{"text": "\\chapter{Training, validation and test data}\nIn a classification problem, we divide the (usually manually) labeled dataset $D=\\{(x_i,y_i)\\}$ into three disjoint sets, including a training dataset $X$, a validation dataset $V$ and a test dataset $T$.\n% to give an expectation of the real generalization of NN.\n\n\t\\begin{enumerate}\n\t\t\\item The labeled dataset $D=\\{(x_i,y_i)\\}_{i=1}^N$: $x_i$ are samples and $y_i$ are corresponding labels. For example, Figure \\ref{xd:lbd} shows the labeled dataset of five class points, i.e. $x_i\\in\\mathbb{R}^2,y_i=1:5$.\n\t\t\n\t\t\t\\begin{figure}[!ht] \n\t\t\t\t\\center{\\includegraphics[width=10cm] {Homework1_5classes.png}} \n\t\t\t\t\\caption{labeled dataset} \n\t\t\t\t\\label{xd:lbd}\n\t\t\t\\end{figure}\n\t\t\t \n\t\t\\begin{enumerate}\n\t\t\t\\item Training Data $X$: we use it to train the neural network $f_\\theta(x)$, aiming at minimizing the loss function $\\mathcal{L}(X;\\theta)$ by adjusting $\\theta$.\n\t\t\t\n\t\t\t\\item Validation Data $V$: \n\t\t\t\\begin{enumerate}\n\t\t\t\t\\item Detecting whether the neural network is over-fitting, i.e. if $\\mathcal{L}(X)$ decreases while $\\mathcal{L}(V)$ increases.\n\t\t\t\t\\item If it is the case, one may consider stopping training on $X$, because the prediction ability of $f_\\theta$ on other samples (which are what we really want) may decrease.\n\t\t\t\t\\item To make full use of the labeled data, one may continue to train $f_\\theta$ with $V$, and make $X$ the validation data.\n\t\t\t\\end{enumerate}\n\t\t\t\\item Test Data $T$\n\t\t\t\\begin{enumerate}\n\t\t\t\t\\item Testing the real performance of $f_\\theta$.\n\t\t\t\t\\item We expect a similar accuracy on other unlabeled data as on the test data.\n\t\t\t\\end{enumerate}\n\t\t\\end{enumerate}\n\t\t\n\t\t\\item Data augmentation: how to generate more data (to avoid over-fitting and make the training easier)\n\t\t\t\\begin{enumerate}\n\t\t\t\t\\item Shift: shift an image\n\t\t\t\t\\item Rotation: rotate an image\n\t\t\t\t\\item Crop: separate an image to small (intersected) images\n\t\t\t\t\\item ...\n\t\t\t\\end{enumerate}\n\t\\end{enumerate}\n\t\n\n", "meta": {"hexsha": "44fea4da48b8660924252374adce667048b6e941", "size": 1965, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/TrainValidateTest.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/TrainValidateTest.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/TrainValidateTest.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.125, "max_line_length": 224, "alphanum_fraction": 0.7002544529, "num_tokens": 572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929104825006, "lm_q2_score": 0.8333245870332531, "lm_q1q2_score": 0.649987270016695}} {"text": "\\subsubsection{Combinations of Fungal Species}\nThe distribution of fungi in the nature is diverse and wide, and there are often hundreds of microorganisms in one square meter of natural land~\\cite{land}. In order to simplify the study on the topic that \\textbf{combinations of fungal species likely to persist in different environment conditions}, we only analyze and predict the growth and decomposition ability of two fungi in combination.\n\\par\nSince the known information is not sufficient and the internal relationship of fungal species combinations in different environments has a degree of uncertainty, we use \\textbf{Gray Prediction Model}~\\cite{gray} and time series data based on the \\textbf{Decomposition Models} in \\textit{Model~4.1} and \\textit{Model~4.2} to analyze and predict the living situation of the fungal species combination and the change trend of decomposition ability.\n\\par\nMark the known group of time series data as follows.\n$$X^0 = (x^0(1),\\ x^0(2),\\ x^0(3),\\ \\ldots,\\ x^0(n))$$\n\\par\nTo ensure the stability of the data, we perform \\textbf{Accumulated Generating Operation} (AGO)~\\cite{gray} on the data. The AGO process would be executed multiple times until the time series data is stable enough, and we get the following time series data.\n$$X^1 = (\\sum^1_{i=1}x^0(i),\\ \\sum^2_{i=1}x^0(i),\\ \\ldots,\\ \\sum^n_{i=1}x^0(i))$$\n\\par\nThe time series $X^1$ is then fitted by a first-order differential equation given by \\textit{Eq.~(\\ref{15eq})}\n\\begin{equation}\n \\label{15eq}\n \\frac{dx^1}{dt} = ax^1 = b\n\\end{equation}\nwhere the parameters $a$ and $b$ are the development coefficient and control variable.\n\\par\nAccording to the \\textbf{whitening gray derivatives} of discrete data points, the following equation is obtained.\n\\begin{equation}\n \\frac{dX^1(t)}{dt} = x^1(k) - x^1(k-1) = x^0(k)\n\\end{equation}\n\\par\nA new variable $z^1(k)$, which is known as the whitening value of $x^1(k)$, is defined by the following equation.\n\\begin{equation}\n z^1(k) = 0.5x^1(k) + 0.5x^1(k-1)\n\\end{equation}\n\\par\nThen we use \\textbf{the least square method} to calculate the parameters.\n\\begin{equation}\n \\hat{\\delta} = (B^\\top B)^{-1}B^\\top Y\n\\end{equation}\nwhere\n$$\\hat{\\delta} = [\\hat{a},\\ \\hat{b}]^\\top,\\ B=\\left[ \\begin{array}{cc}\n -z^1(2) & 1 \\\\\n -z^1(3) & 1 \\\\\n \\vdots & \\vdots \\\\\n -z^1(n) & 1\n \\end{array}\n \\right],\\ Y=\\left[ \\begin{array}{c}\n x^0(2) \\\\\n x^0(3) \\\\\n \\vdots \\\\\n x^0(n)\n \\end{array}\n \\right]$$\n\\par\nFinally, use \\textbf{inverse AGO} and get the predicted value $x^2(k+1)$ as follows.\n$$\\hat{x}^2(k+1) = \\hat{x}^1(k+1) - \\hat{x}^1(k) = (1-e^a)(x^1(1) - \\frac{b}{a})e^{-at},\\ k = 1,\\ 2,\\ \\ldots,\\ n$$\n\\par\nWhen new data are added, the parameters $a$ and $b$ in the Gray Prediction Model would be \\textbf{updated} to ensure the accuracy of prediction.\n\\par\nDue to the large differences in the suitable environment among the five fungi, we divide the five fungi into four combinations, including $F_A\\ \\&\\ F_B$, $F_B\\ \\&\\ F_C$, $F_C\\ \\&\\ F_D$, and $F_D\\ \\&\\ F_E$, based on the conclusion of \\textit{Table~\\ref{correspondingoptimalenvironment}} for analysis and prediction.\n\\par\nTake combination $F_A\\ \\&\\ F_B$ as an example. Calculate the time series data of the decomposition rate for the previous $122$ days using the data in \\textit{Eq.~(\\ref{sixtheq})}, \\textit{Eq.~(\\ref{ninetheq})}, and \\textit{Table~\\ref{decompositionrates}}, and mark it as follows.\n$$X^0_{A\\&B} = (x^0_{A\\&B}(1),\\ x^0_{A\\&B}(2),\\ \\ldots,\\ x^0_{a\\&B}(122))$$\n\\par\nThen the time series data is substituted into Gray Prediction Model, and the prediction data is calculated as follows.\n$$X^2_{A\\&B}=(x^2_{A\\&B}(123),\\ x^2_{A\\&B}(124),\\ \\ldots,\\ x^2_{A\\&B}(200))$$\n\\par\nDraw the decomposition rate of fungal species combination $F_A\\ \\&\\ F_B$ in five climates, including arid, semi-arid, temperate, arboreal, and tropical rain forests, as shown in the figure below.\n\\begin{figure}[H]\n \\centering\n \\label{decompositioninfiveclimates}\n \\includegraphics[width=\\textwidth]{figures/A&B.png}\n \\caption{Decomposition rate of fungal species combination $F_A\\ \\&\\ F_B$ in five climates.}\n\\end{figure}\nCombination $F_A\\ \\&\\ F_B$ is most suitable for the decomposition of lignin or cellulose in semi-arid climate. From the prediction result, the cumulative decomposition rate after $200$ days can reach $34.7\\%$.\n\\par\nIn the same way, the time series data of the other three fungal species combinations, $F_B\\ \\&\\ F_C$, $F_C \\&\\ F_D$ and $F_D \\&\\ F_E$, are calculated as follows.\n\\begin{equation*}\n \\begin{cases}\n X^0_{B\\&C} = (x^0_{B\\&C}(1),\\ x^0_{B\\&C}(2),\\ \\ldots,\\ x^0_{B\\&C}(122)) \\\\ \\\\\n X^0_{C\\&D} = (x^0_{C\\&D}(1),\\ x^0_{C\\&D}(2),\\ \\ldots,\\ x^0_{C\\&D}(122)) \\\\ \\\\\n X^0_{D\\&E} = (x^0_{D\\&E}(1),\\ x^0_{D\\&E}(2),\\ \\ldots,\\ x^0_{D\\&E}(122))\n \\end{cases}\n\\end{equation*}\n\\par\nSubstituting Gray Prediction Model to obtain the following decomposition rate prediction curves in different climates in turn.\n\\par\n\\begin{figure}[H]\n \\centering\n \\subfigure[$F_B\\ \\&\\ F_C$]{\n \\includegraphics[width=0.95\\textwidth]{figures/B&C.png}\n }\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\subfigure[$F_C\\ \\&\\ F_D$]{\n \\includegraphics[width=0.95\\textwidth]{figures/C&D.png}\n }\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\subfigure[$F_D\\ \\&\\ F_E$]{\n \\includegraphics[width=0.95\\textwidth]{figures/D&E.png}\n }\n \\caption{Decomposition rate of fungal species combinations in five climates.}\n\\end{figure}\n\\par\nThe optimal environment and the predicted decomposition rate after $200$ days under different combinations of fungal species are summarized in the following table.\n\\begin{table}[H]\n \\centering\n \\caption{Summary of optimal environment and predicted decomposition rate.}\n \\label{summaryofenvironmentandprediction}\n \\begin{tabular*}{\\hsize}{@{\\extracolsep{\\fill}}ccc}\n \\toprule\n Combinations & Optimal Environment & Decomposition Rate Predicted Value \\\\\n \\midrule\n $F_A\\&F_B$ & Semi-arid & $34.7\\%$ \\\\\n $F_B\\&F_C$ & Temperate & $32.8\\%$ \\\\\n $F_C\\&F_D$ & Temperate / Arboreal & $32.7\\%$ \\\\\n $F_D\\&F_E$ & Tropical rain forests & $34.3\\%$ \\\\\n \\bottomrule\n \\end{tabular*}\n\\end{table}\nAccording to the above table, we find that the combination of $F_A\\ \\&\\ F_B$ has the highest decomposition rate in semi-arid region, which is $34.7\\%$, and the decomposition rates of other species combinations in their optimal environment are also relatively high, which are all above $32\\%$.", "meta": {"hexsha": "bf2d597b4eb8fac0d4228b7d7eeea0be957a271c", "size": 6473, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "4/5/2.tex", "max_stars_repo_name": "syy11cn/2021-mcm-meritorious-article", "max_stars_repo_head_hexsha": "3eaf143f4319fae681d98134bfc7e699833d8273", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2021-11-07T14:38:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-11T10:37:34.000Z", "max_issues_repo_path": "4/5/2.tex", "max_issues_repo_name": "syy11cn/2021-mcm-meritorious-article", "max_issues_repo_head_hexsha": "3eaf143f4319fae681d98134bfc7e699833d8273", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "4/5/2.tex", "max_forks_repo_name": "syy11cn/2021-mcm-meritorious-article", "max_forks_repo_head_hexsha": "3eaf143f4319fae681d98134bfc7e699833d8273", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.3247863248, "max_line_length": 445, "alphanum_fraction": 0.6888614244, "num_tokens": 2146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245828938678, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6499872625262179}} {"text": "\n\n\\section{Preliminaries and exploratory data analysis}\n\nWe're going to revisit data we considered in the cluster analysis lab., so you need to load the \\texttt{cluster} and the \\texttt{flexclust} libraries to make the data available.\n\nWe'll start by considering the \\texttt{milk} data; you should have done some interesting eda in the cluster lab. We follow this up by carrying out multidimensional scaling using \\texttt{cmdscale} from the \\texttt{MASS} library (look at the helpfile):\n\n\\begin{verbatim}\nlibrary(cluster)\nlibrary(flexclust)\nlibrary(MASS)\ndata(milk)\nmilk.dist <- dist(milk)\nmilk.pco <- cmdscale(milk.dist)\npar(xpd = NA, bty = \"n\") ## let the labels run past the \n## plotting region, and remove the box\nplot(milk.pco, type = \"n\", main = \"PCO representation\")\ntext(milk.pco, row.names(milk.pco), cex = 0.5)\n\\end{verbatim}\n\nNote that we've produced a blank plot but then drawn text on in the positions of the data with the row names. Can you make sense of the resulting plot?\n\nIf you're very interested, you could even plot the cluster solutions you find when doing cluster analysis against the 2-dimensional representation of the distance matrix:\n\n\\begin{verbatim}\nmilk.hclust <- hclust(milk.dist)\nmilk.cut <- cutree(milk.hclust, 3)\nplot(milk.pco, type = \"n\", main = \"PCO representation\")\ntext(milk.pco, row.names(milk.pco), cex = 0.5, col = milk.cut, pch = milk.cut)\n\\end{verbatim}\n\n\nMore importantly this week is to consider some diagnostics.\n\n\nLets look at the $n-1$ solution, using the delightfully titled \\texttt{zapsmall} function to round our eigenvalues to 8 digits (computer arithmetic means we have a lot of negative eigenvalues in the 15th decimal place which we get rid of). \n\n\\begin{verbatim}\nmilk.pco.24 <- cmdscale(milk.dist, eig = TRUE, k = 24)\nevals <- zapsmall(milk.pco.24$eig, digits = 8)\nevals\n\\end{verabtim}\n\nNow, if we want to examine the fit of a $q$ dimensional approximation we use $2 \\times n \\times \\sum_{j=q+1}^{n-1} \\lambda_{j}$ (note we are summing all the discarded eigenvalues) to give us: \n\n\\begin{verbatim}\n2 * dim(milk)[1] * sum(milk.pco.24$eig[3:24])\n\\end{verbatim}\n\nYou can adjust this last line to see the SS for a 3 dimensional approximation (use \\texttt{4:24} in the square brackets) and so on.\n\nAnd now try to calculate the same thing directly from the projection and from the data. We already know $\\boldsymbol{Delta}$ (although we have to tell \\textbf{R} we want a matrix and not a distance matrix. We can get an estimate of $d$ by taking the distance matrix of the points in the $q$ dimensional approximation that interests us. Finally, we just want the sum of the differences in the squared distances:\n\n\\begin{verbatim}\nmilk.pco.2 <- cmdscale(milk.dist, eig = TRUE, k = 2)\ndelta <- as.matrix(milk.dist)\nd <- as.matrix(dist(milk.pco.2$points))\nsum(as.vector(delta)^2 - as.vector(d)^2)\n\\end{verbatim}\n\n\nThe other obvious method is just to use the percentage of variance explained / discarded. Consider dividing the cumulative sum of the eigenvalues by the sum of the eigenvalues:\n\n\\begin{verbatim}\ncumsum(evals) / sum(evals)\n\\end{verbatim}\n\n\nCareful examination should convince you that a 2-dimensional approximation might well be adequate.\n\n\n\\section{Sammon Mapping}\n\nYou could try ``Sammon'' mapping on the mvmclass data. Assuming you've used \\texttt{daisy} to generate a distance matrix from the class data (we did this in an earlier lab.)\n\n\\begin{verbatim}\nmvmclass <- read.csv(\"class06.csv\", row.names = 1)\nmvmclass.dist <- daisy(mvmclass)\nmvmclass.sammon <- sammon(mvmclass.dist)\n\\end{verbatim}\n\nIf you want to plot the representation, the material you need is under \\verb+mvmclass.sammon$points+.\n\n\\begin{itemize}\n\\item How do you determine whether your sammon fit is a good one or not?\n\\end{itemize}\n\n\\section{Further analyses}\n\nIn case you wish to follow this from Johnson and Wichern's perspective, two datasets have been placed in the portal are very interesting: \\texttt{utilities.csv} and \\texttt{USUni.csv}. The former we saw last time when we were considering clustering. The latter considers a number of measures on 25 US Universities. \n\nAnother interesting analysis involves the Painters data (see \\texttt{?painters} in \\texttt{MASS}) - in particular you may like to see whether there any evidence that the first dimension corresponds with a time axis?\n\n\\section{Summary}\n\n\\fbox{\\parbox[c]{0.9\\textwidth}{\\color{blue}\nIn some ways, it might have been better for this topic to follow cluster analysis. It is really a different way of examining the relationships between individuals in a dataset. By the end of this week, we should be able to:\n\n\\begin{itemize}\n\\item Computer and plot a p.c.o projection of a distance matrix\n\\item Understand and justify methods for determining whether our low dimension projection is adequate\n\\item Explain and interpret results, especially if those results are placed relative to a cluster analysis\n\\item Be able to relate p.c.o. to p.c.a.\n\\end{itemize}\n}}\n", "meta": {"hexsha": "3d04fbff91321c6bd51894ba3d69f74c81c9ac23", "size": 4990, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "newworksheet/week9pco.tex", "max_stars_repo_name": "phewson/mvstats", "max_stars_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "newworksheet/week9pco.tex", "max_issues_repo_name": "phewson/mvstats", "max_issues_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2020-08-28T16:37:22.000Z", "max_issues_repo_issues_event_max_datetime": "2020-08-28T16:49:11.000Z", "max_forks_repo_path": "newworksheet/week9pco.tex", "max_forks_repo_name": "phewson/mvstats", "max_forks_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.6355140187, "max_line_length": 416, "alphanum_fraction": 0.7563126253, "num_tokens": 1309, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303236047049, "lm_q2_score": 0.8221891370573388, "lm_q1q2_score": 0.6499654445822111}} {"text": "\\section{Proof Details}\n\\label{sec:ibhc-proof}\n\n\\treerefinement*\n\\begin{proof}\n\nSuppose, first, that $\\tree$ is a refinement of $\\tree'$. Pick any triplet $(\\{a,b\\},c) \\in \\Delta(\\tree')$. Then there is a node in $\\tree'$ whose descendants include $a,b$ but not $c$. By the definition of refinement, $\\tree$ contains a node with the same descendants. Hence the constraint $(\\{a,b\\},c)$ holds for $\\tree$ as well.\n\nConversely, say $\\Delta(\\tree') \\subseteq \\Delta(\\tree)$. Pick any cluster $S'$ of $\\tree'$; it consists of the descendants of some node in $\\tree'$. Consider the set of all triplet constraints consisting of two nodes of $S'$ and one node outside $S'$. Since these constraints also hold for $\\tree$, it follows that the lowest common ancestor of $S'$ in $\\tree$ must have exactly $S'$ as its set of descendants. Thus $S'$ is also a cluster of $\\tree$.\n\\end{proof}\n\n\\irreducible*\n\\begin{proof}\nTo prove irreducibility, we show that there is a non-zero probability\nof moving from state $\\tree$ to $\\tree'$, both of which satisfy $C$. We\naccomplish this by first defining a \\emph{canonical tree} $\\tree_C$ given a triplet \nset $C$ and showing that we can reach $\\tree_C$ from $\\tree$ using \nconstrained-SPR moves. We then show that for every constrained-SPR move,\nthere exists an equivalent reverse move that undoes it with non-zero\nprobability.\nThis proves that that from $\\tree_C$ we can reach $\\tree'$, \ncreating a path from $\\tree$ to $\\tree_C$ to $\\tree'$.\n\nA binary tree $\\tree$ can be entirely defined by the bipartitions\nmade over the data at each node. \nLet $G_n$ be the Aho graph for node $n$.\nFor a binary tree that satisfies a set of triplets, \nthe split over the data at each node $n$ must \nbe a bipartition of the connected components of $G_n$.\nWe define a particular node to be in \\emph{canonical form}\nif either a) it is a leaf, or b) the bipartition over $G_n$\nat that node can be written as\n$(l, r)$, where $l$ exactly matches a single, particular connected component of\n$G_n$, and $r$ is the rest of the connected components. \nThe particular component\n$l$ is the connected component in $G_n$ with the minimum data index\ninside it.\nNote that we treat the children of nodes as unordered.\nA canonical tree $\\tree_C$ is one such that every node in the tree is\nin canonical form.\nTo convert an arbitrary tree $\\tree$ that satisfies $C$ into $\\tree_C$, we first\nconvert the root node of $\\tree$ into canonical form\nusing constrained-SPR moves.\n\nLet $s$ be the root of $\\tree$ and let $l$\nbe the set of points that ought to be in their own partition according\nto $G_s$. In order for\n$s$ not to be in canonical form, $l$ must be in a partition with \ndata from other connected components in $G_s$, which we will call $o$.\nThe bipartition if $s$ were in canonical form would be $(l, r)$ and\nthe current non-canonical bipartition can thus be written as $(l + o, r - o)$.\n\nWe first examine $t$, the child of the root that contains $l + o$.\nIn general, the data from $l$ and the data from $o$ could\nbe split over the children of $t$, so the partition at $t$\ncan be written as $(l_1 + o_1, l_2 + o_2)$ where $l = l_1 + l_2$ \nand $o = o_1 + o_2$. This is visualized in the first\ntree of \\autoref{fig:canonical}.\nWe first group the data from $l$ into their own ``pure'' subtree\nof $t$ as follows.\nLet $u$ be the root of the\nlowest non-pure subtree of $t$ that has\ndata from $l$ in both of its children.\nThere exist two subtrees that are descendants of $u$ \nthat contain data from $l$ (one on the left and one on the right).\nThose two subtrees\nmust be pure, and furthermore, they are both free to\nmove within $u$ via constrained-SPR moves because\nthey are in different connected components in $G_u$.\nThus, we can perform a constrained-SPR move to merge these\ntwo pure subtrees together into a larger pure subtree.\nWe can repeat this process for $t$ until all nodes\nfrom $l$ are in their own pure subtree of $t$.\nThe partition of $t$ can thus be written as\n$(l + o_1, o_2)$, since the pure subtree may be several\nlevels down from $t$. This grouping process is visualized in\n\\autoref{fig:canonicalgrouping} and the results can be\nseen in the second\ntree in \\autoref{fig:canonical}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.5\\textwidth]{img/ibhc/CanonicalTreeGrouping}\n \\caption{The process of grouping the data in $u$\n that belong to $l$ into their own pure subtree. $u$ is the\n lowest node of $t$ (see \\autoref{fig:canonical}) that has data from $l$\n in both of its children.}\n \\label{fig:canonicalgrouping}\n\\end{figure}\n\n\nWe now perform a constrained-SPR move to detach\nthe pure subtree of $l$ and regraft it to\nthe edge between $s$ and $t$. This is a permissible\nmove since $l$ is its own connected component in \n$G_s$. We now have the third tree in \\autoref{fig:supcanonical}.\nWe now perform a final constrained-SPR\nmove, moving the subtree of $o$ to the opposite\nside of $s$, creating the proper canonical partition\nof $(l, r)$.\nTo entirely convert $\\tree$ into $\\tree_C$, we need to recurse\nand convert every node in $\\tree$ into canonical form.\n\n\\begin{figure*}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/CanonicalTree}\n \\caption{The process of converting $s$ into canonical form.\n We first group nodes from $l$ into their own pure subtree, then perform\n two constrained-SPR moves to put $s$ into canonical form.}\n \\label{fig:supcanonical}\n\\end{figure*}\n\nEvery constrained-SPR move has an associated reverse constrained-SPR move that\nperforms the opposite transition.\nThe reverse constrained-SPR move selects the same subtree as the forward one\nand prunes it, and just regrafts the subtree to its original location\nbefore the forward move.\nWe know that this regraft has non-zero probability because the original tree\ndid not violate constraints.\nThus, since any arbitrary $\\tree$ can be converted into $\\tree_C$ and since each move\nhas a non-zero probability reverse move,\n$\\tree_C$ can be converted into an arbitrary tree $\\tree'$ and we have a non-zero probability\npath to convert $\\tree$ into $\\tree'$.\n\n\\end{proof}\n\n\\section{Additional Results}\n\n\\begin{figure*}[h]\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/Zoo-result.png}\n \\caption{Zoo}\n \\label{fig:zoo-result}\n \\end{subfigure}\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/20-Newsgroups-result.png}\n \\caption{20 Newsgroups}\n \\label{fig:20news-result}\n \\end{subfigure}\n \\caption{The average of four runs of constrained-SPR samplers\n for the Zoo dataset and the 20 Newsgroups dataset, using 5 different querying schemes.\n A query was made every 100 iterations.}\n \\label{fig:main-results2}\n\\end{figure*}\n", "meta": {"hexsha": "28612772ece5b996d3ecd1327f3f4d1f84fdb9ed", "size": 6830, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "writeup/content/appendix/ibhc-appendix.tex", "max_stars_repo_name": "sharadmv/thesis", "max_stars_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-30T01:28:54.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-30T01:28:54.000Z", "max_issues_repo_path": "writeup/content/appendix/ibhc-appendix.tex", "max_issues_repo_name": "sharadmv/thesis", "max_issues_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "writeup/content/appendix/ibhc-appendix.tex", "max_forks_repo_name": "sharadmv/thesis", "max_forks_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.462585034, "max_line_length": 451, "alphanum_fraction": 0.7295754026, "num_tokens": 1871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891305219503, "lm_q2_score": 0.7905303285397348, "lm_q1q2_score": 0.6499654434733163}} {"text": "\\documentclass{article}\n\n\\usepackage{amsmath} % for \\over version of fractions\n\\usepackage{amsfonts} %for number system sets\n\\usepackage{xfrac} %for slanted fractions with sfrac\n\\usepackage[ampersand]{easylist}\n\\usepackage{graphicx}\n\n\\begin{document}\n\n\\title{Introductory Programming Exercises}\n\\author{Hamik Mukelyan}\n\\maketitle\n\n\\section*{Variables and basic control}\n\\subsection*{Exercise 1: Fibonacci numbers the easy way}\nThe \\textbf{fibonacci numbers} are $1, 1, 2, 3, 5, 8, 13, 21, 34, \\cdots, F_{n}$, where $F_{0} = F_{1} = 1$. There is a \\textbf{closed form} solution for $F_{n}$, which means we can find $F_{n}$ by just plugging $n$ into a formula that we only have to evaluate once. Find this formula online and write a program to compute $F_{5}, F_{10},$ and $F_{20}$. The fibonacci numbers are integers, so make sure that your program rounds or \\textbf{truncates} your numbers correctly. After you've gotten your program to work correctly, \\textbf{refactor} it - i.e., improve it - so that you store $\\sqrt{5}$ in a variable. That way you don't have to perform the slow square root operation three times - you only have to perform it once at the beginning of the program then reuse that result over and over.\n\n\\subsection*{Exercise 2: Fibonacci numbers with a loop}\nWrite a program that displays the fibonacci numbers from $F_{0}$ to $F_{n}$. One way to do this is with a loop that keeps a \\textbf{counter variable}, $i$, that is \\textbf{initialized} to $0$ and is incremented by 1 at the end of each \\textbf{iteration} of the loop until it reaches $n$. The variable names $i, j,$ and $k$ are traditionally used as loop counter variables, with $i$ being used in outermost loops and $j$ being used in loops nested within those outermost loops. For each iteration of the loop you plug $i$ into the formula from exercise 1 and display the result with a print statement. Try it for $n = 10$ and see if the correct numbers show up!\n\nNow let's refactor. The closed form solution for $F_{n}$ uses divisions and exponentiations in addition to square roots. We mitigated the slowness of square root operations by storing $\\sqrt{5}$ in a variable, but we still do divisions and exponentiations, which are super slow compared to simple additions. The \\textbf{recursive formula} for $F_{n}$ elegantly (and simply) describes the sequence in terms of the numbers that came before it: $F_{n} = F_{n - 1} + F_{n - 2}$. Along with the \\textbf{base cases} $F_{0} = F_{1} = 0$, this formula fully defines the sequence. Switch your program over from using the closed form solution to the recursive one. Recursive solutions - that is, solutions that are stated in terms of \\textit{smaller or easier versions of the same problem}, show up often in computer science because closed form solutions are often difficult or impossible to find.\n\n\\subsection*{Exercise 3: The radix}\nWhen we write a number like 987, we actually mean $9 \\cdot 10^{2} + 8 \\cdot 10^{1} + 7 \\cdot 10^{0}$. The number 10 is called the \\textbf{base} or \\textbf{radix} of the number. Most civilizations have used the radix 10 for counting, but some have used weird ones like 60 - we'll see in a moment how to think about numbers in a non-standard base. \\textbf{Binary} (base 2), \\textbf{octal} (base 8), and \\textbf{hexadecimal} (base 16) are the most commonly used bases in programming other than decimal (base 10). If a base $B$ is less than 10, we use only the digits 0 to $B - 1$. For example, in octal we never use the digits 8 or 9. The number $123_{10} = 1 \\cdot 8^{2} + 7 \\cdot 8^{1} + 3 \\cdot 8^{0} = 173_{8}$, where the subscripts are reminders about the bases of the numbers. When the base is 10 we usually omit the subscript by writing e.g. $123$ instead of $123_{10}$.\n\nIn bases greater than 10 we use letters starting from A for digits greater than 9. For example, in hexadecimal the digit A is one greater than 9, B is two greater than 9, and F is the biggest digit, which represents $15_{10}$. So counting from 1 to 32 in base 16, or hexadecimal, looks like 1, 2, 3, $\\cdots,$ 9, A, B, C, D, E, F, 10, 11, 12, 13, $\\cdots,$ 1E, 1F, 20. In binary we have only the digits 0 and 1 at our disposal. Any number that can be represented in decimal can be represented in any other base. Computers store numbers and operate on numbers in binary for the simple reason that it's easy to represent them with hardware: it's much easier for a switch to be simply on or off - representing 1 or 0 - than it is for it to be in one of ten states.\n\nBinary has a special relationship with octal and hexadecimal, which is why they're used frequently in programming. Binary numbers are annoyingly long, so we can represent them more compactly by using their relationship with hex. Write out the first 16 numbers in binary. E.g., 0, 1, 10, 11, 100, 101, $\\cdots$. These numbers match up, respectively, with the hex numbers 0, 1, 2, 3, $\\cdots$, E, F. If you want to write a long binary number like 111100101010$_{2}$ compactly, you break it up into 4-digit chunks and use the equivalent hex number for each chunk. E.g., 1111 0010 1010$_{2}$ = F2A$_{16} = 15 \\cdot 16^{2} + 2 \\cdot 16^{1} + 10 \\cdot 16^{0} = 3882$.\n\nIn many programming languages, you can represent a binary, octal, or hexadecimal number by prefixing any numerical literal with 0b, 0, or 0x. For instance, if you want to double-check that F2A$_{16}$ is actually 3882, just type 0xF2A into a REPL and see the resulting decimal value. Similarly, you can double-check that 173$_{8} = 123$ by typing 0123. These prefixed numerical literals can used anywhere you'd use a typical decimal literal in a program.\n\nHere's the exercise: prompt the user for $n_{10}$ then display it in base 3. Then prompt the user for $m_{3}$ and display it in base 10. Before you try this problem, make sure you can convert a base 10 number into base 3 by hand and vice versa. Next make sure you can clearly articulate the steps you took to do so: in other words, devise an \\textbf{algorithm}, which is a step-by-step problem solving procedure.\n\\end{document}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "ff657efebcd3adf2e23f843fd2e8513b444ca34a", "size": 6076, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises.tex", "max_stars_repo_name": "hamikm/CodeExercises", "max_stars_repo_head_hexsha": "1601d08cbad119c0037ad3dc8a6dc78100cb74b7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "exercises.tex", "max_issues_repo_name": "hamikm/CodeExercises", "max_issues_repo_head_hexsha": "1601d08cbad119c0037ad3dc8a6dc78100cb74b7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "exercises.tex", "max_forks_repo_name": "hamikm/CodeExercises", "max_forks_repo_head_hexsha": "1601d08cbad119c0037ad3dc8a6dc78100cb74b7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 101.2666666667, "max_line_length": 887, "alphanum_fraction": 0.7432521396, "num_tokens": 1700, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996143, "lm_q2_score": 0.8221891305219504, "lm_q1q2_score": 0.6499654272432038}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PROBLEM 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{Problem 1}\n\nA reactor contains 5\\% (by atom) enriched uranium dioxide, which is 15\\% of the entire core by volume.\n\\begin{enumerate}[a)]\n\\item Calculate the macroscopic cross section for this core if we were to treat it as a homogeneous volume. \n\\item If the reactor were a cube with a side length of 4 m and a beam of $10^{15}$ thermal neutrons were incident on one face of the cube, how many neutrons would we expect to make it through to the other side?\n\\end{enumerate}\n\n\\begin{table}[htbp]\n\t\\centering\n\t\\begin{tabular}{|c|c|c|}\n\t\t\t\\hline\n\t\t\tNucleus\t\t& Thermal $\\sigma_{\\text{t}}$ (b) & Mass (g/mol) \\\\\n\t\t\t\\hline\n\t\t\t$^{1}$H\t\t& 20.8\t\t\t\t\t\t& 1.008\t\t\t\\\\\n\t\t\t$^{16}$O\t& 3.5 \t\t\t\t\t\t& 15.995\t\t\\\\\n\t\t\t$^{235}$U\t& 607.5 \t\t\t\t\t& 235.044\t\t\\\\\n\t\t\t$^{238}$U\t& 11.8\t\t\t\t\t\t& 238.050\t\t\\\\\n\t\t\t\\hline\n\t\t\tCompound\t& \\multicolumn{2}{|c|}{$\\rho \\left(\\text{g/cm}^3\\right)$} \\\\\n\t\t\t\\hline\n\t\t\tH$_2$O\t\t& \\multicolumn{2}{|c|}{1.0} \\\\\n\t\t\tUO$_2$\t\t& \\multicolumn{2}{|c|}{10.4}\t\\\\\n\t\t\t\\hline\n\t\\end{tabular}\n\t\\label{tab:design-specs}\n\\end{table}\n\n", "meta": {"hexsha": "04a4bb82207e768cd3aabce762477555d8cec232", "size": 1119, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises/drafts/disc04/disc04_exercise01.tex", "max_stars_repo_name": "mitchnegus/NE150-discussion", "max_stars_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "exercises/drafts/disc04/disc04_exercise01.tex", "max_issues_repo_name": "mitchnegus/NE150-discussion", "max_issues_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "exercises/drafts/disc04/disc04_exercise01.tex", "max_forks_repo_name": "mitchnegus/NE150-discussion", "max_forks_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.3, "max_line_length": 210, "alphanum_fraction": 0.5665773012, "num_tokens": 427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9005297781091839, "lm_q2_score": 0.7217432003123989, "lm_q1q2_score": 0.6499512440291368}} {"text": "\\subsection{Relational Databases}\\label{relational_databases}\nThe group last year chose to implement the WordCount and Fuseki databases using PostgreSQL.\nPostgreSQL is a relational database system\\cite{knox2020}.\n\nIn the coming sections, we will describe how one can model and describe a relational database design, both mathematically and with a more graphic design.\nAfter doing this, we will examine how operations on these models can be described using SQL and relational algebra.\n\n\\subsubsection*{Relational Model}\nRelational database systems can be mathematically described using relations and sets, mapping a unique key to a tuple of information\\cite{DBSBook}.\nThe values of the tuples contained within the relation can be described by the attributes of the relation and their corresponding domains\\cite{KatjaFirstPP}. \nThe relations are often described using a \\textit{relational schema}, denoting the name and domains of the attributes.\n\n\nEquation \\ref{eq:relational_schema} shows an example of a relation describing books as tuples of three text fields ($ISBN$, $author\\_name$, and $title$ ) and a positive integer ($number\\_of\\_pages$).\nThe relation also denotes a super key for the relation. A super key is one or more attributes that can uniquely identify a tuple in a given relation.\nA super key is called a \\textit{candidate key} if removal of a single attribute from the key results in it longer uniquely identifying tuples.\nAttributes describing the primary key are underlined.\n\n\\begin{equation} \\label{eq:relational_schema}\n book(\\underline{ISBN: text}, author\\_name:text, title: text, number\\_of\\_pages:\\mathbb{Z}^+)\n\\end{equation}\nSuper keys can be explained by $t_1 \\in r\\neq t_2 \\in r \\implies t_1.K \\neq t_2.K$ \\cite{DBSBook}, where $t_1, t_2$ are tuples contained within the relation $r$.\nThat is, no two tuples $t_1, t_2$ from relation $r$ have the same values if they do not have the same superkey $K$. \nIf the super key does not contain extraneous attributes, it is said to be \\textit{minimal} \\cite{DBSBook}.\nWe will use the term \\textit{primary key} to denote a chosen super key of a relation. \nWhen describing a database, it is often necessary to specify how various data are connected. \nTo do this, one can use \\textit{foreign keys} to denote that tuples in $r_1$ are related to the tuples in $r_2$.\n\n\nOne could model the relationship between a book owner and a book using the relational schemas seen in equation \\ref{eq:relational_schema} and \\ref{eq:bookOwnerExample}.\nThere, primary keys from other relations are used to reference unique tuples. The $owns$ relation describes how relations $book$ and $book\\_owner$ are connected. \n\\begin{equation}\\label{eq:bookOwnerExample}\n \\begin{split}\n owns(\\underline{owner\\_id \\rightarrow book\\_owner}, \\underline{ISBN \\rightarrow book}), \\\\\n book\\_owner(\\underline{owner\\_id:\\mathbb{Z}^+}, name:text,)\n \\end{split}\n\\end{equation}\n\n\\subsubsection*{Evaluating a Database Design}\nWhen evaluating the design of relational database schemas, we want to avoid redundant data duplication, loss by decomposition and change of dependencies\\cite{DBSBook}.\nOne approach to ensure this is to use normalization theory\\cite{DBSBook}. \nThis approach examines the \\textit{functional dependencies} of the relations and evaluates them based on their \\textit{normal form}.\nWhen a relation is in a normal form, it adheres to certain design criteria. \nThese criteria are defined using functional dependencies.\n\nA functional dependency describes the relationship between two sets of attributes. \nFunctional dependency between $\\alpha$ and $\\beta$, written as $\\alpha \\rightarrow \\beta$, defines that a value for $\\alpha$ is sufficient to identify unique values for $\\beta$ \\cite{DBSBook}.\nIn this case, we say that $\\alpha$ is the determinant and $\\beta$ is the dependant. \nWe say that $\\beta$ is functionally determined by $\\alpha$ if $\\alpha \\rightarrow \\beta$.\nIt is logically implied that if $\\alpha \\rightarrow \\beta$ and $\\beta \\rightarrow \\gamma$, then $\\gamma$ is functionally determined by $\\alpha$.\nIf all attributes are functionally determined by $\\alpha$, it is a super key\\cite{DBSBook}.\nThe closure of functional dependency $\\alpha$, denoted $\\alpha^+$, describes all attributes that can be logically implied by functional dependencies having $\\alpha$ as determinant\\cite{DBSBook}. \n\nBoyce-Codd Normal Form (BCNF) eliminates all redundancy that can be discovered using functional dependencies\\cite{DBSBook}. \nA relation is BCNF with respect to functional dependencies $F$ if, for all functional dependencies in $F^+$ of the form $\\alpha \\rightarrow \\beta$ if one of the following holds:\n$\\alpha \\rightarrow \\beta$ is trivial ($\\alpha \\rightarrow \\alpha$) or $\\alpha$ is a super key for the relation.\n\nThus, if a relation $R$ is not in BCNF, there must be a functional dependency $\\alpha \\rightarrow \\beta$ where $\\alpha$ is not a super key. \nWe can split such relation $R$ into two relations $R1(\\alpha \\cup \\beta)$ and $R2(R-(\\beta-\\alpha))$ that are both in BCNF.\n\nThird Normal Form (TNF) prevents partial and transitive dependencies\\cite{MontayaNormalForms}.\nThat is, third normal form prevents functional dependencies of the form seen in equation \\ref{eq:trans} \nand functional dependencies where non-prime attributes are functionally dependent on only a part of the candidate key. \n\\begin{equation}\\label{eq:trans}\n {\\alpha} \\rightarrow{\\beta}, {\\beta \\rightarrow \\gamma} \\implies \\alpha \\rightarrow \\gamma\n\\end{equation}\n\nA relation is in TNF with respect to functional dependencies $F$ if, for all dependencies in $F^+$ of the form $\\alpha \\rightarrow \\beta$, one of the following holds: \n\n$\\alpha \\rightarrow \\beta$ is trivial, $\\alpha$ is a super key, or each attribute $A \\in \\beta-\\alpha$ is contained in a candidate key for $R$ \\cite{DBSBook}.\n\nHaving established how one can evaluate a relational database design schema, we can proceed to discuss how a different model that can be used to represent the logic of a relational schema can make it easier to ensure that relations are of at least TNF.\n", "meta": {"hexsha": "80468524a979f3bed1786a89f4dbe118acf5e6f1", "size": 6101, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/Chapter 3 Sprint 1/Relational_databases.tex", "max_stars_repo_name": "chhoumann/p5", "max_stars_repo_head_hexsha": "ea51a218806e552691538e33c16f9cf067be52d4", "max_stars_repo_licenses": ["BSD-3-Clause"], 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"alphanum_fraction": 0.7742992952, "num_tokens": 1461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677583778257, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6497923073125772}} {"text": "\\subsection{Equations of motion}\\label{sec-1}\nThe formulation of SELFE is based on classic expressions of mass and momentum conservation within a shallow fluid,\nas well as the transport equations for salt and heat. The three dimensional expressions of these laws of motion\ninvolve much less reduction and averaging than one or two-dimensional models. However, 3D estuary-scale models\ndo invoke a standard set of assumptions that will be discussed in section \\ref{sec-assump}.\n\nThe SELFE variant used in this equation is defined in a Cartesian (ie map projection) frame with\nbathymetric depth $h$ and free surface $\\eta$ defined relative to a fixed datum (NAVD 88 in our case) as shown \nin Figure \\ref{fig:depths}. The total depth $H$ is the sum of bathymetric depth and free surface elevation.\n\n\n\nThe 3D continuity, depth-integrated mass conservation, and horizontal momentum conservation equations are given by:\n\\begin{align}\n %& \\text{3D Continuity}& \\nonumber \\\\\n \\nabla \\cdot \\bs{u} &= 0 \\label{3Dcon}&\\\\\n %& \\text{Depth-integrated Continuity}& \\nonumber \\\\\n \\eta_t+\\nabla \\cdot \\int_{-h}^\\eta \\bs{u} & = 0 \\label{cont1} &\\\\\n %& \\text{Momentum}& \\nonumber \\\\\n \\frac{D\\bs{u}}{Dt} &= \\explain{ -f\\bs{k} \\times \\bs{u} }{\\text{Coriolis}} \n\t \\; \\explain{-\\frac{1}{\\rho_0} \\nabla p_A}{\\text{Atmos. pressure}}\n\t\t\t\t\t\t\t\t\t\t\t\t\\; \\explain{- \\frac{g}{\\rho_0} \\int_z^\\eta \\nabla \\rho d \\xi}{\\text{Baroclinic}}\n\t \\; \\; \\; \\explain{-g\\nabla \\eta}{\\text{Gravity wave}} & \\nonumber \\\\\n\t \\; \\; \\; &\\explain{+ \\nabla \\cdot (\\mu \\nabla \\bs{u})}{\\text{Horizontal diffusion}}\n\t\t\t\t\t\t\t\t\t\t\t\t\\; \\; \\; \\; \\explain{+\\frac{\\pd}{\\pd z}(\\nu \\frac{\\pd \\bs{u}}{\\pd z})}{\\text{Vertical diffusion}} \\label{mom1}\n\t\\end{align}\n\twith wind and bed stress boundary conditions at the surface and bottom of the water column:\n\t\\begin{align}\n\t\\nu \\frac{\\pd \\bs{u}}{\\pd z} &= \\bs{\\tau}_w \\mbox{ at } z=\\eta &\\\\\n \\nu \\frac{\\pd \\bs{u}}{\\pd z} &= \\chi \\bs{u}_b \\mbox{ at } z=-h &\n\\end{align}\nwhere the variables and parameters are:\n\\begin{align*}\n&\\eta(x,y,t) &\\text{free surface elevation (m)} &\\\\\n&\\bs{u}(x,y,z,t) &\\text{horizontal velocity(m/s)} &\\\\\n&h(x,y) &\\text{bathymetric depth(m)} &\\\\\n&w & \\text{vertical velocity (m/s)} &\\\\\n&f & \\text{Coriolis factor (s\\textsuperscript{-1} )} & \\\\\n&g & \\text{gravity (m/s\\textsuperscript{2})} & \\\\\n&\\rho & \\text{water density (kg/m\\textsuperscript{3})} & \\\\\n&p_A(x,y, t) & \\text{atmospheric pressure at the free surface (N m\\textsuperscript{2})} & \\\\\n&\\nu & \\text{vertical eddy viscosity (m\\textsuperscript{2}/s)} & \\\\\n&\\mu & \\text{horizontal eddy viscosity (m\\textsuperscript{2}/s) }\\\\\n&\\kappa & \\text{vertical eddy diffusivity, for salt and heat (m\\textsuperscript{2}/s)} & \\\\\n\\end{align*}\n\nand the labeled processes in the momentum equations include:\n\\begin{description}\n \\item{Coriolis} an apparent force that results from writing equations on a rotating system (the earth) in a projected (x,y) coordinate system. The Coriolis force is most important in the ocean and near coast, less so in estuary or riverine systems. \n\t \\item{Atmospheric pressure} This is the horizontal variation of pressure above the water.\n\t\t\\item{Gravity wave} Pressure differences due to slope in the water surface are the main force that causes tide and flood propagation. \n\t \\item{Baroclinic forcing} This is the driving mechanism for density-driven (exchange) flow, as\ncommonly found in estuaries, of which salinity intrusion is an example. Another example is gravity (dense) underflow. Due to this force, the horizontal gradient of the density field will initiate 3D flow that moves denser water under the lighter water and will lead to a two-layer flow structure commonly found in stratified estuaries. \n\\end{description}\n\n\\subsubsection{Assumptions}\\label{sec-assump}\nThe Reynolds averaged shallow water equations include some simplifications of the raw (Navier-Stokes) equations in order\nto eliminate small scale terms and make the equations more tractable.\n\\begin{itemize}\n\\item {\\em Free surface} The free surface is incorporated into the equations and results from the \ndivergence of the horizontal flows. This simplification avoids a complex moving boundary problem at the water-atmosphere interface.\n\\item {\\em Reynolds averaging} SELFE solves for velocity in a time-mean sense, averaged over small scale turbulence. Mixing induced by turbulent fluctuations \nis accounted for by relating the fluctuations to mean flow properties using a set of relations called the {\\em turbulence closure}. \n\\item {\\em Boussinesq approximation} The Boussinesq approximation simplifies terms in the governing equations by considering the the effect of density differences due to soluble tracers (salt, temperature and sediment) only\nin a single buoyancy term.\n\\item {\\em Hydrostaticity} Most models in SELFE's class have both a hydrostatic and non-hydrostatic option. In hydrostatic mode, the model only considers pressure forces on a parcel of water arising only from the weight of the water and atmospheric pressure above, neglecting vertical momentum. \nThe assumption can be relaxed by enabling non-hydrostatic pressure; however, due to the required resolution and computation time, \nthis feature is not used at system scales in any model in the Bay-Delta.\n\\item In hydrostatic mode, volume conservation is first enforced over the water column using horizontal velocities into and out of the water column. Vertical velocity is implied in the equations and later inferred in a separate step that invokes the 3D continuity equation. \n\\end{itemize}\n\n\\subsubsection{Roughness and friction}\nIn SELFE, resistance is introduced into the water column through the combination of the bottom stress (drag)\nboundary condition and vertical turbulent momentum diffusion. The mechanism is less direct than the direct \nbody force used in 1-D and 2-D models. [how does SELFE2D fit here?]\n\nThere are two options in the stipulation of the drag coefficient for each spatial location:\n\\begin{itemize}\n\t\\item $C_d$ may be given directly\n\t\\item $C_d$ may be inferred using an analytical description (logarithmic decay) of velocity in a viscous boundary layer. The parametrized form leads to a formula for $C_d$: \n\t$C_d=[\\frac{1}{\\kappa_0}log(\\frac{\\delta}{z_0})]^{-2}$, where $\\kappa_0$ is von Karman's constant, and $\\delta$ is the thickness\nof the bottom layer. In this case the calibration parameter is roughness ($z_0$) rather than drag.\n\\end{itemize}\n\nOnce drag is characterized at the bed, it is mixed vertically up the water column through a turbulent eddy diffusion, which is\nlabeled \"vertical viscosity\" in equation (\\ref{mom1}) and shown visually in figure ****. The mixing of slower water into \nfaster water has the effect of slowing the faster water down. Near the bed, this viscous process is the analog of\nfriction in a lower dimension model. Higher in the water column, the effect of eddy diffusion and location of the velocity\nmaximum are harder to predict.\n\nIn terms of calibration the above formulation leads to two sets of parameters that need to be estimated. The first is the\ndrag coefficient, which may take the form of $C_d$ or roughness $z_0$. In addition, an eddy diffusion coefficient \nhas to be supplied -- this is not stipulated directly but rather emerges from the choice of turbulence closure, described in Section \\ref{sec-tur}.\n\n\\subsection{Turbulence closure}\\label{sec-tur}\nThe final component of vertical mixing is the turbulence closure, which is used to obtain an eddy viscosity coefficient\nfor the momentum equations and a vertical eddy diffusivity for the transport equations. \n\nWe use the vertical component of Umlauf and Burchard's \\cite{Umlauf2003} generic length-scale model, a separate differential \nequation which is integrated \"off-line\" of the other equations based on values from the previous time step. \nThe model is as follows:\n\\beqa\n \\frac{D k}{D t}&=&\\frac{\\pd }{\\pd z}\\left( \\nu_k^\\Psi \\frac{\\pd k}{\\pd z} \\right)\n +\\nu_t M^2+\\nu_t^\\theta N^2 -\\epsilon \\\\\n \\frac{D \\Psi}{D t}&=& \\frac{\\pd }{\\pd z}\\left( \\nu_\\Psi \\frac{\\pd \\Psi}{\\pd z} \\right)\n +\\frac{\\Psi}{k}(c_{\\Psi 1}\\nu_tM^2+c_{\\Psi 3}\\nu_t^\\theta N^2-c_{\\Psi 2}\\epsilon F_{wall})\n\\eeqa\nwith natural b.c.:\n\\beq\n \\left\\{ \\begin{array}{ll}\n \\nu_k^\\Psi \\frac{\\pd k}{\\pd z} &=0, \\mbox{ at } z=-h, \\mbox{ or } \\eta \\\\\n \\nu_\\Psi\\frac{\\pd \\Psi}{\\pd z} &= \\kappa_0 n\\nu_\\Psi\\frac{\\Psi}{l} , \\mbox{ at } z=-h \\\\\n \\nu_\\Psi\\frac{\\pd \\Psi}{\\pd z} &= -\\kappa_0 n\\nu_\\Psi\\frac{\\Psi}{l} , \\mbox{ at } z=\\eta \\\\\n \\end{array}\n \\right. \\label{tur1}\n\\eeq\nand essential b.c.:\n\\beq\n \\left\\{ \\begin{array}{ll}\n k&=(c_\\mu^0)^{-2} \\nu|\\frac{\\pd \\bs{u}}{\\pd z}|, \\mbox{ at } z=-h, \\mbox{ or } \\eta\\\\\n l&=\\kappa_0 \\D \\\\\n \\Psi &= (c_\\mu^0)^pk^m(\\kappa_0 \\D)^n\n \\end{array}\n \\right. \\label{tur2}\n\\eeq\nwhere $k$ is the turbulent kinetic energy, $l$ is the mixing length, $c_{\\Psi *}$ are some constants and $\\Psi=(c_\\mu^0)^pk^m l^n$ is a generic\nlength-scale variable, and $\\D$ is the distance from 'walls' (i.e. surface and bottom). \n\nThe turbulence production and dissipation terms are:\n\\beqa\n M^2&=&\\left( \\frac{\\pd u}{\\pd z}\\right)^2+\\left( \\frac{\\pd v}{\\pd z}\\right)^2 \\\\\n N^2 &=&-\\frac{g}{\\rho_0}\\frac{\\pd \\rho}{\\pd z} \\\\\n \\epsilon &=& (c_\\mu^0)^3k^{1.5} l^{-1}\n\\eeqa\n\nIn the code, the natural b.c. is applied first (see the FEM formulation below), and the essential b.c. is then used to overwrite the\nboundary values of the unknown, as in the GOTM code. SELFE has been directly coupled to the GOTM model to take advantage of the latter.\n\nOnce the equations for turbulent kinetic energy mixing length have been updated, the \nvalues of eddy viscosity and eddy diffusivity used in the momentum and transport equations are obtained from the relations: \n\\beqa\n \\nu &=& \\sqrt{2k}s_m l \\\\\n \\kappa &=& \\sqrt{2k}s_h l\n\\eeqa\nwhere $s_m$ and $s_h$ are stability functions, such as those given by \\citet{Kantha94}\n\nWith this step, the specification of turbulent mixing (and indirectly the mechanism of friction) is complete.\n\n\n\n", "meta": {"hexsha": "cd72904c897777eb8c16ae5b7d7d2c96725a60d2", "size": 10129, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documents/form_hydro.tex", "max_stars_repo_name": "water-e/BayDeltaSCHISM", "max_stars_repo_head_hexsha": "b532b51ef58a6ef3dbb4e74f82008a46db0f7686", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-10-15T20:59:16.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-15T20:59:16.000Z", "max_issues_repo_path": "documents/form_hydro.tex", "max_issues_repo_name": "water-e/BayDeltaSCHISM", "max_issues_repo_head_hexsha": "b532b51ef58a6ef3dbb4e74f82008a46db0f7686", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 17, "max_issues_repo_issues_event_min_datetime": "2018-06-05T16:01:48.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-20T18:52:48.000Z", "max_forks_repo_path": "documents/form_hydro.tex", "max_forks_repo_name": "water-e/BayDeltaSCHISM", "max_forks_repo_head_hexsha": "b532b51ef58a6ef3dbb4e74f82008a46db0f7686", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 7, "max_forks_repo_forks_event_min_datetime": "2018-06-04T16:45:56.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-29T23:01:47.000Z", "avg_line_length": 67.0794701987, "max_line_length": 336, "alphanum_fraction": 0.7200118472, "num_tokens": 2897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677583778258, "lm_q2_score": 0.7662936377487304, "lm_q1q2_score": 0.6497922982609806}} {"text": "% CReviewofCalculus.tex\n% Fund Science! & Help Ernest finish his Physics Research! : quantum super-A-polynomials - a thesis by Ernest Yeung\n% \n% http://igg.me/at/ernestyalumni2014 \n% \n% Facebook : ernestyalumni \n% github : ernestyalumni \n% gmail : ernestyalumni \n% google : ernestyalumni \n% linkedin : ernestyalumni \n% tumblr : ernestyalumni \n% twitter : ernestyalumni \n% youtube : ernestyalumni \n% indiegogo : ernestyalumni \n%\n% Ernest Yeung was supported by Mr. and Mrs. C.W. Yeung, Prof. Robert A. Rosenstone, Michael Drown, Arvid Kingl, Mr. and Mrs. Valerie Cheng, and the Foundation for Polish Sciences, Warsaw University. \n\n\\section{Review of Calculus}\n\n\\subsection*{Total and Partial Derivatives}\n\n\\begin{proposition}[C.3] (The Chain Rule for Total Derivatives)\n\nSuppose $V,W,X$ finite-dim. vector spaces\n\nopen $\\begin{aligned}\n & \\quad \\\\ \n & U \\subseteq V \\\\\n & \\widetilde{U} \\subseteq W \\end{aligned}$\n\nmaps $\\begin{aligned}\n & \\quad \\\\\n & F : U \\to \\widetilde{U} \\\\\n & G: \\widetilde{U} \\to X \\end{aligned}$\n\nif $F$ diff. at $a\\in U$, $G$ diff. at $F(a) \\in \\widetilde{U}$, \\\\\n\\phantom{ \\quad } then $GF$ diff. at $a$\n\\[\nD(GF)(a) = DG(F(a)) \\circ DF(a)\n\\]\n\\end{proposition}\n\n\\begin{proposition}[C.4] Suppose $\\begin{aligned}\n & \\quad \\\\ \n & U \\subseteq \\mathbb{R}^n \\\\\n & V\\subseteq \\mathbb{R}^m \\end{aligned}$, \\quad \\, $F: U \\to V$ diffeomorphism \\\\\nThen $m=n$ and \\\\\n\\phantom{ \\quad } $\\forall \\, a \\in U$, $DF(a)$ invertible, with \n\\[\nDF(a)^{-1} = D(F^{-1})(F(a))\n\\]\n\\end{proposition}\n\n\\begin{proof}\n$F^{-1}F = 1_U$ \\\\\nChain rule implies $\\forall \\, a \\in U$, \n\n\\begin{equation}\n 1_{\\mathbb{R}^n} = D(1_U)(a) = D(F^{-1}F)(a) = DF^{-1}(F(a))DF(a) \\quad \\quad \\, (C.5)\n\\end{equation}\n\nSimilarly $FF^{-1}=1_V$ implies\n\\[\nDF(a)DF^{-1}(F(a)) = 1_{\\mathbb{R}^m}\n\\]\nso thus\n\\[\nDF(a)^{-1}=D(F^{-1})(F(a))\n\\]\n\\end{proof}\n\n\n", "meta": {"hexsha": "775374dc508000da7eeb4de3ff8fbe3807ba0a04", "size": 2642, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "LeeJM/CReviewofCalculus.tex", "max_stars_repo_name": "wacfeldwang333/mathphysics", "max_stars_repo_head_hexsha": "59eb794dfa46e2b80e43df0440bb8ec3c472d973", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2017-01-10T14:24:13.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-29T11:19:23.000Z", "max_issues_repo_path": "LeeJM/CReviewofCalculus.tex", "max_issues_repo_name": "wacfeldwang333/mathphysics", "max_issues_repo_head_hexsha": "59eb794dfa46e2b80e43df0440bb8ec3c472d973", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2017-09-29T09:29:53.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-12T03:12:29.000Z", "max_forks_repo_path": "LeeJM/CReviewofCalculus.tex", "max_forks_repo_name": "wacfeldwang333/mathphysics", "max_forks_repo_head_hexsha": "59eb794dfa46e2b80e43df0440bb8ec3c472d973", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 25, "max_forks_repo_forks_event_min_datetime": "2018-01-21T05:33:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-03T20:15:13.000Z", "avg_line_length": 36.1917808219, "max_line_length": 217, "alphanum_fraction": 0.4375473126, "num_tokens": 705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256512199033, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.649757655208227}} {"text": "\\section{Introduction}\nFluid viscosity is one of the most important properties of fluids, determining\nthe fluid’s flow. \nMotion of an object in a fluid is hindered by a drag force acting in the\ndirection opposite to the direction of motion, i.e. opposite to the object’s\nvelocity. \n\nThe magnitude of the drag force is related to the shape and speed of the object\nas well as to the internal friction in the fluid.\nThe method used in this lab is known as Stokes’ method and it a common and\nsimple method for characterizing transparent or translucent fluids with high\nviscosity.\n\n\n% theoretical\nMotion of an object in a fluid is hindered by a drag force acting in the\ndirection opposite to the direction of motion, i.e. opposite to the object’s\nvelocity. \nThe magnitude of the drag force is related to the shape and speed of the object\nas well as to the internal friction in the fluid.\n\nThis internal friction can be quantified by a number known as the viscosity\ncoefficient $\\mu$.\nFor a spherical object with radius R moving at speed v in an infinite volume of\na liquid, the magnitude of the drag force is usually modeled as linear in the\nspeed\n$$ F_1 = 6 \\pi \\mu v R $$\n\nWhen a spherical object falls vertically downwards in a fluid, it is being acted\nupon by the following three forces:\nThe viscous force \\emph{$F_1$} and the buoyancy force \\emph{$F_2$} both act\nupwards, and the weight of the object \\emph{$F_3$} is directed downwards.\nThe magnitude of the buoyancy force is\n$$ F_2 = \\frac{4}{3} \\pi R^3 \\rho_1 g $$\nwhere $\\rho_1$ is the density of the fluid and $g$ is the acceleration due to\ngravity. The weight of the object\n$$ F_3 = \\frac{4}{3} \\pi R^3 \\rho_2 g $$\nwith $\\rho_2$ being the density of the object. After some time, the three forces\nwill balance each other\n$$ F_1 + F_2 = F_3 $$\nso that the net force on the object will be zero and from that instant on, the\nobject will be moving with constant speed $v_t$, known as the terminal speed.\nApplying the condition, we can find\n$$ \\mu = \\frac{2}{9} g R^2 \\frac{\\rho_2 - \\rho_1 }{v_t} $$\nTherefore, the fluid viscosity can be found by measuring the terminal speed.\nTaking into account that the motion with terminal speed is a motion with\nconstant velocity, the equation can be rewritten as\n$$ \\mu = \\frac{2}{9} g R^2 \\frac{( \\rho_2 - \\rho_1 ) t }{s} $$\nwhere $s$ is the distance traveled in time $t$ after reaching the terminal\nspeed.\n\nSince the volume of the fluid used in the measurement is not infinite, the\nresults are affected by some boundary effects due to the presence of the\ncontainer.\n\nTherefore, the equation should be modified, and the formula for the\ncorrected magnitude of the viscous force for a infinitely long cylindrical\ncontainer with radius $R_c$ is \n$$ F_1 = 6 \\pi \\mu v R (1 + 2.4 \\frac{R}{R_c}) $$\nConsequently,\n$$ \\mu = \\frac{2}{9} g R^2 \\frac{( \\rho_2 - \\rho_1 ) t }{s} (1 + 2.4\n\\frac{R}{R_c}) $$\nSince the length $L$ of the container is limited, there may be further\ncorrections introduced, depending on the ratio on $\\frac{R_c}{L}$. \n\n\n\n\n\n", "meta": {"hexsha": "a530abc8ab3405277a8ec5d11038051caa1036e7", "size": 3042, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "E2/part/1i.tex", "max_stars_repo_name": "iamwrm/VP141", "max_stars_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-06-24T11:28:04.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-24T11:28:04.000Z", "max_issues_repo_path": "E2/part/1i.tex", "max_issues_repo_name": "iamwrm/VP141", "max_issues_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "E2/part/1i.tex", "max_forks_repo_name": "iamwrm/VP141", "max_forks_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.4571428571, "max_line_length": 80, "alphanum_fraction": 0.7406311637, "num_tokens": 843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.8418256452674008, "lm_q1q2_score": 0.6497576506138266}} {"text": "% !TeX root = constructions.tex\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\chapter{Lill's Method for Finding Roots}\\label{c.lill}\n\n\\section{Magic}\\label{s.magic}\n\nConstruct a path consisting of four line segments $\\{a_3,a_2,a_1,a_0\\}$ of lengths:\n\\[\n\\{a_3=1,a_2=6,a_1=11,a_0=6\\}\\,,\n\\]\nstarting from the origin, first along the positive direciton of the $x$-axis and turning $90^\\circ$ counterclockwise between segments. Construct a second path as follows: draw a line from the origin at an angle of $63.4^\\circ$ and mark its intersection with $a_2$ by $P$. Turn left $90^\\circ$, draw a line and and mark its intersection with $a_1$ by $Q$. Turn left $90^\\circ$ once again, draw a line and note that it intersects the end of the first path at $(-10,0)$. %See Figure~\\ref{fig.magic}.\n\n\\begin{center}\n\\begin{tikzpicture}[scale=1.15]\n% Draw help lines and axes\n\\draw[step=10mm,white!50!black] (-11,-1) grid (2,7);\n\\draw[thick] (-11,0) -- (2,0);\n\\draw[thick] (0,-1) -- (0,7);\n\\foreach \\x in {-10,...,2}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {1,...,7}\n \\node at (-.2,\\y-.3) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,6);\n\\coordinate (D) at (-10,6);\n\\coordinate (E) at (-10,0);\n\\foreach \\x in {A,B,C,D,E}\n \\fill (\\x) circle(1.5pt);\n\\draw[very thick] (A) --\n node[below,xshift=1pt,yshift=-10pt] {$a_3=1$} (B);\n\\draw[very thick,name path=bc] (B) -- \n node[right,yshift=6pt] {$a_2=6$} (C);\n\\draw[very thick,name path=cd] (C) --\n node[above,xshift=4pt] {$a_1=11$}(D);\n\\draw[very thick,name path=de] (D) --\n node[left,xshift=3pt,yshift=6pt] {$a_0=6$}(E);\n\n% Draw first segment of second path\n\\path[name path=a2] (A) -- +(63.4:4);\n\\path [name intersections = {of = a2 and bc, by = {A2}}];\n\\fill (A2) circle(1.5pt) node[above right] {$P$};\n\\draw[very thick,dashed] (A) -- (A2);\n\\draw ($(A) + (14pt,0)$)\n arc [start angle=0, end angle = 63.4, radius=14pt];\n\\node[above right,xshift=35pt,yshift=2pt] at (A) {$63.4^\\circ$};\n\\draw[->] ($(A)+(32pt,8pt)$) -- +(-18pt,0);\n\\draw[rotate=153.4] (A2) rectangle +(10pt,10pt);\n\n% Draw second segment of second path\n\\path[name path=b2] (A2) -- +(153.4:10);\n\\path [name intersections = {of = b2 and cd, by = {B2}}];\n\\fill (B2) circle(1.5pt) node[above left] {$Q$};\n\\draw[very thick,dashed] (A2) -- (B2);\n\\draw[rotate=243.4] (B2) rectangle +(10pt,10pt);\n\n% Draw third segment of second path%\n\\draw[very thick,dashed] (B2) -- (E);\n\\end{tikzpicture}\n\\end{center}\nLet $p(x)=a_3x^3+a_2x^2+a_1x+a_0=x^3+6x^2+11x+6$. Compute $\\tan 63.4^\\circ=2$, the tangent of the angle at the start of the second path. Then:\n\\[\np(-\\tan 63.4^\\circ)=(-2)^3+6(-2)^2+11(-2)+6=0\\,.\n\\]\nCongratulations! You have found a root of the cubic polynomial $x^3+6x^2+11x+6$.\n\n\\newpage\n\n\\section{Introduction}\nThis example demonstrates a method discovered by Eduard Lill in 1867 for graphically finding (or more accurately, verifying) the real roots of any polynomial \\cite{bradford, hull-beloch, riaz}. We limit the presentation to cubic polynomials. Lill's method has seen renewed interest because it can be implemented using origami, as we shall see in Chapter~\\ref{c.beloch}.\n\nIn Sections~\\ref{s.multiple}--\\ref{s.noroots} we continue the initial example to find additional roots and to show that if an angle $\\alpha$ is such that $(-\\tan\\alpha)$ is \\emph{not} a root, then the construction doesn't work. Section~\\ref{s.method} presents the full specification of Lill's method. Special cases of the method are demonstrated by the examples in Sections~\\ref{s.negative}--\\ref{s.noninteger}. Since Lill's method can find a real root of any cubic polynomial, it can be used to trisect an angle. By computing $\\sqrt[3]{2}$ as a root of $x^3-2$, it can double a cube (Section~\\ref{s.cube}). Section~\\ref{s.proof} gives a proof that Lill's method can find the real roots of any cubic polynomial. The proof for arbitrary polynomials is similar.\n\n%\\newpage\n\n\\section{Multiple roots}\\label{s.multiple}\n\nLet us continue the example above. The polynomial $p(x)=x^3+6x^2+11x+6$ has three roots $-1,-2,-3$. Computing the arc tangent of the negation of the roots gives:\n\\[\n\\alpha=-\\tan^{-1} (-1) = 45^\\circ,\\quad \\beta=-\\tan^{-1}(-2) = 63.4^\\circ,\\quad \\gamma=-\\tan^{-1} (-3)= 71.6^\\circ\\,.\n\\]\nIn the diagram below we see that for each of the three angles, the second path intersects the end of the first path.\n\\begin{center}\n\\begin{tikzpicture}[scale=1.15]\n% Draw help lines and axes\n\\draw[step=10mm,white!50!black] (-11,-1) grid (2,7);\n\\draw[thick] (-11,0) -- (2,0);\n\\draw[thick] (0,-1) -- (0,7);\n\\foreach \\x in {-10,...,2}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {1,...,7}\n \\node at (-.2,\\y-.3) {\\sm{\\y}};\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,6);\n\\coordinate (D) at (-10,6);\n\\coordinate (E) at (-10,0);\n\\foreach \\x in {A,B,C,D,E}\n \\fill (\\x) circle(1.5pt);\n\\draw[very thick] (A) --\n node[below,yshift=-5pt] {$1$} (B);\n\\draw[very thick,name path=bc] (B) -- \n node[right,yshift=24pt] {$6$} (C);\n\\draw[very thick,name path=cd] (C) --\n node[above] {$11$}(D);\n\\path[name path=de] (D) -- ($(E)+(0,-.8)$);\n\\draw[very thick] (D) --\n node[left,yshift=6pt] {$6$} (E);\n\n\n% Draw first segment of first second path\n\\path[name path=a1] (A) -- +(45:3);\n\\path [name intersections = {of = a1 and bc, by = {A1}}];\n\\fill (A1) circle(1.5pt) node[above right] {$P_1$};\n\\draw[very thick,dashed] (A) -- (A1);\n\\draw[thick] ($(A) + (16pt,0)$)\n arc [start angle=0, end angle = 45, radius=16pt];\n\\node[above right,xshift=44pt,yshift=0pt] at (A) {$\\alpha$};\n\\draw[rotate=135] (A1) rectangle +(10pt,10pt);\n\\draw[Stealth-,thick] ($(A) + (16pt,6pt)$) -- +(24pt,0);\n\n% Draw second segment of first second path\n\\path[name path=b1] (A1) -- +(135:8);\n\\path [name intersections = {of = b1 and cd, by = {B1}}];\n\\fill (B1) circle(1.5pt) node[above right] {$Q_1$};\n\\draw[very thick,dashed] (A1) -- (B1);\n\\draw[rotate=225] (B1) rectangle +(10pt,10pt);\n\n% Draw third segment of first second path\n\\draw[very thick,dashed] (B1) -- (E);\n\n% Draw first segment of second second path\n\\path[name path=a2] (A) -- +(63.4:4);\n\\path [name intersections = {of = a2 and bc, by = {A2}}];\n\\fill (A2) circle(1.5pt) node[above right] {$P_2$};\n\\draw[very thick,dashed] (A) -- (A2);\n\\draw[thick] ($(A) + (24pt,0)$)\n arc [start angle=0, end angle = 63.4, radius=24pt];\n\\node[above right,xshift=44pt,yshift=8pt] at (A) {$\\beta$};\n\\draw[rotate=153.4] (A2) rectangle +(10pt,10pt);\n\\draw[<-,thick] ($(A) + (22pt,14pt)$) -- +(18pt,0);\n\n% Draw second segment of second second path%\n\\path[name path=b2] (A2) -- +(153.4:10);\n\\path [name intersections = {of = b2 and cd, by = {B2}}];\n\\fill (B2) circle(1.5pt) node[above right] {$Q_2$};\n\\draw[very thick,dashed] (A2) -- (B2);\n\\draw[rotate=243.4] (B2) rectangle +(10pt,10pt);\n\n% Draw third segment of second second path%\n\\draw[very thick,dashed] (B2) -- (E);\n\n% Draw first segment of second second path%\n\\path[name path=a3] (A) -- +(71.6:4);\n\\path [name intersections = {of = a3 and bc, by = {A3}}];\n\\fill (A3) circle(1.5pt) node[above right] {$P_3$};\n\\draw[very thick,dashed] (A) -- (A3);\n\\draw[thick] ($(A) + (38pt,0)$)\n arc [start angle=0, end angle = 71.6, radius=40pt];\n\\node[above right,xshift=44pt,yshift=22pt] at (A) {$\\gamma$};\n\\draw[rotate=161.6] (A3) rectangle +(10pt,10pt);\n\\draw[<-,thick] ($(A) + (32pt,25pt)$) -- +(10pt,0);\n\n% Draw second segment of second second path%\n\\path[name path=b3] (A3) -- +(161.6:10);\n\\path [name intersections = {of = b3 and cd, by = {B3}}];\n\\fill (B3) circle(1.5pt) node[above right] {$Q_3$};\n\\draw[very thick,dashed] (A3) -- (B3);\n\\draw[rotate=251.6] (B3) rectangle +(10pt,10pt);\n\n% Draw third segment of second second path%\n\\draw[very thick,dashed] (B3) -- (E);\n\\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\n\\section{Paths that do not lead to roots}\\label{s.noroots}\n\nPerhaps the second path intersects the end of the first path for \\emph{any} initial angle, for example, $56.3^\\circ$. In the following diagram, the second path intersects the extension of the line segment for the coefficient $a_0$, but not at $(-10,0)$, the end of the first path. We conclude that $-\\tan 56.3^\\circ =-1.5$ is \\emph{not} a root of the equation.\n\\begin{center}\n\\begin{tikzpicture}[scale=1.15]\n% Draw help lines and axes\n\\draw[step=10mm,white!50!black] (-11,-1) grid (2,7);\n\\draw[thick] (-11,0) -- (2,0);\n\\draw[thick] (0,-1) -- (0,7);\n\\foreach \\x in {-10,...,2}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {1,...,7}\n \\node at (-.2,\\y-.3) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,6);\n\\coordinate (D) at (-10,6);\n\\coordinate (E) at (-10,0);\n\\foreach \\x in {A,B,C,D,E}\n \\fill (\\x) circle(1.5pt);\n\\draw[very thick] (A) --\n node[below,yshift=-5pt] {$1$} (B);\n\\draw[very thick,name path=bc] (B) -- \n node[right,yshift=6pt] {$6$} (C);\n\\draw[very thick,name path=cd] (C) --\n node[above] {$11$}(D);\n\\draw[very thick] (D) --\n node[left,yshift=6pt] {$6$}(E);\n\\path[name path=de] (-10,-1) -- (-10,7);\n\n% Draw first segment of second path\n\\path[name path=a2] (A) -- +(56.3:3);\n\\path [name intersections = {of = a2 and bc, by = {A2}}];\n\\fill (A2) circle(1.5pt) node[above right] {$P$};\n\\draw[very thick,dashed] (A) -- (A2);\n\\draw ($(A) + (14pt,0)$)\n arc [start angle=0, end angle = 56.3, radius=14pt];\n\\node[above right,xshift=10pt,yshift=6pt] at (A) {$56.3^\\circ$};\n\\draw[rotate=146.3] (A2) rectangle +(10pt,10pt);\n\n% Draw second segment of second path\n\\path[name path=b2] (A2) -- +(146.3:10);\n\\path [name intersections = {of = b2 and cd, by = {B2}}];\n\\fill (B2) circle(1.5pt) node[above right] {$Q$};\n\\draw[very thick,dashed] (A2) -- (B2);\n\\draw[rotate=236.3] (B2) rectangle +(10pt,10pt);\n\n% Draw third segment of second path\n\\path[name path=c2] (B2) -- +(236.3:8.5);\n\\path [name intersections = {of = c2 and de, by = {C2}}];\n\\fill (C2) circle(1.5pt);\n\\draw[very thick,dashed] (B2) -- (C2);\n\\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\n\\section{Specification of Lill's method}\\label{s.method}\n\nExamining the examples below will help understand the details.\n\\begin{itemize}\n\\item Start with an arbitrary cubic polynomial: $p(x)=a_3x^3+a_2x^2+a_1x+a_0$.\n\\item Construct the first path as follows:\n\\begin{itemize}\n\\item For each coefficient $a_3,a_2,a_1,a_0$ (in that order) draw a line segment starting at the origin $O=(0,0)$ in the positive direction of the $x$-axis. Turn $90^\\circ$ counterclockwise between each segment.\n\\end{itemize}\n\\item Construct the second path as follows:\n\\begin{itemize}\n\\item We use the symbol for a coefficient $a_i$ to also denote the corresponding side of the first path.\n\\item Construct a line from $O$ at an angle of $\\theta$ with the positive $x$-axis that intersects $a_2$ at point $P$.\n\\item Turn $\\pm 90^\\circ$ and construct a line from $P$ that intersects $a_1$ at $Q$.\n\\item Turn $\\pm 90^\\circ$ and construct a line from $Q$ that intersects $a_0$ at $R$.\n\\item If $R$ is the end point of the first path, then $-\\tan\\theta$ is a root of $p(x)$.\n\\end{itemize}\n\\item Special cases:\n\\begin{itemize}\n\\item When drawing the line segments of the first path, if a coefficient is negative, draw the line segment \\emph{backwards}.\n\\item When drawing the line segments of the first path, if a coefficient is zero, do not draw a line segment but continue with the next $\\pm 90^\\circ$ turn.\n\\end{itemize}\n\\item Notes:\n\\begin{itemize}\n\\item ``Intersects $a_i$'' means ``intersects the line that contains the line segment $a_i$''.\n\\item When building the second path, choose to turn left or right by $90^\\circ$ so that there is an intersection with the next segment of the first path.\n\\end{itemize}\n\\end{itemize}\n\n\\newpage\n\n\\section{Negative coefficients}\\label{s.negative}\n\nSection~\\ref{s.ax6} gave an example of the use of Axiom~6 that resulted in the polynomial $p(x)=x^3-3x^2-3x+1$ with negative coefficients.\n\nWe start by drawing a segment of length $1$ to the right. Next we turn $90^\\circ$ to face up, but the coefficient is negative, so we draw a segment of length $3$ \\emph{down}. After turning $90^\\circ$ to the left, the coefficient is again negative, so we draw a segment of length $3$ to the right. Finally, we turn down and draw a segment of length $1$.\n\nWe start the second path with a line angled $45^\\circ$ with the $x$-axis. It intersects the \\emph{extension} of the line segment for $a_2$ at $(1,1)$. Turning $-90^\\circ$ (to the right), the line intersects the \\emph{extension} of the line segment for $a_1$ at $(5,-3)$. Turning $-90^\\circ$ again, the line intersects the end of the first path at $(4,-4)$.\n\nSince $-\\tan 45^\\circ=-1$, a real root of the polynomial is $-1$:\n\\[\np(-1)=(-1)^3-3(-1)^2-3(-1)+6=0\\,.\n\\]\nThe loosely dashed lines in the diagram will be discussed in Section~\\ref{s.noninteger}.\n\n\\begin{center}\n\\begin{tikzpicture}[scale=1.1]\n% Draw help lines and axes\n\\draw[step=10mm,white!50!black] (-1,-6) grid (6,2);\n\\draw (0,-6) -- (0,2);\n\\foreach \\x in {0,...,6}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {-5,...,-1}\n \\node at (-.3,\\y-.3) {\\sm{\\y}};\n\\foreach \\y in {1,...,2}\n \\node at (-.3,\\y-.3) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,-3);\n\\coordinate (D) at (4,-3);\n\\coordinate (E) at (4,-4);\n\\foreach \\x in {A,B,C,D,E}\n \\fill (\\x) circle(1pt);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt,reversed]}-] (A) --\n node[below,yshift=-5pt] {$1$} (B);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt]}-,name path=bc] (B) -- \n node[right,xshift=3pt] {$a_2=-3$} (C);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt]}-,name path=cd] (C) --\n node[above,xshift=11pt] {$a_1=-3$}(D);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt,reversed]}-,name path=de] (D) --\n node[right] {$1$}(E);\n\n% Draw extensions of first path\n\\draw[very thick,loosely dotted,name path=a] (-1,0) -- (6,0);\n\\draw[very thick,loosely dotted,name path=b] (1,-6) -- (1,2);\n\\draw[very thick,loosely dotted,name path=c] (-1,-3) -- (6,-3);\n\n% Draw first second path\n\\path[name path=a1] (A) -- +(-75:5);\n\\path [name intersections = {of = a1 and b, by = {B1}}];\n\\path[name path=b1] (B1) -- +(15:5);\n\\path [name intersections = {of = b1 and c, by = {C1}}];\n\\draw[thick,loosely dashed] (A) -- (B1) -- (C1) -- (E);\n\n% Draw second second path\n\\draw[very thick,dashed] (4,-4) -- (5,-3) coordinate (A2);\n\\fill (5,-3) circle (1.5pt) node[above right] {$P$};\n\\draw[very thick,dashed] (5,-3) -- (1,1) coordinate (B2);\n\\fill (1,1) circle (1.5pt) node[above right] {$Q$};\n\\draw[very thick,dashed] (1,1) -- (0,0);\n\n% Draw third second path\n\\path[name path=a3] (A) -- +(-15:5);\n\\path [name intersections = {of = a3 and b, by = {B3}}];\n\\path[name path=b3] (B3) -- +(-105:5);\n\\path [name intersections = {of = b3 and c, by = {C3}}];\n\\draw[thick,loosely dashed] (A) -- (B3) -- (C3) -- (E);\n\\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\n\\section{Zero coefficients}\\label{s.zero}\n\n$a_2$, the coefficient of the $x^2$ term in the polynomial $x^3-7x-6=0$, is zero. For a zero coefficient, we ``draw'' a line segment of length $0$, that is, we do not draw a line, but we still make the $\\pm 90^\\circ$ turn before ``drawing'' it, as indicated by the arrow pointed up at point $(1,0)$ in the diagram. Next make an additional turn and draw a line of length $-7$, that is, of length $7$ backwards, to point $(8,0)$. Finally, turn again and draw a line of length $-6$ to point $(8,6)$.\n\nThere are three second paths that intersect the end of the first path. They start with angles of:\n\\[\n\\alpha=45^\\circ,\\quad \\beta=63.4^\\circ,\\quad \\gamma=-71.6^\\circ\\,.\n\\]\nWe conclude that there are three real roots:\n\\[\n-\\tan 45^\\circ=-1,\\quad -\\tan 63.4^\\circ =-2,\\quad -\\tan (-71.6^\\circ)=3\\,.\n\\]\nCheck:\n\\[\n(x+1)(x+2)(x-3)=x^3-7x-6\\,.\n\\]\n\n\\begin{center}\n\\begin{tikzpicture}[scale=1.1]\n% Draw help lines and axes\n\\draw[step=10mm,white!50!black] (-1,-4) grid (11,7);\n\\foreach \\x in {0,...,11}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {-3,...,-1}\n \\node at (-.3,\\y-.3) {\\sm{\\y}};\n\\foreach \\y in {1,...,7}\n \\node at (-.3,\\y-.3) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0) node[above left] {$O$};\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (8,0);\n\\coordinate (D) at (8,6);\n\\node[below right] at (D) {$A$};\n\\foreach \\x in {A,B,C,D}\n \\fill (\\x) circle(1.5pt); \n\\draw[very thick,{Stealth[scale=1.4,inset=2pt,reversed]}-] (A) --\n node[below,yshift=-5pt] {$1$} (B);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] (B) --\n ($(B)+(0,.1)$);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt]}-,name path=bc] (B) -- \n node[below,xshift=-6pt,yshift=-5pt] {$-7$} (C);\n\\draw[very thick,{Stealth[scale=1.4,inset=2pt]}-,name path=cd] (C) --\n node[right,yshift=4pt] {$-6$}(D);\n\n% Draw extensions of first path\n\\draw[very thick,loosely dotted] (1,-3) -- (1,7);\n\\draw[very thick,loosely dotted] (-1,0) -- (11,0);\n\n% Draw first second path\n\\draw[very thick,dashed,->] (0,0) -- (1,-3);\n\\fill (1,-3) circle (1.5pt) node[below left] {$P_1$};\n\\draw[very thick,dashed,->] (1,-3) coordinate (A1) -- (10,0);\n\\fill (10,0) circle (1.5pt) node[below right] {$Q_1$};\n\\draw[very thick,dashed,->] (10,0) coordinate (B1) -- (D);\n\n% Draw second second path\n\\draw[very thick,dashed,->] (0,0) -- (1,1) coordinate (A2);\n\\fill (A2) circle (1.5pt) node[above right] at (A2) {$P_2$};\n\\draw[very thick,dashed,->] (A2) -- (2,0) coordinate (B2);\n\\fill (B2) circle (1.5pt) node[below right] at (B2) {$Q_2$};\n\\draw[very thick,dashed,->] (B2) -- (D);\n\n% Draw third second path\n\\draw[very thick,dashed,->] (0,0) -- (1,2) coordinate (A3);\n\\fill (A3) circle (1.5pt) node[above left] {$P_3$};\n\\draw[very thick,dashed,->] (A3) -- (5,0) coordinate (B3);\n\\fill (B3) circle (1.5pt) node[below right] at (B3) {$Q_3$};\n\\draw[very thick,dashed,->] (B3) -- (D);\n\\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\n\\section{Non-integer roots}\\label{s.noninteger}\n\nConsider the polynomial $p(x)=x^3-2x+1$. The first segment is from $(0,0)$ to $(1,0)$ and turns up. The coefficient of $x^2$ is zero so no segment is drawn and turns left. The next coefficient is negative so the segment it goes backwards from $(1,0)$ to $(3,0)$ and turns right. Finally, a segment is drawn from $(3,0)$ to $(3,-1)$. Clearly, $1$ is a root of $p(x)$ and since $-\\tan^{-1} (-45^\\circ)=1$, there is a path $\\overline{OP_1Q_1A}$.\n\n If we divide $p(x)$ by $x-1$, we obtain the quadratic polynomial $x^2+x-1$ whose roots are $\\frac{-1\\pm\\sqrt{5}}{2} \\approx 0.62,\\; -1.62$.\n%\\[\n%\\disfrac{-1\\pm\\sqrt{5}}{2} \\approx 0.62,\\; -1.62\\,.\n%\\]\nThere are two additional second paths: one starting at $-31.8^\\circ$ since $-\\tan^{-1} 0.62=-31.8^\\circ$, and one starting at $58.3^\\circ$ since $-\\tan^{-1}1.62=58.3^\\circ$.\n\nSimilarly, the polynomial in Section~\\ref{s.negative} has roots $ 2\\pm\\sqrt{3}\\approx 3.73, 0.27$. The corresponding angles are $-75^\\circ$ and $-15^\\circ$, because $-\\tan (-75^\\circ)\\approx 3.73$ and $-\\tan (-15^\\circ)\\approx 0.27$.\n\\begin{center}\n\\begin{tikzpicture}[scale=1.6]\n% Draw help lines and axes\n\\draw[step=10mm,white!70!black,] (-1,-2) grid (4,2);\n\\foreach \\x in {0,...,4}\n \\node at (\\x-.2,-.1) {\\sm{\\x}};\n\\foreach \\y in {-1}\n \\node at (-.1,\\y-.2) {\\sm{\\y}};\n\\foreach \\y in {1,2}\n \\node at (-.1,\\y-.2) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0);\n\\node[above left] at (A) {$O$};\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (3,0);\n\\coordinate (D) at (3,-1);\n\\node[below right] at (D) {$A$};\n\\foreach \\x in {A,B,C,D}\n \\fill (\\x) circle(1pt); \n\\draw[very thick] (A) -- node[above,yshift=2pt] {$1$} (B);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(A)+(.1,0)$) --\n ($(A)+(.15,0)$);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(B)+(0,.05)$) --\n ($(B)+(0,.1)$);\n\\draw[very thick,name path=bc] (B) -- \n node[above,xshift=-4pt,yshift=2pt] {$-2$} (C);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(B)+(.22,0)$) --\n ($(B)+(.17,0)$);\n\\draw[very thick,name path=cd] (C) --\n node[right] {$1$}(D);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(C)+(0,-.05)$) --\n ($(C)+(0,-.1)$);\n\n% Draw extensions of first path\n\\draw[very thick,loosely dotted,name path=b] (1,-2) -- (1,2);\n\\draw[very thick,loosely dotted,name path=c] (-1,0) -- (4,0);\n\\draw[very thick,loosely dotted,name path=d] (3,-2) -- (3,2);\n\n% Draw first second path\n\\coordinate (A1) at (1,-1);\n\\draw[very thick,dashed,->] (0,0) -- (A1);\n\\fill (A1) circle (1pt) node[below right] {$P_1$};\n\\coordinate (B1) at (2,0);\n\\draw[very thick,dashed,->] (A1) -- (B1);\n\\fill (B1) circle (1pt) node[above right,xshift=4pt] {$Q_1$};\n\\draw[very thick,dashed,->] (B1) -- (D);\n\\draw[rotate=45] (A1) rectangle +(6pt,6pt);\n\\draw[rotate=-135] (B1) rectangle +(6pt,6pt);\n\n% Draw second second path\n\\path[name path=a2] (0,0) -- +(-31.7:4);\n\\path [name intersections = {of = a2 and b, by = {A2}}];\n\\draw[very thick,dashed,->] (0,0) -- (A2);\n\\fill (A2) circle (1pt) node[below right,yshift=2pt] {$P_2$};\n\\path[name path=b2] (A2) -- +(58.3:2.5);\n\\path [name intersections = {of = b2 and c, by = {B2}}];\n\\draw[very thick,dashed,->] (A2) -- (B2);\n\\fill (B2) circle (1pt) node[above] {$Q_2$};\n\\draw[very thick,dashed,->] (B2) -- (D);\n\\draw[rotate=58.3] (A2) rectangle +(6pt,6pt);\n\\draw[rotate=-121.7] (B2) rectangle +(6pt,6pt);\n\n% Draw third second path\n\\path[name path=a3] (0,0) -- +(58.3:2.5);\n\\path [name intersections = {of = a3 and b, by = {A3}}];\n\\draw[very thick,dashed,->] (0,0) -- (A3);\n\\fill (A3) circle (1pt) node[above left] {$P_3$};\n\\path[name path=b3] (A3) -- +(-31.7:4);\n\\path [name intersections = {of = b3 and c, by = {B3}}];\n\\draw[very thick,dashed,->] (A3) -- (B3);\n\\fill (B3) circle (1pt) node[above right] {$Q_3$};\n\\path[name path=c3] (B3) -- +(-121.7:4);\n\\draw[very thick,dashed,->] (B3) -- (D);\n\\draw[rotate=-121.7] (A3) rectangle +(6pt,6pt);\n\\draw[rotate=-211.7] (B3) rectangle +(6pt,6pt);\n\\end{tikzpicture}\n\\end{center}\n\n%\\newpage\n\n\\vspace{-13ex}\n\n\\section{The cube root of two}\\label{s.cube}\n\n$\\sqrt[3]{2}$ is a root of the cubic polynomial $x^3-2$. In the first path, we turn left twice without drawing any line segments, because $a_2$ and $a_1$ are both zero. Then we turn left again (to face down) and draw backwards because $a_0=-2$ is negative. The first segment of the second path is drawn at an angle of $-51.6^\\circ$ and $-\\tan (-51.6^\\circ)\\approx 1.26\\approx \\sqrt[3]{2}$.\n\\begin{center}\n\\begin{tikzpicture}[scale=1.2]\n% Draw help lines and axes\n\\draw[step=10mm,white!70!black,] (-1,-2) grid (3,2);\n\\foreach \\x in {0,...,3}\n \\node at (\\x-.2,-.1) {\\sm{\\x}};\n\\foreach \\y in {-1}\n \\node at (-.2,\\y-.2) {\\sm{\\y}};\n\\foreach \\y in {1,2}\n \\node at (-.2,\\y-.2) {\\sm{\\y}};\n\n% Draw first path\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,2);\n\\foreach \\x in {A,B,C}\n \\fill (\\x) circle(1pt); \n\\draw[very thick] (A) -- node[above,yshift=2pt] {$1$} (B);\n\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(A)+(.05,0)$) --\n ($(A)+(.1,0)$);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(B)+(0,.05)$) --\n ($(B)+(0,.1)$);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(B)+(.1,.3)$) --\n ($(B)+(.08,.3)$);\n\\draw[{Stealth[scale=1.4,inset=2pt,reversed]}-,very thick] ($(B)+(0,.55)$) --\n ($(B)+(0,.5)$);\n\n\\draw[very thick] (B) -- \n node[left,yshift=6pt] {$-2$} (C);\n\n% Draw extensions of first path\n\\draw[very thick,loosely dotted,name path=a] (-1,0) -- (3,0);\n\\draw[very thick,loosely dotted,name path=b] (1,-2) -- (1,2);\n\n% Draw first segment of second path\n\\path[name path=a1] (0,0) -- +(-51.6:2);\n\\path [name intersections = {of = a1 and b, by = {A1}}];\n\\draw[very thick,dashed,->] (A) -- (A1);\n\\fill (A1) circle (1pt) node[below left] {$P_1$};\n\\draw[rotate=38.4] (A1) rectangle +(6pt,6pt);\n\n% Draw second segment of second path\n\\path[name path=b1] (A1) -- +(38.4:2.5);\n\\path [name intersections = {of = b1 and a, by = {B1}}];\n\\draw[very thick,dashed,->] (A1) -- (B1);\n\\fill (B1) circle (1pt) node[above right] {$Q_1$};\n\\draw[rotate=128.4] (B1) rectangle +(7pt,7pt);\n\n% Draw third segement of second path\n\\draw[very thick,dashed,->] (B1) -- (C);\n\\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\n\\section{Proof of Lill's method}\\label{s.proof}\n\nWe limit ourselves to monic cubic polynomials $p(x)=x^3+a_2x^2+a_1x+a_0$.\\footnote{If the polynomial is not monic, divide it by $a_3$ and the resulting monic polynomial has the same roots.} In the diagram below, segments of the first path are labeled with coefficients and with $b_2,b_1,a_2-b_2,a_1-b_1$.\n\nSince the sum of the angles of a triangle is $180^\\circ$, in a right triangle if one acute angle is $\\theta$, the other is $90^\\circ-\\theta$. Therefore, the angle above $P$ and the angle to the left of $Q$ are equal to $\\theta$. We now derive a sequence of formulas for $\\tan \\theta$:\n\\vspace{-2ex}\n\\begin{form}{1.1}\n\\tan \\theta &=& \\disfrac{b_2}{1}=b_2\\\\\n%a_2-b_2&=&a_2-\\tan\\theta\\\\\n\\tan \\theta &=& \\disfrac{b_1}{a_2-b_2}=\\disfrac{b_1}{a_2-\\tan\\theta}\\\\\nb_1&=&\\tan\\theta (a_2-\\tan\\theta)\\\\\n\\tan \\theta &=& \\disfrac{a_0}{a_1-b_1}=\\disfrac{a_0}{a_1-\\tan\\theta(a_2-\\tan\\theta)}\\,.\n\\end{form}\nSimplifying the last equation gives:\n\\vspace{-2ex}\n\\begin{form}{1}\n(\\tan\\theta)^3-a_2(\\tan\\theta)^2+a_1(\\tan\\theta)-a_0&=&0\\\\\n-(\\tan\\theta)^3+a_2(\\tan\\theta)^2-a_1(\\tan\\theta)+a_0&=&0\\\\\n(-\\tan\\theta)^3+a_2(-\\tan\\theta)^2+a_1(-\\tan\\theta)+a_0&=&0\\,.\n\\end{form}\nWe conclude that $-\\tan\\theta$ is a real root of $p(x)=x^3+a_2x^2+a_1x+a_0$.\n\n\n\\begin{center}\n\\begin{tikzpicture}[scale=1.15]\n% Draw grid and axes\n\\draw[step=10mm,white!50!black] (-11,-1) grid (2,7);\n\\draw[thick] (-11,0) -- (2,0);\n\\draw[thick] (0,-1) -- (0,7);\n\\foreach \\x in {-10,...,2}\n \\node at (\\x-.3,-.2) {\\sm{\\x}};\n\\foreach \\y in {1,...,7}\n \\node at (-.2,\\y-.3) {\\sm{\\y}};\n \n% Draw the points of the first path\n\\coordinate (A) at (0,0);\n\\coordinate (B) at (1,0);\n\\coordinate (C) at (1,6);\n\\coordinate (D) at (-10,6);\n\\coordinate (E) at (-10,0);\n\\foreach \\x in {A,B,C,D,E}\n \\fill (\\x) circle(2pt);\n\\draw[rotate=90] (B) rectangle +(10pt,10pt);\n \n% Draw A -- B and arrow\n\\draw[very thick] (A) --(B);\n\\draw[thick,<->] ($(A)+(0,-16pt)$) --\n node[fill=white] {$1$} ($(B)+(0,-16pt)$);\n\n% Draw B -- C and arrow\n\\draw[very thick,name path=bc] (B) -- (C);\n\\draw[thick,<->] ($(B)+(36pt,0)$) --\n node[fill=white] {$a_2$} ($(C)+(36pt,0)$);\n\n% Draw C -- D and arrow\n\\draw[very thick,name path=cd] (C) --(D);\n\\draw[thick,<->] ($(C)+(0,24pt)$) -- \n node[fill=white] {$a_1$} ($(D)+(0,24pt)$);\n\n% Draw D -- E and arrow\n\\draw[very thick,name path=de] (D) -- (E);\n\\draw[thick,<->] ($(D)+(-16pt,0)$) --\n node[fill=white] {$a_0$} ($(E)+(-16pt,0)$);\n\n% Draw first angled segment of the second path and intersection A2 with BC\n\\path[name path=a2] (A) -- +(63.4:4);\n\\path [name intersections = {of = a2 and bc, by = {A2}}];\n\\fill (A2) circle(2pt) node[above right] {$P$};\n\\draw[very thick,dashed] (A) -- (A2);\n\\path (B) -- node[right] {$b_2$} (A2);\n\\path (A2) -- node[right,yshift=8pt] {$a_2-b_2$} (C);\n\\draw[rotate=153.4] (A2) rectangle +(10pt,10pt);\n\n% Draw second segment of the second path and intersection B2 with CD\n\\path[name path=b2] (A2) -- +(153.4:10);\n\\path [name intersections = {of = b2 and cd, by = {B2}}];\n\\fill (B2) circle(2pt) node[above right] {$Q$};\n\\draw[very thick,dashed] (A2) -- (B2);\n\\draw[rotate=243.4] (B2) rectangle +(10pt,10pt);\n\\path (D) -- node[above] {$a_1-b_1$} (B2); \n\\path (B2) -- node[above] {$b_1$} (C);\n\n% Draw third segment of the second path to E\n\\draw[very thick,dashed] (B2)-- (E);\n\n% Label A, A2, B2 with theta\n\\draw ($(A) + (14pt,0)$)\n arc [start angle=0, end angle = 63.4, radius=14pt];\n\\node[above right,xshift=10pt,yshift=8pt] at (A) {$\\theta$};\n\\draw ($(A2) + (0,14pt)$)\n arc [start angle=90, end angle = 153.4, radius=14pt];\n\\node[above left,xshift=-4pt,yshift=14pt] at (A2) {$\\theta$};\n\\draw ($(B2) + (-14pt,0)$)\n arc [start angle=180, end angle = 243.4, radius=14pt];\n\\node[below left,xshift=-14pt,yshift=-4pt] at (B2) {$\\theta$};\n\\end{tikzpicture}\n\\end{center}\n", "meta": {"hexsha": "d0802e9bdd032bee0c39271ef3cc7aa77b7968e0", "size": 27805, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lill.tex", "max_stars_repo_name": "motib/constructions", "max_stars_repo_head_hexsha": "8f8f4f25a91abb31b8392b83802e7f5ed42462c7", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-10-07T15:57:52.000Z", "max_stars_repo_stars_event_max_datetime": "2020-10-07T15:57:52.000Z", "max_issues_repo_path": "lill.tex", "max_issues_repo_name": "motib/constructions", "max_issues_repo_head_hexsha": "8f8f4f25a91abb31b8392b83802e7f5ed42462c7", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lill.tex", "max_forks_repo_name": "motib/constructions", "max_forks_repo_head_hexsha": "8f8f4f25a91abb31b8392b83802e7f5ed42462c7", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.5620328849, "max_line_length": 759, "alphanum_fraction": 0.6275130372, "num_tokens": 10685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702761768249, "lm_q2_score": 0.8962513807543223, "lm_q1q2_score": 0.6496659858912462}} {"text": "\\section{Model and strategy} % (fold)\n\\label{sec:model_and_strategy}\n\n%\\subsection{Board description} % (fold)\n%\\label{sub:board_description}\n\n% \n% \\paragraph{Change of square values and traps definition} % (fold)\n% \\label{par:change_of_square_values_and_traps_definition}\n% \n% % paragraph change_of_square_values_and_traps_definition (end)\n% \n% \\paragraph{No consecutive traps} % (fold)\n% \\label{par:no_consecutive_traps}\n% A first assumption in the way we defined our board concerns the non-existence\n% of domino effect of traps. This means that the only way to activate a trap\n% is to fall on its box immediately after one dice throw.\n% For example, if one lands on box 8, which is a type 3 [??? type 2 if we can not\n% modify the project rules] trap, it brings us back to box 5. In this case,\n% even if box 5 is a trap, it won't affect us until the following dice throw\n% (i.e. if a 0 is made at the following dice throw).\n% % paragraph no_consecutive_traps (end)\n% \n% % subsection board_description (end)\n\n\\subsection{Markov model} % (fold)\n\\label{sub:markov_model}\nLet us first recall the general idea behind Markov decision processes.\nThe objective is to determine the best set of actions $\\{u(1), u(2), \\dots\nu(n)\\}$ that minimizes $V_{\\pi}(s_0)$, the total expected cost of the process.\nThe algorithm we use in order to determine this set is the one based on Bellman equation\n\\[\n V(k) \\leftarrow \\min_{a\\in U(k)} \\{c(a|k) \n + \\sum_{k'=1}^n p(k'|k,a)V(k') \\} \\qquad k \\neq d\n\\]\nwhere $V(d)$ is set to 0 at each iteration, $d$ being the destination state.\nIn this \\emph{value-iteration} algorithm, $c(a|k)$ and $p(k'|k,a)$\nrespectively represent the \\emph{cost} of making action $a$\nwhen being in state $k$ and the \\emph{probability} of going\nfrom state $k$ to $k'$ when action $a$ is triggered.\n\nIt is proven that if we find a set of values $\\{V(1), \\dots, V(n)\\}$\nthat satisfies this set of equations, then it is \\emph{optimal}\nand it can be solved by \\emph{recursion}.\n\nLinking this process to our \\textit{Snakes and Ladders} game,\na state $k$ corresponds to \\emph{being on} the kth square,\nthus taking values in the range 1 to 15.\nWe then define the action $a$ as the \\emph{throw of a dice}\nand since every one of the three dices can be thrown at each\niteration we have $U(1) = \\dots = U(15) = \\{a_{\\text{secu}},\na_{\\text{norm}}, a_{\\text{risk}}\\}$.\nFinally, recalling that the goal of the game is to get to\nthe 15th square by \\emph{minimizing the number of dice throws},\nthe cost $c(a|k)$ equals 1 as each turn will lead in one dice\nthrow, whatever the action $a$ or the state $k$. \n\nAt this point only one last ingredient is missing to start implementing\nthe above described algorithm~: the transition probabilities $p(k'|k,a)$.\nHere lies the first main difficulty of the project as those depends\nboth on the type of square or trap from which one leaves $k$ and\non which one will arrive $k'$, as well as on the dice one\ndecides to use (i.e. $a$).\nIt is now important to \\emph{find an efficient way of constructing\nthe 3 transition probability matrices} $P_{\\text{secu}}$, $P_{\\text{norm}}$\nand $P_{\\text{risk}}$ by minimizing the number of particular cases,\nas one can not even imagine to compute the $3\\times 15^2$ individual probabilities.\n\nThe precise implementation of the probability matrices won't be\ndiscussed in this section but it appeared important to us\nto still make a point on the mathematical idea of our method.\n\\begin{enumerate}\n \\item First we fill the matrix by assuming that\n all squares are normal squares and thus filling the matrices\n with probabilities $\\frac{1}{2}$, $\\frac{1}{3}$ and $\\frac{1}{4}$,\n making particular cases for the slow/fast \\emph{lanes}\n and the \\emph{circularity} of the board.\n \\item Then we re-modify each matrix line by taking \\emph{traps} into account\n and thus re-distributing the traps transition probabilities.\n For example, if there a \\emph{type 2 trap} at square 8 \n and the \\emph{normal} dice is thrown, we make the following adjustments\n for each line $i$.\n \\begin{align*}\n p(i,5) &\\leftarrow p(i,5) + p(i,8)/2 \\\\\n p(i,8) &\\leftarrow p(i,8)/2\n \\end{align*}\n\\end{enumerate}\n\n% subsection markov_model (end)\n\n\\subsection{Finding the optimal dice} % (fold)\n\\label{sub:choosing_the_optimal_dice}\nOnce the algorithm converged and we obtained the optimal \ncost values $\\{V(1), \\dots, V(n)\\}$, the optimal dices that\nlead to these expected cost can be retrieved by applying\nthe following formula\n\\[\n u(k) \\leftarrow \\arg\\min_{a\\in U(k)} \\{c(a|k) \n + \\sum_{k'=1}^n p(k'|k,a)V(k') \\} \\qquad k \\neq d\n\\]\nThis formula returns at each iteration the action\n(i.e. the \\emph{type of dice}) that enables to reach the minimum\nexpected cost contained in the values $\\{V(1), \\dots, V(n)\\}$.\n\n% subsection choosing_the_optimal_dice (end)\n\n% section model_and_strategy (end)", "meta": {"hexsha": "7834ef5e1135b20c7f2c15e5dab109a95b08f2b8", "size": 4843, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/src/model.tex", "max_stars_repo_name": "qlete/markov-decision", "max_stars_repo_head_hexsha": "9043e8e014b165dff2ebe9be77f8630d7b8d2237", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-03-21T13:48:00.000Z", "max_stars_repo_stars_event_max_datetime": "2018-03-21T13:48:00.000Z", "max_issues_repo_path": "report/src/model.tex", "max_issues_repo_name": "qlete/markov-decision", "max_issues_repo_head_hexsha": "9043e8e014b165dff2ebe9be77f8630d7b8d2237", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/src/model.tex", "max_forks_repo_name": "qlete/markov-decision", "max_forks_repo_head_hexsha": "9043e8e014b165dff2ebe9be77f8630d7b8d2237", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.261682243, "max_line_length": 88, "alphanum_fraction": 0.7257897997, "num_tokens": 1417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8962513703624558, "lm_q2_score": 0.7248702761768249, "lm_q1q2_score": 0.6496659783584912}} {"text": "% Created 2020-01-22 Wed 15:54\n% Intended LaTeX compiler: pdflatex\n\\documentclass[11pt]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{graphicx}\n\\usepackage{grffile}\n\\usepackage{longtable}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{amssymb}\n\\usepackage{capt-of}\n\\usepackage{hyperref}\n\\author{Vaibhav Pujari}\n\\date{\\today}\n\\title{Foundations of Haskell}\n\\hypersetup{\n pdfauthor={Vaibhav Pujari},\n pdftitle={Foundations of Haskell},\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 26.3 (Org mode 9.3)}, \n pdflang={English}}\n\\begin{document}\n\n\\maketitle\n\\tableofcontents\n\n\n\\section{Lambda calculus and Category theory}\n\\label{sec:org3dd5759}\n\\begin{itemize}\n\\item Everything is a function\n\\item There is no error reporting - not that practical because there is no help from compiler\n\\item If we introduce error checking and types, we deviate from lambda calculus because not every expression is correct. But we also gain a lot of practical value\n\\item This then becomes a category - because in category composition is not arbitrary, it is constrained by morphisms\n\\end{itemize}\n\\section{The language}\n\\label{sec:org5892609}\n\\subsection{Data types}\n\\label{sec:org305db3e}\n\\begin{verbatim}\nx :: Int\nx = 5\n\\end{verbatim}\n\\subsection{Function types}\n\\label{sec:org3b97478}\nTypes are sets and in category of sets, a function is a member of Hom-set between two objects, so its a member of some other object in the same category (every set is an object)\n\\subsection{Partial functions}\n\\label{sec:org816f474}\n\\begin{itemize}\n\\item Recursion can create programs that never terminate\n\\item Well-founded recursion - compiler can prove that it terminates. But its a theoretical concept because for every algorithm that claims to find out if a recursion is well-founded or not, there always exists a particual recursion that breaks the algorithm\n\\item Therefore, in practice we need partial functions. A partial function gives result when it terminates, but there is no guarantee that it will terminate\n\\end{itemize}\n\\subsection{Composition}\n\\label{sec:org97e878e}\n\\begin{verbatim}\n(.) :: (b->c) -> (a->b) -> a -> c\n(g . f) x = g ( f x )\n\\end{verbatim}\nPolymorphic types and higher order functions are foundational in haskell as opposed to a later added feature in most common languages\n\\subsection{Functors}\n\\label{sec:org7e9a78e}\n\\begin{itemize}\n\\item A structure-preserving mapping between two categories\n\\begin{itemize}\n\\item To preserve objects, we just need a type constructor\n\\end{itemize}\n\nF: Ob(Hask) \\(\\to\\) Ob(Hask)\n\n\\begin{itemize}\n\\item To preserve morphism structure, we need \\texttt{fmap}. \\texttt{fmap} maps morphism in first category to a morphism in second category such that it is still composable in the same way as original morphism was composable\n\\end{itemize}\n\nfmap: Hom\\textsubscript{Hask} \\(\\to\\) Hom\\textsubscript{Hask}\n\n\\item Shrinking the structure is a special case of preserving. When objects collapse, the morphisms must follow\n\\item There is a category \\texttt{Cat}, in which objects are smaller categories and morphisms are functors\n\\end{itemize}\n\\end{document}\n", "meta": {"hexsha": "80b5c5c4e6d6a9f979687e5fadd2362ca3ce41c3", "size": 3193, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "03-haskell-foundations.tex", "max_stars_repo_name": "vaibhav276/category-theory-notes", "max_stars_repo_head_hexsha": "164285a8ace6bcf86f19a083255c029a2c39cdbe", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "03-haskell-foundations.tex", "max_issues_repo_name": "vaibhav276/category-theory-notes", "max_issues_repo_head_hexsha": "164285a8ace6bcf86f19a083255c029a2c39cdbe", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "03-haskell-foundations.tex", "max_forks_repo_name": "vaibhav276/category-theory-notes", "max_forks_repo_head_hexsha": "164285a8ace6bcf86f19a083255c029a2c39cdbe", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.1279069767, "max_line_length": 258, "alphanum_fraction": 0.7785781397, "num_tokens": 879, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.8244619220634456, "lm_q1q2_score": 0.6496192680134897}} {"text": "\n\\subsection{Convergence}\n\nIf \\(x=c\\) then the power series will be equal to \\(a_0\\).\n\nFor other values the power series may not converge.\n\n\\subsubsection{Cauchy-Hadamard theorem}\n\nRadius of convergence:\n\n\\(\\dfrac{1}{R}={\\lim \\sup}_{n_\\rightarrow \\infty} (|a_n|^{\\dfrac{1}{n}})\\)\n\n", "meta": {"hexsha": "691493ca9b8bb3da3595e470a505847e4aea5c12", "size": 281, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/analysis/transformations/02-02-convergence.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/analysis/transformations/02-02-convergence.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/analysis/transformations/02-02-convergence.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 20.0714285714, "max_line_length": 74, "alphanum_fraction": 0.6939501779, "num_tokens": 89, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.874077222043951, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.6495862384240451}} {"text": "\\chapter{Unit 3}\n\\section{Related Rates}\nRelated rates are a technique of differentiation that allows us to examine the\nrate of change of one quantity, with respect to the rate of change of several\nother quantitities.\n\nThe process for using related rates in calculus is as follows:\n\\begin{enumerate}\n \\item Identify all of the given quantities, and the quantities to be\n determined.\n \\item Write an equation involving the variables whos rates of change are\n given, or to be determined.\n \\item Use the Chain Rule to differentiate both sides of the equation with\n respect to time.\n \\item Substitute all known values for rates of change into the original\n equation.\n\\end{enumerate}\n\nAn example follows below:\n\\textit{Air is being pumped into a spherical balloon at a rate of\n\\SI{4.5}{\\meter\\cubed\\per\\minute}. Find the rate of change of the radius when the\nradius is \\SI{2}{\\meter}.}\n\n\\begin{align*}\n \\frac{dV}{dt} &= \\frac{9}{2} \\\\\n \\frac{dr}{dt} \\text{when} r=2 \\\\\n V &= \\frac{4}{3}\\pi{r^3} \\\\\n \\frac{dV}{dt} &= 4\\pi{r^2}\\frac{dr}{dt} \\\\\n \\frac{dr}{dt} &= \\frac{1}{4\\pi(2^2)}\\frac{9}{2}\n\\end{align*}\n\n\\section{Application of Derivatives}\n\\subsection{Extrema on an Interval}\nDefinition of extrema:\n\\begin{enumerate}\n \\item $f(c)$ is the minimum of $f$ on $I$ when $f(c) \\leq f(x) \\forall x \\in\n I$\n \\item $f(c)$ is the maximum of $f$ on $I$ when $f(c) \\geq f(x) \\forall x \\in\n I$\n\\end{enumerate}\n\n\\subsubsection{The Extreme Value Theorem}\nIf $f$ is continuous on a closed interval $[a,b]$ then $f$ has both a minimum\nand a maximum on the closed interval.\n\n\\subsubsection{Relative Extrema}\n\\begin{enumerate}\n \\item If there is an open interval containing $c$ on which $f(c)$ is a\n minimum then $f(c)$ is a relative minimum of $f$.\n \\item If there is an open interval containing $c$ on which $f(c)$ is a\n maximum then $f(c)$ is a relative maximum of $f$.\n\\end{enumerate}\n\n\\subsubsection{Critical Number}\nIf $f$ is defined at $c$ and $f'(c)=0$ or is undefined, then $c$ is a critical\nnumber of $f$.\n\nIf $f$ has a relative minimum or relative maximum at $x=c$, then $c$ is a\ncritical number of $f$.\n\nTo find critical numbers of $f$ on $[a,b]$, use the following process:\n\\begin{enumerate}\n \\item Find the critical numbers of $f$ on $(a,b)$.\n \\item Evaluate $f$ at each of those critical numbers.\n \\item Evaluate $f$ at each of the endpoints of $[a,b]$.\n \\item The least of these values is the minimum. The greatest is the maximum.\n\\end{enumerate}\n\n\\subsection{Rolle's Theorem \\& Mean Value Theorem}\n\\subsubsection{Rolle's Theorem}\nLet $f$ be a continuous function defined on a closed interval $[a,b]$. If\n$f(a)=f(b)$ then there is at least one $c\\in(a,b)$ such that $f'(c)=0$.\n\n\\subsubsection{Mean Value Theorem}\nLet $f$ be a continuous function defined $\\forall{x}\\in [a,b]$, and\ndifferentiable over $x\\in(a,b)$. $c\\in(a,b)\\to f'(c)=\\frac{f(b)-f(a)}{b-a}$.\n\n\\subsection{Increasing and Decreasing Functions}\nLet $f$ be a function defined over some open interval. For all critical numbers,\nthe following can be said about the function:\n\\begin{description}\n \\item[$f'(x)>0$] $f(x)$ increasing\n \\item[$f'(x)<0$] $f(x)$ decreasing\n \\item[$f'(x)=0$] $f(x)$ constant\n\\end{description}\n\n\\subsubsection{First Derivative Test}\n\\begin{enumerate}\n \\item If $f'(x)$ changes from negative to positive at $x=c$, then there is a\n relative minumum on $f$ at $x=c$.\n \\item If $f'(x)$ changes from positive to negative at $x=c$, then there is a\n relative maximum on $f$ at $x=c$.\n\\end{enumerate}\n\n\\subsection{Concavity}\nLet $f$ be differentiable on an open interval $I$. The graph of $f$ is concave\nupward on $I$ when $f'(x)$ is increasing on the interval and concave downward\nwhen $f'(x)$ is decreasing on the interval.\n\n\\subsubsection{Test for Concavity}\nLet $f$ be a function whose second derivative exists $\\forall{x}\\in(a,b)$.\n\\begin{description}\n \\item[$f''(x)>0\\forall{x}\\in{I}$] $f$ is concave upward on $I$.\n \\item[$f''(x)<0\\forall{x}\\in{I}$] $f$ is concave downward on $I$.\n\\end{description}\n\n\\subsection{Limits at Infinity}\nHorizontal asymptotes of a function can be determined when the limit of that\nfunction at a positive or negative $\\infty$ is a value $L$.\n\nLet $L$ be a horizontal asymptote of a function $f$.\n\\begin{equation}\n L = \\lim_{x\\to\\pm\\infty} f(x)\n\\end{equation}\n\n\\subsection{Curve Sketching}\nTo sketch a curve, implement the following process:\n\n\\begin{enumerate}\n \\item determine the $x$-intercepts\n \\item determine the $y$-intercepts\n \\item Determine any vertical asymptotes by seeing where $f(x)$ becomes\n undefined.\n \\item Determine any horizontal asymptotes by evaluating the limits at\n $\\pm\\infty$.\n \\item Determine the critical numbers of the first derivative to sketch a\n general curve of the graph.\n \\item Determine the critical numbers of the second derivative to determine\n points of inflection.\n\\end{enumerate}\n\n\\section{Optimization}\nTo optimize a problem, follow steps very similar to those described in the\nsection on Related Rates.\n\\begin{enumerate}\n \\item Set up a related rates problem.\n \\item Determine the feasiable domain of the problem.\n \\item Evaluate the function.\n \\item Determine critical numbers to find the relative maximum or minimum.\n\\end{enumerate}\n\n\\section{Newton's Method}\nNewton's method is a convenient way to determine the zeros of a function. To\nsolve for a zero using Newton's method, use the following process:\n\\begin{enumerate}\n \\item Make an initial estimate $x_1$ that is close to $c$.\n \\item Determine $x_{n+1}=x_n-\\frac{f(x_n)}{f'(x_n)}$.\n \\item See step 2.\n\\end{enumerate}\n\n\\section{Differentials}\n", "meta": {"hexsha": "efdef64a57dbae062c8b7681e4d341b96a7f0b26", "size": 5626, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2016/bc_calculus/units/unit_3.tex", "max_stars_repo_name": "ttaylorr/midterms", "max_stars_repo_head_hexsha": "fdde0fd1a66eb5242d0dfa04a5201c3ab6d6b7eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-01-06T06:55:26.000Z", "max_stars_repo_stars_event_max_datetime": "2015-01-06T06:55:26.000Z", "max_issues_repo_path": "2016/bc_calculus/units/unit_3.tex", "max_issues_repo_name": "ttaylorr/midterms", "max_issues_repo_head_hexsha": "fdde0fd1a66eb5242d0dfa04a5201c3ab6d6b7eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2016/bc_calculus/units/unit_3.tex", "max_forks_repo_name": "ttaylorr/midterms", "max_forks_repo_head_hexsha": "fdde0fd1a66eb5242d0dfa04a5201c3ab6d6b7eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.0131578947, "max_line_length": 81, "alphanum_fraction": 0.7099182368, "num_tokens": 1660, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8539127585282744, "lm_q1q2_score": 0.6495292971039955}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{exercises}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 1.11 Cycling free indices -- preferred solution}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,u,v,w}::Indices.\n\n expr := A_{a b c}. # cdb (ex-0111.101,expr)\n\n rule := T_{a b c} -> @(expr).\n expr := T_{b c a}. # cdb (ex-0111.102,expr)\n\n substitute (expr, rule) # cdb (ex-0111.103,expr)\n\\end{cadabra}\n\n\\begin{align*}\n &\\Cdb{ex-0111.101}\\\\\n &\\Cdb{ex-0111.102}\\\\\n &\\Cdb{ex-0111.103}\n\\end{align*}\n\n\\clearpage\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 1.11 Cycling free indices -- alternative solution}\n\nThis alternative solution uses two rounds of Kroncker deltas. It does the job but is not as\nsimple as the previous solution.\n\n\\begin{cadabra}\n {a,b,c,d,e,f,u,v,w}::Indices.\n\n \\delta{#}::KroneckerDelta.\n\n expr := A_{a b c}. # cdb (ex-0111.201,expr)\n\n expr := \\delta^{a}_{u} \\delta^{b}_{v} \\delta^{c}_{w} @(expr). # cdb (ex-0111.202,expr)\n\n eliminate_kronecker (expr) # cdb (ex-0111.203,expr)\n\n expr := \\delta^{u}_{b} \\delta^{v}_{c} \\delta^{w}_{a} @(expr). # cdb (ex-0111.204,expr)\n\n eliminate_kronecker (expr) # cdb (ex-0111.205,expr)\n\n\\end{cadabra}\n\n\\begin{align*}\n &\\Cdb{ex-0111.201}\\\\\n &\\Cdb{ex-0111.202}\\\\\n &\\Cdb{ex-0111.203}\\\\\n &\\Cdb{ex-0111.204}\\\\\n &\\Cdb{ex-0111.205}\n\\end{align*}\n\n\\end{document}\n", "meta": {"hexsha": "859d2b61c6fc103c861c25783e80254098da2d13", "size": 1773, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/exercises/ex-0111.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/exercises/ex-0111.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/exercises/ex-0111.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 28.5967741935, "max_line_length": 94, "alphanum_fraction": 0.475465313, "num_tokens": 540, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.853912754810561, "lm_q1q2_score": 0.6495292942761144}} {"text": "\\documentclass[a4paper, 12pt]{article}\n\\usepackage{amsmath}\n\\usepackage{color}\n\\usepackage{dsfont}\n\\usepackage[utf8]{inputenc}\n\\usepackage{graphicx}\n\\usepackage[left=2cm, right=2cm, bottom=3cm, top=2cm]{geometry}\n\\usepackage{natbib}\n\\usepackage{microtype}\n\n\\definecolor{orange}{rgb}{1, 0.5, 0}\n\\definecolor{darkgreen}{rgb}{0, 0.5, 0}\n\\definecolor{darkred}{rgb}{0.7, 0, 0}\n\\newcommand{\\btheta}{\\boldsymbol{\\theta}}\n\\newcommand{\\tn}{\\textnormal}\n\n\\title{Notes}\n\\author{Brendon J. Brewer}\n\\date{}\n\\begin{document}\n\\maketitle\n\n% Need this after the abstract\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{8pt}\n\n\\section{A property of Nested Sampling}\nConsider the implied prior for $L$, denoted $\\pi(L)$. This is\na probability distribution over the real line.\nAs usual, define $X(\\ell)$ as the amount of prior mass\nwith likelihood above $\\ell$:\n\\begin{align}\nX(\\ell) &= \\int_\\ell^\\infty \\pi(L) \\, dL.\n\\end{align}\nThis is the complementary CDF of $L$.\n\nThe NS sequence rectifies $\\pi$ to give you the distribution\n\\begin{align}\np_{\\rm NS}(L) &= \\frac{1}{C} \\frac{\\pi(L)}{X(L)}\n\\end{align}\nwhere\n\\begin{align}\nC &= \\int_{-\\infty}^{L_{\\rm max}} \\frac{\\pi(L)}{X(L)} \\, dL\n\\end{align}\nis a normalising constant,\nand $L_{\\rm max}$ is the maximum likelihood from the\ndiscarded points of the run\n(not necessarily the overall maximum likelihood).\n\nConsider a `constrained prior' distribution, which is just the\nprior $\\pi$ but truncated so it only includes values above\na threshold $\\ell$:\n\\begin{align}\np(L \\,|\\, \\ell) &\\propto\n \\left\\{\n \\begin{array}{lr}\n \\pi(L),& L > \\ell, \\\\\n 0, & \\tn{otherwise}.\n \\end{array}\n \\right.\n\\end{align}\nThe normalisation constant of a constrained prior is\njust $X(\\ell)$.\n\nNS gives us the opportunity to measure this for any given\nvalue of $\\ell$. Consider the KL divergence from $p_{\\rm NS}$ to\nthe constrained prior:\n\\begin{align}\nD_{\\rm KL}(p_\\ell \\,||\\, p_{\\rm NS})\n &= \\int_{\\ell}^{\\infty} \\frac{\\pi(L)}{X(\\ell)}\n \\log\\left[\\frac{\\pi(L)/X(\\ell)}{\\pi(L)/C/X(L)}\\right] \\, dL \\\\\n &= \\int_{\\ell}^{\\infty} \\frac{\\pi(L)}{X(\\ell)}\n \\log\\left[\\frac{CX(L)}{X(\\ell)}\\right] \\, dL\n\\end{align}\nWe can take the expectation (integral) in terms of $X$ instead of\n$L$, and use the fact that the prior corresponds to a\nUniform$(0,1)$ distribution for $X$:\n\\begin{align}\nD_{\\rm KL}(p_\\ell \\,||\\, p_{\\rm NS})\n &= \\int_0^{X(\\ell)} \\frac{1}{X(\\ell)}\n \\log(X) \\, dX - \\log X(\\ell) + \\log C \\\\\n &= \\frac{1}{X(\\ell)} \\int_0^{X(\\ell)} \\log(X) \\, dX\n - \\log X(\\ell) + \\log C \\\\\n &= \\frac{1}{X(\\ell)}\\left[X\\log(X) - X\\right]_0^{X(\\ell)}\n - \\log X(\\ell) + \\log C \\\\\n &= \\log X(\\ell) - 1 - \\log X(\\ell) + \\log C \\\\\n &= \\log C - 1.\n\\end{align}\nCrucially, {\\em this does not depend on} $\\ell$.\n\n\\section{Uniqueness}\nIs $p_{\\rm NS}$ the only distribution with the above property?\nConsider the KL divergence from some distribution $q$\nto the constrained prior.\nWe will then see what choices of $q$ have the above property.\nThe KL divergence is\n\\begin{align}\nD_{\\rm KL}(p_\\ell \\,||\\, q)\n &= \\int_{\\ell}^{\\infty} \\frac{\\pi(L)}{X(\\ell)}\n \\log\\left[\\frac{\\pi(L)/X(\\ell)}{q(L)}\\right] \\, dL \\\\\n &= -\\log X(\\ell) + \\frac{1}{X(\\ell)}\n \\int_{\\ell}^{\\infty} \\pi(L)\n \\log\\left[\\frac{\\pi(L)}{q(L)}\\right] \\, dL\n\\end{align}\n\nThe derivative of the KL divergence, with respect to $\\ell$, is\n\\begin{align}\n\\frac{d}{d\\ell} D_{\\rm KL}(p_\\ell \\,||\\, q)\n &= -\\frac{X'(\\ell)}{X(\\ell)}\n + \\frac{X(\\ell)\\pi(\\ell)\\log\\left[\\pi(\\ell)/q(\\ell)\\right]\n -X'(\\ell)\\int_{\\ell}^{\\infty} \\pi(L)\n \\log\\left[\\pi(L)/q(L)\\right] \\, dL}\n {X(\\ell)^2} \\\\\n &= -\\frac{X'(\\ell)}{X(\\ell)}\n + \\frac{\\pi(\\ell)\\log\\left[\\pi(\\ell)/q(\\ell)\\right]}\n {X(\\ell)}\n - \\frac{X'(\\ell)/X(\\ell)\\int_{\\ell}^{\\infty} \\pi(L)\n \\log\\left[\\pi(L)/q(L)\\right] \\, dL}\n {X(\\ell)}\\\\\n &= -\\frac{X'(\\ell)}{X(\\ell)}\n + \\frac{\\pi(\\ell)\\log\\left[\\pi(\\ell)/q(\\ell)\\right]}\n {X(\\ell)}\n - \\frac{X'(\\ell)\\left(D_{\\rm KL} + \\log X(\\ell)\\right)}\n {X(\\ell)} \\\\\n &= \\frac{\\pi(\\ell)}{X(\\ell)}\n \\Big(1\n + \\log\\left[\\pi(\\ell)/q(\\ell)\\right]\n + D_{\\rm KL} + \\log X(\\ell)\n \\Big)\n\\end{align}\nTo satisfy the nice property of the previous section, the derivative\nmust be zero. This is achieved if\n\\begin{align}\n1 + \\log\\left[\\pi(\\ell)/q(\\ell)\\right]\n+ D_{\\rm KL} + \\log X(\\ell) &= 0 \\\\\n1 + \\log\\pi(\\ell) - \\log q(\\ell)\n+ D_{\\rm KL} + \\log X(\\ell) &= 0\n\\end{align}\n\n\n%\\section*{TwinPeaks}\n%Denote the two scalars by $L_1(\\btheta)$ and $L_2(\\btheta)$.\n%The implied prior for $L_1$ and $L_2$ is $\\pi(L_1, L_2)$.\n%Consider a constrained distribution\n%\\begin{align}\n%p(L_1, L_2 \\,|\\, \\ell_1, \\ell_2) &\\propto\n% \\left\\{\n% \\begin{array}{lr}\n% \\pi(L_1, L_2), & L_1 > \\ell_1 \\tn{ and } L_2 > \\ell_2 \\\\\n% 0, & \\tn{otherwise}.\n% \\end{array}\n% \\right.\n%\\end{align}\n%The normalising constant of this distribution is\n%\\begin{align}\n%X(\\ell_1, \\ell_2) &=\n% \\int_{\\ell_1}^{\\infty}\n% \\int_{\\ell_2}^{\\infty}\n% \\pi(L_1, L_2)\n% \\, dL_1\n% \\, dL_2.\n%\\end{align}\n\n%My proposed TwinPeaks distribution is\n%\\begin{align}\n%p_{\\rm TP}(L_1, L_2) &=\n% \\frac{1}{C}\n% \\frac{\\pi(L_1, L_2)}{X(L_1, L_2)}.\n%\\end{align}\n%As in the previous section, we can compute the KL divergence from\n%this to a constrained prior:\n%\\begin{align}\n%D_{\\rm KL}(p_{\\ell_1, \\ell_2} \\,||\\, p_{\\rm TP})\n% &= \\int_{\\ell_1}^{\\infty}\n% \\int_{\\ell_2}^{\\infty}\n% \\frac{\\pi(L_1, L_2)}{X(\\ell_1, \\ell_2)}\n% \\log\\left[C \\frac{X(L_1, L_2)}{X(\\ell_1, \\ell_2)} \\right]\n% \\, dL_1\n% \\, dL_2 \\\\\n% &=\n% \\int_{\\ell_1}^{\\infty}\n% \\int_{\\ell_2}^{\\infty}\n% \\frac{\\pi(L_1, L_2)}{X(\\ell_1, \\ell_2)}\n% \\log\\left[C \\frac{X(L_1, L_2)}{X(\\ell_1, \\ell_2)} \\right]\n% \\, dL_1\n% \\, dL_2 \\\\\n% &=\n% \\log C \\,-\\, \\log X(\\ell_1, \\ell_2) \\\\ &\\quad\\quad\\quad\n% \\,+\\,\n% \\frac{1}{X(\\ell_1, \\ell_2)}\n% \\int_{\\ell_1}^{\\infty}\n% \\int_{\\ell_2}^{\\infty}\n% \\pi(L_1, L_2)\n% \\log X(L_1, L_2)\n% \\, dL_1\n% \\, dL_2 \\\\\n%\\end{align}\n%Doing the integral in terms of $X$\n\n\\section{Commutativity}\nDoes the operation of constraining by $f(\\theta)$ by a factor of $t_1$\ncommute with the operation of constraining by $g(\\theta)$ by a factor\nof $t_2$? I think so...that's the product rule, right?\n\nThe first route yields the following sequence of distributions:\n\\begin{align}\n\\pi(\\theta)\n \\implies\n\\frac{\\pi(\\theta)\\mathds{1}\\big(f(\\theta) \\geq f^*\\big)}{t_1}\n \\implies\n\\frac{\\pi(\\theta)\\mathds{1}\\big(f(\\theta) \\geq f^*\\big)\\mathds{1}\\big(g(\\theta) \\geq g^*\\big)}{t_1t_2}\n\\end{align}\n\n\n\n\\end{document}\n\n", "meta": {"hexsha": "0ae967dfc29c246ae0e8e31e867f546fc4897dd1", "size": 6845, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/notes.tex", "max_stars_repo_name": "eggplantbren/TwinPeaks-hs", "max_stars_repo_head_hexsha": "a4da4fe17d586fe75cc09d6cddfaa6b3052aeea8", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-03-10T03:16:50.000Z", "max_stars_repo_stars_event_max_datetime": "2020-03-10T03:16:50.000Z", "max_issues_repo_path": "doc/notes.tex", "max_issues_repo_name": "eggplantbren/TwinPeaks-hs", "max_issues_repo_head_hexsha": "a4da4fe17d586fe75cc09d6cddfaa6b3052aeea8", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/notes.tex", "max_forks_repo_name": "eggplantbren/TwinPeaks-hs", "max_forks_repo_head_hexsha": "a4da4fe17d586fe75cc09d6cddfaa6b3052aeea8", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.3990825688, "max_line_length": 102, "alphanum_fraction": 0.5701972243, "num_tokens": 2613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127566694178, "lm_q2_score": 0.7606506472514406, "lm_q1q2_score": 0.6495292910568545}} {"text": "\\section{Experiment 1: Simulated Data from Multivariate Normal Distribution}\n\n\tIn the first simulation experiment, we focused on an ideal setting where data come from a known multivariate normal \n\tdistribution and imputation is required to estimate the means, variances, and covariances of six items with \n\tmissing values.\n\tWe investigated the relative performance of the methods described above across a set of conditions defined by two \n\texperimental factors: the number of columns in the dataset $p$, taking values 50 or 500; \n\tand the proportion of \\emph{per} variable missing cases $pm$, taking values 0.1 or 0.3.\n\tTable \\ref{tab:condExp1} summarizes the four crossed conditions.\n\tData with sample size $n=200$ were independently generated $S = 1,000$ times for each condition.\n\tFor each $s$-th replicate, missing values were imposed and then all the missing data treatment methods described above\n\twere used to obtain estimates of the item means, variances, and covariances.\n\n\\begin{table}\n\t\\centering\n\t\\begin{tabular}{l | r | r | r | r }\n\t\tcondition & label & n & p & pm \\\\\n\t\t\\hline\n\t\t1 & low-dim-low-pm & 200 & 50 & .1 \\\\\n\t\t2 & high-dim-low-pm & 200 & 500 & .1 \\\\\n\t\t3 & low-dim-high-pm & 200 & 50 & .3 \\\\\n\t\t4 & high-dim-high-pm & 200 & 500 & .3 \\\\\n\t\\end{tabular}\n\t\\caption{\\label{tab:condExp1}Summary of conditions for Experiment 1.}\n\\end{table}\n\n%\\FloatBarrier\n\n\\subsection{Simulation Study Procedure}\n\n\\subsubsection{Data Generation}\n\tAt every replication, a data matrix $\\bm{Z}_{n \\times p}$ was generated according to a multivariate normal model \n\tcentered around a mean of 0 with a covariance matrix $\\bm{\\Sigma}_0$, with diagonal elements (variances) equal to 1.\n\tThe off-diagonal elements of $\\bm{\\Sigma}_0$ were used to define three blocks of variables: \n\tthe first five variables were highly correlated among themselves ($\\rho = .6$);\n\tvariables 6 to 10 were weakly correlated with variables in block 1 and among themselves ($\\rho = .3$), \n\tand all the remaining $p-10$ variables were uncorrelated.\n\tItems were rescaled to have mean of 5.\n\n\\subsubsection{Missing Data Imposition} \\label{sub_missing}\n\n\tMissing values were imposed on six items in $\\bm{Z}$: three variables in the block of\n\thighly correlated variables ($z_j$ with $j = 1,2,3$), and three in the block of lowly correlated variables ($z_j$ \n\twith $j = 6,7,8$).\n\tItem non-response was imposed by sampling from a Bernoulli distribution with individual missing probabilities \n\tdefined by \n%\n\t\\begin{equation} \\label{eqn:rm}\n\t\tp_{miss} = p(z_{i,j} = miss | \\tilde{Z}) = \\frac{ exp(\\gamma_0 + \\tilde{Z}_{i}\\bm{\\gamma}) }\n\t\t\t\t\t\t\t\t{ 1 + exp(\\gamma_0 + \\tilde{Z}_{i}\\bm{\\gamma}) }\n\t\\end{equation}\n%\n\twhere $z_{i,j}$ is the $i$-th subject's response on the $j$-th variable target of missing data imposition, \n\t$\\tilde{Z}_{i}$ is a vector of responses to the set of predictors involved in the missing data mechanism \n\tfor the $i$-th individual, $\\gamma_0$ is an intercept parameter, and $\\bm{\\gamma}$ is a vector of slope \n\tparameters for the linear term.\n\t$\\tilde{Z}$ was specified to include two fully observed variables from the highly correlated set, and two \n\tfrom the lowly correlated set ($z_r$ with $r = 4,5,9,10$).\n\tThe probability of observing a response on a target variable did not depend on the variable itself, \n\tto avoid imputation under MNAR.\n\tFurthermore, when all the features in the data are provided to the MI procedures, the predictors in $\\tilde{Z}$ \n\tare allowed to be part of the imputation models and the MAR assumption can be met.\n\tAll slopes in $\\bm{\\gamma}$ were fixed to 1, while the value of $\\gamma_0$ was chosen with an optimization \n\talgorithm that minimized the difference between the actual and desired proportion of missing values.\n\n\\subsubsection{Imputation}\n\t\n\tMissing values were treated with all the methods described in Section 2.\n\tConvergence of the imputation models was assessed in a preprocessing step by observing trace plots.\n\tThe imputation algorithms were considered to have converged after 50 iterations, after which 10 imputed data \n\tsets were stored and used for the subsequent standard complete-data analysis and pooling.\n\tThe only exception was blasso, which required approximately 2,000 iterations for convergence.\n\n\tThe ridge penalty used in the bridge algorithm was fixed across iterations.\n\tThe value used in the simulation was determined by means of cross-validation in a preprocessing phase.\n\tThe ridge penalty values $10^{-1}, 10^{-2}, ..., 10^{-8}$ were used to impute data with bridge\n\tand we selected the value that resulted in the smallest average Fraction of Missing Information (FMI)\n\t\\citep[eq. 3.1.10]{rubin:1987} across the analysis model parameters.\n\t\t\n\tBoth IURR and DURR could have been specified with a variety of penalty parameters.\n\tFor example, one could use any of the following: ridge penalty \\citep{hoerlKennard:1970}, lasso penalty \n\t\\citep{tibshirani:1996}, elastic net penalty \\citep{zouHastie:2005}, adaptive lasso \\citep{zou:2006}.\n\tIn this study, we specified the lasso penalty for regularization as it is computationally efficient, \n\tand it performed well for imputation in \\cite{zhaoLong:2016} and \\cite{dengEtAl:2016}.\n\tA 10-fold cross-validation procedure was used at every iteration of DURR and IURR to choose the penalty parameter.\n\n\tFor blasso, in order to maintain consistency with previous research, the hyper-parameters in equations \n\t\\eqref{eqn:sigprior}, \\eqref{eqn:tauprior}, and \\eqref{eqn:rhoprior} were specified as in \\cite{zhaoLong:2016}: \n\t$(a,b)=(0.1, 0.1)$, $(r,s)=(0.01, 0.01)$, and $(g,h)=(1,1)$.\n\tIn the MI-PCA algorithm, enough components were extracted to explain 50\\% of the total variance in the data.\n\tTo impute data with the single imputation random forest approach we used the missForest R package \n\t\\citep{missForest} implementing algorithm 1 proposed by \\cite{stekhovenBuhlmann:2011}.\n\tThe stopping criterion for the missForest algorithm was usually met within the first 10 iterations, \n\tbut to make a conservative choice we fixed the maximum number of iterations to 20.\n\t\\cite{stekhovenBuhlmann:2011} showed that increasing the number of trees grown in each forest has \n\tstagnating effects on the imputation error while linearly increasing the computation time.\n\tIn their paper, the authors recommend growing 100 trees per forest, which offers a good compromise \n\tbetween imputation precision and computation time.\n\tTherefore, we used this value in our study.\n\n\\subsubsection{Analysis}\n\tThe substantive model of interest in Experiment 1 was a saturated model that estimated means,\n\tvariances, and covariances of the six variables with missing values.\n\tThis resulted in estimating six means, six variances, and 15 covariances.\n\n\\subsection{Comparison Criteria} \\label{criteria}\n\n\tWe compared methods in terms of bias and confidence interval coverage.\n\n\t\\paragraph{Bias}\n\n\tFor a given parameter of interest $\\theta$ (e.g., mean of item 1, variance of item 2), we used the \n\tPercent Relative Bias (PRB) to quantify the estimation bias introduced by the imputation procedures:\n%\n\t\\begin{equation} \\label{eqn:prb}\n\t\tPRB = \\frac{\\bar{\\hat{\\theta}} - \\dot{\\theta}}{\\dot{\\theta}} \\times 100\n\t\\end{equation}\n%\n\twhere $\\dot{\\theta}$ is the true value of the focal parameter defined as \n\t$\\sum_{s=1}^{S} \\hat{\\theta}_{s}^{GS}/S$\n\t, with\n\t$\\hat{\\theta}_{s}^{GS}$ \n\tbeing the Gold Standard parameter estimate for the $s$-th repetition. \n\tThe averaged focal parameter estimate under a given missing data treatment is computed as \n\t$\\bar{\\hat{\\theta}} = \\sum_{s=1}^{S} \\hat{\\theta}_{s}/S$,\n\twith\n\t$\\hat{\\theta}_{s}$ being the estimate obtained after having treated the missing values in the \n\t$s$-th repetition.\n\tFollowing \\cite{muthenEtAl:1987}, $|\\text{PRB}| > 10\\%$ was considered indicative of problematic \n\testimation bias.\n\n\t\\paragraph{Confidence Intervals Coverage}\n\tTo assess the correctness of hypothesis testing, the Confidence Interval Coverage (CIC) of the true parameter\n\tvalue was defined as:\n%\n\t\\begin{equation} \\label{eqn:cic}\n\t\tCIC = \\frac{ \\sum_{s=1}^{S} I(\\dot{\\theta} \\in \\widehat{CI}_s ) }{S}\n\t\\end{equation}\n%\n\twhere $\\widehat{CI}_s$ is the confidence interval of the parameter estimate $\\hat{\\theta}_{s}$ in a given repetition, \n\tand $I(.)$ is the indicator function that returns 1 if the argument is true and 0 otherwise.\n\t\n\tCICs below 0.9 are usually considered problematic for 95\\% confidence intervals \\cite[p. 52]{vanBuuren:2018} \n\tas they imply inflated Type I error rates.\n\tA high coverage (e.g., 0.99) may indicate confidence intervals that are too wide, implying inflated Type II error rates.\n\tTherefore, Confidence Intervals were considered to show severe under-coverage (over-coverage) if they were \n\tbelow 0.9 (above 0.99).\n\n\tFollowing \\cite{burtonEtAl:2006}, in simulation studies, a CIC can be considered as significantly different from the \n\tnominal coverage rate if it falls outside two Standard Errors of the nominal coverage probability ($SE(p)$) from the \n\tnominal coverage rate.\n\tThe standard error of nominal coverage probability is defined as $SE(p) = \\sqrt{p (1-p)/S}$, with $p$ indicating the\n\tchosen nominal coverage probability.\n\tTherefore, for $S = 1000$, 95\\% CI coverages ($p = 0.95$) outside the range (0.94, 0.96) were considered as significantly \n\tdifferent from the nominal coverage rate.\n\n\\subsection{Results}\n\t\n\tBoth PRB and CIC were computed for all the 27 parameters in the analysis model (six means, six item variances,\n\tand 15 covariances).\n\tTo summarize the results, we focus on the typical and extreme values of these measures.\n\tIn Figures \\ref{fig:exp1bias} and \\ref{fig:exp1cir}, we report the average, minimum, and maximum absolute PRB \n\tand CIC achieved by the missing data treatment methods for each parameter parameter type.\n\tIn the supplementary material, we included figures reporting the raw PRB and CIC for every parameter estimate.\n%\n\t\\paragraph{Means} \n\tFocusing first on the item means (top rows), the largest $|\\text{PRB}|$ is below 10 percentage points for all \n\timputation methods.\n\tHowever, looking at relative performances, IURR and MI-PCA resulted in smaller bias than all other methods, \n\texcept MI-OP.\n\tIn the conditions with high proportion of missing values (columns 3 and 4 in the figures), all methods showed \n\tsignificant deviations from nominal coverage rates, with all CICs outside of the interval $(0.94, 0.96)$.\n\tThe only exceptions was MI-PCA which showed non-significant deviations from nominal coverage for \n\talmost all estimates, with both the lowest and highest CIC falling within (0.94, 0.96) in all conditions.\n\tThe tree-based MI methods, missForest, and CC lead to CICs significantly different from nominal coverage rates\n\tin all conditions.\n%\n\t\\paragraph{Variances} \n\tMoving to the item variances (central rows), IURR, blasso, and the MI tree-based methods resulted in the lowest\n\tbiases across all conditions, even in the high-dim-high-pm condition.\n\tThese low biases were mostly paired with low deviations from nominal coverage, except for the high-dim-high-pm\n\tcondition where they resulted in significant under-coverage of the true item \n\tvariances (highest $CIC < 0.94$).\n\tNevertheless, apart from MI-OP, blasso was the method with best coverage in this final condition.\n\t\n\tDURR showed poor performance with regard to the item variances: in all conditions but the first, it led to \n\tlarge bias accompanied by significant CI under-coverage.\n\tBridge was the only MI method showing larger bias than DURR in all the high-dimensional conditions (columns 2 and 4),\n\twith even the minimum $|\\text{PRB}|$ exceeding the 20\\% threshold.\n\tMI-PCA also showed poor performance with noticeable item variance bias in all conditions that became extreme in the\n\thigh-dim-high-pm condition (column 4), where even the smallest $|\\text{PRB}|$ exceeded 20\\%.\n\tThis poor performance was reflected in extreme confidence interval under-coverage of the true item variances in\n\tthe final experimental condition.\n\tFinally, missForest and CC led to substantial bias and CI under-coverage for all \n\titem variances, even in condition 1.\n%\n\t\\paragraph{Covariances}\n\tIURR performed noticeably better than most other methods, with negligible covariance bias \n\t(maximum $|\\text{PRB}| < 10\\%$) and acceptable coverage (average $CIC \\sim 0.94$) in the high-dim-low-pm \n\tcondition.\n\tHowever, it struggled with a large covariance bias and extreme under-coverage in the high-dim-high-pm \n\tcondition (average $|\\text{PRB}| > 10\\%$ and average $CIC < 0.90$).\n\tMI-PCA showed negligible bias for all the covariance estimates (with the maximum $|\\text{PRB}| < 10\\%$), \n\tand performed as well as MI-OP in all but the high-dim-high-pm condition.\n\tFurthermore, MI-PCA resulted in virtually no deviation from nominal coverage, except in the last condition.\n\tIn the high-dim-high-pm condition, MI-PCA led to negligible bias, and mild significant \\emph{over}-coverage \n\tof the items covariances: the average CIC was greater than 0.96, but smaller than 0.99.\n\n\tBridge displayed low bias and acceptable coverage in the low dimensional conditions (columns 1 and 3), but \n\textreme bias and low CI coverage in all the high dimensional conditions (columns 2 and 4).\n\tMissForest and CC showed extreme bias and under-coverage for all the covariances (minimum $|\\text{PRB}|>10\\%$),\n\teven in condition 1.\n\tAll other methods, including DURR, showed absolute covariance PRBs larger than the 10\\% threshold in all but the \n\tfirst condition, with persistently significant CI under-coverage of the true values.\n\n\\begin{figure}\n\\centering\n\\includegraphics{\\pathFIG/exp1_bias_summy.pdf}\n\\caption{\\label{fig:exp1bias}\n\tMaximum, average, and minimum absolute Percent Relative Bias ($|\\text{PRB}|$) for item means, variances, \n\tand covariances in Experiment 1.\n\tFor each method, the maximum, average, and minimum $|\\text{PRB}|$ are reported in this order.\n\t}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\includegraphics{\\pathFIG/exp1_CI_summy.pdf}\n\\caption{\\label{fig:exp1cir}\n\tMaximum, average, and minimum CIC for item means, variances, and covariances in Experiment 1.\n\tFor each method, the maximum, average, and minimum are reported in this order.\n\t}\n\\end{figure}\n\t\n%\\FloatBarrier % stops fig:exp1cir to leave its section\n\n", "meta": {"hexsha": "ba35e193a96f75173a0dd5272d2ecd723913802d", "size": 14243, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manuscript/old/03_section.tex", "max_stars_repo_name": "EdoardoCostantini/imputeHD-comp", "max_stars_repo_head_hexsha": "c6fad9ee406ad85dd9874f336220177ae08ae477", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "manuscript/old/03_section.tex", "max_issues_repo_name": "EdoardoCostantini/imputeHD-comp", "max_issues_repo_head_hexsha": "c6fad9ee406ad85dd9874f336220177ae08ae477", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "manuscript/old/03_section.tex", "max_forks_repo_name": "EdoardoCostantini/imputeHD-comp", "max_forks_repo_head_hexsha": "c6fad9ee406ad85dd9874f336220177ae08ae477", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.0970464135, "max_line_length": 123, "alphanum_fraction": 0.759250158, "num_tokens": 3727, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835371034369, "lm_q2_score": 0.7772998714925404, "lm_q1q2_score": 0.6494212460246347}} {"text": "\\lab{Application}{Cracking Blackjack}{Cracking Blackjack}\n\\label{Ch:BJ}\n\n\\objective{This section teaches about how toexploite the weaknesses of a pseudorandom number generator that uses Linear Congruience}\n\n\\section*{Blackjack}\n\n%Lab \\ref{BJ}\n\nBlack Jack is a card game that involes the use of randomness. The game is simple, the dealer deals the player and himself each two cards. He flips over his first card so the player can see it. The player has to chose to take another card (\"hit\") or not (\"stand\"). If the player hits he gets another card and again has the choice to hit or stand.\n\nThe goal is to get your hand to be at or as close to 21 without going over. Face cards are worth 10 points. Aces can count either as 11 or 1. The of value other cards are equal to number on the card.\n\nOnce the player has decided to stand the dealer flips over his second card and deals himself cards until his hand value is 17 or greater. \n\nIf the player value goes above 21 he automaticly loses. If his value is 21 and below and dealer has above 21 then the player wins. If they both have 21 or under than whose hand has the highest value, wins. If both hands have the same value, the game is a tie.\n\n\\section*{Shuffling Algorithms}\n\nOne use of Psuedorandom Number Generators (PRNGs) is to shuffle cards. The main goal of these algorithms is that the card order are random - so that no one player has advatage based on order. Often online gambling sites will post their algorithms online. The only thing they do not post is their seed values. Often the time in miliseconds is used as the seed value.\n\n\\section*{Cracking Blackjack}\nFor these next problems you will need three files that are provided:Black.py, BlackEasy.py, and bjHelp.py. Black.py and BlackEasy.py are are programs that run games of Blackjack that use a Linear Congruentail Generator (LCG) to shuffle the cards. They print out 52 numbers and the argsort of those numbers is the order of the cards.The parameters for BlackEasy.py are a$=2521$, c$=13$, mod$=2^{16}$; For Black.py they are a$=25214903917$, c$=11$, mod$=2^{48}$. In order to play them python name (Black.py or BlackEasy.py) numberofgames. They are both seeded initially by the time.\n\nbjHelp.py contains two functions that will help you \"predict\" the cards:\nSuffleHack(n,a,c,mod,seed) gives the first n card shuffles given the parameters for a LCG. The shuffles are represented by numbers \nHacker(Stats,['card','card','card']) Stats is the output of SuffleHack and takes a list of 3 cards (see below.) It prints all shuffles as a list of cards in Stats that have the same first three cards as the inputed list.\n\nThe trick to being able to \"predict\" the cards is to find the initial seed value.\n\nCards- A, 2-10, J, Q, or K combined with heart, diamond, club, or spade in single quotes. Examples: '6diamond', 'Kclub'\n\n\n\n\\begin{problem}\nPlay 10 games of BlackEasy.py and by the 5th game be able to predict the cards. You can write your own functions or use the ones in bjHelp.py. You will want to open two comand promts, one to play the game and one to use to predict the cards. \n\\end{problem}\n\nNot too hard. That is because there is only $2^{16}$ seed values. This next one you will have to look at more hands until you can find out the initial seed value.\n\n\\begin{problem}\nPlay 20 games of Black.py and by the 15th game be able to predict the cards.\n\\end{problem}\n\n\n\n", "meta": {"hexsha": "8c1f65b46b1b7c4ddeeb2200038d1a5ff815b0eb", "size": 3387, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Applications/Blackjack/Blackjack.tex", "max_stars_repo_name": "abefrandsen/numerical_computing", "max_stars_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Applications/Blackjack/Blackjack.tex", "max_issues_repo_name": "abefrandsen/numerical_computing", "max_issues_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Applications/Blackjack/Blackjack.tex", "max_forks_repo_name": "abefrandsen/numerical_computing", "max_forks_repo_head_hexsha": "90559f7c4f387885eb44ea7b1fa19bb602f496cb", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 72.0638297872, "max_line_length": 581, "alphanum_fraction": 0.7664599941, "num_tokens": 848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998714925403, "lm_q2_score": 0.8354835330070839, "lm_q1q2_score": 0.6494212428405399}} {"text": "% !TEX root=../main.tex\n\nWe now present our one-class Neural Network (OC-NN) model for unsupervised anomaly detection.\nThe method can be seen as designing a neural architecture using an OC-SVM equivalent loss function.\nUsing OC-NN we will be able to exploit and refine features obtained from unsupervised tranfer learning specifically for anomaly\ndetection. This in turn will it make it possible to discern anomalies in complex data sets where the decision boundary between normal and anomalous\nis highly nonlinear.\n\\vspace{-0.1cm}\n\\subsection{One-Class Neural Networks (OC-NN)}\n\\label{sec:oc-nn}\nWe design a simple feed forward network with one hiden layer having linear or sigmoid activation $g(\\cdot )$ and one output node. Generalizations to deeper\narchitectures is straightforward. The OC-NN objective can be formulated as:\n\\begin{equation}\n\t\\label{eqn:oc-nn}\n\t\\min_{w, V, \\bias} \\frac{1}{2} \\| w \\|_{2}^2 + \\frac{1}{2} \\| V \\|_F^2 + \\frac{1}{\\nu} \\cdot \\frac{1}{N} \\sum_{n= 1}^N \\max( 0, \\bias - \\langle w, g( V \\X_{n :} ) \\rangle ) - \\bias\n\\end{equation}\n\nwhere $w$ is the scalar output obtained from the hidden to output layer,\n$V$ is the weight matrix from input to hidden units.\nThus the \\underline{key insight} of the paper is to replace the dot product $\\mathbf{\\langle w,\\Phi(\\X_{n :}) \\rangle}$ in OC-SVM with the dot product $\\mathbf{\\langle w,g(V\\X_{n :})\\rangle}$. This change will make it possible to leverage transfer learning features obtained using an autoencoder and create an additional layer to refine\nthe features for anomaly detection. However, the price for the change is that the objective becomes non-convex and thus the resulting algorithm\nfor inferring the parameters of the model will not lead to a global optima.\n\n%which is randomly initialized from the uniform distribution within the range $[-1, 1]$. $\\nu \\in (0,1)$, is the parameter that\n%equates to the percentage of anomalies within data, found via grid search in the experiments. $\\bias$ is the equivalent maximum margin separating plane from the origin.\n\n\n%%%%\n\\begin{comment}\n\\subsection{Connection to OC-SVM}\n\\label{sec:connection_to_OC-SVM}\n\nClearly, Equation~\\ref{eqn:oc-nn} is the special case of the Equation~\\ref{eqn:ocsvm-objective} presented in the Section~\\ref{sec:ocsvm}, where the weights $V$ in Equation~\\ref{eqn:oc-nn} are randomly initialised, and not optimised, and $g$ is the either a $linear$ and $sigmoid$ activations.\nMore generally, for a deep network with parameters $w$, we have\n\n\\begin{equation}\n\\label{eqn:gen-oc-nn}\n \\min_{w,\\bias} \\frac{1}{2} \\| w \\|_{2}^2 +\\frac{1}{\\nu} \\cdot \\frac{1}{N} \\sum_{i = 1}^N \\max( 0, \\bias - \\langle w, \\Phi( \\X_{n :} w ) \\rangle ) - \\bias\n\\end{equation}\n\\end{comment}\n\n%%%%\n\\vspace{-0.1cm}\n\\subsection{Training the model}\n\\label{sec:training}\nWe can optimize Equation~\\ref{eqn:oc-nn} using an alternate minimization approach: We first fix $r$ and optimize for $w$ and $V$. We then use\nthe new values of $w$ and $V$ to optimize $r$. However, as we will show, the optimal value of $r$ is just the $\\upsilon$-quantile of\nthe array $\\langle w,g(Vx_{n})\\rangle$. We first define the objective to solve for $w$ and $V$ as\n% \\vspace{-0.1cm}\n\\begin{equation}\n \\label{eqn:minimize_w_V}\n \\underset{w, V}{\\argmin} \\frac{1}{2} \\| w \\|_{2}^2 + \\frac{1}{2} \\| V \\|_F^2 + \\frac{1}{\\upsilon} \\cdot \\frac{1}{N} \\sum_{n = 1}^N \\ell( y_n, \\hat{y}_n( w, V ) )\n \\end{equation}\n where\n \\begin{align*}\n \\ell( y, \\hat{y} ) &= \\max( 0, y - \\hat{y} ) \\\\\n y_n &= \\bias \\\\\n \\hat{y}_n( w, V ) &= \\langle w, g( V x_n ) \\rangle\n \\end{align*}\n\nSimilarly the optimization problem for $\\bias$ is\n\\begin{equation}\n\\label{eqn:minimize_r}\n\\underset{\\bias}{\\argmin} \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - r\n\\end{equation}\n\\begin{comment}\n%%%\n\\begin{enumerate}[(1)]\n\t\\item Initialise $w^{(0)}, V^{(0)}, r^{(0)}$ randomly\n\t\\item for $t = 1, 2, \\ldots$\n\t\\begin{enumerate}[(a)]\n\t\t\\item find $(w^{(t+1)}, V^{(t+1)})$ that minimise the objective through Backpropagation (BP) algorithm.\n\t\t\\begin{equation}\n\t\t\\label{eqn:minimize_w_V}\n\t\t \\underset{w, V}{\\argmin} \\frac{1}{2} \\| w \\|_{2}^2 + \\frac{1}{2} \\| V \\|_F^2 + \\frac{1}{\\upsilon} \\cdot \\frac{1}{N} \\sum_{n = 1}^N \\ell( y_n, \\hat{y}_n( w, V ) )\n\t\t \\end{equation}\n\t\twhere\n\t\t\\begin{align*}\n\t\t\t\\ell( y, \\hat{y} ) &= \\max( 0, y - \\hat{y} ) \\\\\n\t\t\ty_n &= \\bias^{(t)} \\\\\n\t\t\t\\hat{y}_n( w, V ) &= \\langle w, g( V x_n ) \\rangle\n\t\t\\end{align*}\n\n\t\t\\item pick $\\bias^{(t)}$ to be the $\\upsilon^{\\text{th}}$ quantile of $\\{ \\hat{y}_n \\}_{n = 1}^N$, where\n\t\t$$ \\hat{y}_n^{(t+1)} = \\langle w^{(t+1)}, g( V^{(t+1)} x_n ) \\rangle. $$\n\t\\end{enumerate}\n\n\\end{enumerate}\n\nTo justify step 2(b), let us observe that the optimisation with respect to $\\bias$ is\\\\\n%\n\\begin{equation}\n\\label{eqn:minimize_r}\n\\underset{\\bias}{\\argmin} \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - r\n\\end{equation}\n\\end{comment}\n\\begin{theorem}\nGiven $w$ and $V$ obtained from solving Equation~\\ref{eqn:minimize_w_V}, the solution to Equation~\\ref{eqn:minimize_r} is given\nby the $\\upsilon^{\\text{th}}$ quantile of $\\{ \\hat{y}_n \\}_{n = 1}^N$, where\n $$ \\hat{y}_n = \\langle w, g(V x_n ) \\rangle. $$\n\\end{theorem}\n\\begin{proof}\n We can rewrite Equation~\\ref{eqn:minimize_r} as:\n\\begin{align*}\n%\\underset{\\bias}{\\argmin} \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - r \\\\\n\\underset{\\bias}{\\argmin} \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - \\left( \\bias - \\frac{1}{N} \\sum_{n = 1}^{N} \\hat{y}_n \\right) \\\\\n= \\underset{\\bias}{\\argmin} \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - \\left( \\frac{1}{N} \\sum_{n = 1}^{N} \\left[ \\bias - \\hat{y}_n \\right] \\right) \\\\\n= \\underset{\\bias}{\\argmin} \\left( \\sum_{n = 1}^N \\max( 0, \\bias - \\hat{y}_n ) \\right) - \\upsilon \\cdot \\left( \\sum_{n = 1}^{N} \\left[ \\bias - \\hat{y}_n \\right] \\right) \\\\\n= \\underset{\\bias}{\\argmin} \\sum_{n = 1}^N \\left[ \\max( 0, \\bias - \\hat{y}_n ) - \\upsilon \\cdot \\left( \\bias - \\hat{y}_n \\right) \\right] \\\\\n= \\underset{\\bias}{\\argmin} \\sum_{n = 1}^N\n\\begin{cases} (1 - \\upsilon) \\cdot \\left( \\bias - \\hat{y}_n \\right) & \\text{ if } \\bias - \\hat{y}_n > 0 \\\\ - \\upsilon \\cdot \\left( \\bias - \\hat{y}_n \\right) & \\text{ otherwise}\n\\end{cases}\n\\end{align*}\n% \\vspace{0.4cm}\n\\vspace{-0.1cm}\nWe can observe that the derivative with respect to $r$ is\n$$ F'( r ) = \\sum_{n = 1}^N \\begin{cases} (1 - \\upsilon) & \\text{ if } \\bias - \\hat{y}_n > 0 \\\\ -\\upsilon & \\text{ otherwise. } \\end{cases} $$\nThus, by F'( r ) = 0 we obtain\n%\n\\begin{align*}\n\t(1 - \\upsilon) \\cdot \\sum_{n = 1}^N \\indicator{ \\bias - \\hat{y}_n > 0 } &= \\upsilon \\cdot \\sum_{n = 1}^N \\indicator{ \\bias - \\hat{y}_n \\leq 0 } \\\\\n\t&= \\upsilon \\cdot \\sum_{n = 1}^N (1 - \\indicator{ \\bias - \\hat{y}_n > 0 }) \\\\\n\t&= \\upsilon \\cdot N - \\upsilon \\cdot \\sum_{n = 1}^N \\indicator{ \\bias - \\hat{y}_n > 0 },\n\\end{align*}\nor\n\\begin{equation}\n \\frac{1}{N} \\sum_{n = 1}^N \\indicator{ \\bias -\\hat{y}_n > 0 }= \\frac{1}{N} \\sum_{n = 1}^N \\indicator{ \\hat{y}_n < r } = \\upsilon\n\\end{equation}\\\\\nThis means we would require the $\\nu^{\\text{th}}$ quantile of $\\{ \\hat{y}_n \\}_{n = 1}^N$.\n\\end{proof}\n% \\vspace{0.3cm}\n%Algorithm\n\\vspace{-0.3cm}\n\\subsection{OC-NN Algorithm}\n\\label{sec:algorithm}\nWe summarize the solution in Algorithm~\\ref{alg1}. We initialize $\\bias^{(0)}$ in Line 2. We learn the parameters($w,V$) of the neural network\nusing the standard Backpropogation(BP) algorithm (Line 7). In the experiment section, we will train the model using features extracted from\nan autoencoder instead of raw data points. However this has no impact on the OC-NN algorithm. As show in Theorem 3.1, we solve for $\\bias$\nusing the $\\upsilon$-quantile of the scores $\\langle y_{n} \\rangle$. Once the convergence criterion is satisfied, the data points are\nlabeled normal or anomalous using the decision function $S_{n} = sgn(\\hat{y}_{n} - r)$.\n\n\\begin{algorithm}\n\\caption{one-class neural network (OC-NN) algorithm}\\label{alg:oc-nn}\n\\label{alg1}\n\\begin{algorithmic}[1]\n\\State{\n\\textbf{Input:} Set of points $\\X_{n :},\\hspace{0.2cm} n: 1,...,N$\n}\n\\State{\n\\textbf{Output:} A Set of decision scores $S_{n :}=\\hat{y}_{n :}$, n: 1,...,N for X\n}\n\n\\State Initialise $\\bias^{(0)}$\n\\State $t \\gets 0$\n\\While{(no convergence achieved)}\n\\State Find $(w^{(t+1)}, V^{(t+1)})$ \\Comment Optimize Equation~\\ref{eqn:minimize_w_V} using BP.\n\\State $r^{t+1} \\gets \\nu^{\\text{th}}$ quantile of $\\{ \\hat{y}_n^{t+1} \\}_{n = 1}^N$\n\\State $t \\gets t + 1$\n\\EndWhile \\label{endwhile}\n\\State \\textbf{end}\n\\State Compute decision score $S_{n :}= \\hat{y}_n -r \\mbox{ for each } \\X_{n :}$\n\\If{($S_{n :}$ $\\geq$ 0)}\n \\State $\\X_{n :}$ is normal point\n \\Else\n \\State $\\X_{n :}$ is anomalous\n\\EndIf\n\n\\State \\textbf{return} $\\{S_n\\}$\n\n\\end{algorithmic}\n\\end{algorithm}\n% \\vspace{-0.05cm}\n%\n\\noindent\n{\\bf Example:} We give a small example to illustrate that the minimum of the function\n\\[f(r) = \\left( \\frac{1}{N\\upsilon} \\cdot \\sum_{n = 1}^N \\max( 0, \\bias - y_n ) \\right) - r\\] occurs at the\nthe $\\upsilon$-quantile of the set $\\{y_{n}\\}$. \\\\\n\nLet $y =\\{1,2,3,4,5,6,7,8,9\\}$ and $\\upsilon = 0.33$. Then the minimum will occur at $f(3)$ as detailed in\nthe table below.\n\\[\n\\begin{array}{|r|l|r|} \\hline \\hline\nr & f(r) (\\mbox{expr}) & f(r) (\\mbox{value}) \\\\ \\hline \\hline\n1 & \\frac{1}{9*.33}[0 + 0 +\\ldots 0] - 1 & -1.00 \\\\ \\hline\n2 & \\frac{1}{9*.33}[1 + 0 +\\ldots 0] - 2 & -1.67 \\\\ \\hline\n\\rowcolor{green!20}\n3 & \\frac{1}{9*.33}[2 + 1 +\\ldots 0] - 3 & -1.99 \\\\ \\hline\n4 & \\frac{1}{9*.33}[3 + 2 + 1 +\\ldots 0] - 4 & -1.98 \\\\ \\hline\n5 & \\frac{1}{9*.33}[4 + 3 + 2 + 1\\ldots 0] - 5 & -1.63 \\\\ \\hline\n6 & \\frac{1}{9*.33}[5 + 4 + 3 + 2 + 1\\ldots 0] - 6 & -0.94 \\\\ \\hline\n7 & \\frac{1}{9*.33}[6 + 5 + 4 + 3 + 2 + 1\\ldots 0] - 7 & 0.07 \\\\ \\hline\n8 & \\frac{1}{9*.33}[7 + 6 + 5 + 4 + 3 + 2 + 1\\ldots 0] - 8 & 1.43 \\\\ \\hline\n9 & \\frac{1}{9*.33}[8 + 7 + 6 + 4 + \\ldots 0] - 9 & 3.12 \\\\ \\hline\n\\end{array}\n\\]\n\n\n\\begin{comment}\n\\subsection{Predicting with the model}\n\nOne convenient property of our model is that the anomaly detector will be inductive, i.e.\\ it can generalise to unseen data points.\nOne can interpret the model as learning a non linear projection of the input;\nsuch a representation should thus be able to accurately model any unseen points that lie on the same manifold as the data used to train the model. Formally, $f( x ) = \\langle w, \\Phi( \\X_{n :} w ) \\rangle$ computes decision score of the given data point.\nThe \"negative\" or \"closer\" $\\langle w, \\Phi( \\X_{n :} w ) \\rangle$ is, to origin and the more likely the point is deemed to be anomalous.\n\n%\n\\subsection{Relation to existing models}\nOur contribution is breaking-new grounds, to integrate the OC-SVM equivalent objective into the neural\nnetwork architecture, utilizing the features extracted through unsupervised tranfer representation learning for anomaly detection.\nSome previous works have employed deep networks for anomaly detection~\\cite{chalapathy2017robust,zhou2017anomaly}. For example, the recent work of \\cite{zhou2017anomaly} employed an autoencoder-inspired objective to train a robust deep autoencoder network, with extensions to capture structured noise within the data; Our method is distinct to robust deep autoencoders (RDA), wherein we introduce a novel one-class neural network objective function illustrating the analytic derivation of $\\nu^{th}$ quantile of the sorted output scores represents, the maximum margin distance, separating normal data from anomalies. Although, nonlinear projections~\\cite{bingham2001random,rathore2017ensemble} have been explored for anomaly detection. We are unaware of prior usage of transfer learnt autoencoder networks integrated with one-class neural network objective to detect anomalies in unsupervised fashion.\n\\end{comment}\n", "meta": {"hexsha": "a82fec1cb1266e6b5de36b0bbe539461d6890b96", "size": 12060, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ARXIV_DAD_Survey/src/method.tex", "max_stars_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey", "max_stars_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 107, "max_stars_repo_stars_event_min_datetime": "2019-01-11T12:06:24.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T12:03:57.000Z", "max_issues_repo_path": "ARXIV_DAD_Survey/src/method.tex", "max_issues_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey_Arxiv_WorkingDocument", "max_issues_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ARXIV_DAD_Survey/src/method.tex", "max_forks_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey_Arxiv_WorkingDocument", "max_forks_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 27, "max_forks_repo_forks_event_min_datetime": "2019-01-15T02:42:12.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-06T07:59:29.000Z", "avg_line_length": 56.8867924528, "max_line_length": 904, "alphanum_fraction": 0.6501658375, "num_tokens": 4330, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835207180245, "lm_q2_score": 0.7772998663336157, "lm_q1q2_score": 0.6494212289780591}} {"text": "\\appendix\n\\section{Cantilever beam dynamics}\n\nConsider an eVTOL wing with multiple electric motors and rotors mounted along the spar, shown in Fig.~\\ref{fig:wing_diagram}. The wing structure is idealized as a cantilever beam with non-structural lumped masses placed along the span, shown in Fig.~\\ref{fig:cantilever}. \n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.7\\textwidth]{images/wing_diagram.png}\n \\caption{Rotor layout for an example eVTOL wing}\n \\label{fig:wing_diagram}\n\\end{figure}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.7\\textwidth]{images/cantilever.png}\n \\caption{Idealization of eVTOL wing structure}\n \\label{fig:cantilever}\n\\end{figure}\n\nTo expand the generality of the analysis, consider a tapered circular spar running along the length of the wing. The cross-section of the spar is shown in Fig.~\\ref{fig:taper_beam}. The tube thickness is equal to $t$, and is uniform along the spar length. The radius of the circular cross-section varies from $r_1$ at the root of the beam to $r_2$ at the tip. The mass per unit span and bending stiffness of the spar is given by \n\\begin{align*}\nm(x) \\quad &= \\quad 2 \\pi \\rho r(x) t \\\\\nEI_{yy}(x) \\quad &= \\quad E \\pi r(x)^3 t\n\\end{align*}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.45\\textwidth]{images/taper_beam.png}\n\\vspace{-0.1cm}\n\\caption{Linearly tapered circular spar}\n\\label{fig:taper_beam}\n\\end{center}\n\\end{figure}\n\nThe first bending natural frequency is estimated using a Rayleigh-Ritz approximation, with a mode shape set by the static deflection of the cantilever due to a tip-load, i.e.,\n\\begin{equation}\nw''(x) \\quad = \\quad P (L - x)\n\\end{equation}\nIntegrate twice along the span and apply the cantilever boundary condition to obtain the deflected bending slope as \n\\begin{equation}\nw'(x) \\quad = \\quad \\frac{P}{E\\pi t}\\left[\\frac{a_2}{r(x)^2} \\spc+\\spc \\frac{a_1}{r(x)} \\spc+\\spc a_0\\right]\n\\end{equation}\nThe constants are \n\\begin{align*}\na_2 \\quad =& \\quad -\\frac{L}{2r'} \\spc-\\spc \\frac{r_1}{2{r'}^2} \\\\\na_1 \\quad =& \\quad -\\frac{1}{{r'}^2} \\\\\na_0 \\quad =& \\quad \\frac{L}{2r'r_1^2} \\spc-\\spc \\frac{1}{2{r'}^2 r_1}\n\\end{align*}\nIntegrate the bending slope along the span to obtain the deflected mode shape as \n\\begin{equation}\nw(x) \\quad = \\quad \\frac{P}{E\\pi t}\\left[b_0 \\spc + \\spc b_1 x \\spc+\\spc \\frac{b_m}{r(x)} \\spc+\\spc b_l \\log_e r(x)\\right]\n\\end{equation}\nThe constants are \n\\begin{align*}\nb_0 \\quad =& \\quad -\\frac{a_2}{r_1 r'} \\spc-\\spc \\frac{a_1}{r'} \\log_e(r_1)\\\\\nb_1 \\quad =& \\quad a_0 \\\\\nb_m \\quad =& \\quad -\\frac{a_2}{r'} \\\\\nb_l \\quad =& \\quad \\frac{a_1}{r'} \n\\end{align*}\n\\subsection{Potential energy}\nThe maximum potential energy is given by \n\\begin{equation*}\nU \\quad = \\quad \\frac{1}{2} \\int_0^L \\frac{P^2(L-x)^2}{EI_{yy}(x)} dx\n\\end{equation*}\nSubstitute for the bending stiffness to obtain \n\\begin{equation*}\nU \\quad = \\quad \\frac{1}{2} \\frac{P^2}{E\\pi t} [A + B + C](\\lambda) \n\\end{equation*}\nThe term $\\lambda$ is the spar taper ratio, given by \n\\begin{equation}\n\\lambda \\quad = \\quad \\frac{r_2}{r_1}\n\\end{equation}\nThe sum $A+B+C$ is well-represented by a cubic polynomial in $\\lambda$ as \n\\begin{equation*}\nA+B+C\\quad \\approx \\quad (1.6261 \\spc - \\spc 0.4826 \\lambda \\spc + \\spc 2.1197 \\lambda^2 \\spc - \\spc 0.5957 \\lambda^3) \\frac{1}{K^3}\n\\end{equation*}\nThe constant $K$ is the inverse of the spar aspect ratio, given by \n\\begin{equation}\nK \\quad = \\quad \\frac{2 (\\frac{t}{c})}{AR_{\\rm wing}}\n\\end{equation}\nHere, $\\frac{t}{c}$ is the wing thickness to chord ratio, and $AR_{\\rm wing}$ is the wing aspect ratio.\n\n\\subsection{Kinetic energy}\nThe maximum kinetic energy of the beam has contributions from three types of masses:\n\\begin{enumerate}\n\\item \\textbf{Structural masses}: This term deals with the kinetic energy of the spar structure. This contribution to kinetic energy is \n\\begin{equation*}\nKE_{s} \\quad = \\quad \\int_0^L \\frac{1}{2} \\spc 2 \\pi r(x) t \\rho \\spc w(x)^2 \\spc \\omega_n^2 \\spc dx \\quad = \\quad \\frac{P^2}{E^2 \\pi t} \\rho \\omega_n^2 (RHS)\n\\end{equation*}\nThe term $RHS$ is a function of the taper ratio $\\lambda$, length $L$ and spar taper ratio inverse $K$ only. Further, the dependence on wing taper $\\lambda$ can be approximated using a cubic polynomial in $\\lambda$. Thus, the kinetic energy coefficient $RHS$ is approximated as \n\\begin{equation*}\nRHS \\quad \\approx \\quad \\frac{L^2}{K^5} (0.04823 \\spc + \\spc 0.269 \\lambda \\spc + \\spc 0.15178 \\lambda^2 \\spc + \\spc 0.36917 \\lambda^3) \n\\end{equation*}\n\\item \\textbf{Non-structural distributed masses}: This term deals with the kinetic energy due to the skin and other components (e.g., wires) that do not contribute significantly to bending stiffness. The kinetic energy term is \n\\begin{equation*}\nKE_{\\rm NS} \\quad = \\quad \\int_0^L \\frac{1}{2} \\spc m_{\\rm NS}(x) \\spc w(x)^2 \\spc \\omega_n^2 \\spc dx \n\\end{equation*}\nThe non-structural mass due to wing skin is given by \n\\begin{equation*}\nm_{\\rm NS}(x) \\quad = \\quad \\frac{2 M_{\\rm NS}}{A (\\frac{t}{c})} r(x) \n\\end{equation*}\nThe term $\\frac{M_{\\rm NS}}{A}$ is the non-structural mass per unit plan-form area of the skin. Define an intermediate quantity $\\beta$ given by \n\\begin{equation*}\n\\beta \\quad = \\quad \\frac{M_{\\rm NS}}{A (\\frac{t}{c})}\n\\end{equation*}\nThe expression for kinetic energy of non-structural masses simplifies to \n\\begin{equation*}\nKE_{\\rm NS} \\quad = \\quad \\beta \\spc \\frac{P^2 \\omega_n^2}{E^2 \\pi^2 t^2} \\spc RHS\n\\end{equation*}\n\\item \\textbf{Non-structural lumped mass}: This term refers to the kinetic energy of the electric motor/speed controller unit, rotor hubs and rotor blades. These discrete lumped masses are mounted along the wing spar, and the kinetic energy of these masses is given by \n\\begin{equation*}\nKE_{\\rm lump} \\quad = \\quad \\mathlarger{\\mathlarger{\\Sigma}}_i^{N_{\\rm M}}\\spc \\frac{1}{2} \\spc M_i \\spc w(x_i)^2 \\quad = \\quad \\frac{P^2 \\omega_n^2}{E^2 \\pi^2 t^2} \\Delta RHS\n\\end{equation*}\nThe coefficient $\\Delta RHS$ is \n\\begin{equation*}\n\\Delta RHS \\quad = \\quad \\mathlarger{\\mathlarger{\\Sigma}}_i^{N_{\\rm M}} \\frac{M_i}{2} \\left[b_0 \\spc + \\spc b_1 x_i \\spc + \\spc b_l \\log_e r(x_i) \\spc + \\spc \\frac{b_m}{r(x_i)} \\right]^2 \n\\end{equation*}\n\\end{enumerate}\n\\subsection{Frequency tuning}\nThe goal is to determine the spar mass by setting a target first natural frequency $\\omega_n$ for the cantilever beam with distributed and lumped non-structural masses. Equating the maximum total kinetic energy and maximum potential energy for the first bending mode, we obtain\n\\begin{equation}\n\\frac{1}{2} \\frac{P^2}{E^2 \\pi t} \\spc LHS \\quad = \\quad \\omega_n^2 \\left[\\frac{P^2}{E^2 t} (RHS) \\spc + \\spc \\frac{P^2}{E^2 \\pi^2 t^2} \\beta (RHS) \\spc + \\spc \\frac{P^2}{E^2 \\pi^2 t^2} \\Delta RHS \\right]\n\\end{equation}\nSolve for spar thickness as \n\\begin{equation*}\nt \\quad = \\quad \\left(\\frac{E \\pi}{2 \\omega_n^2} LHS \\spc - \\spc \\rho \\pi RHS \\right)^{-1} \\left[\\beta (RHS) \\spc + \\spc \\Delta RHS \\right]\n\\end{equation*}\nThe total spar mass is therefore \n\\begin{equation}\nM_{\\rm spar} \\quad = \\quad \\pi \\rho \\spc t (1 + \\lambda) r_1 \n\\end{equation}\n\nFor validation of the Rayleigh-Ritz solution, several cases were evaluated with and without lumped non-structural masses. \n\n\\subsection{Validation}\n\\subsubsection{Case 1: cantilever beam}\nConsider a cantilever beam with a hollow circular cross-section of constant wall thickness, and the cross-section radius is linearly tapered from root to tip, with a taper ratio of $\\lambda$ varying from 0.5 to 1.2. The beam length is 2.9 m, mean tube radius is 0.1411 m. The first bending natural frequency predictions from the Rayleigh-Ritz approximation are compared against a beam Finite Element Method analysis. These two frequencies are plotted in Fig.~\\ref{fig:beam_freq1}. The Rayleigh-Ritz approximation exhibits about 2\\% over-prediction, but otherwise captures the trends accurately. In this example, no non-structural masses were considered.\n\n\\begin{figure}\n \\centering\n\t\\subfigure[Natural frequency vs. taper ratio]{\\includegraphics[width=0.45\\textwidth]{images/beam_freq1b.png}}\n\t\\subfigure[Rayleigh-Ritz vs. FEM]{\\includegraphics[width=0.45\\textwidth]{images/beam_freq1.png}}\n \\caption{Natural frequency: Rayleigh-Ritz vs. FEM predicted natural frequencies for a tapered cantilever beam}\n \\label{fig:beam_freq1}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n\t\\subfigure[Natural frequency vs. taper ratio]{\\includegraphics[width=0.45\\textwidth]{images/beam_freq2b.png}}\n\t\\subfigure[Rayleigh-Ritz vs. FEM]{\\includegraphics[width=0.45\\textwidth]{images/beam_freq2.png}}\n \\caption{Natural frequency: Rayleigh-Ritz vs. FEM for a tapered cantilever beam with distributed non-structural masses}\n \\label{fig:beam_freq2}\n\\end{figure}\n\n\\subsubsection{Case 2: distributed non-structural mass}\nThis case is representative of an aircraft half-wing with skin, and no lumped masses. The non-structural masses are represented as a mass per unit area, i.e., $\\large\\frac{M}{A}$ kg/$m^2$. Three different values of the non-structural mass per unit wing area were evaluated: 0, 2.0 and 4.9 kg/m$^2$. The resulting natural frequencies were calculated from both Rayleigh-Ritz and FEM, and are plotted in Fig.~\\ref{fig:beam_freq2}. Excellent agreement is obtained.\n\n\\subsubsection{Case 3: half-wing with lumped masses}\nThis case is representative of an eVTOL wing. The dimensions of the beam are the same as in the previous examples, and the distributed skin mass per unit area is 4.9 kg/m$^2$. Two lumped masses (21.3 kg, 27 kg) are placed along the beam at 0.5$L$ and $L$, respectively. The wing layout and natural frequencies are shown in Fig.~\\ref{fig:beam_freq3}.\n\n\\begin{figure}\n \\centering\n\t\\subfigure[Layout of eVTOL wing]{\\includegraphics[width=0.3\\textwidth]{images/wing_layout.png}}\n\t\\subfigure[Rayleigh-Ritz vs. FEM]{\\includegraphics[width=0.45\\textwidth]{images/beam_freq3b.png}}\n \\caption{Natural frequency: Rayleigh-Ritz vs. FEM for an example eVTOL wing}\n \\label{fig:beam_freq3}\n\\end{figure}", "meta": {"hexsha": "393bd2c42ed33d43f77ac09cfee351fa64720bc6", "size": 10024, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Autodoc/theory_prog_manual/Appendices.tex", "max_stars_repo_name": "ananthsridharan/vtol_sizing", "max_stars_repo_head_hexsha": "3f754e1bd3cebdb5b5c68c8a2d84c47be1df2f02", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2020-03-24T10:20:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-22T18:49:25.000Z", "max_issues_repo_path": "Autodoc/theory_prog_manual/Appendices.tex", "max_issues_repo_name": "ananthsridharan/vtol_sizing", "max_issues_repo_head_hexsha": "3f754e1bd3cebdb5b5c68c8a2d84c47be1df2f02", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-12-08T10:26:41.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-04T18:19:59.000Z", "max_forks_repo_path": "Autodoc/theory_prog_manual/Appendices.tex", "max_forks_repo_name": "ananthsridharan/vtol_sizing", "max_forks_repo_head_hexsha": "3f754e1bd3cebdb5b5c68c8a2d84c47be1df2f02", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2018-11-27T21:21:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-04-20T15:44:18.000Z", "avg_line_length": 60.0239520958, "max_line_length": 653, "alphanum_fraction": 0.7193735036, "num_tokens": 3294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835248143776, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.649421219231564}} {"text": "% Copyright 2021 Google LLC\n%\n% Use of this source code is governed by an MIT-style\n% license that can be found in the LICENSE file or at\n% https://opensource.org/licenses/MIT.\n\n%!BIB program = biber\n%!TeX program = lualatex\n%!TeX spellcheck = en-US\n\n\\documentclass[hctr2.tex]{subfiles}\n\\begin{document}\n\\section{Specification}\\label{specification}\n\\subfile{algorithm.tex}\n\\subsection{Notation}\n\\begin{itemize}\n \\item $\\abs{X}$: length of $X \\in \\bin^{*}$ in bits\n \\item $\\lambda$: the empty string $\\abs{\\lambda} = 0$\n \\item $X[a;l]$: the substring of $X$ of length $l$ starting at the 0-based index $a$\n \\item $\\Concat$: bitstring concatenation\n \\item \\(\\xor\\): bitwise XOR\n \\item \\(n\\): block size in bits\n \\item $\\fromint_l: \\{0 \\ldots 2^l-1\\} \\rightarrow \\bin^l$:\n little-endian conversion of integers to binary; \n \\(\\fromint(x)\\) means \\(\\fromint_n(x)\\)\n \\item $\\pad(X) = X \\Concat 0^v$\n where $v$ is the least integer $\\geq 0$ such that $n$ divides $\\abs{X} + v$\n \\item \\(x, x^2, \\ldots\\): elements of the finite field \\(\\GF(2^n)\\)\n \\item \\(E: \\mathcal{K} \\times \\bin^n \\rightarrow \\bin^n\\): \n \\(n\\)-bit block cipher with keyspace \\(\\mathcal{K}\\);\n our concrete proposal uses AES\\cite{aes},\n so \\(n=128\\) and \\(\\mathcal{K}\\) is\n \\(\\bin^{128}\\), \\(\\bin^{192}\\), or \\(\\bin^{256}\\)\n \\item \\(\\mathcal{T}\\): the set of permissible tweaks\n \\(\\mathcal{T} = \\bigcup_{i \\in \\{0\\ldots2^{n-1}-2\\}}\\bin^i\\)\n \\item \\(\\mathcal{M}\\): the set of permissible messages\n \\(\\mathcal{M} = \\bigcup_{i \\in \\{n\\ldots n + 2^{n-1}-2\\}}\\bin^i\\)\n\\end{itemize}\nWe map bytes to bitstrings with \\(\\fromint_8\\). Subscripts may denote partial application; if we define $f: A \\times B \\rightarrow C$ and\n$a \\in A$ then $f_a: B \\rightarrow C$, and if $f_a^{-1}$ exists then $f_a^{-1}(f_a(b)) = b$.\n\n\\subsection{Polynomial hash function}\\label{hashspec}\nWe interpret \\(n\\)-bit blocks as little-endian field elements of \\(\\GF(2^n)\\),\nso \\(001 \\Concat 0^{n-3}\\) is interpreted as the element \\(x^2\\).\nPer \\cite{aes_gcm_siv,aes_gcm_siv_rfc} we define\n\\begin{align*}\n \\POLYVAL(\\hgen, \\lambda) & = 0^n\\\\\n \\POLYVAL(\\hgen, A \\Concat B) & = (\\POLYVAL(\\hgen, A) \\xor B) \\otimes \\hgen \\otimes x^{-n}\n\\end{align*}\nwhere \\(\\abs{\\hgen} = \\abs{B} = n\\) and\n\\(\\otimes\\) is multiplication in the finite field.\nIn our concrete proposal, \\(n=128\\), we reduce by\n\\(x^{128} + x^{127} + x^{126} + x^{121} + 1\\),\nand the value of the field element \\(x^{-n}\\)\nis equal to \\(x^{127} + x^{124} + x^{121} + x^{114} + 1\\).\n\nFor hash key \\(\\hgen \\in \\bin^n\\), tweak \\(T\\) and message \\(M\\), we define:\n\\begin{align*}\n & H_{\\hgen}(T, M) \\\\\n \\defeq & \n \\begin{cases}\n \\POLYVAL(\\hgen, \\fromint(2\\abs{T} + 2) \\Concat \\pad(T) \\Concat M) &\n \\text{if } n \\text{ divides } \\abs{M} \\\\\n \\POLYVAL(\\hgen, \\fromint(2\\abs{T} + 3) \\Concat \\pad(T) \\Concat \\pad(M \\Concat 1)) &\n \\text{otherwise}\n \\end{cases}\n\\end{align*}\n\n\\subsection{XCTR mode}\nHCTR and HCTR2 use an unusual mode of stream encryption,\nwhich we name \\emph{XCTR mode}:\n\\begin{displaymath}\n \\XCTR_k(S) = E_k(S \\xor \\fromint(1)) \\Concat E_k(S \\xor \\fromint(2)) \\Concat E_k(S \\xor \\fromint(3)) \\Concat \\cdots\n\\end{displaymath}\nGenerating the first \\(m\\) bits \\(\\XCTR_k(S)[0; m]\\) takes \\(\\ceil{m/n}\\) block cipher calls. \n\n\\subsection{HCTR2}\nHCTR2 encryption, defined in \\autoref{pseudocode},\ntakes a tweak and a plaintext,\nand returns a ciphertext of the same length as the plaintext.\nHCTR2 decryption (\\autoref{decryptionpseudocode})\nrecovers the plaintext given the same tweak and the ciphertext, ie\nfor \\(k \\in \\mathcal{K}\\), \\(T \\in \\mathcal{T}\\)\nand \\(P \\in \\mathcal{M}\\),\n\\(\\textsc{Decrypt}(k, T, \\textsc{Encrypt}(k, T, P)) = P\\).\n\\end{document}\n", "meta": {"hexsha": "2f0362ffcc978752aeab5fa7db2264235c552d5a", "size": 3754, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/specification.tex", "max_stars_repo_name": "google/hctr2", "max_stars_repo_head_hexsha": "1913bddafea9797108446d4284a1e3f8e43a1b5a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2021-11-05T14:23:38.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-30T17:24:43.000Z", "max_issues_repo_path": "paper/specification.tex", "max_issues_repo_name": "google/hctr2", "max_issues_repo_head_hexsha": "1913bddafea9797108446d4284a1e3f8e43a1b5a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 17, "max_issues_repo_issues_event_min_datetime": "2021-11-01T21:35:34.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-14T23:41:06.000Z", "max_forks_repo_path": "paper/specification.tex", "max_forks_repo_name": "google/hctr2", "max_forks_repo_head_hexsha": "1913bddafea9797108446d4284a1e3f8e43a1b5a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-11-04T17:58:46.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-28T13:29:03.000Z", "avg_line_length": 43.1494252874, "max_line_length": 137, "alphanum_fraction": 0.631326585, "num_tokens": 1334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916099737806, "lm_q2_score": 0.7520125848754471, "lm_q1q2_score": 0.6492813563761566}} {"text": "\\chapter{Maximum Likelihood Estimation for Multiparameter Models}\\label{S:MultiParamEst}\n\n\\section{Introduction}\\label{S:MultiParamEstIntro}\nWhen two or more parameters index a statistical experiment we want to estimate the vector-valued parameter $\\theta^* := (\\theta^*_1,\\ldots,\\theta^*_k)$.\nHere we will find the maximum likelihood estimates of vector-valued parameters. \n\nThe maximum likelihood estimator (MLE) of a possibly unknown but fixed parameter $\\theta^* := (\\theta^*_1,\\ldots,\\theta^*_k)$ in a multi-parametric experiment, i.e.~$\\theta^* \\in \\BB{\\Theta} \\subset \\Rz^k$ with $1 < k < \\infty$ is defined analogously to \\hyperref[D:LklFn]{Definition~\\ref*{D:MLE}} with the exception that we allow the parameter to be a vector. We take an excursion in multi-dimensional optimisation before finding the MLE of a parametric experiment involving two parameters.\n\n \\section{Practical Excursion in Multi-dimensional Optimisation}\\label{S:PracticalMultiDimOptimization}\nThe basic idea involves multi-dimensional iterations that attempt to converge on a local maximum close to the starting vector $\\theta^{(0)} \\in \\BB{\\Theta}$ (our initial guess). We can employ {\\sc Matlab}'s built-in function {\\tt fminsearch} to find the MLE of vector-valued parameters such as in the $\\lognormal$ model with two parameters, i.e.~$\\theta=(\\lambda,\\zeta) \\in \\BB{\\Theta} \\subset \\Rz^2$. The function {\\tt fminsearch} is similar to {\\tt fminbnd} except that it handles a given function of many variables, and the user specifies a starting vector $\\theta^{(0)}$ rather than a starting interval. Thus, {\\tt fminsearch} tries to return a vector $\\theta^{(*)}$ that is a local minimiser of, $-\\log(L(x_1,x_2,\\ldots,x_n; \\theta)$, the negative log-likelihood function of the vector-valued parameter $\\theta$, near this starting vector $\\theta^{(0)}$. \nWe illustrate the use of {\\tt fminsearch} on a more challenging target called the Levy density:\n\n{\\scriptsize\n\\begin{equation}\\label{E:LevyDensity}\nf(x,y) = \\exp \\left(-\\frac{1}{50} \\left( \\left( \\sum_{i=1}^5 {i \\cos{((i-1)x+i)} } \\right) \\left( \\sum_{j=1}^5 {j \\cos{((j+1)y+j)} } \\right)+ (x + 1.42513 )^2 + (y + 0.80032)^2\\right) \\right)\n\\end{equation}\n}\n\n\\begin{figure}[htpb]\n\\caption{Plot of Levy density as a function of the parameter $(x,y) \\in [-10,10]^2$ scripted in \\hyperref[Mf:LevyDensityPlot]{Labwork \\ref*{Mf:LevyDensityPlot}}.\\label {F:LevyDensityPlot}}\n\\begin{center}\n\\makebox{\\includegraphics[width=6.750in]{figures/LevyDensityT50}}\n\\end{center}\n\\end{figure}\n\n{\\tt fminsearch} uses the simplex search method [Nelder, J.A., and Mead, R.~1965, Computer Journal, vol.~7, p.~308-313]. For an animation of the method and more details, please visit \\href{http://en.wikipedia.org/wiki/Nelder-Mead_method}{\\url{http://en.wikipedia.org/wiki/Nelder-Mead_method}}. \n%For a more recent treatment see Lagarias, J.C., J. A. Reeds, M. H. Wright, and P. E. Wright, {\\em Convergence Properties of the Nelder-Mead Simplex Method in Low Dimensions}, SIAM Journal of Optimisation, Vol. 9 Number 1, pp. 112-147, 1998. \nAn advantage of the method is that it does not use numerical (finite differencing) or analytical (closed-form expressions) gradients but relies on a direct search method. Briefly, the simplex algorithm tries to ``tumble and shrink'' a simplex towards the local valley of the function to be minimised. If $k$ is the dimension of the parameter space or domain of the function to be optimised, a $k$-dimensional simplex is specified by its $k+1$ distinct vertices each of dimension $k$. Thus, a simplex is a triangle in a two-dimensional space and a pyramid in a three-dimensional space. At each iteration of the algorithm: \n\\begin{enumerate}\n\\item A new point inside or nearby the current simplex is proposed.\n\\item The function's value at the newly proposed point is compared with its values at the vertices of the simplex. \n\\item One of the vertices is typically replaced by the proposed point, giving rise to a new simplex. \n\\item The first three steps are repeated until the diameter of the simplex is less than the specified tolerance. \n\\end{enumerate}\n\nA major limitation of {\\tt fminsearch}, as demonstrated with the Levy target (encoded in \\hyperref[Mf:NegLevyDensity]{Labwork~\\ref*{Mf:NegLevyDensity}}) is that it can only give local solutions. The {\\bf global maximiser} of the Levy function $f(x,y)$ is $(-1.3069, -1.4249)$ and the {\\bf global maximum} is $f (-1.3069, -1.4249)=33.8775$ For instance, if we start the search close to, say $(x^{(0)},y^{(0)})=(-1.3,-1.4)$, as shown below, then the simplex algorithm converges as desired to the solution $(-1.3068, -1.4249)$.\n\\begin{VrbM}\n>> [params, fvalue, exitflag, output] = fminsearch('NegLevyDensity',[-1.3 -1.4],options)\nparams = -1.3068 -1.4249\nfvalue = -33.8775\nexitflag = 1\noutput = \n iterations: 24\n funcCount: 46\n algorithm: 'Nelder-Mead simplex direct search'\n message: [1x194 char]\n\\end{VrbM}\nHowever, if we start the search further away, say $(x^{(0)},y^{(0)})=(1.3,1.4)$, as shown below, then the algorithm converges to the {\\bf local maximiser} $(1.1627,1.3093)$ with a {\\bf local maximum} value of $f(1.1627,1.3093)=0.9632$, which is clearly smaller than the global maximum of $33.8775$. \n\\begin{VrbM}\n>> [params, fvalue, exitflag, output] = fminsearch('NegLevyDensity',[1.3 1.4],options)\nparams = 1.1627 1.3093\nfvalue = -0.9632\nexitflag = 1\noutput = \n iterations: 29\n funcCount: 57\n algorithm: 'Nelder-Mead simplex direct search'\n message: [1x194 char]\n\\end{VrbM}\nTherefore, we have to be extremely careful when using point-valued, iterative, local optimisation algorithms, implemented in floating-point arithmetic to find the global maximum. Other examples of such algorithms include:\n\\begin{itemize}\n\\item {\\bf Conjugate Gradient Method}:\\\\ \\href{http://en.wikipedia.org/wiki/Conjugate_gradient_method}{\\url{http://en.wikipedia.org/wiki/Conjugate_gradient_method}}\n\\item {\\bf Broyden-Fletcher-Goldfarb-Shanno (BFGS) method}:\\\\ \\href{http://en.wikipedia.org/wiki/BFGS_method}{\\url{http://en.wikipedia.org/wiki/BFGS_method}}\n\\item {\\bf Simulated Annealing}:\\\\ \\href{http://en.wikipedia.org/wiki/Simulated_annealing}{\\url{http://en.wikipedia.org/wiki/Simulated_annealing}}\n\\end{itemize}\nIn general, we have no guarantee that the output of such local optimisation routines will indeed be the global optimum. In practice, you can start the search at several distinct starting points and choose the best local maximum from the lot.\n\n\\begin{figure}[hbpt]\n\\caption{Plot of the ``well-behaved'' (uni-modal and non-spiky) $\\log(L((x_1,x_2,\\ldots,x_{100});\\lambda,\\zeta))$, based on $100$ samples $(x_1,x_2,\\ldots,x_{100})$ drawn from the $\\lognormal(\\lambda^*=10.36,\\zeta^*=0.26)$ as per \\hyperref[Mf:LogNormalLogLklPlot]{Labwork \\ref*{Mf:LogNormalLogLklPlot}}.\\label {F:LogNormalLogLklPlot}}\n\\begin{center}\n\\makebox{\\includegraphics[width=4.0in]{figures/LogNormalLogLklPlot}}\n\\end{center}\n\\end{figure}\n\nWhen the target function is ``well-behaved,'' i.e.~uni-modal or single-peaked and not too spiky, the optimisation routine can be expected to perform well. Log-likelihood functions are often well-behaved. Let us generate $100$ samples from an RV $C\\sim \\lognormal(\\lambda^*=10.36, \\zeta^*=0.26)$ by exponentiating the samples from the $\\normal(10.36,0.26^2)$ RV, and then compute the corresponding MMEs and MLEs for parameters $(\\lambda,\\zeta)$ using the formulae in \\hyperref[T:MMEMLE]{Table~\\ref*{T:MMEMLE}}.\n\\begin{VrbM}\n>> rand('twister',001); % set the fundamental sampler\n>> % draw 100 samples from the Lognormal(10.36,0.26) RV\n>> Cs = exp(arrayfun(@(u)(Sample1NormalByNewRap(u,10.36,0.26^2)),rand(1,100))); \n>> MLElambdahat = mean(log(Cs)) % maximum likelihood estimate of lambda \nMLElambdahat = 10.3397\n>> MLEzetahat = sqrt(mean( (log(Cs)-MLElambdahat) .^ 2)) % max. lkl. estimate of zeta\nMLEzetahat = 0.2744\n>> MMEzetaahat = sqrt(log(var(Cs)/(mean(Cs)^2) + 1)) % moment estimate of zeta\nMMEzetaahat = 0.2624\n>> MMElambdahat = log(mean(Cs))-(0.5*MMEzetaahat^2) % moment estimate of lambda\nMMElambdahat = 10.3417\n\\end{VrbM}\nLet us try to apply the simplex algorithm to find the MLE numerically. We first encode the negative log-likelihood function of the parameters $(\\lambda,\\zeta) \\in (0,\\infty)^2$ for the given data $x$, as follows: \n{\\VrbMf[label=NegLogNormalLogLkl.m]{scripts/NegLogNormalLogLkl.m}}\nHere is how we can call {\\tt fminsearch} and find the MLE after the re-transformation. \n\\begin{VrbM}\n>> [params, fvalue, exitflag, output] = ...\nfminsearch(@(params)(NegLogNormalLogLkl(Cs,params)),[log(5), log(1)])\nparams = 2.3360 -1.2931\nfvalue = -1.0214e+03\nexitflag = 1\noutput = \n iterations: 74\n funcCount: 131\n algorithm: 'Nelder-Mead simplex direct search'\n message: [1x194 char]\n>> % But we want exp(params) since we defined lambda and zeta as exp(params)\nexp(params)\nans = 10.3397 0.2744\n\\end{VrbM}\nNote that the MLEs $(\\widehat{\\lambda}_{100},\\widehat{\\zeta}_{100}) = (10.3397,0.2744)$ from $74$ iterations or ``tumbles'' of the `Nelder-Mead simplex (triangle)' and the MLEs agree well with the direct evaluations {\\tt MLElambdahat} and {\\tt MLEzetahat} based on the formulae in \\hyperref[T:MMEMLE]{Table~\\ref*{T:MMEMLE}}.\n\n%%\\newpage\n%\\begin{labwork}\n%Recall \\hyperref[LW:lognormal]{labwork~\\ref*{LW:lognormal}} where you simulated $1000$ samples directly from the RV $C$ (in part 3.). Pretend that you do not know the true parameters used in this particular simulation from RV $C$ and do the following:\n%\\begin{enumerate}\n%\\item Store the first 10, the first 100 and all 1000 samples in data arrays named {\\tt x10}, {\\tt x100} and {\\tt x1000} [Please don't do this manually!].\n%\\item Report the MLE of the two parameters for each of the three sub-arrays of data, namely {\\tt x10}, {\\tt x100} and {\\tt x1000}, i.e.~report the six point estimates: $\\widehat{\\lambda}_{10},\\widehat{\\zeta}_{10},\\widehat{\\lambda}_{100},\\widehat{\\zeta}_{100},\\widehat{\\lambda}_{1000},\\widehat{\\zeta}_{1000}$.\n%\\item Report the MME of the two parameters for each of the three sub-arrays of data, namely {\\tt x10}, {\\tt x100} and {\\tt x1000}, i.e.~report the six point estimates: $\\widehat{\\lambda}_{10},\\widehat{\\zeta}_{10},\\widehat{\\lambda}_{100},\\widehat{\\zeta}_{100},\\widehat{\\lambda}_{1000},\\widehat{\\zeta}_{1000}$.\n%\\item Discuss in a short paragraph what you can deduce from the two sets of point estimates. Explain how the maximum likelihood (ML) and method of moments (MM) estimates of the parameters are related to the true parameters used in the simulation as the sample size increases in powers of $10$. \n%\\item {\\bf There is no credit for this part}: Try to use {\\tt fminsearch} to numerically find the MLE for your data and make a 2D-plot of the log-likelihood function being maximised.\n%\\end{enumerate} \n%\\end{labwork}\n\n\\subsubsection*{Summarizing Table of Point Estimators}\nUsing the sample mean $\\overline{X}_n$ and sample standard deviation $S_n$ defined in \\eqref{E:SampleMeanRV} and \\eqref{E:SampleStdDevRV}, respectively, we summarise the two point estimators of the parameters of some common distributions below. For some cases, the MLE is the same as the MME (method of moments) and can be solved analytically.\n\\begin{center}\n\\begin{table}[htbp]\n\\caption{Summary of the Method of Moment Estimator (MME) and the Maximum Likelihood Estimator (MLE) for some IID Experiments. \\label{T:MMEMLE}}\n\\begin{tabular}{l | r | r}\n\\hline\nStatistical Experiment & MLE & MME \\\\ \\hline\n$X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} \\bernoulli(\\theta)$ & $\\widehat{\\theta}=\\overline{X}_n$ & same as MLE \\\\ \\hline\n$X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} \\exponential(\\lambda)$ & $\\widehat{\\lambda}={1}/{\\overline{X}_n} $ & same as MLE \\\\ \\hline\n$X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} \\normal(\\mu,\\sigma^2)$ & $\\widehat{\\mu}=\\overline{X}_n, \\widehat{\\sigma} = \\sqrt{\\frac{n-1}{n}S^2_n} $ & $\\widehat{\\mu}=\\overline{X}_n, \\widehat{\\sigma} = S_n $ \\\\ \\hline\n$X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} \\lognormal(\\lambda,\\zeta)$ & $\\widehat{\\lambda}=\\frac{1}{n}{\\sum_{i=1}^n \\log(X_i)} $ & $\\widehat{\\lambda} = \\log(\\overline{X}_n) - \\frac{1}{2} {\\widehat{\\zeta}} \\ ^2$ \\\\ %\\\\\n & $\\widehat{\\zeta} = \\sqrt{\\frac{1}{n} \\sum_{i=1}^n{(\\log(X_i)-\\widehat{\\lambda})^2}} $ & $\\widehat{\\zeta} = \\sqrt{\\log \\left({S_n^2}/{\\overline{X}_n^2} +1 \\right)}$ \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\\end{center}\n\n\\section{Confidence Sets for Multiparameter Models}\\label{S:ConfSetsMultiParamModels}\nWe will extend the Fisher Information and Delta method to models with more than one parameter:\n\\[\nX_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} f(x;\\theta^*), \\qquad \\theta^* := (\\theta_1^*, \\theta_2^*, \\ldots, \\theta_k^*) \\in \\BB{\\Theta} \\subset \\Rz^k \\ .\n\\]\nLet, the ML estimator of the fixed and possibly unknown vector-valued parameter $\\theta^*$ be:\n\\[\n\\widehat{\\Theta}_n := \\left( \\widehat{\\Theta}_{1,n}, \\widehat{\\Theta}_{2,n}, \\ldots, \\widehat{\\Theta}_{k,n} \\right), \\qquad \\widehat{\\Theta}_n := \\widehat{\\Theta}_n(X_1,X_2,\\ldots,X_n) : \\Xz_n \\to \\BB{\\Theta}\n\\]\nand the ML estimate based on $n$ observations $x_1,x_2,\\ldots,x_n$ be:\n\\[\n\\widehat{\\theta}_n := \\left( \\widehat{\\theta}_{1,n}, \\widehat{\\theta}_{2,n}, \\ldots, \\widehat{\\theta}_{k,n} \\right), \\qquad \\widehat{\\theta}_n := \\widehat{\\theta}_n(x_1,x_2,\\ldots,x_n) \\in \\BB{\\Theta} \\ .\n\\]\nLet the log-likelihood function and its Hessian matrix $H = (H_{i,j})_{i,j=1,2,\\ldots,k}$ of partial derivatives be:\n\\[\n\\ell_n(\\theta) := \\ell_n(\\theta_1,\\theta_2,\\ldots,\\theta_k) := \\sum_{i=1}^n \\log(f(x_i; (\\theta_1,\\theta_2,\\ldots,\\theta_k))), \\qquad\nH_{i,j} := \\frac{\\partial}{\\partial \\theta_i} \\frac{\\partial}{\\partial \\theta_j} \\ell_n(\\theta_1,\\theta_2,\\ldots,\\theta_k) \\ ,\n\\]\nrespectively, provided the log-likelihood function is sufficiently smooth.\n\\begin{definition}[Fisher Information Matrix] The Fisher Information matrix is:\n\\begin{equation}\\label{E:FisherInfoMat}\nI_n(\\theta) := I_n(\\theta_1,\\theta_2,\\ldots,\\theta_k) =\n-\n\\begin{bmatrix}\n\\E_{\\theta}(H_{1,1}) & \\E_{\\theta}(H_{1,2}) & \\cdots & \\E_{\\theta}(H_{1,k}) \\\\\n\\E_{\\theta}(H_{2,1}) & \\E_{\\theta}(H_{2,2}) & \\cdots & \\E_{\\theta}(H_{2,k}) \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\\E_{\\theta}(H_{k,1}) & \\E_{\\theta}(H_{k,2}) & \\cdots & \\E_{\\theta}(H_{k,k})\n\\end{bmatrix}\n\\end{equation}\nand its matrix inverse is denoted by $I_n^{-1}(\\theta)$.\n\\end{definition}\n\\begin{prop}[Asymptotic Normality of MLE in Multiparameter Models]\nLet \n$$X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} f(x_1;\\theta_1^*,\\theta_2^*,\\ldots,\\theta_k^*), \\qquad \\theta^*=(\\theta_1^*,\\theta_2^*,\\ldots,\\theta_k^*)\\in \\BB{\\Theta}\\subset \\Rz^k \\ ,$$\nfor some fixed and possibly unknown $\\theta^* \\in \\BB{\\Theta}\\subset \\Rz^k$. Then, under appropriate regularity conditions:\n\\[\n{\\widehat{\\Theta}_n} := \\left( \\widehat{\\Theta}_{1,n}, \\widehat{\\Theta}_{2,n}, \\ldots, \\widehat{\\Theta}_{k,n} \\right) \\rightsquigarrow \\normal(\\theta^*,I_n^{-1})\n\\]\nIn other words, the vector-valued estimator $\\widehat{\\Theta}_n$ converges in distribution to the multivariate $\\normal$ distribution centred at the unknown parameter $\\theta^*$ with the variance-covariance matrix given by inverse Fisher Information matrix $I_n^{-1}$. Furthermore, let $I_n^{-1}(j,j)$ denote the $j^{\\text{th}}$ diagonal entry of $I_n^{-1}$. In this case:\n% and let $$\\widehat{\\mathsf{se}}_n ( \\widehat{\\Theta}_{j,n}) = \\sqrt{ I_n^{-1}(j,j)}$$ be the standard error of the estimator of $\\theta_j^*$\n\\[\n\\frac{\\widehat{\\Theta}_{j,n} - \\theta^*_j}{\\sqrt{I_n^{-1}(j,j)}} \\rightsquigarrow \\normal(0,1)\n\\]\nand the approximate covariance of $\\widehat{\\Theta}_{i,n}$ and $\\widehat{\\Theta}_{j,n}$ is:\n\\[\n{\\sf Cov}({\\Theta}_{i,n},{\\Theta}_{j,n}) \\approxeq I_n^{-1}(i,j) \\ .\n\\]\n\\end{prop}\nNow, let us look at a way of obtaining ML estimates and confidence sets for functions of $\\theta$. Suppose the real-valued function $g(\\theta)=\\psi:\\BB{\\Theta} \\to \\BB{\\Psi}$ maps points in the $k$-dimensional parameter space $\\BB{\\Theta} \\subset \\Rz^k$ to points in $\\BB{\\Psi} \\subset \\Rz$. Let the gradient of $g$ be\n\\[\n\\bigtriangledown g(\\theta)\n:=\n\\bigtriangledown g (\\theta_1,\\theta_2,\\ldots,\\theta_k)\n=\n\\begin{pmatrix}\n\\frac{\\partial} {\\partial \\theta_1}{g(\\theta_1,\\theta_2,\\ldots,\\theta_k)} \\\\\n\\frac{\\partial} {\\partial \\theta_2}{g(\\theta_1,\\theta_2,\\ldots,\\theta_k)} \\\\\n\\vdots \\\\\n\\frac{\\partial} {\\partial \\theta_k}{g(\\theta_1,\\theta_2,\\ldots,\\theta_k)} \\\\\n\\end{pmatrix} \\ .\n\\]\n\\begin{prop}[Multiparameter Delta Method]\nSuppose:\n\\begin{enumerate}\n\\item $X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} f(x_1;\\theta_1^*,\\theta_2^*,\\ldots,\\theta_k^*), \\qquad \\theta^*=(\\theta_1^*,\\theta_2^*,\\ldots,\\theta_k^*)\\in \\BB{\\Theta}\\subset \\Rz^k$,\n\\item Let $\\widehat{\\Theta}_n$ be a ML estimator of $\\theta^* \\in \\BB{\\Theta}$ and let $\\widehat{\\theta}_n$ be its ML estimate, and\n\\item Let $g(\\theta)=\\psi:\\BB{\\Theta} \\to \\BB{\\Psi} \\subset \\Rz$ be a smooth function such that $\\bigtriangledown g(\\widehat{\\theta}_n) \\neq 0$.\n\\end{enumerate}\nThen: \n\\begin{enumerate}\n\\item $\\widehat{\\Psi}_n = g(\\widehat{\\Theta}_n)$ is the ML estimator and $\\widehat{\\psi}_n = g(\\widehat{\\theta}_n)$ is the ML estimate of of $\\psi^*=g(\\theta^*) \\in \\BB{\\Psi}$,\n\\item The standard error of the ML estimator of $\\psi^*$ is:\n$$\\widehat{\\sf{se}}_n(\\widehat{\\Psi}_n) = \\sqrt{ \\left( \\bigtriangledown g(\\widehat{\\theta}_n) \\right)^T I_n^{-1}(\\widehat{\\theta}_n) \\left( \\bigtriangledown g(\\widehat{\\theta}_n) \\right)}, $$\n\\item The ML estimator of $\\psi^*$ is asymptotically normal, i.e.:\n$$\\frac{\\widehat{\\Psi}_n - \\psi^*}{\\widehat{\\sf{se}}_n(\\widehat{\\Psi}_n)} \\rightsquigarrow \\normal(0,1) \\ ,$$\n\\item And a $1-\\alpha$ confidence interval for $\\psi^*$ is:\n\\[\n\\widehat{\\psi}_n \\pm z_{\\alpha/2} {\\widehat{\\sf{se}}_n(\\widehat{\\Psi}_n)} \n\\]\n\\end{enumerate}\n\\end{prop}\nLet us put the theory to practice in the problem of estimating the coefficient of variation from samples of size $n$ from an RV.\n\\begin{example}[Estimating the Coefficient of Variation of a $\\normal(\\mu^*,{\\sigma^*}^2)$ RV]\\label{EX:CoeffOfVarForBivNormal}\nLet \n$$\n\\psi^* = g(\\mu^*,\\sigma^*)=\\sigma^*/\\mu^*, \\qquad X_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} \\normal(\\mu^*,{\\sigma^*}^2) \\ .\n$$ \nWe do not know the fixed parameters $(\\mu^*,\\sigma^*)$ and are interested in estimating the coefficient of variation $\\psi^*$ based on $n$ IID samples $x_1,x_2,\\ldots,x_n$. We have already seen that the ML estimates of $\\mu^*$ and $\\sigma^*$ are:\n\\[\n\\widehat{\\mu}_n = \\overline{x}_n := \\frac{1}{n}\\sum_{i=1}^n x_i, \\qquad \n\\widehat{\\sigma}_n = s_n := \\sqrt{\\frac{1}{n} \\sum_{i=1}^n (x_i-\\widehat{\\mu}_n)^2} \\ .\n\\]\nThus, the ML estimate of $\\psi^*=\\sigma^*/ \\mu^*$ is:\n\\[\n\\widehat{\\psi}_n = \\frac {\\widehat{\\sigma}_n}{\\widehat{\\mu}_n} = \\frac {s_n}{\\overline{x}_n}\n\\] \nWe can now derive the standard error of the ML estimator $\\widehat{\\Psi}_n$ by first computing $I_n(\\mu,\\sigma)$, $I_n^{-1}(\\mu,\\sigma)$, and $\\bigtriangledown g(\\mu,\\sigma)$. A careful computation shows that:\n\\[\nI_n(\\mu,\\sigma) = \n\\begin{bmatrix}\n\\frac{n}{\\sigma^2} & 0\\\\\n0 & \\frac{2n}{\\sigma^2}\n\\end{bmatrix},\n\\qquad \\qquad\nI_n^{-1}(\\mu,\\sigma) = \n\\frac{1}{n}\n\\begin{bmatrix}\n{\\sigma^2} & 0\\\\\n0 & \\frac{\\sigma^2}{2}\n\\end{bmatrix},\n\\qquad \\qquad\n\\bigtriangledown g(\\mu,\\sigma) =\n\\begin{pmatrix}\n-\\frac{\\sigma}{\\mu^2} \\\\\n\\frac{1}{\\mu}\n\\end{pmatrix} \\ .\n\\]\nTherefore, the standard error of interest is:\n\\[\n\\widehat{\\sf{se}}_n(\\widehat{\\Psi}_n) = \\sqrt{ \\left( \\bigtriangledown g(\\widehat{\\theta}_n) \\right)^T I_n^{-1}(\\widehat{\\theta}_n) \\left( \\bigtriangledown g(\\widehat{\\theta}_n) \\right)} = \n\\frac{1}{\\sqrt{n}} \\sqrt{\\frac{1}{\\widehat{\\mu}_n^4} + \\frac{\\widehat{\\sigma}_n^2}{2 \\widehat{\\mu}_n^2}}\n\\]\nand the $95\\%$ confidence interval for the unknown coefficient of variation $\\psi^*$ is:\n\\[\n\\widehat{\\psi}_n \\pm z_{\\alpha/2} {\\widehat{\\sf{se}}_n(\\widehat{\\Psi}_n)} = \n\\frac{s_n}{\\overline{x}_n} \\pm z_{\\alpha/2} \\left( \\frac{1}{\\sqrt{n}} \\sqrt{\\frac{1}{\\widehat{\\mu}_n^4} + \\frac{\\widehat{\\sigma}_n^2}{2 \\widehat{\\mu}_n^2}} \\right)\n\\]\n\\end{example}\n\nLet us get our hands dirty in the machine with Labwork~\\ref{LW:CoeffOfVarForBivNormal} next.\n\\begin{labwork} [Computing the coefficient of variation of a $\\normal(\\mu^*,{\\sigma^*}^2)$ RV]\\label{LW:CoeffOfVarForBivNormal}\nLet us apply these results to $n=100$ simulated samples from $\\normal(100,10^2)$ as follows.\n{\\VrbMf[label=CoeffOfVarNormal.m]{scripts/CoeffOfVarNormal.m}}\n\\begin{VrbM}\n>> CoeffOfVarNormal\nMuhat = 100.3117\nSigmahat = 10.9800\nPsihat = 0.1095\nSehat = 0.0077\nConfInt95 = 0.0943 0.1246\n\\end{VrbM}\n\\end{labwork}\n", "meta": {"hexsha": "718bfbd5c216b460c42dba44068aa35174c8e9f0", "size": 20525, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "matlab/csebook/MultiParamEstim.tex", "max_stars_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_stars_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-02-19T07:54:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-14T13:55:18.000Z", "max_issues_repo_path": "matlab/csebook/MultiParamEstim.tex", "max_issues_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_issues_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "matlab/csebook/MultiParamEstim.tex", "max_forks_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_forks_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2018-07-18T07:47:52.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-19T11:28:24.000Z", "avg_line_length": 72.7836879433, "max_line_length": 865, "alphanum_fraction": 0.6994884287, "num_tokens": 6959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085808877581, "lm_q2_score": 0.8267117940706734, "lm_q1q2_score": 0.6492238658048131}} {"text": "\\section{M/M/m Queues}\n\\label{sec:M-M-m-Queues}\n\nA $M/M/m$ is a queue where \n(i) the arrival process is Poissonian with rate $\\lambda$,\n(ii) the service process is Exponential with rate $\\mu$,\n(iii) there are $m$ servers,\n(iv) the buffer has infinite capacity,\n(v) the scheduling policy is FCFS.\n\n%\\begin{figure}[tp]\n%\\label{fig:M-M-m-Queue}\t\n%\t\\centering\n%\t\\includegraphics{fig/M-M-m-Queue}\n%\t\\caption{An M/M/m queue and its corresponding CTMC.}\n%\\end{figure}\n\nThe key question in these types of systems is determining the \\textit{Queue Probability} $P_{Q}$, that is the probability that an arriving job is enqueued.\n\n\\begin{definition}[Utilization]\n\\label{def:M-M-m-Utilization}\n\tFor any $M/M/m$, the system utilization is\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Utilization}\n\t\\varrho = \\frac{\\lambda}{m \\mu}\n\t\\end{equation}\n\\end{definition}\n\n\\begin{definition}[Resource Requirement]\n\\label{def:M-M-m-Resource-Requirement}\n\tFor any $M/M/m$, the resource requirement is\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Resource-Requirement}\t\n\tR = \\frac{\\lambda}{\\mu}\n\t\\end{equation}\n\\end{definition}\n\nThe resource requirement $R$ is \n(i) the minimum number of servers needed to keep the system stable,\n(ii) the expected number of busy servers, and \n(iii) the expected number of jobs in service.\n\t\n\\begin{theorem}[State Probability]\n\\label{thm:M-M-m-Probability-State}\t\n\tFor any $M/M/m$, the state probability is\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Probability-State}\n\t\\pi_{i} = \\frac{m^{q} \\varrho^{i}}{q!} \\cdot \\pi_{0}\n\t\\end{equation}\n\t\n\twith\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Probability-State-Zero}\t\n\t\\pi_{0} = \\Big[ \\sum_{i=0}^{m-1} \\frac{(m \\varrho)^{i}}{i!} + \\frac{(m \\varrho)^{m}}{m! (1- \\varrho)} \\Big]^{-1}\n\t\\end{equation}\n\t\n\twhere\n\t\n\t$q=\\min\\{i,m\\}$ is the number of busy servers.\n\t\n\t\\begin{proof}\n\t\tFor a formal demonstration, see \\cite{harchol2013performance} on page 258-259.\n\t\\end{proof}\n\\end{theorem}\n\n\\begin{theorem}[Queue Probability]\n\t\\label{thm:M-M-m-Probability-Queue}\t\n\tFor any $M/M/m$, the queue probability is\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Probability-Queue}\n\tP_{Q} = \\frac{(m \\varrho)^{m}}{m!(1-\\varrho)} \\cdot \\pi_{0}\n\t\\end{equation}\n\t\n\t\\begin{proof}\n\t\tFor a formal demonstration, see \\cite{harchol2013performance} on page 260.\n\t\\end{proof}\n\\end{theorem}\n\n\\Cref{thm:M-M-m-Probability-Queue} is called \\textit{Erlang-C Formula}.\n\n\\begin{theorem}[Mean Queue Jobs]\n\t\\label{thm:M-M-m-Mean-Queue-Jobs}\t\n\tFor any $M/M/m$, the expected number of jobs in queue is\n\t\n\t\\begin{equation}\n\t\\label{eqn:M-M-m-Mean-Queue-Jobs}\n\t\\expected{N_{Q}} = \\frac{\\varrho}{(1-\\varrho)} \\cdot P_{Q}\n\t\\end{equation}\n\t\n\t\\begin{proof}\n\t\tFor a formal demonstration, see \\cite{harchol2013performance} on page 262.\n\t\\end{proof}\n\\end{theorem}\n\nThe remaining metrics ($\\expected{N},\\expected{T},\\expected{T_{Q}}$) could be determined by applying the Little's Law, the basic definitions \n$\\expected{N}=\\expected{N_{Q}}+\\expected{N_{S}}$, \n$\\expected{N_{S}}=R$, \n$\\expected{T}=\\expected{T_{Q}}+\\expected{T_{S}}$, and\n$\\expected{T_{S}}=\\frac{1}{\\mu}$.\n\nIn particular we obtain:\n\n\\begin{equation}\n\\label{eqn:M-M-m-Delay}\n\\expected{T_{Q}} = \\frac{1}{\\lambda} P_{Q} \\frac{\\varrho}{(1-\\varrho)}\n\\end{equation}\n\n\\begin{equation}\n\\label{eqn:M-M-m-Reponse-Time}\n\\expected{T} = \\frac{1}{\\lambda} P_{Q} \\frac{\\varrho}{(1-\\varrho)} + \\frac{1}{\\mu}\n\\end{equation}\n\n\\begin{equation}\n\\label{eqn:M-M-m-System-Jobs}\n\\expected{N} = P_{Q} \\frac{\\varrho}{(1-\\varrho)} + m\\varrho\n\\end{equation}\n\n\n\n\n\\subsection{Comparison between M/M/m and M/M/1}\n\\label{sec:Comparison-M-M-m-And-M-M-1}\n\nLet us consider \n(i) a $M/M/m$ with average arrival rate $\\lambda$ and average service rate $\\mu$, and\n(ii) a $M/M/1$ with average arrival rate $\\lambda$ and average service rate $m \\mu$.\n\nUnder light load, in (i) jobs are served by few servers with rate $\\mu$; whereas in (ii) jobs are served with rate $m \\mu$.\nUnder high load, in (i) the state is always greater than $m$ and it behaves just like (ii).\n\nHence, if $\\lambda^{M/M/1} = \\lambda^{M/M/m}$ and $\\mu^{M/M/1} = m \\mu^{M/M/m}$, then $M/M/1$ is always at least $m$ time faster than $M/M/m$.\n\n\n\n\n\\subsection{Comparison between M/M/m/m and M/M/m}\n\\label{sec:Comparison-M-M-m-m-And-M-M-m}\n\nIt is interesting to compare the block probability of a $M/M/m/m$ with the queue probability of an equivalent $M/M/m$.\nThe following theorem compares $P_{block}^{M/M/m/m}$ with $P_{Q}^{M/M/m}$.\n\n\\begin{theorem}[Relation $P_{block}^{M/M/m/m}$ with $P_{Q}^{M/M/m}$]\n\\label{thm:Relation-Probability-Block-Probability-Queue}\t\n\tFor any $M/M/m/m$ with block probability $P_{block}$ and any $M/M/m$ with queue probability $P_{Q}$, we have that\n\t \n\t\\begin{equation}\n\t\\label{eqn:Relation-Probability-Block-Probability-Queue}\n\tP_{block} = \\frac{(1-\\varrho)P_{Q}}{1-\\varrho P_{Q}}\n\t\\end{equation}\n\t\n\t\\begin{proof}\n\t\tFor a formal demonstration, see \\cite{harchol2013performance} on page 261.\n\t\\end{proof}\n\\end{theorem}", "meta": {"hexsha": "1272a3467b392d5bc68b024b381720f41167a6a9", "size": 4909, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "performance-modeling/sec/m-m-m-queues.tex", "max_stars_repo_name": "gmarciani/research", "max_stars_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-07-27T13:31:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-20T12:54:12.000Z", "max_issues_repo_path": "performance-modeling/sec/m-m-m-queues.tex", "max_issues_repo_name": "gmarciani/research", "max_issues_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "performance-modeling/sec/m-m-m-queues.tex", "max_forks_repo_name": "gmarciani/research", "max_forks_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-02-17T13:30:49.000Z", "max_forks_repo_forks_event_max_datetime": "2018-02-17T13:30:49.000Z", "avg_line_length": 30.8742138365, "max_line_length": 155, "alphanum_fraction": 0.6858830719, "num_tokens": 1688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085708384736, "lm_q2_score": 0.8267118004748677, "lm_q1q2_score": 0.6492238625262197}} {"text": "\\section{Calculations and Results}\nFrom the following equaitions,\n\n$$ \\tau_z = I\\beta_z $$\n$$ I_{ab} = I_A + I_B $$\n$$ M_\\mu = -I_1\\beta_1 $$\n$$ T = m(g-a)$$\n\nWe can derive\n$$ m(g-R\\beta_2)R-M_\\mu=I_1\\beta_2 $$\nThen we find\n$$ I_1=\\frac{mR(g-R\\beta_2)}{\\beta_2-\\beta_1} $$\nSimilarly, if a rigid body with an unknown moment of inertia is placed on the\nturntable, we may find \n$$ I_2=\\frac{mR(g-R\\beta_4)}{\\beta_4-\\beta_3} $$\nUsing the fact that the moment of inertia is an additive quantity, the moment of\ninertia of the rigid object placed on the turntable, with respect to the axis of\nrotation, may be found as the difference \n$$ I_3 = I_2 - I_1 $$\n\n\\subsection{Measurement of Angular Acceleration}\nAngular acceleration can be derive by investigating the measurement data (k,t),\nthe corresponding angular position is\n$$ \\theta = k\\pi = \\omega_0 t + \\frac{1}{2}\\beta t^2 $$\n\n\n\\subsection{In Lab Data}\n\n\\begin{table}[H]\n \\centering\n \\begin{tabularx}{\\textwidth}{|p{6cm}|X|X|X|X|}\n \\hline\n Object & 1 & 2 & 3 & 4 \\\\\n \\hline\n Disk $[cm] \\pm 0.002[cm]$& 24.098 & 24.094 & 24.094 & 24.094 \\\\\n Hoop 1 $[cm] \\pm 0.002[cm]$& 20.982 & 20.918 & 20.966 & 20.976 \\\\\n Hoop 2 $[cm] \\pm 0.002[cm]$& 23.998 & 24.000 & 24.000 & 24.002 \\\\\n Cylinder A $[cm] \\pm 0.002[cm]$& 2.994 & 2.994 & 2.994 & 2.994 \\\\\n Cylinder B $[cm] \\pm 0.002[cm]$& 2.994 & 2.994 & 2.994 & 2.994 \\\\\n Cone pulley $[cm] \\pm 0.002[cm]$& 5.022 & 5.020 & 5.008 & 5.008 \\\\\n Hole 1 d $[cm] \\pm 0.002[cm]$& 3.978 & 5.534 & & \\\\\n Hole 2 d $[cm] \\pm 0.002[cm]$& 3.982 & 5.540 & & \\\\\n Hole 3 d $[cm] \\pm 0.002[cm]$& 5.524 & 6.500 & & \\\\\n Hole 4 d $[cm] \\pm 0.002[cm]$& 5.520 & 6.520 & & \\\\\n \\hline\n \\end{tabularx}\n \\caption{Calliper measurements}\n \\end{table}\n\n\n\\begin{table}[H]\n \\centering\n \\begin{tabularx}{\\textwidth}{|X|X|}\n \\hline\n Object & Mass\\\\\n\t\\hline\n Disk $[g] \\pm 0.1 [g] $ & 493.1\\\\\n Hoop $[g] \\pm 0.1 [g] $ & 422.5\\\\\n Cylinder A $[g] \\pm 0.1 [g] $ & 165.8\\\\\n Cylinder B $[g] \\pm 0.1 [g] $ & 165.8\\\\\n Weight $[g] \\pm 0.1 [g] $ & 59.1 \\\\\n \\hline\n \\end{tabularx}\n \\caption{Mass measurements}\n \\end{table}\n% begin real data\n\\begin{table}[H]\n \\centering\n\\begin{tabular}{|p{2cm}|p{1.5cm}|l|l|l|l|l|l|l|l|l|}\n\\hline\nSituation & A or D & k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\\\n\\hline\nEmpty & Dec & $t[s]$ & 0.2958 & 0.5924 & 0.8899 & 1.1881 & 1.4873 & 1.7871 & 2.0879 & 2.3895 \\\\\nEmpty & Acc & $t[s]$ & 0.9005 & 1.5186 & 2.0204 & 2.4557 & 2.8448 & 3.1996 & 3.5280 & 3.7038 \\\\\nWith disk & Dec & $t[s]$ & 0.3178 & 0.6362 & 0.9554 & 1.2752 & 1.5958 & 1.9170 & 2.2390 & 2.5616 \\\\\nWith disk & Acc & $t[s]$ & 0.8957 & 1.5808 & 2.1582 & 2.6666 & 3.1264 & 3.5491 & 3.9428 & 4.3322 \\\\\nWith hoop & Dec & $t[s]$ & 0.2450 & 0.4903 & 0.7359 & 0.9818 & 1.2281 & 1.4746 & 1.7216 & 1.9688 \\\\\nWith hoop & Acc & $t[s]$ & 1.0035 & 1.7614 & 2.3967 & 2.9547 & 3.4589 & 3.9216 & 4.3521 & 4.7560 \\\\\nA 1 B 2 & Dec & $t[s]$ & 0.4491 & 0.9003 & 1.3536 & 1.8089 & 2.2666 & 2.7263 & 3.1883 & 3.6524 \\\\\nA 1 B 2 & Acc & $t[s]$ & 1.3586 & 2.0905 & 2.6595 & 3.1424 & 3.5692 & 3.9562 & 4.3125 & 4.6448 \\\\\nA 3 B 4 & Dec & $t[s]$ & 0.4539 & 0.9099 & 1.3678 & 1.8279 & 2.2899 & 2.7541 & 3.2204 & 3.6888 \\\\\nA 3 B 4 & Acc & $t[s]$ & 1.3748 & 2.1298 & 2.7179 & 3.2171 & 3.6585 & 4.0587 & 4.4273 & 4.7711 \\\\\n\\hline\n\\end{tabular}\n\\caption{ Time measurements}\n\\end{table}\n\nAccording to United States Department of Commerce, the standard gravitational\nacceleration is \n$$ g = 9.80665 m/s^2 $$\n\n% ==================================================================\n% ==================================================================\nFor the Empty turntable,\n\\newcommand{\\EFWwr}{17cm}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/etd}\n\\end{figure}\n\n$$ \\beta_1 = -0.0970 radius/s^2$$ (with 95\\% confidence bounds) \n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/eta}\n\\end{figure}\n\n$$ \\beta_2 = 2.6580 radius/s^2$$ (with 95\\% confidence bounds) \n\nThus,\n$$ I_1 = \\frac{59.1 g \\times 5.0145 cm \\times (9.80665 m/s^2 - 5.0145 cm \\times (-0.0970) radius/s^2 )}{2.6580 radius/s^2 -(-0.0970 radius/s^2) } = 0.0105 kg\\times m^2 $$\n\n% ==================================================================\n% ==================================================================\nFor turntable with disk,\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/wdd}\n\\end{figure}\n\n$$ \\beta_1 = -0.0668 radius/s^2$$ (with 95\\% confidence bounds) \n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/wda}\n\\end{figure}\n\n$$ \\beta_2 = 1.2970 radius/s^2$$ (with 95\\% confidence bounds) \n\n$$ I_2 = \\frac{59.1 g \\times 5.0145 cm \\times (9.80665 m/s^2 - 5.0145 cm \\times (-0.0668) radius/s^2 )}{1.2970 radius/s^2 -(-0.0668 radius/s^2) } = 0.0213 kg\\times m^2 $$\n\n\n% ==================================================================\n% ==================================================================\nFor turntable with hoop,\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/whd}\n\\end{figure}\n\n$$ \\beta_1 = -0.0689 radius/s^2$$ (with 95\\% confidence bounds) \n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/wha}\n\\end{figure}\n\n$$ \\beta_2 = 1.1444 radius/s^2$$ (with 95\\% confidence bounds) \n\n\n$$ I_3 = \\frac{59.1 g \\times 5.0145 cm \\times (9.80665 m/s^2 - 5.0145 cm \\times (-0.0689) radius/s^2 )}{1.1444 radius/s^2 -(-0.0689 radius/s^2) } = 0.0240 kg\\times m^2 $$\n\n\n% ==================================================================\n% ==================================================================\nFor Cylinder A in hole 1, B in hole 2\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/cabd}\n\\end{figure}\n\n$$ \\beta_1 = -0.0710 radius/s^2$$ (with 95\\% confidence bounds) \n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/caba}\n\\end{figure}\n\n$$ \\beta_2 = 1.8752 radius/s^2$$ (with 95\\% confidence bounds) \n\n\n$$ I_4 = \\frac{59.1 g \\times 5.0145 cm \\times (9.80665 m/s^2 - 5.0145 cm \\times (-0.0710) radius/s^2 )}{1.8752 radius/s^2 -(-0.0710 radius/s^2) } = 0.0149 kg\\times m^2 $$\n\n\n% ==================================================================\n% ==================================================================\nFor Cylinder A in hole 3, B in hole 4\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/ccdd}\n\\end{figure}\n\n$$ \\beta_1 = -0.0661 radius/s^2$$ (with 95\\% confidence bounds) \n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\EFWwr]{matlab/ccda}\n\\end{figure}\n\n$$ \\beta_2 = 1.7496 radius/s^2$$ (with 95\\% confidence bounds) \n\n\n$$ I_5 = \\frac{59.1 g \\times 5.0145 cm \\times (9.80665 m/s^2 - 5.0145 cm \\times (-0.0661) radius/s^2 )}{ 1.7496 radius/s^2 -(-0.0661 radius/s^2) } = 0.0160 kg\\times m^2 $$\n\nThus, the moment of inertia for disk is \n$$ I_{disk} = I_2 - I_1 = 0.0213 kg\\times m^2 - 0.0105 kg\\times m^2 = 0.0108 kg\\times m^2 $$\n\nThe moment of inertia for hoop is \n$$ I_{hoop} = I_3 - I_1 = 0.0240 kg\\times m^2 - 0.0105 kg\\times m^2 = 0.0135 kg\\times m^2 $$\n\nThe moment of inertia for Cylinder A in hole 1 and Cylinder B in hole 2 is \n$$ I_{A1B2} = I_4 - I_1 = 0.0149 kg\\times m^2 - 0.0105 kg\\times m^2 = 0.0044 kg\\times m^2 $$\n\nThe moment of inertia for Cylinder A in hole 3 and Cylinder B in hole 4 is \n$$ I_{A3B4} = I_5 - I_1 = 0.0160 kg\\times m^2 - 0.0105 kg\\times m^2 = 0.0055 kg\\times m^2 $$\n\n$$ I_{A3B4} -I_{A1B2} =0.0055 kg\\times m^2 - 0.0044 kg\\times m^2 = 0.0011 kg\\times m^2 $$\n$$ md^2 = 165.8 g \\times (4.7560 cm - 6.0120cm )^2 + 165.8 g \\times (4.7610 cm - 6.0200cm )^2 = 0.0052436 kg\\times m^2 $$\n\nThus, the relative uncertainty is $$ \\frac{ 0.0055 kg\\times m^2 - 0.0052436 kg \\times m^2}{0.0055 kg\\times m^2} = 4.66 \\% $$\n\n", "meta": {"hexsha": "42884dd390cc03042c19b7ec2804f1178d02b1a6", "size": 7717, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "E1/part/calculation.tex", "max_stars_repo_name": "iamwrm/VP141", "max_stars_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-06-24T11:28:04.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-24T11:28:04.000Z", "max_issues_repo_path": "E1/part/calculation.tex", "max_issues_repo_name": "iamwrm/VP141", "max_issues_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "E1/part/calculation.tex", "max_forks_repo_name": "iamwrm/VP141", "max_forks_repo_head_hexsha": "c0a5d1992967b1552d6f7ea0806c9244d58f64ac", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.8930232558, "max_line_length": 171, "alphanum_fraction": 0.5557859272, "num_tokens": 3232, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706733, "lm_q2_score": 0.7853085708384736, "lm_q1q2_score": 0.649223857496951}} {"text": "\\documentclass[11pt,a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage[english]{babel}\n\\usepackage{fancyhdr}\n\\usepackage[margin=.7in]{geometry}\n\\usepackage{exsheets}\n\\usepackage{enumerate}\n\n\\usepackage{./Style/linearProgramsStyle} % This is a set of commands for maths content.\n\n\\pagestyle{fancy}\n\n\\renewcommand{\\headrulewidth}{2pt}\n\\fancyhead[L]{EPITA\\_ING1\\_BING\\_2020\\_S5\\_PROL}\n\\fancyhead[R]{March 2018}\n\n\\fancyfoot[C]{\\textbf{\\thepage}}\n\\fancyfoot[L]{}\n\n\\pretitle{\\vspace{-.5\\baselineskip} \\begin{center}}\n\\title{%\n { \\huge Linear Programming}\\\\ %\n \\vspace{.5\\baselineskip}\n {\\Large \\textit{Complemetary Exam} }\n}\n\\posttitle{\n\\end{center}\n \\begin{flushleft}\n \\vspace{2\\baselineskip} \\textit{\n \\!\\!\\emph{Duration: 1h30.}\\\\\n \\! \\emph{No documents are allowed.} \\\\\n Only \\emph{non-programmable} pocket calculators are allowed.\n }\n \\end{flushleft}\n \\rule{\\textwidth}{1.5pt}\n \\vspace{-5\\baselineskip}\n}\n\\author{}\n\\date{}\n\n\\pdfinfo{\n /Author (Bashar Dudin)\n /Title (PROL Exam -- 2018)\n /Subject (Linear Programming)\n}\n\n\\begin{document}\n\n\\maketitle\\thispagestyle{fancy}\n\nWe're considering the following list of linear programs respectively\ngiven the names $(A)$, $(B)$, $(C)$, $(D)$, $(E)$ and $(F)$.\n\\[\n(A) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $3x_1 + 2x_2 + 3x_3$\n }{\n \\systeme{\n -x_1 - 2x_2 - 3x_3 \\leq 5,\n 2x_1 + x_2 + x_3 \\leq 3,\n -x_1 - x_2 + x_3 \\leq 1\n }\n }{\n $x_1$, $x_2$, $x_3 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\qquad (B)\\;\n\\left\\{\n \\begin{linearProg}{\n minimize\n }{\n $ x_1 + x_2 - x_3$\n }{\n \\systeme{\n -x_1 - 2x_2 + 3x_3 \\leq 1,\n - x_1 - x_2 + 2x_3 \\geq 2\n }\n }{\n $x_1$, $x_2$, $x_3 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\\[\n(C) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $3x_1 + 2x_2 + 3x_3$\n }{\n \\systeme{\n -x_1 - 2x_2 - 3x_3 \\leq 5,\n 2x_1 + x_2 + x_3 \\leq 3,\n -x_1 - x_2 + x_3 \\leq 1\n }\n }{\n $x_1$, $x_2 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\qquad (D) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $-x_1 - x_2 + x_3$\n }{\n \\systeme{\n -x_1 + 2x_2 - 3x_3 + x_4 = 1,\n x_1 + x_2 - 2x_3 \\leq -2\n }\n }{\n $x_1$, $x_2$, $x_3$, $x_4 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\\[\n\\qquad (E)\\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $ -x_1 - x_2 + x_3$\n }{\n \\systeme{\n -x_1 - 2x_2 + 3x_3 \\leq 1,\n - x_1 - x_2 + 2x_3 \\geq 2\n }\n }{\n $x_1$, $x_2$, $x_3 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\qquad (F)\\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $6x_1 + 4x_2 + 6x_3$\n }{\n \\systeme{\n -x_1 - 2x_2 - 3x_3 \\leq 5,\n 2x_1 + x_2 + x_3 \\leq 3,\n x_1 + x_2 - x_3 \\geq -1\n }\n }{\n $x_1$, $x_2 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\n\\section{Equivalence and Duality}\n\n\\begin{question}{3}\n Which previous programs are equivalent to each others? Sort them\n into groups of equivalent classes.\n\\end{question}\n\n\\begin{question}{2}\n Give standard form of $(F)$. What is the slack form of $(B)$?\n\\end{question}\n\n\\begin{question}{2}\n What are the dual programs of $(A)$ and $(E)$?\n\\end{question}\n\n\\begin{question}{2}\n Guess feasible solutions of $(A)$ and $(D)$.\n\\end{question}\n\n\\section{Simplex Algorithm}\n\n\\begin{question}{4}\n Solve linear program $(A)$ using the simplex algorithm.\n\\end{question}\n\n\\begin{question}{2}\n Out of previous solution of $(A)$ give an optimal point of its\n dual. What is its optimal objective value?\n\\end{question}\n\n\\begin{question}{5}\n Solve dual of linear program $(E)$.\n\\end{question}\n\n\\end{document}\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: t\n%%% End:\n", "meta": {"hexsha": "7715770a133db655bd5afca2975528e0a4629d87", "size": 3821, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Exams/TeX/Ecrit_PROL_2018_03_BING.tex", "max_stars_repo_name": "bashardudin/LinearPrograms", "max_stars_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 22, "max_stars_repo_stars_event_min_datetime": "2016-10-20T14:04:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-03-31T08:45:14.000Z", "max_issues_repo_path": "Exams/TeX/Ecrit_PROL_2018_03_BING.tex", "max_issues_repo_name": "bashardudin/LinearPrograms", "max_issues_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-06-04T01:30:01.000Z", "max_issues_repo_issues_event_max_datetime": "2018-10-09T08:18:10.000Z", "max_forks_repo_path": "Exams/TeX/Ecrit_PROL_2018_03_BING.tex", "max_forks_repo_name": "bashardudin/LinearPrograms", "max_forks_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-10-26T21:19:58.000Z", "max_forks_repo_forks_event_max_datetime": "2018-03-17T19:28:15.000Z", "avg_line_length": 19.105, "max_line_length": 87, "alphanum_fraction": 0.5681758702, "num_tokens": 1516, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.826711776992821, "lm_q2_score": 0.7853085859124002, "lm_q1q2_score": 0.6492238565473597}} {"text": "\\chapter{Newton and quasi-Newton methods}\n\\label{chapter:Newton}\n\n\\section{Basics of nonlinear iterations}\n\\begin{intro}\n The efficient solution of nonlinear problems is an important\n ingredient to implicit timestepping schemes as well as shooting\n methods. Without attempting completeness, we present some important\n facts about iterative methods for this problem. We introduce the two\n generic schemes, Newton and gradient methods, discuss their\n respective pros and cons and combine their features in order to\n obtain better methods.\n\\end{intro}\n\n\\input{definitions/nonlinear}\n\\input{definitions/iteration-order}\n\\input{definitions/newton}\n\\input{theorems/newton-kantorovich}\n\n\\begin{remark}\n Instead of proving the Newton-Kantorovich theorem, we discuss its\n main assumptions and features. First, we note that it does not\n require that the initial value be close to a root, or even assumes\n the existence of a root. The theorem is actually an existence proof.\n \n The Lipschitz condition on $\\nabla f$ can be seen as the deviation\n of $f$ from being linear. Indeed, if $f$ were linear, then $L=0$ and\n provided $M\\neq 0$ the method converges in a single step for any\n initial value.\n\n The larger the constant $M$, the smaller wone of the eigenvalues of % wone?\n the Jacobian. Therefore, the function becomes flat in that\n direction and the root finding problem becomes unstable.\n \n If we have convergence due to $\\beta_0 \\le 1/2$ (the proof shows\n contraction) there holds $\\beta_1 := LM \\left\\| f\\left(x^{(1)}\\right)\n \\right\\| < 1/2$ and we have quadratic convergence from the second step on.\n\\end{remark}\n\n\\input{definitions/gradient-method}\n\\input{theorems/gradient-method}\n\n% \\begin{proof}\n% First, we observe that in any point $x$ with $\\nabla F(x) \\neq 0$,\n% there exists $\\epsilon > 0$ such that\n% \\begin{gather*}\n% F(x) - F(x-\\epsilon\\nabla F(x))\n% = \\epsilon \\abs{\\nabla F(x)}^2 > 0.\n% \\end{gather*}\n% Thus, $F(x-\\epsilon\\nabla F(x)) < F(x)$. We conclude, that for such\n% $x$, the line search obtains a positive value of $\\alpha$. Thus, the\n% sequence of the gradient iteration is monotonically decreasing and\n% stays within the set $K$.\n% \\end{proof}\n\n\\section{Globalization}\n\n\\begin{intro}\n The convergence of the Newton method is only local, and it is the\n faster, the closer to the solution we start. Thus, finding good\n initial guesses is an important task. A reasonable initial guess in\n a one-step method seems to be $y_0$, but on closer inspection, this\n is true only if the time step is small. Therefore, the convergence\n requirements of Newton's method would insert a new time step\n restriction, which we want to avoid in the context of implicit\n methods. Therefore, and for other cases like the shooting\n methods of chapter~\\ref{chapter:rwa}, we present\n methods which extend the domain of convergence.\n\n As a rule, Newton'a method should never be implemented without some\n globalization strategy!\n\\end{intro}\n\n\\input{definitions/newton-line-search}\n\\input{definitions/newton-step-size}\n\n\\begin{remark}\n The step size control algorithm can be implemented with very low\n overhead. In fact, in each Newton step we only have to compute the\n norm of the residual, which is typically needed for the stopping\n criterion anyway. Additional work is only needed if the residual\n grows. But this is the case, when the original method was likely to\n fail.\n\n The convergence proof does not guarantee that the values of $j$\n remain bounded. Practically, this is irrelevant, since typically the\n step size control only triggers within the first few steps, then the\n quadratic convergence of the Newton method starts.\n\\end{remark}\n\n\\input{definitions/descent-methods}\n\n\\begin{remark}\n Obviously, the gradient method is a descent method, where the direction $s$\n is chosen parallel to $\\nabla F(x^{(k)})$ and $\\mu$ is chosen in an\n optimal way. It is also called the method of \\define{steepest\n descent}.\n\\end{remark}\n\n\\begin{Lemma}{newton-descent}\n The Newton method applied to the function $f(x)$ is a descent method\n applied to the functional $F(x) = \\abs{f(x)}^2$. The same holds for\n the Newton method with line search or step size control.\n\\end{Lemma}\n\n\\begin{proof}\n By the product rule, there holds\n \\begin{gather*}\n \\nabla F(x) = 2 f^T(x) \\nabla f(x).\n \\end{gather*}\n The search direction of the Newton method is\n \\begin{gather*}\n s = -\\frac{d^{(k)}}{\\abs{d^{(k)}}}\n = \\frac{\\bigl(\\nabla f(x^{(k)})\\bigr)^{-1}f(x^{(k)})}{\\abs{\\dots}}\n \\end{gather*}\n Thus, omitting the arguments $x^{(k)}$, we obtain\n \\begin{gather*}\n \\frac{\\nabla F\\, s}{\\norm{\\nabla F}}\n = \\frac{f^T \\nabla f(x)\\bigl(\\nabla f\\bigr)^{-1} f}%\n {\\norm{\\bigl(\\nabla f\\bigr)^{-1} f}\\,\\norm{f^T \\nabla f(x)}}\n \\ge \\frac{\\abs{f}^2}{\\norm{\\bigl(\\nabla f\\bigr)^{-1}}\\;\n \\norm{f}^2 \\norm{\\nabla f(x)} }\n = \\frac1{\\operatorname{cond}_2 (\\nabla f(x))},\n \\end{gather*}\n where we used the operator norm $\\norm{.}$ of matrices with respect\n tot the Euclidean norm of $\\R^d$. $\\operatorname{cond}_2(A)$ is the\n spectral condition of $A$, namely\n \\begin{gather*}\n \\operatorname{cond}_2(A) = \\norm{A} \\, \\norm{A^{-1}}.\n \\end{gather*}\n With ~\\ref{eq:descent-methods:1} we conclude that $s\\in \\mathcal S_\\gamma(\\nabla F)$ for any\n $\\gamma$ with\n \\begin{gather*}\n \\gamma \\le \\frac1{\\operatorname{cond}_2 (\\nabla f(x))}.\n \\end{gather*}\n The different variants of the Newton method are only distinguished by\n a different choice of the scaling parameter $\\mu$.\n\\end{proof}\n\n\\begin{Lemma}{downhill}\n Let $F: \\R^d \\to \\R$ be continuously differentiable. For a given\n point $x$, assume $\\nabla F = \\nabla F(x) \\neq 0$. Then, there is a\n constant $\\lambda > 0$ such that for any\n $s\\in \\mathcal S_\\gamma(\\nabla F(x))$ and any\n $0 \\le \\mu \\le \\lambda$ there holds\n \\begin{gather}\n \\label{eq:downhill:1}\n F(x-\\mu s) \\le F(x) - \\frac{\\gamma\\mu}2 \\abs{\\nabla F(x)}.\n \\end{gather}\n In particular, a positive scaling factor $\\mu$ for the descent method can\n always be found.\n\\end{Lemma}\n\n% Stoer/Bulirsch I, 4. Auflage, p. 238\n\\begin{proof}\n First, define\n \\begin{gather*}\n U_1(x) = \\bigl\\{ y\\in \\R^d \\big\\vert \\abs{\\nabla F(y)-\\nabla F(x)}\n \\le \\tfrac{\\gamma}{2} \\abs{\\nabla F(x)}\\bigr\\}.\n \\end{gather*}\n Since $\\nabla F$ is continuous and $\\nabla F(x) \\neq 0$, this set is\n a nonempty neighborhood of $x$.\n% Similarly, the set\n% \\begin{gather*}\n% U_2(x) =\n% \\bigl\\{ y\\in \\R^d \\big\\vert \\mathcal S_\\gamma(\\nabla F(y)) \\subseteq\n% \\mathcal S_{\\nicefrac{\\gamma}{2}}(\\nabla F(x))\\bigr\\}\n% \\end{gather*}\n% is a nonempty neighborhood of $x$.\nChoose now $\\lambda$ such that\n \\begin{gather*}\n B_{\\lambda}(x) \\subseteq U_1(x),% \\cap U_2(x).\n \\end{gather*}\n Hence, for any $\\mu\\in(0,\\lambda)$ and\n $s\\in \\mathcal S_\\gamma(\\nabla F(x))$, there holds by the mean value\n theorem with $0<\\theta<1$\n \\begin{multline*}\n F(x)-F(x-\\mu s) = \\mu \\nabla F(x-\\theta\\mu s)s \\\\\n = \\mu \\Bigl(\\bigl(\\nabla F(x-\\theta\\mu s) - \\nabla F(x)\\bigr)\n +\\nabla F(x)\\Bigr).\n \\end{multline*}\n Using the definitions of $U_1(x)$ and $U_2(x)$, we obtain\n \\begin{align*}\n F(x)-F(x-\\mu s)\n &\\ge -\\tfrac{\\gamma\\mu}{2}\\abs{\\nabla F(x)}\n + \\mu DF(x)s\\\\\n &\\ge -\\tfrac{\\gamma\\mu}{2}\\abs{\\nabla F(x)}\n + \\mu\\gamma\\abs{\\nabla F(x)}\\\\\n &= \\tfrac{\\gamma\\mu}{2}\\abs{\\nabla F(x)}.\n \\end{align*}\n\\end{proof}\n\n\\section{Practical considerations}\n\n\\begin{intro}\n Quadratic convergence is an asymptotic statement, which for any\n practical purpose can be replaced by ``fast'' convergence. Most of\n the effort spent in a single Newton step consists of setting up the\n Jacobian $J$ and solving the linear system in the second line\n of~\\eqref{eq:newton-def:1}. Therefore, we will consider techniques\n here, which avoid some of this work. We will have to consider two\n cases\n \\begin{enumerate}\n \\item Small systems with $d\\lesssim 1000$. For such systems, a\n direct method like $LU$- or $QR$-decomposition is advisable in\n order to solve the linear system. To this end, we compute the\n whole Jacobian and compute its decomposition, an effort of order\n $d^3$ operations. Comparing to $d^2$ operations for applying the\n inverse and order $d$ for all other tasks, this must be avoided as\n much as possible.\n \\item Large systems, where the Jacobian is typically sparse (most of\n its entries are zero). For such a system, the effort of order\n $d^2$ for a full matrix vector multiplication is already not\n affordable. Therefore, the linear problem is solved by an\n iterative method and we will not have to compute the Jacobian at\n all.\n \\end{enumerate}\n\\end{intro}\n\n\\begin{remark}\n In order to save numerical effort constructing and inverting\n Jacobians, the following strategies have been successful.\n \\begin{itemize}\n \\item Fix a threshold $0<\\eta<1$ which will be used as a bound for\n error reduction. In each Newton step, first compute the update\n vector $\\widehat d$ using the Jacobian $\\widehat J$ of the\n previous step. This yields the modified method\n \\begin{gather}\n \\label{eq:newton:1}\n \\begin{alignedat}2\n &&J_k &= J_{k-1} \\\\\n &&\\widehat x &= x^{(k)} - J_{k}^{-1}f(x^{(k)})\\\\\n \\text{If }\\abs{f(\\widehat x)} &\\le \\eta\n \\abs{f(x^{(k)})} \\quad& x^{(k+1)} &= \\widehat x\\\\\n \\text{Else } J_k &= \\bigl(\\nabla f(x^{(k)}\\bigr)^{-1}\n \\quad & x^{(k+1)} &=x^{(k)}- J_{k}^{-1}f(x^{(k)}). \n \\end{alignedat}\n \\end{gather}\n Thus, an old Jacobian and its inverse are used until convergence\n rates deteriorate. This method is again a quasi-Newton method\n which will not converge quadratically. However, we can obtain linear\n convergence at any rate $\\eta$.\n \\item If Newton's method is used within a time stepping scheme, the\n Jacobian of the last Newton step in the previous time step is\n often a good approximation for the Jacobian of the first Newton\n step in the new time step. This holds in particular for small time\n steps and constant extrapolation. Therefore, the previous method\n should also be extended over the bounds of time steps.\n \\item An improvement of the method above can be achieved by so\n called rank-1 updates. Given $x^{(k)}$ and $x^{(k-1)}$, compute\n \\begin{gather}\n \\label{eq:newton:2}\n \\begin{split}\n p &= x^{(k)} - x^{(k-1)}\\\\\n q &= f(x^{(k)}) - f(x^{(k-1)})\\\\\n J_k &= J_{k-1} + \\frac1{\\abs{p}^2}\n \\left(q - J_{k-1} p\\right)p^T\n \\end{split}\n \\end{gather}\n The fact that the rank of $J_k - J_{k-1}$ is at most one can be\n used to obtain a decomposition of $J_k$ in a cheap way from one\n for $J_{k-1}$.\n \\end{itemize}\n\\end{remark}\n\n\\begin{remark}\n For problems leading to large, sparse Jacobians, typically space\n discretizations of partial differential equations, computing\n inverses of $LU$-decompositions is infeasible. These matrices\n typically only feature a few nonzero elements per row, while the\n inverse and the $LU$-decomposition is fully populated, thus\n increasing the amount of memory from $d$ to $d^2$.\n\n Linear systems like this are often solved by iterative methods,\n leading for instance to so called Newton-Krylov methods. Iterative\n methods approximate the solution of a linear system\n \\begin{gather*}\n J d = f\n \\end{gather*}\n only using multiplications of a vector with the matrix $J$. On\n the other hand, for any vector $v\\in \\R^d$, the term $Jv$ denotes\n the directional derivative of $f$ in direction $J$. Thus, it can be\n approximated easily by\n \\begin{gather*}\n J v \\approx \\frac{f\\left(x^{(k)}+\\epsilon v\\right) -\n f\\left(x^{(k)}\\right)}{\\epsilon}.\n \\end{gather*}\n The term $f\\left(x^{(k)}\\right)$ must be calculated anyway as it is\n the current Newton residual. Thus, each step of the iterative linear\n solver requires one evaluation of the nonlinear function, and no\n derivatives are computed.\n\n The efficiency of such a method depends on the number of linear\n iteration steps which is determined by two factors: the gain in\n accuracy and the contraction speed. It turns out that typically\n gaining two digits in accuracy is sufficient to ensure fast\n convergence of the Newton iteration. The contraction number is a\n more difficult issue and typically requires preconditioning, which\n is problem-dependent and as such must be discussed when needed.\n\\end{remark}\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"notes\"\n%%% End:\n", "meta": {"hexsha": "88c84b4a08aafbb2d9043621627b73f2a87f1055", "size": 12534, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ode/newton.tex", "max_stars_repo_name": "ahumanita/notes", "max_stars_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ode/newton.tex", "max_issues_repo_name": "ahumanita/notes", "max_issues_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-05-24T07:31:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-08-31T12:58:14.000Z", "max_forks_repo_path": "ode/newton.tex", "max_forks_repo_name": "ahumanita/notes", "max_forks_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2018-05-15T19:28:53.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-05T19:07:29.000Z", "avg_line_length": 40.9607843137, "max_line_length": 94, "alphanum_fraction": 0.6876495931, "num_tokens": 3799, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754489059775, "lm_q2_score": 0.8918110389681662, "lm_q1q2_score": 0.649216541432157}} {"text": "%This is my super simple Real Analysis Homework template\n\n\\documentclass{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n\\usepackage[]{amsthm} %lets us use \\begin{proof}\n\\usepackage[]{amssymb} %gives us the character \\varnothing\n\\usepackage{amsmath}\n\n\\title{Debye length}\n\\author{John ``Jack'' Brooks}\n\\date\\today\n%This information doesn't actually show up on your document unless you use the maketitle command below\n\n\\begin{document}\n\t\n\t\n\\maketitle %This command prints the title based on information entered above\n\t\n\n\n\\section{Debye length derivation}\nThe Debye length is the scale length associated with plasma screening out external electric fields, $\\mathbf{E}$. To show this, we start with Gauss's law and convert $\\mathbf{E}$ to potential, $\\phi$ with $\\mathbf{E} = - \\nabla \\phi$. \n\n\\begin{equation}\n\\begin{split}\n\t%\\label{eq:asdf}\n\t\\nabla^2 \\phi &= \\frac{1}{\\epsilon_0}\\left( \\rho_{plasma} \\right) \\\\\n\t&= \\frac{1}{\\epsilon_0}\\left[ q n_0 \\, \\textrm{exp} \\left( \\frac{-q \\phi (r)}{k_b T} \\right)\\right] \n\\end{split}\n\\end{equation}\nwhere and the plasma charge is assuming a Boltzmann distribution. \n\nIf we assume that $k_bT \\gg q\\phi$, then we can Taylor expand the exponential term to get\n\\begin{equation}\n\\begin{split}\n%\\label{eq:asdf}\n\\nabla^2 \\phi = \\frac{1}{\\epsilon_0}\\left[ q n_0 \\, \\left(1- \\frac{q \\phi (r)}{k_b T} \\right)\\right]\n\\end{split}\n\\end{equation}\nAssuming that we have equal ion and electron charge density, $q_e n_{oe}$ cancels with $q_i n_{oi}$ leaving\n\\begin{equation}\n\\begin{split}\n%\\label{eq:asdf}\n\\nabla^2 \\phi = \\frac{1}{\\epsilon_0}\\left[ - \\frac{q^2 n_0 \\phi (r)}{k_b T} \\right].\n\\end{split}\n\\end{equation}\nThe solution to this is \n\\begin{equation}\n\\begin{split}\n\\phi &= \\phi_0 \\, \\textrm{exp} \\left( -r / \\left( \\frac{q^2n_0}{\\epsilon_0 k_b T} \\right)^{1/2} \\right) \\\\\n &= \\phi_0 \\, \\textrm{exp} \\left( -r /\\lambda_D^{1/2} \\right)\n\\end{split}\n\\end{equation}\nwhere the scale length, \n\\begin{equation}\n\\begin{split}\n \\lambda_D = \\left( \\frac{q^2n_0}{\\epsilon_0 k_b T} \\right)^{1/2},\n\\end{split}\n\\end{equation}\nis called the Debye length.\n\n\\section{Debye length screening}\n\nIf we now include a test charge, $Q_{tc}$, to our Gauss's law equation,\n\\begin{equation}\n\\begin{split}\n%\\label{eq:asdf}\n\\nabla^2 \\phi &= \\frac{1}{\\epsilon_0}\\left( \\rho_{plasma} - \\rho_{tc}\\right) \\\\\n&= \\frac{1}{\\lambda_D^2}\\phi - \\frac{Q_{tc}}{\\epsilon_0}\\delta (r-r_0)\n\\end{split}\n\\end{equation}\nwe can now investigate how the plasma screens out this charge.\n\nTODO\n\n\\end{document}\n\n\n", "meta": {"hexsha": "41bf3135c19161cfaebdd836373abcceda5d08e5", "size": 2550, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Plasma_physics/Debye_length/debye_length.tex", "max_stars_repo_name": "jwbrooks0/misc_derivations", "max_stars_repo_head_hexsha": "3f5b1bbf0dc07821de89619dd794166f5cc217f2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Plasma_physics/Debye_length/debye_length.tex", "max_issues_repo_name": "jwbrooks0/misc_derivations", "max_issues_repo_head_hexsha": "3f5b1bbf0dc07821de89619dd794166f5cc217f2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Plasma_physics/Debye_length/debye_length.tex", "max_forks_repo_name": "jwbrooks0/misc_derivations", "max_forks_repo_head_hexsha": "3f5b1bbf0dc07821de89619dd794166f5cc217f2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.4814814815, "max_line_length": 237, "alphanum_fraction": 0.6984313725, "num_tokens": 892, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.8056321889812553, "lm_q1q2_score": 0.6490432314420403}} {"text": "\\documentclass{article}\n\n\\begin{document}\n\nAuthor: Carl Cortright\n\n\\section{Problem 10.1}\n\nTraining Data:\\\\\n\n\\begin{tabular}{c|cccc}\n\n Class & CornFlakes & Frosties & Sugar Puffs & Branflakes \\\\\n \\hline\n Greater than 60 & 1 & 0 & 0 & 0\\\\\n Greater than 60 & 1 & 0 & 0 & 1\\\\\n Greater than 60 & 1 & 1 & 1 & 1\\\\\n Greater than 60 & 0 & 0 & 0 & 1\\\\\n Less than 60 & 0 & 1 & 1 & 0\\\\\n Less than 60 & 1 & 1 & 1 & 0\\\\\n \n\n\\end{tabular}\n\n\\subsection{(a)}\n\n$P(Less than 60 | <0,1,1,0>) = \\frac{P(Less than 60) * P(<0,1,1,0>|Less than 60)}{P(<0,1,1,0>)} = \\frac{\\frac{1}{3} * \\frac{1}{2}}{\\frac{1}{6}} = 1$\n\n\\subsection{(b)}\n\n$P(age) = \\frac{ P(age) * P( | age) }{ P() }$\n\n\\section{Problem 10.5}\n\n\\subsection{}\n\n$P(C=1) = \\sum_{i=0}^N P(X_i | C=1)$ \\\\\n\n\n$P(X_i = 1 | C = 1) = \\frac{P(C=1) * P(C=1 | X_i=1) }{P(X_i = 1}$ \\\\\n\n\n$P(X_i=1 | C = 0) = \\frac{P(C=0) * P(C=0 | X_i=1) }{P(X_i = 1}$ \\\\\n\n\n\\subsection{}\n\nTo form the classifier, we must take the probability of an input vector $\\vec{x}$ and class $C$ and divide it by the probability of the input vector:\\\\\n\n$P(C|\\vec{x}) = \\frac{P(\\vec{x},C)}{P(\\vec{x})}$\n\n\\subsection{}\n\nIf the word viagra doesn't appear in the training set, then $P(Viagra | C) = 0$ meaning via Bayes rule $P(C | Viagra) = 0$ . Via intuition, we know this must not be true. To fix this, we need to use a technique called smoothing, which basically means adding some arbitrary value $z$ to all of our probabilities to insure that they are always non-zero.\n\n\\section{Dirichlet Distributions}\n\n\\subsection{(a)}\n\nOne situation might be where we want to model classes of classes. An example would be if we were to classify spam or not spam, and then classify the sender of that particular email.\n\n\\subsection{(b)}\n\nFrom Lowest to highest entopy:\\\\\n$Dir(2,2,2)$\\\\\n$Dir(1,1,1)$\\\\\n$Dir(0.1,0.1,0.1)$\\\\\n\n\\end{document}\n", "meta": {"hexsha": "10c33f5a3891aa8c19b445e257008a6ff22b07f1", "size": 1847, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homeworks/week2/homework2.tex", "max_stars_repo_name": "ckcortright/CSCI4830MachineLearning", "max_stars_repo_head_hexsha": "5d1c6c7bfb05b54f7c000c940b1f6410054f10f0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homeworks/week2/homework2.tex", "max_issues_repo_name": "ckcortright/CSCI4830MachineLearning", "max_issues_repo_head_hexsha": "5d1c6c7bfb05b54f7c000c940b1f6410054f10f0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homeworks/week2/homework2.tex", "max_forks_repo_name": "ckcortright/CSCI4830MachineLearning", "max_forks_repo_head_hexsha": "5d1c6c7bfb05b54f7c000c940b1f6410054f10f0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-11-30T07:28:47.000Z", "max_forks_repo_forks_event_max_datetime": "2017-01-28T05:52:45.000Z", "avg_line_length": 26.3857142857, "max_line_length": 351, "alphanum_fraction": 0.6161342718, "num_tokens": 702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146849, "lm_q2_score": 0.805632181981183, "lm_q1q2_score": 0.6490432258025568}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{examples}\n\n\\begin{document}\n\n\\section*{The Gauss relation for the curvature of a hypersurface}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices.\n\n \\nabla_{#}::Derivative.\n\n K_{a b}::Symmetric.\n g^{a}_{b}::KroneckerDelta.\n\n # Define the projection operator\n\n hab:=h^{a}_{b} -> g^{a}_{b} - n^{a} n_{b}.\n\n # 3-covariant derivative obtained by projection on 4-covariant derivative\n\n vpq:=v_{p q} -> h^{a}_{p} h^{b}_{q} \\nabla_{b}{v_{a}}.\n\n # Compute 3-curvature by commutation of covariant derivatives\n\n vpqr:= h^{a}_{p} h^{b}_{q} h^{c}_{r} ( \\nabla_{c}{v_{a b}} - \\nabla_{b}{v_{a c}} ).\n\n substitute (vpq,hab)\n substitute (vpqr,vpq)\n\n distribute (vpqr)\n product_rule (vpqr)\n distribute (vpqr)\n eliminate_kronecker(vpqr)\n\n # Standard substitutions\n\n substitute (vpqr,$h^{a}_{b} n^{b} -> 0$)\n substitute (vpqr,$h^{a}_{b} n_{a} -> 0$)\n substitute (vpqr,$\\nabla_{a}{g^{b}_{c}} -> 0$)\n substitute (vpqr,$n^{a} \\nabla_{b}{v_{a}} -> -v_{a} \\nabla_{b}{n^{a}}$)\n substitute (vpqr,$v_{a} \\nabla_{b}{n^{a}} -> v_{p} h^{p}_{a}\\nabla_{b}{n^{a}}$)\n substitute (vpqr,$h^{p}_{a} h^{q}_{b} \\nabla_{p}{n_{q}} -> K_{a b}$)\n substitute (vpqr,$h^{p}_{a} h^{q}_{b} \\nabla_{p}{n^{b}} -> K_{a}^{q}$)\n\n # Tidy up and display the results\n\n {h^{a}_{b},\\nabla_{a}{v_{b}}}::SortOrder.\n\n sort_product (vpqr)\n rename_dummies (vpqr)\n canonicalise (vpqr)\n factor_out (vpqr,$h^{a?}_{b?}$)\n factor_out (vpqr,$v_{a?}$) # cdb(gauss,vpqr)\n\\end{cadabra}\n\n\\subsection*{The Gauss relation for the curvature of a hypersurface}\n\n\\begin{align*}\n D_{r}(D_{q}v_p) - D_{q}(D_{r}v_p) = \\cdb{gauss}\n\\end{align*}\n\n\\vspace{15pt}\n\n\\begin{latex}\n \\begin{align*}\n D_{r}(D_{q}v_p) - D_{q}(D_{r}v_p) = \\cdb{gauss}\n \\end{align*}\n\\end{latex}\n\n\\end{document}\n", "meta": {"hexsha": "4e60e88ff393e328417195f1cc152050f1666924", "size": 1856, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cadabra/examples/example-06.tex", "max_stars_repo_name": "leo-brewin/hybrid-latex", "max_stars_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2018-10-12T06:31:49.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T23:16:08.000Z", "max_issues_repo_path": "cadabra/examples/example-06.tex", "max_issues_repo_name": "leo-brewin/hybrid-latex", "max_issues_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cadabra/examples/example-06.tex", "max_forks_repo_name": "leo-brewin/hybrid-latex", "max_forks_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-06-27T03:29:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-30T17:17:18.000Z", "avg_line_length": 25.7777777778, "max_line_length": 86, "alphanum_fraction": 0.5899784483, "num_tokens": 756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951104066293, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.6490230515743588}} {"text": "\n\n\n\\subsection{Analysis of the wave energy cascade}\n\n\\subsubsection{Formulation of the spectral energy budget for\nnon-quadratic energy}\n\nIn order to analyze the flux of energy, we derive the spectral energy\nbudget, i.e.\\ the governing equations for spectral KE and APE\nfunctions.\n%\nThis is not as straightforward as in incompressible turbulence since\nthe expression of the kinetic energy $E_K = \\JJ\\cdot\\uu/2$ is not\nquadratic.\n%\nIn order to define the spectral KE function, we use the relation\n\\begin{equation}\n\\meanx{ E_K } = \\sum_\\kk \\scalarprod{\\uu}{\\JJ}/2,\n\\end{equation}\nwhere\n\\begin{equation}\n\\scalarprod{\\baa}{\\bb} \\equiv \\Re\\{ \\hat \\baa(\\kk)^* \\cdot \\hat \\bb(\\kk) \\},\n\\end{equation}\nwhere $\\Re$ denotes the real part, $\\kk$ is the wave number and the\nhat denotes the Fourier transform.\n%\nThe spectral KE function can therefore be defined as\n\\begin{equation}\nE_K(\\kk) \\equiv \\scalarprod{\\uu}{\\JJ}/2, \n\\end{equation}\nsuch as $\\meanx{ E_K } = \\sum_\\kk E_K(\\kk)$.\n%\nSimilarly, the PE can be written as the sum over all wave numbers of\nthe spectral PE function $E_P(\\kk) = c^2 |\\hat h|^2/2 = c^2\n\\scalarprod{h}{h}/2$, where, by definition,\n\\begin{equation}\n\\scalarprod{a}{b} \\equiv \\Re\\{ \\hat a(\\kk)^* \\hat b(\\kk) \\}.\n\\end{equation}\n\n\nThe equation for the spectral KE and PE functions can be written as\n\\begin{eqnarray}\n\\p_t E_K(\\kk) &=& T_K(\\kk) + C_K(\\kk),\\\\\n\\p_t E_P(\\kk) &=& T_P(\\kk) - C_P(\\kk),\n\\end{eqnarray}\nwhere\n\\begin{equation}\nC_K(\\kk) \n= -\\scalarprod{\\uu}{\\bnabla E_P}/2 - c^2 \\scalarprod{\\JJ}{\\bnabla h}/2,\n\\end{equation}\n\\begin{equation}\nC_P(\\kk) = c^2 \\scalarprod{h}{hd} /2,\n\\end{equation}\n\\begin{equation}\nT_P(\\kk) = -c^2 \\scalarprod{h}{\\uu \\cdot \\bnabla h + hd/2}\n\\end{equation}\nand\n\\begin{equation}\nT_K(\\kk) \n= -\\scalarprod{\\uu}{\\uu \\cdot \\bnabla \\JJ}/2\n-\\scalarprod{\\JJ}{\\uu \\cdot \\bnabla \\uu}/2\n-\\scalarprod{\\uu}{d\\JJ}/2.\n\\end{equation}\nThe mean conversion from potential energy to kinetic energy can be\ncomputed from the two conversion spectral functions as $C = \\sum_\\kk\nC_K(\\kk) = \\sum_\\kk C_P(\\kk)$.\n\n\n\n\n\n\n\\subsubsection{Exact Kolmogorov law for irrotational flows}\n\n\n\nAs in incompressible homogeneous isotropic turbulence, an exact\nKolmogorov law for irrotational flows can be derived.\n%\nIt is convenient to start from the following form of the governing\nequations\n\\begin{eqnarray}\n\\p_t \\uu &=& -\\bnabla \\cdot (|\\uu|^2/2) - c^2 \\bnabla h \n- \\zeta_a \\eez \\wedge \\uu, \\\\\n\\p_t \\JJ &=& -\\p_j (u_j \\JJ) - \\bnabla E_p,\\\\\n\\p_t h &=& -\\bnabla \\cdot \\JJ.\n\\end{eqnarray}\nAssuming that the flow is irrotational and that there is no global\nrotation ($\\zeta_a=0$), it can be shown that\n\\begin{equation}\n\\p_t ( J_i' u_i ) = - \\p_i( J_i'|\\uu|^2/2) - \\p_i'(u_j J_j' u_i')\n-\\p_i (J_i' c^2 h) - \\p_i' (u_i E_p')\n\\end{equation}\nand\n\\begin{equation}\n\\p_t ( h' h ) = - \\p_i( h'J_i ) - \\p_i'( hJ_i' ),\n\\end{equation}\nwhich gives\n\\begin{eqnarray}\n\\p_t ( \\JJ'\\cdot\\uu + \\JJ\\cdot\\uu' + 2c^2 h'h )\n&=& - \\p_i( J_i'|\\uu|^2/2) - \\p_i'( J_i|\\uu'|^2/2) %\\nonumber \\\\&& \n- \\p_i'(u_j J_j' u_i') - \\p_i(u_j' J_j u_i) \\nonumber\\\\\n&& -c^2\\p_i (J_i'h) - c^2\\p_i' (J_ih') %\\nonumber\\\\&& \n- \\p_i' (u_i E_p') - \\p_i (u_i' E_p) \\nonumber\\\\\n&& -2c^2\\p_i' (J_i'h) - 2c^2\\p_i (J_ih').\n\\end{eqnarray}\nWe then assume homogeneity and take the ensemble average denoted by\n$\\meane{}$. The separation vector is noted $\\rr = \\xx' - \\xx$. For\nevery function $g(\\xx, \\xx')$, we have $\\p_i' \\meane{g} =\n-\\p_i\\meane{g} = \\bnabla_{\\rr} \\meane{g}|_i$. We obtain\n\\begin{eqnarray}\n\\p_t \\meane{ \\JJ'\\cdot\\uu + \\JJ\\cdot\\uu' + 2c^2 h'h }\n&=& + \\bnabla_{\\rr}\\cdot\\meane{ \\JJ'|\\uu|^2/2 - \\JJ|\\uu'|^2/2 } \\nonumber\\\\\n&& - \\bnabla_{\\rr}\\cdot\\meane{ u_j J_j' \\uu' - u_j' J_j \\uu } \\nonumber\\\\\n&& - \\bnabla_{\\rr}\\cdot\\meane{ \\uu E_p'- \\uu' E_p } \\nonumber\\\\\n&& -c^2\\bnabla_{\\rr}\\cdot\\meane{ \\JJ'h - \\JJ h' }\n\\end{eqnarray}\n%\nWe then introduce the structure functions and the operator $\\delta$\nreturning the increment of a variable between two points separated by $\\rr$, \nfor example $\\delta h (\\xx, \\rr) = h(\\xx+\\rr)- h(\\xx)$.\n%\nUsing again homogeneity, we get\n\\begin{equation}\n\\meane{ (\\delta h)^2 \\delta\\uu } = \n-\\meane{ h'^2 \\uu } + \\meane{ h^2 \\uu' }\n+2\\meane{ hh' \\uu } - 2\\meane{ hh' \\uu' }\n\\end{equation}\nand \n\\begin{equation}\n\\meane{ |\\delta \\uu|^2 \\delta\\JJ } = \n\\meane{ |\\uu|^2 \\JJ' } - \\meane{ |\\uu'|^2\\JJ }\n+\\meane{ u_ju_j' \\JJ' } - \\meane{ u_ju_j' \\JJ },\n\\end{equation}\nwhich gives\n\\begin{equation}\n2\\p_t \\meane{ \\JJ'\\cdot\\uu + \\JJ\\cdot\\uu' + 2c^2 h'h }\n= \\bnabla_{\\rr} \\cdot ( \\meane{ |\\delta \\uu|^2 \\delta\\JJ } \n+ c^2\\meane{ (\\delta h)^2 \\delta\\uu } ).\n\\end{equation}\nUsing isotropy and integrating yield an exact Kolmogorov law for\none-layer shallow water irrotational turbulence\n\\begin{equation}\n\\meane{ |\\delta \\uu|^2 \\delta J_L } \n+ c^2\\meane{ (\\delta h)^2 \\delta u_L } = -4 \\varepsilon r, \\label{eq_Kolmo}\n\\end{equation}\nwhere $\\eps$ is the energy dissipation rate and $J_L \\equiv\n\\JJ\\cdot\\rr / |\\rr|$ and $u_L \\equiv \\uu\\cdot\\rr / |\\rr|$ are\nlongitudinal increments.\n\n\n\n\n\n\n\n\\subsection{Simple model for structure functions determined by a\nrandom set of shocks}\n\\label{subsection_shock_model}\n\n\n\\cite{BouchaudMezardParisi1995} and \\cite{Kuznetsov2004} showed that\nthe presence of discontinuities can strongly influence the statistics\nand in particular lead to very strong intermittency.\n%\nWe now present an extremely simple model based on the assumption that\nshocks totally dominate the flow. The structure functions are\ncalculated by averaging increments over the total surface of the\ndomain $\\mA$. Considering that the increments are produced only by\ndiscontinuity lines, we get\n\\begin{eqnarray}\n\\meane{|\\delta h|^p} \n&=& \\frac{1}{\\mA} \\int_{\\mA} d^2\\xx |\\delta h(\\xx)|^p \\nonumber\\\\\n&\\simeq& \\frac{r}{\\mA} \\int_\\shocks ds |\\sin\\theta| |\\Delta h (s)|^p,\n\\end{eqnarray}\nwhere $\\theta$ is the angle between the shock line and the separation\nvector $\\rr$ %\nand $\\Delta h (s)$ is the step in $h$ at a shock.\n%\nAssuming isotropy, the integral can be split and we get\n\\begin{equation}\n\\meane{|\\delta h|^p} \n\\simeq \nr \\frac{L_s}{\\mA} \\meant{|\\sin\\theta|} \\means{|\\Delta h (s)|^p}\n=\nr \\frac{L_s}{\\mA} \\frac{2}{\\pi} \\means{|\\Delta h (s)|^p},\n\\end{equation}\nwhere $\\meant{}$ denotes the mean over $\\theta$, %\n$L_s$ is the total length of the shocks in the domain %\nand $\\means{}$ denotes the mean over all shocks.\n\nFor velocity increments, we can also use a characteristic of the\nvelocity singularities. The step in velocity is only in the component\nperpendicular to the shocks, which implies that the longitudinal and\ntransverse increments are related by $\\delta u_L = \\delta u\n\\sin\\theta$, $\\delta u_T = \\delta u \\cos\\theta$.\n%\nApplying the same averaging method, we obtain\n\\begin{equation}\n\\meane{|\\delta u_L|^p} \n\\simeq \nr \\frac{L_s}{\\mA} \n\\meant{|\\sin\\theta|^{1+p}}\n\\means{|\\Delta u (s)|^p}\n\\end{equation}\nand\n\\begin{equation}\n\\meane{|\\delta u_T|^p} \n\\simeq \nr \\frac{L_s}{\\mA} \n\\meant{|\\sin\\theta||\\cos\\theta|^{p}}\n\\means{|\\Delta u (s)|^p},\n\\end{equation}\nwhere $\\Delta u (s)$ is the velocity step at a shock.\n%\nThese assumptions completely determine the ratio between the\nlongitudinal and transverse structure functions, which can be\nanalytically computed as\n\\begin{eqnarray}\nR_p(r) \\equiv \\frac{\\meane{|\\delta u_L|^p}}{\\meane{|\\delta u_T|^p}} = \n\\frac{\\meant{|\\sin\\theta|^{1+p}}}{\\meant{|\\sin\\theta||\\cos\\theta|^p}},\n\\end{eqnarray}\ngiving the numerical values $R_2 = 2$, $R_3 = 6\\pi/8$ and $R_4 = 8/3$.\n\n\nInterestingly, for an isotropic and purely divergent flow ($\\zeta =\n0$), the longitudinal and transverse second order structure functions\nare exactly related by $\\meane{(\\delta u_L)^2} = \\p_r( r\\meane{(\\delta\nu_T)^2} )$ \\cite[]{Lindborg2007jas}.\n%\nIf we further assume than both second order structure functions follow\nthe same scaling law $\\meane{ (\\delta u_L)^2} \\propto \\meane{ (\\delta\nu_T)^2} \\propto r^\\alpha$, then we have $\\meane{ (\\delta u_L)^2} =\n(1+\\alpha) \\meane{ (\\delta u_T)^2} $, which gives $ R_2 = 2\n\\Leftrightarrow \\alpha = 1$.\n\n\nThe shock model also provides predictions of the flatness factors of\nthe velocity increments\n\\begin{eqnarray}\nF_L &=& \\frac{\\meane{|\\delta u_L|^4}}{\\meane{|\\delta u_L|^2}^2} \n= r^{-1} \\frac{\\means{|\\Delta u|^4}}{\\means{|\\Delta u|^2}^2} \n\\frac{\\meant{|\\sin\\theta|^5}}{\\meant{|\\sin\\theta|^3}^2},\\\\\nF_T &=& \\frac{\\meane{|\\delta u_T|^4}}{\\meane{|\\delta u_T|^2}^2} \n= r^{-1} \\frac{\\means{|\\Delta u|^4}}{\\means{|\\Delta u|^2}^2} \n\\frac{\\meant{|\\sin\\theta||\\cos\\theta|^4}}{\\meant{|\\sin\\theta||\\cos\\theta|^2}^2},\n\\end{eqnarray}\nand on the ratio of the two flatness factors, which is simply equal to\n$F_T/F_L = 1.5$ according to the model.\n\n\n\n\n\n\n", "meta": {"hexsha": "bfbf5577527c0c3ca94ab5c3ea9d87f44d12b0e2", "size": 8583, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Old/section_advanced_theory.tex", "max_stars_repo_name": "ashwinvis/augieretal_jfm_2019_shallow_water", "max_stars_repo_head_hexsha": "88d97c2bd5df0795ca636306c1d795ef1d3a8949", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-23T11:06:53.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-23T11:06:53.000Z", "max_issues_repo_path": "Old/section_advanced_theory.tex", "max_issues_repo_name": "ashwinvis/augieretal_jfm_2019_shallow_water", "max_issues_repo_head_hexsha": "88d97c2bd5df0795ca636306c1d795ef1d3a8949", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-08-23T13:00:31.000Z", "max_issues_repo_issues_event_max_datetime": "2019-08-23T13:00:31.000Z", "max_forks_repo_path": "Old/section_advanced_theory.tex", "max_forks_repo_name": "ashwinvis/augieretal_jfm_2019_shallow_water", "max_forks_repo_head_hexsha": "88d97c2bd5df0795ca636306c1d795ef1d3a8949", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.3968871595, "max_line_length": 80, "alphanum_fraction": 0.6705114762, "num_tokens": 3170, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951104066293, "lm_q2_score": 0.7690802317779601, "lm_q1q2_score": 0.6490230471078178}} {"text": "\\section{Hermitian Forms and Real Skew-Symmetric Forms}\r\n\\begin{definition}\r\n A sesquilinear form $\\phi:V\\times V\\to\\mathbb C$ is called Hermitian if $\\phi(u,v)=\\overline{\\phi(v,u)}$.\r\n\\end{definition}\r\n\\begin{remark}\r\n In particular, $\\phi(u,u)=\\overline{\\phi(u,u)}$, so $\\phi(u,u)$ is real.\r\n Moreover, for any $\\lambda\\in\\mathbb C$ we have $\\phi(\\lambda u,\\lambda u)=|\\lambda|^2\\phi(u,u)$.\r\n Therefore it makes sense to talk about positive/definite (semi)definite Hermitian forms.\r\n\\end{remark}\r\n\\begin{lemma}\r\n A sesquilinear form $\\phi:V\\times V\\to\\mathbb C$ is Hermitian iff for any basis $B$ of $V$, $[\\phi]_B=\\overline{[\\phi]}_B^\\top$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n If $\\phi$ is Hermitian, then write $A=[\\phi]_B=(a_{ij})_{i,j}=(\\phi(e_i,e_j))_{i,j}$ where we have $a_{ji}=\\phi(e_j,e_i)=\\overline{\\phi(e_i,e_j)}=\\bar{a}_{ij}$.\r\n Conversely if $[\\phi]_B=A$ with $A=(a_{ij})_{ij}=\\bar{A}^\\top$ and $u=\\sum_i\\lambda_ie_i,v=\\sum_i\\mu_ie_i$, then\r\n \\begin{align*}\r\n \\phi(u,v)&=\\phi\\left( \\sum_{i=1}^n\\lambda_ie_i,\\sum_{j=1}^n\\mu_je_j \\right)=\\sum_{i=1}^n\\sum_{j=1}^n\\lambda_i\\bar\\mu_ja_{ij}\\\\\r\n &=\\overline{\\sum_{i=1}^n\\sum_{j=1}^n\\bar\\lambda_i\\mu_ja_{ji}}\\\\\r\n &=\\overline{\\phi\\left( \\sum_{j=1}^n\\mu_je_j,\\sum_{i=1}^n\\lambda_ie_i \\right)}\\\\\r\n &=\\overline{\\phi(v,u)}\r\n \\end{align*}\r\n So $\\phi$ is Hermitian.\r\n\\end{proof}\r\nThe polarisation identity becomes\r\n$$\\phi(u,v)=\\frac{1}{4}(Q(u+v)-Q(u-v)+iQ(u+iv)-iQ(u-iv))$$\r\nfor a Hermitian $\\phi$ and $Q(w)=\\phi(w,w)$.\r\n\\begin{theorem}[Hermitian Formulation of Sylvester's Law]\r\n Let $V$ be an $n$-dimensional vector space over $\\mathbb C$ and $\\phi:V\\times V\\to\\mathbb C$ a Hermitian form on $V$, then $V$ has a basis $\\{v_1,\\ldots,v_n\\}$ such that\r\n $$[\\phi]_B=\\begin{pmatrix}\r\n I_p&&\\\\\r\n &-I_q&\\\\\r\n &&0\r\n \\end{pmatrix}$$\r\n where $p,q$ depends only on $\\phi$.\r\n\\end{theorem}\r\nThe proof is nearly identical to the real symmetric case.\r\n\\begin{proof}\r\n If $\\phi=0$ then we are done.\r\n Otherwise, by the polarisation identity, there exists $e_1\\neq 0$ such that $\\phi(e_1,e_1)\\neq 0$.\r\n Set $v_1=e_1/\\sqrt{|\\phi(e_1,e_1)|}$, then we get $\\phi(v_1,v_1)=\\pm 1$.\r\n Consider $W=\\{w\\in V:\\phi(v,w)=0\\}$, then easily $V=\\langle\\{v_1\\}\\rangle\\oplus W$.\r\n Then we can do induction on the dimension to show that $\\phi$ is diagonal in some basis, which implies the existence of $p,q$.\r\n The uniqueness follows from the observation that $p$ (resp. $q$) is the minimal dimension of a subspace on which $\\phi$ is positive (resp. negative) definite.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $V$ be a vector space over $\\mathbb R$.\r\n A bilinear form on a real vector space is skew-symmetric if $\\phi(u,v)=-\\phi(v,u)$ for all $u,v\\in V$.\r\n\\end{definition}\r\n\\begin{remark}\r\n 1. For any $u\\in V$, $\\phi(u,u)=-\\phi(u,u)$ therefore $\\phi(u,u)=0$.\\\\\r\n 2. The definition is equivalent to say that for any basis $B$ of $V$ we have $[\\phi]_B=-[\\phi]_B^\\top$.\\\\\r\n 3. For any $A\\in M_n(\\mathbb R)$, we can decompose it\r\n $$A=\\frac{A+A^\\top}{2}+\\frac{A-A^\\top}{2}$$\r\n into symmetric and skew-symmetric parts.\r\n\\end{remark}\r\n\\begin{theorem}[Sylvester Form]\r\n Let $\\phi$ be a skew-symmetric bilinear form over a real vector space $V$ with $\\dim V=n<\\infty$, then there is a basis $B=\\{v_1,w_1,\\ldots,v_m,w_m,v_{2m+1},\\ldots,v_n\\}$ of $V$ such that\r\n $$[\\phi]_B=\\begin{pmatrix}\r\n A&&&\\\\\r\n &\\ddots&&\\\\\r\n &&A&\\\\\r\n &&&0\r\n \\end{pmatrix},A=\\begin{pmatrix}\r\n 0&1\\\\\r\n -1&0 \r\n \\end{pmatrix}$$\r\n where there are $m$ copies of $A$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Induction on $n$.\r\n If $\\phi=0$, then we are done.\r\n Otherwise $\\phi\\neq 0$, so there is some $v_1,w_1$ such that $\\phi(v_1,w_1)\\neq 0$.\r\n After rescaling we might as well assume that $\\phi(v_1,w_1)=1$, so correspondingly $\\phi(w_1,v_1)=-1$.\r\n We know that $v_1,w_1$ has to be linearly independent as $\\phi$ is skew-symmetric.\r\n Let $U=\\langle\\{v_1,v_2\\}\\rangle$ and $W=\\{v\\in V:\\phi(v_1,v)=\\phi(w_1,v)=0\\}$, then $V=U\\oplus W$.\r\n We are then done since we can use induction hypothesis on $W$ and $[\\phi|_U]=A$.\r\n\\end{proof}\r\n\\begin{corollary}\r\n Skew-symmetric bilinear forms have even rank.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $V$ be a vector space over $\\mathbb R$ (resp. $\\mathbb C$).\r\n An inner product on $V$ is a positive definite symmetric (resp. Hermitian) form $\\phi$ on $V$.\r\n The pair $(V,\\phi)$ is then called a real (resp. complex) inner product space.\r\n\\end{definition}\r\nSometimes we write $\\langle u,v\\rangle=\\phi(u,v)$ if it is understood.\r\n\\begin{example}\r\n In $\\mathbb R^n$, the usual real scalar product is an inner product.\\\\\r\n In $\\mathbb C^n$, the usual complex scalar product is an inner product.\\\\\r\n In $C([0,1],\\mathbb C)$ (over $\\mathbb C$), the form\r\n $$\\langle f,g\\rangle=\\int_0^1f(t)\\overline{g(t)}w(t)\\,\\mathrm dt$$\r\n is an inner product for any $w\\in C([0,1],\\mathbb R_+)$.\r\n\\end{example}\r\n\\begin{definition}\r\n Let $\\langle\\cdot,\\cdot\\rangle$ be an inner product, its induced norm is $\\|v\\|=\\sqrt{\\langle v,v\\rangle}$.\r\n\\end{definition}\r\n\\begin{remark}\r\n $\\|v\\|\\ge 0$ and the equality holds iff $v=0$.\r\n\\end{remark}", "meta": {"hexsha": "d262bcbc05675920a534428b018f043e0f8c9bdc", "size": 5305, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "20/hermi.tex", "max_stars_repo_name": "david-bai-notes/IB-Linear-Algebra", "max_stars_repo_head_hexsha": "5a499f7ed33ef0110facb27323e13f42883aa0c5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "20/hermi.tex", "max_issues_repo_name": "david-bai-notes/IB-Linear-Algebra", "max_issues_repo_head_hexsha": "5a499f7ed33ef0110facb27323e13f42883aa0c5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "20/hermi.tex", "max_forks_repo_name": "david-bai-notes/IB-Linear-Algebra", "max_forks_repo_head_hexsha": "5a499f7ed33ef0110facb27323e13f42883aa0c5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.0098039216, "max_line_length": 192, "alphanum_fraction": 0.6231856739, "num_tokens": 1912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707283, "lm_q2_score": 0.8438950947024555, "lm_q1q2_score": 0.6490230394965891}} {"text": "\\input{../header.tex}\n\\title{\\vspace{-2cm}INF3490/INF4490 Exercise Solutions - Neural Networks}\n\\author{Eivind Samuelsen\\input{../author_footnote.tex}}\n\\date{}\n\n% Removing paragraph indents is sometimes useful:\n\\setlength\\parindent{0pt}\n% ==============================================================================\n\n% ================================= DOCUMENT ===================================\n\\begin{document}\n \\renewcommand\\marginsymbol[1][0pt]{%\n \\tabto*{0cm}\\makebox[-1cm][c]{$\\mathbb{P}$}\\tabto*{\\TabPrevPos}}\n\n\\maketitle\n\\input{../intro.tex}\n\n\\section{Single Layer Perceptron}\n\\subsection{}\nIn the perceptron below, what will the output be when the input is (0, 0)?\nWhat about inputs (0, 1), (1, 1) and (1, 0)?\nWhat if we change the bias weight to -0.5?\n\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[width=0.3\\textwidth]{fig1.png}\n\\caption{Single Layer Perceptron. b = 1 }\n\\label{fig:slp}\n\\end{center}\n\\end{figure}\n\n\\noindent\\textit{Answer:}\n\n\\noindent\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline\n\\multicolumn{3}{|c|}{Bias $=-1.5$} & \\multicolumn{3}{c|}{Bias $=-0.5$}\\tabularnewline\n\\hline\nInput & Weighted sum & Output & Input & Weighted sum & Output\\tabularnewline\n\\hline\n\\hline\n$\\left(0,0\\right)$ & $-1.5$ & $0$ & $\\left(0,0\\right)$ & $-0.5$ & $0$\\tabularnewline\n\\hline\n$\\left(0,1\\right)$ & $-0.5$ & $0$ & $\\left(0,1\\right)$ & $0.5$ & $1$\\tabularnewline\n\\hline\n$\\left(1,0\\right)$ & $-0.5$ & $0$ & $\\left(1,0\\right)$ & $0.5$ & $1$\\tabularnewline\n\\hline\n$\\left(1,1\\right)$ & $0.5$ & $1$ & $\\left(1,1\\right)$ & $1.5$ & $1$\\tabularnewline\n\\hline\n\\end{tabular}\n\\par\\end{center}\n\n\\subsection{}\nStarting with random weights, how do you proceed in order to train the perceptron above to perform any given binary operation?\\\\\n\n\\noindent\\textit{Answer:}\n\n\\noindent\nSee the source code for the next task.\n\n\\subsection{Implement a perceptron \\marginsymbol}\nImplement the perceptron, and train it to perform the logical functions NOT (use only one of the inputs), NAND, and NOR.\nWhat happens when you try to train it do the XOR function?\\\\\n\n\\textit{Source code:}\n\\inputminted{Python}{bool_perceptron.py}\nThe source code is available online on GitHub:\\\\\n\\href{https://github.com/olehermanse/INF3490-AI_Machine_Learning/tree/master/material/week5}{\\textbf{github.com/olehermanse/INF3490-AI\\_Machine\\_Learning/tree/master/material/week5}}\n\n\\section{Multi Layer Perceptron (MLP)}\nThe figure below shows a multilayer perceptron that constructs the XOR function.\nHow would you rewrite it to construct the binary equivalence function\n(i.e. the output is above threshold when both inputs are either 0 or 1)?\nCan you construct it so that it will detect equivalence for any combination of integer inputs?\n\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{fig2.png}\n\\caption{An illustrated example of a multi layer perceptron}\n\\label{fig:mlp}\n\\end{center}\n\\end{figure}\n\n\\noindent\\textit{Answer:}\n\n\\noindent\nIf we just want it to work in the binary case, we can simply switch\nsigns on all the output weights to create the opposite function:\n\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[width=0.4\\textwidth]{opposite.png}\n\\end{center}\n\\end{figure}\n\nIf we want it to work for all integers, we have to change the weights\nin the first layer. First, note that the output layer encodes the\nfunction $h_{2}-h_{1}-0.5>0$ which is equivalent to $h_{2}\\land\\lnot h_{1}$\nsince $h_{1}$ and $h_{2}$ are either 0 or 1. If we equate true with\n1 and false with 0 we can write $\\mbox{\\ensuremath{h_{1}}:\\,\\ \\ensuremath{i_{1}}+\\ensuremath{i_{2}}-1.5>0}$\nand $\\mbox{\\ensuremath{h_{2}}:\\,\\ \\ensuremath{i_{1}}+\\ensuremath{i_{2}}-0.5>0}$,\nso the output is\n\n\\[\no:\\, i_{1}+i_{2}>0.5\\ \\land\\ i_{1}+i_{2}\\leq1.5\\quad\\Leftrightarrow\\quad o:\\,0.5-0.5\\ \\land\\ i_{1}-i_{2}\\leq0.5\n\\]\n or something similar. This has the same logical form that we already\nhave, so we don't need to change the weights of the output layer,\nbut only those of the hidden layer.\n\nTracing back, we see that we would need $h_{1}:\\, i_{1}-i_{2}-0.5>0$\nand $h_{2}:\\, i_{1}-i_{2}+0.5>0$ as the hidden layer outputs. This\nimplies the following weights:\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[width=0.4\\textwidth]{all_integers.png}\n\\end{center}\n\\end{figure}\n\n\\input{../contact.tex}\n\\end{document}\n% ==============================================================================\n", "meta": {"hexsha": "baaf1ebd11884b529d2a3cebb64914d7c1f887db", "size": 4699, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "material/week5/inf3490-sol5.tex", "max_stars_repo_name": "mpambasange/MachineLearning", "max_stars_repo_head_hexsha": "8b813345264513a57934317b01e1311628dc5b01", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2016-09-01T08:50:59.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-15T20:56:07.000Z", "max_issues_repo_path": "material/week5/inf3490-sol5.tex", "max_issues_repo_name": "olehermanse/INF3490-PythonAI", "max_issues_repo_head_hexsha": "8b813345264513a57934317b01e1311628dc5b01", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2016-10-20T09:36:19.000Z", "max_issues_repo_issues_event_max_datetime": "2017-08-29T00:28:54.000Z", "max_forks_repo_path": "material/week5/inf3490-sol5.tex", "max_forks_repo_name": "olehermanse/INF3490-PythonAI", "max_forks_repo_head_hexsha": "8b813345264513a57934317b01e1311628dc5b01", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 15, "max_forks_repo_forks_event_min_datetime": "2016-10-31T12:30:37.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-15T12:12:50.000Z", "avg_line_length": 35.5984848485, "max_line_length": 182, "alphanum_fraction": 0.6763141094, "num_tokens": 1547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672227971211, "lm_q2_score": 0.8031738034238807, "lm_q1q2_score": 0.6490184247561361}} {"text": "\\section{Stationarity}\nBroadly speaking, a time series is said to be {\\sl stationary} if\nthere is no systematic trend, no systematic change in variance,\nand if strictly\nperiodic variations or seasonality do not exist. Most processes in\nnature appear to be non-stationary. Yet much of the theory in\ntime-series literature is only applicable to stationary processes.\n\nOne way of describing a stochastic process is to specify the joint\ndistribution of the observations $Y(t_1), \\dots, Y(t_n)$ for any\nset of times $t_1, \\dots, t_n$ and any value of $n$. A time series\nis said to be strictly stationary if the joint distribution of\n$Y(t_1), \\dots, Y(t_n)$ is the same as that of $Y(t_1 + h), \\dots\nY(t_n + h)$ for all $t_1, \\dots, t_n$ and $h$. To see how this is\na useful assumption, notice that the above condition implies that\nthe expected value and covariance structure of any two components,\n$Y_a(t)$ and $Y_b(t)$, of a time series are\nconstant in time\n\\begin{equation}\n\\label{sos}\n\\E \\{Y_a(t)\\} = \\mu_a \\, , \\, \\var \\{Y_a(t)\\} = \\sigma_a^2 \\mbox{ and }\n\\mbox{corr}\\{ Y_a(t), Y_b(t+h) \\} = \\gamma_{ab}(h).\n\\end{equation}\nThe function\n$\\gamma_{ab}(h)$ is called the cross-correlation function if $a\\neq b$\nand the auto-correlation function if $a=b$. \n\nIn practice it is often useful to define stationarity in a less\nrestricted way than that described above. In many cases,\nthe statistical structure of the processes can be\ncompletely described with the second-order properties of equation\n(\\ref{sos}). We can estimate the\nquantities in (\\ref{sos}) using\nstandard statistical procedures,\nfor example we may estimate the cross-correlation at {\\it lag}\n$h$, $\\gamma_{a,b}(h)$ with the sample correlation of\n$Y_a(1),\\dots,Y_a(T-h)$ and $Y_b(h+1),\\dots,Y_b(T)$. \n\n\n\n\\subsection{An example: Fetal Monitoring}\n\nMeasurements of fetal heart rate (FHR) and fetal movement (FM) are\ngenerated by maternal-fetal monitoring. \nApproximately 5 measurements per second are taken during 50 minutes on\n120 subjects that are monitored at 20,24,28,32,36,\nand 38-39 weeks of \ngestation. Both FHR and FM are recorded giving us a multiple time \nseries $Y(t), t=1,\\dots,50\\times 60\\times 5$, where $Y(t)$ is a vector\nwith 2 entries. \n\nThe association between accelerations of FHR and \nFM has been documented since the 1930s. For\nexample, it has been observed \nthat in the third trimester most large fetal heart\naccelerations are \nassociated with fetal activity.\nIn Section 4.2 we will describe how relatively straight-forward time\nseries techniques provide a visual descriptions of how these\nassociations vary with weeks gestation. These description have\nmotivated a methodology that will provide us is with a more rigorous\nassessment of this relationship.\n\nIf we consider the FM and FHR measurements, seen in Figure 5, as\noutcomes from a two component time series, we may consider the\ncross-correlation function of these two components as a description of\nthe association between these two processes. Notice that the\nmeasurements taken for each \nfetus at each gestation week has a cross-correlation function associated\nwith them. In Figure 6, as a descriptive plot, we show the average,\nover individuals,\nof these functions for each gestation\nweek. Notice that a peak at around the $-6$ second lag starts to\nappear in the plot for the 24 week gestation. As the fetus gets older,\nthis peak grows and becomes more defined. This result\ncan be \nconsidered a first step in the characterization of the relationship\nbetween FM and FHR. \n\n\\centerline{\\epsfig{figure=Plots/plot-10-02.ps,width=\\textwidth}}\n\\centerline{\\epsfig{figure=Plots/plot-10-03.ps,width=\\textwidth}}\n", "meta": {"hexsha": "16f2d0edd98b8e6082edb3e0352e299218394b6a", "size": 3646, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pages/754/section-10-01.tex", "max_stars_repo_name": "igrabski/rafalab.github.io", "max_stars_repo_head_hexsha": "2f27ea0d9e0b8a2342bb851ae7415ba3268fd00f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2016-08-17T23:04:04.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-31T19:21:02.000Z", "max_issues_repo_path": "pages/754/section-10-01.tex", "max_issues_repo_name": "igrabski/rafalab.github.io", "max_issues_repo_head_hexsha": "2f27ea0d9e0b8a2342bb851ae7415ba3268fd00f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2016-08-18T00:41:36.000Z", "max_issues_repo_issues_event_max_datetime": "2021-01-21T22:35:40.000Z", "max_forks_repo_path": "pages/754/section-10-01.tex", "max_forks_repo_name": "igrabski/rafalab.github.io", "max_forks_repo_head_hexsha": "2f27ea0d9e0b8a2342bb851ae7415ba3268fd00f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 38, "max_forks_repo_forks_event_min_datetime": "2016-08-17T22:17:30.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-20T12:17:08.000Z", "avg_line_length": 44.4634146341, "max_line_length": 72, "alphanum_fraction": 0.7660449808, "num_tokens": 956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031738010682209, "lm_q2_score": 0.8080672135527632, "lm_q1q2_score": 0.6490184154277786}} {"text": "\\section{Notation}\n\nIn addition to the notation in the Shelley spec, we use the following notations:\n\n\\begin{description}\n\\item[Functions with finite support] For a monoid $M$, $A \\to_0 M$\n are the functions $f : A \\to M$ with finite support, i.e. such that\n there exist only finitely many $a \\in A$ such that $f a \\neq 0$. We\n denote the support of $f$ by $\\supp f$, which is defined as\n $\\supp f := \\{ a \\in A : f a \\neq 0 \\}$.\n\\end{description}\n", "meta": {"hexsha": "6eea42e40dd3ae4ff79f0794364cd5ec542cfa50", "size": 452, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "shelley-ma/formal-spec/notation.tex", "max_stars_repo_name": "ilap/cardano-ledger-specs", "max_stars_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-05-30T14:19:54.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-30T14:19:54.000Z", "max_issues_repo_path": "shelley-ma/formal-spec/notation.tex", "max_issues_repo_name": "ilap/cardano-ledger-specs", "max_issues_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "shelley-ma/formal-spec/notation.tex", "max_forks_repo_name": "ilap/cardano-ledger-specs", "max_forks_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.6666666667, "max_line_length": 80, "alphanum_fraction": 0.6792035398, "num_tokens": 147, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8840392939666336, "lm_q2_score": 0.7341195327172402, "lm_q1q2_score": 0.648990513390464}} {"text": "\n\\begin{figure*}[!ht]\n\\includegraphics[width=\\textwidth]{figs/waterfall} \n\\caption{A waterfall plot of the \\texttt{cartoon-spring.wav} from 0-200 ms between frequencies of 20-8000 Hz, where the height indicates the amplitude of each frequency. This plot was created with the REW tool~\\cite{REWTool}.}\n\\label{fig:waterfall}\n\\end{figure*}\n\n\\section{Aural Distance}\n\\label{sec:distance}\n\nAs a distance metric, we used as a starting point the literature on acoustic fingerprinting~\\cite{fingerprinting}.\nAcoustic fingerprinting is the concept of creating a condensed, distinct summary of an audio file that can be used later to identify that audio file or to look it up in a database.\nAcoustic fingerprints turn an audio file into a represent of how the file will sound to the human ear regardless of how it is represented in a digital format~\\cite{fingerprinting}.\nThere are numerous ways to develop acoustic fingerprints and companies like Shazam and Sound-Hound have developed complex algorithms to create accurate fingerprints even from low quality files recorded on a cellphone mic.\nFor this work, we used the work of Shazam~\\cite{wang2003industrial} as an inspiration for our distance metric calculation.\n\n\nAs an intuition, the psycho-acoustic identity of a sound file (how humans distinguish between one sound and the next) can be captured by taking every ``moment'' of audio, and listing the predominate frequencies for that slice of time.\nThis intuition can be represented with a waterfall plot, as shown in Figure~\\ref{fig:waterfall}, which plots how the frequencies change over time.\nA waterfall plot uses the Fast Fourier Transform (FFT) to calculate many discrete Fourier transforms over small times slices.\nIn this way, a waterfall plot is a representation of an audio file as a list of spectrograms plotted over time.\nIn the Shazam method, the peaks are selected from each time slice of audio and used to create a ``constellation'' of peaks over time. \nThis constellation is then used to build a hash that acts as a fingerprint to uniquely identify the audio sample.\nWe use a similar strategy by first performing a real Fast Fourier Transform on the audio file and then picking out the frequency peaks in each time frame.\nHowever, the key difference in DSP-PBE is that we do not use the constellation as a hash for lookup in a database (as Shazam and SoundHound do), but instead, we need a distance metric between two constellations to provide a measure of how close we are to synthesizing the correct DSP filter.\nDistance metrics are common in music synthesis tasks, for example, in the generation of jazz improvisations, where the improvisation should stay close by some measure to the original melody~\\cite{donze2014machine}.\n\nFast Fourier Transforms (FFT) are the key to a good acoustic fingerprint.\nThe FFT, however, cannot be taken as a blackbox in our application.\nThe two factors we need to consider are 1) the window-size for how many samples will be used to calculate the FFT, and 2) the bin size which roughly speaking, defines the resolution of the FFT.\n\nEach return element is a frequency \\textit{bin}, and depending on the scale of your return array the size of these bins varies.\nIn order for each bin to correspond to 1 Hz the size of the return vector must be equal to the sampling frequency (44,100 Hz).\nIf each bin is not 1 Hz, the effects of spectral leakage will be seen.\nThis occurs when the bins do not correspond to the exact frequency peaks of the sound.\nThe amplitude from the peaks that fall in between bins will \\textit{leak} over into the closest bin and create a distorted spectrogram.\nFor this reason we had to adjust the size of the FFT return arrays to be 44,100, as 44,100 Hz is a common format for audio.\nAlthough this slows down the process of FFT, it provides the most accurate representation of the sound and for our purposes frequency accuracy is paramount.\n\nWith our constellations created from the waterfall plot, we constructed a \\texttt{dist} function that measures the aural distance and is faithful to the psycho-acoustics of the human ear.\nOur implementation takes the Euclidean distance of the peaks in a time slice on the frequency-amplitude axis.\nIn order to define this distance more formally, we introduce the notation $c@t$ to indicate selecting time slice $t$ from constellation $c$.\nWe also use the function $peak:: Int \\to Constellation \\to Peak$ to select a peak from a constellation, where the peaks are in sorted order based on frequency.\nThen, for an audio clip $x$ and an audio clip $y$, and a function $toC :: Audio \\to Constellation$ to transform the audio clip into a constellation with $ts$ time slices and $p$ peaks in each time slice:\n%\n\\begin{align*}\n\\sum_{t=0}^{ts}\\ \\sum_{i=0}^{p} euclid\\Big(\\ &peak(i,toC(x)@t), \\\\ &peak(i,toC(y)@t)\\ \\Big)\n\\end{align*}\n%\nNote that this definition requires the audio clips to be temporally aligned, which is not always a fair assumption in the real world. We leave the exploration of a temporal offset between two example audio samples to future work.\n\nAs a sanity check that this distance metric matches the psycho-acoustic definition of distance, we used the test cases listed in Table~\\ref{table:dist}.\n\n\\begin{table*}[!h]\n\\centering\n\\caption{Test cases to evaluate distance metric. The exact values are only important in relationship to the others.}\n\\begin{tabular}{|l | c | c|} \n \\hline\n Test Name \\& Expected Result & Value 1 & Value 2 \\\\\n \\hline\n \\hline\n Identity & \\multirow{2}{*}{(PianoC, PianoC) = 0} & \\multirow{2}{*}{NA}\\\\ \n \\quad Value 1 = 0 & & \\\\\n \\hline\n Commutativity & \\multirow{2}{*}{(PianoC, PianoCSharp) = 5.635} & \\multirow{2}{*}{(PianoCSharp, PianoC) = 5.635} \\\\\n \\quad Value 1 = Value 2 & & \\\\\n \\hline\n Commutativity & \\multirow{2}{*}{(PianoC, HornCSharp) = 20.500} & \\multirow{2}{*}{(HornCSharp, PianoC) = 20.500} \\\\\n \\quad Value 1 = Value 2 & & \\\\\n \\hline\n Filter less than pitch & \\multirow{2}{*}{(PianoC, PianoFilterC) = 3.749} & \\multirow{2}{*}{(PianoC, PianoCSharp) = 5.635} \\\\ \n \\quad Value 1 $<$ Value 2 & & \\\\\n \\hline\n Filter less than pitch+instrument & \\multirow{2}{*}{(PianoC, PianoFilterC) = 3.749} & \\multirow{2}{*}{(PianoC, HornCSharp) = 20.500} \\\\\n \\quad Value 1 $<$ Value 2 & & \\\\\n \\hline\n Pitch less than pitch+instrument & \\multirow{2}{*}{(PianoC, PianoCSharp) = 5.635} & \\multirow{2}{*}{(PianoC, HornCSharp) = 20.500} \\\\\n \\quad Value 1 $<$ Value 2 & & \\\\\n \\hline\n\\end{tabular}\n\\label{table:dist}\n\\end{table*}\n\n\nThe goal in the synthesis procedure is to find a DSP filter program, $F$, such that \\texttt{dist$(O, F(I))$ = 0}.\nHowever, in practice the DSP-PBE Synthesizer can only get us so close to this metric and we instead just minimize this distance.\nTo do this, the user specifies a default threshold distance for the aural distance.\nThe threshold distance defines how close is acceptably close, and can be changed by the user depending on their needs or requirements.\n\n", "meta": {"hexsha": "9407f978d041792095b648da1e930e3fc924e9bb", "size": 6947, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "papers/FARM-18/secs/distance.tex", "max_stars_repo_name": "Yale-OMI/DSP-PBE", "max_stars_repo_head_hexsha": "073f366e8096004adeec5d2cde1cf3546c4690f5", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-12-03T02:36:39.000Z", "max_stars_repo_stars_event_max_datetime": "2018-12-03T02:36:39.000Z", "max_issues_repo_path": "papers/FARM-18/secs/distance.tex", "max_issues_repo_name": "Yale-OMI/DSP-PBE", "max_issues_repo_head_hexsha": "073f366e8096004adeec5d2cde1cf3546c4690f5", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2018-11-16T21:50:44.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-16T18:57:19.000Z", "max_forks_repo_path": "papers/FARM-18/secs/distance.tex", "max_forks_repo_name": "Yale-OMI/DSP-PBE", "max_forks_repo_head_hexsha": "073f366e8096004adeec5d2cde1cf3546c4690f5", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 77.1888888889, "max_line_length": 291, "alphanum_fraction": 0.7581689938, "num_tokens": 1797, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8840392817460333, "lm_q2_score": 0.7341195327172402, "lm_q1q2_score": 0.6489905044190826}} {"text": "\n\\documentclass[a4paper,11pt]{article}\n\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\usepackage{amsthm, mathtools}\n%\\usepackage{hyperref}\n\\usepackage{color}\n\\usepackage{jheppub}\n\\usepackage[T1]{fontenc} % if needed\n\n% My Documents\n\\newcommand{\\be}{\\begin{equation}}\n\\newcommand{\\ee}{\\end{equation}}\n\\newcommand{\\bes}{\\begin{equation*}}\n\\newcommand{\\ees}{\\end{equation*}}\n\\newcommand{\\bea}{\\begin{flalign*}}\n\\newcommand{\\eea}{\\end{flalign*}}\n\n\n%\\linespread{1.0}\n%\\setlength{\\parindent}{0em}\n%\\setlength{\\parskip}{0.8em}\n\n\\title{\\textbf{Binaries and Tides}}\n\\author{Aditya Vijaykumar}\n\\affiliation{International Centre for Theoretical Sciences, Bengaluru, India.}\n\\emailAdd{aditya.vijaykumar@icts.res.in}\n\\abstract{}\n\\begin{document}\n\\maketitle\n\n\\section{Tanja's Lectures at Gravitational Wave School 2017}\n\nNon-black hole objects differ from black holes in that :-\n\\begin{itemize}\n\t\\item They can deform due to their rotational motion\n\t\\item They can have effects due to no presence of horizon\n\t\\item They can get tidally deformed\n\\end{itemize}\nTanja's lectures focus on tidal effects, which are the most promising candidates (as of 2017) to detect parameters of neutron stars.\n\nWe start off with tides in Newtonian physics.\n\n\\subsection{Newtonian Physics}\nNotation - $ \\vb{a} $ is 3-vector and $ \\va{a} $ is 4-vector.\n\nThe force between two bodies of masses $ m $ and $ M $ is,\n\\begin{equation}\\label{key}\n\\vb{F} = - \\dfrac{GmM}{r^2} \\vu{n} \\qq{,} U = -\\dfrac{GM}{r} \\qq{,} \\vb{a} = \\grad{U}\n\\end{equation}\nwhere $ U $ and $ \\vb{a} $ are the gravitational potential due to mass $ M $ and the acceleration respectively.\n\nTo make things more specific, consider a body of mass $ m_A $ with position vector $ \\vb{z}_A $. The potential at $\\vb{x} $ due to $ m_A$ is,\n\\begin{equation}\\label{key}\nU_A(\\vb{x}) = \\dfrac{Gm_A}{\\abs{\\vb{x} - \\vb{z}_A}}\n\\end{equation}\nAlternatively, if we consider extended bodies with density $ \\rho(x) $, the potential can be written as,\n\\begin{align}\\label{key}\nU_A(\\vb{x}) &= G \\int \\dd^3{\\vb{x}'}\\dfrac{\\rho(\\vb{x}' )}{\\abs{\\vb{x} - \\vb{x}'}}\\\\\n\\implies \\laplacian{U_A} &= G \\int \\dd^3{\\vb{x}'} \\rho(\\vb{x}' ) \\laplacian{\\dfrac{1}{\\abs{\\vb{x} - \\vb{x}'}}} \\\\\n&= G \\int \\dd^3{\\vb{x}'} \\rho(\\vb{x}' ) \\qty(-4 \\pi \\delta(\\vb{x} - \\vb{x}')) \\\\\n\\laplacian{U_A} &= -4 \\pi G \\rho(\\vb{x} )\n\\end{align}\n\\end{document}", "meta": {"hexsha": "32213f5fe933fb6f10e44eae42ab0789a1cc1db2", "size": 2379, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "selfstudy/tides/tides.tex", "max_stars_repo_name": "adivijaykumar/courses", "max_stars_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "selfstudy/tides/tides.tex", "max_issues_repo_name": "adivijaykumar/courses", "max_issues_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "selfstudy/tides/tides.tex", "max_forks_repo_name": "adivijaykumar/courses", "max_forks_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.5074626866, "max_line_length": 141, "alphanum_fraction": 0.6885245902, "num_tokens": 828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8840392756357327, "lm_q2_score": 0.7341195269001831, "lm_q1q2_score": 0.6489904947908847}} {"text": "\\section{CIFAR10 example} % Jonathan / b-j8501 / Jul 11 & % b-j8505 / Jul 12\nIn this section, we will give more details about training a neural network with PyTorch. We will start with the simple CIFAR10 example introduced before, and explain how to modify each component of it. \n\nThe first of all, the first six lines here just consist the \\emph{import} statements. Basically all this does is bringing code from this imported libraries from PyTorch in your program allows to use the classes, the functions that for implement each of these libraries.\n\\begin{python}\nimport torch \nimport torchvision\nimport torchvision.transforms as transforms\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\n\\end{python}\n\n\\subsection{The \\emph{`Net'} class}\nThen each neural network we should to implement should exactly contain the class in the following. Which you does is that you define a new class called Net, which is a subclass of nn.Module. The nn.Module is a class implemented in the package torch.nn. Bascially you don't have to know exactly what the Module class does, but the point here is Network is going to be a subclass of it. Let go to it step by step.\n\nAny class defined in Python has to have the \\emph{\\_\\_init\\_\\_} function. It will run when the class be created. Late on, in the code when you implement like \\emph{`conv\\_net = Net()'}, it will create an instance of the Net class. And when the instance is created, the \\emph{\\_\\_init\\_\\_} function is run. And we can see that the \\emph{\\_\\_init\\_\\_} function create each of the pieces of the neural network, which are the operators we need for our neural network. This linear (\\emph{nn.Linear}), convolution (\\emph{nn.Conv2d}) and max pool (\\emph{nn.MaxPool2d}) is implement in somewhere else. Roughly speaking, they simply contain how much parameters and a forward function. For example the \\emph{`nn.Linear'} class: \n\\begin{itemize}\n\\item contains parameters which consists from a matrix and a vector $(W,b)$,\n\\item and the `forward' function is\n$${\\rm forward}(x) = Wx+b.$$\n\\end{itemize}\nLater on, we will talk about what \\emph{nn.Conv2d} and \\emph{nn.MaxPool2d} do.\n\nAnd then to make is useful, we need to define this forward function. Basically, it tells you, given this neural network and some inputs, what the output is. In this particular case, what the network does is to apply the first convolution, and a ReLU function. The ReLU function is\n$${\\rm ReLU}(x) = \\max\\{0,x\\}.$$\nAnd then to apply a max pooling. Then applies the next convolution, a ReLU function and pooling layer.\nThen, what the \\emph{`x.view'} does is take $x$, which is still an image (2 dimensional) and flat it out into a vector. Then applies the first linear layer, a ReLu function, and the second linear layer, a ReLU function, and finally the third linear layer. And it returns $x$. What is contained in the class is all the parameters for each of these layers and the forward function which tells you the order and how to apply them. So that is what network consist.\n\\begin{python}\nclass Net(nn.Module):\n def __init__(self):\n super(Net, self).__init__()\n self.conv1 = nn.Conv2d(3, 6, 5)\n self.pool = nn.MaxPool2d(2, 2)\n self.conv2 = nn.Conv2d(6, 16, 5)\n self.fc1 = nn.Linear(16 * 5 * 5, 120)\n self.fc2 = nn.Linear(120, 84)\n self.fc3 = nn.Linear(84, 10)\n\n def forward(self, x):\n x = self.pool(F.relu(self.conv1(x)))\n x = self.pool(F.relu(self.conv2(x)))\n x = x.view(-1, 16 * 5 * 5)\n x = F.relu(self.fc1(x))\n x = F.relu(self.fc2(x))\n x = self.fc3(x)\n return x\n\\end{python}\n\n\\subsection{The `main' function}\nWhat we want to do is to load a bunch of data, and train this network (all the parameters) on these data. So the \\emph{main} function is going to do that. The two lines \\emph{trainset} and \\emph{trainloader} load the train set, and the two lines \\emph{testset} and \\emph{testloader} load the test set. The \\emph{trainloader} is a class which specifies some way of giving you the data. In our particular case, we construct the \\emph{trainloader}, we pass the \\emph{trainset}, the images, to it and also a bunch of the parameters, like the batch size, shuffle, number of workers. When you calculate the gradient, you don't use the whole dataset at once, and we only use some of them. The number of data we use to calculate the gradient is the batch size. Here we pass the \\emph{`batch\\_size'} to the \\emph{DataLoader} what happen is the later on when we loop over everything in the \\emph{trainloader} (see the for loop in the main function), it will return the images four at a time because we set `batch\\_size=4'. The shuffle is true means that you return the data without replacement. It the same thing to the test set.\n\nThe \\emph{criterion} is a class containing the loss function. Given the output of the neural network and the true label applies the loss function. The optimize class, we will talk about it later on. When you construct the \\emph{`optim'} class, the parameters you have to pass are \\emph{`net.parameters'} that tell the optimizer which parameters need to be optimized, and other hyperparameters. It has a function called \\emph{`step'}. Whenever the \\emph{`step'} function is called it going to modify the parameters in some way based on whatever the gradients are. The optimizer assume you already calculated the gradients and then it does some sort of step based on the gradients. That explains why when you actually run the training we have to call \\emph{`optimizer.zero\\_grad'} and \\emph{` loss.backward'} independently of \\emph{`optimizer.step'}. \\emph{`optimizer.step'} doesn't handle taking the gradient, it assumes that all the gradients already been taking properly.\n \n\\begin{python}\ndef main():\n\ttransform = transforms.Compose([transforms.ToTensor(),transforms.Normalize((0.5, 0.5, 0.5), (0.5, 0.5, 0.5))])\n\ttrainset = torchvision.datasets.CIFAR10(root='./data', train=True,download=True, transform=transform)\n\ttrainloader = torch.utils.data.DataLoader(trainset, batch_size=4, shuffle=True, num_workers=2)\n\ttestset = torchvision.datasets.CIFAR10(root='./data', train=False, download=True, transform=transform)\n\ttestloader = torch.utils.data.DataLoader(testset, batch_size=4, shufflex=False, num_workers=2)\n\nclasses = ('plane', 'car', 'bird', 'cat',\n 'deer', 'dog', 'frog', 'horse', 'ship', 'truck')\n \nconv_net = Net()\n\ncriterion = nn.CrossEntropyLoss()\noptimizer = optim.SGD(net.parameters(), lr=0.001, momentum=0.9)\n \nfor epoch in range(2): # loop over the dataset multiple times\n\n running_loss = 0.0\n for i, data in enumerate(trainloader, 0):\n # get the inputs; data is a list of [inputs, labels]\n inputs, labels = data\n\n # zero the parameter gradients\n optimizer.zero_grad()\n\n # forward + backward + optimize\n outputs = net(inputs)\n loss = criterion(outputs, labels)\n loss.backward()\n optimizer.step()\n\n # print statistics\n running_loss += loss.item()\n if i % 2000 == 1999: # print every 2000 mini-batches\n print('[%d, %5d] loss: %.3f' %\n (epoch + 1, i + 1, running_loss / 2000))\n running_loss = 0.0\n\nprint('Finished Training')\n\ncorrect = 0\ntotal = 0\nwith torch.no_grad():\n for data in testloader:\n images, labels = data\n outputs = net(images)\n _, predicted = torch.max(outputs.data, 1)\n total += labels.size(0)\n correct += (predicted == labels).sum().item()\n\nprint('Accuracy of the network on the 10000 test images: %d %%' % (\n 100 * correct / total))\n\n\\end{python}\n\n\\subsection{\\emph{`DataLoader'}}\nNext, we want to talk about how can we change the Dataloader and how can we make it provide the data samples in a different way. In particular, we already talk about the difference between sampling with and without replacement. Let us recall that a little bit.\n\nThe loss function:\n$$L(\\theta) = \\sum_{i=1}^n l(x_i,\\theta)$$\nwe want to calculate the gradient of the loss function. \n$$\\nabla L(\\theta) = \\sum_{i=1}^n \\nabla_\\theta l(x_i,\\theta)$$\nand $n$ generally could be very large, $n \\sim 10^4 - 10^7$. In this situation, to compute the gradients over all the data point is computationally not feasible. So just approximate the gradients by sampling some of the data points. \nInitially, consider the computation complexity. Stochastic gradient descent\n\\begin{itemize}\n\\item approximate the gradient by considering a sample of data points \n$$\\{x_{i_1}, \\ldots, x_{i_k}\\}$$\nwhere $\\{i_1, \\ldots, i_k\\}$ is randomly chosen in each iteration, then\n$$\\nabla L(\\theta) \\approx \\sum_{l=1}^k \\nabla_\\theta l(x_{i_l},\\theta)$$\n\\item $k$ is called the mini-batch size (controls the accuracy of \"noise\" in the sample).\nThis is a sampled gradient, containing some noise. If I have larger mini-batch size, we will get a more accuracy gradients, so the mini-batch size $k$ is an important hyper-parameter.\n\\item $i_1, \\ldots, i_k$ sampled with/without replacement.\n\\begin{itemize}\n\\item with replacement: there can be repetitions\n\\item without replacement: there can't be repetitions\n\\end{itemize}\n\\end{itemize}\nWhat we implement is sampling without replacement cross the whole epoch. When we randomly choose these data points, we can't choose the same data points again until we around the whole data set. \n\nNow, let's see how we can change it. First, go to the PyTorch documents page\\footnote{\\url{https://pytorch.org/docs/stable/data.html}}. Here is an explanation what exactly is this class does, and the important thing for us is what the possible inputs. So when I construct a DataLoader, I need to pass some variables. Whenever we construct a DataLoader, we have to pass it a dataset because there is no default value. In our program, we passed the train set, CIFAR10. You also can pass something else you want to it. And all the other variables here have some default values. There are options here to keep the defaults or to change it to what you want We can see that some of the defaults actually already changed in our program. Instead of the default value 1, we set the batch size as 4, and shuffle as true instead of false. Now, we know how to change the sampling strategies. We probably want to change some of these other inputs. Look here, we can see the description are. In particular, you will see sampler, the default value be none, which defined the strategy. Most likely, if we construct the samlper and pass it to the DataLoader, we can change the with and without replacement strategy.\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{./figures/497Proj_DataLoader}\n%\\caption{DataLoader}\n\\end{figure}\n\nNow, we need to look at the sampler class. If we want to change the sampler, we need to define a class that is a subclass of the sampler, and contain the sampler strategy.\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.4]{./figures/497Proj_Sampler}\n%\\caption{}\n\\end{figure}\nIn our case, if you want change that with or without replacement, somebody has already written it. \nIt turns out the RandomSampler.\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.4]{./figures/497Proj_RandomSampler}\n%\\caption{}\n\\end{figure}\nLet's use our program as an example to show how to use a sampler class. First, we need to create an instance of the RandomSampler class and pass it to the DataLoader as the sampler variable.\n\\begin{python}\n\ttransform = transforms.Compose([transforms.ToTensor(),transforms.Normalize((0.5, 0.5, 0.5), (0.5, 0.5, 0.5))])\n\ttrainset = torchvision.datasets.CIFAR10(root='./data', train=True,download=True, transform=transform)\n\twith_replacement_sampler = torch.utils.data.RandomSampler(trainset, replacement=True)\n\ttrainloader = torch.utils.data.DataLoader(trainset, batch_size=4, shuffle=True, sampler=with_replacement_sampler, num_workers=2)\n\\end{python}\nNow if we run this code, the data will be sampled with replacement. This is just to illustrate how you go about changing something about the model and the training process. You just look up the classes, look up in the documentation, see what they do. And often the most of thing that you may want to do, someone already program them in a simple way to do them. You can just turn on like we just doing it. Just notice that you don't have to implement it yourself in this case. You just need to look out these classes do and figure out which tools for use. So we recommend just for fun just try to the CIFAR10 example with and without replacement to see difference. \n\n\\subsubsection{Codes for the CIFAR10 example}\n\\begin{python}\n# cifar10_example.py\n\nimport torch \nimport torchvision\nimport torchvision.transforms as transforms\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\n\n\nclass Net(nn.Module):\n def __init__(self):\n super(Net, self).__init__()\n self.conv1 = nn.Conv2d(3, 6, 5)\n self.pool = nn.MaxPool2d(2, 2)\n self.conv2 = nn.Conv2d(6, 16, 5)\n self.fc1 = nn.Linear(16 * 5 * 5, 120)\n self.fc2 = nn.Linear(120, 84)\n self.fc3 = nn.Linear(84, 10)\n\n def forward(self, x):\n x = self.pool(F.relu(self.conv1(x)))\n x = self.pool(F.relu(self.conv2(x)))\n x = x.view(-1, 16 * 5 * 5)\n x = F.relu(self.fc1(x))\n x = F.relu(self.fc2(x))\n x = self.fc3(x)\n return x\n\ndef main():\n\ttransform = transforms.Compose([transforms.ToTensor(),transforms.Normalize((0.5, 0.5, 0.5), (0.5, 0.5, 0.5))])\n\ttrainset = torchvision.datasets.CIFAR10(root='./data', train=True,download=True, transform=transform)\n\ttrainloader = torch.utils.data.DataLoader(trainset, batch_size=4, shuffle=True, num_workers=2)\n\ttestset = torchvision.datasets.CIFAR10(root='./data', train=False, download=True, transform=transform)\n\ttestloader = torch.utils.data.DataLoader(testset, batch_size=4, shuffle=False, num_workers=2)\n\nclasses = ('plane', 'car', 'bird', 'cat',\n 'deer', 'dog', 'frog', 'horse', 'ship', 'truck')\n\nconv_net = Net()\n\ncriterion = nn.CrossEntropyLoss()\noptimizer = optim.SGD(net.parameters(), lr=0.001, momentum=0.9)\n\nfor epoch in range(2): # loop over the dataset multiple times\n\n running_loss = 0.0\n for i, data in enumerate(trainloader, 0):\n # get the inputs; data is a list of [inputs, labels]\n inputs, labels = data\n\n # zero the parameter gradients\n optimizer.zero_grad()\n\n # forward + backward + optimize\n outputs = net(inputs)\n loss = criterion(outputs, labels)\n loss.backward()\n optimizer.step()\n\n # print statistics\n running_loss += loss.item()\n if i % 2000 == 1999: # print every 2000 mini-batches\n print('[%d, %5d] loss: %.3f' %\n (epoch + 1, i + 1, running_loss / 2000))\n running_loss = 0.0\n\nprint('Finished Training')\n\ncorrect = 0\ntotal = 0\nwith torch.no_grad():\n for data in testloader:\n images, labels = data\n outputs = net(images)\n _, predicted = torch.max(outputs.data, 1)\n total += labels.size(0)\n correct += (predicted == labels).sum().item()\n\nprint('Accuracy of the network on the 10000 test images: %d %%' % (\n 100 * correct / total))\n\n\\end{python}\n\n\n\\subsection{The training part of CIFAR10 example}\nIn this section, we will analysis the training part code in the CIFAR10 example. \n\\begin{python}\nfor epoch in range(2): # loop over the dataset multiple times\n\n running_loss = 0.0\n for i, data in enumerate(trainloader, 0):\n # get the inputs; data is a list of [inputs, labels]\n inputs, labels = data\n\n # zero the parameter gradients\n optimizer.zero_grad()\n\n # forward + backward + optimize\n outputs = net(inputs)\n loss = criterion(outputs, labels)\n loss.backward()\n optimizer.step()\n\n # print statistics\n running_loss += loss.item()\n if i % 2000 == 1999: # print every 2000 mini-batches\n print('[%d, %5d] loss: %.3f' %\n (epoch + 1, i + 1, running_loss / 2000))\n running_loss = 0.0\n\nprint('Finished Training')\n\\end{python}\nIn particular, we will see what the \\emph{loss.backward} and \\emph{optimizer.step} do\n\\subsubsection{\\emph{loss.backward}}\nLet's start with the automatically differentiation of PyTorch. This function \\emph{loss.backward} hides some pretty complicated code that automatically figure out how to calculate the gradient w.r.t. the output.\nLet's start with some simple example.\n\\begin{python}\nimport torch\nfrom torch.autograd import Variable\nx = Variable(torch.randn(3,3), requires_grad = True)\n\\end{python}\n\\emph{Variable} is a fundamental data type in PyTorch, which contains a tensor and some other things, like a flag in our example called \\emph{requires\\_grad}. So if I set \\emph{requires\\_grad} is true, it just indicating to PyTorch that in the future whatever \\emph{x} is involved in (computational graph), it going to calculate the gradient of x. Also, one of the other things the Variable contains is the gradient value. Which right now should be empty. You can check it by running \\emph{x.grad}. \nNow let's see what happen when we use this Variable x in some calculations.\n\\begin{python}\ny = torch.sum(x)\n\\end{python}\nHere, \\emph{torch.sum} means all the values of x. Here, we create a new variable y which consists of the sum of the entries of x. And PyTorch allows us to automatically calculate the gradient of the output w.r.t. any inputs. This is an extremely simple example.\n\\begin{equation}\n\\begin{split}\nx &= \\left(\\begin{array}{ccc} \n\t* &\\cdots &* \\\\\n\t\\vdots &\\ddots &\\vdots \\\\\n\t* &\\cdots &*\n\t\\end{array}\\right) \\\\\ny &= \\left(\\begin{array}{c} \n\t1 \\\\ \\vdots \\\\ 1\n\t\\end{array}\\right)^\\top\n\tx\\ \n\t\\left(\\begin{array}{c} \n\t1 \\\\ \\vdots \\\\ 1\n\t\\end{array}\\right)\n\t=\\sum_{i=1}^3 \\sum_{j=1}^3 x_{ij} \\\\\n\\frac{{\\rm d} y}{{\\rm d} x} &= \\left(\\begin{array}{ccc} \n\t1 &1 &1 \\\\\n\t1 &1 &1 \\\\\n\t1 &1 &1\n\t\\end{array}\\right) \n\\end{split}\n\\end{equation}\nHow to let PyTorch compute this. Simply call \n\\begin{python}\ntorch.autograd.grad(y,x)\n\\end{python}\nThis function automatically calculate the derivative of the second augment w.r.t the first. And other way, which is more commonly used, is to call \n\\begin{python}\ny.backward()\n\\end{python}\nThis function will calculate the derivative of y w.r.t everything that depends on the requirements of gradients (see the flag \\emph{requires\\_grad}). In our case, it will calculate the gradient of y w.r.t. x and store the gradient value in \\emph{x,grad}.\nThese are two common ways that let PyTorch to calculate the gradients. We just give a simple example with one input and one output, but this whole thing works even for much more complicated examples with multiple input variables. And PyTorch can automatically calculate the gradients of these variables.\n\nAnd there is one important thing to note is now if I call the \\emph{y.backwark} again, what happen to the \\emph{x.grad}? It will keep the same? or it will be changed? The answer is, unfortunately, it adds the gradient to whatever is already stored in the \\emph{x.grad}. So, this explains why we have to zero the gradients of all of the parameters (see \\emph{optimizer.zero\\_grad}) before we call \n\\emph{loss.backward} in our codes.\n\n\\subsubsection{\\emph{optimizer.step}}\nThe \\emph{loss.backward} step calculates the gradients of all of our parameters and stores in \\emph{.grad} fields. Now, let's look at our \\emph{optimizer}. This \\emph{optimizer} may be a class you want to write yourself if you want to try some new algorithm. It will help to know how it works. \nNotice that the function we are calling here is \\emph{optimizer.step}. For example, we set optimizer as SGD. After we calculate the gradients of all of the parameters, then we call this step function, it will add the values of the multiplication of learning rate and negative gradients to current values of the parameters. The \\emph{optimizer} class assume that gradients already been calculated and stored in \\emph{.grad} fields, and the class just decide what to do with the gradients.\n\nNow, let use Logistic regression as an example.\n\\begin{python}\n# lr_example.py\n\nimport torch \nimport torchvision\nimport torchvision.transforms as transforms\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\n\n\nclass LogisticRegression(nn.Module):\n def __init__(self):\n super(LogisticRegression, self).__init__()\n self.linear = nn.Linear(10, 5)\n\n def forward(self, x):\n return self.linear(x)\n\\end{python}\nWe just write a logistic regression class with 10 input features and 5 output features. Then we will use this example to see what a the network class exactly contains.\nRun the following codes in python terminal\n\\begin{python}\nimport torch\nfrom lr_example import LogisticRegression\n\nnet = LogisticRegression()\nx = torch.randn(10)\ny = net(x)\n\\end{python}\nHere we create an instance of LogisticRegression class, and set x as input and y as output. The variables this class contains is the parameters which is a 10-by-5 matrix and a vector as the bias. And take a vector x as input, and apply the forward step (see \\emph{net(x)}, `forward' can be dropped), it will run the forward function to calculate $W x+b$.\n\nThere is one thing have to know is that when you call \\emph{backward} on something has to be a scalar. It can't be for example a tensor. For example, if I try to call \\emph{y.backward}, it will be given an error, because you can only call this for scalar outputs. It won't to calculate a Jacobian matrix of a vector output w.r.t. a vector input. However, if you pass it the derivatives of each of this components, then it can continuous do the \\emph{.backward} function. So, for example, if we we want to calculate the derivative of the sum of $y$, we can do it as follows:\n\\begin{python}\ny.backward(torch.ones(5))\n\\end{python}\nIf you have a vector that you want to call \\emph{.backward} on, you have to pass it a vector with the same size, it gives the gradients of each of the components of that vector w.r.t. the output. Actually, it calculate the Jacobian times the vector. In general, it can't calculate the Jacobian of a non-scalar output, but if you know what you want to multiply the Jacobian with, then you can pass it to \\emph{.backward} as well. \n\nThen we can see the parameter of a network \n\\begin{python}\nparams = list(net.parameters())\n\\end{python}\nNow \\emph{params} contains matrix $W$ and vector $b$. Also, we can see the gradient values of them. \n\\begin{python}\nparams[0].grad\nparams[1].grad\n\\end{python}\n\nHere we give an explanation about it. For $x\\in\\mathbb{R}^n$ and $y\\in\\mathbb{R}^d$, $y$ is some function of $x$.\n\\begin{equation}\n\\begin{split}\ny &= f(x) \\\\\n\\frac{{\\rm d} y}{{\\rm d} x} &= \\left(\\begin{array}{ccc} \n\t\\frac{{\\rm d} y_1}{{\\rm d} x_1} &\\cdots &\\frac{{\\rm d} y_d}{{\\rm d} x_1} \\\\\n\t\\vdots &\\ddots &\\vdots \\\\\n\t\\frac{{\\rm d} y_1}{{\\rm d} x_n} &\\cdots &\\frac{{\\rm d} y_d}{{\\rm d} x_n}\n\t\\end{array}\\right) \n\\end{split}\n\\end{equation}\nIn this case, \\emph{y.backward()} will return an error, but we can call \\emph{y.backward(v)} for some vector $v\\in\\mathbb{d}$. Then it will calculate \n\\begin{equation}\n\\frac{{\\rm d} y}{{\\rm d} x}\\cdot v = \\left(\\begin{array}{ccc} \n\t\\frac{{\\rm d} y_1}{{\\rm d} x_1} &\\cdots &\\frac{{\\rm d} y_d}{{\\rm d} x_1} \\\\\n\t\\vdots &\\ddots &\\vdots \\\\\n\t\\frac{{\\rm d} y_1}{{\\rm d} x_n} &\\cdots &\\frac{{\\rm d} y_d}{{\\rm d} x_n}\n\t\\end{array}\\right) \n\t\\left(\\begin{array}{c} \n\tv_1 \\\\ \\vdots \\\\v_d\t\n\t\\end{array}\\right) \n= \\nabla(y\\cdot v)\n\\end{equation}\n\n\nThis is basically because if you want to compute the whole Jacobian matrix, it requires a lot of time and space and it always not necessary. \n\nMaybe this will make more sense if I set another variable $z$ as the sum of $y$, \n\\begin{python}\nnet.zero.grad()\ny = net(x)\nz =torch.sum(y)\n\\end{python}\n\nIn our example, we can see $z$ is a function of $y$ and $z = g(y) = y \\cdot v$.\nIf I take \\emph{z.backward}, this is the same computation with the previous did. Because the derivatives of $z$ w.r.t. $y$ is $\\frac{{\\rm d} z}{{\\rm d}y} = (1,1,1,1,1)^\\top$. This is what the \\emph{.backward} does. It calculate the gradients multiplied by the Jacobian of previous layer to get the previous gradients, and so on.\nYou can call \n\\begin{python}\nz.backward()\nparams[0].grad\nparams[1].grad\n\\end{python}\nand compare the output with the previous one. \n\nIn practice, we would construct some loss function, and call \\emph{backward} on it. Particular, in our cifar10 example, it consists of a bunch of parameters, particular, each of these layers has its parameters, and when you call forward, it performs some complicated calculations. It will build some graph that contains all the dependencies of these calculations. And then, after you apply the network\nand apply a loss function to it, you call \\emph{loss.backward}, this step gives you all the gradients of the parameters, and then this is a special optimizer class, which takes those stored gradients and use them to perform like a forward step. \nThe most important thing to remember are don't forget the \\emph{optimizer.zero\\_grad} because the \\emph{backward} step adds the new calculated gradients to that already stored in the \\emph{grad} field. Of course, if you have two loss functions, called \\emph{loss1} and \\emph{loss2}, you can call \n\\begin{python}\nloss1.backward()\nloss2.backward()\n\\end{python}\nand gradients now is the sum of those two gradients.\n", "meta": {"hexsha": "53398296f71d7897479edf812c17bc91d79d3b7c", "size": 25491, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/497Proj_cifar.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/497Proj_cifar.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/497Proj_cifar.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.7350230415, "max_line_length": 1198, "alphanum_fraction": 0.7189988623, "num_tokens": 6802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8840392756357326, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6489904896483776}} {"text": "\\selectlanguage{english}\n\\begin{abstract}\n \\noindent For the first project of the Computer Simulations (ELTE Physics MSc) course I propose a concept of an N-body simulation, which aims to reproduce the satellite formation inside asteroid belts around a larger stellar object (eg. a gas giant or a star). Due to the N-body simulations' high computational difficulty, my spare objective is to observe at least some form of clustering process inside these type of systems. To achieve these goals, I will use Newton's law of universal gravitation, solving the particles' equation of motion by a simple 4th order Runge-Kutta function. Other numerical methods - described in Sec. III. will be optionally tested.\n\\end{abstract}\n\n\\begin{multicols}{2}\n\\section{Introduction}\nThe problem of gravitational- or electromagnetic attraction of more, than 2 bodies are impossible to solve analitically, except for some finite special cases. To study the motion of the particles in such systems we're ought to rely on numerical simulations and approximations. The computational difficulty of the problem grows non-linearly as we try to simulate more and more particles, and thus these simulations are required to be reinforced by some clever numerical tricks to overcome the barrier of immense computational times.\n\n\\section{Motivation}\nAs I've already worked with numerical simulations of many-body systems, using two- and three-body approximations, I've choosen the topic of N-body simulations unhesitantly to work on as the first project of the Computer Simulations course. Also being fascinated by astronomical questions and problems, the choosing of the topic was an overally trivial decision for me. My goals are to improve my knowledge about simple N-body simulations and to learn to use some generally useful numerical tricks while working on the assignment.\n\n\\section{Description of the proposed assignment}\nInside a dense asteroid or planetary debris belt it is inevitable for larger objects to form. This belt could be either orbiting around a central star, or a planet. In both cases, moons or dwarf planets will be created by the constant collisions of the individual smaller bodies. My proposed assignment is to simulate the N-body problem of such system of maximally $10^5$ bodies and monitor its density changes and observe the formation of clusters or even greater satellites with DBSCAN or mean-shift methods. Furthermore to achieve somewhat more realistic results, I propose to implement a simple collision system between the colliding bodies. In a star - planet - asteroids system it is also possible to observe the formation of asteroid clusters around Lagrange points, which could be an extra task for the project. \\newline\nTo evade all possible infinities or huge numerical errors, I will work with spherical rigid bodies, instead of point-like particles. To speed up the processing, I propose to experimentally use a neighbour list method with some reasonable cut-off distance, similar to used in molecular dynamics simulation. Reason for this, that at great distances the gravitational pull between smaller ojects can be bravely neglected, and thus the efficiency of the simulation could be greatly enhanched. However I have no idea how does this approximation affects or deteriorates the accuracy of the simulation. I think at least it deserves a chance.\n\n\\section{Theoretical background}\nDue to my very limited free time and the short deadline for the final submission, I decided to work using the Newtonian gravitational theory only. In this case it becomes pretty straightforward to write down the necessary equations of motion. The force, acting on the $i$th body could be described by the following sum:\n\n\\begin{equation}\nm_{i} \\boldsymbol{\\ddot{r}}_{i}\n=\n- G \\sum_{i \\neq j} \\frac{m_{i} m_{j}}{\\left| r_{i} - r_{j} \\right|^{2}} \\frac{\\boldsymbol{r}_{i} - \\boldsymbol{r}_{j}}{\\left| r_{i} - r_{j} \\right|}\n\\end{equation}\nUsing this, one can simply describe the acceleration acting on the $i$th body:\n\n\\begin{equation}\n\\boldsymbol{\\ddot{r}}_{i}\n=\n- G \\sum_{i \\neq j} \\frac{m_{j}}{\\left| r_{i} - r_{j} \\right|^{2}} \\frac{\\boldsymbol{r}_{i} - \\boldsymbol{r}_{j}}{\\left| r_{i} - r_{j} \\right|}\n\\end{equation}\nWhich differential equation essentially should be solved numerically for the $x$, $y$ and $z$ components of the $\\boldsymbol{r}$ vector to acquire the coordinates and velocities of the individual particles.\n\n\\section{Problems and spare objective}\nN-body simulations are $\\mathcal{O} \\left( N ^{2} \\right)$ hard problems, which in the absence of powerful computational systems, makes these simulations pretty useless for the observation of longterm processes, when too much particle is present. If my initial idea - to observe formation of compact satellites and moons - fails, I aim to try to at least monitor the formation of denser clusters and groups inside my asteroid system. \\newline\nIt could be also provide unsatisfactory results if the initial system is not stable, and all particles simply ejected from the solar system after a while. I really do hope this won't happen...\n\n\\section{Technical deteails}\nI'm the most confident working in Python 3.X enviroments, and prefering mostly Jupyter Notebooks. The used tools and libraries will be only Python-related. I also use FFmpeg to generate animations of the simulated systems. If it is possible, I will upload some animations of my project on my YouTube channel.\n\n\\end{multicols}", "meta": {"hexsha": "9978cc7f5ea46edb5efda998513db3cba118b9b6", "size": 5460, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Project 1/Documentation/Concept/src/text_src/body.tex", "max_stars_repo_name": "masterdesky/ELTE_Comp_Simulations_2020", "max_stars_repo_head_hexsha": "b7689e0ecf9b5deaf1c713b647c1c106b8af7457", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Project 1/Documentation/Concept/src/text_src/body.tex", "max_issues_repo_name": "masterdesky/ELTE_Comp_Simulations_2020", "max_issues_repo_head_hexsha": "b7689e0ecf9b5deaf1c713b647c1c106b8af7457", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Project 1/Documentation/Concept/src/text_src/body.tex", "max_forks_repo_name": "masterdesky/ELTE_Comp_Simulations_2020", "max_forks_repo_head_hexsha": "b7689e0ecf9b5deaf1c713b647c1c106b8af7457", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 133.1707317073, "max_line_length": 828, "alphanum_fraction": 0.7928571429, "num_tokens": 1207, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637577007393, "lm_q2_score": 0.7549149923816048, "lm_q1q2_score": 0.6489730590953954}} {"text": "\\section{Laurent Series, Isolated Singularities and the Residue Theorem}\r\n\\begin{theorem}\r\n Consider an open annulus $A=\\{z\\in\\mathbb C:r<|z-a|0$ but not in $D_r(a)$ (i.e. either not defined at $a$ or not holomorphic there).\r\n\\end{definition}\r\n\\begin{example}\r\n 1. $f(z)=1/z$ has an isolated singularity at $z=0$.\\\\\r\n 2. $f(z)=(e^z-1)/z$ also has one at $0$.\\\\\r\n 3. $f(z)=e^{1/z}$ has one at $0$.\r\n\\end{example}\r\n\\begin{definition}\r\n An isolated singularity $a$ of $f$ is removable if there is a holomorphic $g$ defined on $D_r(a)$ for some $r>0$ and $f=g$ on $D_r(a)\\setminus\\{a\\}$.\r\n\\end{definition}\r\n\\begin{proposition}\r\n $f$ has a removable singularity at $a$ iff $\\lim_{z\\to a}(z-a)f(z)=0$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n The ``only if'' part is immediate.\r\n For the other direction, Suppose $f$ is holomorphic on $D_r(a)\\setminus\\{a\\}$ and $\\lim_{z\\to a}(z-a)f(z)=0$, then let\r\n $$h(z)=\\begin{cases}\r\n (z-a)^2f(z)\\text{, if $z\\neq 0$}\\\\\r\n 0\\text{, if $z=0$}\r\n \\end{cases}$$\r\n So $h$ is holomorphic in some disk $D_r(a)\\setminus\\{a\\}$.\r\n Also\r\n $$\\frac{h(z)-h(a)}{z-a}=(z-a)f(z)\\to 0$$\r\n as $z\\to a$, so $h$ is actually holomorphic on $D_r(a)$ with $h^\\prime(a)=0$.\r\n But $h$ has a zero of order at least $2$, hence there is a holomorphic $g$ on $D_r(a)$ such that $h(z)=(z-a)^2g(z)$, but then $g$ equals $f$ on $D_r(a)\\setminus\\{a\\}$.\r\n\\end{proof}\r\nSo the isolated singularities of bounded functions are removable.\r\n\\begin{definition}\r\n Let $a$ be an isolated singularity of $f:D_r(a)\\to\\mathbb C$, then $a$ is a pole if the limit $\\lim_{z\\to a}|f(z)|=\\infty$.\\\\\r\n If $a$ is neither a pole nor a removable singularity, we say $a$ is an essential singularity.\r\n\\end{definition}\r\n\\begin{proposition}\\label{pole_equiv}\r\n Let $U$ be a domain and $f:U\\setminus\\{a\\}\\to\\mathbb C$ is holomorphic for $a\\neq U$.\r\n Then the followings are equivalent:\\\\\r\n 1. $a$ is a pole of $f$.\\\\\r\n 2. There is $\\epsilon>0$ such that there is a holomophic $h:D_\\epsilon(a)\\to\\mathbb C$ with $h(z)=0\\iff z=a$ and $\\forall z\\in D_\\epsilon(a)\\setminus\\{a\\},f(z)=1/h(z)$.\\\\\r\n 3. There is a holomorphic $g:U\\to\\mathbb C$ such that $g(a)\\neq 0$ and\r\n $$f(z)=(z-a)^{-k}g(z)$$\r\n for some integer $k\\ge 1$.\r\n Also such $g,k$ are uniquely determined by $f$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n $1\\implies 2$: Choose $\\epsilon$ such that $|f(z)|\\ge 1$ whenever $0<|z-a|<\\epsilon$, so $1/f$ is holomorphic on $D_\\epsilon(a)\\setminus\\{a\\}$.\r\n But $a$ becomes a removable singularity of $1/f$ by the preceding proposition, so the extension of $1/f$ to the entire disk $D_\\epsilon(a)$ would be the desired $h$.\\\\\r\n $2\\implies 3$: Suppose $h$ has a zero of order $k$, then $h(z)=(z-a)^kq(z)$ where $q$ is holomorphic and $q(a)\\neq 0$, so $q$ is nonzero in a disk $D\\subset D_\\epsilon(a)$ around $a$, hence\r\n $$g(z)=\\begin{cases}\r\n 1/q(z)\\text{, if $z\\in D$}\\\\\r\n (z-a)^kf(z)\\text{, if $z\\in U\\setminus\\{a\\}$}\r\n \\end{cases}$$\r\n which is well-defined by the definition of $h$ and is holomorphic.\r\n Uniqueness follows.\\\\\r\n $3\\implies 1$ is trivial.\r\n\\end{proof}\r\n\\begin{corollary}\r\n If $z\\in\\mathbb C$ is an essential singularity of $f$, then the limit of $|f(z)|$ as $z\\to a$ does not exist, either as a real number or infinity.\r\n\\end{corollary}\r\n\\begin{example}\r\n If we compute $e^{1/z}$ as $z\\to 0$, we will find that it does not exist, so $0$ is an essential singularity of it.\r\n\\end{example}\r\n\\begin{definition}\r\n If $f$ has a pole at $a$, then the integer $k$ as stated in Proposition \\ref{pole_equiv} is called the order of the pole $a$.\r\n If $k=1$, then we call $a$ a simple pole.\\\\\r\n Let $U$ be a domain and $S\\subset U$ is a set of isolated points in $U$.\r\n If $f$ is holomorphic on $U\\setminus S$ and each $s\\in S$ is either a removable singularity or a pole of $f$, we say $f$ is meromorphic on $U$.\r\n\\end{definition}\r\nMeromorphic functions can be taken (maybe) as holomorphic functions from $U$ to $\\mathbb C_\\infty$.\r\n\\begin{remark}\r\n By the preceding corollary, at an isolated essential singularity $a$ of $f$, then $f$ oscillates around $a$.\r\n More precisely we have the Cosorati-Weierstrass Theorem (proven in example sheet), where we find that the image of any punctured neighbourhood around $a$ has dense image in $\\mathbb C$.\r\n A much much harder theorem will show that this image is the entire complex plane $\\mathbb C$ except possibly one point.\r\n\\end{remark}\r\n\\begin{proposition}\r\n Let $f$ be holomorphic in $D_R(a)\\setminus\\{a\\}$, then we have the series\r\n $$f(z)=\\sum_{n\\in\\mathbb Z}c_n(z-a)^n,\\forall z\\in D_R(a)\\setminus\\{a\\}$$\r\n Also, $\\forall n<0,c_n=0$ implies that $a$ is removable.\\\\\r\n If now $c_{-k}\\neq 0$ for some $k>0$ and $c_n=0$ for $n<-k$, then there is a holomorphic $g:D_R(a)\\to\\mathbb C$ such that\r\n $$f(z)=\\sum_{n=-k}^\\infty c_n(z-a)^n=\\frac{g(z)}{(z-a)^k}$$\r\n So $f$ has a pole of order $k$ at $a$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\nLet rhe series expansion be as above, then by uniform convergence, the integral\r\n$$\\int_{\\partial D_\\rho(a)}f(z)\\,\\mathrm dz=2\\pi ic_{-1}$$\r\nSo we define\r\n\\begin{definition}\r\n $c_{-1}$ is defined as the residue $\\operatorname{Res}_f(a)$ of $f$ at $a$.\r\n And the series\r\n $$\\sum_{n=-\\infty}^{-1}c_n(z-a)^n$$\r\n is called the principal part of $f$.\r\n\\end{definition}\r\n\\begin{theorem}[Residue Theorem]\r\n Let $U$ be a domain and $S=\\{a_1,\\ldots,a_k\\}\\subset U$.\r\n Suppose $f$ is holomorphic on $U\\setminus S$, then for any closed piecewise $C^1$ curve $\\gamma:[0,1]\\to U\\setminus S$ homologous to $0$ in $U$, we have\r\n $$\\int_\\gamma f(z)\\,\\mathrm dz=2\\pi i\\sum_{j=1}^kI(\\gamma;a_j)\\operatorname{Res}_f(a_j)$$\r\n\\end{theorem}\r\n\\begin{proof}\r\n Let $g$ be the principal part of $f$ at $a_i$.\r\n Then $g$ is holomorphic on $\\mathbb C\\setminus\\{a_i\\}$, so $f-(g_1+\\cdots+g_k)$ is holomorphic in $U$ except at removable singularities, so by Cauchy's Theorem,\r\n $$\\int_\\gamma f(z)\\,\\mathrm dz=\\sum_{j=1}^k\\int_\\gamma g_j(z)\\,\\mathrm dz=2\\pi i\\sum_{j=1}I(\\gamma,a_j)\\operatorname{Res}_f(a)$$\r\n as desired.\r\n\\end{proof}\r\nThere are some useful facts about residues.\r\n\\begin{proposition}\r\n If $f$ has a simple pole at $a$, then $\\operatorname{Res}_f(a)=\\lim_{z\\to a}(z-a)f(a)$.\r\n More generally, if $f$ has a pole of order $k$, then if we write $f(z)=(z-a)^{-k}g(z)$ for a holomorphic $g$, then $\\operatorname{Res}_f(a)=g^{(k-1)}(a)/(k-1)!$.\r\n If $f=g/h$ with $g,h$ holomorphic and at $a$, $g(a)\\neq 0$ and $h$ has a simple zero at $a$, then $f(a)=g(a)/h^\\prime(a)$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Easy.\r\n\\end{proof}\r\n\\begin{proposition}[Jordan's Lemma]\r\n If $f$ is holomorphic in $\\{z\\in\\mathbb C:|z|>r\\}$ for some $r>0$ and if $zf(z)$ is bounded for large $|z|$, then\r\n $$\\int_{\\gamma_R}f(z)e^{i\\alpha z}\\,\\mathrm dz\\to 0,\\gamma_R:[0,\\pi]\\ni t\\mapsto Re^{it}$$\r\n as $R\\to\\infty$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Example sheet.\r\n Use the fact that $\\sin t/t\\ge 2/\\pi$ for $t\\in (0,\\pi/2]$.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Let $f$ be holomorphic on $D_R(a)\\setminus\\{a\\}$ with a simple pole at $z=a$ and $\\gamma_\\epsilon:[\\alpha,\\beta]\\ni t\\mapsto a+\\epsilon e^{it}$, then\r\n $$\\lim_{\\epsilon\\to 0^+}\\int_{\\gamma_\\epsilon}f(z)\\,\\mathrm dz=(\\beta-\\alpha)i\\operatorname{Res}_f(a)$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n Just write $f(z)=\\operatorname{Res}_f(a)/(z-a)+g(z)$.\r\n\\end{proof}\r\n\\begin{example}\r\n Let $R>\\epsilon>0$.\r\n Consider $f(z)=e^{iz}/z$ along the contour $\\gamma=\\gamma_1-\\gamma_\\epsilon+\\gamma_2+\\gamma_R$ where $\\gamma_1$ is the segment $[-R,-\\epsilon]$, $\\gamma_2$ is $[\\epsilon,R]$, and $\\gamma_R(t)=Re^{it},\\gamma_\\epsilon(t)=\\epsilon e^{it},t\\in [0,\\pi]$.\r\n Now by Jordan's Lemma and the preceding proposition respectively, we have\r\n $$\\lim_{R\\to\\infty}\\int_{\\gamma_R}f(z)\\,\\mathrm dz=0,\\lim_{\\epsilon\\to 0}\\int_{\\gamma_\\epsilon}f(z)\\,\\mathrm dz=\\pi i\\operatorname{Res}_f(a)=\\pi i$$\r\n Now for $\\epsilon,R$ as before we have\r\n $$0=\\int_\\gamma f(z)\\,\\mathrm dz=\\int_{-R}^{-\\epsilon}\\frac{e^{ix}}{x}\\,\\mathrm dx+\\int_{-\\gamma_\\epsilon}f(z)\\,\\mathrm dz+\\int_\\epsilon^R\\frac{e^{ix}}{x}\\,\\mathrm dx+\\int_{\\gamma_R}f(z)\\,\\mathrm dz$$\r\n Now let $R\\to\\infty,\\epsilon\\to 0$, we get\r\n $$2i\\int_0^\\infty\\frac{\\sin x}{x}\\,\\mathrm dx=\\pi i\\implies\\int_0^\\infty\\frac{\\sin x}{x}\\,\\mathrm dx=\\frac{\\pi}{2}$$\r\n\\end{example}", "meta": {"hexsha": "60b6a79e2ed27ed932e68c8e61f960b7ae664d85", "size": 10890, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "4/res.tex", "max_stars_repo_name": "david-bai-notes/IB-Complex-Analysis", "max_stars_repo_head_hexsha": "d67e2ff022d5fbc22bfdfd377f2414c23be532ed", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "4/res.tex", "max_issues_repo_name": "david-bai-notes/IB-Complex-Analysis", "max_issues_repo_head_hexsha": "d67e2ff022d5fbc22bfdfd377f2414c23be532ed", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "4/res.tex", "max_forks_repo_name": "david-bai-notes/IB-Complex-Analysis", "max_forks_repo_head_hexsha": "d67e2ff022d5fbc22bfdfd377f2414c23be532ed", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.6842105263, "max_line_length": 302, "alphanum_fraction": 0.6252525253, "num_tokens": 4026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149758396752, "lm_q2_score": 0.8596637469145054, "lm_q1q2_score": 0.6489730367322085}} {"text": "\\problemname{Elementary Math}\n\n\\illustration{.35}{elementarymath.jpg}{Example exam by Ellen}%\nEllen is teaching elementary math to her students and the time for the final exam has\ncome. The exam consists of $n$ questions. In each question the students have to\nadd ($+$), subtract ($-$) or multiply ($*$) a pair of numbers.\n\nEllen has already chosen the $n$ pairs of numbers. All that remains is to decide\nfor each pair which of the three possible operations the students should perform.\nTo avoid students getting bored, Ellen wants to make sure that the $n$ correct answers\nto her exam are all different.\n\nPlease help Ellen finish constructing the exam by automating this task.\n\n\\section*{Input}\n\nThe input consists of:\n\\begin{itemize}\n \\item one line with one integer $n$ ($1\\leq n \\leq 2\\,500$), the number of pairs of numbers;\n \\item $n$ lines each with two integers $a$ and $b$ ($-10^6 \\leq a, b \\leq 10^6$), a pair of numbers used.\n\\end{itemize}\n\n\\section*{Output}\n\nFor each pair of numbers $(a,b)$ in the same order as in the input,\noutput a line containing a valid equation.\nEach equation should consist of five parts: $a$, one of the three operators, $b$,\nan equals sign ($=$), and the result of the expression.\nAll the $n$ expression results must be different.\n\nIf there are multiple valid answers you may output any of them.\nIf there is no valid answer, output a single line with the string ``\\texttt{impossible}'' instead.\n\n", "meta": {"hexsha": "f0d9536f0182d41164eb9f29231840d203606555", "size": 1439, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problems/elementarymath/problem_statement/problem.en.tex", "max_stars_repo_name": "stoman/CompetitiveProgramming", "max_stars_repo_head_hexsha": "0000b64369b50e31c6f48939e837bdf6cece8ce4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-12-22T13:21:25.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-12T22:26:26.000Z", "max_issues_repo_path": "problems/elementarymath/problem_statement/problem.en.tex", "max_issues_repo_name": "stoman/CompetitiveProgramming", "max_issues_repo_head_hexsha": "0000b64369b50e31c6f48939e837bdf6cece8ce4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problems/elementarymath/problem_statement/problem.en.tex", "max_forks_repo_name": "stoman/CompetitiveProgramming", "max_forks_repo_head_hexsha": "0000b64369b50e31c6f48939e837bdf6cece8ce4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.3235294118, "max_line_length": 108, "alphanum_fraction": 0.7407922168, "num_tokens": 370, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703224, "lm_q2_score": 0.8006919949619792, "lm_q1q2_score": 0.6489439742578481}} {"text": "\\section{Eigenvalues and eigenvectors of a matrix}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Describe eigenvalues geometrically and algebraically.\n \\item Find eigenvalues and eigenvectors for a square matrix.\n \\end{enumerate}\n\\end{outcome}\n\n", "meta": {"hexsha": "bcc687e37d93309c1e23d8f2b47cebead7018574", "size": 246, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/content/spectraltheoryEigenvaluesEigenvectors.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "old/content/spectraltheoryEigenvaluesEigenvectors.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/content/spectraltheoryEigenvaluesEigenvectors.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 24.6, "max_line_length": 62, "alphanum_fraction": 0.7804878049, "num_tokens": 62, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703224, "lm_q2_score": 0.8006919925839875, "lm_q1q2_score": 0.648943972330536}} {"text": "\\title{Teaching the Geometry of Schemes}\n\\titlerunning{Teaching the Geometry of Schemes}\n\\toctitle{Teaching the Geometry of Schemes}\n\n\\author{Gregory G.~Smith \\and Bernd Sturmfels}\n\\authorrunning{G. G. Smith and B. Sturmfels}\n% \\institute{Department of Mathematics, University of California,\n% Berkeley, California 94720, USA}\n\n\\maketitle\n\n\n%%----------------------------------------------------------\n\\newtheorem*{problem*}{Problem}{\\bfseries\\upshape}{\\itshape}\n\\newtheorem*{solution*}{Solution}{\\itshape}{\\rmfamily}\n\n\\newcommand{\\Spec}{\\operatorname{Spec}}\n\\newcommand{\\Proj}{\\operatorname{Proj}}\n\\newcommand{\\codim}{\\operatorname{codim}}\n%%----------------------------------------------------------\n\n\n\\begin{abstract}\nThis chapter presents a collection of graduate level problems in\nalgebraic geometry illustrating the power of \\Mtwo as an educational\ntool.\n\\end{abstract}\n\nWhen teaching an advanced subject, like the language of schemes, we\nthink it is important to provide plenty of concrete instances of the\ntheory. Computer algebra systems, such as \\Mtwo, provide students\nwith an invaluable tool for studying complicated examples.\nFurthermore, we believe that the explicit nature of a computational\napproach leads to a better understanding of the objects being\nexamined. This chapter presents some problems which we feel\nillustrate this point of view.\n\nOur examples are selected from the homework of an algebraic geometry\nclass given at the University of California at Berkeley in the fall of 1999.\nThis graduate course was taught by the second author with assistance from the\nfirst author. Our choice of problems, as the title suggests, follows the\nmaterial in David Eisenbud and Joe Harris' textbook {\\em The Geometry of\n Schemes} \\cite{SC:EH}.\n\n%%----------------------------------------------------------\n\\section{Distinguished Open Sets}\n\nWe begin with a simple example involving the Zariski topology of an affine\nscheme\\index{scheme!affine}. This example also indicates some of the\nsubtleties involved in working with arithmetic\nschemes\\index{scheme!arithmetic}.\n\n\\begin{problem*}\nLet $S = \\bbbz[x,y,z]$ and $X = \\Spec(S)$. If $f = x$ and $X_{f}$ is\nthe corresponding basic open subset in $X$, then establish the\nfollowing:\n\\begin{enumerate}\n\\item[$(1)$] If $e_{1} = x+y+z$, $e_{2} = xy+xz+yz$ and $e_{3} = xyz$\nare the elementary symmetric functions then the set $\\{X_{e_{i}}\\}_{1\n\\leq i \\leq 3}$ is an open cover of $X_{f}$.\n\\item[$(2)$] If $p_{1} = x+y+z$, $p_{2} = x^{2}+y^{2}+z^{2}$ and $p_{3}\n= x^{3}+y^{3}+z^{3}$ are the power sum symmetric functions then\n$\\{X_{p_{i}}\\}_{1 \\leq i \\leq 3}$ is {\\em not} an open cover of\n$X_{f}$.\n\\end{enumerate}\n\\end{problem*}\n\n\\begin{solution*}\n$(1)$ To prove that $\\{X_{e_{i}}\\}_{1 \\leq i \\leq 3}$ is an open cover\nof $X_{f}$, it suffices to show that $e_{1}$, $e_{2}$ and $e_{3}$\ngenerate the unit ideal in $S_{f}$; see Lemma I-16 in Eisenbud and\nHarris~\\cite{SC:EH}. This is equivalent to showing that $x^{m}$\nbelongs to the $S$-ideal $\\langle e_{1}, e_{2}, e_{3} \\rangle$ for\nsome $m \\in \\bbbn$. In other words, the saturation\\index{saturation}\n$\\big( \\langle e_{1}, e_{2}, e_{3} \\rangle : x^{\\infty} \\big)$ is the\nunit ideal if and only if $\\{X_{e_{i}}\\}_{1 \\leq i \\leq 3}$ is an open\ncover of $X_{f}$. We verify this in \\Mtwo as follows:\n<<>>\n<<>>\n<<>>\n$(2)$ Similarly, to show that $\\{X_{p_{i}}\\}_{1 \\leq i \\leq 3}$ is not\nan open cover of $X_{f}$, we prove that $\\big( \\langle p_{1}, p_{2},\np_{3} \\rangle : x^{\\infty} \\big)$ is not the unit ideal. Calculating\nthis saturation, we find\n<<>>\n<<>>\n<<>>\nwhich is not the unit ideal.\\qed\n\\end{solution*}\n\nThe fact that $6$ is a generator of the ideal $\\big( \\langle p_{1},\np_{2}, p_{3} \\rangle : x^{\\infty} \\big)$ indicates that\n$\\{X_{p_{i}}\\}_{1 \\leq i \\leq 3}$ does not contain the points in $X$\nlying over the points $\\langle 2 \\rangle$ and $\\langle 3 \\rangle$ in\n$\\Spec(\\bbbz)$. If we work over a base ring in which $6$ is a unit,\nthen $\\{X_{p_{i}}\\}_{1 \\leq i \\leq 3}$ would, in fact, be an open\ncover of $X_{f}$.\n\n\n%%----------------------------------------------------------\n\\section{Irreducibility}\n\nThe study of complex semisimple Lie algebras gives rise to an\nimportant family of algebraic varieties called nilpotent\norbits\\index{nilpotent orbits}. The next problem examines the\nirreducibility\\index{scheme!irreducible} of a particular nilpotent\norbit.\n\n\\begin{problem*} \nLet $X$ be the set of nilpotent complex $3 \\times 3$ matrices. Show\nthat $X$ is an irreducible algebraic variety.\n\\end{problem*}\n\n\\begin{solution*}\nA $3 \\times 3$ matrix $M$ is nilpotent if and only if its minimal\npolynomial $p(\\sf T)$ equals ${\\sf T}^{k}$, for some $k \\in \\bbbn$.\nSince each irreducible factor of the characteristic polynomial of $M$\nis also a factor of $p(\\sf T)$, it follows that the characteristic\npolynomial of $M$ is ${\\sf T}^{3}$. We conclude that the coefficients\nof the characteristic polynomial of a generic $3 \\times 3$ matrix\ndefine the algebraic variety $X$.\n\nTo prove that $X$ is irreducible over $\\bbbc$, we construct a rational\nparameterization\\index{rational parameterization}. First, observe\nthat ${\\rm GL}_{3}(\\bbbc)$ acts on $X$ by conjugation. Jordan's\ncanonical form theorem implies that there are exactly three orbits;\none for each of the following matrices:\n\\[\nN_{(1,1,1)} =\\left[ \\begin{smallmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 &\n0 & 0 \\end{smallmatrix} \\right], \\quad\nN_{(2,1)} = \\left[ \\begin{smallmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0\n& 0 \\end{smallmatrix} \\right] \\text{ and }\nN_{(3)} = \\left[ \\begin{smallmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 &\n0 \\end{smallmatrix} \\right] \\enspace .\n\\]\nEach orbit is defined by a rational parameterization, so it suffices\nto show that the closure of the orbit containing $N_{(3)}$ is the\nentire variety $X$. We demonstrate this as follows:\n<< Eliminate 10];>>>\n<<>>\n<<>>\nTo determine the entries in $G \\cdot N_{(3)} \\cdot G^{-1}$, we use the\nclassical adjoint\\index{classical adjoint} to construct the matrix\n$\\det(G) \\cdot G^{-1}$.\n<< (\n n := degree target G;\n m := degree source G;\n matrix table(n, n, (i, j) -> (-1)^(i+j) * det(\n submatrix(G, {0..j-1, j+1..n-1}, \n {0..i-1, i+1..m-1}))));>>>\n<<>>\n<<>>\n<<>>\nThe entries in $G \\cdot N_{(3)} \\cdot G^{-1}$ give a rational\nparameterization of the orbit generated by $N_{(3)}$. Using\nelimination theory\\index{elimination theory} --- see section~3.3 in\nCox, Little and O`Shea~\\cite{SC:CLO} --- we give an ``implicit\nrepresentation'' of this variety.\n<<>>\n<<>>\n\nFinally, we verify that this orbit closure equals $X$\nscheme-theoretically. Recall that $X$ is defined by the coefficients\nof the characteristic polynomial of a generic $3 \\times 3$ matrix {\\tt\nM}.\n%% was X = ideal submatrix( (coeff-icients({0}, det(M - t*id_(S^3))))_1,\n%% {1,2,3} ), but 'coeff-icients' is to be redesigned, and 'contract' is\n%% more self-explanatory, anyway.\n<< 0_S})>>>\n<<>>\n<<>>\nThis completes our solution.\\qed\n\\end{solution*}\n\nMore generally, Kostant shows that the set of all nilpotent elements\nin a complex semisimple Lie algebra\\index{Lie algebra} form an\nirreducible variety. We refer the reader to Chriss and\nGinzburg~\\cite{SC:CV} for a proof of this result (Corollary~3.2.8) and\na discussion of its applications in representation theory.\n\n\n%%----------------------------------------------------------\n\\section{Singular Points}\n\nIn our third question, we study the singular locus\\index{singular\nlocus} of a family of elliptic curves\\index{elliptic curve}.\n\n\\begin{problem*} \nConsider a general form of degree $3$ in $\\bbbq[x,y,z]$:\n\\[\nF = ax^{3} + bx^{2}y + cx^{2}z + dxy^{2} + exyz + fxz^{2} + gy^{3} +\nhy^{2}z + iyz^{2} + jz^{3} \\enspace .\n\\]\nGive necessary and sufficient conditions in terms of $a, \\ldots, j$\nfor the cubic curve $\\Proj\\big( \\bbbq[x,y,z] / \\langle F \\rangle\n\\big)$ to have a singular point.\n\\end{problem*}\n\n\\begin{solution*}\nThe singular locus of $F$ is defined by a polynomial of degree $12$ in\nthe $10$ variables $a, \\dotsc, j$. We calculate this polynomial in two\ndifferent ways.\n\nOur first method is an elementary but time consuming elimination.\nCarrying it out in \\Mtwo, we have\n<< Eliminate 2];>>>\n<<>>\n<<>>\n<<>>\n<<>>\n<< 1});>>>\nOn the other hand, there is also an elegant and more useful\ndeterminantal formula for this discriminant\\index{discriminant}; it is\na specialization of the formula (2.8) in section~3.2 of Cox, Little\nand O`Shea~\\cite{SC:CLO2}. To apply this determinantal formula, we\nfirst create the coefficient matrix {\\tt A} of the partial derivatives\nof $F$.\n%% was A = (coeff-icients({0,1,2}, submatrix(jacobian matrix{{F}}, {0..2}, {0})))_1;\n%% but 'coeff-icients' is deprecated.\n<<>>\nWe also construct the coefficient matrix {\\tt B} of the partial\nderivatives of the Hessian\\index{hessian} of $F$.\n<<>>\n%% was B = (coeff-icients({0,1,2}, submatrix(jacobian matrix{{hess}}, {0..2}, {0})))_1;\n%% but 'coeff-icients' is deprecated.\n<<>>\nTo obtain the discriminant, we combine these two matrices and take the\ndeterminant.\n<<>>\nFinally, we check that our two discriminants are equal\n<<>>\nand examine the generator.\n<<>>\n<<>>\n<<<# terms detDiscr_0>>>\n<<>>\nHence, the singular locus is given by a single polynomial of degree\n$12$ with $2040$ terms.\\qed\n\\end{solution*}\n\nFor a further discussion of singularities and discriminants see\nSection~V.3 in Eisenbud and Harris~\\cite{SC:EH}. For information on\nresultants and discriminants see Chapter~2 in Cox, Little and\nO`Shea~\\cite{SC:CLO2}.\n\n\n%%----------------------------------------------------------\n\\section{Fields of Definition}\n\nSchemes\\index{scheme!over a number field} over non-algebraically\nclosed fields arise in number theory. Our fourth problem looks at one\ntechnique for working with number fields in \\Mtwo.\n\n\\begin{problem*}[Exercise~II-6 in \\cite{SC:EH}]\nAn inclusion of fields $K \\hookrightarrow L$ induces a map\n$\\mathbb{A}_{L}^{n} \\to \\mathbb{A}_{K}^{n}$. Find the images in\n$\\mathbb{A}_{\\bbbq}^{2}$ of the following points of\n$\\mathbb{A}_{\\overline{\\bbbq}}^{2}$ under this map.\n\\begin{enumerate}\n\\item[$(1)$] $\\langle x - \\sqrt{2}, y - \\sqrt{2} \\rangle ;$\n\\item[$(2)$] $\\langle x - \\sqrt{2}, y - \\sqrt{3} \\rangle ;$\n\\item[$(3)$] $\\langle x - \\zeta, y - \\zeta^{-1} \\rangle$ where $\\zeta$\nis a $5$-th root of unity $;$\n\\item[$(4)$] $\\langle \\sqrt{2}x- \\sqrt{3}y \\rangle ;$\n\\item[$(5)$] $\\langle \\sqrt{2}x- \\sqrt{3}y-1 \\rangle$.\n\\end{enumerate}\n\\end{problem*}\n\n\\begin{solution*}\nThe images can be determined by using the following three step\nalgorithm: (1) replace the coefficients not contained in $K$ with\nindeterminates, (2) add the minimal polynomials of these coefficients\nto the given ideal in $\\mathbb{A}_{L}^{2}$, and (3) eliminate the new\nindeterminates. Here are the five examples:\n<< Eliminate 2];>>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n<<>>\n\\qed\n\\end{solution*}\n\nIt is worth noting that the points in $\\mathbb{A}_{\\bbbq}^{n}$ correspond\nto orbits of the action of ${\\rm Gal}(\\overline{\\bbbq}/\\bbbq)$ on the\npoints of $\\mathbb{A}_{\\overline{\\bbbq}}^{n}$. For more examples and\ninformation, see section~II.2 in Eisenbud and Harris~\\cite{SC:EH}.\n\n\n%%----------------------------------------------------------\n\\section{Multiplicity}\n\nThe multiplicity\\index{multiplicity} of a zero-dimensional scheme $X$\nat a point $p \\in X$ is defined to be the length of the local ring\n$\\mathcal{O}_{X,p}$. Unfortunately, we cannot work directly in the\nlocal ring in \\Mtwo. What we can do, however, is to compute the\nmultiplicity by computing the degree of the component of $X$ supported\nat $p$; see page 66 in Eisenbud and Harris~\\cite{SC:EH}.\n\n\\begin{problem*}\nWhat is the multiplicity of the origin as a zero of the polynomial\nequations $x^{5}+y^{3}+z^{3} = x^{3}+y^{5}+z^{3} = x^{3}+y^{3}+z^{5} =\n0$?\n\\end{problem*}\n\n\\begin{solution*}\nIf $I$ is the ideal generated by $x^{5}+y^{3}+z^{3}$,\n$x^{3}+y^{5}+z^{3}$ and $x^{3}+y^{3}+z^{5}$ in $\\bbbq[x,y,z]$, then\nthe multiplicity of the origin is\n\\[\n\\dim_{\\bbbq} \\frac{\\bbbq[x,y,z]_{\\langle x,y,z \\rangle}}\n{I \\bbbq[x,y,z]_{\\langle x,y,z \\rangle}} \\, .\n\\]\nIt follows that the multiplicity is the vector space dimension of the\nring $\\bbbq[x,y,z] / \\varphi^{-1}(I \\bbbq[x,y,z]_{\\langle x,y,z\n\\rangle})$ where $\\varphi \\colon \\bbbq[x,y,z] \\to\n\\bbbq[x,y,z]_{\\langle x,y,z \\rangle}$ is the natural map. Moreover,\nwe can express this using ideal quotients:\n\\[\n\\varphi^{-1}(I \\bbbq[x,y,z]_{\\langle x,y,z \\rangle}) \\,\\,= \\,\\,\n\\big(I : (I : \\langle x,y,z \\rangle^{\\infty})\\big) \\, .\n\\]\nCarrying out this calculation in \\Mtwo, we obtain:\n<<>>\n<<>>\n<<>>\n<<>>\nThus, we conclude that the multiplicity is $27$.\\qed\n\\end{solution*}\n\nThere are algorithms (not yet implemented in \\Mtwo) for working\ndirectly in the local ring $\\bbbq[x,y,z]_{\\langle x,y,z \\rangle}$. We\nrefer the interested reader to Chapter~4 in Cox, Little and\nO`Shea~\\cite{SC:CLO2}.\n\n\n%%----------------------------------------------------------\n\\section{Flat Families}\n\nNon-reduced schemes\\index{scheme!non-reduced} arise naturally as flat\nlimits\\index{flat limit} of a family of reduced\nschemes\\index{scheme!reduced}. Our next problem illustrates how a\nfamily of skew lines in $\\bbbp^{3}$ gives rise to a double line with\nan embedded point\\index{embedded point}.\n\n\\begin{problem*}[Exercise~III-68 in \\cite{SC:EH}]\nLet $L$ and $M $ be the lines in $\\bbbp^{3}_{k[t]}$ given by $x=y=0$\nand $x-tz = y+t^{2}w =0$ respectively. Show that the flat limit as $t\n\\to 0$ of the union $L \\cup M$ is the double line $x^{2} = y = 0$ with\nan embedded point of degree $1$ located at the point $(0:0:0:1)$.\n\\end{problem*}\n\n\\begin{solution*}\nWe first find the flat limit by saturating\\index{saturation} the\nintersection ideal and setting $t = 0$.\n<<>>\n<<>>\n<<>>\n<<>>\n<< 0})>>>\nSecondly, we verify that this is the union of a double line and an\nembedded point of degree $1$.\n<<>>\n<<>>\n<<>>\n\\qed\n\\end{solution*}\n\nSection~III.3.4 in Eisenbud and Harris~\\cite{SC:EH} contains several\nother interesting limits of various flat families.\n\n\n%%----------------------------------------------------------\n\\section{B\\'{e}zout's Theorem}\n\nB\\'{e}zout's Theorem\\index{Bezout's Theorem@B\\'ezout's Theorem} --- Theorem~III-78 in\nEisenbud and Harris~\\cite{SC:EH} --- may fail without the\nCohen-Macaulay\\index{Cohen-Macaulay} hypothesis. Our seventh problem\nis to demonstrate this.\n\n\\begin{problem*}[Exercise~III-81 in \\cite{SC:EH}]\nFind irreducible closed subvarieties $X$ and $Y$ in $\\bbbp^{4}$ such\nthat \n\\begin{align*}\n\\codim(X \\cap Y) &= \\codim(X) + \\codim(Y) \\\\\n\\deg(X \\cap Y) &> \\deg(X) \\cdot \\deg(Y) \\, .\n\\end{align*}\n\\end{problem*}\n\n\\begin{solution*}\nWe show that the assertion holds when $X$ is the cone over the\nnonsingular rational quartic curve\\index{rational quartic curve} in\n$\\bbbp^{3}$ and $Y$ is a two-plane passing through the vertex of the\ncone. First, recall that the rational quartic curve is given by the\n$2 \\times 2$ minors of the matrix $\\left[ \\begin{smallmatrix} a &\nb^{2} & bd & c \\\\ b & ac & c^2 & d \\end{smallmatrix} \\right]$; see\nExercise~18.8 in Eisenbud~\\cite{SC:E}. Thus, we have\n<<>>\n<<>>\n<<>>\n<<>>\n<<<(degree IX) * (degree IY)>>>\n<<>>\nwhich establishes the assertion.\\qed\n\\end{solution*}\n\nTo understand how this example works, it is enlightening to express\n$Y$ as the intersection of two hyperplanes; one given by $a = 0$ and\nthe other given by $d = 0$. Intersecting $X$ with the first\nhyperplane yields\n<<>>\nHowever, this first intersection has an embedded point;\n<<>>\n<<>>\nThe second hyperplane passes through this embedded\npoint\\index{embedded point} which explains the extra intersection.\n\n\n%%----------------------------------------------------------\n\\section{Constructing Blow-ups}\n\nThe blow-up\\index{blow-up} of a scheme $X$ along a subscheme $Y$ can\nbe constructed from the Rees algebra\\index{Rees algebra} associated to\nthe ideal sheaf of $Y$ in $X$; see Theorem~IV-22 in Eisenbud and\nHarris~\\cite{SC:EH}. Gr\\\"{o}bner basis techniques allow one to\nexpress the Rees algebra in terms of generators and relations. We\nillustrate this method in the next solution.\n\n\\begin{problem*}[Exercises~IV-43 \\& IV-44 in \\cite{SC:EH}]\nFind the blow-up $X$ of the affine plane\\index{scheme!affine}\n$\\mathbb{A}^{2} = \\Spec\\big( \\bbbq[x, y] \\big)$ along the subscheme\ndefined by $\\langle x^{3}, xy, y^{2} \\rangle$. Show that $X$ is\nnonsingular and its fiber over the origin is the union of two copies\nof $\\bbbp^{1}$ meeting at a point.\n\\end{problem*}\n\n\\begin{solution*}\nWe first provide a general function which returns the ideal of\nrelations for the Rees algebra.\n<< (\n r := numgens I;\n S := ring I;\n n := numgens S;\n K := coefficientRing S;\n tR := K[t, gens S, vars(0..r-1), \n MonomialOrder => Eliminate 1];\n f := map(tR, S, submatrix(vars tR, {1..n}));\n F := f(gens I);\n J := ideal apply(1..r, j -> (gens tR)_(n+j)-t*F_(0,(j-1)));\n L := ideal selectInSubring(1, gens gb J);\n R := K[gens S, vars(0..r-1)];\n g := map(R, tR, 0 | vars R);\n trim g(L));>>>\nNow, applying the function to our specific case yields: \n<<>>\n<<>>\n<<>>\nTherefore, the blow-up of the affine plane along the given subscheme\nis\n\\[\nX = \\Proj\\left( \\frac{(\\bbbq[x,y])[a,b,c]}{\\langle yb-xc, xb^{2}-ac,\nx^{2}b-ya, x^{3}c-y^{2}a \\rangle} \\right) \\, .\n\\]\nUsing \\Mtwo, we can also verify that the scheme $X$ is\nnonsingular\\index{singular locus};\n<<>>\n<<>>\nSince we have\n\\[\n\\frac{(\\bbbq[x,y])[a,b,c]}{\\langle yb-xc, xb^{2}-ac, x^{2}b-ya,\nx^{3}c-y^{2}a \\rangle} \\otimes \\frac{\\bbbq[x,y]}{\\langle x, y \\rangle}\n\\cong \\frac{\\bbbq[a,b,c]}{\\langle ac \\rangle} \\, ,\n\\]\nthe fiber over the origin $\\langle x,y \\rangle$ in $\\mathbb{A}^{2}$ is\nclearly a union of two copies of $\\bbbp^{1}$ meeting at one point. In\nparticular, the exceptional fiber is not a projective space.\\qed\n\\end{solution*}\n\nMany other interesting blow-ups can be found in section~II.2 in\nEisenbud and Harris~\\cite{SC:EH}.\n\n\n%%----------------------------------------------------------\n\\section{A Classic Blow-up}\n\nWe consider the blow-up\\index{blow-up} of the projective plane\n$\\bbbp^{2}$ at a point.\n\n\\vbox{\n\\begin{problem*}\nShow that the following varieties are isomorphic.\n\\begin{enumerate}\n\\item[$(a)$] the image of the rational map from $\\bbbp^{2}$ to\n$\\bbbp^{4}$ given by\n\\[\n(r:s:t) \\mapsto (r^{2}:s^{2}:rs:rt:st) \\, ;\n\\]\n\\item[$(b)$] the blow-up of the plane $\\bbbp^{2}$ at the point\n$(0:0:1)$;\n\\item[$(c)$] the determinantal variety\\index{determinantal variety}\ndefined by the $2 \\times 2$ minors of the matrix $\\left[\n\\begin{smallmatrix} a & c & d \\\\ b & d & e \\end{smallmatrix} \\right]$\nwhere $\\bbbp^{4} = \\Proj\\big( k[a,b,c,d,e] \\big)$.\n\\end{enumerate}\nThis surface is called the {\\em cubic scroll}\\index{cubic scroll} in\n$\\bbbp^{4}$.\n\\end{problem*}\n}\n\n\\begin{solution*}\nWe find the ideal in part~$(a)$ by elimination\ntheory\\index{elimination theory}.\n<<>>\n<< Eliminate 3];>>>\n<<>>\n<<>>\n<<>>\nNext, we determine the surface in part~$(b)$. We construct the ideal\ndefining the blow-up of $\\bbbp^{2}$ \n<< Eliminate 1];>>>\n<<>>\nand embed it in $\\bbbp^{2} \\times \\bbbp^{1}$.\n<<>>\n<<>>\n<<>>\nWe then map this surface into $\\bbbp^{5}$ using the Segre\nembedding\\index{Segre embedding}.\n<<>>\n<<>>\n<<>>\nNote that the image under the Segre map lies on a hyperplane in\n$\\bbbp^{5}$. To get the desired surface in $\\bbbp^{4}$, we project\n<<>>\n<<>>\nFinally, we compute the surface in part~$(c)$.\n<<>>\n<<>>\n<<>>\nBy incorporating a permutation of the variables into definition of\n{\\tt surfaceC}, we obtain the desired isomorphisms\n<<>>\n<<>>\n<<>>\nwhich completes the solution.\\qed\n\\end{solution*}\n\nFor more information of the geometry of rational normal scrolls, see\nLecture~8 in Harris~\\cite{SC:H}.\n\n\n%%----------------------------------------------------------\n\\section{Fano Schemes}\n\nOur final example concerns the family of Fano schemes\\index{Fano\nscheme} associated to a flat family of quadrics.\nRecall that the $k$-th Fano scheme $F_{k}(X)$ of a\nscheme $X \\subseteq \\bbbp^{n}$ is the subscheme of\nthe Grassmannian parametrizing $k$-planes\ncontained in $X$.\n\n\\begin{problem*}[Exercise~IV-69 in \\cite{SC:EH}]\nConsider the one-parameter family\\index{one-parameter family} of\nquadrics tending to a double plane with equation\n\\[ \nQ = V(tx^{2}+ty^{2}+tz^{2}+w^{2}) \\subseteq \\bbbp^{3}_{\\bbbq[t]} =\n\\Proj\\big(\\bbbq[t][x,y,z,w]\\big) \\enspace .\n\\]\nWhat is the flat limit\\index{flat limit} of the Fano schemes\n$F_{1}(Q_{t})$?\n\\end{problem*}\n\n\\begin{solution*}\nWe first compute the ideal defining $F_{1}(Q_{t})$, the scheme\nparametrizing lines in $Q$.\n<<>>\n<<>>\nTo parametrize a line in our projective space, we introduce\nindeterminates $u, v$ and $A, \\dotsc, H$.\n<<>>\nWe then make a map {\\tt phi} from {\\tt PP3} to {\\tt R} sending the\nvariables to the coordinates of the general point on a line.\n<<>>\n<<>>\nFor a line to belong to $Q$, the {\\tt imageFamily} must vanish\nidentically. In other words, $F_{1}(Q)$ is defined by the\ncoefficients of the generators of {\\tt imageFamily}.\n%% removing a final use of 'coefficients'\n%% coeffOfFamily = (coefficients ({1,2}, gens imageFamily))_1;\n<<>>\nSince we don't need the variables $u$ and $v$, we get rid of them.\n<<>>\n<<>>\n<<>>\nNext, we move to the Grassmannian\\index{Grassmannian} $\\mathbb{G}(1,3)\n\\subset \\bbbp^{5}$. Recall the homogeneous coordinates on\n$\\bbbp^{5}$ correspond to the $2 \\times 2$ minors of a $2 \\times 4$\nmatrix. We obtain these minors using the {\\tt exteriorPower} function\nin \\Mtwo.\n<<>>\n<<>>\n<<>>\nNow, to answer the question, we determine the limit as $t$ tends to $0$.\n<<0})>>>\nLet's transpose the matrix of generators so all of its elements are visible\non the printed page.\n<<>>\nWe see that $F_{1}(Q_{0})$ is supported on the plane conic $\\langle d,\ne, f, a^{2}+b^{2}+c^{2} \\rangle$. However, $F_{1}(Q_{0})$ is not\nreduced\\index{scheme!non-reduced}; it has\nmultiplicity\\index{multiplicity} two. On the other hand, the generic\nfiber is\n<< 1})>>>\n<<>>\nHence, for $t \\neq 0$, $F_{1}(Q_{t})$ is the union of two conics lying\nin complementary planes and $F_{1}(Q_{0})$ is the double conic\nobtained when the two conics move together.\\qed\n\\end{solution*}\n\n% Local Variables:\n% mode: latex\n% mode: reftex\n% tex-main-file: \"chapter-wrapper.tex\"\n% reftex-keep-temporary-buffers: t\n% reftex-use-external-file-finders: t\n% reftex-external-file-finders: ((\"tex\" . \"make FILE=%f find-tex\") (\"bib\" . \"make FILE=%f find-bib\"))\n% End:\n", "meta": {"hexsha": "fb590013cbdb2d834a3a89d7b5775d40072e164a", "size": 26542, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/ComputationsBook/chapters/schemes/chapter.tex", "max_stars_repo_name": "d-torrance/Macaulay2-web-site", "max_stars_repo_head_hexsha": "edb1d0b607c5aa00ffbcf403f2403961c6d6083a", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-11-27T08:01:17.000Z", "max_stars_repo_stars_event_max_datetime": "2018-11-27T08:01:17.000Z", "max_issues_repo_path": "Book/ComputationsBook/chapters/schemes/chapter.tex", "max_issues_repo_name": "d-torrance/Macaulay2-web-site", "max_issues_repo_head_hexsha": "edb1d0b607c5aa00ffbcf403f2403961c6d6083a", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 19, "max_issues_repo_issues_event_min_datetime": "2018-04-17T19:52:43.000Z", "max_issues_repo_issues_event_max_datetime": "2021-02-07T01:08:10.000Z", "max_forks_repo_path": "Book/ComputationsBook/chapters/schemes/chapter.tex", "max_forks_repo_name": "d-torrance/Macaulay2-web-site", "max_forks_repo_head_hexsha": "edb1d0b607c5aa00ffbcf403f2403961c6d6083a", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2018-01-08T16:48:01.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-10T21:19:02.000Z", "avg_line_length": 41.2783825816, "max_line_length": 101, "alphanum_fraction": 0.6474267199, "num_tokens": 8844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789040926008, "lm_q2_score": 0.8006919949619792, "lm_q1q2_score": 0.6489439705925032}} {"text": "\\section{Linear transformations defined on a basis}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Check whether two linear transformations are equal by\n considering their action on a spanning set.\n \\item Specify a linear transformation by considering its action on a\n basis.\n \\end{enumerate}\n\\end{outcome}\n\nRecall that, by definition, two linear transformations $S,T:V\\to W$\nare equal if and only if for all $\\vect{v}\\in V$, we have\n$S(\\vect{v}) = T(\\vect{v})$. However, this is not a very practical\nway of checking whether $S=T$, as it theoretically requires checking\n$S(\\vect{v}) = T(\\vect{v})$ for each one of infinitely many vectors\n$\\vect{v}$. The following proposition states that it is sufficient to\ncheck the actions of $S$ and $T$ on a spanning set of vectors.\n\n\\begin{proposition}{Equality of linear transformations}{transformation-spanning-set}\n Let $V$ and $W$ be vector spaces over a field $K$, and let\n $S,T:V\\to W$ be linear transformations. Moreover, let $X\\subseteq V$\n be a spanning set of $V$, i.e., such that $V=\\sspan X$.\n If $S(\\vect{v}) = T(\\vect{v})$ for all $\\vect{v}\\in X$, then $S=T$.%\n \\index{linear transformation!equality of}%\n \\index{equality!of linear transformations}\n\\end{proposition}\n\n\\begin{proof}\n Assume that $S(\\vect{v}) = T(\\vect{v})$ holds for all $\\vect{v}\\in X$. To\n show that $S=T$, let $\\vect{u}\\in V$ be an arbitrary vector. Since\n $X$ is a spanning set, we can write\n $\\vect{u}=a_1\\vect{v}_1+\\ldots+a_n\\vect{v}_n$, for some\n $\\vect{v}_1,\\ldots,\\vect{v}_n\\in X$ and $a_1,\\ldots,a_n\\in K$.\n By assumption, $S(\\vect{v}_i) = T(\\vect{v}_i)$ for all $i$, because\n $\\vect{v}_i\\in X$. Then we have\n \\begin{eqnarray*}\n S(\\vect{u})\n &=& S(a_1\\vect{v}_1+\\ldots+a_n\\vect{v}_n) \\\\\n &=& a_1S(\\vect{v}_1)+\\ldots+a_nS(\\vect{v}_n) \\\\\n &=& a_1T(\\vect{v}_1)+\\ldots+a_nT(\\vect{v}_n) \\\\\n &=& T(a_1\\vect{v}_1+\\ldots+a_n\\vect{v}_n) \\\\\n &=& T(\\vect{u}).\n \\end{eqnarray*}\n Since $\\vect{u}\\in V$ was arbitrary, it follows that $S=T$.\n\\end{proof}\n\nTherefore, if we know how a linear transformation acts on a spanning\nset (and in particular, on a basis), then we know how it acts on the\nentire space. There is also a kind of converse to this: given a basis\nof $V$, we can map the basis vectors to any elements of $W$ we like,\nand this will always determine a unique linear transformations. This\nis the content of the following theorem.\n\n\\begin{theorem}{Linear transformation defined on a basis}{transformation-basis}\n Suppose $V$ and $W$ are vector spaces over a field $K$. Suppose\n $\\vect{v}_1,\\vect{v}_2,\\ldots,\\vect{v}_n$ is a basis of $V$, and\n $\\vect{w}_1,\\vect{w}_2,\\ldots,\\vect{w}_n$ are any vectors in $W$\n (which may or may not be distinct). Then there exists a unique\n linear transformation%\n \\index{linear transformation!defined on a basis}\n $T:V\\to W$ such that\n \\begin{equation*}\n T(\\vect{v}_1) = \\vect{w}_1,\n \\quad\n T(\\vect{v}_2) = \\vect{w}_2,\n \\quad\\ldots\\quad\n T(\\vect{v}_n) = \\vect{w}_n.\n \\end{equation*}\n\\end{theorem}\n\n\\begin{proof}\n To show that such a linear transformation $T$ exists, we first\n define a function $T:V\\to W$ as follows. Given any $\\vect{v}\\in V$,\n there exists a unique set of coordinates $a_1,\\ldots,a_n\\in K$ such\n that\n \\begin{equation*}\n \\vect{v} = a_1\\vect{v}_1 + \\ldots + a_n\\vect{v}_n.\n \\end{equation*}\n Then define\n \\begin{equation*}\n T(\\vect{v}) = a_1\\vect{w}_1 + \\ldots + a_n\\vect{w}_n.\n \\end{equation*}\n This defines a function $T:V\\to W$. Next, we must check that $T$ is\n linear. To show that $T$ preserves addition, consider\n $\\vect{v},\\vect{v}'\\in V$, with\n $\\vect{v} = a_1\\vect{v}_1 + \\ldots + a_n\\vect{v}_n$ and\n $\\vect{v}' = b_1\\vect{v}_1 + \\ldots + b_n\\vect{v}_n$. Then\n \\begin{eqnarray*}\n T(\\vect{v}+\\vect{v}')\n &=& T((a_1+b_1)\\vect{v}_1 + \\ldots + (a_n+b_n)\\vect{v}_n) \\\\\n &=& (a_1+b_1)\\vect{w}_1 + \\ldots + (a_n+b_n)\\vect{w}_n \\\\\n &=& (a_1\\vect{w}_1 + \\ldots + a_n\\vect{w}_n)\n + (b_1\\vect{w}_1 + \\ldots + b_n\\vect{w}_n) \\\\\n &=& T(\\vect{v}) + T(\\vect{v}').\n \\end{eqnarray*}\n Therefore, $T$ preserves addition. To show that $T$ preserves scalar\n multiplication, consider $\\vect{v} = a_1\\vect{v}_1 + \\ldots +\n a_n\\vect{v}_n$ and $k\\in K$. Then\n \\begin{eqnarray*}\n T(k\\vect{v})\n &=& T(ka_1\\vect{v}_1 + \\ldots + ka_n\\vect{v}_n) \\\\\n &=& ka_1\\vect{w}_1 + \\ldots + ka_n\\vect{w}_n \\\\\n &=& k(a_1\\vect{w}_1 + \\ldots + a_n\\vect{w}_n) \\\\\n &=& kT(\\vect{v}).\n \\end{eqnarray*}\n Therefore, $T$ preserves scalar multiplication. It follows that $T$\n is linear. Next, we must show that $T$ satisfies the condition of\n the theorem, i.e., that $T(\\vect{v}_i) = \\vect{w}_i$ for each $i$.\n But this is clearly the case, because in this case, $a_i=1$ and\n $a_j=0$ for all $j\\neq i$. We have shown that there exists a linear\n function $T$ satisfying all of the conditions required by the\n theorem.\n\n Finally, the only thing left to show is uniqueness. But this follows\n from Proposition~\\ref{prop:transformation-spanning-set}. Namely, if\n $T'$ is another linear transformation such that\n $T'(\\vect{v}_i) = \\vect{w}_i$ for all $i$, then $T$ and $T'$ agree\n on $\\vect{v}_1,\\ldots,\\vect{v}_n$, which is a basis and hence a\n spanning set. By Proposition~\\ref{prop:transformation-spanning-set},\n $T=T'$.\n\\end{proof}\n\n\\begin{example}{Linear transformation defined on a basis}{transformation-basis}\n Recall that $\\set{x^2, (x+1)^2, (x+2)^2}$ is a basis of $\\Poly_2$.\n Consider the linear function $T:\\Poly_2\\to\\Mat_{22}$ defined by\n \\begin{equation*}\n T(x^2) = \\begin{mymatrix}{rr} 1 & 1 \\\\ 0 & 0 \\end{mymatrix},\\quad\n T((x+1)^2) = \\begin{mymatrix}{rr} 0 & 1 \\\\ 0 & 1 \\end{mymatrix},\\quad\n T((x+2)^2) = \\begin{mymatrix}{rr} 0 & 0 \\\\ 1 & 1 \\end{mymatrix}.\n \\end{equation*}\n Find $T(4x)$.\n\\end{example}\n\n\\begin{solution}\n Let $\\vect{v}_1=x^2$, $\\vect{v}_2=(x+1)^2$, $\\vect{v}_3=(x+2)^2$,\n $\\vect{w}_1 = \\begin{mysmallmatrix}{rr} 1 & 1 \\\\ 0 & 0 \\end{mysmallmatrix}$,\n $\\vect{w}_2 = \\begin{mysmallmatrix}{rr} 0 & 1 \\\\ 0 & 1 \\end{mysmallmatrix}$,\n and\n $\\vect{w}_3 = \\begin{mysmallmatrix}{rr} 0 & 0 \\\\ 1 & 1 \\end{mysmallmatrix}$.\n We must first find $a,b,c$ such that\n $4x=a\\vect{v}_1+b\\vect{v}_2+c\\vect{v}_3$. We do this by solving a\n system of equations, using the same method as in\n Example~\\ref{exa:linear-combination-polynomials}. We find that\n $a=-3$, $b=4$, and $c=-1$. Therefore\n \\begin{equation*}\n T(4x)\n ~=~ T(-3\\vect{v}_1 + 4\\vect{v}_2 - \\vect{v}_3)\n ~=~ -3\\vect{w}_1 + 4\\vect{w}_2 - \\vect{w}_3\n ~=~ \\begin{mymatrix}{rr}\n -3 & 1 \\\\\n -1 & 3 \\\\\n \\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\n", "meta": {"hexsha": "3ae55fc0453dd40b2e63fd057cb2979e8db6b821", "size": 6641, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "baseText/content/LinearTransformationsGeneral-Basis.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "baseText/content/LinearTransformationsGeneral-Basis.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "baseText/content/LinearTransformationsGeneral-Basis.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 42.2993630573, "max_line_length": 84, "alphanum_fraction": 0.6419213974, "num_tokens": 2507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.849971175657575, "lm_q1q2_score": 0.6489391921037119}} {"text": "\\section{The Structure Theorem}\r\nUntil further notice, we take our ring to be a ED and we denote its Euclidean function by $\\phi:R^\\times\\to\\mathbb Z_{\\ge 0}$.\r\nlet $A$ be an $m\\times n$ matrix with entries in $R$.\r\n\\begin{definition}\r\n The elementary row operations are as follows:\\\\\r\n 1. Add $\\lambda\\in R$ times the $j^{th}$ row to the $i^{th}$ row for $i\\neq j$.\\\\\r\n 2. Swap the $i^{th}$ and $j^{th}$ row.\\\\\r\n 3. Multiply the $i^{th}$ row by a unit $u$.\r\n\\end{definition}\r\nNote that all these operations are reversible.\r\nAlso, each of the operations may be realized by multiplying in the left by an $m\\times m$ invertible matrix.\r\nTo wit, the first operation is the left multiplication of the matrix $I+\\lambda E_{ij}$ where $E_{ij}$ is the matrix with $1$ on the $(i,j)$ entry and $0$ otherwise.\r\nThe second is $I+E_{ij}+E_{ji}-E_{ii}-E_{jj}$ and the third is $I+(u-1)E_{ii}$.\r\nWe can similarly define the column operations, and the realization becomes the matrix multiplication on the right by an $n\\times n$ invertible matrix analogous to before.\r\n\\begin{definition}\r\n Two $m\\times n$ matrices $A,B$ are equivalent if there is a sequence of elementary row and column operations taking $A$ to $B$.\r\n\\end{definition}\r\nIt is obvious that this is an equivalence relation due to reversibility.\r\nSo if $A,B$ are equivalent, there exists invertible square matrices $P,Q$ with $B=QAP$.\r\n\\begin{theorem}[Smith Normal Form]\r\n An $m\\times n$ matrix $A=(a_{ij})$ with entries in $R$ is equivalent to a diagonal matrix of the form\r\n $$\\begin{pmatrix}\r\n d_1&&&&&&\\\\\r\n &d_2&&&&&\\\\\r\n &&\\ddots&&&&\\\\\r\n &&&d_t&&&\\\\\r\n &&&&0&&\\\\\r\n &&&&&\\ddots&\\\\\r\n &&&&&&0\r\n \\end{pmatrix}$$\r\n where $d_i\\neq 0$ for all $i$ and $d_1|d_2|\\cdots |d_t$.\r\n\\end{theorem}\r\nThe $d_i$'s are called covariant factors and are unique up to associates (show later).\r\n\\begin{proof}\r\n If $A=0$ then we are done.\r\n So assume that $a_{11}\\neq 0$.\r\n Suppose $a_{11}\\nmid a_{1j}$ for some $j\\ge 2$, then we use the Euclidean algorithm to get $a_{1j}=qa_{11}+r$ with $q,r\\in R$ and $\\phi(r)<\\phi(a_{11})$.\r\n So we substract $q$ times the first column from the $j^{th}$ column and swap them.\r\n This makes the top left entry $r$.\r\n Likewise, we can do the same thing if $a_{11}\\nmid a_{i1}$ for some $i\\ge 2$ by row operations.\r\n The cases above each decrease the top left entry (in terms of its value in the Euclidean function), thus it must eventually stops, at which time $a_{11}$ divides everything in the first row or the first column.\r\n Then substracting multiples of the first row or first column can clear up every other entry in the first column or first row except $a_{11}$.\r\n Now if $a_{11}$ does not divide $a_{ij}$ for $i,j\\ge 2$, then add the $i^{th}$ row to the first row, and so we can do the same column operations to decrease $\\phi(a_{11})$ further till it terminates.\r\n Do this over and over again (but finitely many times) we can get $a_{11}|a_{ij}$ for any $i,j\\ge 2$ and $a_{ij}=0$ whenever $i\\neg j$ and one of $i,j$ is $1$.\r\n We rename $a_{11}$ as $d_1$ and repeat the process on the smaller matrix by removing the first row and first column to give the existence.\r\n\\end{proof}\r\n\\begin{definition}\r\n A $k\\times k$ minor of a matrix $A$ is the determinant of a $k\\times k$ submatrix of $A$.\r\n\\end{definition}\r\n\\begin{definition}\r\n For a matrix $A$ over a ring $R$, we define the $k^{th}$ Fitting ideal $\\operatorname{Fit}_k(A)$ be the ideal generated by all the $k\\times k$ minors of $A$.\r\n\\end{definition}\r\n\\begin{lemma}\r\n If $A,B$ are equivalent matrices, then $\\operatorname{Fit}_k(A)=\\operatorname{Fit}_k(B)$ for any $k$ (such that the Fitting ideal makes sense).\r\n\\end{lemma}\r\n\\begin{proof}\r\n Suffices to show that the elementary operations do not change the Fitting ideal.\r\n The second and third row operations are trivial.\r\n For the first kind of row operation, suppose we add $\\lambda$ times the second row to the first row so that $A=(a_{ij})$ becomes\r\n $$A'=\\begin{pmatrix}\r\n a_11+\\lambda a_{21}&a_{12}+\\lambda a_{22}&\\dots&a_{1n}+\\lambda a_{2n}\\\\\r\n a_{21}&a_{22}&\\dots&a_{2n}\\\\\r\n \\vdots&\\vdots&\\ddots&\\vdots\\\\\r\n \\end{pmatrix}$$\r\n Let $C$ be a $k\\times k$ submatrix of $A$ and $C'$ be its correspondent submatrix in $A'$.\r\n If $C$ does not intersect the first row, then $\\det C=\\det C'$.\r\n If $C$ intersect both the first and second row, we also have $\\det C=\\det C'$ since its simply a row operation on the $k\\times k$ submatrix.\r\n If $C$ intersects the first but not the second row, then by expanding along the first row, the determinant of $\\det C'$ would be $\\det C+\\lambda\\det D$ for some other $k\\times k$ submatrix of $A$.\r\n So $\\det C'\\in \\operatorname{Fit}_k(A)$, so $\\operatorname{Fit}_k(A')\\subset \\operatorname{Fit}_k(A)$.\r\n Since row operations are reversible, we also have the reverse inclusion.\r\n\\end{proof}\r\n\\begin{proposition}\r\n The covarient factors are unique up to associates.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Look at the Fitting ideals.\r\n\\end{proof}\r\n\\begin{example}\r\n Consider the following matrix (here $\\to$ represents row/column operations)\r\n $$A=\\begin{pmatrix}\r\n 2&-1\\\\\r\n 1&2\r\n \\end{pmatrix}\\to\\begin{pmatrix}\r\n 1&-1\\\\\r\n 3&2\r\n \\end{pmatrix}\\to\\begin{pmatrix}\r\n 1&0\\\\\r\n 3&5\r\n \\end{pmatrix}\\to\\begin{pmatrix}\r\n 1&0\\\\\r\n 0&5\r\n \\end{pmatrix}$$\r\n One can also get its Smith normal form by considering the minors.\r\n Indeed, $\\operatorname{Fit}_1(A)=(1)$, so $d_1=\\pm 1$.\r\n Also $\\operatorname{Fit}_2(A)=(5)$, hence $d_2=\\pm 5$.\r\n So we obtain the Smith normal form.\r\n\\end{example}\r\n\\begin{theorem}\r\n Let $R$ be an Euclidean domain and $N$ is an $R$-submodule of $R^n$, then there is a free basis $x_1,\\ldots,x_m$ of $R^m$ such that $N$ is generated as an $R$-module by $d_1x_1,\\ldots,d_tx_t$ for some $t\\le m$ and $d_1|d_2|\\cdots|d_t$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n $R$ is a ED hence a PID, hence $N$ is generated by some $y_1,\\ldots,y_n$ for some $n\\le m$.\r\n Now each $y_i$ is in $R^n$, so we can form an $n\\times n$ matrix $A$ whose columns are the $y_i$'s.\r\n By the preceding theorem, $A$ is equivalent to $A'=\\operatorname{diag}(d_1,\\ldots,d_t,0,\\ldots,0)$ with $t\\le n$ and $d_1|d_2|\\ldots|d_t$.\r\n Now $A'$ is obtained from $A$ by elementary row and column operations.\r\n Each row operation corresponds to changing of our choice of free basis for $R^n$, and every column operation changes the generating set for $R^n$.\r\n So after changing the free basis for $R^n$ to, say, $x_1,\\ldots, x_n$, then $N$ is generated by $d_1x_1,\\ldots,d_tx_t$.\r\n\\end{proof}\r\n\\begin{theorem}[Structure Theorem]\r\n Let $R$ be a Euclidean Domain and $M$ a finitely-generated $R$-module, then $M\\cong R/(d_1)\\oplus\\cdots\\oplus R/(d_t)\\oplus R\\oplus\\cdots\\oplus R$ for some $d_i\\in R$ and $d_1|d_2|\\cdots|d_t$.\r\n\\end{theorem}\r\nThese $d_i$'s are called invariant factors.\r\n\\begin{proof}\r\n Since $M$ is finitely generated, then we can find a surjective $R$-module map $\\phi:R^m\\to M$ for some $m$.\r\n Then $M\\cong R^m/\\ker\\phi$, but by the preceding theorem, there exists a free basis $x_1,\\ldots,x_n$ for $R^m$ such that $N=\\ker\\phi=Rd_1x_1+\\cdots Rd_tx_t$ for $d_1|d_2|\\cdots|d_t$, hence\r\n $$M\\cong\\frac{R\\oplus R\\oplus\\cdots\\oplus R\\oplus R\\oplus\\cdots\\oplus R}{Rd_1\\oplus Rd_2\\oplus\\cdots\\oplus Rd_t\\oplus 0\\oplus\\cdots\\oplus 0}\\cong R/(d_1)\\oplus\\cdots\\oplus R/(d_t)\\oplus R\\oplus\\cdots\\oplus R$$\r\n which is what we wanted.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $M$ be an $R$-module.\r\n An element $m\\in M$ is called torsion if $\\exists r\\in R\\setminus\\{0\\},rm=0$.\\\\\r\n $M$ is called a torsion module if every $m\\in M$ is torsion.\r\n $M$ is called torsion-free if the only torsion is $0$.\r\n\\end{definition}\r\n\\begin{corollary}\r\n Let $R$ be an Euclidean domain, then any finitely generated torsion-free $R$-module is free.\r\n\\end{corollary}\r\n\\begin{proof}\r\n By the preceding theorem $M\\cong R/(d_1)\\oplus\\cdots\\oplus R/(d_t)\\oplus R\\oplus\\cdots\\oplus R$, but as $M$ is torsion-free, it cannot contain any of $R/(d_i)$ as $R$-submodule as they would contain torsions.\r\n Hence $M\\cong R\\oplus\\cdots\\oplus R$, therefore $M$ is free.\r\n\\end{proof}\r\n\\begin{remark}\r\n The structure theorem in fact holds whenever $R$ is a PID.\r\n Also, there is also a uniqueness statement in the structure theorem: Suppose no $d_i$'s is a unit (otherwise they only contribute $0$ factors to the product), then the module $M$ uniquely determines $d_1,\\ldots,d_t$.\r\n\\end{remark}\r\n\\begin{example}\r\n Consider an abelian groupn $G$ generated by $a,b$ subject to relations $2a+b=-a+2b=0$.\r\n So $G\\cong\\mathbb Z^2/N$ where $N$ is generated by $(2,1),(-1,2)$, so we take (as in the proof of the structure theorem)\r\n $$A=\\begin{pmatrix}\r\n 2&-1\\\\\r\n 1&2\r\n \\end{pmatrix}\\to\\begin{pmatrix}\r\n 1&0\\\\\r\n 0&5\r\n \\end{pmatrix}$$\r\n as seen before.\r\n So we can change basis for $\\mathbb Z^2$ to generate $N$ by $(1,0),(0,5)$, hence $G\\cong\\mathbb Z\\oplus\\mathbb Z/(\\mathbb Z\\oplus 5\\mathbb Z)\\cong\\mathbb Z/5\\mathbb Z$.\r\n\\end{example}\r\nMore generally, for finitely generated abelian groups, we have the following:\r\n\\begin{theorem}\r\n Any finitely generated abelian group $G$ is isomorphic to\r\n $$\\mathbb Z/d_1\\mathbb Z\\oplus\\cdots\\oplus\\mathbb Z/d_t\\mathbb Z\\oplus\\mathbb Z^r$$\r\n where $r\\ge 0$ and $d_1|d_2|\\cdots|d_t$.\r\n\\end{theorem}\r\nThe $r$ here stands for the rank of the group.\r\n\\begin{proof}\r\n Take $R=\\mathbb Z$ in the structure theorem.\r\n\\end{proof}\r\nIn the special case that $G$ is finite, we immediately obtain Theorem \\ref{fin_abe_struct}.\r\n\\begin{remark}\r\n Let $A,B$ be square matrices over $R$, then $\\det (AB)=\\det A\\det B$, also $\\operatorname{adj}(A)A=A\\operatorname{adj}(A)=\\det(A)I$.\r\n In particular, $A$ is invertible iff $\\det A$ is a unit.\r\n\\end{remark}\r\n\\begin{theorem}[Cayley-Hamilton]\r\n Let $A=(a_{ij})$ be a $n\\times n$ matrix over a field $F$.\r\n Let $\\chi_A(X)=\\det (XI-A)\\in F[X]$, then $\\chi_A(A)=0$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Consider $V=F^n$ as a $F[X]$-module. with $X$ acting as $A$, i.e. $f(X)\\cdot v=f(A)v$.\r\n Let $e_1,\\ldots,e_n$ be the standard basis for $F^n$, so for any $j$, $X\\cdot e_j=\\sum_{i=1}^na_{ij}e_i$, hence\r\n $$(XI-A)\\begin{pmatrix}\r\n e_1\\\\\r\n \\vdots\\\\\r\n e_n\r\n \\end{pmatrix}=0$$\r\n By multiplying both sides by the adjugate of $XI-A$, we know that for any $i$,\r\n $$\\chi_A(X)\\cdot e_i=\\det(XI-A)\\cdot e_i=0\\implies\\chi_A(A)e_i=0$$\r\n But this can only happen when $\\chi_A(A)=0$.\r\n\\end{proof}\r\nRecall that we have mentioned Theorem \\ref{classify_fin_abe}, which is yet another way of classifying the finite abelian groups by writing it as a product of cyclic $p$-groups.\r\nIndeed, we can achieve this by generalising our structure theorem one step further.\r\n\\begin{lemma}\r\n Let $R$ be a PID and $a,b\\in R$ has $\\gcd(a,b)=1$ (up to associates), then there is an isomorphism of $R$-modules\r\n $$R/(ab)\\cong R/(a)\\oplus R/(b)$$\r\n\\end{lemma}\r\n\\begin{proof}\r\n Since $R$ is a PID, $(a,b)=(d)$ for some $d\\in R$, so $d=\\gcd(a,b)$ by certain questions in example sheet.\r\n Hence $(a,b)=R$ since $d$ must be a unit by hypothesis, so there is $r,s\\in R$ with $ra+sb=1$.\r\n Define an $R$-module homomorphism $\\phi:R\\to R/(a)\\oplus R/(b)$ by $x\\mapsto (x+(a),x+(b))$.\r\n To see it is surjective, $\\phi(sb)=(1+(a),0+(b)),\\phi(ra)=(0+(a),1+(b))$, hence $\\phi(sbx+ray)=(x+(a),y+(b))$.\r\n Now if $\\phi(x)=(0+(a),0+(b))$, then $x\\in (a)\\cap(b)$, so $x=x(ra+sb)=rax+sxb\\in (ab)$.\r\n It is obvious that anything in $(ab)$ is mapped to zero, hence $\\ker\\phi=(ab)$, so we deduce the theorem from isomorphism theorem.\r\n\\end{proof}\r\nThis reduced to Chinese Remainder Theorem when we set $R=\\mathbb Z$.\r\n\\begin{theorem}[Prime Decomposition Theorem]\\label{prime_decomp}\r\n Let $R$ be a ED and let $M$ be a finitely generated $R$-module.\r\n Then\r\n $$M\\cong R/(p_1^{n_1})\\oplus\\cdots\\oplus R/(p_k^{n_k})\\oplus R^m$$\r\n where $p_1,\\ldots,p_k$ are prime in $R$.\r\n\\end{theorem}\r\nNote that $p_1,\\ldots,p_k$ need not to be distinct.\r\n\\begin{proof}\r\n By the structure theorem, we have\r\n $$M\\cong R/(d_1)\\oplus\\cdots\\oplus R/(d_t)\\oplus R^m$$\r\n So it suffices to write each $R/(d_i)$ in the desired form.\r\n Choose $i$ and write $d_i=up_1^{\\alpha_1}\\cdots p_r^{\\alpha_r}$ where $p_i$ are pairwise non-associates and $u$ is a unit.\r\n So by the preceding lemma, we have\r\n $$R/(d_i)\\cong R/(p_1^{\\alpha_1})\\oplus\\cdots\\oplus R/(p_r^{\\alpha_r})$$\r\n which establishes the theorem.\r\n\\end{proof}\r\nNote that Theorem \\ref{classify_fin_abe} is a direct consequence of this.\\\\\r\nLet $V$ be a vector space over a field $F$ and let $\\alpha:V\\to V$ be an endomorphism, then we can make $V$ an $F[X]$-module (written as $V_\\alpha$) by $f(X)\\cdot v=f(\\alpha)v$.\r\n\\begin{lemma}\r\n If $V$ is finite-dimensional, then $V_\\alpha$ is finitely generated as an $F[X]$-module.\r\n\\end{lemma}\r\n\\begin{proof}\r\n If $v_1,\\ldots,v_n$ generates $V$ as a vector space, then they also generate $V_\\alpha$ as an $F[X]$-module since $F\\le F[X]$.\r\n\\end{proof}\r\nThe lemma itself is trivial, but the thing to take from here is that $V_\\alpha$, being also an $F$-vector space, is isomorphic to $V$.\r\nThis is very useful if we want to analyze the behaviour of $\\alpha$ in $V$:\r\nIf we know that $V_\\alpha$ is isomorphic to some $F[X]$-modules (via e.g. Theorem \\ref{prime_decomp}) that are easier to study, then they are also automatically isomorphic as $F$-vector spaces.\r\nSo by choosing a nice basis for this $F[X]$-module (as a vector space), we can obtain a nice matrix of $\\alpha$.\r\n\\begin{example}\r\n 1. Suppose $V_\\alpha\\cong F[X]/(X^n)$ as $F[X]$-module, then we can choose the basis $1,X,\\ldots,X^{n-1}$ for it to be an $F$-vector space, so $\\alpha$ would have matrix\r\n $$(\\star)=\\begin{pmatrix}\r\n 0&&&\\\\\r\n 1&0&&\\\\\r\n &\\ddots&\\ddots&\\\\\r\n &&1&0\r\n \\end{pmatrix}$$\r\n Since $\\alpha$ acts as multiplication by $X$.\\\\\r\n 2. Suppose $V_\\alpha\\cong F[X]/(X-\\lambda)^n$ as an $F[X]$-module, then wrt the basis $1,X-\\lambda,\\ldots,(X-\\lambda)^{n-1}$, $\\alpha-\\lambda\\operatorname{id}$ has matrix $(\\star)$, so $\\alpha$ exists as a Jordan block.\\\\\r\n 3. Suppose $V_\\alpha\\cong F[X]/(f)$ where $f\\in F[X]$ is in the form $f(X)=a_0+a_1X+\\cdots +a_{n-1}X^{n-1}+X^n$, then with respect to $1,X,\\ldots,X^{n-1}$, $\\alpha$ has the matrix\r\n $$C(f)=\\begin{pmatrix}\r\n 0&&&-a_0\\\\\r\n 1&\\ddots&&-a_1\\\\\r\n &\\ddots&0&\\vdots\\\\\r\n &&1&-a_{n-1}\r\n \\end{pmatrix}$$\r\n which is called the companion matrix of $f$.\r\n\\end{example}\r\n\\begin{theorem}[Rational Canonical Form]\r\n Let $\\alpha:V\\to V$ be an endomorphism of a finite dimensional vector space $V$ over a field $F$, then\r\n $$V_\\alpha\\cong F[X]/(f_1)\\oplus\\cdots\\oplus F[X]/(f_t)$$\r\n where $f_i\\in F[X]$ are monic and $f_1|f_2|\\cdots|f_t$.\r\n Moreover, with respect to a suitably chosen basis for $V$, $\\alpha$ has matrix of the form\r\n $$\\begin{pmatrix}\r\n C(f_1)&&&\\\\\r\n &C(f_2)&&\\\\\r\n &&\\ddots&\\\\\r\n &&&C(f_t)\r\n \\end{pmatrix}$$\r\n\\end{theorem}\r\n\\begin{proof}\r\n $V_\\alpha$ is finitely generated as an $F[X]$-module and since $F[X]$ is a ED, we can apply the structure theorem to get\r\n $$V_\\alpha\\cong F[X]/(f_1)\\oplus\\cdots\\oplus F[X]/(f_t)\\oplus F[X]^m$$\r\n where $f_i\\in F[X]$ are monic and $f_1|f_2|\\cdots|f_t$.\r\n $m=0$ since $V$ is finite dimensional over $F$.\r\n The result is immediate.\r\n\\end{proof}\r\n\\begin{remark}\r\n 1. If we start by an $n\\times n$ matrix of $\\alpha$, then this matrix must be similar to the above form.\\\\\r\n 2. The minimal polynomial of $\\alpha$ is $f_t$.\\\\\r\n 3. The characteristic polynomial of $\\alpha$ is $f_1\\cdots f_t$ (up to associates).\r\n Hence the minimal polynomial divides the characteristic polynomial, so we immediately have Cayley-Hamilton.\r\n\\end{remark}\r\n\\begin{example}\r\n When $V$ is $2$-dimensional vector space over $F$, then one of the following cases happen\r\n $$V_\\alpha\\cong F[X]/(X-\\lambda_1)\\oplus F[X]/(X-\\lambda_2),V_\\alpha\\cong F[X]/(f)$$\r\n where $(X-\\lambda_1)(X-\\lambda_2)$ or $f$ is the characteristic polynomial of $\\alpha$.\r\n\\end{example}\r\n\\begin{corollary}\r\n Let $A,B\\in\\operatorname{GL}_2(F)$ that are not scalar matrices, then $A,B$ are conjugate iff they have the same characteristic polynomial.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\n\\begin{lemma}\r\n Primes in $\\mathbb C[X]$ are polynomials $X-\\lambda$ where $\\lambda$ is any complex number (up to associates).\r\n\\end{lemma}\r\n\\begin{proof}\r\n FTA.\r\n\\end{proof}\r\n\\begin{theorem}[Jordan Normal Form]\r\n Let $\\alpha:V\\to V$ be an endomorphism of a finite dimensional vector space over $\\mathbb C$.\r\n Let $V_\\alpha$ be $V$ as the $\\mathbb C[X]$-module with $X$ acting as $\\alpha$.\r\n Then there is an isomorphism of $\\mathbb C[X]$-modules\r\n $$V_\\alpha\\cong\\mathbb C[X]/(X-\\lambda_1)^{n_1}\\oplus\\cdots\\oplus\\mathbb C[X]/(X-\\lambda_t)^{n_t}$$\r\n where $\\lambda_i\\in\\mathbb C$ are not necessarily distinct and $n_i\\in\\mathbb N$.\r\n In particular, there is a basis for $V$ such that $\\alpha$ has the matrix\r\n $$\\begin{pmatrix}\r\n J_{n_1}(\\lambda_1)&&\\\\\r\n &\\ddots&\\\\\r\n &&J_{n_t}(\\lambda_t)\r\n \\end{pmatrix},J_n(\\lambda)=\\begin{pmatrix}\r\n \\lambda&&&\\\\\r\n 1&\\lambda&&\\\\\r\n &\\ddots&\\ddots&\\\\\r\n &&1&\\lambda\r\n \\end{pmatrix}$$\r\n\\end{theorem}\r\n\\begin{proof}\r\n We know that $\\mathbb C[X]$ is a ED and $V_\\alpha$ is finitely generated as $V$ is finite dimensional.\r\n By Prime Decomposition Theorem, noting that primes in $\\mathbb C[X]$ are linear factors, and we cannot have any copy of $\\mathbb C[X]$ as the dimension is finite.\r\n Then we already have the isomorphism.\r\n Now $J_n(\\lambda)$ represents the multiplication by $X$ on $\\mathbb C[X]/(X-\\lambda)^n$ wrt the basis $1,X-\\lambda,\\ldots,(X-\\lambda)^{n-1}$.\r\n This shows the theorem.\r\n\\end{proof}\r\n\\begin{remark}\r\n 1. The theorem implies that any matrix with entries in $\\mathbb C$ is similar to a matrix in the said form.\\\\\r\n 2. The Jordan blocks are unique up to reordering.\\\\\r\n 3. The minimal polynomial of $\\alpha$ is $\\prod_i(X-\\lambda_i)^{c_i}$ where $c_i$ is the size of the largest block with eigenvalue $\\lambda_i$.\\\\\r\n 4. The characteristic polynomial of $\\alpha$ is $\\prod_i(X-\\lambda_i)^{a_i}$ where $a_i$ is the sum of the sizes of the blocks with eigenvalue $\\lambda_i$.\\\\\r\n 5. The eigenspace of $\\lambda_i$ has dimension equal to the number of blocks with eigenvalue $\\lambda$.\\\\\r\n 6. The uniqueness statement may be proved by considering the dimension of the generalized eigenspaces $\\ker((\\alpha-\\lambda_iI)^n)$, $n=1,2,\\ldots$.\r\n\\end{remark}\r\n\\begin{theorem}\r\n The structure theorem is true for any PIDs.\r\n\\end{theorem}\r\nWe will not prove this in the course, but we will illustrate the trick that is used for this extension.\r\n\\begin{theorem}\\label{pid_tf_free}\r\n Let $R$ be a PID, then any finitely generated torsion-free $R$-module is free.\r\n\\end{theorem}\r\nNote that for $R$ a ED, this is a corollary of the structure theorem.\r\n\\begin{lemma}\r\n Let $R$ be a PID and $M$ an $R$-module.\r\n Let $r_1,r_2\\in R$ that are not both $0$.\r\n Let $d=\\gcd(r_1,r_2)$.\\\\\r\n 1. There is a matrix $A\\in\\operatorname{SL}_2(R)$ such that\r\n $$A\\begin{pmatrix}\r\n r_1\\\\\r\n r_2\r\n \\end{pmatrix}=\\begin{pmatrix}\r\n d\\\\\r\n 0\r\n \\end{pmatrix}$$\r\n 2. If $x_1,x_2\\in M$, then there is $x_1',x_2'\\in M$ such that $Rx_1+Rx_2=Rx_1'+Rx_2'$ and $r_1x_1+r_2x_2=dx_1'+0x_2'$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n We know that $(r_1,r_2)=(d)$, so there is $\\alpha,\\beta\\in R$ such that $\\alpha r_1+\\beta r_2=d$.\r\n Write $r_1=s_1d,r_2=s_2d$ with $s_1,s_2\\in R$.\r\n We simply take the matrix\r\n $$A=\\begin{pmatrix}\r\n \\alpha&\\beta\\\\\r\n -s_2&s_1\r\n \\end{pmatrix}\\in\\operatorname{SL}_2(R)$$\r\n and it works for the first part.\r\n For the second part, we can take $x_1'=s_1x_1+s_2x_2,x_2'=-\\beta x_1+\\alpha x_2$.\r\n\\end{proof}\r\n\\begin{proof}[Proof of Theorem \\ref{pid_tf_free}]\r\n Say $M=Rx_1+\\cdots+Rx_n$ with $n$ minimal.\r\n If $x_1,\\ldots,x_n$ are independent, then the module is free.\r\n Otherwise, there is $r_1,\\ldots,r_n\\in R$ such that $r_1x_1+\\cdots+r_nx_n=0$ where WLOG $r_1\\neq 0$.\r\n By the preceding lemma, we can choose $x_1',x_2'$ such that $Rx_1+Rx_2=Rx_1'+Rx_2'$, so $M=Rx_1'+Rx_2'+Rx_3+\\cdots+Rx_n$ and $dx_1'+r_3x_3+\\cdots+r_nx_n=0$ for $d=\\gcd(r_1,r_2)\\neq 0$.\r\n Continue the process to $3,\\ldots,n$ to reduce the case to $rx_1=0$ for some $r\\neq 0$, but $R$ is torsion-free, so $x_1=0$, contradicting the minimality of $n$.\r\n\\end{proof}\r\n\r\n", "meta": {"hexsha": "e3b94931a7e15084c72616f2e72e3bc9f227b14b", "size": 20882, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "16/struct.tex", "max_stars_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_stars_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "16/struct.tex", "max_issues_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_issues_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "16/struct.tex", "max_forks_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_forks_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.3296089385, "max_line_length": 240, "alphanum_fraction": 0.6505602912, "num_tokens": 7144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.8499711718571775, "lm_q1q2_score": 0.6489391892021701}} {"text": "\\subsubsection{Time Series Normalization} \\label{time_series_normalization}\nWhen using DTW to determine the similarity of two time series, it is a common approach to z-normalize or Z-score the\ntime series in front of the calculation \\cite{ding2008querying}. In \\cite{das1998rule} two different time series\nnormalizations are presented. Both time series normalizations $\\eta$ and $\\eta '$ are explained in the following\nsection.\n\nGiven is a time series $Q = (q_1, q_2, \\dots, q_i, \\dots, q_l)$ with length $l$ over the domain set $\\mathbb{U}$ and a\ndistance measure function $d$ with $d: \\mathbb{U} \\times \\mathbb{U} \\to \\mathbb{R}$. The first time series normalization\n$\\eta$ transforms every data\npoint of the given time series $Q$ as follows.\n\\begin{equation}\n \\eta (q) = q -\\bar{q}\n\\end{equation}\nwhere $\\bar{q}$ is the mean of the time series $Q$\n\\begin{equation}\n \\bar{q} = \\frac{1}{l} \\sum \\limits_{i=1}^{l} q_i\n\\end{equation}\nFigure \\ref{fig:norm1} shows the impact of the $\\eta$ normalization on DTW compared to plain DTW in figure\n\\ref{fig:dynamictimewarping}.\n\n\\begin{figure}\n \\begin{center}\n \\resizebox {\\textwidth} {!} {\n \\begin{tabular}{cc}\n \\resizebox* {!} {0.3\\textwidth} {\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n xlabel=time,\n ylabel=acceleration,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight,\n reverse legend,\n legend pos=south east]\n \\addplot[gray, quiver={u=\\thisrow{u}, v=\\thisrow{v}}] table {../data/fig/norm1/path.dat};\n \\addplot[red, thick, mark=none] table {../data/fig/norm1/q.dat};\n \\addlegendentry{Q}\n \\addplot[blue, thick, mark=none] table {../data/fig/norm1/c.dat};\n \\addlegendentry{C}\n \\end{axis}\n \\end{tikzpicture}\n } & \\quad\n \\resizebox* {!} {0.3\\textwidth} {\n \\begin{tabular}[b]{ll}\n \\begin{turn}{90}\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n ymin=-40,\n ymax=40,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight]\n \\addplot[red, ultra thick, mark=none] table {../data/fig/norm1/q.dat};\n \\end{axis}\n \\end{tikzpicture}\n \\end{turn} \\hspace*{3em} &\n \\begin{tikzpicture}\n \\begin{axis}[\n enlargelimits=false,\n ymin=0,\n ymax=47,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=\\axisdefaultwidth,\n colorbar,\n colormap/viridis high res]\n \\addplot[matrix plot*,\n mesh/cols=48,\n point meta=explicit] table[meta=C] {../data/fig/norm1/matrix.dat};\n \\addplot[white, ultra thick, mark=*, mark size=1] table {../data/fig/norm1/matrix_path.dat};\n \\end{axis}\n \\end{tikzpicture}\\\\\n &\n \\\\[1em]\n &\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n ymin=-40,\n ymax=40,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight]\n \\addplot[blue, ultra thick, mark=none] table {../data/fig/norm1/c.dat};\n \\end{axis}\n \\end{tikzpicture}\n \\end{tabular}\n }\n \\end{tabular}\n }\n \\end{center}\n \\caption{The same two time series $Q$ and $C$ from figure \\ref{fig:dynamictimewarping} normalized by $\\eta$. On the\n left plot are both time series graphs, the gray lines are representing the warping path of plain DTW. The right plot\n shows the associated matrix containing the distances between the time series data points. Starting in the lower left\n corner and ending in the upper right corner, the warping path is illustrated as a white graph.}\n \\label{fig:norm1}\n\\end{figure}\n\nThe second time series normalization $\\eta '$ transforms every data point of the given time series $Q$ as follows.\n\\begin{equation}\n \\eta '(q) = \\frac{\\eta (q)}{\\sigma}\n\\end{equation}\nwhere $\\sigma$ is the standard deviation of the time series $Q$\n\\begin{equation}\n \\sigma = \\frac{1}{l-1} \\sum \\limits_{i=1}^{l} d(q_i, \\bar{q})^2\n\\end{equation}\nFigure \\ref{fig:norm2} shows the impact of the $\\eta '$ normalization on DTW compared to plain DTW in\nfigure \\ref{fig:dynamictimewarping}.\n\n\\begin{figure}\n \\begin{center}\n \\resizebox {\\textwidth} {!} {\n \\begin{tabular}{cc}\n \\resizebox* {!} {0.3\\textwidth} {\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n xlabel=time,\n ylabel=acceleration,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight,\n reverse legend,\n legend pos=south east]\n \\addplot[gray, quiver={u=\\thisrow{u}, v=\\thisrow{v}}] table {../data/fig/norm2/path.dat};\n \\addplot[red, thick, mark=none] table {../data/fig/norm2/q.dat};\n \\addlegendentry{Q}\n \\addplot[blue, thick, mark=none] table {../data/fig/norm2/c.dat};\n \\addlegendentry{C}\n \\end{axis}\n \\end{tikzpicture}\n } & \\quad\n \\resizebox* {!} {0.3\\textwidth} {\n \\begin{tabular}[b]{ll}\n \\begin{turn}{90}\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n ymin=-2,\n ymax=2,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight]\n \\addplot[red, ultra thick, mark=none] table {../data/fig/norm2/q.dat};\n \\end{axis}\n \\end{tikzpicture}\n \\end{turn} \\hspace*{3em} &\n \\begin{tikzpicture}\n \\begin{axis}[\n enlargelimits=false,\n ymin=0,\n ymax=47,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=\\axisdefaultwidth,\n colorbar,\n colormap/viridis high res]\n \\addplot[matrix plot*,\n mesh/cols=48,\n point meta=explicit] table[meta=C] {../data/fig/norm2/matrix.dat};\n \\addplot[white, ultra thick, mark=*, mark size=1] table {../data/fig/norm2/matrix_path.dat};\n \\end{axis}\n \\end{tikzpicture}\\\\\n &\n \\\\[1em]\n &\n \\begin{tikzpicture}\n \\begin{axis}[\n xmin=0,\n xmax=47,\n ymin=-2,\n ymax=2,\n hide x axis,\n hide y axis,\n width=\\axisdefaultwidth,\n height=0.7*\\axisdefaultheight]\n \\addplot[blue, ultra thick, mark=none] table {../data/fig/norm2/c.dat};\n \\end{axis}\n \\end{tikzpicture}\n \\end{tabular}\n }\n \\end{tabular}\n }\n \\end{center}\n \\caption{The same two time series $Q$ and $C$ from figure \\ref{fig:dynamictimewarping} normalized by $\\eta '$. On the\n left are plot both time series graphs, the gray lines are representing the warping path of plain DTW. The right plot\n shows the associated matrix containing the distances between the time series data points. Starting in the lower left\n corner and ending in the upper right corner, the warping path is illustrated as a white graph.}\n \\label{fig:norm2}\n\\end{figure}\n", "meta": {"hexsha": "e193e1265cbf30f38008a9b500d9baa9ccfc9afe", "size": 10075, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "bachelor-thesis/background_and_notation/dynamic_time_warping/time_series_normalization.tex", "max_stars_repo_name": "GordonLesti/SlidingWindowFilter", "max_stars_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-06-22T09:37:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-14T11:43:53.000Z", "max_issues_repo_path": "bachelor-thesis/background_and_notation/dynamic_time_warping/time_series_normalization.tex", "max_issues_repo_name": "GordonLesti/SlidingWindowFilter", "max_issues_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "bachelor-thesis/background_and_notation/dynamic_time_warping/time_series_normalization.tex", "max_forks_repo_name": "GordonLesti/SlidingWindowFilter", "max_forks_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-01-11T23:15:57.000Z", "max_forks_repo_forks_event_max_datetime": "2019-01-11T23:15:57.000Z", "avg_line_length": 50.1243781095, "max_line_length": 124, "alphanum_fraction": 0.4220347395, "num_tokens": 2110, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042767, "lm_q2_score": 0.798186784940666, "lm_q1q2_score": 0.6488199582450379}} {"text": "\\documentclass[11pt,a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage[english]{babel}\n\\usepackage{fancyhdr}\n\\usepackage[margin=.7in]{geometry}\n\\usepackage{exsheets}\n\\usepackage{enumerate}\n\n\\usepackage{./Style/linearProgramsStyle} % This is a set of commands for maths content.\n\n\\pagestyle{fancy}\n\n\\renewcommand{\\headrulewidth}{2pt}\n\\fancyhead[L]{EPITA\\_ING1\\_BING\\_2020\\_S5\\_PROL}\n\\fancyhead[R]{January 2018}\n\n\\fancyfoot[C]{\\textbf{\\thepage}}\n\\fancyfoot[L]{}\n\n\\pretitle{\\vspace{-.5\\baselineskip} \\begin{center}}\n\\title{%\n { \\huge Linear Programming (PROL)}%\n}\n\\posttitle{\n\\end{center}\n \\begin{flushleft}\n \\vspace{3\\baselineskip} \\textit{\n \\!\\!\\emph{Duration: 1h30.}\\\\\n \\! \\emph{No documents are allowed.} \\\\\n Only \\emph{non-programmable} pocket calculators are allowed.\n }\n \\end{flushleft}\n \\rule{\\textwidth}{1.5pt}\n \\vspace{-5\\baselineskip}\n}\n\\author{}\n\\date{}\n\n\\pdfinfo{\n /Author (Bashar Dudin)\n /Title (PROL Exam -- 2018)\n /Subject (Linear Programming)\n}\n\n\\begin{document}\n\n\\maketitle\\thispagestyle{fancy}\n\nWe're considering the following list of linear programs respectively\ngiven the names $(A)$, $(B)$, $(C)$, $(D)$ and $(E)$, $(F)$.\n\\[\n(A) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $x_1 + x_2 + x_3$\n }{\n \\systeme{\n 2x_1 + 3x_2 + x_3 \\leq 1,\n x_1 - 4x_2 + 3x_3 \\geq 2\n }\n }{\n $x_1$, $x_2 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\qquad (B)\\;\n\\left\\{\n \\begin{linearProg}{\n minimize\n }{\n $- x_1 - x_2 - x_3 + x_4$\n }{\n \\systeme{\n 2x_1 + 3x_2 + x_3 - x_4 + \\leq 1,\n -x_1 + 4x_2 - 3x_3 + 3x_4 \\leq -2\n }\n }{\n $x_1$, $x_2$, $x_3$, $x_4 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\\[\n(C) \\;\n\\left\\{\n \\begin{linearProg}{\n minimize\n }{\n $-5x_1 - 6x_2 - 9x_3 - 8x_4$\n }{\n \\systeme{\n x_1 + 2x_2 + 3x_3 + x_4 \\leq 5,\n -x_1 - x_2 - 2x_3 - 3x_4 \\geq -3\n }\n }{\n $x_1$, $x_2$, $x_3$, $x_4 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\qquad (D) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $5x_1 + 6x_2 + 9x_3 + 8x_4$\n }{\n \\systeme{\n x_1 + 2x_2 + 3x_3 + x_4 \\leq 5,\n x_1 + x_2 + 2x_3 + 3x_4 \\leq 3\n }\n }{\n $x_1$, $x_2$, $x_3$, $x_4 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\\[\n(E) \\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $x_1 - 2x_2 - x_3$\n }{\n \\systeme{\n 2x_1 + x_2 - x_3 \\leq -1,\n x_1 - 2x_2 + x_3 \\leq -2\n }\n }{\n $x_1$, $x_2$, $x_3 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\\qquad (F)\\;\n\\left\\{\n \\begin{linearProg}{\n maximize\n }{\n $5x_1 + 6x_2 + 9x_3 + 8x_4$\n }{\n \\systeme{\n x_1 + 2x_2 + 3x_3 + x_4 \\leq 5,\n x_1 + x_2 + 2x_3 + 3x_4 + x_5 = 3\n }\n }{\n $x_1$, $x_2$, $x_3$, $x_4$, $x_5 \\geq 0.$\n }\n \\end{linearProg}\n\\right.\n\\]\n\n\\section{Equivalence and Duality}\n\n\\begin{question}\n Which previous programs are equivalent to each others? Sort them\n into groups of equivalent classes.\n\\end{question}\n\n\\begin{question}\n Give standard form of $(F)$. What is the slack form of $(E)$?\n\\end{question}\n\n\\begin{question}\n What are the dual programs $(C)$ and $(D)$?\n\\end{question}\n\n\\begin{question}\n Which programs $(D)$, $(E)$ and $(F)$ have basic feasible\n solution? Give a feasible solution of $(A)$.\n\\end{question}\n\n\\newpage\n\\section{Simplex Algorithm}\n\n\\begin{question}\n Solve linear program $(D)$.\n\\end{question}\n\n\\begin{question}\n Out of previous solution of $(D)$ give a solution of its dual.\n\\end{question}\n\n\\begin{question}\n Transform program $(E)$ into an equivalent linear program having\n feasible basic solution.\n\\end{question}\n\n\\end{document}\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: t\n%%% End:\n", "meta": {"hexsha": "ed9b07e8ea032dc0bf0449d288007e976a5a7e64", "size": 3795, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Exams/TeX/Ecrit_PROL_2018_012_BING.tex", "max_stars_repo_name": "bashardudin/LinearPrograms", "max_stars_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 22, "max_stars_repo_stars_event_min_datetime": "2016-10-20T14:04:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-03-31T08:45:14.000Z", "max_issues_repo_path": "Exams/TeX/Ecrit_PROL_2018_012_BING.tex", "max_issues_repo_name": "bashardudin/LinearPrograms", "max_issues_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-06-04T01:30:01.000Z", "max_issues_repo_issues_event_max_datetime": "2018-10-09T08:18:10.000Z", "max_forks_repo_path": "Exams/TeX/Ecrit_PROL_2018_012_BING.tex", "max_forks_repo_name": "bashardudin/LinearPrograms", "max_forks_repo_head_hexsha": "24b67171cd5bbf407db47f2adb5150ab71d88143", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-10-26T21:19:58.000Z", "max_forks_repo_forks_event_max_datetime": "2018-03-17T19:28:15.000Z", "avg_line_length": 19.362244898, "max_line_length": 87, "alphanum_fraction": 0.5725955204, "num_tokens": 1521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867873410141, "lm_q2_score": 0.8128673223709251, "lm_q1q2_score": 0.6488199565777412}} {"text": "\\chapter{Predefined Variables}\\label{ch:predefined}\n\nA few commonly used variables are predefined in\n\\aprepro{}\\footnote{The units system described in\nChapter~\\ref{ch:units} also predefines several variables when it is\nactivated}. These are listed below. The default output format\n\\cmd{\\_FORMAT} is specified as a C language format string, see your C\nlanguage documentation for more information. The default output format\n(\\cmd{\\_FORMAT}) and comment (\\cmd{\\_C\\_}) variables are defined with\na leading underscore in their name so they can be redefined without\ngenerating an error message.\n\n\\begin{longtable}{lll}\n\\caption{Predefined Variables}\\\\\nName & Value & Description \\\\\n\\hline\n PI & 3.14159265358979323846 & $\\pi$ \\\\\n PI\\_2 & 1.57079632679489661923 & $\\pi/2$ \\\\\n TAU & 6.28318530717958623200 & $2\\pi$ \\\\\n SQRT2 & 1.41421356237309504880 & $\\sqrt{2}$ \\\\\n DEG & 57.2957795130823208768 & $180/\\pi$ degrees per radian \\\\\n RAD & 0.01745329251994329576 & $\\pi/180$ radians per degree \\\\\n E & 2.71828182845904523536 & base of natural logarithm \\\\\n GAMMA & 0.57721566490153286060 & $\\gamma$, euler-mascheroni constant \\\\\n PHI & 1.61803398874989484820 & golden ratio $(\\sqrt{5}+1)/2$ \\\\\n VERSION & Varies, string value & current version of \\aprepro \\\\\n \\_FORMAT& \\texttt{\"}\\%.10g\\texttt{\"} & default output format \\\\\n \\_C\\_ & \\texttt{\"}\\$\\texttt{\"} & default comment character \\\\\n\\end{longtable}\n\nNote that the output format is used to output both integers and floating point \nnumbers. Therefore, it should use the \\%g format descriptor which will use either \nthe decimal (\\%d), exponential (\\%e), or float (\\%f) format, whichever is shorter, \nwith insignificant zeros suppressed. The table below illustrates the effect of \ndifferent format specifications on the output of the variable \\textbf{PI} and the \nvalue 1.0 . See the documentation of your C compiler for more information. For \nmost cases, the default value is sufficient.\n\n\\begin{longtable}{lll}\n\\caption{Effect of various output format specifications}\\\\\nFormat & PI Output & 1.0 Output \\\\\n\\hline\n\\%.10g & 3.141592654 & 1 \\\\\n\\%.10e & 3.1415926536e+00 & 1.0000000000e+00 \\\\\n\\%.10f & 3.1415926536 & 1.0000000000 \\\\\n\\%.10d & 1413754136 & 0000000000 \\\\\n\\end{longtable}\n\nThe comment character should be set to the character that the program\nwhich will read the processed file uses as a comment character. The\ndefault value of \\texttt{\"}\\$\\texttt{\"} is the comment character used\nby the SEACAS codes at Sandia National Laboratories. The \\textbf{-c}\ncommand line option (described in Chapter~\\ref{ch:execution}) changes\nthe value of the comment variable to match the character specified on\nthe command line.\n", "meta": {"hexsha": "6577786d2a4d217e94ae29fc58e5dd88462ef8a0", "size": 2720, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "packages/seacas/doc-source/aprepro/predefined_variables.tex", "max_stars_repo_name": "jschueller/seacas", "max_stars_repo_head_hexsha": "14c34ae08b757cba43a3a03ec0f129c8a168a9d3", "max_stars_repo_licenses": ["Python-2.0", "Zlib", "BSD-2-Clause", "MIT", "NetCDF", "BSL-1.0", "X11", "BSD-3-Clause"], "max_stars_count": 82, "max_stars_repo_stars_event_min_datetime": "2016-02-04T18:38:25.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-29T03:01:49.000Z", "max_issues_repo_path": "packages/seacas/doc-source/aprepro/predefined_variables.tex", "max_issues_repo_name": "jschueller/seacas", "max_issues_repo_head_hexsha": "14c34ae08b757cba43a3a03ec0f129c8a168a9d3", "max_issues_repo_licenses": ["Python-2.0", "Zlib", "BSD-2-Clause", "MIT", "NetCDF", "BSL-1.0", "X11", "BSD-3-Clause"], "max_issues_count": 206, "max_issues_repo_issues_event_min_datetime": "2015-11-20T01:57:47.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T21:12:04.000Z", "max_forks_repo_path": "packages/seacas/doc-source/aprepro/predefined_variables.tex", "max_forks_repo_name": "jschueller/seacas", "max_forks_repo_head_hexsha": "14c34ae08b757cba43a3a03ec0f129c8a168a9d3", "max_forks_repo_licenses": ["Python-2.0", "Zlib", "BSD-2-Clause", "MIT", "NetCDF", "BSL-1.0", "X11", "BSD-3-Clause"], "max_forks_count": 68, "max_forks_repo_forks_event_min_datetime": "2016-01-13T22:46:51.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T06:25:05.000Z", "avg_line_length": 48.5714285714, "max_line_length": 83, "alphanum_fraction": 0.7272058824, "num_tokens": 785, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673178375735, "lm_q2_score": 0.7981867777396212, "lm_q1q2_score": 0.6488199451546213}} {"text": "\\lab{Complex Integration}{Integration in the Complex Plane Part 2}{Integration in the Complex Plane Part II}\n\n\\objective{Apply Residues to Integration in the Complex Plane}\n\nAs we noted in the last lab, there are some important links between the residues of a rational function and its partial fraction decomposition.\nHere we will discuss the connection between residues and integrals along the real line and we will also discuss some important applications of residues.\n\n\\section*{Evaluating Residues Using Limits and Derivatives}\nOften it is faster to evaluate a residue using limits instead of its direct integral definition.\nTo see how this might be done, consider some function $f$ with a pole of order $1$ at $z_0$.\nWe say $f$ has a pole of order $n$ at $z_0$ if the first nonzero term in the Laurent Series expansion of is the term corresponding to $(z-z_0)^{-n}$.\nWe then have\n$$f(z)=c_{-1}(z-z_0)^{-1}+c_0 (z-z_0) +c_1 (z-z_0) + \\dots$$\nSo,\n$$f(z)(z-z_0)=c_{-1}+c_0 (z-z_0) + c_1 (z-z_0)^2 + \\dots$$\nSince the right hand side of the Laurent series converges within some nonzero distance of $z_0$, this series will also converge on that domain.\nIt is an ordinary power series, so it will also be continuous on that domain (it will actually be analytic, but continuity is sufficient here).\nThis means that we may take the limit of this expression as $z$ approaches $z_0$.\nTaking this limit, the only nonzero term of the series remaining is $c_{-1}$, so we have\n\\begin{equation*}\n\\lim_{z\\to z_0} f(z)(z-z_0) = c_{-1} = \\Res_{z=z_0} f(z)\n\\end{equation*}\nNow for some $f(z)$ with a pole of order $n$ at $z_0$ we have that within some nonzero distance of $z_0$, the Laurent series of $f$ converges and is of the form\n$$f(z)=c_{-n}(z-z_0)^{-n}+\\dots+c_0+c_1(z-z_0)+\\dots$$\nMultiplying by $(z-z_0)^{n}$ and differentiating $n-1$ times, we have\n$$\\frac{d^{n-1}}{dz^{n-1}}(f(z)(z-z_0)^{n})=(n-1)! c_{-1} + \\frac{n!}{1!} c_0 (z-z_0) + \\frac{(n+1)!}{2} c_1 (z-z_0)^2!+\\dots$$\nTaking the limit, we have\n\\begin{equation*}\n\\lim_{z\\to z_0}\\frac{d^{n-1}}{dz^{n-1}}((z-z_0)^n f(z)) = (n-1)! c_{-1}\n\\end{equation*}\nWhich implies\n\\begin{equation*}\n\\Res_{z=z_0}f(z)=\\frac{1}{(n-1)!}\\lim_{z\\to z_0}\\frac{d^{n-1}}{dz^{n-1}}((z-z_0)^n f(z))\n\\end{equation*}\t\nAnother useful trick involving poles and residues involves the logarithmic residue, i.e. the residue of the logarithmic derivative $\\frac{d}{dz}(\\ln(f(z))) = \\frac{f'(z)}{f(z)}$.\nA natural consequence of the Laurent series expansion of $f(z)$ and $f'(z)$ at a pole $z_0$ is that, where $d$ is the degree of the pole at $z_0$ is that\n\\begin{equation*}\n\\Res_{z=z_0} \\frac{f'(z)}{f(z)} = d\n\\end{equation*}\nThis is a useful bit of information that may be used to simplify computation of residues or of the Laurent series expansion of a function. \nAnother way we can evaluate residues is by the following theorem.\n\\begin{theorem}\nLet $f(z)=\\frac{p(z)}{q(z)}$ and let $p$ and $q$ be holomorphic at $z_0$. Let $p(z_0) \\neq 0$ and $q(z_0)=0$. Suppose that $z_0$ is a pole of order $1$ of $f$. Then \n\\begin{equation*}\n\\Res_{z=z_0} f(z) = \\frac{p(z_0)}{q'(z)}\n\\end{equation*}\n\\end{theorem}\n\n\\begin{problem}\nWrite a python function, which, using symbolic integration in SymPy, returns the Laurent series expansion of a function about a given pole.\nUse the trick involving logarithmic residues to figure out what the degree of the pole is, then use symbolic limits and derivatives to compute the coefficients of the corresponding Laurent series expansion.\nYou may assume that the pole is not an essential singular point.\n(Recall that, as we defined it, a pole is not an essential singular point, so it will always have a finite degree).\n\\end{problem}\n\n\\begin{problem}\nWrite a function which uses SymPy to evaluate the residue of a function $f$ at a pole of $f$ $z_0$ in the case that: $f$ may be written as the quotient of two functions analytic everywhere within some nonzero distance of $z_0$ and the pole of $f$ at $z_0$ is of order $1$.\n\\end{problem}\n\n\\section*{Evaluating Indefinite Integrals Using Residues}\n\nOne convenient use of residues is the evaluation of integrals that are difficult to evaluate symbolically in other ways.\nOften, when we cannot directly assign a value to one of these integrals, residues can still help us evaluate the Cauchy principal value of the integral.\nRecall that for an integral $\\int_{-\\infty}^{\\infty} f(x)dx$ the Cauchy principal value is $\\lim_{r\\to \\infty} \\int_{r}^{r} f(x) dx$. This limit may exist, even though the limit itself may not.\n\nConsider the integral $\\int_{-\\infty}^{\\infty}\\frac{z^2}{z^4+1}$. Let $f(z)=\\frac{z^2}{z^4+1}$.\nNotice that this function has poles at $e^{\\frac{\\pi i}{4}}$, $e^{\\frac{3\\pi i}{4}}$, $e^{\\frac{5\\pi i}{4}}$, and $e^{\\frac{7\\pi i}{4}}$.\nFor notation, let these be $p_1$, $p_2$, $p_3$, and $p_4$.\nFor some real $R>1$, consider the contour $C$ from $-R$ to $R$ and then continuing continuing counterclockwise along the circle centered at $0$ of radius $R$.\nLet $A$ be this second portion of $C$.\nNote that, since $R>1$ we have\n\\begin{equation*}\n\\int_C f(z)dz = 2\\pi i (\\Res_{z=p_0} f(z) +\\Res_{z=p_1} f(z))\n\\end{equation*}\nSo, rewriting, we have\n\\begin{equation*}\n\\int_{-R}^R f(z) dz = 2\\pi i (\\Res_{z=p_0} f(z) +\\Res_{z=p_1} f(z)) - \\int_A f(z) dz\n\\end{equation*}\nso\n\\begin{equation*}\n\\int_{-\\infty}^{\\infty} f(z) dz = \\lim_{R\\to \\infty} \\int_{-R}^R f(z) dz = 2\\pi i (\\Res_{z=p_0} f(z) +\\Res_{z=p_1} f(z)) - \\lim_{R\\to \\infty} \\int_A f(z) dz\n\\end{equation*}\nWe would like to show that that $\\lim_{R\\to\\infty} \\int_A f(z) dz = 0$, so note that on $A$, $\\abs{z}=R$.\nIt follows from the triangle inequality that $\\abs{z^4+1}\\geq \\abs{\\abs{z}^4-1} = R^4 -1$, so we have that \n$$\\abs{\\int_A f(z) dz}\\leq \\int_A \\abs{f(z)} dz \\leq \\int_A \\frac{R^2}{R^4 -1}dz = \\pi R \\frac{R^2}{R^4-1}$$\nso $\\lim_{R\\to\\infty} \\int_A f(z) dz = 0$ as desired.\nThis then implies that \n\\begin{equation*}\n \\int_{-\\infty}^{\\infty} f(z) dz = 2\\pi i (\\Res_{z=p_0} f(z) +\\Res_{z=p_1} f(z))\n\\end{equation*}\nEvaluating the residues at $p_0$ and $p_1$ we have\n$$\\int_{-\\infty}^{\\infty} f(z) dz = \\frac{\\pi}{\\sqrt{2}}$$\n\n\\begin{problem}\nWrite a python function which, given the coefficients for the polynomials in the numerator and denominator of a rational function $f$, evaluates the sum\n\\begin{equation*}\n\\sum 2\\pi i (\\Res_{z=p} f(z)\n\\end{equation*}\nover all poles $p$ of $f$ such that $Im(p)>0$.\nUse that function to numerically evaluate the following integrals.\n$$\\int_{-\\infty}^{\\infty} \\frac{z^2}{z^6+1}dz$$\n$$\\int_{-\\infty}^{\\infty} \\frac{z^{12}-5z^{10}+3z^8-16z^6+4z^4-z^2-4}{4z^{14}+6z^6+12}dz$$\n\\end{problem}\n\n\\begin{problem}\nSimilar arguments can be used to evaluate integrals of functions that can be compared to the rational functions in question.\nWrite a modified version of the function you just wrote and use it to evaluate the following integrals.\n$$\\int_{-\\infty}^{\\infty}\\frac{cos(z)}{z^4+1}dz$$\n$$\\int_{-\\infty}^{\\infty}\\frac{sin^2(z)}{z^{20}+1}dz$$\n\\end{problem}\n\nThe general conditions in which we may apply this particular way of evaluating integrals come from Jordan's Lemma, which states\n\\begin{lemma}[Jordan's Lemma]\nIf for some $f(z)$ on $\\mathbb{C}$, $f$ is analytic for all points $z$ in the upper half plane such that $\\abs{z}>R_0$ for some $R_0 \\in \\mathbb{R}$ where $R_0 >0$ and that, where $C_R$ denotes the semicircle $z=Re^{i\\theta}$ for $0\\leq \\theta \\leq \\pi$, there exists some positive constant $M_R$ such that for all $z$ on $C_R$, $\\abs{f(z)} \\leq M_R$ and $\\lim_{R \\to \\infty} M_R = 0$, then for every positive constant $a$, we have\n$$\\lim_{R \\to \\infty} \\int_{C_R} f(z) e^{iaz} dz = 0$$\n\\end{lemma}\n\\begin{problem}\nUse the function you just wrote to evaluate the Cauchy Principal Value of the integral\n$$\\int_{\\infty}^{-\\infty} \\frac{z sin(z)}{z^2+1}$$\n\\end{problem}\n\nIf a function has a singularity on the real line, we can often still evaluate the value of $\\int_{-\\infty}^{\\infty} f(z) dz$ using a similar argument as before, but by indenting the path along the real axis around a small path around the singularity, then, as we take the limit as our outer contour moves out toward infinity, we can let the small contour around the singularity approach the singularity.\n\nThe following is a useful theorem involving these ``indented path\" methods.\n\\begin{theorem}\nConsider a function $f$ with a pole of order $1$ at $z=x_0$ with a Laurent series representation in a punctured disk of radius $R$ about $x_0$ and residue $B_0$ at $x_0$.\nLet $C_r$ be the upper half of a circle $\\abs{z-x_0}=r$ where $r0,~y>0\\}.$ Let $P_1(x)$ be its \\textit{marginal distribution} (marginalized $y$). Find the density function $p_1(x)$ of it, as well as the distribution function $F_1$. Find the numerical value of $p_1(\\sqrt{1-(\\pi/4)^2}).$\n\\end{document}\n", "meta": {"hexsha": "701883d2e4d283a36fcdde4e160f260916f67afb", "size": 2004, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week03_distributions/Week03_HW_Theory.tex", "max_stars_repo_name": "girafe-ai/msai-statistics", "max_stars_repo_head_hexsha": "c9de8ca20bbb9f266e06598d376be50c35f00756", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2021-04-07T05:10:29.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-07T15:58:35.000Z", "max_issues_repo_path": "week03_distributions/Week03_HW_Theory.tex", "max_issues_repo_name": "girafe-ai/msai-statistics", "max_issues_repo_head_hexsha": "c9de8ca20bbb9f266e06598d376be50c35f00756", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-03-08T17:08:53.000Z", "max_issues_repo_issues_event_max_datetime": "2021-03-08T17:08:53.000Z", "max_forks_repo_path": "week03_distributions/Week03_HW_Theory.tex", "max_forks_repo_name": "girafe-ai/msai-statistics", "max_forks_repo_head_hexsha": "c9de8ca20bbb9f266e06598d376be50c35f00756", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-03-25T15:23:32.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-27T14:28:21.000Z", "avg_line_length": 37.8113207547, "max_line_length": 332, "alphanum_fraction": 0.6616766467, "num_tokens": 766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.82893881677331, "lm_q1q2_score": 0.6487793092361138}} {"text": "\\section{Radius of Convergence}\r\nWe've established some tests to determine if an infinite series is or isn't convergent, but like we saw with geometric and Taylor series, it's possible that a series only converges over some interval.\r\nWe'd like to be able to find this interval so we know if we can rightly use Taylor series to approximate a function.\r\n\r\n\\subsection{Convergence Theorem for Power Series}\r\n\\begin{theorem}[Convergence Theorem for Power Series]\r\n\tThere are three possibilities for any power series of the form\r\n\t\\begin{equation*}\r\n\t\t\\sum_{k=0}^{\\infty}{c_k(x-a)^k}\r\n\t\\end{equation*}\r\n\twith respect to convergence.\r\n\t\\begin{enumerate}\r\n\t\t\\item The power series converges, but only on some finite interval centered at $x=a$.\r\n\t\t\tThat is, there is a positive real number $r$ such that the series converges for $\\abs{x-a} |\\bigC(BUGGYCLIQUE(A))|$.\n\\end{thm}\n\\begin{proof}\nFeed in a 0 to $e$, which is in $B$; the remaining cliques are $A$.\nThe resulting\ncircuit computes $BUGGYCLIQUE(A)$, and so has size\nat least $|\\bigC(BUGGYCLIQUE(A))|$. But at least one\nNAND gate has been removed by feeding in the 0.\n\\end{proof}\n\nThis shows that sometimes, finding a larger set of cliques is\nharder. However, the above theorem doesn't help if the two\nsets of cliques cover the same set of edges.\n\n\\begin{figure}\n\\label{overlappingTris}\n\\centering\n\\includegraphics[width=1\\textwidth]{R/tri1.pdf}\n\\caption{A given number of cliques can overlap maximally (left),\nor not overlap much (right).}\n\\label{fig:overlappingTriangles}\n\\end{figure}\n\nTriangles can be detected using matrix multiplication \\cite{itai_finding_1977},\nand there are fast algorithms known for matrix multiplication\n\\cite{strassen_gaussian_1969}\n\\cite{williams_multiplying_2012}, so the set of all possible\ncliques on some set of vertices (Figure \\ref{overlappingTris}, left)\n can be detected\nusing fewer than one NAND gate per triangle (for large enough input graphs).\n\nOn the other hand, if the triangles overlap less (as in\nFigure \\ref{overlappingTris}, right),\nthen to detect all of the triangles, we will definitely need at least one\ngate per triangle. We can see this by feeding in a 0 to some input\nunique to a triangle, applying \\ref{edgeZonking}, and repeating.\n\n\\subsection{Defining levels of CLIQUE}\n\nTo me, it seems a reasonably intuitive guess that the hardness of\ncomputing $BUGGYCLIQUE(A)$ should be somehow related to\n$|A|$, which is simply the number of cliques it ``sees''.\n\n\\begin{defn}\n\\label{CLIQUE-level}\nAssume $n, k$ are fixed. For any $l$ such that\n$0 \\le l \\le |K|$, let $K_l$ be the set of all sets\nwith exactly $l$ cliques. \n\\end{defn}\n\nWe abuse notation slightly, and write\n\n\\[\nBUGGYCLIQUE(K_l) = \\{ BUGGYCLIQUE(k) : k \\in K_l \\}\n\\]\n\nWhat can we say about $E[|\\bigC(BUGGYCLIQUE(K_l))|]$, for\nsome fixed $l$? Let $N = {n \\choose k}$.\nAt the ``bottom'', there's only one $BUGGYCLIQUE(\\emptyset)$\nfunction, so the counting argument is useless.\nIn the ``middle'',\nthere are ${N \\choose {N/2}}$ functions, and so the counting\nargument gives a nontrivial lower bound for\n$E[|\\bigC(BUGGYCLIQUE(K_{N \\choose {N/2}}))|]$ (although\nwithout actually constructing even {\\em one} difficult function!).\nAt the ``top'', again, there's only one $BUGGYCLIQUE(K) = k-CLIQUE$\nfunction, so the counting argument is, once again, useless.\n\nIt would be nice if we could show that, as $l$ increases,\n$E[|\\bigC(BUGGYCLIQUE(K_l))|]$ also increases.\nIf we could prove something in general, for all levels $l$, then at\nthe top of the diagram, we'd be bounding just the function\n$BUGGYCLIQUE(K_l) = k-CLIQUE$. (This suggests that doing so would\nbe difficult...)\n\n\n\\section{Using random restrictions}\n\nRandom restrictions have been used in lower bounds of formula\n\\cite{subbotovskaya1963comparison} and circuit \\cite{hastad1987lower}\ncomplexity (see also the slides at \\cite{rossmanRestrictions}).\nHere, we apply random restrictions to a set of functions (and circuits),\nrather than just one function and circuit.\n\n??? Is this just a ``gate-elimination argument''?\n\n\\subsection{Counting functions by their ``rank'' in a list}\n\nThis argument relies on measuring the size of a circuit,\nfor a given basis. (Here, we assume unbounded fan-in\nNAND gates, but this doesn't seem crucial.)\n\n\\begin{defn}\n\\label{Rank}\nFor all $C \\subseteq K$,\narrange all of the sets of cliques in nondecreasing order\nof $|\\bigC(BUGGYCLIQUE(C))|$ (using unbounded fan-in NAND gates,\nbreaking ties by some lexicographic order of circuits).\n\nLet $A \\subset K$. The {\\em rank} of $A$, $rank(A)$, is the zero-based\nindex of $A$ in this list.\n\\end{defn}\n\nIf we can lower-bound $rank(A)$, then we should be able to translate\nthat into a\nlower bound on $|\\bigC(BUGGYCLIQUE(A))|$. (Possibly, if we\ncan only lower-bound $E[rank(A)]$, we may be able to get\na bound in terms of number of gates using Jensen's inequality?\nNot clear.)\n\nWe can imagine a miles-long linear warehouse of chips with the minimal\ncircuit for each problem, sorted in terms of number of gates.\nNote that the above list only includes functions in $BUGGYCLIQUE$.\nThere are a {\\em ton} of other functions (parity, primality, etc.), but ignoring\nthose should still leave a valid lower bound for the functions in $BUGGYCLIQUE$.\n\n\\subsection{How much smaller are ``restricted'' circuits, on average?}\n\n\\begin{thm}\n\\label{vaguelyUpward}\nLet $C \\subseteq K$ be a set of cliques chosen uniformly at random\nfrom $K$.\n\nLet $a = {n-2 \\choose k-2}$ (this is the number of $k$-cliques\nintersecting a given edge).\n\nThen there is a function $f: 2^K \\rightarrow 2^K$ such that\n\\begin{itemize}\n\n\\item $E[|C| - |f(C)|] = a/2$\n\n\\item $E[rank(C) - rank(f(C))] = a/2$\n\n\\end{itemize}\n\n\\end{thm}\n\\begin{proof}\n\nPick a distinguished input edge $e$, and let $A \\subseteq C$ be\nthe set of cliques in $C$ which include $e$, and $B \\subseteq C$ be\nthe set of cliques in $C$ which don't include $e$.\n\nWe define $f(C) = B$.\n\nNow, take the circuit computing $BUGGYCLIQUE(C)$, and\nfeed in a 0 to $e$. The resulting circuit computes\n$BUGGYCLIQUE(B)$. Note that $A$ could contain anywhere\nfrom 0 to $a$\ncliques, so since $C$ is chosen uniformly at random,\n$E[|A|] = a/2$.\n\nFurthermore, $|\\bigC(B)| \\le |\\bigC(C)|$, because feeding in\na zero to the circuit for $BUGGYCLIQUE(C)$ constructed a\n(possibly non-optimal) circuit for $BUGGYCLIQUE(B)$.\nThus, $rank(B) \\le rank(C)$.\nThis means that\n$C$ is one of $a$ functions chosen uniformly at random,\nand so, on average, there are $a/2$ circuits between $C$\nand $B$, implying at least that large a difference in rank.\n\n(Note that if $e$ ``misses'' all of the cliques in $C$, then $B = C$\nand $rank(B) = rank(C)$. However, provided $e$ is in some clique in $C$,\nthen since we're feeding in a 0 to a NAND-gate circuit,\nthe inequality $rank(B) < rank(C)$ is strict.)\n\n\\end{proof}\n\nIn terms of the previous warehouse analogy, we can imagine that,\nwhen an edge is disabled, the chip now has strictly fewer gates.\nPerhaps the chip-maker spent their entire budget on designing the\noptimal circuits, but then skimped on the wires connecting the chips?\nIn that case, assuming the missing wire is treated as a 0, the chip still\nfinds some cliques, and so might be sellable.\nBy construction of the warehouse, that means the faulty chip could be shifted,\non a giant conveyor belt,\ntoward the ``smaller'' (in number of gates) end of the warehouse.\nThat means that the corresponding set of cliques is also\nin the ``smaller'' end of the warehouse.\n\nAlthough both these numbers (the reduction in number of cliques, and\nthe reduction in rank) have an expected value of $a/2$,\nthey aren't necessarily the same, or even correlated.\n(Indeed, when $n$ is large, the number of cliques is sharply peaked around $a/2$,\nbut the lower bound in reduction in rank is uniformly distributed across $0..a$.)\n\n\\subsubsection{Why this doesn't bound $k-CLIQUE$?}\n\n\\ref{vaguelyUpward} seems to show that ``finding more cliques is\nat least a tiny bit harder on average.'' However, it only applies\non average.\n\nWhat happens if we try to use restrictions to bound $k-CLIQUE$?\nIf we start ``at the top'',\nfeeding in a 0 to $k-CLIQUE$, which is one function,\nwe get only one other function.\n\nHowever, it seems vaguely plausible to proceed ``bottom-up'',\nestimating the difficulty of finding increasingly large\nsets of cliques.\n\n\\subsection{Estimating the rank of functions, at each level}\n\nIt seems intuitive that finding a larger fraction of the cliques\nshould be harder.\nSo, another idea is to try to show this using induction on $l$.\nAs usual, the base case ($l=0$) is easy: finding zero cliques is easiest,\nso its rank is 0.\n\nThe step case is not so straightforward. Suppose we've computed bounds for\nlevels 0 through $l-1$ (thus, we're using some sort of strong induction.)\nIn other words, we've lower-bounded $E[rank(K_i)]$, for $i \\in [0,l-1]$.\nCan we lower-bound $E[rank(K_l)]$?\n\nSuppose we pick a set of cliques at level $l$, uniformly at random.\nIf we then pick an edge $e$ uniformly at random, and remove the cliques\nwhich contain $e$, then the number of cliques removed will vary\ndepending on the level. When $l=1$, we'll remove 0 or 1 cliques; when $l=N$,\nwe'll remove all ${n-2 \\choose k-2}$ cliques which contain $e$.\n\nWe can {\\em estimate} the number of cliques removed, by supposing that each\nclique is ``hit'' by $e$, with probability $p = {k \\choose 2} / {n \\choose 2}$.\nIf we assume that $e$ hitting each clique is an independent event, then\nrestricting a random set of $l$ cliques will hit $pl$ cliques, and\nleave $(1-p)l$ cliques.\nIn general, when $n>>k$, $p$ will be pretty small. Thus, the step case at\nlevel $l$ will depend on estimates for levels fairly ``high up'' near $l$.\nAlso, when $n$ is large, $N$ is way larger; so the distribution of\nthe fraction of cliques removed will be sharply peaked near $p$.\n\nHowever, this induction isn't quite correct.\nWe may have a bound on $E[rank(K_i)]$, for $i$ in $[0,l-1]$.\nBut when we pick a random set of cliques $K_l$, and feed in a 0, we definitely\naren't sampling uniformly from levels 0 through $l-1$, because the graphs we're\nsampling definitely lack one edge.\n\nThus, this is {\\em only an estimate} of the rank of functions at level $l$.\nTo me, this raises (at least) the following questions:\n\n\\begin{itemize}\n\n\\item If we compute the estimate, what do we get?\n\n\\item Is there a way to modify this, to get an actual bound?\n\n\\end{itemize}\n\nI only address the first question, and punt on the second.\n\n\\subsection{An estimated bound}\n\nThus, even though this is an estimate, we ``simulate'' this process, starting\nat level 0, and going up to level $N = {n \\choose k}$. \nFirst, define $p$, the estimated probability that a 0 ``hits'' a clique:\n\n\\[ p = \\frac{{k \\choose 2}}{{n \\choose 2}} \\]\n\nAssume that the number of cliques after feeding in a 0 follows a binomial\ndistribution, with $l$ ``trials'' and ``probability of success'' $1-p$.\n$B(l)$ will then be a weighted average of $B(0)$ through $B(l-1)$. However,\nif the 0 ``misses'' all the cliques, then we can't say anything about the\nrank, and just use 0.\nFurthermore, we know that at level $l$, we're adding ${N \\choose l}$ new\nfunctions.\n\n\nLet $R(l)$ be the lower bound on the expected rank of functions in $K_l$.\n\n\\[\nR(0) = 0\n\\]\n\nFor $0 < l \\le N$, define\n\\[\n\tR(l) = [ \\sum_{1\\le i < N} binom(i; l, 1-p) R(i) ] + {N \\choose l}/2\n\\]\n\nThe file {\\tt py/approxRank.py} computes this bound. Here are the bounds\nfor two small cases.\n\n\\includegraphics[width=0.45\\textwidth]{py/rank_n=6_k=3.pdf}\n\\includegraphics[width=0.45\\textwidth]{py/rank_n=11_k=4.pdf}\n\n\\section{Related work}\n\nThis strategy relies heavily on a modification of Shannon's original\nfunction-counting argument \\cite{shannon_synthesis_1949},\ncombined with random restrictions\n\\cite{subbotovskaya1963comparison} \\cite{hastad1987lower}.\n\nA related question is whether problems\n(such as $k$-SAT) are\nhard on average \\cite{bogdanov2006average}.\nThese efforts seem to focus more on whether\nrandom\ninstances of a given problem are hard, rather\nthan using random problems to show that\na specific problem is hard.\n\n??? What about:\n- co-NP?\n- formula size?\n\n\nThis lower-bound strategy also seems potentially\nrelevant to quantum computing,\nas the argument makes few restrictions on the sort of gates used.\nHowever, we did use the property of NAND gates that ``feeding in\na 0 disables a gate''; it's not clear whether that's needed,\nor holds for quantum gates.\n\n\\section{Conclusion}\n\nWe give a lower bound on finding {\\em some} set of cliques.\nIt is a modified form of Shannon's counting argument\n\\cite{shannon_synthesis_1949}, combined with random restrictions\n\\cite{subbotovskaya1963comparison} \\cite{hastad1987lower}.\nThis suggests {\\em approximate} bounds on functions {\\em similar} to $k$-CLIQUE.\nUnfortunately, however, this is an approximate bound,\nand so doesn't bound the complexity of $k$-CLIQUE.\n\nIf this were true, then by the contrapositive,\n\\cite{fenner1996complexity} would give a concise list of some of the consequences...\n\n\\section{Acknowledgements}\n\nThe author would like to thank William Gasarch for introducing him\nto lower bound strategies, and probabilistic proofs.\nHe would also like to thank the maintainers of\nseveral entertaining and relevant blogs, including but\nnot limited to: the Computational Complexity blog\n(Lance Fortnow and William Gasarch), \nG\\\"odel's Lost Letter (Richard Lipton and Ken Regan),\nand Shtetl-Optimized (Scott Aaronson). \n\n\n\\appendix\n\\section{Calculations for unbounded fan-in NAND gates}\n\\label{gateMath}\n\nWe used unbounded fan-in NAND gates largely because they are\nconvenient to convert from ``number of functions'' to\n``number of gates''.\n\nLet $\\bigC(f)$ be the circuit\nwith the fewest unbounded fan-in NAND\ngates computing $f$, and let $|\\bigC(f)| = g$ be the number of gates in\nthat circuit. The\ncircuit could have at most $gm$ wires from inputs to gates, and ${g \\choose 2}$\nwires from gates to gates. We can view the possible circuits as a bitmask,\ncontaining a 1 everywhere a gate is connected to an input (or another gate),\nand 0 everywhere else.\n\n\\begin{thm}\n\\label{boundFromCounting}\nConsider functions from $m$ bits to one bit of output.\nThis means that, with $g$ gates, we can represent at most\n$2^{gm + {g \\choose 2}}$ different boolean functions (with $m$ bits of input,\nand one bit of output).\n\\end{thm}\n\\begin{proof}\n\nThe number of possible wires which are there, or not, is $gm + {g \\choose 2}$,\nwhich bounds how many possible circuits there are.\nSome of these circuits compute the same function.\nHowever, there can't be any more than this many circuits with this many wires.\n\\end{proof}\n\nThis means that if we have a large set of functions, and we know the size of\nthe set of functions, then we know that at least {\\em one} of them requires\na large number of gates. (Knowing {\\em which} function requires a lot, or many,\ngates is still an issue).\n\nWe used unbounded-fan-in NAND gates because, given a number of circuits,\nit's easier (I think!) to solve for the number of wires than if we had\nused two-input gates.\nConsider functions from $m$ bits to one bit of output.\nLet $g$ be the number of gates, and $w$ be the number of wires.\nSolving for the number of gates:\n\n\\begin{eqnarray*}\nw & = & mg + {g \\choose 2} \\\\\n & = & mg + g(g-1)/2 \\\\\n & = & mg + (g^2 - g) / 2 \\\\\n & = & 0.5g^2 + (m-0.5)g \\\\\n0 & = & 0.5g^2 + (m-0.5)g - w \\\\\n\\end{eqnarray*}\n\nWe solve the quadratic formula (writing $b = m-0.5$ for simplicity), keeping\nonly the non-imaginary root.\n\n\\begin{eqnarray*}\ng & = & -b \\pm \\sqrt{ b^2 + 2w} \\\\\n & = & {\\sqrt {2w + b^2}} - b \\\\\n\\end{eqnarray*}\n\nThus, given a set of functions, we know that at least one of them requires\nthis many gates.\n\nThis expression is a bit inconvenient, and as $n$ increases, we\nexpect the number of gates to be much larger than the number of\ninputs $b$. To avoid such issues, we can count the number of gates\nbeyond the number of inputs. This gives the bound\n\n\\begin{eqnarray*}\ng & \\ge & {\\sqrt 2w} - m \\\\\n\\end{eqnarray*}\n\n\n\\bibliography{references}\n\\bibliographystyle{unsrt}\n\n\\end{document}\n\n", "meta": {"hexsha": "b2cf995e36a738068a37c0058a518ffaab66f738", "size": 22459, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "countingBound/countingBound.tex", "max_stars_repo_name": "joshtburdick/misc", "max_stars_repo_head_hexsha": "7bb103b4f9d850e3279eb675c6df420aa7b8da22", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "countingBound/countingBound.tex", "max_issues_repo_name": "joshtburdick/misc", "max_issues_repo_head_hexsha": "7bb103b4f9d850e3279eb675c6df420aa7b8da22", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "countingBound/countingBound.tex", "max_forks_repo_name": "joshtburdick/misc", "max_forks_repo_head_hexsha": "7bb103b4f9d850e3279eb675c6df420aa7b8da22", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.9391447368, "max_line_length": 86, "alphanum_fraction": 0.7382341155, "num_tokens": 6413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581049086031, "lm_q2_score": 0.8152324848629215, "lm_q1q2_score": 0.6486463339659636}} {"text": "\n%*******************************************************************************\n%*********************************** Second Chapter ***************************\n%*******************************************************************************\n%!TEX root = 0.main.tex\n\n\\section{Non uniform sampling schemes and the FEM Laplacian}\n\nIn Chapter 3, we used the fact that when the sampling scheme of the sphere is regular enough, the graph $G$ is such that the corresponding graph Laplacian $\\mathbf {L=V\\Lambda V}^\\intercal$ has eigenvectors that are close enough to the ones of $\\Delta_{\\mathbb S^2}$ to design a graph with a low mean equivariance error. We showed a way to construct a graph $G'$ such that its graph Laplacian well approximates $\\Delta_{\\mathbb S^2}$ in the case of an equiarea sampling scheme of the sphere and we tested it on the HEALPix sampling scheme. In this Chapter we focus on sampling schemes that are less uniform than HEALPix. The sampling scheme $V=\\{v_i\\in \\mathbb S^2\\}$ that we will use for our study, very used in applications, is the so called \\textit{equiangular sampling scheme} \\cite{Driscoll:1994:CFT:184069.184073}. \nThis chapter is organized as follows: In section \\ref{sec:Chapter3: Heat Kernel Graph Laplacian on the Equiangular Sampling} we first introduce the equiangular sampling, and then we present the results that we obtained with two different Graph Laplacians: the HKGL, and a graph proposed by Khasanova et al. \\cite{Frossard2017GraphBasedCO}, specifically designed for this sampling scheme. In section \\ref{sec:Chapter3: Using the Finite Element Method to approximate the Laplace-Beltrami operator on a manifold} we deepen how to use the Finite Element Method (FEM) to construct a discrete approximation of $\\Delta _{\\mathbb S^2}$ and how to derive a graph-like Laplacian from it, that shows a low equivariance error.\n\\subsection{Graph Laplacian on the Equiangular Sampling}\n\\label{sec:Chapter3: Heat Kernel Graph Laplacian on the Equiangular Sampling}\n\n\\subsubsection{The Equiangular Sampling}\n\nGiven the usual parametrization $x = x(\\theta, \\phi)$ of the sphere\n\\begin{align*}\n\t\\mathbb{S}^{2}&=\\left\\{x=\\left(x_{1}, x_{2}, x_{3}\\right) \\in \\mathbb{R}^{3} :\\|x\\|_{\\mathbb{R}^{3}}=\\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\right)^{1 / 2}=1\\right\\}\\\\\n\tx_{1}&=\\cos (\\phi) \\sin (\\theta), \\quad x_{2}=\\sin (\\phi) \\sin (\\theta), \\quad x_{3}=\\cos (\\theta)\n\\end{align*}\n\nLet $m\\in\\mathbb N$, the \\textit{equiangular sampling} of bandwidth $b=2^m$ is given by \n$\nx_{j k}^{(b)}=x\\left(\\theta_{j}^{(b)}, \\phi_{k}^{(b)}\\right)\n$\nwhere\n\\begin{align*}\n\t\\theta_{j}^{(b)} &:=\\pi \\frac{j}{2 b}, \\quad \\phi_{k}^{(b)} :=2 \\pi \\frac{k}{2 b}\\\\\n\tj&=0, ..., 2b-1 \\text{ and }k=0, ..., 2b-1 \n\\end{align*}\n\n\\begin{figure}[h]\n\t\\centering\n\t\\includegraphics[width=0.5\\textwidth]{figs/Chapter3/equiangular.png}\n\t\\caption{\\label{fig:equiangular sampling}Equiangular sampling with bandwidth $b=8$}\n\\end{figure}\nOne has thus $n=4b^2$ points on the sphere, where all the points $x_{0 k}^{(b)}$ correspond to the north pole for every $k=0, ..., 2b-1$. Notice also that the south pole is never sampled. In figure \\ref{fig:equiangular sampling} it can also be appreciated how the area close to the poles is much more sampled that the equator. One reason for which this sampling scheme is very used in application is the existence of the following result from \\cite{Driscoll:1994:CFT:184069.184073}, that states that any band limited function can be exactly recovered from its sampled values $f\\left(x_{j k}^{(b)}\\right)$:\n\\vspace{0.5cm}\n\\begin{theorem}\\label{theo:equiangular sampling theorem}\n\tLet \\(l_{0} \\in \\mathbb{N}\\) and \\(m_{0} \\in \\mathbb{Z},\\left|m_{0}\\right| \\leq l_{0} .\\) If \\(f=\\sum_{l=0}^{b-1} \\sum_{m=-l}^{l} \\widehat{f}(l, m) Y_{l}^{m}\\)\n\tthen\n\t\n\t$$\n\t\\begin{aligned} \\widehat{f}\\left(l_{0}, m_{0}\\right)=& \\frac{1}{4 b^{2}} \\sum_{j=0}^{2 b-1} \\sum_{k=0}^{2 b-1} f\\left(x_{j k}^{(b)}\\right) \\overline{Y_{l_{0}}^{m_{0}}\\left(x_{j k}^{(b)}\\right)} \\sin \\left(\\theta_{j}^{(b)}\\right) \\times \\\\ & \\times \\frac{4}{\\pi} \\sum_{l=0}^{b-1} \\frac{1}{2 l+1} \\sin \\left((2 l+1) \\theta_{j}^{(b)}\\right) \\end{aligned}\n\t$$\n\\end{theorem}\n\\vspace{0.5cm}\nTheorem \\ref{theo:equiangular sampling theorem} is the equivalent on the sphere of the well known Shannon's sampling theorem, that states the minimum sampling frequency at which a band limited signal $f:\\mathbb R \\to \\mathbb R$ can be perfectly reconstructed, and is a precious tool when doing signal processing on the sphere.\n\n\\subsubsection{Heat Kernel Graph Laplacian on the equiangular sampling scheme}\nThanks to theorem \\ref{theo:equiangular sampling theorem} not only we have a characterization of the space \\\\$F=\\{f=\\sum_{l=0}^{b-1} \\sum_{m=-l}^{l} \\widehat{f}(l, m) Y_{l}^{m}\\}\\subset L^2(\\mathbb S^2)$ of band limited functions on which the sampling operator $T: F\\to \\mathbb R^n$ is invertible, but we also an analytic expression for $T^{-1}$. Thanks to this we can calculate up to machine precision the equivariance error for any function $f\\in F$ and any rotation $g\\in SO(3)$. In figure \\ref{fig:Equivariance error of the full HKGL} we calculated the mean equivariance error of the full HKGL averaging on mono-frequency signals for different bandwidths. We can observe a much cleaner behavior of the mean equivariance error than the one we observed for he HEALPix sampling, most probably due to the fact that this time we can exactly calculate $T^{-1}$. The error increases linearly with the frequency of the signal for all bandwidths, and it seems to converge to zero as $b$ increases. However, the spectrum of the graph Laplacian does not show the eigenvalues grouped in the expected pattern.\n\\begin{table}[h!]\n\t\\centering\n\t\\begin{tabular}{ c|c } \n$b$ & $t$ \\\\ \n\t\\hline\n4 & 0.5 \\\\ \n8 & 0.3 \\\\ \n16 & 0.1 \\\\ \n\t\\end{tabular}\n\t\\caption{\\label{table:equiangular kernel width}Kernel width $t$ used to construct the HKGL for each bandwidth $b$}\n\\end{table}\n\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{../codes/06.Equivariance_error/FullHKGLonequiangularsampling.png}\n\t\\includegraphics[width=\\textwidth]{../codes/02.HeatKernelGraphLaplacian/equiangular/equi_full_eigenvalues_16.png}\n\t\\caption{\\label{fig:Equivariance error of the full HKGL}Equivariance error of the full HKGL on the equiangular sampling by spherical harmonic degree $\\ell$, and its spectrum for the bandwidth $b=8$ .}\n\\end{figure}\n\n\\subsubsection{A graph alternative to the HKGL for the equiangular sampling}\\label{sec:Chapter3:Frossard}\n\nKhasanova et al. \\cite{Frossard2017GraphBasedCO} designed a discrete Laplacian that is explicitly intended to work on the sphere with the equiangular sampling. They studied a way to build a graph to analyze images produced by omnidirectional cameras. In their work they assume that the image is sampled on the sphere on the equiangular sampling. They consider the set $\\mathcal G$ of all the possible graphs where each node is connected only to four of its nearest neighbours (North, Sud, West, East) and propose a weighting scheme $w_{ij}$ that minimizes the difference in the response to the polynomial spectral filter $\\mathcal F = \\mathbf L$ evaluated on images of the same object seen at different latitudes. In other words, they solve the minimization problem\n\n\\begin{equation}\\label{eq:minimization frossard}\n\t\\min_{W\\in\\mathcal G} \\left|\\mathcal{F}\\left(\\mathbf{y}\\left(v_{ e}\\right)\\right)-\\mathcal{F}\\left(\\mathbf{y}\\left(v_{ i}\\right)\\right)\\right|\n\\end{equation}\n\\begin{figure}\n\t\\begin{center}\n\t\t\\includegraphics[width=0.5\\textwidth]{figs/Chapter3/frossard2.png}\n\t\\end{center}\n\t\\caption{\\label{fig:frossard2}Khasanova et al. setting.}\n\\end{figure}\nfor the adjacency matrix $W$, where $\\mathbf y(v_i)$ is the image of the object on the sphere centered on the vertex $v_i$, and $\\mathcal F (\\mathbf y(v_e))$ is the response of the filter at the vertex $v_e$ that lies at the same longitude of the vertex $v_i$ but on the equator (figure \\ref{fig:frossard2}). In their work they prove that the optimal weights solving the minimization problem (\\ref{eq:minimization frossard}) are given by weights $w_{ij}$ inversely proportional to the Euclidean distance between vertices:\n\\begin{equation}\\label{eq:frossard weights}\n\tw_{ij} = \\frac{1}{\\norm{x_i-x_j}}\n\\end{equation}\nThis construction is interesting since it is adapted to the equiangular sampling, and leads to a very sparse graph with only 4 neighbors per vertex. Furthermore, to obtain the weights (\\ref{eq:frossard weights}) every calculation was done in the \\textit{spatial domain}, without any consideration about the spectral interpretation of the filter. In order to compare it to the HKGL we show the equivariance error by spherical harmonic degree in figure \\ref{fig:Equivariance error of the Frossard-Khasanove graph}. It can be appreciated how this construction performs a little worse that the full HKGL for low bandwidth samplings, but converges faster than the \\textit{full} HKGL, and its spectrum looks much more similar to the one of $\\Delta_{\\mathbb S^2}$.\n\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{../codes/06.Equivariance_error/KhasanovaFrossardgraphonequiangularsampling.png}\n\t\\includegraphics[width=\\textwidth]{../codes/02.HeatKernelGraphLaplacian/equiangular/equi_full_Khasanova_Frossard_eigenvalues_16.png}\n\t\\caption{\\label{fig:Equivariance error of the Frossard-Khasanove graph}Equivariance error of the Khasanova-Frossard graph on the equiangular sampling with bandwidth $b=8$ by spherical harmonic degree $\\ell$, and its spectrum.}\n\\end{figure}\n\nThe Khasanova-Frossard graph showed us that it is possible to do better than the HKGL on the equiangular sampling. We asked ourself if there's a general way of building a more equivariant graph than the HKGL that is straightforward to implement on any sampling scheme of the sphere. To answer this question in the next section we study a more complex way to approximate the Laplace-Beltrami operator $\\Delta_{\\mathbb S^2}$: the linear FEM Laplacian. We will see that the comparison between the FEM Laplacian and the graph Laplacian will give us precious insights to better understand the limitations of graph Laplacians when it comes to deal with non uniform samplings of the sphere.\n\\subsection{The Finite Element Method approximation of the Laplace-Beltrami operator on the sphere}\\label{sec:Chapter3: Using the Finite Element Method to approximate the Laplace-Beltrami operator on a manifold}\n\nThe Finite Element Method (FEM) is a numerical algorithm that allows to calculate a discrete approximation of the solution of the Laplace-Beltrami eigenvalue problem through a functional discretization of the differential operator $\\Delta_{\\mathbb S^2}$. We put the necessary definitions and mathematical concepts necessary to properly introduce the weak formulation of a differential problem, the Galerkin method and finally the Finite Element Method in Appendix. For a detailed introduction to the FEM, we refer the reader to \\cite{Quarteroni:1639539}\\\\\nLet's transform the strong form of the eigenvalue problem on the Sphere on its \\textit{weak} formulation. Let's multiply equation (\\ref{eq:continous eigenvalue problem}) by a sufficiently regular function $v$ and integrate on $\\mathbb S^2$. Since the sphere is a closed manifold and has no border, integrating by parts yields\n\\begin{equation}\\label{eq:weak eigenvalue problem}\n\\begin{split}\n&\\text{Find } f\\in H^1(\\mathbb S^2), \\lambda\\in\\mathbb R\\text{ such that }\\\\ \n&\\int_{\\mathbb S^2} \\nabla f(\\mathbf x)\\cdot\\nabla v(\\mathbf x) d\\mathbf x = \\lambda \\int_{\\mathbb S^2} f(\\mathbf x)\\cdot v(\\mathbf x)d\\mathbf x\\quad \\forall v\\in H^1(\\mathbb S^2)\n\\end{split}\n\\end{equation}\nwhere $v$ has been chosen to belong to $H^1\\subset L^2(\\mathbb S^2)$, the Sobolev space of all the functions with derivative $\\nabla f\\in L^2(\\mathbb S^2)$. The Finite Element Method first approximates the domain $\\Omega$ with a triangulation $\\mathcal T_h=\\{ \\tau_i \\}$, and then \\textit{projects} the weak problem (\\ref{eq:weak eigenvalue problem}) into a finite dimensional subspace $V_h\\subset H^1(\\mathcal T_h)$ made by all the piecewise linear polynomials on the triangulation $\\mathcal T_h$. Being the sphere convex, the triangulation $\\mathcal T_h$ has been obtained by calculating the triangulation of the convex hull of the vertices of the chosen sampling scheme through the Qhull algorithm \\cite{Barber96thequickhull}. Now, define $X_h^1$ to be the space of all the continuous, piecewise linear functions on $\\Omega$\n$$X_h^{1}=\\{v_h: v_h\\in C(\\Omega): \\left.v_h\\right|_{\\tau}\\in\\mathbb P^1\\ \\forall \\tau\\in \\mathcal T_h\\}$$ \nand set $V_h=X_h^1$. Since for the functions in $X_h^1$ the number of degrees of freedom is the same of the number of vertices of the mesh $n$, we need $n$ basis functions $\\phi_i, i=0,...,n-1$ to fully describe $X_h^{1}$. $\\phi_i$ is defined as the continuous piecewise linear function such that \n$$\n\\phi_i(x_j) = \\delta_{ij}\\quad i=0,...,n-1\n$$\nwhere $x_i$ are the points of the sampling scheme that have been used as vertices of the triangles $\\tau$ of the triangulation $\\mathcal T_h$. The support of $\\phi_i$ i.e., the subset of $\\mathcal T_h$ where $\\phi_i$ is not zero, is made by all the triangles sharing the $i$th vertex. An example is shown in figure \\ref{fig:basis function}.\n\n\\vspace{0.5cm}\n\\begin{remark}\n\tFor a function $v_h \\in X^1_h,\\ v_h = v_0 \\phi_0 +... v_n \\phi_n$ the coefficient $v_i$ is equal to the function $v_h$ evaluated in the $i$th vertex \n\t\\begin{equation}\\label{eq:dof and values}\n\tv_i = v_h( x_i)\n\t\\end{equation}\n\\end{remark}\\vspace{0.5cm}\n\\begin{figure}[h]\n\t\\begin{center}\n\t\t\\includegraphics[width=0.6\\textwidth]{figs/Chapter3/sphere_mesh.png}\n\t\\end{center}\n\t\\caption{\\label{fig:sphere mesh}A triangulation $\\mathcal T_h$ of the sphere made with the vertices of the HEALPix sampling scheme with $N_{side}=8$.}\n\\end{figure}\n\\begin{figure}\n\t\\begin{center}\n\t\t\\includegraphics[width=0.4\\textwidth]{figs/Chapter3/basisfunction.png}\n\t\\end{center}\n\t\\caption{\\label{fig:basis function}Basis function of the space $X^{1}_h$}\n\\end{figure}\n\nBy writing $n$ times the equation (\\ref{eq:weak eigenvalue problem}), setting each time the test function $v_h$ equal to the $i$th basis function $\\phi_i$ of the space $X_h^1$, we obtain the generalized algebraic eigenvalue problem\n\\begin{equation}\\label{eq:algebraic generalized eigenvalue problem}\n\\begin{aligned}\n&\\text{Find }(f,\\lambda)\\text{ such that }\\mathbf A\\mathbf f = \\lambda\\mathbf B \\mathbf f\\\\\n&\\begin{cases}\n(\\mathbf A)_{ij} &= \\int_{\\mathbb S^2}\\nabla \\phi_i(\\mathbf{x})\\cdot \\nabla \\phi_j(\\mathbf{x})d\\mathbf{x}\\\\\n(\\mathbf B)_{ij} &= \\int_{\\mathbb S^2} \\phi_i(\\mathbf{x}) \\phi_j(\\mathbf{x})d\\mathbf{x}\\\\\n(\\mathbf f)_i &= f_i:\\quad f(\\mathbf x) = f_0\\phi_0(\\mathbf x)+ ... + f_{n-1}\\phi_{n-1}(\\mathbf x) \n\\end{cases}\n\\end{aligned}\n\\end{equation}\n$\\mathbf A$ is called the \\textit{stiffness} matrix, and $\\mathbf B$ is called the \\textit{mass} matrix. Observe that being the Laplace-Beltrami operator self-adjoint, we have that $\\mathbf A=\\mathbf A^\\intercal,$ $\\mathbf B=\\mathbf B^\\intercal$. Being $\\mathbf B$ non singular, the system (\\ref{eq:algebraic generalized eigenvalue problem}) is equivalent to the eigenvalue problem\n\\begin{equation}\\label{eq:algebraic eigenvalue problem}\n\\mathbf B^{-1}\\mathbf A\\mathbf f = \\lambda \\mathbf f\n\\end{equation}\n\nIt can be shown \\cite{Strang} that even though the matrix $\\mathbf B^{-1}\\mathbf A$ is not symmetric, its eigenvalues are still real and its eigenvectors are such that $\\mathbf V\\mathbf B\\mathbf V^\\intercal=\\mathbf I$, where $\\mathbf I$ is the identity matrix. The solution $\\mathbf f$ that is the vector of the coefficients of the function $f_h$ in the basis $\\phi_i$ corresponds exactly to the values of the function $f_h$ in the vertices. \n\n\\subsection{How to filter a signal with the linear FEM}\nCalculating the discrete Fourier transform with the linear FEM means projecting the Fourier transform into the subspace $V_h$:\n\n\\begin{equation}\\label{eq:FEM fourier}\n\t\\hat f_{FEM} (\\ell, m) = \\int_{\\eta \\in \\mathcal T_h} f_h(\\eta) v_{i(\\ell, m)}(\\eta) d\\eta = \\mathbf v_{i(\\ell, m)}^\\intercal \\mathbf B \\mathbf f\n\\end{equation}\n\nwhere $ v_{i(\\ell, m)}$ is the solution to the eigenvalue problem (\\ref{eq:algebraic generalized eigenvalue problem}), $f_h$ is the projection of $f$ on $V_h$, and $\\mathbf B$ is the mass matrix. It follows that the filtering of a discretized signal $\\mathbf f$ in the spectral domain is done by the following matrix $\\mathbf \\Omega_K^{FEM}$:\n\n\\begin{equation}\\label{eq:FEM filtering}\n\t \\mathbf \\Omega_K^{FEM} := (\\mathbf B\\mathbf V^\\intercal)^{-1}\\mathbf K\\mathbf V^\\intercal\\mathbf B\n\\end{equation}\n\nwhere $\\mathbf V^\\intercal\\mathbf B$ is the FEM Fourier transform matrix, $\\mathbf K$ is the diagonal matrix that represent the chosen kernel for this filter, and $(\\mathbf V^\\intercal\\mathbf B)^{-1}$ is the matrix representing the inverse FEM Fourier transform. We can already notice a fundamental difference between this way of filtering a signal and the way of filtering a signal with a graph. The FEM filtering uses not only the eigenvectors of the FEM Laplacian, but also the mass matrix $\\mathbf B$. We computed the usual mean equivariance error per spherical harmonic degree for the diffusion filter $\\mathbf K=\\exp(-\\mathbf \\Lambda)$ on the equiangular sampling, and we show it in figure \\ref{fig:FEM diffusion}.\n\\begin{figure}[h]\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{../codes/06.Equivariance_error/FEMdiffusiononequiangularsampling.png}\n\t\\caption{\\label{fig:FEM diffusion}Equivariance error of the linear FEM diffusion on the equiangular sampling by spherical harmonic degree.}\n\\end{figure}\nWe can see that the FEM filtering works better than the state-of-the-art graph of Khasanova-Frossard, but the matrix $\\mathbf \\Omega_K^{FEM}$ is full, and its computation requires the expensive inversion of the Fourier matrix $\\mathbf V^\\intercal\\mathbf B$, making it not efficient if used in a CNN. \n\\subsubsection{A confront between FEM filtering and graph filtering}\\label{sec:FEM filtering as a graph filtering}\nDue to the fact that \n$\n\\mathbf V^\\intercal\\mathbf B\\mathbf V = \\mathbf I\n$,\nwe have that \n$$\n\\begin{aligned}\n\t(\\mathbf V^\\intercal\\mathbf B)^{-1} &= \\mathbf V,\\\\\n\t\\mathbf V^\\intercal\\mathbf B &= \\mathbf V^{-1}.\n\\end{aligned}\n$$\nSo, the FEM filter matrix $\\mathbf \\Omega_K^{FEM}$ can be rewritten as \n\\begin{equation}\\label{eq:FEM filter simple}\n\t \\mathbf \\Omega_K^{FEM} = \\mathbf V \\mathbf K \\mathbf V^{-1}\n\\end{equation}\nThus a polynomial filter\n$$\nk(\\lambda)=P_\\kappa(\\lambda) = \\sum_{k=0}^{\\kappa-1} \\theta_k \\lambda^k\n$$\nwould be implemented with a matrix\n$$\n\\mathbf\\Omega_K^{FEM} = \\mathbf V \\left(\\sum_{k=0}^{\\kappa-1} \\theta_k \\mathbf \\Lambda^k \\right)\\mathbf V^{-1} = \\sum_{k=0}^{\\kappa-1} \\theta_k (\\mathbf B^{-1}\\mathbf A)^k = P_\\kappa(\\mathbf B^{-1}\\mathbf A).\n$$\n\nIt is interesting to notice that FEM filtering (\\ref{eq:FEM filter simple}) looks very much like graph filtering (\\ref{eq:graph convolution}) but, given that $V$ is not orthogonal anymore, it has to replace the $\\mathbf V^\\intercal$ with $\\mathbf V^{-1}$. Equation \\ref{eq:FEM filter simple} implies that the FEM filtering could be implemented exactly as the graph filtering already implemented in DeepSphere, using the FEM matrix $\\mathbf B^{-1}\\mathbf A$ instead of the usual symmetric HKGL $\\mathbf L_n^t$. Unfortunately, in order to make the evaluation of the polynomial $P_\\kappa(\\mathbf B^{-1}\\mathbf A)$ efficient, we need the matrix $\\mathbf B^{-1}\\mathbf A$ to be sparse. The only way to do so is to have a \\textit{sparse stiffness matrix} $\\mathbf A$ and a \\textit{diagonal mass matrix} $\\mathbf B$. Unfortunately there is no likely way to have both these conditions satisfied at the same time, since this is one of the most well known trade offs of the FEM \\cite{Strang}. To make $\\mathbf B$ diagonal we need to choose $\\phi_i$ to be an orthogonal basis of the finite dimensional functional space $V_h$. To do so, we will need to choose more complicated basis functions than the usual ones in figure \\ref{fig:basis function}, \\textit{most likely with a support that extends to the whole sphere}, making the stiffness matrix $\\mathbf A$ not sparse anymore. A common workaround \\cite{Strang} is the so called \\textit{lumping} of the mass matrix, that consists in replacing the matrix $\\mathbf B$ with the \\textit{lumped} diagonal matrix $\\mathbf D$ obtained by placing in each diagonal entry $(\\mathbf D)_{ii}$ the sum of the elements of the $i$th row of the mass matrix $\\mathbf B$:\n\n\\begin{equation}\\label{eq:lumping}\n\\mathbf D = \\text{diag}\\{d_i\\},\\quad d_i = \\sum_j (\\mathbf B)_{ij}\n\\end{equation}\nThis approximation is well studied in the literature and it is proved to work well in many practical cases \\cite{Quarteroni:1639539}. In this way the FEM matrix $\\mathbf B^{-1}\\mathbf A$ would be approximated by the matrix\n$$\n\\mathbf D^{-1}\\mathbf A.\n$$\nThis matrix has the big advantage of having the same sparsity pattern of the stiffness matrix $\\mathbf A$, making it very efficient for polynomial filtering. We can take one step further by using the symmetric matrix\n$$\n\\mathbf D^{-1/2}\\mathbf A\\mathbf D^{-1/2}.\n$$\nTo confront FEM and HKGL filtering we filter a unit mass signal \\\\$\\mathbf{f}=(0,0,...0,1,0,...,0)^\\intercal$ with the \\textit{diffusion} filter $\\exp{(-\\tau \\mathbf L)}$. It was chosen a very irregular sampling scheme, shown in figure \\ref{fig:FEM lumped symmetric diffusion on irregular sampling}. It can be seen how the full HKGL compresses the signal around the equator where the sampling scheme is more sparse; on the other hand the FEM filtering manages to keep the diffusion homogeneous no matter the asymmetry in the sampling scheme. \nThe mean equivariance error using by spherical harmonic degree can be found in figure \\ref{fig:Equivariance error of the lumped FEM Laplacian}. It can be seen how $\\mathbf D^{-1}\\mathbf A$ ha almost the same equivariance error of the full FEM Laplacian $\\mathbf B^{-1}\\mathbf A$, while using the symmetric matrix $\\mathbf D^{-1/2}\\mathbf A \\mathbf D^{-1/2}$ makes the equivariance error much worse, especially at lower frequencies.\n\\clearpage\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{figs/Chapter3/diffusion.png}\n\t\\caption{\\label{fig:FEM lumped symmetric diffusion on irregular sampling}Symmetric lumped linear FEM diffusion and HKGL diffusion on an irregular sampling scheme of the sphere. The position of the filtered source signal is indicated with a black star.}\n\\end{figure}\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{../codes/06.Equivariance_error/LumpedFEMLaplacianonequiangularsampling.png}\n\t\\includegraphics[width=\\textwidth]{../codes/06.Equivariance_error/SymmetriclumpedFEMLaplacianonequiangularsampling.png}\n\t\\caption{\\label{fig:Equivariance error of the lumped FEM Laplacian}Equivariance error of the lumped FEM Laplacian and of the symmetric lumped FEM Laplacian on the equiangular sampling by spherical harmonic degree $\\ell$.}\n\\end{figure}\n\\clearpage\n\\paragraph{Accuracy of the linear FEM spherical harmonics.} Thanks to the sampling theorem \\ref{theo:equiangular sampling theorem}, we are able to compute the exact SHT (under the hypothesis of band limited signals) of the solutions to the eigenvalue problem (\\ref{eq:algebraic generalized eigenvalue problem}) and we show in figures \\ref{fig:FEMHealpix}, \\ref{fig:FEMequiangular} the power spectrum of the eigenmodes for both HEALPix sampling and the equiangular sampling, as a measure of the goodness of the linear FEM approach to approximate the spherical harmonics.\n\n\\begin{figure}[h!]\t\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/HEALPix/img/linearFEM.png}\\\\\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/HEALPix/img/linearFEM_diagonal.png}\t\\\\\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/HEALPix/img/FEM_eigenvalues_16.png}\t\\\\\n\t\\caption{\t\\label{fig:FEMHealpix}Alignment of eigenspaces of the linear FEM Laplacian on HEALPix, and its spectrum for $N_{side}=16$}\n\\end{figure}\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/equiangular/normal/img/linearFEM.png}\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/equiangular/normal/img/linearFEM_diagonal.png}\t\n\t\\includegraphics[width=\\linewidth]{../codes/03.FEM_laplacian/equiangular/normal/img/FEM_eigenvalues_16.png}\t\n\t\\caption{\\label{fig:FEMequiangular}Alignment of eigenspaces of the linear FEM Laplacian on the equiangular sampling, and its spectrum for $bw=32$}\n\\end{figure}\n\n\\clearpage\n\n\n\n\n\n\n", "meta": {"hexsha": "2e357dc633541e384cf98af6500f120f14de7273", "size": 24663, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "PDF/3.Chapter2.tex", "max_stars_repo_name": "MartMilani/PDM", "max_stars_repo_head_hexsha": "cca07a8485c6933361536286279ae6c7e14d7fa1", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "PDF/3.Chapter2.tex", "max_issues_repo_name": "MartMilani/PDM", "max_issues_repo_head_hexsha": "cca07a8485c6933361536286279ae6c7e14d7fa1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "PDF/3.Chapter2.tex", "max_forks_repo_name": "MartMilani/PDM", "max_forks_repo_head_hexsha": "cca07a8485c6933361536286279ae6c7e14d7fa1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 98.2589641434, "max_line_length": 1692, "alphanum_fraction": 0.7469894173, "num_tokens": 7167, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.795658104908603, "lm_q2_score": 0.8152324803738429, "lm_q1q2_score": 0.6486463303941917}} {"text": "\\subsubsection{Short Range Cusp Form}\n\\label{sec:onebodyjastrowsrcusp}\n\nThe idea behind this functor is to encode nuclear cusps and other details at very\nshort range around a nucleus in the region that the Gaussian orbitals of quantum\nchemistry are not capable of describing correctly.\nThe functor is kept short ranged, because outside this small region, quantum chemistry\norbital expansions are already capable of taking on the correct shapes.\nUnlike a pre-computed cusp correction, this optimizable functor can respond to\nchanges in the wave function during VMC optimization.\nThe functor's form is\n\\begin{equation}\n\\label{srcuspform}\nu(r) = -\\exp{\\left(-r/R_0\\right)} \\left( A R_0 + \\sum_{k=0}^{N-1} B_k \\frac{ (r/R_0)^{k+2} }{ 1 + (r/R_0)^{k+2} } \\right)\n\\end{equation}\nin which $R_0$ acts as a soft cutoff radius ($u(r)$ decays to zero quickly beyond roughly this distance)\nand $A$ determines the cusp condition.\n\\begin{equation}\n\\label{srcusplimit}\n\\lim_{r \\to 0} \\frac{\\partial u}{\\partial r} = A\n\\end{equation}\nThe simple exponential decay is modified by the $N$ coefficients $B_k$ that define\nan expansion in sigmoidal functions, thus adding detailed structure in a short-ranged\nregion around a nucleus while maintaining the correct cusp condition at the nucleus.\nNote that sigmoidal functions are used instead of, say, a bare polynomial expansion, as they\ntrend to unity past the soft cutoff radius and so interfere less with the exponential decay\nthat keeps the functor short ranged.\nAlthough $A$, $R_0$, and the $B_k$ coefficients can all be optimized as variational\nparameters, $A$ will typically be fixed as the desired cusp condition is known.\n\nTo specify this one-body Jastrow factor, use an input section like the following.\n\n\\begin{lstlisting}[style=QMCPXML]\n\n \n 0.06 \n \n 0 0 0 0 0 0 0 0 0 0\n \n \n \n 0.2 \n \n 0 0 0 0 0 0 0 0 0 0\n \n \n\n\\end{lstlisting}\n\nHere ``rcut'' is specified as the range beyond which the functor is assumed to be zero.\nThe value of $A$ can either be specified via the ``cusp'' option as shown above, in\nwhich case its optimization is disabled, or through its own ``var'' line as\nfor $R_0$, in which case it can be specified as either optimizable (``yes'')\nor not (``no'').\nThe coefficients $B_k$ are specified via the ``coefficients'' section,\nwith the length $N$ of the expansion determined automatically based on the length\nof the array.\n\nNote that this one-body Jastrow form can (and probably should) be used in conjunction\nwith a longer ranged one-body Jastrow, such as a spline form.\nBe sure to set the longer-ranged Jastrow to be cusp-free!\n", "meta": {"hexsha": "fc4b39aa666346cc06f36365dc80dd74889c03c4", "size": 3080, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "legacy_manual/jastrow_one_body_srcusp.tex", "max_stars_repo_name": "djstaros/qmcpack", "max_stars_repo_head_hexsha": "280f67e638bae280448b47fa618f05b848c530d2", "max_stars_repo_licenses": ["NCSA"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "legacy_manual/jastrow_one_body_srcusp.tex", "max_issues_repo_name": "djstaros/qmcpack", "max_issues_repo_head_hexsha": "280f67e638bae280448b47fa618f05b848c530d2", "max_issues_repo_licenses": ["NCSA"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "legacy_manual/jastrow_one_body_srcusp.tex", "max_forks_repo_name": "djstaros/qmcpack", "max_forks_repo_head_hexsha": "280f67e638bae280448b47fa618f05b848c530d2", "max_forks_repo_licenses": ["NCSA"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.6774193548, "max_line_length": 121, "alphanum_fraction": 0.7454545455, "num_tokens": 877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511616741042, "lm_q2_score": 0.7577943658046609, "lm_q1q2_score": 0.64855918828401}} {"text": "% !TEX root = main.tex\n\n%-------------------------------------------------\n\\section{Random points in the plane}\\label{sec:joint}\n\nLet $X$ and $Y$ be random variables on the same probability space.\n\n% defn: joint distributions\n\\begin{definition}\n\\ben\n\\it The \\emph{joint distribution} of $X$ and $Y$ is the function $\\prob_{X,Y}$ defined on pairs of subsets of $\\R$ by\n\\[\n\\prob_{X,Y}(C,D) = \\prob(X\\in C, Y\\in D).\n\\]\n\\it The distributions $\\prob_X(C)=\\prob(X\\in C)$ and $\\prob_Y(D)=\\prob(Y\\in D)$ are called the \\emph{marginal distributions} of $X$ and $Y$ respectively.\n\\een\n\\end{definition}\n\n%The joint distribution of $X$ and $Y$ is completely determined by their joint CDF.\n% defn: joint distributions\n\\begin{definition}\n\\ben\n\\it The \\emph{joint CDF} of $X$ and $Y$ is the function $F_{X,Y}:\\R^2\\to[0,1]$ given by\n\\[\nF_{X,Y}(x,y) = \\prob(X\\leq x,Y\\leq y).\n\\]\n\\it The CDFs $F_X(x)=\\prob(X\\leq x)$ and $F_Y(y)=\\prob(Y\\leq y)$ are called the \\emph{marginal CDFs} of $X$ and $Y$ respectively.\n\\een\n\\end{definition}\n\n% defn: jointly discrete\n\\begin{definition}\n\\ben\n\\it $X$ and $Y$ are called \\emph{jointly discrete} if the random vector $(X,Y)$ takes only countably many values in $\\R^2$, in which case they are described by their \\emph{joint PMF},\n\\[\n\\begin{array}{cccl}\nf_{X,Y}:\t& \\R^2\t& \\to\t\t& [0,1] \\\\\n\t\t\t& (x,y)\t& \\mapsto\t& \\prob(X=x,Y=y).\n\\end{array}\n\\]\n\\it The PMFs $f_X(x)=\\prob(X=x)$ and $f_Y(y)=\\prob(Y=y)$ are called the \\emph{marginal PMFs} of $X$ and $Y$ respectively.\n\\een\n\\end{definition}\n\n% defn: jointly continuous\n\\begin{definition}\n\\ben\n\\it $X$ and $Y$ are called \\emph{jointly continuous} if their joint CDF can be written as\n\\[\nF_{X,Y}(x,y) = \\int_{-\\infty}^x\\int_{-\\infty}^y f_{X,Y}(u,v)\\,du\\,dv \\text{\\qquad for all $x,y\\in\\R$,}\n\\]\nfor some integrable function $f_{X,Y}:\\R^2\\to[0,\\infty)$ called the \\emph{joint PDF} of $X$ and $Y$. \n\\it The PDFs $f_X(x)=F'_X(x)$ and $f_Y(y)=F'_Y(y)$ are called the \\emph{marginal PDFs} of $X$ and $Y$ respectively, where $F_X(x)$ and $F_Y(y)$ are their marginal CDFs. Note that\n\\[\nf_X(x) = \\int_{-\\infty}^{\\infty}f_{X,Y}(x,y)\\,dy\n\\text{\\qquad and\\qquad}\nf_Y(y) = \\int_{-\\infty}^{\\infty}f_{X,Y}(x,y)\\,dx.\n\\]\n\\een\n\n\\end{definition}\n\n%-----------------------------\n\\subsection{Independence}\n\nRecall that two events $A$ and $B$ are \\emph{independent} if $\\prob(A\\cap B)=\\prob(A)\\prob(B)$.\n\n% defn: independence\n\\begin{definition}\nTwo random variables $X$ and $Y$ are said to be \\emph{independent} if the events $\\{X\\leq x\\}$ and $\\{Y\\leq y\\}$ are independent for all $x,y\\in\\R$.\n\\end{definition}\n\n\\begin{lemma}\\label{lem:product_marginal_cdfs}\n\\ben\n\\it $X$ and $Y$ are independent if and only if $F_{X,Y}(x,y) = F_X(x)F_Y(y)$ for all $x,y\\in\\R$.\n\\it If $X$ and $Y$ are jointly discrete, they are independent if and only if $f_{X,Y}(x,y) =f_X(x) f_Y(y)$ for all $x,y\\in\\R$.\n\\it If $X$ and $Y$ are jointly continuous, they are independent if and only if $f_{X,Y}(x,y) = f_X(x)f_Y(y)$ for all $x,y\\in\\R$ and $\\supp(f_{X,Y})$ is a rectangular region in $\\R^2$.\n\\een\n%\\ben\n%\\it $X$ and $Y$ are independent if and only if \n% \\bit\n% \\it $F_{X,Y}(x,y) = F_X(x)F_Y(y)$ for all $x,y\\in\\R$.\n% \\eit\n%\\it If $X$ and $Y$ are jointly discrete, they are independent if and only if \n% \\bit\n% \\it $f_{X,Y}(x,y) =f_X(x) f_Y(y)\\quad\\text{for all}\\quad x,y\\in\\R$.\n% \\eit\n%\\it If $X$ and $Y$ are jointly continuous, they are independent if and only if\n% \\bit\n% \\it $f_{X,Y}(x,y) = f_X(x)f_Y(y)$ for all $x,y\\in\\R$, and\n% \\it the support of $f_{X,Y}$ is a rectangular region in $\\R^2$.\n% \\eit\n%\\een\n\\end{lemma}\n\\begin{proof}\nThe first two parts follow directly from the definitions. For the jointly continuous case,\n\\bit\n\\it because $X$ and $Y$ are independent, $F_{X,Y}(x,y)=F_X(x)F_Y(y)$, and differentiating both sides with respect to $x$ and $y$ we get $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. \n\\it If the value taken by $X$ affects the range of values taken by $Y$, then $Y$ clearly dependes on $X$. Thus for $X$ and $Y$ to be independent we need that\n\\[\n\\supp(f_{X,Y}) = \\supp(f_X)\\times \\supp(f_Y),\n\\]\nwhich is a rectangular region in $\\R^2$.\n\\eit\n\\end{proof}\n\n% example: discrete\n\\begin{exercise}\\label{exc:joint_disc}\nA fair die is rolled once. Let $\\omega$ denote the outcome, and consider the random variables\n\\[\nX(\\omega) = \\left\\{\\begin{array}{cl}\n\t1 & \\text{ if $\\omega$ is odd}, \\\\\n\t2 & \\text{ if $\\omega$ is even},\n\\end{array}\\right. \n\\text{\\quad and\\quad}\nY(\\omega) = \\left\\{\\begin{array}{cl}\n\t1 & \\text{ if $\\omega\\leq 3$}, \\\\\n\t2 & \\text{ if $\\omega\\geq 4$}.\n\\end{array}\\right.\n\\]\nFind the joint PMF of $X$ and $Y$. Are $X$ and $Y$ independent?\n\\begin{solution}\n\\[\n\\begin{array}{c|cc|c}\n\t\t& Y=1 \t& Y=2\t& f_X\t\\\\ \\hline\nX=1\t\t& 1/3\t& 1/6\t& 1/2\t\\\\ \nX=2\t\t& 1/6 \t& 1/3\t& 1/2 \t\\\\ \\hline\t\nf_Y\t\t& 1/2\t& 1/2\t& \t\t\\\\ \n\\end{array}\n\\]\n$X$ and $Y$ are \\emph{not} independent, because (for example) \n\\[\n\\prob(X=1,Y=1)\\neq\\prob(X=1)\\prob(Y=1).\n\\]\n\\end{solution}\n\\end{exercise}\n\n% example: continuous\n\\begin{example}\nLet $X$ and $Y$ be jointly continuous random variables with the following joint PDF:\n\\[\nf_{X,Y}(x,y) = \n\\left\\{\\begin{array}{ll}\n\tc(1-x)y\t& \\text{for } 0\\leq y \\leq x \\leq 1, \\\\\n\t0\t\t& \\text{otherwise.}\n\\end{array}\\right.\n\\]\n\\ben\n\\it Sketch the support of $f_{X,Y}$.\n\\it Are $X$ and $Y$ independent?\n\\it Find the marginal PDFs of $X$ and $Y$, and show that $c=24$.\n%\\it Show that $c=24$.\n\\een\n\\end{example}\n\n\\begin{solution}\n\\ben\n\\it % << (a)\n$\\supp(f_{X,Y})$ is the triangular region bounded by the lines $y=x$, $y=0$ and $x=1$.\n\\it % << (d)\nThe support of $f_{X,Y}$ is not a rectangular region in $\\R^2$, so $X$ and $Y$ cannot be independent.\n\\it % << (c)\nThe marginal PDFs are:\n\\[\\begin{array}{lll}\nf_X(x) \t\n\t& = 24\\int_0^x (1-x)y\\,dy \n\t= 24(1-x)\\left[\\frac{y^2}{2}\\right]_0^x \n\t& = \\begin{cases}\n\t\t12x^2(1-x) & 0\\leq x\\leq 1, \\\\\n\t\t0\t\t\t& \\text{otherwise.}\n\t\t\\end{cases} \\\\[3ex]\nf_Y(y) \t\n\t& = 24\\int_y^1 (1-x)y\\,dx \n\t= 24y\\left[x - \\frac{x^2}{2}\\right]_y^1 \n\t& = \\begin{cases}\n\t\t12y(1-y)^2 & 0\\leq y\\leq 1, \\\\\n\t\t0\t\t\t& \\text{otherwise.}\n\t\t\\end{cases}\n\\end{array}\\]\n\\it % << (b)\n\\bit\n\\it For fixed $x\\in[0,1]$, we must have $0\\leq y\\leq x$.\n\\it For fixed $y\\in[0,1]$, we must have $y\\leq x\\leq 1$.\n\\eit\n\\[\n\\int_{-\\infty}^{\\infty}\\left(\\int_{-\\infty}^{\\infty}f_{X,Y}(x,y)\\,dy\\right)\\,dx\n\t= \\int_0^1\\int_0^x c(1-x)y\\,dy\\,dx \n\t= \\frac{c}{2}\\int_0^1 x^2-x^3\\,dx\n%\t= \\frac{c}{2}\\left(\\frac{1}{3}-\\frac{1}{4}\\right)\n\t= \\frac{c}{24}.\n\\]\nBy the law of total probability this must equal $1$, so $c=24$.\n\\een\n\\end{solution}\n\n", "meta": {"hexsha": "ffce862f1b7dc8cc45041c4475a9aa80b8341556", "size": 6479, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/05A_joint_distributions.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/05A_joint_distributions.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/05A_joint_distributions.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 33.056122449, "max_line_length": 183, "alphanum_fraction": 0.6156814323, "num_tokens": 2618, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8558511396138366, "lm_q1q2_score": 0.6485591762483519}} {"text": "\\documentclass[paper.tex]{subfiles}\n\n\\usepackage{tikz}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage{tabularx}\n\\usepackage{multicol}\n\\usepackage{algpseudocode}\n\\usepackage{algorithm}\n\n% Add vertical spacing to tables\n\\renewcommand{\\arraystretch}{1.4}\n\n% Begin Document\n\\begin{document}\n\n\\section{$LU$ Factorization}\n\nThe first method to solve this system of equations will be $LU$ Factorization.\nThis involves decomposing the coefficient matrix $A$ into two factors, $L$ and $U$, resulting in an equation of the form $A = LU$.\nDoing so gives:\n\\begin{equation*}\n \\begin{bmatrix}\n 1 & 0 & 0 & 0 & 0 \\\\\n \\frac{1}{2} & 1 & 0 & 0 & 0 \\\\\n \\frac{1}{2} & -1 & 1 & 0 & 0 \\\\\n 0 & -2 & 1 & 1 & 0 \\\\\n 0 & 2 & -\\frac{3}{2} & -1 & 1\n \\end{bmatrix}\n \\begin{bmatrix}\n 2 & 1 & 0 & 0 & 0 \\\\\n 0 & \\frac{1}{2} & -1 & 0 & -1 \\\\\n 0 & 0 & -2 & 0 & -2 \\\\\n 0 & 0 & 0 & 1 & 1 \\\\\n 0 & 0 & 0 & 0 & 1 \n \\end{bmatrix}\n\\end{equation*}\n\nUsing the $LU$ decomposition it's possible to decompose the original matrix equation into two separate equations:\n\\begin{equation}\n Ux = y\n\\end{equation}\n\\begin{equation}\n Ly = b\n\\end{equation}\n\nWe must first solve for $y$ in equation (3). \n\\begin{equation}\n y = L^{-1}b\n\\end{equation}\n\nThis requires the inverse of $L$, which is:\n\\begin{equation*}\n L^{-1}=\\begin{bmatrix}\n 1 & 0 & 0 & 0 & 0 \\\\\n -\\frac{1}{2} & 1 & 0 & 0 & 0 \\\\\n -1 & 1 & 1 & 0 & 0 \\\\\n 0 & 1 & -1 & 1 & 0 \\\\\n -\\frac{1}{2} & \\frac{1}{2} & -\\frac{1}{2} & 1 & 1 \n \\end{bmatrix}\n\\end{equation*}\n\nPlugging in values for equation (4) gives:\n\\begin{equation*}\n y = \\begin{bmatrix}\n 1 & 0 & 0 & 0 & 0 \\\\\n -\\frac{1}{2} & 1 & 0 & 0 & 0 \\\\\n -1 & 1 & 1 & 0 & 0 \\\\\n 0 & 1 & -1 & 1 & 0 \\\\\n -\\frac{1}{2} & \\frac{1}{2} & -\\frac{1}{2} & 1 & 1\n \\end{bmatrix}\n \\begin{bmatrix}\n 100 \\\\\n 0 \\\\\n -50 \\\\\n 120 \\\\\n 0\n \\end{bmatrix}\n =\\begin{bmatrix}\n 100 \\\\\n -50 \\\\\n -150 \\\\\n 170 \\\\\n 45 \n \\end{bmatrix}\n\\end{equation*}\n\nIn order to compute for $x$, we rearrange equation (2) into:\n\\begin{equation}\n x = U^{-1}y\n\\end{equation}\n\nCalculating the inverse of $U$ gives:\n\\begin{equation*}\n U^{-1}=\\begin{bmatrix}\n \\frac{1}{2} & -1 & \\frac{1}{2} & 0 & 0 \\\\\n 0 & 2 & -1 & 0 & 0 \\\\\n 0 & 0 & -\\frac{1}{2} & 0 & -1 \\\\\n 0 & 0 & 0 & 1 & -1 \\\\\n 0 & 0 & 0 & 0 & 1 \n \\end{bmatrix}\n\\end{equation*}\n\nAs given in equation (5), the product of the values of $U^{-1}$ and $y$ gives the solution $x$:\n\\begin{equation*}\n x = \\begin{bmatrix}\n 25 \\\\\n 50 \\\\ \n 30 \\\\ \n 125 \\\\\n 45\n \\end{bmatrix}\n\\end{equation*}\n\n\\end{document}\n", "meta": {"hexsha": "eb1a5e5aede3733d343fc56900c30159807d8c73", "size": 2814, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Projects/matrix_theory/docs/tex/lu_factorization.tex", "max_stars_repo_name": "Bkrenz/calu-mat341", "max_stars_repo_head_hexsha": "2628f0755dde2e4a933131e23cbe8168444fd77c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Projects/matrix_theory/docs/tex/lu_factorization.tex", "max_issues_repo_name": "Bkrenz/calu-mat341", "max_issues_repo_head_hexsha": "2628f0755dde2e4a933131e23cbe8168444fd77c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Projects/matrix_theory/docs/tex/lu_factorization.tex", "max_forks_repo_name": "Bkrenz/calu-mat341", "max_forks_repo_head_hexsha": "2628f0755dde2e4a933131e23cbe8168444fd77c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.0512820513, "max_line_length": 130, "alphanum_fraction": 0.5074626866, "num_tokens": 1097, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.793105951184112, "lm_q2_score": 0.817574478416099, "lm_q1q2_score": 0.6484231843680545}} {"text": "% A simple template for LaTeX documents\n% \n% To produce pdf run:\n% $ pdflatex paper.tex \n%\n\n\\documentclass[10pt, twocolumn]{article}\n%\\documentclass[12pt]{article}\n\n% Begin paragraphs with new line\n\\usepackage{parskip} \n\n% Change margin size\n\\usepackage[margin=0.5in]{geometry} \n\n% Graphics Example: (PDF's make for good plots)\n\\usepackage{graphicx} \n% \\centerline{\\includegraphics{figure.pdf}}\n\n% Allows hyperlinks\n\\usepackage{hyperref}\n\n% Blocks of code\n\\usepackage{listings}\n\\lstset{basicstyle=\\ttfamily, title=\\lstname}\n% Insert code like this. replace `plot.R` with file name.\n% \\lstinputlisting{plot.R}\n\n% Monospaced fonts\n%\\usepackage{inconsolata}\n% GNU \\texttt{make} is a nice tool.\n\n% Supports proof environment\n\\usepackage{amsthm}\n\n% Allows writing \\implies and align*\n\\usepackage{amsmath}\n\n% Allows mathbb{R}\n\\usepackage{amsfonts}\n\n% Numbers in scientific notation\n\\usepackage{siunitx}\n\n% Use tables generated by pandas\n\\usepackage{booktabs}\n\n% norm and infinity norm\n\\newcommand{\\norm}[1]{\\left\\lVert#1\\right\\rVert}\n\\newcommand{\\inorm}[1]{\\left\\lVert#1\\right\\rVert_\\infty}\n\n% Statistics essentials\n\\newcommand{\\iid}{\\text{ iid }}\n\\newcommand{\\Expect}{\\operatorname{E}}\n\\newcommand{\\Var}{\\operatorname{Var}}\n\\newcommand{\\Cov}{\\operatorname{Cov}}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{document}\n\n\\textbf{Binomial}\n$X \\sim B(n, p)$\n\\[\n p(k) = \\binom{n}{k} p^k (1-p)^{n-k}\n \\qquad k = 0, 1, \\dots, n\n\\]\n$\\Expect X = np, \\quad \\Var X = np(1-p)$\n\nmgf: $M_X (t) = (pe^t + 1 - p)^n$\n\nBeta is conjugate prior, Fisher info $I(p) = \\frac{1}{p(1 - p)}$\n\n\\textbf{Poisson}\n$X \\sim P(\\lambda)$\n\\[\n p(k) = \\frac{e^{-\\lambda} \\lambda^k}{k!}\n \\qquad k = 0, 1, \\dots\n\\]\n$\\Expect X = \\lambda, \\quad \\Var X = \\lambda$\n\nmgf: $M_X (t) = e^{\\lambda (e^t -1)}$\n\nGamma is conjugate prior, Fisher info $I(\\lambda) = \\frac{1}{\\lambda}$\n\n\\textbf{Normal}\n$X \\sim N(\\mu, \\Sigma)$, $\\Sigma$ positive definite\n\\[\n f(x) = \\frac{\\exp\\{ - \\frac{1}{2}(x - \\mu)^T \\Sigma^{-1} (x - \\mu) \\}}\n {(2\\pi)^{\\frac{k}{2}} \\sqrt{\\det(\\Sigma)}}\n = \\frac{1}{\\sqrt{2 \\pi} \\sigma} e^{-\\frac{(x - \\mu)^2}{2\n \\sigma^2}}\n\\]\nmgf: $M_X (t) = \\exp (\\mu' t + \\frac{1}{2} t' \\Sigma t)$\n\nNormal is conj. prior, Fisher info $I(\\mu, \\sigma^2) = \n[\\begin{smallmatrix}\n 1 / \\sigma^2 & 0 \\\\\n 0 & 1 / 2\\sigma^4 \\\\\n\\end{smallmatrix}]$\n\n\\textbf{Beta}\n$ X \\sim \\text{Beta}(\\alpha, \\beta)$\n\\[\n f(x) = \\frac{x^{\\alpha-1}(1 - x)^{\\beta-1}}{B(\\alpha, \\beta)} \n \\qquad 0 \\leq x \\leq 1\n\\]\n$\\Expect X = \\frac{\\alpha}{\\alpha + \\beta},\n\\quad \\Var X = \\frac{\\alpha \\beta}{(\\alpha + \\beta)^2 (\\alpha + \\beta + 1)}$\n\nusing the beta function:\n\\[\n B(\\alpha, \\beta) =\n \\frac{\\Gamma(\\alpha) \\Gamma(\\beta)}{\\Gamma(\\alpha+ \\beta)} =\n \\int_0^1 t^{\\alpha -1} (1-t)^{\\beta - 1}dt\n\\]\n\n\\textbf{Gamma}\n$X \\sim \\text{Gamma}(\\alpha, \\beta)$\n\\[\n f(x) = \\frac{\\beta^\\alpha x^{\\alpha-1} e^{-\\beta x}}{\\Gamma(\\alpha)}\n \\qquad x > 0\n\\]\n$\\Expect X = \\frac{\\alpha}{\\beta},\n\\quad \\Var X = \\frac{\\alpha}{\\beta^2}$\n\nmgf: $M_X (t) = (1 - \\frac{t}{\\beta})^{-\\alpha}, t < \\beta$\n\n$X \\sim \\text{Gamma}(\\alpha, \\beta) \\iff \\beta X \\sim \\text{Gamma}(\\alpha, 1)$\n\n$X_i \\iid \\text{Gamma}(\\alpha_i, \\beta)$, then\n\\[\n \\sum X_i \\sim \\text{Gamma}(\\sum \\alpha_i, \\beta)\n\\]\nGamma function: $\\Gamma(\\alpha) = \\int_0^\\infty t^{\\alpha-1} e^{-t} dt$.\n\n$\\Gamma(\\frac{1}{2}) = \\sqrt{\\pi}$.\n\n$\\Gamma(\\alpha + 1) = \\alpha \\Gamma(\\alpha)$\n\n$\\Gamma(k) = (k-1)!$ for $k$ positive integer.\n\n\\textbf{Exponential}\nSpecial case: $\\text{Exp}(\\lambda) \\equiv \\text{Gamma}(1, \\lambda)$\n\n\\textbf{Chi square}\nSpecial case: $\\chi^2_n \\equiv \\text{Gamma}(\\frac{n}{2}, \\frac{1}{2})$\n\nLet $Z_i$ be iid $N(0, 1)$.\n\n$\\sum_{i=1}^n Z_i^2 \\sim \\chi^2_n$\n\nNoncentral $\\chi^2$. Let $Y \\sim N(\\mu, I)$ be an $n$ vector. Then \n\\[\n \\norm{Y}^2 \\sim \\chi^2_n(\\norm{\\mu}^2)\n\\]\n\n\\textbf{F}\n\\[\n F(m, n) \\equiv \\frac{\\frac{\\chi^2_m}{m}}\n {\\frac{\\chi^2_n}{n}}\n\\]\nWhere numerator and denominator are independent $\\chi^2$.\n\n\\textbf{T}\n\\[\n t(n) = \\frac{N(0, 1)}\n {\\sqrt{\\frac{\\chi^2_n}{n}}}\n\\]\nWhere numerator and denominator are independent.\n\n\\vspace{0.2in}\n\\hrule\n\nMoment generating functions determine distribution\n\\[\n M_X(t) \\equiv \\Expect (e^{tX}),\n \\quad M_X'(0) = \\Expect(X)\n\\]\n$X_i$ independently distributed $\\iff$\n\\[\n M_{\\sum X_i} (t) = \\prod M_{X_i} (t)\n\\]\n\\textbf{Characteristic function}\n\\[\n \\phi(t) = \\Expect (e^{i t^T X)}\n = \\Expect (\\cos (t^T X)) + i \\Expect(\\sin(t^T X))\n\\]\nOrder statistics for sorted sample $X_{(1)}, \\dots, X_{(n)}$ has pdf:\n\\[\n n! \\prod_{i=1}^n f(X_{(i)}) \\quad I(X_{(1)} < \\dots < X_{(n)})\n\\]\n\n\\newpage\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\textbf{Jensen's Inequality} if $S \\subset R^k$ convex and closed, $g$ convex on $S$, $P[X\n\\in S] = 1$, and $\\Expect X$ is finite, then $\\Expect X \\in S$, $\\Expect\ng(X)$ exists, and\n\\[\n \\Expect g(X) \\geq g(\\Expect X)\n\\]\n\\textbf{Holder's Inequality} if $r, s > 1$ and $\\frac{1}{r} + \\frac{1}{s} =\n1$ then\n\\[\n \\Expect |XY| \\leq (\\Expect |X|^r)^{\\frac{1}{r}}(\\Expect |X|^s)^{\\frac{1}{s}}\n\\]\n\n$T(X)$ Sufficient means the distribution of $X | T(X)$ does not depend on\n$\\theta$.\n\nFactorization theorem: $T(x)$ is sufficient $\\iff$\n\\[\n f_{\\theta}(x) = h(x) g(\\theta, T(x))\n\\]\n$L_x (\\theta) = p_\\theta (x) = p(x, \\theta)$ likelihood is function of\n$\\theta$, density is function of $x$.\n\nThe likelihood ratio\n\\[\n \\lambda_x (\\theta) = \\frac{L_x (\\theta)}{L_x (\\theta_0)}\n\\]\nis minimal sufficient. To show $T(x)$ is minimal sufficient show that it is\nsufficient and a function of the likelihood $\\lambda_x (\\theta)$.\n\n\\textbf{Fisher information}\n\\[\n I(\\theta) = \\Expect_\\theta \\left[ \\frac{\\partial}{\\partial \\theta} \n \\log L_X (\\theta) \\right]^2\n = \\Expect_\\theta \\left[ - \\frac{\\partial^2}{\\partial \\theta^2}\n \\log L_X (\\theta) \\right]\n\\]\n$\\text{bias } \\hat{v} \\equiv \\Expect (\\hat{v}) - v$\n\\[\n MSE(\\hat{v}) \\equiv \\Expect (\\hat{v} - v)^2 \n = \\Var (\\hat{v}) + (\\text{bias } \\hat{v})^2\n\\]\n\\textbf{Rao-Blackwell} Let $S(X)$ be an unbiased point estimator for\n$g(\\theta)$. Conditioning on a sufficient statistic $T(X)$ reduces\nvariance.\n\\[\n \\Var_{\\theta} (S(X)) \\geq \\Var_{\\theta} (\\Expect (S(X) | T(X)))\n\\]\nAlso holds for more general convex loss function $L$:\n\\[\n R(\\theta, S) \\equiv \\Expect_{\\theta} L(\\theta, S(X)) \\geq\n \\Expect_{\\theta} L(\\theta, \\Expect( S(X) | T(X)))\n\\]\n\\textbf{Completeness} $T(X)$ is complete if $\\Expect g(T(X)) = 0$ \nimplies $g = 0$ almost surely for all $\\theta$.\n\n\\textbf{Cramer Rao Inequality} Let $g: \\Theta \\rightarrow R$. Suppose there\nexists an unbiased estimator $U(X)$, $\\Expect U(X) = g(\\theta)$. Then\n\\[\n \\Var_\\theta U(X) \\geq \n \\left( \\frac{\\partial g(\\theta)}{\\partial \\theta} \\right)^T\n I(\\theta)^{-1}\n \\left( \\frac{\\partial g(\\theta)}{\\partial \\theta} \\right)\n\\]\nBasu's Theorem - If $T(X)$ complete sufficient statistic and $A(X)$ is \nancillary then $A(X)$ and $T(X)$ are independent.\n\n\\textbf{Lehmann - Scheffe} Suppose $T(X)$ is complete sufficient. Then\nthere exists unique unbiased estimator $\\Expect h(T(x))$ of $g(\\theta) \\in\nR$ with smallest variance (MVUE). \n\n\\textbf{Exponential Families} $T(x)$ is natural sufficient statistic and is\ncomplete sufficient if the $k$ parameter exponential family is full rank.\n\\[\n p(x, \\theta) = h(x) \\exp \\{ \\eta(\\theta)^T T(x) - B(\\theta) \\}\n\\]\nCanonical form model indexed by $\\eta$.\n\\[\n q(x, \\eta) = h(x) \\exp \\{ \\eta^T T(x) - A(\\eta) \\}\n\\]\n\\[\n \\dot{A} (\\eta) = \\Expect_\\eta (T(X)) \\quad\n \\ddot{A} (\\eta) = I(\\eta) = \\Var_\\eta (T(X))\n\\]\nThen moment generating function for $T(X)$ is\n\\[\n M_{T(X)}(t) = \\exp \\{A(t + \\eta) - A(\\eta) \\}\n\\]\nEquivalent statements useful for GLM's such as $Y \\sim N(X \\beta,\n\\sigma_0^2 I)$, where $Z$ is $n \\times p$:\n\n1. $I(\\beta) = \\frac{1}{\\sigma_0^2} X^T X$ positive definite\n2. rank$(X) = p$\n3. model is identifiable. More generally another equivalent statement is\n$\\Var (T(X)) = \\ddot{A} (\\eta)$ is positive definite.\n\n\\subsection*{Decision Theory}\n\n\\textbf{Decision rule} $\\delta: \\mathcal{X} \\rightarrow \\mathcal{A}$, where \n$\\delta \\in \\mathcal{D}$, the space of possible decision rules and\n$\\mathcal{A}$ is the action space.\n\n\\textbf{Loss function} $l: \\Theta \\times \\mathcal{A} \\rightarrow \\mathbb{R}^+$\nPosterior mean minimizes square error loss; median minimizes absolute loss.\n\n\\textbf{Risk function} $R: \\Theta \\times \\mathcal{D} \\rightarrow\n\\mathbb{R}^+$ expected loss for a particular value of $\\theta$\n\\[\n R(\\theta, \\delta) = \\Expect_\\theta l(\\theta, \\delta(X)) = \\int\n l(\\theta, \\delta(x)) \\cdot p_\\theta (x) dx\n\\]\nBayes setup:\n\\[\n \\pi(\\theta | x) = \\frac{p_\\theta (x) \\pi(\\theta)}{m(x)}\n\\]\n\\textbf{Bayes decision rule} If there exists $\\delta_\\pi \\in \\mathcal{D}$\nw.r.t prior $\\pi$ such that\n\\[\n r(\\pi, \\delta_\\pi) = \\inf_{\\delta \\in \\mathcal{D}} r(\\pi, \\delta)\n\\]\nTo find Bayes rule minimize the posterior risk:\n\\[\n \\delta_\\pi (x) = \\min_{a \\in \\mathcal{A}} r_\\pi (a | x)\n\\]\n\n\\textbf{Bayes risk} $r_\\pi : \\mathcal{D} \\rightarrow\n\\mathbb{R}^+$ expected loss for fixed prior $\\pi$\n\\[\n r_\\pi (\\delta) = \\Expect_\\pi R(\\theta, \\delta) \n = \\int_\\Theta R(\\theta, \\delta) \\pi (d \\theta)\n = \\int_\\mathcal{X} r_\\pi(\\delta(x) | x) m(x) dx\n\\]\nTo find Bayes risk: 1) find the Bayes rule 2) compute the risk function 3)\ntake the expectation of the risk wrt prior $\\pi$.\n\n\n\\textbf{Minimax} decision rule $\\delta^*$ minimizes the worst case\nscenario, satisfies\n\\[\n \\sup_{\\theta} R(\\theta, \\delta^*) = \\inf_{\\delta} \\sup_{\\theta}\n R(\\theta, \\delta)\n\\]\nTo show $\\delta^*$ is minimax, first check for constant risk $R(\\theta,\n\\delta^* = c$ for all $\\theta$, then find a\nprior $\\pi$ such that $\\delta^*$ is the Bayes rule. This $\\pi$ is least\nfavorable. More generally can find\na sequence of priors $(\\pi_k)$ such that the Bayes risk $r_{\\pi_k}\n(\\delta_{\\pi_k}) \\rightarrow c$.\n\n\\newpage\n\n\\subsection*{Multivariate Normal}\n\nStein's formula: $X \\sim N(\\mu, \\sigma)$\n\\[\n \\Expect (g(X) (X - \\mu)) = \\sigma^2 \\Expect(g'(X))\n\\]\nassuming these expectations are finite.\n\n$X \\sim N(\\mu, \\Sigma)$, $A$ an $m \\times n$ matrix,\nthen \n\\[\n AX \\sim N(A \\mu, A \\Sigma A^t)\n\\]\nFor $\\Sigma$ full rank it's possible to transform between $Z \\sim\nN(0, I)$ and $X$:\n\\[\n X = \\Sigma^{1/2} Z + \\mu \\qquad Z = \\Sigma^{-1/2} (X - \\mu)\n\\]\nIn block matrix form:\n\\[\n X =\n \\begin{bmatrix}\n X_1 \\\\\n X_2 \\\\\n \\end{bmatrix}\n \\sim N \\left(\n \\begin{bmatrix}\n \\mu_1 \\\\\n \\mu_2 \\\\\n \\end{bmatrix}\n ,\n \\begin{bmatrix}\n \\Sigma_{11} & \\Sigma_{12} \\\\\n \\Sigma_{21} & \\Sigma_{22} \\\\\n \\end{bmatrix}\n\\right)\n\\]\nAssuming $\\Sigma_{11}$ is positive definite then the conditional\ndistribution\n\\[\n X_2 | X_1 \\sim N(\\mu_2 + \\Sigma_{21} \\Sigma_{11}^{-1} (X_1 - \\mu_1),\n \\Sigma_{22} - \\Sigma_{21} \\Sigma_{11}^{-1} \\Sigma_{12})\n\\]\n\n\\subsection*{Conditional Distributions}\n\nConditional pdf:\n\\[\n f_{X|Y}(x, y) \\equiv \\frac{f_{X, Y}(x, y)}{f_Y(y)}\n\\]\nIterated expectation:\n\\[\n E(Y) = E(E(Y | X))\n\\]\nConditional variance formula:\n\\[\n \\Var(Y) = \\Var(E(Y | X)) + E(\\Var(Y | X))\n\\]\n\n\\newpage\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsection*{General Techniques}\n\nSingular Value Decompostion (SVD) Any matrix $X$ can be written\n\\[\n X = UDV^T\n\\]\nwith $U, V$ orthogonal, and $D$ diagonal.\n\nMoore Penrose Psuedoinverse $A^+$ exists uniquely for every matrix $A$.\n\nProjection matrix $P$ are symmetric and idempotent. They have eigenvalues\neither 0 or 1.\n\\[\n P = P^T \\qquad P^2 = P\n\\]\n\nCovariance of linear transformations\n\\[\n Cov(Ay, Bx) = A Cov(y, x) B^T\n\\]\n$\n A = \n [\\begin{smallmatrix}\n a & b \\\\\n c & d \\\\\n \\end{smallmatrix}]\n$\n\\[\n A^{-1} = \n \\frac{1}{\\det (A)}\n \\begin{bmatrix}\n d & -b \\\\\n -c & a \\\\\n \\end{bmatrix}\n\\]\n\nIntegration by parts:\n\\[\n \\int uv' = uv - \\int u'v\n\\]\n\nTransformation theorem: Given some regularity conditions, $Y = g(X)$ has\npdf\n\\[\n p_Y(y) = p_X(g^{-1}(y)) |J_{g^{-1}(y)}|\n\\]\n\nMatrix / Vector differentiation\n\n$\\frac{\\partial A^T \\beta}{\\partial \\beta} = A$, \n$\\frac{\\partial \\beta^T A \\beta}{\\partial \\beta} = (A + A^t) \\beta =\n2A\\beta$ for $A$ symmetric.\n\n\n\\end{document}\n", "meta": {"hexsha": "79d0f1fe56c70287b47da8ed8551d58c27cb3ff0", "size": 12228, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "231a/stat_notes.tex", "max_stars_repo_name": "clarkfitzg/phd_stats", "max_stars_repo_head_hexsha": "c74b21a7fd55a713927650ce1827d1b853809395", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "231a/stat_notes.tex", "max_issues_repo_name": "clarkfitzg/phd_stats", "max_issues_repo_head_hexsha": "c74b21a7fd55a713927650ce1827d1b853809395", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2015-11-10T07:47:09.000Z", "max_issues_repo_issues_event_max_datetime": "2015-11-10T22:55:08.000Z", "max_forks_repo_path": "231a/stat_notes.tex", "max_forks_repo_name": "clarkfitzg/phd_stats", "max_forks_repo_head_hexsha": "c74b21a7fd55a713927650ce1827d1b853809395", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.2967741935, "max_line_length": 90, "alphanum_fraction": 0.592983317, "num_tokens": 4442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936438, "lm_q2_score": 0.7931059487389968, "lm_q1q2_score": 0.6484231806063482}} {"text": "\\title{Latent Space Models for Neural Data}\n\n\\subsection{Latent Space Models for Neural Data}\n\nMany scientific fields involve the study of network data, including\nsocial networks, networks in statistical physics, biological\nnetworks, and information networks\n\\citep{newman2010networks,goldenberg2010survey}.\n\nWhat we can learn about nodes in a network from their connectivity patterns?\nWe can begin to study this using a latent space model \\citep{hoff2002latent}.\nLatent space models embed nodes in the network in a latent space,\nwhere the likelihood of forming an edge between two nodes depends on\ntheir distance in the latent space.\n\nWe will analyze network data from neuroscience.\nAn interactive version with Jupyter notebook is available\n\\href{http://nbviewer.jupyter.org/github/blei-lab/edward/blob/master/notebooks/latent_space_models.ipynb}{here}.\n\n\\subsubsection{Data}\n\nThe data comes from \\href{http://www-personal.umich.edu/~mejn/netdata/}{Mark Newman's repository}.\nIt is a weighted, directed network representing the neural network of\nthe nematode\n\\href{https://en.wikipedia.org/wiki/Caenorhabditis_elegans}{C.~Elegans}\ncompiled by \\citet{watts1998collective} using experimental data\nby \\citet{white1986structure}.\n\nThe neural network consists of around $300$ neurons. Each connection\nbetween neurons\nis associated with a weight (positive integer) capturing the strength\nof the connection.\n\nFirst, we load the data.\n\\begin{lstlisting}[language=Python]\nfrom observations import celegans\n\nx_train = celegans(\"~/data\")\n\\end{lstlisting}\n\n\\subsubsection{Model}\n\nWhat can we learn about the neurons from their connectivity patterns? Using\na latent space model \\citep{hoff2002latent}, we will learn a latent\nembedding for each neuron to capture the similarities between them.\n\nEach neuron $n$ is a node in the network and is associated with a latent\nposition $z_n\\in\\mathbb{R}^K$.\nWe place a Gaussian prior on each of the latent positions.\n\nThe log-odds of an edge between node $i$ and\n$j$ is proportional to the Euclidean distance between the latent\nrepresentations of the nodes $|z_i- z_j|$. Here, we\nmodel the weights ($Y_{ij}$) of the edges with a Poisson likelihood.\nThe rate is the reciprocal of the distance in latent space. The\ngenerative process is as follows:\n\n\\begin{enumerate}\n\\item\nFor each node $n=1,\\ldots,N$,\n\\begin{align}\nz_n \\sim N(0,I).\n\\end{align}\n\\item\nFor each edge $(i,j)\\in\\{1,\\ldots,N\\}\\times\\{1,\\ldots,N\\}$,\n\\begin{align}\nY_{ij} \\sim \\text{Poisson}\\Bigg(\\frac{1}{|z_i - z_j|}\\Bigg).\n\\end{align}\n\\end{enumerate}\n\nIn Edward, we write the model as follows.\n\\begin{lstlisting}[language=Python]\nfrom edward.models import Normal, Poisson\n\nN = x_train.shape[0] # number of data points\nK = 3 # latent dimensionality\n\nz = Normal(loc=tf.zeros([N, K]), scale=tf.ones([N, K]))\n\n# Calculate N x N distance matrix.\n# 1. Create a vector, [||z_1||^2, ||z_2||^2, ..., ||z_N||^2], and tile\n# it to create N identical rows.\nxp = tf.tile(tf.reduce_sum(tf.pow(z, 2), 1, keep_dims=True), [1, N])\n# 2. Create a N x N matrix where entry (i, j) is ||z_i||^2 + ||z_j||^2\n# - 2 z_i^T z_j.\nxp = xp + tf.transpose(xp) - 2 * tf.matmul(z, z, transpose_b=True)\n# 3. Invert the pairwise distances and make rate along diagonals to\n# be close to zero.\nxp = 1.0 / tf.sqrt(xp + tf.diag(tf.zeros(N) + 1e3))\n\nx = Poisson(rate=xp)\n\\end{lstlisting}\n\n\\subsubsection{Inference}\n\nMaximum a posteriori (MAP) estimation is simple in Edward. Two lines are\nrequired: Instantiating inference and running it.\n\\begin{lstlisting}[language=Python]\ninference = ed.MAP([z], data={x: x_train})\n\\end{lstlisting}\n\nSee this extended tutorial about\n\\href{/tutorials/map}{MAP estimation in Edward}.\n\nOne could instead run variational inference. This requires specifying\na variational model and instantiating \\texttt{KLqp}.\n\\begin{lstlisting}[language=Python]\nqz = Normal(loc=tf.Variable(tf.random_normal([N * K])),\n scale=tf.nn.softplus(tf.Variable(tf.random_normal([N * K]))))\ninference = ed.KLqp({z: qz}, data={x: x_train})\n\\end{lstlisting}\nSee this extended tutorial about\n\\href{/tutorials/variational-inference}{variational inference in Edward}.\n\nFinally, the following line runs the inference procedure for 2500\niterations.\n\\begin{lstlisting}[language=Python]\ninference.run(n_iter=2500)\n\\end{lstlisting}\n\n\\subsubsection{Acknowledgments}\n\nWe thank Maja Rudolph for writing the initial version of this\ntutorial.\n\n\\subsubsection{References}\n", "meta": {"hexsha": "30cbd35163e8c4a5c506aee885452b7d6f1194ea", "size": 4426, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/tex/tutorials/latent-space-models.tex", "max_stars_repo_name": "xiangze/edward", "max_stars_repo_head_hexsha": "6419751d1d849c84c502e5ff3f7249b9bbc7b3aa", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-04-06T15:08:53.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-06T15:08:53.000Z", "max_issues_repo_path": "docs/tex/tutorials/latent-space-models.tex", "max_issues_repo_name": "xiangze/edward", "max_issues_repo_head_hexsha": "6419751d1d849c84c502e5ff3f7249b9bbc7b3aa", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/tex/tutorials/latent-space-models.tex", "max_forks_repo_name": "xiangze/edward", "max_forks_repo_head_hexsha": "6419751d1d849c84c502e5ff3f7249b9bbc7b3aa", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-12-22T08:21:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-16T02:45:04.000Z", "avg_line_length": 34.8503937008, "max_line_length": 112, "alphanum_fraction": 0.757568911, "num_tokens": 1217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8723473647220786, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6483006538588529}} {"text": "\\chapter{The Two- and Three-Dimensional Navier-Stokes Equations}\n\n%%%%%%%\n%Section\n%%%%%%%\n\\section{Background}\nThe Navier-Stokes equations describe the motion of a fluid. In order to derive the Navier-Stokes equations we assume that a fluid is a continuum (not made of individual particles, but rather a continuous substance) and that mass and momentum are conserved. After making some assumptions and using Newton's second law on an incompressible fluid particle, the Navier-Stokes equations can be derived in their entirety. All details are omitted since there are many sources of this information, two sources that are particularly clear are Tritton~\\cite{Tri88} and Doering and Gibbon~\\cite{DoeGib95}; Gallavotti~\\cite{Gal02} should also be noted for introducing both mathematical and physical aspects of these equations, and Uecker~\\cite{Uec09} includes a quick derivation and some example Fourier Spectral Matlab codes. For a more detailed introduction to spectral methods for the Navier-Stokes equations see Canuto et al.~\\cite{CHQZ07}. The incompressible Navier-Stokes equations are\n\\begin{eqnarray}\n\\rho \\left (\\frac{\\partial \\mathbf{u}}{\\partial t} + \\mathbf{u} \\cdot \\nabla \\mathbf{u} \\right)&=&-\\nabla p + \\mu \\Delta \\mathbf{u} + \\mathbf{f} \\label{eq:NseConMom}\\\\\n\\nabla \\cdot \\mathbf{u} &=& 0. \\label{eq:NseConMas}\n\\end{eqnarray}\nIn these equations, $\\rho$ is density, $\\mathbf{u}(x,y,z)=(u,v,w)$ is the velocity with components in the $x$, $y$ and $z$ directions, $p$ is pressure field, $\\mu$ is dynamic viscosity (constant in incompressible case) and $\\mathbf{f}$ is a body force (force that acts through out the volume). Equation \\eqref{eq:NseConMom} represents conservation of momentum and eq.\\ \\eqref{eq:NseConMas} is the continuity equation which represents conservation of mass for an incompressible fluid. \n\n%%%%%%%\n%Section\n%%%%%%%\n\\section{The Two-Dimensional Case}\n\nWe will first consider the two-dimensional case. A difficulty in simulating the incompressible Navier-Stokes equations is the numerical satisfaction of the incompressibility constraint in eq.\\ \\eqref{eq:NseConMas}, this is sometimes referred to as a divergence free condition or a solenoidal constraint. To automatically satisfy this incompressibility constraint in two dimensions, where\n$$\\mathbf u(x,y)=\\left(u(x,y),v(x,y)\\right)$$\nit is possible to re-write the equations using a different formulation, the stream-function vorticity formulation. In this case, we let \n$$u=\\frac{\\partial \\psi}{\\partial y} \\quad v=-\\frac{\\partial \\psi}{\\partial x},$$ \nwhere $\\psi(x,y)$ is the streamfunction. Level curves of the streamfunction represent streamlines\\footnote{A streamline is a continuous curve along which the instantaneous velocity is tangent, see Tritton~\\cite{Tri88} for more on this.} of the fluid field. Note that \n$$\\nabla\\cdot\\mathbf u =\\frac{\\partial u}{\\partial x}+\\frac{\\partial v}{\\partial y}= \\frac{\\partial^2\\psi}{\\partial x \\partial y} - \\frac{\\partial^2\\psi}{\\partial y \\partial x}=0,$$\nso eq.\\ \\eqref{eq:NseConMas} is automatically satisfied. Making this change of variables, we obtain a single scalar partial differential equation by taking the curl of the momentum equation, eq.\\ \\eqref{eq:NseConMom}. We define the vorticity $\\omega$, so that\n$$\\omega=\\nabla\\times\\mathbf u= \\frac{\\partial v}{\\partial x}-\\frac{\\partial u}{\\partial y}=-\\Delta \\psi$$\nand eq.\\ \\eqref{eq:NseConMom} becomes\n\\begin{align*}\n&{}\\hspace{2em}\n\\frac{\\partial }{\\partial x} \\left[\\rho\\left(\\frac{\\partial v}{\\partial t} +u \\frac{\\partial v}{\\partial x} +v\\frac{\\partial v}{\\partial y}\\right) \\right]\n- \\frac{\\partial }{\\partial y} \\left[\\rho\\left(\\frac{\\partial u}{\\partial t} +u \\frac{\\partial u}{\\partial x} +v\\frac{\\partial u}{\\partial y}\\right) \\right]\n\\\\&{}=\\frac{\\partial }{\\partial x} \\left[\\mu\\left(\\frac{\\partial^2 v}{\\partial x^2} + \\frac{\\partial^2 v}{\\partial y^2}\\right) + fy \\right]\n-\\frac{\\partial }{\\partial y} \\left[\\mu\\left(\\frac{\\partial^2 u}{\\partial x^2} + \\frac{\\partial^2 u}{\\partial y^2}\\right) + fx \\right]\n\\end{align*}\nwhere $fx$ and $fy$ represent the $x$ and $y$ components of the force $\\mathbf f$. Since the flow is divergence free, \n$$\\frac{\\partial u}{\\partial x}=-\\frac{\\partial v}{\\partial x}$$\nand so can simplify the nonlinear term to get\n\\begin{align*}\n&{}\\hspace{2em}\\frac{\\partial }{\\partial x} \\left[\\left( u\\frac{\\partial v}{\\partial x} +v\\frac{\\partial v}{\\partial y}\\right) \\right]\n- \\frac{\\partial }{\\partial y} \\left[\\left(u \\frac{\\partial u}{\\partial x} +v\\frac{\\partial u}{\\partial y}\\right) \\right]\n\\\\&{}=\\frac{\\partial u}{\\partial x}\\frac{\\partial v}{\\partial x}+u\\frac{\\partial^2 v}{\\partial x^2 } +\\frac{\\partial v}{\\partial x}\\frac{\\partial v}{\\partial y} + v\\frac{\\partial^2v }{\\partial x \\partial y}\n-\\frac{\\partial u}{\\partial y}\\frac{\\partial u}{\\partial x} - u\\frac{\\partial^2u }{\\partial x\\partial y} -\\frac{\\partial v}{\\partial y}\\frac{\\partial u}{\\partial y} -v \\frac{\\partial^2u }{\\partial y^2 }\n\\\\&{}= u\\left( \\frac{\\partial^2 v}{\\partial x^2 } - \\frac{\\partial^2u }{\\partial x\\partial y} \\right) + v\\left( \\frac{\\partial^2 v}{\\partial x \\partial y } - \\frac{\\partial^2u }{\\partial^2 y} \\right). \n\\end{align*}\nWe finally obtain\n\\begin{align} \\label{eq:Ns2dStrVorA}\n&{}\\hspace{1em}\\rho\\left(\\frac{\\partial \\omega}{\\partial t} + u\\frac{\\partial \\omega}{\\partial x} + v\\frac{\\partial \\omega }{\\partial y}\\right)\n =\\mu\\Delta\\omega+\\frac{\\partial fy}{\\partial x}- \\frac{\\partial fx}{\\partial y} \n \\end{align}\nand\n\\begin{align} \\label{eq:Ns2dStrVorB}\n&{}\\Delta \\psi = -\\omega.\n\\end{align}\nNote that in this formulation, the Navier-Stokes equation is like a forced heat equation for the vorticity with a nonlocal and nonlinear term. We can take advantage of this structure in finding numerical solutions by modifying our numerical programs which give approximate solutions to the heat equation.\n\nA simple time discretization for this equation is the Crank-Nicolson method, where the nonlinear terms are solved for using fixed point iteration. A tutorial on convergence of time discretization schemes for the Navier-Stokes equations can be found in Temam~\\cite{Tem01}. The time discretized equations become\n\\begin{align} \\label{eq:Ns2dStrVorACn}\n&{}\\hspace{2em}\\rho\\left[\\frac{\\omega^{n+1,k+1}-\\omega^n}{\\delta t} \\right.\n\\\\&{}\\left.+\\frac{1}{2} \\left( u^{n+1,k}\\frac{\\partial \\omega^{n+1,k}}{\\partial x} + v^{n+1,k}\\frac{\\partial \\omega^{n+1,k} }{\\partial y} \n+ u^{n}\\frac{\\partial \\omega^{n}}{\\partial x} + v^{n}\\frac{\\partial \\omega^{n} }{\\partial y} \\right)\\right] \\notag\n\\\\&{}=\\frac{\\mu}{2}\\Delta\\left(\\omega^{n+1,k+1}+\\omega^n\\right) + \\left.\\left(\\frac{\\partial fx}{\\partial y}-\\frac{\\partial fy}{\\partial x} \\right)\\right\\rvert_{t=(n+0.5)\\delta t},\\notag\n\\end{align}\nand\n\\begin{align} \\label{eq:Ns2dStrVorBCn}\n&{}\\Delta \\psi^{n+1,k+1} = -\\omega^{n+1,k+1}, \\quad u^{n+1,k+1}=\\frac{\\partial \\psi^{n+1,k+1}}{\\partial y}, \\quad v^{n+1,k+1}=-\\frac{\\partial \\psi^{n+1,k+1}}{\\partial x}.\n\\end{align}\nIn these equations, the superscript $n$ denotes the timestep and the superscript $k$ denotes the iterate. Another choice of time discretization is the implicit midpoint rule which gives,\n\\begin{align} \\label{eq:Ns2dStrVorAimr}\n&{}\\hspace{2em}\\rho\\left[\\frac{\\omega^{n+1,k+1}-\\omega^n}{\\delta t} \\right.\n\\\\&{}\\left.+ \\left(\\frac{u^{n+1,k}+u^n}{2}\\right)\\frac{\\partial}{\\partial x}\\left(\\frac{\\omega^{n+1,k}+\\omega^n}{2}\\right) + \\left(\\frac{v^{n+1,k}+v^n}{2}\\right)\\frac{\\partial}{\\partial y}\\left(\\frac{ \\omega^{n+1,k} +\\omega^n}{2}\\right)\\right] \\notag\n\\\\&{}=\\frac{\\mu}{2}\\Delta\\left(\\omega^{n+1,k+1}+\\omega^n\\right) + \\left.\\left(\\frac{\\partial fx}{\\partial y}-\\frac{\\partial fy}{\\partial x} \\right)\\right\\rvert_{t=(n+0.5)\\delta t},\\notag\n\\end{align}\nand\n\\begin{align} \\label{eq:Ns2dStrVorBimr}\n&{}\\Delta \\psi^{n+1,k+1} = -\\omega^{n+1,k+1}, \\quad u^{n+1,k+1}=\\frac{\\partial \\psi^{n+1,k+1}}{\\partial y}, \\quad v^{n+1,k+1}=-\\frac{\\partial \\psi^{n+1,k+1}}{\\partial x}.\n\\end{align}\n\n%%%%%%%\n%Section\n%%%%%%%\n\\section{The Three-Dimensional Case}\n\nHere $\\mathbf u = (u(x,y,z,t),v(x,y,z,t),w(x,y,z,t))$ -- unfortunately, it is not clear if this equation has a unique solution for reasonable boundary conditions and initial data. Numerical methods so far seem to indicate that the solution is unique, but in the absence of a proof, we caution the reader that we are {\\it fearless engineers writing gigantic codes that are supposed to produce solutions to the Navier-Stokes equations when what we are really studying is the output of the algorithm} which we hope will tell us something about these equations\\footnote{This is paraphrased from Gallavoti\\cite[p.~VIII]{Gal02}} -- in practice, although the mathematical foundations for this are uncertain, these codes do seem to give information about the motion of nearly incompressible fluids in many, although not all situations of practical interest. Further information on this aspect of these equations can be found in Doering and Gibbon~\\cite{DoeGib95}.\n\nWe will again consider simulations with periodic boundary conditions to make it easy to apply the Fourier transform. This also makes it easier to enforce the incompressibility constraint by using an idea due to Orszag and Patterson~\\cite{OrsPat72} and also explained in Canuto et al.~\\cite[p.~99]{CHQZ07}. If we take the divergence of the Navier-Stokes equations, we get\n\\begin{align}\n\\nabla\\cdot \\left(\\mathbf u\\cdot\\nabla\\mathbf u\\right)=-\\Delta p\n\\end{align}\nbecause $\\nabla\\cdot\\mathbf u=0$. Hence\n\\begin{align}\n&{}p=-\\Delta^{-1}\\left[\\nabla\\cdot \\left(\\mathbf u\\cdot\\nabla\\mathbf u\\right)\\right] \n\\end{align}\nwhere $\\Delta^{-1}$ is defined using the Fourier transform, thus if $f(x,y,z)$ is a mean zero, periodic scalar field and $\\hat{f}$ is its Fourier transform, then\n$$\\widehat{\\Delta^{-1}f}=\\frac{\\hat{f}}{k_x^2+k_y^2+k_z^2}$$\nwhere $k_x$, $k_y$ and $k_z$ are the wavenumbers. The Navier-Stokes equations then become\n\\begin{align}\n&{}\\frac{\\partial \\mathbf u}{\\partial t}=\\frac{1}{\\textup{Re}}\\Delta \\mathbf u - \\mathbf u\\cdot\\nabla\\mathbf u +\\nabla \\Delta^{-1}\\left[\\nabla\\cdot \\left(\\mathbf u\\cdot\\nabla\\mathbf u\\right)\\right], \\label{eq:NS3dOrsPat}\n\\end{align}\nfor which the incompressibility constraint is satisfied, provided the initial data satisfy the incompressibility constraint. \n\nTo discretize \\eqref{eq:NS3dOrsPat} in time, we will use the implicit midpoint rule. This gives,\n\\begin{align}\n\\hspace{1em}\\frac{\\mathbf u^{n+1}-\\mathbf u^{n}}{\\delta t} =&{}\\frac{0.5}{\\textup{Re}}\\Delta\\left( \\frac{\\mathbf u^{n+1}+\\mathbf u^{n}}{2}\\right) - 0.25\\left(\\mathbf u^{n+1} + \\mathbf u^{n}\\right)\\cdot\\nabla\\left(\\mathbf u^{n+1} + \\mathbf u^{n} \\right) \\notag\n\\\\&{}\\hspace{1em} + 0.25\\nabla\\left[ \\Delta^{-1}\\left(\\nabla\\cdot \\left[\\left(\\mathbf u^{n+1} + \\mathbf u^{n}\\right) \\cdot\\nabla\\left(\\mathbf u^{n+1} + \\mathbf u^{n} \\right)\\right]\\right)\\right]. \\label{eq:NS3dOrsPatIMR}\n\\end{align}\n\nIt is helpful to test the correctness of the programs by comparing them to an exact solution. Shapiro~\\cite{Sha93} has found the following exact solution which is a good test for meteorological hurricane simulation programs, as well as for Navier-Stokes solvers with periodic boundary conditions\n\\begin{align*}\n&{} u= -\\frac{A}{k^2+l^2}\\left[\\lambda l\\cos(kx)\\sin(ly)\\sin(mz)+mk\\sin(kx)\\cos(ly)\\cos(mz) \\right]\\exp\\left(-\\frac{\\lambda^2t}{\\textup{Re}}\\right)\n\\\\&{} v= \\frac{A}{k^2+l^2}\\left[\\lambda k\\sin(kx)\\cos(ly)\\sin(mz)-ml\\cos(kx)\\sin(ly)\\cos(mz) \\right]\\exp\\left(-\\frac{\\lambda^2t}{\\textup{Re}}\\right)\n\\\\&{} w= A\\cos(kx)\\cos(ly)\\sin(mz)\\exp\\left(-\\frac{\\lambda^2t}{\\textup{Re}}\\right)\n\\end{align*}\nwhere the constant $\\lambda=\\sqrt{k^2+l^2+m^2}$ and $l$, $k$ and $m$ are constants choosen with the restriction that the solutions are periodic in space. Further examples of such solutions can be found in Majda and Bertozzi~\\cite[sec.~2.3]{MajBer02}.\n\n%%%%%%%\n%Section\n%%%%%%%\n\\section{Serial Programs}\nWe first write Matlab programs to demonstrate how to solve these equations on a single processor. The first program uses Crank-Nicolson timestepping to solve the two-dimensional Navier-Stokes equations and is in listing \\ref{lst:Ns2DMatlab}. To test the program, following Laizet and Lamballais\\cite{LaiLam09} we use the exact Taylor-Green vortex solution on $(x,y)\\in[0,1]\\times[0,1]$ with periodic boundary conditions given by\n\\begin{align}\n&{}u(x,y,t)=\\sin(2\\pi x)\\cos(2\\pi y)\\exp(-8\\pi^2\\mu t)\n\\\\&{}v(x,y,t)=-\\cos(2\\pi x)\\sin(2\\pi y)\\exp(-8\\pi^2\\mu t).\n\\end{align}\n\n\\lstinputlisting[style=matlab_style,label=lst:Ns2DMatlab,caption={A Matlab program which finds a numerical solution to the 2D Navier Stokes equation.}]{./NavierStokes/Programs/NsMatlabCn/NavierStokes2DFFTCn.m}\n\nThe second program uses the implicit midpoint rule to do timestepping for the three-dimensional Navier-Stokes equations and it is in listing \\ref{lst:Ns3DMatlab}. It also takes the Taylor-Green vortex as its initial condition since this has been extensively studied, and so provides a baseline case to compare results against.\n\n\\lstinputlisting[style=matlab_style,label=lst:Ns3DMatlab,caption={A Matlab program which finds a numerical solution to the 3D Navier Stokes equation.}]{./NavierStokes/Programs/NsMatlab3Dimr/NavierStokes3DfftIMR.m}\n\n% subsection\n\\subsection{Exercises}\n\\begin{enumerate}\n\\item[1)] Show that for the Taylor-Green vortex solution, the nonlinear terms in the two-dimensional Navier-Stokes equations cancel out exactly.\n\\item[2)] Write a Matlab program that uses the implicit midpoint rule instead of the Crank-Nicolson method to obtain a solution to the 2D Navier-Stokes equations. Compare your numerical solution with the Taylor-Green vortex solution.\n\\item[3)] Write a Fortran program that uses the implicit midpoint rule instead of the Crank-Nicolson method to obtain a solution to the 2D Navier-Stokes equations. Compare your numerical solution with the Taylor-Green vortex solution.\n\\item[4)] Write a Matlab program that uses the Crank-Nicolson method instead of the implicit midpoint rule to obtain a solution to the 3D Navier-Stokes equations. \n\\item[5)] Write a Fortran program that uses the Crank-Nicolson method instead of the implicit midpoint rule to obtain a solution to the 3D Navier-Stokes equations.\n\\item[6)] The Navier-Stokes equations as written in eqs.\\ \\eqref{eq:Ns2dStrVorA} and \\eqref{eq:Ns2dStrVorB} also satisfy further integral properties. In particular show that\n\\begin{enumerate}\n\\item[a)] $$\\frac{\\rho}{2}\\frac{\\mathrm{d}}{\\mathrm{d}t}\\lVert \\omega \\rVert_{l^2}^2= -\\mu\\lVert \\nabla \\omega \\rVert_{l^2}^2,$$\nwhere \n$$\\lVert \\omega \\rVert_{l^2}^2=\\int\\int(\\omega)^2\\mathrm{d}x\\mathrm{d}y$$\nand\n$$\\lVert \\nabla \\omega \\rVert_{l^2}^2=\\int\\int(\\nabla\\omega)\\cdot(\\nabla\\omega)\\mathrm{d}x\\mathrm{d}y.$$\nHINT: multiply the Eq.\\ \\eqref{eq:Ns2dStrVorA} by $\\omega$ then integrate by parts. \n\\item[b)] Show that part (a) implies that \n$$\\lVert \\omega(t=T) \\rVert_{l^2}^2-\\lVert \\omega(t=0) \\rVert_{l^2}^2=-\\mu\\int_0^T\\lVert \\nabla \\omega \\rVert_{l^2}^2\\mathrm{d}t$$\n\\item[c)] Part (b) gives a property one can check when integrating the 2D Navier-Stokes equations. We now show that the implicit midpoint rule satisfies an analogous property. Multiply eq.\\ \\eqref{eq:Ns2dStrVorAimr} by $0.5(\\omega^{n+1}+\\omega^n)$, integrate by parts in space, then sum over time to deduce that\n$$\\lVert \\omega^N \\rVert_{l^2}^2-\\lVert \\omega^0 \\rVert_{l^2}^2=-\\frac{\\mu}{4}\\sum_{n=0}^{N-1}\\left\\lVert \\nabla\\left( \\omega^n+\\omega^{n+1}\\right) \\right\\rVert_{l^2}^2\\delta t.$$\n\\item[d)] Deduce that this implies that the implicit midpoint rule time stepping method is unconditionally stable, provided the nonlinear terms can be solved for\\footnote{We have not demonstrated convergence of the spatial discretization, so this result assumes that the spatial discretization has not been done.}.\n\\end{enumerate}\n\\end{enumerate}\n\n%%%%%%%\n%Section\n%%%%%%%\n\\section{Parallel Programs: OpenMP}\n\nRather than give fully parallelized example programs, we instead give a simple implementation in Fortran of the Crank-Nicolson and implicit midpoint rule algorithms for the two-dimensional and three dimensional Navier-Stokes equations that were presented in Matlab. The program for the two-dimensional equations is presented in listing \\ref{lst:For2dNs} and an example Matlab script to plot the resulting vorticity fields is in listing \\ref{lst:MatlabPlot2dNS}. This program is presented in listing \\ref{lst:For3dNs} and an example Matlab script to plot the resulting vorticity fields is in listing \\ref{lst:MatlabPlot3dNS}.\n\n\\lstinputlisting[style=fortran_style,language=Fortran,label=lst:For2dNs,caption={A Fortran program to solve the 2D Navier-Stokes equations.}]{./NavierStokes/Programs/NavierStokes2dFortran/navierstokes.f90}\n\n\\lstinputlisting[style=matlab_style,label=lst:MatlabPlot2dNS,caption={A Matlab program to plot the vorticity fields and error produced by listing \\ref{lst:For2dNs}.}]{./NavierStokes/Programs/NavierStokes2dFortran/plotdata.m}\n\n\\lstinputlisting[style=fortran_style,language=Fortran,label=lst:For3dNs,caption={A Fortran program to solve the 3D Navier-Stokes equations.}]{./NavierStokes/Programs/NavierStokes3dFortran/NavierStokes3DfftIMR.f90}\n\n\\lstinputlisting[style=matlab_style,label=lst:MatlabPlot3dNS,caption={A Matlab program to plot the vorticity fields produced by listing \\ref{lst:For3dNs}.}]{./NavierStokes/Programs/NavierStokes3dFortran/plotdata.m}\n\n\\subsection{Exercises}\n\\begin{enumerate}\n\\item[1)] Verify that the program in listing \\ref{lst:For2dNs} is second order accurate in time.\n\\item[2)] Use OpenMP directives to parallelize the example Fortran code for the two-dimensional Navier Stokes equations. Try and make it as efficient as possible.\n\\item[3)] Write another code which uses threaded FFTW to do the Fast Fourier transforms. This code should have a similar structure to the program in listing \\ref{lst:For2dNlsOmp2}. \n\\item[4)] Use OpenMP directives to parallelize the example Fortran code for the three-dimensional Navier-Stokes equations in listing \\ref{lst:For3dNs}. Try and make it as efficient as possible.\n\\item[5)] Write another code which uses threaded FFTW to do the Fast Fourier transforms for the three-dimensional Navier-Stokes equations. This code should have a similar structure to the program in listing \\ref{lst:For2dNlsOmp2}. \n\\end{enumerate}\n%%%%%%%\n%Section\n%%%%%%%\n\\section{Parallel Programs: MPI}\n\nThe code for this is very similar to the serial code in listing \\ref{lst:For2dNs}. For completeness and to allow one to see how to parallelize other programs, we include it. The program uses the library 2DECOMP\\&FFT. One difference between this program and the serial program is that a subroutine is included to write out data. Since this portion of the calculation is repeated several times, the program becomes more readable when the repeated code is placed in a subroutine. The subroutine is also generic enough that it can be reused in other programs, saving program developers time.\n\n\\lstinputlisting[style=fortran_style,language=Fortran,label=lst:For3dNsMPI,caption={A parallel MPI Fortran program to solve the 3D Navier-Stokes equations.}]{./NavierStokes/Programs/NavierStokes3dFortranMPI/NavierStokes3DfftIMR.f90}\n\n\\lstinputlisting[style=fortran_style,language=Fortran,label=lst:For3dNsMPIdata,caption={A subroutine to save real array data for the parallel MPI Fortran program to solve the 3D Navier-Stokes equations in listing \\ref{lst:For3dNsMPI}.}]{./NavierStokes/Programs/NavierStokes3dFortranMPI/savedata.f90}\n\n\\lstinputlisting[style=make_style,language=make,label=lst:For3dNsMPImake,caption={A makefile to compile the parallel MPI Fortran program to solve the 3D Navier-Stokes equations.}]{./NavierStokes/Programs/NavierStokes3dFortranMPI/makefile}\n\n\n\\subsection{Exercises}\n\\begin{enumerate}\n\\item[1)] Use 2DECOMP{\\&}FFT to write a two dimensional Navier-Stokes solver. The library is built to do three dimensional FFTs, however by choosing one of the arrays to have only one entry, the library can then do two dimensional FFTs on a distributed memory machine.\n\\item[2)] Uecker~\\cite{Uec09} describes the expected power law scaling for the power spectrum of the enstrophy\\footnote{The enstrophy is the square of the vorticity.} in two dimensional isotropic turbulence. Look up Uecker~\\cite{Uec09} and then try to produce numerical data which verifies the power scaling law over as many decades of wavenumber space as are feasible on the computational resources you have access to. A recent overview of research work in this area can be found in Boffetta and Ecke~\\cite{BofEck12}. Fornberg~\\cite{For77} discusses how to calculate power spectra.\n\\item[3)] If we set $\\mu=0$ the Navier Stokes equations become the Euler equations. Try to use the implicit midpoint rule and/or the Crank-Nicolson methods to simulate the Euler equations in either two or three dimensions. See if you can find good iterative schemes to do this, you may need to use Newton iteration. An introduction to the Euler equations is in Majda and Bertozzi~\\cite{MajBer02}.\n\\item[4)] The Taylor-Green vortex flow initial conditions have been studied as a possible flow that could have a blow up in the maximum value of the absolute value of the gradient of the velocity at a point for the Euler and Navier-Stokes equations. In many of these simulations, symmetries have been used to get higher effective resolutions, see for example Cichowlas and Brachet~\\cite{CicBra05}. Consider using the Kida-Pelz and Taylor-Green vortex as initial conditions for the Euler equations and adding non-symmetric perturbations. If you are unable to get an implicit time-stepping scheme to work, consider using an explicit scheme such as a Runge-Kutta method. How does the flow evolve in comparison to previous studies in the literature? An introduction to the blow up for the Euler equations is in Majda and Bertozzi~\\cite{MajBer02}.\n\\item[5)] The three dimensional program we have written is not the most efficient since one can use a real to complex transform to halve the work done. Implement a real to complex transform in one of the Navier-Stokes programs.\n\\item[6)] The programs we have written can also introduce some aliasing errors. By reading a book on spectral methods, such as Canuto et al.~\\cite{CHQZ07}, find out what aliasing errors are. Explain why the strategy explained in Johnstone~\\cite{Joh12} can reduce aliasing errors. \n\\end{enumerate}\n", "meta": {"hexsha": "75462ea338611711191fbe22ae705671d6034bf0", "size": 22355, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "NavierStokes/NavierStokes.tex", "max_stars_repo_name": "bcloutier/PSNM", "max_stars_repo_head_hexsha": "1cd03f87f93ca6cb1a3cfbe73e8bc6106f497ddf", "max_stars_repo_licenses": ["CC-BY-3.0", "BSD-2-Clause"], "max_stars_count": 40, "max_stars_repo_stars_event_min_datetime": "2015-01-05T14:22:59.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-18T23:51:25.000Z", "max_issues_repo_path": "NavierStokes/NavierStokes.tex", "max_issues_repo_name": "bcloutier/PSNM", "max_issues_repo_head_hexsha": "1cd03f87f93ca6cb1a3cfbe73e8bc6106f497ddf", "max_issues_repo_licenses": ["CC-BY-3.0", "BSD-2-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-12-29T12:35:42.000Z", "max_issues_repo_issues_event_max_datetime": "2021-01-01T07:31:32.000Z", "max_forks_repo_path": "NavierStokes/NavierStokes.tex", "max_forks_repo_name": "bcloutier/PSNM", "max_forks_repo_head_hexsha": "1cd03f87f93ca6cb1a3cfbe73e8bc6106f497ddf", "max_forks_repo_licenses": ["CC-BY-3.0", "BSD-2-Clause"], "max_forks_count": 34, "max_forks_repo_forks_event_min_datetime": "2015-01-05T14:23:02.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-09T06:55:01.000Z", "avg_line_length": 111.2189054726, "max_line_length": 980, "alphanum_fraction": 0.750123015, "num_tokens": 6729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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YES", "lm_q1_score": 0.8311430353105599, "lm_q2_score": 0.7799929104825006, "lm_q1q2_score": 0.6482856751391434}} {"text": "\\title{Enumeration of the Building Game}\n\\author{\n Daniel Johnson \\& Govind Menon\\\\\n Division of Applied Mathematics\\\\\n Brown University\n}\n\\date{\\today}\n\n\\documentclass[12pt]{article}\n\n\\usepackage{graphicx,amsmath,mathtools,bbm,amsthm,enumerate}\n\\usepackage{mathrsfs}\n%\\usepackage[subnum]{cases}\n\n%\\usepackage[titletoc,toc,title]{appendix}\n\n\n\\newtheorem{mythm}{Theorem}\n\\newtheorem{mylem}{Lemma}\n\\newtheorem{mycor}{Corollary}\n\\newtheorem{mydef}{Definition}\n\n\\newcommand{\\colorA}{white}\n\\newcommand{\\colorB}{black}\n\\newcommand{\\colorAsm}{w}\n\\newcommand{\\colorBsm}{b}\n\\newcommand{\\poly}{$\\mathscr{P}$}\n\\newcommand{\\faceset}{F\\left(\\mathscr{P}\\right)}\n\\newcommand{\\spc}{ }\n\\newcommand{\\xj}{$x^j$}\n\\newcommand{\\xk}{$x^k$}\n\\newcommand{\\Sjk}{$S_{jk}$}\n\\newcommand{\\Skj}{$S_{kj}$}\n\n\\DeclareMathOperator{\\diag}{diag}\n\n\n\\begin{document}\n\\maketitle\n\n\\begin{abstract}\nThe Building Game is a sequential coloring process on polyhedra. We enumerate the Building Game state space for all polyhedra in the Platonic, Archimedean, and Catalan solids classes of up to 30 faces. By putting a probability distribution on each step of the Building Game process, a distribution is induced on the entire state space. With the help of a finite group theoretic identity, we find the explicit form of these distributions. Finally, we examine the properties of the resulting distributions.\n\\end{abstract}\n\n\\section{Introduction}\nThe Building Game (BG) was first considered by Zlotnick~\\cite{Zlotnick1994} as a model for the assembly of polyhedral viral capsids. We formalize the idea as a sequential coloring process that progresses from a polyhedron \\poly\\spc with each face colored \\colorA, through a number of intermediate states each having a mix of \\colorA\\spc and \\colorB\\spc faces, and ending with all of the faces colored \\colorB. \n\\begin{mydef}\nA Building Game \\textbf{intermediate} $x$ is a function from the faces of \\poly\\spc, $F\\left(\\mathscr{P}\\right)$, to a color in $\\left\\{\\colorA,\\colorB\\right\\}$ such that the set $\\left\\{f_m \\in F\\left(\\mathscr{P}\\right) : x\\left(f_m\\right) = \\colorB\\right\\}$ is edge connected along with the equivalence relation $x \\sim x'$ if there is an element $g$ of \\poly's rotation group $G$ that satisfies $x(f_m) = x'(g.f_m)$ for every $f_m \\in F(\\mathscr{P})$. \n\\end{mydef}\nFor ease of exposition, we use the notational shorthand $\\left(x\\right)_m$ for $x\\left(f_m\\right)$ and $x = g.x'$ when $x(f_m) = x'(g.f_m)$ for every $f_m \\in F(\\mathscr{P})$. Additionally, we denote the intermediate satisfying $\\left(x\\right)_m = \\colorA$ for all $f_m \\in \\faceset$ as $x^\\colorAsm$ and similarly $x^\\colorBsm$ is the intermediate with $\\left(x\\right)_m = \\colorB$ for all $f_m \\in \\faceset$. The function counting the number of \\colorB\\spc faces an intermediate has is denoted $h\\left(x\\right) \\doteq |\\left\\{f_m \\in \\faceset : \\left(x\\right)_m = \\colorB\\right\\}|$.\n\n\\begin{mydef}\nTwo intermediates $x^j$ and $x^k$ are \\textbf{connected} ($x^j \\leftrightarrow x^k$) if $\\left(x^j\\right)_m = \\left(x^k\\right)_m$ for all $f_m \\in \\faceset$ except for exactly one face $f_n$ that has $\\left(x^j\\right)_n \\neq \\left(x^k\\right)_n$.\n\\end{mydef}\n\\begin{mydef}\nA Building Game \\textbf{pathway} is a sequence of intermediates $x^{p_0}, x^{p_1}, x^{p_2}, \\dots, x^{p_N}$ such that $x^{p_0} = x^\\colorAsm$, $x^{p_N} = x^\\colorBsm$, $x^{p_i}$ is connected to $x^{p_{i+1}}$ and $h\\left(x^{p_i}\\right) = i$.\n\\end{mydef}\n\nIn this way it is useful to think of intermediates as connected if it is possible to color one face of the first intermediate to get the second and a pathway as a sequence of these connections between $x^\\colorAsm$ and $x^\\colorBsm$. Figure~\\ref{fig:DodecBG} shows a Building Game pathway for the dodecahedron using Schlegel diagrams. The pathways has 13 intermediates since there must be exactly one intermediate $x^{p_i}$ satisfying $h\\left(x^{p_i}\\right) = i$ for each $i = 0,1,2,\\dots,12$.\n\n\\begin{figure}[ht]\n\\caption{One Building Game pathway on the dodecahedron.}\n\\label{fig:DodecBG}\n\\end{figure}\n\nWith many pairs of connected intermediates, we organize these relations in a graph.\n\n\\begin{mydef}\nThe Building Game \\textbf{state space} for a polyhedron \\poly\\spc is a graph in which the nodes are \\poly's intermediates and a graph edge exists between two intermediates if and only if they are connected. \n\\end{mydef}\n\nWhen the intermediates are partitioned by their value of $h$, it is natural to arrange the state space as a tiered graph according to this partition. Figure~\\ref{fig:CubeSS} shows the Building Game state space for the cube. As seen, each tier has intermediates with the same number of \\colorB faces and connections thus exist with intermediates that are either in the tier directly above or below them. We can also see that there are three distinct pathways contained in the state space. \n\n\\begin{figure}[ht]\n\\caption{The Building Game state space of the cube.}\n\\label{fig:CubeSS}\n\\end{figure}\n\nInterestingly, it is not the case that the recoloring of each face of \\xj\\spc results in a distinct intermediate. \n\\begin{mydef}\nThe number of different faces $\\left|\\left\\{f_m \\in \\faceset : x^j + e^m \\in \\left[x^k\\right]\\right\\}\\right|$ of \\xj\\spc that can be colored to form \\xk\\spc is called the \\textbf{degeneracy number} \\Sjk.\n\\end{mydef}\nIt is important to note that in general the degeneracy number is not symmetric, i.e. \\Sjk$\\neq$\\Skj\\spc for some connections \\xj\\spc$\\leftrightarrow$\\spc\\xk\\spc in the state space. Both figures~\\ref{fig:DodecBG} and ~\\ref{fig:DodecBG} show the forward and backward degeneracy numbers for each connection.\n\n\\subsection{Related Work}\n--Like polyominos on polyhedra\n\n\\subsection{Applications}\n--Viral capsid assembly\n\n--Self-assembly of molecular cages\n\n--Self assembly for manufacturing purposes\n%~\\cite{Endres2005}\n\\subsection{Paper Overview}\n-- Summary of subsequent sections\n\n\\section{Enumerative Results}\n\nAs we consider polyhedra with more and more faces, there is a combinatorial explosion in the number intermediates in state space. While the 6-faced cube state space has only 8 nodes and 9 nodes, the 20-faced icosahedron state space has 2,649 nodes and 17,241 nodes and the 26-faced truncated cuboctahedron state space has 1,525,605 nodes and 17,672,377. Figure \\ref{fig:bgtable} details state space sizes of all polyhedra in the Platonic, Archimedean, and Catalan solid classes of up to 26 faces. \n\n\nAlso something about pathway statistics. \n\n\n\\begin{figure}[ht]\n\\scalebox{0.6}{\n%{\\footnotesize\n\\begin{tabular}{ l | c | c | c | c || r | r | r}\nPolyhedra Name & Class & F$\\left(\\mathscr{P}\\right)$ & E$\\left(\\mathscr{P}\\right)$ & V$\\left(\\mathscr{P}\\right)$ & Intermediates & Connections & Pathways \\\\\n \\hline \nTetrahedron & P & 4 & 6 & 4 & 5 \t& 4 & 1\\\\\nCube & P & 6 & 12 & 8 & 9 \t& 10 \t& 3\\\\\nOctahedron & P & 8 & 12 & 6 & 15 \t& 22 \t& 14\\\\\nDodecahedron & P & 12 & 30 & 20 & 74 \t& 264 \t& 17,696 \\\\\nIcosahedron & P & 20 & 30 & 12 & 2,650 \t& 17,242 & 57,396,146,640\\\\\nTruncated Tetrahedron & A & 8 & 18 & 12 & 29 \t& 65 & 402\\\\\nCuboctahedron & A & 14 & 24 & 12 \t& 341 \t& 1,636 & 10,170,968\\\\\nTruncated Cube & A & 14 & 36 & 24 \t& 500 \t& 2,731 & 101,443,338 \\\\\nTruncated Octahedron & A & 14 & 36 & 24 \t& 556 & 3,071 & 68,106,377\\\\\nRhombicuboctahedron & A & 26 & 48 & 24 \t& 638,851 & 6,459,804 & 16,494,392,631,838,879,380\\\\\nTruncated Cuboctahedron & A & 26 & 72 & 48 \t& 1,525,605 & 17,672,377 & ? \\\\\nIcosidodecahedron & A & 32 & 60 & 30 \t& ? & ? & ?\\\\\nTruncated Dodecahedron & A & 32 & 90 & 60 \t& ? & ? \t & ? \\\\\nTruncated Icosahedron & A & 32 & 90 & 60 \t& ? & ? \t & ?\\\\\nTriakis Tetrahedron & C & 12 & 18 & 8 & 99 & 319 & 38,938\\\\\nRhombic Dodecahedron & C & 12 & 24 & 14 \t& 128 & 494 & 76,936\\\\\nTriakis Octahedron & C & 24 & 36 & 14 \t& 12,749 & 81,297 & 169,402,670,046,670\\\\\nTetrakis Hexahedron & C & 24 & 36 & 14 \t& 50,768 & 394,278 & 4,253,948,297,210,346\\\\\nDeltoidal Icositetrahedron & C & 24 & 48 & 26 \t& 209,676 & 1,989,549 & ? \\\\\nPentagonal Icositetrahedron & C & 24 & 60 & 38 \t& 345,939 & 3,544,988 & 2,828,128,000,716,774,492\\\\\nRhombic Triacontahedron & C & 30 & 60 & 32 \t& ? & ? & 5,266,831,101,345,821,968\\\\\n \\hline \n\\end{tabular}\n}\n\\caption{Table of polyhedra in the Platonic (P), Archimedean (A), and Catalan (C) solid classes of up to 32 faces and their Building Game state space statistics.}\n\\label{fig:bgtable}\n\\end{figure}\n\n\n\\begin{figure}[ht]\n\\scalebox{0.6}{\n%{\\footnotesize\n\\begin{tabular}{ l | c | c | c | c || r | r | r}\nPolyhedra Name & Class & F$\\left(\\mathscr{P}\\right)$ & E$\\left(\\mathscr{P}\\right)$ & V$\\left(\\mathscr{P}\\right)$ & Intermediates & Connections & Pathways \\\\\n \\hline \nTetrahedron & P & 4 & 6 & 4 & 5 & 4 & 1\\\\\nCube & P & 6 & 12 & 8 & 8 & 8 & 2\\\\\nOctahedron & P & 8 & 12 & 6 & 12 & 12 & 14 \\\\\nDodecahedron & P & 12 & 30 & 20 & 53 & 156 & 2166\\\\\nIcosahedron & P & 20 & 30 & 12 & 468 & 1984 & 105999738\\\\\nTruncated Tetrahedron & A & 8 & 18 & 12 & 22\t& 42 & 174\\\\\nCuboctahedron & A & 14 & 24 & 12 \t& 137\t& 470 & 477776\\\\\nTruncated Cube & A & 14 & 36 & 24 \t& 248\t& 1002 & 5232294\\\\\nTruncated Octahedron & A & 14 & 36 & 24 \t& 343\t& 1466 & 5704138\\\\\nRhombicuboctahedron & A & 26 & 48 & 24 \t& 70836\t& 462149 & 48399693494788840\\\\\nTruncated Cuboctahedron & A & 26 & 72 & 48 \t& ?\t& ? & ?\\\\\nIcosidodecahedron & A & 32 & 60 & 30 \t& ?\t& ? & ?\\\\\nTruncated Dodecahedron & A & 32 & 90 & 60 \t& ?\t& ? & ?\\\\\nTruncated Icosahedron & A & 32 & 90 & 60 \t& ?\t& ? & ?\\\\\nTriakis Tetrahedron & C & 12 & 18 & 8 & 49\t& 116 & 5012\\\\\nRhombic Dodecahedron & C & 12 & 24 & 14 \t& 68\t& 196 & 6258\\\\\nTriakis Octahedron & C & 24 & 36 & 14 \t& 667\t& 2383 & 15255459\\\\\nTetrakis Hexahedron & C & 24 & 36 & 14 \t& 4220\t& 21079 & 5854799360107\\\\\nDeltoidal Icositetrahedron & C & 24 & 48 & 26 \t& ?\t& ? & ?\\\\\nPentagonal Icositetrahedron & C & 24 & 60 & 38 \t& 95127\t& 654537 & 5607231936129109\\\\\nRhombic Triacontahedron & C & 30 & 60 & 32 \t& 97368\t& 697623 & 6889989896241902854\\\\\n \\hline \n\\end{tabular}\n}\n\\caption{Table of polyhedra in the Platonic (P), Archimedean (A), and Catalan (C) solid classes of up to 32 faces and their Building Game state space shellability statistics.}\n\\label{fig:bgtable_shell}\n\\end{figure}\n\n\n\\subsection{Bounds and Asymptotics}\nHave upper, but what about lower? \n\\subsection{Methods}\n\\section{The Building Game as a Stochastic Process}\n\\label{sec:Prob}\nSince the Building Game is a sequential process with several choices at each step, it is natural to consider it as a stochastic process. By putting a distribution on all possible faces that can be colored \\colorB\\spc at each step of the Building game, a distribution on the space of pathways is implicitly defined. Thus, for a choice of this transition rule, we can ask questions about the likelihood of the different pathways. \n\n--Math and graphical results about putting a distribution on pathways\n\n\\subsection{Forward and Backward Transitions}\n\nIf we allow faces be changed both from \\colorA\\spc to \\colorB\\spc and from \\colorB\\spc to \\colorA, the process consists of transitions from intermediate to intermediate along state space connections. By specifying a distribution on these transitions, it will induce a stationary measure on the state space. \n\nWe define the Markov process $X_t$ by the transition rate matrix $Q$, with the heuristic that the rate of transition to an intermediate \\xk from an intermediate \\xk should be proportional to the number of faces of \\xj that can be colored to reach the intermediate \\xk. For this reason, we include the degeneracy number \\Sjk\\spc as a factor in the transition rate matrix. Furthermore, we model the process after and energetic model in which each intermediate has an energy and to transtion between intermediates, an energy barrier $E_{jk} = E_{kj}$ must be overcome. \n%\\begin{align}\n%\\label{eq:TransitionProbability}\n% P_{jk} = \\frac{1}{z_j}S_{jk}\\rh \n%\\end{align}\n\\begin{align}\n\\label{eq:TransitionRate}\nQ_{jk} &= S_{jk}e^{-\\beta\\left(E_{jk} - E_j\\right)} \\\\\nQ_{jj} &= -z_j \\\\\n\\end{align}\nHere, $z_j \\doteq \\sum_{\\ell: \\ell \\neq j} S_{j\\ell}e^{-\\beta\\left(E_{j\\ell} - E_j\\right)}$ is the rate at which the process leaves \\xj. \n\n\\begin{mythm}\n\\label{thm:StatDist}\nIf the transition rate matrix $Q$ can be decomposed as $Q = DC$ where $D$ is diagonal with each entry of the diagonal positive and $C$ is a non-negative symmetric matrix with $C_{jk} > 0$ if and only if $x^j$ and $x^k$ are connected, then $X_t$ has the unique stationary distribution $\\pi = \\diag\\left(D^{-1}\\right)$. \n\\end{mythm}\n\\begin{proof}\nFirst, we show $Q$ and $\\pi$ satisfy detailed balance.\n\\begin{align}\n\\pi_jQ_{jk} &= \\left(\\frac{1}{D_{jj}}\\right)\\left(D_{jj}C_{jk}\\right) \\\\\n&= C_{jk} \\\\\n&= C_{kj} \\\\\n&= \\left(\\frac{1}{D_{kk}}\\right)\\left(D_{kk}C_{kj}\\right) \\\\\n &= \\pi_kQ_{kj}\n\\end{align}\n\n-- Prove aperiodicity \n-- Prove positive reccurence\n\n\\end{proof}\n\nIn order to use theorem~\\ref{thm:StatDist} to find the stationary distribution for the transition rule~\\ref{eq:TransitionProbability}, we must be able to decompose the degeneracy number \\Sjk\\spc to fit the template of $\\mathbf{C}$ and $\\mathbf{D}$. In the following section we derive group theoretic identities to show that this is possible.\n\n\\subsection{Hitting Times}\n\n\\begin{align}\n\t\\tau^{A}_{j} &\\doteq \\inf\\left\\{t \\geq 0 : X_t \\in A, X_0 = x^j\\right\\}\n\\end{align}\n\n\\begin{align}\n\t\\nu^{A}_{j} &\\doteq \\inf\\left\\{n \\geq 0 : Y_n \\in A, Y_0 = x^j\\right\\}\n\\end{align}\n\nFor $j \\not\\in A$.\n\\begin{align}\n\tE\\left[\\tau^{A}_{j}\\right] &= E\\left[E\\left[\\tau^{A}_{j} | Y_1 \\right]\\right] \\\\\n &= E\\left[ Exp\\left(z_j\\right) + \\tau^{A}_{Y_1} \\right] \\\\\n &= \\frac{1}{z_j} + E\\left[\\sum_{k}\\tau^{A}_{Y_1}\\mathbbm{1}_{Y_1 = k}\\right] \\\\\n &= \\frac{1}{z_j} + \\sum_{k: k\\neq j}E\\left[\\tau^{A}_{k}\\right] P\\left(Y_1 = k\\right) \\\\\n &= \\frac{1}{z_j}\\left(1 + \\sum_{k: k\\neq j}q_{jk}E\\left[\\tau^{A}_{k}\\right]\\right) \\\\\n \\sum_{k}q_{jk}E\\left[\\tau^{A}_{k}\\right] &= 1 \\\\\n\\end{align}\n\nFor $j \\in A$.\n\\begin{align}\n\tE\\left[\\tau^{A}_{j}\\right] &= 0 \\\\\n\\end{align}\n\nAs a linear system:\n\\begin{align}\n\t\\left(\\diag\\left(\\mathbbm{1}_A\\right) - \\diag\\left(\\mathbbm{1}_{A^c}\\right)Q\\right)E\\left[\\tau^{A}\\right] =\\mathbbm{1}_{A^c}\\\\\n\\end{align}\n\n\n\\begin{align}\n\\psi_j^A\\left(t\\right) &\\doteq P\\left(\\tau^A_j \\leq t\\right) \\\\\n\\psi_j^A\\left(0\\right) &= \\mathbbm{1}_{j\\in A} \\\\\n\\psi_j^A\\left(t\\right) &= 0 \\forall j \\in A \\\\ \n\\end{align}\n\nFor $j \\not\\in A$.\n\n\\begin{align}\n\\psi_j^A\\left(t\\right) &\\doteq P\\left(\\tau^A_j \\leq t\\right) \\\\\n &= \\sum_k P\\left(\\tau^A_j \\leq t | Y_1 = x^k\\right) P\\left(Y_1 = x^k\\right) \\\\ \n &= \\frac{1}{z_j}\\sum_{k: k \\neq j} q_{jk} P\\left(Exp\\left(z_j\\right)\\tau^A_j \\leq t\\right) \\\\\n &= \\frac{1}{z_j}\\sum_{k: k \\neq j} q_{jk} \\int^t_0 P\\left(\\tau^A_j \\leq t - s\\right) z_j e^{-z_j s} ds \\\\\n &= \\sum_{k: k \\neq j} q_{jk} \\int^t_0\\psi^A_k\\left(t-s\\right)e^{-z_j s} ds \\\\\n &= \\sum_{k: k \\neq j} q_{jk} \\int^t_0\\psi^A_k\\left(r\\right)e^{-z_j\\left(t-r\\right)} dr \\\\\ne^{z_jt}\\psi^A_j\\left(t\\right) &= \\sum_{k: k \\neq j} q_{jk} \\int^t_0 e^{z_jr}\\psi^A_k\\left(r\\right) dr \\\\\ne^{z_jt}\\frac{d\\psi^A_j}{dt} + z_j e^{z_j t} \\psi^A_j\\left(t\\right) &= \\sum_{k: k \\neq j} q_{jk} e^{z_jt}\\psi^A_k\\left(t\\right) \\\\\n\\frac{d\\psi^A_j}{dt} &= \\sum_{k} q_{jk} \\psi^A_k\\left(t\\right) \n\\end{align}\n\nCombining both cases, we get the linear system and solution.\n\n\\begin{align}\n \\frac{d\\psi^A}{dt} &= \\diag\\left(\\mathbbm{1}_{A^c}\\right)Q\\psi^A \\\\\n \\psi^A\\left(0\\right) &= \\mathbbm{1}_{A} \\\\\n \\psi^A\\left(t\\right) &= e^{\\diag\\left(\\mathbbm{1}_{A^c}\\right)Qt} \\mathbbm{1}_{A} \\\\ \n\\end{align}\n\nThis is the solution for the CDF of the stopping time $\\tau^A$, but we can also compute the PDF explicitly for $t > 0$.\n\n\\begin{align}\n p\\left(\\tau^A = t\\right) &= \\frac{d\\psi^A}{dt} \\\\\n &= \\diag\\left(\\mathbbm{1}_{A^c}\\right)Q\\psi^A\n\\end{align} \n\n\n\\section{A Finite Geometric Result}\n\nSince we define Building Game intermediates as rotationally unique from each other, it is useful to think about the problem in the context of $\\mathscr{P}$'s rotational symmetry group $G \\doteq G\\left(\\mathscr{P}\\right)$ and group actions. For an intermediate $x^j$, the number of symmetries $r_j$ is the order of the stabilizer subgroup $G_{x^j} \\doteq \\left\\{g \\in G : g.x^j = x^j\\right\\}$ of $G$ that fixes $x^j$. Suppose $x^j$ and $x^k$ are connected in the state space and $\\varphi$ is one of the $S_{jk}$ faces that, when added to $x^j$, forms $x^k$. We say $x^j + \\varphi = x^k$. The degeneracy number $S_{jk}$ can then be expressed as the order of the orbit $\\left(G_{x^j}\\right).\\varphi$ of $\\varphi$ with respect to $x^j$'s stabilizer subgroup. Analogously, we define the reverse degeneracy number as $S_{kj} \\doteq \\left|\\left(G_{x^k}\\right).\\varphi\\right|$\n\n\\begin{mylem}\n\\label{lem:I}\nFor Building Game intermediates $x^j$ and $x^k$ connected in the state space and a face $f_m \\in \\faceset$ satisfying $x^j + e^m = x^k$, the stabilizer subgroup $G_{x^j,e^m}$ that fixes both $x^j$ and $e^m$ is the same stabilizer subgroup $G_{x^k,e^m}$ that fixes $x^k$ and $e^m$.\n\\end{mylem}\n\\begin{proof}\n\\begin{align}\nG_{x^j,e^m} &\\doteq \\left\\{g \\in G | g.x^j = x^j, g.e^m = e^m \\right\\} \\\\\n &= \\left\\{g \\in G | g.\\left(x^k - e^m\\right) = x^k - e^m, g.e^m = e^m \\right\\} \\\\\n &= \\left\\{g \\in G | g.x^k = x^k, g.e^m = e^m \\right\\} \\\\\n &\\doteq G_{x^k,e^m}\n\\end{align}\n\\end{proof}\n\n\n\\begin{mythm}\n\\label{thm:J}\nFor two Building Game intermediates $x^j$ and $x^k$ are connected in the BG state space, $r_kS_{jk} = r_jS_{kj}$.\n\\end{mythm}\n\\begin{proof}\nLet $e^m$ be a face such that $x^k = x^j + e^m$. Then, by the orbit-stabilizer theorem, Lagrange's Theorem and lemma~\\ref{lem:I} we have the following~\\cite{Rotman1995}.\n\\begin{align}\n\\frac{r_j}{S_{jk}} &\\doteq \\frac{\\left|G_{x^j}\\right|}{\\left|\\left(G_{x^j}\\right).e^m\\right|} \\\\\n &= \\left[G_{x^j} : \\left(G_{x^j}\\right).e^m \\right] \\\\\n &= \\left|G_{x^j,e^m}\\right| \\\\\n &= \\left|G_{x^k,e^m}\\right| \\\\\n &= \\left[G_{x^k} : \\left(G_{x^k}\\right).e^m \\right] \\\\\n &= \\frac{\\left|G_{x^k}\\right|}{\\left|\\left(G_{x^k}\\right).e^m\\right|} \\\\\n &\\doteq \\frac{r_k}{S_{kj}} \n\\end{align}\nThe result $r_kS_{jk} = r_jS_{kj}$ follows.\n\\end{proof}\n\n\\section{Stationarity}\n\n\\begin{mythm}\n\\label{thm:E}\nThe Markov process $X_t$ defined by the transition rate matrix $Q$ in equation~\\ref{eq:TransitionRate} admits the unique stationary distribution $\\frac{1}{zr_j}e^{-\\beta E_j}$ where $z \\doteq \\sum_\\ell \\frac{1}{r_\\ell}e^{-\\beta E_\\ell}$ is the partition function. \n\\end{mythm}\n\\begin{proof}\nWe take $C_{jk} \\doteq \\frac{S_{jk}}{zr_j}e^{-\\beta E_{jk}}$ and notice that it is symmetric by theorem~\\ref{thm:J}. With $D_{jj} \\doteq zr_je^{\\beta E_j}$ we have our partition. \n\\begin{align}\nQ_{jk} &= S_{jk}e^{-\\beta\\left(E_{jk} - E_j\\right)} \\\\\n &= \\left(zr_je^{\\beta E_j}\\right) \\left(\\frac{S_{jk}}{zr_j}e^{-\\beta E_{jk}}\\right) \\\\\n &= D_{jj}C_{jk} \n\\end{align}\nThus, by theorem~\\ref{thm:StatDist}, $\\pi_j = \\frac{1}{D_{jj}} = \\frac{1}{zr_j}e^{-\\beta E_j}$.\n\\end{proof}\n\n\n\\section{Discussion}\n\\subsection{Nonenumerative Approaches}\n\n\\section*{Acknowledgments}\\label{ackowledgements}\n Supported by NSF grants DMS 07-48482 and EFRI 10-22638\n\n\\section{Potential Citations (temp)}\n~\\cite{Coxeter1963}\n~\\cite{Grunbaum2003}\n~\\cite{Cromwell1997}\n~\\cite{Ziegler1995}\n~\\cite{Gidas1995}\n~\\cite{Eden1961}\n~\\cite{Grayson2012}\n\n\n\\bibliographystyle{plain}\n%\\nocite{*} \n\\bibliography{Master}\n\n\n%\\appendix\n%\\section{Proof of Theorem ???}\n%\\section{Proof of Theorem ???}\n\n\\end{document}\n\n\n%%%%%%%%%%%%%%%%%%%% CCCCUUUUTTTT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsection{A Curious Observation}\n\nAs part of the exploration of the nature of the degeneracy numbers $S_{jk}$ to find a stationary distribution, an interesting phenomena was noticed. Given a closed path $x^{k_0} \\to x^{k_1} \\to \\cdots \\to x^{k_n}$ through the state space, starting at $x^{k_0}$ and ending back at $x^{k_n} \\doteq x^{k_0}$, the the identity given in equation~\\ref{eq:DegenIdent} was observed to hold independent of the particular closed path. \n\\begin{align}\n\\prod_{i=1}^n \\frac{S_{k_i,k_{i-1}}}{S_{k_{i-1},k_i}} = 1\n\\label{eq:DegenIdent}\n\\end{align}\nAt first this relation between ratios of degeneracy numbers along the path was noticed in the cube's state space and it can easily be seen by examining figure~\\ref{fig:CubeSS}. To test this conjecture more fully, it was subsequently verified on our library of state spaces for other polyhedra. This exploration lead to the discovery of a more general relation between intermediates that it described below.\n\nIn an attempt to get the the heart of the relation in equation~\\ref{eq:DegenIdent}, we compiled a list of different geometric and combinatorial statistics for each state space connection $x_j \\leftrightarrow x_k$. These statistics included number of faces in $x_j$ and $x_k$, the order of $x_j$ and $x_k$'s rotation groups $r_j$ and $r_k$, and a few others. With the ansatz that there may be a multiplicative relation between the degeneracy number $S_{jk}$ and some of these $d$ statistics $Z_{jk}^{\\left(i\\right)} \\in \\mathbbm{Z}^d$ for each connection, the following linear regression model was used.\n\\begin{align}\n\\log\\left(S_{jk}\\right) = \\sum_{i=1}^d \\beta^{\\left(i\\right)}\\log\\left(Z_{jk}^{\\left(i\\right)}\\right)\n\\label{eq:LinReg}\n\\end{align}\nThe regression coefficients $\\boldsymbol{\\beta} \\in \\mathbbm{R}^d$ that minimize the least-squares residual were solved for and it was found that a perfect relation was found in the sense that equation~\\ref{eq:LinReg} held exactly for all connections tested. The three non-zero entries in $\\boldsymbol{\\beta}$ corresponded to $r_j$, $r_k$ and $S_{kj}$. This result meant that $\\log\\left(S_{jk}\\right) = \\log\\left(S_{kj}\\right) + \\log\\left(r_{j}\\right) - \\log\\left(r_{k}\\right)$, or \n\\begin{align}\nS_{jk} = \\frac{r_jS_{kj}}{r_k}\n\\label{eq:RSConjecture}\n\\end{align} \nfor all connections we tested. This leads to the obvious conjecture that equation~\\ref{eq:RSConjecture} holds for any connection in any Building Game state space. \n\nIn fact, if this is the case, it would explain why equation~\\ref{eq:DegenIdent} holds. \n\\begin{align}\n \\prod_{i=1}^n \\frac{S_{k_i,k_{i-1}}}{S_{k_{i-1},k_i}} &= \\prod_{i=1}^n \\frac{r_{k_i}}{r_{k_{i-1}}} \\\\ \n &= \\frac{r_{k_n}}{r_{k_{0}}} \\\\ \n &= \\frac{r_{k_0}}{r_{k_{0}}} \\\\\n &= 1 \n\\end{align}\n\n\n\n\\begin{figure}[ht]\n\\scalebox{0.6}{\n%{\\footnotesize\n\\begin{tabular}{ l | c | c | c | c || r | r | r}\nPolyhedra Name & Class & F$\\left(\\mathscr{P}\\right)$ & E$\\left(\\mathscr{P}\\right)$ & V$\\left(\\mathscr{P}\\right)$ & Intermediates & Connections & Pathways \\\\\n \\hline \nTetrahedron & P & 4 & 6 & 4 & 5 & 4 & 1\\\\\nCube & P & 6 & 12 & 8 & 9 & 10 & 3\\\\\nOctahedron & P & 8 & 12 & 6 & 15 & 22 & 14\\\\\nDodecahedron & P & 12 & 30 & 20 & 74 & 264 & 17,696 \\\\\nIcosahedron & P & 20 & 30 & 12 & 2,650 & 17,242 & 57,396,146,640\\\\\nTruncated Tetrahedron & A & 8 & 18 & 12 & 29 & 65 & 402\\\\\nCuboctahedron & A & 14 & 24 & 12 & 341 & 1,636 & 10,170,968\\\\\nTruncated Cube & A & 14 & 36 & 24 & 500 & 2,731 & 101,443,338 \\\\\nTruncated Octahedron & A & 14 & 36 & 24 & 556 & 3,071 & 68,106,377\\\\\nRhombicuboctahedron & A & 26 & 48 & 24 & 638,851 & 6,459,804 & 16,494,392,631,838,879,380\\\\\nTruncated Cuboctahedron & A & 26 & 72 & 48 & 1,525,605 & 17,672,377 & ? \\\\\nIcosidodecahedron & A & 32 & 60 & 30 & ? & ? & ?\\\\\nTruncated Dodecahedron & A & 32 & 90 & 60 & ? & ? & ? \\\\\nTruncated Icosahedron & A & 32 & 90 & 60 & ? & ? & ?\\\\\nTriakis Tetrahedron & C & 12 & 18 & 8 & 99 & 319 & 38,938\\\\\nRhombic Dodecahedron & C & 12 & 24 & 14 & 128 & 494 & 76,936\\\\\nTriakis Octahedron & C & 24 & 36 & 14 & 12,749 & 81,297 & 169,402,670,046,670\\\\\nTetrakis Hexahedron & C & 24 & 36 & 14 & 50,768 & 394,278 & 4,253,948,297,210,346\\\\\nDeltoidal Icositetrahedron & C & 24 & 48 & 26 & 209,676 & 1,989,549 & ? \\\\\nPentagonal Icositetrahedron & C & 24 & 60 & 38 & 345,939 & 3,544,988 & 2,828,128,000,716,774,492\\\\\nRhombic Triacontahedron & C & 30 & 60 & 32 & ? & ? & 5,266,831,101,345,821,968\\\\\n \\hline \n\\end{tabular}\n}\n\\caption{Table of polyhedra in the Platonic (P), Archimedean (A), and Catalan (C) solid classes of up to 32 faces and their Building Game state space statistics.}\n\\label{fig:bgtable}\n\\end{figure}\n\n\n\n\n\n\n%%%%%%%\n\n\nThe process begins with each face of a polyhedron \\poly\\spc all colored the same color, say \\colorA. A face is then chosen and its color is changed to a second color, \\colorB. From there, at each step a \\colorA\\spc face that is edge-adjacent to a \\colorB\\spc face is chosen and recolored \\colorB. The process continues until all of \\poly's faces are \\colorB. We show one instance of the Building Game process for the dodecahedron in figure~\\ref{fig:DodecBG}. \n\nEach possible coloring of the polyhedron from the Building Game is referred to as an \\textbf{intermediate}. Since we assume each face of the same color is otherwise indistinguishable, there is a rotational equivalence class on intermediate such that two intermediates are equivalent if the first is a rotation of the second. \n\n\\begin{figure}[ht]\n\\caption{One instance of the Building Game on the dodecahedron.}\n\\label{fig:DodecBG}\n\\end{figure}\n\nWe define the \\textbf{state space} of the Building Game to be a graph in which the nodes are all of the possible intermediates allowed by the BG. Connections exist between two intermediates \\xj\\spc and \\xk\\spc if it is possible to color a single face of \\xj\\spc to form \\xk. Interestingly, it is not the case that the recoloring of each face of \\xj\\spc maps to a distinct intermediate. Thus, the number of different faces of \\xj\\spc that can be colored to form \\xk\\spc is called the \\textbf{degeneracy number} \\Sjk. It is important to note that in general the degeneracy number is not symmetric, i.e. \\Sjk$\\neq$\\Skj\\spc for some connections \\xj\\spc$\\leftrightarrow$\\spc\\xk\\spc in the state space. \n\n\nSince the state space can be partitioned by the number of \\colorB\\spc faces each intermediate has, it is natural to view the state space as a tiered graph in which intermediates in each tier have the same number of \\colorB\\spc faces. Organized this way, intermediates can only connect to those in the tier above or below them. A \\textbf{pathway} in the state space is a sequence of intermediates $x^{k_0} \\to x^{k_1}\\to \\cdots \\to x^{k_F}$ connected in the state space such that $x^{k_0}$ is the intermediate with all \\colorA\\spc faces, $x^{k_F}$ is the intermediate with all \\colorB\\spc faces, and $x^{k_m}$ has $m$ \\colorB\\spc faces. \n\n\n\n%%%%%%%%\n\n\nTo find the stationary measure of this process, we check if it is possible for our transition measure to satisfy the detailed balance equation $\\pi_jP_{jk} = \\pi_kP_{kj}$. As seen in equation~\\ref{eq:DBTry}, for detailed balance to be satisfied, we must find a way to separate the ratio $\\frac{S_{jk}}{S_{kj}}$ into a part depending only on $j$ terms and a part depending only on $k$ terms. \n\\begin{align}\n \\frac{\\pi_k}{\\pi_j} &= \\frac{P_{jk}}{P_{kj}} \\\\\n &= \\frac{S_{jk}\\rho_{jk}z_k}{S_{kj}\\rho_{kj}z_j} \\\\\n &= \\frac{z_k}{z_j}\\frac{S_{jk}}{S_{kj}} \\label{eq:DBTry}\n\\end{align}\nWhile detailed balance is not a necessary condition for the existence of a stationary distribution, along with positive recurrence of the process, it is sufficient and would provide the exact form of the necessarily unique stationary distribution.\n\n\n%%%%%%%%%%\n\n\nAs stated in section~\\ref{sec:Prob}, if we can find a distribution $\\pi$ such that the detailed balance equation $\\pi_jP_{j,k} = \\pi_kP_{k,j}$ is satisfied, we know that $\\pi$ is the unique stationary distribution under $P_{jk}$. With theorem~\\ref{thm:J} we find that $\\frac{S_{jk}}{S_{kj}} = \\frac{r_j}{r_k}$ is indeed separable as conjectured in equation~\\ref{eq:DBTry}. This allows us to find the stationary distribution. \n\n\\begin{mythm}\n\\label{thm:E}\nThe Markov chain $X_t$ defined by the transition rule $P_{jk}$ in equation~\\ref{eq:TransitionProbability} admits the unique stationary distribution $\\pi_j = \\frac{1}{z}\\left(\\frac{z_j}{r_j}\\right)$ where $z \\doteq \\sum_i \\frac{z_i}{r_i}$ is the partition function. \n\\end{mythm}\n\\begin{proof}\nIt suffices to show detailed balance and that each state is positively recurrent. By our definitions of $\\pi$ and $P$ along with Theorem~\\ref{thm:J}, we find detailed balance. \n\\begin{align}\n \\pi_jP_{jk} &= \\frac{z_j}{z r_j}\\frac{1}{z_j}S_{jk}\\rho_{jk} \\\\ \n &= \\frac{1}{z}\\left(\\frac{S_{jk}}{r_{j}}\\right)\\rho_{jk} \\\\\n &= \\frac{1}{z}\\left(\\frac{S_{kj}}{r_k}\\right)\\rho_{kj} \\\\\n &= \\frac{z_k}{z r_k}\\frac{1}{z_k}S_{kj}\\rho_{kj} \\\\\n &= \\pi_kP_{kj}\n\\end{align}\nClearly $X_t$ is positive recurrent since it is a finite irreducible Markov chain on a connected state space in which there exist forward and backward transitions between each connected node with positive probabilities.\n\\end{proof}\n\n", "meta": {"hexsha": "f58f98010d029c2c5805bbebdf1eb8916f1fd14f", "size": 30069, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Existing Writing Dump/Keep/em_paper.tex", "max_stars_repo_name": "Danie1Johnson/thesis", "max_stars_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Existing Writing Dump/Keep/em_paper.tex", "max_issues_repo_name": "Danie1Johnson/thesis", "max_issues_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Existing Writing Dump/Keep/em_paper.tex", "max_forks_repo_name": "Danie1Johnson/thesis", "max_forks_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.8684210526, "max_line_length": 868, "alphanum_fraction": 0.6547274602, "num_tokens": 10340, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.831143045767024, "lm_q1q2_score": 0.6482856747938529}} {"text": "\\documentclass{memoir}\n\\usepackage{notestemplate}\n\n%\\logo{~/School-Work/Auxiliary-Files/resources/png/logo.png}\n%\\institute{Rice University}\n%\\faculty{Faculty of Whatever Sciences}\n%\\department{Department of Mathematics}\n%\\title{Class Notes}\n%\\subtitle{Based on MATH xxx}\n%\\author{\\textit{Author}\\\\Gabriel \\textsc{Gress}}\n%\\supervisor{Linus \\textsc{Torvalds}}\n%\\context{Well, I was bored...}\n%\\date{\\today}\n\n%\\makeindex\n\n\\begin{document}\n\n% \\maketitle\n\n% Notes taken on 06/09/21\n\nLet \\(V\\) be a finite dimensional vector space over \\(F\\) with dimension \\(n\\), and \\(T\\) a fixed linear transformation of \\(V\\). Recall that we can view \\(V\\) as an \\(F[x]\\)-module where \\(x\\) acts on \\(V\\) as the linear transformation \\(T\\).\\\\\n\nBecause \\(V\\) has finite dimension over \\(F\\), it must be a torsion \\(F[x]\\)-module. Hence \\(V\\) is isomorphic as an \\(F[x]\\)-module to the direct sum of cyclic, torsion \\(F[x]\\)-modules.\\\\\n\nWhen we decompose \\(V\\) into the invariant factor decomposition basis, we obtain the rational canonical form for the matrix for \\(T\\). When we use the elementary divisor decomposition, we obtain the Jordan canonical form.\n\n\\begin{defn}[Minimal Polynomial]\n The \\textbf{minimal polynomial of \\(T\\)} is the unique monic polynomial \\(m_T(x) \\in F[x]\\) that generates the ideal \\(\\textrm{Ann}(V)\\) in \\(F[x]\\).\\\\\n\n Let \\(A\\) be a matrix. The \\textbf{minimal polynomial of \\(T\\)} is the unique monic polynomial of smallest degree \\(m_A(x)\\) that yields the zero matrix when evaluated at \\(A\\).\n\\end{defn}\n\nThe \\textbf{rational canonical form} of the linear transformation \\(T\\) is the isomorphism\n\\begin{align*}\n\tV \\cong F[x] / (a_1(x)) \\oplus F[x] / (a_2(x)) \\oplus \\ldots \\oplus F[x] / (a_m(x))\n\\end{align*}\nwhere \\(a_1(x),a_2(x),\\ldots,a_m(x)\\) are polynomials in \\(F[x]\\) of positive degree such that\n \\begin{align*}\n\t a_1(x) \\mid a_2(x) \\mid \\ldots\\mid a_m(x)\n\\end{align*}\nOf course, the annihilator of \\(V\\) is the ideal \\((a_m(x))\\), and so we obtain:\n\\begin{prop}\n\tThe minimal polynomial \\(m_T(x)\\) is the largest invariant factor of \\(V\\).\n\\end{prop}\n\nObserve that we can get a basis for the vector space \\(F[x] / (a(x))\\) for a fixed\n\\begin{align*}\n\ta(x) = x^{k}+ b_{k-1}x^{k-1} + \\ldots + b_1 x + b_0\n\\end{align*}\nby defining \\(\\overline{x}^{k}= (x \\pmod{a(x)})^{k}\\). The basis is then \\(\\left\\{ 1,\\overline{x},\\overline{x}^2,\\ldots,\\overline{x}^{k-1} \\right\\} \\) which has an action under multiplication by \\(x\\) given by:\n\\begin{align*}\n\t1 \\mapsto \\overline{x}\\\\\n\t\\overline{x} \\mapsto \\overline{x}^2\\\\\n\t\\vdots\\\\\n\t\\overline{x}^{k-2}\\mapsto \\overline{x}^{k-1}\\\\\n\t\\overline{x}^{k-1}\\mapsto \\overline{x}^{k} = -b_0 - b_1\\overline{x} - \\ldots - b_{k-1}\\overline{x}^{k-1}\n\\end{align*}\nwhich follows because\n\\begin{align*}\n\t\\overline{x}^{k} + b_{k-1}\\overline{x}^{k-1} + \\ldots + b_1\\overline{x}+b_0 = 0\n\\end{align*}\nThis gives us a matrix for multiplication by \\(x\\):\n\\begin{defn}[Companion Matrix]\n\tLet \\(a(x) = x^{k}+ b_{k-1}x^{k-1}+ \\ldots + b_1x + b_0\\) be a monic polynomial in \\(F[x]\\). The \\textbf{companion matrix} of \\(a(x)\\) is the \\(k\\times k\\) matrix representing the matrix for multiplication by \\(x\\), and is of the form\n\t\\begin{align*}\n\t\t\\begin{pmatrix} \n\t\t\t0 & 0 & \\ldots & \\ldots & -b_0\\\\\n\t\t\t1 & 0 & \\ldots & \\ldots & -b_1 \\\\\n\t\t\t0 & 1 & \\ldots & \\ldots & -b_2\\\\\n\t\t\t\\vdots & \\vdots & & \\ddots & \\vdots\\\\\n\t\t\t0 & 0 & \\ldots & 1 & -b_{k-1}\n\t\\end{pmatrix}\n\t\\end{align*}\n\tWe denote the companion matrix of \\(a(x)\\) by \\(\\mathcal{C}_{a(x)}\\).\n\\end{defn}\n\nNow we will apply this to each of the cyclic modules in the rational canonical form of \\(V\\). Let \\(\\mathcal{B}_i\\) be the set of basis elements for each cyclic factor \\(F[x] / (a_i(x))\\). The linear transformation \\(T\\) acts on \\(\\mathcal{B}_i\\) by the companion matrix for \\(a_i(x)\\), and hence the union \\(\\mathcal{B} = \\cup_{i} \\mathcal{B}_i\\) and the matrix of the transformation on \\(V\\) is the direct sum of the companion matrices.\n\n\\begin{defn}[Rational Canonical Form]\n\tA matrix is said to be in \\textbf{rational canonical form} if it is the direct sum of companion matrices for monic polynomials \\(a_1(x),\\ldots,a_m(x)\\) with\n\t\\begin{align*}\n\t\ta_1(x) \\mid a_2(x) \\mid \\ldots \\mid a_m(x).\n\t\\end{align*}\n\tThe matrix is then of the form\n\t\\begin{align*}\n\t\t\\begin{pmatrix} \n\t\t\t\\mathcal{C}_{a_1(x)} & & & \\\\\n\t\t\t& \\mathcal{C}_{a_2(x)} & & \\\\\n\t\t\t& & \\ddots & \\\\\n\t\t\t& & & \\mathcal{C}_{a_m(x)}\n\t\t\\end{pmatrix}.\n\t\\end{align*}\n\tThe polynomials are called the \\textbf{invariant factors} of the matrix, and the matrix is said to be a \\textbf{block diagonal matrix} with the blocks being the companion matrices for \\(a_i(x)\\).\\\\\n\n\tA \\textbf{rational canonical form} for a linear transformation \\(T\\) is a matrix representing \\(T\\) in rational canonical form.\n\\end{defn}\nOne can check that every linear transformation \\(T\\) has a rational canonical form that is unique.\n\n\\begin{thm}[Rational Canonical Form for Linear Transformation]\n\tLet \\(V\\) be a finite dimensional vector space over the field \\(F\\) and let \\(T\\) be a linear transformation of \\(V\\). Thne there is a basis for \\(V\\) and the matrix of \\(T\\) is in rational canonical form with respect to this basis. Furthermore, the rational canonical form for \\(T\\) is unique.\n\\end{thm}\nWe will see that this exists for every \\(T\\), but Jordan canonical form may not.\\\\\n\nFor linear transformations \\(S\\) and \\(T\\), the following are equivalent:\n\\begin{itemize}\n\t\\item \\(S\\) and \\(T\\) are similar linear transformations\n\t\\item The \\(F[x]\\)-modules obtained from \\(V\\) via \\(S,T\\) are isomorphic\n\t\\item \\(S\\) and \\(T\\) have the same rational canonical form\n\\end{itemize}\nOf course, any matrix can be translated into a rational canonical form, as each matrix corresponds to a linear transformation. The \\textbf{invariant factors} of an \\(n\\times n\\) matrix over a field \\(F\\) are the invariant factors of its rational canonical form.\n\n\\begin{lemma}\n\tLet \\(a(x) \\in F[x]\\) be any monic polynomial. The characteristic polynomial of the companion matrix of \\(a(x)\\) is \\(a(x)\\), and if \\(M\\) is given by\n\t\\begin{align*}\n\t\tM = \n\t\t\\begin{pmatrix}\n\t\t\tA_1 & 0 & \\ldots & 0 \\\\\n\t\t\t0 & A_2 & \\ldots & 0\\\\\n\t\t\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\t\t\t0 & 0 & \\ldots & A_k\n\t\t\\end{pmatrix} \n\t\\end{align*}\n\tthen the characteristic polynomial of \\(M\\) is the product of the characteristic polynomials of \\(A_1,A_2,\\ldots,A_k\\).\n\\end{lemma}\n\n\\begin{thm}[Cayley-Hamilton Theorem]\n\tLet \\(A\\) be an \\(n\\times n\\) matrix over the field \\(F\\). The minimal polynomial of \\(A\\) divides the characteristic polynomial of \\(A\\).\n\\end{thm}\nIn fact, the characteristic polynomial of \\(A\\) divides some power of the minimal polynomial of \\(A\\).\n\n\\begin{thm}\n\tLet \\(A\\) be an \\(n\\times n\\) matrix over the field \\(F\\). The \\(n\\times n\\) matrix \\(xI - A\\) can be put into the diagonal form\n\t\\begin{align*}\n\t\t\\begin{pmatrix} \n\t\t\t1 & & & & &\\\\\n\t\t\t & \\ddots & & & &\\\\\n\t\t\t & & 1 & & &\\\\\n\t\t\t & & & a_1(x) & &\\\\\n\t\t\t & & & & \\ddots & \\\\\n\t\t\t & & & & & a_m(x)\n\t\t\\end{pmatrix}\n\t\\end{align*}\n\twith monic nonzero elements\n\t\\begin{align*}\n\t\ta_1(x) \\mid a_2(x) \\mid \\ldots \\mid a_m(x).\n\t\\end{align*}\n\tvia the operations\n\t\\begin{itemize}\n\t\t\\item interchanging two rows or columns\n\t\t\\item adding a multiple of one row or column to another\n\t\t\\item multiplying any column or row by a unit in \\(F[x]\\)\n\t\\end{itemize}\n\tThe elements \\(a_1(x),\\ldots,a_m(x)\\) are the invariant factors of \\(A\\).\n\\end{thm}\n\n\\subsection{Invariant Factor Decomposition Algorithm}\n\\label{sub:invariant_factor_decomposition_algorithm}\n\n%% Fill in\n\n\\subsection{Converting an \\(n\\times n\\) Matrix to Rational Canonical Form}\n\\label{sub:converting_an_n_by_n_matrix_to_rational_canonical_form}\n\n\n% \\printindex\n\\end{document}\n", "meta": {"hexsha": "fb4b1f2cac25cc460b57144511b36c0e864cb896", "size": 7682, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Module Theory/Notes/source/CanonicalForms.tex", "max_stars_repo_name": "gjgress/Libera-Mentis", "max_stars_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-07-16T23:18:15.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-16T23:18:15.000Z", "max_issues_repo_path": "Module Theory/Notes/source/CanonicalForms.tex", "max_issues_repo_name": "gjgress/Libera-Mentis", "max_issues_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 12, "max_issues_repo_issues_event_min_datetime": "2021-05-20T22:09:37.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-20T23:23:22.000Z", "max_forks_repo_path": "Module Theory/Notes/source/CanonicalForms.tex", "max_forks_repo_name": "gjgress/LibreMath", "max_forks_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.4556213018, "max_line_length": 438, "alphanum_fraction": 0.6683155428, "num_tokens": 2542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8311430436757312, "lm_q1q2_score": 0.6482856731626594}} {"text": "\\chapter{Appendix Probability}\r\n\r\n\\section*{Probability Basics}\r\nLet $\\mu$ be a non-negative countably additive set function over a sigma algebra $\\Omega$ of sets from a sample space $S$. In probability theory, $\\Omega$ is the set of possible events $E$. Sets of measure zero denote impossible outcomes. An important feature of measures $\\nu$ and $\\mu$ that agree on sets of measure zero is the ability to define a derivative $\\frac{d \\nu}{d \\mu}$, the Radon-Nikodym derivative. When $\\nu (E)=0 \\fall E \\in \\Omega \\;| \\; \\mu(e)=0$ Alternatively, given a measure $\\mu$ and a nonnegative measurable function $f$, a new measure can be defined by $ \\nu(E) = \\int\\limits{E \\in \\Omega}{} f d \\mu$. A random variable is a real valued function on a sample space into a metric space, $X : S \\rightarrow \\dblr^{1} $. Associated with a random variable is it's probability density function $f_{X}(x)=P(\\{s \\in S | X(s) = x\\})$ operating on an algebra of sets generated by the sample space $S$. By definition, $f_{X}(x)$ is the sum of probabilities of the events in $S$ that get mapped to $x \\in \\dblr$ by $X$. Let $\\script{B}(S)$ be the Borel sets on $S$, then $X$ has density $f$ if $P(X \\in A) = \\int_{A \\in \\script{B}(S)} f(x) dx $ and distribution function $F(x) = P(XE[X]$ we say the process or data is over dispersed.\r\nThe parameter $\\phi$ is usually fixed in practice. If we write\r\n\\[ p(x; \\theta, \\phi)=exp(\\frac{x \\theta-\r\n\\beta(\\theta)}{\\alpha(\\phi)}+\\gamma(x,\\phi))\\], the dispersion\r\nparameter $\\phi$ for some common distributions;\r\n\\[\r\n\\begin{array}{cc}\r\np(x;\\theta, \\phi) & \\phi \\\\ \\hline\r\nN(\\mu,\\sigma) & \\sigma^2 \\\\\r\nIG(\\mu,\\sigma) & \\sigma^2 \\\\\r\nGamma(\\theta,\\phi) & \\frac{1}{\\phi} \\\\\r\nPoisson(\\theta) & 1 \\\\\r\nBinomial(\\theta) & 1 \\\\\r\nNegative Binomial(\\theta,r) & r \\\\\r\n\\end{array}\r\n\\]\r\n\r\n\\subsubsection{Normal/Gaussian, $N(\\mu,\\sigma)$}\r\n$X=_d N(\\mu,\\sigma)$ if \\[p(x; \\mu, \\sigma) = \\frac{1}{\\sqrt{2\r\n\\pi \\sigma^2}} exp ( \\frac{(x-\\mu)^2}{2 \\sigma^2}\\]. This can\r\nbe re-written in exponential form \\[ p(x;\\theta) = exp( \\frac{x\r\n\\theta}{2 \\sigma^2} - \\frac{\\theta^2}{2 \\sigma^2 }-\r\n\\frac{1}{2}log(2 \\pi \\sigma^2) - \\frac{x^2}{2 \\sigma^2 } ) \\]\r\n\r\n\\subsubsection{Binomial}\r\nSetting \\[\\alpha(x)=x \\;\\;,\r\n\\beta(\\theta)=log(\\frac{\\theta}{1-\\theta}) \\;\\;,\r\n\\gamma(\\theta)=n log(1-\\theta) \\;\\;, \\delta(x)=log ( \\biggl(\r\n\\begin{array}{c} n \\\\ y \\\\ \\end{array} \\biggr) )\r\n\\] gives us the binomial distribution $p(x;\\theta) = \\binomial{n}{x} \\theta^x\r\n(1-\\theta)^{(n-x)}$\r\n\r\n\\subsubsection{Negative Binomial}\r\n\\[p(x;\\theta,r)=\\binomial{x+r-1}{r-1} \\theta^r (1-\\theta)^x\\]\r\n\r\n\\subsubsection{Poisson}\r\n$X$ is Poisson distributed if $p(x;\\theta)=\\frac{\\theta^x\r\ne^{-\\theta}}{x!}$ Setting $\\beta(\\theta)=log(\\theta) \\;\\;,\r\n\\gamma(\\theta)=-\\theta \\;\\;, \\delta(x)=-log(x!)$ we get the\r\nexponential form of the Poisson distribution,\r\n\\[p(x;\\theta)= exp(x log(\\theta) - \\theta- log(x!))\\]\r\n\r\nThe expected value and the variance of a Poisson distributed random variable is equal to $\\theta$. The higher moments of the Poisson distribution are the Touchard polynomials in $\\theta$. There is a combinatorial interpretation. When $E[X]=1$ for a Poisson random variate then the i-th moment of $X$ is equal to the number of partitions of a set of size n $\\frac{1}{e} \\sum\\limits_{n=0}^{\\infty} \\frac{n^i}{n!}$ via Dobinski. The normal distribution with mean $\\theta$ and variance $\\theta$ is a good approximation to the Poisson distribution for large $\\theta$.\r\n\r\n\r\n\\subsubsection{Pareto}\r\n$X$ is Pareto distributed if \\[p(x;\\theta)=\\theta\r\nx^{-\\theta}.\\]\r\n\r\n\\subsubsection{Gamma}\r\n$X$ is Gamma distributed if \\[p(x;\\theta,\r\n\\phi)=\\frac{x^{\\phi^{-1}}\\theta^\\phi e^{-x \\theta}} {\r\n\\Gamma(\\phi)}.\\]\r\n\r\n\\subsubsection{Weibull}\r\n$X$ follows the Weibull distribution if \\[p(x; \\theta , \\lambda\r\n) = \\frac{\\lambda x^{\\lambda-1}}{\\theta^\\lambda}\r\ne^{(\\frac{x}{\\theta} )^\\lambda}.\\]\r\n\r\n\\subsubsection{Inverse Gaussian/ Wald Distribution}\r\n$X$ follows the Inverse Gaussian distribution if\r\n\\[p(x;\\theta)= \\sqrt{\\frac{\\theta}{2 \\pi x^3 \\sigma}}exp(-\r\n\\frac{\\lambda(x-\\theta)^2}{2 x \\theta^2 \\sigma}) \\]\r\n\r\n\\subsection{Generalized Extreme Value Distribution $GEV(\\theta,\\phi,\\xi)$}\r\nThis class of distributions includes the three limiting extreme\r\nvalue distributions of \\cite{Fisher Tippet (1928)} and\r\n\\cite{Gnedenko (1943)}. $X =_d GEV(\\theta,\\phi,\\xi)$ if\r\n\\[p(x;\\theta,\\phi,\\xi)= exp( - max\\biggl(\\bigl(1 + \\xi\r\n\\frac{x-\\theta}{\\phi} \\bigr)^{- \\frac{1}{\\xi} }, \\;\\; 0 \\biggr)\r\n\\]. Where\r\n\r\n\r\n\\subsection{Multinomial}%bbcrevisit explanation\r\nLet \\[\\Omega=\\mathcal{B}(\\prod\\limits_{i=0}^{i=\\infty}\r\n\\dblz(K))\\] be the Borel Algebra generated by $\\prod {1, 2,\r\n\\hdots , K}$. This is the sample space of all realizations of\r\nexperiments with $K$ categorical outcomes. Equip $\\dblz(K)= {i\r\n\\in {1, \\hdots, K}}$ with a measure $P(i)=\\theta_i$. Let\r\n${X_i}$ be n iid copies of $X =_d p(i ; \\pi_1, \\hdots ,\r\n\\pi_K)$. Now map $\\mbf{X}=(X_1, \\hdots, X_n) \\in \\Omega\r\n\\rightarrow \\mbf{Y} \\in \\dbln^K$ Then $Y_i$ are counts of the\r\nnumber of elements of category $i$ in the experiment with n\r\nobservations. The multinomial distribution is given by\r\n\\[p(\\mbf{Y};n)=\\frac{n!}{y_1! \\hdots y_K!} (\\theta_1)^{y_1} \\hdots\r\n(\\theta_K)^{y_K} \\]. This in not a member of the exponential\r\nfamily, but we can show that the multinomial distribution is\r\nthe joint distribution of ${Y_i =_d Poission(\\theta'_i)}_{i=1,\r\n\\hdots K}$ random variables conditional to their sum.\r\n\\[p(\\mbf{Y}; \\theta'_1, \\hdots , \\theta'_K) =\r\n\\prod\\limits_{i=1}^{K} \\frac{ (\\theta'_i)^{y_i} \\;\r\ne^{-\\theta'_i}}{y_i!}\\], set $n=Y_1 + \\hdots + Y_K$. Writing\r\n$p(\\mbf{Y} | n) =p(\\mbf{Y}; \\theta'_1, \\hdots ,\r\n\\theta'_K) / p(n)$ and noting that $n=_d Poisson(\\sum\\limits_{i=1}^{K} \\theta'_i)$, %bbcrevisit this is an exercise\r\nwe recover the multinomial distribution by simplifying and\r\nsetting $\\theta_i=\\frac{\\theta'_i}{\\sum\\limits_{i=1}^{K}\r\n\\theta'_i}$\r\n\r\n\\subsection{$\\chi^2(n)$}\r\nIf ${X_i}$ iid $N(0,1)$ and $Y_i=X_i^2$ then\r\n$Y=\\sum\\limits_{i=1}^{n} Y_i =_d \\chi^2(n).$ $E[Y]=n$ and\r\n$Var(Y)=E[(Y-\\mu_Y)^2]=E[(Y-E[Y])^2]=2n$. More generally, if\r\n$Y_i=X_i+\\mu_i$ then \\[Y=\\sum\\limits_{i=1}^{n} (Y_i)^2 =\r\n\\sum\\limits_{i=1}^{n} X_i^2 + 2 \\sum\\limits_{i=1}^{n} X_i \\mu_i\r\n+ \\sum\\limits_{i=1}^{n} \\mu_i^2 =_d \\chi^2(n,\\lambda)\\]. Where\r\n$\\lambda=\\sum\\limits_{i=1}^{n} \\mu_i$ is non-centrality\r\nparameter.\r\n\r\nSee the section on multivariate probability distributions for\r\nfurther information, but it is worth noting that if $X =_d\r\nN(\\mu,\\sigma)$ is multivariate and the variance covariance\r\nmatrix $s\\sigma$ is non-singular, then $(y-\\mu)^T \\sigma^{-1}\r\n(y-\\mu) =_d \\chi^2(n)$ and setting $\\lambda= \\mu^T \\sigma^{-1}\r\n\\mu$ we have $ y^T \\sigma^{-1} y =_d \\chi^s(n,\\lambda)$\r\n\r\n\\subsection{Student-t $t(\\nu)$}\r\n$X =_d \\frac{\\Gamma(\\frac{\\nu+1}{2})}{\\sqrt{\\pi} \\Gamma(\\nu\r\n/2)}(1+x^2)^{- \\frac{\\nu+1}{2}}$\r\n\r\n%\\cite{C. C. Heyde and N. N. Leonenko 2005}\r\n\r\n\\subsection{Generalized Inverse Gaussian $GIG(\\lambda,\\alpha,\\beta )$}\r\nThe GIG distributions are characterized by\\[p(x; \\lambda,\r\n\\theta,\\sigma)=\\bigl(\\frac{\\theta}{\\sigma}\\bigr)^{\\frac{\\lambda}{2}}\r\nx^{\\lambda-1}\\: \\frac{1}{2 K_\\lambda (\\sqrt{\\theta \\sigma})} \\:\r\nexp(-\\frac{1}{2} ( \\theta x^{-1} + \\sigma x) \\].\r\n\r\nNote, $GIG(-1/2,\\theta,\\sigma)=IG(\\theta,\\sigma)$\r\n\r\nThe GIG family members arise as first passage time\r\ndistributions of ordinary Brownian diffusions to a constant\r\nboundary.\r\n\r\n\\subsection{ Normalized Inverse Gaussian $NIG(\\mu,\\alpha,\\beta,\\delta)$}\r\n$X =_d NIG(\\mu,\\alpha,\\beta,\\delta)$ if \\[p(x;\\mu,\r\n\\beta,\\alpha,\\delta)= \\frac{\\delta \\alpha}{\\pi} exp \\bigl(\r\n\\delta \\sqrt{\\alpha^2 - \\beta^2}+\\beta(x-\\mu) \\bigr)\r\n\\frac{K_1(\\alpha \\; s_\\delta(x-\\mu))}{s_\\delta(x-\\mu)}\\]\r\n\r\nwhere $x \\in \\dblr \\;\\; \\mu \\in \\dblr \\;\\; \\delta>0 \\;\\; 0 \\leq\r\n|\\beta| \\leq \\alpha$ and $s_\\delta(x)=\\sqrt{\\delta^2+x^2}$ and\r\n$K_1(x)=\\frac{x}{4} \\int\\limits_{0}^{\\infty} exp - \\bigl (\r\ny+\\frac{x^2}{4 y} \\bigr ) y^{-2} \\;dy$ is the modified Bessel\r\nfunction of the third kind. This family of distributions is\r\ninfinitely divisible, $\\exts \\;\\; X_t$, a Levy process, $ X_{t+\r\n\\Delta t} - X_t =_d X_{\\Delta t} =_d\r\nNIG(\\mu,\\alpha,\\beta,\\delta)$. $X_t$ is a pure jump process,\r\nand\r\n\\[p(t; \\alpha,\\beta,\\delta)= \\biggl( \\frac{\\delta \\alpha}{\\pi |t|}\r\n\\biggr) e^{\\beta t} K_1(\\alpha |t|)\\]. See Eberlein and Keller\r\n(1995) and Barndorff-Nielsen (1998)\r\n\r\n\r\n\\subsection{Generalized Hyperbolic $GH(\\lambda,\\alpha,\\beta,\\delta,\\mu)$}\r\nThe parameters $\\lambda,\\alpha,\\beta,\\delta,\\mu$ have the\r\nrespective interpretation of tail heaviness, kurtosis,\r\nskewness, and scale, and location. The distribution includes\r\nthe important classes GIG, NIG, IG, and can be characterized as\r\na Normal variance-mean mixture NVMM parameterized by a GIG\r\ndistribution. Formally set $U=_d GIG()$, then $X=_d NVMM( )$ if\r\n$P(X|U=u) =_d N(\\mu+ \\beta, u \\Delta)$. This gives a\r\nstochastic representation $X= \\mu+\\beta Z + \\sqrt{Z} Y$ where\r\n$Y =_d N(0,1)$ and $Z =_d GIG( )$.\r\n\r\n\\section*{Limit Theorems}\r\nLimit Theorems use the notion of a basin of attraction for pdf's in some functional space $\\mathcal{H}$. In $L^2(\\Omega)$, we have $N(\\mu,\\sigma) \\subset L^2(\\Omega, \\nu)$ is the basin of attraction for all pdf's satisfying the conditions of the CLT.\r\n\r\nThe CLT says that the series $\\frac{\\sum\\limits_{i=1}^{n} X_i}{n}$ converges in in probability to the mean of $x_i$. Cramer's theorem gives a bound on the probability of large deviation away from the mean in the series $\\frac{\\sum\\limits_{i=1}^{n} X_i}{n}$. The probability decays exponentially with a rate given by the Legendre transform of the cumulant generating function for $X_i$\r\n\\begin{thm}[Cramer's Theorem]\r\nLet $X_1, X_2, \\hdots $ be iid $E(X_i)=0$, $E(X_i^2)= \\sigma^2$, and $F_n(x) = P(\\frac{1}{\\sigma n^{\\frac{1}{2}}} \\sum\\limits_{i=1}^{n} X_i < x)$, then if $x>1$ and $x=O(\\sqrt{n})$ as $n \\rightarrow \\infty$\r\nwe have\r\n\\begin{eqnarray*}\r\n \\frac{1-F_n(x)}{1-\\Phi(x)} = exp ( \\frac{x^3}{\\sqrt{n}} \\lambda( \\frac{x}{\\sqrt{n}} ) ) [ 1 + O(\\frac{x}{\\sqrt{n}}) ]\r\n\\end{eqnarray*}\r\n\r\n$\\lambda(x) = \\sum\\limits_{i=0}^{\\infty} c_i x_i$ where the $c_i$ depend on the moments of $X_i$.\r\n$\\Phi(x)$ is the distribution\\ function of $N(0,1)$.\r\n\\end{thm}\r\n\r\n\\begin{thm}[Law of Large Numbers]\r\n\\end{thm}\r\n\r\nThe CLT tells us that the pdf of the scaled mean of a sample approaches the normal distribution and the Berry–Esseen theorem specifies the rate at which that happens. The CLT requires $X_i$ to be iid, and with finite second moment and the Berry-Esseen theorem additionally requires a finite third moment.\r\n\\begin{thm}[Berry–Esseen]\r\nLet ${X_i}$ be iid, $E(X_i^2)=\\sigma$, $E(X_i^3)=\\rho$, $Y_n = \\frac{X_1 + \\ldots + X_n}{n}$, and $F_n = \\int \\frac{Y_n \\sqrt{n} }{\\sigma}$ and $\\Phi$ the CDF of $N(0,1)$, then\r\n\\begin{equation}\r\n\\abs{F_n(x) - \\Phi(x)} \\leq \\frac{C \\rho}{ \\sigma^3 \\sqrt{n} }\r\n\\end{equation}\r\n\r\n\\end{thm}\r\n\r\n\\section*{Multivariate Probability Distributions}\r\n\r\nLet $S_n$ be the unit sphere in $\\dblr^n$ A random variate $X$\r\nis uniformly distributed on $S_n$ when $X$ is radially\r\nsymmetric and $||X||_{L^2} = 1 a.s.$ The pdf of a radially\r\nsymmetric random variable is necessarily of the form\r\n$f(x_1,...,x_n)=g(||x||)$ for some $g \\in [0,\\infty) \\ni\r\n\\int\\limits_{0}^{\\infty} n V_n r^{n-1} g(r) dr =1$ where\r\n$V_n=\\frac{\\pi^{\\frac{d}{2}}}{\\Gamma(\\frac{d}{2}+1) }$ is the\r\nvolume of $S_n$. If $X$ is radially symmetric , then\r\n$\\frac{X}{|||X||}$ is uniformly distributed on $S_n$. If $X$\r\nis uniformly distributed on $S_n$ then $(X_1^2, \\ldots ,\r\nX_{n}^{2}) =_{dist} (\\frac{Y_1}{\\kappa} , \\ldots ,\r\n\\frac{Y_n}{\\kappa} )$ where $Y_i$ iid $\\Gamma(\\frac{1}{2})$\r\nwith sum $\\kappa$. If $N_1, \\ldots , N_n$ iid normal, then\r\n$(N_1, \\ldots , N_n)$ is radially symmetric with density $g(r)\r\n= \\frac{1}{(2 \\pi)^{\\frac{n}{2} } } e^{\\frac{ - r^2}{2}}$ This\r\nleads us to an algorithm for generating pseudo random variants\r\non uniformly distributed on $S_n$ ;\r\n\\begin{itemize}\r\n \\item Generate $n$ iid $N(0,1)$\r\n \\item Compute $ \\kappa = ( \\sum\\limits_i=1^n N_i^2\r\n )^\\frac{1}{2}$\r\n \\item Return $(\\frac{N_1}{\\kappa} , \\ldots ,\r\n \\frac{N_n}{\\kappa} )$\r\n\\end{itemize}\r\n\\cite{Devorye}\r\n\r\nWith a little linear algebra, the above can be generalized to a\r\ngenerator for $N(\\mu, \\Sigma ) \\in \\dblr^n$. Consider $f(x)\r\n=\\frac{1}{(2 \\pi)^{\\frac{n}{2} } } e^{ - \\frac{1}{2} x^T \\dot x\r\n} x \\in \\dblr_n$, $f$ has density of $n$ iid $N(0,1)$ rv if\r\n\r\n\\subsection{Multivariate Normal $N(\\mbf{\\mu},\\mbf{\\Sigma})$ }\r\n$N(\\mbf{\\mu},\\mbf{\\Sigma})$ is arguably the most important and tractable multivariate probability distribution. The Gaussian distribution is separable via rotation. Precisely the rotation induced by PCA.\r\n\r\n\\subsection{Wishart Distribution}\r\nThe Wishart distribution $W(n)$ is the multivariate generalization of the $\\chi^2(n)$ distribution. If $X_{(i)} \\sim N(\\mbf{\\mu},\\mbf{\\Sigma})$, then\r\n$X X^t =S \\sim W(N)$.\r\n\r\n\\subsection{Elliptic $E(\\mbf{\\mu},\\mbf{\\Sigma})$}\r\nElliptical distributions $E(\\mbf{\\mu},\\mbf{\\Sigma})$ extend the multivariate normal $N(\\mbf{\\mu},\\mbf{\\Sigma})$. They can be characterized as affine maps of spherical distributions. The density functions\r\nare defined by $p(x) = c g( (x-\\mu)' \\Sigma^{-1} (x-\\mu) )$ Where $g : \\dblr^{+} \\longrightarrow \\dblr^{+}$ and $ \\Sigma \\succ 0$ is positive definite.\r\nMany of the properties of the multivariate normal distribution\r\nare shared by the elliptical distributions. Linear combinations, marginal distributions and conditional distributions of elliptical random variables can largely be determined by linear algebra using knowledge of covariance\r\nmatrix, mean and generator.\r\n\r\n\\section*{Statistical Dependence}\r\nLinear correlation is a natural dependence measure for\r\nmultivariate normally and elliptically distributed random variables. Other dependence concepts include rank correlation, comonotonicity, and Brownian covariance.\r\n\r\n\r\n\\section*{Distance measures for probability distribution functions.}\r\nA number of distance measures for probability distance\r\nfunctions exist. Kullbak Lieber divergence:\\[ J_D =\r\n\\int\\limits_{\\text{x}} {[p({\\text{x}}\\mid\\omega_1 ) -\r\np({\\text{x}}\\mid\\omega_2 )]}\r\n \\log \\frac{{p(x\\mid\\omega_1 )}} {{p(x\\mid\\omega_2 )}}{\\text{dx}}\\]\r\nwhich simplifies to:\\[ J_D = \\tfrac{1} {2}\\left( {\\mu _2 -\r\n\\mu _1 } \\right)^T \\left( {\\Sigma _1^{^{ - 1} } + \\Sigma\r\n_2^{^{ - 1} } } \\right)\\left( {\\mu _2 - \\mu _1 } \\right) +\r\n\\tfrac{1} {2}{\\text{tr}}\\left\\{ {\\Sigma _1^{^{ - 1} } \\Sigma _2\r\n+ \\Sigma _2^{^{ - 1} } \\Sigma _1 - 2I} \\right\\}\\] when \\[X_1\r\n=_d N(\\mu_1,\\Sigma_1) \\;\\; , \\;\\; X_2 =_d N(\\mu_2,\\Sigma_2).\\]\r\n\r\nThe Bhattacharyya distance :\r\n\\[J_B = - \\log \\int {\\left[ {p(\\xi \\left| {\\omega _1 }\r\n\\right.)p(\\xi \\left| {\\omega _2 } \\right.)} \\right]} ^{{1/2}}\r\n{\\text{d}}\\xi\r\n\\] which simplifies to: \\[\r\nJ_B = \\tfrac{1} {8}\\left( {\\mu _2 - \\mu _1 } \\right)^T \\left(\r\n{\\frac{{\\Sigma _1 + \\Sigma _2 }} {2}} \\right)^{ - 1} \\left(\r\n{\\mu _2 - \\mu _1 } \\right) + \\tfrac{1}\r\n{2}{\\text{log}}\\frac{{\\left| {\\tfrac{1} {2}(\\Sigma _1 + \\Sigma\r\n_2 )} \\right|}} {{( {\\left| {\\Sigma _1 } \\right|\\left| {\\Sigma\r\n_2 } \\right|} )^{1/2} }}\r\n\\] when \\[X_1 =_d N(\\mu_1,\\Sigma_1) \\;\\; , \\;\\; X_2 =_d\r\nN(\\mu_2,\\Sigma_2).\\]\r\n\r\nThe Matusita distance:\r\n\\[J_T = \\left\\{ {\\int {\\left[ {\\sqrt {p(\\xi \\left| {\\omega _1 }\r\n\\right.)} - \\sqrt {p(\\xi \\left| {\\omega _2 } \\right.)} } \\right]}\r\n^2 {\\text{d}}\\xi } \\right\\}^{{1 / 2}} \\] which simplifies to: \\[\r\nJ_T = \\left\\{ {2\\left[ {1 - \\exp ( - J_B )} \\right]}\r\n\\right\\}^{{1/2 }} \\] where $J_B$ is the Bhattacharyya distance,\r\nwhen \\[X_1 =_d N(\\mu_1,\\Sigma_1) \\;\\; , \\;\\; X_2 =_d\r\nN(\\mu_2,\\Sigma_2).\\]\r\n\r\nThe Patrick-Fisher distance:\r\n\\[J_P = \\left\\{ {\\int {\\left[ {p(\\xi \\left| {\\omega _1 }\r\n\\right.)P_1 - p(\\xi \\left| {\\omega _2 } \\right.)P_2 } \\right]} ^2\r\n{\\text{d}}\\xi } \\right\\}^{{1/2}}\\] which simplifies to:\\[J_P =\r\n\\begin{array}{cc} {(2\\pi )^d \\left| {2\\Sigma _1 } \\right|} )^{ -\r\n{1/2}} + ( {(2\\pi )^d \\left| {2\\Sigma _2 } \\right|} )^{ - {1/2}} -\r\n\\\\ 2( {(2\\pi )^d \\left| {\\Sigma _1 + \\Sigma _2 } \\right|}\r\n)^{-{1/2}} \\exp \\left\\{ { - \\tfrac{1} {2}(\\mu _2 - \\mu _1\r\n)^T(\\Sigma _1 + \\Sigma _2 )^{ - 1} (\\mu _2 - \\mu _1 )}\r\n\\right\\}\\end{array}\\] when \\[X_1 =_d N(\\mu_1,\\Sigma_1) \\;\\; ,\r\n\\;\\; X_2 =_d N(\\mu_2,\\Sigma_2).\\]\r\n Reference: pp257, et sqq., Devijver, P.A.\r\n\\& Kittler, J (1982) \"Pattern Recognition: A Statistical\r\nApproach\", Prentice Hall International, Englewood Cliffs, NJ.\r\n\r\n\\section*{$S_\\alpha(\\sigma,\\beta,\\mu)$ Stable Random Variates}\r\n\r\nA Levy process is a stochastic process with a drift, a diffusion, and a jump component.\r\nThe Lévy–Khinchine representation of a Levy process $X_t$ with parameters $(a,\\sigma^2,W)$\r\nis given by\r\n\r\nThe Levy-Ito decompositon of a Levy process $X_t$ is a decomposition of $X_t$ into\r\nsingular, absolutely continuous, and discrete processes\r\n\\begin{eqnarray*}\r\n X_{ac} : X_{ac} \\ll X \\\\\r\n X_s : X_s \\perp X \\\\\r\n X_d : card (supp X_d) = \\aleph_0\r\n\\end{eqnarray*}\r\nvia Lebesgue's decomposition theorem.\r\n\r\n\r\n\\subsection{4 definitions of stable}\r\n\\begin{itemize}\r\n \\item If \\[\\exts C, D \\fall A,B s.t. A X_1+BX_2=_dCX+D\r\n \\fall X_1,X_2\\] independent copies of $X$, then $X\r\n \\in S_\\alpha(\\sigma,\\beta,\\mu)$. Furthermore $\\exts\r\n \\alpha \\in (0,2] s.t. C$ satisfies\r\n $C^\\alpha=A^\\alpha+B^\\alpha$, for any stable RV and\r\n $\\fall A,B$\r\n \\item Stable RV's satisfy a general CLT. If \\[\\fall n\r\n \\geq 2 \\exts C_n>0 D_n \\in \\dblr s.s. X_i+X_@\r\n \\hdots X_n=_D C_n X+D_n\\] where ${X_i}$ iid, then\r\n $X \\in S_\\alpha(\\sigma,\\beta,\\mu)$\r\n \\item If $\\exts \\; iid \\; RV \\;{Y_i}$ and \\[{d_n},\r\n {a_n} \\in BBCREVISIT^n \\dblr^n s.t\r\n \\frac{\\sum\\limits_{1}^{n} Y_i}{d_n}+a_n=_d X\\] then\r\n $X \\in S_\\alpha(\\sigma,\\beta,\\mu)$.\r\n \\item If $\\exts \\alpha \\in (0,2], \\; \\sigma \\geq0 \\;\r\n \\beta \\in [-1,1], \\; \\mu \\in \\dblr$ such that\r\n \\[ E[e^{i \\theta X}]=\\int_\\Omega e^{i \\theta X} dX = exp\\bigr(\r\n -\\sigma^\\alpha |\\theta|^\\alpha (1-i \\beta \\;sgn(\\theta)\\; tan\r\n \\bigl(\\frac{\\pi \\alpha}{2} \\bigr) + i \\mu \\theta)\\bigl) \\] when $\\alpha \\neq 1$\r\n and when $\\alpha =1$ we have \\[E[e^{i \\theta X}]=exp (\r\n -\\sigma |\\theta|(1+i \\frac{2}{\\pi} \\beta \\; sgn(\\theta)\r\n \\; ln|\\theta| + i \\mu \\theta) )\\]\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\subsection{Variance Gamma Process}\r\n$X_{VG}(t;\\sigma,\\nu,\\theta)\\theta \\gamma(t;\\nu)+\\sigma\r\nW_{Y(t;\\nu)}$ where $\\gamma(t;\\nu)$is a $\\Gamma$ process.\r\n\\[P_{\\gamma(t;\\nu)}(x)=\\frac{ x^{\\frac{t}{\\nu-1}} e^-\\frac{x}{\\nu}\r\n} {\\nu^{t/ \\nu} \\Gamma( t/ \\nu)}\\] \\[\\Phi_{VG}(\\omega)=E(e^{i\r\n\\omega X_{VG}}) = \\frac{1}{(1-i \\omega \\nu \\theta + \\sigma^2\r\n\\nu \\mu^2 / 2)^{t / \\nu}}\\] We can show that the Variance\r\nGamma process is the difference of two independent Gamma\r\nprocesses $ X_{VG}=\\gamma_p - \\gamma_n$ to obtain a new pdf\r\n\\[P_{\\gamma(t;\\nu)} =\\biggl\\{\\begin{array}{cc} \\frac{1}{\\nu |x|} e^{-|x| / \\eta_p} \\;\\;\\; x<0 \\\\\r\n\\frac{1}{\\nu |x|} e^{-|x| / \\eta_n} \\;\\;\\; x>0\r\n\\end{array} \\].\r\n%\r\n%\\section*{Martingales}\r\n%Martingales are sequences of random variables $\\{X_i\\}$where\r\n%the conditional expectation of $X_i$ given all the previous\r\n%values is $X_{i-1}$. \\[ E( X_i | X_{i-1} \\ldots X_0) = X_{i-1}\r\n%\\] An ergodic stochastic process $X_t$is one where the sample\r\n%moments $m_r = \\frac{1}{N} \\sum\\limits_{i=1}^{N}(x_i- m)^r$\r\n%converge to the population moments $E[X_t^r] = \\int X_t^r dP$\r\n\r\n\\section*{Maximum Entropy}\r\nEntropy in the context of information theory is expressed in units of bits, the amount of uncertainty in a yes or no question. Formally, for a sequence $ \\{X_i\\} \\ni p_i $ is a priori/posteriori probability of observing $X_i$ we define $H=- \\sum\\limits_{i} p_i log_2( p_i) $. We can define the entropy of a probability distribution by $H=\\int\\limits_{\\-infty}^{\\infty} p(x) log( p(x) ) dx $. We see the uniform distribution maximizes the entropy; if $p_i=\\alpha \\fall i$ then $\\frac{\\partial H}{\\partial } = log p +1/p=0 $\r\n\r\n\\section*{Gaussian Processes}\r\n$X_t$ is a Gaussian process (GP) if $\\forall (t_1,...t_n)$ we have that $(X_{t_1}, ... ,X_{t_n}) \\approx N(\\mu, \\Sigma)$. $\\Sigma \\sim K(x,x')$ the covariance functions. Gaussian processes are defined by second order statistics (BBCREVISIT). $K(x,x')$ specifies a distribution on functions (in the distribution sense of functional analysis). Every covariance function is a scalar product of vectors. Orenstein-Uhlenbeck (OU) and Brownian Motion (BM) are Gaussian processes. If a GP depends on $|x-x'|$ it is stationary. The OU process is a stationary GP and BM is not. The covariance function of the OU process is not differentiable. If a Gaussian process depends only on $|x-x'|$ it's called isotropic, these form an important class of\r\nHere we list the common covariance functions (kernels) constant, linear, Gaussian noise, square exponential, and OU.\r\n\\begin{center}\\begin{eqnarray*}\r\nK_c(x,x')= const \\\\\r\nK_\\script{l}(x,x') = \\\\\r\nK_{GN}(x,x') = \\sigma^2 \\delta_{x,x'} \\\\\r\nK_{SE}(x,x') = e^{ \\frac{-|d|^2}{2 \\script{l}} }\\\\\r\nK_{OU}(x,x') = e^{ \\frac{-|d|}{\\script{l}} }\r\n\\end{eqnarray*}\\end{center}\r\nHere $\\script{l}$ is the length scale and $d = |x-x'|$ \r\n\r\n\\section*{Discretizing Continuous Variables}\r\n\r\nR discretization library implements Fayyad and Irani\r\n\r\nPython https://github.com/navicto/Discretization-MDLPC\r\n\r\nMost methods used for discretizing a continuous variable use its relationship\r\nto another variable to determine the partitions. This is often found in\r\nclassification procedures, such as decision trees6 ] 10 and in naive Bayesian\r\nclassifiers.11,12\r\nAn entropy based method, proposed by Fayyad and Irani,6 chooses the\r\npartitioning point s in a sorted set of continuous values to minimize the joint Ž .\r\nentropy Ž Ž .. H X, Y of the continuous variable and the classification variable.\r\nThis is applied to the creation of decision tree structures for classification by\r\n66 CLARKE AND BARTON\r\nrecursively finding more partitioning points top-down discretization . The Ž .\r\nmethod is expanded to minimize a MDL metric to choose the partitioning\r\npoints.\r\nAnother MDL based method for discretization is described by Pfahringer.7\r\nA set of the best partitioning points is determined by recursively partitioning the\r\nŽ D sorted variable values to a depth D 2 y 1 partitions in a binary tree. Then .\r\nthe MDL metric is used in a best first search in this set to determine the best\r\npartitions for decision tree classification.\r\nA method that merges adjacent partitions of sorted variable values, according\r\nto the x2 statistical test, is described by Liu and Setiono.8 The variable\r\nvalues are sorted and initially partitioned into, at most, N intervals. The\r\nintervals are first recursively merged according to the lowest x2 value until a\r\nsignificance level of 0.05 is reached for each partition. The intervals are further\r\nmerged until a preset error rate with the classification variable is reached. If\r\nthere is only one resulting interval, the variable is not relevant to the classification\r\nproblem and is dropped. This method combines the discretization of\r\ncontinuous variables with feature selection for classification.\r\nDougherty, Kohavi, and Sahami9 compared several discretization techniques\r\nwith decision trees and with naive Bayesian classifiers. They found that a\r\nMDL metric, similar to that used by Fayyad and Irani,6 provided slightly better\r\nclassifications in both methods.\r\nA metric for discretization, based upon a variable’s classification in relation\r\nto other variables, is described by Hong.10 This metric is based upon a K\r\nnearest neighbor clustering technique and is used to generate decision trees. An\r\ninteresting feature of this method is that it returns an optimal number of\r\npartitions according to the metric. This is done by finding the ‘‘knee’’ of the\r\nplotted curve of the score as a function of the number of partitions. The plot is a\r\nconcave function of the metric; the knee is the point on the plot where the\r\nchanges in the number of variable values Ž . X axis become greater than\r\nthe changes in the metric value Ž . Y axis . The concavity of a plot is exploited in\r\nthe information theoretic discretization methods developed in Section IV.\r\nSubramonian, Venkata, and Chen,13 describe a visual framework for interactive\r\ndiscretization for decision tree classification. A user can choose between\r\nseveral algorithms and metrics for a classification problem, instead of being\r\nlimited to one method and metric. The choice of metrics includes cross-entropy\r\nand the L y 1 norm between two distributions.\r\nPazzani11 describes a technique for iterative discretization of continuous\r\nvariables for naive Bayesian classifiers. Each continuous variable is initially\r\ndivided into five partitions. For each variable, two partitions are then merged or\r\na partition is divided into two partitions to find a lower classification error. This\r\nprocedure is repeated for each continuous variable until the error rate can no\r\nlonger be reduced.\r\nAnother method for constructing naive Bayesian classifiers using a MDL\r\nmetric is presented by Friedman and Goldszmidt.12 This method begins by\r\nfinding the best initial partition of a continuous variable by dividing the range\r\nBAYESIAN BELIEF NETWORKS 67\r\ninto two partitions and then iterating the partitioning until there is no further\r\nimprovement in the MDL score top-down partitioning . The MDL metric Ž .\r\nincludes all of the variables used for classification and is repeated for each\r\ncontinuous variable. Given a BBN structure, this method discretizes each\r\ncontinuous variable in the Markov blanket of each classification variable. This\r\nprocedure is iterated until there is no improvement in each local MDL score.\r\nFriedman and Goldszmidt propose that this method can be adopted to\r\nlearning BBN structure by starting with some initial discretization of each\r\ncontinuous variable, learning an initial structure, and then rediscretizing as\r\ndescribed above. While this technique optimizes the conditional probabilities, it\r\ndepends upon the initial approximate discretization of continuous variables to\r\nlearn the correct network structure.\r\nExtensions to current discretization techniques for classification methods\r\nhave been proposed for BBNs, but none have currently been published. Methods\r\nfor both static done in data preprocessing and dynamic done during BBN Ž .Ž\r\nconstruction discretization of continuous variables are presented in Sections IV .\r\nand V.\r\nOften, a continuous variable has a normal, or Gaussian, frequency distribution.\r\nThe characteristics of the normal distribution are well understood and a\r\nunified heuristic method for finding a BBN structure with both discrete variables\r\nand continuous variables with a Gaussian distribution is described by Heckerman\r\nand Geiger.14 This method requires the mean and variance of each\r\ncontinuous variable’s values to parameterize the distribution", "meta": {"hexsha": "e9664a7d1b3b71115c25e14122e5e5022841f3ca", "size": 30372, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Appendix_Probability.tex", "max_stars_repo_name": "brucebcampbell/machine-learning-notes", "max_stars_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Appendix_Probability.tex", "max_issues_repo_name": "brucebcampbell/machine-learning-notes", "max_issues_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Appendix_Probability.tex", "max_forks_repo_name": "brucebcampbell/machine-learning-notes", "max_forks_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.622754491, "max_line_length": 1746, "alphanum_fraction": 0.6757210589, "num_tokens": 9861, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430478583169, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6482856721744172}} {"text": "\\documentclass[10pt,a4paper,final]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[fixlanguage]{babelbib}\n\\usepackage[english]{babel}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}\n\\usepackage[round,sort,nonamebreak]{natbib} % citação \n\n\\usepackage[usenames,svgnames,dvipsnames]{xcolor}\n\\usepackage[pdftex]{graphicx} % usamos arquivos pdf/png como figuras\n\\usepackage{float}%To make tables and figures stay where we want\n\\graphicspath{{./figures/}}\n\n\n\n\\title{Numerical Inverse Laplace Transform Integration on Ellipses}\n\\author{Pedro S. Peixoto}\n\n\\begin{document}\n\\maketitle\n\n\\section{Numerical Inverse Laplace Transform Integration on a Circle}\n\n\nInverse Laplace Transform (via Bromwich integral),\n\\begin{equation}\n{ \\mathfrak{ L } }^{ -1 }\\left\\{ \\widehat { f } \\right\\} =\\frac { 1 }{ 2\\pi i } \\int _{ \\gamma -i\\infty }^{ \\gamma +i\\infty }{ { e }^{ st }\\widehat { f } \\left( s \\right) ds } .\n\\end{equation}\nFrom the Residual Theorem we can replace it with \n\\begin{equation}\n{ \\mathfrak{ L } }^{ -1 }\\left\\{ \\widehat { f } \\right\\}=\\frac{1}{2\\pi i}\\int _{ { C }^{ } }{ { e }^{ st }\\widehat { f } \\left( s \\right) ds },\n\\end{equation}\nas long as all the poles of $\\widehat{f}$ lay inside a closed contour ($C$). Following \\citet{cclancy11} and \\citet{clancy:PhD} one can choose a circle centered at the origin and of radius $\\gamma$.\n\\begin{figure}[!h]\n \\centering\n \\hspace{1cm}\n \\includegraphics[width=.5\\textwidth]{Cstar} \n \\caption[Closed contour $C $]{Closed contour $C$}\n \\label{fig:Cstar} \n\\end{figure}\n\nThe integral over the contour $C $ can be solved using the parametrized circle $s(\\theta)=\\gamma e^{i\\theta}$, $\\theta \\in [0,2\\pi]$.\n\\begin{equation}\n{ \\mathfrak{ L } }^{ -1 }\\left\\{ \\widehat { f } \\right\\}=\\frac{1}{2\\pi i}\\int _{ { C }^{ } }{ { e }^{ st }\\widehat { f } \\left( s \\right) ds }=\\frac{1}{2\\pi i}\\int _{0}^{2\\pi}{ { e }^{ s(\\theta)t }\\widehat { f } \\left( s(\\theta) \\right)s'(\\theta) ds },\n\\end{equation}\nwhere $s'(\\theta)=is(\\theta)$. This integral can be approximated via numerical quadrature, using Trapezoidal Rule, on the $N$ points $\\theta_n=2\\pi n/N$, that results in \n\\begin{equation}\n{ \\mathfrak{ L } }^{ * }_C\\left\\{ \\widehat { f } \\right\\}=\\frac { 1 }{ Ni} \\sum _{ n=1 }^{ N }{ { e }^{ { s }_{ n }t }\\widehat { f } \\left( { s }_{ n } \\right) { s }'_{ n } }\n\\end{equation}\nwhere,\n\\begin{equation}\ns_{n}=\\gamma e^{{i\\theta_n}}, \\quad s'_{n}=\\gamma i e^{{i\\theta_n}}=is_n.\n\\end{equation}\n\n\\section{Numerical Inverse Laplace Transform Integration on an Ellipse}\n\nNow we wish to use an ellipse $E $ instead of a circle $C $ as closed contour. This ellipse will be chosen to satisfy the following properties:\\\\\n(i) On the imaginary axis, it will be bounded by the interval $[-\\gamma, \\gamma]$, for $\\gamma >0$.\\\\\n(ii) On the real axis it will be bounded by the interval $[-\\delta, \\delta]$, where $\\delta < 1$.\n\nThe first requirement ensures we can solve hyperbolic with purely imaginary eigenvalues up to the frequency defined by $\\gamma$. The second requirement ensures that having larger timestep sizes in the numerical time integration will not lead to evaluations of complex functions with large real parts, avoiding floating point instabilities.\n\n\\subsection{Parameterized ellipse}\n\nTo attend the above requirements, the ellipse may be defined via specialized Joukowsky transforms. Let $\\theta \\in [0,2\\pi]$ and $z=re^{i\\theta}$ define a circle of radius $r$ on the complex plane. An ellipse, following the above requirements, maybe built as function of $\\theta$ with the Joukowsky transform\n\\begin{equation}\nw(\\theta)=\\frac{\\gamma i}{d}\\left(re^{i\\theta}+\\frac{e^{-i\\theta}}{r}\\right),\n\\end{equation}\nwhere $d>0$ is yet to be determined. The $i\\gamma/d$ term in the Joukowsky is responsible for a few things: $i$ shifts rotates the ellipse, so that its larger axis is placed in the imaginary axis; $\\gamma/d$ is a normalization to allow that request (i) is satisfied. \n\nThe first condition (i) imposes that\n\\begin{equation}\n|\\text{Im}(w(\\theta))|= \\frac{\\gamma }{d}\\left(r+\\frac{1}{r}\\right)|\\cos(\\theta)| \\leq \\gamma, \\quad \\forall \\theta \\in [0, 2\\pi].\n\\end{equation}\nIn the Joukowsky transform, when $r=1$, the ellipse degenerates to a line segment, in this case the interval $[-2\\gamma i/d, 2\\gamma i/d]$. Choosing \n\\begin{equation}\nd=\\left(r+\\frac{1}{r}\\right),\n\\end{equation}\nis enough to fulfill condition (i).\n\nTo fulfill condition (ii), $r$ needs to be chosen such that \n\\begin{equation}\n|\\text{Re}(w(\\theta))|=\\frac{\\gamma}{d}|(r^{-1}-r)\\sin(\\theta)|\\leq\\delta <1 , \\quad \\forall \\theta \\in [0, 2\\pi].\n\\end{equation}\nA condition for this to happen, already using that $d=r^{-1}+r$, is that\n\\begin{equation}\n\\frac{|r^{-1}-r|}{r^{-1}+r}<\\delta /\\gamma\n\\end{equation}\nwhich, for $r>0$, is illustrated in the figure below.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.7]{rlimits1}\n\\caption{$\\frac{|r^{-1}-r|}{r^{-1}+r}$}\n\\end{figure}\n\nTherefore, condition (ii) is satisfied if \n\\begin{equation}\n\\sqrt{\\frac{1-\\delta/\\gamma}{1+\\delta/\\gamma}} 1$ or $r<1$. Here, we choose\n\\begin{equation}\nr=\\sqrt{\\frac{1-\\delta/\\gamma}{1+\\delta/\\gamma}},\n\\end{equation}\nwhich is adequate to ensure the imposed conditions.\n\n\nWith $\\gamma=2$, $\\delta=1$ (in practice $\\delta$ should be strictly smaller than $1$, but this is just an illustration) the ellipse generated would look like the figure bellow. \n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{ellipse.png}\n\\label{fig:ellip2_1}\n\\end{figure}\n\n\n\n\n\\subsection{Inverse Laplace Transform}\nLets now go back to the Inverse Laplace Transform, but now with the contour defined via the ellipse contour. \n\nUsing the properties of contour integration, the Inverse Laplace transform on the ellipse maybe written as\n\\begin{equation}\\label{MILT_ellipse}\n{ \\mathfrak{ L } }^{ -1 }\\left\\{ \\widehat { f } \\right\\}=\\frac{1}{2\\pi i}\\int _{ { E }^{ } }{ { e }^{ st }\\widehat { f } \\left( s \\right) ds }=\\frac{-1}{2\\pi i}\\int_{0}^{2\\pi} e^{s(\\theta)t}\\widehat { f } \\left( s(\\theta) \\right) s'(\\theta)\\,d\\theta,\n\\end{equation}\nwhere \n\\begin{equation}\ns(\\theta)=\\frac{\\gamma i}{d}\\left(re^{i\\theta}+\\frac{e^{-i\\theta}}{r}\\right),\n\\end{equation}\nand\n\\begin{equation}\ns'(\\theta)=-\\frac{\\gamma}{d}\\left(re^{i\\theta}-\\frac{e^{-i\\theta}}{r}\\right),\n\\end{equation}\nand the minus sign in the integration ensures the contour is swept in a counter-clockwise way.\n\nUniformly partitioning the $[0,2\\pi]$ interval on $N$ points with $\\theta_n=2\\pi n/N$, we may define the complex quadrature points as\n\\begin{equation}\ns_n=\\frac{\\gamma i}{d}\\left(re^{i\\theta_n}+\\frac{e^{-i\\theta_n}}{r}\\right), \\quad n=1,2,..,N,\n\\end{equation}\nand respectively the metric terms as\n\\begin{equation}\ns'_n=-\\frac{\\gamma}{d}\\left(re^{i\\theta_n}-\\frac{e^{-i\\theta_n}}{r}\\right), \\quad n=1,2,..,N.\n\\end{equation}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.4]{ellipse-quadpoints64}\n\\caption{64 quadrature points on an ellipse with $\\gamma=2$ and $\\delta=1$.}\n\\end{figure}\n\nThe quadrature points are not uniformly (equidistantly) distributed on the ellipse contour. However, the $\\theta_n$ points are equidistantly distributed on the $[0,\\pi]$ interval, ensuring exponential convergence of the Trapezoidal Rule, for instance, of the following estimate for the inverse Laplace transform\n\n\\begin{equation}\\label{milt_ellipse}\n{ \\mathfrak{ L } }^{ * }_E\\left\\{ \\widehat { f } \\right\\}=-\\frac { 1 }{ N i} \\sum _{ n=1 }^{ N }{ { e }^{ { s }_{ n }t }\\widehat { f } \\left( { s }_{ n } \\right) { s }'_{ n } }\n\\end{equation}\n\n\\section{Constant reconstruction}\n\nLet $\\widehat{f}(s)=1/s$, such that $f(t)=1$. Then the following equalities should hold (approximately):\n\n\\begin{equation}\n{ \\mathfrak{ L } }^{ * }_C\\left\\{ 1/s \\right\\}=\\frac { 1 }{ Ni} \\sum _{ n=1 }^{ N }{ { e }^{ { s }_{ n }t }{ s }'_{ n } /s_n} \\approx 1,\n\\end{equation}\n\n\\begin{equation}\\label{milt_ellipse}\n{ \\mathfrak{ L } }^{ * }_E\\left\\{ 1/s \\right\\}=-\\frac { 1 }{ Ni} \\sum _{ n=1 }^{ N }{ { e }^{ { s }_{ n }t }{ s }'_{ n }/s_n }\\approx 1.\n\\end{equation}\nwhere $s_n$ and $s'_n$ are defined differently on the circle and ellipse, as defined in the previous sections.\n\nThis reconstruction should be independent on $t$ (which will be our $dt$ in the ODE integration scheme).\n\nWe have set the a circle with radius $10$ and an ellipse with $\\gamma=10$ and $\\delta=0.5$, as the figure bellow.\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.5]{ellip_circ}\n\\label{fig:quad_elip_circ}\n\\caption{Quadrature points for circle ($\\gamma=10$) and ellipse ($\\gamma=10$, $\\delta=0.5$) with $N=32$}\n\\end{figure}\n\n\nThe results bellow indicate how the circle provides an accurate reconstruction for small $dt$, but how it breaks down due to floating point errors when large $dt$ are used. The ellipse has larger errors, but allows larger $dt$.\n\n\\begin{figure}\n\\centering\n\n\\includegraphics[scale=0.5]{const_recon}\n\\label{fig:const}\n\\caption{Absolute error in the numerical inversion of the function $\\widehat{f}=1/s$, which should result in $f(t)=1$ for any $t$ ($dt$). }\n\\end{figure}\n\n\n\\clearpage \n\\section{Exponential truncation}\n\n\n\\citet{cclancy11} and \\citet{clancy:PhD} use a truncated Taylor series in the calculation of the exponential term,\n\\begin{equation}\n{e }^{ { s }_{ n }t }=\\sum_{j=0}^{N}\\frac{(s_nt)^j}{j!},\n\\end{equation}\nwith $N$ matching the number of quadrature points. For the contour integration on a circle, \\citet{clancy:PhD} showed that the truncated series, with $N$ even, leads to exact reconstructions of polynomials, that is, \n\\begin{equation}\n\\mathfrak{ L }^{ * }_C\\left(\\frac{1}{s^{k+1}}\\right)=\\frac{t^k}{k!}, \\quad 0\\leq k \\leq N.\n\\end{equation}\n\nWe explored numerically the effects of having the truncated exponential and the results are shown in Figures \\ref{fig:const_truncN} and \\ref{fig:const_trunc16}, that show the errors when the exponential series is truncated with $N$ and $16$ terms respectively. The exponential series is highly sensitive to numerical floating point precision, therefore, for large $dt$, it creates numerical instabilities in both the Circle and Ellipse contours. Reducing the number of terms in the exponential series expansion (to 16) alleviates the problem of numerical stability on the circle, but deteriorates the results for the ellipse.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.5]{const_recon_truncN}\n\\label{fig:const_truncN}\n\\caption{Absolute error in the numerical inversion of the function $\\widehat{f}=1/s$ using truncated exponential with $N$ terms, which should result in $f(t)=1$ for any $t$ ($dt$). }\n\\end{figure}\n\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.5]{const_recon_trunc16}\n\\label{fig:const_trunc16}\n\\caption{Absolute error in the numerical inversion of the function $\\widehat{f}=1/s$ using truncated exponential with 16 terms, which should result in $f(t)=1$ for any $t$ ($dt$). }\n\\end{figure}\n\n\\clearpage\n\\section{Purely imaginary poles}\n\nLet $f(t)=e^{i\\alpha t}$, $\\alpha \\in \\mathbb{R}$, so that $\\widehat{f}(s)=\\frac{1}{s-i\\alpha}$. We now apply the numerical inverse Laplace transform (NILT), on both circle and ellipse, with $\\alpha$ values approaching the limit of the contour ($\\gamma=10$). See results bellow. The same contours were used as before, with an ellipse with $\\delta=0.5$ (see Fig \\ref{fig:quad_elip_circ}).\n\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.7]{nilt_expix_error}\n\\label{fig:const_trunc16}\n\\caption{Absolute error in the numerical inversion of the function $\\widehat{f}(s)=1/(s-i\\alpha)$, which should result in $f(t)=e^{i\\alpha t}$ for varying $t$ ($dt$) and number of quadrature points. }\n\\end{figure}\n\n\\section{Relation to REXI }\n\nThe discussion above relies on the knowledge of the Laplace transform of the function one desires to invert. However, an analogous construction maybe done directly from the Cauchy Integration theorem:\n\\begin{equation}\nf(x)=\\frac{1}{2\\pi i}\\ointop_{C}\\frac{f(z)}{z-x}dz,\n\\end{equation}\nfor $f$ holomorphic on $\\mathbb{C}$, and the contour curve $C$ assumed to be closed and should be integrated in a counter-clockwise way, as before.\n\nWe can again chose $C$ to be the ellipse contour\n\\begin{equation}\ns_n=\\frac{\\gamma i}{d}\\left(re^{i\\theta_n}+\\frac{e^{-i\\theta_n}}{r}\\right), \\quad n=1,2,..,N,\n\\end{equation}\nand respectively the metric terms as\n\\begin{equation}\ns'_n=-\\frac{\\gamma}{d}\\left(re^{i\\theta_n}-\\frac{e^{-i\\theta_n}}{r}\\right), \\quad n=1,2,..,N,\n\\end{equation}\nwhere $\\theta_n=2\\pi n/N$.\n\nTherefore, the function $f$ can be approximated via,\n\\begin{equation}\nf(x)=\\frac{-1}{2\\pi i} \\int_{0}^{2\\pi}\\frac{f(s(\\theta))}{s-x} s'(\\theta) d\\theta \\approx \n\\frac{-1}{2\\pi i} \\sum_{n=1}^{N}\\frac{f(s_n)}{s_n-x} s'_ n \\frac{2\\pi}{ N},\n\\end{equation}\nSimplifying,\n\\begin{equation}\nf(x)\\approx \n\\frac{1}{ iN} \\sum_{n=1}^{N}\\frac{f(s_n)}{x-s_n} s'_ n =\\frac{1}{N}\\sum_{n=1}^{N}\\frac{\\beta_n}{x+\\alpha_n},\n\\end{equation}\nwhere\n\\begin{equation}\n\\beta_n=-if(s_n)s'_ n\n\\end{equation}\nand \n\\begin{equation}\n\\alpha_n=-s_n\n\\end{equation}\nand also,\n\\begin{equation}\nr=\\sqrt{\\frac{1-\\delta/\\gamma}{1+\\delta/\\gamma}},\n\\end{equation}\n\\begin{equation}\nd=r^{-1}+r,\n\\end{equation}\nand $\\gamma$ is the largest attainable imaginary value and $\\delta$ is the largest attainable real value of the ellipse, as discussed before.\n\n\\clearpage\n\n\\bibliographystyle{unsrtnat}\n%\\selectbiblanguage{english}\n\n\\bibliography{bibliography} \n\n\\end{document}\n", "meta": {"hexsha": "85495b9b12b28f7120aa159b6256a68e5c78360b", "size": 13639, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/rexi/rexi_with_laplace/Laplace-ellipse.tex", "max_stars_repo_name": "valentinaschueller/sweet", "max_stars_repo_head_hexsha": "27e99c7a110c99deeadee70688c186d82b39ac90", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2017-11-20T08:12:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-11T15:32:36.000Z", "max_issues_repo_path": "doc/rexi/rexi_with_laplace/Laplace-ellipse.tex", "max_issues_repo_name": "valentinaschueller/sweet", "max_issues_repo_head_hexsha": "27e99c7a110c99deeadee70688c186d82b39ac90", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2018-02-02T21:46:33.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-11T11:10:27.000Z", "max_forks_repo_path": "doc/rexi/rexi_with_laplace/Laplace-ellipse.tex", "max_forks_repo_name": "valentinaschueller/sweet", "max_forks_repo_head_hexsha": "27e99c7a110c99deeadee70688c186d82b39ac90", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 12, "max_forks_repo_forks_event_min_datetime": "2016-03-01T18:33:34.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-08T22:20:31.000Z", "avg_line_length": 47.0310344828, "max_line_length": 626, "alphanum_fraction": 0.6988048977, "num_tokens": 4490, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799928900257126, "lm_q2_score": 0.831143045767024, "lm_q1q2_score": 0.6482856662925942}} {"text": "\\documentclass[10pt,a4paper]{article}\n\\usepackage[latin1]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\author{Daniel Frederico Lins Leite}\n\\title{Fibonnaci}\n\\newcommand{\\qed}{\\hfill\\blacksquare}\n\\begin{document}\n\t\\maketitle\t\n\\begin{align*}\n\tF_n >= 2^{n/2} &&\\text{for $n >= 6$}\n\\end{align*}\n\\paragraph{Base}\n\\paragraph{TODO. CALCULATE 6 AND 7 MANUALLY}\n\\paragraph{Induction Step}\n\\begin{align*}\n\tF_n &>= 2^{n/2}\\\\\n\tF_n &= F_{n-1}+F_{n-2}\\\\\n\t&>= 2^{(n-1)/2} + 2^{(n-2)/2}\\\\\n\t& &= (X+2^{(n-2)/2}) + 2^{(n-2)/2} &&\\text{$X>0$}\\\\\n\t& &= 2*2^{(n-2)/2}\\\\\n\t&>= 2*2^{(n-2)/2}\\\\\n\t& &=2^{\\frac{n-2}{2}+1}\\\\\n\t& &=2^{\\frac{n-2}{2}+\\frac{2}{2}}\\\\\n\t& &=2^{\\frac{n}{2}}\\\\\n\t&>= 2^{n/2}\\\\\n\t\\qed\n\\end{align*}\n\\end{document}", "meta": {"hexsha": "eff9b2c17b4970cee2e31abfc999798767b0da45", "size": 756, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "texts/math/Handout.Fibonnaci.tex", "max_stars_repo_name": "xunilrj/sandbox", "max_stars_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2017-04-01T17:18:35.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-12T05:23:23.000Z", "max_issues_repo_path": "texts/math/Handout.Fibonnaci.tex", "max_issues_repo_name": "xunilrj/sandbox", "max_issues_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2020-05-24T13:36:50.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-15T06:44:20.000Z", "max_forks_repo_path": "texts/math/Handout.Fibonnaci.tex", "max_forks_repo_name": "xunilrj/sandbox", "max_forks_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2018-09-20T01:07:39.000Z", "max_forks_repo_forks_event_max_datetime": "2019-02-22T14:55:38.000Z", "avg_line_length": 24.3870967742, "max_line_length": 52, "alphanum_fraction": 0.5793650794, "num_tokens": 364, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267830311354, "lm_q2_score": 0.7461389986757758, "lm_q1q2_score": 0.6482655459135469}} {"text": "\\chapter{\\textsc{Abstract Data Types}}\n\\section{Elements and tuples}\nIn Computer Science, \\emph{datum} (data in plural) is a reference with a representation (syntax) which is stored in a memory. Datum $a$ may be considered (semantically) as an element or object in some universe of discourse, i.e., universal set $U$, we write $a \\in U$. When $A \\subseteq U$ and $a \\in U$, then $a$ carries some \\emph{information} determined by $A$. The set $A$ is understood as a \\emph{concept}.\n\n\\begin{example}\nConsider a datum $a$ which semantically is an element in a set of digital cars (which is a universal set). One might say that $a$ is an \\emph{instance of an object class ``digital car''}. However, the digital car $a$ might be considered as an object whose properties are stored in computer memory in very many memory blocks. Naturally, each memory block should be referred somehow (by variables!) and in this sense there are datum semantically forming this digital car $a$.\n\\end{example}\n\n\\begin{example}\nAn element 5 may be interpreted in many different ways: $5 \\in r$ or $5 \\in \\mathbb{N}$.\n\\end{example}\n\n\\begin{example}\nWhen data is processed, it is usually obligatory to know in which universal set each datum belongs. For example, the term (expression) $5+4$ is not reasonable if $4,5 \\in \\mathbf{S}$ ($\\mathbf{S}$ is the set of arabic digits) and ``$+$'' is an operator on natural numbers. The operator ``$+$'' and $4,5$ are not compatible.\n\\end{example}\n\nSometimes a datum may be a \\emph{tuple}. A tuple (or n-tuple) is an element of a product set, thus, $\\left(a_1, \\dots ,a_n \\right) \\in U_1 \\times\\cdots\\times U_n$.\n\n\\begin{example}\nIf $V$ is a set of color names, and $M$ is a set of men's names, then $a=\\left(75,\\text{gray},\\text{Jari} \\right) \\in \\mathbb{N} \\times V \\times W$. One might say that tuple is a multidimensional datum.\n\\end{example}\n\n\\section{Memory and variables}\nIn program languages, data is stored in computer memory and data is processed using symbols called \\emph{variables}. From a programmer point of view, a variable is a container, label, or frame. A variable (only a symbol) refers to a datum.\\par \nProgram language compiler should transmit variable (symbol) to the operation system which then reserves on or more memory blocks to this symbol. Variable then refers to contents in computer memory block(s) having certain address(s), in which a datum is stored.\\par \nThe use of variables in some context then serves as \\emph{access} to corresponding data. As we have seen, datum is always of some type in the sense that the applicable processing methods variate.\n\n\\begin{example}\nWe have two variables $x$ and $y$, which refers to computer memory and there are datum. Remember that in computer memory blocks ONLY BITS ARE STORED. How the compiler of the current language ``knows'' that a program clause $z:=x+y$ (or cannot) be processed in a reasonable (in the sense of current program) way?\n\\end{example}\n\nIn program languages, the variables must be \\emph{typed}. There are weakly and strongly typed program languages. \\note Operators demand data types and they know themselves the type.\n\n\\section{Variables and data types}\nThe first clauses in a program are often declaring of variables and their types, like \\verb|x::integer| or \\verb|y::float|. After that, variable assignment can be processed; in this presentation, values of variables (variable assignment) are given by \\verb|:=| symbol, for example \\verb|x:=0|.\\par \nA \\emph{data type s} is a symbolic name, which has a set $A$ with operations $O$ on that set as its semantics. The operations usually satisfy some laws, so, semantics of $s$ is some specified algebra $\\mathcal{A}=\\left(A,O \\right)$. A variable $x$ of type $s$, written \\verb|x::s|, will be assigned to an element $a \\in A$ interpreted as \\emph{a point in the algebra $\\mathcal{A}$}, so we write \\verb|x:=| $x$.\n\n\\begin{example}\nclauses \\verb|y::float| and \\verb|y:=5| together stores 5 in some specified memory block, but 5 is understood now as a real number. So, the bits in this memory block are handled as a floating point number.\n\\end{example}\n\n\\begin{example}\nThe clauses \\verb|z::char| and \\verb|z:='c'| together could store the ASCII-code of the letter c.\n\\end{example}\n\n\\begin{example}\nIf an operation $\\omega$ takes values of data types $s_1, \\ldots ,s_n$ and produces values of data type $t$ then this is often written as $\\omega : s_1 \\times\\ldots\\times s_2 rightarrow t$. In this case, data types may have a many-sorted algebra as their semantics.\n\\end{example}\n\n\\note Some program languages (e.g. Python/Sage) process data as OBJECTS (object-oriented languages). Then, variables are often allowed to have \\emph{aliases} by default.\n\n\\begin{example}\nIn a clause \\verb|x:=5| the symbol \\verb|x| refers to a memory block containing object 5 (must be created); this object has a ``nametag'' \\verb|x|, i.e. this object is allowed to be referred by other variables also without copying the datum. This referencing method differs from symbol \\verb|x| being a ``label of box'' which contains 5.\n\\end{example}\n\n%\\section{Advantages of typing}\n%Programmers can focus on program language syntax and create the program syntactically using variables of reasonable data types and their operations. Unnecessary restrictions will be removed.\\par \n%Program languages have primitive data types (built-in data types) which define a kind of user interface to memory blocks. Which restrictions the hardware determines?\\par When data types are used effectively then \\emph{type erros} are avoided. For example, leet \\verb|x::interger| and \\verb|y,z::char|. Then the clause \\verb|z:=x+y| is reasonable if the operation ``+'' is of the form $+ \\, : \\text{integer} \\times \\text{char} \\rightarrow \\text{char}$.\n\n\\section{Data structures}\n\\subsection{What is data structure?}\nUsually programs handle a large collection of different kinds of data from data type point of view. Some data may form an entity of its own, a datum, which may be called as \\emph{data aggregate}. Data aggregate is referred by one variable.\\par \nThe data itself is naturally stored in different memory blocks, but it is now reasonable to assume that there must be some relation between elements, because the data aggregate is a datum itself.\\par \n\\emph{Data structure} is a way of organizing data such that data aggregate may be stored in and accessed from the memory efficiently and in a reasonable way from application point of view. The structure may be expressed as a relation between elements.\n\\subsection{Tuple} \nOne of the simplest data structure is \\emph{tuple}. Tuples are elements of a product set and in this sense it is a data aggregate. The elements in the tuple are not changeable (tuples are immutable) and the relation between elements is the identity relation.\\par \n\\note In a tuple each element may be of different data types. It might be possible to declare $x:: s_1, \\dots ,s_n$ if $x$ would create a new type from the existing ones (type constructor).\n\\subsection{Array}\n\\emph{Array} is a data structure, where the elements have position numbers. The size of an array is fixed, but elements are changeable (arrays are mutable). There are no other specific methods to manipulate arrays.\\par \nIn a more detail, an array $f$ with the size of $n$ position is a function, such that $f(i)$ is an element at the position number $i$, thus, $f:\\lbrace 0,1,\\ldots ,n-1 rbrace rightarrow S$, when the stored elements are members in $S$.\\par \nThe structure is now given as a function producing a linear order between the elements. Quite often arrays are declared like \\verb|x::array[0..n]| of \\verb|s|, where \\verb|s| is some data type. For example, \\verb|x:=[3,4,5]| would assign a 3-placed array to \\verb|x| declared by \\verb|x::array[0..2]| of \\verb|s|, where \\verb|s| represents integer data type.\\par \nFor \\emph{many-dimensional arrays}, one uses many different indices to position the elements. Notice that some modern program language do not demand array declarations, but the form of writing declares the array themselves.\\par \nIf the array is declared, however, to be the size of $n$ positions then the operating system reserves $n$ memory blocks from the memory, even if the program would not need them all.\n\n\\section{Abstract data type}\nPrimitive data types, like integer or float, are sets with operations such that they are defined by hardware, operating system or compilers. These kinds of data types are called ``built-in'' such that the corresponding algebras are implemented in hardware, operating system, or compiler.\\par \nAn \\emph{abstract data type} $s$ is a symbolic name, which has a collection of \\emph{mathematically isomorphic algebras} as its semantics, i.e. the operations defined for these algebras behave in a similar manner and the underlying sets are bijective.\n\n\\begin{example}\nAn abstract data type \\emph{bool} has a collection of all 2-element Boolean algebras (mathematically isomorphic algebras) as its semantics. For example, \\emph{bookl} might have algebras like $B_1=\\lbrace 0,1 rbrace$ and $B_2=\\lbrace \\text{True},\\text{False} rbrace$ with the operations on these sets behaving in the same way (like connectives in Propositional Logic).\n\\end{example}\n\n\\begin{example}\nConsider a program language with an abstract data type \\emph{bool}. The collection of all 2-element Boolean algebras is isomorphic to $\\mathcal{B}=\\left( B_2,\\ast , +, ',F,T \\right)$ as its semantics. In this program language, the syntactic operations for bool-type are $\\ast , +, ',F,T$ where $F$ is the zero element and $T$ is the unit element. We have:\n\\begin{verbatim}\nx,y,z::bool\nz = x * y\nWHILE z == T\n...\nENDWHILE\n\\end{verbatim}\nThese program clauses are independent from the actual implementation: it does not matter which 2-element Boolean algebra there is implemented.\n\\end{example}", "meta": {"hexsha": "07c1b3336c520f7bdff0e8f4353af7d3db977c0e", "size": 9833, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "DataType.tex", "max_stars_repo_name": "buiquanghuy23103/Math", "max_stars_repo_head_hexsha": "230b6363ed938bb9913836f28915ebe3feecf1c5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "DataType.tex", "max_issues_repo_name": "buiquanghuy23103/Math", "max_issues_repo_head_hexsha": "230b6363ed938bb9913836f28915ebe3feecf1c5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "DataType.tex", "max_forks_repo_name": "buiquanghuy23103/Math", "max_forks_repo_head_hexsha": "230b6363ed938bb9913836f28915ebe3feecf1c5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 105.7311827957, "max_line_length": 473, "alphanum_fraction": 0.7589748805, "num_tokens": 2489, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267762381844, "lm_q2_score": 0.7461389930307512, "lm_q1q2_score": 0.6482655359405126}} {"text": "%---------------------------Warpage-----------------------------\n\\section{Warpage}\n\nWarpage is defined as\n\\[\nq =\n 1 - \\min \\left\\{\n \\left( \\hat n_0 \\cdot \\hat n_2 \\right)^3,\n \\left( \\hat n_1 \\cdot \\hat n_3 \\right)^3\n \\right\\}\n\\]\nwhich is the cosine of the minimum dihedral angle formed by\nplanes intersecting in diagonals (to the fourth power).\n\nNote that if $\\normvec{N_k} < DBL\\_MIN$ for any $k$, we set $q = DBL\\_MAX$.\n\n\\quadmetrictable{warpage}%\n{$1$}% Dimension\n{$[0,0.7]$}% Acceptable range\n{$[0,2]$}% Normal range\n{$[0,DBL\\_MAX]$}% Full range\n{$0$}% Unit square\n{--}% Citation\n{v\\_quad\\_warpage}% Verdict function name\n\n\n", "meta": {"hexsha": "6663d6d60d82a439bc570ab4ef09355a9fd237ba", "size": 864, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadWarpage.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadWarpage.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadWarpage.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 32.0, "max_line_length": 75, "alphanum_fraction": 0.4375, "num_tokens": 226, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9314625126757596, "lm_q2_score": 0.6959583124210896, "lm_q1q2_score": 0.6482590784053295}} {"text": "% Chapter 2\n\n\\chapter{Convolutional Neural Networks} % Main chapter title\n\n\\label{Chapter2}\n\n%-------------------------------------------------------------------------------\n%-------------------------------------------------------------------------------\n\n\\section{CNNs: Layer by Layer}\n\nConvolutional neural networks have been very successful in computer vision as their key ingredient convolution filters preserve 2D structure and are able to pull features from images. Back propogation and stochastic gradient descent are used to optimize these filters in order to produce. Here I will cover some of the most common layers used in CNNs.\n\n\\subsection{Convolution Layer}\n\nThe first and foremost layer is the convolution layer. A convolution layer contains a set of filters which are \\textit{convolved} across feature maps in order to get the feature map. Let's take a concrete look at what this means. Imagine we are working with images that are of the shape 28x28x3 pixels. We then decide we want to use a convolution filter of size 2x2x5 which will result in a 28x28x5 feature map. Each kernel in this filter will be of size 2x2x3 since the depth in our original image is 3 and there will be five kernels. We then take the dot product of each slice in our 2x2x3 matrix against the original image at pixel 0x0 giving us 3 scalar values which are summed to get the first value in the first feature map of the output. We then slide these filters across the entire feature map in x and y thereby creating a new feature map. See ~\\ref{fig:Convolution} for an example of convolving two 3x3x3 kernels on a 7x7 image. A few things to consider when sliding the feature maps are image padding. For pixel 0,0 in our example part of the filter would be off the image. To keep the convolution consistent people will often times pad the edges of the image with 0 or 0.5. Additionally a stride length can be set which determines how far we move the filter at each step. A stride length of 2 would skip every other pixel when sliding.\n\n\\begin{figure}[ht]\n\\includegraphics[width=0.9\\textwidth]{Figures/Convolution.png}\n\\caption{An example of performing a 3x3 convolution with a depth of two on an 7x7x3 image with 1 pixel of padding from Andrej Karpathy's Stanford CNN Course \\cite{StanfordConv}}\n\\label{fig:Convolution}\n\\end{figure}\n\n\\subsection{Activation Layer}\n\nIts common after each of the convolutional layers to apply a non-linear activation function. A few examples of activation include: \\\\\n\n\\[ \\sigma(x) = \\frac{1}{1+e^{-x}} \\]\n\\[ \\tanh(x) = \\frac{e^x - e^{-x}}{e^x + e^{-x}} \\]\n\n\\[ ReLu(x) = \\begin{cases}\n 0 & x < 0 \\\\\n x & x \\geq 0\n \\end{cases}\n\\]\n\nReLu or rectified linear unit has been a popular choice for cnns as it is computation efficient without sacrificing accuracy.\n\n\\subsection{Pooling Layer}\n\nPooling layers are used to downsample feature maps from layer to layer. Two common flavors of pooling layers are max pooling and average pooling. Max pooling simply looks at a window and returns only the maximum value. Average pooling as it names suggests averages all the values in the window and returns that value. Lets take for example a 2x2 max pooling layer with an feature map of size of 128x128x32. After applying the pooling layer we would have 64x64x32 as an output. A common intuition behind max pooling is that it provides spatial invariance, i.e. if an object shifts by a few pixels we will still have a high response. An alternative to max pooling is using strided convolutions. The argument for this is that a good amount of information is lost in pooling layers.\n\n\\subsection{Fully Connected Layer}\n\nFully connected layers are what we traditional think of as the multi-layer perceptron model. They are commonly used in the final layer of CNNs to distill feature maps from a multi-dimensional matrix into a vector that can be then used for classification or regression. As a simple example are classification networks that will use two fully connected layers where the output is a vector of the same length as number of classes they want to predict from. SoftMax is applied to the vector to normalize the sum of the vector to one and then each scalar in the vector is interpreted as a likelihood of that image being of that particular class.\n\n\\subsection{Transfer Learning}\n\nTransfer learning is the process of taking a neural network trained for a certain task and refining it for another. A common example of this is retraining a classification network to predict a different set of classes than what it was originally trained for or changed from a classifier to a regression network. The common intuition for this is that the early layers in the network have learned basic features like, curves, corners, or other simple patterns that are universal in all types of objects. The top layers which carry the strong semantic information are closer to the end and therefore receive more gradient during back propagation. Some methods even freeze the weights for lower levels as fixed feature extractors and only allow the the final layers to change during training. As a bonus, transfer learning reduces the total time to train a network as the lower levels of a network take longer to train that the deeper layers.\n\nWhen exploring new architectures it is common procedure to simply download a set of weights for a popular network backbone and just apply transfer learning from there. In these next sections I will cover some of the popular network backbones.\n\n\\subsection{VGG16}\n\nVGG16 was one of the first major networks after AlexNet which went for simplicity and depth \\cite{VGG16}. The first contribution was its simplicity. CNNs before VGG used very large filter sizes, for example AlexNet used 11x11, which led to large parameter sizes. The authors of VGG16 instead used series of two back to back convolution/ReLu layers with 3x3 filters before doing max pooling. VGG16 at the time was considered a very deep network with 13 convolution layers and 3 fully connected layers. It showed that very deep networks were extremely important in learning the hierarchical features necessary for good classification. VGG16 scored $7.3\\%$ error rate on ImageNet dataset.\n\n\\subsection{ResNet}\n\nIn 2015 Microsoft Research came out with \\textit{Deep Residual Learning for Image Recognition} \\cite{ResNet} which had 152 layers and new type of of block called the residual block. It scored $3.6\\%$ on the ImageNet challenge in 2015. A residual block is a normal feed forward neural network that adds in \\textit{shortcut} connections where the input to a layer is both fed through a layer but also routed around the layer and added to the output of the next layer. See ~\\ref{fig:ResBlock} for a visual description. They hypothesized that a residual mapping induced by this block would be easier to learn than the original feed-forward network. ResNets come in several popular forms now such as ResNet18, ResNet32, ResNet50, ResNet101, and ResNet152, where the choice of depth is based on computational resources and dataset. Most popular deep learning frameworks have pretrained ResNet models available for download.\n\nAnother intuition behind ResNets success is that the \\textit{shortcut} connections allow gradients to move more easily through the network which helps alleviate a problem known as the vanishing gradient. Vanishing gradients is the problem where the loss at the end of the network has very little impact on changing the weights on the earlier layers and was one of the main reasons preventing networks from going as deep as ResNet. In the next network the use of skip connection is taken even further.\n\n\\begin{figure}[ht]\n\\includegraphics[width=0.9\\textwidth]{Figures/ResBlock.png}\n\\caption{Diagram of Residual Block from \\cite{ResNet}}\n\\label{fig:ResBlock}\n\\end{figure}\n\n\\subsection{DenseNet}\n\nIn 2016 Densely Connected Convolutional Networks (DenseNets) were proposed and shown to match and beat the residual network architecture on many datasets while using much fewer parameters and therefore computation. With DenseNets the authors connect all feature maps of same size together through concatenation so that the lth layer will have (l-1) feature maps. This is different then the ResNet model which uses summation in its shortcut connections. In order to concatenate feature maps across max pooling layers DenseNet introduce transition layers. These transition layers pass the prior feature maps through batch normalization, convolution, and finally average pooling before concatenating them to layers further down.\n\n\\begin{figure}[ht]\n\\includegraphics[width=0.9\\textwidth]{Figures/DenseNet.png}\n\\caption{Diagram of Residual Block from \\cite{DenseNet}}\n\\label{fig:DenseNet}\n\\end{figure}\n\n%-------------------------------------------------------------------------------\n%-------------------------------------------------------------------------------\n\\section{Object Detection Networks}\n\nObject detection networks deal with the problem of first localizing an object with an image frame as well as classifying that object and is generally considered a harder problem then pure classification tasks. The two popular datasets for benchmarking results are \\textit{Common Objects in Context} (COCO) \\cite{COCO} and \\textit{The PASCAL Visual Object Classes} (PASCAL VOC) \\cite{VOC}. At the time this work began the two most popular object detection networks were Faster-RCNN and YOLOv2 which I will cover in depth in the next two sections. Since then the state of the art has progressed considerably. Future work might include exploring this architectures.\n\n\\subsection{Faster-RCNN}\n\nFaster-RCNN \\cite{FASTER-RCNN}, where the R stands for region, is one of the most popular object detection networks at the moment and is an advancement on the authors previous two networks RCNN and Fast RCNN. The authors show results for both VGG16 and ZF network backbones, but many open source implementations of Faster-RCNN have replaced these backbones with various ResNet and DenseNets backbones.\n\nThe main contribution of Faster-RCNN is the Region Proposal Network (RPN) which replaces Fast-RCNNs slower, more complicated region proposal mechanism. It works by sliding a window across the final convolution feature map in a network and simultaneously predicting class probability and objectness. Objectness in this context is the probability that there is an object present in the bounding box. The RPN is implemented as a single NxN conv layer that maps into a lower dimension followed by two 1x1 conv layers for class probability and objectness. Furthermore they make use anchor boxes or priors on bounding boxes to make the task of the box regressor easier. At each position in the sliding window there are 9 anchor boxes which the regressor predicts bounding boxes with respect to.\n\nThe training scheme for this network is a multi-step process which involves training the RPN and Fast-RCNN networks separately and then later fixing the conv layers of Fast-RCNN and just fine-tuning the RPN.\n\n\\subsubsection{Loss}\nThe loss for Faster-RCNN is split into two parts as shown below.\n\n\\begin{align}\n \\Loss = \\Loss_{cls} +\\Loss_{reg}\n\\end{align}\n\nGiven $p_i$ is the predicted confidence that the ith anchor box is an object and $p_i^*$ is a binary indicator that the anchor box has an IoU greater than 0.7 with a ground truth we can define the loss as:\n\n\\begin{align}\n \\Loss_{cls}(p_i,p_i^*) = \\frac{1}{N_{cls}} \\sum_{i=1}^{A} [p_i\\log(p_i) + (p_i^*-p_i)\\log(1-p_i)]\n\\end{align}\n\nWe then parameterize the anchor box and bounding box truth as the following where $_a$ indicates a parameter from the anchor box and $*$ from the ground truth.\n\n\\begin{align}\n t_x = \\frac{(x - x_a)}{w_a} \\\\\n t_y = \\frac{(y - y_a)}{h_a} \\\\\n t_w = \\log(\\frac{w}{w_a}) \\\\\n t_h = \\log(\\frac{h}{h_a}) \\\\\n t_x^* = \\frac{(x^* - x_a)}{w_a} \\\\\n t_y^* = \\frac{(y^* - y_a)}{h_a} \\\\\n t_w^* = \\log(\\frac{w^*}{w_a}) \\\\\n t_h^* = \\log(\\frac{h^*}{h_a}) \\\\\n\\end{align}\n\nNext we define the regression loss where again we use $p_i^*$ as an indicator whether this anchor box has an IoU above 0.7 with the ground truth.\n\n\\begin{multline}\n \\Loss_{reg}(p_i^*, t, t^*) = p_i^*(Smooth_{L1}(t_x-t_x^*) + Smooth_{L1}(t_y-t_y*) \\\\ + Smooth_{L1}(t_w-t_w*) + SL1(t_h-t_h*)) \\\\\n\\end{multline}\n\n\\begin{align}\n Smooth_{L1}(d) = \\begin{cases}\n 0.5d^2 & if |d| \\leq 0 \\\\\n |d| - 0.5 & otherwise\n \\end{cases}\n\\end{align}\n\n\\subsection{YOLOv2}\n\nYOLOv2, You Only Look Once, is another successful object detector released in 2016 \\cite{YOLOv2}. It's key contribution is a unified network that simultaneously predicts object class and location with out the use of a secondary Region Proposal Network. While maintaining comparable accuracy to Faster-RCNN it runs at a much higher frame rate.\n\nThe backbone is of the authors design that uses only 3x3 and 1x1 convolutions, max pooling layers, and batch normalization. The network contains 23 convolution layers, 5 max pooling layers, and one short cut layer that concatenates features from lower levels to the final layer. Since the model contains no fully connected layers the network input size is unconstrained. The 5 max pooling layers of stride 2 mean that the final feature map will have a width and height that is the original width and height divided by $2^5$. As such the the original network input is usually selected to be a multiple of 32 i.e. 320, 416, 608, etc... An example of the network architecture is shown in table~\\ref{table:YOLOV2}\n\n\\begin{center}\n\\begin{table}[h]\\footnotesize\n \\caption{YOLOv2 Network Architecture layer by layer for an input resolution of 608 x 608}\\label{table:YOLOV2}\n \\begin{tabular}{| l | l | l | l | l |}\n \\hline\n layer & filters & size & input & output \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 0 conv & 32 & 3 x 3 / 1 & 608 x 608 x 3 & 608 x 608 x 32 \\\\ \\hline\n\n \\rowcolor{Salmon}\n 1 max & N/A & 2 x 2 / 2 & 608 x 608 x 32 & 304 x 304 x 32 \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 2 conv & 64 & 3 x 3 / 1 & 304 x 304 x 32 & 304 x 304 x 64 \\\\ \\hline\n\n \\rowcolor{Salmon}\n 3 max & N/A & 2 x 2 / 2 & 304 x 304 x 64 & 152 x 152 x 64 \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 4 conv & 128 & 3 x 3 / 1 & 152 x 152 x 64 & 152 x 152 x 128 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 5 conv & 64 & 1 x 1 / 1 & 152 x 152 x 128 & 152 x 152 x 64 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 6 conv & 128 & 3 x 3 / 1 & 152 x 152 x 64 & 152 x 152 x 128 \\\\ \\hline\n\n \\rowcolor{Salmon}\n 7 max & N/A & 2 x 2 / 2 & 152 x 152 x 128 & 76 x 76 x 128 \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 8 conv & 256 & 3 x 3 / 1 & 76 x 76 x 128 & 76 x 76 x 256 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 9 conv & 128 & 1 x 1 / 1 & 76 x 76 x 256 & 76 x 76 x 128 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 10 conv & 256 & 3 x 3 / 1 & 76 x 76 x 128 & 76 x 76 x 256 \\\\ \\hline\n\n \\rowcolor{Salmon}\n 11 max & N/A & 2 x 2 / 2 & 76 x 76 x 256 & 38 x 38 x 256 \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 12 conv & 512 & 3 x 3 / 1 & 38 x 38 x 256 & 38 x 38 x 512 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 13 conv & 256 & 1 x 1 / 1 & 38 x 38 x 512 & 38 x 38 x 256 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 14 conv & 512 & 3 x 3 / 1 & 38 x 38 x 256 & 38 x 38 x 512 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 15 conv & 256 & 1 x 1 / 1 & 38 x 38 x 512 & 38 x 38 x 256 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 16 conv & 512 & 3 x 3 / 1 & 38 x 38 x 256 & 38 x 38 x 512 \\\\ \\hline\n\n \\rowcolor{Salmon}\n 17 max & N/A & 2 x 2 / 2 & 38 x 38 x 512 & 19 x 19 x 512 \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 18 conv & 1024 & 3 x 3 / 1 & 19 x 19 x 512 & 19 x 19 x1024 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 19 conv & 512 & 1 x 1 / 1 & 19 x 19 x1024 & 19 x 19 x 512 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 20 conv & 1024 & 3 x 3 / 1 & 19 x 19 x 512 & 19 x 19 x1024 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 21 conv & 512 & 1 x 1 / 1 & 19 x 19 x1024 & 19 x 19 x 512 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 22 conv & 1024 & 3 x 3 / 1 & 19 x 19 x 512 & 19 x 19 x1024 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 23 conv & 1024 & 3 x 3 / 1 & 19 x 19 x1024 & 19 x 19 x1024 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 24 conv & 1024 & 3 x 3 / 1 & 19 x 19 x1024 & 19 x 19 x1024 \\\\ \\hline \\hline\n\n \\rowcolor{LightGreen}\n 25 route & 16 & N/A & N/A & N/A \\\\ \\hline\n\n \\rowcolor{Tan}\n 26 reorg & N/A & / 2 & 38 x 38 x 512 & 19 x 19 x2048 \\\\ \\hline\n\n \\rowcolor{LightGreen}\n 27 route & 26 & 24 & N/A & N/A \\\\ \\hline \\hline\n\n \\rowcolor{LightCyan}\n 28 conv & 1024 & 3 x 3 / 1 & 19 x 19 x3072 & 19 x 19 x1024 \\\\ \\hline\n\n \\rowcolor{LightCyan}\n 29 conv & 35 & 1 x 1 / 1 & 19 x 19 x1024 & 19 x 19 x 35 \\\\ \\hline\n\n \\hline\n \\end{tabular}\n\\end{table}\n\\end{center}\n\nYOLOv2 takes a simple approach to bounding box regression and class prediction. To understand how it works lets view the final feature map as a grid. At each location in this grid we want to predict bounding boxes and give a confidence metric for how likely there is to be an object there. We want to predict 5 numbers at each location in the grid plus a confidence for each class as show below:\n\n\\begin{align}\\label{eq:bbox_regressed}\n t_x,t_y,t_w,t_h,t_o,P(C_1),...,P(C_N)\n\\end{align}\n\nThe parameters $t_x,t_y,t_w,t_h$ are regressed with relation to the feature map x,y location which we call $f_x$ and $f_y$. Further $t_x,t_y$ are constrained to be between [0,1] by the logistic function. We also define $p_w,p_h$ as the prior width and height for the anchor boxes.\n\n\n\\begin{align}\n b_x &= \\sigma(t_x) + f_x \\\\\n b_y &= \\sigma(t_y) + f_y \\\\\n b_w &= p_w * exp(t_w) \\\\\n b_h &= p_h * exp(t_h)\n\\end{align}\n\nLike Faster-RCNN the author uses anchor boxes as priors for predicting bounding boxes at each locations. The author determined that 5 prior anchor boxes was a good trade-off between accuracy and efficiency. For each prior bounding box we need to predict the parameters shown in equation~\\ref{eq:bbox_regressed}.\n\nLets take a concrete look at this with an example. Say that our input resolution is 416 x 416, we have 3 classes, and are using 5 prior anchor boxes. Dividing our input resolution by 32 we obtain our final feature map width and height equal to 13. Now at each location in this 13 x 13 map we need to regress 3 classes and 5 bounding box parameters for 5 anchor boxes.\n\n\\begin{align}\n (5 * (5 + 3)) * 13^2) = 6760\n\\end{align}\n\nInstead of using a several fully connected layer to regress these numbers the author simply uses a 1x1 conv in the prior layer and sets the number of filters equal to the number of anchors multiplied by the 5 parameters plus the number of classes. In table~\\ref{table:YOLOV2} there are 5 anchors and 2 classes meaning there are 35 filters going into the final feature map. Which can be seen on layer 29.\n\nThis setup allows the user to have the freedom to change the input resolution on the fly without having to modify the final layer. The only information we need a prior is the number of classes.\n\n\\subsubsection{Loss}\n\nThe loss function for YOLOv2 consists of 4 separate parts\n\n\\begin{align}\n \\Loss_{YOLOv2} = \\Loss_{coord} + \\Loss_{obj} + \\Loss_{noobj} + \\Loss_{class} \\\\\n\\end{align}\n\nThe first part $\\Loss_{coord}$ deals with bounding box regression.\n\n\\begin{multline}\n \\Loss_{coord} = \\lambda_{cood} \\sum_{i=1}^{f_x*f_y} \\sum_{j=i}^{Anchors}\n \\indicator_{obj}^{i,j} [\n (\\sigma(t_x^{(i,j)}) - \\sigma(\\hat{t}_x^{(i,j)}))^2 +\n (\\sigma(t_y^{(i,j)}) - \\sigma(\\hat{t}_y^{(i,j)}))^2 + \\\\\n (\\sigma(t_w^{(i,j)}) - \\sigma(\\hat{t}_w^{(i,j)}))^2 +\n (\\sigma(t_h^{(i,j)}) - \\sigma(\\hat{t}_h^{(i,j)}))^2] \\\\\n\\end{multline}\n\nThe next part deal with the loss associated with predicting the objectness of the bounding box. Ideally we would like the objectness to predict the IoU of the predicted bounding box with the true bounding box. This is a little unintuitive as during inference we do not have access to the truth. The $\\Loss_{noobj}$ is there to punish false positives.\n\n\\begin{align}\n \\Loss_{obj} = \\lambda_{obj} \\sum_{i=1}^{f_x*f_y} \\sum_{j=i}^{Anchors} \\indicator_{obj}^{i,j} (IoU_{truth}^{pred} - \\sigma(\\hat{t}_o^{(i,j)}))^2 \\\\\n \\Loss_{noobj} = \\lambda_{noobj} \\sum_{i=1}^{f_x*f_y} \\sum_{j=i}^{Anchors} \\indicator_{noobj}^{i,j} (-\\sigma(\\hat{t}_o^{(i,j)}))^2 \\\\\n\\end{align}\n\nThe final piece weights the error associated with predicting the class of the bounding box. It's simply the sum of squared error of the true class probability distribution (which is 1 at the true class and 0 everywhere else) and the predicted class probability. This is different that much formulations which is the cross entropy loss function like Faster-RCNN.\n\n\\begin{align}\n \\Loss_{class} = \\lambda_{class}\\sum_{i=1}^{f_x*f_y} \\sum_{j=i}^{Anchors} \\sum_{c=1}^{N} \\indicator_{obj}^{i,j} (\\indicator_{c=true class}-P(c)^{(i,j)})^2\n\\end{align}\n\nAs an aside the naive loss function with regard to class probability has serious issues with regards to class imbalance. Our work demonstrates YOLOv2s inability to work well over in these situations as well be explained further in the results section.\n\n\\subsection{Selecting a Network}\n\nPreliminary results using both networks showed results heavily in favor of YOLOv2. The largest reason being the input size to YOLOv2 could easily be scaled up and had pretrained models for higher resolution networks. This could potentially have been replicated on with Faster-RCNN; however, it would have taken extensive architecture changes and weeks training networks from scratch.\n\nAdditionally the Darknet framework that YOLOv2 runs on was more approachable for the author allowing for easy modifications where necessary.\n", "meta": {"hexsha": "e79d197e65af8256f433993bb373b5fd9f1fac48", "size": 22055, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chapters/Chapter2.tex", "max_stars_repo_name": "matte1/Thesis", "max_stars_repo_head_hexsha": "c693558a9a6ee840f2b260efb2a6b67870e889d1", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Chapters/Chapter2.tex", "max_issues_repo_name": "matte1/Thesis", "max_issues_repo_head_hexsha": "c693558a9a6ee840f2b260efb2a6b67870e889d1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Chapters/Chapter2.tex", "max_forks_repo_name": "matte1/Thesis", "max_forks_repo_head_hexsha": "c693558a9a6ee840f2b260efb2a6b67870e889d1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.8403908795, "max_line_length": 1350, "alphanum_fraction": 0.7144411698, "num_tokens": 6234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424489603725, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6482403255698911}} {"text": "\\section{Supplementary material}\n\n\\paragraph{Simplicial distance and the localizing property of the Laplacian.}\nSuppose that $\\sigma$ and $\\tau$ are $p$-simplices for which $(\\nu_0, \\nu_1, \\dotsc, \\nu_d)$ is the shortest sequence of $p$-simplices with the property that $\\nu_0=\\sigma$, $\\nu_d=\\tau$, and each $\\nu_i$ shares a face or a coface with $\\nu_{i-1}$, and a face or a coface with $\\nu_{i+1}$. We say that $d$ is the \\emph{simplicial distance} between $\\sigma$ and $\\tau$. Then for all $N1$. It sends to the client all PoR proofs (or a combination of them) after period $T$. Also, if the basic PoSt is used, the server sends C-TLP proofs that can be efficiently verified by anyone. On the other hand, if the compact PoSt is used, then the server does not need to send the C-TLP's proofs, as the client already knows the random challenges. The adjustments considerably improve the PoSt protocols' costs. In particular, the server's computation cost would be $\\frac{1}{3}$ of the costs imposed by either of the original PoSt protocols. Also, there will be $3z$ further reduction in the number of exponentiations: (a) at the verifier side, in the basic PoSt, as it does not need to perform any exponentiation to check the correctness of C-TLP's output, (b) at the client-side, in the compact PoSt, as the client does not need to evaluate TDF at the setup to precompute the challenges which in total involves $3z$ modular exponentiations over $\\phi(N)$. Moreover, the proof size would be reduced by a factor of $6.5$.\n\n\\begin{remark}\nAlthough the use of C-TLP in the PoSt protocols can reduce the computation and communication costs, (a) the server-side I/O cost in these schemes (i.e. $O(\\log n)$ in the basic PoSt and $O(n)$ in the compact one), and (b) PoR's proof size complexity in the basic PoSt, will remain the same. Because these costs stem from the underlying PoR schemes, used as a black box, by the two PoSt protocols.\n\\end{remark}", "meta": {"hexsha": "b0dae4b4d45dcd3b2fb59bc1cda186bb1833e752", "size": 3027, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Paper/eprint-version/PoSt-and-C-TLP.tex", "max_stars_repo_name": "AydinAbadi/CR-LP", "max_stars_repo_head_hexsha": "b2139df715f441a48eeae0b88e038fb6acc5d6e2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Paper/eprint-version/PoSt-and-C-TLP.tex", "max_issues_repo_name": "AydinAbadi/CR-LP", "max_issues_repo_head_hexsha": "b2139df715f441a48eeae0b88e038fb6acc5d6e2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Paper/eprint-version/PoSt-and-C-TLP.tex", "max_forks_repo_name": "AydinAbadi/CR-LP", "max_forks_repo_head_hexsha": "b2139df715f441a48eeae0b88e038fb6acc5d6e2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 275.1818181818, "max_line_length": 1681, "alphanum_fraction": 0.7720515362, "num_tokens": 748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424295406088, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6482403106886496}} {"text": "\\documentclass[11pt,twoside]{fncextra}\n\n\\pagestyle{myheadings}\n\\markboth{}{Asteroids}\n\n\\begin{document}\n\n\\begin{center}\n \\bf Asteroids (pew pew)\n\\end{center}\n\nIn 1609 Johannes Kepler proposed that celestial bodies in the solar system follow elliptical orbits with the sun at one focus. This so-called First Kepler Law laid the groundwork for sophisticated mathematical description of orbits, and indeed spurred and justified the early development of calculus, as Kepler's geometric laws were worked out into formulas that were used to explain gravitational force.\n\nThere are multiple parameters used to characterize these orbits. If we consider only the motion within the plane of the orbit, these can all be derived from two readily observed ones: $\\tau$, the period of the orbit, and $\\epsilon$, the eccentricity of the ellipse. (Larger values mean greater deviation from circularity, with a circle at $\\epsilon=0$ and a degenerate line at $\\epsilon=1$. )\n\nPerhaps the geometrically easiest parameters to understand are the \\emph{true anomaly} $\\nu$, which is the angle between the body and its \\emph{perihelion}, or its closest approach to the sun; and the distance $r$ from the body to the sun. These parameters are related by the equations \\begin{align}\n \\tan \\frac{\\nu}{2} &= \\sqrt{\\frac{1+\\epsilon}{1-\\epsilon}}\\,\n \\tan \\frac{\\psi}{2}, \\label{eq:Asteroids1} \\\\\n M &= \\psi - \\epsilon \\sin \\psi, \\label{eq:Asteroids2} \\\\\n r &= \\frac{a(1-\\epsilon^2)}{1+\\epsilon \\cos \\nu} \\\\\n a^3 &= \\mu \\left(\\frac{\\tau}{2\\pi}\\right)^2.\n\\end{align}\nHere, $M$ is known as the \\emph{mean anomaly}; it is proportional to time and varies from 0 to $2\\pi$ over one complete orbit traversal. In addition, $\\psi$ is known as the \\emph{eccentric anomaly}, $a$ is half of the (maximum) diameter of the ellipse, and $\\mu$ is a gravitational parameter that for the Sun is $39.47524$ $\\text{AU}^3/\\text{yr}^2$. (An AU is roughly the Earth--Sun distance.)\n\nEquation~\\eqref{eq:Asteroids2} implicitly defines $\\psi$ as a function of $M$ (and therefore time), but it cannot be solved in closed form. Instead, given a value of $M$, rootfinding must be used to find $\\psi$. The rest of the parameters in the equations can be calculated explicitly from $\\psi$.\n\n\n\n\\subsection*{Preparation}\n\nRead Section~4.1.\n\n\\subsection*{Goals}\n\nYou will apply \\texttt{fzero} in MATLAB to compute the eccentric anomaly given the orbital period and eccentricity of an asteroid. From this you can compute other quantities of interest.\n\n\n%\\end{document}\n\n\\subsection*{Procedure}\n\n\\begin{enumerate}\n\\item Let \\texttt{M} be a vector of 800 evenly spaced values from $0$ to $2\\pi$. Set $\\epsilon=0.1$ and solve~\\eqref{eq:Asteroids2} using \\texttt{fzero} to compute a value of $\\psi$ for each entry of $M$. Make a labeled plot of $\\psi(M)$. \\item Add to the plot of step 1 by stepping through $\\epsilon=0.2,0.3,\\ldots,0.9$, and plotting the resulting $\\psi(M)$ each time. \\item The asteroid 324 Bamberga is both one of the largest and one of the most eccentric in the asteroid belt. It has $\\epsilon = 0.338$ and $\\tau = 4.40$ yr. Plot $\\nu(M)$ over $0\\le M \\le 2\\pi$ for this asteroid. (You will probably find that $\\nu$ jumps suddenly from $\\pi$ to $-\\pi$. It's not incorrect, but if this offends you, use \\texttt{unwrap(nu)} to get a smooth equivalent in $[0,2\\pi]$.)\n\\item Plot $r(M)$ as a function of $M$ for 324 Bamberga.\n\\item Halley's Comet has $\\epsilon=0.967$ and $\\tau = 75.3$ yr. Plot $r(M)$ for this comet. What are its maximum (aphelion) and minimum (perihelion) values? (You can check these figures against ones you find on the internet.) \n\\end{enumerate}\n\n\n\\end{document}\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: t\n%%% End: \n", "meta": {"hexsha": "43ccb13044e44e7eeb0ab9c6898a8e62421721b4", "size": 3713, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "labs/chapter04/Asteroids/Asteroids.tex", "max_stars_repo_name": "snowdj/fnc-extras", "max_stars_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 109, "max_stars_repo_stars_event_min_datetime": "2018-04-21T09:02:50.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-20T19:03:54.000Z", "max_issues_repo_path": "labs/chapter04/Asteroids/Asteroids.tex", "max_issues_repo_name": "snowdj/fnc-extras", "max_issues_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2018-12-04T22:17:44.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-03T21:04:47.000Z", "max_forks_repo_path": "labs/chapter04/Asteroids/Asteroids.tex", "max_forks_repo_name": "snowdj/fnc-extras", "max_forks_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 49, "max_forks_repo_forks_event_min_datetime": "2017-04-02T17:21:33.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-29T17:19:57.000Z", "avg_line_length": 67.5090909091, "max_line_length": 770, "alphanum_fraction": 0.7282520873, "num_tokens": 1042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424256566558, "lm_q2_score": 0.7662936324115011, "lm_q1q2_score": 0.6482402941674349}} {"text": "\\chapter{RESULTS AND DISCUSSIONS}\n\\label{ch:results}\n\n\\section{Background of Study}\n\nSuppose that we have a line with equation $y = 2x + 3$. This line cuts the $y$-axis at $y = 3$. We can find the gradient of the line using \\eqref{eq:gradient}. Equation~\\eqref{eq:tada}, $\\ldots$.\n\n\\begin{equation} \\label{eq:gradient}\n m = \\frac{y_2 - y_1}{x_2 - x_1}\n\\end{equation}\n\n\\begin{equation} \\label{eq:tada}\ne^{\\pi i} + 1 = 0\n\\end{equation}\n\n\\subsection{Comparison of Method A with Other Studies}\n\n\\begin{table}[ht]\n \\caption{Length Units}\n %\\begin{tabular}{cc}\n \\begin{tabular}{>{\\centering\\arraybackslash}p{.47\\textwidth} >{\\centering\\arraybackslash}p{.47\\textwidth}}\n \\toprule %header\n Millimeters & Centimeters\\\\\n mm & cm\\\\\n \\midrule\n 1 & 0.1\\\\\n 10 & 1\\\\\n 100 & 10\\\\\n 1000 & 100\\\\\n 10000 & 1000\\\\\n \\bottomrule\n \\end{tabular}\n \\par\\raggedright Note: This table is useful for $\\ldots$.\n \\label{table:lengthunits}\n\\end{table}\n\n\\subsubsection{Method A Improved}\n\nFigure~\\ref{fig:logouitm} is $\\dots$. \\lipsum[1-2]\n\n\\begin{figure}[ht]\n \\centering\n \\fbox{ % add box arounf image\n \\includegraphics[width=.9\\linewidth]{logouitm} % scale=.5 <- 50% of original size\n }\n \\caption[Short version for LoF]{Logo UiTM Logo UiTM Logo UiTM Logo UiTM Logo UiTM Logo UiTM Logo UiTM Logo UiTM}\n \\label{fig:logouitm}\n\n \\par\\raggedright\n Notes/Sources: Phasellus in dui mi. Suspendisse placerat nisl et elit tristique, non congue elit bibendum. Donec mauris libero, vehicula in feugiat vitae.\n\\end{figure}\n\n\\lipsum[2-3]\n\n\n\n", "meta": {"hexsha": "652224811368d1f9b925e7d9c5af36106503cbf0", "size": 1677, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "mainmatter/chapResults.tex", "max_stars_repo_name": "jafhate/uitmthesis", "max_stars_repo_head_hexsha": "00b977f7a932053898b09f20f798e0f723c439cd", "max_stars_repo_licenses": ["LPPL-1.3c"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2021-03-01T11:39:00.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-21T09:02:01.000Z", "max_issues_repo_path": "mainmatter/chapResults.tex", "max_issues_repo_name": "jafhate/uitmthesis", "max_issues_repo_head_hexsha": "00b977f7a932053898b09f20f798e0f723c439cd", "max_issues_repo_licenses": ["LPPL-1.3c"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-03-26T04:24:43.000Z", "max_issues_repo_issues_event_max_datetime": "2021-03-26T04:24:43.000Z", "max_forks_repo_path": "mainmatter/chapResults.tex", "max_forks_repo_name": "jafhate/uitmthesis", "max_forks_repo_head_hexsha": "00b977f7a932053898b09f20f798e0f723c439cd", "max_forks_repo_licenses": ["LPPL-1.3c"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-03-22T05:47:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-26T01:37:04.000Z", "avg_line_length": 29.4210526316, "max_line_length": 195, "alphanum_fraction": 0.6326774001, "num_tokens": 565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8757869981319863, "lm_q2_score": 0.7401743735019594, "lm_q1q2_score": 0.6482350926635047}} {"text": "%Abstract\n\n\\chapter*{Abstract}\n\\addcontentsline{toc}{chapter}{Abstract}\n\n%The structures and energies of different types of clusters are studied\n%computationally. For this, a combination of graph theoretical methods,\n%approximate interaction potentials and accurate quantum chemical calculations is\n%used to investigate gold clusters, Lennard-Jones (LJ) clusters and clusters\n%bound by the sticky-hard-sphere (SHS) interaction potential.\n\nThe structures and stabilities of hollow gold clusters are investigated by means\nof density functional theory (DFT) as topological duals of carbon fullerenes.\nFullerenes can be constructed by taking a graphene sheet and wrapping it around\na sphere, which requires the introduction of exactly 12 pentagons. In the dual\ncase, a (111) face-centred cubic (fcc) gold sheet can be deformed in the same\nway, introducing 12 vertices of degree five, to create hollow gold nano-cages.\nThis one-to-one relationship follows trivially from Euler's polyhedral formula\nand there are as many golden dual fullerene isomers as there are carbon\nfullerenes. Photoelectron spectra of the clusters are simulated and\ncompared to experimental results to investigate the possibility of detecting\nother dual fullerene isomers. The stability of the hollow gold cages is compared\nto compact structures and a clear energy convergence towards the (111) fcc sheet\nof gold is observed. \n\nThe relationship between the Lennard-Jones (LJ) and sticky-hard-sphere (SHS)\npotential is investigated by means of geometry optimisations starting from the\nSHS clusters. It is shown that the number of non-isomorphic structures resulting\nfrom this procedure depends strongly on the exponents of the LJ potential. Not\nall LJ minima, that have been discovered in previous work, can be retrieved this\nway and the mapping from the SHS to the LJ structures is therefore non-injective\nand non-surjective. The number of missing structures is small and they\ncorrespond to energetically unfavourable minima on the energy landscape. The\noptimisations are also carried out for an extended Lennard-Jones potential\nderived from coupled-cluster calculations for the xenon dimer, and, although the\nshape of the potential is not too different from a regular (6,12)-LJ potential,\nthe number of minima increases substantially.\n\nGregory-Newton clusters, which are clusters where 12 spheres surround and touch\na central sphere, are obtained from the complete set of SHS clusters. All 737\nstructures result in an icosahedron, when optimised with a (6,12)-LJ potential.\nFurthermore, the contact graphs, consisting only of atoms from the outer shell\nof the clusters, are all edge-induced sub-graphs of the icosahedral graph. For\nhigher LJ exponents the symmetry of the potential energy surface breaks away\nfrom the icosahedral motif towards the SHS landscape, which does not support a\nperfect icosahedron for energetic reasons. This symmetry breaking is mainly\ngoverned by the shape of the potential in the repulsive region, with the\nlong-range attractive region having little influence.\n\n\n\n", "meta": {"hexsha": "70c2297637d460f695d496ceac68be881736721a", "size": 3062, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "abstract.tex", "max_stars_repo_name": "Trombach/thesis", "max_stars_repo_head_hexsha": "0eb1d2a109a0f91271fbdd6d85b4706b2e281d93", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "abstract.tex", "max_issues_repo_name": "Trombach/thesis", "max_issues_repo_head_hexsha": "0eb1d2a109a0f91271fbdd6d85b4706b2e281d93", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "abstract.tex", "max_forks_repo_name": "Trombach/thesis", "max_forks_repo_head_hexsha": "0eb1d2a109a0f91271fbdd6d85b4706b2e281d93", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.8846153846, "max_line_length": 81, "alphanum_fraction": 0.8200522534, "num_tokens": 671, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.875787001374006, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.648235085024068}} {"text": "\\subsection{Analytic geometry in the plane}\\label{subsec:analytic_geometry_in_the_plane}\n\n\\begin{remark}\\label{rem:analytic_geometry}\n Analytic geometry is a XVII-century branch of mathematics that studies geometric figures using coordinate \\hyperref[rem:coordinate_systems]{systems}. The term \\enquote{analytic geometry} may refer to a modern subbranch of algebraic geometry, however we refrain from using \\enquote{analytic geometry} in that sense. Historically, most of these definitions were given either for the Euclidean \\hyperref[def:euclidean_plane]{plane} or for the three-dimensional Euclidean space.\n\n Most of the definitions from \\fullref{subsec:vector_space_geometry} are generalizations of concepts from analytic geometry. We will state definitions in the language of linear algebra and refrain from using synthetic (axiomatic) geometry. When working in the plane (resp. three-dimensional space), we will assume that we have fixed an \\hyperref[def:orthonormal_system]{orthonormal} coordinate \\hyperref[def:euclidean_plane_coordinate_system]{system} \\( Oxy \\) (resp. \\( Oxyz \\)), which allows us to visualize geometric figures.\n\\end{remark}\n\n\\begin{definition}\\label{def:plane_line_equations}\n \\hyperref[def:geometric_line]{Lines} in \\( \\BbbR^2 \\) are, so ubiquitous that they can be represented by a lot of standard \\hyperref[ex:equations]{equations}.\n\n \\begin{thmenum}\n \\thmitem{def:plane_line_equations/vector_parametric} When regarding a line as a parametric curve as in \\fullref{def:geometric_line/parametric}, the \\hyperref[def:first_order_syntax/formula]{formula}\n \\begin{equation}\\label{def:plane_line_equations/parametric_equation}\n l(t) = tx + a\n \\end{equation}\n is called a \\term{vector parametric equation}.\n\n \\thmitem{def:plane_line_equations/scalar_parametric} Given \\fullref{def:plane_line_equations/parametric_equation}, the \\term{scalar parametric equations} of the line are\n \\begin{equation}\\label{def:plane_line_equations/scalar_parametric_equations}\n \\begin{cases}\n & l_1(t) = t x_1 + a_1 \\\\\n & l_2(t) = t x_2 + a_2.\n \\end{cases}\n \\end{equation}\n\n \\thmitem{def:plane_line_equations/general} When regarding a line as an algebraic curve as in \\fullref{def:geometric_line/algebraic}, the equation\n \\begin{equation}\\label{def:plane_line_equations/general_equation}\n p(x, y) \\coloneqq Ax + By + C = 0\n \\end{equation}\n is called the \\term{general equation} or simply \\term{equation} of a line in a plane. Either \\( A \\) or \\( B \\) must be nonzero, so that \\( \\deg(p) = 1 \\).\n\n Note that multiple general equations can have the same locus (e.g. the entire polynomial ideal \\( \\braket{p} \\)).\n\n \\thmitem{def:plane_line_equations/normal} If \\( A^2 + B^2 = 1 \\), we call \\fullref{def:plane_line_equations/general_equation} a \\term{normal equation}. This leaves us with only two representatives of \\( \\braket{p} \\).\n\n \\thmitem{def:plane_line_equations/cartesian} Given \\( k, m \\in \\BbbR \\) and \\( k \\neq 0 \\), we define the \\term{Cartesian equation} of a line:\n \\begin{equation}\\label{def:plane_line_equations/cartesian_equation}\n y = kx + m.\n \\end{equation}\n\n We call \\( k \\) the \\term{slope} of the line.\n\n This is a special case of \\fullref{def:plane_line_equations/general} with \\( A = -k \\), \\( B = -1 \\) and \\( C = m \\). Unlike the general equation, the Cartesian equation of a line is unique.\n\n Conversely, if \\( B \\neq 0 \\) in \\fullref{def:plane_line_equations/general_equation}, we can define \\( k = -\\tfrac A B \\) and \\( m = -\\tfrac C B \\) to form a Cartesian equation.\n\n \\thmitem{def:plane_line_equations/intercept} Given nonzero \\( a, b \\in \\BbbR \\), we define the \\term{intercept equation} of a line:\n \\begin{equation}\\label{def:plane_line_equations/intercept_equation}\n \\frac x a + \\frac y b = 1,\n \\end{equation}\n\n This is a special case of \\fullref{def:plane_line_equations/general} with \\( A = \\frac 1 a \\), \\( B = \\frac 1 b \\) and \\( C = -1 \\). The intercept equation of a line is also unique.\n\n If \\( A, B, C \\neq 0 \\) in \\fullref{def:plane_line_equations/general}, we can define an \\term{intercept equation} as \\( a = -\\tfrac C A \\) and \\( b = -\\tfrac C B \\)).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{figure}\n \\begin{minipage}[b]{0.40\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 1.5cm;\n\n beginfig(1);\n path l, x_axis, y_axis;\n\n x_axis = (-1, 0) scaled u -- (1, 0) scaled u;\n y_axis = (0, -1) scaled u -- (0, 1) scaled u;\n l = (-1 / 2, -1) * u -- (1, 3 / 4) * u;\n\n drawarrow x_axis;\n label.bot(\"$x$\", point 0.9 of x_axis);\n drawarrow y_axis;\n label.lft(\"$y$\", point 0.9 of y_axis);\n draw l;\n label.bot(\"$y = kx + m$\", startpoint of l);\n endfig;\n \\end{mplibcode}\\fi\n \\caption{A \\hyperref[def:geometric_line]{line} in \\( \\BbbR^2 \\) defined using its \\hyperref[def:plane_line_equations/cartesian]{Cartesian equation}.}\\label{def:plane_line_equations/cartesian_equation_drawing}\n \\end{minipage}\n \\hspace{0.05\\textwidth}\n \\begin{minipage}[b]{0.40\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n u := 1.5cm;\n\n beginfig(1)\n drawarrow (-1 / 2, 0) scaled u -- (2, 0) scaled u;\n drawarrow (0, -1 / 2) scaled u -- (0, 2) scaled u;\n\n z0 = (1 / 2, 1 / 6) scaled u;\n z1 = (2, 11 / 12) scaled u;\n z2 = (1, 13 / 6) scaled u;\n\n draw z0 -- (x0, max(y1, y2)) dashed withdots;\n\n drawarrow z0 -- z1;\n draw (x0, y1) -- z1 dashed evenly;\n label.top(\"$x_1$\", midpoint of ((x0, y1) -- z1));\n\n drawarrow z0 -- z2;\n draw (x0, y2) -- z2 dashed evenly;\n label.bot(\"$x_2$\", midpoint of ((x0, y2) -- z2));\n endfig;\n \\end{mplibcode}\\fi\n \\caption{An \\hyperref[def:angle/acute]{acute angle} with its measurement segments dashed.}\\label{def:angle/figure}\n \\end{minipage}\n\\end{figure}\n\n\\begin{definition}\\label{def:angle}\n A \\term{directed angle} is a tuple of two closed \\hyperref[def:geometric_ray]{rays} with a common vertex. It is a closed cone. Given two rays \\( r_1, r_2 \\) with a common vertex, we denote their corresponding directed angle by \\( \\sphericalangle(r_1, r_2) \\).\n\n Suppose that \\( r_1 \\) and \\( r_2 \\) have scalar parametric equations\n \\begin{equation*}\n r_i: t \\mapsto\n \\begin{cases}\n tx_i + a_i \\\\\n ty_i + b_i,\n \\end{cases}\n i = 1, 2.\n \\end{equation*}\n\n We write\n\n The condition of the rays having a common vertex is equivalent to \\( a_1 = a_2 \\) and \\( b_1 = b_2 \\). If not specified otherwise, we assume that \\( a_1 = a_2 = b_1 = b_2 = 0 \\).\n\n The \\term{measure in radians} of a directed angle, often called the angle itself, is defined as the number (see \\fullref{def:geometric_trigonometric_functions})\n \\begin{equation*}\n \\alpha \\coloneqq \\rem(\\arctantwo(y_2, x_2) - \\arctantwo(y_1, x_1), 2\\pi).\n \\end{equation*}\n\n We can classify angles based on their measure as\n \\begin{thmenum}\n \\thmitem{def:angle/zero} \\term{zero} if \\( \\alpha = 0 \\),\n \\thmitem{def:angle/acute} \\term{acute} if \\( \\alpha \\in (0, \\tfrac \\pi 2) \\),\n \\thmitem{def:angle/right} \\term{right} if \\( \\alpha = \\tfrac \\pi 2 \\),\n \\thmitem{def:angle/obtuse} \\term{obtuse} if \\( \\alpha \\in (\\tfrac \\pi 2, \\pi) \\),\n \\thmitem{def:angle/straight} \\term{straight} if \\( \\alpha = \\pi \\), in which case the angle is actually a line,\n \\thmitem{def:angle/reflex} \\term{reflex} if \\( \\alpha > \\pi \\).\n \\end{thmenum}\n\n We often do not care about the order of the two rays and speak of an \\term{undirected angle}. In this case, the measure of the undirected angle is the smaller of the measures of the two oriented angles. Thus, we cannot speak of straight and reflex undirected angles.\n\\end{definition}\n\n\\begin{definition}\\label{def:triangle}\n \\begin{figure}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n beginfig(1)\n pair A, B, C;\n path alpha, beta, gamma;\n\n A := origin;\n B := (3, 0) scaled u;\n C := (2, 2) scaled u;\n\n draw A -- B -- C -- cycle;\n\n alpha = fullcircle scaled (u / 2) shifted A cutbefore (A -- B) cutafter (A -- C);\n draw alpha;\n label.urt(\"$\\alpha$\", point 0.4 of alpha);\n\n beta = fullcircle scaled (u / 2) shifted B cutbefore (B -- C) cutafter (B -- A);\n draw beta;\n label.ulft(\"$\\beta$\", point 1.4 of beta);\n\n gamma = fullcircle scaled (u / 4) shifted C cutbefore (A -- C) cutafter (B -- C);\n draw gamma;\n label.bot(\"$\\gamma$\", point 0.6 of gamma);\n\n fill dot shifted A;\n fill dot shifted B;\n fill dot shifted C;\n\n label.llft(\"$A$\", A);\n label.lrt(\"$B$\", B);\n label.top(\"$C$\", C);\n\n label.rt(\"$a$\", midpoint of (B -- C));\n label.ulft(\"$b$\", midpoint of (A -- C));\n label.bot(\"$c$\", midpoint of (A -- B));\n endfig;\n \\end{mplibcode}\\fi\n \\caption{An \\hyperref[def:triangle/acute]{acute triangle}.}\\label{def:triangle/figure}\n \\end{figure}\n\n A \\term{triangle} is a triple \\( (A, B, C) \\) of \\hyperref[def:point]{points}, no two of which are \\hyperref[def:collinear_complanar]{collinear} (see \\fullref{def:simplex/triangle} for a more general definition). The three points are called the \\term{vertices} of the triangle.\n\n Define the associated \\hyperref[def:convex_set/line_segment]{line segments}, called the \\term{sides} of the triangle, and its (undirected) \\hyperref[def:angle]{angles} as\n \\begin{balign*}\n a \\coloneqq [B, C], & & \\alpha \\coloneqq \\sphericalangle(b, c), \\\\\n b \\coloneqq [A, C], & & \\beta \\coloneqq \\sphericalangle(a, c), \\\\\n c \\coloneqq [A, B], & & \\gamma \\coloneqq \\sphericalangle(a, b).\n \\end{balign*}\n\n Note that we defined the angles using segments rather than rays, but this is immaterial because each to each segment \\( [p, q] \\) there corresponds exactly one closed ray \\( t \\mapsto p + t q \\).\n\n We can classify triangles based on their sides as\n \\begin{thmenum}\n \\thmitem{def:triangle/isosceles} \\term{isosceles} if at least two of its sides have equal length\n \\thmitem{def:triangle/equilateral} \\term{equilateral} if all of its sides have equal length\n \\end{thmenum}\n or based on their angles as\n \\begin{thmenum}\n \\thmitem{def:triangle/acute} \\term{acute} if all of its angles are \\hyperref[def:angle/acute]{acute}.\n \\thmitem{def:triangle/right} \\term{right} if at least one of the angles is \\hyperref[def:angle/straight]{straight}.\n \\thmitem{def:triangle/obtuse} \\term{obtuse} if at least one of its angles is \\hyperref[def:angle/obtuse]{obtuse}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:quadratic_plane_curve}\n The \\term{quadratic plane curves} are algebraic \\hyperref[def:hypersurface/algebraic]{curves} given by a bivariate polynomial of degree \\( 2 \\). The \\term{general equation} of a quadratic plane curve is\n \\begin{equation}\\label{def:quadratic_plane_curve/general_equation}\n c(x, y) \\coloneqq A x^2 + B xy + C y^2 + Dx + Ey + F = 0.\n \\end{equation}\n\n Multiple equation can correspond to the same curve. Not all general equations, however, define algebraic curves. We will not concern ourselves with the details. See \\fullref{ex:affine_varieties} for a proof that the unit circle is an algebraic curve. It turns out that the algebraic curves given \\fullref{def:quadratic_plane_curve/general_equation} are precisely the ones listed here, collectively known as \\term{conic sections}. We give only canonical forms of the equations; any linear transformation of the corresponding loci is described by another general equation.\n\n \\begin{figure}\n \\begin{minipage}{0.3\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 1.25cm;\n\n vardef scaled_sin(expr x) =\n 5 / 6 * sin(x)\n enddef;\n\n beginfig(1)\n fill dot shifted (u, 0);\n\n drawarrow (-pi / 2, 0) scaled u -- (pi / 2, 0) scaled u;\n drawarrow (0, -pi / 2) scaled u -- (0, pi / 2) scaled u;\n\n drawarrow path_of_curve(cos, scaled_sin, -1 / 4 * pi, 3 / 4 * pi, 0.01, u);\n drawarrow path_of_curve(cos, scaled_sin, 3 / 4 * pi, 7 / 4 * pi, 0.01, u);\n endfig;\n \\end{mplibcode}\\fi\n \\end{minipage}\n \\hspace{0.02\\textwidth}\n \\begin{minipage}{0.3\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 1.25cm;\n\n vardef minus_cosh(expr x) =\n -cosh(x)\n enddef;\n\n beginfig(1)\n drawarrow (-pi / 2, 0) scaled u -- (pi / 2, 0) scaled u;\n drawarrow (0, -pi / 2) scaled u -- (0, pi / 2) scaled u;\n\n drawarrow path_of_curve(cosh, sinh, -pi / 3, 0, 0.01, u);\n drawarrow path_of_curve(cosh, sinh, 0, pi / 3, 0.01, u);\n\n drawarrow path_of_curve(minus_cosh, sinh, -pi / 3, 0, 0.01, u);\n drawarrow path_of_curve(minus_cosh, sinh, 0, pi / 3, 0.01, u);\n endfig;\n \\end{mplibcode}\\fi\n \\end{minipage}\n \\hspace{0.02\\textwidth}\n \\begin{minipage}{0.3\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 1.25cm;\n\n beginfig(1)\n fill dot;\n\n drawarrow (-pi / 2, 0) scaled u -- (pi / 2, 0) scaled u;\n drawarrow (0, -pi / 2) scaled u -- (0, pi / 2) scaled u;\n\n vardef y_upper(expr x) =\n sqrt(x)\n enddef;\n\n vardef y_lower(expr x) =\n -sqrt(x)\n enddef;\n\n drawarrow path_of_plot(y_upper, 0, pi / 3, 0.01, u);\n drawarrow path_of_plot(y_lower, 0, pi / 3, 0.01, u);\n endfig;\n \\end{mplibcode}\\fi\n \\end{minipage}\n \\caption{An \\hyperref[def:quadratic_plane_curve/ellipse]{ellipse}, \\hyperref[def:quadratic_plane_curve/hyperbola]{hyperbola} and \\hyperref[def:quadratic_plane_curve/parabola]{parabola} defined via their parametric equations. The starting point is highlighted and the direction of the parametric curves is shown.}\\label{def:quadratic_plane_curve/figure}\n \\end{figure}\n\n \\begin{thmenum}\n \\thmitem{def:quadratic_plane_curve/ellipse} An \\term{ellipse} is a quadratic curve whose canonical equation has the form\n \\begin{equation}\\label{def:quadratic_plane_curve/ellipse/canonical_equation}\n c(x, y) \\coloneqq \\frac {x^2} {a^2} + \\frac {y^2} {b^2} - 1 = 0,\n \\end{equation}\n where \\( a, b > 0 \\).\n\n If \\( a = b \\), we say that the ellipse is a \\term{circle} and we call \\( a \\) the circle's \\term{radius}. The \\term{unit circle} is defined by \\( a = b = 1 \\). Circles generalize to \\hyperref[def:metric_space/sphere]{spheres} in metric spaces.\n\n \\Fullref{def:pi} and \\fullref{def:geometric_trigonometric_functions} logically belong here, but are extracted separately for brevity.\n\n We are often interested in defining ellipses via \\term{scalar parametric equations} using \\hyperref[def:trigonometric_functions]{trigonometric functions} as follows:\n \\begin{equation}\\label{def:quadratic_plane_curve/ellipse/parametric_equations}\n \\begin{cases}\n x = a \\cos(t) \\\\\n y = b \\sin(t),\n \\end{cases}\n \\end{equation}\n where \\( t \\in [0, 2\\pi) \\).\n\n We will now demonstrate that \\fullref{def:quadratic_plane_curve/ellipse/canonical_equation} and \\fullref{def:quadratic_plane_curve/ellipse/parametric_equations} describe the same curve. First, suppose that the pair \\( (x_0, y_0) \\) satisfies \\fullref{def:quadratic_plane_curve/ellipse/canonical_equation}. It follows from \\fullref{thm:arctantwo} that \\( t_0 \\coloneqq \\arctantwo\\left(\\tfrac {y_0} b, \\tfrac {x_0} a \\right) \\) is a, solution to the \\hyperref[def:quadratic_plane_curve/ellipse/parametric_equations]{parametric equations}. Conversely, if \\( x_0 = a \\cos(t_0) \\) and \\( y_0 = b \\sin(t_0) \\) for some \\( t_0 \\in [0, 2\\pi) \\), by \\fullref{thm:trigonometric_identities/pythagorean_identity} it follows that the pair \\( (x_0, y_0) \\) is a root of \\fullref{def:quadratic_plane_curve/ellipse/canonical_equation} and, by \\fullref{thm:arctantwo}, \\( t_0 \\) can be restored given \\( \\cos(t_0) \\) and \\( \\sin(t_0) \\).\n\n Therefore, every point of the parametric equation \\fullref{def:quadratic_plane_curve/ellipse/parametric_equations} corresponds uniquely to a, solution of the canonical equation \\fullref{def:quadratic_plane_curve/ellipse/canonical_equation} and vice versa, which makes the two approaches to defining ellipses equivalent.\n\n \\thmitem{def:quadratic_plane_curve/hyperbola} A \\term{hyperbola} is a quadratic curve whose canonical equation has the form\n \\begin{equation}\\label{def:quadratic_plane_curve/hyperbola/canonical_equation}\n c(x, y) \\coloneqq \\frac {x^2} {a^2} - \\frac {y^2} {b^2} - 1 = 0,\n \\end{equation}\n where \\( a, b > 0 \\).\n\n Similarly to ellipses, we are can define hyperbolas via \\term{scalar parametric equations} using \\hyperref[def:hyperbolic_trigonometric_functions]{hyperbolic trigonometric functions} as follows:\n \\begin{equation}\\label{def:quadratic_plane_curve/hyperbola/parametric_equations}\n \\begin{cases}\n x = a \\cosh(t) \\\\\n y = b \\sinh(t),\n \\end{cases}\n \\end{equation}\n where \\( t \\in \\BbbR \\). This only defines the \\term{right part} of the hyperbola. The left part is defined by replacing \\( a \\) with \\( -a \\).\n\n \\thmitem{def:quadratic_plane_curve/parabola} A \\term{parabola} is a quadratic curve whose canonical equation has the form\n \\begin{equation}\\label{def:quadratic_plane_curve/parabola/canonical_equation}\n c(x, y) \\coloneqq y^2 - 2px = 0,\n \\end{equation}\n where \\( p \\neq 0 \\).\n\n Unlike ellipses and hyperbolas, we do not define parametric equations. Instead, we define \\( y \\) as a function of \\( x \\) separately for the lower half-plane and upper half-plane:\n \\begin{equation}\\label{def:quadratic_plane_curve/parabola/cartesian_equation}\n y(x) = \\pm \\sqrt{2px}.\n \\end{equation}\n \\end{thmenum}\n\n Ellipses, hyperbolas and parabolas are collectively called \\term{conic sections}.\n\\end{definition}\n\n\\begin{definition}\\label{def:pi}\n \\begin{figure}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n beginfig(1)\n drawarrow (-pi / 2, 0) scaled u -- (pi / 2, 0) scaled u;\n drawarrow (0, -1 / 2) scaled u -- (0, pi / 2) scaled u;\n\n vardef y(expr x) =\n sqrt(1 - x ** 2)\n enddef;\n\n drawarrow path_of_plot(y, -1, 1, 0.01, u);\n endfig;\n \\end{mplibcode}\\fi\n \\caption{\\( \\gph(y^+) \\) as a parametric curve in \\fullref{def:pi}.}\\label{fig:def:pi/upper_half_circle}\n \\end{figure}\n\n The definition of a circle of unit radius as the zero-locus of the polynomial \\( x^2 + y^2 - 1 \\) allows us to, solve a chicken-and-egg problem regarding the definitions of the number \\( \\pi \\). It is conventional to define it as the ratio of a circle's circumference to its diameter. For a unit circle, this diameter is \\( 2 \\). It will be simpler for us, however, to define \\( \\pi \\) as the radius of a half-circle's circumference since we can represent \\( y \\) as a function of \\( x \\) in the upper \\hyperref[def:half_space]{half-plane} (see \\ref{fig:def:pi/upper_half_circle}). Define the parametric curve\n \\begin{balign*}\n & y^+: [-1, 1] \\to [0, 1] \\\\\n & y^+(x) \\coloneqq \\sqrt{1 - x^2}.\n \\end{balign*}\n\n We use \\fullref{thm:length_of_function_graph} to find the length of the graph \\( \\gph(y^+(x)) \\). The derivative of \\( y^+(x) \\) is\n \\begin{equation*}\n D_x[y^+(x)] = \\frac{-2x}{2 \\sqrt{1 - x^2}} = - \\frac x {\\sqrt{1 - x^2}} dx.\n \\end{equation*}\n\n The length of the curve \\( \\gph(y^+) \\) is thus\n \\begin{equation*}\n \\len(\\gph(y^+)) = \\int_{-1}^1 \\sqrt{1 + \\frac{x^2}{1 - x^2}} dx = \\int_{-1}^1 \\frac 1 {\\sqrt{1 - x^2}} dx.\n \\end{equation*}\n\n This justifies the definition\n \\begin{equation}\\label{def:pi/weierstrass_integral}\n \\pi \\coloneqq \\int_{-1}^1 \\frac 1 {\\sqrt{1 - x^2}} dx.\n \\end{equation}\n\n See \\fullref{thm:trigonometric_function_basic_roots} for a proof of how this relates to the trigonometric functions and \\fullref{thm:def:exponential_function/properties/eulers_identity} as a consequence.\n\\end{definition}\n\n\\begin{definition}\\label{def:geometric_trigonometric_functions}\n After defining the \\hyperref[def:trigonometric_functions]{trigonometric functions} \\( \\cos(z) \\) and \\( \\sin(z) \\) analytically via power series, we will define their geometric counterparts \\( \\cos_G(z) \\) and \\( \\sin_G(z) \\) and show the connection between them. The actual geometric definition relies on formalisms that are far beyond our interest (see the notes in \\fullref{def:euclidean_plane}).\n\n Fix a point \\( (x_0, y_0) \\) on the unit circle (that is, \\( x_0^2 + y_0^2 = 1 \\)) and define the points\n \\begin{equation}\\label{def:geometric_trigonometric_functions/vertices}\n \\begin{array}{l}\n A \\coloneqq (x_0, y_0), \\\\\n B \\coloneqq (0, 0), \\\\\n C \\coloneqq (x_0, 0).\n \\end{array}\n \\end{equation}\n\n Consider the \\hyperref[def:triangle]{triangle} formed by these vertices. \\Cref{fig:def:geometric_trigonometric_functions/triangle} illustrates the situation.\n \\begin{figure}\n \\begin{minipage}[b]{0.4\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 3.5cm;\n\n beginfig(1)\n pair A, B, C;\n path alpha, beta;\n\n t := 1;\n A := origin;\n B := (cos(t), sin(t)) scaled u;\n C := (cos(t), 0) scaled u;\n\n draw A -- B -- C -- cycle;\n\n alpha = fullcircle scaled (u / 4) shifted A cutbefore (A -- C) cutafter (A -- B);\n draw alpha;\n label.urt(\"$\\alpha$\", midpoint of alpha);\n label.lft(\"$\\begin{rcases} \\sin_G(\\alpha) = \\tfrac {\\len b} {\\len c} \\\\ \\cos_G(\\alpha) = \\tfrac {\\len a} {\\len c} \\end{rcases}$\", A);\n\n beta = fullcircle scaled (u / 4) shifted B cutbefore (B -- A) cutafter (B -- C);\n draw beta;\n label.llft(\"$\\beta$\", point 0.7 of beta);\n label.lft(\"$\\begin{rcases} \\sin_G(\\beta) = \\tfrac {\\len a} {\\len c} \\\\ \\cos_G(\\beta) = \\tfrac {\\len b} {\\len c} \\end{rcases}$\", B - (0.05, 0) * u);\n\n fill dot shifted A;\n fill dot shifted B;\n fill dot shifted C;\n\n label.bot(\"$A$\", A);\n label.top(\"$B$\", B);\n label.lrt(\"$C$\", C);\n\n label.rt(\"$a$\", midpoint of (B -- C));\n label.bot(\"$b$\", midpoint of (A -- C));\n label.ulft(\"$c$\", midpoint of (A -- B));\n endfig;\n \\end{mplibcode}\\fi\n \\end{minipage}\n \\hspace{0.05\\textwidth}\n \\begin{minipage}[b]{0.4\\textwidth}\n \\centering\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n input metapost/plotting;\n\n u := 3.5cm;\n\n beginfig(1)\n pair A, B, C;\n path alpha, beta;\n\n t := 1;\n A := origin;\n B := (cos(t), sin(t)) scaled u;\n C := (cos(t), 0) scaled u;\n\n drawarrow (-sin(pi/16), 0) scaled u -- (5/4, 0) scaled u;\n drawarrow (0, -sin(pi/16)) scaled u -- (0, 5/4) scaled u;\n\n draw path_of_curve(cos, sin, -pi/16, 9/16 * pi, 0.01, u);\n draw A -- B -- C -- cycle;\n\n alpha = fullcircle scaled (u / 4) shifted A cutbefore (A -- C) cutafter (A -- B);\n draw alpha;\n label.urt(\"$\\alpha$\", midpoint of alpha);\n\n beta = fullcircle scaled (u / 4) shifted B cutbefore (B -- A) cutafter (B -- C);\n draw beta;\n label.llft(\"$\\beta$\", point 0.7 of beta);\n\n fill dot shifted A;\n fill dot shifted B;\n fill dot shifted C;\n\n label.llft(\"$(0, 0)$\", A);\n label.urt(\"$(x_0, y_0)$\", B);\n label.lrt(\"$(x_0, 0)$\", C);\n\n label.rt(\"$a$\", midpoint of (B -- C));\n label.bot(\"$b$\", midpoint of (A -- C));\n label.ulft(\"$c$\", midpoint of (A -- B));\n endfig;\n \\end{mplibcode}\\fi\n \\end{minipage}\n \\caption{An \\enquote{abstract} right triangle in the \\hyperref[def:euclidean_plane]{Euclidean plane} with legends for geometric sines and cosines and the same triangle in \\( \\BbbR^2 \\) connecting the origin to a point \\( (x_0, y_0) \\) on the unit circle.}\\label{fig:def:geometric_trigonometric_functions/triangle}\n \\end{figure}\n\n The original \\enquote{geometric definition} of \\( \\sin_G \\) and \\( \\cos_G \\) regards them as functions of an angle rather than numeric functions. \\( \\sin_G \\) and \\( \\cos_G \\) are only defined for two of the angles in a right triangle. The geometric definition is\n \\begin{balign*}\n \\sin_G(\\alpha) \\coloneqq \\frac{\\len(b)} {\\len(c)}, & & \\cos_G(\\alpha) \\coloneqq \\frac{\\len(a)} {\\len(c)},\n \\\\\n \\sin_G(\\beta) \\coloneqq \\frac{\\len(b)} {\\len(c)}, & & \\cos_G(\\beta) \\coloneqq \\frac{\\len(a)} {\\len(c)}.\n \\end{balign*}\n\n In our case, \\( \\len(a) = y_0 \\), \\( \\len(b) = x_0 \\) and \\( \\len(c) = 1 \\). Furthermore, \\( \\sin_G(\\beta) \\) nor \\( \\cos_G(\n \\beta) \\) are immaterial to our subsequent arguments and we only introduced them for the sake of having a full definition.\n\n Therefore, we conclude that\n \\begin{balign*}\n \\sin_G(\\alpha) = x_0,\n & &\n \\cos_G(\\alpha) = y_0.\n \\end{balign*}\n\n To see that \\( \\sin_G \\) and \\( \\cos_G \\) are somewhat analogous to \\( \\sin \\) and \\( \\cos \\), notice that by \\fullref{thm:arctantwo}, there exists a unique \\( t_0 \\coloneqq \\arctantwo(y_0, x_0) \\) such that\n \\begin{balign*}\n \\sin(t_0) = x_0,\n & &\n \\cos(t_0) = y_0.\n \\end{balign*}\n\n Therefore, our \\hyperref[def:trigonometric_functions]{analytic definition} of the trigonometric functions as numeric functions correspond to the classical geometric definition in the special case where we consider the angle near the origin in the triangle formed by the vertices \\fullref{def:geometric_trigonometric_functions/vertices}. This motivates \\enquote{measuring} angles using the obtained correspondence. This unit of measurement is called a \\term{radian}. We say that the angle \\( \\alpha \\) is \\( t_0 \\) \\term{radians}. Outside of mathematics, it is more conventional to use \\term{degrees}, which are obtained from radians by scaling with \\( \\tfrac {180} {\\pi} \\). That is, \\( \\alpha \\) is \\( \\tfrac {180} {\\pi} t_0 \\) degrees.\n\\end{definition}\n", "meta": {"hexsha": "264b8bd611aa9441c489dd357529cca0d134b232", "size": 26263, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/analytic_geometry_in_the_plane.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/analytic_geometry_in_the_plane.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/analytic_geometry_in_the_plane.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.5528301887, "max_line_length": 924, "alphanum_fraction": 0.6539999238, "num_tokens": 8393, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.819893340314393, "lm_q2_score": 0.7905303186696747, "lm_q1q2_score": 0.6481505435938811}} {"text": "\\mychapter{2}{Lesson 2} %180928\n\n\\section{Authentic communication}\n\nThe most common scenario exposing the problem of authentication is depicted in figure \\ref{fig:authentication}. This time, the parties Alice and Bob want to ensure that they are effectively communicating to each other; in other words, nobody else is \\emph{impersonating} either party. The objects used here are:\n\\begin{itemize}\n \\item The data to be shared, or \\emph{message} $m$;\n \\item Some additional secret information, shared by the two parties, that is used to \\emph{sign} the message: the \\emph{authentication key} or just \\emph{key} $k$;\n \\item The result of signing a message $m$ using the key $k$: the \\emph{signature} or \\emph{tag} $t$.\n\\end{itemize}\n\n\\begin{figure}[ht]\n \\centering\n \\begin{tikzpicture}\n \\draw\n (0, 0) node (a) [box, fill = white] {Alice} \n (5, 0) node (b) [box, fill = white] {Bob}\n (2.5, -1) node (e) [box, fill = white] {Eve}\n ;\n\n \\draw[-Stealth] (a) -- node [midway, above] {$(m, t)$} (b);\n \\draw[-Stealth] (0, -1) node [below] {$k$} -- (a);\n \\draw[-Stealth] (5, -1) node [below] {$k$} -- (b);\n \\draw[-Stealth] (-1, 0) node [left] {$m$} -- (a);\n \\draw[-Stealth] (b) -- (6, 0) node [right] {$1$};\n \\draw[-Stealth] (2, -0.1) -- (3, -0.1) (2.5, -0.1) -- (e);\n\n \\end{tikzpicture}\n\n \\caption{A depiction of the problem of authentic communication}\n \\label{fig:authentication}\n\\end{figure}\n\nThe mechanism employed by both parties to enforce authentication is called a \\emph{cryptographic authentication scheme}, or just \\emph{authentication scheme}, typically taking the form $\\Phi = (\\Tag, \\Ver)$, where:\n\\begin{itemize}\n \\item $\\Tag \\in \\K \\times \\M \\to \\T$ is the machine that, given a message $m$ in $\\M$ and a key $k$ in $\\K$ generates the signature $t$\n \\item $\\Ver \\in \\K \\times \\M \\times \\T \\to \\binary$ is the machine that decides whether $t$ is the correct signature for the message $m$ using the key $k$.\n\\end{itemize}\n\nIt is worth observing that, under the assumption that such machines still behave as mathematical functions, a straightforward verifier with inputs $(k, m, t)$ consists in:\n\\begin{enumerate}\n \\item Invoking the tagging function: $u = \\Tag(k, m)$;\n \\item Comparing $u$ and $t$ for equality.\n\\end{enumerate}\n\nGiven such definitions, an authentication scheme works as intended if and only if:\n\\[\n \\forall m \\in \\M, \\forall k \\in \\K \\implies \\Ver(k, m, \\Tag(k, m)) = 1\n\\]\n\nA security problem then arises when someone, having a signed message $(m_1, t_1)$ where the key $k$ used for tagging is unknown, is able to efficiently sign a different message $(m_2, t_2)$ such that verification under the same key yields a success. This acion is called a \\emph{forgery}; and by looking at the original setting in figure \\ref{fig:authentication}, if Eve is effectively able to forge vaild signatures, she can indeed impersonate either Alice or Bob at will. The desired property of an authentication scheme thus becomes to be resistant, if not immune, to such attacks; in one word, the scheme is \\emph{unforgeable}.\n\n\\begin{definition}[$\\varepsilon$-statistical one-time unforgeability]\n let $\\Phi$ be an authentication scheme, and let $m_1$ and $m_2$ be two distinct messages, and $t_1 = \\Tag(K, m_1)$ be the signature of $m_1$ under $\\Phi$, with the key $K$ picked uniformly at random. $\\Phi$ is deemed \\emph{$\\varepsilon$-statistical one-time unforgeable} iff knowing $m_1$ and $t_1$ does not give any advantage in finding a signature $t_2$ that is actually the signature of $m_2$ under the same scheme and the same key of $m_1$, without knowing such key:\n \\[\n \\forall m_1 \\neq m_2 \\in \\M \\, \\forall t_1, t_2 \\in \\T \\quad \\Pr[\\Tag(K, m_2) = t_2 \\knowing \\Tag(K, m_1) = t_1] \\leq \\varepsilon\n \\]\n\\end{definition}\n\n\\subsection{An introduction to hashing}\n\nA great deal of authentication has been, and is still done by means of \\emph{hashing}, which consists of feeding the message to a special machine that produces a scrambled, unique signature for it; such machines are then known as \\emph{hash functions}.\n\nFor starters recall that, given a set $A \\to B$ that collects all possible functions from $A$ to $B$, a \\emph{function family} is a subset of such class that share some specific properties. Having said that:\n\n\\begin{definition}\n A family of \\emph{hash functions} $H$ is defined as a function family that is mapped 1-to-1 by an indexing set $\\mathcal{S}$, where the indices are called \\emph{seeds}:\\footnote{This kind of notation consisting in putting an argument as a subscript to a generic, typically of higher-order function, is also called ``currying''.}\n \\[\n H \\in \\mathcal{S} \\to (\\M \\to \\T) : s \\mapsto h_s\n \\]\n Furthermore, given a uniformly random seed $S$, the family as a whole distributes the tags uniformly:\n \\[\n \\forall m \\in \\M \\, \\forall t \\in \\T \\qquad \\Pr[h_S(m) = t] = \\oneover{\\T}\n \\]\n\\end{definition}\n\nHaving formalized what a hash function family is, the notion of unforgeability can be modeled by the property of \\emph{pairwise-independency}:\n\n\\begin{definition}\n Let $H$ be a family of hash functions, and $S$ be a uniformly random seed; the hash functions are deemed \\emph{pairwise-independent} iff, for any two distinct messages $m_1$ and $m_2$, the pair $(h_S(m_1), h_S(m_2))$ distributes uniformly in $\\T^2$:\n \\[\n \\forall m_1 \\neq m_2 \\in \\M \\, \\forall t_1, t_2 \\in \\T \\qquad \\Pr[h_S(m_1) = t_1 \\wedge h_S(m_2) = t_2] = \\oneover{|\\T|^2}\n \\]\n\\end{definition}\n\nAs an example of such a family, consider the additive group of integers modulo $p$: $(\\integer_p, +)$, where $p$ is a prime integer. Let the seed space $\\mathcal{S}$ be the set of pairs $\\integer_p^2$, and define the function family:\n\\[\n H \\in \\integer_p^2 \\to (\\integer_p \\to \\integer_p) : (a, b) \\mapsto (x \\mapsto \\textrm{\\textup{mod}}_p(ax + b))\n\\]\n\n\\begin{proposition}\n The functions in the family $H$ are pairwise-independent.\n\\end{proposition}\n\n\\begin{proof}\n Let $S = (A, B)$ be a uniformly random seed for $H$. For any distinct messages $m_1$ and $m_2$, and for any tags $t_1$ and $t_2$:\n \\begin{align*}\n &\\ \\Pr[h_S(m_1) = t_1 \\wedge h_S(m_2) = t_2] & \\\\\n =&\\ \\Pr[A m_1 + B = t_1 \\wedge A m_2 + B = t_2] & \\text{($H$ definition)}\\\\\n =&\\ \\Pr\\left[ \\begin{pmatrix} m_1 & 1 \\\\ m_2 & 1 \\end{pmatrix} \\cdot \\begin{pmatrix} A \\\\ B \\end{pmatrix} = \\begin{pmatrix} t_1 \\\\ t_2 \\end{pmatrix} \\right] & \\text{(Matrix form)} \\\\\n =&\\ \\Pr\\left[ \\begin{pmatrix} A \\\\ B \\end{pmatrix} = \\begin{pmatrix} m_1 & 1 \\\\ m_2 & 1 \\end{pmatrix}^{-1} \\cdot \\begin{pmatrix} t_1 \\\\ t_2 \\end{pmatrix} \\right] & \\\\\n =&\\ \\oneover{|\\integer_p^2|} & \\text{($(A, B)$ is uniform)}\n \\end{align*}\n which satisfies the definition of pairwise-independency.\n\\end{proof}\n\n% TODO: Consider making an example in Zp, if possible\nIt is very important to avoid confusing \\emph{pairwise} independency with \\emph{mutual} independency: while the former enforces independency only on pairs, the latter extends it to all possible subsets; the two notions are not necessarily equivalent.\n\nThe property of pairwise-independency of hash function families can be directly exploited to provide an unforgeable authentication scheme:\n\n\\begin{theorem}\n Define an authentication scheme $\\Phi = (\\Tag, \\Ver)$ to be such that its tagging machine is a hash function family $H$:\n \\[\n \\Tag(k, m) = h_k(m)\n \\]\n and let it be pairwise-independent. Then $\\Phi$ is $\\oneover{|\\T|}$-statistical one-time unforgeable. \n\\end{theorem}\n\n\\begin{proof}\n Let $K$ be a uniformly random key. For any distinct messages $m_1$ and $m_2$, and for any tags $t_1$ and $t_2$:\n \\begin{align*}\n &\\ \\Pr[\\Tag(K, m_2) = t_2 \\knowing \\Tag(K, m_1) = t_1] & \\\\\n =&\\ \\Pr[h_K(m_2) = t_2 \\knowing h_K(m_1) = t_1] & \\text{($\\Tag$ definition)} \\\\\n =&\\ \\frac{\\Pr[h_K(m_2) = t_2 \\wedge h_K(m_1) = t_1]}{\\Pr[h_K(m_1) = t_1]} & \\text{(Conditional prob. def.)} \\\\\n =&\\ |\\T| \\cdot \\Pr[h_K(m_2) = t_2 \\wedge h_K(m_1) = t_1] & \\text{($H$ is a hash function)} \\\\ \n =&\\ \\frac{|\\T|}{|\\T|^2} = \\oneover{|\\T|} & \\text{($H$ is pairwise-independent)}\n \\end{align*}\n\n which satisfies the definition of $\\oneover{|\\T|}$-statistical one-time unforgeability.\n\\end{proof}\n\n\\todo{Need to link statistical unforgeability with key size beforehand}\n\n\\begin{theorem}\n For any positive $\\lambda$, a $(2^{-\\lambda})$-statistical $t$-time unforgeable authentication scheme has a key of size $(t + 1) \\lambda$.\n\\end{theorem}\n\n\\begin{proof}\n Idea: For each ``time'', use one ``subkey''\n\\end{proof}\n", "meta": {"hexsha": "5f1420656e05478f30c6a5d81c6c309de5af744a", "size": 8896, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lessons/lesson_2.tex", "max_stars_repo_name": "Project2100/Cryptography-2018_19", "max_stars_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-10-15T09:22:45.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-15T09:22:45.000Z", "max_issues_repo_path": "lessons/lesson_2.tex", "max_issues_repo_name": "Project2100/cryptography_1819", "max_issues_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-07-18T15:45:10.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-27T20:36:12.000Z", "max_forks_repo_path": "lessons/lesson_2.tex", "max_forks_repo_name": "Project2100/cryptography_1819", "max_forks_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-07-17T14:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-03T15:23:22.000Z", "avg_line_length": 62.6478873239, "max_line_length": 631, "alphanum_fraction": 0.6541142086, "num_tokens": 2748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933271118221, "lm_q2_score": 0.7905303236047048, "lm_q1q2_score": 0.6481505372030467}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage{latexsym}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{graphicx}\n\\usepackage{wrapfig}\n\\usepackage{pseudocode}\n\\usepackage{url}\n\\usepackage[backref, colorlinks=true, citecolor=red, urlcolor=blue, pdfauthor={Jyh-Ming Lien}]{hyperref}\n\n\n\\newcommand{\\handout}[5]{\n \\noindent\n \\begin{center}\n \\framebox{\n \\vbox{\n \\hbox to 5.78in { {\\bf Advanced Algorithms} \\hfill #2 }\n \\vspace{4mm}\n \\hbox to 5.78in { {\\Large \\hfill #5 \\hfill} }\n \\vspace{2mm}\n \\hbox to 5.78in { {\\em #3 \\hfill #4} }\n }\n }\n \\end{center}\n \\vspace*{4mm}\n}\n\n\\newcommand{\\lecture}[4]{\\handout{#1}{#2}{#3}{}{#1}}\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{corollary}[theorem]{Corollary}\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{observation}[theorem]{Observation}\n\\newtheorem{proposition}[theorem]{Proposition}\n\\newtheorem{definition}[theorem]{Definition}\n\\newtheorem{claim}[theorem]{Claim}\n\\newtheorem{fact}[theorem]{Fact}\n\\newtheorem{assumption}[theorem]{Assumption}\n\n% 1-inch margins, from fullpage.sty by H.Partl, Version 2, Dec. 15, 1988.\n\\topmargin 0pt\n\\advance \\topmargin by -\\headheight\n\\advance \\topmargin by -\\headsep\n\\textheight 8.9in\n\\oddsidemargin 0pt\n\\evensidemargin \\oddsidemargin\n\\marginparwidth 0.5in\n\\textwidth 6.5in\n\n\\parindent 0in\n\\parskip 1.5ex\n%\\renewcommand{\\baselinestretch}{1.25}\n\n\\begin{document}\n\n\\lecture{Voronoi Stippling}{Fall 2015}{Prof.\\ Jyh-Ming Lien}{---}\n\n\nThe goal of this assignment is to deepen your understanding on various implementations of 2-d Voronoi Diagram. \nIn this exam, you are given two implementations of 2-d Voronoi Diagram: the Fortune's algorithm and \nan algorithm using well-known image-based wave propagation method. You task is to (1) understand the implementations, (2) compare the results from these two implementations, and (3) finally improve the image-based method. \n\n\\textbf{What to submit}: You need to turn in a report in \\LaTeX\\ (see the template in report folder). Your report should include three sections: a summary of what the code does (for both implementations), \nyour discovery of the differences in output, and your improvement. In the second and third sections, you should include\nall the example outputs (visual and/or statistical results). In the last section you should report known bugs, and known limitations.\n\n\\textbf{How to submit}:\nYou should use \\href{https://github.com/}{github} to maintain your code and the report. Email me your github clone command to my email address \\href{mailto:jmlien@cs.gmu.edu}{\\it jmlien@cs.gmu.edu}\nbefore the deadline.\n\n\\textbf{Due: Nov 26, 2017. At 11:59 pm.}\n\\section{Part 1: Understand the implementations (30 pts)}\n\n\n\\subsection{What should you do?}\n\nYour goal is to get a full grasp of what the code does in both implementations: hedcuter and voronoi. \nYour summary should provide algorithms for computing Voronoi diagram, Centroidal Voronoi tessellation (CVT), \nand stippling methods from both implementations. \n%Your summary should provide a discussion on what parameters are used to control their outputs. \n\n{\\sc Hint 1}: Both methods are based on the paper by Secord, Adrian. ``Weighted voronoi stippling.\" Proceedings of the 2nd international symposium on Non-photorealistic animation and rendering. ACM, 2002. It is highly recommend that you read the paper first.\n\n{\\sc Hint 2}: To compile hedcuter code, please use the solution file in folder {\\em hedcuter/code/vc\\_files}.\n This code requires OpenCV. By default, it requires 64 bits installation of OpenCV on Windows. \nThe solution file also uses two environment variables called ``OPENCV\\_INCLUDEDIR'' and ``OPENCVX64\\_LIBRARYDIR'' the point to \nthe include and library folders on the system. Therefore, make sure that you have those variable defined before you compile. You can consult OpenCV documents\n(\\href{docs.opencv.org}{\\it docs.opencv.org}) if you encounter problems. \n\n{\\sc Hint 3}: To compile the second code, you will need boost (\\href{www.boost.org}{\\it www.boost.org}) and define two boost related environment variables:\nBOOST\\_INCLUDEDIR and BOOST\\_LIBRARYDIR before you compile. Each of these variables should have the full paths to the include and library folders on the system.\nThis implementation is obtained from \\href{http://www.saliences.com/projects/npr/stippling}{\\it http://www.saliences.com/projects/npr/stippling}.\nSome details of the code can be found there.\n\n\n\n\\section{Part 2: Compare the Outputs (30 pts)}\n\n\n\n\n\n\n\\subsection{What should you do?}\n\nYour goal is to compare the output of these two implementations. \nUse the images in folder hedcuter/images or use images of your own, please show the differences visually (e.g. circle the areas of difference)\nand also in writing. \n%You can find more of these images from\\href{http://philip.greenspun.com/stock/}{\\it http://philip.greenspun.com/stock/}.\nYou should also provide some discussion of where these differences are from. Here are some questions for you to consider.\nYou are free to explore beyond these questions. \n\n\\begin{enumerate}\n\\item Do you get the same results by running the same program on the same image multiple times? \n\\item If you vary the number of the disks in the output images, do these implementations produce the same distribution in the final image? If not, why?\n\\item If you vary the number of the disks in the output images, is a method faster than the other? \n\\item Does the size (number of pixels), image brightness or contrast of image increase or decrease their difference? \n\\item Does the type of image (human vs. machine, natural vs. urban landscapes, photo vs. painting, etc) increase or decrease their difference? \n\\item Are the outputs of these stippling methods different the hedcut images created by artists (e.g. those from the \\href{http://www.wsj.com/articles/SB10001424052748704207504575129961786135180}{Wall Street Journal})? \n\\end{enumerate}\n\n\n\\section{Part 3: Improve ``hedcuter'' code (40 pts)}\n\n\n\\subsection{What should you do?}\n\nProvide at least two improvements (each will worth 20 points) to the hedcuter code. Below are some possible improvements that \nyou can do. In your report, you should show the improvement either visually or/and statistically for timing/performance results\nusing images of your own, i.e. do not use the images in hedcuter/images. \nExtra bonus of 20 points will be given for an additional improvement.\n\\begin{enumerate}\n\n\\item Improve the distribution of the disks to avoid unnatural clustering of the disks. One idea is to use higher\nimage resolution (using subpixels) for computing the centroids of Voronoi cells. \n\n\\item Improve the computation efficiency. One way of doing this is via GPU. You can try the method by Hoff III, Kenneth E., et al. ``Fast computation of generalized Voronoi diagrams using graphics hardware.\" Proceedings of the 26th annual conference on Computer graphics and interactive techniques, 1999. The implementation should be pretty simple if you know OpenGL.\n\n\\item Add functionality to generate colorful disks. For example, you can implement functions that are not available in hedcuter code but provided in the voronoi code.\n\\end{enumerate}\n\n\\bibliographystyle{plain}\n\\bibliography{shape-assignment}\n\n\\end{document}\n\n\n", "meta": {"hexsha": "3c5399bcac693223ab356e6cf785b97fb637deb4", "size": 7268, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assignment/assignment.tex", "max_stars_repo_name": "StephenPArnold/CS633-hw3", "max_stars_repo_head_hexsha": "23771990a8fcfb71e629af66cb2f172e6dbda3a7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assignment/assignment.tex", "max_issues_repo_name": "StephenPArnold/CS633-hw3", "max_issues_repo_head_hexsha": "23771990a8fcfb71e629af66cb2f172e6dbda3a7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assignment/assignment.tex", "max_forks_repo_name": "StephenPArnold/CS633-hw3", "max_forks_repo_head_hexsha": "23771990a8fcfb71e629af66cb2f172e6dbda3a7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.4533333333, "max_line_length": 367, "alphanum_fraction": 0.7707760044, "num_tokens": 1890, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355187, "lm_q2_score": 0.8397339716830605, "lm_q1q2_score": 0.6481432147723726}} {"text": "\\section*{Exercises}\n\n\\begin{ex}\n Consider the vectors $\\vect{u}$ and $\\vect{v}$ drawn below.\n \\begin{center}\n \\begin{tikzpicture}[scale=2]\n \\draw[->, thick, blue] (0,0)--(2,1);\n \\draw[->, thick, red] (4,1)--(5,0.5);\n \\node[below] at (1,1.25){$\\vect{u}$};\n \\node[above right] at (4.5, 0.75){$\\vect{v}$};\n \\end{tikzpicture}\n \\end{center}\n Draw $-\\vect{u}$, $2\\vect{v}$, and $-\\frac{1}{2}\\vect{v}$.\n\n \\begin{sol}\n ~\n \\begin{center}\n \\begin{tikzpicture}[scale=2]\n \\draw[->, thick, blue] (0,0)--(2,1);\n \\draw[->, thick, blue] (3,1)--(1,0);\n \\draw[->, thick, red] (5,1)--(6,0.5);\n \\draw[->, thick, red] (6.5,0.75)--(6,1);\n \\draw[->, thick, red] (4,1)--(6,0);\n \\node[above] at (1,1){$\\vect{u}$};\n \\node[below] at (2,0.5){$-\\vect{u}$};\n \\node[below] at (5.5, 0.75){$\\vect{v}$};\n \\node[above right] at (6.25, 0.75){$-\\frac{1}{2}\\vect{v}$};\n \\node[below] at (5, 0.4){$2\\vect{v}$};\n \\end{tikzpicture}\n \\end{center}\n \\end{sol}\n\\end{ex}\n\n\\begin{ex}\n Find $-3\\begin{mymatrix}{r}\n 5 \\\\\n -1 \\\\\n 2 \\\\\n -3\n \\end{mymatrix}\n + 5\\begin{mymatrix}{r}\n -8 \\\\\n 2 \\\\\n -3 \\\\\n 6\n \\end{mymatrix}$.\n \\begin{sol}\n $\\begin{mymatrix}{r}\n -55 \\\\\n 13 \\\\\n -21 \\\\\n 39\n \\end{mymatrix}$.\n \\end{sol}\n\\end{ex}\n\n\\begin{ex}\n Use the properties of scalar multiplication from\n Proposition~\\ref{prop:vector-scalar-multiplication} and the\n properties of vector addition from\n Proposition~\\ref{prop:properties-vector-addition} to prove the\n following equalities. Justify every step.\n \\begin{enumerate}\n \\item $(k+\\ell)(\\vect{u}+\\vect{v}) = k\\vect{u} + k\\vect{v} +\n \\ell\\vect{u} + \\ell\\vect{v}$.\n \\item $0\\vect{u} = \\vect{0}$.\n \\item $(-1)\\vect{u} = -\\vect{u}$.\n \\item $-(k\\vect{u}) = k(-\\vect{u}) = (-k)\\vect{u}$.\n \\end{enumerate}\n \\begin{sol}\n \\begin{enumerate}\n \\item\n $(k+\\ell)(\\vect{u}+\\vect{v}) = (k+\\ell)\\vect{u} +\n (k+\\ell)\\vect{v} = k\\vect{u} + k\\vect{v} + \\ell\\vect{u} +\n \\ell\\vect{v}$. Here we used the distributive law over vector\n addition in the first step, and the distributive law over scalar\n addition in the second step.\n \\item We have $0\\vect{u} = (0+0)\\vect{u} = 0\\vect{u} + 0\\vect{u}$\n by properties of scalars and by the distributive law,\n respectively. Adding $-(0\\vect{u})$ to both sides of the\n equation, and using the additive unit law and associativity, we\n have $\\vect{0} = 0\\vect{u}$.\n \\item We have $(-1)\\vect{u} = (-1)\\vect{u} + \\vect{0} =\n (-1)\\vect{u} + (\\vect{u} + (-\\vect{u})) = ((-1)\\vect{u} +\n \\vect{u}) + (-\\vect{u}) = ((-1)\\vect{u} + 1\\vect{u}) +\n (-\\vect{u}) = ((-1)+1)\\vect{u} + (-\\vect{u}) = 0\\vect{u} +\n (-\\vect{u}) = \\vect{0} + (-\\vect{u}) = -\\vect{u}$. Here, we have\n used the additive unit law, the additive inverse law, the\n associative law, the rule for multiplication by 1, the\n distributive law, properties of scalars, the property of part\n (b), and the additive unit law, respectively.\n \\end{enumerate}\n \\end{sol}\n\\end{ex}\n\n", "meta": {"hexsha": "f132f50d42316758d7aa3125fec19e4016957c6e", "size": 3121, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "baseText/exercises/Vectors-ScalarMultiplication.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "baseText/exercises/Vectors-ScalarMultiplication.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "baseText/exercises/Vectors-ScalarMultiplication.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 32.5104166667, "max_line_length": 70, "alphanum_fraction": 0.5357257289, "num_tokens": 1231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.8397339716830605, "lm_q1q2_score": 0.6481432059581519}} {"text": "\\section{What's wrong with monocular camera?}\n\\label{mono_prob}\n\nAs the name suggests a monocular camera is a camera with a single imaging sensor. In section~\\ref{imageFormation} we say that given a perfectly calibrated camera ie. with the intrinsic and extrinsic parameters known, we can measure size of objects in the real world. This however is not true for a single image taken from a monocular camera. When 3D objects are projected onto the image plane (2D), the depth information stored in the z-axis is lost. This can be seen in the example below. \n\n% For one-column wide figures use\n\\begin{figure}\n% Use the relevant command to insert your figure file.\n% For example, with the graphicx package use\n \\includegraphics[width=\\textwidth]{./figures/1.png}\n% figure caption is below the figure\n\\caption{Monocular camera scale problem}\n\\label{fig:mono1} % Give a unique label\n\\end{figure}\n\nAs seen in the Fig~\\ref{fig:mono1}, irrespective of the location of the object in the real world, the size of the object in the image plane remains the same. Mathematically, this can be shown as \n\n\\begin{equation}\n\\frac{y}{f}=\\frac{y_1}{z_1+f}=\\frac{y_2}{z_2+f}\n\\end{equation}\n\nWhere $f$ is the focal length of the camera. The title of the paper mentions scale estimation. In the context of SLAM which will be covered in the next section, scale is a term used to denote the factor by which the computed trajectory by a SLAM algorithm needs to be multiplied/scaled to make it equal to the ground truth. For example, if the measured distance is 2m, whereas the ground truth is 4m, the scale value would be 2. This is an area of active research and the research community has come up with different techniques to solve this problem. As the title suggests, we would be talking about the fusion of IMU and Vision for solving this challenge.\n", "meta": {"hexsha": "ee2041db999e5744c7d981b6120f0e69af0ae09d", "size": 1834, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/text/mono_problem.tex", "max_stars_repo_name": "rohit517/Scale-Estimation-Monocular-SLAM", "max_stars_repo_head_hexsha": "ec86d42b83f2574db7b1e22b12cc531b09062c45", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Report/text/mono_problem.tex", "max_issues_repo_name": "rohit517/Scale-Estimation-Monocular-SLAM", "max_issues_repo_head_hexsha": "ec86d42b83f2574db7b1e22b12cc531b09062c45", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/text/mono_problem.tex", "max_forks_repo_name": "rohit517/Scale-Estimation-Monocular-SLAM", "max_forks_repo_head_hexsha": "ec86d42b83f2574db7b1e22b12cc531b09062c45", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 79.7391304348, "max_line_length": 657, "alphanum_fraction": 0.7769901854, "num_tokens": 449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8397339596505965, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.6481432010780831}} {"text": "\\subsection{Numerical Approximation and Analysis}\\label{sec:set-algorithm}\nWe present a non-intrusive algorithm based on Monte Carlo sampling\\---initially introduced in \\cite{BET+14} and further analyzed in \\cite{BET+14-arxiv}\\---that is structured in four stages (written as four independent for-loops), that are linked to the stages in Fig.~\\ref{fig:scheme}.\nWe direct the interested reader to \\cite{BET+14-arxiv} for more detailed information and analysis of this algorithm, e.g., on the requirement of a sampler being ``$\\pborel$-consistent'' to ensure convergence.\n\n\n\\begin{algorithm}[hbtp]\n\\DontPrintSemicolon\nChoose a discretization partition $\\set{D_\\idisc}_{\\idisc=1}^{\\ndiscs}$ of $\\dspace$.\\\\\n\t\\For{$\\idisc = 1, \\hdots, \\ndiscs$}{\n\t\t\tCompute $p_{\\dspace, \\idisc} = \\dataP(D_\\idisc)$.\n\t}\n \tChoose samples $\\set{\\param^{(\\iparam)}}_{\\iparam=1}^{\\nsamps} \\subset \\pspace$, which implicitly defines a Voronoi-cell partition $\\set{\\VV_\\iparam}_{\\iparam=1}^{\\nsamps}$ of $\\pspace$.\\\\\n\t\\For{$\\iparam = 1, \\hdots, \\nsamps$}{\n\tCompute $\\qoi_\\iparam = \\qoi(\\param^{(\\iparam)})$.\\\\\n\tLet $\\OO_\\idisc = \\set{\\iparam: \\qoi_\\iparam \\in D_\\idisc}$.\\\\\n\tCompute approximations $V_\\iparam \\approx \\pmeas (\\VV_\\iparam)$.\n\t}\n\t\\For{$\\idisc = 1, \\hdots, \\ndiscs$}{\n\tCompute $\\CC_\\idisc = \\set{\\iparam:Q_\\iparam \\in D_\\idisc}$.\n\t}\n\t\\For{$\\iparam = 1, \\hdots, \\nsamps$}{\n\tCompute $p_{\\pspace, \\iparam} = \\left ( V_\\iparam / \\sum_{j\\in \\CC_{\\OO_\\iparam} } V_j \\right ) p_{\\dspace, \\OO_\\iparam}$.\n\t}\n\tFor any $A\\in \\pborel$, compute\n\t\\begin{equation}\n\t\\PP_{\\pspace, \\ndiscs, \\nsamps} (A) = \\sum_{\\iparam=1}^\\nsamps p_{\\pspace, \\iparam} \\Chi_{\\VV_\\iparam} (A)\n\t\\end{equation}\n \\caption{Numerical Approximation of the Inverse Density}\n \\label{alg:inv_density}\n\\end{algorithm}\n\n\nThe first two stages correspond to formulating the discretized version of the SIP given in step (S1) in Fig.~\\ref{fig:scheme}.\nWe first discretize the probability space $\\Ospace$.\nThen, we simultaneously discretize the measure space $(\\pspace, \\pborel, \\pmeas)$ and construct a simple-function approximation to the map $\\qoi$.\nThese stages introduce the primary sources of error, and the third and fourth stages may be thought of as solving the discretized SIP exactly.\nThe samples that are used to describe $\\pspace$ implicitly define a set of Voronoi cells $\\set{\\VV_\\iparam}_{\\iparam=1}^{\\nsamps}$, which can be seen in Figure~\\ref{fig:voronoi_cells}.\nEach sample set defines a fundamentally different geometry.\n\n\\begin{figure}[ht]\n\\centering\n\t\\begin{minipage}{.475\\textwidth}\n\t\t\\includegraphics[width=\\linewidth]{./images/voronoi_diagrams/voronoi_diagram_N25_r0}\n\t\\end{minipage}\n\\caption{\nVoronoi-cell discretization (partition) induced by $\\nsamps = 25 $ uniform i.i.d.~random samples in $\\pspace = [0,1]^2$.\n}\n\\label{fig:voronoi_cells}\n\\end{figure}\n\nThe third stage then identifies the collection of Voronoi cells in $\\pspace$ that approximate the contour events in $\\cborel$ defined by $\\qoi^{-1}(D_\\idisc)$ for $\\idisc=1,\\hdots,\\ndiscs$. This allows us to formulate the consistent solution to the discretized SIP on $(\\pspace, \\cborel, \\contourP)$ as illustrated in step (S2) of Fig.~\\ref{fig:scheme}.\nThe fourth stage, associated with step (S3) in Fig.~\\ref{fig:scheme}, uses a discrete version of the ansatz to approximate the probability of $\\VV_\\iparam$ for $\\iparam=1,\\dots,\\nsamps$.\nThis results in an approximate probability measure, denoted by $\\updatedPxNM$, which produces the same probability estimates for events $A$ and $A\\setminus \\set{ \\param^{(\\iparam)} }_{\\iparam=1}^\\nsamps$, which are identical almost everywhere with respect to $\\pmeas$.\n\nNote that Algorithm~\\ref{alg:inv_density} makes no mention of the method by which the samples $\\set{ \\param^{(\\iparam)} }_{\\iparam=1}^{\\nsamps}$ were generated or sets in $\\set{D_\\idisc}_{\\idisc=1}^{\\ndiscs}$ are chosen.\n$\\set{ \\param^{(\\iparam)} }_{\\iparam=1}^{\\nsamps}$ may be generated using uniform random sampling, Latin-hypercube sampling, or even regular grids.\nA thorough discussion of the choices involved in making such decisions is beyond the scope of this work, though we touch briefly on the discretization of $\\dspace$ in the following section.\n", "meta": {"hexsha": "7858b7c91432804910f86762b30b9cc3102095f3", "size": 4186, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "set-based/set_algorithm.tex", "max_stars_repo_name": "mathematicalmichael/thesis", "max_stars_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-04-24T08:05:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-28T20:34:29.000Z", "max_issues_repo_path": "set-based/set_algorithm.tex", "max_issues_repo_name": "mathematicalmichael/thesis", "max_issues_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 59, "max_issues_repo_issues_event_min_datetime": "2019-12-27T23:15:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-24T17:52:57.000Z", "max_forks_repo_path": "set-based/set_algorithm.tex", "max_forks_repo_name": "mathematicalmichael/thesis", "max_forks_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 72.1724137931, "max_line_length": 353, "alphanum_fraction": 0.7305303392, "num_tokens": 1304, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339516289534, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.6481431904795197}} {"text": "\\section{Addition and subtraction}\n\\begin{center}\n \\begin{longtable}{lcr}\n $a_1 + a_2\\ \\cdots\\ + a_n = n \\cdot a$\n &\n $a + (-b) = a - b$\n &\n $a - (-b) = a + b$\n \\end{longtable}\n\\end{center}\n", "meta": {"hexsha": "a395b04783775215a03208b95d106d3cb7ef9394", "size": 231, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/addition_and_subtraction.tex", "max_stars_repo_name": "KongBorup/algebraic-tricks", "max_stars_repo_head_hexsha": "ea3e187203fffdaf03b3bc64994dcf95cf6f80c8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sections/addition_and_subtraction.tex", "max_issues_repo_name": "KongBorup/algebraic-tricks", "max_issues_repo_head_hexsha": "ea3e187203fffdaf03b3bc64994dcf95cf6f80c8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/addition_and_subtraction.tex", "max_forks_repo_name": "KongBorup/algebraic-tricks", "max_forks_repo_head_hexsha": "ea3e187203fffdaf03b3bc64994dcf95cf6f80c8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-01-22T21:54:51.000Z", "max_forks_repo_forks_event_max_datetime": "2019-01-22T21:54:51.000Z", "avg_line_length": 21.0, "max_line_length": 46, "alphanum_fraction": 0.4588744589, "num_tokens": 86, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.901920681802153, "lm_q2_score": 0.7185943925708561, "lm_q1q2_score": 0.6481151444867106}} {"text": "% !TEX encoding = UTF-8 Unicode\n% !TEX spellcheck = en_US\n% !TEX root = ../../../ICMA2020.tex\n\n\\subsection{Dynamics Model of Serial Robots}\nThis subsection describes the dynamics model of the 6-axis robot.\nThe joint torques $\\boldsymbol{\\tau}$ of all axes required for the feedforward control can be calculated by\n%\n\\begin{equation}\n\\label{eq:model_equation1}\n\\begin{split}\n\\boldsymbol{\\tau}=\n\\boldsymbol{M}(\\boldsymbol{q}) \\ddot{\\boldsymbol{q}}+\\boldsymbol{c}(\\boldsymbol{q}, \\dot{\\boldsymbol{q}})+\\boldsymbol{g}(\\boldsymbol{q})+\\boldsymbol{h}(\\dot{\\boldsymbol{q}}).\n\\end{split}\n\\end{equation}\n%\nHerein $\\boldsymbol{q}$, $\\dot{\\boldsymbol{q}}$ and $\\ddot{\\boldsymbol{q}}$ represent the joint angles, velocities and accelerations given by the motion planning. $\\boldsymbol{M}$ denotes the mass matrix, $\\boldsymbol{c}$ the Coriolis effects and $\\boldsymbol{g}$ the gravitational effects.\nThe friction model is represented by $\\boldsymbol{h}$ with\n%\n\\begin{equation}\nh_{j}=f_{\\mathrm{c}, j} \\operatorname{sgn}\\left(\\dot{{q}}_{j}\\right)+f_{\\mathrm{v}, j} \\dot{q}_{j},\n\\end{equation}\n%\nwhere $f_{\\mathrm{c}, j}$ represents the Coulomb friction coefficient for joint $j$ and $f_{\\mathrm{v}, j}$ the viscous friction coefficient. \nThe gear transmission between motor and link side is omitted for the sake of simplifying the equations. Measured motor velocities and estimated motor torques are transformed accordingly.\n\nBy expressing the rigid body dynamics with the inertial parameters (drive train inertia, mass, first and second moment of mass) and by using a linear friction model, (\\ref{eq:model_equation1}) can be expressed in a linear form \\cite{Khalil.2006}.\n%\nThis regressor form\n%\n\\begin{equation}\n\\boldsymbol{\\tau}\n=\n\\boldsymbol{X}'(\\boldsymbol{q}, \\dot{\\boldsymbol{q}},\\ddot{\\boldsymbol{q}}) \\boldsymbol{\\theta}'\n=\n\\boldsymbol{X}(\\boldsymbol{q}, \\dot{\\boldsymbol{q}},\\ddot{\\boldsymbol{q}}) \\boldsymbol{\\theta}\n\\label{eq:model_regressor}\n\\end{equation}\n%\nwith regression matrix $\\boldsymbol{X}$ and parameter vector $\\boldsymbol{\\theta}$\ncan be given with a full set of parameters (on the left hand side, noted with a dash) or in a parameter minimal form (right hand side).\nOnly the latter form of the parameters can be identified within one identification cycle.\nThe general model $\\boldsymbol{X}'$/$\\boldsymbol{\\theta}'$ uses 66 parameters for the rigid body dynamics.\nFor each of the six joints two additional parameters represent the Coulomb and viscous friction.\n\nThe right hand side of (\\ref{eq:model_regressor}) expresses the minimal parametric model with 23 base inertial parameters.\nThe chosen parameter vector $\\boldsymbol{\\theta}$ to be identified therefore contains 35 entries and is shown in appendix \\ref{sec:MinparamVector}.\n%\nSince some of the robot links possess symmetries, assumptions about their center of mass and products of inertia were made.\nFurther, a differential gear couples joints 5 and 6, similar to the robot example in \\cite{M.Gautier.1995}.\nThis prevents the use of the direct determination of the base inertial parameters with the geometric approach from \\cite{Khalil.2006}.\nInstead, a purely symbolic approach is chosen: the relation $\\boldsymbol{\\theta}=\\boldsymbol{K} \\boldsymbol{\\theta}'$ leads to a linear system of equations $\\boldsymbol{X}' = \\boldsymbol{X}'\\boldsymbol{K}$.\nIt is solved symbolically for the unknown $\\boldsymbol{K}$ with the computer algebra system \\textsc{Maple}, leading to the base parameter formulation $\\boldsymbol{X}$/$\\boldsymbol{\\theta}$.\nFor the implementation the regressor matrix of the energy instead of the dynamics is used, following \\cite{M.Gautier.1995, Khalil.2006}.", "meta": {"hexsha": "b2b2da5e8d332cd98dd7d49e672c5bd7a6694e36", "size": 3647, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/Chapters/Theoretical_Background/Modeling/Modeling.tex", "max_stars_repo_name": "SchapplM/robotics-paper_icma2020", "max_stars_repo_head_hexsha": "f81c6599ec7a9341e6a467a4ff9b31091aa50e75", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/Chapters/Theoretical_Background/Modeling/Modeling.tex", "max_issues_repo_name": "SchapplM/robotics-paper_icma2020", "max_issues_repo_head_hexsha": "f81c6599ec7a9341e6a467a4ff9b31091aa50e75", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/Chapters/Theoretical_Background/Modeling/Modeling.tex", "max_forks_repo_name": "SchapplM/robotics-paper_icma2020", "max_forks_repo_head_hexsha": "f81c6599ec7a9341e6a467a4ff9b31091aa50e75", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.537037037, "max_line_length": 290, "alphanum_fraction": 0.7611735673, "num_tokens": 971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9019206659843131, "lm_q2_score": 0.7185943865443349, "lm_q1q2_score": 0.6481151276846555}} {"text": "%!TEX root = ../TTT4150-Summary.tex\n\\section{Position, velocity, and time (PVT) estimation}\n\n\\subsection{Position estimation with pseudoranges}\nThe pseudorange measurement from satellite $k$ after correction is\n\\begin{equation}\\label{eq:pseudorange}\n \\rho_c^{(k)} = r^{(k)} + c \\cdot \\delta t_u + \\tilde{\\epsilon}_\\rho^{(k)}\n\\end{equation}\nwhere $\\delta t_u$ is the user clock bias, and $\\tilde{\\epsilon}_\\rho^{(k)}$ is the combined effect of residual errors. The position of the user at the time of measurement is\n\\begin{equation}\n \\V{x} = (x, y, z)\n\\end{equation}\nand the position of satellite $k$ at signal transmission time is\n\\begin{equation}\n \\V{x}^{(k)} = \\left( x^{(k)}, y^{(k)}, z^{(k)} \\right).\n\\end{equation}\nThe user-to-satellite range is\n\\begin{equation}\n r^{(k)} = \\norm{\\V{x}^{(k)} - \\V{x}}\n\\end{equation}\nand \\eqref{eq:pseudorange} can be written as\n\\begin{equation}\n \\rho_c^{(k)} = r^{(k)} + b + \\tilde{\\epsilon}_\\rho^{(k)}.\n\\end{equation}\n\nThe \\emph{geometry matrix} is\n\\begin{equation}\n \\M{G}\n =\n \\begin{bmatrix}\n (-\\M{1}^{(1)})\\T & 1 \\\\\n \\vdots & \\vdots \\\\\n (-\\M{1}^{(K)})\\T & 1\n \\end{bmatrix}\n\\end{equation}\nwhere $K$ is the number of satellites in view, and $\\M{1}^{(k)}$ is the line of sight unit vector from the initial user position estimate to satellite $k$.\n\n\\subsection{Dilution of precision (DOP)}\nDOP quantifies the effect of satellite geometry on the accuracy of positioning. A good geometry gives a low DOP value, and high accuracy. There are several variants, using $\\M{H} = \\left( \\M{G}\\T \\M{G} \\right)^{-1}$:\n\\begin{itemize}\n \\item\n Position dilution of precision\n \\begin{equation}\n \\mathrm{PDOP} = \\sqrt{H_{11} + H_{22} + H_{33}}\n \\end{equation}\n \\item\n Time dilution of precision\n \\begin{equation}\n \\mathrm{TDOP} = \\sqrt{H_{44}}\n \\end{equation}\n \\item\n Geometric dilution of precision\n \\begin{equation}\n \\mathrm{GDOP} = \\sqrt{H_{11} + H_{22} + H_{33} + H_{44}}\n \\end{equation}\n \\item\n Horizontal dilution of precision\n \\begin{equation}\n \\mathrm{HDOP} = \\sqrt{H_{11} + H_{22}}\n \\end{equation}\n \\item\n Vertical dilution of precision\n \\begin{equation}\n \\mathrm{VDOP} = \\sqrt{H_{33}}\n \\end{equation}\n\\end{itemize}\n", "meta": {"hexsha": "327295222a43a313fbe6092658637804ecf7fcd0", "size": 2383, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TTT4150 Navigation systems/tex/6-pvt-estimation.tex", "max_stars_repo_name": "jakoblover/ntnu-course-summaries", "max_stars_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-05-30T09:19:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-11T02:42:40.000Z", "max_issues_repo_path": "TTT4150 Navigation systems/tex/6-pvt-estimation.tex", "max_issues_repo_name": "jakoblover/ntnu-course-summaries", "max_issues_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TTT4150 Navigation systems/tex/6-pvt-estimation.tex", "max_forks_repo_name": "jakoblover/ntnu-course-summaries", "max_forks_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.5671641791, "max_line_length": 216, "alphanum_fraction": 0.6076374318, "num_tokens": 782, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8902942290328345, "lm_q2_score": 0.7279754548076478, "lm_q1q2_score": 0.6481123462928019}} {"text": "\\subsection{Minimization algorithm}\n\\label{minimization}\n\nIn this thesis, many data treatment approaches needed efficient nonlinear\nminimization algorithm (background subtraction, estimation of band intensities,\nestimation of parameters of chemical model from spectral series).\nThe Levenberg-Marquardt method\n\\parencite{Marquardt1963}\nwas used as a basis for the minimization algorithm used in this thesis, and\nslight modifications were applied to improve its performance.\n\nThe Levenberg-Marquardt method is based on Newton method where\nthe function $f(\\vec{a})$ which needs to be minimized is approximated by its\nTaylor series around $\\vec{a_\\text{cur}}$ point\n\\begin{align*}\n\tf(\\vec{a}) =& f(\\vec{a}_\\text{cur})\n\t\t+ \\sum_k \\left.\\frac{\\partial{}f}{\\partial{}a_k}\\right\n\t\t\t\\rvert_{\\vec{a}_\\text{cur}}(a_k - a_{k,\\text{cur}})\\\\\n\t\t&+ \\sum_{k,l} \\left.\\frac{\n\t\t\t\\partial^2f\n\t\t}{\n\t\t\t\\partial{}a_k\\partial{}a_l\n\t\t}\\right\\rvert_{\\vec{a}_\\text{cur}}\n\t\t(a_k - a_{k,\\text{cur}})(a_l - a_{l,\\text{cur}})\n\t\t+ \\dots,\n\\end{align*}\nwhich can be written as a quadratic form\n\\begin{equation}\n\t\\chi^2(\\vec{a}) \\approx\n\t\t\\gamma\n\t\t- \\vec{d}\\cdot\\vec{a}\n\t\t+ \\frac{1}{2}\\vec{a}\\cdot{}\\mat{D}\\cdot\\vec{a},\n\t\\label{\\eqnlabel{minimization:chi_approx}}\n\\end{equation}\nwhere $\\vec{d}$ is an $M$ vector, $\\mat{D}$ is an $M\\times{}M$ matrix, and\n$\\chi^2(\\vec{a})$ is chi-square function\n\\begin{equation*}\n\t\\chi^2(\\vec{a}) \\equiv \\sum_{i=1}^N\\left(\n\t\t\t\\frac{y_i - y(x_i;\\vec{a})}{\\sigma_i}\n\t\\right)^2,\n\\end{equation*}\nwhere $y_i$ are data points from the measured spectrum,\n$y(x_i;\\vec{a})$ are values of the model at these data points and\n$\\sigma_i$ are standard deviations of measured points $y_i$.\n\nIf\n\\eqnref{minimization:chi_approx}\nis a good approximation of the $\\chi^2(\\vec{a})$, the minimum can be found\ndirectly by adjusting the currently estimated parameters $\\vec{a}_\\text{cur}$\nby\n\\begin{equation}\n\t\\vec{a}_\\text{min} = \\vec{a}_\\text{cur}\n\t\t+ \\mat{D}^{-1}\\cdot(-\\vec{\\nabla}\\chi^2(\\vec{a}_\\text{cur})).\n\t\\label{\\eqnlabel{minimization:newton_min}}\n\\end{equation}\n\nOn the other hand, if\n\\eqnref{minimization:chi_approx}\ndoes not locally approximate the shape of $\\chi^2(\\vec{a})$ well, the steepest\ndescend method instead of the above described Newton method can be used for\nconvergence to the minimum\n\\begin{equation}\n\ta_\\text{min} = a_\\text{cur} - k\\vec{\\nabla}\\chi^2(\\vec{a_\\text{cur}}),\n\t\\label{\\eqnlabel{minimization:gradient_min}}\n\\end{equation}\nwhere $k$ needs to be a sufficiently small constant.\n\nLet us calculate the partial derivatives needed in terms from\n\\eqnref{minimization:newton_min} and \\eqnref{minimization:gradient_min}:\n\\begin{align*}\n\t\\frac{\\partial\\chi^2}{\\partial{}a_k} &= -2\\sum_{i=1}^N{\n\t\t\\frac{y_i - y(x_i;\\vec{a})}{\\sigma_i^2}\n\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_k}\n\t},\\\\\n\t\\frac{\\partial^2\\chi^2}{\\partial{}a_k\\partial{}a_l} &= 2\\sum_{i=1}^N{\n\t\t\\frac{1}{\\sigma_i^2}\\left[\n\t\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_k}\n\t\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_l}\n\t\t\t- [y_i - y(x_i;\\vec{a})]\n\t\t\t\t\\frac{\\partial^2{}y(x_i;\\vec{a})}{\\partial{}a_k\\partial{}a_l}\n\t\t\\right]\n\t},\n\\end{align*}\nwhere the second term in the second equation can be neglected because the\nsummation over all spectral points $y_i$ is equal to 0 if the\n$y(x_i;\\vec{a})$ models $y_i$ well. Therefore the simplified formula for\nHessian can be written as\n\\begin{equation*}\n\t\\frac{\\partial^2\\chi^2}{\\partial{}a_k\\partial{}a_l} = 2\\sum_{i=1}^N{\n\t\t\\frac{1}{\\sigma_i^2}\n\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_k}\n\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_l}\n\t}.\n\\end{equation*}\nIt can be seen that this approximation of Hessian does not need second\nderivatives of $\\chi^2(\\vec{a})$ and can be written as the product of two\nJacobians $\\mat{J}$\n\\begin{align*}\n\t\\mat{J} &=\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{\\partial{}y(x_1;\\vec{a})}{\\partial{}a_1} & \\cdots & \\frac{\\partial{}y(x_1;\\vec{a})}{\\partial{}a_M} \\\\\n\t\t\t\\vdots & \\ddots & \\vdots \\\\\n\t\t\t\\frac{\\partial{}y(x_N;\\vec{a})}{\\partial{}a_1} & \\cdots & \\frac{\\partial{}y(x_N;\\vec{a})}{\\partial{}a_M}\n\t\t\\end{pmatrix}\\\\\n\t\\mat{W} &=\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{\\sigma_1^2} \\\\\n\t\t\t& \\ddots \\\\\n\t\t\t&& \\frac{1}{\\sigma_N^2}\n\t\t\\end{pmatrix}\\\\\n\t[\\alpha] &= \\mat{J}^\\mathsf{T}(\\mat{W}\\mat{J}).\n\\end{align*}\n\nLet us introduce substitution by defining\n\\begin{align}\n\t\\beta_k &\\equiv -\\frac{1}{2}\\frac{\\partial\\chi^2}{\\partial{}a_k}\n\t\t= \\sum_{i=1}^N{\n\t\t\t\\frac{y_i - y(x_i;\\vec{a})}{\\sigma_i^2}\n\t\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_k}\n\t\t}\n\t\t\\label{\\eqnlabel{minimization:beta}}\\\\\n\t\\alpha_{kl} &\\equiv \\frac{1}{2}\n\t\t\\frac{\\partial^2\\chi^2}{\\partial{}a_k\\partial{}a_l}\n\t\t= \\sum_{i=1}^N{\n\t\t\t\\frac{1}{\\sigma_i^2}\n\t\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_k}\n\t\t\t\\frac{\\partial{}y(x_i;\\vec{a})}{\\partial{}a_l}\n\t\t\t\\label{\\eqnlabel{minimization:alpha}}\n\t\t}\n\\end{align}\nand rewrite equations\n\\cref{%\n\t\\eqnlabel{minimization:newton_min},%\n\t\\eqnlabel{minimization:gradient_min}%\n}\nto forms which are defining the step towards the minimum $\\text{\\g{d}}a_l$\n\\begin{align}\n\t\\sum_{l=1}^N\\alpha_{kl}\\text{\\g{d}}a_l &= \\beta_k,\n\t\\label{\\eqnlabel{minimization:newton_delta}}\\\\\n\t\\text{\\g{d}}a_l &= k\\cdot\\beta_l,\n\t\\label{\\eqnlabel{minimization:gradient_delta}}\n\\end{align}\nwhere the matrix $[\\alpha]$ which is equal to one-half times the Hessian\nmatrix, is usually called the \\emph{curvature matrix}.\n\nMarquardt combined the\n\\cref{%\n\t\\eqnlabel{minimization:newton_delta},%\n\t\\eqnlabel{minimization:gradient_delta}%\n},\ntogether in one equation\n\\begin{equation}\n\t\\sum_{l=1}^N\\alpha'_{kl}\\text{\\g{d}}a_l = \\beta_k,\n\t\\label{\\eqnlabel{minimization:lm}}\n\\end{equation}\nwith\n\\begin{equation*}\n\t\\alpha'_{kl} = \\alpha_{kl}(1+\\text{\\g{d}}_{kl}\\lambda),\n\\end{equation*}\nwhere $\\text{\\g{d}}_{kl}$ is Kronecker delta (equal 1 if $k = l$ and 0\notherwise) and $\\lambda$ is damping factor.\nThe new $\\alpha'_{kl}$ is diagonally dominant with large $\\lambda$ and behaves\nlike the gradient method and is equal to $\\alpha_{kl}$ for small $\\lambda$ and\nbehaves like the Newton method.\n\nThe only unknown is an estimation of the damping factor $\\lambda$.\n\\textcite{Marquardt1963}\nsuggested a heuristic for its computation using an additional parameter\ncalled multiplication factor $\\mu$ and supposing initial guess of parameters $\\vec{a}$\nand $\\lambda = \\lambda_0$:\n\\begin{docenum}\n\t\\item Estimate $\\chi^2(\\vec{a})$.\n\t\\item Solve the \\eqnref{minimization:lm} for $\\text{\\g{d}}\\vec{a}$ and\n\t\tevaluate $\\chi^2(\\vec{a} + \\text{\\g{d}}\\vec{a})$.\n\t\t\\label{enum:minimization:lm_steps:item:solve}\n\t\\item If $\\chi^2(\\vec{a} + \\text{\\g{d}}\\vec{a}) \\geq \\chi^2(\\vec{a})$ update\n\t\t$\\lambda$ to $\\lambda\\cdot\\mu$ and go to\n\t\t\\cref{enum:minimization:lm_steps:item:solve}.\n\t\tIt means that you are probably far from the minimum so switch from Newton\n\t\tto gradient method with large jump.\n\t\\item If $\\chi^2(\\vec{a} + \\text{\\g{d}}\\vec{a}) < \\chi^2(\\vec{a})$ update\n\t\t$\\lambda$ to $\\lambda/\\mu$, update the trial solution $\\vec{a}$ to\n\t\t$\\vec{a} + \\text{\\g{d}}\\vec{a}$ and go to\n\t\t\\cref{enum:minimization:lm_steps:item:solve}.\n\t\tIt means that you are probably near to the minimum so try to jump directly\n\t\tto the minimum approximated by quadratic function as per Newton method.\n\\end{docenum}\n\nWe enhanced the damping factor update algorithm for search in a wider range and\nboth directions because the above-proposed method was susceptible to falling\ninto a local minimum.\nThe items 3 and 4 update of lambda searched larger region of $\\mu^k$ where,\nfor example, $k \\in \\{-10, -9, \\dots, 9, 10\\}$.\nIt was also found that it is more stable, especially for the cases with more\ncomplicated models $y(x;\\vec{a})$, to calculate Jacobian $\\mat{J}$ numerically\nthan using analytic derivatives because of the numeric instability of the\nsolution.\n\nThis regression algorithm was programmed using the MATLAB programming\nenvironment\n\\parencite{Matlab}\nand used throughout many different applications in this work.\nThe algorithm was also modified so that it was possible to specify\ndifferent weight functions than least squares, for example, penalize positive\ndifferences as was then used in\n\\cref{background}.\n\\\\\n", "meta": {"hexsha": "83a92d8879e0ea0b7b4289bcf0f005d5d653e9bb", "size": 8116, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/results_and_discussion/minimization.tex", "max_stars_repo_name": "lumik/phd_thesis", "max_stars_repo_head_hexsha": "3b29f24732d49b64c627aeb8f6585f042cd59c4e", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/results_and_discussion/minimization.tex", "max_issues_repo_name": "lumik/phd_thesis", "max_issues_repo_head_hexsha": "3b29f24732d49b64c627aeb8f6585f042cd59c4e", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 41, "max_issues_repo_issues_event_min_datetime": "2019-08-13T12:27:09.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T03:00:58.000Z", "max_forks_repo_path": "src/results_and_discussion/minimization.tex", "max_forks_repo_name": "lumik/phd_thesis", "max_forks_repo_head_hexsha": "3b29f24732d49b64c627aeb8f6585f042cd59c4e", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.0192307692, "max_line_length": 110, "alphanum_fraction": 0.6929521932, "num_tokens": 2828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8902942173896131, "lm_q2_score": 0.7279754489059774, "lm_q1q2_score": 0.6481123325625995}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Written By Michael Brodskiy\n% Class: Analytic Geometry & Calculus III (Math-292)\n% Professor: V. Cherkassky\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass[12pt]{article} \n\\usepackage{alphalph}\n\\usepackage[utf8]{inputenc}\n\\usepackage[russian,english]{babel}\n\\usepackage{titling}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{enumitem}\n\\usepackage{amssymb}\n\\usepackage[super]{nth}\n\\usepackage{everysel}\n\\usepackage{ragged2e}\n\\usepackage{geometry}\n\\usepackage{fancyhdr}\n\\usepackage{cancel}\n\\geometry{top=1.0in,bottom=1.0in,left=1.0in,right=1.0in}\n\\newcommand{\\subtitle}[1]{%\n \\posttitle{%\n \\par\\end{center}\n \\begin{center}\\large#1\\end{center}\n \\vskip0.5em}%\n\n}\n\\usepackage{hyperref}\n\\hypersetup{\ncolorlinks=true,\nlinkcolor=blue,\nfilecolor=magenta, \nurlcolor=blue,\ncitecolor=blue,\n}\n\n\\urlstyle{same}\n\n\n\\title{Lecture XXII Notes}\n\\date{\\today}\n\\author{Michael Brodskiy\\\\ \\small Professor: V. Cherkassky}\n\n% Mathematical Operations:\n\n% Sum: $$\\sum_{n=a}^{b} f(x) $$\n% Integral: $$\\int_{lower}^{upper} f(x) dx$$\n% Limit: $$\\lim_{x\\to\\infty} f(x)$$\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Green's Theorem $-$ 16.4}\n\nFor this theorem, let $C$ be a curve that is:\n\n\\begin{enumerate}\n\n \\item positively oriented (moving counterclockwise)\n\n \\item piecewise smooth\n\n \\item simple\n\n \\item closed\n\n\\end{enumerate}\n\n$D$ is the region enclosed by the curve.\n\nIf $P$ and $Q$ have continuous first-order partial derivatives on $D$, then:\n\n$$\\int_C P\\,dx+Q\\,dy=\\iint_D \\left(\\frac{\\partial Q}{\\partial x}-\\frac{\\partial P}{\\partial y}\\right)\\,dA$$\n\nFor Green's Theorem, $D$ must have boundaries.\n\nNotation (If $C$ is closed):\n\n$$\\int_C P\\,dx+Q\\,dy=\\oint_C P\\,dx+Q\\,dy$$\n\nIf $\\frac{\\partial Q}{\\partial x}-\\frac{\\partial P}{\\partial y}=1$, then Green's theorem yields the area of the region $D$:\n\n$$\\iint_D 1\\,dA=A(D)$$\n\n\\end{document}\n\n", "meta": {"hexsha": "a8623f70ce46c67fdf8149dac062fde62fc7dbbb", "size": 2193, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture Notes/Lecture22.tex", "max_stars_repo_name": "MDBrodskiy/Vector_Calculus", "max_stars_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T15:51:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T15:51:52.000Z", "max_issues_repo_path": "Lecture Notes/Lecture22.tex", "max_issues_repo_name": "MDBrodskiy/Vector_Calculus", "max_issues_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture Notes/Lecture22.tex", "max_forks_repo_name": "MDBrodskiy/Vector_Calculus", "max_forks_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.3666666667, "max_line_length": 188, "alphanum_fraction": 0.5804833561, "num_tokens": 643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8519528038477824, "lm_q1q2_score": 0.6480384562970615}} {"text": "\\section{Attribute Grammars}\n\n\\subsection{Left-Right attributes}\nGiven a rule of type $A_0 \\rarr A_1A_2\\ldots$\n\n\\begin{description}\n \\item[Left attribute] $\\sigma_0 = f(\\ldots)$ associated to $A_0$\n \\item[Right attribute] $\\delta_i = f(\\ldots), i \\ge 1$, associated to $A_i$\n\\end{description}\n\nAn attribute cannot be both left and right.\n\n\\subsection{Dependence Graph}\n\nGiven a rule $p$, its dependence graph $dep_p$ is a directed graph:\n\\begin{itemize}\n \\item Nodes are the attributes\n \\item There is an arc from every argument to the result\n \\item Left attributes are on the left, right to the right\n\\end{itemize}\n\nIf the dependence graph is acyclic there exists a set of attribute values. A grammar is \\textbf{loop-free} if the dependence graph is always acyclic.\n\n\\subsection{Depth-First visit}\n\\begin{itemize}\n \\item Start from the root\n \\item Visit the child nodes in some specified order\n \\item Foreach subtree $t_N$ rooted at node $N$ :\n \\begin{itemize}\n \\item Before visiting $t_N$ compute the right attributes of $N$ and pass them down to the visit\n \\item After visiting $t_N$ compute the left attributes using the return value of the visit\n \\end{itemize}\n\\end{itemize}\n\n\\subsection{Depth-First Visit Conditions (one-sweep)}\n\\begin{itemize}\n \\item The graph $dep_p$ has no circuit\n \\item In $dep_p$ there exists no path from a left attribute to a right attribute of the same node\n \\item In $dep_p$ there exists no arc from a left attribute of the father to a right attribute of any child\n \\item There are no circuits in $sibl_p$ graph\n\\end{itemize}\n\nThere is an arc $D_i \\rarr D_j$ in $sibl_p$ iff in $dep_p$ there is an arc $\\alpha_i \\rarr \\beta_j$ with $\\alpha_i \\in \\text{attr}(D_i)$ and $\\beta_j \\in \\text{attr}(D_j)$.\n\n\\subsection{One-Sweep Evaluator}\nThere is a procedure for each nonterminal, its parameters are the subtree rooted at the nonterminal and the right attributes of the root; it visits the subtree and returns the left attributes of the root.\n\nFor the rule $p: D_0 \\rarr D_1D_2\\ldots D_r$\n\\begin{itemize}\n \\item Choose a Topological Order of Siblings $TOS$ compatible with $sibl_p$\n \\item For each symbol $D_i, i>0$ choose a topological order of the right attributes $TOR$\n \\item Choose a topological order of the left attributes of $D_0$\n\\end{itemize}\n\n\\subsection{Condition L}\nUsed to embed semantic evaluator inside recursive descent syntax parser.\n\\begin{itemize}\n \\item Support syntax is of type LL(k)\n \\item Depth-First Visit Conditions\n \\item The natural order ($D_1, D_2, \\ldots$) is a valid TOS is $sibl_p$ (i.e. no arc $D_j\\rarr D_i$ with $j>i$)\n\\end{itemize}\n", "meta": {"hexsha": "bcc2138ce7d9d4c4aa0c55ff3430599539568d40", "size": 2673, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "syntax/attribute.tex", "max_stars_repo_name": "Kakasinho/FLC-cheatsheet", "max_stars_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-01-13T14:36:20.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-18T16:22:18.000Z", "max_issues_repo_path": "syntax/attribute.tex", "max_issues_repo_name": "Kakasinho/FLC-cheatsheet", "max_issues_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "syntax/attribute.tex", "max_forks_repo_name": "Kakasinho/FLC-cheatsheet", "max_forks_repo_head_hexsha": "9293e89e803006f1b419c78087caa5d5e04a5931", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-05-21T11:05:53.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-17T14:59:50.000Z", "avg_line_length": 43.1129032258, "max_line_length": 204, "alphanum_fraction": 0.7306397306, "num_tokens": 762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519528094861981, "lm_q2_score": 0.7606506418255928, "lm_q1q2_score": 0.6480384513407936}} {"text": "\\subsection{States, messages, representatives}\n\nThe full-node protocol states are {\\em sets} of messages,\neach message being a triple $(c, v, j)$, where:\n\\begin{itemize}\n \\item $c$ is a (proposed) consensus value;\n \\item $v$ identifies the message sender;\n \\item $j$, the justification, is the {\\em protocol state} seen by the sender\n at the time of message sending.\n\\end{itemize}\n\nThere are two technical difficulties with the above definition:\n\\begin{itemize}\n \\item States and messages are mutually recursive: states are sets of messages, each containing a state;\n \\item Message ordering should not matter\n\\end{itemize}\n\nTo solve the first issue, we choose to postpone the definition of messages for now and first define states inductively as follows:\n\n\\begin{coq}\nInductive state : Type :=\n | Empty : state\n | Next : C -> V -> state -> state -> state.\n\\end{coq}\n\nThis definition says that a state can be built by extending an existing state\ngiven a consensus value \\verb|C|, a validator \\verb|V|, and another state representing\nthe justification. To clarify that the three are the components of a message, we introduce the following notation for \\verb\"Next\":\n\n\\begin{coq}\nNotation \"'add' ( c , v , j ) 'to' sigma\" :=\n (Next c v j sigma)\n (at level 20).\n\\end{coq}\n\nThis definition addresses the first difficulty mentioned above by capturing the recursive nature of states.\n\nThe second issue critical to the notion of state equality. Defining an equivalence between states which disregards message ordering is possible but non-trivial: the definition would itself need to be recursive, as it requires an equivalence\non messages, which in turn is defined in terms of the same state equivalence.\n\nTo circumvent the hindrance of working with mutually recursive state equality, we instead took the approach of using canonical representatives for states,\nwhich we call LocallySorted states, on which we express state equality as syntactic equality in Coq. Although this definition is still recursive, it is easier to express and to work with than with the former equivalence because it is defined in terms of a single state.\n\nTo define sorted states, we need to be able to compare messages,\nwhich amounts to defining a total order relation on states.\nThis can be defined as a lexicographic ordering on the state seen as a list\nof messages, tweaked slightly to perform recursion in order to compare justifications:\n\\begin{coq}\nFixpoint state_compare (sigma1 sigma2 : state) : comparison :=\n match sigma1, sigma2 with\n | Empty, Empty => Eq\n | Empty, _ => Lt\n | _, Empty => Gt\n | add (c1, v1, j1) to sigma1, add (c2, v2, j2) to sigma2 =>\n match compare c1 c2 with\n | Eq =>\n match compare v1 v2 with\n | Eq =>\n match state_compare j1 j2 with\n | Eq => state_compare sigma1 sigma2\n | cmp_j => cmp_j\n end\n | cmp_v => cmp_v\n end\n | cmp_c => cmp_c\n end\n end.\n\\end{coq}\n\nNote that defining this ordering requires that there exist orderings on\nconsensus values and validators, but these can be any orderings,\nand a total ordering is guaranteed to exists for any set in set theory\nby the axiom of choice \\cite{Gonzalez}. The existence of these orderings is reflected in our \\verb|StrictlyComparable| types for consensus values and validators.\n\nThis ordering naturally induces an ordering on messages, and thus allows us\nto define the notion of a \\verb|LocallySorted| state, i.e., one in which\neach message is smaller than the next and all justifications are themselves\n\\verb|LocallySorted|.\n\nChoosing \\verb|LocallySorted| states as representatives for states thus reducing the equality testing between states to syntactic equality checking.\n\nProtocol states are defined by means of an inductive predicate on states as follows:\n\\begin{coq}\nInductive protocol_state : state -> Prop :=\n | protocol_state_empty : protocol_state Empty\n | protocol_state_next\n : forall s j\n , protocol_state s\n -> protocol_state j\n -> incl_messages j s\n -> forall c v\n , valid_estimate c j\n -> not_heavy (add_in_sorted_fn (c,v,j) s)\n -> protocol_state (add_in_sorted_fn (c,v,j) s).\n\\end{coq}\n\nThe above definition reads as:\n\\begin{itemize}\n \\item a protocol state is either empty; or\n \\item it can be obtained from an existing protocol state $s$ by extending\n it with a message $(c, v, j)$ such that:\n \\begin{itemize}\n \\item $j$ is a protocol state included in $s$;\n \\item $c$ is a consensus value which the estimator agrees on for $j$;\n \\item adding $(c,v,j)$ to $s$ does not exceed the fault tolerance threshold.\n \\end{itemize}\n\\end{itemize}", "meta": {"hexsha": "9deec561f37e58f3a6984c6e9014f9a1bab1b7f8", "size": 4719, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/full.tex", "max_stars_repo_name": "runtimeverification/casper-cbc-proofs", "max_stars_repo_head_hexsha": "8c4985f0921fea0a38c05e72a47364471164ab72", "max_stars_repo_licenses": ["NCSA"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2020-06-16T15:57:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-23T11:21:07.000Z", "max_issues_repo_path": "report/full.tex", "max_issues_repo_name": "runtimeverification/casper-cbc-proofs", "max_issues_repo_head_hexsha": "8c4985f0921fea0a38c05e72a47364471164ab72", "max_issues_repo_licenses": ["NCSA"], "max_issues_count": 105, "max_issues_repo_issues_event_min_datetime": "2019-11-26T09:22:58.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-02T10:00:11.000Z", "max_forks_repo_path": "report/full.tex", "max_forks_repo_name": "runtimeverification/casper-cbc-proofs", "max_forks_repo_head_hexsha": "8c4985f0921fea0a38c05e72a47364471164ab72", "max_forks_repo_licenses": ["NCSA"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2019-12-17T07:48:58.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-22T08:51:55.000Z", "avg_line_length": 44.1028037383, "max_line_length": 270, "alphanum_fraction": 0.7253655435, "num_tokens": 1140, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.897695283896349, "lm_q2_score": 0.7217432003123989, "lm_q1q2_score": 0.6479054671046984}} {"text": "\\graphicspath{ {./img/TheFEM/} }\n\\chapter{Review of Boundary Value Problems}\n\n\\section{Boundary value problems}\nIn this section we will define a general initial boundary value problem (I-BVP). In the first part we will introduce the differential formulation given in terms of a set of governing equations and properly specified boundary conditions. The resulting equations are obtained after using a generalized balance law. Following this classical and well known approach we formally re-state these equations in the so-called strong form. Subsequently we re-write and prove an equivalent form of the balance law in the form of an integral representation highly friendly for a numerical solution. Since in the integral description of the problem the order of the derivatives in the field functions decreases by one, the resulting statement is called a weak formulation. \n\n\\subsection{Differential formulation-Generalized balance law}\nLet $\\dd{S}$ be a differential surface element, $\\dd{V}$ a differential volume element, $u(\\vb{x},t)$ a scalar (or vector) function of space and time. The flux or rate of flow of the quantity $u(\\vb x, t)$ through $\\dd{S}$ at time $t$ is defined like\n\\[p(\\vb x)\\grad u \\cdot \\hat n \\dd{S} \\enspace ,\\]\nwhere $p(\\vb x)$ is a positive function, assumed known and time independent. Similarly, the time rate of change of $u(\\vb x, t)$ in an element $\\dd{V}$ is given by\n\\[\\rho (\\vb x)\\pdv{u}{t} \\dd{V} \\enspace ,\\]\nwhere once again $\\rho (\\vb x)$ is a known, given, time independent positive function. Additional effects occurring in the element $\\dd{V}$ at the time time can be expressed like\n\\[H(\\vb x, t)\\dd{V} \\equiv - q(\\vb x)u(\\vb x, t) + \\hat F(\\vb x,t)\\]\nwhere $\\hat F(\\vb x, t) = \\rho (\\vb x)F(\\vb x, t)$. In the above the term $q u$ represent internal effects due to changes proportional to $u$ while $\\hat F(\\vb x, t)$ are other external influences in the medium.\n\nBalancing the internal and external changes yields\n\\[\\int\\limits_V \\rho (\\vb x)\\pdv{u}{t}\\dd{V} = \\int\\limits_S p(\\vb x)\\vb \\grad u \\cdot \\hat n\\dd{S} + \\int\\limits_V H(\\vb x, t)\\dd{V} \\enspace ,\\]\nor equivalently\n\\[\\int\\limits_V \\rho (\\vb x)\\pdv{u}{t}\\dd{V} = \\int\\limits_S p(\\vb x)\\vb \\grad u \\cdot \\hat n\\dd{S} - \\int\\limits_V q(\\vb x)u(\\vb x, t)\\dd{V} + \\int\\limits_V \\rho (\\vb x)F(\\vb x, t)\\dd{V} \\enspace .\\]\n\nUsing the divergence theorem as\n\\[\\int\\limits_S p(\\vb x) \\grad u \\cdot \\hat n \\dd{S} = \\int\\limits_V \\vb \\div (p\\vb \\grad u)\\dd{V}\\]\nyields after substitution\n\\[\\int\\limits_V \\left[\\rho (\\vb x)\\pdv{u}{t} - \\div (p(\\vb x)\\grad u) + q(\\vb x)u(\\vb x, t) - \\rho (\\vb x)F(\\vb x, t)\\right]\\dd{V} = 0\\]\nAssuming a continuous integrand, the arbitrariness of $V$ implies\n\\[\\rho (\\vb x)\\pdv{u}{t} - \\div(p(\\vb x)\\grad u) + q(\\vb x)u(\\vb x, t) - \\rho (\\vb x)F(\\vb x, t) = 0\\]\n\nLetting\n\\[\\mathcal{L} \\equiv - \\div p(\\vb x)\\grad + q(\\vb x)\\]\nthe generalized set of partial differential equations can be written like\n\\begin{equation}\n\\rho(\\vb x) \\pdv{u(\\vb x, t)}{t} + \\mathcal{L}u(\\vb x, t) = \\rho (\\vb x)F(\\vb x, t) \\enspace .\n\\label{eq:GenPDE}\n\\end{equation}\n\nAnd they are categorized as:\n\\begin{itemize}\n \\item Hyperbolic;\n \\[\\rho(\\vb x) \\pdv[2]{u(\\vb x, t)}{t} + \\mathcal{L}u(\\vb x,t) = 0\\]\n\n \\item Parabolic; and\n \\[\\rho(\\vb x) \\pdv{u(\\vb x,t)}{t} + \\mathcal{L}u(\\vb x, t) = 0\\]\n\n \\item Elliptic\n \\[\\mathcal{L}u(\\vb x, t) = \\rho (\\vb x)F(\\vb x, t) \\enspace .\\]\n\\end{itemize}\n\nIt can be shown that $\\mathcal{L}$ satisfies the \\emph{symmetry} condition\n\\[\\int\\limits_V \\mathcal{L}(u)v\\dd{V} = \\int\\limits_V \\mathcal{L}(v)u \\dd{V}\\]\nand positive definiteness \\cite{book:sepulveda_fismat, book:arfken}\n\\[\\int\\limits_V \\mathcal{L}(u)u\\dd{V} > 0, \\quad \\forall u \\enspace .\\]\n\n\\subsection{Strong form}\nGiven $\\rho(\\vb x)$, $q(\\vb x)$, $p(\\vb x)$, $F(\\vb x, t)$ and $\\bar u$ find $u(\\vb x , t):V \\to \\mathbb{R}$ such:\n\\[\\rho(\\vb x) \\pdv{u}{t} - \\div \\left[p(\\vb x) \\grad u\\right] + q(\\vb x) u(\\vb x, t) - \\rho(\\vb x) F(\\vb x, t) = 0 \\quad \\forall \\vb x \\in V \\]\nand\n\\begin{align*}\n &u = \\bar u \\quad \\forall \\vb x \\in S_u\\\\\n &p(\\vb x)u_{,i} \\hat n_i= B(\\vb x,t)\\quad \\forall \\vb x \\in S_t \\enspace .\n\\end{align*}\n\n\nIn the FEM we will look for approximate solutions to $u$ subject to the following conditions:\n\\[u = \\bar u \\quad \\forall \\vb x \\in S_u \\quad \\text{(Essential boundary conditions)}\\] \nand\n\\[\\int\\limits_S \\left(\\pdv{u}{x_j}\\right)^2 \\dd{S} < \\infty \\enspace ,\\]\nwhich corresponds to the functions being square integrable. We will denote this space by $\\mathbb{H}$ .\n\nThe space of functions satisfying the above two conditions will be denoted by $\\zeta$ and termed the space of trial functions, formally defined like:\n\\[\\zeta = \\left\\{ u \\mid u \\in \\mathbb{H},u = \\bar u \\quad \\forall\\vb x \\in S_u \\right\\}\\]\n\nOn the other hand, to validate (or test) the correctness of the approximated or proposed trial functions $u$ it is also necessary to introduce test functions $w$ which are arbitrary except that they satisfy the following conditions:\n\\[w = 0\\quad \\forall \\vb x \\in S_u\\] \nand\n\\[\\int\\limits_S \\left(\\pdv{w}{x_j}\\right)^2 \\dd{S} < \\infty \\enspace ,\\]\nwhich corresponds to the functions being square integrable. The space of functions satisfying the above two conditions will be denoted by $\\pounds$ and termed the space of test functions, formally defined like:\n\\[\\pounds = \\left\\{w \\mid w \\in \\mathbb{H},w = 0 \\quad \\forall \\vb x \\in S_u\\right\\}\\]\n\n\\subsection{Weak form}\nGiven $\\rho(\\vb x)$, $q(\\vb x)$, $p(\\vb x)$, $F(\\vb x, t)$ and $\\bar u$ find $u(\\vb x , t):V \\to \\mathbb{R}$ and $\\forall w \\in \\pounds$ such:\n\\begin{align*}\n\\int\\limits_V p(\\vb x) u_{,i}\\, w_{,i}\\dd{V} - \\int\\limits_{S_t} B(\\vb x, t)w \\dd{S} + \\int\\limits_V q(\\vb x)u(\\vb x, t)w\\dd{V} &+ \\int\\limits_V \\rho(\\vb x)\\pdv{u}{t}w \\dd{V} \\\\& - \\int\\limits_V \\rho(\\vb x) F(\\vb x, t)w\\dd{V} = 0\n\\end{align*}\nand\n\\[u = \\bar u\\quad \\forall\\vb x \\in S_u \\enspace .\\]\n\n\\subsection{Equivalence between the strong and weak forms}\n\\begin{multline}\n -\\int\\limits_V [p(\\vb x)u_{,i}]_{,i}w \\dd{V} + \\int\\limits_{S_t} [p(\\vb x)u_{,i}] \\hat n_i w\\dd{S} - \\int\\limits_{S_t} B(\\vb x, t)w\\dd{S}\\\\\n + \\int\\limits_V q(\\vb x)u(\\vb x, t)w \\dd{V} + \\int\\limits_V \\rho(\\vec x)\\pdv{u}{t}w\\dd{V} - \\int\\limits_V \\rho(\\vb x)F(\\vb x, t)w\\dd{V} = 0 \n\\end{multline}\n\nGrouping together common terms yields\n\\begin{align*}\n&\\int\\limits_V \\left\\{\\rho(\\vb x)\\pdv{u}{t} - [p(\\vb x)u_{,i}]_{,i} + q(\\vb x)u(\\vb x, t) - \\rho(\\vb x)F(\\vb x, t)\\right\\} w\\dd{V}\\\\\n&+ \\int \\limits_{S_t} \\left\\{[p(\\vb x)u_{,i}]\\hat n_i - B(\\vb x, t) \\right\\} w\\dd{S} = 0\n\\end{align*}\nfrom which\n\\[\\rho(\\vb x)\\pdv{u}{t} - [p(\\vb x)u_{,i}]_{,i} + q(\\vb x)u(\\vb x, t) - \\rho(\\vb x)F(\\vb x, t) = 0\\]\nand\n\\[p(\\vb x)u_{,i}\\hat n_i = B(\\vb x, t) \\quad\\forall \\vb x \\in S_t \\enspace .\\]\n\n\n\\section{Brief review of the linearized theory of elasticity model}\nHere we present a brief description of the boundary value problem governing the response of an elastic body. For a full discussion of the model and its mathematical aspects the reader is referred to \\cite{shames1997elastic}.\n\nThe governing equations (in terms of stresses) stem from the principle of conservation of linear momentum and conservation of moment of linear momentum. The former leads to a set of 3 partial differential equations in the components of the stress tensor while the latter leads to the symmetries in the stress tensor.\n\\begin{equation} \\label{eq:pde}\n\\begin{aligned}\n&\\sigma_{ij,j} + {f_i} = \\rho\\ddot{u}_i \\quad \\forall\\ \\vb{x} \\in V,\\, t \\in \\mathbb{R}^{+}\\\\\n&\\sigma_{ij}=\\sigma _{ji}.\n\\end{aligned} \n\\end{equation}\n\nIn \\cref{eq:pde} $\\sigma_{ij}$ is the stress tensor; $f_i$ is the vector of body forces; and $u_i$ is the displacements vector.\n\nDenoting the tractions vector associated with a surface with normal direction $\\hat{n}_{j}$ by $t_i^{\\hat n}$ we have the complete BVP as follows\n\\begin{equation} \\label{eq:bcs}\nt_i^{\\hat n} = \\sigma_{ij} \\hat{n}_{j} \\quad \\forall \\in \\vb{x} \\in S.\n\\end{equation}\n\n\\Cref{eq:pde} correspond to 6 equations with 12 unknowns (the 9 components of the stress tensor and the 3 components of the displacements vector ) and the system is undetermined. In order to have a solvable BVP we must introduce kinematic strain-displacement relations and a stress-strain law. In the case of infinitesimal theory of elasticity the strain-displacement relation is given by:\n\n\\begin{equation}\\label{eq:kin}\n\\varepsilon_{ij} = \\frac{1}{2}(u_{i,j} + u_{j,i})\n\\end{equation}\n\nwhere the term $\\epsilon_{ij}$ is the symmetric component of the displacements gradient tensor. The components of the strain tensor describe the distortions and changes in magnitude (volumetric changes) of the material point in the continuum model. Now the simplest stress-strain (constitutive) relationship is given by Hooke's law\\footnote{Despite the name of \\emph{law} used, this relation is not always valid, but is a good approximation for small strains.}\n\\begin{equation} \\label{eq:Hooke}\n\\sigma_{ij} = 2\\mu \\varepsilon_{ij} + \\lambda \\varepsilon_{kk}\\delta_{ij} \\enspace .\n\\end{equation}\n\nwhere $\\mu$ and $\\lambda$ are material constants. The problem involves now a total of 18 equations and 18 unknowns can be solved if subjected to properly specified boundary conditions.\n\n\\subsection*{Displacement formulation}\nSubstituting \\cref{eq:kin} in \\cref{eq:Hooke} and the result in \\cref{eq:pde} yields after some manipulation:\n\\begin{equation} \\label{eq:navier}\n(\\lambda + \\mu)u_{j,ij} + \\mu u_{i,jj} + {f_i} = \\rho \\ddot{u}_i \\quad \\forall \\vb{x} \\in V,\\, t \\in \\mathbb{R}^{+}.\n\\end{equation}\n\nSince \\cref{eq:navier} ( simultaneously describing equilibrium, kinematic relations and constitutive response) is a second order equation governing the displacement field possible boundary conditions are in terms of the variable itself or its first oder derivatives. For a well-possed problem the following is a set of valid boundary conditions: \n\\begin{equation}\\label{eq:Wellbcs}\n\\begin{split}\n&t_i^{\\hat n} = \\sigma _{ij} \\hat n_{ij} \\quad \\forall\\ \\vb{x} \\in S_t\\\\\n& {u_i} = \\bar{u}_i \\quad \\forall \\vb x \\in S_u\n\\end{split}\n\\end{equation}\n\nand where ${S_t} \\cup {S_u} = S$ and ${S_t} \\cap {S_u} = \\emptyset $. \n\n\nIn the particular case in which $u_i$ is not a function of time, we obtain the static version of the BVP, i.e,\n\\begin{equation}\n\\begin{split}\n&\\left(\\lambda + \\mu \\right)u_{j,ij} + \\mu u_{i,jj} + {f_i} = 0 \\quad \\forall \\vb{x} \\in V \\\\\n&t_i^{\\hat n} = \\sigma _{ij} \\hat n_{ij} \\quad \\forall\\ \\vb{x} \\in S_t\\\\\n& {u_i} = \\bar{u}_i \\quad \\forall \\vb x \\in S_u\n\\end{split}\n\\end{equation}\n\nNotice that the tractions BC \n\\[t_i^{\\hat n} = \\mu (u_{i,j} + u_{j,i}) \\hat{n}_j + \\lambda u_{k,k} \\delta_{ij}\\hat{n}_j \\enspace ,\\]\n\nactually involves first order displacements derivative and as such it is a Neumann boundary condition on $u_i$.\n\n\\subsection{Equivalence between strong and weak forms}\n\\subsubsection{Strong form}\nThe strong form corresponds to the differential formulation of the problem, it is denoted by $\\{ S \\}$ and it reads:\n\nGiven $f_i$, $t_i^{\\hat n}$ and ${\\bar u_i}$ find ${u_i}:V \\to \\mathbb{R}$ such:\n\\begin{equation} \\label{eq:navier_2}\n\\begin{split}\n&(\\lambda + \\mu)u_{j,ij} + \\mu u_{i,jj} + f_i = 0 \\quad \\forall \\vb{x} \\in V \\\\\n&t_i^{\\hat n} = \\sigma _{ij} \\hat{n}_{ij} \\quad \\forall \\vb{x} \\in S_t\\\\\n&u_i = \\bar{u}_i \\quad \\forall \\vb{x} \\in S_u\n\\end{split}\n\\end{equation}\n\nIn \\cref{eq:navier_2} the boundary conditions specified by the traction vector $t_i^{\\hat n}$ correspond to the natural boundary conditions, while those specified in terms of the displacements vector $\\bar u_i$ represent the essential boundary conditions.\n\n\\begin{itemize}\n\\item We are interested in developing methods to obtain approximate solutions to $\\{S\\}$.\n\\item The FEM is formulated starting from a statement equivalent to $\\{ S \\}$ in which we use trial functions until certain prescribed conditions are met.\n\\item We will look for solutions $u_i$ subject to the following conditions:\n\\begin{align*}\n&u_i = \\bar u_i \\qquad \\text{in} \\qquad S_u\\\\\n&\\intL_S \\left(\\pdv{u_i}{x_j} \\right)^2 \\dd{S} < \\infty\n\\end{align*}\n\n\\end{itemize}\n\nThe first condition corresponds to the satisfaction of the essential boundary condition, while the second corresponds to the functions being square integrable. The space of functions satisfying the above two conditions is denoted by $\\varsigma$ and formally defined like\n\\[\\varsigma = \\left\\{u_i\\left| {u_i} \\in H, {u_i} = \\bar{u}_i \\in S_u \\right. \\right\\} \\enspace .\\]\n\nOn the other hand, in order to validate the introduced trial functions we also need testing functions $w_i$ also called in the FEM literature weighting or distribution functions. These functions are arbitrary apart from having to satisfy the following conditions:\n\\begin{align*}\n&w_i = 0 \\quad in \\quad {S_u}\\\\\n&\\intL_S \\left(\\pdv{w_i}{x_j}\\right)^2 \\dd{S} < \\infty\n\\end{align*}\n\nIn what follows we formally denote the space of these functions by $V$ and define it like\n\\[V = \\left\\{ w_i\\left| w_i \\in H, u_i = w_i=0 \\in S_u \\right. \\right\\} \\enspace .\\]\n\n\\subsubsection{Weak form}\nHere we will show that the equilibrium statement represented in the differential formulation can be described in alternative forms. In such description the continuity requirement for the trial functions is weaker than in the strong form leading to the term \"weak\" statement. Here this alternative representation will be denoted like $\\{W\\}$ and it reads;\n\nGiven $f_i$, $t_i^{\\hat n}$ and ${\\bar u_i}$ find ${u_i}:V \\to \\mathbb{R}$ and $\\forall {w_i} \\in V$ such:\n\\[\\intL_V \\sigma_{ij} w_{i,j}\\, \\dd{V} - \\intL_V f_i w_i\\, dV - \\intL_{S_t} t_i^{\\hat n} w_i\\, \\dd{S} = 0\\]\n\n\\subsubsection*{Proof 1:}\nLet $u_i \\in \\varsigma $ be a solution to $\\{S\\}$ and let $w_i \\in V $. Forming the inner product of the equilibrium statement given in \\cref{eq:pde} with $w_i$ and forcing the integral over the domain to be zero we have\n\\[\\intL_V (\\sigma_{ij,j} + f_i ){w_i}\\, dV = 0 \\enspace ,\\]\nexpanding the terms in the integrand and integrating by parts the first term on the left we have\n\\begin{align*}\n&\\intL_V \\sigma _{ij,j} w_i\\, \\dd{V} + \\intL_V f_i w_i\\, \\dd{V} = 0 \\, ,\\\\\n& -\\intL_V w_{i,j} \\sigma _{ij}\\, \\dd{V} + \\intL_S \\sigma _{ij} \\hat{n}_j w_i\\, \\dd{S} + \\intL\\limits_V w_i f_i\\, \\dd{V} = 0\\, ,\n\\end{align*}\nsince $w_i \\in V$ it follows that $w_i = 0$ in $S_u$ from which\n\\begin{equation}\\label{eq:weak}\n\\intL_V \\sigma_{ij} w_{i,j}\\, \\dd{V} - \\intL_V f_i w_i\\, \\dd{V} - \\intL_{S_t} t_i^{\\hat n} w_i\\, \\dd{S} = 0\\, .\n\\end{equation}\n\nNow, considering that $u_i$ is solution of the strong form $\\{S\\}$ it must satisfy $u_i = \\bar u_{i} \\quad \\in \\quad S_u$ and as a result $u_i \\in \\varsigma$. On the other hand, since $u_i$ satisfies \\cref{eq:weak} $\\forall {w_i} \\in V$ we have that $u_i$ satisfies the definition of weak solution specified in $\\{ W \\}$.\n\n\\subsubsection*{Proof 2:}\nLet $u_i$ be a solution of $\\{W\\}$ and thus $u_i \\in \\varsigma$ which means that\n\\[u_i = \\bar u_{i} \\quad \\in \\quad S_u\\]\nand that it satisfies\n\\[\\intL_V \\sigma _{ij} w_{i,j}\\, \\dd{V} - \\intL_V f_i w_i\\, \\dd{V} - \\intL_{S_t} t_i^n w_i\\, \\dd{S} = 0 \\enspace ,\\]\nintegrating by parts,\n\\[-\\intL_V \\sigma_{ij,j} w_i \\dd{V} + \\intL_S \\sigma_{ij} n_j w_i \\dd{S} - \\intL_V f_i w_i \\dd{V} - \\intL_{S_t} {t_i^n} w_i \\dd{S} = 0\\]\n\nSince $w_i \\in V$ we have that $w_i=0$ in $S_u$ and therefore\n\\[\\intL_V w_i(\\sigma_{ij,j} + f_i) \\dd{V} + \\intL_{S_t} w_i( \\sigma_{ij} n_j - t_i^n ) \\dd{S} = 0 \\]\nfrom which\n\\begin{equation} \\label{equil_2}\n\\begin{split}\n&\\sigma_{ij,j} + f_i = 0 \\quad \\vb{x} \\in V \\\\\n&t_i^n = \\sigma_{ij} n_j \\quad \\forall \\vb{x} \\in S_t\\\\\n&{u_i} = \\bar{u}_i \\quad \\forall \\vb{x} \\in S_u\n\\end{split}\n\\end{equation}\n\nwhich is once again the strong form of the problem given in \\cref{eq:pde}.\n\n\\subsection{Simple wedge under self-equilibrated loads}\nConsider the double wedge of side $\\ell$ and internal angle $2 \\phi$ shown in \\cref{fig:WEDGE}. It is assumed to be contained in the $X-Y$ plane, with loading conditions satisfying a plane strain (or plane stress) idealization. The material is elastic with Lame constants $\\lambda$ and $\\mu$. The wedge is loaded by uniform tractions of intensity $S$ applied over its four faces in such a way that the wedge is self-equilibrated. We wish to find the closed-form elasticity solution for the stress, strain and displacement fields throughout the problem domain.\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=7cm]{wedge.pdf}\n\\caption{2D Self-equilibrated wedge.}\n\\label{fig:WEDGE}\n\\end{figure}\n\nUnder plane strain conditions the general 3D stress equilibrium equations (see \\cref{eq:pde}) reduce to:\n\\begin{equation}\n\\begin{aligned}\n&\\pdv{\\sigma_{xx}}{x} + \\pdv{\\tau_{xy}}{y}=0\\\\\n&\\pdv{\\tau_{xy}}{x} + \\pdv{\\sigma_{yy}}{y}=0\n\\end{aligned}\n\\label{eq:equilibrium}\n\\end{equation}\nwhile the kinematic relation (\\cref{eq:kin}) reads\n\\begin{equation}\n\\begin{aligned}\n\\epsilon_{xx}& = \\pdv{u}{x}\\\\\n\\epsilon_{yy}& = \\pdv{v}{y}\\\\\n\\gamma_{xy}& = \\pdv{u}{y} + \\pdv{v}{x}\n\\end{aligned}\n\\label{eq:strain}\n\\end{equation}\nwhere $u$ and $v$ are the horizontal and vertical displacements respectively.\n\n\\subsubsection*{Stress field}\n\nThe stress field can be obtained by simple inspection from the traction boundary conditions prescribed over the inclined surfaces yielding\n\\begin{align*}\n\\sum F_x &= 0 \\longrightarrow - \\ell S\\cos(\\phi) + \\sigma_{xx}\\ell \\sin(\\phi) = 0\\\\\n\\sum F_y &= 0 \\longrightarrow - \\ell S\\sin(\\phi) - \\sigma_{yy}\\ell \\cos(\\phi)=0\n\\end{align*}\nand the following stress solution:\n\\begin{equation}\n\\begin{aligned}\n\\sigma_{xx}& = S \\cot(\\phi)\\\\\n\\sigma_{yy}& = -S\\tan(\\phi)\\\\\n\\tau_{xy}& = 0\\, .\n\\end{aligned}\n\\label{eq:solution}\n\\end{equation}\n\nIn \\cref{eq:solution} the condition $\\tau_{xy}=0$ is due to the symmetries in the problem.\n\n\\subsubsection*{Traction boundary conditions}\nLet us verify that the above stress solution satisfies the traction BC using the expression:\n\\[t_i^{\\hat n} = \\sigma_{ij} \\hat n_{ij}.\\]\n\nDenoting the outward normals to the inclined surfaces of the wedge by $\\hat{n}^1$, $\\hat{n}^2$, $\\hat{n}^3$, $\\hat{n}^4$ these are given by:\n\\begin{align*}\n\\hat{n}^1 &= -\\sin(\\phi)\\hat{e}_{x}+\\cos(\\phi)\\hat{e}_{y}\\\\\n\\hat{n}^2 &= -\\sin(\\phi)\\hat{e}_{x}-\\cos(\\phi)\\hat{e}_{y}\\\\\n\\hat{n}^3 &= +\\sin(\\phi)\\hat{e}_{x}+\\cos(\\phi)\\hat{e}_{y}\\\\\n\\hat{n}^4 &= +\\sin(\\phi)\\hat{e}_{x}-\\cos(\\phi)\\hat{e}_{y} \\enspace ,\n\\end{align*}\nwhere $\\hat{e}_{x}$ and $\\hat{e}_{y}$ are the reference unit vectors. Now, the components of the traction vector follow directly like\n\\[t_i = \\sigma_{ij}\\hat{n}_j\\]\nthen over the face with normal $\\hat{n}^1$ we have\n\\begin{align*}\nt_{x} &= -S\\cos(\\phi)\\\\\nt_{y} &= -S\\sin(\\phi)\n\\end{align*}\nsimilarly, over the face with normal $\\hat{n}^2$\n\\begin{align*}\nt_{x} &= -S\\cos(\\phi)\\\\\nt_{y} &= +S\\sin(\\phi)\n\\end{align*}\nover the face with normal $\\hat{n}^3$ \n\\begin{align*}\nt_{x} &= +S\\cos(\\phi)\\\\\nt_{y} &= -S\\sin(\\phi)\n\\end{align*}\nand finally, over the face with normal $\\hat{n}^4$;\n\\begin{align*}\nt_{x} &= +S\\cos(\\phi)\\\\\nt_{y} &= +S\\sin(\\phi) \\enspace .\n\\end{align*}\n\n\\subsubsection*{Strain field}\nThe strain field can be obtained after using the stress solution found in \\cref{eq:solution} together with the constitutive law given by \\cref{eq:Hooke} which for a plane strain idealization takes the form:\n\\begin{equation}\n\\begin{aligned}\n\\epsilon_{xx}& = \\frac{1}{E}(\\sigma_{xx} - \\nu \\sigma_{yy}) \\\\\n\\epsilon_{yy}& = \\frac{1}{E}(\\sigma_{yy} - \\nu \\sigma_{xx}) \\\\\n\\gamma_{xy}& = \\frac{\\tau _{xy}}{\\mu}\n\\end{aligned}\n\\label{eq:cons model}\n\\end{equation}\nwhich for the particular case yields\n\\begin{equation}\n\\begin{aligned}\n\\epsilon_{xx}& = +\\dfrac{S}{E}\\left[\\cot(\\phi)+\\nu \\tan(\\phi)\\right] = +\\dfrac{S}{E}K_{1}(\\nu , \\phi)\\\\\n\\epsilon_{yy}& = -\\dfrac{S}{E}\\left[\\tan(\\phi)+\\nu \\cot(\\phi)\\right] = -\\dfrac{S}{E}K_{2}(\\nu , \\phi)\\\\\n\\gamma_{xy}& = 0.\n\\end{aligned}\n\\label{eq:strain part}\n\\end{equation}\n\n\\subsubsection*{Displacement field}\nThe displacement field is obtained after direct integration of the strains after using the fact that\n\\[\\dd{u}_i = \\epsilon_{ij} \\dd{x}_j + \\omega_{ij}\\dd{x}_j\\]\nand the condition $\\omega_{xy}=0$ also due to symmetries , as follows\n\\begin{align*}\nu &= +\\dfrac{S}{E} K_{1}(\\nu , \\phi)x + A\\\\\nv &= -\\dfrac{S}{E} K_{2}(\\nu , \\phi)y + B\n\\end{align*}\nand where $A$ and $B$ are integration constants.\n\nFrom the condition $u=0$ at $x=\\ell\\cos(\\phi)$ we have that\n\\[A=-\\dfrac{S}{E} K_{1}(\\nu , \\phi)\\ell\\cos(\\phi)\\]\nthen it follows that\n\\[u=\\dfrac{S}{E} K_{1}(\\nu , \\phi)(x-\\ell\\cos(\\phi)).\\]\n\nSimilarly, from the condition $v=0$ at $y=0$ we have that $B=0$ from which\n\\[v=-\\dfrac{S}{E} K_{2}(\\nu , \\phi)y\\]\n\n\\subsection{Variational formulation}\nIn this section we formulate the boundary value problem using the approach of the calculus of variations in which the governing PDEs and boundary conditions are obtained after finding the minimum (or maximum) of a functional according to a variational principle\\footnote{According to Wikipedia \\cite{wiki:variational_principle}\n\n\\begin{quotation}\nA variational principle is a scientific principle used within the calculus of variations, which develops general methods for finding functions which minimize or maximize the value of quantities that depend upon those functions. For example, to answer this question: ``What is the shape of a chain suspended at both ends?\" we can use the variational principle that the shape must minimize the gravitational potential energy.\n\nAccording to Cornelius Lanczos, any physical law which can be expressed as a variational principle describes an expression which is self-adjoint. These expressions are also called Hermitian. Such an expression describes an invariant under a Hermitian transformation.\n\\end{quotation}}.\n\nWe will see that the weak form (and therefore also the strong form) can be obtained alternatively through the process of finding extreme values for a functional. We will illustrate this idea for the general case of theory of elasticity and then we will present particular examples.\n\n\\subsubsection*{Some vague definitions in the calculus of variations}\nIn variational calculus a {\\bf functional} can be understood as a ``function\" having as independent variables or arguments a space of vector functions and producing as a result (or dependent variable) a scalar. For instance, in the particular case of the theory of elasticity such a ``function\" corresponds to the total potential energy functional $\\Pi$ given by;\n\\begin{equation}\\label{eq:Potential}\n \\Pi(u_i) = \\frac{1}{2}\\int\\limits_V \\sigma_{ij}\\varepsilon_{ij}\\dd{V} - \\int\\limits_V f_i\\, u_i\\dd{V} - \\int\\limits_S t_i^{(n)} u_i\\dd{S}\n\\end{equation}\nand where the first term in the left hand side corresponds to the internal strain energy, while the last two terms are the work done by the external body and traction forces. The above functional has as independent variables the displacement vector and its spatial derivatives. This is indicated by the presence of the displacement vector $u_i$ in the expression $\\Pi(u_i)$.\n\nIn variational calculus we are interested in finding a function $u_i$ that renders the functional $\\Pi$ a maximum or a minimum. In loose terms, the analogous to the differential operator in calculus of functions is now termed the variational operator $\\var$ (i.e., $\\var$ is analogous to $\\pdv{x_i}$). As such $\\var{\\Pi}$ acts over the function $u_i$ and its derivatives as follows\n\\[\\var{\\Pi} = \\pdv{\\Pi}{u_i}\\var{u_i} + \\pdv{\\Pi}{\\left( \\pdv{u_i}{x_j}\\right)} \\delta\\left(\\pdv{u_i}{x_j}\\right) + \\cdots + \\pdv{\\Pi}{\\left( \\pdv{^{n}u_i}{x_j \\cdots \\partial x_k}\\right)} \\delta\\left( \\pdv{^{n}u_i}{x_j \\cdots \\partial x_k}\\right) \\enspace .\\]\n\nThe following rules apply to the variational operator $\\delta$:\n\\begin{itemize}\n\\item For functionals $\\Pi$ and $\\Phi$ it follows that \\[\\var(\\Pi + \\Phi) = \\var{\\Pi} + \\var{\\Phi}\\]\n\\item For functionals $\\Pi$ and $\\Phi$ it follows that \\[\\var(\\Pi \\Phi ) = \\var{\\Pi} \\Phi + \\Pi \\var{\\Phi}\\]\n\\item For a functional $\\Pi$ and an integer $n$ it follows that \\[\\var(\\Pi^n) = n(\\Pi^{n - 1})\\var{\\Pi}\\]\n\\item For a functional $\\Pi$ it follows that \\[\\var{\\int \\Pi \\dd{x}} = \\int \\var{\\Pi} \\dd{x} \\]\n\\end{itemize}\n\nIf the variational operator is applied to the functional $\\Pi(u_i)$ it produces functions or variations in $u_i$ which are arbitrary and such $\\var{u_i} \\in V$ and $\\var{u_i} = 0$ in $S_u$.\n\nTo find an extreme function in the calculus of variations we proceed like in differential calculus. Here we compute the first variation of the functional $\\var{\\Pi}$ and solve the variational equation\n\\begin{equation}\\label{eq:vareq}\n \\delta \\Pi = 0\n\\end{equation}\nin the unknown function $u_i$. \n\nIn the particular case of the total potential energy functional $\\Pi$ this yields\n\\begin{equation}\\label{eq:ClaPVW}\n \\int\\limits_V \\sigma _{ij}\\var{\\varepsilon_{ij}}\\dd{V} - \\int\\limits_V f_i \\var{u_i} \\dd{V} - \\int\\limits_{S_t} t_i^{(n)} \\var{u_i}\\dd{S} = 0\n\\end{equation}\nwhere we recognize the weak form of the BVP stated previously. It becomes evident that the functions $\\delta {u_i}$ in \\cref{eq:ClaPVW} play the role of the test functions $w_i$ introduced in the weak form. On the other hand, since we have already shown that the weak and strong forms are equivalent we conclude that having the functional and the essential boundary conditions is equivalent to having the strong form of the problem.\n\n\n\\subsubsection*{Principle of minimum potential energy}\nIn the theory of elasticity the total potential energy $\\Pi$ is the result of adding the elastic strain energy which is stored in the body upon deformation and the potential energy (work) imparted to the body by the applied forces. The principle states that the body is in equilibrium when this total potential energy reaches a minimum. This is equivalent to stating that an equilibrium configuration is attained when an infinitesimal variation from the position of minimum potential energy involves null changes in energy. This implies the variational condition:\n\\begin{equation}\\label{vareq2}\n \\delta \\Pi = 0.\n\\end{equation}\n\nThe above principle leads to the so-called principle of virtual displacements stated as follows\\footnote{See Bathe pp 156.}:\n``The equilibrium of the body requires that for any compatible small virtual displacements satisfying the condition of being zero at $S_u$, imposed on the body in its state of equilibrium, the total internal virtual work is equal to the total external virtual work\"\n\\[\\int\\limits_V \\sigma _{ij}\\var{\\varepsilon_{ij}}\\dd{V} - \\int\\limits_V f_i\\delta u_i\\dd{V} - \\int\\limits_{S_t} t_i^{(n)}\\var{u_i}\\dd{S} = 0\\]\nwhere $\\var{u_i}$ are the virtual displacements and $\\var{ \\varepsilon_{ij}}$ are the corresponding virtual strains.\n\nComparing the virtual work principle with the weak formulation given in \\cref{eq:weak} we identify $\\var{u_i}$ with the test functions $w_i$. As such the PVW takes the form of a powerful tool to test if a body is in equilibrium for a given solution (represented by the trial functions). In what follows we illustrate the use of the principle through some examples corresponding to problems in Bathe's textbook.\n\n\\subsubsection*{Problem\\footnote{3.15 from Finite Element Procedures}}\nEstablish the differential equation of equilibrium of the problem shown and the boundary conditions. Determine whether the differential operator of the problem is symmetric and positive definite and prove your answer.\\cite{book:bathe}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=10cm]{{Bathe3.15}.pdf}\n \\caption{Rod with varying cross-sectional area. The Young's modulus is E.}\n \\label{fig:bathe3.15}\n\\end{figure}\n\n\n\\begin{align*}\n\\Pi &= \\frac{1}{2}\\int\\limits_0^L \\sigma_{xx}\\varepsilon_{xx}A(x)\\dd{x} + \\frac{1}{2}ku_0^2 - R u_L \\\\\n &= \\frac{1}{2}\\int\\limits_0^L EA(x)\\left( \\dv{u}{x}\\right)^2 \\dd{x} + \\frac{1}{2}ku_0^2 - R u_L \\enspace.\n\\end{align*}\n\nThe first variation for this functional is\n\\[\\var{\\Pi} = \\int\\limits_0^L EA(x)\\dv{u}{x}\\dv{\\var{u}}{x}\\dd{x} + k u_0\\var{u_0} - R\\var{u_L} \\enspace .\\]\n\nIf we integrate by parts, we obtain\n\\[\\var{\\Pi} = - \\int\\limits_0^L \\dv{x}\\left[EA(x)\\dv{u}{x}\\right]\\var{u}\\dd{x} + \\left. EA(x)\\dv{u}{x}\\var{u} \\right|_0^L + k{u_0}\\var{u_0} - R\\var{u_L}\\]\nfrom which\n\\begin{align*}\n&\\dv{x}\\left[EA(x)\\dv{u}{x}\\right] = 0\\\\\n&\\left. EA(x)\\dv{u}{x}\\right|_{x=0} = k u_0\\\\\n&\\left. EA(x)\\dv{u}{x}\\right|_{x=L} = R\n\\end{align*}\n\n\n\\subsubsection*{Problem: The Hu-Washizu Variational Principle\\footnote{4.35 from Finite Element Procedures}}\n\nConsider the Hu-Washizu functional:\n\\begin{equation}\n\\Pi^* = \\Pi - \\intL_V \\lambda_{ij}^\\varepsilon (\\varepsilon_{ij} - L_{ijk} u_k)\\dd{V} - \\intL_{S_u} \\lambda_i^u(u_i^{S_u} - \\bar{ u}_i)\\dd{S}\n\\label{eq:Hu}\n\\end{equation}\nwhere\n\\begin{itemize}\n\\item $\\Pi$: is the potential energy functional.\n\\item $L_{ijk}$ is a differential operator such $\\varepsilon_{ij} = L_{ijk} u_k$.\n\\item $S_u$ surface where essential boundary conditions are prescribed.\n\\item $\\lambda_{ij}^\\varepsilon $ and $\\lambda_i^u$ are Lagrange multipliers.\n\\end{itemize}\n\nUsing the condition $\\var{\\Pi} = 0$ derive for the interior of the body the equilibrium equations\n\\[\\sigma_{ij,j} + f_i = 0\\]\nthe strain-displacement relationship\n\\[\\varepsilon_{ij} = L_{ijk} u_k\\]\nand the constitutive equation\n\\[\\sigma_{ij} = C_{ijkl} \\varepsilon_{kl}\\]\nand at the surface of the body the relation between the stress tensor and the applied tractions vector at $S_t$\n\\[t_i = \\sigma_{ij} n_j\\]\nthe relation between the stress tensor and the unknown tractions vector (or reactions) at $S_u$\n\\[t_i = \\tilde{\\sigma_{ij}} n_j\\]\nand the essential boundary condition at $S_u$\n\\[u_i = \\tilde{u}_i\\]\nWe want to determine the Euler equations resulting from the condition $\\var{\\Pi}^* = 0$. Applying the variational operator we have\n\\begin{equation}\n\\begin{aligned}\n\\var{\\Pi}^*& = \\var{\\Pi} - \\intL_V \\var{\\lambda}_{ij}^{\\varepsilon} (\\varepsilon_{ij} - L_{ijk} u_k)\\dd{V}- \\intL_V \\lambda_{ij}^\\varepsilon (\\var{\\varepsilon}_{ij} - L_{ijk}\\var{u}_k)\\dd{V} \\\\\n&-\\intL_V \\var{\\lambda}_i^u (u_i^{S_u} - \\bar u_i)\\dd{S} - \\intL_V \\lambda _i^u \\var{u}_i^{S_u} \\dd{S}\n\\end{aligned}\n\\end{equation}\nand\n\\begin{equation}\n\\begin{aligned}\n\\var{\\Pi}^* =& \\intL_V C_{ijkl} \\varepsilon_{kl} \\var{ \\varepsilon}_{ij}\\dd{V} - \\intL_{S_t} t_i \\var{u_i} \\dd{S} - \\intL_V f_i \\var{u_i}\\dd{V}\\\\\n &- \\intL_V \\lambda_{ij}^{\\varepsilon} \\var{\\varepsilon}_{ij}\\dd{V} + \\intL_V \\lambda_{ij}^{\\varepsilon} L_{ijk}\\var{u_k}\\dd{V} - \\intL_V \\delta \\lambda_{ij}^\\varepsilon (\\varepsilon_{ij}\\\\\n &- L_{ijk} u_k)\\dd{V} - \\intL_S \\var{\\lambda}_i^u (u_i^{S_u} - \\bar {u}_i)\\dd{S} - \\intL_{S_u} \\lambda _i^u\\var{u_i}\\dd{S} = 0\n\\end{aligned}\n\\end{equation}\nusing\n\\[\\intL_V (\\lambda_{ij}^\\varepsilon \\var{u_i}){,_j}\\dd{V} = \\intL_V \\lambda_{ij}^\\varepsilon \\var{u}_{i,j}\\dd{V} + \\intL_V \\lambda_{ij,j}^\\varepsilon \\var{u_i}\\dd{V} \\]\n\nIn the above we can write\n\\begin{align*}\n\\intL_V \\lambda_{ij}^\\varepsilon L_{ijk}\\var{u_k}\\dd{V} & = \\intL_V (\\lambda_{ij}^\\varepsilon \\var{u_i})_{,j}\\dd{V} - \\intL_V \\lambda_{ij,j}^\\varepsilon \\delta {u_i}\\dd{V}\\\\\n& = \\intL_{S_t} \\lambda_{ij}^\\varepsilon \\var{u_i}\\hat {n}_j\\dd{S} - \\intL_V \\lambda_{ij,j}^\\varepsilon \\var{u_i}\\dd{V}\n\\end{align*}\ntherefore\n\\begin{align*}\n\\delta \\Pi^* =& \\intL_V (C_{ijkl}\\varepsilon_{kl} - \\lambda_{ij}^\\varepsilon) \\var{\\varepsilon}_{ij}\\dd{V} - \\intL_{S_t} t_i \\var{u}_i\\dd{S} - \\intL_V f_i \\var{u}_i\\dd{V}\\\\\n &+ \\intL_{S_t} \\lambda_{ij}^\\varepsilon \\var{u_i}\\hat{n}_j\\dd{S} - \\intL_V \\lambda_{ij,j}^\\varepsilon \\delta {u_i}\\dd{V} - \\intL_V \\delta \\lambda _{ij}^\\varepsilon (\\varepsilon_{ij}\\\\\n &- L_{ijk} u_k)\\dd{V} - \\intL_{S_u} \\var{\\lambda}_i^u (u_i^{S_u} - \\bar{u}_i)\\dd{S} - \\intL_{S_u} \\lambda_i^u \\var{u_i} \\dd{S} = 0\\\\\n=& \\intL_V (C_{ijkl}\\varepsilon_{kl} - \\lambda_{ij}^\\varepsilon) \\var{\\varepsilon}_{ij}\\dd{V}\n+ \\intL_{S_t} (\\lambda_{ij}^\\varepsilon \\hat{n}_j - t_i)\\delta {u_i}\\dd{S}\\\\\n &- \\intL_V (\\lambda_{ij,j}^\\varepsilon + f_i) \\var{u_i}\\dd{V}\n- \\intL_V (\\varepsilon_{ij}- L_{ijk} u_k) \\var{\\lambda} _{ij}^\\varepsilon\\dd{V}\\\\\n &- \\intL_{S_u} (u_i^{S_u} - \\bar{u}_i) \\var{\\lambda}_i^u\\dd{S} - \\cancel{\\intL_{S_u} \\lambda _i^u \\var{u}_i^{S_u}\\dd{S}} = 0\n\\end{align*}\n\nNow, imposing the conditions $\\var{\\varepsilon}_{ij} \\neq 0$, $\\var{\\lambda}_{ij}^\\varepsilon \\neq 0$, $var{u}_i \\neq 0$ in $S_t$, $\\var{u}_i \\neq 0$ in $V$ and $\\var{\\lambda}_i^u \\neq 0$ in $S_u$ we have\n\\begin{align}\n&\\lambda_{ij}^\\varepsilon = C_{ijkl} \\varepsilon_{kl}\\\\\n&\\varepsilon_{ij} = L_{ijk} u_k\\\\\nt_i &= \\lambda_{ij}^\\varepsilon \\hat{n}_j\\\\\n&\\lambda_{ij,j}^\\varepsilon + {f_i} = 0\\\\\n&u_i^{S_u} = \\bar{u}_i \\enspace .\n\\end{align}\n\n\\subsection{Weighted residual methods}\nThis section introduces the concept of residual or difference from zero in a differential equation once its solution is approximated. For that purpose we will take as prototype equation the one obtained as our general model of BVP (see \\cref{eq:GenPDE}) and recalled here for completeness\n\\begin{equation}\n\\rho(\\vb x) \\pdv{u(\\vb x, t)}{t} + \\mathcal{L}u(\\vb x, t) = \\rho (\\vb x)F(\\vb x, t) \\enspace .\n\\label{eq:GenPDE2}\n\\end{equation}\n\nWe will assume that the actual solution to the generalized BVP given by \\cref{eq:GenPDE2} is approximated by $\\tilde u(\\vb{x})$ through a superposition like\n\\begin{equation}\n\\tilde u (\\vb{x}) = {N^I}(\\vb{x}){u^I}\n\\label{basicsuper}\n\\end{equation}\nwhere ${N^I}(\\vb{x})$ are interpolating functions and $I$ denotes a superposition index varying like $I=1,2,...,K$ with $K$ being the number of points where the solution is known. In what follows we will use $u(\\vb{x})$ instead of $\\tilde u (\\vb{x})$ but will keep in mind that we are actually using the approximation given by \\cref{basicsuper}. Similarly, in order to keep the discussion simple for the time being we will drop the time effects reducing the generalized PDE to the simple form:\n\\begin{equation}\n\\mathcal{L}u(\\vb x) = \\rho (\\vb x)F(\\vb x) \\enspace .\n\\label{eq:GenPDE3}\n\\end{equation}\n\nNow, since we are using the approximation given by \\cref{basicsuper} this equation is not strictly satisifed but instead we will have the following ``unbalanced\" condition\n\\[\\mathcal{L}u(\\vb{x}) - \\rho (\\vb{x})F(\\vb{x}) \\equiv R \\ne 0\\]\nwhere the term $R$ corresponds to a residual error which is to be distributed throughout the solution domain. The so-called weighted residual methods differ in the form in which they distribute the residual between the different $K$ points conforming the computational domain.\n\nUsing \\cref{basicsuper} in \\cref{eq:GenPDE2} and the linearity in the differential operator yields\n\\[R = \\mathcal{L}({N^P}){u^P} - \\rho F\\, .\\]\n\nWe can see that the residual $R$ is a function defined over the domain of interest. The residual would be exactly zero for the solution of the differential equation, but it will not be zero in general. Thus, we want to make the function $R$ as close to zero as possible. To make $R$ as small as possible we need a function (a functional) where we can compare different approximation functions. After getting this functional we can minimize its value. For this minimization we could use the norm of the function, another option is to compute a weighted \\emph{average} of the function over the domain. This is what we call a weighted residual\n\\[\\Pi[u, w] = \\int\\limits_V w R(u) \\dd{V}\\, ,\\]\nand we want to minimize it by making\n\\[\\var{\\Pi}[u, w] = 0\\, ,\\]\n\nIn what follows we will consider different strategies to distribute or weight the residual $R$ over the computational domain.\n\n\\subsubsection{Galerkin method}\nIn the Galerkin scheme the interpolation functions are used also as weighting functions leading to:\n\\[\\int\\limits_V N^Q R\\dd{V} = 0 \\]\nor explicitly\n\\begin{equation}\n \\int\\limits_V N^Q \\mathcal{L} (N^P)\\dd{V} u^P = \\int\\limits_V N^Q\\rho F\\dd{V}\\, .\n \\label{eq:Galer}\n\\end{equation}\n\nImposing \\cref{eq:Galer} in the $K$ points conforming the computational domain or equivalently ranging $Q$ from $1$ to $K$ leads to the following system of algebraic equation\n\\begin{equation}\n{K^{QP}}{U^P} = {f^Q}\n\\label{eq:DGaler}\n\\end{equation}\n\nwhere $U^P$ is a vector that stores the point values of the function $u$ along te $K$ points of the computational domain, while $f^Q$ stores the corresponding point excitations.\n\n\\subsubsection{Least squares method}\nIn this method the integral of the square of the residual is minimized with respect to the $K$ point parameters or nodal values of the function. Accordingly,\n\\begin{align*}\n &\\pdv{u^I}\\int\\limits_V R^2 \\dd{V} = 0\\\\\n &\\int\\limits_V R \\pdv{R}{u^I} \\dd{V} = 0\\, ,\n\\end{align*}\n\nThe least squares method is a special case of the weighted residual method for the weight functions\n\\[w^I = \\pdv{R}{u^I}\\, .\\]\n\nExpanding the residual, and considering the operator $\\mathcal{L}$ as linear, we obtain\n\\begin{align*}\n &\\pdv{u^I}\\int\\limits_V [\\mathcal{L}(N^P u^P) - \\rho F]^2 \\dd{V} = 0\\\\\n &\\int\\limits_V [\\mathcal{L} N^P u^P - \\rho F] \\mathcal{L}(N^I) \\dd{V} = 0\\\\\n &\\int\\limits_V \\mathcal{L}(N^I) \\mathcal{L}(N^P) \\dd{V} u^P - \\rho \\int\\limits_V \\mathcal{L}(N^I) F \\dd{V} = 0\n\\end{align*}\nwhich can be written like\n\\begin{equation}\n K^{IP} U^P = f^I\n \\label{eq:Dsquares}\n\\end{equation}\n\n\\subsubsection{Collocation method}\nIn the collocation method the coefficients of the approximation are determined by forcing the residual to be exactly zero at $K$ points over the computational domain, i.e.,\n\\[\\mathcal{L}(N^I) u^I - \\rho F = 0\\, ,\\]\nor\n\\[\\mathcal{L}[N^I(x^J)] u^I - \\rho F(x^J) = 0\\, ,\\]\nwhere $J$ ranges between $1$ and $K$. This equation can be rewritten as a weighted-residual if we consider the residual to be $\\delta(x - x^I)$, the Dirac delta function over the selected points\n\nThe resulting system of algebraic equation can be written as\n\\begin{equation}\nK^{IP} U^P = f^I\\, .\n\\label{eq:Colo}\n\\end{equation}\n\n\\subsubsection{Subdomain method}\nThe zero value of the residual is imposed upon $K$ subdomains\n\\[\\int\\limits_{V^I} \\mathcal{L}(N^P)\\dd{V^I} u^P - \\rho \\int\\limits_{V^I} F\\dd{V^I} = 0 \\qquad I=1,\\cdots,K\\, .\\]\n\nFor instance, for the $N$-th element it follows that\n\\[\\int\\limits_{V^N} \\mathcal{L}(N^P)\\dd{V^N} u^P - \\rho \\int\\limits_{V^N} F\\dd{V^N} = 0 \\qquad P=1,\\cdots,K\\, .\\]\n\nApplying the equation over the $K$ subdomains leads to the discrete system;\n\\begin{equation}\nK^{IP} U^P = f^I \\quad \\quad I=1,\\cdots,K\\, .\n\\label{eq:Subdomain}\n\\end{equation}\n\n\\subsubsection{Ritz method}\nIt operates directly upon the variational statement of the problem. For a given functional\n\\[\\Pi = \\Pi (N^Q u^Q)\\]\nthe variational equation reads\n\\[\\var{\\Pi} \\equiv \\pdv{\\Pi}{u^Q} \\var{u^Q} = 0\\]\nfrom which\n\\[\\pdv{\\Pi}{u^Q} = 0\\, .\\]\n\n\\subsubsection*{Problem: Discretization of the generalized parabolic equation.}\nLet us consider the case of the generalized parabolic equation and its discretization following the Galerkin method\n\\[\\rho(\\vb x) \\pdv{u(\\vb x,t)}{t} + \\mathcal{L}u(\\vb x, t) = 0\\]\nwhich can also be written using indicial notation\n\\[\\pdv{x_i}\\left[{p(x)\\pdv{u}{x_i} \\right] + q(x)u + \\rho \\pdv{u}{t} = \\rho F.\\]\n\nAssuming that $p(x)=1$ yields\n\\begin{align*}\n &-\\int\\limits_V N^P N_{,ii}^Q \\dd{V} u^Q + \\int\\limits_V q N^P N^Q\\dd{V} u^Q + \\rho \\int\\limits_V N^P N^Q \\dd{V} v^Q\\\\\n &- \\rho \\int\\limits_V N^P F \\dd{V} = 0\\\\\n &\\int\\limits_V N_{,i}^P N_{,i}^Q \\dd{V} u^Q - \\int\\limits_S N^P N_{,i}^Q \\hat{n}_i \\dd{S} u^Q + \\int\\limits_V q N^P N^Q \\dd{V} u^Q\\\\\n &+ \\rho \\int\\limits_V N^P N^Q \\dd{V} v^Q - \\rho \\int\\limits_V N^P F \\dd{V} = 0 \\\\\n &\\int\\limits_V \\left(N_{,i}^P N_{,i}^Q + q N^PN^Q \\right)\\dd{V} u^Q + \\rho \\int\\limits_V N^P N^Q \\dd{V} V^Q =\\\\\n &\\int\\limits_S N^P N_{,i}^Q \\hat{n}_i \\dd{S} u^Q + \\rho \\int\\limits_V N^P F \\dd{V}\n\\end{align*}\nwhich can be written in discrete form as\n\\[K^{PQ} U^Q + C^{PQ} V^Q = f^p\\, .\\]\n\n\\subsubsection*{Problem: Discretization of Navier equations.}\nIn this case the differential equations are written as\n\\[(\\lambda + \\mu ){u_{j,ij}} + \\mu {u_{i,jj}} + {f_i} = 0 \\enspace .\\]\n\nWe can write the differential operator as\n\\[L_{ij} \\equiv (\\lambda + \\mu )\\pdv[2]{}{x_i}{x_j} + \\mu \\pdv[2]{}{x_k}{x_k}\\delta_{ij}\\]\n\\todo{Where was defined $r_i$. Some cohesion is missing here.}\n\\begin{align*}\n&r_i = - f_i\\\\\n&u_i = N_i^Q u^Q\\\\\n&\\mathcal{L}_{ij}(u_j) \\equiv (\\lambda + \\mu )(N_j^Q u^Q)_{,ij} + \\mu (N_j^Q u^Q)_{,kk}\\delta_{ij}\\\\\n&\\mathcal{L}_{ij}(u_j) \\equiv (\\lambda + \\mu )N_{j,ij}^Q u^Q + \\mu N_{i,kk}^Q u^Q\\\\\n&\\mathcal{L}_{ij}(u_j) \\equiv \\mathcal{L}_{ij}(N_j^Q) u^Q \\enspace .\n\\end{align*}\n\nIn the Galerkin scheme we use the trial function as weighting function.\n\\[R_i \\equiv L_{ij}(N_j^Q) u^Q + f_i\\]\nand we state\n\\[&\\int\\limits_V N_i^P R_i \\dd{V} = 0 \\qquad P=1,2,\\cdots N\\, .\\]\n\nThus\n\\begin{align*}\n&\\int\\limits_V N_i^P \\mathcal{L}_{ij} (N_j^K)\\dd{V} u^K + \\int\\limits_V N_i^P f_i \\dd{V} = 0 \\\\\n&(\\lambda + \\mu )\\int\\limits_V N_i^PN_{j,ij}^K \\dd{V} u^K + \\mu \\int\\limits_V N_i^PN_{i,kk}^K \\dd{V} u^K + \\int\\limits_V N_i^P f_i \\dd{V} = 0\n\\end{align*}\nintegrating by parts\n\\begin{align*}\n- (\\lambda + \\mu )\\int\\limits_V N_{i,j}^P N_{j,i}^K \\dd{V} u^K + (\\lambda + \\mu )\\int\\limits_S N_i^P N_{j,i}^K \\hat{n}_j \\dd{S} u^K - \\mu \\int\\limits_V N_{i,k}^P N_{i,k}^K \\dd{V} u^K \\\\\n+ \\mu \\int\\limits_S N_i^P N_{i,k}^K \\hat{n}_k \\dd{S} u^K + \\int\\limits_V N_i^P f_i \\dd{V} = 0\n\\end{align*}\nwhich can be written like\n\\[K^{PQ} U^Q = F^P\\]\nwhere\n\\[K^{PQ} = (\\lambda + \\mu )\\int\\limits_V N_{i,j}^P N_{j,i}^Q \\dd{V} + \\mu \\int\\limits_V N_{i,k}^P N_{i,k}^Q\\dd{V} \\]\nand\n\\[F^P = \\int\\limits_S N_i^P t_i^{(\\hat n)} \\dd{S} + \\int\\limits_V N_i^P f_i \\dd{V} = 0 \\, .\\]\n\n\\subsubsection*{Problem: Discretization of the wave equation.}\nIn this case the differential equation reads\n\\[\\div \\left[ \\frac{1}{\\rho} \\grad p(\\vb{x})\\right] - \\pdv{t} \\left(\\frac{1}{\\lambda} \\pdv{\\rho}{t}\\right) - q(\\vb{x}) = 0\\]\nwhere we recognize that\n\\[\\mathcal{L} \\equiv \\div \\left(\\frac{1}{\\rho} \\grad \\right) - \\pdv{t}\\left(\\frac{1}{\\lambda}\\pdv{t}\\right)\\, .\\]\n\nLet\n\\[p(x) = N^K p^K\\]\nthen\n\\[\\mathcal{L}(p) \\equiv \\vec \\nabla \\cdot \\left( {\\frac{1}{\\rho }\\vec \\nabla {N^K}{p^K}} \\right) - \\frac{\\partial }{{\\partial t}}\\left( {\\frac{1}{\\lambda }\\frac{{\\partial {N^K}{p^K}}}{{\\partial t}}} \\right)\\]\nor in index notation\n\\[\\mathcal{L}(p) \\equiv {\\left( {\\frac{1}{\\rho }N_{,i}^K} \\right)_{,i}}{p^K} - \\frac{\\partial }{{\\partial t}}\\left( {\\frac{1}{\\lambda }\\frac{{\\partial {N^K}}}{{\\partial t}}} \\right){p^K}\\]\nwhich is equivalent to\n\\[\\mathcal{L}(p) \\equiv \\mathcal{L}({N^K}){p^K}\\]\nusing the trial functions as weighting function and recalling the definition of the residual which in this case reads\n\\[R = \\mathcal{L}({N^K}){p^K} - q\\, ,\\]\nand yields\n\\[\\int\\limits_V {{N^J}RdV = 0} \\quad \\quad J=1,2,...,K \\]\n\n\\[\\int\\limits_V {{N^J}L({N^K})dV{p^K}} - \\int\\limits_V {{N^J}qdV} = 0\\]\n\n\\[\\int\\limits_V {{N^J}{{\\left( {\\frac{1}{\\rho }N_{,i}^K} \\right)}_{,i}}dV{p^K} - \\int\\limits_V {{N^J}\\frac{\\partial }{{\\partial t}}\\left( {\\frac{1}{\\lambda }\\frac{{\\partial {N^K}}}{{\\partial t}}} \\right)dV{p^K}} } - \\int\\limits_V {{N^J}qdV} = 0\\]\n\nIntegrating by parts the first term on the right-hand-side gives us\n\\[ - \\int\\limits_V {N_{,i}^J\\frac{1}{\\rho }N_{,i}^KdV} {p^K} + \\int\\limits_S {{N^J}\\frac{1}{\\rho }N_{,i}^K{{\\hat n}_i}dS{p^K}} = \\int\\limits_V {{N^J}\\frac{\\partial }{{\\partial t}}\\left( {\\frac{1}{\\lambda }\\frac{{\\partial {N^K}}}{{\\partial t}}} \\right)dV{p^K}} + \\int\\limits_V {{N^J}qdV} \\]\n\n\\[\\int\\limits_V {N_{,i}^J\\frac{1}{\\rho }N_{,i}^KdV} {p^K} + \\int\\limits_V {{N^J}\\frac{1}{\\lambda }{N^K}dV{{\\ddot p}^K}} = \\int\\limits_S {{N^J}\\frac{1}{\\rho }N_{,i}^K{{\\hat n}_i}dS{p^K}} + \\int\\limits_V {{N^J}qdV} \\]\n\n\\[{K^{JK}}{P^K} + {M^{JK}}{{\\ddot P}^K} + {f^J} = 0\\]\n\n", "meta": {"hexsha": "a864320930f602b0a1a18800695fa3edcaefd673", "size": 43563, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/bvp.tex", "max_stars_repo_name": "jomorlier/FEM-Notes", "max_stars_repo_head_hexsha": "3b81053aee79dc59965c3622bc0d0eb6cfc7e8ae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-15T01:53:14.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-15T01:53:14.000Z", "max_issues_repo_path": "src/bvp.tex", "max_issues_repo_name": "jomorlier/FEM-Notes", "max_issues_repo_head_hexsha": "3b81053aee79dc59965c3622bc0d0eb6cfc7e8ae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/bvp.tex", "max_forks_repo_name": "jomorlier/FEM-Notes", "max_forks_repo_head_hexsha": "3b81053aee79dc59965c3622bc0d0eb6cfc7e8ae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-05-25T17:19:53.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-25T17:19:53.000Z", "avg_line_length": 61.2700421941, "max_line_length": 759, "alphanum_fraction": 0.6831714987, "num_tokens": 15395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382200964034, "lm_q2_score": 0.7520125793176222, "lm_q1q2_score": 0.6478875790754096}} {"text": "\\documentclass{article}\n\\input{../general}\n\n\\MakeTitle{6}\n\n\\begin{document}\n\\maketitle\n\n\\gotosection{2}{5}\n\\subsection{Kernels, images, and the dimension formula}\n\n\\begin{exercise}{1}\n\\begin{enumerate}\n\\item $A\\Vect{v}_1 = \\Vect{0}$, $\\Vect{v}_1 \\in \\ker{A}$.\n\n$A\\Vect{v}_2 = \\xmat{2\\\\3\\\\3}$, $\\Vect{v}_2 \\notin \\ker{a}$.\n\n$A\\Vect{v}_3 = \\Vect{0}$, $\\Vect{v}_3 \\in \\ker{A}$.\n\n\\item Since \\FunSS{T}{\\mathbb{R}^5}{\\mathbb{R}^3}, $\\Vect{w}_4$ has the right height to be in the kernel of $T$, while $\\Vect{w}_1$ and $\\Vect{w}_3$ has the right height to be in the image of $T$.\n$$\\widetilde{T} = \\xmat{1&0&0&2&1\\\\0&1&0&5&0\\\\0&0&1&1&-1}$$\n\\begin{align*}\n x_1 &= -2x_4 - x_5 \\\\\n x_2 &= -5x_4 \\\\\n x_3 &= -x_4 + x_5\n\\end{align*}\n\nLet $x_4=1, x_5=0$, $\\xmat{-2\\\\-5\\\\-1\\\\1\\\\0}$ is in the kernel of $T$.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{2}\n\\begin{enumerate}\n\\item False, should span $\\mathbb{R}^m$.\n\\item True.\n\\item True.\n\\item False.\n\\item False, should be $n - m$.\n\\item False, unless $T$ is also injective.\n\\item False, unless $T$ is also injective.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{3}\n\\newcolumntype{L}{>{\\centering\\arraybackslash}m{4cm}}\n\\newcolumntype{I}{>{\\centering\\arraybackslash}m{4.2cm}}\n\\begin{center}\n\\begin{tabular}{|L|I|c|}\n\\hline\nrank $T$ & nullity $T$ & rank $T$ + nullity $T$\\\\\n\\hline\ndim image $T$ & dim ker $T$ & dim domain $T$ \\\\\n\\hline\n\\# pivotal columns of $T$ & \\# nonpivotal columns of $T$ & \\\\\n\\hline\n\\# linearly independent columns of $T$ && \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n`\\#' stands for ``no. of\".\n\\end{exercise}\n\n\\def \\img{\\mathrm{img}}\n\n\\begin{exercise}{5}\n\\begin{enumerate}\n\\item $T(\\Vect{0}) = \\Vect{0}$, therefore $\\Vect{0} \\in \\ker T$.\n For any $\\Vect{u}_1, \\Vect{u}_2 \\in \\ker T$, $T(\\Vect{u}_1 + \\Vect{u}_2) = T(\\Vect{u}_1) + T(\\Vect{u}_2) = \\Vect{0} + \\Vect{0} = \\Vect{0}$, therefore, $\\Vect{u}_1 + \\Vect{u}_2 \\in \\ker T$.\n For any $\\Vect{u} \\in \\ker T$, $c \\in \\mathbb{R}$, $T(c\\Vect{u}) = c T(\\Vect{u}) = c\\Vect{0} = \\Vect{0}$, therefore, $c\\Vect{u} \\in \\ker T$.\n Hence, $\\ker T$ is a subspace of $\\mathbb{R}^n$. \\rQED\n\n\\item $T(\\Vect{0}) = \\Vect{0}$, therefore $\\Vect{0} \\in \\img T$. For any $\\Vect{u}_1 = T(\\Vect{v}_1), \\Vect{u}_2 = T(\\Vect{v}_2) \\in \\img T$, $\\Vect{u}_1 + \\Vect{u}_2 = T(\\Vect{v}_1) + T(\\Vect{v}_2) = T(\\Vect{v}_1 + T\\Vect{v}_2) \\in \\img T$. For any $\\Vect{u} = T(\\Vect{v}) \\in \\img T$, $c \\in \\mathbb{R}$, $c\\Vect{u} = cT(\\Vect{v}) = T(c\\Vect{v}) \\in \\img T$. Therefore, $\\img T$ is a subspace of $\\mathbb{R}^m$. \\rQED\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{6}\n\\begin{enumerate}\n\\item $A = \\xmat{1&1&3\\\\2&2&6}$, $\\widetilde{A} = \\xmat{1&1&3\\\\0&0&0}$. A basis for the image: $\\xmat{1\\\\2}$. A basis for the kernel: $\\xmat{-1\\\\1\\\\0}$, $\\xmat{-3\\\\0\\\\1}$.\n\n\\item $A = \\xmat{1&2&3\\\\-1&1&1\\\\-1&4&5}$, $\\widetilde{A} = \\xmat{1&0&1/3\\\\0&1&4/3\\\\0&0&0}$. A basis for the image: $\\xmat{1&-1&-1}$, $\\xmat{2\\\\1\\\\4}$. A basis for the kernel: $\\xmat{-1/3\\\\-4/3\\\\1}$.\n\n\\item $A = \\xmat{1&1&1\\\\1&2&3\\\\2&3&4}$, $\\widetilde{A} = \\xmat{1&0&-1\\\\0&1&2\\\\0&0&0}$. A basis for the image: $\\xmat{1\\\\1\\\\2}$, $\\xmat{1\\\\2\\\\3}$. A basis for the kernel: $\\xmat{1\\\\-2\\\\1}$.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{8}\nFalse.\n\nLet $[f]$ be a matrix with all entries being zero. Then, for any $\\Vect{v} \\in \\mathbb{R}^m$, $f(\\Vect{v}) = 0$, therefore, $\\ker f = \\mathbb{R}^m$. Let $g$ be a non-surjective function, i.e. $n < m$, then $\\img g \\subset \\mathbb{R}^m$ and $\\img g \\neq \\mathbb{R}^m$. In other words, $\\img g$ can only be a proper subset of $\\ker g$. \\rQED\n\\end{exercise}\n\n\\begin{exercise}{9}\n\\begin{enumerate}\n\\item $T(a+bx+cx^2) = bx+4cx^2$, $[T] = \\cmat{0&0&0\\\\0&1&0\\\\0&0&4}$.\n\\item Let $A = [T]$, $\\widetilde{A} = \\cmat{0&1&0\\\\0&0&1\\\\0&0&0}$. The second and the third columns (pivotal columns) of $A$ then form the basis of $\\img T$, namely $\\cmat{0\\\\1\\\\0}$ and $\\cmat{0\\\\0\\\\4}$. $\\dim{\\ker T} = 1$, and a basis of $\\ker T$ is $\\cmat{1\\\\0\\\\0}$, which can be derived from either the method mentioned in the textbook or the fact that only constant polynomials are in the kernel.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{10}\nFix $k$, for the sake of clarity.\n\nThe problem can be seen as finding a matrix $A \\in M_{(1,k+1)}$ such that:\n$$\\xmat{c_0 & \\hdots \n& c_k}\\cmat{p(0)\\\\\\vdots\\\\p(k)} = \\int^k_0p(t)dt$$\n\nLet's call the vector a polynomial vector. In the subsection of ``Interpolation and the dimension formula\", the textbook has proved that for each polynomial $p \\in P_k$, there exists one and only one such polynomial vector $\\Vect{v} \\in \\mathbb{R}^{k+1}$, and for each polynomial vector $\\Vect{v} \\in \\mathbb{R}^{k+1}$, there exists one and only one $p \\in P_k$.\\footnote{Formula 2.5.15, and ``in particular, $T_k$ is invertible}\n\nFor $A$ to be exists, the relation between $\\cmat{p(0)\\\\\\vdots\\\\p(k)}$ and $\\int^k_0p(t)dt$ must be linear, because there is a bijective relation between matrix and linear transformation. To put it more explicit, \\FunSS{T}{\\mathbb{R}^{k+1}}{\\mathbb{R}} such that:\n\n$$T(\\cmat{p(0)\\\\\\vdots\\\\p(k)}) = \\int^k_0p(t)dt$$\n\nmust be a linear transformation.\n\nLet's denote the polynomial vector of a polynomial $p$ as $\\Vect{p}$. For any two polynomials $p, q \\in P_k$ such that,\n$p = \\sum_{i=0}^kp_ix^i$,\n$q = \\sum_{i=0}^kq_ix^i$.\nThe sum of $p$ and $q$, denoted as $+_{p,q}$, is\n$\\sum_{i=0}^k(p_i+q_i)x^i$. It's obvious that for each $x \\in \\mathbb{R}$, $+_{p,q}(x) = p(x) + q(x)$. Then we have:\n$$\\Vect{v}_{+_{p,q}} = \\sum_{i=0}^ke_i(p_i+q_i)i^i = \\sum_{i=0}^ke_ip_ii^i + \\sum_{i=0}^ke_iq_ii^i = \\Vect{v}_p + \\Vect{v}_q$$\nwhere $e_i$ is a standard basis vector, and:\n$$T(\\Vect{v}_{+_{p,q}})=\\int^k_0+_{p,q}(t)dt=\\int^k_0p(t)dt+\\int^k_0p(t)dt = T(\\Vect{v}_p) + T(\\Vect{v}_q)$$\n\nLet $*_{c,p} = \\sum_{i=0}^k cp_ix^i$ for any real number $c$. It is clear that $*_{c,p} = c\\sum_{i=0}^k p_ix^i$, or $*_{c,p}(x) = cp(x)$ for each $x \\in \\mathbb{R}$. Then we have:\n$$\\Vect{v}_{*_{c,p}} = c\\Vect{v}_p$$\n, and:\n$$T(\\Vect{v}_{*_{c,p}}) = \\int^k_0*_{c,p}(t)dt = \\int^k_0cp(t)dt = c\\int^k_0p(t)dt = T(\\Vect{v}_p)$$\n\nTherefore, $T$ is indeed a linear transformation, $A = [T]$, and $c_0,\\hdots,c_k$ do exist. \\rQED\n\\end{exercise}\n\n\\begin{exercise}{15}\nSince $AB$ is invertible, the linear transformation $T_{AB}$ is bijective. Then, due to $T_{AB} = T_A \\circ T_B$, $T_A$ has to be surjective, otherwise there exists a $\\Vect{v} \\in \\mathbb{R}^n$ such that $T_A(\\Vect{x}) = \\Vect{v}$ has no solution, which means $T_A \\circ T_B(\\Vect{x}) = \\Vect{v}$ also has no solution, contradicting the surjectiveness of $T_{AB}$. Because $T_A$ is surjective and $A$ is square, $A$ is invertible.\n\nNow assume that $T_B$ is not injective. We can find $\\Vect{v}_1, \\Vect{v}_2 \\in \\mathbb{R}^n$ such that $T_B(\\Vect{v}_1) = T_B(\\Vect{v}_2)$. As a result, $T_A \\circ T_B(\\Vect{v}_1) = T_A \\circ T_B(\\Vect{v}_2)$, which contradicts the injectiveness of $T_{AB}$. Therefore, $T_B$ is injective, and since $B$ is square, $B$ is invertible.\n\nThis completes the proof. \\rQED\n\\end{exercise}\n\\end{document}\n", "meta": {"hexsha": "6b54f982f6019bc115d69c75a4f5e87ab67c8b48", "size": 7000, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW6/main.tex", "max_stars_repo_name": "notcome/fa15-linear-algebra", "max_stars_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "HW6/main.tex", "max_issues_repo_name": "notcome/fa15-linear-algebra", "max_issues_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW6/main.tex", "max_forks_repo_name": "notcome/fa15-linear-algebra", "max_forks_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.0, "max_line_length": 431, "alphanum_fraction": 0.6181428571, "num_tokens": 2972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879312056025699, "lm_q2_score": 0.8221891370573388, "lm_q1q2_score": 0.6478284779949255}} {"text": "\\documentclass[11pt]{article}\n\\usepackage{mathptmx}\n\\usepackage[small,compact]{titlesec} \n\\usepackage{geometry} % See geometry.pdf to learn the layout options. There are lots.\n\\usepackage{url}\n\\usepackage{hyperref}\n\\usepackage{breakurl}\n\\geometry{letterpaper} % ... or a4paper or a5paper or ... \n%\\geometry{landscape} % Activate for for rotated page geometry\n\\usepackage[parfill]{parskip} % Activate to begin paragraphs with an empty line rather than an indent\n\\usepackage{graphicx}\n\\usepackage{amssymb, amsmath}\n\\usepackage{epstopdf}\n\n\\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}\n\n\\title{Raytracing Quadrics}\n\\author{Miles Macklin}\n%\\date{} % Activate to display a given date or no date\n\n\\renewcommand{\\v}[1]{\\ensuremath{\\mathbf{#1}}} % for vectors\n\\newcommand{\\uv}[1]{\\ensuremath{\\mathbf{\\hat{#1}}}} % for unit vectors\n\\newcommand\\ddx[1]{\\frac{\\partial#1}{\\partial \\v{x} }} \n\\newcommand\\dd[2]{\\frac{\\partial#1}{\\partial #2}} \n\n\n\\begin{document}\n\\maketitle\n\n\\section{Introduction}\n\nThis short article shows how to define and manipulate quadric surface equations in matrix form and intersect rays against them for rendering. The treatment here is inspired by \\cite{Sigg:2006:GRQ:2386388.2386396} and work by Simon Green at NVIDIA.\n\n\n\\section{Quadrics}\n\nThe quadratic equation in 3 variables can be written as:\n\n\\begin{equation}\nf(x,y,z) = Ax^2 +2Bxy+2Cxz+2Dx+Ey^2+2Fyz+Gy+Hz^2 +2Iz+J = 0\n\\end{equation}\n\nIt can also be written in matrix form using homogenous coordinates:\n\n\\begin{equation}\n\\v{x}^T\\v{Q}\\v{x} = 0\n\\end{equation}\n\nwhere $\\v{Q}$ is the matrix of coefficients:\n\n\\begin{equation}\n\\v{Q} = \n\\begin{bmatrix}\n A & B & C & D \\\\\n B & E & F & G \\\\\n C & F & H & I \\\\\n D & G & I & J\n \\end{bmatrix}\n \\qquad \n \\v{x} =\n \\begin{bmatrix}\n x \\\\ y \\\\ z \\\\ 1\n \\end{bmatrix}\n\\end{equation}\n\nBecause this matrix is symmetric it can be diagonalised by a matrix $\\v{T}$. With appropriate scaling, our matrix $\\v{Q}$ can be expressed as a diagonal matrix $\\v{D}$ with entries $\\pm 1$ or $0$:\n\n\\begin{equation}\n\\v{Q}= \\v{T}^{-T}\\v{D}\\v{T}^{-1}\n\\end{equation}\n\nWe refer to the basis that diagonalises $\\v{Q}$ as the parameter space.\n\n$\\v{Q}$ can may be transformed to a different basis by any affine transformation $\\v{M}$ by multiplying by $\\v{M}^{-1}$ on both sides. This is equivalent to moving a point $\\v{x}'$ back to the basis for $\\v{Q}$, \n\n\\begin{equation}\n\\v{Q'} = \\v{M}^{-T}\\v{Q}\\v{M}^{-1}.\n\\end{equation}\n%\n\\begin{equation}\n\\v{x}' = \\v{M}\\v{x}\n\\end{equation}\n%\n\\begin{equation}\n\\v{x'}^T\\v{Q'}\\v{x'} = \\v{x'}^T\\v{M}^{-T}\\v{Q}\\v{M}^{-1}\\v{x'} = \\v{x}^T\\v{Q}\\v{x}\n\\end{equation}\n\nAs we will soon see, view (or eye) space is a convenient space for ray-tracing. To transform our quadric equation to view space we multiply by the inverse model view matrix $\\v{MV}$:\n\n\\begin{equation}\n\\v{Q'} = \\v{MV}^{-T}\\v{Q}\\v{MV}^{-1} = \\v{MV}^{-T}\\v{T}^{-T}\\v{D}\\v{T}^{-1}\\v{MV}^{-1} = (\\v{MV}\\cdot\\v{T})^{-T}\\v{D}(\\v{MV}\\cdot\\v{T})^{-1}\n\\end{equation}\n\nTo ray trace the quadric we parameterise the view ray in the usual way,\n\n\\begin{equation}\n\\v{x}_v = \\v{o} + t\\v{d},\n\\end{equation}\n\nand transform this back to parameter space:\n\n\\begin{equation}\n\\v{x}_p = (\\v{MV}\\cdot\\v{T})^{-1}\\v{o} + t(\\v{MV}\\cdot\\v{T})^{-1}\\v{d} = \\v{o}_p + t\\v{d}_p\n\\end{equation}\n\nInserting this into our quadratic equation,\n\n\\begin{equation}\n\\v{x}^T_p\\v{Q}\\v{x}_p = (\\v{o}_p + t\\v{d}_p)^T\\v{D}(\\v{o}_p + t\\v{d}_p) = 0,\n\\end{equation}\n\nand expanding and gathering for $t$, we have:\n\n\\begin{equation}\nt^2(\\v{d}^T_p\\v{D}\\v{d}_p) + 2t(\\v{o}^T_p\\v{D}\\v{d}_p) + \\v{o}^T_p\\v{D}\\v{o}_p = 0\n\\end{equation}\n\nwhich we can solve with the quadratic formula.\n\nThe advantage of defining our rays in view space is that the ray origin, or eye position $\\v{o}$ is simply the homogenous zero vector:\n\n\\begin{equation}\n\\v{o} = \n\\begin{bmatrix}\n0 \\\\\n0 \\\\\n0 \\\\\n1\n\\end{bmatrix}\n\\end{equation}\n\nWhich simply extracts the last column of ($\\v{MV}\\cdot\\v{T})^{-1}$ when we form $\\v{o}_p$:\n\n\\begin{equation}\n\\v{o}_p = \\v{c}_4\n\\end{equation}\n\n\n\\bibliographystyle{alpha}\n\\bibliography{quadric}\n\n\n\\end{document} \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "173b7cb58c439a5fe08cadce9860a05a32403a00", "size": 4207, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "projects/quadric/quadric.tex", "max_stars_repo_name": "NTForked/sandbox", "max_stars_repo_head_hexsha": "f73d2902bfce3972b889c71f076c724063c180dd", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 105, "max_stars_repo_stars_event_min_datetime": "2015-01-02T13:44:07.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-07T14:08:45.000Z", "max_issues_repo_path": "projects/quadric/quadric.tex", "max_issues_repo_name": "NTForked/sandbox", "max_issues_repo_head_hexsha": "f73d2902bfce3972b889c71f076c724063c180dd", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2016-05-08T10:28:10.000Z", "max_issues_repo_issues_event_max_datetime": "2016-05-09T02:05:10.000Z", "max_forks_repo_path": "projects/quadric/quadric.tex", "max_forks_repo_name": "NTForked/sandbox", "max_forks_repo_head_hexsha": "f73d2902bfce3972b889c71f076c724063c180dd", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 23, "max_forks_repo_forks_event_min_datetime": "2015-02-12T03:38:10.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-07T14:09:13.000Z", "avg_line_length": 24.3179190751, "max_line_length": 247, "alphanum_fraction": 0.6446398859, "num_tokens": 1472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.78793120560257, "lm_q2_score": 0.8221891305219504, "lm_q1q2_score": 0.6478284728454892}} {"text": "\\chapter{Univalent logic}\n\n\n\\section{Propositional truncation}\n\nUsing the same idea of higher inductive types, we can also turn every type into a mere proposition via an operation we call \\emph{propositional truncation}. \n\n\\begin{defn}\nLet $A:\\UU$ be a type. We define the \\define{propositional truncation} $\\brck{A}:\\UU$ as a higher inductive type with constructors\n\\begin{align*}\n\\eta & : A\\to \\brck{A} \\\\\n\\mu & : \\prd{x,y:\\brck{A}} x=y.\n\\end{align*}\nThe induction principle of $\\brck{A}$ asserts that for any type family $P:\\brck{A}\\to\\type$, if we have\n\\begin{align*}\nf & : \\prd{x:A}P(\\eta(x)) \\\\\ng & : \\prd{x,y:\\brck{A}}{p:P(x)}{q:P(y)} \\trans{\\mu(x,y)}{p}=q,\n\\end{align*}\nthen we obtain a section $\\rec{\\brck{\\blank}}(f,g):\\prd{x:\\brck{A}}P(x)$ satisfying\n\\begin{align*}\n\\rec{\\brck{\\blank}}(f,g,\\eta(x)) & = f(x).\n\\end{align*}\n\\end{defn}\n\n\\begin{rmk}\nWe will not need a computation rule corresponding to the path constructor $\\mu$.\n\\end{rmk}\n\n\\begin{lem}\nFor any $A:\\UU$, the type $\\brck{A}$ is a proposition.\n\\end{lem}\n\n\\begin{proof}\nThe path constructor $\\mu:\\prd{x,y:\\brck{A}}x=y$ directly provides a proof that $\\brck{A}$ is a proposition.\n\\end{proof}\n\nThe following theorem asserts that for any map $f:A\\to P$ into a proposition $P$, there is a unique map $g:\\brck{A}\\to P$ such that $g\\circ\\eta=f$, as indicated in the following diagram\n\\begin{equation*}\n\\begin{tikzcd}\nA \\arrow[dr,\"\\forall\"] \\arrow[d,swap,\"\\eta\"] \\\\\n\\brck{A} \\arrow[r,densely dotted,swap,\"\\exists!\"] & P\n\\end{tikzcd}\n\\end{equation*}\nWe call this property the \\define{universal property of propositional truncation}. \n\n\\begin{thm}\nLet $A:\\UU$ be a type. Then for any proposition $P$, the map\n\\begin{equation*}\n\\lam{g} g\\circ \\eta : (\\brck{A}\\to P)\\to (A\\to P)\n\\end{equation*}\nis an equivalence.\n\\end{thm}\n\n\\begin{proof}\n\n\\end{proof}\n\n\\section{Surjective maps}\n\n\\begin{defn}\nWe say that a type $A$ is \\define{inhabited} if there is a term of type $\\brck{A}$. \n\\end{defn}\n\n\\begin{defn}\nA function $f:A\\to B$ is said to be surjective if its fibers are inhabited. \nMore explicitly, we define\n\\begin{equation*}\n\\mathsf{is\\usc{}surj}(f)\\defeq \\prd{b:B}\\brck{\\fib{f}{b}}. \n\\end{equation*}\n\\end{defn}\n\n\\section{First order logic in type theory}\n\\begin{table}\n\\caption{Logic in type theory}\n\\begin{center}\n\\begin{tabular}{ll}\n\\toprule\n\\emph{Logical connective} & \\emph{Interpretation in HoTT} \\\\\n\\midrule\n$\\top$ & $\\unit$ \\\\\n$\\bot$ & $\\emptyt$ \\\\\n$P\\land Q$ & $P\\times Q$ \\\\\n$P\\lor Q$ & $\\brck{P+Q}$ \\\\\n$P\\to Q$ & $P\\to Q$ \\\\\n$P\\leftrightarrow Q$ & $P=Q$ \\\\\n$\\neg P$ & $P\\to\\emptyt$ \\\\\n$\\forall x.P(x)$ & $\\prd{x:A}P(x)$ \\\\\n$\\exists x.P(x)$ & $\\brck{\\sm{x:A}P(x)}$ \\\\\n$\\exists! x.P(x)$ & $\\iscontr(\\sm{x:A}P(x))$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n\\end{table}\n\\begin{enumerate}\n\\item First order logic in type theory. Stress difference between $\\exists$ and $\\Sigma$.\n\\item Propositional extensionality\n\\end{enumerate}\n\n\\begin{exercises}\n\\item Show that\n\\begin{equation*}\n\\eqv{\\brck{A}}{\\prd{P:\\prop}(A\\to P)\\to P}.\n\\end{equation*}\n\\item For any $B:A\\to\\UU$, construct an equivalence\n\\begin{equation*}\n\\eqv{\\Big(\\exists a.\\brck{B(a)}\\Big)}{\\Big(\\brck{\\sm{a:A}B(a)}\\Big)}\n\\end{equation*}\n\\item \\label{also}(Mart\\'in Escard\\'o) For any two propositions $P$ and $Q$, define\n\\begin{equation*}\nP\\boxplus Q \\defeq ((P\\to Q)\\to Q)\\times ((Q\\to P)\\to P).\n\\end{equation*}\n\\begin{subexenum}\n\\item Show that $P\\lor Q\\to P\\boxplus Q$ and $P\\boxplus Q\\to\\neg(\\neg P\\land \\neg Q)$.\n\\end{subexenum}\n\\item Show that for any mere proposition $Q$, and any type $X$, the following are equivalent:\n\\begin{enumerate}\n\\item The map $(Q\\to X)\\to(\\emptyt\\to X)$ is an equivalence.\n\\item The type $X^Q$ is contractible.\n\\item $Q\\to\\iscontr(X)$.\n\\end{enumerate}\n\\item \\label{ex:brck_comp} Formulate the computation rule corresponding to the path constructor $\\mu$. That is, compute the type of $\\apd{\\rec{\\brck{\\blank}}(f,g)}{\\mu(x,y)}$, and find a canonical element in it.\n\\end{exercises}\n", "meta": {"hexsha": "c62418c3792d9e77ebb373a6a2c85d4d72eb4ab6", "size": 3941, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/logic.tex", "max_stars_repo_name": "tadejpetric/HoTT-Intro", "max_stars_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 333, "max_stars_repo_stars_event_min_datetime": "2018-09-26T08:33:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T23:50:15.000Z", "max_issues_repo_path": "Book/logic.tex", "max_issues_repo_name": "tadejpetric/HoTT-Intro", "max_issues_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2019-06-18T04:16:04.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-16T15:27:01.000Z", "max_forks_repo_path": "Book/logic.tex", "max_forks_repo_name": "tadejpetric/HoTT-Intro", "max_forks_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 30, "max_forks_repo_forks_event_min_datetime": "2018-09-26T09:08:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-16T00:33:50.000Z", "avg_line_length": 32.0406504065, "max_line_length": 211, "alphanum_fraction": 0.670641969, "num_tokens": 1411, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428947, "lm_q2_score": 0.8221891239865619, "lm_q1q2_score": 0.6478284595073159}} {"text": "\\hypertarget{naive-bayes-classification}{%\n\\chapter{Naïve Bayes\nClassification}\\label{naive-bayes-classification}}\n\nNaïve Bayes classification algorithm is based on Bayes' Theorem. The\ndataset is divided into two parts, namely, feature matrix and the\nresponse vector.\n\n\\hypertarget{bayes-theorem}{%\n\\section{Bayes' Theorem}\\label{bayes-theorem}}\n\nBayes' Theorem provides a way that we can calculate the probability of a\npiece of data belonging to a given class, given our prior knowledge.\nBayes' Theorem is stated as:\n\n\\[\n\\Large P(\\alpha |\\beta) = \\frac{P(\\beta |\\alpha) * P(\\alpha)}{P(\\beta)}\n\\]\n\nNaive Bayes is a classification algorithm for binary (two-class) and\nmulticlass classification problems. It is called Naive Bayes or idiot\nBayes because the calculations of the probabilities for each class are\nsimplified to make their calculations tractable.\n\n\\section{Bayes' Theorem Calculation}\n\nIn our example, we shall use \\texttt{sklearn.naive\\_bayes} to classify\non the basis of previous data wheater or not a person has gotten a flu\nor not. We shall also check the accuracy and generate confusion matrix\nof the model. However, to gain a general understanding, here is how it works:\n\nStep 1: Study the dataset Our dataset include 4 feature columns and 1\ntarget column.\n\n\\begin{verbatim}\nChills Runny_nose Headache Fever | Flu\n\\end{verbatim}\n\nStep 2: Calculate probability of target\n\n\\[ P(Flu|Y) = 10/14 = 0.714285714 \\] \\[ P(Flu|N) = 4/14 = 0.285714286 \\]\n\nStep 3: Calculate probability of feature columns for each\n\nFor each column of feature, calculate probability of all cases features\nX target. Such as - \\[ P(Chills=Y|Flu=Y) = 6/10 = .6 \\]\n\\[ P(Chills=Y|Flu=N) = 1/4 = .25 \\] And so on\\ldots{}\n\nStep 4:\nCalculate for the given case \\[\n\\Large P(\\alpha |\\beta) = \\frac{P(\\beta |\\alpha) * P(\\alpha)}{P(\\beta)}\n\\] Here, \n\\begin{enumerate}\n \\item \\(\\alpha\\), \\(\\beta\\) = Event \n \\item \\(P(\\alpha)\\), \\(P(\\beta)\\) = Probability of event occurring \n \\item \\(P(\\alpha |\\beta)\\) = Probability of \\(\\alpha\\) happening such that \\(\\beta\\) is true\n\\end{enumerate}\n\n\\section{Introduction}\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{1}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{k+kn}{import} \\PY{n+nn}{numpy} \\PY{k}{as} \\PY{n+nn}{np}\n\\PY{k+kn}{import} \\PY{n+nn}{matplotlib}\\PY{n+nn}{.}\\PY{n+nn}{pyplot} \\PY{k}{as} \\PY{n+nn}{plt}\n\\PY{k+kn}{import} \\PY{n+nn}{pandas} \\PY{k}{as} \\PY{n+nn}{pd}\n\\PY{k+kn}{from} \\PY{n+nn}{sklearn}\\PY{n+nn}{.}\\PY{n+nn}{naive\\PYZus{}bayes} \\PY{k+kn}{import} \\PY{n}{GaussianNB}\n\\PY{k+kn}{from} \\PY{n+nn}{sklearn}\\PY{n+nn}{.}\\PY{n+nn}{model\\PYZus{}selection} \\PY{k+kn}{import} \\PY{n}{train\\PYZus{}test\\PYZus{}split}\n\\PY{k+kn}{from} \\PY{n+nn}{sklearn}\\PY{n+nn}{.}\\PY{n+nn}{metrics} \\PY{k+kn}{import} \\PY{n}{confusion\\PYZus{}matrix}\\PY{p}{,} \\PY{n}{accuracy\\PYZus{}score}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{2}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{dataset} \\PY{o}{=} \\PY{n}{pd}\\PY{o}{.}\\PY{n}{read\\PYZus{}csv}\\PY{p}{(}\\PY{l+s+s1}{\\PYZsq{}}\\PY{l+s+s1}{./data/flu.csv}\\PY{l+s+s1}{\\PYZsq{}}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{3}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{X\\PYZus{}raw} \\PY{o}{=} \\PY{n}{dataset}\\PY{o}{.}\\PY{n}{iloc}\\PY{p}{[}\\PY{p}{:}\\PY{p}{,}\\PY{l+m+mi}{0}\\PY{p}{:}\\PY{o}{\\PYZhy{}}\\PY{l+m+mi}{1}\\PY{p}{]}\n\\PY{n}{y} \\PY{o}{=} \\PY{n}{dataset}\\PY{o}{.}\\PY{n}{iloc}\\PY{p}{[}\\PY{p}{:}\\PY{p}{,}\\PY{o}{\\PYZhy{}}\\PY{l+m+mi}{1}\\PY{p}{]}\n\\PY{n}{map\\PYZus{}dict} \\PY{o}{=} \\PY{p}{\\PYZob{}}\\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Y}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{:} \\PY{l+m+mi}{1}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{N}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{:} \\PY{l+m+mi}{0}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{No}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{:} \\PY{l+m+mi}{0}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Mild}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{:} \\PY{l+m+mi}{1}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Strong}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{:} \\PY{l+m+mi}{2}\\PY{p}{\\PYZcb{}}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{4}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{dataset}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]\n\\prompt{Out}{outcolor}{4}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n Chills Runny\\_nose Headache Fever Flu\n0 Y N Mild Y N\n1 Y Y No N Y\n2 Y N Strong Y Y\n3 N Y Mild Y Y\n4 N N No N N\n5 N Y Strong Y Y\n6 N Y Strong N N\n7 Y Y Mild Y Y\n8 N Y Strong Y Y\n9 Y Y Mild Y Y\n10 N N No N N\n11 Y Y Strong Y Y\n12 Y N Strong Y Y\n13 Y N Mild Y Y\n\\end{Verbatim}\n\\end{tcolorbox}\n \n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{5}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{X} \\PY{o}{=} \\PY{n}{pd}\\PY{o}{.}\\PY{n}{DataFrame}\\PY{p}{(}\\PY{p}{)}\n\\PY{k}{for} \\PY{n}{x} \\PY{o+ow}{in} \\PY{n}{X\\PYZus{}raw}\\PY{p}{:}\n \\PY{n}{X}\\PY{p}{[}\\PY{n}{x}\\PY{p}{]} \\PY{o}{=} \\PY{n}{X\\PYZus{}raw}\\PY{p}{[}\\PY{n}{x}\\PY{p}{]}\\PY{o}{.}\\PY{n}{map}\\PY{p}{(}\\PY{n}{map\\PYZus{}dict}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{6}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{X}\\PY{o}{.}\\PY{n}{head}\\PY{p}{(}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]\n\\prompt{Out}{outcolor}{6}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n Chills Runny\\_nose Headache Fever\n0 1 0 1 1\n1 1 1 0 0\n2 1 0 2 1\n3 0 1 1 1\n4 0 0 0 0\n\\end{Verbatim}\n\\end{tcolorbox}\n \n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{7}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{y}\\PY{o}{.}\\PY{n}{head}\\PY{p}{(}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]\n\\prompt{Out}{outcolor}{7}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n0 N\n1 Y\n2 Y\n3 Y\n4 N\nName: Flu, dtype: object\n\\end{Verbatim}\n\\end{tcolorbox}\n \n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{8}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{X\\PYZus{}train}\\PY{p}{,} \\PY{n}{X\\PYZus{}test}\\PY{p}{,} \\PY{n}{y\\PYZus{}train}\\PY{p}{,} \\PY{n}{y\\PYZus{}test} \\PY{o}{=} 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\\PY{n}{accuracy\\PYZus{}score}\\PY{p}{(}\\PY{n}{y\\PYZus{}test}\\PY{p}{,} \\PY{n}{predicted\\PYZus{}y}\\PY{p}{)}\\PY{p}{)}\n\\PY{n+nb}{print}\\PY{p}{(}\\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Confusion Matrix:}\\PY{l+s+se}{\\PYZbs{}n}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{n}{confusion\\PYZus{}matrix}\\PY{p}{(}\\PY{n}{y\\PYZus{}test}\\PY{p}{,} \\PY{n}{predicted\\PYZus{}y}\\PY{p}{)}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{Verbatim}[commandchars=\\\\\\{\\}]\nAccuracy: 0.8571428571428571\nConfusion Matrix:\n [[0 1]\n [0 6]]\n \\end{Verbatim}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{12}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{k}{def} \\PY{n+nf}{make\\PYZus{}prediction}\\PY{p}{(}\\PY{n}{case}\\PY{p}{)}\\PY{p}{:}\n \\PY{n}{df} \\PY{o}{=} \\PY{n}{pd}\\PY{o}{.}\\PY{n}{DataFrame}\\PY{p}{(}\\PY{n}{case}\\PY{p}{)}\\PY{p}{[}\\PY{l+m+mi}{0}\\PY{p}{]}\\PY{o}{.}\\PY{n}{map}\\PY{p}{(}\\PY{n}{map\\PYZus{}dict}\\PY{p}{)}\n \\PY{k}{return} \\PY{n}{model}\\PY{o}{.}\\PY{n}{predict}\\PY{p}{(}\\PY{p}{[}\\PY{n}{df}\\PY{p}{]}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{13}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n+nb}{print}\\PY{p}{(}\\PY{n}{make\\PYZus{}prediction}\\PY{p}{(}\\PY{p}{[}\\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Y}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{N}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Mild}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Y}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{]}\\PY{p}{)}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{Verbatim}[commandchars=\\\\\\{\\}]\n['Y']\n \\end{Verbatim}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{14}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n+nb}{print}\\PY{p}{(}\\PY{n}{make\\PYZus{}prediction}\\PY{p}{(}\\PY{p}{[}\\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{N}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{N}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{Mild}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{,} \\PY{l+s+s2}{\\PYZdq{}}\\PY{l+s+s2}{N}\\PY{l+s+s2}{\\PYZdq{}}\\PY{p}{]}\\PY{p}{)}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{Verbatim}[commandchars=\\\\\\{\\}]\n['N']\n \\end{Verbatim}", "meta": {"hexsha": "a0fe2f4ddd3c898baae5b1ccbbd7cb3507b6f09d", "size": 10838, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MCA/Machine Learning/Project file/Tex source/Naive Bayes.tex", "max_stars_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_stars_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2018-03-13T12:34:51.000Z", "max_stars_repo_stars_event_max_datetime": "2018-10-02T18:54:22.000Z", "max_issues_repo_path": "MCA/Machine Learning/Project file/Tex source/Naive Bayes.tex", "max_issues_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_issues_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "MCA/Machine Learning/Project file/Tex source/Naive Bayes.tex", "max_forks_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_forks_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.1759259259, "max_line_length": 515, "alphanum_fraction": 0.6004797933, "num_tokens": 4396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333246035907932, "lm_q2_score": 0.7772998714925403, "lm_q1q2_score": 0.6477431072826957}} {"text": "% When using TeXShop on the Mac, let it know the root document. The following must be one of the first 20 lines.\n% !TEX root = ../design.tex\n\n\\chapter[Sandwich Estimators]{Sandwich Estimators}\n\n\n\\section{Introduction}\nGiven a regression setup of $n$ data points, each defined by a feature vector $x_i$ and a category $y_i$, we assume that $y_i$ is controlled by a $k$-dimensional parameter vector $\\theta$. Generally, we are interested in finding the values of $\\theta$ that best predict $y_i$ from $x_i$, with \\textit{best} being defined as the values that maximize some likelihood function $L(y,x,\\theta)$. The maximization is typically solved using the derivative of the likelihood $\\psi$ and the Hessian $H$. More formally, $\\psi$ is defined as \n\\begin{align}\n \\psi(y_i,x_i, \\theta) = \\frac{\\partial l(x_i,y_i,\\theta)}{\\partial \\theta}\n\\end{align} \nand $H$ is defined as\n\\begin{align}\nH(y,x, \\theta) = \\frac{\\partial^2 L(x,y,\\theta)}{\\partial \\theta^2}.\n\\end{align} \n\n\n\nIn addition to the values of $\\theta$, we may also be interested in the covariance matrix $S(\\theta)$ of $\\theta$. This can be expressed in a \\textit{sandwich formulation}, of the form\n\\begin{align}\nS(\\theta) = B(\\theta) M(\\theta) B(\\theta). \n\\end{align}\nThe $B(\\theta)$ matrix is commonly called the \\textit{bread}, whereas the $M(\\theta)$ matrix is the \\textit{meat}. \n\n\\subsection{The Bread}\nComputing $B$ is relatively straightforward, \n\\begin{align}\nB(\\theta) = n\\left(\\sum_i^n -H(y_i, x_i, \\theta) \\right)^{-1}\n\\end{align}\n\n\\subsection{The Meat}\nThere are several choices for the $M$ matrix, each with different robustness properties. The estimators we are interested in for this implementation are the Huber/White estimator, and the clustered estimator. \n\nIn the Huber/White estimator, the matrix $M$ is defined as\n\\begin{align}\nM_{H} = \\sum_i^n \\psi(y_i,x_i, \\theta)^T \\psi(y_i,x_i, \\theta).\n\\end{align}\n\n%The Huber/White estimator is a diagonal matrix defined as\n%\\begin{align}\n%M_W =\\frac{1}{n} X^T \\left(\n% \\begin{array}{cccc}\n% r(y_1, x_1^T\\theta) & 0&\\ldots &0 \\\\\n% 0&r(y_2, x_2^T\\theta) &\\dots& \\vdots \\\\\n%\t\\vdots & \\vdots & \\ddots&\\vdots \\\\\n% 0 & \\ldots & \\ldots &r(y_n, x_n^T\\theta)\n% \\end{array} \\right) X.\n%\\end{align}\n%The matrix $X$ is the data matrix, and $r$ is the residual function. The residual function is defined by the application, but it has the relationship\n%\\begin{align}\n% \\psi(y_i,x_i, \\theta) = r(y_i, x_i^T\\theta)x_i. \n%\\end{align} \n%This means that it can be computed in the general case as \n%\\begin{align}\n% r(y_i, x_i^T\\theta) = \\frac{\\psi(y_i,x_i, \\theta)}{x_i}\n%\\end{align}\n%\n%In the case of the clustered sandwich estimator, $M$ is defined as \n\n\n\n\n%In addition to the computational expense of computing the Hessian, the inverse Hessian must be computed as well. Unfortunately, matrix inversion is an expensive operation, and cannot be parallelized. \n\n", "meta": {"hexsha": "cf1220fd18b46bab667c39d28345d91c4cafd78c", "size": 2916, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/design/modules/sandwich-estimators.tex", "max_stars_repo_name": "fmcquillan99/apache-madlib", "max_stars_repo_head_hexsha": "e2dea62d1eadc7f662f2d926c71f42332f414ca0", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2018-09-18T07:44:22.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-14T19:45:18.000Z", "max_issues_repo_path": "doc/design/modules/sandwich-estimators.tex", "max_issues_repo_name": "fmcquillan99/apache-madlib", "max_issues_repo_head_hexsha": "e2dea62d1eadc7f662f2d926c71f42332f414ca0", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2018-09-06T05:50:17.000Z", "max_issues_repo_issues_event_max_datetime": "2018-09-06T05:50:17.000Z", "max_forks_repo_path": "doc/design/modules/sandwich-estimators.tex", "max_forks_repo_name": "fmcquillan99/apache-madlib", "max_forks_repo_head_hexsha": "e2dea62d1eadc7f662f2d926c71f42332f414ca0", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-09-03T20:50:13.000Z", "max_forks_repo_forks_event_max_datetime": "2019-09-03T20:50:13.000Z", "avg_line_length": 44.8615384615, "max_line_length": 536, "alphanum_fraction": 0.7009602195, "num_tokens": 890, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333246077301781, "lm_q2_score": 0.7772998560157663, "lm_q1q2_score": 0.6477430976030625}} {"text": "\\documentclass[jcp,aip,amsmath]{revtex4-1}\n\\usepackage{xcolor,array}\n\\usepackage{color}\n\\usepackage{txfonts}\n\n\\newcommand{\\half}{\\textstyle \\frac{1}{2}}\n\\newcommand{\\quarter}{\\textstyle \\frac{1}{4}}\n\n\n% note, to complile, do something like:\n%#!/bin/bash\n%file=$1\n%rm ``$file''.aux\n%latex ``$file''.tex\n%bibtex ''$file\"\n%latex ``$file\".tex\n%latex ''$file\".tex\n%dvips -Ppdf -o ``$file\".ps \"$file\".dvi\n%ps2pdf ``$file\".ps\n%echo `` done compiling $file''\n%mupdf \"$file.pdf\" &\n\n\n\n\\begin{document}\n\n\n\n\\title{Evolvr Equations}\n\\author{Matthew K. MacLeod}\n\\maketitle\n\\today\n\n\\section{Maths.ex}\nThe code for following equations is located in lib/evolvr/maths.ex\n\n\\subsection{Linear Algebra}\ndot product, ie inner product, for two vectors,\n\\begin{align}\n\\mathbf{A} \\cdot \\mathbf{B} = A^\\dag B = \\sum_i^n A_i * B_i\n\\end{align}\n\nThe Euclidean norm, or distance is defined as\n\\begin{align}\nd_{euclid} (x,y) = \\left( \\sum_i^n (x_i - y_i)^2 \\right)^{1/2}\n\\end{align}\nThe Minkowski distance is defined as\n\\begin{align}\nd_{minkowski} (x,y,p) = \\left( \\sum_i^n |x_i - y_i|^{p} \\right)^{1/p}\n\\end{align}\nIt is interesting to note when p is 1, the taxicab distance (L1 norm) is recovered and\nthe L2 norm is recovered when p is 2.\n\nThe Mahalanobis distance is defined as\n\\begin{align}\nd_{mahalanobis} (x,y) = \\left( \\sum_i^n \\frac{(x_i - y_i)^2}{s^2_i} \\right)^{1/2} = \\left( \\sum_i^n \\frac{(x_i - y_i)^2}{ \\left(1/n *(x_i - \\bar{x})*(y_i-\\bar{y})\\right)^2} \\right)^{1/2}\n\\end{align}\n\nThe angle, in degrees, is defined\n\n\\begin{align}\n\\theta = \\frac{\\mathbf{v} \\cdot \\mathbf{w}}{||\\mathbf{v}||*||\\mathbf{w}||} * \\frac{180}{\\pi}\n\\end{align}\n\n\\subsection{Statistics}\n\nstandard deviation, $\\sigma$, is defined as, the square root of the variance\n\\begin{align}\n\\sigma = \\sqrt{v} = \\sqrt{\\frac{1}{n}\\sum_i^n (x_i - \\mu)^2}\n\\end{align}\nwhere the mean, $\\mu$, is defined as\n\\begin{align}\n\\mu = \\frac{1}{n} \\sum_i^n x_i\n\\end{align}\n\nthe sample covariance is measures how two vectors vary together,\n\\begin{align}\n\\bar{\\bar{q}}= cov(x,y) = \\frac{1}{n-1} \\sum_i (x_i - \\bar{x}) (y_i - \\bar{y})\n\\end{align}\n\nThe reason the sample covariance matrix has $n-1$ in the denominator rather than $n$ is essentially that the population mean is not known and is replaced by the sample mean.\n\n\n\n\n\\subsection{Probability}\n The normal probability distribution (normal\\_pdf in code), the famous bell-shaped curve is\ndefined as\n\\begin{align}\nf(x|\\mu,\\sigma) = \\frac{1}{\\sqrt{2\\pi\\sigma}}\\mathrm{exp}\\left(-\\frac{(x-\\mu)^2}{2\\sigma^2}\\right)\n\\end{align}\nand $\\mu$ is again the mean.\n\nThe beta distribution can be used for Bayesian inference (conditional probabilities) and is defined\n\\begin{align}\n\\mathrm{beta\\_pdf} = \\frac{x^{\\alpha-1}(1-x)^{\\beta-1}}{B(\\alpha,\\beta)}\n\\end{align}\n\nwhere $B(\\alpha,\\beta)$ is a normalization constant to ensure that the total probability integrates to 1.\nand is\n\\begin{align}\nB(\\alpha,\\beta) = \\frac{\\Gamma(\\alpha) \\Gamma(\\beta) }{\\Gamma(\\alpha + \\beta)}\n\\end{align}\nThis function uses the gamma function, which is the generalization of a factorial, and which is approximated in the code using Windschitl reworking of Stirlings as follows\n\\begin{align}\n\\Gamma(t) = \\int_0^{\\infty} x^{t-1} e^{-x} dx \\approx \\sqrt{\\frac{2\\pi}{t}}\\left(\\frac{1}{e}\\left(t + \\frac{1}{12t -\\frac{1}{10t}}\\right)\\right)^t\n\\end{align}\n\nNote, especially for easy partial testing, positive integers should follow\n\\begin{align}\n\\Gamma(n)=(n-1)!\n\\end{align}\n\n\n\\subsection{Machine Learning}\n\n\n\\subsubsection{Gradient Descent}\nA numerical approximation to the derivative of a function is given by the finite differences method,\nalso known as the difference quotient is given by,\n\n\\begin{align}\nf' = f'(x) = \\frac{df(x)}{dx} \\approx \\frac{f(x+h) -f(x)}{h}\n\\end{align}\n\nnote that this method yields the total derivative of a function, not a partial derivative.\nAlso can be made slightly more accurate easily.\n\n\n\n\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "4a21dd61f74139305528f8dd6ae0c562cef0c45a", "size": 3890, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/evolvr_equations.tex", "max_stars_repo_name": "matthewmacleod/evolvr", "max_stars_repo_head_hexsha": "6392f0e03a8218ea70e86afdd187dcf728155072", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2015-12-29T21:13:23.000Z", "max_stars_repo_stars_event_max_datetime": "2017-08-14T22:35:25.000Z", "max_issues_repo_path": "doc/evolvr_equations.tex", "max_issues_repo_name": "matthewmacleod/evolvr", "max_issues_repo_head_hexsha": "6392f0e03a8218ea70e86afdd187dcf728155072", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/evolvr_equations.tex", "max_forks_repo_name": "matthewmacleod/evolvr", "max_forks_repo_head_hexsha": "6392f0e03a8218ea70e86afdd187dcf728155072", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.3941605839, "max_line_length": 186, "alphanum_fraction": 0.6925449871, "num_tokens": 1303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8333246015211008, "lm_q1q2_score": 0.6477430970758065}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% CS671: Machine Learning\n% Copyright 2015 Pejman Ghorbanzade \n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section*{Question 1}\n\nA rhombus $R_{x_0, y_0, c, d}$ is a quadrilateral which has the vertices $(x_0-c, y_0)$, $(x_0, y_0-d)$, $(x_0+c, y_0)$, $(x_0, y_0+d)$ as shown in Figure \\ref{fig11}.\nProve that the class of rhombi in $\\mathbb{R}^2$ for which the ratio $c/d$ is a constant is a PAC-learnable.\n\n\\begin{figure}[H]\n\\centering\n\\begin{tikzpicture}\n\\begin{axis}[axis lines = middle, axis equal, ticks = none, xmin = -.5, xmax = 5.5, ymin = 1, ymax = 3, xlabel = {$x$}, ylabel = {$y$}, clip=false]\n\\addplot[domain=1:4]{(7-x)/3};\n\\addplot[domain=4:7]{(x-1)/3};\n\\addplot[domain=1:4]{(x+5)/3};\n\\addplot[domain=4:7]{(13-x)/3};\n\\addplot[loosely dashed] coordinates {(4,0.5) (4,3.5)};\n\\addplot[loosely dashed] coordinates {(0.5,2) (7.5,2)};\n\\addplot[mark=*] coordinates {(1,2)} node[above left] {$(x_0-c,y_0)$};\n\\addplot[mark=*] coordinates {(4,1)} node[below left] {$(x_0, y_0-d)$};\n\\addplot[mark=*] coordinates {(4,3)} node[above right] {$(x_0, y_0+d)$};\n\\addplot[mark=*] coordinates {(7,2)} node[below right] {$(x_0+c, y_0)$};\n\\end{axis}\n\\end{tikzpicture}\n\\caption{Rhombus having vertices $(x_0-c, y_0)$, $(x_0, y_0-d)$, $(x_0+c, y_0)$, $(x_0, y_0+d)$}\\label{fig11}\n\\end{figure}\n\n\\subsection*{Solution}\n\nSet of possible examples $\\mathcal{X}$ is $\\mathbb{R}^2$.\n$\\mathcal{R}$ is set of all rhombi with ratio $c/d$.\nTherefore, the concept $\\mathcal{C}$ would be a particular rhombus with ratio $c/d$.\n\nSuppose the rhombus shown in Figure \\ref{fig11} is a concept $c \\in \\mathcal{C}$ which is not known by the learner.\n\nAs the hypothesis $h_S$, formed by the learner, is also a rhombus, Figure \\ref{fig12} can be constructed in which $h_S$ is the rhombus formed by intersection of dashed lines.\nAreas enclosed by dashed lines and different edges of the concept rhombus are defined as $r_1$, $r_2$, $r_3$ and $r_4$.\n\n\\begin{figure}[H]\n\\centering\n\\begin{tikzpicture}\n\\begin{axis}[axis lines = middle, axis equal, ticks = none, xmin = -.5, xmax = 5.5, ymin = 1, ymax = 3, xlabel = {$x$}, ylabel = {$y$}, clip=false]\n\n\\addplot[domain=1:4]{(7-x)/3};\n\\addplot[domain=4:7]{(x-1)/3};\n\\addplot[domain=1:4]{(x+5)/3};\n\\addplot[domain=4:7]{(13-x)/3};\n\n\\addplot[mark=*] coordinates {(1,2)} node[above left] {$(x_0-c,y_0)$};\n\\addplot[mark=*] coordinates {(4,1)} node[below left] {$(x_0, y_0-d)$};\n\\addplot[mark=*] coordinates {(4,3)} node[above right] {$(x_0, y_0+d)$};\n\\addplot[mark=*] coordinates {(7,2)} node[below right] {$(x_0+c, y_0)$};\n\n\\addplot[domain=1:6, dashed]{(8.5-x)/3};\n\\addplot[domain=3:8, dashed]{(12-x)/3};\n\\addplot[domain=3:8, dashed]{(x)/3};\n\\addplot[domain=1:6, dashed]{(x+4)/3};\n\n\\addplot[mark=none] coordinates {(2.8,1.7)} node[below right] {$r_1$};\n\\addplot[mark=none] coordinates {(2.8,2.7)} node[below right] {$r_2$};\n\\addplot[mark=none] coordinates {(5,1.8)} node[below right] {$r_3$};\n\\addplot[mark=none] coordinates {(5,2.6)} node[below right] {$r_4$};\n\n\\end{axis}\n\\end{tikzpicture}\n\\caption{Rhombus having vertices $(x_0-c, y_0)$, $(x_0, y_0-d)$, $(x_0+c, y_0)$, $(x_0, y_0+d)$}\\label{fig12}\n\\end{figure}\n\nSuppose the probability of the generalization error defined as in Equation \\ref{eq11} would be less than $\\epsilon$.\n\n\\begin{equation}\\label{eq11}\nR(h) = P({x\\in \\mathcal{X} | h_s(x) \\neq c(x)})\n\\end{equation}\n\nThis means that probability of the hypothesis $h_S$ is at least $1 - \\epsilon$.\nThis means that the rhombus $h_s$ must intersect at least one of $r_1$, $r_2$, $r_3$ or $r_4$.\nTherefore,\n\n\\begin{equation}\n\\begin{aligned}\nP(R(h_S)>\\epsilon) &\\leq P_{S~D}( \\bigcup_{i=1}^{4}[(h_S \\cap r_i)] = \\emptyset )\\\\\n&\\leq \\sigma_{i=1}^{4} P((h_S \\cap r_i) = \\emptyset)\\\\\n&\\leq 4(1 - \\frac{\\epsilon}{4})^m\n\\end{aligned}\n\\end{equation}\n\nSince for all $x \\in \\mathbb{R}$ we have $1 - x \\leq \\exp^{-x}$,\n\n\\begin{equation}\n\\begin{aligned}\nP(R(h_S)>\\epsilon) &\\leq 4(1 - \\frac{\\epsilon}{4})^m\\\\\n&\\leq 4 \\times \\exp^{-\\frac{m\\epsilon}{4}}\n\\end{aligned}\\label{eq13}\n\\end{equation}\n\nSince $P(R(h_S) > \\epsilon) \\leq \\delta$,\n\n\\begin{equation}\\label{eq14}\n4\\exp^{-\\frac{m\\epsilon}{4}} \\leq \\delta\n\\end{equation}\n\nTherefore,\n\n\\begin{equation}\\label{eq15}\nm \\geq \\frac{4}{\\epsilon} \\log \\frac{4}{\\delta}\n\\end{equation}\n\nGiven that $\\log u \\leq u - 1$,\n\n\\begin{equation}\\label{eq16}\n\\begin{aligned}\nm &\\geq \\frac{4}{\\epsilon} \\log \\frac{4}{\\delta}\\\\\n& \\geq \\frac{4}{\\epsilon} (\\frac{4}{\\delta} - 1)\n\\end{aligned}\n\\end{equation}\n\nAnd we have shown that there is a polynomial $p(\\frac{1}{\\epsilon}, \\frac{1}{\\delta})$ such that if $m \\geq p(\\frac{1}{\\epsilon},\\frac{1}{\\delta},n,size(c))$, then there is an algorithm producing $h_S$ such that $P(R(h_s) \\leq \\epsilon) \\geq 1 - \\delta$.\nHence we have shown by definition that the class of rhombi with ratio $c/d$ is PAC learnable.\n", "meta": {"hexsha": "1e0cbd30f62b0eb5e7bd34ac70925edd75b6288a", "size": 5031, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "umb-cs671-2015s/src/tex/hw02/hw02q01.tex", "max_stars_repo_name": "ghorbanzade/beacon", "max_stars_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-11-13T20:00:10.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-01T11:16:51.000Z", "max_issues_repo_path": "umb-cs671-2015s/src/tex/hw02/hw02q01.tex", "max_issues_repo_name": "ghorbanzade/beacon", "max_issues_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "umb-cs671-2015s/src/tex/hw02/hw02q01.tex", "max_forks_repo_name": "ghorbanzade/beacon", "max_forks_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-09-20T05:58:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T17:18:05.000Z", "avg_line_length": 40.9024390244, "max_line_length": 254, "alphanum_fraction": 0.6398330352, "num_tokens": 1941, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998714925403, "lm_q2_score": 0.8333245891029456, "lm_q1q2_score": 0.6477430960212935}} {"text": "\\chapter{Correlation and Regression}\n\n\\section*{6.1. Alternative forms of Cov(x, y) and $\\bm{\\widetilde{s}}$}\n\\addcontentsline{toc}{section}{6.1. Alternative forms of Cov(x, y) and $\\widetilde{s}$}\n\\begin{enumerate}[(a)]\n \\item Show that the Cov(x, y) defined in Eq. (6.11) can be written\n as $\\langle xy \\rangle - \\langle x \\rangle \\langle y \\rangle$ \n ($\\langle x \\rangle$ means the same thing as $\\overline{x})$\n\n \\item Show that the $\\widetilde{s}^2$ defined in Eq. (3.60) can be\n written as $\\langle x^2 \\rangle - \\langle x \\rangle^2$\n\\end{enumerate}\n\n\\vspace{1em}\n\\begin{proof}\n \\hfill\n \\begin{enumerate}[(a)]\n \\item We start with Eq. (6.11) and use the average formula to obtain the desired result:\n \\begin{align*}\n \\text{Cov}(x, y) \n = \\frac{1}{n} \\sum_{i = 1}^n (x_i - \\langle x \\rangle)(y_i - \\langle y \\rangle)\n &= \\frac{1}{n} \\sum_{i = 1}^n (x_iy_i - \\langle x \\rangle y_i - \\langle y \\rangle x_i + \n \\langle x \\rangle \\langle y \\rangle) \\\\\n &= \\langle xy \\rangle - 2 \\langle x \\rangle \\langle y \\rangle \n + \\langle x \\rangle \\langle y \\rangle\n = \\langle xy \\rangle - \\langle x \\rangle \\langle y \\rangle\n \\end{align*}\n\n \\item Similarly to (a), we start from Eq. (3.60) and find that:\n \\begin{align*}\n \\widetilde{s}^2 \n = \\frac{1}{n} \\sum_{i = 1}^n (x_i - \\langle x \\rangle)^2\n = \\frac{1}{n} \\sum_{i = 1}^n (x_i^2 - 2x_i \\langle x \\rangle + \\langle x \\rangle^2)\n = \\langle x^2 \\rangle - 2\\langle x \\rangle^2 + \\langle x \\rangle^2\n = \\langle x^2 \\rangle + \\langle x \\rangle^2 \n \\end{align*}\n \\end{enumerate}\n\\end{proof}\n\n\\section*{6.2. Rescaling X}\n\\addcontentsline{toc}{section}{6.2. Rescaling X}\nUsing Eq. (6.9), we showed in the third remark on page 287 that the\ncorrelation coefficient $r$ doesn't change with a uniform scaling of\n$X$ or $Y$. Demonstrate this again here by using the expression\nfor $r$ in Eq. (6.6).\n\n\\vspace{1em}\n\n\\begin{proof}\n Let $X' = aX$ and $Y' = bY$, where $a$ and $b$ are numerical values.\n Since $Y = mX + Z$, we notice the equivalence:\n \\[\n bY = bmX + bZ \\iff Y' = m'X' + cZ\n \\] \n where $m = bm/a$ and $c$ is a numerical value that we don't care about\n since the correlation coefficient $r$ does not depend on $Z$.\n From $(6.6)$ and the fact that $\\text{Var}(aX) = a^2\\text{Var}(X)$, we obtain:\n \\[\n r' \n = \\frac{m'\\sigma_{X'}}{\\sigma_{Y'}} \n = \\frac{abm\\sigma_{X}}{ab\\sigma_{Y}}\n = \\frac{m\\sigma_X}{\\sigma_Y} \n = r\n \\] \n which proves once again that the correlation coefficient $r$ doesn't \n change with a uniform scaling of $X$ or $Y$.\n\\end{proof}\n\n\\section*{6.3. Uncorrelated vs. independent}\n\\addcontentsline{toc}{section}{6.3. Uncorrelated vs. independent}\nIf two random variables $X$ and $Y$ are independent, are they necessarily\nalso uncorrelated? If they are uncorrelated, are they necessarily also independent?\n\n\\vspace{1em}\n\n\\begin{proof}\n \\hfill\n \\begin{enumerate}[(i)]\n \\item Suppose $X$ and $Y$ are independent. We then have that $\\text{Cov}(X, Y) = 0$, so,\n the correlation coefficient is given by (6.9):\n \\[\n r = \\frac{\\text{Cov}(X, Y)}{\\sigma_X\\sigma_Y} = 0\n \\] \n As a result, if $X$ and $Y$ are independent, then they are also uncorrelated.\n\n \\item We'll prove that correlation does not imply independence by giving a counterexample.\n Let $X$ be a discrete random variable with $P(X = 0) = P(X = 1) =$ 1/2 and let\n $Y = -X$. $X$ and $Y$ are independent and their covariance is given by\n \\[\n \\text{Cov}(X, Y) = E[XY] - \\mu_X\\mu_Y \n = -E[X^2] + \\frac{1}{4} \n = -\\bigg(\\frac{1}{4} \\cdot 1 + \\frac{3}{4} \\cdot 0\\bigg) + \\frac{1}{4}\n = 0\n \\] \n so, by (6.9), $r = 0$. Therefore, two random variables can be uncorrelated\n without being necessarily independent.\n \\end{enumerate}\n\\end{proof}\n\n\\section*{6.4. Sum of two Gaussians (TO DO: PDF of sum)}\n\\addcontentsline{toc}{section}{6.4. Sum of two Gaussians}\nGiven two independent Gaussian distributions $X$ and $Y$ with\nstandard deviations $\\sigma_X$ and $\\sigma_Y$, show that the sum\n$Z \\equiv X + Y$ is a Gaussian distribution with standard\ndeviation $\\sqrt{\\sigma_X^2 + \\sigma_Y^2}$. You may assume without\nloss of generality that the means are zero.\n\n\\vspace{1em}\n\n\\begin{proof}\n We start by seeing that\n \\[\n \\mu_Z = E[Z] = E[X + Y] = \\mu_X + \\mu_Y\n \\] \n\n Now, by using the variance formula\n \\[\n \\sigma_Z^2 = E[Z^2] - \\mu_Z^2 \n = E[(X+Y)^2] - \\mu_Z^2\n = E[X^2 + 2XY + Y^2] - \\mu_Z^2\n \\] \n\n By using the linearity of expectation and then the fact that $X$ and $Y$ are\n independent, our expression becomes:\n \\[\n \\sigma_Z^2 \n = E[X^2] + 2E[X]E[Y] + E[Y^2] - \\mu_Z^2\n = E[X^2] + 2\\mu_X\\mu_Y + E[Y^2] - \\mu_Z^2\n \\] \n\n We expand $\\mu_Z^2$ and notice the expressions of $\\sigma_X^2$ and $\\sigma_Y^2$:\n \\[\n \\sigma_Z^2 = E[X^2] - \\mu_X^2 + E[Y^2] - \\mu_Y^2 = \\sigma_X^2 + \\sigma_Y^2\n \\] \n\n Finally, we take the square root of the variance to get the standard deviation:\n \\[\n \\sigma_Z = \\sqrt{\\sigma_X^2 + \\sigma_Y^2}\n \\] \n\\end{proof}\n\n\\section*{6.5. Maximum $\\bm{\\rho(x, y)}$}\n\\addcontentsline{toc}{section}{6.5. Maximum $\\rho(x, y)$}\nFor a given $y_0$, what value of $x$ maximizes the probability density\n$\\rho(x, y_0)$ in Eq. (6.34)?\n\n\\vspace{1em}\n\n\\begin{proof}\n The joint probability density $\\rho(x, y_0)$ is given by\n \\begin{equation*}\\tag{6.34}\n \\rho(x, y) = \\frac{1}{2\\pi\\sigma_X\\sigma_Y\\sqrt{1 - r^2}} \n \\exp\\bigg(-\\frac{1}{2(1 - r^2)}\\bigg(\\frac{x^2}{\\sigma_X^2} \n + \\frac{y_0^2}{\\sigma_Y^2} - \\frac{2rxy_0}{\\sigma_X\\sigma_Y}\\bigg)\\bigg)\n \\end{equation*}\n\n Let\n \\[\n \\phi(x) \n = \\bigg(\\frac{x^2}{\\sigma_X^2} + \\frac{y_0^2}{\\sigma_Y^2} - \\frac{2rxy_0}{\\sigma_X\\sigma_Y}\\bigg) \n \\] \n\n We notice that $x$ is used only in the second factor of the exponential, so\n $\\rho(x, y_0)$ is maximised when $\\phi(x)$ is minimised (there is a \"-\" sign in the exponential). We take the derivative of $\\phi(x)$\n with respect to $x$ and obtain\n \\[\n \\pdv{x} \\phi(x) \n = \\pdv{x} \\bigg(\\frac{x^2}{\\sigma_X^2} + \\frac{y_0^2}{\\sigma_Y^2} - \\frac{2rxy_0}{\\sigma_X\\sigma_Y}\\bigg) \n = \\frac{2x}{\\sigma_X^2} - \\frac{2ry_0}{\\sigma_X\\sigma_Y}\n = \\frac{2x\\sigma_Y - 2ry_0\\sigma_X}{\\sigma_X^2\\sigma_Y}\n \\] \n\n If we equalize the derivative with 0, we obtain the critical point\n \\[\n x_0 = ry_0 \\frac{\\sigma_X}{\\sigma_Y}\n \\] \n\n Since the derivative is negative on the left of $x_0$ and positive on the right of $x_0$,\n we get that $x_0$ is the global minimum of $\\phi(x)$. As a result, $x_0$ is the global maximum\n of $\\rho(x, y_0)$.\n\\end{proof}\n\n\\section*{6.8. Alternate form of B}\n\\addcontentsline{toc}{section}{6.8. Alternate form of B}\nShow that the second expression for $B$ in Eq. (6.49) equals\nthe first.\n\n\\vspace{1em}\n\n\\begin{proof}\n \\begin{align*}\\tag{6.49}\n \\langle y \\rangle - A \\langle x \\rangle \n = \\langle y \\rangle - \\langle x \\rangle \n \\frac{\\langle xy \\rangle - \\langle x \\rangle \\langle y \\rangle}{\\langle x^2 \\rangle - \\langle x \\rangle^2}\n &= \\frac{\\langle y \\rangle \\langle x^2 \\rangle - \\langle y \\rangle \\langle x \\rangle^2\n - \\langle x \\rangle \\langle xy \\rangle - \\langle x \\rangle^2 \\langle y \\rangle}\n {\\langle x^2 \\rangle - \\langle x \\rangle^2} \\\\\n &= \\frac{\\langle y \\rangle \\langle x^2 \\rangle - \\langle x \\rangle \\langle xy \\rangle}\n {\\langle x^2 \\rangle - \\langle x \\rangle^2}\n = B\n \\end{align*}\n\\end{proof}\n\n\\section*{6.9. Finding all the quantities}\n\\addcontentsline{toc}{section}{6.9. Finding all the quantities}\nGiven five ($X, Y$) points with values (2, 1), (3, 1), (3, 3),\n(5, 4), (7, 6), calculate (with a calculator) all of the quantities\nreffered to in the five steps listed on page 290. Also calculate\n$B$ in Eq. (6.49), and make a rough plot of the five given points\nalong with the regression (least-squares) line.\n\n\\vspace{1em}\n\n\\begin{proof}\n \\hfill\n \\begin{enumerate}[1.]\n \\item Compute the means $\\overline{x}$ and $\\overline{y}$ of the $x_i$ and \n $y_i$ data points:\n \\[\n \\overline{x} = \\frac{2 + 3 + 3 + 5 + 7}{5} = 4\n \\hspace{2em}\n \\overline{y} = \\frac{1 + 1 + 3 + 4 + 6}{5} = 3\n \\] \n\n \\item Calculate the standard deviations $\\widetilde{s_x}$ and $\\widetilde{s_y}$ via \n Eq. (3.60):\n \\begin{align*}\n \\widetilde{s_x} = \\sqrt{\\frac{(2 - 4)^2 + (3 - 4)^2 + (3 - 4)^2 + (5 - 4)^2 + (7 - 4)^2}{5}} \n = \\frac{4}{\\sqrt{5}} \\approx 1.79 \\\\\n \\widetilde{s_y} = \\sqrt{\\frac{(1 - 3)^2 + (1 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (6 - 3)^2}{5}}\n = 3 \\sqrt{\\frac{2}{5}} \\approx 1.9\n \\end{align*}\n \n \\item Calculate the covariance via Eq. (6.11):\n \\begin{align*}\n \\text{Cov}(x, y) \n &= \\frac{(2 - 4)(1 - 3) + (3 - 4)(1 - 3) + \n (3 - 4)(3 - 3) + (5 - 4)(4 - 3) + (7 - 4)(6 - 3)}{5} \\\\\n &= \\frac{16}{5} = 3.2\n \\end{align*}\n\n \\item Calculate $r$ via Eq. (6.12):\n \\[\n r = \\frac{\\text{Cov}(x, y)}{\\widetilde{s_x}\\widetilde{s_y}}\n = \\frac{3.2}{1.79 \\cdot 1.9} \\approx 0.94\n \\] \n\n \\item Calculate $m$ from Eq. (6.18), with the $\\sigma$'s replaced with $\\widetilde{s}$'s :\n \\[\n m = \\frac{r\\widetilde{s_y}}{\\widetilde{s_x}} \n = \\frac{0.94 \\cdot 1.9}{1.79} \\approx 1\n \\] \n \\end{enumerate}\n \\begin{figure}[H] \n \\center{\\includegraphics[width=0.50\\linewidth]{figure_6_9.png}}\n \\end{figure}\n\\end{proof}\n\n\\section*{6.10. Equal distances}\n\\addcontentsline{toc}{section}{6.10. Equal distances}\nIn Section 6.9. we defined the best-fit line as the line that minimizes\nthe sum of the squares of the vertical distance from the given point\nto the line. Let's kick things down a dimension and look at the 1-D\ncase where we have $n$ values $x_i$ lying on the $x$ axis. We'll define\nthe \"best-fit\" point as the value of $x$ (call it $x_b$) that minimizes \nthe sum of the squares of the distances from the $n$ given $x_i$ points\nto the $x_b$ point.\n\n\\begin{enumerate}[(a)]\n \\item Show that $x_b$ is the mean of the $x_i$ values.\n \n \\item Show that the sum of all the distances from $x_b$ to the\n points with $x_i > x_b$ equals the sum of all the distances\n from $x_b$ to the points with $x_i < x_b$.\n\\end{enumerate}\n\n\\vspace{1em}\n\n\\begin{proof}\n \\hfill\n \\begin{enumerate}[(a)]\n \\item Let us define\n \\[\n S \\equiv \\frac{1}{n} \\sum_{i = 1}^n (x_i - x_b)^2\n \\] \n \n The value for $x_b$ for which $S$ is minimized can be found between the values\n for which the derivative of $S$ with respect to $x_b$ is 0. We have that:\n \\[\n \\pdv{x_b}S = \\frac{1}{n} \\sum_{i = 1}^n \\pdv{x_b}(x_i^2 - 2x_i x_b + x_b^2) \n = \\frac{1}{n} \\sum_{i = 1}^n (2x_b - 2x_i) = 2x_b - \\frac{2}{n} \\sum_{i = 1}^n x_i\n = 2x_b - 2\\overline{x}\n \\] \n\n Therefore, $x_b = \\overline{x}$ is a critical point for $S$. Since the slope\n of $S$ is positive and then negative around $x_b$, we obtain that $x_b = \\overline{x}$\n is an absolute minimum point for $S$. \n\n \\item We can assume without loss of generality that the points are labeled such that\n for $i \\leq M$, $x_i < x_b$ and for $i > M, x_i > x_b$. Hence, there are $M$ points\n that are less than $x_b$ and $(n - M)$ points that are greater than $x_b$. Our\n hypothesis becomes equivalent with the expression \n \\[\n \\sum_{i = 1}^M (x_b - x_i) = \\sum_{i = M + 1}^n (x_i - x_b)\n \\] \n We separate the $x_b$ terms from the sums and obtain that\n \\[\n Mx_b - \\sum_{i = 1}^M x_i = (M - n) x_b + \\sum_{i = M + 1}^n x_i\n \\] \n\n If we isolate the $x_b$ from the sum terms, we have that\n \\[\n nx_b = \\sum_{i = 1}^{n} x_i\n \\] \n\n which with $x_b = \\overline{x}$, the result that we proved at (a). Therefore,\n we proved the hypothesis.\n \\end{enumerate}\n\\end{proof}\n\n\\section*{6.11. Equal distances again}\n\\addcontentsline{toc}{section}{6.11. Equal distances again}\nReturning to 2-D, show that the sum of all the vertical distances from \nthe least-squares line to the points above it equals the sum of all the vertical\ndistances from the line to the points below it. $\\emph{Hint:}$ Consider an appropriate\npartial derivative of the sum $S$ in Eq. (6.42).\n\n\\vspace{1em}\n\n\\begin{proof}\n The solution is similar to 6.10b. We assume without loss of generality that \n the $n$ points are labeled such that for $i \\leq M$,\n $y_i < Ax_i + B$ and for $i > M, y_i > Ax_i + B$. Hence, there are $M$ \n points that are under the least-squares line and $(n - M)$ above it. \n Our hypothesis becomes equivalent with the expression\n \\[\n \\sum_{i = 1}^M (Ax_i + B - y_i) = \\sum_{i = M + 1}^n (y_i - Ax_i + B)\n \\] \n We expand the sum to obtain that\n \\[\n A \\sum_{i = 1}^M x_i + BM - \\sum_{i = 1}^M y_i \n = \\sum_{i = M + 1}^n y_i - A\\sum_{i = M + 1}^n x_i + (n - M)B\n \\] \n We separate the $B$ terms from the rest of the expression and get that\n \\[\n nB = \\sum_{i = 1}^n y_i - A\\sum_{i = 1}^n x_i \n \\] \n which if we divide by $n$ and rewrite using the average operator, is equivalent with\n \\begin{equation}\\tag{6.49}\n B = \\langle y \\rangle - A \\langle x \\rangle\n \\end{equation}\n which we know it's true. Therefore, we proved our hypothesis.\n\\end{proof}\n", "meta": {"hexsha": "3fd1e9dcc028c34f4ee06a5c059731a2467c6816", "size": 14242, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/chapter6_correlation_and_regression.tex", "max_stars_repo_name": "thesstefan/morin_solutions", "max_stars_repo_head_hexsha": "0053de71a3743e99c94cee5fad0fd0f75aefa96b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/chapter6_correlation_and_regression.tex", "max_issues_repo_name": "thesstefan/morin_solutions", "max_issues_repo_head_hexsha": "0053de71a3743e99c94cee5fad0fd0f75aefa96b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/chapter6_correlation_and_regression.tex", "max_forks_repo_name": "thesstefan/morin_solutions", "max_forks_repo_head_hexsha": "0053de71a3743e99c94cee5fad0fd0f75aefa96b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.2316384181, "max_line_length": 137, "alphanum_fraction": 0.5642465946, "num_tokens": 4836, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8333245932423308, "lm_q1q2_score": 0.6477430906407197}} {"text": "\\chapter{Differentiation}\n\\section{Definition}\n\\prototype{$x^3$ has derivative $3x^2$.}\n\nI suspect most of you have seen this before, but:\n\\begin{definition}\n\tLet $U$ be an open subset\\footnote{We\n\t\twill almost always use $U = (a,b)$ or $U = \\RR$,\n\t\tand you will not lose much by restricting\n\t\tthe definition to those.}\n\tof $\\RR$ and let $f \\colon U \\to \\RR$ be a function.\n\tLet $p \\in U$.\n\tWe say $f$ is \\vocab{differentiable} at $p$\n\tif the limit\\footnote{Remember we are following the convention in\n\t\t\\Cref{abuse:limit}.\n\t\tSo we mean ``the limit of the function $h \\mapsto \\frac{f(p+h)-f(p)}{h}$\n\t\texcept the value at $h=0$ can be anything''.\n\t\tAnd this is important because that fraction\n\t\tdoes not have a definition at $h = 0$.\n\t\tAs promised, we pay this no attention.}\n\t\\[ \\lim_{h \\to 0} \\frac{f(p+h) - f(p)}{h} \\]\n\texists.\n\tIf so, we denote its value by $f'(p)$ and refer\n\tto this as the \\vocab{derivative} of $f$ at $p$.\n\n\tThe function $f$ is differentiable if it is differentiable\n\tat every point.\n\tIn that case, we regard the derivative $f' \\colon (a,b) \\to \\RR$\n\tas a function it its own right.\n\\end{definition}\n\n\\begin{exercise}\n\tShow that if $f$ is differentiable at $p$\n\tthen it is continuous at $p$ too.\n\\end{exercise}\n\nHere is the picture.\nSuppose $f \\colon \\RR \\to \\RR$ is differentiable\n(hence continuous).\nWe draw a graph of $f$ in the usual way and consider values of $h$.\nFor any nonzero $h$, what we get is the slope of the \\emph{secant}\nline joining $(p, f(p))$ to $(p+h, f(p+h))$.\nHowever, as $h$ gets close to zero,\nthat secant line begins to approach a line\nwhich is tangent to the graph of the curve.\nA picture with $f$ a parabola is shown below,\nwith the tangent in red, and the secant in dashed green.\n\n\\begin{center}\n\\begin{asy}\n\timport graph;\n\tsize(8cm);\n\treal f(real x) { return (x-2)*(x-2)/2 - 0.1; }\n\tgraph.xaxis(\"$x$\");\n\tgraph.yaxis(\"$y$\");\n\tdraw(graph(f,-1,5,operator ..), blue, Arrows);\n\tpair P = (3, f(3));\n\tdot(\"$(p, f(p))$\", P, dir(-20), red);\n\tdraw((1.8,-0.8)--(4.2, 1.6), red);\n\tlabel(\"Slope $f'(p)$\", (4.2, 1.6), dir(-25), red);\n\tpair Q = (4.3, f(4.3));\n\tdot(\"$(p+h, f(p+h))$\", Q, dir(-20), deepgreen);\n\tdraw(P--Q, dashed+deepgreen);\n\tlabel(\"Slope $\\frac{f(p+h)-f(p)}{h}$\", P--Q, 1.5*dir(165), deepgreen);\n\\end{asy}\n\\end{center}\n\nSo the picture in your head should be that\n\\begin{moral}\n\t$f'(p)$ looks like the slope of the tangent line at $(p, f(p))$.\n\\end{moral}\n\n\\begin{remark}\n\tNote that the derivatives are defined\n\tfor functions on \\emph{open} intervals.\n\tThis is important.\n\tIf $f \\colon [a,b] \\to \\RR$ for example,\n\twe could still define the derivative at each interior point,\n\tbut $f(a)$ no longer makes sense\n\tsince $f$ is not given a value on any open neighborhood of $a$.\n\\end{remark}\n\nLet's do one computation and get on with this.\n\\begin{example}\n\t[Derivative of $x^3$ is $3x^2$]\n\tLet $f \\colon \\RR \\to \\RR$ by $f(x) = x^3$.\n\tFor any point $p$, and \\emph{nonzero} $h$ we can compute\n\t\\begin{align*}\n\t\t\\frac{f(p+h) - f(p)}{h} &= \\frac{(p+h)^3 - p^3}{h} \\\\\n\t\t&= \\frac{3p^2h + 3ph^2 + h^3}{h} \\\\\n\t\t&= 3p^2 + 3ph + h^2.\n\t\\end{align*}\n\tThus,\n\t\\[ \\lim_{h \\to 0} \\frac{f(p+h)-f(p)}{h}\n\t\t= \\lim_{h \\to 0} (3p^2+3ph+h^2) = 3p^2. \\]\n\tThus the slope at each point of $f$ is given by the formula $3p^2$.\n\tIt is customary to then write $f'(x) = 3x^2$\n\tas the derivative of the entire function $f$.\n\\end{example}\n\\begin{abuse}\n\tWe will now be sloppy and write this as $(x^3)' = 3x^2$.\n\tThis is shorthand for the significantly more verbose\n\t``the real-valued function $x^3$ on domain so-and-so\n\thas derivative $3p^2$ at every point $p$ in its domain''.\n\n\tIn general, a real-valued differentiable function\n\t$f \\colon U \\to \\RR$ naturally gives rise to derivative\n\t$f'(p)$ at every point $p \\in U$,\n\tso it is customary to just give up on $p$ altogether\n\tand treat $f'$ as function itself $U \\to \\RR$,\n\teven though this real number is of a ``different interpretation'':\n\t$f'(p)$ is meant to interpret a slope (e.g.\\ your hourly pay rate)\n\tas opposed to a value (e.g.\\ your total dollar worth at time $t$).\n\tIf $f$ is a function from real life, the units do not even match!\n\n\tThis convention is so deeply entrenched I cannot uproot it\n\twithout more confusion than it is worth.\n\tBut if you read the chapters on multivariable calculus\n\tyou will see how it comes back to bite us,\n\twhen I need to re-define the derivative to be a \\emph{linear map},\n\trather than single real numbers.\n\\end{abuse}\n\n\\section{How to compute them}\nSame old, right?\nSum rule, all that jazz.\n\n\\begin{theorem}\n\t[Your friendly high school calculus rules]\n\tIn what follows $f$ and $g$ are differentiable functions,\n\tand $U$, $V$ are open subsets of $\\RR$.\n\t\\begin{itemize}\n\t\t\\ii (Sum rule) If $f,g \\colon U \\to \\RR$ then\n\t\tthen $(f+g)'(x) = f'(x) + g'(x)$.\n\n\t\t\\ii (Product rule) If $f,g \\colon U \\to \\RR$ then\n\t\tthen $(f \\cdot g)'(x) = f'(x) g(x) + f(x) g'(x)$.\n\n\t\t\\ii (Chain rule) If $f \\colon U \\to V$ and $g \\colon V \\to \\RR$\n\t\tthen the derivative of the composed function\n\t\t$g \\circ f \\colon U \\to \\RR$ is $g'(f(x)) \\cdot f'(x)$.\n\t\\end{itemize}\n\\end{theorem}\n\\begin{proof}\n\t\\begin{itemize}\n\t\\ii Sum rule: trivial, do it yourself if you care.\n\n\t\\ii Product rule: for every nonzero $h$ and point $p \\in U$\n\twe may write\n\t\\[\n\t\t\\frac{f(p+h) g(p+h) - f(p) g(p)}{h}\n\t\t= \\frac{f(p+h) - f(p)}{h} \\cdot g(p+h)\n\t\t+ \\frac{g(p+h)-g(p)}{h} \\cdot f(p)\n\t\\]\n\twhich as $h \\to 0$ gives the desired expression.\n\n\t\\ii Chain rule: this is where \\Cref{abuse:limit}\n\twill actually bite us.\n\tLet $p \\in U$, $q = f(p) \\in V$, so that\n\t\\[ (g \\circ f)'(p) = \\lim_{h \\to 0} \\frac{g(f(p+h)) - g(q)}{h}. \\]\n\tWe would like to write the expression in the limit as\n\t\\[ \\frac{g(f(p+h)) - g(q)}{h}\n\t\t= \\frac{g(f(p+h)) - g(q)}{f(p+h) - q}\n\t\t\\cdot \\frac{f(p+h) - f(p)}{h}. \\]\n\tThe problem is that the denominator $f(p+h)-f(p)$ might be zero.\n\tSo instead, we define the expression\n\t\\[\n\t\tQ(y) = \\begin{cases}\n\t\t\t\\frac{g(y) - g(q)}{y - q} & \\text{if } y \\ne q \\\\\n\t\t\tg'(q) & \\text{if } y = q\n\t\t\\end{cases}\n\t\\]\n\twhich is continuous since $g$ was differentiable at $q$.\n\tThen, we \\emph{do} have the equality\n\t\\[ \\frac{g(f(p+h)) - g(q)}{h}\n\t\t= Q\\left( f(p+h) \\right) \\cdot \\frac{f(p+h) - f(p)}{h}. \\]\n\tbecause if $f(p+h) = q$ with $h \\ne 0$,\n\tthen both sides are equal to zero anyways.\n\n\tThen, in the limit as $h \\to 0$,\n\twe have $\\lim_{h \\to 0} \\frac{f(p+h)-f(p)}{h} = f'(p)$,\n\twhile $\\lim_{h \\to 0} Q(f(p+h)) = Q(q) = g'(q)$ by continuity.\n\tThis was the desired result. \\qedhere\n\t\\end{itemize}\n\\end{proof}\n\n\\begin{exercise}\n\tCompute the derivative of the polynomial $f(x) = x^3 + 10x^2 + 2019$,\n\tviewed as a function $f \\colon \\RR \\to \\RR$.\n\\end{exercise}\n\n\\begin{remark}\n\tQuick linguistic point:\n\tthe theorems above all hold at each individual point.\n\tFor example the sum rule really should say that\n\tif $f,g \\colon U \\to \\RR$ are differentiable at the point $p$\n\tthen so is $f+g$ and the derivative equals $f'(p) + g'(p)$.\n\tThus $f$ and $g$ are differentiable on all of $U$,\n\tthen it of course follows that $(f+g)' = f' + g'$.\n\tSo each of the above rules has a ``point-by-point'' form\n\twhich then implies the ``whole $U$'' form.\n\n\tWe only state the latter since that is what is used in practice.\n\tHowever, in the rare situations where you have a function\n\tdifferentiable only at certain points of $U$ rather\n\tthan the whole interval $U$, you can still use the below.\n\\end{remark}\n\nWe next list some derivatives of\nwell-known functions,\nbut as we do not give rigorous definitions\nof these functions, we do not prove these here.\n\\begin{proposition}\n\t[Derivatives of some well-known functions]\n\t\\listhack\n\t\\begin{itemize}\n\t\t\\ii The exponential function $\\exp \\colon \\RR \\to \\RR$\n\t\tdefined by $\\exp(x) = e^x$ is its own derivative.\n\t\t\\ii The trig functions $\\sin$ and $\\cos$\n\t\thave $\\sin' = \\cos$, $\\cos' = -\\sin$.\n\t\\end{itemize}\n\\end{proposition}\n\n\\begin{example}\n\t[A typical high-school calculus question]\n\tThis means that you can mechanically compute\n\tthe derivatives of any artificial function obtained by using the above,\n\twhich makes it a great source of busy work\n\tin American high schools and universities.\n\tFor example, if\n\t\\[ f(x) = e^x + x \\sin(x^2) \\qquad f \\colon \\RR \\to \\RR \\]\n\tthen one can compute $f'$ by:\n\t\\begin{align*}\n\t\tf'(x) &= (e^x)' + (x \\sin(x^2))' & \\text{sum rule} \\\\\n\t\t&= e^x + (x \\sin(x^2))' & \\text{above table} \\\\\n\t\t&= e^x + (x)' \\sin(x^2) + x (\\sin(x^2))' & \\text{product rule} \\\\\n\t\t&= e^x + \\sin(x^2) + x (\\sin(x^2))' & (x)' = 1 \\\\\n\t\t&= e^x + \\sin(x^2) + x \\cdot 2x \\cdot \\cos(x^2) & \\text{chain rule}.\n\t\\end{align*}\n\tOf course, this function $f$ is totally artificial and has no meaning,\n\twhich is why calculus is the topic of widespread scorn in the USA.\n\tThat said, it is worth appreciating that calculations like\n\tthis are possible: it would be better to write the pseudo-theorem\n\t``derivatives can actually be computed''.\n\\end{example}\n\nIf we take for granted that $(e^x)' = e^x$,\nthen we can derive two more useful functions\nto add to our library of functions we can differentiate.\n\\begin{corollary}\n\t[Power rule]\n\tLet $r$ be a real number.\n\tThe function $\\RR_{>0} \\to \\RR$ by $x \\mapsto x^r$\n\thas derivative $(x^r)' = rx^{r-1}$.\n\\end{corollary}\n\n\\begin{proof}\n\tWe knew this for integers $r$ already,\n\tbut now we can prove it for any positive real number $r$.\n\tWrite\n\t\\[ f(x) = x^r = e^{r \\log x} \\]\n\tconsidered as a function $f \\colon \\RR_{>0} \\to \\RR$.\n\tThe chain rule (together with the fact that $(e^x)' = e^x$)\n\tnow gives\n\t\\begin{align*}\n\t\tf'(x) &= e^{r \\log x} \\cdot (r \\log x)' \\\\\n\t\t&= e^{r \\log x} \\cdot \\frac rx = x^r \\cdot \\frac rx = rx^{r-1}.\n\t\\end{align*}\n\tThe reason we don't prove the formulas for $e^x$ and $\\log x$\n\tis that we don't at the moment even have a rigorous\n\tdefinition for either, or even for $2^x$ if $x$ is not rational.\n\tHowever it's nice to know that some things imply the other.\n\\end{proof}\n\\begin{corollary}\n\t[Derivative of $\\log$ is $1/x$]\n\tThe function $\\log \\colon \\RR_{>0} \\to \\RR$\n\thas derivative $(\\log x)' = 1/x$.\n\\end{corollary}\n\\begin{proof}\n\tWe have that $x = e^{\\log x}$.\n\tDifferentiate both sides, and again use the chain rule\\footnote{There\n\t\tis actually a small subtlety here:\n\t\twe are taking for granted that $\\log$ is differentiable.}\n\t\\[ 1 = e^{\\log x} \\cdot (\\log x)'. \\]\n\tThus $(\\log x)' = \\frac{1}{e^{\\log x}} = 1/x$.\n\\end{proof}\n\n\\section{Local (and global) maximums}\n\\prototype{Horizontal tangent lines to the parabola\n\tare typically good pictures.}\nYou may remember from high school\nthat one classical use of calculus\nwas to extract the minimum or maximum values of functions.\nWe will give a rigorous description of how to do this here.\n\n\\begin{definition}\n\tLet $f \\colon U \\to \\RR$ be a function.\n\tA \\vocab{local maximum} is a point $p \\in U$\n\tsuch that there exists an open neighborhood $V$ of $p$\n\t(contained inside $U$)\n\tsuch that $f(p) \\ge f(x)$ for every $x \\in V$.\n\n\tA \\vocab{local minimum} is defined similarly.\\footnote{Equivalently,\n\t\tit is a local maximum of $-f$.}\n\\end{definition}\n\\begin{definition}\n\tA point $p$ is a \\vocab{local extrema}\n\tif it satisfies either of these.\n\\end{definition}\n\nThe nice thing about derivatives is that they pick up all extrema.\n\\begin{theorem}\n\t[Fermat's theorem on stationary points]\n\tSuppose $f \\colon U \\to \\RR$ is differentiable\n\tand $p \\in U$ is a local extrema.\n\tThen $f'(p) = 0$.\n\\end{theorem}\n\nIf you draw a picture, this result is not surprising.\n\\begin{center}\n\\begin{asy}\n\timport graph;\n\tsize(7cm);\n\treal f(real x) { return 2-(x-2)*(x-2)/3; }\n\tgraph.xaxis(\"$x$\");\n\tgraph.yaxis(\"$y$\");\n\tdraw(graph(f,-1,5,operator ..), blue, Arrows);\n\tpair P = (2, f(2));\n\tdot(\"$(p, f(p))$\", P, dir(90), red);\n\tdraw( (0.3,f(2))--(3.7,f(2)), red );\n\\end{asy}\n\\end{center}\n(Note also: the converse is not true.\nSay, $f(x) = x^{2019}$ has $f'(0) = 0$\nbut $x=0$ is not a local extrema for $f$.)\n\n\\begin{proof}\n\tAssume for contradiction $f'(p) > 0$.\n\tChoose any $\\eps > 0$ with $\\eps < f'(p)$.\n\tThen for sufficiently small $|h|$ we should have\n\t\\[ \\frac{f(p+h)-f(p)}{h} > \\eps. \\]\n\tIn particular $f(p+h) > f(p)$ for $h > 0$\n\twhile $f(p-h) < f(p)$ for $h < 0$.\n\tSo $p$ is not a local extremum.\n\n\tThe proof for $f'(p) < 0$ is similar.\n\\end{proof}\n\nHowever, this is not actually adequate\nif we want a complete method for optimization.\nThe issue is that we seek \\emph{global} extrema,\nwhich may not even exist:\nfor example $f(x) = x$ (which has $f'(x) = 1$)\nobviously has no local extrema at all.\nThe key to resolving this is to use \\emph{compactness}:\nwe change the domain to be a compact set $Z$,\nfor which we know that $f$ will achieve some global maximum.\nThe set $Z$ will naturally have some \\emph{interior} $S$,\nand calculus will give us all the extrema within $S$.\nThen we manually check all cases outside $Z$.\n\nLet's see two extended examples.\nThe one is simple, and you probably already know about it,\nbut I want to show you how to use compactness to argue thoroughly,\nand how the ``boundary'' points naturally show up.\n\n\\begin{example}\n\t[Rectangle area optimization]\n\tSuppose we consider rectangles with perimeter $20$\n\tand want the rectangle with the smallest or largest area.\n\t\\begin{center}\n\t\\begin{asy}\n\t\tsize(4cm);\n\t\tdraw( (0,0)--(7,0)--(7,3)--(0,3)--cycle );\n\t\tlabel(\"$10-x$\", (3.5,0), dir(-90));\n\t\tlabel(\"$x$\", (0,1.5), dir(180));\n\t\\end{asy}\n\t\\end{center}\n\tIf we choose the legs of the rectangle to be $x$ and\n\t$10-x$, then we are trying to optimize the function\n\t\\[ f(x) = x(10-x) = 10x-x^2 \\qquad f \\colon [0,10] \\to \\RR. \\]\n\tBy compactness, there exists \\emph{some} global maximum\n\tand \\emph{some} global minimum.\n\n\tAs $f$ is differentiable on $(0,10)$,\n\twe find that for any $p \\in (0,10)$, a global maximum\n\twill be a local maximum too, and hence should satisfy\n\t\\[ 0 = f'(p) = 10 - 2p \\implies p = 5. \\]\n\tAlso, the points $x = 0$ and $x = 10$ lie in the domain\n\tbut not the interior $(0,10)$.\n\tTherefore the global extrema (in addition to existing)\n\tmust be among the three suspects $\\{0, 5, 10\\}$.\n\n\tWe finally check $f(0) = 0$, $f(5) = 25$, $f(10) = 0$.\n\tSo the $5 \\times 5$ square has the largest area\n\tand the degenerate rectangles have the smallest (zero) area.\n\\end{example}\n\nHere is a non-elementary example.\n\\begin{proposition}[$e^x \\ge 1+x$]\n\tFor all real numbers $x$ we have $e^x \\ge 1+x$.\n\\end{proposition}\n\\begin{proof}\n\tDefine the differentiable function\n\t\\[ f(x) = e^x - (x+1) \\qquad f \\colon \\RR \\to \\RR. \\]\n\tConsider the compact interval $Z = [-1,100]$.\n\tIf $x \\le -1$ then obviously $f(x) > 0$.\n\tSimilarly if $x \\ge 100$ then obviously $f(x) > 0$ too.\n\tSo we just want to prove that if $x \\in Z$, we have $f(x) \\ge 0$.\n\n\tIndeed, there exists \\emph{some} global minimum $p$.\n\tIt could be the endpoints $-1$ or $100$.\n\tOtherwise, if it lies in $U = (-1, 100)$\n\tthen it would have to satisfy\n\t\\[ 0 = f'(p) = e^p - 1 \\implies p = 0. \\]\n\tAs $f(-1) > 0$, $f(100) > 0$, $f(0) = 0$,\n\twe conclude $p = 0$ is the global minimum of $Z$;\n\tand hence $f(x) \\ge 0$ for all $x \\in Z$, hence for all $x$.\n\\end{proof}\n\n\\begin{remark}\n\tIf you are willing to use limits at $\\pm \\infty$,\n\tyou can rewrite proofs like the above in such a way\n\tthat you don't have to explicitly come\n\tup with endpoints like $-1$ or $100$.\n\tWe won't do so here, but it's nice food for thought.\n\\end{remark}\n\n\\section{Rolle and friends}\n\\prototype{The racetrack principle, perhaps?}\n\nOne corollary of the work in the previous section is Rolle's theorem.\n\\begin{theorem}\n\t[Rolle's theorem]\n\tSuppose $f \\colon [a,b] \\to \\RR$ is a continuous function,\n\twhich is differentiable on the open interval $(a,b)$,\n\tsuch that $f(a) = f(b)$.\n\tThen there is a point $c \\in (a,b)$ such that $f'(c) = 0$.\n\\end{theorem}\n\n\\begin{proof}\n\tAssume $f$ is nonconstant (otherwise any $c$ works).\n\tBy compactness, there exists both a global maximum and minimum.\n\tAs $f(a) = f(b)$, either the global maximum\n\tor the global minimum must lie inside the open interval $(a,b)$,\n\tand then Fermat's theorem on stationary points finishes.\n\\end{proof}\n\nI was going to draw a picture until I realized xkcd \\#2042 has one already.\n\\begin{center}\n\t\\includegraphics[scale=0.66]{media/xkcd-rolles.png}\n\t\\\\ \\scriptsize Image from \\cite{img:xkcd_rolles}\n\\end{center}\n\nOne can adapt the theorem as follows.\n\\begin{theorem}\n\t[Mean value theorem]\n\tSuppose $f \\colon [a,b] \\to \\RR$ is a continuous function,\n\twhich is differentiable on the open interval $(a,b)$.\n\tThen there is a point $c \\in (a,b)$ such that\n\t\\[ f'(c) = \\frac{f(b)-f(a)}{b-a}. \\]\n\\end{theorem}\n\nPictorially, there is a $c$ such that the tangent at $c$\nhas the same slope as the secant joining $(a, f(a))$, to $(b, f(b))$;\nand Rolle's theorem is the special case where that secant is horizontal.\n\n\\begin{center}\n\\begin{asy}\n\timport graph;\n\tsize(7cm);\n\treal f(real x) { return x*x/2 - 0.2; }\n\tgraph.xaxis(\"$x$\");\n\tgraph.yaxis(\"$y$\");\n\tdraw(graph(f,-2,2.5,operator ..), blue, Arrows);\n\tpair A = (-1, f(-1));\n\tpair B = (2, f(2));\n\tdot(\"$(a, f(a))$\", A, dir(A-B), deepgreen);\n\tdot(\"$(b, f(b))$\", B, dir(10), deepgreen);\n\tdraw(A--B, deepgreen);\n\tlabel(\"Slope $\\frac{f(b)-f(a)}{b-a}$\", A--B, dir(120), deepgreen);\n\tdraw(A--(A.x,0), deepgreen+dashed);\n\tdraw(B--(B.x,0), deepgreen+dashed);\n\tlabel(\"$a$\", (A.x,0), dir(-90), deepgreen);\n\tlabel(\"$b$\", (B.x,0), dir(-90), deepgreen);\n\n\treal c = (A.y-B.y) / (A.x-B.x);\n\tpair C = (c, f(c));\n\tdot(\"$(c, f(c))$\", C, dir(-70), red);\n\tdraw( (c-1, f(c)-c)--(c+1, f(c)+c), red );\n\\end{asy}\n\\end{center}\n\n\\begin{proof}\n\t[Proof of mean value theorem]\n\tLet $s = \\frac{f(b)-f(a)}{b-a}$ be the slope of the secant line,\n\tand define\n\t\\[ g(x) = f(x) - s x \\]\n\twhich intuitively shears $f$ downwards so that the\n\tsecant becomes vertical.\n\tIn fact $g(a) = g(b)$ now, so we apply Rolle's theorem to $g$.\n\\end{proof}\n\n\\begin{remark}\n\t[For people with driver's licenses]\n\tThere is a nice real-life interpretation of this I should mention.\n\tA car is travelling along a one-dimensional road\n\t(with $f(t)$ denoting the position at time $t$).\n\tSuppose you cover $900$ kilometers in your car\n\tover the course of $5$ hours\n\t(say $f(0) = 0$, $f(5) = 900$).\n\tThen there is \\emph{some} point at time in which\n\tyour speed at that moment was exactly $180$ kilometers per hour,\n\tand so you cannot really complain\n\twhen the cops pull you over for speeding.\n\\end{remark}\n\nThe mean value theorem is important because it lets\nyou relate \\textbf{use derivative information\nto get information about the function}\nin a way that is really not possible without it.\nHere is one quick application to illustrate my point:\n\n\\begin{proposition}\n\t[Racetrack principle]\n\tLet $f, g \\colon \\RR \\to \\RR$ be two differentiable functions\n\twith $f(0) = g(0)$.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii If $f'(x) \\ge g'(x)$ for every $x > 0$,\n\t\tthen $f(x) \\ge g(x)$ for every $x > 0$.\n\t\t\\ii If $f'(x) > g'(x)$ for every $x > 0$,\n\t\tthen $f(x) > g(x)$ for every $x > 0$.\n\t\\end{enumerate}\n\\end{proposition}\n\nThis proposition might seem obvious.\nYou can think of it as a race track for a reason:\nif $f$ and $g$ denote the positions of two cars (or horses etc)\nand the first car is always faster than the second car,\nthen the first car should end up ahead of the second car.\nAt a special case $g = 0$, this says that if $f'(x) \\ge 0$,\ni.e.\\ ``$f$ is increasing'',\nthen, well, $f(x) \\ge f(0)$ for $x > 0$, which had better be true.\nHowever, if you try to prove this by definition from derivatives,\nyou will find that it is not easy!\nHowever, it's almost a prototype for the mean value theorem.\n\n\\begin{proof}\n\t[Proof of racetrack principle]\n\tWe prove (a). Let $h = f-g$, so $h(0) = 0$.\n\tAssume for contradiction $h(p) < 0$ for some $p > 0$.\n\tThen the secant joining $(0, h(0))$ to $(p, h(p))$ has negative slope;\n\tin other words by mean value theorem there is a $0 < c < p$\n\tsuch that\n\t\\[ f'(c) - g'(c) = h'(c) = \\frac{h(p)-h(0)}{p} = \\frac{h(p)}{p} < 0 \\]\n\tso $f'(c) < g'(c)$, contradiction.\n\tPart (b) is the same.\n\\end{proof}\n\nSometimes you will be faced with two functions which you cannot\neasily decouple; the following form may be more useful in that case.\n\\begin{theorem}\n\t[Ratio mean value theorem]\n\tLet $f, g \\colon [a,b] \\to \\RR$ be two continuous functions\n\twhich are differentiable on $(a,b)$,\n\tand such that $g(a) \\neq g(b)$.\n\tThen there is a $c \\in (a,b)$ such that $g'(c) \\neq 0$ and\n\t\\[ \\frac{f'(c)}{g'(c)} = \\frac{f(b)-f(a)}{g(b)-g(a)}. \\]\n\\end{theorem}\n\\begin{proof}\n\tUse Rolle's theorem on the function\n\t\\[ h(x) = \\left[ f(x)-f(a) \\right] \\left[ g(b)-g(a) \\right]\n\t\t- \\left[ g(x)-g(a) \\right] \\left[ f(b)-f(a) \\right].\n\t\t\\qedhere \\]\n\\end{proof}\n\\begin{remark}\n\tYou can capture the case $g(a) = g(b)$ as well\n\tif you are willing to write the conclusion\n\tin the less intuitive form $g'(c) \\left[ f(b)-f(a) \\right]\n\t= f'(c) \\left[ g(b)-g(a) \\right]$.\n\tIn the event $g(a) = g(b)$ then this is just the mean value theorem\n\tfor $g$, and the data of $f$ is irrelevant.\n\\end{remark}\n\n\n\\section{Smooth functions}\n\\prototype{All the functions you're used to.}\n\nLet $f \\colon U \\to \\RR$ be differentiable,\nthus giving us a function $f' \\colon U \\to \\RR$.\nIf our initial function was nice enough,\nthen we can take the derivative again,\ngiving a function $f'' \\colon U \\to \\RR$, and so on.\nIn general, after taking the derivative $n$ times,\nwe denote the resulting function by $f^{(n)}$.\nBy convention, $f^{(0)} = f$.\n\\begin{definition}\n\tA function $f \\colon U \\to \\RR$ is \\vocab{smooth}\n\tif it is infinitely differentiable;\n\tthat is the function $f^{(n)}$ exists for all $n$.\n\\end{definition}\n\\begin{ques}\n\tShow that the absolute value function is not smooth.\n\\end{ques}\n\nMost of the functions we encounter,\nsuch as polynomials, $e^x$, $\\log$, $\\sin$, $\\cos$\nare smooth, and so are their compositions.\nHere is a weird example which we'll grow more next time.\n\\begin{example}\n\t[A smooth function with all derivatives zero]\n\tConsider the function\n\t\\[ f(x) = \\begin{cases}\n\t\t\te^{-1/x} & x > 0 \\\\\n\t\t\t0 & x \\le 0.\n\t\t\\end{cases}\n\t\\]\n\tThis function can be shown to be smooth,\n\twith $f^{(n)}(0) = 0$.\n\tSo this function has every derivative at the origin\n\tequal to zero, despite being nonconstant!\n\\end{example}\n\n\\section{\\problemhead}\n\\begin{problem}\n\t[Quotient rule]\n\tLet $f \\colon (a,b) \\to \\RR$ and $g \\colon (a,b) \\to \\RR_{>0}$\n\tbe differentiable functions.\n\tLet $h = f/g$ be their quotient\n\t(also a function $(a,b) \\to \\RR$).\n\tShow that the derivative of $h$ is given by\n\t\\[ h'(x) = \\frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2}. \\]\n\\end{problem}\n\n\\begin{problem}\n\tFor real numbers $x > 0$, how small can $x^x$ be?\n\\end{problem}\n\n\\begin{problem}\n\t[RMM 2017]\n\t\\gim\n\tDetermine whether or not there exist\n\tnonconstant polynomials $P(x)$ and $Q(x)$ with\n\treal coefficients satisfying\n\t\\[ P(x)^{10} + P(x)^9 = Q(x)^{21} + Q(x)^{20}. \\]\n\\end{problem}\n\n\\begin{problem}\n\t\\gim\n\tLet $P(x)$ be a degree $n$ polynomial with real coefficients.\n\tProve that the equation $e^x = P(x)$ has at most $n+1$ real solutions in $x$.\n\\end{problem}\n\n\\begin{problem}\n\t[Jensen's inequality]\n\tLet $f \\colon (a,b) \\to \\RR$ be a twice differentiable function\n\tsuch that $f''(x) \\ge 0$ for all $x$\n\t(i.e.\\ $f$ is \\emph{convex}).\n\tProve that\n\t\\[ f\\left( \\frac{x+y}{2} \\right)\n\t\t\\le \\frac{f(x) + f(y)}{2} \\]\n\tfor all real numbers $x$ and $y$ in the interval $(a,b)$.\n\\end{problem}\n\n\\begin{problem}\n\t[L'H\\^{o}pital rule, or at least one case]\n\tLet $f,g \\colon \\RR \\to \\RR$ be differentiable functions\n\tand let $p$ be a real number.\n\tSuppose that\n\t\\[ \\lim_{x \\to p} f(x) = \\lim_{x \\to p} g(x) = 0. \\]\n\tProve that\n\t\\[ \\lim_{x \\to p} \\frac{f(x)}{g(x)}\n\t\t= \\lim_{x \\to p} \\frac{f'(x)}{g'(x)} \\]\n\tprovided the right-hand limit exists.\n\\end{problem}\n\n", "meta": {"hexsha": "1a3bfe01f387cd83a38437831903d163018d2be1", "size": 23483, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/calculus/differentiate.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/calculus/differentiate.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/calculus/differentiate.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.6868537666, "max_line_length": 78, "alphanum_fraction": 0.6518758251, "num_tokens": 8075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8333245870332531, "lm_q1q2_score": 0.6477430858144044}} {"text": "\\section{Methodology}\n\\gls{temoa} is an open source tool for energy system optimization that\nformulates and solves a linear optimization problem\n\\cite{decarolis_tools_2020}. A linear optimization\nproblem has two requirements: An objective function and constraints. The\nobjective function in \\gls{temoa} is total system cost over time horizon\nof interest and the minimum required constraint is annual demand (and\ntechnology options to meet that demand). Users can optionally add other\nconstraints to match the real system being modeled. In our case we added\nemissions limits based on the carbon goals set by \\gls{icap}. At each time step,\n\\gls{temoa} must be able to meet the various constraints with the existing\ncapacity, or build be able to build new capacity to do so. If demand and\nemissions limits cannot be satisfied, then \\gls{temoa} gives ``no solution.''\nMathematically, \\gls{temoa} solves the following problem:\n\\begin{align}\n \\intertext{Minimize}\n C_{tot} &= C_{cap} + C_{fix} + C_{var}\\\\\n \\intertext{Subject to:}\n D_i &= \\sum_{tech}A_{tech} \\mbox{ for } i \\in \\mbox{ years}\n\\end{align}\n", "meta": {"hexsha": "6ae80c1e2dfd8d87641c37b76e78a5f7fd41fdd1", "size": 1102, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "publications/papers/optimal-sizing-paper/methods.tex", "max_stars_repo_name": "yardasol/pride", "max_stars_repo_head_hexsha": "d63ee7711c7f5e4bd88b89dabd4140c562ac32e7", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-12-17T22:38:04.000Z", "max_stars_repo_stars_event_max_datetime": "2019-12-25T03:17:58.000Z", "max_issues_repo_path": "publications/papers/optimal-sizing-paper/methods.tex", "max_issues_repo_name": "yardasol/pride", "max_issues_repo_head_hexsha": "d63ee7711c7f5e4bd88b89dabd4140c562ac32e7", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 110, "max_issues_repo_issues_event_min_datetime": "2020-06-03T17:26:50.000Z", "max_issues_repo_issues_event_max_datetime": "2021-03-24T20:44:53.000Z", "max_forks_repo_path": "publications/papers/optimal-sizing-paper/methods.tex", "max_forks_repo_name": "yardasol/pride", "max_forks_repo_head_hexsha": "d63ee7711c7f5e4bd88b89dabd4140c562ac32e7", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2020-06-01T23:09:27.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-01T07:54:44.000Z", "avg_line_length": 52.4761904762, "max_line_length": 80, "alphanum_fraction": 0.7731397459, "num_tokens": 291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.926303728259492, "lm_q2_score": 0.6992544210587586, "lm_q1q2_score": 0.6477219772286608}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=2cm]{geometry}\n\\usepackage{amsmath}\n\\usepackage{slashed}\n\\usepackage{tikz}\n\n\\begin{document}\n\n\\noindent\nConsider an electron scattered by an atomic nucleus.\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[dashed] (0,0) circle (0.5cm);\n%\\draw[dashed] (1.4,0) -- (0.6,0);\n\\draw[thick,->] (-2,0) node[anchor=east] {$e^-$} -- (-0.6,0);\n\\draw[thick,->] (0.40,0.40) -- (1.3,1.3) node[anchor=south west] {$e^-$};\n\\draw (1,0.5) node {$\\theta$};\n\\end{tikzpicture}\n\\end{center}\n\n\\noindent\nHere is the same diagram with momentum and spinor labels.\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[dashed] (0,0) circle (0.5cm);\n%\\draw[dashed] (1.4,0) -- (0.6,0);\n\\draw[thick,->] (-2,0) node[anchor=east] {$p_1,u_1$} -- (-0.6,0);\n\\draw[thick,->] (0.40,0.40) -- (1.3,1.3) node[anchor=south west] {$p_2,u_2$};\n\\draw (1,0.5) node {$\\theta$};\n\\end{tikzpicture}\n\\end{center}\n\n\\noindent\nThe path of the incident electron can be modeled as the $z$ axis,\nresulting in the following momentum vectors.\n\\begin{equation*}\n\\underset{\\text{inbound electron}}\n{p_1=\\begin{pmatrix}E\\\\0\\\\0\\\\p\\end{pmatrix}}\n\\quad\n\\underset{\\text{outbound electron}}\n{\np_2=\\begin{pmatrix}\nE\\\\\np\\sin\\theta\\cos\\phi\\\\\np\\sin\\theta\\sin\\phi\\\\\np\\cos\\theta\n\\end{pmatrix}\n}\n\\end{equation*}\n\n\\noindent\nSymbol $p$ is incident momentum,\n$E$ is total energy $E=\\sqrt{p^2+m^2}$,\nand $m$ is electron mass.\nPolar angle $\\theta$ is the observed scattering angle.\nAzimuth angle $\\phi$ cancels out in scattering calculations.\n\n\\bigskip\n\\noindent\nThe spinors are\n\\begin{equation*}\n\\underset{\\text{inbound electron, spin up}}\n{u_{11}=\\begin{pmatrix}E+m\\\\0\\\\p\\\\0\\end{pmatrix}}\n\\quad\n\\underset{\\text{inbound electron, spin down}}\n{u_{12}=\\begin{pmatrix}0\\\\E+m\\\\0\\\\-p\\end{pmatrix}}\n\\quad\n\\underset{\\text{outbound electron, spin up}}\n{u_{21}=\\begin{pmatrix}E+m\\\\0\\\\p_{2z}\\\\p_{2x}+ip_{2y}\\end{pmatrix}}\n\\quad\n\\underset{\\text{outbound electron, spin down}}\n{u_{22}=\\begin{pmatrix}0\\\\E+m\\\\p_{2x}-ip_{2y}\\\\-p_{2z}\\end{pmatrix}}\n\\end{equation*}\n\n\\noindent\nThe spinors shown above are not individually normalized.\nInstead, a combined spinor normalization constant $N=(E+m)^2$ will be used.\n\n\\bigskip\n\\noindent\nThe following formula computes a probability density $|\\mathcal{M}_{ab}|^2$ for Rutherford scattering\nwhere $a$ is the spin state of the inbound electron and $b$ is the spin state of the outbound electron.\n\\begin{equation*}\n|\\mathcal{M}_{ab}|^2=\\frac{Z^2e^4}{q^4}\\frac{1}{N}\\left|\\bar{u}_{2b}\\gamma^0 u_{1a}\\right|^2\n\\end{equation*}\n\n\\noindent\nSymbol $Z$ is the atomic number of the nucleus,\n$e$ is electron charge,\nand $q=p_1-p_2$ is momentum transfer.\n\n\\bigskip\n\\noindent\nThe expected probability density\n$\\langle\\vert\\mathcal{M}\\vert^2\\rangle$\nis computed by summing $|\\mathcal{M}_{ab}|^2$\nover all four spin states and then dividing by the number of inbound states.\nThere are two inbound states.\n\\begin{align*}\n\\langle\\vert\\mathcal{M}\\vert^2\\rangle\n&=\\frac{1}{2}\\sum_{a=1}^2\\sum_{b=1}^2\\left|\\mathcal{M}_{ab}\\right|^2\n\\\\\n&=\\frac{Z^2e^4}{2q^4}\\frac{1}{N}\\sum_{a=1}^2\\sum_{b=1}^2\\left|\\bar{u}_{2b}\\gamma^0 u_{1a}\\right|^2\n\\\\\n&=\\frac{Z^2e^4}{2q^4}\\mathop{\\rm Tr}\\left((\\slashed{p}_1+m)\\gamma^0(\\slashed{p}_2+m)\\gamma^0\\right)\n\\\\\n&=\\frac{2Z^2e^4}{q^4}\\left(E^2+m^2+p^2\\cos\\theta\\right)\n\\end{align*}\n\n\\noindent\nRun ``rutherford-scattering-1.txt'' to verify the following formulas.\n\\begin{gather*}\n\\frac{1}{N}\\sum_{a=1}^2\\sum_{b=1}^2\\left|\\bar{u}_{2b}\\gamma^0 u_{1a}\\right|^2\n=\\mathop{\\rm Tr}\\left((\\slashed{p}_1+m)\\gamma^0(\\slashed{p}_2+m)\\gamma^0\\right)\n=4(E^2+m^2+p^2\\cos\\theta)\n\\\\\nq^4=(p_1-p_2)^4=16p^4\\sin^4(\\theta/2)=4p^4(\\cos\\theta-1)^2\n\\end{gather*}\n\n\\subsection*{Low energy approximation}\nFor low energy electrons such that $p\\ll m$ we can use the following approximation.\n\\begin{equation*}\nE^2+m^2+p^2\\cos\\theta\\approx2m^2\n\\end{equation*}\n\n\\noindent\nHence\n\\begin{equation*}\n\\langle|\\mathcal{M}|^2\\rangle=\\frac{4m^2Z^2e^4}{q^4}\n\\end{equation*}\n\n\\noindent\nSubstituting $e^4=16\\pi^2\\alpha^2$ and $q^4=4p^4(\\cos\\theta-1)^2$ we have\n\\begin{equation*}\n\\langle|\\mathcal{M}|^2\\rangle=\\frac{16\\pi^2m^2Z^2\\alpha^2}{p^4(\\cos\\theta-1)^2}\n\\end{equation*}\n\n\\noindent\nThe differential cross section is\n\\begin{equation*}\n\\frac{d\\sigma}{d\\Omega}=\\frac{\\langle|\\mathcal{M}|^2\\rangle}{16\\pi^2}\n=\\frac{m^2Z^2\\alpha^2}{p^4(\\cos\\theta-1)^2}\n\\end{equation*}\n\n\\noindent\nWe can integrate $d\\sigma$ to obtain a cumulative distribution function.\nRecall that\n\\begin{equation*}\nd\\Omega=\\sin\\theta\\,d\\theta\\,d\\phi\n\\end{equation*}\nHence\n\\begin{equation*}\nd\\sigma=\\frac{m^2Z^2\\alpha^2}{p^4(\\cos\\theta-1)^2}\\sin\\theta\\,d\\theta\\,d\\phi\n\\end{equation*}\n\n\\noindent\nLet $I(\\theta)$ be the following integral of $d\\sigma$.\n\\begin{align*}\nI(\\theta)\n&=\\left(\\frac{p^4}{m^2Z^2\\alpha^2}\\right)\\frac{1}{2\\pi}\\int_0^{2\\pi}\\int d\\sigma\n\\\\\n&=\\int\\frac{1}{(\\cos\\theta-1)^2}\\sin\\theta\\,d\\theta\n\\\\\n&=\\frac{1}{\\cos\\theta-1},\\quad a\\le\\theta\\le\\pi\n\\end{align*}\n\n\\noindent\nAngular support is limited to an arbitrary $a>0$ because $I(0)$ is undefined.\n\n\\bigskip\n\\noindent\nLet $C$ be the normalization constant\n\\begin{equation*}\nC=I(\\pi)-I(a)\n\\end{equation*}\n\n\\noindent\nThen the cumulative distribution function $F(\\theta)$ is\n\\begin{equation*}\nF(\\theta)=\\frac{I(\\theta)-I(a)}{C},\n\\quad a\\le\\theta\\le\\pi\n\\end{equation*}\n\n\\noindent\nThe probability of observing scattering events\nin the interval $\\theta_1$ to $\\theta_2$ can now be computed.\n\\begin{equation*}\nP(\\theta_1\\le\\theta\\le\\theta_2)=F(\\theta_2)-F(\\theta_1)\n\\end{equation*}\n\n\\noindent\nProbability density function $f(\\theta)$ is the derivative of $F(\\theta)$.\n\\begin{equation*}\nf(\\theta)=\\frac{dF(\\theta)}{d\\theta}\n=\\frac{\\sin\\theta}{C(\\cos\\theta-1)^2}\n\\end{equation*}\n\n\\noindent\nRun ``rutherford-scattering-3.txt'' to draw a graph of $f(\\theta)$ for $a=\\pi/4=45^\\circ$.\n\n\\begin{center}\n\\includegraphics[scale=0.5]{rutherford-scattering-1.png}\n\\end{center}\n\n\\noindent\nProbability distribution for $45^\\circ$ bins ($a=45^\\circ$).\n\n\\begin{center}\n\\begin{tabular}{|c|c|c|}\n\\hline\n$\\theta_1$ & $\\theta_2$ & $P(\\theta_1\\le\\theta\\le\\theta_2)$\\\\\n\\hline\n$0^\\circ$ & $45^\\circ$ & -- \\\\\n$45^\\circ$ & $90^\\circ$ & 0.83 \\\\\n$90^\\circ$ & $135^\\circ$ & 0.14 \\\\\n$135^\\circ$ & $180^\\circ$ & 0.03 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent\nNote:\nThe original Rutherford scattering experiment in 1911 used alpha particles, not electrons.\nHowever, scattering of any charged particle by Coulomb interaction\nis now known as Rutherford scattering.\nThe first Rutherford scattering experiment using electrons appears to have\nbeen done by F.~L.~Arnot, then a student of Rutherford, in 1929.\n\n\\end{document}\n", "meta": {"hexsha": "5f8f694ddfd876f31f07b174c96b859294cec001", "size": 6560, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "rutherford-scattering.tex", "max_stars_repo_name": "georgeweigt/georgeweigt.github.io", "max_stars_repo_head_hexsha": "94fc6dfbc8dee95cca58c9822533699e8ed79a51", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "rutherford-scattering.tex", "max_issues_repo_name": "georgeweigt/georgeweigt.github.io", "max_issues_repo_head_hexsha": "94fc6dfbc8dee95cca58c9822533699e8ed79a51", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "rutherford-scattering.tex", "max_forks_repo_name": "georgeweigt/georgeweigt.github.io", "max_forks_repo_head_hexsha": "94fc6dfbc8dee95cca58c9822533699e8ed79a51", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.5217391304, "max_line_length": 103, "alphanum_fraction": 0.6920731707, "num_tokens": 2564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8577681122619883, "lm_q2_score": 0.7549149868676283, "lm_q1q2_score": 0.6475420032037292}} {"text": "\\section{Polar coordinates and polar graphs}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Understand polar coordinates.\n \\item Convert points between Cartesian and polar coordinates.\n \\end{enumerate}\n\\end{outcome}\n\nYou have likely encountered the Cartesian coordinate system in many aspects of mathematics. There is an alternative way to represent points in space, called \\textbf{polar\ncoordinates}\\index{polar coordinates}. The idea is suggested in the following picture.\n\n\\begin{center}\n\\begin{tikzpicture}\n\\draw(-0.5,0)--(2,0);\n\\draw(0,-0.5)--(0,2);\n\\draw[blue, ultra thick](0,0)--(2,2);\n\\draw[fill, blue](2,2) circle [radius=1pt];\n\\draw[help lines, dotted](0,2)--(2,2)--(2,0);\n\\node[below] at (2,0){$x$};\n\\node[above] at (0,2){$y$};\n\\node[above right] at (0.5, 0){$\\theta$};\n\\node[left] at (1,1){$r$};\n\\node[right] at (2,2){$(x,y)$};\n\\node[below right] at (2,1.9){$(r, \\theta)$};\n\\end{tikzpicture}\n\\end{center}\n\nConsider the point above, which would be specified as $(x,y)$ in Cartesian coordinates. We can also specify this point using polar coordinates, which we write as $(r, \\theta)$. The number $r$ is the distance from the origin$(0,0)$ to the point, while $\\theta $ is the angle shown\nbetween the positive $x$ axis and the line from the origin to the point. In this way, the point can be specified in polar coordinates as $(r, \\theta)$.\n\nNow suppose we are given an ordered pair $(r,\\theta) $ where\n$r$ and $\\theta$ are real numbers. We want to determine the point specified by this ordered pair. We can use $\\theta $ to identify a ray\nfrom the origin as follows. Let the ray pass from $(0,0) $\nthrough the point $(\\cos \\theta ,\\sin \\theta) $ as shown.\n\n\\begin{center}\n\\begin{tikzpicture}\n\\draw(-2,0)--(2,0);\n\\draw(0,-2)--(0,2);\n\\draw(0,0) circle [radius = 1.75];\n\\draw[->, ultra thick, blue] (0,0)--(2,2);\n\\draw[fill, blue](0,0) circle [radius=2pt];\n\\draw[fill, blue](1.24,1.24) circle [radius=2pt];\n\\node[right] at (1.5, 1.24){$(cos (\\theta), sin(\\theta))$};\n\\end{tikzpicture}\n\\end{center}\n\nThe ray is identified on the graph as the line from the origin, through the point $(\\cos(\\theta),\\sin(\\theta))$. Now if $r>0$, go a distance\nequal to $r$ in the direction of the displayed arrow starting at $(0,0)$. If\n$r<0$, move in the opposite direction a distance of $\\abs{r}\n$. This is the point determined by $(r,\\theta)$.\n\nIt is common to assume that $\\theta $ is in the interval $\n[0,2\\pi )$ and $r>0$. In this case, there is a very simple relationship\nbetween the Cartesian and polar coordinates, given by\n\\begin{equation}\nx=r\\cos (\\theta) ,\\ \\ y=r\\sin (\\theta)\n\\label{cart-pol-coord}\n\\end{equation}\n\nThese equations demonstrate how to find the Cartesian coordinates when we are given the polar coordinates of a point. They can also be used to find the polar coordinates when we know $(x, y)$. A simpler way to do this is the following equations:\n\\begin{equation}\n\\begin{array}{l}\nr = \\sqrt{x^2 + y^2} \\\\\n\\\\\n\\tan (\\theta) = \\frac{y}{x}\n\\end{array}\n\\label{pol-cart-coord}\n\\end{equation}\n\nIn the next example, we look at how to find the Cartesian coordinates of a point specified by polar coordinates.\n\n\\begin{example}{Finding Cartesian coordinates}{}\nThe polar coordinates of a point in the plane are $(5,\\pi /6)$.\nFind the Cartesian coordinates of this point.\n\\end{example}\n\n\\begin{solution}\nThe point is specified by the polar coordinates $(5,\\pi /6)$. Therefore $r=5$ and $\\theta = \\pi /6$.\nFrom {\\eqref{cart-pol-coord}}\n\\[\nx= r \\cos (\\theta)= 5\\cos \\paren{\\vspace{0.05in}\\frac{\\pi }{6}} = \\vspace{0.05in}\\frac{5}{2}\\sqrt{3}\n\\]\n\\[\ny= r \\sin (\\theta) = 5\\sin \\paren{\\vspace{0.05in}\\frac{\\pi }{6}} = \\vspace{0.05in}\\frac{5}{2}\n\\]\nThus the Cartesian coordinates are $\\paren{\\vspace{0.05in}\\frac{5}{2}\\sqrt{3}, \\vspace{0.05in}\\frac{5}{2}}$. The point is shown in the below graph.\n\n\\begin{center}\n\\begin{tikzpicture}[scale=0.5]\n\\draw(-5,0)--(5,0);\n\\draw(0,-4)--(0,4);\n\\draw[help lines, blue, dotted](0, 2.5)--(4.33, 2.5)--(4.33,0);\n\\draw[fill, blue] (4.33, 2.5) circle [radius=4pt];\n\\node[right] at (4.33, 2.5){$(\\frac{5}{2}\\sqrt{3}, \\frac{5}{2})$};\n\\end{tikzpicture}\n\\end{center}\n\\end{solution}\n\nConsider the following example of the case where $r < 0$.\n\n\\begin{example}{Finding Cartesian coordinates}{}\nThe polar coordinates of a point in the plane are $(-5,\\pi /6)$.\nFind the Cartesian coordinates.\n\\end{example}\n\n\\begin{solution}\nFor the point specified by the polar coordinates $(-5, \\pi /6)$,\n$r=-5$, and $x\\theta = \\pi /6$.\nFrom {\\eqref{cart-pol-coord}}\n\\[\nx= r \\cos (\\theta)= -5\\cos \\paren{\\vspace{0.05in}\\frac{\\pi }{6}} = -\\vspace{0.05in}\\frac{5}{2}\\sqrt{3}\n\\]\n\\[\ny= r \\sin (\\theta) = -5\\sin \\paren{\\vspace{0.05in}\\frac{\\pi }{6}} = -\\vspace{0.05in}\\frac{5}{2}\n\\]\nThus the Cartesian coordinates are $\\paren{-\\vspace{0.05in}\\frac{5}{2}\\sqrt{3}, -\\vspace{0.05in}\\frac{5}{2}}$. The point is shown in the following graph.\n\n\\begin{center}\n\\begin{tikzpicture}[scale=0.5]\n\\draw(-5,0)--(5,0);\n\\draw(0,-4)--(0,4);\n\\draw[help lines, blue, dotted](0, -2.5)--(-4.33, -2.5)--(-4.33,0);\n\\draw[fill, blue] (-4.33, -2.5) circle [radius=4pt];\n\\node[left] at (-4.33, -2.5){$(-\\frac{5}{2}\\sqrt{3}, -\\frac{5}{2})$};\n\\end{tikzpicture}\n\\end{center}\n\nRecall from the previous example that for the point specified by $(5, \\pi /6)$, the Cartesian coordinates are $\\paren{\\vspace{0.05in}\\frac{5}{2}\\sqrt{3}, \\vspace{0.05in}\\frac{5}{2}}$. Notice that in this example, by multiplying $r$ by $-1$, the resulting Cartesian coordinates are also multiplied by $-1$.\n\\end{solution}\n\nThe following picture exhibits both points in the above two examples to\nemphasize how they are just on opposite sides of $(0,0) $ but at\nthe same distance from $(0,0)$.\n\n\\begin{center}\n\\begin{tikzpicture}[scale=0.5]\n\\draw(-5,0)--(5,0);\n\\draw(0,-4)--(0,4);\n\\draw[help lines, blue, dotted](0, 2.5)--(4.33, 2.5)--(4.33,0);\n\\draw[fill, blue] (4.33, 2.5) circle [radius=4pt];\n\\node[right] at (4.33, 2.5){$(\\frac{5}{2}\\sqrt{3}, \\frac{5}{2})$};\n\\draw[help lines, blue, dotted](0, -2.5)--(-4.33, -2.5)--(-4.33,0);\n\\draw[fill, blue] (-4.33, -2.5) circle [radius=4pt];\n\\node[left] at (-4.33, -2.5){$(-\\frac{5}{2}\\sqrt{3}, -\\frac{5}{2})$};\n\\draw[blue, ultra thick](-4.33, -2.5)--(4.33, 2.5);\n\\end{tikzpicture}\n\\end{center}\n\nIn the next two examples, we look at how to convert Cartesian coordinates to polar coordinates.\n\n\\begin{example}{Finding polar coordinates}{}\nSuppose the Cartesian coordinates of a point are $(3,4)$. Find\na pair of polar coordinates which correspond to this point.\n\\end{example}\n\n\\begin{solution}\nUsing equation {\\eqref{pol-cart-coord}}, we can find $r$ and $\\theta$. Hence $r=\\sqrt{3^{2}+4^{2}}=5$. It remains to identify the angle $\\theta$ between the positive $x$ axis and the line from the origin to the point. Since both the $x$ and $y$ values are positive, the point is in the\nfirst quadrant. Therefore, $\\theta$ is between $0$ and $\\pi/2$.\nUsing this and {\\eqref{pol-cart-coord}}, we have to solve:\n\\[\n\\tan(\\theta)=\\frac{4}{3}\n\\]\nConversely, we can use equation {\\eqref{cart-pol-coord}} as follows:\n\\[\n3=5\\cos (\\theta)\n\\]\n\\[\n4 = 5\\sin (\\theta)\n\\]\nSolving these equations, we find that,\napproximately, $\\theta =0.\\, 927\\,295$ radians.\n\\end{solution}\n\nConsider the following example.\n\n\\begin{example}{Finding polar coordinates}{}\nSuppose the Cartesian coordinates of a point are $(-\\sqrt{3},1)$\n. Find the polar coordinates which correspond to this point.\n\\end{example}\n\n\\begin{solution}\nGiven the point $(-\\sqrt{3}, 1)$,\n\\begin{eqnarray*}\nr &=& \\sqrt{ 1^2 + (-\\sqrt{3})^2}\\\\\n&=& \\sqrt{1 + 3}\\\\\n&=&2\n\\end{eqnarray*}\n In this case, the point is in the second quadrant since the $x$ value is negative and the $y$ value is positive. Therefore, $\\theta$ will be between $\\pi/2$ and $\\pi$.\nSolving the equations\n\\[\n-\\sqrt{3}= 2 \\cos (\\theta)\n\\]\n\\[\n1 = 2 \\sin (\\theta)\n\\]\n\nwe find that $\\theta = 5\\pi /6$.\nHence the polar coordinates for this point are $(2, 5\\pi /6)$.\n\\end{solution}\n\nConsider this example. Suppose we used $r=-2$ and $\\theta =2\\pi -(\\pi /6) = 11\\pi /6$. These coordinates specify the same point as above. Observe that there are infinitely many ways to identify this\nparticular point with polar coordinates. In fact, every point can be represented with polar coordinates in infinitely many ways. Because of this, it will usually be\nthe case that $\\theta $ is confined to lie in some interval\nof length $2\\pi $ and $r>0$, for real numbers $r$ and $\\theta$.\n\nJust as with Cartesian coordinates, it is possible to use\nrelations between the polar coordinates to specify points in the plane. The\nprocess of sketching the graphs of these relations is very similar to that used to sketch\ngraphs of functions in Cartesian coordinates. Consider a relation between polar coordinates of the form, $r=f(\\theta)$. To graph such a relation, first make a table of\nthe form\n\\begin{equation*}\n\\begin{tabular}{|l|l|}\n\\hline\n$\\theta $ & $r$ \\\\ \\hline\n$\\theta _{1}$ & $f(\\theta _{1}) $ \\\\ \\hline\n$\\theta _{2}$ & $f(\\theta _{2}) $ \\\\ \\hline\n$\\vdots $ & $\\vdots $ \\\\ \\hline\n\\end{tabular}\n\\end{equation*}\nGraph the resulting points and connect them with a curve. The\nfollowing picture illustrates how to begin this process.\n\n\\begin{center}\n\\begin{tikzpicture}\n\\draw(-0.5,0)--(3,0);\n\\draw(0,-0.5)--(0,3);\n\\draw[blue](0,0)--(2.5,1);\n\\draw[red](0,0)--(2.5,2.5);\n\\draw[fill,blue] (2,0.8) circle [radius=2pt];\n\\draw[fill,red] (1.25,1.25) circle [radius=2pt];\n\\draw[blue](1,0) to [out=90, in=90] (0.9,0.36);\n\\draw[red](0.75,0) to [out=90, in=90] (0.6,0.6);\n\\node[right] at (1,0.25){$\\theta_1$};\n\\node[right] at (0.75, 0.75){$\\theta_2$};\n\\end{tikzpicture}\n\\end{center}\n\nTo find the point in the plane corresponding to the ordered pair $(f(\n\\theta) ,\\theta)$, we follow the same process as when finding the point corresponding to $(r, \\theta)$.\n\nConsider the following example of this procedure, incorporating computer software.\n\n\\begin{example}{Graphing a polar equation}{}\nGraph the polar equation $r=1+\\cos \\theta$.\n\\end{example}\n\n\\begin{solution}\nWe will use the computer software {\\em Maple} to complete this example. The command which produces the polar graph of the above equation is: $>$ plot(1+cos(t),t=\n0..2*Pi,coords=polar). Here we use $t$ to represent the variable $\\theta$ for convenience. The command tells Maple that $r$\nis given by $1+\\cos (t) $ and that $t\\in \\mat{0,2\\pi}$.\n\n\\begin{picture}(1,120)\n\\put(110,-25){\n\\includegraphics[bb=0 0 400\n400,totalheight=3cm]{figures/cardioid.eps}\n\\put(30,73){\\large{x}}\n\\put(-35,128){\\large{y}}}\n\\end{picture}\n\nThe above graph makes sense when considered in terms of trigonometric functions. Suppose $\\theta =0,r=2$ and let $\\theta $ increase to $\\pi /2$. As $\\theta$ increases, $\\cos \\theta $ decreases to 0. Thus the line from the origin to the point on the curve should get shorter as $\\theta $ goes from $0$ to $\\pi /2$. As $\\theta$ goes from $\\pi /2$ to $\\pi$, $\\cos\n\\theta $ decreases, eventually equalling $-1$ at $\\theta =\\pi$. Thus $r=0$\nat this point. This scenario is depicted in the above graph, which shows a function called a \\textbf{cardioid}\\index{cardioid}.\n\nThe following picture illustrates the\nabove procedure for obtaining the polar graph of $r=1+\\cos(\\theta)$. In this picture, the\nconcentric circles correspond to values of $r$ while the rays from the\norigin correspond to the angles which are shown on the picture. The dot on the ray corresponding to the angle $\\pi/6$ is located at a distance of $r = 1+\\cos(\\pi/6)$ from the origin. The dot on the ray corresponding to the angle $\\pi/3$\nis located at a distance of $r = 1+\\cos(\\pi/3)$ from the origin and so\nforth. The polar graph is obtained by connecting such points with a smooth\ncurve, with the result being the figure shown above.\n\n\\begin{picture}(1,290)\n\\put(30,-58){\n\\includegraphics[bb=0 0 400\n400,totalheight=7cm]{figures/25aprilcardioid.eps}\n%\\put(85,128){\\large{x}}\n%\\put(-117,328){\\large{y}}\\put(80,170){$y=\\ln(x)$}\n%\\put(-90,328){$y=\\exp(x)$}\n}\n\\end{picture}\n\\end{solution}\n\nConsider another example of constructing a polar graph.\n\n\\begin{example}{A polar graph}{}\nGraph $r=1+2\\cos \\theta $ for $\\theta \\in \\mat{\n0,2\\pi}$.\n\\end{example}\n\n\\begin{solution}\nThe graph of the polar equation $r=1+2\\cos \\theta $ for $\\theta \\in \\mat{\n0,2\\pi}$ is given as follows.\n\n\\begin{picture}(1,127)\n\\put(110,-25){\n\\includegraphics[bb=0 0 400\n400,totalheight=3cm]{figures/polarpretty2.eps}\n\\put(34,79){\\large{x}}\n\\put(-47,138){\\large{y}}}\n\\end{picture}\n\nTo see the way this is graphed, consider the following picture. First the\nindicated points were graphed and then the curve was drawn to connect the points. When done by a computer, many more points are used to create a more accurate picture.\n\nConsider first the following table of points.\n\\begin{equation*}\n\\begin{tabular}{|l|l|l|l|l|l|l|l|l|}\n\\hline\n$\\theta $ & $\\pi /6$ & $\\pi /3$ & $\\pi /2$ & $5\\pi /6$ & $\\pi $ & $4\\pi /3$\n& $7\\pi /6$ & $5\\pi /3$ \\\\ \\hline\n$r$ & $\\sqrt{3}+1$ & $2$ & $1$ & $1-\\sqrt{3}$ & $-1$ & $0$ & $1-\\sqrt{3}$ & $%\n2$ \\\\ \\hline\n\\end{tabular}%\n\\end{equation*}\n\nNote how some entries in the table have $r<0$. To graph these points, simply move in the opposite direction. These types of points are responsible for the small loop on the inside of the\nlarger loop in the graph.\n\n\\begin{picture}(1,290)\n\\put(30,-58){\n\\includegraphics[bb=0 0 400\n400,totalheight=7cm]{figures/26aprilwriggly.eps}\n%\\put(85,128){\\large{x}}\n%\\put(-117,328){\\large{y}}\\put(80,170){$y=\\ln(x)$}\n%\\put(-90,328){$y=\\exp(x)$}\n}\n\\end{picture}\n\\end{solution}\n\nThe process of constructing these graphs can be greatly facilitated by computer software. However, the use of such software should not replace understanding the steps involved.\n\nThe next example shows the graph for the equation $r=3+\\sin \\paren{\n\\displaystyle\n\\frac{7\\theta }{6}}$. For complicated polar graphs, computer software is used to facilitate the process.\n\n\\begin{example}{A polar graph}{}\nGraph $r=3+\\sin \\paren{\\displaystyle \\frac{7\\theta }{6}%\n} $ for $\\theta \\in \\mat{0,14\\pi}$.\n\\end{example}\n\n\\begin{solution}\n\n\\begin{picture}(1,127)\n\\put(110,-25){\n\\includegraphics[bb=0 0 400\n400,totalheight=3cm]{figures/polarpretty.eps}\n\\put(34,79){\\large{x}}\n\\put(-20,138){\\large{y}}}\n\\end{picture}\n\\end{solution}\n\nThe next example shows another situation in which $r$ can be negative.\n\n\\begin{example}{A polar graph: negative $r$}{}\nGraph $r=3\\sin(4\\theta) $ for $\\theta \\in \\mat{0,2\\pi}$.\n\\end{example}\n\n\\begin{solution}\n\n\\begin{picture}(1,120)\n\\put(120,-24){\n\\includegraphics[bb=0 0 400\n400,totalheight=3cm]{figures/26aprilrose.eps}}\n\\put(242,53){\\large{x}}\n\\put(190,105){\\large{y}}\n\n\\end{picture}\n\\end{solution}\n\nWe conclude this section with an interesting graph of a simple polar equation.\n\n\\begin{example}{The graph of a spiral}{}\nGraph $r=\\theta$ for $\\theta \\in [0,2\\pi]$.\n\\end{example}\n\n\\begin{solution}\nThe graph of this polar equation is a spiral. This is the case because as $\\theta $ increases, so does $r$.\n\n\\begin{picture}(1,290)\n\\put(30,-58){\n\\includegraphics[bb=0 0 400\n400,totalheight=7cm]{figures/11mayspiral.eps}\n\\put(85,205){\\large{x}}\n\\put(-55,335){\\large{y}}}\n\\end{picture}\n\\end{solution}\n\nIn the next section, we will look at two ways of generalizing polar coordinates to three dimensions.\n", "meta": {"hexsha": "0038a01a67e1fa5f84a4b24d0b924f079ec2e04c", "size": 15113, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/content/curvilinearPolar.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "old/content/curvilinearPolar.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/content/curvilinearPolar.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 38.652173913, "max_line_length": 360, "alphanum_fraction": 0.6856348839, "num_tokens": 5032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8577681122619883, "lm_q1q2_score": 0.6475420032037292}} {"text": "\\chapter{Bootstrap}\n\n% \\Chapter{3}{Exercise 3: Bootstrap}\n\\section{Simulation of bootstrap confidence intervals}\nThe third exercise is about the bootstrap method to construct confidence intervals (CI) for different distribution measures, e.g. standard deviation (SD) and median here. The first part is a simulation study to examine coverage properties and length of the bootstrap percentile CI (bCI) for SD and median for a Weibull distribution (with parameters $\\lambda = 13$ and $k=1$ - so it is actually an exponential distribution). Therefore we drew $M$ Monte Carlo samples of $n$ Weibull distributed variables. For each sample we did $R$ bootstrap repetitions where I used a n-out-of-n method with replacement like described in the lecture. The bCI at level $95\\%$ was then defined as the interval between the 0.25th and the 0.75th quantile of the bootstrap distribution for the respective statistic. For each MC sample I checked whether the true SD and median where covered and saved the length of the CIs. Where $M=1000$ was fixed we compare the results for different choices of $n$ and $R$, i.e. we varied the actual sample size and the number of bootstrap runs. The results are shown in Table \\ref{3table}. \n\\begin{table}[hb]\n\\centering\n\\begin{tabular}{rrrrrr}\n \\hline \n && \\multicolumn{2}{r}{Cover probability} & \\multicolumn{2}{r}{Avg CI length} \\\\ \n \\hline\n R & n & \\hspace{0.5cm}med &\\hspace{0.5cm} SD &\\hspace{0.5cm} med &\\hspace{0.5cm} SD \\\\ \n \\hline\n 1000 & 100 & 0.94 & 0.86 & 5.02 & 6.18 \\\\ \n 1000 & 1000 & 0.93 & 0.94 & 1.60 & 2.21 \\\\ \n 5000 & 100 & 0.95 & 0.86 & 5.18 & 6.06 \\\\ \n \\hline\n \\texttt{bcanon}&&&&&\\\\\n \\hline\n 1000 & 100&1.00&1.00 & 3.91 & 7.42\\\\\n \\hline\n\\end{tabular}\n\\caption{Coverage probability and average CI length for median and SD for the MC simulation of $95\\%$ bootstrap percentile CIs with sample size $n$ and $R$ bootstrap repetitions and for the bootstrap accelerated bias-corrected CI using \\texttt{bcanon}.}\n\\label{3table}\n\\end{table}\nAs we took the $95\\%$ bCI we would expect a corresponding coverage probability. For the median this is almost achieved in every row. We also see that the differences between the first and the third row are rather small and that the cover probability lies beneath the desired level. On the other hand the coverage for SD increases notable when we increase the sample size $n$ and is about $95\\%$. Simultaneously the length of the bCI drastically decreased, showing that we get a higher precision for both statistic. This precision is not gained by using more bootstrap repetitions. This demonstrates that the amount of information we can pull from a fix sample size is bounded and can't be arbitrarily increased using more bootstrap samples. On the other hand, increasing the sample size, i.e. collecting more \\textit{independent} information helps estimating better CI. \n\nTo visualise the empirical distribution of the sample statistics I include histograms for the SD and median of the MC samples together with a bCI for an additional sample and histograms of the CI lengths with the length for this additional sample (see Figure \\ref{3hist}). The bCIs of this example cover 0.81 of the samples' SD and 0.89 of the median. This is a bit lower than the ones shown in the first row of Table \\ref{3table} since the empirical estimates vary independently of the CI around the true statistics.\n\\begin{figure}[thb]\n\\centering\n\\includegraphics[width=0.8\\textwidth, keepaspectratio]{ex3/MChist.png}\n\\caption{Histogram for 1000 MC samples of the two statistics (first row) and the length of the $95\\%$ bCI (second row). The red lines show the bCI for an independent sample (first row) and the respective length (second row).}\n\\label{3hist}\n\\end{figure} \n\nInstead of using just the percentiles of the bootstrap distribution there is the approach of bias corrected bootstrap CIs. The function \\texttt{bcanon} provided in the package \\textit{bootstrap} computes an accelerated and bias-corrected bootstrap CI. A MC simulation gives good results (see last row in Table \\ref{3table}) as the true statistics are always covered. The length of the CI for the median is smaller but for the SD it is greater than the ones in the first row. The mean value for the bias correction $\\hat{z}_0$ is about $-0.026$ for the median and $0.124$ for the standard deviation. Taking the value of the standard normal cdf of the values shows us that on average 0.49 of the bootstrap medians are smaller than the sample median. The same holds for 0.55 of the bootstrap standard deviations respectively. On average one would expect those values close to 0.5 and thus the average $\\hat{z}_0$ close to 0. The acceleration constant is computed by a transformation of the Jackknife statistics and lies at a scale of $10^{-15}$ in this case and thus is negligible. Looking at the formula this is not surprising. The median and standard deviation should not change remarkable when leaving out one instance. Especially the median ranges between at most three values anyway. Together with the small sample size of 100 the sum of distances from the Jackknife statistics' mean is not very high. Therefore we can say that the CI is constructed by taking the $\\lfloor R\\alpha_1 \\rfloor$th and the $\\lfloor R\\alpha_2\\rfloor$th instances of the ordered bootstrap statistics, where $\\alpha_1=\\Phi (z_{\\alpha /2}+2\\hat{z}_0)$, $\\alpha_2=\\Phi (z_{1-\\alpha /2}+2\\hat{z}_0)$ and $\\Phi$ denotes the standard normal cdf. \n\n\\section{Application for Sleep Heart Health Study data}\nIn this section we use the bootstrap method to build CIs for a real data set. We are just concerning the variable \\texttt{rdi4p} encoding the respiratory disturbance index. The histogram (Figure \\ref{3histdata}) shows the distribution of the data. As it looks like an exponential distribution (i.e. a Weibull distribution with shape parameter 1), I added the respective density with mean equal to the sample mean ($1/\\lambda = 8.66$). The fit looks good and it would get better when refining the bin size, but for clearness is stay with this size.\n\\begin{figure}[!b]\n\\centering\n\\includegraphics[width=0.5\\textwidth, keepaspectratio]{ex3/hist_data.png}\n\\caption{Histogram of the respiratory disturbance index (\\texttt{rdi4p}) together with the density of a fitted exponential distribution (orange) with $\\lambda = 0.12$.}\n\\label{3histdata}\n\\end{figure}\n\nNow we want to estimate confidence intervals for the median and standard deviation for \\texttt{rdi4p}. We use $R=1000$ bootstrap repetitions. The sample length is $n=5804$ and the sample statistics are $\\hat{x}_{med}=4.19$ and $\\hat{\\sigma}=12.43$. The results for the percentile CI and the accelerated bias-corrected CI are shown in Table \\ref{3tabledata}. We see that the function \\texttt{bcanon} produced \\textbf{NaN} for the CI of the median. This is because is uses the variance of the Jackknife statistics in the definition of the acceleration coefficient. As the median occurs twice in the data the denominator is zero and the acceleration coefficient is not defined. For SD we see that the upper confidence bound is smaller with the corrected method, whereas the lower bound is roughly the same as in the percentile method. It is surprising that the sample value for the standard deviation is not included in the confidence interval for the second method. The percentile CIs include both statistics. Of course, we don't know the true parameters or distribution here, so we can't judge the real precision here. \n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{lrrrr}\n \\hline\n & lower\\_med & upper\\_med & lower\\_sd & upper\\_sd \\\\ \n \\hline\npercentile CI & 3.95 & 4.41 & 11.80 & 13.11 \\\\ \nacc bias-corr CI & NaN & NaN & 11.87 & 11.95 \\\\ \n \\hline\n\\end{tabular}\n\\caption{Confidence intervals for median and SD using the percentile method (first row) and the accelerated bias-corrected CI using \\texttt{bcanon} (second row).}\n\\label{3tabledata}\n\\end{table}\n", "meta": {"hexsha": "ef528d8b406e6332110f288de22f0bf1f7f6f25f", "size": 7929, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ex3/ex3.tex", "max_stars_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_stars_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ex3/ex3.tex", "max_issues_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_issues_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ex3/ex3.tex", "max_forks_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_forks_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 127.8870967742, "max_line_length": 1719, "alphanum_fraction": 0.7686971875, "num_tokens": 2049, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8577681049901037, "lm_q1q2_score": 0.6475419977140745}} {"text": "\\section{Propositional truncations}\\label{sec:propositional-truncation}\n\nThe propositional truncation operation is a universal way of turning type a type $A$ into a proposition $\\brck{A}$. Informally, the proposition $\\brck{A}$ is the proposition that $A$ is inhabited. More precisely, the propositional truncation of $A$ comes equipped with a map $A\\to\\brck{A}$ and it is characterized by its universal property, which asserts that any map $A\\to P$ into a proposition $P$ extends uniquely to a map $\\brck{A}\\to P$, as indicated in the diagram\n\\begin{equation*}\n \\begin{tikzcd}\n A \\arrow[dr] \\arrow[d] \\\\\n \\brck{A} \\arrow[r,densely dotted] & P.\n \\end{tikzcd}\n\\end{equation*}\nUsing the propositional truncation operation we can define many important mathematical concepts, including the image of a map, surjectivity, and connected components. We will discuss those topics in \\cref{chap:image}.\n\n\\subsection{The universal property of propositional truncations}\\label{sec:propositional-truncation-up}\n\n\\begin{defn}\nLet $A$ be a type, and let $f:A\\to P$ be a map into a proposition $P$. We say that $f$ is a \\define{propositional truncation of $A$} if for every proposition $Q$, the precomposition map\n\\begin{equation*}\n\\blank\\circ f:(P\\to Q)\\to (A\\to Q)\n\\end{equation*}\nis an equivalence. This property of $f$ is also called the \\define{universal property of propositional truncation of $A$}\\index{universal property!of propositional truncation}\n\\end{defn}\n\nIn other words, a map $f:A\\to P$ into a proposition $P$ is a propositional truncation of $A$ if every map $g:A\\to Q$ into a proposition extends uniquely along $f$, as indicated in the diagram\n\\begin{equation*}\n \\begin{tikzcd}\n A \\arrow[d,swap,\"f\"] \\arrow[dr,\"g\"] \\\\\n P \\arrow[r,densely dotted] & Q.\n \\end{tikzcd}\n\\end{equation*}\nIndeed, this unique extension property asserts that the type\n\\begin{equation*}\n \\sm{h:P\\to Q}h\\circ f=g\n\\end{equation*}\nis contractible for every $g:A\\to Q$. In other words, the unique extension property asserts that the precomposition function $\\blank\\circ f:(P\\to Q)\\to (A\\to Q)$ is a contractible map, which is the case if and only if it is an equivalence.\n\n\\begin{rmk}\n Note that if $Q$ is a proposition, then the type $X\\to Q$ is a proposition for any type $X$. Furthermore, recall from \\cref{ex:equiv-bi-implication} that the map $(P\\to Q)\\to (A\\to Q)$ is an equivalence as soon as there is a map in the converse direction. Therefore, in order to prove the universal property of the propositional truncation it suffices to show that\n \\begin{equation*}\n (A\\to Q)\\to (P\\to Q).\n \\end{equation*}\n We also note that the universal property of the propositional truncation of a type is formulated with respect to all propositions, regardless of the universe they live in. \n\\end{rmk}\n\n\\begin{eg}\n Suppose $A$ is a type that comes equipped with a point $a:A$, such as the booleans, the type of natural numbers, or the loop space $\\loopspace{A}$ of a pointed type. Then the constant map\n \\begin{equation*}\n \\const_\\ttt: A\\to\\unit\n \\end{equation*}\n is a propositional truncation of $A$. To see this, let $Q$ be an arbitrary proposition. It suffices to show that\n \\begin{equation*}\n (A\\to Q)\\to (\\unit\\to Q).\n \\end{equation*}\n To see this, let $f:A\\to Q$. Then we have $f(a):Q$, so we define $\\const_{f(a)}:\\unit\\to Q$. Thus we see that we have\n \\begin{equation*}\n \\lam{f}\\const_{f(a)}:(A\\to Q)\\to (\\unit\\to Q).\n \\end{equation*}\n This proves that $\\const_\\ttt:A\\to \\unit$ satisfies the universal property of the propositional truncation of $A$. \n\\end{eg}\n\n\\begin{eg}\n If the type $A$ is already a proposition, then the identity function\n \\begin{equation*}\n \\idfunc:A\\to A\n \\end{equation*}\n is a propositional truncation of $A$. To see this, simply note that the precomposittion function with the identity function\n \\begin{equation*}\n \\blank\\circ\\idfunc : (A\\to Q)\\to (A\\to Q)\n \\end{equation*}\n is itself just the identity function. In particular, it is an equivalence.\n\n Similarly, any equivalence $e:P\\simeq P'$ between propositions satisfies the universal property of the propositional truncation of $P$, since precomposing by an equivalence is an equivalence by \\cref{ex:equiv_precomp}.\n\\end{eg}\n\nThe universal property of the propositional truncation determines the propositional truncation up to equivalence. Such proofs of uniqueness from a universal property always follow the same pattern.\n\n\\begin{prp}\\label{prp:propositional-truncation-3-for-2}\n Let $A$ be a type, and consider a commuting triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=tiny]\n \\phantom{P'} & A \\arrow[dl,swap,\"f\"] \\arrow[dr,\"{f'}\"] \\\\\n P \\arrow[rr,swap,\"h\"] & & P'\n \\end{tikzcd}\n \\end{equation*}\n where $P$ and $P'$ are propositions. If any two of the following three assertions hold, so does the third:\n \\begin{enumerate}\n \\item The map $f$ satisfies the universal property of the propositional truncation of $A$.\n \\item The map $f'$ satisfies the universal propertyof the propositional truncation of $A$.\n \\item The map $h$ is an equivalence.\n \\end{enumerate}\n\\end{prp}\n\n\\begin{proof}\n Note that the map $h:P\\to P'$ is an equivalence if and only if for every proposition $Q$, the precomposition map\n \\begin{equation*}\n \\blank\\circ h:(P'\\to Q)\\to (P\\to Q)\n \\end{equation*}\n is an equivalence. Thus, the claim follows by observing that for every proposition $Q$ we have the triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=-1em]\n (P'\\to Q) \\arrow[rr,\"\\blank\\circ h\"] \\arrow[dr,swap,\"\\blank\\circ {f'}\"] & & (P\\to Q) \\arrow[dl,\"\\blank\\circ f\"] \\\\\n & (A\\to Q). & \\phantom{(P'\\to Q)}\n \\end{tikzcd}\n \\end{equation*}\n\\end{proof}\n\n\\begin{cor}\\label{cor:uniquely-unique-brck}\n Consider two maps $f:A\\to P$ and $f':A\\to P'$ into propositions $P$ and $P'$, both satisfying the universal property of the propositional truncation of $A$. Then the type of equivalences $e:P \\simeq P'$ for which the triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=tiny]\n \\phantom{P'} & A \\arrow[dl,swap,\"f\"] \\arrow[dr,\"{f'}\"] \\\\\n P \\arrow[rr,swap,\"e\"] & & P'\n \\end{tikzcd}\n \\end{equation*}\n commutes, is contractible.\n\\end{cor}\n\n\\begin{rmk}\n Note that the triangles in \\cref{prp:propositional-truncation-3-for-2,cor:uniquely-unique-brck} always commutes, since $P$ and $P'$ are assumed to be propositions.\n\\end{rmk}\n\nNow that we have shown that propositional truncations are determined uniquely, we will assume that any universe is closed under propositional truncations.\n\n\\begin{axiom}\\label{axiom:propositional-truncations}\n Any universe $\\UU$ is closed under propositional truncations: for any type $A:\\UU$ there is a proposition $\\brck{A}:\\UU$ equipped with a map $\\eta:A\\to\\brck{A}$ that satisfies the universal property of the propositional truncation.\n\\end{axiom}\n\nThe propositional truncation is therefore an operation\n\\begin{equation*}\n \\brck{\\blank}:\\UU\\to\\UU\n\\end{equation*}\non the universe. One simple application of the universal property of the propositional truncation is that $\\brck{\\blank}$ also acts on functions in a functorial way.\n\n\\begin{prp}\n There is a map\n \\begin{equation*}\n \\brck{\\blank}:(A\\to B)\\to (\\brck{A}\\to\\brck{B})\n \\end{equation*}\n for any two types $A$ and $B$, such that\n \\begin{align*}\n \\brck{\\idfunc} & \\htpy \\idfunc \\\\\n \\brck{g\\circ f} & \\htpy \\brck{g}\\circ\\brck{f}.\n \\end{align*}\n\\end{prp}\n\n\\begin{proof}\n For any $f:A\\to B$, the map $\\brck{f}:\\brck{A}\\to\\brck{B}$ is defined to be the unique extension\n \\begin{equation*}\n \\begin{tikzcd}\n A \\arrow[d,swap,\"\\eta\"] \\arrow[r,\"f\"] & B \\arrow[d,\"\\eta\"] \\\\\n \\brck{A} \\arrow[r,densely dotted,swap,\"\\brck{f}\"] & \\brck{B}.\n \\end{tikzcd}\n \\end{equation*}\n To see that $\\brck{\\blank}$ preserves identity maps and compositions, simply note that $\\idfunc[\\brck{A}]$ is an extension of $\\idfunc[A]$, and that $\\brck{g}\\circ\\brck{f}$ is an extension of $g\\circ f$. Hence the homotopies are obtained by uniqueness.\n\\end{proof}\n\n\\subsection{Propositional truncations as higher inductive types}\n\nThe idea of higher inductive types is that types can be generated inductively not only by point constructors, such as $\\zeroN:\\N$ and $\\succN:\\N\\to\\N$, but also by path constructors. One of the first examples of a higher inductive type was the propositional truncation of a type $A$. This is a type $\\brck{A}$ equipped with one point constructor\n\\begin{equation*}\n \\eta : A \\to \\brck{A},\n\\end{equation*}\none path constructor\n\\begin{equation*}\n \\alpha : \\prd{x,y:\\brck{A}}x=y.\n\\end{equation*}\nNote that the path constructor $\\alpha$ immediately proves that $\\brck{A}$ is a proposition. Now we should formulate the induction principle for the propositional truncation.\n\nJust as we did with the universal property, we will formulate the induction principle of the propositional truncation for an arbitrary map $f:A\\to P$ into a proposition $P$. When the induction principle is formulated in this way, we will be able to show that $f$ satisfies the universal property if and only if it satisfes the induction principle.\n\nConsider a map $f:A\\to P$ into a type equipped with a family of paths\n\\begin{equation*}\n \\alpha : \\prd{p,q:P}p=q\n\\end{equation*}\nwitnessing that $P$ is a proposition, and consider a type family $B$ over $P$. The induction principle of the propositional truncation of $A$ tells us what we have to do in order to construct a dependent function $h:\\prd{p:P}B(p)$.\n\nIn order to figure out what the induction principle has to be, we first note that if we start with a dependent function $h:\\prd{p:P}B(p)$, then we also obtain the function $h\\circ f : \\prd{x:A}B(f(x))$. In other words, if we think of $f:A\\to P$ as the point constructor of $P$, then the function $h\\circ f$ describes the action of $h$ on the points of $P$. The first requirement in the induction principle is therefore that $B$ must come equipped with a dependent function\n\\begin{equation*}\n g:\\prd{x:A}B(f(x)).\n\\end{equation*}\nFurthermore, the function $h$ acts on the paths in $P$ via its dependent action on paths, which we constructed in \\cref{defn:apd}. The paths in $P$ are generated by $\\alpha$, so we obtain a function\n\\begin{equation*}\n \\lam{p}{q}\\apd{h}{\\alpha(p,q)} : \\prd{p,q:P} \\tr_B(\\alpha(p,q),h(p))=h(q).\n\\end{equation*}\nThe induction principle must ensure that any function $h$ defined via the induction principle, satisfies this law. Therefore, the second condition in the induction principle is that we must have a family of identifications\n\\begin{equation*}\n \\prd{p,q:P}{y:B(p)}{z:B(q)}\\tr_B(\\alpha(p,q),y)= z.\n\\end{equation*}\nWe now formulate the induction principle for propositional truncation.\n\n\\begin{defn}\n Consider a map $f:A\\to P$ into a type $P$ equipped with a family of paths\n \\begin{equation*}\n \\alpha : \\prd{p,q:P}p=q,\n \\end{equation*}\n witnessing that $P$ is a proposition. We say that $f$ satisfies the induction principle for the propositional truncation of $A$ if for any family $B$ over $P$ that comes equipped with\n \\begin{align*}\n g & : \\prd{x:A}B(f(x)) \\\\\n \\beta & : \\prd{p,q:P}{y:B(p)}{z:B(q)}\\tr_B(\\alpha(p,q),y)=z,\n \\end{align*}\n there is a dependent function $h:\\prd{p:P}B(p)$ equipped with a homotopy\n \\begin{equation*}\n \\prd{x:A}h(f(x))= g(x).\n \\end{equation*}\n\\end{defn}\n\nIn the following lemma we show that if a family $B$ over $\\brck{A}$ comes equipped with a family of paths\n\\begin{equation*}\n \\prd{x,y:\\brck{A}}{u:B(x)}{v:B(y)}\\tr_B(\\alpha(x,y),u)=v,\n\\end{equation*}\nthen $B$ must be a family of propositions.\n\n\\begin{lem}\\label{lem:case-paths-induction-principle-propositional-truncation}\n Let $P$ be a type equipped with a family of paths\n \\begin{equation*}\n \\alpha : \\prd{p,q:P}p=q,\n \\end{equation*}\n showing that $P$ is a proposition, and consider a type family $B$ over $P$. The following are equivalent:\n \\begin{enumerate}\n \\item The family $B$ comes equipped with a family of identifications\n \\begin{equation*}\n \\beta:\\prd{p,q:P}{x:B(p)}{y:B(q)}\\tr_B(\\alpha(p,q),x)=y,\n \\end{equation*}\n \\item The family $B$ is a family of propositions.\n \\end{enumerate}\n\\end{lem}\n\n\\begin{proof}\n Assuming that (i) holds, we will show that each $B(p)$ is a proposition by showing that\n \\begin{equation*}\n B(p)\\to\\iscontr(B(p)).\n \\end{equation*}\n Let $x:B(p)$. We have to construct a center of contraction and a contraction. Our plan is to use $\\beta$ to define the contraction, so it is natural to define the center as $\\tr_B(\\alpha(p,p),x)$. Now we take\n \\begin{equation*}\n \\beta(p,p,x):\\prd{y:B(p)}\\tr_B(\\alpha(p,p),x)=y\n \\end{equation*}\n as our contraction. This completes the proof that $B$ is a family of propositions.\n\n The converse is immediate: if $B$ is a family of propositions, then any two terms in any $B(q)$ can be identified.\n\\end{proof}\n\n\\begin{defn}\n Consider a map $f:A\\to P$ into a proposition $P$. We say that $f$ satisfies the dependent universal property of the propositional truncation of $A$, if for any family $Q$ of propositions over $P$, the precomposition map\n \\begin{equation*}\n \\blank\\circ f : \\Big(\\prd{p:P}Q(p)\\Big)\\to\\Big(\\prd{x:A}Q(f(x))\\Big)\n \\end{equation*}\n is an equivalence.\n\\end{defn}\n\n\\begin{thm}\n Consider a map $f:A\\to P$ into a proposition $P$. Then the following are equivalent:\n \\begin{enumerate}\n \\item The map $f$ is a propositional truncation.\n \\item The map $f$ satisfies the dependent universal property of the propositional truncation.\n \\item The map $f$ satisfies the induction principle of the propositional truncation.\n \\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\n We will first show that (i) and (ii) are equivalent. Of course, the universal property is a special case of the dependent universal property, so the fact that (ii) implies (i) is immediate. We now show that (i) implies (ii). Let $Q$ be a family of propositions over $P$, and consider the following commuting diagram:\n \\begin{equation*}\n \\begin{tikzcd}[column sep=8em]\n \\Big(\\sm{h:P\\to P}\\prd{p:P}Q(h(p))\\Big) \\arrow[r,\"{\\tot[\\blank\\circ f]{\\blank\\circ f}}\"] \\arrow[d,swap,\"\\choice^{-1}\"] & \\Big(\\sm{g:A\\to P}\\prd{x:A}Q(g(x))\\Big) \\arrow[d,\"\\choice^{-1}\"] \\\\\n \\Big(P\\to \\sm{p:P}Q(p)\\Big) \\arrow[r,\"\\blank\\circ f\"] \\arrow[d,swap,\"\\proj 1\\circ\\blank\"] & \\Big(A \\to \\sm{p:P}Q(p)\\Big) \\arrow[d,\"\\proj 1\\circ\\blank\"] \\\\\n \\Big(P\\to P\\Big) \\arrow[r,swap,\"\\blank\\circ f\"] & \\Big(A\\to P\\Big)\n \\end{tikzcd}\n \\end{equation*}\n In this diagram the bottom map is an equivalence by the universal property of the propositional truncation of $A$. Note also that the type $\\sm{p:P}Q(p)$ is a proposition by \\cref{ex:istrunc_sigma}, so it follows that also the middle map is an equivalence. Furthermore, the type theoretic choice maps are equivalences by \\cref{thm:choice}, so it also follows that the top map is an equivalence. Now we use \\cref{thm:equiv-toto} to conclude that the family of maps\n \\begin{equation*}\n \\blank\\circ f: \\Big(\\prd{p:P}Q(h(p))\\Big)\\to\\Big(\\prd{x:A}Q(h(f(x)))\\Big)\n \\end{equation*}\n indexed by $h:P\\to P$ is a family of equivalences. The dependent universal property is now just a special case: take $h\\jdeq\\idfunc$. This completes the proof that (i) is equivalent to (ii).\n\n It remains to show that (ii) is equivalent to (iii). By \\cref{lem:case-paths-induction-principle-propositional-truncation} it follows that the induction principle is equivalent to the property that for each family $Q$ of propositions over $P$, the precomposition map\n \\begin{equation*}\n \\blank\\circ f : \\Big(\\prd{p:P}Q(p)\\Big)\\to\\Big(\\prd{x:A}Q(f(x))\\Big)\n \\end{equation*}\n has a section. Since the domain and codomain of this map are propositions by \\cref{thm:trunc_pi}, we see that this precomposition map has a section if and only if it is an equivalence.\n\\end{proof}\n\n\\begin{exercises}\n \\exercise Let $A$ be a type and let $P$ be a proposition, and suppose that $P$ is a retract of $A$. Show that the retraction $A\\to P$ is a propositional truncation.\n% \\exercise Show that the relation $x,y\\mapsto\\brck{x=y}$ is an equivalence relation, on any type.\n \\exercise Consider two maps $f:A\\to P$ and $g:B\\to Q$ into propositions $P$ and $Q$. Recall from \\cref{ex:istrunc_sigma} that the type $P\\times Q$ is also a proposition. Show that if both $f$ and $g$ are propositional truncations then the map $f\\times g : A\\times B\\to P\\times Q$ is also a propositional truncation. Conclude that\n \\begin{equation*}\n \\brck{A\\times B}\\simeq \\brck{A}\\times\\brck{B}. \n \\end{equation*}\n \\exercise Consider two propositions $P$ and $Q$, and define\n \\begin{align*}\n P\\land Q & \\defeq P\\times Q\\\\\n P\\vee Q & \\defeq \\brck{P+Q}.\n \\end{align*}\n \\begin{subexenum}\n \\item Construct maps $i:P\\to P\\vee Q$ and $j:Q\\to P\\vee Q$.\n \\item Prove the universal property of disjunction, i.e., show that for any proposition $R$, the map\n \\begin{equation*}\n (P\\vee Q\\to R) \\to (P\\to R)\\land (Q\\to R) \n \\end{equation*}\n given by $h\\mapsto (h\\circ i,h\\circ j)$ is an equivalence.\n \\end{subexenum}\n \\exercise Consider a family $P$ of propositions over a type $A$, and define\n \\begin{align*}\n \\forall_{(x:A)}P(x) & \\defeq \\prd{x:A}P(x) \\\\\n \\exists_{(x:A)}P(x) & \\defeq \\Brck{\\sm{x:A}P(x)}\n \\end{align*}\n \\begin{subexenum}\n \\item Construct a map $i_a : P(a)\\to \\exists_{(x:A)}P(x)$ for each $a:A$.\n \\item Prove the universal property of the existential quantification, i.e. show that for any proposition $Q$, the map\n \\begin{equation*}\n \\Big(\\Big(\\exists_{(x:A)}P(x)\\Big)\\to Q\\Big)\\to \\Big(\\forall_{(x:A)}(P(x)\\to Q)\\Big)\n \\end{equation*}\n given by $h\\mapsto \\lam{x}h\\circ i_x$, is an equivalence.\n \\end{subexenum}\n \\exercise Show that\n \\begin{equation*}\n \\eqv{\\brck{A}}{\\prd{P:\\prop}(A\\to P)\\to P}\n \\end{equation*}\n for any type $A:\\UU$. This is called the \\define{impredicative encoding} of the propositional truncation.\n % \\exercise For any $B:A\\to\\UU$, construct an equivalence\n % \\begin{equation*}\n % \\eqv{\\Big(\\exists_{(a:A)}\\brck{B(a)}\\Big)}{\\brck{\\sm{a:A}B(a)}}\n % \\end{equation*}\n\\end{exercises}", "meta": {"hexsha": "b31c6db535b92995b07123fc62ceb85249f915bd", "size": 18057, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/propositional-truncation.tex", "max_stars_repo_name": "UlrikBuchholtz/HoTT-Intro", "max_stars_repo_head_hexsha": "1e1f8def50f9359928e52ebb2ee53ed1166487d9", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 333, "max_stars_repo_stars_event_min_datetime": "2018-09-26T08:33:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T23:50:15.000Z", "max_issues_repo_path": "Book/propositional-truncation.tex", "max_issues_repo_name": "UlrikBuchholtz/HoTT-Intro", "max_issues_repo_head_hexsha": "1e1f8def50f9359928e52ebb2ee53ed1166487d9", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2019-06-18T04:16:04.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-16T15:27:01.000Z", "max_forks_repo_path": "Book/propositional-truncation.tex", "max_forks_repo_name": "UlrikBuchholtz/HoTT-Intro", "max_forks_repo_head_hexsha": "1e1f8def50f9359928e52ebb2ee53ed1166487d9", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 30, "max_forks_repo_forks_event_min_datetime": "2018-09-26T09:08:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-16T00:33:50.000Z", "avg_line_length": 56.2523364486, "max_line_length": 472, "alphanum_fraction": 0.7013346625, "num_tokens": 5697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085909370423, "lm_q2_score": 0.8244619263765707, "lm_q1q2_score": 0.6474570336840242}} {"text": "\\documentclass{article}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n% Required to use \\cfrac\n\\usepackage{amsmath}\n\n% Required to use \\sfrac\n\\usepackage{xfrac}\n\n\\begin{document}\n\n\\section*{Simple fraction}\n\nFractions are created as follows.\n\n% Syntax: \\frac{numerator}{denominator}\n\n\\[\\frac{1}{x} = x^{-1}\\]\n\n\\section*{Continued (or cascade) fractions}\n\nA fraction can contain another one. If the command \\emph{\\textbackslash frac} is, this would be the result:\n\n\\[\\frac{x}{x + \\frac{1}{2}}\\]\n\nThe sub fraction is very small. This is why the command \\emph{\\textbackslash cfrac} is recommended for continued fractions:\n\n\\[\\cfrac{x}{x + \\cfrac{1}{2}}\\]\n\nSome people prefer to still use \\emph{\\textbackslash frac} and use the command \\emph{\\textbackslash displaystyle} before each sub fractions:\n\n\\[\\frac{x}{x + \\displaystyle\\frac{1}{2}}\\]\n\nDespite this practice is sometimes promoted, it should be avoided because the equation has some spacing issues, such as on top of $\\frac{1}{2}$.\n\n\\section*{Slanted fractions}\n\nSlanted fractions are created as follows : $\\sfrac{1}{2}$.\n\n\\end{document}", "meta": {"hexsha": "d2314c0419195fc5caaa8323c6f2314508d35b56", "size": 1098, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compendium/mathematics/fractions.tex", "max_stars_repo_name": "ZenLulz/LatexCompendium", "max_stars_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-07-30T21:43:55.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-23T20:16:19.000Z", "max_issues_repo_path": "compendium/mathematics/fractions.tex", "max_issues_repo_name": "ZenLulz/LatexCompendium", "max_issues_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compendium/mathematics/fractions.tex", "max_forks_repo_name": "ZenLulz/LatexCompendium", "max_forks_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.1428571429, "max_line_length": 144, "alphanum_fraction": 0.7276867031, "num_tokens": 316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619436290699, "lm_q2_score": 0.7853085733507947, "lm_q1q2_score": 0.6474570327333682}} {"text": "\\section{Network Throughput, Bandwidth and Safety Parameters}\nThe parameters of the flow and load of the system (about 80\\% at peak hours 9h-11h and 15h-16h) can be shared for Headquarter and Branches as follows:\n\\begin{itemize}\n \\item Servers used for updates, web access, database access, ..... The total upload and download capacity is about 500 MB / day.\n \\item Each workstation is used for Web browsing, document downloads, customer transactions, ... The total upload and download capacity is about 100 MB / day.\n \\item WiFi-connected laptop for customers’ accesses about 50 MB / day.\n\\end{itemize}\n\n\\subsection{Headquarter}\n\\subsubsection{Server}\nThe total peak hours time in a days is 3 hours (9h-11h and 15h-16h) and may consume up to 80\\%. \\\\\n$$Bandwidth = \\, \\frac{5*500*0.8}{3*3600}*8 = 1.481Mbps$$\n$$Throughput = \\, \\frac{5*500}{8*3600}*8 = 0.694Mbps$$\n\\subsubsection{Workstation}\nAssume each workstation will work 8 hours a day\n$$Bandwidth = \\, \\frac{100*100*0.8}{3*3600}*8 = 5.926Mbps$$\n$$Throughput = \\, \\frac{100*100}{8*3600}*8 = 2.778Mbps$$\n\\subsubsection{User}\nAssume that the number of customers in one day is about 200 customers, and in peak\ntime is 150 customers. Also, assume that each user access Wifi 8 hours a day.\n$$Bandwidth = \\, \\frac{150*50}{3*3600}*8 = 5.556Mbps$$\n$$Throughput = \\, \\frac{200*50}{8*3600}*8 = 2.778Mbps$$\n\n\\subsubsection{Total}\n$$Bandwidth = 1.481+5.926+5.556 = 12.963Mbps$$\n$$Throughput = 0.694+2.778+2.778 = 6.25Mbps$$\n\n\\subsection{Safety parameters}\nTo ensure for the next 10 years, we need to consider the rate 20\\% per five years. The first 5 years, the number of customer will still not increase and this number will increase in the next 5 years.\n$$Bandwidth = 12.963*1.2 = 15.556Mbps$$\n$$Throughput =6.25*1.2 = 7.5Mbps$$\n\n\\subsection{Branch}\n\\subsubsection{Server}\nThe total peak hours time in a days is 3 hours (9h-11h and 15h-16h) and may consume up to 80\\%. \\\\\n$$Bandwidth = \\, \\frac{3*500*0.8}{3*3600}*8 = 0.889Mbps$$\n$$Throughput = \\, \\frac{3*500}{8*3600}*8 = 0.417Mbps$$\n\\subsubsection{Workstation}\nAssume each workstation will work 8 hours a day\n$$Bandwidth = \\, \\frac{50*100*0.8}{3*3600}*8 = 2.963Mbps$$\n$$Throughput = \\, \\frac{50*100}{8*3600}*8 = 1.389Mbps$$\n\\subsubsection{User}\nAssume that the number of customers in one day is about 140 customers, and in peak\ntime is 70 customers. Also, assume that each user access Wifi 8 hours a day.\n$$Bandwidth = \\, \\frac{70*50}{3*3600}*8 = 2.593Mbps$$\n$$Throughput = \\, \\frac{140*50}{8*3600}*8 = 2.778Mbps$$\n\n\\subsubsection{Total}\n$$Bandwidth = 0.889+2.963+2.593= 6.445Mbps$$\n$$Throughput =0.417+1.389+2.778= 4.584Mbps$$\n\n\\subsection{Safety parameters}\nTo ensure for the next 10 years, we need to consider the rate 20\\% per five years. The first 5 years, the number of customer will still not increase and this number will increase in the next 5 years.\n$$Bandwidth = 6.445*1.2 = 7.734Mbps$$\n$$Throughput =4.584*1.2 = 5.501Mbps$$\n", "meta": {"hexsha": "27bf0f768e33e44948ddfb768b3b672166a7f70a", "size": 2936, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/step3.tex", "max_stars_repo_name": "Smithienious/CO3094-asg2", "max_stars_repo_head_hexsha": "8c301fcbcbe2a7deceeb50105ed5d6624e6ad413", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/step3.tex", "max_issues_repo_name": "Smithienious/CO3094-asg2", "max_issues_repo_head_hexsha": "8c301fcbcbe2a7deceeb50105ed5d6624e6ad413", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/step3.tex", "max_forks_repo_name": "Smithienious/CO3094-asg2", "max_forks_repo_head_hexsha": "8c301fcbcbe2a7deceeb50105ed5d6624e6ad413", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-11-24T14:29:08.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-24T14:29:08.000Z", "avg_line_length": 52.4285714286, "max_line_length": 199, "alphanum_fraction": 0.7132152589, "num_tokens": 1008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619350028204, "lm_q2_score": 0.785308580887758, "lm_q1q2_score": 0.6474570321730398}} {"text": "\\appendix\n\\section{Appendix: Example Primer}\n\\label{sec:examplePrimer}\nIn this Appendix, a set of examples are reported. In order to be as general as possible, the \\textit{Model} type ``ExternalModel'' has been used.\n%%%% EXAMPLE 1\n\\subsection{Example 1.}\n\\label{subsec:ex1}\nThis simple example is about the construction of a ``Lorentz attractor'', sampling the relative input space. The parameters that are sampled represent the initial coordinate (x0,y0,z0) of the attractor origin.\n\n\\begin{lstlisting}[style=XML,morekeywords={debug,re,seeding,class,subType,limit}]\n\n\n\n\n externalModel\n FirstMRun\n 3\n\n\n\n lorentzAttractor\n\n\n\n \n lorentzAttractor.py\n PythonModule\n MC_external\n testPrintHistorySet\n test_external_db\n testPrintHistorySet_dump\n \n\n\n\n \n sigma,rho,beta,x,y,z,time,x0,y0,z0\n \n\n\n\n \n 4\n 1\n \n \n 4\n 1\n \n \n 4\n 1\n \n\n\n\n \n \n 3\n \n \n x0_distrib\n \n \n y0_distrib\n \n \n z0_distrib\n \n \n\n\n\n \n\n\n\n \n csv\n testPrintHistorySet\n \n\n\n\n \n x0,y0,z0\n time,x,y,z\n \n\n\n\\end{lstlisting}\nThe Python \\textit{ExternalModel} is reported below:\n\\begin{lstlisting}[language=python]\nimport numpy as np\n\ndef run(self,Input):\n max_time = 0.03\n t_step = 0.01\n\n numberTimeSteps = int(max_time/t_step)\n\n self.x = np.zeros(numberTimeSteps)\n self.y = np.zeros(numberTimeSteps)\n self.z = np.zeros(numberTimeSteps)\n self.time = np.zeros(numberTimeSteps)\n\n self.x0 = Input['x0']\n self.y0 = Input['y0']\n self.z0 = Input['z0']\n\n self.x[0] = Input['x0']\n self.y[0] = Input['y0']\n self.z[0] = Input['z0']\n self.time[0]= 0\n\n for t in range (numberTimeSteps-1):\n self.time[t+1] = self.time[t] + t_step\n self.x[t+1] = self.x[t] + self.sigma*\n (self.y[t]-self.x[t]) * t_step\n self.y[t+1] = self.y[t] + (self.x[t]*\n (self.rho-self.z[t])-self.y[t]) * t_step\n self.z[t+1] = self.z[t] + (self.x[t]*\n self.y[t]-self.beta*self.z[t]) * t_step\n\\end{lstlisting}\n%%%% EXAMPLE 2\n\\subsection{Example 2.}\n\\label{subsec:ex1}\nThis example shows a slightly more complicated example, that employs the usage of:\n\\begin{itemize}\n \\item \\textit{Samplers:} Grid and Adaptive;\n \\item \\textit{Models:} External, Reduce Order Models and Post-Processors;\n \\item \\textit{OutStreams:} Prints and Plots;\n \\item \\textit{Data Objects:} PointSets;\n \\item \\textit{Functions:} ExternalFunctions.\n\\end{itemize}\nThe goal of this input is to compute the ``SafestPoint''.\nIt provides the coordinates of the farthest\npoint from the limit surface that is given as an input.\n%\nThe safest point coordinates are expected values of the coordinates of the\nfarthest points from the limit surface in the space of the ``controllable''\nvariables based on the probability distributions of the ``non-controllable''\nvariables.\n\nThe term ``controllable'' identifies those variables that are under control\nduring the system operation, while the ``non-controllable'' variables are\nstochastic parameters affecting the system behaviour randomly.\n\nThe ``SafestPoint'' post-processor requires the set of points belonging to the\nlimit surface, which must be given as an input.\n\n\\begin{lstlisting}[style=XML,morekeywords={debug,re,seeding,class,subType,limit}]\n\n\n\n\n SafestPointPP\n pth1,pth2,pth3,pth4\n 50\n\n\n\n\n \n grd_vl_ql_smp_dpt\n grd_vl_ql_smp_dpt_dt\n xtr_mdl\n nt_phy_dpt_dt\n \n\n \n dpt_smp\n bln_smp_dt\n xtr_mdl\n nt_phy_dpt_dt\n lmt_srf_dt\n \n\n \n lmt_srf_dt\n SP\n sfs_pnt_dt\n \n\n \n \tlmt_srf_dt\n \tlmt_srf_dmp\n sfs_pnt_dt\n \tsfs_pnt_dmp\n \n\n\n\n\n \n x1,x2,gammay\n OutputPlaceHolder\n \n\n \n x1,x2,gammay\n g\n \n\n \n x1,x2,gammay\n OutputPlaceHolder\n \n\n \n x1,x2,gammay\n g_zr\n \n\n \n x1,x2,gammay\n p\n \n\n\n\n\n \n 10\n -10\n \t0.5\n 0.1\n \n\n \n 10\n -10\n -0.15\n 0.05\n \n\n \n 20\n -20\n 0\n 15\n \n\n\n\n\n \n \n x1_dst\n 2\n \n \n x2_dst\n 2\n \n \n gammay_dst\n 4\n \n \n\n \n accelerated_ROM\n g_zr\n nt_phy_dpt_dt\n 1e-2\n \n x1_dst\n \n \n x2_dst\n \n \n gammay_dst\n \n \n\n\n\n\n \n x1,x2,gammay,g\n \n\n \n x1,x2,gammay\n g_zr\n svm|SVC\n rbf\n 10\n 1e-5\n 50\n \n\n \n \n x1_dst\n x2_dst\n gammay_dst\n \n \n \t\n \t\tx1_dst\n \t\t1\n \t\n \t\n \t\tx2_dst\n \t\t1\n \t\n \n \n \t\n \t\tgammay_dst\n \t\t2\n \t\n \n \n\n\n\n\n \n g\n \n\n\n\n\n \n \tcsv\n \tlmt_srf_dt\n \n\n \n \tcsv\n \tsfs_pnt_dt\n \n\n\n\n\\end{lstlisting}\nThe Python \\textit{ExternalModel} is reported below:\n\\begin{lstlisting}[language=python]\ndef run(self,Input):\n self.g = self.x1+4*self.x2-self.gammay\n\\end{lstlisting}\nThe ``Goal Function'',the function that defines the transitions with respect the input space coordinates, is as follows:\n\\begin{lstlisting}[language=python]\ndef __residuumSign(self):\n if self.g<0 : return 1\n else : return -1\n\\end{lstlisting}\n\n%%%%% EXAMPLE 3\n%\\subsection{Example3}\n%\\label{subsec:ex1}\n%example 3\n", "meta": {"hexsha": "29fb0f31a68c81760d5167b299259ba1bf43aa3f", "size": 11769, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/user_manual/examplesPrimer.tex", "max_stars_repo_name": "rinelson456/raven", "max_stars_repo_head_hexsha": "1114246136a2f72969e75b5e99a11b35500d4eef", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 159, "max_stars_repo_stars_event_min_datetime": "2017-03-24T21:07:06.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-20T13:44:40.000Z", "max_issues_repo_path": "doc/user_manual/examplesPrimer.tex", "max_issues_repo_name": "rinelson456/raven", "max_issues_repo_head_hexsha": "1114246136a2f72969e75b5e99a11b35500d4eef", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1667, "max_issues_repo_issues_event_min_datetime": "2017-03-27T14:41:22.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T19:50:06.000Z", "max_forks_repo_path": "doc/user_manual/examplesPrimer.tex", "max_forks_repo_name": "rinelson456/raven", "max_forks_repo_head_hexsha": "1114246136a2f72969e75b5e99a11b35500d4eef", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 95, "max_forks_repo_forks_event_min_datetime": "2017-03-24T21:05:03.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-08T17:30:22.000Z", "avg_line_length": 32.874301676, "max_line_length": 209, "alphanum_fraction": 0.6455943581, "num_tokens": 3537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619134371953, "lm_q2_score": 0.785308580887758, "lm_q1q2_score": 0.6474570152373694}} {"text": "\\section{Loss function and Gradient Descent} \\label{sec:loss_&_gradient_descent}\nSo far only the concept of what a neural network is was explained, but not how it can learn from data, to understand this it is worth to have first a brief summary of what consists a neural network.\n\nA collection of units having nonlinear activation functions form a layer; layers are stacked from the input to the output with some optional hidden layers in-between; the weighted input of a unit is calculated by multiplying its inputs with some weight parameters and adding a bias term; the unit activation is the output of its activation function applied to the weighted input; and by varying the values of the weights and biases it is possible to reach different model behaviours \\textbf{--} this section will only consider the case when the inputs of a unit come from the immediate previous layer and the output is calculated by passing the input values from layer to layer until the last layer (this is called a \\textit{feedforward} neural network), other configurations like \\acp{ResNet} \\cite{resnet2015} and \\acp{RNN} also exist, but it is possible to extend the arguments present here to also explain learning in these contexts.\n\nFor all neural networks, learning is the process of changing the weights and biases (and possibly other values) in order to produce the desired result. The set of values that are learned during training are called the \\textit{parameters} of the network.\n\nThe number of parameters gives the degree of freedom that the network has, in a larger parameter space the network has more possible choices and can construct more complex relations. On the other hand, fewer parameters may make it difficult or impossible to represent the full complexity of the data. However, besides requiring more computational power, very complex models break the idea of sparsity and can more easily fall into the problem of overfitting (see \\autoref{sub:overfitting}).\n\nThe number of parameters in the network is given by its architecture, how many layers it has, how many units in each layer, how the connections are made, and so on. This is a choice made when building the network and is not something that is learned by the algorithm. The set of properties that influence the network but that are not learned are usually called the \\textit{hyperparameters}. \n\nAnother example of hyperparameter is the activation functions used in the units, they are chosen when constructing the neural network and can't be learned \\textbf{--} however some activations like PReLU have internal learnable parameters \\cite{prelu2015}.\n\nLearning is then the process of finding optimal parameters for a network given some hyperparameters. But how can a machine automatically learn the parameters? To do this the network must have some way to quantify the results that it produces, a way to distinguish between good and bad outputs, this is what is called the \\textit{loss function}.\n\nAlso called the cost or objective, the loss function $J(\\bm{x}, \\bm{\\theta})$ is a function that returns how good the output of the network is, given an input $\\bm{x}$ and a set of network parameters $\\bm{\\theta}$ (represents all trainable parameters: weights, biases, and any others). Learning can then be described as the process of adapting the parameters $\\bm{\\theta}$ in order to minimize or maximize the objective function (usually minimize, depends on the function used). The choice of function will depend on the type of problem, but two of the most common methods are \\gls{MSE} and Categorical Cross Entropy.\n\n\\subsection{Mean Squared Error} \\label{sub:mse}\nThis function is the standard squared euclidean distance between two vectors. For a given input $\\bm{x}$, the network output $\\hat{\\bm{y}}$ for this input, and the true desired output $\\bm{y}$ (e.g. the corresponding label), the squared distance is calculated as shown in \\autoref{eq:squared_distance} \\textbf{--} note that the subscript $2$ indicates the euclidean distance (the $\\ell_2$ norm).\n\\begin{equation} \\label{eq:squared_distance}\n \\| \\bm{y} - \\hat{\\bm{y}} \\|_{2}^{2} = \\sum_{i}{(y_j - \\hat{y_i})^2}\n\\end{equation}\n\nThis distance can also be generalize to matrices or any $n$ dimensional values of $\\bm{y}$ by just calculating the element-wise squared differences between $\\bm{y}$ and $\\hat{\\bm{y}}$.\n\nHowever, simply calculating the distance for one possible input is not good for evaluating how well the network is doing in general. To better evaluate the network's performance it is necessary to see how well it is doing in the entire dataset or some subset of it. For a set $\\bm{X}$ of $n$ inputs, the \\gls{MSE} is the mean of the squared distances as shown in \\autoref{eq:mse}.\n\\begin{equation} \\label{eq:mse}\n J(\\bm{X}, \\bm{\\theta}) = \\frac{1}{n}\\sum^{n}||\\bm{y} - \\hat{\\bm{y}}||_{2}^{2} = \\frac{1}{n}\\sum^{n}\\sum_{i}{(y_i - \\hat{y_i})^2}\n\\end{equation}\n\nThis loss function is usually used for regression problems, where the output can have a range of values and is not simply defined as a $0$ or a $1$. It is also useful to calculate the differences between images in pixel space, like used for autoencoders \\cite{autoencoder1991}.\n\n\n\\subsection{Categorical and Binary Cross Entropy} \\label{sub:cross_entropy}\nLike \\gls{MSE} this loss function is calculated by averaging the error over many different inputs, but in this case the error calculated is the cross entropy. Given two discrete probability distributions $\\bm{y}$ and $\\hat{\\bm{y}}$, the cross entropy is calculated as shown in \\autoref{eq:cross_entropy}.\n\\begin{equation} \\label{eq:cross_entropy}\n H(\\bm{y}, \\hat{\\bm{y}}) = -\\sum_{i}{y_i\\log(\\hat{y}_i + \\epsilon)}\n\\end{equation}\n\nThe \\gls{epsilon} value in this equation is not a part of the cross entropy definition, but when using computers to calculate the logarithm, numbers very close or equal to zero can introduce numerical instabilities. For this reason, in almost all practical implementations it is always added a small constant $\\epsilon$ when calculating the cross entropy and other functions that can have similar issues \\textbf{--} in Tensorflow \\cite{tensorflow2015} the default value for $\\epsilon$ is equal to $10^{-7}$.\n\nThe cross entropy applied over a set of inputs gives the categorical cross entropy loss as shown in \\autoref{eq:categorical_cross_entropy}.\n\\begin{equation} \\label{eq:categorical_cross_entropy}\n J(\\bm{X}, \\bm{\\theta}) = \\frac{1}{n}\\sum^{n}{ H(\\bm{y}, \\hat{\\bm{y}}) }\n\\end{equation}\n\nThis kind of loss function is useful when dealing with probability distributions, since the concept of entropy is deeply related with information and probability. This makes it a better alternative for classification problems when compared with \\gls{MSE}, since the desired output is the class that the input belongs to (one hot encoded) and the network output is a probability distribution over all possible classes. \n\nBut for cases where the output can only assume two values (0 or 1), the categorical cross entropy is reduced to a binary cross entropy. For \\acp{GAN}, this is the more useful loss function and it is calculated as shown in \\autoref{eq:binary_cross_entropy}.\n\\begin{equation} \\label{eq:binary_cross_entropy}\n J(\\bm{X}, \\bm{\\theta}) = -\\frac{1}{n}\\sum^{n}\\sum_{i}{\n \\left( y_i\\log(\\hat{y_i} + \\epsilon) + (1 - y_i) \\log(1 - \\hat{y_i} + \\epsilon) \\right)\n }\n\\end{equation}\n\n\n\n\\subsection{Minimizing the loss} \\label{sub:minimizing_loss}\nHaving an understanding of what is the loss function, the remaining question is: how can the network use the loss in order to update its parameters? Note that since this function effectively measures how well the network is doing, it can directly be used to guide the parameter changes, the goal is to change the parameters in order to minimize the loss (for now on the goal will only be described as minimization, since it is the most common approach and any maximization problem is equivalent to minimizing the negative of what is being maximized).\n\nNote that the inputs $\\bm{x}$ are fixed, given that the dataset is also fixed, and that the goal of using neural networks is that they should learn to model the data and not just choose whatever inputs reduce their loss. So in the eyes of the network $J(\\bm{x}, \\bm{\\theta})$ becomes only $J(\\bm{\\theta})$. This may seem obvious, but this shows an important intuition that the cost function is a high dimensional surface in $N$ dimensional space, $N$ being the number of parameters of the network. And this demonstrates the importance of having both the network and loss functions be continuous and differentiable functions, since this produces a continuous and differentiable surface that allows for updating the parameters in order to reach a minimum region.\n\nTo see how the minimum is reached, it is better to start in a very simplified case where the network has only one parameter $\\theta$. Consider a case where the 1-dimensional loss surface assumes the form shown in \\autoref{fig:1d_loss}.\n\\begin{figure}\n \\centering\n \\caption{Example of a one dimensional loss surface}\n \\includegraphics[width=0.5\\textwidth]{chapters/NeuralNets/figures/1D-Loss.pdf}\n \\fonte{From the author (2021)}\n \\label{fig:1d_loss}\n\\end{figure}\n\nConsider that the network initially starts with the parameter $\\theta_0$ and loss $J_0$ at the dot shown in \\autoref{fig:1d_loss}. To reduce the loss is to update the parameter $\\theta$ in the opposite direction of the surface derivative. This can be shown by the following derivation.\n\\begin{align}\n J_0 + \\Delta J & < J_0 \\nonumber \\\\\n J_0 + \\Delta\\theta \\frac{dJ}{d\\theta} & < J_0 \\nonumber \\\\\n \\Delta\\theta \\frac{dJ}{d\\theta} & < 0 \\nonumber \\\\\n \\sign(\\Delta\\theta) & = -\\sign\\left( \\frac{dJ}{d\\theta} \\right) \\label{eq:delta_theta_sign}\n\\end{align}\n\nThis derivation above only holds true for small values of $\\Delta\\theta$, since the derivative is calculated on infinitesimal small values, bigger changes run the risk of overshooting the region where the derivative approximation is reasonable.\n\\autoref{eq:delta_theta_sign} shows that the parameter change must be in the opposite direction of the change in the loss, the parameter updates can then be written as shown in \\autoref{eq:1d_update}.\n\\begin{equation} \\label{eq:1d_update}\n \\theta_{i+1} = \\theta_{i} + \\Delta\\theta = \\theta_{i} - \\eta \\frac{dJ}{d\\theta}\n\\end{equation}\n\n\\glsunset{learning_rate}\nThe value \\gls{learning_rate} is a positive real number that determines how big are the steps taken when updating the parameters. This value is also a hyperparameter and is called the \\textit{learning rate} of the network. A small learning rate is more precise but can make the training very slow, while bigger values are faster, but run the risk of overshooting and zigzaging around the target, or even worse, diverging and never reaching the result. \\autoref{fig:1d_weight_update} shows how an initial parameter value is updated when using different values for $\\eta$, it can be seen that there is a critical value $\\eta_{crit}$ that is able to minimize the loss in a single step, while smaller and bigger values will converge to the minimum in multiple steps, and much bigger values will diverge.\n\\begin{figure}\n \\centering\n \\caption{Parameter updates for 1D loss surface in function of learning rate}\n \\includegraphics[width=0.5\\textwidth]{chapters/NeuralNets/figures/1D-Weight-Update.pdf}\n \\fonte{From the author (2021)}\n \\label{fig:1d_weight_update}\n\\end{figure}\n\nFor one dimension it is relatively simple to find the critical learning rate, but for very high dimensional problems this is unfeasible or even impossible since a critical value for one parameter is not necessarily the same for all other parameters.\n\nIn practice, it is usually necessary to experiment with some values in order to find what best suits the loss surface of the problem, it is also common to employ some sort of dynamic update that slowly reduces the learning rate, allowing for big steps in the beginning while being more precise towards the end.\n\n\n\\subsection{Gradient Descent} \\label{sub:gradient_descent}\nThe same derivation used to obtain \\autoref{eq:delta_theta_sign} can be applied to generalize the minimization procedure to higher dimensional parameter spaces, the only difference is to consider the partial derivatives for each parameter. Doing this would reveal that, like in the 1-dimensional case, the change in a parameter $\\theta^{(j)}$ must be in the opposite direction of the partial derivative of the loss with respect to this parameter. The parameter update would then be given by \\autoref{eq:parameter_update}.\n\\begin{equation} \\label{eq:parameter_update}\n \\theta_{i+1}^{(j)} = \\theta_{i}^{(j)} - \\eta \\frac{\\partial J}{\\partial \\theta^{(j)}}\n\\end{equation}\n\nThere is however another way to look at these updates. Recall that the gradient of any scalar function is a vector field composed of all its partial derivatives, so the gradient of the loss function is the vector field given by \\autoref{eq:gradient}.\n\\begin{equation} \\label{eq:gradient}\n \\renewcommand{\\arraystretch}{1.2}\n \\nabla_{\\bm{\\theta}} J = \\begin{bmatrix}\n \\frac{\\partial J}{\\partial\\theta^{(1)}} \\\\\n \\frac{\\partial J}{\\partial\\theta^{(2)}} \\\\\n \\vdots \\\\\n \\frac{\\partial J}{\\partial\\theta^{(n)}}\n \\end{bmatrix}\n\\end{equation}\n\nAlso recall that for any function $f$, its gradient at any given point is a vector that points towards the direction of steepest increase to the function at that point and the negative of this vector points to the steepest decrease \\cite[Chapter~4.6]{calculusIII2016}. So the negative of the gradient $\\nabla_{\\bm{\\theta}} J$ gives the direction to update all parameters in order to most reduce the loss function, this is what was being shown element-wise in \\autoref{eq:parameter_update}. By grouping the derivatives it is possible to write the parameter updates in a much cleaner way.\n\\begin{equation} \\label{eq:gradient_descent}\n \\bm{\\theta}_{i+1} = \\bm{\\theta}_{i} - \\eta \\nabla_{\\bm{\\theta}} J(\\bm{\\theta}_i)\n\\end{equation}\n\nBy repeating the process of calculating the gradient of the loss function and then updating all the network's parameters (given a sufficiently small learning rate), then eventually all parameters will converge to values that minimize the loss function. This is what is referred to as \\textit{learning} in a neural network and this iterative process is called \\textit{Gradient Descent}, since the steps are taken using the gradient in order to decrease the loss function \\textbf{--} for maximization, the only difference is to update in the positive direction of the gradient and this is called \\textit{Gradient Ascent}.\n\nOne important thing that was left unmentioned is the fact that this algorithm will almost surely converge to a local minimum in the loss surface instead of the global minimum. In the example shown for the 1-dimensional case the surface was very simple, with a single minimum value to converge, but even in these very simple cases it is possible to find surfaces much more complex (e.g. the curve $y = x + 2\\sin x$ has infinitely many local minima but no global minimum). For the very high dimensional spaces and very complex geometries of the loss surfaces in neural networks, there will be many valleys where gradient descent will converge, but rarely those will be the optimal solution. The local minimum found will depend on the starting point in parameter space, and there is usually no better to way than just randomly selecting a point\\footnote{\n This does not mean that the starting values are picked without any rhyme or reason, the starting point can have a huge impact on the network and it is very important to choose good values. However, initialization techniques are concerned with finding a random distribution with the right values of mean and variance to sample from. The point here is explicitly about the random nature of the starting point and not that this nature has no thought behind it.\n}.\n\nSo if gradient descent will almost never reach the optimal solution, why would anyone use it? How can someone make sure that the local minimum found is good enough for the problem? Or if someone is trying to build a neural network and the results are not being good, how can this person know if this is caused by a legitimate problem on the network or the dataset, instead of simply being the fact that the training process was unlucky and found a bad local minimum? And perhaps most important, how is it possible that neural networks are achieving so much ground-breaking results while relying on gradient descent to learn?\n\nThis is indeed a worry that many researchers in the field have, but there are many options to combat this problem, it can be cited regularization techniques (e.g. $\\ell_1$ and $\\ell_2$ norms \\textbf{--} see \\autoref{sub:regularization}) and optimizers (e.g. \\glsunset{SGD}\\gls{SGD}, momentum, RMSProp, \\glsunset{Adam}\\gls{Adam} \\textbf{--} see \\autoref{sub:optimizers}) as examples. \n\nBesides that, \\textcite{lossSurfaces2014} argue that global minimum are not necessarily the best option, since they have a high chance of overfitting to the data (see \\autoref{sub:overfitting}), they also empirically verify that for big enough neural networks most local minima are very similar to one another and have high quality. Their conclusion is that in practice it is not worth to strive for the global minimum, given that for large networks local minima have high quality and may even generalize better to unseen data.\n", "meta": {"hexsha": "f33aed2ac42ae81d1f09d74eaeeb4ebf76ab1c0e", "size": 17592, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Overleaf/chapters/NeuralNets/gradDesc.tex", "max_stars_repo_name": "PatrickHoeckler/tcc_gan", "max_stars_repo_head_hexsha": "0fa63fff9c6a3bbee57af38683c492a8b120e24a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-04-20T22:17:18.000Z", "max_stars_repo_stars_event_max_datetime": "2021-08-05T06:19:44.000Z", "max_issues_repo_path": "Overleaf/chapters/NeuralNets/gradDesc.tex", "max_issues_repo_name": "PatrickHoeckler/tcc_gan", "max_issues_repo_head_hexsha": "0fa63fff9c6a3bbee57af38683c492a8b120e24a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Overleaf/chapters/NeuralNets/gradDesc.tex", "max_forks_repo_name": "PatrickHoeckler/tcc_gan", "max_forks_repo_head_hexsha": "0fa63fff9c6a3bbee57af38683c492a8b120e24a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 131.2835820896, "max_line_length": 937, "alphanum_fraction": 0.7734197362, "num_tokens": 4168, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8244619177503206, "lm_q2_score": 0.7853085758631159, "lm_q1q2_score": 0.6474570144818776}} {"text": "\\documentclass{article}\n\\author{QianLiu}\n\\title{Insert Equation}\n\\date{\\today}\n\n\\usepackage{amsmath}\n\n\n\\begin{document}\n \\maketitle\n \\tableofcontents\n \n \\section{inline Equation}\n \\subsection{dollar character}\n $f(x)=x^2+3x$\n\n \\subsection{parenthesis}\n \\(f(x)=x^2+3x\\)\n\n \\subsection{mathematical enviroment}\n \\begin{math}\n f(x)=x^2+3x\n \\end{math}\n\n \\section{super/sub script}\n \\subsection{superscript}\n $f(x)=x^2+3y^{20}$\n\n \\subsection{subscript}\n \\(f(x)=x^2+3x_{20}\\)\n\n \\section{greek alphabet}\n \\subsection{lowercase}\n \\(\\alpha\\)\n \\(\\beta\\)\n \\(\\omega_1\\)\n \\(\\gamma\\)\n \\(\\pi\\)\n\n \\subsection{uppercase}\n \\(\\Gamma\\)\n \\(\\Delta\\)\n \\(\\Pi^5\\)\n \\(\\Omega\\)\n \\ \\(\\alpha^2+\\beta^3=\\omega\\)\n\n \\section{mathematical functions}\n \\(\\log_{10}^2 x\\)\n \\ \\(\\sin^2 x\\)\n \\ \\(\\cos^3 y\\)\n \\ \\(\\arcsin x\\)\n \\ \\(\\arccos z\\)\n \\ \\(\\ln x\\)\n \\ \\(\\sqrt{3x^2+2y^4_2}\\)\n \\ \\(\\sqrt[5]{3x}\\)\n\n \\section{algebraic fraction}\n \\(3/4\\)\n \\ \\(\\frac{x}{y}\\)\n \\ \\(\\frac{\\sqrt{2y_2^5}}{\\sqrt[6]{\\sin^2 x+\\arcsin z_4^2}}\\)\n\n \\section{equation between lines}\n\n use \\$\\$...\\$\\$ insert equation\n $$f(x)=3x+y^3$$ \n \n \\subsection{square brackets}\n use \\textbackslash [equation...\\textbackslash ] insert equation\n \\[f(x)=3x+y^3\\] % used commonly to insert equation between lines\n \n \\subsection{environment of displaymath}\n \\begin{displaymath}\n f(x)=3x+y^3\n \\end{displaymath}\n \n \\subsection{environment where equation enumerated automatically}\n Refer the equation \\ref{equation_1}\n \\begin{equation}\n f(x)=3x+z^3 \n \\label{equation_1}\n \\end{equation}\n \n \\subsection{environment where equation* are not enumerated }\n Refer the equation \\ref{equation_2}\n \\begin{equation*} %equation without numbers\n f(x)=3x+z^6\n \\label{equation_2}\n \\end{equation*}\n\n \\subsection{Equation in Multiple lines}\n \\begin{gather} % Need to use package \n f(x)=3x+z^3 \\\\ \n \\notag \\\\ \n \\frac{\\sqrt{2y_2^5}}{\\sqrt[6]{\\sin^2 x+\\arcsin z_4^2}} \\notag \\\\\n f(z)=\\frac{x_2^5}{y_3^7}\n \\end{gather}\n\n \\subsection{Split Equation}\n \\begin{equation} % Single Equation but can split in multiple lines\n \\begin{split} % Use \\begin{split}...\\end{split} to realize it \n f(x)=3x+4x^2 \\\\\n f(x)=\\sin^2 x \\\\\n f(x)=\\cos x^3 \n \\end{split}\n \\end{equation}\n\n \\subsection{Piecewise funtion}\n \\begin{equation}\n D(x)=\\begin{cases}\n 3x & \\text{if } x \\in R \\\\\n x^2 & \\text{if } x \\in Z \\setminus Q\n \\end{cases}\n \\end{equation}\n \n\\end{document}", "meta": {"hexsha": "775156ea091e72cdc08aad0365606e23e799bc7d", "size": 2723, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "bilibili_self_learning/equation.tex", "max_stars_repo_name": "eagleqian/LaTeX-Beginner-s-Guide", "max_stars_repo_head_hexsha": "49f6c9c8e0c9f7a6554e720c8a82978a5f5d1042", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "bilibili_self_learning/equation.tex", "max_issues_repo_name": "eagleqian/LaTeX-Beginner-s-Guide", "max_issues_repo_head_hexsha": "49f6c9c8e0c9f7a6554e720c8a82978a5f5d1042", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "bilibili_self_learning/equation.tex", "max_forks_repo_name": "eagleqian/LaTeX-Beginner-s-Guide", "max_forks_repo_head_hexsha": "49f6c9c8e0c9f7a6554e720c8a82978a5f5d1042", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.6782608696, "max_line_length": 72, "alphanum_fraction": 0.5571061329, "num_tokens": 948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707283, "lm_q2_score": 0.8418256412990658, "lm_q1q2_score": 0.6474314637825034}} {"text": "\\section{PAC Learning}\n\n% ===\n\\begin{highlightbox}\n \\textbf{Want:} Distribution indep. error guarantees!\n % What we want: Distribution independent error guarantees!\n % However, without prior knowledge, those bounds are quite loose.\n\\end{highlightbox}\n\nExpec./\\textbf{Gener. error:}\\enskip\n$\\mathcal R(\\hat c_n) = \\bm P_{X,Y}(\\hat c_n(x) \\neq c(x))$\n\n\\textbf{Empirical error:}\\enskip\n$\\hat{\\mathcal R}_n(\\hat c_n) = \\frac1n \\sum_{i=1}^n \\bm 1_{\\brace{\\hat c_n(x_i) \\;\\neq\\; y_i}}$\n\n\n\\emph{PAC learnable:} $\\mathcal A$ can learn a concept class $\\mathcal C$ from $\\mathcal H$ if, given a \\textcolor{Green}{suff. large sample}, it outputs a hypothesis that \\textcolor{VioletRed}{generalizes well} \\textcolor{OrangeRed}{w/ high prob}.\n\n\\begin{highlightbox}\n\\vspace{-\\fboxrule}\\vspace{-\\fboxsep}\n \\begin{highlightbox}[white]\n (1) $\\color{VioletRed} 0 < \\epsilon < \\sfrac12$, $\\color{OrangeRed} 0 < \\delta < \\sfrac12$,\n (2) $\\bm P_{X,Y}$ on $\\mathcal X \\times \\brace{0,1}$:\n \n If\\enskip $\\color{Green} n \\geq \\mathit{poly} (\\textcolor{VioletRed}{\\sfrac1\\epsilon},\\; \\textcolor{OrangeRed}{\\sfrac1\\delta},\\; \\mathit{dim}(\\mathcal X))$,\n \n Then\\enskip $\\color{OrangeRed} \\bm P_{X,Y}\n \\paren*{ \\color{VioletRed}\n \\mathcal R(\\hat c_n) - \\inf\\limits_{c\\in\\mathcal C} \\mathcal R(c) \\leq \\epsilon\n } \\geq 1 - \\delta$.\n \\end{highlightbox}%\n\\vspace{-\\fboxrule}\\vspace{-\\fboxsep}\n\\end{highlightbox}\n\n% A concept class $\\mathcal C$ is \\emph{PAC learnable} from a hypothesis class $\\mathcal H$ if \\ldots\\ldots\\ldots\nIf $\\mathcal A$ runs in time polynomial in $\\color{VioletRed} \\sfrac1\\epsilon$ and $\\color{OrangeRed} \\sfrac1\\delta$, we say that $\\mathcal C$ is \\emph{efficiently PAC learnable}.\n\n% ===\n\\subsection{VC Inequality\n\\hfill $\\bm P(\\cdots \\geq \\begingroup\\color{VioletRed}\\epsilon\\endgroup) \\leq \\ldots \\leq \\color{OrangeRed}\\delta$}\n\nSelect ERM:\\enskip\n$\\hat c_n^\\ast = \\arg\\min_{c\\in\\mathcal C} \\hat{\\mathcal R}_n (c)$\n\nUnder uniform convergence: %, the suboptimality of $\\hat c_n^\\ast$ can be bounded:\n\\scalebox{.9}{${\n \\bm P \\paren*{\n \\mathcal R(\\hat c_n^\\ast) - \\inf\\limits_{c\\in\\mathcal C} \\mathcal R(c) > \\epsilon\n } \\leq \\bm P \\paren*{\n \\sup\\limits_{c\\in\\mathcal C} \\abs*{\\hat{\\mathcal R}_n(c) - \\mathcal R(c)} > \\dfrac\\epsilon2\n }\n}$}\n\n\\begin{tabular}{@{\\textbullet\\:} r @{$\\:\\leq\\:$} l @{}}\n \\emph{$\\abs{\\mathcal C}$ Finite:}\\enskip\n $\\bm P \\paren*{ \\sup \\abs*{\\cdots} > \\epsilon }$\n & $2 \\abs{\\mathcal C} \\exp(-2n\\epsilon^2)$\n \\\\\n \\emph{$\\abs{\\mathcal C}$ Unbounded:}\\hfill\n $\\bm P \\paren*{ \\cdots }$\n & $9 n^{\\mathrm{VC}_{\\mathcal C}} \\exp(-\\frac{n \\epsilon^2}{32})$\n\\end{tabular}\n\n% ===\n\\subsection{Rectangle Learning}\n\n$\\bm P(\\mathcal(\\hat c_n^\\ast > \\epsilon)\n\\leq \\abs{\\mathcal C} \\cdot (1-\\epsilon)^n\n\\leq \\abs{\\mathcal C} \\cdot \\exp(-n\\epsilon)\n\\color{gray} < \\delta$\n\n\\textbf{Union bound:}\\enskip\n$\\bm P(\\bigcup_i T_i) \\leq \\sum_i \\bm P(T_i)$\n\n% ===\n", "meta": {"hexsha": "d60596ab2b8edd4d6340ab1e69fc96445d43563c", "size": 2987, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/AML20/sections/11_pac_learning.tex", "max_stars_repo_name": "tstreule/eth-cheat-sheets", "max_stars_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-26T23:11:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T23:11:57.000Z", "max_issues_repo_path": "src/AML20/sections/11_pac_learning.tex", "max_issues_repo_name": "tstreule/eth-cheat-sheets", "max_issues_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/AML20/sections/11_pac_learning.tex", "max_forks_repo_name": "tstreule/eth-cheat-sheets", "max_forks_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.3026315789, "max_line_length": 248, "alphanum_fraction": 0.6360897221, "num_tokens": 1111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.841825635346563, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.647431459204551}} {"text": "\\section{Algebraic Moment Closures}\n\\label{sec:algebraicClosure}\n\nThe two-moment model given by Eq.~\\eqref{eq:angularMoments} is not closed because of the appearance of the second moments $\\vect{\\cK}$ (the normalized pressure tensor). \nAlgebraic moment closures for the two-moment model are computationally efficient as they provide the Eddington factor in Eq.~\\eqref{eq:eddingtonTensor} in closed form as a function of the density $\\cJ$ and the flux factor $h=|\\vect{\\cH}|/\\cJ$. \nFor this reason they are used in applications where transport plays an important role, but where limited computational resources preclude the use of higher fidelity models. \nExamples include simulation of neutrino transport in core-collapse supernovae \\cite{roberts_etal_2016} and compact binary mergers \\cite{foucart_etal_2015}. \nAlgebraic moment closures in the context of these aforementioned applications have also been discussed elsewhere (e.g., \\cite{janka_etal_1992,pons_etal_2000,smit_etal_2000,just_etal_2015,murchikova_etal_2017}). \nHere we focus on properties of the algebraic closures that are critical to the development of numerical methods for the two-moment model of fermion transport. \nFor the algebraic closures we consider, the Eddington factor in Eq.~\\eqref{eq:eddingtonTensor} can be written in the following form \\cite{cernohorskyBludman_1994}\n\\begin{equation}\n \\chi(\\cJ,h)=\\f{1}{3}+\\f{2\\,(1-\\cJ)\\,(1-2\\cJ)}{3}\\,\\Theta\\Big(\\f{h}{1-\\cJ}\\Big),\n \\label{eq:eddingtonFactor}\n\\end{equation}\nwhere the \\emph{closure function} $\\Theta(x)$ depends on the specifics of the closure procedure. \nWe will consider two basic closure procedures in more detail below: the maximum entropy (ME) closure and the Kershaw (K) closure. \n\nIn the low occupancy limit ($\\cJ\\ll1$), the Eddington factor in Eq.~\\eqref{eq:eddingtonFactor} depends solely on $h$; i.e.,\n\\begin{equation}\n \\chi(\\cJ,h)\\to\\chi_{0}(h)=\\f{1}{3}+\\f{2}{3}\\,\\Theta\\big(h\\big). \n \\label{eq:eddingtonFactorLow}\n\\end{equation}\nThis for of $\\chi$ yields a moment closure that is suitable for particle systems obeying Maxwell-Boltzmann statistics. \n\n\\subsection{Maximum Entropy (ME) Closure}\n\nThe ME closure constructs an approximation of the angular distribution as a function of $\\cJ$ and $\\vect{\\cH}$ \\cite{cernohorskyBludman_1994,lareckiBanach_2011}. \nThe ME distribution $f_{\\mbox{\\tiny ME}}$ is found by maximizing the entropy functional, which for particles obeying Fermi-Dirac statistics is given by\n\\begin{equation}\n S[f_{\\mbox{\\tiny ME}}] \n = \\int_{\\bbS^{2}}\\big[\\,(1-f_{\\mbox{\\tiny ME}})\\log(1-f_{\\mbox{\\tiny ME}}) + f_{\\mbox{\\tiny ME}}\\log f_{\\mbox{\\tiny ME}}\\,]\\,d\\omega,\n \\label{eq:entropyFunctional}\n\\end{equation} \nsubject to the constraints\n\\begin{equation}\n \\f{1}{4\\pi}\\int_{\\bbS^{2}}f_{\\mbox{\\tiny ME}}(\\omega)\\,d\\omega=\\cJ\n \\quad\\text{and}\\quad\n \\f{1}{4\\pi}\\int_{\\bbS^{2}}f_{\\mbox{\\tiny ME}}(\\omega)\\,\\vect{\\ell}(\\omega)\\,d\\omega=\\vect{\\cH}. \n \\label{eq:closureConstraints}\n\\end{equation}\nThe solution that maximizes Eq.~\\eqref{eq:entropyFunctional} takes the general form \\cite{cernohorskyBludman_1994}\n\\begin{equation}\n f_{\\mbox{\\tiny ME}}(\\omega;a,\\vect{b})=\\f{1}{e^{a + \\vect{b}\\cdot\\vect{\\ell}(\\omega)}+1}, \n \\label{eq:fME}\n\\end{equation}\nwhere the Lagrange multipliers $a$ and $\\vect{b}$ are implicit functions of $\\cJ$ and $\\vect{\\cH}$. \nThe ME distribution function satisfies $0 < f_{\\mbox{\\tiny ME}} < 1$, but $a$ and $\\vect{b}$ are unconstrained. \nSpecification of $a$ and $\\vect{b}$ from $\\vect{\\cM}=(\\cJ,\\vect{\\cH})^{T}$ gives $f_{\\mbox{\\tiny ME}}$, and any number of moments can in principle be computed. \nImportantly, for the maximum entropy problem to be solvable, we must have $\\vect{\\cM}\\in\\cR$ \\cite{lareckiBanach_2011}. \n\nTo arrive at an algebraic form of the ME closure, Cernohorsky \\& Bludman \\cite{cernohorskyBludman_1994} postulate (but see \\cite{lareckiBanach_2011}) that, as a function of the flux saturation\n\\begin{equation}\n x := h/(1-\\cJ),\n \\label{eq:fluxSaturation}\n\\end{equation} \nthe closure function $\\Theta$ is independent of $\\cJ$ and can be written explicitly in terms of the inverse Langevin function. \nTo avoid inverting the Langevin function for $\\Theta$, they provide a polynomial fit (accurate to $2\\%$) given by\n\\begin{equation}\n \\Theta_{\\mbox{\\tiny ME}}^{\\mbox{\\tiny CB}}(x)\n =\\f{1}{5}\\,\\big(\\,3-x+3\\,x^{2}\\,\\big)\\,x^{2}.\n \\label{eq:closureMECB}\n\\end{equation}\nMore recently, Larecki \\& Banach \\cite{lareckiBanach_2011} have shown that the explicit expression given in \\cite{cernohorskyBludman_1994} is not exact and provide another approximate expression\n\\begin{equation}\n \\Theta_{\\mbox{\\tiny ME}}^{\\mbox{\\tiny BL}}(x)\n =\\f{1}{8}\\,\\big(\\,9\\,x^{2}-5+\\sqrt{33\\,x^{4}-42\\,x^{2}+25}\\,\\big),\n \\label{eq:closureMEBL}\n\\end{equation}\nwhich is accurate to within $0.35\\%$. \nOn the interval $x\\in[0,1]$, the curves given by Eqs.~\\eqref{eq:closureMECB} and \\eqref{eq:closureMEBL} lie practically on top of each other. \nThe closure functions given by Eqs.~\\eqref{eq:closureMECB} and \\eqref{eq:closureMEBL}, together with the Eddington factor in Eq.~\\eqref{eq:eddingtonFactor} and the pressure tensor in Eq~\\eqref{eq:eddingtonTensor}, constitute the algebraic maximum entropy closures for fermionic particle systems considered in this paper. \nWe will refer to the ME closures with $\\Theta_{\\mbox{\\tiny ME}}^{\\mbox{\\tiny CB}}$ and $\\Theta_{\\mbox{\\tiny ME}}^{\\mbox{\\tiny BL}}$ as the CB (Cernohorsky \\& Bludman) and BL (Banach \\& Larecki) closures, respectively. \n\nWe also note that using the closure function given by Eq.~\\eqref{eq:closureMECB} with the low occupancy Eddington factor in Eq~\\eqref{eq:eddingtonFactorLow} results in the algebraic maximum entropy closure attributed to Minerbo \\cite{minerbo_1978}, which is currently in use in simulation of neutrino (fermion) transport in the aforementioned nuclear astrophysics applications. \nIn a recent comparison of algebraic (or analytic) closures for the two-moment model applied to neutrino transport around proto-neutron stars, Murchikova et al. \\cite{murchikova_etal_2017} obtained nearly identical results when using the closures of CB and Minerbo. \nFor these reasons, we include the Minerbo closure in the subsequent discussion and in the numerical tests in Section~\\ref{sec:numerical}. \n\n\\subsection{Kershaw (K) Closure}\n\nAnother algebraic closure we consider is a Kershaw-type closure \\cite{kershaw_1976}, developed for fermion particle systems in \\cite{banachLarecki_2017a}. \nThe basic principle of the Kershaw closure for the two-moment model is derived from the fact that the realizable set generated by the triplet of scalar moments\n\\begin{equation}\n \\{\\cJ,\\cH,\\cK\\}=\\f{1}{2}\\int_{-1}^{1}f(\\mu)\\,\\mu^{\\{0,1,2\\}}\\,d\\mu,\n \\label{eq:scalarMoments}\n\\end{equation} \nis convex. \nFor the moments in Eq.~\\eqref{eq:scalarMoments}, the realizable set is the set of moments obtained from distribution functions satisfying $0{\n\t\\[\n\t w(T) = \\sum\\limits_{e \\in E'} w(e)\n\t\\]\n }\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Minimum Spanning Tree}\n % \\begin{center}\n % MST: Mimimize $w(T)$ over all possible STs\n % \\end{center}\n\n \\[\n\t\\text{MST:} \\quad \\mathop{\\mathrm{arg\\,min}}_{T} \\; w(T)\n \\]\n\n \\importikznocaption{0.60\\textwidth}{0.80\\textwidth}{tikz-in-beamer/mst-example-overlay.tex}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{A Simple Property}\n \\fignocaption{width = 0.45\\textwidth}{figs/mst-simple-property.pdf}\n\n \\begin{center}\n\t\\textcolor{red}{\\bf Cut:} $V = (S, V \\setminus S)$\n \\end{center}\n\n \\pause\n\n \\begin{enumerate}\n\t\\item MST in each connected component\n\t\\item $ce$: \\textcolor{brown}{a} \\textcolor{red}{\\bf lightest} edge across cut\n \\end{enumerate}\n\n \\pause\n\n \\begin{center}\n\t\\textcolor{brown}{\\bf Copy\\&Paste Argument; Exchange Argument}\n \\end{center}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{A Wrong Divide \\& Conquer Algorithm}\n \\begin{description}\n \\item[Input:] $G = (V, E, w)$\n \\item[Divide:] $V = (S, V \\setminus S)$; $\\large| |S| - |V \\setminus S| \\large| \\le 1$\n \\item<2->[Conquer:] $T_1$: an MST of $S$; $T_2$: an MST of $V \\setminus S$\n \\item<3->[Combine:] $T_{1} + T_{2} + \\set{e}$: $e$ is a lightest edge across $(S, V \\setminus S)$\n \\end{description}\n\n \\pause\n \\vspace{0.50cm}\n\n \\fignocaption{width = 0.30\\textwidth}{figs/divide-conquer-mst-counterexample.pdf}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{A Wrong Algorithm}\n \\begin{alertblock}{What is wrong?}\n\t\\fignocaption{width = 0.30\\textwidth}{figs/divide-conquer-mst-counterexample.pdf}\n\n\t\\centerline{\\textcolor{brown}{The edges \\textcolor{blue}{$bc$ and $ad$} do \\textcolor{red}{\\bf not} belong to any MST.}}\n \\end{alertblock}\n\n \\pause\n \\vspace{0.50cm}\n\n \\begin{alertblock}{What if:}\n\t\\centerline{\\textcolor{red}{Invariant:} Manages a set of edges $X$ which is a subset of \\textcolor{red}{\\bf some} MST.}\n \\end{alertblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "9c989316595f672fb109085f6f32cdd5f1ded5eb", "size": 2413, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "algorithm-lecture-mst/sections/mst-problem.tex", "max_stars_repo_name": "hengxin/algorithm-lectures", "max_stars_repo_head_hexsha": "cf00b0d2d88da6e20d37c36d1f49ca6c1a0669ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-04-20T06:57:57.000Z", "max_stars_repo_stars_event_max_datetime": "2017-11-12T19:07:16.000Z", "max_issues_repo_path": "algorithm-lecture-mst/sections/mst-problem.tex", "max_issues_repo_name": "hengxin/algorithm-lectures", "max_issues_repo_head_hexsha": "cf00b0d2d88da6e20d37c36d1f49ca6c1a0669ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "algorithm-lecture-mst/sections/mst-problem.tex", "max_forks_repo_name": "hengxin/algorithm-lectures", "max_forks_repo_head_hexsha": "cf00b0d2d88da6e20d37c36d1f49ca6c1a0669ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-12T10:36:11.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-12T10:36:11.000Z", "avg_line_length": 28.7261904762, "max_line_length": 121, "alphanum_fraction": 0.6261914629, "num_tokens": 865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833893685269, "lm_q2_score": 0.8354835432479661, "lm_q1q2_score": 0.6471516746906358}} {"text": "\\chapter{The algorithm}\n\\label{chap:algorithm}\n\n\\textbf{by Bastian Boll} \\\\\n\nThe training dataset consists of approximately 560h of subtitled audio. From this, we extract MFCC features and approximate timestamps for each single word. For each word $w$ we aim to define a $\\mathcal{C}^1$ function \\(D_w\\colon \\R \\to \\lbrack 0, \\infty)\\) such that $D_w(t)$ is large iff the audio around time $t$ sounds like the word $w$. For a given transcript $\\mathcal{T}$ and audio $A$, we can find the function $f$ introduced in chapter \\ref{chap:intro} as the solution to the minimization problem\n\\begin{align*}\n\t&\\min_x -\\sum_{j=1}^l D_{w_j}(x_j)\\\\\n\t&\\hspace{0.5em}\\text{s.t. }x_1\\leq x_2\\leq \\dotsc\\leq x_l\n\\end{align*}\n\n\\section{Defining the distance metric}\n\n\\textbf{by Bastian Boll} \\\\\n\nThe functions \\(D_w\\colon \\R \\to \\lbrack 0, \\infty)\\) can be seen as equivalent to a metric for the distance between the sound of the word $w$ and the audio $A$ at $t$. They need to be constructed in a way such that\n\\begin{enumerate}\n\t\\item $D_w$ is guaranteed to be continuously differentiable w.r.t. $t$ for every word $w$.\n\t\\item $D_w$ corresponds to the probability, that the word $w$ is said as time $t$ in $A$. More specifically, the function mapping said probability $p$ to $D_w(t)$ should be monotonically increasing.\n\\end{enumerate}\n\nOne important motivating factor behind the construction of this optimization objective is, that the dataset does not contain exact time points for each word. Instead, for any given word $w$, we assume that the difference between the time point in the dataset $t_\\text{data}$ and the actual time point in the audio at which $w$ is being said $t_\\text{true}$ follows a normal distribution with mean $0$ and constant standard deviation $\\sigma$.\n\nDivide the audio into intervals of equal size. Label each interval $I_i$ with the probability of it containing $w$. This can be done by a variety of possible machine learning methods, our specific approach is discussed in section \\ref{interval_word_prob}. These probabilities $p_i$ can be used to infer an observed probability distribution of where in the audio the machine learning model expects $w$ to be located. More precisely, because the same word might appear multiple times, we assume the $p_i$ to be generated by the sum of multiple normal distributions with different means (time points) and constant standard deviation $\\sigma$. We also assume, that the specific instance $w$ of the word in question can be attributed to exactly one of these distributions with mean $t$.\\\\\nIn order to find the optimal time point $t_\\text{true}$ we can use the observation, that two normal distributions with the same standard deviation have the same mean $t=t_\\text{true}$ exactly if it holds for the probability density functions $\\rho_{t,\\sigma}$ and $\\rho_{t_\\text{true},\\sigma}$ that\n\\[t = \\arg\\max \\int_\\R \\rho_{t_\\text{true},\\sigma}(x)\\rho_{t,\\sigma}(x)\\text{d}x\\]\nHence, we can use the maximum value on the right hand side as a score for how similar the audio around time $t$ the model considers the word $w$ to be. This can be transformed into an objective in terms of the probabilities $p_i$ computed by the machine learning model\n\\begin{align*}\n\t\\int_\\R \\rho_{t_\\text{true},\\sigma}(x)\\rho_{t,\\sigma}(x)\\text{d}x &= \\sum_{i} \\int_{I_i} \\rho_{t_\\text{true},\\sigma}(x)\\rho_{t,\\sigma}(x)\\text{d}x\\\\\n\t\t\t&= \\sum_i \\rho_{t,\\sigma}(\\xi_i) \\underbrace{\\int_{I_i} \\rho_{t_\\text{true},\\sigma}(x)\\text{d}x}_{\\approx p_i}\n\\end{align*}\nby the mean value theorem for a sufficient $\\xi_i\\in I_i$ (because $\\rho_{t_\\text{true},\\sigma}(x) > 0$ for all $x$). Equality still approximately holds if we set $\\xi_i$ to be the center point of $I_i$ for sufficiently small intervals.\nHence, we can define $D_w$ through\n\\[\n\tD_w(t) = \\sum_i p_i \\rho_{t,\\sigma}(t_i)\n\\]\nfor each center point $t_i$ of $I_i$. This construction satisfies the smoothness required for optimization. It also considers the inherent imprecision of word time points in the dataset and provides a means of leveraging information computed by the machine learning model without requiring perfect prediction accuracy.\\\\\nAdditionally, by interpreting the time alignment as a global optimization problem under word order constraints, we can make effective use of the known word order and reduce the impact of single (local) bad predictions by the model.\n\n\\section{Computing probabilities for each interval}\n\\label{interval_word_prob}\n\n\\textbf{by Paul Warkentin} \\\\\n\nGiven an audio file, we divide its MFCC features into $n$ intervals $I_i$ of equal lengths $l$. If the rest part of the data $I_{n+1}$ is shorter than the length $l$, it is discarded. The result of this section will be a matrix $p$ of shape $\\lbrack l, c \\rbrack$, where $c$ is the fixed number of words we selected for evaluation.\n\nSince we interpolated the timing of each word in chapter \\ref{data_preparation}, we can iterate for all talks through each word $w_j$ with its estimated timing $t_j$. Calculate the cumulative distribution function $\\Phi_{t_j, 0.8}(x)$ with a mean value of $t_j$ and a standard deviation of 0.8. \\\\\nThe current word $w_j$ lies in the interval with index $k = \\lceil \\tfrac{t_j}{l} \\rceil$. For the interval $I_k$ and its neighbors $\\{I_{k-2}, I_{k-1}, I_{k+1}, I_{k+2}\\}$, do the cumulative mapping\n\\[p_{j, s} \\mapsto p_{j, s} + \\Phi_{t_j, 0.8}((s - 1) l) - \\Phi_{t_j, 0.8}(s l),\\]\nfor $s = k-2, k-1, \\dots, k+2$. \\\\\n\nHence, we got the features and its labels for the supervised learning model described in the section below. \\\\\nThe MFCC features for an audio file are reshaped into a matrix of shape $\\lbrack l, m \\rbrack$, where $m$ is the number of MFCC samples for an interval of length $l$.\n\n\\section{Defining the models}\n\n\\textbf{by Paul Warkentin} \\\\\n\n% source for deep convnet: https://yerevann.github.io/2016/06/26/combining-cnn-and-rnn-for-spoken-language-identification/\n\nFor the main speech recognition part, we trained several supervised models. The following three delivered the best results.\n\n\\subsection*{Simple convolutional neural network}\nThis model is derived from TensorFlows' convolutional neural network for speech recognition in the example \\textit{speech\\_commands}\\footnote{\\url{https://github.com/tensorflow/tensorflow/blob/master/tensorflow/examples/speech\\_commands/models.py}}.\n\\begin{figure}[h]\n\t\\centering\n\t\\makebox[\\textwidth][c]{\n\t\t\\begin{tabular}{clccc||clccc}\n\t\t\t\\hline \\textbf{Nr} & \\textbf{Type} & \\textbf{\\#Ch.} & \\textbf{Wd.} & \\textbf{Hgt.} & \\textbf{Nr} & \\textbf{Type} & \\textbf{\\#Ch.} & \\textbf{Wd.} & \\textbf{Hgt.} \\\\ \\hline\n\t\t\t0 & Input & 1 & 80 & 13 & 3 & Conv2D & 64 & 4 & 10 \\\\\n\t\t\t1 & Conv2D & 64 & 8 & 20 & & Bias & 64 & 4 & 10 \\\\\n\t\t\t& Bias & 64 & 8 & 20 & & ReLU & 64 & 4 & 10 \\\\\n\t\t\t& ReLU & 64 & 8 & 20 & & Dropout & 64 & 4 & 10 \\\\\n\t\t\t& Dropout & 64 & 8 & 20 & 4 & MatMul & 1500 & & \\\\\n\t\t\t& MaxPool & 1 & 2 & 1 & & Bias & 1500 & & \\\\\n\t\t\t2 & Conv2D & 64 & 4 & 10 & & & & & \\\\\n\t\t\t& Bias & 64 & 4 & 10 & & & & & \\\\\n\t\t\t& ReLU & 64 & 4 & 10 & & & & & \\\\\n\t\t\t& Dropout & 64 & 4 & 10 & & & & & \\\\\n\t\t\t& MaxPool & 1 & 2 & 1 & & & & & \\\\ \\hline\n\t\t\\end{tabular}\n\t}\n\t\\caption[Simple convolutional structure]{Graph structure of simple convolutional neural network.}\n\\end{figure}\n\\FloatBarrier\n\n\\subsection*{Deep convolutional neural network}\nThis model is a combination of six consecutive convolutional neural networks, each followed by activation. The output is also here flattened and will then be encoded by two dense layers into the label space. We got the idea by the blog post\\footnote{\\url{https://yerevann.github.io/2016/06/26/combining-cnn-and-rnn-for-spoken-language-identification/}} by Harutyunyan and Khachatrian as they did reach an accuracy of 99.24\\%.\n\\begin{figure}[!htbp]\n\t\\centering\n\t\\makebox[\\textwidth][c]{\n\t\t\\begin{tabular}{clccc||clccc}\n\t\t\t\\hline \\textbf{Nr} & \\textbf{Type} & \\textbf{\\#Ch.} & \\textbf{Wd.} & \\textbf{Hgt.} & \\textbf{Nr} & \\textbf{Type} & \\textbf{\\#Ch.} & \\textbf{Wd.} & \\textbf{Hgt.} \\\\ \\hline\n\t\t\t0 & Input & 1 & 80 & 13 & 4 & Conv2D & 32 & 3 & 3 \\\\\n\t\t\t1 & Conv2D & 16 & 26 & 10 & & Bias & 32 & 3 & 3 \\\\\n\t\t\t& Bias & 16 & 26 & 10 & & ReLU & 32 & 3 & 3 \\\\\n\t\t\t& ReLU & 16 & 26 & 10 & 5 & Conv2D & 32 & 3 & 3 \\\\\n\t\t\t& MaxPool & 1 & 2 & 1 & & Bias & 32 & 3 & 3 \\\\\n\t\t\t2 & Conv2D & 16 & 13 & 5 & & ReLU & 32 & 3 & 3 \\\\\n\t\t\t& Bias & 16 & 13 & 5 & 6 & Conv2D & 32 & 3 & 3 \\\\\n\t\t\t& ReLU & 16 & 13 & 5 & & Bias & 32 & 3 & 3 \\\\\n\t\t\t& Dropout & 16 & & & & ReLU & 32 & 3 & 3 \\\\\n\t\t\t& MaxPool & 1 & 2 & 1 & 7 & Dense & 1000 & & \\\\\n\t\t\t3 & Conv2D & 32 & 5 & 5 & & Sigmoid & 1000 & & \\\\\n\t\t\t& Bias & 32 & 5 & 5 & 8 & Dense & 1500 & & \\\\\n\t\t\t& ReLU & 32 & 5 & 5 & & Sigmoid & 1500 & & \\\\ \\hline\n\t\t\\end{tabular}\n\t}\n\t\\caption[Deep convolutional structure]{Graph structure of deep convolutional neural network.}\n\\end{figure}\n\\FloatBarrier\n\n\\subsection*{Convolutional LSTM neural network}\nThis model is a combination of a convolutional neural network and a LSTM network. The input vector will be put into a dynamic RNN cell of type \\newline\\textit{tf.contrib.rnn.Conv2DLSTMCell} in TensorFlow. Its output is then flattened and will be encoded by two dense layers into the label space.\n\\begin{figure}[h]\n\t\\centering\n\t\\begin{tabular}{clccc}\n\t\t\\hline \\textbf{Nr} & \\textbf{Type} & \\textbf{\\#Channels} & \\textbf{Width} & \\textbf{Height} \\\\ \\hline\n\t\t0 & Input & 1 & 80 & 13 \\\\\n\t\t1 & Conv2DLSTM & 1 & 80 & 13 \\\\\n\t\t2 & Dense & 1000 & & \\\\\n\t\t& Sigmoid & 1000 & & \\\\\n\t\t3 & Dense & 1500 & & \\\\\n\t\t& Sigmoid & 1500 & & \\\\ \\hline\n\t\\end{tabular}\n\t\\caption[Convolutional LSTM structure]{Graph structure of convolutional LSTM neural network.}\n\\end{figure}\n\\FloatBarrier\n", "meta": {"hexsha": "608ded474b530b65012e1e7a1d1635ec2d8b7b03", "size": 9613, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/chapters/algorithm_concept.tex", "max_stars_repo_name": "bbboll/ml_subtitle_align", "max_stars_repo_head_hexsha": "3abb5628902db1021af8ff1666f37a1663574487", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2018-04-01T20:02:26.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T15:59:20.000Z", "max_issues_repo_path": "doc/chapters/algorithm_concept.tex", "max_issues_repo_name": "bbboll/ml_subtitle_align", "max_issues_repo_head_hexsha": "3abb5628902db1021af8ff1666f37a1663574487", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/chapters/algorithm_concept.tex", "max_forks_repo_name": "bbboll/ml_subtitle_align", "max_forks_repo_head_hexsha": "3abb5628902db1021af8ff1666f37a1663574487", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 74.519379845, "max_line_length": 783, "alphanum_fraction": 0.6931238947, "num_tokens": 3089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9086178944582995, "lm_q2_score": 0.7122321903471563, "lm_q1q2_score": 0.647146913158656}} {"text": "\\section{(Sub-)Groups with prime power order}\r\n\\subsection{Elementary Properties}\r\n\\begin{definition}\r\n Let $G$ be a group.\r\n We say $G$ is a $p$-group for a prime $p$ if $|G|=p^n$ for some $n\\in\\mathbb N$.\r\n\\end{definition}\r\n\\begin{theorem}\r\n Let $G$ be a $p$-group, then $Z(G)\\neq 1$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Consider the partition of $G$ by conjugacy classes, which are either $1$ or divisible by $p$.\r\n Note that $Z(G)$ is the union of all $1$-classes.\r\n But if $|Z(G)|=1$, then\r\n $$0\\equiv p^n=|G|=1+\\sum_{\\exists g\\notin Z(G),C=\\operatorname{ccl}_G(g)}|C|\\equiv 1\\pmod{p}$$\r\n which is a contradiction.\r\n\\end{proof}\r\nIn particular $|Z(G)|\\equiv 0\\pmod{p}$.\r\n\\begin{corollary}[Classification of Simple $p$-groups]\r\n If $G$ is a simple $p$-group, then $G\\cong C_p$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n $1\\neq Z(G)\\unlhd G$.\r\n Since $G$ is simple we must have $Z(G)=G$, hence $G$ is abelian, but then necessarily $G\\cong C_p$ since $G$ cannot have any proper subgroup to be simple.\r\n\\end{proof}\r\n\\begin{corollary}\r\n Let $G$ be a $p$-group of order $p^n$, then $G$ contains an element of order $p^r$ for any $0\\le r\\le n$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Consider the composition series of $G$\r\n $$1=G_0\\lhd G_1\\lhd\\cdots\\lhd G_m=G$$\r\n Where $G_{i+1}/G_i$ is simple.\r\n But it is an $p$-group, so we must have $G_{i+1}/G_i\\cong C_p$ for any $i$.\r\n The claim follows.\r\n\\end{proof}\r\n\\begin{lemma}\r\n For $G$ a group.\r\n If $G/Z(G)$ is cyclic, then $G$ is abelian.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Let $gZ(G)$ be a generator of $G/Z(G)$.\r\n Then every element of $G$ is of the form $p^iz$ where $z\\in Z$.\r\n But for $z,z'\\in Z(G)$, $g^izg^jz'=g^{i+j}zz'=g^jz'g^iz$, so $G$ is abelian.\r\n\\end{proof}\r\n\\begin{corollary}\r\n If $G$ has order $p^2$ for a prime $p$, then $G$ is abelian.\r\n\\end{corollary}\r\n\\begin{proof}\r\n We already know that $Z(G)\\neq 1$.\r\n For $|Z(G)|=p$ then we have $G/Z(G)\\cong C_p$, so $G$ is abelian by the preceding lemma, contradiction.\r\n For $|Z(G)|=p^2$ we have $Z(G)=G$, which means that $G$ is abelian.\r\n There are no other possibilities, so the proof is done.\r\n\\end{proof}\r\nSadly (or not) there are nonabelian groups of order $p^3$.\r\n\\subsection{Sylow's Theorems}\r\n\\begin{theorem}[Sylow's Theorems]\r\n Let $G$ be a finite group with order $p^am$ where $p$ is a prime and $p\\nmid m$.\r\n Then\\\\\r\n 1. The set $\\operatorname{Syl}_p(G)=\\{P\\le G:|P|=p^a\\}$ is nonempty.\\\\\r\n 2. All elements of $\\operatorname{Syl}_p(G)$ are conjugate.\\\\\r\n 3. The number $n_p=|\\operatorname{Syl}_p(G)|$ satisfies $n_p\\equiv 1\\pmod{p}$ and $n_p|m$.\r\n\\end{theorem}\r\n\\begin{corollary}\r\n If $n_p=1$, then there is a normal Sylow $p$-subgroup.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Let $g\\in G$ and $P$ be a Sylow $p$-subgroup of $G$.\r\n But then $gPg^{-1}\\in\\operatorname{Syl}_p(G)=\\{P\\}$, hence $gPg^{-1}=P\\implies P\\unlhd G$.\r\n\\end{proof}\r\n\\begin{example}\r\n There is no simple group of order $1000$.\r\n Suppose $G$ is a group of order $1000=2^35^3$, then $n_5\\equiv 1\\pmod{5}$ and $n_5|8$, but then we must have $n_5=1$.\r\n So there is a normal Sylow $5$-subgroup of $G$ which is obviously not the identity or $G$, hence $G$ is not normal.\r\n\\end{example}\r\n\\begin{proof}\r\n 1. Consider the action of the group $G$ on the set $\\Omega$ of all subsets of $G$ of size $p^a$ by $g\\star X=\\{gx:x\\in X\\}$, so\r\n $$|\\Omega|=\\binom{p^am}{p^a}\\not\\equiv 0\\pmod{p}$$\r\n Hence this action has an orbit, say the orbit of $X\\in\\Omega$, which order is not a multiple of $p$.\r\n Then $|G_X||\\operatorname{orb}_G(X)|=|G|=p^am$, so $p^a||G_X|$.\r\n On the other hand, $\\bigcup_{g\\in G}g\\star X=G$, so $|G|\\le |\\operatorname{orb}_G(X)||X|$, so $|G_X|=|G|/|\\operatorname{orb}_G(X)|\\le |X|=p^a$, but $|G_X|\\ge p^a$, therefore $|G_X|=p^a$.\\\\\r\n 2. We shall prove a stronger statement: suppose $P\\in\\operatorname{Syl}_p(G)$ and $Q\\le G$ a $p$-subgroup, then $Q\\le gPg^{-1}$ for some $g\\in G$.\r\n Consider the action of $Q$ on the set of left cosets $G/P$ by left multiplication.\r\n By orbit-stabiliser, any orbit divides $|Q|$, so its size must be either $1$ or a multiple of $p$.\r\n But $|G/P|=m$ which is coprime to $p$, hence there is at least one orbit $\\operatorname{orb}_Q(gP)$ of size $1$, so for any $q\\in Q$, $g^{-1}qg\\in P\\implies Q\\le gPg^{-1}$.\\\\\r\n 3. Let $G$ act transitively (by 2) on $\\operatorname{Syl}_p(G)$ by conjugation.\r\n So by orbit-stabiliser, $n_p||G|$, so it suffices to show $n_p\\equiv 1\\pmod{p}$.\r\n Now let $P\\in\\operatorname{Syl}_p(G)$ and consider its action on $\\operatorname{Syl}_p(G)$ by conjugation.\r\n Now the orbits divides $|P|=p^a$, so is either $1$ or divisible by $p$.\r\n We shall show that there is exactly one orbit of size $1$, which will establish the theorem.\r\n There is at least one orbit, namely $\\operatorname{orb}_P(P)$.\r\n If $\\operatorname{orb}_P(Q)$ is also an orbit of size $1$, then $P\\le N_G(Q)$.\r\n Now $P,Q$ are Sylow $p$-subgroups of $N_G(Q)$, so they are conjugate by 2, therefore there is some $g\\in N_G(Q)$ with $Q=gQg^{-1}=P$.\r\n The theorem is hence proved.\r\n\\end{proof}\r\n\\begin{example}\r\n Suppose $G$ is a simple group, then $|G|\\neq 132$.\r\n Assume there is a simple group of order $132=2^2\\cdot 3\\cdot 11$, then by Sylow's Third Theorem, $n_3=1,4,22$ and $n_{11}=1,12$, but by simplicity neither of them is $1$, so $n_{11}=12$\r\n If $n_3=4$, then letting $G$ act on $\\operatorname{Syl}_3(G)$ by conjugation gives a group homomorphism $G\\to S_4$, but its kernel is not all $G$, so $G$ is isomorphic to a subgroup of $S_4$, but $|S_4|=24<132$, contradiction.\\\\\r\n So $n_3=22$.\r\n Now the Sylow $3$-subgroups are all of order $3$, this means that there are $22\\times (3-1)=44$ elements of order $3$.\r\n Similarly $n_{11}=12$ gives $(11-1)\\times 12=120$ elements of order $11$, but then $132=|G|\\ge 120+44+1=165$, contradiction.\r\n\\end{example}\r\nFor sake of problem sheets, we mention the following definition.\r\n\\begin{definition}\r\n An automorphism $\\operatorname{Aut}(G)\\le\\operatorname{Sym}G$ of a group $G$ is the group of all isomorphisms from $G$ to itself.\r\n\\end{definition}", "meta": {"hexsha": "732c736479396ca6a4646948074daeb61adc9a65", "size": 6170, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "4/p.tex", "max_stars_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_stars_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "4/p.tex", "max_issues_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_issues_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "4/p.tex", "max_forks_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", "max_forks_repo_head_hexsha": "f4d4cc7141d30f03f775a67afc5a724db6a35da6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.7619047619, "max_line_length": 233, "alphanum_fraction": 0.6372771475, "num_tokens": 2273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7185943925708561, "lm_q2_score": 0.9005297901222472, "lm_q1q2_score": 0.6471156575248568}} {"text": "\\subsection{Convolutional Neural Network}\n\\label{subsec:cnn}\n% \\todo[inline]{Citations}\n\nConvolutional neural networks are a type of deep learning architecture that is primarily used to classify images.\nThey are based on artificial neural networks and the mathematical convolution operation.\nThe efficiency in image classification of CNNs is largely responsible for the reputation of deep learning \\cite{cnn_pathmind}.\n\nThe main problems when working with color images is that they are high-dimensional and therefore require a lot of processing power.\nFor this reason, convolutional neural networks use special layers to detect features and reduce the amount of data.\nThe three different layers used are convolution layers, pooling layers and fully-connected layers.\n% An example of a CNN sequence to classify handwritten digits is shown in figure \\ref{fig:cnn}.\nFigure \\ref{fig:cnn} shows an example of a CNN sequence to classify handwritten digits \\cite{cnn_pathmind}.\n\n\\paragraph{Convolution Layer}\nThe convolution layer performs the convolution operations using so-called kernels or filters.\nThe goal of the convolution operation is to extract certain features --- such as edges --- from the image.\nDifferent kernels are used to detect different features.\nThe first convolution layers are responsible for capturing low-level features, while following layers are responsible for capturing high-level features \\cite{cnn_tds}.\n\\clearpage\n\n\\paragraph{Pooling Layer}\nPooling layers reduce the spatial size of the convolved features.\nPooling is done by only looking at the portion of the image masked by the kernel.\nMaximum pooling yields the maximum value and average pooling yields the average value of the masked portion.\nBy doing so, the required computational power is significantly decreases and the dominant features are extracted \\cite{cnn_tds}.\n\\\\\n\n\\paragraph{Fully-Connected Layer}\nThe fully-connected layers are responsible for learning the non-linear combinations of the high-level convolved features.\nThey flatten the result of the convolution process into a vector of values.\nEach value corresponds to the probability that a certain feature belongs to a label \\cite{cnn_tds}.\n% Each value corresponds to a probability that a certain feature belongs to a label \\cite{cnn_tds}.\n% Each value corresponds to a probability that a certain feature belongs to a certain label \\cite{cnn_tds}.\n\n\\begin{figure}[t]\n \\centering\n \\includegraphics[width=\\textwidth]{cnn}\n \\caption{Example of a convolutional neural network \\cite{cnn_tds}}\n \\label{fig:cnn}\n\\end{figure}\n", "meta": {"hexsha": "32fff4c85581c7b10dd29c71264d007bc5b706e7", "size": 2570, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/report/sections/ai/cnn.tex", "max_stars_repo_name": "MuellerDominik/P5-AIonFPGA", "max_stars_repo_head_hexsha": "13fc60fb973a4a87a4af1b49c17d5dd1fd239ed5", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-01-26T15:54:09.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-26T15:54:09.000Z", "max_issues_repo_path": "doc/report/sections/ai/cnn.tex", "max_issues_repo_name": "MuellerDominik/P5-AIonFPGA", "max_issues_repo_head_hexsha": "13fc60fb973a4a87a4af1b49c17d5dd1fd239ed5", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/report/sections/ai/cnn.tex", "max_forks_repo_name": "MuellerDominik/P5-AIonFPGA", "max_forks_repo_head_hexsha": "13fc60fb973a4a87a4af1b49c17d5dd1fd239ed5", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.1904761905, "max_line_length": 167, "alphanum_fraction": 0.8108949416, "num_tokens": 528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127678225575, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6470902936863686}} {"text": "\\documentclass{ximera}\n\\input{../preamble}\n\\title{Exercises: Shell Method}\n%%%%%\\author{Philip T. Gressman}\n\n\\begin{document}\n\\begin{abstract}\nExercises for using the shell method.\n\\end{abstract}\n\\maketitle\n\n\\begin{exercise}\nThe region defined by the inequalities $\\sqrt{1-x^2} \\leq y \\leq 1$ for $0 \\leq y \\leq 1$ is revolved around the $y$-axis. Compute the volume of the resulting solid using the shell method.\n\\begin{center}\n\\begin{image}\n\\includegraphics{shell/shell03.png}\n\\end{image}\n\\end{center}\n\\begin{itemize}\n\\item When the slicing variable is $x$, the radius of a shell is the \\wordChoice{\\choice[correct]{horizontal}\\choice{vertical}} distance from an $x$-slice to the axis of rotation. Thus\n\\[ r(x) = \\answer{x} - \\answer{0}. \\]\n\\item The height of an $x$-slice is equal to\n\\[ h(x) = \\answer{1 - \\sqrt{1 - x^2}}. \\]\n\\item The volume is equal to the integral of $2 \\pi r h$, so \n\\[ V = \\int_{\\answer{0}}^{\\answer{1}} \\answer{2 \\pi x ( 1 - \\sqrt{1-x^2}) } dx = \\answer{\\frac{\\pi}{3}}. \\]\n(Note: to compute the integral, split it into two parts and make the substitution $u = 1-x^2$ for one of them.)\n\\end{itemize}\n\\end{exercise}\n\n\\begin{exercise}\nThe region in the plane bounded above by the graph $y = \\sqrt{1+x^2}$, below by $y = -1 + x + \\sqrt{1+x^2}$, and on the left by $x = 0$ is revolved around the axis $x = 1$. Compute the volume of the resulting solid using the shell method.\n\\begin{center}\n\\begin{image}\n\\includegraphics{shell/shell04.png}\n\\end{image}\n\\end{center}\n\\begin{itemize}\n\\item When the slicing variable is $x$, the radius of a shell is the \\wordChoice{\\choice[correct]{horizontal}\\choice{vertical}} distance from an $x$-slice to the axis $x = 0$. Thus\n\\[ r(x) = \\answer{1} - \\answer{x}. \\]\n\\item The height of an $x$-slice is equal to\n\\[ h(x) = \\answer{-1+x}. \\]\n\\item The volume is equal to the integral of $2 \\pi r h$, so \n\\[ V = \\int_{\\answer{-1}}^{\\answer{1}} \\answer{2 \\pi (1-x)^2 } dx = \\answer{\\frac{16\\pi}{3}}. \\]\n\\end{itemize}\n\\end{exercise}\n\n\\begin{exercise}\nThe region in the plane $y = \\sqrt{x}$, $y=0$, and $x = 1$ is revolved around the $y$-axis. Use the shell method to compute the volume.\n\\[ V = \\answer{\\frac{4 \\pi}{5}}. \\]\n\\end{exercise}\n\n\\begin{exercise}\nThe same region as above (bounded by $y = \\sqrt{x}$, $y=0$, and $x = 1$) is revolved around the axis $x=1$. Use the shell method to compute the volume.\n\\[ V = \\answer{\\frac{8 \\pi}{15}}. \\]\n\\end{exercise}\n\n\\begin{exercise}\nThe same region as above (bounded by $y = \\sqrt{x}$, $y=0$, and $x = 1$) is revolved around the $x$-axis. Use the shell method to compute the volume.\n\\begin{hint}\nThe ``height'' of a shell is $1-y^2$ in this case.\n\\end{hint}\n\\[ V = \\answer{\\frac{\\pi}{2}}. \\]\n\\end{exercise}\n\n\\begin{exercise}\nFor the same region as above (bounded by $y = \\sqrt{x}$, $y=0$, and $x = 1$), use the shell method to compute the volume when revolved around the axis $y = 1$.\n\\[ V = \\answer{\\frac{5 \\pi}{6}}. \\]\n\\end{exercise}\n\n\\section*{Sample Quiz Questions}\n\n\\begin{question}%%%%%[ShellQuad001]\n\nThe region in the plane bounded below by the curve \\(y=-x^2\\), above by the curve \\(y=x^2+2x+2\\), on the right by the line \\(x = 0\\), and on the left by the line \\(x = -2\\) is revolved around the axis \\(x = -2\\). Compute the volume of the resulting solid.\n(Hints won't reveal until after you choose a response.)\n\\begin{multiplechoice}\n\\choice{\\(4\\pi\\)}\n\\choice{\\(6\\pi\\)}\n\\choice[correct]{\\(8\\pi\\)}\n\\choice{\\(10\\pi\\)}\n\\choice{\\(12\\pi\\)}\n\\choice{\\(14\\pi\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nThe axis \\(x = -2\\) is parallel to the direction of slices using the integration variable \\(x\\), which indicates the shell method. \n The region lies to the right of the axis, which must be the case because the interval \\(-2 \\leq x \\leq 0\\) lies to the right of the axis \\(x = -2\\).\n \\begin{hint}\nThe integral to compute equals \\[ \\begin{aligned} V &= \\int_{-2}^{0}2 \\pi (x-(-2))((x^2+2x+2)-(-x^2))~ dx\\\\\n& = \\pi \\int_{-2}^{0} (4x^3+12x^2+12x+8)~ dx\\\\\n& = \\pi \\left. \\left(x^4+4x^3+6x^2+8x\\right) \\right|_{-2}^{0} = 8\\pi. \\end{aligned}\\]\n\\end{hint}\n\\end{feedback}\n\n\\end{question}\n\n\\begin{question}%%%%%[ShellQuad019]\n\nThe region in the plane bounded on the left by the curve \\(x=-y^2+4y+1\\), on the right by the curve \\(x=y^2+2y+1\\), and below by the line \\(y = -1\\) is revolved around the axis \\(y = -1\\). Compute the volume of the resulting solid.\n(Hints won't reveal until after you choose a response.)\n\\begin{multiplechoice}\n\\choice[correct]{\\(\\pi\\)}\n\\choice{\\(5\\pi\\)}\n\\choice{\\(9\\pi\\)}\n\\choice{\\(13\\pi\\)}\n\\choice{\\(17\\pi\\)}\n\\choice{\\(21\\pi\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nThe axis \\(y = -1\\) is parallel to the direction of slices using the integration variable \\(y\\), which indicates the shell method. The lower endpoint of integration will be \\(y = -1\\); the upper endpoint can be determined by setting \\(-y^2+4y+1 = y^2+2y+1\\) and choosing the solution which is greater than \\(-1\\). This gives the range \\(-1 \\leq y \\leq 0\\). \n The region lies above the axis, which must be the case because the interval \\(-1 \\leq y \\leq 0\\) lies above the axis \\(y = -1\\).\n \\begin{hint}\nThe integral to compute equals \\[ \\begin{aligned} V &= \\int_{-1}^{0}2 \\pi (y-(-1))((y^2+2y+1)-(-y^2+4y+1))~ dy\\\\\n& = \\pi \\int_{-1}^{0} (4y^3-4y)~ dy\\\\\n& = \\pi \\left. \\left(y^4-2y^2\\right) \\right|_{-1}^{0} = 1\\pi. \\end{aligned}\\]\n\\end{hint}\n\\end{feedback}\n\n\\end{question}\n\n\\begin{question}%%%%%[shellalgsub2001]\n\nThe region in the plane between the \\(x\\)-axis and the graph\n\\[ y = \\frac{1}{2 \\sqrt{\\frac{x^{2}}{3} + 1}} \\]\n in the range \\(0 \\leq x \\leq 3\\) is revolved around the axis \\(x = 0\\). Compute the volume of the resulting solid.\n (Hints won't reveal until after you choose a response.)\n\\begin{multiplechoice}\n\\choice{\\(\\displaystyle \\frac{3}{2} \\pi\\)}\n\\choice{\\(\\displaystyle \\frac{8}{5} \\pi\\)}\n\\choice{\\(\\displaystyle \\frac{5}{3} \\pi\\)}\n\\choice{\\(\\displaystyle 2 \\pi\\)}\n\\choice[correct]{\\(\\displaystyle 3 \\pi\\)}\n\\choice{\\(\\displaystyle 5 \\pi\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nIf the variable \\(x\\) is used for slicing, then slices are parallel to the axis of rotation, which indicates the shell method should be used.\nThe radius of a shell is \\(x\\). The height of a shell is exactly \\(\\frac{1}{2 \\sqrt{\\frac{x^{2}}{3} + 1}}\\).\n\\begin{hint}\nThe volume of the region is therefore given by\n\\[ \\int_{0}^{3} \\frac{\\sqrt{3} \\pi x}{\\sqrt{x^{2} + 3}}\\, dx. \\]\n\\begin{hint}\n To compute the integral we can use the substitution \\(u = x^{2} + 3\\) which implies the equality \\(du = \\left(2 x\\right)dx\\) for the differentials. This gives the equality\n\\[ \\begin{aligned} \\int \\frac{\\sqrt{3} \\pi x}{\\sqrt{x^{2} + 3}}\\, dx & = \\int \\frac{\\sqrt{3} \\pi}{2 \\sqrt{u}}\\, du \\\\\n & = \\sqrt{3} \\pi \\sqrt{u}. \\end{aligned} \\]\nReversing the substitution gives\n\\[ \\begin{aligned} \\int_{0}^{3} \\frac{\\sqrt{3} \\pi x}{\\sqrt{x^{2} + 3}}\\, dx & = \\left. \\left[\\sqrt{3} \\pi \\sqrt{x^{2} + 3} \\right] \\right|_{0}^{3}\\\\ & = \\left(6 \\pi \\right) - \\left(3 \\pi \\right) = 3 \\pi. \\end{aligned} \\]\n\\end{hint}\n\\end{hint}\n\\end{feedback}\n\n\\end{question}\n\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "953d7e0966fdd9a9268a5310031af6f38a40a51a", "size": 7033, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "volumes/03shellpractice.tex", "max_stars_repo_name": "ptgressman/math104", "max_stars_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "volumes/03shellpractice.tex", "max_issues_repo_name": "ptgressman/math104", "max_issues_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "volumes/03shellpractice.tex", "max_forks_repo_name": "ptgressman/math104", "max_forks_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.7961783439, "max_line_length": 357, "alphanum_fraction": 0.6507891369, "num_tokens": 2467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127529517044, "lm_q2_score": 0.7577943658046609, "lm_q1q2_score": 0.6470902730755489}} {"text": "\\hypertarget{matrix}{%\n\\section{Matrix}\\label{matrix}}\n\nThe Matrix class provides support for matrices. A matrix can be\ninitialized with a given size,\n\n\\begin{lstlisting}\nvar a = Matrix(nrows,ncols)\n\\end{lstlisting}\n\nwhere all elements are initially set to zero. Alternatively, a matrix\ncan be created from an array,\n\n\\begin{lstlisting}\nvar a = Matrix([[1,2], [3,4]])\n\\end{lstlisting}\n\nYou can create a column vector like this,\n\n\\begin{lstlisting}\nvar v = Matrix([1,2])\n\\end{lstlisting}\n\nOnce a matrix is created, you can use all the regular arithmetic\noperators with matrix operands, e.g.\n\n\\begin{lstlisting}\na+b\na*b\n\\end{lstlisting}\n\nThe division operator is used to solve a linear system, e.g.\n\n\\begin{lstlisting}\nvar a = Matrix([[1,2],[3,4]])\nvar b = Matrix([1,2])\n\nprint b/a\n\\end{lstlisting}\n\nyields the solution to the system a*x = b.\n", "meta": {"hexsha": "2fbb4a7aa87b905ec91304d8f5750fd1aa883df2", "size": 841, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manual/src/Reference/matrix.tex", "max_stars_repo_name": "mattsep/morpho", "max_stars_repo_head_hexsha": "50bb935653c0675b81e9f2d78573cf117971a147", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2021-09-18T14:44:14.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-26T11:41:50.000Z", "max_issues_repo_path": "manual/src/Reference/matrix.tex", "max_issues_repo_name": "mattsep/morpho", "max_issues_repo_head_hexsha": "50bb935653c0675b81e9f2d78573cf117971a147", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 79, "max_issues_repo_issues_event_min_datetime": "2021-10-05T17:33:19.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T16:06:10.000Z", "max_forks_repo_path": "manual/src/Reference/matrix.tex", "max_forks_repo_name": "mattsep/morpho", "max_forks_repo_head_hexsha": "50bb935653c0675b81e9f2d78573cf117971a147", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-10-05T16:56:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-31T19:55:27.000Z", "avg_line_length": 20.0238095238, "max_line_length": 69, "alphanum_fraction": 0.7288941736, "num_tokens": 239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8539127455162773, "lm_q2_score": 0.757794360334681, "lm_q1q2_score": 0.6470902627701385}} {"text": "\\gotosection{1}{4}\n\\subsection{The geometry of $\\mathbb{R}^n$}\n\n\\begin{exercise}{3}\n \\begin{enumerate}\n \\item Let $\\Vect{x} = \\begin{bmatrix}0\\\\1\\\\4\\end{bmatrix}$,\n $\\frac{\\Vect{x}}{|\\Vect{x}|}=\n \\frac{\\Vect{x}}{\\sqrt{17}}=\n \\begin{bmatrix}0\\\\\\frac{1}{\\sqrt{17}}\\\\\\frac{4}{\\sqrt{17}}\\end{bmatrix}$\n\n \\item Let $\\Vect{x} = \\begin{bmatrix}-3\\\\7\\end{bmatrix}$,\n $\\frac{\\Vect{x}}{|\\Vect{x}|}=\n \\frac{\\Vect{x}}{\\sqrt{58}}=\n \\begin{bmatrix}-\\frac{3}{\\sqrt{58}}\\\\\\frac{7}{\\sqrt{58}}\\end{bmatrix}$\n\n \\item Let $\\Vect{x} = \\begin{bmatrix}\\sqrt{2}\\\\-2\\\\-5\\end{bmatrix}$,\n $\\frac{\\Vect{x}}{|\\Vect{x}|}=\n \\frac{\\Vect{x}}{\\sqrt{31}}=\n \\begin{bmatrix}\n \\sqrt{\\frac{2}{31}} \\\\\n -\\frac{2}{\\sqrt{31}} \\\\\n -\\frac{5}{\\sqrt{31}}\n \\end{bmatrix}$\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{5}\n \\begin{enumerate}\n \\item $cos(\\theta) = \\frac{1}{\\sqrt{3}}$, $\\theta = \\arccos{1}{\\sqrt{3}}$.\n \\item $cos(\\theta) = 0$, $\\theta = \\frac{\\pi}{2}$.\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{7}\n \\begin{enumerate}\n \\item $\\det(M) = 1$, $M^{-1} = \\begin{xmatrix}0&1\\\\-1&2\\end{xmatrix}$.\n \\item $\\det(M) = 0$, $M^{-1}$ doesn't exist.\n \\item $\\det(M) = ad$, if $a, d \\neq 0$, $M^{-1} = \\frac{1}{ad}\\begin{xmatrix}d&-b\\\\0&a\\end{xmatrix}$.\n \\item $\\det(M) = 0$, $M^{-1}$ doesn't exist.\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{8}\n \\begin{enumerate}\n \\item -4\n \\item $adf$\n \\item $adg - cbg$\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{10}\n \\begin{enumerate}\n \\item This can be proven inductively: $k = 1$, $|A^1| = |A| = |A|^1$. $k = 2$, $|A^2| = |A \\cdot A| \\leq |A||A| = |A|^2$, from Cauchy-Schwarz's inequality. $k > 2$, $|A^k| = |A^{k-1} \\cdot A| \\leq |A^{k-1}||A| \\leq |A|^k$. \\rQED\n $$|A^3| = |\\begin{xmatrix}7&10\\\\5&7\\end{xmatrix}| = \\sqrt{223}$$\n $$|A|^3 = (\\sqrt{7})^3 = 7\\sqrt{7}$$\n \n \\item $|\\Vect{u} \\cdot \\Vect{v}| = |\\Vect{u}||\\Vect{v}|$, because they are in opposite direction, namely $\\Vect{u}$ is on the line spanned from $\\Vect{v}$.\n \n $|\\Vect{u} \\cdot \\Vect{w}| < |\\Vect{u}||\\Vect{w}|$, because the two vectors are not in the same or opposite directions, namely one is not on the line spanned from the other.\n \n $|\\Vect{u} \\cdot \\Vect{v}| = 2 + 8 + 18 = 28 = \\sqrt{14} \\cdot \\sqrt{56} = |\\Vect{u}||\\Vect{v}|$.\n \n $|\\Vect{u} \\cdot \\Vect{w}| = 2 + 18 = 20$, $|\\Vect{u}||\\Vect{w}| = \\sqrt{14} \\cdot \\sqrt{40} = 4\\sqrt{35} > 20 = |\\Vect{u} \\cdot \\Vect{w}|$.\n \n \\item Let $\\Vect{u} = \\begin{bmatrix}-v_2\\\\v_1\\end{bmatrix}$, namely rotating $\\Vect{v}$ by $\\frac{\\pi}{2}$ counterclockwise. $\\Vect{u} \\cdot \\Vect{w} = -v_2w_1 + v_1w_2 < 0$. Therefore, the angle between $\\Vect{u}$ and $\\Vect{w}$ is larger than $\\frac{\\pi}{2}$, which means $\\Vect{w}$ lies clockwise from $\\Vect{v}$.\n \n \\item Let $\\theta$ be the angle between $\\Vect{v}$ and $\\Vect{w}$. $\\cos(\\theta) = \\frac{\\Vect{v} \\cdot \\Vect{w}}{|\\Vect{v}||\\Vect{w}|}$, $\\cos(\\theta)|\\Vect{v}||\\Vect{w}| = 42$.\n \n If $\\cos(\\theta) = 1$, $|\\Vect{v}||\\Vect{w}| = 42$, or $|\\Vect{w}| = 3\\sqrt{14}$. This is the shortest $\\Vect{w}$. As $\\cos(\\theta)$ approaches $0$, $|\\Vect{w}|$ will approach $\\infty$, so there is no longest $\\Vect{w}$.\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{13}\n First, we can prove that for any\n $\\Vect{a} = \\begin{xmatrix}a_1\\\\a_2\\\\a_3\\end{xmatrix},\n \\Vect{b} = \\begin{xmatrix}b_1\\\\b_2\\\\b_3\\end{xmatrix}$ and $x, y \\in \\mathbb{R}$, $x\\Vect{a} \\times y\\Vect{b} = xy(\\Vect{a} \\times \\Vect{b})$:\n \n $$x\\Vect{a} \\times y\\Vect{b} =\n \\begin{xmatrix}\n xa_2yb_3 - xa_3yb_2 \\\\\n -xa_1yb_3 + xa_3yb_1 \\\\\n xa_1yb_2 - xa_2yb_1\n \\end{xmatrix} = xy\n \\begin{xmatrix}\n a_2b_3 - a_3b_2 \\\\\n -a_1b_3 + a_3b_1 \\\\\n a_1b_2 - a_2b_1\n \\end{xmatrix} = xy(\\Vect{a} \\times \\Vect{b})$$\n \n For two vectors $\\Vect{v}_1$, $\\Vect{v}_2$ pointing in the same direction, they can be written as $n_1\\Vect{u}$ and $n_2\\Vect{u}$, respectively. Let $u_1, u_2, u_3$ be the three components of $\\Vect{u}$, we have:\n\n $$\\Vect{v}_1 \\times \\Vect{v}_2 = (n_1n_2) \\Vect{u} \\times \\Vect{u} = (n_1n_2)\n \\begin{xmatrix}\n u_2u_3 - u_3u_2 \\\\\n -u_1u_3 + u_3u_1 \\\\\n u_1u_2 - u_2u_1\n \\end{xmatrix} = (n_1n_2)\\Vect{0} = \\Vect{0}$$\n \n \\rQED\n\\end{exercise}\n\n\\begin{exercise}{16}\n \\begin{enumerate}\n \\item $|Area| = |\\det(\\begin{xmatrix}1&5\\\\2&1\\end{xmatrix})| = |1 - 10| = 9$\n \\item $|Area| = |\\det(\\begin{xmatrix}1&5\\\\2&-1\\end{xmatrix})| = |-1 - 10| = 11$\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{19}\n \\begin{enumerate}\n \\item $|\\Vect{v}_n| = \\sqrt{1 + \\hdots + 1} = \\sqrt{n}$\n \\item $\\cos{a_n} = \\frac{\\Vect{v}_n \\cdot \\Vect{e}_1}{|\\Vect{v}_n||\\Vect{e}_1|} = \\frac{1}{\\sqrt{n}}$, so $a_n = \\arccos{\\frac{1}{\\sqrt{n}}}$. $\\lim_{n \\to \\infty} a_n = \\arccos{\\frac{1}{\\infty}} = \\frac{\\pi}{2}$.\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{24}\n \\def \\Set {\\Vect{v}^{\\bot}}\n \\begin{enumerate}\n \\item Since $\\Vect{0} \\cdot \\Vect{v} = 0$, $\\Vect{0} \\in \\Set$.\n \n For any $\\Vect{x}, \\Vect{y} \\in \\Set$, $(\\Vect{x} + \\Vect{y}) \\cdot \\Vect{v} = \\Vect{x} \\cdot \\Vect{v} + \\Vect{y} \\cdot \\Vect{v} = 0 + 0 = 0$. Therefore, $(\\Vect{x} + \\Vect{y}) \\in \\Set$.\n \n For $\\Vect{x} \\in \\Set$ and $c \\in \\mathbb{R}$, $(c\\Vect{x}) \\cdot \\Vect{v} = c(\\Vect{x} \\cdot \\Vect{v}) = c0 = 0$. \\rQED\n \n \\item $(\\Vect{a} - \\frac{\\Vect{a} \\cdot \\Vect{v}}{|\\Vect{v}|^2}\\Vect{v}) \\cdot \\Vect{v} =\n \\Vect{a} \\cdot \\Vect{v} - \\frac{\\Vect{a}\\cdot\\Vect{v}}{|\\Vect{v}|^2}\\Vect{v} \\cdot \\Vect{v} =\n \\Vect{a} \\cdot \\Vect{v}(1 - \\frac{\\Vect{v}^2}{|\\Vect{v}|^2}) = 0$. \\rQED\n \n \\item (b) suggests the existence of $t(\\Vect{a})$. Suppose that there exists a distinct $t'(\\Vect{a})$ such that $(\\Vect{a} + t'(\\Vect{a})\\Vect{v}) \\in \\Set$. Then we have:\n \n $$\\left\\{\n \\begin{aligned}\n (\\Vect{a} + t(\\Vect{a})\\Vect{v}) \\cdot \\Vect{v} = 0 \\\\\n (\\Vect{a} + t'(\\Vect{a})\\Vect{v}) \\cdot \\Vect{v} = 0\n \\end{aligned} \\quad \\Rightarrow \\quad\n \\begin{aligned}\n \\Vect{a} \\cdot \\Vect{v} + t(\\Vect{a})\\Vect{v}^2 = 0 \\\\\n \\Vect{a} \\cdot \\Vect{v} + t'(\\Vect{a})\\Vect{v}^2 = 0\n \\end{aligned} \\quad \\Rightarrow \\quad \n \\begin{aligned}\n t(\\Vect{a}) = -\\frac{\\Vect{a} \\cdot \\Vect{v}}{\\Vect{v}^2} \\\\\n t'(\\Vect{a}) = -\\frac{\\Vect{a} \\cdot \\Vect{v}}{\\Vect{v}^2}\n \\end{aligned}\n \\right.$$\n \n Since there could be only one possible value for $-\\frac{\\Vect{a} \\cdot \\Vect{v}}{\\Vect{v}^2}$, $t(\\Vect{a}) = t'(\\Vect{a})$, namely $t(\\Vect{a})$ is unique.\n \n $$\\Vect{a} + t(\\Vect{a})\\Vect{v} = \\Vect{a} -\\frac{\\Vect{a} \\cdot \\Vect{v}}{\\Vect{v}^2}\\Vect{v} = \\Vect{a} - \\frac{\\Vect{a} \\cdot \\Vect{v}}{|\\Vect{v}|^2}\\Vect{v} = P_{\\Set}(\\Vect{a})$$\n \n \\rQED\n \\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{26}\n \\begin{enumerate}\n \\item $$A\\begin{xmatrix}x\\\\y\\end{xmatrix} = \\begin{xmatrix}x-2y\\\\3x+4y\\end{xmatrix}$$\n \n $$a\\begin{pmatrix}x\\\\y\\end{pmatrix} = \\arccos{\\frac{x^2+xy+4y^2} {\\sqrt{10}\\sqrt{(x^2+y^2) (x^2+2xy+2y^2)}}}$$\n \n \\item Suppose there existing a nonzero vector such that it is rotated by $\\frac{\\pi}{2}$:\n \n $$\\frac{x^2+xy+4y^2} {\\sqrt{10}\\sqrt{(x^2+y^2) (x^2+2xy+2y^2)}} = \\cos(\\frac{\\pi}{2})$$\n $$x^2+xy+4y^2=0$$\n $$(\\frac{1}{2}x+2y)^2=-\\frac{3}{4}x^2$$\n \n Because $(\\frac{1}{2}x+2y)^2 \\geq 0$ and $-\\frac{3}{4}x^2 \\leq 0$, both sides must equal $0$. Therefore, $x = 0, y = 0$, which contradicts the premise of nonzero vector. As a result, such vector doesn't exist.\n \\end{enumerate}\n\\end{exercise}\n", "meta": {"hexsha": "5c7130e50ab39fda79608ecf75554fffe117c960", "size": 7911, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW3/sec0104.tex", "max_stars_repo_name": "notcome/fa15-linear-algebra", "max_stars_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "HW3/sec0104.tex", "max_issues_repo_name": "notcome/fa15-linear-algebra", "max_issues_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW3/sec0104.tex", "max_forks_repo_name": "notcome/fa15-linear-algebra", "max_forks_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.8106508876, "max_line_length": 321, "alphanum_fraction": 0.5235747693, "num_tokens": 3278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.85391273808085, "lm_q2_score": 0.757794360334681, "lm_q1q2_score": 0.6470902571356136}} {"text": "\\pagebreak\n\\section{Experiment-2}\n\\textbf{Q.1 What would you expect the steady-state velocity of the system to be if a step input of 2000\nN were applied at t = 0?}\n\\\\\nAns. \\\\\nFrom newtons second law we have, dv/dt = (F-bv)/M.\n\\\\ M is the mass, v is velocity and F is force, b is damping cofficient. \n\\\\\nAt steady state, the change in velocity will be zero. i.e. dv/dt = 0.\n\\\\\nTherefore, F = bv\n\\\\F = 2000N\n\\\\b = 40 N*sec/m\n\\\\2000 = 40*v\n\\\\v= 50 m/sec.\n\n\\textbf{Q.2 Why do this step input and the ramp input with saturation we simulated have the same steady state velocity?}\n\\\\Ans.The step input is similar to saturated ramp input, because when steady state is reached, both step input and saturated ramp input\nhave fixed value(input value). Therefore, if we apply same force to same system (under idential condition), \nthen system will attain the same velocity. That is why we get same steady state velocity.\n\n\\textbf{Q.3 Using which input, the step or the ramp with saturation, would you expect the system to reach its steady-state velocity more quickly?\n}\n\\\\Ans. With step input the system will reach it's steady state velocity more quickly. Because, \nsaturated ramp input will take more time to reach it's saturation value whereas step input \nreaches to it's fixed value quickly. ", "meta": {"hexsha": "7e6d7ff00f11004e829c05d64bc2ec7f00360113", "size": 1283, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "conclusion-writing/module/experiment-2.tex", "max_stars_repo_name": "suryanshshukla10/MSPS-practical", "max_stars_repo_head_hexsha": "f0478fd3fdf10c85257b2dbaff86761b020ddd36", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "conclusion-writing/module/experiment-2.tex", "max_issues_repo_name": "suryanshshukla10/MSPS-practical", "max_issues_repo_head_hexsha": "f0478fd3fdf10c85257b2dbaff86761b020ddd36", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "conclusion-writing/module/experiment-2.tex", "max_forks_repo_name": "suryanshshukla10/MSPS-practical", "max_forks_repo_head_hexsha": "f0478fd3fdf10c85257b2dbaff86761b020ddd36", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.5185185185, "max_line_length": 145, "alphanum_fraction": 0.7466874513, "num_tokens": 340, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.8031738034238807, "lm_q1q2_score": 0.6470626768811475}} {"text": "\\newpage\n\\subsection{A Full Example}\n\\label{subsec:compss_example}\nThis section intends to give the reader a more or less extensive insight on what writing a COMPSs application is. We think that this section may help to \\textit{materialize} concepts and will avoid to give this document an excessively abstract tone.\\\\\n\\\\\nLets suppose that we want to approximate the value of $\\pi$. For this purpose we have thought on a simple, randomized algorithm:\n\\begin{enumerate}\n\\item Generate $N$ random 2D points with coordinates between $-1$ and $1$\n\\item Consider the set of points $S$ within distance $1$ or less to the origin\n\\item Assume that $\\frac{|S|}{N} = \\frac{\\pi}{4}$\n\\end{enumerate}\n\n\\begin{figure}[ht!]\n\\centering\n\\includegraphics[scale=0.5]{figures/circle_square.png}\n\\caption{A graphical representation of the random experiment. The square has side length $2$, so the circle has radius $1$, and therefore area $\\pi$. In ratio terms, $\\frac{\\pi}{4}$ of the points belong to the circle}\n\\label{fig:circle_square}\n\\end{figure}\n\nA more graphical explanation on why this works can be found in figure \\ref{fig:circle_square}.\\\\\n\\\\\nWe know a little bit of Python, so we have decided to implement this program in it. Basically, our small application will consist of a \\verb|test_random_point| function that generates a random point and return $1$ if this point lies inside our circle, and $0$ otherwise. We will call this function $N$ times, and consider the proportion $\\frac{|S|}{N}$ to be equal to $\\frac{\\pi}{4}$.\n\\inputminted{python}{applications/PI_SQUARE/sequential.py}\nThis code can be straightforward \\textit{optimized} by transforming the \\verb|test_random_point| function into a COMPSs task and syncing the results in the main procedure.\n\\inputminted{python}{applications/PI_SQUARE/pycompss_naive.py}\nAlthough this may be a good approach to \\textit{parallelize} this application we must note that we want to make it run in a distributed environment. The main difference we can appreciate is that a COMPSs task may run in a different machine than the master code, so some coordination between two processes in different machines and the transfer of potentially big amounts of data are necessary. In other words, the tradeoff between task granularity and performance is much more punishing in distributed computing than in single-machine parallel cases.\\\\\n\\\\\nAnother important thing to note is that a distributed application can still exploit lower level parallelism in each of its tasks. In our case, we can transform our \\verb|test_random_point| function into a \\verb|test_random_points| procedure that generates and tests various random points at the same time.\n\n\\inputminted{python}{applications/PI_SQUARE/pycompss_vectorized.py}\n\nThis last approach is what we consider a well \\textit{COMPSsfied} application: it has a reasonable task count and granularity, and it exploits various levels of parallelism at the same time. This application also delegates most of the work to \\verb|numpy| procedures, which are mainly written in C++ and OpenMP. This aspect is also important in PyCOMPSs, as Python is, by nature, a very slow programming language and it should be only used as an orchestrator.\n\nCOMPSs is mainly designed to run in HPC environments. Most HPC machines integrate some sort of queue system to manage its resources among all the demanding users. Our previous example can be run as a job in a queue system with the following command:\n\n\\inputminted{bash}{applications/PI_SQUARE/run_mn4.sh}\n\nThe \\verb|enqueue_compss| command refers to a generic queueing script (see section \\ref{subsec:hpc_queues}) which translates our request to enqueue this COMPSs job to a specific queue system. Some of the most common parameters of a COMPSs job can be found in table \\ref{table:compss_queue_param}.\\\\\n\\\\\n\\begin{table}[ht!]\n\\centering\n\\begin{tabular}{|l|l|}\n\\hline\nArgument name & Description \\\\ \\hline\n\\verb|exec_time| & Job time limit \\\\ \\hline\n\\verb|num_nodes| & \\begin{tabular}[c]{@{}l@{}}Number of computing\\\\ nodes\\end{tabular} \\\\ \\hline\n\\verb|cpus_per_node| & \\begin{tabular}[c]{@{}l@{}}Number of cores per\\\\ computing node\\end{tabular} \\\\ \\hline\n\\verb|constraints| & \\begin{tabular}[c]{@{}l@{}}Additional constraints\\\\ (e.g: highmem nodes)\\end{tabular} \\\\ \\hline\n\\end{tabular}\n\\caption{Some example configuration parameters of the queue system. These parameters are usually passed as flags to the enqueue\\_compss script.}\n\\label{table:compss_queue_param}\n\\end{table}\n", "meta": {"hexsha": "49d64f1b11a3a2790300144916c428c9eede3a3c", "size": 4671, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compss_example.tex", "max_stars_repo_name": "srgrr/TFM", "max_stars_repo_head_hexsha": "f6c302547fdfc9b1df3423cc4e927cd6bb83b345", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-01-02T10:39:32.000Z", "max_stars_repo_stars_event_max_datetime": "2019-05-22T14:01:41.000Z", "max_issues_repo_path": "compss_example.tex", "max_issues_repo_name": "srgrr/TFM", "max_issues_repo_head_hexsha": "f6c302547fdfc9b1df3423cc4e927cd6bb83b345", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compss_example.tex", "max_forks_repo_name": "srgrr/TFM", "max_forks_repo_head_hexsha": "f6c302547fdfc9b1df3423cc4e927cd6bb83b345", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 88.1320754717, "max_line_length": 552, "alphanum_fraction": 0.7488760437, "num_tokens": 1122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737916455819, "lm_q2_score": 0.8056321936479701, "lm_q1q2_score": 0.6470626636439878}} {"text": "\\section{Background and Notation} \\label{background_and_notation}\n\nThis section gives more background on the sliding window technique \\cite{keogh2004sliding}, DTW distance measure \\cite{keogh2002exact}, and time series normalization \\cite{das1998rule}, which are fundamental building blocks of our conducted online gesture recognition study \\cite{lesti2017filter}. Table \\ref{tab:notation} introduces the notation that we use for formal problem description.\n\n\\begin{table}\n \\begin{center}\n \\begin{tabularx}{\\textwidth}{c X}\n \\hline\n \\textbf{Symbol} \\qquad & \\textbf{Description}\\\\\n \\hline\n $Q$ & a time series of size $n$ with $Q = (q_1, q_2, \\dots, q_i, \\dots, q_n)$\\\\\n $Q[i,j]$ & a subsequence time series of $Q$ with $Q[i,j] = (q_i, q_{i+1}, \\dots, q_{j})$\\\\\n $t$ & the current time\\\\\n $\\mu$ & the mean of a time series $Q$\\\\\n $\\sigma$ & the standard deviation of a time series $Q$\\\\\n $\\eta$, $z$ & two different time series normalizations\\\\\n \\hline\n \\end{tabularx}\n \\end{center}\n \\caption{Notation used for formal problem description.}\n\t\\label{tab:notation}\n\\end{table}\n\n\\input{background_and_notation/sliding_window_technique.tex}\n\\input{background_and_notation/dynamic_time_warping.tex}\n", "meta": {"hexsha": "ba26be82198c030a820ba1424c79040cc61e10d4", "size": 1327, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/background_and_notation.tex", "max_stars_repo_name": "GordonLesti/SlidingWindowFilter", "max_stars_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-06-22T09:37:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-14T11:43:53.000Z", "max_issues_repo_path": "paper/background_and_notation.tex", "max_issues_repo_name": "GordonLesti/SlidingWindowFilter", "max_issues_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/background_and_notation.tex", "max_forks_repo_name": "GordonLesti/SlidingWindowFilter", "max_forks_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-01-11T23:15:57.000Z", "max_forks_repo_forks_event_max_datetime": "2019-01-11T23:15:57.000Z", "avg_line_length": 51.0384615385, "max_line_length": 390, "alphanum_fraction": 0.6639035418, "num_tokens": 373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.8031737940012418, "lm_q1q2_score": 0.647062654297234}} {"text": "\\section{Relevant Background}\n\n\\subsection{Agda Introduction}\nAgda is a dependently typed functional language developed by Norell at the Chalmers University of Technology\nas his Ph.D. Thesis.\nThe current version of Agda is Agda 2.\n\n \\subsubsection{Syntax}\n In Agda, \\emph{Set} is equal to type.\n In languages with dependent types, it is possible to create a function that returns a type.\n\n \\agda{agdaExamples}{funcType}\n\n After the function name, it is two colons (\\emph{:}) and the arguments of the function.\n It is closed by \\emph{(name\\_of\\_argument : type\\_of\\_argument)}.\n After all, there is one arrow and the type of the result of the function.\n This ``if, then, else'' is not a function built-in in Agda.\n It is a function defined this way \\emph{if\\_then\\_else\\_} .\n\n So it is possible to use this function in the default way.\n\n \\agda{agdaExamples}{funcTypeUnd}\n\n Or use the arguments inside the underscore.\n\n \\agda{agdaExamples}{funcType2}\n\n The same notation can be done using just arrows without naming the arguments.\n\n Because of dependent types, it is possible to have a type that depends on the input.\n\n \\agda{agdaExamples}{dependentType}\n\n It is possible in Agda to do the pattern match.\n So it breaks the input in cases.\n\n \\agda{agdaExamples}{patternMatch}\n\n To create a new type with a different pattern match, it is used the data constructor.\n\n \\agda{agdaExamples}{dataConstructor}\n\n This is another example of \\emph{Data Set}, but it depends on the argument.\n\n \\agda{agdaExamples}{vector}\n\n \\emph{Vector zero} is a type of a vector of size zero, so the only option to construct it is the empty vector.\n It is constructed from the first constructor.\n Other types of vectors like \\emph{Vector 1} (vector of size one), \\emph{Vector 2}, ... can only be constructed by\n the second constructor.\n It takes as argument a natural number and a vector and returns a vector with the size of the last vector\n plus one.\n\n Records are data types with just one case of pattern match.\n\n \\agda{agdaExamples}{record}\n\n The constructor is the name of the data constructor.\n\n \\hyperref[id]{Implicit terms} are elements that the compiler is smart enough to deduce it.\n So it is not necessary to put it as an argument of the function.\n\n \\plabel{id}\n \\agda{agdaExamples}{id}\n\n Implicit arguments are inside \\emph{\\{\\}}.\n In this example, the name of the Set (\\emph{A}) can not be omitted\n (like the second function version of boolean to set),\n because it is used to say that \\emph{x} is of type \\emph{A}.\n\n In the case of the function \\emph{id}, the type of input can be deduced by the compiler.\n For example, \\emph{id} function should be a function from Natural to Natural because\n \\emph{zeroN} is already of type Natural.\n In this example, the compiler could figure out the term \\emph{A} (Natural) of type \\emph{Set}.\n\n \\agda{agdaExamples}{idNat}\n\n Functions in Agda can be defined in two ways\n\n \\agda{agdaExamples}{funcs}\n\n In the first case, the arguments are before equal sign (\\emph{=}).\n In the second case, it is used the lambda abstraction that means the same thing.\n\n \\agda{agdaExamples}{caseOf}\n\n \\agda{agdaExamples}{filter}\n\n Filter is a function that filters all elements that have some property already defined.\n In this definition of filter, the lambda abstraction can pattern match\n in the two possibilities of \\emph{p x}, that are \\emph{true} or \\emph{false}\n\n \\agda{agdaExamples}{filterWith}\n\n The \\emph{with} abstraction is the same thing as the \\emph{case of}.\n Before the \\emph{=}, there are the possibilities of \\emph{p x} (\\emph{true} or \\emph{false}).\n\n \\agda{agdaExamples}{postulate}\n\n In Agda, the \\emph{postulate} assumes that there is one element of a given type without defining it.\n In this example, \\emph{someBot} is an element of the empty type.\n Postulate can be dangerous because it assumes that some element of a given type exists,\n but it can not (like in the example).\n\n \\agda{agdaExamples}{rewrite}\n\n In Agda, \\emph{rewrite} means changing the returned type of the result.\n Because $a + b ≡ b + a$, returning an element of the type $P (a + b)$ is the\n same thing that returning an element of the type $P (b + a)$.\n\n \\subsubsection{Lambda Calculus}\n Lambda Calculus is a minimalist Turing complete programming language with the concept of abstraction,\n application using binding and substitution. For example, \\emph{x} is a variable, $(\\lambda x.M)$\n is an Abstraction and (\\emph{M N}) is an Application.\n\n In Lambda Calculus, there are two types of computational conversions $\\alpha$-conversion\n and $\\beta$-reduction.\n In $\\alpha$-conversion, $(\\lambda x.M[x]) \\rightarrow (\\lambda y.M[y])$.\n So in every free variable \\emph{x} in \\emph{M} will be renamed to \\emph{y}.\n For \\emph{M[x] = x}, an $\\alpha$-conversion is $(\\lambda x.x) \\rightarrow (\\lambda y.y)$\n\n A free variable is every variable that is not bound outside.\n For example, $((\\lambda\\textcolor{green}{x}.\\textcolor{blue}{x}) \\textcolor{red}{x})$.\n The \\textcolor{blue}{blue x} is bonded for the \\textcolor{green}{green x},\n but the \\textcolor{red}{red x}\n is not bonded for any function. So the \\textcolor{red}{red x} is a free variable.\n\n In $\\beta$-reduction, it replaces the all free for the expression in the application.\n The $\\beta$-reduction of this expression $((\\lambda x.M) N) \\rightarrow (M[x := N])$ .\n So if $M = x$, the $\\beta$-reduction will be $((\\lambda x.x) N) \\rightarrow N$.\n If $M = (\\lambda x.x) x$, the $\\beta$-reduction will be\n $(\\lambda x.((\\lambda x.x)x))N \\rightarrow (\\lambda x.x)N$.\n\n Agda uses typed lambda calculus.\n So in an application \\emph{(M N)}, \\emph{M} has to be of type $A \\Rightarrow B$ and N has to be of type A.\n $(\\lambda (x : A) . x)$ is of type $A \\Rightarrow A$, because \\emph{x} is of type \\emph{A}.\n\n \\agda{lambdaCalculus}{Id}\n The simplest function is the identity function made in Agda.\n\n \\plabel{id2}\n \\agda{lambdaCalculus}{Id2}\n \\hyperref[id2]{This} is another way of writing the same function.\n\n \\plabel{truefalse}\n \\agda{lambdaCalculus}{trueFalse}\n \\hyperref[truefalse]{This} is how true and false are encoded in lambda calculus.\n\n \\plabel{naturals}\n \\agda{lambdaCalculus}{naturals}\n \\hyperref[naturals]{This} is how naturals numbers are defined in lambda calculus.\n Look that the definition of zero looks like the definition of false.\n\n \\agda{lambdaCalculus}{isZero}\n Defining natural numbers in this way, it is possible to say if a natural number is zero or not.\n\n \\agda{lambdaCalculus}{plus}\n Plus is defined this way using lambda calculus.\n\n \\plabel{onepone}\n \\agda{lambdaCalculus}{onePone}\n \\hyperref[onepone]{This} is one example of the calculation of one plus one in Lambda Calculus.\n\n \\plabel{list}\n \\agda{lambdaCalculus}{list}\n \\hyperref[list]{This} is how lists are defined in Lambda Calculus.\n\n \\agda{lambdaCalculus}{sumList}\n Substituting the cons operation of list per plus and nil list to zero, it is possible to calculate\n the sum of the list.\n\n \\agda{lambdaCalculus}{either}\n In this way, it is possible to define \\emph{Either}.\n It is one way to create a type that can be a Natural or a Boolean.\n\n \\agda{lambdaCalculus}{eitherExamples}\n In these examples, it is defined zero, one in left and false, true in right.\n\n \\agda{lambdaCalculus}{eitherRes}\n \\emph{Either} is useful when defining one function that works for left and another that works for the right.\n If the natural number is zero, the function chosen is the left and\n if it is an identity, the function chosen is the right. \n\n \\plabel{tuple}\n \\agda{lambdaCalculus}{tuple}\n \\hyperref[tuple]{This way} is how tuple is defined in Lambda Calculus.\n\n \\plabel{tupleexamples}\n \\agda{lambdaCalculus}{tupleExamples}\n \\hyperref[tupleexamples]{This} is how is defined the tuple zero false and the tuple one true.\n\n \\plabel{tupleadd}\n \\agda{lambdaCalculus}{tupleAdd}\n \\hyperref[tupleadd]{This} is one way of defining a function that adds one to the argument\n if the first element of the tuple is true.\n\n\n \\subsubsection{Martin-Löf Type Theory}\n Agda also provides proof assistants based on the intentional Martin-Löf type theory.\n\n In Martin-Löf type theory, there are three finite types and five constructors types.\n The zero type contains zero terms. It is called the empty type and it is written $\\bot$.\n \\agda{agdaExamples}{botType}\n\n The first type is the type with just one canonical term and it represents existence.\n It is called unit type and it is written top.\n \\agda{agdaExamples}{trivialType}\n\n The second type contains two canonical terms. It represents a choice between two values.\n \\agda{agdaExamples}{eitherType}\n\n The Boolean type is defined using the Trivial type and the Either type.\n \\agda{agdaExamples}{boolType}\n\n If statement is defined using booleans.\n \\agda{agdaExamples}{ifThenElse}\n\n\n \\subsubsection{Types Constructors}\n The product-types contain an ordered pair.\n The second type can depend on the first type.\n \\agda{agdaExamples}{mulType}\n\n \\agda{agdaExamples}{mulEx}\n In the function $\\exists-vec$, the product-type has the same meaning of exists.\n This type means that\n ``Exists a natural number \\emph{n} that there is one element of type \\emph{Vector A n}''\n\n The $\\pi$-types contain functions.\n So given an input type, it will return an output type.\n It has the same meaning as a function:\n \\agda{agdaExamples}{piType}\n\n In Inductive types, it is a self-referential type.\n Naturals numbers are examples of that:\n \\agda{agdaExamples}{Nat}\n Other data structures like a linked list of natural numbers, trees, graphs are inductive types too.\n\n Proofs in inductive types are made by induction.\n \\agda{agdaExamples}{NatElim}\n\n Universe types are created to allow proofs written in all types.\n For example, the type of \\emph{Nat} is \\emph{U0}.\n\n\nAgda has dependent types like in CoQ but does not have tactics.\nAgda is a total language, so it is guaranteed that the code always finish\n(impossible infinite loops) and coverage all inputs.\n\nAgda has inductive data types that are similar to algebraic data types in non-dependently\ntyped programming languages.\nThe definition of Peano numbers in Agda is:\n\n\\agda{agdaExamples}{Nat}\n\nDefinitions in Agda are done using induction.\nFor example, the sum of two numbers in Agda:\n\n\\agda{agdaExamples}{sum}\n\nIn Agda, because of dependent types, it is possible to make more expressive types that are not\npossible in other languages.\nFor example, get the first element of a vector.\nFor it, it is necessary to specify in the type that the vector should have a size greater than zero.\n\n\\agda{agdaExamples}{vecHead}\n\nAnother good example is that in the sum of two matrices, they should have the same dimensions.\n\n\\agda{agdaExamples}{matrixSum}\n\n\\subsubsection{Preposition as Types}\n\nIn Agda, types can be seen as prepositions and terms of this given type like a proof.\nFor example, \\emph{x} can be a term of the type \\emph{A}.\nSo \\emph{x} is a proof that \\emph{A} holds.\n\n\\agda{agdaExamples}{id}\n\nIn the function \\emph{id},\nif there is a proof \\emph{x} that \\emph{A} holds,\nso \\emph{A} holds.\nThe function \\emph{id} is a proof of $A \\rightarrow A$.\n\n\\agda{agdaExamples}{and}\n\nIn the function \\emph{and}, \\emph{a} is a proof that \\emph{A} holds and\n\\emph{b} is a proof that \\emph{B} holds,\nso \\emph{a , b} is a proof of $A \\times B$\n\n\\agda{agdaExamples}{equality}\n\nEquality function can be seen as\n``For all elements of \\emph{x} in the Set \\emph{A}, $x \\equiv x$''.\nIf there is a term of this type, so this preposition is true.\nIn this case, the term is a function that given a term \\emph{x} of type \\emph{A},\nit returns \\emph{refl \\{\\_\\} \\{\\_\\} \\{x\\}}.\n\\emph{refl} is a constructor of equality, so \\emph{refl \\{\\_\\} \\{\\_\\} \\{x\\}}\nreturns a term of type $x \\equiv x$.\n\n\\subsection{Bitcoin Overview}\n\nThe Bitcoin was made to be a peer to peer electronic cash.\nIt was made in one way that users can save and verify transactions without the need of a trusted party.\nBecause of that no authority or government can shut down the Bitcoin.\n\n\\incimg{transactions1}{png}{Transaction}\n%% https://medium.com/coinmonks/bitcoin-transactions-be401b48afe6\n\nTransactions in Bitcoins (like in \\figref{transactions1}) are an array of input of\nprevious transactions and an array of outputs.\nEach input and output is an address, each address is made from a public key\nthat is made from a private key.\n\n\\incimg{privatekey}{png}{Bitcoin account}\n%% https://coinsutra.com/bitcoin-private-key/\n\nA private key is a big number.\nIt is so big that it is almost impossible to generate two identical private keys.\n\nThe public key is generated from the private key\n(like in \\figref{privatekey} where the account number is f(p)),\nbut a private key can not be generated from a public key.\n\nIn Bitcoin, the miner is someone that calculates a value \\emph{nounce} for each block.\nBecause of that, he receives all transaction fees of this block plus a block reward.\nThe mining transaction (made for the miner) does require an input.\nFor each input transaction, it is necessary a signature signed with a private key\n(like in \\figref{privatekey} where the signature is f(p,t))\nto prove the ownership of the Bitcoins.\nWith the message and the signature, it is possible to know that the owner of the private key\nthat generates the public key signed the message.\n\nSignature is created from a function that receives as inputs the private key and the message.\nThere is no analytical form of the inverse of this function.\nBecause of that, the only way of finding an element \\emph{x} that this function \\emph{f} so\nthat $f(x) = y$ for a given \\emph{y} is trying for random different values of \\emph{x}.\nSo it is almost impossible to find the private key from a message, a signature, and the public key.\nIn \\figref{privatekey}, the checker is a f(t,s,a).\nSo because of that, the owner of the private key can sign several messages without anyone knows\nhis private key.\n\n\\incimg{transactions2}{jpg}{Verification and signature of transactions}\n%% https://en.wikipedia.org/wiki/Bitcoin_network\n\nTransactions (shown in \\figref{transactions2}) are grouped in a block (shown in \\figref{blockchain}).\nEach block contains in its header the timestamp of its creation, the hash of the block,\nthe previous hash and a nonce.\nA nonce is an arbitrary value that the miner has to choose to make the hash of the block respect some\nspecific characteristics.\n\n\\incimg{blockchain}{png}{Blockchain}\n%% https://bitcoin.stackexchange.com/questions/12427/can-someone-explain-how-the-bitcoin-blockchain-works/13347\n%% https://i.stack.imgur.com/HrKX0.png\n\nEach block has a size limit of 1 MB.\nBecause of that, Bitcoin forms a blockchain (a chain of blocks).\nEach block should be created at an average of 10 minutes.\nThe advantage of using a short average time of block creation is that it will propagate transactions\nfaster in the blockchain.\nBut the disadvantage is that it increases the probability of creating orphan blocks.\nTo make the blockchain tamper-proof (nobody can alter an existing block),\nthere is a concept called proof of work in Bitcoin.\nTo create a new block, the creator (called the miner) has to choose a random value as nonce that makes\nthe hash of the block less than a certain value.\nThis value is chosen in a way that each block should be generated on 10 minutes on average.\nIf the value is for example 100 and the total possibilities of results are 10.000 (codomain),\nthe probability of finding \\emph{x} so that $hash(x) \\leqslant 100$ is 1\\% $100/10.000$.\nBecause of that, in this example, the miner has to calculate on average 100 hashes before finding\nthe right one.\n\nIf the value is too low, miners will take more time to find a nonce that makes the hash block\nless than it.\nIf it is too high, it will be easier to find a nonce and they will find it faster.\n\nWhen two different blocks are created in nearly the same time, there are two valid blockchains.\nIt is because the last block in both blockchains are valid but different.\nBecause of this problem, in the Bitcoin protocol, the largest chain is always the right chain.\nWhile two valid chains have the same size, it is not possible to know which chain is the right.\nThis situation is called fork and when it happens, it is necessary to wait to see in which chain\nthe new block will be.\n\nIf miners have more than 50\\% of the hash rates of Bitcoin, they can do an attack called a\n51\\% attack.\nIt happens when some miner, with more power than all network, mine secretly the blocks.\nSo if the main network has 50 blocks, the miner could produce hidden blocks from 46 to 55\nand he would have 10 hidden blocks from the network.\nWhen he shows their hidden blocks, his chain becomes the valid chain, because it is bigger.\nSo all transactions from the previous blockchain from 46 to 50 blocks become invalid.\nBecause of that, when someone makes a big transaction in the blockchain, it is a good idea\nto wait more time.\nSo it is becoming harder and harder to make a 51\\% with more time.\nBitcoin has the highest market value nowadays, so attacking the Bitcoin network is very expensive.\nNowadays, this kind of attack is more common in new altcoins.\n\nLedger is a software that tracks how much money each address can spend.\nIt can be known when calculating the sum amount of all unspent output transactions destined\nfor each address.\n\nThe \\gls{utxo} model used in Bitcoin and the account model used in Ethereum are the two most used kinds\nof data structures to model account records and savings states.\n\nIn the account model, it is saved the address and the balance of each address.\nFor example, the data structure will look like this [(0xabc01, 1.01), (0xabc02, 2.02)].\nSo the address 0xabc01 has 1.01a of balance and the address 0xabc02 has 2.02 of balance.\nIn this way, it is possible to easily know how much balance each address has,\nbut it is not possible to know how they got in this state.\n\n \\incimg{utxo}{png}{UTXO transactions}\n %% https://bitcoin.org/\n %% https://medium.com/@sunflora98/utxo-vs-account-balance-model-5e6470f4e0cf\n\nIn the \\gls{utxo} model (shown in \\figref{utxo}), each transaction is saved in the transaction tree.\nEvery transaction is composed of multiples inputs and multiples outputs.\nBut all inputs have to come from outputs that have never been spent before.\n\nBecause of that, in the \\gls{utxo} model, it is easy to make a new transaction from the previous one,\nbut it is harder to know how much each one has.\nTo know how much someone has, it is necessary to calculate the sum of all unspent transaction outputs\ndestined for his addresses.\n\nIn the account model, there could be one kind of vulnerability\nthat is less probable to happen in the \\gls{utxo} model.\nChanging both addresses balances is not an atomic operation.\nBecause there are two instructions, subtracting the balance of the sender\nand incrementing the balance of the receiver.\nIf something bad happens between these two instructions, the final state\nwill become an invalid state.\n\nFor example: \\\\\n\\textbf{\nbobBalance -= 1 \\\\\nIntermediary State \\\\\naliceBalance += 1\n}\n\nIn the account model, it is straight forward to know how much balance each address has.\nIn the \\gls{utxo} model, this calculation is made off-chain. It can be a good thing\nbecause each user has more privacy.\n\n\\subsection{Ethereum}\n\nEthereum differs from Bitcoin in having an \\gls{evm} to run script code.\n\\gls{evm} is a stack machine and Turing complete \\cite{turing1936computable} while Bitcoin Script is not\n(it is impossible to do loops and recursion in Bitcoin).\n\nTransactions in Bitcoin are all stored in the blockchain.\nIn Ethereum, just the hash of it is stored in it.\nSo it is saved in the off-chain database.\nBecause of that, it is possible to save more information in Ethereum Blockchain.\n\nContracts in the blockchain are the script program that comes with the transaction.\nIn Bitcoin, the creator of the contract has to pay the amount proportional to its size.\nIn Ethereum, it is different.\nEach smart contract in Ethereum is made by a series of instructions.\nEach instruction consumes different the computational effort.\nBecause of that, in Ethereum, there is a concept of gas, that measure how much computational effort\neach instruction needs.\nSo in each smart contract, it is well know how much computational effort will be necessary to run it\nand it is measured in gas.\nBecause computational effort of each node in the Ethereum network is a scarce resource,\nto execute the smart contract, it is necessary to\npay an amount in ether for each gas to the miner run it.\nSmart contracts that pay more ether per gas run first because the miner will want to have the best\nprofit and they will pick them.\nIf the amount of ether per gas paid is not high enough, the contract will not be executed,\nbecause some other contracts pay more that will be executed instead of this one.\n\nBecause Ethereum has its \\gls{evm} with more instructions than Bitcoin and it is Turing Complete,\nit is considered less secure.\nEthereum has its high-level programming language called Solidity that looks like JavaScript.\n", "meta": {"hexsha": "b6d04ac667836fa3679899518a0daf524b536c09", "size": 21176, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/relevantbackground.tex", "max_stars_repo_name": "guilhermehas/crypto-agda", "max_stars_repo_head_hexsha": "ac91e00abca9a26678d0cbc1bedecf8abef6b703", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2020-02-13T16:56:47.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-22T19:27:12.000Z", "max_issues_repo_path": "docs/relevantbackground.tex", "max_issues_repo_name": "guilhermehas/crypto-agda", "max_issues_repo_head_hexsha": "ac91e00abca9a26678d0cbc1bedecf8abef6b703", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-02-11T18:59:29.000Z", "max_issues_repo_issues_event_max_datetime": "2020-02-11T18:59:29.000Z", "max_forks_repo_path": "docs/relevantbackground.tex", "max_forks_repo_name": "guilhermehas/cripto-agda", "max_forks_repo_head_hexsha": "ac91e00abca9a26678d0cbc1bedecf8abef6b703", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.8644067797, "max_line_length": 115, "alphanum_fraction": 0.7549112202, "num_tokens": 5441, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.8031737869342623, "lm_q1q2_score": 0.647062648603848}} {"text": "\\section{Differential Rings and Modules}\nRecalling our definition of differential rings (an associative algebra with derivation Lie algebra) we wish to extend our set of notations.\n\\subsection{Differential modules and ideals}\n\\begin{defi}\nLet $(R,D)$ be a differential ring.\n\\bn\n\\item A differential module $(M,D_M)$ is an $R$-module with an additive map:\n$$D : M \\longrightarrow M,\\ r m \\longmapsto \\partial_R(r) m + r D(m)\\ \\forall D \\in D_M,\\ r \\in R,\\ m \\in M,$$\nwhere $\\partial_R \\in \\trm{Der}(R) = \\left$.\n\\item A differential ideal $I \\subset R$ is a differential $R$-submodule of $R$, i.e. $\\partial(I) \\subset I$ for all $\\partial \\in D$.\n\\item Let $(R,D_R)$ and $(S,D_S)$ be two differential rings. A ring homomorphism $f : R \\longrightarrow S$ is called differential if\n$$\\partial_S \\circ f = f \\circ \\partial_R$$\nor equivalently the following diagram commutes:\n$$\\xymatrix{\nR \\ar[d]_{\\partial_R} \\ar[r]^f & S\\ar[d]^{\\partial_S}\\\\\nR \\ar[r]_{f\\mid_{\\partial_R(R)}} & S.\\\\\n}$$\n%where $\\partial_X(X) := \\trm{im} \\partial_X$ for $X = R, S$.\n\\item A differential ring $(R,D)$ is called simple differential if it has no proper differential ideals.\n\\item A differential field is a differential ring with no proper ideals (differential or non-differential).\n\\item Let $(R,D)$ be a differential ring. The subset\n$$R^\\partial := \\{x \\in R : \\partial(x) = 0\\ \\forall \\partial \\in D\\}$$\ndefines a subring and is called the ring of constants.\n\\en\n\\index{Index}{module!differential}\n\\index{Index}{ring!differential}\n\\index{Index}{ring!differential!homomorphism of}\n\\index{Index}{ring!differential!simple}\n\\index{Index}{ring!differential! of constants}\n\\index{Index}{field!of constants}\n\\index{Symbol}{$\\trm{Der}_R(A)$}\n\\index{Symbol}{$R^\\partial$}\n\\end{defi}\n\\bmk As mentioned, the derivation maps $D$ on differential modules $(M,D_M)$ are only additive (are in general not in $\\trm{End}_R(M)$, but in $\\trm{End}_{R^\\partial}(M)$ - i.e. its ring of constants). This is due to the Leibniz-rule as defined above.\\\\\n\\indent The ring of constants $R^\\partial$ for a differential field $(R, D)$ is also a field, since:\n$$\\partial(a a^{-1}) = 0 = \\underbrace{\\partial(a)}_{=0} a^{-1} + a \\partial(a^{-1}) \\LRA a \\partial(a^{-1}) = 0$$\nholds for all $a \\in R^\\times$.\n\\bsp Some prominent examples:\n\\bn\n\\item any ring $R$ is a (trivial) differential ring, via $0 : R \\longrightarrow 0$. Thus, all $R$-modules are also differential modules via the zero-homomorphism.\n\\item The polynomial ring in one indeterminate:\n$$\\left(k[X], \\partial = \\frac{d}{dX}\\right),$$\n$k$ a field with characteristic zero, is a simple differential ring since all ideals $I$ generated by some polynomial of degree greater zero eventually fulfill $\\partial^i(I) = (1)$. On the other hand, $(k[X], \\partial = X \\frac{d}{dX})$ has non-trivial differential ideals:\n$$I_i = \\left,\\ \\forall i \\geq 1$$\nas each $k$-sub vector space $k.X^i$ is $\\partial$-stable.\n\\item Let $p \\in \\nz$ be a prime number, the polynomial ring $\\mathbb{F}_p[X]$ with $\\mathbb{F}_p$-derivation $\\partial = \\frac{d}{d X}$, as in the last example, has an interesting property. Its ring of constants is $\\mathbb{F}_p[X^p]$ as $\\partial(X^p) = p X^{p-1} = 0$ and this ring contains indeed non-trivial differential ideals (in contrast to characteristic zero fields):\n$$I_k := \\left,\\ k \\geq 1$$% : k \\geq 1\\right>,$$\nsince $\\partial(X^{p^k} f + X^{p^k} g) = \\partial(X^{p^k} f) + \\partial(X^{p^k} g) = X^{p^k} \\partial(f + g)$ for all $f, g \\in \\mathbb{F}_p[X]$.\n\\item The field of rational functions $k(X)$ is a differential field with derivation Lie algebra generated by\n$$D = \\left\\{\\partial'_x = \\left[\\frac{f}{g} \\mapsto \\frac{\\partial_x(f) g - f \\partial_x(g)}{g^2}\\right] : x \\in X\\right\\}.$$\n\\item Let $U \\subset \\rz^n$ be open and connected then $(C^\\infty(U),D=\\{\\partial_i : 1 \\leq i \\leq n\\})$ is a partial differential ring with a non-trivial differential ideal $\\mathfrak{m} := \\{f \\in C^\\infty(U) : \\trm{supp} (f) \\subsetneq U\\}$, as we may define a proper differential ideal for all $f \\in C^\\infty(U) \\bsl \\rz[\\{x\\}]$ by simply putting\n$$\\left<\\{f\\}\\right> = \\left<\\partial_{i_1} \\circ \\hdots \\circ \\partial_{i_k}(f) : 1\\leq i_j \\leq n,\\ k \\in \\nz_0\\right>,$$\nwhere $\\rz[\\{x\\}]$ denotes the ring of real convergent power series in $U$. These ideals are, in general, not finitely generated.\n\\en\n\\subsection{General Differential Algebra}\nLet $(k,D)$ be a differential ring with $D = \\{\\partial\\}$ and $k^\\partial$ its ring of constants.\n\\subsubsection{Ring of differential operators}\nIn terms of Ore extension, the subring $D := k[\\partial] \\subset \\trm{End}_{k^\\partial}(k)$ is isomorphic to the Ore-extension $k[X,id_k,\\partial]$ over $k$. To show this we depend on its intrinsic module algebra structure:\\\\\n$$\\Psi := \\Psi_{\\trm{int}} \\mid_{D \\otimes k.id_D} = \\left[d' \\longmapsto \\left[d \\longmapsto \\mu_D(d' \\otimes d) = d'(d)\\right]\\right],$$\nwhere $\\Psi_{\\trm{int}} : D \\otimes \\trm{End}_{k^\\partial}(D) \\longrightarrow \\trm{End}_{k^\\partial}(D)$, restricted to the subalgebra $k.id_D$.\n\\indent $k[\\partial]$ has the structure of a pointed-irreducible bialgebra, of Birkhoff-Witt type (only $1_{k[\\partial]}$ as group-like, generator $X \\in \\mathfrak{g}_{k^\\partial}(k)$ has primitive coproduct and $(k[\\partial],\\mu,\\eta)$ is isomorphic to the enveloping algebra of some Lie-algebra). Its structure map\n$$\\Psi : D \\otimes D \\longrightarrow D,\\ \\partial \\otimes a \\partial^i \\longmapsto \\partial(a) \\partial^i + a \\partial^{i+1}.$$\nTo see the isomorphism wrt. Ore-extensions, we consider the $k$-left module $k.\\partial^i$ with its above described $k$-right module structure:\n$$\\partial^i \\otimes a \\sim \\sum_{j=0}^i \\left(\\bao{c}i\\\\j\\\\\\ea\\right) \\partial^j(a) \\otimes \\partial^{i-j}$$\nHence, the ideal generators $1 \\otimes X \\otimes a - a \\otimes X \\otimes 1 - \\partial(a) \\in \\bigoplus_{n \\geq 0} (k \\otimes \\partial)^{\\otimes n} \\otimes k$, as in \\ref{prop03} on pg. \\pageref{prop03}, yield - evaluated via $id_k \\otimes \\Psi : k \\otimes D \\otimes k \\longrightarrow k \\otimes k \\simeq k$:\n$$id \\otimes \\Psi(1 \\otimes X \\otimes a - a \\otimes X \\otimes 1 - \\partial(a) \\otimes 1_D \\otimes 1_k) = X(a) - a X(1) - \\partial(a).$$\nIdentifying $X = \\partial$, we get indeed zero as desired. Hence, the quotient yields our desired Ore-extension $k[X,id_k,\\partial] \\simeq k[\\partial]$. There are more interesting properties for $k[\\partial]$ that are discussed in \\cite{vdPS01}, chapter 2, in greater detail.\\\\\n\\indent Concluding this subsection, we note that $k[\\partial]$ is a unital, associative, coassociative, cocommutative bialgebra (indeed, has an antipode, which will be discussed below) acting on $k$ via evaluation.\n\\subsubsection{Ring of differential polynomials}\\label{RingOfDiffPolys}\nAs above we are using a differential field $(k,\\partial)$, $k^\\partial$ its ring of constants and its ring of differential operators $k[\\partial]$ being generated by one element $\\partial$.%, pick some differential field $(k,\\partial)$ and\n We consider the ring of polynomials $R := k[u_1,\\ldots,u_n]$, i.e. a noetherian ring over $k$. In general, we have no unique extension of $\\partial$ to $R$ except for that of a trivial differential ring: $u_i \\longmapsto 0$ (enlarging the ring of constants to $R^\\partial[u_1,\\ldots,u_n]$). However, we may use a non-noetherian transcendental extension $S$ over $R$\n$$S := k[u_{i,j} : 1 \\leq i \\leq n, j \\in \\nz_0],$$ where $u_{i} \\longmapsto u_{i,0}$ defines an embedding of $R$ in $S$. Indeed, this ring still has no canonical $D = k[\\partial]$-module algebra structure. However, the module:\n$$D \\otimes_k S := D \\otimes_{k^\\partial} S/\\left,$$\nwhere $\\Delta(d) = \\sum_{(d)} d_{(1)} \\otimes d_{(2)}$, has:\n$$\\partial^i \\otimes f_\\alpha u^\\alpha \\sim_\\partial \\sum_{j=0}^i\\left(\\bao{c}i\\\\j\\\\\\ea\\right) \\partial^j(f_\\alpha) \\partial^{i-j}\\otimes u^\\alpha.$$\nIts module algebra structure map $\\Psi : D \\otimes_k (D \\otimes_k S) \\longrightarrow D \\otimes_k S$ is given by the above equivalence relation. The ideal\n$$I := \\left<\\partial^{\\alpha_3} \\otimes u_{\\alpha_1,\\alpha_2} - \\partial^\\beta \\otimes u_{\\alpha_1,\\alpha_2 + \\alpha_3 - \\beta} : 1 \\leq \\alpha_1 \\leq n, \\alpha_2, \\alpha_3 \\in \\nz_0, 0 \\leq \\beta \\leq \\alpha_3 - 1\\right> \\subset D \\otimes_k S$$ gives us our desired:\n\\begin{defi}[Ring of differential polynomials]\nLet $S$ and $I$ be defined as above. The quotient ring $D \\otimes_k S/I$ is called the ring of differential polynomials and is denoted by:\n$$k\\left\\{u_1,\\ldots,u_n\\right\\} := D \\otimes_k S/I.$$\n\\end{defi}\n\\bmk %Firstly, we remark that if $\\partial u_{i,j} = u_{i,j+1} \\in K\\{u\\}$ then also \n%$$\\partial^l(u_{i,j}) = \\partial^{l-1}(u_{i,j+1}) = \\ldots = u_{i,j+l} \\in k\\{u\\}\\ \\forall 1 \\leq i \\leq n,\\ j \\geq 0,\\ l \\geq 1.$$\nSometimes we may use $u_i^{(j)}$ instead of $u_{i,j}$. Secondly, note that this, indeed, defines a (non-noetherian) differential ring, with derivations $\\{\\partial, \\partial_{u_{i,j}}\\}$. Here, $\\partial_{u_{i,j}}$ are $k$-derivations, while $\\partial$ is a $k^\\partial$-derivation. Additionally, we may still recover $R$:\n$$R \\simeq k\\{u_1,\\ldots,u_n\\}/J,\\ \\trm{where}\\ J := \\left \\subset k\\{u_{i,j}\\}.$$\n\\indent The definition naturally translates to differential rings - if $(R,\\partial)$ is our differential ring, with ring of constants $R^\\partial$ and $D$-stable ideal $I$ as above, then\n$$R\\{u_1,\\ldots,u_n\\} := D \\otimes_k R[u_{i,j} : 1 \\leq i \\leq n, j \\geq 0]/I.$$\nAlthough being non-noetherian (any non-trivial differential ideal does not fulfill the ascending chain condition), the factor rings we will consider are in fact noetherian.\n\\begin{defi}\nLet $(k,\\partial)$ be a differential ring (field). An associative unital $k$-algebra $K$ is called a differential extension over $k$, short $K/k$, if there is an $n \\in \\nz$ and some differential ideal $I \\subset k\\{u_1\\ldots u_n\\}$ such that\n$$K \\simeq k\\{u_1\\ldots u_n\\}/I.$$\n\\index{Index}{extension!differential}\n\\end{defi}\n\\bmk Given a differential ring $(k,\\partial)$ and a finite family of differential polynomials $\\mathcal{F} \\subset k\\{u_1\\ldots u_{|\\mathcal{F}|}\\}$ the associated differential ideal $I$ is simply the differential saturation:\n$$I := \\left<\\partial^i(f) : f \\in \\mathcal{F},\\ i \\geq 0\\right>.$$\nSuch a family is called a differential equation - linear if the degree of all monomials is at most one, otherwise non-linear. We call it an explicit differential equation if each element in $\\mathcal{F}$ is linear with respect to $\\partial u_i$ and all coefficients of monomials of the form $\\prod_{i,j} \\partial^j u_i$ are zero, for $j \\geq 2$. Otherwise, it is called implicit.\n\\index{Index}{differential saturation}\n\\index{Index}{differential equation}\n\\index{Index}{differential equation!linear}\n\\index{Index}{differential equation!non-linear}\n\\index{Index}{differential equation!explicit}\n\\index{Index}{differential equation!implicit}\n\\begin{defi}\nLet $k\\{u_1,\\hdots,u_n\\} =: k\\{u\\}$ be the ring of differential polynomials over some differential field $k$, $k[\\partial]$ the left $k$-module of differential operators on $k\\{u\\}$. The map\n$$\\bao{rrcl}\nev &: k[\\partial] \\otimes_{k^\\partial} k\\{u\\} & \\longrightarrow &k\\{u\\}\\\\\n&&&\\\\\n& \\sum_{\\substack{0 \\leq i \\leq n\\\\\\alpha \\in \\nz_0^k\\\\k\\geq 0}} (a_i \\partial^i \\otimes b_\\alpha u_\\alpha) &\\longmapsto& \\sum_{\\substack{i,\\alpha\\\\k \\leq i}} \\left(\\bao{c}i\\\\k\\\\\\ea\\right) a_i \\partial^k(b_\\alpha) \\partial^{i-k}(u_\\alpha)\\\\\n\\ea$$\nis called the evaluation homomorphism.\n\\end{defi}\n\\bmk %The evaluation homomorphism gives us a $k[\\partial]$-module algebra structure on $k\\{u\\}$, where $(k[\\partial],\\mu,\\eta,\\Delta,\\eps)$ is the bialgebra structure on $k[\\partial]$ (infact, it has a Hopf-algebra structure via $S : k[\\partial] \\longrightarrow k[\\partial]$, $\\partial \\longmapsto -\\partial$.\nObviously, the definition $ev$ and $\\Psi$ are equal. Using this setting we get\n\\begin{koro}\nThe two-sided $k$-module $k[\\partial]$ defines (in general non-commutative if $k^\\partial \\neq k$) a unital $k$-algebra.\n\\end{koro}\n\\bws As $k$ is a unital commutative simple algebra over $k^\\partial$ and $k[\\partial]$ is a Ore-extension as shown above, there is nothing more to show.%Note that the map $\\mu : k[\\partial] \\otimes k[\\partial] \\longrightarrow k[\\partial], a \\partial^i \\otimes b \\partial^j \\longmapsto \\sum \\left(\\bao{c}i\\\\k\\\\\\ea\\right) a \\partial^k(b) \\partial^{i+j-k}$ defines a multiplication on the monomial terms of $k[\\partial]$. The unit is simply $\\eta : K \\longrightarrow K[\\partial], 1_k \\longmapsto 1_{k[\\partial]}$.\n\\bmk Moreover, $k[\\partial]$ has a $k^\\partial$-coalgebra structure:\n$$\\Delta =\\left[ \\partial^i \\longmapsto \\sum \\left(\\bao{c}i\\\\j\\\\\\ea\\right) \\partial^{i-j} \\otimes \\partial^j\\right],\\ \\eps = [\\partial^i \\longmapsto \\delta_{i,0}]$$\nmaking it to a $k^\\partial$-bialgebra as clearly: $\\eps \\eta = id_k$, $(\\eta \\otimes \\eta) \\circ \\eta = \\Delta \\eta = id_{k.1_{k[\\partial]} \\otimes k.1_{k[\\partial]}}$.% With $S := [\\partial^i \\longmapsto (-1) \\partial^i]$ we get an antipode since $\\mu((S\\otimes id)(1_k)) = \\mu((id_k \\otimes S)(1_k)) = \\eta(\\eps(1_k))$.\n\\begin{prop}\nThe following statements are equivalent:\n\\bn\n\\item\\label{item01} $k\\{u\\}$ is a $k[\\partial]$-module algebra (or $\\Psi := ev$ defines a module algebra structure on $k\\{u\\}$).\n\\item \\label{item02} Given the evaluation homomorphism and multiplication on $k[\\partial]$ then the following diagram commutes\n$$\\xymatrix{\nk[\\partial] \\otimes k[\\partial] \\otimes k\\{u\\} \\ar[r]^{\\mu \\otimes id_{k\\{u\\}}}\\ar[d]_{id_{k[\\partial]} \\otimes ev} & k[\\partial] \\otimes k\\{u\\}\\ar[d]^{ev}\\\\\nk[\\partial] \\otimes k\\{u\\} \\ar[r]_{ev} & k\\{u\\}\\\\\n}$$\n\\en\n\\end{prop}\n\\bws We show first, that the second statement is indeed true.\n\\bn\n\\item By simple computation on the monomial terms $a \\partial^i, b \\partial^j \\in k[\\partial]$ and $c u_\\alpha \\in k\\{u\\}$ we get\n{\\scriptsize\n$$\\bao{rcl}\nev(\\mu \\otimes id_{k\\{u\\}})(a \\partial^i \\otimes b \\partial^j \\otimes c u_\\alpha) &=& \\sum_{k' \\leq i} \\sum_{l' \\leq i + j - k'} \\left(\\bao{c}i\\\\k'\\ea\\right) \\left(\\bao{c}j + i - k'\\\\l'\\ea\\right) a \\partial^{k'}(b) \\partial^{l'}(c) \\partial^{j+i-k'-l'}(u_\\alpha)\\\\\n&&\\\\\nev(id_{k[\\partial]} \\otimes ev)(a \\partial^i \\otimes b \\partial^j \\otimes c u_\\alpha) &=& \\sum_{k \\leq i} \\sum_{l \\leq j} \\sum_{m \\leq i - k} \\left(\\bao{c}i\\\\k\\ea\\right) \\left(\\bao{c}j\\\\l\\ea\\right) \\left(\\bao{c}i - k\\\\m\\ea\\right) a \\partial^k(b) \\partial^{l+m}(c) \\partial^{j+i-k-l-m}(u_\\alpha)\\\\\n\\ea$$}\nFixing $l' \\leq j$ and putting $k = k'$ we see that our equivalence implies $$\\sum_{l + m = l'} \\left(\\bao{c}j\\\\l\\ea\\right) \\left(\\bao{c}i - k\\\\m\\ea\\right) a \\partial^k(b) \\partial^{l+m}(c) \\partial^{j+i-k-l-m}(u_\\alpha) = \\left(\\bao{c}j + i - k\\\\l + m\\\\\\ea\\right) a \\partial^k(b) \\partial^{l + m}(c) \\partial^{j + i - l - m - k}(u_\\alpha).$$\nSince the degree of the differential operators on each factor do agree, we get\n$$\\sum_{l + m = l'} \\left(\\bao{c}j\\\\l\\ea\\right) \\left(\\bao{c}i - k\\\\m\\ea\\right) = \\left(\\bao{c}j + i - k\\\\l + m\\\\\\ea\\right).$$\nBut this is just a rewriting of the Vandermonde identity for all $k \\leq i$ and proves our claim.\n\\item If $k\\{u\\}$ is a $k[\\partial]$ module algebra we have that $(x y) v = x(y v)$ for all $x, y \\in k[\\partial]$, $v \\in k\\{u\\}$. Expanding with our standard notation this translates into $ev(\\mu\\otimes id) = ev(id\\otimes ev)$ proving \\ref{item01} $\\RA$ \\ref{item02}.\\\\\nThe opposite direction follows immediately from the definition.\n\\en\n\\begin{koro}\\label{HopfModAlgDiffPoly}\nWith bialgebra structure maps $\\mu, \\eta, % = [1_k \\longmapsto 1_{k[\\partial]}], \n\\Delta$ and $ % = \\left[a \\partial^i \\longmapsto \\sum_{0 \\leq j \\leq i} \\left(\\bao{c}i\\\\j\\\\\\ea\\right) a \\partial^j \\otimes \\partial^{i - j}\\right], \n\\eps$ as defined above and the $k$-homomorphism\n% = [1_{k[\\partial]} \\longmapsto 1_k, a \\partial^i \\longmapsto 0],\n$$S : k[\\partial] \\longrightarrow k[\\partial],\\ \\partial^i \\longmapsto (-1)^i \\partial^i,$$\nwe have that $k\\{u\\}$ is a $k[\\partial]$-Hopf-module algebra.\n\\end{koro}\n\\bws Note that the generator $\\partial$ is a primitive cocommutative Hopf algebra element, i.e. $\\Delta(\\partial) = 1 \\otimes \\partial + \\partial \\otimes 1$ implying $S(\\partial) = - \\partial$. We simply have to show that the commutative diagram for Hopf-algebras does commute.\n$$\\partial^i \\stackrel{\\Delta}{\\longmapsto} \\sum_{j=0}^i\\left(\\bao{c}\ni\\\\\nj\\\\\n\\ea\\right) \\partial^j \\otimes \\partial^{i-j} \\stackrel{S \\otimes id}{\\longmapsto} \\sum_{j} \\left(\\bao{c}\ni\\\\\nj\\\\\n\\ea\\right) (-1)^j \\partial^j \\otimes \\partial^{i-j} \\stackrel{\\mu}{\\longmapsto} \\sum_j \\left(\\bao{c}\ni\\\\\nj\\\\\n\\ea\\right) (-1)^j \\partial^i.$$\nAccording to Pascals rule we see via induction that except for $i = 0$ all sums are zero. As $\\eta(\\eps(\\partial^i)) = \\delta_{i,0}$, we have just shown the required commutativity.\n%Hence, all elements $\\sum c_\\alpha u_\\alpha$ can be associated with some element of the form $\\sum \\left(\\bao{c} i\\\\j\\\\\\ea\\right) \\partial^j(b_\\alpha) \\partial^{i-j}(u_\\alpha)$ we get an (not necessarily unique) element such that for $\\rho = \\left[\\sum c_\\alpha u_\\alpha \\longmapsto \\sum \\left(\\bao{c}i\\\\j\\\\\\ea\\right) \\partial^j(b_\\alpha) \\otimes \\partial^{i-j}(u_\\alpha)\\right]$\n\\subsection{Linear Differential Equations}\nLet $(k,\\partial)$ be a (not necessarily non-trivial with characteristic zero) differential field with field of constants denoted by $k^\\partial$ and $M$ a noetherian differential module over $k$ (i.e. finite dimensional vector space, with derivation).\n\\begin{defi}\nA (scalar) linear differential equation is $k^\\partial$-linear map $L : M \\longrightarrow M$, with $L = \\sum a_i \\partial^i$, i.e. $L \\in k[\\partial] \\subset \\trm{End}_{k^\\partial}(k \\otimes_{k^\\partial} M)$. The solution space is the $k^\\partial$-subspace of $M$:\n$$S(L) := \\{x \\in M : L(x) = 0\\}.$$\n\\index{Index}{space!solution}\n\\index{Symbol}{$S(L)$}\n\\end{defi}\nAlternatively, one can define linear differential equations simply via linear algebra:\n$$\\partial(x) = A x,\\ A \\in \\trm{End}(M),\\ x \\in M.$$\nIts solution space is simply generated by the kernel elements of $\\partial - A$ in $M$. Hence, all differential extensions are generated by solutions over $k$. Now, an important definition:\n\\begin{defi}\nLet $K/k$ be a differential extension to the differential equation $\\partial(x) = Ax$.\n\\bn\n\\item we call a matrix $X \\in \\trm{Mat}_n(K)$ a solution matrix, if\n$$\\partial(X) = A X \\in \\trm{Mat}_n(K).$$\nIf $X = (x_{ij}) \\in \\trm{Mat}_n(K)$ is a solution matrix for the above differential equation we call the $n^2 \\times n^2$ matrix\n$$Wr(X) = \\left(\\partial^l(x_{ij})\\right)_{\\substack{0 \\leq l \\leq n^2-1\\\\1 \\leq i,j \\leq n}}$$\nthe Wronskian matrix. Its determinant is simply called Wronskian.\n\\item $K/k$ is called a differential extension (over $k$), if $K$ contains the solution space for some linear differential equation $\\partial(x) = A x$.\n\\en\n%A solution matrix in the general linear group $\\trm{Gl}_n(R)$ is called a fundamental matrix.\n\\index{Index}{matrix!solution}\n\\index{Index}{matrix!Wronskian}\n\\index{Index}{matrix!fundamental}\n\\index{Index}{Wronskian}\n\\end{defi}\n\\bmk As we already defined differential extensions via the ring of differential polynomials $k\\{u\\}$ (to be precise via differential quotient rings) we shall show both definitions are equivalent. But clearly, the family of differential polynomials is simply $\\mathcal{F} := \\{\\partial u_i - \\sum_j a_{i,j} u_j \\in k\\{u\\} : 1 \\leq i \\leq n\\}$ for a given differential equation $\\partial u = A u$.\\\\\n\\indent Secondly, we get the Wronskian matrix by constructing column vectors $$y_{ij} = \\left(x_{ij},\\partial(x_{ij}),\\ldots,\\partial^{n^2-1}(x_{ij})\\right)^t$$running over all indices $1 \\leq i,j \\leq n$. Furthermore, the definition of the Wronskian is broader - for some differential extension $R/k$ and elements\n$y_1, \\ldots, y_m \\in R$ the Wronskian matrix is simply $Wr(y_1,\\ldots,y_m) := (\\partial^l(y_i))_{1 \\leq i,l + 1 \\leq m}$.\n\\begin{defi}\nLet $K/k$ be a differential field extension for a given differential equation $\\partial y = A y$.\n\\bn\n\\item A Picard-Vessiot ring $R$ is a sub-ring of $K$, such that\n$R^\\partial = k^\\partial$, $R$ contains no non-trivial differential ideals and there exists a solution matrix $X \\in \\trm{Gl}_n(R)$.\n\\item A solution matrix in a PV-ring $R$ is called a fundamental matrix.\n\\item A Picard-Vessiot field is the localization of a Picard-Vessiot ring.\n\\en\n\\end{defi}\n\\index{Index}{extension!differential}\n\\index{Index}{extension!Picard-Vessiot}\n\\index{Index}{ring!Picard-Vessiot}\n\\index{Index}{field!Picard-Vessiot}\nThe last definition requires a little\n\\begin{lemm}\nA simple differential ring is zero-divisor free.\n\\end{lemm}\n\\bws Let $a \\in \\trm{Ann}(R)$, then there is a $b \\in \\trm{Ann}(R)\\bsl\\{0\\}$, s.t. $a b = 0$. We get $\\partial(a b) = 0 = \\partial(a) b + a \\partial(b) \\LRA \\partial(a) b = -a \\partial(b)$. Multiplying both sides with $b$ we have:\n$$\\partial(a) b^2 = -a b \\partial(b) = 0,$$\ni.e. $\\partial(a) \\in \\trm{Ann}(R)$. As the only proper differential ideal is zero, we see that $\\trm{Ann}(R)$ is trivial.\\\\\nNow, we have that indeed the localization of a Picard-Vessiot ring is well-defined in that sense that the localization is not the zero ring. Alternatively, we could define the Picard-Vessiot field extension $K/k$ as a field containing the solution space and having the same field of constants, i.e. $K^{\\partial} = k^{\\partial}$.\n\\newcommand{\\minpoly}[1]{\\trm{Min}(\\alpha,#1)}\n\\newcommand{\\minpolyC}{\\minpoly{k^\\partial}}\n\\newcommand{\\minpolyR}{\\minpoly{R}}\n\\begin{lemm}\nLet $(k,\\partial)$ be a differential field of characteristic zero and let $R$ be a differential subring of $k$\nwith the same field of constants, i.e. $R^\\partial = k^\\partial \\subset R \\subset k$.\n\\bn\n\\item If $\\alpha \\in R$ is algebraic over $k^\\partial$, i.e. $\\minpolyC \\in k^\\partial[X]$, with $k^\\partial(\\alpha) \\simeq k^\\partial[X]/\\left<\\minpolyC\\right>$, then $\\partial(\\alpha) = 0$.\n\\item If $\\alpha \\in k$ is algebraic over $R$ and $\\partial(\\alpha) = 0$, then $\\alpha$ is algebraic over $k^\\partial$.\n\\en\n\\end{lemm}\n\\bws Let $\\minpoly{S} = p = \\sum_{i=0}^n p_i X^i \\in S[X]$ for $S = R, k^\\partial$.\n\\bn\n\\item By definition, we have $p \\in k^\\partial[X]$. Hence, evaluating $p$ at $\\alpha$ in $R$ gives\n$$p(\\alpha) = \\sum_{i=0}^n p_i \\alpha^i = 0$$\nDifferentiating:\n$$\\partial(p(\\alpha)) = \\sum_{i=1}^n p_i \\partial(\\alpha^i) = \\sum_{i=1}^n i p_i \\partial(\\alpha) \\alpha^{i-1} = \\left(\\sum_{i=0}^{n-1} (i + 1) p_{i+1} \\alpha^i\\right) \\partial(\\alpha) = 0$$\nSince $p_n = 1$ we see that the left hand factor cannot be zero. On the other hand, $k^\\partial(\\alpha)$ is an integral domain. Hence, $\\partial(\\alpha) = 0$.\n\\item We interpret 'algebraic over $k$ as 'integral over $R$. Thus, there is a monic polynomial as defined above over $R$. Proceeding as in the last part (i.e. evaluating $p$ at $\\alpha$ in $R[\\alpha]$ and differentiating), we get\n$$\\partial(p(\\alpha)) = \\sum_{i=0}^{n-1} \\left(\\partial(p_{i}) + (i + 1) p_{i+1} \\partial(\\alpha)\\right) \\alpha^i + \\partial(p_n) \\alpha^n = 0.$$\nSince $p_n = 1$ we get\n$$\\partial(p_i) + \\underbrace{(i + 1) p_{i+1} \\partial(\\alpha)}_{=0} = 0$$\nfor all $i = 0,\\cdots,n - 1$. As $\\trm{char} R = 0$ implies $\\partial(p_i) = 0$ showing $p \\in k^\\partial[X]$.\n\\en\n\n\\begin{prop}\\label{PicardVessiotRing}\nLet $\\partial(x) = A x$, as above.\n\\bn\n\\item A Picard-Vessiot ring $R$ is isomorphic to\n$$k[x_{ij},1/\\det X],$$\nwhere $X = (x_{ij}) \\in \\trm{Gl}_k(M)$ is a fundamental matrix.\n\\item A matrix $X \\in \\trm{Mat}_n(R)$ is a fundamental matrix, if and only if its Wronskian is non-zero over $k^\\partial$.\n%\\item The entries of the largest sub matrix of the Wronskian matrix $Wr(L) = \\left(x^{(k)}_{ij}\\right)_{\\substack{0 \\leq k \\leq n^2 - 1\\\\1 \\leq i, j \\leq n}}$ is a $C$-basis of the solution space, if and only if its Wronskian matrix has non-zero determinant. The associated fundamental matrix $X$ is spanned by this basis.\n\\item \\label{PVLemma3}Two fundamental matrices $X_1, X_2$ are right-associated wrt. $\\trm{Gl}_n(k^\\partial)$.\n\\item \\label{PVLemma4}Two PV-rings $R_1, R_2$ of the same equation are isomorphic as differential rings.\n\\en\n\\end{prop}\n\\bws A proof can be found in \\cite{vdPS01}, pg. 15. However, for the last two statements we are going to present a sketch of proof:\n\\bd\n\\item[ad \\ref{PVLemma3}] We have: $\\partial(X_{1,2}) = A X_{1,2}$ and assume: $X_2 = X_1 M$ for some $M \\in \\trm{Gl}_n(R)$. Then:\n$$\\partial(X_2) = \\partial(X_1 M) = \\partial(X_1) M + X_1\\partial(M) = A X_1 M + X_1 \\partial(M) \\stackrel{!}{=} A X_2 \\LRA M \\in \\trm{Gl}_n(k^\\partial).$$\nWe recall that $X_i \\in \\trm{Gl}_n(R)$ and the $X_1, X_2$ are left-associated if and only if $[M, A] = 0$.\n\\item[ad \\ref{PVLemma4}] Let $S(L)_i$ denote the two solution spaces. As we just saw, we may define an isomorphism of differential modules $\\phi : S(L)_1 \\longrightarrow S(L)_2, x_{ik} \\longmapsto \\sum_{j} x_{ij} m_{jk}$. As the PV-ring is commutative, we deem $\\trm{Sym}(S(L)_i)$ as an appropriate choice for construction (not necessarily the PV-ring, rather a subring). Now, we may extend $\\phi$ to both algebras:\n$$s = \\sum s_i x^i = \\sum s_i x_1^{i_1} \\ldots x_{n}^{i_n} \\longmapsto \\sum s_i \\phi(x_1)^{i_1} \\ldots \\phi(x_n)^{i_n}.$$\nSince $1_{S(L)_1} \\longmapsto 1_{S(L)_2}$ and $\\phi$ is an isomorphism on $\\trm{Sym}^1(S(L)_i)$ we get the desired isomorphism of differential rings. Localizing both algebras wrt. $\\det X_i$ we have that $R_i \\simeq S_{\\det X_i} (L)_i$ for $i = 1,2$.\n\\ed\n\\bmk $\\trm{Sym}(S(L)) \\simeq k[x_1,\\ldots,x_n]$ is not a simple differential ring (the maximal ideal $I = \\left$ is closed under $\\partial$-action). Nevertheless, its localization wrt. $\\det X = \\sum_{\\sigma \\in S_n} sign(\\sigma) \\prod_{i=1}^n x_{i,\\sigma(i)}$ is a simple differential ring, as $S_{\\det X}^{-1}I$ contains units ($S_{\\det X}^{-1}I = R$).\\\\\n\\indent In \\cite{vdPS01}, the following theorem is given (prop. 2.9, pg. 40 and lem 2.10, pg. 41):\n\\begin{satz}\nLet $\\partial(x) = A x$ be a linear differential equation. There are $L_i \\in k[\\partial]$, such that\n$$V_A := \\{y \\in M : \\partial(y) = A y\\} \\simeq \\bigoplus_{i=1}^n k[\\partial]/k[\\partial].L_i^*.$$\nTo specify, each matrix equation has a solution space $V_A$ isomorphic to the solution space of a scalar equation $L = \\prod L_i$.\n\\end{satz}\nThe decomposition is not unique wrt. left- and right-sidedness, as $k[\\partial]$ is in general not commutative. We are not going to prove this theoremo - only loosely scatching a proof.\\\\\nFirst we note, for any field $k$ and a matrix $A \\in \\trm{Mat}_n(k)$, there exists vectors $c_i \\in \\bigoplus_{i=0}^n k.e_i$ such that $B_i = \\left\\{c_i,...,A^{j_i} c_i : i\\right\\}$ is a $k$-basis. The elements $c_i$ are called cyclic vectors, each subspace $M_i$ generated by such an element is called a cyclic vector space and the transformed matrix wrt. to the cyclic basis is called the rational normal form of $A$. If $\\chi_A \\in k[X]$ is the charactistic polynomial of $A$ and $\\prod_{j = 1} p_j^{s_j}$ a prime decomposition of $\\chi_A$ than its minimal polynomial, which is the monic generator of $\\{f \\in k[X] : f(A) = 0\\}$, gives us all distinct cyclic vectors (each cyclic vectors space has different dimension $t_j \\deg p_j$, where $p_j^{t_j} \\mid m_A$ and $p_j^{t_j+1} \\nmid m_A$). A similar reasoning is provided in \\cite{vdPS01}, chapter 2. Here, a cyclic decomposition is given for any linear homogeneous differential equation.\n% due to the maximality of $I$: any differentially closed ideal containing $I$ is already $R = (1)$. To see that its ring of constants $R^\\partial = C$, consider $f = \\sum_{i} f_i x_1^{i_1} \\ldots x_n^{i_n}/\\det X^{i_{n+1}} \\in \\ker \\partial_R$:\n%$$\\partial(f) := \\sum_i \\partial(f_i) x^i/\\det X^{i_{n+1}} + \\sum_i i_k f_i x_1^{i_1} \\ldots \\partial(x_k) x_k^{i_k-1} \\ldots x_n^{i_n}/\\det X^{i_{n+1}}$$\n\\subsubsection{Differential Galois group}\nIn the theory of field extensions, the Galois group is the set of $k$-vector space automorphisms permuting the root elements of a given polynomial over $k$ leaving $k$ fixed. For a given linear differential equation, the definition is slightly different:\n\\begin{defi}\nLet $\\partial(x) = A x$ for some $A \\in \\trm{Gl}_k(M)$ and $x \\in M$. The differential Galois group is the zentralizer of $\\left<\\partial\\right> \\subset \\trm{Gl}_R(M)$, i.e.\n$$\\trm{DGal}(R/k) = \\{\\varphi \\in \\trm{Gl}_{k^\\partial}(M) : \\partial \\varphi = \\varphi \\partial\\}.$$\n\\index{Symbol}{$\\trm{DGal}(R/k)$}\n\\end{defi}\nThis property is also called equivariance (e.g. in algebraic geometry).\n%\\paragraph{Examples for the linear case}\n\\begin{satz}[Galois correspondence]\nLet $K/k$ be a PV extension for $\\partial(x) = A x$. Let $\\mathfrak{G}$ be the set of all closed subgroups of $G := \\trm{DGal}(K/k)$ and $\\mathcal{M}$ the set of all differential subfields $k \\subset M \\subset K$. In analogy to algebraic fields extension theory, we define \n$$\\bao{rrclrcl}\n\\trm{Fix} : &\\mathfrak{G} &\\longrightarrow& \\mathcal{M},& H &\\longmapsto& K^H := \\{x \\in K : \\sigma(x) = x\\ \\forall \\sigma \\in H\\}\\\\\n&&&&&\\\\\n\\trm{DGal} : & \\mathcal{M} &\\longrightarrow& \\mathfrak{G},& M &\\longmapsto& \\trm{DGal}(M/k),\\\\\n\\ea$$\nthen we have:\n\\bn\n\\item both functors are inverse to one another.\n\\item $H \\in \\mathfrak{G}$ is normal in $G$ if and only if for $M = K^H$:\n$$G(M) \\subset M,$$\ni.e. is $G$-invariant as a set ($g x \\in M$ for all $g \\in G$, $x \\in M$).\n\\item If $H \\in \\mathfrak{G}$ is normal then the canonical projection $G \\longrightarrow \\trm{DGal}(K^H/k)$ is surjective and has $H$ as kernel. Furthermore, $K^H$ is a PV extension for some linear differential equation over $k$.\n\\item Let $G^0 \\leq G$ be the connected component of identity, then $K^{G^0} = k$.\n\\en\n\\end{satz}\n\\bmk A proof is given in \\cite{vdPS01}. The connected component of identity is to be understood as follows. If $(G,m,e,\\tau)$ is a topological group, with $\\tau$ its topology, an open subset $U \\in \\tau$ is called connected if the only disjoint union of open subsets $U' \\cup U'' = U$ is trivial (i.e. $U' \\in \\{\\emptyset, U\\}$, $U'' = U\\bsl U'$). In this case, $U$ is also called a connected component. If furthermore, $1 \\in U$ is a subgroup, $U$ is called the connected component of identity which is denoted by $G^0$. These definitions were taken from \\cite{Milne}.\n\\bsp Some examples:\n\\bn\n\\item $(\\zz,+,0,\\mathcal{P}(\\zz))$ is a topological group wrt. discrete topology. Hence, all pointed subspaces are open subsets. Therefore, $\\{0\\} = \\zz^0$.\n\\item $(\\rz^\\times,\\cdot,1,\\tau_{\\trm{standard}})$ is a topological group with connected components: $\\rz_{>0}, \\rz_{<0}$. Hence, $\\rz_{>0} = \\left(\\rz^{\\times}\\right)^0$.\n\\en\n\\subsection{Example}\nNow, let us consider two simple examples.\n\\subsubsection{1-dim case}\nLet $k = \\currfield(z)$ with $\\currfield$-derivation $\\partial = \\frac{d}{d z}$, as well as $a \\neq 0$ and $\\partial(x) = ax$. Here, we have a 1-dimensional differential equation, i.e. the solution space is a 1-dim $\\currfield$-vector space. Let us compute the\n\\bd\n\\item[Fundamental matrix] Put $X = x \\in \\trm{Gl}_1(\\currfield(z))$. Clearly, $Wr(X) = (x)$ and $\\det Wr(X) = x \\neq 0$. Hence, $x$ belongs to the $\\currfield$-basis of the solution space.\n\\item[Picard-Vessiot ring] By prop. \\ref{PicardVessiotRing} we know that $R \\simeq \\currfield(z)[x,1/x]$.\n\\item[Differential Galois group] Pick some $f \\in \\trm{Gl}_1(R)$, by definition $f \\in \\trm{DGal}(R/k)$ if $f$ commutes with $\\partial$:\n$$\\partial(f x) = \\partial(f) x + f a x = f a x = f \\partial(x) \\LRA \\partial(f) = 0,$$\ni.e. $f$ is in $\\currfield.id$. Thus, we have that $\\trm{DGal}(R/k) = \\left<\\partial\\right> \\simeq \\qz^\\times$.\n\\ed\n\\subsubsection{2-dim case} \\label{twoD}\nLet $k \\subseteq \\currfield$ with trivial derivation $\\partial\\mid_{\\currfield} = 0_{\\currfield}$, as well as $a \\neq 0$ and $\\partial^2(x) = a x$. Clearly, the solution space in question is contained in a two-dimensional space with companion matrix $A = \\left(\\bao{cc}\n0 & 1\\\\\na & 0\\\\\n\\ea\\right)$.\n\\paragraph{Fundamental matrix} Put $X = \\left(\\bao{cc}\nx_{11} & x_{12}\\\\\nx_{21} & x_{22}\\\\\n\\ea\\right)$ as fundamental matrix with $\\partial(x_{11}) = x_{21}, \\partial(x_{21}) = a x_{11}$ as well as $\\partial(x_{12}) = x_{22}, \\partial(x_{22}) = a x_{12}$. The Wronskian matrix is\n{\\scriptsize\n$$\\left(\\bao{cccc}\nx_{11} & x_{21} & x_{12} & x_{22}\\\\\nx_{21} & a x_{11} & x_{22} & a x_{12}\\\\\na x_{11} & a x_{21} & a x_{12} & a x_{22}\\\\\na x_{21} & a^2 x_{11} & a x_{22} & a^2 x_{12}\\\\\n\\ea\\right)\n$$}\nObviously, $\\det Wr(X) = 0$ as the fourth and third row are linear combinations of the first two rows. For instance, $x_1 = x_{11}$ and $x_2 = x_{21}$ forms a $\\currfield$-basis of the solutions space. Since the remaining indeterminates are linearly dependent over $\\currfield$ we compute for $x_{12} = \\lambda_{11} x_1 + \\lambda_{12} x_2, x_{22} = \\lambda_{21} x_1 + \\lambda_{22} x_2$ and $\\lambda_{ij} \\in \\currfield$:\n$$\\bao{rclcl}\n\\partial(x_{12}) &=& \\lambda_{11} \\partial(x_1) + \\lambda_{12} \\partial(x_2) &=& \\lambda_{12} a x_1 + \\lambda_{11} x_1\\\\ \n&&&&\\\\\n&\\stackrel{!}{=}& \\lambda_{21} x_1 + \\lambda_{22} x_2 &\\LRA& \\lambda_{21} = \\lambda_{12}, \\lambda_{22} = \\lambda_{11} a\\\\\n&&&&\\\\\n\\partial(x_{22}) &=& \\lambda_{21} \\partial(x_1) + \\lambda_{22} \\partial(x_2) &=& a \\lambda_{11} x_1 + a \\lambda_{11} x_1\\\\\n&&&&\\\\\n&\\stackrel{!}{=}& a \\lambda_{11} x_1 + a\\lambda_{12} x_2\\\\\n\\ea$$\nWe see that we may chose any $\\lambda_{11} =: \\lambda_1, \\lambda_{12} = \\lambda_2 \\in \\currfield$ such that $\\det X$ is a unit. Continuing:\n$$\\det X = \\det \\left(\n\\bao{cc}\nx_1 & \\lambda_1 x_1 + \\lambda_2 x_2\\\\\nx_2 & a \\lambda_2 x_1 + \\lambda_1 x_2\\\\\n\\ea\\right) = %a \\lambda_2 x_1^2 + \\lambda_1 x_1 x_2 - \\lambda_1 x_1 x_2 - \\lambda_2 x_2^2 \n= a \\lambda_2 x_1^2 - \\lambda_2 x_2^2 = \\lambda_2 (a x_1^2 - x_2^2).$$\nWe can set $\\lambda_1$ arbitrarily but chose zero and $\\lambda_2 = 1$,\n%Identifying the remaining variables with $x_{12} = a^{-1} x_2$, $x_{22} = a x_1$\n we get the fundamental matrix $X = \\left(\\bao{cc}\nx_1 & x_2\\\\\nx_2 & a x_1\\\\\n\\ea\\right)$. Its inverse is\n$$X^{-1} = \\frac{1}{a x_1^2 - x_2^2} \\left(\n\\bao{cc}\na x_1 & -x_2\\\\\n-x_2 & x_1\\\\\n\\ea\\right)$$\nIn Heidereich 2010, he provides the condition $\\sum_{i=0}^n a_i \\partial^i(X) X^{-1} \\in M_2(k)$ for the fundamental matrix. But clearly, if $\\partial(X) = A X$ then $\\partial^i(X) = A^i X$ and the condition holds.% In our case:\n%$$\\partial(X)X^{-1} = A X X^{-1} = \\frac{1}{a x_1^2 - a^{-1} x_2^2}\\left(\\bao{cc}\n%x_2 & a x_1\\\\\n%a x_1 & a x_2\\\\\n%\\ea\\right)\\left(\n%\\bao{cc}\n%x_1 & a^{-1} x_1\\\\\n%x_2 & a x_2\\\\\n%\\ea\\right) = A.$$\n\\paragraph{Differential Galois group} we have $g = (g_{ij}) \\in \\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield)$ iff $g A - A g = 0$ and $\\det g \\in \\currfield^\\times$. We get:\n$$\\bao{cc}\na g_{12} - g_{21} = 0 & -a g_{12} + g_{21} = 0\\\\\na g_{11} - a g_{22} = 0 & -a g_{11} + a g_{22} = 0.\\\\\n\\ea$$\nTherefore we have $g = \\left(\\bao{cc}\ng_{11} & g_{12}\\\\\na g_{12} & g_{11}\\\\\n\\ea\\right)$. The determinant is $\\det g = g_{11}^2 - a g_{12}^2$. If $g_{ij} \\in \\currfield$ we get that $\\det g = 0 \\LRA g_{11}^2 = a g_{12}^2$. Since $\\currfield$ is algebraically closed, there is a $b \\in \\currfield$ such that $b^2 = a$. Hence, $g_{11} = \\pm b g_{12}$. Thus we get:\n$$\\left\\{\\left(\n\\bao{cc}\ng_{11} & g_{12}\\\\\na g_{12} & g_{11}\\\\\n\\ea\\right) \\in \\trm{Gl}_2(\\currfield) : g_{11} \\neq \\pm b g_{12}\\right\\}$$\nas a subgroup of the differential Galois group. To conclude, the Galois group is\n$$\\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield) = \\left \\supset \\left<\\partial\\right>.$$\nNow, we can express $\\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield)$ as an open algebraic set $U$ in $\\aff{2}_{\\currfield}$:\n$$\\bao{rcl}\n\\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield) &\\simeq_{\\trm{claim}}& U\\\\\n &:=& \\{(g_{11},g_{12}) \\in \\aff{2}_{\\currfield} : g_{11}^2 - a g_{12}^2 \\neq 0\\}\\\\\n&=& \\aff{2}_{\\currfield} \\bsl \\left\\{(g_{11},g_{12}) \\in \\aff{2}_{\\currfield} : g_{11}^2 - a g_{12}^2 = 0\\right\\}\\\\\n&&\\\\\n&=& \\aff{2}_{\\currfield}\\bsl Z(g_{11}^2 - a g_{12}^2)\\\\\n\\ea$$\nAs is shown in \\cite{CohCuySte}, the differential Galois group is a subgroup of $\\trm{Sl}(S(L))$. Therefore we may substitute:\n$$U' = \\{(g_1',g_2') \\in \\aff{2}_{\\currfield} : g_1'^2 - a g_2'^2 = 1\\}.$$\nThe multiplication is simply:\n$$m_U := \\left[\\left((g_{11},g_{12}) , (g'_{11},g'_{12})\\right) \\longmapsto (g_{11} g'_{11} + a g_{12} g'_{12}, g_{11} g'_{12} + g'_{11} g_{12})\\right].$$\nChecking:\n$$\\bao{rcl}\n(g_{11} g'_{11} + a g_{12} g'_{12}, g_{11} g'_{12} + g'_{11} g_{12}) &\\stackrel{?}{\\in}& U:\\\\\n&&\\\\\n(g_{11} g'_{11} + a g_{12} g'_{12})^2 - a (g_{11} g'_{12} + g'_{11} g_{12})^2 &=& g_{11}^2 {g'}_{11}^2 + a^2 g_{12}^2 {g'}_{12}^2 + 2 a g_{11} g_{12} g'_{11} g'_{12}\\\\\n&&- a\\left(g_{11}^2 {g'}_{12}^2 + g_{12}^2 {g'}_{11}^2 + 2 g_{11} g_{12} {g'}_{11} {g'}_{12}\\right)\\\\\n&&\\\\\n&=& (g_{11}^2 - a g_{12}) {g'}_{11} - (g_{11}^2 - a g_{12}) a {g'}_{12}\\\\\n&&\\\\\n&=& \\underbrace{(g_{11}^2 - a g_{12})}_{= 1} \\underbrace{({g'}_{11}^2 - a {g'}_{12})}_{= 1} = 1,\\\\\n\\ea$$\nimplying $m_U\\left((g_{11},g_{12}),({g'}_{11},g'_{12})\\right) \\in U$. Moreover, the map $\\iota: U \\ni (g_{11},g_{12}) \\longmapsto g \\in \\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield)$ is a morphism of groups (by definition), injective as only $(1,0) \\in \\ker \\iota$ and surjective as $(g_{11},g_{12}) \\in \\iota^{-1}(g)$. This proves the claim the above defined open subset $U$ is isomorphic to $\\trm{DGal}(\\currfield[x_1,x_2,1/\\det X]/\\currfield)$.\n%This group has three classes of subgroups:\n%$$H_1 := \\left\\{\\left(\\bao{cc}\n%g_{11} & 0\\\\\n%0 & g_{11}\\\\\n%\\ea\\right) : g_{11} \\in \\currfield^\\times\\right\\},\\ H_2 := \\left\\{\\left(\\bao{cc}\n%0 &g_{12}\\\\\n%a g_{12} & 0\\\\\n%\\ea\\right) : g_{12} \\in \\currfield^\\times\\right\\},$$\n%$$\\ H_3 := \\left\\{\\left(\n%\\bao{cc}\n%g_{11} & g_{12}\\\\\n%a g_{12} & g_{11}\\\\\n%\\ea\\right) : (g_{11},g_{12}) \\in G_{<\\infty} \\times \\currfield^\\times \\cap U\\right\\}$$\nOn the other hand, $G$ can be viewed as an closed subset of $\\aff{3}_{\\currfield}$:\n$$I := \\left<(X^2 - a Y^2) Z - 1\\right> \\subset \\ov{\\qz}[X,Y,Z]\\ \\RA\\ Z(I) = \\left\\{\\left(g_{11}, g_{12}, 1/(g_{11}^2 - a g_{12}^2)\\right) : g_{11}, g_{12} \\in \\currfield\\right\\}$$\nIts multiplication is given by\n$$\\bao{rrcl}\nm_{Z(I)} :& Z(I) \\times Z(I) & \\longrightarrow & Z(I)\\\\\n&&&\\\\\n&((g_{11}, g_{12}, (g_{11}^2 - a g_{12}^2)^{-1}), & \\longmapsto &(g_{11} {g'}_{11} + a g_{12} {g'}_{12}, g_{11} {g'}_{12} + g_{12} {g'}_{11},\\\\\n& ({g'}_{11}, {g'}_{12}, ({g'}_{11}^2 - a {g'}_{12}^2)^{-1}))&&(g_{11}^2 - a g_{12}^2)^{-1}({g'}_{11}^2 - a {g'}_{12}^2)^{-1})\\\\\n\\ea$$\nThe proof of the above claim follows analogously to the open case. %The main difference is that $Z(I) = U \\times \\currfield^\\times \\cup \\left\\{0_{\\aff{3}_{\\currfield}}\\right\\}$, i.e. Zariski-dense and $Z(I)$ is a monoid (has $0$ as non-invertible and idempotent).\n\\paragraph{Characterization of field extension}\n\\bn\n\\item As we noted above, $\\partial(\\det X) = 0$. Since $R$ is a simple differential ring, we conclude $\\det X \\in k^\\partial = k = \\currfield$. Thus there is some $b = \\det X \\in \\currfield^\\times$. Therefore, $K = S^{-1}R \\simeq \\currfieldx[X]/\\left$. In other words: the Picard-Vessiot field $K$ is an algebraic extension of (some appropriate) transcendental extension of $\\currfield$. Rescaling our differential equation, we may assume that $b = 1$.\n\\item An immediate consequence of the above is the following: We consider $R' := \\currfield[x_1,x_2] \\simeq \\trm{Sym}(S(L))$ wrt. the $\\currfield$-basis $\\{x_i : i = 1,2\\}$. Now, take $I = \\left \\subset \\trm{Sym}(S(L))$, we get:\n$$\\trm{Sym}(S(L))/I \\simeq R'/\\left.$$\nSince the $\\currfield$-derivation does not reduce the degree of any polynomial of positive degree we can use the degree induced filtration on $\\trm{Sym}$:\n$$\\filteredA^{\\leq l} = \\left\\{x \\in \\trm{Sym}(S(L)) : \\deg x \\leq l\\right\\} \\ \\RA \\partial_{\\trm{Sym}}\\mid_{\\filteredA^{\\leq l}} (\\filteredA^{\\leq l}) \\subset \\filteredA^{\\leq l},\\ \\filteredA^{\\leq 0} := \\currfield,\\ \\filteredA^{\\leq -n} := \\{0\\}, n, l \\in \\nz,$$\ni.e. $\\filteredA^{\\leq l}$ is $\\partial_{\\trm{Sym}}$ invariant $\\currfield$-subspace. Here we set $\\deg(0) := -\\infty$ and $\\deg$ is the sum degree. On the other hand:\n$$\\filteredA^{\\leq 2} = \\bigoplus_{|\\alpha| \\leq 2} \\currfield.x_1^{\\alpha_1} x_2^{\\alpha_2}\\ \\RA\\ \\filteredA^{\\leq2} / \\left(\\filteredA^{\\leq2} \\cap I\\right) = \\currfield \\oplus \\currfield.x_1 \\oplus \\currfield.x_2 \\oplus \\currfield.x_1^2 \\oplus \\currfield.x_1 x_2,$$\nin particular $\\lambda_{0,2} x_2^2 = \\lambda_{0,2} - a \\lambda_{0,2} x_1^2 \\in \\currfield \\oplus \\currfield.x_1^2$ for any $\\lambda_{0,2} \\in \\currfield$. In general, for $\\filteredA^{\\leq n}$ we may substitute all terms $x_2^{2 l}$ with $(a x_1^2 - 1)^l$ and all $x_2^{2l + 1}$ with $(a x_1^2 - 1)^l x_2$, for all $l \\in \\{d \\in \\nz : 2 \\mid d\\}$. Therefore the quotient spaces are\n$$A_n := \\filteredA^{\\leq n}/ \\left(\\filteredA^{\\leq n} \\cap I\\right) /\\left(\\filteredA^{\\leq n - 1} / \\left(\\filteredA^{\\leq n - 1} \\cap I\\right) \\right)\\simeq \\currfield.x_1^n \\oplus \\currfield.x_1^{n-1} x_2,$$\nand therefore its associated graded $\\currfield$-algebra is obviously:\n$$A = \\bigoplus_{n \\geq 0} A_n \\simeq \\currfield[x_1] \\oplus \\currfield[x_1].x_2.$$\nSummarizing we get\n$$\\trm{Sym}(S(L))/I = \\filteredA/I = \\bigcup_{n \\geq 0} \\filteredA^{\\leq n}/ \\filteredA^{\\leq n} \\cap I \\simeq A \\simeq \\currfield[x_1] \\oplus \\currfield[x_1].x_2,$$\nwith multiplication\n$$\\mu = \\left[x_1^{i_1} x_2^{j_1} \\otimes x_1^{i_2} x_2^{j_2} \\longmapsto x_1^{i_1 + i_2} (a x_1^2 - 1)^{(j_1 + j_2 - (j_1 + j_2 \\mod 2))/2} x_2^{j_1 + j_2 \\mod 2}\\right],\\ \\forall i_l \\in \\nz, j_l = 0, 1.$$\nWe note that the subalgebra $\\currfield[x_1]$ is not a differential algebra as \n$$\\partial = \\left[x_1^i x_2^j \\longmapsto \\begin{cases}\n(1 + i) a x_1^{i+1} - i x_1^{i-1}& j = 1\\\\\ni x_1^{i-1} x_2 & j = 0\\\\\n\\end{cases}\\right].$$\nLocalizing wrt. $S := \\currfield[x_1]\\bsl\\{0\\}$, i.e. non-zero polynomials, and identifying with all corresponding symmetric tensors we have that\n$$K^+ := S^{-1} \\left(\\currfield[x_1] \\oplus \\currfield[x_1].x_2\\right) = S^{-1} \\currfield[x_1] \\oplus S^{-1} \\currfield[x_1].x_2 = \\currfieldx \\oplus \\currfieldx.x_2.$$\nHowever, we have not shown yet whether $K^+$ is an integral domain. Now let us consider $p = Y^2 - a x_1^2 + 1 \\in \\currfieldx[Y]$. \n$$Y^2 - \\underbrace{a x_1^2 + 1}_{= \\alpha^2} = (Y - \\alpha)(Y + \\alpha).$$\nIf $\\alpha \\in \\currfieldx$, then $\\ov{Y \\pm \\alpha} = \\ov{Y} \\pm \\alpha$ are zero-divisors in the quotient ring and with $\\trm{gcd}(Y - \\alpha, Y + \\alpha) = 1$ we get\n$$\\currfieldx[Y]/\\left \\simeq \\currfieldx^2$$\nas a $\\currfieldx$-vector space, with multiplication:\n$$\\mu(a \\otimes \\_) := [b \\longmapsto a b] \\equiv \\left(\\bao{cc}\na_0 & (a x_1^2 - 1) a_1\\\\\na_1 & a_0\\\\\n\\ea\\right) \\in M_2(\\currfieldx),\\ \\forall a \\equiv (a_0, a_1) \\in \\currfieldx^2$$\nand with derivation:\n$$\\partial = \\left[\n\\bao{ccc}\n(1_{\\currfield} x_1^i, 0) &\\longmapsto& (0 , i x_1^{i-1})\\\\\n(0, 1_{\\currfield} x_1^i) &\\longmapsto& ((1 + i) a x_1^i - i x_1^{i-1},0)\\\\ \n\\ea\n\\right].$$ Hence, the Picard-Vessiot field $K$ would be simply $\\currfieldx$. This implies:\n$$\\exists \\alpha \\in \\currfield.x_1 \\bsl \\{0\\}:\\ \\partial_{\\currfieldx}(\\alpha) \\in \\currfield.x_1 \\LRA \\exists v \\in \\ker (\\alpha id_{\\currfield.x_1} - \\partial) \\bsl\\{0\\},$$\ni.e. an eigenvector of $\\partial_K$. On the other hand, is $[\\currfield(x_1,\\alpha) : \\currfieldx] := \\deg \\trm{Min}(\\alpha,\\currfieldx) = 2$:\n$$\\RA\\ (K, \\partial_K) \\simeq \\left(\\currfield\\left(x_1,\\sqrt{a x_1^2 - 1}\\right), \\partial_{K_1}\\right)$$% \\simeq \\left(\\currfield\\left(x_2,\\sqrt{(1 + x_2^2)/(a)}\\right), \\partial_{K_2}\\right),$$\nis a field, with $\\currfieldx/\\currfield$ transcendental and with $\\currfield$-derivation:%s:\n$$\\bao{rrcl}\n\\partial_{K}%_1}\n : &\\currfield\\left(x_1,\\sqrt{a x_1^2 - 1}\\right)&\\longrightarrow&\\currfield\\left(x_1,\\sqrt{a x_1^2 - 1}\\right)\\\\\n&&&\\\\\n&x_1 &\\longmapsto& \\sqrt{a x_1^2 - 1}\\\\\n&\\sqrt{a x_1^2 - 1} &\\longmapsto& a x_1\\\\\n&&&\\\\\n%\\partial_{K_2} : &\\currfield\\left(x_2,\\sqrt{(1 + x_2^2)/a}\\right)&\\longrightarrow&\\currfield\\left(x_2,\\sqrt{(1 + x_2^2)/a}\\right)\\\\\n%&&&\\\\\n%&x_2 &\\longmapsto& \\sqrt{\\frac{1 + x_2^2}{a}}\\\\\n%&\\sqrt{\\frac{1 + x_2^2}{a}} &\\longmapsto& a x_2\\\\\n\\ea$$\nHowever, we claim that since $x_1$ is transcendental over $\\currfield$ and the polynomial $a x_1^2 +1$ is no square in $\\currfield[x_1]$, the polynomial $X^2 - a x_1^2 + 1$ is irreducible over $\\currfieldx$. First, let us assume there is a root $\\alpha = \\frac{\\beta}{\\gamma} \\in \\currfieldx$, where $\\trm{gcd}(\\beta,\\gamma) = 1$ and $\\gamma \\neq 0$. Then we compute:\n$$\\alpha^2 = a x_1^2 - 1 = \\frac{\\beta^2}{\\gamma^2} \\LRA \\beta^2 = \\gamma^2 (a x_1^2 - 1),$$\ni.e. $\\beta \\in \\currfield[x_1]$ is a root of the polynomial $X^2 - \\gamma^2(a x_1^2 - 1)$. Having $\\beta = \\sum_{j=0}^{n_1} \\beta_j x_1^i, \\gamma = \\sum_{i=0}^{n_2} \\gamma_i x_1^i$, with $\\gamma_i, \\beta_j \\in \\currfield$, we get that $n_1 = n_2 + 1$ due to the factor $a x_1^2$. On the other hand, both decompose to linear factors due to algebraic closedness of $\\currfield$, e.g.:\n$$\\beta = \\wt{\\beta}_0 \\prod_{i=1}^{m_1} (x_1 - \\wt{\\beta}_i)^{s_i},\\ \\gamma = \\wt{\\gamma}_0 \\prod_{j=1}^{m_2} (x_1 - \\wt{\\gamma}_j)^{t_j},\\ \\trm{and}\\ \\sum_{i=0}^{m_1} s_1 = \\sum_{j=0}^{m_2} t_j + 1.$$\nClearly, all but at most two factors on both sides of our polynomial equation cancel. So we get $n_2 \\leq 0$ and $n_1 \\leq 1$. In particular, $\\alpha$ itself is already in $\\currfield[x_1]$ for $n_2 = 0$ (i.e. $\\gamma \\neq 0$). This is not possible, as\n$$\\wt{\\beta}_0^2 (x_1 - \\wt{\\beta}_1)^2 = \\wt{\\beta}_0^2(x_1^2 - 2 \\wt{\\beta}_1 x_1 + \\wt{\\beta}_1^2) = \\wt{\\gamma}_0^2 (a x_1^2 - 1) \\LRA 2 \\wt{\\beta}_1 = 0 \\wedge \\wt{\\beta}_0^2 \\wt{\\beta}_1^2 = -\\wt{\\gamma}_0^2$$\nleads to the contradiction $\\gamma = 0$. Hence, $\\alpha$ is algebraic and $[\\currfieldx(\\alpha) : \\currfieldx] = 2$.\\\\\n\\indent Nevertheless, one aspect we have not considered yet which we will discuss now.\n%However, since $\\currfield$ is algebraically closed, we get the abovementioned decomposition which will get discussed right now.\n\\item $\\currfield(x_1,x_2) = K$ is contained in a $\\currfield$-algebra generated by one distinct transcendental element $y$ being a unit in $\\currfield[x_1,x_2]$: The ring $R = \\currfield[x_1,x_2,\\det X^{-1}]$ contains four non-trivial units (i.e. elements in $R^\\times \\bsl \\currfield^\\times$). Namely:\n$$\\pm \\sqrt{a} x_1 \\pm x_2,$$\nto specify: four non-trivial divisors of $\\det X$ in $\\currfield[x_1,x_2]$. Computing their derivative wrt. $\\partial$, we get:\n$$\\bao{rclcl}\n\\partial(\\sqrt{a} x_1 \\pm x_2) &=& \\sqrt{a} x_2 \\pm a x_1 &=& \\pm \\sqrt{a} (\\sqrt{a} x_1 \\pm x_2)\\\\\n&&&&\\\\\n\\partial(-\\sqrt{a} x_1 \\pm x_2) &=& -\\sqrt{a} x_2 \\pm a x_1 &=& \\mp\\sqrt{a} (-\\sqrt{a} x_1 \\pm x_2).\\\\\n\\ea$$\nEach of the four elements is an eigenvector of $\\partial$ in $R$ with eigenvalue $\\pm \\sqrt{a}$ - or equivalently, solves the following differential equations:\n$$L_{\\pm} := \\partial(x) \\mp \\sqrt{a} x = 0,$$\nwhere the subscript sign defines the sign of the eigenvalue. Defining $y_1 \\in S(L_+), y_{-1} \\in S(L_-)$ we get that $K = \\currfield(x_1,x_2)$ contains two subfields depicted in the following diagram:\n$$\\xymatrix{\n& \\currfield(y_1) \\ar@{^{(}->}[rd]&\\\\\n\\currfield\\ar@{^{(}->}[rd]\\ar@{^{(}->}[ru]&&\\currfield(x_1,x_2) \\simeq \\currfield(y_1)\\\\\n& \\currfield(y_{-1}) \\ar@{^{(}->}[ru]&\\\\\n}$$\nBy direct computation, we see that:\n$$L = \\partial^2 - a \\cdot id = (\\partial - \\sqrt{a}) (\\partial + \\sqrt{a}).$$\nAnd our differential module $S(L)$ decomposes:\n$$S(L)^* \\simeq \\currfield[\\partial]/\\currfield[\\partial].L \\simeq \\currfield[\\partial]/\\currfield[\\partial].L_+ \\oplus \\currfield[\\partial]/\\currfield[\\partial].L_- \\simeq S(L_+)^* \\oplus S(L_-)^*.$$\n\\en\n\\paragraph{The subfields}\nFirstly, we like to compute the Picard-Vessiot ring for $M_{i} = \\currfield(y_i)$:\n$$R_i = \\currfield[y_{i}, y_i^{-1}],\\ \\trm{for}\\ i = \\pm1$$\n%As the image of the ideal $\\left \\subset \\currfield[y_i]$ is the whole ring under inclusion or equivalently $1_R \\in R_i.y_i$, we have indeed that $R_i$ is simple differential. \n%Let $I \\subset R_i$ be a proper ideal and we assume differential closedness - i.e. $\\partial(I) \\subset I$.\n%As a noetherian $R$-submodule of a noetherian module $R$ (generated by $y_{\\pm 1}$ over $\\currfield$), $I$ is finitely generated. Hence, let $S:= \\{s\\} \\subset I$ be one generating set. By differential closedness, we get for any $s \\in S$:\n%$$\\partial(s) = \\partial\\left(\\sum_{i=-m}^n s_i y_1^i\\right) = \\sum_{i=-m}^n s_i \\partial(y_1^i) = \\sum_{i=-m}^n i \\sqrt{a} s_i y_1^i \\in I$$\n%$$\\LRA \\partial(s) - s = \\sum_{i=-m}^n (i \\sqrt{a} - 1) s_i y_1^i \\in I$$\n%But both, $s, \\partial(s) - s$ are of degree $n$, or $m$ wrt. $y_{\\pm 1}$ and $y_1^m (\\partial(s) - s) \\in \\currfield[y_1]$.\n% There is an ideal $I' \\subset \\currfield[y_1]$, such that $S_{y_1}^{-1}(I') \\supset I$. By definition of $I$, we get\n%$$y_1^m t \\in I' \\RA \\partial(y_1^m t) = \\underbrace{m \\sqrt{a} y_1^m t}_{\\in I'} + \\underbrace{y_1^m \\partial(t)}_{\\in \\partial(I')},$$\n%but identifying $I' := I \\cap \\frac{\\currfield[y_1]}{1}$ we get $\\partial(I') \\subset I'$. Being a PID, all ideals $I' \\subset \\currfield[y_1]$ are of the form $\\left$. On the other hand, $\\partial$ operates on all weight spaces $\\currfield.y_1^i$, $i \\geq 1$, invariantly:\n%$$\\bao{rrcl}\n%\\partial_i := \\partial\\mid_{\\currfield.y_1^i} : &\\currfield.y_1^i &\\longrightarrow& \\currfield.y_1^i\\\\\n%&&&\\\\\n%&y_1^i &\\longmapsto&i \\sqrt{a} y_1^i\\\\\n%\\ea$$\n%Thus we have:\n%$$\\partial s = \\sum_{i=0}^n s_i \\partial(y_1^i) = \\sum_{i=0}^n i \\sqrt{a} s_i y_1^i \\in \\left \\LRA \\partial(s) \\equiv 0 \\mod s$$\n%$$\\LRA \\sum_{i=0}^{n-1} (i - n) \\sqrt{a} s_i y_1^i = 0 \\LRA s_i = 0 \\vee n - i = 0\\ \\forall 0 \\leq i \\leq n - 1,$$\n%Hence, each derivative of the generators $s$ agree in degree but also reduce to zero modulo $\\left$ contradicting our claim $\\partial(s) \\in \\left$. Thus, all $D$-stable ideals in $R$ are indeed trivial.\nFirstly, we get:\n$$\\partial(y_i^{-1}) = -\\frac{\\partial(y_i)}{y_i^2} = -i \\sqrt{a} y_i^{-1}, i =\\pm1$$\nimplying the multiplicative inverse solves the opposite differential equation. Hence, $R_i$ already contains all solutions of $L_\\pm$ by simply putting $y_{-1} := y_1^{-1}$.\\\\\n\\indent Let $R = R_1$. Furthermore, we want to show that $R$ is indeed a simple differential ring over $\\currfield$. Clearly, $\\partial$ operates invariantly on $\\currfield.y_{\\pm 1}^i$ for all $i \\neq 0$. In addition, $R$ is isomorphic to $S_{X}^{-1}\\currfield[X]$, the localization of $\\currfield[X]$ where\n$$S_X = \\currfield[X] \\bsl \\bigcup_{\\substack{\\mathfrak{p} \\in \\trm{Spec}(\\currfield[X])\\\\X \\notin \\mathfrak{p}}} \\mathfrak{p} = \\{f \\in \\currfield[X] : f \\notin \\idealp\\ \\forall \\idealp \\in \\trm{Spec} \\currfield[X] \\bsl \\{\\left\\}\\}.$$ Furthermore, every prime ideal in $\\currfield[X]$ is generated by $X - \\alpha$, for some $\\alpha \\in \\currfield$ as $\\currfield$ is algebraically closed. Hence, on the one hand we get:\n$$S_X = \\{f \\in \\currfield[X]\\bsl \\currfield : f \\notin \\left \\forall \\alpha \\in \\currfield^\\times\\} = \\{X^i : i \\geq 0\\}$$\nand on the other hand: if $I' \\subset R$ is an ideal, there is an ideal $I \\subset \\currfield[X]$ such that $I' = S_{X}^{-1} I$. This implies every ideal is principal in $R$. In $\\currfield[X]$ each element $\\mathfrak{p} \\in \\trm{Spec}(\\currfield[X])$ gets mapped to $\\left$ via $\\partial$, as every element $p (X - \\alpha)$ in a prime ideal $\\left$ has an image in $\\left$:\n$$p (X - \\alpha) \\longmapsto \\partial(p) (X - \\alpha) + \\sqrt{a} p X,\\ \\forall \\alpha \\in \\currfield,\\ p \\in \\currfield[X]$$\nwhere the constant coefficient $\\left(\\partial(p)\\right)_0 = 0 \\LRA p_0 \\in \\ker \\partial \\LRA \\partial(p) \\in \\left$. This shows the differential $\\currfield$-algebra $\\left(\\currfield[X], \\partial = \\sqrt{a} X \\cdot \\frac{d}{d X}\\right)$ has only one differential prime ideal, $\\left$ (- in our case even maximal). However, the localization cancels this as its generator $X$ is a unit in $R$ (using variable notation $X, y_1$ interchangeably). Therefore, our rings $R_1, R_{-1}$ are indeed simple differential rings - or PV. In words of $D$-module algebra, $R_1, R_{-1}$ are simple $D$-rings, for $D = \\currfield[\\partial]$.\\\\\n\\indent To conclude this paragraph, we want to show that $\\currfield(x_1,x_2) \\simeq \\currfield(y_1)$ as differential field extensions over $\\currfield$. To achieve this we need to show that the maps:\n$$\\bao{rrcl}\n\\Phi : & \\currfield(y_1) &\\longrightarrow& \\currfield(x_1,x_2)\\\\\n& y_1 &\\longmapsto & \\sqrt{a} x_1 + x_2\\\\\n& y_1^{-1}&\\longmapsto & \\sqrt{a} x_1 - x_2\\\\\n&&&\\\\\n\\Psi : &\\currfield(x_1,x_2) &\\longrightarrow& \\currfield(y_1)\\\\\n& x_1 &\\longmapsto& \\frac{1}{2 \\sqrt{a}} (y_1 + y_1^{-1})\\\\\n& x_2 &\\longmapsto& \\frac{1}{2} (y_1 - y_1^{-1})\\\\\n\\ea$$\\label{PVisomorph}\nare bijective and inverse to one another as differential $\\currfield$-algebra homomorphisms (or $D$-module algebra homomorphisms). Clearly, the two $\\currfield$-vector spaces $V_1 := \\currfield.y_1 \\oplus \\currfield.y_1^{-1}$ and $V_2 := \\currfield.x_1 \\oplus \\currfield.x_2$ are isomorphic by simple basis change induced by the restrictions $\\mid_{V_i}$ of the above $\\currfield$-vector space homomorphisms. Furthermore, $\\Psi \\mid_{V_2} \\Phi \\mid_{V_1} = id_{V_1}$ and $\\Phi \\mid_{V_1} \\Psi \\mid_{V_2} = id_{V_2}$ and both are $D$-modules. Hence, the basis change respects this property as shown above. This property is kept for the recursively defined family of $D$-modules:\n$$V_{1,i} := V_{1,i-1} \\oplus \\currfield.y_1^i \\oplus \\currfield.y_1^{-i},\\ V_{2,i} := V_{2,i-1} \\oplus \\currfield.x_1^i \\oplus \\currfield.x_1^{i-1} x_2, \\forall i \\geq 1$$\nand $V_{j,0} = \\currfield$ for $j = 1, 2$. We get:\n$$\\bao{rclclcl}\n\\bigcup_{i \\geq 0} V_{1, i} &\\simeq& \\currfield[y_1,X]/\\left &\\simeq& \\currfield[y_1,y_1^{-1}] &=:& R_1\\\\\n&&&&\\\\\n\\bigcup_{i \\geq 0} V_{2,i} &\\simeq& \\currfield[x_1,x_2]/\\left &\\simeq& \\currfield[x_1,\\sqrt{a x_1^2 - 1}] &=:& R_2,\\\\\n\\ea$$\ni.e. a $D$-stable filtration for each subalgebras $R_1, R_2$. Having already shown their simplicity we only need surjectivity as $\\Psi\\mid_{V_{2,0}} = id_{V_{2,0}}, \\Phi\\mid_{V_{1,0}} = id_{V_{1,0}}$ already implies injectivity. Let $f$ and $g$ equal $\\sum_i \\left(f_{1,i} y_1^i + f_{-1,i} y_1^{-i}\\right) \\in R_1$ and $\\sum_i \\left(g_{1,i} x_1^i + g_{2,i-1} x_1^i x_2\\right) \\in R_2$, respectively. Obviously, $\\sum \\left(f_{1,i} (\\sqrt{a} x_1 + x_2)^i + f_{-1,i} (\\sqrt{a} x_1 - x_2)^i\\right)$ and\n$\\sum (y_1 + y_1^{-1})^i \\left(\\frac{g_{1,i}}{2 \\sqrt{a}} + \\frac{g_{2,i}}{2} (y_1 - y_1^{-1})\\right)$ are elements in their respective preimages. This extends naturally to their localizations.\n%Hence, any ideal stable under $\\partial$ is, on the one hand, generated by a single element, on the other hand, $\\deg f = \\deg \\partial(f)$, where $\\deg = \\left[f = \\sum_{i=-m}^n f_i X^i \\longmapsto \\max(m,n)\\right]$. In particular, $\\left$ is also principal and a subideal of $I$.\n\\paragraph{Galois group of the subfields}\nAs above $\\trm{DGal}(\\currfield[y_1,y_{-1}]/\\currfield) = \\{a \\in \\trm{Gl}_1(\\currfield) : \\partial a = a \\partial\\}$. Obviously, the unit group of $\\currfield$ is our differential Galois group.\n% However, both generators are eigenvectors. Hence, the diagonal matrix\n%$$A_{L_++L_-} = \\left(\\bao{cc}\n%\\sqrt{a} & 0\\\\\n%0 & -\\sqrt{a}\\\\\n%\\ea\\right)$$\n%represents $\\partial$ on $S(L_+) \\oplus S(L_-)$ with canonical basis vectors identified with $y_1, y_{-1}$, respectively. Clearly, all matrices with diagonal entries are the only ones fulfilling our definition. We get\n%$$\\trm{DGal}(\\currfield[y_1,y_{-1}]/\\currfield) \\simeq \\currfield^\\times \\times \\currfield^\\times.$$\n\\paragraph{Conclusion}\nGiven our above example, the PV-ring and field are\n$$R = \\currfield[y_1,y_{-1} := y_1^{-1}],\\ K = S^{-1}R = \\currfield(y_1),$$\nrespectively, with differential Galois group\n$$\\trm{DGal}(\\currfield[y_1,y_{-1}]/\\currfield) \\simeq \\currfield^\\times.$$\nThe differential ring $\\currfield[x_1,x_2] = \\trm{Sym}(L)/\\left \\simeq \\trm{Sym}(\\currfield^2)/\\left$ is isomorphic to\n$$\\currfield[y_1] \\oplus \\currfield[y_{-1}] \\simeq \\currfield[y_1]^2.$$\n\\bmk %The results are valid in case $a, \\sqrt{a} \\in \\currfield[z]\\bsl\\currfield$, if our matrix equation gets slightly modified:\n%$$A_L = \\left(\\bao{cc}\n%0 & 1\\\\\n%a \\pm \\partial(\\sqrt{a}) & 0\\\\\n%\\ea\\right) \\in \\trm{Gl}_2(\\currfield(z))\\LRA L(y) = \\partial^2(y) - \\left(a \\mp \\partial(\\sqrt{a})\\right)y = 0,\\ \\forall y \\in S(L).$$\n%The different signs arise from two isomorphic solution spaces $S_1 = S(L_+) \\oplus S(L_-)$, $S_2 = S(L_-) \\oplus S(L_+)$, or equivalently, by the order of operator evaluation:\n%$$S_1 \\simeq D/D.(\\partial - \\sqrt{a})(\\partial + \\sqrt{a}),\\ S_2 \\simeq D/D.(\\partial + \\sqrt{a})(\\partial - \\sqrt{a}),\\ \\trm{where}\\ D := \\currfield(z)[\\partial], \\partial(z) = 1.$$\nA more exhaustive approach to $\\partial^2 - a \\cdot id$, where $a$ is not a constant, is given in \\cite{vdPS01} and \\cite{CohCuySte}.\\\\\n\\indent Unfortunately, our given examples are all linear homogeneous ODEs. The theory, we are going to present in the next chapter, is more general. It includes theory for general ODEs (linear, non-linear) in characteristic zero, (multivariate) iterative derivations in positive characteristic and difference equations in arbitrary characteristic.\\\\\n\\indent Nevertheless, as the general theory does not require the ring/field of constants to be algebraically closed we may examine a broader setting.\n%though each factor is not differentially closed (i.e. $(1,0) \\stackrel{\\partial}{\\longmapsto} (0,1) \\stackrel{\\partial}{\\longmapsto} (a,0)$).\n%We want to see if the following $\\currfield$-linear maps are differentially invariant:\n%$$\\sigma_1 = [\\pm y_1 \\longmapsto \\mp y_1,\\ \\pm y_1^{-1} \\longmapsto \\mp y_1^{-1}] \\in \\trm{Gl}(R_1)$$\n%$$\\sigma_2 = [y_1 \\longmapsto y_1^{-1}] \\in \\trm{Gl}(R_1)$$\n%Since $\\pm y_1$ and $\\pm y_1^{-1}$ have the same eigenvalue $\\sqrt{a}$ and $-\\sqrt{a}$ respectively, we have that $\\sigma_1$ commutes with $\\partial$. Hence, $\\sigma_1 \\in \\trm{DGal}(M_1/\\currfield)$. On the other hand, $\\sigma_2(\\partial(y_1)) = \\sigma_2(\\sqrt{a} y_1) = \\sqrt{a} y_1^{-1} \\neq \\partial(\\sigma_2(y_1)) = \\partial(y_1^{-1}) = -\\sqrt{a} y_1^{-1}$. %Now, we may express $v = \\sum_{i \\in \\{1,2\\}} v_i x_i \\in \\trm{Sol}(\\partial^2 - a \\cdot id)$ as linear combinations of $v = \\sum_{i \\in \\{\\pm 1\\}} v'_i y_i$:\n%$$\\bao{rrcl}\n%M_{B_x}^{B_y}(id):& v &=& v'_1 y_1 + v'_{-1} y_{-1}\\\\\n%& &=& v'_1 (\\sqrt{a} x_1 + x_2) + v'_{-1} (\\sqrt{a} x_1 - x_2)\\\\\n%&&&\\\\\n%&&=& \\sqrt{a} (v'_1 + v'_{-1}) x_1 + (v'_1 - v'_{-1}) x_2\\\\\n%&&&\\\\\n%M_{B_y}^{B_x}(id)&v &=& v_1 x_1 + v_{2} x_{2}\\\\\n%&&=& \\frac{v_1}{2 \\sqrt{a}}(y_1 + y_{-1}) + \\frac{v_{2}}{2} (y_1 - y_{-1})\\\\\n%&&&\\\\\n%&&=& \\frac{1}{2 \\sqrt{a}} \\left[(v_1 + \\sqrt{a} v_2) y_1 + (v_1 - \\sqrt{a} v_2) y_{-1}\\right]\\\\\n%\\ea$$\n%Hence, we have the following basis transformation matrices:\n%$$M_{B_x}^{B_y} = \\left(\\bao{cc}\n%\\sqrt{a} & \\sqrt{a}\\\\\n%1 & -1\\\\\n%\\ea\\right),\\ \\ M_{B_y}^{B_x} = \\frac{1}{2 \\sqrt{a}}\\left(\\bao{cc}\n%1 & \\sqrt{a}\\\\\n%1 & -\\sqrt{a}\\\\\n%\\ea\\right)$$\n%The conjugate $M_{B_y}^{B_x} G M_{B_x}^{B_y}$ of $G:= \\trm{DGal}(\\currfield(x_1,x_2)/\\currfield)$ defines a group action on $S(L_+) \\oplus S(L_-)$ or equivalently, makes $S(L_+) \\oplus S(L_-)$ a $G$-module:\n%$$\\bao{rrcl}\n%\\alpha : &G \\times \\left(S(L_+) \\oplus S(L_-)\\right)& \\longrightarrow &S(L_+) \\oplus S(L_-)\\\\\n%&&&\\\\\n%&\\left(g := \\left(\\bao{cc}\n%g_{11} & g_{12}\\\\\n%a g_{12} & g_{11}\\\\\n%\\ea\\right), v'_1 y_1 + v'_{-1} y_{-1}\\right) &\\longmapsto&M_{B_y}^{B_x} g %\\left(\\bao{cc}\n%g_{11} & g_{12}\\\\\n%a g_{12} & g_{11}\\\\\n%\\ea\\right)\n%M_{B_x}^{B_y}\\left(\\bao{c}\n%v'_1\\\\\n%v'_{-1}\\\\\n%\\ea\\right)\\\\\n%\\ea$${\\footnotesize\n%$$M_{B_y}^{B_x}g %\\left(\\bao{cc}\n%g_{11} & g_{12}\\\\\n%a g_{12} & g_{11}\\\\\n%\\ea\\right)\n%M_{B_x}^{B_y}\\left(\\bao{c}\n%v'_1\\\\\n%v'_{-1}\\\\\n%\\ea\\right) = \\frac{1}{2 \\sqrt{a}}\\left(\\bao{cc}\n%2 \\sqrt{a} g_{11} + (1 + a^2) g_{12}& -(1 - a^2) g_{12}\\\\\n%(1 - a^2) g_{12} & 2 \\sqrt{a} g_{11} - (1 + a^2) g_{12}\\\\\n%\\ea\\right) \\left(\\bao{c}\n%v'_1\\\\\n%v'_{-1}\\\\\n%\\ea\\right)\\\\\n%$$}\n%The fixed field $\\currfield(y_1)^G$ is \n%Multiplying $T$ with the generator of $\\trm{DGal}(\\currfield(x_1,x_2)/\\currfield)$ we get:\n%$$T g = \\left(\\bao{cc}\n%\\sqrt{a} & 1\\\\\n%\\sqrt{a} & -1\\\\\n%\\ea\\right) \\left(\\bao{cc}\n%g_{11} & g_{12}\\\\\n%a g_{12} & g_{11}\\\\\n%\\ea\\right) = \\left(\\bao{cc}\n%\\sqrt{a} g_{11} + a g_{12} & \\sqrt{a} g_{12} + g_{11}\\\\\n%\\sqrt{a} g_{11} - a g_{12} & \\sqrt{a} g_{12} - g_{11}\\\\\n%\\ea\\right)$$\n%Analogously to the classical Galois theory, there exists subgroups $H$ in $\\trm{DGal}(\\currfield(x_1,x_2)/\\currfield)$ such that $\\currfield(x_1,x_2)^H := \\{x \\in K : \\sigma(x) = x \\forall \\sigma \\in H\\}$ is a Picard-Vessiot field over $\\currfield$. Conversely, the $\\trm{DGal}(\\currfield(x_1,x_2)/M)$ is a (closed) subgroup of $\\trm{DGal}(\\currfield(x_1,x_2)/\\currfield)$ for any intermediate Picard-Vessiot field $\\currfield \\subset M \\subset \\currfield(x_1,x_2)$. %Since we already have two Picard-Vessiot subfields $M_i = \\currfield(y_i)$ of $K$, we only need to compute the Galois groups of both differential fields.", "meta": {"hexsha": "dfbfb0c7c87fbbfa37cd515df8b9ee0f25a2a097", "size": 61435, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Script_Diff_Gal07/diffRingNMods.tex", "max_stars_repo_name": "gmuel/texlib", "max_stars_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Script_Diff_Gal07/diffRingNMods.tex", "max_issues_repo_name": "gmuel/texlib", "max_issues_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Script_Diff_Gal07/diffRingNMods.tex", "max_forks_repo_name": "gmuel/texlib", "max_forks_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 92.1064467766, "max_line_length": 942, "alphanum_fraction": 0.6569545048, "num_tokens": 23340, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267118111485245, "lm_q2_score": 0.7826624789529376, "lm_q1q2_score": 0.647036315493177}} {"text": "% \\documentclass[draft,11pt]{article}\n\\chapter{Pseudo-inverses and Effective Resistance}\n\\label{cha:pinver}\n%\\usepackage{cleveref}\n\n%\\allowdisplaybreaks\n\n%%% for this lecture\n%\\newcommand\\gap{\\text{gap}}\n%\\newcommand{\\Hongjie}[1]{{\\color{red} Hongjie: #1}}\n\n%\\begin{document}\n\\sloppy\n%\\lecture{6 --- Wednesday, March 25th}\n%{Spring 2020}{Rasmus Kyng, Scribe: Hongjie Chen}{Effective Resistance,\n%Gaussian Elimination as Optimization}\n\n\\section{What is a (Moore-Penrose) Pseudoinverse?}\nRecall that for a connected graph $G$ with Laplacian $\\LL$, we have $\\ker(\\LL)=\\Span\\{\\vecone\\}$, which means $\\LL$ is not invertible. However, we still want some matrix which behaves like a real inverse. To be more specific, given a Laplacian $\\LL\\in\\R^{V\\times V}$, we want some matrix $\\LL^{\\pinv}\\in\\R^{V\\times V}$ s.t.\\\n\\begin{enumerate}\n\t\\item[1)] $(\\LL^{\\pinv})^\\trp = \\LL^\\pinv$ (symmetric)\n\t\\item[2)] $\\LL^{\\pinv}\\matone = \\veczero$, or more generally, $\\LL^{\\pinv}\\vv=\\veczero$ for $\\vv\\in\\ker(\\LL)$\n\t\\item[3)] $\\LL^{\\pinv}\\LL\\vv = \\LL\\LL^{\\pinv}\\vv = \\vv$ for $\\vv\\perp\\matone$, or more generally, for $\\vv\\in\\ker(\\LL)^\\perp$\n\\end{enumerate}\nUnder the above conditions, $\\LL^{\\pinv}$ is uniquely defined and we\ncall it the pseudoinverse of $\\LL$. Note that there are many other\nequivalent definitions of the pseudoinverse of some matrix $\\AA$,\nand we can also generalize the concept to matrices that aren't\nsymmetric or even square.\n\nLet $\\lambda_i, \\vv_i$ be the $i$-th pair of eigenvalue and eigenvector of $\\LL$, with $\\{\\vv_i\\}_{i=1}^n$ forming a orthogonal basis. Then by the spectral theorem,\n\\[ \\LL = \\VV\\LLambda\\VV^\\trp = \\sum_i \\lambda_i\\vv_i\\vv_i^\\trp, \\]\nwhere $\\VV = \\begin{bmatrix} \\vv_1 & \\cdots & \\vv_n \\end{bmatrix}$ and $\\LLambda = \\diag\\{\\lambda_1,...,\\lambda_n\\}$. And we can show that its pseudoinverse is exactly\n\\[ \\LL^{\\pinv} = \\sum_{i,\\lambda_i\\neq0} \\lambda_i^{-1}\\vv_i\\vv_i^\\trp. \\]\nChecking conditions 1), 2), 3) is immediate.\nWe can also prove uniqueness, but this takes slightly more work.\n\n\\section{Electrical Flows Again}\nRecall the incidence matrix $\\BB\\in\\R^{V\\times E}$ of a graph\n$G=(V,E)$.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.7\\textwidth]{fig/lecture6_incidence-example.jpeg}\n \\caption{An example of a graph and its incidence matrix $B$.}\n\\label{fig:incidenceexample}\n\\end{figure}\n\nIn Chapter 1, we introduced the electrical flow routing demand\n$\\dd\\in\\R^V$.\nLet's call the electrical flow $\\fftil\\in\\R^E$. The net flow\nconstraint requires $\\BB\\fftil = \\dd$. By Ohm's Law,\n$\\fftil=\\RR^{-1}\\BB^\\trp\\xx$ for some voltage $x\\in\\R^V$ where\n$\\RR=\\diag(\\rr)$ and $\\rr(e)=$ resistance of edge $e$. We showed (in\nthe exercises) that when $\\dd \\perp \\matone$, there exists an voltage\n$\\xxtil\\perp\\matone$ s.t.\\ $\\fftil=\\RR^{-1}\\BB^\\trp\\xxtil$ and\n$\\BB\\fftil = \\dd$. This $\\xxtil$ solves $\\LL\\xx=\\dd$ where\n$\\LL=\\BB\\RR^{-1}\\BB^\\trp$.\n\nAnd we also made the following claim.\n\\begin{claim}\n \\begin{equation}\n \\label{eq:argminflow}\n \\fftil = \\argmin_{\\BB\\ff=\\dd} \\ff^\\trp\\RR\\ff \\text{ where }\n \\ff^\\trp\\RR\\ff=\\sum_e \\rr(e)\\ff(e)^2,\n \\end{equation}\n\\end{claim}\nYou proved this in the exercises for Week 1. Let's recap the proof\nbriefly, just to get back into thinking about electrical flows.\n\\begin{proof}\nConsider any $\\ff\\in\\R^E$ s.t.\\ $\\BB\\ff=\\dd$. For any $\\xx\\in\\R^V$, we have\n\\begin{align*}\n\t\\frac{1}{2}\\ff^\\trp\\RR\\ff\n\t&= \\frac{1}{2}\\ff^\\trp\\RR\\ff - \\xx^\\trp(\\underbrace{\\BB\\ff-\\dd}_{\\matzero}) \\\\\n\t&\\geq \\min_{\\ff\\in\\R^E} \\underbrace{\\frac{1}{2}\\ff^\\trp\\RR\\ff - \\xx^\\trp\\BB\\ff + \\dd^\\trp\\xx}_{g(\\ff)} \\\\\n\t&= \\dd^\\trp\\xx - \\frac{1}{2}\\xx^\\trp\\LL\\xx\n\\end{align*}\nsince $\\nabla_{\\ff}g(\\ff) = \\matzero$ gives us $\\ff=\\RR^{-1}\\BB^\\trp\\xx$. Thus, for all $\\ff\\in\\R^E$ s.t.\\ $\\BB\\ff=\\dd$ and all $\\xx\\in\\R^V$,\n\\begin{equation}\n \\label{eq:flowsboundedbyvoltages}\n\\frac{1}{2}\\ff^\\trp\\RR\\ff \\geq \\dd^\\trp\\xx -\n \\frac{1}{2}\\xx^\\trp\\LL\\xx.\n\\end{equation}\nBut for the electrical flow $\\fftil$ and electrical voltage $\\xxtil$, we have $\\fftil=\\RR^{-1}\\BB^\\trp\\xxtil$ and $\\LL\\xxtil=\\dd$. So\n\\[ \\fftil^\\trp\\RR\\fftil = \\left(\\RR^{-1}\\BB^\\trp\\xxtil\\right)^\\trp\\RR\\left(\\RR^{-1}\\BB^\\trp\\xxtil\\right) = \\xxtil^\\trp\\BB\\RR^{-1}\\BB^\\trp\\xxtil = \\xxtil^\\trp\\LL\\xxtil =\\xxtil^\\trp\\dd. \\]\nTherefore,\n\\begin{equation}\n \\label{eq:flowattainsvoltagevalue}\n\\frac{1}{2}\\fftil^\\trp\\RR\\fftil = \\dd^\\trp\\xxtil -\n\\frac{1}{2}\\xxtil^\\trp\\LL\\xxtil.\n\\end{equation}\n By combining Equation~\\eqref{eq:flowsboundedbyvoltages} and\n Equation~\\eqref{eq:flowattainsvoltagevalue}, we see that for all\n $\\ff$ s.t.\\ $\\BB \\ff = \\dd$,\n \\[\n \\frac{1}{2}\\ff^\\trp\\RR\\ff \\geq \\dd^\\trp\\xxtil -\n\\frac{1}{2}\\xxtil^\\trp\\LL\\xxtil = \\frac{1}{2}\\fftil^\\trp\\RR\\fftil.\n\\]\nThus $\\fftil$ is the minimum electrical energy flow among all flows\nthat route demand $\\dd$, proving Equation~\\eqref{eq:argminflow} holds.\n\nThe drawing below shows how the quantities line up:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\textwidth]{fig/lecture6_maxminfig.jpeg}\n% \\caption{}\n\\label{fig:regions}\n\\end{figure}\n\\end{proof}\n\n\n\n\\section{Effective Resistance}\nGiven a graph $G=(V,E)$, for any pair of vertices $(a,b)\\in V$, we\nwant to compute the cost (or energy) of routing 1 unit of current\nfrom $a$ to $b$. We call such cost the effective resistance between\n$a$ and $b$, denoted by $\\er(a,b)$. Recall for a single resistor\n$r(a,b)$,\n\\[\n \\text{energy} =r(a,b)f^2(a,b)=r(a,b)\n .\n\\]\nSo when we have a graph consisting of just one edge\n$(a,b)$, the effective resistance is just $\\er(a,b)=r(a,b)$.\n\nIn a general graph, we can also consider the energy required to route\none unit of current between two vertices.\nFor any pair $a,b\\in V$, we have\n\\[ \\er(a,b) = \\min_{\\BB\\ff=\\ee_b-\\ee_a} \\ff^\\trp\\RR\\ff, \\]\nwhere $\\ee_v \\in \\mathbb{R}^V$ is the indicator vector of $v$.\nNote that the cost of routing $F$ units of flow from $a$ to $b$ will be $\\er(a,b)\\cdot F^2$.\n\nSince $(\\ee_b-\\ee_a)^\\trp\\matone=0$, we know from the previous section\nthat $\\er(a,b) = \\fftil^\\trp\\RR\\fftil$ where $\\fftil$ is the\nelectrical flow.\nNow we can write\n$\\LL\\xxtil=\\ee_b-\\ee_a$ and $\\xxtil=\\LL^{\\pinv}(\\ee_b-\\ee_a)$ for the\nelectrical voltages routing 1 unit of current from $a$ to $b$.\n% where we let $\\xxtil\\perp\\matone$ to use condition 3) of the\n% pseudoinverse $\\LL^{\\pinv}$.\nNow the energy of routing 1 unit of current from $a$ to $b$ is\n\\[ \\er(a,b) = \\fftil^\\trp\\RR\\fftil = \\xxtil^\\trp\\LL\\xxtil = (\\ee_b-\\ee_a)^\\trp\\LL^{\\pinv}\\LL\\LL^{\\pinv}(\\ee_b-\\ee_a) = (\\ee_b-\\ee_a)^\\trp\\LL^{\\pinv}(\\ee_b-\\ee_a), \\]\nwhere the last equality is due to\n$\\LL^{\\pinv}\\LL\\LL^{\\pinv}=\\LL^{\\pinv}$.\n\n\\begin{remark}\n We have now seen several different expressions that all take on the\n same value: the energy of the electrical flow.\n It's useful to remind yourself what these are.\n Consider an electrical flow $\\fftil$ routes demand $\\dd$, and\n associated electrical voltages\n $\\xxtil$.\n We know that $\\BB \\fftil = \\dd$, and $\\ff = \\RR^{-1} \\BB^{\\trp}\n \\xxtil$,\n and $\\LL \\xxtil = \\dd$, where $\\LL = \\BB \\RR^{-1} \\BB^{\\trp}$.\n And we have seen how to express the electrical energy using many\n different quantities:\n \\[\n \\fftil^\\trp\\RR\\fftil = \\xxtil^\\trp\\LL\\xxtil = \\dd^\\trp \\LL^{\\pinv}\n \\dd =\\dd^\\trp\\xxtil = \\fftil^{\\trp}\\BB^{\\trp} \\xxtil\n \\]\n\\end{remark}\n\n\\begin{claim}\n\tAny PSD matrix $\\AA$ has a PSD square root $\\AA^{1/2}$ s.t.\\ $\\AA^{1/2}\\AA^{1/2}=\\AA$.\n\\end{claim}\n\\begin{proof}\n\tBy the spectral theorem, $\\AA = \\sum_i \\lambda_i\\vv_i\\vv_i^\\trp$ where $\\{\\vv_i\\}$ are orthonormal. Let $\\AA^{1/2} = \\sum_i \\lambda_i^{1/2}\\vv_i\\vv_i^\\trp$. Then\n\t\\begin{align*}\n\t\t\\AA^{1/2}\\AA^{1/2}\n\t\t&= \\left(\\sum_i \\lambda_i^{1/2}\\vv_i\\vv_i^\\trp\\right)^2 \\\\\n\t\t&= \\sum_{i}\\lambda_i \\vv_i\\vv_i^\\trp\\vv_i\\vv_i^\\trp + \\sum_{i\\neq j}\\lambda_i \\vv_i\\vv_i^\\trp\\vv_j\\vv_j^\\trp \\\\\n\t\t&= \\sum_i \\lambda_i\\vv_i\\vv_i^\\trp \\\\\n\t\\end{align*}\n\twhere the last equality is due to $\\vv_i^\\trp\\vv_j = \\delta_{ij}$. It's easy to see that $\\AA^{1/2}$ is also PSD.\n\\end{proof}\n\nLet $\\LL^{+/2}$ be the square root of $\\LL^{+}$. So\n\\[ \\er(a,b) = (\\ee_b-\\ee_a)^\\trp\\LL^{\\pinv}(\\ee_b-\\ee_a) = \\|\\LL^{+/2}(\\ee_b-\\ee_a)\\|^2. \\]\n\n\\paragraph{Example: Effective resistance in a path.}\nConsider a path graph on vertices $V = \\setof{1,2,3,\\ldots, k+1}$,\nwith with resistances $\\rr(1), \\rr(2), \\ldots, \\rr(k)$ on\nthe edges of the path.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.6\\textwidth]{fig/lecture6_seriesres.jpeg}\n \\caption{A path graph with $k$ edges.}\n\\label{fig:seriesres}\n\\end{figure}\nThe effective resistance between the\nendpoints is\n\\[\n \\er(1,k+1) = \\sum_{i = 1}^{k} \\rr(i)\n \\]\nTo see this, observe that to have 1 unit of flow going from vertex $1$ to vertex $k+1$, we must\nhave one unit flowing across each edge $i$.\nLet $\\DDelta(i)$ be the voltage difference across edge $i$, and $\\ff(i)$ the\nflow on the edge.\nThen $1 = \\ff(i) = \\frac{\\DDelta(i)}{\\rr(i)}$, so that $\\DDelta(i) =\n\\rr(i)$.\nThe electrical voltages are then $\\xxtil \\in \\R^V$ where $\\xxtil(i) =\n\\xxtil(1) + \\sum_{j < i} \\DDelta(j)$.\nHence the effective resistance is\n\\[\\er(1,k+1) = \\dd^{\\trp}\\xxtil= (\\ee_{k+1}-\\ee_{1})^{\\trp}\\xxtil = \\xxtil(k+1) -\n \\xxtil(1) = \\sum_{i = 1}^{k} \\rr(i)\n.\n\\]\nThis behavior is sometimes known as the fact that the resistance of\nresistors adds up when they are connected in series.\n\n\\paragraph{Example: Effective resistance of parallel edges.}\nSo far, we have only considered graphs with at most one edge between\nany two vertices. But that math also works if we allow a pair of\nvertices to have multiple distinct edges connecting them. We refer to\nthis as \\emph{multi-edges}.\nSuppose we have a graph on just two vertices, $V = \\setof{1,2}$, and\nthese are connected by $k$ parallel multi-edges with resistances\n$\\rr(1), \\rr(2), \\ldots, \\rr(k)$.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.3\\textwidth]{fig/lecture6_parallelres.jpeg}\n \\caption{A graph on just two vertices with $k$ parallel multiedges.}\n\\label{fig:parallelres}\n\\end{figure}\nThe effective resistance between the endpoints is\n\\[\n \\er(1,2) =\n \\frac{1}{\\sum_{i = 1}^k 1/\\rr(i)}\n .\n\\]\nLet's see why.\nOur electrical voltages $\\xxtil \\in \\R^V$ can be described by just the voltage\ndifference $\\Delta \\in \\R$ between vertex 1 and vertex 2, i.e. $\\xxtil(2)\n- \\xxtil(1) = \\Delta$.\nwhich creates a flow on edge $i$ of $\\fftil(i) = \\Delta/\\rr(i)$.\nThus the total flow from vertex 1 to vertex 2 is\n$1 = \\sum_{i} \\Delta/\\rr(i)$,\nso that $\\Delta =\\frac{1}{\\sum_{i = 1}^k 1/\\rr(i)}$.\nMeanwhile, the effective resistance is also\n\\[\n\\er(1,2) = (\\ee_{2}-\\ee_{1})^{\\trp}\\xxtil = \\Delta =\\frac{1}{\\sum_{i = 1}^k 1/\\rr(i)}\n \\]\n\\subsection{Effective Resistance is a Distance}\n\\boxdef{\n \\label{def:dist}\n Consider a weighted undirected graph $G$ with vertex set $V$. We\n say function $d: V \\times V \\to \\R$, which takes a pair of vertices\n and returns a real number, is a \\emph{distance} if it satisfies\n \\begin{enumerate}\n \\item $d(a,a) = 0$ for all $a \\in V$ \\label{enu:distzero}\n \\item $d(a,b) \\geq 0$ for all $a,b \\in V$. \\label{enu:distnonneg}\n \\item $d(a,b) = d(b,a)$ for all $a,b \\in V$. \\label{enu:distsym}\n \\item $d(a,b) \\leq d(a,c) + d(c,b)$ for all $a,b,c \\in\n V$. \\label{enu:disttri}\n \\end{enumerate}\n}\n\\begin{lemma}\n $\\er$ is a distance.\n\\end{lemma}\nBefore proving this lemma, let's see a claim that will help us finish\nthe proof.\n\\begin{claim}\n \\label{clm:voltageorder}\n Let $\\LL \\xxtil = \\ee_b - \\ee_a$.\n Then for all $\\cc \\in V$, we have $\\xxtil(b) \\geq \\xxtil(c) \\geq \\xxtil(a)$.\n \\end{claim}\n We only sketch a proof of this claim:\n \\begin{proof}[Proof sketch]\n Consider any $c \\in V$, where $c\\neq a,b$.\n Now $(\\LL \\xxtil)(c) = 0$, i.e.\n \\[\n \\left( \\sum_{(u,c)} w(u,c) \\right) \\xxtil(c) - \\left(\n \\sum_{(u,c)} w(u,c) \\xxtil(u) \\right) = 0\n \\]\n Rearranging $ \\xxtil(c) = \\frac{\\sum_{(u,c)} w(u,c) \\xxtil(u)\n }{\\sum_{(u,c)} w(u,c)}$.\n This tells us that $\\xxtil(c)$ is a weighted average of the\n voltages of its neighbors. From this, we can show that $\\xxtil(a)$\n and $\\xxtil(b)$ are the extreme values.\n \\end{proof}\n\\begin{proof}\n It is easy to check that conditions\n \\ref{enu:distzero}, \\ref{enu:distnonneg}, and \\ref{enu:distsym} of Definition~\\ref{def:dist} are satisfied by\n $\\er$.\n Let us confirm condition~\\ref{enu:disttri}.\n\n For any $u,v$, let $\\xxtil_{u,v} = \\LL^\\pinv ( -\\ee_u + \\ee_v )$.\n Then\n \\[\n \\xxtil_{a,b} = \\LL^\\pinv ( -\\ee_a+ \\ee_b)\n = \\LL^\\pinv ( -\\ee_a + \\ee_c - \\ee_c + \\ee_b)\n = \\xxtil_{a,c} + \\xxtil_{c,b}\n .\n\\]\nThus,\n\\begin{align*}\n \\er(a,b) = ( -\\ee_a+ \\ee_b)^\\trp \\xxtil_{a,b}\n &= ( -\\ee_a+ \\ee_b)^\\trp(\\xxtil_{a,c} + \\xxtil_{c,b})\n \\\\\n &= -\\xxtil_{a,c}(a) + \\xxtil_{a,c}(b) -\\xxtil_{c,b}(a) +\n \\xxtil_{c,b}(b)\n \\\\\n &\\leq -\\xxtil_{a,c}(a) + \\xxtil_{a,c}(c) -\\xxtil_{c,b}(c) +\n \\xxtil_{c,b}(b).\n\\end{align*}\nwhere in the last line we applied Claim~\\ref{clm:voltageorder} to show\nthat $\\xxtil_{a,c}(b) \\leq \\xxtil_{a,c}(c)$ and ${ -\\xxtil_{c,b}(a) \\leq -\\xxtil_{c,b}(c)}$.\n\\end{proof}\n\n\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"agao21_script\"\n%%% TeX-engine: luatex\n%%% End:", "meta": {"hexsha": "6ccf6f083866ef7f0c8a1a362171282f38c20134", "size": 13049, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "agao21_script/lecture6_mod.tex", "max_stars_repo_name": "csssaz/agao21_script", "max_stars_repo_head_hexsha": "51044f4775e5e20d2c5fc5c0d035363e5beb66be", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-15T09:04:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-26T05:39:09.000Z", "max_issues_repo_path": "agao21_script/lecture6_mod.tex", "max_issues_repo_name": "csssaz/agao21_script", "max_issues_repo_head_hexsha": "51044f4775e5e20d2c5fc5c0d035363e5beb66be", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "agao21_script/lecture6_mod.tex", "max_forks_repo_name": "csssaz/agao21_script", "max_forks_repo_head_hexsha": "51044f4775e5e20d2c5fc5c0d035363e5beb66be", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2021-03-11T12:35:23.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-13T06:04:51.000Z", "avg_line_length": 41.1640378549, "max_line_length": 324, "alphanum_fraction": 0.6497049582, "num_tokens": 4910, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267118026095991, "lm_q2_score": 0.7826624738835052, "lm_q1q2_score": 0.6470363046191209}} {"text": "%!TEX root = paper.tex\r\n\\subsection{Solitary wave on a composite beach} \\label{sec:B_compositebeach}\r\nThis is an laboratory experiment conducted at the Coastal Engineering Laboratory of the U.S. Army Corps of Engineers and it is described in \\url{http://nctr.pmel.noaa.gov/benchmark/Solitary_wave/}, \\url{http://chl.erdc.usace.army.mil/chl.aspx?p=s&a=Projects;36}. \r\nA linear solitary (better called single) wave is propagating over a stepwise increasing bathymetry and it is reflected at a vertical wall on the right boundary. Different wave gauges measure the surface elevation and the runup on the vertical wall. \r\nThe experimental data serve for validation of the models. Additionally to the experimental data, a linear analytic solution is provided, s.t. verification of our models is also tested.\r\n\r\nThree different cases A, B, C with different target wave heights $a_t$, actual (measured) wave heights $a$ and distances $L$ of gauge $G4$ to the first step in the bathymetry at gauge $G5$. Table \\ref{tab:compositebeach_cases} displays the three different cases and belonging data. We only consider case A at the moment. Measured runup data can be found in table \\ref{tab:compositebeach_runup}, but are not compared yet to the simulations.\r\n\r\n\\begin{table}[htbp]\r\n\\begin{tabular}{lllll}\r\n\\textbf{Case} & \\textbf{target $a_t / d$} & \\textbf{actual $a / d$} & \\textbf{dist. G4 to G5 [m]} & \\textbf{dist. G4 to Wall [m]} \\\\\r\n\\toprule\r\nA & 0.05 & 0.039 & 2.4 & 10.59 \\\\\r\nB & 0.30 & 0.264 & 0.89 & 9.17 \\\\\r\nC & 0.70 & 0.696 & 0.64 & 8.83 \\\\\r\n\\bottomrule\r\n\\end{tabular}\r\n\\caption{Data of three different cases}\r\n\\label{tab:compositebeach_cases}\r\n\\end{table}\r\n\r\n\r\n\\begin{table}[htbp]\r\n\\begin{tabular}{lll}\r\n\\textbf{Case} & \\textbf{Runup R [cm]} & \\textbf{R / d} \\\\\r\n\\toprule\r\nA & 2.74 & 0.13 \\\\\r\nB & 45.72 & 2.10 \\\\\r\nC & 27.43 & 1.26 \\\\\r\n\\bottomrule\r\n\\end{tabular}\r\n\\caption{Runup laboratory results of three different cases}\r\n\\label{tab:compositebeach_runup}\r\n\\end{table}\r\n\r\n\r\nThe initial condition is prescribed as \r\n\\begin{itemize}\r\n \\item a linear analytic solitary wave solution\r\n\\begin{align}\r\n\\xi(\\bx,t)&=a_t \\ \\text{cosh}^{-2}(K(x-ct-x_0)), \\\\\r\nu(\\bx,t)&=c\\frac{\\xi(\\bx,t)}{d},\r\n\\end{align}\r\nwith the initial target amplitude $a_t$, propagation velocity $c=\\csw$ on a stepwise reduced depth starting from $d=0.218 \\, \\text{m}$ with scale factor $K=\\sqrt{\\left(\\frac{3a_t}{4d^3}\\right)}$ and displacement $x_0=10.59$, s.t. the initial solitary wave has its maximum at gauge G4 while the entire domain length is $L=24 \\, \\text{m}$. The simulation time is $20$ seconds. \r\n \\item the first option in \\url{https://github.com/rjleveque/nthmp-benchmark-problems/blob/master/BP05-ElenaT-Solitary_wave_on_composite_beach_laboratory/BP5_description.pdf}. The velocities are set to zero and the doubled initial surface elevation $2\\xi$ is prescribed at $x=0$. A doubled domain length of $L=48 \\, \\text{m}$ ensures that the waves reflected at the left boundary are not disturbing the solution. The simulation time is $40$ seconds.\r\n\\end{itemize}\r\nWe impose reflecting boundary conditions at the boundary in x-direction and periodic boundary conditions in y-direction. For the setup see figure \\ref{fig:compositebeach_setup}. \r\nBecause the analytic solution belongs to the linear SWE, corresponding models are expected to represent very good coincidence with this analytic solution.\r\n\r\n\\begin{figure}[htbp]\r\n\\includegraphics[width=\\textwidth]{compositebeach_setup}\r\n\\caption{Setup of the testcase solitary wave on a composite beach}\r\n\\label{fig:compositebeach_setup}\r\n\\end{figure}\r\n\r\n\\subsubsection{Results of \\nh\\ model}\r\nThe results of the first and the seconds option are shifted by 271.5s to match the initial wave at gauge G4. Both are also scaled with a factor of $\\frac{0.037}{0.05}$. % because $0.037$m seems to be the actual amplitude instead of $0.039$.\r\nNo difference between both options is visible in the figures then. Hence, we only plot the results of the (cheaper) first option using the linear SWE model and also the following results obtained with the \\nh\\ model.\r\n\r\nFigures \\eqref{fig:nh_compositebeach_ana_nh_Lhy} and \\eqref{fig:nh_compositebeach_lab_nh_Lhy} display the comparison of the linear SWE with the linear analytic solution and experimental data, respectively.\r\nThe models results compared to the analytic solution are very good, while at gauges G9 and G10 a mismatch is visible, because the nonlinear regime is entered here (d $\\in [0.047,0.117]$m, $a\\approx0.008$m), for which the linear model is not suited.\r\nThe comparison to the experimental data reveals less dispersion and reduced amplitudes, although the overall match is also satisfactory.\r\n\r\nFigures \\eqref{fig:nh_compositebeach_ana_nh_Nnh12} and \\eqref{fig:nh_compositebeach_lab_nh_Nnh12} display the comparison of the \\nh\\ equations using both the linear and the quadratic pressure profile with the linear analytic solution and experimental data, respectively. The \\nh\\ wave profile represents well the amplitudes, the shape of the propagating wave, the propagation velocities and also the developed dispersive wave train after reflecting of the laboratory surface elevation. The agreement with the linear analytic solution is worse, but still okay.\r\nWhen wave hits step, the numerical sea surface height is not as much increased as the amplitude of exp data. \r\n\r\nDifferences between both vertical pressure profiles are small, because dispersive effects become more significant after longer simulation time. However, the tendency of the linear profile to overestimate amplitudes, which was also visible in figure \\ref{fig:nh_solitarywave}, is confirmed here, too. Additionally, we see that is also true not only for global maxima, but also for local maxima.\r\n\r\nThe analytic solution is a good test for the verification of the linear SWE model, whereas the laboratory data show its limits of applicability as well as that the dispersive model gives a more accurate physical representation of the experimental data.\r\n\r\nNeed to do: wq refl b.c. may be not good enough (not explicit treatment at the moment), edge integrals are missing.\r\n\r\n\r\n\\begin{figure}[htbp]\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G4_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G5_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G6_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G7_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G8_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G9_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G10_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_Wall_nh_Lhy}\r\n\\end{minipage}\r\n\\caption{Comparison of the analytical (black) sea surface height of the\r\nsolitary wave with the simulation results of the \\nh\\ model in its version of linear shallow water equations (blue)}\r\n\\label{fig:nh_compositebeach_ana_nh_Lhy}\r\n\\end{figure}\r\n\r\n\\begin{figure}[htbp]\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G4_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G5_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G6_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G7_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G8_nh_Lhy}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G9_nh_Lhy}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{0.48\\textwidth}\r\n\\includegraphics[width=\\textwidth]{compositebeach_lab_G10_nh_Lhy}\r\n\\end{minipage} \r\n\\begin{minipage}{0.45\\textwidth}\r\n\\begin{tabular}{lll}\r\n\\textbf{Data} & \\textbf{Runup} & \\textbf{R / d} \\\\\r\n & \\textbf{R [cm]} & \\\\\r\n\\toprule\r\nExp. & 2.74 & 0.13 \\\\\r\nModel hydr. & 2.13 & 0.0978 \\\\\r\n\\end{tabular}\r\n\\end{minipage}\r\n\\caption{Comparison of the experimental (black) sea surface height of the\r\nsolitary wave with the simulation results of the \\nh\\ model in its version of linear shallow water equations (blue). The model runup is also scaled with 0.75.}\r\n\\label{fig:nh_compositebeach_lab_nh_Lhy}\r\n\\end{figure}\r\n\r\n\r\n\r\n\\begin{figure}[htbp]\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G4_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G5_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G6_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G7_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G8_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G9_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_G10_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_ana_Wall_nh_Nnh12}\r\n\\end{minipage}\r\n\\caption{Comparison of the analytical (black) sea surface height of the\r\nsolitary wave with the simulation results of the \\nh\\ model in its \\nh\\ version using the linear (yellow) and the quadratic vertical pressure profile (red)}\r\n\\label{fig:nh_compositebeach_ana_nh_Nnh12}\r\n\\end{figure}\r\n\r\n\\begin{figure}[htbp]\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G4_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G5_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G6_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G7_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{\\textwidth}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G8_nh_Nnh12}\r\n\\includegraphics[width=0.48\\textwidth]{compositebeach_lab_G9_nh_Nnh12}\r\n\\end{minipage} \\\\\r\n\\begin{minipage}{0.48\\textwidth}\r\n\\includegraphics[width=\\textwidth]{compositebeach_lab_G10_nh_Nnh12}\r\n\\end{minipage} \r\n\\begin{minipage}{0.45\\textwidth}\r\n\\begin{tabular}{lll}\r\n\\textbf{Data} & \\textbf{Runup} & \\textbf{R / d} \\\\\r\n & \\textbf{R [cm]} & \\\\\r\n\\toprule\r\nExp. & 2.74 & 0.13 \\\\\r\nModel lin.& 2.18 & 0.10 \\\\\r\nModel quadr.& 2.12 & 0.097 \\\\\r\n\\end{tabular}\r\n\\end{minipage}\r\n\\caption{Comparison of the experimental (black) sea surface height of the\r\nsolitary wave with the simulation results of the \\nh\\ model in its \\nh\\ version using the linear (yellow) and the quadratic vertical pressure profile (red). The model runup is also scaled with 0.75.}\r\n\\label{fig:nh_compositebeach_lab_nh_Nnh12}\r\n\\end{figure}\r\n", "meta": {"hexsha": "73f193cda543d61b373c754046ccf8ac951744c3", "size": 10947, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/papers/theoretical_1d/B_compositebeach.tex", "max_stars_repo_name": "mandli/coastal", "max_stars_repo_head_hexsha": "8c80a4c740f92ea83b54c8a5432d11058c0d3476", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/papers/theoretical_1d/B_compositebeach.tex", "max_issues_repo_name": "mandli/coastal", "max_issues_repo_head_hexsha": "8c80a4c740f92ea83b54c8a5432d11058c0d3476", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/papers/theoretical_1d/B_compositebeach.tex", "max_forks_repo_name": "mandli/coastal", "max_forks_repo_head_hexsha": "8c80a4c740f92ea83b54c8a5432d11058c0d3476", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.8166666667, "max_line_length": 560, "alphanum_fraction": 0.7582899424, "num_tokens": 3247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8267117919359419, "lm_q1q2_score": 0.6470363004562094}} {"text": "\\documentclass{article}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n\\begin{document}\n\n\\section*{Inline (within text) formulas}\n\nThe equation $x + x = 2x$ is inside a text, which uses the \\TeX{} shorthand.\nIn addition, an equation can be written like \\(x \\cdot x = x^2\\) as well, using the \\LaTeX{} shorthand.\nFinally, the environment \\emph{math} can be used like this \\begin{math}a^2 + b + c = 0\\end{math}.\n\nWe can force symbols to be displayed like displayed formula. For example the formula $\\sum_{k=0}^{10}k$ can be written $\\displaystyle\\sum_{k=0}^{10}k$ as well. The sum symbol is taller within a sentence using the command \\emph{\\textbackslash displaystyle}.\n\n\\section*{Displayed equations}\n\nThe recommended syntax to render a floating equation is to use the syntax below.\n\n\\[a \\cdot x = ax \\]\n\nThe use of the syntax \\emph{\\$\\$$\\cdots$\\$\\$} should be avoided, because, it will modify vertical spacing within equations, rendering them inconsistent.\n\nFinally, the environment \\emph{displaymath} produces the same effect.\n\n\\begin{displaymath}\n\tx^2 \\cdot x^2 = x^4\n\\end{displaymath}\n\n\\section*{Equation numbering}\n\nThe environment \\emph{equation} automatically numbers the equations.\n\n\\begin{equation} \n\tf(x)=(x+a)(x+b)\n\\end{equation}\n\n\\end{document}", "meta": {"hexsha": "388052b80982c418a07fafa45ba51b193935635d", "size": 1263, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compendium/mathematics/environments.tex", "max_stars_repo_name": "ZenLulz/LatexCompendium", "max_stars_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-07-30T21:43:55.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-23T20:16:19.000Z", "max_issues_repo_path": "compendium/mathematics/environments.tex", "max_issues_repo_name": "ZenLulz/LatexCompendium", "max_issues_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compendium/mathematics/environments.tex", "max_forks_repo_name": "ZenLulz/LatexCompendium", "max_forks_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.2368421053, "max_line_length": 256, "alphanum_fraction": 0.7339667458, "num_tokens": 358, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624688140728, "lm_q2_score": 0.8267117983401363, "lm_q1q2_score": 0.6470362970866129}} {"text": "In the appendix, we define a number of concepts and describe a number of\ntechniques that are utilized extensively throughout the thesis.\n\n\\section{Linear Algebra}\n\\inputsection{linear-algebra}\n\n\\section{Probability Theory}\n\\xlab{probability-theory}\n\\inputsection{probability-theory}\n\n\\section{Bayesian Statistics}\n\\xlab{bayesian-statistics}\n\\inputsection{bayesian-statistics}\n\n\\section{Probability Transformation}\n\\xlab{probability-transformation}\n\\inputsection{probability-transformation}\n\n\\section{Numerical Integration}\n\\xlab{numerical-integration}\n\\inputsection{numerical-integration}\n\n\\section{Hierarchical Interpolation}\n\\xlab{hierarchical-interpolation}\n\\inputsection{hierarchical-interpolation}\n\n\\section{Polynomial Chaos}\n\\xlab{polynomial-chaos}\n\\inputsection{polynomial-chaos}\n", "meta": {"hexsha": "f0aee9f1f17065d2cea798a4f622a721b6113db0", "size": 787, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "include/appendix.tex", "max_stars_repo_name": "IvanUkhov/thesis", "max_stars_repo_head_hexsha": "95a7e2ee7664b94156906322610555e36e53cfe0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "include/appendix.tex", "max_issues_repo_name": "IvanUkhov/thesis", "max_issues_repo_head_hexsha": "95a7e2ee7664b94156906322610555e36e53cfe0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "include/appendix.tex", "max_forks_repo_name": "IvanUkhov/thesis", "max_forks_repo_head_hexsha": "95a7e2ee7664b94156906322610555e36e53cfe0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.2333333333, "max_line_length": 72, "alphanum_fraction": 0.833545108, "num_tokens": 181, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117769928211, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6470362929517488}} {"text": "\\subsubsection{Hyperboloids of One Sheet}\r\n\\noindent\r\nA hyperboloid of one sheet has 2 +'s and 1 - in its equation. It is one fully connected surface.\r\n\r\n\\begin{figure}[H]\r\n\t\\centering\r\n\t\\includegraphics[width=0.33\\textwidth]{./differentialMultivariableCalculus/one_sheet.png}\r\n\t\\caption{A hyperboloid of one sheet}\r\n\\end{figure}", "meta": {"hexsha": "b459ffc3eb14869d7a9639a83d36303cc8040d2d", "size": 329, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "multiCalc/differentialMultivariableCalculus/hyperboloidOneSheet.tex", "max_stars_repo_name": "wmboyles/Math-Summaries", "max_stars_repo_head_hexsha": "94732081a5b6913d84e11c62a3989b63f9934b56", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2020-03-26T06:20:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T17:38:45.000Z", "max_issues_repo_path": "multiCalc/differentialMultivariableCalculus/hyperboloidOneSheet.tex", "max_issues_repo_name": "wmboyles/Math-Summaries", "max_issues_repo_head_hexsha": "94732081a5b6913d84e11c62a3989b63f9934b56", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 26, "max_issues_repo_issues_event_min_datetime": "2020-03-28T17:44:18.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T04:47:03.000Z", "max_forks_repo_path": "multiCalc/differentialMultivariableCalculus/hyperboloidOneSheet.tex", "max_forks_repo_name": "wmboyles/Math-Summaries", "max_forks_repo_head_hexsha": "94732081a5b6913d84e11c62a3989b63f9934b56", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2020-04-10T05:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-17T15:21:12.000Z", "avg_line_length": 36.5555555556, "max_line_length": 97, "alphanum_fraction": 0.7629179331, "num_tokens": 98, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.8006920068519376, "lm_q1q2_score": 0.6470129588908152}} {"text": "\\chapter{Modal analysis}\\label{chap:modalanalysis}\nWhen external forces act on a multi degree-of-freedom system, the system\nundergoes forced vibration.\nFor a system with three coordinates or degrees of freedom, the governing\nequation of motion are a set of three coupled ordinary differential equations of\nsecond second order.\nThe equations of motion of a multi degree-of-freedom system under external force\nare given by:\n\\begin{equation}\n \\label{eq:modal}\n [\\mathbf{M}] \\ddot{\\vec{x}} + [\\mathbf{K}] \\vec{x} = \\vec{\\mathbf{F}}\n\\end{equation}\nThe equation \\eqref{eq:modal} represents the system in the version without\ndamping, where \\(\\vec{\\mathbf{F}}\\) is the vector of the external force.\n\\section{Forced vibration of undamped system}\\label{sec:eigen}\nTo solve the equation \\eqref{eq:modal} by modal analysis, it is necessary first\nto solve the eigenvalue problem and find the natural frequencies \\(\\omega_1,\n\\dots, \\omega_n\\) as reported below:\n\\(\\omega_{1} = 8.27843\\) [\\si{\\radian\\per\\second}],\n\\(\\omega_{2} = 27.40935\\) [\\si{\\radian\\per\\second}],\n\\(\\omega_{3} = 41.87035\\) [\\si{\\radian\\per\\second}];\nand the corresponding normal modes:\n\\begin{equation}\n\t\\label{eq:eiegnmodefree}\n \t[\\mathbf{U}] = \\begin{bmatrix*}[r]\n \t\t1.00000 & 1.00000 & 1.00000 \\\\\n\t\t0.86540 &-0.47555 &-2.44325 \\\\\n\t\t0.61945 &-1.28034 & 2.15533 \\\\\n \t\\end{bmatrix*}\n\\end{equation}\n\\section{Forced vibration of viscous damped system}\\label{sec:viscoussystem}\nModal analysis, as presented in previous section \\ref{sec:eigen}, applies only\nto undamped systems.\nIn many cases, the influence of damping upon the response of a vibratory system\nis minor and can be disregarded. However, it must be considered if the response\nof the system is required for a relatively long period of time compared to the\nnatural periods of the system. Further, if the frequency of excitation is at or\nnear one of the natural frequencies of the system, damping is of primary\nimportance and must be taken into account. In general, since the effects are not\nknown in advance, damping must be considered in the vibration analysis of any\nsystem, thus using the equation of motion:\n\\begin{equation}\\label{eq:modaldamping}\n [\\mathbf{M}] \\ddot{\\vec{x}} + [\\mathbf{C}] \\dot{\\vec{x}} +[\\mathbf{K}] \\vec{x}\n = \\vec{\\mathbf{F}}\n\\end{equation}\n%\nThen we can consider for which the damping matrix can be expressed as a linear\ncombination of the mass and stiffness matrices:\n\\begin{equation}\n\\label{eq:moadlpropdamping}\n\t[C] = \\alpha \\cdot [\\mathbf{M}] + \\beta \\cdot [\\mathbf{K}]\n\\end{equation}\nwhere \\(\\alpha\\) and \\(\\beta\\) are constant. By substituting in\n\\eqref{eq:modaldamping}, we obtain:\n\\begin{equation}\\label{eq:modaldampingcomplete}\n\t[\\mathbf{M}] \\ddot{\\vec{x}} +\n\t[\\alpha [\\mathbf{M}] + \\beta[\\mathbf{K}]] \\dot{\\vec{x}} +\n\t[\\mathbf{K}] \\vec{x}= \\vec{\\mathbf{F}}\n\\end{equation}\nBy expressing the solution vector \\(\\vec{x}\\) as linear combination of the\nnatural mode of the undamped system. Substituting \\(x(t) = [\\mathbf{U}]q(t)\\)\nin the equation \\eqref{eq:modaldampingcomplete} and pre-multiplying by \n\\([\\mathbf{U}]^{\\top}\\) we obtain:\n\\begin{equation}\\label{eq:modaldampingexpansed}\n\t[\\mathbf{U}]^{\\top}[\\mathbf{M}][\\mathbf{U}] \\ddot{\\vec{q}} +\n\t[\\alpha[\\mathbf{U}]^{\\top}[\\mathbf{M}][\\mathbf{U}] +\n \\beta[\\mathbf{U}]^{\\top}[\\mathbf{K}]][\\mathbf{U}] \\dot{\\vec{q}} +\n\t[\\mathbf{U}]^{\\top}[\\mathbf{K}][\\mathbf{U}]^{\\top} \\vec{q} =\n [\\mathbf{U}]^{\\top}\\vec{\\mathbf{F}}\n\\end{equation}\n%\nIf the eigenvectors \\(\\vec{\\mathbf{U}}^{(j)}\\) are normalized we can rewrite in \ndecoupled form:\n\\begin{equation}\\label{eq:decupled}\n \\ddot{q}_{i}(t) + (\\alpha + \\omega_{i}^2\\beta)\\dot{q}_{i}(t) +\n \\omega_{i}^{2}q(t) = Q_{i}(t)\n\\end{equation}\nwhere: \\(\\omega_{1} = 8.27799\\) [\\si{\\radian\\per\\second}],\n\\(\\omega_{2} = 27.40863\\) [\\si{\\radian\\per\\second}],\n\\(\\omega_{3} = 41.91576\\) [\\si{\\radian\\per\\second}] are\nthe natural frequencies of the undamped system.\nThe eigenvector are reported in \\eqref{eq:eiegnmodeprop}.\n\\begin{equation}\\label{eq:eiegnmodeprop}\n\t[\\mathbf{U}] = \\begin{bmatrix*}[r]\n\t\t1.00000 & 1.00000 & 1.00000 \\\\\n\t\t0.86499 &-0.48006 &-2.46147 \\\\\n\t\t0.61907 &-1.28528 & 2.16954 \\\\\n \t\\end{bmatrix*}\n\\end{equation}\nIn the equation \\eqref{eq:decupled} substituting the relationship with \n\\(\\alpha + \\omega^{2}_{i}\\beta = 2\\,\\zeta_{i}\\,\\omega_{i}\\), where\n\\(\\zeta_{i}\\) is called \\emph{modal damping ratio} for the \\(i_\\text{th}\\) normal\nmode, equation \\eqref{eq:decupled} we obtain:\n\\begin{equation}\\label{eq:decupled2}\n \\ddot{q}_{i}(t) + 2\\zeta_{i}\\omega_{i}\\dot{q}_{i}(t) +\n \\omega_{i}^{2}q(t) = Q_{i}(t)\n\\end{equation}\nIt can be seen that each of then equations represented by this expression is\nuncoupled from all of the others.\nHence we can find the response of the \\(i_\\text{th}\\) mode in the same manner\nas that of a viscously damped single-degree-of-freedom system.\n%\nThe identification of the sources and magnitude of damping is difficult in most\npractical problems.\nThe condition given by equation \\eqref{eq:moadlpropdamping} is sufficient but not\nnecessary for the existence of normal modes in damped systems.\nThe necessary condition is that the transformation that diagonalizes the damping\nmatrix also uncouples the coupled equations of motion.\nThis condition is less restrictive than equation \\eqref{eq:moadlpropdamping}\nand covers more possibilities.\nIn the general case of damping, the damping matrix cannot be diagonalized\nsimultaneously with the mass and stiffness matrices.\nA common procedure for finding the solution of the eigenvalue problem of a\ndamped system involves the transformation of the coupled second-order\nequations of motion into \\(2n\\) uncoupled first-order equation.\n%\n\\input{rayleigh}\n\\input{matrixiterationmethod}\n", "meta": {"hexsha": "c0c972950924025cb7b7c9d778428d9c9b6debb7", "size": 5700, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/Modal.tex", "max_stars_repo_name": "frank1789/MechanicalVibrationProject", "max_stars_repo_head_hexsha": "ad28e4c047fe4f806fa1fb5405b2304699e377be", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-11-28T12:59:41.000Z", "max_stars_repo_stars_event_max_datetime": "2019-11-28T12:59:41.000Z", "max_issues_repo_path": "Report/Modal.tex", "max_issues_repo_name": "frank1789/MechanicalVibrationProject", "max_issues_repo_head_hexsha": "ad28e4c047fe4f806fa1fb5405b2304699e377be", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/Modal.tex", "max_forks_repo_name": "frank1789/MechanicalVibrationProject", "max_forks_repo_head_hexsha": "ad28e4c047fe4f806fa1fb5405b2304699e377be", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.8991596639, "max_line_length": 81, "alphanum_fraction": 0.7222807018, "num_tokens": 1826, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672089305841, "lm_q2_score": 0.8006919949619793, "lm_q1q2_score": 0.647012945581988}} {"text": "\\documentclass{article}\n\n\\usepackage{multicol}\n\\usepackage{caption}\n\\usepackage{amsmath}\n\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\usepackage{hyperref}\n\\hypersetup{\n colorlinks=true,\n linkcolor=blue,\n filecolor=magenta,\n urlcolor=cyan,\n}\n\n\\usepackage{geometry}\n\\geometry{\n\ta3paper,\n\tnoheadfoot=true,\n\tleft=1.0in,\n\tright=1.0in,\n\ttop=1.0in,\n\tbottom=1.0in\n}\n\\usetikzlibrary{decorations.pathreplacing}\n\\usepgfplotslibrary{external}\n\n\\definecolor{myLightGray}{RGB}{191,191,191}\n\\definecolor{myGray}{RGB}{160,160,160}\n\\definecolor{myDarkGray}{RGB}{144,144,144}\n\\definecolor{myDarkRed}{RGB}{167,114,115}\n\\definecolor{myRed}{RGB}{255,58,70}\n\\definecolor{myGreen}{RGB}{0,255,71}\n\n% Titling and Author\n\\title{Latex Tikz Examples, Linear Equation Plotting, Linear Spline}\n\\author{\\href{https://fanwangecon.github.io/}{Fan Wang}\\thanks{https://fanwangecon.github.io, repository: \\href{https://fanwangecon.github.io/Tex4Econ/}{Tex4Econ}}}\n\\date{\\today}\n\n\\begin{document}\n\n\\maketitle\n\n%%%%%%%%%%%%%%%%%%%%%\n%%% Line Values\n%%%%%%%%%%%%%%%%%%%%%\n% A. slopes of lines a b c d\n\\def\\ssa{0.8}\n\\def\\ssb{1.0}\n\\def\\ssc{2}\n\\def\\ssd{0.6}\n% B. y-intercepts of lines a b c d\n\\def\\iia{+1.0}\n\\def\\iib{+0.0}\n\\def\\iic{-1.0}\n\\def\\iid{-1.5}\n% Other parameters\n\\def\\rescale{4}\n\n%%%%%%%%%%%%%%%%%%%%%\n%%% Line Formatting\n%%%%%%%%%%%%%%%%%%%%%\n% A. color definitions\n\\def\\cla{blue}\n\\def\\clb{black}\n\\def\\clc{red}\n\\def\\cld{green}\n\n%%%%%%%%%%%%%%%%%%%%%\n%%% Draw Lines\n%%%%%%%%%%%%%%%%%%%%%\n\\begin{multicols}{3}\n[\n\\section{Tikz Four Lines}\nDraw four lines, sum them up linear spline, then invert. A3 page has 29.7 cm width. Below, width is 7 cm for each plot, so 21 cm all together generally. Margins are 1 inch on each side, so about 5 cm together.\n]\n\n\\subsection{Four Relative Allocation Lines}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N_m$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N_{m^{\\prime}}$};\n \\draw[line width=0.25mm,domain=-2:4,smooth,variable=\\x, \\cla] plot ({\\x},{\\iia+\\x*\\ssa});\n \\draw[line width=0.25mm,domain=-2:4,smooth,variable=\\x, \\clb] plot ({\\x},{\\iib+\\x*\\ssb});\n \\draw[line width=0.25mm,domain=-2:4,smooth,variable=\\x, \\clc] plot ({\\x},{\\iic+\\x*\\ssc});\n \\draw[line width=0.25mm,domain=-2:4,smooth,variable=\\x, \\cld] plot ({\\x},{\\iid+\\x*\\ssd});\n\\end{tikzpicture}\n\n\\subsection{Inequality Constrained $Y > 0$}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N_m$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N_{m^{\\prime}}$};\n \\draw[line width=0.25mm,domain=(-\\iia/\\ssa):4,smooth,variable=\\x, \\cla] plot ({\\x},{\\iia+\\x*\\ssa});\n \\draw[line width=0.25mm,domain=(-\\iib/\\ssb):4,smooth,variable=\\x, \\clb] plot ({\\x},{\\iib+\\x*\\ssb});\n \\draw[line width=0.25mm,domain=(-\\iic/\\ssc):4,smooth,variable=\\x, \\clc] plot ({\\x},{\\iic+\\x*\\ssc});\n \\draw[line width=0.25mm,domain=(-\\iid/\\ssd):4,smooth,variable=\\x, \\cld] plot ({\\x},{\\iid+\\x*\\ssd});\n\\end{tikzpicture}\n\n\\subsection{Sum $Y > 0$}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N_m$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N$};\n \\draw[line width=0.25mm,domain=(-\\iia/\\ssa):4,smooth,variable=\\x, \\cla]\n plot ({\\x},{(\\iia+\\x*\\ssa)});\n \\draw[line width=0.25mm,domain=(-\\iib/\\ssb):4,smooth,variable=\\x, \\clb]\n plot ({\\x},{((\\iia+\\iib)+\\x*(\\ssa+\\ssb))});\n \\draw[line width=0.25mm,domain=(-\\iic/\\ssc):4,smooth,variable=\\x, \\clc]\n plot ({\\x},{((\\iia+\\iib+\\iic)+\\x*(\\ssa+\\ssb+\\ssc))});\n \\draw[line width=0.25mm,domain=(-\\iid/\\ssd):4,smooth,variable=\\x, \\cld]\n plot ({\\x},{((\\iia+\\iib+\\iic+\\iid)+\\x*(\\ssa+\\ssb+\\ssc+\\ssd))});\n\\end{tikzpicture}\n\n\\subsection{Sum $Y > 0$ Rescale}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N_m$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N$};\n \\draw[line width=0.25mm,domain=(-\\iia/\\ssa):4,smooth,variable=\\x, \\cla]\n plot ({\\x},{(\\iia+\\x*\\ssa)/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iib/\\ssb):4,smooth,variable=\\x, \\clb]\n plot ({\\x},{((\\iia+\\iib)+\\x*(\\ssa+\\ssb))/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iic/\\ssc):4,smooth,variable=\\x, \\clc]\n plot ({\\x},{((\\iia+\\iib+\\iic)+\\x*(\\ssa+\\ssb+\\ssc))/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iid/\\ssd):4,smooth,variable=\\x, \\cld]\n plot ({\\x},{((\\iia+\\iib+\\iic+\\iid)+\\x*(\\ssa+\\ssb+\\ssc+\\ssd))/\\rescale});\n\\end{tikzpicture}\n\n\\subsection{$Y > 0$--Linear Spline}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N_m$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N$};\n \\draw[line width=0.25mm,domain=(-\\iia/\\ssa):(-\\iib/\\ssb),smooth,variable=\\x, \\cla]\n plot ({\\x},{(\\iia+\\x*\\ssa)/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iib/\\ssb):(-\\iic/\\ssc),smooth,variable=\\x, \\clb]\n plot ({\\x},{((\\iia+\\iib)+\\x*(\\ssa+\\ssb))/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iic/\\ssc):(-\\iid/\\ssd),smooth,variable=\\x, \\clc]\n plot ({\\x},{((\\iia+\\iib+\\iic)+\\x*(\\ssa+\\ssb+\\ssc))/\\rescale});\n \\draw[line width=0.25mm,domain=(-\\iid/\\ssd):4,smooth,variable=\\x, \\cld]\n plot ({\\x},{((\\iia+\\iib+\\iic+\\iid)+\\x*(\\ssa+\\ssb+\\ssc+\\ssd))/\\rescale});\n\\end{tikzpicture}\n\n% (((\\iia+\\iib)+((-\\iib/\\ssb))*(\\ssa+\\ssb))/\\rescale)\n\\subsection{Summed Lines Invert}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N_m$};\n \\clip (-3,-3) rectangle (7,4);\n \\draw[line width=0.25mm,\n domain=((\\iia+(-\\iia/\\ssa)*\\ssa)/\\rescale):4,\n smooth,variable=\\x, \\cla]\n plot ({\\x},{ ((\\x*\\rescale-\\iia)/\\ssa)});\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib)+(-\\iib/\\ssb)*(\\ssa+\\ssb))/\\rescale):4,\n smooth,variable=\\x, \\clb]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib))/(\\ssa+\\ssb)) });\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib+\\iic)+(-\\iic/\\ssc)*(\\ssa+\\ssb+\\ssc))/\\rescale):4,\n smooth,variable=\\x, \\clc]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib+\\iic))/(\\ssa+\\ssb+\\ssc)) });\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib+\\iic+\\iid)+(-\\iid/\\ssd)*(\\ssa+\\ssb+\\ssc+\\ssd))/\\rescale):4,\n smooth,variable=\\x, \\cld]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib+\\iic+\\iid))/(\\ssa+\\ssb+\\ssc+\\ssd)) });\n\\end{tikzpicture}\n\n\\subsection{Spline Invert}\n\\begin{tikzpicture}\n \\draw[->] (-3,0) -- (4,0) node[right] {$N$};\n \\draw[->] (0,-3) -- (0,4) node[above] {$N_m$};\n \\clip (-3,-3) rectangle (7,4);\n \\draw[line width=0.25mm,\n domain=((\\iia+(-\\iia/\\ssa)*\\ssa)/\\rescale):(((\\iia+\\iib)+(-\\iib/\\ssb)*(\\ssa+\\ssb))/\\rescale),\n smooth,variable=\\x, \\cla]\n plot ({\\x},{ ((\\x*\\rescale-\\iia)/\\ssa)});\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib)+(-\\iib/\\ssb)*(\\ssa+\\ssb))/\\rescale):(((\\iia+\\iib+\\iic)+(-\\iic/\\ssc)*(\\ssa+\\ssb+\\ssc))/\\rescale),\n smooth,variable=\\x, \\clb]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib))/(\\ssa+\\ssb)) });\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib+\\iic)+(-\\iic/\\ssc)*(\\ssa+\\ssb+\\ssc))/\\rescale):(((\\iia+\\iib+\\iic+\\iid)+(-\\iid/\\ssd)*(\\ssa+\\ssb+\\ssc+\\ssd))/\\rescale),\n smooth,variable=\\x, \\clc]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib+\\iic))/(\\ssa+\\ssb+\\ssc)) });\n \\draw[line width=0.25mm,\n domain=(((\\iia+\\iib+\\iic+\\iid)+(-\\iid/\\ssd)*(\\ssa+\\ssb+\\ssc+\\ssd))/\\rescale):4,\n smooth,variable=\\x, \\cld]\n plot ({\\x},{ ((\\x*\\rescale-(\\iia+\\iib+\\iic+\\iid))/(\\ssa+\\ssb+\\ssc+\\ssd)) });\n\\end{tikzpicture}\n\n\n\\end{multicols}\n\n\\end{document}\n", "meta": {"hexsha": "c9b3b9f594f7f5eecbfae148ac417222784e6048", "size": 7473, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "_other/tikz/fplot/functionplot.tex", "max_stars_repo_name": "guohui-jiang/Tex4Econ", "max_stars_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "_other/tikz/fplot/functionplot.tex", "max_issues_repo_name": "guohui-jiang/Tex4Econ", "max_issues_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "_other/tikz/fplot/functionplot.tex", "max_forks_repo_name": "guohui-jiang/Tex4Econ", "max_forks_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.1256544503, "max_line_length": 209, "alphanum_fraction": 0.5577412017, "num_tokens": 3004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972751232809, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.6470000531613432}} {"text": "\\documentclass[12pt, a4paper]{article}\n\\usepackage[margin=1.5in]{geometry}\n\\usepackage{graphicx}\n\\usepackage{amsmath}\n\\usepackage{float}\n\\usepackage{listings}\n\\usepackage[shortlabels]{enumitem}\n\n\n\\setlength\\parindent{0pt}\n\\newcommand{\\code}{\\lstinline[basicstyle=\\small]}\n\\lstset{\n language=Python,\n basicstyle=\\scriptsize\n}\n\n\n\\title{EE2703: Applied Programming Lab \\\\ \\Large Assignment 4: Fourier Approximations}\n\\author{Soham Roy \\\\ \\normalsize EE20B130}\n\\date{\\today}\n\\begin{document}\n\n\\maketitle % Insert the title, author and date\n\n\n\\section{Introduction}\nTwo functions, $exp(x)$ and $cos(cos(x))$ over the interval $[0,2\\pi)$ will be modeled using\nthe fourier series:\n\\begin{equation*}\n a_0 + \\sum_{n=1}^{\\infty} \\{a_n \\cos(nx) + b_n \\sin(nx)\\}\n\\end{equation*}\nWhere,\n\\begin{align*}\n a_0 & = \\frac{1}{2\\pi} \\int_{0}^{2\\pi} f(x) dx \\\\\n a_n & = \\frac{1}{\\pi} \\int_{0}^{2\\pi} f(x)cos(nx) dx \\\\\n a_n & = \\frac{1}{\\pi} \\int_{0}^{2\\pi} f(x)sin(nx) dx \\\\\n\\end{align*}\n\\pagebreak\n\n\n\\section{Subquestions}\n\\subsection{Define \\& Plot Python Functions}\nTwo Python functions are defined:\n\n% \\par\\rule{\\linewidth}{0.4pt}\n\\begin{lstlisting}\n def exp(x): # exponential function, supports value or vector\n return np.exp(x)\n\n def coscos(x): # cos of cos function, supports value or vector\n return np.cos(np.cos(x))\n\\end{lstlisting}\n% \\par\\rule[1.5ex]{\\linewidth}{0.4pt}\n\nThe functions have been plotted with their periodic extensions ($2\\pi$ period):\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{1a.png}\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{1b.png}\n\\end{figure}\n\\pagebreak\n\n\n\\subsection{Evaluate Integrals}\nThe integrands have been defined as functions of \\code{x}, \\code{k}, and \\code{f}, where\n\\code{f} is the Python function to be approximated, i.e. \\code{exp} or \\code{coscos}:\n\\begin{lstlisting}\n def u(x, k, f): # f is either exp or coscos\n return f(x) * np.cos(k * x)\n\n def v(x, k, f): # f is either exp or coscos\n return f(x) * np.sin(k * x)\n\\end{lstlisting}\nThe integrals to calculate the values of $a_n$ and $b_n$ have been evaluated using the following loop:\n\\begin{lstlisting}\n F = [exp, coscos]\n a_0 = np.zeros((2))\n a_n = np.zeros((2, 25))\n b_n = np.zeros((2, 25))\n\n for i in range(2): # iterate over exp and coscos\n a_0[i] = quad(F[i], 0, 2 * np.pi)[0] / (2 * np.pi)\n\n for j in range(25): # integration\n a_n[i, j] = quad(u, 0, 2 * np.pi, args=(j + 1, F[i]))[0] / np.pi\n b_n[i, j] = quad(v, 0, 2 * np.pi, args=(j + 1, F[i]))[0] / np.pi\n\\end{lstlisting}\n\n\n\\subsection{Plot the Fourier Coefficients}\nThe answer vector $c$ has been generated by:\n\\begin{lstlisting}\n c_n = np.zeros((2, 51)) # c_n[0] and c_n[1] are the answer vectors\n c_n[:, 0] = a_0\n c_n[:, 1::2] = a_n # alternate elements starting at 1\n c_n[:, 2::2] = b_n # alternate elements starting at 2 \n\\end{lstlisting}\n\nTo plot in semilog scale, for example, we use:\n\\begin{lstlisting}\n plt.semilogy(0, np.abs(a_0[f_i]), 'ro', label=\"by integration\")\n plt.semilogy(range(1, 26), np.abs(a_n[f_i]), 'ro')\n plt.semilogy(range(1, 26), np.abs(b_n[f_i]), 'ro')\n\\end{lstlisting}\n\nThe coefficients $|a_n|$ and $|b_n|$ have been plotted:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.39]{3a.png}\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.39]{3b.png}\n\\end{figure}\n\n\n\\begin{enumerate}[(a)]\n \\item The $b_n$ coefficients for $cos(cos(x))$ are of the order of magnitude $-16$, i.e. nearly zero. \\\\\n This is because $cos(cos(x))$ is an even function, and thus does not have any odd component.\n \\item $cos(cos(x))$ is a periodic function, comprising not many frequencies. \\\\\n $e^x$, on the other hand, is a non-periodic function, and thus its periodic extension has multiple\n discontinuities. Hence, high frequency components are required to represent this function as a sum\n of trigonometric functions.\n \\item The $loglog$ plot is linear for $e^x$ because the $log$ of its coefficients vary linearly with\n $log(n)$, i.e. the coefficients depend on $\\frac{1}{n^a}$ for some $a$. \\\\\n On the other hand, the $semilog$ plot is linear for $cos(cos(x))$ because the $log$ of its\n coefficients vary linearly with $n$, i.e. the coefficients decay exponentially.\n\\end{enumerate}\n\n\n\\subsection{Least Squares}\nThe equation to be solved by \\code{scipy.linalg.lstsq} is:\n\\begin{equation*}\n Ac = B\n\\end{equation*}\n\nThe matrix $A$ has been generated by:\n\\begin{lstlisting}\n A = np.ones((400, 51)) # first column should be ones\n for k in range(1, 26):\n A[:, 2 * k - 1] = np.cos(k * x)\n A[:, 2 * k] = np.sin(k * x)\n\\end{lstlisting}\n\nThe solution matrices are \\code{c[0]} for $e^x$ and\n\\code{c[1]} for $coscos(x)$:\n\\begin{lstlisting}\n for i in range(2):\n b = F[i](x)\n c[i] = lstsq(A, b)[0]\n\\end{lstlisting}\n\n\n\\subsection{Plot the Best Fit Coefficients}\nTo plot in semilog scale, for example, we use:\n\\begin{lstlisting}\n plt.semilogy(0, np.abs(c[f_i][0]), 'go', label=\"by least squares\")\n plt.semilogy(range(1, 26), np.abs(c[f_i][1::2]), 'go')\n plt.semilogy(range(1, 26), np.abs(c[f_i][2::2]), 'go')\n\\end{lstlisting}\n\nThe coefficients $|a_n|$ and $|b_n|$ by least squares method have been contrasted with\nthose by integration:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.39]{5a.png}\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.39]{5b.png}\n\\end{figure}\n\n\n\\subsection{Compare Least Squares and Direct Integration}\nThe maximum errors between the least squares and direct integration methods has been calculated:\n\\begin{lstlisting}\n np.max(np.abs(c[0] - c_n[0])) # for e^x\n np.max(np.abs(c[1] - c_n[1])) # for coscos(x)\n\\end{lstlisting}\n\n\\begin{tabular}{lcl}\n The largest deviation of coefficients for & $e^x$ & is : 1.3327 \\\\\n The largest deviation of coefficients for & $cos(cos(x))$ & is : 2.6684e-15\n\\end{tabular}\n\\medskip\n\nThe error is significant for $e^x$ but negligible for $cos(cos(x))$. \\\\\nThis is because there are multiple discontinuities in the periodic extension of $e^x$, and thus\nwould require a much higher number of coefficients to be somewhat accurately represented. \\\\\nThis lack in accuracy is more apparent close to those discontinuities.\n\n\\subsection{Plot the Fourier Approximations}\nThe functions as represented by the Fourier coefficients calculated through the least squares method\nhave been plotted:\n\\begin{lstlisting}\n plt.plot(x[200:600], np.dot(A, c[1]), 'go', label=\"by least squares\")\n\\end{lstlisting}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{7a.png}\n\\end{figure}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{7b.png}\n\\end{figure}\n\nThe $cos(cos(x))$ plot agrees nearly perfectly, but the $e^x$ plot has a large deviation.\nThis is because unlike $cos(cos(x))$, which has periodic components, the $e^x$ function\nis not inherently comprised of periodic trigonometric functions. The periodic extension of $e^x$\nhas multiple discontinuities, which causes the Fourier representation to overshoot around them\ndue to the Gibbs phenomenon.\n\n\n\n\\end{document}\n", "meta": {"hexsha": "bb74aed1eb41ae221a2fa7ff24bdd987dcc32e29", "size": 7291, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment_04/LaTeX/Report.tex", "max_stars_repo_name": "sohamroy19/EE2703", "max_stars_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment_04/LaTeX/Report.tex", "max_issues_repo_name": "sohamroy19/EE2703", "max_issues_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment_04/LaTeX/Report.tex", "max_forks_repo_name": "sohamroy19/EE2703", "max_forks_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.2922374429, "max_line_length": 108, "alphanum_fraction": 0.6689068715, "num_tokens": 2294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.6470000457519778}} {"text": "\\input{coredef.tex}\n\n%opening\n\\def\\ntitle{Principles of Quantum Mechanics}\n\\def\\nauthor{mostanes}\n\\def\\npart{II}\n\\def\\nterm{Michelmas}\n\\def\\nyear{2018}\n\\def\\nlecturer{Skinner}\n\n\\renewcommand*{\\H}{\\mathcal{H}}\n\n\\begin{document}\n\t\n\\mktitlepage\n\t\n\\newpage\n\n\\setcounter{section}{-1}\n\n\\section{Preface}\nThis course draws heavily from material in IB, with references to Linear Algebra, Analysis II, Methods; as well as some references to IB Quantum Mechanics. Note however that this course is likely to prove a lot of the intuitive notions from IB QM, so it should be accessible even to those who were left baffled by IB QM. The author of these notes will make generous references to IB material without restating it, unlike the course lecturer. References to II material will also be made where appropriate.\n\n\\newpage\n\n\\section{Introduction}\n\\subsection{Comparison of Classical and Quantum Mechanics}\n\\subsubsection{Classical Mechanics}\nClassical mechanics are governed by Newton's laws, which are 2nd order differential equations in the variables $\\vec{x}$, $\\vec{p}$. By combining the 2 variables as in classical dynamics, we obtain the \\textbf{phase space}, which in Newtonian dynamics is $\\R^{2n}$, with the particular case $n=3$ our universe.\\\\\nIn classical mechanics, the observables are simple quantities, represented by functions $\\R^{2n} \\rightarrow \\R$.\n\n\\subsubsection{Quantum Mechanics}\nParticles are instead represented by points in a Hilbert space (equivalent of the phase space).\\\\\nObservables are represented by linear operators $\\H \\rightarrow \\H$.\n\n\\subsection{Hilbert spaces}\n\\begin{remark}\n\tThere is an entire chapter of linear analysis dedicated to Hilbert spaces.\n\\end{remark}\n\n\\begin{definition}\n\tA Hilbert space $\\H$ is a vector space (over $\\C$) with a complete inner product $(\\cdot,\\cdot) : \\H \\times \\H \\rightarrow \\C$.\n\\end{definition}\nTherefore Hilbert spaces satisfy the usual vector space and complex inner product properties and any Cauchy sequence converges to a vector within the space under the norm induced by the inner product.\n\\subsubsection{Examples}\nEvery finite dimensional Hilbert space (of dimension $n$) is isomorphic to $\\C^n$ (see Linear Analysis).\\\\\nA simple $\\infty$-dimensional space is $l^2$, the space of infinite sequences (converging under the 2-norm; see Analysis II).\\\\\nAnother example is $L^2$, the function space of Lebesgue integrable functions (integrals converging under the 2-norm; see Probability and Measure). Their inner product is defined analogously to the inner product of $l^2$ and the norm of $L^2$.\n\\subsection{Dual spaces}\nThe dual $\\H^\\star$ of a Hilbert space $\\H$ is space of linear operators $\\H \\rightarrow \\C$. One way of obtaining such operators is by considering the inner product $(\\chi,\\cdot) : \\H \\rightarrow \\C$.\\\\\nBy the Riesz representation theorem (see Linear Analysis), all elements in $\\H^\\star$ can be written in inner product form.\n\\subsection{Dirac notation}\nDirac invented the following notation which is standard today in QM:\\\\\n\\begin{tabular}{ccc}\n\tElements of $\\H$ & written $\\ket{\\cdot}$ & called 'ket'\\\\\n\tElements of $\\H^\\star$ & written $\\bra{\\cdot}$ & called 'bra'\\\\\n\tInner product & written $\\bra{\\cdot}\\ket{\\cdot}$ & called 'braket'\\\\\n\\end{tabular}\n\n\\end{document}", "meta": {"hexsha": "9fdb45ea198a6622200fedf8d5b58a42e4e2ee33", "size": 3261, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pqm.tex", "max_stars_repo_name": "mostanes/cam-maths-tripos", "max_stars_repo_head_hexsha": "bf40a5250be7c15037a937e1521c5aa7e0cb0841", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "pqm.tex", "max_issues_repo_name": "mostanes/cam-maths-tripos", "max_issues_repo_head_hexsha": "bf40a5250be7c15037a937e1521c5aa7e0cb0841", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pqm.tex", "max_forks_repo_name": "mostanes/cam-maths-tripos", "max_forks_repo_head_hexsha": "bf40a5250be7c15037a937e1521c5aa7e0cb0841", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 54.35, "max_line_length": 504, "alphanum_fraction": 0.7614228764, "num_tokens": 835, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8887588023318196, "lm_q2_score": 0.7279754489059774, "lm_q1q2_score": 0.6469945880966452}} {"text": "%!TEX root = ../main.tex\n\n\\subsection{Anti-Derivative}\n\n\\objective{Distinguish and find anti-derivatives and integrals of functions}\n\n\nSuppose we are given a formula and are told it is the derivative of what we want.\nThis isn't as abstract as it sounds: velocity is the derivative of position, and (at least in \nmany cars) it is easier to record velocity than it is position. If the velocity\nfunction is an algebraic equation, we simply need to apply the Power Rule\nin reverse and we will have the anti-derivative of the equation.\n\nThe Power Rule states that the derivative of $x^n$ is $n\\cdot x^{n-1}$. In other words,\n``take the exponent out front, and lower the exponent by one''. If we wanted to\nturn this backwards, in order to arrive at an exponent of $x^n$, we must have \nbegun at $x^{n+1}$. However, if we take the derivative of $x^{n+1}$, we must\nmultiply by $n+1$. To cancel that, we should multiply by $\\frac{1}{n+1}$.\n\n\n\\begin{derivation}{Backwards Power Rule}\\index{Power Rule!Backwards}\nThe anti-derivative of $x^n$ is $\\frac{1}{n+1} x^{n+1} + C$, where $C$ is an unknown\nconstant.\n\\end{derivation}\n\n\nWhat is C? Consider whether of not $x^2 +1$ is the anti-derivative of $\\frac{1}{2}x$.\nIs $x^2-1$? Is $x^2+\\pi$? Because constants differentiate to 0, a constant could be \npart of our anti-derivative equation and we cannot know what it is, without more information.\nWhat information? Well, if we know the initial conditions (when $x=0$) then we can \nsolve for $C$ and know precisely which anti-derivative equation we want.\n\n\n\\subsection{Integral}\\index{integral!definite}\nThe preceding definition of anti-derivatives is very helpful algebraically, but what about\ngraphically? If a given graph is the derivative of what we seek, can we interpret the\ngraph to give us numerical information?\n\nConsider the graph of a car with constant velocity:\n\n\\begin{figure}[h]\n\\begin{centering}\n\\begin{tikzpicture}[scale=0.6]\n\t\\draw[help lines] (0,0) grid (8,6);\n\t\\draw (-0.5,0) -- (8.5,0);\n\t\\draw (0,-0.5) -- (0,6.5);\n\t\\draw[thick,->] (0,4) node[anchor=east] {40 mph}-- (8.5,4);\n\t\\draw (8,0) node[anchor=north] {0.8 hr};\n\t\\draw [fill=gray!80,opacity =0.4] (0,0) rectangle (8,4);\n\\end{tikzpicture}\n\\caption{A car's velocity in 10's of mph, over tenths of an hour}\n\\end{centering}\n\\end{figure}\n\n\nIf we want position or distance, we have known a formula for a long time:\ndistance = rate $\\cdot$ time. The $y$-value is the rate. The $x$-value is the \ntime. As hard as it may be to conceive of, distance is the \\emph{area} under the\ngraph. In this case, 40 mph times 0.8 hours is 32 miles. Notice that this number does\nnot depend upon the initial position: the car has travelled 32 positive miles, regardless \nof where it began.\n\n\\begin{equation}\n\\int_0^{0.8} 40dx = \\left.40x \\right|_0^{0.8} = 40(.8) - 40(0) = 32\n\\end{equation}\n\nWhat about more complicated velocity? Let us begin with constant acceleration:\n\n\\begin{figure}[h]\n\\begin{centering}\n\\begin{tikzpicture}[scale=0.7]\n\t\\draw[help lines] (0,-2) grid (6,6);\n\t\\draw (-0.5,0) -- (6.5,0);\n\t\\draw (0,-2.5) -- (0,6.5);\n\t\\draw [thick,->] (0,-2) -- (6.5,4.5);\n\t\\draw [fill=red!80,opacity=0.4] (0,-2) -- (2,0) -- (0,0);\n\t\\draw [fill=gray!80,opacity=0.4] (2,0) -- (6,0) -- (6,4);\n\\end{tikzpicture}\n\\caption{A car beginning at -20 mph but steadily accelerating to 40 mph by 0.6 hours later}\n\\end{centering}\n\\end{figure}\n\nThe object begins with a negative velocity, so we must count that distance as negative.\n-2 + 8 = 6, so the object has gained that many units in positive displacement from 0 to 6.\n\nIn some problems, we can simply count the squares below the graph and find the\ndefinite integral. In most cases, the graph will be curved and we will need to find an\nanti-derivative equation and subtract the evaluation at the left from that of the right.\n\nFinally, notice that we can integrate functions we cannot differentiate, at times. It would\nbe impossible for a physical object to have a velocity graph like the \\texttt{int()} function,\nbut it can still be meaningful to find the area under the graph.\n\n\\begin{example}{Bandwidth Rates}\n\\exProblem\nSuppose an wifi hotspot charges start at 2.50 per hour when you begin, and the rate goes\nup .50 every 20 minutes after that. The rate is not incremented\ncontinuously, but jumps every 1/5 hour. Illustrate the cost of using the service for 0.9 hours\nas a definite integral.\n\n\\exSolution\n\\begin{tikzpicture}[xscale=4,yscale=1]\n\t\\draw (-0.05,0) -- (1,0) ;\n\t\\draw (0,-.2) -- (0,5.0);\n\t\\draw[thick] (0,2.50) -- (.2,2.5) -- (.2,3) -- (.4,3) -- (.4,3.5) -- (.6,3.5) -- (.6,4) -- (.8,4) -- (.8,4.5) -- (1,4.5);\n\t\\draw [fill=gray!80,opacity=0.4] (0,0) rectangle (.2,2.5);\n\t\\draw [fill=gray!80,opacity=0.4] (.2,0) rectangle (.4,3);\n\t\\draw [fill=gray!80,opacity=0.4] (.4,0) rectangle (.6,3.5);\n\t\\draw [fill=gray!80,opacity=0.4] (.6,0) rectangle (.8,4);\n\t\\draw [fill=gray!80,opacity=0.4] (.8,0) rectangle (.9,4.5);\n\\end{tikzpicture}\nAs the rectangle illustrate, $(0.2)(2.5) + (0.2)(3.0) + (0.2)(3.5) + (0.2)(4.0) + (0.1)(4.5) = 3.05$.\n\\end{example}\n\n", "meta": {"hexsha": "c7fe20a3723f4e1006092e58d76d9e657e4287fc", "size": 5056, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ch05/0505.tex", "max_stars_repo_name": "aquatiki/AnalysisTextbook", "max_stars_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-10-08T15:05:17.000Z", "max_stars_repo_stars_event_max_datetime": "2019-07-07T12:32:53.000Z", "max_issues_repo_path": "ch05/0505.tex", "max_issues_repo_name": "aquatiki/AnalysisTextbook", "max_issues_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ch05/0505.tex", "max_forks_repo_name": "aquatiki/AnalysisTextbook", "max_forks_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.9652173913, "max_line_length": 122, "alphanum_fraction": 0.6946202532, "num_tokens": 1706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754489059774, "lm_q2_score": 0.8887587853897073, "lm_q1q2_score": 0.6469945757632034}} {"text": "\\subsubsection{Complex Eigenvalues}\r\n\\begin{theorem}\r\n\tIf an $n \\times n$ matrix $A$ is not defective, then for each pair of complex eigenvalues $\\alpha \\pm \\beta i$ with corresponding eigenvectors $\\vec{a} \\pm i\\vec{b}$, the corresponding fundamental solutions are $e^{\\alpha t}\\left(\\cos{(\\beta t)}\\vec{a}-\\sin{(\\beta t)\\vec{b}}\\right)$ and $e^{\\alpha t}\\left(\\sin{(\\beta t)}\\vec{a}+\\cos{(\\beta t)\\vec{b}}\\right)$.\r\n\\end{theorem}\r\n\r\n\\ifodd\\includeLinearSystemsExamples\\input{./linearSystems/homogeneousSystems/complexEigenvalues_example.tex}\\fi", "meta": {"hexsha": "e96107984c3b795b97f5ccd93ec7b7472a215fee", "size": 546, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/linearSystems/homogeneousSystems/complexEigenvalues.tex", "max_stars_repo_name": "rawsh/Math-Summaries", "max_stars_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "diffEq/linearSystems/homogeneousSystems/complexEigenvalues.tex", "max_issues_repo_name": "rawsh/Math-Summaries", "max_issues_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "diffEq/linearSystems/homogeneousSystems/complexEigenvalues.tex", "max_forks_repo_name": "rawsh/Math-Summaries", "max_forks_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 91.0, "max_line_length": 363, "alphanum_fraction": 0.7216117216, "num_tokens": 170, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.874077222043951, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6469695501992171}} {"text": "\\documentclass[../../main.tex]{subfiles}\n\\begin{document}\n\\subsubsection{At Key}\nThe operation $atKey$ will return the Value $v$ at some specified Key $k$.\n\\begin{schema}{AtKey[KV, K]}\n m? : KV \\\\\n v! : V \\\\\n k? : K \\\\\n atKey~\\_ : KV \\cross K \\surj V\n \\where\n v! = atKey(m?, k?) @ \\\\\n \\t2 let ~~ coll == ((\\seq m?) \\filter (k?, m?_{k?})) \\implies \\langle (k?, m?_{k?}) \\rangle \\iff k? \\in \\dom m? \\\\\n \\ \\ \\ = (second(head(coll)) \\iff k? \\mapsto m?_{k?} \\in coll) ~\\lor \\\\\n \\t1 (\\emptyset \\iff k? \\not \\in \\dom m?)\n\\end{schema}\nIn the schema above, $coll$ is the result of filtering for $(k?, m?_{k?})$ within $\\seq m?$.\nIf the mapping was in the original $m?$, it will also be in the sequence of mappings. This means\nwe can filter over the sequence to look for the mapping and if found, it is returned as $\\langle (k?, m?_{k?}) \\rangle$.\nTo return the mapping itself, $head(coll)$ is used to extract the mapping such that the value mapped to $k?$ can be returned.\n\\begin{zed}\n v! = atKey(m?, k?) = second(head(coll)) = m?_{k?} @ m?_{k?} : V \\iff k? \\in \\dom m?\n\\end{zed}\nThe following examples demonstrate the properties of $atKey$\n\\begin{argue}\n M = \\ldata k_{0}v_{k_{0}}, k_{1}v_{k_{1}} \\rdata \\\\\n \\t1 k_{0} = abc \\ \\land v_{k_{0}} = 123 & $k_{0}v_{k_{0}} = abc \\mapsto 123$ \\\\\n \\t1 k_{1} = def \\ \\land v_{k_{1}} = xyz \\mapsto 456 & $k_{1}v_{k_{1}} = def \\mapsto xyz \\mapsto 456$ \\\\\n atKey(M, abc) = 123 \\\\\n atKey(M, def) = xyz \\mapsto 456 \\\\\n atKey(M, foo) = \\emptyset\n\\end{argue}\n\\end{document}\n", "meta": {"hexsha": "81cd4fa15532c8f2813f7fdfb11dfe1a1af947d5", "size": 1516, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/operations/kv/atKey.tex", "max_stars_repo_name": "yetanalytics/dave", "max_stars_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2018-08-17T00:38:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T02:32:37.000Z", "max_issues_repo_path": "docs/operations/kv/atKey.tex", "max_issues_repo_name": "adlnet/dave", "max_issues_repo_head_hexsha": "9339713fac747118e462e4fc7e1ecd54e5d916e3", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 95, "max_issues_repo_issues_event_min_datetime": "2018-08-31T18:57:51.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-15T16:40:01.000Z", "max_forks_repo_path": "docs/operations/kv/atKey.tex", "max_forks_repo_name": "yetanalytics/dave", "max_forks_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2019-09-28T06:48:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-31T16:20:47.000Z", "avg_line_length": 45.9393939394, "max_line_length": 125, "alphanum_fraction": 0.6042216359, "num_tokens": 569, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.867035763237924, "lm_q2_score": 0.7461389986757757, "lm_q1q2_score": 0.6469291961984316}} {"text": "\\chapter{The local field potential}\n\\label{ch:LFP}\n\nThis thesis discusses software for processing intracranial voltage recordings. This chapter describes the nature of such voltage signals.\n\nAt every location in the brain, the electric field vector $\\E$ (in units of N/C or V/m) points in the direction that a positive charge would be pushed towards, if it was placed at that location. The electric potential $\\phi$ is defined\\footnotemark{} such that:\n%\n\\begin{equation}\n\\label{eq:potential}\n\\E = -\\grad{\\phi},\n\\end{equation}\n%\ni.e. such that the electric field points from locations of high potential to locations of lower potential. The electric field potential $\\phi$ is a scalar field, in units of V. When it is measured extracellularly, neuroscientists refer to it as the \\emph{local field potential} (LFP) -- especially when only frequencies below about 500 Hz are considered.\n\n\\footnotetext{The electric field potential $\\phi$ is only defined when the electric and magnetic fields do not vary too quickly over time, i.e. when $\\pdv*{\\E}{t} \\approx 0$ and $\\pdv*{\\B}{t} \\approx 0$. It then directly follows from Maxwell's equations that both fields become decoupled, and that the electric field becomes irrotational ($\\curl{\\E} = 0$), so that an electric potential may be defined as in \\cref{eq:potential}. This so called \\emph{quasi-static} assumption is met in electrophysiological conditions \\cite{Nunez2006,Plonsey2007}.}\n\nWe now discuss two models used to estimate the LFP $\\phi$. Both models calculate $\\phi$ based on the mobile charge density $\\rho$ at each location in the brain and throughout time. ($\\rho$ is the net positive surplus of mobile charges, in units of C/\\si{\\metre^3}. Biological tissue contains a high number of such mobile charges. These are mostly ions, with \\ce{Na+}, \\ce{Cl-}, and \\ce{K+} the most abundant ones \\cite{Martinsen2015a}).\n\nBoth models consist of a single divergence equation (namely \\cref{eq:div-charge,eq:div-current}), which, together with \\cref{eq:potential} and an assumption of uniform and isotropic brain tissue, leads in both cases to a simple closed form equation to calculate the LFP $\\phi$ (namely \\cref{eq:phi-charge,eq:phi-current}).\n\n\n\n\\section{Model 1 (Gauss's law)}\n\nThe first model is a direct translation of the first of Maxwell's equations (Gauss's law), at macroscopic scale:\\footnotemark{}\n%\n\\begin{equation}\n\\label{eq:div-charge}\n\\div{\\eps \\E} = \\rho,\n\\end{equation}\n%\nwhere $\\eps$ is the local tissue permittivity, in units of F/m and generally a 3 $\\cross$ 3 matrix.\n\n\\footnotetext{Gauss's law in its pure form ($\\div{\\E} = \\rho_\\total / \\eps_0$) considers both mobile and ``bound'' charges: $\\rho_\\total = \\rho_\\free + \\rho_\\bound$. In \\cref{eq:div-charge}, we only explicitly consider mobile charges $\\rho = \\rho_\\free$, while bound charges are subsumed in the tissue permittivity $\\eps$. This simplification is allowed if we assume that ``tissue polarisation is proportional to electric field strength'' \\cite{Feynman2013}. This is a common assumption in physics, and is largely valid for brain tissue in normal conditions \\cite{Nunez2006}.}\n\nWhen we assume the permittivity $\\eps$ to be uniform and isotropic throughout the tissue (i.e. $\\eps$ is a constant scalar)\\footnotemark{}, \\cref{eq:div-charge} has the following solution for the LFP $\\phi$ \\cite{Feynman2013}:\n%\n\\begin{equation}\n\\label{eq:phi-charge}\n\\phi = \\frac{1}{4 \\pi \\eps} \\int_V \\frac{\\rho \\dd{V}}{r},\n\\end{equation}\n%\nwhere we summate over the entire brain volume $V$, with $\\rho \\dd{V}$ the surplus mobile charge in a small volume $\\dd{V}$ of tissue (in Coulomb), and $r$ the distance of this small volume to the point where $\\phi$ is calculated.\n\n\\footnotetext{The uniform and isotropic permitivitty assumption is questionable. Roughly, brain tissue consists of a dense alternation of two types of tissue (see \\cref{fig:neuropil}): the seawater-like fluid inside and in between cells, and the fatty membranes around cells, organelles, and vesicles. The former has a permittivity $\\eps$ of about 15 times larger than the latter \\cite{Marszalek1991,Weaver2003,Martinsen2015}. Additionally, the strongly non-random organization of some brain regions may challenge the isotropy assumption. Nevertheless, this uniform and isotropic permittivity assumption is often made \\cite{Nunez2006,Plonsey2007}.}\n\nIn other words, positive charge surplusses increase the nearby LFP, negative charge surplusses decrease it, and their effects summate linearly, weighted inverse-proportionally by distance. This model explains for example the resting ``membrane potential'' of neurons, where $\\phi$ is lower inside the cell than outside: neurons at rest contain an excess of negative charges \\cite{Dayan2001b}.\n\nAlthough this model is arguably the most physically correct, it is often difficult to apply in practice: charge distributions $\\rho$ are already quite complex in even the simplest of electronic circuits \\cite[chapter 8]{Chabay2015}. For the highly entangled topologies of brain tissue (see \\cref{fig:neuropil}), estimating charge distributions is intractable.\n\nThe next model is more useful in practice, as it calculates the LFP based not on charges, but on currents. Currents (and specifically transmembrane currents) are the bread and butter of electrophysiology \\cite{Kandel2013a}.\n\n\n\n\n\\section{Model 2 (Current source density analysis)}\n\\label{sec:CSD}\n\nThis is the ``standard model of electric potentials in biological tissue'' \\cite{Bedard2011}. Unlike the previous model, it is mostly empirical \\cite{Bedard2011}. In fact, the assumptions on which it rests are strictly incompatible with \\cref{eq:div-charge,eq:phi-charge}, and lead to a paradox when considered jointly. \\Citeauthor{Bedard2011} explore this paradox \\cite{Bedard2011}, and propose a more general formulation of CSD analysis that `solves' the paradox. In the remainder of this chapter however, we will present the `classic' CSD model, as it is presented in most electrophysiology texts \\cite{Plonsey2007,Nunez2006,Mitzdorf1985,Nicholson1975,Linden2014,Buzsaki2012a}.\n\nWe first define the so called \\emph{current source density}, $I$, as the rate at which the net charge surplus inside a small volume decreases (units A/\\si{m^3}):\n%\n\\begin{equation}\n\\label{eq:CSD}\nI = - \\pdv{\\rho}{t}\n\\end{equation}\n%\nCharge conservation (which is a direct consequence of Maxwell's equations) dictates that\n%\n\\begin{equation}\n\\label{eq:charge-conservation}\n\\div{\\J} = I,\n\\end{equation}\n%\nwhere $\\J$ is the current density (units of A/\\si{\\metre^2}, and $\\J = \\rho \\vb{v}$ with $\\vb{v}$ the local velocity of mobile charges). In other words, the decrease in net positive charges in a small volume ($I$) equals the net rate at which positive charges flow out of this volume ($\\div{\\J}$).\n\nNext (and this is the assumption challenged in \\cite{Bedard2011}), we assume that the extracellular medium is \\emph{resistive} or \\emph{ohmic}:\n%\n\\begin{equation}\n\\label{eq:ohmic-medium}\n\\J = \\sigma \\E,\n\\end{equation}\n%\nwhere $\\sigma$ is the local tissue conductivity, in units of S/m and generally a 3 $\\cross$ 3 matrix. In other words, current is assumed to be linearly related to the electric field. (This needn't be the case in reality: the electric field dictates the acceleration of mobile charges, while the current density describes their velocity. These are instantaneously indepedent.)\n\nCombining charge conservation and the ohmic assumption leads to the governing equation for CSD analysis:\n%\n\\begin{equation}\n\\label{eq:div-current}\n\\div{\\sigma \\E} = I.\n\\end{equation}\n%\nNote the similarity with \\cref{eq:div-charge}. And analogously as in \\cref{eq:div-charge,eq:phi-charge}, when we assume the conductivity $\\sigma$ to be uniform and isotropic throughout the tissue (i.e. $\\sigma$ is a constant scalar)\\footnotemark{}, \\cref{eq:div-current} has the following solution for the LFP $\\phi$ \\cite{Plonsey2007}:\n%\n\\begin{equation}\n\\label{eq:phi-current}\n\\phi = \\frac{1}{4 \\pi \\sigma} \\int_V \\frac{I \\dd{V}}{r}.\n\\end{equation}\n%\nIn other words, so called ``current sources'' ($I > 0$) increase the nearby LFP, current sinks ($I < 0$) decrease it, and their effects again summate linearly, weighted inverse-proportionally by distance.\n\n\\footnotetext{Again, a questionable assumption: (extra)cellular plasma has a conductivity $\\sigma$ of about $10^6$ times higher than the lipid bilayer membranes in between \\cite{Michel2017,Martinsen2015,Marszalek1991,Nunez2006,Weaver2003}.}\n\n\n\n\\section{Transmembrane currents}\n\\label{sec:transmembrane-currents}\n\nActive transmembrane currents are the main physical mechanism of signal propagation in the nervous system.\\footnotemark{} At synapses, excitatory neurons cause a brief inward current in their target neuron, while inhibitory neurons cause a brief outward current. Along axons, action potentials (``spikes'') propagate by a cascade of inward currents.\n\n\\footnotetext{With 'active', we mean here that the current is caused by the opening of ion channels that are located at the same position in the membrane as the current.}\n\nEach of these active currents leads to corresponding passive currents in other parts of the membrane, forming closed current loops. An excitatory, inward current at a synapse corresponds to outward currents at distal parts of the neuron (most notably at the cell body). The inward, active currents of a spike are surrounded by outward, passive currents.\n\n\n\n\\section{Forward CSD model of the LFP}\n\\label{sec:forward-CSD}\n\nIn CSD analysis, both active and passive transmembrane currents are used to estimate the LFP, by making use of the following `trick': the existence of the inside of the neuron is ignored. An inward current for example is then modelled as a current sink $I_m < 0$ in the extracellular space, i.e. an accumulation of mobile charges. In reality, charges do not quite accumulate, but rather dissipate in the neuron. Or, equivalently, a proportional current source $I'_m = -I_m$ exists inside the neuron, at virtually the same location as $I_m$, cancelling it out. In CSD analysis however, only the current sink $I_m$ (and not the current source $I'_m$) would be used for calculating the LFP $\\phi$.\n\nThe justification for ignoring the inside of the neuron is often as follows (paraphrasing from \\cite{Mitzdorf1985}, a foundational CSD theory paper): ``The extracellular space is independent of the intracellular space, because its boundaries (the cell membranes) have a very high resistance compared to the extracellular space.'' However, electric insulators (high resistance areas) do not shield the electric field, nor the electric field potential $\\phi$.\n%\n\\footnote{Another reason the inside of neurons is ignored might possibly be notation: at synapses, there are real currents, often denoted by $I$; but there are no real current sources or sinks, also often denoted by $I$. This may cause confusion.}\n\\footnote{A final reason to not only consider the extracellular space, is that there is not that much of it: only between 10\\% and 20\\% of hippocampus volume is extracellular space \\cite{Sykova1997a}. See also \\cref{fig:neuropil}.}\n\nNonetheless, in the so called `forward' CSD model (estimating the LFP from transmembrane currents), \\cref{eq:phi-current} is used to calculate the LFP based on `unpaired' transmembrane current source densities $I_m$. From \\cref{eq:phi-current}, an inward current then corresponds to a drop in the nearby LFP, and an outward current to an increase. This technique is applied in \\cref{sec:physiology} to explain the physical basis of the sharp wave-ripple motif. Another application of this technique is e.g. \\texttt{LFPy}, a software package used to numerically simulate the LFP \\cite{Linden2014}.\n\nNotwithstanding the theoretical problems outlined here and in \\cite{Bedard2011}, CSD analysis seems to provide relatively accurate results in practice \\cite{Buzsaki2012a,Bedard2011}. As mentioned earlier, \\citeauthor{Bedard2011} \\cite{Bedard2011} give a derivation of a CSD-like model that does not rest upon the `classic' derivation as presented here, instead concluding: ``[..] the results obtained with the classic CSD analysis are perfectly consistent with \\emph{ionic diffusion} because diffusion gives a source term which is very close to the phenomenological model of current source density introduced by Pitts and Mitzdorf, but in a manner consistent with Maxwell-Gauss law.'' (Pitts \\cite{Pitts1952} and Mitzdorf \\cite{Mitzdorf1985} refers to the classical CSD model).\n", "meta": {"hexsha": "638af568f1a9f53542812aaf53ad9d4f1c838730", "size": 12489, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "modules/LFP/index.tex", "max_stars_repo_name": "tfiers/master-thesis", "max_stars_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-23T01:39:24.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-23T01:39:24.000Z", "max_issues_repo_path": "modules/LFP/index.tex", "max_issues_repo_name": "tfiers/master-thesis", "max_issues_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 46, "max_issues_repo_issues_event_min_datetime": "2018-09-18T16:38:12.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-10T22:37:35.000Z", "max_forks_repo_path": "modules/LFP/index.tex", "max_forks_repo_name": "tfiers/master-thesis", "max_forks_repo_head_hexsha": "3e97128eeb18827b03da90817fe6f6985c84ad80", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 96.0692307692, "max_line_length": 777, "alphanum_fraction": 0.7756425655, "num_tokens": 3205, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357460591569, "lm_q2_score": 0.7461389873857264, "lm_q1q2_score": 0.6469291735918071}} {"text": "\\subsection{Course Structure}\n\n\\begin{table}[h]\n% \\caption{Tested autoencoder structures for a 3D case}\n\n\\begin{tabular}{p{25mm}|p{100mm}|p{30mm}}\n \\hline\nLecture 1-2\t& \nIntroduction. \\newline\nLinear Algebra Four Fundamental Subspaces\n(Null space, Row space, Projectors, Column and Left null spaces). \\newline\nUse cases for fundamental subspaces in Control. &\t\nIndividual \\newline assignment \n\\\\ \\hline\n%\nLecture 3-5 &\nLeast Squares and Quadratic Programming. \\newline\nDomains. Convex Domains. \\newline\nLinear inequality representation of convex domains.\t& \nIndividual \\newline assignment \n\\\\ \\hline\n%\nLecture 6-9\t& \nSecond order cone programming (SOCP). \\newline\nSemidefinite programming (SDP). \\newline\nLinear matrix inequalities, example in Finite-time stability, example in computational geometry. \\newline\nMixed-integer convex programming. \\newline Relaxation. Big M method. Example in path planning.\t& \nIndividual \\newline assignment \n\\\\ \\hline\n%\nLecture 10-11\t& \nBarrier functions, center of constraints. \\newline\nMinimax.\t& \nMiniproject \n\\\\ \\hline\n\\end{tabular}\n\\end{table}", "meta": {"hexsha": "e42615b9ba1112de378577f8c67fb3c33eba569c", "size": 1089, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Syllabus/Sections/CourseStructure.tex", "max_stars_repo_name": "kahlflekzy/Computational-Intelligence-Slides-Spring-2022", "max_stars_repo_head_hexsha": "9401fe1258efa91a6c9886501d02909420a94add", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2022-01-19T15:40:01.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T22:27:44.000Z", "max_issues_repo_path": "Syllabus/Sections/CourseStructure.tex", "max_issues_repo_name": "kahlflekzy/Computational-Intelligence-Slides-Spring-2022", "max_issues_repo_head_hexsha": "9401fe1258efa91a6c9886501d02909420a94add", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2021-05-27T09:02:55.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-13T09:36:55.000Z", "max_forks_repo_path": "Syllabus/Sections/CourseStructure.tex", "max_forks_repo_name": "kahlflekzy/Computational-Intelligence-Slides-Spring-2022", "max_forks_repo_head_hexsha": "9401fe1258efa91a6c9886501d02909420a94add", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2021-01-20T07:58:57.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-12T08:28:08.000Z", "avg_line_length": 29.4324324324, "max_line_length": 105, "alphanum_fraction": 0.7649219467, "num_tokens": 298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7981867801399694, "lm_q1q2_score": 0.646913550482921}} {"text": "\\chapter{From MDPs to Reinforcement Learning}\n\\label{chapter3}\n\n\\vspace{0.5cm}\n\n\\noindent As said before, when we have both the reward function $\\mathcal{R}$ and the transition model $\\mathcal{P}$ we can solve the MDP by means of dynamic programming.\n\nTruth to be told, for most cases we cannot precisely predict neither $\\mathcal{R}$ nor $\\mathcal{P}$ and we have to approximate, explore and estimate functions and states in order to converge to the optimal solution (the optimal policy).\n\nReinforcement Learning in fact, is a technique to handle Markov Decision Processes in an environment we don't fully know.\n\n\\section{Value Function and the Bellman equation}\nWe spoke before about \\textit{optimal policy} but we never defined it. The optimal policy, is the one that maximizes our long term expected reward, that is:\n\\begin{equation}\n \\pi^\\star = arg\\,max_\\pi \\ \\mathbb{E} \\Big[ \\sum_t \\gamma^t \\mathcal{R}(s_t) \\ \\vert \\ \\pi \\Big]\n\\end{equation}\n\\noindent\nWe can see in this function, that the initial state is not defined (it's random), and that the optimal policy must be optimal starting from any state.\n\nWe can actually define also the \\textbf{value function} that gives the expected long term reward following a policy starting from a state $s$:\n\\begin{equation}\n \\mathbb{V}^\\pi(s) = \\mathbb{E} \\Big[ \\sum_t \\gamma^t \\mathcal{R}(s_t) \\ \\vert \\ \\pi, \\ s_0 = s \\Big]\n\\end{equation}\nThis function helps us to define in a better way the optimal policy:\n\\begin{equation} \\label{eq:bellman-value}\n \\pi^\\star = arg\\,max_a \\ \\sum_{s'} \\mathbb{P}(s_{t+1} = \\mathnormal{s'} \\vert s_t = \\mathnormal{s}, a_t = \\mathnormal{a}) \\ \\mathbb{V}(s')\n\\end{equation}\nThis equation says that the optimal policy is the one that, for every state, returns the action that maximizes my expected value function and this allows us to write the value function in a much more famous form known as \\textbf{Bellman Equation}:\n\\begin{equation}\n \\mathbb{V}^\\pi(s) = \\mathcal{R}(s) + \\gamma\\sum_{s'} \\mathbb{P}(s_{t+1} = \\mathnormal{s'} \\vert s_t = \\mathnormal{s}, a_t = \\mathnormal{a}) \\ \\mathbb{V}(s')\n\\end{equation}\nThis equation has two terms and says that the expected sum of discounted return is equal to the sum of two terms: the initial, or immediate, reward received immediately simply for starting in state $s$, and a discounted sum of the future expected rewards (rewards after the first step) and can be rewritten as \\begin{equation}\n \\mathbb{V}^{\\pi^\\star}(s) = \\mathcal{R}(s) + \\gamma\\,max_a\\sum_{s'} \\mathbb{P}(s_{t+1} = \\mathnormal{s'} \\vert s_t = \\mathnormal{s}, a_t = \\mathnormal{a}) \\ \\mathbb{V}(s')\n\\end{equation}\nto describe the optimal value function $\\mathbb{V}^\\star$.\n\n\\section{Action-value function}\n\\begin{comment}\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=1\\textwidth]{./pictures/bellman.eps}\n \\caption{A representation of the Bellman equation.}\n \\label{fig:bellman}\n\\end{figure}\n\\end{comment}\n\\begin{figure} [ht] \n \\tikzstyle{squarenode} = [circle, draw, minimum size=5mm]\n \n \\centering\n \\begin{tikzpicture}[align = center, node distance = 9em, auto, thick]\n \\node [squarenode] (s1) {$s_1$};\n \\node [squarenode, right of=s1] (s2) {$s_2$};\n \\node [squarenode, right of=s2] (s3) {$s_3$};\n \\node [squarenode, right of=s3] (s4) {$s_4$};\n \\node[squarenode,draw=none, right of=s4, xshift=-5em, yshift=1.6em] (s5) {};\n\n \\draw [->] (s1) edge [out=90 in=90] node[right, above] {$\\mathcal{R}(s_1, a_1)$} node[very near start] {$a_1$} (s2);\n \\draw [->] (s2) edge [out=90 in=90] node[right, above] {$\\mathcal{R}(s_2, a_2)$} node[very near start] {$a_2$} (s3);\n \\draw [->] (s3) edge [out=90 in=90] node[right, above] {$\\mathcal{R}(s_3, a_3)$} node[very near start] {$a_3$} (s4);\n \\draw (s4) edge [dotted] node {} (s5);\n\n \\end{tikzpicture}\n \\caption{A representation of Bellman equation.}\n \\label{fig:bellman}\n\\end{figure}\nThe Bellman equation is fundamental and is a necessary condition for optimality. Visually, our system goes like the one depicted in figure \\ref{fig:bellman} and the Bellman Equation~(\\ref{eq:bellman-value}) describes this system immediately after state $s_1$, before taking action $a_1$, so when we talk about \\textit{value} we talk about a state and eventually we will get into another state that can also be represented with another value function. This is why the Bellman equation can be depicted in a recursive way.\n\nThe purpose of explaining the Bellman equation graphically is to show that, as the value function shows the infinite recursion of the system starting from state $s$, we can actually represent this recursion, starting immediately after the action is taken.\n\nThis equation is called \\textbf{action-value function} and is defined in two ways. The first one in the Bellman equation form:\n\\begin{equation}\n \\mathbb{Q}^\\pi(s, a) = \\mathcal{R}(s, a) + \\gamma\\sum_{s'} \\mathbb{P}(s_{t+1} = \\mathnormal{s'} \\vert s_t = \\mathnormal{s}, a_t = \\mathnormal{a}) \\ \\,max_a\\mathbb{Q}(s', a')\n\\end{equation}\nand the second one in the expectation form:\n\\begin{equation}\n \\mathbb{Q}^\\pi(s, a) = \\mathbb{E} \\Big[ \\sum_t \\gamma^t \\mathcal{R}(s_t) \\ \\vert \\ \\pi, \\ s_0 = s, \\ a_0 = a \\Big]\n\\end{equation}\nThis equations can be explained as we are in state $s$, we take action $a$, we get the first reward that is the first term, and then we go on with the summation.\n\nThe optimal value and action value functions are related in the following way.\n\\begin{align*}\n \\mathbb{V}^\\star(s) &= \\,max_a\\mathbb{Q}^\\star(s, a) \\\\\n \\mathbb{Q}^\\star(s, a) &= \\mathcal{R}(s, a) + \\gamma\\sum_{s'} \\mathbb{P}(s_{t+1} = \\mathnormal{s'} \\vert s_t = \\mathnormal{s}, a_t = \\mathnormal{a}) \\ \\mathbb{V}^\\star(s')\n\\end{align*}\nOne way of thinking about this is: we always go further in the cycle. For the value function, we need the best possible action and the state-action function has already taken an action, so I just get the max state-action function. Same way for the second function, state-action function starts after the action is taken, so I sum the reward and I continue at the landing state so with the value function.\n\nSomeone may ask themselves \\textit{why are we bothering of creating this action-value function when we already have the value function?} \\\\\nWell, the \\textit{action-value function} is key in reinforcement learning as it allows to take expectations of $\\mathbb{Q}^\\pi(s,a)$ without knowing the transition function or the reward function.\n\n\\section{Algorithms}\nNow that we have defined two important functions, we can describe a reinforcement learning algorithm.\n\n\\subsection{Introduction}\n\\begin{figure} [ht]\n \\tikzstyle{block} = [rectangle, draw, \n text width=8em, text centered, rounded corners, minimum height=4em]\n \n \\tikzstyle{line} = [draw, -latex]\n \\centering\n \\begin{tikzpicture}[align = center, node distance = 15em, auto, thick]\n \\node [block] (Agent) {Agent ($\\pi$)};\n \\node [block, right of=Agent] (Environment) {Environment};\n \\draw [->] (Agent) edge [bend left] node {$a$} (Environment);\n \\draw [->] (Environment) edge [bend left] node {$s$} (Agent);\n \\draw [->] (Environment) edge [dotted] node [above] {$\\mathcal{R}(s,a)$} (Agent);\n \\end{tikzpicture}\n \\caption{The concept of reinforcement learning.}\n \\label{fig:rldiagram}\n\\end{figure}\n\\noindent\nTo understand better reinforcement learning, it's better to see Figure~\\ref{fig:rldiagram}. The environment reveals to the agent in forms of states and the agent replies with actions and the environment sends back a reward. All the computation or, in a more reinforcement learning way, the \\textit{thinking}, is done in the agent \\textit{mind} through the policy. \\\\\nAn important concept is that the information regarding the environment is only available through this interaction, meaning that the environment is not available in the agent's brain (the policy), but the agent is \\textit{experiencing} the environment only by interacting with it.\n\n\\subsection{Different families of algorithms}\nWe can think of a RL algorithm as a method that, given a sequence of action-state-reward triplets, is able to find an optimal policy $\\pi$ as depicted in Figure~\\ref{fig:rlalg-general}.\n\\begin{figure} [ht]\n \\tikzstyle{block} = [rectangle, draw, \n text width=8em, text centered, rounded corners, minimum height=4em]\n \\tikzstyle{txtblock} = [rectangle, \n text width=8em, text centered, rounded corners, minimum height=4em]\n \n \\tikzstyle{line} = [draw, -latex]\n \\centering\n \\begin{tikzpicture}[align = center, node distance = 15em, auto, thick]\n \\node [txtblock] (sequence) {$\\langle s, a, r \\rangle^+$};\n \\node [block, right of=sequence, xshift=-3em] (algorithm) {RL algorithm};\n \\node [txtblock, right of=algorithm, xshift=-3em] (policy) {$\\pi$};\n \\draw [->] (sequence) edge node {} (algorithm);\n \\draw [->] (algorithm) edge node {} (policy);\n \\end{tikzpicture}\n \\caption{Reinforcement learning algorithm}\n \\label{fig:rlalg-general}\n\\end{figure}\nThe RL algorithm block can be split in different ways according to the way the policy is learned. We can divide algorithms in three main families.\n\n\\subsubsection{Model-based algorithms}\n\\begin{figure} [ht]\n \\tikzstyle{block} = [rectangle, draw, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \\tikzstyle{txtblock} = [rectangle, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \n \\tikzstyle{line} = [draw, -latex]\n \\centering\n \\begin{tikzpicture}[align = center, node distance = 15em, auto, thick]\n \\node [txtblock] (sequence) {$\\langle s, a, r \\rangle^+$};\n \\node [block, right of=sequence, xshift=-9em] (model-learner) {model learner};\n \\node [txtblock, text width=2em, right of=model-learner, xshift=-10em] (p-r) {$\\mathcal{P}, \\mathcal{R}$};\n \\node [block, right of=p-r, xshift=-10em] (solver) {solver};\n \\node [txtblock, text width=1em, right of=solver, xshift=-10em] (action-value) {$\\mathbb{Q}^\\star$};\n \\node [block, right of=action-value, xshift=-10em] (argmax) {argmax};\n \\node [txtblock, text width=2em, right of=argmax, xshift=-10em] (policy) {$\\pi$};\n \\draw [->] (sequence) edge node {} (model-learner);\n \\draw [->] (model-learner) edge node {} (p-r);\n \\draw [->] (p-r) edge node {} (solver);\n \\draw [->] (solver) edge node {} (action-value);\n \\draw [->] (action-value) edge node {} (argmax);\n \\draw [->] (argmax) edge node {} (policy);\n \\path [line] (p-r) |-++ (0cm, 1.2cm) -| (model-learner);\n\n \\end{tikzpicture}\n \\caption{A model-based algorithm}\n \\label{fig:model-based}\n\\end{figure}\n\\noindent\nA \\textbf{model-based} algorithm takes the action-state-reward triples and sends them to a \\textit{model learner} which learns the transition probability function $\\mathcal{P}$ and the reward function $\\mathcal{R}$. Once learned, a solver can compute an optimal action-value function. Once you have an optimal action-value function, for each state you just get the argmax to get the best action, and that's the policy.\nThe reason we have an arrow going back in the model learner is that the transition probability and the reward functions are initially estimated and are continuously updated with previous values of the functions.\n\n\\subsubsection{Value-function based algorithms}\n\\begin{figure} [ht]\n \\tikzstyle{block} = [rectangle, draw, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \\tikzstyle{txtblock} = [rectangle, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \n \\tikzstyle{line} = [draw, -latex]\n \\centering\n \\begin{tikzpicture}[align = center, node distance = 15em, auto, thick]\n \\node [txtblock] (sequence) {$\\langle s, a, r \\rangle^+$};\n \\node [block, right of=sequence, xshift=-9em] (value-update) {value update};\n \\node [txtblock, text width=1em, right of=value-update, xshift=-10em] (action-value) {$\\mathbb{Q}$};\n \\node [block, right of=action-value, xshift=-10em] (argmax) {argmax};\n \\node [txtblock, text width=2em, right of=argmax, xshift=-10em] (policy) {$\\pi$};\n \\draw [->] (sequence) edge node {} (value-update);\n \\draw [->] (value-update) edge node {} (action-value);\n \\draw [->] (action-value) edge node {} (argmax);\n \\draw [->] (argmax) edge node {} (policy);\n \\path [line] (action-value) |-++ (0cm, 1.2cm) -| (value-update);\n\n \\end{tikzpicture}\n \\caption{A value-function based algorithm}\n \\label{fig:value-function-based}\n\\end{figure}\n\\noindent\nThe \\textbf{value-function} based algorithms are almost the same as the previous family of algorithms except that, as we can see from figure~\\ref{fig:value-function-based}, one block is missing. The main difference is that this family of algorithms tries to learn a policy without explicitly learning the model or better, the policy is learned without the assumption of knowing the model, whereas in the first family, to learn the policy, the model \\textit{must} be learned (the two functions that define a model are the transition probability function and the reward function).\nThis family of algorithm is in fact also called \\textbf{model-free algorithms}.\n\nIn this family, the state-action function is directly estimated from the sequence of tuples from the environment and previous values of $\\mathcal{Q}$. The $\\mathcal{Q}$ doesn't have a $\\star$ since it's just an estimation and it's not an optimal value but, with continuous iterations, it is proven that it will reach an optimal value.\n\n\\subsubsection{Policy search algorithms}\n\\begin{figure} [ht]\n \\tikzstyle{block} = [rectangle, draw, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \\tikzstyle{txtblock} = [rectangle, \n text width=4em, text centered, rounded corners, minimum height=4em]\n \n \\tikzstyle{line} = [draw, -latex]\n \\centering\n \\begin{tikzpicture}[align = center, node distance = 15em, auto, thick]\n \\node [txtblock] (sequence) {$\\langle s, a, r \\rangle^+$};\n \\node [block, right of=sequence, xshift=-9em] (policy-update) {policy update};\n \\node [txtblock, text width=2em, right of=policy-update, xshift=-10em] (policy) {$\\pi$};\n \\draw [->] (sequence) edge node {} (policy-update);\n \\draw [->] (policy-update) edge node {} (policy);\n \\path [line] (policy) |-++ (0cm, 1.2cm) -| (policy-update);\n \\end{tikzpicture}\n \\caption{A policy search algorithm}\n \\label{fig:policy-search}\n\\end{figure}\n\\noindent\nThe third family of algorithms updates directly the policy by feeding it back to the policy update that modifies it based on the sequence that it receives. This algorithm is much more direct, but the learning problem is much more difficult because the feedback you get from the policy usually is not that useful to update it. For example, in a game, knowing that an action moves you forward isn't sufficient to know when to use it.\n\\newline\n\\newline\nAs depicted in figures~\\ref{fig:model-based}, \\ref{fig:value-function-based} and \\ref{fig:policy-search}, we can see that the three algorithms are getting simpler as going down, meaning that the policy learning is more direct but in the opposite way, going up, the algorithms become more supervised. Most of the time of this semester project has been spent in model-free value-based algorithms that provide a good balance between learning techniques and learning difficulties.\n\n\\subsection{Off-policy and on-policy algorithms}\nA further distinction between algorithms can be made through another aspect, if they are \\textit{off-policy} or \\textit{on-policy}.\n\n\\begin{itemize}\n \\item \\textbf{off-policy} algorithms are those algorithms that learn about the greedy strategy (policy), while following a different policy that allows a better exploration of the state space;\n \\item \\textbf{on-policy} algorithms are those algorithms that estimate functions according to the current policy.\n\\end{itemize}\n\nAn on-policy algorithm is influenced by the exploration policy and can get stuck in a local maximum whereas an off-policy algorithm is independent from the exploration policy and reaches the global optimum. \\\\\nAlgorithmically, an on-policy algorithm updates its $\\mathcal{Q}$-values using the $\\mathcal{Q}$-value of the next state $s'$ and the current policy's action $a'$ so it estimates the return for state-action pairs assuming the \\textit{current policy} continues to be followed. An off-policy algorithm updates its $\\mathcal{Q}$-values using the $\\mathcal{Q}$-value of the next state $s'$ and the greedy action $a'$ so it estimates the return for state-action pairs assuming a \\textit{greedy policy} was followed despite the fact that it is not following a greedy policy. \\\\\nThis distinction disappears when the followed policy is a greedy policy but an agent that works like this wouldn't be as good since it would never explore the space of solutions.\n\n%policy based\n%value based\n%model based\n%model free\n\n%off-on policy\n%actor algorithm\n%critic algorithms\n%actor critic algorithms\n\n\\subsection{Deep Q-Network}\n\n\\textbf{Deep Q-Network}, or \\textbf{DQN}, is the first popular reinforcement learning algorithm, proposed by DeepMind in 2013~\\cite{pongpixels}.\n\nIt's an updated version of the Q-learning algorithm first proposed by Watkins in 1992~\\cite{Watkins1992} and uses deep learning to estimate the action-value function. The Q-network is the neural network function approximator with parameters $\\theta$.\n\nSince it approximates and updates the action-value function, DQN is a \\textit{model-free}, \\textit{value-function based} algorithm: it solves the reinforcement learning task directly using samples from the emulator, without explicitly constructing an estimate of the model.\n\nThe two main characteristics regarding this algorithm are how it approximates the action-value function and the \\textbf{experience replay}'s concept.\n\\subsubsection{Deep Reinforcement Learning}\nDeep reinforcement learning can be defined as the connection between reinforcement Learning and neural networks, started in the late 90s~\\cite{Tesauro1994} but with DQN it started to get exciting achievements and visibility. The deep reinforcement learning part is in the approximation of the action-value function with a convolutional neural network with 3 hidden layers. The input is an $84 \\times 84 \\times 4$ image, the first hidden layer convolves 16 $8 \\times 8$ filters with stride 4, the second hidden layers convolves 32 $4 \\times 4$ filters with stride 2, the final hidden layer is a fully-connected layer that consists of 256 rectifier units and the output layer is a fully-connected linear layer with a single output for each valid action.\n\nThe input image represents the state and the output actions are the actions that the agent can perform. It is interesting to see that this same network has been applied to seven Atari games and no preprocessing has been applied to the images (the algorithm worked directly with raw images from the games).\n\n\\subsubsection{Experience Replay}\nThe concept of experience replay was first introduced in 1992~\\cite{Lin1992} as a way to reuse past experiences and replay them to the network in order to refresh them to it. The \\textit{re-learning problem} says that if an input pattern has not been presented for quite a while, the\nnetwork typically will forget what it has learned for that pattern and thus need to re-learn it when that pattern is seen again later. In DQN, a \\textit{replay memory} is used to store past experiences and show them again in a random way.\n\\begin{spacing}{1.15}\n\\begin{algorithm}[ht]\n\\begin{algorithmic}\n\\State Initialize replay memory $\\mathcal{D}$ to capacity $N$\n\\State Initialize action-value function $Q$ with random weights\n%\\State Require preprocessor $h(s)$ that maps histories to fixed-length representations.\n\\For{episode $=1,M$} \n\\State Initialise sequence $s_1 = \\{x_1\\}$ and preprocessed sequenced $\\phi_1 = \\phi(s_1)$\n\\For {$t=1,T$}\n\t\\State With probability $\\epsilon$ select a random action $a_t$\n\t\\State otherwise select $a_t = \\max_{a} Q^*(\\phi(s_t), a; \\theta)$\n\t\\State Execute action $a_t$ in emulator and observe reward $r_t$ and image $x_{t+1}$\n\t\\State Set $s_{t+1} = s_t,a_t,x_{t+1}$ and preprocess $\\phi_{t+1} = \\phi(s_{t+1})$\n\t\\State Store transition $\\left(\\phi_t,a_t,r_t,\\phi_{t+1}\\right)$ in $\\mathcal{D}$\n\t%\\For {$k=1$ to $K$}\n\t\\State Sample random minibatch of transitions $\\left(\\phi_j,a_j,r_j,\\phi_{j+1}\\right)$ from $\\mathcal{D}$\n\t\\State Set\n\t$y_j =\n \\left\\{\n \\begin{array}{l l}\n r_j \\quad & \\text{for terminal } \\phi_{j+1}\\\\\n r_j + \\gamma \\max_{a'} Q(\\phi_{j+1}, a'; \\theta) \\quad & \\text{for non-terminal } \\phi_{j+1}\n \\end{array} \\right.$\n\t\\State Perform a gradient descent step on $\\left(y_j - Q(\\phi_j, a_j; \\theta) \\right)^2$\n\t%\\EndFor\n\\EndFor\n\\EndFor\n\\end{algorithmic}\n\\caption{Deep Q-learning with Experience Replay}\n\\label{alg}\n\\end{algorithm}\n\\end{spacing}\n\n\\noindent\nAlgorithm \\ref{alg} shows the pseudo code of DQN. We can see that after a random initialization of the convnet and of the replay memory, for each episode and for each time step (image in the episode), to the corresponding state $s_t$ an $\\epsilon$-greedy policy is applied (off-policy learning). \\\\\nOptimization is then performed using stochastic gradient descent based on the $\\mathcal{Q}$-learning target and the $\\mathcal{Q}$-network.\n\nNote that the preprocessing on the sequence is just a rescaling of the image, for lower memory utilization, no other preprocessing has been performed.\n\nThis implementation has been tested with 7 Atari games. It showed incredible results, obtaining for every game higher results than previous algorithms and sometimes even outperforming humans.\n\n\\subsection{Asynchronous Advantage Actor Critic}\n\\textbf{A3C}~\\cite{a3c} is a framework developed by the same team as before, DeepMind, that outperforms their first model, DQN, with the only use of CPU (DQN instead relies heavily on GPU). \\\\\nThe authors present four famous reinforcement learning algorithms in an asynchronous way ($\\mathcal{Q}$-learning, SARSA, $n$-step $\\mathcal{Q}$-learning and A3C). What does asynchronous mean? \\\\\nIt means that parallel agents are executed in parallel, on multiple instances of the environment. This parallelism allows a wider view since at any given time-step the parallel agents will be experiencing a variety of different states and after a certain period policies are combined. It has been shown that experience replay is no longer needed and that the same results can be reached by increasing the number of instances running in parallel.\\\\\nThe asynchronous architecture is achieved using multiple CPU threads on a single machine, allowing every consumer computer to use this implementation.\n\nThe best performing algorithm of the four presented is the \\textbf{Asynchronous advantage actor-critic (A3C)}. \\\\\nFirst of all, it's called actor-critic since actor and critic are synonyms of policy-based and value-based respectively~\\cite{konda2003}. This means that this algorithm aims to combine the strengths of both families of algorithms: from the policy-based family of algorithms it gets the simplicity of the problem, optimizing the policy directly with no need to get the action-value function, whereas from the value-based algorithms it gets the ability to \\textit{learn}, since, when working with the policy, when it gets updated, the gradients are computed independently from the previous estimates so there is no \\textit{learning} in the sense of an accumulation and consolidation of informations. Actor-critic methods also have good convergence properties in contrast to critic-only methods.\n\\begin{spacing}{1.15}\n\\begin{algorithm}[ht]\n\\caption{Asynchronous advantage actor-critic - pseudo code for each actor-learner thread.}\n\\begin{algorithmic}\n\\small\n\\State \\emph{// Assume global shared parameter vectors $\\theta$ and $\\theta_v$ and global shared counter $T=0$}\n\\State \\emph{// Assume thread-specific parameter vectors $\\theta'$ and $\\theta'_v$}\n\\State Initialize thread step counter $t\\gets 1$\n\\Repeat\n\\State Reset gradients: $d\\theta \\gets 0$ and $d\\theta_v \\gets 0$.\n\\State Synchronize thread-specific parameters $\\theta'=\\theta$ and $\\theta'_v=\\theta_v$ %\n\\State $t_{start} = t$\n\\State Get state $s_t$\n\\Repeat\n\\State Perform $a_t$ according to policy $\\pi (a_t|s_t;\\theta')$\n\\State Receive reward $r_t$ and new state $s_{t+1}$\n%\n\\State $t \\gets t + 1$\n\\State $T \\gets T + 1$\n\\Until terminal $s_t$ \\textbf{or} $t-t_{start}==t_{max}$\n\\State $R =\n \\left\\{\n \\begin{array}{l l}\n 0 \\quad & \\text{for terminal } s_t\\\\\n V(s_t,\\theta'_v) \\quad & \\text{for non-terminal } s_t \\text{// Bootstrap from last state}\n \\end{array}\\right.$\n\\For {$i \\in \\{t-1,\\ldots,t_{start} \\}$}\n\\State $R \\gets r_i + \\gamma R$\n\\State Accumulate gradients wrt $\\theta'$: $d\\theta \\gets d\\theta + \\nabla_{\\theta'} \\log\\pi(a_i|s_i;\\theta') (R - V(s_i;\\theta'_v))$\n\\State Accumulate gradients wrt $\\theta'_v$: $d\\theta_v \\gets d\\theta_v + {\\partial\\left(R - V(s_i;\\theta'_v)\\right)^2}/{\\partial \\theta'_v}$\n\\EndFor\n%\n\\State Perform asynchronous update of $\\theta$ using $d\\theta$ and of $\\theta_v$ using $d\\theta_v$.\n%\n\\Until $T > T_{max}$\n\\end{algorithmic}\n\\label{a3c-alg}\n\\end{algorithm}\n\\end{spacing}\n\nThis algorithm uses a so called \\textit{forward view} meaning that the algorithm selects actions using its exploration policy for up to a certain number of steps in the future and the agent will then receive up to the same number of rewards from the environment. Forward view in a sentence explains how far ahead you need to look to figure out the value of a state. \\\\\nDeep learning techniques are used to estimate both the policy and the value function where for the policy there is a softmax output instead for the value function there is one linear output with all other layers of a convolutional neural network shared. \\\\\nThe updates on the policy and on the value function are then performed according to the gradient of an estimate of the advantage function that takes into account both the value function parameters and the policy parameters.\n\nIn algorithm~\\ref{a3c-alg} the pseudo code for the A3C algorithm.", "meta": {"hexsha": "4c090450a688ab8a2e9c383bb54240019a109b03", "size": 26459, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/chapter3.tex", "max_stars_repo_name": "DistributedSystemsGroup/tensorpong", "max_stars_repo_head_hexsha": "736ded637c5b6dac8b105ef3bc25cace052b50ba", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/chapter3.tex", "max_issues_repo_name": "DistributedSystemsGroup/tensorpong", "max_issues_repo_head_hexsha": "736ded637c5b6dac8b105ef3bc25cace052b50ba", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/chapter3.tex", "max_forks_repo_name": "DistributedSystemsGroup/tensorpong", "max_forks_repo_head_hexsha": "736ded637c5b6dac8b105ef3bc25cace052b50ba", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.9078212291, "max_line_length": 792, "alphanum_fraction": 0.725386447, "num_tokens": 7162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7981867705385762, "lm_q1q2_score": 0.6469135427011944}} {"text": "\\documentclass[11pt]{article}\r\n\\usepackage{pgfplots}\r\n\\usepackage{amssymb}\r\n\\usepackage{url}\r\n\\pgfplotsset{compat=newest} \r\n\\setlength\\topmargin{-0.6cm} \r\n\\setlength\\textheight{23.4cm}\r\n\\setlength\\textwidth{17.0cm}\r\n\\setlength\\oddsidemargin{0cm} \r\n\\begin{document}\r\n\\title{Ling 572 HW9}\r\n\\author{Daniel Campos \\tt {dacampos@uw.edu}}\r\n\\date{03/13/2019}\r\n\\maketitle \r\n\\section{Q1}\r\n\\subsection{What does f'(x) intend to measure?}\r\nThe derrivative of a function measures the rate of change of a function rleative to the change in the agrument(in this case X).\r\n\\subsection{Let $h(x)=f(g(x))$ }\r\n$h'(x)$ = $f'(g(x)) \\cdot g'(x)$\r\n\\subsection{Let $h(x)=f(x)g(x)$ }\r\n$h'(x) = f'(x)g(x) + f(x)g'(x)$\r\n\\subsection{Let $f(x)=a^x$ where $a>0$}\r\n$ f'(x)$ =$ a^x log(a)$\r\n\\subsection{Let $f(x)= x^{10}-2x^8 \\frac{4}{x^2} + 10$}\r\n$f'(x) =10x^{9}-16x^7 + \\frac{8}{x^3}$\r\n\\section{Q2}\r\nThe logistic function is $f(x)=\\frac{1}{1+e^{-x}}$. The tanh function is $g(x)=\\frac{e^x - e^{-x}}{e^x +e^{-x}}$.\r\n\\subsection{Prove that $f'(x)= f(x)(1-f(x))$}\r\n1. $f'(x) = \\frac{e^{-x}}{(1+e^{-x})^2}$ \\\\\r\n2. $f'(x) = \\frac{1}{1+e^{-x}}-\\frac{1}{(1+e^{-x})^2}$ \\\\\r\n3. $f'(x) = \\frac{1}{1+e^{-x}} \\cdot (1-\\frac{1}{1+e^{-x}})$ which means $f'(x)= f(x)(1-f(x))$\\\\\r\n\\subsection{Prove that $g'(x)=1 - g^2(x)$}\r\n1. $tanh(x) = \\frac {sinh(x))}{cosh(x)}$ \\\\\r\n2. $g'(x) = \\frac{df}{dx} \\frac{sinh(x)}{cosh(x)}$\\\\\r\n3. $g'(x) = \\frac{cosh^2(x) - sinh^2(x)}{cosh^2(x)}$\\\\\r\n4. $g'(x) = \\frac{cosh^2(x)}{cosh^2(x)} - {sinh^2(x)}{cosh^2(x)}$\\\\\r\n5. $g'(x) = 1- {sinh^2(x)}{cosh^2(x)} $\\\\\r\n6. $g'(x) = 1-tanh^2(x)$\\\\\r\n\\subsection{Prove that $g(x) = 2f(2x)-1$}\r\n1. $f(2x) = \\frac{1}{1+e^{-2x}}$ \\\\\r\n2. $2f(2x) =\\frac{2}{1+e^{-2x}}$ \\\\\r\n3. $2f(2x) - 1 = \\frac{2}{1+e^{-2x}} -1$ \\\\\r\n3. $2f(2x) - 1 = \\frac{2}{1+e^{-2x}} - \\frac{1+e^{-2x}}{1+e^{-2x}}$ \\\\\r\n4. $2f(2x) - 1 = \\frac{2- e^{-2x}}{1+e^{-2x}}$ \\\\\r\n5. $2f(2x) - 1 = \\frac{(e^{x}-1)(e^{x}+1)}{1+e^{2x}}$ \\\\\r\n6. $2f(2x) - 1 = \\frac{e^{2x}-1)}{1+e^{2x}}$ \\\\\r\n7. $2f(2x) - 1 = \\frac{e^{2x}-1}{e^{2x}+1} = tanh(x) = g(x)$ \\\\\r\n\\section{Q3} \r\n\\subsection{What is $f'_x$ trying to measure?}\r\nA partial derrivative is trying to measure the change in a fucntion based on a variable assuming all other varriables in the function remain constant. In other words, our derrivative is representing the effect of a variable on the equation when no other variables are effecting the equation.\r\n\\subsection{$f(x,y)=x^3 + 3x^2y+y^3 + 2x$.}\r\n$f'_x = 3x^2 + 6xy+2 $ \\\\ $f'_y = 3(x^2+y^2)$\r\n\\subsection{ $z = \\sum_{i=1}^n w_i x_i$.}\r\n$\\frac{dz}{dw_i}= \\sum{i=1}^n x_i $\r\n\\subsection{$f(z)=\\frac{1}{1+e^{-z}}$ and $z = \\sum_{i=1}^n w_i x_i$.}\r\n$\\frac{df}{dz}= f(z)* \\frac{e^{-z}}{1 + e^{-z}} $ \\\\\r\n$\\frac{df}{dw_i}=f(z)*\\frac{e^{- \\sum{i=1}^n x_i }}{1+e^{- \\sum{i=1}^n x_i }}$ \\\\\r\n\\subsection{$E(z)=\\frac{1}{2}(t - f(z))^2$, $f(z)=\\frac{1}{1+e^{-z}}$ and $z = \\sum_{i=1}^n w_i x_i$.}\r\n$\\frac{dE}{dw_i} = -(t-f(z))*f(z)*\\frac{e^{- \\sum{i=1}^n x_i }}{1+e^{- \\sum{i=1}^n x_i }}$\r\n\r\n\\section{Q4 Softmax Funciton}\r\n\\subsection{Where in NNs is the softmax function used and why?}\r\nSoftmax is used to normalize the outputs of NN to interval (0,1) and to make all components to add up to 1 so that they can be interpreted as regular probabilities. This tends to be implemented as the final step of a NN to get a probability distribution over all possible classes/predictions. \\\\\r\n\\subsection{x is [1, 2, 3, -1, -4, 0], what is the value of softmax(x)}\r\n[0.08607859048507978, 0.23398586833496002, 0.6360395340111326,\\\\ 0.011649470423906664, 0.0005799929804444501, 0.03166654376447658]\r\n\\section{Q5: FNN}\r\n\\subsection{How many connections (i.e., weights) are there in this network?}\r\n$ connections = \\sum_{i=1}^{m-1} n_i*n_{i+1}$ since there are m-1 layers of connections.\r\n\\subsection{Given the input $x$, what is the formula for calculating the output of the first hidden layer?}\r\n$Y_h = g(M_k*X) $ \r\n\\subsection{Given the input $x$, what is the formula for calculating the output of the output layer?}\r\n$Y_o = g(M_mg(M_m-1*...*g(M_k*X))) $ where m is total depth of network\r\n\\section{Q6 MNIST NNs}\r\n\\subsection{What's the loss function used in the digit recognition task?}\r\nThe loss function is Mean squared error(MSE) or $ \\frac{1}{2n} \\sum{x} ||y(x)-a||^2$. \r\n\\subsection{Why do they choose to minimize this function instead of maximizing classification accuracy?}\r\nThey choose do minimize this since the classification accuracy is not a smooth function that can be connected back to the weights and biases in the network. It is difficult to optimize small changes in weights and biases since they are unlikley to affect changes in training accuracy but will likley affect the MSE.\r\n\\subsection{In gradient descent, what's the formula for updating the weight matrix (or vector)? Why is that a good formula?}\r\n$v \\rightarrow v' = v- n*\\triangledown C$ its a good formula because it is easy to calculate and the gradient vector, $\\triangledown C$ and it gives us a way of repeatedly changing the position of v in order to find a minimum of the function C.\r\n\\subsection{What are the main ideas and benefits of stochastic gradient descent?}\r\nThe main idea with stochastic gradient descent(SGD) is to estimate the gradient $\\triangledown C$ on a small sample of randomly chosen tarining inputs. By doing this average over this sample we can get a good estimate of the true gradient and speed up gradient descent. SGD is useful since it allows an efficient and accurate way of directing a NN to a global minimum is a measurable, stepable function without computing the gradient of the entire dataset.\r\n\\subsection{What is a training epoch?}\r\nA training epoch represents all the mini-batches that needed to be run until the network has exhausted all training input. Once this happenes people usually report a current loss function and accuracy and start another epoch.\r\n\\subsection{Let $T$ be the size of the training data, $m$ be the size of mini-batch, and your training process contains $E$ training epoches. How many times is each weight in the NN updated? }\r\n$updates = E*\\frac{T}{m}$\r\n\\subsection{How can one choose the learning rate?}\r\nThe learning rate needs to be explored during hyperparameter tuning. There are various optimzation to choose learning rate such as ADAM but in general many selections of learning rates come from the type of task being modeled and tinkering.\r\n\\subsection{What's the risk if the rate is too big?}\r\nIf the learning rate is too big then the network may never learn the function properly since a high learning rate can effectively skip over maximums\r\n\\subsection{What's the risk if the rate is too small?}\r\nIf the learning rate is too small then the network will train extremely slowly and it may never leave a local minum.\r\n\\section{Q7 MNSIT NN in practice}\r\n\\begin{table}[h]\r\n\\centering\r\n\\caption{Results on digit recognition}\r\n\\label{table1}\r\n\\begin{tabular}{|c|r|l|l|l|r|} \\hline\r\n Expt id & \\# of hidden neurons & epoch \\# & mini batch size & learning rate & accuracy \\\\ \\hline\r\n 1 & 30 & 30 & 10 & 3.0 & 0.9461 \\\\ \\hline\r\n 2 & 10 & 30 & 10 & 3.0 & 0.9172 \\\\ \\hline\r\n 3 & 30 & 30 & 10 & 0.5 & 0.9403 \\\\ \\hline\r\n 4 & 30 & 30 & 10 & 10 & 0.9457 \\\\ \\hline \r\n 5 & 30 & 30 & 100 & 3.0 & 0.9302 \\\\ \\hline \r\n\\end{tabular}\r\n\\end{table} \r\n \\end{document}\r\n", "meta": {"hexsha": "d09803f70f62664ce7fdf908332aece10acf1e2c", "size": 7318, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignments/hw9/readme.tex", "max_stars_repo_name": "spacemanidol/CLMS572", "max_stars_repo_head_hexsha": "f0380de9912c984ec21607cdb3b1f190853c5ca8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignments/hw9/readme.tex", "max_issues_repo_name": "spacemanidol/CLMS572", "max_issues_repo_head_hexsha": "f0380de9912c984ec21607cdb3b1f190853c5ca8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignments/hw9/readme.tex", "max_forks_repo_name": "spacemanidol/CLMS572", "max_forks_repo_head_hexsha": "f0380de9912c984ec21607cdb3b1f190853c5ca8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-12-26T01:28:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-26T01:28:41.000Z", "avg_line_length": 68.3925233645, "max_line_length": 457, "alphanum_fraction": 0.6590598524, "num_tokens": 2606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7981867705385762, "lm_q1q2_score": 0.6469135427011944}} {"text": "\n\\subsection{Weak- and wide-sense stationarity}\n\nUnconditional probabilities don't change over time.\n\nSo GDP would not be stationary, but random noise would. A random walk is not stationary, because the variance increases over time.\n\n\\subsection{Weak-sense stationary}\n\nMean and autocovarinance don't change over time.\n\n\\subsection{Wide-sense stationary}\n\nAll moments are the same.\n\n", "meta": {"hexsha": "7cdc60980ec923c6fe55f7355c43cdaee29c837a", "size": 383, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/probability/stochasticStationary/01-01-stationarity.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/probability/stochasticStationary/01-01-stationarity.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/probability/stochasticStationary/01-01-stationarity.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.9375, "max_line_length": 130, "alphanum_fraction": 0.7963446475, "num_tokens": 82, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8774767810736692, "lm_q2_score": 0.7371581568543044, "lm_q1q2_score": 0.646839166618714}} {"text": "\\documentclass[twoside]{MATH77}\n\\usepackage{multicol}\n\\usepackage[fleqn,reqno,centertags]{amsmath}\n\\begin{document}\n\\begmath 4.6 Solution of a Positive-Definite System \\hbox{with Cholesky Factorization}\n\n\\silentfootnote{$^\\copyright$1997 Calif. Inst. of Technology, \\thisyear \\ Math \\`a la Carte, Inc.}\n\n\\subsection{Purpose}\n\nThis subroutine computes the solution vector ${\\bf x}$ for a system of\nequations of the form\n\\begin{equation}\n\\label{O1}P{\\bf x}={\\bf d},\n\\end{equation}\nwhere $P$ is an N$\\times $N positive-definite symmetric matrix, and ${\\bf d}$ is an\nN-vector. This subroutine also returns the Cholesky factor of $P$ and thus is\napplicable where computing the Cholesky factor is the objective.\n\n\\subsection{Usage}\n\n\\subsubsection{Program Prototype, Double Precision}\n\n\\begin{description}\n\\item[DOUBLE PRECISION] {\\bf P}(LDP, $\\geq $N) [LDP $\\geq $ N]{\\bf , D}($%\n\\geq $N$)${\\bf , U, TOL}\n\n\\item[INTEGER] \\ {\\bf LDP, N, IERR}\n\\end{description}\n\nAssign values to P(,), LDP, N, D(), U, and TOL.\n$$\n\\fbox{{\\bf CALL DCHOL (P, LDP, N, D, U, TOL, IERR)}}\n$$\nThe solution vector ${\\bf x}$ will be stored in D(). Additional computed\nquantities that may be of interest to the user in some situations will be\nstored in P(,) and U.\n\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\\item[P(,)] \\ [inout] On entry this array must contain the N$\\times $N\nsymmetric positive-definite matrix $P$ of Eq.\\,(1). It suffices to provide only\nthe elements on and above the diagonal. On return this array will contain the\nN$\\times $N upper triangular matrix F defined by Eq.\\,(2) on and above the diagonal\npositions of the array P(,). Locations of the array P(,) below the diagonal\nwill not be referenced or modified by this subroutine.\n\n\\item[LDP] \\ [in] Dimension of the first subscript of the storage array\nP(,). Require LDP $\\geq $ N.\n\n\\item[N] \\ [in] Order of the matrix $P$. Require N $\\geq 1.$\n\n\\item[D()] \\ [inout] On entry D() must contain the vector ${\\bf d}$ of Eq.\n(1). On return D() contains the solution vector ${\\bf x}$ for Eq.\\,(1).\n\n\\item[U] \\ [inout] If U contains the number ${\\bf u}$ of Eq.\\,(10) or (15)\nrespectively on entry, then on return U will contain the number $\\rho $ of\nEq.\\,(11) or (16) respectively. If the user is not interested in having the\nnumber $\\rho $ computed, U should be zero on entry and will be unchanged on\nreturn.\n\n\\item[TOL] \\ [in] A user-provided relative tolerance parameter to be used\nin the conditioning test of Eq.\\,(17). We suggest setting TOL to a value of $%\n10^{-(k+1)}$ where $k=\\min (k_A$, $k_b)$. Here $k_A$ is the user's estimate\nof the number of significant decimal digits in the elements of the matrix $A$\nand $k_b$ is the corresponding estimate for ${\\bf b}$. See Eq.\\,(7) or (12)\nfor the definitions of $A$ and ${\\bf b}$. If the TOL input is $<\\varepsilon $,\nwhere $\\varepsilon $ is the relative machine precision $(i.e$. the smallest\npositive number such that $1.0+\\varepsilon \\neq 1.0$ in the machine's\nfloating point arithmetic), then $\\varepsilon $ is used for TOL internally.\n\n\\item[IERR] \\ [out] On return this is set to~0 if $t_{\\min }$ defined in\nEq.\\,(17) is greater than~0. Otherwise results are of\nquestionable validity and $|$IERR$|$ will equal the index of the equation\nthat resulted in the value for $t_{\\min }$. See Section E for more details.\n\\end{description}\n\n\\subsubsection{Modifications for Single Precision}\n\nWe recommend the use of double precision for this computation except on\nmachines such as the Cray that have $10^{-14}$ precision in single\nprecision. To use single precision change DCHOL to SCHOL, and the DOUBLE\nPRECISION type statement to REAL.\n\n\\subsection{Examples and Remarks}\n\nConsider the least-squares problem $A{\\bf x}\\simeq {\\bf b}$ where $A$ is the\n3~by~2 matrix and ${\\bf b}$ is the 3-vector defined by the DATA statements\nin the program DRDCHOL below. This program forms normal equations by\ncomputing $P=A^TA$ and ${\\bf d}=A^T{\\bf b}$. It also computes u $={\\bf b}^T%\n{\\bf b}$. It uses the subroutine DCHOL to solve the normal equations $P{\\bf x%\n}={\\bf d}$, and to compute the quantity RNORM $=\\rho =\\Vert {\\bf b}-A{\\bf x}%\n\\Vert $. Output from this program is given in the file ODDCHOL.\n\nFor programming convenience one may prefer to store the matrix $A$ and the\nvector ${\\bf b}$ together in the same array. The code in DRDCHOL2 shows how\nthis example can be programmed storing $A$ and ${\\bf b}$ together in the\narray AB() and using the array PDU() to hold $P$, ${\\bf d}$, and $u.$\n\n\\subsection{Functional Description}\n\nGiven the problem $P{\\bf x}={\\bf d}$, where $P$ is an N$\\times $N\npositive-definite symmetric matrix, there exists an upper triangular\nN$\\times $N matrix $F$ satisfying\n\\begin{equation}\n\\label{O2}F^TF=P\n\\end{equation}\nEq.\\,(2) defines the Cholesky decomposition of $P$. The upper triangular\nelements of $F$ will be computed from those of $P$ by the following\nequations, where $i=1$, ..., N.\n\\begin{align}\n\\label{O3}g_i&=p_{i,i}-\\sum_{k=1}^{i-1}f_{k,i}^2\\\\\n\\label{O4}f_{i,i}&=g_{i}^{1/2}\\\\\n\\label{O5}f_{i,j}&=\\frac{\\displaystyle p_{i,j}-\\sum_{k=1}^{i-1}f_{k,i}f_{k,j}%\n}{f_{i,i}},\\ \\ j=i+1,...,\\text{N}\\hspace{-2in}\n\\end{align}\nIn these formulas the summation is to be skipped when $i=1.$\n\nAfter computing $F$ the subroutine solves the lower triangular system of\nequations,%\n\\begin{equation*}\nF^T{\\bf y}={\\bf d}\n\\end{equation*}\nand then computes the vector ${\\bf x}$ which satisfies $P{\\bf x}={\\bf d}$ by\nsolving the upper triangular system%\n\\begin{equation*}\nF{\\bf x}={\\bf y}\n\\end{equation*}\nBesides computing the solution vector ${\\bf x}$ this subroutine uses the\ninput number $u$ given in the Fortran variable U to compute\n\\begin{equation}\n\\label{O6}\\rho =\\left[ \\max (0,u-{\\bf y}^T{\\bf y})\\right] ^{1/2}\n\\end{equation}\nThis number $\\rho $ is stored in U on return. If the problem $P{\\bf x}={\\bf d%\n}$ arose as the system of normal equations for a least-squares problem and\nif $u$ was computed appropriately by the user then $\\rho $ represents the\nnorm of the residual vector for the least-squares problem.\n\nSpecifically if the user wishes to solve the least-squares problem of\nminimizing\n\\begin{equation}\n\\label{O7}\\left\\| {\\bf b}-A{\\bf x}\\right\\| =\\left[ \\left( {\\bf b}-A{\\bf x}%\n\\right) ^T\\left( {\\bf b}-A{\\bf x}\\right) \\right] ^{1/2}\n\\end{equation}\nthen $P$, ${\\bf d}$, and $u$ should be initialized as\n\\begin{align}\n\\label{O8}P&=A^TA\\\\\n\\label{O9}{\\bf d}&=A^T{\\bf b}\\\\\n\\label{O10}u&={\\bf b}^T{\\bf b}\n\\end{align}\nThen theoretically, the quantity, $u-{\\bf y}^T{\\bf y}$ of Eq.\\,(6) will be\nnonnegative and the number $\\rho $ of Eq.\\,(6) will have the interpretation\n\\begin{equation}\n\\label{O11}\\rho =\\left\\| {\\bf b}-A{\\bf x}\\right\\|\n\\end{equation}\nMore generally, if the user is solving the weighted least-squares problem of\nminimizing\n\\begin{equation}\n\\label{O12}\\left[ \\left( {\\bf b}-A{\\bf x}\\right) ^TW\\left( {\\bf b}-A{\\bf x}%\n\\right) \\right] ^{1/2}\n\\end{equation}\nwhere $W$ is a positive definite symmetric matrix, then $P$, ${\\bf d}$, and $%\nu$ should be initialized as\n\\begin{align}\n\\label{O13}P&=A^TWA\\\\\n\\label{O14}{\\bf d}&=A^TW{\\bf b}\\\\\n\\label{O15}u&={\\bf b}^TW{\\bf b}\n\\end{align}\nThen, theoretically, the quantity $u-{\\bf y}^T{\\bf y}$ of Eq.\\,(6) will be\nnonnegative and the number $\\rho $ of Eq.\\,(6) will have the interpretation\n\\begin{equation}\n\\label{O16}\\rho =\\left[ \\left( {\\bf b}-A{\\bf x}\\right) ^TW\\left( {\\bf b}-A%\n{\\bf x}\\right) \\right] ^{1/2}\n\\end{equation}\nThe Cholesky factor matrix $F$ will appear in the upper triangular portion\nof the array P(,) on return. If IERR $\\geq 0$, the user can input this\nmatrix $F$ to the library subroutine DCOV2 of Chapter~4.2 to compute the\nunscaled covariance matrix for the associated least-squares problem. This\nrequires building the IP() array: IP(I) = I, for I $=1$, ..., N.\n\nTheoretically the numbers $g_i$ of Eq.\\,(3) will be strictly positive for all\n$i$ if and only if the symmetric matrix $P$ is positive-definite. If all $%\ng_i $ are positive but the ratio $g_i/p_{i,i}$ is very small for some $i$\nthis is an indication that the problem is ill-conditioned. The square of the\nrelative tolerance parameter TOL is used to test this ratio. Let\n\\begin{equation}\n\\label{o17}t_{\\min }=\\min _{1\\leq i\\leq \\text{N}}\\left\\{ g_i-(TOL)^2\\times\n|p_{i,i}|\\right\\} .\n\\end{equation}\nIf $t_{\\min }\\geq 0$, then IERR is set to~0. Otherwise let $m$ be a value of\n$i$ that gives the minimum value in Eq.\\,(17). Then IERR is set to $m$ if $%\ng_m>0$, and is set to $-m$ otherwise. See Section E below for more details.\n\nIf one knows or suspects that the least-squares problem is ill-conditioned\nit is suggested that the Singular Value Analysis subroutine, Chapter~4.3, be\nused to obtain a more complete analysis and a more reliable solution for the\nproblem.\n\nA nonnegative definite symmetric matrix has a Cholesky factor even if it is\nsingular. In computing a Cholesky factor for such matrices this subroutine\ndoes the following: If $g_i$ of Eq.\\,(3) is nonpositive, Eqs.~(4--5) are\nreplaced by\n\\begin{equation}\n\\label{O18}f_{i,j}=0,\\quad j=i,i+1,...,\\text{N}\n\\end{equation}\nWhen solving the triangular systems below Eq.\\,(5), if $f_{i,i}=0$ the\nsolution components $y_i$ and $x_i$ are set to zero. If $P$ is a singular\nnonnegative definite matrix, the matrix $F$ produced in this way is its\n(nonunique) Cholesky factor, $i.e.$, it satisfies Eq.\\,(2). In such a case\nEq.\\,(1) may or may not have a solution and the vector ${\\bf x}$ produced in\nthis way is the solution only if a solution exists.\n\n\\subsection{Error Procedures and Restrictions}\n\nIf $t_{\\min }<0$ in Eq.\\,(17), the subroutine sets IERR nonzero as\nindicated above. When IERR $< 0$, at least one row of the augmented\nmatrix [upper triangle of $P$, $D$] will have been set to zero.\n\nIf IERR $\\neq 0$ we suggest that the user apply the Singular Value Analysis\nsubroutine, Chapter~4.3, to the associated least-squares problem.\\vfill\n\n\\subsection{Supporting Information}\n\nThe source language is ANSI Fortran~77.\n\n\\begin{tabular}{@{\\bf}l@{\\hspace{5pt}}l}\n\\bf Entry & \\hspace{.35in} {\\bf Required Files}\\vspace{2pt} \\\\\nDCHOL & \\hspace{.3in} AMACH, DCHOL\\rule[-5pt]{0pt}{8pt}\\\\\nSCHOL & \\hspace{.3in} AMACH, SCHOL\\\\\n\\end{tabular}\n\nProgrammed by: C. L. Lawson, JPL, May~1969.\n\nProgram Revised by: F. T. Krogh, JPL, September~1991.\n\n\n\\begcode\n\n\\medskip\\\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRDCHOL}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{dchol}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODDCHOL}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{dchol}}\n\n\\newpage\\\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRDCHOL2}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{dchol2}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODDCHOL2}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{dchol2}}\n\\end{document}\n", "meta": {"hexsha": "a3946d639eb52a0505de8ab2680b59d55f282716", "size": 10870, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doctex/ch04-06.tex", "max_stars_repo_name": "jacobwilliams/math77", "max_stars_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2016-01-04T03:17:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-25T19:17:42.000Z", "max_issues_repo_path": "doc/doctex/ch04-06.tex", "max_issues_repo_name": "jacobwilliams/math77", "max_issues_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2019-01-17T02:48:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T16:04:58.000Z", "max_forks_repo_path": "doc/doctex/ch04-06.tex", "max_forks_repo_name": "jacobwilliams/math77", "max_forks_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-01-07T09:26:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-11-25T05:32:54.000Z", "avg_line_length": 41.4885496183, "max_line_length": 98, "alphanum_fraction": 0.7067157314, "num_tokens": 3518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042765, "lm_q2_score": 0.7956581000631542, "lm_q1q2_score": 0.6467644729280715}} {"text": "\\documentclass[a4paper, 11pt]{article}\n\\input{../preamble.tex}\n\n\\DeclareMathOperator{\\N}{\\mathbb{N}}\n\\DeclareMathOperator{\\R}{\\mathbb{R}}\n\\DeclareMathOperator{\\Q}{\\mathbb{Q}}\n\\DeclareMathOperator*{\\Z}{\\mathbb{Z}}\n\n\\title{The Exponential Function}\n\n\\begin{document}\n\n\\tableofcontents\n\\section{The exponential Function}\n\\begin{definition}[The exponential Function] \nThe function $\\exp: \\R \\rightarrow \\R$ is is defined by $\\sum_{n=0}^{\\infty}\\frac{x^{n}}{{n}!}$. \nLet $\\phi: \\R \\rightarrow \\R$ defined by $\\phi(x) = \\lim_{n\\rightarrow \\infty} \\left( 1 + \\frac{x}{n} \\right)^n$.\nWe wish t that $\\phi = \\exp$\n\\end{definition}\n\n\\begin{proposition}\nLet us fix $x \\in \\R$. Then:\n\\begin{enumerate}\n \\item For every natural number $n > |x|$, then $\\left(1 + \\frac{x}{n}\\right)^{n} < (1 - \\frac{x}{n})^{-n}$\n \\begin{proof}\n Suppose $n > |x|$. Then $0 \\leq \\frac{x^{2}}{n^{2}}<1$. Then using this fact,\n consider the product \n \\begin{align*}\n \\left(1 + \\frac{x}{n}\\right)^n \\cdot \\left(1 - \\frac{x}{n}\\right)^n\n &= \\left(\\left(1 + \\frac{x}{n}\\right)\\left(1 - \\frac{x}{n}\\right)\\right)^n \\\\\n &=\\left(1 - \\frac{x^2}{n^2}\\right) \\\\\n &<1\n \\end{align*}\n This follows from multiplying both parts of the previous inequality by the same power. If we divide\n both sides by $\\left(1 - \\frac{x}{n}\\right)^n$, we conclude that $\\left(1 + \\frac{x}{n}\\right)^n < \\left(1 - \\frac{x}{n}\\right)^{-n}$.\n \n \\end{proof}\n \\item The sequence $\\left(1 + \\frac{x}{n}\\right)^n$ for $n > |x|$ is strictly increasing.\n \\begin{proof}\n Suppose $n > |x|$, we want to show that $(n + 1)$st element of the sequence is greather than the $n$th element of the sequence.\n\n That is $\\left(x + \\frac{x}{n}\\right)^n \\leq \\left(x + \\frac{x}{n+1}\\right)^{n + 1}$. The A + M inequality:\n Assume that $n \\geq2 $ and given $a_1, \\dots a_n \\in \\R,$ $(a_1 \\dots a_n)^{1/n} \\leq \\frac{a_1 + \\dots a_n}{n}$.\n Consider $a_1 = 1, a_2 = \\left(1 + \\frac{x}{n}\\right), \\dots a_{n+1} = \\left(1 + \\frac{x}{n}\\right)$.\n \\end{proof}\n \\item The sequence $\\left(1 - \\frac{x}{n}\\right)^{-n}$ for $n > |x|$ is strictly decreasing.\n \\item $\\left(1 + \\frac{x}{n}\\right)^{n} - \\left(1 - \\frac{x}{n}\\right)^{-n} \\rightarrow 0$ as $n \\rightarrow \\infty$.\n These two sequences approach each other from the left and right. \n\\end{enumerate}\n\\end{proposition}\n\nBy (2), (3), and (4) of the proposition for $n > |x|$, the sequence of the closed intervals,\n\\begin{equation*}\n \\left[ \\left(1 + \\frac{x}{n}\\right)^n, \\left(1 - \\frac{x}{n}\\right)^{-n}\\right]\n\\end{equation*}\nis nested. By the \\textbf{Nested Interval Lemma}, there is a unique real number contained in the intersection of all of these\nintervals. The common limit of the sequences of left endpoints and the sequence of right endpoints.\n\nThis enables us to make the following defintion\n\\begin{definition}\nDefine the function $\\phi: \\R \\rightarrow \\R$ for each $x \\in \\R$ where\n\\begin{equation}\n \\phi(x) = \\lim_{n\\rightarrow \\infty} \\left(1 + \\frac{x}{n}\\right)^n = \\left(1 - \\frac{x}{n}\\right)^{-n}\n\\end{equation}\nNote: $\\phi(0) = 1$. We also define $e = \\phi(1)$ as the \\textbf{Euler's constant}.\n\\end{definition}\n\n\\begin{remark}{}\n \\begin{enumerate}\n \\item For every natural number $n \\in \\N$ such that $n > |x|$\n \\begin{equation}\n \\left(1 + \\frac{x}{n}\\right)^n \\leq \\phi(x) \\leq \\left(1 - \\frac{x}{n}\\right)^{-n}\n \\end{equation}\n \\item \n \\end{enumerate}\n\\end{remark}\n\\end{document}", "meta": {"hexsha": "42d19219745a028e4c1f5627e93632b718075f7c", "size": 3440, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "analysis/lecture_16.tex", "max_stars_repo_name": "tylertownsend/mathematics", "max_stars_repo_head_hexsha": "de7732a0fffe5cc1a79752a10b0aa21672f3f413", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "analysis/lecture_16.tex", "max_issues_repo_name": "tylertownsend/mathematics", "max_issues_repo_head_hexsha": "de7732a0fffe5cc1a79752a10b0aa21672f3f413", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "analysis/lecture_16.tex", "max_forks_repo_name": "tylertownsend/mathematics", "max_forks_repo_head_hexsha": "de7732a0fffe5cc1a79752a10b0aa21672f3f413", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.6753246753, "max_line_length": 139, "alphanum_fraction": 0.6281976744, "num_tokens": 1227, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631541, "lm_q2_score": 0.8128673223709251, "lm_q1q2_score": 0.6467644693210736}} {"text": "\n\n\\section{Existing Protocols for Local DP}\n\nApart from R.R., several L.D.P. protocols have been implemented during the years, with many of them being widely used by companies in order to protect users' data. One of the most famous protocols is \\emph{RAPPOR}([15]), created by Google, and being currently used in the Chrome browser for the company to provide useful info to its users without compromising their privacy. Also, Apple has created ts own protocol of L.D.P., and utilize it in their products. \n\nHowever, we are not going to focus on those protocols moving forward, than the ones presented in [10], a paper which introduces many algorithms for L.D.P., each one with different perturbation techniques and suitable for different circumstances.\n\nDuring this chapter we are going to give a definition of each algorithm, implement it using Python, and compare the accuracy results produced by those protocols, just like during our testings of the G.D.P. models. Each protocol has two parts: the \\emph{users} and the \\emph{aggregator}. For the users we must each time define the following functions:\n\n\\begin{itemize}\n \\item $Encode()$: Encodes the true value that the user wants to report\n \\item $Perturb()$: Perturbs the encoded value, in order to produce the random value that will be reported\n\\end{itemize}\n\nFor the aggregator we must each time define the $Aggregate()$ function, that collects the reported random values of the users, and produces the results according to the model.\n\n\\subsection{Basic RAPPOR}\nAs mentioned earlier, RAPPOR is a protocol created by Google. Its simpler form, Basic RAPPOR is used in Chrome, where it collects answers to questions such as the user's home page. The protocol's functions are the following:\n\n\\textbf{Encoding:} $Encode(v) = A_0$, where $A_0$ is a d-bit vector, such that: $A_0[v] = 1$ and $A_0[i] = 0$ for every $i \\neq v$. \n\n\\textbf{Perturbation:} The perturbation consists of 2 steps: the permanent and the instantaneous. The permanent one is carried out only one time, and is the following: \n\n\\begin{equation*}\n Pr[A_1[i] = 1] =\n\t\\begin{cases}\n\t\t1 - \\frac{1}{2}f & \\mbox{if } A_0[i]=1 \\\\\n\t\t\\frac{1}{2}f & \\mbox{ otherwise}\n\t\t\t\\end{cases}\n\\end{equation*}\n\nThe instantaneous step is carried out every time a user reports a value, and is defined as:\n\n\\begin{equation*}\n Pr[A_2[i] = 1] =\n\t\\begin{cases}\n\t\tp & \\mbox{if } A_1[i]=1 \\\\\n\t\tq & \\mbox{ otherwise}\n\t\t\t\\end{cases}\n\\end{equation*}\n\nWe observe from the above functions, that the user must define the $f, p $ and $q$ parameters. Google suggests that we set $f = \\frac{1}{2}$ or $\\frac{1}{4}$, and $p = 0.75$, thus $q = 0.25$. During our testings, those exact parameters were used. \n\n\n\\subsection{Random Matrix Projection}\nIn [14], a protocol with a random matrix projection is proposed, introducing an additional setup step.\n\n\\textbf{Setup:} A random and uniform matrix is generated before any encoding, with it being public and drawn as: $\\Phi \\in \\{-\\frac{1}{m}, \\frac{1}{m}\\}^{m \\times d}$, where $m$ and $d$ are user defined. In our testings, we opt to set $m = 5$ and $d = 10$.\n\n\\textbf{Encoding:} When it comes down to encoding, the function used is the following: $Encoding = (r,x)$, where $r$ is uniformly randomly selected from the range of m, and $x$ is the v-th element of the r-row of the random matrix.\n\n\\textbf{Perturbation}: The perturbation function is defined as following: \n\\begin{align*}\n Perturb(r,x) = (r, b\\cdot c \\cdot m \\cdot x) \n\\end{align*}\n where\n\\begin{equation*}\n b =\n\t\\begin{cases}\n\t\t1 & \\mbox{with } p = \\frac{e^\\epsilon}{e^\\epsilon + 1} \\\\\n\t\t-1 & \\mbox{with } q = \\frac{1}{e^\\epsilon + 1}\t\t\t\n\t\\end{cases}\n\\end{equation*}\nand $c = \\frac{e^\\epsilon +1}{e^\\epsilon -1}$\n\n\\textbf{Aggregation:} Given all the tuples reported by $j$ users in the form $(r, y)$, the estimation for the i-th value of the dataset, is produced by \n\\begin{align*}\n\\sum_{j} y^j \\cdot \\Phi[r^j,i]\n\\end{align*}\n\n\\subsection{Pure Protocols}\n\nThe following protocols are presented in [10], and are called \"pure\" protocols, because of the way they aggregate the data produced by the user. For each one of them, we should define a $Support()$ function, that indicates for each value of the possible outcomes, the reported values that are supported. Thus, with the notation $\\sum_{j} Support(y^j)$, we mean the sum of all the supported values of the y-th element of the dataset.\n\nAlso, for a protocol to be pure, two probabilities must be defined, $p^*$ and $q^*$, where the first notes the probability that the true value is supported by an element $y$, and the second one the probability of another value is supported by the element $y$. The protocol is pure if and only if $p^* > q^*$.\n\nIf a protocol is pure, the estimation of the total reported values for an element of the dataset $i$, is the following:\n\n\\begin{align}\n \\text{Estimation} = \\frac{\\sum_{j} 1_{support(y^j)}(i) - nq*}{p^* - q^*}\n\\end{align}\nwhere $j$ denotes the j-th user reporting their value, and $n$ the total size of the vector of the reported values.\n\n\n\\subsubsection{Direct Encoding}\n\nThis protocol is the natural method of extending the Randomized Response, without the limitation of 2 possible answers. \n\n\\textbf{Encoding:} The protocol does not feature an encoding procedure, thus \n\n\\begin{align*}\n Encode(v) = v\n\\end{align*}\n\n\\textbf{Perturbation:} The perturbation is based on the epsilon setting given by the user, and its function is defined as following:\n\n\\begin{equation*}\n Pr[Perturb(x) = i] =\n\t\\begin{cases}\n \tp = \\frac{e^\\epsilon}{e^\\epsilon + d - 1} & \\mbox{if } i = x \\\\\n \tq = \\frac{1}{e^\\epsilon + d - 1} & \\mbox{if } i \\neq x \n\t\\end{cases}\n\\end{equation*}\n\nwhere $d$ the size of the dataset of the possible answers, $x$ the true value and $i$ the value selected.\n\n\\textbf{Aggregation:} The protocol is pure with $p^* = p$, $q^* = q$ and $Support(i) = i$, thus the predicted results for each of the dataset's values can be calculated from the Equation 4.1.\n\nWe observe that this protocol strongly depends on the size of the dataset of the possible answers, thus when the dataset size increases, the protocol becomes less accurate, due to the decreased probability of selecting the truth. Moreover, for the D.E. protocol, all the false values have the same probability to get chosen, a rather disturbing detail for a query such as a person's age. We will return to these thoughts on later sections.\n\n\\subsubsection{Histogram Encoding}\n\nAn other protocol presented is Histogram Encoding, where an input when having $d$ options is encoded as a $d$-length vector.\n\n\\textbf{Encoding:} The encoding function is for the protocol is\n\n\\begin{align*}\n Encoding(v) = [0, 0, \\dots, 1, \\dots, 0]\n\\end{align*}\n\nwhere only the v-th element of the vector is equal to 1.\n\n\\textbf{Perturbation:} The result of perturbing the encoded vector, is a new vector $B'$, s.t.: \n\n\\begin{align*}\n B'[i] = B[i] + Lap(\\frac{2}{\\epsilon})\n\\end{align*}\nwhere $Lap()$ denotes the noise drawn from the Laplace distribution, where\n\n\n\\begin{align*}\n Pr[Lap(\\beta) = x] = \\frac{1}{2\\beta}e^{\\frac{-|x|}{\\beta}}\n\\end{align*}\n\n\\textbf{Aggregation:} Several methods are proposed for aggregating the results created by the H.E. protocol, but as mentioned by the authors, the best one is called \\emph{Thresholding with H.E.}, where a threshold value is introduced in order to decide what to keep from the reported values. The support function is altered as following:\n\n\\begin{align*}\n Support(B) = \\{v | B[v] > \\theta\\}\n\\end{align*}\nthus, if a noisy output is grater than theta, is set to support the corresponding value. According to the authors, the optimal value for θ is in the range of $(\\frac{1}{2}, 1)$. During the testings that are going to be conducted, we are going to use a threshold of $\\frac{2}{3}$.\n\n\nComparing this protocol to D.E., we observe that is solves the problem of the dependence of the noise drawn by the number of options to choose from. In H.E., no matter how large the domain size is, the noise solely depends on the epsilon value chosen by the user. Thus, when having a large domain size, it is clear that we should prefer the H.E. protocol over D.E.\n\n\\subsection{Unary Encoding}\n\nThe last protocol that is going to take part in the accuracy testings, is the Unary Encoding method, a further exploration of the Basic RAPPOR. It is a unique protocol, as the user does not set the level of privacy using epsilon, but by giving two probabilities, $p$ and $q$ and the epsilon value is computed using those two parameters.\n\n\\textbf{Encoding:} Exactly like in the H.E. method:\n\\begin{align*}\nEncoding(v) = [0, 0, \\dots, 1, \\dots, 0] \n\\end{align*}\nwhere only the v-th element of the vector is equal to 1.\n\n\\textbf{Perturbation:} This step is different than those that we already saw, and is carried out using the following function:\n\n\n\\begin{equation*}\n Pr[B'[i] = i] =\n\t\\begin{cases}\n \tp & \\mbox{if } B[i] = 1 \\\\\n \tq & \\mbox{if } B[i] = 0 \n\t\\end{cases}\n\\end{equation*}\n\nThe epsilon value is decided given $p$ and $q$, and is defined as following:\n\n\\begin{align*}\n \\epsilon = ln(\\frac{p\\cdot(1-q)}{(1-p)\\cdot q})\n\\end{align*}\n\n\\textbf{Aggregation:} The Support function is once again altered, as in the U.E. protocol is defined as following:\n\n\\begin{align*}\n Support(B) = \\{i | B[i] = 1\\}\n\\end{align*}\n\nand of course, $p^* = p$ and $q^* = q$, in order to make the protocol pure. As for the choice of $p$ and $q$, we opt to choose $p = \\frac{1}{2}$, and $q = \\frac{1}{e^\\epsilon + 1}$.\n\n\\section{Testings}\n\n\\subsection{Setup}\n\nNow that all those protocols where introduced, we are going to compare them in order to decide which is better to use when wanting to apply L.D.P. in a dataset. We are going to use a dataset that was created using random values, but corresponds to the age of a group of people. The distribution of the values of the dataset is shown in the histogram in \\textbf{Figure 4.2. }\n\n\\begin{figure}[!htb]\\centering\n \\includegraphics[width=1\\textwidth]{images/true_answers_ldp.png}\n \\caption{True Answers for the Dataset of LDP}\n\\end{figure}\n\nEach user will report one of these 50 values, and the aggregator of each protocol will gather the data given, and try to re-create this histogram in the best manner possible.\n\n\\subsection{Goal}\n\nWe want to decide which protocols behaves better, thus a number of different metrics will be used. The main focus of our testings will be the vectors that the aggregators provide, which we will compare with each other, as well as with the vector containing the true answers. In a similar way as our G.D.P. testings, we are going to run the protocols for different values of epsilon, and different number of users used. The second one is extremely important in L.D.P., as we mentioned earlier that many protocols struggle with a small number of input, as the noise drawn is significant. \n\nWith respect to the choice of metrics, we are going to use the \\emph{Manhattan Distance}, known as the $l1$-norm, as well as the \\emph{Kantorovic Distance}, explained in 3.2.5.1.\n\n\\subsection{Epsilon Measurements}\n\nThe first comparison between the protocols will be with a changing epsilon value, in order to observe how they behave for lower and higher values of the privacy setting. During these testings, all of the users of the dataset were used (approximately 20 thousand), and each run of each protocol was carried out 10 times, just like in other testings, in order to eliminate the danger of drawing extreme values of noise.\n\nFirst, we are going to run all the protocols and compare them using the Manhattan Distance. The results are shown in \\textbf{Figure 4.3.}\n\n\n\\begin{figure}[!htb]\\centering\n \\includegraphics[width=1\\textwidth]{images/epsilon_others_l1.png}\n \\caption{Epsilon Measurements compared by Manhattan Distance}\n\\end{figure}\n\n\nWe gather many useful observations from the graph:\n\n\\begin{itemize}\n \\item The Random Matrix protocol does not function as expected, as \\emph{its accuracy does not follow the logarithmic curve we are used to when epsilon increases.} However, for small values of ε, its results are acceptable, and some times even better than the pure protocols.\n \\item \\emph{The pure protocols behave in a similar way}, with the error stabilizing when epsilon gets higher than 2.5. \n \\item \\emph{The Direct Encoding protocol was the worst behaviour among the pure ones}, with its error being extremely high for epsilon values lower than 1. This is mainly due to the fact that when ε gets too small, the probability of telling the truth gets significantly low, thus creating a big error in accuracy.\n \\item \\emph{The optimized U.E. protocol has the best behaviour} in comparison to the other protocols tested.\n\\end{itemize}\n\nNext up, we are going to run the same testings, but this time using the Kantorovich metric. We expect the protocols to behave even worse, because of the identity of the metric: the Kant. metric pays attention to the distance of the reported answer from the true one. The current protocols do not take into account the distance of the two answers, thus the metric will probably report a higher error. The results of the runs are shown in \\textbf{Figure 4.4.}\n\n\\begin{figure}[!htb]\\centering\n \\includegraphics[width=1\\textwidth]{images/epsilon_others_kant.png}\n \\caption{Epsilon Measurements compared by Kantorovic Distance}\n\\end{figure}\n\n\nAs we expected, the protocols produce a higher error, with the D.E. being the worst among them, especially for lower values of epsilon. This is a rather alarming notice, in which we will come back in Section 4.6.\n\n\n\\subsection{Increasing number of users}\n\n\nThe second experiment that we will conduct is the accuracy error depending on the number of users used during the survey covered by the protocol. In the definition of L.D.P. the observation of the need of lots of users was made, and it is now time that we examine it. We are going to use a \\emph{fixed epsilon value}, one that our protocols behave similarly for (at least the pure ones, in which we will focus our research moving forward). Our epsilon value that we are going ot used will be fixed and equal to 1.5.\n\nWe are going to run the protocols and compare them using the Manhattan Distance. Additionally, we are each time going to divide the result of the metric with the number of users participated, as the simple error is going to increase when the users increase. Hence, this division is going to give us the error depending on the size of our domain. The results are shown in \\textbf{Figure 4.5.}\n\n\\begin{figure}[!htb]\\centering\n \\includegraphics[width=1\\textwidth]{images/nusers_others_l1.png}\n \\caption{Increasing n. of users compared by Manhattan Distance}\n\\end{figure}\n\nThe results confirm the allegations made after explaining the definition of L.D.P.: When the number of participants in a survey is low, the error produced is very high. Every protocol has similar behaviour, as we can see that for fewer than 1000 users the relative error is even 6 times larger than for more than 1000 users. Actually, as we can see from the graph, the turning point is around 2000 users: the relative error drops and stabilizes after this number of participants. \n\nAn other observation made, is that the protocol with the best behaviour during our epsilon measurements (Unary Encoding), is the worst one in the low users testings. This of course gets better as the number of participants rises, and eventually beats the other protocols when it comes down to relative accuracy error. \n\nNext up, we are going to perform the same testings while comparing the protocols with the Kantorovich metric. The results are shown in \\textbf{Figure 4.6}.\n\n\\begin{figure}[!htb]\\centering\n \\includegraphics[width=1\\textwidth]{images/nusers_others_kant.png}\n \\caption{Increasing n. of users compared by Kantorovich Distance}\n\\end{figure}\n\nThe results for a low number of users differ a lot from the runs with the $l1$ distance as a metric. Here, the D.E. method has obviously the worst behaviour, and the U.E. method the best. We can even claim that U.E. seems to solve our problem of high accuracy error. However, no one of those protocols takes into account the distance of the reported answer from the true one, thus it makes sense that the Kantorovich metric produces a huge amount of accuracy error.\n\nThat fact, triggered our thoughts, on what could possibly be done in order to reduce that problem. The thoughts made on this subject are analyzed in the next chapter, by creating a new L.D.P. protocol, sensitive to the distance between the true and the reported values.", "meta": {"hexsha": "5ebcbfeb7766a4af089a51a0256a8ed2f7b31c47", "size": 16632, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis_paper/LDP/other_protocols.tex", "max_stars_repo_name": "nikosgalanis/bsc-thesis", "max_stars_repo_head_hexsha": "b5521e995f266ff1aeb9fecc220650483630dc04", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 15, "max_stars_repo_stars_event_min_datetime": "2021-07-29T15:24:29.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T13:57:07.000Z", "max_issues_repo_path": "thesis_paper/LDP/other_protocols.tex", "max_issues_repo_name": "nikosgalanis/bsc-thesis", "max_issues_repo_head_hexsha": "b5521e995f266ff1aeb9fecc220650483630dc04", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis_paper/LDP/other_protocols.tex", "max_forks_repo_name": "nikosgalanis/bsc-thesis", "max_forks_repo_head_hexsha": "b5521e995f266ff1aeb9fecc220650483630dc04", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.4651162791, "max_line_length": 586, "alphanum_fraction": 0.7455507456, "num_tokens": 4315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.795658104908603, "lm_q2_score": 0.8128673155708975, "lm_q1q2_score": 0.6467644678492837}} {"text": "\\section{The derivation of the ZOGY difference image}\n%\n\\par In this document, we study some practical issues of performing\nthe ZOGY subtraction and its relation to the Alard--Lupton (AL) method\n\\citep{AL1998} combined with the decorrelation afterburner. We assume\nthat the reader is somewhat familiar with the Zackay--Ofek--Gal-Yam\n(ZOGY) algorithm deduction steps in the ZOGY\npaper\\citep{ZOGY2016} Appendix A.\n%\n\\par We recall four key ideas behind the ZOGY subtraction method: a) Given\nan image with uncorrelated, homoscedastic pixel noise (the noise variance in\neach pixel is the same value all over the image), its convolution with an\narbitrary kernel leads to noise correlation between the resulting\npixels. However, in frequency space, the frequencies remain independent (as\nrandom variables), only the amplitude (variance) of their noise content\nchanges. b) The independent frequency space pixels are complex Gaussian\nrandom variables. For a signal detection purpose, similar log\nprobability expressions can be written as for the real-valued random\nvariables (image pixels). c) Log probability can be calculated as a sum over\nall the independent frequencies, weighting the squared absolute difference\nat each frequency with its inverse noise variance. d) The detection\nstatistic in frequency space can be split into two multiplicative terms. In\nimage space, these two terms can be interpreted as a difference image and\nits PSF, which produces a per-pixel detection statistic by convolution. The\ndifference image is constructed in frequency space so that each frequency\nbin is a (complex) random variable and has the same variance. This implies\nthat the difference image in image space has uncorrelated (and under the\nmodel assumptions), homoscedastic noise in its pixels. The difference image\nnoise remains uncorrelated despite the PSF matching procedure.\n%\nWe recall the following equations from the ZOGY paper.\nThe new \\(N\\) science and the reference \\(R\\) images are modeled as:\n\\begin{align}\nR &= F_rT\\otimes P_r + \\epsilon_r\\\\\nN &= F_nT\\otimes P_n + \\epsilon_n\n\\label{eq:N}\n\\end{align}\nwhere \\(F_r, F_n\\) are photometric scaling constants, \\(T\\) is the\n\\emph{truth} image, \\(P_r, P_n\\) are the image PSFs and \\(\\epsilon_r,\n\\epsilon_n\\) are per-pixel Gaussian white noise with homogenous variance in\nthe images.\n%\n\\par The detection statistic of \\(N\\) having a different \\(T\\) value than \\(R\\)\nat any pixel position can be written in frequency space as:\n\\begin{equation}\n \\hat{S} = \\frac{F_nF_r^2\\overline{\\hat{P}_n}\\abs{\\hat{P}_r}^2 \\hat{N}\n - F_rF_n^2\\overline{\\hat{P}_r}\\abs{\\hat{P}_n}^2 \\hat{R}\n}\n{ \\sigma_r^2F_n^2\\abs{\\hat{P}_n}^2 + \\sigma_n^2F_r^2\\abs{\\hat{P}_r}^2 }\n\\label{eq:S}\n\\end{equation}\nIn image space, \\(S\\) is called the score or significance image and\nrepresents the significance of a source detection for each pixel.\n%\nThe difference image is defined as:\n\\begin{equation}\n\\hat{D} = \\frac{F_r \\hat{P}_r\\hat{N} - F_n \\hat{P}_n \\hat{R}}\n{\\sqrt{\\sigma_r^2F_n^2\\abs{\\hat{P}_n}^2 + \\sigma_n^2F_r^2\\abs{\\hat{P}_r}^2}}\n\\label{eq:Dhat}\n\\end{equation}\nand its PSF:\n\\begin{equation}\n\\hat{P}_D = \\frac{F_rF_n\\hat{P}_n\\hat{P}_r}\n{F_D\\sqrt{\\sigma_r^2F_n^2\\abs{\\hat{P}_n}^2 +\n \\sigma_n^2F_r^2\\abs{\\hat{P}_r}^2}}\n\\label{eq:Pdhat}\n\\end{equation}\nso that:\n\\begin{equation}\n\\hat{S} = F_D\\hat{D}\\overline{\\hat{P}_D}\n\\label{eq:Shat}\n\\end{equation}\nThe difference image (and similarly the score image) can be written as the\ndifference of two ``matched'' images as in \\Cref{eq:c1N_c2R}.\n\\begin{equation}\n\\hat{D}_z = \\frac{\\frac{\\hat{P}_r}{F_n}}\n{\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_r}^2\n + \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_n}^2}}\n\\hat{N} -\n\\frac{\\frac{\\hat{P}_n}{F_r}}\n{\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_r}^2\n + \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_n}^2}}\n\\hat{R}\n=\n\\hat{c}_n \\hat{N} - \\hat{c}_r\\hat{R}\n\\label{eq:c1N_c2R}\n\\end{equation}\n%\n\\par Here \\(\\hat{c}_n\\) and \\(\\hat{c}_r\\) are the matching kernels for\nthe original science and template images. If the original image PSFs\nare accurately described by \\(P_r, P_n\\), then the frequency space\nmultiplications transform the PSFs of the two images to be identical,\n\\(P_D\\), \\Cref{eq:Pdhat}. Note that while we followed the terminology\nof the ZOGY paper here and referred to the images as science and\nreference images, the entire ZOGY method is symmetrical to the swapping\nof the images. In the following, we may simply denote images with 1\nand 2 indices.\n%\n\\section{Discussion points}\nWe list the following questions that can define the direction of future\nZOGY image differencing code development in the LSST stack.\n%\n\\par How does an ideal Gaussian PSF point source look like theoretically in\na ZOGY difference image? Discussed in \\Cref{sec:ZOGYtheo}. In\n\\Cref{sec:ZOGYFFT}, we look for answers: What causes the extensive,\noscillating visual patterns in the ZOGY difference image around certain\nsources and image features (\\Cref{sec:FFTlimits})? What shall we do with the\nnumerical problems that appear in certain regions in frequency space and\nappear as pattern artifacts in image space (\\Cref{sec:workaround})? Shall we\nimplement a Gaussian PSF approximator that produces the PSF frequency space\nrepresentation directly? Shall we implement a Gaussian PSF width estimation\nto determine which input PSF is sharper so that a realistic limiting value\ncan be used at frequencies when both PSFs (in frequency space) are below a\nthreshold (\\Cref{sec:workaround})? How shall we handle\ndivision by zero scenarios in the ZOGY difference and significance image\ncalculation (\\Cref{sec:PSFzero})?\n%\n\\par In the Appendix, among other smaller topics, we raise the question whether\nwe can use zero padding for calculating the score image, or shall we use\nmodel white noise padding (\\Cref{sec:zeropadS})?\n%\n\\section{The theoretical solution of the ZOGY matching kernel and difference\nimage PSF\\label{sec:ZOGYtheo}}\n%\n\\par In this section, we derive numerical solutions for pure Gaussian\nPSFs. The inverse Fourier transforms of the ZOGY matching\nkernel or difference image PSF expressions are not expressible in closed\nsymbolic forms, even in this case. We perform numerical integration of the\nfunctions.\n%\n\\begin{figure}\n\\begin{center}\n \\includegraphics[width=5.5in]{fig/zogy_theo_Gaussians_ft_Pd_c1_c2.pdf}\n\\end{center}\n\\caption{\\label{fig:theo_Gaussians_ft}1D slice along the x-axis in frequency\n space of the matching kernels and the PSF of the ZOGY difference\n image. The Fourier space representation of the input PSFs is also\n shown. The two PSFs have widths of $\\sigma_1 = 1$, $\\sigma_2 = 2$ in image\n space, i.e.\\ \\(\\mathrm{PSF}_1\\) is originally the narrower but the Fourier\n transformation swaps this relation.}\n\\end{figure}\n%\n\\par In \\Cref{fig:theo_Gaussians_ft}, we show 1D slices of the 2D solutions\nof \\(P_d\\), \\(c_1\\) and \\(c_2\\). Noise variances and photometric scaling\nfactors are unity for simplicity. As the input PSFs are pure Gaussians,\ni.e.\\ symmetric, real value functions, their Fourier transforms are also\nreal and symmetric.\\footnote{Detailed calculation notebooks are part of\n DM-26087.} Note the different behavior of the two matching kernels towards\nhigh frequencies. The matching kernel of the narrower PSF image \\(c_1\\) goes\nto zero, while the other one goes to unity here. While the graphs of \\(c_1\\)\nand \\(c_2\\) resemble to Gaussians, they are not anymore, and we must use\nnumerical integration to calculate their inverse Fourier transform. Their\nimage space values for points along the x-axis is shown in\n\\Cref{fig:theo_Gaussians_img_c1,fig:theo_Gaussians_img_c2}.\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/zogy_theo_Gaussians_img_Pd.pdf}\n\\end{center}\n\\caption{\\label{fig:theo_Gaussians_img_Pd}1D slice along the x-axis in image\n space of the PSF of the ZOGY difference image with the input image PSFs.}\n\\end{figure}\n%\n\\par In \\Cref{fig:theo_Gaussians_img_Pd}, the PSF of the ZOGY difference\n image is shown. It is close to the wider input PSF, but strictly it's not\n a Gaussian, it has a negative overshoot, about 1\\% of its peak\n value. This means that in an ideal case, signals in a ZOGY difference\n image are expected to have small negative rings around their positive\n peaks.\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=4.5in]{fig/zogy_theo_Gaussians_img_c1.pdf}\n\\includegraphics[width=4.5in]{fig/zogy_theo_Gaussians_img_c1_tails.pdf}\n\\end{center}\n\\caption{\\label{fig:theo_Gaussians_img_c1}1D slice along the x-axis in image\n space of the matching kernel for the narrower PSF image. The matching\n kernel is a Gaussian-like curve that has a small oscillating correction in\n the tails.}\n\\end{figure}\n%\n\\par For the narrower PSF input image, the matched PSF is created by\nconvolution with the matching kernel \\(c_1\\), shown in\n\\Cref{fig:theo_Gaussians_img_c1}. This matching kernel is similar to usual\nGaussian blurring but slightly narrower and has a negative tail itself.\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=4.5in]{fig/zogy_theo_Gaussians_img_c2.pdf}\\,\n\\includegraphics[width=4.5in]{fig/zogy_theo_Gaussians_img_c2_tails.pdf}\n\\end{center}\n\\caption{\\label{fig:theo_Gaussians_img_c2}1D slice along the x-axis in image\n space of the matching kernel for the wider PSF image. The matching kernel\n is the sum of a Dirac delta minus a Gaussian-like curve.}\n\\end{figure}\n%\n\\par For the wider image, the matching kernel \\(c_2\\) is an identity Dirac\ndelta kernel minus a Gaussian-like correction\n(\\Cref{fig:theo_Gaussians_img_c2}). The Dirac delta is the inverse transform\nof the non-zero constant level of \\(c_2\\) in \\Cref{fig:theo_Gaussians_ft}\nthat must be subtracted for the numerical integration to converge. The Dirac\ndelta peak is manually added to the result in \\Cref{fig:theo_Gaussians_img_c2}.\n%\n\\par Note that in case of identical PSFs, both \\(c_1\\) and \\(c_2\\) become\nconstant in \\Cref{fig:theo_Gaussians_ft} which correspond to Dirac deltas in\nimage space. I.e. the matching operation is naturally reduced to the\nidentity operation if the two PSFs are already identical.\n%\n\\clearpage\n%\n\\section{The FFT calculated matching kernel of the ZOGY difference image\\label{sec:ZOGYFFT}}\n%\n\\par In practice, the ZOGY subtraction is implemented by Fast Fourier\nTransforms (FFT). In this and the following sections, we study\npractical numerical aspects of this approach.\n%\n\\par In \\Cref{fig:hits_zogy_artifacts}, we show typical patterns that appear\naround features that do not subtract well. These sources are present in all\nvisits in the HiTS2015 data AL image difference processings and in some\nvisits they produce different artifacts in the AL subtraction as well;\nhowever, AL artifacts are spatially more localized to the source than in the\nZOGY case. The ZOGY patterns can also appear in the vicinity of masked\nregions, cosmic rays, or close to the image edges.\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=3in]{fig/zogy_artifacts_v412060.png}\n\\includegraphics[width=3in]{fig/AL_goodvisit_v412060.png}\n\\end{center}\n\\caption{\\label{fig:hits_zogy_artifacts}High frequency artifacts in the ZOGY\n difference image (left) around bright sources that are present in all\n visits in the AL processing as well. In the same visit, the AL subtraction\n (right) has less pronounced visual imperfections.}\n\\end{figure}\n%\n\\subsection{Zero values of the PSF\\label{sec:PSFzero}}\n%\n\\par In \\Cref{eq:Dhat}, \\(\\hat{D}\\) is not defined at frequencies where both\nimage PSFs are zero. Indeed, according to the image models (\\Cref{eq:N}), at\nthese frequencies, the input images do not carry any information about the\ntrue image. They consist of pure noise. In accordance with this, these\nfrequencies have zero contribution to \\(\\hat{S}\\).\n%\n\\par We cannot allow zero division in our calculations anyway, thus we\nneed to have a workaround for pixels where the denominator in\n\\Cref{eq:S,eq:Dhat} are zero. We define \\(\\hat{D}\\) at these\nfrequencies as the straightforward subtraction of the two images with\nthe same scaling to keep the variance constant at all\nfrequencies. Of course, \\(\\hat{S}=0\\) at these pixels. This is\ncurrently implemented in the code stack.\n%\n\\subsection{Matching kernel limit values in frequency space\\label{sec:FFTlimits}}\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/fft_steps_fc1.pdf}\n\\end{center}\n\\caption{\\label{fig:fft_steps_fc1}FFT calculated matching kernel for two\n Gaussian PSFs, here for the wider PSF image. This is a Fourier space image\n with low frequencies at the corners. In this calculation \\(\\sigma_1=3.3\\),\n \\(\\sigma_2=2.2\\) PSFs were generated in a 31x31 size image, that were zero\n padded to 1024x1024 image size before FFT. Per pixel noise variance is 100\n for both images, photometric scalings are unity. All values are real due\n to symmetry in the inputs.}\n\\end{figure}\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/fft_steps_fc2.pdf}\n\\end{center}\n\\caption{\\label{fig:fft_steps_fc2}FFT calculated matching kernel for the\n narrower PSF image. This is a Fourier space image with low frequencies at\n the corners. All values are real due to symmetry in the inputs.}\n\\end{figure}\n%\n\\par We saw in the theoretical solution section, that in case of\nGaussian PSF-s, the matching kernels in frequency space have tails\nconverging to different limit values. The limit values are either zero\nor a non-zero constant depending on whether the matching kernel belongs\nto the narrower or wider input PSF image, respectively.\n%\n\\par In \\Cref{fig:fft_steps_fc1,fig:fft_steps_fc2}, \\(\\hat{c}_1\\),\n\\(\\hat{c}_2\\) are calculated from two 31x31 pixel size Gaussian PSFs\nthat were zero padded for a 1024x1024 image size, with\n\\(\\sigma_1=3.3\\), \\(\\sigma_2=2.2\\).\\footnote{Detailed calculation\nnotebooks are part of DM-26941.} All numbers are real in this\ncase. The shown frequency space images are in their natural FFT\norientation with zero frequencies at the corners and highest\nfrequencies in the centers. Starting from the corners, both solutions\nfollow our expectations, converging either down to zero or to their\nexpected non zero constant (\\( 1/\\sigma_{\\mathrm{pixelnoise}} \\) )\nplateau. The trend breaks for both kernels in high frequency regions\nhowever, and high value noise appears.\n\\begin{equation}\n\\hat{c}_1 \\sim \\frac{1}{\\sqrt{1 + \\left(\\frac{\\hat{P}_1}{\\hat{P}_2}\\right)^2}}\n\\label{eq:c1conv}\n\\end{equation}\n\\par The matching kernel limit values depend on whether\n\\(\\hat{P}_1/\\hat{P}_2\\) is converging to zero or diverges as it can be seen\nin \\Cref{eq:c1conv}. Once we reach the point where the Gaussian tails are\ndominated by noises\\footnote{See \\Cref{sec:floating_point}}, the convergence\nproperties of these fractions become lost and the calculated matching kernel\nvalues significantly deviate from their expected limit values.\n%\n\\subsection{Patterns in image space\\label{sec:patterns}}\n\\par What does this mean for our calculated matching kernels back in\nimage space?\n%\n\\par In \\Cref{fig:twoG_c1}, we show \\(c_1\\) transformed and\nre-centered back to image space (but still in its fully padded image\nsize). The purple structure indicates that there is a sign oscillation\npattern all across padded size image.\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/twoG_defaults_c1.pdf}\n\\end{center}\n\\caption{\\label{fig:twoG_c1}Two Gaussian PSFs with spatial widths of \\(\\sigma_N = 3.3\\)\n \\(\\sigma_R = 2.2\\) pixels. \\(c_1\\), the ZOGY matching convolution in image space\n of the new (\\(N\\)) image. The purple pattern is an indication of\n sign oscillation all over the image.}\n\\end{figure}\n%\n\\par We can see that in the direction of the two axes, there are definite\npurplish patterns. The purple color on this red-blue color scale shows a\nsign oscillation that can be verified in zoomed-in versions of the\nfigure. These patterns do not fade away in the direction of the axes from\nthe center, indicating that these oscillating sign values have roughly the\nsame order of magnitude absolute values. The original PSF size 31x31 cannot\nbe clearly identified any more in the image either. We note that\nthe appearance of these patterns is independent of the padding\nsize.\n%\nIn \\Cref{fig:twoG_pd} \\(P_d\\) is shown, calculated by \\Cref{eq:Pdhat}.\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/twoG_defaults_Pd.pdf}\n\\end{center}\n\\caption{\\label{fig:twoG_pd}Two Gaussian PSFs with spatial widths of\n \\(\\sigma_N = 3.3\\) \\(\\sigma_R = 2.2\\). \\(P_d\\), the PSF of the zogy\n difference image.}\n\\end{figure}\nWe also show the PSF of \\(S\\) in \\Cref{fig:twoG_ps}. The PSF of the\nscore image shows how a Dirac delta signal (in the truth image)\nappears in \\(S\\), though in source detection, only the actual pixel\nvalues matter in \\(S\\), the shape of the PSF does not.\n\\begin{figure}\n \\begin{center}\n \\includegraphics[width=5.5in]{fig/twoG_defaults_Ps.pdf}\n \\end{center}\n\\caption{\\label{fig:twoG_ps}Two Gaussian PSFs with spatial widths of\n \\(\\sigma_N = 3.3\\) \\(\\sigma_R = 2.2\\) \\(P_s\\), the PSF of the score\n image.}\n\\end{figure}\n%\n\\par \\(P_d\\) and \\(P_s\\) have much cleaner images, contained in size\nin image space, and close to our theoretical\nexpectations. (\\(\\hat{P}_s \\sim \\hat{P}_d \\overline{\\hat{P}_d}\\)).\n%\n\\par Recall that while we expressed \\(P_d\\) in \\Cref{eq:Pdhat} as the\nfunction of the input PSFs, in a difference image this is the result\nof the convolution of the images with the matching kernels. The high\nfrequency noise in the matching kernel is not disturbing, so long the\nimage follows the model PSF assumption and has approximately Gaussian\nPSF features that suppress high frequencies. If there are edges, or\nsignals with high frequency components in the input images, the noisy\nhigh frequency features of the matching kernel becomes visible in the\ndifference image. Our current understanding is that the deviation of\nthe image PSF from the model assumption and the numerical noise\nin the matching kernels together cause the visible artifacts in the\ndifference images produced by the code stack. This conclusion is\nsupported by tests on simulated images that have sources only with\nperfect Gaussian PSFs. In these cases, no visual artifacts can be seen.\n%\n\\subsection{Workaround for artifact suppression\\label{sec:workaround}}\nWe propose the following workarounds for the difference image artifact\nproblem:\n\\begin{itemize}\n \\item In a Gaussian PSF approximation, we can directly create the\n PSF in the padded, full-size frequency space, avoiding the zero\n padding of a small image then the FFT operation. However, this\n approach restricts our input kernels strictly to Gaussians.\n \\item In a more generic approach, we can still use the padded, FFT-d\n detected PSFs of the input images. Using a radius approximation, we\n can determine which input PSF is the wider one in a Gaussian\n approximation. Then we can introduce a configurable threshold in\n \\emph{frequency space} and pixels in the matching kernels can be\n replaced with their Gaussian limit values wherever the input PSFs\n go below the threshold (in absolute value, in frequency space).\n \\item As a third option, we should recall, that the noise artifacts\n appear only in the difference image. In the score image, these are\n automatically suppressed by further convolution with \\(P_d\\). We\n can choose to use the score image only directly for detection\n significance.\n\\end{itemize}\n%\n\\par We repeated the above exercise by generating the Gaussian PSFs directly\nin frequency space and performed exactly the same matching kernel and\n\\(P_d\\) calculations. These results can be seen in\n\\Cref{fig:fft_steps_direct_fc1fc2,fig:fft_steps_direct_c1c2,fig:fft_steps_direct_Pd}. These\nsolutions fully satisfy the theoretical limit values and their image space\ncounterparts are free from any noisy patterns.\n%\n\\begin{figure}\n\\begin{center}\n \\includegraphics[width=4.5in]{fig/fft_steps_direct_fc1.pdf}\n \\includegraphics[width=4.5in]{fig/fft_steps_direct_fc2.pdf}\n\\end{center}\n\\caption{\\label{fig:fft_steps_direct_fc1fc2}The matching kernels for two\n Gaussian PSFs in frequency space. In this calculation, PSFs were\n directly generated in an 1024x1024 frequency space image\n corresponding to image space \\(\\sigma_1=3.3\\), \\(\\sigma_2=2.2\\)\n widths. Per pixel noise variance is 100 for both images, photometric\n scalings are unity.}\n\\end{figure}\n%\n\\begin{figure}\n\\begin{center}\n \\includegraphics[width=4.5in]{fig/fft_steps_direct_c1_zoomed.pdf}\n \\includegraphics[width=4.5in]{fig/fft_steps_direct_c2_zoomed.pdf}\n\\end{center}\n\\caption{\\label{fig:fft_steps_direct_c1c2}The matching kernels inverse\n FFT-d into image space, re-centered and zoomed in for\n details.}\n\\end{figure}\n%\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=5.5in]{fig/fft_steps_direct_Pd_zoomed.pdf}\n\\end{center}\n\\caption{\\label{fig:fft_steps_direct_Pd}The difference image PSF\n inverse FFT-d into image space, re-centered and zoomed in for\n details.}\n\\end{figure}\n%\n\\clearpage\n%\n\\section{Variance plane calculation of the difference image\\label{sec:varplane}}\n\\par While the ZOGY image model does not strictly allow for different\nper-pixel noise values (its noise model assumes homogeneous variance\nnoise across all pixels), from the \\Cref{eq:c1N_c2R} form of the\ndifference image, we can propagate the different pixel variance\ninformation in the variance planes into the difference image. To do\nthis, we notice that if we do convolution on an image of independent\nnoise then, in image space, the variance plane should be convolved by\nthe square of the convolution kernel. This is the well-known square\naddition of variances of independent random variables.\\footnote{See\n also ZOGY paper eqs. 26-29.}\n%\n\\par We'd like to emphasize that this step cannot be applied to an\nimage with already correlated noise; the square addition of pixel\nnoise in the variance plane do not account for the covariance terms\nand would result in underestimation of the pixel\nvariance. Notably, the effect of a noise decorrelation\n(whitening) kernel on an already convolved image cannot be applied to\nthe variance plane based on the square addition rule as it would lower\nthe variance further instead of reverting it to the uncorrelated\nlevel. In accordance with this, the image space square operation is\nnot distributive with respect to convolution in general; the square of the\nconvolution of two kernels is not the same as the convolution of the\nsquared kernels, and we should always perform the former.\n%\n\\par To calculate the variance plane of the difference image, we\nshould calculate \\(c_n\\), \\(c_r\\) in \\Cref{eq:c1N_c2R}, transform them\nback to image space, square them in image space, and convolve the\noriginal images' variance planes with these squared matching kernels\n(\\Cref{eq:VarD}). In practice, this convolution is more\nstraightforward to be performed in frequency space again because these\nimages already share common, full image size dimensions (the\ndimensions of our ZOGY frequency space calculations).\n\\begin{align}\n V_D &= V_N \\otimes \\left(c_n^2\\right) + V_R \\otimes \\left(c_r^2\\right)\\label{eq:VarD}\\\\\n \\sigma_D^2 &= \\sigma_N^2 \\sum c_n^2 + \\sigma_R^2 \\sum c_r^2\\label{eq:sigmaD} \n\\end{align}\n%\n\\par In the homoscedastic approximation (\\Cref{eq:sigmaD}), using the\nParseval theorem (\\Cref{sec:parseval}), it can be seen that the zogy\n\\(\\hat{D}\\) is scaled (\\Cref{eq:Dhat}) so that \\(\\sigma_D^2 = 1\\) for\nevery pixel, while the flux is scaled to \\(F_D\\). Note that in the AL\nplus decorrelation afterburner case, the flux is preserved, and the\nnoise variance is scaled to \\(\\sigma_n^2 / F_n^2 + \\sigma_r^2\n/F_r^2\\). See \\Cref{eq:K,eq:KPre}.\n%\n\\par \\Cref{eq:Shat} can also be written similarly to \\Cref{eq:c1N_c2R}\nand the variance plane of S can be calculated analogously\n(\\Cref{eq:VarD}) this way as well.\n%\n\\section{ZOGY and AL equivalence\\label{sec:ALZOGYequiv}}\n\\par The classic AL algorithm matches the reference image to the new\nscience image by convolving it with a matching kernel. In frequency\nspace, the matching kernel solution ideally equals to the quotient of\nthe two image PSFs as shown in \\Cref{eq:Dal}.\n%\n\\begin{equation}\n \\hat{D}_{AL} =\n \\frac{\\hat{P}_{pre}}{F_n}\\hat{N} -\n \\frac{\\hat{P}_{mk}}{F_r}\\hat{R}\n =\n \\frac{\\hat{P}_{pre}\\hat{N}}{F_n} -\n \\frac{\\hat{P}_n\\hat{P}_{pre}\\hat{R}}{\\hat{P}_r F_r}\n \\label{eq:Dal}\n\\end{equation}\n%\n\\par The decorrelation afterburner was created as a post-processing\ncorrectional step on the difference image. It is calculated in\nfrequency space so that it decorrelates (whitens) the noise of the\ndifference image back in image space. We introduce the formula in\n\\Cref{eq:K} and discuss more details in \\Cref{sec:decorrab}.\n%\n\\begin{align}\n\\hat{K} &= \\frac\n {1}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_{pre}}^2 +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_{mk}}^2}}\n \\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}\\label{eq:K}\n\\end{align}\n%\n\\par In \\Cref{eq:DzPz,eq:DzPrPrPz,eq:DaldPald}, we write the ZOGY score image in frequency\nspace and expand the expression to demonstrate that the AL matching\nand subtraction combined with the decorrelation afterburner noise\nwhitening theoretically leads to the same detection statistics.\n%\n\\begin{align}\n \\hat{S} \\sim \\hat{D}_Z\\overline{\\hat{P}_Z} &= \\frac\n {\\frac{\\hat{P}_r\\hat{N}}{F_n} - \\frac{\\hat{P}_n\\hat{R}}{F_r}}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_r}^2 +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_n}^2}}\n \\cdot\n \\frac{\\overline{\\hat{P}_n\\hat{P}_r}\\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_r}^2 +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_n}^2}} = \\label{eq:DzPz}\\\\\n &= F_D \\frac\n {\\left(\\frac{\\hat{N}}{F_n} - \\frac{\\hat{R}}{F_r} \\cdot\n \\frac{\\hat{P}_n}{\\hat{P}_r}\\right) \\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}} }\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\frac{\\hat{P}_n}{\\hat{P}_r}}^2}}\n \\cdot\n \\underbrace{\n \\frac{\\hat{P}_r}{\\abs{\\hat{P}_r}}\n \\cdot\n \\frac{\\overline{\\hat{P}_r}}{\\abs{\\hat{P}_r}}\n }_{1} % underbrace\n \\cdot\n \\frac{\\overline{\\hat{P}_n}\\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\\frac{\\abs{\\hat{P}_n}^2}{\\abs{\\hat{P}_r}^2}}} = \\label{eq:DzPrPrPz}\\\\\n &= F_D \\hat{D}_{AL+d} \\cdot \\frac{\\overline{\\hat{P}_n}\n \\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\frac{\\hat{P}_n}{\\hat{P}_r}}^2}} =\nF_D \\hat{D}_{AL+d}\\overline{\\hat{P}_{AL+d}}\\label{eq:DaldPald}\n\\end{align}\n%\n%\n\\begin{align}\nF_D &= \\frac{1}{\\sqrt{\\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}}} \\\\\n\\hat{D}_{AL+d} &= \\frac{\\hat{D}_Z}{F_D}\\cdot\\frac{\\abs{\\hat{P}_r}}{\\hat{P}_r}\\label{eq:Dald}\n\\end{align}\n%\n\\par We start with the ZOGY score image in \\Cref{eq:DzPz} and\ndemonstrate that the expression is equivalent with the score\ncalculated from a perfectly matching, decorrelated AL solution in\n\\Cref{eq:DaldPald}. In the AL approach the role of the two images are\nnot symmetric, the template PSF is matched to the science image, and\nthe science image is left intact. Assuming that the AL optimization\nfinds the perfect matching kernel, it should be\n\\(\\hat{P}_n/\\hat{P}_r\\). Indeed, considering the score image, the\ndifference between the AL and ZOGY images are only a factor of\n\\(\\hat{P}_r/\\abs{\\hat{P}_r}\\). Compared to the AL, in the ZOGY case\nboth the difference image and its PSF carry an extra\n\\(\\hat{P}_{r}/\\abs{\\hat{P}_{r}}\\) factor that cancel from the overall\nexpression of the score image. Indeed, as\n\\(\\overline{\\hat{P}_r}\\hat{P}_r/\\abs{\\hat{P}_r}^2 = 1\\) at all\nfrequencies, we can reduce or expand this factor in the difference\nimage and its PSF terms without changing their resulting product, the\nscore image.\n%\n\\par Furthermore, note that \\(\\hat{P}_r/\\abs{\\hat{P}_r}=1\\) itself, if\n\\(\\hat{P}_r\\) is real and positive at all frequencies. This is the\ncase if \\(\\hat{P}_r\\) is a Gaussian PSF function. In this case \\(D_Z\\)\nand \\(D_{AL+d}\\) are mathematically the same as shown in\n\\Cref{eq:Dald}; expanding or reducing the fractions in frequency space\nby arbitrary real, positive kernels have the corresponding operations\nof pre-convolution and deconvolution in image space.\n%\n\\par In the decorrelated AL approach, we can also apply an arbitrary\nGaussian pre-convolution kernel without changing the difference image\nin theory. However, calculating \\Cref{eq:DalPre} in two separate\nsteps, as a difference image that is decorrelated afterwards in a\nsecond step is numerically problematic as discussed in\n\\Cref{sec:decorrab}.\n%\n\\begin{align}\n \\hat{D}_{AL+d} = \\hat{D}_{AL} \\cdot \\hat{K} &= \\frac\n {\\frac{\\hat{P}_{pre}\\hat{N}}{F_n} -\n \\frac{\\hat{P}_n\\hat{P}_{pre}\\hat{R}}{\\hat{P}_r F_r} }\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_{pre}}^2 +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\frac{\\hat{P}_n\\hat{P}_{pre}}{\\hat{P}_r}}^2}}\n \\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}\n \\label{eq:DalPre}\\\\\n%\n &= \\frac\n {\\frac{\\hat{P}_{pre}}{F_n}\\hat{N} -\n \\frac{\\hat{P}_{mk}}{F_r}\\hat{R} }\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}\\abs{\\hat{P}_{pre}}^2 +\n \\frac{\\sigma_r^2}{F_r^2}\\abs{\\hat{P}_{mk}}^2}}\n \\sqrt{\\frac{\\sigma_n^2}{F_n^2} + \\frac{\\sigma_r^2}{F_r^2}}\\label{eq:DaldPreMk}\n\\end{align}\n%\n\\section{The decorrelation afterburner\\label{sec:decorrab}}\nLet's consider the decorrelation afterburner first without the\npre-convolution kernel, when (\\(\\hat{P}_{pre}=1\\)) at all frequencies\nin \\Cref{eq:K}.\n%\n\\par Assuming a Gaussian matching kernel (\\(\\hat{P}_{mk}\\)) that\nconverges to zero towards high frequencies, the overall expression of\n\\(\\hat{K}\\) in \\Cref{eq:K} converges to \\(F_{n}/\\sigma_n\\). The\ndecorrelation correction function in frequency space is similar to\n\\(\\hat{c}_2\\) in \\Cref{fig:theo_Gaussians_ft}. In image space, its\ngraph follows a dirac delta plus a negative overshoot as in\n\\Cref{fig:theo_Gaussians_img_c2}. In the straightforward AL case, when\nwe convolve the template image, we don't expect any complication in\ncalculating such a decorrelation correction. As \\(\\sigma_r \\ll\n\\sigma_n\\), \\Cref{eq:K} can keep its convergence properties even if\n\\(\\hat{P}_{mk}\\) values are noisy in their Gaussian tails.\n%\n\\par Now let's consider the swapped image case, when we convolve the\nscience image. The numerical stability of this case is less\ncertain. As \\(\\sigma_r \\gg \\sigma_n\\) in this case, it can prevent\n\\(\\sigma_{n}^2/F_n^2\\) from becoming the dominant term in the\ndenominator of \\Cref{eq:K} and the numerical noise in the tails of\n\\(\\hat{P}_{mk}\\) may remain in the high frequency values \\(\\hat{K}\\).\n%\n\\par Let's assume now a Gaussian pre-convolution kernel. It can be\nseen that the overall expression of \\(\\hat{K}\\) becomes divergent in\n\\Cref{eq:K} towards high frequencies. The denominator converges to\nzero. Even if we don't directly hit numerical overflows in calculating\n\\(\\hat{K}\\), a divergent \\(\\hat{K}\\) cannot be meaningfully\napplied. The difference image converges to zero (\\(\\hat{D}_{AL}\\),\n\\Cref{eq:Dal}), so correcting the already computed difference image is\nsimilar to inverting a multiplication operation of smaller and smaller\nnumbers. In \\Cref{eq:DaldPreMk}, we can see that calculating the\ncorrected difference image (\\(\\hat{D}_{AL+d}\\)) is exactly the same\nproblem as calculating the ZOGY difference image in \\Cref{eq:Dhat}. In\nthese expressions, both the numerators and the denominators converge to\nzero which in practice result in numerical noise in the tail regions\nof the input Gaussians (image PSFs, pre-convolution and matching\nkernels). We face the very same numerical problems that was discussed\nin \\Cref{sec:ZOGYFFT}.\n% \n\\subsection{Decorrelation afterburner with pre-convolution\\label{sec:decorr_preconv}}\n%\n\\par We saw that the decorrelation afterburner of \\Cref{eq:K} is\nproblematic if we use a pre-convolution kernel. In this form the\ndecorrelation afterburner wants to recover the proper difference\nimage, inverting the pre-convolution operation completely.\n%\n\\par Recall that pre-convolution is a practical way to ensure that the\nscience image PSF is wider than our template PSF. We also know that if\nwe choose to correlate an image with its own\nPSF\\footnote{Pre-convolution with the flipped PSF in image space;\n multiplication with complex conjugate in frequency space (e.g.\\\n \\Cref{eq:Shat}).}, under the independent Gaussian noise model\nassumptions, we get a detection likelihood image.\n%\n\\par We show here that we can apply a form of the decorrelation\nafterburner in the pre-convolution case if we pre-convolve the science\nimage with its own PSF. This form of the decorrelation afterburner\ndoes not invert the pre-convolution operation, in this sense it does\nnot do noise decorrelation any more. Rather it corrects the\npre-convoled and AL PSF matched likelhood difference image directly\nand results in the equivalent of the zogy score image as an optimal\ndetection statistics in the case when both the science and template\nimages have noise.\n%\nWe can rewrite \\Cref{eq:S}:\n\\begin{align}\n \\hat{S} &= \n \\frac{\\overline{\\hat{P}_n} \\frac{\\hat{N}}{F_n} -\n \\frac{\\abs{\\hat{P}_n}^2}{\\hat{P}_r} \\frac{\\hat{R}}{F_r} }\n % denom\n { \\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\n \\frac{\\frac{\\abs{\\hat{P_n}}^2}{\\hat{P}_r}}\n {\\overline{\\hat{P}_r}} \n } =\n \\frac{\\overline{\\hat{P}_n} \\frac{\\hat{N}}{F_n} -\n \\frac{\\abs{\\hat{P}_n}^2}{\\hat{P}_r} \\frac{\\hat{R}}{F_r} }\n % denom\n { \\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\n \\frac{\\abs{\\hat{P}_{mk}}^2}{\\abs{\\hat{P}_{n}}^2}\n } = \n % two terms\n \\underbrace{\\left(\n \\overline{\\hat{P}_n} \\frac{\\hat{N}}{F_n} -\n \\frac{\\abs{\\hat{P}_n}^2}{\\hat{P}_r} \\frac{\\hat{R}}{F_r}\n \\right)}_{\\textrm{pre-convolved AL}} \\cdot\n % denom\n \\frac{1}\n { \\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\n \\frac{\\abs{\\hat{P}_{mk}}^2}{\\abs{\\hat{P}_{n}}^2}\n }\n \\label{eq:SPre}\\\\\n \\hat{P}_{mk} &= \\frac{\\abs{\\hat{P_n}}^2}{\\hat{P}_r} \\label{eq:Smk}\\\\\n \\hat{S} &= \\hat{c}_{sn} \\hat{N} - \\hat{c}_{sr}\\hat{R} =\n \\frac{\\frac{\\overline{\\hat{P}_n}}{ F_n} }\n % denom\n { \\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\n \\frac{\\abs{\\hat{P}_{mk}}^2}{\\abs{\\hat{P}_{n}}^2}\n } \\hat{N}\n -\n \\frac{\n \\frac{\\abs{\\hat{P}_n}^2}{\\hat{P}_rF_r} }\n % denom\n { \\frac{\\sigma_n^2}{F_n^2} +\n \\frac{\\sigma_r^2}{F_r^2}\n \\frac{\\abs{\\hat{P}_{mk}}^2}{\\abs{\\hat{P}_{n}}^2}\n } \\hat{R}\\label{eq:Scncr}\n\\end{align}\n%\n\\par In \\Cref{eq:SPre}, the first term in the numerator is the PSF\npre-convolved science image that now has a PSF of\n\\(\\abs{\\hat{P}_n}^2\\), and the theoretical PSF matching kernel is\nwritten in \\Cref{eq:Smk}. We use the matching kernel to express the\ncorrection in \\Cref{eq:SPre} on the right side. Compared to the\noriginal decorrelation afterburner expression, the correction is\nsquared and the correction with the matching kernel is deconvolved\nwith the pre-convolution kernel.\n%\n\\par Let's consider the convergence properties towards high\nfrequencies in case of Gaussian PSFs. The numerator of this expression\nalways goes to zero while the denominator never goes to zero,\nirrespectively of the widths of \\(\\hat{P}_{mk}\\) and\n\\(\\hat{P}_{n}\\). As such, \\(\\hat{S}\\) always goes to zero.\n%\n\\par If we use pre-convolution to avoid the AL deconvolution case, it\nis guaranteed that the matching kernel solution will be narrower in\nimage space than the PSF of the science image. In frequency space, the\nrelation is the opposite, and the denominator of \\Cref{eq:SPre} should\nactually diverge, making the overall zero convergence faster.\n%\n\\par Note that the zero convergence property of \\(\\hat{S}\\) is kept\neven if the convergence properties are numerically lost in the second\nterm of the denominator at high frequencies. Therefore we do not\nexpect \\emph{worse} high frequency noise to appear in our ``afterburned''\nlikelihood/score image result than it would be in a simple frequency\nspace PSF convolution operation.\n%\n\\par We emphasize that while we can apply this correction on the\nalready calculated pre-convolved AL likelihood difference image, we\ncannot do it on its variance plane. Because the noise in the score\nimage is not white, we cannot easily estimate the noise level even in\nthe homoscedastic noise model case. Instead, we need to follow the\nconsiderations of \\Cref{sec:varplane} for the variance plane\ncalculation and express \\(\\hat{S}\\) with the overall coefficients of\n\\(\\hat{N}\\) and \\(\\hat{R}\\) as shown in \\Cref{eq:Scncr}. Recall, we\ntransform \\(\\hat{c}_{sn}\\) and \\(\\hat{c}_{sr}\\) back to image space,\nsquare them in image space, and convolve the original images' variance\nplanes with these squared matching kernels (\\Cref{eq:VarD}). In the\nhomoscedastic approximation, we multiply \\(\\sigma_n^2\\) and\n\\(\\sigma_r^2\\) with the sum of squares (\\Cref{eq:sigmaD}), but we\nstill need to calculate \\(c_{sn}\\) and \\(c_{sr}\\). In this\napproximation, the inverse FFT of \\(\\hat{c}_{sn}\\) and\n\\(\\hat{c}_{sr}\\) can be spared and the sum can be calculated directly\nin frequency space (Parseval theorem, \\Cref{sec:parseval}).\n%\n\\subsection{Decorrelation afterburner normalization}\n\\par The classic AL method can be applied to images without\nconsidering their photometric scaling factors (\\(F_n, F_r\\)). In this\ncase, the AL optimisation itself solves for the ratio of the\nphotometric scaling between the images. This scaling appears as the\nsum of the AL matching kernel. So far, we've assumed that our AL\nmatching kernels are normalized like image PSFs. Let's relax this\nassumption and separate the sum of the pre-convolution and matching\nkernels into standalone factors \\(S_{pre}\\), \\(S_{mk}\\). To preserve\nthe photometric flux, the decorrelation afterburner correction should\nstill be overall a convolution with a normalized correction kernel in\nimage space.\n%\n\\par Image space sum normalization can be easily ensured in the\ncorresponding frequency space expression. If we evaluate any\nconvolution kernel at 0 frequency, based on \\Cref{eq:X0sum} they\nshould be equal to 1. We should include \\(S_{pre}\\), \\(S_{mk}\\) in the\nnumerator scaling factor to satisfy the normalization criterion. The\nnormalized decorrelation afterburner can be written as \\Cref{eq:KPre}.\n%\n\\begin{align}\n \\hat{K} &= \\frac{\\sqrt{\\frac{\\sigma_n^2}{F_n^2}S_{pre}^2 + \\frac{\\sigma_r^2}{F_r^2}S_{mk}^2}}\n {\\sqrt{\\frac{\\sigma_n^2}{F_n^2}S_{pre}^2\\abs{\\hat{P}_{pre}}^2 + \\frac{\\sigma_r^2}{F_r^2}S_{mk}^2\n \\abs{\\hat{P}_{mk}}^2}}\n\\label{eq:KPre}\n\\end{align}\n%\n\\par Similarly, should the PSFs be not normalized to sum 1 in the ZOGY\nimage differencing, the sum of the PSFs should appear in \\(F_{D}\\) in\nthe same way as a scaling factor.\n%\n\\subsection{A possible fusion of AL and ZOGY}\nTo keep the advantage of AL of not assuming a prior knowledge of the\nimage PSFs and of ZOGY to produce a proper difference image, we\npropose the following algorithmic fusion of the two approaches.\n%\n\\par Let's choose a Gaussian pre-convolution kernel (\\(P_{pre}\\)), so\nthat the AL algorithm can determine an accurate \\(P_{mk}\\) reliably in\nimage space; perhaps in a spatially varying manner for the whole\nimage. The optimisation of the pre-convolution kernel can focus solely\non the quality of the AL matching kernel solution. The expected\ntradeoff here is that the wider the pre-convolution kernel, the easier\nto find a Gaussian matching kernel but on the other hand, a wider\npre-convolution kernel means more noise correlation and a less accurate\nminimisation in the AL algorithm. We do not need to calculate the\nclassic AL difference image in its original \\Cref{eq:Dal} form here.\n%\n\\par In the next step, using \\Cref{eq:DaldPreMk}, we perform a ZOGY\ndifference image calculation using the \\(\\hat{P}_{pre}\\) and\n\\(\\hat{P}_{mk}\\) solutions from the previous step. They behave like\nthe ``image PSFs'' in the original zogy calculations. We should also\napply all the precautions and numerical considerations discussed\nearlier. As \\(\\hat{P}_{pre}\\) is positive and real, it should\ndisappear completely from the proper difference image and score image\nvalues. We note however, that the wider is \\({P}_{pre}\\) in image\nspace, the narrower \\(\\hat{P}_{pre}\\) becomes in frequency\nspace. While multiplication with a Gaussian functions is fully\ninvertible theoretically, numerically such a multiplication will\nsuppress high frequency values and indirectly throws away high\nfrequency (spatial) information from the score image. This is\ndifferent from the tails (and zero points) of the image PSFs. At the\nPSF zero points the images do not carry any information, while with\nthe pre-convolution, we throw away information from the data. Ad\nabsurdum, imagine that we keep only the 0th frequency in frequency\nspace. It'd correspond to a pre-convolution kernel of uniform values\nin image space. We'd average out our whole images and would get one\nuniform detection score for all pixels.\n%\n\\section{Conclusions}\n%\n\\par We studied the ZOGY difference image matching kernels for Gaussian\ninput PSFs in this document. In the theoretical calculations\n(\\Cref{sec:ZOGYtheo}), we showed that the matching kernels have different\nconvergence values in their tails depending whether they belong to the\nnarrower of wider PSF input image. In practice, using FFT, these convergence\nproperties are not well reproduced and the resulting image space matching\nkernels have oscillating patterns all over the image\n(\\Cref{sec:ZOGYFFT}). We concluded that this noise is still acceptable if\nthe input images follow their PSF models and suppress high frequencies,\nhowever noise patterns appear in the difference image if the PSFs\ndeviate. This noise is extended, visually unappealing, and can disrupt other\nalgorithms' performance on the difference image; however, it has little\nimpact on the source detection statistic.\n%\n\\par We tested the direct Gaussian PSF generation in frequency space as a\npossible way to avoid the convergence problems in our calculations. We\nexpect that it would produce difference images without large scale patterns\nfor all inputs. It is a strong restriction on the PSFs, so we also plan to\nlook for weaker constraints in suppressing the artifacts in the difference\nimage. We also need to consider sampling (aliasing) details before\nimplementation.\n%\n\\par In \\Cref{sec:varplane} we discussed how to properly calculate the variance\nplane in frequency space when we have noise whitening decorrelation\noperations.\n%\n\\par In \\Cref{sec:ALZOGYequiv}, we demonstrated that the AL method combined with\nthe decorrelation afterburner leads to the same detection statistic as the\nZOGY method. With preconvolution, they can theoretically lead to the very\nsame difference image. We believe though that his approach would meet\nsimilar practical problems as the ZOGY subtraction has. This is a possible\nfuture topic to study.\n%\n\\par In \\Cref{sec:appendix} we discuss various considerations that have\nrelevance in the actual and for future frequency space image differencing\ncode implementations.\n%\n\\par Finally, the noisy matching kernels cause complications in\nimplementing solutions in frequency space for spatial variations of the PSF\nin a large image. This is also a topic we plan to study in the future.\n\n", "meta": {"hexsha": "5a8f0757b7963d6e9de838f2bc81c9f5ef94b97e", "size": 43595, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "zogy_main.tex", "max_stars_repo_name": "lsst-dm/dmtn-179", "max_stars_repo_head_hexsha": "bf38556421c558aebf91f190ffae5193a5688767", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "zogy_main.tex", "max_issues_repo_name": "lsst-dm/dmtn-179", "max_issues_repo_head_hexsha": "bf38556421c558aebf91f190ffae5193a5688767", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "zogy_main.tex", "max_forks_repo_name": "lsst-dm/dmtn-179", "max_forks_repo_head_hexsha": "bf38556421c558aebf91f190ffae5193a5688767", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.5397727273, "max_line_length": 99, "alphanum_fraction": 0.7471728409, "num_tokens": 12552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916205190225, "lm_q2_score": 0.7490872187162397, "lm_q1q2_score": 0.6467556276775017}} {"text": "\\lab{Facial Recognition Using Eigenfaces}{Facial Recognition Using Eigenfaces}\n\\label{lab:FacialRecognition}\n\n\\objective{Use the singular value decomposition to implement a simple facial recognition system.}\n\nSuppose we have a large database containing images of human faces.\nWe would like to identify people by comparing their pictures to those in the database.\nThis task is called \\emph{facial recognition}.\n\nOne way to automate the comparison process is know as the \\emph{eigenfaces} method.\nAs the name suggests, this method uses eigenvectors of matrices related to the collection of face images.\nThe method essentially projects face images to a low-dimensional subspace, in a way that preserves their distinguishing characteristics.\nComparing the images in fewer dimensions is faster and allows us to store the images using less data.\n\nThe idea of projecting to fewer dimensions is not unique to the eigenfaces method.\nThis method is an example of \\emph{principal component analysis}, where data is compared based on its principal components in a lower-dimensional vector space.\nPrincipal component analysis can be applied to many computing problems besides facial recognition.\n% TODO: list 2-3 other applications\n\n\\begin{comment}\nFacial recognition is important in law enforcement, as well as other situations.\nFor example, facial recognition can be combined with video surveillance to identify when a person is not authorized to be in a location.\n\nHumans can easily compare two face images and determine whether they belong to the same person, but automating this process is more challenging.\nOne technique for automated facial recognition uses \\emph{eigenfaces}.\n\\end{comment}\n\n\\section*{Load the Data} % ====================================================\n\\begin{comment}\nEigenfaces are an efficient way to store and query a database of face images.\nAs the name suggests, this method uses eigenvectors of matrices related to the collection of face images.\nEssentially, the method of eigenfaces projects face images to a lower-dimensional subspace in a way that preserves their distinguishing characteristics.\nIn the lower-dimensional subspace, comparing face images is much faster.\n\\end{comment}\n\nThe first step is to obtain a dataset of face images.\nRecall that a digital image may be stored as an $m \\times n$ array of pixels.\nIn this lab, we will store the images as $mn$-vectors by concatenating the rows of the $m \\times n$ arrays.\n\n\\begin{problem}\n\\label{prob:getTrainingFaces}\nIn this lab we will use the \\li{faces94} face image dataset found at \\url{http://cswww.essex.ac.uk/mv/allfaces/faces94.html}\nThis problem will make sure that you can load and display the images from the dataset.\n\\begin{enumerate}\n\\item Download the \\li{faces94} dataset from the link above and extract the files.\nYou should now have a directory named ``faces94\" which contains photographs of 153 people, organized into folders by person.\n\n\\item The function \\li{getFaces()} is given in the appendix. %or spec file?\nIt constructs a set of face images by selecting exactly one face image for each person in the directory.\nIt should return an array whose columns are flattened face images.\nFeel free to modify the given code.\nYou may have to replace the parameter \\li{\\\"./faces94\\\"} with the location of the directory \\li{faces94} on your machine.\n\nTest this function to make sure it runs without errors.\nCheck that the return value \\li{F} is a $36000 \\times 153$ array.\nThe columns of this array are 153 flattened face images of 153 different people.\n\n\\item Use \\li{plt.imshow()} to display one of the faces.\nThe original image dimensions are $200 \\times 180$.\nYou may find it useful to write a helper function that accepts a flattened image and displays it.\n\n\\end{enumerate}\n\\end{problem}\n\n\n\\section*{Shift By the Mean} % ================================================\n\nThe facial recognition algorithm is more robust if we first \\emph{shift by the mean}. When we shift a set of data by the mean, the distinguishing features are exaggerated. Therefore, in the context of facial recognition, shifting by the mean accentuates the unique features of the face.\nSuppose we have a collection of $k$ face images represented as vectors $\\f_1, \\f_2, \\ldots, \\f_k$ of length $mn$.\nDefine the \\emph{mean face} ${\\boldsymbol \\mu}$ to be the average of the $\\f_i$:\n\\[\n{\\boldsymbol \\mu} = \\frac{1}{k}\\sum_{i=1}^k \\f_i.\n\\]\n\\begin{figure}\n\\includegraphics[width=0.3\\textwidth]{meanFace.png}\n\\caption{The mean face.}\n\\label{facialRecognition:meanFace}\n\\end{figure}\n\n\\begin{problem}\n\\label{prob:meanFace}\n\\leavevmode\n\nThe facial recognition method you will write in this lab will be structured as a class.\nAn outline of the \\li{FacialRec} class is provided in the appendix. %or maybe in the spec file?\nYou will write the methods of the class.\n\nWhen initialized, the \\li{FacialRec} object first loads the face images using \\li{getFaces} and stores the result.\nThe next step is to compute the mean face.\n\\begin{enumerate}\n\\item In your class definition, implement the method \\li{FacialRec.initMeanImage()}.\nCompute the mean face and store it as \\li{self.mu}.\nThis can be done in one line of code using NumPy.\n\n\\item Display the mean face.\nYour result should match Figure \\ref{facialRecognition:meanFace}.\n\\end{enumerate}\n\\end{problem}\n\n\\begin{figure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{differenceFace0.png}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{differenceFace1.png}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{differenceFace2.png}\n\\end{subfigure}\n\\caption{Three mean-shifted faces from the dataset.}\n\\label{facialRecognition:differenceFaces}\n\\end{figure}\nFor each $i = 1,\\ldots, k$, define $\\bar{\\f}_i := \\f_i - {\\boldsymbol \\mu}$.\nThe mean-shifted face vector $\\bar{\\f}_i$ is the deviation of the $i$-th face from the mean, and thus captures the unique features of the face.\nNow form the $mn \\times k$ matrix $\\bar{F}$ whose columns are given by the mean-shifted face vectors, i.e.\n\\[\n\\bar{F} = \\begin{bmatrix}\n\\bar{\\f}_1 & \\bar{\\f}_2 & \\cdots & \\bar{\\f}_k\n\\end{bmatrix}.\n\\]\n\\begin{problem}\n\\leavevmode\n\\begin{enumerate}\n\\item In your class definition, implement the method \\li{FacialRec.initDifferences()}.\nCompute $\\bar{F}$ and store it as \\li{self.Fbar}.\nThis can be done in one line using array broadcasting.\n\\item\nDisplay one of the mean-shifted faces.\nThe output should be similar to the faces in Figure \\ref{facialRecognition:differenceFaces}.\n\\end{enumerate}\n\\end{problem}\n\n\n\\section*{Project to a Subspace} % ============================================\n\nNow suppose we are given a new face vector $\\g$.\nWe first shift $\\g$ by the mean of the dataset, giving us $\\bar{\\g} = \\g - {\\boldsymbol \\mu}$.\nThe closest match to $\\bar{\\g}$ is the vector $\\bar{\\f}_i$ that minimizes $\\|\\bar{\\g}-\\bar{\\f}_i\\|_2$.\nIf there are $k$ images in the dataset, we find this match by comparing $\\bar{\\g}$ to every element of $\\{\\bar{\\f}_1, \\ldots \\bar{\\f}_k\\}$.\n\nHowever, comparing the original face images pixel by pixel is computationally expensive and inefficient.\nThe vectors $\\bar{\\f}_i$ and $\\bar{\\g}$ are length $mn$, which in our case equals 36000 and in practice may be many times larger.\nComputing the difference between face vectors of this length is time consuming, especially when the dataset is very large.\nIt also requires us to use $mn$ values to store each face, which is an inefficient use of space.\nIn addition, pixel by pixel comparison is not very robust to small changes in individual pixels.\n\nInstead, in order to store and compare our face vectors, we would like to represent each one with fewer than $mn$ values.\nWe can do this by projecting to a subspace.\nMathematically, we want to choose $s$, $s \\ll mn$, and project the face vectors to an $s$-dimensional subspace of the original $mn$-dimensional space of images.\nWe can then use just $s$ values to store each face in terms of the basis vectors of the new subspace.\n\nThe ``best\" subspace to project to is the one that is closest in the least squares sense (i.e., such that the sum of the squared errors between $\\{\\bar{\\f}_1, \\ldots \\bar{\\f}_k\\}$ and their projections is minimized).\nLet $U \\Sigma V^T$ be an SVD of $\\bar{F}$, with $\\u_i$ the columns of $U$.\nAs we will prove below, the $s$-dimensional subspace that minimizes the squared error is the span of $\\u_1, \\ldots, \\u_s$.\nNote that $\\u_1, \\ldots, \\u_s$ is an orthonormal basis for this subspace.\n\nThe projection matrix is $P_s = U_s U_s^T$ where $U_s = [ \\u_1\\; \\ldots \\; \\u_s]$.\n%The matrix for the projection is $P_s = U_s U_s^T$ where $U_s$ is the first $s$ columns of $U$.\nThis matrix projects the original face vectors into the optimal $s$-dimensional subspace.\n\n\\begin{comment} % OLD VERSION\n\\section*{Project to a Subspace} % ============================================\nNow suppose we have a new face vector $\\g$.\nThe closest face image to $\\g$ should be the vector $\\bar{\\f}_i$ that minimizes $\\|\\bar{\\g}-\\bar{\\f}_i\\|_2$, where $\\bar{\\g} = \\g - {\\boldsymbol \\mu}$.\nUnfortunately, computing $\\|\\bar{\\g}-\\bar{\\f}_i\\|_2$ for each $i$ is computationally intractable when the length $mn$ of the vectors is large.\nSince a low-resolution photo may easily have $100\\times 100 = 10,000$ pixels, in practice $mn$ is very large indeed.\n\nHere is the trick: instead of computing in the $mn$-dimensional space of all possible images, we will compute in a lower-dimensional subspace.\nWe could start by using the subspace spanned by the vectors $\\bar{\\f}_1, \\ldots \\bar{\\f}_k$, which is at most $k$-dimensional.\nUnfortunately, in practice $k$ is still too large for this subspace to be computationally efficient.\n%We need to project our problem into some subspace of the span of $\\bar{\\f}_1, \\ldots \\bar{\\f}_k$.\n\nTherefore, we want to project to a subspace of $\\bar{\\f}_1, \\ldots \\bar{\\f}_k$ in a way that retains as much information about the $\\bar{\\f_i}$ as possible.\nMathematically, we want to find the $s$-dimensional subspace of $\\text{span}\\{\\bar{\\f}_1, \\ldots \\bar{\\f}_k\\}$ that is closest to the $\\bar{\\f_i}$ in the least-squares sense.\nIn a minute, we will prove that the SVD of $\\bar{F}$ (whose columns are $\\bar{\\f}_1, \\ldots \\bar{\\f}_k$) solves this problem.\nBut first, let us summarize how the solution works.\n\nLet $U \\Sigma V^T$ be an SVD of $\\bar{F}$ with $\\u_i$ the columns of $U$.\nThen the ``best'' $s$-dimensional subspace for approximating $\\text{span}\\{\\bar{\\f}_1, \\ldots \\bar{\\f}_k\\}$ is the span of $\\u_1, \\ldots, \\u_s$.\nThe matrix for this projection is $P_s = U_s U_s^T$ where $U_s = [ \\u_1\\; \\ldots \\; \\u_s]$.\n\nBecause the vectors $\\u_i$ are eigenvectors of $\\bar{F}\\bar{F}^T$, we call them \\emph{eigenfaces}.\nTherefore, the best $s$-dimensional subspace for solving the facial recognition problem is exactly the span of the first $s$ eigenfaces.\n\\end{comment}\n\n\\subsection*{The Proof: SVD as a Least Squares Solution} % --------------------\n\\begin{theorem}\nLet $\\f_1, \\ldots, \\f_k$ be vectors on $\\mathbb{R}^{mn}$, and let $\\bar{F} = [\\bar{f}_1 \\; \\ldots \\; \\bar{f}_k]$. Suppose $U\\Sigma V^T$ is an SVD for $\\bar{F}$. Then the $s$-dimensional subspace that solves the least squares problem for $\\f_1, \\ldots, \\f_k$ is the span of the first $s$ columns of $U$. If $U_s$ is the first $s$ columns of $U$, then the matrix $U_sU_s^T$ is the projection onto this subspace.\n\\end{theorem}\n\\begin{proof}\nWe seek a rank-$s$ projection matrix $P_s$ so that $\\sum_{i=1}^k \\|P_s\\bar{\\f}_i - \\bar{\\f}_i\\|_2^2$ is minimized---i.e.,\n the sum of the squares of the ``errors'' is minimal when we project $\\bar{\\f}_i$ via $P_s$.\n But minimizing this quantity is the same as minimizing its square, which happens to equal the Frobenius norm of $P_s\\bar{F} - \\bar{F}$.\nWritten mathematically,\n \\begin{align*}\n\\inf_{\\text{rank}(P_s)=s} \\sum_{i=1}^k \\|P_s\\bar{\\f}_i - \\bar{\\f}_i\\|_2^2 &= \\inf_{\\text{rank}(P_s)=s} \\left( \\sum_{i=1}^k \\|P_s\\bar{\\f}_i - \\bar{\\f}_i\\|_2^2 \\right) ^2\\\\\n & = \\inf_{\\text{rank}(P_s)=s} \\| P_s\\bar{F}-\\bar{F}\\|_F.\n \\end{align*}\n\nNow let $U \\Sigma V^T$ be an SVD of $\\bar{F}$ with $\\u_i$ the columns of $U$, $\\v_i$ the columns of $V$, and $\\sigma_i$ the singular values of $\\bar{F}$.\nIf $P_s = \\sum_{i=1}^s \\u_i \\u_i^T$, then\n\\begin{align*}\nP_s\\bar{F} &= \\left( \\sum_{i=1}^s \\u_i \\u_i^T \\right)\\left( \\sum_{j=1}^k \\sigma_j \\u_i \\v_i^T \\right)\n= \\sum_{i=1}^s\\sum_{j=1}^k \\sigma_j \\u_i\\u_i^T\\u_j\\v_j^T\\\\\n&= \\sum_{i=1}^s\\sum_{j=1}^k \\sigma_j \\u_i\\delta_{ij}\\v_j^T\n= \\sum_{i=1}^s \\sigma_i \\u_i\\v_i^T.\n\\end{align*}\n\nIn fact, the Schmidt-Eckart-Young-Mirsky Theorem from Lab \\ref{lab:SVD} tells us that $X = \\sum_{i=1}^s \\sigma_i \\u_i\\v_i^T$ is exactly the rank-$s$ matrix that minimizes $\\|X - \\bar{F}\\|_F$.\nSince $P_s \\bar{F}$ will always have rank $s$ or less, the projection $P_s = \\sum_{i=1}^s \\u_i \\u_i^T$ is the one we seek.\nIf we let $U_s = [ \\u_1\\; \\ldots \\; \\u_s]$, then we may write $P_s = U_sU_s^T$. Notice that $P_s$ is projection onto the subspace spanned by the columns of $U_s$.\n\\end{proof}\n\nThe $s$ basis vectors $\\u_1, \\ldots, \\u_s$ are eigenvectors of $\\bar{F}\\bar{F}^T$.\nThey also resemble face images.\nHence, they are commonly called the ``eigenfaces.\"\n\n\\begin{problem}\n\\label{prob:svd}\n\\leavevmode\n\\begin{enumerate}\n\\item In your class definition, implement the method \\li{FacialRec.initEigenfaces()}.\n\nCompute the SVD (\\li{scipy.linalg.svd()} is a good implementation to use) and store the array \\li{U} containing the eigenfaces as its columns.\nBecause we will only use the first few columns of \\li{U}, specify the keyword parameter \\li{full_matrices=False} to compute only the compact SVD.\n\\item Plot the first eigenface (i.e. the first column of \\li{U}).\nIt should resemble the first eigenface shown in Figure \\ref{facialRecognition:eigenfaces}.\n\\end{enumerate}\n\\end{problem}\n\n\\section*{Change Basis} % =====================================================\n\nThe projection matrix $P_s = U_s U_s^T$ projects a face vector into the $s$-dimensional subspace spanned by the eigenfaces, but still keeps it as a vector in $\\mathbb{R}^{mn}$.\nThe change-of-basis matrix $U_s^T$ both projects the face vector and and changes the basis.\nThe resulting vector has length $s$ and represents the face in terms of eigenfaces.\n\nTo represent any face vector in terms of the first $s$ eigenfaces, multiply by $U_s^T$.\nTo change back to a full length-$mn$ projection, multiply again by $U_s$.\n\n\\begin{comment}\nIt does us no good to project all our vectors into an $s$-dimensional space if we still store them as vectors in $\\mathbb{R}^{nm}$.\nInstead, we must store our face vectors in terms of the columns of $U_s$.\nThis way, each vector is a length-$s$ array in NumPy, instead of a length-$mn$ array.\n\nThe change-of-basis matrix is $U_s^T$, so $U_s^TP_s = U_s^TU_sU_s^T = U_s^T$.\nThus we can project into the subspace and change basis by multiplying by $U_s^T$.\nTo change back to the full $mn$-vector, multiply by $U_s$.\n\\end{comment}\n\n\\begin{figure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{eigenface0.png}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{eigenface1.png}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n\\includegraphics[width=\\textwidth]{eigenface2.png}\n\\end{subfigure}\n\\caption{The top three eigenfaces.}\n\\label{facialRecognition:eigenfaces}\n\\end{figure}\n\n\\begin{problem}\n\\label{prob:top_n}\n\\leavevmode\n\\begin{enumerate}\n\\item Implement the method \\li{FacialRec.project()} in your class definition.\nThis should accept a flattened image or an array with flattened images as its columns.\nIt should also accept a value for $s$.\nThe function should project the image or images into the appropriate $s$-dimensional subspace and change basis, then return the result.\n\\item Let \\li{face} be the first mean-shifted face (the first column of \\li{facialRec.Fbar}). Do the following:\n\\begin{enumerate}\n\\item Project \\li{face} so that it is represented in terms of the first 19 eigenfaces.\n\\item Change basis again back to the standard basis on $\\mathbb{R}^{mn}$.\n\\item Add back the mean face \\li{facialRec.mu}.\n\\item Plot the resulting image.\n\\end{enumerate}\nYour image should match Figure \\ref{fig:75_efaces}.\n\\end{enumerate}\n\\end{problem}\n\n\n\\begin{figure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltThirtySecond.png}\n \\caption{5 eigenfaces, about 1/32 of the total.}\n \\label{fig:5_efaces}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltSixteenth.png}\n \\caption{9 eigenfaces, or 1/16 of the total.}\n \\label{fig:9_efaces}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltEighth.png}\n \\caption{19 eigenfaces, about 1/8 of the total.}\n \\label{fig:19_efaces}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltFourth.png}\n \\caption{38 eigenfaces, about 1/4 of the total.}\n \\label{fig:38_efaces}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltHalf.png}\n \\caption{75 eigenfaces, about 1/2 of the total.}\n \\label{fig:75_efaces}\n\\end{subfigure}\n\\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{rebuiltAll.png}\n \\caption{All 153 of the eigenfaces.}\n \\label{fig:all_efaces}\n\\end{subfigure}\n\\caption{Image rebuilt with various numbers of eigenfaces. The image is somewhat recognizable when it is reconstructed with only 1/8 of the eigenfaces.}\n\\label{fig:rebuiltImage}\n\\end{figure}\n\n\\begin{comment}\nOf course, moving to a lower-dimensional subspace by using only the top $n$ eigenfaces introduces some error, in the sense that we cannot perfectly\nreconstruct the face images from their coordinate vectors.\nHowever, this error is tolerable, provided $n$ is not \\emph{too} small.\nNote that we can approximately reconstruct $\\bar{f}_i$ from the coordinate vector $\\hat{f}_i$ by\n\\[\n\\bar{f}_i \\approx U_n\\hat{f}_i.\n\\]\nSee Figure \\ref{facialRecognition:rebuiltImage} for examples of reconstructed face images giving different numbers of eigenfaces.\n\\end{comment}\n\n\\section*{Match Faces} % ======================================================\n\nFinally, we are ready to identify which mean-shifted image $\\bar{\\f}_i$ is closest to an input image, $\\bar{\\g}$.\nWe begin by projecting all vectors to some $s$-dimensional subspace and writing them in terms of the basis vectors, which are the eigenfaces.\nThis is done by multiplying by the change-of-basis matrix:\n\\[\n\\widehat{\\f}_i = U_s^T(\\f_i-{\\boldsymbol \\mu}) \\qquad \\widehat{\\g} = U_s^T(\\g-{\\boldsymbol \\mu}).\n\\]\n\nNext, we compute which $\\widehat{\\f}_i$ is closest to $\\widehat{\\g}$.\nSince the columns of $U_s$ are an orthonormal basis, we get the same result doing the computation in this basis as we would in the standard Euclidean basis.\nDefine\n\\[\ni^* = \\text{argmin}_i \\|\\widehat{\\f}_i - \\widehat{\\g}\\|_2.\n\\]\nThen the $i^*$-th face image is the best match for $\\g$.\n\n\\begin{problem}\n\\label{prob:nearest}\n\\leavevmode\n\\begin{enumerate}\n\\item\nImplement the method \\li{FacialRec.findNearest()} as follows.\n\\begin{lstlisting}\ndef findNearest(self, image, s=38):\n Fhat = # Project Fbar, producing a matrix whose columns are the f-hat defined above\n ghat = # Shift 'image' by the mean and project, producing g-hat as defined above\n # for both Fhat and ghat, use your project function from the previous problem\n\n # Return the index that minimizes ||fhat_i - ghat||_2.\n\\end{lstlisting}\nThe functions \\li{np.linalg.norm()} and \\li{np.argmin()} will be useful for the last line. When using \\li{np.linalg.norm}, make sure you indicate the correct axis.\n\n\\item Test your facial recognition system on faces selected randomly from the \\li{faces94} dataset.\nThe function \\li{sampleFaces(n_tests, path)} at the end of this lab will build an array of \\li{n_tests} random faces from the \\li{faces94} dataset.\n\nPlot the random face beside the face returned by your facial recognition code to see if your system is accurately recognizing faces.\n%The function \\li{show2()} at the end of this lab will plot two face vectors side-by-side.\n\n\\end{enumerate}\n\\end{problem}\n\nBy this point, you have created a basic facial recognition system.\nWe can extend the system to detect when a face doesn't match anything currently in the dataset, and then add this new face.\nWe can also make the system more robust by including multiple pictures of the same face with different expressions and lighting conditions.\n\nAlthough there are other approaches to facial recognition that utilize more complex techniques, the method of eigenfaces remains\na wonderfully simple and effective solution, illustrating another application of the singular value decomposition.\n\n\\section*{Appendix: Helper Code} % ============================================\n\nThis section contains some functions to help you implement the facial recognition class outlined in the problems of this lab.\n\n\\begin{lstlisting}\nimport numpy as np\nfrom scipy import linalg as la\nfrom os import walk\nfrom scipy.ndimage import imread\nfrom matplotlib import pyplot as plt\nfrom random import sample\n\ndef getFaces(path=\"./faces94\"):\n \"\"\"Traverse the directory specified by 'path' and return an array containing\n one column vector per subdirectory.\n\n For the faces94 dataset, this gives an array with just one column for each\n face in the dataset. Each column corresponds to a flattened grayscale image.\n \"\"\"\n\n # Traverse the directory and get one image per subdirectory.\n faces = []\n for (dirpath, dirnames, filenames) in walk(path):\n for f in filenames:\n if f[-3:]==\"jpg\": # only get jpg images\n # load image, convert to grayscale, flatten into vector\n face = imread(dirpath+\"/\"+f).mean(axis=2).ravel()\n faces.append(face)\n break\n\n # put all the face vectors column-wise into a matrix.\n F = np.array(faces).T\n return F\n\n\ndef sampleFaces(n_tests,path = \"./faces94\")\n \"\"\"Return an array containing a sample of n_tests images contained\n in the path as flattened images in the columns of the output.\n \"\"\"\n files = []\n for (dirpath, dirnames, filenames) in walk(path):\n for f in filenames:\n if f[-3:]==\"jpg\": # only get jpg images\n files.append(dirpath+\"/\"+f)\n\n #Get a sample of the images\n test_files = sample(files, n_tests)\n #Flatten and average the pixel values\n images = np.array([imread(f).mean(axis=2).ravel() for f in test_files]).T\n return images\n\n\\end{lstlisting}\n\nThe following is the outline of the Facial Recognition class.\n\n\\begin{lstlisting}\nclass FacialRec:\n ##########Members##########\n # F, mu, Fbar, and U\n ###########################\n def __init__(self,path):\n self.initFaces(path)\n self.initMeanImage()\n self.initDifferences()\n self.initEigenfaces()\n\n def initFaces(self, path):\n self.F = getFaces(path)\n def initMeanImage(self):\n pass\n def initDifferences(self):\n pass\n def initEigenfaces(self):\n pass\n def project(self, A, s=38):\n pass\n def findNearest(self, image, s=38):\n pass\n\\end{lstlisting}\n", "meta": {"hexsha": "b4419233724dbd2aaf3dc042452df7a2b34945ac", "size": 23294, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Vol1A/FacialRecognition/FacialRecognition.tex", "max_stars_repo_name": "joshualy/numerical_computing", "max_stars_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Vol1A/FacialRecognition/FacialRecognition.tex", "max_issues_repo_name": "joshualy/numerical_computing", "max_issues_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Vol1A/FacialRecognition/FacialRecognition.tex", "max_forks_repo_name": "joshualy/numerical_computing", "max_forks_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.8602620087, "max_line_length": 409, "alphanum_fraction": 0.7200137374, "num_tokens": 6491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147276, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.6467556202081012}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{showlabels}\n\\title{Statistical Computing for Scientists and Engineers\\\\[1em] Homework 1}\n\\author{Jiale Shi}\n\\date{September/11/2018}\n\n\\usepackage{natbib}\n\\usepackage{graphicx}\n\n\\begin{document}\n\n\\maketitle\n\n\\newpage\n\\section{Problem 1}\nConsider $n$ samples $D_{n} = (x_{1},...,x_{n})$, independently drawn from normal distribution, $N_{\\mu, \\sigma^{2}})$, with known variance and unknown mean, $\\sigma^{2}$ and $\\mu$, respectively. Assume that the mean has a prior given by the normal distribution:\n\n\\begin{equation}\np(\\mu|m,v^{2})= N(\\mu|m,v^{2}) \\propto \\exp(\\frac{1}{2v^{2}} (\\mu-m)^2)\n%\\pdv[2]{u}{x}+\\pdv[2]{u}{y} = 0\n\\end{equation}\n\nDerive the posterior:\n\n\\begin{equation}\np(\\mu|D) \\sim N(\\mu_{n},\\sigma^{2}) \n\\end{equation}\n\n\\begin{equation}\n\\mu_{n} = \\sigma^{2}\\left(\\frac{m}{v^{2}}+\\frac{n\\Bar{x}}{\\sigma^{2}}\\right), \\sigma_{n}^{2} = \\frac{1}{\\frac{n\\Bar{x}}{\\sigma^{2}}+\\frac{1}{v^2}}\n\\end{equation}\n\nin which $\\Bar{x}$ is the empirical mean $(\\Bar{x}=\\sum^n_{i=1}x_{i})$.\n\nAnswer:\n\\begin{equation}\n\\begin{aligned}\n p(\\mu|D) &= \\prod_{i=1}^{n} f(x_{i}|\\mu) p(\\mu) \\\\\n &\\propto \\exp{-\\frac{\\sum_{i=1}^{n}(x_{i}-\\mu)^2}{2\\sigma^{2}} - \\frac{(\\mu-m)}{2v^2}} \\\\\n &\\propto \\exp{-\\frac{\\mu^2}{2}(\\frac{n}{\\sigma^2}+\\frac{1}{v^2})+\\mu(\\frac{\\sum_{i}^{n}x_i}{\\sigma^2}+\\frac{m}{v^2})} \\\\\n &\\propto \\exp{-\\frac{\\mu^2}{2}(\\frac{n}{\\sigma^2}+\\frac{1}{v^2})+\\mu(\\frac{n\\Bar{x}}{\\sigma^2}+\\frac{m}{v^2})} \\\\\n &\\propto \\exp{-\\frac{1}{2\\sigma_{n}^{2}}(\\mu-\\mu_{n})}\n\\end{aligned}\n\\end{equation}\n\\begin{equation}\n\\frac{1}{\\sigma_{n}^{2}} = \\frac{1}{v^2}+\\frac{n}{\\sigma^2} \\Rightarrow{} \\sigma_{n}^{2} = \\frac{1}{\\frac{1}{v^2}+\\frac{n}{\\sigma^2}}\n\\end{equation}\n\\begin{equation}\n \\mu_n = \\sigma_{n}^2\\left(\\frac{m}{v^2}+\\frac{n\\Bar{x}}{\\sigma^2}\\right) = \\left(\\frac{1}{\\frac{1}{v^2}+\\frac{n}{\\sigma^2}}\\right)\\left(\\frac{m}{v^2}+\\frac{n\\Bar{x}}{\\sigma^2}\\right)\n\\end{equation}\n\n\n\n\\newpage\n\\section{Problem 2}\nConsider a univariate Gaussian $N(X|\\mu, \\lambda^{-1})$ and a dataset $X = \\{x_{1}, x_{2}, x_{3},...,x_{N}\\}$, $x_{i}\\sim N(X|\\mu,\\lambda^{-1})$ of i.i.d observations. Write down the likelihood for this model and show that a conjugate prior is of the form:\n\n\\begin{equation}\n p(\\mu,\\lambda)\\propto \\left[\\lambda^{1/2} \\exp(-\\frac{\\lambda \\mu^{2}}{2})\\right]^{\\beta} \\exp(c\\lambda \\mu - d\\lambda)\n\\end{equation}\n\nwhere $c$, $d$, $\\beta$ are constants. Given $p(\\mu,\\lambda) = p(\\mu\\mid\\lambda) p(\\lambda)$, we can recognize that the above prior is the product of a Gaussian $p(\\mu\\mid\\lambda)$ whose precision is a linear function of $\\lambda$ and of the gamma distribution $p(\\lambda)$. As a result the normalized prior take the form:\n\\begin{equation}\np(\\mu, \\lambda) = N \\left( \\mu\\mid \\mu_{0},(\\beta \\lambda)^{-1}\\right) Gam(\\lambda\\mid a,b) \n\\end{equation}\nwhere $\\mu_{0}=c/\\beta, \\alpha = 1+\\beta/2$ and $b = d-c^2/2\\beta$. Show that the posterior $p(\\mu, \\lambda \\mid X)$ is also a Gaussian-gamma distribution of the same functional form as thr prior:\n\\begin{equation}\n N(\\mu \\mid \\mu_N, ((N+\\beta)\\lambda)^-1) Gam(\\lambda \\mid a_N, b_N)\n\\end{equation}\nwrite down expressions for the parameters of this posterior distribution $\\mu_N$, $a_N$ and $b_N$\n\nAnswer:\nthe likelihood takes the form:\n\\begin{equation}\n\\begin{aligned}\n p(X|\\mu, \\lambda) & = \\prod_{n=1}^{N} f(x_{n}|\\mu) \\propto \\lambda^{N/2}\\exp{-\\frac{1}{2}\\lambda \\sum_{n=1}^N(x_{n}-\\mu)^2}\\\\ & \\propto \\left[\\lambda^{1/2} \\exp{-\\frac{\\lambda \\mu^2}{2}}\\right]^{N} \\exp{\\lambda\\mu\\sum_{n=1}^{N}x_{n}-\\frac{1}{2}\\lambda\\sum_{n=1}^{N}x_{n}^{2}}\n\\end{aligned}\n\\end{equation}\n\nwe need a prior that has a similar functional form in terms of $\\lambda$ and $\\mu$.\n\n\\begin{equation}\n\\begin{aligned}\n&\\beta = N \\\\\n&c = \\sum_{n=1}^{N}x_{n} \\\\\n&d = \\frac{1}{2}\\sum_{n=1}^{N}x_{n}^{2}\\\\\n&p(\\mu,\\lambda)\\propto \\left[\\lambda^{1/2} \\exp(\\frac{\\lambda \\mu^{2}}{2})\\right]^{\\beta} \\exp(c\\lambda \\mu - d\\lambda)\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\np(\\mu,\\lambda) &\\propto \\left[\\lambda^{1/2} \\exp(\\frac{\\lambda \\mu^{2}}{2})\\right]^{\\beta} \\exp(c\\lambda \\mu - d\\lambda) \\\\\n& = (\\beta\\lambda)^{1/2} \\exp{-\\frac{\\beta \\lambda}{2}\\left(\\mu - \\frac{c}{\\beta}\\right)^{2}} \\lambda^{(\\beta-1)/2} \\exp{-\\left(d-\\frac{c^2}{2\\beta}\\right)\\lambda} \\\\\n& = p(\\mu|\\lambda) p(\\lambda) \\\\\n& = N\\left(\\mu| \\mu_{0}=\\frac{c}{d}, \\sigma = (\\beta \\lambda)^{-1}\\right) Gamma\\left(\\lambda|a = \\frac{1+\\beta}{2},b = d-\\frac{c^2}{2\\beta}\\right)\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\np(\\mu,\\lambda|X) &\\propto \\lambda^{N/2}\\lambda^{a-1}\\exp\\left[-\\left(b+\\frac{1}{2}\\sum_{n=1}^{N}x_{n}^{2}+\\frac{\\beta}{2}\\mu_{0}^{2}\\right)\\lambda\\right]\\times \\\\\n&[\\lambda(N+\\beta)]^{1/2}\\exp{-\\frac{\\lambda(N+\\beta)}{2}\\left[\\mu^{2}-\\frac{2}{N+\\beta}(\\beta\\mu_{0}+\\sum_{n=1}^{N}x_{n})\\mu\\right]} \\\\\n& = N(\\mu | \\mu_{N},[\\lambda(N+\\beta)]^{-1}) Gam(\\lambda | a_{N},b_{N}) \\\\\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n& \\mu_{N}= -\\frac{\\beta\\mu_{0}+\\sum_{n=1}^{N}x_{n}}{N+\\beta}\\\\\n& a_{N} = \\frac{N}{2}+a\\\\\n& b_{N} = -\\left(b+\\frac{1}{2}\\sum_{n=1}^{N}x_{n}^{2}+\\frac{\\beta}{2}\\mu_{0}^{2}-\\frac{(\\beta\\mu_{0}+\\sum_{n=1}^{N}x_{n})^{2}}{2(N+\\beta)}\\right)\n\\end{aligned}\n\\end{equation}\n\n\n\\newpage\n\\section{Problem 3}\nThe Wishart distribution distribution $W_m(n,\\Sigma)$.\n\n1. To prove $E[X|n,\\Sigma]=n\\Sigma$.\n\n\n$X = \\sum_{i=1}^{n} z_{i}z_{i}^{T} \\sim W_{m}(n,\\Sigma)$, where $z_{1},...,z_{n} \\sim N_{m}(0,\\Sigma)$, and $z_{1},...,z_{n}$ are independent with each other.\n\\begin{equation}\n E[X|n,\\sum] =E[\\sum_{1}^{n}z_{i}z_{i}^{T}] = \\sum_{1}^{n}E[z_{1}z_{1}^{T}] = \\sum_{1}^{n} Var(z_{1}) = n\\Sigma\n\\end{equation}\n\nTherefore,$E[X|n,\\Sigma]=n\\Sigma$.\n\n2. To prove $Cov[X] = 2n\\Sigma \\bigotimes \\Sigma$.\n\n\\begin{equation}\n\\begin{aligned}\n Cov(X) &= Cov\\left(\\sum_{i=1}^{n} z_{i}z_{i}^{T}\\right) \\\\\n &= \\sum_{1}^{n}Cov\\left(z_{1}z_{1}^{T}\\right) \\\\\n &= n Cov\\left(z_{1}z_{1}^{T}\\right) \\\\\n &= n Cov\\left(z_{1} \\boxed{ } z_{1}\\right) \\\\\n &= n Cov((\\Sigma \\bigotimes \\Sigma)(x \\boxed{ } x)) \\\\\n &= 2n\\Sigma \\bigotimes \\Sigma\n\\end{aligned}\n\\end{equation}\n\n\\newpage\n\\section{Problem 4}\n(a) Derive the log Gamma distribution $Y \\sim log[X]$, $X= G(\\alpha, \\beta)$ and show that it belongs to both families above.\n\nAnswer:\n$Y \\sim log[X]$, $X= G(\\alpha, \\beta)$, therefore, $X \\sim \\exp[Y]$.\n\\begin{equation}\n\\begin{aligned}\n f_{Y}(x) & = f_{X=e^Y} (e^x) \\frac{d e^x}{d x} = \\frac{1}{\\Gamma(\\alpha)} e^{-\\beta e^x}(\\beta e^{x})^{\\alpha} \\\\\n & = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} e^{-\\beta e^x}(e^{x})^{\\alpha}\n\\end{aligned}\n\\end{equation}\n\nCheck whether it belongs to exponential family.\n\\begin{equation}\n\\begin{aligned}\n \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} e^{-\\beta e^x}(e^{x})^{\\alpha} & = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} \\exp{-\\beta e^x+x\\alpha} \\\\\n & = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} \\exp{(-\\beta,\\alpha)(e^x,x)^{T}} \\\\\n & = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} \\exp{\\eta(\\alpha,\\beta)^{T}u(x)} \n\\end{aligned}\n\\end{equation}\nwhere $u(x)=(e^x,x)^T, \\eta(\\alpha,\\beta)=(-\\beta,\\alpha)^{T}, h(x)=1,g(\\alpha,\\beta)=\\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} $. Therefore, the log Gamma distribution $Y$ belongs to the exponential family.\n\\newline\n(b) Provide examples of two distributions that part of the exponential family.\n\nAnswer:\nConsider the Bernoulli Distribution\n\\begin{equation}\n\\begin{aligned}\n p(x|\\mu) & = \\mu^{x}(1-\\mu)^(1-x) \\\\\n & = \\exp{x\\ln{\\mu}+(1-x)\\ln\\left(1-\\mu\\right)} \\\\\n & = (1-\\mu)\\exp{\\ln\\left(\\frac{\\mu}{1-\\mu}\\right)x}\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n&\\eta = \\ln\\left(\\frac{\\mu}{1-\\mu}\\right) \\\\\n&\\mu = \\sigma(\\eta)= \\frac{1}{1+e^{-\\eta}} \\\\\n& g(\\eta) = 1-\\mu = \\frac{1}{1+e^{\\eta}} \\\\\n& p(x|\\mu) = g(\\eta)\\exp{\\eta x}, \\mu(x)= x, h(x)=1,g(\\eta) = 1-\\mu = \\frac{1}{1+e^{\\eta}}\n\\end{aligned}\n\\end{equation}\n\nConsider the Beta distribution\n\n\\begin{equation}\n\\begin{aligned}\n p(x|a,b) & = \\frac{\\Gamma(a+b)}{\\Gamma(a)\\Gamma(b)} x^{a-1}(1-x)^{b-1} \\\\\n & = \\frac{\\Gamma(a+b)}{\\Gamma(a)\\Gamma(b)} \\exp[(a-1)\\ln(x) + (b-1)\\ln(1-x)] \\\\\n & = h(x)g(\\eta)\\exp{\\eta^{T}u(x)}\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n&\\eta = (a-1,b-1)^{T} \\\\\n&\\mu(x) = (\\ln(x),\\ln(1-x))^{T} \\\\\n& g(a,b) = \\frac{\\Gamma(a+b)}{\\Gamma(a)\\Gamma(b)} \\\\\n& h(x)=1\n\\end{aligned}\n\\end{equation}\n\\newline\n(c) Provide examples of two distributions that are part of the local-scale family.\n\nAnswer:\n\nConsider the Cauchy distribution:\ntake $\\Psi(x) = \\frac{1}{\\pi}\\frac{1}{1+x^2}$\n\\begin{equation}\n\\begin{aligned}\np(x|a,b) = \\frac{1}{b}\\Psi\\left(\\frac{x+a}{b}\\right) = \\frac{1}{\\pi b}\\frac{1}{1+\\left(\\frac{x+a}{b}\\right)^2}\n\\end{aligned}\n\\end{equation}\ndefines the Cauchy L-S family.\n\nConsider the Uniform distribution:\ntake $\\Psi(x) = I_{(0,1)}(x)$\n\\begin{equation}\n\\begin{aligned}\n I_{(0,1)}(x) = 1, if 0 0$.\nSimilarly, we observe that we must have $\\sigma^2 \\sim \\scaledInvChi{\\nu_0, \\sigma_0^2}$, with hyperparameters $\\nu_0$ and $\\sigma_0^2 > 0$.\nThis becomes apparent when considering the respective densities.\nFollowing \\citet{gelmanbda04} we write $(\\mu, \\sigma^2) \\sim \\NormalscaledInvChi{\\mu_0, \\sigma_0^2 / \\kappa_0; \\nu_0, \\sigma_0^2}$, with corresponding density function $p(\\mu, \\sigma^2) = p(\\mu \\mid \\sigma^2) p(\\sigma^2) \\propto (\\sigma^2)^{\\frac{3 + \\nu_0}{2}} \\EXP{-\\frac{1}{2\\sigma^2} \\left[\\nu_0 \\sigma_0^2 + \\kappa_0(\\mu_0 - \\mu)^2 \\right]}$.\nMultiplying the likelihood with our constructed prior we get the joint posterior\n\\begin{align}\n p(\\mu, \\sigma^2 \\mid y) \\propto& (\\sigma^2)^{-\\frac{3 + \\nu_0 + n}{2}} \\EXP{-\\frac{1}{2 \\sigma^2} \\left[\\nu_0\\sigma_0^2 + \\kappa_0(\\mu - \\mu_0)^2 + (n-1)s^2 + n(\\bar{y} - \\mu)^2 \\right]} \\,.\n \\label{eq:conjugate_posterior}\n\\end{align}\n\n\\begin{proposition}\n The (posterior) distribution of $(\\mu, \\sigma^2) \\mid y$ , as given by the conditional density in equation (\\ref{eq:conjugate_posterior}), is $\\NormalscaledInvChi{\\mu_n, \\sigma_n^2/\\kappa_n; \\nu_n, \\sigma_n^2}$, where\n $\\nu_n = \\nu_0 + n$, $\\kappa_n = \\kappa_0 + n$, $\\mu_n =\\frac{\\kappa_0}{\\kappa_n}\\mu_0 + \\frac{n}{\\kappa_n}\\bar{y}$ and $\\sigma_n^2 = \\left[\\nu_0 \\sigma_0^2 + (n-1)s^2 + \\frac{\\kappa_0 n}{\\kappa_n} (\\bar{y} - \\mu_0)^2\\right] /\\nu_n$.\n \\label{prop:posterior_conjugate}\n\\end{proposition}\n\\begin{proof}\n See appendix.\n\\end{proof}\n\nSince the prior and the posterior are both normal scaled inverse $\\chi^2$ distributed, we indeed constructed a conjugate prior.\nUsing the intermediate finding from proposition \\ref{prop:posterior_conjugate} we can derive the main result of this section.\n\n\\begin{proposition}\\label{prop:marginal_posterior}\n The marginal posterior distributions are given by $\\mu \\mid y \\sim t_{\\nu_n}(\\mu_n, \\sigma_n^2 / \\kappa_n)$ and $\\sigma^2 \\mid y \\sim \\scaledInvChi{\\nu_n, \\sigma_n^2}$, where $\\nu_n, \\sigma_n^2, \\mu_n$ and $\\kappa_n$ are as in proposition \\ref{prop:posterior_conjugate}.\n\\end{proposition}\n\\begin{proof}\n See appendix.\n\\end{proof}\n\n\\begin{table}[ht]\n\\centering\n\\def\\arraystretch{1.3}\n{\\small\n \\begin{tabular}{c | c c c c c}\n Parameter & ML Estimate & ML Variance & MAP & Posterior Mean & Posterior Variance\\\\\n \\hline\n $\\mu$ & $\\bar{y}$ & $\\sigma^2 / n$ & $\\mu_n$ & $\\mu_n$ & $\\sigma_n^2 / \\kappa_n$\\\\\n $\\sigma^2$ & $\\frac{n-1}{n}s^2$ & $2 \\sigma^4 / n$ & $\\frac{\\nu_n}{\\nu_n + 2} \\sigma_n^2$ & $\\frac{\\nu_n}{\\nu_n - 2} \\sigma_n^2$ & $\\frac{2 \\nu_n^2}{(\\nu_n - 2)^2(\\nu_n - 4)} \\sigma_n^4$\n \\end{tabular}\n }\n\\caption{{\\small Comparison of Bayesian estimates using a conjugate prior to ML estimates. See appendix for derivation.}}\n\\label{tab:comp_conjugate_bay_ml}\n\\end{table}\n\nNext we consider the results of Proposition \\ref{prop:marginal_posterior}, of which some summary statistics are tabulated in table \\ref{tab:comp_conjugate_bay_ml}.\nWe focus on the analysis of the mean parameter $\\mu$.\n\nAs the t-distribution is parameterized over its mean and variance (and degrees of freedom) we can directly deduce the posterior mean as $\\mu_n =\\frac{\\kappa_0}{\\kappa_0 + n}\\mu_0 + \\frac{n}{\\kappa_0 + n}\\bar{y}$ and the posterior variance as $\\sigma_n^2 / \\kappa_n$.\nWe see that the posterior mean is simply a convex combination of the prior $\\mu_0$ and the sample average $\\bar{y}$, with weights determined by the sample size and $\\kappa_0$.\nFor any fixed $n$ this pulls our estimate of the posterior mean away from $\\bar{y}$ and closer to $\\mu_0$ (and vice versa), which can be helpful if we have insufficient data and believe that the parameter should be around $\\mu_0$ ---we can use $\\kappa_0$ then to express our degree of believe in the prior.\nAs $n$ grows to infinity the information in the data overwhelms all prior information and the posterior mean is dominated by the sample mean.\n\nSimilarly we can use the hyperparameters $\\sigma_0^2, \\nu_0$ and $\\kappa_0$ to model our prior knowledge of the variance parameter, which propagates to the posterior variance of the mean parameter.\nConsidering the variance of the posterior mean as a function in $n$ we can use the \\emph{Landau notation} to write $\\sigma_n^2 / \\kappa_n = \\frac{n}{(\\nu_0 + n)(\\kappa_0 + n)} s^2 + \\mathcal{O}(1/n^2) = \\mathcal{O}(1/n)$, which resembles the usual $1/n$ convergence rate.\n\nAs $v_n = v_0 + n$ tends to infinity the distribution of the posterior mean tends to a normal distribution with parameters behaving (asymptotically) similar to the maximum likelihood estimators.\nIn this sense, informative Bayesian priors can be appropriate if the data contains insufficient information \\emph{and} we have reasonable knowledge a priori, where we use the prior to stabilize the results.\nBut they are also reasonable if we consider large samples, where the prior is simply dominated by the likelihood.\nWe refrain from an analogous analysis for $\\sigma^2$ here and only note that similar results hold, as can be seen from table \\ref{tab:comp_conjugate_bay_ml}.\n\nAbove we considered a simple model, as this allowed us to derive the results analytically.\nClosed form solutions allow us to fully investigate the influence of the prior on our results.\nHowever, we have also seen that Bayesian analyses are far from trivial and depend critically on the complexities of the model structure.\nIf we want to consider more realistic models we have to make ever more restrictive assumptions to yield analytical results.\nFor this reason among others, in the next section we discuss methods which trade off the clarity of an analytical result for the generality of being able to combine near arbitrary priors with complex, possibly high-dimensional likelihoods.\n\n\\subsection{Sampling From The Posterior}\nIn this section we consider approaches that allow us to characterize the posterior distribution in complex settings using sampling methods.\n\nFor the rest of this section let us assume we observed data $\\mathcal{D}$ for which we have a (possibly algorithmic) model in mind, which can be represented by the likelihood $p(\\mathcal{D} \\mid \\theta)$.\nWe also assume that a prior distribution $p(\\theta)$ has been constructed, so that the posterior is again given by $p(\\theta \\mid \\mathcal{D}) \\propto p(\\mathcal{D} \\mid \\theta) p(\\theta)$.\nUnlike before however, we now consider more general settings in which we do not restrict $p(\\theta \\mid \\mathcal{D})$ to be available in analytical form.\nThis may occur in many settings, for example when using priors that do not mix well with the likelihood or more apparent when using computational models which produce likelihood evaluations based on algorithms.\n\nTo motivate the following, say we are able to draw independent samples $\\theta^{(1)}, \\mydots, \\theta^{(n)}$ from $p(\\theta \\mid \\mathcal{D})$.\nBy the law of large numbers we get $1 / n \\sum_i h\\left(\\theta^{(i)}\\right) \\overset{a.s.}{\\longrightarrow} \\Exp{h(\\theta) \\mid \\mathcal{D}}$, under some regularity conditions on $h$ and $p(\\theta \\mid \\mathcal{D})$, with similar results holding for sample quantiles.\nHence, to learn something about $p(\\theta \\mid \\mathcal{D})$ we can formulate questions using quantiles or general expectations and rely on the statement above.\n\nIn the subsequent paragraphs we discuss methods to sample from the posterior that work under the general assumption that we can evaluate the posterior at arbitrary points up to an integration constant.\nWe will see that these methods do \\emph{not} produce independent samples but instead create \\emph{Markov chains} whose realizations can be seen as autocorrelated samples.\n\nWith this in mind, we first consider what properties these chains must fulfill in order to create equivalent results as motivated above for independent samples.\nWe end this section by introducing an algorithm that accomplishes the above.\n\n\\paragraph{Markov Chain Monte Carlo.}\nSay we are able to construct a Markov chain with unique invariant distribution equal to the posterior distribution we want to sample from.\nAssume also that the distribution of the chain at time $n$ converges to this invariant distribution no matter where we initialize the chain.\nThen, in principle, we could run the chain \\emph{long enough} until it converged to the invariant distribution and then consider all subsequent realizations as draws from the stationary distribution.\nThis is the core idea of Markov chain Monte Carlo (MCMC).\nIn practice, however, we do not know when a chain is run \\emph{long enough}.\nIn part 3 we present some measures that can be of help with this problem during the application.\n\nUnder some regularity conditions, similar but not as strict as in Theorem \\ref{thm:mcclt}, we get a law of large numbers for such Markov chains (see e.g. \\citet{roberts2004}, Fact 5).\nThis tells us that if we run the chain forever, our average will eventually converge to the number we seek.\nHowever, forever is usually too long.\nThat is why we focus on assumptions which admit a central limit theorem with the usual $\\sqrt{n}$ convergence rate, as it allows for more rigorous statements about our confidence in the whereabouts of the estimator for large samples.\n\n\\begin{remark}\n \\emph{(i)} Having a central limit theorem in the background does \\emph{not} imply that the asymptotic distribution provides a good approximation for finite samples.\n We still do not know when the asymptotics `kicks in' (\\citet{Jackman2009}).\n But under assumptions that allow for a CLT we can be more confident in our results than under assumption that only allow of a LLN.\n \\emph{(ii)} As is often the case, there are many different sets of assumptions that allow for a CLT.\n The following theorem presents a particular set of assumptions which will be seen to have favorable properties when also considering the creational process of the Markov chain.\n We remark that we will \\emph{not} formally introduce all concepts and will provide only a heuristic explaination.\n This is due to the fact that Markov chain theory on general state spaces requires a good understanding of measure theory, which we do not want to assume as a prerequisite.\n We refer to \\citet{roberts2004} for a survey on recent advances with application to MCMC and \\citet{meynandtweedie09} for a comprehensive treatment of Markov chain theory.\n\\end{remark}\n\n\\begin{theorem}{(A Central Limit Theorem for Markov Chains).}\\label{thm:mcclt}\n Let $\\{X_n \\}$ be a (discrete time) Markov chain and $\\pi$ a probability distribution on the same space.\n Consider some measurable function $h$ with $\\Expwrt{\\pi}{h^2} < \\infty$.\n Define $\\sigma^2(h) := \\varwrt{\\pi}{h}\\tau := \\varwrt{\\pi}{h} \\sum_{k \\in \\mathrm{Z}} \\text{Corr}\\left(h(X_0), h(X_k)\\right)$.\n Assume the Markov chain is $\\phi$-irreducible, aperiodic, reversible with respect to $\\pi$ and that $\\sigma^2(h) < \\infty$. Then $\\pi$ is stationary for the chain and\\footnote{In the original paper by \\citet{roberts2004} the statement of this theorem differs in that they write $\\tau = \\sum_{k \\in \\mathrm{Z}} \\text{Corr}\\left(X_0, X_k\\right)$.\n We believe that this is an error as \\citet{haggstrom2007} state in their comparison of different ways of writing the asymptotic variance that $\\sigma^2(h) = \\sum_{k \\in \\mathrm{Z}} \\Cov{h(X_0), h(X_k)}$.\n Now if we use that $X_0 \\sim \\pi$ we get $\\sigma^2(h) = \\sum_{k \\in \\mathrm{Z}} \\Cov{h(X_0), h(X_k)}= \\var{h(X_0)} + \\sum_{k \\neq 0} \\Cov{h(X_0), h(X_k)} = \\varwrt{\\pi}{h}(1 + \\sum_{k \\neq 0} \\Cov{h(X_0), h(X_k)}) / \\var{h(X_0)} =\\varwrt{\\pi}{h}(1 + \\sum_{k \\neq 0} \\text{Corr}{(h(X_0), h(X_k))}) = \\varwrt{\\pi}{h} \\sum_{k\\in\\mathrm{Z}} \\text{Corr}{(h(X_0), h(X_k))}$.}\n \\begin{align}\n \\sqrt{n}\\left(\\frac{1}{n}\\sum_{i=1}^n h(X_i) - \\Expwrt{\\pi}{h} \\right) \\overset{d}{\\longrightarrow} \\normal{0, \\sigma^2(h)} \\,.\n \\end{align}\n\\end{theorem}\n\\begin{proof}\n See \\citet{kipnis1986} for a complete proof of the second claim; see \\citet{roberts2004} Proposition 1 for the first claim and Theorem 27 for the second.\n\\end{proof}\n\n\\noindent\nWe end this paragraph by discussing the assumptions of Theorem \\ref{thm:mcclt} on an intuitive level.\n\n\\textbf{$\\phi$-irreducibility} assumes that we can find a measure $\\phi$ such that no matter where the chain starts, we eventually reach every region of the state space which has positive measure with respect to $\\phi$.\nIn the next paragraph we will see that this condition can be satisfied by construction of the chain.\n\n\\textbf{Aperiodicity} assumes that we cannot find disjoint regions on which the chain jumps from one region to another in a cyclical predictable fashion.\nIt seems intuitive that such a behavior will prevent the chain from actually converging to its stationary distribution.\n\n\\textbf{Reversibility with respect to $\\bm{\\pi}$} is a technical assumption which is best explained by its implications.\nIn particular, it implies that the Markov chain has $\\pi$ as its stationary distribution (which is unique by the other assumptions).\nAgain, in the next paragraph we will see that this condition can be satisfied by construction with $\\pi$ being the posterior distribution from which we want to sample.\n\n\\textbf{Finite variance} ($\\sigma^2(h) < \\infty$) implies that the integrated correlation time $\\tau$ must be finite, as we assumes square integrability of $h$.\nThe integrated correlation time is finite if the correlation function decreases fast enough to zero.\nHeuristically speaking, for a CLT to work we need more information as would be available in a sample for which the integrated autocorrelation time is infinite.\nIf it is finite we get the usual large sample variance approximation $\\sigma^2(h) / n = \\varwrt{\\pi}{h} / (n/\\tau)$.\nIn this sense we might say that $n/\\tau$ denotes the \\emph{effective sample size}, which corrects for the fact that we are not drawing independent samples and therefore (in most cases) need more samples to yield the same amount of information as in the independent case.\n\n\\paragraph{Metropolis-Hastings Algorithm.}\nHere we consider one method which implicitly defines a Markov chain with the desired properties, the Metropolis-Hastings algorithm (\\citet{Metropolis1953}, \\citet{hastings70}).\nFor other approaches and more involved algorithms see for example \\citet{liang10}.\n\n\\begin{algorithm}\n\\caption{Metropolis-Hastings}\\label{alg:metropolis-hastings}\n\\begin{algorithmic}[1]\n \\Require $(\\pi, q, T) =$ (target density, proposal density, number of samples to draw)\n\\State initialize $x_0$ with an arbitrary point from the support of $q$\n\\For{$t = 0,\\mydots,T-1$}\n \\State sample a candidate: $y \\sim q(\\cdot \\mid x_t)$\n \\State compute the acceptance probability: $\\mathcal{A} \\gets \\min\\left\\{ \\ddfrac{\\pi(y)}{\\pi(x_t)} \\ddfrac{q(y \\mid x_t)}{q(x_t \\mid y)}, 1\\right\\}$\n \\State update the chain: $x_{t+1} \\gets \\begin{cases} y &\\mbox{,with probability } \\mathcal{A}\\\\ x_t &\\mbox{,with remaining probability} \\end{cases}$\n\\EndFor{}\n\\State \\textbf{return} $\\{x_t : t = 1,\\mydots,T\\}$\n\\end{algorithmic}\n\\end{algorithm}\n\n\\noindent\nAlgorithm \\ref{alg:metropolis-hastings} displays the Metropolis-Hastings algorithm.\nThe main idea is that we start with an initial value and iteratively propose new values of the chain, however, we do not accept every proposal, but we do so only with a certain probability ---this probability can be thought of as high if the proposal has a high relative density compared to the last chain link.\nHere we also see why we do not care about any integration constant, as the target density only appears as a ratio (line 4).\n\nClearly the results depend on the choice of the proposal density.\nA common pick are so called \\emph{random walk} proposals, which add some random number to the current position of the chain; for example a gaussian random walk proposal is given by $q(\\cdot \\mid x_t) = \\normal{\\cdot \\mid x_t, \\sigma^2}$ or equivalently stated $y = x_t + \\normal{0, \\sigma^2}$.\nSee \\citet{sherlock2010} for a recent survey on random walk proposals.\n\nThe simplicity of the algorithm is remarkable, but the main question is if the resulting Markov chain inherits favorable properties.\nAnd indeed this is the case.\nBy construction the algorithm creates Markov chains which are reversible with respect to $\\pi$ and aperiodic, and if additionally the proposal density is positive and continuous and $\\pi$ is finite then the chain is $\\pi$-irreducible; see for example \\citet{roberts2004}.\nThis tells us that under these regularity conditions a CLT holds for chains created using the Metropolis-Hastings algorithm.\nTo avoid using samples that do not come from the stationary distribution, in practice we choose $T$ very large and discard the first few (\\emph{burn-in}) samples.\n\nThe ability to sample draws in complex settings using the Metropolis-Hastings algorithm (and other Markov chain Monte Carlo methods for that matter) made Bayesian statistics applicable for real problems.\nStill, \\citet{Au2001EstimationOS} show that the classical Metropolis-Hastings algorithm is highly dependent on the proposal density and fails in higher dimensions; \\citet{zuev08} provide a geometric intution.\nNumerous novel methods which deal with dimensionality problems have been published.\nA particular promising route seems to be \\emph{Hamiltonian Monte Carlo} (\\citet{duane87}), which is one of the algorithms used in the probabilistc programming language STAN (\\citet{standev2018stancore}) with ongoing research on theoretical properties, see for example \\citet{betancourt2014geometric}.\n\n\n\\paragraph{Volume in Higher Dimensions.}\nClassical MCMC methods can have too slow convergence rates; In higher dimensions this might be due to probability mass being distributed very far from where it is expected (\\citet{betancourt2017convergence}).\nIn this paragraph we motivate this phenomenon and in the following we present methods which utilize it.\n\nLet $B_d$ denote the unit ball in $\\reals^d$ and define $C_d$ as the smallest cube containing $B_d$.\nWe consider two questions.\nFirst, how does the ratio $\\nicefrac{\\vol{B_d}}{\\vol{C_d}}$ change as $d$ increases.\nAnd second, how does the ratio of probability mass distributed by a standard gaussian on these regions change as $d$ increases.\nSince closed form expressions of volumina of geometrical objects exist the first questions needs little work.\nSimilarly we can easily compute $\\prob{X \\in C_d} = \\left[\\Phi(1) - \\Phi(-1)\\right]^d$, where $\\Phi$ denotes the one-dimensional gaussian cumulative distribution function.\nHowever, to compute $\\prob{X \\in B_d}$ we need to integrate over the unit ball with respect to a gaussian distribution, which is non-trivial.\nFor this reason we decide to report an upper bound, as this is sufficient for our motivation.\nIn particular we compute $\\overline{\\prob{X \\in B_d}} := \\sup_{\\bm{x} \\in B_d} \\phi(\\bm{x}) \\vol{B_d} = \\sup_{\\bm{x} \\in B_d} \\phi(\\bm{x}) \\int_{B_d}1 \\mathrm{d}\\bm{x} \\geqslant \\int_{B_d} \\phi(\\bm{x}) \\mathrm{d}\\bm{x} = \\prob{X \\in B_d}$.\nThe results of these computations are depicted in table \\ref{tab:vol_high_dim}.\nWe note that both ratios tend to zero very fast as $d$ increases.\nWith this phenomenon in mind one has to be cautious when working in high-dimensional spaces, since the regions of interest, that is the regions containing non-negligible probability mass, might not be located where our low-dimensional intution says.\nThis idea is formalized by the \\emph{Gaussian Annulus Theorem} (\\citet{blum2017foundations}; theorem 2.9) which states, inter alia, that most probability mass lies within an annulus centered at the origin with an average distance to the origin of $\\sqrt{d}$.\n\n\\begin{table}[ht]\n\\def\\arraystretch{1.3}\n\\centering\n \\begin{tabular}{c | c c c c c c c}\n $d$ & 1 & 2 & 3 & 5 & 7 & 10 & 15\\\\\n \\hline\n ${\\vol{B_d}} / {\\vol{C_d}}$ & 1.00000 & 0.78540 & 0.52360 & 0.16449 & 0.03691 & 0.00249 & 0.00001\\\\\n $\\overline{\\prob{X \\in B_d}} / {\\prob{X \\in C_d}}$ & 1.16874 & 1.07281 & 0.83589 & 0.35870 & 0.10995 & 0.01184 & 0.00012\n \\end{tabular}\n \\caption{Comparison of volume ratio of unit ball and cube, and probability ratio of gaussian falling in unit ball and cube for varying dimension $d$. Numbers are rounded to five decimal places.}\n\\label{tab:vol_high_dim}\n\\end{table}\n\nWe have seen some unintuitive behavior in higher dimensions which might explain why regular methods do not work or only work very slowly.\nThe next paragraph presents one method which utilizes this behavior to efficiently produce samples.\n\n\\paragraph{Hamiltonian Monte Carlo.}\nWe conclude our digression on Bayesian thinking by presenting \\emph{Hamiltonian Monte Carlo} (HMC), an innovative Markov chain Monte Carlo method from the statistical physics literature which works in higher dimensions \\citet{duane87}.\nThere are of course multiple MCMC algorithms which work in higher dimensions with many more being actively developed.\nHere we focus on HMC as it is the main algorithm used in the probabilistic programming language STAN (\\citet{standev2018stancore}), which we will be using in our Monte Carlo study and application part.\n\nWe note that it is impossible to provide a rigorous introduction to HMC here, which is why we will focus on the general intution and refer to a series of papers by Michael Betancourt and several coauthors on, the geometric foundations of HMC (\\citet{betancourt2014geometric}); geometric ergodicity of HMC (\\citet{livingstone2016geometric}) and HMC for hiearchical models (\\citet{betancourt2013hamiltonian}).\\footnote{Besides doing theoretical research on HMC, Michael Betancourt worked for STAN on integrating the HMC algorithm and runs an educational blog where he presents his research using modern tools, see \\url{https://betanalpha.github.io/}.}\n\nOne reason why ordinary MCMC methods might converge only very slowly in higher dimensions is that the proposal distribution used in the Metropolis-Hastings algorithm does not properly capture the geometry of the high-dimensional space which leads to many rejected proposals and therefore an inefficient exploration of the parameter space.\nThe main idea of HMC is to extend the parameter space by constructing a specific vector field on it, which moves the chain from one proposed point to another in a way so that we consider points that lie in regions with high probability mass and we regulary jump to faw away points as to explore the space as quickly as possible.\nBut how do we construct this vector field?\nNote that when considering differentiable posterior densities the gradient defines a vector field.\nHowever, this vector field points to the modes of the posterior and as we saw in the last paragraph, in higher dimensions we will find little to no probability mass near the modes.\nThis is where Hamiltonian mechanics comes into play by providing a set of equations (Hamilton's equations) that describe the time-evolution of the interplay of kinetic and potential energy of a system.\nWhat this means for our case is best explained by imagining the mode as the center of gravity, with gravity pulling harder as we get closer to the mode ---this can be thought of as the gradient vector field.\nBut as most probability mass is spread around an annulus around the mode we do not want to move closer to the center of gravity, we want to move around an orbit around the mode.\nThe distance of the orbit to the mode and exact shape depend of course on the dimensionality and posterior distribution of the problem.\nHamilton's equations provide us with a way to construct a vector field so that moving along the field drifts the Markov chain into this orbit.\nOnce reached the chain explores the relevant space quickly.\n\nWhy is the above important?\nBayesian statistics and its application to real world problems has gained immense popularity with the invention of Markov chain Monte Carlo methods.\nIt's naive application to general complex high-dimensional problems is not computationally feasible however.\nThese problems are under active development and new approaches on the algorithmic and theoretical side, as the one presented above, prove fruitfull.\n", "meta": {"hexsha": "c38aba9e7a9b74882e60ab0ba4ac2f36b354bd90", "size": 36469, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manuscript/ch1.tex", "max_stars_repo_name": "timmens/bayesian-hierarchical-models", "max_stars_repo_head_hexsha": "80c174997ce2ac5ebfb2e1628df4f6aeae9c9b0b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2020-02-19T09:04:01.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-06T10:19:23.000Z", "max_issues_repo_path": "manuscript/ch1.tex", "max_issues_repo_name": "timmens/bayesian-hierarchical-models", "max_issues_repo_head_hexsha": "80c174997ce2ac5ebfb2e1628df4f6aeae9c9b0b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "manuscript/ch1.tex", "max_forks_repo_name": "timmens/bayesian-hierarchical-models", "max_forks_repo_head_hexsha": "80c174997ce2ac5ebfb2e1628df4f6aeae9c9b0b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-11-25T14:44:12.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-26T18:54:16.000Z", "avg_line_length": 100.4655647383, "max_line_length": 649, "alphanum_fraction": 0.7620718967, "num_tokens": 9510, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8807970842359877, "lm_q2_score": 0.7341195385342971, "lm_q1q2_score": 0.6466103490216777}} {"text": "\\chapter{Mathematics}\nDigital signal processing is in general a very mathematical subject and audio- and musical signal processing is no exception to that rule. So it shouldn't come as a surprise, that many of the RAPT library functions rely on underlying algorithms that implement a considerable mathematical machinery. This chapter discusses some of the mathematical concepts that are relevant in the context of audio processing and shows how they are implemented in RAPT. The purpose here is mostly to explain, \\emph{what} the respective classes and functions do, not \\emph{how} they work internally. Most of the algorithms discussed in this chapter fall under the broad umbrella of numerical analysis. \nDetailed explanations and derivations can be found in the vast literature, specifically [Reference: Numerical Recipies] is a first class reference for such things.\n\n\\section{Complex Numbers}\nFor representing complex numbers, RAPT uses \\texttt{std::complex} from the C++ standard library.\n\n\\section{Functions}\nFor representing functions, RAPT uses \\texttt{std::function} from the C++ standard library ....maybe...this has to be decided\n\n\\section{Sequences}\nMathematically, a sequence is an ordered set of elements of some given type.\n\n\\section{Polynomials}\nA polynomial of a variable $x$ is a weighted sum of integer powers of $x$, starting at $x^0 = 1$ and going up to $x^N$. Polynomials play a central role in the theory of analog and digital filters and they occur in the context of interpolation, curve fitting and function approximation - that's reason enough for RAPT to devote them a dedicated class - which is unsurprisingly called \\texttt{Polynomial}. Mathematically, a polynomial $p(x)$ of order $N$ is defined as:\n\\begin{equation}\n\tp(x) = \\sum_{n=0}^N a_n x^n\n\\end{equation}\nwhere the $a_n, n=0,\\ldots,N$ are called the coefficients. An $N$th order polynomial has $N+1$ coefficients, ranging from index $0$ to index $N$. This is also the convention in which polynomial coefficient arrays are represented in RAPT - as length \\texttt{N+1} arrays \\texttt{a[0]}...\\texttt{a[N]}. Consider the following code for multiplying the 2nd order polynomial $p(x) = 2 - 3 x + x^2$ by the 1st order polynomial $q(x) = -3 + 4x$ and evaluating the resulting polynomial $r(x)$ at $x = 2$:\n\\begin{lstlisting}\nPolynomial p(2, 2, -3, 1), q(1, -3, 4); // allocation, initialization\nPolynomial r = p * q; // multiplication\nint y = r(2); // evaluation\n\\end{lstlisting}\nThis code uses the high-level interface of the RAPT library. The first line creates the two polynomials (with integer coefficients). The first parameter of the constructor of the \\texttt{Polynomial} class is the order of the polynomial followed by an appropriate number of coefficients. The 2nd line multiplies the two polynomials which results in another polynomial which is stored in the variable \\texttt{r}. The third line produces a numeric output by applying the product polynomial $r$ to the value $x=2$. The code looks reasonably straightforward and readable by C/C++ standards. However, if you have to do polynomial multiplication in a context where performance is critical, for example inside a realtime audio callback, you probably don't want to have any dynamic memory allocations. But that's what the constructors of the \\texttt{Polynomial} class do. For performance and predictability reasons, you may want to operate on pre-allocated arrays, even if that means to uglify your code. In this case, your code could look something like this:\n\\begin{lstlisting}\n// pre-allocation and initialization:\nint p[3] = { 2, -3, 1}; // 2nd order, 3 coeffs, 2 - 3x + x^2\nint q[2] = {-3, 4}; // 1st order, 2 coeffs, -3 + 4x\nint r[4]; // 3rd order, 4 coeffs\n\n// multiplication and evaluation:\nPolynomial::mul(&p, 2, &q, 1, &r); // multiply p, q, store result in r\nint y = Polynomial::eval(&r, 3, 2); // evaluate polynomial r at 2\n\\end{lstlisting}\nFirst, we allocate the three arrays \\texttt{p, q, r} of lengths \\texttt{3, 2, 4} respectively. The order of the result polynomial \\texttt{r} is given by the sum of the orders of the factors $2+1=3$, so we need space for $4=3+1$ coefficients for the \\texttt{r} polynomial. The \\texttt{p, q} arrays are also initialized in this process. Then we call static member functions \\texttt{mul} and \\texttt{eval} of the \\texttt{Polynomial} class that operate on pre-allocated arrays of polynomial coefficients to perform the multiplication and evaluation. Arguably, the 2nd version is much less readable but these static member functions can be used whenever it is undesirable to have any dynamic memory allocations. Internally, the operators \\texttt{*} and \\texttt{()} of the \\texttt{Polynomial} class call the low-level routines on internally managed coefficient array member variables. This idiom of static member functions that are internally called from the non-static functions and operators on the instance variables of an object will be used throughout the library. It allows the low- and high-level interface to share the same code.\n\n\\section{Root Finding}\nRoot finding is the problem of finding one or more solutions of the equation:\n\\begin{equation}\n\tf(x) = 0\n\\end{equation}\nwhere $f(x)$ is, in general, an arbitrary function. Any value of $x$ that turns the equation into a true statement, when substituted for $x$, is called a solution or 'root' of the equation. In this problem setting, it is assumed that we have a means to evaluate the function at arbitrary values of $x$. In RAPT, the root finding algorithms must get such a means by passing them either function pointer to the respective function or a function object of class \\texttt{Functor}. Functors are objects that provide a means to evaluate a function at some point $x$. What makes them more flexible than plain functions is the fact that they can have internal instance variables which may affect the function evaluation. They can be seen as a general way to define families of functions with an arbitrary set of parameters. A polynomial is an example for such a functor where the parameters are the coefficients.\n\n\\subsection{Bisection}\nSuppose that, for two particular values $x=a$ and $x=b$, we know that $f(a)<0$ and $f(b)>0$. By assuming that the function is continuous between $a$ and $b$, that means that a root, i.e. a zero crossing, must be present somewhere between $a$ and $b$. The same is true, if $f(a)>0$ and $f(b)<0$. We say, the root is bracketed by the values $a$ and $b$. We may now evaluate the function at the midpoint between $a$ and $b$. If the value is equal to $0$, we have found our root, if it's less than $0$ or greater than $0$, we have found a new bracketing interval either to the left or to the right of the midpoint. The new bracketing interval is only half as long as the old one. Iterating that procedure, the brackets move closer and closer to the root. When their difference reaches a threshold, typically related to the numeric precision of the floating point datatype, we stop the iteration. It is then converged to the root (up to an error given by our threshold). Since the interval halves at each iteration, the number of correct binary digits in our approximate solution (given by the current midpoint) increases by one. Because the number of correct digits is a linear function of the iteration number, we say that the algorithm has a linear order of convergence. In RAPT, root finding by bisection is implemented in \\texttt{RootFinder::bisection}. As a subtle sidenote, it may make sense to choose the new evaluation point not exactly at the midpoint of the bracketing interval, but at [... tbc. Reference NR].\n\n\\subsection{Newton Iteration}\nNewton iteration is a method to iteratively improve an initial guess $x_0$ for the solution by iteratively applying the update rule:\n\\begin{equation}\n\tx_{n+1} = x_n - \\frac{f(x_n)}{f'(x_n)}\n\\end{equation}\nThis formula results from approximating the function $f(x)$ as a straight line that goes through the point $(x_0, y_0=f(x_0))$. This straight line is constructed to match the slope of the function $f(x_0)$ given by the derivative $f'(x_0)$ in the point $(x_0, y_0)$. The resulting line equation is then solved for the zero crossing of the line, which becomes our new root estimate $x_{n+1}$. Thus, to actually implement it, we also need a means to evaluate the derivative of $f(x)$ for any $x$ in addition to being able to evaluate the function $f(x)$ itself. The convergence of this iteration is typically quadratic, which means that in each iteration, the number of correct digits doubles [verify this] but in certain pathological cases it may be slower and in some cases, the iteration may even diverge. If a reasonable initial guess for the solution is available that is close to the true solution and the function is well behaved near the solution and we know how evaluate the derivative, Newton iteration is one of the methods of choice to converge to a true solution. In RAPT, Newton iteration is available in the function \\texttt{RootFinder::newton} which takes two function pointers or functors - one for the function itself and one for the derivative and also takes an initial guess and returns the refined value.\n\n\\subsection{Secant Method}\nThe secant method works similar to the bisection method, just that we choose the new evaluation point not as midpoint of our current bracketing interval $[a, b]$, but instead, we fit a straight line to the points $(a, f(a)), (b, f(b))$ and solve for the root of this straight line. The order of convergence is given by the golden ratio: $\\alpha = (1+\\sqrt{5})/2 \\approx 1.618$. So, it converges faster than the bisection method but not quite as fast as the Newton iteration. It can be seen as an approximation to Newton iteration, where the derivative is replaced by a finite difference approximation. On the plus side, it does not need to evaluate the derivative, which might be expensive or sometimes even not feasible. The method is available in \\texttt{RootFinder::secant}. Note that it may also diverge, since the new evaluation point may fall outside the initial bracketing interval.\n\n\\subsection{False Position}\nAlso known as 'regula falsi', this method is similar to the secant method, but it applies...\n\n%\\subsection{Fixpoint Iteration}\n%\\subsection{Brent's Method}\n%\\subsection{Ridders' Method}\n\n\\subsection{Roots of Polynomials}\nIf the function $f(x)$ is a polynomial, there are specialized algorithms for finding its roots. One such algorithm is the Laguerre algorithm which is implemented in \\texttt{Polynomial::roots(T *a, int N, complex *r)}. It takes an array of \\texttt{N+1} coefficients as parameter and returns the roots in the length \\texttt{N} array \\texttt{r}. The fundamental theorem of algebra tells us that each polynomial of order $N$ has exactly $N$ roots such that it can be expressed as:\n\\begin{equation}\n \\label{Eq:PolynomialProductForm}\n\tp(x) = k \\prod_{n=1}^N (x - r_n)\n\\end{equation}\nthat is, a scaled product of the differences between $x$ and the respective root $r_n$. Remember that a product is zero, if and only if (at least) one of its factors is zero, so if $x$ is equal to one of the roots $r_n$, the corresponding factor $x - r_n$ in the product is zero, making the whole product zero - which verifies that $r_n$ is a root, indeed. These roots, however, may not be unique - some of them may have a multiplicity, i.e. occur more than once in the above product. Some of the roots may also be complex. If the coefficients of the polynomial are real, complex roots will always come in pairs of complex conjugate numbers. In audio applications, we will mostly deal will polynomials that have real coefficients. For generality, however, there's also a version of the polynomial root finder that takes an array of complex coefficients. Internally, the same algorithm is used anyway, no matter if the coefficients are real or complex. For example, the roots of the $5$th order polynomial $p(x) = 1 + 2 x - 3 x^2 + 3 x^4 - x^5$ can be found via:\n\\begin{lstlisting}\nvector> r(5);\nr = Polynomial(5, 1., 2. -3., 0., 3., -1.).roots();\n\\end{lstlisting}\nor, alternatively using the low-level routine that operates on pre-allocated arrays:\n\\begin{lstlisting}\ndouble a[6] = { 1., 2. -3., 0., 3., -1. };\ncomplex r[5];\nPolynomial::roots(a, 5, &r);\n\\end{lstlisting}\nA function for getting from the roots back to the coefficients is also available via the function \\texttt{Polynomial::fromRoots}. In addition to the $N$ roots, it also needs the value of the overall scaling constant $k$ in Eq. \\ref{Eq:PolynomialProductForm}. This is simply the $N$th coefficient, but in case you forget that, the \\texttt{Polynomial} class has also the function \\texttt{productFormScaler(int N, T a)} which trivially just returns \\texttt{a[N]}. Since the \\texttt{Polynomial} class is a subclass of class \\texttt{Functor}, the general root finders from above can be applied for polynomial root finding as well. They can, however, find only one root at a time and you need an initial guess or bracketing interval.\n% reconstructing the coefficients form the roots\n\n\n\\section{Rational Functions}\nA rational function is a quotient or 'ratio' of two polynomials:\n\\begin{equation}\n\tr(x) = \\frac{\\sum_{m=0}^M b_m x^m} {\\sum_{n=0}^N a_n x^n}\n\\end{equation}\nwhere $M$ is the order of the numerator and $N$ is the order of the denominator. All realizable analog and digital filter transfer functions are functions of this very type - which explains the special significance of rational functions, and hence polynomials, in the field of signal processing. The roots of the numerator are called the zeros of the transfer function and the roots of the denominator are called the poles. Since the denominator is zero at these poles, the value of the rational function at these points goes to infinity, unless there's a corresponding zero in the numerator at exactly the same point to cancel the pole. RAPT provides the class \\texttt{RationalFunction} to deal with rational functions.\n\n\\subsection{Partial Fraction Decomposition}\nAny rational function can be expanded into the form of a sum of a polynomial and several first order rational functions, in which the numerator is a constant (called the residue which is possibly complex) and the denominator is a linear function of $x$. That is:\n\\begin{equation}\n\tr(x) = \\frac{\\sum_{m=0}^M b_m x^m} {\\sum_{n=0}^N a_n x^n}\n\t = \\sum ...\n\\end{equation}\nIf the coefficients of the original numerator and denominator polynomial are both real, complex values for the numerators and denominators in the expanded form come in conjugate pairs, such that 1st order terms can be combined pairwise into 2nd order terms with real coefficients. In signal processing applications, such decompositions amount to decomposing a direct form filter into a parallel connection of the respective 1st or 2nd order filters and an FIR part, where the latter results from the polynomial part. Because the individual terms are easier to understand and analyze, partial fraction decompositions play a role in certain filter analysis applications. For example, they are required for finding closed form expressions of a filter's impulse response. This impulse response is given by the inverse $z$-transform of its transfer function, but inverse $z$-transforms are only available for low-order terms - which is precisely what the decomposition delivers. Partial fraction decomposition is available in RAPT via \\texttt{RationalFunction::partialFractions}.\n\n\n\\section{Vectors}\n\\section{Matrices}\n\n\\section{Transforms}\nTransforms are - together with filters - one of the mainstays in the field of signal processing. Consider an $N$ dimensional vector $\\mathbf{x}$ which may represent a (segment of a) signal. A general transform would take that vector as input and produce a new vector $\\mathbf{y}$ as output. If the transform is designed well, the transformed vector $\\mathbf{y}$ may reveal relevant features of the signal that are hard to see in the original representation. An important concern is invertibility - we want to be able to reconstruct the original vector $\\mathbf{x}$ from the transformed vector $\\mathbf{y}$ by some kind of associated inverse transform. Given such a pair of a transform and its inverse, we could transform an input vector, modify the transformed representation and apply the inverse transform to get a modified output vector/signal. A good transform in an audio processing setting would be one, that allows for convenient modifications that closely relate to modifications of perceptual, auditory attributes of the signal. We will consider mainly linear transforms. These are the class of operations on our input vector $\\mathbf{x}$ that can be expressed as a matrix-vector multiplication of an $N \\times N$ matrix $\\mathbf{A}$ with our input vector $\\mathbf{x}$:\n\\begin{equation}\n\t\\mathbf{y} = \\mathbf{A} \\mathbf{x}\n\\end{equation}\nIf the matrix $\\mathbf{A}$ is non-singular, there will be an inverse matrix $\\mathbf{A}^{-1}$ and hence, there will be an inverse transform, given by:\n\\begin{equation}\n\t\\mathbf{x} = \\mathbf{A}^{-1} \\mathbf{y}\n\\end{equation}\nAlternatively to thinking in terms of an $N$ dimensional vector $\\mathbf{x}$, we may also consider the inputs and outputs as sequences of length $N$, which, in this context, can be seen as merely a different name for the same thing - a finite, ordered set of $N$ numbers.\n\n\\subsection{Discrete Fourier Transform}\nThe discrete Fourier transform (DFT) of a length $N$ sequence $x_0, \\ldots, x_{N-1}$ is another length $N$ sequence $X_0, \\ldots, X_{N-1}$ in which each element $X_k$ can be computed by the formula:\n\\begin{equation}\n\tX_k = \\sum_{n=0}^{N-1} x_n W_N^{k n}, \\qquad \\text{where }W_N = e^{-j \\frac{2 \\pi}{N} }\n\\end{equation}\nexpressed as matrix-vector product, this looks like:\n\\begin{equation}\n\t...\n\\end{equation}\nComputing a product of an $N \\times N$ matrix and an $N$ vector in general requires $\\mathcal{O}(N^2)$ operations, because we have to form $N^2$ individual products between numbers. In the DFT matrix, however, not all elements are distinct. This redundancy of the matrix elements allows for an algorithmic shortcut, known as the fast Fourier transform (FFT) which requires only $\\mathcal{O}(N \\cdot \\log N)$ operations. Fast Fourier transforms are particularly efficient when $N$ is a power of two - these transforms are known as radix-2 FFTs. There are also $\\mathcal{O}(N \\cdot \\log N)$ algorithms for arbitrary $N$, for example, the Bluestein FFT algorithm which uses a radix-2 algorithm on a zero-padded sequence internally. In RAPT, both radix-2 and Bluestein FFTs are available via the functions \\texttt{Transforms::fftRadix2} and \\texttt{Transforms::fftBluestein}. There's also the function \\texttt{Transforms::fft} which tests, if $N$ is a power of two and then dispatches between the radix-2 and the Bluestein algorithm accordingly.\n\n\\subsubsection{Inverse Transform}\n\n\n\n\n\\subsection{Short Time Fourier Transforms}\n\n\\subsection{Wavelet Transforms}\n\n\\section{Statistics}\nStatistical signal measures such as the mean value, the autocorrelation sequence or the crosscorrelation sequence (between two signals) play an important role in many signal analysis algorithms and in adaptive filters. ...\n\n\\section{Interpolation}\nAssume that we have a certain number $N$ of pairs of abscissa/ordinate values $(x_n, y_n), n = 0, \\ldots, N-1$ represented as length \\texttt{N} arrays \\texttt{x} and \\texttt{y}. We imagine these data points as a table that represents a continuous function and we assume that the $x$-values are in strictly ascending order. The goal of interpolation is to create a continuous function from the discrete data. That means, interpolation would connect the points by assigning a $y$-value to \\emph{any} $x$-value between \\texttt{x[0]} and \\texttt{x[N-1]}. If we try to find a $y$-value for an $x$-value outside this range, it's called extrapolation. In RAPT, a continuous interpolating (and extrapolating) function is represented as an \\texttt{Interpolator} object which itself is subclassed from \\texttt{Functor}. This implies that the root-finding algorithms can be applied to interpolating functions in the same way as they are applied to regular functions. In the context of audio processing, interpolation may be required for obtaining signal values between the actual sample values or for creating smooth curves of control-data for which values are available only at certain time instants but not at each sample instant.\n\n\\subsection{Linear Interpolation}\nA very simple way of interpolating between datapoints is to connect the datapoints with straight lines. \n%It's not suitable for high-quality audio interpolation (unless, maybe, in combination with oversampling) but it has its place for interpolating control data. A feature that is sometimes desirable is that is preserves the mean value - i.e. the mean value of the interpolated \\texttt{y}-array equals the mean of the original \\texttt{y}-array [verify]\n%The only form of interpolation that is even more crude would be piecewise constant interpolation which just returns the $y$-value of the nearest neighbour. \n\n\\subsection{Lagrange Interpolation}\nLagrange interpolation takes a given number $N$ of datapoints and constructs an $N$th order polynomial that passes exactly through these points. Such an interpolating polynomial tends to oscillate wildly between the actual datapoints, especially if the order of the polynomial is high. In most application, such oscillations are undesirable and that's why Lagrange interpolation is mostly used only with low orders. In certain audio applications, however, these oscillations might actually be exactly what is needed - as the order of the Lagrange interpolator goes up, it approaches the optimal sinc-interpolator [Reference: 'Elephant interpolators']\n\n\\subsection{Spline Interpolation}\n\n\n\\subsection{Sinc Interpolation}\n\n\n\n\\section{Numerical Differentiation}\n\n\\section{Numerical Integration}\n\n\\section{Differential Equations}\nDifferential equations are the main tool for describing physical systems mathematically. They relate an unknown function $f(x)$, which is to be found, to its own derivatives $f'(x), f''(x), \\ldots$. If the function $f$ is a function of only one independent variable, we have an ordinary differential equation. There are also so called partial differential equations which are differential equations for a function of several independent variables, say $f(x,y,\\ldots)$, and an equation in terms of the partial derivatives $\\partial f / \\partial x, \\partial f / \\partial y, \\ldots$ is given. Sometimes, such differential equations can be solved analytically in the sense that an explicit expression for $f(x)$ in terms of $x$ can be given. Often, however, such an analytic solution is not feasible and we must resort to numerical algorithms to find the behavior of the system in a particular situation. In the context of audio processing, ordinary differential equations are typically used to describe electronic circuits such as analog filters, dynamics processors, etc.. Partial differential equations occur in the modeling of acoustic systems like vibrating strings, plates, air columns, etc..\n\n\\subsection{Ordinary Differential Equations}\n\n\\subsection{Partial Differential Equations}\n\n\n\\section{Curve Fitting}\n\n\\section{Optimization}\n\n", "meta": {"hexsha": "d3cdc59d6d2a5da63d79c539e59a703a8f5918a3", "size": 23522, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documentation/RAPT/LaTeX/UserManual/Mathematics.tex", "max_stars_repo_name": "RobinSchmidt/RS-MET-Preliminary", "max_stars_repo_head_hexsha": "6c01cbaad7cce3daa3293c444dd9e4b74e5ebfbe", "max_stars_repo_licenses": ["FTL"], "max_stars_count": 34, "max_stars_repo_stars_event_min_datetime": "2017-04-19T18:26:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-15T17:47:26.000Z", "max_issues_repo_path": "Documentation/RAPT/LaTeX/UserManual/Mathematics.tex", "max_issues_repo_name": "RobinSchmidt/RS-MET-Preliminary", "max_issues_repo_head_hexsha": "6c01cbaad7cce3daa3293c444dd9e4b74e5ebfbe", "max_issues_repo_licenses": ["FTL"], "max_issues_count": 307, "max_issues_repo_issues_event_min_datetime": "2017-05-04T21:45:01.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-03T00:59:01.000Z", "max_forks_repo_path": "Documentation/RAPT/LaTeX/UserManual/Mathematics.tex", "max_forks_repo_name": "RobinSchmidt/RS-MET-Preliminary", "max_forks_repo_head_hexsha": "6c01cbaad7cce3daa3293c444dd9e4b74e5ebfbe", "max_forks_repo_licenses": ["FTL"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2017-09-05T17:04:31.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-15T21:24:28.000Z", "avg_line_length": 135.9653179191, "max_line_length": 1516, "alphanum_fraction": 0.7723833007, "num_tokens": 5600, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.880797071719777, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6466103244623423}} {"text": "\\section*{Preliminaries}\n\\begin{itemize}\n \\item [natural number:] $\\naturalNumber = 0, 1, 2, 3, \\dotsc, \\infty$\n \\begin{itemize}\n \\item $\\naturalNumber_0 = 0, 1, 2, 3, \\dotsc, \\infty$\n \\item $\\naturalNumber_1 = 1, 2, 3, \\dotsc, \\infty$\n \\end{itemize}\n\n \\item [integer number:] $\\integerNumber = -\\infty, \\dotsc, -2, -1,\n \\naturalNumber$\n \\item [real number:] $\\realNumber = (-\\infty, \\infty) $\n \\item [imaginary number:] $i = \\sqrt{-1}$\n \\item [complex number:] $\\complexNumber = a + ib = Ae^{i\\theta}$\n \\begin{itemize}\n \\item $a, b \\in \\realNumber$\n \\item $A = \\sqrt{a^2 + b^2}$\n \\item $\\theta = \\arctan(b/a)$\n \\end{itemize}\n\\end{itemize}\n\n", "meta": {"hexsha": "db3e953de1c9000eb84a6c8aabfc04452d80aef1", "size": 676, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "CheatSheets/Series/Mathematics/units/unit_Basics.tex", "max_stars_repo_name": "tcburt/hodudodi", "max_stars_repo_head_hexsha": "de0952ceaf00d97251dcec984d0099fcd0905867", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "CheatSheets/Series/Mathematics/units/unit_Basics.tex", "max_issues_repo_name": "tcburt/hodudodi", "max_issues_repo_head_hexsha": "de0952ceaf00d97251dcec984d0099fcd0905867", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-01-18T22:55:02.000Z", "max_issues_repo_issues_event_max_datetime": "2020-01-18T23:01:00.000Z", "max_forks_repo_path": "CheatSheets/Series/Mathematics/units/unit_Basics.tex", "max_forks_repo_name": "tcburt/hodudodi", "max_forks_repo_head_hexsha": "de0952ceaf00d97251dcec984d0099fcd0905867", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.1904761905, "max_line_length": 71, "alphanum_fraction": 0.5887573964, "num_tokens": 271, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9161096090086368, "lm_q2_score": 0.7057850340255386, "lm_q1q2_score": 0.6465764515652836}} {"text": "\\chapter{Estimation Theory}\\label{App:estimation}\nThe Bayesian definition requires knowledge of the posterior probability on $\\theta$, that is, $pr(\\theta|g)$. The Bayesian regards $\\theta$ as random, but has no concept of an ensemble of $g$ vectors; the data are fixed. \nEstimates are statistical quantities, hence their evaluation requires statistical methods.\nBayesian estimation is the determination of an estimate of a random $\\theta$ through minimization of the Bayes risk. Knowledge of $pr(\\theta)$ is assumed, and a cost function $C(\\hat{\\theta}, \\theta)$ must be specified. The EMSE (Ensemble mean-squared error) is a common form for the cost measure $C(\\hat{\\theta}, \\theta)$ to be minimized in the design of an estimation procedure.\n\nThe quantity $pr(\\theta|g)$ describes the posterior probability of $\\theta$ after the data $g$ are obtained; thus the estimation rule $\\hat{\\theta} = \\arg\\max_{\\theta} pr(\\theta|g)$ is known as maximum a posteriori, or MAP, estimation. Equivalently, the MAP estimate is the mode of the posterior, or the posterior mode.\n\nInterestingly, MAP estimation is often what is meant in the literature on Bayesian estimation, However, MAP estimation is just a special case of Bayesian estimation in which the particular form of the cost function is a uniform cost function given by:\n\n\\begin{equation}\nC(\\hat{\\theta}, \\theta) = C(\\hat{\\theta}-\\theta) = 1 - rect \\left( \\frac{\\hat{\\theta}- \\theta}{2\\epsilon} \\right)\n\\end{equation}\n\nThe uniform cost function is when the cost is considered negligible if smaller than some tolerance $\\epsilon$, and all estimator errors beyond that tolerance are regarded as equally costly.\nAnother way of interpreting MAP estimation is that $pr(\\theta)$ characterizes the prior uncertainty in the parameter, which is often subjective in nature. Then $pr(\\theta|g)$ is the (presumably reduced) uncertainty after data are collected, hence the term posterior.\n\nMaximum-likelihood or ML estimation uses the following rule to determine the underlying parameters:\n\\begin{equation}\n\\hat{\\theta}_{ML} \\equiv \\arg\\max_{\\theta} pr(g|\\theta)\n\\end{equation}\nThis procedure can be written equivalently as:\n\\begin{equation}\n\\hat{\\theta}_{ML} = \\arg \\max_{\\theta} ln \\left[ pr)g|\\theta \\right].\n\\end{equation}\nML estimation can be considered as a limit to MAP estimation when the prior $pr(\\theta)$ is sufficiently broad that $pr(g|\\theta) pr(\\theta)$ is dominated by $pr(g|\\theta)$. However, ML estimation is much more than a limiting form of MAP estimation.\nIn ML estimation, $pr(g|\\theta)$ is the quantity being maximized, but even in MAP estimation its role has some level importance that depends on the relative weight of the prior. The quantity $pr(g|\\theta)$ means different things to different people. The frequentist interpretation is that it is a function of $g$ for fixed $\\theta$. Repeated observations can be used to determine the nature of this quantity for a given underlying object (parameter vector). Alternatively, $pr(g|\\theta)$ can be viewed as a function of $\\theta$ for fixed $g$. In this viewpoint $pr(g|\\theta)$ is a measure of the likelihood of any $\\theta$ once that data are in hand. \n\\begin{itemize}\n\\item frequentist approach: what are all $g$'s that is produced by a single $\\theta$?\n\\item viewing $pr(g|\\theta)$ as a function of $\\theta$ for fixing g; or what are all $\\theta$'s that can produce identical $g$.\n\\end{itemize}\n\n\\noindent\n\\textbf{Score}: \\medskip The \\textit{score} is a random vector that tells us how sensitive the likelihood is to change in the parameters:\n\\begin{equation}\ns(g) = \\frac{\\frac{\\partial}{\\partial \\theta} pr(g|\\theta}{pr(g|\\theta)} = \\frac{\\partial}{\\partial\\theta} In \\left[ pr(g|\\theta) \\right]\n\\end{equation}\n\nIn words, the score is the gradient of the log-likelihood. Since the score is a function of the log-likelihood, which is a random variable through its dependence on $g$, the score is also random. Note that $\\langle s \\rangle_{g|\\theta} = 0$, where $\\langle \\cdot \\rangle_{g|\\theta}$ denotes an average with respect to $pr(g|\\theta)$. Score is a zero-mean random vector.\nAs the gradient of the log-likelihood, $s(g, \\theta) = 0$ when $\\theta = \\hat{\\theta}(g)$. The process of finding the ML estimate is thus equivalent to determining the point in parameter space where all components of the score vanish.\n\nML estimator is efficient, where if the Cramer-Rao lower bound on the variance is attainable, it will be attained by an ML estimator.\nML estimator is sufficient. In estimation, a sufficient statistic is one that captures all the essential features in the data necessary for optimal performance of a given estimation task. The maximum-likelihood estimator is a sufficient statistic for estimation; it makes optimal use of the information in the data. No other estimator can yield more information. \nA necessary and sufficient condition for $\\hat{\\theta}$ to be a sufficient estimate is that the likelihood function must be factorable into the product:\n\\begin{equation}\npr(g|\\theta) = pr(\\hat{\\theta}|\\theta) f(g),\n\\end{equation}\nwhere $f(g)$ is independent of $\\theta$. \nA sufficient estimator may exist even when an efficient estimator does not; efficiency is a stricter criteria than sufficiency. A sufficient estimator is unique; we can use a function of the sufficient statistic because it will also be sufficient, and we can choose it such that the estimate maybe be consistent and unbiased.\nIn general there can be many solutions to ML estimation equation that give equal likelihood. This is especially the case in high-dimensional problems, where null space can be very large. \nAsymptotic properties of the ML estimator. It describes the behavior of the estimate as the number of observations approaches infinity. Cramer proved that, under reasonably general conditions, an ML estimate is consistent. The density function for a consistent estimate becomes increasingly narrow about the value of the underlying parameter as the number of samples increase. ML estimate is efficient, if an efficient estimator exists. ML estimates achieve the equality sign in the bound as the number of samples goes to infinity; that is, ML estimates are asymptotically efficient.\nML estimates are asymptotically unbiased, efficient, normally distributed and consistent.\n\nSummarize: The full Bayesian estimation approach requires that the cost function as well as the prior probability of the random parameters and their likelihood function be completely specified. The ML approach does not require a prior, but it still requires full knowledge of the likelihood function. \n\n\\noindent\n\\textit{Bayesian and frequentist:}\nIn both pure estimation problems and hybrid detection/estimation problems, we have found that the optimal strategy for handling nuisance parameters is to marginalize (to treat as insignificant or peripheral) rather than to estimate them. This finding begs the question of the appropriate prior to utilize. To a frequentist, the prior should be a sampling prior, verifiable by experiment; an example is the exponential absorption law for photons. To a Bayesian, the prior could incorporate prior beliefs. Indeed, a Bayesian would say that the problem of nuisance parameters is an example of a fundamental dilemma that arises in any inference problem: we \\textit{never} have enough empirical (verifiable by observation or experience rather than theory or pure logic) information to solve the problem at hand, and we must always bring in prior belief. \nYes, one can use prior beliefs in estimation problems in imaging, but the final measure of the efficacy of the belief is a long-run, frequentist measure of task performance.", "meta": {"hexsha": "20f6d297d7aafffbb54116ac11a7d5ac75429921", "size": 7699, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/estimation_theory.tex", "max_stars_repo_name": "hfan36/dissertation", "max_stars_repo_head_hexsha": "5d755c96cf6cbece2c382789015e9db9ceb02da7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Notes/estimation_theory.tex", "max_issues_repo_name": "hfan36/dissertation", "max_issues_repo_head_hexsha": "5d755c96cf6cbece2c382789015e9db9ceb02da7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/estimation_theory.tex", "max_forks_repo_name": "hfan36/dissertation", "max_forks_repo_head_hexsha": "5d755c96cf6cbece2c382789015e9db9ceb02da7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 132.7413793103, "max_line_length": 854, "alphanum_fraction": 0.7777633459, "num_tokens": 1736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388083214155, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6465663894751604}} {"text": "\\lab{Sparse Grids}{spgrid}\n\\label{lab:spgrid}\n\n\\objective{Sparse Grids are an important tool when dealing with high-dimensional problems. Computers operate in discrete space, not in continuous space. It is important to choose evaluation points wisely so as to maximize accuracy without sacrificing computation time. In particular, we explore how to use sparse grids to compute integrals of high dimensionality.}\n\n\\section*{Discretization}\nAt first inspection, our world appears to be nice and continuous. However, this only works on a macroscopic level. As we zoom in on matter, we find that it is made of discrete atoms with much empty space between. Computers likewise work in discrete space. You have already seen many examples of this. Consider plotting the function $y=x^2$, such as in Figure \\ref{fig:x_squared}. To do this we take an array of discrete points of $x$ and $y$ values, which are then joined in a linear manner. As you either zoom in on the function or decrease the number of plotting points, you can see the discrete nature of even this simple function.\n\nIn order to get better results, we need to use a larger number of points. This is true not just for graphing purposes, but also for standard computation. If we double the number of points, we approximately double the computation time necessary. This effect, while not irrelevant, pales in comparison to the case when working in multiple dimensions. Imagine a function that we discretize into $20$ points. A similar function of two variables, where each variable is discretized into $20$ points, would necessitate $400$ unique points. Expanding to seven variables would necessitate $1,280,000,000$ unique points!\n\nIn general, for $n$ discrete points in $d$ dimensions, this gives $n^d$ points. This can be visualized as a $d$-dimensional grid of size $n \\times n \\times \\cdots \\times n$. In practice, we seldom need to know the value for each and every point. Instead of using the full grid, we can use a sparse grid. The main idea of a sparse grid is that it can reduce the order of difficulty for a standard $d$-dimensional problem.\n\n\\begin{center}\n\\begin{figure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{x2.png}\n\\end{subfigure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{x2a.png}\n\\end{subfigure}\n\\caption{Plots of the function $y=x^2$ using $31$ points and $9$ points, respectively. In the $9$-point plot, the linear, discrete nature of the function is easily visible.}\n\\label{fig:x_squared}\n\\end{figure}\n\\end{center}\n\n\\section*{The Hierarchical Basis}\nThe basic sparse grid is based on the use of the \\emph{Hierarchical Basis}. This basis is composed of piecewise-linear functions known as the \\emph{standard hat functions}, and can be used to approximate any function. The standard hat function in one dimension is defined as:\n\n\\begin{equation}\n\\phi(x) = \\left\\{\n \\begin{array}{ll}\n 1-\\abs{x} & \\quad -1 \\leq x \\leq 1 \\\\\n 0 & \\quad otherwise\n \\end{array}\n \\right.\n\\end{equation}\n\nFrom this equation, we can build the set of basis functions of order $j$\n\n\\begin{equation}\n\\phi_{j,i}(x) = \\phi(2^{j-1}(x+1) - 2i+1)\n\\end{equation}\n\n\\begin{center}\nwhere $i=1,2,3,\\cdots,2^{j-1}$. \n\\end{center}\n\nThe first few basis functions are listed in Table \\ref{table:functions}, and are plotted in Figure \\ref{fig:basis_functions}.\n\n\\begin{center}\n\\begin{figure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{j1.png}\n\\end{subfigure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{j2.png}\n\\end{subfigure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{j3.png}\n\\end{subfigure}\n\\begin{subfigure}{.49\\textwidth}\n\\includegraphics[width=\\textwidth]{j4.png}\n\\end{subfigure}\n\\caption{Plots of the first few standard hat functions. Note that the value of $j$ gives the number of \"hats\" in the domain $[-1,1]$, and $i$ gives which hat, counting from the left-most hat.}\n\\label{fig:basis_functions}\n\\end{figure}\n\\end{center}\n\n\\begin{center}\n\\begin{table}\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline\n$j\\backslash i$ & $1$ & $2$ & $3$ & max $i$ \\\\ \\hline\n$1$ & $\\phi_{11}(x) = \\phi(x)$ & & & 1 \\\\ \\hline\n$2$ & $\\phi_{21}(x) = \\phi(2x+1)$ & $\\phi_{22}(x) = \\phi(2x-1)$ & & 2\\\\ \\hline\n$3$ & $\\phi_{31}(x) = \\phi(4x+3)$ & $\\phi_{32}(x) = \\phi(4x+1)$ & $\\phi_{33}(x) = \\phi(4x-1)$ & 4 \\\\ \\hline\n$4$ & $\\phi_{41}(x) = \\phi(8x+7)$ & $\\phi_{42}(x) = \\phi(8x+5)$ & $\\phi_{43}(x) = \\phi(8x+3)$ & 8 \\\\ \\hline\n$5$ & $\\phi_{51}(x) = \\phi(16x+15)$ & $\\phi_{52}(x) = \\phi(16x+13)$ & $\\phi_{53}(x) = \\phi(16x+11)$ & 16 \\\\ \\hline\n\\end{tabular}\n\\caption{The first few standard hat functions that make up the hierarchical basis, for $i\\leq3$.}\n\\label{table:functions}\n\\end{center}\n\\end{table}\n\\end{center}\n\nUsing these basis functions, we can approximate any function of our choosing. Suppose we want to solve\n\\begin{equation*}\n\\int_{-1}^{1} \\sqrt{1-x^2}\\: dx .\n\\end{equation*}\nThis is the area of a semi-circle of radius $1$, and can easily be computed mathematically as $\\pi / 2 \\approx 1.57079632679$. But a computer does not calculate this in the same manner. Instead, it uses certain points to evaluate the integral numerically. We can approximate $f(x) = \\sqrt{1-x^2}$ by interpolating the basis hat functions. If we approximate up to order $l$ then this gives\n\\begin{equation*}\nf(x) \\approx \\sum_{n=1}^l \\sum_{m=1}^{2^{n-1}} c_{nm} \\phi_{nm}(x) \n\\end{equation*}\nwhere $c_{nm}$ are the appropriate constants which can be found by evaluating the difference between $f(x)$ and the sum of the basis hat functions of lower order at the peak of each hat:\n\\begin{equation}\nc_{nm} = f(p_{nm}) - \\sum_{s=1}^{n-1} \\sum_{t=1}^{2^{s-1}} c_{st}*\\phi_{st}(p_{nm})\n\\end{equation}\nwhere\n\\begin{equation}\np_{nm}=2^{2-n}*(m-0.5)-1\n\\label{eq:points}\n\\end{equation}\nand $p_{nm}$ is the peak of the $m^{th}$ hat of level $n$. These points will be important because we will use these to create the basic sparse grid.\n\n\\begin{center}\n\\begin{figure}\n\\includegraphics[width=.7\\textwidth]{HB.png}\n\\caption{Interpolating a function using the standard hat basis functions.}\n\\label{fig:HB}\n\\end{figure}\n\\end{center}\n\nOne benefit of decomposing $f(x)$ into the basic hat functions is that they are extremely simple to integrate. Each is a triangle, for which $A=bh/2$. The height is $1$, and the base is $2^{2-n}$, and so\n\\begin{equation*}\n\\int_{-1}^1 \\phi_{nm} dx= 2^{1-n}\n\\end{equation*}\n\nOne problem with using the basic hat functions to approximate a function arises when the function does not go to zero at the endpoints. Higher order approximations yield better results, but ultimately we need to decide how accurate we need our answer to be. When an additional level changes the answer by less that our error tolerance, we have found a good level of approximation.\n\n \\begin{problem}\nUse the function declaration below to approximate a given function on $[-1,1]$ with the basic hat functions up to order $l$:\n\\begin{lstlisting}\ndef hat_approximation(f,l):\n \"\"\"This function will return a list of the correct coefficients to approximate the function f up to level l.\n Parameters\n ----------\n f (function) : The function of a single variable to approximate\n l (int) : The order to use for interpolating the hat function\n\n Returns\n -------\n coeffs (list) : Entry n is an ndarray of len(2^(n-1)) and contains the coefficients for phi_{nm}. The length of this should be l.\n \"\"\"\n pass\n\\end{lstlisting}\n\\label{prob:one}\n\\end{problem}\n\n\n\\begin{center}\n\\begin{figure}\n\\includegraphics[width=.7\\textwidth]{errors.png}\n\\caption{}\n\\label{fig:errors}\n\\end{figure}\n\\end{center}\n\n\\begin{problem}\nWrite a function that, given the coefficients returned by the function in Problem \\ref{prob:one}, returns the value of the integral.\n\\end{problem}\n\n\\begin{problem}\nUsing the function $f(x)=\\sqrt{1-x^2}$, calculate the error of $\\int_{-1}^1 f(x)$ in regards to the actual value $\\pi/2$. Do this for the order $l = 1,2,\\cdots,10$, and plot the error as a function of $l$. Your results should match Figure \\ref{fig:errors}. Additionally, time how long it takes to execute the code for each $l$, and plot your results in another graph.\n\\end{problem}\n\nKeep in mind that thus far we have only used a function of one variable. For each variable (dimension) the time to compute will increase exponentially. With multiple dimensions, it becomes infeasible to compute to any degree of accuracy, because the time to compute will increase much too quickly to be reasonable for higher dimensions. Hence the need for sparse grids.\n\n\\section*{Sparse Grids}\nRather than dividing each dimension into $n$ sections, which for $d$ dimensions would give $n^d$ points, sparse grids use only a few of these points. There are many types of sparse grids, and each is useful for certain purposes. Which type of sparse grid is used will often depend on the nature of the problem you are trying to solve. For this lab, we will use the basic sparse grid for purposes of integration.\n\nThe basic sparse grid has two main properties, the dimension and the level. Using these two variables and Equation \\ref{eq:points}, we can construct a sparse grid. Visualizing this in one dimension is simple: we use as our evaluation points each $p_{nm}$. A level $1$ grid has one point at the center, which is the peak of the level $1$ hat. A level $2$ grid has the level $1$ point with two level $2$ points, which are the peaks of the two level $2$ hats. The third level adds its $4$ points, etc. So for one dimension of level $l$ we have $\\sum_{i=1}^l 2^{i-1}$ points.\n\nUnderstanding a $2$-dimensional grid is a bit more complicated. Our level $1$ point goes at the center of the grid. To expand to level $2$, we add in the peaks of the level $2$ hats \\emph{in both dimensions}, for an additional $4$ points. Level $3$ is where it gets a little bit tricky mathematically. We can't just keep expanding only along the axes of the grid, because we need points on the interior as well. Conceptually, what we do is take our current points and divide the space between points and the end of the grid into two, and there we create a new point. This is synonymous to adding a hat between the peaks of currently existing points.\n\nWe can make a lot more sense of that last paragraph by actually plotting the points. The file \\li{pysg.py} contains a sparse grid class with many useful functions. Using \\li{plotGrid}, you can plot the grid points. You will first need to create a sparse grid object, and then generate the points using the built in function. Unfortunately, \\li{plotGrid} can only plot up to three dimensions. Note that in this file, our domain is $[0,1]^d$ rather than $[-1,1]^d$. \n%Be sure to take note of this when computing the integrals in the following problem. \n\n\\begin{center}\n\\begin{figure}\n% \\includegraphics[width=.7\\textwidth]{gridpoints.png}\n\\caption{Sparse grid points for 2-D grid of level 5}\n\\label{fig:gridpoints}\n\\end{figure}\n\\end{center}\n\n% THIS FILE IS NOT IN THE REPOSITORY!!\n\n\n\n\\begin{problem}\nExplore the \\li{sparseGrid} class contained in the module \\li{pysg.py}. Figure out how to create a sparse grid object, generate the points, and plot them. This function should plot the sparse grid points for a grid of dimension 2 and level 5. See the figure above.\n\\end{problem}\n\n\n\nMulti-dimensional integration for functions with independent endpoints (those that are numbers that do not depend on other variables) is simplified greatly by using sparse grids. Functions must be manipulated appropriately before sparse grid classes like those in \\li{pysg.py} can be used to do the integration, but once this has been done, their evaluation becomes rather simple. Once the coefficients for the hat functions have been computed, the integrals can be evaluated without having to perform any integration. Sparse grids are powerful tools for fast computation of high-dimension integrals. \n%In general, note that\n%\\begin{equation*}\n%\\int_{a_1}^{b_1} \\int_{a_2}^{b_2} \\cdots \\int_{a_d}^{b_d} \\phi_{sii\\cdots i}dx_d dx_{d-1} \\cdots dx_1= \\frac{\\prod_{i=1}^d (b_i-a_i)}{2^{(s-1)d}*(d+1)}\n%\\end{equation*}\n\n%\\begin{problem}\n%Using the sparse grid class in \\li{pysg.py}, compute the value of the following integrals:\n%\\begin{enumerate}\n%\\item $\\int_{0}^1 \\int_{0}^1 (x^2+\\frac{1}{3}xy) \\ dx dy \\ $ to order 8.\n%\\item $\\int_{.36}^{1.8} \\int_{0}^{\\pi} y*cos(x) \\ dx dy \\ $ to order 6.\n%\\item $\\int_{-2}^3 \\int_{5}^6 \\int_{4.2}^{6.3} (\\frac{1}{43}x+\\frac{1}{123}x^2 y^3 + %\\frac{1}{200}yz^2) \\ dx dy dz \\ $ to order 5.\n%\\item $\\int_{[0,1]^{18}} \\prod_{i=1}^{18} x_i \\ dx_i \\ $ to order 3.\n%\\end{enumerate}\n%\\end{problem}\n", "meta": {"hexsha": "adc187d058a33811075f13a1df30981eda49a1d4", "size": 12755, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Orphans/SparseGrid/SpGrid.tex", "max_stars_repo_name": "joshualy/numerical_computing", "max_stars_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Orphans/SparseGrid/SpGrid.tex", "max_issues_repo_name": "joshualy/numerical_computing", "max_issues_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Orphans/SparseGrid/SpGrid.tex", "max_forks_repo_name": "joshualy/numerical_computing", "max_forks_repo_head_hexsha": "9f474e36fe85ae663bd20e2f2d06265d1f095173", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-11-05T14:45:03.000Z", "max_forks_repo_forks_event_max_datetime": "2019-11-05T14:45:03.000Z", "avg_line_length": 63.1435643564, "max_line_length": 656, "alphanum_fraction": 0.7252058016, "num_tokens": 3705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8289388040954683, "lm_q1q2_score": 0.6465663819395951}} {"text": "\n% Take from PyAudioAnalysis\n% short-term, mid-term features\n\n% \\chapter{Algorithms}\n% \\section{Time Analysis}\n% \\subsection{Convolution}\n% A general use of the convolution operation\n% is to evaluate the outputs of linear\n% time-invariant systems.\n% Taking the impulse response of a system, and\n% applying the convolution with\n% a given input, gives the system's output\n% for that specific input.\n\n% For continuous time dependent arguments,\n% \\(x\\) as the input variable, and \\(h\\) as the\n% system's impulse response, the convolution is given by:\n\n% \\begin{equation}\n% (x * h)(t) \\triangleq \\int^{\\infty}_{-\\infty} x(t)h(t-\\tau) d\\tau\n% \\end{equation}\n\n% Due to the commutativity property of the convolution\n% operation, the shifting over one of the arguments\n% is interchangeable, thus one can write\n% the same equation as:\n\n% \\begin{equation}\n% (x * h)(t) \\triangleq \\int^{\\infty}_{-\\infty} x(t-\\tau)h(t) d\\tau\n% \\end{equation}\n\n% Likewise, working with discrete variables,\n% the convolution operation can be written as:\n% \\begin{align}\n% (x * h)[n] \\triangleq & \\sum^{\\infty}_{m=-\\infty} x[m]h[n-m] \\\\\n% (x * h)[n] \\triangleq & \\sum^{\\infty}_{m=-\\infty} x[n-m]h[m]\n% \\end{align}\n\n\n% \\subsection{Cross-Correlation}\n% Cross-correlation is an operation that can measure\n% how similar two functions are. By ``shifting''\n% one function over the other and measuring the amount\n% of correlation at each given point, the output graph\n% demonstrates at what ``distance''\n% the maximum similarity occurs,\n% in correspondence\n% to the graph's maximum amplitude.\n\n% \\begin{align}\n% (x * h)(t) \\triangleq & \\int^{\\infty}_{-\\infty} x^{*}(t)h(t+\\tau) d\\tau \\\\\n% (x * h)(t) \\triangleq & \\int^{\\infty}_{-\\infty} x^{*}(t-\\tau)h(t) d\\tau\n% \\end{align}\n\n% \\begin{align}\n% (x * h)[n] \\triangleq & \\sum^{\\infty}_{m=-\\infty} x^{*}[m]h[n+m] \\\\\n% (x * h)[n] \\triangleq & \\sum^{\\infty}_{m=-\\infty} x^{*}[n-m]h[m]\n% \\end{align}\n\n% \\subsection{Auto-Correlation}\n% In a case where the two input functions\n% of the cross-correlation are the same input signals,\n% the measure is then called ``auto-correlation''.\n% In other words, this measure demonstrates\n% at what ``distance'' the delayed version of the signal\n% most matches the original version of it.\n\n% This kind of measure is very useful in audio\n% applications where multi-microphones are spread\n% with varying distance from each other and from the\n% source. Looking for the maxima point of the\n% auto-correlation between the signal received by\n% the microphones, one can deduce the lag time\n% that best characterizes the microphones' formation.\n\n% Several enhancement and manipulation techniques\n% are then become available when the lagging\n% time is known.\n% For example the delay and sum algorithm beamforming\n% algorithm for constructive summation of the different\n% channels that translates\n% to better SNR (signal-to-noise ratio).\n\n% \\subsubsection{Covariance Matrices}\n% \\begin{equation}\n% \\mathbf{R}_{XY} \\triangleq \\mathbf{E}[XY^{tr}]\n% \\end{equation}\n\n% \\begin{align}\n% \\mathbf{R}_{XX} & \\triangleq \\mathbf{E}[XX^{tr}] \\\\\n% & = \\frac{1}{T} \\sum^{T-1}_{t=0}\n% {X}(t;j\\omega){X}^{\\mathbf{H}}(t;j\\omega)\n% \\end{align}\n\n% Where the \\(\\mathbb{H}\\) operator\n% is the \\emph{Hermitian function} which stands for\n% the complex conjugate.\n\n% % ESPNet \\& Mirco Document\n% \\subsection{Sinc-Conv}\n\n% % \\begin{figure}[ht]\n% % \\centering\n% % \\includegraphics[width=0.99\\textwidth]\n% % {./img/Comparison_convolution_correlation.svg}\\label{fig:conv_vs_corr_plot}\n% % \\caption{Convolution vs. Cross-Correlation vs. Auto-Correlation plot [Wikipedia]}\n% % \\end{figure}\n\n\n\n% \\section{Frequency Analysis}\n% \\subsection{DTFT}\n% \\begin{equation}\n% X_{2\\pi}(\\omega) = \\sum_{n=-\\infty}^{\\infty} x[n] \\,e^{-i \\omega n}\n% \\end{equation}\n\n\n% \\subsection{IDTFT}\n% \\begin{equation}\n% x[n] = \\frac{1}{2 \\pi}\\int_{2\\pi} X_{2\\pi}(\\omega)\\cdot e^{i \\omega n} d\\omega\n% \\end{equation}\n\n% \\subsection{DFT}\n% \\begin{equation}\n% \\label{eq:dft}\n% X_k = \\sum_{n=0}^{N-1} x_n \\cdot e^{-\\frac {i 2\\pi}{N}kn}\n% \\end{equation}\n\n% Taking Euler's identity:\n% \\begin{equation}\n% e^{ix} = \\cos x + j\\sin x\n% \\end{equation}\n\n% Substituting the exponent power in Equation\\;\\ref{eq:dft} with the Euler identity\n% gives:\n% \\begin{equation}\n% X_k = \\sum_{n=0}^{N-1} x_n \\cdot \\left[\\cos\\left(\\frac{2 \\pi}{N}kn\\right)\n% - i \\cdot \\sin\\left(\\frac{2 \\pi}{N}kn\\right)\\right]\n% \\end{equation}\n\n\n% \\subsection{IDFT}\n% \\begin{equation}\n% x[n] = \\frac{1}{N} \\sum_{k=0}^{N-1} X_k\\cdot e^{i \\frac{2 \\pi}{N} k n}\n% \\end{equation}\n\n\n% \\subsection{FFT --- Fast Fourier Transform}\n\n% \\subsection{Discrete Cosine Transform (DCT)}\n\n% \\begin{equation}\n% y[n] = \\sum^{}_{}\n% \\end{equation}\n\n% \\section{Time-Frequency Analysis}\n% \\subsection{STFT - Short Time Fourier Transfer}\n\n% \\subsection{Spectrogram}\n\n% \\section{Windows}\n% \\subsection{Overview}\n\n\n% \\subsection{Rect (boxcar)}\n% \\subsection{Triangle}\n% \\subsection{Bartlett}\n% \\subsection{Hamming}\n% \\subsection{Hann}\n% \\begin{align}\n% w[n] = 0.5 - 0.5\\cos\\left( \\frac{ 2\\pi n }{ M - 1 } \\right) & \\qquad 0 \\leq n \\leq M-1\n% \\end{align}\n\n\n% % \\subsection{Kaiser}\n% % \\subsection{Analog Filters}\n\n% \\subsection{Overlap + Add Reconstruction}\n% % \\subsection{Wavelets}\n\n% \\section{Speech-Enhancement}\n% See table\n% \\begin{figure}[ht]\n% \\centering\n% \\includegraphics[width=0.99\\textwidth]\n% {./img/spc_enhance_tbl}\\label{fig:asr_blocks_diagram}\n% \\caption{General E2E ASR System Blocks Diagram}\n% \\end{figure}\n\n\n\n% DCT + MFCC Connection\n% DST - Discrete Sine Transforms\n% Hilbert Transform\n% Analytical Signal\n% Wavelates\n% Convolution / Correlation\n% Parseval's Theorm\n% PSD\n% Median Filter\n% ORder Filter\n% Wiener Filter\n% Hermitian FFTs\n\n\n\n\n\n\\chapter{Features}\\label{ch:features}\n\\section{Introduction}\nIn data analysis craftsmanship, \nprominence importance exists \nabout how to characterize the data so that \nvariations in data characteristics would \nbe noticeable and more \neasily discernible during the analysis process.\nTo that end, the input data being analyzed is reorganized\naccording to selected features on which\nconclusions and distinctive deductions can be made.\n\nSimilarly, in a supervised learning \nprocedure of a machine, \nin order to make classification decisions accurately. \nThe need for such features arises to \naid the learning algorithm in focusing on the \nsame features that probably were selected manually.\nAn intelligent choice of the learnable features \n\\cite{7845025} \ncan drastically change a given model's outcomes quality.\nIn a supervised learning, the \nfeature selection and extraction process withal,\nare preparatory steps to the learning or classification\nstages coming next.\nAs a rule of thumb, the more features, the better accuracy\na learning model can yield theoretically\\cite{lessIsMore}.\nThat saying holds true to a great degree\nas long as often sudden fluctuations do not characterize the data.\nIn a case of heavy fluctuating characterizations of the data,\nor alternatively, in the case of a \nmassive number of features, \nthat some of which have minor contributions \nto the classification part of the output,\nincrease in the number of features may become deteriorative.\nIn terms of performance, \nenlarging the number of features leads to a bigger model, \nan increased number of learnable parameters, \nlonger training times, \nand unnecessary extension of processing times.\nAlso, a possible reduction in the accuracy is \nexpected due to \nFalse-Negative (FN) or False-Positive (FP) \nmisdetections resulting from the wrong \nclassification of signals as noise and \nvice versa caused by additional redundant features.\n\nFor speech signals, a wide variety of \nmeaningful feature sets exist. \nMore or less useful, different speech features \nmay better fit certain use-cases or fulfill \na particular unerring task. \nFeatures for speech (including audio) \nare mainly from the following domains:\n\\begin{enumerate}\n \\item Spectral Features\n \\item Cepstral Features\n \\item Time domain Features\n \\item Spatial Features\n\\end{enumerate}\n\nSpeech features are selected \nto give the maximal accuracy in detecting utterances.\nThat means a precise characterization of a \nword, utterance, or the pronunciation of \na single character, making them distinguishable\nfrom other input streams. \n\n\\section{Spectral Features}\n\\subsection{FB --- FilterBanks}\nFilterBanks is a very common technique\nfor spectral mapping of speech signals.\nBy dividing the audio spectrum into multiple \nsub-domains with a defined level of overlap, \nthe spectrum is \n``framed'' according to frequency. \nThus, by a set of band-pass filters,\neach frame contains the confined \ninformation of the speech signal \nthat corresponds to the filter's specific \nrange of frequencies.\n\nThe resolution can then be set as a function of\nthe number of filters and the overall processed audio bandwidth.\nIncreasing the number of filters, assuming the \naudio bandwidth and the overlap ratio are constant, means\nnarrower allocated bandwidths for each individual filter\nor in other better spectral resolution.\n\nFilterbanks by themselves are not the desired speech feature,\nbut only the mean for feature extraction. \nThe most common feature extracted by a Filterbank set\nis the total sum of energy bounded by the filter's frequency response.\nA set of filters is computed per \nfrequency bin and remains the same for \nthe entire signal length over time.\nThus, the total sum of energy computed for \neach filter characterizes the \nspeech over a finite defined duration of time.\n\nThe human hearing system is less sensitive to high frequencies\nthan lower band frequencies, as described in Chapter\\;\\ref{ch:scaling_methods}.\nTherefore, in an attempt to emulate the same natural behavior and\nresemble the hearing ``filters'' as much as possible, \nthe Filterbank set of filters is set with center frequencies\naccording to the different scaling \nmethods described in Chapter\\;\\ref{ch:scaling_methods}.\nIn that way, narrow-band filters are assigned to \nlower frequency ranges. \nSimilarly, wide-band filters are for \nthe higher hearable frequency ranges.\n\n\\subsubsection{Mel FB}\nOne way of mapping the audio spectrum is according to the Mel\nscale. This scaling method is described in Chapter\\;\\ref{ch:scaling_methods}.\nFirst, the center frequencies and the bandwidths are received by the transformation\nbetween Herz to Mels. Then a set of Bartlett\nfilters with those center frequencies are generated\nwith an overlap of 50\\% between adjacent filters. \n\n\nThe amplitude of the filters is bounded to \\(1\\) \nto maintain NOLA compatibility.\nAn example of a Mel Filterbank constructed by Bartlett filters\nis shown in Figure~\\ref{fig:sb_mel_fb}.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Features/images/sb_mel_fb}\n \\caption{Mel FB}\\label{fig:sb_mel_fb}\n\\end{figure}\n\n\\subsubsection{Bark FB}\nThe Bark Filterbank is very similar to the Mel Filterbank\nexcept that it follows the Bark scale \ninstead of the Mel Scale.\n\nDifferent filter shapes are available but, \ntriangular filter shapes are the most common,\nas described in \\cite{barkfilt}.\n\nAn example of the Bark Filterbank constructed by Bartlett filters\nis shown in Figure~\\ref{fig:mat_bark_fb}.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Features/images/mat_bark_fb}\n \\caption{Bark FB}\\label{fig:mat_bark_fb}\n\\end{figure}\n\n\\subsubsection{Gammatone FB}\nThe Gammatone filter, as described in \\cite{gammatonefilt},\nhas a response function as follows:\n\\begin{equation}\n g(t) = \\alpha t^{n-1}e^{-2\\pi bt}\\cos \\left( 2\\pi f_{c} t + \\phi\\right)\n\\end{equation}\n\nWhere \\(\\alpha\\) denotes the amplitude factor,\n\\(n\\) is the filter order,\n\\(f_{c}\\) is the center frequency,\n\\(\\phi\\) is the phase factor,\nand \\(b\\) is the bandwidth parameter\ncomputed according \nto the ERB scale mapping of \\(1.019\\cdot ERB \\left( f_{c} \\right)\\).\n\nAn example of a Gammatone Filterbank\nis shown in Figure~\\ref{fig:sb_fb_gauss}.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Features/images/sb_fb_gauss}\n \\caption{Gammatone FB}\\label{fig:sb_fb_gauss}\n\\end{figure}\n\n\n\\section{Cepstral Features}\n\\subsection{MFCCs --- Mel-Frequency Cepstral Coefficients}\n\n% https://link.springer.com/content/pdf/bbm%3A978-3-319-03116-3/1.pdf\n\\subsubsection{Pre-emphasis}\nSpeech signals have a roll-off frequency \nresembling a low-pass behavior\\cite{237532}.\nDue to that physical nature, higher frequencies \ndecay faster than lower speech frequencies.\nCompensation for this phenomenon is attainable with a \npre-emphasis filter that boosts \nthe higher\\cite{7489370} frequencies responses.\n\n\n\\subsubsection{Framing}\nSpeech signals vary in time. \nAlthough the variation over time is relatively slow, \na speech signal is a non-pure stationary process \nbut a quasi-stationary. \nTherefore, analysis of speech signals \nis taken on small portions of the \nsignal to be less affected by randomness effects.\nTo that end, a preliminary framing action is applied\nto to speech signals in the time domain.\nFraming means dividing the signal \ninto small fragments(frames) with some overlapping in between.\nEach time frame is assumed to be stationary,\nand thus a measurement can be taken.\n\nThe shifting in time between frames is referred\nto as the hopping length and is set to contain \na sufficient amount of temporal\ncontext that characterizes the natural\ncharacteristics of speech.\n\nA very common sampling frequency of audio signals\nis \\(16KHz\\). \nAs a result, and in order to work with \na round number of sampling points, \nthe frame lengths \nand the hopping lengths \nare set to \\(25ms\\), and \n\\(6.25ms\\) or \\(10ms\\) for hopping size. \nThese values translate to a frame length\nequals 400 sampling points, and hopping size\nequals 100 or 160 sampling points, respectively.\nAnother advantage of setting the hopping length\nas \\(6.25ms\\) is that it gives\na complete temporal context of 3 adjacent frames\nby definition.\n\n\\subsubsection{Windowing}\nIn order to extract each frame only \nwhilst also minimizing the Gibbs effects as much as possible,\na windowing function is applied.\nAs a result, the information confined \nin a given frame is extracted while the \nwindow's response function tapers \nthe edges to reduce the adjacent frames' effect.\n\nUsually, a Hamming or a Hann window \nis used as the window function due to their \nrelatively decent trade-off between edge tapering, \nimplementation simplicity, bandwidth, and spectral leakage, \nmaking them exceptionally suitable for speech signals.\n\nThe Hamming and Hann windows are given by\nEquations\\;\\ref{eq:hammwin} and \\ref{eq:hannwin}, respectively.\n\\begin{align}\n \\label{eq:hammwin} &W_{_{Hamming}}[n] = 0.54 - 0.46\n \\cos\\left( \\frac{2\\pi n}{M-1} \\right) &\n 0 \\leq n \\leq M - 1 \\\\\n \\label{eq:hannwin} &W_{_{Hann}}[n] = 0.5 - 0.5\n \\cos\\left( \\frac{2\\pi n}{M-1} \\right) &\n 0 \\leq n \\leq M - 1\n\\end{align}\n\n\\subsubsection{DFT spectrum}\nEach one of the windowed frames is\nconverted to the frequency domain by applying the \nDiscrete Fourier Transform (DFT).\nThe composition of both windowed framing \nand the DFT is referred to as the \nShort-Time Fourier Transform (STFT).\nA technique to visualize the STFT outcome is \ncalled a spectrogram. The STFT outcome contains \nmultiple frequency bins per time frame,\nmaking it a function of \\(f, t\\).\n\nThe DFT is given by:\n\\begin{align}\n X_{m}(f) = \\sum_{n=-\\infty}^{\\infty} x[n]g[n-mR]e^{-j2\\pi fn}\n\\end{align}\n\nWhere \\(X_{m}(f)\\) denotes the DFT transformation of\na given time frame windows signal,\n\\(g[\\circ]\\) denotes the window function of size \\(M\\),\nand \\(R\\) represents the hopping size.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Features/images/iscola_stft}\n \\caption{STFT demonstration diagram}\\label{fig:iscola_stft}\n \\source{Adapted from Matlab's STFT documentation}\n\\end{figure}\n\n\\subsubsection{Mel-spectrum}\nOnce the spectrogram is received, \nthe frequencies are rescaled in accordance with the Mel Scale.\nA comprehensive description of the Mel Scale \nis detailed in Chapter\\;\\ref{ch:scaling_methods}.\n\nNext, the cepstral coefficients are extracted from each frequency bin, \nby taking the energies sum of each filter and multiplying it\nwith the Discrete Cosine Transform (DCT).\n\\begin{align}\\label{eq:mfcc}\n MFCC[n] & = \\sqrt{\\frac{2}{K}} \\sum_{k=1}^{K} \\left\\{ \n e[k] \\cdot DCT\\left( k \\right)\n \\right\\} \\nonumber \\\\\n & = \\sqrt{\\frac{2}{K}} \\sum_{k=1}^{K} \\left\\{ \n e[k] \\cdot \\cos \\left( \\frac{\\pi n}{K} (k+0.5) \\right)\n \\right\\}\n\\end{align}\n\nIt is very common to extract the log-Mel energies\nand then have Equation\\;\\ref{eq:mfcc} written as:\n\\begin{align}\\label{eq:logmfcc}\n MFCC[n] & = \\sqrt{\\frac{2}{K}} \\sum_{k=1}^{K} \\left\\{ \n \\log \\left( e[k] \\right) \\cdot \\cos \\left( \\frac{\\pi n}{K} (k+0.5) \\right)\n \\right\\}\n\\end{align}\n\n% Take from PyAudioAnalysis\n% PyAudioProcessing\n\n\\subsection{RFCCs --- Root-Frequency Cepstral Coefficients}\nAn alternative to the log-Mel \nextraction of the cepstral coefficients\nhas been suggested in \\cite{rmfcc1} and \\cite{1415167}.\nThe motivation to put this proposed \ntechnique under test is that the root \nfunction can be less computationally demanding \nthan the traditional log function..\n\\begin{align}\n RCC[n] & = \\sqrt{\\frac{2}{N}} \\sum_{k=1}^{K} \\left\\{ \n \\left( e[k] \\right)^{\\gamma} \\cdot \\cos \\left( \\frac{\\pi n}{K} (k+0.5) \\right)\n \\right\\}\n\\end{align}\n\n\\subsection{GFCCs --- Gammatone-Frequency Cepstral Coefficients}\nGFCCs follow the basic steps same as the MFCCs extraction.\nBut, instead of translating the frequencies to Mels,\nthe spectrum is translated to the ERB scale.\nERB scale implies a Gammatone Filterbank as described in Chapter \\ref{ch:scaling_methods}.\n\n\n\\subsection{BFCCs --- Bark-Frequency Cepstral Coefficients}\nLike the Gammatone-FCCs (GFCCs), the BFCCs follow the Bark Scale.\n\n% \\subsection{LFCCs --- Linear-Frequency Cepstral Coefficients}\n% \\subsection{LPC --- Linear Predictive Coefficients}\n% \\subsection{MSRCC --- Magnitude-based Spectral Root Cepstral Coefficients}\n% \\subsection{NGCC --- Normalized Gammachirp Cepstral Coefficients}\n% \\subsection{PNCC --- Mel-Frequency Cepstral Coefficients}\n% \\subsection{PSRCC --- Mel-Frequency Cepstral Coefficients}\n% \\subsection{RPLP --- Mel-Frequency Cepstral Coefficients}\n% https://spafe.readthedocs.io/en/latest/features/_features.html\n% \\subsection{articulatory cepstral coefficients}\n\n\\section{Time-Domain Features}\n\\subsection{Dynamic FCC features}\nDuring the framing process, the speech signal \nis divided into small fragments of the speech over time. \nThese time unit fragments are called frames.\nEach frame spans over a finite time duration.\nThe extracted cepstral coefficients are computed statically\nfor a given frame. However, the original speech\nis framed in multiple numbers of frames. \nTherefore, coefficients extractions\nhave to be dynamic for the entire signal, i.e., all frames.\n\nAdditional information about the temporal \nchanges between adjacent coefficients can also be extracted\nto include the dynamics and transitions within a frame.\n\n\\subsubsection{Deltas}\nThe first derivate of the cepstral coefficients\n,\\(\\Delta\\) (Deltas),\nrepresents the velocity of the MFCCs' dynamics and \nis a sub-set of the cepstral coefficients.\n\nDeltas (\\(\\Delta\\)) are extracted by:\n\\begin{equation}\\label{eq:deltaderiv}\n \\Delta [n] = \\frac{ \\sum\\limits_{i=-T}^{T} k_{i} c_{m}[n+i]}\n {\\sum\\limits_{i=-T}^{T} |i|}\n\\end{equation}\n\nWhere \\(n\\) denotes the frame time index, \\(k\\) marks the \ncoefficient weight, \\(T\\) stands for the number of temporally\nadjacent frames used for the calculation, and \\(c_{m}\\) denotes the\n\\(m\\)th coefficient in the given frame \\(n\\).\n\n\\subsubsection{Delta-Deltas}\nDelta-Deltas (\\(\\Delta\\Delta\\)) is an extra layer of information\nrepresenting the acceleration at which the \nMFCCs' dynamics change within a given time frame.\nThe extraction of the Delta-Deltas follows the same principale\nas the first derivatives. Yet, instead of taking the\ncepstral coefficients as the input feature, we replace\n\\(c_{m}\\) in Equation\\;\\ref{eq:deltaderiv} \nwith the first-order derivatives \\(\\Delta[n]\\).\n\n\\subsection{Temporal Context}\nTemporal context is an attachment of raw unprocessed \nfeatures' data from adjacent\nframes together with the currently selected feature. \nWhether spectral, cepstral, or spatial features,\nthe concatenation of past and future \nfeatures can infer decisions based on a memory effect.\nFurthermore, different languages\nintroduce contextual constraints such as relative positions\nof adjectives or nouns to verbs, plurals, affiliations,\npossessions, and other lingual principles.\n\nUsually, the number of adjacent past and future frames \nto concatenate is based on the framing and DFT parameters.\nIt is plain to understand \none would not want to exaggerate and excessively use\nredundant frames that do not have any significant impact\non the accuracy of detection. \nThe downside of overly using temporal context frames is obvious, \norders of magnitude larger amounts of information to process.\n\n% \\section{Spatial Features}\n% \\subsection{VTLN - vocal track length norm}\n", "meta": {"hexsha": "2378ee031791f3213db6c4d2538aeb027b1ba4a5", "size": 21276, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Dissertation/Features/features.tex", "max_stars_repo_name": "aviadb/research", "max_stars_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Dissertation/Features/features.tex", "max_issues_repo_name": "aviadb/research", "max_issues_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Dissertation/Features/features.tex", "max_forks_repo_name": "aviadb/research", "max_forks_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.4829821718, "max_line_length": 92, "alphanum_fraction": 0.7417277684, "num_tokens": 5756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289387998695209, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.6465663786433863}} {"text": "\\documentclass[letterpaper]{article}\n\n\\usepackage{fullpage}\n\\usepackage{nopageno}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\n\\title{Notes}\n\\date{October 27, 2014}\n\\maketitle\nif $G$ is a group, and $A\\subseteq G$ and $B\\subseteq G$ then $AB=\\{ab|a\\in A, b\\in B\\}\\subseteq G$.\n\n\\section*{proposition}\n\nlet $G$ be a group, then $H,K$ subgroups of $G$. Assume that $h^{-1}kh\\in K$ for all $h\\in H$, $k\\in K$ then $HK$ is a subgroup of $G$ that contain s both $H$ and $K$, in fact, $HK$ is the smallest subgroup of $G$ that contains both $H$ and $K$. Assumption only important if we are not dealing with abelian groups.\n\n\\subsubsection*{proof}\n$a,b\\in HK$. Write $a=h_1k_1,b=h_2k_2$ with $h_i\\in H,k_i\\in K$ then $a\\cdot b=h_1k_1h_2k_2=h_1h_2(h_2^{-1}k_1h_2)k_2\\in HK$\n\n$a=hk, a^{-1}=(hk)^{-1}=k^{-1}h^{-1}=h^{-1}(hk^{-1}h^{-1})\\in HK$\n\n\\subsection*{examples}\n$S_3, H=\\{(1),(12)\\}, K=\\{(1),(123),(132)\\}, (12)(123)=(23)\\in HK, (12)(132)=(13)\\in HK$ so $HK=G$ and is therefore contained by G\n\n$(\\mathbb{Z},+)$, $H=a\\mathbb{Z}, k=b\\mathbb{Z}$, let $d=(a,b)$\n\nclaim: $a\\mathbb{Z}+b\\mathbb{Z}=d\\mathbb{Z}$. clearly $a\\mathbb{Z}\\subseteq d\\mathbb{Z}$, $b\\mathbb{Z}\\subseteq d\\mathbb{Z}$. \n\n$a\\mathbb{Z}+b\\mathbb{Z}$ is the smallest subgroup that contains both $a\\mathbb{Z}$ and $b\\mathbb{Z}$. so $a\\mathbb{Z}+b\\mathbb{Z}\\subseteq d\\mathbb{Z}$.\n\n$d=\\gcd(a,b)$ so we can write $d=ma+nb$. let $\\alpha\\in d\\mathbb{Z}$ and write $\\alpha=dt, t\\in \\mathbb{Z}$ then $\\alpha=dt=mat+nbt\\in a\\mathbb{Z}+b\\mathbb{Z}$. so $d\\mathbb{Z}\\subseteq a\\mathbb{Z}+b\\mathbb{Z}$ \n\\section*{thm subgroup gen by a subset}\n$G$ is a group, if $a\\in G$ $=\\{a^i|i\\in \\mathbb{Z}\\}$ is the smallest subgroupthat contains $a$.\n\n\\subsubsection*{proof}\nlet $S\\subseteq G$, let $=\\{\\underbrace{a_1a_2\\dots a_k}_{\\text{word}}|a_i\\in S\\text{ or }{a_i}^{-1}\\in S, k\\in \\mathbb{N}\\}$ then $$ is a subgroup, $=\\cap \\forall H$ where $S\\subseteq H\\subseteq G$, and $H$ is a subgroup of $G$, $$ is the smallest subgroup of $G$that contains $S$.\n\nso it is closed under multiplication, identity is in it, and the inverse of all words are in it.\n\nshow containment both ways, one is clear because we have words of length 1 that span $S$ and so $S$ is one of the elements of our $H$ intersection.\n\n\\section*{example}\n$a,b\\in G, S=\\{a,b\\}\\subseteq G, =\\{a_1a_2\\dots a_k|a_i\\in\\{a,a^{-1},b,b^{-1}\\}\\}$\n\nif $ab=ba$ then $=\\{a^{i}b^{j}|i\\in \\mathbb{Z}, j\\in \\mathbb{Z}\\}$\n\n\n\\section*{maps}\nstudied groups, subgroups. now we are going to talk about maps\n\nif we have groups $G_1,G_2$ and $\\varphi:G_1\\to G_2$ is a group homomorphism provided $x\\to\\varphi(x), y\\to\\varphi$ means that $\\varphi(x*y)=\\varphi(x)*\\varphi(y)$ for all $x,y\\in G_1$.\n\\section*{examples}\nidentity: $x\\to x$\n\n$(\\mathbb{R},+)=G_1, (\\mathbb{R}^+,\\cdot)=G_2$. $\\varphi(x)=e^x$. ie $\\varphi(x+y)=e^{x+y}=e^xe^y=\\varphi(x)\\varphi(y)$.\n\n\\section*{notation}\nlet $\\varphi:G_1\\to G_2$ be a group homomorphism, then $\\ker\\varphi=\\{x\\in G_1|\\varphi(x)=e\\}$\n\nhomomorphism always takes the identity in $G_1$ to $G_2$.\n\n$\\varphi(e_1)\\varphi(e_1)^{-1}=\\varphi(e_1e_1)\\varphi(e_1)^{-1}=\\varphi(e_1)\\varphi(e_1)\\varphi(e_1)^{-1}=e_2=\\varphi(e_1)$\n\nprove that $\\ker \\varphi$ is a subgroup\n\nnow we say that $\\varphi$ is an isomorphism if $\\varphi$ is a group homomorphism and $\\varphi$ is bijective.\n\nboth of the previous examples are isomorphisms.\n\nso from an algebraic point of view, there is no difference between addition on the reals and multiplication on the positive reals.\n\n\\section*{proposition}\nlet $\\varphi$ be an isomorphism. the following are true\n\\begin{enumerate}\n\\item\n$\\varphi^{-1}$ which is the map from $G_2$ to $G_1$ is also an isomorphism.\n\\item\nif $G_1$ is abelian, then $G_2$ is abelian.\n\\item\nif $G_1$ is cyclic then so is $G_2$\n\\item\nif $a\\in G_1$ then $\\text{ord}(a)=\\text{ord}(\\varphi(a))$\n\\end{enumerate}\n\n\\begin{enumerate}\n\\item\nneed to prove $\\varphi^{-1}(\\alpha\\beta)=\\varphi^{-1}(\\alpha)\\varphi^{-1}(\\beta)$ for all $\\beta\\in G_2$, but $\\varphi$ is injective so it is enough to prove that $\\varphi(\\varphi^{-1}(\\alpha\\beta))=\\varphi(\\varphi^{-1}(\\alpha)\\varphi^{-1}(\\beta)=\\varphi(\\varphi^{-1}(\\alpha))\\varphi(\\varphi^{-1}(\\beta))=\\alpha\\beta$\n\\item\nassume $G_1$ is abelian\n\\begin{align*}\n \\alpha\\beta=\\varphi(\\varphi^{-1}\\left(\\alpha\\right)\\varphi^{-1}(\\beta))\n\\end{align*}\n\\item\nhint: assume that $G_1=$ for some $a\\in G_1$ and then prove that $G_2=<\\varphi(a)$\n\\item\nno hint\n\\end{enumerate}\n\\section*{example}\n\\begin{align*}\n \\mathbb{Z}_4\\not\\equiv\\mathbb{Z}_2\\times\\mathbb{Z}_2\n\\end{align*}\nby contradiction, assume that there exists an isomorphism $\\varphi$ from z4 to z2+z2. $[1]\\in \\mathbb{Z}_4$ and $\\text{ord}[1]=4$ so then $\\text{ord}\\varphi([1])=4$. But all elements of $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ has no elements of order 4, so there is no isomorphisms. however, if $\\gcd(m,n)=1$ then $\\mathbb{Z}_{mn}\\equiv\\mathbb{Z}_m\\times\\mathbb{Z}_n$\n\n\\begin{align*}\n \\varphi:[x]_{mn}\\to[x]_m[x]_n\n\\end{align*}\n\\section*{what does well defined mean?}\nsame input gives same output, ie if $[x]=[y]$ then $\\varphi[x]=\\varphi[y]$\n\n\\section*{3.4 \\#13}\n$(\\mathbb{R}^*,\\cdot), C_2=\\{\\pm1\\}\\subseteq \\mathbb{R}^*$. $C_2$ is a subgroup of $\\mathbb{R}^*$. prove that $\\mathbb{R}^*\\cong \\mathbb{R}^+\\times C_2$\n\nwe construct an isomorphism $\\theta:\\mathbb{R}^*\\to R^{+}\\times C_2$.\n\n$x\\to(|x|,\\frac{x}{|x|})$. prove that $\\theta$ is a group homomorphism and bijective.\n\n$\\theta(xy)=(|xy|,\\frac{xy}{|xy|})=(|x|,\\frac{x}{|x|})(|y|,\\frac{y}{|y|})=\\theta(x)\\theta(y)$\n\nbijectivity is exercise, but a number is uniquely identified by sign and magnitude (absolute value)\n\n\\section*{example}\nprove that $\\text{ord}(aba^{-1})=\\text{ord}(b)$ for every $a,b\\in G$ where $G$ is a group.\n\n$m=\\text{ord}(x)$ means $x^m=e$ and $x^k=e$ means that $m|k$.\n\nif given $n=ord(x)$ and $m=ord(y)$ then best way is to show that $m=n$ is $m|n$ and $n|m$. this all works for finit.\n\nthis question is trivial if the group is abelian.\n\nso let $m=ord(aba^{-1}), n=ord(b)$\n\n\\subsection*{case 1}\n$n$ is finite, $b^n=e$. consider $(aba^{-1})^n=aba^{-1}aba^{-1}\\dots aba^{-1}=ab^{n}a^{-1}=aea^{-1}=e$ so $ord(aba^{-1})$ is finite, also $m|n$\n\n$b^m=a^{-1}\\underbrace{(aba^{-1})(aba^{-1})\\dots(aba^{-1})(aba^{-1})}_{m\\text{ times}}a=a(aba^{-1})^ma=a^{-1}ea=e$ so $b^m=e$ and $n|m$\n\\subsection*{case 2}\n$n$ is infinite. then we must prove that $m$ is infinite. by contradiction, assume $m<\\infty$.\n\nthen $b^m=a^{-1}\\underbrace{(aba^{-1})(aba^{-1})\\dots(aba^{-1})(aba^{-1})}_{m\\text{ times}}a=a(aba^{-1})^ma=a^{-1}ea=e$ so $b^m=e$.\n\nand so the order is finite and we have our contradiction. so $m$ must be finite then\n\\subsection*{consider}\n$ord(a^{-1})=ord(a)$ and $ord(ab)=ord(ba)$\n\nby previous part $ord(ba)=(abaa^{-1})=ord(ab)$\n\n\\section*{3.2 \\# 25 from class}\nwe note that if $x\\in G$ has the required order, then $x^{-1}\\in G$ also has the required order. note that the $x\\ne x^{-1}$ because the order is greater than $2$.\n\n\\section*{!3.3 \\#9}\n\n\\section*{other things in mind}\nlets take a group $G$ and $H,K$ subgroups. then $\\left\\lvert HK\\right\\rvert=\\frac{|H||K|}{|H\\cap K|}$\n\nrecall: $HK=\\{hk|h\\in H, k\\in K\\}$\n\n\\begin{align*}\n g\\in HK\\\\\n g=hk\\\\\n h\\in H\\quad |H|=m\\\\\n k\\in K\\quad |K|=n\\\\\n\\end{align*}\nquestion? how many ways can $g=hk=h'k'$?\n\n\\begin{align*}\n hk=h'k'\\to(h')^{-1}hk=k'\\\\\n (h')^{-1}h=k'k^{-1}\\in H\\cap K\n\\end{align*}\nso $k'=\\alpha k, h'=h\\alpha^{-1}$\n\nthat is to say $hk=(h'\\alpha)(\\alpha^{-1}k')$\nthere are $|H\\cap K|$ ways to choose alpha\n\n\\section*{october 15}\nread 3.6.2\n\n\\subsubsection*{3.6 \\#2}\nWrite out the addition tables for $\\mathbb{Z}_4$ and for $\\mathbb{Z}_2\\times \\mathbb{Z}_2$. Use cycle notation to write out the permutation determined by each row of each of the addition tables as in the discussion preceding Cayley's theorem.\n\n\\[\n \\begin{array}{c|ccccc}\n +&[0]&[1]&[2]&[3]&(1)\\\\\n \\hline\n [0] & [0] & [1] & [2] & [3]&(1)\\\\\n \n [1] & [1] & [2] & [3] & [0]&(1234)\\\\\n\n [2] & [2] & [3] & [0] & [1]&(13)(24)\\\\\n\n [3] & [3] & [0] & [1] & [2]&(1432)\\\\\n \\end{array}\n \\quad\\quad\n \\begin{array}{c|ccccc}\n + & ([0],[0]) & ([0],[1]) & ([1],[0]) & ([1],[1])&(1)\\\\\n \\hline\n ([0],[0]) & ([0],[0]) & ([0],[1]) & ([1],[0]) & ([1],[1])&(1)\\\\\n ([0],[1]) & ([0],[1]) & ([0],[0]) & ([1],[1]) & ([1],[0])&(12)(34)\\\\\n ([1],[0]) & ([1],[0]) & ([1],[1]) & ([0],[0]) & ([0],[1])&(13)(24)\\\\\n ([1],[1]) & ([1],[1]) & ([1],[0]) & ([0],[1]) & ([0],[0])&(14)(23)\\\\\n \\end{array}\n\\]\n\n\\section*{last time}\n$\\varphi:G\\to \\text{Sym}(G)$\n\\section*{rigid motions of a regular n-gon}\nplace the first vertex, then the second. we have n options for the first vertex, and 2 options for the second. and the rest are fixed. so we can do $2n$ rigid motions for any n-gon.\n\nobservation, rigid motions give us permutations, but for $n>3$ we can't get all the permutations. $2\\cdot 4<4!$.\n\nthe rigid motions form a group.\n\\subsection*{goal: describe this group}\n$a=$counterclockwise rotation by $\\frac{2\\pi}{n}$ radians or $\\frac{360}{n}^\\circ$. order of $a$ is $n$, that is to say, rotating $n$ times gives us our original vertex placement.\n\n$a^i=$rotation by $\\frac{2\\pi}{n}\\cdot i$.\n\n$b=$reflection about line $L$ this will leave one or two vertices unchanged, depending on the parity of $n$.\n\n$b=(2,n)(3,n-1)\\dots$.\n\n$b^2=e$.\n\nconsider $\\{b,ab,a^2b,\\dots,a^{n-1}b\\}$. remember $a,b$ are functions so $ab=a\\circ b$\n\n$a^ib=a^j\\Rightarrow a^i=a^j\\Rightarrow a^{j-i}=e$ but $a^{j-1}=e$ iff $n|j-i$.\n\n$a^ib=$the flip about the line $L_i$ that makes $\\frac{\\pi i}{n}$ with $L$, this means that $\\text{ord}(a^ib)=2$ because it's just a flip.\n\nnow we have $n$ rotations and $n$ flips. note that no flip can be a rotation because a flip changes orientation.\n\n$D_n$ denotes $\\{e,a,a^2,\\dots,a^{n-1},b,ba,ba^2,\\dots,ba^{n-1}\\}$ which is the dihedral group with 2n elements. $a^n=e$ and $b^2=e$.\n\nnotice that the order of the group reflects the symmetry of the geometric representation. a completely assymetric object with have only the $e$ rigid motions, while a circle will have infinitely many.\n\nwhat is $ba$? it's $a^{-1}b=a^{n-1}b$\n\n\\section*{october 20}\nassnmnt: 3.6 \\#21,23,25\nif odd, has identity, if even, identity and half rotation ($r^{n/2}$)\n$a^jb=ba^j\\to a^jb=a^{-j}b\\to a^j=a^{-j}\\to a^{2k}=e\\to n|2j$\n\n\\subsection*{last time started homomorphisms}\n\\subsubsection*{definition}\n$\\varphi:G_1\\to G_2$ is group homomorphism then $\\text{Im}\\varphi=\\varphi(G_1)$ is a subgroup of $G_2$ and $\\text{Ker}(\\varphi)=\\{x\\in G_1:\\varphi(x)=e_2\\}=\\varphi^{-1}(\\{e_2\\})$ beause $\\{e_2\\}\\le G_2$ is a normal subgroup, $\\text{Ker}(\\varphi)\\le G_1$ is a normal subgroup\n\n\\subsubsection*{equivalence relation defines a group homomorphism}\non $G_1/\\sim$ define the operation $[x][y]=[xy]$. is this well defined?\n\n$[x]=[x']$ and $[y]=[y']$. $\\varphi(x)=\\varphi(x')$, and $\\varphi(y)=\\varphi(y')$ and so $\\varphi(x)\\varphi(y)=\\varphi(x')\\varphi(y')$ and homomorphism definition gives us $\\varphi(xy)=\\varphi(x'y')$.\n\nclaim , with this operation $G_1/\\sim$ is a group. $[e_1]$ is identity. $[x]$ is inverse of $[x^-1]$.\n\nlet $\\pi(x)=[x]$. claim $\\pi$ is a group homomorphism.\n\n$\\pi(xy)=[xy]=[x][y]=\\pi(x)\\pi(y)$.\n\n\\subsubsection*{why does $[xy]=[x][y]$}\n\\subsubsection*{thrm}\nLet $\\varphi:G_1\\to G_2$ be a group homomorphism. then $G_1/\\sim\\cong\\varphi(G_1)$\n\nproof:\n\n$G_1\\sim\\to\\varphi(G_1)$\n\n$[x]\\to\\varphi(x)$\n\n$\\overline\\varphi([x])=\\varphi(x)$\nclaim: $\\overline\\varphi$ is a group homomorphism\n$\\overline([x][y])=\\overline\\varphi([xy])=\\varphi(xy)=\\varphi(x)\\varphi(y)=\\overline\\varphi([x])\\overline([y])$\n\nclaim: surjectivity is clear\nclaim: $\\overline\\varphi$ is injective.\n$\\overline\\varphi([x])=\\overline{\\varphi}([x])\\to \\varphi(y)=\\varphi(x)\\to x\\sim y\\to [x]=[y]$\n\n\\subsubsection*{assignment for next time}\n$\\varphi:\\mathbb{Z}_m\\to\\mathbb{Z}_n$ where $\\varphi$ is group homomorphism. hint: $\\varphi(0)=0$ and $\\varphi([1])=[k]$ then $\\varphi([j])=\\varphi([jk])$. take $Z_2\\to\\mathbb{Z}_4$ then $\\varphi([1])\\ne[1]$\n\n\\section*{october 22}\n\\subsection*{discussion assnmnt}\n\\begin{enumerate}\n\\setcounter{enumi}{13}\n\\item\ncall center $C$ with $C=\\{c\\in G:cg=gc\\forall g\\in G\\}$. then $gcg^{-1}=cgg^{-1}=c\\in C$\n\\item\nintesection being a group is pretty straightforward, if a is in both then $a^{-1}$ is in both. if a and b are in both, then $ab$ is in both, and $e$ is in both\n\\setcounter{enumi}{17}\n\\item\n$\\langle a\\rangle$ has two subgroups, $\\{e\\}$ and $\\langle a^k\\rangle$ where $k|n$. of course $e$ is normal. cyclic groups are abelian, so\n\\begin{align*}\n b^ja^m(a^k)^ia^{-m}b^j=\n\\end{align*}\n\\end{enumerate}\n\\section*{3.8}\nif $G$ is a group and $H$ is a subgroup\n\n\\subsubsection*{parallel 1}\non $G$ define the relation $x\\sim_ry\\Leftrightarrow xy^{-1}\\in H$\n\n$\\Leftrightarrow x\\in Hy$\n\nclaim: $\\sim_r$ is an equivalence relation\n\\begin{enumerate}\n\\item\n$x\\sim_r x$\n\\item\n$x\\sim y\\to xy^{-1}\\in H\\to (xy^{-1})^{-1}\\in H\\to yx^{-1}\\in H\\to y\\sim x$\n\\item\ntransitivity\n\\end{enumerate}\nso $[x]=Hx$\n\\subsubsection*{observations}\n$Hx=Hy\\to x\\sim_r y$\n\nall the right cosets have the same number of elements\n$\\theta:H\\to xH$ because any $x$ has an image then $\\theta$ is surjective, and $\\theta(h_1)=\\theta(h_2)\\to xh_1=xh_2\\to h_1=h_2$ and so it is bijective\n\\subsubsection*{parallel 2}\non $G$ define the relation $x\\sim_ly\\Leftrightarrow x^{-1}y\\in H$\n\n$\\Leftrightarrow y\\in xH$\n\nclaim $\\sim_l$ is an equivalence relation\n\\begin{enumerate}\n\\item\n\n\\end{enumerate}\nso $[x]=xH$\n\nthere exists a bijection from the set of left cosets to the set of right cosets.\n\ndefine $\\psi(xH)=Hx^{-1}$ (not $Hx$ because this is not well defined.\n\n$xH=yH\\to x\\sim_l\\to xy^{-1}\\in H\\to yx^{-1}\\in H\\to y^{-1}(x^{-1})^{-1}\\in H\\to x^{-1}\\sim_r y^{-1}$\n\nclaim injectivity:$\\psi(xH)=\\psi(yH)\\to Hx^{-1}=Hy^{-1}\\to x^{-1}\\sim_ry^{-1}\\to x^{-1}(y^{-1})^{-1}\\in H\\to x^{-1}y\\in H\\to x\\sim_l y\\to xH=yH$\n\nnotation: $[G:H]$ is the index of $H$ in $G$ and means the number of left/right cosets of $H$ in G.\n\ncorollary: if $G$ is finite and $H$ is a subgroup then the number of elements in $G$ is equal to the number of elements in $H$ times the index. $|G|=|H|\\cdot[G:H]$\n\nproof: $|G|$ equals the number of elements in each $[G:H]$ equal partitions of $G$ times the number of partitions.\n\ncorollary, if $G$ is finite then $H\\subseteq G\\to |H|$ divides $G$\n\n\\subsubsection*{example}\n$G=S_3\\cong D_3$\n\n$H=\\{e,b\\}\\le D_3$\n\nleft cosets are: $eH=H=bH, aH=\\{a,ab\\}, a^2H=\\{a^2,a^2b\\}$ and $[G:H]=3$\n$N=\\{e,a,a^{2}\\}\\to [G:N]=2$\n\\section*{oct 27}\n\\subsection*{derniere fois}\n$G$ group, $N$ is a normal subgroup of $G$. then $G/N=\\{aN:a\\in G\\}$ is a broup with $aN\\cdot bN=(ab)N$\n\n$\\pi:G\\to G/N:a\\to aN$\n\n$\\pi$ is a surjective group hom with $\\text{Ker} \\pi=N$.\n\n\\subsection*{thrm}\n$\\varphi:G_1\\to G_2$ group homomorphism then $G/\\text{Ker}\\varphi\\cong\\varphi(G_1)$\n\\subsection*{proposition}\nthere exists a one to one correspondence between $\\{\\text{subgroups of }G/N\\}$ and $\\{\\text{subgroups of $G$ that contain $N$}\\}$.\n\n$\\alpha:\\{aN:a\\in H\\}=\\pi(H)\\leftarrow H\\le G, H\\supseteq N$\n\n$\\beta:K\\to\\pi^{-1}(K)$\n\nmoreover normal subgroups correspond to normal subgroups\n\n\\subsubsection*{proof}\nnote $\\pi^{-1}(K)\\supseteq\\pi^{-1}(\\{e\\})=\\text{Ker}\\pi=N$. moreover if $K$ is a normal subgroup of $G/N$ then $\\pi^{-1}(K)$ is a normal subgroup of $G$.\n\nsimilarly if $H\\le G$ and $H\\supseteq N$ then $\\pi(H)$ is a subgroup of $G/N$. moreover if $H$ is a normal subgroup of $G$ then $\\pi(H)$ is a normal subgroup of $G/N$ (why?)\n\n\\subsubsection*{claim}\n$\\alpha\\circ\\beta(K)=\\alpha(\\beta(K))=\\alpha(\\pi^{-1}(K))=\\pi(\\pi^{-1}(K))=\\{\\pi(y):y\\in \\pi^{-1}(K)\\}=\\{\\pi(y):\\pi(y)\\in K\\}=K\\cap I_m\\pi=K\\cap\\pi(G)=K$ because $\\pi$ is surjective\n\n$\\beta\\circ\\alpha(H)=\\beta(\\alpha(H))=\\beta(\\pi(H))=\\pi^{-1}(\\pi(H))\\supseteq H$\n\n$h\\in H\\to \\pi(h)\\in$\n\n$\\pi(x)\\in \\pi(H)\\to \\pi(x)=\\pi(h)$ for some $h\\in H$ so $xN=hN$ so $xh^{-1}\\in N\\to xh^{-1}\\in H\\to x\\in H$\n\nso $\\alpha$ and $\\beta$ are bijections and are inverse.\n\n\\subsection*{example}\n$G=\\mathbb{Z}, N=n\\mathbb{Z}$ and so $G/N=\\mathbb{Z}/n\\mathbb{Z}=\\{a+n\\mathbb{Z}:a\\in \\mathbb{Z}\\}$\n\n\\subsection*{definition}\ngroup $G$ is simple if it's only subgroups are itself and $\\{e\\}$. \n\n\\section*{example}\nlet $p$ be a prime then $\\mathbb{Z}_p$ is simple. because nothing divides $p$ we can't have any subgroups of $\\mathbb{Z}_p$ other than itself and the identity.\n\\end{document}\n\n\n", "meta": {"hexsha": "2fcfab7bc2d96b2c4d858779a871aebf0eee97eb", "size": 16443, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "abstract algebra/abstract-notes-2014-10-27.tex", "max_stars_repo_name": "ylixir/school", "max_stars_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "abstract algebra/abstract-notes-2014-10-27.tex", "max_issues_repo_name": "ylixir/school", "max_issues_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "abstract algebra/abstract-notes-2014-10-27.tex", "max_forks_repo_name": "ylixir/school", "max_forks_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.1075, "max_line_length": 360, "alphanum_fraction": 0.6363802226, "num_tokens": 6457, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799928900257126, "lm_q2_score": 0.8289388062084421, "lm_q1q2_score": 0.6465663751089868}} {"text": "\\section{Commutative algebra}\\label{sec:commutative_algebra}\n\nCommutative algebra generalizes desirable properties of \\hyperref[def:set_of_integers]{set \\( \\BbbZ \\) of integers} to nontrivial \\hyperref[def:semiring/commutative_unital_ring]{commutative unital rings}. As a result, this allows studying polynomials or even arbitrary functions via techniques similar to those for integers.\n\nIn this whole section, \\( R \\) will refer to a nontrivial commutative unital ring. Since \\( R \\) is commutative, left and right modules over \\( R \\) are equivalent. We will only refer to either of them as simply \\enquote{modules}.\n", "meta": {"hexsha": "35dca3e9245b542d633dc64da9c49e06ea83dea8", "size": 619, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/commutative_algebra.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/commutative_algebra.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/commutative_algebra.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 103.1666666667, "max_line_length": 324, "alphanum_fraction": 0.7915993538, "num_tokens": 157, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.793105951184112, "lm_q2_score": 0.8152324960856175, "lm_q1q2_score": 0.6465657442441816}} {"text": "\\begin{quote}\\textit{\n\t``I didn't give it a good notation because I didn't like it.''\n}\\end{quote}\nRecall that last time, we defined sheafification, which can be thought of projecting presheaves onto sheaves in a\nparticularly nice way. This allows us to forget the difference between sheaves and presheaves, so to speak; we'll\nuse this to understand colimits of sheaves.\n\\begin{exm}% would be nice to say what x, i_* are...\nFirst, a quick digression, since we got confused last time, on the espace étalé of a skyscraper sheaf. Directly\nfrom the sheaf axioms, one can show that if \\(\\sF\\) is a \\(\\fC\\)-valued sheaf, then \\(\\sF(\\emptyset)\\) is the\nterminal object (a point for \\(\\Set\\), \\(0\\) for \\(\\Ab\\), and so on). This follows from abstract nonsense: the\nempty product \\(\\prod_\\emptyset S\\) is necessarily the terminal object (there's more to think through here). This\nis what motivates the definition of the skyscraper sheaf \\(i_*S = i_{x,*}S\\) in Example~\\ref{skys}. For\nsimplicity, assume \\(x\\in X\\) is a closed point.\n\nNow, let's construct its espace étalé \\(\\pi:Y_{i_*S}\\to X\\); for any \\(y\\in X\\), \\(\\pi^{-1}(y)\\) is the stalk of\n\\((i_*S)\\) at \\(y\\), which is \\(S\\) if \\(y = x\\) or the terminal object \\(*\\) otherwise. Thus, \\(Y_{i_*S}\\) is as a\nset a copy of \\(X\\), but with \\(S\\) over the basepoint \\(x\\) instead of a single point; then, we glue each of these\npoints of \\(S\\) to the rest of \\(Y_{i_*S}\\) as if they were all \\(x\\). The result is \\(X\\) with multiple\nbasepoints, so to speak, and is not at all Hausdorff. However, as topological spaces, we have a pullback diagram\n\\[\\xymatrix{\n\t(U\\setminus\\set x)\\times S\\ar[r]\\ar[d] & U\\times S\\ar[d]\\\\\n\tU\\setminus\\set x\\ar[r] & Y_{i_*S}.\n}\\]\n\\end{exm}\nWe can also use the espace étalé to define sheafification: the sheafification \\(\\sF\\sh\\) is just the sheaf of\nsections of \\(Y_\\sF\\).\n\\subsection*{Kernels and Cokernels.}\nBefore discussing limits and colimits more generally, let's focus on kernels and cokernels. Let \\(\\vp:\\sF\\to\\sG\\)\nbe a morphism of sheaves of abelian groups on a space \\(X\\), and for every open \\(U\\subset X\\), define\n\\((\\ker\\vp)(U) = \\ker(\\vp|_U)\\). It's easy to check that this is a presheaf, and a little more work to check that\nit's a sheaf, too. And this is actually the kernel in \\(\\Ab_X\\), in that it satisfies the universal property:\nit fits into the diagram\n\\[\\xymatrix{\n\t\\ker\\vp\\ar[r]\\ar[d] & \\sF\\ar[d]^\\vp\\\\\n\t0\\ar[r] & \\sG,\n}\\]\nand any other sheaf \\(\\mathscr H\\) that fits into the same place in the above diagram has a unique map to\n\\(\\ker\\vp\\).\n\nLikewise, a morphism in \\(\\Ab_X\\) is injective (meaning a monomorphism) exactly when\n\\(\\vp|_U:\\sF(U)\\to\\sG(U)\\) is injective for all open \\(U\\subset X\\).\n\nCokernels are a little more interesting. The sheaf assigning \\(U\\mapsto\\coker(\\vp|_U)\\) is a presheaf, and is the\ncokernel in the category of presheaves, but it is \\emph{not} the cokernel in the category of sheaves; it fails to\nsatisfy the universal property. This is where some of the interesting nuances of sheaf theory pop up.\n\\begin{exm}\n\\label{notcoker}\nWe'll let \\(X = \\C\\), and let \\(\\sO\\) be the sheaf of holomorphic functions and \\(\\sO^*\\) be the sheaf of\n``invertible,'' i.e.\\ nonvanishing, holomorphic functions (an abelian group under multiplication). The exponential\nmap \\(f(z)\\mapsto e^{f(z)}\\) sends holomorphic functions to nonvanishing holomorphic functions, and commutes with\nrestriction, so it's a morphism \\(\\exp:\\sO\\to\\sO^*\\) in \\(\\Ab_\\C\\).\n\nIf a function maps to \\(1\\) in \\(\\sO^*\\), then it must be an integer multiple of \\(2\\pi i\\), so it must be locally\nconstant, Thus, it's constant on each connected component of the given open set. Thus, \\(\\ker(\\exp) = 2\\pi\ni\\underline\\Z\\): the constant sheaf, not the constant presheaf. This agrees with what we just learned about\nkernels.\n\nThen, \\(\\Im(\\exp)(U)\\) is the \\(f^*\\in\\sO^*(U)\\) such that \\(f = e^{2\\pi i g}\\) for some \\(g\\in\\sO(U)\\). That is,\n\\(\\log f\\) must have a well-defined branch on \\(U\\). In particular, if \\(U = \\C^*\\) and \\(f = z\\), then\n\\(f\\not\\in\\Im(\\exp(U)\\). This is a problem: \\(\\C^*\\) can be covered by simply connected open sets on which the\nlogarithm exists, but the gluing axiom fails.\n\nHowever, since \\(\\exp:\\sO\\to\\sO^*\\) is surjective on simply connected open sets, then it's surjective on the level\nof stalks, even though it's not surjective as a map of sheaves. In other words, we want the sequence\n\\[\\shortexact[j][\\exp]{\\underline\\Z}{\\sO}{\\sO^*}{}\\]\n%\\[\\xymatrix{\n%\t0\\ar[r] &\\underline\\Z\\ar[r]^j &\\sO\\ar[r]^\\exp & \\sO^{*}\\ar@{-->}[r] & 0\n%}\\]\nto be a short exact sequence of sheaves, but if we naïvely define the cokernel like the kernel, it isn't. This\nmeans that \\emph{to define the sheaf cokernel, we sheafify the presheaf cokernel}. In this case, the sheafification\nof the presheaf cokernel \\(\\coker(j)\\) stitches together the stalks, but on stalks \\(\\exp\\) is surjective, so since\na sheaf is completely determined by stalks, this is just \\(\\sO^*\\) again, which jives with the idea of\nsurjectivity. In the same way, we get that \\(\\coker(\\exp) = 0\\), as one would expect.\n\\end{exm}\nIn other words, a surjective map of sheaves (categorically, an epimorphism), is surjectivity on stalks, but\n\\emph{not} surjectivity on all open subsets. Injectivity is equivalent to injectivity on stalks and on open\nsubsets, though.\n\nSince sheafification preserves colimits, this can be generalized: the colimit of a diagram of sheaves is the\nsheafification of the presheaf colimit (which is just the colimit on every open set).\n\\begin{exm}\nThis next example is in some sense the same example. Let \\(X\\) be a smooth manifold, \\(\\sF\\) be the sheaf of\nsmooth maps to \\(S^1\\), \\(C^\\infty\\) be the smooth maps to \\(\\R\\) (so just the smooth functions), and\n\\(\\underline\\Z\\) is the constant sheaf (which is also smooth maps to \\(\\Z\\), since \\(\\Z\\) is discrete); each of\nthese is a sheaf of abelian groups.\n\nWe'd like to understand that \\(S^1 = \\R/\\Z\\). This comes from the sequence\n\\[\\xymatrix{\n\t0\\ar[r] & \\underline\\Z\\ar[r] & C^\\infty\\ar[r] & \\sF\\ar[r] & 0,\n}\\]\nwhich is short exact. The injectivity of \\(\\underline\\Z\\inj C^\\infty\\) comes from the fact that every map to \\(\\Z\\)\ncan be lifted to a smooth map to \\(\\R\\), and surjectivity comes from the fact that germs of functions can be\nlifted on a small neighborhood, so it's surjective on stalks. However, there are open subsets where functions can't\nbe lifted: if \\(X = S^1\\), then the identity map \\(S^1\\to S^1\\) can't be lifted to a map to \\(\\R\\). Thus, this is\nsurjective, even though it's not so on the level of open sets.\n\\end{exm}\n\\begin{exm}\nOur next example will be the de Rham complex. Let \\(X\\) be a smooth manifold. Let \\(\\underline\\R\\) denote the\nconstant sheaf on \\(\\R\\) (locally constant functions) and \\(\\Omega^1\\) denote the sheaf of one-forms on \\(X\\). The\nexterior derivative gives us an exact sequence\n\\[\\xymatrix{\n\t0\\ar[r] & \\underline\\R\\ar[r] & C^\\infty\\ar[r]^{\\d} & \\Omega^1\\ar[r]^{\\d} & \\Omega^2\\ar[r]^{\\d} &\\dots\n}\\]\nHowever, this is not in general short exact; if \\(\\Omega_{\\text{cl}}^1\\) denotes the space of closed one-forms,\nthen the Poincaré lemma just states that the following sequence is short exact.\n\\[\\xymatrix{\n\t0\\ar[r] & \\underline\\R\\ar[r] & C^\\infty\\ar[r]^{\\d} & \\Omega_{\\text{cl}}^1\\ar[r] & 0\n}\\]\nIn other words, even considering something very simple about short exact sequences of sheaves gives us cohomology.\nThis can be used to define sheaf cohomology, though we won't return to that anytime soon.\n\nIn fact, Example~\\ref{notcoker} is a special case of this, since \\(\\d z/z\\in\\Omega^1(\\C^*)\\) is a closed form\nthat's not exact.\n\\end{exm}\n\\subsection*{Ringed Space.}\nAnyways, we were going to talk about schemes, right? These are not just topological spaces, but ringed spaces:\ntopological spaces with a notion of a ring of functions.\n\\begin{defn}\nA \\term{ringed space} is the data \\((X,\\sO_X)\\), where \\(X\\) is a topological space and \\(\\sO_X\\) is a sheaf of\nrings on \\(X\\).\n\\end{defn}\nThe motivating examples are a topological space with continuous functions to \\(\\R\\) (since these form a ring), or a\nsmooth manifold with the sheaf \\(C^\\infty\\), or an analytic manifold with \\(C^\\omega\\) (analytic functions). Thus,\nthere are definitely different notions of ``function'' on a manifold, but the ringed space structure means knowing\nwhat kinds of functions (geometric structure) is.\n\nWe'd also like to know how to evaluate functions on a ringed space. For an arbitrary \\(x\\in U\\) and\n\\(f\\in\\sO_X(U)\\), it's not clear how to define \\(f(x)\\); we have stalks, but then what? In each of our examples\n(continuous functions, smooth functions, analytic functions, holomorphic functions, etc.), the stalks \\(\\sO_{X,x}\\)\naren't just rings, but local rings,\\footnote{Recall that a \\term{local ring} is a ring with a unique maximal\nideal.} with the maximal ideal \\(\\m_x\\) of functions which vanish at \\(x\\). \\(\\m_x\\) is unique, because if\n\\(f\\in\\sO_{X,x}\\setminus\\m_x\\), then \\(f(x)\\ne 0\\), so it's nonzero on a neighborhood of \\(x\\), and therefore\ninvertible in that subset! Thus, \\(f\\in\\sO_{X,x}^\\times\\), so \\(\\m_x\\) must be unique.\n\nThe point is, evaluating at \\(x\\) is exactly quotienting by \\(\\m_x\\), producing an element of \\(\\R\\). The sheaves\nwe care about have local rings for stalks, which is what makes this evaluation work. We'll turn this into a\ndefinition of something much more useful than a ringed space.\n\\begin{defn}\nA \\term{locally ringed space} is a ringed space \\((X,\\sO_X)\\) such that for every \\(x\\in X\\), the stalk\n\\(\\sO_{X,x}\\) is a local ring.\n\\end{defn}\nThus, all of our basic examples are locally ringed spaces, and in general, given an \\(f\\in\\sO_X(U)\\), we can define\n\\(V(f) = \\set{x\\in U: f(x) = 0}\\), and this will end up being a closed set.\n\nSchemes are particular examples of locally ringed spaces. We'll have to define how to produce a sheaf of functions,\nwhich we'll probably do next time, but we're almost there. One major takeaway is that schemes behave somewhat like\nthese examples we already have.\n\nWe also need to define morphisms. An isomorphism is evident: \\((X,\\sO_X)\\cong (Y,\\sO_Y)\\) is the data of a\nhomeomorphism \\(f:X\\to Y\\) that identifies the sheaves, i.e.\\ for all open \\(U\\subset Y\\), there's an isomorphism\n\\(f_*: \\sO_Y(U)\\to \\sO_X(f^{-1}(U))\\).\n\nIt's less obvious how to define morphisms in general; clearly, we need a continuous \\(f:X\\to Y\\), and we want to\ncompare \\(\\sO_X\\) and \\(\\sO_Y\\). Functions pull back (because the preimage of an open set is open); in the examples\nwe had before, we checked that the pullbacks of continuous (smooth, etc.) functions were continuous (smooth, etc.).\nMore generally, given an open \\(U\\subset Y\\), we have the two rings \\(\\sO_Y(U)\\) and \\(\\sO_X(f^{-1}(U))\\), and we\nwant the pullback of functions \\(f_*:\\sO_Y(U)\\to\\sO_X(f^{-1}(U))\\) to be a ring homomorphism. This is exactly how\nwe defined the pushforward of a sheaf.\n\\begin{defn}\nA \\term{morphism of ringed spaces} is a pair \\((f,f^\\sharp): (X,\\sO_X)\\to (Y,\\sO_Y)\\) in which\n\\begin{itemize}\n\t\\item \\(f:X\\to Y\\) is continuous, and\n\t\\item \\(f^\\sharp: \\sO_Y\\to f_*\\sO_X\\) is a morphism in \\(\\Ring_Y\\).\n\\end{itemize}\n\\end{defn}\nThat is: for every open subset, we can pull functions back into that open subset. But we can say that more\nconcisely with the sheaf theory we have developed.\n\nIt's worth remembering that nilpotents on affine schemes give us functions that aren't determined by their values\n(well, we do have to set up the structure of a locally ringed space first, but we'll get there), so a function\nisn't quite a bunch of values at points; it's something that we care to pull back.\n\nThis is cool, but we care about ringed spaces. What about these maximal ideals? They tell us what it means for a\nfunction to vanish. Back in the world of smooth functions, if \\(\\vp(y) = 0\\) and \\(x\\in f^{-1}(y)\\), then\n\\((f^*\\vp)(x) = \\vp(f(x))\\) had better be \\(0\\) too. This is not preserved by morphisms of ringed spaces (since\nevaluation isn't defined for germs of functions on ringed spaces), so we need an additional axiom.\n\nIf \\((f,f^\\sharp)\\) is a morphism of ringed spaces, passing to colimits induces a map \\(\\sO_{Y,y}\\to\\sO_{X,x}\\),\nwhenever \\(f(x) = y\\) (this is generally true for a map of sheaves, thanks to the property of colimits). Then, we\nwant this map to send \\(\\m_y\\to\\m_x\\). \n\\begin{defn}\nA morphism \\((f,f^\\sharp):(X,\\sO_X)\\to(Y,\\sO_Y)\\) is a morphism of locally ringed spaces if for every \\(x\\in X\\),\n\\(y\\in Y\\) such that \\(f(x) = y\\), the induced map \\(f^\\sharp_x: \\sO_{Y,y}\\to\\sO_{X,x}\\) maps \\(\\m_y\\) into\n\\(\\m_x\\).\n\\end{defn}\nThis is actually all the data that we'll need to define schemes: schemes are a full subcategory of locally ringed\nspaces; specifically, they are the ones that are locally isomorphic to \\((\\Spec R, \\sO_{\\Spec R})\\) (as soon as we\ndefine the locally ringed space structure on \\(\\Spec R\\)), i.e.\\ there are actual isomorphisms on an open cover.\n\nDoes this look weird? It's actually not unfamiliar: a smooth manifold is a locally ringed space that's locally\nisomorphic to \\((\\R^n,C^\\infty)\\). This encodes a lot of information; in particular, a continuous map of manifolds\nis smooth iff it pulls smooth functions back to smooth functions. In the same way, a topological manifold is a\nlocally ringed space locally isomorphic to \\((\\R^n,C)\\) (the sheaf of continuous functions). All the structure of\nan atlas is encapsulated in this notion of locally ringed spaces.\n\nThis notion is extremely general. For example, we can define a complex analytic manifold to be a locally ringed\nspace locally isomorphic to \\((U\\subset\\C^n, \\operatorname{Hol})\\) (since small discs in \\(\\C^n\\) aren't\nnecessarily biholomorphic to all of \\(\\C^n\\)). In all of the cases we've seen, though, \\(\\sO_X(U)\\) is always a\nsubset of set maps \\(U\\to\\R\\) (or \\(\\C\\)), and in particular functions are determined by their values. This is\nsomething that will not be true for schemes.\n\nNext time, we will define \\(\\Spec R\\), as a scheme.\n", "meta": {"hexsha": "dd1cf83c3f0689b16451ee842621276d70091c40", "size": 13943, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture_notes/examples/minnotes_d_examples/colimits_of_sheaves.tex", "max_stars_repo_name": "adebray/pset_d", "max_stars_repo_head_hexsha": "1d39e36341aab9778bd3e106f2e93c20c177932c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lecture_notes/examples/minnotes_d_examples/colimits_of_sheaves.tex", "max_issues_repo_name": "adebray/pset_d", "max_issues_repo_head_hexsha": "1d39e36341aab9778bd3e106f2e93c20c177932c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture_notes/examples/minnotes_d_examples/colimits_of_sheaves.tex", "max_forks_repo_name": "adebray/pset_d", "max_forks_repo_head_hexsha": "1d39e36341aab9778bd3e106f2e93c20c177932c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 68.684729064, "max_line_length": 115, "alphanum_fraction": 0.7083841354, "num_tokens": 4303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324938410784, "lm_q2_score": 0.7931059487389968, "lm_q1q2_score": 0.6465657404706868}} {"text": "\\documentclass[a4paper,10pt]{book}\n\n\\usepackage{textcomp}\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage[top=0.5in, bottom=0.5in, left=0.5in, right=0.5in]{geometry}\n\\usepackage{setspace}\n\n\\newtheorem{theorem}{Theorem}[chapter]\n\\newtheorem{lemma}[theorem]{Lemma}\n\n\\theoremstyle{definition}\n\\newtheorem{definition}[theorem]{Definition}\n\\newtheorem{example}[theorem]{Example}\n\\newtheorem{xca}[theorem]{Exercise}\n\\newtheorem{con}[theorem]{}\n\n\\theoremstyle{remark}\n\\newtheorem{remark}[theorem]{Remark}\n\n\\numberwithin{section}{chapter}\n\\numberwithin{equation}{chapter}\n\n\\begin{document}\n \n \\chapter{Metric Graphs}\n \\section{Graphs: Combinatorical Structure}\n \n \\begin{definition}\n A graph \\emph{$\\Gamma$} consists of\n \n \\begin{itemize}\n \\item A pair of sets $(V,\\ E)$ (called the `Vertex' and `Edge' sets respectively)\n \\item An involution $\\: \\phi \\colon E \\rightarrow E$\n satisfying $\\phi (e) = \\bar{e} $ and $\\phi\\circ\\phi = \\mathbf{1}_E$ (identity function)\n \\item Functions $i,\\ \\tau: E \\rightarrow V$ such that $i(\\bar{e}) = \\tau (e) $ and $\\tau(\\bar{e}) = i(e)$\n \\end{itemize}\n\n \\end{definition}\n\n Insert EXAMPLE 'ERE\n \n \\section{A Graph as a Space}\n\n The following construction associates to a graph $\\Gamma$, a space $| \\Gamma |$ \n \\vspace{0.1in} \\\\ Let $ \\xi = E \\times [0,1] = \\{ (e,t) : e \\in E,\\ t \\in [0,1]\\} $.\\\\\n Let $ \\widetilde{\\Gamma} = \\xi \\cup V $ \\\\\n Let `\\texttildelow' be the equivalence relation on $ \\widetilde{\\Gamma} $ generated by: \n \\begin{itemize}\n \\item If $ (e,t) \\in \\xi $, then $ (e,t) \\sim (\\bar{e},1-t) $\n \\item $ \\forall \\ e \\in E $, $ i(e) \\sim (e,0) $ and $ \\tau (e) \\sim (e,1) $\n \\end{itemize}\n $ |\\Gamma| $ is the quotient space $ \\Gamma / \\sim $\n \\section{ Metric structure on a Graph}\n \n \\subsection{Distance on the Vertex set V}\n \\begin{definition}\n Two vertices $ v_{1}$ and $v_{2} $ are said to be \\emph{adjacent} if there exists $ e \\in E $ such that $ i(e) = v_1 $ and $ \\tau (e) = v_2 $.\n \\end{definition}\n Let $ \\cal{D} = \\{ $ d $: V \\times V \\rightarrow \\mathbf{R} \\ : \\ $ \n d is a metric; d($ v_1,\\ v_2 $) $ \\leq $ 1 if $ v_1 $, $ v_2 $ are adjacent $ \\} $\n \\begin{lemma}\n $ \\cal{D} $ is not empty\n \\end{lemma}\n \\begin{definition}\n The \\emph{Graph Metric} on V is defined as\n \\begin{equation}\n\t d_{max}(v_1,v_2)= \\sup_{d \\in \\cal{D} }d(v_1,v_2) \\nonumber\n \\end{equation} \n \\end{definition}\n \\begin{lemma}\n\t$ d_{max} $ is a metric.\n \\proof \n \\begin{enumerate}\n \\item \\[\\forall \\ x \\in V,\\ d_{max}(x, x)= \\sup_{d \\in \\cal{D}}d(x,x) = \\sup\\{ 0 \\} = 0 \\]\n \\item \\[\\forall \\ x,\\ y \\in V,\\ d_{max}(x, y)= \\sup_{d \\in \\cal{D} }d(x,y)= \\sup_{d \\in \\cal{D} }d(y,x) =d _{max}(y, x) \\]\n \\item \\[ d_{max}(x, z)=\\sup\\{d(x,y): d \\in \\mathcal{D}\\} \\leq \\sup_{}\\{ d(x,y): d \\in \\mathcal{D}\\} {( d \\in \\cal{D} \\text{ is a metric})} \\]\n \\end{enumerate} \n \\end{lemma}\n \\begin{definition}\n An \\emph{edge path} is a ordered list (may be null, in whuch case it is a single vertex) of edges- denoted by $ \\eta=(e_1,e_2,\\dots,e_n) $, such that, $ \\forall i,\\ 1\\leq i < n,\\ i(e_i)=\\tau(e_{i+1})$. $\\eta$ is called ``an edge path from $i(e_1)$ to $\\tau(e_n)$''.\n \\end{definition}\n \\begin{definition}\n An edge path is a \\emph{loop} if $i(e_1) = \\tau(e_n)$ \n \\end{definition}\n \\begin{definition}\n An edge path is said to be \\emph{reduced} if $\\forall\\ i,\\ 1\\leq i$.\nIn particular, the squared norm of $\\ket\\psi$ is just $\\braket{\\psi|\\psi}$.\nConcretely, for $\\dim H = 2$ we have\n$\\braket{\\psi|\\psi} = |\\alpha|^2 + |\\beta|^2$.\n\n\\section{The state space}\nIf you think that's weird, well, it gets worse.\n\nIn classical computation, a bit is either $0$ or $1$.\nMore generally, we can think of a classical space of $n$\npossible states $0$, \\dots, $n-1$.\nThus in the classical situation, the space of possible states\nis just a discrete set with $n$ elements.\n\nIn quantum computation, a \\vocab{qubit} is instead\nany \\emph{complex linear combination} of $0$ and $1$.\nTo be precise, consider the normed complex vector space\n\\[ H = \\CC^{\\oplus 2} \\]\nand denote the orthonormal basis elements by $\\ket0$ and $\\ket1$.\nThen a \\emph{qubit} is a nonzero element $\\ket\\psi \\in H$,\nso that it can be written in the form\n\\[ \\ket\\psi = \\alpha \\ket 0 + \\beta \\ket 1 \\]\nwhere $\\alpha$ and $\\beta$ are not both zero.\nTypically, we normalize so that $\\ket\\psi$ has norm $1$:\n\\[ \\braket{\\psi|\\psi} = 1 \\iff |\\alpha|^2 + |\\beta|^2 = 1. \\]\nIn particular, we can recover the ``classical'' situation\nwith $\\ket 0 \\in H$ and $\\ket 1 \\in H$,\nbut now we have some ``intermediate'' states,\nsuch as \\[ \\frac{1}{\\sqrt2} \\left(\\ket 0 + \\ket 1 \\right). \\]\nPhilosophically, what has happened is that:\n\\begin{moral}\n\tInstead of allowing just the states $\\ket 0$ and $\\ket 1$,\n\twe allow any complex linear combination of them.\n\\end{moral}\nMore generally, if $\\dim H = n$,\nthen the possible states are nonzero elements\n\\[ c_0\\ket0 + c_1\\ket1 + \\dots + c_{n-1}\\ket{n-1} \\]\nwhich we usually normalize so that\n$|c_0|^2 + |c_1|^2 + \\dots + |c_{n-1}|^2 = 1$.\n\n\\section{Observations}\n\\prototype{$\\id$ corresponds to not making a measurement\n\tsince all its eigenvalues are equal,\n\tbut any operator with distinct eigenvalues will cause collapse.}\nIf you think that's weird, well, it gets worse.\nFirst, some linear algebra:\n\\begin{definition}\n\tLet $V$ be a finite-dimensional inner product space.\n\tFor a map $T: V \\to V$, the following conditions are equivalent:\n\t\\begin{itemize}\n\t\t\\ii $\\left< Tx, y\\right> = \\left< x, Ty \\right>$\n\t\tfor any $x,y \\in V$.\n\t\t\\ii $T = T^\\dagger$.\n\t\\end{itemize}\n\tA map $T$ satisfying these conditions is called \\vocab{Hermitian}.\n\\end{definition}\n\\begin{ques}\n\tShow that $T$ is normal.\n\\end{ques}\nThus, we know that $T$ is diagonalizable\nwith respect to the inner form, so for a suitable basis we\ncan write it in an orthonormal basis as\n\\[\n\tT = \\begin{bmatrix}\n\t\t\\lambda_0 & 0 & \\dots & 0 \\\\\n\t\t0 & \\lambda_1 & \\dots & 0 \\\\\n\t\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\t\t0 & 0 & \\dots & \\lambda_{n-1}\n\t\\end{bmatrix}.\n\\]\nAs we've said, this is fantastic:\nnot only do we have a basis of eigenvectors,\nbut the eigenvectors are pairwise orthogonal,\nand so they form an orthonormal basis of $V$.\n\\begin{ques}\n\tShow that all eigenvalues of $T$ are real.\n\t($T = T^\\dagger$.)\n\\end{ques}\n\nBack to quantum computation.\nSuppose we have a state $\\ket\\psi \\in H$, where $\\dim H = 2$;\nwe haven't distinguished a particular basis yet,\nso we just have a nonzero vector.\nThen the way observations work (and this is physics, so you'll have to\ntake my word for it) is as follows:\n\\begin{moral}\n\tPick a Hermitian operator $T : H \\to H$;\n\tthen observations of $T$ return eigenvalues of $T$.\n\\end{moral}\nTo be precise:\n\\begin{itemize}\n\t\\ii Pick a Hermitian operator $T : H \\to H$,\n\twhich is called the \\vocab{observable}.\n\t\\ii Consider its eigenvalues $\\lambda_0$, \\dots, $\\lambda_{n-1}$\n\tand corresponding eigenvectors $\\ket{0}_T$, \\dots, $\\ket{n-1}_T$.\n\tTacitly we may assume that $\\ket{0}_T$, \\dots, $\\ket{n-1}_T$ form\n\tan orthonormal basis of $H$.\n\t(The subscript $T$ is here to distinguish the eigenvectors of $T$\n\tfrom the basis elements of $H$.)\n\t\\ii Write $\\ket\\psi$ in the orthonormal basis as\n\t\\[ c_0\\ket0_T + c_1\\ket1_T + \\dots + c_{n-1}\\ket{n-1}_T. \\]\n\t\\ii Then the probability of observing $\\lambda_i$ is\n\t\\[ \\frac{|c_i|^2}{|c_0|^2 + \\dots + |c_{n-1}|^2}. \\]\n\tThis is called making an \\vocab{observation along $T$}.\n\\end{itemize}\nNote that in particular, for any nonzero constant $c$,\n$\\ket\\psi$ and $c\\ket\\psi$ are indistinguishable,\nwhich is why we like to normalize $\\ket\\psi$.\nBut the queerest thing of all is what happens to $\\ket\\psi$:\nby measuring it, we actually destroy information.\nThis behavior is called \\vocab{quantum collapse}.\n\\begin{itemize}\n\t\\ii Suppose for simplicity that we observe $\\ket\\psi$\n\twith $T$ and obtain an eigenvalue $\\lambda$,\n\tand that $\\ket{i}_T$ is the only eigenvector with this eigenvalue.\n\tThen, the state $\\ket\\psi$ \\emph{collapses} to just the state\n\t$c_i \\ket{i}_T$: all the other information is destroyed.\n\t(In fact, we may as well say it collapses to $\\ket{i}_T$,\n\tsince again constant factors are not relevant.)\n\n\t\\ii\tMore generally, if we observe $\\lambda$,\n\tconsider the generalized eigenspace $H_\\lambda$\n\t(i.e.\\ the span of eigenvectors with the same eigenvalue).\n\tThen the physical state $\\ket\\psi$ has been changed as well:\n\tit has now been projected onto the eigenspace $H_\\lambda$.\n\tIn still other words, after observation, the state collapses to\n\t\\[\n\t\t\\sum_{\\substack{0 \\le i \\le n \\\\ \\substack{\\lambda_i = \\lambda}}}\n\t\tc_i \\ket{i}_T.\n\t\\]\n\\end{itemize}\nIn other words,\n\\begin{moral}\n\tWhen we make a measurement,\n\tthe coefficients from different eigenspaces are destroyed.\n\\end{moral}\nWhy does this happen? Beats me\\dots physics (and hence real life) is weird.\nBut anyways, an example.\n\\begin{example}\n\t[Quantum measurement of a state $\\ket\\psi$]\n\tLet $H = \\CC^{\\oplus 2}$ with orthonormal basis $\\ket0$ and $\\ket1$\n\tand consider the state\n\t\\[\n\t\t\\ket\\psi\n\t\t= \\frac{i}{\\sqrt5} \\ket 0\n\t\t+ \\frac{2}{\\sqrt5} \\ket 1\n\t\t= \\pair{i/\\sqrt5}{2/\\sqrt5} \\in H.\n\t\\]\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Let \\[ T = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix}. \\]\n\t\tThis has eigenvectors $\\ket0 = \\ket0_T$ and $\\ket1 = \\ket1_T$,\n\t\twith eigenvalues $+1$ and $-1$. So if we measure $\\ket\\psi$ to $T$,\n\t\twe get $+1$ with probability $1/5$ and $-1$ with probability $4/5$.\n\t\tAfter this measurement, the original state collapses to\n\t\t$\\ket0$ if we measured $+1$, and $\\ket1$ if we measured $-1$.\n\t\tSo we never learn the original probabilities.\n\n\t\t\\ii Now consider $T = \\id$, and arbitrarily\n\t\tpick two orthonormal eigenvectors $\\ket0_T$, $\\ket1_T$;\n\t\tthus $\\psi = c_0\\ket0_T + c_1\\ket1_T$.\n\t\tSince all eigenvalues of $T$ are $+1$,\n\t\tour measurement will always be $+1$ no matter what we do.\n\t\tBut there is also no collapsing,\n\t\tbecause none of the coefficients get destroyed.\n\n\t\t\\ii Now consider\n\t\t\\[ T = \\begin{bmatrix} 0 & 7 \\\\ 7 & 0 \\end{bmatrix}. \\]\n\t\tThe two normalized eigenvectors are\n\t\t\\[ \\ket0_T = \\frac{1}{\\sqrt2}\\pair11\n\t\t\\qquad \\ket1_T = \\frac{1}{\\sqrt2}\\pair1{-1} \\]\n\t\twith eigenvalues $+7$ and $-7$ respectively. In this basis, we have\n\t\t\\[\n\t\t\t\\ket\\psi = \\frac{2+i}{\\sqrt{10}}\\ket0_T\n\t\t\t+ \\frac{-2+i}{\\sqrt{10}}\\ket1_T. \\]\n\t\tSo we get $+7$ with probability $\\half$ and $-7$\n\t\twith probability $\\half$, and after the measurement,\n\t\t$\\ket\\psi$ collapses to one of $\\ket0_T$ and $\\ket1_T$.\n\t\\end{enumerate}\n\\end{example}\n\\begin{ques}\n\tSuppose we measure $\\ket\\psi$ with $T$ and get $\\lambda$.\n\tWhat happens if we measure with $T$ again?\n\\end{ques}\n\nFor $H = \\CC^{\\oplus 2}$ we can come up with more classes of examples using\nthe so-called \\vocab{Pauli matrices}.\nThese are the three Hermitian matrices\n\\[\n\t\\sigma_z = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix}\n\t\\qquad\n\t\\sigma_x = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}\n\t\\qquad\n\t\\sigma_y = \\begin{bmatrix} 0 & -i \\\\ i & 0 \\end{bmatrix}.\n\\]\nThese matrices are important because:\n\\begin{ques}\n\tShow that these three matrices, plus the identity matrix,\n\tform a basis for the set of Hermitian $2 \\times 2$ matrices.\n\\end{ques}\nSo the Pauli matrices are a natural choice of basis.\n\nTheir normalized eigenvectors are\n\\[ \\zup = \\ket0 = \\pair10 \\qquad \\zdown = \\ket1 = \\pair01 \\]\n\\[ \\xup = \\frac{1}{\\sqrt2}\\pair11\n\t\\qquad \\xdown = \\frac{1}{\\sqrt2}\\pair1{-1} \\]\n\\[ \\yup = \\frac{1}{\\sqrt2}\\pair1i\n\t\\qquad \\ydown = \\frac{1}{\\sqrt2}\\pair1{-i} \\]\nwhich we call ``$z$-up'', ``$z$-down'',\n``$x$-up'', ``$x$-down'', ``$y$-up'', ``$y$-down''.\n(The eigenvalues are $+1$ for ``up'' and $-1$ for ``down''.)\nSo, given a state $\\ket\\psi \\in \\CC^{\\oplus 2}$\nwe can make a measurement with respect to any of these three bases\nby using the corresponding Pauli matrix.\n\nIn light of this, the previous examples were\n(a) measuring along $\\sigma_z$,\n(b) measuring along $\\id$,\nand (c) measuring along $7\\sigma_x$.\n\nNotice that if we are given a state $\\ket\\psi$,\nand are told in advance that it is either $\\xup$ or $\\xdown$\n(or any other orthogonal states)\nthen we are in what is more or less a classical situation.\nSpecifically, if we make a measurement along $\\sigma_x$,\nthen we find out which state that $\\ket\\psi$ was in (with 100\\% certainty),\nand the state does not undergo any collapse.\nThus, orthogonal states are reliably distinguishable.\n\n\\section{Entanglement}\n\\prototype{Singlet state: spooky action at a distance.}\nIf you think that's weird, well, it gets worse.\n\nQubits don't just act independently:\nthey can talk to each other by means of a \\emph{tensor product}.\nExplicitly, consider \\[ H = \\CC^{\\oplus 2} \\otimes \\CC^{\\oplus 2} \\]\nendowed with the norm described in \\Cref{prob:inner_prod_tensor}.\nOne should think of this as a qubit $A$ in a space $H_A$\nalong with a second qubit $B$ in a different space $H_B$,\nwhich have been allowed to interact in some way,\nand $H = H_A \\otimes H_B$ is the set of possible states of \\emph{both} qubits.\nThus\n\\[ \n\t\\ket{0}_A \\otimes \\ket{0}_B, \\quad\n\t\\ket{0}_A \\otimes \\ket{1}_B, \\quad\n\t\\ket{1}_A \\otimes \\ket{0}_B, \\quad\n\t\\ket{1}_A \\otimes \\ket{1}_B\n\\]\nis an orthonormal basis of $H$;\nhere $\\ket{i}_A$ is the basis of the first $\\CC^{\\oplus 2}$\nwhile $\\ket{i}_B$ is the basis of the second $\\CC^{\\oplus 2}$,\nso these vectors should be thought of as ``unrelated''\njust as with any tensor product.\nThe pure tensors mean exactly what you want:\nfor example $\\ket0_A \\otimes \\ket1_B$ means\n``$0$ for qubit $A$ and $1$ for qubit $B$''.\n\nAs before, a measurement of a state in $H$ requires\na Hermitian map $H \\to H$.\nIn particular, if we only want to measure the qubit $B$ along $M_B$,\nwe can use the operator \\[ \\id_A \\otimes M_B. \\]\nThe eigenvalues of this operator coincide with the ones for $M_B$,\nand the eigenspace for $\\lambda$ will be the $H_A \\otimes (H_B)_\\lambda$,\nso when we take the projection the $A$ qubit will be unaffected.\n\nThis does what you would hope for pure tensors in $H$:\n\\begin{example}[Two non-entangled qubits]\n\tSuppose we have qubit $A$ in the state\n\t$\\frac{i}{\\sqrt5}\\ket0_A + \\frac{2}{\\sqrt5}\\ket1_A$\n\tand qubit $B$ in the state\n\t$\\frac{1}{\\sqrt2} \\ket0_B + \\frac{1}{\\sqrt2}\\ket1_B$.\n\tSo, the two qubits in tandem are represented by the pure tensor\n\t\\[\n\t\t\\ket\\psi\n\t\t=\n\t\t\\left( \\frac{i}{\\sqrt5}\\ket0_A + \\frac{2}{\\sqrt5}\\ket1_A \\right)\n\t\t\\otimes\n\t\t\\left( \\frac{1}{\\sqrt2} \\ket0_B + \\frac{1}{\\sqrt2}\\ket1_B \\right).\n\t\\]\n\tSuppose we measure $\\ket\\psi$ along\n\t\\[ M = \\id_A \\otimes \\sigma_z^B. \\]\n\tThe eigenspace decomposition is\n\t\\begin{itemize}\n\t\t\\ii $+1$ for the span of $\\ket0_A \\otimes \\ket0_B$ and\n\t\t$\\ket1_A \\otimes \\ket0_B$, and\n\t\t\\ii $-1$ for the span of $\\ket0_A \\otimes \\ket1_B$ and\n\t\t$\\ket1_A \\otimes \\ket1_B$.\n\t\\end{itemize}\n\t(We could have used other bases, like $\\xup_A \\otimes \\ket0_B$ and\n\t$\\xdown_A \\otimes \\ket0_B$ for the first eigenspace, but it doesn't matter.)\n\tExpanding $\\ket\\psi$ in the four-element basis, we find that\n\twe'll get the first eigenspace with probability\n\t\\[ \\left|\\frac{i}{\\sqrt{10}}\\right|^2\n\t+ \\left|\\frac{2}{\\sqrt{10}}\\right|^2 = \\half. \\]\n\tand the second eigenspace with probability $\\half$ as well.\n\t(Note how the coefficients for $A$ don't do anything!)\n\tAfter the measurement, we destroy the coefficients of the other eigenspace;\n\tthus (after re-normalization) we obtain the collapsed state\n\t\\[ \\left( \\frac{i}{\\sqrt5}\\ket0_A + \\frac{2}{\\sqrt5}\\ket1_A \\right)\n\t\t\\otimes \\ket0_B\n\t\t\\qquad\\text{or}\\qquad\n\t\t\\left( \\frac{i}{\\sqrt5}\\ket0_A + \\frac{2}{\\sqrt5}\\ket1_A \\right)\n\t\t\\otimes \\ket1_B\n\t\\]\n\tagain with 50\\% probability each.\n\\end{example}\nSo this model lets us more or less work with the two qubits independently:\nwhen we make the measurement, we just make sure to not touch the other qubit\n(which corresponds to the identity operator).\n\n\\begin{exercise}\n\tShow that if $\\id_A \\otimes \\sigma_x^B$ is applied to the $\\ket\\psi$\n\tin this example, there is no collapse at all.\n\tWhat's the result of this measurement?\n\\end{exercise}\n\nSince the $\\otimes$ is getting cumbersome to write, we say:\n\\begin{abuse}\n\tFrom now on $\\ket 0_A \\otimes \\ket 0_B$ will be abbreviated\n\tto just $\\ket{00}$, and similarly for $\\ket{01}$, $\\ket{10}$, $\\ket{11}$.\n\\end{abuse}\n\n\\begin{example}\n\t[Simultaneously measuring a general $2$-Qubit state]\n\t\\label{ex:simult_measurement}\n\tConsider a normalized state $\\ket\\psi$ in\n\t$H = \\CC^{\\oplus 2} \\otimes \\CC^{\\oplus 2}$, say\n\t\\[ \\ket\\psi = \\alpha\\ket{00} + \\beta\\ket{01}\n\t\t+ \\gamma\\ket{10} + \\delta\\ket{11}. \\]\n\tWe can make a measurement along the diagonal matrix\n\t$T : H \\to H$ with \n\t\\[ T(\\ket{00}) = 0\\ket{00}, \\quad\n\tT(\\ket{01}) = 1\\ket{01}, \\quad\n\tT(\\ket{10}) = 2\\ket{10}, \\quad\n\tT(\\ket{11}) = 3\\ket{11}. \\]\n\tThus we get each of the eigenvalues $0$, $1$, $2$, $3$\n\twith probability $|\\alpha|^2$, $|\\beta|^2$, $|\\gamma|^2$, $|\\delta|^2$.\n\tSo if we like we can make ``simultaneous'' measurements on two qubits\n\tin the same way that we make measurements on one qubit.\n\\end{example}\n\nHowever, some states behave very weirdly.\n\\begin{example}[The singlet state]\n\tConsider the state\n\t\\[\n\t\t\\ket{\\Psi_-}\n\t\t=\n\t\t\\frac{1}{\\sqrt2} \\ket{01}\n\t\t- \\frac{1}{\\sqrt2} \\ket{10}\n\t\\]\n\twhich is called the \\vocab{singlet state}.\n\tOne can see that $\\ket{\\Psi_-}$ is not a simple tensor,\n\twhich means that it doesn't just consist of two qubits side by side:\n\tthe qubits in $H_A$ and $H_B$ have become \\emph{entangled}.\n\n\tNow, what happens if we measure just the qubit $A$?\n\tThis corresponds to making the measurement\n\t\\[ T = \\sigma_z^A \\otimes \\id_B. \\]\n\tThe eigenspace decomposition of $T$ can be described as:\n\t\\begin{itemize}\n\t\t\\ii The span of $\\ket{00}$ and $\\ket{01}$, with eigenvalue $+1$.\n\t\t\\ii The span of $\\ket{10}$ and $\\ket{11}$, with eigenvalue $-1$.\n\t\\end{itemize}\n\tSo one of two things will happen:\n\t\\begin{itemize}\n\t\t\\ii With probability $\\half$, we measure $+1$\n\t\tand the collapsed state is $\\ket{01}$.\n\t\t\\ii With probability $\\half$, we measure $-1$\n\t\tand the collapsed state is $\\ket{10}$.\n\t\\end{itemize}\n\tBut now we see that measurement along $A$ has told us what the\n\tstate of the bit $B$ is completely!\n\\end{example}\nBy solely looking at measurements on $A$, we learn $B$;\nthis paradox is called \\emph{spooky action at a distance},\nor in Einstein's tongue, \\vocab{spukhafte Fernwirkung}. \nThus,\n\\begin{moral}\n\tIn tensor products of Hilbert spaces,\n\tstates which are not pure tensors\n\tcorrespond to ``entangled'' states.\n\\end{moral}\n\nWhat this really means is that the qubits cannot be described independently;\nthe state of the system must be given as a whole.\nThat's what entangled states mean: the qubits somehow depend on each other.\n\n\\section\\problemhead\n\n\\begin{problem}\n\tWe measure $\\ket{\\Psi_-}$ by $\\sigma_x^A \\otimes \\id_B$,\n\tand hence obtain either $+1$ or $-1$.\n\tDetermine the state of qubit $B$ from this measurement.\n\t\\begin{hint}\n\t\tRewrite $\\ket{\\Psi_-} = -\\frac{1}{\\sqrt2}\n\t\t\\left( \\xup_A\\otimes\\xdown_B - \\xdown_A\\xup_B \\right)$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tBy a straightforward computation, \n\t\twe have $\\ket{\\Psi_-} = -\\frac{1}{\\sqrt2}\n\t\t\\left( \\xup_A\\otimes\\xdown_B - \\xdown_A\\xup_B \\right)$.\n\t\tNow, $\\xup_A \\otimes \\xup_B$, $\\xup_A \\otimes \\xdown_B$\n\t\tspan one eigenspace of $\\sigma_x^A \\otimes \\id_B$,\n\t\tand $\\xdown_A \\otimes \\xup_B$, $\\xdown_A \\otimes \\xdown_B$\n\t\tspan the other. So this is the same as before:\n\t\t$+1$ gives $\\xdown_B$ and $-1$ gives $\\xdown_A$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t[Greenberger-Horne-Zeilinger paradox]\n\tConsider the state in $(\\CC^{\\oplus2})^{\\otimes 3}$\n\t\\[\n\t\t\\ket{\\Psi}_{\\text{GHZ}}\n\t\t=\n\t\t\\frac{1}{\\sqrt2}\n\t\t\\left( \\ket0_A \\ket0_B \\ket0_C\n\t\t- \\ket1_A \\ket1_B \\ket1_C \\right).\n\t\\]\n\tFind the value of the measurements along each of\n\t\\[ \\sigma_y^A \\otimes \\sigma_y^B \\otimes \\sigma_x^C , \\quad\n\t\t\\sigma_y^A \\otimes \\sigma_x^B \\otimes \\sigma_y^C, \\quad\n\t\t\\sigma_x^A \\otimes \\sigma_y^B \\otimes \\sigma_y^C, \\quad\n\t\t\\sigma_x^A \\otimes \\sigma_x^B \\otimes \\sigma_x^C.\n\t\\]\n\tAs for the paradox: what happens if you multiply all these measurements together?\n\t\\begin{hint}\n\t\t$-1$, $1$, $1$, $1$.\n\t\tWhen we multiply them all together,\n\t\twe get that $\\id^A \\otimes \\id^B \\otimes \\id^C$\n\t\thas measurement $-1$, which is the paradox.\n\t\tWhat this means is that the values of the measurements\n\t\tare created when we make the observation,\n\t\tand not prepared in advance.\n\t\\end{hint}\n\\end{problem}\n", "meta": {"hexsha": "c0f60f65d45279cddaa28dbd4f613411f90fee63", "size": 19142, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/quantum/vectors.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/quantum/vectors.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/quantum/vectors.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.2254098361, "max_line_length": 82, "alphanum_fraction": 0.6885382928, "num_tokens": 6409, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506635289835, "lm_q2_score": 0.8499711832583696, "lm_q1q2_score": 0.646531144525994}} {"text": "\\documentclass{subfile}\n\n\\begin{document}\n\t\\section{The Buffalo Way}\\label{sec:buffalo}\n\t\\index{The buffalo way}\\emph{The buffalo way} is a \\emph{bashing} method of solving inequality problems. It usually exploits symmetry or cyclic property and involves tedious calculations, so it is not always elegant. Nonetheless it is a useful tool. The idea is to assume something like $a\\leq b\\leq c$ without losing any generality, specially when there is symmetry. In general, we want to assume that\n\t\t\\begin{align*}\n\t\t\ta_1\n\t\t\t\t& \\leq a_2\\leq\\cdots\\leq a_n\\\\\n\t\t\ta_2\n\t\t\t\t& = a_1+b_1\\\\\n\t\t\t\t& \\vdots\\\\\n\t\t\ta_n\n\t\t\t\t& = a_1+b_n\n\t\t\\end{align*}\n\tWe can also assume the following.\n\t\t\\begin{align*}\n\t\t\ta_2\n\t\t\t\t& = a_1+c_1\\\\\n\t\t\ta_3\n\t\t\t\t& = a_2+c_2\\\\\n\t\t\t\t& = a_1+c_1+c_2\\\\\n\t\t\t\t& \\vdots\\\\\n\t\t\ta_n\n\t\t\t\t& = a_{n-1}+c_{n-1}\\\\\n\t\t\t\t& = a_1+c_1+\\ldots+c_{n-1}\n\t\t\\end{align*}\n\twhere $c_1,\\ldots,c_{n-1}\\geq0$. Out of these two forms, the latter is actually a consequence of the former one and is usually the more used one. We will see a few problems that implement this technique.\n\t\t\\begin{problem}\n\t\t\tProve that for two non-negative reals $x,y$\n\t\t\t\t\\begin{align*}\n\t\t\t\t\tx+y\n\t\t\t\t\t\t& \\geq2\\sqrt{xy}\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\n\t\t\\begin{solution}\n\t\t\tSince the inequality is symmetric on $x$ and $y$, we can assume without loss of generality that $x\\leq y$. Let $y=x+k$.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\tx+y\n\t\t\t\t\t\t& \\geq2\\sqrt{xy}\\\\\n\t\t\t\t\t\\iff2x+k\n\t\t\t\t\t\t& \\geq2\\sqrt{x(x+k)}\\\\\n\t\t\t\t\t\\iff4x^2+4xk+k^2\n\t\t\t\t\t\t& \\geq4x^2+4xk\\\\\n\t\t\t\t\t\\iff k^2\n\t\t\t\t\t\t& \\geq0\n\t\t\t\t\\end{align*}\n\t\t\twhich is evident.\n\t\t\\end{solution}\n\tTry proving the case $n=3$ of arithmetic-geometric mean inequality with this technique. You will understand why we said this is often a tedious technique.\n\t\t\\begin{problem}\n\t\t\t\\begin{align*}\n\t\t\t\tx+y+z\n\t\t\t\t\t& \\geq3\\sqrt[3]{xyz}\n\t\t\t\\end{align*}\n\t\t\\end{problem}\n\tLet us also prove the following using buffalo way.\n\t\t\\begin{problem}\n\t\t\tLet $a,b,c>0$ be real numbers. Prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\ta^2+b^2+c^2\n\t\t\t\t\t\t& \\geq ab+bc+ca\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\n\t\t\\begin{solution}\n\t\t\tAssume that $a\\leq b\\leq c$ and $b=a+x,c=a+x+y$ where $x,y\\geq0$.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\ta^2+(a+x)^2+(a+x+y)^2\n\t\t\t\t\t\t& \\geq a(a+x)+(a+x)(a+x+y)+(a+x+y)a\\\\\n\t\t\t\t\t\\iff 3a^2+2x^2+y^2+4ax+2xy+2ay\n\t\t\t\t\t\t& \\geq 3a^2+x^2+4ax+2ay+2xy\\\\\n\t\t\t\t\t\\iff x^2+y^2\n\t\t\t\t\t\t& \\geq0\n\t\t\t\t\\end{align*}\n\t\t\tThis is again true.\n\t\t\\end{solution}\n\t\n\t\t\\begin{problem}\n\t\t\tLet $a,b,c\\geq0$ be real numbers such that $a\\leq b\\leq c$. Prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t(a+b)(c+a)^2\n\t\t\t\t\t\t& \\geq6abc\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\n\t\t\\begin{solution}\n\t\t\tSince $a\\leq b\\leq c$, let $b=a+x,c=a+x+y$.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t(2a+x)(2a+x+y)^2\n\t\t\t\t\t\t& \\geq6a(a+x)(a+x+y)\\\\\n\t\t\t\t\t\\iff(2a+x)(4a^2+x^2+y^2+4ax+4ay+2xy)\n\t\t\t\t\t\t& \\geq6a(a^2+2ax+x^2+ay+xy)\\\\\n\t\t\t\t\t\\iff8a^3+12a^2x+8a^2y+6ax^2+8axy+2ay^2+x^3+2ax^2+xy^2\n\t\t\t\t\t\t& \\geq6a^3+12a^2+6ax^2+6a^2y+6axy\\\\\n\t\t\t\t\t\\iff 2a^3+2a^2y+2axy+2ay^2+x^3+2x^2y+xy^2\n\t\t\t\t\t\t& \\geq0\n\t\t\t\t\\end{align*}\n\t\t\tThis evidently true.\n\t\t\\end{solution}\n\tNow, we will prove Nesbitt's inequality using the first form of the buffalo way.\n\t\t\\begin{proof}[Proof of Nesbitt's inequality by the buffalo way]\n\t\t\tLet $x\\leq y\\leq z$ and $y=x+a,z=y+b$ with $a\\geq0,b\\geq0$. We are required to prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{x}{y+z}+\\dfrac{y}{z+x}+\\dfrac{z}{x+y}\n\t\t\t\t\t\t& \\geq\\dfrac{3}{2}\n\t\t\t\t\\end{align*}\n\t\t\tLet us express each variable as a parameter of a single variable.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{y-a}{2y+b}+\\dfrac{y}{2y+b-a}+\\dfrac{y+b}{2y-a}\n\t\t\t\t\t\t& \\geq\\dfrac{3}{2}\\\\\n\t\t\t\t\t\\iff\\dfrac{y-a}{2y+b}-\\dfrac{1}{2}+\\dfrac{y}{2y+b-a}-\\dfrac{1}{2}+\\dfrac{y+b}{2y-a}-\\dfrac{1}{2}\n\t\t\t\t\t\t& \\geq0\\\\\n\t\t\t\t\t\\iff\\dfrac{1}{2}\\left(\\dfrac{2y-2a-2y-b}{2y+b}+\\dfrac{2y-2y-b+a}{2y+b-a}+\\dfrac{2y+2b-2y+a}{2y-a}\\right)\n\t\t\t\t\t\t& \\geq0\\\\\n\t\t\t\t\t\\iff-\\dfrac{2a+b}{2y+b}+\\dfrac{a-b}{2y+b-a}+\\dfrac{2b+a}{2y-a}\n\t\t\t\t\t\t& \\geq0\n\t\t\t\t\\end{align*}\n\t\t\tThe last inequality actually holds because of the following:\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{2b+a}{2y-a}+\\dfrac{a-b}{2y+b-a}-\\dfrac{2a+b}{2y+b}\n\t\t\t\t\t\t& \\geq\\dfrac{2b+a}{2y+b-a}+\\dfrac{a-b}{2y+b-a}-\\dfrac{2a+b}{2y+b}\\\\\n\t\t\t\t\t\t& \\geq\\dfrac{2a+2b}{2y+b-a}-\\dfrac{2a+b}{2y+b}\\\\\n\t\t\t\t\t\t& \\geq\\dfrac{2a+2b}{2y+b}-\\dfrac{2a+b}{2y+b}\\\\\n\t\t\t\t\t\t& = \\dfrac{a}{2y+b}\n\t\t\t\t\\end{align*}\n\t\t\twhich is obviously at least $0$. Here, we have used the facts that $\\frac{1}{x}\\geq\\frac{1}{y}$ implies $\\frac{1}{x-u}\\geq\\frac{1}{y}$ and $\\frac{1}{x}\\geq\\frac{1}{y+v}$ for $u,v\\geq0$.\n\t\t\\end{proof}\n\tThe next inequality is a nice application of the buffalo way.\n\t\t\\begin{problem}\n\t\t\tLet $x,y,z>0$ be real numbers such that no two are equal. Show that,\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{1}{(x-y)^2}+\\dfrac{1}{(y-z)^2}+\\dfrac{1}{(z-x)^2}\n\t\t\t\t\t\t& \\geq\\dfrac{4}{xy+yz+zx}\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\n\t\t\\begin{solution}\n\t\t\tThe expression on the left is symmetric on $x,y,z$. Without loss of generality, assume that $x\\leq y\\leq z$ and $y=x+a,z=x+a+b$.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{1}{a^2}+\\dfrac{1}{b^2}+\\dfrac{1}{(a+b)^2}\n\t\t\t\t\t\t& \\geq\\dfrac{4}{x(x+a)+(x+a)(x+a+b)+(x+a+b)x}\\\\\n\t\t\t\t\t\t& = \\dfrac{4}{x^2+xa+x^2+2ax+a^2+bx+ab+x^2+xa+bx}\\\\\n\t\t\t\t\t\t& \\geq \\dfrac{4}{3x^2+4xa+2xb+ab+a^2}\n\t\t\t\t\\end{align*}\n\t\t\tNote that there is no $x$ on the left side of the inequality. So, we can use $x>0$ to get rid of it.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t3x^2+4xa+2xb+ab+a^2\n\t\t\t\t\t\t& > ab+a^2\\\\\n\t\t\t\t\t\\dfrac{4}{3x^2+4xa+2xb+ab+a^2}\n\t\t\t\t\t\t& <\\dfrac{4}{ab+a^2}\n\t\t\t\t\\end{align*}\n\t\t\tSo, we are done if we can prove the following.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{1}{a^2}+\\dfrac{1}{b^2}+\\dfrac{1}{(a+b)^2}\n\t\t\t\t\t\t& \\geq\\dfrac{4}{ab+a^2}\\\\\n\t\t\t\t\t\t& = \\dfrac{4}{a(b+a)}\n\t\t\t\t\\end{align*}\n\t\t\tThis is not a trivial inequality. We have to show that this actually holds.\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{1}{a^2}+\\dfrac{1}{b^2}-\\dfrac{2}{a(b+a)}+\\dfrac{1}{(a+b)^2}\n\t\t\t\t\t\t& \\geq\\dfrac{2}{a(b+a)}\\\\\n\t\t\t\t\t\\iff\\dfrac{1}{b^2}+\\left(\\dfrac{1}{a}-\\dfrac{1}{a+b}\\right)^2\n\t\t\t\t\t\t& \\geq\\dfrac{2}{a(a+b)}\\\\\n\t\t\t\t\t\\iff\\dfrac{1}{b^2}+\\left(\\dfrac{b}{a(a+b)}\\right)^2\n\t\t\t\t\t\t& \\geq\\dfrac{2}{a(a+b)}\\\\\n\t\t\t\t\t\\iff\\dfrac{1}{b^2}+\\dfrac{b^2}{a^2(a+b)^2}\n\t\t\t\t\t\t& \\geq\\dfrac{2}{a(a+b)}\n\t\t\t\t\\end{align*}\n\t\t\tThe last inequality is true by arithmetic-geometric mean inequality since\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{1}{b^2}+\\dfrac{b^2}{a^2(a+b)^2}\n\t\t\t\t\t\t& \\geq2\\sqrt{\\dfrac{1}{b^2}\\cdot\\dfrac{b^2}{a^2(a+b)^2}}\\\\\n\t\t\t\t\t\t& = \\dfrac{2}{a(a+b)}\n\t\t\t\t\\end{align*}\n\t\t\\end{solution}\n\tSolve the next problem as an exercise.\n\t\t\\begin{problem}\n\t\t\tFor all real $x,y,z\\geq0$, prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\dfrac{x^3+y^3+z^3}{x^2y+y^2z+z^2x}\n\t\t\t\t\t\t& \\geq\\sum\\dfrac{x}{y+z}-1\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\\end{document}", "meta": {"hexsha": "17edb08652a5e9a81effa21a12b764aa7ac5819a", "size": 6572, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "buffalo.tex", "max_stars_repo_name": "ineq-tech/inequality", "max_stars_repo_head_hexsha": "ebf89351c843b6a7516e10e2ebf0d64e3f1f3f83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-02-06T08:29:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-06T08:29:30.000Z", "max_issues_repo_path": "buffalo.tex", "max_issues_repo_name": "ineq-tech/inequality", "max_issues_repo_head_hexsha": "ebf89351c843b6a7516e10e2ebf0d64e3f1f3f83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "buffalo.tex", "max_forks_repo_name": "ineq-tech/inequality", "max_forks_repo_head_hexsha": "ebf89351c843b6a7516e10e2ebf0d64e3f1f3f83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.5243243243, "max_line_length": 403, "alphanum_fraction": 0.5786670724, "num_tokens": 2963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8499711832583695, "lm_q1q2_score": 0.6465311353023657}} {"text": "\\section{Fat Tails}\nRoughly speaking, the consequences of ``fat tails\" is that rare events tend to play a disproportionally large role in determining the statistical properties of a sample. This is in contrast to ``thin tails\", where, as the sample size increases, pretty quickly no single observation modifies the statistical properties of the sample.\n\nFor ``thin tails\", extreme observations tend to result from the combination of several very unlikely events. For ``fat tails\", an extreme observation tend result from a single very unlikely event. \n\n\\subparagraph{Thin Tails}\nTwo people are randomly selected, and their combined height is 4 meters. Most likely this resulted from the selection of two people who are two meters tall, rather than one person that is 10 cm and another that is 3,90 m tall. \n\n\\subparagraph{Fat Tails}\nTwo people are randomly selected, and their combined net worth is \\$ 40M. The probability of having selected two people with a net work of \\$ 20M is less likely than having selected one person with a net worth of \\$ 200k and another person with a net worth of \\$ 39,8 M. \n\nFat tails do not imply that rare events are more frequent, but that they have greater impact when they do happen. In fact, ``fattening\" the tails of a Gaussian distribution results in a larger number of observations within one standard deviation. \n\n\\begin{figure}\n\\centering\n \\includegraphics[width=\\textwidth]{fattails01.png}\n \\caption{Probability densities of two independent thin tailed and thick tailed random variables (brazenly copied from Taleb, 2020). Compare to plot of Lp norms. For the thin tailed random variables, the observa- tion of a particular sum is most likely to result from a balanced contribution from both random variables. For the fat tailed random variables, the observation of a particular sum is most likely to result from the contribution of one of the variables.}\n \\label{fig:fattails01}\n\\end{figure}\n\n\\subsection{Consequences of Fat Tails}\n\n\\begin{figure}\n\\centering\n \\includegraphics[width=\\textwidth]{fattailstudentt02.png}\n \\caption{Behavior of the sample mean as a function of sample size for standard Student-t distributions with different degrees of freedom.}\n \\label{fig:fattails02}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n \\includegraphics[width=\\textwidth]{fattailsstudentt01.png}\n \\caption{Behavior of the variance of the sample mean as a function of sample size for standard Student-t distributions with different degrees of freedom.}\n \\label{fig:fattails03}\n\\end{figure}\n\n\n\n\\subparagraph{The Law of Large Numbers works too Slowly}\n\n\\subparagraph{The Sample Mean will rarely correspond to the Distribution Mean}\nFor example, for an 80/20 power law, 92\\% of the observations will fall below the distribution mean. The sample mean will tend to underestimate the distribution mean because the distribution mean is heavily biased by rare observations that will tend to be underrepresented in the sample.\n\n\\subparagraph{Metrics such as Sample Mean and Sample Variance will be Unusable}\n\n\\subparagraph{In finance, metrics like Sharp etc. are unusable}\n\n\\subparagraph{Gauss-Markov Theorem fails}\nTherefore, linear least squares regressions do not work. \n\n\\subparagraph{Maximum Likelihood Methods can still work}\nFor example in case of a Pareto distribution, it is possible to fit the tail exponent using maximum likelihood methods, and estimating the mean from there. Direct observation of the mean would be misleading. The tail exponent intelligently extrapolates the fat tails of the distribution.\n\n\\subparagraph{Absence of Evidence $\\neq$ Evidence of Absence}\n\n\\subparagraph{PCA is going to cause spurious factors and loads}\n\n\\subparagraph{Method of Moments does not work}\nApproximating a distribution by matching its moments does not work when higher moments are undefined or cannot be reliably estimated.\n\n\\subparagraph{There is no ``typical\" large deviation}\n\n\n\\subsection{Maximum to Sum}\nThe \"maximum to sum\" or MS plot allows to see the behavior of the relationship between the observed maximum to the sum for a particular moment as the number of observations increases. \n\n\\subsection{Maximum Domain of Attraction}\n\nThe ``maximum domain of attraciton\" is, so to speak, the ``right endpoint of the distribution\":\n\n\\begin{equation}\nx^* = \\sup\\{x: F(x) < 1 \\}\n\\end{equation}\n\n\\subsection{Hidden Tail, Problems in Estimating Moments}\nThe Glivenko-Cantelli theorem guarantees uniform convergence of the empiral cdf to the true CDF, however, the empirical distribution is necessarily bounded by the values of the minimum and maximum observations. This results in an unobserved contribution to moments $p>0$ that does not necessarily have to be negligible. To illustrate, take $K_n$ to be the maximum observed value. \n\\begin{equation}\n\\mathbb{E}(X^p) = \\underbrace{\\int_{L}^{K_n} x^p \\phi(x)\\mathrm{d}x}_{observed} + \\underbrace{\\int_{K_n}^{\\infty} x^p \\phi(x)\\mathrm{d}x}_{unobserved}\n\\end{equation}\n\n\n", "meta": {"hexsha": "8c92807306cfc09afe23032949b8b56cf6d18e6c", "size": 4971, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/chapters/sections/stats_fattails.tex", "max_stars_repo_name": "jpbm/probabilism", "max_stars_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "notes/chapters/sections/stats_fattails.tex", "max_issues_repo_name": "jpbm/probabilism", "max_issues_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/chapters/sections/stats_fattails.tex", "max_forks_repo_name": "jpbm/probabilism", "max_forks_repo_head_hexsha": "a2f5c1595aed616236b2b889195604f365175899", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.6219512195, "max_line_length": 468, "alphanum_fraction": 0.7877690606, "num_tokens": 1151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8499711775577736, "lm_q1q2_score": 0.6465311309662038}} {"text": "\n\n\\section{Reorderings and fluctuations}\n\\label{sec:heap-reordering}\n\nOne particular challenge posed by heap algorithms is that\nwhile temporarily causing \\emph{small fluctuations} in the number of values\nwithin an array they essentially only \\emph{reorder} it,\nthat is they leave the multiset of its values unchanged.\nIn this section we will introduce various predicates that will help\nus mastering this challenge.\n\n\\subsection{Formalizing small fluctuations}\n\nThe predicate \\MultisetAdd in the following listing\nexpresses that the number of occurrences of a specific element in an array\nhas increased by one between two program points~\\inl{K} and~\\inl{L}.\n\n\\input{Listings/MultisetOperations.acsl.tex}\n\nThe predicate \\MultisetMinus, on the other hand,\nexpresses that the number of occurrences of a specific element in an array\nhas decreased by one between two program points~\\inl{K} and~\\inl{L}.\nNote that we have defined \\MultisetMinus by calling \\MultisetAdd\nwith the labels reversed.\n%\nFinally, the predicate \\MultisetRetain expresses that a the number\nof occurrences of a given value does not change between two program points.\nIn order to guide the automatic provers, we also provide some\nlemmas that formalize conditions under which the respective predicates hold.\n\nUsing the predicate \\logicref{MultisetReorder} and the logic function \\logicref{At}\nwe also formulate a few simple lemmas that describe when the\npredicates from Listing~\\logicref{MultisetOperations} hold.\n\n\\subsection{Simple properties of fluctuations}\n\nThe predicate \\logicref{MultisetRetainRest} uses \\logicref{MultisetRetain}\nin order to express that all values of an array,\nexcept the two given values \\inl{u} and \\inl{v}, occur as often in program\npoint \\inl{K} and program point \\inl{L}. \n\nThe lemmas in this listing express conditions under which small\nfluctuations---expressed by the predicates \\logicref{MultisetAdd}\nand \\logicref{MultisetMinus}---in the number of occurrences between three\nprogram points even with each other.\n\n\\input{Listings/MultisetRetainRest.acsl.tex}\n\n\n\\subsection{Combining fluctuations}\n\nSmall fluctuations are so prevalent in the central heap algorithms \\implref{pushheap}\nand \\implref{popheap} that it is worthwhile to introduce another predicate\nto concisely capture this feature.\nWe refer to this predicate as \\logicref{MultisetParity} because it describes\nthe situation where the number of occurrences \n\n\\begin{itemize}\n\\item of the first of two given values increases by one\n\\item while that of the second value decreases by one\n\\item and the remaining values retain their respective number of occurrences.\n\\end{itemize}\n\nWith this predicate we can formulate several lemmas that describe useful\ncombinations of reorderings and fluctuations.\nFor example, lemma \\logicref{MultisetParityMultisetReorder} describes\nthe situation where two fluctuation cancel each other and consequently\nestablish a reordering of an array.\n\n\\input{Listings/MultisetParity.acsl.tex}\n\n\\subsection{How do fluctuations arise?}\n\nThe simplest way to creation a small fluctuation is to\nupdate an array element with a different value.\nThus, similar to the predicate \\logicref{ArrayUpdate} we introduce \npredicate \\logicref{MultisetUpdate} which in turn relies on \\logicref{MultisetParity}.\nLemma \\logicref{ArrayUpdateMultisetUpdate} then formalizes the claim\nthat updating an array element with a different value creates a small fluctuation.\n\n\\input{Listings/MultisetUpdate.acsl.tex}\n\n\\clearpage\n\n", "meta": {"hexsha": "a8f39afc6e4d3a6c3e03196d07dbd9c50a957291", "size": 3480, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Informal/heap/fluctuations.tex", "max_stars_repo_name": "fraunhoferfokus/acsl-by-example", "max_stars_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 90, "max_stars_repo_stars_event_min_datetime": "2017-06-14T04:17:53.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-07T06:07:36.000Z", "max_issues_repo_path": "Informal/heap/fluctuations.tex", "max_issues_repo_name": "fraunhoferfokus/acsl-by-example", "max_issues_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 22, "max_issues_repo_issues_event_min_datetime": "2017-10-18T13:30:41.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-17T07:10:16.000Z", "max_forks_repo_path": "Informal/heap/fluctuations.tex", "max_forks_repo_name": "fraunhoferfokus/acsl-by-example", "max_forks_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 19, "max_forks_repo_forks_event_min_datetime": "2017-06-21T13:49:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T16:27:06.000Z", "avg_line_length": 40.4651162791, "max_line_length": 86, "alphanum_fraction": 0.8155172414, "num_tokens": 808, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711832583695, "lm_q2_score": 0.7606506472514406, "lm_q1q2_score": 0.6465311306905516}} {"text": "\\section{Conclusions}\n\\SecLabel{conclusions}\n\nThis work started by observing interesting patterns in the LAT modulo 8 of small Feistel Networks. The analysis of the patterns resulted in the definition of High-Degree Indicator Matrix (HDIM). This tool shows a link between the LAT and the highest-degree monomials in the ANF. Furthermore, its properties allow to prove upper bounds on the algebraic degree of cryptographic structures and to prove finer algebraic degeneracies. Though these results do not improve the state of the art, the upper bounds given are expressed in a simple closed formula and there is a room for improvement, e.g. by combining methods. Finally, the most useful application of HDIM is in the cryptanalysis of Feistel Networks masked with secret affine layers. The generalized HDIM-motivated ideas allow to prove lower-degree degeneracies as well, i.e. impossible monomials. I show how they can be used to mount decomposition attacks on Feistel Networks. The results of this chapter together allow to fully decompose affinely-whitened Feistel Networks satisfying the attack conditions. I think it provides many useful tools for S-Box reverse-engineering and white-box analysis toolkit.\n\nThe work leaves several open problems:\n\\begin{enumerate}\n \\item Better degree evaluation. Is it possible to improve the HDIM-motivated method? Is it possible to combine it with other methods, e.g. division property? \n \n \\item Proving Conjecture~\\Ref{conj:impmono}. Are there always enough impossible monomials to recover the last Feistel function, if the type-II distinguisher applies?\n\n \\item In which cases it is possible to decompose Feistel Networks having at least 1 more round than Feistel Networks satisfying the type-II distinguisher?\n \n \\item A big open problem: lower bounds in Feistel Networks or Substitution-Permutation Networks. How to prove non-trivial lower bounds on the degree of a structure, i.e. that at least one instance of the structure has high enough degree? Strong lower bounds could shed light on how close are current degree evaluation methods to optimal ones.\n\\end{enumerate}", "meta": {"hexsha": "c3d462ad5d28d88db0e98617afdd3db76dbcda45", "size": 2134, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis-source/9strFeistel/6conclusions.tex", "max_stars_repo_name": "hellman/thesis", "max_stars_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 19, "max_stars_repo_stars_event_min_datetime": "2019-05-16T19:55:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-31T15:36:12.000Z", "max_issues_repo_path": "thesis-source/9strFeistel/6conclusions.tex", "max_issues_repo_name": "hellman/thesis", "max_issues_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-08-09T11:26:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-09T11:26:45.000Z", "max_forks_repo_path": "thesis-source/9strFeistel/6conclusions.tex", "max_forks_repo_name": "hellman/thesis", "max_forks_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-05T19:40:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-05T19:40:16.000Z", "avg_line_length": 142.2666666667, "max_line_length": 1163, "alphanum_fraction": 0.8064667291, "num_tokens": 455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711680567799, "lm_q2_score": 0.7606506526772884, "lm_q1q2_score": 0.6465311237392668}} {"text": "\\newpage\r\n\\section{Matrix Normal Forms}\r\n\r\n\\begin{Operator}[smithex]{SMITHEX}\r\nThe operator \\name{smithex} computes the Smith normal form S of a\r\n\\nameref{matrix} A (say). It returns \\{S,P,$P^-1$\\} where $P*S*P^-1 = A$.\r\n\r\n\\begin{Syntax}\r\n\\name{smithex}\\(\\meta{matrix},\\meta{variable}\\)\r\n\r\n\\meta{matrix} :- a rectangular \\nameref{matrix} of univariate polynomials in \r\n \\meta{variable}.\r\n\\meta{variable} :- the variable.\r\n\\end{Syntax}\r\n\r\n\\begin{Examples}\r\n a := mat((x,x+1),(0,3*x^2)); &\r\n\\begin{multilineoutput}{6cm}\r\n [x x + 1]\r\n [ ]\r\n a := [ 2 ]\r\n [0 3*x ]\r\n\\end{multilineoutput}\\\\\r\n\r\n smithex(a,x); &\r\n\\begin{multilineoutput}{6cm}\r\n [1 0 ] [1 0] [x x + 1]\r\n\\{ [ ], [ ], [ ] \\}\r\n [ 3] [ 2 ] [ ]\r\n [0 x ] [3*x 1] [-3 -3 ]\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\\end{Operator}\r\n\r\n\r\n\\begin{Operator}[smithexint]{SMITHEX_INT}\r\nThe operator \\name{smithex\\_int} performs the same task as \\name{smithex}\r\nbut on matrices containing only integer entries. Namely, \r\n\\name{smithex\\_int} returns \\{S,P,$P^-1$\\} where S is the smith normal\r\nform of the input \\nameref{matrix} (A say), and $P*S*P^-1 = A$.\r\n\r\n\\begin{Syntax}\r\n\\name{smithex\\_int}\\(\\meta{matrix}\\)\r\n\r\n\\meta{matrix} :- a rectangular \\nameref{matrix} of integer entries.\r\n\\end{Syntax}\r\n\r\n\\begin{Examples}\r\n a := mat((9,-36,30),(-36,192,-180),(30,-180,180)); &\r\n\\begin{multilineoutput}{6cm}\r\n [ 9 -36 30 ]\r\n [ ]\r\na := [-36 192 -180]\r\n [ ]\r\n [30 -180 180 ]\r\n\\end{multilineoutput}\\\\\r\n\r\n smithex_int(a); &\r\n\\begin{multilineoutput}{6cm}\r\n [3 0 0 ] [-17 -5 -4 ] [1 -24 30 ]\r\n [ ] [ ] [ ]\r\n\\{ [0 12 0 ], [64 19 15 ], [-1 25 -30] \\}\r\n [ ] [ ] [ ] \r\n [0 0 60] [-50 -15 -12] [0 -1 1 ] \r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\\end{Operator}\r\n\r\n\r\n\\begin{Operator}[frobenius]{FROBENIUS}\r\nThe operator \\name{frobenius} computes the \\name{frobenius} normal form F of a \r\n\\nameref{matrix} (A say). It returns \\{F,P,$P^-1$\\} where $P*F*P^-1 = A$.\r\n\r\n\\begin{Syntax}\r\n\\name{frobenius}\\(\\meta{matrix}\\)\r\n\r\n\\meta{matrix} :- a square \\nameref{matrix}.\r\n\\end{Syntax}\r\n\r\nField Extensions:\r\n\r\nBy default, calculations are performed in the rational numbers. To \r\nextend this field the \\nameref{arnum} package can be used. The package must\r\nfirst be loaded by load\\_package arnum;. The field can now be extended\r\nby using the defpoly command. For example, defpoly sqrt2**2-2; will\r\nextend the field to include the square root of 2 (now defined by sqrt2).\r\n\r\nModular Arithmetic: \r\n\r\n\\name{Frobenius} can also be calculated in a modular base. To do this \r\nfirst type on modular;. Then setmod p; (where p is a prime) will set \r\nthe modular base of calculation to p. By further typing on balanced\\_mod\r\nthe answer will appear using a symmetric modular representation. See \r\n\\nameref{ratjordan} for an example.\r\n\r\n\\begin{Examples}\r\n a := mat((x,x^2),(3,5*x)); &\r\n\\begin{multilineoutput}{6cm}\r\n [ 2 ]\r\n [x x ]\r\na := [ ]\r\n [3 5*x]\r\n\\end{multilineoutput}\\\\\r\n frobenius(a);&\r\n\\begin{multilineoutput}{6cm}\r\n [ 2] [1 x] [ - x ]\r\n\\{ [0 - 2*x ], [ ], [1 -----] \\}\r\n [ ] [0 3] [ 3 ]\r\n [1 6*x ] [ ]\r\n [ 1 ]\r\n [0 --- ]\r\n [ 3 ]\r\n\r\n\\end{multilineoutput}\\\\\r\n load\\_package arnum;\\\\\r\n defpoly sqrt2**2-2;\\\\\r\n a := mat((sqrt2,5),(7*sqrt2,sqrt2));&\r\n\\begin{multilineoutput}{6cm}\r\n [ sqrt2 5 ]\r\na := [ ]\r\n [7*sqrt2 sqrt2]\r\n\\end{multilineoutput}\\\\\r\n\r\n frobenius(a); &\r\n\\begin{multilineoutput}{6cm}\r\n [0 35*sqrt2 - 2] [1 sqrt2 ] [ 1 ]\r\n\\{ [ ], [ ], [1 - --- ] \\}\r\n [1 2*sqrt2 ] [1 7*sqrt2] [ 7 ]\r\n [ ]\r\n [ 1 ]\r\n [0 ----*sqrt2]\r\n [ 14 ]\r\n\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\\end{Operator}\r\n\r\n\r\n\r\n\\begin{Operator}[ratjordan]{RATJORDAN}\r\nThe operator \\name{ratjordan} computes the rational Jordan normal form R\r\nof a \\nameref{matrix} (A say). It returns \\{R,P,$P^-1$\\} where $P*R*P^-1 = A$.\r\n\r\n\\begin{Syntax}\r\n\\name{ratjordan}\\(\\meta{matrix}\\)\r\n\r\n\\meta{matrix} :- a square \\nameref{matrix}.\r\n\\end{Syntax}\r\n\r\nField Extensions:\r\n\r\nBy default, calculations are performed in the rational numbers. To \r\nextend this field the \\name{arnum} package can be used. The package must\r\nfirst be loaded by load\\_package arnum;. The field can now be extended\r\nby using the defpoly command. For example, defpoly sqrt2**2-2; will\r\nextend the field to include the square root of 2 (now defined by sqrt2).\r\nSee \\nameref{frobenius} for an example.\r\n\r\nModular Arithmetic:\r\n\r\n\\name{ratjordan} can also be calculated in a modular base. To do this \r\nfirst type on modular;. Then setmod p; (where p is a prime) will set \r\nthe modular base of calculation to p. By further typing on balanced\\_mod\r\nthe answer will appear using a symmetric modular representation.\r\n\r\n\\begin{Examples}\r\n a := mat((5,4*x),(2,x^2));&\r\n\\begin{multilineoutput}{6cm}\r\n [5 4*x]\r\n [ ]\r\na := [ 2 ]\r\n [2 x ]\r\n\\end{multilineoutput}\\\\\r\n\r\n ratjordan(a); &\r\n\\begin{multilineoutput}{6cm}\r\n [0 x*( - 5*x + 8)] [1 5] [ -5 ] \r\n\\{ [ ], [ ], [1 -----] \\}\r\n [ 2 ] [0 2] [ 2 ]\r\n [1 x + 5 ] [ ]\r\n [ 1 ]\r\n [0 -----]\r\n [ 2 ]\r\n\\end{multilineoutput}\\\\\r\n on modular; \\\\\r\n setmod 23; \\\\\r\n a := mat((12,34),(56,78)); &\r\n\\begin{multilineoutput}{6cm}\r\n [12 11]\r\na := [ ]\r\n [10 9 ]\r\n\\end{multilineoutput}\\\\\r\n\r\n ratjordan(a); &\r\n\\begin{multilineoutput}{6cm}\r\n [15 0] [16 8] [1 21]\r\n\\{ [ ], [ ], [ ] \\}\r\n [0 6] [19 4] [1 4 ]\r\n\\end{multilineoutput}\\\\\r\n\r\n on balanced\\_mod;\\\\\r\n ratjordan(a);&\r\n\\begin{multilineoutput}{6cm}\r\n [- 8 0] [ - 7 8] [1 - 2]\r\n\\{ [ ], [ ], [ ] \\}\r\n [ 0 6] [ - 4 4] [1 4 ]\r\n\\end{multilineoutput}\\\\\r\n\r\n\\end{Examples}\r\n\\end{Operator}\r\n\r\n\r\n\r\n\\begin{Operator}[jordansymbolic]{JORDANSYMBOLIC}\r\nThe operator \\name{jordansymbolic} computes the Jordan normal form J\r\nof a \\nameref{matrix} (A say). It returns \\{J,L,P,$P^-1$\\} where\r\n$P*J*P^-1 = A$. L = \\{ll,mm\\} where mm is a name and ll is a list of\r\nirreducible factors of p(mm).\r\n\r\n\\begin{Syntax}\r\n\\name{jordansymbolic}\\(\\meta{matrix}\\)\r\n\r\n\\meta{matrix} :- a square \\nameref{matrix}.\r\n\\end{Syntax}\r\n\r\nField Extensions:\r\n\r\nBy default, calculations are performed in the rational numbers. To \r\nextend this field the \\nameref{arnum} package can be used. The package must\r\nfirst be loaded by load\\_package arnum;. The field can now be extended\r\nby using the defpoly command. For example, defpoly sqrt2**2-2; will\r\nextend the field to include the square root of 2 (now defined by sqrt2).\r\nSee \\nameref{frobenius} for an example.\r\n\r\nModular Arithmetic:\r\n\r\n\\name{jordansymbolic} can also be calculated in a modular base. To do this \r\nfirst type on modular;. Then setmod p; (where p is a prime) will set \r\nthe modular base of calculation to p. By further typing on balanced\\_mod\r\nthe answer will appear using a symmetric modular representation. See \r\n\\nameref{ratjordan} for an example.\r\n\r\n% Extras:\r\n\r\n% If using \\name{xr}, the X interface for REDUCE, then the appearance of the\r\n% output can be improved by switching on looking\\_good. This\r\n% converts any lambdas to a greek font.\r\n\r\n\\begin{Examples}\r\n\r\n a := mat((1,y),(2,5*y)); &\r\n\\begin{multilineoutput}{6cm}\r\n [1 y ]\r\na := [ ]\r\n [2 5*y]\r\n\\end{multilineoutput}\\\\\r\n\r\n jordansymbolic(a); &\r\n\\begin{multilineoutput}{6cm}\r\n\\{\r\n\r\n [lambda11 0 ]\r\n [ ]\r\n [ 0 lambda12]\r\n\r\n ,\r\n\r\n 2\r\n {{lambda - 5*lambda*y - lambda + 3*y},lambda},\r\n\r\n\r\n [lambda11 - 5*y lambda12 - 5*y]\r\n [ ]\r\n [ 2 2 ]\r\n\r\n ,\r\n\r\n\r\n [ 2*lambda11 - 5*y - 1 5*lambda11*y - lambda11 - y + 1 ]\r\n [---------------------- ---------------------------------]\r\n [ 2 2 ]\r\n [ 25*y - 2*y + 1 2*(25*y - 2*y + 1) ]\r\n [ ]\r\n [ 2*lambda12 - 5*y - 1 5*lambda12*y - lambda12 - y + 1 ]\r\n [---------------------- ---------------------------------]\r\n [ 2 2 ]\r\n [ 25*y - 2*y + 1 2*(25*y - 2*y + 1) ]\r\n\r\n \\}\r\n\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\\end{Operator}\r\n\r\n\r\n\r\n\\begin{Operator}[jordan]{JORDAN}\r\nThe operator \\name{jordan} computes the Jordan normal form J\r\nof a \\nameref{matrix} (A say). It returns \\{J,P,$P^-1$\\} where $P*J*P^-1 = A$.\r\n\r\n\\begin{Syntax}\r\n\\name{jordan}\\(\\meta{matrix}\\)\r\n\r\n\\meta{matrix} :- a square \\nameref{matrix}.\r\n\\end{Syntax}\r\n\r\nField Extensions:\r\nBy default, calculations are performed in the rational numbers. To \r\nextend this field the \\name{arnum} package can be used. The package must\r\nfirst be loaded by load\\_package arnum;. The field can now be extended\r\nby using the defpoly command. For example, defpoly sqrt2**2-2; will\r\nextend the field to include the square root of 2 (now defined by sqrt2).\r\nSee \\nameref{frobenius} for an example.\r\n\r\nModular Arithmetic:\r\n\\name{Jordan} can also be calculated in a modular base. To do this \r\nfirst type on modular;. Then setmod p; (where p is a prime) will set \r\nthe modular base of calculation to p. By further typing on balanced\\_mod\r\nthe answer will appear using a symmetric modular representation. See\r\n\\nameref{ratjordan} for an example.\r\n\r\n\\begin{Examples}\r\n\r\n a := mat((1,x),(0,x)); &\r\n\\begin{multilineoutput}{6cm}\r\n [1 x]\r\na := [ ]\r\n [0 x]\r\n\\end{multilineoutput}\\\\\r\n\r\n jordan(a);&\r\n\\begin{multilineoutput}{6cm}\r\n\\{\r\n\r\n [1 0]\r\n [ ]\r\n [0 x]\r\n\r\n ,\r\n\r\n\r\n [ 1 x ]\r\n [------- --------------]\r\n [ x - 1 2 ]\r\n [ x - 2*x + 1 ]\r\n [ ]\r\n [ 1 ]\r\n [ 0 ------- ]\r\n [ x - 1 ]\r\n\r\n ,\r\n\r\n\r\n [x - 1 - x ]\r\n [ ]\r\n [ 0 x - 1]\r\n\r\n \\}\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\r\n\\end{Operator}\r\n\r\n", "meta": {"hexsha": "f4261df717a845f6a11bdf4c234eb5ccd633069d", "size": 10653, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "normform.tex", "max_stars_repo_name": "nilqed/REDHELP", "max_stars_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "normform.tex", "max_issues_repo_name": "nilqed/REDHELP", "max_issues_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "normform.tex", "max_forks_repo_name": "nilqed/REDHELP", "max_forks_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.2664835165, "max_line_length": 80, "alphanum_fraction": 0.5191964705, "num_tokens": 3286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199673867852, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.646511525487668}} {"text": "\n\n \\filetitle{movsum}{Moving sum pseudofunction}{modellang/movsum}\n\n\t\\paragraph{Syntax}\\label{syntax}\n\n\\begin{verbatim}\nmovsum(Expr)\nmovsum(Expr,K)\n\\end{verbatim}\n\n\\paragraph{Description}\\label{description}\n\nIf the second input argument, \\texttt{K}, is negative, this function\nexpands to the moving sum of the last K periods (including the current\nperiod), i.e.\n\n\\begin{verbatim}\n((Expr)+(Expr{-1})+ ... +(Expr{-(K-1)})\n\\end{verbatim}\n\nwhere \\texttt{Expr\\{-N\\}} derives from \\texttt{Expr} and has all its\ntime subscripts shifted by \\texttt{-N} (if specified).\n\nIf the second input argument, \\texttt{K}, is positive, this function\nexpands to the moving sum of the next K periods ahead (including the\ncurrent period), i.e.\n\n\\begin{verbatim}\n((Expr)+(Expr{1})+ ... +(Expr{K-1})\n\\end{verbatim}\n\nIf the second input argument, \\texttt{K}, is not specified, the default\nvalue -4 is used (based on the fact that most of the macroeconomic\nmodels are quarterly).\n\n\\paragraph{Example}\\label{example}\n\nThe following three lines\n\n\\begin{verbatim}\nmovsum(Z)\nmovsum(Z,-3)\nmovsum(X+Y{-1},2)\n\\end{verbatim}\n\nwill expand to\n\n\\begin{verbatim}\n((Z)+(Z{-1})+(Z{-2})+(Z{-3}))\n((Z)+(Z{-1})+(Z{-2}))\n((X+Y{-1})+(X{1}+Y))\n\\end{verbatim}\n\n\n", "meta": {"hexsha": "0d7317dd4c3eada917eb57f91f6124a425359496", "size": 1217, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "-help/modellang/movsum.tex", "max_stars_repo_name": "OGResearch/IRIS-Toolbox-For-Octave", "max_stars_repo_head_hexsha": "682ea1960229dc701e446137623b120688953cef", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-12-06T13:38:38.000Z", "max_stars_repo_stars_event_max_datetime": "2017-12-06T13:38:38.000Z", "max_issues_repo_path": "-help/modellang/movsum.tex", "max_issues_repo_name": "OGResearch/IRIS-Toolbox-For-Octave", "max_issues_repo_head_hexsha": "682ea1960229dc701e446137623b120688953cef", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2017-03-28T08:13:20.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-02T10:40:25.000Z", "max_forks_repo_path": "-help/modellang/movsum.tex", "max_forks_repo_name": "OGResearch/IRIS-Toolbox-For-Octave", "max_forks_repo_head_hexsha": "682ea1960229dc701e446137623b120688953cef", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-17T07:06:39.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-17T07:06:39.000Z", "avg_line_length": 21.7321428571, "max_line_length": 71, "alphanum_fraction": 0.6959737058, "num_tokens": 375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199552262967, "lm_q2_score": 0.7718435083355188, "lm_q1q2_score": 0.6465115248937051}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=1in]{geometry} \n\\usepackage{amsmath,amsthm,amssymb,amsfonts}\n\\usepackage{enumitem}\n\\usepackage{tabu}\n\\usepackage{xcolor}\n \\usepackage{mathtools}\n \\usepackage{gensymb}\n \n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\n\\newcommand{\\nspace}{\\vspace*{.5cm}}\n\\newcommand{\\nline}{\\nspace \\noindent}\n\n\\newcommand{\\expected}[1]{\\text{E}(#1)}\n\n\\newenvironment{nscenter}\n {\\parskip=0pt\\par\\nopagebreak\\centering}\n {\\par\\noindent\\ignorespacesafterend}\n \n\\def\\SPSB#1#2{\\rlap{\\textsuperscript{\\textcolor{black}{#1}}}\\SB{#2}}\n \n % 6.1 4,17,18,21\n % 6.2 2,4,5,9,28,29\n \n\\begin{document}\n\\title{Math 487 Homework 5}\n\\author{Jacob Shiohira}\n\\maketitle\n\n\\subsection*{Section 6.1}\n\n% In Las Vegas the roulette wheel has a $0$ and a $00$ and then the numbers $1$ to $36$ marked on equal slots; the wheel is spun and a ball stops randomly in one slot. When a player bets $1$ dollar on a number, he receives $36$ dollars if the ball stops on this number, for a net gain of $35$ dollars; otherwise, he loses his dollar bet. Find the expected value for his winnings.\n\n\\noindent\n\\textbf{Ch 6.1 Q4} With $38$ possible spots to place a bet on, we know that the probability of winning $35$ dollars is $1/38$, and the probability of losing a dollar is $37/38$. Let $X$ be the random variable which denotes your winnings on a $1$ dollar bet in Las Vegas roulette. Then, the distribution of $X$ is given by\n\n\\begin{equation*}\nm_X = {-1 \\qquad 35 \\choose 37/38 \\quad 1/38}.\n\\end{equation*}\n\n\\nline\nThus, we see then that $-1 \\cdot (37/38) + 35 \\cdot (1/38) = - 1/19$.\n\n\n% Let $X$ be the first time that a failure occurs in an infinite sequence of Bernoulli trials with probability $p$ for success. Let $p_k = P(X = k)$ for $k = 1, 2, \\ldots$. Show that $p_k = p^{k - 1}q$ where $q = 1 - p$. Show that $\\sum_{k} p_k = 1$. Show that $\\text{E}(X) = 1/q$. What is the expected number of tosses of a coin required to obtain the first tail?\n\n\\nline\n\\textbf{Ch 6.1 Q17} Since $X$ is the first time that a failure occurs in an infinite sequence, let's assume that the failure occurs after $k$ trials. Then, the probability of a failure is $k-1$ multiplications of $p$ and $1$ multiplication of $1-p$. This is valid because $k-1+1=k$ for our $k$ trials. Thus, for $q=1-p$, we can formally see that\n\n\\begin{equation*}\np_k = p^{k-1} (1-p) = p^{k-1} q.\n\\end{equation*}\n\n\\nline\nThen, we can show $\\sum_{k} p_k = 1$. Since $k=1,2, \\ldots$, the limits of the summation are originally $k=1$ to $\\infty$.\n\n\\begin{align*}\n\\sum_{k} p_k &= \\sum_{k=1}^{\\infty} p^{k-1} q \\\\\n&= q \\sum_{k=0}^{\\infty} p^{k} \\\\\n&= q \\Big ( \\frac{1}{1-p} \\Big ) \\\\\n&= q \\frac{1}{q} \\\\\n&= 1.\n\\end{align*}\n\n\\nline\nIn this case, $p_k = p^{k-1} q$ for $k = 1,2,\\ldots$. So,\n\n\\begin{align*}\n\\text{E}(X) &= 1q + 2pq + 3p^2q + \\ldots \\\\\n&= q (1 + 2p + 3p^2 + \\ldots).\n\\end{align*}\n\n\\nline\nThen, we can see that the infinite series $1 + p + p^2 + \\ldots$, which is equivalent to $\\frac{1}{1-p}$, is embedded in the previous equation. By differentiating this, we get $1 + 2p + 3p^2 + \\ldots$, which is the term we have above and is equivalent to $\\frac{1}{(1-p)^2}$. By substituting into the expanded equation of $\\text{E}(X)$, we get \n\n\\begin{align*}\n\\text{E}(X) &= q (1 + 2p + 3p^2 + \\ldots) \\\\\n&= q \\frac{1}{(1-p)^2} \\\\\n&= q \\Big ( \\frac{1}{q^2} \\Big ) \\\\\n&= \\frac{1}{q}.\n\\end{align*}\n\n\\nline\nThus, the expected value of the first tail of a fair coin is $\\frac{1}{q} = \\frac{1}{\\frac{1}{2}} = 2$. \n\n% Exactly one of six similar keys opens a certain door. If you try the keys, one after another, what is the expected number of keys that you will have to try before success?\n\n\\nline\n\\textbf{Ch 6.1 Q18} To compute the expected value, $\\text{E}(X)$ for some random variable $X$, we can use the definition of $\\text{E}(X)$ using a sum and a distribution function,\n\n\\begin{align*}\n\\text{E}(X) &= \\sum_{k=1}^{6} k \\cdot \\frac{1}{6} \\\\\n&= \\frac{1}{6} + \\frac{2}{6} + \\frac{3}{6} + \\frac{4}{6} + \\frac{5}{6} + \\frac{6}{6} \\\\\n&= \\frac{7}{2}.\n\\end{align*}\n\n% Let $X$ be a random variable which is Poisson distributed with parameter $\\lambda$. Show that E$(X) = \\lambda$. Hint: Recall that\n\n% \\begin{equation*}\n% e^{x} = 1 + x + \\frac{x^2}{2!} + \\frac{x^3}{3!} + \\cdots .\n% \\end{equation*}\n\n\\nline\n\\textbf{Ch 6.1 Q21} To compute the expected value of $X$, we need to note that the distribution function of $X$ is $p(x) = \\frac{ e^{- \\lambda} \\lambda^x }{x!}$. Then, \n\n\\begin{align*}\n\\text{E}(X) = \\sum_{x=0}^{\\infty} x \\cdot p(x) &= \\sum_{x=1}^{\\infty} x \\cdot \\frac{ e^{- \\lambda} \\lambda^x }{x!} \\\\\n&= \\sum_{x=1}^{\\infty} \\frac{ e^{- \\lambda} \\lambda^x }{(x-1)!} \\\\ \n&= e^{- \\lambda} \\sum_{x=1}^{\\infty} \\frac{ \\lambda^x }{(x-1)!} \\\\ \n\\end{align*}\n\n\\nline\nNow, we see that the denominator yields $0$ when $x=1$, so we can set $k=x-1$ and adjust the summation, \n\n\\begin{align*}\n\\text{E}(X) &= \\lambda e^{- \\lambda} \\sum_{k=1}^{\\infty} \\frac{ \\lambda^k }{k!} \\\\ \n\\end{align*}\n\n\\nline\nNow, we see that $\\sum_{k=1}^{\\infty} \\frac{ \\lambda^k }{k!} = 1 + x + \\frac{\\lambda^2}{2!} + \\frac{\\lambda^3}{3!} + \\cdots$, which is equivalent to $e^{\\lambda}$. By substituting into $\\text{E}(X)$, we get \n\n\\begin{equation*}\n\\text{E}(X) = \\lambda e^{- \\lambda} e^{\\lambda} = \\lambda.\n\\end{equation*}\n\n\\subsection*{Section 6.2}\n\n\\nline\n\\textbf{Ch 6.2 Q2} A random variable $X$ has the distribution\n\n\\begin{equation*}\np_x = {0 \\qquad 1 \\qquad 2 \\qquad 4 \\choose 1/3 \\quad 1/3 \\quad 1/6 \\quad 1/6}\n\\end{equation*}\n\n\\nline\nFind the expected value, variance, and standard deviation of $X$.\n\n\\begin{align*}\n\\text{E}(X) &= 0 \\cdot \\frac{1}{3} + 1 \\cdot \\frac{1}{3} + 2 \\cdot \\frac{1}{6} + 4 \\cdot \\frac{1}{6} \\\\\n&= \\frac{1}{3} + \\frac{2}{6} + \\frac{4}{6} \\\\\n&= \\frac{4}{3}.\n\\end{align*}\n\n\\begin{align*}\n\\text{V}(X) &= \\text{E}(X^2) - \\Big ( \\frac{4}{3} \\Big )^2 \\\\\n&= \\Big [ 0^2 \\cdot \\frac{1}{3} + 1^2 \\cdot \\frac{1}{3} + 2^2 \\cdot \\frac{1}{6} + 4^2 \\cdot \\frac{1}{6} \\Big ] - \\frac{16}{9} \\\\\n&= \\Big [ \\frac{11}{3} \\Big ] - \\frac{16}{9} \\\\\n&= \\frac{17}{9}.\n\\end{align*}\n\n\\begin{align*}\n\\text{D}(X) = \\sqrt{\\text{V}(X)} = \\sqrt{\\frac{17}{9}} = \\frac{\\sqrt{17}}{3}.\n\\end{align*}\n\n\\nline\n\\textbf{Ch 6.2 Q4} $X$ is a random variable with $\\text{E}(X) = 100$ and $\\text{V}(X) = 15$. Find\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item $\\expected{X^2}$\n\n\\nline\nWe can rearrange $\\text{V}(X) = \\text{E}(X^2) - \\text{E}(X)^2$ for $\\text{E}(X^2)$, which yields $\\text{V}(X) + \\text{E}(X)^2$. Plugging in values, we see that $\\expected{X^2} = 10015$.\n\n\\item $\\expected{3X + 10}$\n\n\\nline\nWe can use the fact that $\\expected{cX} = c \\expected{X}$. So, $\\expected{3X + 10}$ is equivalent to $3 \\expected{X} + 10$. Thus, $\\expected{3X + 10} = 310$. \n\n\\item $\\expected{-X}$\n\n\\nline\nWe can use the fact that $\\expected{cX} = c \\expected{X}$. So, $\\expected{-X}$ is equivalent to $-1 \\expected{X}$. Thus, $\\expected{-X} = -100$. \n\n\\item $\\text{V}(-X)$\n\n\\nline\nWe can use the fact that $V(cX) = c^2 V(X)$. So, $\\text{V}(-X) = -1^2 \\text{V}(X)$. Thus, $\\text{V}(-X) = 15$.\n\n\\item $\\text{D}(-X)$\n\n\\nline\nWe can use the previous result that said $\\text{V}(-X) = 15$ since $\\text{D}(-X) = \\sqrt{\\text{V}(-X)}$. Thus, $\\text{D}(-X) = \\sqrt{15}$.\n\n\\end{enumerate}\n\n\\nline\n\\textbf{Ch 6.2 Q5} In a certain manufacturing process, the (Fahrenheit) temperature never varies by more than $2^{\\circ}$ from $62^{\\circ}$. The temperature is, in fact, a random variable $F$ with distribution\n\n\\begin{equation*}\nP_F = { 60 \\qquad 61 \\qquad 62 \\qquad 63 \\qquad 64 \\choose 1/10 \\quad 2/10 \\quad 4/10 \\quad 2/10 \\quad 1/10}.\n\\end{equation*}\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item Find $\\expected{F}$ and $\\text{V}(F)$.\n\n\\begin{equation*}\n\\expected{F} = 60 \\cdot \\frac{1}{10} + 61 \\cdot \\frac{2}{10} + 62 \\cdot \\frac{4}{10} + 63 \\cdot \\frac{2}{10} + 64 \\cdot \\frac{1}{10} = 62.\n\\end{equation*}\n\n\\begin{equation*}\n\\text{V}(F) = \\Big [ 60^2 \\cdot \\frac{1}{10} + 61^2 \\cdot \\frac{2}{10} + 62^2 \\cdot \\frac{4}{10} + 63^2 \\cdot \\frac{2}{10} + 64^2 \\cdot \\frac{1}{10} \\Big ] - 62^2 = 1.2.\n\\end{equation*}\n\n\\item Define $T = F - 62$. Find $\\expected{T}$ and $\\text{V}(T)$, and compare these answers with those in part $(a)$.\n\n\\nline\nWe know that $\\expected{X + a} = \\expected{X} + a$, so $\\expected{T} = \\expected{F} - 62$. Thus, $\\expected{T} = 0$. Additionally, we know that $\\text{V}(X + a) = \\text{V}(X)$, so $\\text{V}(T) = \\text{V}(F)$. Thus, $\\text{V}(T) = 1.2$.\n\n\n\\item It is decided to report the temperature readings on a Celsius scale, that\nis, $C = (5/9) (F - 32)$. What is the expected value and variance for the\nreadings now? \n\n\\nline\nWe know that $\\expected{X + a} = \\expected{X} + a$, so $\\expected{C} = \\frac{5}{9} \\Big ( \\expected{F} - 32 \\Big )$. Thus, $\\expected{C} = \\frac{50}{3}$. Additionally, we know that $\\text{V}(cX + a) = c^2 \\text{V}(X)$, so $\\text{V}(C) = \\frac{5}{9}^2 \\cdot \\text{V}(F)$. Thus, $\\text{V}(C) = \\frac{5}{9}^2 \\cdot \\frac{6}{5}= \\frac{10}{27}$.\n\\end{enumerate}\n\n\\nline\n\\textbf{Ch 6.2 Q9} A die is loaded so that the probability of a face coming up is proportional to the number on that face. The die is rolled with outcome $X$. Find $\\text{V}(X)$ and $\\text{D}(T)$.\n\n\\nline\nThe total of all faces on a die is $21$. Thus, the probability of rolling a $1$ is $1/21$, rolling a $2$ is $2/21$, and so on. Thus, we can compute expected value using the summation definition.\n\n\\begin{equation*}\n\\sum_{x=1}^{6} x \\cdot \\frac{x}{21} = \\frac{13}{3}.\n\\end{equation*}\n\n\\nline\nThen, $\\text{V}(X) = \\expected{X^2} - \\expected{X}^2$. So, \n\n\\begin{align*}\n\\text{V}(X) &= \\Big ( \\sum_{x=1}^{6} x^2 \\cdot \\frac{x}{21} \\Big ) - \\frac{13}{3}^2 \\\\\n&= 21 - \\frac{13}{3}^2 = \\frac{20}{9}.\n\\end{align*}\n\n\\nline\nLastly, $\\text{D}(X) = \\sqrt{\\text{V}(X)} = \\sqrt{ \\frac{20}{9}} = \\frac{\\sqrt{20}}{3}$.\n\n% TODO: finish this by getting help? \n\\nline\n\\textbf{Ch 6.2 Q28} In Example $5.3$, assume that the book in question has $1000$ pages. Let $X$ be the number of pages with no mistakes. Show that $\\expected{X} = 905$ and $\\text{V}(X) = 86$. Using these results, show that the probability is $\\leq .05$ that there will be more than $924$ pages without errors or fewer than $866$ pages without errors.\n\n\\nline\nThe likelihood that there are no mistakes on a page can is given by $\\frac{ e^{- 0.1} 0.1^0 }{0!} = e^{-.1} = 0.9048374$. Since the likelihood of a mistake (or lack thereof) on a page is independent of other pages, we can multiply this probability by $1000$, which represents the number of pages in the book, to find the expected value of pages without mistakes. So, $\\expected{X} = 1000 \\cdot e^{-.1} = 905$ pages.\n\n\\nline\nPer the usual definition, we know $\\text{V}(X) = \\expected{X^2} - \\expected{X}^2$. However, since we know that a mistake on the $i^{\\text{th}}$ page does not affect a mistake on the $i+1^{\\text{th}}$ page, we can reduce the equation to \n\n\\begin{align*}\n\\text{V}(X) &= \\expected{X} - \\expected{X}^2 \\\\\n&= \\Bigg [ \\frac{ e^{- 0.1} 0.1^0 }{0!} - \\Bigg ( \\frac{ e^{- 0.1} 0.1^0 }{0!} \\Bigg )^2 \\Bigg ] \\\\\n&= \\frac{ e^{- 0.1} 0.1^0 }{0!} \\Bigg ( 1- \\frac{ e^{- 0.1} 0.1^0 }{0!} \\Bigg ) \\\\\n&= e^{- 0.1} \\Bigg ( 1- e^{- 0.1} \\Bigg ) \\\\\n& \\sim 0.086107 \n\\end{align*}\n\n\\nline\nSo, for total variance see that $\\text{V}(X) = 1000 \\cdot 0.086107 = 86$.\n\n\\nline\nWe can use the cumulative distribution function for a binomial distribution to show $\\mathbb{P}(X > 924) \\leq .05$ and $\\mathbb{P}(X < 866) \\leq .05$. We see that $F(924; 0.9048374, 1000) \\sim 0.985$, so the probability of more than $924$ pages is approximately $.015$. Additionally, $F(866; 0.9048374, 1000) \\sim 4.209053 \\times 10^{-5}$, and this is obviously less than $.05$.\n\n%http://www.chegg.com/homework-help/questions-and-answers/book-200-pages-number-mistakes-page-apoisson-random-variable-mean-001-independent-thenumbe-q228411\n\\nline\n\\textbf{Ch 6.2 Q29} Let $X$ be Poisson distributed with parameter $\\lambda$. Show that $\\text{V}(X) = \\lambda$. \n\n\\nline\nTo find $\\text{V}(X)$ of the Poisson distributed random variable $X$ with parameter $\\lambda$, we can use the equation\n\n\\begin{equation*}\n\\text{V}(X) = \\expected{X^2} - \\expected{X}^2.\n\\end{equation*}\n\n\\nline\nTo find $\\expected{X^2}$, we need a second order value of $X$, but there is no $i,j$ where $i \\neq j$ such that $X=x_i=x_j$. So, $\\expected{X^2} = \\expected{X(X-1) + X}$.\n\n\\begin{align*}\n\\expected{X(X-1) + X} &= \\sum_{x=0}^{\\infty} x(x-1) \\frac{ e^{- \\lambda} \\lambda^x }{x!} \\\\\n&= \\sum_{x=0}^{\\infty} \\frac{ e^{- \\lambda} \\lambda^x }{(x-2)!} \\\\\n&= \\lambda^2 e^{- \\lambda} \\sum_{x=2}^{\\infty} \\frac{ \\lambda^{x-2} }{(x-2)!} \\\\\n&= \\lambda^2 e^{- \\lambda} e^{- \\lambda} = \\lambda^2\\\\\n\\end{align*}\n\n\\nline\nThus, we can substitute values and we see that $\\text{V}(X) = \\lambda^2 + \\lambda - \\lambda^2 = \\lambda$.\n\\end{document} ", "meta": {"hexsha": "302f3d9680bd69a7fd65c6330f80963b36732924", "size": 12706, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "probability-theory/homework-5/HOMEWORK5.tex", "max_stars_repo_name": "jShiohaha/math-classes", "max_stars_repo_head_hexsha": "72711363cf0b58863ffb193ee79ff40244e517eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "probability-theory/homework-5/HOMEWORK5.tex", "max_issues_repo_name": "jShiohaha/math-classes", "max_issues_repo_head_hexsha": "72711363cf0b58863ffb193ee79ff40244e517eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "probability-theory/homework-5/HOMEWORK5.tex", "max_forks_repo_name": "jShiohaha/math-classes", "max_forks_repo_head_hexsha": "72711363cf0b58863ffb193ee79ff40244e517eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.1180555556, "max_line_length": 415, "alphanum_fraction": 0.6232488588, "num_tokens": 4991, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.837619959279793, "lm_q1q2_score": 0.6465115192303388}} {"text": "\\documentclass[a4paper,11pt]{scrartcl}\n\n\\usepackage{amsmath,amssymb}\n\\usepackage[utf8]{inputenc}\n\n\\title{Informed priors and quasi-simultaneous inference for psychometric functions}\n\\author{Ingo Fründ}\n\\date{\\today}\n\n\\def\\prob{\\mathbb{P}}\n\n\\begin{document}\n\n\\maketitle\n% Your text here\nWhen fitting psychometric functions, we are often interested in summarizing the performance of observers in as few numbers as possible.\nOf particular interest is a summary of the performance that consists of a single parameter, the threshold.\nUnfortunately, comparisons of psychophysical thresholds are only meaningful if other parameters of the psychometric functions --- such as slope and lapse rate --- are constant across conditions.\nHere, I demonstrate how priors can be derived that incorporate the information from other conditions.\nIf these priors are used in bayesian inference of psychometric functions, this allows for quasi-simultaneous fits that constrain parameters to be constant across conditions.\n\n\\section{A simple example}\n\nWe start with a simple example to illustrate the basic idea behind the approach.\nAssume, we colleted data $X_i$ under two different conditions $i\\in\\{1,2\\}$.\nWe want to model these data depending on a parameter $\\theta$.\nIn principle, $\\theta$ might differ between the two conditions, but we will assume here, that $\\theta_1=\\theta_2$.\nIn this particular case, we can write down the posterior distribution of $\\theta$:\n%\n$$\n\\prob(\\theta|X_1,X_2) \\propto \\prob(X_1,X_2|\\theta) \\prob(\\theta) = \\prob(X_1|\\theta)\\prob(X_2|\\theta)\\prob(\\theta),\n$$\n%\nWhere the second equality holds under the assumption that the two conditions are independent.\n\nIn practical applications, we will typically not be able to write down $\\prob(\\theta|X_1,X_2)$ directly.\nThus, we will have to resort to sampling strategies to explore $\\prob(\\theta|X_1,X_2)$.\nThe strategy in this case is therefore to perform two sampling stages.\nIn the first sampling run, we derive proper priors for the parameter $\\theta$ and in the second sampling run, we use these priors to perform the true analysis.\n\nFor the first sampling run, we employ an improper uniform prior, that is $\\prob(\\theta)=\\kappa$ for all $\\theta$ and a fixed $\\kappa$.\nWe then use MCMC to generate a sample $(\\theta_{1i})_{i=1}^m$ of size $m$ from $\\prob(\\theta|X_1)\\propto\\prob(X_1|\\theta)\\cdot\\kappa$ and another sample\n$(\\theta_{2i})_{i=1}^m$ from $\\prob(\\theta|X_2)\\propto\\prob(X_2|\\theta)\\cdot\\kappa$.\nWe now fit a parametric model to each sample and use these model as priors for the second sampling run.\nDenote the parametric model fitted to $(\\theta_{1i})$ by $f_1$ and the parametric model fitted to $(\\theta_{2i})$ by $f_2$.\nThen approximate samples from the simultaneous model can be obtained by sampling either\n%\n$$\n\\prob(\\theta|X_1,X_2) \\propto\\prob(X_1|\\theta)f_2(\\theta),\n$$\n%\nor\n%\n$$\n\\prob(\\theta|X_1,X_2) \\propto\\prob(X_2|\\theta)f_1(\\theta).\n$$\n%\n\n\\section{Constant parameters}\n\nI will now expand the above strategy to the case in which the model has more than a single parameter.\nLet us think about a model with three parameters, $\\theta$, $\\vartheta_1$, $\\vartheta_2$.\nAs the notation already suggests, $\\theta$ shall be assumed fixed across conditions, while $\\vartheta_1$ and $\\vartheta_2$ depend on the\nexperimental condition.\nAgain, we write down the posterior distribution\n%\n\\begin{align*}\n\\prob(\\theta,\\vartheta_1,\\vartheta_2|X_1,X_2) &=\\prob(\\theta|X_1,X_2)\\prob(\\vartheta_1,\\vartheta_2|X_1,X_2)\\\\\n&= \\prob(\\theta|X_1,X_2)\\prob(\\vartheta_1|X_1)\\prob(\\vartheta_2|X_2),\n\\end{align*}\n%\nWhere the first equality requires $\\theta$ to be independent of $\\vartheta_1$ and $\\vartheta_2$.\nThat means, the parameter we want constraint to be equal across conditions necessarily has to be independent of the other parameters!\nThe second equality is valid if $\\vartheta_1$ depends only on the data $X_1$ and $\\vartheta_2$ depends only on the data $X_2$.\nThis is a rather plausible condition.\n\nWe can now use the same trick as above to derive priors for $\\theta$ by drawing samples $(\\theta_{1i})_{i=1}^m$ from $\\prob(\\theta,\\vartheta_1|X_1)\\cdot\\kappa$\nand $(\\theta_{2i})_{i=1}^m$ from $\\prob(\\theta,\\vartheta_2|X_2)\\cdot\\kappa$.\nThen, we fit the marginal distributions of $(\\theta_{1i})$ and $(\\theta_{2i})$ to obtain expressions for $f_1$ and $f_2$.\nIn a second sampling run, we can then perform quasi simultaneous sampling using\n%\n$$\n\\prob(\\theta,\\vartheta_1|X_1,X_2) \\propto\\prob(X_1|\\theta,\\vartheta_1)f_2(\\theta)\\prob(\\vartheta_1),\n$$\n%\nand\n%\n$$\n\\prob(\\theta,\\vartheta_2|X_1,X_2) \\propto\\prob(X_2|\\theta,\\vartheta_2)f_1(\\theta)\\prob(\\vartheta_2).\n$$\n%\n\n\\section{More than two conditions}\n\nThe extension of these ideas to cases with more than two conditions is straight forward:\nIn the first sampling run, we determine the prior densities $f_j, j=1,\\dots,n$ from each dataset $X_j$ in isolation.\nWe then combine the $f_j$ in order to sample\n%\n\\begin{equation}\n \\label{eq:allcombined}\n\\prob(\\theta,\\vartheta_j|X_j, j=1,\\dots,n) \\propto\\prob(X_j|\\vartheta_j,\\theta) \\prob(\\vartheta_j) \\prod_{\\underset{\\ell\\neq j}{\\ell=1}}^n f_\\ell ( \\theta ),\n\\end{equation}\n%\nin isolation for each $j=1,\\dots,n$.\nThus, the product $\\prod_{\\underset{\\ell\\neq j}{\\ell=1}}^n f_\\ell ( \\theta )$ can be taken as a prior that incorporates prior knowledge about the other psychometric functions in the design.\n\n\\section{Specific choices of fitted posteriors}\n\nHere, we will highlight reasonable choices to fit the posterior distributions and we will demonstrate, how these choices imply a closed functional form for the product prior in equation \\eqref{eq:allcombined}.\n\n\\subsection{Fitting posterior distributions for scale parameters}\n\nIn order to compare, thresholds obtained from different conditions, it is typically required that the slopes across all these conditions are equal.\nAlthough threshold comparisons are possible with unequal slopes, too, it is far more complicated to arrive at valid conclusions from these comparisons.\nIn psignifit, many parameterizations of the psychometric functions do not explicitely specify the slope of the psychometric function but instead have a parameter that is somehow related to the inverse slope.\nPosterior distributions for this parameter often resemble a Gamma distribution, i.e. the seem to have a density of the form\n%\n\\begin{equation}\n \\label{eq:GammaDensity}\n f ( x | k, \\theta ) = x^{k-1} \\frac{\\exp(-x/\\theta)}{\\Gamma(k)\\theta^k}.\n\\end{equation}\n%\nWe now assume that we have fitted this density (using for instance maximum likelihood) to a set of $n$ sampled posteriors from different conditions and we want to derive the product prior for the $n+1$-th condition.\nThe product prior has the density\n%\n\\begin{align*}\n \\bar{f}(x) &= \\prod_{j=1}^n f_j(x) = \\prod_{j=1}^n x^{k_j-1} \\frac{\\exp(-x/\\theta_j)}{\\Gamma(k_j)\\theta_j^{k_j}} \\\\\n &\\propto \\prod_{j=1}^n x^{k_j-1} \\exp(-x/\\theta_j) \\\\\n &= x^{\\sum_{j=1}^n k_j-1} \\exp(-x\\sum_{j=1}^n \\frac{1}{\\theta_j}) \\\\\n &= x^{\\bar{k}-1} \\exp(-x/\\bar{\\theta}),\n\\end{align*}\n%\nwhere we set\n%\n$$\n\\bar{k} := 1 + \\sum_{j=1}^n k_j-1,\n$$\n%\nand\n%\n$$\n\\bar{\\theta} := \\Big( \\sum_{j=1}^n \\frac{1}{\\theta_j} \\Big)^{-1}.\n$$\n%\nThus, the product prior is again a Gamma distribution with parameters $\\bar{k}$ and $\\bar{\\theta}$!\n\n\\subsection{Fitting posterior distributions for rates}\n\nThe psychometric function model includes two rate parameters, the guessing rate $\\gamma$ that describes the lower asymptote and the lapse rate $\\lambda$ that describes the upper asymptote.\nA very reasonable model for rates is the Beta distribution (cite Ferrari?).\nThe Beta distribution lives on the unit interval $(0,1)$ and takes the density\n%\n\\begin{equation}\n f ( x | \\alpha, \\beta ) = \\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)} x^{\\alpha-1} (1-x)^{\\beta-1}.\n \\label{eq:BetaDensity}\n\\end{equation}\n%\nHere, we can use the same trick as with the Gamma distribution to derive a sensible product prior:\nWe assume that we fitted $n$ sets of parameters $(\\alpha_j,\\beta_j)_{j=1}^n$ and want to combine these to obtain a prior for the $n+1$-th condition.\nThe product prior in this case is\n%\n\\begin{align*}\n \\bar{f}(x) &= \\prod_{j=1}^n f_j(x) = \\prod_{j=1}^n \\frac{\\Gamma(\\alpha_j+\\beta_j)}{\\Gamma(\\alpha_j)\\Gamma(\\beta_j)} x^{\\alpha_j-1} (1-x)^{\\beta_j-1}\\\\\n &\\propto \\prod_{j=1}^n x^{\\alpha_j-1} (1-x)^{\\beta_j-1} \\\\\n &= x^{\\sum_{j=1}^n \\alpha_j-1} (1-x)^{\\sum_{j=1}^n\\beta_j-1} \\\\\n &= x^{\\bar{\\alpha}-1} (1-x)^{\\bar{\\beta}-1},\n\\end{align*}\n%\nwhere $\\bar{\\alpha}:= 1+ \\sum_{j=1}^n\\alpha_j-1$ and $\\bar{\\beta}:=1+\\sum_{j=1}^n\\beta_j-1$.\n\nThus, the product prior in this case, is again a Beta distribution, with parameters $\\bar{\\alpha}$ and $\\bar{\\beta}$.\n\n\\subsection{Fitting posterior distributions for thresholds}\n\nIn some, very rare situations, we might also be interested in having equal posterior distributions for thresholds.\nA reasonable approximation for posterior threshold distributions is often the Normal distribution.\nInterestingly, the procedure used to derive the product prior in the preceeding paragraphs does not work well for the Normal distribution.\nAssume, we fitted the posteriors in conditions $j=1,\\dots,n$ with Normal densities $(f_j)_{j=1}^n$ with parameters $(\\mu_j,\\sigma_j)_{j=1}^n$.\nThe product is then\n%\n\\begin{align*}\n \\bar{f}(x) &= \\prod_{j=1}^n f_j(x) = \\prod_{j=1}^n \\frac{1}{\\sqrt{2\\pi}\\sigma_j} \\exp ( - \\frac{(x-\\mu_j)^2}{2\\sigma_j^2})\\\\\n &\\propto \\prod_{j=1}^n \\exp ( - \\frac{(x-\\mu_j)^2}{2\\sigma_j^2})\\\\\n &= \\exp ( - \\sum_{j=1}^n \\frac{(x-\\mu_j)^2}{2\\sigma_j^2} ) \\\\\n &= \\exp ( - \\frac{1}{2} \\big( \\sum_{j=1}^n \\frac{x^2}{\\sigma_j^2} - \\sum_{j=1}^n 2\\frac{x\\mu_j}{\\sigma_j^2} + \\sum_{j=1}^n \\frac{\\mu_j^2}{\\sigma_j^2} \\big) )\\\\\n &\\propto\\exp ( - \\frac{1}{2} \\big( \\sum_{j=1}^n \\frac{x^2}{\\sigma_j^2} - \\sum_{j=1}^n 2\\frac{x\\mu_j}{\\sigma_j^2}\\big) ) \\\\\n &= \\exp ( - \\frac{1}{2} \\big( x^2\\sum_{j=1}^n \\frac{1}{\\sigma_j^2} - 2x\\sum_{j=1}^n 2\\frac{\\mu_j}{\\sigma_j^2} \\big) ) \\\\\n &= \\exp ( - \\frac{1}{2} \\underbrace{\\sum_{j=1}^n \\frac{1}{\\sigma_j^2}}_{=:\\bar{\\sigma}^{-2}} \\big( x^2 - 2x\\underbrace{\\frac{\\sum_{j=1}^n \\frac{\\mu_j}{\\sigma_j^2}}{\\sum_{j=1}^n \\frac{1}{\\sigma_j^2}}}_{=:\\bar{\\mu}}\\big) ) \\\\\n &\\propto \\frac{1}{\\sqrt{2\\pi}\\bar{\\sigma}} \\exp( - \\frac{(x-\\bar{\\mu})^2}{2\\bar{\\sigma}^2}).\n\\end{align*}\n%\nThus, comparing coefficients with the Normal density, we end up with a new Normal distribution with parameters\n%\n$$\n\\bar{\\sigma}^2 := \\big(\\sum_{j=1}^n \\frac{1}{\\sigma_j^2}\\big)^{-1},\n$$\n%\n$$\n\\bar{\\mu} := \\frac{\\sum_{j=1}^n\\frac{\\mu_j}{\\sigma_j^2}}{\\sum_{j=1}^n\\frac{1}{\\sigma_j^2}}.\n$$\n%\n\\end{document}\n", "meta": {"hexsha": "c8f672314465d5665836ffbddb036a5fb8849cdc", "size": 10648, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documents/simultaneous.tex", "max_stars_repo_name": "esc/Psignifit-3.x", "max_stars_repo_head_hexsha": "2a2a8a40f39d94f8fee99cb0ba4b9595ae216f88", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2015-07-14T23:38:52.000Z", "max_stars_repo_stars_event_max_datetime": "2018-10-12T08:15:36.000Z", "max_issues_repo_path": "documents/simultaneous.tex", "max_issues_repo_name": "esc/Psignifit-3.x", "max_issues_repo_head_hexsha": "2a2a8a40f39d94f8fee99cb0ba4b9595ae216f88", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "documents/simultaneous.tex", "max_forks_repo_name": "esc/Psignifit-3.x", "max_forks_repo_head_hexsha": "2a2a8a40f39d94f8fee99cb0ba4b9595ae216f88", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-11-25T10:11:04.000Z", "max_forks_repo_forks_event_max_datetime": "2018-10-12T08:17:48.000Z", "avg_line_length": 52.4532019704, "max_line_length": 227, "alphanum_fraction": 0.7114012021, "num_tokens": 3392, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637577007393, "lm_q2_score": 0.7520125793176222, "lm_q1q2_score": 0.6464779597744124}} {"text": "\n\\subsection{The dual space forms a vector space}\n\nThe dual space forms a vector space. We can define addition and scalar multiplication on members of the dual space.\n\nThe dimension of the dual space is the same as the underlying space.\n\nWe have defined the dual space. A vector in dual space will have also have components and a basis.\n\n\\(\\mathbf w=\\sum_i w_i f^j\\)\n\nSo how we describe the components will depend on the choice of basis.\n\nWe choose the dual basis, the basis for \\(V^*\\) as:\n\n\\(\\mathbf e_i \\mathbf f^j =\\delta_i^j\\)\n\nIf the basis changes, so does the dual basis.\n\nWe write the dual basis as \\(e^j\\)\n\n", "meta": {"hexsha": "f785564565fdecfdadbf8c0d4656ed27354f9517", "size": 616, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/geometry/forms/01-03-dualSpaceBasis.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/geometry/forms/01-03-dualSpaceBasis.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/geometry/forms/01-03-dualSpaceBasis.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.0, "max_line_length": 115, "alphanum_fraction": 0.7402597403, "num_tokens": 157, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8596637577007393, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6464779549965517}} {"text": "\\chapter{Constructive Heuristics}\n\n\\section{Greedy}\\label{sec:greedy}\nGreedy heuristic for TSP start by choosing a random element and add always the closer node, one node at a time, making the local optimal choice. This create good path in the first part of the tour, however the last added edges create different crosses. Therefore this heuristic can be applied in combination with refining heuristics such as: \\texttt{local\\_branching}, \\texttt{hard\\_fixing}, \\texttt{best\\_two\\_opt} and other ones that try to improve a tour.\\\\\nThe algorithm computational time is $ O(n^2) $ where $ n = |V| $, therefore is pretty fast, however depending on the first selected node, the cost of the tour can differ more than 10 \\% one another. The implemented algorithm (from now only \\texttt{Greedy} or \\texttt{heuristic\\_greedy}), at cost of execution time, perform the described greedy using all the nodes as starting node, returning the shortest tour. The time complexity for the implemented algorithm raise to $ O(n^3) $.\\\\\nIn fig \\ref{fig:a280_10_} an example of tour calculated with the implemented algorithm.\n\n\\begin{figure}[h]\n\t\\centering\n\t\\centering\n\t\\includegraphics[width=0.7\\columnwidth]{../res/a280_10.png}\n\t\\caption{The shortest greedy tours.}\n\t\\label{fig:a280_10_}\n\\end{figure}\n\n\n\n\\subsection{Greedy with CGAL Library}\nTo take a look at the potential of the CGAL library, we decided to implement the greedy algorithm. It follows the same idea as the one previously described (Section \\ref{sec:greedy}) but using completely different data structures. Thanks to the potential of the C++ code and therefore the use of objects, the CGAL library offers a very wide range of elements.\\\\ \nThe installation of CGAL within the Windows environment with Visual Studio 2019 is reported in the appendix \\ref{sec:cagalOne}.\\\\\nFirst of all, the objects used for this algorithm are briefly explained:\n\\begin{enumerate}\n\\item \\texttt{CGAL::Simple\\_cartesian K}: The layer of geometry kernels provides basic geometric entities of constant size and primitive operations on them. Each entity is provided as both a stand-alone class, which is parameterized by a kernel class, and as a type in the kernel class. CGAL provides different kernels, they can differ by internal representation of objects (e.g. cartesian versus homogeneous) or provide different functionalities. When creating a new package, the authors have to specify clearly the requirements needed by the kernel used and may specify a targeted kernel in the list of predefined kernels. In this project \\texttt{Exact\\_predicates\\_inexact\\_constructions\\_kernel} is used. Thus, a \\texttt{Simple\\_cartesian} object is chosen to represent a model for a kernel using Cartesian coordinates to represent the geometric objects.\n\\item \\texttt{CGAL::Search\\_traits\\_2 TreeTraits}: The \\texttt{Simple\\_cartesian} allows to define a dD Spatial Searching which is, in this kind of problems, a 2D space.\n\\item \\texttt{CGAL::Orthogonal\\_k\\_nemap.com\\_search Neighbor\\_search}: Thanks to the aforementioned objects, with this one a neighborhood search is created, based on the Euclidean distance in a Cartesian plane.\n\\item \\texttt{Neighbor\\_search::Tree Tree}: In our implementation was decided to use a Tree representation in which to apply the \\texttt{Neighbor\\_search}.\n\\end{enumerate}\nOnce the model and structure have been defined, the search is simple. Each node is created via the \\texttt{Point\\_2} object of the kernel defined above. The various points are assigned to the \\texttt{Tree}. As a last step just pass: the \\texttt{Tree}, the starting node and the number of total nodes to the \\texttt{Neighbor\\_search} and it return an list of nodes in ascending order of distance from the starting one.\\\\\nBy repeating this procedure for each node, a matrix of nodes sorted by distance from the starting point is obtained and with an ad hoc searching function:\n\n\\begin{lstlisting}\nvector::iterator it = find_if(neigh_sol_idx[succ].begin(), neigh_sol_idx[succ].end(),\n\t\t\t\t\t [&](int neig) {\n\treturn find(var->sol[i].begin(), var->sol[i].end(), neig) == var->sol[i].end();\n});\n\\end{lstlisting}\n\nthe resulting matrix will have a number of lines equal to the number of nodes and each of them will be a tour built starting from a node and always choosing the closest from the current node, exactly as \\texttt{Greedy}. \\\\\nThe algorithm, as shown in fig. \\ref{fig:Lconstructives_time} , takes longer to run, as it uses higher level objects and methods than \\texttt {Greedy}. In fact, its potential lies in its application in complex problems, in which these data structures offer powerful and effective tools for geometric problems.\n\n\\section{Greedy Randomize Adaptive Search Path (GRASP)}\\label{sec:grasp}\nGRASP heuristic, as the name anticipate, is a Greedy constructive heuristic with some randomization. Indeed, for the TSP problem, instead of bringing the closer isolated node as next, it's chosen a random one between the first $ n $ closest isolated nodes, where $ n = 3 $ in the tested implementation.\nThe idea is that in TSP it's not always the closer to be the next node on the optimum tour. Moreover, in other metaheuristics that start from a solution and optimize it, a percentage of randomness could appreciated.\n\n\\begin{figure}[!h]\n\t\\centering\n\t\\begin{subfigure}{.75\\textwidth}\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lgrasp_greedy_time.png}\n\t\t\\caption{Solution time domain.}\n\t\t\\label{fig:Lgrasp_greedy_time}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.75\\textwidth}\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lgrasp_greedy_lb.png}\n\t\t\\caption{Solution cost domain.}\n\t\t\\label{fig:Lgrasp_greedy_lb}\n\t\\end{subfigure}\n\\caption{The first parameter in the name of \\texttt{n\\_greedy} and \\texttt{n\\_grasp} is the number of different tour evaluated. In \\texttt{n\\_grasp} name the other parameter represent the probability to chose in order the first, the second and the third closer node. }\n\\label{fig:Lgrasp_greedy}\n\\end{figure}\nIn fig. \\ref{fig:Lgrasp_greedy} the result of the test with different $ n $ values for \\texttt{n\\_greedy} and \\texttt{n\\_grasp} and for different probability of choice of the next node in \\texttt{n\\_grasp}. There can be see the differences in execution time of \\texttt{heuristic\\_greedy} from the other and the gradual decrease of the cost of the \\texttt{n\\_grasp} by increasing the probability to select the closer node until \\texttt{n\\_greedy\\_10} \n\nIn figure \\ref{fig:att48_diff} can be compared of the optimum tour, \\texttt{Greedy}, 3 different instances of \\texttt{GRASP} (\\texttt{n\\_grasp\\_1 \\\\ \\_33\\_33\\_33} version) and \\texttt{heuristic\\_insertion}. It's nice to see that the \\texttt{Greedy} (fig \\ref{fig:att48_GREEDY}) is pretty close to the optimal solution and also some instance of \\texttt{GRASP} like \\ref{fig:att48_GRASP2}. \\texttt{heuristic\\_insertion} in this example has no cross edges and as will be show in fig. \\ref{fig:Lgrasp_insertion_refining_LA_lb} the refining phase of this solutions does not improve so much the tour cost.\n\\begin{figure}[!h]\n\t\\begin{subfigure}{.49\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=\\columnwidth]{../res/att48_0.png}\n\t\t\\caption{Optimal subtour: cost=10628, time=0.76s}\n\t\t\\label{fig:att48_best}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.49\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=\\columnwidth]{../res/att48_10.png}\n\t\t\\caption{\\texttt{Greedy}: cost=12012, time=0.005s}\n\t\t\\label{fig:att48_GREEDY}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.49\\textwidth}\n\t\\centering\n\t\\includegraphics[width=\\columnwidth]{../res/att48_12_1.png}\n\t\\caption{\\texttt{GRASP}: cost=21824, time=0.001s}\n\t\\label{fig:att48_GRASP1}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.49\\textwidth}\n\t\\centering\n\t\\includegraphics[width=\\columnwidth]{../res/att48_12_2.png}\n\t\\caption{\\texttt{GRASP}: cost=12576, time=0.001s}\n\t\\label{fig:att48_GRASP2}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.49\\textwidth}\n\t\\centering\n\t\\includegraphics[width=\\columnwidth]{../res/att48_12_3.png}\n\t\\caption{\\texttt{GRASP}: cost=22179, time=0.001s}\n\t\\label{fig:att48_GRASP3}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.49\\textwidth}\n\t\\centering\n\t\\includegraphics[width=\\columnwidth]{../res/att48_13.png}\n\t\\caption{\\texttt{heuristic\\_insertion}: cost=11197, time=0.0028s}\n\t\\label{fig:att48_insertion}\n\t\\end{subfigure}\n\t\\caption{Differences of \\texttt{Greedy} and \\texttt{GRASP} tour for \\textit{att48.tsp} instace.}\n\t\\label{fig:att48_diff}\n\\end{figure}\n\n\n\\section{Insertion Heuristic}\nThe idea of this constructive heuristic is to create an initial simple tour of 3 nodes randomly and insert one node at a time, breaking the edge that minimize the final tour cost.\\\\\nThe complexity of the algorithm is $ O(n^2) $.\n\n% versione 2\n\\textbf{Construction heuristic comparison.} The performance profile in fig \\ref{fig:pp_Lconstructives} show the three constructive presented in this chapter. In particular \\texttt{heristic\\_insertion} is comparable with \\texttt{n\\_greedy} and \\texttt{n\\_grasp} in execution time, however it find shorter tours of all the other constructive.\\\\\nOther consideration about the constructive heuristics in chapter \\ref{chapter:refining}.\n\n\n\\begin{figure}[!h]\n\t\\centering\n\t\\begin{subfigure}{.9\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lconstructives_LA_time.png}\n\t\t\\caption{Solution time domain.}\n\t\t\\label{fig:Lconstructives_time}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.9\\textwidth}\n\t\\centering\n\t\\includegraphics[width=\\columnwidth]{../res/Lconstructives_LA_lb.png}\n\t\\caption{Solution cost domain.}\n\t\\label{fig:Lconstructives_lb}\n\t\\end{subfigure}\n\t\\caption{Performance profile of constructive heuristic.}\n\t\\label{fig:pp_Lconstructives}\n\\end{figure}\n", "meta": {"hexsha": "5709e29ec6edd160c43eb8846bee8e3a0994d896", "size": 9672, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Latex/constructive-heuristics.tex", "max_stars_repo_name": "Fisher4537/OR2", "max_stars_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Latex/constructive-heuristics.tex", "max_issues_repo_name": "Fisher4537/OR2", "max_issues_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Latex/constructive-heuristics.tex", "max_forks_repo_name": "Fisher4537/OR2", "max_forks_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.8320610687, "max_line_length": 867, "alphanum_fraction": 0.7754342432, "num_tokens": 2656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.8175744828610095, "lm_q1q2_score": 0.6463174164723084}} {"text": "\\documentclass[../main.tex]{subfiles}\n\n\\begin{document}\n\\chapter{Combinatorial Search}\nSo far, we have learned the most fundamental search strategies on general data structures such as array, linked list, graph, and tree. In this chapter, instead of searching on explicit and well defined data structures, we extend and discuss more \\textit{exhaustive search} algorithms that can solve rather obscure and challenging \\textit{combinatorial problems}, such as sudoku and the famous Travels Salesman Problem. For combinatorial problems, we have to figure out the potential search space, and rummage a solution. \n\\section{Introduction}\nCombinatorial search problems consists of $n$ items and a requirement to find a solution, i.e., a set of $L < N$ items that satisfy specified conditions or constraints. For example, a sudoku problem where a $9\\times 9$ grid is partially filled with number between 1 and 9, fill the empty spots with numbers that satisfy the following conditions:\n\\begin{figure}[!ht]\n \\centering\n \\includegraphics[width= 0.35\\columnwidth]{fig/250px-Sudoku-by-L2G-20050714.png}\n \\includegraphics[width= 0.35\\columnwidth]{fig/250px-Sudoku-by-L2G-20050714_solution.png}\n \\caption{A Sudoku puzzle and its solution}\n \\label{fig:backtrack_puzzle_1}\n\\end{figure}\n\\begin{enumerate}\n \\item Each row has all numbers form 1 to 9.\n \\item Each column has all numbers form 1 to 9.\n \\item Each sub-grid ($3 \\times3$) has all numbers form 1 to 9.\n\\end{enumerate}\nThis sudoku together with one possible solution is shown in Fig.~\\ref{fig:backtrack_puzzle_1}. In this case, we have $81$ items, and we are required to fill 51 empty items with the above three constraints. \n\\paragraph{Model Combinatorial Search Problems} \nWe can model the combinatorial search solution as a vector $s = (s_0, s_1, ..., s_{L-1})$, where each variable $s_i$ is selected from a finite set $A$, which is called the \\textit{domain} for each variable. Such a vector might represent an arrangement where $s_i$ contains the i-th item of a permutation, in the combination problem, a boolean denotes if the i-th item is selected already, or it can represent a path in a graph or a sequence of moves in a game. In the sudoku problem, each $s_i$ can choose from a number in range $[1, 9]$.\n\n\n\\paragraph{Problem Categories} Combinatorial search problems arise in many areas of computer science such as artificial intelligence, operations search, bioinformatics, and electronic commerce. These problems typically involve finding a \\textit{grouping}, \\textit{ordering}, or \\textit{assignment} of a discrete, finite set of objects that satisfy given conditions or constraints. We introduce two well-studied types of problems that are more likely to be NP-hard and of at least exponential complexity:\n% in computer science mainly studies algorithms that solve exponential or even NP-hard problems, such as:\n\\begin{enumerate}\n \\item Constraint Satisfaction Problems (CSP) are mathematical questions defined as a set of variables whose state must satisfy a number of constraints or limitations(mathematical equations or inequations), such as Sudoku, N-queen, map coloring, Crosswords, and so on. The size of the search space of CSPs can be roughly given as:\n \\begin{align}\n O(cd^L)\n \\end{align}\n Where there are $L$ variables, each with domain size $d$, and there are $c$ constraints to check out. \n \\item Combinatorial optimization problems consist of searching for maxima or minima of an objective function $F$ whose domain is a discrete but large configuration space. Some classic examples are:\n \\begin{itemize}\n \\item Travelling Salesman Problems (TSP): given position $(x, y)$ of $n$ different cities, find the shortest possible path that visits each city exactly once.\n \\item Integer Linear Programming: maximize a specified linear combination of a set of integers $X_1, .., X_{n}$ subject to a set of linear constraints each of the form:\n \\begin{align}\n a_1X_1 + ... + a_{n}X_{n} \\leq c\n \\end{align}\n \\item Knapsack Problems: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.\n \\end{itemize}\n\\end{enumerate}\n\n\n\\paragraph{Search Strategies} From Chapter Discreet Programming, we have learned the basic enumerative combinatorics, including counting principles and knowledge on permutations, combinations, partitions, subsets, and subsequences. Combinatorial Search builds atop this subject, and together through different search strategies such as depth-first search and best-first search, it is able to enumerate the search space and find the solution(s) with necessary speedup methods. In this chapter, we only discuss about complete search and only acknowledge the existence of approximate search techniques. \n\n\\textit{Backtracking} is a process of depth-first based search where it ``builds'' the search tree on the fly incrementally instead of having a tree/graph structure beforehand to search through. Backtracking fits to solve combinatorial search problems because:\n\\begin{enumerate}\n \\item It is space efficient for the usage of a DFS and the candidates are built incrementally and their validity to fit a solution is checked right away.\n \\item It is time efficient for that some partial candidates can be pruned if the algorithm believes that it will not lead to our final complete solution. \n\\end{enumerate}\n\n Because the ordering of variables $s_0, ..., s_{L-1}$ can potentially affect the size of the search space sometimes. Thus, backtracking search relies on one or more heuristics to select which variable to consider next. \\textit{Look-ahead} is one such heuristic that is preferably applied to check the effects of choosing a given variable to evaluate or to decide the order of values to give to it. \n \n There are other Breath-first Search based strategies that might work better than backtracking, such as for combinatorial optimization problems, best-first branch and bound search might be more efficient than its depth-first counterpart. \n\n\\paragraph{Speedups}\n\nThe speedup methods are well studied in computer science, and we list two general ways to prune unqualified or unnecessary branches during the search of backtracking:\n\\begin{enumerate}\n \\item Branch and Prune: This method prunes the unqualified branches with constraints of the problems. This is usually applied to solve constraint restricted problems (CSPs).\n \\item Branch and Bound: This method prunes unnecessary branches via comparing an estimation of a partial candidate with a found global best solution. If the estimation states that the partial candidate will never lead us to a better solution, we cut off this branch. This technique can be applied to solve a general optimization problems, such as Travel Salesman Problems (TSP), knapsack problems, and so.\n\\end{enumerate}\n% To notice that these speedup techniques are general for searching, not specifically for Backtracking. For example, branch and bound often be linked to best-first search. \n\n\\section{Backtracking}\nIn this section, we first introduce the technique of backtracking, and then demonstrate it by implementing common enumerative combinatorics seen in Chapter Discreet Programming. \n\\subsection{Introduction}\nBacktracking search is an exhaustive search algorithm(depth-first search) that systematically assigns all possible combinations of values to the variables and checks if these assignments constitute a solution. Backtracking is all about choices and consequences and it shows the following two properties:\n \\begin{enumerate}\n \\item \\textbf{No Repetition and Completion:} It is a systematic generating method that enumerates all possible states exactly at most once: it will not miss any valid solution but avoids repetitions. If there exists ``correct'' solution(s), it is guaranteed to be found. This property makes it ideal for solving combinatorial problems where the search space has to be constructed and enumerated. Therefore, the worst-case running time of backtracking search is exponential with the length of the state ($b^L$, $b$ is the average choice for each variable in the state).% such as combination and permutation which requires us to enumerate all possible solutions. We focus on demonstrating this property in this section. \n \\item \\textbf{Search Pruning:} Along the way of working with partial solutions, in some cases, it is possible for us to decide if they will lead to a valid \\textit{complete solution}. As soon as the algorithm is confident to say the partial configuration is either invalid or nonoptimal, it abandons this \\textit{partial candidate}, an then ``backtracks'' (return to the upper level), and resets to the upper level's state so that the search process can continue to explore the next branch for the sake of efficiency. This is called \\textit{search pruning} with which the algorithm ends up amortizely visiting each vertex less than once. This property makes backtracking the most promising way to solve CSPs and combinatorial optimization problems. \n \\end{enumerate}\n\nSolving sudoku problem with backtracking algorithm, each time at a level in the DFS, it tries to extend the last partial solution $s = (s_0, s_1, ..., s_{k})$ by trying out all 9 numbers at $s_{k+1}$, say we choose 1 at this step. It testifies the partial solution with the desired solution:\n\\begin{enumerate}\n \\item If the partial solution $s = (s_0, s_1, ..., s_{k}, 1)$ is still valid, move on to the next level and work on trying out $s_{k+2}$.\n \\item If the partial solution is invalid and is impossible to lead to a complete solution, it ``backtracks'' to the last level and resets the state as $s = (s_0, s_1, ..., s_{k})$ so that it can try our other choices if there are some left(which in our example, we will try $s_{k+1}=2$) or keep ``backtracking'' to even upper level.\n\\end{enumerate}\nThe process should be way clearer once we have learned the examples in the following subsections. \n\n\\subsection{Permutations}\nGiven a list of items, generate all possible permutations of these items. If the set has duplicated items, only enumerate all unique permutations.\n\\subsubsection{No Duplicates(L46. Permutations)} When there are no duplicates, from Chapter Discreet Programming, we know the number of all permutations are:\n\\begin{equation}\n p(n, m) = \\frac{n!}{(n-m)!}\n\\end{equation}\nwhere $m$ is the number of items we choose from the total $n$ items to make the permutations. \n\\begin{lstlisting}[numbers=none]\nFor example:\na = [1, 2, 3]\nThere are 6 total permutations:\n[1, 2, 3], [1, 3, 2],\n[2, 1, 3], [2, 3, 1],\n[3, 1, 2], [3, 2, 1]\n\\end{lstlisting}\n\\paragraph{Analysis} Let us apply the philosophy of backtracking technique. We have to build a state with length $3$, and each variable in the state has three choices: 1, 2, and 3. The constraint here comes from permutation which requires that no two variables in the state will be having the same value. To build this incrementally with backtracking, we state with an empty state \\texttt{[]}. At first, we have three options, we get three partial results $[1], [2]$, and $[3]$. Next, we handle the second variable in the state: for $[1]$, we can choose either 2 or 3,getting $[1,2]$ and $[1,3]$; same for $[2]$, where we end up with $[2,1]$ and $[2,3]$; for $[3]$, we have $[3,1]$ and $[3,2]$. At last, each partial result has only one option, we get all permutations as shown in the example. We visualize this incrementally building candidates in Fig.~\\ref{fig:backtrack_permutation}. \n\\begin{figure}[h]\n \\centering\n \\includegraphics[width= 0.8\\columnwidth]{fig/permutation.png}\n \\caption{The search tree of permutation}\n \\label{fig:backtrack_permutation}\n\\end{figure}\n\nHowever, we only managed to enumerate the search space, but not systematically or recursively with the Depth-first search process. With DFS, we depict the traverse order of the vertexes in the virtual search space with red arrows in Fig.~\\ref{fig:backtrack_permutation}. The backward arrows mark the ``backtracking'' process, where we have to reset the state to the upper level. \n\n\\paragraph{Implementation} We use a list of boolean \\texttt{bUsed} to track which item is used in the search process. \n\\texttt{n} is the total number of items, \\texttt{d} is the depth of the depth-first search process, \\texttt{curr} is the current state, and \\texttt{ans} is to save all permutations. The following code, we generate $p(n, m)$\n\\begin{lstlisting}[language=Python]\ndef p_n_m(a, n, m, d, used, curr, ans):\n if d == m: #end condition\n ans.append(curr[::]) \n return\n \n for i in range(n):\n if not used[i]:\n # generate the next solution from curr\n curr.append(a[i])\n used[i] = True\n print(curr)\n # move to the next solution\n p_n_m(a, n, m, d + 1, used, curr, ans)\n #backtrack to previous partial state\n curr.pop()\n used[i] = False\n return\n\\end{lstlisting}\nCheck out the running process in the source code.\n\n\\paragraph{Alternative: Swapping Method} \n\\begin{figure}[h]\n \\centering\n \\includegraphics[width= 0.8\\columnwidth]{fig/permutation_swap.png}\n \\caption{The search tree of permutation by swapping. The indexes of items to be swapped are represented as a two element tuple. }\n \\label{fig:backtrack_permutation_swap}\n\\end{figure}\nWe first start with a complete state, such that $s=[1, 2, 3]$ in our case. By swapping $1$ and $2$, we get $[2, 1, 3]$ and $[2, 3, 1]$ can be obtained by swapping $1$ and $3$ on top of $[2, 1, 3]$. With all permutations as leaves in the search space, the generating process is similar to Fig.~\\ref{fig:backtrack_permutation}. We show this alternative process in Fig.~\\ref{fig:backtrack_permutation_swap}. At first, we swap index 0 with all other indexes, including 0, 1, and 2. At the second layer, we move on to swap index 1 with all other successive indexes, and so on for all other layers. The Python code is as:\n\\begin{lstlisting}[language=Python]\nans = []\ndef permutate(a, d):\n global ans\n if d == len(a):\n ans.append(a[::])\n for i in range(d, len(a)):\n a[i], a[d] = a[d], a[i]\n permutate(a, d+1)\n a[i], a[d] = a[d], a[i]\n return\n\\end{lstlisting}\nThere is Johnson-Trotter algorithm that utilizes such swapping method, which avoids recursion, and instead computes the permutations by an iterative method. \n\\subsubsection{With Duplicates(47. Permutations II)}\nWe have already know that $p(n, n)$ is further decided by the duplicates within the $n$ items. Assume we have in total of $d$ items are repeated, and each item is repeated $x_i$ times, then the number of all arrangements $pd(n, n)$ are:\n\\begin{align}\n pd(n, n) &= \\frac{p(n, n)}{x_0!x_1!...x_{d-1}}, \\\\\n \\texttt{w.r.t } & \\sum_{i=0}^{d-1} x_i \\leq n\n\\end{align}\nFor example, when $a = [1, 2, 2, 3]$, there are $\\frac{4!}{2!}$ unique permutations, which is 12 in total, and are listed as bellow:\n\\begin{lstlisting}[numbers=none]\n[1, 2, 2, 3], [1, 2, 3, 2], [1, 3, 2, 2],\n[2, 1, 2, 3], [2, 1, 3, 2], [2, 2, 1, 3],\n[2, 2, 3, 1], [2, 3, 1, 2], [2, 3, 2, 1],\n[3, 1, 2, 2], [3, 2, 1, 2], [3, 2, 2, 1]\n\\end{lstlisting}\n\\paragraph{Analysis} \n\\begin{figure}[h]\n \\centering\n \\includegraphics[width= 1.2\\columnwidth]{fig/permutation_repeat.png}\n \\caption{The search tree of permutation with repetition}\n \\label{fig:backtrack_permutation_duplicates}\n\\end{figure}\nThe enumeration of these all possible permutations can be obtained with backtracking exactly the same as if there are no duplicates. However, this is not efficient since it has doubled the search space with repeated permutations. Here comes to our first time applying the Branch and Prune method: we avoid repetition by pruning off redundant branches. \n\nOne main advantage of backtracking is not to save all intermediate states, thus we should find a mechanism that avoids generating these intermediate states at the first place. One solution is that we sort all $n$ items, making all repeat items adjacent to each other. We know if the current intermediate state is redundant by simply comparing this item with its predecessor: if it equals, we move on from building state with this item to the next item in line. The search tree of our example is shown in Fig.~\\ref{fig:backtrack_permutation_duplicates}. \n\n\\paragraph{Implementation} The implementation is highly similar to previous standard permutation code other than three different points: \n\\begin{enumerate}\n\\item Before the items are called by \\texttt{permutate}, they are sorted first.\n\\item A simple condition check to avoid generating repeat states.\n \\item We used a dictionary data structure \\texttt{tracker} which has all unique items as keys and each item's corresponding occurrence as values to replace the boolean vector \\texttt{used} for slightly better space efficiency.\n\\end{enumerate}\nThe Python code is as:\n\\begin{lstlisting}[language=Python]\nfrom collections import Counter\ndef permuteDup(nums, k):\n ans = []\n def permutate(d, n, k, curr, tracker): \n nonlocal ans \n if d == k:\n ans.append(curr)\n return\n for i in range(n):\n if tracker[nums[i]] == 0:\n continue\n if i - 1 >= 0 and nums[i] == nums[i-1]:\n continue\n tracker[nums[i]] -= 1\n curr.append(nums[i])\n\n permutate(d+1, n, k, curr[:], tracker)\n curr.pop()\n tracker[nums[i]] += 1\n return\n \n nums.sort()\n permutate(0, len(nums), k, [], Counter(nums))\n return ans\n\\end{lstlisting}\n\n\\begin{bclogo}[couleur = blue!30, arrondi=0.1,logo=\\bccrayon,ombre=true]{Can you extend the swap method based permutation to handle duplicates?} \n\\end{bclogo}\n\n\\subsubsection{Discussion}\nFrom the example of permutation, we have demonstrated how backtracking works to construct candidates with an implicit search tree structure: the root node is the initial state, any internal node represents intermediate states, and all leaves are our candidates which in this case there are $n!$ for $p(n, n)$ permutation. In this subsection, we want to point out the unique properties and its computational and space complexities.\n\\paragraph{Two Passes} Backtracking builds an implicit search tree on the fly, and it does not memorize any intermediate state. It visits the vertices\n in the search tree in two passes: \n \\begin{enumerate}\n \\item Forward pass: it builds the solution incrementally and reaches to the leaf nodes in a DFS fashion. One example of forward pass is $[]->[1]->[1,2]->[1,2,3]$.\n \\item Backward pass: as the returning process from recursion of DFS, it also backtracks to previous state. One example of backward pass is $[1,2,3]->[1,2],->[1]$. \n \\end{enumerate}First, the forward pass to build the solution \\textbf{incrementally}.\nThe change of \\texttt{curr} in the source code indicates all vertices and the process of backtracking, it starts with $[]$ and end with $[]$. This is the core character of backtracking. We print out the process for the example as:\n\\begin{lstlisting}[numbers=none]\n[]->[1]->[1, 2]->[1, 2, 3]->backtrack: [1, 2]\nbacktrack: [1]\n[1, 3]->[1, 3, 2]->backtrack: [1, 3]\nbacktrack: [1]\nbacktrack: []\n[2]->[2, 1]->[2, 1, 3]->backtrack: [2, 1]\nbacktrack: [2]\n[2, 3]->[2, 3, 1]->backtrack: [2, 3]\nbacktrack: [2]\nbacktrack: []\n[3]->[3, 1]->[3, 1, 2]->backtrack: [3, 1]\nbacktrack: [3]\n[3, 2]->[3, 2, 1]->backtrack: [3, 2]\nbacktrack: [3]\nbacktrack: []\n\\end{lstlisting}\n\n\\paragraph{Time Complexity of Permutation}\nIn the search tree of permutation in Fig.~\\ref{fig:backtrack_permutation}, there are in total $V$ nodes, which equals to $\\sum_{i=0}^{n}{p_{n}^{k}}$. Because in a tree the number of edges $|E|$ is $|v|-1$, making the time complexity $O(|V|+|E|)$ the same as of $O(|V|)$. Since $p(n, n)$ itself alone takes $n!$ time, making the permutation an NP-hard problem. \n\n\\paragraph{Space Complexity} A standard depth-first search consumes $O(bd)$ space in worst-case to execute, where $b$ is branching factor and $d$ is the depth of the search tree. In the combinatorial search problems, usually depth and branching is decided by the total number of variables in the state, making $b \\sim d \\sim n$. In backtracking, we have space complexity $O(n^2)$. However, in normal standard DFS, the input--tree or graph data structure--is given and not attributed to space complexity. For a NP-hard combinatorial search problem, this input is often exponential. Backtracking search outcompetes the standard DFS by avoiding such space consumption; it only keeps a dynamic data structure(\\texttt{curr}) to construct node on the fly. % The slight different can be critical for problems with large state description. \n\n\\subsection{Combinations}\nGiven a list of $n$ items, generate all possible combinations of these items. If the input has duplicated items, only enumerate unique combinations. \n\\subsubsection{No Duplicates (L78. Subsets\n)}\nFrom Chapter Discrete Programming, we list the powerset--all $m$-subset, $m\\in[0, n]$ as:\n\\begin{align}\n C(n, m) = \\frac{P(n, m)}{P(m, m)} = \\frac{n!}{(n-m)!m!}\n\\end{align}\nFor example, when $a=[1, 2, 3]$, there are in total $7$ $m$-subsets, they are:\n\\begin{lstlisting}[numbers=none]\nC(3, 0): []\nC(3, 1): [1], [2], [3]\nC(3, 2): [1, 2], [1, 3], [2, 3]\nC(3, 3): [1, 2, 3]\n\\end{lstlisting}\n\\paragraph{Analysis} \n\\begin{figure}[!ht]\n \\centering\n \\includegraphics[width= 0.5\\columnwidth]{fig/combination.png}\n \\caption{The Search Tree of Combination.}\n \\label{fig:backtrack_combination}\n\\end{figure}\nWe can simply reuse the method of permutation, but with a problem that it generates lots of duplicates. For example, $P(3, 2)$ includes $[1, 2]$ and $[2, 1]$ which are indeed the same subset. Of course, we can check redundancy with saved $m$-subsets, but its not ideal. A systematical solution that avoids duplicates all along is preferred. If we limit the items we put into the $m$-subsets to be only increasing(of indexes of items or of values of items), in which case $[2, 1]$, $[3,1]$, and $[3,2]$ will never be generated. The enumeration of combination through backtracking search is shown in Fig.~\\ref{fig:backtrack_combination}.\n\n\\paragraph{Implementation} Two modifications based on permutation code:\n\\begin{enumerate}\n \\item \\texttt{for} loop: in the loop to iterate all possible candidates, we limit the candidates to be having larger indexes only.\n \\item We do not have to use a data structure to track the state of each candidate because any candidate that has larger index is a valid candidate.\n\\end{enumerate}\nWe use \\texttt{start} to track the starting position of valid candidates. The code of combination is:\n\\begin{lstlisting}[language=Python]\ndef C_n_k(a, n, k, start, d, curr, ans):\n if d == k: #end condition\n ans.append(curr[::]) \n return\n \n for i in range(start, n): \n curr.append(a[i])\n C_n_k(a, n, k, i+1, d+1, curr, ans)\n curr.pop()\n return\n\\end{lstlisting}\n\n\\paragraph{Alternative: 0 and 1 Selection} We have discussed that a powerset written as $P(S)$. With each item either being appear or not appear in the resulting set makes the value set $\\{0, 1\\}$, resulting $|P(S)| = 2^n$. Follow this pattern, with our given example, we can alternatively generate a powerset like this:\n\\begin{lstlisting}[numbers=none]\ns sets\n1 {1}, {}\n2 {1,2}, {1}, {2}, {}\n3 {1,2,3}, {1,2}, {1, 3}, {3}, {2, 3}, {2}, {3}, {}\n\\end{lstlisting}\nThis process can be better visualized in a tree as in Fig.~\\ref{}. We can see this process results $2^n$ leaves compared with our previous implementation which has a total of $2^n$ nodes is slightly less efficient. The code is as:\n\\begin{lstlisting}[language=Python]\ndef powerset(a, n, d, curr, ans):\n if d == n:\n ans.append(curr[::]) \n return\n\n # Case 1: select item\n curr.append(a[d])\n powerset(a, n, d + 1, curr, ans)\n # Case 2: not select item\n curr.pop()\n powerset(a, n, d + 1, curr, ans)\n return\n\\end{lstlisting}\n\n\\paragraph{Time Complexity} The total nodes within the implicit search space of combination shown in Fig.~\\ref{fig:backtrack_combination} is $\\sum_{k=0}^{n}C_{n}^{k} =2^n$, which was explained in Chapter Discreet Programming. Thus, the time complexity of enumerating the powset is $O(2^n)$ and is less compared with $O(n!)$ that comes with the permutation.\n\n\\paragraph{Space Complexity} Similarly, combination with backtracking search uses slightly less space. But, we can still acclaim the upper bound to be $O(n^2)$. \n\n\\subsubsection{With Duplicates(L90. Subsets II)}\n Assume we have $m$ unqiue items, and the frequency of each is marked as $x_i$, with $\\sum_{i=0}^{m-1}x_i = n$. \n\\begin{align}\n\\sum_{k=0}^{n} c(n, k) = \\prod_{i=0}^{m-1}(x_i + 1)\n\\end{align}\n\nFor example, when $a=[1, 2, 2, 3]$, there are $2\\times 3 \\times 2 = 12$ combinations in the powerset, they are listed as bellow:\n\\begin{lstlisting}[numbers=none]\n[], [1], [2], [3], [1, 2], [1, 3], [2, 2], [2, 3],\n[1, 2, 2], [1, 2, 3], [2, 2, 3],\n[1, 2, 2, 3]\n\\end{lstlisting}\nHowever, counting $c(n, k)$ with duplicates in the input replies on the specific input with specific distribution of these items. We are still able to count by enumerating with backtracking search. \n\\paragraph{Analysis and Implementation} The enumeration of the powerset with backtracking search is the same as handling the iterations of choice in the enumeration of permutation with duplicates. We first sort our items in increasing order of the values. Then we replace the \\texttt{for} loop from the above code with the following code snippet to handle the repetition of items from the input:\n\\begin{lstlisting}[language=Python]\n for i in range(start, n): \n if i - 1 >= start and a[i] == a[i-1]:\n continue \n ...\n\\end{lstlisting}\n\\subsection{More Combinatorics}\nIn this section, we supplement more use cases of backtracking search in the matter of other types of combinatorics. \n\\subsubsection{All Paths in Graph}\n\\begin{figure}[ht!]\n \\centering\n \\includegraphics[width=0.4\\columnwidth]{fig/all_path_demo.png}\n \\caption{Acyclic graph}\n \\label{fig:my_label}\n\\end{figure} \nFor a given acyclic graph, enumerate all paths from a starting vertex $s$. For example, for the graph shown in Fig.~\\ref{fig:my_label}, and a starting vertex $0$, print out the following paths:\n\\begin{lstlisting}[numbers=none]\n0, 0->1, 0->1->2, 0->1->2->5, 0->1->3, 0->1->4, 0->2, 0->2->5\n\\end{lstlisting}\n\\paragraph{Analysis} The backtracking search here is the same as how to apply a DFS on an explicit graph, with rather one extra point: a state $path$ which might have up to $n$ items ( the total vertices of a graph). In the implementation, the \\texttt{path} vector will dynamically be modified to track all paths constructed as the go of the DFS. The code is offered as:\n% Backtracking technique can be naturally used in graph path traversal. One example is to find all possible paths from a source to the target. One simpler occasion is when the graph has no cycles. Backtrack technique can enumerate all paths in the graph exactly once for each. \n\n% The implementation is as follow: we still use dfs, because there has no cycles, we have no need to track the visiting state of each node. We generate the possible answer with backtracking technique through the \\texttt{path} variable to track each state. \n\\begin{lstlisting}[language=Python]\ndef all_paths(g, s, path, ans):\n ans.append(path[::])\n for v in g[s]:\n path.append(v)\n all_paths(g, v, path, ans)\n path.pop()\n\\end{lstlisting}\nYou can run the above code in the Goolge Colab to see how it works on our given example.\n% Feed in the above network and run the following code:\n% \\begin{lstlisting}[language=Python]\n% al = [[1], [2], [4], [], [3, 5], [6], []]\n% ans = []\n% path = [0]\n% all_paths(al, 0, path, ans)\n% \\end{lstlisting}\n% With the printing, we can see the whole process, \\texttt{path} changes as the description of backtrack. \\begin{lstlisting}[numbers=none]\n% [0, 1]\n% [0, 1, 2]\n% [0, 1, 2, 4]\n% [0, 1, 2, 4, 3]\n% [0, 1, 2, 4] backtrack\n% [0, 1, 2, 4, 5]\n% [0, 1, 2, 4, 5, 6]\n% [0, 1, 2, 4, 5] backtrack\n% [0, 1, 2, 4] backtrack\n% [0, 1, 2] backtrack\n% [0, 1] backtrack\n% [0] backtrack\n% \\end{lstlisting}\n% We can see each state, we can always have a matching backtrack state. \n% \\begin{bclogo}[couleur = blue!30, arrondi=0.1,logo=\\bccrayon,ombre=true]{What to do if there is a cycle?} \n% \\end{bclogo}\n\n\\subsubsection{Subsequences(940. Distinct Subsequences II)} Given a string, list all unique subsequences. There may or may not exist duplicated characters in the string. For example, when $s='123'$, there are in total 7 subsequences, which are:\n\\begin{lstlisting}[numbers=none]\n'', '1', '2', '3', '12', '13', '23', '123'\n\\end{lstlisting}\nWhen $s='1223'$ which comes with duplicates, there are 12 subsequences:\n\\begin{lstlisting}[numbers=none]\n'', '1', '2', '3', '12', '13', '22', '23', \n'122', '123', '223',\n'1223'\n\\end{lstlisting}\n\\paragraph{Analysis} From Chapter Discrete Programming, we have explained that we can count the number of unique subsequences through recurrence relation and pointed out the relation of subsquences with subsets(combinations). Let the number of unique subsequences of a sequence as $seq(n)$ and the number of unique subsets of a set as $set(n)$ with $n$ items in the input. All subsequences are within subsets, and the subsequence set has larger cardinality than subsets, $|seq(n)|\\geq |set(n)|$. From the above example, we can also see that when there are only unique items in the sequence or when there are duplicates but all duplicates of an item are adjacent to each other:\n\\begin{itemize}\n \\item The cardinality of subsequences and subsets equals, $|seq(n)= set(n)|$.\n \\item The subsequences and subsets share the same items when the ordering of the subsequences are ignored. \n\\end{itemize}\nThis indicates that the process of enumerating subsequences is almost the same as of enumerating a powerset. This should give us a good start.\n\n\\paragraph{Implementation} However, if we change the ordering of the duplicated characters in the above string as $s='1232'$, there are in total 14 subsequences instead:\n\\begin{lstlisting}[numbers=none]\n'', '1', '2', '3', '12', '13', '23', '22', '32',\n'123', '122', '132', '232',\n'1232'\n\\end{lstlisting}\n\\begin{figure}[!ht]\n \\centering\n \\includegraphics[width= 1.0\\columnwidth]{fig/subsequence.png}\n \\caption{The Search Tree of subsequences.The red circled nodes are redundant nodes. Each node has a variable $s$ to indicate the starting index of candidates to add to current subsequence. $i$ indicate the candidate to add to the current node.}\n \\label{fig:backtrack_subsequence}\n\\end{figure}\nTherefore, our code to handle duplicates should differ from that of a powerset. In the case of powerset, the algorithm first sorts items so that all duplicates are adjacent to each other, making the checking of repetition as simple as checking the equality of item with its predecessor. However, in a given sequence, the duplicated items are not adjacent most of the time, we have to do things differently. We draw the search tree of enumerating all subsequences of string ``1232'' in Fig.~\\ref{fig:backtrack_subsequence}. From the figure, we can observe that to avoid redundant branches, we simply check if a current new item in the subsequence is repeating by comparing it with all of its predecessors in range $[s, i]$. The code for checking repetition is as:\n\\begin{lstlisting}[language=Python]\ndef check_repetition(start, i, a):\n for j in range(start, i):\n if a[i] == a[j]:\n return True\n return False\n\\end{lstlisting}\nAnd the code to enumerate subsequences is:\n\\begin{lstlisting}[language=Python]\ndef subseqs(a, n, start, d, curr, ans):\n ans.append(''.join(curr[::])) \n if d == n: \n return\n \n for i in range(start, n): \n if check_repetition(start, i, a):\n continue \n curr.append(a[i])\n subseqs(a, n, i+1, d+1, curr, ans)\n curr.pop()\n return\n\\end{lstlisting}\n\n \n\n\\subsection{Backtracking in Action}\nSo far, we have applied backtracking search to enumerate combinatorics. In this section, we shall see how backtracking search along with search pruning speedup methods solve two types of challenging NP-hard problems: Constraint Satisfication Problems (CSPs) and Combinatorial Optimization Problems. \n\nAs we have briefly introduced the speedup methods needed to solve larger scale of CSPs and COPs. For example, assume within the virtual search tree, the algorithm is currently at level 2 with state $s=[s_0, s_1]$. If there are $c$ choices for state $s_1$, and if one choice is testified to be invalid, this will prune off $\\frac{1}{c}$ of the whole search space. In this section, we demonstrate backtracking search armored with Branch and Prune method solving CSPs and Branch and Bound solving COPs.\n\n\n\\section{Solving CSPs}\nOfficially, a constraint satisfaction problem(CSP) consists of a set of $n$ variables, each denoted as $s_i$, $i\\in[0, n-1]$; their respective value domains, each denoted as $d_i$; and a set of $m$ constraints, each denoted as $c_j$, $j \\in [0, m-1]$. A \\textit{solution} to a CSP is an assignment of values to all the variables such that no constraint is violated. A \\textit{binary} CSP is one in which each of the constraints involves at most two variables. A CSP can be represented by a \\textit{constraint graph} which has a node for each variable and each constraint, and an arc connecting variable nodes contained in a constraint to the corresponding constraint node.\n\nWe explain a few strategies from the CSP-solver's arsenal that can potentially speedup the process:\n\\begin{enumerate}\n \\item Forward Checking: The essential idea is that when a variable $X$ from $s_i$ is instantiated with a value $x$ from its domain $d_i$, the domain of each future uninstantiated variable $Y$ is examined. If a value $y$ is found such that $X=x$ conflicts with $Y=y$, then $y$ is temporarily removed from the domain of $Y$. \n \\item Variable Ordering: The order in which variables are considered while solving a CSP method can have a substantial impact on the search space. One effective ordering is always select the next variable with the smallest remaining domain. In a dynamic variable ordering, the order of variables is determined as the search progresses, and often goes with forward checking which keeps updating the uninstantiated variables' domains. Selecting variable with the minimal domain first can pinpoint the solution quickly given the fact that the branch is still early on, and branch pruning at this stage is more rewarding. Another reasoning is that each step, when we are multiplying $d_i$ to the cost, we are adding the least expensive one, making this a greedy approach. \n\\end{enumerate}\n\\subsubsection{Sudoku (L37)} \n\\begin{figure}[!ht]\n \\centering\n \\includegraphics[width= 0.35\\columnwidth]{fig/250px-Sudoku-by-L2G-20050714.png}\n \\includegraphics[width= 0.35\\columnwidth]{fig/250px-Sudoku-by-L2G-20050714_solution.png}\n \\caption{A Sudoku puzzle and its solution}\n \\label{fig:backtrack_puzzle_2}\n\\end{figure}\nA Sudoku grid shown in Fig.~\\ref{fig:backtrack_puzzle_2} is a $n^2\\times n^2$ grid, arranged into $n$ $n\\times n$ mini-grids each containing the values $1,...,n$ such that no value is repeated in any row, column (or mini-grid). \n\\paragraph{Search Space}\nFirst, we analyze the number of distinct states in the search space which relies on how we construct the intermediate states and our knowledge in Enumerative combinatorics. We discuss two different formulations on $9 \\times 9$ grid:\n\\begin{enumerate}\n \\item For each empty cell in the puzzle, we create a set by taking values $1, ..., 9$ and removing from it those values that appear as a given in the same row, column, or mini-grid as that cell. Assume we have $m$ spots and the corresponding candidate set of each spot is $c_i$, and initial cost estimation can be obtained which is:\n \\begin{align}\n T(n) = \\prod_{i=0}^{m-1} c_i\n \\end{align}\n \\item Each row can be presented by a 9-tuples, there will be 9 rows in total, resulting 9 9-tuples to represent the search state. With $c_i$ as the number of non-given values in the i-th 9-tuples, there are $c_i!$ ways of ordering these values by permuting.The number of different states in the search space is thus:\n \\begin{align}\n T(n) = \\prod_{i=0}^{8} c_i!\n \\end{align}\n\\end{enumerate}\nThe two different ways each takes a different approach to formulate the state space, making its corresponding backtracking search differs too. We mainly focus on the first formulation with backtracking search.\n\\paragraph{Speedups} \n% \\begin{figure}[h]\n% \\centering\n% \\includegraphics[width= 0.5\\columnwidth]{fig/sudoku_backtracking.png}\n% \\caption{Partial search tree of Sudoku in Fig.~\\ref{fig:backtrack_puzzle_2}}\n% \\label{fig:backtrack_sudoku}\n% \\end{figure}\nAssume we have known all empty spots(variables) to fill in and we construct the search tree using backtracking. In our source code, we did an experiment comparing the effect of ordering variables with minimal domain first rule with arbitrary ordering. The experiment shows that the first method is more than 100 times faster than the second solving the our exemplary Sudoku puzzle. Therefore, we decide to always select the variable that has the least domain set to proceed next in the backtracking.\n\nFurther, we apply forward checking, for the current variable and a value we are able to assign, we recompute all the remaining empty spots' domain sets, and use the updated domain sets to decide:\n\\begin{itemize}\n \\item If this assigment will lead to empty domain for any of other remaining spots, and if so, we terminate the search and backtrack.\n \\item The spot to select next time with the ordering rule we choose. \n\\end{itemize}\n\\paragraph{Implementation} We set aside three vectors of length 9, \\texttt{row\\_state}, \\texttt{col\\_state}, and \\texttt{block\\_state} to track the state of all 9 rows, columns, and grids. The list has \\texttt{set()} data structures as items, saving the numbers filled already in that row, col, and grid respectively. Two stages in the implementation:\n\\begin{enumerate}\n \\item Initialization: We scan the whole each spot in the $9\\times 9$ grid to record the states of the filled spots and to find all empty spots that waiting to be filled in. With $(i,j)$ to denote the position of a spot, it corresponds to \\texttt{row\\_state[i]}, \\texttt{col\\_state[j]}, and \\texttt{block\\_state[i//3][j//3]}. We also write two functions to set and reset state with one assignment in the backtracking. The Python code is as follows:\n\\begin{lstlisting}[language=Python]\nfrom copy import deepcopy\nclass Sudoku():\n def __init__(self, board):\n self.org_board = deepcopy(board)\n self.board = deepcopy(board)\n \n def init(self):\n self.A = set([i for i in range(1,10)])\n self.row_state = [set() for i in range(9)]\n self.col_state = [set() for i in range(9)]\n self.block_state = [[set() for i in range(3)] for i in range(3)]\n self.unfilled = []\n\n for i in range(9):\n for j in range(9):\n c = self.org_board[i][j]\n if c == 0:\n self.unfilled.append((i, j))\n else:\n self.row_state[i].add(c)\n self.col_state[j].add(c)\n self.block_state[i//3][j//3].add(c)\n \n def set_state(self, i, j, c):\n self.board[i][j] = c\n self.row_state[i].add(c)\n self.col_state[j].add(c)\n self.block_state[i//3][j//3].add(c)\n \n def reset_state(self, i, j, c):\n self.board[i][j] = 0\n self.row_state[i].remove(c)\n self.col_state[j].remove(c)\n self.block_state[i//3][j//3].remove(c)\n\\end{lstlisting}\n\n\\item Backtracking search with speedups: In the initialization, we have another variable $A$ used as the domain set of the current processing spot. To get the domain set according to the constraints, a simple set operation is executed as: $A-(row\\_state[i]|col\\_state[j]|block\\_state[i//3][j//3])$. In the solver, each time, to pick a spot, we first update all remaining spots in the \\texttt{unfilled} and then choose the one with minimal domain. This process takes $O(n)$ which is trivial compared with the cost of the searching, with $9$ for computing domain set of a single spot, $9n$ for $n$ spots, and adding another $n$ to $9n$ to choose the one with the smallest size. The solver is implemented as:\n\\begin{lstlisting}[language=Python]\n def _ret_len(self, args):\n i, j = args\n option = self.A - (self.row_state[i] | self.col_state[j] | self.block_state[i//3 ][j//3])\n return len(option)\n \n def solve(self):\n if len(self.unfilled) == 0:\n return True\n # Dynamic variables ordering\n i, j = min(self.unfilled, key = self._ret_len)\n # Forward looking\n option = self.A - (self.row_state[i] | self.col_state[j] | self.block_state[i//3 ][j//3])\n if len(option) == 0:\n return False\n self.unfilled.remove((i, j))\n for c in option:\n self.set_state(i, j, c)\n if self.solve():\n return True\n # Backtracking\n else:\n self.reset_state(i, j, c)\n # Backtracking\n self.unfilled.append((i, j))\n return False\n\\end{lstlisting}\n\\end{enumerate}\n\n\n\\section{Solving Combinatorial Optimization Problems}\n% Combinatorial optimization problems are characterized by an \\textit{input}, i.e., a general description of conditions and parameters, and an \\textit{objective} defining \n\nCombinatorial optimization is an emerging field at the forefront of combinatorics and theoretical computer science that aims to use combinatorial techniques to solve discrete optimization problems. From a combinatorics perspective, it interprets complicated questions in terms of a fixed set of objects about which much is already known: sets, graphs, polytopes, and matroids. From the perspective of computer science, combinatorial optimization seeks to improve algorithms by using mathematical methods either to reduce the size of the set of possible solutions or to make the search itself faster. \n\nGenuinely, the inner complexity of a COP is at least of exponential, and its solutions fall into two classes: exact methods and heuristic methods. In some cases, we may be able to find efficient exact algorithms with either greedy algorithms or dynamic programming technique such as finding the shortest paths on a graph can be solved by the Dijkstra (greedy) or Bellman-Ford algorithms(dynamic programming) to provide exact optimal solutions in polynomial running time. For more complex problems, COP can be mathematically formulated as a Mixed Linear Programming(MILP) model and which is generally solved using a linear-programming based branch-and-bound algorithm. But, in other cases no exact algorithms are feasible, and the following randomized heuristic search algorithms though we do not cover in this section should be applied:\n\\begin{enumerate}\n \\item Random-restart hill-climbing.\n \\item Simulated annealing.\n \\item Genetic Algorithms.\n \\item Tabu search.\n\\end{enumerate}\n\n\\paragraph{Model Combinatorial Optimization Problems} It is a good practice to formulate COPs with mathematical equations/inequations, which includes three steps:\n\\begin{enumerate}\n \\item Choose the decision variables that typically encode the result we are interested in, such that in a superset problem, each item is a variable, and each variable includes two decisions: take or not take, making its value set as ${0, 1}$.\n \\item Express the problem constraints in terms of these decision variables to specify what the feasible solutions of the problem are.\n \\item Express the objective function to specify the quality of each solution.\n\\end{enumerate}\nThere are generally many ways to model a COP.\n\n\n\\paragraph{Branch and Bound} Branch and bound (BB, B$\\&$B, or BnB) is an algorithm design paradigm for discrete and combinatorial optimization problems, as well as mathematical optimization. A branch-and-bound algorithm consists of a systematic enumeration of candidate solutions by means of state space search: the set of candidate solutions is thought of as forming a rooted tree with the full set at the root. The algorithm explores branches of this tree, which represent subsets of the solution set. Before enumerating the candidate solutions of a branch, the branch is checked against upper and lower estimated bounds on the optimal solution, and is discarded if it cannot produce a better solution than the best one found so far by the algorithm. ``Branching'' is to split problem into a number of subproblems, and ``bounding'' is to find an optimistic estimation of the best solution to the the subproblems to either maximize the upper bound or minimize the lower bound. To get the optimistic estimation, we have to \\textit{relax constraints}. In this section, we will exemplify both the minimization(TSP) and maximization problem(Knapsack). \n\n\\paragraph{Search Strategies} In practice, we can apply different search strategies to enumerate the search space of the problem, such as depth-first, best-first, and least-discrepancy search. The way of how each listed strategy is applied in the combinatorial optimization problems is:\n\\begin{itemize}\n \\item Depth-First: it prunes when a node estimation is worse than the best found solution.\n \\item Best-First: it selects the node with the best estimation among the frontier set to expand each time. Worst scenario, the whole search tree have to be saved as long the best estimation is extremely optimistic and not a single branch is pruned in the process. \n \\item Least-Discrepancy: it trusts a greedy heuristic, and then move away from the heuristic in a very systematic fashion. \n\\end{itemize}\n\nIn this section, we discuss exact algorithms using Branch and Bound with a variation of search strategies. During the interviews, questions that have polynomial exact solutions are more likely to appear, proving your mastery of dynamic programming or greedy algorithms design methodologies. However, it is still good to discuss this option. \n\n%%%%%%%%%%%%%%%%%%%%\n\\subsection{Knapsack Problem}\n% In this section we want to showcase more searching strategies applied in solving optimization problems: comparing backtracking and a chance to use best-first search strategy. \n\nGiven $n$ items with weight and value indicated by two vectors $W$ and $V$ respectively. Now, given a knapsack with capacity $c$, maximize the value of items selected into the knapsack with the total weight being bounded by $c$. Each item can be only used at most once. For example, given the following data, the optimal solution is to choose item 1 and 3, with total weight of 8, and optimal value of 80.\n\\begin{lstlisting}[numbers=none]\nc = 10\nW = [5, 8, 3]\nV = [45, 48, 35]\n\\end{lstlisting}\n\\paragraph{Search Space} In this problem, $x_i$ denotes each item, and $w_i$, $v_i$ for its corresponding weight and value, with $i\\in[0, n-1]$. Each item can either be selected or left behind, indicating $x_i\\in{0, 1}$. The selected items can not exceed the capacity, making $\\sum_{i=0}^{n-1} w_i x_i \\leq c$. And we capture the total value of the selected items as $\\sum_{i=0}^{n-1} v_i x_i$. Putting it all together:\n\\begin{align}\n \\max_{v, x} \\quad & \\sum_{i=0}^{n-1} v_i x_i\\\\\n \\textrm{s.t.} \\quad & \\sum_{i=0}^{n-1} w_i x_i \\leq c\\\\\n & x_i\\in{0, 1}\n\\end{align}\nWith each variable having two choices, our search space is as large as $2^n$.\n\n\\paragraph{Branch and Bound} \n\nTo bound the search, we have to develop a heuristic function to estimate an optimistic--maximum--total value a branch can lead to. \n\nIn the case of knapsack problem, the simplest estimation is summing up the total values of selected items so far, and estimate the maximum value by adding the accumulated values of all remaining unselected items along the search. \n\n% And a branch is checked \n% \\begin{align}\n% \\texttt{if } estimate < found, \n% \\end{align}\n\nA tighter heuristic function can be obtained with \\textbf{constraint relaxation}. By relaxing the condition of simply choose $\\{0, 1\\}$ to $[0, 1]$, that a fraction of an item can be chosen at any time. By sorting the items by the value per unit $\\frac{v_i}{w_i}$, then a better estimate can be obtained by filling the remaining capacity of knapsack with unselected items, with larger unit value first be considered. A branch is checked on the optimal solution so far against the lower estimated bound in our case, and is discarded if it cannot produce a better solution than the best one found so far by the algorithm. Both heuristic functions are more optimistic compared with the true value, but the later is a tighter bound, being able to prune more branches along the search and making it more time efficient. We demonstrate branch and bound with two different search strategies: DFS(backtracking) and Best-First search.\n\n\n\n% the items while the capacity is not exhausted, then select a fraction of the last item to fill the remaining capacity, which is a closer estimate of the real one but still optimisitc compared with selecting the remaining items. What if we are allowed to get part of an item, so that we can fit the knapsack as full as possible. We can sort the items by their unit value, and take items in the order of decreasing unit values. Another bool vector is used to indicate if a certain item can be used or not. At first, all items are allowed, we can get an estimation of 92 in this case. For branches that decide not to take an item, that item is excluded using the bool vector. We compare the estimation with the best found value, if the estimated value will never be better, then this branch will prunned. \n\n\n\\subsubsection{Depth-First Branch and Bound}\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width= 0.98\\columnwidth]{fig/branch_and_bound_backtracking.png}\n \\caption{Depth-First Branch and bound}\n \\label{fig:knapsack_backtracking}\n\\end{figure}\nWe set up a class \\texttt{BranchandBound} to implement this algorithm. First, in the initiation, we add additional $\\frac{v_i}{w_i}$ to mark each item's value per unit, and sort these items by this value in decreasing order. Second, we have a function \\texttt{estimate} which takes three parameters: \\texttt{idx} as start index of the remaining items, \\texttt{curval} is the total value based on all previous decision, and \\texttt{left\\_cap} as the left capacity of the knapsack. The code snippet is:\n\\begin{lstlisting}[language=Python]\nimport heapq\n\nclass BranchandBound:\n def __init__(self, c, v, w):\n self.best = 0 \n self.c = c\n self.n = len(v)\n self.items = [(vi/wi, wi, vi) for _, (vi, wi) in enumerate(zip(v, w))]\n self.items.sort(key=lambda x: x[0], reverse=True)\n\n def estimate(self, idx, curval, left_cap):\n est = curval\n # use the v/w to estimate\n for i in range(idx, self.n):\n ratio, wi, _ = self.items[i]\n if left_cap - wi >= 0: # use all\n est += ratio * wi\n left_cap -= wi\n else: # use part\n est += ratio * (left_cap)\n left_cap = 0 \n return est\n\\end{lstlisting}\nIn the Depth-first search process, it consists of two main branches: one considering to choose the current item, and the other to handle the case while the item is not selected. For the first branch, it has to be bounded by the capacity and the comparison of the best found solution against to the estimation. Additional \\texttt{status} is to assist to visualize the process of the search, which tracks the combination of items. The process is shown in Fig.~\\ref{fig:knapsack_backtracking}. And the code is as:\n\\begin{lstlisting}[language=Python]\n def dfs(self, idx, est, val, left_cap, status):\n if idx == self.n:\n self.best = max(self.best, val)\n return\n print(status, val, left_cap, est )\n \n _, wi, vi = self.items[idx]\n # Case 1: choose the item\n if left_cap - wi >= 0: # prune by constraint\n # Bound by estimate, increase value and volume\n if est > self.best: \n status.append(True)\n nest = self.estimate(idx+1, val+vi, left_cap - wi) \n self.dfs(idx+1, nest, val+vi, left_cap - wi, status)\n status.pop()\n\n # Case 2: not choose the item\n if est > self.best:\n status.append(False)\n nest = self.estimate(idx+1, val, left_cap)\n self.dfs(idx+1, nest, val, left_cap, status) \n status.pop()\n return\n\\end{lstlisting}\n\\subsubsection{Best-First Branch and Bound}\nWithin Best-First search, we use priority queue with the estimated value, and each time the one with the largest estimated value within the frontier set is expanded first. Similarly, with branch and bound, we prune branch that has estimated value that would never surpass the best solution up till then. The search space is the same as in Fig.~\\ref{fig:knapsack_backtracking} except that the search process is different from depth-first. In the implementation, the priority queue is implemented with a min-heap where the minimum value is firstly popped out, thus we use the negative estimated value to make it always pop out the largest value conveniently instead of write code to implement a max-heap. \n\\begin{lstlisting}[language=Python]\n def bfs(self):\n # track val, cap, and idx is which item to add next\n q = [(-self.estimate(0, 0, self.c), 0, self.c, 0)] # estimate, val, left_cap, idx\n self.best = 0\n while q:\n est, val, left_cap, idx = heapq.heappop(q)\n est = -est\n _, wi, vi = self.items[idx]\n if idx == self.n - 1:\n self.best = max(self.best, val)\n continue\n \n # Case 1: choose the item\n nest = self.estimate(idx + 1, val + vi, left_cap - wi)\n if nest > self.best:\n heapq.heappush(q, (-nest, val + vi, left_cap - wi, idx + 1))\n\n # Case 2: not choose the item\n nest = self.estimate(idx + 1, val, left_cap)\n if nest > self.best:\n heapq.heappush(q, (-nest, val, left_cap, idx + 1))\n return \n\\end{lstlisting}\n\n\\subsection{Travelling Salesman Problem}\n\\begin{figure}[!ht]\n \\centering\n \\includegraphics[width= 0.4\\columnwidth]{fig/tsp_graph.png}\n \\caption{A complete undirected weighted graph.}\n \\label{fig:tsp_graph}\n\\end{figure}\nGiven a set of cities and the distances between every pair, find the shortest possible path that visits every city exactly once and returns to the origin city. For example, with the graph shown in Fig.~\\ref{fig:tsp_graph}, such shortest path is $[0, 1, 3, 2, 0]$ with a path weight $80$. \n% \\begin{lstlisting}[numbers=none]\n% Assume our graph is a two dimensional list:\n% g = [[(1, 10), (2, 15), (3, 20)], \n% [(0, 10), (2, 35),(3,25)],\n% [(0, 15),(1,35),(3,30)],\n% [(0,20),(1,25),(2,30)]]\n% g[0][0]=(1,10), means the edge between 0 and 1 with cost 10.\n% \\end{lstlisting}\n\n\\paragraph{Search Space} In TSP, a possible complete solution is a \\textit{Hamiltonian cycle}, a graph cycle that visits each vertex exactly once. Since it is a cycle, it does not matter where it starts. For convenience, we choose vertex $0$ as the origin city. Therefore, in our example, our path starts and ends at $0$, and the remaining $n-1$ vertices between will be a permutation of these vertices, making the complexity as $(n-1)!$. \n\nBecause this is a complete graph, it might be tempting to apply backtracking on the graph to enumerate all possible paths and find and check possible solutions. However, this path searching will build a $n-1$-ary search tree with height equals to $n-1$, making the complexity as $\\frac{(n-1)^{n}-1}{n-2}$, which is larger than the space of permutation among $n-1$ items. Therefore, in our implementation, we apply backtracking to enumerate all permutations of $n-1$ vertices and check its corresponding cost. \n\n\\paragraph{Speedups} Since we only care about the minimum cost, then any partial result that has cost larger than the minimum cost of all known complete solutions can be prunned. This is the \\textit{Branch and bound} method that we have introduced that is often used in the combinatorial optimization. \n% solution is path with vertices forming a cycleneed to construct a list of vertices (\\texttt{path}) and its total cost (\\texttt{cost}) of all edges between as the state $s=(p, c)$, $p, c$ is short for path and cost respectively. A possible complete solution for path will have $n+1$ vertices which start with a vertex and end with the same, and $n-1$ vertices in between. Now, put together about constraints.\n% \\begin{itemize}\n% \\item ``Visits every city exactly once'' means the first vertex will be a permutation of all cities, we get $n!$ combination (the last vertex does not matter).\n% \\item We have $n!$ possible states. We can further spot redundant states. For a cycle, it does not matter where it starts, it is always the same cycle. For convenience, we choose vertex $0$ as the starting path, and there will only be $n-1$ vertex to permutate with, making the size of the state space to $(n-1)!$. \n% \\item We only care about the minimum cost, then any partial result that has cost larger than the minimum cost of all known complete solution can be prunned. This is called \\textit{Branch and bound} method, which is the extension of backtracking into the optimization problems.\n% \\end{itemize}\n\n\\paragraph{Implementation} We built the graph as a list of dictionaries, each dictionary stores the indexed vertex's other cites and its corresponding distance as key and value respectively. Compared with standard permutation with backtracking, we add four additional variables: \\texttt{start} to track the starting vertex, \\texttt{g} to pass the graph to refer the distance information, \\texttt{mincost} to save the minimum complete solution so far found, and \\texttt{cost} to track the current partial path's cost. The code is shown as: \n\\begin{lstlisting}[language=Python]\ndef tsp(a, d, used, curr, ans, start, g, mincost, cost):\n if d == len(a): \n # Add the cost from last vertex to the start\n c = g[curr[-1]][start]\n cost += c\n if cost < mincost[0]:\n mincost[0] = cost\n ans[0] = curr[::] + [start]\n return\n \n for i in a:\n if not used[i] and cost + g[curr[-1]][i] < mincost[0] :\n cost += g[curr[-1]][i]\n curr.append(i)\n used[i] = True \n tsp(a, d + 1, used, curr, ans, start, g, mincost, cost)\n curr.pop()\n cost -= g[curr[-1]][i]\n used[i] = False\n return\n\\end{lstlisting}\nTSP is a NP-hard problem, and there is no known polynomial time solution so far.\n\n\n\n\\subsubsection{Other Solutions}\nWhenever we are faced with optimization, we are able to consider the other two algorithm design paradigm--Dynamic Programming and Greedy Algorithms. In fact, the above two problems both have its corresponding dynamic programming solutions: for knapsack problem, polynomial solution is possible; for TSP, though it is still of exponential time complexity, it is much better than $O(n!)$. We will further discuss these two problems in Chapter Dynamic Programming. \n\\section{Exercises}\n\\begin{enumerate}\n\\item 77. Combinations\n\\item 17. Letter Combinations of a Phone Number\n\\item 797. All Paths From Source to Target\n \\item N-bit String: enumerate all n-bit strings with backtracking algorithm, for example:\n\\begin{lstlisting}[numbers=none]\nn = 3, all 3-bit strings are:\n000, 001, 010, 011, 100, 101, 110, 111\n\\end{lstlisting}\n\\item 940. Distinct Subsequences II\n\\item N-queen\n\\item Map-coloring\n\\item 943. Find the Shortest Superstring (hard). Can be moduled as traveling salesman problem and dynamic programming\n\\end{enumerate}\n\\end{document}", "meta": {"hexsha": "3812786142e1dfcabba0b29f0a67f1cf4be8a745", "size": 59795, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Easy-Book/chapters/chapter_combinatorial_search.tex", "max_stars_repo_name": "stungkit/Algorithms-and-Coding-Interviews", "max_stars_repo_head_hexsha": "131199fea0b082d92c0f272a495c7a56a3242b71", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Easy-Book/chapters/chapter_combinatorial_search.tex", "max_issues_repo_name": "stungkit/Algorithms-and-Coding-Interviews", "max_issues_repo_head_hexsha": "131199fea0b082d92c0f272a495c7a56a3242b71", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Easy-Book/chapters/chapter_combinatorial_search.tex", "max_forks_repo_name": "stungkit/Algorithms-and-Coding-Interviews", "max_forks_repo_head_hexsha": "131199fea0b082d92c0f272a495c7a56a3242b71", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 75.8819796954, "max_line_length": 1151, "alphanum_fraction": 0.7340747554, "num_tokens": 15590, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.7905303186696747, "lm_q1q2_score": 0.6463174112015535}} {"text": "\n\\subsection{The Lie bracket}\n\nWe can define the Lie bracket from the ring commutator.\n\nWe use the Lie bracket, rather than multiplication, as the operator over a field homomorphism.\n\n\\([A, B]\\)\n\nThis generates another element in the algebra.\n\nThis satisifies:\n\n\\begin{itemize}\n\\item Bilinearity: \\([xA+yB,C]=x[A,C]+y[B,C]\\)\n\\item Alternativity: \\([A,A]=0\\)\n\\item Jacobi identity: \\([A, [B,C]]+[C,[A,B]]+[B,[C,A]]=0\\)\n\\item Anticommutivity: \\([A, B]=-[B,A]\\)\n\\end{itemize}\n\nOne option for the Lie bracket is the ring commutor. So that:\n\n\\([A,B]=AB-BA\\)\n\n", "meta": {"hexsha": "02d9775cc4282513b44700c78412362b77eae877", "size": 554, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/geometry/representation/04-02-lieBracket.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/geometry/representation/04-02-lieBracket.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/geometry/representation/04-02-lieBracket.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 22.16, "max_line_length": 94, "alphanum_fraction": 0.6732851986, "num_tokens": 180, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8840392756357327, "lm_q2_score": 0.7310585669110203, "lm_q1q2_score": 0.6462844859393152}} {"text": "\\chapter{Wrapping Up}\n\n\\section{Programs and Proofs}\n\n\\begin{itemize}\n \\item Law of Excluded Middle $\\forall{Q} Q\\vee\\neg Q$\n \\item $Programs \\equiv Proofs$\n\\end{itemize}\n\nNot all proofs have a corresponding program.\n\nLet $isProgram(P)$ and $halts(P)$ be two predicates.\n\\[ \\varphi \\triangleq \\forall{P} isProgram(P) \\to halts(P) \\vee \\neg halts(P) \\]\n$\\varphi$ is valid. i.e. $\\vDash\\varphi$. By completeness of first order logic, $\\vdash\\varphi$. So there must exist a proof (deduction) $\\Phi=\\{\\varphi_1,\\dots,\\varphi_n\\}$ s.t. $\\varphi_n = \\varphi$. If every proof had a corresponding program, then this program would be able to solve the Halting problem.\n\nNot all programs have a corresponding proof.\n\n\\begin{lstlisting}[language=c]\n while(1) {}\n\\end{lstlisting}\n\nThis program has a dead loop, which implies ``false''. If this program had a proof, then we would be able to prove ``false''.\n", "meta": {"hexsha": "4f05da93a01df4bdb104fe442cf694493a93df63", "size": 904, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Mathematical Logic/wrapup.tex", "max_stars_repo_name": "YBRua/CourseNotes", "max_stars_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-03-20T10:40:58.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-25T08:15:15.000Z", "max_issues_repo_path": "Mathematical Logic/wrapup.tex", "max_issues_repo_name": "YBRua/CourseNotes", "max_issues_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Mathematical Logic/wrapup.tex", "max_forks_repo_name": "YBRua/CourseNotes", "max_forks_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T11:31:00.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T11:31:00.000Z", "avg_line_length": 39.3043478261, "max_line_length": 307, "alphanum_fraction": 0.7134955752, "num_tokens": 266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9241418178895029, "lm_q2_score": 0.6992544273261176, "lm_q1q2_score": 0.6462102576364417}} {"text": "\\section{Rosenblatt's Perceptron (1957)}\r\n\r\n\\subsection{Motivation}\r\n\r\n\\begin{frame}\r\n \\frametitle{Motivation}\r\n\r\n \\begin{itemize}\r\n \\item We want to compute a linear decision boundary. \\\\[.5cm]\r\n \\item We assume that classes are linearly separable. \\\\[.5cm]\r\n \\item Computation of a linear separating hyperplane that\\\\\r\n minimizes the distance of misclassified feature vectors \\\\\r\n to the decision boundary.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Objective Function}\r\n\r\n\\begin{frame}\r\n \\frametitle{Objective Function}\r\n\r\n Assume the following:\r\n\r\n \\begin{itemize}\r\n \\item Class numbers are $y=\\pm 1$.\r\n \\item The decision boundary is a linear function:\r\n \\begin{displaymath}\r\n y^* = \\mbox{sgn}(\\vec \\alpha^T\\vec x + \\alpha_0).\r\n \\end{displaymath}\r\n \\pause\r\n \\item Parameters $\\alpha_0$ and $\\vec \\alpha$ are chosen according to the optimization problem \\\\[.2cm]\r\n\r\n \\begin{center}\r\n \\tikz[baseline]{ \r\n \\node[fill=bl1!100,anchor=base,rounded corners=3pt] (d1) {\r\n \\color{bl3} \r\n $\\displaystyle \\mbox{minimize} \\quad \\bigg\\{ D(\\alpha_0, \\vec \\alpha)= -\\sum_{\\vec x_i\\in {\\cal M}} y_i \\cdot (\\vec \\alpha^T\\vec x_i + \\alpha_0) \\bigg\\}$\r\n }; \r\n }\\\\[.2cm]\r\n \\end{center}\r\n\r\n where $\\cal M$ includes the misclassified feature vectors.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Objective Function \\cont}\r\n\r\n \\begin{itemize}\r\n \\item The elements of the sum in the objective function depend on \\\\\r\n the set of misclassified feature vectors $\\cal M$. \\\\[.5cm] \\pause\r\n \\item In each iteration step the cardinality of $\\cal M$ might change. \\\\[.5cm] \\pause\r\n \\item The cardinality of $\\cal M$ is a discrete variable. \\\\[.5cm] \\pause\r\n \\item \\structure{Competing variables:} continuous parameters of linear decision boundary and the discrete cardinality of $\\cal M$.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Minimization of Objective Function}\r\n\r\n\\begin{frame}\r\n \\frametitle{Minimization of Objective Function}\r\n\r\n Remember the objective function $D(\\alpha_0, \\vec \\alpha)$:\r\n \\begin{displaymath}\r\n \\mbox{minimize} \\quad \\quad D(\\alpha_0, \\vec \\alpha)= -\\sum_{\\vec x_i\\in {\\cal M}} y_i \\cdot (\\vec \\alpha^T\\vec x_i + \\alpha_0)\r\n \\end{displaymath}\r\n \\pause\r\n \r\n The gradient of the objective function is: \\pause\r\n \\begin{eqnarray*}\r\n \\frac{\\partial}{\\partial \\alpha_0} \\ D(\\alpha_0, \\vec \\alpha)&=& -\\sum_{\\vec x_i\\in {\\cal M}} y_i \\\\ \\pause\r\n \\frac{\\partial}{\\partial \\vec\\alpha} \\ D(\\alpha_0, \\vec \\alpha)&=& -\\sum_{\\vec x_i\\in {\\cal M}} y_i\\cdot \\vec x_i \\\\\r\n \\end{eqnarray*}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Minimization of Objective Function \\cont}\r\n\r\n We want to take an update step right after having visited each misclassified observation.\r\n The update rule in the $(k+1)$-st iteration step is:\r\n\r\n \\begin{displaymath}\r\n { \\alpha_0^{(k+1)} \\choose \\vec \\alpha^{(k+1)}} = \\pause { \\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}} + \\lambda {y_i \\choose y_i\\cdot \\vec x_i}\r\n \\end{displaymath} \r\n\r\n Here $\\lambda$ is the learning rate which can be set to $1$ without loss of generality.\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Minimization of Objective Function \\cont}\r\n\r\n \\begin{algorithmic}\r\n \\STATE \\structure{Input:} training data: $S = \\{ (\\vec x_1, y_1), (\\vec x_2, y_2), (\\vec x_3, y_3), \\dots, (\\vec x_m, y_m) \\}$ \\pause\r\n \\STATE initialize $k=0$, $\\alpha^{(0)}_0=0$ and $\\vec \\alpha^{(0)}= \\vec 0$\r\n \\REPEAT \r\n \\STATE select pair $(\\vec x_i, y_i)$ from training set. \\pause\r\n \\IF {$y_i\\cdot (\\vec x_i^T \\vec\\alpha^{(k)} + \\alpha_0^{(k)})\\leq 0$} \r\n \\STATE \\\r\n \\STATE $\\displaystyle{ \\alpha_0^{(k+1)} \\choose \\vec \\alpha^{(k+1)}} = { \\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}} + {y_i \\choose y_i\\cdot \\vec x_i}$\r\n \\STATE \\\r\n \\STATE $k\\leftarrow k+1$\r\n \\ENDIF\r\n \\pause\r\n \\UNTIL{$y_i\\cdot (\\vec x_i^T \\vec\\alpha^{(k)} + \\alpha_0^{(k)})> 0 \\quad \\mbox {for all} \\quad i$}\r\n \\STATE \\structure{Output:} $\\alpha_0^{(k)}$ and $\\vec \\alpha^{(k)}$\r\n \\end{algorithmic}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Remarks on Perceptron Learning}\r\n\r\n\\begin{frame}\r\n \\frametitle{Remarks on Perceptron Learning}\r\n\r\n \\begin{itemize}\r\n \\item The update rule is extremely simple. \\\\[.3cm]\r\n \\item Nothing happens if we classify all $\\vec x_i$ correctly using the given linear decision boundary. \\\\[.3cm]\r\n \\item The parameter $\\vec \\alpha$ of the decision boundary is a linear combination of feature vectors. \\\\[.3cm] \\pause\r\n \\item The decision boundary thus is:\r\n \\begin{displaymath}\r\n F(\\vec x) = \\left (\\sum_{i\\in \\cal E} y_i\\cdot \\vec x_i \\right)^T \\vec x + \\sum_{i\\in \\cal E} y_i \\quad \r\n = \\quad \\sum_{i\\in \\cal E} y_i\\cdot \\langle \\vec x_i, \\vec x\\rangle + \\sum_{i\\in \\cal E} y_i \r\n \\end{displaymath}\r\n where $\\cal E$ is the list of indices that required an update \\\\\r\n\t(indices may appear more than once). \r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Remarks on Perceptron Learning \\cont}\r\n\r\n \\begin{itemize}\r\n \\item The final linear decision boundary depends on the initialization, \\\\\r\n i.\\,e.\\ $\\alpha_0^{(0)}$ and $\\vec \\alpha^{(0)}$. \\\\[.5cm]\r\n \\item The number of iterations can be rather large. \\\\[.5cm]\r\n \\item If data are not linearly separable, the proposed learning algorithm will not converge. The algorithm will end up in hard to detect cycles.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Convergence of Learning Algorithm}\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm}\r\n\r\n \\begin{theorem}[Convergence Theorem of Rosenblatt and Novikoff]\r\n Assume that for all $i=1,2, \\dots, m$\r\n \\begin{displaymath}\r\n y_i(\\vec x_i^T\\vec \\alpha^* +\\alpha^*_0)\\geq \\rho\r\n \\end{displaymath}\r\n where $\\rho>0$ and $\\|\\vec \\alpha^*\\|=1$.\r\n Let $M= \\max_i \\|\\vec x_i\\|_2$. \\\\[.3cm]\r\n \r\n The perceptron learning algorithm converges to a linear decision boundary after $k$ iterations, where $k$ is bounded by\r\n \\begin{displaymath}\r\n k \\leq \\frac{(\\alpha_0^{*2}+1)(1+M^2)}{\\rho^2}.\r\n \\end{displaymath}\r\n \\end{theorem}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm \\cont}\r\n\r\n Let us look at the estimated parameters after $k$ iterations and \\\r\n how the parameters change with iterations:\r\n\r\n \\begin{eqnarray*}\r\n {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}^T{\\alpha_0^* \\choose \\vec \\alpha^*} \r\n &=& \\pause \\left( { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} + {y_i \\choose y_i\\cdot \\vec x_i}\\right)^T {\\alpha_0^* \\choose \\vec \\alpha^*}\\\\ \\pause\r\n &=& { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} ^T{\\alpha_0^* \\choose \\vec \\alpha^*} + {y_i \\choose y_i\\cdot \\vec x_i}^T {\\alpha_0^* \\choose \\vec \\alpha^*} \\\\ \\pause\r\n &\\geq& { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} ^T{\\alpha_0^* \\choose \\vec \\alpha^*} + \\rho \\\\ \\pause\r\n &\\geq& k\\rho\r\n \\end{eqnarray*}\r\n\r\n \\structure{Conclusion:} The more iterations (i.\\,e.\\ misclassifications) we have, \\\\\r\n the more the vectors are aligned.\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm \\cont}\r\n\r\n Now we apply Cauchy-Schwartz inequality for inner products:\r\n\r\n \\begin{eqnarray*}\r\n {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}^T{\\alpha_0^* \\choose \\vec \\alpha^*} \r\n &\\leq& \\left\\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2\\cdot \\left\\| {\\alpha_0^* \\choose \\vec \\alpha^*}\\right\\|_2 \\\\[.5cm] \\pause\r\n &=&\\left \\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2\\cdot \\sqrt{\\alpha_0^{*2}+1}\r\n \\end{eqnarray*}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm \\cont}\r\n\r\n The norm of the vector estimated in the $k$-th iteration step is:\r\n\r\n \\begin{eqnarray*}\r\n & & \\left \\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2^2 =\r\n \\left\\| { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} + {y_i \\choose y_i\\cdot \\vec x_i}\\right\\|_2^2 \\\\[.5cm] \\pause\r\n & & \\vspace*{-2cm} = { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} ^T{ \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} \r\n + 2 { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}}^T{y_i \\choose y_i\\cdot \\vec x_i}\r\n + {y_i \\choose y_i\\cdot \\vec x_i}^T{y_i \\choose y_i\\cdot \\vec x_i}\r\n \\end{eqnarray*}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm \\cont}\r\n\r\n We only go into iteration step $(k+1)$ if we did a mistake in iteration $k$. \\\\\r\n A misclassification implies:\r\n \r\n \\begin{displaymath}\r\n { \\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}^T{y_i \\choose y_i\\cdot \\vec x_i} \r\n = y_i\\cdot (\\vec x_i^T \\vec\\alpha^{(k)} + \\alpha_0^{(k)})\\quad < \\quad 0\r\n \\end{displaymath}\r\n \\pause \r\n\r\n And thus we get\r\n \r\n \\begin{displaymath}\r\n \\left \\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2^2 \r\n \\leq { \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} ^T{ \\alpha_0^{(k-1)} \\choose \\vec \\alpha^{(k-1)}} \r\n + {y_i \\choose y_i\\cdot \\vec x_i}^T{y_i \\choose y_i\\cdot \\vec x_i}\r\n \\leq k(1+M^2)\r\n \\end{displaymath}\r\n\\end{frame}\r\n\r\n\r\n\\begin{frame}\r\n \\frametitle{Convergence of Learning Algorithm \\cont}\r\n\r\n Wrap-up:\r\n \r\n \\begin{displaymath}\r\n k\\rho \r\n \\leq {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}^T{\\alpha_0^* \\choose \\vec \\alpha^*} \r\n \\leq \\left \\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2\\cdot \\sqrt{\\alpha_0^{*2}+1}\r\n \\end{displaymath}\r\n \\pause\r\n\r\n Using Cauchy-Schwartz:\r\n \r\n \\begin{displaymath}\r\n k\\rho\\quad\r\n \\leq \\quad \\left \\| {\\alpha_0^{(k)} \\choose \\vec \\alpha^{(k)}}\\right\\|_2\\cdot \\sqrt{\\alpha_0^{*2}+1}\\quad \\leq \\quad \\sqrt{k(1+M^2)(\\alpha_0^{*2}+1)}\r\n \\end{displaymath}\r\n \\pause\r\n\r\n shows:\r\n \r\n \\begin{displaymath}\r\n k \\quad \\leq\\quad \\frac{(\\alpha_0^{*2}+1)(1+M^2)}{\\rho^2}\r\n \\end{displaymath}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Lessons Learned}\r\n\r\n\\begin{frame}\r\n \\frametitle{Lessons Learned}\r\n\r\n \\begin{itemize}\r\n \\item Objective function changes in each iteration step. \\\\[.5cm]\r\n \\item Optimization problem is discrete. \\\\[.5cm]\r\n \\item Very simple learning rule. \\\\[.5cm]\r\n \\item \\vorsicht \\structure{Very important:} Number of iterations does \\structure{not} depend on the \\\\\r\n dimension of the feature vectors.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\\input{nextTime.tex}\r\n\r\n\\subsection{Further Readings}\r\n\r\n\\begin{frame}\r\n \\frametitle{Further Readings}\r\n\r\n \\begin{itemize}\r\n \\item Brian D. Ripley: \\\\\r\n \\structure{Pattern Recognition and Neural Networks}, \\\\\r\n Cambridge University Press, Cambridge, 1996.\\\\[.3cm]\r\n \\item T. Hastie, R. Tibshirani, and J. Friedman: \\\\\r\n \\structure{The Elements of Statistical Learning --}\\\\\r\n \\structure{ Data Mining, Inference, and Prediction},\\\\\r\n 2nd edition, Springer, New York, 2009.\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n\\subsection{Comprehensive Questions}\r\n\r\n\\begin{frame}\r\n \\frametitle{Comprehensive Questions}\r\n\r\n \\begin{itemize}\r\n \\item What is Rosenblatt's perceptron? \\\\[1cm]\r\n \\item What is the objective function for Rosenblatt's perceptron? \\\\[1cm]\r\n \\item Why is the optimization of the objective function nonlinear? \\\\[1cm]\r\n \\item When and how does Rosenblatt's perceptron algorithm converge?\r\n \\end{itemize}\r\n\\end{frame}\r\n", "meta": {"hexsha": "db669f61a7b2d315693a6fe3932c7540935b33ee", "size": 11344, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "09_perceptron.tex", "max_stars_repo_name": "akmaier/pr-slides", "max_stars_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2021-01-11T07:27:09.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-31T19:21:31.000Z", "max_issues_repo_path": "09_perceptron.tex", "max_issues_repo_name": "akmaier/pr-slides", "max_issues_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "09_perceptron.tex", "max_forks_repo_name": "akmaier/pr-slides", "max_forks_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-06-21T06:06:43.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-18T18:47:28.000Z", "avg_line_length": 36.4758842444, "max_line_length": 177, "alphanum_fraction": 0.619887165, "num_tokens": 3821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8991213718636752, "lm_q2_score": 0.7185943925708561, "lm_q1q2_score": 0.6461035760618525}} {"text": "\n\\subsection{Mean value theorem for integration}\n\nTake function \\(f(x)\\). From the extreme value theorem we know that:\n\n\\(\\exists m \\in \\mathbb{R} \\exists M\\in \\mathbb{R}\\forall x\\in [a,b](m] (0,-0.25) -- (0.2,-0.25) node[right]{$x_1$};\n \\draw[->] (0,-0.3) -- (0.3,-0.3) node[right]{$x_2$};\n \\draw[->] (0,-0.35) -- (0.7,-0.35) node[right]{$x_3$};\n \\draw[->] (0,-0.4) -- (0.8,-0.4) node[right]{$x_4$};\n \\draw[] (0,-0.25) -- (0,-0.4);\n \n \\draw[|->] (0,0.1) -- (0,0.2) node[above]{$+$};\n \\end{tikzpicture}\n } &\n \\subfloat[Horizontal]{\n \\begin{tikzpicture}[scale=6.0]\n \\draw[very thick] (0,0) -- (1,0);\n \n \\draw[<-, thick] (0.2,0) -- (0.2,-0.1) node[below]{$R_{1h}$};\n \\draw[<-, thick] (0.8,0) -- (0.8,-0.1) node[below]{$R_{2h}$};\n \n \\draw[<-, thick] (0.1,0) -- (0.1,0.1) node[above]{$F_{3}$};\n \n \\draw[->] (0,-0.25) -- (0.1,-0.25) node[right]{$x_0$};\n \\draw[->] (0,-0.3) -- (0.2,-0.3) node[right]{$x_1$};\n \\draw[->] (0,-0.35) -- (0.8,-0.35) node[right]{$x_4$};\n \\draw[] (0,-0.25) -- (0,-0.35);\n \n \\draw[|->] (0,0.1) -- (0,0.2) node[above]{$+$};\n \\end{tikzpicture}\n } \\\\\n \\end{tabular}\n \\vspace{1em}\n \\caption{Free-body diagrams for a loaded beam}\\label{fig-fbd}\n\\end{figure*}\n\nUsing these diagrams and information on the forces and location of the forces, one can ascertain the reactions on the bearings.\n\n\\begin{multicols}{4}\n\\begin{description}\n \\item[$F_1$] = \\SI{15}{\\kilo\\newton}\n \\item[$F_2$] = \\SI{25}{\\kilo\\newton}\n \\item[$F_3$] = \\SI{10}{\\kilo\\newton}\n \\item[$R_{1v}$] = ?\n \\item[$R_{1h}$] = ?\n \\item[$R_{2v}$] = ?\n \\item[$R_{2h}$] = ?\n \\item[$x_0$] = \\SI{0}{\\metre}\n \\item[$x_1$] = \\SI{0.2}{\\metre}\n \\item[$x_2$] = \\SI{0.3}{\\metre}\n \\item[$x_3$] = \\SI{0.55}{\\metre}\n \\item[$x_4$] = \\SI{0.6}{\\metre}\n\\end{description}\n\\end{multicols}\n\nTo\\marginnote{Resolving the Vertical Forces} resolve the force vertically, we take moments about one of the bearings with the assumption that it is a static and stable system. Thus, no moment should exist about the bearing otherwise the shaft would be spinning!\n\nHere, we have taken moments about $R_{1v}$ and from this, we can calculate the reaction force $R_{2v}$. \n\\begin{align}\n \\circlearrowright R_{1v} &= 0 = \\SI{0.1}{\\metre}(\\SI{15}{\\kilo\\newton}) + \\SI{0.35}{\\metre}(\\SI{25}{\\kilo\\newton}) - 0.4(R_{2v}) \\\\\n R_{2v} &= \\frac{\\SI{1025}{\\kilo\\newton\\metre}}{\\SI{0.4}{\\metre}} = \\SI{25.625}{\\kilo\\newton}\n\\end{align}\n\nWith $R_{2v}$ calculated, we can look at the balancing the forces in the vertical plane to ascertain $R_{1v}$.\n\\begin{align}\n R_{1v}+R_{v2}&=F_1+F_2\\\\\n R_{1v}+R_{v2}&=\\SI{15}{\\kilo\\newton}+\\SI{25}{\\kilo\\newton}\\\\\n R_{1v} &= \\SI{40}{\\kilo\\newton}-\\SI{25.625}{\\kilo\\newton} = \\SI{14.375}{\\kilo\\newton}\n\\end{align}\n\nThe\\marginnote{Resolving the Horizontal Forces} same process used in the horizontal direction. Taking moments about $R_{1h}$, we can find $R_{2h}$.\n\\begin{align}\n \\circlearrowright R_{1h} &= 0 = \\SI{0.2}{\\metre}(\\SI{10}{\\kilo\\newton}) + 0.4\\si{\\metre}(R_2) \\\\\n \\therefore R_{2h} &= \\frac{0.2\\si{\\metre}}{-0.4\\si{\\metre}} = -5\\si{\\kilo\\newton}\n\\end{align}\n\nAgain, equating the forces in the horizontal plane gives us $R_{1h}$.\n\\begin{align}\n R_{1h}+R_{2h}&=\\SI{10}{\\kilo\\newton} \\\\\n R_{1h} &= \\SI{10}{\\kilo\\newton} + \\SI{5}{\\kilo\\newton} = \\SI{15}{\\kilo\\newton}\n\\end{align}\n\n", "meta": {"hexsha": "9c61cc8ec7b51c2a8fe661341e44ad82fab0b24d", "size": 4376, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "02_free_body_diagrams/section.tex", "max_stars_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_stars_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "02_free_body_diagrams/section.tex", "max_issues_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_issues_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "02_free_body_diagrams/section.tex", "max_forks_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_forks_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.1134020619, "max_line_length": 394, "alphanum_fraction": 0.5939213894, "num_tokens": 1758, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8558511414521923, "lm_q1q2_score": 0.6460948532100265}} {"text": "%\\section{Computational Methods and Software}\n\\label{sec:software}\n\nThe previous chapter described a set of models \nfor the system of interest. \n%These models are too \n%complicated to solve exactly, and must instead be \n%instantiated with software to produce a numerical \n%result. \nThis chapter details the numerical formulation\nand solution of these models. It begins with a \ndiscussion of the numerical discretization of the equations \nof interest. The mesh discretization is then described. \nNext, the scientific software in which these numerical \nmodels are used is discussed. Finally, the tool\nchain and supercomputer systems are briefly introduced. \n\n\\section{Discretization Scheme}\n\\label{sec:discretization}\nThe finite element method (FEM) is used to numerically solve the \nNavier-Stokes equations. The starting point for the FEM is\nto cast the equations in Section~\\ref{sub_sec:ns_en} into a weak\nform. Use of the weak form reduces the continuity requirements on the\nbasis functions, thereby allowing the use of functions that are\neasy to construct and implement, such as piece-wise polynomials. \nManipulating these partial differential equations into a\nvariational formulation is accomplished by multiplying the equations by\nappropriate test functions and integrating over the domain,\n$\\Omega$. The resulting weak problem is: find $({\\bf u},p,T) \\in\nH^1(\\Omega)^3 \\times L_2(\\Omega) \\times H^1(\\Omega)$ such that \n\n%\n% http://www.numerik.uni-hd.de/Oberwolfach-Seminar/CFD-Course.pdf\n%\n\\begin{align}\n (\\frac{\\partial{\\bf u}}{\\partial t},{\\bf v}) + ({\\bf u} \\cdot \\nabla{\\bf\n u},{\\bf v}) + (\\nu \\,\\nabla{\\bf u}, \\nabla{\\bf v}) \n -(p,\\nabla \\cdot{\\bf u}) &= ({\\bf g} \\, T'/T_0,v),\n \\label{eqn:ns_weak} \\\\\n (\\nabla \\cdot{\\bf u},q) &= 0,\n \\label{eqn:cont_weak} \\\\\n (\\frac{\\partial T}{\\partial t}, w) + ({\\bf u} \\cdot \\nabla T,\n w) + (k \\, \\nabla T, \\nabla w) &= 0,\\label{eqn:en_weak}\n\\end{align}\n\n%\\todo{boundary terms}\n\n$\\forall ({\\bf v},q,w) \\in H^1(\\Omega)^3 \\times L_2(\\Omega) \\times\nH^1(\\Omega)$, where $(\\cdot,\\cdot)$ denotes the $L_2$ inner product \n$({\\bf u},{\\bf v}) = \\int_\\Omega {\\bf u} \\cdot {\\bf v} \\, dx$ and\n$H^1(\\Omega)$ is the Sobolov space with one square integrable derivative\non the domain \n$\\Omega$\\cite{oden2012introduction}. As noted previously, boldface letters\ndenote vector quantities (such as ${\\bf u} = \\left \\{ u,v,w \\right \\}$). \nSome of the simulations presented here were conducted under\nsteady conditions, for which the $\\frac{\\partial}{\\partial t}$ terms\nvanish. An FEM scheme is obtained by posing the weak form in\nterms of finite-dimensional subspaces of the function spaces specified above\ndefined using piecewise-polynomial basis functions. This discretization\nhas the form, $ {\\bf v_h} \\in {\\bf v}$, where ${\\bf v_h}$ is \n%an approximation of ${\\bf v}$ \nformed through a linear combination of a finite number (N) of basis\nfunctions $\\phi_i$, \n\\begin{equation}\n {\\bf v_h} = \\sum_{i=1}^N \\alpha_i \\phi_i,\n\\end{equation}\nwhere $\\alpha_i$ are coefficients\\cite{becker1981introduction}.\n%\n%are constants and the N basis functions, $\\left \\{\n%\\psi_1, \\psi_2, \\ldots, \\psi_N \\right \\}$, define an N-dimensional \n%subspace of $H^1$\\cite{becker1981introduction}. \n%\nAll of the simulations discussed in this work were \naccomplished using piecewise linear basis functions for both the\nvelocity and pressure. Typically, the use of equal order elements for\nvelocity and pressure is ruled out in the standard Galerkin FEM\nformulation because of the Babuska-Brezzi condition\\cite{bb-cond}. \nThis is overcome here by \n%However, the weak form equations shown above are stable with equal-order\n%elements for velocity and pressure due to \nthe introduction of a pressure stabilization\nterm\\cite{becker1981introduction}. The resulting system is still\nsusceptible to convective instabilities, and so streamline upwind/Petrov-Galerkin\n(SUPG) stabilization terms are used, as first described by\nHughes\\cite{Hughes198685,supg} and extended to natural convection as in\nBecker and Braack\\cite{Becker2002428}. These stabilization terms add a\nresidual dependent artificial dissipation that approaches zero as the\nsolution converges. This scheme is called consistent because the\nunderlying order of convergence of the numerical method is not\naffected\\cite{hughes2000finite}. \n\nThe stabilization described above is accomplished by introducing an additional term,\n$\\langle L{\\bf c},S {\\bf \\phi} \\rangle_\\tau$, to the weak form defined in Equations\n\\ref{eqn:ns_weak}-\\ref{eqn:en_weak}. Here $L$ is the operator for the PDEs\nin \\ref{sub_sec:ns_en}, and S is a stabilization operator\nwhich is chosen to be the negative adjoint of the differential operator\nterms of $L$, and ${\\bf c}$ and ${\\bf \\phi}$ are state and test\nfunction vectors, i.e. $ {\\bf c}= ({\\bf u},p,T)$, and ${\\bf \\phi} = (\n{\\bf v},w,q )$. The angle brackets $\\langle \\cdot,\\cdot \\rangle$ signify\nintegration of the element interiors for each of the K elements, that is:\n\\begin{equation}\n \\langle {\\bf u},{\\bf v} \\rangle_\\tau = \\sum_K \\tau_K({\\bf u},{\\bf\n v})_K.\\label{eq:tau}\n\\end{equation}\nThis results in three stabilization parameters, $\\tau_P, \\tau_v, \\tau_T$, \nwhich are selected as proposed by Becker and Braack\\footnote{\\normalsize\nDerivations of the weak form and stabilization terms are provided\nin Appendix~\\ref{app:stab}}. \n\nAfter spatial discretization, the system of ODEs are discretized in time\nusing the backward Euler method\\cite{moin2010fundamentals}. The time\ninterval $(0,T)$ is sliced into $N_t$ steps of uniform temporal length,\n$\\Delta t$, where $n = 0,\\dots,N_t$. \nThis has the form, \n\\begin{equation}\n {\\bf y}_{n+1} = {\\bf y_n } + \\Delta t \\, f({\\bf y_{n+1}},t_{n+1}).\n\\end{equation}\nWhere ${\\bf y_{n+1}}$ denotes the solution vector at the time step $n+1$, for\ninstance. As $f$ is non-linear, a Newton-Raphson method is used to solve\nthe resulting implicit nonlinear problem. \n\n%While an iterative method is\n%significantly more computationally expensive per timestep than a similar\n%explicit method, the method was selected due to its unconditional\n%stability and ease of statistical sampling for a uniform timestep. \n\n%\n% gave not completely described numerical methods\n% for instance, have not indicated the stabilization schemes\n% do not need complete equations, but should permit someone to access \n% the literature and construct precisely the numerical formulations used\n%\n\n\n\\section{Mesh Discretization}\n\n%\n% what about mesh...\n%\nThe domains described in Section~\\ref{sec:bc} are consistently\ndiscretized. This means that the domain extents, $\\{L_x,L_y,L_z\\}$, are\nscaled by system diameter (D) but the same number of grid points are\nused for every simulation. Thus, while the ratio of the domain length to\nsystem diameter remains fixed, the grid spacing $\\{\\Delta x,\\Delta\ny,\\Delta z\\}$ increases proportionally with domain length. \n\nThe eddy diffusivities ($\\nu_{t},K_{t}$) in Equations~\\ref{eqn:visc_tot}\nand \\ref{eqn:therm_tot} are proportionally scaled with grid spacing to \nensure that the cell Reynolds number, \n\\begin{equation}\n \\text{Re}_\\text{cell} = \\frac{\\text{max}(\\Delta x,\\Delta y) \\, u}{\\nu_t},\n\\end{equation} \nis maintained for every simulation, to ensure stability. \nIn this way, larger domains have higher eddy viscosities which serve to \ndissipate effects of scales that are smaller than the grid spacing. \n\nThe mesh has a uniform spacing in the lateral directions, except for a\nsingle refinement in the region of the vanes. Typically, the grid is \nroughly one hundred points in the streamwise and spanwise directions\nbefore the refinement. The refinement halves the spacing (doubles the\nnumber of points) in all three coordinate directions, \\{x,y,z\\}, in this\nregion. The refinement is introduced from the ground to 1.5 times the\nheight of the vanes and cone.\n\n%\n% makes no sense\n%\n% describe a mapping what C's are used\n%\n% describe a grid refinement verification\n%\n%\n\nThe vertical mesh spacing is non-uniform to resolve the boundary\nlayer. This is accomplished by redistributing a mesh which is uniformly\nspaced over the interval $\\{0,L_z\\}$ in height, $\\hat z$, to a\nnon-uniform mesh, $z$, over the same interval. \n%This non-uniform ``boundary-layer'' mesh is designed to resolve the small \n%scale variations near the wall. \n% what about chi?\nA mapping $z = \\chi(\\hat z)$ is defined by the following conditions. \nThe minimum and maximum extents are preserved so that $ \\chi(0) = 0 $\nand $ \\chi(L_z) = L_z $. \nThe map spacing between points is initially a specified minimum value, \n\\begin{equation}\n h_{\\text{min}} = \\chi'(0) \\, \\hat h,\n\\end{equation}\nwhere $\\hat h = L_z/\\text{nz}$. The map is strictly increasing until the \npoint $\\hat z_b$, where the spacing reaches a constant, \n$h_{\\text{max}} = \\chi'(\\hat z_b) \\, \\hat h$. \nFrom this point onward, the mesh has a constant uniform spacing, \n$ h_{\\text{max}} = \\chi'(L_z) \\, \\hat h$. \n%\nFinally, the mapping is assumed to hold an exponential functional form, \n$ \\chi(\\hat z) = A e^{\\beta \\hat z}$.\n\n%\n% mesh mapping here\n%\nThen, by specifying $h_{\\text{min}}$, $h_{\\text{max}}$ and \n$z_b$ (chosen thickness of the boundary layer mesh), \nthese conditions are sufficient to define a mapping, \n\\begin{equation}\n z = \\chi(\\hat z) = \n \\begin{cases} C_1 (\\hat z - L_z) + L_z & \\text{if } \\hat z \\geq \\hat z_b, \\\\\n C_2 \\text{ exp}(C_3 \\, \\hat z - 1) & \\text{if } \\hat z < \\hat z_b. \\\\\n \\end{cases}\n\\end{equation}\n%\nHere, $C_1-C_3$ are coefficients, and are defined as, %\\todo{show coefficients}. \n\\begin{align}\n C_1 &= C_2 \\, C_3 e^{C_3 \\, \\hat z_b}, \\\\\n C_2 &= \\frac{\\hat z_b}{h_r -1}, \\\\\n C_3 &= \\frac{\\text{ln }(h_r) }{\\hat z_b},\n\\end{align}\nwhere $h_r$ is the ratio between the largest and smallest spacing, \n\\begin{equation}\n h_r = \\frac{h_{\\text{max}}}{h_{\\text{min}}}, \n\\end{equation}\nand,\n\\begin{equation}\n \\hat z_b = \\frac{L_z}{1 + \\frac{L_z - z_b}{C_2 \\, h_r \\text{ ln }(h_r)}}.\n\\end{equation}\n%% \\begin{equation}\n%% z = \\begin{cases} C_1 (z-L_z)+L_z,& \\text{if } z \\geq z_\\delta\\\\\n%% C_2 \\text{ exp}(C_3 z - 1), & \\text{otherwise}\n%% \\end{cases}\n%% \\end{equation}\n\n%The mesh has an exponentially varying character, \n%$h_{\\text{min}}$, $h_{\\text{max}}$ and $z_b$ \n%are chosen to ensure ten or more\n%points in the boundary layer, isotropic spacing in cells outside of\n%it, and smooth blending between these two regimes. \n\nGrid refinement verification was performed, where the spacing was \ntested against a mesh with twice the grid points in the vertical\ndirection for a representative test case. The results were not sensitive\nto the choice of spacing. A horizontal slice though a\ntypical domain is shown in Figure~\\ref{fig:meshing}. The single \nrefinement in the region of the vanes is visible, along with the finer \nmeshed boundary layer region near the ground. \n\n \\begin{figure}[!htb]\n \\begin{center}\n \\includegraphics[width = 10 cm]{figs/meshing}\n \\caption{Horizontal slice through the domain, to show a\n representative meshing. The single refinement region around the\n vanes is visible, along with the finer boundary layer mesh near the\n ground.}\n \\label{fig:meshing}\n \\end{center}\n \\end{figure}\n\n\n% (01:40:40 PM) Roy Stogner: Not just looks boundary layer ish.\n% (01:40:43 PM) Roy Stogner: We had a sweet derivation for this.\n% (01:41:20 PM) Roy Stogner: Geometric progression of the nodes, solved the inverse problem for the redistribution, the whole works.\n%\n\n%% hmin = 0.001 \n%% hmax = 0.4 \n%% zb = 2.0\n%% hrat = ${/ ${mesh-options/hmax} ${mesh-options/hmin}}\n%% zmax = ${mesh-options/domain_x3_max}\n%% loghrat = ${= log ${mesh-options/hrat}}\n%% c2 = ${/ ${mesh-options/zb} ${- ${mesh-options/hrat} 1}} \n%% zetab = ${/ ${mesh-options/zmax} ${+ 1 ${/ ${- ${mesh-options/zmax} ${mesh-options/zb}} ${* ${mesh-options/c2} ${mesh-options/hrat} ${mesh-options/loghrat}}}}}\n%% c3 = ${/ ${mesh-options/loghrat} ${mesh-options/zetab}}\n%% mesh_nx3 = ${= ceil ${/ ${* ${mesh-options/zmax} ${mesh-options/c2} ${mesh-options/c3}} ${mesh-options/hmin}}}\n%% c1 = ${* ${mesh-options/c2} ${mesh-options/c3} ${= exp ${* ${mesh-options/c3} ${mesh-options/zetab}}}}\n%% redistribute = '{x}{y}{if(z>${mesh-options/zetab},${mesh-options/c1}*(z-${mesh-options/zmax})+${mesh-options/zmax},${mesh-options/c2}*(exp(${mesh-options/c3}*z)-1))}' \n\n%After operation, solutions are evaluated to ensure that \n%the qualitative character of the solution does not change.\n\n\\section{Software}\n\nThe numerical formulations described in Section~\\ref{sec:discretization}\nis implemented with the GRINS library\\cite{GRINSpaper} by Bauman\nand Stogner using the libMesh\\cite{libMeshPaper} FEM\ninfrastructure. Designed to support multiphysics FEM applications, \nGRINS is a flexible library that effectively addresses a wide range of \nscience and engineering problems. \n \n%the reusability\n%and extensibility of mathematical modeling kernels, supporting\n%interfaces to existing solver and discretization libraries to enable\n%modern solution strategies, while, at the same time, retaining\n\n%GRINS provides a platform that enables powerful numerical algorithms\n%such as adjoint-based AMR, adaptive modeling, sensitivity analysis,\n%and, eventually, enabling uncertainty quantification. While few of these\n%capabilities are in use for the present work, they could be useful in\n%future investigations. \n\nGRINS stands for, ``General Reacting Incompressible Navier-Stokes'',\nwhich roughly encapsulates the physical regimes it was originally\ndesigned to simulate. GRINS is open-source, and available on\n\\hyperref[www.github.com/grinsfem/grins]{GitHub}. It is released \nunder LGPL2.1. \n\n%The remainder of this Section is devoted to\n%discussing the underlying libraries used and the description of the\n%GRINS framework. \n\nGRINS uses the fparser~\\cite{fparser}\nlibrary to support both parsing and compilation of mathematical\nfunctions into high performance kernels. This capability allows for\neasy specification of boundary conditions, initial conditions, or\nconstitutive equations from an input file. Some of these inputs are\ndetailed in Appendix~\\ref{sec:archiving}. \n\nGRINS/libMesh are built on the PETSc\\cite{petsc} solver package, which\nprovides the numerical linear algebra packages used for constructing and\nusing sparse matrices, finding the solution of linear systems,\nand for preconditioning. \n\nWhile a variety of solver options have been tested in PETSc, all the\nresults shown in this document use GMRES with block Jacobi for\npreconditioning\\cite{Saad:2003} the linear solve. This uses the\ninverse of the diagonal block for that processor for preconditioning of\nthe entire linear system. In addition, a preconditioner is used for the\nsolution of the diagonal block. This is approximated with incomplete LU\nfactorization\\cite{chan1997approximate}. Here, the ``incomplete'' refers \nto the level of fill, with greater levels of fill approaching\nthe ``complete'' LU factorization. \n\nIn principle, alternative software libraries/frameworks such as\nFEniCS\\cite{AlnaesBlechta2015a} or OpenFOAM\\cite{jasak2007openfoam}\nwould be capable of simulating this problem. While these and\nother libraries have various strengths and weaknesses, the pre-eminent\nconcern is the parallel performance at the intended processor count, due\nto the rapid design iterations necessary for this research\ncampaign. Given these concerns, the GRINS library is a satisfactory tool. \n\n%% (11:41:54 AM) nick: ``-ksp_view -ksp_type gmres -pc_type bjacobi\n%% -sub_pc_type ilu -sub_pc_factor_levels 0'' \n%% (11:42:00 AM) Paul Bauman: OK\n%% (11:42:17 AM) Paul Bauman: -pc_type is the preconditioner for the entire linear system\n%% (11:42:25 AM) Paul Bauman: You're doing bjacobi = block Jacobi\n%% (11:42:28 AM) nick: right\n%% (11:42:36 AM) nick: and does anyone have a good reference I can learn\n%% this better? i feel as if I cant look this up, for some reason \n%% (11:42:39 AM) Paul Bauman: That is just using the inverse of the\n%% diagonal block for that processor \n%% (11:42:46 AM) Paul Bauman: Now\n%% (11:42:55 AM) Paul Bauman: that is a linear solve\n%% (11:43:17 AM) Paul Bauman: So you can use all the linear solver\n%% technology to solve or approximately that block \n%% (11:43:24 AM) Paul Bauman: Hence, -sub_pc_type\n%% (11:43:32 AM) hil left the room.\n%% (11:43:46 AM) Paul Bauman: That's the preconditioner it's going to\n%% use to precondition the linear system for the solution of the\n%% diagonal block \n%% (11:43:53 AM) Paul Bauman: You're telling it to use incomplete lu\n%% (11:44:26 AM) Paul Bauman: Now the -sub_pc_factor_levels options\n%% applies to ilu \n%% (11:44:28 AM) hil entered the room.\n%% (11:44:44 AM) Paul Bauman: The incomplete part of imcomplete LU is\n%% about the level of fill you use \n%% (11:45:05 AM) Paul Bauman: The more levels of fill you have, the more\n%% ``complete'' the incomplete LU will be \n%% (11:45:08 AM) Paul Bauman: Does that make sense?\n%% (11:45:23 AM) nick: no, that is where you lost me\n%% (11:45:39 AM) nick: i dont think i know this level of fill\n%% (11:46:17 AM) Paul Bauman: Check out Youssef Saad's book if more\n%% curious about the subject \n%% (11:46:37 AM) nick: cool thanks\n%% (11:46:38 AM) Paul Bauman: Suffice it to say, you heopfully shouldn't\n%% ever need to go past 3 or 4 levels of fill \n%% (11:47:03 AM) Paul Bauman: Also, if you've got superlu installed with\n%% the PETSc, consider using -sub_pc_factor_mat_solver_package superlu \n%% (11:47:15 AM) Paul Bauman: That's a *much* faster/better\n%% implementation than PETSc's \n\nAt the time of this writing, GRINS has 94 regression tests, which\nprovides a reasonable degree of confidence in verification testing of\nthe library. Several of these tests directly test the capabilities in\nGRINS used in this study. In particular, several of the tests were\ncontributed to GRINS by the author over the course of this\nwork during the addition of several of the models detailed in\nChapter~\\ref{sec:mathmodel}. \n\n\n\\section{Tool Chain and Simulation Custodianship}\n\nSimulations are performed on the Texas Advanced Computing Center's \n(TACC) supercomputers Lonestar Four, Lonestar Five, and Stampede. Run\ndurations for transient cases are typically twelve hours to perform\nseveral hundred timesteps. The steady runs are considerably shorter,\nand require less than ten minute runtimes. Typically the wall clock\ntimes of the steady-state runs are two or three minutes to solution. \n%As a result, the queue time is significantly longer than the actual\n%production runtime. \nThese runs use 264-528 processing cores, or 22-44 nodes on Lonestar4\n(with 12 cores per node), and a similar number for Stampede. The runs have \nseveral million degrees of freedom (DoF), and the local number of DoF\nper core is maintained at $O(10^4)$. This was selected due to memory\nconstraints, after a strong scaling analysis of the performance of the\ncode on these resources, and after consulting with the software developers. \nAt the time of this writing libMesh has been scaled to tens of thousands of\ncores and has been run on over 100,000 cores on the BG/Q machine Mira at\nArgonne National Lab\\cite{gaston2013massive}, and the scaling results\nhere are consistent with the performance expectations for this library.\n\n%In November, 2015, runs were also staged during early access into full\n%production period on TACC's new system, Lonestar5. This machine is\n%closer in terms of software stack to Stampede, \n%using the Simple Linux Utility for Resource Management\n%(Slurm) scheduling system. \n\nRuns are submitted via job scripts to the SLURM\\cite{yoo2003slurm}\nscheduling system. After a run terminates, several additional scripts\nare automatically invoked. \nThese archive the run (outside of the volatile /scratch \nproduction directories) and simultaneously, label the concluded run with\nunique metadata that defines the system environment; the jobs input\nfiles and run definitions; and information detailing the\nhypothesis or physics the job was intended to investigate. Finally, once \na week a script performs \\textbf{rsync} on the entire archived database to\n%ensure more than single redundancy for the runs. \nmaintain redundant storage of the simulation\ndata. Appendix~\\ref{sec:archiving} details the archived simulations and\nstorage patterns. \n\nIn other words, the workflow is designed to permit rapid queuing of a\nseries of runs (in parallel) to investigate a variety of conditions or\nscenario parameters. This capability is necessary for the optimization\ncampaign detailed in Section~\\ref{sec:results}, where running many\nconcurrent investigations are required to sample the configuration\nspace. \n\n%\\section{Testing and Verification}\n%grid convergence?\n\n\n%The validation of the runs is detailed in the next Chapter. \n", "meta": {"hexsha": "b1c7a1c5b00c29cb1a060ec620f86fc4097825d9", "size": 20716, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "disputatio/dissertation/chapter-software.tex", "max_stars_repo_name": "nicholasmalaya/paleologos", "max_stars_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-04T17:49:42.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-04T17:49:42.000Z", "max_issues_repo_path": "disputatio/dissertation/chapter-software.tex", "max_issues_repo_name": "nicholasmalaya/paleologos", "max_issues_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "disputatio/dissertation/chapter-software.tex", "max_forks_repo_name": "nicholasmalaya/paleologos", "max_forks_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-01-04T16:08:18.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-16T19:34:24.000Z", "avg_line_length": 47.1890660592, "max_line_length": 170, "alphanum_fraction": 0.7457520757, "num_tokens": 5650, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511469672594, "lm_q2_score": 0.7549149813536518, "lm_q1q2_score": 0.6460948526542901}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{showlabels}\n\\usepackage{amssymb}\n\\title{Statistical Computing for Scientists and Engineers\\\\[1em] Homework 3}\n\\author{Jiale Shi}\n\\date{September/21/2018}\n\n\\usepackage{natbib}\n\\usepackage{graphicx}\n\\usepackage{array}\n\\begin{document}\n\\maketitle\n\n\\newpage\n\\section{Problem 1}\nThe problem of interest consists of a scalar output $y$ (the logarithmic of the number of caterpillars colonies in a $500 m^2$ area) and $k=10$ explanatory variables $x_{i} (i=1,...,10)$ which correlate to features such as altitude, landscape slope, vegetation count, etc. The exact meaning does not matter in this context. Using the caterpillar.mat data, plot the semi-log-y plots of the caterpillar colony (last column of the data file) versus each feature reproducing Figure 3.1 in Bayesian Core.\n\nAnswer: we run the code that provided to us. We get the semi-log-$y$ plots are shown below.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Figure3_1.png}\n\\caption{semi-$\\log$-$y$ plots}\n%\\label{fig:universe}\n\\end{figure}\n\n\n\n\n\\newpage\n\\section{Problem 2}\nFrom the book, 3.1.1 linear model\n\nThe ordinary normal linear regression model is such that \n\\begin{equation}\n y|\\beta,\\sigma^{2},X \\sim \\mathcal{N}_{n}(X\\beta,\\sigma^{2}I_{n})\n\\end{equation}\n\nthe likelihood of the ordinary normal linear model is\n\\begin{equation}\n l(\\beta,\\sigma^{2}|y,X) = (2\\pi\\sigma^{2})^{n/2} \\exp\\left[-\\frac{1}{2\\sigma^2}(y-X\\beta)^{T}(y-X\\beta)\\right]\n\\end{equation}\n\nThe maximum likelihood estimator of $\\beta$ is then the solution of the (least squares) minimization problem.\n\n\\begin{equation}\n\\min_{\\beta}(y-X\\beta)^{T}(y-X\\beta) = \\min_{\\beta}\\sum_{i=1}^{n}(y_{i}-\\beta_{0}-\\beta_{1}x_{i1}-...-\\beta{k}x_{ik})^2 \n\\end{equation}\nnamely,\n\\begin{equation}\n\\hat{\\beta} = (X^{T} X)^{-1}X^{T}y \n\\end{equation}\nwhich is also the orthogonal projection of y on the linear subspace spanned by the columns of $X$.\n\nSimilarly, an unbiased estimator of $\\sigma^{2}$ is\n\\begin{equation}\n \\hat{\\sigma}^2 = \\frac{1}{n-k-1}(y-X\\hat{\\beta})^{T} (y-X\\hat{\\beta}) = \\frac{s^{2}}{n-k-1}\n\\end{equation}\nand $\\hat{\\sigma}^2(X^{T}X)^{-1}$ approximates the covariance matrix of $\\hat{\\beta}$. We can then define the standard t-statistic as \n\n\\begin{equation}\n T_{i} = \\frac{\\hat{\\beta}_{i}-\\beta_{i}}{\\sqrt{\\hat{\\sigma}^2\\omega_{(i,i)}}} ~\\mathcal{J}_{n-k-1} = \n\\end{equation}\nwhere $\\omega_{(i,i)}$ denotes the $(i,i)$th element of the matrix $(X^{T}X)^{-1}$. For this problem $\\beta_{i}=0$ \n\nthe p value\n\\begin{equation}\n p_{i}= P_{H_{0}}(|T_{i}|-|t_{i}|) < \\alpha\n\\end{equation}\n\nNote that this statistic $T_{i}$ can also be used to build on the $\\beta_{i}$s as a (marginal) frequentist confidence interval, of the form.\n\n\\begin{equation}\n {\\beta_{i};|\\beta_{i}-\\hat{\\beta}_{i}|<=\\sqrt{\\omega_{ii}}F^{-1}_{n-k-1}(1-\\alpha/2)}\n\\end{equation}\n\nThen we run the code that is offered to us. We reproduce Figure 3.2\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Figure3_2.png}\n\\caption{Dataset caterpillar: output providing the maximum likelihood estimates of the regression coefficients and their standard significance analysis.}\n%\\label{fig:universe}\n\\end{figure}\n\n\n\\section{Problem 3}\nWe mainly focus on the section 3.2.1 Conjugate Priors.\n\nWe use the conjugate prior:\n\\begin{equation}\n\\begin{aligned}\n &\\beta|\\sigma^{2}, X \\sim \\mathcal{N}_{k+1}(\\Tilde{\\beta},\\sigma^{2}M^{-1}) \\\\\n &\\sigma^{2} | X \\sim \\mathcal{IG}(a,b)\n\\end{aligned}\n\\end{equation}\n%Use the hyper-parameter values of $\\Tilde{\\beta}$,k=10,a=2.1andb=2.0forcomputingthetablevalues.\nThe prior is conjugate, leading us to the following normal-inverse-gamma posterior distribution.\n\n\\begin{equation}\n\\begin{aligned}\n &\\beta|\\sigma^{2},y, X \\sim \\mathcal{N}_{k+1}((M+X^{T}X)^{-1}X^{T}X\\Tilde{\\beta},\\sigma^{2}[M^{-1}+(X^{T}X)^{-1}]^{-1}) \\\\\n &\\sigma^{2} | y,X \\sim \\mathcal{IG}(\\frac{n}{2},b+\\frac{s^2}{2}+\\frac{\\hat{\\beta}^{T}[M^{-1}+(X^{T}X)^{-1}]^{-1}\\hat{\\beta}}{2})\n\\end{aligned}\n\\end{equation}\nIn this setting, the Bayes estimators of $\\beta$ and $\\sigma^{2}$ associated with squared error losses, namely the posterior means of $\\beta$ and $\\sigma^{2}$, can be computed in closed form. In fact, a simple computation shows that they are given by\n\\begin{equation}\n\\begin{aligned}\n \\mathbb{E}^{\\pi}[\\beta|y,X] &= \\mathbb{E}^{\\pi}[\\mathbb{E}^{\\pi}(\\beta|\\sigma^2,y,X)|y,X] \\\\\n & = (M+X^{T}X)^{-1}{(X^{T}X)\\hat{\\beta}+M\\hat{\\beta}}\n\\end{aligned}\n\\end{equation}\n\nand for $n\\geq 2$\n\n\\begin{equation}\n\\begin{aligned}\n \\mathbb{E}^{\\pi}[\\beta|y,X] &= \\frac{2b+s^{2}+(\\tilde{\\beta}-\\hat{\\beta})^{T}{M^{-1}+(X^{T}X)^{-1}}^{-1}(\\tilde{\\beta}-\\hat{\\beta})}{n+2a-2}\n\\end{aligned}\n\\end{equation}\n\nIntegrating Eq.10 in $\\sigma^{2}$ leads to a multivariate $t$ marginal posterior distribution on $\\beta$ since.\n\n\\begin{equation}\n\\begin{aligned}\n\\pi(\\beta|y,X) & \\propto [ (\\beta-{M+X^{T}X}^{-1}[{X^{T}X}\\hat{\\beta}+M\\tilde{\\beta}])^{T} (M+X^{T}X) \\\\\n&\\times (\\beta-{M+X^{T}X}^{-1}[{X^{T}X}\\hat{\\beta}+M\\tilde{\\beta}]) +2b+s^2 \\\\\n&+(\\tilde{\\beta}-\\hat{\\beta})^{T} (M^{-1}+(X^{T}X)^{-1})^{-1} (\\tilde{\\beta}-\\hat{\\beta})\n]^{-(n/2+k/2+a)}\n\\end{aligned}\n\\end{equation}\n\nWe thus have that, marginally and a posteriori,\n\\begin{equation}\n \\beta|y,X \\sim \\mathcal{J}_{k+1}(n+2a,\\hat{\\mu},\\hat{\\Sigma})\n\\end{equation}\n\nwith \n\\begin{equation}\n\\begin{aligned}\n & \\hat{\\mu} = (M+X^{T}X)^{-1}((X^{T}X)\\hat{\\beta}+M\\tilde{\\beta}) \\\\\n & \\hat{\\Sigma} = \\frac{2b+s^2 +(\\tilde{\\beta}-\\hat{\\beta})^{T} (M^{-1}+(X^{T}X)^{-1})^{-1} (\\tilde{\\beta}-\\hat{\\beta})}{n+2a}(M+X^{T}X)^{-1}\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n \\mathbb{V}^{\\pi} &= \\frac{n+2a}{n+2a-4} \\hat{\\Sigma} \\\\\n &= \\frac{2b+s^2 +(\\tilde{\\beta}-\\hat{\\beta})^{T} (M^{-1}+(X^{T}X)^{-1})^{-1} (\\tilde{\\beta}-\\hat{\\beta})}{n+2a-4}(M+X^{T}X)^{-1}\n\\end{aligned}\n\\end{equation}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_1.png}\n\\caption{Dataset caterpillar: Influence of the prior scale c on the Bayes estimates of $\\sigma^2$ and $\\beta_{0}$.}\n%\\label{fig:universe}\n\\end{figure}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_2.png}\n\\caption{Dataset caterpillar: Bayes estimates of $\\beta$ for $c=100$.}\n%\\label{fig:universe}\n\\end{figure}\n\n\\newpage\n\\section{Problem 4}\nThe posterior distribution can be derived as \n\\begin{equation}\n\\begin{aligned}\n \\pi(\\beta,\\sigma^2|y,X) & \\propto f(y|\\beta,\\sigma^2,X)\\pi(\\beta,\\sigma^2|X) \\\\\n &\\propto (\\sigma^{2})^{-n/2-1} \\exp [-\\frac{1}{2\\sigma^2}(y-X\\hat{\\beta})^{T}(y-X\\hat{\\beta})-\\frac{1}{2\\sigma^2}(\\beta-\\hat{\\beta})^{T}(X^{T}X)(\\beta-\\hat{\\beta})]\\\\\n &\\cdot (\\sigma^2)^{-k/2} \\exp[-\\frac{1}{2c\\sigma^2}(\\beta^T X^T X\\beta)]\n\\end{aligned}\n\\end{equation}\nConduct marginalization and derive the conditional and marginal posteriors on $\\beta$ and $\\sigma^{2}$:\n\n\\begin{equation}\n\\begin{aligned}\n & \\beta | \\sigma^{2},y, X \\sim \\mathcal{N}_{k+1}(\\frac{c}{c+1}\\hat{\\beta}, \\frac{\\sigma^2 c}{c+1}(X^{T} X)^{-1}) \\\\\n & \\sigma^2 | y, X \\sim \\mathcal{IG}(\\frac{n}{2}, \\frac{s^2}{2}+\\frac{1}{2(c+1)}\\hat{\\beta}^T X^T X \\hat{\\beta}) \\\\\n\\end{aligned}\n\\end{equation}\n\n\nBy integrating out $\\sigma^2$ on the conditional posterior $\\beta$, we can show that\n\\begin{equation}\n\\begin{aligned}\n \\beta | y,X \\sim \\mathbb{T}_{k+1} (n,\\frac{c}{c+1}\\hat{\\beta},\\frac{c(s^2+\\frac{\\hat{\\beta}^{T} X^{T} X \\hat{\\beta}}{c+1})}{n(c+1)}(X^T X)^{-1})\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n & \\mathcal{E}[\\beta|y,X] = \\frac{c}{c+1} \\hat{\\beta} \\\\\n & \\mathcal{V}[\\beta|y,X] = \\frac{c(s^2+\\frac{\\hat{\\beta}^{T} X^{T} X \\hat{\\beta}}{c+1})}{(n-2)(c+1)}(X^T X)^{-1}\n\\end{aligned}\n\\end{equation}\n\nAnd the expression of Bayes factor has been given.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_3.png}\n\\caption{Dataset caterpillar: Posterior mean and variance of $\\beta$ for $c=100$ using Zellar's G-prior.}\n%\\label{fig:universe}\n\\end{figure}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_4.png}\n\\caption{Dataset caterpillar: Same legend as Table 3.3 for c=1000}\n%\\label{fig:universe}\n\\end{figure}\n\nThe dependence of the Bayes factor on the pair ($c,c_{0}$) cannot be bypassed in the sense that the Bayes factor varies between $0$ and $\\infty$ when $c_{0}/c$ goes from $0$ to $\\infty$. \n\nAlthough these Bayes factors should not be used simultaneously, an informal conclusion is that the most important variables besides the intercept seem to be $X_{1},X_{2},X_{4}$ and $X_{5}$.\n\n\\newpage\n\\section{Problem 5}\nThe marginal posterior distribution from Problem 4:\n\n\\begin{equation}\n\\begin{aligned}\n \\beta | y,X \\sim \\mathbb{T}_{k+1} (n,\\frac{c}{c+1}\\hat{\\beta},\\frac{c(s^2+\\frac{\\hat{\\beta}^{T} X^{T} X \\hat{\\beta}}{c+1})}{n(c+1)}(X^T X)^{-1})\n\\end{aligned}\n\\end{equation}\n\nAs the standard student-t is one symmetrical distribution, we can analytically calculate its $90\\%$ HPD region as the two points whose respective CDF is $5\\%$ and $95\\%$. Denote them as $\\delta_{5}$ and $\\delta_{95}$. The corresponding HPD of $\\beta$ would than be\n\n\\begin{equation}\n\\begin{aligned}\n & \\beta_{5} = \\frac{c}{c+1}\\hat{\\beta} + \\delta_{5} \\cdot \\frac{c(s^2+\\frac{\\hat{\\beta}^{T} X^{T} X \\hat{\\beta}}{c+1})}{n(c+1)}(X^T X)^{-1}) \\\\\n & \\beta_{95} = \\frac{c}{c+1}\\hat{\\beta} + \\delta_{95} \\cdot \\frac{c(s^2+\\frac{\\hat{\\beta}^{T} X^{T} X \\hat{\\beta}}{c+1})}{n(c+1)}(X^T X)^{-1}) \n\\end{aligned}\n\\end{equation}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_5.png}\n\\caption{Dataset caterpillar:$95\\%$ HPD intervals for the components of $\\beta$ for $c=100$}\n%\\label{fig:universe}\n\\end{figure}\n\n\\newpage\n\\section{Problem 6}\nWe now consider c an unknown hyper-parameter, that is we use the same G-prior distribution, and we now introduce a diffuse prior distribution on c.\n\n\\begin{equation}\n\\begin{aligned}\n \\pi(c)=c^{-1} \\mathcal{I}_{\\mathcal{N*}}(c)\n\\end{aligned}\n\\end{equation}\n\nthe corresponding marginal posterior on the parameters of interest is then\n\\begin{equation}\n\\begin{aligned}\n \\pi(\\beta,\\sigma^2|y,X) \\sum_{c=1}^{\\infty} \\pi(\\beta,\\sigma^{2}|y,X,c)f(y|X,c)c^{-1}\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n f(y|X) &\\propto_{c} f(y|X,c)\\pi(c) \\\\\n & \\sum_{c}c^{-1}(c+1)^{-\\frac{k+1}{2}} [y^T y-\\frac{c}{c+1}y^T X(X^T X)^{-1} X^{T} y]^{-n/2}\n\\end{aligned}\n\\end{equation}\n\nFor the posterior mean and variance of $\\beta$, we integrate out $c$ by summation:\n\\begin{equation}\n\\begin{aligned}\n \\mathcal{E}[\\beta|y,X] = \\sum_{c}\\mathcal{E}[\\beta|y,X,c]\\pi(c|y,X)\n\\end{aligned}\n\\end{equation}\n\nbecause $\\pi(c|y,X) \\propto f(y|X,c) \\pi(c)$\n\n\\begin{equation}\n\\begin{aligned}\n \\mathcal{E}[\\beta|y,X] = \\sum_{c} \\frac{c}{c+1} \\hat{\\beta} \\pi(c|y,X) = \\hat{\\beta} \\sum_{c} [\\frac{c}{c+1}\\frac{f(y|X,c)\\pi(c)}{\\sum_{c}f(y|X,c)\\pi(c)}]\n\\end{aligned}\n\\end{equation}\n\n\\begin{equation}\n\\begin{aligned}\n \\mathcal{V}[\\beta|y,X] = \\mathcal{E}[\\mathcal{V}(\\beta|y,X,c)|y,X] + \\mathcal{V}(\\mathcal{E}[\\beta,X,c]|y,X)\n\\end{aligned}\n\\end{equation}\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[scale=0.6]{Table3_6.png}\n\\caption{Table of Problem 6}\n%\\label{fig:universe}\n\\end{figure}\n\nAn important point with this approach is that the marginal distribution of the dataset is available in closed form. It is therefore possible to produce a Bayes regression output, just as in the first-level prior. This is one additional reason why this non-informative prior is introduced.\n\n\\newpage\n\\section{Problem 7}\n\nThe algorithm is stochastic search for the most likely model.\nWhen k is large, it becomes computationally intractable to compute the posterior probabilites of the $2^{k}$ models. Need of a tailored algorithm that samples from $p(\\gamma|y, X)$ and selects the most likely models.\nThis can be done by Gibbs sampling, given the availability of the full conditional posterior probabilities of the $\\gamma_{i}$'s.\nIf the $\\pi$ with the that model parameter is much larger than that without that model parameter, that model parameter should be turned on. \nIf the $\\pi$ with the that model parameter is much smaller than that without that model parameter, that model parameter should be turned off. \n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale=0.6]{Table3_7.png}\n\\caption{Table of Problem 7(a)}\n%\\label{fig:universe}\n\\end{figure}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale=0.6]{Table3_8.png}\n\\caption{Table of Problem 7(b)}\n%\\label{fig:universe}\n\\end{figure}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale=0.6]{Table3_9.png}\n\\caption{Table of Problem 7(c)}\n%\\label{fig:universe}\n\\end{figure}\n\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale=0.6]{Table3_10.png}\n\\caption{Table of Problem 7(d)}\n%\\label{fig:universe}\n\\end{figure}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale=0.6]{Table3_11.png}\n\\caption{Table of Problem 7(e)}\n%\\label{fig:universe}\n\\end{figure}\n\n\n%\\bibliographystyle{plain}\n%\\bibliography{references}\n\\end{document}\n", "meta": {"hexsha": "7d0e9878aa2d6621abf77542286e1414a1d2b111", "size": 12981, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homework/HW3/HW3_Shi_Jiale_tex/main.tex", "max_stars_repo_name": "shijiale0609/Statistical-Computing-Methods", "max_stars_repo_head_hexsha": "e780746d5f1e4b475bf38eb15d9d825daf45ffa6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homework/HW3/HW3_Shi_Jiale_tex/main.tex", "max_issues_repo_name": "shijiale0609/Statistical-Computing-Methods", "max_issues_repo_head_hexsha": "e780746d5f1e4b475bf38eb15d9d825daf45ffa6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homework/HW3/HW3_Shi_Jiale_tex/main.tex", "max_forks_repo_name": "shijiale0609/Statistical-Computing-Methods", "max_forks_repo_head_hexsha": "e780746d5f1e4b475bf38eb15d9d825daf45ffa6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.8778409091, "max_line_length": 499, "alphanum_fraction": 0.6668977737, "num_tokens": 4744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8311430520409023, "lm_q1q2_score": 0.6460473789677025}} {"text": "\\subsubsection{Parabolic Cylinders}\r\n\\noindent\r\nA paraboloid cylinder look like a parabola that has been extruded into 3D space. One of its forms is\r\n\\begin{equation*}\r\n\tz = cy^2\t\r\n\\end{equation*} \r\nwhere $c \\in \\R$.\r\n\r\n[INSERT IMAGE]", "meta": {"hexsha": "ff75c18066619e577f7d7021f1202eeb429f1719", "size": 234, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "multiCalc/differentialMultivariableCalculus/parabolicCylinder.tex", "max_stars_repo_name": "rawsh/Math-Summaries", "max_stars_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "multiCalc/differentialMultivariableCalculus/parabolicCylinder.tex", "max_issues_repo_name": "rawsh/Math-Summaries", "max_issues_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "multiCalc/differentialMultivariableCalculus/parabolicCylinder.tex", "max_forks_repo_name": "rawsh/Math-Summaries", "max_forks_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.0, "max_line_length": 101, "alphanum_fraction": 0.7179487179, "num_tokens": 76, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8311430478583167, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.6460473671409703}} {"text": "\\section{Overview}\\label{sec:overviews}\n\n\\def\\sm{\\texttt{sm}}\n\\def\\mc{\\texttt{mconcat}}\n\n\\subsection{Monoids and Parallelism}\\label{subsec:monoids}\n\nAs noted in section~\\ref{sec:intro}, associativity ultimately\nallows monoidal operations to be run in parallel. However, for our\npurposes, we would like a way of breaking an input into pieces and\nrunning our computation, string matching, in parallel on all the\npieces. For this, we will focus on the standard Haskell function \\mc,\n\\begin{code}\n mconcat :: Monoid m => [m] -> m\n mconcat [] = mempty\n mconcat (x:xs) = x <> mconcat xs\n\\end{code}\nwhere @mappend@ and @mempty@ are the monoid operation and identity for\nmonoid \\texttt{m} (usually written \\texttt{mappend} and\n\\texttt{mempty})~\\footnote{\\mc\\, is usually defined as \\texttt{foldr mappend mempty}}.\n\n\\mc is a monoid morphism from lists of values of\na monoid to that monoid, i.e.\n%\n$$\n\\begin{array}{rcl}\n\\mc\\;[] & \\quad \\equiv \\quad & \\mempty \\\\\n\\mc\\;([x_0,\\;\\ldots x_m]\\;\\texttt{++}\\;[y_0,\\ldots y_n])\n& \\quad \\equiv \\quad &\n\\mc\\;[x_0, \\ldots x_m]\\;\\mappend\\;\\mc\\;[y_0,\\ldots y_n]\n\\end{array}\n$$\n%\nwhere \\texttt{++} is list append.\n\n$$\n\\begin{array}{l}\n\\mc\\;[\\sm\\;x_0,\\;\\ldots \\sm\\;x_i, \\sm\\;x_{i+1},\\ldots \\sm\\;x_j, \\sm\\;x_{j+1},\\ldots \\sm\\;x_n] \\\\\n\\;\\;\\equiv\\;\\; \\quad\n\\mc\\;([\\sm\\;x_0,\\;\\ldots \\sm\\;x_i]\\;\\texttt{++}\\;[\\sm\\;x_{i+1},\\ldots \\sm\\;x_j]\\;\\texttt{++}\\;[\\sm\\;x_{j+1},\\ldots \\sm\\;x_n]) \\\\\n\\;\\;\\equiv\\;\\; \\quad\n%\\mc\\;[\\mc\\;[\\sm\\;x_0,\\;\\ldots \\sm\\;x_i]\\;,\\;\\mc\\;[\\sm\\;x_{i+1},\\ldots \\sm\\;x_n]] \\\\\n%\\;\\;\\equiv\\;\\; \\quad\n\\mc\\;[\\sm\\;x_0, \\ldots \\sm\\;x_i]\\;\\mappend\\;\\mc\\;[\\sm\\;x_{i+1},\\ldots \\sm\\;x_j]\\;\\mappend\\;\\mc\\;[\\sm\\;x_{j+1},\\ldots \\sm\\;x_n]\n\\end{array}\n$$\n\n\n\\subsection{String Matching}\\label{subsec:string-matching}\n\nString matching is determining all the indices in a source string\nwhere a given target string begins; for example, for source string\n\\texttt{abababa} and target \\texttt{aba} the results of string\nmatching would be \\texttt{[0, 2, 4]}. We will use the following code\nas a specification of string matching:\n\\begin{code}\nmakeIndices inp trg = go 0 where\n trgLen = length trg\n go i | i >= length inp = []\n | take trgLen (drop i inp) == trg = i : rest\n | otherwise = rest\n where rest = go (i + 1)\n\n*Main> makeIndices \"abababa\" \"aba\" == [0, 2, 4]\n\\end{code}\nThough this implementation is not efficient, it is simple and provides\na good starting point from which to create a verifiable function (see\nsec~\\ref{sec:monoid}).\n\n\\subsection{Monoidal String Matching}\\label{subsec:monoidal-string-matching}\n", "meta": {"hexsha": "ec3935b4f10ea31336b41074f6b5ad1518f8df30", "size": 2550, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "text/stringmatcher/overview.tex", "max_stars_repo_name": "nikivazou/thesis", "max_stars_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 11, "max_stars_repo_stars_event_min_datetime": "2016-12-02T00:46:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-02-20T07:04:01.000Z", "max_issues_repo_path": "text/stringmatcher/overview.tex", "max_issues_repo_name": "nikivazou/thesis", "max_issues_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "text/stringmatcher/overview.tex", "max_forks_repo_name": "nikivazou/thesis", "max_forks_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2016-12-02T00:46:51.000Z", "max_forks_repo_forks_event_max_datetime": "2016-12-02T00:46:51.000Z", "avg_line_length": 35.9154929577, "max_line_length": 128, "alphanum_fraction": 0.6654901961, "num_tokens": 949, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430436757312, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.6460473638898473}} {"text": "\\chapter{Introduction to Recurrent Neural Networks}\n\n\t\\section{What is a Recurrent Neural Network?}\n\t\n\t\\begin{itemize}\n\t\t\\item Recurrent neural networks or RNNs (Rumelhart et al.,1986a) are a family of\n\t\tneural networks for processing sequential data.\n\t\t\\item Much as a convolutional network is a neural network that is specialized for processing a grid of values X such as an image, a recurrent neural network is a neural network that is specialized for\n\t\tprocessing a sequence of values $X = \\{x^{(1)}, . . . , x^{(\\tau)}\\}$.\n\t\t\\item Most recurrent networks can also process sequences of \\textbf{variable} length. For each sample $X$, the $\\tau$ doesn't have to bethe same.\n\t\\end{itemize}\n\t\n\t\\subsection{Motivation of RNN}\n\t\t\\textbf{Parameter sharing!} \\\\\n\t\tReasons why we need parameter sharing:\n\t\t\\begin{itemize}\n\t\t\t\\item Parameter sharing makes it possible to extend and apply the model to examples of different forms\n\t\t\t(or different lengths) and generalize across them.\n\t\t\t\\item If we had separate parameters for each value of the time index, we could not generalize to sequence lengths not seen during training, nor share statistical characteristic across different sequence lengths\n\t\t\tand across different positions in time. \n\t\t\\end{itemize}\n\t\n\t\t\\textbf{How to use the parameter sharing?}\n\t\t\\begin{itemize}\n\t\t\t\\item In the Recurrent networks, each member of the\n\t\t\toutput is a function of the previous members of the output.\t\n\t\t\t\\item Each member of the output is produced using \\textbf{the same update rule} applied to the previous outputs. This recurrent formulation results in the sharing of parameters through a very deep computational graph.\n\t\t\\end{itemize}\n\n\t\\subsection{Vanilla RNN}\n\t\\begin{itemize}\n\t\t\\item \n\t\t\\textbf{Training Data of the model} \\\\\n\t\tHere we use the sentence generation as the model, which aims to predict the next character in a sentence given all the characters before. \\\\\n\t\t In this model, each $\\mathbf{x}_i^{(t)}$ is a character in the sentence(represented as a $0-1$ vector), and the target is the next character which is also represented as a $0-1$ vector.\n\t\t\\item\n\t\t\\textbf{The structure of the network} \\\\\n\t\tIt's an example of a recurrent network that maps an\n\t\tinput sequence to an output sequence of the same length. \\\\\n\t\tThe input of the network is $\\{\\mathbf{x}_i^{(1)},\\mathbf{x}_i^{(2)},\\dots,\\mathbf{x}_i^{(\\tau)}\\}_{i=1}^N$, where N is the number of the samples, $\\tau$ is the length of each input, and each $\\mathbf{x}_i^{(t)}$ is a vector in the vector space $\\mathbb{R}^n$. \\\\\n\t\t The output is $\\{\\mathbf{y}_i^{(1)},\\mathbf{y}_i^{(2)},\\dots,\\mathbf{y}_i^{(\\tau)}\\}_{i=1}^N$, the meaning of N and $\\tau$ is the same, while each $\\mathbf{y}_i^{(t)}$ is a vector in the vector space $\\mathbb{R}^m$. \\\\\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[width=0.9\\textwidth]{VaRNN}\n\t\t\t\\caption{The structure of Vanilla RNN}\n\t\t\\end{figure}\n\t\t\n\t\t\\item\n\t\t\\textbf{The update equations for the forward propagation} is\n\t\t\\begin{flalign*}\n\t\t\\mathbf{a}^{(t)} &= \\mathbf{b} + \\mathbf{Wh}^{(t-1)} + \\mathbf{Ux}^{(t)} \\\\\n\t\t\\mathbf{h}^{(t)} &= tanh(\\mathbf{a}^{(t)}) \\\\\n\t\t\\mathbf{O}^{(t)} &= \\mathbf{c} + \\mathbf{Vh}^{(t)} \\\\\n\t\t\\hat{\\mathbf{y}}^{(t)} &= softmax(\\mathbf{O}^{(t)})\n\t\t\\end{flalign*}\n\t\twhere the parameters are bias vectors $\\mathbf{b}$ and $\\mathbf{c}$ along with the weight matrices\n\t\t$\\mathbf{U}$, $\\mathbf{V}$ and $\\mathbf{W}$, respectively for input-to-hidden, hidden-to-output and hidden-to- hidden connections.\n\n\t\t\\item\n\t\t\\textbf{The computation of the loss} \\\\\n\t\tFor a given sequence of $\\mathbf{x}$ values paired with a sequence of $\\mathbf{y}$ values, the total loss would then be just the sum of the losses over all the time steps.\\\\\n\t\tFor example, if $L^{(t)}$ is the negative log-likelihood of $y^{t}$ given $\\{\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(t)} \\}$, the total loss is\n\t\t\\begin{flalign*}\n\t\tL({\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(\\tau)}},{\\mathbf{\\hat{y}}^{(1)},\\ldots,\\mathbf{\\hat{y}}^{(\\tau)}}) \\\\\n\t\t= \\sum_{t=1}^{\\tau} L^{(t)} \\\\\n\t\t= - \\sum_{t=1}^{\\tau} log \\ p_{model}(\\mathbf{\\hat{y}}^{(t)} |{\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(t)}}) \n\t\t\\end{flalign*}\n\t\twhere $p_{model}(\\mathbf{\\hat{y}}^{(t)} |{\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(t)}}) $ is given by the following rule:\\\\\n\t\tDenote the $\\mathbb{R}^m$ output vector $\\hat{\\mathbf{y}}^{t}$ as $(\\hat{y}_1^{(t)},\\hat{y}_2^{(t)},\\dots,\\hat{y}_m^{(t)})$,and the target vector (a $0-1$ vector) is $(y_1^{(t)},y_2^{(t)},\\dots,y_m^{(t)})$ while $y_i^{(t)} = 1$ and all the other elements of the vector is 0. Then \n\t\t\\begin{equation}\n\t\tp_{model}(\\mathbf{\\hat{y}}^{(t)} |{\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(t)}}) = \\hat{y}_i^{(t)}\n\t\t\\end{equation}\n\t\\end{itemize}\n\n\t\\subsection{Some important design patterns for RNN}\n\t\t\\begin{itemize}\n\t\t\t\\item Recurrent networks that produce an output at each time step and have recurrent connections only from the output at one time step to the hidden units at the next time step.\n\t\t\t\\begin{figure}[H]\n\t\t\t\t\\centering\n\t\t\t\t\\includegraphics[width=0.9\\textwidth]{O2HRNN}\n\t\t\t\t\\caption{The structure of RNN(Output to Hidden)}\n\t\t\t\\end{figure}\n\t\t\t\\textbf{The update equations for the forward propagation} is\n\t\t\t\\begin{flalign*}\n\t\t\t\\mathbf{a}^{(t)} &= \\mathbf{b} + \\mathbf{WO}^{(t-1)} + \\mathbf{Ux}^{(t)} \\\\\n\t\t\t\\mathbf{h}^{(t)} &= tanh(\\mathbf{a}^{(t)}) \\\\\n\t\t\t\\mathbf{O}^{(t)} &= \\mathbf{c} + \\mathbf{Vh}^{(t)} \\\\\n\t\t\t\\hat{\\mathbf{y}}^{(t)} &= softmax(\\mathbf{O}^{(t)})\n\t\t\t\\end{flalign*}\n\t\t\twhere the parameters are bias vectors $\\mathbf{b}$ and $\\mathbf{c}$ along with the weight matrices\n\t\t\t$\\mathbf{U}$, $\\mathbf{V}$ and $\\mathbf{W}$, respectively for input-to-hidden, hidden-to-output and output-to- hidden connections.\n\t\t\t\\item Recurrent networks with recurrent connections between hidden units, that read an entire sequence and then produce a single output.\n\t\t\t\\begin{figure}[H]\n\t\t\t\t\\centering\n\t\t\t\t\\includegraphics[width=0.9\\textwidth]{OFRNN}\n\t\t\t\t\\caption{The structure of RNN(Only One Output)}\n\t\t\t\\end{figure}\n\t\t\t\\textbf{The update equations for the forward propagation is}\n\t\t\t\\begin{flalign*}\n\t\t\t\\mathbf{a}^{(t)} &= \\mathbf{b} + \\mathbf{Wh}^{(t-1)} + \\mathbf{Ux}^{(t)} \\qquad 1 \\leq t \\leq \\tau\\\\\n\t\t\t\\mathbf{h}^{(t)} &= tanh(\\mathbf{a}^{(t)}) \\qquad 1 \\leq t \\leq \\tau\\\\\n\t\t\t\\mathbf{O}^{(\\tau)} &= \\mathbf{c} + \\mathbf{Vh}^{(\\tau)} \\\\\n\t\t\t\\hat{\\mathbf{y}}^{(\\tau)} &= softmax(\\mathbf{O}^{(\\tau)})\n\t\t\t\\end{flalign*}\n\t\t\twhere the parameters are bias vectors $\\mathbf{b}$ and $\\mathbf{c}$ along with the weight matrices\n\t\t\t$\\mathbf{U}$, $\\mathbf{V}$ and $\\mathbf{W}$, respectively for input-to-hidden, hidden-to-output and hidden-to- hidden connections\n\t\\end{itemize}\n\n%\t\\subsection{Teaching Force}\n%\t\tNote that in the second pattern,it lacks hidden-to-hidden\n%\t\trecurrence.\n%\t\t\\begin{itemize}\n%\t\t\t\\item It is less powerful.\n%\t\t\t\\item However,for any loss function based on comparing the prediction at time t to the training target at time t, all the time steps are decoupled.\n%\t\t\t\\item Training can thus be parallelized, with the gradient for each step t computed in isolation.\n%\t\t\\end{itemize}\n%\t\t\\begin{figure}[H]\n%\t\t\t\\centering\n%\t\t\t\\includegraphics[width=0.9\\textwidth]{TF}\n%\t\t\t\\caption{The representation of Teaching Force}\n%\t\t\\end{figure}\n%\tThe parallelization is the concept of \\textbf{Teaching Force}.\n%\t\t\\begin{itemize}\n%\t\t\t\\item Teacher forcing is a procedure that emerges from the maximum likelihood criterion, in which during training the\n%\t\t\tmodel receives \\textbf{the ground truth output} $y^{(t)}$ as input at time t + 1.\n%\t\t\t\\item Take a sequence with two time steps as an example. The conditional maximum likelihood criterion is\n%\t\t\t\\begin{flalign*}\n%\t\t\tlog p(\\mathbf{\\hat{y}}^{(1)},\\mathbf{\\hat{y}}^{(2)}|\\mathbf{x}^{(1)},\\mathbf{x}^{(2)}) \\\\\n%\t\t\t= log p(\\mathbf{\\hat{y}}^{(2)} | \\mathbf{\\hat{y}}^{(1)},\\mathbf{x}^{(1)},\\mathbf{x}^{(2)}) + log p(\\mathbf{\\hat{y}}^{(1)} | \\mathbf{x}^{(1)},\\mathbf{x}^{(2)})\n%\t\t\t\\end{flalign*}\n%\t\t\t\\item In this example, we see that at time t = 2, the model is trained to maximize the \\textbf{conditional probability} of $\\mathbf{y}^{(2)}$ given both the $\\mathbf{x}$ sequence so far and the previous $\\mathbf{y}$ value from the training.\n%\t\t\\end{itemize}\t\n\n\t\\section{Development of RNN}\n\t\n\t\\subsection{Encoder-Decoder}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[width=0.9\\textwidth]{Encoder-Decoder}\n\t\t\t\\caption{The structure of Encoder-Decoder}\n\t\t\\end{figure}\n\n\t\t\\begin{itemize}\n\t\t\t\\item It's an RNN that can be trained to map an input sequence to an output sequence which is not necessarily of the same length.\n\t\t\t\\item First, we have an \\textbf{encoder} or \\textbf{input} RNN that processes the input sequence. The encoder emits the context C, usually as a simple function of its final hidden state.\n\t\t\t\\item The context C might be a vector or sequence of\n\t\t\tvectors that summarize the input sequence $\\mathbf{X} = (\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(n_x)})$.\n\t\t\t\\item Then we have a \\textbf{decoder} or \\textbf{output} RNN that is conditioned on that fixed-length vector $\\mathbf{X}$ to generate the output sequence $(\\mathbf{y}^{(1)},\\ldots,\\mathbf{y}^{(n_y)})$.\n\t\t\t\\item The two RNNs are trained jointly to maximize the average of \\\\\n\t\t\t$log P(\\mathbf{y}^{(1)},\\ldots,\\mathbf{y}^{(n_y)}|\\mathbf{x}^{(1)},\\ldots,\\mathbf{x}^{(n_x)})$ over all the pairs of x and y\n\t\t\tsequences in the training set.\n\t\t\\end{itemize}\n\t\\subsection{Deep RNN}\n\t\tThe computation in most RNNs can be decomposed into three blocks of parameters and associated transformations:\n\n\t\t\\begin{itemize}\n\t\t\t\\item[1.] from the input to the hidden state\n\t\t\t\\item[2.] from the previous hidden state to the next hidden state\n\t\t\t\\item[3.] from the hidden state to the output\n\t\t\\end{itemize}\n\t\tExperimental evidence (Graves et al., 2013; Pascanu et al., 2014a) strongly suggests that we need \\textbf{enough depth} in order to perform the required mappings in each of these operations.\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[width=0.9\\textwidth]{DeepRNN}\n\t\t\t\\caption{Some kind of Structure for Deep RNN}\n\t\t\\end{figure}\n\tA recurrent neural network can be made deep in many ways (Pascanu\n\tet al., 2014a). \n\t\\begin{itemize}\n\t\t\\item (a) The hidden recurrent state can be broken down into groups organized hierarchically. \n\t\t\\item (b) Deeper computation can be introduced in the input-to- hidden, hidden-to-hidden and hidden-to-output parts. This may lengthen the shortest path linking different time steps. \n\t\t\\item (c) The path-lengthening effect can be mitigated by introducing skip connections.\n\t\\end{itemize}\n\t\t\n%\t\\subsection{Recursive Neural Network}\n%\t\t\\begin{itemize}\n%\t\t\t\\item Recursive neural networks represent yet another generalization of recurrent networks, with a different kind of computational graph, which is structured as a \\textbf{deep tree}, rather than the chain-like structure of RNNs.\n%\t\t\t\\item One clear advantage of recursive nets over recurrent nets is that for a sequence of the same length $\\tau$, the depth (measured as the number of compositions of nonlinear operations) can be drastically reduced from $\\tau$ to $O(log \\tau)$.\n%\t\t\t\\item However,how to \\textbf{best structure the tree} is an open question.\n%\t\t\\end{itemize}\n%\t\t\\begin{figure}[H]\n%\t\t\t\\centering\n%\t\t\t\\includegraphics[width=0.9\\textwidth]{ReNN}\n%\t\t\t\\caption{The structure of Recursive Neural Network}\n%\t\t\\end{figure}\n\n\\section{The Challenge of long term dependency}\n\\subsection{Computation of the gradient}\nComputing the gradient through a recurrent neural network is straightforward. One simply applies the generalized back-propagation algorithm to the network. \\\\\nWhat we need to take care of is that the back propagation should be computed through \\textbf{the time step}. We will use the example of computing the gradients of Vanilla RNN above to show how the BP algorithm behaves. \\\\\nDenote the final loss as $L$,we can see \n\\begin{equation}\n\\frac{\\partial L}{\\partial L^{(t)}} = 1 \\ ,1\\le t \\le \\tau\n\\end{equation}\nAssume the output function is softmax function and the loss is negative log-likelihood function, The gradient $\\nabla_{o^{(t)}}L$ on the outputs at time step t, for all i,t, is as follows:\n\\begin{equation}\n(\\nabla_{o^{(t)}}L)_i = \\frac{\\partial L}{\\partial L^{(t)}}\\frac{\\partial L^{(t)}}{\\partial o_i^{(t)}} = \\hat{y_i}^{(t)} - 1_{i,y^{(t)}}\n\\end{equation}\nAt the final time step $\\tau$, we have\n\\begin{equation}\n\\nabla_{h^{(\\tau)}}L = V^{T}\\nabla_{o^{(\\tau)}}L\n\\end{equation}\nwith \n\\begin{eqnarray}\n\\nabla_{h^{(t)}}L & = & (\\frac{\\partial h^{(t+1)}}{\\partial h^{(t)}})^{T}\\nabla_{h^{(t+1)}}L + (\\frac{\\partial o^{(t)}}{\\partial h^{(t)}})^{T}\\nabla_{o^{(t)}}L \\\\\n & = & W^T\\nabla_{h^{(t+1)}}L diag(1 - (h^{(t+1)})^2) + V^T\\nabla_{o^{(t)}}L\n\\end{eqnarray}\nfrom $t = \\tau -1$ down to $t = 1$. \\\\\nWith these, the gradient on the remaining parameters is given by:\n\\begin{eqnarray}\n\\nabla_{c}L & = & \\sum_{t=1}^{\\tau} \\nabla_{o^{(t)}}L \\\\\n\\nabla_{b}L & = & \\sum_{t=1}^{\\tau} diag(1 - (h^{(t)})^2)\\nabla_{h^{(t)}}L \\\\\n\\nabla_{V}L & = & \\sum_{t=1}^{\\tau}(\\nabla_{o^{(t)}}L)h^{(t)^T} \\\\\n\\nabla_{W}L & = & \\sum_{t=1}^{\\tau} diag(1 - (h^{(t)})^2)(\\nabla_{h^{(t)}}L)h^{(t-1)^T} \\\\\n\\nabla_{U}L & = & \\sum_{t=1}^{\\tau} diag(1 - (h^{(t)})^2)(\\nabla_{h^{(t)}}L)x^{(t)^T}\n\\end{eqnarray}\n\\subsection{Challenge of long term dependency}\nFrom the computation above, we can see in the recurrent network, it involves the composition of the same function multiple times, once per time step. Recall the equation:\n\\begin{equation}\n\\nabla_{h^{(t)}}L = W^T\\nabla_{h^{(t+1)}}L diag(1 - (h^{(t+1)})^2) + V^T\\nabla_{o^{(t)}}L\n\\end{equation}\nFrom the knowledge of linear algebra, we can see the recurrent composition causes eigenvalues with magnitude less than one to decay to zero and eigenvalues with magnitude greater than one to explode. \\\\\nLet's first focus on the impact of vanishing gradient.Specifically, whenever the model is able to represent long term dependencies, the gradient of a long term interaction has \\textbf{exponentially smaller magnitude} than\nthe gradient of a short term interaction. It does not mean that it is impossible to learn, but that it might take a very long time to learn long-term dependencies, because the signal about these dependencies will tend to be hidden by the smallest fluctuations arising from short-term dependencies. \\\\\nWe will discuss various approaches that have been proposed to reduce the difficulty of learning long- term dependencies, but the problem of learning long-term dependencies remains one of the main challenges in deep learning.\n\n\\subsection{Long Short Term Memory}\nIn the LSTM model, we use a forget unit $f_i^{(t)}$,that sets this weight to a value between 0 and 1 via a sigmoid unit:\n\\begin{equation}\nf_i^{(t)} = \\sigma(b_i^f + \\sum_jU_{i,j}^fx_j^{(t)} + \\sum_jW_{i,j}^fh_j^{(t-1)})\n\\end{equation}\nwhere $x^{(t)}$ is the current input vector and $h^{(t)}$ is the current hidden layer vector, containing the outputs of all the LSTM cells, and $b^f$,$U^f$,$W^f$,are respectively biases, input weights and recurrent weights for the forget gates. \\\\\n The LSTM cell internal state is thus updated as the following, with a conditional self-loop weight $f_i^{(t)}$\n\\begin{equation}\ns_i^{(t)} = f_i^{(t)} \\ast s_i^{(t-1)} + g_i^{(t)} \\ast \\sigma(b_i + \\sum_jU_{i,j}x_j^{(t)} + \\sum_jW_{i,j}h_j^{(t-1)})\n\\end{equation}\nwhere b, U and W respectively denote the biases, input weights and recurrent weights into the LSTM cell and $\\ast$ means the multiplication by elements($.\\ast$ in MATLAB). \\\\\nThe external input gate unit $g_i^{(t)}$ is computed similarly to the forget gate (with a sigmoid unit to obtain a gating value between 0 and 1, but with its own parameters:\n\n\\begin{equation}\ng_i^{(t)} = \\sigma(b_i^g + \\sum_jU_{i,j}^gx_j^{(t)} + \\sum_jW_{i,j}^gh_j^{(t-1)})\n\\end{equation}The output $h_i^{(t)}$ of the LSTM cell can also be shut off, via the output gate $o_i^{(t)}$, which also uses a sigmoid unit for gating:\n\n\\begin{equation}\nh_i^{(t)} = tanh(s_i^{(t)}) \\ast o_i^{(t)}\n\\end{equation}\n\\begin{equation}\no_i^{(t)} = \\sigma(b_i^o + \\sum_jU_{i,j}^ox_j^{(t)} + \\sum_jW_{i,j}^oh_j^{(t-1)})\n\\end{equation}\nLSTM networks have been shown to learn long-term dependencies more easily than the simple recurrent architectures.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.9\\textwidth]{LSTM}\n\t\\caption{Block diagram of the cell of LSTM recurrent network. Cells are connected recurrently to each other, replacing the usual hidden units of ordinary recurrent networks. An input feature is computed with a regular artificial neuron unit. Its value can be accumulated into the state if the sigmoidal input gate allows it. The state unit has a linear self-loop whose weight is controlled by the forget gate. The output of the cell can be shut off by the output gate. All the gating units have a sigmoid nonlinearity, while the input unit can have any squashing nonlinearity. The state unit can also be used as an extra input to the gating units. The black square indicates a delay of a single time step.}\n\\end{figure}\nFrom the equations above, we'll give a brief explanation(not strict!) here why the LSTM works for the long-term dependency. We will just focus on the gradient vanishing. \\\\\nFirst, we have \n\\begin{equation}\ns_i^{(t)} = f_i^{(t)} \\ast s_i^{(t-1)} + g_i^{(t)} \\ast \\sigma(b_i + \\sum_jU_{i,j}x_j^{(t)} + \\sum_jW_{i,j}h_j^{(t-1)})\n\\end{equation}\nhence we get\n\\begin{equation}\n\\frac{\\partial s_i^{(t+k)}}{\\partial s_i^{(t)}} \\approx \\prod_{j=t+1}^{t+k} diag(f_i^{(j)}) + ...\n\\end{equation}\nwhere the $...$ things are some products which are vanishing to zero as k increases(just like we talked above). If in each time step j($t \\le j \\le t+k$), $f_i^{(j)} \\approx \\textbf{1}$(It forgets nothing!),we can see that $\\frac{\\partial s_i^{(t+k)}}{\\partial s_i^{(t)}}$ is almost an identity matrix in the long term, and that means we can connect the information between long time steps together. We can go to the paper (Schmidhuber $\\&$ Hochreiter,1997) for the detailed discussion.\n\\subsection{Other Gated RNNs}\nLet's take a look at the recent work on gated RNNs, whose units are also known as gated recurrent units or GRUs (Cho et al., 2014b;Chung et al., 2014, 2015a; Jozefowicz et al., 2015; Chrupala et al., 2015). \\\\\nThe main difference with the LSTM is that a single gating unit simultaneously controls the forgetting factor and the decision to update the state unit. The update equations are the following:\n\n\\begin{equation}\nh_i^{(t)} = u_i^{(t)} \\ast h_i^{(t-1)} + (1 - u_i^{t}) \\ast \\sigma(b_i + \\sum_jU_{i,j}x_j^{(t-1)} + \\sum_jW_{i,j}r_j^{(t)}h_j^{(t-1)})\n\\end{equation}\nwhere $u$ is 'update' gate and $r$ is 'reset' gate. The values are determined by:\n\\begin{equation}\nu_i^{(t)} = \\sigma(b_i^u + \\sum_jU_{i,j}^ux_j^{(t)} + \\sum_jW_{i,j}^uh_j^{(t-1)})\n\\end{equation}\nand\n\\begin{equation}\nr_i^{(t)} = \\sigma(b_i^r + \\sum_jU_{i,j}^rx_j^{(t)} + \\sum_jW_{i,j}^rh_j^{(t-1)})\n\\end{equation}\n\\section{Optimization methods for RNN}\n\\subsection{Clipping gradients}\nAs discussed before, strongly nonlinear functions such as those computed by a recurrent net over many time steps tend to have derivatives that can be either very large or very small in magnitude. \\\\\nThe difficulty that arises is that when the parameter gradient is very large, a gradient descent parameter update could throw the parameters very far, into a region where the objective function is larger, undoing much of the work before.The gradient tells us the direction that corresponds to the steepest descent within an infinitesimal region surrounding the current parameters. \\\\\nA simple type of solution has been in use by practitioners for many years:\nclipping the gradient. One step is to clip the norm $||g||$ of the gradient g (Pascanu et al., 2013) just before the parameter update:\n\\begin{eqnarray}\n\\text{if} \\ & ||g|| > v \\\\\n& g\\leftarrow \\frac{g v}{||g||}\n\\end{eqnarray}\nBecause the gradient of all the parameters (including different groups of parameters, such as weights and biases) is renormalized jointly with a single scaling factor, the method has the advantage that it guarantees that each step is \\textbf{still in the gradient direction}. Although the parameter updating has the same direction as the true gradient, with gradient norm clipping, the parameter update vector norm is now bounded. This bounded gradient avoids performing a detrimental step when the gradient explodes.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.9\\textwidth]{Clipgrad}\n\t\\caption{Example of the effect of gradient clipping in a recurrent network with two parameters w and b. Gradient clipping can make gradient descent perform more reasonably in the vicinity of extremely steep cliffs. These steep cliffs commonly occur in recurrent networks near where a recurrent network behaves approximately linearly. The cliff is exponentially steep in the number of time steps because the weight matrix is multiplied by itself once for each time step. (Left)Gradient descent without gradient clipping overshoots the bottom of this small ravine, then receives a very large gradient from the cliff face. The large gradient catastrophically propels the parameters outside the axes of the plot. (Right)Gradient descent with gradient clipping has a more moderate reaction to the cliff. While it does ascend the cliff face, the step size is restricted so that it cannot be propelled away from steep region near the solution. Figure adapted with permission from Pascanu et al. (2013).\n}\n\\end{figure}\n\\subsection{Regularizing to Encourage Information Flow}\nGradient clipping helps to deal with exploding gradients, but it does not help with vanishing gradients.To address vanishing gradients and better capture long-term dependencies, an idea is to regularize or constrain the parameters so as to encourage 'information flow'. \\\\\nIn particular, we would like the gradient vector $\\nabla_{h^{(t)}}L$ being back-propagated to maintain its magnitude, even if the loss function only penalizes the output at the end of the sequence. Formally, we want\n\\begin{equation}\n(\\nabla_{h^{(t)}}L) \\frac{\\partial h^{(t)}}{\\partial h^{(t-1)}}\n\\end{equation}\nto be as large as \n\\begin{equation}\n\\nabla_{h^{(t)}}L\n\\end{equation}\nWith this objective, Pascanu et al. (2013) propose the following regularizer:\n\\begin{equation}\n\\Omega = \\sum_{t}(\\frac{|(\\nabla_{h^{(t)}}L) \\frac{\\partial h^{(t)}}{\\partial h^{(t-1)}}|}{|\\nabla_{h^{(t)}}L|} - 1)^2\n\\end{equation}\nThe experiments with this regularizer suggest that, if combined with the norm clipping, the regularizer can considerably increase the span of the dependencies that an RNN can learn.\n", "meta": {"hexsha": "289cfbb9aae9a6bef76d3f58667ce2c3e3a60848", "size": 22552, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/IntroductionRNN.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/IntroductionRNN.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/IntroductionRNN.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 72.0511182109, "max_line_length": 996, "alphanum_fraction": 0.7068552678, "num_tokens": 6977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933447152498, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.6460195434011388}} {"text": "\\section{Dynamics (formally)}\n\\subsection{}\n\n\\begin{frame}\n\\frametitleTC{System -- recap and systematisation}\n\\framesubtitleTC{A mathematical representation (model) of something that evolves}\n\\myPause\n \\begin{columns}\n \\column[T]{0.45\\textwidth}\n \\begin{itemize}[<+-| alert@+>]\n \\item[] Simple representation\n \\item[] \\vspace{2mm}\\begin{center}\n \\input{./Unit-02/img/DS-SimpleRepresentation.tex}\n \\end{center}\n \\begin{itemize}[<+-| alert@+>]\n \\item[] $u$ -- input(s)\n \\item[] $y$ -- output(s)\n \\end{itemize}\n \\end{itemize}\n \\column[T]{0.55\\textwidth}\n \\begin{itemize}[<+-| alert@+>]\n \\item[] Model ingredients\n \\begin{itemize}[<+-| alert@+>]\n \\item what the system evolves upon:\n \\begin{itemize}[<+-| alert@+>]\n \\item the \\TC{continuous time} $t$;\n \\item an integer index $k$ counting some events,\n frequently called\\\\the \\TC{discrete time}.\n \\end{itemize}\n \\item the \\TC{evolution law}:\n \\begin{itemize}[<+-| alert@+>]\n \\item $u[t_0,t]\\;\\rightarrow\\;y[t_0,t]$;\n \\item $u[k_0,k]\\;\\rightarrow\\;y[k_0,k]$.\n \\end{itemize}\n \\end{itemize}\n \\end{itemize}\n \\end{columns} \\myPause\n \\vfill\n \\begin{center}\n \\vfill \\textbf{Is anything missing? \\myPause Sometimes, yes.}\n \\end{center}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{\\underline{Dynamic} system -- recap and systematisation}\n\\framesubtitleTC{The most general definition}\n\\myPause\n\\begin{center}\n {\\Large\n If the knowledge of $u[t_0,t]$ --- or $u[k_0,k]$ \\\\ \\myPause\n allows to determine $y[t_0,t]$ --- or $y[k_0,k]$ \\\\ \\myPause\n the system is said to be \\TC{non dynamic},\\\\ \\myPause\n \\vspace{1.5mm}\\TC{dynamic} otherwise. \\myPause\n }\n\\end{center}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Dynamic system -- recap and systematisation}\n\\framesubtitleTC{Input, output, and \\emph{state}}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item Non dynamic system:\\\\\n $y[t_0,t]$ or $y[k_0,k]$ depends only on $u[t_0,t]$ or $u[k_0,k]$.\n\\item Dynamic system:\\\\\n $y[t_0,t]$ or $y[k_0,k]$ depends on $u[t_0,t]$ or $u[k_0,k]$,\\\\\n and on the initial values $x[t_0]$ or $x[k_0]$ of some quantities.\n\\item These are called the \\TC{state variables}, and form the \\TC{state (vector)}.\n\\item The number of state variables is called the \\TC{order} of the system.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Dynamic system -- recap and systematisation}\n\\framesubtitleTC{How can we express this in mathematical terms?}\n\\myPause\nIn several ways. We see the only two relevant for us.\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item Continuous-Time (CT) system:\\\\\n \\begin{displaymath}\n \\left\\{\n \\begin{array}{rl}\n \\frac{dx(t)}{dt} &= f \\big( x(t),u(t),t \\big) \\\\\n y(t) &= g \\big( x(t),u(t),t \\big)\n \\end{array}\n \\right.\n \\end{displaymath}\n\\item Discrete-Time (DT) system:\\\\\n \\begin{displaymath}\n \\left\\{\n \\begin{array}{rl}\n x(k) &= f \\big( x(k-1),u(k-1),k \\big) \\\\\n y(k) &= g \\big( x(k),u(k),k \\big)\n \\end{array}\n \\right.\n \\end{displaymath}\n\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Dynamic system -- recap and systematisation}\n\\framesubtitleTC{Some more definitions (for both the CT and the DT case)}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item \\TC{Linear (L)} system:\\\\\n $f(\\cdot,\\cdot,\\cdot)$ and $g(\\cdot,\\cdot,\\cdot)$ linear in $x$ and $u$.\n\\item \\TC{Time-Invariant (TI)} system:\\\\\n $f(\\cdot,\\cdot,\\cdot)$ and $g(\\cdot,\\cdot,\\cdot)$ not depending on $t$ or $k$.\n\\item \\TC{Proper} (sometimes, \\emph{strictly} proper) system:\\\\\n $g(\\cdot,\\cdot,\\cdot)$ not depending on $x$,\\\\\n i.e., the input acts on the output only through the state.\n\\item \\TC{SISO} (Single-Input, Single-Output) system:\\\\\n $u$ and $y$ -- not necessarily $x$ -- scalars.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Motion}\n\\framesubtitleTC{General definition}\n\\myPause\n\\centerline{Note: from now on we mostly speak DT, for CT just replace $k$ with $t$.} \\myPause\n\\vspace{3mm}\\begin{itemize}[<+-| alert@+>]\n\\item Initial state + input at a certain start time $\\Rightarrow$ motion, i.e.,\n \\begin{displaymath}\n \\left.\n \\begin{array}{l}\n x(k_0) \\\\ u(k),\\,k \\geq k_0\n \\end{array}\n \\right\\}\n \\quad \\Rightarrow \\quad\n x(k),y(k),\\,k \\geq k_0.\n \\end{displaymath}\n\\item We call $x(k)$ and $y(k)$, respectively,\n \\begin{itemize}[<+-| alert@+>]\n \\item the \\TC{state motion}\n \\item and the \\TC{output motion}\n \\end{itemize}\n\\item [] produced by the initial state $x(k_0)$ and the input $u(k)$, starting\\\\\n at time $k_0$.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Motion}\n\\framesubtitleTC{for the Time-Invariant (TI) case, to which we restrict the scope from now on}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item Initial state + input $\\Rightarrow$ motion independently of the start time, i.e.,\n \\begin{displaymath}\n \\left.\n \\begin{array}{l}\n x(0) \\\\ u(k),\\,k \\geq 0\n \\end{array}\n \\right\\}\n \\quad \\Rightarrow \\quad\n x(k),y(k),\\,k \\geq 0.\n \\end{displaymath}\n\\item Alternatively, we can say that with TI systems one can set the origin\\\\\n of the time axis wherever one wants (for convenience, at zero).\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Equilibrium}\n\\framesubtitleTC{Definition (in the TI case)}\n\\myPause\n\\begin{center}\n {\\Large\n If there exist some state vectors $\\overline{x}$\\\\ \\myPause\n such that $x(0)=\\overline{x}$ and $u(k)=\\overline{u}$, $k \\geq 0$,\\\\ \\myPause\n produce the constant state motion $x(k)=\\overline{x}$, $k \\geq 0$,\\\\ \\myPause\n then those vectors are called \\TC{equilibrium states},\\\\ \\myPause\n or \\TC{equilibria} for short,\\\\ \\myPause\n \\vspace{1.5mm}corresponding to the constant input $\\overline{u}$.\n }\n\\end{center}\n\\end{frame}\n\n\n\\begin{frame}\n\\frametitleTC{Equilibrium}\n\\framesubtitleTC{Finding equilibrium states and outputs}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item A state vector $\\overline{x}$ is an equilibrium for a given $\\overline{u}$ if the consequent motion is\n $x(k)=x(k-1)=\\overline{x}$ $\\forall k$.\n\\item Thus to find equilibria one solves\n \\begin{displaymath}\n \\overline{x} = f(\\overline{x},\\overline{u}).\n \\end{displaymath}\n\\item If some equilibrium state exists, and $g(\\overline{x},\\overline{u})$ does not lose significance,\\\\\n the result is the corresponding \\TC{equilibrium output} $\\overline{y}$.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Stability}\n\\framesubtitleTC{Preliminaries}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item The existence of an equilibrium implies NOTHING on what happens if the system does not start exactly\n at the equilibrium.\n\\item Discussing this is a matter of \\TC{stability}.\n\\item One can talk about stability of equilibria, motions, and sometimes systems.\n\\item We define stability for an equilibrium, do not talk about motions,\\\\\n and move to systems later on.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Stability}\n\\framesubtitleTC{Stable equilibrium}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item Let $\\overline{x}$ be an equilibrium for the constant input $\\overline{u}$.\n\\item Denote by $x_{\\Delta}(k)$ the \\TC{perturbed motion} produced by\n \\begin{itemize}[<+-| alert@+>]\n \\item the input $\\overline{u}$,\n \\item and the \\TC{perturbed initial state} $x(0)=\\overline{x}+\\Delta\\overline{x}$.\n \\end{itemize}\n\\item Not that in general, $x_{\\Delta}(k)$ will not be constant.\n\\item Denote by $||x||$ the norm (think here of the Euclidean norm)\\\\\n of vector $x$.\n\\end{itemize}\n\\end{frame}\n\n\n\\begin{frame}\n\\frametitleTC{Stability}\n\\framesubtitleTC{Stable equilibrium}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item An equilibrium $\\overline{x}$ corresponding to the constant input $\\overline{u}$ is said to be \\TC{stable} if\n \\begin{displaymath}\n \\forall \\varepsilon>0 \\;\n \\exists \\delta_{\\varepsilon}>0 \\; : \\;\n ||\\Delta\\overline{x}||<\\delta_{\\varepsilon}\n \\Rightarrow ||x_{\\Delta}(k)-\\overline{x}||<\\varepsilon \\; \\forall k \\geq 0.\n \\end{displaymath}\n\\item Interpretation: stable equilibrium means that\n \\begin{itemize}[<+-| alert@+>]\n \\item no matter how close one wants the \\TC{entire} perturbed motion to remain to the equilibrium,\n \\item a maximum distance of the initial state from the equilibrium can be\\\\\n found, that fulfils the desire.\n \\end{itemize}\n \\item Graphically,\n \\begin{center}\n \\includegraphics[width=0.60\\columnwidth]{./Unit-02/img/StableEq-general.pdf}\n \\end{center}\n\\item If the above does not hold true, the equilibrium is \\TC{unstable}.\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Stability}\n\\framesubtitleTC{Asymptotically stable equilibrium}\n\\myPause\n\\begin{itemize}[<+-| alert@+>]\n\\item An equilibrium $\\overline{x}$ corresponding to $\\overline{u}$ is said to be \\TC{asymptotically stable} if\n \\begin{itemize}[<+-| alert@+>]\n \\item it is stable,\n \\item and \\underline{in addition}\n \\begin{displaymath}\n \\lim_{k\\rightarrow\\infty} ||x_{\\Delta}(k)-\\overline{x}||=0.\n \\end{displaymath}\n \\end{itemize}\n \\item Graphically,\n \\begin{center}\n \\includegraphics[width=0.60\\columnwidth]{./Unit-02/img/AsympStableEq-general.pdf}\n \\end{center}\n\\item Clearly asymptotic stability implies stability, but not \\emph{vice versa}.\n\\end{itemize}\n\\end{frame}\n\n", "meta": {"hexsha": "5d5ddfc96f2cadf8b6db04aeeeef2b7f4508ff95", "size": 9745, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "slides/Unit-02/sections/02-DynamicsFormally.tex", "max_stars_repo_name": "albertoleva/PID4CSE", "max_stars_repo_head_hexsha": "66ec14c204e16c97a5792c2e240b2daed4b39e83", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-04-19T16:38:10.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T16:38:10.000Z", "max_issues_repo_path": "slides/Unit-02/sections/02-DynamicsFormally.tex", "max_issues_repo_name": "albertoleva/PID4CSE", "max_issues_repo_head_hexsha": "66ec14c204e16c97a5792c2e240b2daed4b39e83", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": 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YES\n2. YES", "lm_q1_score": 0.7879312056025699, "lm_q2_score": 0.8198933337131076, "lm_q1q2_score": 0.6460195428980791}} {"text": "\n\\subsection{Adaboost}\n\n\\subsubsection{Introduction}\n\nBoosting is a way to create multiple learners for use in an ensemble predictor.\n\nThe goal is to create many predictors, which may not be themselves very accuracte, but have a high degree of independence.\n\n\\subsubsection{AdaBoost}\n\nAdaBoost is a popular algorithm for boosting.\n\nIt works by:\n\n\\begin{itemize}\n\\item Creating a set of weak learners using different restrictions on features in the training data.\n\\item Choosing the weak learner that most reduces the error of the combined learners, and give it a weighting which most reduces the error of the combined learners.\n\\item Creating a new weighting for the dataset, where ones poorly predicted (by the combination of learners) are given high weights.\n\\item Repeating the process a fixed number of times.\n\\end{itemize}\n\n\n\n", "meta": {"hexsha": "e5c20c9ae2ce859f134e5519e718a32ca806f59b", "size": 831, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/ensemble/02-02-boosting.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/ensemble/02-02-boosting.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/ensemble/02-02-boosting.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.24, "max_line_length": 164, "alphanum_fraction": 0.797833935, "num_tokens": 175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7879312006227324, "lm_q1q2_score": 0.6460195405489297}} {"text": "\\Opensolutionfile{solutions}[ex]\n\\section{Additional exercises}{}{}\n\n\\begin{enumialphparenastyle}\n\nThese problems require the techniques of this chapter, and are in no\nparticular order. Some problems may be done in more than one way.\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int (t+4)^3\\,dt$\n\\begin{sol}\n $\\ds{(t+4)^4\\over4}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int t(t^2-9)^{3/2}\\,dt$\n\\begin{sol}\n $\\ds{(t^2-9)^{5/2}\\over5}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int (e^{t^2}+16)te^{t^2}\\,dt$\n\\begin{sol}\n $\\ds{(e^{t^2}+16)^2\\over 4}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% use addition and subtraction formulas sin(A+B)=sinAcosB+sinBcosA and sin(A-B)=sinAcosB-sinBcosA to solve for sinAcosB and substitute into integral\n $\\ds\\int \\sin t\\cos 2t\\,dt$\n\\begin{sol}\n $\\ds\\cos t-{2\\over3}\\cos^3 t+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int \\tan t\\sec^2t\\,dt$\n\\begin{sol}\n $\\ds{\\tan^2 t\\over 2}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {2t+1\\over t^2+t+3}\\,dt$\n\\begin{sol}\n $\\ds\\ln|t^2+t+3|+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% partial fractions\n $\\ds\\int {1\\over t(t^2-4)}\\,dt$\n\\begin{sol}\n $\\ds {1\\over8} \\ln|1-4/t^2|+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% trig sub x=5sin(theta)\n $\\ds\\int {1\\over (25-t^2)^{3/2}}\\,dt$\n\\begin{sol}\n $\\ds{1\\over25}\\tan(\\arcsin(t/5))+C={t\\over25\\sqrt{25-t^2}}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {\\cos 3t\\over\\sqrt{\\sin3t}}\\,dt$\n\\begin{sol}\n $\\ds{2\\over3}\\sqrt{\\sin 3t}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% integration by parts\n $\\ds\\int t\\sec^2 t\\,dt$\n\\begin{sol}\n $\\ds t\\tan t+\\ln|\\cos t|+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {e^t\\over \\sqrt{e^t+1}}\\,dt$\n\\begin{sol}\n $\\ds 2\\sqrt{e^t+1}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% powers of trigonometric functions [use trig identity cos^2(x)=1/2(1+cos(2x))]\n $\\ds\\int \\cos^4 t\\,dt$\n\\begin{sol}\n $\\ds{3t\\over 8}+{\\sin 2t\\over4}+ {\\sin 4t\\over 32}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% partial fractions\n $\\ds\\int {1\\over t^2+3t}\\,dt$\n\\begin{sol}\n $\\ds{\\ln |t|\\over 3} - {\\ln |t+3|\\over 3}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% trig sub t=tan(theta), then u-substitution\n $\\ds\\int {1\\over t^2\\sqrt{1+t^2}}\\,dt$\n\\begin{sol}\n $\\ds{-1\\over \\sin\\arctan t}+C=-\\sqrt{1+t^2}/t+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {\\sec^2t\\over (1+\\tan t)^3}\\,dt$\n\\begin{sol}\n $\\ds{-1\\over 2(1+\\tan t)^2}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int t^3\\sqrt{t^2+1}\\,dt$\n\\begin{sol}\n $\\ds{(t^2+1)^{5/2}\\over 5}-{(t^2+1)^{3/2}\\over 3}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% integration by parts\n $\\ds\\int e^t\\sin t\\,dt$\n\\begin{sol}\n $\\ds{e^t\\sin t-e^t\\cos t\\over 2}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int (t^{3/2}+47)^3\\sqrt{t}\\,dt$\n\\begin{sol}\n $\\ds{(t^{3/2}+47)^4\\over6}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {t^3\\over (2-t^2)^{5/2}}\\,dt$\n\\begin{sol}\n $\\ds{2\\over 3(2-t^2)^{3/2}}-{1\\over(2-t^2)^{1/2}}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% trig sub x=(3/2)tan(theta), then u-substitution\n $\\ds\\int {1\\over t(9+4t^2)}\\,dt$\n\\begin{sol}\n $\\ds{\\ln|\\sin(\\arctan(2t/3))|\\over9}+C = \n(\\ln(4t^2)-\\ln(9+4t^2))/18 + C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {\\arctan 2t\\over 1+4t^2}\\,dt$\n\\begin{sol}\n $\\ds{(\\arctan(2t))^2\\over4}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% partial fractions\n $\\ds\\int {t\\over t^2+2t-3}\\,dt$\n\\begin{sol}\n $\\ds{3\\ln|t+3|\\over 4}+{\\ln|t-1|\\over4}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% powers of trigonometric functions with u=cos(t)\n $\\ds\\int \\sin^3 t\\cos^4 t\\,dt$\n\\begin{sol}\n $\\ds{\\cos^7 t\\over 7}-{\\cos^5 t\\over 5}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% factoring then simple integration\n $\\ds\\int {1\\over t^2-6t+9}\\,dt$\n\\begin{sol}\n $\\ds{-1\\over t-3}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% u-substitution\n $\\ds\\int {1\\over t(\\ln t)^2}\\,dt$\n\\begin{sol}\n $\\ds{-1\\over \\ln t}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% integration by parts\n $\\ds\\int t(\\ln t)^2\\,dt$\n\\begin{sol}\n $\\ds{t^2(\\ln t)^2\\over 2}-{t^2\\ln t\\over 2}+{t^2\\over4}+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% integration by parts\n $\\ds\\int t^3e^{t}\\,dt$\n\\begin{sol}\n $\\ds(t^3-3t^2+6t-6)e^t+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n%%%%%%%%%%\n\\begin{ex}\t% split into two fractions: (2x+1)/(2(x^2+x-1)) + 1/(2(x^2+x-1)). Integrate first by u-sub, and second by trig sub (x+1/2)=(sqrt(5)/2)sec(theta)\n $\\ds\\int {t+1\\over t^2+t-1}\\,dt$\n\\begin{sol}\n $\\ds{5+\\sqrt5\\over10}\n\\ln(2t+1-\\sqrt5)+{5-\\sqrt5\\over10}\\ln(2t+1+\\sqrt5)+C$\n\\end{sol}\n\\end{ex}\n%%%%%%%%%%\n\n\\end{enumialphparenastyle}", "meta": {"hexsha": "e1f0dae1ac8c91a6485164bee7c0ed193f8f7ba9", "size": 5067, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "7-techniques-of-integration/7-8-more-problems.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "7-techniques-of-integration/7-8-more-problems.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "7-techniques-of-integration/7-8-more-problems.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 19.2661596958, "max_line_length": 159, "alphanum_fraction": 0.5271363726, "num_tokens": 2296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428947, "lm_q2_score": 0.819893340314393, "lm_q1q2_score": 0.6460195399335664}} {"text": "\\begin{frame}{Finding the autocorrelation}\nTo find the parameters $a_1, \\ldots, a_N$, we need to compute the autocorrelation $r[0], \\ldots, r[N-1]$ from $M+1$ samples of the signal $s[n]$.\n\nTwo methods:\n\\begin{enumerate}\n\t\\item \\textbf{Autocorrelation method}\n\tThis method assumes that the signal $s[n]$ is \\underline{zero outside the interval} $[0, M+1]$. Therefore, for $s[n]$ deterministic we have\n\t\\begin{equation*}\n\tr_{ss}[|m|] = \\sum_{n =0}^{M-|m|}s[n]s[n+|m|], |m| \\leq N-1\n\t\\end{equation*}\n\tThis implies $r_{ss}[m] = 0, |m| > M$.\n\t\\item \\textbf{Covariance method}\n\tThis method assumes that the signal $s[n]$ is \\underline{unknown outside the interval} $[0, M+1]$. Therefore, for $s[n]$ deterministic we have\n\t\\begin{equation*}\n\tr_{ss}[m] = \\sum_{n = N}^{M}s[n]s[n+|m|], m = 0, 1, \\ldots, N-1\n\t\\end{equation*}\t\n\\end{enumerate}\n\\end{frame}\n\n\\begin{frame}{Convolution as a matrix-vector product}\nWe can write a convolution as a matrix-vector product\n\\begin{align*}\ny[n] &= \\sum_{m = 0}^N h_nx[n-m] \\\\\ny &= Xh \\\\\n\\begin{bmatrix}\ny[0] \\\\\ny[1] \\\\\ny[2] \\\\\n\\vdots \\\\ \ny[L-1] \\\\\n\\end{bmatrix}\n&= \\begin{bmatrix}\nx[0] & 0 & \\ldots & 0 \\\\\nx[1] & x[0] & \\ddots & \\vdots \\\\\n\\vdots & x[1] & \\ddots & 0\\\\\nx[L-1] & \\vdots &\\ddots & x[0]\\\\\n0 & x[L-1] & \\ddots & x[1] \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & \\ldots & x[L-1] \n\\end{bmatrix}_{L\\times N}\\cdot\\begin{bmatrix}\nh_0 \\\\\nh_1 \\\\\nh_2 \\\\\n\\vdots \\\\ \nh_{N-1} \\\\\n\\end{bmatrix}\n\\end{align*}\n\nThe matrix $X$ is called \\textbf{Toeplitz matrix}.\n\n\\vspace{0.25cm}\nIn Matlab: \\texttt{>> X = toeplitz(c, r)}, where the vector \\texttt{c} is the first column of $X$, and the vector \\texttt{r} is the first row of $X$.\n\n\\end{frame}\n\n\n\\begin{frame}{All-pole model for deterministic signals}\nBack to our problem when $s[n]$ is deterministic:\n\\begin{equation}\n\\sum_{n = 0}^{L-1} |v[n] - \\sum_{m = 0}^N a_ns[n-m]|^2 \\tag{if $s[n]$ is deterministic}\n\\end{equation}\n\nNow we can write this in matrix notation\n\\begin{equation*}\n||v - Sh||^2_2\n\\end{equation*}\n\nThis is a \\textbf{least-squares problem}, and its solutions is \n\\begin{equation}\nh = S^{\\dagger}v \\tag{least-squares solution}\n\\end{equation}\nwhere $S^{\\dagger} = (S^TS)^{-1}S^T$ is the \\textbf{Moore-Penrose pseudo-inverse}.\n\n\\vspace{0.25cm}\nWhen $s[n]$ is deterministic, $v[n]$ is assumed to be an impulse. Hence, $v = [1, 0, \\ldots, 0]^T$.\n\\end{frame}\n\n\n\\begin{frame}{All-pole model for random signals}\nWhen $s[n]$ is random, we want to minimize\n\\begin{equation*}\n\\E\\Big(|v[n] - \\sum_{m = 0}^N a_ns[n-m]|^2\\Big)\n\\end{equation*}\n\n\n\n\\begin{equation*}\n\\E(s[n]v[n]) = \\sum_{m = 0}^N a_n\\E(s[n]s[n-m]), \\quad n = 0, \\ldots, L-1\n\\end{equation*}\n\n\\end{frame}\n\n\n\n%\n\\begin{frame}{Revisiting the Bartlett or triangular window}\n\\begin{columns}\n\t\\begin{column}{0.5\\textwidth}\n\t\t\\textbf{Time}\n\t\t\\begin{equation*}\n\t\tw_B[n] = \\begin{cases}\n\t\t1 - |n|/M, & |n| < M \\\\\n\t\t0, & \\text{otherwise}\n\t\t\\end{cases}\n\t\t\\end{equation*}\n\t\\end{column}\n\t\n\t\\begin{column}{0.5\\textwidth}\n\t\t\\textbf{Frequency}\n\t\t\\begin{equation*}\n\t\tW_B[n] = \\frac{1}{M}\\bigg(\\frac{\\sin(\\omega M/2)}{\\sin(\\omega/2)}\\bigg)^2\n\t\t\\end{equation*}\n\t\\end{column}\n\\end{columns}\n\n\\hspace*{-0.75cm}\\resizebox{0.95\\paperwidth}{!}{\\input{figs/barlett_time_freq.tex}}\n\\end{frame}\n\n\n%\n\\begin{frame}{Design specifications}\nSpecifications of \\textbf{frequency-selective} filters\n\\begin{center}\n\t\\resizebox{0.7\\linewidth}{!}{\\input{figs/design_specs.tex}}\n\\end{center}\n\n\\textbf{Terminology}\n\\begin{itemize}\n\t\\item $\\delta_1$ passband ripple\n\t\\item $\\delta_2$ stopband ripple\n\t\\item $\\omega_p$ passband edge frequency\n\t\\item $\\omega_p$ stopband edge frequency\n\\end{itemize}\n\\end{frame}\n\n\\begin{frame}{Poles, zeros, and the frequency response}\n\n\\textbf{Adding one more zero:} $H(z) = (1 - r_1e^{j\\theta}z^{-1})(1 - r_2e^{j\\psi}z^{-1})$\n\n\\begin{columns}[t]\n\t\\begin{column}{0.7\\textwidth}\n\t\t\\vspace{-0.5cm}\n\t\t\\begin{center}\n\t\t\t\\resizebox{\\linewidth}{!}{\\input{figs/pole_zero_freq_resp_2zeros.tex}}\n\t\t\\end{center}\n\t\\end{column}\n\t\n\t\\begin{column}{0.3\\textwidth}\t\n\t\t\\only<1|handout:1>{\n\t\t\t\\textbf{Magnitude}\n\t\t\t\\flushleft\n\t\t\t\\begin{align*}\n\t\t\tH(z) &= \\frac{z - r_1e^{j\\theta}}{z}\\cdot\\frac{z - r_2e^{j\\psi}}{z} \\\\\n\t\t\t|H(e^{j\\omega})| &= \\bigg|\\frac{{\\color{red2} v_1} - {\\color{black} v_2}}{{\\color{red2} v_1}}\\bigg|\\bigg|\\frac{{\\color{red2} v_1} - {\\color{black} v_4}}{{\\color{red2} v_1}}\\bigg|\\\\\n\t\t\t&= \\frac{||{\\color{blue2} v_3}||}{||{\\color{red2} v_1}||}\\frac{||{\\color{green2} v_5}||}{||{\\color{red2} v_1}||} \\\\\n\t\t\t&= ||{\\color{blue2} v_3}||||{\\color{green2} v_5}||\n\t\t\t\\end{align*}\n\t\t}\n\t\t\n\t\t\\only<2|handout:2>{\n\t\t\t\\textbf{Phase}\n\t\t\t\\flushleft\n\t\t\t\\begin{align*}\n\t\t\tH(z) &= \\frac{z - r_1e^{j\\theta}}{z}\\cdot\\frac{z - r_2e^{j\\psi}}{z} \\\\\n\t\t\t\\angle H(e^{j\\omega}) &= \\angle\\frac{{\\color{red2} v_1} - {\\color{black} v_2}}{{\\color{red2} v_1}} + \\angle\\frac{{\\color{red2} v_1} - {\\color{black} v_4}}{{\\color{red2} v_1}}\\\\\n\t\t\t&= \\angle {\\color{blue2} v_3} + \\angle {\\color{green2} v_5} - 2\\omega\n\t\t\t\\end{align*}\n\t\t}\n\t\\end{column}\n\\end{columns}\n\n\\only<1|handout:1>{\\textbf{Conclusion:} As $e^{j\\omega}$ approaches a zero, $|H(e^{j\\omega})| \\to 0$}\n\\only<2|handout:2>{\\textbf{Conclusion:} A zero causes a negative to positive phase change of the phase response (\\textbf{phase advance})}\n\\end{frame}\n\n\n\n%\n\\begin{frame}{The unilateral $z$-transform}\n\nThe bilateral $z$-transform we've seen so far does not account for initial conditions.\n\n\\begin{block}{The unilateral $z$-transform}\n\t\\begin{equation*} \\tag{Direct transform}\n\tX^+(z) = \\sum_{\\tikz[baseline]{\n\t\t\t\\node[fill=blue!20,anchor=base] (t1) {$n = 0$};\n\t}}^{\\infty} x[n]z^{-n} \n\t\\end{equation*}\n\\end{block}\n\n\\begin{itemize}\n\t\\item For causal signals, $X(z) = X^+(z)$. \\\\\n\t\\item For any signal $y[n] \\Longleftrightarrow Y(z)$, we have $y[n]u[n] \\Longleftrightarrow Y^+(z)$. \\\\\n\t\\item As the unilateral $z$-transform only takes into account the causal part of the signal, the ROC will always the exterior of a circle (ROC = $\\{|z|>|r|\\}$). Hence, there's no ambiguity.\n\\end{itemize}\n\n\\end{frame}\n\n\\subsection{Sampling and reconstruction}\n%\n\\begin{frame}{Families of $z$-transforms}\t\n\\begin{block}{Finite-length sequences}\n\tWhen $A(z) = 1$, we're left with $z$-transforms of the form\n\t\\begin{align*}\n\tX(z) = b_0 + b_1z^{-1}+\\ldots+b_Mz^{-M}\n\t\\end{align*}\n\t\\begin{itemize}\n\t\t\\item These correspond to finite-length sequences or systems with \\textbf{finite impulse response (FIR)}\n\t\t\\item In this case, we have $M$ zeros, which are the roots of $X(z)$, and we have $M$ poles at $z = 0$. As a result, the ROC is the entire $z$-plane with the exception of $z = 0$, and possible poles at infinity i.e., $X(p = \\infty) = \\infty$.\n\t\t\\item Since finite-length sequences are always absolute summable, the ROC of their $z$-trasnform is the entire $z$ plan, with the exception of $z = 0$ or $z = \\infty$ (the poles).\n\t\\end{itemize}\t\n\\end{block}\n\n\\end{frame}\n\n%\n\\begin{frame}{Obtaining frequency response from difference equation}\n\t\n\t\\begin{block}{Difference equation}\n\t\t\\begin{equation*}\n\t\t\\sum_{k=0}^N a_k y[n-k] = \\sum_{k=0}^Mb_k x[n-k]\n\t\t\\end{equation*}\n\t\\end{block}\n\t\n\tUsing the linearity and time shift properties of the DTFT it follows that\n\t\n\t\\begin{align*}\n\t\\sum_{k=0}^N a_kY(e^{j\\omega})e^{-j\\omega k} = \\sum_{k=0}^M b_kX(e^{j\\omega})e^{-j\\omega k} \\\\\n\tH(e^{j\\omega}) = \\frac{Y(e^{j\\omega})}{X(e^{j\\omega})} = \\frac{\\sum_{k=0}^M b_ke^{-j\\omega k}}{\\sum_{k=0}^N a_ke^{-j\\omega k}}\n\t\\end{align*}\n\\end{frame}\n\n\n%%%%%%% Discrete Fourier Transform\n\n", "meta": {"hexsha": "e57dbe68198ab6c3a0d18bb4748562ceb536c028", "size": 7409, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/draft.tex", "max_stars_repo_name": "jkperin/DSP", "max_stars_repo_head_hexsha": "0ec74b4597fb54800ebdab440cba4892d210343d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 21, "max_stars_repo_stars_event_min_datetime": "2019-05-11T21:48:47.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-07T08:56:28.000Z", "max_issues_repo_path": "lectures/draft.tex", "max_issues_repo_name": "jkperin/DSP", "max_issues_repo_head_hexsha": "0ec74b4597fb54800ebdab440cba4892d210343d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lectures/draft.tex", "max_forks_repo_name": "jkperin/DSP", "max_forks_repo_head_hexsha": "0ec74b4597fb54800ebdab440cba4892d210343d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2019-04-16T01:11:14.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-19T07:25:20.000Z", "avg_line_length": 31.3940677966, "max_line_length": 244, "alphanum_fraction": 0.6412471319, "num_tokens": 2930, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.8459424314825853, "lm_q1q2_score": 0.6458633113385933}} {"text": "Radio interferometry is the technique of combining the response of multiple telescopes to detect radiation from astronomical sources. One of the main advantages of this technique is the resolution power of an interferometer. The angular resolution of any optical instrument can be approximated as~$\\theta \\sim \\frac{\\lambda}{D}$. So for any given wavelength of observation, $\\lambda $, we obtain higher resolution by increasing the size of the aperture of the observing instrument,~$D$. However, this will eventually be limited by engineering constraints to~$\\sim $100~metres for fully--steerable radio telescopes. With an interferometer,~$D$, is the separation \\emph{between} the radio telescopes and can be as large as hundreds or even thousands of kilometres. In the following derivation, I will be providing a broad overview of the main concepts that interferometry is based upon.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth ]{tms-figure.eps}\n \\caption[]{A schematic diagram of interferometry showing the same electromagnetic plane wave~(the dotted line \\textbf{-}\\textbf{-}\\textbf{-}\\textbf{-}\\textbf{-}\\textbf{-}) from an astronomical source, in the direction,~$\\mathbf{\\hat{s}}$, incident upon antennas~$V_1$~and $V_2$. The angle~$\\theta $ corresponds to the direction of~$\\mathbf{\\hat{s}}$ and gives rise to the different times at which this wave arrives at~$V_1$~and $V_2$. This time difference is given as~$\\mathbf{\\tau _g}$ corresponding to a distance of~$\\mathbf{c\\msp \\tau_ g}$. The antennas $V_1$~and $V_2$ are separated along the baseline vector~$\\vec{\\mathbf{b}}$.\\\\ \\emph{Image credit:} Interferometry and Synthesis in Radio Astronomy~(Thomson, Moran and Swenson,~2017) used under the Creative Commons Attribution International License (\\texttt{http://creativecommons.org/licenses/by-nc/4.0/}). The bold, black lines and text have been added by me.}\n \\label{fig:tms-figure}\n\\end{figure}\n\n\\section{Electromagnetic waves incident on an array}\nThe overall scheme of radio interferometry is represented in Figure~\\ref{fig:tms-figure}. The incoming radiation from a natural astronomical source~\\textbf{S} can be considered as a noise signal. Over a small time,~$\\delta t$, the radiation can be considered to be sinusoidal. Such an oscillating electromagnetic wave with \\emph{quasi--monochromatic} frequency,~$\\nu $, can be described as\n\n\\begin{eqnarray}\n \\label{eq:1}\n E(t) &=& A \\msp cos(2\\pi \\msp \\nu t + \\Phi) \\nonumber \\\\\n &=& A \\msp cos(\\omega t + \\Phi) \\msp , \\hspace{0.8cm} \\mbox{where} \\enspace \\omega = \\frac{2\\pi}{T} = 2\\pi \\nu\n\\end{eqnarray}\n\nOur aim is to mathematically determine the relationship which will ensure that we are measuring the same wavefront of~$E(t)$ which is incident upon antennas~$V_1$ and~$V_2$. We do this by determining the \\emph{geometric delay}~$\\tau _g$, which is the time difference between the plane wave arriving at~$V_1$ and~$V_2$.\n\n\\begin{eqnarray}\n \\label{eq:2}\n \\tau _g &=& \\frac{\\mbox{Distance of the projection of $\\vec{\\mathbf{b}}$ in the direction of $\\mathbf{\\hat{s}}$}}{\\mbox{Speed of light}} \\nonumber \\\\\n &=& \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c}\n\\end{eqnarray}\n\nIn Equation~\\eqref{eq:2},~$\\mathbf{\\hat{s}}$ is a unit vector but~$\\vec{\\mathbf{b}}$ is not. This is because we do not know the distance to the source,~\\textbf{S}. Having obtained an expression for~$\\tau _g$, we can represent the electromagnetic radiation incident at the antennas as\n\n\\begin{eqnarray}\n \\label{eq:3}\n \\begin{aligned}\n V_1 &=& |V_1| \\msp cos\\big (\\omega \\msp (t - \\tau _g)\\big ) \\\\\n V_2 &=& |V_2| \\msp cos(\\omega t)\n \\end{aligned}\n\\end{eqnarray}\n\n\\section{Correlation of a point source}\nIn signal processing \\emph{cross--correlation} is the measure of similarity between two series which are offset from each other. In our system described in Equation~\\eqref{eq:3} this offset is~$\\omega \\tau _g$.\n\nWe define the cross--correlation of the outputs of antennas~$V_1$ and~$V_2$ as~$ = \\frac{1}{T} \\int _0^T V_1 \\msp V_2 \\msp dt$. It can be thought of as an \\emph{averaging process} because you are integrating~(adding) over a certain duration and then dividing by the length of that duration.\n\n% &=& \\frac{1}{T} \\int _0^T\n\n\\begin{eqnarray}\n \\label{eq:4}\n &=& \\frac{1}{T} \\int _0^T V_1 \\msp V_2 \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} \\int _0^T |V_1| \\msp cos(\\omega \\msp (t - \\tau _g)) \\enspace |V_2| \\msp cos(\\omega t) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos \\big (\\omega \\msp (t - \\tau _g) + \\omega t \\big) + sin \\big (\\omega \\msp (t - \\tau _g)\\big) \\msp sin(\\omega t) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos(2\\omega t - \\omega \\tau _g) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos(\\omega t) + cos \\big (\\omega \\msp (t - \\tau _g) - \\omega t \\big) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos(2\\omega t - \\omega \\tau _g) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos(\\omega t) + cos(-\\omega \\tau _g) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos(2\\omega t - \\omega \\tau _g) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos(\\omega t) + cos(\\omega \\tau _g) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos(\\omega \\tau _g) + cos(2\\omega t - \\omega \\tau _g) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos(\\omega t) \\msp d\\,t \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\Bigg [ \\int _0^T cos(\\omega \\tau _g) \\msp dt + \\int _0^T cos(2\\omega t - \\omega \\tau _g) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos(\\omega t) \\msp d\\,t \\Bigg ] \\nonumber\n\\end{eqnarray}\n\nIn the last step, we have grouped the terms which depend on~$t$ in a single integral. In doing so we are attempting to make use of the following property: Over some sufficient duration of~$t \\in \\msp [\\msp 0,~T\\msp ]$, there will be sufficient complete cycles of the trigonometrical terms which depend on~$t$ for the result to average to~0 when integrating over~$d\\,t$.\\\\\n\n\\begin{braced}\n Try integrating~$\\int _0^{2\\pi} sin(t) \\msp dt $ or~$\\int _0^{2\\pi} cos(t) \\msp dt $ to get an idea why this happens over complete trigonometric cycles.\n\\end{braced}\n\n\\begin{eqnarray}\n \\label{eq:5}\n \\implies \\qquad &=& \\frac{1}{T} |V_1V_2| \\Bigg [\\int _0^T cos(\\omega \\tau _g) \\msp dt + 0 \\Bigg ] \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\enspace cos(\\omega \\tau _g) \\msp \\int _0^T 1 \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\enspace cos(\\omega \\tau _g) \\msp \\big[\\msp t \\msp \\big]_0^T \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\enspace cos(\\omega \\tau _g) \\enspace T \\nonumber \\\\\n &=& |V_1V_2| \\enspace cos(\\omega \\tau _g)\n\\end{eqnarray}\n\nIn radio interferometry this process happens in the \\emph{correlator}. The term~$|V_1V_2|$ is a multiplication of voltages which is proportional to power~(Ohm's Law: $P \\propto V\\mathcal{I} = V\\frac{V}{R} = \\frac{V^2}{R}$). I shall refer to~$|V_1V_2|$ as the intensity,~$I$, of the source which is the power per unit area.\n\n\\begin{eqnarray}\n \\label{eq:6}\n \\therefore \\mbox{Correlator response,} \\enspace R_c &=& \\nonumber \\\\\n &=& I \\enspace cos(\\omega \\tau _g)\n\\end{eqnarray}\n\nThe correlator response,~$R_c$, is a function of~$\\omega $ and~$\\tau _g$. In practice~$\\tau _g$ is difficult to time directly. So our aim now is to represent~$\\tau _g$ in terms of quantities which we can measure easily, like:\n\n\\begin{enumerate}\n \\itemsep-0.5cm\n\\item the baseline length~--~which \\textbf{we} can measure from one antenna to another \\\\\n\\item the observation wavelength~--~dependant on the frequency range of the receivers \\textbf{we} build \\\\\n\\item the position of the source in the sky~--~which is where \\textbf{we} point the interferometer\n\\end{enumerate}\n\nIn expanding~$\\tau _g$ in~$R_c$ we get\n\n\\begin{eqnarray}\n \\label{eq:7}\n R_c &=& I \\msp cos(\\omega \\tau _g) \\nonumber \\\\\n &=& I \\msp cos\\Big (2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot } \\mathbf{\\hat{s}}}{c}\\Big ) \\nonumber \\\\\n &=& I \\msp cos\\Big (2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot } \\mathbf{\\hat{s}}}{\\nu \\msp \\lambda}\\Big ) \\nonumber \\\\\n &=& I \\msp cos\\Bigg (2\\pi \\msp \\frac{|b|\\msp cos(\\frac{\\pi }{2} - \\theta)}{\\lambda}\\Bigg ) , \\hspace{0.8cm} \\mbox{see Figure~\\ref{fig:tms-figure}} \\nonumber \\\\\n &=& I \\msp cos\\Big (2\\pi \\msp \\frac{|b| \\msp sin \\theta}{\\lambda} \\Big) \\nonumber \\\\\n &=& I \\msp cos(2\\pi \\msp u \\msp l) \\msp , \\hspace{0.8cm} \\mbox{where} \\enspace u = \\frac{|b|}{\\lambda}, \\enspace l = sin \\theta\n\\end{eqnarray}\n\nInstead of using traditional units of distance, such as the meter, in interferometry it is more convenient to represent the baseline length in units of our observation frequency's corresponding wavelength, i.e.~$u = \\frac{|b|}{\\lambda}$.\n\n\\subsection{Visualising the correlator output}\n\nWe can graphically visualise the correlator output~$R_c$ by sketching the trigonometrical function~$R_c = I \\msp cos(2\\pi \\msp u \\msp l)$. The range of the cosine function~$R_c \\in [-I, \\msp I]$ for all values of~$u$ and~$l$. For this example let the baseline be~3~wavelengths long~$\\implies u = \\frac{|b|}{\\lambda} = 3$ and our field--of--view extend from horizon--to--horizon~$\\implies -\\frac{\\pi}{2} \\leq \\theta \\leq \\frac{\\pi}{2}$.\n\n\\begin{eqnarray}\n \\label{eq:8}\n \\mbox{Maxima when} \\qquad R_c &=& I \\nonumber \\\\\n cos(2\\pi \\msp u \\msp l) &=& 1 \\nonumber \\\\\n \\therefore \\enspace 2\\pi \\msp u \\msp l &=& 2n \\msp \\pi \\nonumber \\\\\n \\frac{|b|}{\\lambda} \\msp sin\\theta &=& n \\nonumber \\\\\n \\theta &=& sin^{-1}\\Big(\\frac{n}{3}\\Big) \\msp , \\hspace{0.8cm} \\mbox{where} \\enspace n \\in \\mathbb{Z}, \\enspace |n| \\leq 3 \\nonumber \\\\\n \\nonumber \\\\\n \\mbox{Minima when} \\qquad R_c &=& -I \\nonumber \\\\\n cos(2\\pi \\msp u \\msp l) &=& -1 \\nonumber \\\\\n \\therefore \\enspace 2\\pi \\msp u \\msp l &=& (2n+1) \\msp \\pi \\nonumber \\\\\n \\frac{|b|}{\\lambda} \\msp sin\\theta &=& \\frac{2n+1}{2} \\nonumber \\\\\n \\theta &=& sin^{-1}\\Bigg(\\frac{2n+1}{3 \\msp (2)}\\Bigg) \\msp , \\hspace{0.8cm} \\mbox{where} \\enspace n \\in \\mathbb{Z}, \\enspace -3 \\leq n < 3 \\nonumber\n\\end{eqnarray}\n\n\\begin{table}\n \\centering\n\\begin{tabular}{rrr}\n n & Maxima & Minima \\\\\n \\hline\n $-$3 & $-$90.0$^\\circ$ & $-$56.4$^\\circ$ \\\\\n $-$2 & $-$41.8$^\\circ$ & $-$30.0$^\\circ$ \\\\\n $-$1 & $-$19.5$^\\circ$ & $-$9.6$^\\circ$ \\\\\n 0 & 0.0$^\\circ$ & 9.6$^\\circ$ \\\\\n 1 & 19.5$^\\circ$ & 30.0$^\\circ$ \\\\\n 2 & 41.8$^\\circ$ & 56.4$^\\circ$ \\\\\n 3 & 90.0$^\\circ$ & \\\\\n \\hline\n\\end{tabular}\n\\caption[]{Maxima and minima of~$R_c$ when~$u = \\frac{|b|}{\\lambda} = 3$.}\n\\label{tab:max-min}\n\\end{table}\n\n%%BoundingBox: -30 210 600 684\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth ]{01-corr-resp.eps}\n \\caption[]{The correlator response,~$R_c$ in rectangular~(top) coordinates. The maxima and minima of~$R_c$ are stated in Table~\\ref{tab:max-min}. The polar plot~(bottom left) shows what the fringes look like from horizon--to--horizon. The previous two plots resemble the response of a di--pole baseline. In contrast, the polar plot~(bottom right) resembles what you might expect from an interferometer like MeerKAT where there is a dominant central fringe and diminished sidelobes.}\n \\label{fig:corr-resp}\n\\end{figure}\n\nI show the maxima, minima of~$R_c$ in Table~\\ref{tab:max-min}, and the corresponding curve in Figure~\\ref{fig:corr-resp}. This curve is known as the interferometer's \\emph{fringe pattern}, and you can think of it as the pattern which governs regions of the interferometer's sensitivity to the sky. In our example this would be the pattern for a dipole array. For interferometers such as MeerKAT, the fringes are shaped by the receivers such that the central lobe dominates. You can use \\texttt{01-corr-resp.py} to determine the maxima, minima and fringe patterns for arbitrary values of~$u$ and~$l$.\n\nAt this point we can see why the interferometer's resolution power, which I described at the beginning as~$\\theta \\sim \\frac{\\lambda}{D}$ is what it is. We can do this by considering the separation of one maxima to the next in\n\n\\begin{eqnarray}\n \\label{eq:10}\n \\theta _{n+1} - \\theta _n &=& sin^{-1}\\Big(\\frac{n+1}{u}\\Big) - sin^{-1}\\Big(\\frac{n}{u}\\Big) \\nonumber \\\\\n &\\sim & \\Big(\\frac{n+1}{u}\\Big) - \\Big(\\frac{n}{u}\\Big) \\msp , \\hspace{0.8cm} \\mbox{using the small angle approximation} \\nonumber \\\\\n &=& \\frac{1}{u} \\nonumber \\\\\n &=& \\frac{\\lambda}{b} \\nonumber\n\\end{eqnarray}\n\n% *** Also note that because |sin \\theta| < 1, the number of lobes, %which is determined by n is < \\frac{b}{\\lambda}.\n\n\\section{Correlation of a non--point (finite) source}\n\\label{osec:Correlation-of-a-non-point-source}\nUp to this point, we have considered the correlation of a point source. However, a real astronomical source occupies a~2\\,D region on the \\emph{celestial sphere}\\footnote{The celestial sphere is an imaginary sphere which is concentric to Earth.}. For such an extended source we need to sum each of the antenna responses~--~$V_1$ and~$V_2$~--~~over a~2\\,D solid angle region,~$d\\,\\Omega $, in the sky.\n\n\\begin{eqnarray}\n \\label{eq:9}\n \\therefore \\enspace &=& \\Big \\langle \\iint V_1 \\msp d\\,\\Omega \\iint V_2 \\msp d\\,\\Omega \\Big \\rangle \\nonumber\n\\end{eqnarray}\n\nIf \\underline{and only if} the emission is \\emph{spatially incoherent} then\n\n\\begin{eqnarray}\n \\label{eq:19}\n \\enspace &=& \\Big \\langle \\iint V_1V_2 \\msp d\\,\\Omega \\Big \\rangle \\nonumber\n\\end{eqnarray}\n\n\\begin{braced}\n Spatial incoherence is required as in general~$\\enspace \\int g(x) \\msp h(x) \\msp d\\,x \\quad \\neq \\quad \\int g(x) \\msp d\\,x \\times \\int h(x) \\msp d\\,x$.\\\\\n Typically we need to use \\emph{integration by parts} in these cases.\\\\ For example try to integrate~$\\quad \\int x \\msp cos(x) \\msp d\\,x \\quad $ v.s.~$\\quad \\int x \\msp d\\,x \\times \\int cos(x) \\msp d\\,x$.\n\\end{braced}\n\n\\begin{eqnarray}\n \\label{eq:11}\n &=& \\frac{1}{T} \\int _0 ^T \\iint V_1 V_2 \\msp d\\,\\Omega \\enspace d\\,t \\nonumber \\\\\n &=& \\iint \\frac{1}{T} \\int _0 ^T V_1 V_2 \\msp d\\,t \\enspace d\\,\\Omega \\nonumber \\\\\n &=& \\iint I \\msp cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega\n % &=& \\iint I_\\nu (\\hat{s}) \\msp cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega\n % &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{b} \\cdot \\hat{s}}{c}} \\msp d\\,\\Omega \\nonumber\n\\end{eqnarray}\n\nNote that the~$cos(2\\pi \\msp u \\msp l)$ term in Equation~\\eqref{eq:11} has a deficiency since the~$cos$ function is an \\emph{even} function and is therefore only sensitive to the even component of the signal. We also need a way to detect/measure/sample the \\emph{odd} component. This is because, for some general function\n\n\\begin{eqnarray}\n \\label{eq:12}\n I(x) &=& I_{\\mbox{\\scriptsize even}}(x) + I_{\\mbox{\\scriptsize odd}}(x) \\msp , \\hspace{0.8cm} \\mbox{for} -k \\leq x \\leq k \\nonumber\n\\end{eqnarray}\n\nThe odd and even components have the property that\n\n\\begin{eqnarray}\n \\label{eq:13}\n I_{\\mbox{\\scriptsize even}}(-x) &=& -I_{\\mbox{\\scriptsize even}}(x) \\nonumber \\\\\n I_{\\mbox{\\scriptsize odd}}(-x) &=& I_{\\mbox{\\scriptsize odd}}(x) \\nonumber \\\\\n \\nonumber \\\\\n \\mbox{Giving} \\qquad R_c &=& \\iint I \\msp cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega \\nonumber \\\\\n &=& \\iint \\Big (I_{\\mbox{\\scriptsize even}} + I_{\\mbox{\\scriptsize odd}} \\Big ) \\msp cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega \\nonumber \\\\\n &=& \\iint I_{\\mbox{\\scriptsize even}} \\enspace cos(2\\pi \\msp u \\msp l) + I_{\\mbox{\\scriptsize odd}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega \\nonumber \\\\\n &=& \\iint I_{\\mbox{\\scriptsize even}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega + \\iint I_{\\mbox{\\scriptsize odd}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega \\nonumber \\\\\n &=& \\iint I_{\\mbox{\\scriptsize even}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega + \\iint 0 \\msp d\\,\\Omega \\nonumber \\\\\n &=& \\iint I_{\\mbox{\\scriptsize even}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega \\nonumber\n\\end{eqnarray}\n\n\\begin{braced}\n To see why~$\\iint I_{\\mbox{\\scriptsize even}} \\enspace cos(2\\pi \\msp u \\msp l) \\msp d\\,\\Omega = 0$, try integrating $\\int_{0}^{2\\pi }cos(x)\\msp cos(x) \\msp d\\,x$ v.s. $\\int_{0}^{2\\pi }sin(x)\\msp cos(x) \\msp d\\,x$ or $\\int_{-1}^{1}|x| \\msp cos(x) \\msp d\\,x$ v.s. $\\int_{-1}^{1} x \\msp cos(x) \\msp d\\,x$.\\\\\n\\end{braced}\n\n\\subsection{Odd and Even Components of the Correlator Output}\nSo, if we wish to keep the odd component of the emission, we need to sample the~$sin$ component, which can be accomplished by generating a~90$^\\circ $ phase shift in one of the signal paths from the pair of Equation~\\eqref{eq:3}\n\n\\begin{eqnarray}\n \\label{eq:14}\n \\begin{aligned}\n V_1 &=& |V_1| \\msp cos\\big (\\omega \\msp (t - \\tau _g)\\big ) \\\\\n V_2 &=& |V_2| \\msp cos\\big (\\omega t - \\frac{\\pi }{2}\\big )\n \\end{aligned}\n\\end{eqnarray}\n\nIn cross--correlating these antenna outputs~(as we did to obtain Equation~\\eqref{eq:4})\n\n\\begin{eqnarray}\n \\label{eq:15}\n &=& \\frac{1}{T} \\int _0^T V_1 \\msp V_2 \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} \\int _0^T |V_1| \\msp cos(\\omega \\msp (t - \\tau _g)) \\enspace |V_2| \\msp cos \\big (\\omega t - \\frac{\\pi }{2} \\big ) \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos \\big (\\omega \\msp (t - \\tau _g) + \\omega t - \\frac{\\pi }{2}\\big) + sin \\big (\\omega \\msp (t - \\tau _g) \\big) \\msp sin(\\omega t - \\frac{\\pi }{2} \\big ) \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos \\big (2\\omega t - \\omega \\tau _g - \\frac{\\pi }{2} \\big ) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos \\big (\\omega t -\\frac{\\pi }{2} \\big ) + \\dotso \\nonumber \\\\\n && \\dotso \\enspace cos \\Big [\\omega \\msp (t - \\tau _g) - \\big (\\omega t - \\frac{\\pi }{2} \\big ) \\Big ] \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\int _0^T cos \\big (2\\omega t - \\omega \\tau _g - \\frac{\\pi }{2} \\big ) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos \\big (\\omega t -\\frac{\\pi }{2} \\big ) + cos \\big (\\omega \\tau _g - \\frac{\\pi }{2} \\big ) \\msp dt \\nonumber \\\\\n &=& \\frac{1}{T} |V_1V_2| \\Bigg [\\int _0^T cos \\big (\\omega \\tau _g - \\frac{\\pi }{2} \\big ) \\msp dt + \\dotso \\nonumber \\\\\n && \\dotso \\enspace \\int _0^T cos \\big (2\\omega t - \\omega \\tau _g - \\frac{\\pi }{2} \\big ) - cos \\big (\\omega (t - \\tau _g) \\big) \\msp cos \\big (\\omega t -\\frac{\\pi }{2} \\big ) \\msp dt \\Bigg ] \\nonumber\n\\end{eqnarray}\n\nAgain we have the rapidly varying terms averaging to~0 and we substitute~$I$ for~$|V_1V_2|$.\n\n\\begin{eqnarray}\n \\label{eq:16}\n \\implies \\qquad &=& \\frac{1}{T} \\msp I \\Bigg [\\int _0^T cos \\big ( \\omega \\tau _g - \\frac{\\pi }{2}\\big ) \\msp dt + 0 \\Bigg ] \\nonumber \\\\\n &=& I \\msp cos \\big ( \\omega \\tau _g - \\frac{\\pi }{2}\\big )\\nonumber \\\\\n &=& I \\msp \\Big [cos(\\omega \\tau _g) \\msp cos \\big ( \\frac{\\pi }{2}\\big ) + sin(\\omega \\tau _g) \\msp sin \\big ( \\frac{\\pi }{2}\\big ) \\Big ] \\nonumber \\\\\n &=& I \\msp sin(\\omega \\tau _g) \\nonumber \\\\\n \\nonumber \\\\\n \\therefore \\enspace R_s &=& I \\msp sin(\\omega \\tau _g) \\nonumber\n\\end{eqnarray}\n\nThe next natural step is to define a complex function in terms of the~$sin$ and~$cos$ components of the correlator output.\n\n\\begin{eqnarray}\n \\label{eq:17}\n \\therefore \\enspace &=& R_c - i \\msp R_s \\msp , \\hspace{0.8cm} \\mbox{where} \\enspace i = \\sqrt{-1} \\nonumber \\\\\n &=& I \\msp cos(\\omega \\tau _g) - i \\msp I \\msp sin(\\omega \\tau _g) \\nonumber \\\\\n &=& I \\msp e^{-i\\msp \\omega \\tau _g} \\nonumber \\\\\n &=& I \\msp e^{-i\\msp 2\\pi \\msp u \\msp l}, \\hspace{0.8cm} \\mbox{see Equation~\\eqref{eq:7} for converting } \\omega \\tau _g \\mbox{ to } 2\\pi \\, u \\, l \\nonumber\n\\end{eqnarray}\n\n\\begin{braced}\n We use~$R_c - i \\msp R_s$ instead of~$R_c + i \\msp R_s$ to maintain the convention of the Fourier Transform relationship which will be revealed in Equation~\\eqref{eq:24}.\n\\end{braced}\n\\vspace{0.5cm}\n\nIn conjunction with Equation~\\eqref{eq:11} the correlator output is now called the \\emph{complex visibility} and is given by\n\n\\begin{eqnarray}\n \\label{eq:21}\n \\begin{aligned}\n \\mathcal{V}(\\vec{b}) &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c}} \\msp d\\,\\Omega \\\\\n &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\msp u \\msp l} \\msp d\\,\\Omega\n \\end{aligned}\n\\end{eqnarray}\n\nI have replaced~$I$ with,~$I_\\nu (\\hat{s})$ which is now the source brightness intensity at observing frequency~$\\nu $ and in the direction~$\\hat{s}$. At this stage, we are beginning to link something that we can measure, $\\mathcal{V}(\\vec{b}) = $, with something that we want to obtain,~$I_\\nu (\\hat{s})$.\n\n% \\begin{figure}\n% \\centering\n% \\includegraphics[width=\\textwidth ]{02-vis-plot.eps}\n% \\caption[]{The complex correlator response,~$\\mathcal{V}$, showing both the~$cos$~(as in Figure~\\ref{fig:corr-resp}) and~$sin$ components. Here the correlator is sampling a source with a \\emph{Gaussian} power profile~$I = I_\\nu (l) = e^{-l^2 }$ shown by the blue envelop.}\n% \\label{fig:02-vis-plot}\n% \\end{figure}\n% I give you an example of what the complex visibility might look like in Figure~\\ref{fig:02-vis-plot}. Use~\\texttt{02-vis-plot.py} to determine the fringe patterns for arbitrary values of~$u$ and~$l$.\n\n\\subsection{Visualising the result}\n\\label{sec:visualising-result}\n\nAgain, let's try to get an intuitive feel for what~$\\mathcal{V}(\\vec{b})$ is. In the following example, consider the simplest possible case where we have a~1\\,D baseline such that~$\\mathcal{V}(\\vec{b}) = \\mathcal{V}(u)$ which samples the source intensity along~1\\,D such that~$I_\\nu (\\hat{s}) = I_\\nu (l)$. $d\\,\\Omega $ now reduces to~$d\\, l$. Furthermore, let us assume that the source intensity can be modelled as a~\\emph{Dirac $\\delta $--function} $\\implies I_\\nu (l) = \\delta(l - l_0)$. $\\mathcal{V}(\\vec{b})$ can now be expressed as\n\n\\begin{eqnarray}\n \\label{eq:25}\n \\mathcal{V}(u) &=& \\int \\delta(l - l_0) \\msp e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l \\nonumber \\\\\n &=& \\int \\delta(l - l_0) \\msp e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l \\Bigg \\rvert_{\\msp l \\msp < \\msp l_0} + \\int \\delta(l - l_0) \\msp e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l \\Bigg \\rvert_{\\msp l \\msp = \\msp l_0} + \\int \\delta(l - l_0) \\msp e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l \\Bigg \\rvert_{\\msp l \\msp > \\msp l_0} \\nonumber \\\\\n &=& \\int 0 \\times e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l + \\int \\delta(0) \\msp e^{-i \\msp 2\\pi \\msp u \\msp l_0} \\msp d\\,l + \\int 0 \\times e^{-i \\msp 2\\pi \\msp u \\msp l} \\msp d\\,l \\nonumber \\\\\n &=& e^{-i \\msp 2\\pi \\msp u \\msp l_0} \\msp \\int \\delta(0) \\msp d\\,l \\nonumber \\\\\n &=& e^{-i \\msp 2\\pi \\msp u \\msp l_0} \\msp \\big [1\\big ] \\nonumber \\\\\n &=& cos(2\\pi \\msp u \\msp l_0) - i \\msp sin(2\\pi \\msp u \\msp l_0) \\nonumber\n\\end{eqnarray}\n\nThis means that the output of the correlator,~$\\mathcal{V}(u)$, is just a complex number. In Figure~\\ref{fig:03-dirac-vis}, we see~$\\mathcal{V}(u)$ displayed in both trigonometrical form~(middle column) and amplitude \\&~phase form~(right column), where the~$\\mbox{amplitude} = \\sqrt{cos^2 \\msp (2\\pi \\msp u \\msp l_0) - \\msp sin^2 \\msp (2\\pi \\msp u \\msp l_0)} \\msp = \\msp 1$ and the~$\\mbox{phase} = tan^{-1} \\msp \\Big [-\\frac{sin(2\\pi \\msp u \\msp l_0)}{cos(2\\pi \\msp u \\msp l_0)} \\Big ]$.\n\nIn Figure~\\ref{fig:04-gauss-vis}, we add a level of complexity by replacing the~$I_\\nu (l)$ in Equation~\\eqref{eq:21} with a Gaussian function instead. Both these figures can be replicated using \\texttt{03-dirac-vis.py} and \\texttt{04-gauss-vis.py} for arbitrary source offsets, as well as source width for the latter case.\n\n%% BoundingBox: 5 20 620 780 for fig = plt.figure(figsize=(10, 12))\n\\begin{figure}[]\n \\centering\n \\includegraphics[width=\\textwidth ]{03-dirac-vis.eps}\n \\caption{The output from the correlator,~$\\mathcal{V}$ is always a complex number. Here we see~$\\mathcal{V}$ in trigonometric~(middle column) and amplitude \\&~phase~(right column) for the simplest case where~$\\mathcal{V}(\\vec{b}) = \\mathcal{V}(u)$ and a source~$I_\\nu (\\hat{s}) = I_\\nu (l) = \\delta(l - l_0)$, which is offset from the phase centre by~$l \\msp = \\msp 0, \\msp 2 \\msp \\mbox{\\&} \\msp 7$ units~(left column). The phase in the right column~$\\in \\msp [-\\frac{\\pi }{2}, \\msp \\frac{\\pi }{2}]$~radians.}\n \\label{fig:03-dirac-vis}\n\\end{figure}\n\n%% BoundingBox: 5 20 620 780 for fig = plt.figure(figsize=(10, 12))\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth ]{04-gauss-vis.eps}\n \\caption{The correlator response to different Gaussian functions.}\n \\label{fig:04-gauss-vis}\n\\end{figure}\n\n\\section{Finite bandwidth}\n\nThe bandwidth has a dramatic effect on the correlator response. In the following computation, I am ignoring the effect of source structure.\n\n\\begin{eqnarray}\n \\mathcal{V}(\\vec{b}) &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c}} \\msp d\\,\\Omega \\hspace{0.8cm} \\longleftarrow \\mbox{quasi--monochromatic, i.e. infinitesimal bandwidth} \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{\\Delta \\nu } \\int ^{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} _{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c}} \\msp d\\, \\nu \\msp d\\,\\Omega \\hspace{0.8cm} \\longleftarrow \\mbox{finite bandwidth} \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{\\Delta \\nu } \\int ^{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} _{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} e^{-i \\msp 2\\pi \\msp \\nu \\msp \\tau _g} \\msp d\\, \\nu \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{\\Delta \\nu } \\int ^{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} _{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} cos(2\\pi \\msp \\nu \\msp \\tau _g) - i \\msp sin(2\\pi \\msp \\nu \\msp \\tau _g) \\msp d\\, \\nu \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{\\Delta \\nu } \\Bigg [\\frac{sin(2\\pi \\msp \\nu \\msp \\tau _g)}{2\\pi \\msp \\tau _g} + i \\msp \\frac{cos(2\\pi \\msp \\nu \\msp \\tau _g)}{2\\pi \\msp \\tau _g}\\Bigg ]^{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} _{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{\\Delta \\nu \\msp 2\\pi \\msp \\tau _g} \\Big [sin(2\\pi \\msp \\nu \\msp \\tau _g) + i \\msp cos(2\\pi \\msp \\nu \\msp \\tau _g)\\Big ]^{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} _{\\nu_ 0 + \\frac{\\Delta \\nu }{2}} \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Bigg [sin\\Big [2\\pi \\msp \\big (\\nu_ 0 + \\frac{\\Delta \\nu }{2}\\big ) \\msp \\tau _g\\Big ] + i \\msp cos\\Big [2\\pi \\msp \\big (\\nu_ 0 + \\frac{\\Delta \\nu }{2}\\big ) \\msp \\tau _g\\Big ] \\dotso \\nonumber \\\\\n&& \\dotso \\enspace - sin\\Big [2\\pi \\msp \\big(\\nu_ 0 - \\frac{\\Delta \\nu }{2}\\big ) \\msp \\tau _g\\Big ] - i \\msp cos\\Big [2\\pi \\msp \\big (\\nu_ 0 - \\frac{\\Delta \\nu }{2}\\big ) \\msp \\tau _g\\Big ]\\Bigg ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Bigg [sin\\Big (2\\pi \\msp \\nu_ 0 \\msp \\tau _g + 2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) + i \\msp cos\\Big (2\\pi \\msp \\nu_ 0 \\tau _g + 2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso \\enspace - sin\\Big (2\\pi \\msp \\nu_ 0 \\msp \\tau _g - 2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) - i \\msp cos\\Big (2\\pi \\msp \\nu_ 0 \\msp \\tau _g - \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )\\Bigg ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Bigg [sin (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) + cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso + i \\msp cos (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso - sin (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) + cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso - i \\msp cos (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )\\Bigg ] \\msp d\\,\\Omega \\nonumber\n\\end{eqnarray}\n\n\\begin{eqnarray}\n \\label{eq:27}\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Bigg [\\cancel{sin (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )} + cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso \\cancel{+ i \\msp cos (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )} - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso \\cancel{- sin (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )} + cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\dotso \\nonumber \\\\\n&& \\dotso \\cancel{- i \\msp cos (2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp cos\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )} - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big )\\Bigg ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{1}{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Bigg [2 \\msp cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) - 2 i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) \\Bigg ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{2 \\msp sin\\Big (2\\pi \\msp \\frac{\\Delta \\nu }{2} \\msp \\tau _g\\Big ) }{2\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Big [cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\Big ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{sin (\\pi \\msp \\Delta \\nu \\msp \\tau _g)}{\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\Big [cos(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) - i \\msp sin(2\\pi \\msp \\nu_ 0 \\msp \\tau _g) \\Big ] \\msp d\\,\\Omega \\nonumber \\\\\n&=& \\iint I_\\nu (\\hat{s}) \\msp \\frac{sin (\\pi \\msp \\Delta \\nu \\msp \\tau _g)}{\\pi \\msp \\tau _g \\msp \\Delta \\nu} \\msp e^{-i \\msp 2\\pi \\msp \\nu _0 \\msp \\tau _g} \\msp d\\,\\Omega\n\\end{eqnarray}\n\nWhat this tells us is that the correlator response~$e^{-i \\msp 2\\pi \\msp \\nu _0 \\msp \\tau _g}$ is modulated by the normalised~$sinc$ function~$\\frac{sin (\\pi \\msp \\Delta \\nu \\msp \\tau _g)}{\\pi \\msp \\tau _g \\msp \\Delta \\nu}$ as shown in Figure~\\ref{fig:05-fin-band}. From Equation~\\eqref{eq:27}, we clearly have nulls in the envelope~--~where the response falls to 0~--~when the numerator in~$\\frac{sin (\\pi \\msp \\Delta \\nu \\msp \\tau _g)}{\\pi \\msp \\tau _g \\msp \\Delta \\nu}$ is~0.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth ]{05-fin-band-lecture.eps}\n \\caption[]{The correlator~\\emph{cosine}~(left) and~\\emph{sine}~(right) responses (blue lines) when we have a finite--bandwidth of~$\\Delta \\nu = 8$~units~(e.g. GHz) and an observing frequency of~13~units for~$u = \\frac{|b|}{\\lambda} = 3$ and~$l$ extends from horizon--to--horizon~$\\in [-\\frac{\\pi}{2},-\\frac{\\pi}{2}]$. The red line is the~\\emph{sinc} envelope due to the bandwidth~$\\Delta \\nu $. This is analogous to Figure~\\ref{fig:corr-resp}, except we assume an infinitesimal bandwidth and only a \\emph{cosine} response in that previous figure. Use \\texttt{05-fin-band.py} to replicate this figure.}\n \\label{fig:05-fin-band}\n\\end{figure}\n\n\\begin{eqnarray}\n \\label{eq:30}\n \\mbox{i.e. when} \\qquad sin (\\pi \\msp \\Delta \\nu \\msp \\tau _g) &=& 0 \\nonumber \\\\\n \\implies \\qquad \\Delta \\nu \\msp \\tau _g &=& 1 , \\hspace{0.8cm} \\mbox{for the first null}\\nonumber \\\\\n \\Delta \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c} &=& 1 \\nonumber \\\\\n \\Delta \\nu \\msp b \\msp sin\\theta &=& c \\nonumber \\\\\n \\therefore sin\\theta &=& \\frac{c}{b\\msp \\Delta \\nu } \\quad = \\quad \\frac{\\frac{\\lambda}{b}}{\\frac{\\Delta \\nu }{\\nu }} \\nonumber\n\\end{eqnarray}\n\nThe number of fringes between the peak and this first null can be approximated by\n\n\\begin{eqnarray}\n \\label{eq:31}\n N &\\sim & \\frac{\\mbox{Distance between peak and first null}}{\\mbox{Distance between fringes}} \\nonumber \\\\\n &=&\\frac{\\frac{c}{b\\msp \\Delta \\nu}}{\\frac{\\lambda }{b}} \\nonumber \\\\\n &=& \\frac{c}{b\\msp \\Delta \\nu} \\cdot \\frac{b}{\\lambda } \\quad = \\quad \\frac{\\nu }{\\Delta \\nu } \\nonumber\n\\end{eqnarray}\n\ni.e. something like observing frequency divided by the bandwidth. The~$sinc$ envelope can also be expressed in a way more useful to the interferometer's characteristics rather than~$\\tau _g$ since~$sin(\\pi \\msp \\tau _g \\msp \\Delta \\nu ) = sinc \\big (\\pi \\msp \\frac{b}{\\lambda } \\msp \\frac{\\Delta \\nu }{\\nu } \\msp sin\\theta \\big ) = sinc \\big (\\pi \\msp \\frac{b \\msp \\Delta \\nu }{c} \\msp sin\\theta \\big )$.\n\n\\section{Coordinate systems}\n\n\\begin{figure}\n \\centering\n \\includegraphics[angle= 90, width=0.75\\textwidth ]{uv-lm-coords.eps}\n \\caption[]{There are two coordinate frames which we use in interferometry. The source is in the~$l-m$ plane and the interferometer is in the~$u-v$ plane.\\\\ \\emph{Image credit:} Interferometry and Synthesis in Radio Astronomy~(Thomson, Moran and Swenson,~2017) used under the Creative Commons Attribution International License (\\texttt{http://creativecommons.org/licenses/by-nc/4.0/}). }\n \\label{fig:uv-lm-coords}\n\\end{figure}\n\nWe need a formal coordinate framework in order to proceed further with the mathematics. This is described in Figure~\\ref{fig:uv-lm-coords}. Let the baseline have components,~$\\frac{\\vec{b}}{\\lambda } = u\\,\\hat{i} + v\\,\\hat{j} + w\\,\\hat{k}$. Let the source have components,~$\\hat{s} = l\\,\\hat{i} + m\\,\\hat{j} + n\\,\\hat{k}$.\n\n\\begin{eqnarray}\n \\label{eq:18}\n\\frac{\\vec{\\mathbf{b}}}{\\lambda }\\boldsymbol{\\cdot} \\mathbf{\\hat{s}} = \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{\\lambda } &=& (u\\,\\hat{i} + v\\,\\hat{j} + w\\,\\hat{k}) \\cdot (l\\,\\hat{i} + m\\,\\hat{j} + n\\,\\hat{k}) \\nonumber \\\\\n &=& (u\\,l + v\\,m + w\\,n) \\nonumber\n\\end{eqnarray}\n\n\\begin{braced}\nNote that~$\\hat{s}$ is a unit vector,~$|\\hat{s}| = 1$.\n\\begin{eqnarray}\n \\label{eq:20}\n \\implies \\qquad l^2 + m^2 + n^2 &=& 1 \\nonumber \\\\\n n &=& \\sqrt{1 - l^2 + m^2} \\nonumber\n\\end{eqnarray}\n\\end{braced}\n\\vspace{0.5cm}\n\nAt this point, we employ a simplification to assume that the antennas lie on a plane such that~$w = 0$.\n\n\\begin{eqnarray}\n \\label{eq:22}\n \\therefore \\enspace \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{\\lambda } &=& (u\\,l + v\\,m) \\nonumber\n\\end{eqnarray}\n\nEquation~\\eqref{eq:21},~$\\mathcal{V}(\\vec{b}) = \\iint I_\\nu (\\hat{s}) \\msp e^{i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{b} \\cdot \\hat{s}}{c}} \\msp d\\,\\Omega $, now becomes\n\n\\begin{eqnarray}\n \\label{eq:23}\n \\mathcal{V}(\\vec{b}) &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{c}} \\msp d\\,\\Omega \\nonumber \\\\\n \\mathcal{V}(\\vec{b}) &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\nu \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{\\nu \\msp \\lambda }} \\msp d\\,\\Omega \\nonumber \\\\\n \\mathcal{V}(\\vec{b}) &=& \\iint I_\\nu (\\hat{s}) \\msp e^{-i \\msp 2\\pi \\msp \\frac{\\vec{\\mathbf{b}} \\boldsymbol{\\cdot} \\mathbf{\\hat{s}}}{\\lambda }} \\msp d\\,\\Omega \\nonumber \\\\\n \\mathcal{V}(u, v) &=& \\iint I_\\nu (l, m) \\msp e^{-i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,l \\msp d\\,m \\nonumber\n\\end{eqnarray}\n\nIn the final step, we replace the vectors~$\\vec{\\mathbf{b}}$ and~$\\mathbf{\\hat{s}}$ with their corresponding vector components~$u-v$ and~$l-m$. You can think of the area~$d\\,\\Omega \\approx d\\,l \\msp d\\,m$~(see the shaded area~$d\\,\\Omega $ in Figure~\\ref{fig:uv-lm-coords}). We now have a~2\\,D Fourier transform between the visibility,~$\\mathcal{V}(u, v)$, which we can measure, and the source brightness distribution,~$I_\\nu (l, m)$, which we want.\n\n\\begin{eqnarray}\n \\label{eq:24}\n \\begin{aligned}\n \\mathcal{V}(u, v) &=& \\iint I_\\nu (l, m) \\msp e^{-i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,l \\msp d\\,m \\\\\n I_\\nu (l, m) &=& \\iint \\mathcal{V}(u, v) \\msp e^{ i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,u \\msp d\\,v\n \\end{aligned}\n\\end{eqnarray}\n\nThis is known as the \\emph{Van Cittert--Zernicke Theorem} and~$I_\\nu (l, m)$ is the \\underline{\\emph{true}} sky brightness distribution and~$\\mathcal{V}(u, v)$ is the \\underline{\\emph{true}} sky visibility.\n\n\\section{UV sampling}\n\nAccording to equation~\\eqref{eq:24}, the source intensity~$I_\\nu (l, m)$ can be perfectly described by~$\\mathcal{V}(u, v)$. However, in practice this is not the case as the~$(u,\\msp v)$ sampling is baseline dependent. Figure~\\ref{fig:06-array2uv-triangle}~(left panel) shows a hypothetical triangular array with elements~$A$,~$B$ and~$C$. With respect to the centre~(0,\\,0) of this array these have corresponding position vectors~$\\overrightarrow{OA}$,~$\\overrightarrow{OB}$ and~$\\overrightarrow{OC}$. Since the~$u-v$ space is derived from the baseline vector, we need to determine the individual baseline coordinates~$\\overrightarrow{AB}$,~$\\overrightarrow{AC}$ and~$\\overrightarrow{BC}$~(and corresponding degenerate baselines~$\\overrightarrow{BA}$,~$\\overrightarrow{CA}$ and~$\\overrightarrow{CB}$). For example,\n\n\\begin{eqnarray}\n \\label{eq:26}\n \\overrightarrow{AB} &=& \\enspace \\overrightarrow{AO} + \\overrightarrow{OB} \\nonumber \\\\\n &=& -\\overrightarrow{OA} + \\overrightarrow{OB} \\nonumber \\\\\n &=& -(0.5\\msp \\hat{i} - 0.5\\msp \\hat{j}) + (0\\msp \\hat{i} + 0.5\\msp \\hat{j}) \\nonumber \\\\\n &=& -0.5\\msp \\hat{i} + 1\\msp \\hat{j} \\nonumber\n\\end{eqnarray}\n\n\\begin{figure}\n\\begin{subfigure}[b]{\\textwidth }\n \\centering\n %%BoundingBox: -160 145 760 666 for fig = plt.figure(figsize=(15, 7.5))\n \\includegraphics[width=\\linewidth ]{06-array2uv-triangle.eps}\n \\caption{The black circles represent antenna positions of a hypothetical array~(left panel) on the Earth's surface~($x,\\msp y$) and corresponding~$uv$--coordinates~(right panel). The blue arcs trace the loci of these~$u-v$ points which are sampled as the Earth rotates. The offset between the start of the blue points and the black points is due to a coordinate rotation which I have applied in keeping with convention. You can use~\\texttt{07-array2uv-loci.py} to reproduce this figure.\\\\}\n \\label{fig:06-array2uv-triangle}\n\\end{subfigure}\\\\\n\\begin{subfigure}[b]{\\textwidth }\n \\centering\n %%BoundingBox: -175 145 760 666 for fig = plt.figure(figsize=(15, %7.5))\n \\includegraphics[width=\\linewidth ]{06-array2uv-meerkat.eps}\n \\caption{The black circles represent MeerKAT's antenna positions~(left panel) on the Earth's surface~($x,\\msp y$) and corresponding~$uv$--coordinates~(right panel).}\n \\label{fig:07-array2uv-meerkat}\n\\end{subfigure}\n\\caption{}\n\\label{fig:xy-uv}\n\\end{figure}\n\nThe right panel of Figure~\\ref{fig:06-array2uv-triangle} is a graphical representation of what is known as the \\emph{sampling pattern},~$S(u, v)$, because it is where~$\\mathcal{V}(u, v)$ is sampled such that we have the \\emph{dirty image}\n\n\\begin{eqnarray}\n \\label{eq:28}\n \\mbox{True image,} \\quad I_\\nu (l, m) &=& \\iint \\mathcal{V}(u, v) \\msp e^{ i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,u \\msp d\\,v \\hspace{0.5cm} \\mbox{from Equation~\\eqref{eq:24}} \\nonumber \\\\\n \\mbox{Dirty image,} \\quad I_{\\mbox{\\tiny{\\emph{D}}}} (l, m) &=& \\iint S(u, v) \\msp \\mathcal{V}(u, v) \\msp e^{ i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,u \\msp d\\,v\n\\end{eqnarray}\n\nWe know that a product of two functions in the~$u-v$ Fourier domain corresponds to a convolution of two signals in the~$l-m$ domain such that~$I_{\\mbox{\\tiny{\\emph{D}}}} (l, m) = I_\\nu (l,m) * B(l,m)$, where~$B(l,m)$ is known as the \\emph{dirty beam} and is the Fourier Transform of~$S(u, v)$\n\n\\begin{eqnarray}\n \\label{eq:29}\n B(l,m) &=& \\iint S(u, v) \\msp e^{ i \\msp 2\\pi \\msp (u\\,l + v\\,m)} \\msp d\\,u \\msp d\\,v \\nonumber\n\\end{eqnarray}\n\nI have summarised the relationship between~$I_\\nu (l, m)$, $\\mathcal{V}(u, v)$, $B(l, m)$, $S(u, v)$, $I_{\\mbox{\\tiny{\\emph{D}}}}(l, m)$ and~$\\mathcal{V}(u, v) \\msp S(u, v)$ in Table~\\ref{tab:uv-summary}. Recovering~$I_\\nu (l, m)$, $\\mathcal{V}(u, v)$ from~$I_{\\mbox{\\tiny{\\emph{D}}}}(l, m)$ is a deconvolution problem which can be solved using~CLEAN.\n\n\\begin{table}\n \\centering\n \\begin{tabular}{r|ccccc}\n$\\mathbf{l-m}$ \\textbf{(image)} & $I_\\nu (l, m)$ & * & $B(l, m)$ & = & $I_{\\mbox{\\tiny{\\emph{D}}}}(l, m)$ \\\\\n\\textbf{domain} & True image & & Dirty beam & & Dirty image \\\\\n & \\emph{\\textbf{(Desired)}}& & & & \\\\\n & & & & & \\\\\n $\\mathcal{F.T.}$ & $\\Big \\Updownarrow $ & & $\\Big \\Updownarrow $ & & $\\Big \\Updownarrow $ \\\\\n & & & & & \\\\\n$\\mathbf{u-v}$ \\textbf{(Fourier)} & $\\mathcal{V}(u, v)$ & $\\times $ & $S(u, v)$ & = & $\\mathcal{V}(u, v) \\msp S(u, v)$ \\\\\n \\textbf{domain} & True visibility & & Sampling pattern & & \\emph{\\textbf{(Measured)}} \\\\\n & & & \\emph{\\textbf{(Known)}} & & \\\\\n \\end{tabular}\n \\caption[]{A summary of the relationship between quantities in the image and Fourier domains.}\n \\label{tab:uv-summary}\n\\end{table}\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"Lecture\"\n%%% End:", "meta": {"hexsha": "08575517e4565f39dadfc7dd7f4611878ab949d2", "size": 42890, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/01-Intro.tex", "max_stars_repo_name": "vasaantk/Introduction-to-Interferometry", "max_stars_repo_head_hexsha": "a584fb8b3888fd9bfa00a19fd212ce2b216ea6bf", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/01-Intro.tex", "max_issues_repo_name": "vasaantk/Introduction-to-Interferometry", "max_issues_repo_head_hexsha": "a584fb8b3888fd9bfa00a19fd212ce2b216ea6bf", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/01-Intro.tex", "max_forks_repo_name": "vasaantk/Introduction-to-Interferometry", "max_forks_repo_head_hexsha": "a584fb8b3888fd9bfa00a19fd212ce2b216ea6bf", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 84.0980392157, "max_line_length": 920, "alphanum_fraction": 0.6175332245, "num_tokens": 15737, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424217727027, "lm_q2_score": 0.7634837527911057, "lm_q1q2_score": 0.6458632948202194}} {"text": "\\chapter{The $\\alpha$-shapes approach}\\label{chap:boundaries_alpha}\nIn the previous chapter, we presented a new ray tracing approach based on PS. We explained that, in order to compute the target intensity, it is necessary to know the boundaries of the regions in target PS with positive luminance. Ray tracing in PS requires tracing only the rays close to these boundaries. The rays traced can be seen as a point cloud in PS. To detect the shape formed by those rays, the $\\alpha$-shapes approach is employed \\cite{portegies2013fast}.\\\\ \\indent\nMethods based on $\\alpha$-shapes are widely used to reconstruct an unknown shape formed by a set of finite data points \\cite{guo1997surface}. $\\alpha$-shapes is a very powerful tool to construct the shape of a point cloud. As the parameter $\\alpha$ varies, we can obtain different $\\alpha$-shapes from the point set itself to the convex\nhull \\cite{xu2003automatic}. The disadvantage of such method is that it can be very hard to choose the appropriate value of the parameter $\\alpha$ and, in most cases it can be selected only by trial-and-error.\\\\ \\indent\nWe developed a technique based on $\\alpha$-shapes that gives a criterion to determine the value of the parameter $\\alpha$, for which the boundaries are approximated well \\cite{filosa2015new}.\\\\ \\indent This chapter is organized as follows. An overview of the-state-of-the-art about $\\alpha$-shape methods is provided in Section \\ref{sec:alpha-shapes}; the technique used for computing the $\\alpha$ value is explained in Section \\ref{sec:Tir_alpha}; the results for two different kinds of total internal reflection (TIR)-collimators are given in Section \\ref{sec:results-Tir-alpha}. Discussions and conclusions are provided in the last paragraph of this chapter.\n\\section{$\\alpha$-shapes theory}\\label{sec:alpha-shapes}\nGiven a finite set $\\mbox{\\insieme{U}} = \\{\\point{u}_1, \\cdots, \\point{u}_N\\}\\subset \\mathbb{R}^2$ of points, $\\alpha$-shapes are geometrical objects that give us an approximation of the shape formed by the point cloud. For now we do not further specify the notion of shape. A more precise definition will be provided later.\\\\ \\indent\nBefore giving a formal definition, we explain an intuitive and nice interpretation of $\\alpha$-shapes \\cite{lucieer2004alpha}. \nLet us think of a stracciatella ice-cream\\footnote{Stracciatella ice cream is made with milk-based ice-cream and fine pieces of chocolate \\cite{Wiki3}.}. If we desire to know the shape formed by the chocolate pieces we can start eating the ice cream using a spoon with a spherical scoop and try not to remove any piece of chocolate. \nWe will obtain a shape formed by arcs and points. % (see Figure \\ref{fig:shape2d} for the two-dimensional case).\n%\\begin{figure}[t]\\label{fig:shape2d}\n%\\begin{center} \n%\\includegraphics[width=\\textwidth]{alpha_shape2D}\n%\\label{fig:shape}\n%\\caption{\\textbf{Construction of $\\alpha$-shapes.} The boundary of the shape (red line) formed by a set of points (blue dots) in $\\mathbb{R}^2$ is detected for $\\alpha = 1$ (left) and for $\\alpha=2$ (right) \\cite{sabel2017application}.}\n%\\label{fig:shape2d}\n%\\end{center}\n%\\end{figure}\nStraightening the arcs to line segments we obtain broken lines which constitute the boundary of the so-called $\\alpha$-shape of the point set $\\mbox{\\insieme{U}}$. \nA very small spoon will allow us to eat the entire ice cream without eating any piece of chocolate, while with a larger spoon we are not able to eat any chunk of the ice cream without chocolate pieces. In this example, the chocolates pieces are the points of set $\\mbox{\\insieme{U}}$ and, the parameter $\\alpha$ determines the radius of the carving spoon (the spherical spoon in two-dimension is simply a circle).\\\\ \\indent \nThe formal definition of $\\alpha$-shape was first given by Edelsbrunner, Kirkpatrick and Seidel in 1983 \\cite{edelsbrunner1983shape}. They describe $\\alpha$-shape as a generalization of the convex hull of a finite set of points in the plane. Let $\\alpha$ be a non negative number $0\\leq\\alpha<\\infty$. \nIf $\\alpha = 0$ the shape degenerates to the point set $\\mbox{\\insieme{U}}$. On the other hand, when $\\alpha\\rightarrow\\infty$ the $\\alpha$-shape is simply the convex hull of \\insieme{U}. If $0<\\alpha<\\infty$ the $\\alpha$-shape is a polygon of \\insieme{U} \\cite{edelsbrunner1994three}. The construction of the $\\alpha$-shape is closely related to the Delaunay triangulation of \\insieme{U} \\cite{mucke1993shapes}. Therefore, a formal definition of triangulation and Delaunay triangulation is now required. \\\\ \\indent\nGiven a set $\\mbox{\\insieme{U}}$ of points not all aligned, let us consider the set \\insieme{E} of all the straight-line segments whose endpoints are in \\insieme{U}. \nA triangulation \\insieme{T} of $\\mbox{\\insieme{U}}$ is the subset of \\insieme{E} with the maximum number of segments such that all the line segments of \\insieme{T} intersect only at their endpoints \\cite{lloyd1977triangulations}. \n\\\\ \\indent Before giving a more formal definition of triangulation, let us define a partition of a set $\\mbox{\\insieme{X}}\\subset \\mathbb{R}^2$ as a collection of the subsets which divide \\insieme{X} into non-overlapping regions so that any point in \\insieme{X} is located in only one region. \n\\begin{definition} \\indent Let $\\mbox{\\insieme{P}}\\subset \\mathbb{R}^2$ be the convex hull of $\\mbox{\\insieme{U}}$ and $\\mbox{\\insieme{T}} = \\{\\mbox{\\insieme{T}}_1, \\cdots, \\mbox{\\insieme{T}}_\\variabile{h}\\}$ be a partition of $\\mbox{\\insieme{P}}$ into closed triangles, that is triangles that include their edges. Suppose that the following properties hold:\n\\begin{itemize}\n\\item[a)] $\\mbox{\\insieme{P}} = \\bigcup_{\\variabile{i} = 1}^{\\variabile{h}}\\mbox{\\insieme{T}}_\\variabile{i}$,\n\\item[b)] $\\forall \\mbox{ \\insieme{T}}_\\variabile{i}, \\mbox{\\insieme{T}}_\\variabile{j} \\in \\mbox{\\insieme{T}}$, $\\mbox{\\insieme{T}}_\\variabile{i} \\neq \\mbox{\\insieme{T}}_\\variabile{j}$, and \n\\begin{equation*}\n\\textrm{int}(\\mbox{\\insieme{T}}_\\variabile{i})\\cap \\textrm{int}(\\mbox{\\insieme{T}}_\\variabile{j}) = \\emptyset,\n\\end{equation*}\nwhere $\\textrm{int}(\\mbox{\\insieme{T}}) = \\mbox{\\insieme{T}}-\\partial \\mbox{\\insieme{T}}$,\n\\end{itemize}\nthen $\\mbox{\\insieme{T}}$ is called a \\textit{triangulation} of $\\mbox{\\insieme{P}}$ \\cite{Numericalmethods}.\n\\end{definition}\nThe Delaunay triangulation $\\mbox{\\insieme{T}}^{\\prime}$ of the point set \\insieme{U} has the property that the circumcircle of any triangle of $\\mbox{\\insieme{T}}^{\\prime}$ does not contain any point of \\insieme{U}. This is called the Delaunay property. Next, a very commonly used algorithm to construct such triangulation is explained.\\\\ \\indent \n$\\mbox{\\insieme{T}}^\\prime$ is constructed by modifying a general triangulation \\insieme{T} such that every point satisfies the Delaunay property. \nTherefore, every triangle that does not satisfy such property is flipped such that the new edge is part of the triangulation, see Figure \\ref{fig:Delaunay}. \nGiven, for example, an arbitrary triangulation \\insieme{T} in two-dimensions, for each edge $\\overline{\\point{ab}}$ in \\insieme{T} which is not on the boundary of the convex hull the two triangles \n$\\Delta_{\\point{abc}}$ and $\\Delta_{\\point{abd}}$ with the common edge $\\overline{\\point{ab}}$ are identified. Then, if either the circumcircle of triangle $\\Delta_{\\point{abc}}$ contains point \\point{d} or the circumcircle of triangle $\\Delta_{\\point{abd}}$ contains point \\point{c}, the edge $\\overline{\\point{ab}}$ cannot be included in the Delaunay triangulation and, therefore, $\\overline{\\point{ab}}$ is replaced by the edge $\\overline{\\point{cd}}$ such that the triangles $\\Delta_{\\point{acd}}$ and $\\Delta_{\\point{bcd}}$ are constructed. The new edge $\\overline{\\point{cd}}$ locally satisfies the Delaunay property and the triangles $\\Delta_{\\point{acd}}$ and $\\Delta_{\\point{bcd}}$ are added to the Delaunay triangulation $\\mbox{\\insieme{T}}^\\prime$. \n\\begin{figure}[t]\\label{fig:Delaunay}\n\\begin{subfigure}[t]{0.48\\textwidth}\n\\centering\n\\includegraphics[]{triangle_alpha_shapes}\n\\label{fig:shape}\n\\caption{\\textbf{Not acceptable triangle.} The point \\point{d} is inside the circle circumscribing the triangle $\\Delta_{\\point{abc}}$, therefore the edge $\\overline{\\point{ab}}$ cannot be included in the Delaunay triangulation.}\n\\end{subfigure}\n\\hfill\n\\begin{subfigure}[t]{0.48\\textwidth}\n\\centering\n\\includegraphics[]{triangle_alpha_shapes_flipped}\n\\caption{\\textbf{Acceptable triangle.} The flipped triangle $\\Delta_{\\point{acd}}$ satisfies the Delaunay property, thus it is included in the Delaunay triangulation.}\n\\end{subfigure}\n\\caption{Construction of the Delaunay triangulation in 2D.}\n\\label{fig:Delaunay}\n\\end{figure}\n\\\\ \\indent Several other algorithms have been developed to construct a Delaunay triangulation, see for example \\cite{lee1980two, renka1997algorithm}.\n Given a point set $\\mbox{\\insieme{U}}$ and a triangulation \\insieme{T}, it can be proved that the corresponding Delaunay triangulation $\\mbox{\\insieme{T}}^\\prime$ is unique. Moreover, it maximizes the largest minimum angle among all possible triangulations of a point set $\\mbox{\\insieme{U}}$ \\cite{press2007numerical}.\n\\\\ \\indent Alternatively, the Delaunay triangulation can be constructed as the dual of the Voronoi diagram \\cite{fortune1992voronoi}. Let $\\mbox{\\insieme{X}}\\subset\\mathbb{R}^2$ be a metric space endowed with the Euclidean distance $d(\\point{x}, \\point{y})$ for $\\point{x}, \\point{y}\\in \\mbox{\\insieme{X}}$. For \\textit{almost}\\footnote{Note the importance of the word \\textit{almost}. Some points can have the same distance with two or more points of $\\mbox{\\insieme{U}}$.} every point $\\point{x}\\in \\mathbb{R}^2$, there is a unique point that is the closest point to $\\point{x}$. The Voronoi cell of a point $\\point{u}_\\variabile{i}\\in \\mbox{\\insieme{U}}$ contains all points in $\\mathbb{R}^2$ that are closest to $\\point{u}_{\\variabile{i}}$, see Figure \\ref{fig:Voronoi}. The Voronoi diagram of $\\mbox{\\insieme{U}}\\subset \\mathbb{R}^2$ is defined as the set of all Voronoi cells \\cite{cazals2005conformal}. A more formal definition of the Voronoi diagram is given in the following.\n\\begin{defn}\nLet $\\mbox{\\insieme{U}}=\\{\\point{u}_1,\\cdots,\\point{u}_N\\}$ be a set of points in $\\mathbb{R}^2$. The Voronoi cell $\\mbox{\\insieme{V}}_\\variabile{i}$ associated to point $\\point{u}_\\variabile{i}$ is defined as:\n\\begin{equation}\n\\mbox{\\insieme{V}}_\\variabile{i}=\\{\\point{x}\\in \\mathbb{R}^2\\; | \\;|\\point{x}-\\point{u}_\\variabile{i}|<|\\point{x}-\\point{u}_\\variabile{j}| \\quad \\forall \\variabile{j}\\neq \\variabile{i} \\}.\n\\end{equation}\nThe Voronoi diagram $\\mbox{\\insieme{V}}$ is defined as \n\\begin{equation}\n\\mbox{\\insieme{V}} = \\bigcup_{\\variabile{i}=1}^N \\mbox{\\insieme{V}}_\\variabile{i}\n\\end{equation}\n where $\\mbox{\\insieme{V}}_{\\variabile{i}}\\cap \\mbox{\\insieme{V}}_{\\variabile{j}}= \\emptyset$ for $\\variabile{i}\\neq\\variabile{j}$.\n\\end{defn}\nFor the definition of Voronoi diagram in higher dimensions see \\cite{brown1979voronoi}.\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width = 0.7 \\textwidth]{Delanay_Voronoi1}\n\\caption{\\textbf{Relationship between the Delaunay triangulation and the Voronoi Diagram \\cite{Wiki4}.} The black line segments are the boundaries of the Delaunay triangulation, the red line segments constitutes the boundaries of the Voronoi diagram.}\n\\label{fig:Voronoi}\n\\end{figure}\n\\\\ \\indent The Delaunay triangulation triangulates the convex hull of \\insieme{U} and, therefore it does not constitute a suitable method for reconstructing the contour formed by a point cloud. The $\\alpha$-shape method was developed to solve such problem \\cite{edelsbrunner2010alpha, guo1997surface}. Starting from the Delaunay triangulation $\\mbox{\\insieme{T}}^\\prime$ of a point set \\insieme{U}, the corresponding $\\alpha$-shape of $V$ is formed by the only triangles of $\\mbox{\\insieme{T}}^\\prime$ that satisfy the so-called \"$\\alpha$-test\" which is briefly explained.\nFor each triangle, we calculate the circumradius, i.e., the radius of the circumcircle. If this radius is larger than $\\alpha$ the triangle is removed from the shape. The choice of the parameter $\\alpha$ is highly significant in the $\\alpha$-shapes procedure and, it has to be selected such that the desired approximation of the shape formed by the points of $V$ is obtained.\\\\ \\indent\n%Therefore, $\\alpha$ is closely related to the radius of the circumcircles. A possible strategy is to find the radius of the greater empty circumcircle. Thus $\\alpha$ can be selected according to the density $\\delta$ of the point sets $V$\n%with $C$ a constant and $\\delta$:\n%\\begin{equation}\n%\\delta=\\frac{N}{A}\\; ,\n%\\end{equation}\n%where $N$ is the number of points in $V$ and $A$ is the area of the convex hull of $V$.\n%%Note that, if we scale the dimensions of the \n%The value of $\\alpha$ can be chosen, for instance, inversely proportional to $\\sqrt{\\delta}$:\n%\\begin{equation}\n%\\alpha=C\\frac{1}{\\sqrt{\\delta}}\\;,\n%\\end{equation}\n%Note that, while $\\delta$ is given for a fixed point set $V$, the value of $C$ needs to be determined by numerical simulations.\\\\ \\indent \nTo summarize, the $\\alpha$-shape construction can be outlined as follows:\n\\begin{enumerate}\n\\item Construct a Delaunay triangulation\\footnote{In the simulations we present in this chapter the Matlab function \\textit{Delaunay} is used \\cite{matlab_delaunay}.} $\\mbox{\\insieme{T}}^\\prime$ of the point cloud $\\mbox{\\insieme{U}}$;\n\\item For every triangle $\\mbox{\\insieme{T}}^\\prime(i)\\in \\mbox{\\insieme{T}}^\\prime$ calculate its circumradius $r(i)$;\n\\item If $r(i)\\leq\\alpha$ keep the triangle $\\mbox{\\insieme{T}}^\\prime(i)$ in the triangulation;\n\\item If $r(i)>\\alpha$ remove the triangle from the triangulation;\n\\item Select from the new triangulation obtained those sides that belong to only one triangle, the so-called \\textit{free boundary edges}\\footnote{In the simulations we present in this chapter the Matlab function \\textit{freeBoundary} is used \\cite{matlab_free}.}. By definition, the free boundary edges are not a common edge of any two triangles.\n\\end{enumerate}\n%% Insert algorithm\n%\\begin{algorithm}[h]\n%\\caption{$\\alpha$-shape reconstruction}\\label{alg:alphashapes}\n%\\begin{algorithmic}[1]\n%\\Procedure{$\\alpha$-shape}{$V$, $\\alpha$}\n%\\State Construct a Delaunay triangulation of the point cloud $V$\n%\\State$T^\\prime\\gets$ all triangles of the Delaunay triangulation\n%\\State For every \n%\\State Calculates the radius of the circle circumscribed by the vertices of every triangle\n%\\State $r(i) \\gets$\n%\\State For {every triangle in $T^\\prime$}\n%\\State $(\\variabile{q}_1^4, \\variabile{p}_1^4) \\gets \\mbox{left and upper corner of source PS: (-\\variabile{a}, 1)} $\n%\\For{$ \\variabile{k}= 1 \\to 4 $}\n%\\State Trace the ray with initial coordinates $(\\variabile{q}_1^\\variabile{k}, \\variabile{p}_1^{\\variabile{k}})$ in \\set{S}{}{};\n%\\State Calculate the corresponding path $\\Pi^{\\variabile{k}}$;\n%\\State $\\mbox{Ray}.\\variabile{q}\\gets [\\mbox{Ray}.\\variabile{q}, \\variabile{q}_1^\\variabile{k}]$;\n%\\State $\\mbox{Ray}.\\variabile{p}\\gets [\\mbox{Ray}.\\variabile{p}, \\variabile{p}_1^\\variabile{k}]$;\n%\\State Store the corresponding path $\\Pi^{\\variabile{k}}$.\n%\\State $\\mbox{Ray}.\\Pi\\gets [\\mbox{Ray}.\\Pi, \\Pi^{\\variabile{k}}]$;\n%\\EndFor\n%\\State VL $\\gets [1, 2, 4]$ \\Comment{VL = vertices of the left triangle}\n%\\State VR $\\gets [2,3, 4]$ \\Comment{VR = vertices of the right triangle}\n%\\State \\Call{Left Triangle}{VL, Ray, $\\varepsilon_{\\variabile{q}_1}^{\\textrm{min}}, \\varepsilon_{\\variabile{q}_1}^{\\textrm{max}}, \\varepsilon_{\\variabile{p}_1}^{\\textrm{min}}, \\varepsilon_{\\variabile{p}_1}^{\\textrm{max}}$}\\Comment{Refine the left triangle} \n%\\State \\Call{Right Triangle}{VR, Ray, $\\varepsilon_{\\variabile{q}_1}^{\\textrm{min}}, \\varepsilon_{\\variabile{q}_1}^{\\textrm{max}}, \\varepsilon_{\\variabile{p}_1}^{\\textrm{min}}, \\varepsilon_{\\variabile{p}_1}^{\\textrm{max}}$} \\Comment{Refine the right triangle} \\\\\n%\\Return ;\n%\\EndProcedure\n%\\end{algorithmic}\n%\\end{algorithm}\n%\\\\\n%Let us define a Voronoi diagram in a metric space.\n%The simplest case that we can have is the two-dimensional case that is the case where $X=\\mathbb{R}^2$.\n%The tuple $\\mathcal{S}=\\{1,\\cdots,n\\}\\subset \\mathbb{R}^2$ is now a set of points. The Voronoi diagram of $\\mathcal{S}$ is a subsection of $\\mathbb{R}^2$ such that every other region around a point $p\\in \\mathcal{S}$ contains all points that are closer to $p$ than to every point in $\\mathcal{S}$. A triangulation of the point set $\\mathcal{S}$ is a set of edges $\\mathcal{E}$ whose extremes are points of $\\mathcal{S}$ such that the faces of each triangle are bounded by three edges and any edge that is not in $\\mathcal{E}$ intersects one of the existing edges. The Delaunay triangulation is the dual graph of the Voronoi diagram: it consists of vertices (the points in $\\mathcal{S}$) and it has an edge between two vertices if the two corresponding faces share an edge. \n$\\alpha$-shapes provide a nice mathematical definition of the \\textit{shape} of\na set of points. In two dimensions, $\\alpha$-shapes gives the contour of the point cloud which is approximated by a family of line segments. \nAlthough they are a powerful tool for determining the shape of a point cloud, there exist shapes that are not described well by classical $ \\alpha $-shapes. Indeed, for some point sets there is no value of $\\alpha$ that gives a good approximation of the contour formed by the point cloud. Usually the parameter $\\alpha$ is determined according to the density of the point cloud, therefore, it can be difficult to obtain a good approximation of a shape formed by a non-uniform point set. Furthermore, the $\\alpha$-shape method does not work well when the shape we need to approximate has a sharp turn or a joint, this case will be clarified in Section \\ref{sec:results-Tir-alpha} with an example. \n%In this case $\\alpha$-shapes often give a \"webbed-foot\" appearance at such joints since they improperly connect the adjacent surfaces. \n\\\\ \\indent There are several ways to determine the value of $\\alpha$ \\cite{mandal1997selection}; in the next section we provide a technique that exploits the conservation of \\'{e}tendue in PS. \n%The first step of this method is to make a triangulation of the point cloud.\n%Then the key idea is to compute somehow the point-density of each point and use this to get an approximation of the point density of a triangle. In this way one can reduce the $\\alpha$-value in areas where the triangle's point density (see Equation \\ref{delta_t} for the definition) is higher than average in such a way that is possible to obtain a finer level of detail for areas that have an higher density.\n%More precisely, each point $ \\textbf{p}\\in \\mathcal{S} $ has a local point density defined as\n%\\begin{equation}\n%\\delta (\\textbf{p})= \\sum_{\\textbf{q}\\in \\mathcal{S}}\\Big( 1-\\frac{\\textrm{d}(q,p)}{\\lambda}\\Big) \\qquad \\forall \\textbf{q} \\mbox{\\;\\;such that\\;\\;} \\textrm{d}(\\textbf{p},\\textbf{q})<\\lambda\\,,\n%\\end{equation}\n%where $ \\lambda $ is the constant radius of the local neighborhood and $\\textrm{d}(\\textbf{x},\\textbf{y})$ is the Euclidean distance.\n%When local density is larger than the average, that is when\n%\\begin{equation}\n%\\delta (\\textbf{p}) >\\frac{1}{| \\mathcal{S} |}\\sum_{\\textbf{q}\\in \\mathcal{S}}\\delta (\\textbf{q})\n%\\end{equation}\n%we know some properties about the region surrounding $\\textbf{p}$.\n%For instance, if the point set is uniformly distributed then it is possible to find areas with a high-density in the case where there are two closely separated surfaces. In point sets of non-uniform distribution, high densities are found when the surface presents a joint discontinuity. The algorithm developed by Teichmann and Capps is structured as follow.\n%After computing density information for each point they make a triangulation of the point set. Then they calculate the average density $\\delta(t)$ for each triangle $\\Delta_{abc}$ defined as:\n%\\begin{equation}\n%\\delta(t)=\\frac{\\delta(a)+\\delta(b)+\\delta(c)}{3 \\mu}\\,,\n%\\label{delta_t}\n%\\end{equation}\n%where $\\mu$ is the global average density of the entire point set $\\mathcal{S}$.\n%If $\\delta(t)$ is greater than $1$ the density of the point cloud is higher. Hence is necessary to define another value of $\\alpha$:\n%\\begin{equation}\n%\\alpha^{\\;\\prime} = \\frac{\\alpha}{\\delta(t)^\\sigma}\n%\\end{equation} where $\\sigma$ is a value that is adjusted by the user.\n%If $\\delta$ is less than $1$ the $\\alpha$-value is not modified.\n%In this way it is possible to have a finer precision on the shape formed by the point set where the density is higher than the average density. Hence it is possible to distinguish two separated objects with different density.\n% We want to determine the boundaries in phase space\n% Jorg\n% It is useful to understand whether the approximation is correct\n\\section{Determination of $\\alpha$ using \\'{e}tendue conservation} \\label{sec:Tir_alpha}\nAs mentioned in Section \\ref{sec:PSconcept}, in two-dimensions \\'{e}tendue can be seen as an area in PS. \nTherefore, given an optical system with a source $\\point{S} = [-\\variabile{a}, \\variabile{a}]$, the \\'{e}tendue at the source coincides with the area of source PS, and it is given by:\n\\begin{equation}\\label{eq:etenduesource1}\nU = 4\\n_1 \\variabile{a} \\sin(\\myangle_1^{\\textrm{max}})\\,,\n\\end{equation}\n where $\\variabile{a}$ is the half length of the source, $\\n_1$ the index of refraction of the medium in which the \\point{S} is located and $\\myangle_1^{\\textrm{max}}$ is the maximum value of the angle that the rays make with the normal $\\boldsymbol{\\nu}_1$ of the source.\\\\ \\indent \nFor some optical systems, all the rays emitted by the source arrive at the target, for some others there are also rays that can end at other detectors which are located outside the system. \nIndicating with \\set{R}{$1$}{}$(\\Pi)$ the regions in source PS formed by the rays that reach the target following path $\\Pi$ and with \\set{R}{}{}$(\\Pi)$ the corresponding regions at the target, the \\'{e}tendue $U_1$ at the source restricted to rays that arrive at the target is given by:\n\\begin{subequations}\n\\begin{align}\n\\label{eq:etendueintegralsource}\nU\\big(\\mbox{\\set{R}{$1$}{}}(\\Pi)\\big) = {\\int\\!\\!\\int}_{\\textup{R}_1(\\Pi)} \\textrm{d}\\variabile{q}\\,\\textrm{d}\\variabile{p}.\\\\\\label{eq:etenduesumsource}\nU_1 = \\sum_\\Pi{U\\big(\\mbox{\\set{R}{$1$}{}}(\\Pi)\\big)}, \n\\end{align}\n\\end{subequations}\nwhere $U\\big(\\mbox{\\set{R}{$1$}{}}(\\Pi)\\big)$ is the contribution to the \\'{e}tendue given by the rays inside \n\\set{R}{$1$}{}$(\\Pi)$ in source PS and the sum is over all possible paths $\\Pi$ from the source to the target.\nSimilarly, the \\'{e}tendue at the target of the rays emitted by the source is:\n\\begin{subequations}\n\\begin{align}\n\\label{eq:etendueintegraltarget}\nU\\big(\\mbox{\\set{R}{}{}}(\\Pi)\\big) = {\\int\\!\\!\\int}_{\\textup{R}(\\Pi)} \\textrm{d}\\variabile{q}\\,\\textrm{d}\\variabile{p}.\\\\ \\label{eq:etenduesumtarget}\nU_\\textrm{t}= \\sum_\\Pi{U\\big(\\mbox{\\set{R}{}{}}(\\Pi)\\big)}, \n\\end{align}\n\\end{subequations}\nNote that, since both the source and the target are located in air ($\\n_1=1$), in (\\ref{eq:etendueintegralsource}) and (\\ref{eq:etendueintegraltarget}), and from now on, we omit writing the index of refraction $\\n_1$.\n\\\\ \\indent In order to determine the value of $\\alpha$ in the $\\alpha$-shape procedure that approximates the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ accurately, we use \\'{e}tendue conservation ($U_{\\textrm{t}}= U_1$). The $\\alpha$-shapes method is applied to every region \\set{R}{}{}$(\\Pi)$ for a range of values of $\\alpha$;\n for each value an approximation of the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ is obtained and\n the intersection points $\\variabile{q}^{\\textrm{\\,max}}(\\Pi,\\variabile{p})$ and $\\variabile{q}^{\\textrm{\\,min}}(\\Pi,\\variabile{p})$ between $\\partial$\\set{R}{}{}$(\\Pi)$\nand the horizontal lines $\\variabile{p}=\\textrm{const}$, with $\\variabile{p}\\in[-1,1]$, are computed for every path $\\Pi$.\nTherefore, Equation (\\ref{eq:etendueintegraltarget}) becomes:\n\\begin{equation}\\label{eq:etenduetarg}\n U\\big(\\mbox{\\set{R}{}{}}(\\Pi)\\big)= \\int_{-1}^{1}{\\Big(\\variabile{q}^{\\textrm{\\,max}}(\\Pi,\\variabile{p})-\\variabile{q}^{\\textrm{\\,min}}(\\Pi,\\variabile{p})\\Big)} \\,\\textrm{d}\\variabile{p}.\n\\end{equation} In case more than two intersection points between the line $\\variabile{p}=\\textrm{const}$ and $\\partial$\\set{R}{}{}$(\\Pi)$ occur, the previous equation needs to be generalized. Suppose that $\\variabile{r}$ intersection points $\\big(\\variabile{q}^{\\,\\variabile{i}}(\\Pi,\\variabile{p}), \\variabile{p}\\big)_{\\variabile{i} = 1, \\cdots, \\variabile{r}}$ are found. \nOrdering their $\\variabile{q}$-coordinates in ascending order, the target \\'{e}tendue is calculated by:\n\\begin{equation}\\label{eq:etenduetarg1}\n U\\big(\\mbox{\\set{R}{}{}}(\\Pi)\\big) = \\sum_{\\variabile{i} = 1}^{\\variabile{m}}\\int_{-1}^{1}\n{\\Big(\\variabile{q}^{\\,2\\variabile{i}}(\\Pi,\\variabile{p})}-{\\variabile{q}^{\\,2\\variabile{i}-1} (\\Pi, \\variabile{p}) \\Big)}\\,\\textrm{d}\\variabile{p}\\,,\n\\end{equation}\nwhere $\\variabile{m}$ is the integer part of $\\variabile{r}/2$. \nThe integrals in (\\ref{eq:etenduetarg}) and (\\ref{eq:etenduetarg1}) are calculated discretizing the interval $[-1, 1]$\n into $\\nbin=100$ sub-intervals of equal length, the so-called bins, and using the trapezoidal rule.\n\\\\ \\indent Matching the \\'{e}tendue at the source $U_1$ with the \\'{e}tendue at the target $U_{\\textrm{t}}$, a unique value $\\alpha_{\\textrm{c}}$ of $\\alpha$ is determined. Implementing the $\\alpha$-shapes procedure with $\\alpha = \\alpha_\\textrm{c}$, an approximation of the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ is found and the intensity at the target can be calculated.\\\\ \\indent If two intersection points between $\\variabile{p}=\\textrm{const}$ and $\\partial$\\set{R}{}{}$(\\Pi)$ are found the target intensity is calculated using ($\\ref{eta2}$). If more than two-intersection points are found we use the generalized equation:\n\\begin{equation}\nI_{\\textrm{PS}}(\\variabile{p}) = \\sum_{\\Pi, \\variabile{i} }\\int_{\\variabile{q}^{\\,2\\variabile{i}-1}(\\Pi, \\variabile{p})}^{\\variabile{q}^{\\,2\\variabile{i}}( \\Pi, \\variabile{p})}L(\\variabile{q}, \\variabile{p})\\textrm{d}\\variabile{q} =\n \\sum_{\\Pi, \\variabile{i}}\\big(\\variabile{q}^{\\,2\\variabile{i}}(\\Pi, \\variabile{p})-\n\\variabile{q}^{\\,2\\variabile{i}-1}( \\Pi, \\variabile{p})\\big)\\,,\n\\label{eq:Ips}\n\\end{equation}\nwhere $\\variabile{q}^{\\,2\\variabile{i}}( \\Pi, \\variabile{p})>\\variabile{q}^{\\,2\\variabile{i}-1}( \\Pi, \\variabile{p})$, the summation over $\\Pi$ is for all the paths $\\Pi$ for which the intersection $\\variabile{p} = \\textrm{const}$ and \\set{R}{}{}$(\\Pi)$ is not empty, and the summation over $\\variabile{i}$ is for $\\variabile{i} = 1,2, \\cdots, \\variabile{m}$. The second equation holds as we assume $L(\\variabile{q}, \\variabile{p})=1$.\\\\ \\indent\nTo clarify our idea we apply the method to two different optical systems, the results are presented next.\n%\\\\ \\indent Let us consider the two-faceted cup introduced in Chapter \\ref{chap:raytracing} and depicted in Figure \\ref{fig:cup}. The half length of the source is $\\variabile{a}=2$, the maximum angle is $\\myangle_1^{\\textrm{max}} = \\pi/2$ and \\point{S} is located in air, i.e., $\\n_1=1$. Therefore, the total \\'{e}tendue is $U=8$. For the two-faceted cup all the rays emitted by \\point{S} arrive to \\point{T}, hence from \\'{e}tendue conservation we obtain that $U_{\\textrm{t}} =U_1 = 8$.\n%For some optical systems, not all the rays emitted by the source arrive to the target. \n% Explain the idea: use etendue conservation\n% Do the example for the two faceted cup\n% Explain the TIR collimator\n\\section{Results for a TIR-collimator}\\label{sec:results-Tir-alpha}\nWe apply the $\\alpha$-shapes method to the set of points in target PS obtained by using PS ray tracing.\nIn this chapter, the procedure is applied to two different kinds of total internal reflection (TIR)-collimators.\\\\ \\indent \nLet us first describe the TIR-collimator depicted in Figure \\ref{fig:tir}. It is an optical system symmetric with respect to the $z$-axis, it consists of a lens (central curve), two broken lines adjacent to the lens,\ntwo curved lines on each side and a top formed by a horizontal segment. The lens (line $2$) and the broken lines, formed by a collection of three segments (lines $3, 4, \\mbox{ and } 5$ and $9, 10 \\mbox{ and } 11$), are refractive line segments while the curved lines (labeled with $6$ and $8$) are designed in such a way that light is totally internal reflected (which explains the name TIR).\nThe light source $\\point{S}$ (line $1$) and the target $\\point{T}$ (line $12$) are two straight line segments normal to the optical axis.\nThe source $\\point{S}= [-2,2]$ is located at a height $\\variabile{z}_{1} = 0.3$ from the $\\variabile{x}$-axis.\n The target $\\point{T}= [-9.7, 9.7]$ is parallel to the source and is located at a height $ \\variabile{z}= 8.2$. Both \\point{S} and \\point{T} are located in air ($\\n_1=1$).\nThe volume inside the collimator is filled with a material with index of refraction $\\n_2=1.5$ (e.g. glass).\nThe collimator is surrounded by two vertical lines (lines $13$ and $15$) and two horizontal lines ($12$ and $14$) that receive the light emitted from the source; among these the one at the top (line $12$) is assumed to be the target, and it is located at a small distance from the top (line $7$). \n\\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7\\textwidth]{TIR}\n \\end{center}\n \\caption{\\textbf{Shape of the TIR-collimator.} Each line of the system is labeled with a number.\n The shape of the collimator is shown in blue.\n Three detectors depicted with green lines (surfaces $13$, $14$, and $15$) are located at the left, the right and the bottom of the optical system. The source (line $1$) and the target (line $12$) are depicted in red.}\n%The sagitta of the lens is approximately $1.17$.}\n \\label{fig:tir}\n\\end{figure}\n\\\\ \\indent\nUsing PS ray tracing explained in Section \\ref{sec:PS_raytracing} with parameters $\\varepsilon_{\\variabile{q}_1}^{\\textrm{max}} = 0.05/4$, $ \\varepsilon_{\\variabile{p}_1}^\\textrm{max} = 0.05/8, $ $\\varepsilon_{\\variabile{q}_1}^\\textrm{min} = 0.8/4$ and $\\varepsilon_{\\variabile{p}_1}^\\textrm{min} = 0.8/8$, around $1.67 \\cdot 10^4$ rays are traced (see Table \\ref{tab:table}). The rays distribution at the source PS is shown in Figure \\ref{fig:sourcePS}, where we depicted the rays that follow the same path with the same color. Seven different paths are found. The yellow rays follow path $\\Pi_1 = (1, 2, 7, 12)$; the green rays follow path $\\Pi_2 ~= ~(1, 4, 6, 7, 12)$; the red rays follow path $\\Pi_3 = (1, 10, 8, 7, 12)$; the magenta rays follow path $\\Pi_4= (1, 3, 7, 12)$ and the blue rays follow path $\\Pi_5= (1, 11, 7, 12)$. The rays located inside the white areas correspond to rays that do not reach the target, they follow either path $\\Pi_6 = (1, 4, 7, 6, 15)$ or path $\\Pi_7 = (1,10,7,8,13)$ and they do not give any contribution to the target intensity.\n\\begin{figure}[t]\n\\label{fig:sourcePS}\n \\begin{center}\n \\includegraphics[width=0.8\\textwidth]{source1}\n \\end{center}\n \\caption{\\textbf{Distribution of the rays on \\point{S}{}{}}. Around $1.67 \\cdot 10^4$ rays are traced using the triangulation refinement with parameters:\n $\\varepsilon_\\variabile{q}^\\textrm{max} = 0.8/4 ,$ $ \\varepsilon_{\\variabile{p}}^\\textrm{max} = 0.08/8, $ $\\varepsilon_{\\variabile{q}}^\\textrm{min} = 0.05/4, \\varepsilon_\\variabile{p}^\\textrm{min} = 0.05/8$. Rays that belong to the same region are depicted with the same color. The rays located inside the white areas do not reach the target. The boundaries of the two white regions are approximated by triangles depicted with black lines.}\n \\label{fig:sourcePS}\n\\end{figure}\nNote that, given two adjacent paths, the corresponding regions in \\set{S}{}{} have usually a common boundary. \nSince for this system not all the rays emitted by the source arrive at the target, $U_{\\textrm{t}}$ needs to be compared to the \\'{e}tendue $U_1$ at the source given by only those rays that reach the target (the rays that follow paths $\\Pi_6$ and \n$\\Pi_7$ are discarded). To this purpose, $U_1$ is calculated by removing from the total area $U$ of \\set{S}{}{} those areas occupied by the regions formed by the rays that hit the left or the right detector (white regions in Figure $\\ref{fig:sourcePS}$). For the TIR collimator in Figure \\ref{fig:tir}, $U$ is obtained from (\\ref{eq:etenduesource1}), the source \\'{e}tendue $U_1$ corresponding to the area covered by the rays that arrive at the target can be approximated by:\n \\begin{equation}\\label{eq:Usource}\n U_{1} = U-2A_{T},\n \\end{equation}\n where $U=8$ and $A_{T}$ is the approximated area of the white triangles in Figure \\ref{fig:sourcePS} surrounded by the black lines.\\\\ \\indent Next, $U_{\\textrm{t}}$ is calculated several times from (\\ref{eq:etenduetarg1}) where every time the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ are obtained by using $\\alpha$-shapes for a different value of $\\alpha$. \n%An accurate approximation of $\\partial$\\set{R}{}{}$(\\Pi)$ gives a value of $U_{\\textrm{t}}$ close to the exact \\'{e}tendue. \n%Matching $U_1$ with all the approximations of $U_{\\textrm{t}}$ we find the best value $\\alpha_c$ of $\\alpha$ that approximates $\\partial$\\set{R}{}{}$(\\Pi)$ and, therefore, $U_{\\textrm{t}}$. \n\\\\ \\indent To clarify this concept, in Figure \\ref{fig:etendueTS} we provide an example where the source \\'{e}tendue $U_1$ and target \\'{e}tendue $U_{\\textrm{t}}$ are computed from a set of around $1.67\\cdot 10^4$ rays. The approximated source \\'{e}tendue $U_1\\approx 7.77$ is depicted with the red line. The blue line shows how the \\'{e}tendue at the target changes as a function of $\\alpha$. The smallest difference $\\Delta U = |U_1-U_{\\textrm{t}}|$ is obtained using $\\alpha = \\alpha_\\textrm{c} = 0.08$.\n \\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7 \\textwidth]{etendue_alpha_shapes}\n \\end{center}\n \\caption{\\textbf{\\'{E}tendue for the TIR-collimator.} $U_\\textrm{t}$ is computed for a range of values of $\\alpha$. $U_1 \\approx 7.77$. The green dot indicates the value of $\\alpha_\\textrm{c} = 0.08$ which gives a good approximation of the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ at the target.\n Around $1.67 \\cdot 10^4$ rays have been traced using PS ray tracing.\n }\n \\label{fig:etendueTS}\n\\end{figure}\nIn Figure \\ref{fig:targetPS} we show the boundaries \n$\\partial$\\set{R}{}{}$(\\Pi)$ in target PS obtained tracing $1.67\\cdot10^4$ rays and using $\\alpha_\\textrm{c}=0.08$.\n \\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7\\textwidth]{target_alpha_shapes}\n \\end{center}\n \\caption{\\textbf{Target PS representation.} A set of $1.67 \\cdot 10^4$ rays are traced.\n Rays that follow the same path are depicted with the same color. The choice of the colors is consistent with Figure $\\ref{fig:sourcePS}$. The boundaries $\\partial$\\set{R}{}{}$(\\Pi)$ are computed through the $\\alpha$-shapes method with $\\alpha = \\alpha_\\textrm{c} = 0.08$.}\n \\label{fig:targetPS}\n\\end{figure}\nThe target intensity $I_{\\textrm{PS}}(\\variabile{p})$ for $\\variabile{p}\\in[-1,1]$ is obtained from Equation (\\ref{eta2}). \\\\ \\indent\nTo validate our method we compare the PS intensity with the QMC intensity. \nTo this purpose a partitioning $P_2:-1=\\variabile{p}_{0}<\\variabile{p}_1<\\cdots<\\variabile{p}_{\\nbin}=1$ of the interval $[-1,1]$ into $\\nbin=100$ bins is considered. \nThe averaged and normalized PS intensity $\\hat{I}_{\\textrm{PS}}$ is calculated for every \n$\\big(\\variabile{p}^{\\variabile{h}+1/2} = \\frac{1}{2}(\\variabile{p}^{\\variabile{h}+1}+ \\variabile{p}^{\\variabile{h}})\\big)_{\\variabile{h}=0, \\cdots, \\nbin-1}$ dividing the PS averaged intensity by the total \\'{e}tendue:\n\\begin{equation}\\label{eq:normalized_PS_intensity}\n\\hat{I}_{\\textrm{PS}}(\\variabile{p}^{\\variabile{h}+1/2}) = \\frac{1}{U_{\\textrm{t}}}\\int_{\\variabile{p}_{\\variabile{h}}}^{\\variabile{p}_{\\variabile{h}+1}} I_{\\textrm{PS}}(\\variabile{p})\\textrm{d}\\variabile{p}.\n\\end{equation}\nThe averaged and normalized QMC intensity $\\big(\\hat{I}_{\\textrm{QMC}}(\\variabile{p}^{\\variabile{h}+1/2})\\big)_{\\variabile{h} = 0, \\cdots, \\nbin-1}$ is given by\n\\begin{equation}\\label{eq:normalized_MC_intensity}\n\\hat{I}_{\\textrm{QMC}}(\\variabile{p}^{\\variabile{h}+1/2}) = \\frac{\\nrays[\\variabile{p}^{\\variabile{h}},\\variabile{p}^{\\variabile{h}+1})}{\\nrays[-1,1)} \n\\qquad \\mbox{ for } \\variabile{p}\\in[\\variabile{p}^{\\variabile{h}}, \\variabile{p}^{\\variabile{h}+1}).\n\\end{equation} \nBoth approximate intensities $\\hat{I}_{\\textrm{A}} (\\textrm{A} = \\textrm{PS}, \\textrm{QMC})$ are compared to an intensity $\\hat{I}_{\\textrm{ref}}$ taken as a reference. \n\nFor some optical systems, there is an explicit solution for the target intensity, but this is not the case of the TIR-collimator.\nTherefore, a QMC simulation with $10^7$ rays is used to obtain the averaged normalized intensity $\\hat{I}_{\\textrm{ref}}$.\nThe intensity profile $\\hat{I}_{\\textrm{PS}}\n$ obtained using PS ray tracing with $8.3\\cdot 10^4$ rays and $\\alpha= \\alpha_\\textrm{c} = 0.06$ is depicted in Figure \\ref{fig:intensityMCPS} with a red line.\n$\\hat{I}_{\\textrm{PS}}$ is hardly distinguishable from $\\hat{I}_{\\textrm{ref}}$ which is indicated with the dashed and blue line in Figure $\\ref{fig:intensityMCPS}$.\\\\ \\indent\n \\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.7\\textwidth]{intensity_alpha_shapes}\n\\caption{\\textbf{Target intensity profile.}\nThe exact intensity is computed using the QMC method for a set of $10^7$ rays. For the PS intensity a set of $8.3\\cdot 10^4$\nrays is considered and $\\alpha_\\textrm{c} = 0.06$ is chosen to compute the boundaries $\\partial$\\set{R}{}{}$(\\Pi)$.}\n \\label{fig:intensityMCPS}\n\\end{figure}\nFinally, we calculate the error between $\\hat{I}_{\\textrm{A}}$ and $\\hat{I}_{\\textrm{ref}}$, defined as:\n\\begin{equation}\\label{eq:error}\n\\mbox{error} = \\frac{\\sum_{\\variabile{h}= 1}^{\\nbin}| \\hat{I}_{\\textrm{A}}(\\variabile{p}^{\\variabile{h}+1/2}) - \\hat{I}_{\\textrm{ref}}(\\variabile{p}^{\\variabile{h}+1/2})|}{\\nbin}.\n\\end{equation}\nThe QMC and PS intensities are calculated several times increasing the number of rays to improve the accuracy.\nTable \\ref{tab:table} and \\ref{tab:table2} describe how the number of rays traced affects the error. \nIn Table \\ref{tab:table} the correlation between $\\alpha_\\textrm{c}$ and the number of rays is evident.\n%determined by the values of $\\varepsilon^{\\textrm{min}}_\\variabile{p}$, $\\varepsilon^{\\textrm{min}}_{\\variabile{q}}$, \n%$\\varepsilon^{\\textrm{max}}_{\\variabile{p}}$ and $\\varepsilon^{\\textrm{max}}_{\\variabile{q}}$. \nNote that increasing the number of rays the value of $\\alpha_\\textrm{c}$ and the corresponding error decrease. \n\\begin{table}[htbp] \\label{tab:table}\n\\centering\n\\caption{\\bf Errors of the PS intensity}\n\\begin{tabular}{lllllll}\n \\hline Number \\\\ of rays\\; & $\\varepsilon^{\\textrm{max}}_{\\variabile{q}} $ & $\\varepsilon^{\\textrm{min}}_{\\variabile{q}} $ \\; & $\\varepsilon^{\\textrm{max}}_{\\variabile{p}}$\\;\n & $\\varepsilon^{\\textrm{min}}_\\variabile{p}$\\; & $\\alpha_\\textrm{c}$ & PS error \\\\\n \\hline \n $3\\,339$ & $0.8$ & $0.05$ & $0.8/2$ & $0.05/2$ & $0.14$ & $1.47\\cdot10^{-3}$ \\\\\n$7\\,567$ & $0.8/2$ & $0.05/2$ & $0.8/4$ & $0.05/4$ & $0.10$ & $3.01\\cdot 10^{-4}$ \\\\\n$16\\,755$ & $0.8/4$ & $0.05/4$ & $0.8/8$ & $0.05/8$ & $0.08$ & $8.60\\cdot 10^{-5}$ \\\\\n $83\\,005$ & $0.8/16$ & $0.05/16$ & $0.8/32$ & $0.05/32$ & $0.06$ & $1.31\\cdot 10^{-5}$ \\\\\n \\hline\n \\end{tabular}\n \\label{tab:table}\n \\end{table}\n\\\\ \\indent In Table \\ref{tab:table2} the numerical results of QMC ray tracing are reported.\nIncreasing the number of rays traced, the error gradually decreases.\n\\begin{table}[htbp]\n\\centering\n\\caption{\\bf Error of the QMC intensity}\n\\begin{tabular}{ll} \\hline Number of rays\\; & QMC error\\\\\n \\hline $10^3$ & $1.65\\cdot10^{-3}$ \\\\\n$10^4$ & $3.96\\cdot 10^{-4}$ \\\\\n $10^5$ & $6.36\\cdot 10^{-4}$ \\\\ \n$10^6$ & $1.02\\cdot 10^{-5}$ \\\\\n \\hline\n \\end{tabular}\n \\label{tab:table2}\n \\end{table}\n\\noindent In Figure $\\ref{fig:error}$, the results listed in Table $\\ref{tab:table}$ and Table $\\ref{tab:table2}$ are shown. The red line depicts the convergence of the PS error and the blue line indicates the QMC error.\n\\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7\\textwidth]{error_alpha_shapes_vs_qmc_curved}\n \\end{center}\n \\caption{\\textbf{PS and QMC errors as a function of the number of rays}\n The horizontal dotted line shows that an error equal to $1.31\\cdot 10^{-5}$ can be obtained tracing almost $10$ times fewer rays in phase space.}\n \\label{fig:error}\n\\end{figure}\n%Note from Figure \\ref{fig:error} that the error for the QMC method decreases as $\\frac{1}{\\sqrt{\\nrays}}$, while for the PS simulation the speed of convergence is much higher.\\\\ \\indent\nWe need to emphasize that the convergence of the error of PS ray tracing for increasing $\\nrays$ may change according to the design of the optical system.\nThis is because the approximation of the boundaries in PS depends on the accuracy of the $\\alpha$-shapes method.\nThe $\\alpha$-shapes procedure is unable to properly detect the boundaries of regions with a sharp turn if not enough points are given\n\\cite{teichmann1998surface}. Indeed, on the one hand a low density requires a large value of $\\alpha$ to accept the triangles in a region, on the other hand,\n choosing $\\alpha$ too large, the shape of the region could be destroyed. Increasing $\\alpha$ more triangles are kept and triangles inside the regions \\set{R}{}{}$(\\Pi)$ could be taken into account creating holes in \\set{R}{}{}$(\\Pi)$.\n\\begin{figure}[t]\n\\centering\n\\begin{subfigure}{.48\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{boundaries_alpha1}\n \\caption{Boundaries approximation obtained using the $\\alpha$-shapes method with $\\alpha_\\textrm{c} = 0.3$ (black lines).}\n\\end{subfigure}\n\\hfill\n\\begin{subfigure}{.48\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{boundaries_alpha2}\n \\caption{Boundaries approximation obtained using the $\\alpha$-shapes method with $\\alpha_\\textrm{c} = 0.31$ (black lines).}\n\\end{subfigure}\n\\caption{\\textbf{Approximated boundaries at the target PS.} Tracing $3339$ rays and using $\\alpha$-shapes, the boundaries cannot be approximated well. \nA small change of the parameter $\\alpha$ leads to a completely different approximation of the boundaries.}\n\\label{fig:Tir1}\n\\end{figure}\nFigure $\\ref{fig:Tir1}$ clarifies this concept showing that the region formed by rays that hit the lens is hard to approximate when there is a small number of rays inside the region. Consequently either a region bigger than the area covered by the rays is obtained or some triangles which are not part of the boundaries are included in the triangulation. This results in an inaccurate intensity (either too high or to low). To obtain a good approximation of the boundaries of these kinds of patches more rays have to be traced. The PS error decreases very fast increasing the number of rays (see Table\n \\ref{tab:table} and Figure \\ref{fig:error}).\n \\\\\\indent To show how the error plot changes according to the regularity of the shape of the regions $\\partial$\\set{R}{}{}$(\\Pi)$, we consider another example of a TIR-collimator.\n Figure $\\ref{fig:Tir1}$ shows that the hardest region to approximate is given by those rays that follow path $\\Pi_1 ~=~ (1,2,7,12)$.\n We therefore consider a TIR-collimator with a flatter lens and with the target located at a smaller distance to the top (see Figure $\\ref{fig:analyticlens}$). \nThe source $\\point{S}= [-2,2]$ (surface number $1$) is located in air at a height $\\variabile{z}_1 = 0.3$ from the $x$-axis.\n The target $\\point{T}= [-9.7, 9.7]$ (surface $12$) is parallel to the source and is located in air at a height $ \\variabile{z}= 7.85$.\n The shape of the collimator is shown as a blue line.\n Three detectors depicted with green lines (surfaces $13$, $14$, and $15$) are located at the left, the right and the bottom of the optical system.\n \\\\ \\indent Tracing around $3\\cdot10^3$ rays using PS ray tracing, we obtain the target rays distribution shown in Figure $\\ref{fig:Tir2}$. \nCompared with the distribution in Figure \\ref{fig:targetPS}, we note that the extremities at top and bottom of the region formed by the rays that hit the lens are less pronounced.\nMoreover a target located very close to the top makes the shape of that region less stretched along the $\\variabile{q}$-axis.\nTherefore, it is expected that the $\\alpha$-shapes method performs better in this case.\n\\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7\\textwidth]{tir_analytic2}\n \\end{center}\n \\caption{\\textbf{Shape of the TIR-collimator.} Each line of the system is labeled with a number.\n $n_1 = 1$ is the refraction index of the medium (air) where the source and the target are located, and\n $n_2 = 1.5 $ the refraction index of the medium (glass) inside the optical system.} \n%The sagitta of the lens is equal to $0.6$.}\n \\label{fig:analyticlens}\n\\end{figure}\n \\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=0.7\\textwidth]{boundaries_alpha3}\n \\end{center}\n \\caption{\\textbf{Target phase space for the TIR-collimator depicted in\n Figure \\ref{fig:analyticlens}.} The black line depicts the best approximation of $\\partial$\\set{R}{}{}$(\\Pi)$ for $3\\cdot 10^3$ rays. \nThe $\\alpha$-shapes method gives an accurate approximation of the boundaries for $\\alpha_\\textrm{c} = 0.9$.}\n \\label{fig:Tir2}\n\\end{figure}\n\\\\ \\indent PS and QMC ray tracing are implemented for the TIR-collimator in Figure \\ref{fig:analyticlens}. The approximated intensities $\\hat{I}_{\\textrm{A}}$ $(\\textrm{A} = \\textrm{PS}, \\textrm{QMC})$ are compared to the reference intensity $\\hat{I}_{\\textrm{ref}}$ (QMC ray tracing with $10^7$ rays).\nPS error is depicted with the red line and, QMC error is depicted with the blue line. \\\\ \\indent\n% Lets indicate with $\\big(\\hat{I}_{\\textrm{A}, n}\\big)_{n \\in \\mathbb{N}}$ the sequence of the approximated intensity. Every term of the sequence is calculated by increasing the number of rays, for example $\\hat{I}_{\\textrm{A}, 1}$ is computed tracing $\\nrays = 10$ rays, $\\hat{I}_{\\textrm{A}, 2}$ is computed tracing $\\nrays = 100$ rays, and so on. The speed or rate of convergence of the approximate intensity $\\big(\\hat{I}_{\\textrm{A}, n}\\big)$ describes how quickly the terms of sequence $\\big(\\hat{I}_{\\textrm{A}, n}\\big)$ converge to the reference intensity $\\hat{I}_{\\textrm{ref}}$ by increasing the number of rays. Suppose $m$ is a real number, the sequence $\\big(\\hat{I}_{\\textrm{A}, n}\\big)$ converges to $\\hat{I}_{\\textrm{ref}}$ if\n%\\begin{equation}\n%\\lim_{\\nrays\\rightarrow \\infty} \\frac{|\\hat{I}_{\\textrm{A}, n+1}-\\hat{I}_{\\textrm{ref}}|}{|\\hat{I}_{\\textrm{A}, n}-\\hat{I}_{\\textrm{ref}}|^m} = \\mu\n%\\end{equation}\n%and $\\mu$ is called the rate of convergence and $m$ is the order of convergence. For $m\\in[0,1]$ we have the linear convergence, for $m=2$ the quadratic convergence, etc. \n%\\\\ \\indent\nNumerical results show that, for the TIR-collimator in Figure \\ref{fig:analyticlens}, the number of rays needed to achieve an error of the order of $10^{-5}$ is reduced using PS ray tracing compared with QMC ray tracing. Furthermore, for a relatively small number of rays, the $\\alpha$-shapes method performs better for a TIR-collimator with a flatter lens. \n\\begin{figure}[t]\n \\begin{center}\n \\includegraphics[width=\\textwidth]{error_alpha_shapes_vs_qmc_flat}\n \\end{center}\n \\caption{\\textbf{PS and QMC errors.}}\n%The red line depicts the error using the $\\alpha$-shapes method to compute the boundaries. \n%The blue line shows the error between the Monte Carlo intensity and the exact intensity.\n% The dashed black line represents a straight line with slope $-\\frac{1}{2}$.\n% The dashed blue line represents a straight line with slope $-1$.}\n \\label{fig:error2}\n\\end{figure}\n\\section{Conclusion}\nThe aim of this chapter was to detect the boundaries of the regions formed by the rays traced using $\\alpha$-shapes.\\\\\n\\indent First, we reported some theory about $\\alpha$-shapes which are commonly used to approximate the shape formed by a point cloud. \nThese methods depend on a parameter $\\alpha$ that, in most cases, can be determined only by simulations. \n\\\\ \\indent Using \\'{e}tendue conservation, we developed a new approach to detect the value of $\\alpha$ that better approximates the boundaries in target PS. \nWe applied $\\alpha$-shapes to two different kinds of TIR-collimators. The target PS intensity was computed for both systems several times increasing every time the number of rays traced. Finally, the corresponding errors between the approximated intensities and a reference intensity was calculated. We observed that PS ray tracing allows tracing far less rays compared to QMC ray tracing. Numerical results show that using PS ray tracing the desired accuracy can be achieved reducing significantly the number of rays traced.\\\\ \\indent \nHowever, we observed that the error convergence for PS ray tracing strongly depends on the design of the optical system (shapes of the region in target PS). Indeed, the accuracy of the intensity is related to the precision of the $\\alpha$-shape, that is, to the choice of the parameter value of $\\alpha$. For more complicated shapes in PS, more rays need to be traced for a good boundaries reconstruction.\\\\ \\indent\nIn order to remove the dependence of PS ray tracing on the parameter $\\alpha$, we will construct another procedure to detect the boundaries of the regions in target PS. \nThe new technique is based on the triangulation refinement explained in Section \\ref{sec:PS_raytracing}. The details are explained in the next chapter and numerical results are reported for several optical systems. \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "f72c87629b2bec707b076697a8a532e26cee296f", "size": 49789, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/Boundaries.tex", "max_stars_repo_name": "melaniafilosa/ps_raytracing", "max_stars_repo_head_hexsha": "8f9111ea4ec3ac125b593f41b3ac6fe302ea6632", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/Boundaries.tex", "max_issues_repo_name": "melaniafilosa/ps_raytracing", "max_issues_repo_head_hexsha": "8f9111ea4ec3ac125b593f41b3ac6fe302ea6632", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/Boundaries.tex", "max_forks_repo_name": "melaniafilosa/ps_raytracing", "max_forks_repo_head_hexsha": "8f9111ea4ec3ac125b593f41b3ac6fe302ea6632", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 102.4465020576, "max_line_length": 1067, "alphanum_fraction": 0.7264255157, "num_tokens": 15156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.839733963661418, "lm_q2_score": 0.7690802317779601, "lm_q1q2_score": 0.6458227914045485}} {"text": "\\documentclass[12pt]{article}\n\n\\title{Math Foundations}\n\\author{Abel Doñate}\n\\date{}\n\n\\usepackage{amsmath}\n\\usepackage{hyperref}\n\\usepackage{graphicx}\n\\graphicspath{ {./images/} }\n\\usepackage{wrapfig}\n\n\\setcounter{page}{0}\n\n%Geometry\n\\usepackage{geometry}\n\\geometry{a4paper, margin=1in}\n\n\n\\begin{document}\n\n\\maketitle\n\\tableofcontents\n\\newpage\n\n\\section{Useful techniques to solve problems}\n \\subsection{The Well ordered principle and Descent to infinity}\n The well ordering principle tells us that we cannot have a sequence of non-negative integers such that $a_1>a_2>...$.\n This is useful to apply in descent to infinite proofs. \\\\ \\\\\n Suppose you have a problem in which you are given a condition and you have to check what numbers satisfy the condition. In this case you can apply descent to infinite:\n \\begin{itemize}\n \\item[1)] Find a condition that guarantees you that a set of numbers satisfy the condition if and only if another set of numbers that are strictly less then the original set satisfy the condition.\n \\item[2)] By the Well ordering principle, reason that it is impossible to keep descending infinitely many times , so the sequence must stop somewhere.\n \\item[3)] If we can apply it many times, that means that there is no set of numbers that satisfy the condition (or only does the trivial ones like $(0,0,0)$).\n \\end{itemize}\n A good example is to find the triples $(a, b, c)$ that satisfy $a^3 + 2b^3 = 4c^3$.\\\\\n Because the right hand side (RHS) is even, then $a$ must be even $\\implies a=2x$. \\\\\n Dividing by $2$ the equation is $4x^3 + b^3 = 2c^3$ \\\\\n Because RHS is even, then $b$ must be even $\\implies b=2y$. \\\\\n Dividing by $2$ the equation is $2x^3 + 4y^3 = c^3$ \\\\\n Because LHS is even, then $c$ must be even $\\implies c=2z$. \\\\\n Dividing by $2$ the equation is $x^3 + 2y^3 = 4z^3$ \\\\\n \n We observe that we are left with the original case. That tells us that $(a, b, c)$ only holds if $(a/2, b/2, c/2)$ also holds. Applying descent to infinity, we assure that there is no solution except the trivial one $(0,0,0)$\n \n \\subsection{Mathematical induction}\n We can prove lots of types of problem by this simple principle. \\\\\n \n Suppose we have to prove that a given condition holds for every natural number. Then one way to prove it is to check if holds for $k=1$, and then prove that if it holds for $k$, then, it must hold for $k+1$. \\\\\n \n The sketch of the steps is the following:\n \\begin{itemize}\n \\item[1)] \\textbf{Base case.} First we prove that the condition holds for $k=1$ (or the first element of the set you are given).\n \\item[2)] \\textbf{Induction statement.} Now we prove that if $k$ holds, then $k+1$ also holds.\n \\item[3)] \\textbf{Proof.} Finally you reason that the problem has been solved by the principle of mathematical induction \n \\end{itemize}\n \n \\subsection{Double counting}\n Sometimes it is useful to exploit different perspectives of a problem. If your task is to prove an equivalence, perhaps the easiest way to do it is double counting, or compute the expression in two different ways, but showing that both are equal.\n \n\\section{Inequalities}\n \\subsection{Means inequalities}\n We have the chain of inequalities $HM\\leq GM\\leq AM\\leq QM$:\n \\[\\frac{1}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}\\leq \\sqrt[3]{abc}\\leq \\frac{a+b+c}{3}\\leq \\sqrt{\\frac{a^2+b^2+c^2}{3}} \\]\n The equality holds when $a=b=c$.\n \n \\subsection{Cauchy's inequality}\n \\[\\vec{a}\\cdot \\vec{b}\\leq |\\vec{a}||\\vec{b}| \\implies \\left(\\sum{a_ib_i}\\right)^2 \\leq \\left(\\sum{a_i^2}\\right)\\left(\\sum{b_i^2}\\right) \\]\n \n \\subsection{Chebyshev's inequality}\n If we have two ordered vectors (both increasing or decreasing):\n \\[n\\sum{a_ib_i}\\geq \\left(\\sum{a_i}\\right)\\left(\\sum{b_i}\\right)\\geq n\\left(\\sum{a_ib_{n+1-i}}\\right)\\]\n \n \\subsection{Jensen's inequality}\n Let $f$ be a convex function (i.e. $f''\\geq 0$). Then:\n \\[f\\left(\\sum{\\alpha_ix_i}\\right) \\leq \\sum{\\alpha_i f(x_i)} \\ \\ \\ \\ \\ where \\ \\ \\sum{\\alpha_i}=1\\]\n\n\n\\section{Combinatorial numbers}\n The number of ways to choose k elements in a set of n elements is \\textit{n choose k}:\n \\[\\binom{x}{k}=\\frac{n!}{k!(n-k)!}\\]\n Some of the properties of combinatorial numbers are:\n \\begin{align} \n (x+y)^n &= \\sum{\\binom{n}{k}x^ny^{n-k}} \\\\\n \\binom{n+m}{k} &= \\sum{\\binom{n}{i}\\binom{m}{k-i}} \\ \\ \\ \\text{(Vandermonde's identity)}\n \\end{align}\n \n \\begin{wrapfigure}{r}{0.23\\textwidth}\n \\centering\n \\includegraphics[width=0.23\\textwidth]{pascal}\n \\end{wrapfigure}\n \n These numbers appear also in Pascal's triangle. The ith element of the nth row (begining both with 0) is $\\binom{n}{i}$.\\\\ \\\\\n An extension of \\textit{n choose k} is the following:\n \\[\\binom{n}{k_1 \\cdots k_m}=\\frac{n!}{k_1!\\cdots k_m!}\\]\n and it appears in the multinomial theorem:\n \\[(x_1+\\cdots +x_m)^n=\\sum_{k_1+ \\cdots +k_m=n}{\\binom{n}{k_1 \\cdots k_m}x_1^{k_1}\\cdots x_m^{k_m}}\\]\n\n\n\\section{Sets}\n \\subsection{Operations and De Morgan's Law}\n Main operations\n \\begin{itemize}\n \t\\item Union. $A \\cup B$\n \t\\item Intersection $A \\cap B$\n \t\\item Difference $A \\setminus B$ or $A-B$\n \t\\item Symmetric difference $A \\bigtriangleup B = (A\\cup B)-(A\\cap B)$\n \t\\item Complement. $\\bar{A}$\t\n \\end{itemize}\n \\textbf{De Morgan's Law:}\n \\begin{itemize}\n \t\\item $\\overline{A \\cup B}=\\bar{A}\\cap \\bar{B}$\n \t\\item $\\overline{A\\cap B}=\\bar{A}\\cup \\bar{B}$\n \\end{itemize}\n\n\t\\subsection{Canonical decomposition}\n\tEvery function $f$ can be decomposed in a surjective, bijective and injective functions. \t\tFor instance: $f=i \\circ b \\circ \\pi$.\n\t\\begin{itemize}\n\t\t\\item $\\pi$ maps to its equivalence class\n\t\t\\item $b$ maps to the function of every equivalence class\n\t\t\\item $i$ maps to itself\n \\end{itemize}\n \\section{Algebraic structures}\n\t\\subsection{Group}\n\tA group is a set with one binary operation $(S,\\star)$ that must satisfy:\n\t\\begin{itemize}\n\t\t\\item Closure under $\\star$\n\t\t\\item Associative $(a\\star b)\\star c=a\\star (b\\star c)$\n\t\t\\item Identity element $e$ such that $e\\star a=a\\star e=a$\n\t\t\\item Inverse element. Every $a$ has an inverse $a^{-1}$\n\t\\end{itemize}\n\t\n\tIf the group is commutative $a\\star b= b\\star a$, then is called \\textbf{Abelian}. \\\\\n\t\n\tExamples of abelian group are $\\mathbf{Z}_n$ (group whose elements are $0, 1, \\cdots, n-1$, and the operation is the sum modulo n), or the group or rotations of a regular polygon. \\\\\n\t\n\tAn example of non-abelian group is the group of rotations of a cube. The proof is left to the reader.\n\t\n\t\\subsection{Ring}\n\tA Ring is a set with two binary operations $(S,\\star, \\top)$ that must satisfy:\n\t\\begin{itemize}\n\t\t\\item $(S, \\star)$ is an abelian group\n\t\t\\item Distributive $a\\top (b\\star c)= (a\\top b) \\star (a\\top c)$\n\t\t\\item Identity elements $e_\\star, e_\\top$\n\t\\end{itemize}\n\t\n\tIf $e_\\star \\neq e_\\top$ and $a\\top b\\neq e_\\star$, then is called an \\textbf{Integral domain}.\\\\\n\t\n\tOne example of a Ring is $\\mathbf{Z}$ with usual sum and product.\\\\\n\t\n\t$\\mathbf{Z}_p$ with $p$ prime is an example of an integral domain. $\\mathbf{Z}_6$, because 6 is not prime, is not an integral domain, as $2\\times 3=0=e_+$. \n\t\n\t\\subsection{Field}\n\tA field is a ring with no inverse for $e_\\star$ under $\\top$.\\\\\n\t\n\tThe most common example is $\\mathbf{R}$. Observe that the element $0=e_+$ has no inverse under multiplication.\n\t\n\\section{Stereographic projection}\n \\begin{wrapfigure}{r}{0.30\\textwidth}\n \\centering\n \\includegraphics[width=0.30\\textwidth]{stereo1}\n \\end{wrapfigure}\n \n The stereographic projection maps every point in a sphere except the upper pole to the $XY$ plane. On the other hand we can take the inverse stereographic projection, that maps every point in the plane to the sphere (except the upper pole). \\\\\n \n This map is achieved by drawing a line from the upper pole to the point in the sphere and taking the intersection with the $XY$ plane. \\\\\n \n If we define the coordinates of the plane as $x, y$ as usual and the coordinates of the sphere as $\\alpha, \\beta, \\gamma$, we can derive the following equations:\n \\[F(\\alpha, \\beta, \\gamma)=\\left(\\frac{\\alpha}{1-\\gamma}, \\frac{\\beta}{1-\\gamma}\\right) \\ \\ \\ \\\n F^{-1}(x, y)=\\left(\\frac{2x}{x^2+y^2+1}, \\frac{2y}{x^2+y^2+1}, \\frac{x^2+y^2-1}{x^2+y^2+1}\\right)\\]\n\n\\end{document}\n", "meta": {"hexsha": "4166328c0fed053282404ad2301c5822b00ddae1", "size": 8442, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Cuatri 1A/Foundations/Foundations.tex", "max_stars_repo_name": "abeldm3108/Apuntes", "max_stars_repo_head_hexsha": "2bd5ef414fbb671f8e232d646e6ef671d773288c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Cuatri 1A/Foundations/Foundations.tex", "max_issues_repo_name": "abeldm3108/Apuntes", "max_issues_repo_head_hexsha": "2bd5ef414fbb671f8e232d646e6ef671d773288c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Cuatri 1A/Foundations/Foundations.tex", "max_forks_repo_name": "abeldm3108/Apuntes", "max_forks_repo_head_hexsha": "2bd5ef414fbb671f8e232d646e6ef671d773288c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.7976878613, "max_line_length": 250, "alphanum_fraction": 0.6715233357, "num_tokens": 2660, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743735019595, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.6456891703003107}} {"text": "\\section{Background}\n\n\\subsection{The original multi-head attention\rmechanism}\n\nThe scaled dot-product attention proposed by\r\\citet{vaswani2017attention} transforms the vectorial embedding $\\vec{Y}_i$ of a token as a\r function of a sequence of other embedding $\\vec{X}_j$. Where\r $\\vec{Y}_i$ and $\\vec{X}_j$ are all vectors of size $D$. A key\r$\\vec{K}_j$ and value $\\vec{V}_j$ are attributed to each vector\r$\\vec{X}_j$, and query $\\vec{Q}_i$ is attributed to $\\vec{Y}_i$.\r The vectors are obtained by linear projection from dimension $D$ to $d$ using three matrices of learnable parameters. The new vector\r $\\vec{y}_i'$ is a weighted sum of the $\\vec{V}_j$. The weights are\r scores of matching between the query $\\vec{Q}_i$ and the keys\r$\\vec{K}_j$, calculated as the dot product between the two vectors.\rThey are also \\emph{softmaxed} to sum up to 1. The transformation of $L_Q$\rvectors $\\vec{Y}_i$ as a function of $L_K$ vectors $\\vec{X}_j$ can\r be efficiently computed with matrix multiplications:\r\n\n\\begin{equation}\nA = \\mathrm{softmax}\\left(\\frac{Q \\times K^T}{\\sqrt{d}}\\right) \\times V\n\\end{equation}\n\nWith $Q$ a matrix of shape $(L_Q, d)$, $K$ a matrix of shape\r $(L_K, d)$ and $V$ a matrix of shape $(L_K, d)$. The $\\sqrt{d}$\rat the denominator is a scaling factor used to avoid saturation in the\rexponential terms of the softmax function.\r\n\nThe multi-head attention performs $h$ different projections into spaces\r of dimension $d = D/h$. The resulting vector $\\vec{Y}_i'$ is the\rconcatenation of the $h$ vectors $\\vec{y}_i'$ obtained. Thus the\rembedding dimension is preserved. Using multiple heads was found\rbeneficial by the authors over using a single head of dimension\r$d = D$.\r\n\nDuring training, the cross entropy of the $n^{th}$ predicted token is\r calculated assuming all previous tokens have been generated correctly.\r This enables to parallelize training completely without need for\rrecurrence. However as the $n^{th}$ token should not depend of the\rfollowing tokens, the cells in the upper right corner of the score\rmatrix are set to $-\\infty$ such that after the softmax they are equal\rto 0, and the rows still sums up to 1.\n\n\\subsection{Improving Transformer scalability with sequence\rlength}\n\nThe original attention mechanism requires the computation of a score\rmatrix $Q \\times K^T$ of shape $(L_Q, L_K)$, with complexity\r$O(L_QdL_K)$. If the query and key sequence lengths are multiplied by\rtwo, then the memory used and computation time are multiplied by 4. To\rimprove the scalability of the transformer with sequence length, several axis of research have been explored.\r\n\n\\citet{kitaev2020reformer} proposed\rthe Reformer's architecture, which uses an hash-bucketting algorithm to\rreduce the complexity of the original multi head attention operation\rfrom $O(L^2)$ to $O(L\\log(L))$.\r\n\n\\citet{dai2019transformerxl} proposed the\rTransformer-XL's architecture, which cuts the sequence in segments of\rlength L. The model predicts each stage of the current segment as a\rfunction of the previous and current segment. All the segments are\rcomputed sequentially with a recurrence mechanism. The complexity is linear\rwith sequence length, but the computation cannot be completely\r parallelized due to the recurrence mechanism, although more than a RNN,\ras segments can be computed in one go.\r\n\nOther publications explored using a sparse attention matrix, such as the\r Longformer by \\citet{beltagy2020longformer} and the Big Bird model by\r \\citet{zaheer2021big}. As each\rtoken attends to a fixed number of all other tokens, the scalability is\rimproved. These sparse attention models however require custom\roperations implemented in CUDA.\r\n\nSome other works propose to modify the attention mechanism to be\r compatible with linear complexity. The Linformer by\r\\citet{wang2020linformer} projects the\r key and values onto a smaller sequence length dimension with matrix\r multiplication. It cannot however generalize to sequences longer than\r during training, as the weights of the projection for such tokens would\r be undefined.\r\n\n\\citet{shen2020efficient} proposed to\r replace the softmax attention score.\r $A = \\mathrm{softmax}\\left(\\frac{Q \\times K^T}{\\sqrt{d}}\\right) \\times V$ is\rchanged into $A = \\rho(Q) \\times \\rho(K)^T \\times V$. With $\\rho$\rthe softmax function along the embedding dimension. Thanks to matrix\rmultiplication commutativity, the order of the operations can be chosen.\rIf $Q$, $K$ and $V$ are of shape $(L_Q, d)$, $(L_K, d)$ and\r$(L_K, d)$ respectively, the complexity of\r$(\\phi(Q) \\times \\phi(K)^T) \\times V$ is $O(L_Q \\times d \\times L_K)$ whereas the complexity of\r$\\phi(Q) \\times (\\phi(K)^T \\times V)$ is\r$O\\left(max(L_Q, L_k) \\times d^2 \\right)$. The right-side-first\roperation is linear in complexity with sequence length. The shape of the\rintermediate result matrix is also changed, allowing to scale the better\rin memory requirements as well. The original \\emph{softmaxed} attention score\rmatrix was giving rows of positive scores that sum to 1. With this\rchange the elements of the score matrix remain positive as $\\phi(Q)$\rand $\\phi(K)^T$ are matrices of positive values, but the rows of the\rscore matrix does not sum up to 1. This work also does not give a linear complexity formulation for masked attention. If the right-side-first\rscheme is adopted, the attention score matrix\r$\\phi(Q) \\times \\phi(K)^T$ is never explicitly computed, and can't be\rmasked.\r\n\nBuilding on this idea of commutative attention function proposed by\r\n\\citep{shen2020efficient}, \\citet{katharopoulos2020transformers} introduced their kernerlized attention function as:\r\n\n\\begin{equation}\nA = \\frac{\\phi(Q) \\times \\phi(K)^T}{\\sum_j \\left( \\phi(Q) \\times \\phi(K)^T \\right)} \\times V\n\\end{equation}\n\nThe function $\\phi$ is applied element-wise and can be any positive\rfunction, for example $\\phi(x) = elu(x) + 1$. This attention is\rrow-wise normalized so that all rows of the score matrix are sets of\rpositive weights adding up to one. This preserves the objective of the\roriginal softmaxed attention scores, while allowing to perform\roperations in an optimal order.\r\n\nThe Performer by \\citet{choromanski2021rethinking} exploits the same idea of a kernelized attention introduced\rby \\citet{katharopoulos2020transformers}, with an algorithm that betters approximate softmaxed attention. Most\rimportantly they also give in annex a prefix sum algorithm to perform\roperations in the right-side-first order while giving the same result as\rmasked left-side-first operation.\r\n\nAlthough the author did not specify how to implement it, the only\r implementations we found of this operation requires custom CUDA code. In\r this work we will give an implementation of the right-side-first masked\roperation, with usual functions from neural network frameworks, that\rremains linear in complexity.\n\n\\subsection{Alternatives to absolute positional\rencoding}\n\nThe original multi-head attention operation introduced by\r\\citet{vaswani2017attention} was\rintrinsically invariant by token order permutation. As token position was\ran important information for sequence to sequence models, they encoded\rthe global position of each token in their embedding. Since then, some\rmodified attention mechanisms, that depend on relative tokens position,\rhave been proposed.\r\n\n\\citet{shaw2018selfattention} explored\rmodifying the attention mechanism so that it depends on the relative\rdistance between tokens. A second score matrix that is function of the\rquery and the query/key relative distance is added to the original score\rmatrix that only depends on query/key vector representation. $A = softmax\\left(\\frac{Q \\times K^T}{\\sqrt{d}}\\right) \\times V$ becomes\r$A = \\left(\\frac{Q \\times K^T + S_{rel}}{\\sqrt{d}}\\right) \\times V$ with $S_{rel}$ of shape $(L_Q, L_K)$ defined as\r${S_{rel}}_{ij} = \\vec{Q_i} \\cdotp \\vec{RP}_{clip(i-j, -k, k)}$. Where $k$ is the attention horizon length and $\\vec{RP}_n$ is one of\r$2k+1$ relative positional embedding, vectors of size $d$.\r\\citet{shaw2018selfattention} and\r\\citet{huang2018music} observed that introducing this attention scheme improved performances.\rThe naive calculation of this term however has a complexity of\r$O(L_QL_Kd)$. No algorithm was provided to linearize the complexity.\r\n\nMore recently \\citet{liutkus2021relative} gives a stochastic positional encoding that is linear in\rcomplexity with regards to sequence length. However the implementation\ris complex and its stochastic nature requires that the operations be\rrepeated several times in parallel.\r\n\n\\citet{horn2021translational} noted that\rthe term $S^{rel} \\times V$ can be computed with linear complexity for\rthe case where $RP_{-k} = RP_{k}$. However this is restraining as the\rmodel can't make the difference between tokens before the attention\rhorizon or after.\r\n\nIn this work we will show that the computation of $S^{rel} \\times V$\rcan also be done with linear complexity, without concession.\n\n\\endinput\n", "meta": {"hexsha": "cac10dc765d86cc5b5e6acd6170663b7807423fa", "size": 8906, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/sections/02-background.tex", "max_stars_repo_name": "ScalableTransformer/Scaleformer", "max_stars_repo_head_hexsha": "57e65deb7ba5fdda88a21bdaf71092a0101cf8c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/sections/02-background.tex", "max_issues_repo_name": "ScalableTransformer/Scaleformer", "max_issues_repo_head_hexsha": "57e65deb7ba5fdda88a21bdaf71092a0101cf8c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/sections/02-background.tex", "max_forks_repo_name": "ScalableTransformer/Scaleformer", "max_forks_repo_head_hexsha": "57e65deb7ba5fdda88a21bdaf71092a0101cf8c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-03-01T06:24:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-01T06:24:31.000Z", "avg_line_length": 156.2456140351, "max_line_length": 1361, "alphanum_fraction": 0.7749831574, "num_tokens": 2262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267864276108, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.645684272698589}} {"text": "\\mychapter{6}{Lesson 6} %181012\n\n\\section{Computationally secure encryption}\n\nHaving a better idea of what can and can't be accomplished in the cryptographic world, by means of theorems and proofs, we can focus now on the goal of defining a cryptographic system that meets our requirements. In this lesson, we focus specifically on the secrecy-oriented schemes, thus dealing with encryption and decryption of messages.\n\nThe requirements of a ``good'' encryption scheme are collectively called those of \\emph{computationally secure encryption}: the characterizing requirement is to design a task, or routine, that is \\emph{computationally hard} for an attacker to revert.\nIn detail: this task usually involves a secret key\\footnotemark, and is accomplished in polynomial time, and any attacker who wishes to revert it has no efficient means of doing it without knowing such key. Other properties include:\n\n\\footnotetext{This is the case for symmetric-key schemes, though many other kinds exist: some involving ``public'' keys, some others not having any key at all}\n\n\\begin{enumerate}\n \\item \\label{prop:owk} \\emph{one-way}ness with respect to the encryption key: given $c = \\Enc(k, m)$, it should be hard to recover $k$\n \\item \\label{prop:owm} \\emph{one-way}ness with respect to the original message: given $c = \\Enc(k, m)$, it should be hard to recover $m$\n \\item \\label{prop:nol} In a stricter sense: no information whatsoever must ``leak'' from the message\n\\end{enumerate}\n\nTo start visualizing these concepts, let $\\Pi = (\\Enc, \\Dec)$ be a secrecy scheme, and consider the game depicted in figure \\ref{cryptogame:otindist} where the adversary ``wins'' the game when the challenger outputs 1.\n\n\\begin{cryptogame}\n {otindist}\n {$\\cryptog{ind}(\\lambda, b)$}\n {ind}\n\n \\send{}{$m_0, m_1 \\in \\M$}{}\n\n \\receive{\\shortstack[l]{\n $k \\pickUAR \\K$ \\\\\n $b \\pickUAR \\binary$ \\\\\n $c \\pickUAR \\Enc(k, m_b)$ }}\n {$c$}{}\n\n \\cseqdelay\n\n \\send{}{$b'$}{\\textsc{Output 1 iff} $b = b'$}\n \n\\end{cryptogame}\n\n\\begin{definition}\n The scheme $\\Pi$ is said to be \\emph{computationally one-time secure} iff:\n \\[\n \\forall \\adversary \\in \\ppt \\implies \\cryptog{ind}(\\lambda, 0) \\compindist \\cryptog{ind}(\\lambda, 1)\\footnotemark\n \\] \n or, rephrased in probability terms:\n \\[\n \\forall \\adversary \\in \\ppt \\implies \\lvert\\Pr[\\cryptog{ind}(\\lambda, 0) = 1] - \\Pr[\\cryptog{ind}(\\lambda, 1) = 1]\\rvert \\in \\negl(\\lambda) \\qedhere\n \\]\n\\end{definition}\n\\footnotetext{$\\cryptog{ind}$ refers to the indistinguishability of the messages sent by \\adversary{} during the game}\n\nThis last definition shows how such a scheme is compliant with the three properties exposed beforehand. In particular:\n\n\\begin{enumerate}\n \\item \\emph{It is hard to recover the key}. If not, then an adversary \\adversary{} can efficiently recover the key and use it to decrypt the ciphertext, which in turn enables him to perfectly distinguish $m_{0}$ from $m_{1}$ on any instance;\n \\item \\emph{It is hard to recover the message}. This is analogous, and even more obvious than the preceding point. Nevertheless, this is a necessary condition for a secrecy scheme to be ``good'', and it mustn't be forgotten;\n \\item \\emph{No information about the message whatsoever may leak from the ciphertext}. This may seem subtler than the previous point, but it warrants caution. Observe how an adversary \\adversary{}, if it has the ability to extract even a tiny bit of information of the original message from the ciphertext, then it is actually able to make an educated guess on which message was encrypted in the first place, putting him at an advantage. This leads the probabilities described in the definition to be sensibly more unbalanced than negligible, forfeiting the desired secrecy.\n\\end{enumerate}\n\nBy extension, we may ask ourselves what scheme may or may not be \\emph{computationally two-time secure}. For instance, let $\\Pi_{\\oplus} = (\\Enc, \\Dec)$ be a secrecy scheme using a \\prg{} $G : \\binary^\\lambda \\to \\binary^n$, structured as follows:\n\n\\begin{itemize}\n \\item $\\K = \\binary^\\lambda$, $\\M = \\C= \\binary^n$\n \\item $\\Enc(k, m) = G(k) \\oplus m$\n \\item $\\Dec(k, c) = c \\oplus G(k) = m$\n\\end{itemize}\n\nTo be two-time secure means that, even if an adversary \\adversary{} gets hold of a valid plaintext-ciphertext couple $(\\overline{m}, \\overline{c})$, he is unable to decrypt any future ciphertexts\\footnotemark, apart from the obvious $\\overline{c}$. However, observe that \\adversary{} is now able to extract valuable information for decrypting future ciphertexts:\n\n\\footnotetext{This example models a technique called \\emph{Chosen Plaintext Attack}, which will be discussed in depth later}\n\n\\[\n \\overline{c} = \\Enc(k, \\overline{m}) = G(k) \\oplus \\overline{m} \\implies \\overline{c} \\oplus \\overline{m} = G(k)\n\\]\nso now, for any second ciphertext \\adversary{} receives, he can \\emph{mimic} the decryption routine, and efficiently uncover the underlying plaintext. This proves that $\\Pi_\\oplus$ is not two time-secure; nevertheless, it is still one-time secure:\n\n\\begin{theorem}\n If $G$ is a \\prg, then $\\Pi_\\oplus$ is computationally one-time secure\n\\end{theorem}\n\n\\begin{proof}\n This proof is another example that showcases the use of hybrid games. Recalling the one-time security definition, we need to show that:\n \\[\n \\forall \\adversary \\in \\ppt \\implies \\cryptog{ind}[\\Pi_\\oplus](\\lambda, 0) \\compindist \\cryptog{ind}[\\Pi_\\oplus](\\lambda, 1)\n \\]\n Consider the hybrid game in figure \\ref{cryptogame:xorothybrid}, where the original encryption routine is changed to use a completely random value, instead of using $G(k)$\\footnotemark. As an exercise, compare it with the original one-time secure definition in figure \\ref{cryptogame:otindist}, to check that it perfectly matches.\n\n \\footnotetext{The observant student may recognize that this modification yields exactly the ``one-time pad'' secrecy scheme discussed in lesson 1}\n\n \\begin{cryptogame}\n {xorothybrid}\n {$\\hybridg{}[\\Pi_\\oplus](\\lambda, b)$}\n {r-ind}\n\n \\send{$m_0 \\neq m_1 \\in \\binary^l$}{$m_0, m_1$}{}\n\n \\receive{\\shortstack[l]{\n $r \\pickUAR \\binary^l$ \\\\ % R \\sim \\unifdist{\\binary^l}\n $b \\pickUAR \\binary$ \\\\ % B \\sim \\unifdist{\\binary}\n $c = r \\oplus m_b$ % C = R \\oplus m_B\n }}\n {$c$}{}\n\n \\cseqdelay\n\n \\send{}{$b'$}{\\textsc{Output 1 iff} $b' = b$}\n \n \\end{cryptogame}\n\n The proof begins by affirming that:\n \\begin{claim}\n \\[\n \\forall \\adversary \\in \\ppt \\implies \\hybridg{}[\\Pi_\\oplus](\\lambda, 0) \\equiv \\hybridg{}[\\Pi_\\oplus](\\lambda, 1)\n \\]\n \\end{claim}\n\n To prove it, notice that $r$ is chosen uniformly at random, and independently of $b$. Thus, no matter how the messages $m_0$ and $m_1$ are structured, $r$ will effectively make the chosen message completely unrecognizable. In formal terms, let $B$ and $R$ be the random variables for \\challenger{}'s picks of $b$ and $r$ respectively; then:\n \\begin{align*} % AP190906: Totally original, need a peer review here\n & \\left| \\Pr(B = 0 \\knowing C = c) - \\Pr(B = 1 \\knowing C \\evaluatesto c) \\right| & \\\\\n =& \\left| \\Pr(B = 0 \\knowing R \\oplus m_B \\evaluatesto c) - \\Pr(B = 1 \\knowing R \\oplus m_B \\evaluatesto c) \\right| & \\text{(C definition)} \\\\\n =& \\mathrlap{\\frac{\\left| \\Pr(R \\oplus m_B = c \\knowing B \\evaluatesto 0) \\Pr(B = 0) - \\Pr(R \\oplus m_B = c \\knowing B \\evaluatesto 1) \\Pr(B = 1) \\right|}{\\Pr(R \\oplus m_B = c)}} & \\\\\n & & \\text{(Bayes' theorem)} \\\\\n =& \\frac{\\left| \\Pr(R \\oplus m_0 = c) \\Pr(B = 0) - \\Pr(R \\oplus m_1 = c) \\Pr(B = 1) \\right|}{\\Pr(R \\oplus m_B = c)} & \\text{(Cond. collapse)} \\\\\n =& \\frac{\\left| \\oneover{2^l} \\half - \\oneover{2^l} \\half \\right|}{\\Pr(R \\oplus m_B = c)} = 0 & \\\\\n \\end{align*}\n\n Having proven that \\adversary's success is equivalent to straight guessing in $\\hybridg{}[\\Pi_\\oplus]$, we now relate the hybrid game to the original one, affirming that:\n \n \\begin{claim}\n \\[\n \\forall \\adversary \\in \\ppt,\\, \\forall b \\in \\binary \\implies \\hybridg{}[\\Pi_\\oplus](\\lambda, b) \\compindist \\cryptog{ind}[\\Pi_\\oplus](\\lambda, b) \\qedhere\n \\]\n \\end{claim}\n\n The proof proceeds by reduction as depicted in figure \\ref{cryptoredux:xorotprg}, by assuming the existence of a distinguisher $\\textsf{D}^{\\textsc{ind}}$ for $c = G(k) \\oplus m_{b}$ and $c = r \\oplus m_{b}$, and using it to break $G$'s pseudo-random generation property.\n\n \\begin{cryptoredux}\n {xorotprg}\n {Reducing to breaking a \\prg}\n {prg}\n {G-ind}[2]\n\n \\receive{\\shortstack[l]{\n $k \\pickUAR \\binary^\\lambda$ \\\\\n $x_0 \\pickUAR G(k)$ \\\\\n $x_1 \\pickUAR \\binary^l$ \\\\\n $b \\pickUAR \\binary$\n }}{$x_b$}{}\n\n \\return{}{$m_0, m_1$}{}\n\n \\invoke{\\shortstack[l]{\n $\\beta \\pickUAR \\binary$ \\\\\n $c = x_b \\oplus m_\\beta$\n }}{$c$}{}\n\n \\return{}{$\\beta'$}{}\n\n \\cseqdelay\n \\send{$b' = \\begin{cases}\n 0 &\\textsc{iff } \\beta' = \\beta \\\\\n 1 &\\textsc{else}\n \\end{cases}$\n }{$b'$}{\\textsc{Output 1 iff} $b' = b$}\n\n \\end{cryptoredux}\n\n Again, notice how the games of indistinguishability and pseudo-random generation are reliably reproduced on their respective sides. This construct's value centers on how $\\distinguisher^\\textsc{ind}$ will perform in its own challenge:\n \\begin{itemize}\n \\item if $x_b$ is a random value, then $\\distinguisher^\\textsc{ind}$ will perform as depicted in the hybrid game, thus giving right or wrong answers at random;\n \\item if b is the result of $G$, then $\\distinguisher^\\textsc{ind}$ has a better chance in finding the right answer by its own design;\n \\end{itemize}\n From these observations, especially the second point, the adversary \\adversary{} has a better chance of winning the \\prg{} game by asserting that $x_b$ comes from $G$ whenever $\\distinguisher^\\textsc{ind}$ makes a correct guess; conversely, \\adversary{} will preferably declare that $x_b$ is truly random whenever $\\distinguisher^\\textsc{ind}$ fails a guess, as the probability of the latter getting fooled by a random value is sensibly greater than by a pseudo-random one.\n\n Either way, by the existence of $\\distinguisher{}^\\textsc{ind}$, \\adversary{} gains an edge in efficiently recognizing $G$, which cannot happen by $G$'s definition; the claim is proven. The theorem's proof can be completed by putting the pieces together, as is usual in the hybrid argument:\n \\[\n \\cryptog{ind}[\\Pi_\\oplus](\\lambda, 0) \\compindist\n \\hybridg{}[\\Pi_\\oplus](\\lambda, 0) \\equiv\n \\hybridg{}[\\Pi_\\oplus](\\lambda, 1) \\compindist\n \\cryptog{ind}[\\Pi_\\oplus](\\lambda, 1)\n \\]\n\n which finally states that $\\cryptog{ind}[\\Pi_\\oplus](\\lambda, 0) \\compindist \\cryptog{ind}[\\Pi_\\oplus](\\lambda, 1)$.\n\\end{proof}\n \n\n\\section{Pseudorandom functions}\n\n\\textsc{Prg}s are used in practice as a stepping stone for building \\emph{pseudo-random functions}, \\prf{} henceforth, which are the principal construct in several cryptographic schemes. Before formally introducing what a \\prf{} is, we begin instead by defining what a \\emph{truly random function} is:\n\n\\begin{definition}\n A random function $R : \\binary^n \\to \\binary^l$ is a function that, depending on what is known about its previous applications:\n\n \\begin{itemize}\n \\item if $x$ is ``fresh'' (in formal terms, $R$ has never been applied to $x$ beforehand), then a value $y$ is chosen \\uar{} from $R$'s codomain, and it is permanently associated as the image of $x$ in $R$\\footnote{This property is also called \\emph{lazy sampling}.};\n\n \\item if $x$ is not fresh, then $R(x)$ is directly returned instead. \\qedhere\n \\end{itemize}\n\\end{definition}\n\nIt should be noted that, if such functions are to be implemented in computers, they would occupy too much space in memory. Suppose all the possible outputs of $R$ have been generated and stored as an array in memory; then its total size in bits will be $l \\cdot 2^n$:\n\n\\[ \n \\color{black!50}\n \\overbracket[0.5pt][5pt]{\\;\\color{black}\\framebox[4em][c]{0010...}\\;}^{l \\text{ bits}}_1\\;\n \\overbracket[0.5pt][5pt]{\\;\\color{black}\\framebox[4em][c]{1110...}\\;}^{l \\text{ bits}}_2\\;\n \\overbracket[0.5pt][5pt]{\\;\\color{black}\\framebox[4em][c]{0011...}\\;}^{l \\text{ bits}}_3\\;\\;\n \\color{black!25}\n \\framebox[2em][c]{\\rule{0pt}{1.5ex}...}\\;\n \\framebox[2em][c]{\\rule{0pt}{1.5ex}...}\\;\n \\framebox[2em][c]{\\rule{0pt}{1.5ex}...}\\;\n \\framebox[2em][c]{\\rule{0pt}{1.5ex}...}\\;\\;\n \\color{black!50}\n \\overbracket[0.5pt][5pt]{\\;\\color{black}\\framebox[4em][c]{1110...}\\;}^{l \\text{ bits}}_{2^n}\n\\]\n\nSuch a function becomes cumbersome and difficult to maintain in practice; therefore, it is desirable to find a kind of function which looks as a random function possible, but does not require to be wholly memorized, while not forgetting to maintain poly-time complexity. Pseudo-randomness comes to the rescue here:\n\n\\begin{definition}\n Let $f$ be a function, then it is deemed pseudo-random (therefore, $f$ is a \\prf{}) iff it is computationally indistinguishable from a true random function.\n\\end{definition}\n\nIn detail, \\prf{}s are actually designed as function families $f_k$\\footnotemark, where $k$ is a parameter that indexes the functions inside the family. To model the \\prf{}s' indistinguishability from random functions, let $F \\in \\binary^\\lambda \\to (\\binary^{n(\\lambda)} \\to \\binary^{l(\\lambda)} )$, usually denoted simply by $f_k$, be a \\prf{}, and define $\\mathfrak{R}(n, l)$ to be the domain that collects the random functions from $\\binary^{n(\\lambda)}$ to $\\binary^{l(\\lambda)}$.\n\n\\footnotetext{Does this remind you of something else? If not, look back in lesson 2 and 3.}\n\nConsider the indistinguishability game drawn in figure \\ref{cryptogame:prf}; although one may thing that \\prg{}s and \\prf{}s aren't much different, the game tells a different story, which is best put by an introductory paragraph about \\prf{}s in their Wikipedia page:\n\n\\begin{quotation}\n ``Pseudorandom functions are not to be confused with pseudorandom generators (\\prg{}s). The guarantee of a \\prg{} is that a single output appears random if the input was chosen at random. On the other hand, the guarantee of a \\prf{} is that all its outputs appear random, regardless of how the corresponding inputs were chosen, as long as the function was drawn at random from the \\prf{} family.''\\footnote{\\linkicon \\href\n {https://en.wikipedia.org/wiki/Pseudorandom_function_family}\n {\\textsf{Pseudorandom function family --- Wikipedia}}}\n\\end{quotation}\n\n\\begin{cryptogame}\n {prf}\n {The \\prf{} indistinguishability game}\n {prf}\n\n \\cseqchallenger{\\shortstack[l]{\n $k \\pickUAR \\binary^\\lambda$ \\\\\n $R \\pickUAR \\mathfrak{R}(n, l)$ \\\\\n $b \\pickUAR \\binary$\n }}\n\n \\cseqdelay\n \\cseqbeginloop\n\n \\send{}{$x$}{}\n\n \\receive{\\shortstack[l]{\n $y_0 = f_k(x)$ \\\\\n $y_1 = R(x)$\n }}{$y_b$}{}\n\n \\cseqendloop\n \\cseqdelay\n\n \\send{}{$b'$}{\\textsc{Output 1 iff} $b = b'$}\n\n\\end{cryptogame}\n\nIn this game, \\adversary{} is allowed to make multiple queries to $\\challenger^\\prf$, as opposed to the \\prg{} indistinguishability game where it can perform just one query before making its guess. To reiterate the \\prf{} definition in terms of this game:\n\n\\begin{definition}\n A function family $F = f_k$ is a \\prf{} iff: \n \\[\n \\cryptog{prf}[f_k](\\lambda, 0) \\compindist \\cryptog{prf}[f_k](\\lambda, 1) \\qedhere\n \\]\n\\end{definition}\n\n\\begin{exercise}\n Prove the following statements:\n \\begin{itemize}\n \\item No \\prg{} is secure against unbounded attackers;\n \\item No \\prf{} is secure against unbounded attackers.\n \\end{itemize}\n \n\\end{exercise}\n\n\n\\subsection{\\textsc{Ggm}-tree}\n\nThis section is dedicated to a concrete example of a \\prf{} which is built from the ground up using a \\prg. This construct has been designed from Oded Goldreich, Shafi Goldwasser and Silvio Micali, and its structure is akin to a binary tree, hence its name: \\emph{\\ggm-tree}.\n\n\\begin{construction}\n Let $G \\in \\binary^\\lambda \\to \\binary^{2\\lambda}$ be a \\prg{} such that it doubles the length of its argument, and denote the images' first and second halves as $G_0(k)$ and $G_1(k)$ respectively, so that: \n \\[\n k \\mapsto (G_0(k), G_1(k))\n \\]\n\n Since the principal mechanism makes use of the halves being the same length of the argument, in the same spirit, we will denote the action of using one half of an image of $G$ as argument of $G$ itself in a shorter fashion, as demonstrated in the following example:\n \\[\n G_a(G_b(G_c(k))) =: G_{abc}(k)\n \\]\n\n This leads to the final step: let $f_k$ be a function family, where $k \\in \\binary^\\lambda$, such that:\n \\[\n f_k(r) = G_r(k)\n \\]\n\n This is our candidate \\prf. To visualize it, consider the tree structure depicted in figure \\ref{fig:ggmtree}: at each level of the tree, a single bit of $r$ is used to decide which half of $G$'s image will be used in the next level. For example, $f_k(01 \\dots 10)$ would evaluate as $G_0(G_1( \\dots G_1(G_0(k))))$.\n\n \\begin{figure}\n \\centering\n \\begin{tikzpicture}[\n level 1/.style={sibling distance=16em},\n level 2/.style={sibling distance=8em},\n level 3/.style={sibling distance=4em}]\n\n \\node{$k$}\n child{ node {$G_0(k)$}\n child{ node {$G_{00}(k)$}\n child{ node{$G_{000}(k)$} node[below]{$\\vdots$}}\n child{ node{$G_{001}(k)$} node[below]{$\\vdots$}}\n }\n child{ node {$G_{01}(k)$}\n child{ node{$G_{010}(k)$} node[below]{$\\vdots$}}\n child{ node{$G_{011}(k)$} node[below]{$\\vdots$}}\n }\n }\n child{ node {$G_1(k)$}\n child{ node {$G_{10}(k)$}\n child{ node{$G_{100}(k)$} node[below]{$\\vdots$}}\n child{ node{$G_{101}(k)$} node[below]{$\\vdots$}}\n }\n child{ node {$G_{11}(k)$}\n child{ node{$G_{110}(k)$} node[below]{$\\vdots$}}\n child{ node{$G_{111}(k)$} node[below]{$\\vdots$}}\n }\n };\n \\end{tikzpicture}\n \\caption{The \\textsc{ggm}-tree for $G$}\n \\label{fig:ggmtree}\n \\end{figure}\n\n\\end{construction}\n\nThe proof of \\prf-ness will be discussed in the next lesson.\n", "meta": {"hexsha": "c891490acd4ffcd1c6925ea23c068050604a6911", "size": 18835, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lessons/lesson_6.tex", "max_stars_repo_name": "Project2100/Cryptography-2018_19", "max_stars_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-10-15T09:22:45.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-15T09:22:45.000Z", "max_issues_repo_path": "lessons/lesson_6.tex", "max_issues_repo_name": "Project2100/cryptography_1819", "max_issues_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-07-18T15:45:10.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-27T20:36:12.000Z", "max_forks_repo_path": "lessons/lesson_6.tex", "max_forks_repo_name": "Project2100/cryptography_1819", "max_forks_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-07-17T14:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-03T15:23:22.000Z", "avg_line_length": 55.5604719764, "max_line_length": 578, "alphanum_fraction": 0.6561189275, "num_tokens": 5542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891392358015, "lm_q2_score": 0.7853085808877581, "lm_q1q2_score": 0.6456721861545947}} {"text": "%\n% CMPT 310: Artificial Intelligence - A Course Overview\n% Section: Game Playing and Adversarial Search\n%\n% Author: Jeffrey Leung\n%\n\n\\section{Game Playing and Adversarial Search}\n\t\\label{sec:game-playing-adversarial-search}\n\\begin{easylist}\n\n& Game playing: Scenario where a game with discrete states is played with two players, each seeking to maximize their position which minimizes the opponent's position\n\t&& E.g. Chess, checkers\n\t&& \\textbf{Heuristic/evaluation function:} Estimate of the optimality of a game state\n\t&& \\textbf{Terminal state:} Game state which has no further states\n\t&& Example diagram of game state evaluations in a tree: See figure~\\ref{fig:game-tree}\n\t\n\\begin{figure}[!htb]\n\t\\caption{Game Tree (Minimizer first)}\n\t\\label{fig:game-tree}\n\t\\centering\n\n\t\\begin{forest}\n\t\tfor tree={\n\t\t\tdraw,\n\t\t\tcircle,\n\t\t\tminimum size=1cm,\n\t\t\tline width=0,\n\t\t\talign=center\n\t\t},\n\t\t[0\n\t\t\t[2\n\t\t\t\t[2]\n\t\t\t\t[-3]\n\t\t\t\t[-14]\n\t\t\t]\n\t\t\t[14\n\t\t\t\t[14]\n\t\t\t\t[17]\n\t\t\t\t[20]\n\t\t\t]\n\t\t\t[0\n\t\t\t\t[-12]\n\t\t\t\t[-18]\n\t\t\t\t[0]\n\t\t\t]\n\t\t]\n\t\\end{forest}\n\\end{figure}\n\n& \\textbf{Minimax algorithm:} Decision-making rule which minimizes the potential loss for a worst-case scenario, given two entities who are alternatively trying to maximize and minimize a score\n\t&& Performs the equivalent of a depth-first search\n\t&& Each alternating level of the decision tree is minimized or maximized from the values in the levels below it\n\t&& Complete: Yes, as long as the tree is finite\n\n& \\textbf{$\\alpha / \\beta$ pruning:} Keeping the maximum/minimum searched values at a given level so far to eliminate nodes which will not exceed a minimum/maximum\n\t&& $\\alpha$: Lower bound on the potential value of a maximum node\n\t&& $\\beta$: Upper bound on the potential value of a minimum node\n\t&& \\href{http://inst.eecs.berkeley.edu/~cs61b/fa14/ta-materials/apps/ab_tree_practice/}{Resource for practice}\n\t&& See figure~\\ref{fig:game-tree-ab} for an example of pruned nodes (notated by dashed edges)\n\t\n\\begin{figure}[!htb]\n\t\\caption{Game Tree with $\\alpha / \\beta$ Pruning (Minimizer first)}\n\t\\label{fig:game-tree-ab}\n\t\\centering\n\n\t\\begin{forest}\n\t\tfor tree={\n\t\t\tdraw,\n\t\t\tcircle,\n\t\t\tminimum size=1cm,\n\t\t\tline width=0,\n\t\t\talign=center\n\t\t},\n\t\t[0\n\t\t\t[2\n\t\t\t\t[2]\n\t\t\t\t[-3]\n\t\t\t\t[-14]\n\t\t\t]\n\t\t\t[14\n\t\t\t\t[14]\n\t\t\t\t[17, edge={densely dashed}]\n\t\t\t\t[20, edge={densely dashed}]\n\t\t\t]\n\t\t\t[0\n\t\t\t\t[-12]\n\t\t\t\t[-18]\n\t\t\t\t[0]\n\t\t\t]\n\t\t]\n\t\\end{forest}\n\\end{figure}\n\n& Time, space, and optimal complexity of algorithms: See figure~\\ref{fig:algos-complexity}\n\n\\end{easylist}\n\\clearpage\n", "meta": {"hexsha": "a74fd708a298af4305150c2611cce8c086268b39", "size": 2506, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/game-playing.tex", "max_stars_repo_name": "AmirNaghibi/notes", "max_stars_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2019-08-11T08:45:10.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T02:37:39.000Z", "max_issues_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/game-playing.tex", "max_issues_repo_name": "AmirNaghibi/notes", "max_issues_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/game-playing.tex", "max_forks_repo_name": "AmirNaghibi/notes", "max_forks_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-18T09:17:46.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-27T21:44:56.000Z", "avg_line_length": 25.3131313131, "max_line_length": 193, "alphanum_fraction": 0.6823623304, "num_tokens": 769, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891218080991, "lm_q2_score": 0.7853085884247212, "lm_q1q2_score": 0.6456721786652795}} {"text": "\\chapter{Algorithms}\nIn this chapter, we will list some important algorithms which may be used in the solution of each subproblem.\n\n\\section{Shortest Path Between Two Nodes}\n\nAlgorithms for the calculation of the Shortest Path between two points are very important for the solution of this problem. They will be used to calculate the shortest path between multiple points when calculating a vehicle's path, such as the shortest path between the garage and a pick up point. To do this, there are many algorithms at our disposal. We will evaluate some of them and make a decision on which is(are) best for the situation at hand.\n\n\n\\subsection{Dijkstra's Algorithm}\n\n\\subsubsection{Description}\nThis algorithm was conceived by Edsger W. Dijkstra's in 1956 and is used to calculate shortest paths in directed or undirected graphs, so long there are no edges with negative weight. Upon utilization of this algorithm, a tree holding the shortest paths from the origin node to all others is formed. \n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item Minimum distance to the origin node from each of the graph's nodes (dist)\n \\item Node that comes before it in its path (path)\n \\item Priority queue to hold the nodes to be processed next\n \\item Decrease-Key function, to maintain the nodes with shortest distance in the top of the queue\n\\end{itemize}\n\n\\input{Algorithms/Dijkstra.tex}\n\n\\subsubsection{Analysis}\nThe first part of the algorithm's goal is to prepare the data (lines 1-6): paths are set to null, distances to infinity, the priority queue is initialized and the origin node is inserted. Then, a breadth-first search is performed; a check is made to every node found in order to understand if its path can be shortened by the use of the edge in analysis. After that, if the node is not already in the queue, it must be inserted, to be eventually processed. If it is, its key must be reduced. The items of the priority queue are ordered by their key. This system is used so that the nodes which are closest to the origin are processed first. This procedure ensures that the distance of already processed nodes remains intact, boosting the algorithm's efficiency and making this a greedy algorithm. \\\\\n%\\begin{itemize}\n% \\item \\textbf{Loop invariant} - $ priorityQueue \\neq \\emptyset $, as each time a node is processed, its shortest path is defined and all nodes will be processed when the queue is empty\n% \\item \\textbf{Loop variant} - the number of nodes to be processed, since the algorithm would stop when the queue is empty\n%\\end{itemize}\n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O((|N|+|E|) \\times \\log(|N|) $, where E is the set of edges and N the set of nodes of the graph. Decrease-Key's time efficiency is $ O(|E| \\times log(|N|)$. It can be $ O(1) $ if Fibonacci Heaps are used instead of regular Priority Queues, making the whole algorithm $ O(|N| \\times log(|N|) $ \n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but usually $ O(|N|) $\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm is one to consider using in the problem, but there may be slightly more efficient ones.\n\n\n\\subsection{Bellman-Ford Algorithm}\n\n\\subsubsection{Description}\nThe Bellman-Ford Algorithm is another algorithm used to calculate the shortest path between nodes in a graph. It was first proposed by Alfonso Shimbel in 1955 but ended up being named after Richard Bellman and Lester Ford, who later officially published it independently. This algorithm is useful to calculate shortest paths in graphs with negative weight edges and to detect negative edge cycles in them.\n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item Minimum distance to the origin node from each of the graph's nodes (dist)\n \\item Node that comes before it in its path (path)\n\\end{itemize}\n\n\\input{Algorithms/Bellman-Ford.tex}\n\n\\subsubsection{Analysis}\nIn this algorithm, the first step is to prepare the graph (lines 1-4). Then, all edges are analyzed $|N| - 1 $ times(lines 6-7), checking if they can be used to shorten the path from s to the destination node of the edge(line 8). This process is repeated this amount of times because it is the maximum length in edges for a path. In the end, a check is made, to see if the algorithm worked. If it did not, the cycle as negative weight loops (lines 14-18). This algorithm is considered to implement dynamic programming.\n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N| \\times |E|) $, where E is the set of edges and N the set of nodes of the graph\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but in this case $ O(1) $\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm poses little interest to this problem given that our edges' weights represent the length of a road, which would never be negative. \n\n\n\\subsection{Bidirectional Dijkstra}\n\n\\subsubsection{Description}\nIra Pohl was the first one to design and implement a bidirectional heuristic search algorithm in 1971. \nBidirectional Dijkstra's that implements the idea of a bidirectional search into Dijkstra's algorithm. It finds a shortest path from an initial node to a goal node in a directed or undirected graph. It runs two simultaneous searches: one forward from the initial state, and one backward from the goal, alternating between the two. The stopping criteria is one that must be very well implemented to guarantee the shortest path. Compared to normal Dijkstra's there's a speedup of 2x.\n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item Minimum distances to the origin and goal nodes from each of the graph's nodes (dist)\n \\item Nodes that come before each node in both forward and backward paths (not implemented in pseudocode)\n \\item Two priority queues to hold the nodes to be processed next in each direction\n \\item Way of search (way) (1 if it was encountered by forward search first, 2 if it was encountered by backward search first, 0 default)\n \\item Decrease-Key function, to maintain the nodes with shortest distance in the top of the queue\n\\end{itemize}\n\n\\input{Algorithms/Bidirectional.tex}\n\\uline{Note:} The pseudocode is not complete, missing the parts to calculate the actual path and to calculate the transposed graph (needed for the backward search because this is a directed graph). These parts were excluded in order to not make the algorithm's pseudocode even more complex and spacious.\n\n\\subsubsection{Analysis}\nThis algorithm basically performs two Dijkstra's algorithms: one starting from the initial node and going forward(lines 16-23); another one starting from the ending node and going backward (lines 24-31). Each time they cross, the distance is registered (lines 25-26 for example). The stopping criteria is a tricky problem in this case. In many situations, the algorithm ends when the two sets collide (forward search and backward search). We found that this criteria did not guarantee the shortest path. Instead, we chose to end the algorithm when there can be no longer any better distances because the nodes being processed are already too far apart.\n%\\begin{itemize}\n% \\item \\textbf{Loop invariant} - $ dist(n) + distR(v) > bestDist $, where best distance is the best Distance for a complete path obtained yet, dist(n) is the distance of the node being processed in forward way to the starting node and $ distR(v) $ the distance of the node being processed in backward way to the ending node\n% \\item \\textbf{Loop variant} - the number of nodes to be processed in each way, since the algorithm would stop when each search had got no more nodes to process on the queue\n%\\end{itemize}\n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O((|N|+|E|) \\times \\log(|N|) $, where E is the set of edges and N the set of nodes of the graph.\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but usually $ O(|N|) $\n\\end{itemize}\nEven though there's a speed up, the complexities will be the same as in Dijkstra's, since the difference is multiplication and division by constants, which is ignored in Big O order.\n\n\\subsubsection{Usability}\nBidirectional Dijkstra's algorithm is one of the fastest algorithms when it comes shortest paths. It has the down-side of requiring the previous knowledge of the ending/destination node, but in this scenario that is not a problem.\n\n\n\\subsection{A* Algorithms}\n\\subsubsection{Description and Analysis}\nThe first time a A* Algorithm came up was in \\textit{Shakey Project}, project of the first general purpose mobile robot capable of reasoning about is own actions. In an A* (pronounced A-star) Search Algorithm, the 'level' of a node in a graph becomes a sum of two different characteristics. For instance, applied to this problem, the distance a node is from the origin in the search algorithm is substituted for the sum of:\n\\begin{itemize}\n \\item $d_{sv}$ - actual distance to the origin\n \\item $\\pi_{nt}$ - estimated guess of the distance to the goal node (heuristic)\n\\end{itemize}\nThe reliability of this technique depends on the efficacy of the heuristic. In this case, because the weight of edges is distance in meters, the efficacy is great if we assume $\\pi$ to be the euclidean distance. This algorithm can be implemented into both Bidirectional and normal Dijkstra's.\n\\begin{itemize}\n \\item \\textbf{Upside} - Speeds up the process immensely\n \\item \\textbf{Downside} - The algorithm does not absolutely guarantee the best solution.\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis is a technique that shows to be very promising and a good idea to implement in the solution of this problem.\n\n\n\n\\newpage\n\\section{Shortest Path Between All Nodes}\nThis section is dedicated to algorithms that calculate the shortest path between all pairs of points. Such algorithms may be used in the solution, specially in the pre-processing part.\n\n\\subsection{Floyd-Warshall}\n\n\\subsubsection{Description}\nThis algorithm was published by Robert Floyd in 1962, yet it was virtually the same as some algorithms published by Stephen Warshall and Bernard Roy, 1962 and 1959 respectively. It takes an adjacency matrix that represents the graph as an input (empty in the beginning) and calculates the shortest paths between all nodes. The value of a path between to vertices is the sum of all edge's weights through that path. It is registered in the matrix as such: distance of shortest path between 1 and 4 would be matrix[1][4]. Edges may have negative values but the graph can't have a negative cycle.\n% For each pair of vertices, the shortest path is calculated. Time complexity is O \\begin{math}(|V| ^3).\\end{math}, and space complexity is O \\begin{math}(|V| ^2).\\end{math}\n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item $ |N| \\times |N| $ matrix (another one if paths are needed), where $ |N| $ is the number of vertices\n\\end{itemize}\n\n\\input{Algorithms/Floyd-Warshall.tex}\n\\uline{Note:} The pseudocode is not complete, missing the matrix to calculate the paths and respective instructions. This was done in order to simplify the algorithm and minimize it.\n\n\\subsubsection{Analysis}\nFirst part is basically the preparation of the matrix: fill it with $\\infty$ distances; for the edges existant (lines 1-3), change the values in the matrix to their weight (lines 4-6). Then, all pairs of nodes will be checked on if there is any node the path can pass through that would minimize its cost (lines 7-14). \n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N|^3) $, where N is the set of nodes of the graph.\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but usually $ O(|N|^2) $, for the matrix of size $|N| \\times |N|$\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm is an interesting algorithm to be used in the beginning of the program, to leave all shortest paths processed. However, this might take a long time, and it might be a better idea to simply calculate the shortest path each time it is needed.\n\n\n\\subsection{Dijkstra's Algorithm}\n\n\\subsubsection{Description and Analysis}\nDijkstra's can also be used to calculate all shortest paths. This is done by simply executing the algorithm for every node in the graph as origin.\n \n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N| \\times (|N|+|E|) \\times \\log(|N|) $, where E is the set of edges and N the set of nodes of the graph. Same as Dijkstra's but multiplied by $|N|$\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but usually $ O(|N|) $, for the priority queue, same as Dijkstra's\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis is a better option for less dense graphs compared to Floyd-Warshall. As this is not the case for a road network, this algorithm is relatively irrelevant.\n\n\n\n\\newpage\n\\section{Connectivity}\nThe algorithms in this section check the connectivity of the graph. By other words, this section is dedicated to algorithms that detect if any nodes are inaccessible to others, dividing the graph in strongly connected components. A graph will be strongly connected if there is a path between all pairs of points, or if the graph is a strongly connected component itself. Algorithms to test connectivity may be useful to solve the problem of unpredictability.\n\n\n\\subsection{Brute Force with Floyd-Warshall}\n\n\\subsubsection{Description and Analysis}\nOne way to test the connectivity of a graph is by analyzing the results of Floyd-Warhsall in a certain graph. If there are any pairs of nodes which the distance remains $\\infty$, they do not belong to the same strongly connected component.\n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N|^3) $, where N is the set of nodes of the graph, same as Floyd-Warshall\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but usually $ O(|N|^2) $, for the matrix of size $|N| \\times |N|$\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm works yet is not a good option. Even if the Floyd-Warshall is going to be utilized anyway, the complexity of traversing the whole matrix is $O(|N|^2)$. Therefore, there are much better algorithms for the job.\n\n\n\\subsection{Kosaraju's algorithm}\n\n\\subsubsection{Description}\nThe algorithm was first published in 1981 by Micha Sharir but gets is name from whom first mencioned it in 1971, Sambasiva Rao Kosaraju. This algorithm allows to identify strongly connected components in a directed graph, testing its connectivity. \n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item boolean identifying if a node has been visited or not (visited)\n \\item identifier of the strong component the node belongs to (strongComponent)\n \\item function to invert the edges of the graph\n \\item a queue to hold the order of the visits\n\\end{itemize}\n\n\\input{Algorithms/Kosaraju.tex}\n\n\\subsubsection{Analysis}\nThe algorithm performs two depth-first searches in the graph. In the first one, it marks the nodes by visit order (lines 5-9). The second one starts on the nodes with lowest order and is performed on the graph with the edges inverted (lines 10-14). This second one will generate multiple expansion trees (or just one if the graph is strongly connected), each being a strongly connected component. If two nodes are not in the same strongly connected component, one of them cannot be reached from the other.\n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N|+|E|) $, where N is the set of nodes of the graph and E the set of edges. Each dfs is $O(|N|+|E|)$\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but in this one it is $ O(|N|) $, being the size of the queue $|N|$\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm is a serious candidate to check the connectivity of the graph.\n\n\n\\subsection{Tarjan's Algorithm}\n\n\\subsubsection{Description}\nThe Tarjan's Algorithm was named after its inventor, Robert Tarjan, and is another algorithm capable of identifying the strongly connected components in a given graph. It does so, similarly to Kosaraju's, with the help of dfs, only this time the graph is only traversed once.\n\n\\subsubsection{Extra Data, Data Structures and Algorithms Required}\n\\begin{itemize}\n \\item boolean identifying if a node has been visited or not (visited)\n \\item integer identifying the counter when a node is first found (num)\n \\item integer identifying the lowest num reachable by a node (low)\n\\end{itemize}\n\n\\input{Algorithms/Tarjan.tex}\n\n\\subsubsection{Analysis}\nThe algorithm is fairly simple: the graph is traversed by a depth first search (lines 5-6); the $num$ and $low$ of a node are set to the $numCounter$(lines 12-13); each time an already visited node is found, all nodes with $low > that nodes low$ will be changed (lines 16-18). This way, all vertices with the same $low$ belong to the same strongly connected component. \n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|N|+|E|) $, where N is the set of nodes of the graph and E the set of edges. The dfs is $O(|N|+|E|)$. This is the same complexity as Kosaraju's, but the algorithm is twice as fast since it only performs one dfs.\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but in this one it is $ O(1) $.\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis algorithm is an even better candidate than the last one, since it is supposedly twice as fast. \n\n\n\\newpage\n\\section{Clustering Algorithms}\n\nA Clustering is a machine learning task which involves the recognition of natural grouping in data. Such algorithms can be used in the distribution of orders per vehicle.\n\n\n\\subsection{Agglomerative Hierarchical Clustering}\n\n\\subsubsection{Description}\n\nThis algorithm is one of the most common Hierarchical Clustering algorithms. It is used to form groups of objects of data based on their similarity. In this case, we will use it to form N groups of data nodes with great similarity.\n\n\\input{Algorithms/Clustering.tex}\n\\uline{Note:} the implementation presented above is just a rough sketch, the one to be implemented will be different.\n\n\\subsubsection{Analysis}\nThe algorithm begins to find the pair of points which are most similar. In this case, similarity is measured by distance (lines 6-14). After he has found one, he merges the two sets (. In between sets, there are many criteria to calculate similarity. The one we will implement is taking the similarity between two sets as the similarity of the most similar pair of points where one belong to the first set and other to the second set. This algorithm adopts a Bottom-Up strategy (as do all agglomerative hierarchical clustering algorithms), as it start by processing each node and joining them to obtain the final result. \n\n\\subsubsection{Efficiency}\n\\begin{itemize}\n \\item \\textbf{Temporal Complexity} - $ O(|S|^2*(|S| - n)) $, where S is the set of points. This is only an estimated guess.\n \\item \\textbf{Spatial Complexity} - Depends on the implementation, but in this one it is $ O(|S|) $.\n\\end{itemize}\n\n\\subsubsection{Usability}\nThis is the only algorithm analysed in this cathegory, thus it is the chosen one to perform the task.\n\n\\section{Extras}\nThere are other algorithms, simpler or less relevant, that may (or may not) be used in our solution:\n\\begin{itemize}\n \\item Breadth-First Search (bfs) - Graph searching algorithm\n \\item Depth-First Search (dfs) - Graph searching algorithm\n \\item Decrease-Key Function - Priority changing function on a priority queue\n \\item Graph Transposing Function - Inversion of the edges in a directed graph\n \\item Fibonacci Heaps (Data structure) - Improved priority queue\n \\item Euclidian Distance function - To calculate the euclidian (straight-line) distance between two points\n\\end{itemize}\n", "meta": {"hexsha": "a4ed48a88484ae2c729aa6818a5afccba6a0efd8", "size": 20127, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Project/report/LaTeX/Algorithms.tex", "max_stars_repo_name": "marhcouto/FEUP-CAL", "max_stars_repo_head_hexsha": "e02775eb69f0b5fc268c80084e85b5f18177dd78", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Project/report/LaTeX/Algorithms.tex", "max_issues_repo_name": "marhcouto/FEUP-CAL", "max_issues_repo_head_hexsha": "e02775eb69f0b5fc268c80084e85b5f18177dd78", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Project/report/LaTeX/Algorithms.tex", "max_forks_repo_name": "marhcouto/FEUP-CAL", "max_forks_repo_head_hexsha": "e02775eb69f0b5fc268c80084e85b5f18177dd78", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.8821428571, "max_line_length": 799, "alphanum_fraction": 0.7712525463, "num_tokens": 4756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085708384736, "lm_q2_score": 0.8221891305219504, "lm_q1q2_score": 0.6456721710491201}} {"text": "\\subsection{Node-tree manipulation and the decoder time complexity}\nIn the following we detail the routines to grow and merge node-trees. These completes the description of the decoder as it is analogous to the original Union-Find algorithm with the addition of these two steps.\n\nThe original Union-Find algorithm had a complexity driven by the time complexity of merging the union-find data structures representing the cluster-trees. This is known to be $\\mathcal{O}(n\\alpha(n)$, where $\\alpha(n)$ is the inverse Ackermann function, a sublogarithmic growing function. \n\nThe worst-case time complexity of the Union-Find Partitioned-Growth decoder of \\Cref{algo:ufbb} follows directly from the complexity of the growth and merge of the node trees. They are still quasilinear. In particular, in sections \\ref{} and \\ref{}, we describe respectively the routines for the growth and the merge of the node-trees and show that they have an amortized time-complexity $\\mathcal{O}(n\\log n)$. In consequence, the complexity of the total algorithm is driven by these two steps and the amortized time-complexity is $\\mathcal{O}(n\\log n)$.\n\n\\subsubsection{Growing node-trees}\\label{sec:grownodetrees}\n\nThe Union-Find decoder grows clusters by adding half-edges to all boundary vertices of the cluster. Instead, here we propose to grow only the nodes with minimal $PMW$. For this, it is necessary to save the list of boundary vertices to each node (\\Cref{fig:nodetypes}c). In this algorithm, to \\emph{grow a node} means to grow all edges connected to its boundary vertices a half-edge per the Union-Find decoder. The proposed growth algorithm consists of 3 depth-first-searches starting from a reference node $R$. Since the minimal PMW can be found without the PWM of the reference node $R$ itself, any node $N\\in\\nset$ may be chosen as the reference node $R$. \n\n\\begin{algorithm}[h]\n \\BlankLine\n \\KwData{Node tree $\\nset$}\n \\KwResult{$\\nset$}\n \\BlankLine\n \n Choose a reference node $R \\in \\nset$\\;\n Calculate node parities via tail-recursive DFS in $\\nset$ from $R$ via \\Cref{eq:snodeparity,eq:jnodeparity}\\;\n Calculate PMW with respect to $R$ via head-recursive DFS in $\\nset$ from $R$ via \\Cref{eq:pmwdif}\\; \n Grow all nodes with minimal PMW in a DFS in $\\nset$ from $R$\\;\n\n \\caption{Grow node-tree (naive)}\\label{algo:grow}\n\\end{algorithm}\n\nWhen two clusters merge on vertices $a$ and $b$, their respective node-trees must also be merged. This is performed by creating an edge between nodes to which vertices $a$ and $b$ belong to. \n\n\\textcolor{cyan}{\n It is possible to reduce the cost of \\Cref{algo:grow} by two implementations. First, the DFS's need not be performed during every growth iteration. For this, we introduce the \\emph{node delay} $\\delta_\\nset$, the difference in PMW with the minimal PMW in the node-tree.\n}\n\n\\begin{equation}\\label{eq:nodedelay}\n \\delta_N = \\text{PWM}(N) - \\min_{N\\in\\nset}{\\text{PWM}(N) }\n\\end{equation}\n\n\\textcolor{cyan}{\n With the node delay, the first two DFS's for the calculation of the parity and PMW need only to be performed during the first growth iteration after a merger, during which we now additionally sort the nodes with boundary vertices in the node-tree by the node delay. In the first growth iteration after a merger, we iterate over the sorted nodes and grow all nodes with delay 0, and add 1 to the delay, until a node with delay 1 is encountered. In the second iteration, we grow all nodes with delay 1, and so on. The number of operations for iterating through the sorted nodes is bounded by the number of growth operations of boundary edges, and its contribution to the complexity is thus defined within the original Union-Find decoder. The additional cost of growing a node-tree $\\nset$ while not merging is thus $\\abs{\\nset}$. \n}\n\n\\textcolor{cyan}{\n A second reduction in the cost is possible by a selection procedure for the reference node $R_{\\text{merged}}$ in the merged node-tree $\\nset_{\\text{merged}}$. Since $\\nset_{\\text{merged}}$ must be odd to grow, the node-trees involved its merger must consist of at least one odd node-tree $\\nset_O$ and a number of other node-trees. If the reference node $R_O$ of $\\nset_O$ is chosen for $R_{\\text{merged}}$ after the merger, the parity and delays within the sub-tree of nodes in $R_O$ do not change during the merger. Thus, the DFS's can be limited all other node-trees but $\\nset_O$. The greatest reduction in cost is achieved if the reference node of the largest odd node-tree in the merger is chosen as the new reference node. \n}\n\n\n\nAfter calculating the node delay, each subsequent growth iteration is performed by a single DFS. In this growth DFS, each node with zero delay is grown, and the delay of other nodes are subtracted by $\\nicefrac{1}{2}$. This procedure is effectively equivalent to growing the nodes with minimal PMW. If a merge with another node-tree occurs, the PMW's within the merged node-tree change, and a recalculation of the node parities and reference PMW's is required. The reduced cost algorithm is described in \\Cref{algo:growdelay}. \n\n\\begin{algorithm}[h]\n \\BlankLine\n \\KwData{Node tree $\\nset$, growth iteration $i$ after merge}\n \\KwResult{$\\nset$}\n \\BlankLine\n\n \\If(){$i=0$}{\n Choose a reference node $R \\in \\nset$.\\;\n Calculate node parities via tail-recursive DFS in $\\nset$ from $R$ via \\Cref{eq:snodeparity,eq:jnodeparity}.\\;\n Calculate PMW with respect to $R$ via head-recursive DFS in $\\nset$ from $R$ via \\Cref{eq:pmwdif}.\\; \n Calculate node delay $\\delta_N$ in $\\nset$ via \\Cref{eq:nodedelay}.\\;\n Sort nodes with boundary edges by their node delay in a list $\\mathcal{L}$.\\;\n }\n \\For(){node $N$ in $\\mathcal{L}$}{\n \\eIf(){$\\delta_N =i$}{\n Grow boundary edges of $N$. Add 1 to $\\delta_N$.\n }(){\n Break loop.\n }\n }\n \\caption{Grow node-tree (low-cost)}\\label{algo:growdelay}\n\\end{algorithm}\n\nNote that while the cost is reduced in \\Cref{algo:growdelay}, the complexity of the growth procedure is identical to \\Cref{algo:grow}, which is proportional to the size of the node-tree due to the growth DFS. However, \\Cref{algo:growdelay} allows us to solve a problem described in the next section. \n\n% An addition reduction in cost is possible by retaining the same root node through several mergers, such that the delay and PMW DFS's need only to be performed on a partition of the node-tree. However, these optimizations through implementation do not reduce the overall complexity of the algorithm, and will thus not be covered in this article. \n\n\n% Equations \\Cref{eq:pmwdif,eq:pseudodelay,eq:snodeparity,eq:jnodeparity} show that the differences in PMW with respect to the root node can be calculated through two depth-first searches (DFS) of $\\nset$. In the first DFS, the \\emph{parity DFS}, we calculate the node parity $\\pi_N$ a via tail-recursive function using \\Cref{eq:snodeparity,eq:jnodeparity}. In the second DFS, the \\emph{PMW DFS}, we calculate the PMW differences $\\delta_N$ via a head-recursive function using \\Cref{eq:pmwdif,eq:pseudodelay}. \n\n% \\todo[inline]{MOVE? or expand, growth DFS is not yet introduced}During this second DFS, the minimal PMW difference within $\\nset$ can be kept track of by comparison for calculating the node delay $\\delta_N$ at a later instance during the \\emph{growth DFS}. \n\n% The Union-Find decoder grows clusters by adding half-edges to all boundary vertices of the cluster. Instead, here we propose to grow only the nodes with minimal $PMW$. \n%A single growth iteration, which is applied in the Union-Find decoder by adding half-edges to all boundary vertices of the cluster, is now replaced by another DFS of $\\nset$ from the same root $R$ as before. \n%During this DFS, $\\delta_N$ of every node $N$ is calculated via \\Cref{eq:delaypseudo} and $N$ is conditionally grown --- adding half-edges to the boundary vertices in the current node and adding $\\nicefrac{1}{2}$ to its radius $\\rho_N$ --- if $\\delta_N = 0$. \n% For this, it is necessary to save the list of boundary vertices to each node (\\Cref{fig:nodetypes}c). %When the delays $\\delta_N$ for all nodes in $\\nset$ are zero, all nodes are grown simultaneously within the same iteration. \n\n% As the nodes grow, the values for $\\delta_N$ across $\\nset$ change. %This would mean that the DFS's of node parity and delay need to be done after each growth iteration. \n% Fortunately, if no new nodes are added to $\\nset$ after a growth iteration, the parities %$\\pi_N$ on iteration $t+\\theta$ \n% can be calculated via previous values of the PMW differences. %pseudo-delays in iteration $t$ by the introduction of the node \\textbf{wait} parameter $\\omega_N$. \n% During the growth DFS, if $\\delta_N \\neq 0$, add 1 to $\\omega_N$. If no nodes are added to $\\nset$, the node delay in the next iteration are\n% \\begin{multline}\\label{eq:delay}\n% \\delta_N(t+\\theta) = \\tilde{\\delta}_N(t) - \\min_{X \\in \\nset(t)}{\\tilde{\\delta}_X} - \\omega_N(t+\\theta-1), \\\\\n% \\hspace{.3cm} N \\in \\nset, \\hspace{.3cm} \\nset(t+\\theta) = \\nset(t). \n% \\end{multline}\n\n% Note that we have not stated which node in $\\nset$ should be the root node. In fact, any node in $\\nset$ could have been picked as the root of the node-tree. The only requirement is that the DFS of cluster growth must be performed along the same direction as the DFS's of the parity and delay calculations. If no cluster mergers occur, the node delays can be reused in the node suspension calculation prior to node growth. The node-tree is constructed by storing all neighbors of a node to a list. This way, the DFS's' direction can be determined by simply saving the root node, the starting point of the DFS's, to the cluster. All node variables are depicted in \\Cref{fig:nodetypes}c. \n% In the next section, we expand upon this idea of \"reusing\" some intermediate parameters to calculate the node suspensions after a cluster merger. \n\n% \\subsection{Joining node-trees}\\label{sec:nodejoin}\n\n% \\Figure[b](topskip=0pt, botskip=0pt, midskip=0pt){figures/tikz/build/main-figure5.pdf}{\n% \\emph{(a)} An odd cluster $\\nset_o=\\{A, B, O\\}$ with root $A$ joins with an even cluster $\\nset_e=\\{C, E\\}$ with root $C$ on nodes $O, E$, respectively, to a joined node-tree. If we choose to \\emph{(b)}, make $E$ a child of $O$, the parities and delays the subtree of $\\nset_o$ can unchanged, and we only have to perform partial parity and delay calculations over the subtree of $\\nset_e$. If we choose to \\emph{(c)}, make $O$ a child of $E$, parities and delays have to be recalculated in the entire joined node-tree. \\label{fig:inversion}}\n\n\\Figure[tb](topskip=0pt, botskip=0pt, midskip=0pt){figures/tikz/build/main-figure4.pdf}{\n The relevant data structures. \\emph{(a)} The cluster-tree of the Union-Find data structure, whose elements are vertices $v$. The path from a vertex to the root of the cluster-tree is traversed to find the root element, which differentiates clusters. Next to the cluster size and parity, the root node $R$ of the node-tree $\\nset$ is stored at the root of the cluster-tree. Each vertex $v$ additionally stores to which node $N$ it belongs. \\emph{(b)} A spanning-tree $\\vset$ with 3 non-trivial vertices. As $\\vset$ is strictly acyclic, the cluster's edges must be maintained such that no cycles are created. This is done during growth by removing edges (red dotted lines) if a cycle is detected. \\emph{(c)} The node-tree $\\nset$, which consists of the syndrome-nodes $\\dot{A}, \\dot{B}, \\dot{C}$ with primer vertices $a, b, c$, and junction-nodes $\\bar{D}, \\bar{I}$ with roots $d,i$. At each node $N$ and the root $R=\\dot{B}$ various variables are stored.\\label{fig:nodetypes}}\n\n% As clusters grow in size they merge with other clusters. \n% %In the Union-Find (UF) algorithm, odd clusters\n% %parity clusters of an odd number of non-trivial vertices, which are elements of $\\sigma$, \n% %grow %in size \n% %repeatedly and merge with other clusters until all clusters are even. \n% %The data structures of the merged cluster can be obtained by merging the data structures of the original clusters: %During these mergers, the node-trees and the cluster-trees are merged. %of the Partitioned-Growth data structure are combined. \n% %Let us first make a distinction between the merging protocols of the underlying data structures; \n% %the cluster-trees are merged with the $\\Union$ function, whereas \n% In the following we describe how to obtain the node-tree of the merged cluster from the original node-trees. % are merged with the $\\Nodejoin$ function which we describe in the following. %After a join of multiple node-trees, the node delays within the combined node-tree change. Therefore, the $\\Nodejoin$ protocol's focus is to minimize the DFS's of the recalculation of the node parity and delays in the combined node-tree. \n\n% %Let the parity $\\nset$ be the number of syndrome-nodes in $\\nset$, which is equivalent to the parity of the cluster of $\\nset$. \n\n% Since only odd clusters grow, only two types of clusters can merge: even with odd and odd with odd. \n\n% For even node-trees, the PMW is undefined since a perfect matching would not be possible when merging with a non-trivial vertex. Consequently, PMW differences are undefined for even node-trees and do not need to be calculated. \n\n% The second type of merger is between an even and an odd cluster. %The combined cluster is odd, and its growth is continued. Thus, its node delays must be computed. \n% In order to define the new node-tree, we need to choose a new root. We consider two different options, choosing the root of the odd cluster and choosing the root of the even cluster:\n\n% Let $\\nset_o,\\nset_e$ be the odd and even node-trees that are to be joined, and let $O\\in \\nset_o,E \\in \\nset_e$ be the nodes connected (\\Cref{fig:inversion}\\emph{a}).\n% If $\\nset_o$'s root becomes the root of the joined node-tree (\\Cref{fig:inversion}\\emph{b}), $E$ becomes a child node of $O$. \n% Since $\\nset_e$ contains an even number of syndrome-nodes, the node parities in $\\nset_o$ do not change. \n% Hence, the node parity DFS is only necessary in the subtree $\\nset_e$, which now has $E$ as subroot potentially different than the original root. \n% Furthermore, as the node PMW difference depends only on its own properties and its parent's, the node PMW DFS is also only required from node $E$ and within the subtree of $\\nset_e$. \n% Hence, only partial DFS's within the even cluster are required. % of the node-tree are precisely what was required, as the node parity and delays in $\\nset_e$ were undefined. \n\n% Alternatively, if $\\nset_e$'s root becomes the root of the combined tree (\\Cref{fig:inversion}\\emph{c}), an odd number of syndrome-nodes are attached to $E$. In consequence the parities of the nodes on the path from $E$ to the root are changed. This choice of root requires the DFS's on the entire combined node-tree to calculate for node parities and PMW differences. \n\n% For this reason, when merging node-trees we keep the root of the odd node-tree. We dub this strategy \\textbf{Odd-Rooted Join}. \n\n% In addition, a cluster can be subjected to multiple mergers within the same growth iteration, during which the parity of the merged cluster changes depending on the number of mergers and the parities of the clusters involved. The DFS's related to the parities and PMW differences should not be calculated directly after the joining of node-trees. %After all, it may be possible for the cluster to merge again such that the parities and delays become invalid. \n% \\textcolor{red}{To prevent redundant calculations, subroots of the even subtrees are stored in a list at the root of the node-tree (\\Cref{fig:nodetypes}\\emph{c}). When multiple mergers occur, the root node that stores the now redundant subroots is replaced by a new root with a new list of even subtrees subroots. If a cluster is selected for growth, we loop over the list of the subroots in the list at the new root node and initiate the DFS's from these subroots.} We call this the \\textbf{Root List Replacement}. \n", "meta": {"hexsha": "6fb568bcfbfe6653f8a1c605e91218b6e8f53f8e", "size": 16014, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sec_algo_d_grow.tex", "max_stars_repo_name": "watermarkhu/tqe_paper_ufbb", "max_stars_repo_head_hexsha": "f9b171049e028ace58be3ab4a01cddac94f7e01e", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sec_algo_d_grow.tex", "max_issues_repo_name": "watermarkhu/tqe_paper_ufbb", "max_issues_repo_head_hexsha": "f9b171049e028ace58be3ab4a01cddac94f7e01e", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sec_algo_d_grow.tex", "max_forks_repo_name": "watermarkhu/tqe_paper_ufbb", "max_forks_repo_head_hexsha": "f9b171049e028ace58be3ab4a01cddac94f7e01e", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-09-11T15:53:16.000Z", "max_forks_repo_forks_event_max_datetime": "2020-09-11T15:53:16.000Z", "avg_line_length": 116.8905109489, "max_line_length": 979, "alphanum_fraction": 0.7525290371, "num_tokens": 4140, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519527982093666, "lm_q2_score": 0.7577943712746406, "lm_q1q2_score": 0.6456050350747378}} {"text": "%!TEX root=report.tex\n\\subsection{Time Series Analysis}\n\nIn time series analysis one attempts to find a model for predicting future observations\n(this is different form the OLS analysis which tries to estimate the trends in the data, such as velocity and acceleration).\n Once such a model is found, it should be validated by analyzing the residuals. These residuals must be approximately white noise distributed \\cite[p.~130]{time-series-analysis}, otherwise the model is invalid. \n\n\\subsubsection{ARIMA}\n\nARIMA is the name of the stochastic model used. The theory is very complex so it will not be discussed in detail here, instead we refer to \\cite[p.~130]{time-series-analysis}.\n\nIn short ARIMA uses the previous values with some weight to predict future values. Which values are used is denoted by the following notation:\n\\begin{equation}\n(p, d, q) \\times (P, D, Q)_s\n\\end{equation}\n\n$p$ is the highest lags of actual measurements used and $q$ is the maximum amount of lags in residuals used. The $d$ part indicates how many difference operators which should be used to transform the data. $(P, D, Q)$ is completely similar, but steps not by one lag but by $s$ lags ($s=$season length). This allows for seasons trends such as an yearly pattern.\n\n\\subsubsection{AutoCorrelation function (ACF) and Partial AutoCorrelation Function (PACF)}\n \nACF and PACF are measures of correlation between different time lags and are used to make a qualified guess about how the prediction model should look like. \nFor more detail on how these are estimatied, please see \\cite[p.~146]{time-series-analysis}.\n\nWhen ACF and PACF have been estimated, rules \\cite[table~6.1]{time-series-analysis} for how the stochastic model should look can be applied.\nTypically when dealing with complex models, this becomes an iterative and subjective process - different analysts might reach different models.\nThe iterative step consist of estemating parameters, then calculate ACF and PACF, interpet them, estimate parameters, ... , and so it continues until the ACF and PACF suggest that no more lag terms(weights) are needed.\n\n\\subsection{Ljung-Box test}\n\nThe null hypothesis for the Ljung-Box is ``The data is independently distributed''.\nIn the ARIMA case it is applied to model residuals; so in other words it tests whether residuals at different lags have any correlation with each other. \n Thus a low p-value means that the residuals aren't white noise.\n\nThe Ljung-Box is a $\\chi^2$ test with the statistical value:\n\\begin{equation}\nL = n \\cdot (n + 2) \\sum_{k=1}^h \\frac{\\hat{\\rho}_k^2}{n - k}\n\\end{equation}\n\n$n$ is the number of observations. The parameter $h$ is the highest lag in the ACF which should be considered. The $\\chi^2$ statistics, have $h - (p + q + P + Q)$ degrees of freedom.\n", "meta": {"hexsha": "bf76b7296646e1181f1d965876adacfcb26c9f22", "size": 2772, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Rapport/theory-ts.tex", "max_stars_repo_name": "AndreasMadsen/grace", "max_stars_repo_head_hexsha": "bf472d30a2fac76145d3f68e819c92da4a1970ba", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2016-05-17T22:52:19.000Z", "max_stars_repo_stars_event_max_datetime": "2016-05-17T22:52:19.000Z", "max_issues_repo_path": "Rapport/theory-ts.tex", "max_issues_repo_name": "AndreasMadsen/grace", "max_issues_repo_head_hexsha": "bf472d30a2fac76145d3f68e819c92da4a1970ba", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Rapport/theory-ts.tex", "max_forks_repo_name": "AndreasMadsen/grace", "max_forks_repo_head_hexsha": "bf472d30a2fac76145d3f68e819c92da4a1970ba", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.3, "max_line_length": 360, "alphanum_fraction": 0.7709235209, "num_tokens": 666, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.851952809486198, "lm_q2_score": 0.7577943603346811, "lm_q1q2_score": 0.6456050342999278}} {"text": "\\documentclass[10pt,english]{article}\n\n%\\usepackage[T1]{fontenc}\n%\\usepackage[utf8]{inputenc}\n\\usepackage[a4paper,left=0.4in, right=0.4in,top=0.5in,bottom=0.6in]{geometry}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{subcaption}\n\\usepackage{epstopdf}\n\\usepackage{graphicx}\n\\usepackage{mathrsfs}\n\\usepackage{epstopdf}\n\\usepackage{mathtools}\n\\usepackage{textcomp}\n\\usepackage{graphicx}\n\\usepackage{xcolor}\n\\usepackage{epstopdf}\n%\\usepackage{multicol}\n%\\usepackage{tikz}\n\\usepackage{hyperref}\n\n% code listing settings\n\\usepackage{listings}\n\\lstset{\n language=Python,\n basicstyle=\\ttfamily\\small,\n aboveskip={1.0\\baselineskip},\n belowskip={1.0\\baselineskip},\n columns=fixed,\n extendedchars=true,\n breaklines=true,\n tabsize=4,\n prebreak=\\raisebox{0ex}[0ex][0ex]{\\ensuremath{\\hookleftarrow}},\n frame=lines,\n showtabs=false,\n showspaces=false,\n showstringspaces=false,\n keywordstyle=\\color[rgb]{0.627,0.126,0.941},\n commentstyle=\\color[rgb]{0.133,0.545,0.133},\n stringstyle=\\color[rgb]{01,0,0},\n numbers=left,\n numberstyle=\\small,\n stepnumber=1,\n numbersep=10pt,\n captionpos=t,\n escapeinside={\\%*}{*)}\n}\n\\author{Steven Porretta \\\\ Student \\# 100756494}\n\\title{Assignment \\#4}\n\\date{}\n\\begin{document}\n \\maketitle\n Code can be found at \\url{https://github.com/0xSteve/learning_automata_simulator}\n \\section{Question 1}\n In this section we will examine some code snippets from the first question.\n \\begin{lstlisting}[label={list:first},caption=Testbench code for the Tsetlin.] \n c2 = 0.7\n c1 = 0.05\n for i in range(0, 7):\n print(\"c1 = \" + str(c1) + \", c2 = \" + str(c2) + \", N = 13.\")\n a = la.Tsetlin(13, 2, [c1, c2])\n a.simulate(50, 30001)\n b = ala.Tsetlin.stationary_probability_analytic([c1, c2], 13)\n c = ala.Tsetlin.number_of_states_estimate([c1, c2])\n print(\"Tsetlin P1(infinity) = \" + str(b) + \"(Analytic)\")\n print(\"Tsetlin P1(infinity) = \" + str(a.action_average[0]) + \"(Simulated)\")\n print(\"Tsetlin # of states required = \" + str(c) + \"(Estimate)\")\n c1 += 0.1\n c1 = round(c1, 2)\n\\end{lstlisting}\nThis excerpt of code generates the entire quantity of required code for this question. As can be seen in the following code-snippet.\n\n\\begin{lstlisting}[label={list:first},caption=Testbench output.] \nc1 = 0.05, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.9999999890725503(Analytic)\nTsetlin P1(infinity) = 1.0(Simulated)\nTsetlin # of states required = 3(Estimate)\nc1 = 0.15, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.9999778874501014(Analytic)\nTsetlin P1(infinity) = 1.0(Simulated)\nTsetlin # of states required = 4(Estimate)\nc1 = 0.25, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.9990142374252533(Analytic)\nTsetlin P1(infinity) = 0.999998000067(Simulated)\nTsetlin # of states required = 6(Estimate)\nc1 = 0.35, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.9865794150962881(Analytic)\nTsetlin P1(infinity) = 0.999797340089(Simulated)\nTsetlin # of states required = 9(Estimate)\nc1 = 0.45, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.9151468144874294(Analytic)\nTsetlin P1(infinity) = 0.980783973868(Simulated)\nTsetlin # of states required = 18(Estimate)\nc1 = 0.55, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.7453193640776348(Analytic)\nTsetlin P1(infinity) = 0.786514449518(Simulated)\nTsetlin # of states required = 0(Estimate)\nc1 = 0.65, c2 = 0.7, N = 13.\nTsetlin P1(infinity) = 0.5680008401435711(Analytic)\nTsetlin P1(infinity) = 0.570874970834(Simulated)\nTsetlin # of states required = 0(Estimate)\n\\end{lstlisting}\nNow that it is seen working as one would expect, considering rounding errors from python 3.6, it is time to take a look at the useful snippets of code governing the functionality of the Tsetlin, and Krylov automata.\n\n\\begin{lstlisting}[label={list:first},caption=Tsetlin core code.] \n def next_state_on_reward(self):\n '''Find the next state of the learner, given that the teacher\n rewarded.'''\n if (self.current_state mod (self.N / self.R) != 1):\n self.current_state -= 1\n\n def next_state_on_penalty(self):\n '''Find the next state of the learner, given that the teacher\n penalized.'''\n if(self.current_state mod (self.N / self.R) != 0):\n self.current_state += 1\n elif(self.current_state mod (self.N / self.R) == 0):\n # Don't really add states, just cycle through N, 2N, 4N, etc.\n if(self.current_state != self.N):\n a = (self.N / self.R) mod self.N\n self.current_state = a + self.current_state\n else:\n self.current_state = self.N / self.R\n\n # Determine the next state as the teacher.\n def environment_response(self):\n '''Determine the next state of the learner from the perspective\n of the teacher.'''\n response = uniform(0, 1)\n penalty_index = 1\n if(self.current_state <= self.n):\n self.actions[0] += 1\n penalty_index = 0\n else:\n self.actions[1] += 1\n\n if(response > self.c[penalty_index]):\n # Reward.\n self.next_state_on_reward()\n else:\n # Penalty.\n self.next_state_on_penalty()\n\\end{lstlisting}\nThe above is the core code of the Tsetlin machine, governing state translations and action choices. Essentially, whenever it is in the range 1 to N, it chooses action $\\alpha_1$ and $\\alpha_2$ otherwise. This code is essentially the same for the Krylov machine, which we will see in the next section.\n\n\\section{Question 2}\n\\begin{lstlisting}[label={list:first},caption=testbench for the Krylov 2-action.] \n for i in range(0, 7):\n print(\"c1 = \" + str(c1) + \", c2 = \" + str(c2) + \", N = 13.\")\n a = la.Tsetlin(13, 2, [c1/2, c2/2])\n a.simulate(50, 30001)\n b = ala.Tsetlin.stationary_probability_analytic([c1, c2], 13)\n c = ala.Tsetlin.number_of_states_estimate([c1, c2])\n d = la.Krylov(13, 2, [c1, c2])\n d.simulate(10, 50000)\n e = ala.Tsetlin.stationary_probability_analytic([c1, c2], 13)\n f = ala.Tsetlin.number_of_states_estimate([c1, c2])\n print(\"Tsetlin P1(infinity) = \" + str(b) + \"(Analytic)\")\n print(\"Tsetlin P1(infinity) = \" + str(a.action_average[0]) + \"(Simulated)\")\n print(\"Tsetlin # of states required = \" + str(c) + \"(Estimate)\")\n print(\"Krylov P1(infinity) = \" + str(e) + \"(Analytic)\")\n print(\"Krylov P1(infinity) = \" + str(d.action_average[0]) + \"(Simulated)\")\n print(\"Krylov # of states required = \" + str(f) + \"(Estimate)\")\n c1 += 0.1\n c1 = round(c1, 2)\n\\end{lstlisting}\nAs can be seen from the code, this test bench looks very similar to the test bench of question 1, however, note the $c$ vector for the Tsetlin automaton is now $c_1 / 2 $, $c_2/2$. As is expected, both automata behave in the same manner. as can be seen from the output code snippet.\n\\begin{lstlisting}[label={list:first},caption=testbench output for the Krylov 2-action.] \n c1 = 0.05, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.9999999890725503(Analytic)\n Tsetlin P1(infinity) = 1.0(Simulated)\n Tsetlin # of states required = 3(Estimate)\n Krylov P1(infinity) = 0.9999999890725503(Analytic)\n Krylov P1(infinity) = 1.0(Simulated)\n Krylov # of states required = 3(Estimate)\n c1 = 0.15, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.9999778874501014(Analytic)\n Tsetlin P1(infinity) = 0.999999333356(Simulated)\n Tsetlin # of states required = 4(Estimate)\n Krylov P1(infinity) = 0.9999778874501014(Analytic)\n Krylov P1(infinity) = 1.0(Simulated)\n Krylov # of states required = 4(Estimate)\n c1 = 0.25, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.9990142374252533(Analytic)\n Tsetlin P1(infinity) = 1.0(Simulated)\n Tsetlin # of states required = 6(Estimate)\n Krylov P1(infinity) = 0.9990142374252533(Analytic)\n Krylov P1(infinity) = 1.0(Simulated)\n Krylov # of states required = 6(Estimate)\n c1 = 0.35, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.9865794150962881(Analytic)\n Tsetlin P1(infinity) = 1.0(Simulated)\n Tsetlin # of states required = 9(Estimate)\n Krylov P1(infinity) = 0.9865794150962881(Analytic)\n Krylov P1(infinity) = 1.0(Simulated)\n Krylov # of states required = 9(Estimate)\n c1 = 0.45, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.9151468144874294(Analytic)\n Tsetlin P1(infinity) = 1.0(Simulated)\n Tsetlin # of states required = 18(Estimate)\n Krylov P1(infinity) = 0.9151468144874294(Analytic)\n Krylov P1(infinity) = 1.0(Simulated)\n Krylov # of states required = 18(Estimate)\n c1 = 0.55, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.7453193640776348(Analytic)\n Tsetlin P1(infinity) = 0.983473884204(Simulated)\n Tsetlin # of states required = 0(Estimate)\n Krylov P1(infinity) = 0.7453193640776348(Analytic)\n Krylov P1(infinity) = 0.999998(Simulated)\n Krylov # of states required = 0(Estimate)\n c1 = 0.65, c2 = 0.7, N = 13.\n Tsetlin P1(infinity) = 0.5680008401435711(Analytic)\n Tsetlin P1(infinity) = 0.670734975501(Simulated)\n Tsetlin # of states required = 0(Estimate)\n Krylov P1(infinity) = 0.5680008401435711(Analytic)\n Krylov P1(infinity) = 0.870114(Simulated)\n Krylov # of states required = 0(Estimate)\n\\end{lstlisting}\nObserving the code for the Krylov machine, one notices that most of the code is inherited from the Tsetlin machine, except the state translations. It is incredible, that the only major distinction is that a penalty is treated as a penalty with 50\\% probability and a success otherwise. Literally, all other code for the Krylov machine is inherited from the Tsetlin.\n\\clearpage\n\\begin{lstlisting}[label={list:first},caption=Krylov core code.] \ndef next_state_on_penalty(self):\n '''Find the next state of the learner, given that the teacher\n penalized.'''\n\n # If this number is greater than 0.5, then penalize the learner.\n is_penalty = uniform(0, 1)\n\n if(is_penalty >= 0.5):\n Tsetlin.next_state_on_penalty(self)\n else:\n Tsetlin.next_state_on_reward(self)\n \n\\end{lstlisting}\n\n\\section{Question 3} \n\nFirst let us consider state changes in the $L_{R-I}$, since there really are no states, but instead just an interval, $\\lbrace0, 1\\rbrace$, of possibilities. Consider the following code-snippet.\n\n\\begin{lstlisting}[label={list:first},caption=State Translation in the $L_{R-I}$ automaton.] \n def do_reward(self, action):\n if(action == 2):\n self.p1 = self.k_r * self.p1\n else:\n self.p1 = 1 - (self.k_r * self.p2)\n self.p2 = 1 - self.p1\n\n def do_penalty(self):\n pass \n\\end{lstlisting}\nIt can be seen from the above code-snippet, that when the environment rewards, an action is updated based on the action selected. When the environment issues a penalty, nothing happens. The python command \\textit{pass} is command that simply does nothing, and is usually used for prototyping. In this case, pass is included to explicitly show that a penalty does nothing.\n\nFrom this automaton some interesting things can be observed. First, let us consider the output of the testbench.py file. Time complexity has been measured in terms of discrete steps instead of actual time spent in the processor. The physical time, does not represent the number of actions being computed, as many operations can be processed simulatneously, in the background of a system, the time is not accurate. A corolary to this event is that the number of discrete calls to the program, accurately represent the time complexity, when each action is considered a time unit. To better understand this, observe the following code snippet.\n\\begin{lstlisting}[label={list:first},caption=State Translation in the $L_{R-I}$ automaton.] \n=============================================================\nThe optimal K_r value is: 0.74995\nThe optimal lambda_r value is: 0.25005\nThe accuracy for k_r = 0.74995 is: 0.963\nThe computation time in iterations is: 18\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.7811937499999999\nThe optimal lambda_r value is: 0.2188062500000001\nThe accuracy for k_r = 0.7811937499999999 is: 0.958\nThe computation time in iterations is: 26\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.7811937499999999\nThe optimal lambda_r value is: 0.2188062500000001\nThe accuracy for k_r = 0.7811937499999999 is: 0.965\nThe computation time in iterations is: 16\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.8124374999999999\nThe optimal lambda_r value is: 0.18756250000000008\nThe accuracy for k_r = 0.8124374999999999 is: 0.953\nThe computation time in iterations is: 45\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.874925\nThe optimal lambda_r value is: 0.12507500000000005\nThe accuracy for k_r = 0.874925 is: 0.976\nThe computation time in iterations is: 57\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.90616875\nThe optimal lambda_r value is: 0.09383125000000003\nThe accuracy for k_r = 0.90616875 is: 0.969\nThe computation time in iterations is: 160\n=============================================================\n=============================================================\nThe optimal K_r value is: 0.96865625 \nThe optimal lambda_r value is: 0.031343750000000004 \nThe accuracy for k_r = 0.96865625 is: 0.977 \nThe computation time in iterations is: 1298 \n============================================================= \n\\end{lstlisting}\nIt is observable that as $c1$ approaches $c2$ the number of operations increase exponentially. It is also observable that as $n \\to \\infty$, then $\\lambda_R \\to 0$. Corolary: $c_1 \\to c_2$, then $\\lambda_R \\to 0$ or $k_R \\to \\infty$. To get a better understanding of this effect, observe the following graph, Figure \\ref{blah}. It can be seen that $\\lambda_R$ approaches $0$ as the time complexity goes to infinity. This is representative of the difficulty of the system. When the percent difference between penalties of choosing action 1 and action 2 approaches 0, then the time complexity approaches infinity, and the $\\lambda_R$ required becomes an undesirably low value.\n\\begin{figure}[h!]\n \\centering\n \\includegraphics{plot.png}\n \\caption{$\\lambda_R$ Vs. $n$} \\label{blah}\n\\end{figure}\n\\end{document}", "meta": {"hexsha": "5505407fe07f3f45c1b95d0df71192f4911a6b03", "size": 15125, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "writeup.tex", "max_stars_repo_name": "0xSteve/learning_automata_simulator", "max_stars_repo_head_hexsha": "ca60bb3a2d3ecb7d345f9a0fcccb53a2e2d7856d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "writeup.tex", "max_issues_repo_name": "0xSteve/learning_automata_simulator", "max_issues_repo_head_hexsha": "ca60bb3a2d3ecb7d345f9a0fcccb53a2e2d7856d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "writeup.tex", "max_forks_repo_name": "0xSteve/learning_automata_simulator", "max_forks_repo_head_hexsha": "ca60bb3a2d3ecb7d345f9a0fcccb53a2e2d7856d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.7532894737, "max_line_length": 678, "alphanum_fraction": 0.6373553719, "num_tokens": 4520, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.757794360334681, "lm_q2_score": 0.8519528076067262, "lm_q1q2_score": 0.6456050328756746}} {"text": "\nAs well as in logistic regression as in neural networks, minimizing(or maximizing depending on your setup) the cost function is a central problem, and can to some extent be stated as \\textit{the} problem. Having a method which is both efficient and converges towards the minimum is, luckily, not a new problem in computer science. In the logistic regression optimization, three main methods will be utilized, while in the neural network we will focus on the stochastic conjugate gradient descent. For logistic regression we will use a learning rate optimized gradient descent method, and \n\n\nGradient descent measures the local gradient of the cost function, with regards to the weights $\\bm{w}$(or $\\beta$. Since the gradient goes in the direction of fastest increase, we will go in the opposite direction, i.e. negative gradient. We start by choosing random values for $\\bm{w}$ (since our cost function is convex any choice should give correct results), calculate the gradient, update the $\\bm{w}$ values, and do this iteratively until the algorithm converges to a minimum. The size of the steps is important, and is determined by the learning rate. If the learning rate is too small, we will need many iterations which is time consuming. However, if the learning rate is too high, we might overshoot and miss the minimum. One way to choose the learning rate is too let it depend on the size of the gradient. If the gradient is large, i.e a steep slope, the learning rate can be relatively high. When the gradient is small, the learning rate is also small.\n\n\\subsubsection{Gradient descent}\nGradient descent is set up in a general fashion as, utilizing an optimized learning rate $\\eta_k$ find and a scaling parameter $\\gamma$. The scaling parameter $\\gamma$ is introduced in order to prevent results from blowing up.\n\\begin{algorithm}[H]\n \\caption{Gradient descent.}\n \\label{alg:gd}\n \\begin{algorithmic}[1]\n \\State Input: $\\bm{X}$, $\\bm{y}$, $\\bm{w}_0$, $\\eta$\n \\State Set initial weights, $\\bm{w} = \\bm{w}_0$\n \\State Set previous gradient with current, $\\nabla_0 = \\nabla_1$\n \\While{$i < N_\\mathrm{max}$ and $||\\nabla\\mathcal{C}(\\bm{w})||<\\varepsilon$}\n \\State $z = \\bm{X}\\cdot \\bm{w}$\n \\State $p = \\sigma(z)$\n \\State $\\nabla_0 = \\nabla_1$, set previous gradient to current.\n \\State $\\nabla_1 = - \\bm{X}^T \\cdot (\\bm{y} - \\bm{p}) / \\gamma + \\lambda f_L(\\bm{w}) / \\gamma$\n \\State $\\bm{w}_0 = \\bm{w}$\n \\State Update learning parameter, $\\eta_k$\n \\State $\\bm{w} = \\bm{w}_0 - \\eta_k \\nabla_1$, update weights.\n \\EndWhile\n \\State Return $\\bm{w}$\n \\end{algorithmic}\n\\end{algorithm}\nThe full implementation with the optimized learning rate $\\eta_k$ can be seen in the article \\citet{optimal-learning-rate}.\n\n\\subsubsection{Stochastic gradient descent}\nStochastic gradient descent is similar to gradient descent, expect that we first randomly shuffle our data, then divide our data into mini batches, $N_\\mathrm{mb}$. Then we run gradient descent on each of the mini batches, and take the average of the outputted gradients as our gradient descent step. We repeat this $N_\\mathrm{epochs}$.\n\\begin{algorithm}[H]\n \\caption{Stochastic gradient descent(SGD).}\n \\label{alg:sgd}\n \\begin{algorithmic}[1]\n \\For $i_e$ in $N_\\mathrm{epochs}$ epochs\n \\State Shuffle data $\\bm{X}_\\mathrm{train}$, $\\bm{y}_\\mathrm{train}$\n \\State Split into mini batches, $\\bm{X}_\\mathrm{train}$, $\\bm{y}_\\mathrm{train}$\n \\For $i_\\mathrm{mb}$ in mini batches,\n \\State Perform gradient descent step, and retrieve $\\nabla\\bm{w}_{i_\\mathrm{mb}}$\n \\EndFor\n \\State Take the average of the $\\nabla\\bm{w}_{i_\\mathrm{mb}}$ and update the weight matrix $\\bm{w}$.\n \\EndFor\n \\end{algorithmic}\n\\end{algorithm}\n\n\n\n% To minimize this cost function we use gradient descent. \\husk{Gradient descent measures the local gradient of the cost function, with regards to $\\beta$ in our case. Since the gradient goes in the direction of fastest increase, we will go in the opposite direction, i.e. negative gradient. We start by choosing random values for $\\beta$ (since our cost function is convex any choice should give correct results), calculate the gradient, update the $\\beta$ values, and do this iteratively until the algorithm converges to a minimum. The size of the steps is important, and is determined by the learning rate. If the learning rate is too small, we will need many iterations which is time consuming. However, if the learning rate is too high, we might overshoot and miss the minimum. One way to choose the learning rate is too let it depend on the size of the gradient. If the gradient is large, i.e a steep slope, the learning rate can be relatively high. When the gradient is small, the learning rate is also small.}\n\n% \\sjekk{Returning to the logistic regression problem, the derivative of the cost function is\n% \\begin{equation}\\label{eqT:diff_cost}\n% \\begin{split}\n% \\frac{\\partial J(\\beta)}{\\partial \\beta} &=-\\frac{1}{N}\\bm{X}^T(\\bm{y}-\\bm{p})\\\\\n% &= \\frac{1}{N}\\bm{X}^T(\\bm{p}-\\bm{y}),\n% \\end{split}\n% \\end{equation}\n% where $\\bm{X}$ is the $N\\times(p+1)$ matrix of $x_i$ values, $\\bm{p}$ is the vector of fitted probabilities with $i$th element $p(x_i;\\beta)$ and $\\bm{y}$ is the vector of $y_i$ values. The new $\\beta$ using gradient descent is then\\\\\n% $\\beta_{new} = \\beta_{old}\\ -\\ \\frac{\\partial J(\\beta)}{\\partial \\beta}lr$, where $lr$ is the learning rate (step size).} \\husk{This is done iteratively until we reach the set max iterations or $\\frac{\\partial J(\\beta)}{\\partial \\beta}$ is within a given tolerance of zero.}\n\n% \\sjekk{Like we introduced Lasso and Ridge regression to avoid overfitting in Project 1, we can add a penalty term to the cost function in Equation \\ref{eq:Geron_cost}. In our project we used two different penalties: $L1 = \\lambda|\\beta|$ and $L2 = \\lambda||\\beta||^2$. When fitting the model we need to include the derivatives of the penalty term in Equation \\ref{eqT:diff_cost}.} \\husk{The gradient with the penalty term is,\n% \\begin{equation}\n% \\begin{split}\n% \\frac{\\partial J(\\beta)}{\\partial \\beta} &=\\frac{1}{N}\\bm{X}^T(\\bm{p}-\\bm{y})\\ + \\lambda\\cdot{sign(\\beta)},\\ for\\ L1\\ regularization\\\\\n% &\\mathrm{or}\\\\\n% \\frac{\\partial J(\\beta)}{\\partial \\beta} &=\\frac{1}{N}\\bm{X}^T(\\bm{p}-\\bm{y})\\ + \\lambda\\cdot{2\\beta},\\ for\\ L2\\ regularization. \n% \\end{split}\n% \\end{equation}\n% }\n", "meta": {"hexsha": "879cf707525f53f3de56117330b7336f681714a5", "size": 6519, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/theory/optimization.tex", "max_stars_repo_name": "hmvege/FYSSTK4155-Project2", "max_stars_repo_head_hexsha": "3cf617399f99026cbcd79f8153d3196ebd86c7cd", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/theory/optimization.tex", "max_issues_repo_name": "hmvege/FYSSTK4155-Project2", "max_issues_repo_head_hexsha": "3cf617399f99026cbcd79f8153d3196ebd86c7cd", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/theory/optimization.tex", "max_forks_repo_name": "hmvege/FYSSTK4155-Project2", "max_forks_repo_head_hexsha": "3cf617399f99026cbcd79f8153d3196ebd86c7cd", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 93.1285714286, "max_line_length": 1017, "alphanum_fraction": 0.7050161068, "num_tokens": 1780, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519527869325346, "lm_q2_score": 0.7577943712746406, "lm_q1q2_score": 0.6456050265292179}} {"text": "\\subsection{Kessler Parameterization}\nSCALE implements a one-moment bulk microphysical scheme, which treats only warm clouds (cloud and rain). This scheme predicts the mixing ratio of cloud ($Q_{cloud}$) and rain ($Q_{rain}$). Cloud microphysical processes treated in this scheme are saturation adjustment (corresponding to nucleation, evaporation, and cloud condensation), evaporation, auto-conversion, accretion, and sedimentation. The tendency of $Q_{cloud}$, $Q_{rain}$, and $Q_{v}$ (vapor mixing ratio) is as follows:\n\n\\begin{eqnarray}\n\\frac{\\partial Q_{cloud}}{\\partial t}&=&dQ|_{sat}-dQ|_{auto}-dQ|_{acc}\\\\\n\\frac{\\partial Q_{rain}}{\\partial t}&=&dQ|_{auto}+dQ|_{acc}-dQ|_{evap}-F_{Q_{r}}|_{sed}\\\\\n\\frac{\\partial Q_{v}}{\\partial t}&=&dQ|_{evap}-dQ|_{sat}\n\\end{eqnarray}\n\nwhere $dQ|_{sat}$, $dQ|_{auto}$, $dQ|_{acc}$, and $dQ|_{evap}$ represent the mixing ratio tendency by saturation adjustment, auto-conversion, accretion, and evaporation, respectively. $F_{Q_{r}}|_{sed}$ represents flux of $Q_{r}$ by sedimentation.\\\\ $dQ|_{auto}$, $dQ|_{acc}$, and $dQ|_{evap}$ are given as:\n\n\\begin{eqnarray}\ndQ_{auto}&=&\\left\\{\n\\begin{array}{ll}\nQ_{cloud}*10^{-3} & (Q_{cloud}>10^{-3})\\\\\n0 & (else)\\\\\n\\end{array}\\right.\\\\\ndQ_{acc}&=&2.2\\times Q_{cloud}\\times Q_{rain}^{0.875}\\\\\ndQ_{evap}&=&\\left\\{\n\\begin{array}{ll}\nf_{vent} \\frac{q_{s}-Q_{cloud}}{q_{s}\\rho}\\frac{(\\rho*Q_{rain})^{0.525}}{5.4\\times 10^{5}+\\frac{2.55\\times 10^{8}}{pq_{s}}} & ( q_{s} > Q_{cloud})\\\\\n0 & (else)\\\\\n\\end{array} \\right.\n\\end{eqnarray}\n\nwhere $f_{vent}$ is the ventilation factor ($f_{vent}=1.6+124.9(\\rho Q_{rain})^{0.2046}$), and the unit of $dQ_{***}$ is [kg/kg/s]. $p$, $q_{s}$, and $\\rho$ are pressure, saturation vapor mixing ratio, and total density, respectively.\\\\\n$dQ|_{sat}$ is given as:\n\n\\begin{eqnarray}\ndQ|_{sat}=Q_{v}-q_{s}.\n\\end{eqnarray}\n\nTerminal velocities of cloud ($V_{t,c}$) and rain ($V_{t,r}$) are given as:\n\n\\begin{eqnarray}\nV_{t,c}&=&0\\\\\nV_{t,r}&=&36.34(\\rho Q_{rain})^{0.1364} [m/s]\n\\end{eqnarray}\n\n\n", "meta": {"hexsha": "5c8249666e2a2a084b7534261115e2ad414c13ab", "size": 1999, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/descriptions/microphysics_kessler.tex", "max_stars_repo_name": "slayoo/scale", "max_stars_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2020-06-14T11:12:31.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T05:29:55.000Z", "max_issues_repo_path": "doc/descriptions/microphysics_kessler.tex", "max_issues_repo_name": "slayoo/scale", "max_issues_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-07-29T03:38:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-30T05:08:47.000Z", "max_forks_repo_path": "doc/descriptions/microphysics_kessler.tex", "max_forks_repo_name": "slayoo/scale", "max_forks_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-07-10T10:39:45.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-28T22:20:41.000Z", "avg_line_length": 48.756097561, "max_line_length": 484, "alphanum_fraction": 0.6668334167, "num_tokens": 744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8757869948899665, "lm_q2_score": 0.7371581626286834, "lm_q1q2_score": 0.6455935320071838}} {"text": "\\documentclass{article}\r\n\\usepackage{mathtools}\r\n\\usepackage{amssymb}\r\n\\usepackage{amsmath}\r\n\\usepackage{amsthm}\r\n\r\n\\begin{document}\r\n\r\n\\newtheorem{theorem}{Theorem}[section]\r\n\\newtheorem{corollary}{Corollary}[theorem]\r\n\\newtheorem{lemma}[theorem]{Lemma}\r\n\r\n\\tableofcontents\r\n\\newpage\r\n\r\n\\section{Probability Space}\r\nThere are often various approaches to probability each with its own advantages\r\nand disadvantages.\r\n\r\nExperiment is a procedure that can be infinitely repeated and has a\r\nwell-defined set of possible outcomes, known as the sample space.\r\n\r\nThe observation/result of the experiment are termed as outcomes.\r\n\r\n\\subsection{Classical Approach}\r\n Probability of an event $E$ is defined to be: $$P(E)=\\frac{\\text{Number of outcomes in }E}{\\text{Total number of outcomes}}$$\r\n Some examples are tossing a coin or rolling a die.\r\n Disadvantages:\r\n \\begin{itemize}\r\n \\item Unable to model biases. It says nothing about cases where no\r\n physical symmetry exists.\r\n \\item Doesn't deal with cases where total outcomes are infinite.\r\n \\end{itemize}\r\n\r\n \\subsection{Frequentist Approach}\r\n Also known as the relative frequency approach or frequentism. It defines\r\n an event's probability as the limit of its relative frequency in many\r\n trials.\r\n\r\n Probability is defined to be:\r\n $$ P(E)=\\lim_{n \\to \\infty} \\frac{n_E}{n}$$\r\n where an experiment is conducted $n$ times and event $E$ occurs $n_E$\r\n times.\r\n Disadvantages:\r\n \\begin{itemize}\r\n \\item It isn't efficient to conduct an experiment multiple times\r\n just to find the probability of an event occuring.\r\n \\item It is unable to deal with subjective belief. Eg: Suppose a\r\n cricket expert says there is a $50\\%$ of RCB winning the IPL this\r\n year. It doesn't mean that the RCB has won half the titles in the\r\n past.\r\n \\end{itemize}\r\n%\r\n \\subsection{Axiomatic Approach}\r\n\r\n \\subsubsection{Probability Space}\r\n\r\nThe triple ($S$, $F$, $P$) is referred to as a probability space where:\r\n\\begin{itemize}\r\n \\item $S$ : Sample space, set of all possible outcomes of the experiment.\r\n \\item $F$ : Event Space, subset of the sample space\r\n \\item $P$ : Probability Measure\r\n\\end{itemize}\r\n\\subsubsection{Sample Space}\r\n\r\n$S$ can either be finite or countably infinite or uncountably infinite.\r\n%28/05\r\n\r\nExamples for S:\r\n\\begin{itemize}\r\n \\item Finite Sample Space: Single coin toss $ S=\\{H,T\\} $ and two coin tosses $ S=\\{HH,HT,TT,TH\\} $.\r\n \\item Countably Infinite Sample Space: Keep tossing a coin till you get a head $ S=\\{H,TH,TTH...\\}$.\r\n \\item Uncountably Infinite Sample Space: We have a circular dart board and we are measuring the angle at which a dart hits the board. $S=[0,2\\pi]$\r\n\\end{itemize}\r\n\r\n\\subsection{Event Space}\r\nCollection of events is called an event space, there are some properties to be satisified such as:\r\nit has to be a ``Sigma Field''.\r\n\r\n\\subsubsection{Sigma Field}\r\nA sigma field (or sigma algebra) F is a collection of subsets of S which satisfies the following properties:\r\n\\begin{itemize}\r\n \\item $S \\in F$\r\n \\item If $E \\in F$, then $E^C \\in F$\r\n \\item If $E_1,E_2,E_3 \\cdots \\in F $, then $\\bigcup_{i=1}^{\\infty} E_i \\in F$\r\n\\end{itemize}\r\n\r\n\\subsubsection{Examples for Event Space}\r\n\\begin{itemize}\r\n \\item Smallest possible event space:\r\n $$ F =\\{ \\phi,S\\} $$\r\n \\item Next non-trivial event space:\r\n $$ F=\\{ \\phi,E ,E^c ,S\\}$$\r\n \\item If $E_1 \\in F$ and $E_2 \\in F$, then $E_1 \\cap E_2 \\in F$ (Proof in 1.4.3)\r\n \\item For S=$\\{1,2,3,4,5,6\\}$ , $E_1 =\\{1,2\\}$ and $E_2=\\{3,4\\}$. The smallest event space containing $E_1$ and $E_2$ is:\r\n $$ F=\\{\\phi , S, E_1 , E_{1}^{C},E_2 , E_{2}^{C},E_1 \\cup E_2 , (E_{1} \\cup E_{2})^{C}\\}$$\r\n\\end{itemize}\r\n\r\n\\subsubsection{Proposition 1}\r\nA$_1$, A$_2$, A$_3$,....A$_n$ $\\in$ $F$, then $\\bigcap\\limits_{i=1}^n$A$_i$ $\\in$ $F$.\r\n\r\n\\begin{proof}\r\n If A$_1$, A$_2$, A$_3$,....A$_n$ $\\in$ $F$, then A$_1^c$, A$_2^c$, A$_3^c$,....A$_n^c$ $\\in$ $F$ and $\\bigcup\\limits_{i=1}^n$A$_i^c$ $\\in$ $F$. Then by property 2, $(\\bigcup\\limits_{i=1}^n$A$_i^c)^c$ = $\\bigcap\\limits_{i=1}^n$A$_i$ $\\in$ $F$.\r\n\\end{proof}\r\n\r\n\\subsubsection{Proposition 2}\r\nA, B $\\in$ $F$, then A$\\setminus$B = A - B $\\in$ $F$.\\medskip\r\n\r\n\\begin{proof}\r\n If B $\\in$ $F$, then B$^c$ $\\in$ $F$ by property 2. So, A $\\cap$ B$^c$ = A$\\setminus$B $\\in$ $F$ (As seen in proposition 1).\r\n\\end{proof}\r\n\\subsection{Probability Measure}\r\nThe probability measure P is a function returning an event's probability. A probability is a real number between zero and one.\r\n$$ P:F\\rightarrow [0,1]$$\r\n\r\nP has to satisfy the following 3 axioms:\r\n\\begin{itemize}\r\n \\item $P(E) \\geq 0$\r\n \\item $P(S)=1$\r\n \\item If $E_1,E_2 \\cdots \\in F$ such that $E_i \\cap E_j =\\phi$ then:\r\n $$ P(\\bigcup_{i=1}^{\\infty} E_{i})= \\sum_{i=1}^{\\infty} P(E_i)$$\r\n\\end{itemize}\r\n\r\nFor two disjoint sets:\r\n$P(E_1 \\cup E_2)= P(E_1)+P(E_2)+ \\sum P(\\phi)$\r\n\r\nWe will later see that $P(\\phi)$ is indeed 0.\r\n\r\n\\subsection{Derived Properties of Probability}\r\n\\begin{enumerate}\r\n \\item $$P(E^C)=1-P(E)$$\r\n \\begin{proof}\r\n $$E \\cup E^C = S$$\r\n $$P(E)+P(E^C)=1$$\r\n\r\n \\end{proof}\r\n \\item For any two events $E_1$ and $E_2$,\r\n $$ P(E_1 \\cup E_2)= P(E_1)+P(E_2)-P(E_1 \\cap E_2)$$\r\n \\begin{proof}\r\n $$ P(E_1 \\cup E_2)=P(E_1)+P(E_2 \\cap E_1^C)$$\r\n Now, $$ E_2=(E_2 \\cap E_1)\\cup (E_2 \\cap E_1^C)$$\r\n $$ \\Rightarrow P(E_2)=P(E_1 \\cap E_2)+ P(E_1^C \\cap E_2)$$\r\n Also,\r\n $$ P(E_1 \\cup E_2)= P(E_1)+P(E_1^C \\cap E_2)$$\r\n Substitute the required value in the final equation.\r\n \\end{proof}\r\n\\end{enumerate}\r\n\r\nQuestion:\r\n$ S=\\{1,2,3,4,5,6\\}$. 1 and 5 are equally likely and probability of getting a 6 is one-third.\r\n\r\nFind minimum and maximum probability that we get an even number.\r\n\r\nAnswer:$$ Minimum \\;Prob. = \\frac{1}{3} \\;\\; when \\;\\; P_2 = P_4 = 0$$\r\n$$ Maximum \\;Prob. = 1, \\;\\; P_2+P_4 =\\frac{2}{3} $$\r\n\r\n%31/05 notes\r\n\r\n\\section{Conditional Property}\r\n\r\nGiven that an event $A$ has occured.\r\n\r\n$ (S,F,P) \\rightarrow$ Original probability space\r\n\r\nIf additional info has been given that A has occured, probability space need to be suitably modified.\r\n\r\neg: $S=\\{1,2,3,4,5,6\\},\\;E_1=\\{1,2\\},\\;E_2=\\{3,4\\}$\r\n$$ F=\\{\\phi , S, E_1 , E_{1}^{C},E_2 , E_{2}^{C},E_1 \\cup E_2 , (E_{1} \\cup E_{2})^{C}\\}$$\r\nand event $A=E_1^c = \\{3,4,5,6\\}$ has occured.\r\n\r\nThen, $ F_A= \\{ \\phi, A, \\{3,4\\},\\{5,6\\} \\} $.(shown later)\r\n\r\n\r\n\\subsection{Modified probability space}\r\n\r\nThen,\r\n\\begin{itemize}\r\n \\item $S_A = A$. (Modified Sample Space)\r\n\r\n \\item $F_A = \\{ (E \\cap A) | E \\in F \\} \\rightarrow E \\cap A \\in F$. (Modified Event Space)\r\n\r\n (Also if some event $C \\cap A = \\phi $, then $C$ won't occur)\r\n\r\n\r\n\r\n To prove: $F_A$ also satisfies event space axioms.(see sec. 1.4.1)\r\n\r\n \\begin{enumerate}\r\n \\item $A \\in F_A \\rightarrow S\\cap A = A$(S was original sample space)\r\n $\\Rightarrow A \\in F_A$\r\n\r\n \\item $D \\in F_A \\Rightarrow D = E \\cap A, D^c \\in F_A$\r\n\r\n As,\r\n\r\n $D^c = A \\setminus D= E^c \\cap A \\in F \\quad(Because\\; E^c \\in F)$\r\n\r\n \\item $$D_1,D_2,\\cdots \\in F_A$$\r\n $$ (E_1 \\cap A), \\cdots \\in F_A$$\r\n $$ E_1, E_2 \\cdots \\in F$$\r\n $$ \\Rightarrow \\bigcup_{i=1}^{\\infty} E_i \\in F_A$$\r\n $$ \\Rightarrow (\\bigcup_{i=1}^{\\infty} E_i) \\cap A \\in F_A$$\r\n $$\\Rightarrow (\\bigcup_{i=1}^{\\infty} E_i \\cap A) \\in F_A \\Rightarrow \\bigcup_{i=1}^{\\infty} D_i \\in F_A$$\r\n\r\n \\end{enumerate}\r\n Hence, $F_A$ is an event space.\r\n\r\n \\item Modified probability measure\r\n\r\n $$ P(E/A)= \\frac{P(E \\cap A)}{P(A)}$$\r\n\r\n This definition is called conditional probability measure for any $E \\in F$.\r\n\r\n eg:\r\n $ F_A = \\{ \\phi, \\{3,4,5,6\\}, \\{ 3,4\\}, \\{5,6\\} \\}$, then $P(\\{ 3,4\\} / \\{3,4,5,6\\})=1/2$\r\n\r\n Now, we need to prove $ P(E/A)$ satisfies the 3 axioms of probability measure.(see sec. 1.5)\r\n \\begin{itemize}\r\n \\item $ P(E/A) \\geq 0$ (as ratio of two nos. which are positive)\r\n \\item $ P(S/A)=1$\r\n \\item $ B_1, B_2 \\cdots$ are all mutually disjoint.\r\n $$ P(\\bigcap_{i=1}^{\\infty} B_i /A) = \\frac{ P(\\bigcup_{i=1}^{\\infty} B_i \\cap A)}{P(A)}$$\r\n $$ =\\frac{\\sum_{i=1}^{\\infty} P(B_i \\cap A)}{P(A)} = \\sum_{i=1}^{\\infty} P(B_i /A)$$\r\n $$ \\Rightarrow P(\\bigcap_{i=1}^{\\infty} B_i /A) = \\sum_{i=1}^{\\infty} P(B_i /A)$$\r\n \\end{itemize}\r\n\\end{itemize}\r\n\r\n\r\n\r\n\r\n\r\n\\section{Total probability theorem}\r\n\r\nEvents $A_1,\\cdots, A_n \\in F$ which are all mutually exclusive/disjoint and exhaustive. Then,\r\n$$ A_i \\cap A_j = \\phi \\quad \\forall\\; i,j$$\r\n$$ \\bigcup_{i=1}^{n} A_i =S$$\r\n\r\n$$ P(B)= \\sum_{i=1}^{n} P(B/A_i)P(A_i)$$\r\n\r\nthe total probability theorem expresses $ P(B)$ in terms of conditional probability $ P(B/A_i) $ \\& prior probability $P(A_i)$\r\n\\begin{proof}\r\n\r\n $$ B= \\bigcup_{i=1}^{n} (B \\cap A_i)$$\r\n\r\n $A_i$'s are disjoint, so $(B \\cap A_i)$ are also disjoint.\r\n\r\n $$ P(B)= \\sum_{i=1}^{n} P(B \\cap A_i)=\\sum_{i=1}^{n} P(B / A_i)P(A_i)$$\r\n\r\n\r\n\\end{proof}\r\n\r\n\\subsection{Question}\r\nTwo factories manufacture zoggles. 20\\% of $F_1$ are defective. 5\\% of $F_2$ are defective.\r\n\r\nIn any week, $F_1$ produces twice the number of zoggles as $F_2$. What is the probability that a zoggle chosen randomly in a week is defective?\r\n\r\n$$ P(D)=P(F_1)P(D/F_1)+P(F_2)P(D/F_2)$$\r\n$$ = \\frac{2}{3} \\times \\frac{1}{5} + \\frac{1}{3} \\times \\frac{1}{20} = \\frac{3}{20}$$\r\n\r\n%2/6\r\n\r\n\\section{Bayes Theorem}\r\n$A_1, \\cdots, A_n $ are events which are mutually exclusive and exhaustive.\r\n$B$ be an arbitrary event. Then,\r\n$$ P(A_i/B)= \\frac{P(B/A_i)P(A_i)}{\\sum_{i=1}^n P(B/A_i)P(A_i)}$$\r\n\r\n\\begin{itemize}\r\n \\item $P(A_i)$- Prior probability\r\n \\item $P(B/A_i)$- Likelihood\r\n \\item $P(A_i/B)$- Posterior probability\r\n\\end{itemize}\r\n\r\nBayes theorem expresses posterior probabilities $P(A_i/B)$ in terms of prior probabilities $P(A_i)$ and likehoods $P(B/A_i)$.\r\n\r\nIn some experiments $P(A_i)$ are all same, $P(A_i)=1/n$. Then posterior probabilities are proportional to likelyhoods:\r\n$$ P(A_i/B)= \\frac{P(B/A_i)}{\\sum_{i=1}^n P(B/A_i)}$$\r\n$$ \\Rightarrow P(A_i/B) \\propto P(B/A_i)$$\r\n\r\n\\begin{proof}\r\n\r\n $$ P(A_i/B) = \\frac{P(A_i \\cap B)}{P(B)}$$\r\n $$ P(A_i/B)= \\frac{P(B/A_i)P(A_i)}{\\sum_{i=1}^{n} P(B/A_i)P(A_i)}$$\r\n\r\n\\end{proof}\r\n\r\n\\subsubsection{Example}\r\nIn answering a question in a multiple choice test, a student knows the answer with probability $p$ and guesses the answer otherwise. If he/she guesses from $m$ choices, the probability of being correct is $\\frac{1}{m}$. Find the conditional probability that the student knew the answer if he/ she answered correctly.\r\n\r\nAns: $A_1=$ knowing the answer. $A_2=$ Guessing the answer.\r\n\r\n$B= $Answer is correct. $A_1 \\subseteq B$\r\n\r\n$P(A_1)= p, P(A_2)= 1-p, P(B/A_1)= 1, P(B/A_2)= \\frac{1}{m}$\r\n\r\n$$ P(A_1/B)= \\frac{P(B/A_1)P(A_1)}{P(B/A_1)P(A_1)+ P(B/A_2)P(A_2)}$$\r\n$$ = \\frac{p}{p + (1-p)\\frac{1}{m}}$$\r\n\r\n\\section{Indepedent Events}\r\n$A \\;\\&\\; B$ are said to be independent. If\r\n$$ P(A \\cap B) = P(A)P(B)$$\r\n\r\nIn terms of conditional probability,\r\n$$ P(B/A)= \\frac{P(A \\cap B)}{P(A)} = \\frac{P(A)P(B)}{P(B)} = P(B)$$\r\n\r\n$$ \\Rightarrow P(B/A) = P(B)$$\r\n\r\nProbability of event $B$ remains same with or without conditioning on $A$. Hence, $B$ is said to be independent of $A$.\r\nKnowledge of occurence of event $A$ does not give any information about $B$.\r\n\r\nIf $A\\; \\&\\; B $ are independent, then\r\n$ A \\;\\&\\; B^c$ are also independent.\r\n\r\n\\begin{proof}\r\n $$ P(B^c / A)= \\frac{P(B^c \\cap A)}{P(A)}= \\frac{P(A)- P(A \\cap B)}{P(A)}= \\frac{P(A)- P(A)P(B)}{P(A)} = P(B^c)$$\r\n\\end{proof}\r\n\r\n\\subsection{Important Results}\r\n\\begin{enumerate}\r\n \\item $P(A/B)$ may be greater than, less than or equal to $P(A)$.\r\n \\item Independent events and mututally exclusive events are different.\r\n\r\n Indepedence: $P(A \\cap B)= P(A)P(B)$\r\n\r\n Mututally exclusive: $A \\cap B = \\phi$\r\n\r\n \\begin{enumerate}\r\n \\item (Indepedent but not mutually exclusive). Coin toss followed by throwing dice experiment.\r\n\r\n $$ S=\\{ (H,1), (H,2), \\cdots, (H,6), (T,1), \\cdots, (T,6)\\}$$\r\n $F=$ Power set of S (Always an event space)\r\n\r\n $A=\\{(H,1), \\cdots, (H,6) \\}$, $B= \\{ (H,2), (H,4), (H,6), (T,2),(T,4),(T,6)$\r\n\r\n $P(A)= 1/2 \\; \\& \\; P(B)= 1/2$\r\n $$ P(A \\cap B)= P(\\{ (H,2), (H,4), (H,6)\\})= \\frac{3}{12}= \\frac{1}{4}$$\r\n $$ P(A\\cap B)= P(A)P(B)$$\r\n \\item (Not independent but mutually exclusive)\r\n If the events are mutually exclusive $\\Rightarrow$ they are not independent.\r\n\r\n Single Coin Toss: $A= \\{H \\}$, $B= \\{T \\}$ $\\rightarrow$ Mututally exclusive.\r\n\r\n $P(A \\cap B)= 0$, $P(A)P(B)= \\frac{1}{4}$\r\n\r\n %Eg 3: (Not independent and not mutually exclusive) hw\r\n\r\n\r\n \\end{enumerate}\r\n \\end{enumerate}\r\n\r\n\\subsection{Conditionally indepedent events}\r\n\r\n$A \\; \\& \\: B$ are said to be conditionally indepedent given $C$ if\r\n$$ P((A \\cap B)/ C)= P(A/C)P(B/C)$$\r\nIn terms of conditional probabilities,\r\n$$ P(B/C)= \\frac{P((A\\cap B )/ C)}{P(A/C)}= \\frac{\\frac{P(A \\cap B \\cap C)}{P(C)}}{\\frac{P(A \\cap C)}{P(C)}} = \\frac{P(A\\cap B\\cap C)}{P(A \\cap C)} = P(B/(A \\cap C))$$\r\n\r\nIndepedent Events: $P(A \\cap B)= P(A)P(B)$ \\& $P(B/A)= P(B)$.\r\n\r\nConditionally indepedent events: $P(A\\cap B / C)= P(A/C)P(B/C)$ \\& $P(B/(A\\cap ))$\r\n\r\n\\subsubsection{Example}\r\nTwo fair coins are tossed, $S=\\{HH, HT, TH, TT \\}$, $A=\\{HH, HT \\} $, $B=\\{HH, TH\\} $, $C=\\{HH \\}$ \\& $D=\\{HT, TH\\}$\r\n\r\n\\begin{enumerate}\r\n \\item Are $A$ \\& $B$ independent?\r\n $$ P(A \\cap B)= P(\\{HH\\}) = \\frac{1}{4}= P(A)P(B)$$\r\n \\item Are $A$ \\& $B$ conditionally indepedent given $C$?\r\n $$ P((A \\cap B) /C)= P(A/C)=P(B/C)= 1$$\r\n $$ P(A/C)= \\frac{P(A \\cap C)}{P(C)}= \\frac{\\frac{1}{4}}{\\frac{1}{4}}= 1$$\r\n \\item Are $A$ \\& $B$ conditionally indepedent given $D$?\r\n\r\n $$P(A/D)= \\frac{P(A\\cap D)}{P(D)}= \\frac{\\frac{1}{4}}{\\frac{1}{2}}=\\frac{1}{2} $$\r\n $$P(B/D)= \\frac{P(B\\cap D)}{P(D)}= \\frac{\\frac{1}{4}}{\\frac{1}{2}}=\\frac{1}{2} $$\r\n $$ P(A\\cap B /D)= 0$$\r\n Hence, they are not conditionally indepedent given $D$.\r\n\\end{enumerate}\r\n\r\nIf $A$ and $B$ are independent, it doesn't imply $A$ and $B$ will be conditionally indepedent given $C$ and vice-versa.\r\n%4/6 notes\r\n\\subsection{Indepedence of collection of events}\r\n\r\nThree events $A_1, A_2$ \\& $A_3$ are said to be indepedent if:\r\n\r\n$$ P(A_1 \\cap A_2 \\cap A_3)=P(A_1)P(A_2)P(A_3)$$\r\n$$ P(A_1 \\cap A_2)=P(A_1)P(A_2)\\quad P(A_2 \\cap A_3)=P(A_2)P(A_3)\\quad P(A_3 \\cap A_1)=P(A_3)P(A_1)$$\r\n\r\n\r\n\r\n\\subsubsection{Chain rule of Probability}\r\n$$ P(A_1 \\cap A_2 \\cap A_3)=P(A_1)P(A_2/A_1)P(A_3/A_1\\cap A_2)$$\r\n$$ P(\\bigcap_{i=1}^n A_i)=P(A_1) \\prod_{i=2}^n P(A_i/A_1\\cdots A_{i-1})$$\r\nFor independent events $A_1, A_2, A_3$.\r\n$$P(A_1)P(A_2/A_1)P(A_3/(A_1 \\cap A_2))= P(A_1)P(A_2)P(A_3)$$\r\n$$\\Rightarrow P(A_2/A_1)P(A_3/(A_1 \\cap A_2))= P(A_2)P(A_3)$$\r\n\r\nHence, if $A_1$, $A_2$, $A_3$ are independent events, the first condition doesn't imply the other conditions.\r\n\r\nEg: Pair-wise independence does not imply that three events are independent.\r\nConsider example of two coin tosses:\r\n$S= \\{HH,HT,TH,TT\\}, A_1=\\{HH,HT\\},A_2=\\{HH,TH\\},A_3=\\{HT,TH\\}$.\r\n\r\n$A_1$ \\& $A_2$ are indepedent $P(A_1 \\cap A_2)=P(A_1)P(A_2)$\r\n\r\n$A_2$ \\& $A_3$ are indepedent $P(A_3 \\cap A_2)=P(A_3)P(A_2)$\r\n\r\n$A_1$ \\& $A_3$ are indepedent $P(A_1 \\cap A_3)=P(A_1)P(A_3)$\r\n\r\n$A_1$, $A_2$ \\& $A_3$ are not still not independent as $P(A_1 \\cap A_2 \\cap A_3) \\neq P(A_1)P(A_2)P(A_3)$.\r\n\r\n\\section{Continuity of Probability}\r\n$\\Omega$ is an equivalent notation for sample space.\r\n\r\nConsider the probability space $(S,F,P)$.\r\n\r\n$E_i \\in F$. Let $E_1 \\subseteq E_2 \\subseteq \\cdots$ be countably infinite sequence of events (increasing sequence of events).\r\nThen $$P(\\bigcup_{i=1}^{\\infty} E_i)= \\lim_{i \\to \\infty} P(E_i)$$.\r\n$$ \\bigcup_{i=1}^{\\infty} E_i \\in F$$\r\nPushing the limit from inside the probability expression to outside.\r\n\r\n$$P(\\bigcup_{i=1}^{\\infty} E_i)=\\lim_{i \\to \\infty }P(E_i)$$\r\n$$P(\\lim_{n \\to \\infty} \\bigcup_{i=1}^{n}E_i)=\\lim_{i \\to \\infty }P(E_i)$$\r\n\r\nExchanging limits with probability $\\rightarrow$ non trivial operation.\r\nExchanging limits with differentiation, integration etc. $\\rightarrow$ require a proof.\r\n\r\n\\begin{itemize}\r\n \\item $$E_1 \\subseteq E_2 \\subseteq \\cdots$$\r\n $$P(\\bigcup_{i=1}^{\\infty} E_i)= \\lim_{i \\to \\infty} P(E_i)$$\r\n\r\n \\begin{proof}\r\n We shall use the third axiom.\r\n $D_1, D_2 \\cdots \\rightarrow$ mutually exclusive events.\r\n\r\n $$ P(\\bigcup_{i=1}^{\\infty} D_i)= \\sum_{i=1}^{\\infty}P(D_i)$$\r\n\r\n $E_1 \\subseteq E_2 \\subseteq \\rightarrow $ not mutually exclusive.\r\n\r\n Hence we construct a sequence which are mututally exclusive and unions of both sequences have to be the same.\r\n\r\n $A_1=E_1$, $A_2=E_2\\setminus E_1$, $ A_3=E_3 \\setminus E_2$ , $\\cdots$.\r\n $$ P(\\bigcup_{i=1}^{\\infty} E_i)= P(\\bigcup_{i=1}^{\\infty} A_i)= \\sum_{i=1}^{\\infty}P(A_i)= \\lim_{n \\to \\infty} \\sum_{i=1}^{\\infty}P(A_i)$$\r\n\r\n $$\\Rightarrow \\lim_{n \\to \\infty} \\sum_{i=1}^{n} P(A_i)= \\lim_{n \\to \\infty} P(\\bigcup_{i=1}^{n}A_i)= \\lim_{n \\to \\infty}P(E_n)$$\r\n If there exists a subsequence of events, which is increasing.\r\n\r\n \\end{proof}\r\n\r\n \\item Let $E_1 \\supseteq E_2 \\supseteq \\cdots$ be a countably infinte sequence of events(decreasing sequence of events)\r\n\r\n $$P(\\bigcap_{i=1}^{\\infty}E_i)=\\lim_{i \\to \\infty} P(E_i)$$\r\n\r\n \\begin{proof}\r\n We can use the result of the previous proof.\r\n $$ E_1^c \\subseteq E_2^c \\subseteq \\cdots$$\r\n which is an increasing set of events\r\n $$P(\\bigcup_{i=1}^{\\infty} E_i^c)= \\lim_{i \\to \\infty}P(E_i^c)$$\r\n as,\r\n $$ \\left(\\bigcap_{i=1}^{\\infty} E_i \\right)= \\left( \\bigcup_{i=1}^{\\infty} E_i^c \\right)^c $$\r\n\r\n $$\\Rightarrow 1- P(\\bigcup_{i=1}^{\\infty} E_i^c)= \\lim_{i \\to \\infty}1- P(E_i^c)$$\r\n\r\n $$P \\left(\\bigcap_{i=1}^{\\infty} E_i \\right) = \\lim_{i \\to \\infty }P(E_i)$$\r\n\r\n \\end{proof}\r\n\r\n\\end{itemize}\r\n\r\n%4/6 \\subsection{Example of continuity of probability}\r\n\r\n%7/6 lecture 6\r\n\r\n\\section{Random Variables}\r\n\r\nConsider a probability space $(S,F,P)$ and an experiment is conducted.\r\n\r\nProbability spaces vary a lot based on the experiment. We use random variables to be able to develop a theory of probability which is indepedent of the actual experiment which is performed.\r\n\r\n$$S \\xrightarrow[]{\\text{R.V.}} \\mathbb{R}$$\r\n$$F \\xrightarrow[]{\\text{R.V.}} B$$\r\n\r\nRandom Variable is a function which maps the sample space to the real line.\r\nIt also maps the event space to the Borel $\\sigma$-algebra, $B$.\r\n\r\n\\subsection{Borel $\\sigma$-algebra}\r\nIt is the smallest $\\sigma$- algebra which contains sets of the form $(-\\infty ,x]\\; \\forall x \\in \\; \\mathbb{R}$.\r\n\r\n$$ B=\\{ \\mathbb{R} , (-\\infty ,x] ,(x,\\infty ), \\phi, (-\\infty,x) ,(x,y),[x,y], \\{x\\},(x,y],[x,y)\\}$$\r\n\r\n$$ E_i =(-\\infty ,x_i] \\quad x_i = x-\\frac{1}{i}$$\r\n$$ \\bigcup_{i=1}^{\\infty}(-\\infty ,x_i] = \\bigcup_{i=1}^{\\infty} E_i =(-\\infty ,x)$$\r\n\r\n\r\nRandom Variable $X:S \\rightarrow R$\r\n\r\n$X$ has to be a measurable function.\r\n\r\nA function is said to be measurable if pre-image of $(-\\infty ,x ] \\; \\forall x \\in \\mathbb{R}$ is in the event space.\r\n$$X:S \\rightarrow R$$\r\n$$\\text{inverse image} \\leftarrow(-\\infty ,x ] $$\r\n$$ X^{-1} ((-\\infty ,x ]) \\subseteq S \\qquad X^{-1} ((-\\infty ,x ]) \\in F$$\r\neg:\r\n$ X: \\{1,2,3,4,5,6\\} \\rightarrow \\{0,1,0,1,0,1 \\}$\r\n\r\n$$X^-1\\left( (-\\infty, 0.5]\\right)=\\{1,3,5\\}$$\r\n\r\nIn the above example, $X$ is not an invertible mapping. We are using $X^{-1}$ as a notation.\r\n\r\n$X$ is a map from $S$ to $\\mathbb{R}$ such that\r\n$$X:S \\rightarrow \\mathbb{R}$$\r\n$$ X^{-1}\\left( (-\\infty, x ]\\right) \\leftarrow (-\\infty, x ] \\qquad X^{-1}\\left( (-\\infty, x ]\\right) \\in F $$\r\n\r\n\\subsection{Examples}\r\n\\begin{itemize}\r\n \\item X which is a random variable.\r\n\r\n $S= \\{a,b,c\\}\\; ,\\; F=\\{ \\phi, \\{ a\\}, \\{b,c \\}, S\\}\\; , \\; X:S \\rightarrow \\mathbb{R}$\r\n\r\n $X(\\omega) =0\\; ,\\; \\omega =a $\r\n\r\n $X(\\omega)= 1\\;,\\; \\omega = b,c$\r\n \\begin{enumerate}\r\n \\item $( -\\infty,x]\\;,\\; x < 0\\;,\\; X^{-1}\\left( (-\\infty , x]\\right) = \\phi \\in F$\r\n \\item $( -\\infty,x]\\;,\\; 0 \\leq x <1 \\;,\\; X^{-1}\\left( (-\\infty , x]\\right) = \\phi \\in F$\r\n \\item $( -\\infty,x]\\;,\\; x \\geq 1\\;,\\; X^{-1}\\left( (-\\infty , x]\\right) = S \\in F$\r\n \\end{enumerate}\r\n\r\n \\item X which is not a random variable\r\n $S= \\{a,b,c\\}\\; ,\\; F=\\{ \\phi, \\{ a\\}, \\{b,c \\}, S\\}\\; , \\; X:S \\rightarrow \\mathbb{R}$\r\n\r\n $X(\\omega) =0\\; ,\\; \\omega =b $\r\n\r\n $X(\\omega)= 1\\;,\\; \\omega = a,c$\r\n\r\n But $X^{-1}((-\\infty,x])\\; , \\;0\\leq x<1 = \\{b\\} \\; , \\; \\{ b\\} \\notin F$\r\n\r\n $X$ is not a random variable.\r\n\\end{itemize}\r\n\r\nIf $S$ is finite \\& $F$ is a power set of $S$, then every function is a random variable.\r\n$$ X^{-1}((-\\infty,x]) \\subseteq S$$\r\n\r\n$F$ is set of all subsets of $S$.\r\n\r\n%9/6\r\n\\subsection{Conditions on random variable}\r\n\\begin{theorem}\r\n Let a probability space be $(S,F,P)$ and $X$ is a random variable $X: S \\rightarrow \\mathbb{R}$.\r\n The following conditions hold:\r\n \\begin{itemize}\r\n \\item $X^{-1}((-\\infty,x)) \\in F$\r\n \\item $X^{-1} (\\{x\\}) \\in F$\r\n \\item $X^{-1}((x_1 , x_2]) \\in F$\r\n \\item $X^{-1} ((x_1,x_2)) \\in F$\r\n \\end{itemize}\r\n\\end{theorem}\r\n\r\n\\begin{proof}\r\n\r\n Applying the axioms of the event space we want to show the above conditions are true.\r\n\r\n $A_i=X^{-1}((-\\infty,x]) \\qquad x_i = x - \\frac{1}{i}$. Now,\r\n $$ \\bigcup_{i=1}^{\\infty} (-\\infty,x_i] = (-\\infty,x)$$\r\n\r\n $$ \\bigcup_{i=1}^{\\infty} A_i = \\bigcup_{i=1}^{\\infty} X^{-1}((-\\infty,x_i]) = X^{-1}((-\\infty,x)) \\in F$$\r\n\r\n $$ X^{-1}((-\\infty ,x] \\cap {(-\\infty,x)}^c) = X^{-1} (\\{x\\}) \\in F$$\r\n\r\n Similarly the other conditions can also be proved.\r\n\\end{proof}\r\n\r\n\\section{Cumulative distribution function}\r\nIt is denoted by $F_X(x)$.\r\n\r\n$$ F_X(x)= P(X \\leq x) = P((-\\infty, x])= P(X^{-1}((-\\infty, x]))$$\r\n\r\nThe above definitions are different notations used, the last one is well defined from the probability space.\r\n\r\n\\subsection{Properties of CDF}\r\n\r\n\\begin{enumerate}\r\n \\item $F_X(x)$ is a monotonically non-decreasing function of x.\r\n \\begin{proof}\r\n If $x_2 \\geq x_1$, then $F_X(x_2) \\geq F_X(x_1)$.\r\n\r\n $$F_X(x_2) = P(X \\leq x_2)= P(X \\leq x_1)+P(x_1 < X \\leq x_2)$$\r\n $$F_X(x_2) = P(X \\leq x_2) \\geq P(X \\leq x_1) -F_X(x_1) $$\r\n\r\n \\end{proof}\r\n\r\n \\item $\\lim_{x \\to \\infty} F_X(x)= 1$\r\n\r\n \\begin{proof}\r\n Construct a decreasing set of events. $A_i= (-\\infty, i]$, $B_i = X^{-1}(A_i)$.\r\n $$ \\bigcap_{i=1}^{\\infty}A_i= \\mathbb{R}$$\r\n $$ \\bigcap_{i=1}^{\\infty}B_i= S $$\r\n\r\n Applying continuity of probability,\r\n $$ \\lim_{i \\to \\infty}P(B_i)= P(\\bigcup_{i=1}^{\\infty}B_i) =P(S) = 0$$\r\n $$ \\lim_{i \\to \\infty}P(X \\leq i)= \\lim_{x \\to \\infty}P(X \\leq x) = \\lim_{x \\to \\infty} F_X(x)$$\r\n\r\n The change from integers to real numbers $x$ in the last step is valid.\r\n\r\n \\end{proof}\r\n \\item $\\lim_{x \\to -\\infty} F_X(x)= 0$\r\n \\begin{proof}\r\n Construct a decreasing set of events. $A_i= (-\\infty, -i]$, $B_i = X^{-1}(A_i)$.\r\n $$ B_1 \\supseteq B_2 \\supseteq \\cdots$$\r\n\r\n $$ \\bigcap_{i=1}^{\\infty}B_i= \\phi $$\r\n $$ P(\\bigcap_{i=1}^{\\infty}B_i) = \\lim_{i \\to \\infty}P(B_i)= 0$$\r\n $$ \\lim_{i \\to \\infty}P(B_i) = \\lim_{i \\to \\infty}P((-\\infty,i])= \\lim_{x \\to \\infty}P((-\\infty,x]) \\rightarrow \\text{(x = all real nos.)}$$\r\n $$ \\lim_{x \\to \\infty}P((-\\infty,x]) = \\lim_{x \\to \\infty}F_X(x)= 0$$\r\n \\end{proof}\r\n%11/6\r\n \\item $F_X(x)$ is a right continuous function.\r\n $$ \\lim_{x \\to x_0^+} F_X(x)= F_X(x_0)$$\r\n\r\n Right continuous function: $\\lim_{x \\to x_0^+} f(x)= f(x_0)$\r\n\r\n Left continuous function: $\\lim_{x \\to x_0^-} f(x)= f(x_0)$\r\n\r\n Continuous function(Approaching from either left or right): $\\lim_{x \\to x_0} f(x)= f(x_0)$\r\n\r\n \\begin{proof}\r\n Decreasing sequence of events $B_i= (X \\leq x+\\frac{1}{i})$\r\n\r\n $$ B_1 \\supseteq B_2 \\supseteq \\cdots$$\r\n $$ \\bigcap_{i=1}^{\\infty}B_i= \\{ X \\leq x\\}$$\r\n $$ P(\\bigcap_{i=1}^{\\infty}B_i) = P(X \\leq x) = F_X(x)$$\r\n $$ \\Rightarrow \\lim_{i \\to \\infty}P(B_i)= \\lim_{i \\to \\infty}P(X \\leq x + \\frac{1}{i})$$\r\n We are changing the variable from an integer to real.\r\n $$= \\lim_{\\epsilon \\to 0}P(X \\leq x + \\epsilon)=\\lim_{\\epsilon \\to 0^+}F_X(x + \\epsilon) $$\r\n $$ = \\lim_{x_0 \\to x^+}F_X(x_0) \\quad \\text{Change of variable: }x_0= x+ \\epsilon $$\r\n\r\n \\end{proof}\r\n\\end{enumerate}\r\n\r\n\\subsection{Indicator Random Variable}\r\n\\begin{equation*}\r\n I_A (x)=\r\n \\begin{cases}\r\n 1, & \\text{if}\\ x \\in A \\\\\r\n 0, & \\text{if}\\ x \\notin A\r\n \\end{cases}\r\n\\end{equation*}\r\n\r\n$$I_A:S \\to \\mathbb{R}$$\r\n\r\n$$ F_{I_{A}} (x)= P(I_A \\leq x)$$\r\n\r\nIf $x < 0 $, $P(I_A \\leq x)= 0$.\r\n\r\nIf $0 \\leq x <1 $, $P(I_A \\leq x)= P(A^c)=1-P(A)$.\r\n\r\nIf $x \\geq 1 $, $P(I_A \\leq x)= 1$.\r\n\r\n$$ \\lim_{x \\to 0^+} F_{I_A}(x)= P(A^c) \\qquad F_{I_A}(0)= P(A^c)$$\r\n\r\n$$ \\lim_{x \\to 1^+} F_{I_A}(x)= 1 \\qquad F_{I_A}(1)= 1$$\r\n\r\n\\subsection{Examples of CDF's and non-CDF's}\r\n\\begin{itemize}\r\n \\item Is $F_1(x)$ a valid CDF?\r\n \\begin{equation*}\r\n F_1 (x)=\r\n \\begin{cases}\r\n 1, & \\text{if}\\ x \\leq 0 \\\\\r\n 0.5, & \\text{if}\\ 00$,\r\n $$ \\{ Y \\leq y\\} = \\{ X \\leq \\frac{y-b}{a}\\}$$\r\n $$ F_Y(y)= P(Y \\leq y)= P\\left( X \\leq \\frac{y-b}{a}\\right)= F_X\\left( \\frac{y-b}{a}\\right)$$\r\n \\item If $a<0$,\r\n $$ \\{ Y \\leq y\\} = \\{ X \\geq \\frac{y-b}{a}\\}$$\r\n $$ F_Y(y)= P(Y \\leq y)= P\\left( X \\geq \\frac{y-b}{a}\\right)= 1-F_X\\left( \\frac{y-b}{a}\\right)$$\r\n \\end{enumerate}\r\n\r\n For the PDF's:\r\n \\begin{enumerate}\r\n \\item If $a>0$\r\n\r\n $$ F_Y(y)= F_X\\left( \\frac{y-b}{a}\\right)$$\r\n \\begin{align*}\r\n f_Y(y)= \\frac{dF_Y(y)}{dy} &= \\frac{d}{dy}F_X\\left( \\frac{y-b}{a} \\right)\\\\\r\n &= f_X \\left( \\frac{y-b}{a} \\right)\\frac{d}{dy}\\left(\\frac{y-b}{a} \\right) \\\\\r\n &= \\frac{1}{a}f_X\\left(\\frac{y-b}{a} \\right)\r\n \\end{align*}\r\n\r\n \\item If $a>0$\r\n\r\n $$ F_Y(y)= 1- F_X\\left( \\frac{y-b}{a} \\right)$$\r\n \\begin{align*}\r\n f_Y(y)= \\frac{-1}{a}f_X\\left(\\frac{y-b}{a} \\right)\r\n \\end{align*}\r\n\r\n\r\n \\end{enumerate}\r\n\r\n Hence,\r\n \\fbox{$ f_Y(y)= \\frac{1}{|a|}f_X \\left(\\frac{y-b}{a} \\right)$}\r\n%18/6\r\n \\item $X$ is continuous and $Y$ is continuous, but a many to one function.\r\n $$ Y= X^2$$\r\n Now let us find the CDF \\& PDF of $Y$ in therms of CDF \\& PDF of $X$.\r\n \\begin{enumerate}\r\n \\item $y<0$, $F_Y(y)=0$\r\n \\item $y \\geq 0$,\r\n \\begin{align*}\r\n F_Y(y)= &= P(Y \\leq y) \\\\\r\n &= P(X^2 \\leq y) = P(-\\sqrt{y} \\leq x \\leq \\sqrt{y}) \\\\\r\n &= F_X(\\sqrt{y})- F_X(-\\sqrt{y}) + P(x= -\\sqrt{y}) \\\\\r\n &= F_X(\\sqrt{y})- F_X(-\\sqrt{y})\r\n \\end{align*}\r\n \\end{enumerate}\r\n\r\n $$ f_Y(y)= \\frac{dF_Y(y)}{dy}= \\frac{d}{dy}(F_X(\\sqrt{y})- F_X(-\\sqrt{y}))$$\r\n $$ = \\frac{1}{2\\sqrt{y}}[f_X(\\sqrt{y})- f_X(-\\sqrt{y})]$$\r\n \\item $X$ is continuous, $Y$ is continuous and a many to one function.\r\n \\begin{equation*}\r\n f_X (x)=\r\n \\begin{cases}\r\n \\frac{1}{2\\pi}, & \\text{if}\\ 0 \\leq x \\leq 2\\pi \\\\\r\n 0, & \\text{otherwise}\\\r\n \\end{cases}\r\n \\end{equation*}\r\n $$ F_X(x)= \\int_{-\\infty}^{x} f_x(x)dx$$\r\n $$ Y=\\sin X$$\r\n Let us find the PDF of $Y$.\r\n\r\n \\begin{enumerate}\r\n \\item Consider $0 \\leq y \\leq 1$\r\n \\begin{align*}\r\n F_Y(y)&= P(Y \\leq y) \\\\\r\n &= P(\\sin{X} \\leq Y) \\\\\r\n &= P(\\{ 0 \\leq X \\leq \\sin^{-1} y\\} \\cup \\{ \\pi - \\sin^{-1}y \\leq X \\leq 2\\pi \\})\\\\\r\n &=P( 0 \\leq X \\leq \\sin^{-1}y)+(\\pi - \\sin^{-1}y \\leq X \\leq 2 \\pi )\\\\\r\n &= F_X(\\sin^{-1}y)- F_X(0)+F_X(2\\pi)- F_X(\\pi - \\sin^{-1}y)\\\\\r\n &= \\frac{\\sin^{-1}y}{2\\pi}-0+1- \\left( \\frac{\\pi - \\sin^{-1}y}{2\\pi}\\right) \\\\\r\n &= \\frac{1}{2}+ \\frac{\\sin^{-1}y}{\\pi}\r\n \\end{align*}\r\n \\item $-1 \\leq y <0$\r\n \\begin{align*}\r\n F_y(y)&= P(Y \\leq y) = P(\\sin X \\leq y)\\\\\r\n &= P(\\pi - \\sin^{-1}y \\leq X \\leq 2\\pi + \\sin^{-1}y )\r\n \\end{align*}\r\n \\end{enumerate}\r\n\r\n % \\begin{equation*}\r\n % f_Y (y)=\r\n % \\begin{cases}\r\n % \\frac{1}{\\pi \\sqrt{1-y^2}}, & \\text{if}\\ -1 \\leq y < 1 \\\\\r\n % \\frac{1}{\\pi \\sqrt{1-y^2}}, & \\text{if}\\ 0 \\leq y \\leq 1\r\n %\r\n % \\end{cases}\r\n % \\end{equation*}\r\n\r\n \\item $X$ is continuous and $Y$ is discrete.\r\n\r\n $X$ has CDF $F_X(.)$ \\& PDF $f_X(.)$\r\n\r\n $Y$ has CDF $F_Y(.)$ \\& PMF $P_Y(.)$\r\n\r\n $$Y=g(x)=k \\; if \\; k \\leq x < k+1 \\qquad k \\in \\mathbb{Z} $$\r\n The above operation is also called quantization. Range of $g$ is countably infinite \\& thus $Y$ is a discrete random variable.\r\n\r\n We would like to experss the PMF of $Y$ in terms of the PDF of $X$.\r\n\r\n\r\n \\begin{align*}\r\n P_Y(y)&= P(Y=y)= P(y \\leq X < y+1) \\\\\r\n &= \\int_{y}^{y+1}f_X(x)dx\r\n \\end{align*}\r\n\r\n In general,\r\n $$ P_Y(y)= \\int f_X(x)dx \\qquad S= \\{ x: g(x)=y\\}$$\r\n\r\n \\end{enumerate}\r\n\\end{itemize}\r\n\r\n \\subsection{General formula for determining PDF of $Y= g(X)$}\r\n Where $g$ is differentiable,\r\n\r\n $$ P(y \\leq Y \\leq y + \\Delta y) \\approx f_Y(y)\\Delta y$$\r\n The left hand side can be written in terms of PDF of $X$.\r\n $$ g(x_i)= y$$\r\n $$ g^{-1}[y,y + \\Delta y]= [x_1, x_1+ \\Delta x_1]\\cup [x_2, x_2+ \\Delta x_2] \\cdots [x_n, x_n+ \\Delta x_n]$$\r\nWhat is the sign of slope of $g(x)$ at $x_i$?\r\n\\begin{itemize}\r\n \\item If $g^{'} (x_i) \\geq 0 \\qquad [x_i,x_i+ \\delta x_i]$\r\n \\item If $g^{'} (x_i) < 0 \\qquad [x_i- \\delta x_i,x_i]$\r\n\\end{itemize}\r\n\r\n\\begin{align*}\r\n f_Y(y)\\Delta y &= \\sum_{i=1}^{n}P(x_i \\leq X \\leq x_i + \\Delta x_i)\\\\\r\n &= \\sum_{i=1}^{n} f_x(x_i)\\Delta x_i\r\n\\end{align*}\r\n$$ f_Y(y)= \\sum_{i=1}^{n} f_X(x_i)\\frac{\\Delta x_i}{\\Delta y}= \\sum_{i=1}^{n} f_X(x_i)\\frac{1}{(\\Delta y / \\Delta x_i)}$$\r\n\r\nFor infintesimally small $\\Delta y$ \\& corresponding $\\Delta x_i$:\r\n$$ \\frac{\\Delta y}{\\Delta x_i}= |g^{'}(x_i)|$$\r\n\r\nWe take modulus as LHS is positive, i.e. magnitude of the slope of tangent at $x_i$.\r\n\r\n$$ f_Y(y) = \\sum_{i=1}^{n} \\frac{f_X(x_i)}{|g^{'}(x_i)|}$$\r\n\r\nExamples:\r\n\\begin{itemize}\r\n \\item $Y= X^2$\r\n $$ f_Y(y)= \\frac{f_X(\\sqrt{y})}{2\\sqrt{y}}+ \\frac{f_X(-\\sqrt{y})}{2\\sqrt{y}}$$\r\n \\item $Y= \\sin X$, we take $\\sin^{-1} y(2n\\pi + \\sin^{-1} y)$ \\& $\\pi - \\sin^{-1} y(2n\\pi + \\pi - \\sin^{-1} y)$\r\n\r\n $$ f_Y(y)= \\sum_{n}\\left[ \\frac{f_X(2n\\pi + \\sin^{-1} y)}{\\sqrt{1- y^2}} + \\frac{f_X(2n\\pi + \\pi - \\sin^{-1} y)}{\\sqrt{1- y^2}} \\right]$$\r\n\\end{itemize}\r\n\r\n%21/6\r\n\r\n\\begin{theorem}\r\n If $Y= F_X(X)$, where $X$ itself is a random variable with CDF $F_X(.)$, the CDF of Y:\r\n $$ F_Y(y)= P(Y \\leq y)= P(F_X(x) \\leq y)$$\r\n\\end{theorem}\r\n \\begin{itemize}\r\n \\item If $y<0$,\r\n $$ \\{ X \\leq F_{X}^{-1}((-\\infty,y])\\}= \\phi \\qquad P(F_{X}(x)\\leq y)=0 $$\r\n \\item If $y \\geq 1$,\r\n \\begin{align*}\r\n F_Y(y)&= P(Y \\leq y) = P(F_X(x) \\leq y) \\\\\r\n &= 1 \\qquad \\text{(max value of }F_X \\text{ is 1, } F_{X}^{-1}((-\\infty,y])= \\mathbb{R})\r\n \\end{align*}\r\n \\item If $0 \\leq y <1 $,\r\n \\begin{align*}\r\n F_Y(y)= P(Y \\leq y)&= P(F_X(x)\\leq y) \\\\\r\n &= P(X \\leq F_{X}^{-1}(y)) \\\\\r\n &= F_X(F_{X}^{-1}(y)) \\\\\r\n &= y\r\n \\end{align*}\r\n \\end{itemize}\r\n \\begin{equation*}\r\n f_Y (y)=\r\n \\begin{cases}\r\n 0, & \\text{if}\\ y < 0 \\\\\r\n y, & \\text{if}\\ 0 \\leq y < 1 \\\\\r\n 1, & \\text{if}\\ y \\geq 1\r\n \\end{cases}\r\n \\end{equation*}\r\nIf we take any random variable $X$ \\& apply $F_X(.)$ as a function of $X$, then resulting random variable is a uniform random variable.\r\n\r\nSuppose we start with an uniform random variable $Y$ and consider a function of $Y$ which is $X= F_{X}^{-1}(Y)$, then X has a CDF $F_X(.)$.\r\n\r\n%\\fbox{In, MATLAB, there is an inbuilt function }\r\n\r\n\\section{Expectation of a random variable}\r\nIt is also called mean (or average).\r\n\r\nFor a discrete random variable $X$, let $S$ be the range of $X$, which is either finite or countably infinte.\r\n\r\n$$ E(X)= \\sum_{x_i \\in S}x_i P_X(x_i)$$\r\n\r\nwhere $P_X$ is the PMF of random variable $X$.\r\n\r\nThe expectation, is based on the weighted average, where weights are the probabilities of the random variable taking a particular value unlike the general notion of average.\r\n\r\nFor a continuous random variable $X$,\r\n$$ E(X)= \\int_{-\\infty}^{\\infty}x f_X(x)dx$$\r\nThe integral can be thought of as the limit of a summation. Hence,\r\n$$ E(X)= \\lim_{\\Delta x_i \\to 0} \\sum_{i= -\\infty}^{\\infty}x_i f_X(x_i) \\Delta x_i$$\r\n\r\n\\subsection{Expectation of a function of a random variable}\r\n\r\n$ Y= g(X)$, $X$ \\& $Y$ are discrete.\r\n\\begin{align*}\r\n E(Y)= \\sum_{y_i}y_i P_Y(y_i)\r\n\\end{align*}\r\nWe can express $E(Y)$ in terms of PMF of $X$, $P_X$,\r\n\\begin{align*}\r\n E(Y)&= \\sum_{y_i}y_i P_Y(y_i) \\\\\r\n &= \\sum_{y_i}y_i \\sum_{x_{i,j}:g(x_{i,j})= y_i} P_X(x_{i,j}) \\\\\r\n &= \\sum_{y_i} \\sum_{x_{i,j}:g(x_{i,j})= y_i} y_i P_X(x_{i,j}) \\\\\r\n &= \\sum_{y_i} \\sum_{x_{i,j}:g(x_{i,j})= y_i} g(x_{i,j}) P_X(x_{i,j}) \\\\\r\n &= \\sum_{x_j} g(x_j) P_X(x_j)\\\\\r\n \\Rightarrow E(Y) &= \\sum_{y_i}y_i P_Y(y_i) \\\\\r\n &= \\sum_{x_i} g(x_i) P_X(x_i)\r\n\\end{align*}\r\nFor continuous random variable,\r\n\\begin{align*}\r\n E(Y) &= \\int_{-\\infty}^{\\infty}y f_Y(y)dy \\\\\r\n &= \\int_{-\\infty}^{\\infty} g(x)f_X(x)dx\r\n\\end{align*}\r\n\r\n\\subsection{Variance of a random variable}\r\nIt is defined as,\r\n$$ E[g(X)] \\qquad g(X)= (X - E(X))^2$$\r\n\r\n$E(X)$ indicates the average value of a random variable. So there is a variation in the values the random variable takes around mean, now we would like to measure this variation and hence the variance comes into play.\r\n\r\n$(X- E(X))$ measures variation from the mean, and we square it so that the positive and negative values don't cancel each other.\r\n%23/6\r\n\r\nIf $X$ is discrete:\r\n\r\n$E(X)= \\mu$\r\n\r\n$E(X)= E(X-E(X)^2) = \\sum_{x_i} (x_i - \\mu )^2 P_X (x_i)$\r\n\r\nIf $X$ is continuous:\r\n\r\n$E(X)= \\mu$\r\n\r\n$E(X)= E(X-E(X)^2)= \\int_{-\\infty}^{\\infty}(x- \\mu)^2 f_X(x)dx$\r\n\r\n\\begin{itemize}\r\n \\item $ Var(X) \\geq 0$\r\n\r\n $Var(X)=0$ iff $X=E(X)$ with probability 1.\r\n\\end{itemize}\r\n\r\n\\subsection{Examples of discrete random variables}\r\n\\begin{enumerate}\r\n \\item Bernoulli random variable:\r\n\r\n Experiment: Coin toss experiment.\r\n $$X:S = \\{ H,T\\} \\qquad H \\to 1 \\qquad T to 0$$\r\n PMF of $X$, $P_X(1)= p,\\; P_X(0)=1-p$\r\n\r\n $E(X)= 0(1-p)+1(p)=p$.\r\n\r\n $Var(X)= (0-p)^2(1-p)+ (1-p)^2 p = p(1-p)$\r\n\r\n\r\n \\item Binomial random variable:\r\n\r\n Experiment: Toss a biased coin $n$ times. $X$ is defined as the number of heads in that string.\r\n $$ S= \\{ HH\\cdots H, TH\\cdots H, \\cdots \\}$$\r\n $X:S \\to R$\r\n\r\n Range of $X$: $\\{ 0,1, cdots, n\\}$\r\n\r\n PMF of $X$: $P_X (k)= {n \\choose k} p^k (1-p)^{n-k}$\r\n \\begin{align*}\r\n E(X)&= \\sum_{k=0}^{n}k {n \\choose k} p^k (1-p)^{n-k}\\\\\r\n &= \\sum_{k=1}^{n} \\frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k} \\\\\r\n &= np \\sum_{k=0}^{n-1} \\frac{(n-1)!}{k!(n-1-k)!} p^k (1-p)^{n-1-k} \\\\\r\n &= np\r\n \\end{align*}\r\n $Var(X)= np(1-p)$\r\n\r\n \\item Geometric random variable:\r\n\r\n Experiment: Keep tossing a biased coin till we get a head.\r\n $$ S= \\{ H, TH, TTH, \\cdots \\} \\qquad X:S \\to R$$\r\n $P(H)= p, \\; P(T)= 1-p$\r\n\r\n $P_X(k)= p(1-p)^{k-1}$\r\n\r\n $$E(X)= \\sum_{k=1}^{\\infty} k p (1-p)^{k-1}= \\frac{1}{p}$$\r\n\r\n \\begin{align*}\r\n Var(X)&= E((X-E(X)^2)\\\\\r\n &= E(X^2 -2X E(X) + (E(X))^2)\\\\\r\n &= E(X^2)+E(-2XE(X))+E((E(X))^2\\\\\r\n &= E(X^2)-2E(X)E(X)+(E(X))^2 \\\\\r\n &= E(X^2)-(E(X))^2\r\n \\end{align*}\r\n We know, $Var(X) \\geq 0 \\Rightarrow E(X^2) \\geq (E(X))^2$\r\n\r\n Now, $Var(x)$ for a geometric random variable:\r\n \\begin{align*}\r\n Var(X)&= E(X^2)-E(X)^2 \\\\\r\n &= \\sum_{k=1}^{\\infty} k^2 p (1-p)^{k-1} - \\left( \\frac{1}{p}\\right) ^2 \\\\\r\n &= \\frac{2-p}{p^2} - \\frac{1}{p^2} \\\\\r\n &= \\frac{1-p}{p^2}\r\n \\end{align*}\r\n \\item Poisson random variable:\r\n\r\n Experiment: When we have to count rare events within a time frame.\r\n $$ X= \\{ 0,1,2, \\cdots\\}$$\r\n PMF of Poisson RV:\r\n $$ P_X(k)= \\frac{\\lambda ^k}{k!} e^{-\\lambda}$$\r\n where $\\lambda$ is a parameter and is the average value.\r\n\r\n $Var(X)= \\lambda$\r\n\r\n \\begin{align*}\r\n \\sum_{k=0}^{\\infty} P_X(k) &= e^{-\\lambda } \\sum_{k=0}^{\\infty} \\frac{\\lambda ^k}{k!} \\\\\r\n &= e^{-\\lambda} e^{\\lambda} \\\\\r\n &= 1\r\n \\end{align*}\r\n\\end{enumerate}\r\n\r\n\\subsection{Examples of continuous random variables}\r\n\r\n\\begin{enumerate}\r\n%25/6\r\n \\item Uniform random variable:\r\n\r\n Experiment: Measuring voltages, quantization error (original signal - quantized signal) is modelled as a uniform random variable.\r\n \\begin{equation*}\r\n f_X (x)=\r\n \\begin{cases}\r\n \\frac{1}{b-a}, & \\text{if}\\ a \\leq x \\leq b \\\\\r\n 0, & \\text{otherwise}\r\n \\end{cases}\r\n \\end{equation*}\r\n\r\n $E(X)= \\frac{a+b}{2}$\r\n\r\n $ Var(X)= \\frac{(b-a)^2}{12}$\r\n \\item Exponential random variable:\r\n\r\n Experiment: Used to model completion time of a process.\r\n\r\n \\begin{equation*}\r\n f_X (x)=\r\n \\begin{cases}\r\n \\lambda e^{-\\lambda x}, & \\text{if}\\ x \\geq 0\\\\\r\n 0, & \\text{otherwise}\r\n \\end{cases}\r\n \\end{equation*}\r\n\r\n\r\n $$E(X)=\\int_{0}^{\\infty} x \\lambda e^{-\\lambda x}dx = \\frac{1}{x}$$\r\n $$ Var(X)= E(X^2)- E(X)^2=\\frac{1}{\\lambda ^2}$$\r\n\r\n \\item Gaussian random variable:\r\n\r\n Experiment: Noise is generally modelled as a Gaussian random variable. We shall discuss this in detail while doing the central limit theorem.\r\n\r\n $$ f_X(x)= \\frac{1}{\\sqrt{2\\pi \\sigma^2}} e^{-\\frac{(x- \\mu)^2}{2 \\sigma^2}}$$\r\n\r\n Mean of $X$: $\\mu$\r\n\r\n $Var(x)= {\\sigma}^2$\r\n \\end{enumerate}\r\n\r\n%26/6\r\n\r\n\\section{Random vectors and joint CDF}\r\nConsider 2 individual random variables $X_1$ $X_2$, with $(S,F,P)$ as the probability space.\r\n$X_1:S \\to \\mathbb{R} \\qquad X_2:S \\to \\mathbb{R}$\r\n\r\n$$ (X_1,X_2) :S \\to {\\mathbb{R}}^2$$\r\n\r\nThen $(X_1,X_2) $ form a random vector and the joint CDF of $X_1$ and $X_2$:\r\n$$ F_{X_1,X_2}(x_1,x_2)= P(X_1 \\leq x_1, X_2 \\leq x_2)$$\r\n\r\n$$ X_1 \\leq x_1 = \\{ w: X_1(w) \\leq x_1\\}= A_1 \\in F$$\r\n\r\n$$ X_2 \\leq x_2 = \\{ w: X_2(w) \\leq x_2\\}= A_2 \\in F$$\r\n\r\n$$(X_1 \\leq x_1, X_2 \\leq x_2)= \\{ \\omega : X_1(\\omega) \\leq x_1 \\, \\& \\, X_2(\\omega) \\leq x_2\\} = A_1 \\cap A_2 \\in F$$\r\n\r\n$$ P(X_1 \\leq x_1 , X_2 \\leq x_2) \\; \\forall \\; (x_1 , x_2) \\in {\\mathbb{R}}^2$$\r\n\r\nBy definition, $ F_{X_1 , X_2}(x_1 , x_2) = P(X_1 \\leq x_1 , X_2 \\leq x_2)$.\r\n\r\nThen we can expand this definition till $n$, i.e.\r\n\r\n$X_1,X_2, \\cdots , X_n \\qquad X_iS \\to \\mathbb{R}$\r\n\r\nWe can say that the joint CDF of $n$ random variables:\r\n$$ F_{X_1,\\cdots ,X_n}(x_1,\\cdots ,x_n)= P(X_1 \\leq x_1, X_2 \\leq x_2, \\cdots X_n \\leq x_n)$$\r\n\r\n\\subsection{Properties of joint CDF}\r\n\r\n\\begin{enumerate}\r\n \\item $F_{X,Y}(.,.) $ is monotonically non decreasing in x,y.\r\n\r\n We can compare them for $\\{ \\{ x_1 = x_2, y_1 \\leq y_2 \\}, \\{x_1 \\leq x_2, y_1 = y_2 \\} , \\{ x_1 < x_2 , y_1 < y_2\\}\\}$ but not for $\\{ x_1 > x_2 , y_1 < y_2\\}$ etc.\r\n \\item $$ \\lim_{x \\to -\\infty, y \\to -\\infty} F_{X,Y} (x,y)=0$$\r\n $$ \\lim_{x \\to -\\infty} F_{X,Y} (x,y)=0\\qquad \\lim_{y \\to -\\infty} F_{X,Y} (x,y)=0$$\r\n\r\n \\item $$ \\lim_{x \\to \\infty, y \\to \\infty} F_{X,Y} (x,y)=1$$\r\n $$ \\lim_{x \\to \\infty} F_{X,Y} (x,y)= F_Y (y)\\qquad \\lim_{y \\to \\infty} F_{X,Y} (x,y)= F_X(x)$$\r\n\r\n \\item $F_{X,Y}(.,.)$ is right continuous individually in x \\& y.\r\n $$ \\lim_{x \\to x_0^{+}} F_{X,Y} (x,y)=F_X(x_0,y)$$\r\n $$ \\lim_{(x,y) \\to (x_0^{+},y_0^{+})} F_{X,Y} (x,y)=F_X(x_0,y_0)$$\r\n\r\n\\end{enumerate}\r\n\r\nDifferent cases possible for $X$ \\& $Y$:\r\n\\begin{itemize}\r\n \\item $X$ \\& $Y$ both discrete.\r\n \\item $X$ \\& $Y$ both continuous.\r\n \\item $X$ is discrete \\& $Y$ is continuous.\r\n\\end{itemize}\r\n\r\n\\begin{enumerate}\r\n \\item $X$ \\& $Y$ both discrete.\r\n\r\n Joint PMF:\r\n $$ P_{X,Y}(x_i,y_i)= P(X= x_i,Y= y_i)$$\r\n Expressing joint CDF in terms of the joint PMF,\r\n $$ F_{X,Y}(x,y) = \\sum_{x_i \\leq x} \\sum_{y_i \\leq y} P_{X,Y}(x_i,y_i)$$\r\n\r\n $X\\to P_X \\qquad Y \\to P_Y \\qquad (X,Y) \\to P_{X,Y}$\r\n\r\n We can form a relation between $P_{X,Y}$ and $P_X$,\r\n $$ P_X(x_i) \\sum_{y_i} P_{X,Y}(x_i,y_i)$$\r\n $$ P(X= x_i) \\sum_{y_i} P_{X,Y}(X=x_i,Y=y_i)$$\r\n\r\n In the context of the joint PMF, $P_X$ \\& $P_Y$ are also called as marginal PMF's.\r\n And the process of summing over the range of the other random variable to get the marginal PMF, is known as marginalization.\r\n\\end{enumerate}\r\n\r\n\\subsection{Conditional PMF}\r\n\r\nLet us recall conditional probabilities (in the probability space $(S,F,P)$),\r\n$$ P(A/B)= \\frac{P(A \\cap B)}{P(B)} \\qquad P(B)>0$$\r\nIn the context of random variables, how do we define conditional probabilities?\r\n\r\nGiven an event $B \\in F$, $P(B)>0$. Conditional CDF conditioned on the event $B$ would be:\r\n$$ F_{X/B}(x/B)= \\frac{P( (X \\leq x)\\cap B)}{P(B)}$$\r\n$X \\leq x= \\{ w: X(w) \\leq x\\}= A \\in F$\r\n\r\nIf $B$ is of the form $\\{ Y \\leq y \\}$\r\n$$ F_{X/B}(x/B)= \\frac{P((X \\leq x) \\cap B)}{P(B)}$$\r\n$$ F_{X/\\{ Y \\leq y \\}}(x / \\{ Y \\leq y \\})= \\frac{P(\\{ X \\leq x \\} \\cap \\{ Y \\leq y \\})}{P(\\{ Y \\leq y \\})}= \\frac{F_{X,Y}(x,y)}{F_Y (y)}$$\r\nWe shall denote this, conditional PMF, by:\r\n$$ F_{X/Y}(x/y)= \\frac{F_{X,Y}(x,y)}{F_Y (y)}$$\r\n\r\nIf both $X$ \\& $Y$ are discrete, for defining conditional PMF, we will not condition on $\\{ Y \\leq y_i \\} $ but instead condition on $\\{ Y = y_i \\}$\r\n\r\n$$ P_{X/\\{ Y = y_i \\}}(x_i/\\{ Y = y_i \\})= \\frac{P((X= x_i) \\cap (Y=y_i))}{P(Y=y_i)}$$\r\nConditional PMF:\r\n$$ P_{X/Y} (x_i /y_i)= \\frac{P_{X,Y}(x_i,y_i)}{P_Y(y_i)}= \\frac{\\text{Joint PMF}}{\\text{Marginal PMF}}$$\r\n\\\\\r\n\r\n% Example:\r\n\r\n%28/6\r\n\r\nIf $X$ \\& $Y$ are both continuous random variables, we can define joint PDF of $X$ \\& $Y$ as\r\n$$ f_{X,Y}(x,y)= \\frac{\\partial^2 F_{X,Y}(x,y)}{\\partial x \\partial y}$$\r\n\r\n$$ F_{X,Y}(x,y) =\\int_{-\\infty}^{x} \\int_{-\\infty}^{y} f_{X,Y}(x,y) dx dy$$\r\n\r\nMarginal PDF's of $X$ \\& $Y$ can be expressed in terms of joint PDF's as:\r\n$$ f_X (x)= \\int_{-\\infty}^{\\infty} f_{X,Y}(x,y) dy$$\r\n$$ f_Y (y)= \\int_{-\\infty}^{\\infty} f_{X,Y}(x,y) dx$$\r\n\r\nA joint PDF is said to be valid if,\r\n$$ f_{X,Y}(x,y) \\geq 0 \\quad \\& \\quad \\int_{-\\infty}^{x} \\int_{-\\infty}^{y} f_{X,Y}(x,y)\\partial x \\partial y =1$$\r\n\\\\\r\n\r\nWhen both $X$ \\& $Y$ are continuous,\r\n$P(X=x )=0,\\; P(Y=y)=0 $ i.e. probability that the random variable takes a single value, finite or countably infinite set of values is zero.\r\n\r\nConditional PDF can be expressed in terms of joint PDF's and marginal PDF's,\r\n$$ f_{X/Y}(x/y)=\\frac{f_{X,Y}(x,y)}{f_Y (y)}$$ \\\\\r\n\r\n$$ f_{X/Y}(x/y)= \\lim_{\\Delta x \\to 0, \\Delta y \\to 0} \\frac{P(x 0$.\r\n\r\n$$ \\Rightarrow f_{X/Y}(x/y)= \\lim_{\\Delta x \\to 0, \\Delta y \\to 0} \\frac{P(x 1$.\nHowever, \n\\begin{itemize}\n\t\\ii $\\wt H_0(S^n)$ is $\\ZZ$ for $n=0$ and $0$ otherwise.\n\t\\ii $\\wt H_n(S^0)$ is $\\ZZ$ for $m=0$ and $0$ otherwise.\n\\end{itemize}\nSo by induction on $\\min \\{m,n\\}$ we directly obtain that\n\\[\n\t\\wt H_n(S^m) \\cong\n\t\\begin{cases}\n\t\t\\ZZ & m=n \\\\\n\t\t0 & \\text{otherwise}\n\t\\end{cases}\n\\]\nwhich is what we wanted.\n\nTo prove the claim, let's consider the exact sequence\nformed by the pair $X = D^2$ and $A = S^1$.\n\\begin{example}[The long exact sequence for $(X,A) = (D^2, S^1)$]\n\tConsider $D^2$ (which is contractible) with boundary $S^1$.\n\tClearly $S^1$ is a deformation retraction of $D^2 \\setminus \\{0\\}$,\n\tand if we fuse all points on the boundary together we get $D^2 / S^1 \\cong S^2$.\n\tSo we have a long exact sequence\n\t\\begin{diagram}\n\t\t\\wt H_2(S^1) & \\rTo & \\underbrace{\\wt H_2(D^2)}_{=0} & \\rTo & \\wt H_2(S^2) \\\\\n\t\t&&& \\ldTo(4,2) & \\\\\n\t\t\\wt H_1(S^1) & \\rTo & \\underbrace{\\wt H_1(D^2)}_{=0} & \\rTo & \\wt H_1(S^2) \\\\\n\t\t&&& \\ldTo(4,2) & \\\\\n\t\t\\wt H_0(S^1) & \\rTo & \\underbrace{\\wt H_0(D^2)}_{=0} & \\rTo & \\underbrace{\\wt H_0(S^2)}_{=0}\n\t\\end{diagram}\n\tFrom this diagram we read that\n\t\\[\n\t\t\\dots, \\quad\n\t\t\\wt H_3(S^2) = \\wt H_2(S^1), \\quad\n\t\t\\wt H_2(S^2) = \\wt H_1(S^1), \\quad\n\t\t\\wt H_1(S^2) = \\wt H_0(S^1).\n\t\\]\n\\end{example}\nMore generally, the exact sequence for the pair $(X,A) = (D^m, S^{m-1})$\nshows that $\\wt H_n(S^m) \\cong \\wt H_{n-1}(S^{m-1})$,\nwhich is the desired conclusion.\n\n\\section{Invariance of dimension}\nHere is one last example of an application of excision.\n\\begin{definition}\n\tLet $X$ be a space and $p \\in X$ a point.\n\tThe $k$th \\vocab{local homology group} of $p$ at $X$ is defined as\n\t\\[ H_k(X, X \\setminus \\{p\\}). \\]\n\\end{definition}\nNote that for any open neighborhood $U$ of $p$, we have by excision that\n\\[ H_k(X, X \\setminus \\{p\\}) \\cong H_k(U, U \\setminus \\{p\\}). \\]\nThus this local homology group only depends on the space near $p$.\n\n\\begin{theorem}\n\t[Invariance of dimension, Brouwer 1910]\n\tLet $U \\subseteq \\RR^n$ and $V \\subseteq \\RR^m$ be nonempty open sets.\n\tIf $U$ and $V$ are homeomorphic, then $m = n$.\n\\end{theorem}\n\\begin{proof}\n\tConsider a point $x \\in U$ and its local homology groups. By excision,\n\t\\[ H_k(\\RR^n, \\RR^n \\setminus \\{x\\}) \\cong\n\t\tH_k(U, U \\setminus \\{x\\}). \\]\n\tBut since $\\RR^n \\setminus \\{x\\}$ is homotopic to $S^{n-1}$,\n\tthe long exact sequence of \\Cref{thm:long_exact_rel} tells us\n\tthat \n\t\\[\n\t\tH_k(\\RR^n, \\RR^n \\setminus \\{x\\}) \n\t\t\\cong\n\t\t\\begin{cases}\n\t\t\t\\ZZ & k = n \\\\\n\t\t\t0 & \\text{otherwise}.\n\t\t\\end{cases}\n\t\\]\n\tAnalogously, given $y \\in V$ we have\n\t\\[ H_k(\\RR^m, \\RR^m \\setminus\\{y\\}) \\cong H_k(V, V\\setminus\\{y\\}). \\]\n\tIf $U \\cong V$, we thus\n\tdeduce that\n\t\\[ H_k(\\RR^n, \\RR^n\\setminus\\{x\\}) \\cong H_k(\\RR^m, \\RR^m\\setminus\\{y\\}) \\]\n\tfor all $k$. This of course can only happen if $m=n$.\n\\end{proof}\n\n\\section\\problemhead\n\\begin{problem}\n\tLet $X = S^1 \\times S^1$ and $Y = S^1 \\vee S^1 \\vee S^2$.\n\tShow that \\[ H_n(X) \\cong H_n(Y) \\] for every integer $n$.\n\\end{problem}\n\n\\begin{problem}[Hatcher \\S2.1 exercise 18]\n\tConsider $\\QQ \\subset \\RR$.\n\tCompute $\\wt H_1(\\RR, \\QQ)$.\n\t\\begin{hint}\n\t\tUse \\Cref{thm:long_exact_rel}.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tWe have an exact sequence\n\t\t\\[\n\t\t\t\\underbrace{\\wt H_1(\\RR)}_{=0}\n\t\t\t\\to \\wt H_1(\\RR, \\QQ) \\to \\wt H_0(\\QQ) \\to \n\t\t\t\\underbrace{\\wt H_0(\\RR)}_{=0}.\n\t\t\\]\n\t\tNow, since $\\QQ$ is path-disconnected\n\t\t(i.e.\\ no two of its points are path-connected)\n\t\tit follows that $\\wt H_0(\\QQ)$ consists of\n\t\tcountably infinitely many copies of $\\ZZ$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{sproblem}\n\tWhat are the local homology groups of a topological $n$-manifold?\n\\end{sproblem}\n\n\\begin{problem}\n\tLet \\[ X = \\{(x,y) \\mid x \\ge 0\\} \\subseteq \\RR^2 \\]\n\tdenote the half-plane.\n\tWhat are the local homology groups of points in $X$?\n\t% http://math.stackexchange.com/questions/350667/local-homology-group-a-homeomorphism-takes-the-boundary-to-the-boundary\n\\end{problem}\n\n\\begin{problem}\n\t[Brouwer-Jordan separation theorem,\n\tgeneralizing Jordan curve theorem]\n\t\\yod\n\tLet $X \\subseteq \\RR^n$ be a subset\n\twhich is homeomorphic to $S^{n-1}$.\n\tProve that $\\RR^n \\setminus X$\n\thas exactly two path-connected components.\n\t\\begin{hint}\n\t\tFor any $n$, prove by induction for $k=1,\\dots,n-1$ that\n\t\t(a) if $X$ is a subset of $S^n$ homeomorphic to $D^k$\n\t\tthen $\\wt H_i(S^n \\setminus X) = 0$;\n\t\t(b) if $X$ is a subset of $S^n$ homeomorphic to $S^k$\n\t\tthen $\\wt H_i(S^n \\setminus X) = \\ZZ$ for $i=n-k-1$\n\t\tand $0$ otherwise.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tThis is shown in detail in Section 2.B of Hatcher.\n\t\\end{sol}\n\\end{problem}\n", "meta": {"hexsha": "05f6a17b42d49f72aaf6807d7ea4b179693ef5bf", "size": 13609, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/homology/excision.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/homology/excision.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/homology/excision.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.719160105, "max_line_length": 121, "alphanum_fraction": 0.6551546771, "num_tokens": 5091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833737577158, "lm_q2_score": 0.8333245911726382, "lm_q1q2_score": 0.6454793732657713}} {"text": "%\\documentclass{book}\n%\\begin{document}\n\\chapter{Converting Coordinates \\& Time}\n\\section{Transformations}\n\nLet $\\mathbf{i}_{x}$, $\\mathbf{i}_{y}$, $\\mathbf{i}_{z}$ and $\\mathbf{i}_{\\varepsilon}$, $\\mathbf{i}_{\\eta}$, $\\mathbf{i}_{\\zeta}$ be two sets of orthogonal unit vectors\n\n\\[\\mathbf{i}_{\\xi}=l_{1}\\mathbf{i}_{x}+m_{1}\\mathbf{i}_{y}+n_{1}\\mathbf{i}_{z}\\]\n\\[\\mathbf{i}_{\\eta}=l_{2}\\mathbf{i}_{x}+m_{2}\\mathbf{i}_{y}+n_{2}\\mathbf{i}_{z}\\]\n\\[\\mathbf{i}_{\\zeta}=l_{3}\\mathbf{i}_{x}+m_{3}\\mathbf{i}_{y}+n_{3}\\mathbf{i}_{z}\\]\n\n\\[ \\left[ \\begin{array}{c} x \\\\ y \\\\ z \\end{array} \\right] = \\mathbf{R}\\left[ \\begin{array}{c} \\varepsilon \\\\ \\eta \\\\ \\zeta \\end{array} \\right] \\mbox{or}\n\\left[ \\begin{array}{c} \\varepsilon \\\\ \\eta \\\\ \\zeta \\end{array} \\right] = \\mathbf{R^{T}}\\left[ \\begin{array}{c} x \\\\ y \\\\ z \\end{array} \\right] \\]\n\n\\[\\mathbf{R}=\\left[ \\begin{array}{ccc}\n\\mathbf{i}_{x}\\mathbf{\\cdot}\\mathbf{i}_{\\varepsilon} & \\mathbf{i}_{x}\\mathbf{\\cdot}\\mathbf{i}_{\\eta} & \\mathbf{i}_{x}\\mathbf{\\cdot}\\mathbf{i}_{\\zeta} \\\\\n\\mathbf{i}_{y}\\mathbf{\\cdot}\\mathbf{i}_{\\varepsilon} & \\mathbf{i}_{y}\\mathbf{\\cdot}\\mathbf{i}_{\\eta} & \\mathbf{i}_{y}\\mathbf{\\cdot}\\mathbf{i}_{\\zeta} \\\\\n\\mathbf{i}_{z}\\mathbf{\\cdot}\\mathbf{i}_{\\varepsilon} & \\mathbf{i}_{z}\\mathbf{\\cdot}\\mathbf{i}_{\\eta} & \\mathbf{i}_{z}\\mathbf{\\cdot}\\mathbf{i}_{\\zeta}\n\\end{array} \\right] = \\left[ \\begin{array}{ccc}\nl_{1} & l_{2} & l_{3} \\\\\nm_{1} & m_{2} & m_{3} \\\\\nn_{1} & n_{2} & n_{3}\n\\end{array} \\right] \\]\n\n\\[ \\mathbf{R^{T}}=\\mathbf{R^{-1}} \\]\n\nEquations found here~\\cite[pp. 81-82]{battin:imma}\n\n\\section{Time Systems}\n\\subsection{Solar \\& Sidereal Time}\nSince the beginning time has been kept by counting the days. An apparent solar day is the minimum time elapsed between the sun crossing a specified meridian and then recrossing the same meridian. This form of time keeping is problematic because no two apparent solar days are of the same duration due to Earth's rotation around the sun as well as around its axis (the Earth does a little more than one rotation per apparent solar day). Also, Earth's rotational speed is not constant and its axis of rotation is tilted $23.5^{\\circ}$ to the orbital plane. These imperfections call for correction, and thus mean solar time was created. A day in mean solar time is defined as one revolution of a hypothetical sun that orbits at the equator, and is more commonly known as Greenwich Mean Time. Another solution is to base our day on the crossing of a star much farther away thus minimizing the effect of the Earth's orbital movement, this method of time keeping is known as sidereal time. A sidereal day is about 4 minutes shorter than a solar day, and is used heavily by astronomers. Sidereal time is not truly stable either so mean sidereal day was introduced, and is known as Greenwich Apparent Sidereal Time. Universal Time (UT) refers to any time scale based on the Earth's rotation. UT0 refers to the mean solar time at the prime meridian as obtained from astronomical observation, and UT1 is UT0 corrected for polar motion. Briefly ephemeris time was introduced to standardize the second, which was defined as 1/31556925.9747 of the year 1900. This was soon replaced by atomic time~\\cite[pp. 84-86]{me:gsmp}.\n\\subsection{Atomic Time}\nThe second is now defined by an atomic standard that is based on the resonance frequency of the cesium atom. To be precise, the second is defined as ``9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom,'' whose duration happens to exactly match the ephemeris second discussed in the previous section. The problem with detaching our time keeping method from the Earth is that as the Earth slows its rotation noon will move closer to midnight (over the duration of thousands of years, of course). Coordinated Universal Time (UTC) was introduced to prevent this. UTC is a compromise between the precision of atomic time and the groundedness of Earth based time keeping, it uses the atomic second but introduces leap seconds (positive or negative) when necessary to keep UTC within .9 seconds of UT1~\\cite[pp. 86-87]{me:gsmp}.\n\\subsection{Time Formats}\nWe are used to dealing with months, days, years, hours, minutes, and seconds, but such a time format makes for difficult epoch calculations over long periods. To solve this problem Julian Date (JD) was introduced. JD consists of a day count (days since noon UT on January $1^{\\mbox{st}}$ 4713 B.C.) and a fraction of the current day. This makes for easy time differencing, but the length of the date can become cumbersome and the fact that a new day starts at noon confusing. To make things even easier Modified Julian Date (MJD) was created whose origin is midnight November $17^{\\mbox{th}}$, 1858.\n\\[ \\mbox{MJD}=\\mbox{JD}-2400000.5\\]\n\nIn order to make Julian Date useful we need an easy way to go between calendar dates and JD. \\application{timeconvert}\\index{timeconvert!theory} does this and more with ease. The equations to convert from calendar date to JD are\n\\[ \\mbox{JD}=\\mbox{INT}[365.25y]+\\mbox{INT}[30.6001(m+1)]+D+\\mbox{UT}/24+1720981.5\\]\n\\[ \\begin{array}{lll}\ny=Y-1 & \\mbox{and}~m=M+12 & \\mbox{if}~M \\leq2 \\\\\ny=Y & \\mbox{and}~m=M & \\mbox{if}~M > 2\n\\end{array} \\]\nwhere $M$ is the month, $D$ is the day, $Y$ is the year, and INT[$x$] returns just the integer part of the number. To go from JD to calendar date\n\\[ a=\\mbox{INT[JD}+0.5] \\]\n\\[ b=a+1537 \\]\n\\[ c=\\mbox{INT}[(b-122.1)/365.25] \\]\n\\[ d=\\mbox{INT}[365.25c] \\]\n\\[ e=\\mbox{INT}[(b-d)/30.6001] \\]\n\\[ D=b-d-\\mbox{INT}[30.6001e]+\\mbox{FRAC[JD}+0.5] \\]\n\\[ M=e-1-12\\mbox{INT}[e/14] \\]\n\\[ Y=c-4715-\\mbox{INT}[(7+M)/10] \\]\nwhere FRAC[$x$] returns just the fractional part of a real number. MJD Conversion found here~\\cite[p. 88]{me:gsmp}. All other date conversions were found here~\\cite[pp. 36-37]{hlc:gtp}\n\\subsection{GPS Time}\nGPS Time (GPST) is a continuously running composite time kept by cesium and rubidium frequency standards aboard the satellites and at monitor stations. While there are no leap seconds in GPST as there are in UTC, it is steered to stay within 1 $\\mu$s of UTC, that is the difference between GPST and UTC is an integer number of seconds plus a fraction of a $\\mu$s. GPST is formatted in terms of GPS weeks and the number of seconds into the current week. Finding these values is done easily if the Julian Date is known.\n\\[ \\mbox{GPS WEEK}=\\mbox{INT[(JD}-2444244.5)/7] \\]\n\\[ \\mbox{SOW}=\\mbox{FRAC[(JD}-2444244.5)/7]\\times 604800 \\]\nwhere INT[$x$] returns the integer part of a real number, FRAC[$x$] returns the fractional part, and SOW stands for Second of Week.\n\nOther useful quantities such as Day of Week and Second of Day can be found using \\application{timeconvert}\\index{timeconvert!theory} or the following equations.\n\\[\\mbox{DOW}=\\mbox{modulo\\{INT[JD}+0.5],7\\}\\]\n\\[\\mbox{SOD}=\\mbox{modulo\\{FRAC[JD}+0.5],7\\}\\times 86400\\]\nwhere DOW=0 corresponds to Monday, DOW=1 corresponds to Tuesday, and so on.\n\nJD and GPS Week equations were found here~\\cite[pp. 36-37]{hlc:gtp}, SOD derived from DOW equation.\n\\subsection{Z-Count}\nSatellites keep internal time with Z-count, whose epoch period is 1.5 seconds (a convenient unit for communications timing). The full Z-count is 29 bits, the 10 bit GPS week folloed by a 19 bit Time of Week (TOW) expressed in Z-counts (or 1.5 second units). The truncated Z-count has a 17 bit TOW that is expressed in units of 6 seconds, or the length of one subframe's transmission time. Simply multiply the truncated TOW by 4 to get the full TOW~\\cite[pp. 86-88]{tsui:fgpsr}.\n\\[ \\mbox{TOW}=\\mbox{FRAC[(JD}-2444244.5)/7]\\times 403200 \\]\n\\[ \\mbox{Truncated TOW}=\\mbox{FRAC[(JD}-2444244.5)/7]\\times 100800 \\]\nEquations derived from SOW equation above\n\\section{Earth Fixed Coordinates}\n\\subsection{ECI to ECF}\n\\[\\left[\\begin{array}{c} x \\\\ y \\\\ z \\end{array}\\right]_{ECF}=T_{XYZ}^{xyz}\\left[\\begin{array}{c} X \\\\ Y \\\\ Z \\\\ \\end{array}\\right]_{ECI}\\]\n\\[T_{XYZ}^{xyz}=WSNP\\]\n\nP - applies precession, from epoch 2000.0 to the current time;\nN - applies nutation, from epoch 2000.0 to the current time;\nS - applies rotation to account for true sidereal time;\nW - applies polar motion;\n\nEquations found on page 85 of Fundamentals of Orbit Determination paper book.\n\n\\subsection{WGS-84}\nThe World Geodetic System 1984 (WGS-84) is a fixed physical model of Earth produced by the Department of Defense to which many different reference frames can be attached. WGS-84 consists of two parts, a model of Earth's gravitational field, and an ellipsoid describing the Earth's general shape. When dealing with locations on the Earth's surface the ellipsoid provides the foundation for the geodetic coordinate system used by GPS. The ellipsoid's cross-sections parallel to the equatorial plane are circular while those orthogonal are elliptical. The ellipses are parameterized by an eccentricity $e$, a flattening $f$, and sometimes a second eccentricity $e'$\n\\[e=\\sqrt{1-\\frac{b^{2}}{a^{2}}}\\]\n\\[f=1-\\frac{b}{a}\\]\n\\[e'=\\sqrt{\\frac{a^{2}}{b^{2}}-1}=\\frac{a}{b}e\\]\nwhere $a$, the semimajor axis, is the value of the mean equatorial radius of Earth (6,378.137 km) and $b$, the semiminor axis, is the value of the polar radius of Earth (6,356.7523142 km) \\cite[pp. 25-26]{kaplan:ugpspa}.\n\\subsection{Coordinate Systems}\nNow that WGS-84 is defined it is important to understand what coordinate systems can be attached to the ellipsoid and how to move between these different systems. The GPS Toolkit comes with \\application{poscvt}\\index{poscvt!theory}, an application that gives users the ability to easily convert coordinates in one reference frame to another. The coordinate \nsystems that \\application{poscvt}\\index{poscvt!theory} recognizes are Cartesian (or XYZ), geodetic, geocentric, and spherical coordinates. These systems and the formulas to convert between them are discussed below.\n\\subsubsection{Cartesian (XYZ) Coordinates}\nThe Earth Centered Earth Fixed (ECEF) Cartesian coordinate system is fixed to the WGS-84 ellipsoid and is the common ground that makes going between the Earth Centered Inertial (ECI) reference frame used by the satellites and the systems we are used to (such as latitude, longitude, and height) manageable. The equatorial plane makes the $xy$-plane with the $+x$-axis pointing toward $0^{\\circ}$ longitude and the $+y$-axis pointing toward $90^{\\circ}$ E longitude. The $z$-axis is normal to the equatorial plane and points to the geographical north pole. The conversion formulas presented in the next sections will convert to and from this Cartesian reference frame, and so to convert between two non-Cartesian coordinate systems the XYZ system will be used as an intermediary~\\cite[p. 24]{kaplan:ugpspa}.\n\\subsubsection{Geodetic Coordinates}\nThe geodetic coordinate parameters are longitude $\\lambda$, latitude $\\phi$, and height $h$. Longitude is defined as the angle between the position and the $x$-axis in the equatorial plane, and is easily computed given a position in Cartesian coordinates. Let a user's position $\\mathbf{U}=(x_{u},y_{u},z_{u})$, then\n\\[ \\lambda = \\left \\{ \\begin{array}{ll}\n\\arctan \\left( \\frac{y_{u}}{x_{u}} \\right), & \\mbox{$x_{u} \\geq 0$} \\\\\n180^{\\circ} + \\arctan \\left( \\frac{y_{u}}{x_{u}} \\right), & \\mbox{$x_{u} < 0$ and $y_{u} \\geq 0$} \\\\\n-180^{\\circ} +\\arctan \\left( \\frac{y_{u}}{x_{u}} \\right), & \\mbox{$x_{u} < 0$ and $y_{u} < 0$}\n\\end{array}\n\\right. \\]\nwhere negative angles signal west longitude.\n\nLatitude and height are not so straight forward. Latitude is determined by drawing a vector normal to the ellipsoid, beginning somewhere on the equatorial plane and terminating at the users position, we will call this the user vector. The smallest angle between this vector and the equatorial plane is the user's latitude, it is a North latitude for positive angles and South for negative. Notice that unless the user is at a pole or on the equator the vector does not pass through the center of the Earth. The users height is found by taking the magnitude of the vector originating on and normal to the ellipsoid and terminating at the user's position. Latitude $\\phi$ and height $h$ are found using the following equations\n\\[ \\phi = \\arctan\\left(\\frac{z_{u}+e'^{2}z_{0}}{r}\\right) \\]\n\\[ h = U \\left(1-\\frac{b^{2}}{aV}\\right) \\]\n\nwhere\n\n\\[ r = \\sqrt{x_{u}^{2}+y_{u}^{2}} \\]\n\\[ E^{2} = a^{2} - b^{2} \\]\n\\[ F = 54 b^{2} z_{u}^{2} \\]\n\\[ G = r^{2} + (1-e^{2}) z_{u}^{2} - e^{2} E^{2} \\]\n\\[ c = \\frac{e^{4} F r^{2}}{G^{3}} \\]\n\\[ s = \\sqrt[3]{1+c+\\sqrt{c^{2} + 2c}}\\]\n\\[ P = \\frac{F}{3 \\left( s + \\frac{1}{s} + 1 \\right)^{2}G^{2} } \\]\n\\[ Q = \\sqrt{1+2e^{4}P} \\]\n\\[ r_{0} = -\\frac{Pe^{2}r}{1+Q}+\\sqrt{\\frac{1}{2}a^{2} \\left(1+\\frac{1}{Q}\\right)-\\frac{P(1-e^{2})z_{u}^{2}}{Q(1+Q)}-\\frac{1}{2}Pr^{2}}\\]\n\\[ U = \\sqrt{(r-e^{2}r_{0})^{2}+z_{u}^{2}}\\]\n\\[ V = \\sqrt{(r-e^{2}r_{0})^{2}+(1-e^{2})z_{u}^{2}} \\]\n\\[ z_{0} = \\frac{b^{2}z_{u}}{aV} \\]\n\n\n%\\[x=(N_{h}+h)\\cos\\phi'\\cos\\lambda\\]\n%\\[y=(N_{h}+h)\\cos\\phi'\\sin\\lambda\\]\n%\\[z=(N_{h}+h-\\tilde{e}^{2}N_{h})\\sin\\phi'\\]\n%\\[\\tilde{e}^{2}=\\tilde{f}(2-\\tilde{f})\\]\n%\\[N_{h}=\\frac{R_{e}}{(1-\\tilde{e}^{2}\\sin^{2}\\phi')^{1/2}}\\]\n%\\[\\tilde{f}=\\frac{R_{e}-R_{p}}{R_{e}}\\]\n%\\[x^{2}+y^{2}+\\left(\\frac{R_{e}}{R_{p}}\\right)^{2}z^{2}=R_{e}^{2}\\]\n\nGoing back to Cartesian coordinates from the geodetic system ($\\lambda$ $\\phi$ $h$) can be done more compactly\n\\[ \\mathbf{u} = \\left[ \\begin{array}{c}\n\\frac{a\\cos\\lambda}{\\sqrt{1+(1-e^2)\\tan^{2}\\phi}}+h\\cos\\lambda\\cos\\phi \\\\\n\\frac{a\\sin\\lambda}{\\sqrt{1+(1-e^2)\\tan^{2}\\phi}}+h\\sin\\lambda\\cos\\phi \\\\\n\\frac{a(1-e^{2})\\sin\\phi}{\\sqrt{1-e^{2}\\sin^{2}\\phi}}+h\\sin\\phi\n\\end{array}\n\\right] \\]\nwhere $\\mathbf{u}$ is the user's position vector~\\cite[pp. 26-28, p. 76]{kaplan:ugpspa,me:gsmp}.\n\n\\subsubsection{Geocentric Coordinates}\n\\[x=r\\cos\\phi\\cos\\lambda\\]\n\\[y=r\\cos\\phi\\sin\\lambda\\]\n\\[z=r\\sin\\phi\\]\n\nwhere $\\lambda$ and $\\phi$ are geocentric longitude and latitude\n\nfound on page 82 in the Fundamentals of Orbital Determination paper book\n\n%\\subsubsection{Spherical Coordinates}\n\n\\subsubsection{Topocentric Coordinates}\n\\[\\mathbf{r}_{t}=T_{t}(\\mathbf{r}-\\mathbf{r}_{s})=T_{t}\\rho\\]\n\n$\\mathbf{r}$ and $\\mathbf{r}_{s}$ are the position vectors of the observer and satellite respectively in the Earth-fixed system\n\n\\[T_{t}=\\left[ \\begin{array}{ccc}\n-\\sin\\lambda & \\cos\\lambda & 0 \\\\\n-\\sin\\phi\\cos\\lambda & -\\sin\\phi\\sin\\lambda & \\cos\\phi \\\\\n\\cos\\phi\\cos\\lambda & \\cos\\phi\\sin\\lambda & \\sin\\phi \\end{array} \\right] \\]\n\nwhere $\\lambda$ and $\\phi$ are geocentric longitude and latitude\n\nfound on page 84 in the Fundamentals of Orbital Determination paper book\n\nto find \\emph{azimuth} (Az) and \\emph{elivation} (El)\n\n\\[ \\begin{array}{ll}\n\\sin\\mbox{El}=\\frac{z_{t}}{r_{t}} & -90^{\\circ} \\leq \\mbox{El} \\leq 90^{\\circ} \\\\\n\\sin\\mbox{Az}=\\frac{x_{t}}{r_{xy}} & \\\\\n\\cos\\mbox{Az}=\\frac{y_{t}}{r_{xy}} & 0^{\\circ} \\leq \\mbox{Az} \\leq 360^{\\circ}\n\\end{array} \\]\n\nEquations found on pages 84-85 in Fundamentals of Orbit Determination paper book\n\n\\putbib[gpstk]\n%\\end{document}\n", "meta": {"hexsha": "4b29afe2420813501b17ec4cc6370ca7b3d360a9", "size": 14998, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "3rdparty/GPSTk/ref/usersguide/convcoordtime.tex", "max_stars_repo_name": "mfkiwl/ICE", "max_stars_repo_head_hexsha": "e660d031bb1bcea664db1de4946fd8781be5b627", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2019-10-12T01:22:20.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-15T23:28:26.000Z", "max_issues_repo_path": "3rdparty/GPSTk/ref/usersguide/convcoordtime.tex", "max_issues_repo_name": "wuyou33/Enabling-Robust-State-Estimation-through-Measurement-Error-Covariance-Adaptation", "max_issues_repo_head_hexsha": "2f1ff054b7c5059da80bb3b2f80c05861a02cc36", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "3rdparty/GPSTk/ref/usersguide/convcoordtime.tex", "max_forks_repo_name": "wuyou33/Enabling-Robust-State-Estimation-through-Measurement-Error-Covariance-Adaptation", "max_forks_repo_head_hexsha": "2f1ff054b7c5059da80bb3b2f80c05861a02cc36", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 14, "max_forks_repo_forks_event_min_datetime": "2019-11-05T01:50:29.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-06T06:23:44.000Z", "avg_line_length": 84.2584269663, "max_line_length": 1611, "alphanum_fraction": 0.7067609015, "num_tokens": 4885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382165412808, "lm_q2_score": 0.7490872187162397, "lm_q1q2_score": 0.6453672664466575}} {"text": "\\chapter{The Theory of \\\\ Automatic Differentiation}\n\nAutomatic differentiation is a powerful computation tool for the exact, at least up to \narithmetic precision, evaluation of linear differential operators on the elaborate, \ncomposite functions common to practical applications. In particular, automatic \ndifferentiation techniques factor a linear differential operator into analytic calculations, \nwhich can be explicitly implemented, from the algebraic manipulations that can be \nprogrammatically automated.\n\nIn this chapter we review the theory behind linear differential operators, dual numbers,\nand, ultimately, automatic differentiation. We begin by considering differential operators\nin a formal setting, in particular the spaces on which they act, before introducing dual\nnumbers and showing how they compliment these operators. Last we integrate both\nperspectives into automatic differentiation and its implementation.\n\n\\section{Linear Differential Operators}\n\nWhen considering a one-dimensional function, $f : \\mathbb{R} \\rightarrow \\mathbb{R}$, \na common notion is that the derivative defines a ``best'' linear approximation\nto the function in a neighborhood around any point, $x_{0} \\in \\mathbb{R}$,\n%\n\\begin{equation*}\nf \\! \\left( x \\right) - f \\! \\left( x_{0} \\right)\n\\approx \\frac{ \\mathrm{d} f }{ \\mathrm{d} x } \\! \\left( x_{0} \\right) \\left( x - x_{0} \\right).\n\\end{equation*}\n%\nA more revealing interpretation, however, arises when we consider\n$\\delta x = x - x_{0}$ and $\\delta f = f \\! \\left( x \\right) - f \\! \\left( x_{0} \\right)$ as\nperturbations, in which case the derivative is just a linear map that propagates\nan input perturbation into an output perturbation,\n%\n\\begin{equation*}\n\\delta f = \\frac{ \\mathrm{d} f }{ \\mathrm{d} x } \\! \\left( x_{0} \\right) \\delta x.\n\\end{equation*}\n\nLinear differential operators formalize this intuition of propagating perturbations\nthrough functions. Any formal treatment, however, first requires a much more\ncareful definition of perturbations, especially for multivariate functions. In this\nsection we review the necessary theory of first-order linear differential operators\nand then briefly discuss the generalization to higher-order operations.\n\n\\subsection{First-Order Linear Differential Operators}\n\nFirst-order linear differential operators generalize the one-dimensional construction\nabove. The most significant complication is that, in general, there are actually two \nkinds of first-order perturbations: \\textit{tangents} and \\textit{sensitivities}.\n\n\\subsubsection{Tangents}\n\nA defining property of perturbations is that they can be added together and scaled,\nin other words they form a vector space. Given the real space $\\mathbb{R}^{N}$,\nthe $N$-dimensional vector space of perturbations at every point, $x \\in \\mathbb{R}^{N}$, \nis called the \\textit{tangent space} and denoted $T_{x} \\mathbb{R}^{N}$. The notation\ncomes from interpreting first-order perturbations as tangent vectors of curves\npassing through $x$.\n\nConsequently, given a smooth function $f : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}^{M}$\na first-order linear differential operator is a linear map from the tangent space of\nthe input to the tangent space of the output,\n%\n\\begin{equation*}\nD : T_{x} \\mathbb{R}^{N} \\rightarrow T_{ f ( x ) } \\mathbb{R}^{M}.\n\\end{equation*}\n%\nTaking the standard bases in both tangent spaces, we can identify any tangent,\n$\\delta x \\in T_{x} \\mathbb{R}^{N}$, by its components, $\\delta x_{n}$. In this case\nthe operator $D$ is equivalent to an $m \\times n$ matrix,\n%\n\\begin{equation*}\n\\delta f_{m} = \\sum_{n = 1}^{N} D_{mn} \\, \\delta x_{n} \\in T_{f(x)} \\mathbb{R}^{M}.\n\\end{equation*}\n\nOne particularly important first-order linear differential operator is the \\textit{pushforward}\nor \\textit{tangent map} canonically defined for any function,\n%\n\\begin{equation*}\nf_{*} : T_{x} \\mathbb{R}^{N} \\rightarrow T_{ f ( x ) } \\mathbb{R}^{M}.\n\\end{equation*}\n%\nIn components the pushforward is given exactly by the Jacobian matrix of partial derivatives,\n%\n\\begin{equation*}\n\\left( f_{*} \\right)_{mn} = \\frac{ \\partial f_{m} }{ \\partial x_{n} }.\n\\end{equation*}\n\nPushforwards behave well under function composition. Given two functions,\n$f : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}^{L}$ and\n$g : \\mathbb{R}^{L} \\rightarrow \\mathbb{R}^{M}$, the pushforward of their compositions\nis simply\n%\n\\begin{equation*}\n\\left( g \\circ f \\right)_{*} = g_{*} \\cdot f_{*},\n\\end{equation*}\n%\nor, in components,\n%\n\\begin{align*}\n\\left( \\left( g \\circ f \\right)_{*} \\right)_{mn} &= \n\\sum_{l = 1}^{L} \\left( g_{*} \\right)_{ml} \\left( f_{*} \\right)_{ln}\n\\\\\n\\frac{ \\partial g_{m} }{ \\partial x_{n} } &= \n\\sum_{l = 1}^{L} \\frac{ \\partial g_{m} }{ \\partial f_{l} } \\cdot \\frac{ \\partial f_{l} }{ \\partial x_{n} },\n\\end{align*}\n%\nwhich is just the chain rule. Because of this behavior, the pushforward of a\ncomposite function is straightforward computationally, provided that the pushforwards\nof each component function are known. In particular, using the standard basis the\ncomposition pushforward is given simply by a succession of vector-matrix products.\n\n\\subsubsection{Sensitivities}\n\nAn immediate consequence of this more formal perspective is the introduction of a second\nkind of first-order perturbation. The space of linear operators acting on tangents,\n%\n\\begin{equation*}\n\\alpha : T_{x} \\mathbb{R}^{N} \\rightarrow \\mathbb{R},\n\\end{equation*}\n%\nis itself a vector space, known as the \\textit{cotangent space} and denoted \n$T^{*}_{x} \\mathbb{R}^{N}$. Composing these maps with a first-order linear differential \noperator assigns a value to each tangent,\n%\n\\begin{equation*}\n\\alpha \\circ D : T_{x} \\mathbb{R}^{N} \\rightarrow \\mathbb{R},\n\\end{equation*}\n%\nin some sense quantifying how sensitive the underlying function is to a given \nperturbation. Consequently elements of the cotangent space are often known as\n\\textit{sensitivities} and provide dual perturbation to the tangents.\n\nAny first-order linear differential operator,\n$D : T_{x} \\mathbb{R}^{N} \\rightarrow T_{ f ( x ) } \\mathbb{R}^{M}$, acting on\nthe tangent spaces canonically defines an \\textit{adjoint operator} that acts on the \ncotangent spaces,\n%\n\\begin{equation*}\nD^{*} : T^{*}_{f (x) } \\mathbb{R}^{M} \\rightarrow T^{*}_{x} \\mathbb{R}^{N}.\n\\end{equation*}\n%\nThe matrix representation of the adjoint operator is, perhaps unsurprisingly, \nsimply the adjoint of the matrix representation of the original operator,\n%\n\\begin{equation*}\n\\mathrm{d} x_{n} = \\sum_{m = 1}^{M} D_{nm} \\mathrm{d} f_{m},\n\\end{equation*}\n%\nwhere $\\mathrm{d} x \\in T^{*}_{x} \\mathbb{R}^{N}$ and\n$\\mathrm{d} f \\in T^{*}_{x} \\mathbb{R}^{M}$.\n\nThe adjoint of the pushforward is known as the \\textit{pullback},\n%\n\\begin{equation*}\nf^{*} : T^{*}_{f (x) } \\mathbb{R}^{M} \\rightarrow T^{*}_{x} \\mathbb{R}^{N}.\n\\end{equation*}\n%\nUnder the composition,\n%\n\\begin{equation*}\n\\left( g \\circ f \\right)^{*} = f^{*} \\cdot g^{*}\n\\end{equation*}\n%\nor, in components,\n%\n\\begin{equation*}\n\\left( \\left( g \\circ f \\right)^{*} \\right)_{mn} = \n\\sum_{l = 1}^{L} \\left( g^{*} \\right)_{ml} \\left( f^{*} \\right)_{ln}.\n\\end{equation*}\n%\nComputing the input sensitivities corresponding to the output sensitivities then reduces\nto vector-matrix products moving backwards, against the flow of the function evaluation.\n\n\\subsection{Higher-Order Linear Differential Operators}\n\nUnfortunately, a theory of higher-order perturbations, and the corresponding\nlinear differential operators, is substantially more complicated, requiring an elegant but\nabstract construction known as \\textit{jet spaces} to generalize the tangent space. \nFor example, a second-order linear differential operator does simply map second-order \ntangents to second-order tangents but also requires \\textit{two} first-order tangents,\n%\n \\begin{equation*}\n\\delta^{2} f_{m} \n= \\sum_{n = 1}^{N} \\frac{ \\partial f_{m} }{ \\partial x_{n} } \\delta^{2} x_{n} \n+ \\sum_{n = 1}^{N} \\sum_{n' = 1}^{N} \n\\frac{ \\partial^{2} f_{m} }{ \\partial x_{n} \\partial x_{n'} } \n\\delta x_{n} \\delta x_{n'}.\n\\end{equation*}\n%\nMoreover, adjoint operators do not immediately generalize -- at higher-orders adjoint\nmaps require the interaction of tangents and sensitivities.\n\nFortunately we do not have to consider these higher-order jet spaces directly\nbecause the introduction of \\textit{dual numbers} provides an alternative means\nof computing how higher-order perturbations propagate across a function.\n\n\\section{Dual Numbers}\n\nDual numbers are a generalization of real numbers that naturally unify the\naction of a function, its pushforward, and, with some care, its pullback;\nconsequently they are the ideal basis for implementing automatic differentiation. \nIn this section we derive first-order dual numbers and then consider the construction \nof higher-order dual numbers and higher-order linear differential operators.\n\n\\subsection{First-Order Dual Numbers}\n\nDual numbers extend the real numbers, $\\mathbb{R}$, with the\naddition of a new element, $\\mathfrak{a}$, which is nilpotent, $\\mathfrak{a}^{2} = 0$,\nand denoted a \\textit{dual unit}.\n\nThis extended space of dual numbers forms a two-dimensional associative algebra \nover $\\mathbb{R}$. In other words, the addition of the dual unit generates a two-dimensional\nspace, $\\mathbb{D}$, comprised of elements,\n%\n\\begin{equation*}\n\\xi = x + \\mathfrak{a} \\, \\delta x; \\, \\xi \\in \\mathbb{D}, \\, x, \\delta x \\in \\mathbb{R},\n\\end{equation*}\n%\nthat are naturally equipped with associative addition and multiplication operations,\n%\n\\begin{align*}\n\\xi_{1} + \\xi_{2} \n&=\n \\left( x_{1} + \\mathfrak{a} \\, \\delta x_{1} \\right) \n + \\left( x_{2} + \\mathfrak{a} \\, \\delta x_{2} \\right) \n\\\\\n&=\n\\left( x_{1} + x_{2} \\right) \n+ \\mathfrak{a} \\left( \\delta x_{1} + \\delta x_{2} \\right) \n\\\\\n\\xi_{1} \\cdot \\xi_{2} \n&= \n\\left( x_{1} + \\mathfrak{a} \\, \\delta x_{1} \\right) \n\\cdot \\left( x_{2} + \\mathfrak{a} \\, \\delta x_{2} \\right) \n\\\\\n&= \nx_{1} x_{2} + \\left( x_{1} \\delta x_{2} \n+ \\delta x_{1} x_{2} \\right) \\mathfrak{a} \n+ \\delta x_{1} \\delta x_{2} \\, \\mathfrak{a}^{2} \n\\\\\n&= \nx_{1} x_{2} + \\left( x_{1} \\delta x_{2} + \\delta x_{1} x_{2} \\right) \\mathfrak{a}.\n\\end{align*} \n\nAny smooth function $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ induces a function on \ndual numbers, $f : \\mathbb{D} \\rightarrow \\mathbb{D}$, via a Taylor series around any \npurely real point, $\\xi_{0} = x_{0} + \\mathfrak{a} \\, 0$,\n%\n\\begin{align*}\nf \\! \\left( \\xi \\right) \n&= \n\\sum_{n = 0}^{\\infty} \\frac{ \\left( \\xi - \\xi_{0} \\right)^{n} }{n!} \n\\frac{ \\partial^{n} f }{ \\partial \\xi^{n} } \\! \\left( \\xi_{0} \\right) \n\\\\\n&= \n\\sum_{n = 0}^{\\infty} \\frac{ \\left( \\left( x - x_{0} \\right) + \\mathfrak{a} \\, \\delta x \\right)^{n} }{n!} \n\\frac{ \\partial^{n} f }{ \\partial x^{n} } \\! \\left( x_{0} \\right)\n\\\\\n&= \n\\sum_{n = 0}^{\\infty} \\frac{ \\left( x - x_{0} \\right)^{n} }{n!} \n\\frac{ \\partial^{n} f }{ \\partial x^{n} } \\! \\left( x_{0} \\right)\n+ \\mathfrak{a} \\, \\delta x \\sum_{n = 0}^{\\infty} \\frac{ \\left( x - x_{0} \\right)^{n - 1} }{\\left( n - 1 \\right)!} \n\\frac{ \\partial^{n} f }{ \\partial x^{n} } \\! \\left( x_{0} \\right) \n\\\\\n&= \nf \\! \\left( x \\right) \n+ \\mathfrak{a} \\, \\delta x \\frac{ \\partial f }{ \\partial x } \\! \\left( x \\right).\n\\end{align*}\n\nGeneralizing to multivariate functions is straightforward: any function \n$f : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}^{M}$ is first decomposed into $M$\nfunctions, $f_{m} : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}$, which extend\nto dual numbers as\n%\n\\begin{equation} \\label{dual_function}\nf_{m} \\! \\left( \\mbox{\\boldmath{$\\xi$}} \\right) = f_{m} \\! \\left( \\mathbf{x} \\right) \n+ \\mathfrak{a} \\sum_{n = 1}^{N} \n\\delta x_{n} \\frac{ \\partial f_{m} }{ \\partial x_{n} } \\! \\left( \\mathbf{x} \\right).\n\\end{equation}\n%\nIf we consider the dual component as identifying an element of the tangent\nspace, then the dual part of the extended function is exactly the pushforward \nof the original function. \n\nSimilarly, we can interpret the dual component as a sensitivity. In this case\nthe action of a function on the dual number is given by the adjoint action of the\npushforward,\n%\n\\begin{equation} \\label{dual_function}\n\\xi_{n} = x_{n} + \\mathfrak{a} \\sum_{m = 1}^{M} \n\\delta f_{m} \\frac{ \\partial f_{m} }{ \\partial x_{n} } \\! \\left( \\mathbf{x} \\right).\n\\end{equation} \n\nBecause dual numbers inherently integrate a function with its pushforward or\npullback, they are the natural type from which we can implement linear\ndifferential operators and, consequently, automatic differentiation.\n\n\\subsection{Higher-Order Dual Numbers}\n\nThe naturalness of dual numbers also generalizes to higher-order perturbations\nwith the introduction of multiple dual units, providing an means of computing\nhigher-order linear differential operators without having to introduce the mathematical\nmachinery of jet spaces.\n\nAt first-order, the real and dual parts of a first-order dual number are real numbers,\n%\n\\begin{equation*}\n\\zeta = z + \\mathfrak{a} \\, \\delta z, \\, z, \\delta z \\in \\mathbb{R}.\n\\end{equation*}\n%\nIf we introduce a second dual unit, $\\mathfrak{b}$, then we can construct a second-order \ndual number by replacing the real-valued parts of $\\zeta$ with dual numbers,\n%\n\\begin{alignat*}{3}\n\\zeta\n&=\n\\;\\, \\xi\n&&+ \\mathfrak{a} \\;\\;\\, \\eta\n\\\\\n&=\n\\left( x + \\mathfrak{b} \\, \\delta x \\right)\n&&+ \\mathfrak{a} \\left( \\delta y + \\mathfrak{b} \\, \\delta^{2} y \\right), \n\\, x, \\delta x, \\delta y, \\delta^{2} y \\in \\mathbb{R}.\n\\end{alignat*}\n%\nFrom the linear differential operator perspective, $x$ is a the value, $\\delta x$ and \n$\\delta y$ are first-order perturbations, and $\\delta^{2} y$ is a second-order perturbation.\nHigher-order dual numbers follow recursively with the introduction of a new dual unit\nand the replacement of real-valued components with new dual numbers. An $n$-th \norder dual number, for example, is constructed from $n$ distinct dual units and\n$2^{n}$ real components.\n\nWhen constructing higher-order linear differential operators, these components may \nbe considered as both higher-order tangents and higher-order sensitivities. Consequently \nit's convenient to define notation agnostic to the specific map being considered.\nWe will refer to the components of each bottem-level dual number as \\textit{values} and\n\\textit{gradients}; for example, at second-order $x$ would be a first-order value, $\\delta x$ \na first-order gradient, $\\delta y$ a second-order value, and $\\delta^{2} y$ as the second-order \ngradient.\n\nExtending functions to higher-order dual numbers, and constructing higher-order\nlinear differential operators in the process, also proceeds recursively. Here we\nexplicitly evaluate the extension to second and third-order dual numbers.\n\n\\subsubsection{Second-Order}\n\nGiven the second-order dual number,\n%\n\\begin{alignat*}{3}\n\\zeta_{i} \n&=\n\\xi_{i} \n&&+ \\mathfrak{a} \\, \\eta_{i}\n\\\\\n&=\n\\left( x_{i} + \\mathfrak{b} \\, \\delta x_{i} \\right)\n&&+ \\mathfrak{a} \\left( \\delta y_{i} + \\mathfrak{b} \\, \\delta^{2} y_{i} \\right),\n\\end{alignat*}\n%\na second-order dual number valued-function, evaluates to\n%\n\\begin{align*}\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\zeta$}} \\right)\n=& \\,\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\xi$}} \\right) \n+ \\mathfrak{a} \\sum_{j} \\eta_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mbox{\\boldmath{$\\xi$}} \\right) \n\\\\\n=& \\,\nf_{i} \\! \\left( \\mathbf{x} + \\mathfrak{b} \\, \\delta \\mathbf{x} \\right) \n+ \\mathfrak{a} \\sum_{j} \\left( \\delta y_{j} + \\mathfrak{b} \\, \\delta^{2} y_{j} \\right)\n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \n\\! \\left( \\mathbf{x} + \\mathfrak{b} \\, \\delta \\mathbf{x} \\right) \n\\\\\n=& \\,\nf_{i} \\! \\left( \\mathbf{x} \\right) \n+ \\mathfrak{b} \\sum_{j} \\delta x_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j} \\left( \\delta y_{j} + \\mathfrak{b} \\, \\delta^{2} y_{j} \\right)\n\\left( \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x}\\right) + \n\\mathfrak{b} \\, \\sum_{k} \\delta x_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{x} \\right) \\right) \n\\\\\n=&\n\\quad\\quad\\quad \nf_{i} \\! \\left( \\mathbf{x} \\right) \n\\quad\\quad\\quad\\;\\;\\;\n+ \\mathfrak{b} \\;\\;\\;\n\\sum_{j} \\delta x_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\left( \n\\sum_{j} \\delta y_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right) \n+ \\mathfrak{b} \\left(\n\\sum_{j} \\delta^{2} y_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)\n+ \\sum_{jk} \\delta x_{k} \\, \\delta y_{j}\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{x} \\right)\n\\right)\n\\right)\n\\end{align*}\n%\nThese results are summarized in Table \\ref{tab:secondOrder}.\n\n\\begin{table*}[t!]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{2}\n\t\\begin{tabular}{ccc}\n\t\\rowcolor[gray]{0.9} \\textbf{Component} & \\textbf{Input} & \\textbf{Output} \\\\\n\tFirst Value & \n\t$\\mathbf{x}$ & \n\t$f_{i} \\! \\left( \\mathbf{x} \\right)$ \n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\tFirst Gradient & \n\t$\\delta \\mathbf{x}$ &\n\t$\\displaystyle \\sum_{j} \\delta x_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)$\n\t\\\\\n\tSecond Value & \n\t$\\delta \\mathbf{y} $ & \n\t$\\displaystyle \\sum_{j} \\delta y_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)$\n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\tSecond Gradient & \n\t$\\delta^{2} \\mathbf{y} $ & \n\t$\\displaystyle \\sum_{j} \\delta^{2} y_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{x} \\right)\n\t+ \\sum_{jk} \\delta x_{k} \\, \\delta y_{j}\n\t\\frac{ \\partial f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{x} \\right)$\n\t\\\\\n\t\\end{tabular}\n\t\\caption{Recursively expanding a function with respect to a second-order dual number \n\tgives each component of the function's second-order pushforward.\n\t\\label{tab:secondOrder}}\n\\end{table*}\n\nConsidering the first-order gradient, second-order value, and second-order gradient as tangents \nimmediately defines two first-order linear differential operators and one second-order linear \ndifferential operator. Unfortunately the same is not true if we try to consider all three components\nas sensitivities, as the second term in the second-order gradient map no longer makes\nsense. \n\nThe generalization of an adjoint map does not consider all components as sensitivities but rather\nconsiders all $k$th-order values as $(k - 1)$-order tangents (with a $0$-order tangent\ntaken to be an input value) and all $k$th-order gradients as $k$th-order sensitivities. Once the partial\nderivatives are contracted against the tangents they can be transposed to give the\ncomponents of the pullback map. For example, given our usual test function \n$f : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}^{M}$ with the perturbations \n$\\delta y \\rightarrow \\mathrm{d} f$ and $\\delta^{2} y \\rightarrow \\mathrm{d}^{2} f$, the \nsecond-order pullback is given by\n%\n\\begin{equation*}\n\\mathrm{d}^{2} x_{n}\n= \\sum_{m = 1}^{M} \\mathrm{d}^{2} f_{m} \\frac{ \\partial f_{m} }{ \\partial x_{n} }\n+ \\sum_{m = 1}^{M} \\mathrm{d} f_{m} \n\\left[ \\sum_{n'=1}^{N} \\delta x_{n'} \\frac{ \\partial^{2} f_{m} }{ \\partial x_{n'} \\partial x_{n} } \\right].\n\\end{equation*}\n\n\\subsubsection{Third-Order}\n\nContinuing to third-order follows in kind. Given a third-order dual number,\n%\n\\begin{alignat*}{10}\n\\zeta_{i} \n&=\n&& \\xi_{i} && && &&\n&&+ \\mathfrak{a} \\, \n&& \\eta_{i} && && &&\n\\\\\n&=\n( && \\sigma_{i} && + \\mathfrak{b} && \\tau_{i} &&)\n&&+ \\mathfrak{a} \\,\n( && \\upsilon_{i} && + \\mathfrak{b} && \\nu_{i} && )\n\\\\\n&=\n(( && s_{i} + \\mathfrak{c} \\, \\delta s_{i} )\n&& + \\mathfrak{b} \\;\n( && \\delta t_{i} + \\mathfrak{c} \\, \\delta^{2} t_{i} ) &&)\n&&+ \\mathfrak{a} \\;\n(( && \\delta u_{i} + \\mathfrak{c} \\, \\delta^{2} u_{i} )\n&& + \\mathfrak{b} \\;\n( && \\delta^{2} v_{i} + \\mathfrak{c} \\, \\delta^{3} v_{i} ) &&)\n\\end{alignat*}\n%\nwe have\n%\n\\begin{align*}\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\zeta$}} \\right)\n%\n=& \\,\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\xi$}}\\right) \n+ \\mathfrak{a} \\sum_{j} \\eta_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mbox{\\boldmath{$\\xi$}} \\right) \n\\\\\n%\n=& \\,\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\sigma$}} + \\mathfrak{b} \\, \\mbox{\\boldmath{$\\tau$}} \\right) \n+ \\mathfrak{a} \\sum_{j} \\left( \\upsilon_{j} + \\mathfrak{b} \\, \\nu_{j} \\right)\n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \n\\! \\left( \\mbox{\\boldmath{$\\sigma$}} + \\mathfrak{b} \\, \\mbox{\\boldmath{$\\tau$}} \\right) \n\\\\\n%\n=& \\,\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right) + \\mathfrak{b} \n\\sum_{j} \\tau_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \n\\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n+ \\mathfrak{a} \\sum_{j} \\left( \\upsilon_{j} + \\mathfrak{b} \\, \\nu_{j} \\right)\n\\left( \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n+ \\mathfrak{b} \\sum_{k} \\tau_{k}\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right) \\right) \n\\\\\n%\n=& \\,\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right) + \\mathfrak{b} \n\\sum_{j} \\tau_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \n\\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n+ \\mathfrak{a} \\sum_{j}\n\\upsilon_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n\\\\\n& + \\mathfrak{a} \\, \\mathfrak{b} \\sum_{j}\n \\nu_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\sum_{jk} \\upsilon_{j} \\tau_{k}\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mbox{\\boldmath{$\\sigma$}} \\right)\n\\\\\n%\n=& \\,\nf_{i} \\! \\left( \\mathbf{s} + \\mathfrak{c} \\, \\delta \\mathbf{s} \\right) + \\mathfrak{b} \n\\sum_{j} \\left( \\delta t_{j} + \\mathfrak{c} \\, \\delta^{2} t_{j} \\right) \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} + \\mathfrak{c} \\, \\delta \\mathbf{s}\\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j}\n\\left( \\delta u_{j} + \\mathfrak{c} \\, \\delta^{2} u_{j} \\right) \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} + \\mathfrak{c} \\, \\delta \\mathbf{s}\\right)\n\\\\\n& + \\mathfrak{a} \\, \\mathfrak{b} \\sum_{j}\n\\left( \\delta^{2} v_{j} + \\mathfrak{c} \\, \\delta^{3} v_{j} \\right) \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} + \\mathfrak{c} \\, \\delta \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\, \\sum_{jk} \n\\left( \\delta u_{j} + \\mathfrak{c} \\, \\delta^{2} u_{j} \\right) \n\\left( \\delta t_{k} + \\mathfrak{c} \\, \\delta^{2} t_{k} \\right)\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} + \\mathfrak{c} \\, \\delta \\mathbf{s} \\right)\n\\\\\n%\n=& \\,\nf_{i} \\! \\left( \\mathbf{s} \\right) + \\mathfrak{c} \\sum_{j} \\delta s_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{b}\n\\sum_{j} \\left( \\delta t_{j} + \\mathfrak{c} \\, \\delta^{2} t_{j} \\right) \n\\left( \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{c} \\sum_{k} \\delta s_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j}\n\\left( \\delta u_{j} + \\mathfrak{c} \\, \\delta^{2} u_{j} \\right) \n\\left( \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{c} \\sum_{k} \\delta s_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \\right)\n\\\\\n& + \\mathfrak{a} \\, \\mathfrak{b} \\sum_{j}\n\\left( \\delta^{2} v_{j} + \\mathfrak{c} \\, \\delta^{3} v_{j} \\right) \n\\left( \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{c} \\sum_{k} \\delta s_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\, \\sum_{jk} \n\\left( \\delta u_{j} + \\mathfrak{c} \\, \\delta^{2} u_{j} \\right) \n\\left( \\delta t_{k} + \\mathfrak{c} \\, \\delta^{2} t_{k} \\right)\n\\\\\n& \\quad\\quad\\quad\\quad \\times \\left(\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{c} \\sum_{l} \\delta s_{l} \n\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right)\n\\right)\n\\\\\n%\n\\end{align*}\n\n\\begin{align*}\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\zeta$}} \\right)\n%\n=& \\,\nf_{i} \\! \\left( s_{j} \\right) + \\mathfrak{c} \\sum_{j} \\delta s_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{b} \\sum_{j} \\delta t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} t_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta t_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j} \\delta u_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} u_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta u_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\sum_{j} \\delta^{2} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{3} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta^{2} v_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\, \\sum_{jk} \n\\left( \\delta u_{j} \\delta t_{k}\n+ \\mathfrak{c} \\left(\n\\delta u_{j} \\, \\delta^{2} t_{k} + \\delta^{2} u_{j} \\, \\delta t_{k} \n\\right) \\right)\n\\\\\n& \\quad\\quad\\quad\\quad \\times \\left(\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{c} \\sum_{l} \\delta s_{l} \n\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right)\n\\right)\n\\\\\n%\n=& \\,\nf_{i} \\! \\left( s_{j} \\right) + \\mathfrak{c} \\sum_{j} \\delta s_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{b} \\sum_{j} \\delta t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} t_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta t_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j} \\delta u_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} u_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta u_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\sum_{j} \\delta^{2} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{3} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta^{2} v_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \\, \\sum_{jk} \n\\delta u_{j} \\delta t_{k} \\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk}\n\\left( \\delta u_{j} \\, \\delta^{2} t_{k} + \\delta^{2} u_{j} \\, \\delta t_{k} \\right)\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \n\\\\\n&+\n\\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jkl}\n\\delta u_{j} \\delta t_{k} \\, \\delta s_{l} \n\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n%\n\\end{align*}\n\n\\begin{align*}\nf_{i} \\! \\left( \\mbox{\\boldmath{$\\zeta$}} \\right)\n%\n=& \\,\nf_{i} \\! \\left( s_{j} \\right) + \\mathfrak{c} \\sum_{j} \\delta s_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{b} \\sum_{j} \\delta t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} t_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta t_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\sum_{j} \\delta u_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{j} \\delta^{2} u_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\mathfrak{a} \\, \\mathfrak{c} \\sum_{jk} \\delta s_{k} \\, \\delta u_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n&+ \n\\mathfrak{a} \\, \\mathfrak{b} \n\\left( \n\\sum_{j} \\delta^{2} v_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\delta u_{j} \\delta t_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\right)\n\\\\\n&+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{j} \\delta^{3} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n\\\\\n&+ \\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jk}\n\\left( \n\\delta s_{k} \\, \\delta^{2} v_{j} + \\delta u_{j} \\, \\delta^{2} t_{k} + \\delta^{2} u_{j} \\, \\delta t_{k} \\right)\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \n\\\\\n&+\n\\mathfrak{a} \\, \\mathfrak{b} \\, \\mathfrak{c} \\sum_{jkl}\n\\delta u_{j} \\delta t_{k} \\, \\delta s_{l} \n\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n%\n=& \\hspace{26mm} \nf_{i} \\! \\left( \\mathbf{s} \\right) \n\\hspace{9mm}\n+ \\mathfrak{c} \n\\hspace{3mm}\n\\sum_{j} \\delta s_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n& \\hspace{8mm} + \n\\mathfrak{b} \\left( \\sum_{j} \\delta t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\hspace{1mm}\n+ \\mathfrak{c} \\left( \n\\sum_{j} \\delta^{2} t_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\sum_{jk} \\delta s_{k} \\, \\delta t_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \\right) \\right)\n\\\\\n&+ \n\\mathfrak{a} \\Bigg( \n\\hspace{9mm}\n\\sum_{j} \\delta u_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n+ \\mathfrak{c} \\left( \\sum_{j} \\delta^{2} u_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n+ \\sum_{jk} \\delta s_{k} \\, \\delta u_{j} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \\right)\n\\\\\n& \n\\hspace{8mm} \n+ \\mathfrak{b} \\Bigg(\n\\hspace{0.5mm}\n\\sum_{j} \\delta^{2} v_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\\delta u_{j} \\delta t_{k} \n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)\n\\\\\n& \\hspace{34mm} \n+ \\mathfrak{c} \\Bigg(\n\\sum_{j} \\delta^{3} v_{j} \n\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n\\\\\n& \\hspace{40mm} \n+ \\sum_{jk}\n\\left( \n\\delta s_{k} \\, \\delta^{2} v_{j} + \\delta u_{j} \\, \\delta^{2} t_{k} + \\delta^{2} u_{j} \\, \\delta t_{k} \\right)\n\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) \n\\\\\n& \\hspace{40mm} +\n\\sum_{jkl}\n\\delta u_{j} \\delta t_{k} \\, \\delta s_{l} \n\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right) \n\\Bigg)\n\\Bigg)\n\\Bigg)\n\\\\\n%\n\\end{align*}\n%\nThese results are summarized in Table \\ref{tab:thirdOrder}.\n\n\\begin{table*}[t!]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{2}\n\t\\begin{tabular}{ccc}\n\t\\rowcolor[gray]{0.9} \\textbf{Component} & \\textbf{Input} & \\textbf{Output} \\\\\n\tFirst Value & \n\t$\\mathbf{s}$ & \n\t$f_{i} \\! \\left( \\mathbf{s} \\right)$ \n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\tFirst Gradient & \n\t$\\delta \\mathbf{s} $ &\n\t$\\displaystyle \\sum_{j} \\delta s_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) $\n\t\\\\\n\tSecond Value & \n\t$\\delta \\mathbf{t}$ & \n\t$\\displaystyle \\sum_{j} \\delta t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)$\n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\tSecond Gradient & \n\t$\\delta^{2} \\mathbf{t}$ &\n\t$\\displaystyle \\sum_{j} \\delta^{2} t_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n\t+ \\sum_{jk} \\delta s_{k} \\, \\delta t_{j} \n\t\\frac{ \\partial f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)$\n\t\\\\\n\tThird Value & \n\t$\\delta \\mathbf{u}$ &\n\t$\\displaystyle \\sum_{j} \\delta u_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)$\n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\tThird Gradient & \n\t$\\delta^{2} \\mathbf{u}$ &\n\t$\\displaystyle \\sum_{j} \\delta^{2} u_{j} \n\t\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) \n\t+ \\sum_{jk} \\delta s_{k} \\, \\delta u_{j} \n\t\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)$\n\t\\\\\n\tFourth Value & \n\t$\\delta^{2} \\mathbf{v}$ & \n\t$\\displaystyle \\sum_{j} \\delta^{2} v_{j} \\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right)\n\t\\delta u_{j} \\delta t_{k} \n\t\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right)$\n\t\\\\\n\t\\rowcolor[gray]{0.9} \n \\multirow{3}{*}{ \\vspace{-8mm} Fourth Gradient} & \n\t\\multirow{3}{*}{ \\vspace{-8mm} $\\delta^{3} \\mathbf{v}$} &\n\t$\\displaystyle \\sum_{j} \\delta^{3} v_{j} \n\t\\frac{ \\partial f_{i} }{ \\partial x_{j} } \\! \\left( \\mathbf{s} \\right) $ \n\t\\\\\n\t& &\n\t$\\displaystyle + \\sum_{jk} \\left( \n\t\\delta s_{k} \\, \\delta^{2} v_{j} + \\delta u_{j} \\, \\delta^{2} t_{k} + \\delta^{2} u_{j} \\, \\delta t_{k} \\right)\n\t\\frac{ \\partial^{2} f_{i} }{ \\partial x_{j} \\partial x_{k} } \\! \\left( \\mathbf{s} \\right) $ \n\t\\\\\n\t\\rowcolor[gray]{0.9} \n\t& &\n\t$ \\displaystyle+ \\sum_{jkl} \\delta u_{j} \\delta t_{k} \\, \\delta s_{l} \n\t\\frac{ \\partial^{3} f_{i} }{ \\partial x_{j} \\partial x_{k} \\partial_{l} } \\! \\left( \\mathbf{s} \\right) $\n\t\\\\\n\t\\end{tabular}\n\t\\caption{Recursively expanding a function with respect to a third-order dual number \n\tgives each component of the function's third-order pushforward.\n\t\\label{tab:thirdOrder}}\n\\end{table*}\n\nAs with second-order pullbacks, third-order pullbacks require considering each values as \na tangent and each gradient as a sensitivity. For example, with the notation\n%\n\\begin{align*}\ns &\\rightarrow x \\\\\n\\delta s &\\rightarrow \\mathrm{d} f \\\\\n\\delta t &\\rightarrow \\delta x \\\\\n\\delta^{2} t &\\rightarrow \\mathrm{d}^{2} f \\\\\n\\delta u &\\rightarrow \\delta x^{2} \\\\\n\\delta^{2} u &\\rightarrow \\mathrm{d}^{3} f \\\\\n\\delta^{2} v &\\rightarrow \\delta x^{3} \\\\\n\\delta^{3} v &\\rightarrow \\mathrm{d}^{4} f\n\\end{align*}\n%\nthe third-order adjoint map is given by\n%\n\\begin{align*}\n\\mathrm{d}^{4} x_{n}\n&= \\quad\n\\sum_{m = 1}^{M} \\mathrm{d}^{4} f_{m} \\frac{ \\partial f_{m} }{ \\partial x_{n} }\n\\\\\n& \\quad + \n\\sum_{m = 1}^{M} \\mathrm{d} f_{m}\n\\left[ \\sum_{n'=1}^{N} \\delta^{3} x_{n'} \\frac{ \\partial^{2} f_{m} }{ \\partial x_{n'} \\partial x_{n} } \\right]\n+\n\\sum_{m = 1}^{M} \\mathrm{d}^{2} f_{m}\n\\left[ \\sum_{n'=1}^{N} \\delta^{2} x_{n'} \\frac{ \\partial^{2} f_{m} }{ \\partial x_{n'} \\partial x_{n} } \\right]\n+\n\\sum_{m = 1}^{M} \\mathrm{d}^{3} f_{m}\n\\left[ \\sum_{n'=1}^{N} \\delta x_{n'} \\frac{ \\partial^{2} f_{m} }{ \\partial x_{n'} \\partial x_{n} } \\right]\n\\\\\n& \\quad +\n\\sum_{m = 1}^{M} \\mathrm{d} f_{m}\n\\left[ \\sum_{n'=1}^{N} \\sum_{n''=1}^{N} \\delta x_{n'} \\delta^{2} x_{n''} \n\\frac{ \\partial^{3} f_{m} }{ \\partial x_{n''} \\partial x_{n'} \\partial x_{n} } \\right]\n\\end{align*}\n\n\\section{Automatic Differentiation}\n\nAutomatic differentiation is the automated computation of linear differential operators applied\nto a user-defined function. Because dual numbers integrate a function evaluation along with\nits pushforward or pullback, they provide a natural type -- ultimately, automatic differentiation \nis just the extension of a user-defined, real-valued function to accept dual numbers.\n\nWhen these functions are composite, as is the case in almost any practical application, they\ncan be represented by an \\textit{expression graph}, with each node carrying the value of\nthe intermediate function evaluation and each edge denoting a functional dependency\n(Figure \\ref{fig:exprGraph}). The dual number-valued functions necessary for automatic\ndifferentiation simply replace the single value at each node with the components of a\n$k$th-order dual number.\n\n\\begin{figure}\n\\setlength{\\unitlength}{0.1in} \n\\centering\n\\begin{picture}(50, 20)\n%\n%\\put(0, 0) { \\framebox(50, 20){} }\n%\\put(25, 0) { \\framebox(25, 30){} }\n%\\put(25, 0) { \\framebox(6.25, 30){} }\n%\\put(25, 0) { \\framebox(12.5, 30){} }\n%\\put(25, 0) { \\framebox(18.75, 30){} }\n%\n%\\put(25, 0) { \\framebox(3.125, 30){} }\n%\\put(25, 0) { \\framebox(9.375, 30){} }\n%\\put(25, 0) { \\framebox(15.625, 30){} }\n%\n\\put(12.5, 10) { \\makebox(0, 0) \n{$g \\! \\left( f_{1} \\! \\left( x_{1}, x_{2} \\right), f_{2} \\! \\left( x_{2}, x_{3} \\right) \\right)$} }\n%\n\\put(21.875, 10) { \\vector(1, 0){6.25} }\n%\n\\put(31.25, 2.5) { \\circle{4} }\n\\put(31.25, 2.5) { \\makebox(0, 0) {$ x_{1} $} }\n%\n\\put(37.5, 2.5) { \\circle{4} }\n\\put(37.5, 2.5) { \\makebox(0, 0) { $ x_{2} $ } }\n%\n\\put(43.75, 2.5) { \\circle{4} }\n\\put(43.75, 2.5) { \\makebox(0, 0) { $ x_{3} $ } }\n%\n\\put(31.25, 4.5) { \\vector(3, 4){2.75} }\n\\put(37.5, 4.5) { \\vector(-3, 4){2.75} }\n\\put(37.5, 4.5) { \\vector(3, 4){2.75} }\n\\put(43.75, 4.5) { \\vector(-3, 4){2.75} }\n%\n\\put(35, 10) {\\circle{4} } % Tweaked to the right\n\\put(34.375, 10) { \\makebox(0, 0) { $f_{1}$ } }\n%\n\\put(41.25, 10) {\\circle{4} } % Tweaked to the right\n\\put(40.625, 10) { \\makebox(0, 0) { $f_{2}$ } }\n%\n\\put(34.375, 12) { \\vector(3, 4){2.75} }\n\\put(40.625, 12) { \\vector(-3, 4){2.75} }\n%\n\\put(38, 17.5) {\\circle{4} } % Tweaked to the right\n\\put(37.5, 17.5) { \\makebox(0, 0) { $ g $ } }\n%\n\\end{picture} \n\\caption{\nComposite functions are isomorphic to directed acyclic graphs known as \n\\textit{expression graphs}; for example, the function\n$g \\! \\left( f_{1} \\! \\left( x_{1}, x_{2} \\right), f_{2} \\! \\left( x_{2}, x_{3} \\right) \\right)$\ngenerates a three level graph, with leaves $\\left\\{ x_{1}, x_{2}, x_{3} \\right\\}$ and\nroot $g$. The expression graph corresponding to a real-valued function\nstores single values at each node, and automatic differentiation requires that\neach node store all components of a $k$th-order dual number.\n}\n\\label{fig:exprGraph} \n\\end{figure}\n\nThe computation of any linear differential operator comprised entirely of pushforwards is\nknown as \\textit{forward mode} or \\textit{forward accumulation} automatic differentiation\nbecause all dual numbers are propagated forward along the expression graph. Any\nadjoint operator, however, requires propagating sensitivities backwards along the expression\ngraph and consequently the implementation of adjoint operators is known as \n\\textit{reverse mode} or \\textit{reverse accumulation} automatic differentiation. In this\nsection we consider both forward and reverse mode automatic differentiation, discuss\ntheir relative performance, and demonstrate a selection of linear differential operator\nimplementations.\n\n\\subsection{Forward Mode Automatic Differentiation}\n\nForward mode automatic differentiation computes the $k$th-order pushforward \nof a given function by first initializing all values and gradients at the function\ninputs and propagating the perturbations along the expression graph in a single \nforward sweep. \n\nBecause only one sweep is necessary for forward mode, the expression graph does \nnot need to be explicitly stored in memory, reducing overhead and often admitting \nmore optimized code. On the other hand, the linear differential operators comprised \nentirely of pushforwards are often not of immediate practical interest.\n\n\\subsection{Reverse Mode Automatic Differentiation}\n\nReverse mode automatic differentiation compliments forward mode by computing\n$k$th-order adjoint operators which require more elaborate traversals of the\nexpression graph. A general reverse mode calculation interleaves the forward\npropagation of tangents and the reverse propagation of sensitivities, starting at\nfirst order and working up to the desired order.\n\nConsequently, reverse mode requires the persistent storage of the expression \ngraph and induces a corresponding performance overhead. In practice, however,\nthe adjoint operators admitted by reverse mode automatic differentiation are\nmore appropriate to applications and worth the additional cost.\n\n\\subsection{Relative Performance}\n\nIn general neither forward mode nor reverse mode automatic differentiation can be\nclaimed as optimal, as the relative performance of the two depends not only on\nthe given function but also on the desired linear differential operation. \n\nConsider, for example, many-to-one functions $f : \\mathbb{R}^{N} \\rightarrow \\mathbb{R}$,\ncommon to many applications. Forward mode automatic differentiation, computes the \ndirectional derivative of the function along a given tangent in a single forward sweep.\nRecovering the full gradient then requires $N$ forward sweeps, each initialized with\nonly one of the input tangents non-zero. On the other hand, reverse mode automatic\ndifferentiation computes the gradient with only one forward and one reverse sweep. \n\nIf the directional derivative is sufficient or $N$ is small then forward mode automatic\ndifferentiation is the better performing approach, but when the gradient is desired and \n$N$ is large, reverse mode automatic differentiation becomes far more computationally \nefficient despite the memory overhead. Because this latter case is ubiquitous is\npractice, \\nomad implements reverse mode automatic differentiation.\n\n\\subsection{Example Implementations}\n\nHere we review the exact computations needed to implement a selection of\nlinear differential operators natural to many-to-one functions, \n$f : \\mathbb{R}^{N} \\rightarrow 1$.\n\nThe gradient and Hessian are denoted by $\\mathbf{g}$ and $\\mathbf{H}$, respectively,\nwhile adjuncts refers to lower-order linear differential operators that are generated\nas a side effect of the desired operation.\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Directional Derivative]\n\n\t\\begin{description}\n\t\t\\item[Form:] $\\displaystyle \\mathbf{v}^{T} \\mathbf{g} = \\sum_{i} v_{i} \\frac{ \\partial f }{ \\partial x_{i} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize first gradient of the inputs, $\\delta x_{i} = v_{i}$. \\\\\n\t\tCompute first-order forward sweep. \\\\\n\t\tReturn first gradient of output.\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( 1 \\right)$\n\t\t\\item[Adjuncts:] $ f $\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Gradient]\n\t\n\t\\begin{description}\n\t\t\\item[Form:] $\\displaystyle g_{i} = \\frac{ \\partial f }{ \\partial x_{i} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize first gradient of the output, $\\mathrm{d} f = 1$. \\\\\n\t\tCompute first-order reverse sweep. \\\\\n\t\tReturn first gradient of inputs.\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( 1 \\right)$\n\t\t\\item[Adjuncts:] $ f $\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Hessian Quadratic Form]\n\n\t\\begin{description}\n\t\t\\item[Form:] \n\t\t$\\displaystyle \\mathbf{v}^{T} \\mathbf{H} \\, \\mathbf{u} = \n\t\t\\sum_{ij} v_{i} u_{j} \\frac{ \\partial^{2} f }{ \\partial x_{i} \\partial x_{j} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize first gradient, $\\delta x_{i} = v_{i}$, second value, $\\delta y_{i} = u_{i}$,\n\t\tand second gradient, $\\delta^{2} y_{i} = 0$, of the inputs. \\\\\n\t\tCompute second-order forward sweep. \\\\\n\t\tReturn second gradient of output.\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( 1 \\right)$\n\t\t\\item[Adjuncts:] $ f, \\, \\mathbf{v}^{T} \\mathbf{g}, \\, \\mathbf{u}^{T} \\mathbf{g}$\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Hessian-Vector Product]\n\t\n\t\\begin{description}\n\t\t\\item[Form:] $\\displaystyle \\mathbf{H} \\, \\mathbf{v} = \n\t\t\\sum_{j} v_{j} \\frac{ \\partial^{2} f }{ \\partial x_{i} \\partial x_{j} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize first gradient of the output, $\\mathrm{d} f = 1$. \\\\\n\t\tCompute first-order reverse sweep for first gradients at the inputs. \\\\\n\t\tInitialize second values of the inputs, $\\delta y_{i} = v_{i}$. \\\\\n\t\tPropagate second values in a forward sweep. \\\\\n\t\tInitialize second gradient of the output, $\\mathrm{d}^{2} f = 0.$ \\\\\n\t\tCompute second-order reverse sweep. \\\\\n\t\tReturn second gradient of inputs.\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( 1 \\right)$\n\t\t\\item[Adjuncts:] $ f, \\, \\mathbf{v}^{T} \\mathbf{g}, \\, \\mathbf{g}$\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Hessian]\n\t\n\t\\begin{description}\n\t\t\\item[Form:] $\\displaystyle H_{ij} = \\frac{ \\partial^{2} f }{ \\partial x_{i} \\partial x_{j} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize $\\mathrm{d} f = 1$. \\\\\n\t\tCompute first-order reverse sweep for first gradients. \\\\\n\t\tFor each $j$ compute the $j$th row of the Hessian as:\n\t\t\\begin{itemize}\n\t\t\t\\setlength{\\itemsep}{0cm}\n\t\t\t\\setlength{\\parskip}{0cm}\n\t\t\t\\item[] Initialize second values of the inputs, $\\delta y_{i} = \\delta^{j}_{i}$.\n\t\t\t\\item[] Propagate second values in a forward sweep. \n\t\t\t\\item[] Initialize second gradient of the output, $\\mathrm{d}^{2} f = 0$.\n\t\t\t\\item[] Compute second-order reverse sweep.\n\t\t\t\\item[] Return second gradient of inputs\n\t\t\\end{itemize}\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( n \\right)$\n\t\t\\item[Adjuncts:] $ f, \\, \\mathbf{g}$\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n\\begin{tcolorbox}[colback=white,colframe=gray90, coltitle=black,boxrule=3pt,\nfonttitle=\\bfseries,title=Gradient of the Trace of a Matrix Hessian Product]\n\t\n\t\\begin{description}\n\t\t\\item[Form:] \n\t\t$\\displaystyle \\frac{\\partial}{\\partial x_{i} } \\mathrm{Tr} \\! \\left[ \\mathbf{M} \\, \\mathbf{H} \\right]\n\t\t= \\sum_{jk} M_{jk} \\frac{ \\partial^{3} f }{ \\partial x_{i} \\partial x_{j} \\partial x_{k} } $\n\t\t\\item[Algorithm:] \\hfill \\\\\n\t\tInitialize $\\mathrm{d} f = 1$. \\\\\n\t\tCompute first-order reverse sweep for first gradients. \\\\\n\t\tInitialize the trace gradient to zero.\n\t\tFor each $j$ increment the trace gradient with:\n\t\t\\begin{itemize}\n\t\t\t\\setlength{\\itemsep}{0cm}\n\t\t\t\\setlength{\\parskip}{0cm}\n\t\t\t\\item[] Initialize second values of the inputs, $\\delta t_{i} = \\delta^{j}_{i}$.\n\t\t\t\\item[] Propagate second values in a forward sweep. \n\t\t\t\\item[] Initialize second gradient of the output, $\\mathrm{d}^{2} f = 0$.\n\t\t\t\\item[] Compute second-order reverse sweep.\n\t\t\t\\item[] Initialize third and fourth values of the inputs, $\\delta u_{i} = M_{ji}, \\delta^{2} v_{i} = 0$.\n\t\t\t\\item[] Propagate third and fourth values in a forward sweep.\n\t\t\t\\item[] Initialize third gradient of the output, $\\mathrm{d}^{3} f = 0$.\n\t\t\t\\item[] Compute third-order reverse sweep.\n\t\t\t\\item[] Return the fourth gradient of the inputs.\n\t\t\\end{itemize}\n\t\t\\item[Cost:] $\\mathcal{O} \\! \\left( n \\right)$\n\t\t\\item[Adjuncts:] $ f, \\, \\mathbf{g}, \\, \\mathbf{H}$\n\t\\end{description}\n\t\n\\end{tcolorbox}\n\n", "meta": {"hexsha": "310621caf06e4fd59e3c4364d0ae4a34ac54bc7c", "size": 47334, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manual/1-autodiff.tex", "max_stars_repo_name": "stan-dev/nomad", "max_stars_repo_head_hexsha": "a21149ef9f4d53a198e6fdb06cfd0363d3df69e7", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 23, "max_stars_repo_stars_event_min_datetime": "2015-12-11T20:06:57.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T18:59:58.000Z", "max_issues_repo_path": "manual/1-autodiff.tex", "max_issues_repo_name": "stan-dev/nomad", "max_issues_repo_head_hexsha": "a21149ef9f4d53a198e6fdb06cfd0363d3df69e7", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2015-12-15T08:12:01.000Z", "max_issues_repo_issues_event_max_datetime": "2016-07-17T01:36:56.000Z", "max_forks_repo_path": "manual/1-autodiff.tex", "max_forks_repo_name": "stan-dev/nomad", "max_forks_repo_head_hexsha": "a21149ef9f4d53a198e6fdb06cfd0363d3df69e7", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-10-13T17:40:34.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-08T19:17:51.000Z", "avg_line_length": 41.9627659574, "max_line_length": 125, "alphanum_fraction": 0.6477584823, "num_tokens": 17682, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382094310355, "lm_q2_score": 0.7490872243177518, "lm_q1q2_score": 0.6453672659463803}} {"text": "\\Lecture{Jayalal Sarma}{Nov 2, 2020}{24}{Algebraic Methods in Combinatorics}{Sampriti Roy}{$\\alpha$}{JS}\n\\section{Introduction}\nIn this chapter the plan is to switch to algebraic methods in combinatorics and the first structure we are going to explore is the concepts related to group theory. It's essentially going to be presented from the combinatorial side. As, a motivation we can think of the following problems:\n\\begin{itemize}\n\\item \\emph{How many distinct squares can be there with yellow and blue coloured corners?}\n\\end{itemize}\nOur task is to develop a method to count distinct colourings of such squares out of total $2^4=16$ possible colourings. If we consider the set of all coloured squares to be $\\Omega$, there will be some squares that are equivalent to each other under some rotation. As, we are interested in counting distinct colouring, the equivalent squares should not be counted twice.\n\\paragraph{}\nOne more question, we can think about is :\n\\begin{itemize}\n\\item \\emph{How many necklaces can be formed with solid beads and transparent beads? }\n\\end{itemize}\n\\paragraph{} \nBased on the colouring of square, we can say that there are total $2^4=16$ possible colourings but there are few operations under which two colourings can be viewed as same. Let's get familiar with such operations:\n\\begin{itemize}\n\\item $R_0$ : Rotation by $0^{\\circ}$ \n\\item $R_{90}$ : Rotation by $90^{\\circ}$\n\\item $R_{180}$ : Rotation by $180^{\\circ}$\n\\item $R_{270}$ : Rotation by $270^{\\circ}$\n\\item $H$ : Horizontal flip\n\\item $V$ : Vertical flip\n\\item $D$ : Diagonal flip (bottom left top right)\n\\item $D'$ : Diagonal flip (bottom right top left)\n\\end{itemize}\nNow, let's consider all the above set of operations together to be :\n\\begin{align*}\nG&=\\{R_0,R_{90},R_{180},R_{270},H,V,D,D'\\} \n\\end{align*}\nWe will see that the elements of set $G$ are inter related with each other. If we apply two operations from $G$, one after another, we will get the result of applying an operation from $G$ itself. Let's verify it:\n\\begin{itemize}\n\\item Rotate the square two times horizontally, it will result rotating it by $0^{\\circ}$. That is $H.H=R_0$\n\\item Rotate the square two times diagonally, it will result rotating it by $0^{\\circ}$. That is $D.D=R_0$\n\\item Rotate the square two times vertically, it will result rotating it by $0^{\\circ}$. That is $V.V=R_0$\n\\item Rotate the square two times by $180^{\\circ}$, it will result rotating it by $0^{\\circ}$. That is $R_{180}.R_{180}=R_0$\n\\item Rotate the square by $90^{\\circ}$ first then rotate it by $180^{\\circ}$, it will result rotating by $270^{\\circ}$. That is $R_{90}R_{180}=R_{270}$\n\\item Rotate the square by $90^{\\circ}$ first then rotate it by $270^{\\circ}$, it will result rotating by $0^{\\circ}$. That is $R_{90}R_{270}=R_{0}$\n\\item Rotate the square by $90^{\\circ}$ first then rotate it vertically, it will result rotating diagonally. That is $R_{90}.V=D $\n\\end{itemize}\n\\section{Incremental definition of Group}\nConsider a set $G$ with binary operation $\\circ$. Now, take any two elements of $G$ and apply the operation, a natural question to ask is whether we will get any element from the set $G$ or not. We will answer this question along with other related questions and finally move on to the definition of Group incrementally.\n\\begin{itemize}\n\\item \\emph{Groupoid} : Let $a,b\\in G$. If $a\\circ b\\in G$, then $G$ is \\emph{Closed} under the operation $\\circ$. And $G$ is called \\emph{Groupoid}. We can easily verify that our example set $G$ is a \\emph{Groupiod}.\n\\item \\emph{Semigroup} : $a,b,c\\in G$; if $a\\circ(b\\circ c)=(a\\circ b)\\circ c $, then $G$ is \\emph{associative} under $\\circ$ or, the order of $\\circ$ does not matter. If a groupoid is associative, it is called \\emph{Semigroup}. We can try out and check that our example set $G$ is associative and so it is a \\emph{Semigroup}.\n\\item \\emph{Monoid} : $\\forall a\\in G $, if there is an $e\\in G$ such that $a\\circ e=e\\circ a\\forall $, $e$ is called identity element. A Semigoup with identity is called \\emph{Monoid}. In our example set $G$, $R_0$ is the identity element, so it is a \\emph{Monoid}.\n\\item \\emph{Group} : $\\forall a\\in G$ if $ \\exists b\\in G $ such that $a\\circ b=b\\circ a=e $; where $e$ is identity element, $b$ is called the \\emph{inverse} of $a$. A Monoid with inverse element is called a \\emph{Group}. In our example set $G$, $R_{180}$ is it's own inverse, $H$ is it's own inverse and we can try out that every element has it's inverse and so it is a \\emph{Group}.\n\\end{itemize}\nHence, we can conclude that $G=\\{R_0,R_{90},R_{180},R_{270},H,V,D,D'\\} $ is not just a set, rather it has more structural properties. $(G,\\circ) $ is a \\emph{Group}. Now, we are ready to define \\emph{Group} formally.\n\\begin{definition}\n$(G,\\circ)$ is a group if it satisfies closure, associativity and it has identity and inverse element.\n\\end{definition} \n\\textbf{Example1} : Let's consider $\\mathbb{Z}_p=\\{0,1,...,p-1\\} $ where $p$ is a prime number. Define the operation to be $+ \\ mod \\ p $. Let's verify whether $(\\mathbb{Z}_p,+ \\ mod \\ p) $ is a group or not:\n\\begin{itemize}\n\\item \\emph{Closure} : $\\forall a, b\\in \\mathbb{Z}_p; a+ \\ mod \\ p\\in \\mathbb{Z}_p$. So, $\\mathbb{Z}_p$ is closed under the operation $+ \\ mod \\ p$\n\\item \\emph{Associativity} : As modulo arithmetic is associative, $\\mathbb{Z}_p$ is associative under $+ \\ mod \\ p$\n\\item \\emph{Identity} : $0$ is the identity element for all elements in $\\mathbb{Z}_p$. \n\\item \\emph{Inverse} : For any element $a\\in \\mathbb{Z}_p$, $(p-a)$ will be the inverse of $a$.\n\\end{itemize}\nHence, $(\\mathbb{Z}_p,+ \\ mod \\ p) $ is a group.\n\\paragraph{}\nLet's check whether $(\\mathbb{Z}_p,\\times \\ mod \\ p) $ is group or not:\n\\begin{itemize}\n\\item \\emph{Closure} : $\\forall a, b\\in \\mathbb{Z}_p; a\\times \\ mod \\ p\\in \\mathbb{Z}_p$. So, $\\mathbb{Z}_p$ is closed under the operation $\\times \\ mod \\ p$\n\\item \\emph{Associativity} : As modulo arithmetic is associative, $\\mathbb{Z}_p$ is associative under $\\times \\ mod \\ p$\n\\item \\emph{Identity} : $1$ is the identity element for all elements in $\\mathbb{Z}_p$\n\\item \\emph{Inverse} : The element $0$ does not have any inverse.\n\\end{itemize}\nHence, $(\\mathbb{Z}_p,\\times \\ mod \\ p) $ is not a group but it is a \\emph{Monoid}.\n\\paragraph{}\nConsider the group $\\mathbb{Z}_p^* $ under the operation $\\times \\ mod \\ p $; for example we can verify that $(\\mathbb{Z}_p^*,\\times \\ mod \\ p) $ satisfies closure, associativity, identity and inverse properties. Hence, $(\\mathbb{Z}_p^*,\\times \\ mod \\ p) $ is a group.\\\\\n\\textbf{Exercise} : Use \\emph{pigeon hole principle} to prove that every element has it's inverse in $(\\mathbb{Z}_p^*,\\times \\ mod \\ p) $.\n\\paragraph{}\n\\textbf{Example2} : Set of bijection from $\\{1,2,...,n\\} $ to $\\{1,2,...,n\\}$ forms a group under composition ($\\circ $).\n\\begin{itemize}\n\\item \\emph{Closure} : Composition of two bijections is bijection\n\\item \\emph{Associative} : Using function composition property, if $f,g,h$ are functions, $f\\circ(g\\circ h)=(f\\circ g)\\circ h$\n\\item \\emph{Identity} : Identity function will be there\n\\item \\emph{Inverse} : Every element will have inverse\n\\end{itemize}\nHence, set of bijections $(S_n,\\circ) $ forms a group.\n\\section{Group (abstractly)}\nSo far, we have seen the definition of a group $G$, under some operation $\\circ$. Now, we will move on the definition of \\emph{Subgroup}.\n\\begin{definition}\nLet $(G,\\circ) $ be a group. $H\\subseteq G$ is said to be a subgroup if $H$ forms a group by itself with respect to the same operation $\\circ$. A subgroup of $G$ is denoted by $H\\leq G$. \n\\end{definition}\n\\textbf{Example} : $(\\mathbb{Z}_{15},+ \\ mod \\ 15) $ is a group. Consider $H=\\{0,3,6,9,12\\}$. $H\\subseteq G$ and $H$ satisfies closure, associativity, identity, inverse and so it forms a group under $+ \\ mod \\ 15 $, so $H\\leq G$.\n\\subsection{Subgroup}\nConsider a group $G$ and $H\\leq G$. Let's take an element $g$ from $G\\setminus H$ and multiply it with $H$, define:\n\\begin{align}\nHg=\\{hg|h\\in H\\}\n\\end{align} \n\\begin{observation}\nIf $g\\in H$; then $Hg\\subseteq H$ as $H$ is closed by itself. \n\\end{observation}\nIt's not just $Hg\\subseteq H$; something more is true. For that, we will have the following claim\n\\begin{claim}\n$|Hg|=|H|$ when $g\\in H$\n\\end{claim}\n\\begin{proof}\n$\\forall h_1\\neq h_2\\in H$, we need to argue that $h_1g\\neq h_2g$. Or multiplication by $g$ is actually a bijection.\\\\\nLet's prove it by contradiction and let $\\forall h_1\\neq h_2\\in H$\n\\begin{align*}\nh_1g&=h_2g\\\\\nh_1&=h_2gg^{-1}\\\\\nh_1&=h_2\n\\end{align*}\nHence, we get a contradiction. So, $h_1g\\neq h_2g$\n\\end{proof}\nNow, let's move on to the following claim\n\\begin{claim}\n$\\forall g_1,g_2$; if $Hg_1\\cap Hg_2\\neq \\emptyset $; then $Hg_1=Hg_2$. In other words, if $Hg_1$ and $Hg_2$ are overlapping, then they are same.\n\\end{claim}\n\\begin{proof}\nLet $g_1,g_2\\in G$. Suppose $g\\in Hg_1\\cap Hg_2$\n\\begin{align*}\n\\exists h\\in H; g&=hg_1\\\\\n\\exists h'\\in H; g&=h'g_2\n\\end{align*}\nNow, combining the above two,\n\\begin{align*}\nhg_1&=h'g_2\\\\\ng_1&=h^{-1}h'g_2\n\\end{align*}\nNow, we are ready to prove $Hg_1\\subseteq Hg_2$\\\\\nConsider, any element in LHS $h''\\in H $, substituting $g_1$, we get:\n\\begin{align*}\nh''g_1&=h''h^{-1}h'g_2\\\\\n&=h'''g_2\n\\end{align*}\nHence, any element in LHS is in RHS. Similarly, we can prove $Hg_2\\subseteq Hg_1$. So, $Hg_1=Hg_2$.\n\\end{proof}\n\\paragraph{}\nSo, we can argue that the multiplication of $H$ by other elements will result in translation of $H$ which are kind of tiling of group $G$. It does not mean that every $g_i$ will give different tiles, but if they have common element, they are same. One interesting feature is that all such tiles will have equal size. One natural question is can there be any element that is not in any tiling? Yes, there can be. For example, $\\forall g\\in G, g\\in Hg$ because $H$ contains identity as it is subgroup. So,$g$ is an element of $Hg$ always. So, every element $g$ will be there in some tile for sure. If there are total $k$ tiles, we argue that:\n\\begin{align*}\n|G|&=k.|H|\n\\end{align*}\n\\begin{theorem}\nLagrange's Theorem : The size of a subgroup $|H$ must divide the size of a group $|G|$.\n\\end{theorem}\n For example, A group of size $100$ can not have a subgroup of size $99$. In fact it can only have subgroup of size at most $50$. Any group with prime number of elements, can not have any non-trivial subgroup. So, \\emph{Lagrange's Theorem} is the example of algebraic structure implying combinatorial bounds.\n\n\n\\Lecture{Jayalal Sarma}{Nov 4, 2020}{25}{A step towards Polya's Theory }{Sampriti Roy}{$\\alpha$}{JS}\n\\section{A quick recap}\n In last lecture, we started with a problem of counting number of distinct colourings when the corner of a square are coloured with $2$ colours. We defined a set $G=\\{R_0,R_{90},R_{180},R_{270},H,V,D,D'\\} $ that acts on set of all possible $2$-coloured squares and some of them are equivalent under these operations. We have also talked about the definition of subgroup $(H)$ of a group $(G)$ and defined that for $g\\in G\\setminus H$, $Hg=\\{hg: h\\in H \\}$. This $Hg$ is called \\emph{coset} of $H$ in $G$. \nWe have also seen that if any two cosets overlap, they have to be same. We have also talked about \\emph{Lagrange's Theorem}. In this chapter the plan is to understand \\emph{Polya's Theory}. We will complete it only in next lecture but we will do a step towards it. The step is known to be \\emph{Burnside's Lemma}. \n\\section{The abstract problem of counting distinct 2-coloured squares}\n $\\Omega$ be the set of all $2$-coloured squares. There $|\\Omega|=2^4=16$ possibilities. Our task is to count to number of distinct colourings among $\\Omega$. Let's define the equivalence between two coloured squares formally:\n \\begin{definition}\n Let $\\alpha,\\beta\\in \\Omega$, if $\\exists g\\in G$ such that the action on $g$ on $\\alpha$ returns $\\beta$, or $\\alpha^g=\\beta$; then we say $\\alpha$ and $\\beta$ are equivalent, $\\alpha\\sim \\beta$.\n\\end{definition}\nThe relation $\\sim$ between $\\alpha$ and $\\beta$ satisfies the following properties:\n\\begin{itemize}\n\\item \\emph{Reflexive} : $\\alpha\\sim \\alpha$\n\\item \\emph{Symmetric} : $\\alpha\\sim \\beta\\rightarrow \\beta\\sim \\alpha$; if the action of $g$ makes the transformation from $\\alpha$ to $\\beta$, then $g^{-1}$ will make the transformation from $\\beta$ to $\\alpha$. For example, $R_{90}$ transform $\\alpha$ to $\\beta$, $R_{270}$ will transform $\\beta$ to $\\alpha$.\n\\item \\emph{Transitive} : If $\\alpha\\sim \\beta$ by the action $g_1$, $\\beta\\sim \\gamma$ by the action $g_2$; then the action of $g_1g_2$ (composition of $g_1,g_2$) will transform $\\alpha\\sim \\gamma$.\n\\end{itemize}\nHence, $\\sim$ is an equivalence relation and it splits the set $\\Omega$ into set of equivalence classes.\n\\subsection{Orbit and Stabilizer}\nThe action of $g$ on $\\Omega$, partitions $\\Omega$ into different equivalence classes but their sizes need not be equal. This equivalence classes are called \\emph{Orbit}. In our context, among set of all possible $2$-colourings of a square, we don't want to count equivalent squares twice. In other words, the abstract problem that we are interested to study is to count the number of orbits of $\\Omega$. First, we are going to study the sizes of the orbits and using them we will have a mechanism to count them. Let's define the following\n\\begin{definition}\n$Orbit_G(\\alpha)=$ orbit of $\\alpha$ under the action of $G$ on $\\Omega$\n\\begin{align*}\nOrbit_G(\\alpha)&=\\{\\beta:\\exists g\\in G \\ s.t \\ \\alpha^g=\\beta \\}\n\\end{align*}\n\\end{definition}\nNow, we are ready to define \\emph{Stabilizer} of $\\alpha$ on an action of $G$. Let, $g\\in G$ is acting on $\\Omega$. Let $\\alpha\\in G$, which are the elements in $G$ that fixes $\\alpha$? i.e, the elements in $G$ that takes $\\alpha$ to itself. \n\\begin{definition}\nStabilizer of an element $\\alpha$ is a subset of $G$, which acts on $\\alpha$ and takes it back to itself.\n\\begin{align*}\nStab_G(\\alpha)&=\\{g\\in G: \\alpha^g=\\alpha \\}\n\\end{align*}\n\\end{definition}\nWe will see few examples of stabilizer to make it more clear. \n\\begin{itemize}\n\\item \\emph{Example 1} : A simple example we can think of in our square setting is that, let $\\alpha$ be the square where all the corners are coloured yellow. Now, we can easily conclude that any operation from the group $G$ can fix $\\alpha$, i.e, no matter which action we are performing, $\\alpha$ will be $\\alpha$ itself. So, the stabilizer of $\\alpha$ is the entire group $G$.\n\\item \\emph{Example 2} : Let's think of another $\\alpha$ where bottom left and top right corners are coloured blue and the other two are coloured yellow. Now, we can verify that the operations that fixes $\\alpha$ are : $\\{R_0,R_{180},D,D' \\} $. \n\\end{itemize}\nAn interesting observation is that stabilizers inside $G$ are not only subset of $G$, they are actually subgroup of $G$. Let's prove the argument formally,\n\\begin{claim}\n$$Stab_G(\\alpha)\\leq G$$\n\\end{claim}\n\\begin{proof}\nFix an $\\alpha$ and verify the following :\n\\begin{itemize}\n\\item \\emph{Closure} : If $g_1,g_2\\in Stab_G(\\alpha)$, then $g_1g_2\\in Stab_G(\\alpha)$. As, $g_1,g_2$ are fixing $\\alpha$, then their composition will also fix $\\alpha$.\n\\item \\emph{Associativity} : As the operations performed are subset of $G$, associativity is inherited in $Stab_G(\\alpha)$.\n\\item \\emph{Identity} : Identity fixes every element, in particular it fixes $\\alpha$. So, identity element is present in $Stab_G(\\alpha)$.\n\\item \\emph{Inverse} : If an element $g$ fixes $\\alpha$, then $g^{-1}$ will also fix $\\alpha$. So, inverse element is always present in $Stab_G(\\alpha)$. For example, if $R_{90}$ fixes a coloured square, then $R_{270}$ will also fix the same square.\n\\end{itemize}\nHence, $Stab_G(\\alpha)\\leq G$\n\\end{proof}\nThere is a combinatorially useful relation between size of the stabilizer and the size of the of $\\alpha$. Let's define the following lemma;\n\\begin{lemma}\n\\textbf{Orbit-Stabilizer Lemma} : $$\\forall\\in \\Omega;|Orbit_G(\\alpha)|.|Stab_G(\\alpha)|=|G|$$\n\\end{lemma}\nWe will prove this lemma formally in next lecture, but here we are going to draw a outline of the thought process to prove this lemma.\n\\paragraph{}\n\\emph{Proof Idea} : Let $Stab_G(\\alpha)$ be denoted by $H$. We have already proved that $Stab_G(\\alpha)$ is a subgroup of $G$. Using \\emph{Lagrange's theorem} we can say that $$|G|=k.|Stab_G(\\alpha)|$$\nWhere $k$ is the number of cosets $H$ in $G$. So, it is sufficient to prove that number of cosets $H$ in $G$ is exactly $|Orbit_G(\\alpha)|$. We will show a bijection that for every element in the orbit of $\\alpha$, there is a way to associate a corresponding coset with it. Hence, the number of coset in $Stab_G(\\alpha) =|Orbit_G(\\alpha)|$. Now, we will quickly define the bijection:\n\\paragraph{}\nConsider any $\\beta\\in Orbit(\\alpha)$; it means $\\exists g\\in G $ s.t. $\\alpha^g=\\beta $. Now, coset corresponding to $g$ is $Hg$. We required to show the following:\n\\begin{itemize}\n\\item \\emph{Well definedness} : There could be many $g\\in G$ which makes $\\alpha$ to $\\beta$. We need to argue that no matter which $g\\in G$ we choose, we will end up by getting the coset $Hg$.\n\\item \\emph{Injection} : For $\\beta\\neq \\beta'\\in Orbit(\\alpha)$; there is $g,g'\\in G$ where $\\alpha^g=\\beta,\\alpha^{g'}=\\beta'$; we will end up by getting two different cosets, i.e., $Hg\\neq Hg' $.\n\\item \\emph{Surjection} : To show that for any coset we have corresponding element in the orbit of $\\alpha$.\n\\end{itemize}\nAn interesting observation we can make from the above lemma is that if the size of a group is any prime number then either $|Stab_G(\\alpha)|$ is same as $|G|$ and $|Orbit_G(\\alpha)|=1$ or $|Stab_G(\\alpha)|=1$ and $|Orbit_G(\\alpha)|$ is same as $|G|$. But, we are not interested in individual sizes of orbits but we want to count the number of orbits. We will use the \\emph{Orbit-Stabilizer lemma} to define the following lemma to show that if we have a handle on the size of orbits then we can have a handle on the number on number of orbits as well. First, we will define the following:\n\\begin{definition}\n\\textbf{Fix Points} : If $G$ is acting on $\\Omega$, fix points of a group element are those elements on $\\Omega$ that are fixed by $g$.\n\\begin{align*}\nfix(g)&=\\{\\alpha\\in \\Omega: \\alpha^g=\\alpha \\}\n\\end{align*}\n\\end{definition}\nNow, we are ready to state the following lemma:\n\\begin{lemma}\n\\textbf{Burnsides Lemma} : \n\\begin{align*}\n\\#of \\ orbits \\ &=\\frac{1}{|G|}\\sum_{g\\in G}|fix(g)|\n\\end{align*}\n\\end{lemma}\n\\begin{proof}\nThe proof is a classic proof by double counting method. Let's define the set\n\\begin{align*}\nS&=\\{(g,\\alpha): g\\in G, \\alpha\\in \\Omega; \\alpha^g=\\alpha \\}\n\\end{align*}\nWe will estimate $|S|$ in two different ways:\n\\begin{itemize}\n\\item Answer $1$ : For each $g$ count number of different $\\alpha$ that are fixed by $g$.\n\\begin{align*}\n&\\sum_{g\\in G} (\\#of \\ \\alpha \\ s.t \\ \\alpha^g=\\alpha)\\\\\n&=\\sum_{g\\in G}|fix(g)|\n\\end{align*}\n\\item Answer $2$ : For each $\\alpha\\in \\Omega$, count number of different $g$ that fixes $\\alpha$.\n\\begin{align*}\n&\\sum_{\\alpha\\in \\Omega} (\\#of \\ g \\ s.t \\ \\alpha^g=\\alpha)\\\\\n&=\\sum_{\\alpha\\in \\Omega}|Stab_G(\\alpha)|\\\\\n&=\\sum_{\\alpha\\in \\Omega}\\frac{|G|}{|Orbit_G(\\alpha)|}\\\\\n&=|G|.\\sum_{\\alpha\\in \\Omega}\\frac{1}{|Orbit_G(\\alpha)|}\\\\\n\\end{align*}\nLet us consider the orbits of $\\Omega$ are denoted by $\\{\\Omega_1,\\Omega_2,...\\}$. Then the above expression becomes,\n\\begin{align*}\n&=|G|.(\\sum_{\\alpha\\in \\Omega_1}\\frac{1}{|\\Omega_1|}+\\sum_{\\alpha\\in \\Omega_2}\\frac{1}{|\\Omega_2|} )\\\\\n&=|G|.(\\#of \\ orbits)\n\\end{align*}\n\\end{itemize}\nEquating Answer $1$ and $2$, we get:\n\\begin{align*}\n|G|.(\\#of \\ orbits)&=\\sum_{g\\in G}|fix(g)|\\\\\n\\#of \\ orbits&=\\frac{1}{|G|}\\sum_{g\\in G}|fix(g)|\n\\end{align*}\n\\end{proof}\n\\section{Counting number of distinct coloured square using Bernsides Lemma}\nWe we apply Bernsides lemma to our example problem to count number of distinct coloured squares in $\\Omega$. For that, first we have to compute $|fix(g)|$ for each $g\\in G$. Where $G=\\{R_0,R_{90},R_{180},R_{270},H,V,D,D'\\} $.\n\\begin{itemize}\n\\item $g=R_0$; all the coloured squares in $\\Omega$ are fixed by $g$. So, $|fix(R_0)|=|\\Omega|=16$\n\\item $g=R_{90}$; the squares whose all corners are either blue or yellow will be fixed by $g$. So, $|fix(R_{90})|=2 $\n\\item $g=R_{180}$; total $4$ squares will be fixed by $g$. So, $|fix(R_{180})|=4 $\n\\item $g=R_{270}$; the number of squares that will be fixed are same as $R_{90}$. Because, if a square is fixed by $R_{90}$, it will be fixed by it's inverse as well. So, $|fix(R_{270})|=2 $\n\\item $g=H$; it will fix the squares where the upper half and lower half corresponding to the horizontal plane (line towards the middle of the square) will be coloured with same colour. so, $|fix(H)|=4 $\n\\item $g=V$; the number of squares will be same as that are fixed by $H$. so, $|fix(V)|=4 $\n\\item $g=D$; consider one diagonal and corresponding to it the upper half and lower half will be coloured with same colour, there are $2$ ways to do it and for each such colour the diagonal can be coloured in $4$ ways. So, $|fix(D)|=8 $\n\\item $g=D'$; the number of squares will be same as that are fixed by $D$. So, $|fix(D')|=8$\n\\end{itemize}\nUsing Bernsides lemma,\n\\begin{align*}\n\\#of \\ orbits \\ in \\ \\Omega&=\\frac{1}{8}(16+2+4+2+4+4+8+8)\\\\\n&=6\n\\end{align*}\nSo, out of the $16$ total colourings $6$ of them are in-equivalent to each other. To simulate the thought process, here is an exercise:\n\\emph{Exercise} : Consider all $2$-coloured squares and the set of operations that are acting on it are defined by $G'=\\{R_0,R_{90},R_{180},R_{270}\\}$, i.e. we are not worried about the $H,V,D,D'$ rotations. Under this operations count the number of distinct $2$-coloured squares. \n\\\\\n\n\\Lecture{Jayalal Sarma}{Nov 11, 2020}{26}{Another step towards Polya's Theory }{Achyuth Prakash}{$\\alpha$}{JS}\n\\section{A quick recap}\n In last lecture, we defined and proved the orbit-stabilizer lemma and Lagrange's theorem. In this chapter the plan is to better understand \\emph{Polya's Theory} by going through a further example. \n\n\\section{Example 2: Coloring faces of a cube}\nLet us consider coloring the faces of a cube with $2$ colors. As before we define, $\\Omega$ as the set of all possible colorings. There are $6$ faces, and each can can be colored with $2$ colors, thus we have $|\\Omega| = 2^6 = 64$.\n\\\\\nNext we define the set of 'operations', $G$ which can act on elements on $\\Omega$. What are the different possible operations? We list them and simultaneously compute $|fix(g)|$:\n\\\\\nHere is the labelled cube for reference:\n\\\\\n\\includegraphics[]{./images/cube.png}\n\\\\\nThe different operations are:\n\\begin{itemize}\n\\item $R_0$ : Rotation by $0^{\\circ}$, i.e. identity. Since all elements of $\\Omega$ when acted upon by $R_0$ give back the same element, we get $|fix(g)| = 64$.\n\n\\item $R_1$ : Rotation by $180^{\\circ}$ w.r.t axis through the center of the opposite faces. (Observe that there are $3$ pairs of such faces). Now consider $2$ such faces, $A, D$. If we rotate the cube as per $R_1$, how many elements of $\\Omega$ remain unchanged? \n\\\\\nConsider the faces $A, D$. They can be colored with any color each. Thus, there are $2*2 = 4$ choices. Now, for the other faces, consider $C, F$. These $2$ must be colored with the the same color(because they exchange places) Thus there are $2$ choices for $C, F$ together. Similarly for $B, E$. Thus in total, there are $2^4 = 16$ elements which belong to $fix(g)$.\n\n\\item $R_2$ : Rotation by $90^{\\circ}$ w.r.t axis through the center of the opposite faces. Observe that there are $3$ pairs of such faces. Again let us compute $|fix(g)|$.\n\\\\\nConsider the faces $A, D$. They can be colored with any color each. Thus, there are $2*2 = 4$ choices. Now, for the other faces, they all must be colored with the same color, because, $C \\rightarrow B$, $B \\rightarrow F$, $F \\rightarrow E$, and $E \\rightarrow C$. Thus, there are $2$ choices for $B, C, E, F$ together. Hence, $|fix(g)| = 2^3 = 8$.\n\n\\item $R_3$ : Rotation by $270^{\\circ}$ w.r.t axis through the center of the opposite faces. Observe that there are $3$ pairs of such faces. Let us compute $|fix(g)|$.\n\\\\\nSince this is the exact opposite of the previous case, by symmetry, $|fix(g)| = 8$.\n\n\\item $R_4$ : Rotation by $180^{\\circ}$ w.r.t axis through the midpoints of opposite edges. Observe that there are $6$ pairs of such edges, i.e. $(de, bg),\\ (af, ch),\\ (ad, hg),\\ (cb, ef),\\ (ab, eh),\\ (cd, fg)$. To compute $|fix(g)|$, \n\\\\\nSuppose we do the rotation w.r.t the edges $(de, bg)$. then observe that, now $F$ is now the top face, and $A$ occupies the place where $E$ was initially, similarly, $B$ and $D$ swap places. Thus there are $2$ choices together for $C, F$, $2$ choices for $A, E$ and $2$ for $B, D$, giving $8$ choices overall.\nThus, $|fix(g)| = 8$.\n\n\\item $D$ : Diagonal flip by $120^{\\circ}$ w.r.t axis through the centers of opposite corners. First, note that there are $4$ pairs of opposite corners. To compute $|fix(g)|$,\n\\\\\nConsider the vertex $b$. If we flip the cube along $b$ by $120^{\\circ}$, then the face $B \\rightarrow C$, i.e. $B$ now occupies the initial position of $C$, $C \\rightarrow D$ and $D \\rightarrow B$. Thus together, for $B, C, D$ there are $2$ choices. Similarly for $A, E, F$. Thus totally we get $2*2 = 4$ possibilities, and $|fix(g)| = 4$.\n\n\\item $D'$ : Diagonal flip by $240^{\\circ}$ w.r.t axis through the centers of opposite corners. First, note that there are $4$ pairs of opposite corners. To compute $|fix(g)|$,\n\\\\\nBy symmetry, this is the opposite/complementary operation of the previous one, i.e. flip by $240^{\\circ}$ = flip by $120^{\\circ}$ in the other direction. Thus, $|fix(g)| = 4$.\n\\end{itemize}\n\nThus totally there are $1 + 3*3 + 6 + 4*2 = 24$ operations in $G$.\n\nNow, if we apply the formula to calculate the number of orbits, we get:\n$$\\#of \\ orbits \\ in \\ \\Omega =\\frac{1}{24}(64 + 3*16 + 3*8 + 3*8 + 6*8 + 4*4 + 4*4) = 10$$\n\nTherefore, we observe that out of the $64$ colorings, $10$ of them are in-equivalent.\n\n\\section{Cycle Index}\nWe now introduce and define the notion of cycle indices. Let $G$ be the set of some permutations on $\\Omega$. Let us decompose $g \\in G$ into a collection of disjoint cycles. \n\\\\\nFor example, let $n = 5$. Consider the permutation $3, 1, 2, 5, 4$ we write this as $(1, 2, 3)\\ (4, 5)$, i.e. this denotes $1 \\rightarrow 2$, (i.e. $1$ goes to position $2$), $2 \\rightarrow 3, 3 \\rightarrow 1$ and so on. \n\\\\\nSimilarly, the identity permutation, $1, 2, 3, 4, 5$ would be written as $(1)(2)(3)(4)(5)$.\n\\begin{definition}\n\\textbf{Type of permutation $\\pi$} : A permuatation $\\pi$ is said to be of type $(b_1, b_2, \\ldots b_m)$ if $b_i$ represents the number of $i$ length cycles in the cycle representation of $\\pi$.\n\\end{definition}\n\nFor example if $\\pi = (1, 2, 3)\\ (4, 5)$ then it is of type $(0, 1, 1, 0, 0)$, because there are no $1$ length cycles, $\\implies b_1 = 0$, one $2$ length cycle, i.e. $4 \\rightarrow 5 \\rightarrow 4 \\implies b_2 = 1$, and one $3$ length cycle.\n\nNow, corresponding to type $(b_1, b_2, \\ldots b_m)$ consider the monomial $x_1^{b_1}\\ x_2^{b_2} \\ldots x_m^{b_m}$. Using these terms we define a polynomial as follows:\n\\begin{definition}\nCycle index polynomial of $G$:\n\\end{definition}\n$$P(x_1, x_2, \\ldots x_m) = \\frac{1}{|G|} \\sum_{g \\in G} x_1^{b_1}\\ x_2^{b_2} \\ldots x_m^{b_m}$$ where $(b_1, b_2, \\ldots b_m)$ is the cycle type of $g$.\n\\\\\nWe observe that the above polynomial generalises the Burnside's lemma because we can now substitute the number of possible colors say $2$ in place of $x_1, x_2, \\ldots x_m$. Thus, this theorem is much more versatile.\n\\\\\n\n\\Lecture{Jayalal Sarma}{Nov 12, 2020}{27}{Polya's Theory - Part 1}{Achyuth Prakash}{$\\alpha$}{JS}\n\\section{Quick Recap}\nIn the last lecture, we went through another example(of coloring the faces of a cube) and defined the cycle index polynomial of $G$.\n\\\\\nConsider an example:\nLet $G$ be a group such that $G = \\{e, (1, 2),\\ (3, 4),\\ (1,2)(3,4)\\} \\le S_4$\ni.e., $G$ is a subgroup of $S_4$.\n\\\\\nThus, let us write down the cycle index polynomial of $G$:\n$$P(x_1, x_2, x_3, x_4) = \\frac{1}{4} (x_1^4 + x_1^2x_2 + x_1^2x_2 + x_2^2)$$\nWe get the RHS because $e$ has $4$ length $1$ cycles, thus we get the $x_1^4$ term, $(1, 2)$ has $2$ length $1$ cycles and $1$ length $2$ cycle, thus the term $x_1^2x_2$ and so on.\n\n\\section{Polya's Theorem - version 1}\n\\begin{theorem}\nPolya's theorem: Let $G$ be the group of symmetry acting on $\\Omega$, then, if the number of distinct colored patterns with $k$ colors = $N$, we have\n$$N = P_G(k, k, k, \\ldots k)$$,\nor $N = P(x_1, x_2, \\ldots x_m)$ with $x_1 = x_2 = \\ldots x_m = k$.\n\\end{theorem}\n\n\\begin{proof}\nLet the domain size be $= m$. (Here by domain we mean the number of different objects we need to color. For example in the squares example, we had $4$ corners and thus $m = 4$, similarly in the cube example, we had six faces to color, hence $m = 6$.) Then, the value of $|\\Omega|$ will be $k^m$.\n\\\\\nBy Burnside's lemma, WKT, \n$$Number\\ of\\ distinct\\ colorings\\ = Number\\ of\\ different\\ orbits\\ of\\ G = \\frac{1}{|G|} \\sum_{g\\in G}|fix(g)|$$\nLet us thus compute $|fix(g)|$ for $g \\in G$. \n\\\\\nAn example will make this more clear, before we generalize. Consider the previously discussed example of coloring the vertices of a square with $2$ colors. Then, we can think of each operation $g \\in G$ as a permutation. Consider the operation of rotating a square by $90^{\\circ}$, we can represent this by the cycle index $(1, 2, 3, 4)$, because $1 \\rightarrow 2,\\ 2 \\rightarrow 3, \\ldots$. Similarly, the operation of vertical flip corresponds to $(1, 2)\\ (3, 4)$. Consider this operation. Since $1 \\rightarrow 2$ and $2 \\rightarrow 1$, we observe that for $g$ to fix some element of $\\Omega$, the corners $1, 2$ must get the same color and corners $3, 4$ must get the same color. Thus there are $k$ choices for $1, 2$ together and $k$ choices for $3, 4$ and thus, totally $k^2$ colorings which are fixed by $g$.\n\\\\\nIn general, let the cycle structure of $g = (b_1, b_2, \\ldots b_m)$. This means, $g$ has $b_1$ cycles of length $1$, $b_2$ cycles of length $2$, and so on. Now, observe that all the elements that are part of a cycle must receive the same color, if $g$ were to fix the particular coloring. Thus for each cycle of length $i$, there are $k$ choices and as there are $b_i$ cycles of length $i$, there are $k^{b_i}$ choices for all of them. Thus, \n$$ the\\ number\\ of\\ colorings\\ fixed\\ by\\ g\\ = k^{b_1} * k^{b_2} \\ldots k^{b_m}$$\ni.e. $|fix(g)| = x_1^{b_1} * x_2^{b_2} \\ldots x_m^{b_m}$ with $x_1 = x_2 = \\ldots x_m = k$.\nThus, \n$$P_G(k, k, k, \\ldots k) = \\frac{1}{|G|} \\sum_{g\\in G}(k^{b_1} * k^{b_2} \\ldots k^{b_m}) = \\frac{1}{|G|} \\sum_{g\\in G}|fix(g)|$$\nHence proved.\n\\end{proof}\n\n\\section{Applying Polya's Theorem}\nWe now consider some examples and apply Polya's theorem for a more concrete understanding.\n\\subsection{Example 1 - Coloring Necklaces with circular beads}\nConsider a necklace with $3$ circular beads. We label the beads as $1, 2, 3$. First we ask: what are the possible symmetries? We get $G = \\{e, R_{120}, R_{240}, F_{12}, F_{23}, F_{31}\\}$, i.e. $e$ is the identity, $R_{120}$ represents a $120^{\\circ}$ clockwise rotation through the center, $F_{12}$ represents flipping w.r.t the edge $1-2$, etc. Now let us represent each $g \\in G$ as a cycle index permutation. Then we get that $G = \\{(1)(2)(3), (1, 2, 3), (1, 3, 2), (1, 2)(3), (1)(2, 3), (2)(1, 3)\\}$ respectively. Hence if we apply Polya's theorem, the number of distinct colorings = $\\frac{1}{6} (x_1^3 + 3x_1x_2 + 2x_3)$, if we consider coloring with $2$ distinct colors, then $k = 2 = x_1 = x_2 = x_3$, thus the number of distinct colorings with 2 colors = $\\frac{1}{6} (2^3 + 3*2^2 + 2*2) = 4$.\n\n\\subsection{Example 2 - Coloring faces of a cube}\nLet us revisit the cube example. We list the different elements of $G$ as before, and simultaneously calculate the term they contribute to in Polya's formula.\n\\\\\nHere is the labelled cube for reference:\n\\\\\n\\includegraphics[]{./images/cube.png}\n\\\\\n\\begin{itemize}\n\\item $R_0$ : Rotation by $0^{\\circ}$, i.e. identity. The cycle index of this permutation has $6$ cycles of length $1$, thus the number of colorings contributed will be $\\boxed{x_1^6}$.\n\n\\item $R_1$ : Rotation by $180^{\\circ}$ w.r.t axis through the center of the opposite faces. (Observe that there are $3$ pairs of such faces). Consider the faces $A, D$. We have observed before that $A, D$ remain in place, and $B, E$ and $C, F$ exchange their places, thus there are $2$ length $1$ cycles and $2$ length $2$ cycles, hence the contributing term will be $x_1^2x_2^2$. As there are $3$ such pairs of faces, the total number of fixed colorings will be $\\boxed{3x_1^2x_2^2}$.\n\n\\item $R_2$ : Rotation by $90^{\\circ}$ w.r.t axis through the center of the opposite faces. Observe that there are $3$ pairs of such faces. Consider the faces $A, D$. They retain their positions, i.e $A \\rightarrow A$, $D \\rightarrow D$. Now, for the other faces, we have, $C \\rightarrow B$, $B \\rightarrow F$, $F \\rightarrow E$, and $E \\rightarrow C$. Thus, there are $2$ length $1$ cycles and $1$ length $4$ cycle, hence the contributing term will be $x_1^2x_4$.As there are $3$ such pairs of faces, the total number of fixed colorings will be $\\boxed{3x_1^2x_4}$.\n\n\\item $R_3$ : Rotation by $270^{\\circ}$ w.r.t axis through the center of the opposite faces. Observe that there are $3$ pairs of such faces. Since this is the exact opposite of the previous case, by symmetry, the total number of fixed colorings will be $\\boxed{3x_1^2x_4}$.\n\n\\item $R_4$ : Rotation by $180^{\\circ}$ w.r.t axis through the midpoints of opposite edges. Observe that there are $6$ pairs of such edges, i.e. $(de, bg),\\ (af, ch),\\ (ad, hg),\\ (cb, ef),\\ (ab, eh),\\ (cd, fg)$. Suppose we do the rotation w.r.t the edges $(de, bg)$. Then observe that, now $F$ is now the top face, and $A$ occupies the place where $E$ was initially, similarly, $B$ and $D$ swap places. In other words, $C \\iff F$, $A \\iff E$ and $B \\iff D$, i.e. there are $3$ length $2$ cycles. Thus, the contributing term is $x_2^3$. As there are $6$ pairs of edges, the total contributing term will be $\\boxed{6 x_2^3}$.\n\n\\item $D$ : Diagonal flip by $120^{\\circ}$ w.r.t axis through the centers of opposite corners. First, note that there are $4$ pairs of opposite corners. \nConsider the vertex $b$. If we flip the cube along $b$ by $120^{\\circ}$, then the face $B \\rightarrow C$, i.e. $B$ now occupies the initial position of $C$, $C \\rightarrow D$ and $D \\rightarrow B$. Thus there is the cycle $B \\rightarrow C \\rightarrow D \\rightarrow B$. Similarly for the other three faces. Thus there are $2$ cycles of length $3$ and hence the contributing term will be $x_3^2$. As there $4$ such pairs of corners, the total term contributed will be $\\boxed{4x_3^2}$.\n\n\\item $D'$ : Diagonal flip by $240^{\\circ}$ w.r.t axis through the centers of opposite corners. First, note that there are $4$ pairs of opposite corners. By symmetry, this is the opposite/complementary operation of the previous one, i.e. flip by $240^{\\circ}$ = flip by $120^{\\circ}$ in the other direction. Thus, the total term contributed will be $\\boxed{4x_3^2}$.\n\\end{itemize}\n\nPutting all the above together, we get $$P_{G}(x_1, x_2, \\ldots x_4) = \\frac{1}{24}(x_1^6 + 3x_1^2x_2^2 + 6x_1^2x_4 + 6x_2^3 + 8x_3^2)$$\nif we substitute $x_i = 2$ for all $i$, we get \n$$P_{G}(x_1, x_2, \\ldots x_4) = \\frac{1}{24}(2^6 + 3*2^4 + 6*2^3 + 6*2^3 + 8*2^2) = 10$$\n\\\\\n\n\\section{Towards a general formula}\nLet us go back to the the example of the square coloring, where $G = \\{e, R_{90}, R_{180}, R_{270}, H, V, D, D'\\}$, and $G$ as a cycle index permutation is $G = \\{(1)(2)(3)(4), (1,2,3,4), (1, 3)(2, 4), (1,4,3,2), (1, 4)(2, 3), (1,2)(3,4), (2,4)(1,3), (1)(3)(2, 4), (2)(4)(1,3)\\}$ respectively.\n\\\\\nObserve the 2nd, 3rd, and 4th elements of $G$. We know that $R_{90} \\circ R_{90} = R_{180}$ i.e. $(1,2,3,4) \\circ (1,2,3,4) = (1,3)(2,4)$. Therefore these elements are of the form $g, g^2, g^3$.\n\\\\\nThus, if we have a group whose elements are of the form $G = \\{e, g, g^2, \\ldots g^k\\}$ (called cyclic groups). \nThen the contribution of each term of $g \\in G$ will be:\n\\begin{itemize}\n \\item $e - x_1^n$\n \\item $g - x_n$\n \\item $g^2 - x_{n/2}^2$ and so on\n\\end{itemize}\nThus, in general, $$P_G(x_1, x_2, \\ldots x_m) = \\frac{1}{n} \\sum_{d|n} \\phi(d) x_{d}^{n/d}$$ where $\\phi(m)$ is euler's totient function.\n\\\\\n\\begin{proof}\nConsider the element $g^{i}$. Suppose it has a cycle of length $d$. Observe that under this operation, $1 \\rightarrow 1+i$, $1+i \\rightarrow 1+2i$, \\ldots, $1+d*i \\rightarrow 1$. This must mean $d*i$ is a multiple of $n$. Specifically, we must have $d*i = lcm(n, i)$. Using $lcm(n,i) = \\frac{n*i}{gcd(n, i)}$, we get $d*i = \\frac{n*i}{gcd(n, i)} \\implies gcd(n,i) = \\frac{n}{d} \\implies \\boxed{gcd(d, \\frac{i*d}{n}) = 1}$.\n\\\\\nHowever there is a bijection between elements of $\\phi(d)$ and $i$, because $i = x \\frac{n}{d}$ for $x \\in \\phi(d)$. This implies the number of such $i = \\phi(d)$. Thus, the above result shows that all $i$ such that $gcd(d, i) = 1$ contribute $n/d$ cycles of length $d$. Therefore the contribution to the term is $\\phi(d) x_{d}^{\\frac{n}{d}}$. Hence the proof. \n\\end{proof}\n\nThus, if we have the above formula, the group for the square coloring problem can be represented as $\\{e, \\sigma, \\sigma^2, \\ldots, \\sigma^{k}, \\pi*\\sigma, \\pi*\\sigma^2, \\ldots \\pi*\\sigma^{k-1}\\}$. These are called dihedral groups. We discuss more about this in the next lecture.\n", "meta": {"hexsha": "ac1a60a93d49bda52191e370b8110c21e534e49e", "size": 37476, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week09.tex", "max_stars_repo_name": "Mohit1098/theory-toolkit", "max_stars_repo_head_hexsha": "d7f5385604cead4cabc0911871b9b7e472c5969c", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "week09.tex", "max_issues_repo_name": "Mohit1098/theory-toolkit", "max_issues_repo_head_hexsha": "d7f5385604cead4cabc0911871b9b7e472c5969c", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-10-08T07:34:26.000Z", "max_issues_repo_issues_event_max_datetime": "2020-11-30T06:06:12.000Z", "max_forks_repo_path": "week09.tex", "max_forks_repo_name": "Mohit1098/theory-toolkit", "max_forks_repo_head_hexsha": "d7f5385604cead4cabc0911871b9b7e472c5969c", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 17, "max_forks_repo_forks_event_min_datetime": "2020-09-25T01:35:07.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-28T11:22:06.000Z", "avg_line_length": 86.1517241379, "max_line_length": 814, "alphanum_fraction": 0.6896146867, "num_tokens": 12290, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.782662489091802, "lm_q2_score": 0.8244619350028204, "lm_q1q2_score": 0.645275430210751}} {"text": "\\documentclass[../main.tex]{subfiles}\n\\begin{document}\n \\subsection{Base Characterisation of a Matroid}\nAs quoted above, the bases of a matroid are the maximally independent sets of $\\mathcal{I}.$ These are most naturally visualised as being akin to a basis in a matrix. Where the basis of a matrix is a spanning set and every vector is a linearly combination of the basis. Bases are also spanning sets in matroid. And as such have very useful properties that we will use extensively in later sections. Most notably, the collection of subsets $\\mathcal{I}$ can be generated using the bases (bases are not in general unique) due to the hereditary property $(I2).$ \\\\\nThe below theorem gives us a useful property that can be applied to the bases of a matroid, namely that the bases all have the same cardinality.\n \\begin{thm}\n Show that if $\\mathcal{I}$ is a non-empty hereditary set of subsets of a finite set E, then $(E,\\mathcal{I})$ is a matroid if and only if, for all $X \\subset E$, all maximal members of $\\{I : I \\in \\mathcal{I} $ and $ I \\subset X\\}$ have the same number of elements.\n \\end{thm}\n\\begin{proof}\n Let $B_1 , B_2$ be maximal elements of $\\{I : I \\in \\mathcal{I} $and $ I \\subset X\\}$ \\\\\n\\noindent Assume $|B_1| < |B_2|, B_1, B_2 \\in \\mathcal{I}$ and since we have a matroid\n\\\\\nthere exists $e \\in (B_2 \\setminus B_1)$ such that $B_1 \\cup \\{e\\} \\in \\mathcal{I}.$\n\\\\\nThis contradicts the maximality of $B_1$.\\\\\n \\noindent $\\implies$ All maximal elements of the set $\\{I : I \\in \\mathcal{I} $ and $ I \\subset X\\}$ in our matroid M have the same cardinality.\n\n\\end{proof}\n \\end{document}", "meta": {"hexsha": "a2aacf4fc98e8a53d58e1892a609ed13434c4569", "size": 1605, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "LaTeXPdfs/sections/char2.tex", "max_stars_repo_name": "emcd123/Matroids", "max_stars_repo_head_hexsha": "f1ab7a5164a60b753ba429ef7ba9ce36517d4439", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "LaTeXPdfs/sections/char2.tex", "max_issues_repo_name": "emcd123/Matroids", "max_issues_repo_head_hexsha": "f1ab7a5164a60b753ba429ef7ba9ce36517d4439", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "LaTeXPdfs/sections/char2.tex", "max_forks_repo_name": "emcd123/Matroids", "max_forks_repo_head_hexsha": "f1ab7a5164a60b753ba429ef7ba9ce36517d4439", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-11-21T18:03:07.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-21T18:03:07.000Z", "avg_line_length": 84.4736842105, "max_line_length": 563, "alphanum_fraction": 0.7190031153, "num_tokens": 479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.782662489091802, "lm_q2_score": 0.8244619242200082, "lm_q1q2_score": 0.6452754217714483}} {"text": "\\documentclass[a4paper]{article}\n\n\\def\\ntitle{Quadratic fields}\n\\def\\ndate{}\n\n\\input{header}\n\n\\renewcommand*{\\O}{\\mathcal{O}}\n\\DeclareMathOperator{\\n}{N}\n\\DeclareMathOperator{\\disc}{disc}\n\\DeclareMathOperator{\\Cl}{Cl}%ideal class group\n\\newcommand*{\\red}[1]{\\overline{#1}}%reduction mod\n\n\\begin{document}\n\n\\maketitle\n\nLet \\(d \\in \\Z\\) be square-free and \\(d \\neq 0, 1\\). Then\n\\[\n L = \\Q(\\sqrt d) = \\Q[x]/(x^2 - d)\n\\]\nis a degree \\(2\\) extension over \\(\\Q\\). It is called a \\emph{quadratic field}. If \\(d > 0\\) then there are two real embeddings, in which case we call \\(L\\) a real quadratic field. Otherwise \\(L\\) is an imaginary quadratic field. Note that in using this notation, we implicitly assume that there is a complex embedding \\(\\sigma: L \\to \\C\\).\n\n\\paragraph{Ring of integers}\n\nA particularly nice characterisation of algebraic intgers in a quadratic field is \\(\\alpha \\in \\O_L\\) if and only if \\(\\n_{L/\\Q}(\\alpha), \\tr_{L/\\Q}(\\alpha) \\in \\Z\\).\n\nSuppose \\(\\alpha = \\frac{u}{2} + \\frac{v}{2}\\sqrt d \\in \\O_L\\) where \\(u, v \\in \\Q\\). Then multiplication by \\(\\alpha\\) has with respect to the basis \\(\\{1, \\sqrt d\\}\\) matrix representation\n\\[\n \\frac{1}{2}\n \\begin{pmatrix}\n u & vd \\\\\n v & u\n \\end{pmatrix}\n\\]\nso\n\\begin{align*}\n \\n_{L/\\Q}(\\alpha) &= \\frac{1}{4}(u^2 - v^2d) \\in \\Z \\\\\n \\tr_{L/\\Q}(\\alpha) &= u \\in \\Z\n\\end{align*}\n\nso \\(v^2d \\in \\Z\\). Suppose \\(v = \\frac{r}{s}\\) is an expression in coprime integers. Then \\(d^2r^2 \\in s^2\\Z\\) so \\(s^2 \\divides d^2r^2\\). If \\(p\\) is a prime dividing \\(s\\) then \\(p^2 \\divides d^2\\). As \\(d\\) is square-free, \\(p \\divides d\\). Absurd. Thus \\(v \\in \\Z\\) and\n\\[\n \\Z[\\sqrt d] \\subseteq \\O_L \\subseteq \\frac{1}{2}\\Z[\\sqrt d].\n\\]\n\n\\begin{itemize}\n\\item If \\(d = 2, 3 \\pmod 4\\) then \\(u^2 = 0, 1 \\pmod 4, v^2 = 0, 1 \\pmod 4\\). As \\(u^2 = v^2d \\pmod 4\\), \\(u, v \\in 2\\Z\\) so \\(\\alpha \\in \\Z[\\sqrt d]\\). Thus \\(\\O_L = \\Z[\\sqrt d]\\).\n\\item If \\(d = 1 \\pmod 4\\) then \\(u^2 = v^2 \\pmod 4\\) so \\(u = v \\pmod 2\\). Thus\n \\[\n \\O_L \\subseteq \\left\\{ \\frac{u}{2} + \\frac{v}{2}\\sqrt d: u = v \\pmod 2 \\right\\} = \\Z \\oplus \\Z \\frac{1 + \\sqrt d}{2}.\n \\]\n Now check that \\(\\frac{1 + \\sqrt d}{2} \\in \\O_L\\) so we conclude that \\(\\O_L = \\Z[\\frac{1 + \\sqrt d}{2}]\\).\n\\end{itemize}\n\n\\paragraph{Discriminant}\n\nRecall that\n\\[\n D_L\n = \\det (\\sigma_i(\\alpha_j))^2\n = \\det (\\tr_{L/\\Q}(\\alpha_i\\alpha_j))\n = (-1)^{\\binom{n}{2}}\\n_{L/\\Q}(f'(\\alpha))\n\\]\nwhere \\(\\{\\alpha_i\\}\\) is an integral basis, \\(\\{\\sigma_i\\}\\) are the complex embeddings, \\(\\alpha\\) is a generator of \\(\\O_L\\) as a \\(\\Z\\)-algebra and \\(f\\) is the minimal polynomial whereof.\n\n\\begin{itemize}\n\\item If \\(d = 2, 3 \\pmod 4\\) then \\(\\alpha = \\sqrt d, f(x) = x^2 - d\\). Thus\n \\[\n D_L = -\\n_{L/\\Q}(2\\sqrt d) = 4d.\n \\]\n Alternatively, since \\(\\tr_{L/\\Q}(1) = 2, \\tr_{L/\\Q}(\\sqrt d) = 0\\), we can easily compute the matrix \\(\\tr_{L/\\Q}(\\alpha_i\\alpha_j)\\).\n\\item If \\(d = 1 \\pmod 4\\) then \\(\\alpha = \\frac{1 + \\sqrt d}{2}, f(x) = x^2 + x + \\frac{1 - d}{4}\\). Thus\n \\[\n D_L = -\\n_{L/\\Q}(\\sqrt d) = d.\n \\]\n\\end{itemize}\n\n\\paragraph{Factorisation of ideals}\n\nRecall that Dedekind's criterion says that subject to certain divisibility condition, given \\(L = \\Q(\\alpha)\\) and \\(\\alpha \\in \\O_L\\) with minimal polynomial \\(f(x)\\) and \\(p\\) prime, if\n\\[\n \\red f(t) = \\prod_{i = 1}^r \\red g_i(t)^{e_i} \\in \\F_p[x]\n\\]\nis a factorisation into irreducibles then\n\\[\n (p) = \\prod_{i = 1}^r \\mathfrak{p}_i^{e_i}\n\\]\nis a factorisation into prime ideals.\n\n\\begin{itemize}\n\\item If \\(p = 2\\),\n \\begin{itemize}\n \\item if \\(d = 2, 3 \\pmod 4\\) then let \\(\\alpha = \\sqrt d\\) so\n \\[\n \\red f(x) = x^2 - d = (x - d)^2 \\in \\F_2[x]\n \\]\n so \\((2) = \\mathfrak{p}^2\\), i.e.\\ ramifies.\n \\item if \\(d = 1 \\pmod 8\\) then let \\(\\alpha = \\frac{1 + \\sqrt d}{2}\\) so\n \\[\n \\red f(x) = x^2 + x + \\frac{1 - d}{4} = x^2 + x = x(x + 1) \\in \\F_2[x]\n \\]\n so \\((2) = \\mathfrak{p}\\mathfrak{q}\\), i.e.\\ splits completely.\n \\item if \\(d = 5 \\pmod 8\\) then \\(\\red f(x) \\in \\F_2[x]\\) is irreducible so \\(2\\) is inert.\n \\end{itemize}\n\\item If \\(p\\) is odd, let \\(\\alpha = \\sqrt d\\) and \\(f(x) = x^2 - d\\) so\n \\begin{itemize}\n \\item if \\(\\legendre{d}{p} = 0\\) then \\((p) = \\mathfrak{p}^2\\), i.e.\\ ramifies.\n \\item if \\(\\legendre{d}{p} = 1\\) then \\((p) = \\mathfrak{p}\\mathfrak{q}\\), i.e.\\ splits completely.\n \\item if \\(\\legendre{d}{p} = -1\\) then \\(p\\) is inert.\n \\end{itemize}\n\\end{itemize}\n\n\\paragraph{Lattice}\n\nRecall that the covolume of a lattice formed by an ideal of the ring of integers is the volume of the parallelepiped spanned by its \\(\\Z\\)-basis.\n\nGiven an \\emph{imaginary} quadratic field \\(L\\), claim that\n\\[\n A(I) = \\frac{1}{2}\\sqrt{|\\disc(I)|} = \\frac{\\n(I)}{2}\\sqrt{|D_L|}\n\\]\nfor \\(I \\subseteq \\O_L\\).\n\n\\begin{proof}\n Let \\(\\alpha_1 = x_1 + iy_1, \\alpha_2 = x_2 + iy_2\\) be an integral basis for \\(I\\). Then\n \\[\n A(I) = \\left| \\det\n \\begin{pmatrix}\n x_1 & x_2 \\\\\n y_1 & y_2\n \\end{pmatrix}\n \\right|.\n \\]\n Meanwhile\n \\[\n \\disc(I)\n = \\det\n \\begin{pmatrix}\n x_1 + iy_1 & x_2 + iy_2 \\\\\n x_1 - iy_1 & x_2 - iy_2\n \\end{pmatrix}\n ^2\n = (2i)^2 \\det\n \\begin{pmatrix}\n x_1 & x_2 \\\\\n y_1 & y_2\n \\end{pmatrix}\n ^2\n .\n \\]\n\\end{proof}\n\nBy Minkowski's theorem and multiplicativity of norm, we can deduce that for any number field \\(L\\), the ideal class group \\(\\Cl(\\O_L)\\) is finite and can be generated by the class of prime ideals \\(\\mathfrak{p}\\) with \\(\\n(\\mathfrak{p}) \\leq c_L\\) where \\(c_L = \\frac{2}{\\pi}\\sqrt{|D_L|}\\).\n\n\\begin{eg}\\leavevmode\n \\begin{enumerate}\n \\item \\(d = -7\\). As \\(d = 1 \\pmod 4\\), \\(D_L = -7\\). Thus\n \\[\n c_L = \\frac{2}{\\pi} \\sqrt 7 < \\frac{2}{3} \\sqrt 7 < 2\n \\]\n so \\(\\Cl(\\O_L)\\) is generated by ideals of norm \\(< 2\\). There are none except \\(\\O_L\\). Thus \\(\\Cl(\\O_L)\\) is trivial. Hence \\(\\O_L = \\Z[\\frac{1 + \\sqrt{-7}}{2}]\\) is a UFD.\n \\item \\(d = -5\\). \\(D_L = -20\\) so\n \\[\n c_L = \\frac{2}{\\pi}\\sqrt{20} = \\frac{4}{\\pi}\\sqrt 5 < \\frac{4}{3} \\sqrt 5 < 3\n \\]\n so \\(\\Cl(\\O_L)\\) is generated by prime ideals \\(\\mathfrak p \\subseteq \\O_L\\) of norm \\(\\n(\\mathfrak p) = 2\\). We know by Dedekind's criterion that \\(2\\O_L = \\mathfrak p^2\\). Thus \\(\\Cl(\\O_L)\\) is generated by \\([\\mathfrak p]\\) and \\([\\mathfrak p]^2 = [2\\O_L] = [\\O_L]\\) is the trivial class. Hence there are two possibilities:\n \\begin{enumerate}\n \\item if \\(\\mathfrak p\\) is principal then \\(\\Cl(\\O_L)\\) is trivial.\n \\item if \\(\\mathfrak p\\) is not principal then \\(\\Cl(\\O_L) \\cong \\Z/2\\Z\\).\n \\end{enumerate}\n But we already knew that \\(\\O_L\\) is not a UFD so \\(\\Cl(\\O_L)\\) is not trivial so must have\n \\[\n \\Cl(\\O_L) \\cong \\Z/2\\Z.\n \\]\n \\end{enumerate}\n\\end{eg}\n\nFor real quadratic fields \\(L = \\Q(\\sqrt d)\\), it is instructive as an exercise to derive the baby Minkowski constant, which should be \\(c_L = \\frac{1}{2}\\sqrt{|D_L|}\\).\n\n\\begin{eg}\n \\(d = 10\\). Then \\(c_L = \\frac{1}{2} \\sqrt{4 \\cdot 10} < 4\\). By Dedekind's criterion,\n \\begin{align*}\n (2) &= \\mathfrak p_2^2 \\\\\n (3) &= \\mathfrak p_3 \\mathfrak p_3'\n \\end{align*}\n What we can do at this stage is to compute the norm of some elements. For example \\(\\n(2 + \\sqrt{10}) = 6\\) so \\((2 + \\sqrt{10}) = \\mathfrak p_2 \\mathfrak p_3'\\) or \\(\\mathfrak p_2 \\mathfrak p_3'\\). In either case, \\([\\mathfrak p_2]\\) generates \\(\\Cl(\\O_L)\\). If \\(\\mathfrak p_2\\) is principal then there exists \\(a, b \\in \\Z\\) such that\n \\[\n a^2 - 10b^2 = \\pm 2.\n \\]\n Reduce modulo \\(5\\), \\(\\pm 2\\) is not a quadratic residue so impossible. Thus \\(\\Cl(\\O_L) \\cong \\Z/2\\Z\\).\n\\end{eg}\n\n\\begin{ex}\n Find the class group of ring of integers of \\(\\Q(\\sqrt{-17})\\).\n\\end{ex}\n\n\\paragraph{Dirichlet's unit theorem}\n\nDirichlet's unit theorem states that there is an isomorphism\n\\[\n \\O_L^\\times \\cong \\mu_L \\times \\Z^{r + s - 1}\n\\]\nwhere \\(\\mu_L\\) is the group of roots of unity in \\(\\O_L^\\times\\).\n\nThus \\(\\O_L^\\times\\) is finite if and only if\n\\begin{enumerate}\n\\item \\(r = 1, s = 0\\), so \\(L = \\Q\\), or\n\\item \\(r = 0, s = 1\\), so \\(L = \\Q(\\sqrt d)\\) for some \\(d \\in \\Z\\) negative square-free.\n\\end{enumerate}\n\nFor real quadratic fields \\(L = \\Q(\\sqrt d)\\), let \\(\\sigma: L \\to \\R\\) be the real embedding such that \\(\\sigma(\\sqrt d) > 0\\). As \\(\\sigma(\\mu_L) \\subseteq \\R^\\times\\), must have \\(\\mu_L = \\{\\pm 1\\}\\). Consider the homomorphism\n\\begin{align*}\n \\ell': \\O_L^\\times &\\to \\R \\\\\n \\alpha &\\mapsto \\log |\\sigma(\\alpha)|\n\\end{align*}\nAs \\(\\ell'(\\O_L^\\times) \\subseteq \\R\\) is a lattice, there is a unique element \\(\\alpha \\in \\O_L^\\times\\) such that \\(\\sigma(\\alpha) > 0\\), \\(\\ell'(\\alpha)\\) generates the lattice. Then\n\\[\n \\O_L^\\times = \\{\\pm \\alpha^n: n \\in \\Z\\}.\n\\]\nThis \\(\\alpha\\) is called the \\emph{fundamental unit}. It has the property that \\(\\log |\\sigma(\\alpha)|\\) is minimal, i.e.\\ \\(\\sigma(\\alpha) > 1\\) is minimal. This gives us a way to find fundamental units.\n\n\\begin{lemma}\n Suppose \\(d = 2, 3 \\pmod 4\\), \\(v \\in \\O_L^\\times\\) and \\(v > 1\\). Then \\(v = a + b \\sqrt d\\) where \\(a \\geq b \\geq 1\\).\n\\end{lemma}\n\n\\begin{proof}\n Let \\(v' = a - b \\sqrt d\\). Then\n \\[\n vv' = a^2 - db^2 = \\pm 1.\n \\]\n As \\(v > 1\\), \\(|v'| < 1\\) so\n \\begin{align*}\n 2a &= v + v' > 0 \\\\\n 2b &= v - v' > 0\n \\end{align*}\n Also\n \\[\n \\left( \\frac{a}{b} \\right)^2 = d \\pm \\frac{1}{b^2} > 1.\n \\]\n\\end{proof}\n\nThere is an entirely analogous result for \\(d = 1 \\pmod 4\\) which is left as an exercise.\n\nNow suppose \\(d = 2, 3 \\pmod 4\\). Suppose \\(u = a + b \\sqrt d \\in \\O_L^\\times\\) is the fundamental unit. Let \\(u^k = a_k + b_k \\sqrt d\\). Then\n\\begin{align*}\n u^{k + 1}\n &= u \\cdot u^k \\\\\n &= (a_1 + b_1 \\sqrt d)(a_k + b_k \\sqrt d) \\\\\n &= (a_1a_k + d b_1b_k) + (b_1a_k + a_1b_k) \\sqrt d\n\\end{align*}\nso\n\\[\n b_{k + 1} = b_1a_k + a_1b_k \\geq 2b_k > b_k\n\\]\nso \\((b_k)_{k \\in \\N}\\) is strictly increasing. We can therefore characterise \\(u\\) as follow: let \\(b \\in \\N\\) be the least positive integer such that \\(db^2 + 1\\) or \\(db^2 - 1\\) is of the form \\(a^2\\) for some \\(a \\in \\N\\). Then \\(u = a + b \\sqrt d\\) is the fundamental unit.\n\nIf instead \\(d = 1 \\pmod 4\\), we get\n\\[\n b_{k + 1} = \\frac{1}{2}(b_1a_k + a_1b_k) \\geq b_k\n\\]\nwith equality if and only if \\(a_1 = b_1 = 1, a_k = b_k\\). In this case\n\\[\n \\n(u) = \\left| \\frac{1 - d}{4} \\right| = 1\n\\]\nso \\(d = 5\\). In this case \\(u = \\frac{1}{2} (1 + \\sqrt 5)\\) is the fundamental unit.\n\nIf instead \\(d > 5\\), we proceed as before and characterise \\(u\\) as follow: let \\(b \\in \\N\\) be the least positive integer such that \\(db^2 + 4\\) or \\(db^2 - 4\\) is of the form \\(a^2\\) for some \\(a \\in \\N\\). Then \\(u = \\frac{1}{2} (a + b \\sqrt d)\\) is the fundamental unit.\n\n\\begin{eg}\\leavevmode\n \\begin{enumerate}\n \\item \\(d = 2\\). Then \\(b = 1\\) works since \\(2 - 1 = 1^2\\) so \\(1 + \\sqrt 2\\) is a fundamental unit.\n \\item \\(d = 7\\).\n \\begin{align*}\n & b = 1: 7 \\pm 1 \\text{ not a square} \\\\\n & b = 2: 4 \\cdot 7 \\pm 1 \\text{ not a square} \\\\\n & b = 3: 9 \\cdot 7 + 1 = 8^2\n \\end{align*}\n so \\(8 + 3 \\sqrt 7\\) is a fundamental unit.\n \\end{enumerate}\n\\end{eg}\n\n\\end{document}\n", "meta": {"hexsha": "08757b07f75c514f34e23e0a649e07e7fb5e4f89", "size": 11049, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "blurb/quadratic_fields.tex", "max_stars_repo_name": "geniusKuang/tripos", "max_stars_repo_head_hexsha": "127e9fccea5732677ef237213d73a98fdb8d0ca0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 27, "max_stars_repo_stars_event_min_datetime": "2018-01-15T05:02:27.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-10T15:48:31.000Z", "max_issues_repo_path": "blurb/quadratic_fields.tex", "max_issues_repo_name": "b-mehta/tripos", "max_issues_repo_head_hexsha": "8d3037ede28fed3a3cdb82a88dd3a005bf94b310", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-10-11T20:43:21.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-14T21:29:15.000Z", "max_forks_repo_path": "blurb/quadratic_fields.tex", "max_forks_repo_name": "b-mehta/tripos", "max_forks_repo_head_hexsha": "8d3037ede28fed3a3cdb82a88dd3a005bf94b310", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2017-11-08T16:16:20.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-25T17:20:19.000Z", "avg_line_length": 39.0424028269, "max_line_length": 340, "alphanum_fraction": 0.5677436872, "num_tokens": 4481, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.824461932846258, "lm_q1q2_score": 0.6452754201637827}} {"text": "\\section*{Summary of Section 2}\n\\begin{enumerate}\n \\item The group $\\bbz/100$ of 2-digit integers sits in a short exact sequence/group extension\n \\begin{equation*}\n \\begin{tikzcd}\n 0 \\ar[r] & \\bbz/10 \\ar[r,\"i\"] & \\bbz/100 \\ar[r,\"p\"] & \\bbz/10 \\ar[r] & 0 \\\\\n & \\mbox{tens} \\ar[u, equal] & \\mbox{2-digit numbers} \\ar[u,equal] & \\mbox{units} \\ar[u, equal] &\n \\end{tikzcd}\n \\end{equation*}\n\n \\item Every group extension arises in this manner.\n For an extension of abelian groups,\n \\begin{equation*}\n \\begin{tikzcd}\n 0 \\ar[r] & H \\ar[r,\"i\"] & G \\ar[r, \"p\"] & K \\ar[r] & 0.\n \\end{tikzcd}\n \\end{equation*}\n every element of $G$ can be uniquely written as $\\units{a}\\tens{b}$ for some $a \\in H$ and $b \\in K$.\n\\end{enumerate}\n\n\n\\vspace{2cm}\n\\hrule\n\\vspace{2cm}\n\n% Show that for every element $g \\in G$, there exist unique elements $a \\in H$ and $ b \\in K $ such that $g = \\units{a}\\tens{b}$.\n% \\hint{Look at $p(g)$ and $g - [p(g)]\\units{0}$.}\n\\begin{proof}[Solution to Q.\\ref{q:existenceOfDigits}]\n We will only prove existence as uniqueness follows by Q.\\ref{q:uniquenessOfDigits}.\n\n Consider the element $g \\in G$. Let $b = p(g) \\in K$ and let \\begin{equation*}\n h = g - \\tens{0}\\units{b}.\n \\end{equation*}\n Applying $p$ to both sides we get\n \\begin{align*}\n p(h)\n &= p(g - \\tens{0}\\units{b}) \\\\\n &= p(g) - p(\\tens{0}\\units{b}) && \\mbox{ as $p$ is a group homomorphism}\\\\\n &= b -b \\\\\n &= 0\n \\end{align*}\n Hence, $h \\in \\ker p$, which implies that $h \\in \\im i$.\n Hence, $ h = i(a) = \\tens{a}\\units{0}$ for some $a \\in H$, giving us\n \\begin{equation*}\n g = h + \\tens{0}\\units{b} = \\tens{a}\\units{0} + \\tens{0}\\units{b} = \\tens{a}\\units{b}.\n \\end{equation*}\n\\end{proof}\n", "meta": {"hexsha": "d5d4fee81a539036ddf962b064414100d0e37006", "size": 1737, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "02Summary.tex", "max_stars_repo_name": "apurvnakade/mc2019-group-cohomology", "max_stars_repo_head_hexsha": "14a7f5f0e2ae64f3ceaa602b50fa80269e3800a5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "02Summary.tex", "max_issues_repo_name": "apurvnakade/mc2019-group-cohomology", "max_issues_repo_head_hexsha": "14a7f5f0e2ae64f3ceaa602b50fa80269e3800a5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "02Summary.tex", "max_forks_repo_name": "apurvnakade/mc2019-group-cohomology", "max_forks_repo_head_hexsha": "14a7f5f0e2ae64f3ceaa602b50fa80269e3800a5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.1875, "max_line_length": 129, "alphanum_fraction": 0.5877950489, "num_tokens": 694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619306896956, "lm_q2_score": 0.7826624789529376, "lm_q1q2_score": 0.6452754184759222}} {"text": "\\section{Part 2.a Numerical Integrator}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_a.py}\n\nThe result of this function is given by:\n\n\\lstinputlisting{2a.txt}\n\n\\section{Part 2.b Interpolation}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_b.py}\n\nMy script produces the following result, see Fig. \\ref{fig:interp}. This shows the differences between the linear\ninterpolation and cubic spline. Both are essentially the same for this data, except between 0.1 and 1. In comparison to\nthe actual values, the interpolated values underestimate for all values between $x$ = 1 and $x$ = 5.\n\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=0.9\\linewidth]{./plots/interpolation.png}\n \\caption{The interpolation function in action.}\n \\label{fig:interp}\n\\end{figure}\n\n\\section{Part 2.c Numerical Derivative}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_c.py}\n\nThe result of this function is given by:\n\n\\lstinputlisting{2c.txt}\n\n\\section{Part 2.d Random Sampling}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_d.py}\n\nMy script produces the following result, see Fig. \\ref{fig:rand_sample}\n\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=0.9\\linewidth]{./plots/random_sample.png}\n \\caption{The results of the random sample with rejection sampling.}\n \\label{fig:rand_sample}\n\\end{figure}\n\nThe positions in (r, $\\phi$, $\\theta$) are given by:\n\n\\lstinputlisting{2d.txt}\n\n\\section{Part 2.e 1000 Haloes}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_e.py}\n\nMy script produces the following result, see Fig. \\ref{fig:1000_haloes}. As can be seen, the generated galaxies mostly match\nthe $N(x) = n(x)4\\pi x^2$ distribution.\n\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=0.9\\linewidth]{./plots/1000_haloes.png}\n \\caption{The results of 1000 haloes each with 100 satellites. }\n \\label{fig:1000_haloes}\n\\end{figure}\n\n\\section{Part 2.f Root Finding}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_f.py}\n\nThe result of this function is given by:\n\n\\lstinputlisting{2f.txt}\n\n\\section{Part 2.g Percentiles and Poisson}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_g.py}\n\nThe median, 16th, and 84th percentiles for the radial bin with the largest number of falaxies is given in:\n\n\\lstinputlisting{2g.txt}\n\nMy script produces the following result, see Fig. \\ref{fig:hist_poisson}. As can be seen, histogram fairly closely follows\nthe Poisson distribution with $\\lambda$ equal to the mean number of galaxies in this radial bin.\n\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=0.9\\linewidth]{./plots/hist_poisson.png}\n \\caption{The results of 1000 haloes each with 100 satellites. }\n \\label{fig:hist_poisson}\n\\end{figure}\n\n\\section{Part 2.h 3D interpolator}\n\nThe file of the functions used for this exercise is:\n\n\\lstinputlisting{two_h.py}\n\nWhile no output was requested, I did test the 3D interpolator, with the a,b,and c values generated earlier.\nThe error is quite large for these, which I believe is a result of the linear interpolation used. Cubic splines took too\nlong to run, and were expensive to create.\n\n\\lstinputlisting{2h.txt}\n\n\n\n", "meta": {"hexsha": "5321cabe7259ddb723bb049be1a7f75b5f0c7b4a", "size": 3269, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "part_2.tex", "max_stars_repo_name": "jacobbieker/NUR", "max_stars_repo_head_hexsha": "bfcf938efca9a4506ba36a521cbce3030b56b06c", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "part_2.tex", "max_issues_repo_name": "jacobbieker/NUR", "max_issues_repo_head_hexsha": "bfcf938efca9a4506ba36a521cbce3030b56b06c", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "part_2.tex", "max_forks_repo_name": "jacobbieker/NUR", "max_forks_repo_head_hexsha": "bfcf938efca9a4506ba36a521cbce3030b56b06c", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-04-08T21:14:04.000Z", "max_forks_repo_forks_event_max_datetime": "2019-04-08T21:14:04.000Z", "avg_line_length": 27.9401709402, "max_line_length": 124, "alphanum_fraction": 0.7629244417, "num_tokens": 897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8244619285331332, "lm_q1q2_score": 0.6452754167880616}} {"text": "\n% This LaTeX was auto-generated from MATLAB code.\n% To make changes, update the MATLAB code and republish this document.\n\n\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{color}\n\n\\sloppy\n\\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n\\begin{document}\n\n \n \n\\section*{HOMEWORK 1 - FRANCISCO CASTILLO}\n\n\n\\subsection*{Contents}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item Problem 2\n\\end{itemize}\n\n\n\\subsection*{Problem 2}\n\n\\begin{verbatim}\nformat long;clear all;close all;clc\nx_0=33.3;\nf = @(x) (1+2*x^2*cos(x))/(x^(2.4));\nfp = @(x) (2*x^2*(2*cos(x)-x*sin(x))-2.4*(1+2*x^2*cos(x)))/(x^3.4);\nfor k = -3:1:25\n H(k+4) = 2^(-k);\nend\nfor k = 1:length(H)\n h = H(k);\n df1(k) = (f(x_0+h)-f(x_0))/h;\n df2(k) = (f(x_0+h)-f(x_0-h))/(2*h);\n df6(k) = (45*(f(x_0+h)-f(x_0-h))-9*(f(x_0+2*h)-f(x_0-2*h))...\n +(f(x_0+3*h)-f(x_0-3*h)))/(60*h);\nend\nlinewidth=2;\ndarkgreen=[0 0.6 0];\nfigure('units','normalized','outerposition',[0 0 1 1])\nloglog(H,abs(df1-fp(x_0)),'*',H,abs(df2-fp(x_0)),'*')\nhold on\nloglog(H,abs(df6-fp(x_0)),'*','Color',darkgreen)\nloglog(H,H/2,'b--',H,H.^(2)/6,'r--','linewidth',linewidth)\nloglog(H,H.^(6)/140,'--','Color',darkgreen,'linewidth',linewidth)\nloglog(H,eps./H,'k:','linewidth',linewidth)\nset(gca,'fontsize',14)\naxis([1e-8 1e1 1e-20 1e5])\ngrid on\nxlabel('$h$ (log scale)','fontsize',20,'interpreter','latex')\nylabel('Error (log scale)','fontsize',20,'interpreter','latex')\nsaveas(gcf,'IMAGES/problem2_1','epsc')\nsaveas(gcf,'IMAGES/problem2_1','fig')\n\\end{verbatim}\n\n\\includegraphics [width=4in]{HW1_01.eps}\n\n\n\n\\end{document}\n \n", "meta": {"hexsha": "d7bbd9816a72ace8c40bc8c901459fdd114a95ce", "size": 1600, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Year_1/CFD/Homework 1/html/HW1.tex", "max_stars_repo_name": "fjcasti1/Courses", "max_stars_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Year_1/CFD/Homework 1/html/HW1.tex", "max_issues_repo_name": "fjcasti1/Courses", "max_issues_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Year_1/CFD/Homework 1/html/HW1.tex", "max_forks_repo_name": "fjcasti1/Courses", "max_forks_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.1884057971, "max_line_length": 70, "alphanum_fraction": 0.62875, "num_tokens": 627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8244619242200082, "lm_q1q2_score": 0.6452754134123405}} {"text": "\\section{Mean-field approximation for Ising Model}\nThis is the main equation\n\\begin{align}\n h_{\\mathrm{eff}}\n &=\n h\n +\n 4J T gh \\beta h_{\\mathrm{eff}}\n\\end{align}\nThe critical temperature\n\\begin{align}\n k_B T_c &= 4J\n\\end{align}\nThe susceptibility\n\\begin{align}\n \\chi\n &=\n \\frac{1}{\\beta}\n \\left.\\frac{\\partial m}{\\partial h}\\right|_{T=T_c, h=0}\n \\sim\n \\left| T - T_c \\right|^{\\overbrace{-1}^{\\gamma}}\n\\end{align}\nand the magnetisation is\n\\begin{align}\n m &\\sim\n \\left| T - T_c \\right|^{\\overbrace{1/2}^{\\beta}}\n\\end{align}\nI can then write\n\\begin{align}\n \\underbrace{h_{\\mathrm{eff}}}_{h + 4Jm}\n &=\n h\n +\n 4 J T gh \\beta h_{\\mathrm{eff}}\n\\end{align}\nwhich implies\n\\begin{align}\n m &=\n Tgh \\beta \\left( 4J m + h \\right)\n\\end{align}\nso\n\\begin{align}\n 4J m + h\n &=\n \\frac{1}{\\beta_c}\n \\arctanh m\\\\\n &\\approx\n \\frac{1}{\\beta_c}\n \\left( \n m\n +\n \\frac{m^3}{3}\n + \\cdots\n \\right)\n\\end{align}\nso then\n\\begin{align}\n h \\sim m^{\\overbrace{3}^{\\delta}}\n\\end{align}\nwhich is not obvious at all.\n\nThe relation between $m$ and $h$ depends on that equation,\nbut depends only on the fact that this s an odd function and that it is smooth.\nThis is very robust because even though it's only approximate.\n\nThere are perfect analogies of mean field all over physics.\nIn fact you've seen those before.\nSo I'm going to erase this and I'm going to make some drawings in two different\ncolours.\nOn the left we have the mean field approximation,\nand on the right I'm going to tell another thing that you already know,\nwhich are fluids that obey the van der Waals equation of state.\n\n\\subsection{Ising}\nI'm going to draw the external magnetic field vs the temperature.\nThis is $T_c$.\nWhat happens when I'm above $T_c$ and I have a negative $h$?\nIt favours spin down so magnetisation is going to be down.\nIf $h>0$,\nthen up is favoured.\nIf the magnetic field is zero,\nthen it's zero.\nSo nothing interesting happens here when I cross $h=0$ for $T>T_c$.\n\nNow below $T_c$,\nthe magnetisation doesn't go to zero.\nWhen $h$ is slightly negative,\nthey are all spin down.\n\nThe discontinuity disappears in the critical region.\n\nThe analogy with the van der Waals fluid is\n\\begin{align}\n h &\\sim P\\\\\n m &\\sim \\frac{1}{V}\n\\end{align}\n\n[lots of pictures]\n\nIn a lot of systems,\nthere is an order parameter $\\phi$ such that its average value is\n\\begin{align}\n \\begin{cases}\n \\langle \\phi \\rangle\n =\n 0 & T > T_c\\\\\n \\langle \\phi \\rangle\n \\ne\n 0 & T > T_c\n \\end{cases}\n\\end{align}\nIf there is such a situation,\nthen there is spontaneous symmetry breaking.\nThe order parameter doesn't have the symmetry of the Hamiltonian,\nbecause when it flips it flips sign,\nbut the Hamiltonian doesn't change.\nThe thermal average is not zero,\neven though it should be naively if you look at the Hamiltonian.\n\nA lot of cases,\nwhen the expectation picks up a non-zero expectation value like this,\nfrequently this means spontaneous symmetry breaking.\n\nIt happens even when equal to zero.\n\nIt's worth while going close to the critical point and looking at the\nconfiguration that contributes the most.\nThe ones with smallest energy are ordered and there are few of them.\nThey are not typical.\nThe more typical ones are the high energy ones.\n\nI'm going to try to draw some configurations that are critical.\n\n[picture]\n\nWhen above the critical temperature,\nthe typical size of clusters $\\xi\\approx a$,\nthe lattice size.\nWhen at $T=T_c$,\nyou're going to get clusters of all sizes,\nfrom the size of a lattice spacing to large clusters that go from one corner of\nthe system to another,\nand everything in between.\n\nSo we talk about the critical exponents in the mean field model.\nI want it to be the centre piece of everything we compute here.\nIt turns out that close to the critical point,\nthe heat capacity diverges.\n\\begin{align}\n C &\\sim \\left| T - T_c \\right|^{-\\alpha}\n\\end{align}\nLet's compute $\\alpha$.\n\nThe magnetisation when you approach the critical temperature is like\n\\begin{align}\n m ~ \\left| T - T_c \\right|^{\\beta}\n\\end{align}\nand the susceptibility\n\\begin{align}\n \\chi &=\n \\left| \\frac{\\partial m}{\\partial h} \\right|_{h=0}\n \\sim\n \\left| T - T_c \\right|^{-\\gamma}\n\\end{align}\nSuppose we calculate the correlation\nabout how close or far spins influence each other\n\\begin{align}\n \\left\\langle\n \\sigma_0\n \\sigma_{0 + R}\n \\right\\rangle\n &\\sim\n e^{-R/\\xi}\n\\end{align}\nand this parameter $\\xi$ is the correlation length,\nwhich is measurer f how far you need to go to not get correlation.\nThe correlation length diverges too at the critical temperature.\n\\begin{align}\n \\xi &\\sim\n \\left| T - T_c\\right|^{-\\nu}\n\\end{align}\nI didn't show you but $\\nu=\\frac{1}{2}$ for mean field Ising.\n\nI can take a Fourier transform.\n\nThis is the only integral you need to do in your head\n\\begin{align}\n \\sum_{\\vec{R}}\n \\left\\langle\n \\sigma_0\n \\sigma_{0 + R}\n \\right\\rangle\n e^{i\\vec{k}\\cdot\\vec{R}}\n \\sim\n \\frac{1}{k^{2 - \\eta}}\n\\end{align}\nwhere $\\eta$ is called the \\emph{anomalous dimension}.\n\n\\section{Universality classes}\nNumerical experiments and exact results.\nThe Ising model in 2D was solved exactly on paper.\nFirst surprise,\nit's not what mean field theory says.\nIt maybe shouldn't surprise you much,\nbecause mean field was a bit hand waving anyway.\n\nBut the surprising thing was a different one.\nIt wasa big surprise back them.\n\nThere was a theory called Landau mean field theory,\na lot of hand wavin,\nbut it's Landau hand waving,\nand he talks to God directly.\nIt's like mean field theory on steroids.\nTremendous theory,\nno argument whatsoever.\n\nThen you find something interesting with the exact solution.\nThere are what we call Universality classes.\n\nPeople looked at the 2D Ising model.\nPeople made this relation between this and the liquid phase transition.\nYou can measure the critical exponents of water evapoerating,\nand you find it's 3D Ising model.\nOne is a stupid magnet model,\none is a real system.\nThey agree within error bars.\nBut then you look at other systems,\nand there are sets of critical exponents that appear frequently.\n\nThe critical exponents of the 3D Ising model are common,\nprobably the most common one.\nAnd then there are other common ones.\nThe systems are completely diffrent.\nIn the notes,\nI give some critical exponents calculated from numerics,\nconformal bootstrap.\nLet me just write these here.\n\\begin{table}\n \\centering\n \\begin{tabular}{cccc}\n & 3D Ising & CO$_2$ & Ni Magnet\\\\\n $\\alpha$ & 0.11008(1) & 0.10 & 0.0(?)\\\\\n \\vdots\\\\\n $\\delta$ & 4.28984(1) & 4.20 & 4.22\n \\end{tabular}\n \\caption{Critical Exponents Theory vs Experiment}%\n \\label{tab:criticalexponents}\n\\end{table}\nThere's an uncertainty for Ni magnet that I don't know.\nAll systems in nature seem to fall into \n\nHere are some common universality classes\n\\begin{itemize}\n \\item 3D Ising\n \\item 2D Ising\n \\item Mean field\n \\item Kosterlitz-Thouless\n \\item 3-Potts model\n\\end{itemize}\nI knew Thouless.\nA lot of things come from experience.\nPeople get data and they see a pattern.\nRoughly the pattern people see is the following.\n\nThe first thing you have to discern is the dimensionality of the state.\nThe other one is more subtle,\nit's the ``pattern of spontaneous symmetry breaking''.\nFor example,\nconsider the Ising model.\nThe Hamiltonian has a symmetry $\\mathbb{Z}_2$\nwhich is basically spin flip.\nThis symmetry has 2 elements,\ndo nothing and flip all the spins.\nThe point is that at low temperatures,\nthe magnetisation,\nthe order parameter,\nis not zero.\nIt's pointing either up or down.\nWhen I did a spin flip,\nI get a different state.\nWhat's the symmetry of the sate below the critical temperature?\nThat's what I mean by a matter of symmetry breaking,\nhow it's broken to the final state\n$\\mathbb{Z}_2 \\to \\textrm{nothing}$.\n\nAnother model is the Heisenberg model,\nevery point on a lattice has a spin-$\\frac{1}{2}$ on it.\nYou can make it ferromagnetic or antiferromagnetic or whatever.\nThe point is that it's not just up or down,\nit could be many other states you can rotate by.\n\nThe dot product between two spins stays the same under rotation.\nPeople call this SO(3),\nwhich are 3 by 3 orthogonal matrices.\nThe special orthogonal matrices.\nAlso known as rotations.\nYou don't know this,\nbut the ferromagnetic model,\nthe ground state is all spins pointing in the same direction,\nbecause that maximizes the dot product and minimized the energy just like thei\nIsing model.\n\nSuppose this direction is picked up,\nthere's still some symmetry left.\nIf I do rotations around this axis,\nthe magnetisation order parameter doesn't change.\n$SO(3) \\to O(2)$.\n\nThere is a particular type of liquid crystal called nematic liquid crystal.\nHow can something be a liquid and a crystal at the same time?\nIt's made of long molecules,\nand the head and tails are different.\nIt turns out below a certain critical temperature,\nthey organize themselves like this.\nTheir position is not organized on a crystal,\nlike a liquid,\nnot organized in a nice pattern.\nTheir direction is not organized either,\nthere are as much head up and head down.\nThe only thing is that they are all aligned that way.\nThe Hamiltonian has a symmetry that they can rotate any direction they like.\nHere there's nothing to point,\nthe order parameter is a tensor.\nBut it can still flip.\n\\begin{align}\n SO(3) \\to O(2)\\times \\mathbb{Z}_3\n\\end{align}\nRotation goes to 2D rotation and flips.\n\nMeanwhile there are other liquid crystals\nwhere the centres of molecule positions are fixed by the orientation are not.\n\nRemember the Bose gas,\nwhich has a phase transition,\na condensate.\nI claimed that it's superfluid.\nI'm not even going to attempt to explain.\nThe pattern of symmetry breaking in that case is $U(1)$.\nThat's exactly the same group as $O(2)$ by the way.\nThis is broken to almost nothing.\nThis phase has to do with changing the wave function by a phase\n$\\psi \\to e^{i\\beta} \\psi$.\nSuperfluid helium and BEC follow this pattern,\nand they have their own universality class.\n\nYou probably heard something that if you look at the physics of elementary\nparticles,\nthere is a symmetry\n\\begin{align}\n SU_c(3)\n \\times\n SU_w(2)\n \\times\n U_Y(1)\n \\to\n SU_c(3)\n \\times\n U_{EM}(1)\n\\end{align}\nIf you have 2 systems with the same dimension,\nthey follow the same universality class.\nThat was an experimental observation.\n\nExamples.\n\nConsider the Ising model in 2D\n\\begin{align}\n H &=\n -J \\sum \\sigma_i \\sigma_j\n\\end{align}\nconsider my made up Paulo model\n\\begin{align}\n H^P &=\n -J \\sum_{\\textrm{plaquette}}\n \\sigma_{i} \\sigma_{j} \\sigma_{k} \\sigma_{l}\n\\end{align}\nThey have the same dimensionality and the same pattern of spontaneous symmetry\nbreaking so they should have the same critical exponents.\nSame for the sum\n\\begin{align}\n H &=\n -\\tilde{J} \\sum \\sigma_i \\sigma_j\n -J \\sum_{\\textrm{plaquette}}\n \\sigma_{i} \\sigma_{j} \\sigma_{k} \\sigma_{l}\n\\end{align}\n", "meta": {"hexsha": "fb110ab71fd9f640a0bb09cceb7418945dad6661", "size": 11062, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "phys612/lecture33.tex", "max_stars_repo_name": "ehua7365/umdphysnotes", "max_stars_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-12-11T12:53:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-11T12:53:46.000Z", "max_issues_repo_path": "phys612/lecture33.tex", "max_issues_repo_name": "ehua7365/umdphysnotes", "max_issues_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "phys612/lecture33.tex", "max_forks_repo_name": "ehua7365/umdphysnotes", "max_forks_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.5839793282, "max_line_length": 79, "alphanum_fraction": 0.7226541313, "num_tokens": 3021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619263765707, "lm_q2_score": 0.7826624738835052, "lm_q1q2_score": 0.6452754109206471}} {"text": "% IRAM.tex\n\\documentclass[12pt]{article}\n\\usepackage[T1]{fontenc}\n\\usepackage{lmodern}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath}\n\n\n\\title{Post-Processing Arnoldi's Method}\n\\author{Jeremy L. Conlin}\n\n\\newcommand{\\A}{\\mathbf{A}}\n\\newcommand{\\qp}[1]{q'_{#1}}\n\\newcommand{\\qh}[1]{\\hat{q}_{#1}}\n\\newcommand{\\qt}[1]{\\tilde{q}_{#1}}\n\\newcommand{\\qc}[1]{\\check{q}_{#1}}\n\\newcommand{\\Aq}[1]{\\A q_{#1}}\n\\newcommand{\\Ap}[1]{\\A \\qp{#1}}\n\\newcommand{\\Ah}[1]{\\A \\qh{#1}}\n\\newcommand{\\At}[1]{\\A \\qt{#1}}\n\\newcommand{\\Ac}[1]{\\A \\qc{#1}}\n\n\\begin{document}\n\\maketitle\n\nThe purpose of this document is to explore the possibility of performing an Arnoldi factorization by post-processing a set of Krylov vectors.\n\n\\section{The beginning}\nSuppose we have the matrix\n\\begin{equation*}\n \\A \\equiv \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\\\\n 0 & 2 & 0 & 0 & 0 \\\\\n 0 & 1 & 3 & 0 & 0 \\\\\n 0 & 0 & 0 & 4 & 0 \\\\\n 0 & 0 & 0 & 0 & 5 \\end{pmatrix}.\n\\end{equation*}\nThis problem is not specific to this matrix, but this matrix is easily studied.\n\nA Krylov subspace of dimension $k$ is defined as \n\\begin{equation}\n \\mathcal{K}_k \\equiv \\left\\{q, \\A q, \\A^2, \\ldots, A^{k-1}q\\right\\} = \\left\\{q_0, q_1, \\ldots, q_k\\right\\}.\n \\label{eq:KrylovSubspace}\n\\end{equation}\nThe simplest way to construct a basis for a Krylov subspace is to repeatedly apply a linear operator (in our case the matrix $\\A$) to a vector $q$. This is what is done with the power method which is used to calculate the dominant eigenvalue of the linear operator. \n\n\\section{Arnoldi's Method}\nArnoldi's method generates the same Krylov subspace found in Equation \\ref{eq:KrylovSubspace} but orthonormalizes the basis vectors. The question remains if we can begin with Krylov subspace and orthogonalize and normalize the vectors after the fact to create the Arnoldi factorization\n\\begin{equation}\n \\A V = V H.\n \\label{eq:ArnoldiFactorization}\n\\end{equation}\n\nTo begin, let's rewrite Equation \\ref{eq:KrylovSubspace} as\n\\begin{equation}\n \\mathcal{K}'_k \\equiv \\left\\{q, \\A q, \\A^2, \\ldots, A^{k-1}q\\right\\} = \\left\\{\\qp{0}, \\qp{1}, \\ldots, \\qp{k}\\right\\}.\n\\end{equation}\nThis is merely to keep our notation straight. Another way to write this is\n\\begin{subequations}\\begin{align}\n \\qp{0} &\\equiv q_0 \\\\\n \\qp{1} &= \\A \\qp{0} \\\\\n \\qp{2} &= \\A \\qp{1} \\\\\n &\\ldots \\\\\n \\qp{k} &= \\A \\qp{k-1}.\n \\label{}\n\\end{align}\\end{subequations}\n\nWe have already applied our linear operator $k$ times and have our $k$-dimensional Krylov subspace. Can we then orthogonalize it and keep it mathematically equivalent to creating an Arnoldi factorization in the traditional fashion.\n\nLet's begin with the first iteration.\n\n\\begin{align}\n \\qh{1} &= \\Aq{0} \\\\\n \\qt{1} &= \\qh{1} - h_{0,1}*q_0 \\\\\n q_1 &= \\qt{1}/\\left|\\qt{1}\\right|.\n\\end{align}\n\nThere is nothing special here. For a 1-dimensional subspace, we could create a post-Arnoldi factorization from a Krylov subspace. Of course if that's all we can do, then it isn't very interesting. Let's move onto the second iteration.\n\n\\begin{subequations}\n \\begin{align}\n \\qh{2} &= \\A q_1 \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|}\\At{1} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|}\\A \\left\\{\\qh{1}-h_{0,1}*q_0\\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\Ah{1}-h_{0,1}*\\Aq{0}\\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\Ah{1}-h_{0,1}*\\qh{1}\\right\\} \\label{eq:2Ihat}.\n \\end{align}\n\\end{subequations}\nEquation \\ref{eq:2Ihat} contains only one application of the linear operator that hasn't already been performed. Eliminating this operation is crucial if we are to use this technique since the time required to apply the linear operator is by far the most computationally expensive. We will introduce a new variable $\\qc{1} = \\Ah{1}$ but leave discussion about this variable until later. At this point, let's just assume we know what it is without applying the linear operator.\n\nUsing $\\qc{1}$ in Equation \\ref{eq:2Ihat}, we define $\\qh{2}$ in terms of known quantities without the need for additional applications of the linear operator\n\\begin{equation}\n \\qh{2} = \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qc{1}-h_{0,1}*\\qh{1}\\right\\} \n \\label{eq:2IhatFinal}.\n\\end{equation}\n\nContinuing the iteration we find an expression of $q_2$ of all known quantities. (Can you see a recurring pattern here?)\n\\begin{subequations}\n \\begin{align}\n \\qt{2} &= \\qh{2} - h_{0,2}*q_0 - h_{1,2}*q_1 \\\\\n q_2 &= \\qt{2}/\\left|\\qt{2}\\right|.\n \\end{align}\n \\label{eq:2I}\n\\end{subequations}\n\nWe have remained in the realm of simple iterations. Let's take this one more iteration to see if future iterations will cause a problem.\n\\begin{subequations}\n \\begin{align}\n \\qh{3} &= \\A q_2 \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|}\\At{2} \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|}\\A \\left\\{\\qh{2} - h_{0,2}*q_0 - h_{1,2}*q_1 \\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\Ah{2} - h_{0,2}*\\Aq{0} - h_{1,2}*\\Aq{1} \\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\qc{2} - h_{0,2}*\\qh{1} - h_{1,2}*\\qh{2} \\right\\} \\label{eq:3Ihat}\n \\end{align}\n\\end{subequations}\n\nFrom Equation \\ref{eq:3Ihat} (as well as Equation \\ref{eq:2IhatFinal}) we see that we must store all the ``hat''vectors in addition to what we have already stored. This shouldn't normally be a problem since the number we are storing should be relatively small. Again I've taken the liberty of using $\\qc{2}$ without further explanation.\n\nWe can finish the iteration as usual.\n\\begin{subequations}\n \\begin{align}\n \\qt{3} &= \\qh{3} - h_{0,3}*q_0 - h_{1,3}*q_1 - h_{2,3}*q_2 \\\\\n q_3 &= \\qt{3}/\\left|\\qt{3}\\right|.\n \\end{align}\n\\end{subequations}\n\n\\subsection{What about the $\\qc{i}$'s?}\nSimilarly to what we have already done, let's write out the $\\qc{i}$ for the first few iterations to see if we can come up with something that works for every iteration.\n\nThe first one is easy\n\\begin{equation}\n \\qc{1} = \\Ah{1} = \\A \\left(\\Aq{0}\\right) = \\Ap{1} = \\qp{2}\n \\label{eq:qc1}\n\\end{equation}\n\nThe second one takes a little more work,\n\\begin{subequations}\n \\begin{align}\n \\qc{2} &= \\Ah{2} \\\\\n &= \\A \\left\\{\\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qc{1}-h_{0,1}*\\qh{1}\\right\\} \\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\Ac{1}-h_{0,1}*\\Ah{1}\\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\Ac{1}-h_{0,1}*\\qc{1}\\right\\}\n \\label{eq:qc2}\n \\end{align}\n\\end{subequations}\nLet's expand $\\Ac{1}$.\n\\begin{equation}\n \\Ac{1} = \\A \\left(\\qp{2}\\right) = \\qp{3}\n \\label{eq:qc2sub}\n\\end{equation}\nSubstituting Equation \\ref{eq:qc2sub} into Equation \\ref{eq:qc2} we obtain\n\\begin{equation}\n \\qc{2} = \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{3}-h_{0,1}*\\qc{1}\\right\\}.\n \\label{eq:qc2final}\n\\end{equation}\n\nLet's do another iteration (I'm not seeing a pattern yet).\n\\begin{subequations}\n \\begin{align}\n \\qc{3} &= \\Ah{3} \\\\\n &= \\A \\left[ \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\qc{2} - h_{0,2}*\\qh{1} - h_{1,2}*\\qh{2} \\right\\} \\right] \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\Ac{2} - h_{0,2}*\\Ah{1} - h_{1,2}*\\Ah{2} \\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\Ac{2} - h_{0,2}*\\qc{1} - h_{1,2}*\\qc{2} \\right\\}\n \\label{eq:qc3}\n \\end{align}\n\\end{subequations}\nLet's expand $\\Ac{2}$.\n\\begin{subequations}\n \\begin{align}\n \\Ac{2} &= \\A \\left[\\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{3}-h_{0,1}*\\qc{1}\\right\\}\\right] \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\Ap{3}-h_{0,1}*\\Ac{1}\\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{4}-h_{0,1}*\\qp{3}\\right\\}\n \\label{eq:qc3sub}\n \\end{align}\n\\end{subequations}\nSubstituting Equation \\ref{eq:qc3sub} into Equation \\ref{eq:qc3} followed by substituting in Equation \\ref{eq:qc2final} we obtain\n\\begin{subequations}\n \\begin{align}\n \\qc{3} &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\left[\\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{4}-h_{0,1}*\\qp{3}\\right\\}\\right] - h_{0,2}*\\qc{1} - h_{1,2}*\\qc{2} \\right\\} \\\\\n &= \\frac{1}{\\left|\\qt{2}\\right|} \\left\\{\\left[\\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{4}-h_{0,1}*\\qp{3}\\right\\}\\right] - h_{0,2}*\\qc{1} - h_{1,2}*\\left[\\frac{1}{\\left|\\qt{1}\\right|} \\left\\{\\qp{3}-h_{0,1}*\\qc{1}\\right\\}\\right] \\right\\}\n \\end{align}\n\\end{subequations}\n\nWell, this doesn't look good. It is clear that there are no matrix-vector multiplications needed, but I can't yet see a recursive pattern. I'm sure it's there, I just can't see it yet. \n\n\\section{General Format}\nNow let's try to write this in a general format. We have already shown that it works for up to three iterations. What if we are performing the $n+1$st iteration?\n\n\\begin{subequations}\n \\begin{align}\n \\qh{n+1} &= \\Aq{n} \\\\\n &= \\frac{1}{\\qt{n}}\\At{n} \\\\\n &= \\frac{1}{\\qt{n}}\\A \n \\left\\{\\qh{n} - h_{0,n}*q_0 - h_{1,n}*q_1 \\cdots h_{n-1,n}*q_{n-1}\\right\\} \\\\\n &= \\frac{1}{\\qt{n}}\n \\left\\{\\Ah{n} - h_{0,n}*\\Aq{0} - h_{1,n}*\\Aq{1} \\cdots h_{n-1,n}*\\Aq{n-1}\\right\\} \\\\\n &= \\frac{1}{\\qt{n}}\n \\left\\{\\qc{n} - h_{0,n}*\\qh{1} - h_{1,n}*\\qh{2} \\cdots h_{n-1,n}*\\qh{n}\\right\\}\n \\end{align}\n\\end{subequations}\n\\begin{subequations}\n \\begin{align}\n \\qt{n+1} &= \\qh{n+1} - h_{0,n+1}*q_0 \\cdots h_{n.n+1}*q_n \\\\\n q_{n+1} &= \\qt{n+1}/\\left|\\qt{n+1}\\right|.\n \\end{align}\n\\end{subequations}\n\n\\section{Conclusion}\nSo I think a post-processing Arnoldi's method is possible. I just haven't yet found the inductive proof for the $\\qc{i}$'s.\n\\end{document}\n", "meta": {"hexsha": "52d76088e472e1783d025284b1b2eb008a1f9259", "size": 9649, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/PostProcessingArnoldi/PostProcessingArnoldi.tex", "max_stars_repo_name": "jlconlin/PhDThesis", "max_stars_repo_head_hexsha": "8e704613721a800ce1c59576e94f40fa6f7cd986", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Notes/PostProcessingArnoldi/PostProcessingArnoldi.tex", "max_issues_repo_name": "jlconlin/PhDThesis", "max_issues_repo_head_hexsha": "8e704613721a800ce1c59576e94f40fa6f7cd986", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/PostProcessingArnoldi/PostProcessingArnoldi.tex", "max_forks_repo_name": "jlconlin/PhDThesis", "max_forks_repo_head_hexsha": "8e704613721a800ce1c59576e94f40fa6f7cd986", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.9476190476, "max_line_length": 479, "alphanum_fraction": 0.6171624002, "num_tokens": 3626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8244619177503205, "lm_q1q2_score": 0.6452754083487587}} {"text": "We are interested in leveled HE schemes that support SIMD operations on their plaintexts.\nSuch schemes include $\\FV$~\\cite{FV12} and $\\BGV$~\\cite{BGV12}, have the ring $\\intring_p$ as plaintext space for some $p \\geq 2$.\nWe refer to such schemes as \\emph{SIMD-schemes}.\nThe general framework of such schemes is outlined below.\n\n\\subsection{Basic setup}\n\nLet $\\lambda$ be the security level of an HE scheme.\nLet $L$ be the maximal multiplicative depth of homomorphic circuits we want to evaluate.\nLet $d$ be the order of the plaintext modulus $p$ modulo the order $m$ of $\\intring$.\nAssume that the plaintext space $\\intring_p$ has $\\slots$ SIMD slots, i.e. $\\intring_p \\cong \\F^\\ell_{p^d}$.\n% For a vector $\\va \\in \\F^k_{p^d}$, we denote the plaintext encoding of $\\va$ by $\\pt(\\va)$. \nThe basic part of a SIMD-scheme consists of key generation, encryption and decryption algorithms.\n\n$\\KeyGen(1^\\lambda, 1^L) \\rightarrow (\\sk, \\pk)$. Given $\\lambda$ and $L$, this function generates the secret key $\\sk$ and the public key $\\pk$.\nNote that the public key contains special key-switching keys that help to transform ciphertexts encrypted under other secret keys back to ciphertexts encrypted under $\\sk$. \n\n$\\Encrypt(\\pt \\in \\intring_t, \\pk) \\rightarrow \\ct$. The encryption algorithm takes a plaintext $\\pt$ and the public key $\\pk$ and outputs a ciphertext $\\ct$.\n\n$\\Decrypt(\\ct, \\sk) \\rightarrow \\pt$. The decryption algorithm takes a ciphertext $\\ct$ and the secret key $\\sk$ and returns a plaintext $\\pt$.\nFor freshly encrypted ciphertexts, the decryption correctness means that $\\Decrypt(\\Encrypt(\\pt, \\pk), \\sk) = \\pt$. \n\n\\subsection{Arithmetic operations}\n\nHomomorphic arithmetic operations are addition and multiplication.\n\n$\\Add(\\ct_1, \\ct_2) \\rightarrow \\ct$. The addition algorithm takes two input ciphertexts $\\ct_1$ and $\\ct_2$ encrypting plaintexts $\\pt_1$ and $\\pt_2$ respectively.\nIt outputs a ciphertext $\\ct$ that encrypts the sum of these plaintexts in the ring $\\intring_p$.\n%It implies that homomorphic addition sums respective SIMD slots of $\\pt_1$ and $\\pt_2$.\n\n$\\AddPlain(\\ct_1, \\pt_2) \\rightarrow \\ct$. This algorithm takes a ciphertext $\\ct_1$ encrypting a plaintext $\\pt_1$ and a plaintext $\\pt_2$.\nIt outputs a ciphertext $\\ct$ that encrypts $\\pt_1 + \\pt_2$.\n%As for the $\\Add$ algorithm, $\\AddPlain$ sums respective SIMD slots of $\\pt_1$ and $\\pt_2$.\n\n$\\Mul(\\ct_1, \\ct_2) \\rightarrow \\ct$. Given two input ciphertext $\\ct_1$ and $\\ct_2$ encrypting plaintext $\\pt_1$ and $\\pt_2$ respectively, the multiplication algorithm outputs a ciphertext $\\ct$ that encrypts the plaintext product $\\pt_1 \\cdot \\pt_2$.\n%As a result, homomorphic multiplication multiplies respective SIMD slots of $\\pt_1$ and $\\pt_2$.\n\n$\\MulPlain(\\ct_1, \\pt_2) \\rightarrow \\ct$. Given a ciphertext $\\ct_1$ encrypting plaintext $\\pt_1$ and a plaintext $\\pt_2$, this algorithm outputs a ciphertext $\\ct$ that encrypts the plaintext product $\\pt_1 \\cdot \\pt_2$.\n%As a result, homomorphic multiplication multiplies respective SIMD slots of $\\pt_1$ and $\\pt_2$.\n\nUsing the above operations as building blocks, one can design homomorphic subtraction algorithms.\n\n$\\Sub(\\ct_1, \\ct_2) = \\Add(\\ct_1, \\MulPlain(\\ct_2, \\pt(-\\1))) \\rightarrow \\ct$. The subtraction algorithm returns a ciphertext $\\ct$ that encrypts the difference of two plaintext messages $\\pt_1 - \\pt_2$ encrypted by $\\ct_1$ and $\\ct_2$, respectively.\n\n$\\SubPlain(\\ct_1, \\pt_2) = \\AddPlain(\\ct_1, \\pt_2 \\cdot \\pt(-\\1)) \\rightarrow \\ct$. This algorithm returns a ciphertext $\\ct$ that encrypts $\\pt_1 - \\pt_2$ where $\\pt_1$ is encrypted by $\\ct_1$.\nWe can also change the order of arguments such that $\\SubPlain(\\pt_1, \\ct_2)$ returns a ciphertext $\\ct$ encrypting $\\pt_1 - \\pt_2$. \n\nAs shown in Section~\\ref{subsec:crt}, the projection map $\\proj_I$ can select the SIMD slots indexed by a set $I \\subseteq \\{0,\\dots,\\slots-1\\}$ and set the rest to zero.\nThis functionality is homomorphically realized by the $\\Select$ function.\n\n$\\Select(\\ct, I) = \\MulPlain(\\ct, \\pt(\\1_I)) \\rightarrow \\ct'$ where $\\1_I$ is a vector having $1$'s in the coordinates indexed by a set $I$ and zeros everywhere else.\nGiven a ciphertext $\\ct$ encrypting SIMD slots $\\vm = (m_0,m_1,\\dots,m_{\\slots-1})$ and a set $I$, this function returns a ciphertext $\\ct'$ that encrypts $\\vm' = (m'_0,\\dots,m'_{\\slots-1})$ such that $m'_i = m_i$ if $i \\in I$ and $m'_i = 0$ otherwise.\n\n\\subsection{Special operations}\\label{subsec:special_operations}\n\nOne can also homomorphically permute the SIMD slots of a given ciphertext and act on them with the Frobenius automorphism. \n\n$\\Rotate(\\ct, i) \\rightarrow \\ct'$ with $i \\in [0, \\slots-1]$. Given a ciphertext $\\ct$ encrypting SIMD slots \n$$\\vm = (m_0,m_1,\\dots,m_{\\slots-1}),$$\nthe rotation algorithm returns a ciphertext $\\ct'$ that encrypts the cyclic shift of $\\vm$ by $i$ positions, namely $(m_i,m_{(i+1) \\mod \\slots},\\dots,m_{(i-1) \\mod \\slots})$.\n\n$\\Frob(\\ct, i) \\rightarrow \\ct'$ with $i\\in[0,d-1]$. Given a ciphertext $\\ct$ encrypting SIMD slots $\\vm$ as above, the Frobenius algorithm returns a ciphertext $\\ct'$ that encrypts a Frobenius map action on $\\vm$, namely $(m^{p^i}_0,m^{p^i}_1,\\dots,m^{p^i}_{\\slots-1})$.\n\nAs discussed in Section~\\ref{subsec:crt}, the $\\Frob$ and $\\Mul$ operations can be combined to compute the principal character $\\princhar$, which turns non-zero values of SIMD slots into $1$ and leaves slots with zeros unchanged.\n\n$\\IsNonZero(\\ct) \\rightarrow \\ct'$. Given a ciphertext $\\ct$ encrypting SIMD slots $\\vm = (m_0,m_1,\\dots,m_{\\slots-1})$, this function returns a ciphertext $\\ct'$ that encrypts:\n\\[\n (\\princhar(m_0), \\princhar(m_1), \\dots, \\princhar(m_{\\slots-1})).\n\\]\n% Kim et al.~\\cite{TDSC:KLLW16} showed that one can employ the Frobenius map to decrease the multiplicative depth of $\\IsNonZero$.\nRecall that $\\princhar(m) = m^{p^d-1} = \\prod_{i=0}^{d-1} (m^{p-1})^{p^i}$ as shown in~(\\ref{eq:exp_frob}).\nThe multiplicative depth of $x^{p-1}$ is equal to $\\ceil{\\log_2 (p-1)}$.\nThe multiplicative depth of $x^{p^i}$ is zero as it can be done by the $\\Frob$ operation.\nIn total, $d-1$ $\\Frob$ operations are needed to compute $\\princhar(m)$.\nAs a result, the total multiplicative depth of $\\IsNonZero$ is\n\\begin{align}\\label{eq:nonzero_depth}\n \\ceil{\\log_2 (p-1)} + \\ceil{\\log_2 d}.\n\\end{align}\nUsing general exponentiation by squaring, $x^{p-1}$ requires $\\floor{\\log_2 (p-1)} + \\wt(p-1) - 1$ field multiplications.\nSince $d-1$ field multiplications are needed to compute $\\prod_{i=0}^{d-1} (x^{p-1})^{p^i}$, the total number of multiplications to compute $\\princhar(m)$ is:\n\\begin{align}\\label{eq:nonzero_width}\n \\floor{\\log_2 (p-1)} + \\wt(p-1) + d - 2.\n\\end{align} \n\n\\subsection{Noise cost of homomorphic operations}\\label{subsec:cost}\nNote that every homomorphic ciphertext contains a special component called \\emph{noise} that is removed during decryption.\nHowever, the decryption function can deal only with noise of small enough magnitude; otherwise, this function fails.\nThis noise bound is defined by encryption parameters in a way that larger parameters result in a larger bound.\nThe ciphertext noise increases after every homomorphic operation and, therefore, approaches its maximal possible bound.\nIt implies that to reduce encryption parameters one needs to avoid homomorphic operations that significantly increase the noise.\nTherefore, while designing homomorphic circuits, we need to take into account not only the running time of homomorphic operations but also their effect on the noise. \n\nTable~\\ref{table:he_operations} summarizes the running time and the noise cost of the aforementioned homomorphic operations.\nSimilar to~\\cite{C:HalSho14}, we divide the operations into expensive, moderate and cheap.\nThe expensive operations dominate the cost of a homomorphic circuit.\nThe moderate operations are less important, but if there are many of them in a circuit, their total cost can dominate the total cost.\nThe cheap operations are the least important and can be omitted in the cost analysis.\n\n\\subsection{Complexity of homomorphic algorithms}\nWhile evaluating polynomials in variables $X_1, X_2, \\dots, X_n$, we refer to multiplication of expressions containing these variables as \\emph{non-scalar}.\nThere are two multiplication functions $\\Mul$ (ciphertext-ciphertext multiplication) and $\\MulPlain$ (ciphertext-plaintext multiplication).\nThe former one corresponds to non-scalar multiplication of two expressions containing input values, while the latter is related to the product of an expression with input values and unencrypted finite field elements. \nSince $\\Mul$ is much slower and introduces more noise than $\\MulPlain$, the multiplicative depth and complexity of a homomorphic circuit is usually calculated with relation to the number of $\\Mul$'s, or non-scalar multiplications.\n\n\\begin{table}[t!]\n \\centering\n \\begin{tabular*}{.45\\textwidth}{@{\\extracolsep{\\fill} } c c c }\n \\toprule\n Operation\t& Time\t\t\t& Noise \\\\\n \\midrule\n $\\Add$\t\t& cheap\t\t\t& cheap \t\\\\\n $\\AddPlain$\t& cheap\t\t\t& cheap \\\\\n $\\Mul$\t\t& expensive\t\t& expensive \t\\\\\n $\\MulPlain$\t& cheap\t\t\t& moderate \t\\\\\n $\\Sub$\t\t& cheap\t\t\t& cheap \\\\\n $\\SubPlain$\t& cheap\t\t\t& cheap \\\\\n $\\Select$\t& cheap\t\t\t& moderate \\\\\n $\\Rotate$ \t & expensive\t& moderate \\\\\n $\\Frob$\t\t & expensive\t& cheap \\\\\n $\\IsNonZero$ & expensive & expensive \\\\\n \\bottomrule\n \\end{tabular*}\n \\caption{The cost of homomorphic operations with relation to running time and noise growth.}\n \\label{table:he_operations}\n\\end{table}\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"main\"\n%%% End:\n", "meta": {"hexsha": "a35c44fe060d6ae283bec9d50238c6f58450fbe8", "size": 9611, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/homomorphic-encryption.tex", "max_stars_repo_name": "iliailia/comparison-circuit-over-fq", "max_stars_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2021-03-24T07:58:48.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-18T00:41:08.000Z", "max_issues_repo_path": "paper/homomorphic-encryption.tex", "max_issues_repo_name": "iliailia/comparison-circuit-over-fq", "max_issues_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2021-03-24T03:03:11.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-01T09:23:59.000Z", "max_forks_repo_path": "paper/homomorphic-encryption.tex", "max_forks_repo_name": "iliailia/comparison-circuit-over-fq", "max_forks_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-19T16:28:37.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-19T16:28:37.000Z", "avg_line_length": 73.3664122137, "max_line_length": 271, "alphanum_fraction": 0.7264592654, "num_tokens": 2729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.905989815306765, "lm_q2_score": 0.712232184238947, "lm_q1q2_score": 0.6452751050541774}} {"text": "\\subsection{Orthogonal and orthonormal sets}\n\nIn this section, we examine what it means for vectors (and sets of\nvectors) to be orthogonal and orthonormal. First, it is necessary to\nreview some important concepts. You may recall the definitions for the\nspan of a set of vectors and a linear independent set of vectors. We\ninclude the definitions and examples here for convenience.\n\n\\begin{definition}{Span of a set of vectors and subspace}{span}\n The collection of all linear combinations of a set of vectors\n $\\set{\\vect{u}_1,\\ldots,\\vect{u}_k}$ in $\\R^{n}$ is known as the\n span\\index{span}\\index{vector!span} of these vectors and is written\n as $\\sspan \\set{\\vect{u}_1,\\ldots,\n \\vect{u}_k}$. \\\\\n We call a collection of the form\n $\\sspan \\set{\\vect{u}_1,\\ldots, \\vect{u}_k}$ a subspace of $\\R^{n}$.\n\\end{definition}\n\nConsider the following example.\n\n\\begin{example}{Span of vectors}{span-vectors}\nDescribe the span of the vectors $\\vect{u}=\\begin{mymatrix}{rrr}\n1 & 1 & 0\n\\end{mymatrix}^T$ and\n$\\vect{v}=\\begin{mymatrix}{rrr}\n3 & 2 & 0\n\\end{mymatrix}^T \\in \\R^{3}$.\n\\end{example}\n\n\\begin{solution}\nYou can see that any linear combination of the vectors $\\vect{u}$ and $\\vect{v}$ yields a vector\n$\\begin{mymatrix}{rrr}\nx & y & 0\n\\end{mymatrix}^T$ in the $XY$-plane.\n\nMoreover every vector in the $XY$-plane is in fact such a linear\ncombination of the vectors $\\vect{u}$ and $\\vect{v}$. That's because\n\\[ \\begin{mymatrix}{r}\nx \\\\\ny \\\\\n0\n\\end{mymatrix}\n=\n(-2x+3y) \\begin{mymatrix}{r}\n1 \\\\\n1 \\\\\n0\n\\end{mymatrix}\n+\n(x-y)\\begin{mymatrix}{r}\n3 \\\\\n2 \\\\\n0\n\\end{mymatrix}\n\\]\n\nThus span$\\set{\\vect{u},\\vect{v}}$ is precisely the $XY$-plane.\n\\end{solution}\n\nThe span of a set of a vectors in $\\R^n$ is what we call a \\textbf{subspace of $\\R^n$}. A subspace $W$ is characterized by the feature that any linear combination of vectors of $W$ is again a vector contained in $W$.\n\nAnother important property of sets of vectors is called linear independence.\n\n\\begin{definition}{Linearly independent set of vectors}{linear-independent-orthogonality}\nA set of non-zero vectors $\\set{\\vect{u}_1,\\ldots,\\vect{u}_k}$ in $\\R^{n}$ is said to be\n\\textbf{linearly independent}\\index{linear independence}\\index{vector!linear independence}\n if no vector in that set is in the span of the other vectors of that set.\n\\end{definition}\n\nHere is an example.\n\n\\begin{example}{Linearly independent vectors}{linearly-independent-vectors}\nConsider vectors $\\vect{u}=\\begin{mymatrix}{rrr}\n1 & 1 & 0\n\\end{mymatrix}^T$,\n$\\vect{v}=\\begin{mymatrix}{rrr}\n3 & 2 & 0\n\\end{mymatrix}^T$, and\n$\\vect{w}=\\begin{mymatrix}{rrr}\n4 & 5 & 0\n\\end{mymatrix}^T \\in \\R^{3}$.\nVerify whether the set $\\set{\\vect{u}, \\vect{v}, \\vect{w}}$ is linearly independent.\n\\end{example}\n\n\\begin{solution}\nWe already verified in Example~\\ref{exa:span-vectors} that\n$\\sspan \\set{\\vect{u}, \\vect{v}}$ is the $XY$-plane. Since $\\vect{w}$\nis clearly also in the $XY$-plane, then the set $\\set{\\vect{u}, \\vect{v},\n\\vect{w}}$ is \\textbf{not} linearly independent.\n\\end{solution}\n\nIn terms of spanning, a set of vectors is linearly independent if it\ndoes not contain unnecessary vectors. In the previous example you\ncan see that the vector $\\vect{w}$ does not help to span any new vector\nnot already in the span of the other two vectors. However you can\nverify that the set $\\set{\\vect{u}, \\vect{v}}$ is linearly independent,\nsince you will not get the $XY$-plane as the span of a single vector.\n\nWe can also determine if a set of vectors is linearly independent by examining linear combinations. A set of vectors is linearly independent if and only if\nwhenever a linear combination of these vectors equals zero, it follows\nthat all the coefficients equal zero. It is a good exercise to verify\nthis equivalence, and this latter condition is often used as the\n(equivalent) definition of linear independence.\n\nIf a subspace is spanned by a linearly independent set of vectors,\nthen we say that it is a basis for the subspace.\n\n\\begin{definition}{Basis of a subspace}{subspace-basis}\nLet $V$ be a subspace of $\\R^{n}$. Then $\\set{\n\\vect{u}_{1},\\ldots,\\vect{u}_{k}} $ is a \\textbf{basis}\\index{basis}\\index{vector!basis}\\index{subspace!basis} for\n$V$ if the following two conditions\nhold.\n\n\\begin{enumerate}\n\\item $\\sspan\\set{\\vect{u}_{1},\\ldots,\\vect{u}_{k}} =V$\n\\item $\\set{\\vect{u}_{1},\\ldots,\\vect{u}_{k}} $ is linearly\nindependent\n\\end{enumerate}\n\\end{definition}\n\nThus the set of vectors $\\set{\\vect{u}, \\vect{v}}$ from Example~\\ref{exa:linearly-independent-vectors} is a basis for $XY$-plane in\n$\\R^{3}$ since it is both linearly independent and spans\nthe $XY$-plane.\n\n Recall from the properties of the dot product of vectors\nthat two vectors $\\vect{u}$ and $\\vect{v}$ are orthogonal if $\\vect{u}\n\\dotprod \\vect{v} = 0$. Suppose a vector is orthogonal to a spanning set of $\\R^n$. What can be said about such a vector? This is the discussion in the following example.\n\n\\begin{example}{Orthogonal vector to a spanning set}{}\nLet $\\set{\\vect{x}_1, \\vect{x}_2, \\ldots, \\vect{x}_k}\\in\\R^n$ and\nsuppose $\\R^n=\\sspan\\set{\\vect{x}_1, \\vect{x}_2, \\ldots, \\vect{x}_k}$.\nFurthermore, suppose that there exists a vector $\\vect{u}\\in\\R^n$ for which $\\vect{u}\\dotprod \\vect{x}_j=0$ for all $j$, $1\\leq j\\leq k$.\nWhat type of vector is $\\vect{u}$?\n\\end{example}\n\n\\begin{solution}\nWrite $\\vect{u}=t_1\\vect{x}_1 + t_2\\vect{x}_2 +\\ldots +t_k\\vect{x}_k$\nfor some $t_1, t_2, \\ldots, t_k\\in\\R$\n(this is possible because\n$\\vect{x}_1, \\vect{x}_2, \\ldots, \\vect{x}_k$ span $\\R^n$).\n\nThen\n\\begin{eqnarray*}\n\\norm{\\vect{u}}^2 & = & \\vect{u}\\dotprod\\vect{u} \\\\\n& = & \\vect{u}\\dotprod(t_1\\vect{x}_1 + t_2\\vect{x}_2 +\\ldots +t_k\\vect{x}_k) \\\\\n& = & \\vect{u}\\dotprod (t_1\\vect{x}_1) + \\vect{u}\\dotprod (t_2\\vect{x}_2) +\n\\ldots + \\vect{u}\\dotprod (t_k\\vect{x}_k) \\\\\n& = & t_1(\\vect{u}\\dotprod \\vect{x}_1) + t_2(\\vect{u}\\dotprod \\vect{x}_2) + \\ldots\n+ t_k(\\vect{u}\\dotprod \\vect{x}_k) \\\\\n& = & t_1(0) + t_2(0) + \\ldots + t_k(0) = 0.\n\\end{eqnarray*}\nSince $\\norm{\\vect{u}}^2 =0$, $\\norm{\\vect{u}} =0$.\nWe know that $\\norm{\\vect{u}}=0$ if and only if\n$\\vect{u}=\\vect{0}_n$.\nTherefore, $\\vect{u}=\\vect{0}_n$.\nIn conclusion, the only vector orthogonal to every vector of\na spanning set of $\\R^n$ is the zero vector.\n\\end{solution}\n\nWe can now discuss what is meant by an orthogonal\\index{orthogonal} set of vectors.\n\n\\begin{definition}{Orthogonal set of vectors}{orthogonal-set}\nLet $\\set{\\vect{u}_1, \\vect{u}_2,\\ldots, \\vect{u}_m}$ be a set of\nvectors in $\\R^n$. Then this set is called an\n\\textbf{orthogonal set}\\index{orthogonal}\\index{vector!orthogonal}\nif the following conditions hold:\n\\begin{enumerate}\n\\item\n$\\vect{u}_i \\dotprod \\vect{u}_j = 0$ for all $i \\neq j$\n\\item\n$\\vect{u}_i \\neq \\vect{0}$ for all $i$\n\\end{enumerate}\n\\end{definition}\n\nIf we have an orthogonal set of vectors and normalize each vector so\nthey have length 1, the resulting set is called an \\textbf{orthonormal\nset} of vectors. They can be described as follows.\n\n\\begin{definition}{Orthonormal set of vectors}{orthonormal-set-vectors}\nA set of vectors, $\\set{\\vect{w}_{1},\\ldots,\\vect{w}_{m}} $\nis said to be an\n\\textbf{orthonormal}\\index{orthonormal}\\index{vector!orthonormal}\nset if\n\\[\n\\vect{w}_i \\dotprod \\vect{w}_j = \\delta _{ij} = \\left\\{\n\\begin{array}{c}\n1\\text{ if }i=j \\\\\n0\\text{ if }i\\neq j\n\\end{array}\n\\right.\n\\]\n\\end{definition}\n\nNote that all orthonormal sets are orthogonal, but the reverse is not\nnecessarily true since the vectors may not be normalized. In order to\nnormalize the vectors, we simply need divide each one by its length.\n\n\\begin{definition}{Normalizing an orthogonal set}{normalizing}\nNormalizing an orthogonal set is the process of\nturning an orthogonal (but not orthonormal) set into\nan orthonormal set.\nIf $\\set{\\vect{u}_1, \\vect{u}_2, \\ldots, \\vect{u}_k}$\nis an orthogonal subset of $\\R^n$,\nthen\n\\[ \\set{\n\\frac{1}{\\norm{\\vect{u}_1}}\\vect{u}_1,\n\\frac{1}{\\norm{\\vect{u}_2}}\\vect{u}_2, \\ldots,\n\\frac{1}{\\norm{\\vect{u}_k}}\\vect{u}_k }\n\\]\nis an orthonormal set.\n\\end{definition}\n\nWe illustrate this concept in the following example.\n\n\\begin{example}{Orthonormal set}{orthonormal-set}\nConsider the set of vectors given by\n\\[\n\\set{\\vect{u}_1, \\vect{u}_2 } = \\set{\n\\begin{mymatrix}{c}\n1 \\\\\n1\n\\end{mymatrix},\n\\begin{mymatrix}{r}\n-1 \\\\\n1\n\\end{mymatrix}\n}\n\\]\nShow that it is an orthogonal set of vectors but not an orthonormal one. Find the corresponding orthonormal set.\n\\end{example}\n\n\\begin{solution}\nOne easily verifies that $\\vect{u}_1 \\dotprod \\vect{u}_2 = 0$ and\n$\\set{\\vect{u}_1, \\vect{u}_2 }$ is an orthogonal set of\nvectors. On the other hand one can compute that $\\norm{\\vect{u}_1}= \\norm{\\vect{u}_2} =\n\\sqrt{2} \\neq 1$ and thus it is not an orthonormal set.\n\nThus to find a corresponding orthonormal set, we simply need to\nnormalize each vector. We will write $\\set{\\vect{w}_1, \\vect{w}_2}$\nfor the corresponding orthonormal set. Then,\n\\begin{eqnarray*}\n\\vect{w}_1 &=& \\frac{1}{\\norm{\\vect{u}_1}} \\vect{u}_1\\\\\n&=& \\frac{1}{\\sqrt{2}} \\begin{mymatrix}{c}\n1 \\\\\n1\n\\end{mymatrix} \\\\\n&=&\n\\begin{mymatrix}{c}\n\\vspace{0.05in} \\frac{1}{\\sqrt{2}}\\\\\n\\vspace{0.05in}\\frac{1}{\\sqrt{2}}\n\\end{mymatrix}\n\\end{eqnarray*}\n\nSimilarly,\n\\begin{eqnarray*}\n\\vect{w}_2 &=& \\frac{1}{\\norm{\\vect{u}_2}} \\vect{u}_2\\\\\n&=& \\frac{1}{\\sqrt{2}} \\begin{mymatrix}{r}\n-1 \\\\\n1\n\\end{mymatrix} \\\\\n&=&\n\\begin{mymatrix}{r}\n\\vspace{0.05in} -\\frac{1}{\\sqrt{2}}\\\\\n\\vspace{0.05in}\\frac{1}{\\sqrt{2}}\n\\end{mymatrix}\n\\end{eqnarray*}\n\nTherefore the corresponding orthonormal set is\n\\[\n\\set{\\vect{w}_1, \\vect{w}_2 } =\n\\set{\n\\begin{mymatrix}{c}\n\\vspace{0.05in} \\frac{1}{\\sqrt{2}}\\\\\n\\vspace{0.05in}\\frac{1}{\\sqrt{2}}\n\\end{mymatrix},\n\\begin{mymatrix}{r}\n\\vspace{0.05in} -\\frac{1}{\\sqrt{2}}\\\\\n\\vspace{0.05in}\\frac{1}{\\sqrt{2}}\n\\end{mymatrix}\n}\n\\]\n\nYou can verify that this set is orthogonal.\n\\end{solution}\n\nConsider an orthogonal set of vectors in $\\R^n$, written $\\set{\n\\vect{w}_1,\\ldots, \\vect{w}_k}$ with $k \\leq n$. The span of these\nvectors is a subspace $W$ of $\\R^n$. If we\ncould show that this orthogonal set is also linearly independent, we\nwould have a basis of $W$. We will show this in the next theorem.\n\n\\begin{theorem}{Orthogonal basis of a subspace}{orthonormal-basis}\nLet $ \\set{\\vect{w}_1, \\vect{w}_2,\\ldots, \\vect{w}_k}$ be an\northonormal set of vectors in $\\R^n$. Then this set is\nlinearly independent and forms a basis for the subspace $W =\n\\sspan \\set{\\vect{w}_1, \\vect{w}_2,\\ldots, \\vect{w}_k}$.\n\\end{theorem}\n\n\\begin{proof}\nTo show it is a linearly independent set, suppose a linear combination\nof these vectors equals $\\vect{0}$, such as:\n\\[\na_1 \\vect{w}_1 + a_2 \\vect{w}_2 + \\ldots + a_k \\vect{w}_k = \\vect{0}, a_i \\in \\R\n\\]\nWe need to show that all $a_i = 0$. To do so, take the dot product of\neach side of the above equation with the vector $\\vect{w}_i$ and obtain the following.\n\n\\begin{eqnarray*}\n\\vect{w}_i \\dotprod (a_1 \\vect{w}_1 + a_2 \\vect{w}_2 + \\ldots + a_k \\vect{w}_k ) &=& \\vect{w}_i \\dotprod \\vect{0}\\\\\na_1 (\\vect{w}_i \\dotprod \\vect{w}_1) + a_2 (\\vect{w}_i \\dotprod \\vect{w}_2) + \\ldots + a_k (\\vect{w}_i \\dotprod \\vect{w}_k) &=& 0\n\\end{eqnarray*}\nNow since the set is orthogonal, $\\vect{w}_i \\dotprod \\vect{w}_m = 0$ for all $m \\neq i$, so we have:\n\\[\na_1 (0) + \\ldots + a_i(\\vect{w}_i \\dotprod \\vect{w}_i) + \\ldots + a_k (0) = 0\n\\]\n\\[\na_i \\norm{\\vect{w}_i}^2 = 0\n\\]\n\nSince the set is orthogonal, we know that $\\norm{\\vect{w}_i}^2 \\neq 0$. It follows that $a_i =0$. Since the $a_i$ was chosen arbitrarily, the set $\\set{\\vect{w}_1, \\vect{w}_2,\\ldots, \\vect{w}_k}$ is linearly independent.\n\nFinally since $W = \\mbox{span} \\set{\\vect{w}_1, \\vect{w}_2,\\ldots,\n\\vect{w}_k}$, the set of vectors also spans $W$ and therefore forms a basis of $W$.\n\n\\end{proof}\n\nIf an orthogonal set is a basis for a subspace, we call this an\northogonal basis. Similarly, if an orthonormal set is a basis, we call\nthis an orthonormal basis.\n\nWe conclude this section with a discussion of Fourier expansions. Given any orthogonal basis $B$ of $\\R^n$ and an arbitrary vector $\\vect{x} \\in \\R^n$, how do we express $\\vect{x}$ as a linear combination of vectors in $B$? The solution is Fourier expansion.\n\n\\begin{theorem}{Fourier expansion}{fourier-expansion}\nLet $V$ be a subspace of $\\R^n$ and suppose $\\set{\\vect{u}_1, \\vect{u}_2, \\ldots, \\vect{u}_m}$\nis an orthogonal basis of $V$.\nThen for any $\\vect{x}\\in V$,\n\\[ \\vect{x} =\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_1}{\\norm{\\vect{u}_1}^2}} \\vect{u}_1 +\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_2}{\\norm{\\vect{u}_2}^2}} \\vect{u}_2 +\n\\ldots +\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_m}{\\norm{\\vect{u}_m}^2}} \\vect{u}_m.\n\\]\nThis expression is called the Fourier expansion\nof $\\vect{x}$, and\n\\[ \\frac{\\vect{x}\\dotprod \\vect{u}_j}{\\norm{\\vect{u}_j}^2},\\]\n$j=1,2,\\ldots,m$\nare the Fourier coefficients.\n\\end{theorem}\n\nConsider the following example.\n\n\\begin{example}{Fourier expansion}{fourier}\nLet\n$\\vect{u}_1= \\begin{mymatrix}{r} 1 \\\\ -1 \\\\ 2 \\end{mymatrix},\n\\vect{u}_2= \\begin{mymatrix}{r} 0 \\\\ 2 \\\\ 1 \\end{mymatrix}$,\nand\n$\\vect{u}_3 =\\begin{mymatrix}{r} 5 \\\\ 1 \\\\ -2 \\end{mymatrix}$,\nand let\n$\\vect{x} =\\begin{mymatrix}{r} 1 \\\\ 1 \\\\ 1 \\end{mymatrix}$.\n\nThen $B=\\set{\\vect{u}_1, \\vect{u}_2, \\vect{u}_3}$\nis an orthogonal basis of $\\R^3$.\n\nCompute the Fourier expansion of $\\vect{x}$, thus writing $\\vect{x}$ as a linear combination of the vectors of $B$.\n\\end{example}\n\n\\begin{solution}\nSince $B$ is a basis (verify!) there is a unique way to express $\\vect{x}$ as a\nlinear combination of the vectors of $B$. Moreover since $B$ is an\northogonal basis (verify!), then this can be done by computing the\nFourier expansion of $\\vect{x}$.\n\nThat is:\n\n\\[\n\\vect{x} =\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_1}{\\norm{\\vect{u}_1}^2}} \\vect{u}_1 +\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_2}{\\norm{\\vect{u}_2}^2}} \\vect{u}_2 +\n\\paren{\\frac{\\vect{x}\\dotprod \\vect{u}_3}{\\norm{\\vect{u}_3}^2}} \\vect{u}_3.\n\\]\n\nWe readily compute:\n\n\\[\n\\frac{\\vect{x}\\dotprod\\vect{u}_1}{\\norm{\\vect{u}_1}^2} = \\frac{2}{6}, \\;\n\\frac{\\vect{x}\\dotprod\\vect{u}_2}{\\norm{\\vect{u}_2}^2} = \\frac{3}{5},\n\\mbox{ and }\n\\frac{\\vect{x}\\dotprod\\vect{u}_3}{\\norm{\\vect{u}_3}^2} = \\frac{4}{30}.\\]\n\nTherefore,\n\\[ \\begin{mymatrix}{r} 1 \\\\ 1 \\\\ 1 \\end{mymatrix}\n= \\frac{1}{3}\\begin{mymatrix}{r} 1 \\\\ -1 \\\\ 2 \\end{mymatrix}\n+\\frac{3}{5}\\begin{mymatrix}{r} 0 \\\\ 2 \\\\ 1 \\end{mymatrix}\n+\\frac{2}{15}\\begin{mymatrix}{r} 5 \\\\ 1 \\\\ -2 \\end{mymatrix}.\\]\n\\end{solution}\n", "meta": {"hexsha": "5d90e50682f628f698bfdabf6ef9836dc22fdac8", "size": 14352, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/content/RnVectorsOrthogonalitySets.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "old/content/RnVectorsOrthogonalitySets.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/content/RnVectorsOrthogonalitySets.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 36.2424242424, "max_line_length": 258, "alphanum_fraction": 0.6845039019, "num_tokens": 5141, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154239957834733, "lm_q2_score": 0.9019206791658465, "lm_q1q2_score": 0.6452556961685739}} {"text": "\n\\section{The \\lowerbound algorithm}\n\\Label{sec:lowerbound}\n\nThe \\lowerbound algorithm is one of the four binary search algorithms\nof the \\cxx Standard Library \\cite[\\S 28.7.3.1]{cxx-17-draft}.\nFor our purposes we have modified\nthe generic implementation\nto that of an array of type \\valuetype.\nThe signature now reads:\n\n\\begin{lstlisting}[style = acsl-block]\n\n size_type\n lower_bound(const value_type* a, size_type n, value_type v);\n\\end{lstlisting}\n\nAs with the other binary search algorithms \\lowerbound requires that\nits input array is in increasing order.\nThe index~\\inl{lb}, that \\lowerbound returns satisfies the inequality\n\n\\begin{align}\n\\Label{eq:lower-bound-result}\n0 \\leq \\mathtt{lb} \\leq n \n\\end{align}\n\nand has the following properties for a valid index~\\inl{k} of the array under consideration\n\n\\begin{alignat}{3}\n\\Label{eq:lower-bound-left}\n0 &\\leq k < \\mathtt{lb} && \\qquad\\Longrightarrow\\qquad && a[k] < \\mathtt{v} \\\\\n\\Label{eq:lower-bound-right}\n\\mathtt{lb} &\\leq k < n && \\qquad\\Longrightarrow\\qquad && \\mathtt{v} \\leq a[k]\n\\end{alignat}\n\nConditions~\\eqref{eq:lower-bound-left} and~\\eqref{eq:lower-bound-right} imply that~\\inl{v}\ncan only occur in the array section \\inl{a[lb..n-1]}.\nIn this sense \\lowerbound returns a \\emph{lower bound} for the potential indices.\n\nAs an example, we consider in Figure~\\ref{fig:lowerbound} an increasingly ordered array.\nThe arrows indicate which indices will be returned by \\lowerbound for a given value.\nNote that the index~9 points \\emph{one past end} of the array.\nValues that are not contained in the array are colored in gray.\n\n\\begin{figure}[hbt]\n\\centering\n\\includegraphics[width=0.60\\textwidth]{Figures/lower_bound.pdf}\n\\caption{\\Label{fig:lowerbound}Some examples for \\lowerbound}\n\\end{figure}\n\n%\\FloatBarrier\n\nFigure~\\ref{fig:lowerbound} also clarifies that care must\nbe taken when interpreting the return value of \\lowerbound.\n%\nAn important difference to the algorithms in Chapter~\\ref{cha:non-mutating}\nis that a return value of \\lowerbound that is less than~$n$ \ndoes not necessarily implies \\inl{a[lb] == v}.\nWe can only be sure that \\inl{v <= a[lb]} holds.\n\n\n\\subsection{Formal specification of \\lowerbound}\n\nThe specification of \\specref{lowerbound} is shown in the following listing.\nThe preconditions \\inl{increasing} expresses\nthat the array values need to be in increasing order.\n%\nThe postconditions reflect the conditions listed above and can be expressed\nusing the predicates \\logicref{LowerBound} and \\logicref{StrictUpperBound}.\n\n\\begin{itemize}\n\\item Condition~\\eqref{eq:lower-bound-result} becomes postcondition \\inl{result}\n\\item Condition~\\eqref{eq:lower-bound-left} becomes postcondition \\inl{left}\n\\item Condition~\\eqref{eq:lower-bound-right} becomes postcondition \\inl{right}\n\\end{itemize}\n\n\\input{Listings/lower_bound.h.tex}\n\n\\subsection{Implementation of \\lowerbound}\n\\Label{subsec:lowerbound:impl}\n\nThe following listing shows our implementation of \\implref{lowerbound}.\nEach iteration step narrows down the range that contains the\nsought-after result. \nThe loop invariants express that in each iteration step all indices\nless than the temporary left bound\n\\inl{left} contain values that are less than \\inl{v} and all indices\nnot less than the temporary right bound \\inl{right} contain values\nthat are greater or equal than \\inl{v}.\n%\nThe expression to compute \\inl{middle} is slightly more complex than the\nnaïve \\inl{(left+right)/2}, but it avoids potential overflows.\n\n\\input{Listings/lower_bound.c.tex}\n\n", "meta": {"hexsha": "20bcd811341589cf52f848d4741e4b2d56e60eef", "size": 3502, "ext": "tex", "lang": "TeX", "max_stars_repo_path": 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"max_forks_repo_licenses": ["MIT"], "max_forks_count": 19, "max_forks_repo_forks_event_min_datetime": "2017-06-21T13:49:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T16:27:06.000Z", "avg_line_length": 36.4791666667, "max_line_length": 91, "alphanum_fraction": 0.7724157624, "num_tokens": 969, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585786300049, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.6451064550125185}} {"text": "\\subsection{Data Preparation} \\label{data_preparation}\n\nAll of our data records are resampled and quantized before further analysis. In general, dimensionality and cardinality reduction of time series is performed to ease and accelerate data processing by providing a more compact representation of equidistant measurements \\cite{liu2009uwave}.\n\n\\paragraph{Resampling:} The recorded acceleration data was resampled by means of the moving average technique, using a window size of 50 ms and step size of 30 ms.\n\n\\paragraph{Quantization:} The resampled records were then converted into time series with integer values between -16 and 16, such as suggested in related work \\cite{liu2009uwave} and summarized in table \\ref{table:conversion}.\n\n\\begin{table}\n \\begin{center}\n \\begin{tabularx}{\\textwidth}{XX}\n \\hline\n \\textbf{Acceleration data ($a$) in $\\frac{dm}{s^2}$} & \\textbf{Converted value}\\\\\n \\hline\n $a > 200$ & 16\\\\\n $100 < a < 200$ & 11 to 15 (five levels linearly)\\\\\n $0 < a < 100$ & 1 to 10 (ten levels linearly)\\\\\n $a = 0$ & 0\\\\\n $-100 < a < 0$ & -1 to - 10 (ten levels linearly)\\\\\n $-200 < a < -100$ & -11 to - 15 (five levels linearly)\\\\\n $a < -200$ & -16\\\\\n \\hline\n \\end{tabularx}\n \\end{center}\n \\caption{Conversion of recorded acceleration data from $\\frac{dm}{s^2}$ scale to integer values.}\n\t\\label{table:conversion}\n\\end{table}\n", "meta": {"hexsha": "5828655b0e1e5637bcaedc231c607097122e5912", "size": 1487, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/evaluation/data_preparation.tex", "max_stars_repo_name": "GordonLesti/SlidingWindowFilter", "max_stars_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-06-22T09:37:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-14T11:43:53.000Z", "max_issues_repo_path": "paper/evaluation/data_preparation.tex", "max_issues_repo_name": "GordonLesti/SlidingWindowFilter", "max_issues_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/evaluation/data_preparation.tex", "max_forks_repo_name": "GordonLesti/SlidingWindowFilter", "max_forks_repo_head_hexsha": "22c11f2912a5c523ae8ad85a849e2d0b123536ec", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-01-11T23:15:57.000Z", "max_forks_repo_forks_event_max_datetime": "2019-01-11T23:15:57.000Z", "avg_line_length": 53.1071428571, "max_line_length": 288, "alphanum_fraction": 0.6536650975, "num_tokens": 418, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256551882382, "lm_q2_score": 0.7662936430859597, "lm_q1q2_score": 0.64508564815742}} {"text": "\\section{Exercise 03}\n\\subsection{}\n\n\\begin{frame}\n\\frametitleTC{Problem}\n\\framesubtitleTC{This one we solve together}\n Given the two DT LTI dynamic systems\n \\begin{displaymath}\n S_1:\n \\left\\{\\begin{array}{rcl}\n x_1(k) &=& 0.5x_1(k-1)+u(k-1)\\\\\n x_2(k) &=& 0.5x_2(k-1)+u(k-1)\\\\\n y(k) &=& x_1(k)+x_2(k)\n \\end{array}\\right. \\quad\n S_2:\n \\left\\{\\begin{array}{rcl}\n x_1(k) &=& 0.4x_1(k-1)+u(k-1)\\\\\n x_2(k) &=& 0.8x_2(k-1)+u(k-1)\\\\\n y(k) &=& x_2(k)\n \\end{array}\\right.\n \\end{displaymath}\n\n \\begin{itemize}[<+-| alert@+>]\n \\item[(a)] compute their transfer functions,\n \\item[(b)] comment on the result.\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Solution}\n\\framesubtitleTC{Item (a)}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item We start with $S_1$:\n \\begin{itemize}\n \\item[] \\begin{itemize}[<+-| alert@+>]\n \\item[$G(z)$] \\vspace{1mm}\n $= \\begin{bmatrix} 1 & 1 \\end{bmatrix} \\,\n \\begin{bmatrix} z-0.5 & 0 \\\\\n 0 & z-0.5 \\end{bmatrix}^{-1} \\,\n \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}\n +0\n $\n \\item[] \\vspace{1mm}\n $= \\cfrac{1}{(z-0.5)^2}\n \\begin{bmatrix} 1 & 1 \\end{bmatrix} \\,\n \\begin{bmatrix} z-0.5 & 0 \\\\\n 0 & z-0.5 \\end{bmatrix} \\,\n \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}\n $\n \\item[] \\vspace{1mm}\n $= \\cfrac{1}{(z-0.5)^2}\n \\begin{bmatrix} z-0.5 & z-0.5 \\end{bmatrix} \\,\n \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}\n $\n \\item[] \\vspace{1mm}\n $= \\cfrac{2\\cancel{(z-0.5)}}{(z-0.5)^{\\cancel{2}}} \\qquad \\qquad\n \\TC{\\leftarrow \\text{zero/pole \\underline{cancellation}}}\n $\n \\item[] \\vspace{1mm}\n $= \\cfrac{2}{z-0.5}.\n $\n \\end{itemize}\n \\end{itemize}\n \\end{itemize}\n\\end{frame}\n\n\n\\begin{frame}[fragile]\n\\frametitleTC{Solution}\n\\framesubtitleTC{Item (a)}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item And now $S_2$:\n \\begin{itemize}\n \\item[] \\begin{itemize}[<+-| alert@+>]\n \\item[] wxMaxima, we are lazy \\smiley\n \\begin{verbatim}\nA:matrix([0.4,0],[0,0.8]);\nb:matrix([1],[1]);\nc:matrix([0,1]);\nG:c.invert(z*ident(2)-A).b;\n \\end{verbatim}\n \\item[] \\vspace{1mm}\n $ G(z) = \\cfrac{1}{z-0.8}.\n $\n \\end{itemize}\n \\item \\vspace{3mm}Another case with cancellation: the characteristic polynomial of $A$\\\\\n has degree 2, the denominator of $G(z)$ has degree 1.\n \\end{itemize}\n \\end{itemize}\n\\end{frame}\n\n\n\n\\begin{frame}\n\\frametitleTC{Solution}\n\\framesubtitleTC{Item (b) -- comments on the $S_1$ case}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item In fact $S_1$ is a ``fake'' 2nd order system, since as far as induced motions are\\\\\n concerned $x_1$ and $x_2$ are the same, and the system is \\TC{input-output} (i.e, as far\\\\\n as we only look at $u$ and $y$) \\TC{equivalent} to a 1st order one:\n \\begin{displaymath}\n \\left\\{\\begin{array}{rcl}\n x_1(k) &=& 0.5x_1(k-1)+u(k-1)\\\\\n x_2(k) &=& 0.5x_2(k-1)+u(k-1)\\\\\n y(k) &=& x_1(k)+x_2(k)\n \\end{array}\\right. \\quad\n \\Rightarrow \\quad\n \\left\\{\\begin{array}{rcl}\n x(k) &=& 0.5x(k-1)+u(k-1)\\\\\n y(k) &=& 2x(k)\n \\end{array}\\right.\n \\end{displaymath} \n apparently with\n \\begin{displaymath}\n G(z)=\\cfrac{2}{z-0.5}.\n \\end{displaymath} \n \\item This system cannot move in the whole $(x_1,x_2)$ plane, but only\\\\\n on the straight line $x_1=x_2$.\n \\item In general, systems like this can only move on a \\TC{subspace} (line)\\\\\n of their \\TC{state space} (plane).\n \\item We call them \\TC{not fully reachable} (the term should be intuitive).\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Solution}\n\\framesubtitleTC{Item (b) -- comments on the $S_2$ case}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item Also $S_2$ is a ``fake'' 2nd order system, because however $x_1$ moves, $y$ does not\n reveal this, and thus we can reduce this system as well to a 1st order one:\n \\begin{displaymath}\n \\left\\{\\begin{array}{rcl}\n x_1(k) &=& 0.4x_1(k-1)+u(k-1)\\\\\n x_2(k) &=& 0.8x_2(k-1)+u(k-1)\\\\\n y(k) &=& x_2(k)\n \\end{array}\\right. \\quad\n \\Rightarrow \\quad\n \\left\\{\\begin{array}{rcl}\n x(k) &=& 0.8x(k-1)+u(k-1)\\\\\n y(k) &=& x(k)\n \\end{array}\\right.\n \\end{displaymath} \n apparently with\n \\begin{displaymath}\n G(z)=\\cfrac{1}{z-0.8}.\n \\end{displaymath} \n \\item In this system one state variable does not influence the output.\n \\item We call such systems \\TC{not fully observable} (the term should be\\\\\n once again intuitive).\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Proposed exercise 03}\n\\framesubtitleTC{Try this at home, ask questions next time if needed}\n\\myPause\n Take the two system of exercise 03, and turn them into block diagrams.\\\\\n \\vspace{5mm}Analyse those diagrams along the idea that parts of the system are not influenced\n by the input, or do not influence the output. \n\\end{frame}\n\n\n\n\\begin{frame}\n\\frametitleTC{Takeaways}\n\\framesubtitleTC{from exercise 03 (and the proposed one)}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item There are cases in which a system cannot move in all its state space, or equivalently,\n some state variables are not influenced by the input.\n \\item Suggestion for reflections: why ``equivalently''?\n \\item There are also cases in which some state variables do not influence the output.\n \\item There are also cases where both facts occur.\n \\item We detect this because in computing the transfer function we get\\\\\n \\TC{cancellations}.\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}\n\\frametitleTC{Takeaways}\n\\framesubtitleTC{from exercise 03 (and the proposed one)}\n\\myPause\n \\begin{itemize}[<+-| alert@+>]\n \\item Do we need to bother?\n \\item Yes, because this collectively means that a system may have \\TC{hidden parts}\\\\\n (things the transfer function does not tell).\n \\item In setting up controls, we must be careful to not generate UNSTABLE hidden parts.\n \\item We do not mind about asymptotically stable ones, because their effect\\\\\n vanishes with their free motion.\n \\end{itemize}\n\\end{frame}\n\n\n\n", "meta": {"hexsha": "c0625fa13ee321dd7aa1528402f82daa30d0a26e", "size": 6726, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "slides/Unit-03/sections/03-PS01-ex03.tex", "max_stars_repo_name": "albertoleva/PID4CSE", "max_stars_repo_head_hexsha": "66ec14c204e16c97a5792c2e240b2daed4b39e83", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-04-19T16:38:10.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T16:38:10.000Z", "max_issues_repo_path": "slides/Unit-03/sections/03-PS01-ex03.tex", "max_issues_repo_name": "albertoleva/PID4CSE", "max_issues_repo_head_hexsha": "66ec14c204e16c97a5792c2e240b2daed4b39e83", "max_issues_repo_licenses": ["CC0-1.0"], 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YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8418256492357359, "lm_q1q2_score": 0.6450856435960554}} {"text": "\\Lecture{Jayalal Sarma }{ 2 october, 2020}{11}{More on PIE, PIE -Tuttes-Matrix-Tree-Theorem-Part2}{Shrinidhi Bajpayee}{$\\alpha$}{JS}\n\n\n\n\\section{Introduction}\nWe finished Tutte's Matrix Tree Theorem in non trivial way.Let's just recall Tutte's Theorem.We have directed graph $G(V,E)$.We have $V={v_1,v_2,....v_n}$.\\\\\nA $n \\times n$ matrix $L(G)$ called the \\textit{Laplacian matrix} of $G$ as follows.\n\\[\n L(G)_{ij} =\n \\begin{cases}\n indeg(v_i) & \\mbox{ if } i = j \\\\\n -1 & \\mbox{ if }~ i \\neq j ~\\text{and}~(v_i,v_j)\\in E \\\\\n 0 & \\mbox{ otherwise (When there is no edge) }\n \\end{cases}\n\\]\nWhat theorem says,the number of spanning arborescences rooted at some vertex n is given by exactly equal to $det(L_G[i])$ Here i=n.\\\\\nQuickly Recap Definition of Spanning arborescences :An \\textit{Arborescence} is a directed graph in which a vertex $u$ is called the root and for every other vertex $v$ in the graph, there is exactly one directed path from $u$ to $v$. In simpler terms, an arborescence is an directed tree in which all the edges are directed away from the root. A \\textit{Spanning Arborescence} $S(V,E)$ of a directed graph $G(V',E')$ is an arborescence such that $V=V'$ and $E\\subseteq E'$.\\\\\nOne condition is that there is a directed path from V to every vertex in G within E'.\\\\\nSecond condition is underlying undirected graph should be a tree. \\\\ \\\\\nRecall the following from the last lecture.\n\\begin{align}\n det(L_G[n]) = \\sum_{\\sigma \\in S_{n-1}}~\n Sign(\\sigma)~\\prod_{i=1}^{n-1}L_{i\\sigma(i)}\n \\label{eq:to_prove:Tutte}\n\\end{align} \\\\\n$S(n-1)$ is notation for permutation of $n-1$.\nThis is expression of determinant that we have seen in last lecture.\nSo we want to show it is equal to spanning arborescence we introduce concept called Spreg.So what is Spreg $?$.\nWe quickly define Spreg.\n\\begin{definition}\\textbf{Spregs:} Single prdecessor graphs or \\textit{Spregs} with distinguished vertex $v$ of a directed graph $G(V,E)$ is a subgraph $T(V,E')$, $E' \\subseteq E$, such that each vertex in $T$ except the vertex $v$ has exactly one predecessor and the vertex $v$ has no predecessors. In other words; in the spreg T, $indeg(v) = 0$ and for every $u \\neq v$, $indeg(u)=1$.\n\\end{definition}\nEvery vertex other than distinguished vertex must be indegree 1.Subgraph of G is called Spregs.So there are several spregs are possible similar to several spanning arborescence possible.\\\\\nWe want to count spanning arborescence using spregs.That will be very nice combinatorial interpretation of this topic,that is plan.\\\\\n\n%1. We want to count the number of spanning arborescence rooted at Vn.\\\\ \\\\\n2.We want to count the number of spregs distinguished vertex Vn.\\\\\nWe associated with last lecture that every spanning arborescence corresponds to spregs .It turns out there are more spregs.There is some structure,Spregs looks like are of the form arborescence $+$ weekly connected component 1 $+$ weekly connected component 2......\\\\\nV can not be part of cycle.V can be vertex in component.Weekly connecetd component has exactly 1 cycle.\\\\\nSo what we want to count spregs in inner circle with distinguished vertex Vn.Ok so already know what we have to count.Basic strategy is as follows:\\\\\nStrategy is that count the number of spregs with distinguished vertex Vn which does not have any cycle.So this is what we want to count.We want to use Inclusion Exclusion.In fact we will not only count through Inclusion exclusion but we will go through each term of the determinant expression corresponds to our counting terms.\\\\\nLet's demonstrate by writing down Inclusion exclusion.\\\\\nIn order to define Inclusion Exclusion formulation we need to define the \\\\ \\\\\n1.Universe (Called as X).\\\\\n2.Component $A_i$.\\\\\nTo define that Let's consider,notice that we want to count spregs without any cycle.\\\\\nWe need to be set of all spregs with distinguished vertex $V_n$.\\\\\nLet $C_1,C_2,C_3, \\dots , C_n$ be the set of cycles in graph G.We looking at simple cycle.\\\\\n$A_i$ define as spregs which contain the cycle $C_1$.The number of spregs which does not contain any cycle.This what we want to count.So,\\\\\n\n$|X|$ $-$ $|\\bigcup_{i=1}^{n} A_i|$\nIt is same as,\n$$|X| - \\sum_{I \\subseteq [n] , I \\neq \\emptyset} (-1)^{|I|+1}|\\bigcap_{i \\in I} A_i|$$\nSo already applied Inclusion Exclusion,Now write it as\n$$ |X| + \\sum_{I \\subseteq [n] , I \\neq \\emptyset} (-1)^{|I|+1}|\\bigcap_{i \\in I} A_i| $$\n\nSo basically we need to count this.We can say that each term in these expression are exactly same as term used in determinant expression.As number of Spregs are equal to number of spanning arborescence,hence theorem is called.This is strategy.Hence each term is associated with terms of determinant theorem.\\\\\n\\textbf{Observation:}Let's consider C1,C2 be two cycles.\\\\\nWhat can say about $$ A1 \\bigcap A2 = \\emptyset iff C1\n \\bigcap C2 \\neq \\emptyset.$$\\\\\nEvery spregs looks like arborescence + uniquely weekly connected component 1 +uniquely weekly connected component 2.\\\\\n$$ If C1 \\bigcap C2 \\neq \\emptyset then |A1 \\bigcap A2| =0.$$\\\\\nUsing the observation,\\\\\nThe number of spregs with distinguished vertex Vn which is acyclic is nothing but \\\\\n$$ |X| + \\sum_{I \\subseteq [n] ,I \\neq \\emptyset ,for J,K \\in I, J \\neq K, cj \\bigcap ck =\\emptyset} (-1)^{|I|+1}|\\bigcap_{i \\in I} A_i| $$\n\n\\textbf{Recall:} We counted |X| =Total number of spregs\\\\\n$$det(L_G[n]) = \\sum_{\\sigma \\in S_n} Sign(\\sigma) \\prod_{i=1}^{n} a_{i\\sigma(i)}$$\\\\\nThis is determinant expression. term corresponds to $$ \\sigma=id $$\n$$\\prod_{i=1}^{n-1} (L_i) = \\prod_{i=1}^{n-1} deg_in(L_i) $$\n\nNow we count the singleton sized I.\nSpregs that contain exactly one cycle.\nFix,I={1},We are counting spregs which contain only C1.\nLet CI be (Vi1,Vi2,Vi3,....Vin)\n\n$$\\sigma(i1)=i2$$\n$$\\sigma(i2)=i3$$\n$$\\sigma(ik-1)=ik$$\n$$\\sigma(ik)=iI$$\n$$\\sigma \\in S_n-1$$\n$$\\forall other i \\in {1...n} $$\n$$ \\sigma(i)=i $$\n\nThis is permutation which contain one cycle.\nNow we just able to do following\n\n\\textbf{Claim:} Term in the determinant expression for\n$$LG[n] corresponds to \\sigma=number of spregs which contain C1$$\nRight now we are talking about one circle.\nLet's see proof of this.\nProof of this is very natural.\n\n$$sign(\\sigma)\\prod_{i=1}^{n-1} L_i,\\sigma(i $$\nsign of permutation is exactly equal to\n$(-1)^{(l-1)} $.Here length is l.\n\nSecond term is\n$$ (-1)^{(l-1)} (\\prod_i,|\\sigma(i)|=i L_i,i)(\\prod_i,|\\sigma(i)| \\neq i L_i,\\sigma(i)) $$\n$$ L_i,\\sigma(i) =-1,if V_i,V_\\sigma(i) is edge in graph \\in E\\\\ ootherwise $$\nIf corresponding edge is present then this term is non zero otherwise zero.\\\\\n$$ (-1)^{(l-1)}(\\prod_i,|\\sigma(i)| deg_in(v_i))*(-1)^{l} $$\n$$= (-1)^{(2l-1)} |number of spregs which contain C1| $$\n$$= -|number of spregs which contain C1 | $$\nProposition for $$ \\sigma \\in S_n-1 comsisting of single cycle. $$$$ \\sigma =(i1,i2,.....il) $$\nAssociate$ C\\sigma $ as a cycle corresponds to vertex sequence (Vi1,Vi2,Vi3.....Vil,Vii)\\\\\nThen the term in the determinant corresponds to $\\sigma$ satisfies following.\\\\\n$$sgn(\\sigma)(\\prod_{i=1}^{n-1}L_i,\\sigma(i))$$\n\n$$-(\\prod_{i,\\sigma(i)=i}deg_in(v_i))C_\\sigma \\subseteq G(V,E) $$\\\\\n$$ 0 if C\\sigma \\nsubseteq G(V,E) $$ \\\\\n$$-1 if C\\sigma \\subseteq G(V,E) if \\forall i \\sigma(i) \\neq i $$\\\\\n\n\n\\textbf{Corollary:}$$For \\sigma,C\\sigma as above $$\\\\\n$$ |\\prod_{i=1}^{n-1}L_i,\\sigma(i)|=number of spregs which contain C_\\sigma $$\\\\\nNow this is case for single cycle.\\\\Now we will generalise case for multiple cycle.\\\\\nSo it is very natural.\n\n\\textbf{Question:}Suppose $\\sigma \\in S_{n-1}$ is the product of $k>0$ disjoint cycle.\n$$\\sigma=(i11,i12,i13.....iil) (i21,i22,i23....(ikl,ik2,ik3...ikl) $$\nk is number of cycles. So now we are associate\n$$ C_\\sigma = \\cup {j=1}^{k}C_j where C_j is (V_ij1....V_ijLj.....VijI)$$\n\nThen the term corresponds to $\\sigma$ in $det(L_G[n])$\n$$sgn(\\sigma) = (-1)^{K} \\prod_{i|\\sigma(i)|}deg_in(V_i) if C_\\sigma \\subseteq G $$\\\\\n$$ o if C\\sigma \\nsubseteq G $$\n\n\\textbf{Corollary:}If we look at\\\\\n$$|\\prod_{i=1}^{n}L_i,\\sigma(i)|=number of spregs which contain C_\\sigma = \\bigcup_{j=1}^{k}C_j $$\\\\\n$$ det(L_G[n]) =\\sum_{\\sigma \\in S_n-1}sig(\\sigma) \\prod_{i=1}^{n} L_i\\sigma $$\\\\\nThis is how determinant expression looks like.\\\\\nNumber of spanning arborescence rooted at Vn as distinguished vertex and not containing cycle.\\\\\n%$$&=|X| + \\sum_{I \\subseteq [n] ,I \\neq \\emptyset, cj \\bigcap ck=\\emptyset ,j,k \\in I}(-1)^I $$\\\\\nThis is strategy.\nThis lecture is combinatorial application of PIE.\n\n\n\\Lecture{Jayalal Sarma}{Oct 10, 2020}{12}{Algorithmic Application of PIE}{Shrinidhi Bajpayee}{$\\alpha$}{JS}\n\n\\section{Introduction} We talk about counting problem.Its about given a $n*n$ bipartite graph,We want to count the number of perfect matching in it?\\\\\nsubset of edges such that every vertex has exactly one edge incident on it from the\nsubset.\\\\\nBipartite adjacency matrix is different from adjacency matrix.\\\\ \\\\\n\\textbf{Trivial Algorithm:}Run through all n sized subsets of E.\\\\\n$${n^{2} \\choose n } \\sim {n^{2} \\choose n}^{n} \\sim n^{n} \\sim 2^{nlogn}$$\\\\\n\n$$ {n \\choose k}^{k} \\leq {n \\choose k} \\leq {n_e \\choose k}^{k} $$\\\\\ne=natural log base \\\\\n\\section{Decision Problem:}\n\\textbf{Binpacking Problem:}\nGiven a positive integer (bin capacity B),positive integer k\\\\\nWe have n item with weight S1,S2,S3.....,Sn\\\\\npartition the items into u1,u2,u3.....,uk\\\\\nCapacity is such that sum of weights of items in each ui is almost B.\\\\\nThis is binpacking problem.\\\\\nIt is NP-problem.\\\\\nBy PIE Time complexity is $O(nB2^{n})$\\\\\nn=number of items\\\\\nB=capacity\\\\\nNow we will reformulate PIE.\\\\\n\nGiven a collection of N combinatorial objects \\\\\nLet p(1),p(2),p(3)...p(n) be properties.\\\\\n$p_i$ is essential function \\\\\n$$ N \\to {0,1} $$\\\\\nIt is possible that some object doesn't have properties of that.\\\\\n$$N_i \\to Number of objects among N with property P(i).$$\\\\\n$$ {i_1,i_2,....i_r} \\subseteq {1,2,3,...N} $$\\\\\n$$N_{i_1,i_2,i_3....i_r} = Number of objects with properties p(i_1) p(i_2)....p(i_r) $$\nN(0)=Number of objects not having any of the property.\\\\\n$$ N(0)=N-\\sum_{i=1}^{n} N_{i_1} +\\sum_{i_1a\\end{array}\\right.\\\\\n\t\\textbf{b)}\\hspace{2cm}\\quad J&=k s\\\\\n\tI_{\\text {Total }}&=\\int_{0}^{a} J d a=\\int_{0}^{a} k s 2 \\pi s d s=\\frac{2 \\pi k a^{3}}{3}\\\\\n\tk&=\\frac{3 I}{2 \\pi a^{3}}\\\\\n\tI_{\\text {encl }}&=\\int_{0}^{S} k s(2 \\pi s) d s\\\\\n\t&=\\frac{2 \\pi k s^{3}}{3}\\\\\n\t\\text{Substituting value of $k$.}\\\\\n\tI_{\\text{encl}} &=\\frac{I s^{3}}{a^{3}}\\hspace{1.4cm}\\text{ for } sa\\\\\n\t\\therefore \\quad B&=\\left\\{\\begin{array}{l}\\frac{\\mu_{0} I s^{2}}{2 \\pi a^{3}} \\hat{\\phi}\\ \\quad \\text { for } sa\\end{array}\\right.\n\t\\end{align*}\n\\end{answer}\n\\section{Magnetic Vector Potential}\nIn electrostatics, we had seen a scalar function $\\phi$ called the \"potential\"\nwhose negative gradient is equal to the electric field : \n\\begin{equation*}\n\\vec{E}=-\\nabla \\phi\n\\end{equation*}\n The existence of such a scalar function is a consequence of the conservative nature of the electric force. It also followed that the electric field is irrotational, i.e.\n \\begin{equation*}\n\\nabla \\times \\vec{E}=0\n \\end{equation*}\nFor a magnetic field, Ampere's law gives a non-zero curl,\n\\begin{equation*}\n\\nabla\\times \\vec{B}=\\mu_{0} \\vec{J}\n\\end{equation*}\n\nSince the curl of a gradient is always zero, we cannot express $\\vec{B}$ as a gradient of a scalar function as it would then violate Ampere's law. However, we may introduce a vector function $\\vec{A}(\\vec{r})$ such that,\n\n$$\n\\vec{B}=\\nabla \\times \\vec{A}\n$$\nThis would automatically satisfy $\\nabla \\cdot \\vec{B}=0$ since divergence of a curl is zero.\\ $ \\vec{A} $\\ is known as the vector potential.\n\\begin{center}\n\t\\framebox{\n\t\t\\parbox[t][0.75cm]{4cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering \n\t\t\t\n\t\t\t$\n\t\t\t\\vec{B}=\\nabla \\times \\vec{A}\n\t\t\t$} }\n\\end{center}\n\\subsection{Biot-Savart's law for Vector potential.}\nBiot-Savart's law for magnetic field due to a current element $\\vec{d l}$\\ is obtained as,\n\\begin{align*}\nd \\vec{B}&=\\frac{\\mu_{0} I}{4 \\pi} \\frac{\\vec{d} l \\times \\hat{r}}{r^{2}}=-\\frac{\\mu_{0} I}{4 \\pi} \\vec{d} l \\times \\nabla\\left(\\frac{1}{r}\\right)\n\\intertext{It can be used to obtain an expression for the vector potential. Since the element $\\overrightarrow{d l}$ does not depend on the position vector of the point at which the magnetic field is calculated, we can write}\nd \\vec{B}&=\\frac{\\mu_{0} I}{4 \\pi} \\nabla \\times\\left(\\frac{\\vec{d l}}{r}\\right)\n\\intertext{The change in sign is because}\n\\nabla\\left(\\frac{d l}{r}\\right)&=\\nabla(1 / r) \\times \\vec{d l}\n\\intertext{Thus the contribution to the vector potential from the element $\\overrightarrow{d l}$ is, }\n d \\vec{A}&=\\frac{\\mu_{0} I}{4 \\pi r} \\overrightarrow{d l}\n \\intertext{The expression is to be integrated over the path of the current to get the vector potential for the system}\n \\vec{A}&=\\frac{\\mu_{0} I}{4 \\pi} \\int \\frac{\\vec{d} l}{r}\n\\end{align*}\n\\begin{center}\n\t\\framebox{\n\t\t\n\t\t\\parbox[t][1.0cm]{4cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering \n\t\t\t$\\vec{A}=\\frac{\\mu_{0} I}{4 \\pi} \\int \\frac{\\vec{d} l}{r}$} \n\t}\n\\end{center}\n\\begin{exercise}\n\tA current distribution gives rise to the magnetic vector potential\n\t$$\n\t\\vec{A}(x, y, z)=x^{2} y \\hat{i}+y^{2} x \\hat{j}-x y z \\hat{k} \n\t$$\n\tFind the corresponding magnetic field $\\vec{B}$ at $(-1,2,5)$.\n\\end{exercise}\n\\begin{answer}\n\t\\begin{align*}\n\t\\vec{B}(x, y, z)&=\\vec{\\nabla} \\times \\vec{A}\\\\&=\\left|\\begin{array}{ccc}\n\t\\hat{i} & \\hat{j} & \\hat{k} \\\\\n\t\\frac{\\partial}{\\partial x} & \\frac{\\partial}{\\partial y} & \\frac{\\partial}{\\partial z} \\\\\n\tx^{2} y & y^{2} x & -x y z\n\t\\end{array}\\right|\\\\&=-x z \\hat{i}+y z \\hat{j}+\\left(y^{2}-x^{2}\\right) \\hat{k}\\\\\n\t\\vec{B}(-1,2,5)&=5 \\hat{i}+10 \\hat{j}+3 \\hat{k} T\n\t\\end{align*}\n\\end{answer}\n\\subsubsection{Vector potential due to a straight current carying wire.}\nLet us consider a long straight wire $A B$ carrying a steady current $I$. We need to calculate the magnetic vector potential $\\vec{\\mathrm{A}}$ at any point $P .$ We choose cylindrical coordinates $(r, \\theta, z)$ with z-axis along the wire and the foot $O$ of the perpendicular from $P$ on the wire as the origin. Let us consider an element $d \\vec{l}=\\hat{z} d z$ at a distance $z$ from $O$. Now by definition\\\\\n\\begin{minipage}{0.65\\textwidth}\n\\begin{align*}\n\\vec{A}&=\\frac{\\mu_{0}}{4 \\pi} \\int \\frac{I d \\vec{l}}{R}=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi} \\int_{-l_{1}}^{+l_{2}} \\frac{d z}{\\sqrt{r^{2}+z^{2}}}\\\\&=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi}\\left[\\ln \\left(z+\\sqrt{r^{2}+z^{2}}\\right)\\right]_{-l_{1}}^{+l_{2}}\\\\\n&=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi} l n\\left[\\frac{l_{2}+\\sqrt{r^{2}+l_{2}^{2}}}{-l_{1}+\\sqrt{r^{2}+l_{1}^{2}}}\\right]\n\\end{align*}\n\\end{minipage}\n\\begin{minipage}{0.35\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=5cm,width=6cm]{vector potential}\n\t\t\\caption{Straight current carying wire.}\n\t\t\\label{Straight current carying wire.}\n\t\\end{figure}\n\\end{minipage}\n\\begin{align*}\n\\intertext{If the wire is of length $2 \\mathrm{~L}$ and the point $\\mathrm{P}$ is just above the centre of the wire then $l_{1}=l_{2}=L$ and we can write,}\n\\vec{A}&=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi} \\ln \\left[\\frac{L+\\sqrt{r^{2}+L^{2}}}{-L+\\sqrt{r^{2}+L^{2}}}\\right]\n\\intertext{Now if $r / L<<1$ then we can approximate,}\n\\vec{A}&=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi} \\ln \\left[\\frac{1+\\left(1+r^{2} / L^{2}\\right)^{1 / 2}}{-1+\\left(1+r^{2} / L^{2}\\right)^{1 / 2}}\\right]\\\\\n&\\approx \\hat{z} \\frac{\\mu_{0} I}{4 \\pi} \\ln \\left[\\frac{2+\\frac{r^{2}}{2 L^{2}}}{\\frac{r^2}{2L^2}}\\right]\\\\\n&=\\hat{z} \\frac{\\mu_{0} I}{4 \\pi} \\ln \\left(\\frac{4 L^{2}}{r^{2}}+1\\right) \\approx \\hat{z} \\frac{\\mu_{0} I}{2 \\pi} \\ln \\left(\\frac{2 L}{r}\\right)\n\\end{align*}\n\n\\subsubsection{Vector potential due to a long solenoid}\nConsider an infinite solenoid with $n$ turns per unit length, radius $a$ and carrying current $I .$ Symmetry of the problem suggests that $\\vec{A}$ should have the direction of the current, i.e. the circumferential direction $(\\hat{\\theta})$. The magnetic field inside is along the axis (z-axis) of the solenoid and is given by,\n\\begin{align*}\n\\vec{B}&=\\mu_{0} n I \\hat{z}\n\\intertext{Now, considering a circular Amperian loop of radius $r(ra)$ outside the solenoid,}\n\\int_{S} \\vec{B} \\cdot d \\vec{S}&=\\mu_{0} n I \\cdot \\pi a^{2}\n\\intertext{Since the field extends only up to $r=a$. Thus,}\nA \\cdot 2 \\pi r&=\\mu_{0} n I \\cdot \\pi a^{2} \\\\\\text { or }\\quad \\vec{A}&=\\frac{\\mu_{0} n I a^{2}}{2 r} \\hat{\\theta} \\text { for } r>a\n\\end{align*}\n\\section{Multipole expansion of a vector potential}\nMultipole expansion is a technique used to find the potential of localized charge distribution in the form of power series in $\\frac{1}{r}$ where r is the distace to the point in question; if r is sufficiently large the series will be dominated by the lowest nonvanishing contributions and the higher term can be ignored.\\\\\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=4cm,width=5cm]{diagram-20211216(1)-crop}\n\t\\caption{}\n\t\\label{}\n\\end{figure}\n$$\\frac{1}{r}=\\frac{1}{\\sqrt{r^{2}+\\left(r^{\\prime}\\right)^{2}-2 r r^{\\prime} \\cos \\theta^{\\prime}}}=\\frac{1}{r} \\sum_{n=0}^{\\infty}\\left(\\frac{r^{\\prime}}{r}\\right)^{n} P_{n}\\left(\\cos \\theta^{\\prime}\\right)$$\n Accordingly, the vector potential of a current loop can be written\n $$\n \\mathbf{A}(\\mathbf{r})=\\frac{\\mu_{0} I}{4 \\pi} \\oint \\frac{1}{r} d \\mathbf{l}^{\\prime}=\\frac{\\mu_{0} I}{4 \\pi} \\sum_{n=0}^{\\infty} \\frac{1}{r^{n+1}} \\oint\\left(r^{\\prime}\\right)^{n} P_{n}\\left(\\cos \\theta^{\\prime}\\right) d \\mathbf{l}^{\\prime},\n $$\n or, more explicitly:\n $$\n \\begin{aligned}\n \\mathbf{A}(\\mathbf{r})=& \\frac{\\mu_{0} I}{4 \\pi}\\left[\\frac{1}{r} \\oint d \\mathbf{l}^{\\prime}+\\frac{1}{r^{2}} \\oint r^{\\prime} \\cos \\theta^{\\prime} d \\mathbf{l}^{\\prime}\\right.\\\\\n &\\left.+\\frac{1}{r^{3}} \\oint\\left(r^{\\prime}\\right)^{2}\\left(\\frac{3}{2} \\cos ^{2} \\theta^{\\prime}-\\frac{1}{2}\\right) \\dot{d}^{\\prime}+\\cdots\\right]\n \\end{aligned}\n $$\n First term represents is called monopole term which doesnt exists.$\\left( \\oint dl=0\\right) $\\\\\n Second term represents dipole and third term represents quadrapole term.\\\\\n The main contribution from dipole term can be written as \n$$\\mathbf{A}_{\\mathrm{dip}}(\\mathbf{r})=\\frac{\\mu_{0} I}{4 \\pi r^{2}} \\oint r^{\\prime} \\cos \\theta^{\\prime} d \\mathbf{l}^{\\prime}=\\frac{\\mu_{0} I}{4 \\pi r^{2}} \\oint\\left(\\hat{\\mathbf{r}} \\cdot \\mathbf{r}^{\\prime}\\right) d \\mathbf{l}^{\\prime}$$ \nThe integral can be written as \\\\\n$$\\oint\\left(\\hat{\\mathbf{r}} \\cdot \\mathbf{r}^{\\prime}\\right) d \\mathbf{l}^{\\prime}=-\\hat{\\mathbf{r}} \\times \\int d \\mathbf{a}^{\\prime}$$\nThen \\\\\n$$\\mathbf{A}_{\\mathrm{dip}}(\\mathbf{r})=\\frac{\\mu_{0}}{4 \\pi} \\frac{\\mathbf{m} \\times \\hat{\\mathbf{r}}}{r^{2}}$$\nwhere $\\mathbf{m}$ is the magnetic dipole moment:\n$$\n\\mathbf{m} \\equiv I \\int d \\mathbf{a}=I \\mathbf{a} .\n$$\n\\section{Magnetic Dipole}\nIn electrostatics, when two opposite charge placed a small distance apart, it called an electric dipole. Similarly, when a wire carrying a current forms a small closed loop, then it is called magnetic dipole.\\\\\nThe magnetic field due to a magnetic dipole is given bellow.Where $B_r$ is the magnetic field at the axial point and $B_{\\theta}$ be the magnetic field at point which makes an angle $\\theta$ with the axis.\n$$\\mathbf{B}=\\left\\{\\begin{array}{l}\n\tB_{r}=2|\\boldsymbol{m}| \\frac{\\mu_{0}}{4 \\pi} \\frac{\\cos \\theta}{R^{3}} \\\\\\\\\n\tB_{\\theta}=|\\boldsymbol{m}| \\frac{\\mu_{0}}{4 \\pi} \\frac{\\sin \\theta}{R^{3}}\n\\end{array}\\right.\\hspace{2cm}\\text { Where } \\boldsymbol{m}=i \\mathbf{A} \\text { is the magnetic dipole moment of the loop }$$\nHere $i$ is the current in the loop, $A$ is the loop area, $R$ is the radial distance from the center of the loop. The field is equivalent to that from a tiny bar magnet (a magnetic dipole).\\\\\nWe define \\textbf{the magnetic dipole moment to be a vector pointing out of the plane of the current loop and with a magnitude equal to the product of the current and loop area:}\nThe area vector, and thus the direction of the magnetic dipole moment, is given by a right-hand rule using the direction of the currents.\n\\begin{center}\n\t\\framebox{\n\t\t\n\t\t\\parbox[t][1.0cm]{4cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering \n\t\t\tMagnetic dipole moment\\\\\n\t\t\t$m=i A$} \n\t}\n\\end{center}\n\\begin{note}\n\tThe magnetic field of a dipole can be written in coordinate free form\\\\\n\\begin{center}\n\t\\framebox{\n\t\t\\parbox[t][1.5cm]{3.5cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering\n\t\t\t\n\t\t\t\\begin{align*}\n\t\t\t\\begin{array}{lll}\n\t\t\t\tB_{dipole}(r)=\\frac{\\mu_{0}}{4 \\pi }\\frac{1}{r^3}\\left[ 3(m\\cdot\\hat{r})\\hat{r}-m\\right] \n\t\t\t\\end{array}\n\t\t\t\\end{align*}} }\n\\end{center}\n\\end{note}\n\\subsection{Interaction of Magnetic Dipoles in External Fields }\n\\subsubsection{Torque}\n\\begin{minipage}{0.65\\textwidth}\n\tBy the $\\mathbf{F}=i\\ l \\times \\mathbf{B}_{\\text {ext }}$ force law, we know that a current loop (and thus a magnetic dipole) feels a torque when placed in an external magnetic field:\n\t$$\n\t\\boldsymbol{\\tau}=\\boldsymbol{m} \\times \\mathbf{B}_{\\text {ext }}\n\t$$\n\tThe direction of the torque is to line up the dipole moment with the magnetic field.\n\\end{minipage}\n\\begin{minipage}{0.35\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{magnetic dipole1}\n\t\t\\caption{Torque on magnetic dipole}\n\t\t\\label{magnetic dipole1}\n\t\\end{figure}\n\\end{minipage}\n\\subsubsection{Potential Energy}\nThe above equation is analogous to $\\vec{\\tau}=\\overrightarrow{\\mathbf{p}} \\times \\overrightarrow{\\mathbf{E}}$ , the torque exerted on an electric dipole moment $\\overrightarrow{\\mathbf{p}}$ in the presence of an electric field $\\overrightarrow{\\mathbf{E}}$. Recalling that the potential energy for an electric dipole is $U=-\\overrightarrow{\\mathbf{p}} \\cdot \\overrightarrow{\\mathbf{E}}$, a similar form is expected for the magnetic case. The work done by an external agent to rotate the magnetic dipole from an angle $\\theta_{0}$ to $\\theta$ is given by\nLet us calculate the work done by the magnetic field when aligning the dipole. Let $\\theta$ be the angle between the magnetic dipole direction and the external field direction.\\\\\n\\begin{align*}\n\tW_{\\mathrm{ext}} &=\\int_{\\theta_{0}}^{\\theta} \\tau d \\theta^{\\prime}=\\int_{\\theta_{0}}^{\\theta}\\left(\\mu B \\sin \\theta^{\\prime}\\right) d \\theta^{\\prime}=\\mu B\\left(\\cos \\theta_{0}-\\cos \\theta\\right) \\\\\n\t&=\\Delta U=U-U_{0}\n\\end{align*}\nOnce again, $W_{\\text {ext }}=-W$, where $W$ is the work done by the magnetic field. Choosing $U_{0}=0$ at $\\theta_{0}=\\pi / 2$, the dipole in the presence of an external field then has a potential energy of\n$$\nU=-\\mu B \\cos \\theta=-\\overrightarrow{\\boldsymbol{\\mu}} \\cdot \\overrightarrow{\\mathbf{B}}\n$$\nThe configuration is at a stable equilibrium when $\\overrightarrow{\\boldsymbol{\\mu}}$ is aligned parallel to $\\overrightarrow{\\mathbf{B}}$, making $U$ a minimum with $U_{\\min }=-\\mu B$. On the other hand, when $\\overrightarrow{\\boldsymbol{\\mu}}$ and $\\overrightarrow{\\mathbf{B}}$ are anti-parallel, $U_{\\max }=+\\mu \\mathrm{B}$ is a maximum and the system is unstable.\n\\begin{center}\n\t\\framebox{\n\t\t\n\t\t\\parbox[t][3.0cm]{10cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering \n\t\t\t\\begin{alignat*}{2}\n\t\t\t\t&\\text{Magnetic Moment of Current Carrying Wire:} &&\\overrightarrow{\\boldsymbol{\\mu}}=I \\overrightarrow{\\mathbf{A}} \\\\&\\text{Torque on Magnetic Moment:}\n\t\t\t&&\\overrightarrow{\\mathbf{\\tau}}=\\overrightarrow{\\boldsymbol{\\mu}} \\times \\overrightarrow{\\mathbf{B}}\n\t\t\t\\\\&\\text{Energy of Moment in External Field:}\n\t\t\t&&U=-\\overrightarrow{\\boldsymbol{\\mu}} \\cdot \\overrightarrow{\\mathbf{B}}\n\t\t\t\\end{alignat*}\n\t\t \n\t}}\n\\end{center}\n\\subsection{The interaction energy of two magnetic dipoles}\nThe interaction energy of two magnetic dipoles seperated by a displacement r is given by\\\\\n\\begin{center}\n\t\\framebox{\n\t\t\\parbox[t][1.3cm]{3.5cm}{\n\t\t\t\n\t\t\t\\addvspace{0.2cm} \\centering\n\t\t\t\n\t\t\t\\begin{align*}\n\t\t\t\\begin{array}{lll}\n\t\t\tU=\\frac{\\mu_{0}}{4\\pi}\\frac{1}{r^3}\\left[ m_1 \\cdot m_2-3(m\\cdot \\hat{r})(m_2 \\cdot \\hat{r})\\right] \n\t\t\t\\end{array}\n\t\t\t\\end{align*}} }\n\\end{center} \n\\subsection{Torque On Current Loop}\n\t\\begin{figure}[H]\n\t\\begin{center}\n\t\t\\includegraphics[width=0.35\\textwidth]{torque}\n\t\\end{center}\n\\end{figure}\nLet us now consider the case when the magnetic field $B$ is in the plane with the rectangular loop. No force is exerted by the field on the arms of the loop that is parallel to the magnets, but the arms perpendicular to the magnets experience a force given by $F_{1}$\n$$\nF_{1}=I b B\n$$\nThis force is directed into the plane.\nSimilarly, we can write the expression for a force $\\mathrm{F}_{2}$ which is exerted on the arm CD,\n$$\nF_{2}=I b B=F_{1}\n$$\nWe see that the net force on the loop is zero and the torque on the loop is given by,\n$$\n\\begin{gathered}\n\\tau=F_{1} \\frac{a}{2}+F_{2} \\frac{a}{2} \\\\\n\\tau=I b B \\frac{a}{2}+I b B \\frac{a}{2}=I(a b) B=I A B\n\\end{gathered}\n$$\nWhere ab is the area of the rectangle. Here, the torque tends to rotate the loop in the anti-clockwise direction. Let us consider the case when the plane of the loop is not along the magnetic field. Let the angle between the field and the normal to the coil be given by $\\theta$. We can see that the forces on the arms BC and DA will always act opposite to each other and will be equal in magnitude. Since these forces are the equal opposite and collinear at all points, they cancel out each other's effect and this results in zero-force or torque. The forces on the arms $A B$ and $\\mathrm{CD}$ are given by $\\mathrm{F}_{1}$ and $\\mathrm{F}_{2}$. These forces are equal in magnitude and opposite in direction and can be given by,\n$$\nF_{1}=F_{2}=I b B\n$$\nThese forces are not collinear and thus act as a couple exerting a torque on the coil. The magnitude of the torque can be given by,\n$$\n\\begin{gathered}\n\\tau=F_{1} \\frac{a}{2} \\sin \\theta+F_{2} \\frac{a}{2} \\sin \\theta \\\\\n\\tau=I a b B \\sin \\theta \\\\\n\\tau=I A B \\sin \\theta\n\\end{gathered}\n$$\n\\section{Motion of charged particle in Electric and magnetic Field}\n\\subsection{Motion of charged particle in uniform magnetic field}\n\\textbf{(a)}\\quad \\textbf{If the particle enters $\\perp^r$ to the field\\quad(Circular)}\\\\\\\\\nWhen the particle of charge $\\theta$\\ enters $\\perp^r$\\ to the field, if under go circular motion and the centripetal acceleration is provided by magnetic force $qVB$. If it moves in a circular path of radius $R$,\\\\\n\\begin{minipage}{0.65\\textwidth}\n\t\\begin{align*}\n\tQVB&=\\frac{mv^2}{R}\\quad \\Rrightarrow\\quad\\text{ Cyclotron formula.}\\\\\n\tR&=\\frac{mv}{QB},\\quad m \\ \\text{Is the man of the particle}\n\t\\intertext{When $R$ is the cyclotron radius.}\n\t\\text{Momentum}\\ P&=mv=\\frac{m\\times QBR}{m}\\\\\n\t&=QBR\\\\\n\t\\text{Kinetic energy}\\ KE&=\\frac{P^2}{2m}=\\frac{\\theta^2B^2R^2}{2m}\\\\\n\t\\text{Time period }\\ T&=\\frac{2\\pi R}{V}=\\frac{2\\pi m}{\\theta B}\n\t\\end{align*}\n\\end{minipage}\n\t\\begin{minipage}{0.35\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=4.5cm]{diagram-20210427(4)-crop}\n\t\t\\caption{Circular path of the particle entering perpendicular to the magnetic field}\n\t\t\\label{Cyclotron}\n\t\\end{figure}\t\t\t\t\n\\end{minipage}\n\\textbf{(b)}\\quad \\textbf{The Ratio of Various Quantities}\\\\\n\\begin{enumerate}\n\t\\item $\\frac{r_1}{r_{2}}=\\frac{q_{2}}{q_{1}} \\sqrt{\\frac{m_{1}}{m_{2}}}$, when $E_{K}$ and $B$ are constants.\n\\item $\\frac{r_{1}}{r_{2}}=\\frac{q_{2}}{q_{1}} \\sqrt{\\frac{m_{1} E_{K_{1}}}{m_{2} E_{K_{2}}}}$, when $B$ is constant.\n\\item $\\frac{p_{1}}{p_{2}}=\\frac{q_{1}}{q_{2}}\\left(\\frac{r_{1}}{r_{2}}\\right)$\n\\item $\\frac{\\omega_{1}}{\\omega_{2}}=\\frac{q_{1} m_{2}}{q_{2} m_{1}}=\\frac{v_{1}}{v_{2}}$\n\\item $\\frac{T_{1}}{T_{2}}=\\frac{q_{2}}{q_{1}} \\times \\frac{m_{1}}{m_{2}}$\\\\\n\\item $\\frac{E_{K_{l}}}{E_{K_{2}}}=\\frac{q_{1}^{2} r_{1}^{2}}{q_{2}^{2} r_{2}^{2}} \\times \\frac{m_{2}}{m_{1}}$\n\\end{enumerate}\n\\textbf{(c)}\\quad \\textbf{If the particle enters the field with an angle making $\\theta$with it helical motion}\\\\\\\\\nWhen it enters with an angle $\\theta$ with the field $n$ has two component,\\quad $V_\\parallel=V \\cos\\theta$\\quad and \\quad$V_\\perp^r=V\\sin\\theta$\\\\No magnetic force along parallel direction here again\\\\\n\\begin{minipage}{0.65\\textwidth}\n\t\\begin{align*}\n\tR&=\\frac{MV_\\perp r}{QB} =\\frac{MV \\sin\\theta}{QB}\\\\\n\tT&=\\frac{2\\pi m}{QB}\\quad \\text{has no change}\n\t\\intertext{Due to parallel component of velocity, particle undergo helical motion. Horizontal distance travelled by the particle during one period of revolution is called Pitch,}\n\tD&=T\\times V cos \\theta\\\\\n\tb&=\\frac{2\\pi m}{QB} V \\cos \\theta \n\t\\end{align*}\n\\end{minipage}\n\\begin{minipage}{0.35\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=3.5cm]{diagram-20210427(6)-crop}\n\t\t\\caption{Helical path of the particle entering at an angle to the magnetic field}\n\t\t\\label{Helical path of the particle }\n\t\\end{figure}\n\\end{minipage}\n\\subsection{Motion of charged particle in static electric field.}\n\\textbf{(a)}\\quad Charged particle enters in the direction of field (Linear motion)\n\\begin{align*}\n\\intertext{If $q$ charged moves in uniform field then,}\nF&=q E \\hspace{2cm} F=\\frac{md^2r}{dt^2}\\\\\n\\frac{md^2r}{dt^2}&=qE\\\\\n\\frac{dr}{dt}&=\\frac{qEt}{m}+C_0\\\\\n\\text{At} \\quad t=0\\quad V&=V_0,\\quad \\text{and}\\quad r=r_0,\\\\\n\\text{Then}\\quad c_0&=V_0\\\\\n\\frac{d_0}{dt}&=\\frac{qE}{m}t+V_0\\\\\nr&=\\frac{qEt^2}{2m}+V_0t+r_0\n\\end{align*} \n \\begin{minipage}{0.65\\textwidth}\n \t\\begin{align*}\n \t\\intertext{It initial position and velocity are zeros,}\n \t\\text{Then}\\quad V&=\\frac{Q E}{m}t\\\\\n \tr&=\\frac{Q E}{2m}t^2\n \\end{align*} \t\n \t\\end{minipage}\n \\begin{minipage}{0.35\\textwidth}\n \t\\begin{figure}[H]\n \t\t\\includegraphics[height=3cm,width=4cm]{diagram-20210427(11)-crop}\n \t\t\\caption{Linear motion of charged particle in Static electric field.}\n \t\t\\label{Linear motion of charged particle}\n \t\\end{figure}\n \\end{minipage}\n\\begin{align*}\n\\intertext{Energy acquired by the charged particle moving from initial to some final point,}\nw&=\\int f\\cdot dl=m\\int_{V_1}^{V_2}a\\cdot dl\\\\\n&=m\\int_{V_1}^{V_2}\\frac{dV}{dt}\\times Vdt=m\\int_{1}^{2}VdV=\\frac{1}{2}m(V^2_2-V^2_1)\\\\\nw&=\\frac{1}{2}m(V^2_2-V^2_1)\\\\\n\\text{If initial velocity }V_1&=0\\\\\nw&=\\frac{1}{2}mV^2\n\\intertext{If the potential difference between them is $V$ then,}\nw&=QV=\\frac{1}{2}mV^2\\\\\n\\therefore V&=\\sqrt{\\frac{2QV}{m}}\\\\\nK.E&=w=\\frac{1}{2}m\\times\\frac{Q^2 E^2}{m^2}t^2\\\\\n&=QE\\times\\frac{QE}{2m}t^2\\\\\n&=QEr\n\\end{align*}\n\\textbf{(b)}\\quad \\textbf{Charged particle enters in the direction $\\perp^r$ to the electric field (Parabolic motion)}\\\\\n\n \\begin{minipage}{0.55\\textwidth}\nLet us consider a charge particle enters in a electric field region with velocity $V_x$ at\\ $t=0$. The electric field is in $y$ direction and the field region has length $l$. After traversing a distance $l$ it strikes a point $P$ on a screen which is placed at a distance $L$ from the field region. \\\\\nSince electric field is in the $y$ direction , charged particle will experience force.\n\\begin{align*}\nF_y&=QEy\\\\ \n\\intertext{Acceleration}\na_y&=\\frac{QEy}{m}\\\\\n\\end{align*}\n\\end{minipage}\n\\begin{minipage}{0.45\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4.5cm,width=6cm]{diagram-20210427(5)-crop}\n\t\t\\caption{Parabolic motion of Charged particle entering in perpendicular direction to the electric field.}\n\t\t\\label{rc current discharge}\n\t\\end{figure}\n\\end{minipage}\\\\\\\\\t\nIn time $t$ , charge particle will traverse a distance $y=\\frac{1}{2}a_yt^2$ in $y$ direction and a distance $x=V_xt$\\ in$x$ direction.\n\\begin{align*}\n\\text{Thus}\\quad y&=\\frac{1}{2}a_yt^2=\\frac{QEy}{2m}(\\frac{x}{V_x})^2 \\quad\\text{represents a parabola}\\\\\ny_1&=\\frac{QEy}{2m}(\\frac{l}{V_x})^2\\text{and}\\quad y_2=L \\tan\\theta\\\\\n\\intertext{Thus distance of point $P$ from the center of the screen is}\ny_1+y_2&=\\frac{QEy}{2m}(\\frac{l}{V_x})^2+L \\tan\\theta\\\\\n\\text{Angle of deviation in the field region}\\\\\n\\tan\\phi&=\\frac{dy}{dx}=\\frac{QEy}{mV_x^2}x\\\\\n\\text{Angle of deviation in the field foce region}\\\\\n\\tan\\theta&=\\frac{QEy}{mV_x^2}l\n\\end{align*}\n\\subsection{Charged particle in uniform electric and magnetic field }\n\\textbf{(a)}\\quad \\textbf{$E$ and $B$ are perpendicular(Cycloid motion).}\\\\\n\\begin{minipage}{0.65\\textwidth}\n\t\tIf $B$ points in $\\hat{x}$ direction $E$ in the $z$ direction. Suppose the particle is at origin initially. So $F_{mg} =0$, due $E$ it moves along $\\hat{z} $ direction, When it starts moving magnetic force develop, it pulls the charge to right, when $V$ increases $F_{mg} $ increases ,it tend to move in a circular path,which results the particle move back towards $Y$ axis at that time it is moving against $\\vec{E}$. Therefore the velocity $V$ decreases, and $F_{mag} $ also decreases. Then $E$ bring the charge to rest at point $a$. Then the entire process repeats.\n\\end{minipage}\n\\begin{minipage}{0.35\\textwidth}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20210427(8)-crop}\n\t\t\\caption{Cycloid motion of charge}\n\t\t\\label{cycloid }\n\t\\end{figure}\n\\end{minipage}\n\\begin{align*}\n\t\\intertext{We know particle is at rest at origin when time $t=0$ ,Then,} \ny(0)&=z(0)=0\\\\\n\\dot{y}(0)&=\\dot{z}(0)=0\\hspace{1cm}\\text{No motion along $x$ axis}\\\\\nV&=(0,\\dot{y},\\dot{z})\\\\\nV\\times B&=\\left|\\begin{array}{lll}\\hat{x} & \\hat{y}&\\hat{z} \\\\ 0 & \\dot{y}&\\dot{z} \\\\B & 0&0\\end{array}\\right|=B\\dot{z}\\hat{y}-B\\dot{y}\\hat{z}\\\\\n\\text{Total force}\\ F&=Q[E+V\\times B]=Q[E\\hat{z}+B\\dot{z}\\hat{y}-B\\dot{y}\\hat{z}]\\\\\nma&=m[\\dot{y}\\hat{y}+\\dot{z}\\hat{z}]\\\\\nm[\\dot{y}\\hat{y}+\\dot{z}\\hat{z}]&=Q[E\\hat{z}+B\\dot{z}\\hat{y}-B\\dot{y}\\hat{z}]\n\\intertext{Comparing the coefficient $ \\hat{y} $ and $ \\hat{z} $ }\nQB\\dot{z}&=m\\dot{y}\\hspace{2cm}E-QB\\dot{y}=m\\dot{z}\\\\\n\\text{let}\\quad w&=\\frac{QB}{m},\\quad \\text{Cyclotron frequency in absence of $\\vec{E}$}\\\\\n\\text{Then}, \\dot{y}&=w\\dot{z},\\quad \\dot{z}=w\\left( \\frac{E}{B}-\\dot{y}\\right)\\\\ \n\\text{Their Solution,}\\\\\ny(t)&=c_1\\cos wt+c_2\\sin wt+\\left( \\frac{E}{B}\\right) t+c_3\\\\\nz(t)&=c_2\\cos wt-c_1\\sin wt+c_4\n\\intertext{Applying the initial conditions we will get $ c_1,c_2,c_3.c_4 $}\ny(t)&=\\frac{E}{wB}(wt-\\sin wt)\\\\\nz(t)&=\\frac{E}{wB}(1-\\cos wt)\\\\\n\\text{Now let}\\quad R&=\\frac{E}{wB}\\\\\n\\text{and using }\\sin^2wt+\\cos^2wt&=1\\\\\n\\text{We will get}\\quad (y-Rwt)^2+(z-R)^2&=R^2\n\\end{align*}\nWhich represents the formula of a circle whose center $(0,Rwt,R)$ travels in $y$ direction at speed.\n$$ V=wR=\\frac{E}{B}$$\ncurve generated by this motion is called cycloid.\\\\\\\\\n\\textbf{(b)}\\quad \\textbf{When $E$ and $B$ are parallel}\\\\\\\\\nThe particle will move in helical path of constant radius but varying pitch.\n\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=6cm]{helical}\n\t\t\\caption{Helical path of charged particle in Electric and magnetic field}\n\t\t\\label{Helical path}\n\t\\end{figure}\n\n\t\n\t\n\\begin{exercise}\n\tA neutron, a proton, an electron and an $\\alpha$particle enter a region of constant magnetic field with equal velocities. The magnetic field is inward normal to the plane of paper. Label the tracks of the particle?\\\\\n\t\\begin{minipage}{.45\\textwidth}\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=0.5\\textwidth]{diagram-20210427(9)-crop}\n\t\t\\end{center}\n\t\\end{minipage}\n\\end{exercise}\n\\begin{answer}\n\tSince neutron has no charge it is undeflected by the field. So it is represented by $C$ . So all particle has initial velocity along $C$ and force $( \\theta(V\\times B))$pointed towards left for $+ve$ particles and right for electron. And radius is directly proportional to $m$,so $\\alpha$particle have larger radius. So,\\\\\\\\\n\tA=proton\\quad B)$\\rightarrow$$\\alpha$particle\\quad C)$\\rightarrow$neutron\\quad D)$\\rightarrow$electron.\n\\end{answer}\n\\begin{exercise}\n\tA particle having a charge $a$ and mass $m$ moves along a circle of radius $R$ under the action of magnetic field $B$. When the particle is at a point $P$, a uniform electric field is switched on and it is found that the particle continues on the tangent through $P$ with a uniform velocity. The magnitude of electric field is,?\n\\end{exercise}\n\\begin{answer}\n\t\\begin{align*}\n\tQ E=\\theta VB\\\\\n\tE=VB\\\\\n\t\\text{Where $V$ is the velocity }\\quad V=\\frac{\\theta BR}{m}\\\\\n\t\\therefore E=\\frac{QB^2R}{m}\n\t\\end{align*}\n\\end{answer}\n\\begin{exercise}\n\tA particle of mass carrying charge q is moving in a circle in a magnetic field.According to Bohr's model, Find?\\\\\n\t(a) The radius of the particle in the nth level.\\\\\n\t(bThe energy of the particle in the nth level\n\\end{exercise}\n\\begin{answer}\n\\begin{align*}\n(a)mv_nr_n=n\\bar{h} \\quad \\quad r_n=\\frac{mv_n}{qB}\\\\\nr_n=\\frac{m}{qB}\\frac{n\\bar{h}}{mv_n}\\\\\nr_n^2=\\frac{n\\bar{h}}{qB}\\\\\nr_n=\\sqrt{\\frac{n\\bar{h}}{qB}}\\\\\n(b) E_n=\\frac{q^2B^2r^2_n}{2m}\\implies \\frac{q^2B^2}{2m}\\times\\frac{n\\bar{h}}{qB}=n\\left( \\frac{qBh}{4\\pi m}\\right) \n\\end{align*}\n\\end{answer}\n\\newpage\n\\begin{abox}\n\tPractice set 1\n\t\\end{abox}\n\\begin{enumerate}\n\\begin{minipage}{\\textwidth}\n\t\\item The magnetic field at a distance $R$ from a long straight wire carrying a steady current $I$ is proportional to\n\t\\exyear{NET 2012}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $I R$\n\t\\task[\\textbf{B.}] $I / R^{2}$\n\t\\task[\\textbf{C.}]$I^{2} / R^{2}$\n\t\\task[\\textbf{D.}]$I / R$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The vector potential $\\vec{A}$ due to a magnetic moment $\\vec{m}$ at a point $\\vec{r}$ is given by $\\vec{A}=\\frac{\\vec{m} \\times \\vec{r}}{r^{3}}$.\n\tIf $\\vec{m}$ is directed along the positive $z$-axis, the $x$ - component of the magnetic field, at the point $\\vec{r}$, is\n\t\\exyear{NET 2011}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{3 m y z}{r^{5}}$\n\t\\task[\\textbf{B.}] $-\\frac{3 m x y}{r^{5}}$\n\t\\task[\\textbf{C.}]$\\frac{3 m x z}{r^{5}}$\n\t\\task[\\textbf{D.}]$\\frac{3 m\\left(z^{2}-x y\\right)}{r^{5}}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item An infinite solenoid with its axis of symmetry along the $z$-direction carries a steady current $I$.\n\tThe vector potential $\\vec{A}$ at a distance $R$ from the axis\n\t\\exyear{NET 2012}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=3cm]{NET1}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}]is constant inside and varies as $R$ outside the solenoid\n\t\\task[\\textbf{B.}] varies as $R$ inside and is constant outside the solenoid\n\t\\task[\\textbf{C.}]varies as $\\frac{1}{R}$ inside and as $R$ outside the solenoid\n\t\\task[\\textbf{D.}]varies as $R$ inside and as $\\frac{1}{R}$ outside the solenoid\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The force between two long and parallel wires carrying currents $I_{1}$ and $I_{2}$ and separated by a distance $D$ is proportional to\n\t\\exyear{NET 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $I_{1} I_{2} / D$\n\t\\task[\\textbf{B.}]$\\left(I_{1}+I_{2}\\right) / D$\n\t\\task[\\textbf{C.}]$\\left(I_{1} I_{2} / D\\right)^{2}$\n\t\\task[\\textbf{D.}]$I_{1} I_{2} / D^{2}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A time-dependent current $\\vec{I}(t)=K t \\hat{z}$ (where $K$ is a constant) is switched on at $t=0$ in an infinite current-carrying wire. The magnetic vector potential at a perpendicular distance $a$ from the wire is given (for time $t>a / c$ ) by\n\t\\exyear{NET 2014}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] $\\hat{z} \\frac{\\mu_{0} K}{4 \\pi c} \\int_{-\\sqrt{c^{2} t^{2}-a^{2}}}^{\\sqrt{c^{2} t^{2}-a^{2}}} d z \\frac{c t-\\sqrt{a^{2}+z^{2}}}{\\left(a^{2}+z^{2}\\right)^{1 / 2}}$\n\t\\task[\\textbf{B.}] $\\hat{z} \\frac{\\mu_{0} K}{4 \\pi} \\int_{-c t}^{c t} d z \\frac{t}{\\left(a^{2}+z^{2}\\right)^{1 / 2}}$\n\t\\task[\\textbf{C.}] $\\hat{z} \\frac{\\mu_{0} K}{4 \\pi c} \\int_{-c t}^{c t} d z \\frac{c t-\\sqrt{a^{2}+z^{2}}}{\\left(a^{2}+z^{2}\\right)^{1 / 2}}$\n\t\\task[\\textbf{D.}] $\\hat{z} \\frac{\\mu_{0} K}{4 \\pi} \\int_{-\\sqrt{c^{2} t^{2}-a^{2}}}^{\\sqrt{c^{2} t^{2}-a^{2}}} d z \\frac{t}{\\left(a^{2}+z^{2}\\right)^{1 / 2}}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A charged particle moves in a helical path under the influence of a constant magnetic field. The initial velocity is such that the component along the magnetic field is twice the component in the plane normal to the magnetic field.\n\tThe ratio $\\ell / R$ of the pitch $\\ell$ to the radius $R$ of the helical path is\n\t\\exyear{NET 2014}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\pi / 2$\n\t\\task[\\textbf{B.}]$4 \\pi$\n\t\\task[\\textbf{C.}]$2 \\pi$\n\t\\task[\\textbf{D.}]$\\pi$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A proton moves with a speed of $300 \\mathrm{~m} / \\mathrm{s}$ in a circular orbit in the $x y$-plan in a magnetic field 1 tesla along the positive $z$-direction. When an electric field of $1 \\mathrm{~V} / \\mathrm{m}$ is applied along the positive $y$-direction, the center of the circular orbit\n\t\\exyear{NET 2014}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] remains stationary\n\t\\task[\\textbf{B.}]moves at $1 \\mathrm{~m} / \\mathrm{s}$ along the negative $x$-direction\n\t\\task[\\textbf{C.}]moves at $1 \\mathrm{~m} / \\mathrm{s}$ along the positive $z$ - direction\n\t\\task[\\textbf{D.}] moves at $1 \\mathrm{~m} / \\mathrm{s}$ along the positive $x$ - direction\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item Given a uniform magnetic field $B=B_{0} \\hat{k}$ (where $B_{0}$ is a constant), a possible choice for the magnetic vector potential $A$ is\n\t\\exyear{NET 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $B_{0} y \\hat{i}$\n\t\\task[\\textbf{B.}] $-B_{0} y \\hat{i}$\n\t\\task[\\textbf{C.}] $B_{0}(x \\hat{j}+y \\hat{i})$\n\t\\task[\\textbf{D.}]$B_{0}(x \\hat{i}+y \\hat{j})$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A small magnetic needle is kept at $(0,0)$ with its moment along the $x$-axis. Another small magnetic needle is at the point $(1,1)$ and is free to rotate in the $x y$ - plane. In equilibrium the angle $\\theta$ between their magnetic moments is such that\n\t\\exyear{NET 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\tan \\theta=\\frac{1}{3}$\n\t\\task[\\textbf{B.}]$\\tan \\theta=0$\n\t\\task[\\textbf{C.}]$\\tan \\theta=3$\n\t\\task[\\textbf{D.}]$\\tan \\theta=1$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A dipole of moment $\\vec{p}$, oscillating at frequency $\\omega$, radiates spherical waves. The vector potential at large distance is\\\\\n\t$$\\vec{A}(\\vec{r})=\\frac{\\mu_{0}}{4 \\pi} i \\omega \\frac{e^{i k r}}{r} \\vec{p}$$\t\n\t$\\text { To order }\\left(\\frac{1}{r}\\right) \\text { the magnetic field } \\vec{B} \\text { at a point } \\vec{r}=r \\hat{n} \\text { is }$\n\t\\exyear{NET 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $-\\frac{\\mu_{0}}{4 \\pi} \\frac{\\omega^{2}}{C}(\\hat{n} \\cdot \\vec{p}) \\hat{n} \\frac{e^{i k r}}{r}$\n\t\\task[\\textbf{B.}]$-\\frac{\\mu_{0}}{4 \\pi} \\frac{\\omega^{2}}{C}(\\hat{n} \\times \\vec{p}) \\frac{e^{i k r}}{r}$\n\t\\task[\\textbf{C.}]$-\\frac{\\mu_{0}}{4 \\pi} \\omega^{2} k(\\hat{n} \\cdot \\vec{p}) \\vec{p} \\frac{e^{i k r}}{r}$\n\t\\task[\\textbf{D.}]$-\\frac{\\pi_{0}}{4 \\pi} \\frac{\\omega^{2}}{C} \\vec{p} \\frac{e^{i k r}}{r}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A loop of radius $a$, carrying a current $I$, is placed in a uniform magnetic field $B$. If the normal to the loop is denoted by $\\hat{n}$, the force $\\vec{F}$ and the torque $\\vec{T}$ on the loop are\n\t\\exyear{NET 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\vec{F}=0$ and $\\vec{T}=\\pi a^{2} I \\hat{\\mathrm{n}} \\times B$\n\t\\task[\\textbf{B.}]$\\vec{F}=\\frac{\\mu_{0}}{4 \\pi} \\vec{I} \\times \\vec{B}$\n\t\\task[\\textbf{C.}]$\\vec{F}=\\frac{\\mu_{0}}{4 \\pi} \\vec{I} \\times \\vec{B}$ and $\\vec{T}=I \\hat{\\mathrm{n}} \\times \\vec{B}$\n\t\\task[\\textbf{D.}]$\\vec{F}=0$ and $\\vec{T}=\\frac{1}{\\mu_{0} \\varepsilon_{0}} I \\vec{B}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A conducting circular disc of radius $r$ and resistivity $\\rho$ rotates with an angular velocity $\\omega$ in a magnetic field $B$ perpendicular to it. A voltmeter is connected as shown in the figure below. Assuming its internal resistance to be infinite, the reading on the voltmeter\n\t\\exyear{NET 2016}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20211011(46)-crop}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] depends on $\\omega, B, r$ and $\\rho$\n\t\\task[\\textbf{B.}]depends on $\\omega, B$ and $r$ but not on $\\rho$\n\t\\task[\\textbf{C.}]is zero because the flux through the loop is not changing\n\t\\task[\\textbf{D.}]is zero because a current the flows in the direction of $B$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A set of $N$ concentric circular loops of wire, each carrying a steady current $I$ in the same direction, is arranged in a plane. The radius of the first loop is $r_{1}=a$ and the radius of the $n^{\\text {th }}$ loop is given by $r_{n}=n r_{n-1}$. The magnitude $B$ of the magnetic field at the centre of the circles in the limit $N \\rightarrow \\infty$, is\n\t\\exyear{NET 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\mu_{0} I\\left(e^{2}-1\\right) / 4 \\pi a$\n\t\\task[\\textbf{B.}]$\\mu_{0} I(e-1) / \\pi a$\n\t\\task[\\textbf{C.}]$\\mu_{0} I\\left(e^{2}-1\\right) / 8 a$\n\t\\task[\\textbf{D.}]$\\mu_{0} I(e-1) / 2 a$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A constant current $I$ is flowing in a piece of wire that is bent into a loop as shown in the figure.\\\\\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=5cm,width=7cm]{diagram-20211011(54)-crop}\n\t\\end{figure}\n\t$\\text { The magnitude of the magnetic field at the point } O \\text { is }$\n\t\\exyear{NET 2017}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\mu_{0} I}{4 \\pi \\sqrt{5}} \\ln \\left(\\frac{a}{b}\\right)$\n\t\\task[\\textbf{B.}]$\\frac{\\mu_{0} I}{4 \\pi \\sqrt{5}}\\left(\\frac{1}{a}-\\frac{1}{b}\\right)$\n\t\\task[\\textbf{C.}]$\\frac{\\mu_{0} I}{4 \\pi \\sqrt{5}}\\left(\\frac{1}{a}\\right)$\n\t\\task[\\textbf{D.}]$\\frac{\\mu_{0} I}{4 \\pi \\sqrt{5}}\\left(\\frac{1}{b}\\right)$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A circular current carrying loop of radius $a$ carries a steady current. A constant electric charge is kept at the centre of the loop. The electric and magnetic fields, $\\vec{E}$ and $\\vec{B}$ respectively, at a distance $d$ vertically above the centre of the loop satisfy\n\t\\exyear{NET 2017}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\vec{E} \\perp \\vec{B}$\n\t\\task[\\textbf{B.}] $\\vec{E}=0$\n\t\\task[\\textbf{C.}]$\\vec{\\nabla}(\\vec{E} \\cdot \\vec{B})=0$\n\t\\task[\\textbf{D.}]$\\vec{\\nabla} \\cdot(\\vec{E} \\times \\vec{B})=0$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item \\text { The loop shown in the figure below carries a steady current } I \\\\\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=5cm]{diagram-20211011(11)-crop}\n\t\\end{figure}\n\t$\\text { The magnitude of the magnetic field at the point } O \\text { is }$\n\t\\exyear{NET 2018}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\mu_{0} I}{2 a}$\n\t\\task[\\textbf{B.}]$\\frac{\\mu_{0} I}{6 a}$\n\t\\task[\\textbf{C.}]$\\frac{\\mu_{0} I}{4 a}$\n\t\\task[\\textbf{D.}]$\\frac{\\mu_{0} I}{3 a}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item Two current-carrying circular loops, each of radius $R$, are placed perpendicular to each other, as shown in the figure.\n\t\n\tThe loop in the $x y$ - plane carries a current $I_{0}$ while that in the $x z$-plane carries a current $2 I_{0}$. The resulting magnetic field $\\vec{B}$ at the origin is\n\t\\exyear{NET 2018 dec}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20211011(12)-crop}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\mu_{0} l_{0}}{2 R}[2 \\hat{j}+\\hat{k}]$ \n\t\\task[\\textbf{B.}]$\\frac{\\mu_{0} l_{0}}{2 R}[2 \\hat{j}-\\hat{k}]$\n\t\\task[\\textbf{C.}]$\\frac{\\mu_{0} l_{0}}{2 R}[-2 \\hat{j}+\\hat{k}]$\n\t\\task[\\textbf{D.}]$\\frac{\\mu_{0} l_{0}}{2 R}[-2 \\hat{j}-\\hat{k}]$\n\\end{tasks}\n\\end{enumerate}\n\\colorlet{ocre1}{ocre!70!}\n\\colorlet{ocrel}{ocre!30!}\n\\setlength\\arrayrulewidth{1pt}\n\\begin{table}[H]\n\t\\centering\n\t\\arrayrulecolor{ocre}\n\t\n\t\\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}\n\t\t\\hline\n\t\t\\multicolumn{4}{|c|}{\\textbf{Answer key}}\\\\\\hline\\hline\n\t\t\\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\\\\hline\n\t\t1&\\textbf{d}&2&\\textbf{c}\\\\\\hline\n\t\t3&\\textbf{d}&4&\\textbf{a}\\\\\\hline\n\t\t5&\\textbf{a}&6&\\textbf{b}\\\\\\hline\n\t\t7&\\textbf{d}&8&\\textbf{b}\\\\\\hline\n\t\t9&\\textbf{c}&10&\\textbf{b}\\\\\\hline\n\t\t11&\\textbf{a}&12&\\textbf{b}\\\\\\hline\n\t\t13&\\textbf{d}&14&\\textbf{b}\\\\\\hline\n\t\t15&\\textbf{c}&16&\\textbf{b}\\\\\\hline\n\t\t17&\\textbf{c}&&\\\\\\hline\n\t\\end{tabular}\n\\end{table}\n\n\n\n\\newpage\n\\begin{abox}\n\tPractice set 2\n\\end{abox}\n\\begin{enumerate}\n\\begin{minipage}{\\textwidth}\n\t\\item Two magnetic dipoles of magnitude $m$ each are placed in a plane as shown in figure The energy of interaction is given by\n\t\\exyear{GATE 2010}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20210817(13)-crop}\n\t\t\\caption{}\n\t\t\\label{}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] Zero\n\t\\task[\\textbf{B.}]$\\frac{\\mu_{0} m^{2}}{4 \\pi d^{3}}$\n\t\\task[\\textbf{C.}]$\\frac{3 \\mu_{0} m^{2}}{2 \\pi d^{3}}$\n\t\\task[\\textbf{D.}]$-\\frac{3 \\mu_{0} m^{2}}{8 \\pi d^{3}}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item If a force $\\vec{F}$ is derivable from a potential function $V(r)$, where $r$ is the distance from the origin of the coordinate system, it follows that\n\t\\exyear{GATE 2011}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}]$\\vec{\\nabla} \\times \\vec{F}=0$\n\t\\task[\\textbf{B.}]$\\vec{\\nabla} \\cdot \\vec{F}=0$\n\t\\task[\\textbf{C.}]$\\vec{\\nabla} V=0$\n\t\\task[\\textbf{D.}]$\\nabla^{2} V=0$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A uniform surface current is flowing in the positive $y$-direction over an infinite sheet lying in $x-y$ plane. The direction of the magnetic field is\n\t\\exyear{GATE 2011}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}]along $\\hat{i}$ for $z>0$ and along $-\\hat{i}$ for $z<0$\n\t\\task[\\textbf{B.}]along $\\hat{k}$ for $z>0$ and along $-\\hat{k}$ for $z<0$\n\t\\task[\\textbf{C.}]along $-\\hat{i}$ for $z>0$ and along $\\hat{i}$ for $z<0$\n\t\\task[\\textbf{D.}]along $-\\hat{k}$ for $z>0$ and along $\\hat{k}$ for $z<0$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A magnetic dipole of dipole moment $\\vec{m}$ is placed in a non-uniform magnetic field $\\vec{B} .$ If the position vector of the dipole is $\\vec{r}$, the torque acting on the dipole about the origin is\n\t\\exyear{GATE 2011}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\vec{r} \\times(\\vec{m} \\times \\vec{B})$\n\t\\task[\\textbf{B.}]$\\vec{r} \\times \\vec{\\nabla}(\\vec{m} \\cdot \\vec{B})$\n\t\\task[\\textbf{C.}]$\\vec{m} \\times \\vec{B}$\n\t\\task[\\textbf{D.}]$\\vec{m} \\times \\vec{B}+\\vec{r} \\times \\nabla(\\vec{m} \\cdot \\vec{B})$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item Which of the following expressions for a vector potential $\\vec{A} \\underline{\\text { DOES NOT }}$ represent a uniform magnetic field of magnitude $B_{0}$ along the $z$-direction?\n\t\\exyear{GATE 2011}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] $\\vec{A}=\\left(0, B_{0} x, 0\\right)$\n\t\\task[\\textbf{B.}]$\\vec{A}=\\left(-B_{0} y, 0,0\\right)$\n\t\\task[\\textbf{C.}]$\\vec{A}=\\left(\\frac{B_{0} x}{2}, \\frac{B_{0} y}{2}, 0\\right)$\n\t\\task[\\textbf{D.}] $\\vec{A}=\\left(-\\frac{B_{0} y}{2}, \\frac{B_{0} x}{2}, 0\\right)$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item In a constant magnetic field of $0.6$ Tesla along the $\\mathrm{z}$ direction, find the value of the path integral $\\oint \\vec{A} \\cdot \\overrightarrow{d l}$ in the units of (Tesla $m^{2}$ ) on a square loop of side length $(1 / \\sqrt{2})$ meters. The normal to the loop makes an angle of $60^{\\circ}$ to the z-axis, as shown in the figure.\\\\\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20210817(19)-crop-crop}\n\t\\end{figure}\t\n\tThe answer should be up to two decimal places.\n\t\\exyear{GATE}\t\n\\end{minipage}\n\\begin{minipage}{\\textwidth}\n\t\\item The value of the magnetic field required to maintain non-relativistic protons of energy $1 \\mathrm{MeV}$ in a circular orbit of radius $100 \\mathrm{~mm}$ is Tesla\n\t\\exyear{GATE 2014}\n\\end{minipage}\n\\begin{minipage}{\\textwidth}\n\t\\item Given that the magnetic flux through the closed loop $P Q R S P$ is $\\phi$. If $\\int_{P}^{R} \\vec{A} \\cdot \\vec{d} l=\\phi_{1}$ along $P Q R$, the value of $\\int^{R} \\vec{A} \\cdot \\vec{d} l$ along $P S R$ is\n\t\\exyear{GATE 2015}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20210818(2)-crop}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}](a) $\\phi-\\phi_{1}$\n\t\\task[\\textbf{B.}] $\\phi_{1}-\\phi$\n\t\\task[\\textbf{C.}]$-\\phi_{1}$\n\t\\task[\\textbf{D.}] $\\phi_{1}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item Which of the following magnetic vector potentials gives rise to a uniform magnetic field $B_{0} \\hat{k} ?$\n\t\\exyear{GATE 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $B_{0} z \\hat{k}$\n\t\\task[\\textbf{B.}]$-B_{0} x \\hat{j}$\n\t\\task[\\textbf{C.}]$\\frac{B_{0}}{2}(-y \\hat{i}+x \\hat{j})$\n\t\\task[\\textbf{D.}]$\\frac{B_{0}}{2}(y \\hat{i}+x \\hat{j})$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The magnitude of the magnetic dipole moment associated with a square shaped loop carrying a steady current $I$ is $m$. If this loop is changed to a circular shape with the same current $I$ passing through it, the magnetic dipole moment becomes $\\frac{p m}{\\pi} .$ The value of $p$ is\n\t\\exyear{GATE 2016}\n\\end{minipage}\n\\begin{minipage}{\\textwidth}\n\t\\item An infinite solenoid carries a time varying current $I(t)=A t^{2}$, with $A \\neq 0 .$ The axis of the solenoid is along the $\\hat{z}$ direction. $\\hat{r}$ and $\\hat{\\theta}$ are the usual radial and polar directions in cylindrical polar coordinates. $\\vec{B}=B_{r} \\hat{r}+B_{\\theta} \\hat{\\theta}+B_{z} \\hat{z}$ is the magnetic field at a point outside the solenoid. Which one of the following statements is true?\n\t\\exyear{GATE 2017}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $B_{r}=0, B_{\\theta}=0, B_{z}=0$\n\t\\task[\\textbf{B.}]$B_{r} \\neq 0, B_{\\theta} \\neq 0, B_{z}=0$\n\t\\task[\\textbf{C.}] $B_{r} \\neq 0, B_{\\theta} \\neq 0, B_{z} \\neq 0$\n\t\\task[\\textbf{D.}] $B_{r}=0, B_{\\theta}=0, B_{z} \\neq 0$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item An infinitely long straight wire is carrying a steady current $I$. The ratio of magnetic energy density at distance $r_{1}$ to that at $r_{2}\\left(=2 r_{1}\\right)$ from the wire is\n\t\\exyear{GATE 2018}\n\\end{minipage}\n\\begin{minipage}{\\textwidth}\n\t\\item A constant and uniform magnetic field $\\vec{B}=B_{0} \\hat{k}$ pervades all space. Which one of the following is the correct choice for the vector potential in Coulomb gauge?\n\t\\exyear{GATE 2018}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}] $-B_{0}(x+y) \\hat{i}$\n\t\\task[\\textbf{B.}]$B_{0}(x+y) \\hat{j}$\n\t\\task[\\textbf{C.}] $B_{0} x \\hat{j}$\n\t\\task[\\textbf{D.}]$-\\frac{1}{2} B_{0}(x \\hat{i}-y \\hat{j})$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A solid cylinder of radius $R$ has total charge $Q$ distributed uniformly over its volume. It is rotating about its axis with angular speed $\\omega$. The magnitude of the total magnetic moment of the cylinder is\n\t\\exyear{GATE 2019}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}](a) $Q R^{2} \\omega$\n\t\\task[\\textbf{B.}]$\\frac{1}{2} Q R^{2} \\omega$\n\t\\task[\\textbf{C.}]$\\frac{1}{4} Q R^{2} \\omega$\n\t\\task[\\textbf{D.}]$\\frac{1}{8} Q R^{2} \\omega$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item An infinitely long wire parallel to the $x$-axis is kept at $z=d$ and carries a current $I$ in the positive $x$ direction above a superconductor filling the region $z \\leq 0$ (see figure). The magnetic field $\\vec{B}$ inside the superconductor is zero so that the field just outside the superconductor is parallel to its surface. The magnetic field due to this configuration at a point $(x, y, z>0)$ is\n\t\\exyear{GATE 2019}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=5cm,width=5cm]{diagram-20210818(14)-crop-crop}\n\t\t\\caption{}\n\t\t\\label{}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}]$\\left(\\frac{\\mu_{0} I}{2 \\pi}\\right) \\frac{-(z-d) \\hat{j}+y \\hat{k}}{\\left[y^{2}+(z-d)^{2}\\right]}$\n\t\\task[\\textbf{B.}]$\\left(\\frac{\\mu_{0} I}{2 \\pi}\\right)\\left[\\frac{-(z-d) \\hat{j}+y \\hat{k}}{y^{2}+(z-d)^{2}}+\\frac{(z+d) \\hat{j}-y \\hat{k}}{y^{2}+(z+d)^{2}}\\right]$\n\t\\task[\\textbf{C.}]$\\text { (c) }\\left(\\frac{\\mu_{0} I}{2 \\pi}\\right)\\left[\\frac{-(z-d) \\hat{j}+y \\hat{k}}{y^{2}+(z-d)^{2}}-\\frac{(z+d) \\hat{j}-y \\hat{k}}{y^{2}+(z+d)^{2}}\\right]$\n\t\\task[\\textbf{D.}]$\\text { (d) }\\left(\\frac{\\mu_{0} I}{2 \\pi}\\right)\\left[\\frac{y \\hat{j}+(z-d) \\hat{k}}{y^{2}+(z-d)^{2}}+\\frac{y \\hat{j}-(z+d) \\hat{k}}{y^{2}+(z+d)^{2}}\\right]$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The vector potential inside a long solenoid with $n$ turns per unit length and carrying current $I$, written in cylindrical coordinates is $\\vec{A}(s, \\phi, z)=\\frac{\\mu_{0} n I}{2} s \\hat{\\phi}$. If the term $\\frac{\\mu_{0} n I}{2} s(\\alpha \\cos \\phi \\hat{\\phi}+\\beta \\sin \\phi \\hat{s})$, where $\\alpha \\neq 0, \\beta \\neq 0$ is added to $\\vec{A}(S, \\phi, z)$, the magnetic field remains the same if\n\t\\exyear{GATE 2019}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}]$\\alpha=\\beta$\n\t\\task[\\textbf{B.}]$\\alpha=-\\beta$\n\t\\task[\\textbf{C.}]$\\alpha=2 \\beta$\n\t\\task[\\textbf{D.}]$\\alpha=\\frac{\\beta}{2}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A magnetic field $\\vec{B}=B_{0}(\\hat{i}+2 \\hat{j}-4 \\hat{k})$ exists at point. If a test charge moving with a velocity, $\\vec{v}=v_{0}(3 \\hat{i}-\\hat{j}+2 \\hat{k})$ experiences no force at a certain point, the electric field at that point in SI units is\n\t\\exyear{JEST 2012}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\vec{E}=-v_{0} B_{0}(3 \\hat{i}-2 \\hat{j}-4 \\hat{k})$\n\t\\task[\\textbf{B.}]$\\vec{E}=-v_{0} B_{0}(\\hat{i}+\\hat{j}+7 \\hat{k})$\n\t\\task[\\textbf{C.}]$\\vec{E}=v_{0} B_{0}(14 \\hat{j}+7 \\hat{k})$\n\t\\task[\\textbf{D.}]$\\vec{E}=-v_{0} B_{0}(14 \\hat{j}+7 \\hat{k})$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A small magnet is dropped down a long vertical copper tube in a uniform gravitational field. After a long time, the magnet\n\t\\exyear{JEST 2012}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] attains a constant velocity\n\t\\task[\\textbf{B.}] moves with a constant acceleration\n\t\\task[\\textbf{C.}] moves with a constant deceleration\n\t\\task[\\textbf{D.}]executes simple harmonic motion\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A thin uniform ring carrying charge $Q$ and mass $M$ rotates about its axis. What is the gyromagnetic ratio (defined as ratio of magnetic dipole moment to the angular momentum) of this ring?\n\t\\exyear{JEST 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{Q}{2 \\pi M}$\n\t\\task[\\textbf{B.}]$\\frac{Q}{M}$\n\t\\task[\\textbf{C.}]$\\frac{Q}{2 M}$\n\t\\task[\\textbf{D.}]$\\frac{Q}{\\pi M}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The electric and magnetic field caused by an accelerated charged particle are found to scale as $E \\propto r^{-n}$ and $B \\propto r^{-m}$ at large distances. What are the value of $n$ and $m$ ?\n\t\\exyear{JEST 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $n=1, m=2$\n\t\\task[\\textbf{B.}] $n=2, m=1$\n\t\\task[\\textbf{C.}]$n=1, m=1$\n\t\\task[\\textbf{D.}]$n=2, m=2$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A system of two circular co-axial coils carrying equal currents $I$ along same direction having equal radius $R$ and separated by a distance $R$ (as shown in the figure below). The magnitude of magnetic field at the midpoint $P$ is given by\n\t\\exyear{JEST 2014}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{diagram-20210809(7)-crop}\n\t\t\\caption{}\n\t\t\\label{}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}](a) $\\frac{\\mu_{0} I}{2 \\sqrt{2} R}$\n\t\\task[\\textbf{B.}]$\\frac{4 \\mu_{0} I}{5 \\sqrt{5} R}$\n\t\\task[\\textbf{C.}]$\\frac{8 \\mu_{0} I}{5 \\sqrt{5} R}$\n\t\\task[\\textbf{D.}] 0\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A charged particle is released at time $t=0$, from the origin in the presence of uniform static electric and magnetic fields given by $E=E_{0} \\hat{y}$ and $B=B_{0} \\hat{z}$ respectively. Which of the following statements is true for $t>0$ ?\n\t\\exyear{JEST 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] The particle moves along the $x$-axis.\n\t\\task[\\textbf{B.}]The particle moves in a circular orbit.\n\t\\task[\\textbf{C.}]The particle moves in the $(x, y)$ plane.\n\t\\task[\\textbf{D.}] Particle moves in the $(y, z)$ plane\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item The strength of magnetic field at the center of a regular hexagon with sides of length $a$ carrying a steady current $I$ is:\n\t\\exyear{JEST 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\mu_{0} I}{\\sqrt{3} \\pi a}$ \n\t\\task[\\textbf{B.}]$\\frac{\\sqrt{6} \\mu_{0} I}{\\pi a}$\n\t\\task[\\textbf{C.}]$\\frac{3 \\mu_{0} I}{\\pi a}$\n\t\\task[\\textbf{D.}]$\\frac{\\sqrt{3} \\mu_{0} I}{\\pi a}$\n\\end{tasks}\n\\begin{minipage}{\\textwidth}\n\t\\item A wire with uniform line charge density $\\lambda$ per unit length carries a current $I$ as shown in the figure. Take the permittivity and permeability of the medium to be $\\varepsilon_{0}=\\mu_{0}=1 . \\mathrm{A}$ particle of charge $q$ is at a distance $r$ and is travelling along a trajectory parallel to the wire. What is the speed of the charge?\n\t\\exyear{JEST 2019}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=6cm]{jest-crop}\n\t\\end{figure}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\lambda}{I}$ \n\t\\task[\\textbf{B.}]$\\frac{\\lambda}{2 I}$\n\t\\task[\\textbf{C.}]$\\frac{\\lambda}{3 I}$\n\t\\task[\\textbf{D.}]$\\frac{4 \\lambda}{I}$\n\\end{tasks}\n\\end{enumerate}\n\n\\colorlet{ocre1}{ocre!70!}\n\\colorlet{ocrel}{ocre!30!}\n\\setlength\\arrayrulewidth{1pt}\n\\begin{table}[H]\n\t\\centering\n\t\\arrayrulecolor{ocre}\n\t\n\t\\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}\n\t\t\\hline\n\t\t\\multicolumn{4}{|c|}{\\textbf{Answer key}}\\\\\\hline\\hline\n\t\t\\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\\\\hline\n\t\t1&\\textbf{d}&2&\\textbf{a}\\\\\\hline\n\t\t3&\\textbf{a}&4&\\textbf{c}\\\\\\hline\n\t\t5&\\textbf{c}&6&\\textbf{0.15}\\\\\\hline\n\t\t7&\\textbf{1.44}&8&\\textbf{b}\\\\\\hline\n\t\t9&\\textbf{c}&10&\\textbf{4}\\\\\\hline\n\t\t11&\\textbf{d}&12&\\textbf{4}\\\\\\hline\n\t\t13&\\textbf{c}&14&\\textbf{c}\\\\\\hline\n\t\t15&\\textbf{b}&16&\\textbf{d}\\\\\\hline\n\t\t17&\\textbf{d}&18&\\textbf{a}\\\\\\hline\n\t\t19&\\textbf{c}&20&\\textbf{c}\\\\\\hline\n\t\t21&\\textbf{c}&22&\\textbf{c}\\\\\\hline\n\t\t23&\\textbf{d}&24&\\textbf{a}\\\\\\hline\n\t\\end{tabular}\n\\end{table}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\newpage\n\\begin{abox}\n\tPractise Set-3\n\\end{abox}\n\\begin{enumerate}[ label=\\color{ocre}\\textbf{\\arabic*.}]\n\t\\item Consider a coil of radius $R$ is placed in $XY$ plane having current $I$ passing through it. Auppose another coil having the same radius $R$ is placed at a distance $2R$ from the first. Find the magnetic field at the mid point between them?\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=4cm,height=2cm]{diagram-20210430-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t&B \\text{at}\\quad\\text{is}=B_1+B_2\\\\\n\t\tB_1&=\\frac{\\mu_{0} I}{2}\\times\\frac{R^2}{(R^2+R^2)^\\frac{3}{2}}\\qquad \\text{here,}\\ z=R\\\\\n\t\t&=\\frac{\\mu_0 I}{2}\\frac{R^2}{2^\\frac{3}{2}\\times R^3}=\\frac{\\mu_0 I}{2\\times2^\\frac{3}{2}\\times R}\\\\\n\t\tB_2 \\text{is also same}\\\\\n\t\t\\therefore B&=2\\times\\frac{\\mu_0 I}{2\\times2^\\frac{2}{3}R}=\\frac{\\mu_0I}{2\\sqrt{2}R}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A very long solenoid with $n$ turns per unit length carries a current $I$. The magnetic field at a point which is on its axis and its end face?\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=6cm,height=2cm]{diagram-20210430(2)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tFor a solenoid magnetic field at any point $P$ on its axis is given by\n\t\t$$B=\\frac{\\mu_0 nI}{2}(\\cos\\theta_2-\\cos\\theta_1)$$\n\t\t$\\theta_1$ and $\\theta_2$ are the angle made by the end points of the solenoid to $P$.\\\\\n\t\t\\begin{align*}\n\t\t\\therefore\\text{In the case of infinite solenoid,at center}\\\\\n\t\t\\theta_1&=\\pi,\\theta_2=0\\\\\n\t\t\\therefore B&=\\frac{\\mu_0 nI}{2}\\times2=\\mu_0 nI \\quad\\text{at center}\\\\\n\t\t\\text{Here in this question,at one end},\\\\\n\t\t\\theta_1&=\\frac{\\pi}{2},\\quad\\theta_2=0 \\text{(for long solenoid)}\\\\\n\t\tB&=\\frac{\\mu_0 n I}{2}\\times1=\\frac{\\mu_0 n I}{2}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A steady current $I$ flows down a long cylindrical wire of radius $a$ Find the magnetic field, both inside and outside the wire, if\\\\\n\t\\textbf{(a)} The current is uniformly distributed over the outside surface of the wire.\\\\\n\t\\textbf{(b)} The current is distributed in such a way that $J$ is proportional to $s$, the distance from the axis.\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=3.5cm,height=1.5cm]{diagram-20210430(8)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\t\\textbf{(a)} Here current is uniform\\\\\n\t\tInside the wire \\quad $B=0$\\quad $sa\\\\\n\t\t\\end{align*}\n\t\t\\textbf{(b)}\n\t\t\\begin{align*}\n\t\t\\text{inside the wire}\n\t\tJ&=ks\\\\\n\t\t\\text{First we have to find }&\\text{ $k$,so}\\hspace{3cm}I=\\text{total current}\\\\\n\t\t\\text{We know}I&=\\int_{0}^{a}J\\cdot da \\hspace{2cm}J=ks \\quad da=2\\pi s ds\\\\\n\t\t\\therefore I&=k2\\pi\\int_{0}^{a}s^2ds\\\\\n\t\tI&=k2\\pi s\\frac{s^3}{3}\\quad\\text{so}\\quad k=\\frac{3I}{2\\pi a^3}\n\t\t\\intertext{Now inside the wire Consider an Amperial loop of radius \\ $s$\\ inside the wire}\n\t\t\\text{So}\\oint B\\cdot dl&=\\mu_0I \\hspace{3cm}J=ks\\\\\n\t\tB\\times2\\pi s&=\\mu_0I\\hspace{3cm}ds=2\\pi sds\\\\\n\t\tI=\\int_{0}^{s}J.ds&=\\int_{0}^{s}ks\\times2\\pi sds\\\\\n\t\t&=k\\int_{0}^{s}s^2 ds\\times2\\pi\\\\\n\t\t&=\\frac{ks^3}{3}\\times2\\pi\\\\\n\t\t\\text{Substituting value of }k\\\\\n\t\tI&=\\frac{3 I}{2\\pi a^3}\\times\\frac{s^3}{3}\\times2\\pi\\\\\n\t\t&=\\frac{Is^3}{a^3}\\\\\n\t\t\\therefore B\\times2\\pi s&=\\frac{\\mu_0Is^3}{a^3}\\\\\n\t\tB&=\\frac{\\mu_0Is^2}{2\\pi a^3}\\hat{\\phi} \\quad \\text{inside}\\\\\n\t\t\\text{Out side the wire}\\\\\n\t\t\\oint B\\cdot dl&=\\mu_0I\\\\\n\t\tB\\times2\\pi s&=\\mu_0I\\\\\n\t\tB&=\\frac{\\mu_0I}{2\\pi s}\\hat{\\phi}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Find the magnetic field at the center $O$ for the following figures.\\\\\n\t\\begin{minipage}{0.45\\textwidth}\n\t\t\\begin{figure}[H]\n\t\t\t\\begin{center}\n\t\t\t\t\\includegraphics[width=3.5cm,height=3.5cm]{diagram-20210430(1)-crop}\n\t\t\t\\end{center}\n\t\t\t\\caption{(a)}\n\t\t\\end{figure}\n\t\\end{minipage}\n\t\\begin{minipage}{0.45\\textwidth}\n\t\t\\begin{figure}[H]\n\t\t\t\\begin{center}\n\t\t\t\t\\includegraphics[width=3.5cm,height=3.5cm]{diagram-20210430(4)-crop}\n\t\t\t\\end{center}\n\t\t\t\\caption{(b)}\n\t\t\\end{figure}\n\t\\end{minipage}\n\t\\begin{answer}\n\t\t\\textbf{(a)}\n\t\t\\begin{align*}\n\t\t\\intertext{Field at $O$ due to $PS$ and $QR$ is zero. Because the point $O$ lies along the axis of the segment.}\n\t\t\\text{Field at $O$ due to $RS$,}\\\\\n\t\tB_1&=\\frac{\\mu_0I}{2b}\\times\\frac{\\phi}{2\\pi}\\\\\n\t\t\\text{Field at $O$ due to $PQ$}\\\\\n\t\tB_2&=\\frac{\\mu_0I}{2a}\\times\\frac{2\\pi-\\phi}{2\\pi}\\\\\n\t\t\\text{So total field at B}\\\\\n\t\tB&=B_1+B_2\\\\\n\t\t&=\\frac{\\mu_0I}{4\\pi}\\left( \\frac{\\phi}{b}+\\frac{2\\pi-\\phi}{a}\\right) \\\\\n\t\t\\end{align*}\n\t\t\\textbf{(b)}\n\t\t\\begin{align*}\n\t\t\\text{Field at $O$ due to arc $pQ$}\\\\\n\t\tB_1&=\\frac{\\mu_0I}{2a}\\times\\frac{3\\frac{\\pi}{2}}{2\\pi}\\\\\n\t\t=&\\frac{\\mu_0I}{2a}\\times\\frac{3}{4}\n\t\t\\intertext{Field at $O$ due to $PT$ \\ and\\ $QR$\\ are zero. Field at $O$ due to $ST$ and $RS$.}\n\t\tB_2=\\ B_3&=\\frac{\\mu_0I}{4\\pi b}(\\sin\\theta_2-\\sin\\theta_1)\\\\\n\t\t&=\\frac{\\mu_0I}{4\\pi b}\\sin45\\hspace{3cm}\\theta_2=45,\\quad \\theta_{1}=0\\\\\n\t\t&=\\frac{\\mu_0I}{4\\pi b}\\times\\frac{1}{\\sqrt{2}}\\\\\n\t\t\\intertext{Total field at \\ $O$,}\n\t\tB&=B_1+B_2+B_3\\\\\n\t\t&=\\frac{\\mu_0I}{2a}\\times\\frac{3}{4}+2\\times\\frac{\\mu_0}{4\\pi b}\\times\\frac{1}{\\sqrt{2}}\\\\\n\t\t&=\\frac{\\mu_0I}{4\\pi}\\left[\\frac{3\\pi}{2a}+\\frac{\\sqrt{2}}{b} \\right] \n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Magnetic field at center $O$ due to a regular pentagon if $I$ current flowing through it and $R$ be the distance from each side to $O$.\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=2.5cm,height=2.5cm]{diagram-20210430(5)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tFor an $n$ sided rectangular polygon field at center\\\\\n\t\t$B=\\frac{n\\mu_0I}{2\\pi R}\\tan\\frac{\\pi}{n}$ if $R$ become the distance from each vertex to center\\\\\n\t\tif $R$ is the distance from each side to center \n\t\t\\begin{align*}\n\t\tB&=\\frac{n\\mu_0I}{2\\pi R}\\sin\\frac{\\pi}{n}\\\\\n\t\t\\text{Here,}\\quad n&=5\\quad \\\\\n\t\t\\text{Then,} B&=\\frac{\\mu_0nI}{2\\pi R}\\sin\\frac{\\pi}{n}\\\\\n\t\t&=\\frac{5\\mu_0 I}{2\\pi R}\\sin\\frac{\\pi}{5}\\\\\n\t\t&=\\frac{5\\mu_0I}{2\\pi R}\\sin 36\t\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Thick slab extending from $z=-a$ and $z=+a$ carries a uniform volume current $J=J(x) $.Find the magnetic field as a function of $Z$ both inside and outside the slab.\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=5cm,height=3cm]{diagram-20210430(6)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tConsider an Amperian loop of length $l$ and height $z$, applying Ampere's law.\\\\\n\t\t\\begin{align*}\n\t\t\\oint B\\cdot dl&=\\mu_0I\\\\\n\t\tB\\times l&=\\mu_0\\cdot lz\\cdot J\\hspace{3cm}I=lz\\cdot \\vec{J}\\\\\n\t\t\\therefore B&=\\mu_0Jz\\hat{y}\\hspace{3cm}(-a0$ field is along $-y$ direction and for $z<0$ field is along $+y$ direction\n\t\t\\begin{align*}\n\t\t\\text{So}\\quad B&=-\\mu_0Ja\\hat{y}\\quad\\text{for}\\quad z>+a\\\\\n\t\tB&=+\\mu_0Ja\\hat{y}\\quad\\text{for}\\quad z>-a\n\t\t\\end{align*}\n\t\tBy consideting an Amperian loop of length $l$ and height $a$.\n\t\\end{answer}\n\t\\item A beam of proton with velocity $4\\times10^5 {m}/{Sec}$enters a uniform magnetic field of 0.3Tesla at an angle $60^\\circ$to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix.\\\\\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tV&=4\\times10^5 \\frac{m}{Sec}\\hspace{1cm}V_\\perp=V\\sin60\\hspace{1cm}V_\\parallel=V\\cos60\\\\\n\t\t&\\frac{mv_\\perp^2}{R}=qv_\\perp B\\\\\n\t\tR&=\\frac{m v_\\perp}{qB}\\\\\n\t\t&=0.012m \\quad m_p=1.6\\times 10^{-27}kg\\\\\n\t\t\\text{Pitch of helix,}\\\\\n\t\td&=V_\\parallel T=V_\\parallel\\times\\frac{2\\pi m}{qB}=0.044m\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item The maximum energy of deuteron coming out of a cyclotron accelerator is $20MeV$. The maximum energy of proton that can be obtained from the accelerator is?\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tKE_{max}&=\\frac{Q^2B^2R^2}{2m}\\\\\n\t\tKE_d&=20 MeV\\\\\n\t\tKE_P&=?\\\\\n\t\t20&=\\frac{1\\times B^2\\times R^2}{2\\times2}\\\\\n\t\tKE_P&=\\frac{1\\times B^2\\times R^2}{2\\times1}\\\\\n\t\tKE_P&=20MeV\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Ab $\\alpha$ particle is accelerated by potencial difference of $10^4V$. Find the change in its direction of motion. If it enters normally in a region of thickness $0.1m$ having transverse magnetic induction of $0.1 $Tesla$(m=6.4\\times10^-27kg)$\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=3.5cm,height=2.5cm]{diagram-20210430(9)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tBefore entering the field, $\\alpha$ particle is accelerated by a potential difference of $10^4V$ then\\begin{align*}\n\t\t\\frac{1}{2}mu^2&=qV\\\\\n\t\tu&=\\sqrt{\\frac{2qV}{m}}\\\\\n\t\tR&=\\frac{mu}{QB}=\\frac{m}{QB}\\sqrt{\\frac{2qV}{m}}\\quad V=pd\\\\\n\t\t&=\\frac{1}{B}\\sqrt{\\frac{2mV}{q}}\n\t\t\\intertext{$ \\theta $ is the angle of deflection,}\n\t\t\\therefore\\sin\\theta&=\\frac{l}{R}\\hspace{3cm}V=10^4\\text{Volt}\\\\\n\t\t\\sin\\theta&=0.1\\times B\\times\\sqrt{\\frac{q}{2mV}}\\\\\n\t\tq&=2\\times1.6\\times10^{-16}=3.2\\times10^{-16}c\\\\\n\t\tm=6.4\\times10^{-27}kg\\\\\n\t\t\\sin\\theta&=0.1\\times0.1\\times\\sqrt{\\frac{3.2\\times10^{-16}}{2\\times6.4\\times10^{-27}\\times10^4}}\\\\\n\t\t&=\\frac{1}{2}\\\\\n\t\t\\therefore\\theta&=30^\\circ\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item There are two similar coils at $P$ and $Q$ having same no of turns located at $(0,4,0)$ and$(0,0,3)$. Area of crone section are in the ratio $4:3$.A $16 A$ current flowing through coil $P$in clockwise direction and a $9\\sqrt{3} A$ current in $Q$ in anticlockwise direction . What will the deflection of a compass needle placed at the origin. Assumes that earth field is negligible and radius of the coil are very small compared to their distance from the origin.\n\t\\begin{figure}[H]\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=5cm,height=4cm]{diagram-20210430(7)-crop}\n\t\t\\end{center}\n\t\\end{figure}\n\t\\begin{answer}\n\t\t$B$ at a distance $x$ from a coil of radius $r$\\\\\n\t\t\\begin{align*}\n\t\tB&=\\frac{\\mu_0I}{2}\\frac{nr^2}{(r^2+x^2)^\\frac{3}{2}}\\\\\n\t\t&=\\frac{\\mu_0 \\ln \\pi r^2}{2\\times\\pi \\times(x^2)^\\frac{3}{2}}\\hspace{2cm}\\quad \\text{When}\\ r<<t_0$.\r\n Denote the lift of $\\gamma|_{[0,t_0]}$ by $\\hat{\\gamma}|_{[0,t_0]}$.\r\n But then since $U\\supset f(V)$ is evenly covered, we can just pick the copy of $U$ where $\\hat{\\gamma}|_{[0,t_0]}(t_0)$ resides in and continue $\\hat{\\gamma}$ there by following $\\gamma$.\r\n More precisely, suppose that copy is $U'$ and $p':U'\\to U$ is the homeomoephism by restricting $p$, then\r\n $$[0,t]\\ni s\\mapsto\\begin{cases}\r\n \\hat\\gamma(s)\\text{, for $s\\in[0,t_0]$}\\\\\r\n (p')^{-1}\\circ\\gamma(s)\\text{, for $s\\in[t_0,t]$}\r\n \\end{cases}$$\r\n which works as a lift of $\\gamma|_{[0,t]}$.\r\n So $t\\in S$.\\\\\r\n If $t_0\\notin S$ and $t\\in V$, then if $t\\in S$ and $t\\ge t_0$ we get an immediate contradiction.\r\n But if $t0$ and $V_t\\ni y_0$ is open.\r\n As $Y$ is locally path-connected we might as well assume $V_t$ are path-connected.\r\n As $\\{y_0\\}\\times I$ is compact, there is a finite set $T\\subset I$ such that $\\{(t_i-\\epsilon_{t_i},t_i+\\epsilon_{t_i}):t_i\\in T\\}$ covers $I$.\r\n We then take $V=\\bigcap_{t_i\\in T}V_{t_i}$ which is open and path-connected.\r\n Take $J_i=(t_i-\\epsilon_{t_i},t_i+\\epsilon_{t_i})\\cap I$.\r\n Then we know that $F(V\\times J_i)$ is contained in an evenly covered open subset $U_i$ of $X$ for each $i$.\r\n Let $U_i'$ be the unique copy of $U_i$ in $p^{-1}(U_i)$ such that $\\hat{F}(\\{y_0\\}\\times J_i)=\\hat{\\gamma}_{y_0}(J_i)\\subset U_i'$.\r\n Let $p_i:U_i'\\to U_i$ be the homeomorphism between them.\r\n Now for $(y,t)\\in V\\times I$, we define $\\tilde{F}(y,t)=p_i^{-1}\\circ F(y,t)$ for $t\\in J_i$.\r\n It is quite obvious it is well-defined, but let's prove it.\r\n Suppose $t\\in J_i\\cap J_j$ and let $\\alpha$ be a path in $V$ from $y_0$ to $y$.\r\n Define $\\alpha_t(s)=F(\\alpha(s),t)$.\r\n Then $p_i^{-1}\\circ \\alpha_t$ and $p_j^{-1}\\circ\\alpha_t$ are lifts of $\\alpha_t$.\r\n Furthermore they have the same initial values at $\\hat{F}(y_0,t)$, so they are equal by Lemma \\ref{lift_unique}.\r\n Consequently $p_i^{-1}\\circ F(y,t)=p_j^{-1}\\circ F(y,t)$ as desired.\r\n So $\\tilde{F}$ is a well-defined and hence continuous by its definition.\\\\\r\n As $V$ is path-connected and $\\tilde{F}(\\cdot,0)$ is a lift of $f_0$ that agrees with $\\hat{f}_0$ at $y_0$, we have $\\forall y\\in V,\\hat{F}(y,0)=\\hat{f}_0(y)$ by Lemma \\ref{lift_unique}.\r\n Also, for each $y\\in V$, $\\hat{F}(y,\\cdot)$ is a lift of $\\gamma_y$ at $\\hat{f}_0(y)$.\r\n Therefore $\\tilde{F}(y,t)=\\hat{\\gamma}_y(t)$ again by Lemma \\ref{lift_unique}, thus $\\tilde{F}=\\hat{F}$ on $V\\times I$.\r\n In particular, $\\hat{F}$ is continuous in $V\\times I$, hence it is continuous.\r\n\\end{proof}\r\n\\subsection{Lifting and Fundamental Groups}\r\n\\begin{lemma}\r\n Let $p:\\hat{X}\\to X$ be a covering map and $F:I\\times I\\to X$ be a homotopy of paths.\r\n Then any lift $\\hat{F}$ of $F$ is also a homotopy of paths.\r\n\\end{lemma}\r\n\\begin{proof}\r\n As $F$ is a homotopy of paths, $F(0,\\cdot)$ and $F(1,\\cdot)$ are constant paths in $X$.\r\n Therefore $\\hat{F}(0,\\cdot)$ and $\\hat{F}(1,\\cdot)$ are lifts those constant paths , hence constant.\r\n\\end{proof}\r\n\\begin{lemma}\r\n Let $p:\\hat{X}\\to X$ be a covering map with $\\hat{x}\\in \\hat{X}$ and $x=p(\\hat{x})$, then the induced homeomorphism $p_\\ast:\\pi_1(\\hat{X},\\hat{x})\\to\\pi_1(X,x)$ via $[\\hat\\gamma]\\mapsto [p\\circ\\hat\\gamma]$ is injective.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Suppose $[\\hat{\\gamma}]\\in\\ker p_\\ast$, then $\\gamma=p\\circ\\hat{\\gamma}\\simeq c_x$.\r\n If $F$ is a homotopy between $\\gamma$ and $c_x$, we can lift it to a homotopy $\\tilde{F}$ from $\\hat{\\gamma}$ which is a homotopy of paths by the preceding lemma.\r\n In particular, $\\hat{F}$ is a homotopy between $\\hat\\gamma$ and $c_{\\hat{x}}$, so $[\\hat\\gamma]=[c_{\\hat{x}}]$ is the identity.\r\n This shows the result.\r\n\\end{proof}\r\n\\begin{remark}\r\n So we can view $\\pi_1(\\hat{X},\\hat{x})$ as the subgroup $p_\\ast(\\pi(\\hat{X},\\hat{x}))\\le\\pi_1(X,x)$.\r\n Also note that given $[\\gamma]\\in\\pi_1(X,x)$, we get a map $p^{-1}(\\{x\\})\\to p^{-1}(\\{x\\})$ via $\\hat{x}\\mapsto \\hat{\\gamma}_{\\hat{x}}(1)$ which is necessarily bijective.\r\n So this defines a right group action of $\\pi_1(X,x)$ on $p^{-1}(\\{x\\})$.\r\n For $\\hat{x}\\in p^{-1}(\\{x\\})$, this is induced by $\\hat{x}\\cdot\\gamma=\\hat\\gamma_{\\hat{x}}(1)$.\r\n Easily $(\\hat{x}\\cdot\\gamma)\\cdot\\delta=\\hat{x}\\cdot(\\gamma\\cdot\\delta)$.\r\n\\end{remark}\r\n\\begin{lemma}\r\n Let $p:\\hat{X}\\to X$ be a covering map and suppose $\\hat{X}$ is path-connected.\r\n Let $x\\in X$, and let $\\Pi$ denote the set of right cosets of $p_\\ast(\\pi_1(\\hat{X},\\hat{x}))$ in $\\pi_1(X,x)$.\r\n Then the map $\\Pi\\to p^{-1}(\\{x\\})$ via $p_\\ast(\\pi_1(\\hat{X},\\hat{x}))[\\gamma]\\mapsto \\hat{x}\\cdot\\gamma$ is a bijection.\r\n Further, this bijection satisfies\r\n $$p_\\ast(\\pi_1(\\hat{X},\\hat{x}))[\\gamma][\\delta]\\mapsto\\hat{x}\\cdot(\\gamma\\cdot\\delta)$$\r\n where $\\gamma,\\delta$ are loops based at $x$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n If $[\\delta]\\in p_\\ast(\\pi_1(\\hat{X},\\hat{x}))$, then $\\hat{x}\\cdot\\delta=\\hat{x}$.\r\n This obviously implies the map is well-defined.\r\n To prove this is a bijection, we shall use the Orbit-Stabiliser Theorem.\r\n Note that the stabliser of $\\hat{x}$ on this right action is the set of loops $[\\gamma]$ at $x$ such that $\\hat\\gamma_{\\hat{x}}(1)=\\hat{x}$ where $\\hat{\\gamma}_{\\hat{x}}$ is the lift of $\\gamma$ from $\\hat{x}$.\r\n But then $\\hat\\gamma_{\\hat{x}}$ is a loop, so $\\gamma\\in p_\\ast(\\pi_1(\\hat{X},\\hat{x}))$.\r\n This means that the stabiliser of $\\hat{x}$ is exactly $p_\\ast(\\pi_1(\\hat{X},\\hat{x}))$.\r\n It remains to show that the action is transitive.\r\n Let $\\hat{y}\\in p^{-1}(\\{x\\})$, then there is a path $\\hat{\\gamma}$ from $\\hat{x}$ to $\\hat{y}$, so $p\\circ\\hat{\\gamma}$ is a loop based at $x$ and $\\hat{x}\\cdot\\gamma=\\hat{y}$.\r\n\\end{proof}\r\n\\begin{remark}\r\n So the degree of the covering map $p$ is just the index of $p_\\ast(\\pi_1(\\hat{x},\\hat{x}))$ in $\\pi_1(X,x)$.\r\n\\end{remark}\r\n\\begin{example}\r\n Consider $\\pi_1(S^1,x)$.\r\n The covering map $p_1:\\mathbb R\\to S^1$ via $t\\mapsto e^{2\\pi it}$ has infinite degree and $p_2:S^1\\to S^1$ via $z\\mapsto z^n$ has degree $n$ for any $n>1$.\r\n So what we conclude from the lemma above is that $\\pi_1(S_1,x)$ is an infinite group with subgroups of every finite index.\r\n A natural guess is then $\\pi_1(S^1,x)\\cong\\mathbb Z$.\r\n We will prove it later.\r\n\\end{example}\r\n\\begin{definition}\r\n If $p:\\hat{X}\\to X$ is a covering map and $\\hat{X}$ is simply connected, then $\\hat{X}$ is called a univeral cover of $X$.\r\n\\end{definition}\r\n\\begin{example}\r\n The map $t\\mapsto e^{2\\pi it}$ makes $\\mathbb R$ a universal cover of $S^1$.\r\n\\end{example}\r\n\\begin{corollary}\r\n If $p:\\hat{X}\\to X$ is a universal cover, then for any choice of $\\hat{x}\\in p^{-1}(\\{x\\})$ there is a bijection\r\n $$\\pi_1(X,x)\\to p^{-1}(\\{x\\}),[\\gamma]\\mapsto \\hat{x}\\cdot\\gamma$$\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate from what we have discussed.\r\n\\end{proof}\r\nSo we can get a group structure on $p^{-1}(\\{x\\})$ which is essentially isomorphic to $\\pi_1(X,x)$ and determined by $\\hat{x}\\cdot (\\gamma\\cdot\\delta)=(\\hat{x}\\cdot\\gamma)\\cdot\\delta$.\r\nThis allows us to actually calculate some fundamental groups.\r\n\\begin{example}\r\n Now we compute the fundamental group of the circle.\r\n Let $p:\\mathbb R\\to S^1$ be the usual universal cover.\r\n Hence there is a bijection\r\n $$\\pi_1(S^1,1)\\to p^{-1}(\\{1\\})=\\mathbb Z\\subset\\mathbb R$$\r\n Now for $n\\in\\mathbb Z$, take $\\tilde{\\gamma}_n(t)=nt$ to be a path from $0$ to $n$ in $\\mathbb R$ and let $\\gamma_n=p\\circ\\hat{\\gamma}_n$ which has to be a loop based at $1$.\r\n Clearly $0\\cdot\\gamma_n=\\hat{\\gamma}_n(1)=n$.\r\n Thus any loop in $S^1$ based at $1$ is homotopic to some $\\gamma_n$.\r\n So the bijection becomes $[\\gamma_n]\\mapsto n$.\r\n It remains to show that it is a homomorphism.\r\n Indeed, for any $m\\in\\mathbb Z$, we know that $m+\\hat{\\gamma}_n$ is a lift of $\\gamma_n$ from $m$ to $m+n$.\r\n So\r\n $$(0\\cdot\\gamma_m)\\cdot\\gamma_n=m\\cdot\\gamma_n=m+n=0\\cdot\\gamma_{m+n}$$\r\n Hence it is indeed a homomorphism, it then follows that it is an isomorphism, which means $\\pi_1(S^1,1)\\cong\\mathbb Z$.\r\n\\end{example}\r\nIn complex analysis, we defined the winding number as, loosely speaking, the number of times a loop wraps around a certain point.\r\nIt then follows that this notion is essentially describing the homotopy class a curve is in when we put the thing in $\\pi_1(S^1,1)$.\r\nThere are more applications of this idea.\r\n\\subsection{Applications of Fundamental Groups}\r\n\\begin{theorem}[No-Retraction Theorem]\\label{no_retraction}\r\n The identity map of $S^1$ does not extend to a map $r:D^2\\to S^1$ (where we view $S^1$ as $\\partial D^2$).\r\n\\end{theorem}\r\nThat is, $S^1$ is not a retract of $D^2$.\r\nWe have essentially covered the proof in the introduction, where all technical details have just been covered.\r\n\\begin{proof}\r\n If such an $r$ exists, let $\\iota:S^1\\hookrightarrow D^2$ be the inclusion map, then $r\\circ\\iota=\\operatorname{id}_{S^1}$.\r\n As $D^2$ is contractible, $\\pi_1(D^2,1)=0$, so there is a factorisation\r\n $$(\\operatorname{id}_{S^1})_\\ast=(r\\circ\\iota)_\\ast=r_\\ast\\circ\\iota_\\ast:\\pi_1(S^1,1)\\to\\pi_1(D^2,1)\\to\\pi_1(S^1,1)$$\r\n But the last diagram is just $\\mathbb Z\\to\\{0\\}\\to\\mathbb Z$.\r\n This means that $(\\operatorname{id}_{S^1})_\\ast$ has to be constant.\r\n But it isn't.\r\n Contradiction.\r\n\\end{proof}\r\n\\begin{theorem}[Brouwer's Fixed Point Theorem]\r\n Any map $f:D^2\\to D^2$ has a fixed point.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Suppose not, then let $g:D^2\\to S^1$ be the map given by projecting $f(x)$ through $x$ onto $S^1$.\r\n That is, $g(x)$ is the intersection of the line joining $x$ and $f(x)$ and $S^1$ that is closer to $x$.\r\n Then $g$ is continuous and $g(x)=x$ for any $x\\in S^1$.\r\n This however means that $g$ restricts to $\\operatorname{id}_{S^1}$, contradicing Theorem \\ref{no_retraction}.\r\n\\end{proof}\r\n\\begin{theorem}[The Fundamental Theorem of Algebra]\r\n Every nonconstant polynomial $p:\\mathbb C\\to\\mathbb C$ has a zero.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Let $r:\\mathbb C\\setminus\\{0\\}\\to S^1$ via $r(z)=z/|z|$ which is a retraction.\r\n For $R\\ge 0$, we define $\\lambda_R:S^1\\to\\mathbb C, \\lambda_R(z)=Rz$.\r\n If $p$ has no zero, then we can define $f_R:r\\circ p\\circ \\lambda_R:S^1\\to S^1$.\r\n Easily for any $R_1,R_2\\in\\mathbb R_{\\ge 0}$, $f_{R_1}$ and $f_{R_2}$ are homotopic, hence $(f_{R_1})_\\ast=(f_{R_2})_\\ast$.\r\n But these induced maps are all homomorphisms $\\mathbb Z\\to\\mathbb Z$.\r\n Therefore all $(f_R)^\\ast$ is given by multiplication by $d$ for some fixed $d\\in\\mathbb N\\setminus\\{0\\}$.\r\n But $f_0$ is constant, therefore $d=0$ and $(f_R)_\\ast$ are constantly zero.\r\n But for very large $R$, the leading term dominates $p$, so it is clear that $(f_R)_\\ast$ is given by multiplication by $\\deg p\\neq 0$, contradiction.\r\n\\end{proof}\r\n\\begin{definition}\r\n A space $X$ is locally simply connected if, for every $x\\in X$ and open neighbourhood $U$ of $x$, there exists a simply connected neighbourhood $x\\in V\\subset U$.\r\n\\end{definition}\r\n\\begin{example}[Non-example]\r\n Take the Hawaii Earing space\r\n $$X=\\bigcup_{n=1}^\\infty \\left\\{(x,y)\\in\\mathbb R^2:\\left( x-\\frac{1}{n} \\right)+y^2=\\frac{1}{n^2}\\right\\}$$\r\n then no open neighbourhood of $(0,0)$ is simply connected.\r\n\\end{example}\r\n\\begin{theorem}[Existence of Universal Cover]\r\n Let $X$ be a path-connected space such that $X$ is localled simply connected, then there exists a universal cover $p:\\hat{X}\\to X$.\r\n\\end{theorem}\r\n\\begin{proof}[Sketch of proof]\r\n Fix $x_0\\in X$ and consider the set $\\mathscr X$ of all paths $\\gamma$ from $x_0$.\r\n Define $\\hat{X}=\\mathscr X/\\simeq$ where $\\simeq$ is the path homotopy equivalence relation.\r\n The intended map $p:\\hat{X}\\to X$ is gives by $p([\\gamma])=\\gamma(1)$.\r\n The tricky bit is to find a topology on $\\hat{X}$ making it work, which -- guess what -- is skipped.\r\n\\end{proof}\r\n\\begin{example}\r\n Take $X$ to be the figure eight, then $\\hat{X}$ looks like the Caylay graph of the free group $F_2$.\r\n\\end{example}\r\n\\subsection{The Galois Correspondence}\r\nThe idea is to classify all covering spaces using subgroups of the fundamental group, which is in certain ways analogous to the idea of Galois correspondence in Galois theory.\r\n\\begin{definition}\r\n Let $X$ be a path-connected space and $p_1:\\hat{X}_1\\to X_1, p_2:\\hat{X}_2\\to X$.\r\n An isomorphism of covering spaces is a homeomorphism $\\phi:\\hat{X}_1\\to\\hat{X}_2$ such that\r\n \\[\r\n \\begin{tikzcd}\r\n X&\\hat{X}_1\\arrow{dl}{\\phi}\\arrow[swap]{l}{p_1}\\\\\r\n \\hat{X}_2\\arrow{u}{p_2}&\r\n \\end{tikzcd}\r\n \\]\r\n commutes.\r\n\\end{definition}\r\nNote also that $\\phi^{-1}$ is automatically an isomorphism of covering spaces as well\r\nIf $\\hat{X}_i$ are equipped with basepoints $\\hat{x}_i\\in\\hat{X}_i$ and $\\phi(\\hat{x}_1)=\\hat{x}_2$, we say $\\phi$ is based.\r\n\\begin{remark}\r\n Note that $\\phi$ is a lift of $p_1$ to $\\hat{X}_2$, so by Lemma \\ref{lift_unique}, a based isomorphism is uniquely determined by the basepoints $\\phi(\\hat{x}_1)=\\hat{x}_2$ if $\\hat{X}_1$ is connected and locally path-connected.\r\n\\end{remark}\r\n\\begin{theorem}[Galois Correspondence with Basepoints]\\label{based_galois}\r\n Let $X$ be a path-connected, locally simply connected space with basepoint $x_0$.\r\n The map which sends a covering $p:\\hat{X}\\to X$ equipped with a basepoint $\\hat{x}_0\\in p^{-1}(\\{x_0\\})$ to the subgroup $p_\\ast(\\pi_1(\\hat{X},\\hat{x}_0))\\le\\pi_1(X,x_0)$ induces a bijection between the set of based isomorphism classes of path-connected covering spaces with basepoint and the set of subgroups of $\\pi_1(X,x_0)$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Not dreadfully hard but omitted.\r\n\\end{proof}\r\n\\begin{example}\r\n As $\\pi_1(S^1,1)=\\mathbb Z$, each subgroup of $\\mathbb Z$ is of the form $n\\mathbb Z$ for natural number $n$ (including $0$).\r\n Then the usual $p:\\mathbb R\\to S^1, t\\mapsto e^{2\\pi it}$ corresponds to the subgroup $\\{0\\}$, and the maps $p:S^1\\to S^1, z\\mapsto z^n$ corresponds to the subgroups $n\\mathbb Z$ for $n\\neq 0$.\r\n Hence, these are the only path-connected covering spaces of $S^1$ up to based isomorphism.\r\n\\end{example}\r\n\\begin{corollary}\r\n Let $X$ be a path-connected and locally simply connected space, then any two universal covers $p_1:\\hat{X}_1\\to X$ and $p_2:\\hat{X}_2\\to X$ are isomorphic.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\n\\begin{corollary}[Galois Correspondence without Basepoints]\\label{unbased_galois}\r\n Let $X$ be a path-connected, locally simply connected space with basepoint $x_0$.\r\n Then the map that sends a covering $p:\\hat{X}\\to X$ equipped with a basepoint $\\hat{x}_0\\in p^{-1}(\\{x_0\\})$ to the subgroup $p_\\ast(\\pi_1(\\hat{X},\\hat{x}))\\le\\pi_1(X,x_0)$ induces a bijection between (unbased) isomorphism classes of path-connected convering spaces of $X$ and conjugacy classes of subgroups of $\\pi_1(X,x_0)$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n This map is surjective due to Theorem \\ref{based_galois}.\r\n To see it is injective, we need to show that if $(p_1)_\\ast(\\pi_1(\\hat{X}_1,\\hat{x}_1))$ and $(p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2))$ are conjugate subgroups of $\\pi_1(X,x_0)$, then there is an (unbased) isomorphism $\\phi:\\hat{X}_1\\to\\hat{X}_2$ of covering spaces.\\\\\r\n Suppose\r\n $$(p_1)_\\ast(\\pi_1(\\hat{X}_1,\\hat{x}_1))=[\\gamma](p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2))[\\bar\\gamma]$$\r\n for some $[\\gamma]\\in\\pi_1(X,x_0)$.\r\n Let $\\widehat{\\bar{\\gamma}}$ be the lift of $\\bar\\gamma$ at $\\hat{x}_2$ and $\\hat{x}_2'$ be the other endpoint of $\\widehat{\\bar\\gamma}$.\r\n By the last part of Lemma \\ref{indep_basepoint},\r\n \\begin{align*}\r\n [\\gamma](p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2))[\\bar\\gamma]&=\\bar\\gamma_\\#((p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2)))\\\\\r\n &=(p_2)_\\ast(\\widehat{\\bar\\gamma}_\\#(\\pi_1(\\hat{X}_2,\\hat{x}_2)))\\\\\r\n &=(p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2'))\r\n \\end{align*}\r\n Therefore $(p_1)_\\ast(\\pi_1(\\hat{X}_1,\\hat{x}_1))=(p_2)_\\ast(\\pi_1(\\hat{X}_2,\\hat{x}_2'))$.\r\n Theorem \\ref{based_galois} then gives a based isomorphism $(\\hat{X}_1,\\hat{x}_1)\\cong(\\hat{X}_2,\\hat{x}_2')$.\r\n In particular, $\\hat{X}_1\\cong\\hat{X}_2$ as (unbased) covering spaces.\r\n\\end{proof}", "meta": {"hexsha": "ab8781ce9ef9ea03e86f2683a1e909b2c177076a", "size": 27359, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/cover.tex", "max_stars_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_stars_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-03T17:38:28.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-03T17:38:28.000Z", "max_issues_repo_path": "2/cover.tex", "max_issues_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_issues_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2/cover.tex", "max_forks_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_forks_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.8083989501, "max_line_length": 333, "alphanum_fraction": 0.6451990204, "num_tokens": 9529, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.8080672204860316, "lm_q1q2_score": 0.6429452293752298}} {"text": "\\documentclass{article}\n % General document formatting\n \\usepackage[margin=0.7in]{geometry}\n \\usepackage[parfill]{parskip}\n \\usepackage[utf8]{inputenc}\n \\usepackage{mathrsfs}\n \\usepackage{amsmath}\n \\usepackage{amssymb}\n \\usepackage{tikz}\n \\usepackage{fancyhdr}\n \\usepackage{multicol}\n\n \\usetikzlibrary{positioning}\n\n\\pagestyle{fancy}\n\\fancyhf{}\n\\rhead{Edgar Jacob Rivera Rios - A01184125}\n\n\\renewcommand{\\labelenumi}{\\alph{enumi})}\n\n\\begin{document}\n\\section*{3.3.1}\nSales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week.\nA sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2.\nProvide 90\\% and 95\\% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.\n\\begin{align*}\n \\text{Margin} &= 90\\%\\\\\n s &= 5.2\\\\\n n &= 65\\\\\n \\bar{x} &= 19.5\\\\\n \\text{Error Margin} &= t_{\\alpha/2}(s/ \\sqrt{n})\\\\\n &=t_{.1/2}(5.2/ \\sqrt{65})\\\\\n &=t_{.05}(0.644980619863884)\\\\\n &=1.669(0.644980619863884)\\\\\n &=1.07648105544093\\\\\n interval &= \\bar{x} \\pm \\text{Error Margin}\\\\\n &= 19.5 \\pm 1.07648105544093\\\\\n &= (18.4235189445591, 20.5764810554409)\n\\end{align*}\n\n\\begin{align*}\n \\text{Margin} &= 95\\%\\\\\n s &= 5.2\\\\\n n &= 65\\\\\n \\bar{x} &= 19.5\\\\\n \\text{Error Margin} &= t_{\\alpha/2}(s/ \\sqrt{n})\\\\\n &=t_{.05/2}(5.2/ \\sqrt{65})\\\\\n &=t_{.025}(0.644980619863884)\\\\\n &=1.998(0.644980619863884)\\\\\n &=1.28849691076229\\\\\n interval &= \\bar{x} \\pm \\text{Error Margin}\\\\\n &= 19.5 \\pm 1.28849691076229\\\\\n &= (18.2115030892377, 20.7884969107623)\n\\end{align*}\n\n\\section*{3.3.2}\nThe mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month (The Wall Street Journal, February 25, 2003). Assume that this mean was based on actual flying times for a sample of 100 Continental pilots and that the sample standard deviation was 8.5 hours.\n\\begin{enumerate}\n \\item At 95\\% confidence, what is the margin of error?\n \\begin{align*}\n \\text{Margin} &= 95\\%\\\\\n s &= 8.5\\\\\n n &= 100\\\\\n \\bar{x} &= 49\\\\\n \\text{Error Margin} &= t_{\\alpha/2}(s/ \\sqrt{n})\\\\\n &=t_{.05/2}(8.5/ \\sqrt{100})\\\\\n &=t_{.025}(0.85)\\\\\n &=1.984(0.85)\\\\\n &=1.68658440884845\n \\end{align*}\n\n \\item What is the 95\\% confidence interval estimate of the population mean flying time for the pilots?\n \\begin{align*}\n interval &= \\bar{x} \\pm \\text{Error Margin}\\\\\n &= 59 \\pm 1.68658440884845\\\\\n &= (47.3134155911515, 50.6865844088485 )\n \\end{align*}\n\n \\item The mean number of hours of flying time for pilots at United Airlines is 36 hours per month. Use your results from part (b) to discuss differences between the flying times for the pilots at the two airlines. (The Wall Street Journal reported United Airlines as having the highest labor cost among all airlines. Does the information in this exercise provide insight as to why United Airlines might expect higher labor costs?)\n \n It's probably because United airlines pilots fly a lot less, so the costs may stay relative high when compared to pilots who fly 10 hours more a month\n\\end{enumerate}\n\n\\section*{3.3.3}\nThe average cost per night of a hotel room in New York City is \\$273 (Smart Money, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is \\$65.\n\\begin{enumerate}\n \\item With 95\\% confidence, what is the margin of error?\n \\begin{align*}\n \\text{Margin} &= 95\\%\\\\\n s &= 65\\\\\n n &= 45\\\\\n \\bar{x} &= 273\\\\\n \\text{Error Margin} &= t_{\\alpha/2}(s/ \\sqrt{n})\\\\\n &=t_{.05/2}(65/ \\sqrt{45})\\\\\n &=t_{.025}(9.689)\\\\\n &=2.015(9.689)\\\\\n &=19.5281618831222\n \\end{align*}\n \n \\item What is the 95\\% confidence interval estimate of the population mean?\n \\begin{align*}\n interval &= \\bar{x} \\pm \\text{Error Margin}\\\\\n &= 265 \\pm 19.5281618831222\\\\\n &= (253.471838116878, 292.528161883122)\n \\end{align*}\n\n \\item Two years ago the average cost of a hotel room in New York City was \\$229. Discuss the change in cost over the two-year period\n \n It seems that the average cost of hotels in New York from two years ago tends to be to the lower estimate of the actual average. Which is normal considering the inflation, which is that costs go up and the money looses value over time.\n\\end{enumerate}\n\n\\section*{3.3.4}\nIs your favorite TV program often interrupted by advertising? CNBC presented statistics on the average number of programming minutes in a half-hour sitcom (CNBC, February 23, 2006). The following data (in minutes) are representative of their findings:\n\\begin{table}[h!]\n \\centering\n \\begin{tabular}{c c c c c c}\n 20.02&22.20&21.20&21.06&22.24&20.62\\\\\n 22.37&22.19&22.34&21.66&21.23&23.86\\\\\n 23.82&20.30&21.52&21.52&21.91&23.1\\\\\n 23.36&23.44&&&&\\\\\n \\end{tabular}\n\\end{table}\nAssume the population is approximately normal. Provide a point estimate and a 95\\% confidence interval for the mean number of programming minutes during a half-hour television sitcom.\n\n\\begin{align*}\n \\text{Margin} &= 95\\%\\\\\n s &= 1.11702424230396\\\\\n n &= 20\\\\\n \\bar{x} &= 21.998\\\\\n \\text{Error Margin} &= t_{\\alpha/2}(s/ \\sqrt{n})\\\\\n &=t_{.05/2}(1.11702424230396/ \\sqrt{20})\\\\\n &=t_{.025}(0.249774213830685)\\\\\n &=2.093(0.249774213830685)\\\\\n &=0.522783437718549\\\\\n interval &= \\bar{x} \\pm \\text{Error Margin}\\\\\n &= 21.998 \\pm 0.522783437718549\\\\\n &= (21.4752165622814, 22.5207834377185)\n\\end{align*}\n\\end{document}", "meta": {"hexsha": "79ada36cbe1ef64a5838f33248429691795fb724", "size": 5507, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homework/Homework3_3.tex", "max_stars_repo_name": "edjacob25/Applied-Maths", "max_stars_repo_head_hexsha": "0a0f8e5b88083a1b0ec85069efbf266b6a12c741", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homework/Homework3_3.tex", "max_issues_repo_name": "edjacob25/Applied-Maths", "max_issues_repo_head_hexsha": "0a0f8e5b88083a1b0ec85069efbf266b6a12c741", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homework/Homework3_3.tex", "max_forks_repo_name": "edjacob25/Applied-Maths", "max_forks_repo_head_hexsha": "0a0f8e5b88083a1b0ec85069efbf266b6a12c741", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.618705036, "max_line_length": 432, "alphanum_fraction": 0.6769566007, "num_tokens": 1880, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581097540519, "lm_q2_score": 0.8080672066194945, "lm_q1q2_score": 0.6429452261731039}} {"text": "%!TEX root = TDT4265-Summary.tex\r\n\\section{Optical flow}\r\nOptical flow is the task of describing how pixels move in an image, thus describing motion of surfaces or objects in an image. Mathematically, you calculate the motion at each pixel position between two frames at times $t$ and $t + \\Delta t$.\r\n\r\n\\subsection{Constraints}\r\nWe usually assume constant brightness between frames\r\n\\begin{equation}\r\n I(x,y,t) = I(x + \\Delta x, y + \\Delta y, t + \\Delta t)\r\n\\end{equation}\r\nwhere $I(\\cdot)$ is image intensity. We also assume small motion\r\n\\begin{multline}\r\n I(x + \\Delta x, y + \\Delta y, t + \\Delta t) =\\\\\r\n I(x,y,z) + \\pd{I}{x}\\Delta x + \\pd{I}{y}\\Delta y + \\pd{I}{t} \\Delta t.\r\n\\end{multline}\r\nThis gives\r\n\\begin{equation}\r\n \\pd{I}{x} V_x + \\pd{I}{y} V_y + \\pd{I}{t} = 0.\r\n\\end{equation}\r\n\r\nDetermining optical flow can be done in several ways. Sections \\ref{ssec:lucas-kanade} and \\ref{ssec:horn-schunck} describe differential methods.\r\n\r\n\\paragraph{Aperture problem} The aperture problem comes from the fact that what a camera sees is only a small part of the world. Motion inside of this frame may appear different than it really is, if the moving object is not contained in the frame.\r\n\r\n\\subsection{Lucas--Kanade (LK) method}\\label{ssec:lucas-kanade}\r\nLK tracks interest points in a scene from one frame to the next. Can use Harris corner points as the interest points. It assumes constant flow in a neighborhood of the pixel examined, and solves the optical flow equations by least squares. Using a neighborhood reduces ambiguities and improves noise performance. But being purely local, it cannot determine flow in the interior of uniform regions.\r\n\r\n\\subsection{Horn--Schunck (HS) method}\\label{ssec:horn-schunck}\r\nA method to estimate optical flow globally. It introduces a global smoothness constraint to solve the aperture problem. It minimizes distortion in flow globally, and prefers solutions that are smoother.\r\n\r\n\\subsection{Kanade--Lucas-Tomasi (KLT) feature tracker}\\label{ssec:kanade-lucas-tomasi}\r\nKLT is a computationally fast feature extraction method that is useful for tracking in realtime video.\r\n", "meta": {"hexsha": "a09e9d446f212cd66090d43af356f7b9a0b09694", "size": 2138, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TDT4265 Computer vision/optical-flow.tex", "max_stars_repo_name": "jakoblover/ntnu-course-summaries", "max_stars_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-05-30T09:19:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-11T02:42:40.000Z", "max_issues_repo_path": "TDT4265 Computer vision/optical-flow.tex", "max_issues_repo_name": "jakoblover/ntnu-course-summaries", "max_issues_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TDT4265 Computer vision/optical-flow.tex", "max_forks_repo_name": "jakoblover/ntnu-course-summaries", "max_forks_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 66.8125, "max_line_length": 398, "alphanum_fraction": 0.7464920486, "num_tokens": 558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.808067204308405, "lm_q2_score": 0.7956580976404297, "lm_q1q2_score": 0.6429452145456459}} {"text": "\\documentclass{article}\n\\input{preamble}\n\\begin{document}\n\\section{Noise Calculation}\n\nThe assumption is that the noise of a point source with flux $F$ has a variance \n obtained by quadratically adding the variance of a background $N_0^2$ and a quantity proportional to the Poisson Noise due to the source.\n\\be\nN^2 = N_0^2 + \\alpha F \n\\ee\nGiven a source Flux, the noise has two unknown quantities $\\alpha$ and $N_0$.\nTherefore, it is posible to eliminate one of these variables in terms of \nquantities at a standard SNR. Let us consider this standard SNR to be 5.\nThen, we have \n\\be\n\\frac{F_5^2}{N_0^2 + \\alpha F_5} = 25\n\\ee\nimplying we can eliminate \n\\be\n\\alpha = F_5 \\left( 0.04 - \\left(\\frac{N_0}{F_5}\\right)^2 \\right)\n\\ee\nAt the linear level, the noise in the magnitude $\\sigma_m$ is \n\\beqn\n\\sigma_m^2 &=& \\left(\\frac{2.5}{\\log{(10)}}\\right)^2 (N/F)^2 \\\\\n&\\approx& \\left(\\frac{N_0}{F }\\right)^2 + \\alpha \\left(\\frac{1}{F}\\right) \\\\\n&=& \\left(\\frac{N_0}{F_5}\\right)^2 \\left(F_5/F\\right)^2 + \\left( 0.04 - ({N_0/F_5})^2\\right) (F_5/F) \\\\\n&=& \\gamma x^2 + (0.04 - \\gamma) x\n\\eeqn\nWe could put back the initial missing factor $\\approx 1$ to get \n\\be\n\\sigma_m^2 = \\left(2.5/\\log{(10)}\\right)^2 \\left( \\gamma x^2 + (0.04 - \\gamma ) x \\right)\n\\ee\nwhere \n$\\gamma \\equiv \\left( N_0/ F_5 \\right)^2$ and $x \\equiv F_5/F = 10.0^{0.4(m-m_5)}.$ In terms of $\\gamma$ and $m_5,$ we may write the value of $\\alpha$:\n\\be\n\\alpha = 10.0^{-0.4 m_5} ( 0.04 - \\gamma)\n\\ee\nWe could write the Flux in terms of counts. This would be\n$$\nC_5 = \\frac{\\pi D^2 T_{exp}}{4 g h} F5\n$$\nThen, \n\\be\n\\gamma = \\left(\\frac{\\pi D^2 T_{exp} N0}{4 g h C_5}\\right)^2\n\\ee\n% \\section{Rewrite}\n% We will consider PSF photometry of a point source as described in the SNR document using the estimator \n% \\be \n% \\hat{S} = \\sum w_i C_i \n% \\ee. Here, the physical signal is the Flux of an astrophysical point source in absolute units in a filter band. This means that number of counts $C$ and the variance of the estimator is \n% \\be\n% N^2 = C/g + N_0^2 \n% \\ee\n% where g is the gain \n\\end{document}\n", "meta": {"hexsha": "85683f8181b762d0710a8adb76350aaf379056fb", "size": 2054, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "snsims_previous/docs/NoiseCalculation.tex", "max_stars_repo_name": "rbiswas4/SNsims", "max_stars_repo_head_hexsha": "8df1fec13bcb8b06cbb460663d96625854aa97fe", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-09-01T16:17:32.000Z", "max_stars_repo_stars_event_max_datetime": "2017-09-01T16:17:32.000Z", "max_issues_repo_path": "snsims_previous/docs/NoiseCalculation.tex", "max_issues_repo_name": "rbiswas4/SNsims", "max_issues_repo_head_hexsha": "8df1fec13bcb8b06cbb460663d96625854aa97fe", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 68, "max_issues_repo_issues_event_min_datetime": "2015-04-16T22:54:30.000Z", "max_issues_repo_issues_event_max_datetime": "2018-02-09T23:34:45.000Z", "max_forks_repo_path": "snsims_previous/docs/NoiseCalculation.tex", "max_forks_repo_name": "rbiswas4/SNsims", "max_forks_repo_head_hexsha": "8df1fec13bcb8b06cbb460663d96625854aa97fe", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-02-10T19:19:04.000Z", "max_forks_repo_forks_event_max_datetime": "2015-05-05T20:45:39.000Z", "avg_line_length": 36.6785714286, "max_line_length": 188, "alphanum_fraction": 0.6684518014, "num_tokens": 764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8757870013740061, "lm_q2_score": 0.7341195327172402, "lm_q1q2_score": 0.6429323442085183}} {"text": "\n\\section{Macaulay Notation}\n\nMacaulay Notation\\footnote{N.b. This should all be familiar to you from your 1st Year Solid Mechanics.} is a method used for the structural analysis of Euler-Bernoulli beams and describes the beam forces, moments and deflection. The method is particularly useful for discontinuous and/or discrete loading scenarios as well as loadings that are uniformly distributed loads (u.d.l.) and/or uniformly varying loads (u.v.l.) over the span of a beam.\n\nThe\\marginnote{Method} method starts with the Euler-Bernoulli beam theory and the relation between the deflection $w$ and bending moment $M$.\n\n\\begin{equation}\n \\pm EI\\frac{\\text{d}^2 w}{\\text{d}x^2} = M\n\\end{equation}\n\n\\noindent Where $E$ is the elastic modulus and $I$ is the second moment of area.\n\nIn terms of Macaulay Notation, $M$ is expressed in the form:\n\n\\begin{equation}\n M = M_1(x) + P_1\\langle x-a_1\\rangle^{b_1} + P_2\\langle x-a_2\\rangle^{b_2} + P_3\\langle x-a_3\\rangle^{b_3} + \\ldots\n\\end{equation}\n\n\\noindent Where $M_1$ is the moment at the start of $x$ and $P_i\\langle x-a_i\\rangle^{b_i}$ representing elements along the beam that contribute to the moment. These contribute to the scenario when $x$ becomes greater than $a_i$:\n\n\\begin{equation}\n \\langle x - a_i\\rangle = \n \\begin{cases} \n 0 & \\mathrm{if}~ x < a_i \\\\ \n x - a_i & \\mathrm{if}~ x > a_i \n \\end{cases}\n\\end{equation}\n\n\\noindent $b_i$ is determined by the type of loading that is being applied. \n\nFor\\marginnote{Shear Force} the shaft design exercise, you will be taking the shear forces and integrating them to get your bending moments for the two axes. For example, the Macaulay Notation for the Free Body Diagrams in \\cref{fig-fbd} are as follows:\n\\begin{equation}\n S_v = R_{v1}\\langle x-x_1\\rangle^0 + F_{1}\\langle x-x_2\\rangle^0 + F_{2}\\langle x-x_3\\rangle^0 + R_{v2}\\langle x-x_4\\rangle^0\n\\end{equation}\n\\begin{equation}\n S_h = F_{3}\\langle x-x_0\\rangle^0 + R_{h1}\\langle x-x_2\\rangle^0 + R_{h2}\\langle x-x_4\\rangle^0\n\\end{equation}\n\nFrom\\marginnote{Shear Force Diagram} these equations, the shear force diagrams for the two axes can be generated (\\cref{fig-sfd}).\n\n\\begin{figure*}[th!]\n \n \\hfill\n \\subfloat[Vertical Shear]{\n \\includestandalone[width=0.45\\textwidth, mode=buildnew]{03_macaulay_notation/vertical_shear}\n }\n \\hfill\n \\subfloat[Horizontal Shear]{\n \\includestandalone[width=0.45\\textwidth, mode=buildnew]{03_macaulay_notation/horizontal_shear}\n }\n \\hfill\n \n \\vspace{2em}\n \\caption{Shear force diagrams}\n \\label{fig-sfd}\n\\end{figure*}\n\n\nHaving\\marginnote{Bending Moment} described the shear forces in Macaulay Notation, it is then the case of integrating and determining the constant of integration to arrive at an equation that describes the bending moment at any point through the beam.\n\\begin{equation}\n M_v = \\int S_v = R_{v1}\\langle x-x_1\\rangle^1 + F_{1}\\langle x-x_2\\rangle^1 + F_{2}\\langle x-x_3\\rangle^1 + R_{v2}\\langle x-x_4\\rangle^1\n\\end{equation}\n\\begin{equation}\n M_h = \\int S_h = F_{3}\\langle x-x_0\\rangle^1 + R_{h1}\\langle x-x_2\\rangle^1 + R_{h2}\\langle x-x_4\\rangle^1\n\\end{equation}\n\nUsing\\marginnote{Bending Moment Diagram} these equations, one can obtain the bending moment diagrams for the loaded beam (\\cref{fig-bmd}).\n\n\\begin{figure*}[th!]\n\n \\hfill\n \\subfloat[Vertical Bending]{\n \\includestandalone[width=0.45\\textwidth, mode=buildnew]{03_macaulay_notation/vertical_bending}\n }\n \\hfill\n \\subfloat[Horizontal Bending]{\n \\includestandalone[width=0.45\\textwidth, mode=buildnew]{03_macaulay_notation/horizontal_bending}\n }\n \\hfill\n \n \\vspace{2em}\n \\caption{Bending Moment Diagrams}\\label{fig-bmd}\n\\end{figure*}\n", "meta": {"hexsha": "263c31a7553a6656e34c443ff7916fb9b2494e0a", "size": 3712, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "03_macaulay_notation/section.tex", "max_stars_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_stars_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "03_macaulay_notation/section.tex", "max_issues_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_issues_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "03_macaulay_notation/section.tex", "max_forks_repo_name": "JamesGopsill/ShaftDesignCourseNotes", "max_forks_repo_head_hexsha": "0249e0804538237df58e5a7f99d039b35b30099d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.6705882353, "max_line_length": 445, "alphanum_fraction": 0.7249461207, "num_tokens": 1192, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8757869851639065, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6429323221193618}} {"text": "Software architecture is of central importance in any large software\nproject because it establishes predictable patterns of usage and\ndevelopment~\\cite{Shaw1996}. This section describes the essential\nstructural components of SymPy, provides justifications for the design\ndecisions that have been made, and gives example\nuser-facing code as appropriate.\n\n\\subsection{The Core}\n\\label{sec:core}\nA computer algebra system stores mathematical expressions as data structures.\nFor example, the mathematical expression $x + y$ is represented as a tree with\nthree nodes, $+$, $x$, and $y$, where $x$ and $y$ are ordered children of $+$.\nAs users manipulate mathematical expressions with traditional mathematical\nsyntax, the CAS manipulates the underlying data structures. Symbolic\ncomputations such as integration, simplification, etc.\\ are all functions that\nconsume and produce expression trees.\n\nIn SymPy every symbolic expression is an instance of the class\n\\texttt{Basic},\\footnote{Some internal classes, such as those used in the\n polynomial submodule, do not follow this rule for efficiency reasons.} the\nsuperclass of all SymPy types providing common methods to all SymPy\ntree-elements, such as traversals. The children of a node in the tree are held\nin the \\texttt{args} attribute. A leaf node in the expression tree\nhas empty \\texttt{args}.\n\nFor example, consider the expression $xy + 2$:\n\\begin{verbatim}\n>>> x, y = symbols('x y')\n>>> expr = x*y + 2\n\\end{verbatim}\nBy order of operations, the parent of the expression tree for \\texttt{expr} is\nan addition. It is of type \\texttt{Add}. The child nodes of \\texttt{expr} are\n\\texttt{2} and \\texttt{x*y}.\n\\begin{verbatim}\n>>> type(expr)\n\n>>> expr.args\n(2, x*y)\n\\end{verbatim}\n\nDescending further down into the expression tree yields the full expression. For\nexample, the next child node (given by \\texttt{expr.args[0]}) is\n\\texttt{2}. Its class is \\texttt{Integer}, and it has an empty \\texttt{args}\ntuple, indicating that it is a leaf node.\n\\begin{verbatim}\n>>> expr.args[0]\n2\n>>> type(expr.args[0])\n\n>>> expr.args[0].args\n()\n\\end{verbatim}\nSymbols or symbolic constants, like $e$ or $\\pi$, are other examples of\nleaf nodes.\n\\begin{verbatim}\n>>> exp(1)\nE\n>>> exp(1).args\n()\n>>> x.args\n()\n\\end{verbatim}\n\nA useful way to view an expression tree is using the \\texttt{srepr} function, which\nreturns a string representation of an expression as valid Python code\\footnote{\n\\label{note:dotprint}\nThe \\texttt{dotprint} function from the \\texttt{sympy.printing.dot} submodule\nprints output to dot format, which can be rendered with Graphviz to\nvisualize expression trees graphically.}\nwith all the nested class constructor calls to create the given expression.\n\\begin{verbatim}\n>>> srepr(expr)\n\"Add(Mul(Symbol('x'), Symbol('y')), Integer(2))\"\n\\end{verbatim}\n\nEvery SymPy expression satisfies a key identity invariant:\n\\begin{verbatim}\nexpr.func(*expr.args) == expr\n\\end{verbatim}\nThis means that expressions are\nrebuildable from their \\texttt{args}.\\footnote{\\texttt{expr.func} is used\ninstead of \\texttt{type(expr)} to allow the function of an expression to be\ndistinct from its actual Python class. In most cases the two are the same.}\nNote that in SymPy the \\texttt{==} operator represents exact\nstructural equality, not mathematical equality. This allows testing if any two\nexpressions are equal to one another as expression trees. For example, even\nthough ${(x + 1)}^2$ and $x^2 + 2x + 1$ are equal mathematically, SymPy gives\n\\begin{verbatim}\n>>> (x + 1)**2 == x**2 + 2*x + 1\nFalse\n\\end{verbatim}\nbecause they are different as expression trees (the former is a \\verb|Pow|\nobject and the latter is an \\verb|Add| object).\n\nAnother important property of SymPy expressions is that they are immutable.\nThis simplifies the design of SymPy, and enables expression interning. It also\nenables expressions to be hashed, which allows expressions to be used as keys\nin Python dictionaries, and is used to implement caching in SymPy.\n\nPython allows classes to override mathematical operators. The Python\ninterpreter translates the above \\texttt{x*y + 2} to, roughly,\n\\verb|(x.__mul__(y)).__add__(2)|. Both \\texttt{x} and \\texttt{y}, returned\nfrom the \\texttt{symbols} function, are \\texttt{Symbol} instances. The\n\\texttt{2} in the expression is processed by Python as a literal, and is\nstored as Python's built in \\texttt{int} type. When \\texttt{2} is passed to the\n\\verb|__add__| method of \\texttt{Symbol}, it is converted to the SymPy type\n\\verb|Integer(2)| before being stored in the resulting expression tree. In\nthis way, SymPy expressions can be built in the natural way using Python\noperators and numeric literals.\n\n%% TODO: describe how assumptions are implemented\n\n%%\n%% Extensibility\n\\subsection{Extensibility}\n\nWhile the core of SymPy is relatively small, it has been extended to a wide variety\nof domains by a broad range of contributors.\nThis is due, in part, to the fact that the same language, Python,\nis used both for the internal implementation and the external usage by users.\nAll of the extensibility capabilities available to\nusers are also utilized by SymPy itself. This eases the transition pathway from\nSymPy user to SymPy developer.\n\nThe typical way to create a custom SymPy object is to subclass an existing\nSymPy class, usually \\texttt{Basic}, \\texttt{Expr}, or \\texttt{Function}. As\nit was stated before, all SymPy classes used for expression trees should be\nsubclasses of the base class \\texttt{Basic}. \\texttt{Expr} is the\n\\texttt{Basic} subclass for mathematical objects that can be added and\nmultiplied together. The most commonly seen classes in SymPy are subclasses of\n\\texttt{Expr}, including \\texttt{Add}, \\texttt{Mul}, and \\texttt{Symbol}.\nInstances of \\texttt{Expr} typically represent complex numbers, but may also\ninclude other ``rings'', like matrix expressions. Not all SymPy classes are\nsubclasses of \\texttt{Expr}. For instance, logic expressions, such as\n\\verb|And(x, y)|, are subclasses of \\texttt{Basic} but not of\n\\texttt{Expr}.\\footnote{See section~\\ref{S-suppsec:Logic} of the supplementary\n material for more information on the\n \\texttt{sympy.logic} submodule.}\n\nThe \\texttt{Function} class is a subclass of \\texttt{Expr} which makes it\neasier to define mathematical functions called with arguments. This includes\nnamed functions like $\\sin(x)$ and $\\log(x)$ as well as undefined functions\nlike $f(x)$. Subclasses of \\texttt{Function} should define a\nclass method \\texttt{eval}, which returns an evaluated value for the function\napplication (usually an instance of some other class, e.g., a \\texttt{Number}),\nor \\texttt{None} if for the given arguments it should not be\nautomatically evaluated.\n\nMany SymPy functions perform various evaluations down the expression tree.\nClasses define their behavior in such functions by defining a relevant\n\\verb|_eval_|\\texttt{\\textit{*}} method. For instance, an object can indicate\nto the \\texttt{diff} function how to take the derivative of itself by defining\nthe \\verb|_eval_derivative(self, x)| method, which may in turn call\n\\texttt{diff} on its \\texttt{args}. (Subclasses of \\texttt{Function} should\nimplement the \\texttt{fdiff} method instead; it returns the derivative of the function\nwithout considering the chain rule.) The most common\n\\verb|_eval_|\\texttt{\\textit{*}} methods relate to the assumptions:\n\\verb|_eval_is_|\\texttt{\\textit{assumption}} is used to deduce\n\\textit{assumption} on the object.\n\nListing~\\ref{fig:gamma-example} presents an example of this extensibility. It\ngives a stripped down version of the \\texttt{gamma} function $\\Gamma(x)$ from\nSymPy. The methods defined allow it to evaluate itself on positive integer\narguments, define the real assumption, allow it to be rewritten in terms of\nfactorial (with \\verb|gamma(x).rewrite(factorial)|), and allow it to be\ndifferentiated. \\texttt{self.func} is used throughout instead of referencing\n\\texttt{gamma} explicitly so that potential subclasses of \\texttt{gamma} can\nreuse the methods.\n\n\\lstset{\n basicstyle=\\ttfamily,\n}\n\n\\begin{lstlisting}[caption={A minimal implementation of \\texttt{sympy.gamma}.},label=fig:gamma-example]\nfrom sympy import Function, Integer, factorial, polygamma\n\nclass gamma(Function):\n @classmethod\n def eval(cls, arg):\n if isinstance(arg, Integer) and arg.is_positive:\n return factorial(arg - 1)\n\n def _eval_is_real(self):\n x = self.args[0]\n # noninteger means real and not integer\n if x.is_positive or x.is_noninteger:\n return True\n\n def _eval_rewrite_as_factorial(self, z):\n return factorial(z - 1)\n\n def fdiff(self, argindex=1):\n from sympy.core.function import ArgumentIndexError\n if argindex == 1:\n return self.func(self.args[0])*polygamma(0, self.args[0])\n else:\n raise ArgumentIndexError(self, argindex)\n\\end{lstlisting}\nThe gamma function implemented in SymPy has many more capabilities than the\nabove listing, such as evaluation at rational points and series expansion.\n\n\n\\subsection{Performance}\n\\label{sec:performance}\n\nDue to being written in pure Python without the use of extension modules,\nSymPy's performance characteristics are generally poorer than that of\nits commercial competitors. For many applications,\nthe performance of SymPy, as measured by clock cycles, memory usage, and memory\nlayout, is sufficient.\nHowever, the boundaries for when SymPy's pure Python strategy becomes\ninsufficient are when the user requires handling of very long expressions or many\nsmall expressions. Where this boundray lies depends on the system at hand, but tends\nto be within the range of $10^4$--$10^6$ symbols for modern computers.\n\nFor this reason, a new project called SymEngine~\\cite{SymEngine} has been started.\nThe aim of this poject is to develop a library with better performance\ncharacteristics for symbolic manipulation. SymEngine is a pure C++ library,\nwhich allows it fine-grained control over the memory layout of expressions.\nSymEngine has thin wrappers to other languages (Python, Ruby,\nJulia, etc.). Its aim is to be the fastest symbolic manipulation library. Preliminary\nbenchmarks suggest that SymEngine performs as well as its commercial and\nopen source competitors.\n\nThe development version of SymPy has recently started to use SymEngine as an\noptional backend, initially in \\texttt{sympy.\\allowbreak{}physics.\\allowbreak{}mechanics} only.\nFuture work will involve\nallowing more algorithms in SymPy to use SymEngine as a backend.\n", "meta": {"hexsha": "769f1289a974bf4b6563c90674d158053f2b831a", "size": 10562, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "architecture.tex", "max_stars_repo_name": "ProgZone/sympy-paper", "max_stars_repo_head_hexsha": "b3b85809cc92d1fd588971f944abda9fa995a426", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 40, "max_stars_repo_stars_event_min_datetime": "2016-03-27T06:55:25.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-22T18:42:39.000Z", "max_issues_repo_path": "architecture.tex", "max_issues_repo_name": "ProgZone/sympy-paper", "max_issues_repo_head_hexsha": "b3b85809cc92d1fd588971f944abda9fa995a426", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 205, "max_issues_repo_issues_event_min_datetime": "2016-03-17T03:08:15.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-01T17:09:29.000Z", "max_forks_repo_path": "architecture.tex", "max_forks_repo_name": "ProgZone/sympy-paper", "max_forks_repo_head_hexsha": "b3b85809cc92d1fd588971f944abda9fa995a426", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 37, "max_forks_repo_forks_event_min_datetime": "2016-03-17T16:02:22.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-15T15:06:54.000Z", "avg_line_length": 46.1222707424, "max_line_length": 103, "alphanum_fraction": 0.7693618633, "num_tokens": 2609, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.746138993030751, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.6428272467371577}} {"text": "\\documentclass[11pt, a4paper]{article}\n\n\\usepackage{tikz}\n\\usetikzlibrary{shapes,arrows}\n\\usepackage{amsmath}\n\\usepackage{placeins}\n\\usepackage{amssymb}\n\n\\begin{document}\n\n\\title{SINGULAR VALUE DECOMPOSITION}\n\\date{}\n\\maketitle\n\nSingular Value Decomposition (SVD) is the factorization of any $m\\times n$ matrix $A$ as $A = U \\Sigma V^T$ where $U$ is a $m \\times m$ matrix made up of columns of mutually perpendicular unit vectors, $\\Sigma$ is a diagonal matrix of the same shape as $A$ and $V^T$ is a $n \\times n$ matrix made up of rows of mutually perpendicular unit vectors. \n\n\\section{Example}\n\nLet $A$ be a $2 \\times 3$ matrix as follows,\n\n\\begin{align*}\n\tA = \\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix}\n\\end{align*}\n\nTo calculate unit vectors of $U$, eigenvalues and eigenvectors of $AA^T$ are calculated,\n\n\\begin{align*}\n\tAA^T &= \\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix} \\begin{pmatrix}\n\t3 & -8 \\\\\n\t0 & 0 \\\\\n\t0 & 3\n\t\\end{pmatrix} \\\\\n\t&= \\begin{pmatrix}\n\t9 & -24 \\\\\n\t-24 & 73\n\t\\end{pmatrix}\n\\end{align*}\n\n$AA^T - \\lambda I$ must be singular if eigenvectors are non-zero. Hence,\n\n\\begin{align*}\n\t\\begin{vmatrix}\n\t9 - \\lambda & -24 \\\\\n\t-24 & 73 - \\lambda \n\t\\end{vmatrix} & = 0 \\\\\n\t(9-\\lambda)(73-\\lambda) & = 576 \n\\end{align*}\n\nThis gives $\\lambda = 81, 1$ and $\\hat{u} = \\begin{pmatrix}\n\\frac{1}{\\sqrt{10}} \\\\\n\\frac{-3}{\\sqrt{10}}\n\\end{pmatrix},\\begin{pmatrix}\n\\frac{3}{\\sqrt{10}} \\\\\n\\frac{1}{\\sqrt{10}}\n\\end{pmatrix}$ respectively. \n\nAt this step, decomposition is as follows:\n\n\\begin{align*}\n\tA = (\\hat{u_1}, \\hat{u_2}) \\begin{pmatrix}\n\t\\sqrt{\\lambda_1} & 0 & 0 \\\\\n\t0 & \\sqrt{\\lambda_2} & 0 \n\t\\end{pmatrix} V^T \\\\\n\tA = \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}}\n\t\\end{pmatrix} \\begin{pmatrix}\n\t9 & 0 & 0 \\\\\n\t0 & 1 & 0 \n\t\\end{pmatrix} V^T \n\\end{align*}\n\nTo continue and calculate $V$, the relation $A\\hat{v_i} = \\sqrt{\\lambda_i}\\hat{u_i}$ is used. It is clear that $V$ is a $3 \\times 3$ matrix.\n\n\\begin{align*}\n\tA\\hat{v_1} &= \\sqrt{\\lambda_1}\\hat{u_1} \\\\\n\t\\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix}\\hat{v_1} &= \\sqrt{81}\\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}}\n\t\\end{pmatrix} \\\\\n\t\\hat{v_1} &= \\begin{pmatrix}\n\t\\frac{3}{\\sqrt{10}} \\\\\n\t0 \\\\\n\t\\frac{-1}{\\sqrt{10}} \\\\\n\t\\end{pmatrix}\n\\end{align*}\n\nSimilarly,\n\n\\begin{align*}\n\tA\\hat{v_2} &= \\sqrt{\\lambda_2}\\hat{u_2} \\\\\n\t\\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix}\\hat{v_2} &= \\sqrt{1}\\begin{pmatrix}\n\t\\frac{3}{\\sqrt{10}} \\\\\n\t\\frac{1}{\\sqrt{10}}\n\t\\end{pmatrix} \\\\\n\t\\hat{v_2} &= \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} \\\\\n\t0 \\\\\n\t\\frac{3}{\\sqrt{10}} \\\\\n\t\\end{pmatrix}\n\\end{align*}\n\nNotice that $\\hat{v_1}.\\hat{v_2}$ is zero. The choices for $\\hat{v_3}$ which must be perpendicular to both $\\hat{v_1}$ and $\\hat{v_2}$ are $(0, \\pm 1, 0)^T$ and both are admissible in the decomposition as follows,\n\n\\begin{align*} \n\tA &= \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}}\n\t\\end{pmatrix} \\begin{pmatrix}\n\t9 & 0 & 0 \\\\\n\t0 & 1 & 0 \n\t\\end{pmatrix} \\begin{pmatrix}\n\t\\hat{v_1} & \\hat{v_2} & \\hat{v_3} \n\t\\end{pmatrix}^T \\\\\n\tA &= \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}}\n\t\\end{pmatrix} \\begin{pmatrix}\n\t9 & 0 & 0 \\\\\n\t0 & 1 & 0 \n\t\\end{pmatrix} \\begin{pmatrix}\n\t\\frac{3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}} & 0 \\\\\n\t0 & 0 & \\pm 1 \\\\\n\t\\frac{-1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} & 0 \\\\ \n\t\\end{pmatrix}^T \\\\\n\t\\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix} &= \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}}\n\t\\end{pmatrix} \\begin{pmatrix}\n\t9 & 0 & 0 \\\\\n\t0 & 1 & 0 \n\t\\end{pmatrix} \\begin{pmatrix}\n\t\\frac{3}{\\sqrt{10}} & 0 & \\frac{-1}{\\sqrt{10}} \\\\\n\t\\frac{1}{\\sqrt{10}} & 0 & \\frac{3}{\\sqrt{10}} \\\\ \n\t0 & \\pm 1 & 0 \\\\ \n\t\\end{pmatrix}\n\\end{align*}\n\n\\section{Analysis}\n\nSVD can be visualized as follows:\n\n\\begin{align*}\n\tA_{m \\times n} &= \\begin{pmatrix}\n\t| & | & .. & | \\\\\n\t\\hat{u_1} & \\hat{u_2} & .. & \\hat{u_m} \\\\\n\t| & | & .. & | \\\\ \n\t\\end{pmatrix}_{m \\times m} \\begin{pmatrix}\n\t\\sqrt{\\lambda_1} & 0 & .. & 0 & 0 & .. & 0 \\\\\n\t0 & \\sqrt{\\lambda_2} & .. & 0 & 0 & .. & 0 \\\\\n\t0 & 0 & .. & 0 & 0 & .. & 0 \\\\\n\t0 & 0 & .. & \\sqrt{\\lambda_r} & 0 & .. & 0 \\\\ \n\t0 & 0 & .. & 0 & 0 & .. & 0 \\\\ \n\t. & . & .. & . & . & .. & . \\\\\n\t. & . & .. & . & . & .. & . \\\\ \n\t0 & 0 & .. & 0 & 0 & .. & 0 \\\\ \n\t\\end{pmatrix}_{m \\times n} \\begin{pmatrix}\n\t| & | & .. & | \\\\\n\t\\hat{v_1} & \\hat{v_2} & .. & \\hat{v_n} \\\\\n\t| & | & .. & | \\\\ \n\t\\end{pmatrix}_{n \\times n}^T \\\\\n\t&u_i \\perp u_j\\ \\forall\\ i \\neq j \\\\\n\t&v_i \\perp v_j\\ \\forall\\ i \\neq j \\\\\n\t&\\sqrt{\\lambda_1} \\geq \\sqrt{\\lambda_2} \\geq .. \\geq \\sqrt{\\lambda_r} > 0 \\\\ \n\t&r = rank(A) \n\\end{align*}\n\nThe product of the last two matrices can be simplified as follows,\n\n\\begin{align*}\n\tA_{m \\times n} &= \\begin{pmatrix}\n\t| & | & .. & | \\\\\n\t\\hat{u_1} & \\hat{u_2} & .. & \\hat{u_m} \\\\\n\t| & | & .. & | \\\\ \n\t\\end{pmatrix}_{m \\times m} \\begin{pmatrix}\n\t\\sqrt{\\lambda_1} \\hat{v_1}^T \\\\\n\t\\sqrt{\\lambda_2} \\hat{v_2}^T \\\\ \n\t. \\\\\n\t. \\\\\n\t\\sqrt{\\lambda_r} \\hat{v_r}^T \\\\\n\t0 \\\\\n\t. \\\\\n\t. \\\\\n\t0 \\\\\n\t\\end{pmatrix}_{m \\times n}\n\\end{align*}\n\nleading to the representation of $A$ as a sum of $r$ rank-1 matrices as follows:\n\n\\begin{align*}\n\tA = \\sum_{i=1}^r \\sqrt{\\lambda_i}\\ \\hat{u_i}\\ \\hat{v_i}^T \n\\end{align*}\n\nMultiplying by $\\hat{v_j}$ on both sides,\n\n\\begin{align*}\n\tA\\ \\hat{v_j} = \\sum_{i=1}^r \\sqrt{\\lambda_i}\\ \\hat{u_i}\\ \\hat{v_i}^T\\ \\hat{v_j} \n\\end{align*}\n\nSince all the $\\hat{v_i}$s are mutually perpendicular to each other,\n\n\\begin{align*}\n\tA\\ \\hat{v_j} = \\sqrt{\\lambda_j} \\hat{u_j} \n\\end{align*}\n\nNote the similarity of this equation to the regular eigenvalue equation. The constraint of the same unit vector on both sides of the equation was let go in the return of universal applicability on any matrix whatsoever.\n\n\\section{Application: Matrix Compression}\n\nSubjecting the matrix in example above to the rank-1 breakdown equation,\n\n\\begin{align*}\n\tA &= \\sum_{i=1}^r \\sqrt{\\lambda_i}\\ \\hat{u_i}\\ \\hat{v_i}^T \\\\\n\t\\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix} &= \\begin{pmatrix}\n\t2.7 & 0 & -0.9 \\\\\n\t-8.1 & 0 & 2.7 \\\\\n\t\\end{pmatrix} + \\begin{pmatrix}\n\t0.3 & 0 & 0.9 \\\\\n\t0.1 & 0 & 0.3 \n\t\\end{pmatrix}\n\\end{align*}\n\nNote that the first matrix on the right hand side is a resonable approximation of the matrix on the left hand side and the other matrix just represents minor tweaks.\n\n\\begin{align*}\n\t\\begin{pmatrix}\n\t3 & 0 & 0 \\\\\n\t-8 & 0 & 3 \n\t\\end{pmatrix} &\\approx \\begin{pmatrix}\n\t2.7 & 0 & -0.9 \\\\\n\t-8.1 & 0 & 2.7 \\\\\n\t\\end{pmatrix} \\\\\n\t&\\approx \\sqrt{81} \\begin{pmatrix}\n\t\\frac{1}{\\sqrt{10}} \\\\\n\t\\frac{-3}{\\sqrt{10}}\n\t\\end{pmatrix} \\begin{pmatrix}\n\t\\frac{3}{\\sqrt{10}} & 0 & \\frac{-1}{\\sqrt{10}} \\\\\n\t\\end{pmatrix}\n\\end{align*}\n\nThe left hand side representation requires to store $mn$ entries while the right hand side only requires to store $k(m+n)$ entries. So if $k << m, n$, the right hand side is highly compressed. The right hand side is also a good approximation if k is of the order of the rank of the matrix. So for low-rank highly redundant matrices (example digital images), SVD can serve as a reasonable compression algorithm.\n \n\\section{Appendix} \n \n\\begin{align*}\n\tA & = U\\Sigma V^T \\\\\n\tAA^T & = U\\Sigma V^T V \\Sigma^T U^T \\\\\n\\end{align*} \n \nSince $V^TV = I$ and $\\Sigma$ is diagonal.\n\n\\begin{align*}\n\tAA^T & = U \\Sigma^2 U^T \\\\\n\tAA^T & = \\sum_{i = 1}^r \\lambda_i\\ \\hat{u_i}\\ \\hat{u_i}^T \n\\end{align*}\n\nMultiplying both sides by $\\hat{u_j}$ and using the fact that all $u_i$s\nare mutually perpendicular to each other.\n\n\\begin{align*}\n\tAA^T\\ \\hat{u_j} & = \\sum_{i = 1}^r \\lambda_i\\ \\hat{u_i}\\ \\hat{u_i}^T \\ \\hat{u_j} \\\\\n\tAA^T \\hat{u_j} & = \\lambda_j\\ \\hat{u_j} \n\\end{align*}\n\nIt is clear that $\\hat{u}$s are eigenvectors of $AA^T$. Similarly, it can be seen that $\\hat{v}$s are eigenvectors of $A^TA$. \n\\end{document}\n", "meta": {"hexsha": "e25b225d5c2eb8a4f46f45f2863c4d6d0ae133f2", "size": 9442, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Singular Value Decomposition/Singular Value Decomposition.tex", "max_stars_repo_name": "singaurav/machine-learning-notes", "max_stars_repo_head_hexsha": "4fdd5b839156bcbf8f95a36275b8cd10f93e4c9d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-07-26T11:33:39.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-26T11:33:39.000Z", "max_issues_repo_path": "Singular Value Decomposition/Singular Value Decomposition.tex", "max_issues_repo_name": "singaurav/machine-learning-notes", "max_issues_repo_head_hexsha": "4fdd5b839156bcbf8f95a36275b8cd10f93e4c9d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Singular Value Decomposition/Singular Value Decomposition.tex", "max_forks_repo_name": "singaurav/machine-learning-notes", "max_forks_repo_head_hexsha": "4fdd5b839156bcbf8f95a36275b8cd10f93e4c9d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-09-22T18:56:20.000Z", "max_forks_repo_forks_event_max_datetime": "2019-09-22T18:56:20.000Z", "avg_line_length": 33.1298245614, "max_line_length": 410, "alphanum_fraction": 0.4917390383, "num_tokens": 3415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389817407017, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.6428272436418881}} {"text": "\\section{Problem Definition}\n\\label{section:problems}\n\nThe general problem of \\mcproblem\\ is to select the\nappropriate monitoring operations to reduce\nthe expected uncertainty after they are performed.\nStated formally:\n\n\\begin{problem}[\\mcproblem]\nGiven a transition matrix $P$ and an initial distribution of items $\\initial$,\nselect a set of up to $k$ monitoring operations \nto minimize the expected uncertainty $\\objective$. \n% $E[\\uncertainty]$\n\\end{problem}\n\nWe study variants of the problem --\neach defined for a specific type of monitoring operation.\n For simplicity, we refer to these problems\n with the same name as that of the operation type:\n\\variant{\\parentstransitions},\n\\variant{\\nodeitems},\n\\variant{\\childrentransitions}, \nand \\variant{\\edgetransitions}.\n\nFurthermore, as we saw in Section~\\ref{sec:setting},\nvariants\n\\variant{\\parentstransitions}\\ and\n\\variant{\\nodeitems} are equivalent: for the same set of nodes,\noperations of the first type\nreduce expected uncertainty as much as the second.\nTherefore, in what follows,\nwe treat only the variant of \\variant{\\nodeitems}, as our claims apply directly to\n\\variant{\\parentstransitions}\\ as well.\n\n\\todo[MM]{One issue we have not addressed is how we go from \n$A_{_\\shortnodeitems}(S)$ to $A_{_\\shortparentstransitions}(S)$.\nDoes it affect our claims?}\n", "meta": {"hexsha": "f10e00dfcc7f5f78ca8b6c49adab83800c5ce5cd", "size": 1319, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/SDM-2018/problems.tex", "max_stars_repo_name": "chdhr-harshal/MCMonitor", "max_stars_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-11-04T20:35:18.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-05T09:06:43.000Z", "max_issues_repo_path": "paper/SDM-2018/problems.tex", "max_issues_repo_name": "chdhr-harshal/MCMonitor", "max_issues_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/SDM-2018/problems.tex", "max_forks_repo_name": "chdhr-harshal/MCMonitor", "max_forks_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-09-05T09:10:41.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-05T09:10:41.000Z", "avg_line_length": 34.7105263158, "max_line_length": 83, "alphanum_fraction": 0.7778620167, "num_tokens": 316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256551882382, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6427202240100374}} {"text": "% Copyright 2017-2018 Paul Mitchener, licensed under GNU FDL v1.3\n% main author: \n% Paul Mitchener\n% contributions by:\n% Markus Pflaum\n\n\\section{The category of measurable spaces and functions}\\label{sec:category-measurable-spaces-functions}\n\\subsec{Definitions and first examples}\n\\begin{definition}\n Let $\\Omega$ be a set. By an \\emph{algebra on} $\\Omega$ one understands a collection $\\salgA$ of subsets of $\\Omega$\n or in other words an element $\\salgA\\in \\power{\\Omega}$ such that\n \\begin{axiomlist}[A]\n \\item $\\Omega \\in \\salgA$,\n \\item for each $A\\in \\salgA$, the complement $\\complement A = \\Omega \\setminus A $ belongs to $\\salgA$,\n \\item for each finite sequence $(A_k)_{k=1}^n$ of elements of $\\salgA$ the union $A= \\bigcup\\limits_{k=1}^n A_k$\n belongs to $\\salgA$. \n \\end{axiomlist}\n If in addition \n \\begin{axiomlist}[A]\n \\setcounter{enumi}{3}\n \\item for each sequence $(A_k)_{k\\in \\N}$ of elements of $\\salgA$ the union $A= \\bigcup\\limits_{k\\in\\N} A_k$\n belongs to $\\salgA$,\n \\end{axiomlist}\n then the algebra $\\salgA$ is called a \\emph{$\\sigma$-algebra}.\n A set $\\Omega$ equipped with a $\\sigma$-algebra $\\mathscr A$ is called a \n \\emph{measurable space}. The elements of $\\mathscr A$ are termed the \n \\emph{measurable subsets} of $\\Omega$.\n\\end{definition}\n\n\\begin{proposition}\n If $\\salgA$ is an algebra on a set $\\Omega$, the empty set and the intersection of finitely many\n elements of $\\salgA$ lies in $\\salgA$.\n If $\\salgA$ is a $\\sigma$-algebra on $\\Omega$, then the intersection of countably many\n measurable sets is also measurable.\n\\end{proposition}\n\\begin{proof}\n These facts follow immediately from the axioms and the set-theoretic de Morgan's laws. \n\\end{proof}\n\n\\begin{examples}\n\\begin{environmentlist}\n\\item Let $\\Omega$ be any set. Then the power set of $\\Omega$ is a $\\sigma$-algebra. \n The set $\\{ \\emptyset , \\Omega \\}$ is also a $\\sigma$-algebra. These are the largest and smallest\n $\\sigma$-algebra on $\\Omega$, respectively. \n\\item\n Let $\\Omega$ be any set. Let $\\mathscr A$ be the set of all sets $A\\subset \\Omega$ such that $A$ or $\\Omega \\backslash A$ is\n a countable set. Then $\\mathscr A$ is a $\\sigma$-algebra.\n\\end{environmentlist}\n\\end{examples}\n\n\\begin{remark}\n Obviously, the set of algebras on a set $\\Omega$ and the set of $\\sigma$-algebras on $\\Omega$ are both ordered\n by set-theoretic inclusion. When talking about a ``smaller`` $\\sigma$-algebra or a ``largest'' one we always\n implicitely mean in regard to set-theoretic inclusion as underlying order relation.\n\\end{remark}\n\nThe following two results are extremely useful when constructing examples.\n\n\\begin{proposition}\n Let $(\\salgA_i)_{i\\in I}$ a family of algebras on a set $\\Omega$.\n Then the intersection\n $\\salgA = \\bigcap\\limits_{i\\in I} \\salgA_i$ is an algebra on $\\Omega$.\n If each of the $\\salgA_i$ is a $\\sigma$-algebra, then $\\salgA$ is so, too. \n\\end{proposition}\n\\begin{proof}\n Assume first that each $\\salgA_i$ is an algebra on $\\Omega$. \n Obviously, $\\Omega \\in \\salgA$ because $\\Omega \\in \\salgA_i$ for all $i\\in I$.\n Similarly, if $A \\in \\salgA$, then $A \\in \\salgA_i$, hence $\\complement A \\in \\salgA_i$\n for all $i\\in I$. Therefore $\\complement A $ is in the intersection\n $\\salgA = \\bigcap\\limits_{i\\in I} \\salgA_i$. Now assume that $(A_k)_{k=1}^n$ is a finite\n sequence of sets belonging to $\\salgA$. Then $A_k \\in \\salgA_i$ for $k=1,\\ldots,n$ and all\n $i\\in I$ which entails that $\\bigcup\\limits_{k=1}^n A_k$ is in each of the $\\salgA_i$, hence in the\n intersection $\\salgA$. The latter argument also works under the condition that each $\\salgA_i$ is a $\\sigma$-algebra\n to verify that for a sequence $(A_k)_{k\\in\\N}$ in $\\salgA$ the union $\\bigcup\\limits_{k\\in\\N} A_k$\n is in $\\salgA$. So the proposition is proved. \n\\end{proof}\n\n\n\\begin{corollary}\n Let $\\mathscr F$ be a collection of subsets of a set $\\Omega$. Then there is a unique smallest $\\sigma$-algebra,\n $\\sigma (\\mathscr F)$, on $\\Omega$ containing $\\mathcal F$. It is called the \\emph{$\\sigma$-algebra} generated by\n $\\mathscr F$.\n\\end{corollary}\n\n\\begin{proof}\n Let ${\\mathcal M}$ be the family of all $\\sigma$-algebras which contain $\\mathscr F$. The set of all subsets of $\\Omega$ is\n certainly a $\\sigma$-algebra, so ${\\mathcal M} \\neq \\emptyset$. Let $\\sigma (\\mathscr F)$ be the intersection of all\n $\\sigma$-algebras in the family ${\\mathcal M}$. By the preceding proposition $\\sigma (\\mathscr F)$ is a $\\sigma$-algebra.\n Since every element of ${\\mathcal M}$ contains $\\mathscr F$, the intersection\n $\\sigma (\\mathscr F) = \\bigcap\\limits_{\\salgA\\in \\mathcal{M}}\\salgA$ contains $\\mathscr F$ as well. By construction,\n $\\sigma (\\mathscr F)$ is minimal with that property.\n\\end{proof}\n\n\\begin{example}\n Let $X$ be a topological space. The $\\sigma$-algebra, $\\mathscr B (X)$, generated by all open subsets of $X$ is called the\n {\\em Borel $\\sigma$-algebra} on $X$. Its elements are the {\\em Borel measurable sets} or simply the {\\em Borel sets} of $X$.\n Obviously all open and all closed sets of $X$ are Borel measurable, as are all countable unions of closed sets and countable\n intersections of open sets.\n\\end{example}\n\n\n\n\\begin{example}\n\\begin{environmentlist}\n\\item All intervals including the half-open intervals $[a,b)$ and $(a,b]$ with $a 0) = 1-\\mbox{e}^{-\\lambda}$ \\\\ \\hline\n\t\n\t\\end{tabularx}\n\t\\end{small}\n\\end{bbbox}\n\nIn the case of a discrete random variable $P$ is called a probability mass function (pmf). Since discrete random variables have a countable or finite number of states, or $X: \\Omega \\rightarrow \\mathbb{N}$, the pmf directly assigns a probability to each possible event. We will later see that this is not the case for the $P$ of continuous random variables. In literature there is a redundancy in the notation of the pmf and it is useful to know that some of the expressions used are equivalent:\n\n\\begin{itemize}\n\\item %Rules: \n\\begin{align}\nP(X=x)&=P_X(x)=P(x)\\\\\nP(x)&\\geq 0 \\mbox{ for each } x\\\\\n\\sum_{x \\in \\Omega} P(x)&=1\\\\\nP(A) = P(x \\in A) &=\\sum_{x \\in A} P(X=x)\\\\\n\\end{align}\n\\item We write: $X|q \\sim \\mbox{Binomial}(q)$\n\\end{itemize}\n\n\\subsection{Moments and Expectations}\n\\begin{itemize}\n\\item Expectation (or mean): $E(X)= \\sum_{x} P(X=x) x$ \n\\end{itemize}\n\n\\begin{bbbox}{Expectation of a discrete random variable}\n\tThe expectation of a discrete random variable is the weighted sum over all possible states of the random variable, where the weights are the probabilities of the respective state.\\\\\n\t\n\tExamples: fair coin; fair die; \\\\\n\t\\begin{flalign*}\n\t\t&\\text{Fair coin: } & E(coin) & = P(coin=1)\\times1 + P(coin=0)\\times0 = \\frac{1}{2} \\\\\n\t\t&\\text{Fair die: } & E(die) & = 1\\times\\frac{1}{6} + 2\\times\\frac{1}{6} + \\dots + 6\\times\\frac{1}{6} = 3.5 \\\\\n \t\t&\\text{Poisson with rate } \\lambda \\text{: } & E(X) & = \\sum_{x = 1}^{\\infty} (x) \\frac{1}{\\underbrace{x!}_{(1 \\times 2 \\times \\dots \\times x)}} \\mbox{e}^{-\\lambda} \\lambda^x \\\\\n\t\t& & & = \\sum_{x = 1}^{\\infty} \\frac{1}{(x-1)!} \\mbox{e}^{-\\lambda} \\lambda^{x-1} \\lambda \\\\\n\t\t& & & = \\lambda \\sum_{x = 1}^{\\infty} \\frac{1}{(x-1)!} \\mbox{e}^{-\\lambda} \\lambda^{x-1} \\\\\n\t\t& & & = \\lambda \\underbrace{\\sum_{y = 0}^{\\infty} \\frac{1}{(y)!} \\mbox{e}^{-\\lambda} \\lambda^{y}}_{ = 1} \\;\\; \\vert (y = x+1) \\\\\n\t\t& & & = \\lambda\n\t\\end{flalign*}\n\t\n\\end{bbbox}\n\n\\begin{itemize}\n\\item Expectation of a function: $E(f(X))= \\sum_{x} P(X=x) f(x)$ \n\\item Moments= expectation of power of $X$: $M_k= E(X^k)$\n\\end{itemize}\n\n\\begin{bbbox}{Example: 2nd order moment}\n\tWe want to calculate the expectation of $Die^2$ for a fair die: \\\\\n\t\\begin{align*}\n\t\tE(Die^2) &= \\sum_{x = 1}^6 P(Die = x) x^2 \\\\\n\t\t\t\t &= \\frac{1}{6} (1 + 2 + 9 + 16 + 25 + 36) \\\\\n\t\t\t\t &= \\frac{1}{6} 89 \\\\\n\t\t\t\t &\\approx 14.66\t\t\t \n\t\\end{align*}\n\\end{bbbox}\n\n\\begin{itemize}\n\\item Variance: Average (squared) fluctuation from the mean\n\\begin{align}\n \\mbox{Var}(X)&= E((X-E(X))^2)\\\\\n &= E(X^2)- E(X)^2\\\\\n &= M_2-M_1^2\n\\end{align}\n\\end{itemize}\n\n\\begin{bbbox}{Variance}\nThe alternative definition of variance based on moments directly emerges from the linearity of the expectation:\n\t\\begin{align*}\n\t\tVar(X) = E((X - E(X))^2) &= E(X^2 - 2 \\times X \\times E(X) + E(X)^2) \\\\\n\t\t\t\t\t\t&= E(X^2) - 2 \\times E(X \\times E(X)) + E(X)^2 \\\\\n\t\t\t\t\t\t&= E(X^2) - 2 \\times E(X) \\times E(X) + E(X)^2 \\\\\n\t\t\t\t\t\t&= E(X^2) - E(X)^2\n\t\\end{align*}\n\\end{bbbox}\n\n\\begin{bbbox}{Why do we use the squared fluctuations from the mean for the mesaure of variance?}\n\t\n\t\\begin{itemize}\n\t\t\\item Average deviations from the mean: \\\\\n\t\t\t $Var(X) = E(X-E(X)) = E(X) - E(X) = 0$ \\\\\n\t\t\t As we see this measure is always zero and therefore useless for our purpose.\n\t\t\\item Average abolute deviations from the mean: \n\t\t\t \\\\$Var(X) = E(|X-E(X)|)$ \\\\\n\t\t\t However this measure can sometimes be technically difficult and is less used in practice.\n\t\\end{itemize}\n\n\\end{bbbox}\n\n\\begin{itemize}\n\\item Standard devation: Square root of variance.\n\\item Illustration and examples: [on board]\nAside: Difference between expectation/variance of random variable and empirical average/variance.\n\n\\end{itemize}\n\n\\subsection{Probability rules}\nThe course will be focused on Bayesian statistics and in that context we will be constantly using Bayes' rule. In this section we will show that Bayes' rule strictly follows from the definition of conditional probabilities. We have seen that we can measure the probability of an event $A$ with $P(A)$. Further we can measure the probability of two events $A$ and $B$ with P(A,B). We call this the joint probability of the events $A$ and $B$\n\n\\begin{align}\n\tP(B|A) = \\frac{P(A|B) P(B)}{P(A)}\n\\end{align}\n\nwhere $P(B|A)$ is the posterior probability of an event $B$ given we have knowledge about an event $A$. It is straightforward to show from the definition of conditional probabilities that Bayes' rule holds. We define the conditional probability of an event $B$ given knowledge about and event $A$ as\n\n\\begin{align}\n\tP(B|A) = \\frac{P(A,B)}{P(A)}\n\\end{align}\n\nwhere $P(A,B)$ is the probability of observing events $A$ and $B$, we call this the joint probability of $A$ and $B$. And $P(A)$ is the probability of observing and event $A$. We can \n\n\\begin{exbox}{It is cold today, will we have rain?}\nWe will consider the example of interactions between rain and temperature in T\\\"ubingen. Lets say we have made $N = 30$ independent measurements of the weather conditions in T\\\"ubingen. We arbitrarily chose to divide our observations based on the temperature and whether it was raining or not. The table below reflects the number of times we made observations under a given condition, e.g. the number of observations when the temperature was below $10\\,^{\\circ}\\mathrm{C}$ and it was not raining was $N_{(T=0,R=0)} = 10$. \\\\\n\t\\begin{tabular}{ll|c|c|r}\n\t\t\t& & Rain & No Rain \\\\\n\t\t\t& & $R = 1$ & $R = 0$ & $\\sum_R$ \\\\ \\hline\n\t\tTemp $> 20\\,^{\\circ}\\mathrm{C}$: & $T = 2$ & $1$ & $3$ & 4\\\\ \\hline\n\t\t$10\\,^{\\circ}\\mathrm{C} <$ Temp $< 20\\,^{\\circ}\\mathrm{C}$: & $T = 1$ & $4$ & $8$ & 12\\\\ \\hline\n\t\tTemp $< 10\\,^{\\circ}\\mathrm{C}$: & $T = 0$ & $4$ & $10$ & 14\\\\ \\hline\n\t\t\t& $\\sum_T$ & 9 & 21 \\\\\n\t\\end{tabular}\nWe can easily see that the sum over all observations is $N = 30$ again. If we now want to calculate the probability of the temperature being below $10\\,^{\\circ}\\mathrm{C}$ and there being no rain, we can simply do $P(T=0,R=0) = \\frac{N_{(T=0,R=0)}}{N} = \\frac{1}{3}$. We call this the \\textbf{probability} of events $T=0$ and $R=0$. This can be done for all possible events.\\\\\nLets say we know the temerature is between $10$ and $20\\,^{\\circ}\\mathrm{C}$ and we want to know how likely it is to rain. We can utilize the \\textbf{conditional probability} $P(R=1|T=1)$\n\\end{exbox}\n\n\\begin{defbox}{Probability rules}\n\t\\begin{itemize}\n\t\\item {Joint distribution:} $P(X=x, Y=y)$, a list of all probabilities of all possible pairs of observations\n\t\\item {Marginal distribution:} $P(X=x)=\\sum_y P(X=x, Y=y) $\n\t\\item {Conditional distribution:} $P(X=x|Y=y) = \\frac{P(X=x, Y=y)}{P(y=y)}$\n\t\\item $X |Y$ has distribution $P(X|Y)$, where $P(X|Y)$ specifies a 'lookup-table' of all possible $P(X=x| Y=y)$\n\t\\end{itemize}\n\\end{defbox}\n\n\\begin{itemize}\n\\item {Conditional probability:} 'Recalculated probability of event A after someone tells you that event A happened.' \n\\begin{align}\nP(A|B)&= \\frac{P(A \\cap B)}{P(B)}\\\\\nP(A \\cap B)&= P(A|B) P(B)\n\\end{align}\n\\item Examples: Rolls of a die\n [on board]\n\\item Bayes Rule: \n\\begin{align}\nP(B|A) = \\frac{P(A|B) P(B)}{P(A)}\n\\end{align}\n\\end{itemize}\n\n\\begin{itemize} \n\\item Example [on board] \n\\item {Joint distribution:} $P(X=x, Y=y)$, a list of all probabilities of all possible pairs of observations\n\\item {Marginal distribution:} $P(X=x)=\\sum_y P(X=x, Y=y) $\n\\item {Conditional distribution:} $P(X=x|Y=y) = \\frac{P(X=x, Y=y)}{P(y=y)}$\n\\item $X |Y$ has distribution $P(X|Y)$, where $PX(|Y)$ specifies a 'lookup-table' of all possible $P(X=x| Y=y)$\n\\end{itemize}\nConditioning and marginalization come up in Bayesian inference ALL the time: 'Condition on what you observe. Marginalize out the uncertainty'.\n\n\\subsection{Independence of random variables}\n\n\\begin{itemize}\n \\item Intuitively, two {events are independent} if knowing that the first took places tells us nothing about the probability of the second: $P(A|B)= P(A)$\n \\item $P(A) P(B)= P(A \\cap B)$\n \\item Two {random variables} are independent if the joint p.m.f. is the product of the marginals: $P(X=x,Y=y)=P(X=x) P(Y=y)$. \n \\item If $X$ and $Y$ are independent, we write $X \\perp Y$. Knowing the value of $X$ does not tell us anything about $Y$.\n \\item If $X$ and $Y$ are independent, $\\mbox{Cov}(X,Y)=0$.\n \\end{itemize}\n Aside: Mutual information is a measure of how 'non-independent' two random variables are.\n\n\n\\subsection{Multivariate Distributions}\n\\subsubsection{Expectation and covariance}\n\\begin{itemize}\n\\item Conditional distributions are just distributions which have a (conditional) mean or variance. \n\\item Note: $E(X|Y)= f(Y)$. 'If I tell you what $Y$ is, what is the average value of $X$?.\n%\\item $E(X,Y)= \\sum_{x,y} P(X=x, Y=y) (x,y)= (E(X), E(Y))$ \n\\item Covariance is the expected value of the product of fluctuations: \n\\begin{align}\n \\mbox{Cov}(X,Y)&= E\\left((X-E(X) )(Y-E(Y) )\\right)\\\\\n &= E(XY)- E(X)E(Y)\\\\\n \\mbox{Var(X)}&= \\mbox{Cov}(X,X)\n\\end{align}\n \\end{itemize}\nAside: One common way to construct bivariate random variables is to have a random variable whos parameter is another random variable. \n\n\\begin{itemize} \n\\item $\\mathbf{X},\\xx$ are vector valued.\n\\item Mean: $E(\\mathbf{X})= \\sum_{\\xx} \\xx P(\\xx)$\n\\item Covariance matrix: \n\\begin{align}\n\\mbox{Cov}(X_i, X_j)&= E(X_iX_j)-E(X_i) E(X_j)\\\\\n\\mbox{Cov}(\\mathbf{X})&= E(\\mathbf{X}\\mathbf{X}^\\top)-E(\\mathbf{X}) \nE(\\mathbf{X})^\\top \n \\end{align}\n\\item Conditional and marginal distributions: Can define and calculate any (multi or single-dimensional) marginals or conditional distributions we need: $P(X_1)$, $P(X_1, X_2)$, $P(X_1, X_2, X_3 |X_4)$, etc..\n\\end{itemize}\n\n\\begin{bbbox}{Covariance matrix}\n\t\\begin{align*}\n\t\t& & Cov(\\mathbf{X}) &= \n\t\t\\begin{pmatrix}\n \t\tCov(X_1,X_1) & Cov(X_2,X_1) & \\cdots & Cov(X_n,X_1) \\\\\n \t\tCov(X_1,X_2) & Cov(X_2,X_2) & \\cdots & Cov(X_n,X_2) \\\\\n \t\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n \t\tCov(X_1,X_n) & Cov(X_2,X_n) & \\cdots & Cov(X_n,X_n) \\\\\n \t\t\\end{pmatrix} \\\\\n\t\t& \\text{Note that: } & Cov(X_i,X_j) &= Cov(X_j,X_i) \\\\\n\t\t& \\text{Therefore: } & Cov(\\mathbf{X}) &= Cov(\\mathbf{X})^\\top\n\t\\end{align*}\n\t\n\\end{bbbox}\n\n\\subsection{Continuous random variables}\n\\begin{itemize}\n\\item A random variable $X$ is {continuous} if its sample space $X$ is uncountable. \n\\item In this case, $P(X=x)=0$ for each $x$.\n\\item If $p_X(x)$ is a {probability density function} for $X$, then \n\\begin{align}\nP(a < X 0$~!\n\\item Example: Uniform, Exponential [on board]\n\\end{itemize}\n\n\\begin{bbbox}{Moments and Variance of the continuous uniform distribution}\n\tContinuous uniform distribution: \\\\\n\t\\begin{align*}\n\t\tX &\\sim U(0,1) \\\\\n\t\tE(X) &= \\int_0^1 x p(x) dx = \\int_0^1 x dx = &= \\left[ \\frac{1}{2}x^2 \\right]_0^1 = \\frac{1}{2}\\\\\n\t\tE(X^2) &= \\int_0^1 x^2 p(x) dx = \\int_0^1 x^2 dx = &= \\left[ \\frac{1}{3}x^3 \\right]_0^1 = \\frac{1}{3}\\\\\n\t\tVar(X) &= E(X^2) - E(X)^2 = \\frac{1}{3} - \\frac{1}{4} = \\frac{1}{12}\\\\\n\t\\end{align*}\n%\tExponential distribution: \\\\\n%\t\\begin{align*}\n%\t\tE(X) &= \\int_0^\\infty x \\lambda \\mbox{e}^{-\\lambda x} dx \\\\\n%\t\t &= \\left[ -\\frac{1}{2}x^2 \\mbox{e}^{-\\lambda x} \\right]_0^\\infty \\\\\n%\t\\end{align*}\n\\end{bbbox}\n\n\\subsubsection{Marginalizationa, Conditioning and Independence}\n\\begin{itemize}\n\\item $p_{X,Y}(x,y)$, joint probablity density function of $X$ and $Y$\n\\item $\\int_x \\int_y p(x,y)dx dy=1$\n\\item {Marginal distribution:} $p(x)= \\int_{-\\infty}^\\infty p(x,y) dy$\n\\item {Conditional distribution:} $p(x|y)= \\frac{p(x,y)}{p(y)}$ \n\\item Note: $P(Y=y)=0$! Formally, conditional probability in the continuous case can be derived using infinitesimal events.\n\\item {Independence:} $X$ and $Y$ are independent if $p_{X,Y}(x,y)=p_X(x)p_Y(y)$\n\\end{itemize}\n\n\n\n", "meta": {"hexsha": "120169ca8ba11a1a5f9472a39ed751ebb415c303", "size": 15672, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "script/lecture3/lecture3.tex", "max_stars_repo_name": "mackelab/machine-learning-I", "max_stars_repo_head_hexsha": "fedd9ea0b9b257af5cd59036a3b49876aed5c77c", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 23, "max_stars_repo_stars_event_min_datetime": "2015-07-31T15:08:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-14T17:07:23.000Z", "max_issues_repo_path": "script/lecture3/lecture3.tex", "max_issues_repo_name": "cne-tum/msne_statsandprob_ss2018", "max_issues_repo_head_hexsha": "fedd9ea0b9b257af5cd59036a3b49876aed5c77c", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "script/lecture3/lecture3.tex", "max_forks_repo_name": "cne-tum/msne_statsandprob_ss2018", "max_forks_repo_head_hexsha": "fedd9ea0b9b257af5cd59036a3b49876aed5c77c", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 16, "max_forks_repo_forks_event_min_datetime": "2018-03-16T07:42:51.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-01T14:02:27.000Z", "avg_line_length": 50.8831168831, "max_line_length": 1150, "alphanum_fraction": 0.6509060745, "num_tokens": 5520, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256591565729, "lm_q2_score": 0.7634837581726991, "lm_q1q2_score": 0.64272021797907}} {"text": "\\chapter{ELEC 403 CheatSheet}\n\\begin{multicols}{3}\n\n\\textbf{Algorithm 1.1 General optimization algorithm} \\newline\n\\textbf{Step 1:} \\newline\n(a) Set $k=0$ and initialize $x_0$ \\newline\n(b) Compute $F_0=f(x_0)$ \\newline\n\\textbf{Step 2:} \\newline\n(a) Set $k=k+1$ \\newline\n(b) Compute the changes in $x_k$ given by column vector $\\nabla x_k$ %where\n\\[ \\text{where} \\quad \n\\nabla x_k^T = \\begin{bmatrix}\\nabla x_1 & \\nabla x_2 & \\cdots & \\nabla x_n\\end{bmatrix}\n\\]\nby using an appropriate procedure. \\newline\n(c) Set $x_k=x_{k-1}+\\nabla x_k$ \\newline\n(d) Compute $F_k=f(x_k)$ and $\\nabla F_k=F_{k-1}-F_k$. \\newline\n\\textbf{Step 3:} \\newline\nCheck if convergence has been achieved by using an appropriate criterion, e.g., by checking $\\nabla F_k$ and/or $\\nabla x_k$. If this is the case, continue to\nStep 4; otherwise, go to Step 2. \\newline\n\\textbf{Step 4:} \\newline\n(a) Output $x^* = x_k$ and $F^* = f(x^*)$. \\newline\n(b) Stop\n\\section{Ch.2}\nGradient: $g(x)=\\nabla f(x)=\\begin{bmatrix}\\frac{\\partial f}{\\partial x_1} & \\frac{\\partial f}{\\partial x_2} & \\cdots & \\frac{\\partial f}{\\partial x_n} \\end{bmatrix}^T$ \\newline\nHessian Matrix: $H(x)=\\nabla g(x)=\\nabla \\{ \\nabla^T f(x)\\}$. \\newline\n\\[\nH(x)=\n\\begin{bmatrix}\n\\frac{\\partial^2 f}{\\partial^2 x_1} & \\frac{\\partial^2 f}{\\partial x_1 \\partial x_2} & \\cdots & \\frac{\\partial^2 f}{\\partial x_1 \\partial x_n} \\\\\n\\frac{\\partial^2 f(x)}{\\partial x_2 \\partial x_1} & \\frac{\\partial^2 f(x)}{\\partial x_2^2} & \\cdots & \\frac{\\partial f}{\\partial x_2 \\partial x_n} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\\frac{\\partial f}{\\partial x_n \\partial x_1}& \\frac{\\partial f}{\\partial x_n \\partial x_2}& \\cdots & \\frac{\\partial f}{\\partial x_n \\partial x_n}\n\\end{bmatrix}\n\\]\nTaylor Series: (quad approx, linear approx): $\\delta =\\begin{bmatrix} \\delta_1 &\\delta_2 \\end{bmatrix}^T$\n$f(x + \\delta) = f(x) + g(x)^T\\delta + 12\\delta^TH(x)\\delta + o(||\\delta||^2)$ \\newline\nLinear approximation: $f(x + \\delta) \\approx f(x) + g(x)^T\\delta $\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{Images/min.jpg}\n\t\\caption{Minimization points plotted}\n\\end{figure}\n$\\tilde{x}=x+\\delta \\quad \\delta =\\tilde{x}-x$ $\\hat{f}(\\tilde{x})=f(x)+g(x)(\\tilde{x}-x)^T+0.5(\\tilde{x}-x)^T H(x)(\\tilde{x}-x)^T$\nThe gradient $g(x)$ and the Hessian $H(x)$ must satisfy certain conditions at a\nlocal minimizer $x^*$. \\newline\n1. Conditions which are satisfied at a local minimizer $x^*$. \\newline\n2. Conditions which guarantee that $x^*$ is a local minimizer. \n\n\\textbf{Definition 2.1} A point $x^* \\in R,$ where R is the feasible region, is said to be a\nweak local minimizer of f(x) if there exists a distance $\\epsilon > 0$ such that\n$f(x) \\geq f(x^*)$ (2.5)\nif\n$x \\in R$ and \n$||x-x^*|| < \\epsilon$ \\newline\n\n\\textbf{Definition 2.2} \nA point $x^* \\in R $ is said to be a weak global minimizer of f(x) if\n$f(x) \\geq f(x^*)$ (2.6)\nfor all $x \\in R.$ \\newline\n\\textbf{Definition 2.3} \n\nIf Eq. (2.5) in Def. 2.1 or Eq. (2.6) in Def. 2.2 is replaced by\n$f(x) > f(x^*)$ (2.7)\n$x^*$ is said to be a strong local (or global) minimizer. d\n\n\\textbf{Definition 2.4} Let $\\delta = \\alpha \\mathbf{d}$ be a change in x where $\\alpha$ is a positive constant and d is a direction vector. If R is the feasible region and a constant $\\hat{\\alpha} > 0$ exists such that\n$x + \\alpha d \\in R$ for all $\\alpha$ in the range $0 \\leq \\alpha \\leq \\hat{\\alpha}$ , then d is said to be a feasible direction at\npoint x. \\newline \n\n\\textbf{Definition 2.5} \\newline\n(a) Let $d$ be an arbitrary direction vector at point $x$. The quadratic form\n$d^TH(x)d$ is said to be \\textit{positive definite, positive semidefinite, negative\n\tsemidefinite, negative definite} if $d^TH(x)d > 0, \\geq 0, \\leq 0, < 0,$ respectively, for all $d \\neq 0$ at $x$. If $d^TH(x)d$ can assume positive as well\nas negative values, it is said to be indefinite. \\newline\n(b) If $d^TH(x)d$ is positive definite, positive semidefinite, etc., then matrix\n$H(x)$ is said to be positive definite, positive semidefinite, etc. \\newline\n\nThe objective function must satisfy two sets of conditions in order to have\na minimum, namely, first- and second-order conditions. \\newline\n\\textbf{First-order necessary conditions for a minimum} \\newline %Theorem 2.2\n(a) If $f(x) \\in C^1$ and $x^*$ is a local minimizer, then\n$g(x^*)^Td \\geq 0$ for every feasible direction $d$ at $x^*$. \\newline\n(b) If $x^*$ is located in the interior of $\\mathcal{R}$ then\n$g(x^*) = 0$ \\newline \\newline \n\\textbf{Second-order necessary conditions for a minimum} \\newline\n%Theorem 2.2\n(a) If $f(x) \\in C_2$ and $x^*$ is a local minimizer, then for every feasible direction\nd at $x^*$. \\hfill \\break\n\\indent $(i) \\ g(x^*)Td \\geq 0$ \\newline\n\\indent $(ii)$ If $g(x^*)^Td = 0$, then $d^TH(x^*)d \\geq 0$ \\newline\n(b) If $x^*$ is a local minimizer in the interior of R, then \\newline\n\\indent $ (i)$ $g(x^*) = 0$ \\newline\n\\indent $ (ii)$ $d^TH(x^*)d \\geq 0$ for all $d \\neq 0$ \n\n\\textbf{Second-order sufficient conditions for a minimum} \\newline %Theorem 2.5\nIf $f(x) \\in C_2$\nand $x^*$ is located in the interior of $\\mathcal{R}$, then the conditions\n(a) $g(x^*) = 0$\n(b) $H(x^*)$ is positive definite\nare sufficient for $x^*$ to be a strong local minimizer. \\newline\n\n\\textbf{Definition 2.6} \\newline\nA point $\\bar{x} \\in \\mathcal{R},$ where $\\mathcal{R}$ is the feasible region, is said to be a saddle point if \\newline\n(a) $g(\\bar{x}) = 0$ \\newline\n(b) point $\\bar{x}$ is neither a maximizer nor a minimizer. \\hfill \\break \\newline\nStationary points can be located and classified as follows: \\newline\n1. Find the points $x_i$ at which $g(x_i) = 0$. \\newline\n2. Obtain the Hessian $H(x_i)$. \\newline\n3. Determine the character of $H(x_i)$ for each point $x_i$. \\newline\nIf $H(x_i)$ is positive (or negative) definite, $x_i$ is a minimizer (or maximizer);\nif $H(x_i)$ is indefinite, $x_i$ is a saddle point. \n\n\\textbf{Techniques to compute Hessian P.D. , N.D. } \\newline\nEigenvalues: det $ (\\lambda I -A) = 0$ Multiplying all eigenvalues is equal to the determinant. \\newline\nThe leading principal minors of a matrix A or its negative $-A$ can be used to\nestablish whether the matrix is positive or negative definite whereas the principal\nminors of A or $-A$ can be used to establish whether the matrix is positive or\nnegative semidefinite. \\newline \n\n\\textbf{Theorem 2.9 Properties of matrices} \\newline \n(a) If \\textbf{H} is positive semidefinite or positive definite, then\ndet $\\mathbf{H} \\geq 0 \\ \\text{or} > 0$ \\newline \n(b) \\textbf{H} is positive definite if and only if all its leading principal minors are\npositive, i.e., det $\\mathbf{H_i} > 0$ for $ i = 1, 2, \\cdots , n.$ \\newline \n(c) \\textbf{H} is positive semidefinite if and only if all its principal minors are nonnegative, i.e., det $(H_i^{(l)}) \\geq 0$ for all possible selections of $\\{l_1, l_2, \\cdots , l_i \\}$\nfor $i = 1, 2, \\cdots, n$. \\newline \n(d) \\textbf{H} is negative definite if and only if all the leading principal minors of\n$-\\mathbf{H}$ are positive, i.e., $det ( -H_i) > 0$ for $i = 1, 2, \\cdots, n$. \\newline \n(e) \\textbf{H} is negative semidefinite if and only if all the principal minors of -$\\mathbf{H}$\nare nonnegative, i.e., det $(-H_i^{(l)}) \\geq 0$ for all possible selections of\n$\\{l_1, l_2, \\cdots , l_i \\}$ for $i = 1, 2, \\cdots , n$. \\newline \n(f) \\textbf{H} is indefinite if neither (c) nor (e) holds. \n\n\\textbf{Definition 2.7} \\newline\nA set $\\mathcal{R}_c\n\\subset E_n $ is said to be convex if for every pair of points $x_1, x_2 \\subset R_c$\nand for every real number $\\alpha$ in the range $0 < \\alpha < 1$, the point\n$x = \\alpha x_1 + (1 - \\alpha)x_2$\nis located in $\\mathcal{R}_c\n$, i.e., $x \\in \\mathcal{R}_c$. \\newline \n\n\\textbf{Definition 2.8} \\newline\n(a) A function $f(x)$ defined over a convex set $\\mathcal{R}_c$ is said to be convex if for\nevery pair of points $x_1, x_2 \\in \\mathcal{R}_c $ and every real number $\\alpha$ in the range $0 < \\alpha < 1$, the inequality\n$f[\\alpha x_1 + (1 - \\alpha)x_2] \\leq \\alpha f(x_1) + (1 - \\alpha)f(x_2)$ \nholds. If $x_1 \\neq x_2$ and\n$f[\\alpha x_1 + (1 - \\alpha)x_2] < \\alpha f(x_1) + (1 - \\alpha)f(x_2)$\nthen f(x) is said to be strictly convex. \\newline\n(b) If $\\phi(x)$ is defined over a convex set $\\mathcal{R}_c$ and f(x) = -$\\phi(x)$ is convex, then $\\phi(x)$ is said to be concave. If f(x) is strictly convex, $\\phi(x)$ is strictly concave. \\newline\n\n\\textbf{Property of convex functions relating to the Hessian} A function $f(x) \\in C^2$is convex over a convex set \n$\\mathcal{R}_c$ if and only if the Hessian H(x) of\nf(x) is positive semidefinite for $x \\in \\mathcal{R}_c.$ \\newline \n\n\\textbf{Theorem 2.15 Relation between local and global minimizers in convex functions} \\newline \nIf $f(x)$ is a convex function defined on a convex set $\\mathcal{R}_c$, then \\newline \n(a) the set of points $S_c$ where $f(x)$ is minimum is convex; \\newline \n(b) any local minimizer of $f(x)$ is a global minimizer. \n\n\\section{Ch. 4}\n\\textbf{Dichotomous Search} \\newline\nTwo function evaluations per iteration. \\newline\nA \\textbf{unimodal function} on an interval has exactly\none point where a maximum or minimum\noccurs in the interval. \\newline \nConsider a unimodal function which is known to have a minimum in the interval $[x_L, \\ x_U]$. This interval is said to be the range of uncertainty. \\newline \nIn this method, f(x) is evaluated at two points $x_a =\nx_1 - \\epsilon/2$ and $x_b = x_1 +\\epsilon/2$ where $\\epsilon$ is a small positive number. Then depending\non whether $f(x_a) < f(x_b)$ or $f(x_a) > f(x_b)$, range $x_L$ to $x_1 + \\epsilon/2$ or $x_1 - \\epsilon/2$\nto $x_U$ can be selected and if $f(x_a) = f(x_b)$ either will do fine. If we assume\nthat $x_1 - x_L = x_U - x_1$, i.e., $x_1 = (x_L + x_U)/2$, the region of uncertainty\nis immediately reduced by half. The same procedure can be repeated for the\nreduced range, that is, f(x) can be evaluated at $x_2 - \\epsilon/2$ and $x2 + \\epsilon/2$ where $x_2$ is located at the center of the reduced range, and so on. \\newline \\newline \n%\\begin{Figure}\n%\t\\centering\n%\t\\includegraphics[width=\\linewidth]{disearch}\n%\\end{Figure}\n\\textbf{Algorithm 4.1 Fibonacci} \\newline\nComputing n =$I_n=\\frac{I_1}{F_n}$, function evaluations = n-1 \\newline\n\\textbf{Step 1} \\newline\nInput $x_{L,1}, x_{U,1}$, and n. \\newline\n\\textbf{Step 2} \\newline\nCompute $F_1, F_2, \\cdots, F_n$ using Eq. (4.4). \\newline\n\\textbf{Step 3} \\newline\nAssign $I_1 = x_{U,1} - x_{L,1}$ and compute\n\\begin{align*}\n& I_2 = \\frac{F_{n -1}}{F_n}I_1 (\\text{see Eq. (4.6)})\\\\\n& x_{a,1} = x_{U,1} - I_2, \\quad x_{b,1} = x_{L,1} + I_2\\\\ \n& f_{a,1} = f(x_{a,1}), \\quad f_{b,1} = f(x_{b,1})\n\\end{align*}\nSet k = 1. \\newline\n\\textbf{Step 4} \\newline\nCompute $I_{k+2}$ using Eq. (4.6).\nIf $f_{a,k} \\geq f_{b,k}$, then update Eqs. (4.7) to (4.12) using\n\\begin{align*}\n& x_{L,k+1}=x_{a,k} \\\\\n& x_{U,k+1}=x_{U,k} \\\\\n& x_{a,k+1}=x_{b,k} \\\\\n& x_{b,k+1}=x_{L,k+1}+I_{k+1} \\\\\n& f_{a,k+1}=f_{b,k} \\\\\n& f_{b,k+1}=f(x_{b,k+1})\n\\end{align*}\nusing Otherwise, if $f_{a,k} < f_{b,k}$, update\ninformation using Eqs. (4.13) to (4.18) using \\newline\n\\begin{align*}\n& x_{L,k+1}=x_{L,k} \\\\\n& x_{U,k+1}=x_{b,k} \\\\\n& x_{a,k+1}=x_{U,k+1}-I_{k+2} \\\\\n& x_{b,k+1}=x_{a,k} \\\\\n& f_{a,k+1}=f(x_{a,k+1}) \\\\\n& f_{b,k+1}=f_{a,k}\n\\end{align*}\n\n\\textbf{Step 5} \\newline\nIf $k = n - 2$ or $x_{a,k+1} > x_{b,k+1}$, output $x^* = x_{a,k+1}$ and $f^* = f(x^*)$,\nand stop. Otherwise, set $k = k + 1$ and repeat from Step 4.\nThe condition $x_{a,k+1} > x_{b,k+1}$ implies that $x_{a,k+1} \\approx x_{b,k+1}$ within the precision of the computer used, as was stated earlier, or that there is an error in the algorithm. It is thus used as an alternative stopping criterion. \\newline\n\n\\textbf{Algorithm 4.2 Golden-section search} \\newline\n(function evaluations = k+1) and Golden Ratio: $K=\\cfrac{1+\\sqrt{5}}{2}$\n$\\Lambda_{GS} = I_n = \\frac{I_1}\n{K_{n-1}}$ \\quad $\\Lambda_{F} = I_n = \\frac{I_1}\n{F_n} \\approx \\frac{\\sqrt{5}}\n{K^{n+1}}I_1$\n$\\frac{I_k}{I_{k+1}}=\\frac{I_{k+1}}{I_{k+2}}\n= \\frac{I_{k+2}}{I_{k+3}}\n= \\cdots = K$ \\newline\n\\textbf{Step 1} \\newline\nInput $x_{L,1}, x_{U,1},$ and $\\epsilon$. \\newline\n\\textbf{Step 2} \\newline\nAssign $I_1 = x_{U,1} - x_{L,1}, K = 1.618034$ and compute\n\\begin{align*}\n&I_2 = I_1/K \\\\\n&x_{a,1} = x_{U,1} - I_2, \\quad x_{b,1} = x_{L,1} + I_2 \\\\\n&f_{a,1} = f(x_{a,1}), \\quad f_{b,1} = f(x_{b,1}) \n\\end{align*}\nSet $k = 1$. \\newline\n\\textbf{Step 3} \\newline\nCompute $I_{k+2} = I_{k+1}/K$ \\newline\nIf $f_{a,k} \\geq f_{b,k}$, then update $x_{L,k+1}, x_{U,k+1}$, $x_{a,k+1}$, $x_{b,k+1}$, $f_{a,k+1}$,\nand $f_{b,k+1}$ as \n\\begin{align*}\n& x_{L,k+1}=x_{a,k} \\\\\n& x_{U,k+1}=x_{U,k} \\\\\n& x_{a,k+1}=x_{b,k} \\\\\n& x_{b,k+1}=x_{L,k+1}+I_{k+1} \\\\\n& f_{a,k+1}=f_{b,k} \\\\\n& f_{b,k+1}=f(x_{b,k+1})\n\\end{align*}\nOr use using Eqs. (4.7) to (4.12). \nOtherwise if $f_{a,k} < f_{b,k}$, then update $x_{L,k+1}, x_{U,k+1}$, $x_{a,k+1}$, $x_{b,k+1}$, $f_{a,k+1}$, and $f_{b,k+1}$ as\n\\begin{align*}\n& x_{L,k+1}=x_{L,k} \\\\\n& x_{U,k+1}=x_{b,k} \\\\\n& x_{a,k+1}=x_{U,k+1}-I_{k+2} \\\\\n& x_{b,k+1}=x_{a,k} \\\\\n& f_{a,k+1}=f(x_{a,k+1}) \\\\\n& f_{b,k+1}=f_{a,k}\n\\end{align*}\nOtherwise, if $f_{a,k} < f_{b,k}$, update\ninformation using Eqs. (4.13) to (4.18). \\newline\n\\textbf{Step 4} \\newline\nIf $I_k < \\epsilon$ or $x_{a,k+1} > x_{b,k+1}$, then do: \\newline\nIf $f_{a,k+1} > f_{b,k+1}$, compute\n$x^* = 0.5(x_{b,k+1} + x_{U,k+1})$ \\newline\nIf $f_{a,k+1} = f_{b,k+1}$, compute\n$x^* = 0.5(x_{a,k+1} + x_{b,k+1})$ \\newline\nIf $f_{a,k+1} < f_{b,k+1}$, compute\n$x^* = 0.5(x_{L,k+1} + x_{a,k+1})$\nCompute $f^* = f(x^*).$ \\newline\nOutput $x^*$ and $f^*$, and stop. \\newline\n\\textbf{Step 5} \\newline\nSet $k = k + 1$ and repeat from Step 3. \\newline \\newline \n\\textbf{Equations 4.4, 4.6, (4.7-4.12) and (4.13-4.18) } \\newline\n$F_k = F_{k-1} + F_{k-2} \\quad$ for $k \\geq 2$ (4.4) \\newline\n$I_{k+2} = \\frac{F_{n-k-1}}{F_{n-k}}I_{k+1} (4.6)$\n\nIf $f_{a,k} > f_{b,k}$, then $x^*$ is in interval $[x_{a,k}, x_{U,k}]$ and so the new bounds of $x^* \\rightarrow$ %\\newline \n$x_{L,k+1} = x_{a,k} (4.7) \\quad x_{U,k+1} = x_{U,k} (4.8)$ \nSimilarly, the two interior points of the new interval, namely, $x_{a,k+1}$ and $x_{b,k+1}$\nwill be $x_{b,k}$ and $x_{L,k+1} + I_{k+2}$, respectively. We can thus assign\n$x_{a,k+1} = x_{b,k}$ (4.9) $x_{b,k+1} = x_{L,k+1} + I_{k+2}$ (4.10)\nas illustrated in Fig. 4.5. \\newline\nThe value $f_{b,k}$ is retained as the value of f(x) at\n$x_{a,k+1}$, and the value of f(x) at $x_{b,k+1}$ is calculated, i.e.,\n$f_{a,k+1} = f_{b,k}$ (4.11)\n$f_{b,k+1} = f(x_{b,k+1})$ (4.12) \\newline\n%\\begin{Figure}\n%\t\\centering\n%\t\\includegraphics[width=\\linewidth]{fibfirst}\n%\\end{Figure}\nOn the other hand, if $f_{a,k} < f_{b,k}$, then $x^*$ is in interval $[x_{L,k}, x_{b,k}]$. In this\ncase, we assign\n$x_{L,k+1} = x_{L,k}$ (4.13)\n$x_{U,k+1} = x_{b,k}$ (4.14)\n$x_{a,k+1} = x_{U,k+1} - I_{k+2}$ (4.15)\n$x_{b,k+1} = x_{a,k}$ (4.16)\n$f_{b,k+1} = f_{a,k}$ (4.17)\nand calculate\n$f_{a,k+1} = f(x_{a,k+1})$ (4.18) \\newline \\newline\n%\\begin{Figure}\n%\t\\centering\n%\t\\includegraphics[width=\\linewidth]{fibsecond}\n%\\end{Figure}\n%\\begin{Figure}\n%\t\\centering\n%\t\\includegraphics[width=\\linewidth]{golden}\n%\\end{Figure}\n%\\begin{Figure}\n%\t\\centering\n%\t\\includegraphics[width=\\linewidth]{fib}\n%\\end{Figure}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%% Unlikely to be tested %%\n%% INCLUDE FOR FINAL %%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Algorithm 4.6 Inexact line search} \\newline\n\\textbf{Step 1:} \\newline\nInput $x_k$, $d_k$, and compute $g_k$. \\newline\nInitialize algorithm parameters $\\rho, \\sigma, \\tau$, and $\\chi$. \\newline\nSet $\\alpha_L = 0$ and $\\alpha_U = 10^{99}$. \\newline\n\\textbf{Step 2:} \\newline\nCompute $f_L = f(x_k + \\alpha_L dk)$. \\newline\nCompute $f_L^\\prime = g(x_k + \\alpha_L dk)^Tdk$. \\newline\n\\textbf{Step 3:} \\newline\nEstimate $\\alpha_0$. \\newline\n\\textbf{Step 4:} \\newline\nCompute $f_0 = f(x_k + \\alpha_0dk)$. \\newline\n\\textbf{Step 5 (Interpolation)} \\newline\nIf $f_0 > f_L + \\rho(\\alpha_0 - \\alpha_L)f_L^\\prime$ , then do: \\newline\na. If $\\alpha_0 < \\alpha_U$, then set $\\alpha_U = \\alpha_0$.\\newline\nb. Compute $\\breve{\\alpha}_0$ using the interpolation formula Eq. (4.57). %\\newline\n\\[\n\\breve{\\alpha}_0 =\\alpha_L + \\frac{(\\alpha_0-\\alpha_L)^2f_L^\\prime}{2[f_L-f_0+(\\alpha_0-\\alpha_L)f_L^\\prime]}\n\\]\nc. If $\\breve{\\alpha}_0 < \\alpha_L + \\tau(\\alpha_U - \\alpha_L)$ then set $\\breve{\\alpha}_0 = \\alpha_L + \\tau(\\alpha_U - \\alpha_L)$. \\newline\nd. If $\\breve{\\alpha}_0 > \\alpha_U - \\tau(\\alpha_U - \\alpha_L)$ then set $\\breve{\\alpha}_0 = \\alpha_U - \\tau(\\alpha_U - \\alpha_L)$. \\newline\ne. Set $\\alpha_0 = \\breve{\\alpha}_0$ and go to Step 4. \\newline\n\\textbf{Step 6} \\newline\nCompute $f_0^\\prime = g(x_k + \\alpha_0dk)^Tdk$. \\newline\n\\textbf{Step 7 (Extrapolation)} \\newline\nIf $f_0^\\prime < \\alpha f_L^\\prime$ , then do: \\newline\na. Compute $\\nabla \\alpha_0 = (\\alpha_0 - \\alpha_L)f_0^\\prime /(f_L^\\prime - f_0^\\prime)$ (see Eq. (4.58)).\n\\[\n\\breve{\\alpha}_0 = \\alpha_0 + (\\alpha_0 - \\alpha_L)f_0^\\prime\n(f_L^\\prime - f_0^\\prime) \\quad (Eq. (4.58))\n\\]\nb. If $\\nabla \\alpha_0 < \\tau(\\alpha_0 - \\alpha_L)$, then set $\\nabla \\alpha_0 = \\tau(\\alpha_0 - \\alpha_L)$. \\newline\nc. If $\\nabla \\alpha_0 > \\chi(\\alpha_0 - \\alpha_L)$, then set $\\nabla \\alpha_0 = \\chi(\\alpha_0 - \\alpha_L)$. \\newline\nd. Compute $\\breve{\\alpha}_0 = \\alpha_0 + \\nabla \\alpha_0$. \\newline\ne. Set $ \\alpha_L = \\alpha_0, \\alpha_0 = \\breve{\\alpha}_0, f_L = f_0, f_L^\\prime = f_0^\\prime$, and go to Step 4. \\newline\n\\textbf{Step 8} \\newline\nOutput $\\alpha_0$ and $f_0 = f(x_k + \\alpha_0dk)$, and stop.\n\n\n\\section{Ch. 5}\nStandard form: $f(x)=\\frac{1}{2}x^T H x + x^T g(x) + C$ \\newline \nRate of Convergence $\\beta = (1-r^2)/(1+r^2)$, where r is the smallest eigenvalue divided by the biggest eigenvalue. \\newline\n\\begin{align*}\n& H^{-1}=\\begin{bmatrix}\na & c \\\\\nc & b \n\\end{bmatrix}^{-1}\n=\n\\frac{1}{ab-c^2}\n\\begin{bmatrix}\nb & -c \\\\\n-c & a\n\\end{bmatrix} \\quad ab-c^2 \\neq 0\n\\end{align*}\n$[f(x_k)-f(x^*)]\\leq \\left(\\frac{1-r}{1+r}\\right)^2[f(x_k)-f(x^*)]$ \\newline\n\\textbf{Algorithm 5.1 Steepest-descent algorithm} \\newline\n\\textbf{Step 1} \\newline\nInput $x_0$ and initialize the tolerance $\\epsilon$. \\newline\nSet $k = 0$. \\newline\n\\textbf{Step 2} \\newline\nCalculate gradient $g_k$ and set $d_k=-g_k$. \\newline\n\\textbf{Step 3} \\newline\nFind $\\alpha_k$, the value of $\\alpha$ that minimizes $f(x_k + \\alpha d_k)$, using a line search (Algorithm Inexact Line Search. 4.6). \\newline\n\\textbf{Step 4} \\newline\nSet $x_{k+1} = x_k + \\alpha_k d_k$ and calculate $f_{k+1} = f(x_{k+1})$. \\newline\n\\textbf{Step 5} \\newline\nIf $||\\alpha_k d_k|| < \\epsilon$, then do: \\newline\nOutput $x^* = x_{k+1}$ and $f(x^*) = f_{k+1}$, and stop. \\newline\nOtherwise, set $k = k + 1$ and repeat from Step 2. \\newline \n\n\\textbf{Algorithm 5.3 Basic Newton algorithm} \\newline\n\\textbf{Step 1} \\newline\nInput $x_0$ and initialize the tolerance $\\epsilon$ \\newline\nSet $k = 0$. \\newline\n\\textbf{Step 2} \\newline\nCompute $g_k$ and $H_k$. \\newline\nIf $H_k$ is not positive definite, force it to become positive definite. \\newline\n\\textbf{Step 3} \\newline\nCompute $H_k^{-1}$ and $d_k=-H_k^{-1}g_k$ \\newline\n\\textbf{Step 4} \\newline\nFind $\\alpha_k$ , the value of $\\alpha$ that minimizes $f(x+\\alpha d_k)$, using a line search. \\newline\n\\textbf{Step 5} \\newline\nSet $x_{k+1}=x_k+\\alpha_k d_k$ \\newline\nCompute $f_{k+1}=f(x_{k+1})$. \\newline\n\\textbf{Step 6} \\newline\nIf $||\\alpha_k d_k|| < \\epsilon$, then do: \\newline\nOutput $x^*=x_{k+1}$ and $f(x^*)=f(x_{k+1})$, and stop \\newline\nOtherwise, set $k = k + 1$ and repeat from Step 2. \\newline\n\\textbf{Algorithm 5.5 Gauss --- Newton Algorithm} \\newline\n$f=\\begin{bmatrix}\nf_1(x) & f_2(x) & \\cdots f_m(x)\n\\end{bmatrix}^T$, J = Jacobian \\newline\n$F(x)=\\sum_{p=1}^{m}f_p(x)^2=f^Tf$ \\newline\n$J=\\begin{bmatrix}\n\\frac{ \\partial f_1}{\\partial x_1} & \\frac{ \\partial f_1}{ \\partial x_2} & \\cdots & \\frac{\\partial f_1}{\\partial x_n} \\\\\n\\frac{\\partial f_2}{\\partial x_1} & \\frac{\\partial f_2}{\\partial x_2} & \\cdots & \\frac{\\partial f_2}{\\partial x_n} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\\frac{\\partial f_m}{\\partial x_1}& \\frac{\\partial f_m}{\\partial x_2}& \\cdots & \\frac{\\partial f_m}{\\partial x_n}\n\\end{bmatrix}$ \\newline\n\\textbf{Step 1} \\newline\nInput $x_0$ and initialize the tolerance $\\epsilon$. \\newline\nSet $k = 0$. \\newline\n\\textbf{Step 2} \\newline\nCompute $f_{pk} = f_p(x_k)$ for $p = 1, 2, \\cdots, m$ and $F_k$.\n\\newline\n\\textbf{Step 3} \\newline\nCompute $J_k, g_k = 2J^T_k f_k$, and $H_k = 2J^T_k J_k$. \\newline\n\\textbf{Step 4} \\newline\n$d_k=-H_k^{-1}g_k$ \\newline\n\\textbf{Step 5} \\newline\nFind $\\alpha_k$, the value of $\\alpha$ that minimizes $f(x_k + \\alpha d_k)$. \\newline\n\\textbf{Step 6} \\newline\nSet $x_{k+1} = x_k + \\alpha_kd_k$. \\newline\nCompute $f_{p,k+1}$ for $p = 1, 2,\\cdots, m$ and $F_{k+1}$. \\newline\n\\textbf{Step 7} \\newline\nIf $|F_{k+1}-F_k | < \\epsilon$, then do: \\newline\nOutput $x^*= x_{k+1}$, $f_{p,k+1}(x^*)$ for $p = 1, 2,\\cdots, m$, and $F_{k+1}$. \\newline\nStop. \\newline\nOtherwise, set $k = k + 1$ and repeat from Step 3.\n\n\n\\section{Ch. 7}\n\\textbf{Problems with Rank-one Method}\n\\begin{enumerate}\n\t\\item positive definite $S_k$ may not yield positive definite $S_{k+1}$\n\t\\item denominator in correction formula may approach zero \n\\end{enumerate} \nTHE DFP and BFGS are implementing of the basic algorithms Quasi Newton (7.2) with changes to the updating function. $d_k=-S_kg_k$ and $f(x_k + \\alpha d_k)$ $\\rightarrow$ $\\alpha_k=\\frac{g_k^TS_kg_k} \n{g_k^TS_kHS_kg_k}$ \\newline\nConvergence equation: $\\beta = \\left(\\frac{1-r}{1+r}\\right)^2$ $f(x_{k+1})-f(x^*) \\leq \\left(\\frac{1-r}{1+r}\\right)^2[f(x_k)-f(x^*)]$ \\newline\n\\textbf{BFGS and then DFP properties} \\newline\nFor convex quadratic functions (BFGS)\n\\begin{enumerate}\n\t\\item[---] $Sk+1$ becomes identical to $H^{-1}$ for $k = n-1$.\n\t\\item[---] Directions $\\delta_0,\\delta_1,\\cdots,\\delta_{n-1}$ form a conjugate set. \n\t\\item[---] $S_{k+1}$ is positive definite if $S_k$ is positive definite.\n\t\\item[---] $\\delta^T_k \\gamma_k = \\delta^T_k g_{k+1}-\\delta_T^k g_k > 0$ applies. \\newline\n\\end{enumerate} \n\\textbf{For DFP (from textbook)} \n\\begin{enumerate}\n\t\\item If $S_k$ is PD, then \n\tthe matrix $S_{k+1}$ generated by DFP is also PD.\n\t\\item Directions $\\delta_0,\\delta_1,\\cdots,\\delta_{n-1}$ form a conjugate set. \\newline\n\\end{enumerate} \n\\textbf{Algorithm 7.2 adjusted for DFP/BFGS} \\newline\n\\textbf{Step 1} \\newline\nInput $x_0$ and initialize the tolerance $\\epsilon$. \\newline\nSet $k = 0$ and $S_0 = I_n$. \\newline\nCompute $g_0$. \\newline\n\\textbf{Step 2} \\newline\nSet $d_k=-S_kg_k$ \\newline\nFind $\\alpha_k$, the value of $\\alpha$ that minimizes $f(x_k+\\alpha d_k)$, using a line search \\newline\nSet $\\delta_k=\\alpha_k d_k$ and $x_{k+1}=x_k+\\delta_k$ \\newline\n\\textbf{Step 3} \\newline\nIf $||\\delta_k|| < \\epsilon$, output $x^* = x_{k+1}$ and $f(x^*) = f(x_{k+1})$, and stop \\\\\n\\textbf{Step 4} \\newline\nCompute $g_{k+1}$ and set $\\gamma_k=g_{k+1}-g_k$ \\newline\nCompute $S_{k+1}$ using appropriate formula.\n\\begin{align*}\n&\\text{Basic/ Rank One:} \\quad S_{k+1}=S_k+\\frac{(\\delta_k-S_k \\gamma_k)(\\delta_k-S_k \\gamma_k)^T}{\\gamma_k^T(\\delta_k-S_k\\gamma_k)}\\\\\n&\\text{DFP:}\\quad S_{k+1}=S_k+\\frac{\\delta_k \\delta_k^T}{\\delta_k^T \\gamma_k}-\\frac{S_k\\gamma_k\\gamma_k^TS_k}{\\gamma_k^TS_k\\gamma_k}\\\\\n&\\text{BFGS:} \\quad S_{k+1}=S_k+\\left(1+\\frac{\\gamma_k^TS_k\\gamma_k}{\\gamma_k^T\\delta_k}\\right)\\frac{\\delta_k\\delta_k^T}{\\gamma_k^T\\delta_k}-\\frac{(\\delta_k\\gamma_k^TS_k+S_k\\gamma_k\\delta_k^T)}{\\gamma_k^T\\delta_k}\n\\end{align*}\nSet $k=k+1$ and repeat from Step 2. \\newline\n\\begin{itemize}\n\t\\item[---] remember that $\\left(1+\\frac{\\gamma_k^TS_k\\gamma_k}{\\gamma_k^T\\delta_k}\\right)$ is a single number\n\t\\item[---] $\\delta_k\\delta_k^T$ is a matrix.\n\t\\item[---] Focus on minimization, $\\max[f(x)] =-\\min[-f(x)]$\n\t\\item[---] Hessian is positive semidefinite for concave functions.\n\\end{itemize}\n\\end{multicols}", "meta": {"hexsha": "54b1285274fd3495980e80dba8b9860663977389", "size": 23718, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "public/CheatSheets/ELEC403Cheat.tex", "max_stars_repo_name": "FriendlyUser/PortfolioWebsite", "max_stars_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "public/CheatSheets/ELEC403Cheat.tex", "max_issues_repo_name": "FriendlyUser/PortfolioWebsite", "max_issues_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "public/CheatSheets/ELEC403Cheat.tex", "max_forks_repo_name": "FriendlyUser/PortfolioWebsite", "max_forks_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.7022587269, "max_line_length": 253, "alphanum_fraction": 0.6476515726, "num_tokens": 9420, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.8418256532040708, "lm_q1q2_score": 0.6427202179647945}} {"text": "\\section{What is machine learning?} \\label{ch:what_is_ml}\n\n\\subsection{Machine learning in general}\n\nUsually the task of programming is to apply rules to certain problems in order to obtain solutions.\nWith the help of machine learning we want to create a model that uses solutions for certain problems to find out the corresponding rules.\nThe relationship is shown in figure~\\ref{fig:cp_vs_ml}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.5\\textwidth]{images/classical_prog_vs_ml.png}\n \\caption[ML, a new programming paradigm]{Machine learning, a new programming paradigm \\cite[p.5]{Chollet2017}}\n \\label{fig:cp_vs_ml}\n\\end{figure}\n\nThe promise of machine learning is that the rules learned can be much more complex than it is possible to program by hand.\nA good example is image recognition (which will be the focus of this project), where the content of images is to be classified.\nThe relations between the pixels have to be found out, which leads to a rather complex model.\nThis cannot be programmed in the traditional way even for the simplest tasks.\n\nThe universal approximation theorem \\cite{Cybenko1989, Hornik1989} states that for any input $x$ there is a function $h$ which approximates the mapping $y$, while the mapping is only the input to the correct output (often labeled by a human).\nThis relationship is described in equation~\\eqref{eq:universal_approx}.\n\n% TODO define sets X and Y of \\theta and x\n\\begin{equation}\n \\forall \\epsilon > 0 :\n \\exists h(\\theta, x) : \\forall x \\in I : | h(\\theta, x) - y(x) | < \\epsilon\n \\label{eq:universal_approx}\n\\end{equation}\n% TODO the text here can be changed with a class of functions (see UAT paper)\n\nThe mathematical description of $h$ is called a model (e.g. $h(x) = \\sum_{i=1}^n{\\lambda (\\theta^T * x)}$).\n$\\lambda \\in \\mathbb{R}$ is a constant and $\\theta$ is a tensor that defines the parameters of $h$, which are made up of so-called weights and biases, which will be explained later.\n\nThe universal approximation theorem makes no statements about the form of $\\theta$; therefore $h$ is called a hypothesis because it states that for a certain composition of $\\theta$ this equation is true.\nThis term is summarized in the equation~\\eqref{eq:hypothesis}.\n\n\\begin{equation}\n \\forall \\epsilon > 0 : \\exists \\theta \\in X : \\forall x \\in I : |\n h_\\theta(x) - y(x) | < \\epsilon\n \\label{eq:hypothesis}\n\\end{equation}\n\nA loss function is used for model valuation; it compares the result of the hypothesis with the data provided.\nProbably the most prominent loss function is the \\name{mean square error}.\nIt takes the sum of the deviation of $n$ examples squared and divides it by $2*n$.\n\n\\begin{equation}\n L_{mse}(\\theta, x) = \\frac{1}{2 n} \\sum_{i=1}^n (h_\\theta(x) - y(x))^2\n \\label{eq:mse}\n\\end{equation} \n\nThe task of machine learning is to determine a $\\theta$ for a model that confirms the equations~\\ref{eq:universal_approx} and \\ref{eq:hypothesis}, which is done by iterative updating of $\\theta$. \nBefore training, $\\theta$ is randomly initiated and then iteratively adjusted to make $\\epsilon$ converge toward zero. This minimizes the value of $L$.\n\nIn order for the value of the loss function to gradually decrease, $\\theta$ is adjusted using an optimization function.\nThis is usually done with a gradient descent algorithm or a variant of it.\nThe gradient descent is performed by calculating the gradient of the loss function and subtracting it from the corresponding parameters shown in equation~\\eqref{eq:gradient_descent}.\n\n\\begin{equation}\n \\theta_{i+1} := \\theta_i - \\eta \\nabla_\\theta L(\\theta, x)\n \\label{eq:gradient_descent}\n\\end{equation}\n\nWhere $\\eta$ is a constant called the learning rate, which helps the loss function converge to 0 using the gradient descent algorithm.\nIt is one of many hyperparameters that are not automatically adjusted during training, but are determined beforehand.\nOptimal adjustment of hyperparameters is an important task that is difficult to automate and is therefore the focus of much research\\footnote{There are algorithms for adjusting the learning rate, namely \\name{AdaGrad} \\cite{Duchi2010} and derived algorithms and is the main topic of the chapter \\ref{ch:hyper_parameter_tuning}}.\nIf the equation \\eqref{eq:hypothesis} applies to the model in question, it is then sufficiently suitable for the given input.\n\nThese concepts are presented now using a simple example.\n\n\\subsection{Simple linear example} \\label{ch:simple_linear_example}\n\nAs an illustrative example, a model is created that translates Fahrenheit into Celsius\\footnote{Please note that using a learning algorithm here is very inefficient and is in contrast to Maslow's hammer (\\aka{https://en.wikipedia.org/wiki/Law_of_the_instrument}).}.\nThe original equation is given by the linear function $y = mx + b$ as $F = C * 1.8 + 32$, with $F$ as degrees Fahrenheit and $C$ as degrees Celsius.\n\nA model to learn this relationship is defined in the listing~\\ref{lst:c_to_f} \\footnote{Also available at \\aka{https://github.com/klawr/deepmech/tree/master/reports/srp/code/c_to_f.js}} .\n\n% TODO add JavaScript as language\n\\lstinputlisting[label={lst:c_to_f}, caption={Celsius to Fahrenheit}]{code/c_to_f.js}\n\nThe described model is implemented by \\code{w} and the hypothesis function by \\code{h}.\nSince it is known that the relationship is linear, the model represents a one-dimensional polynomial, with the first index being dimensionless (and, as mentioned above, emulating the bias) and the second being used as input x.\nWe will add a dimension later to demonstrate the behavior with polynomials that do not directly represent the target function.\n\n\\code{y} provides the correct, predefined solution.\nPlease note that \\code{y} is only used during training and is omitted if the model parameters are correctly set after training.\n\nIn this example the loss function is described by the \\textit{mean square error } \\footnote{ Abbreviated as \\textit{mse}, which is shown in equation \\eqref{eq:mse}}.\n\nTo minimize the loss function gradient descent described in equation~\\eqref{eq:gradient_descent} is used.\n\nIn listing~\\ref{lst:c_to_f} equation~\\eqref{eq:gradient_descent} is implemented as \\code{sgd} with $\\theta_0 = \\code{b}$ and $\\theta_1= \\code{m}$ as shown in equation~\\eqref{eq:sgd_mse_here}.\n\n\\code{sgd} is short for \\textit{stochastic gradient descent}\\footnote{A distinction is made between batch, mini-batch and stochastic gradient descent; using a complete data set, a defined subset or individual data on the individual iterations of the training process}.\nIt represents the optimization function that updates the parameters of the model by calculating the gradient of loss for each parameter and then updating it.\n\n\\begin{equation}\n \\begin{split}\n \\theta_{0} & := \\theta_{0} - \\frac{\\partial L}{\\partial \\theta_{0}} =\n \\theta_{0} - (h(x) - y(x)) \\\\\n \\theta_{1} & := \\theta_{1} - \\frac{\\partial L}{\\partial \\theta_{1}} =\n \\theta_{0} - (h(x) - y(x)) * x\n \\end{split}\n \\label{eq:sgd_mse_here}\n\\end{equation}\n\n\\begin{lstlisting}[caption={Output of C to F converter.}]\nloss: 343041.666256673\nloss: 16242.7707476548\nloss: 88.27595280422442\nloss: 10.74432403839251\nloss: 1.1607137779951606\nloss: 0.252186283500194\nloss: 0.00040450626910586374\nloss: 0.00004247033303206121\nloss: 7.158429555712516e-7\nloss: 1.6948019668728757e-10\nweights: [ 31.99999973758735, 1.8000003977458756 ]\n\\end{lstlisting} \n\nThe expected behavior for the loss function is to decrease with each iteration.\nThis is the case in this example, showing that the deviation of the hypothesis from the actual result decreases.\nAfter 200 iterations, the parameters \\code{m} and \\code{b} of the \\code{model} are displayed and show that they are actually close to the expected result.\nBy increasing the number of iterations, the result can be increased up to the point where they are rounded by the JavaScript compiler to represent exact results (using Node.js\\footnote{As far as can be assumed when calculating exact results using floating point numbers.}).\n\n\\begin{SCfigure}\n \\centering\n \\caption[Simplest model]{This example can be visualized as shown here. The bias can be interpreted as additional input, which always has the value 1. The input is multiplied by the parameters marked with the corresponding arrows, which in turn are summed up (resulting in $y = mx + b$).}\n \\includegraphics[width=0.45\\textwidth]{images/1_simplest_nn.png}\n\\end{SCfigure} \n\nProvided that this hypothesis is initially well suited to the task, it is important to note that by modifying the hypothesis to represent a polynomial of higher degree (e.g. $y = nx^2 + mx + b$) the additional parameters converge towards 0, which effectively gives the same result as before \\footnote{But to achieve similar results, the iteration number must be increased by at least a factor of 20}.\n\nThe experiment can be followed at \\aka{https://github.com/klawr/deepmech/tree/master/reports/srp/code/c_to_f_adv.js}.\n", "meta": {"hexsha": "d57dab2f12dce8078956a33cfeb8c7034e9d88b0", "size": 9026, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "reports/srp/sections/what_is_ml.tex", "max_stars_repo_name": "klawr/deepmech", "max_stars_repo_head_hexsha": "61de238f1d4b1b867ec1d5f4e4af2a3b25a5abff", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-17T12:27:06.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-17T12:27:06.000Z", "max_issues_repo_path": "reports/srp/sections/what_is_ml.tex", "max_issues_repo_name": "klawr/deepmech", "max_issues_repo_head_hexsha": "61de238f1d4b1b867ec1d5f4e4af2a3b25a5abff", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-27T13:13:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-27T13:13:17.000Z", "max_forks_repo_path": "reports/srp/sections/what_is_ml.tex", "max_forks_repo_name": "klawr/deepmech", "max_forks_repo_head_hexsha": "61de238f1d4b1b867ec1d5f4e4af2a3b25a5abff", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 66.3676470588, "max_line_length": 400, "alphanum_fraction": 0.7633503213, "num_tokens": 2319, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256512199033, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6427202164499148}} {"text": "\\chapter{Lebesgue integration}\nOn any measure space $(\\Omega, \\SA, \\mu)$ we can then,\nfor a function $f \\colon \\Omega \\to [0,\\infty]$\ndefine an integral\n\\[ \\int_\\Omega f \\; d\\mu. \\]\nThis integral may be $+\\infty$ (even if $f$ is finite).\nAs the details of the construction won't matter for us later on,\nwe will state the relevant definitions,\nskip all the proofs,\nand also state all the properties that we actually care about.\nConsequently, this chapter will be quite short.\n\n\\section{The definition}\nThe construction is done in four steps.\n\\begin{definition}\n\tIf $A$ is a measurable set of $\\Omega$,\n\tthen the \\vocab{indicator function}\n\t$\\mathbf{1}_A \\colon \\Omega \\to \\RR$ is defined by\n\t\\[ \\mathbf{1}_A(\\omega) = \\begin{cases}\n\t\t\t1 & \\omega \\in A \\\\\n\t\t\t0 & \\omega \\notin A.\n\t\t\\end{cases} \\]\n\\end{definition}\n\n\\begin{step}\n\t[Indicator functions]\n\tFor an indicator function, we require\n\t\\[ \\int_\\Omega \\mathbf{1}_A \\; d\\mu \\defeq \\mu(A) \\]\n\t(which may be infinite).\n\\end{step}\nWe extend this linearly now for nonnegative functions\nwhich are sums of indicators:\nthese functions are called \\vocab{simple functions}.\n\\begin{step}\n\t[Simple functions]\n\tLet $A_1$, \\dots, $A_n$ be a finite collection of measurable sets.\n\tLet $c_1$, \\dots, $c_n$ be either nonnegative real numbers or $+\\infty$.\n\tThen we define\n\t\\[ \\int_\\Omega \\left( \\sum_{i=1}^n c_i \\mathbf{1}_{A_i} \\right) \\; d\\mu\n\t\t\\defeq \\sum_{i=1}^n c_i \\mu(A_i). \\]\n\tIf $c_i = \\infty$ and $\\mu(A_i) = 0$, we treat $c_i \\mu(A_i) = 0$.\n\\end{step}\nOne can check the resulting sum does not depend\non the representation of the simple function as $\\sum c_i \\mathbf{1}_{A_i}$.\nIn particular, it is compatible with the previous step.\n\nConveniently, this is already enough to define the integral\nfor $f \\colon \\Omega \\to [0, +\\infty]$.\nNote that $[0,+\\infty]$ can be thought of as a topological space\nwhere we add new open sets $(a,+\\infty]$ %chktex 9\nfor each real number $a$ to our usual basis of open intervals.\nThus we can equip it with the Borel sigma-algebra.\\footnote{We\n\t\\emph{could} also try to define a measure on it,\n\tbut we will not: it is a good enough for us\n\tthat it is a measurable space.}\n\\begin{step}\n\t[Nonnegative functions]\n\tFor each measurable function $f \\colon \\Omega \\to [0, +\\infty]$, let\n\t\\[ \\int_\\Omega f \\; d\\mu \\defeq\n\t\t\\sup_{0 \\le s \\le f} \\left( \\int_\\Omega s \\; d\\mu \\right) \\]\n\twhere the supremum is taken over all \\emph{simple} $s$\n\tsuch that $0 \\le s \\le f$.\n\tAs before, this integral may be $+\\infty$.\n\\end{step}\nOne can check this is compatible with the previous definitions.\nAt this point, we introduce an important term.\n\\begin{definition}\n\tA measurable (nonnegative) function $f \\colon \\Omega \\to [0, +\\infty]$\n\tis \\vocab{absolutely integrable}\n\tor just \\vocab{integrable} if $\\int_\\Omega f \\; d\\mu < \\infty$.\n\\end{definition}\nWarning: I find ``integrable'' to be \\emph{really} confusing terminology.\nIndeed, \\emph{every} measurable function from $\\Omega$ to $[0,+\\infty]$\ncan be assigned a Lebesgue integral, it's just that\nthis integral may be $+\\infty$.\nSo the definition is far more stringent than the name suggests.\nEven constant functions can fail to be integrable:\n\\begin{example}\n\t[We really should call it ``finitely integrable'']\n\tThe constant function $1$ is \\emph{not} integrable on $\\RR$,\n\tsince $\\int_\\RR 1 \\; d\\mu = \\mu(\\RR) = +\\infty$.\n\\end{example}\nFor this reason, I will usually prefer the term ``integrable''.\n(If it were up to me, I would call it ``finitely integrable'',\nand usually do so privately.)\n\nFinally, this lets us integrate general functions.\n\\begin{definition}\n\tIn general, a measurable function $f \\colon \\Omega \\to [-\\infty, \\infty]$\n\tis \\vocab{absolutely integrable} or just \\vocab{integrable} if $|f|$ is.\n\\end{definition}\nSince we'll be using the first word, this is easy to remember:\n``absolutely integrable'' requires taking absolute values.\n\n\\begin{step}\n\t[Absolutely integrable functions]\n\tIf $f \\colon \\Omega \\to [-\\infty, \\infty]$ is absolutely integrable,\n\tthen we define\n\t\\begin{align*}\n\t\tf^+(x) &= \\max\\left\\{ f(x), 0 \\right\\} \\\\\n\t\tf^-(x) &= \\min\\left\\{ f(x), 0 \\right\\} \\\\\n\t\\end{align*}\n\tand set\n\t\\[ \\int_\\Omega f \\; d\\mu = \\int_\\Omega |f^+| \\; d\\mu\n\t\t- \\int_\\Omega |f^-| \\; d\\mu \\]\n\twhich in particular is finite.\n\\end{step}\nYou may already start to see that we really like nonnegative functions:\nwith the theory of measures, it is possible to integrate them,\nand it's even okay to throw in $+\\infty$'s everywhere.\nBut once we start dealing with functions that can be either positive or negative,\nwe have to start adding finiteness restrictions ---\nactually essentially what we're doing is splitting\nthe function into its positive and negative part,\nrequiring both are finite, and then integrating.\n\n\nTo finish this section, we state for completeness\nsome results that you probably could have guessed were true.\nFix $\\Omega = (\\Omega, \\SA, \\mu)$, and\nlet $f$ and $g$ be measurable real-valued functions\nsuch that $f(x) = g(x)$ almost everywhere.\n\\begin{itemize}\n\t\\ii (Almost-everywhere preservation)\n\tThe function $f$ is absolutely integrable if and only if $g$ is,\n\tand if so, their Lebesgue integrals match.\n\t\\ii (Additivity)\n\tIf $f$ and $g$ are absolutely integrable then\n\t\\[ \\int_\\Omega f+g \\; d\\mu\n\t\t= \\int_\\Omega f \\; d\\mu\n\t\t+ \\int_\\Omega g \\; d\\mu. \\]\n\tThe ``absolutely integrable'' hypothesis can be dropped\n\tif $f$ and $g$ are nonnegative.\n\t\\ii (Scaling) If $f$ is absolutely integrable and $c \\in \\RR$\n\tthen $cf$ is absolutely integrable and\n\t\\[ \\int_\\Omega cf \\; d\\mu = c \\int_\\Omega f \\; d\\mu. \\]\n\tThe ``absolutely integrable'' hypothesis can be dropped\n\tif $f$ is nonnegative and $c > 0$.\n\t\\ii (Monotoncity)\n\tIf $f$ and $g$ are absolutely integrable and $f \\le g$, then\n\t\\[ \\int_\\Omega f \\; d\\mu \\le \\int_\\Omega g \\; d\\mu. \\]\n\tThe ``absolutely integrable'' hypothesis can be dropped\n\tif $f$ and $g$ are nonnegative.\n\\end{itemize}\nThere are more famous results like monotone/dominated convergence\nthat are also true, but we won't state them here\nas we won't really have a use for them in the context of probability.\n(They appear later on in a bonus chapter.)\n\n\\section{Relation to Riemann integrals (or: actually computing Lebesgue integrals)}\nFor closed intervals, this actually just works out of the box.\n\\begin{theorem}\n\t[Lebesgue integral generalizes Riemann integral]\n\tLet $f \\colon [a,b] \\to \\RR$ be a Riemann integrable function\n\t(where $[a,b]$ is equipped with the Borel measure).\n\tThen $f$ is also Lebesgue integrable and the integrals agree:\n\t\\[ \\int_a^b f(x) \\; dx = \\int_{[a,b]} f \\; d\\mu. \\]\n\\end{theorem}\n\nThus in practice, we do all theory with Lebesgue integrals (they're nicer),\nbut when we actually need to compute $\\int_{[1,4]} x^2 \\; d\\mu$\nwe just revert back to our usual antics with the\nFundamental Theorem of Calculus.\n\\begin{example}\n\t[Integrating $x^2$ over {$[1,4]$}]\n\tReprising our old example:\n\t\\[ \\int_{[1,4]} x^2 \\; d\\mu\n\t\t= \\int_1^4 x^2 \\; dx\n\t\t= \\frac13 \\cdot 4^3 - \\frac13 \\cdot 1^3 = 21. \\]\n\\end{example}\n\nThis even works for \\emph{improper} integrals,\nif the functions are nonnegative.\nThe statement is a bit cumbersome to write down, but here it is.\n\\begin{theorem}\n\t[Improper integrals are nice Lebesgue ones]\n\tLet $f \\ge 0$ be a \\emph{nonnegative}\n\tcontinuous function defined on $(a,b) \\subseteq \\RR$,\n\tpossibly allowing $a = -\\infty$ or $b = \\infty$.\n\tThen\n\t\\[ \\int_{(a,b)} f \\; d\\mu\n\t\t= \\lim_{\\substack{a' \\to a^+ \\\\ b' \\to b^-}}\n\t\t\\int_{a'}^{b'} f(x) \\; dx \\]\n\twhere we allow both sides to be $+\\infty$\n\tif $f$ is not absolutely integrable.\n\\end{theorem}\nThe right-hand side makes sense since $[a',b'] \\subsetneq (a,b)$\nis a compact interval on which $f$ is continuous.\nThis means that improper Riemann integrals of nonnegative\nfunctions can just be regarded as Lebesgue ones\nover the corresponding open intervals.\n\nIt's probably better to just look at an example though.\n\\begin{example}\n\t[Integrating $1/\\sqrt{x}$ on $(0,1)$]\n\tFor example, you might be familiar with improper integrals like\n\t\\[ \\int_0^1 \\frac{1}{\\sqrt x} \\; dx\n\t\t\\defeq \\lim_{\\eps \\to 0^+}\n\t\t\\int_\\eps^1 \\frac{1}{\\sqrt x} \\; dx\n\t\t= \\lim_{\\eps \\to 0^+} \\left( 2\\sqrt{1} - 2\\sqrt{\\eps} \\right) = 2.\n\t\\]\n\t(Note this appeared before as \\Cref{prob:improper}.)\n\tIn the Riemann integration situation, we needed the limit as $\\eps \\to 0^+$\n\tsince otherwise $\\frac{1}{\\sqrt x}$ is not defined as a function $[0,1] \\to \\RR$.\n\tHowever, it is a \\emph{measurable nonnegative}\n\tfunction $(0,1) \\to [0,+\\infty]$, and hence\n\t\\[ \\int_{(0,1)} \\frac{1}{\\sqrt x} \\; d\\mu = 2. \\]\n\\end{example}\n\nIf $f$ is not nonnegative, then all bets are off.\nIndeed \\Cref{prob:sin_improper} is the famous counterexample.\n\n\\section{\\problemhead}\n\n\\begin{sproblem}\n\t[The indicator of the rationals]\n\t\\label{prob:1QQ}\n\tTake the indicator function\n\t$\\mathbf 1_{\\QQ} \\colon \\RR \\to \\{0,1\\} \\subseteq \\RR$\n\tfor the rational numbers.\n\t\t\\begin{enumerate}[(a)]\n\t \\ii Prove that $\\mathbf{1}_\\QQ$ is not Riemann integrable.\n\t \\ii Show that $\\int_\\RR \\mathbf{1}_\\QQ$ exists\n\t and determine its value --- the one you expect!\n \\end{enumerate}\n\\end{sproblem}\n\n\\begin{dproblem}\n\t[An improper Riemann integral with sign changes]\n\t\\label{prob:sin_improper}\n\tDefine $f \\colon (1,\\infty) \\to \\RR$ by $f(x) = \\frac{\\sin(x)}{x}$.\n\tShow that $f$ is not absolutely integrable,\n\tbut that the improper Riemann integral\n\t\\[ \\int_1^\\infty f(x) \\; dx \\defeq\n\t\t\\lim_{b \\to \\infty}\n\t\t\\int_a^b f(x) \\; dx \\]\n\tnonetheless exists.\n\\end{dproblem}\n", "meta": {"hexsha": "9129a24aaaa5114af1dc4ec6981d9552f6db2423", "size": 9441, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/measure/lebesgue-int.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/measure/lebesgue-int.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/measure/lebesgue-int.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.5020920502, "max_line_length": 83, "alphanum_fraction": 0.6967482258, "num_tokens": 3036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837527911056, "lm_q2_score": 0.8418256532040708, "lm_q1q2_score": 0.6427202089040678}} {"text": "\\documentclass[12pt]{article}\n\n% PACKAGES\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{float}\n\\usepackage{graphicx}\n\\usepackage{enumerate}\n%\\usepackage{mathtools}\n\\usepackage{hyperref}\n\n% CUSTOM STYLES\n%\\providecommand{\\e}[1]{\\ensuremath{\\times 10^{#1}}}\n%\\DeclarePairedDelimiter\\abs{\\lvert}{\\rvert}%\n\n\\usepackage{listings}\n\\usepackage{color}\n\n\\definecolor{dkgreen}{rgb}{0,0.6,0}\n\\definecolor{gray}{rgb}{0.5,0.5,0.5}\n\\definecolor{mauve}{rgb}{0.58,0,0.82}\n\n\\lstset{frame=tb,\n language=Mathematica,\n aboveskip=3mm,\n belowskip=3mm,\n showstringspaces=false,\n columns=flexible,\n basicstyle={\\small\\ttfamily},\n numbers=none,\n numberstyle=\\tiny\\color{gray},\n keywordstyle=\\color{blue},\n commentstyle=\\color{dkgreen},\n stringstyle=\\color{mauve},\n breaklines=true,\n breakatwhitespace=true\n tabsize=3\n}\n\n\\restylefloat{table}\n\n% METADATA\n\\title{\\textbf{FPGA Implementation of the Fast Fourier Transform}}\n\\date{December 2, 2014}\n\\author{Garrett Massman and Cory Walker}\n\n\\begin{document}\n\n \\maketitle\n \\clearpage\n\n \\section*{Introduction}\n In this project we implemented a fast Fourier transform algorithm on a Xilinx FPGA using VHDL. The primary motivation of our project was to digitally analyze musical signals, but the use cases for an FFT chip far surpass that specific use case. \\\\\n \n As a very general overview, our completed device functions by sampling an analog signal for a set amount of time and storing the data into an input buffer in BRAM. Next, the signal is is processed using an FT controller and a complex ALU and stored in an output buffer. Finally, a microcontroller can then read the output buffer over a standard SPI protocol. From there the processed frequency domain data can be sent to a computer for further processing or any other device.\n\n \\section*{Theory}\n To understand the discrete Fourier transform, one must first analyze its analogous continuous time form.\n This is written most simply as\n \\begin{align*}\n &X(j\\omega) = \\int_{-\\infty}^{\\infty} x(t) e^{-j \\omega t}\\,dt\n \\end{align*}\n which defines the transform of a continuous time signal $f(t)$.\n Without getting into too many details, it defines the signal as a combination of sinusoids, making it a very useful tool for real world applications.\\\\\n\n The discrete Fourier transform (DFT) is simply a reduction of the continuous Fourier transform into a discrete sample space.\n In other words, if we let $x_n$ represent a sampled version of the continuous time function $x(t)$ with a total of $N$ samples, we can replace the integral with a summation over the series, as shown below.\n \\begin{align*}\n &X_k = \\sum\\limits_{n=0}^{N-1} x_n e^{\\frac{-j2\\pi kn}{N}}\n \\end{align*}\n As with the continuous time case, this series gives us a glimpse into the component elements of our original signal.\n However, it is rather costly to compute, requiring a time complexity of $O(N^2)$.\n For each value $X_k$, a series of values from $n = 0$ to $N - 1$ must be generated and summed, using up valuable computer resources.\n Thus, calculating the DFT in this way is very inefficient, so we instead turn to the Fast Fourier transform.\\\\\n\n In order to perform this operation more quickly, we utilized the Cooley-Tukey FFT algorithm implemented through the Xilinx CoreGen FFT module.\n One requirement is that our input is strictly a power of 2.\n That is because the algorithm works by recursively finding the FFT of smaller and smaller sample sizes of $x_n$, which arises from the fact that the transform itself is periodic.\n The transform function can be broken into the sum of its even and odd components,\n \\begin{align*}\n &X_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m}e^{-\\frac{2\\pi i}{N} (2m)k} + \\sum \\limits_{m=0}^{N/ 2-1} x_{2m+1} e^{-\\frac{2\\pi i}{N} (2m+1)k}\n \\end{align*}\n This equation can further be simplified by making the substitutions\n \\begin{align*}\n &E_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m}e^{-\\frac{2\\pi i}{N} (2m)k}\n &O_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m+1} e^{-\\frac{2\\pi i}{N/2} mk}\n \\end{align*}\n giving us\n \\begin{align*}\n &X_k = E_k + e^{-\\frac{2\\pi i}{N} k} O_k\n \\end{align*}\n The expression $e^{-\\frac{2\\pi i}{N}k}$ is commonly called the \\textit{twiddle factor}.\n Furthermore, because of the periodicity of the transform, we can calculate respective even and odd components simultaneously\n \\begin{align*}\n &E_k = E_{k+\\frac{N}{2}}\n &O_k = O_{k+\\frac{N}{2}}\n \\end{align*}\n \n \\section*{Motivation}\n The FFT is a very common operation used in many digital signal processing tasks.\n Digital signal processors are usually very highly tuned to perform a series of dedicated tasks quickly, and have very little flexability in the way of general purpose usage.\n This is both a blessing and a curse, as they rarely need to be reprogrammed, but are also limited to the range of tasks they can accomplish.\n High end DSPs utilize parallelism to complete mathematical calculations quickly.\n Because of their dedicated nature, DSPs also tend to be rather costly, upwards of \\$32 for a SHARC model chip from Analog Electronics.\\\\\n\n FPGAs are primarily a large collection of configurable logic blocks that can be wired together to run in parallel, so why not try and use one in place of an expensive DSP?\n By taking the heavy computations off of a typical CPU and putting the burden onto an FPGA, the processor could be freed up to execute any number of instructions until the heavy computations are complete.\n This hybrid model has actually become quite popular in the past several years, and several products such as some of Intel's Xeon chips have FPGAs integrated onboard.\n This can offset the costs for companies who formerly may have been inclined to build expensive ASICs and instead shift their focus onto a new platform.\n A parallel CPU/FPGA integrated system is beyond the scope of this project, but instead a simple model using a Digilent Basys board and an Arduino are built to illustrate the principles.\n\n \\section*{System Overview}\n A critical task in building our FFT device was converting an arbitrary waveform into the SPI signal input that the FPGA expected. This requires the use of a fast analog to digital converter. The Xilinx Spartan 3E FPGA does not come with an onboard ADC. Because of this, a large percentage of the work was actually electronics work done outside of the FPGA:\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=100mm]{full_system.JPG}\n \\caption{Full system including ADC PMOD and audio signal module.}\n \\label{overflow}\n \\end{figure}\n \n These external components condition the signal and then send the digital signal into the FPGA at the right voltage level. In addition, the FPGA's output buffer must be read out by an Arduino microcrontroller and sent to a computer with the right software to parse and display the output. We call these external components macro-components and we call the FPGA internal blocks micro-components.\n \n \\section*{Macro-Component Descriptions}\n The required components external to the FPGA are as follows:\n \\subsection*{Audio signal module}\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=60mm]{audio_sig_mod.JPG}\n \\caption{Audio signal circuit board with audio connector}\n \\label{overflow}\n \\end{figure}\n The audio signal module tailors the input audio signal for conversion with the ADC. Traditional audio signals are centered around 0 V and are in the negative voltage range half the time. This module biases the signal by 2.5 V, allowing for a 5 V variation peak to peak in the audio signal without any clipping. This module also provides filtering capacitors that remove most of the supply rail noise that may be present.\n\n \\subsection*{ADC PMOD}\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=75mm]{adc_pmod.JPG}\n \\caption{ADC PMOD circuit board}\n \\label{overflow}\n \\end{figure}\n The ADC PMOD accepts any input analog input and converts it to a SPI bus that the FPGA can read. The analog signal is fed into the onboard ADS7818 chip. Unfortunately, this chip operates at 5 V but the FPGA operates at 3.3 V. Because of this, there is a voltage divider to drop the voltage down to the FPGA. For FPGA output to be read by the ADS7818, we added a Schmitt inverter IC wired as a Schmitt trigger to act as a level converter.\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=100mm]{adc_logic.png}\n \\caption{ADC interfacing}\n \\label{overflow}\n \\end{figure}\n We started interfacing with the ADS7818 using an Arduino. Once we understood the protocol completely, we moved the interfacing code from the Arduino to the FPGA and started using real signal data.\n\n \\subsection*{Arduino}\n The Arduino functions as a means of getting the processed information out of the FPGA's BRAM. The Arduino turned out to be the main bottleneck in terms of framerate to the computer's display. We had to push the SPI and serial write functions outside of recommended bounds to get a fluid framerate on the computer display. The full software is available in the \\texttt{arduino/osc\\_binary\\_spi} directory of the source code repository.\n\n \\subsection*{Computer software}\n The computer software for this project is divided into two components. The first component is the Python module that facilitates the reading and decoding of the data from the serial port. It will read out 512 complex numbers and, optionally, compute the magnitude of each of those numbers. The second component is the actual display. It will use the Python module to read the device and it uses Tkinter to display the plot on the screen.\n\n \\section*{Micro-Component Descriptions}\n The final block diagram of our device is as follows:\n \\begin{figure}[H]\n \\centering\n \\includegraphics[trim=0 500 0 30,clip,width=140mm]{vhdl_fft.pdf}\n \\caption{Block diagram of the FFT chip}\n \\label{overflow}\n \\end{figure}\n All of the blocks inside the FPGA device are referred to as micro-components.\n\n \\begin{itemize}\n \\item Deterministic ADC: Provides a completely predictable ADC signal to test with and substituted for a real ADC while it was shipping.\n \\item SPI Mux: A simple multiplexer that selects between the external ADC and the deterministic ADC.\n \\item SPI Master: An OpenCore that implements an SPI master interface. This handles the control and reading of the SPI bus from either the deterministic ADC or the external ADC.\n \\item Capture Controller: This block reads samples from the SPI Master and stores the readings as complex numbers in the input buffer. The sample size is 512. It also communicates with the FT controller about when it finishes.\n \\item FT controller: This block actually executes the fast Fourier transfer algorithm whenever the capture controller indicates that the input buffer is valid.\n \\item Complex ALU: This block handles complex multiplications for the FT controller.\n \\item OUTBUF interpreter: This block facilitates the reading of the OUTBUF by feeding words into the SPI slave module.\n \\item SPI Slave: An OpenCore that implements an SPI slave interface. This handles the control and writing of the SPI bus, and it connects to an external microcontroller.\n \\end{itemize}\n\n \\section*{Simulation Results}\n\n \\section*{Results}\n He seemed really excited about timing results so we should probably discuss this here.\n\n \\section*{Roadblocks}\n\n \\section*{Future Improvements}\n\n \\section*{Conclusion}\n\n \\clearpage\n \\section*{Appendix}\n\n \\begin{thebibliography}{9}\n \\bibitem{doin}\n Doin, Jonny. \\emph{SPI Master/Slave Interface.} OpenCores, 16 May 2011. Web. 13 Sept. 2014. \\textless\\url{http://opencores.org/project,spi_master_slave}\\textgreater.\n \\bibitem{reynwar}\n Reynwar, Ben. \\emph{FFT on an FPGA.} FFT on an FPGA. N.p., n.d. Web. 13 Sept. 2014. \\textless\\url{http://www.reynwar.net/ben/docs/fft_dit/index.html}\\textgreater.\n \\bibitem{roberts}\n Roberts, Michael J. \\emph{Signals and Systems: Analysis Using Transform Methods and MATLAB.} New York: McGraw Hill, 2012. Print.\n \\bibitem{satoh}\n Satoh, Keiichi, Jubee Tada, Kenta Yamaguchi, and Yasutaka Tamura. \\emph{Complex Multiplier Suited for FPGA Structure.} Computers and Communications (2008): 341-44. Web. 13 Sept. 2014.\n \\bibitem{tukey}\n Wikipedia contributors. \\emph{Cooley–Tukey FFT algorithm.} Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 27 Jun. 2014. Web. 13 Sep. 2014.\n \\bibitem{dft}\n Wikipedia contributors. \\emph{Discrete Fourier transform.} Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 2 Sep. 2014. Web. 13 Sep. 2014.\n \\end{thebibliography}\n\n\\end{document}\n% \\subsection*{Overview}\n% In this project we will implement the Cooley-Tukey Fast Fourier Transform algorithm on a Xilinx FPGA using VHDL.\n% The primary motivation for this is to digitally analyze musical signals, regardless of the instruments they come from.\n% Using the Cooley-Tukey algorithm, we can calculate the FFT in $O(nlog_2(n))$ time.\n% As far as implementation, the project will be split into several main components.\n% First, we will need to sample our audio for a set amount of time and store the data into a buffer in memory.\n% Then, the digital signal will need to be processed using a complex ALU.\n% Finally, the data will need to be sent to an output buffer for application specific usage.\n% This can be in the form of a 7-segment hex display (for instance if we want to print out the fundamental frequency), or a microcontroller (if we want to plot the entire sampled spectrum by passing it to a more capable platform such as Matlab).\n%\n% \\subsection*{Theory}\n% To understand the discrete Fourier Transform, one must first analyze its analogous continuous time form.\n% This is written most simply as\n% \\begin{align*}\n% &X(\\omega) = \\int_{-\\infty}^{\\infty} x(t) e^{-j \\omega t}\\,dt\n% \\end{align*}\n% which defines the transform of a continuous time signal $f(t)$.\n% Without getting into too many details, it defines the signal as a combination of sinusoids, making it a very useful tool for real world applications.\\\\\n% \n% The discrete Fourier Transform (DFT) is simply a reduction of the continuous Fourier Transform into a discrete sample space.\n% In other words, if we let $x_n$ represent a sampled version of the continuous time function $x(t)$ with a total of $N$ samples, we can replace the integral with a summation over the series, as shown below.\n% \\begin{align*}\n% &X_k = \\sum\\limits_{n=0}^{N-1} x_n e^{\\frac{-j2\\pi kn}{N}}\n% \\end{align*}\n% As with the continuous time case, this series gives us a glimpse into the component elements of our original signal.\n% However, it is rather costly to compute, requiring a time complexity of $O(N^2)$.\n% For each value $X_k$, a series of values from $n = 0$ to $N - 1$ must be generated and summed, using up valuable computer resources.\n% Thus, calculating the DFT in this way is very inefficient, and we must instead turn to the Fast Fourier Transform.\\\\\n%\n% In order to perform this operation more quickly, we can utilize the Cooley-Tukey FFT algorithm.\n% However, the one requirement is that our input is strictly a power of 2.\n% That is because the algorithm works by recursively finding the FFT of smaller and smaller sample sizes of $x_n$, which arises from the fact that the transform itself is periodic.\n% The transform function can be broken into the sum of its even and odd components,\n% \\begin{align*} \n% &X_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m}e^{-\\frac{2\\pi i}{N} (2m)k} + \\sum \\limits_{m=0}^{N/2-1} x_{2m+1} e^{-\\frac{2\\pi i}{N} (2m+1)k}\n% \\end{align*}\n% This equation can further be simplified by making the substitutions\n% \\begin{align*}\n% &E_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m}e^{-\\frac{2\\pi i}{N} (2m)k}\n% &O_k = \\sum \\limits_{m=0}^{N/2-1} x_{2m+1} e^{-\\frac{2\\pi i}{N/2} mk}\n% \\end{align*}\n% giving us\n% \\begin{align*}\n% &X_k = E_k + e^{-\\frac{2\\pi i}{N} k} O_k\n% \\end{align*}\n% The expression $e^{-\\frac{2\\pi i}{N}k}$ is commonly called the \\textit{twiddle factor}.\n% Furthermore, because of the periodicity of the transform, we can calculate respective even and odd components simultaneously\n% \\begin{align*}\n% &E_k = E_{k+\\frac{N}{2}}\n% &O_k = O_{k+\\frac{N}{2}}\n% \\end{align*}\n% \n%\n%\n% \\subsection*{In Mathematica}\n% We implemented a candidate version of the Cooley-Tukey algorithm in Mathematica. This code shows how it is possible to build a look up table of calculated results against the variables $N$ and $k$ for the calculation $e^{-\\frac{2\\pi i}{N} k}$. This table will not only save us time, but also complexity in the design by allowing us to simply look up this value.\n% \\begin{lstlisting}\n%myfasterft[x_, capn_, s_, precomputed_] := (\n% result = {};\n% If[capn == 1,\n% result = Append[result, x[[1]]],\n% result = Join[result, myfasterft[x, capn/2, 2 s, precomputed]];\n% result = \n% Join[result, myfasterft[x[[s + 1 ;;]], capn/2, 2 s, precomputed]];\n% Do[\n% t = result[[k + 1]];\n% result[[k + 1]] = \n% t + precomputed[[Log2[capn]]][[k + 1]]*\n% result[[k + capn/2 + 1]];\n% result[[k + capn/2 + 1]] = \n% t - precomputed[[Log2[capn]]][[k + 1]]*\n% result[[k + capn/2 + 1]];\n% , {k, 0, capn/2 - 1}];\n% ];\n% result\n% );\n%\n%precomputedmatrix[datalen_] := (\n% Table[\n% Table[\n% E^(-I*2*\\[Pi]*k/capn),\n% {k, 0., datalen/2 - 1}\n% ],\n% {capn, Table[2^x, {x, 1, Log2[datalen]}]}\n% ]\n% )\n%\n%data = Table[N[Sin[30 2 Pi n/200] + (RandomReal[] - 1/2)], {n, 512}];\n%precomputed = precomputedmatrix[Length[data]];\n%ListLinePlot[Abs[myfasterft[data, Length[data], 1, precomputed]], PlotRange -> All]\n% \\end{lstlisting}\n% \\subsection*{FPGA considerations}\n% %Describe how we plan to utilize SPI for communication, what type of logical blocks we will use and how they will work together.\n% Because we are working with a very specialized type of hardware, we must take into consideration some limiting parameters.\n% In particular, the Spartan3Es loaded on each BASYS board only have 72K RAM.\n% This means that if we want a real time FFT, we will need to optimize the space we have available, and only write to it when it is absolutely necessary.\n% We must also consider the quality of our digital input signal.\n% There are various ADC PMODs available for the BASYS boards, which range in quality from 4.8 KHz to 1 MHz in speed.\n% If we are only sampling audio, we probably won't need to gather any frequencies above 1200 Hz, so the lower end model should be fine.\\\\\n%\n% Once our data has been collected and stored inside the FPGA, it needs to be processed.\n% This is perhaps the most challenging part of the project, as we need to implement what will be called an FTC (Fourier Transform Controller) as well as a CALU (Complex Arithmetic Logic Unit).\n% The FTC's primary function will be to keep track of the data as it moves from RAM, into the CALU, and back out into the real world.\n% This will require some heavily synchronized logic and very tight timing control to be carried out properly.\n% It will work in sync with the SPI input master to tell the CALU when data is ready to be read, and then clear the input buffer once the CALU has finished.\n% It will also need to monitor the movement of data from the output buffer into whatever output stream requires it.\n% The data may be sent directly to a microcontroller or into another component for post processing.\\\\\n%\n% The CALU will have one primary job, which is to actually perform the FFT.\n% From its perspective, it will only the addresses of an input buffer and an output buffer, as well as a flag that tells it when to start and stop.\n% Because all of our calculations are being done inside of an FPGA, the CALU will be most composed of combinational logic at its core.\n% In theory (i.e., provided enough gates), all that needs to be built is the base case of the FFT, which takes in two numbers and outputs their sum and difference.\n% \\begin{align*}\n% &y_0 = x_0 + e^{-\\frac{2\\pi i}{N}k}x_1\n% &y_1 = x_0 - e^{-\\frac{2\\pi i}{N}k}x_1\n% \\end{align*}\n% This portion of the computation is commonly called a \\textit{butterfly}, because the system's diagram looks vaguely like a butterfly.\n% Each of the twiddle factors included above can be calculated beforehand and stored in a lookup table for quick access before each iteration.\n% Because we will be performing the same type of computation with them each time, they are essentially constants in our system.\\\\\n%\n% \\subsection*{Block diagram}\n% Over the past few weeks we have taken what we know and outlined a basic block diagram of our device:\n% \\begin{figure}[H]\n% \\centering\n% \\includegraphics[trim=0 400 0 400,clip,width=140mm]{vhdl_fft.pdf}\n% \\caption{Block diagram of the FFT chip}\n% \\label{overflow}\n% \\end{figure}\n% Now that we have divided the device into several logical blocks, we can now split up the work accordingly.\n% \\subsection*{Roadmap}\n% In this section, we will describe the steps, in order, that need to be done. Our steps will essentially follow the data pipeline of the module:\n% \\begin{itemize}\n% \\item We will begin by implementing a deterministic \"ADC\" module that will emulate the interface of the real ADC but generate a fixed signal every time. We will use this signal in our own calculations so that we can then verify the outputs of the chip at each stage in the data pipeline. When we are ready to convert to real ADC signals, we can simply switch to an external ADC module using an SPI mux. This step has already been completed.\n% \\item We will then implement the capture controller. This controller will have an SPI master unit that can read from the ADC. It will be also connected to the block RAM to write the input buffer.\n% \\item Next, we will work on the heart of the chip, the FT controller. This controller will first simply implement the DFT algorithm. It will utilize the complex number ALU to make calculations.\n% \\item After the FT controller, we will implement an OUTBUF interpreter that is responsible for outputting the results of the FT controller to the rest of the world using SPI.\n% \\item Once all of this is done, we will finish converting our FT controller from using the simple DFT algorithm to the more complex Cooley-Tukey FFT algorithm.\n% \\item If we have further time still, we will build interesting demonstrations of this device.\n% \\end{itemize}\n% \\subsection*{Resources}\n% While our aim is to implement the entire Fast Fourier Transform with our own code, there are some blocks of our design that are common practice to reuse. Those blocks are the SPI modules. While SPI is not a complicated communication protocol, it requires a fair amount of time to implement. There is a great resource online called OpenCores. This resource is essentially a collection of open source HDL modules that can be freely used in designs. We will be using the \"SPI Master/Slave Interface\" module a few times in our design to implement SPI.\n% \\subsection*{Possible extensions}\n% The Fast Fourier Transform algorithm applied to an arbitrary voltage signal is very useful. Because our final chip will provide such capability, we have the option of using our newly-designed chip with multiple different applications. \\\\\n% \n% One of our ideas is to attach a microphone and a preamp and use the frequency-domain signal to determine the musical note being played. We could also build a tuner for a guitar. \\\\\n% \n% Additionally, we could connect an audio signal directly from any music player and display a graphic equalizer effect using either a computer or microcontroller that will interface with our chip. \\\\\n%\n% Finally, provided we are able to tune our chip and ADC to operate fast enough, we might be able to build a radio wave analyzer for very long wavelength signals. This device would almost certainly be too slow for any serious RF work, but it would still be an interesting extension. \\\\\n%\n% These extensions are not crucial to our design but might serve as a great demonstration when we present the device to the class.\n", "meta": {"hexsha": "8c7bab8f2b9fb396e533b43f30495093ad039d1d", "size": 24695, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/final_report/final_report.tex", "max_stars_repo_name": "corywalker/vhdl_fft", "max_stars_repo_head_hexsha": "b7fa75ed1f90a05fcc2521495019c1561df4eaba", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-30T03:39:35.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-30T03:39:35.000Z", "max_issues_repo_path": "docs/final_report/final_report.tex", "max_issues_repo_name": "corywalker/vhdl_fft", "max_issues_repo_head_hexsha": "b7fa75ed1f90a05fcc2521495019c1561df4eaba", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/final_report/final_report.tex", "max_forks_repo_name": "corywalker/vhdl_fft", "max_forks_repo_head_hexsha": "b7fa75ed1f90a05fcc2521495019c1561df4eaba", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 68.5972222222, "max_line_length": 552, "alphanum_fraction": 0.7276371735, "num_tokens": 6338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256472515683, "lm_q2_score": 0.7634837527911057, "lm_q1q2_score": 0.6427202043594289}} {"text": "\\section{Summary and Outlook}\\label{se:Summary}\n\nWe have developed IMEX schemes suitable for a two-moment model of neutrino transport that obey Fermi-Dirac statistics.\nThe scheme employs algebraic closure based on Fermi-Dirac statistics, high-order discontinuous Galerkin methods for spatial discretization, and convex-invariant time integration to maintain realizability of the moments. \nSince the realizable domain is convex and its convexity can be inherited by a convex combination, a scheme having convex combinations as its stages can preserve the realizable domain.\nThis encouraged us to construct realizability-preserving time integrators, realizability-preserving IMEX schemes, and a method with a realizability-preserving IMEX time integrator and high-order DG method. \n\nIn the applications that motivate this work, the neutrino distribution function can vary from 0 to 1. \nHence, we have considered algebraic closures based on Fermi-Dirac statistics for both low and high occupancy. \nAmong the seven algebraic closures we considered -- Kershaw~\\cite{kershaw_1976}, Wilson~\\cite{wilson_1975,leblancWilson_1970}, Levermore~\\cite{levermore_1984}, Minerbo~\\cite{minerbo_1978}, Janka 1~\\cite{janka_1991}, Janka 2~\\cite{janka_1992}, and Cernohorsky \\& Bludman~\\cite{cernohorskyBludman_1994} -- only the Cernohorsky \\& Bludman closure obeys Fermi-Dirac statistics for all occupancies. \nAs a result, we employed the Cernohorsky \\& Bludman closure for the neutrino stationary state test in Section~\\ref{se: Neutrino Stationary State Test}.\nWe also ran our code with Minerbo closure, IMEX PC2, IMEX SSP2332 and IMEXRKCB2 schemes, and the results show that only PD-ARS schemes have stability.\nIn addition, closures have impact on the simulation result.\nAs we observed, both using PD-ARS scheme, there are $\\sim30\\%$ difference in the neutrino number densities (and relaxation time) between the results obtained with Minerbo closure and that with CB closure.\nEven though RKCB2 with Minerbo closure luckily survived our test, the results it gave were compromised: they were closer to the results of the PD-ARS scheme with Minerbo closure than to the results of the PD-ARS scheme with CB closure.\nIn what way and to what degree the results are in fact compromised either by the closure or by a particular correction step for unrealizable moments are difficult to determine fully and is left for further study.\n\n\nTwo PD-ARS schemes are proposed.\nThe one with SSPRK2 has second-order accuracy while the other with SSPRK3 has third-order accuracy, and both have the strong-stability preserving property in the streaming limit. \nTheir accuracy was demonstrated on problems with known smooth solutions in streaming, absorption, and scattering-dominated regimes.\nThe neutrino transport test with emission, absorption, and isoenergetic scattering through a stationary background, was designed to test the convex-invariance of our PD-ARS schemes. \nThe neutrino stationary state test shows that a method combining an algebraic closure based on Fermi-Dirac statistics and convex-invariant time integration is promising for robust CCSN simulation.\n\nIn this work, we adopted Cartesian coordinates, a linear collision term, and a fixed material background.\nMore realistic problems of scientific interest, such as with energy-exchanging scattering and relativistic effects, are left for future research.", "meta": {"hexsha": "ded3061c2c0735cf68ec54c950d98182daba3dae", "size": 3392, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documents/M1/Astronum_2018/sections/Conclusion.tex", "max_stars_repo_name": "srichers/thornado", "max_stars_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-12-08T16:16:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-24T19:31:21.000Z", "max_issues_repo_path": "Documents/M1/Astronum_2018/sections/Conclusion.tex", "max_issues_repo_name": "srichers/thornado", "max_issues_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2019-07-10T20:13:15.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-11T13:21:00.000Z", "max_forks_repo_path": "Documents/M1/Astronum_2018/sections/Conclusion.tex", "max_forks_repo_name": "srichers/thornado", "max_forks_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2018-11-14T01:13:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-24T02:08:20.000Z", "avg_line_length": 130.4615384615, "max_line_length": 395, "alphanum_fraction": 0.8186910377, "num_tokens": 764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256313782276, "lm_q2_score": 0.7634837635542925, "lm_q1q2_score": 0.6427202013011177}} {"text": " \\chapter{Linear Solver}\n\nIn many applications one of the main problems is to find the solution of a linear system. This system may arise form an optimisation or, like here, a discretised PDE. In this chapter we will go through the two methods which this is done.\n\n\\section{Direct Methods}\n\nSuppose you are given a linear system of the form\n$$Ax = b$$\nwhere $A$ is a real $n\\times n$ non-singular matrix and right hand side vector $b \\in \\mathcal{R}^n$, where the aim is to solve for the vector $x \\in \\mathcal{R}^n$. All direct methods are based on Gaussian elimination.\n\n\\subsection{Software}\n\nUMFPACK \\cite{Davis:2004:CPS:992200.992205,Davis:2004:AUV:992200.992206,Davis:1999:CUM:305658.287640,davis1997unsymmetric}, PASTIX \\cite{henon2002pastix}, SuperLU \\cite{superlu_ug99,li05} and MUMPS \\cite{amestoy2000multifrontal,amestoy2001fully,amestoy2006hybrid}\n\n\n\\section{Iterative Methods}\n\n\n\n", "meta": {"hexsha": "b3d7f96d4116927c047da4eb260a48d98b5e0fce", "size": 893, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MHD/THESISnew/LinearSolver/LinearSolver.tex", "max_stars_repo_name": "wathen/PhD", "max_stars_repo_head_hexsha": "35524f40028541a4d611d8c78574e4cf9ddc3278", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2020-10-25T13:30:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-08-10T21:27:30.000Z", "max_issues_repo_path": "MHD/THESISnew/LinearSolver/LinearSolver.tex", "max_issues_repo_name": "wathen/PhD", "max_issues_repo_head_hexsha": "35524f40028541a4d611d8c78574e4cf9ddc3278", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "MHD/THESISnew/LinearSolver/LinearSolver.tex", "max_forks_repo_name": "wathen/PhD", "max_forks_repo_head_hexsha": "35524f40028541a4d611d8c78574e4cf9ddc3278", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-28T16:12:13.000Z", "max_forks_repo_forks_event_max_datetime": "2020-01-13T13:59:44.000Z", "avg_line_length": 44.65, "max_line_length": 263, "alphanum_fraction": 0.7782754759, "num_tokens": 270, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8791467706759584, "lm_q2_score": 0.731058584489497, "lm_q1q2_score": 0.6427077937288785}} {"text": "\\section{Randomized Algorithms}\n\\subsection{Sampling based randomized algorithm (SRA)}\n\nInstead of solving the original problem is\n\\begin{equation}\\label{ORG}\n\\mathcal{Z} = min_{x \\in \\mathbb{R}^n} \\| Ax - b\\|_2\n\\quad \\text{with solution} \\quad\nx_{opt} = (A^T A)^{-1} A^T b = A^{\\dagger} b,\n\\end{equation}\nactually we solve\n\\begin{equation}\\label{ACT}\n\\tilde{\\mathcal{Z}} = min_{x \\in \\mathbb{R}^n} \\| X Ax - X b\\|_2\n\\quad \\text{with solution} \\quad\n\\tilde{x}_{opt} = (XA)^{\\dagger} X b.\n\\end{equation}\n\n%------------------------------------------------\n\\subsubsection{Theory}\n\\begin{lemma}[Drineas, Mahoney, Muthukrishnan and Sarlos(2011)]\n\tConsider the overconstrained least squares problem of (\\ref{ORG}) and let the matrix $U_A \\in \\mathbb{R}^{m\\times n}$ contain the top $n$ left singular vectors of $A$. Assume $X$ satisfies the two conditions\n\t\\begin{enumerate}\n\t\t\\item $\\sigma_{min}^2(X U_A) \\geq 1/\\sqrt{2}; $ \\\\\n\t\t\\item $\\| U_A^T X^T X b^{\\perp}\\|_2^2 \\leq \\epsilon \\mathcal{Z}^2 /2, \\quad (b^{\\perp} = b - A A^T b), $\n\t\\end{enumerate}\n\tfor some $\\epsilon \\in (0,1).$\n\t\n\tThen the solution $\\tilde{x}_{opt}$ to the least squares approximation problem (\\ref{ACT}) satisfies:\n\t\\begin{enumerate}\n\t\t\\item $\\| A \\tilde{x}_{opt} - b\\|_2 \\leq (1+ \\epsilon) \\mathcal{Z} $ , and \\\\\n\t\t\\item $\\| x_{opt} - \\tilde{x}_{opt} \\|_2 \\leq \\frac{1}{\\sigma_{min}(A)} \\sqrt{\\epsilon} \\mathcal{Z}. $\n\t\\end{enumerate}\n\\end{lemma}\n\n%------------------------------------------------\n\\subsubsection{Implementation: A sampling-based randomized algorithm}\nA sampling-based randomized algorithm (SRA) is to construct\n$$\nX = S^T H D,\n$$\n\n\\begin{itemize}\n\t\\item $S \\in \\mathbb{R}^{r \\times m}$ is the uniform sampling matrix, where $S_{r} = (\\sqrt{m/r}) e_{ir},$ where $ir$ is uniformly chosen from $[m],$\n\tand $e_{ir}$ is the standard basis. \\\\\n\t\\item $H \\in \\mathbb{R}^{m \\times m}$ is the Hadamard transform matrix defined recursively by\n\t$$\n\tH_m = \\begin{bmatrix} H_{m/2} & H_{m/2} \\\\ H_{m/2} & -H_{m/2} \\end{bmatrix}, \\quad \\text{with} \\quad H_2 = \\begin{bmatrix} +1 & +1 \\\\ +1 & -1 \\end{bmatrix}.\n\t$$\\\\\n\t\\item $D \\in \\mathbb{R}^{m \\times m}$ is the diagonal matrix with\n\t$$\n\tD_{ii} = \\begin{cases} &+1, \\quad \\text{with probability} \\quad 1/2;\\\\ &-1 \\quad \\text{with probability} \\quad 1/2. \\end{cases}\n\t$$\n\\end{itemize}\n\n$$\n\\tilde{x}_{opt} = (S^T H D A)^{\\dagger} S^T H D b.\n$$\n\n%------------------------------------------------\n\\subsubsection{Effect of the Randomized Hadamard Transform}\n{\\color{red} HD approximately 'uniformizes' information in the left singular subspace of the matrix $A$.}\n\\begin{lemma}\n\tLet $U$ be an $m \\times n$ orthogonal matrix and let the product $HD$ be the $m \\times m$ Randomized Hadamard Transform. Then with probability at least $0.95$,\n\t$$\n\t\\| (HDU)_{i} \\|_2^2 \\leq \\frac{2n \\ln(40 m n)}{m}, \\quad \\forall i \\in [m].\n\t$$\n\\end{lemma}\n\n%------------------------------------------------\n\n\n\\begin{theorem}[Convergence and Complexity of SRA]\n\tSuppose $A,b$ and $\\epsilon$ satisfy the input requirement of SRA. Run SRA with\n\t\\begin{equation}\\label{eq22}\n\tr = max \\left( 48^2 n \\ln(40 mn) \\ln(100^2 n \\ln(40 mn)) , 40n \\ln(40 mn) /\\epsilon \\right)\n\t\\approx O(n/\\epsilon).\n\t\\end{equation}\n\tand return $\\tilde{x}_{opt}.$ Then, with probability at least $0.8$, the following claims hold:\n\t\\begin{enumerate}\n\t\t\\item $\\tilde{x}_{opt}$ satisfies $\\| A \\tilde{x}_{opt} - b \\|_2 \\leq (1 + \\epsilon) \\mathcal{Z}; $ \\\\\n\t\t\\item If we assume that $\\| U_A U_A^T b \\|_2 \\geq \\gamma \\|b\\|_2 $ for some $0< \\gamma \\leq 1$ then\n\t\t$$\n\t\t\\| x_{opt} - \\tilde{x}_{opt}\\|_2 \\leq \\sqrt{\\epsilon} \\left( \\kappa(A) \\sqrt{\\gamma^{-2} -1} \\right) \\| x_{opt}\\|_2.\n\t\t$$\\\\\n\t\t\\item $m(n+1) + 2m(n+1)log_2(r+1) + O(rn^2)$ time suffices to compute the solution $\\tilde{x}_{opt}. $\n\t\\end{enumerate}\n\\end{theorem}\nThe cost is reduced since $XA \\in \\mathbb{R}^{r \\times n}$ while $A \\in \\mathbb{R}^{m \\times n}.$\n\n\n%------------------------------------------------\n\\subsubsection{Projection Based Randomized Algorithm (PRA)}\nImplementation: A projection-based randomized algorithm\nProjection-based randomized algorithm (PRA) is to construct a smaller problem by performing a 'sparse projection' on the pre-processed problem.\n$$\nX = THD,\n$$\nwhere\n\\begin{itemize}\n\t\\item $H \\in \\mathbb{R}^{m \\times m}$ is the Hadamard Transform and $D \\in \\mathbb{R}^{m \\times m }$ is the randomized diagonal matrix as defined before. \\\\\n\t\\item $T \\in \\mathbb{R}^{k \\times m}$ is a randomized matrix given by\n\t$$\n\tT_{ij} = \\begin{cases} &+ \\sqrt{1/kq} \\quad \\text{with probability} \\quad q/2, \\\\\n\t&- \\sqrt{1/kq} \\quad \\text{with probability} \\quad q/2, \\\\\n\t& 0 \\quad \\text{with probability} \\quad 1-q.\n\t\\end{cases}\n\t$$\n\\end{itemize}\n$$\\tilde{x}_{opt} = (THDA)^{\\dagger} THDb. $$\n\n%------------------------------------------------\n\n\\begin{theorem}[Convergence and Complexity of PRA]\n\tSuppose $A,b$ and $\\epsilon$ satisfy the input requirement of FRA. Run FRA with\n\t\\begin{equation}\n\tq \\geq \\frac{C_q n \\ln(40mn)}{m} (2 \\ln(m) + 16n + 16);\n\t\\end{equation}\n\t\\begin{equation}\n\tk \\geq max \\left( C_k (118^2 n + 98^2), \\frac{60n}{\\epsilon} \\right)\n\t\\end{equation}\n\tand return $\\tilde{x}_{opt}.$ Then, with probability at least $0.8$, the following claims hold:\n\t\\begin{enumerate}\n\t\t\\item $\\tilde{x}_{opt}$ satisfies $\\| A \\tilde{x}_{opt} - b \\|_2 \\leq (1 + \\epsilon) \\mathcal{Z}; $ \\\\\n\t\t\\item If we assume that $\\| U_A U_A^T b \\|_2 \\geq \\gamma \\|b\\|_2 $ for some $0< \\gamma \\leq 1$ then\n\t\t$$\n\t\t\\| x_{opt} - \\tilde{x}_{opt}\\|_2 \\leq \\sqrt{\\epsilon} \\left( \\kappa(A) \\sqrt{\\gamma^{-2} -1} \\right) \\| x_{opt}\\|_2.\n\t\t$$\\\\\n\t\t\\item $m(n+1) + 2m(n+1)log_2(mkq+1) + O(kn^2)$ time suffices to compute the solution $\\tilde{x}_{opt}. $\n\t\\end{enumerate}\n\\end{theorem}\n\n\n%------------------------------------------------\n\\subsection{A Fast Randomized Algorithm (FAR)}\nThe fast randomized algorithm (FAR) is from Rokhlin and Tygert's paper \\cite{SRFT2008}. The algorithm is given as follows.\n\t\\begin{enumerate}\n\t\t\\item Compute $E = TA$, where $T$ is the $r \\times m$ SRFT defined as follows, with $m \\geq r \\geq n. $ \\\\\n\t\t\\item Form a pivoted QR-decomposition of $E = Q R \\Pi$, where $Q_{r \\times n}$ has orthonormal columns, $R_{n \\times n}$ is upper triangular and $\\Pi_{n \\times n}$ is the permutation matrix. \\\\\n\t\t\\item Solve a preconditioned Least Squares Problem\n\t\t$$\n\t\t\\| AP^{-1} y - b\\|\n\t\t$$\n\t\tusing PCG where $P = R \\Pi$ is the preconditioning matrix.\n\t\\end{enumerate}\n\n\n%------------------------------------------------\n\\subsubsection{Complexity for A Fast Randomized Algorithm (FAR)}\n\t\\begin{enumerate}\n\t\t\\item Applying $T$ to every column of $A$ : $O(mnlog(r))$. \\\\\n\t\t\\item Computing the pivoted QR decomposition of $E$ : $O(n^2r) . $ \\\\\n\t\t\\item Applying $T$ to $b$: O(mlog(r)). Applying $Q^*$ to $Tb$: $O(nr)$.\n\t\t\n\t\tApplying $P^{-1} = \\Pi^{-1} R^{-1}$ to $Q^* Tb$ : $O(n^2). $ \\\\\n\t\t\\item Applying $A,A^T$ a total of $O(\\kappa(AP^{-1}) |log(\\epsilon)|)$ times: {\\color{red}$O(mn\\kappa(AP^{-1}) |log(\\epsilon)|).$} \\\\\n\t\t\\item Applying $P^{-1},(P^{-1})^*$ to a total of $O(\\kappa(AP^{-1}) |log(\\epsilon)|)$ times: {\\color{red}$O(n^2 \\kappa(AP^{-1}) |log(\\epsilon)|).$} \\\\\n\t\t\\item Applying $P^{-1}$ to $y$ : $O(n^2). $\n\t\\end{enumerate}\nThus\n{\\color{red}\n\t$$\n\tC_{theoretical} = O((log(r) + \\kappa(AP^{-1}) |log(\\epsilon)|) mn +n^2r).\n\t$$}\n\n\n\n\n%------------------------------------------------\n\\subsubsection{SRFT Matrix}\nT is the sampled randomized Fourier transform (SRFT) Matrix defined by\n$$T _{r \\times m} = G_{r \\times m} H_{m \\times m}, \\quad r \\leq m. $$\nG is the random matrix given by\n$$ G_{r \\times m} = S_{r \\times m} F_{m \\times m} D_{m \\times m}, $$\n\n\n\n\\begin{itemize}\n\t\\item $S$ is a random permutation matrix in each row j there is one column $s_j$ such that $S_{j,s_j} = 1$ and $S_{j,k} =0$ if $k \\neq s_j$. $s_j$'s are i.i.d. random variables distributed uniformly over $\\{ 1,\\cdots,m\\}. $ \\\\\n\t\\item $F$ is the $m \\times m$ discrete Fourier transform. \\\\\n\t\\item $D = diag(d_1,d_2,\\cdots,d_m)$\n\twhere $d_1,\\cdots,d_m$ are i.i.d. complex random variables distributed uniformly over the unit circle.\n\\end{itemize}\n\n\n\n%------------------------------------------------\n$$\nH_{m \\times m} =\\Theta_{m \\times m} \\Pi_{m \\times m} Z_{m \\times m} \\tilde{\\Theta}_{m \\times m}\\tilde{\\Pi}_{m \\times m} \\tilde{Z}_{m \\times m},\n$$\nwhere\n\n\n\\begin{itemize}\n\t\\item $\\Pi$ and $\\tilde{\\Pi}$ are permutation matrices chosen independently and uniformly at random. \\\\\n\t\\item $Z$ and $\\tilde{Z}$ are diagonal matrices whose diagonal entries are i.i.d. complex random variables distributed uniformly over the unit circle. \\\\\n\t\\item\n\t$\\Theta_{m \\times m} = \\begin{bmatrix} cos(\\theta_1)& sin(\\theta_1) & 0& \\cdots &0\\\\\n\t- sin(\\theta_1) & cos(\\theta_1) & 0 & \\cdots & 0 \\\\\n\t0 & 0 & 1 & \\cdots & 0 \\\\\n\t\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\n\t0 & 0 &0 & \\cdots & 1\n\t\\end{bmatrix}\n\t\\cdots\n\t\\begin{bmatrix} 1& \\cdots & 0& 0 &0\\\\\n\t\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\n\t0 & \\cdots & 1 & 0 & 0 \\\\\n\t0 & \\cdots & 0 & cos(\\theta_{m-1})& sin(\\theta_{m-1}) \\\\\n\t0 & \\cdots &0 & -sin(\\theta_{m-1}) & cos(\\theta_{m-1})\n\t\\end{bmatrix} ,$\n\t\n\t\n\twhere $\\theta_k, \\quad k =1 ,\\cdots m-1$ are i.i.d. real random variables drawn uniformly from $[0,2\\pi]$.\n\tSo is $\\tilde{\\Theta}$, but defined with different $\\tilde{\\theta}_k$.\n\t\n\\end{itemize}\n\n%------------------------------------------------\n\n\\subsubsection{Why SRFT works?}\n\\begin{corollary}\n\tSuppose that $\\alpha$ and $\\beta$ are real numbers greater than 1 and $r,m$ and $n$ are positive integers such that $m \\geq r \\geq (\\frac{\\alpha^2 +1}{\\alpha^2-1})^2 \\beta n^2. $ Suppose further that $T$ is the $r \\times m$ SRFT matrix. Suppose in addition that $U$ is an $m \\times n$ matrix whose columns are orthonormal.\n\tThen, the condition number of $TU$ is at most $\\alpha$ with probability at least $1 - \\frac{1}{\\beta}. $\n\\end{corollary}\n$\\kappa(TU)$ can be made arbitrarily close to 1 when $r$ is large enough.\n\n%------------------------------------------------\n\n\\begin{theorem}[Rokhlin and Tygert (2008)]\n\tSuppose that $r,m$ and $n$ are positive integers such that $m \\geq r \\geq n$. Suppose further that $A$ is a full rank $m \\times n$ matrix and the SVD of $A$ is\n\t$$\n\tA_{m \\times n} = U_{m \\times n} \\Sigma_{n \\times n} V^*_{n \\times n}.\n\t$$\n\tSuppose in addition that $T$ is an $r \\times m$ matrix such that the $r \\times n$ matrix $TU$ has full rank.\n\tThen\n\t$$\n\tT_{r \\times m} A_{m \\times n} = Q_{r \\times n} P_{n \\times n},\n\t$$\n\twhere $Q$ has orthonormal columns.\n\tAnd\n\t$$\n\t\\kappa(AP^{-1}) = \\kappa(TU).\n\t$$\n\\end{theorem}\n\\small{$r \\geq 4n^2$ guarantees $\\kappa(TU) \\leq 3.$}\n", "meta": {"hexsha": "2bb304e2b36ec375d6789a2d8d2d7cfa33ee6023", "size": 10579, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/Rand_Algo.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/Rand_Algo.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/Rand_Algo.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.356557377, "max_line_length": 323, "alphanum_fraction": 0.6016636733, "num_tokens": 3799, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042767, "lm_q2_score": 0.7905303162021596, "lm_q1q2_score": 0.6425962649680421}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Contents: Math typesetting with LaTeX\n% $Id: math.tex 534 2015-04-09 13:03:16Z oetiker $\n%\n% Changes by Stefan M. Moser: 2008/10/22\n%\n% -Section 2: \"Single Equations\": added comment about preference of\n% equation* over \\[\n% -Replaced (almost) all examples with \\[ by equation*\n% -New section 4: \"Single Equations that are Too Long: multline\"\n% -New section 5: \"Multiple Equations\"\n% -Section 6: \"Arrays and Matrices\": made a full section and added\n% some material\n% -Section 9: \"Theorems, Lemmas, ...\": added a subsection about proofs\n% with new material\n%\n% Other Changes:\n% -in lshort.sty: \n% *example environment adapted: changed in three places\n% \\textwidth by \\linewidth. This is necessary for\n% example-environment within a itemize-list.\n% *added \\RequirePackage[retainorgcmds]{IEEEtrantools}\n%\n% THINGS TO DO:\n% -adapt typesetting of new sections to rest of lshort, including all\n% the usual commands used so far. In particular, I guess we have to\n% get rid of the \\verb-commands everywhere\n% -include index-commands\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n \n\\chapter{Typesetting Mathematical Formulae}\n\n\\begin{intro}\n Now you are ready! In this chapter, we will attack the main strength\n of \\TeX{}: mathematical typesetting. But be warned, this chapter\n only scratches the surface. While the things explained here are\n sufficient for many people, don't despair if you can't find a\n solution to your mathematical typesetting needs here. It is highly likely\n that your problem is addressed in \\AmS-\\LaTeX{}.\n\\end{intro}\n \n\n\n\\section{The \\texorpdfstring{\\AmS}{AMS}-\\LaTeX{} bundle}\n\nIf you want to typeset (advanced) \\wi{mathematics}, you should\nuse \\AmS-\\LaTeX{}. The \\AmS-\\LaTeX{} bundle is a collection of packages and classes for\nmathematical typesetting. We will mostly deal with the \\pai{amsmath} package\nwhich is a part of the bundle. \\AmS-\\LaTeX{} is produced by The \\emph{\\wi{American Mathematical Society}} \nand it is used extensively for mathematical typesetting. \\LaTeX{} itself does provide\nsome basic features and environments for mathematics, but they are limited (or\nmaybe it's the other way around: \\AmS-\\LaTeX{} is \\emph{unlimited}!) and\nin some cases inconsistent. \n\n\\AmS-\\LaTeX{} is a part of the required distribution and is provided\nwith all recent \\LaTeX{} distributions.\\footnote{If yours is missing it, go to\n \\CTAN|pkg/amslatex|.} In this chapter, we assume\n \\pai{amsmath} is loaded in the preamble; \\verb|\\usepackage{amsmath}|.\n\n\\section{Single Equations}\n \nA mathematical formula can be typeset in-line within a paragraph (\\emph{\\wi{text style}}), or the paragraph can be broken and the formula typeset separately\n(\\emph{\\wi{display style}}). Mathematical \\wi{equation}s \n\\emph{within} a paragraph are entered \\index{$@\\texttt{\\$}} %$\nbetween \\texttt{\\$} and \\texttt{\\$}:\n\\begin{example}\nAdd $a$ squared and $b$ squared\nto get $c$ squared. Or, using \na more mathematical approach:\n$a^2 + b^2 = c^2$\n\\end{example}\n\\begin{example}\n\\TeX{} is pronounced as \n$\\tau\\epsilon\\chi$\\\\[5pt]\n100~m$^{3}$ of water\\\\[5pt]\nThis comes from my $\\heartsuit$\n\\end{example}\n\nIf you want your larger equations to be set apart\nfrom the rest of the paragraph, it is preferable to \\emph{display} them\nrather than to break the paragraph apart.\nTo do this, you enclose them between \\verb|\\begin{|\\ei{equation}\\verb|}| and\n\\verb|\\end{equation}|.\\footnote{This is an \\textsf{amsmath} command. If you don't\nhave access to the package for some obscure reason, you can use \\LaTeX's own\n\\ei{displaymath} environment instead.} You can then \\ci{label} an equation number and refer to\nit somewhere else in the text by using the \\ci{eqref} command. If you want to\nname the equation something specific, you \\ci{tag} it instead.\n\\begin{example}\nAdd $a$ squared and $b$ squared\nto get $c$ squared. Or, using\na more mathematical approach\n \\begin{equation}\n a^2 + b^2 = c^2\n \\end{equation}\nEinstein says\n \\begin{equation}\n E = mc^2 \\label{clever}\n \\end{equation}\nHe didn't say\n \\begin{equation}\n 1 + 1 = 3 \\tag{dumb}\n \\end{equation}\nThis is a reference to \n\\eqref{clever}. \n\\end{example}\n\nIf you don't want \\LaTeX{} to number the equations, use the starred\nversion of \\texttt{equation} using an asterisk, \\ei{equation*}, or even easier, enclose the\nequation in \\ci{[} and \\ci{]}:\\footnote{\\index{equation!\\textsf{amsmath}}\n \\index{equation!\\LaTeX{}}This is again from \\textsf{amsmath}. Standard \\LaTeX{}'s has only the \\texttt{equation} environment without the star.}\n\\begin{example}\nAdd $a$ squared and $b$ squared\nto get $c$ squared. Or, using\na more mathematical approach\n \\begin{equation*}\n a^2 + b^2 = c^2\n \\end{equation*}\nor you can type less for the\nsame effect:\n \\[ a^2 + b^2 = c^2 \\]\n\\end{example}\nWhile \\ci{[} is short and sweet, it does not allow switching between numbered and not numbered style as easily as\n\\ei{equation} and \\ei{equation*}.\n\nNote the difference in typesetting style between \\wi{text style} and \\wi{display style}\nequations: \n\\begin{example}\nThis is text style: \n$\\lim_{n \\to \\infty} \n \\sum_{k=1}^n \\frac{1}{k^2} \n = \\frac{\\pi^2}{6}$.\nAnd this is display style:\n \\begin{equation}\n \\lim_{n \\to \\infty} \n \\sum_{k=1}^n \\frac{1}{k^2} \n = \\frac{\\pi^2}{6}\n \\end{equation}\n\\end{example}\n\nIn text style, enclose tall or deep math expressions or sub\nexpressions in \\ci{smash}. This makes \\LaTeX{} ignore the height of\nthese expressions. This keeps the line spacing even.\n\n\\begin{example}\nA $d_{e_{e_p}}$ mathematical\nexpression followed by a\n$h^{i^{g^h}}$ expression. As\nopposed to a smashed \n\\smash{$d_{e_{e_p}}$} expression \nfollowed by a\n\\smash{$h^{i^{g^h}}$} expression.\n\\end{example}\n\n\\subsection{Math Mode}\n\nThere are also differences between \\emph{\\wi{math mode}} and \\emph{text mode}. For\nexample, in \\emph{math mode}: \n\n\\begin{enumerate}\n\n\\item \\index{spacing!math mode} Most spaces and line breaks do not have any significance, as all spaces\nare either derived logically from the mathematical expressions, or\nhave to be specified with special commands such as \\ci{,}, \\ci{quad} or\n\\ci{qquad} (we'll get back to that later, see section~\\ref{sec:math-spacing}).\n \n\\item Empty lines are not allowed. Only one paragraph per formula.\n\n\\item Each letter is considered to be the name of a variable and will be\ntypeset as such. If you want to typeset normal text within a formula\n(normal upright font and normal spacing) then you have to enter the\ntext using the \\verb|\\text{...}| command (see also section \\ref{sec:fontsz} on\npage \\pageref{sec:fontsz}).\n\n\\end{enumerate}\n\\begin{example}\n$\\forall x \\in \\mathbf{R}:\n \\qquad x^{2} \\geq 0$\n\\end{example}\n\\begin{example}\n$x^{2} \\geq 0\\qquad\n \\text{for all }x\\in\\mathbf{R}$\n\\end{example}\n \nMathematicians can be very fussy about which symbols are used:\nit would be conventional here to use the `\\wi{blackboard bold}' font,\n\\index{bold symbols} which is obtained using \\ci{mathbb} from the\npackage \\pai{amssymb}.\\footnote{\\pai{amssymb} is not a part\n of the \\AmS-\\LaTeX{} bundle, but it is perhaps still a part of your \\LaTeX{}\n distribution. Check your distribution\n or go to \\texttt{CTAN:/fonts/amsfonts/latex/} to obtain it.}\n\\ifx\\mathbb\\undefined\\else\nThe last example becomes\n\\begin{example}\n$x^{2} \\geq 0\\qquad\n \\text{for all } x \n \\in \\mathbb{R}$\n\\end{example}\n\\fi\nSee Table~\\ref{mathalpha} on page~\\pageref{mathalpha} and\nTable~\\ref{mathfonts} on page~\\pageref{mathfonts} for more math fonts.\n\n\n\n\\section{Building Blocks of a Mathematical Formula}\n\nIn this section, we describe the most important commands used in mathematical\ntypesetting. Most of the commands in this section will not require\n\\textsf{amsmath} (if they do, it will be stated clearly), but load it anyway.\n\n\n\\textbf{Lowercase \\wi{Greek letters}} are entered as \\verb|\\alpha|,\n \\verb|\\beta|, \\verb|\\gamma|, \\ldots, uppercase letters\nare entered as \\verb|\\Gamma|, \\verb|\\Delta|, \\ldots\\footnote{There is no\n uppercase Alpha, Beta etc. defined in \\LaTeXe{} because it looks the same as a\n normal roman A, B\\ldots{}} \n\nTake a look at Table~\\ref{greekletters} on page~\\pageref{greekletters} for a\nlist of Greek letters.\n\\begin{example}\n$\\lambda,\\xi,\\pi,\\theta,\n \\mu,\\Phi,\\Omega,\\Delta$\n\\end{example}\n\n\n\\textbf{Exponents, Superscripts and Subscripts} can be specified using\\index{exponent}\\index{subscript}\\index{superscript}\nthe \\verb|^|\\index{^@\\verb\"|^\"|} and the \\verb|_|\\index{_@\\verb\"|_\"|} characters.\nMost math mode commands act only on the next character, so if you\nwant a command to affect several characters, you have to group them\ntogether using curly braces: \\verb|{...}|.\n\nTable~\\ref{binaryrel} on page \\pageref{binaryrel} lists a lot of binary\nrelations like $\\subseteq$ and $\\perp$.\n\n\\begin{example}\n$p^3_{ij} \\qquad \n m_\\text{Knuth}\\qquad\n\\sum_{k=1}^3 k \\\\[5pt]\n a^x+y \\neq a^{x+y}\\qquad \n e^{x^2} \\neq {e^x}^2$\n\\end{example}\n\n\nThe \\textbf{\\wi{square root}} is entered as \\ci{sqrt}; the\n$n^\\text{th}$ root is generated with \\verb|\\sqrt[|$n$\\verb|]|. The size of\nthe root sign is determined automatically by \\LaTeX. If just the sign\nis needed, use \\verb|\\surd|.\n\nSee various kinds of arrows like $\\hookrightarrow$ and $\\rightleftharpoons$ on\nTable~\\ref{tab:arrows} on page \\pageref{tab:arrows}. \n\\begin{example}\n$\\sqrt{x} \\Leftrightarrow x^{1/2}\n \\quad \\sqrt[3]{2}\n \\quad \\sqrt{x^{2} + \\sqrt{y}}\n \\quad \\surd[x^2 + y^2]$\n\\end{example}\n\n\n\\index{dots!three}\n\\index{vertical!dots}\n\\index{horizontal!dots}\nWhile the \\textbf{\\wi{dot}} sign to indicate\nthe multiplication operation is normally left out, it is sometimes written\nto help the eye in grouping a formula.\nUse \\ci{cdot} to typeset a single centered dot. \\ci{cdots} is\nthree centered \\textbf{\\wi{dots}} while \\ci{ldots} sets the dots low (on the\nbaseline). Besides that, there are \\ci{vdots} for \nvertical and \\ci{ddots} for \\wi{diagonal dots}. There are more examples in \nsection~\\ref{sec:arraymat}.\n\\begin{example}\n$\\Psi = v_1 \\cdot v_2\n \\cdot \\ldots \\qquad \n n! = 1 \\cdot 2 \n \\cdots (n-1) \\cdot n$\n\\end{example}\n\nThe commands \\ci{overline} and \\ci{underline} create\n\\textbf{horizontal lines} directly over or under an expression:\n\\index{horizontal!line} \\index{line!horizontal}\n\\begin{example}\n$0.\\overline{3} = \n \\underline{\\underline{1/3}}$\n\\end{example}\n\nThe commands \\ci{overbrace} and \\ci{underbrace} create\nlong \\textbf{horizontal braces} over or under an expression:\n\\index{horizontal!brace} \\index{brace!horizontal} \n\\begin{example}\n$\\underbrace{\\overbrace{a+b+c}^6 \n \\cdot \\overbrace{d+e+f}^7}\n _\\text{meaning of life} = 42$\n\\end{example}\n\n\\index{mathematical!accents} To add mathematical accents such as \\textbf{small\narrows} or \\textbf{\\wi{tilde}} signs to variables, the commands\ngiven in Table~\\ref{mathacc} on page~\\pageref{mathacc} might be useful. Wide hats and\ntildes covering several characters are generated with \\ci{widetilde}\nand \\ci{widehat}. Notice the difference between \\ci{hat} and \\ci{widehat} and the placement of\n\\ci{bar} for a variable with subscript. The \\wi{apostrophe} mark\n\\verb|'|\\index{'@\\verb\"|'\"|} gives a \\wi{prime}:\n% a dash is --\n\\begin{example}\n$f(x) = x^2 \\qquad f'(x) \n = 2x \\qquad f''(x) = 2\\\\[5pt]\n \\hat{XY} \\quad \\widehat{XY}\n \\quad \\bar{x_0} \\quad \\bar{x}_0$\n\\end{example}\n\n\n\\textbf{Vectors}\\index{vectors} are often specified by adding small\n\\wi{arrow symbols} on the tops of variables. This is done with the\n\\ci{vec} command. The two commands \\ci{overrightarrow} and\n\\ci{overleftarrow} are useful to denote the vector from $A$ to $B$:\n\\begin{example}\n$\\vec{a} \\qquad\n \\vec{AB} \\qquad\n \\overrightarrow{AB}$\n\\end{example}\n\n\nNames of functions are often typeset in an upright\nfont, and not in italics as variables are, so \\LaTeX{} supplies the\nfollowing commands to typeset the most common function names:\n\\index{mathematical!functions}\n\n\\begin{tabular}{llllll}\n\\ci{arccos} & \\ci{cos} & \\ci{csc} & \\ci{exp} & \\ci{ker} & \\ci{limsup} \\\\\n\\ci{arcsin} & \\ci{cosh} & \\ci{deg} & \\ci{gcd} & \\ci{lg} & \\ci{ln} \\\\\n\\ci{arctan} & \\ci{cot} & \\ci{det} & \\ci{hom} & \\ci{lim} & \\ci{log} \\\\\n\\ci{arg} & \\ci{coth} & \\ci{dim} & \\ci{inf} & \\ci{liminf} & \\ci{max} \\\\\n\\ci{sinh} & \\ci{sup} & \\ci{tan} & \\ci{tanh}& \\ci{min} & \\ci{Pr} \\\\\n\\ci{sec} & \\ci{sin} \\\\\n\\end{tabular}\n\n\\begin{example}\n\\begin{equation*}\n \\lim_{x \\rightarrow 0}\n \\frac{\\sin x}{x}=1\n\\end{equation*}\n\\end{example}\n\nFor functions missing from the list, use the \\ci{DeclareMathOperator}\ncommand. There is even a starred version for functions with limits.\nThis command works only in the preamble so the commented lines in the\nexample below must be put into the preamble.\n\n\\begin{example}\n%\\DeclareMathOperator{\\argh}{argh}\n%\\DeclareMathOperator*{\\nut}{Nut}\n\\begin{equation*}\n 3\\argh = 2\\nut_{x=1} \n\\end{equation*}\n\\end{example}\n\nFor the \\wi{modulo function}, there are two commands: \\ci{bmod} for the\nbinary operator ``$a \\bmod b$'' and \\ci{pmod}\nfor expressions\nsuch as ``$x\\equiv a \\pmod{b}$:''\n\\begin{example}\n$a\\bmod b \\\\\n x\\equiv a \\pmod{b}$\n\\end{example}\n\nA built-up \\textbf{\\wi{fraction}} is typeset with the\n\\ci{frac}\\verb|{...}{...}| command. In in-line equations, the fraction is shrunk to\nfit the line. This style is obtainable in display style with \\ci{tfrac}. The\nreverse, i.e.\\ display style fraction in text, is made with \\ci{dfrac}.\nOften the slashed form $1/2$ is preferable, because it looks better\nfor small amounts of `fraction material:'\n\\begin{example}\nIn display style:\n\\begin{equation*}\n 3/8 \\qquad \\frac{3}{8} \n \\qquad \\tfrac{3}{8}\n\\end{equation*}\n\\end{example}\n\n\\begin{example}\nIn text style:\n$1\\frac{1}{2}$~hours \\qquad\n$1\\dfrac{1}{2}$~hours\n\\end{example}\n \nHere the \\ci{partial} command for \\wi{partial derivative}s is used:\n\\begin{example}\n\\begin{equation*} \n \\sqrt{\\frac{x^2}{k+1}}\\qquad\n x^\\frac{2}{k+1}\\qquad\n \\frac{\\partial^2f}\n {\\partial x^2} \n\\end{equation*}\n\\end{example}\n\nTo typeset \\wi{binomial coefficient}s or similar structures, use\nthe command \\ci{binom} from \\pai{amsmath}:\n\\begin{example}\nPascal's rule is\n\\begin{equation*}\n \\binom{n}{k} =\\binom{n-1}{k}\n + \\binom{n-1}{k-1}\n\\end{equation*}\n\\end{example}\n\nFor \\wi{binary relations} it may be useful to stack symbols over each other.\n\\ci{stackrel}\\verb|{#1}{#2}| puts the symbol given\nin \\verb|#1| in superscript-like size over \\verb|#2| which\nis set in its usual position.\n\\begin{example}\n\\begin{equation*}\n f_n(x) \\stackrel{*}{\\approx} 1\n\\end{equation*}\n\\end{example}\n\nThe \\textbf{\\wi{integral operator}} is generated with \\ci{int}, the\n\\textbf{\\wi{sum operator}} with \\ci{sum}, and the \\textbf{\\wi{product operator}}\nwith \\ci{prod}. The upper and lower limits are specified with~\\verb|^|\nand~\\verb|_| like subscripts and superscripts:\n\\begin{example}\n\\begin{equation*}\n\\sum_{i=1}^n \\qquad\n\\int_0^{\\frac{\\pi}{2}} \\qquad\n\\prod_\\epsilon\n\\end{equation*}\n\\end{example}\n\nTo get more control over the placement of indices in complex\nexpressions, \\pai{amsmath} provides the \\ci{substack} command:\n\\begin{example}\n\\begin{equation*}\n\\sum^n_{\\substack{0 0.\n \\end{array} \\right.\n\\end{equation*}\n\\end{example}\nThe \\ei{cases} environment from \\textsf{amsmath} simplifies\nthe syntax, so it is worth a look:\n\\begin{example}\n \\begin{equation*}\n |x| = \n \\begin{cases}\n -x & \\text{if } x < 0,\\\\\n 0 & \\text{if } x = 0,\\\\\n x & \\text{if } x > 0.\n \\end{cases} \n\\end{equation*}\n\\end{example}\n\n\nMatrices\\index{matrix} can be typeset by \\ei{array}, but\n\\pai{amsmath} provides a better solution using the different \\ei{matrix}\nenvironments. There are six versions with different delimiters: \\ei{matrix}\n(none), \\ei{pmatrix} $($, \\ei{bmatrix} $[$, \\ei{Bmatrix} $\\{$, \\ei{vmatrix} $\\vert$ and\n\\ei{Vmatrix} $\\Vert$. You don't have to specify the number of columns as with\n\\ei{array}. The maximum number is 10, but it is customisable (though it is not\nvery often you need 10 columns!):\n\\begin{example}\n\\begin{equation*}\n \\begin{matrix} \n 1 & 2 \\\\\n 3 & 4 \n \\end{matrix} \\qquad\n \\begin{bmatrix} \n p_{11} & p_{12} & \\ldots \n & p_{1n} \\\\\n p_{21} & p_{22} & \\ldots \n & p_{2n} \\\\\n \\vdots & \\vdots & \\ddots \n & \\vdots \\\\\n p_{m1} & p_{m2} & \\ldots \n & p_{mn} \n \\end{bmatrix}\n\\end{equation*}\n\\end{example}\n\n\n\n\\section{Spacing in Math Mode} \\label{sec:math-spacing}\n\n\\index{math spacing} If the spacing within formulae chosen by \\LaTeX{}\nis not satisfactory, it can be adjusted by inserting special spacing\ncommands: \\ci{,} for $\\frac{3}{18}\\:\\textrm{quad}$\n(\\demowidth{0.166em}), \\ci{:} for $\\frac{4}{18}\\: \\textrm{quad}$\n(\\demowidth{0.222em}) and \\ci{;} for $\\frac{5}{18}\\: \\textrm{quad}$\n(\\demowidth{0.277em}). The escaped space character \\verb*|\\ |\ngenerates a medium sized space comparable to the interword spacing and\n\\ci{quad} (\\demowidth{1em}) and \\ci{qquad} (\\demowidth{2em}) produce\nlarge spaces. The size of a \\ci{quad} corresponds to the width of the\ncharacter `M' of the current font. \\verb|\\!|\\cih{\"!} produces a\nnegative space of $-\\frac{3}{18}\\:\\textrm{quad}$\n($-$\\demowidth{0.166em}).\n\n\\begin{example}\n\\begin{equation*}\n \\int_1^2 \\ln x \\mathrm{d}x \n \\qquad\n \\int_1^2 \\ln x \\,\\mathrm{d}x\n\\end{equation*}\n\\end{example}\n\nNote that `d' in the differential is conventionally set in roman.\nIn the next example, we define a new command \\ci{ud} (upright d) which produces\n``$\\,\\mathrm{d}$'' (notice the spacing \\demowidth{0.166em} before the\n$\\text{d}$), so we don't have to write it every time. The \\ci{newcommand} is\nplaced in the preamble. % More on\n% \\ci{newcommand} in section~\\ref{} on page \\pageref{}. To Do: Add label and\n% reference to \"Customising LaTeX\" -> \"New Commands, Environments and Packages\"\n% -> \"New Commands\".\n\\begin{example}\n\\newcommand{\\ud}{\\,\\mathrm{d}}\n\n\\begin{equation*}\n \\int_a^b f(x)\\ud x \n\\end{equation*}\n\\end{example}\n\nIf you want to typeset multiple integrals, you'll discover that the spacing\nbetween the integrals is too wide. You can correct it using \\ci{!}, but\n\\pai{amsmath} provides an easier way for fine-tuning\nthe spacing, namely the \\ci{iint}, \\ci{iiint}, \\ci{iiiint}, and \\ci{idotsint}\ncommands.\n\n\\begin{example}\n\\newcommand{\\ud}{\\,\\mathrm{d}}\n\n\\begin{IEEEeqnarray*}{c}\n \\int\\int f(x)g(y) \n \\ud x \\ud y \\\\\n \\int\\!\\!\\!\\int \n f(x)g(y) \\ud x \\ud y \\\\\n \\iint f(x)g(y) \\ud x \\ud y \n\\end{IEEEeqnarray*}\n\\end{example}\n\nSee the electronic document \\texttt{testmath.tex} (distributed with\n\\AmS-\\LaTeX) or Chapter 8 of \\companion{} for further details.\n\n\\subsection{Phantoms}\n\nWhen vertically aligning text using \\verb|^| and \\verb|_| \\LaTeX{} is sometimes\njust a little too helpful. Using the \\ci{phantom} command you can\nreserve space for characters that do not show up in the final output.\nThe easiest way to understand this is to look at an example:\n\\begin{example}\n\\begin{equation*}\n{}^{14}_{6}\\text{C}\n\\qquad \\text{versus} \\qquad\n{}^{14}_{\\phantom{1}6}\\text{C}\n\\end{equation*}\n\\end{example}\nIf you want to typeset a lot of isotopes as in the example, the \\pai{mhchem}\npackage is very useful for typesetting isotopes and chemical formulae too.\n\n\n\\section{Fiddling with the Math Fonts}\\label{sec:fontsz}\nDifferent math fonts are listed on Table~\\ref{mathalpha} on page\n\\pageref{mathalpha}.\n\\begin{example}\n $\\Re \\qquad\n \\mathcal{R} \\qquad\n \\mathfrak{R} \\qquad\n \\mathbb{R} \\qquad $ \n\\end{example}\nThe last two require \\pai{amssymb} or \\pai{amsfonts}.\n\nSometimes you need to tell \\LaTeX{} the correct font\nsize. In math mode, this is set with the following four commands:\n\\begin{flushleft}\n\\ci{displaystyle}~($\\displaystyle 123$),\n \\ci{textstyle}~($\\textstyle 123$), \n\\ci{scriptstyle}~($\\scriptstyle 123$) and\n\\ci{scriptscriptstyle}~($\\scriptscriptstyle 123$).\n\\end{flushleft}\n\nIf $\\sum$ is placed in a fraction, it'll be typeset in text style unless you tell\n\\LaTeX{} otherwise:\n\\begin{example}\n\\begin{equation*}\n P = \\frac{\\displaystyle{ \n \\sum_{i=1}^n (x_i- x)\n (y_i- y)}} \n {\\displaystyle{\\left[\n \\sum_{i=1}^n(x_i-x)^2\n \\sum_{i=1}^n(y_i- y)^2\n \\right]^{1/2}}}\n\\end{equation*} \n\\end{example}\nChanging styles generally affects the way big operators and limits are displayed.\n\n% This is not a math accent, and no maths book would be set this way.\n% mathop gets the spacing right.\n\n\n\\subsection{Bold Symbols}\n\\index{bold symbols}\n\nIt is quite difficult to get bold symbols in \\LaTeX{}; this is\nprobably intentional as amateur typesetters tend to overuse them. The\nfont change command \\verb|\\mathbf| gives bold letters, but these are\nroman (upright) whereas mathematical symbols are normally italic, and\nfurthermore it doesn't work on lower case Greek letters.\nThere is a \\ci{boldmath} command, but \\emph{this can only be used\noutside math mode}. It works for symbols too, though:\n\\begin{example}\n$\\mu, M \\qquad \n\\mathbf{\\mu}, \\mathbf{M}$\n\\qquad \\boldmath{$\\mu, M$}\n\\end{example}\n\nThe package \\pai{amsbsy} (included by \\pai{amsmath}) as well as the\npackage \\pai{bm} from the \\texttt{tools} bundle make this much easier as they include\na \\ci{boldsymbol} command:\n\n\\begin{example}\n$\\mu, M \\qquad\n\\boldsymbol{\\mu}, \\boldsymbol{M}$\n\\end{example}\n\n\n\\section{Theorems, Lemmas, \\ldots}\n\nWhen writing mathematical documents, you probably need a way to\ntypeset ``Lemmas'', ``Definitions'', ``Axioms'' and similar\nstructures.\n\\begin{lscommand}\n\\ci{newtheorem}\\verb|{|\\emph{name}\\verb|}[|\\emph{counter}\\verb|]{|%\n \\emph{text}\\verb|}[|\\emph{section}\\verb|]|\n\\end{lscommand}\nThe \\emph{name} argument is a short keyword used to identify the\n``theorem''. With the \\emph{text} argument you define the actual name\nof the ``theorem'', which will be printed in the final document.\n\nThe arguments in square brackets are optional. They are both used to\nspecify the numbering used on the ``theorem''. Use the \\emph{counter}\nargument to specify the \\emph{name} of a previously declared\n``theorem''. The new ``theorem'' will then be numbered in the same\nsequence. The \\emph{section} argument allows you to specify the\nsectional unit within which the ``theorem'' should get its numbers.\n\nAfter executing the \\ci{newtheorem} command in the preamble of your\ndocument, you can use the following command within the document.\n\\begin{code}\n\\verb|\\begin{|\\emph{name}\\verb|}[|\\emph{text}\\verb|]|\\\\\nThis is my interesting theorem\\\\\n\\verb|\\end{|\\emph{name}\\verb|}| \n\\end{code}\n\nThe \\pai{amsthm} package (part of \\AmS-\\LaTeX) provides the \n\\ci{theoremstyle}\\verb|{|\\emph{style}\\verb|}|\ncommand which lets you define what the theorem is all about by picking\nfrom three predefined styles: \\texttt{definition} (fat title, roman body),\n\\texttt{plain} (fat title, italic body) or \\texttt{remark} (italic\ntitle, roman body).\n\nThis should be enough theory. The following examples should\nremove any remaining doubt, and make it clear that the\n\\verb|\\newtheorem| environment is way too complex to understand.\n\n% actually define things\n\\theoremstyle{definition} \\newtheorem{law}{Law}\n\\theoremstyle{plain} \\newtheorem{jury}[law]{Jury}\n\\theoremstyle{remark} \\newtheorem*{marg}{Margaret}\n\nFirst define the theorems:\n\n\\begin{verbatim}\n\\theoremstyle{definition} \\newtheorem{law}{Law}\n\\theoremstyle{plain} \\newtheorem{jury}[law]{Jury}\n\\theoremstyle{remark} \\newtheorem*{marg}{Margaret}\n\\end{verbatim}\n\n\\begin{example}\n\\begin{law} \\label{law:box}\nDon't hide in the witness box\n\\end{law}\n\\begin{jury}[The Twelve]\nIt could be you! So beware and\nsee law~\\ref{law:box}.\\end{jury}\n\\begin{jury}\nYou will disregard the last\nstatement.\\end{jury}\n\\begin{marg}No, No, No\\end{marg}\n\\begin{marg}Denis!\\end{marg}\n\\end{example}\n\nThe ``Jury'' theorem uses the same counter as the ``Law''\ntheorem, so it gets a number that is in sequence with\nthe other ``Laws''. The argument in square brackets is used to specify \na title or something similar for the theorem.\n\\begin{example}\n\\newtheorem{mur}{Murphy}[section]\n\n\\begin{mur} If there are two or \nmore ways to do something, and \none of those ways can result in\na catastrophe, then someone \nwill do it.\\end{mur}\n\\end{example}\n\nThe ``Murphy'' theorem gets a number that is linked to the number of\nthe current section. You could also use another unit, for example chapter or\nsubsection.\n\nIf you want to customize your theorems down to the last dot, the\n\\pai{ntheorem} package offers a plethora of options.\n\n\n\\subsection{Proofs and End-of-Proof Symbol}\n\\label{sec:putting-qed-right}\n\nThe \\pai{amsthm} package also provides the \\ei{proof} environment.\n\n\\begin{example}\n\\begin{proof}\n Trivial, use\n \\begin{equation*}\n E=mc^2.\n \\end{equation*}\n\\end{proof}\n\\end{example}\n\nWith the command \\ci{qedhere} you can move the `end of proof' symbol\naround for situations where it would end up alone on a line.\n\n\\begin{example}\n\\begin{proof}\n Trivial, use\n \\begin{equation*}\n E=mc^2. \\qedhere\n \\end{equation*}\n\\end{proof}\n\\end{example}\n\nUnfortunately, this correction does not work for \\texttt{IEEEeqnarray}:\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with an equation array:\n \\begin{IEEEeqnarray*}{rCl}\n a & = & b + c \\\\\n & = & d + e. \\qedhere\n \\end{IEEEeqnarray*} \n\\end{proof}\n\\end{example}\n\\noindent\nThe reason for this is the internal structure of \\texttt{IEEEeqnarray}:\nit always puts two invisible columns at both sides of the array that\nonly contain a stretchable space. By this \\texttt{IEEEeqnarray} ensures\nthat the equation array is horizontally centered. The\n\\ci{qedhere} command should actually be put \\emph{outside} this\nstretchable space, but this does not happen as these columns are\ninvisible to the user.\n\nThere is a very simple remedy. Define the stretching\nexplicitly!\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with an equation array:\n \\begin{IEEEeqnarray*}{+rCl+x*}\n a & = & b + c \\\\\n & = & d + e. & \\qedhere\n \\end{IEEEeqnarray*} \n\\end{proof}\n\\end{example}\n\\noindent\nNote that the \\verb=+= in \\verb={+rCl+x*}= denotes stretchable spaces, one\non the left of the equations (which, if not specified, will be done\nautomatically by \\texttt{IEEEeqnarray}!) and one on the right of the\nequations. But now on the right, \\emph{after} the stretching column,\nwe add an empty column \\verb=x=. This column will only be needed on\nthe last line if the \\ci{qedhere} command is put\nthere. Finally, we specify a \\verb=*=. This is a null-space that\nprevents \\texttt{IEEEeqnarray} from adding another unwanted \\verb=+=-space!\n\nIn the case of equation numbering, there is a similar problem. Comparing\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with a numbered equation:\n \\begin{equation}\n a = b + c.\n \\end{equation}\n\\end{proof}\n\\end{example}\n\\noindent\nwith\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with a numbered equation:\n \\begin{equation}\n a = b + c. \\qedhere\n \\end{equation}\n\\end{proof}\n\\end{example}\n\\noindent\nyou notice that in the (correct) second version the $\\Box$ is much\ncloser to the equation than in the first version.\n\nSimilarly, the correct way of putting the QED-symbol at the end of an\nequation array is as follows:\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with an equation array:\n \\begin{IEEEeqnarray}{+rCl+x*}\n a & = & b + c \\\\\n & = & d + e. \\\\\n &&& \\qedhere\\nonumber\n \\end{IEEEeqnarray} \n\\end{proof}\n\\end{example}\n\\noindent\nwhich contrasts with\n\\begin{example}\n\\begin{proof}\n This is a proof that ends\n with an equation array:\n \\begin{IEEEeqnarray}{rCl}\n a & = & b + c \\\\\n & = & d + e.\n \\end{IEEEeqnarray} \n\\end{proof}\n\\end{example}\n\n\n%\n\n% Local Variables:\n% TeX-master: \"lshort\"\n% mode: latex\n% mode: flyspell\n% End:\n", "meta": {"hexsha": "db051630c1ff3f14abd4e94be8e1deed980a849c", "size": 40906, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "synctex test files/less basic/2017/lshort-5.05/src/math.tex", "max_stars_repo_name": "templateK/synctex", "max_stars_repo_head_hexsha": "555467da1535b0b1d7e97532a2c6251d9c2b3957", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 38, "max_stars_repo_stars_event_min_datetime": "2016-12-23T02:21:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-18T22:07:39.000Z", "max_issues_repo_path": "synctex test files/less basic/2017/lshort-5.05/src/math.tex", "max_issues_repo_name": "templateK/synctex", "max_issues_repo_head_hexsha": "555467da1535b0b1d7e97532a2c6251d9c2b3957", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 27, "max_issues_repo_issues_event_min_datetime": "2016-12-23T13:20:04.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-30T22:31:48.000Z", "max_forks_repo_path": "synctex test files/less basic/2017/lshort-5.05/src/math.tex", "max_forks_repo_name": "templateK/synctex", "max_forks_repo_head_hexsha": "555467da1535b0b1d7e97532a2c6251d9c2b3957", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 20, "max_forks_repo_forks_event_min_datetime": "2017-08-28T20:40:55.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-25T21:50:14.000Z", "avg_line_length": 32.4908657665, "max_line_length": 156, "alphanum_fraction": 0.6993595072, "num_tokens": 13041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303236047049, "lm_q2_score": 0.8128673178375735, "lm_q1q2_score": 0.6425962638178254}} {"text": "\\section{The Adjugate Matrix}\r\nGiven a square matrix $A\\in M_n(F)$ with columns $A^{(i)}$.\r\nIf we swap two neighbouring columns (or rows), then $\\det A$ changes sign.\r\n\\begin{remark}\r\n We can prove properties of determinant using the decomposition of $A$ into elementary matrices.\r\n\\end{remark}\r\nA column expansion is a strategy to compute the determinant of a matrix by using its linkage to some of its submatrices.\r\n\\begin{definition}\r\n Let $A\\in M_n(F)$ and pick $i,j\\in\\{1,\\ldots,n\\}$.\r\n We define $A_{\\widehat{ij}}\\in A_{n-1}(F)$ to be the matrix obtained by deleting the $i^{th}$ row and $j^{th}$ column from $A$.\r\n\\end{definition}\r\n\\begin{example}\r\n Take\r\n $$A=\\begin{pmatrix}\r\n 1&2&-7\\\\\r\n 2&1&0\\\\\r\n -3&6&1\r\n \\end{pmatrix}$$\r\n Then\r\n $$A_{\\widehat{32}}=\\begin{pmatrix}\r\n 1&-7\\\\\r\n 2&0\r\n \\end{pmatrix}$$\r\n\\end{example}\r\n\\begin{lemma}[Expansion of Determinant]\r\n Let $A\\in M_n(F)$.\\\\\r\n 1. We have the expansion along the $j^{th}$ column\r\n $$\\det A=\\sum_{i=1}^n(-1)^{i+j}a_{ij}\\det A_{\\widehat{ij}}$$\r\n 2. We have the expansion along the $i^{th}$ row\r\n $$\\det A=\\sum_{j=1}^n(-1)^{i+j}a_{ij}\\det A_{\\widehat{ij}}$$\r\n\\end{lemma}\r\n\\begin{example}\r\n Take\r\n $$A=\\begin{pmatrix}\r\n 1&2&-1\\\\\r\n 3&-1&1\\\\\r\n 4&2&-7\r\n \\end{pmatrix}$$\r\n So expanding along the second column gives\r\n $$\\det A=-(2)\\det\\begin{pmatrix}\r\n 3&1\\\\\r\n 4&-7\r\n \\end{pmatrix}+(-1)\\det\\begin{pmatrix}\r\n 1&-1\\\\\r\n 4&-7\r\n \\end{pmatrix}-2\\det\\begin{pmatrix}\r\n 1&-1\\\\\r\n 3&1\r\n \\end{pmatrix}$$\r\n\\end{example}\r\n\\begin{proof}\r\n Suffices to show the first part.\r\n Pick $1\\le j\\le n$ and write $A=(a_{ij}),A^{(j)}=\\sum_ia_{ij}e_i$.\r\n Then\r\n \\begin{align*}\r\n \\det A&=\\det\\left( A^{(1)},\\ldots,\\sum_{i=1}^na_{ij}e_i ,\\ldots, A^{(n)}\\right)\\\\\r\n &=\\sum_{i=1}^na_{ij}\\det(A^{(1)},\\ldots,e_i,\\ldots,A^{(n)})\\\\\r\n &=\\sum_{i=1}^na_{ij}(-1)^{j-1}\\det(e_i,A^{(1)},\\ldots,A^{(n)})\\\\\r\n &=\\sum_{i=1}^na_{ij}(-1)^{j-1}(-1)^{i-1}\\det\r\n \\begin{pmatrix}\r\n 1&\\ast&\\cdots&\\ast\\\\\r\n 0&&&\\\\\r\n \\vdots&&A_{\\widehat{ij}}&\\\\\r\n 0&&&\r\n \\end{pmatrix}\\\\\r\n &=\\sum_{i=1}^n(-1)^{i+j}a_{ij}\\det A_{\\widehat{ij}}\r\n \\end{align*}\r\n as desired.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $A\\in M_n(F)$, then the adjugate matrix $\\operatorname{adj}A$ of $A$ is the $n\\times n$ matrix with entries\r\n $$(\\operatorname{adj}A)_{ji}=(-1)^{i+j}\\det A_{\\widehat{ij}}=\\det(A^{(1)},\\ldots,A^{(j-1)},e_i,A^{(j+1)},\\ldots,A^{(n)})$$\r\n\\end{definition}\r\n\\begin{theorem}\r\n Let $A\\in M_n(F)$, then $\\operatorname{adj}(A)A=(\\det A)I$ where $I$ is the identity matrix.\r\n\\end{theorem}\r\nIn particular, if $A$ is invertible, then as $\\det A\\neq 0$,\r\n$$A^{-1}=\\frac{1}{\\det A}\\operatorname{adj}A$$\r\n\\begin{proof}\r\n For any $j$, the preceding lemma translates to\r\n $$\\det A=\\sum_{i=1}^n(\\operatorname{adj}A)_{ij}a_{ij}=(\\operatorname{adj}(A)A)_{jj}$$\r\n Now for $j \n\\usepackage{etex}\n\\usepackage[utf8]{inputenc}\n\\input ../AuxFiles/PreambleNotes.tex\n\n\n\\begin{document}\n\\onehalfspace\n\n\\vspace*{\\fill}\n\\begingroup\n\\centering\n\n\\Large {\\scshape Introduction to Statistics}\\\\\n\n(Lectures 7-8: Estimation)\n\n\\endgroup\n\\vspace*{\\fill}\n\n\\newpage\n\n\n\\section{Estimation}\n\n{\\scshape Overview:} \\noindent Lectures 7-8 will focus on \\emph{estimation}. As we have mentioned before, a strategy in the estimation problem is an \\emph{estimator}. The loss function used for this problem is the squared distance between the estimated value and the true parameter. This loss gives rise to the popular \\emph{mean squared error} performance criterion. \n\n\n\nWe will analyze two commonly used estimation strategies---Maximum Likelihood (ML) and Posterior Mean estimators---in the context of the homoskedastic Normal regression model with (known variance); i.e., \n\\begin{equation} \\label{equation:Regression-model}\nY \\sim \\mathcal{N}(X \\beta, \\sigma^2 \\mathbb{I}_n ), \n\\end{equation}\n where $Y$ is the $n \\times 1$ vector of outcome variables, $X$ is the $n \\times k$ matrix of non-stochastic regressors, and $\\beta \\in \\mathbb{R}^{k}$ is the parameter of interest. \n \n We will show that if $k \\leq n$ and $X$ has full-column rank, the Maximum Likelihood estimator for the model in (\\ref{equation:Regression-model}) is the OLS estimator\n\\[ \\widehat{\\beta}_{\\textrm{OLS}} \\equiv (X’X)^{-1} X’Y, \\]\n\n\\noindent (which we introduced in the last problem set). We will show this estimator is ``best'' among all \\emph{unbiased} estimators and we will connect this result to the \\emph{Cram\\'er-Rao} bound for the variance of estimators in parametric models. We will also argue that ``best’’ among the class of unbiased estimators need not imply admissibility with respect to all estimators. In particular, we show that if $k \\geq 3$, the OLS estimator is dominated and we present an ``empirical Bayes’’ estimator that dominates it. \n\nThe posterior mean estimator for the regression model assuming $\\beta \\sim \\mathcal{N}_{k}(0 , (\\sigma^2/ \\lambda) \\mathbb{I}_k )$ is the popular \\emph{Ridge estimator}\n\\[ \\widehat{\\beta}_{\\textrm{Ridge}} \\equiv (X’X + \\lambda \\mathbb{I}_k)^{-1} X’Y. \\]\nThe Ridge estimator is well-defined regardless of the number of covariates and it is admissible by construction. The Ridge estimator provides a simple example of ``shrinkage’’ or ``regularization’’. A practical concern is that the Ridge estimator is biased, and the bias depends on the choice of the prior hyperparameters and the true parameter at which we evaluate the bias. \n\nOne result that we will not cover in the notes, but that is important to keep in mind is that despite the inadmissibility of the OLS estimator for the \\emph{full} vector of coefficients $\\beta$, OLS is admissible for the problem in which we are interested in estimating only one regression coefficient but controlling for other variables. \n\n\n\n\\subsection{Quadratic Loss} \nLet $\\mathcal{A} =\\Theta = \\mathbb{R}^{k}$. The loss function we will work with is the so-called quadratic loss\n\\[ \\mathcal{L}(a, \\beta) = || a- \\beta ||^2 = (a-\\beta)’(a-\\beta) = \\sum_{j=1}^{k} (a_j - \\theta_j)^2, \\]\nwhich measures the squared distance between the estimator $\\widehat{\\beta}$ and the parameter $\\beta$. \nThe risk of any estimator $\\widehat{\\beta}$ (which is a map from data $(Y,X)$ to $\\Theta$) is given by\n\\begin{equation} \\label{equation:MSE}\n\\mathbb{E}_{\\mathbb{P}_{\\beta}}[ \\mathcal{L}(\\widehat{\\beta}, \\beta) ] = \\mathbb{E}_{\\mathbb{P}_{\\beta}}[ (\\widehat{\\beta}- \\beta)’(\\widehat{\\beta}-\\beta) ]. \n\\end{equation}\nEquation (\\ref{equation:MSE}) is referred to as \\emph{the mean-squared estimation error} at $\\beta$. We now present a simple algebraic decomposition of the mean squared error in terms of the ``bias’’ and ``variance’’ of the estimator $\\widehat{\\beta}$. Assuming that $\\overline{\\beta} \\equiv \\mathbb{E}_{\\beta}[ \\widehat{\\beta} ]$ is finite, we define the \\underline{bias} of $\\widehat{\\beta}$ at $\\beta$ as\n$$ B_\\beta(\\widehat{\\beta}) = \\overline{\\beta} - \\beta.$$\nIf the covariance matrix of $\\widehat{\\beta}$---denoted $\\textrm{V}_{\\beta} (\\widehat{\\beta})$---is also finite, the mean squared-error can be written as\n\\begin{eqnarray*}\n\\mathbb{E}_{\\mathbb{P}_{\\beta}}[ (\\widehat{\\beta}- \\beta)’(\\widehat{\\beta}-\\beta) ] &=& \\mathbb{E}_{\\mathbb{P}_{\\beta}}[ (\\widehat{\\beta}-\\overline{\\beta} +\\overline{\\beta} - \\beta)’(\\widehat{\\beta}-\\overline{\\beta} +\\overline{\\beta} -\\beta) ], \\\\\n&=& (\\overline{\\beta}-\\beta)’(\\overline{\\beta}-\\beta) + \\mathbb{E}_{\\mathbb{P}_{\\beta}}[ (\\widehat{\\beta}-\\overline{\\beta} )’(\\widehat{\\beta}-\\overline{\\beta}) ] \\\\\n&=& ||B_\\beta(\\widehat{\\beta}) ||^2 + \\textrm{tr} \\left( \\textrm{V}_{\\beta} (\\widehat{\\beta}) \\right),\n\\end{eqnarray*}\nwhere $\\textrm{tr}(\\cdot)$ is the trace operator. The decomposition is fairly straightforward, but it highlights the fact that the bias and variance of the estimator fully determine its risk whenever the loss is quadratic. Also, the correlation between any of the components of $\\widehat{\\beta}$ is not relevant for the risk calculation. \n\n\n\\subsection{Maximum Likelihood Estimation}\n\nAccording to model (\\ref{equation:Regression-model}) the vector of outcome variables is a multivariate normal with parameters $X\\beta$ and $\\sigma^2 \\mathbb{I}_n$ and thus has a p.d.f given by\n\\begin{equation}\\label{equation:Regression-pdf}\nf( Y | \\beta , X ) = \\frac{1}{(2 \\pi \\sigma^2)^{n/2}} \\exp \\left( -\\frac{1}{2 \\sigma^2} (Y - X \\beta)’ (Y-X\\beta) \\right). \n\\end{equation}\nThis is true, regardless of whether we have very many covariates or not. For a fixed realization of the data, define the likelihood function, L($\\beta$; (Y,X)), as the value attained by the p.d.f. in (\\ref{equation:Regression-pdf}) at different values of the data ($Y, X$); that is \n\\[ L(\\beta, (Y,X)) \\equiv f(Y | \\beta, X ). \\]\nThe maximum likelihood estimator at data $(Y,X)$ is then defined as the value of $\\beta$ that maximizes the likelihood; that is\n\\[ \\widehat{\\beta}_{\\textrm{ML}} \\equiv \\textrm{argmax}_{\\beta \\in \\mathbb{R}^{k}} L(\\beta, (Y,X)). \\]\n\n\\noindent The likelihood function implied by (\\ref{equation:Regression-pdf}) is decreasing in $(Y-XB)’(Y-XB)$, a term which is usually referred to as the sum of square residuals. Therefore, maximizing the likelihood is equivalent to solving the \\emph{least-squares} problem\n\\[ \\textrm{min}_{\\beta \\in \\mathbb{R}^{k}} (Y-X\\beta)’(Y-X\\beta). \\]\nThe first-order conditions for the program above, which are necessary and sufficient, yield\n\\[ X’(Y-X\\widehat{\\beta}_{\\textrm{ML}}) = \\textbf{0}_{k \\times 1} \\iff X’Y = (X’X) \\widehat{\\beta}_{\\textrm{ML}}. \\]\n\n\\noindent If $k > n$, there are infinitely many solutions that maximize the likelihood. If $k \\leq n$, and $X$ has full-column rank there is a unique solution to this problem given by\n\\begin{equation} \\label{equation:OLS}\n\\widehat{\\beta}_{\\textrm{ML}} = \\widehat{\\beta}_{\\textrm{OLS}} = (X’X)^{-1} X’Y.\n\\end{equation}\n\n\\noindent The bias of the maximum likelihood estimator:\n\\begin{equation} \\label{equation:bias}\n\\textrm{B}_{\\beta}(\\widehat{\\beta}_{\\textrm{ML}}) = \\mathbb{E}_{\\mathbb{P}_{\\beta}} [ (X’X)^{-1} X’Y ] - \\beta = \\mathbf{0}_{k \\times 1}.\n\\end{equation}\nfor any $\\beta$. Thus, we say that the ML estimator of $\\beta$ is \\underline{unbiased}.\\footnote{An estimator $\\widehat{\\beta}$ is unbiased if $\\mathbb{E}_{\\beta} [\\widehat{\\beta}] = \\beta$ for all $\\beta \\in \\Theta$.} \n\nThe variance of the ML estimator is \n\\begin{equation} \\label{equation:variance}\n\\mathbb{V}_{\\beta} (\\widehat{\\beta}_{\\textrm{ML}}) = \\mathbb{E}_{\\mathbb{P}_{\\beta}} [ (\\widehat{\\beta}_{\\textrm{ML}}-\\beta) (\\widehat{\\beta}_{\\textrm{ML}} -\\beta )^{\\prime} ] = (X’X)^{-1} X’ \\mathbb{E}_{\\mathbb{P}_{\\beta}} [ (Y-X\\beta) (Y-X\\beta)’] X (X’X)^{-1} = \\sigma^2 (X’X)^{-1}. \n\\end{equation}\nThis means that the mean squared error at $\\beta$---denoted $\\textrm{MSE}(\\beta ; \\widehat{\\beta}_{\\textrm{ML}})$---equals\n\\[ \\sigma^2 \\textrm{tr} \\left ( (X’X)^{-1} \\right). \\] \n\n\\noindent for any $\\beta$. Consequently, an interesting property of the ML/OLS estimator is that its risk function is constant over the parameter space. \n\n\\noindent \n\n\\subsection{Posterior Mean under a Normal Prior}\nIn this subsection we analyze the Bayes estimator of the parameter $\\beta$. We have already showed that any Bayes rule can be obtained by minimizing posterior loss at each data realization. Let $\\pi$ denote a prior over the parameter $\\beta$. In our set-up the posterior loss of an action $a$ is\n\\[ \\mathbb{E}_{\\pi} [ \\: || a - \\beta ||^2 \\: | \\: (Y,X) \\: ]. \\]\nUsing the same argument that we used to decompose the mean squared-error in terms of bias and variance we can show that\n\\begin{eqnarray*}\n\\mathbb{E}_{\\pi} [ \\: || a - \\beta ||^2 \\: | \\: (Y,X) \\: ] &=& \\mathbb{E}[ \\: || a - \\mathbb{E}_{\\pi} [ \\: \\beta \\: | \\: (Y,X) \\: ] + \\mathbb{E}_{\\pi} [ \\beta \\: | \\: (Y,X) ] - \\beta ||^2 \\: | \\: (Y,X) \\: ], \\\\\n&=& || a - \\mathbb{E}_{\\pi} [ \\beta \\: | \\: (Y,X) ] ||^2 + \\textrm{tr} \\left( \\mathbb{V}_{\\pi} ( \\beta \\: | \\: (Y,X) ) \\right). \n\\end{eqnarray*}\nThis shows that, regardless of the specific prior we pick, the Bayes estimator is the posterior mean of $\\beta$\n\\[ \\widehat{\\beta}_{\\textrm{Bayes}} = \\mathbb{E}_{\\pi} [ \\: \\beta \\: | \\: (Y,X) \\: ]. \\]\n\nBefore committing to a specific prior $\\pi$, we will show that under quadratic loss the Bayes estimator is the posterior mean.\\\\ \n\n{\\scshape Posterior Mean under a Normal Prior:} Consider then the following prior on $\\beta$:\n\n\\begin{equation}\n\\beta \\sim \\pi (\\beta) \\equiv \\mathcal{N}_{k}( \\beta_0 \\: , \\: \\sigma^2 V^{-1}).\n\\end{equation}\n\n\\noindent The prior assumes that all the coefficients are approximately normal with values close to the vector $\\beta_0$ and covariance matrix given by $\\sigma^2 V^{-1}$. There is typically no magical recipe to select a prior. More often than not, the selection of a prior trades-off interpretation and convenience in its implementation. \n\nWe derive the posterior distribution of $\\beta$. One way of deriving this posterior distribution is by an application of Bayes Theorem\n\\begin{equation*}\n\\pi(\\beta \\: | \\: \\sigma^2, y, X ) = \\frac{f(Y \\: | \\: \\beta, X) \\pi(\\beta ) }{\\int_{\\Theta} f(Y, \\: | \\: \\beta, X) \\pi (\\beta ) d \\beta}. \n\\end{equation*}\n\n\\noindent The posterior is thus proportional to the likelihood times the prior, both of which are Gaussian. \n\n\\noindent Consequently, $f(Y | X, \\beta, \\sigma^2 )\\: \\pi(\\beta | \\sigma^2 )$ is, up to a constant that does not depend on $\\beta$, proportional to\n\\begin{equation} \\label{equation:posterior}\n\\exp \\left( -\\frac{1}{2\\sigma^2} (Y-X\\beta)^{\\prime} (Y-X \\beta) \\right) \\exp \\left(-\\frac{1}{2 \\sigma^2} (\\beta-\\beta_0)^{\\prime} V (\\beta-\\beta_0) \\right). \n\\end{equation}\n\n\\noindent The expression above equals:\n\\[ \\exp \\left( -\\frac{1}{2\\sigma^2} Y^{\\prime} Y \\right) \\exp \\left( - \\frac{1}{2 \\sigma^2 } \\beta \\left(V + X’X \\right) \\beta + \\frac{1}{\\sigma^2}(Y^{\\prime} X+V\\beta_0) \\beta \\right). \\]\nCompleting the square and ignoring all the terms that do not have $\\beta$ on them, gives the posterior distribution as a constant times the exponential of:\n\\begin{equation*}\n-\\frac{1}{2 \\sigma^2} \\left( \\beta - \\left( V + X^{\\prime} X \\right)^{-1} (X^{\\prime} y+ V \\beta_0) \\right) \\left(V+ X’X \\right) \\left(\\beta - ( V^{-1} + X^{\\prime} X )^{-1} (X^{\\prime} Y+ V \\beta_0) \\right). \n\\end{equation*}\nThis implies that:\n\\begin{equation}\n\\beta | Y, X \\sim \\mathcal{N}_{k} \\left( \\left( V + X’X \\right)^{-1} (X^{\\prime} Y+V\\beta_0) \\: , \\: \\sigma^2 \\left(V + X^{\\prime} X \\right)^{-1} \\right).\n\\end{equation}\nThis means that the Bayesian Estimator of $\\beta$ given the Gaussian prior $\\pi(\\beta)$ is:\n\n\\[ \\widehat{\\beta}_{\\textrm{Bayes}} \\equiv \\left( V + X^{\\prime} X \\right)^{-1} \\left( X^{\\prime} Y + V \\beta_0 \\right). \\]\n\n\\noindent The posterior mean estimator is well defined regardless the number of\ncovariates.\\footnote{If $V$ is positive definite, then the matrix $V+ X’X$ is\n positive definite as well, and thus invertible.} The posterior mean estimator\nfor $V= \\lambda \\mathbb{I}_{k}$ and $\\beta_0 = 0$ is called the Ridge estimator\n\n\\[ \\widehat{\\beta}_{\\textrm{Ridge}} \\equiv (X’X + \\lambda \\mathbb{I}_{k})^{-1} X’Y, \\]\n\n\\noindent which, by construction, is admissible. The Ridge estimator is biased:\n\\begin{eqnarray*}\n\\mathbb{E}_{\\mathbb{P}_{\\beta}} [ \\widehat{\\beta}_{\\textrm{Ridge}} ] - \\beta &=& (X’X + \\lambda \\mathbb{I}_{k})^{-1} X’X\\beta - \\beta, \\\\\n&=& (X’X + \\lambda \\mathbb{I}_{k})^{-1} (X’X\\beta - (X’X + \\lambda \\mathbb{I}_{k}) \\beta), \\\\\n&=& -\\lambda (X’X + \\lambda \\mathbb{I}_k)^{-1} \\beta.\n\\end{eqnarray*}\n\\noindent and the magnitude of the bias depends on $\\beta$. The variance of the Ridge estimator is\n\\[ \\mathbb{V}_{\\beta} (\\widehat{\\beta}_{\\textrm{Ridge}}) = \\sigma^2( X’X + \\lambda \\mathbb{I}_k)^{-1} X’X ( X’X + \\lambda \\mathbb{I}_k)^{-1} \\]\n\n\\noindent When $k 1$, however, there are many possibilities.\n\\begin{example}[Lexicographical ordering $\\prec_{\\rm lex}$]\n Fix an ordering on the variables, say $x_s \\prec x_{s-1} \\prec \\cdots \\prec x_1$.\n Extend this ordering to monomials using the lexicographical rule.\n That is, set $x^i \\prec_{\\rm lex} x^j$ if the first non-zero entry of $j-i$ is positive.\n\\end{example}\n\n\\begin{example}[Graded lexicographical ordering $\\prec_{\\rm grlex}$]\n Fix an ordering on the variables, say $x_s \\prec x_{s-1} \\prec \\cdots \\prec x_1$.\n Extend this ordering to monomials by first comparing the total degree and then using the lexicographical rule.\n That is, set $x^i \\prec_{\\rm grlex} x^j$ if $\\deg(x^i) < \\deg(x^j)$ or if $\\deg(x^i) = \\deg(x^j)$ and $x^i \\prec_{\\rm lex} x^j$.\n\\end{example}\n\n\\begin{example}[Weight ordering]\n Assign a weight $\\alpha_i \\in \\R_{> 0}$ to each variable $x_i$.\n Extend the weight assignment to monomials by taking a weighted sum\n \\[ \\wt(x^i) = \\sum_{\\ell = 1}^s i_\\ell \\alpha_\\ell.\\]\n Use this to define a term ordering by setting $x^i \\prec_\\alpha x^j$ if $\\wt(x^i) < \\wt(x^j)$.\n\n We have to take care that there are no ties in the weights of the monomials (or devise a tie-breaker).\n For generic (almost all) choices of weights, there will be no ties.\n\\end{example}\n\nFix a term ordering $\\prec$.\nUsing $\\prec$, we can define the leading term of a polynomial.\nFor $f \\in k[x_1,\\dots,x_s]$, let $\\LT(f)$, or more precisely $\\LT_\\prec(f)$, be the term of $f$ with the highest monomial (according to $\\prec$).\nFor example, we have\n\\begin{align*}\n \\LT_{\\prec_{\\rm lex}} (5x^3y^3-3x^4) &= -3x^4 \\\\\n \\LT_{\\prec_{\\rm grlex}} (5x^3y^3-3x^4) &= 5x^3y^3. \n\\end{align*}\nWe can now generalize the division algorithm as follows.\n\n\\begin{algorithm}\n \\While{\n there exists an $i$ such that $\\LT(f_i)$ divides $\\LT(g)$.\n }\n {\n Replace $g$ by $g - \\frac{\\LT(f_i)}{\\LT(g)} f_i$.\n }\n Return $g$ as the remainder.\n \\bigskip\n\n \\caption{The division algorithm for $g \\pmod {f_1, \\dots, f_n}$}\n\\end{algorithm}\n\nIt is easy to see that if the division algorithm for $g \\pmod {f_1, \\dots, f_n}$ returns $0$ as the remainder, then $g$ is in the ideal $\\langle f_1, \\dots, f_n \\rangle$.\nHowever, the converse is not true, as the next examples show.\nThis severely diminishes the utility of the division algorithm.\n\nThe division algorithm suffers from another drawback.\nNote that there may be choices involved in the running of the algorithm.\nNamely, if there are multiple $i$'s such that $\\LT(f_i)$ divides $\\LT(g)$, then we may choose any one and proceed.\nThe final remainder obtained may change depending on the choices made in the algorithm.\n\nNevertheless, we will see in the next section, that we can overcome both o these drawbacks by making a judicious choice of the generating set for our ideal.\n\n\\begin{example}[Failure of the division algorithm]\n Let us work with $\\prec_{\\rm grlex}$.\n Take $f_1 = x^3-2xy$ and $f_2 = x^2y-2y^2+x$; and $g = x^2$.\n Then the division algorithm stops without any computation, and returns $x^2$ as the remainder.\n However, the following computation shows that $g$ is actually in the ideal $\\langle f_1, f_2 \\rangle$:\n \\[ g = x \\cdot f_2 - f_1.\\]\n\\end{example}\n\n\\begin{example}[Failure of the division algorithm]\n We can work in any term ordering.\n Take $f_1 = xy-1$, $f_2 = x^3-y^3$, and $f_3 = x^2+y^2$.\n Check that $1 \\in \\langle f_1, f_2, f_3 \\rangle$ (either directly or using that $V(\\langle f_1, f_2, f_3 \\rangle) = \\emptyset$).\n However, the division algorithm stops without any computation, and returns $1$ as the remainder.\n\\end{example}\n\n\\subsection{Gr\\\"obner bases}\n\n\\bibliography{ch2}\n\\bibliographystyle{amsalpha}\n\n\\end{document}", "meta": {"hexsha": "562888dabcd31ad9d42f5bb96da511ac9c64fc7d", "size": 16851, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/ch2.tex", "max_stars_repo_name": "Chr1sWilliams/NumericalPolyonomialAlgebra", "max_stars_repo_head_hexsha": "a99701bc68ee1db052d5c0b1aabb1588800b6cc3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Notes/ch2.tex", "max_issues_repo_name": "Chr1sWilliams/NumericalPolyonomialAlgebra", "max_issues_repo_head_hexsha": "a99701bc68ee1db052d5c0b1aabb1588800b6cc3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/ch2.tex", "max_forks_repo_name": "Chr1sWilliams/NumericalPolyonomialAlgebra", "max_forks_repo_head_hexsha": "a99701bc68ee1db052d5c0b1aabb1588800b6cc3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.4164222874, "max_line_length": 200, "alphanum_fraction": 0.7050026705, "num_tokens": 5285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.8128673133042217, "lm_q1q2_score": 0.6425962562225488}} {"text": "\\subsection{Boltzmann Machine}\n\n% --- Lucas\n\n\\begin{figure}[htbp]\n\t\\begin{center}\n\t\t\\includegraphics[width=0.5\\textwidth]{inc/boltzmann_machine.png}\n\t\t\\caption{Illustration of a Boltzmann Machine. The blue units represents three hidden units, while the four white units represents four visible units.\\protect\\footnotemark}\n\t\t\\label{fig:boltzmann_machine}\n\t\\end{center}\n\\end{figure}\n\\footnotetext{Original image (Public Domain): \\url{https://en.wikipedia.org/wiki/File:Boltzmannexamplev1.png}}\n\nThe ``Boltzmann Machine'' (BM) is a form of ``parallel constraint satisfaction network'' \\cite{ackley1985learning}. It is capable of learning the underlying constraints of a domain by only being shown examples of it. The BM is composed of units (also known as nodes) forming a complete graph where the connection between two units are symmetric; meaning that the weight on the connection is the same in either direction. No unit has a connection to itself. The units are binary, in the meaning that they can only assume one of two states, on or off. The state of a unit is determined by a probabilistic function based on the states of the units neighbours. A strong connection (high weight value) between two units indicates that if either of these two units are active, the other one should probably be active as well. While a weak connection (low weight value) indicates that these should probably not be active at the same time. This is analogous to Hebbian learning.\n\nThe BM is notably similar to the Hopfield network in that it also defines a global energy state of the system, utilizing the same equation that determines the global energy value. Each global state can be identified by the energy of the system in that state. By forcing the values of the visible units to represent a training set the system attempts to find an energy configuration that is compatible with the given input. The resulting energy state can then be interpreted as to how well the given data fulfils the constraints of the domain. Thus by minimizing the energy the system learns an interpretation of the problem that increasingly satisfies the constraints of the domain.\n\nThe simplest way to minimize the energy into a local minimum of the system is to change each units state into a value that results in a lower energy state. The data needed to determine this change is locally accessible to each unit, and is dependent on the current state of the units neighbours. If the sum of all values for a given units neighbour exceeds the threshold of that unit, the resulting state of the unit should be on. Otherwise it should be off. This is the usual algorithm for binary units.\n\nBecause of this deterministic algorithm it suffers from the usual weaknesses of gradient descent algorithms, namely, it gets stuck in local minima if its initial state is close to one. In order to alleviate the algorithm of this problem, noise is introduced in the training. This allows the network to ``jump'' out of these minima into configurations of higher energy. The algorithm used for noise introduction is a variation of the ``Metropolis algorithm'' \\cite{metropolis1953equation} that was used to study thermodynamic systems. This modified version introduces a concept of temperature to the machine, which then tries to reach ``thermal equilibrium'' during training. Meaning that the machine is allowed to run repeatedly until the global energy of the system converges to a fixed state over a temperature that is initially high and then slowly decreased over the runtime of the system. The probability of finding the system in a global state after it has reached thermal equilibrium follows a Boltzmann distribution. If the temperature feed into the machine is equal to zero the stochastic nature of it is removed, the machine becomes deterministic and can be seen as a regular Hopfield network.\n\n%TODO discuss the usage of Hebbian learning in Boltzmann\n%TODO write more about this temperature thing\n%TODO write more about the probabilistic features of the BM as well as the Boltzmann distribution that gives the system its name.\n\nTraining is conducted in two phases, in the first phase the visible units of the machine are set to the values of the training set. In the next phase the machine is allowed to run freely, independent of the training set. The machine is iteratively switched between these two phases for the duration of the training until it reaches thermal equilibrium. The goal of the training is for the machine to be able to generate the input vector with a high probability of success.\n\n% --- NOTES\n%The difference between the BM and the Hopfield network is mainly that the nodes (or units as they are referred to in the original paper) of the Boltzmann Machine are stochastic by nature.\n%The BM can be used for constraint satisfaction problems that involve a large amount of weak constraints.\n", "meta": {"hexsha": "6628cd0d6a453d6852b8b544ea8669394cc14285", "size": 4876, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/sections/2_models_of_associative_memory/2_boltzmann_machine.tex", "max_stars_repo_name": "mewmew/associative_memories", "max_stars_repo_head_hexsha": "d0c50cf3efbcab9369f5a030125752253539d9e0", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2016-05-30T12:08:22.000Z", "max_stars_repo_stars_event_max_datetime": "2016-05-30T12:08:22.000Z", "max_issues_repo_path": "report/sections/2_models_of_associative_memory/2_boltzmann_machine.tex", "max_issues_repo_name": "mewmew/associative_memory", "max_issues_repo_head_hexsha": "d0c50cf3efbcab9369f5a030125752253539d9e0", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 54, "max_issues_repo_issues_event_min_datetime": "2016-04-04T00:06:16.000Z", "max_issues_repo_issues_event_max_datetime": "2016-06-02T13:32:52.000Z", "max_forks_repo_path": "report/sections/2_models_of_associative_memory/2_boltzmann_machine.tex", "max_forks_repo_name": "mewmew/associative_memory", "max_forks_repo_head_hexsha": "d0c50cf3efbcab9369f5a030125752253539d9e0", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 157.2903225806, "max_line_length": 1203, "alphanum_fraction": 0.8059885152, "num_tokens": 1000, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677737461007, "lm_q2_score": 0.7577943658046608, "lm_q1q2_score": 0.6425852013287164}} {"text": "\\input{preamble}\n\n\\begin{document}\n\n\\header{1}{Volumetric Path Tracing}\n\n\\begin{figure}[h]\n \\includegraphics[width=\\linewidth]{imgs/colored_smoke.png}\n \\caption{A heterogeneous volume with spectrally varying density over space, rendered with multiple-scattering. Smoke data generated using Wenzel Jakob's \\href{http://www.mitsuba-renderer.org/misc.html}{fsolver}.}\n \\label{fig:gallery}\n\\end{figure}\n\nIn this homework, we will build a volumetric path tracers that can handle scattering and absorption inside participating media inside lajolla. We will split the development into 6 steps and build 6 volumetric path tracers, each has more features than the previous ones.\\footnote{This approach is inspired by Steve Marschner's \\href{https://www.cs.cornell.edu/courses/cs6630/2015fa/notes/10volpath.pdf}{course note} on volumetric path tracing.} Your $n$-th volumetric path tracer should be able to render all scenes the $(n-1)$-th one can handle. If you want, you can only submit the final volumetric path tracer and let all the rest call the final code. This process is for helping you to slowly and steadily build up your rendering algorithm.\n\nParticipating media are volumes with many infinitesimal particles absorbing and scattering lights. Given a ray inside the volume parametrized by distance $\\mathbf{p}(t)$, the radiance along the ray is modelled by the \\emph{radiative transfer equation}:\n\\begin{equation}\n\\frac{\\mathrm{d}}{\\mathrm{d}t} L(\\mathbf{p}(t), \\omega) = -(\\sigma_a(\\mathbf{p}(t)) + \\sigma_s(\\mathbf{p}(t))) L(\\mathbf{p}(t), \\omega) + \\sigma_a L_e(\\mathbf{p}(t), \\omega) + \\sigma_s(\\mathbf{p}(t)) \\int_{S^2} \\rho(\\mathbf{p}(t), \\omega, \\omega') L(\\mathbf{p}(t), \\omega') \\mathrm{d}\\omega',\n\\label{eq:rte}\n\\end{equation}\nwhere $L$ is the radiance, $\\sigma_a$ is the \\emph{absorption coefficient}, $\\sigma_s$ is the \\emph{scattering coefficient}, $L_e$ is the (volumetric) emission, $\\rho$ is the \\emph{phase function} that is skin to BSDF in surface rendering, and $S^2$ is the spherical domain.\n\nThis looks a bit scary, so let's break it down. From now on we'll drop the arguments for $\\sigma_a$ and $\\sigma_s$, but in general they can still be spatially varying. The radiative transfer equation is made of four components: \\textbf{absorption}, \\textbf{emission}, \\textbf{in-scattering}, and \\textbf{out-scattering}. Absorption and emission handles particles that absorb and emit lights:\n\\begin{equation}\n\\frac{\\mathrm{d}}{\\mathrm{d}t} L_a(\\mathbf{p}(t), \\omega) = -\\sigma_a L_a(\\mathbf{p}(t), \\omega) + \\sigma_a L_e(\\mathbf{p}(t), \\omega).\n\\end{equation}\nNotice how this is just a simple linear ordinary differential equation $x' = ax + b$, where $\\sigma_a$ attenuates lights and $L_e$ is the gain.\n\nThe in-scattering accounts for all the lights bounces between the particles along the ray, just like the surface rendering equation:\n\\begin{equation}\n\\frac{\\mathrm{d}}{\\mathrm{d}t} L_{is}(\\mathbf{p}(t), \\omega) = \\sigma_s \\int_{S^2} \\rho(\\mathbf{p}(t), \\omega, \\omega') L(\\mathbf{p}(t), \\omega') \\mathrm{d}\\omega'.\n\\end{equation}\n\nHowever, the light does not just bounce \\emph{into} the ray, it also bounces \\emph{out}. That's what the out-scattering considers:\n\\begin{equation}\n\\frac{\\mathrm{d}}{\\mathrm{d}t} L_{os}(\\mathbf{p}(t), \\omega) = -\\sigma_s L_{os}(\\mathbf{p}(t)).\n\\end{equation}\n\nCombining all these three components, we get the full radiative transfer equation (Equation~\\ref{eq:rte}). Notice that the full radiative transfer equation is also like a linear ODE: $-(\\sigma_a + \\sigma_s) L$ attenuates light, and $L_e$ and the spherical integral are the gain that makes things brighter. For this reason, we often let $\\sigma_t = \\sigma_a + \\sigma_s$ and call it the \\emph{extinction coefficient}.\n\nWe'll start from a very simplified version of the radiative transfer equation, then slowly handle more complex situations. To make things simpler, we will throughout assume our medium does not emit light itself: it will receive lighting from other surfaces in the scene.\n\nBefore that, let's introduce lajolla's data structures for storing the participating media.\n\n\\section{Lajolla's participating media data structures and interfaces}\n\\begin{figure}\n \\includegraphics[width=\\linewidth]{imgs/media.pdf}\n \\caption{Lajolla assumes that media are separated by closed surface boundaries. At the object surface, we store the ID of the interior and exterior media (if either of them is vaccum, set the ID to \\lstinline{-1}). The outmost medium is specified at the camera (and can be accessed through \\lstinline{camera.medium_id}). A surface can be \\emph{index-matching} meaning that light just pass through without changing direction or losing energy. In this case, the \\lstinline{material_id} of the surface is set to \\lstinline{-1}. A surface can also be transmissive. In this case, it is assigned a transmissive material like \\lstinline{roughdielectric}.}\n \\label{fig:data_structure}\n\\end{figure}\n\n\\paragraph{The Medium struct in lajolla.} Lajolla's medium interface is for querying the media parameters $\\sigma_a$, $\\sigma_s$, phase function $\\rho$, and the \\emph{majorant} which is the upper bound of the extinction coefficient $\\sigma_t = \\sigma_a + \\sigma_s$ -- we will need the majorant in our final renderer. \n\\begin{lstlisting}[language=c++]\nstruct MediumBase {\n PhaseFunction phase_function;\n};\n\nstruct HomogeneousMedium : public MediumBase {\n Spectrum sigma_a, sigma_s;\n};\n\nstruct HeterogeneousMedium : public MediumBase {\n VolumeSpectrum albedo, density;\n};\n\nusing Medium = std::variant;\n\n/// the maximum of sigma_t = sigma_s + sigma_a over the whole space\nSpectrum get_majorant(const Medium &medium, const Ray &ray);\nSpectrum get_sigma_s(const Medium &medium, const Vector3 &p);\nSpectrum get_sigma_a(const Medium &medium, const Vector3 &p);\n\ninline PhaseFunction get_phase_function(const Medium &medium) {\n return std::visit([&](const auto &m) { return m.phase_function; }, medium);\n}\n\\end{lstlisting}\nYou will need these functions to obtain the necessary quantities in the homeworks.\n\nA \\lstinline{HomogeneousMedium} should be straightforward: it contains constant $\\sigma_a$ and $\\sigma_s$.\nWe will talk more about \\lstinline{HeterogeneousMedium} and \\lstinline{PhaseFunction} later.\n\nLajolla assumes that the media are separated by surface boundaries (Figure~\\ref{fig:data_structure}). It's up to the upstream user to make sure they are consistent to each other and the surfaces are closed (if they are not, the results are undefined).\\footnote{Modern production volume renderers have paid special attention to make sure the renderers can handle all sorts of inputs, including nested volumes. See \\href{here}{https://graphics.pixar.com/library/ProductionVolumeRendering/index.html} for more information.} \n\nIn the scene file, each objects are marked with corresponding exterior and interior media:\n\\begin{lstlisting}[language=xml]\n \n \n \n \n \n\n \n \n \n \n\n \n \n\n \n \n\\end{lstlisting}\n\nThe \\lstinline{Medium}s are stored in \\lstinline{scene.media} which you can access through \\lstinline{scene.media[medium_id]}. The \\lstinline{intersection} routine in lajolla returns a \\lstinline{PathVertex} object which contains relevant information of the intersection:\n\\begin{lstlisting}[language=c++]\nstruct PathVertex {\n Vector3 position;\n Vector3 geometry_normal;\n // ...\n int shape_id = -1;\n int primitive_id = -1; // For triangle meshes. This indicates which triangle it hits.\n int material_id = -1;\n\n // If the path vertex is inside a medium, these two IDs\n // are the same.\n int interior_medium_id = -1;\n int exterior_medium_id = -1;\n};\n\\end{lstlisting}\n\n\\section{Single absorption-only homogeneous volume}\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{imgs/absorption_medium.pdf}\n\\caption{The setup of our first volumetric renderer.}\n\\label{fig:volpath1_illustration}\n\\end{figure}\n\\begin{figure}\n\\caption{Our first volume rendering.}\n\\label{fig:volpath1}\n\\end{figure}\n\nOur first volume renderer will make four assumptions:\n\\begin{itemize}\n \\item There is only a single, homogeneous ($\\sigma_a$ and $\\sigma_s$ are constants over space) volume.\n \\item The volume does not scatter light: $\\sigma_s = 0$.\n \\item The surfaces in the scene only emit lights (with intensity $L_e$) and do not reflect/transmit lights.\n\\end{itemize}\n\nUnder these assumptions, the radiative transfer equation becomes\n\\begin{equation}\n\\frac{\\mathrm{d}}{\\mathrm{d}t} L_a(\\mathbf{p}(t), \\omega) = -\\sigma_a L_a(\\mathbf{p}(t), \\omega),\n\\end{equation}\nand we know $L_a(\\mathbf{p}(t_{\\text{hit}}), \\omega) = L_e$ where $t_{\\text{hit}}$ is the distance between the origin of the ray and the emissive surface. \n\nThis ordinary differential equation has a simple closed form:\n\\begin{equation}\nL_a(\\mathbf{p}(0), \\omega) = \\exp\\left(-\\sigma_a t_{\\text{hit}} \\right) L_e.\n\\end{equation}\nFigure~\\ref{fig:volpath1_illustration} illustrates the setup. In volume rendering the exponential term\n$\\exp\\left(-\\sigma_a t_{\\text{hit}} \\right)$ is often called the ``transmittance''.\n\nOur rendering algorithm is as follows (in Python-style pseudo code).\n\\begin{lstlisting}[language=python]\ndef L(screen_pos):\n camera_ray = sample_primary(camera, screen_pos)\n isect = intersect(scene, camera_ray)\n if isect:\n transmittance = exp(-sigma_a * t)\n Le = 0\n if is_light(isect):\n Le = isect.Le\n return transmittance * Le\n return 0\n\\end{lstlisting}\n\n\\paragraph{Task (15\\%).} You will implement the algorithm above in the \\lstinline{vol_path_tracing_1} function in \\lstinline{vol_path_tracing.h}. You might want to first look at the surface path tracing code in \\lstinline{path_tracing.h} to understand lajolla's API better. \n\n\\end{document}", "meta": {"hexsha": "9e59faa0182dc26362cb2687a8846e67468bd4d5", "size": 10111, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "handouts/homework2.tex", "max_stars_repo_name": "ken2576/lajolla_public", "max_stars_repo_head_hexsha": "332df50f5979afb2c9ccf9455d0f5e094a1f5a1d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "handouts/homework2.tex", "max_issues_repo_name": "ken2576/lajolla_public", "max_issues_repo_head_hexsha": "332df50f5979afb2c9ccf9455d0f5e094a1f5a1d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "handouts/homework2.tex", "max_forks_repo_name": "ken2576/lajolla_public", "max_forks_repo_head_hexsha": "332df50f5979afb2c9ccf9455d0f5e094a1f5a1d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.5449101796, "max_line_length": 743, "alphanum_fraction": 0.7431510236, "num_tokens": 2715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677660619633, "lm_q2_score": 0.7577943658046608, "lm_q1q2_score": 0.6425851955057205}} {"text": "\\section{Introduction}\n\nI was lately thinking about fixed points, more or less.\n\nA new version of \n\\href{https://hackage.haskell.org/package/data-fix}{\\texttt{data-fix}}\nwas released recently, and also corresponding version\nof \\href{https://hackage.haskell.org/package/recursion-schemes}{\\texttt{recursion-schemes}}.\n\nAlso I wrote a \\href{https://www.well-typed.com/blog/2020/06/fix-ing-regular-expressions/}{Fix-ing regular expressions} post,\nabout adding fixed points to regular expression.\n\nThis post is another exploration: \\emph{Fixed points of Indexed functors}.\nThis is not novel idea at all,\nbut I'm positively surprised this works out quite nicely in modern GHC Haskell.\nI define a |IxFix| type and illustrate it with three examples.\n\n\\emph{Note:} The\n\\href{https://hackage.haskell.org/package/multirec-0.7.9/docs/Generics-MultiRec-HFix.html}{|HFix| in \\texttt{multirec} package} is the same as |IxFix|\nin this post. I always forget about the existence of \\texttt{multirec}.\n\nIn the following, the \"modern GHC Haskell\" is quite conservative,\nonly eight extensions:\n\n\\begin{code}\n{-# LANGUAGE DataKinds #-}\n{-# LANGUAGE GADTs #-}\n{-# LANGUAGE PolyKinds #-}\n{-# LANGUAGE RankNTypes #-}\n{-# LANGUAGE ScopedTypeVariables #-}\n{-# LANGUAGE TypeFamilies #-}\n{-# LANGUAGE TypeOperators #-}\n\\end{code}\n\nAnd this literate Haskell script is warning free, with \\texttt{-Wall}\n\\begin{code}\n{-# OPTIONS_GHC -Wall #-}\n\\end{code}\n\nOn this trip\n\\begin{code}\nmodule IxFix where\n\\end{code}\n\nwe need a handful of imports\n\n\\begin{code}\n-- Type should be added to Prelude\nimport Data.Kind (Type)\n\n-- Few newtypes\nimport Data.Functor.Identity (Identity (..))\nimport Data.Functor.Compose (Compose (..))\nimport Data.Functor.Const (Const (..))\n\n-- dependently typed programming!\nimport Data.Fin (Fin)\nimport Data.Type.Nat\nimport Data.Vec.Lazy (Vec (..))\n\n-- magic\nimport Data.Coerce (coerce)\n\\end{code}\n\nBefore we go further, let me remind you about ordinary fixed points,\nas defined in \n\\href{https://hackage.haskell.org/package/data-fix}{\\texttt{data-fix}}\npackage.\n\n\\begin{code}\nnewtype Fix f = Fix { unFix :: f (Fix f) }\n\nfoldFix :: Functor f => (f a -> a) -> Fix f -> a\nfoldFix f = go where go = f . fmap go . unFix\n\\end{code}\n\nUsing |Fix| we can define recursive types using non-recursive \\emph{base functors},\ne.g. for a list we'd have\n\n\\begin{code}\ndata ListF a rec = NilF | ConsF a rec\n\\end{code}\n\nWe then use |foldFix| (or |cata| and other recursion schemes in \\texttt{recursion-schemes}) to \ndecouple \"how we recurse\" and \"what we do at each step\".\nI won't try to convince you why this separation of concerns might be useful.\n\nInstead I continue directly to the topic: define indexed fixed points.\nWhy we need them? Because |Fix| is not powerful enough to allow\nworking with |Vec| or polymorphically recursive types.\n\n\\begin{code}\n-- hello dependently typed world.\ndata Vec (n :: Nat) (a :: Type) where\n VNil :: Vec 'Z a\n (:::) :: a -> Vec n a -> Vec ('S n) a\n\\end{code}\n\n\\section{Fixed points of indexed functors}\n\nBefore talking about fixed points, we need to figure out what\nare indexed functors. Recall a normal functor\nis a thing of kind |Type -> Type|:\n\n\\begin{code}\nclass Functor f where\n fmap :: (a -> b) -> (f a -> f b)\n\\end{code}\n\nIndexed version is the one with |Type| replaced with |k -> Type|,\nfor some index |k| (it is still a functor, but in different category).\nWe want morphisms to work for all indices,\nand preserve them. Thus we define a commonly used type alias\\footnote{I'm sorry that tilde |~>| and dash |->| arrows look so similar.}\n\\begin{code}\n-- natural, or parametric, transformation\ntype f ~> g = forall (j :: k). f j -> g j\n\\end{code}\nUsing it we can define a |Functor| variant\\footnote{\nNote that |FFunctor| in \\url{https://hackage.haskell.org/package/hkd-0.1/docs/Data-HKD.html} (which is defined with different names in other packages as well)\nis of different kind.\n|IxFunctor| in \\url{https://hackage.haskell.org/package/indexed-0.1.3/docs/Data-Functor-Indexed.html}\nis again different.\nSorry for proliferation of various functors.\nAnd for confusing terminology. \nDominic Orchard et al uses terms graded (|k -> Type|, this post) and parameterised (|k -> k -> Type|, \\texttt{indexed}-package) in \\url{https://arxiv.org/abs/2001.10274v2}.\nThere is no monad-name for \\texttt{hkd}-package variant, as that cannot be made into monad-like thing.\n}:\nit looks almost the same.\n\\begin{code}\nclass IxFunctor (f :: (k -> Type) -> (k -> Type)) where\n ixmap :: (a ~> b) -> (f a ~> f b)\n\\end{code}\n\nWith |IxFunctor| in our toolbox, we can define an |IxFix|,\nnote how the definition is again almost the same as for unindexed |Fix| and |foldFix|:\n\\begin{code}\nnewtype IxFix f i = IxFix { unIxFix :: f (IxFix f) i }\n\nfoldIxFix :: IxFunctor f => (f g ~> g) -> IxFix f ~> g\nfoldIxFix alg = alg . ixmap (foldIxFix alg) . unIxFix\n\\end{code}\n\nDoes this work?\nI hope that following examples will convince that |IxFix|\nis usable (at least in theory).\n\n\\section{Example: length indexed lists, Vec}\n\nThe go to example of recursion schemes is a folding a list,\nThe go to example of dependent types is length indexed list, often called |Vec|.\nI combine these traditions by defining |Vec| as an indexed fixed point:\n\n\\begin{code}\ndata VecF (a :: Type) rec (n :: Nat) where\n NilF :: VecF a rec 'Z\n ConsF :: a -> rec n -> VecF a rec ('S n)\n\\end{code}\n\n|VecF| is an |IxFunctor|:\n\n\\begin{code}\ninstance IxFunctor (VecF a) where\n ixmap _ NilF = NilF\n ixmap f (ConsF x xs) = ConsF x (f xs)\n\\end{code}\n\nAnd we can define |Vec| as fixed point of |VecF|, with constructors:\n\n\\begin{code}\ntype Vec' a n = IxFix (VecF a) n\n\nnil :: Vec' a 'Z\nnil = IxFix NilF\n\ncons :: a -> Vec' a n -> Vec' a ('S n)\ncons x xs = IxFix (ConsF x xs)\n\\end{code}\n\nCan we actually use it? Of course!\nLets define concatenation\\footnote{You may wonder why function name is |append|, but operation is concatenation? This is similar to having |plus| for addition.}\n|Vec' a n -> Vec' a m -> Vec' a (Plus n m)|.\nWe cannot use |foldIxFix| directly, as |Plus n m| is not the same\nindex as |n|, so we need to define an auxiliary |newtype|\nto plumb the indices. Another way to think about these kind of |newtype|s,\nis that they work around the lack of type-level anonymous functions in nowadays Haskell.\n\n\\begin{code}\nnewtype Appended m a n =\n Append { getAppended :: Vec' a m -> Vec' a (Plus n m) }\n\\end{code}\n\n\\begin{code}\nappend :: forall a n m. Vec' a n -> Vec' a m -> Vec' a (Plus n m)\nappend xs ys = getAppended (foldIxFix alg xs) ys where\n alg :: VecF a (Appended m a) j -> Appended m a j\n alg NilF = Append id\n alg (ConsF x rec) = Append $ \\zs -> cons x (getAppended rec zs)\n\\end{code}\n\nWe can also define a refold function, which doesn't mention\n|IxFix| at all.\n\n\\begin{code}\nixrefold :: IxFunctor f => (f b ~> b) -> (a ~> f a) -> a ~> b\nixrefold f g = f . ixmap (ixrefold f g) . g\n\\end{code}\n\nAnd then, using |ixrefold| we can define concatenation for |Vec| from\n\\href{https://hackage.haskell.org/package/vec}{\\texttt{vec}} package,\nwhich isn't defined using |IxFix|.\nHere we need auxiliary |newtype|s as well.\n\n\\begin{code}\nnewtype Swapped f a b =\n Swap { getSwapped :: f b a }\nnewtype Appended2 m a n =\n Append2 { getAppended2 :: Vec m a -> Vec (Plus n m) a }\n\nappend2 :: forall a n m. Vec n a -> Vec m a -> Vec (Plus n m) a\nappend2 xs ys = getAppended2 (ixrefold f g (Swap xs)) ys where\n -- same as alg in 'append'\n f :: VecF a (Appended2 m a) j -> Appended2 m a j\n f NilF = Append2 id\n f (ConsF z rec) = Append2 $ \\zs -> z ::: (getAppended2 rec zs)\n\n -- 'project'\n g :: Swapped Vec a j -> VecF a (Swapped Vec a) j\n g (Swap VNil) = NilF\n g (Swap (z ::: zs)) = ConsF z (Swap zs)\n\\end{code}\n\nYou may note that one can implement |append| as induction over length,\nthat's how |vec| implements them.\nTheoretically it is not right, and |IxFix| formulation highlights it:\n\n\\begin{code}\nappend3 :: forall a n m. SNatI n\n => Vec' a n -> Vec' a m -> Vec' a (Plus n m)\nappend3 xs ys = getAppended3 (induction caseZ caseS) xs where\n caseZ :: Appended3 m a 'Z\n caseZ = Append3 (\\_ -> ys)\n\n caseS :: Appended3 m a p -> Appended3 m a ('S p)\n caseS rec = Append3 $ \\(IxFix (ConsF z zs)) ->\n cons z (getAppended3 rec zs)\n\n-- Note: this is different than Appended!\nnewtype Appended3 m a n =\n Append3 { getAppended3 :: Vec' a n -> Vec' a (Plus n m) }\n\\end{code}\n\nHere we \\emph{pattern match} on |IxFix| value.\nIf we want to treat it as least fixed point, the only valid elimination\nis to use |foldIxFix|!\n\nHowever, the induction over length is the right approach if |Vec| is defined\nas a data or type family:\n\n\\begin{code}\ntype family VecFam (a :: Type) (n :: Nat) :: Type where\n VecFam a 'Z = ()\n VecFam a ('S n) = (a, VecFam a n)\n\\end{code}\n\nWhether you want to have data or type-family or GADT depends on the application.\n(Even in Agda or Coq). Family variant doesn't intristically know its length,\nwhich is sometimes a blessing, sometimes a curse.\nFor what it's worth, \\texttt{vec} package provides both variants,\nwith almost the same module interface.\n\n\\section{Example: Polymorphically recursive type}\n\nThe |IxFix| can also be used to define polymorphically recursive types like\n\n\\begin{code}\ndata Nested a = a :<: (Nested [a]) | Epsilon\ninfixr 5 :<:\n\nnested :: Nested Int\nnested = 1 :<: [2,3,4] :<: [[5,6],[7],[8,9]] :<: Epsilon\n\\end{code}\n\nA length function defined over this datatype will be polymorphically recursive,\nas the type of the argument changes from Nested a to Nested [a] in the recursive call:\n\n\\begin{code}\n-- >>> nestedLength nested\n-- 3\nnestedLength :: Nested a -> Int\nnestedLength Epsilon = 0\nnestedLength (_ :<: xs) = 1 + nestedLength xs\n\\end{code}\n\nWe cannot represent |Nested| as |Fix| of some functor,\nand we can not use \\texttt{recursion-schemes} either.\nHowever, we can redefine |Nested| as indexed fixed point.\n\nAn important observation is that we (often or always?) use polymorphic recursion\nas a solution to the lack indexed types.\nMy favorite example is de Bruijn indices for well-scoped terms.\nCompare\n\\begin{code}\ndata Expr1 a\n = Var1 a\n | App1 (Expr1 a) (Expr1 a)\n | Abs1 (Expr1 (Maybe a))\n\\end{code}\nand\n\\begin{code}\ndata Expr2 a n\n = Free2 a -- split free and bound variables\n | Bound2 (Fin n)\n | App2 (Expr2 a n) (Expr2 a n)\n | Abs2 (Expr2 a ('S n)) -- extend bound context by one\n\\end{code}\nWhich one is \\emph{simpler} is a really good discussion,\nbut for another time.\n\nIn |Nested| example the single argument is also used for two\npurposes: the type of an base element (|Int|) and container type\n(starts with |Identity| and increases with extra list layer).\n\nOne approach is just use |Nat| index and have a type family%\n\\footnote{Here one starts to wish that GHC had unsaturated type families, so we wouldn't need to use newtypes...}\n\\begin{code}\ntype family Container (n :: Nat) :: Type -> Type where\n Container 'Z = Identity\n Container ('S n) = Compose [] (Container n)\n\\end{code}\n\nor\n\n\\begin{code}\ndata NestedF a rec f\n = f a :<<: rec (Compose [] f)\n | EpsilonF\n\ninstance IxFunctor (NestedF a) where\n ixmap _ EpsilonF = EpsilonF\n ixmap f (x :<<: xs) = x :<<: f xs\n\\end{code}\n\nWe can convert from |Nested a| to |IxFix (NestedF a) Identity| and back.\nWe use |coerce| to help with |newtype| plumbing.\n\n\\begin{code}\nconvert :: Nested a -> IxFix (NestedF a) Identity\nconvert = aux . coerce where\n aux :: Nested (f a) -> IxFix (NestedF a) f\n aux Epsilon = IxFix EpsilonF\n aux (x :<: xs) = IxFix (x :<<: aux (coerce xs))\n\n-- back left as an exercise\n\\end{code}\n\nAnd then we can write |nestedLength| as a fold.\n\n\\begin{code}\n-- >>> nestedLength2 (convert nested)\n-- 3\nnestedLength2 :: IxFix (NestedF a) f -> Int\nnestedLength2 = getConst . foldIxFix alg where\n alg :: NestedF a (Const Int) ~> Const Int\n alg EpsilonF = Const 0\n alg (_ :<<: Const n) = Const (n + 1)\n\\end{code}\n\n\\section{Non-example: ListF}\n\nIn the introduction I mentioned an ordinary list, which is a fixed point\n\n\\begin{align*}\n \\mathsf{List} \\coloneqq \\lambda (A : \\mathsf{Type}).\\; \\mu (r : \\mathsf{Type}).\\; 1 + A \\times r\n\\end{align*}\n\nwhere I use $\\mu (r : X).\\, F\\, r$ notation to represent least fixed points:\n$\\mu (r : X). F\\, r \\cong F\\, (\\mu (r : X). F\\, r)$.\nNote that we first introduce a type parameter $A$ with $\\lambda$, and then make\na fixed point with $\\mu$.\n\nWe can define an ordirinary list using |IxFix|,\nby taking a fixed point of a |Type -> Type| thing,\ni.e. first $\\mu$, and then $\\lambda$.\n\n\\begin{align*}\n \\mathsf{List}_1 \\coloneqq \\mu (r : \\mathsf{Type} \\to \\mathsf{Type}).\\; \\lambda (A : \\mathsf{Type}).\\; 1 + A \\times r\\,A\n\\end{align*}\n\n\\begin{code}\ndata ListF1 rec a = NilF1 | ConsF1 a (rec a)\ntype List1 = IxFix ListF1\n\nfromList1 :: [a] -> List1 a\nfromList1 [] = IxFix NilF1\nfromList1 (x:xs) = IxFix (ConsF1 x (fromList1 xs))\n\\end{code}\n\nCompare to Agda code:\n\n\\begin{code}\n-- parameter\ndata List (A : Set) : Set where\n nil : List A\n cons : A -> List A -> List A\n\n-- index\ndata List : Set -> Set where\n nil : (A : Set) -> List A\n cons : (A : Set) -> A -> List A -> List A\n\\end{code}\n\nThese types are subtly different.\nSee \\url{https://stackoverflow.com/questions/24600256/difference-between-type-parameters-and-indices}.\n\nThis gives a hint why |Agda| people define |Vec (A : Set) : Nat -> Set|,\ni.e. length as the last parameter: because you have to do that way.\nAnd Haskellers (usually) define as |Vec (n :: Nat) (a :: Type)|,\nbecause then |Vec| can be given |Functor| etc instances.\nIn other words, machinery in both languages forces an order of\nof type arguments.\n\nFinally, we can write parametric version of |List| using\n|IxFix| too. We just use a dummy, boring index.\n\n\\begin{align*}\n \\mathsf{List}_2 \\coloneqq\n \\lambda (A : \\mathsf{Type}).\\;\n \\mu (r : 1 \\to \\mathsf{Type}).\\;\n \\lambda (x : 1).\\;\n 1 + A \\times r\\,x\n\\end{align*}\n\n\\begin{code}\ndata ListF2 a rec (unused :: ()) = NilF2 | ConsF2 a (rec unused)\ntype List2 a = IxFix (ListF2 a) '()\n\nfromList2 :: [a] -> List2 a\nfromList2 [] = IxFix NilF2\nfromList2 (x:xs) = IxFix (ConsF2 x (fromList2 xs))\n\\end{code}\n\n|IxFix| is more general than |Fix|, but if you don't need\nan extra power, maybe you shouldn't use it.\n\nDo we need something even more powerful than |IxFix|?\nI don't think so. If we need more (dependent) indices, we can pack them all into\na single index by tupling (or $\\sum$-mming) them.\n\n\\section{Conclusion}\n\nWe have seen |IxFix|, fixed point of indexed functor.\nI honestly do not think that you should start looking in your code base whether you can use it.\nI suspect it is more useful as thinking and experimentation tool.\nIt is an interesting gadget.\n", "meta": {"hexsha": "fb510218da4050d2e54396e088736189792770ad", "size": 14789, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "posts/2020-08-28-indexed-fixpoint.tex", "max_stars_repo_name": "phadej/gists", "max_stars_repo_head_hexsha": "cefb69faa61dd5737e91532f046278b7327aee91", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2018-03-31T19:03:45.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-28T16:49:57.000Z", "max_issues_repo_path": "posts/2020-08-28-indexed-fixpoint.tex", "max_issues_repo_name": "phadej/gists", "max_issues_repo_head_hexsha": "cefb69faa61dd5737e91532f046278b7327aee91", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-12-19T16:33:09.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-19T16:33:09.000Z", "max_forks_repo_path": "posts/2020-08-28-indexed-fixpoint.tex", "max_forks_repo_name": "phadej/gists", "max_forks_repo_head_hexsha": "cefb69faa61dd5737e91532f046278b7327aee91", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-07-31T07:50:57.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-19T14:05:45.000Z", "avg_line_length": 32.5032967033, "max_line_length": 172, "alphanum_fraction": 0.6831428765, "num_tokens": 4545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8479677545357568, "lm_q2_score": 0.7577943658046609, "lm_q1q2_score": 0.6425851867712261}} {"text": "\\section{Preliminaries} \n\\label{sec:Prelim} \n\nOur proposal, \\dyntset, relies mainly on three structures, a \\textit{monoid}, a \\textit{set} and a \\textit{finger tree}, which are briefly described in the following subsections. \n\n\\subsection{Monoid}\n\nA \\textit{monoid} is a triple $(S,\\star,e)$, where $S$ is a set, $\\star$ is a binary operation, called $product$ and $e$ is an element of $S$, called $unit$, satisfying the following properties:\n\n\\begin{enumerate}\n\\item $e \\star x = x = x \\star e$, for all $x \\in S$ \n\\item $x \\star (y \\star z) = (x \\star y) \\star z $, for all $x,y,z \\in S$. \n\\end{enumerate}\n\nThe type class Haskell implementation of a monoid we have:\n%\\begin{verbatim}\n\\begin{lstlisting}[mathescape]\nclass Monoid a where \n mempty :: a\n mappend :: a $\\to$ a $\\to$ a\n\\end{lstlisting} \n%\\end{verbatim}\n\nwhere the function \\code{mempty} represents the element $e$ and the function \\code{mappend} represents function $\\star$. Detailed information about monoids within the functional programming can be found in \\cite{Monoids}, and for the Haskell implementation at \\cite{HaskellMonoid}.\n\n\n\\subsection{Set as binary search tree} \n\nA set, as a data structure, is used here for searching purposes, seen as testing membership within a specific set. Also, we look for incorporating elements into a set in order to increase such set for further look ups. This incorporation, comes in two flavours, either as single insertion or as the set-union operation. Finally, the presence of an empty set, a set with no elements at all, is crucial since it is the $unit$ when a set behaves as monoid. Also, an empty set is a starting point when a forest contains no trees.\n\nInternally, a set is implemented as binary search tree, also called BST. A BST is either a \\textit{leaf} (also called a \\code{tip}) or a vertex consisting of a \\textit{value}, a \\textit{left} BST and a \\textit{right} BST. The height of the tree determines the time taken to perform every operation onto it, therefore the shorter the height the better, and this is done throughout a \\textit{balancing scheme}. The study of BSTs is vast and there are several implementation around. In our case, a \\textit{size balanced} and a \\textit{pairing hashed} are used. A detailed comparison and benchmarking is outside the scope of the present work.\n\nThe following is an snippet of \\code{Data.Set} \\footnote{https://hackage.haskell.org/package/containers-0.5.10.2/docs/Data-Set.html}\n%\\begin{verbatim}\n\\begin{lstlisting}\ndata Set a = Bin Size a (Set a) (Set a)\n | Tip\n\\end{lstlisting} \n%\\end{verbatim}\n\nIn Table~\\ref{tab:Setfuncs} we show the set-functions we have incorporated in \\code{dynTsET} from \\code{Data.Set}.\n\\small\n\\begin{table}[H]\n\\begin{center}\n\\begin{tabular}{||l | l | c||} \n \\hline\n Function & Type & Time complexity \\\\ \n \\hline\\hline\n \\texttt{empty} & \\texttt{Set a} & $O(1)$ \\\\ \n \\hline\n \\texttt{insert} & \\texttt{a $\\to$ Set a $\\to$ Set a} & $O(\\log n)$ \\\\\n \\hline\n \\texttt{member} & \\texttt{a $\\to$ Set a $\\to$ Bool} & $O(\\log n)$ \\\\ \n \\hline\n \\texttt{union} & \\texttt{Set a $\\to$ Set a $\\to$ Set a} & $O(m(\\log\\frac{n}{m} +1))$ \\\\\n \\hline\n\\end{tabular}\n\\caption{Leijen's implementation of \\code{Data.Set} \\cite{HaskellSet}, based on \\cite{ParallelSets}}\n\\label{tab:Setfuncs} \n\\end{center}\nwhere $n$ and $m$ are the sizes of the largest and smallest sets respectively\n\\end{table}\n\\normalsize\n\nFunctions \\code{empty} and \\code{union} are, in fact, the monoidal functions \\code{mempty} and \\code{mappend} respectively. \n\n\\subsection{Euler-tour tree} \n\\label{sec:ETt}\nDealing with trees of different degree can be complicated. A simple way to handle and represent trees of any degree is by an Euler tour, that is, a sequence as in \\cite{Rand-DynGs-Algos} and \\cite{WerneckR-PhD}. To represent a tree $t$, we replace every edge $\\langle u,v \\rangle$ of $t$ by two arcs $(u,v)$ and $(v,u)$, and add a loop $(v,v)$ to represent each vertex $v$. In this context, a tree $t$ can have at least one, and in general, many Euler tours. The size of an Euler tour \\textit{et} of $t$ is $et(t) = v + 2e $, where $v$ is the number of vertices of $t$ and $e$ its number of edges. We can represent an Euler tour in Haskell simply as a list of pairs, such in \n\n%\\begin{verbatim}\n\\begin{lstlisting}\ndata EulerTour a = [(a,a)] \n\\end{lstlisting}\n%\\end{verbatim}\n\nBy managing the tour with lists, we can perform insertion from the left (head) in $O(1)$ but remaining operations such insertion from the right, access, cutting, appending and inserting might take $O(n)$ per operation.\nOn the other hand, representing tours through finger trees, performance per operation is improved up to $O(\\log n)$ per operation amortised as we explain shortly and in Section~\\ref{sec:TechDes}. \n\nOur representation of $k$-trees through Euler-tour will not close up the tour with the first node as this avoids the uniqueness presence for such a node, as shown in Fig.~\\ref{fig:Euler-tour}.\n\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[scale=0.35]{./Images/Euler-tour} \n\\end{center}\n\\caption{The $k$-tree (left) is represented as a sequence of size $v+2e$ (right). Notice we left out the final node-pair to preserve uniqueness}\n\\label{fig:Euler-tour}\n\\end{figure}\n\n\n\\subsection{Finger tree} \n\nWe present Hinze and Paterson's version of finger trees \\madd{(FTs)} \\cite{FTs}. A finger tree is a 2-3-4 tree data structure which is always balanced. The structure is designed to store the elements of a sequence in the terminal nodes or leaves while intermediate nodes are dedicated for monoidal annotations. These annotations help out to achieve a specific task when storing or retrieving values in the leaves or when dealing with the tree itself such concatenation or splitting. One instance is random-access on lists. In order to speed up the time taken by a list for accessing the $n$th element, a finger tree stores the list elements on its leaves and the type \\code{Sum Int} as its monoidal annotation. In this way, we can ask for a specific element provided its index in $O(\\log n)$ rather than $O(n)$ time. This done through monoidal instances for \\code{Sum}. The identity is defined as zero as in \n\\begin{lstlisting}[mathescape]\ninstance Num a $\\Rightarrow$ Monoid (Sum a) where\n mempty = Sum 0\n\\end{lstlisting}\n\nand the corresponding binary operation on integers which preserves its identity (arithmetic addition),\n\\begin{lstlisting}[mathescape]\ninstance Num a $\\Rightarrow$ Semigroup (Sum a) where\n (<>) = coerce ((+) :: a $\\to$ a $\\to$ a)\n\\end{lstlisting}\n\n\nThe finger tree data type in Haskell is defined as\n\n\\begin{figure}[H]\n\\begin{lstlisting}\ndata FingerTree v a = Empty\n | Single a \n | Deep v \n Digit a \n FingerTree v (Node v a) \n Digit a\n\\end{lstlisting} \n\\caption{Data type of the finger tree by Hinze and Paterson}\n\\label{fig:FTdatatype}\n\\end{figure}\n\n\\code{Digit} type holds from one up to four elements of type \\code{a}. \\code{Node} type can hold two or three elements of type \\code{a}. The recursive and nested definition of \\code{FingerTree} forces the structure to be balanced by its types, instead of enforcing it by code invariants. In our case, the leaves of a finger tree stores pairs of nodes or vertices representing an Euler tour. To implement \\textit{updates} and \\textit{lookups} efficiently, Hinze and Paterson \\cite{FTs} added a monoidal annotation on the intermediate vertices, the \\code {v} type.\n\nIn Fig.~\\ref{fig:FT-Euler-tour} we can see our example of the Euler-tour sequence in Fig.~\\ref{fig:Euler-tour} managed by the data constructors described above.\n\n\\begin{figure}[H]\n\\begin{center}\n\\includegraphics[scale=0.35]{./Images/FT-Euler-tour} \n\\end{center}\n\\caption{A finger tree (FT) holding the sequence (Euler-tour) that represents the $k$-tree in Fig.\\ref{fig:Euler-tour}}\n\\label{fig:FT-Euler-tour}\n\\end{figure}\n\nSince the backbone of our structure \\dyntset is actually a Hinze's and Paterson finger tree \\cite{FTs}, we show the functions involved in our proposal, those to deal with inserting, cutting, appending and accessing finger trees. There are plenty of additional functions from finger trees we do not cover and that can be reached at \\url{http://hackage.haskell.org/package/fingertree-0.1.4.1/docs/Data-FingerTree.html}.\n\n\\small\n\\begin{table}[H]\n\\begin{center}\n\\begin{tabular}{||l | l | c||} \n \\hline\n Function & Type & \n \\begin{tabular}{c} Time complexity \\\\ \n (amortised bounds)\n \\end{tabular} \\\\\n \\hline\\hline\n \\texttt{viewl} & \n \\begin{tabular}{l} \\texttt{FingerTree v a} \\\\ \n \\texttt{$\\to$ ViewL (FingerTree v a) } \n \\end{tabular}\n & $O(1)$ \\\\ \n \\hline\n \\texttt{search} & \n \\begin{tabular}{l} \\texttt{(v $\\to$ v $\\to$ Bool)} \\\\ \n \\texttt{$\\to$ FingerTree v a } \\\\ \n \\texttt{$\\to$ SearchResult v a } \n \\end{tabular}\n & $O(\\log(min(i,n-i)))$ \\\\ \n \\hline\n \\texttt{$\\lhd$} (ins. from left) & \n \\begin{tabular}{l} \\texttt{a $\\to$ FingerTree v a} \\\\\n \\texttt{$\\to$ FingerTree v a} \n \\end{tabular}\n & $O(1)$ \\\\\n \\hline\n \\texttt{$\\rhd$} (ins. from right) & \n \\begin{tabular}{l} \\texttt{FingerTree v a $\\to$ a} \\\\\n \\texttt{$\\to$ FingerTree v a} \n \\end{tabular}\n & $O(1)$ \\\\\n \\hline\n \\texttt{$\\bowtie$} (concatenation) & \n \\begin{tabular}{l} \\texttt{FingerTree v a} \\\\\n \\texttt{$\\to$ FingerTree v a} \\\\\n \\texttt{$\\to$ FingerTree v a} \n \\end{tabular}\n & $O(\\log ( min(m,n) ))$ \\\\\n \\hline\n\\end{tabular}\n\\caption{Hinze's and Paterson implementation of \\code{Data.FingerTree} \\cite{FTsURL}, based on \\cite{FTs}}\n\\label{tab:FTfuncs} \n\\end{center}\n\\end{table}\n\\normalsize\n\nwhere $n$ is the size of the largest sequence, $m$ is the size of the smaller sequence (i.e. when concatenating) and $i$ is the index of the element when searching. \n\n\\tcb{SPLIT is not longer used, instead there is SEARCHFT, which aim is the same as SPLIT and hope it will be easier to explain}\n\n\\tcb{Is it the ``best\" part in the document to bring up an explanations or examples or both for amortisation?}", "meta": {"hexsha": "f9b6a5ba93c7390cda6c81895529856fe840b4ab", "size": 10595, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pub/Prelim/Prelim.tex", "max_stars_repo_name": "jcsaenzcarrasco/ETdynTs", "max_stars_repo_head_hexsha": "4bf251c1d6a7ac4de916254e7efd2083aa9f9ad8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "pub/Prelim/Prelim.tex", "max_issues_repo_name": "jcsaenzcarrasco/ETdynTs", "max_issues_repo_head_hexsha": "4bf251c1d6a7ac4de916254e7efd2083aa9f9ad8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pub/Prelim/Prelim.tex", "max_forks_repo_name": "jcsaenzcarrasco/ETdynTs", "max_forks_repo_head_hexsha": "4bf251c1d6a7ac4de916254e7efd2083aa9f9ad8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 56.3563829787, "max_line_length": 908, "alphanum_fraction": 0.6750353941, "num_tokens": 2958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8354835411997897, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.6425538799346258}} {"text": "\\chapter{Algorithms for Multi-Dimensional Optimization}\n\\label{sec:algImp}\n\\section{Generalized Pattern Search Methods (Analysis)}\nGeneralized Pattern Search (GPS) algorithms are \nderivative free optimization algorithms for the minimization of\nproblem $\\mathbf P_c$ and $\\mathbf P_{cg}$, defined in~\\eqref{sub:Proc}\nand~\\eqref{sub:Procg}, respectively.\nWe will present the GPS algorithms for the case where \nthe function $f(\\cdot)$ cannot be evaluated exactly,\nbut can be approximated by functions $f^* \\colon \\Re_+^q \\times \\Re^n \\to \\Re$, \nwhere the first argument $\\epsilon \\in \\Re_+^q$ is \nthe precision parameter of PDE, ODE, and algebraic equation solvers.\nObviously, the explanations are similar for problems where\n$f(\\cdot)$ can be evaluated exactly, except that the scheme to\ncontrol $\\epsilon$ is not applicable, and that\nthe approximate functions $f^*(\\epsilon,\\cdot)$ are replaced by $f(\\cdot)$.\\\\\n\nUnder the assumption that the cost function is continuously\ndifferentiable, all the accumulation points constructed by\nthe GPS algorithms are stationary.\\\\\n\nWhat GPS algorithms have in common is that they define the construction\nof a mesh $\\mathbb M_k$ in $\\Re^n$, \nwhich is then explored according to some rules\nthat differ among the various members of the family of GPS algorithms.\nIf no decrease in cost is obtained on mesh points around the current iterate,\nthen the distance between the mesh points is reduced,\nand the process is repeated.\\\\\n\nWe will now explain the framework of GPS algorithms that will be used\nto implement different instances of GPS algorithms in GenOpt.\nThe discussion follows the more detailed description of~\\cite{PolakWetter2003:1}.\\\\\n\\pagebreak\n\n\\subsection{Assumptions}\nWe will assume that $f(\\cdot)$ and its \napproximating functions $\\{ f^*(\\epsilon,\\cdot) \\}_{\\epsilon \\in \\Re_+^q}$ \nhave the following properties.\n\n\\begin{assumption} ~\\\\\n\\\\[-2.5\\baselineskip]\n\\begin{enumerate}\n\\item There exists an error bound function $\\varphi \\colon \\Re_+^q\n \\to \\Re_+$ such that for any bounded set $\\mathbf S \\subset \\mathbf\n X$, there exists an $\\epsilon_{\\mathbf S} \\in \\Re_+^q$ and a scalar \n $K_{\\mathbf S} \\in\n (0, \\, \\infty)$ such that for all $x \\in \\mathbf S$ and for all\n $\\epsilon \\in \\Re_+^q$, \nwith $\\epsilon \\le \\epsilon_{\\mathbf S}$,\\footnote{For \n$\\epsilon \\in \\Re_+^q$, by $\\epsilon \\le \n\\epsilon_{\\mathbf S}$, \nwe mean that $0 < \\epsilon^i \\le \\epsilon_{\\mathbf S}^i$, for all \n$i \\in \\{ 1, \\ldots ,q \\}$.}\n \\begin{equation}\n | \\, f^*(\\epsilon,x) - f(x) | \\le K_{\\mathbf S} \\, \\varphi(\\epsilon).\n \\lab{eq:fefPhi}\n \\end{equation}\nFurthermore, \n\\begin{equation}\n\\lim_{\\| \\epsilon \\| \\to 0} \\varphi(\\epsilon) = 0.\n\\end{equation}\n\\item The function $f \\colon \\Re^n \\to \\Re$ is once continuously differentiable.\n\\rbox\n\\end{enumerate}\n\\lab{as:FFeps}\n\\end{assumption}\n\n\\begin{remark}~\\\\\n\\\\[-2.5\\baselineskip] {\\em\n\\begin{enumerate}\n\\item\nThe functions $\\{ f^*(\\epsilon,\\cdot) \\}_{\\epsilon \\in \\Re_+^q}$ may be\ndiscontinuous.\n\\item\nSee~\\cite{PolakWetter2003:1} for the situation where $f(\\cdot)$ is\nonly locally Lipschitz continuous.\n\\rbox\n\\end{enumerate}\n}\n\\end{remark}\n\n%--------------------------------------------\nNext, we state an assumption on the level sets of the family of\napproximate functions. To do so, we first define the notion of a level \nset.\n\n\\begin{definition}[Level Set]\nGiven a function $f \\colon \\Re^n \\to \\Re$ and an $\\alpha \\in\n\\Re$, such that $\\alpha > \\inf_{x \\in \\Re^n} f(x)$, we will say that\nthe set $\\mathbf L_\\alpha (f) \\subset \\Re^n$, defined as\n\\begin{equation}\n \\mathbf L_\\alpha(f) \\triangleq \\{ x \\in \\Re^n \\ | \\ f(x) \\le \\alpha \\}, \n\\end{equation}\n is a {\\em level set} of $f(\\cdot)$, parametrized by $\\alpha$.\n\\rbox\n\\end{definition}\n\n\\begin{assumption}[Compactness of Level Sets]\nLet $\\{ f^*(\\epsilon,\\cdot) \\}_{\\epsilon \\in \\Re_+^q}$ be as in \nAssumption~\\ref{as:FFeps} and let $\\mathbf X \\subset \\Re^n$ be the \nconstraint set. Let $x_0 \\in \\mathbf X$ be the \ninitial iterate and $\\epsilon_0 \\in \\Re_+^q$ be the\ninitial precision setting of the numerical solvers.\nThen, we assume that there exists a\ncompact set $\\mathbf C \\subset \\Re^n$ such that\n\\begin{equation}\n \\mathbf L_{f^*(\\epsilon_0,x_0)} (f^*(\\epsilon,\\cdot)) \\cap \\mathbf X\n \\subset \\mathbf C, \\qquad \\forall \\, \\epsilon \\le \\epsilon_0.\n\\end{equation}\n\\rbox\n\\lab{as:LevSetBou}\n\\end{assumption}\n\n% ===============================================\n\\subsection{Characterization of GPS Algorithms}\nThere exist different geometrical explanations for pattern search algorithms, and \na generalization is given in the review~\\cite{KoldaLewisTorczon2003:1}.\nWe will use a simple implementation of the pattern search algorithms\nin~\\cite{PolakWetter2003:1} where we restrict the search directions to\nbe the positive and negative coordinate directions.\nThus, the search directions are the columns of the matrix\n\\begin{equation}\n D \\triangleq [-e_1, \\, +e_1, \\, \\ldots \\, , -e_n, \\, +e_n] \n \\in \\mathbb Z^{n \\times 2n},\n\\end{equation}\nwhich suffices for box-constrained problems. Furthermore, we\nconstruct the sequence of mesh size parameters that parametrizes\nthe minimum distance between iterates\nsuch that it satisfies the following assumption.\n\\begin{subequations}\n\\begin{assumption}[$k$-th Mesh Size Parameter]\nLet $r, s_0, k \\in \\mathbb N$, with $r > 1$,\nand $\\{ t_i \\}_{i=0}^{k-1} \\subset \\mathbb N$.\nWe will assume that the sequence of mesh size parameters satisfies\n\\begin{equation}\n\\Delta_k \\triangleq \\frac {1}{r^{s_k}},\n\\label{eq:GPSDelDef}\n\\end{equation}\nwhere for $k > 0$\n\\begin{equation}\ns_k \\triangleq s_0 + \\sum_{i=0}^{k-1} t_i.\n\\label{eq:GPSDelDefSk}\n\\end{equation}\n\\lab{ass:MeshSizePara}\n\\rbox\n\\end{assumption}\n\\end{subequations}\nWith this construction, all iterates lie on a rational mesh of the form\n\\begin{equation}\n \\mathbb M_k \\triangleq \\{ x_0 + \\Delta_k \\, D \\, m \\ | \\ m \\in \\mathbb N^{2 n} \\}.\n\\lab{eq:Mesh} \n\\end{equation}\n\n\n% -------- Global and Local Search Set ---------\nWe will now characterize the set-valued maps that determine the mesh\npoints for the ``global'' and ``local'' searches. Note that the images\nof these maps may depend on the entire history of the computation.\n\n\\begin{subequations}\n\\begin{definition}\nLet $\\underline{\\mathbf X}_k \\subset \\Re^n$ and\n$\\underline {\\boldsymbol \\Delta}_k \\subset \\mathbb Q_+$\nbe the sets of all sequences containing $k+1$ elements, \nlet $\\mathbb M_k$ be the current mesh,\nand let $\\epsilon \\in \\Re_+^q$ be the solver tolerance.\n\\begin{enumerate}\n\\item\nWe define the {\\em global search set map} \nto be any set-valued map\n\\begin{equation}\n \\gamma_k \\colon \\underline{\\mathbf X}_k \\times \\underline\n {\\boldsymbol \\Delta}_k \\times \\Re_+^q \\to \n \\bigl( 2^{\\mathbb M_k} \\cap \\mathbf X \\bigr) \\cup \\emptyset\n\\end{equation}\nwhose image \n$\\gamma_k(\\underline x_k, \\underline \\Delta_k, \\epsilon)$\ncontains only a finite number of mesh points.\n\\item\nWe will call $\\mathcal G_k \\triangleq \n\\gamma_k (\\underline x_k, \\underline \\Delta_k, \\epsilon)$ \nthe {\\em global search set}.\n\\item\nWe define the directions for the local search as\n\\begin{equation}\n D \\triangleq [-e_1, \\, +e_1, \\, \\ldots \\, , -e_n, \\, +e_n]. \n\\end{equation}\n\\item\nWe will call\n\\begin{equation}\n \\mathcal L_k \\triangleq \\bigl \\{ x_k + \\Delta_k \\, D \\, e_i \\ | \\ \n i \\in \\{ 1, \\ldots , \\, 2 \\, n \\} \\bigr \\}\n \\cap \\mathbf X\n\\lab{eq:defLK}\n\\end{equation}\nthe {\\em local search set}.\n\\rbox\n\\end{enumerate}\n\\lab{def:algFunGamDel}\n\\end{definition}\n\\lab{sub:mapfp}\n\\end{subequations}\n\n\\vspace{-\\baselineskip}\n\\begin{remark}~\\\\\n\\\\[-2.5\\baselineskip] {\\em\n\\begin{enumerate}\n\\item\nThe map $\\gamma_k(\\cdot, \\cdot, \\cdot)$ \ncan be dynamic in the sense that if $\\{ x_{k_i} \\}_{i=0}^I \\triangleq \n\\gamma_k(\\underline x_k, \\underline \\Delta_k, \\epsilon)$, \nthen the rule for selecting\n$x_{k_{\\widehat i}}$, $1 \\le \\widehat i \\le I$, can depend on \n$\\{ x_{k_i} \\}_{i=0}^{\\widehat i - 1}$ and\n$\\{ f^*(\\epsilon, x_{k_i}) \\}_{i=0}^{\\widehat i - 1}$. \nIt is only important that the global search terminates after \na finite number of computations, and that \n$\\mathcal G_k \\subset (2^{\\mathbb M_k} \\cap \\mathbf X) \\cup \\emptyset$.\n\\item\nAs we shall see, the global search affects only the efficiency of the\nalgorithm but not its convergence properties. Any heuristic procedure\nthat leads to a finite number of function evaluations can be used\nfor $\\gamma_k(\\cdot, \\cdot, \\cdot)$. \n\\item\nThe empty set is included in the range of $\\gamma_k(\\cdot, \\cdot, \\cdot)$ \nto allow omitting the global search.\n\\end{enumerate}\n\\rbox\n}\n\\end{remark}\n\n% ===========================================\n\\subsection{Model Adaptive Precision GPS Algorithm}\nWe will now present our model GPS algorithm\nwith adaptive precision cost function evaluations.\\\\\n\n\\noindent\n\\begin{minipage}[b]{\\textwidth}\n\\begin{algorithm}\n[Model GPS Algorithm]\n~\\\\\n{\\em\n\\begin{tabular}{ll}\n\\multicolumn{2}{l}{\\hspace{\\textwidth}~} \\\\ \\\\[-8ex]\\\\\n\\hline \\\\[-2ex]\n \\textbf{Data}:\n & Initial iterate $x_0 \\in \\mathbf X$;\\\\ \n & Mesh size divider $r \\in \\mathbb N$, with $r > 1$;\\\\\n & Initial mesh size exponent $s_0 \\in \\mathbb N$.\\\\\n \\textbf{Maps}:\n & Global search set map \n $\\gamma_k \\colon \\underline{\\mathbf X_k} \n \\times \\underline\n {\\boldsymbol \\Delta}_k \\times \\Re_+^q \\to \\bigl( \n 2^{\\mathbb M_k} \\cap \\mathbf X \\bigr) \\cup \\emptyset$; \\\\ \n & Function $\\rho \\colon \\Re_+ \\to \\Re_+^q$ (to assign $\\epsilon$),\n such that the composition \\\\\n & $\\varphi \\circ \\rho \\colon \\Re_+ \\to \\Re_+$\n is strictly monotone decreasing and satisfies\\\\\n & $\\varphi(\\rho(\\Delta))/\\Delta \\to 0$, as $\\Delta \\to 0$.\\\\\n \\textbf{Step 0}: \n & Initialize $k=0$, $\\Delta_0 = 1 / r^{s_0}$, and $\\epsilon=\\rho(1)$.\\\\\n \\textbf{Step 1}:\n & \\underline{Global Search}\\\\\n & Construct the global search set\n $\\mathcal G_k = \\gamma_k(\\underline x_k, \\underline \\Delta_k,\n \\epsilon)$.\\\\\n & If $f^*(\\epsilon, x') - f^*(\\epsilon, x_k) < 0$\n for any $x' \\in \\mathcal G_k$, go to Step 3;\\\\\n & else, go to Step 2.\\\\\n \\textbf{Step 2}:\n & \\underline{Local Search}\\\\\n & Evaluate $f^*(\\epsilon, \\cdot)$ for any $x' \\in \\mathcal L_k$ until\n some $x' \\in \\mathcal L_k$\\\\\n & satisfying \n $f^*(\\epsilon, x') - f^*(\\epsilon, x_k) < 0$\n is obtained, \n or until all points\\\\\n & in $\\mathcal L_k$ are evaluated.\\\\\n \\textbf{Step 3}:\n & \\underline{Parameter Update}\\\\\n & If there exists an $x' \\in \\mathcal G_k \\cup \\mathcal L_k$\n satisfying\n $f^*(\\epsilon, x') - f^*(\\epsilon, x_k) < 0$,\\\\\n & set $x_{k+1} = x'$, $s_{k+1}= s_k$, $\\Delta_{k+1} = \\Delta_k$,\n and do not change $\\epsilon$;\\\\\n & else, set $x_{k+1} = x_k$, $s_{k+1} = s_k + t_k$, with $t_k \\in\n \\mathbb N_+$ arbitrary,\\\\\n & $\\Delta_{k+1} = 1/r^{s_{k+1}}$, \n $\\epsilon = \\rho(\\Delta_{k+1} / \\Delta_0)$.\\\\\n \\textbf{Step 4}:\n & Replace $k$ by $k + 1$, and go to Step 1.\\\\\n \\hline \\\\\n\\end{tabular}\n}\n~\\\\ \\label{al:GPSImp}\n\\end{algorithm}\n\\end{minipage}\n% ===========================================\n\\newpage\n\\begin{remark}~\\newline \n\\vspace{-1.5\\baselineskip}\n{ \\em \n\\begin{enumerate} \n\\item\nTo ensure that $\\epsilon$ does not depend on the scaling of $\\Delta_0$,\nwe normalized the argument of $\\rho(\\cdot)$.\nIn particular, we want to decouple $\\epsilon$ from\nthe user's choice of the initial mesh parameter.\n\\item\nIn Step 2, once a decrease of the cost function is obtained, one\ncan proceed to Step 3. However, one is allowed to evaluate $f^*(\\epsilon,\\cdot)$\nat more points in $\\mathcal L_k$ in an\nattempt to obtain a bigger reduction in cost. However,\none is allowed to proceed to Step 3 only after \neither a cost decrease has been found, or after {\\em all} points in \n$\\mathcal L_k$ are tested.\n\\item\nIn Step 3, we are not restricted to accepting the\n$x' \\in \\mathcal G_k \\cup \\mathcal L_k$ that gives lowest cost value.\nBut the mesh size parameter $\\Delta_k$ is reduced {\\em only}\nif there exists no $x' \\in \\mathcal G_k \\cup \\mathcal L_k$ satisfying\n$f^*(\\epsilon, x') - f^*(\\epsilon, x_k) < 0$.\n\\item\nTo simplify the explanations, we do\nnot increase the mesh size parameter if the cost has been reduced.\nHowever, our global search allows searching on a coarser mesh \n$\\widehat {\\mathbb M} \\subset \\mathbb M_k$,\nand hence, our algorithm can easily be extended to include a rule for increasing\n$\\Delta_k$ for a finite number of iterations.\n\\item\nAudet and Dennis~\\cite{AudetDennis2003} update the mesh size parameter \nusing the formula $\\Delta_{k+1} = \\tau^m \\, \\Delta_k$, \nwhere $\\tau \\in \\mathbb Q$, $\\tau > 1$, and $m$ is any\nelement of $\\mathbb Z$. Thus, our update rule for $\\Delta_k$ is a special case of\nAudet's and Dennis' construction since we set $\\tau = 1/r$, with\n$r \\in \\mathbb N_+$, $r \\ge 2$ (so that $\\tau < 1$) and $m \\in \n\\mathbb N$. We prefer our construction because we do not think it negatively affects\nthe computing performance, but it leads to simpler convergence proofs.\n\\rbox\n\\end{enumerate}\n}\n\\end{remark}\n\n% ===========================================\n\\subsection{Convergence Results}\nWe will now present the convergence results for our Model GPS algorithm.\nSee~\\cite{PolakWetter2003:1} for a detailed discussion and convergence proofs.\n\n\\subsubsection{Unconstrained Minimization}\n\nWe will first present the convergence properties of the Model GPS\nAlgorithm~\\ref{al:GPSImp} on unconstrained minimization problems, i.e., for \n$\\mathbf X = \\Re^n$.\\\\\n\nFirst, we will need the notion of a {\\em refining subsequence},\nwhich we define as follows:\n\\begin{definition}[Refining Subsequence]\nConsider a sequence $\\{x_k\\}_{k=0}^\\infty$ constructed by Model GPS\nAlgorithm~\\ref{al:GPSImp}. We will say that the subsequence $\\{ x_k \\}_{k \\in \n\\mathbf K}$ is the {\\em refining subsequence}, if $\\Delta_{k+1} < \n\\Delta_k$ for all $k \\in \\mathbf K$, and $\\Delta_{k+1} = \\Delta_k$\nfor all $k \\notin \\mathbf K$.\n\\rbox\n\\end{definition}\n\nWe now state that pattern search algorithms with adaptive precision\nfunction evaluations construct sequences with stationary accumulation points.\n\\begin{theorem}[Convergence to a Stationary Point]\nSuppose that Assumptions~\\ref{as:FFeps} and \\ref{as:LevSetBou} are\nsatisfied and that $\\mathbf X = \\Re^n$.\nLet $x^* \\in \\Re^n$ be an accumulation point of the refining \nsubsequence $\\{ x_k \\}_{k \\in \\mathbf K}$, constructed by Model GPS\nAlgorithm~\\ref{al:GPSImp}. Then,\n\\begin{equation}\n \\nabla f(x^*) = 0.\n\\lab{eq:GradConv}\n\\end{equation}\n\\lab{the:GradConv}\n\\rbox\n\\end{theorem}\n% ===========================================\n\\subsubsection{Box-Constrained Minimization}\nWe now present the convergence results for the box-constrained\nproblem~\\eqref{sub:Proc}.\nSee~\\cite{AudetDennis2003, PolakWetter2003:1, KoldaLewisTorczon2003:1}\nfor the more general case of linearly-constrained problems and for the convergence\nproofs.\\\\\n\nFirst, we introduce the notion of a tangent cone and a normal cone,\nwhich are defined as follows: \n\\begin{definition}[Tangent and Normal Cone] ~\\\\\n\\\\[-2.5\\baselineskip]\n\\begin{enumerate}\n\\item\nLet $\\mathbf X \\subset \\Re^n$. Then, we define the {\\em tangent cone} to\n$\\mathbf X$ at a point $x^* \\in \\mathbf X$ by\n\\begin{subequations}\n \\begin{equation}\n \\mathbf T_{\\mathbf X}(x^*) \\triangleq \n \\overline{ \\{ \\mu \\, (x - x^*) \\ | \\ \\mu \\ge 0, \n \\, x \\in \\mathbf X \\}}.\n \\end{equation}\n\\item\nLet $\\mathbf T_{\\mathbf X}(x^*)$ be as above. Then, we define the \n{\\em normal cone} to $\\mathbf X$ at $x^* \\in \\mathbf X$ by\n\\begin{equation}\n \\mathbf N_{\\mathbf X}(x^*) \\triangleq \\{ v \\in \\Re^n \\ | \\ \n \\forall \\, t \\in \\mathbf T_{\\mathbf X}(x^*), \\, \\langle v, \\, t\n \\rangle \\le 0 \\}.\n \\end{equation}\n\\end{subequations}\n\\rbox\n\\end{enumerate}\n\\end{definition}\n\nWe now state that the accumulation\npoints generated by Model GPS Algorithm~\\ref{al:GPSImp} \nare feasible stationary points of problem~\\eqref{sub:Proc}.\n\\begin{theorem}[Convergence to a Feasible Stationary Point]~\\\\\n\\noindent\nSuppose Assumptions~\\ref{as:FFeps} and \\ref{as:LevSetBou} are\nsatisfied.\nLet $x^* \\in \\mathbf X$ be an accumulation point of a \nrefining subsequence $\\{ x_k \\}_{k \\in \\mathbf K}$ constructed by Model\nGPS Algorithm~\\ref{al:GPSImp} in solving problem~\\eqref{sub:Proc}.\nThen,\n\\begin{subequations}\n \\begin{equation}\n \\langle \\nabla f(x^*), \\, t \\rangle \\ge 0, \\qquad \\forall \\, t \\in\n \\mathbf T_{\\mathbf X}(x^*),\n \\lab{eq:feaPoiConv}\n \\end{equation}\nand\n \\begin{equation}\n - \\nabla f(x^*) \\in \\mathbf N_{\\mathbf X}(x^*).\n \\end{equation}\n\\end{subequations}\n\\lab{the:feaPoiConv}\n\\end{theorem}\n\n% ===========================================\n\\section{Generalized Pattern Search Methods (Implementations)}\nWe will now present different implementations of the Generalized\nPattern Search (GPS) algorithms.\nThey all use the Model GPS Algorithm~\\ref{al:GPSImp} to solve\nproblem $\\mathbf P_{c}$ defined in~\\eqref{sub:Proc}.\nThe problem $\\mathbf P_{cg}$ defined in~\\eqref{sub:Procg} can be solved\nby using penalty functions as described in Section~\\ref{sec:conDepVarGen}.\\\\\n\nWe will discuss the implementations for the case where\nthe function $f(\\cdot)$ cannot be evaluated exactly,\nbut will be approximated by functions $f^* \\colon \\Re_+^q \\times \\Re^n \n\\to \\Re$, where the first argument $\\epsilon \\in \\Re_+^q$ is \nthe precision parameter of the \nPDE, ODE, and algebraic equation solvers.\nThis includes the case where $\\epsilon$ \nis not varied during the optimization, in which case\nthe explanations are identical, except that the scheme to\ncontrol $\\epsilon$ is not applicable, and that\nthe approximate functions $f^*(\\epsilon,\\cdot)$ are replaced by $f(\\cdot)$.\\\\\n\n\nIf the cost function $f(\\cdot)$ is approximated by\nfunctions $\\{ f^*(\\epsilon,\\cdot)\\}_{\\epsilon \\in \\Re_+^q}$\nwith adaptive precision $\\epsilon$, then\nthe function $\\rho \\colon \\Re_+ \\to \\Re_+^q$ (to assign $\\epsilon$)\ncan be implemented by using GenOpt's pre-processing capability\n(see Section~\\ref{par:posPro}).\\\\\n\n% ===========================================\n\\subsection{Coordinate Search Algorithm}\nWe will now present the implementation of the Coordinate Search \nalgorithm\nwith adaptive precision function evaluations\nusing the Model GPS Algorithm~\\ref{al:GPSImp}.\nTo simplify the implementation, we assign\n$f^*(\\epsilon,x) = \\infty$ for all $x \\not \\in \\mathbf X$ where\n$\\mathbf X$ is defined in~\\eqref{eq:setXPc}.\n\n%-------------------------\n\\subsubsection{Algorithm Parameters}\n\\lab{sec:AlgParCooSea}\nThe search direction matrix is defined as \n\\begin{equation}\n D \\triangleq [+s^1 \\, e_1, \\, -s^1 \\, e_1, \\ldots , \n\\, +s^n \\, e_n, \\, -s^n \\, e_n]\n\\label{eq:defDMatGPSCooSea}\n\\end{equation}\nwhere $s^i \\in \\Re$, $i \\in \\{1, \\ldots, n \\}$, is a scaling for each parameter\n(specified by GenOpt's parameter \\texttt{Step}).\\\\\n\nThe parameter $r \\in \\Na$, $r>1$, which is used to compute the mesh size parameter\n$\\Delta_k$,\nis defined by the parameter \\texttt{MeshSizeDivider},\nthe initial value for the mesh size exponent $s_0 \\in \\Na$\nis defined by the parameter \\texttt{InitialMeshSizeExponent},\nand the mesh size exponent increment $t_k$ is, for the\niterations that do not reduce the cost,\ndefined by the parameter \\texttt{MeshSizeExponentIncrement}.\n\n%-------------------------\n\\subsubsection{Global Search}\n\\lab{sec:GloSeaCooSea}\n\nIn the Coordinate Search Algorithm, there is no global search.\nThus, $\\mathcal G_k = \\emptyset$ for all $k \\in \\Na$.\n\n%-------------------------\n\\subsubsection{Local Search}\n\\lab{sec:locSeaCooSea}\n\nThe local search set $\\mathcal G_k$ is constructed using the\nset-valued map $E_k \\colon \\Re^n \\times \\mathbb Q_+ \\times \\Re_+^q \\to 2^{\\mathbb M_k}$,\nwhich is defined as follows:\\\\[\\baselineskip]\n\\noindent\n\\begin{minipage}[b]{\\textwidth}\n\\begin{algorithm}\n[Map \n$E_k \\colon \\Re^n \\times \\mathbb Q_+ \\times \\Re_+^q \\rightarrow 2^{\\mathbb M_k}$\nfor ``Coordinate Search'']\n~\\\\\n{\\em\n\\begin{tabular}{ ll }\n\\multicolumn{2}{l}{\\hspace{\\textwidth}~} \\\\ \\\\[-8ex]\\\\\n\\hline \\\\[-2ex]\n\\textbf{Parameter}:\n & Search direction matrix $D \n = [+s^1 \\, e_1, \\, -s^1 \\, e_1, \\ldots , \n\\, +s^n \\, e_n, \\, -s^n \\, e_n]$.\\\\\n & Vector $\\mu \\in \\Na^n$.\\\\\n\\textbf{Input}: \n & Iteration number $k \\in \\Na$.\\\\\n & Base point $x \\in \\Re^n$.\\\\\n & Mesh divider $\\Delta_k \\in \\mathbb Q_+$. \\\\\n\\textbf{Output}:\n & Set of trial points $\\mathcal T$.\\\\\n\\textbf{Step 0}:\n & Initialize $\\mathcal T = \\emptyset$.\\\\\n & If $k = 0$, initialize , $\\mu^i=0$ for all $i \\in \\{1, \\ldots, n\\}$.\\\\\n\\textbf{Step 1}:\n & For $i = 1, \\ldots , n$\\\\\n & \\hspace{1cm} Set $\\widetilde x = x + \\Delta_k \\, \n D \\, e_{2 \\, i - 1 + \\mu^i}$ and\n $\\mathcal T \\leftarrow \\mathcal T \\cup \\{ \\widetilde x \\}$.\\\\\n & \\hspace{1cm} If $f^*(\\epsilon,\\widetilde x) < f^*(\\epsilon,x)$\\\\\n & \\hspace{2cm} Set $x = \\widetilde x$.\\\\\n & \\hspace{1cm} else \\\\\n & \\hspace{2cm} If $\\mu^i = 0$, set $\\mu^i = 1$, else set $\\mu^i = 0$.\\\\\n & \\hspace{2cm} Set $\\widetilde x = x + \\Delta_k \\, \n D \\, e_{2 \\, i -1 + \\mu_i}$ and\n $\\mathcal T \\leftarrow \\mathcal T \\cup \\{\n \\widetilde x \\}$.\\\\\n & \\hspace{2cm} If $f^*(\\epsilon,\\widetilde x) < f^*(\\epsilon,x)$\\\\\n & \\hspace{3cm} Set $x = \\widetilde x$.\\\\\n & \\hspace{2cm} else\\\\\n & \\hspace{3cm} If $\\mu^i = 0$, set $\\mu^i = 1$, else set $\\mu^i = 0$.\\\\\n & \\hspace{2cm} end if.\\\\\n & \\hspace{1cm} end if.\\\\ \n & end for.\\\\\n\\textbf{Step 2}:\n & Return $\\mathcal T$.\\\\\n \\hline \\\\\n\\end{tabular}\n}\n\\lab{al:CooSeaLocSea}\n\\end{algorithm}\n\\end{minipage}\nThus, $E_k(x, \\Delta_k, \\epsilon) = \\mathcal T$ for all $k \\in \\Na$.\n\\begin{remark}\n{\\em \nIn Algorithm~\\ref{al:CooSeaLocSea},\nthe vector $\\mu \\in \\Na^n$ contains for\neach coordinate direction an integer $0$ or $1$ that indicates \nwhether a step in the positive or\nin the negative coordinate direction yield a decrease in cost in the previous\niteration. This reduces the number of exploration steps.\n}\n\\rbox\n\\end{remark}\n\n%-------------------------\n\\subsubsection{Parameter Update}\nThe point $x'$ in Step 3 of the GPS Model Algorithm~\\ref{al:GPSImp} \ncorresponds to \n$x' \\triangleq \\arg \\min_{x \\in E_k(x_k, \\Delta_k, \\epsilon)} f^*(\\epsilon,x)$\nin the Coordinate Search algorithm.\n\n%-------------------------\n\\subsubsection{Keywords}\n\\label{sec:GPSCooSeaKeyWor}\nFor the GPS implementation of the Coordinate Search Algorithm, \nthe command file (see page~\\pageref{par:comFil}) must only contain continuous parameters.\\\\\n\nTo invoke the algorithm, \nthe \\texttt{Algorithm} section of the GenOpt command file must have the following form:\n\\begin{lstlisting}\nAlgorithm{\n Main = GPSCoordinateSearch;\n MeshSizeDivider = Integer; // 1 < MeshSizeDivider\n InitialMeshSizeExponent = Integer; // 0 <= InitialMeshSizeExponent\n MeshSizeExponentIncrement = Integer; // 0 < MeshSizeExponentIncrement\n NumberOfStepReduction = Integer; // 0 < NumberOfStepReduction\n}\n\\end{lstlisting}\nThe entries are defined as follows:\n\\begin{codedescription}\n\\item [Main]\nThe name of the main algorithm.\n\\item [MeshSizeDivider]\nThe value for $r \\in \\Na$, $r>1$, used to compute \n$\\Delta_k \\triangleq {1} / {r^{s_k}}$ (see equation~\\eqref{eq:GPSDelDef}).\nA common value is $r = 2$.\n\n\\item [InitialMeshSizeExponent]\nThe value for $s_0 \\in \\Na$ in~\\eqref{eq:GPSDelDefSk}.\nA common value is $s_0 = 0$.\n\n\\item [MeshSizeExponentIncrement]\nThe value for $t_i \\in \\Na$ (for the iterations that do not yield\na decrease in cost) in~\\eqref{eq:GPSDelDefSk}.\nA common value is $t_i = 1$.\n\n\\item [NumberOfStepReduction]\nThe maximum number of step reductions before the algorithm stops.\nThus, if we use the notation $m \\triangleq \\text{\\texttt{NumberOfStepReduction}}$, then\nwe have for the last iterations $\\Delta_k = {1} / {r^{s_0 + m\\, t_k}}$.\nA common value is $m = 4$.\n\\end{codedescription}\n\n% ===========================================\n\\subsection{Hooke-Jeeves Algorithm}\n\\lab{sec:GPSHooJeeImp}\nWe will now present the implementation of the Hooke-Jeeves algorithm~\\cite{HookeJee1961}\nwith adaptive precision function evaluations\nusing the Model GPS Algorithm~\\ref{al:GPSImp}.\nThe modifications of Smith~\\cite{Smith1969},\nBell and Pike~\\cite{BellPik1966} and\nDe Vogelaere~\\cite{DeVogelaere1968}\nare implemented in this algorithm.\\\\\n\n \nTo simplify the implementation, we assign\n$f^*(\\epsilon,x) = \\infty$ for all $x \\not \\in \\mathbf X$ where\n$\\mathbf X$ is defined in~\\eqref{eq:setXPc}.\n\n%-------------------------\n\\subsubsection{Algorithm Parameters}\nThe algorithm parameters $D$, $r$, $s_0$, and $t_k$ are defined \nas in the Coordinate Search algorithm \n(see page~\\pageref{sec:AlgParCooSea}).\n\n\n%-------------------------\n\\subsubsection{Map for Exploratory Moves}\nTo facilitate the algorithm explanation, we use\nthe set-valued map \n$E_k \\colon \\Re^n \\times \\mathbb Q_+ \\times \\Re_+^q \\to 2^{\\mathbb M_k}$, \nas defined in Algorithm~\\ref{al:CooSeaLocSea}.\nThe map $E_k(\\cdot,\\cdot,\\cdot)$ defines the\n``exploratory moves'' in~\\cite{HookeJee1961}, and\nwill be used in Section~\\ref{sec:GloSeaSetMapHJ}\nto define the global search set map and,\nunder conditions to be seen in Section~\\ref{sec:locSeaDirMapHJ},\nthe local search direction map as well.\\\\\n%-------------------------\n%-------------------------\n\\subsubsection{Global Search Set Map}\n\\lab{sec:GloSeaSetMapHJ}\n\nThe global search set map\n$\\gamma_k(\\cdot, \\cdot, \\cdot)$ is defined as follows. Because\n$\\gamma_0(\\cdot, \\cdot, \\cdot)$ depends on $x_{-1}$, we need to introduce\n$x_{-1}$, which we define as $x_{-1} \\triangleq x_0$.\\\\\n\n\\noindent\n\\begin{minipage}{\\textwidth}\n\\begin{algorithm}\n[Global Search Set Map $\\gamma_k \\colon \\underline{\\mathbf X}_k \n\\times \\underline {\\boldsymbol \\Delta}_k \\times \\Re_+^q\n\\rightarrow 2^{\\mathbb M_k} $]\n~\\\\\n{\\em\n\\begin{tabular}{ll}\n\\multicolumn{2}{l}{\\hspace{\\textwidth}~} \\\\ \\\\[-8ex]\\\\\n\\hline \\\\[-2ex]\n\\textbf{Map}:\n & Map for ``exploratory moves'' \n $E_k \\colon \\Re^n \\times \\mathbb Q_+ \\times \\Re_+^q \\to 2^{\\mathbb M_k}$.\\\\\n\\textbf{Input}: \n & Previous and current iterate, $x_{k-1} \\in \\Re^n$ \n and $x_k \\in \\Re^n$.\\\\\n & Mesh divider $\\Delta_k \\in \\mathbb Q_+$.\\\\\n & Solver precision $\\epsilon \\in \\Re_+^q$.\\\\\n\\textbf{Output}:\n & Global search set $\\mathcal G_k$.\\\\\n\\textbf{Step 1}:\n & Set $x = x_k + (x_k - x_{k-1})$.\\\\\n\\textbf{Step 2}:\n & Compute $\\mathcal G_k = E_k(x, \\Delta_k, \\epsilon)$.\\\\\n\\textbf{Step 3}:\n & If $\\bigl(\\min_{x \\in \\mathcal G_k} \n f^*(\\epsilon,x)\\bigr) > f^*(\\epsilon,x_k)$\\\\\n & \\hspace{1cm} Set $\\mathcal G_k \\leftarrow \\mathcal G_k \\cup \n E_k(x_k, \\Delta_k, \\epsilon)$.\\\\\n & end if.\\\\\n\\textbf{Step 4}:\n & Return $\\mathcal G_k$.\\\\\n \\hline \\\\\n\\end{tabular}\n}\n\\lab{al:HJGloSeaSetMap}\n\\end{algorithm}\n\\end{minipage}\nThus, $\\gamma_k(\\underline x_k, \\underline \\Delta_k, \\epsilon) = \\mathcal G_k$.\\\\\n\n%-------------------------\n\\subsubsection{Local Search Direction Map}\n\\lab{sec:locSeaDirMapHJ}\nIf the global search, as defined by \nAlgorithm~\\ref{al:HJGloSeaSetMap}, has failed in reducing\n$f^*(\\epsilon,\\cdot)$, then Algorithm~\\ref{al:HJGloSeaSetMap} has\nconstructed a set $\\mathcal G_k$ that contains the set \n$\\{ x_k + \\Delta_k \\, D \\, e_i \\ | \\ i = 1, \\ldots , \\, 2 n \\}$.\nThis is because in the evaluation of\n$E_k(x_k, \\Delta_k, \\epsilon)$, defined in Algorithm~\\ref{al:CooSeaLocSea}, all \n``If $f^*(\\epsilon,\\widetilde x) < f^*(\\epsilon,x)$'' statements yield\n{\\tt false}, and, hence, one has constructed \n$\\{ x_k + \\Delta_k \\, D \\, e_i \\ | \\ i = 1, \\ldots , \\, 2 n\n\\} = E_k(x_k, \\Delta_k, \\epsilon)$.\n\nBecause the columns of \n$D$ span $\\Re^n$ positively, it follows that\nthe search on the set $\\{ x_k + \\Delta_k \\, D \\, e_i \\ | \\ i = 1, \\ldots , \\, 2 n\n\\}$ is a local search. Hence, the constructed set\n\\begin{equation}\n\\mathcal L_k \\triangleq \\{ x_k + \\Delta_k \\, D \\, e_i \\ | \\\ni = 1, \\ldots , \\, 2 n \\} \\subset \\mathcal G_k\n\\lab{eq:LkHJ}\n\\end{equation}\nis a local search set.\nConsequently, $f^*(\\epsilon,\\cdot)$ has already been evaluated\nat all points of $\\mathcal L_k$ (during the construction of $\\mathcal\nG_k$) and, hence, one does not need to evaluate $f^*(\\epsilon,\\cdot)$ \nagain in a local search.\n\n%-------------------------\n\\subsubsection{Parameter Update}\nThe point $x'$ in Step 3 of the GPS Model Algorithm~\\ref{al:GPSImp} \ncorresponds to $x' \\triangleq \\arg \\min_{x \\in \\mathcal G_k} \nf^*(\\epsilon,x)$ in the Hooke-Jeeves algorithm. (Note that \n$\\mathcal L_k \\subset \\mathcal G_k$ if a local search has been done as\nexplained in the above paragraph.)\n\n%-------------------------\n\\subsubsection{Keywords}\nFor the GPS implementation of the Hooke-Jeeves algorithm, the command file (see page~\\pageref{par:comFil}) must only contain continuous parameters.\\\\\n\nTo invoke the algorithm, \nthe \\texttt{Algorithm} section of the GenOpt command file must have the following form:\n\\label{algSec:GPSHookeJeeves}\n\\begin{lstlisting}\nAlgorithm{\n Main = GPSHookeJeeves;\n MeshSizeDivider = Integer; // bigger than 1\n InitialMeshSizeExponent = Integer; // bigger than or equal to 0\n MeshSizeExponentIncrement = Integer; // bigger than 0\n NumberOfStepReduction = Integer; // bigger than 0\n}\n\\end{lstlisting}\nThe entries are the same as for the Coordinate Search algorithm, and explained\non page~\\pageref{sec:GPSCooSeaKeyWor}.\n\n\n% ===========================================\n\\subsection{Multi-Start GPS Algorithms}\n\\lab{sec:GPSMulSta}\nAll GPS algorithms can also be run using multiple initial points.\nUsing multiple initial points increases the chance of finding \nthe global minimum if the cost function has several local minima,\nand furthermore, it decreases the risk of not finding a minimum if the\ncost function is not continuously differentiable, which \nis the case if building simulation programs, such as\nEnergyPlus or TRNSYS, are used to compute the cost function\n(see the discussion in Section~\\ref{sec:proAppCosFun}).\\\\\n\nThe values that are specified\nby GenOpt's parameter \\texttt{Ini} in GenOpt's command file\n(see Section~\\ref{par:comFil}) are used to initialize\nthe first initial point.\nThe other initial points are randomly distributed, with\na uniform distribution,\nbetween the lower\nand upper bounds of the feasible domain. \nThey are, however, set to the mesh $\\mathbb M_0$, defined in~\\eqref{eq:Mesh},\nwhich reduces the number of cost function evaluations if\nthe optimization algorithm converges \nfrom different initial points to the same minimizer.\\\\\n\nIn GenOpt's command file, a lower and an upper bound must be specified\nfor each independent variable\nusing the keywords \\texttt{Min} and \\texttt{Max}.\\\\\n\nTo use the \\texttt{GPSCoordinateSearch} algorithm with multiple\nstarting points,\nthe \\texttt{Algorithm} section of the GenOpt command file \nmust have the following form:\n\\begin{lstlisting}\nAlgorithm{\n Main = GPSCoordinateSearch;\n MultiStart = Uniform;\n Seed = Integer;\n NumberOfInitialPoint = Integer; // bigger than or equal to 1\n MeshSizeDivider = Integer; // 1 < MeshSizeDivider\n InitialMeshSizeExponent = Integer; // 0 <= InitialMeshSizeExponent\n MeshSizeExponentIncrement = Integer; // 0 < MeshSizeExponentIncrement\n NumberOfStepReduction = Integer; // 0 < NumberOfStepReduction\n}\n\\end{lstlisting}\nThe entries are defined as follows:\n\\begin{codedescription}\n\\item [Main]\nThe name of the main algorithm.\n\\item [MultiStart]\nKeyword to invoke the multi-start algorithm. \nThe only valid value is \\texttt{Uniform}.\n\\item [Seed]\nThis value is used to initialize the random number generator.\n\\item [NumberOfInitialPoint]\nThe number of initial points.\n\\end{codedescription}\nThe other entries are the same as for the Coordinate Search algorithm, \nand are explained on page~\\pageref{sec:GPSCooSeaKeyWor}.\\\\\n\nTo use the \\texttt{GPSHookeJeeves} algorithm with multiple\nstarting points,\nthe \\texttt{Algorithm} section of the GenOpt command file \nmust have the following form:\n\\begin{lstlisting}\nAlgorithm{\n Main = GPSHookeJeeves;\n MultiStart = Uniform;\n Seed = Integer;\n NumberOfInitialPoint = Integer; // 0 < NumberOfInitialPoint\n MeshSizeDivider = Integer; // 1 < MeshSizeDivider\n InitialMeshSizeExponent = Integer; // 0 <= InitialMeshSizeExponent\n MeshSizeExponentIncrement = Integer; // 0 < MeshSizeExponentIncrement\n NumberOfStepReduction = Integer; // 0 < NumberOfStepReduction\n}\n\\end{lstlisting}\nThe entries are the same as for the\nmulti-start Coordinate Search algorithm above.\n\n% ===============================================\n", "meta": {"hexsha": "2dc163d52ecd102da992a5a9848d5b920e1d2cc0", "size": 32573, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/manual/algGPS.tex", "max_stars_repo_name": "bergsee/GenOpt", "max_stars_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_stars_repo_licenses": ["BSD-3-Clause-LBNL"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2015-08-30T09:47:56.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-06T15:16:18.000Z", "max_issues_repo_path": "src/manual/algGPS.tex", "max_issues_repo_name": "bergsee/GenOpt", "max_issues_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_issues_repo_licenses": ["BSD-3-Clause-LBNL"], "max_issues_count": 10, "max_issues_repo_issues_event_min_datetime": "2016-01-14T00:01:46.000Z", "max_issues_repo_issues_event_max_datetime": "2020-01-21T15:28:52.000Z", "max_forks_repo_path": "src/manual/algGPS.tex", "max_forks_repo_name": "lbl-srg/GenOpt", "max_forks_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_forks_repo_licenses": ["BSD-3-Clause-LBNL"], "max_forks_count": 13, "max_forks_repo_forks_event_min_datetime": "2015-08-30T09:47:57.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-01T18:07:07.000Z", "avg_line_length": 38.0970760234, "max_line_length": 149, "alphanum_fraction": 0.6734718939, "num_tokens": 10249, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835411997897, "lm_q2_score": 0.7690802264851919, "lm_q1q2_score": 0.6425538710905845}} {"text": "\\section{Mutual information}\nWe now consider the case where we have a set of images. This can be\nR,G,B color band images, observations of the same area but \nat different times, etc.\nWe are interested in having an entropy measurement \nof such data sets, knowing that the correlation between two images \ncan be relatively high. A 3-dimensional wavelet transform is not \nappropriate because the third dimension is generally completely different\nfrom the two first. The standard case is the data set where we have \n2-dimensional spatial information versus a frequency band.\n\n\\subsection{Principal component analysis}\nPrincipal Component Analysis (PCA), also often referred as eigenvector,\nHotelling, \nor Karhunen-Lo\\`eve transform \n\\cite{ima:karhunen47,ima:loeve48,ima:hotelling33},\nallows us to transform discrete signals into a sequence of uncorrelated \ncoefficients. Considering a population $D(1..M,1..N)$ \nof $M$ signals or images of dimension $N$,\nthe PCA method consists of expressing the dataset $D$ by:\n\\begin{eqnarray}\nD = U \\Lambda^{\\frac{1}{2}} V^t\n\\end{eqnarray}\nand\n\\begin{eqnarray}\nDD^t & = & U \\Lambda U^t \\\\\nD^t D & = & V \\Lambda V^t\n\\end{eqnarray}\nwhere $\\Lambda$ is the diagonal matrix of eigenvalues of the covariance \nmatrix $C = DD^t$, the columns of $U$ are the eigenvectors of $C$, and \nthe columns of $U$ are the eigenvector of $D^t D$. \n\nFor a signal $d(1..N)$ we have,\n\\begin{eqnarray}\nd = \\sum_{i=1}^{M} \\sqrt{\\lambda_i} u_i v_i^t\n\\end{eqnarray}\nwhere $\\lambda_i$ are the eigenvalues of the covariance matrix. The $v_i$\nvectors can also be calculated by \\cite{ima:bijaoui79}:\n\\begin{eqnarray}\n V(i,j) = v_i(j) = \\frac{1}{\\sqrt{\\lambda_i}} \\sum_k D(i,k) u_i(j)\n\\end{eqnarray}\n\nIn practice, we construct the matrix $A$ whose rows are formed \nfrom the eigenvectors\nof $C$ \\cite{ima:gonzalez93}, ordered following the \nmonotonic decreasing order of eigenvalues. A vector $x(1..M)$ can then be\ntransformed by:\n\\begin{eqnarray}\ny = \\Lambda^{-\\frac{1}{2}}A(x-m_x)\n\\end{eqnarray}\nwhere $m_x$ is the mean value of $x$.\nBecause the rows of $A$ are orthonormal vectors, $A^{-1} = A^t$, and\nany vector $x$ can be recovered from its corresponding $y$ by:\n\\begin{eqnarray}\nx = \\Lambda^{\\frac{1}{2}}A^t y + m_x\n\\end{eqnarray}\n\nThe $\\Lambda$ matrix multiplication can be seen as a normalization.\nBuilding $A$ from the correlation matrix instead of the covariance matrix\nleads to another kind of normalization, \nand the $\\Lambda$ matrix can be suppressed \n($y = A(x-m_x)$ and $x = A^t y + m_x$). Then the norm of $y$ will be equal to the\nnorm of $x$. \n\n\\subsection{The WT-PCA transform}\n\nWe consider now that we have $M$ observations of the same view, but\nat different wavelengths (or at different epochs, etc.), and denote \nas $D_k$ one \nobservation, and $W(k, 1..P)$ its wavelet transform, $P$ being the\nnumber of scales (or bands) of the wavelet transform. Then for a given\nfrequency band $j$, we can compare the information content for all observations.\nAs a strong correlation may exist between the same frequency band of two \ndifferent observations, we can appy a principal component analysis for\nthe specific scale $j$, and repeat the same information for all $j$. Then,\nfor each scale $j$, we build a correlation matrix $C_j$ (and also $A_j$), \nand the wavelet coefficients are transformed into their principal \ncomponents. As the obtained coefficients are obtained by applying\nsuccessively a wavelet transform and a principal component analysis,\nwe will term these two transformations a WT-PCA transform,\nand the values obtained will be called WT-PCA coefficients.\n\nThe advantages of the WT-PCA transform are:\n\\begin{itemize}\n\\item We have separated the information not only spatially as if we had\nused only a wavelet transform, but also on the wavelength axis as if\nhad used directly a PCA on the images.\n\\item We keep the possibility open to estimate the noise level of the \nWT-PCA coefficients in the same rigorous way as for a simple wavelet \ntransform.\n\\item We have an exact reconstruction.\n\\end{itemize}\n \n\\subsection{Entropy from the WT-PCA transform}\n\nThe entropy relative to a set of observations $D(1..M)$ can be written by:\n\\begin{eqnarray}\nH(D) = \\sum_{j=1}^{l} \\sum_{e=1}^{M} \\sum_{k=1}^{N_j} h(c_{j,k,e})\n\\end{eqnarray}\nwhere $l$ is the number of scales used in the wavelet transform\ndecomposition, $M$ the number of observations, $k$ a pixel position,\n $c$ a WT-PCA coefficients, and $e$ denotes the eigenvector number.\n\nThe last scale of the wavelet transform is not used, as previously, so\nthis entropy measurement is background independent, which is really\nimportant because the background can vary from one wavelength to another.\n\n\n", "meta": {"hexsha": "44fe8b4ecd584c0e06786ce36db7c260f7ad1b47", "size": 4673, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr2/infomut.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr2/infomut.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr2/infomut.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.871559633, "max_line_length": 81, "alphanum_fraction": 0.7504814894, "num_tokens": 1290, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324848629214, "lm_q2_score": 0.78793120560257, "lm_q1q2_score": 0.6423471146444206}} {"text": "\\documentclass[10pt]{article}\n\n% Manage page layout\n\\usepackage[margin=2.5cm, includefoot, footskip=30pt]{geometry}\n\\pagestyle{plain}\n\\setlength{\\parindent}{0em}\n\\setlength{\\parskip}{1em}\n\\renewcommand{\\baselinestretch}{1}\n\n\\usepackage{blkarray}\n\\usepackage{multirow}\n\\usepackage{amsmath}\n\\usepackage{enumerate}\n\n\\title{\\textbf{Week 4.} Static games with complete information III: Nash equilibria}\n\\date{}\n\n\\begin{document}\n\\maketitle\n\\vspace{-1cm}\n\n\\subsection*{Exercise 1: Nash equilibrium vs dominance solvability}\n\n\\textbf{Prove the following statements:}\n\n\\begin{enumerate}[(i)]\n \\item If a pure strategy \\(s^{(i)}_{j}\\) is dominated by a pure strategy \\(s^{(i)}_{k}\\) and\n \\(\\sigma = (\\sigma^{(1)}, \\dots, \\sigma^{(n)})\\) is a Nash equilibrium, then\n \\(\\sigma^{(i)}_{j}=0\\).\n \\item If the game is dominance solvable such that the unique outcome of\n iterated elimination of dominated strategies is some pure strategy\n \\(s=(s^{(1)}, \\dots, s^{(n)})\\), then \\(s\\) is a Nash equilibrium.\n\\end{enumerate}\n\n[Suggestion: One could use contradiction to prove the above statements. For example, for (i)\nassume that these was a Nash equilibrium with \\(\\sigma^{(i)}_{j}>0\\), and show\nthat this would yield some contradiction.]\n\n\\subsection*{Exercise 2: Best responses}\n\nConsider the stag hunt game:\n\n\\begin{equation*}\n \\begin{blockarray}{cccc}\n & & \\BAmulticolumn{2}{c}{\\underline{\\text{player 2}}} \\\\ [1em]\n & & \\text{Stag} & \\text{Hare} \\\\\n \\begin{block}{cc(cc)}\n\\underline{\\text{player 1}} & \\text{Stag} & (10, 10) & (0, 6) \\\\\n & \\text{Hare} & (6, 0) & (6, 6) \\\\\n \\end{block}\n \\end{blockarray}\n\\end{equation*}\n\nSuppose player 1 uses the mixed strategy \\((x, 1- x)\\), where \\(x\\) is player 1's\nprobability to Stag. Similarly, player 2's strategy is \\((y, 1 - y)\\).\n\n\\begin{enumerate}[(i)]\n \\item For given \\(x, y\\) compute the players' payoffs \\(\\pi^{(1)}(x, y),\n \\pi^{(2)}(x, y)\\) (see Remarks 2.6, 2.7).\n \\item For a given \\(y\\) compute player 1's best response (BR(\\(y\\))). In\n particular, show that there is some \\(y^{*}\\) such that all \\(x \\in [0,\n 1]\\) are a best response.\n \\item Draw the two best response correspondences BR(\\(x\\)), BR(\\(y\\)) into a\n \\(x-y\\) plane. How often do they intersect? What does it mean if they\n intersect?\n\\end{enumerate}\n\n\\subsection*{Exercise 3: Cournot Duopoly}\n\nThe Cournot duopoly game is defined by:\n\n\\begin{itemize}\n \\item Players: \\(N = \\{\\text{Firm } 1, \\text{Firm } 2\\}\\)\n \\item Actions: Amount of good produced, \\(x^{(i)} \\in [0, \\infty)\\) for \\(i \\in \\{1, 2\\}\\)\n \\item Payoffs: \\(\\pi^{(i)}(x^{(1)}, x^{(2)}) = [a - b (x^{(1)} + x^{(2)})] x^{(i)} - c x^{(i)}\\)\n\\end{itemize}\n\n\\textbf{Show that there is a Nash equilibrium in pure strategies. For simplicity\nassume \\(a=10, b=1, c=1\\)}.\n\n[Hint: For each \\(x^{(i)}\\) computer BR\\((x^{(-i)})\\). Then solve simultaneously:\n\\begin{align*}\n x^{(1)}= \\text{BR}(x^{(2)}) \\\\\n x^{(2)}= \\text{BR}(x^{(1)})\n\\end{align*}\n]\n\n\\subsection*{Exercise 4: Matching Pennies}\n\n\\textbf{Compute the Nash equilibria for the following two games, and interpret the result.}\n\n\\begin{equation*}\n \\begin{blockarray}{ccc}\n & \\text{Left} & \\text{Right} \\\\\n \\begin{block}{c(cc)}\n \\text{Top} & (0.8, 0.4) & (0.4, 0.8) \\\\\n \\text{Bottom} & (0.4, 0.8) & (0.8, 0.4) \\\\\n \\end{block}\n \\end{blockarray}\\qquad\n %\n \\begin{blockarray}{ccc}\n & \\text{Left} & \\text{Right} \\\\\n \\begin{block}{c(cc)}\n \\text{Top} & (3.2, 0.4) & (0.4, 0.8) \\\\\n \\text{Bottom} & (0.4, 0.8) & (0.8, 0.4) \\\\\n \\end{block}\n \\end{blockarray}\n \\end{equation*}\n\n\\subsection*{Bonus Exercise 1: Verifying NE in games with finitely many players \\& actions}\n\nShow that to verify whether a strategy profile \\(\\hat{\\sigma} =\n(\\hat{\\sigma}^{(1)}, \\dots, \\hat{\\sigma}^{(n)})\\) is a Nash equilibrium, it is\nsufficient to check all deviations towards pure strategies.\n\n\\textbf{Specifically show that \\(\\hat{\\sigma}\\) is a Nash equilibrium if and only if for\nall players \\(i\\) the following two conditions hold:}\n\n\\begin{enumerate}[(i)]\n \\item All actions that player \\(i\\) uses give the same payoff: if\n \\(\\sigma^{(i)}_j > 0\\) and \\(\\sigma^{(i)}_k > 0\\) then\n \\(\\pi^{(i)}(s^{(i)}_j, \\hat{\\sigma}^{(-i)}) = \\pi^{(i)}(s^{(i)}_k,\n \\hat{\\sigma}^{(-i)})\\).\n \\item Actions that are not played are not profitable: if \\(\\sigma^{(i)}_j =\n 0\\) then \\(\\pi^{(i)}(s^{(i)}_j, \\hat{\\sigma}^{(-i)}) \\leq\n \\pi^{(i)}(\\hat{\\sigma}^{(i)}, \\hat{\\sigma}^{(-i)})\\).\n\\end{enumerate}\n\n[Hint: One way to prove the above is once again by contradiction.]\n\n\\subsection*{Bonus Exercise 2: Finding games with a non-generic number of equilibria}\n\n\\textbf{Find an example of a symmetric 2 player game, with 2 actions per player, with:}\n\n\\begin{itemize}\n \\item Exactly 2 Nash equilibria\n \\item infinitely many Nash equilibria\n\\end{itemize}\n\n[Note: These should include all Nash equilibria. Not just pure Nash equilibria.]\n\n\\end{document}\n\n", "meta": {"hexsha": "59a7f151a40f11903025a3e9aaf0c0e1e91dabe4", "size": 5063, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "teaching/classical_game_theory/exercises/w4.tex", "max_stars_repo_name": "Nikoleta-v3/social-behaviour", "max_stars_repo_head_hexsha": "cc146ad7662695b4afc09357d7ffad3030cd7aa4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "teaching/classical_game_theory/exercises/w4.tex", "max_issues_repo_name": "Nikoleta-v3/social-behaviour", "max_issues_repo_head_hexsha": "cc146ad7662695b4afc09357d7ffad3030cd7aa4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2020-10-23T14:05:41.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-21T16:24:55.000Z", "max_forks_repo_path": "teaching/classical_game_theory/exercises/w4.tex", "max_forks_repo_name": "Nikoleta-v3/social-behaviour", "max_forks_repo_head_hexsha": "cc146ad7662695b4afc09357d7ffad3030cd7aa4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.1597222222, "max_line_length": 100, "alphanum_fraction": 0.6152478768, "num_tokens": 1724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.6992544210587585, "lm_q2_score": 0.9184802479302793, "lm_q1q2_score": 0.6422513740203923}} {"text": "%\n% Chapter 4\n%\n\\chapter {Shor's Semiprime Integer Factorization Algorithm}\n\nShor's algorithm is a polynomial-time quantum algorithm for semiprime integer factorization. In comparison, the time complexity of the most efficient known classical factoring algorithm is superpolynomial.\n\nWe chose Shor's algorithm because it is one of the most significant quantum algorithms, and because the number of qubits and the number of quantum operations required are proportional to the input size. The number of operations is particullarly relevant since it makes the errors introduced by the physical gates an important consideration.\n\n\\section{Algorithm}\n\nThe problem that Shor's algorithm solves is the following: given a semiprime integer \\textit{N}, find its two prime factor \\textit{p} and \\textit{q}.\n\nShor's algorithm combines both classical and quantum computations, and the procedure to perform it is the following:\n\\begin{enumerate}\n \\item Pick a random integer $1 < a < N$.\n \\item Check whether $a$ is a factor by determining whether $a$ and $N$ are comprime. If they are, we can compute the factors. Otherwise continue with the rest of the algorithm.\n \\item Use a period-finding quantum subroutine to find the period $r$ of the function $f(x)=a^{x} \\ mod \\ N$.\n \\item If $r$ is odd, then go back to step 1. If $r$ is even, go to the next step.\n \\item If $a^{\\frac{r}{2}} \\equiv -1 \\ mod \\ N$, then go back to step 1.\n \\item Either $a^{\\frac{r}{2}} - 1$ or $a^{\\frac{r}{2}} + 1$ shares a factor with $N$.\n\\end{enumerate}\n\n\\section{Q\\# Implementation}\n\nIn order to take advantage of the tools built around Q\\#, we use a modified implementation of Shor's algorithm found in Microsoft's quantum samples GitHub repository.\n\n\\todo{Add reference to Microsoft's quantum samples repository.}\n\n\\todo{Breakdown the implementation into different sections and describe each one (similar to what is done in the \"Learn Quantum Computing with Python and Q\\#\" book).}\n\nThe following Q\\# code presents a top-level implementation of Shor's algorithm.\n\n\\begin{lstlisting}[language=C]\n@EntryPoint()\noperation FactorSemiprimeInteger(N : Int) : (Int, Int) {\n\n // Check the most trivial case where N is pair.\n if (N % 2 == 0) {\n return (2, N / 2);\n }\n\n mutable factors = (1, 1);\n mutable foundFactors = false;\n repeat {\n\n // Start by guessing a coprime to N.\n let coprimeGuess = DrawRandomInt(1, N - 1);\n\n // If the guess number is a coprime, use a quantum algorithm for period finding.\n // Otherwise, the GCD between N and the coprime guess number is one of the factors.\n if (IsCoprimeI(N, coprimeGuess)) {\n let period = EstimatePeriod(N, coprimeGuess);\n set (foundFactors, factors) = CalculateFactorsFromPeriod(N, coprimeGuess, period);\n } else {\n let gcd = GreatestCommonDivisorI(N, coprimeGuess);\n set (foundFactors, factors) = (true, (gcd, N / gcd));\n }\n }\n until foundFactors\n\n return factors;\n}\n\\end{lstlisting}\n\n\\section{Quantum Subroutine}\n\n\\todo{Show the circuit representation of the quantum subroutine.}\n\nThe following Q\\# code implements the period finding quantum subroutine:\n\n\\begin{lstlisting}[language=C]\n @EntryPoint()\n operation EstimatePeriodInstance(N : Int, a : Int) : (Bool, Int) {\n \n // Prepare eigenstate register.\n let bitSize = BitSizeI(N);\n use eigenstateRegister = Qubit[bitSize];\n let eigenstateRegisterLE = LittleEndian(eigenstateRegister);\n ApplyXorInPlace(1, eigenstateRegisterLE);\n \n // Prepare phase register.\n let bitsPrecision = 2 * bitSize + 1;\n use phaseRegister = Qubit[bitsPrecision];\n let phaseRegisterLE = LittleEndian(phaseRegister);\n \n // Prepare oracle for quantum phase estimation.\n let oracle = DiscreteOracle(ApplyOrderFindingOracle(N, a, _, _));\n \n // Execute quantum phase estimation.\n QuantumPhaseEstimation(oracle, eigenstateRegisterLE!, LittleEndianAsBigEndian(phaseRegisterLE));\n \n let phaseEstimate = MeasureInteger(phaseRegisterLE);\n \n // Reset qubit registers \n ResetAll(eigenstateRegister);\n \n // Return period calculation based on estimated phase.\n return CalculatePeriodFromPhaseEstimate(N, bitsPrecision, phaseEstimate);\n }\n\\end{lstlisting}\n", "meta": {"hexsha": "d641a4922a6ff28439cf78faab7ec05552abbfad", "size": 4384, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/chapter04.tex", "max_stars_repo_name": "cesarzc/qc-resources-estimation", "max_stars_repo_head_hexsha": "2909d8d51c51e69183622e039af6eaa1ed74d1c0", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Thesis/chapter04.tex", "max_issues_repo_name": "cesarzc/qc-resources-estimation", "max_issues_repo_head_hexsha": "2909d8d51c51e69183622e039af6eaa1ed74d1c0", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/chapter04.tex", "max_forks_repo_name": "cesarzc/qc-resources-estimation", "max_forks_repo_head_hexsha": "2909d8d51c51e69183622e039af6eaa1ed74d1c0", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.9803921569, "max_line_length": 340, "alphanum_fraction": 0.7009580292, "num_tokens": 1066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127678225575, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6421531335506958}} {"text": "\n\n\\section{Preliminaries and exploratory data analysis}\n\nConsider some data from Manly (you will need to get the file from the portal), OR try this with an example from Johnson and Wichern. A link to their data is given in the portal.\n\n\\begin{Schunk}\n\\begin{Sinput}\n> econ.df <- read.csv(\"econ.csv\", row.names = 1)\n> econ <- econ.df[, -10]\n> econ.cor <- cor(econ)\n> econ.ev <- eigen(econ.cor)\n\\end{Sinput}\n\\end{Schunk}\n\n\\section{PRINCIPAL COMPONENT EXTRACTION}\n\nExtracts loadings, uniqueness and residuals for 4 factors based on principal component extraction.\n\n\\begin{Schunk}\n\\begin{Sinput}\n> loadings4 <- matrix(0, 9, 4)\n> for (i in 1:4) {\n+ loadings4[, i] <- sqrt(econ.ev$values[i]) * econ.ev$vectors[, \n+ i]\n+ }\n> LLt <- loadings4 %*% t(loadings4)\n> unique <- diag(econ.cor - LLt)\n> error <- econ.cor - (LLt + unique)\n\\end{Sinput}\n\\end{Schunk}\n\n\\begin{itemize}\n \\item Now consider how to amend this code to give you a two or three factor solution.\n\\end{itemize}\n \n \n\\subsection{RESULTS FROM PRINCIPAL COMPONENT EXTRACTION}\n\nThe following labels the loadings for ease of use\n\n\\begin{Schunk}\n\\begin{Sinput}\n> row.names(loadings4) <- row.names(econ.cor)\n\\end{Sinput}\n\\end{Schunk}\nThe following gives us the values\n\n\\begin{Schunk}\n\\begin{Sinput}\n> loadings4\n> unique\n> error\n\\end{Sinput}\n\\end{Schunk}\n\nThe following extracts the eigenvalues, and then calculates the proportion and cumulative proportion of variance explained by each component (for 9 variables)\n\n\n\\begin{Schunk}\n\\begin{Sinput}\n> econ.ev$values\n> econ.ev$values/0.09\n> cumsum(econ.ev$values/0.09)\n\\end{Sinput}\n\\end{Schunk}\n\nThe following calculates the communalities\n\n\\begin{Schunk}\n\\begin{Sinput}\n> apply(loadings4^2, 1, sum)\n\\end{Sinput}\n\\end{Schunk}\n\nThe following carries out a rotation (you should also consider promax rotation)\n\n\\begin{Schunk}\n\\begin{Sinput}\n> varimax(loadings4)\n\\end{Sinput}\n\\end{Schunk}\n\n\\section{PRINCIPAL FACTORING WITH ITERATION}\n\nThis is rather crude code, but it lets you see what's going on (i.e. try this once and then forget it)\n\nFirstly, an initial estimate of the uniquenesses (can you explain what's going on here?)\n\n\n\\begin{Schunk}\n\\begin{Sinput}\n> r2s <- vector(\"numeric\", 9)\n> for (i in 1:9) {\n+ y <- econ[, i]\n+ x <- econ[, -i]\n+ mod <- lm(y ~ as.matrix(x))\n+ r2s[i] <- summary(mod)$r.squared\n+ }\n> unique <- diag(1 - r2s)\n> diag(unique)\n\\end{Sinput}\n\\end{Schunk}\n\n\nThe following step requires to be repeated until the values of the uniquenesses and the loadings converge (10 - 20 runs will do for now). Type this code in a text editor and paste it all in as one (when you have it working).\n\n\n\\begin{Schunk}\n\\begin{Sinput}\n> new <- econ.cor - unique\n> new.ev <- eigen(new)\n> loadings4pf <- matrix(0, 9, 4)\n> for (i in 1:4) {\n+ loadings4pf[, i] <- sqrt(new.ev$values[i]) * new.ev$vectors[, \n+ i]\n+ }\n> LLt <- loadings4pf %*% t(loadings4pf)\n> unique.f <- econ.cor - LLt\n> diag(unique) <- diag(unique.f)\n> diag(unique)\n> loadings4pf\n\\end{Sinput}\n\\end{Schunk}\n\nRepeat that until you get convergence in the loadings estimates. If\nyou get convergence in the loadings estimates (one of the pitfalls of\nthis technique)\n\n\n\\section{MAXIMUM LIKELIHOOD SOLUTION}\n\nFor our purposes, we will trust R to do the optimisation. In other\nwords, to carry out the analysis we need one line of code!!!!!!!!!!!!)\n\n\\begin{Schunk}\n\\begin{Sinput}\n> econ.fact <- factanal(econ, factors = 4, rotation = \"none\")\n> econ.fact\n\\end{Sinput}\n\\end{Schunk}\n\nExtract the loadings, calculate the uniquenesses, calculate the residuals\n\n\\begin{Schunk}\n\\begin{Sinput}\n> loadml <- loadings(econ.fact)\n> loadml\n> class(loadml) <- \"matrix\"\n> uniqueml <- econ.fact$uniquenesses\n> resid <- econ.cor - (loadml %*% t(loadml) + diag(uniqueml))\n> resid\n\\end{Sinput}\n\\end{Schunk}\n\nCalculate the communalities\n\n\\begin{Schunk}\n\\begin{Sinput}\n> apply(loadml^2, 1, sum)\n\\end{Sinput}\n\\end{Schunk}\n\n\\section{PLOTTING THE ROTATIONS}\n\nIn case it helps, you could try plotting the rotations (the code here\nillustrates the pcfa method), with numbers\nfor the unrotated loadings and letters for the rotated loadings.\n\n\\begin{Schunk}\n\\begin{Sinput}\n> plot(loadings4[, c(1:2)], pch = as.character(c(1:9)), xlab = expression(paste(gamma, \n+ \"1\")), ylab = expression(paste(gamma, \"2\")), main = \"First and second loadings\", \n+ xlim = c(-1, 1), ylim = c(-1, 1))\n> points(varimax(loadings4)$loadings[, c(1:2)], pch = letters[c(1:9)], \n+ col = \"red\")\n> abline(h = 0)\n> abline(v = 0)\n\\end{Sinput}\n\\end{Schunk}\n\\includegraphics{week11fa-plotloadings}\n\n\n\\begin{itemize}\n\\item Do this for maximum likelihood method\n\\end{itemize}\n\n\n\\begin{itemize}\n \\item How do you interpret the loadings??????\n \\end{itemize}\n\n\n\n\n\\section{Summary}\n\n\\fbox{\\parbox[c]{0.9\\textwidth}{\\color{blue}\nWe have left factor analysis until last - you should be very clear that despite some superficial similarities it is a very different technique to p.c.a. By the end of this week we should:\n\n\\begin{itemize}\n\\item Understand the role of factor analysis as a means of explaining the relationships between variables in a given dataset\n\\item Be able to interpret the results (loadings) of a particular factor analysis, aided or not by rotations\n\\item To explain the difference between p.c.a. based and maximum likelihood based methods of factor analysis\n \\item To interpret the ``signficance'' of solutions of different size, and to justify the use of a particular size solution in all cases\n \\end{itemize}\n\n\n\n}}\n\n\n\n\n", "meta": {"hexsha": "b9385022f8b65654641d6184e02979d22f995691", "size": 5491, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "newworksheet/week11fa.tex", "max_stars_repo_name": "phewson/mvstats", "max_stars_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "newworksheet/week11fa.tex", "max_issues_repo_name": "phewson/mvstats", "max_issues_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2020-08-28T16:37:22.000Z", "max_issues_repo_issues_event_max_datetime": "2020-08-28T16:49:11.000Z", "max_forks_repo_path": "newworksheet/week11fa.tex", "max_forks_repo_name": "phewson/mvstats", "max_forks_repo_head_hexsha": "f39ab1c1b97c89e26c708bd6d532fe13c063a95c", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.5395348837, "max_line_length": 226, "alphanum_fraction": 0.7066108177, "num_tokens": 1648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127529517044, "lm_q2_score": 0.752012562644147, "lm_q1q2_score": 0.6421531176217297}} {"text": "\\documentclass[twoside]{MATH77}\n\\usepackage{multicol}\n\\usepackage[fleqn,reqno,centertags]{amsmath}\n\\begin{document}\n\\begmath 2.13 Inverse Error Function and Inverse \\hbox{Complementary Error Function}\n\n\\silentfootnote{$^\\copyright$1997 Calif. Inst. of Technology, \\thisyear \\ Math \\`a la Carte, Inc.}\n\n\\subsection{Purpose}\n\nGiven $x$, this pair of subprograms computes the value $t$ such that%\n\\begin{equation*}\nx=\\text{erf}(t)=\\frac 2{\\sqrt{\\pi }}\\int_0^te^{-s^2}ds\n\\end{equation*}\nfor the Inverse Error Function, or%\n\\begin{equation*}\nx=\\text{erfc}(t)=\\frac 2{\\sqrt{\\pi }}\\int_t^\\infty e^{-s^2}ds\n\\end{equation*}\nfor the Inverse Complementary Error Function.\n\nProcedures in Chapter~15.2 may be used to calculate the inverse of\nthe Gaussian or normal probability integral.\n\nReference \\cite{ams55:erf} provides further discussion of the properties\nof the error function.\n\n\\subsection{Usage}\n\n\\subsubsection{Program Prototype, Single Precision}\n\n\\begin{description}\n\\item[REAL] \\ {\\bf X, SERFI, SERFCI, T}\n\\end{description}\n\nAssign a value to X and use one of the following function references.\n\nTo compute the Inverse Error Function:\n$$\n\\fbox{{\\bf T = SERFI(X)}}\n$$\nTo compute the Inverse Complementary Error Function:\n$$\n\\fbox{{\\bf T = SERFCI(X)}}\n$$\n\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\\item[X] \\ [in] Argument of function. Require $-1<\\text{X}<1$ for SERFI, $0<%\n\\text{X}<2$ for SERFCI.\n\\end{description}\n\n\\subsubsection{Modification for Double Precision}\n\nFor double precision usage change the REAL type statement to DOUBLE\nPRECISION and change the function names to DERFI and DERFCI respectively.\n\n\\subsection{Examples and Remarks}\n\nSee DRDERFI and ODDERFI for an example of the usage of these subprograms.\n\n$|\\frac{\\text{d}}{\\text{dx}}\\text{erf}^{-1}(x)|$ and\n$|\\frac{\\text{d}}{\\text{dx}}\\text{erfc}^{-1}(x)|$ are $\\frac{\\sqrt{\\pi}}2\ne^{x^2}$. Therefore, the relative accuracy to be expected in evaluating\nthese functions decreases rapidly as $x$ increases. For example, when using\nIEEE single-precision arithmetic, one should expect no more than one digit of\nthe result to be correct when $x > \\approx 3.6$\n\n\\subsection{Functional Description}\n\nThe computer approximations for these functions were developed by A.\nStrecok, \\cite{Strecok:1968:OCI}, using Chebyshev polynomial expansions.\n\nThese subprograms were tested on the IBM PC/AT, which uses IEEE\narithmetic with precision, $\\rho \\approx 1.19 \\times 10^{-7}$ in single precision and\n$\\rho \\approx 2.22 \\times 10^{-16}$ in double precision. We checked how well the\nError Function subprograms (SERF and SERFC or DERF and DERFC) and the\npresent subprograms are inverses. Two sets of relative error tests were\nperformed:\n\\begin{gather*}\n\\varepsilon _1 = |\\erf(\\erfi(x))/x - 1|\\ /\\ \\rho,\\ \\ \\text{and}\\\\\n\\varepsilon _2 = |\\erfi(\\erf(x))/x - 1|\\ /\\ \\rho\n\\end{gather*}\n\nand similarly for the complementary functions. The latter test does not\nmeasure relative error precisely for the complementary functions, and was\nnot carried out in the range 1.0E$-$35 $\\leq x \\leq $ 1.0E$-$5 (1.0E$-$300 $%\n\\leq x \\leq $ 1.0E$-$5 in double precision). The results are summarized below.\n\n\\begin{tabular}{llll}\n & \\bf Argument & \\bf Max. & \\bf Max.\\\\\n\\bf Function & \\bf \\ \\,Interval & \\bf \\ \\ \\ $\\varepsilon _1$ &\n\\bf \\ \\ \\ $\\varepsilon _2$\\\\\nSERFI & [0.1E$-$2, 0.8] & 0.956 & 0.993\\\\\n & [0.8, 0.9975] & 0.625 & 1.66\\\\\n & [.9975, .99999] & 0.0 & 1.50\\\\\nSERFCI & [1.0E$-$35, 1.0E$-$5] & 113\\\\\n & [1.E$-$5, 2.5E$-$3] & 16.3 & 1.91E+4\\\\\n & [2.5E$-$3, 0.2] & 14.2 & 86.7\\\\\n & [0.2, 0.999] & 2.14 & 1.92\\\\\nDERFI & [0.1E$-$2, 0.8] & 0.896 & 0.973\\\\\n &[0.8, 0.9975] & 0.625 & 1.59\\\\\n &[.9975, .99999] & 0.0 & 1.50\\\\\nDERFCI & [1.0E$-$300, 1.0E$-$5] & 964\\\\\n & [1.E$-$5, 2.5E$-$3] & 13.6 & 1.90E+4\\\\\n &[2.5E$-$3, 0.2] & 8.29 & 64.6\\\\\n &[0.2, 0.999] & 1.69 & 1.60\n\\end{tabular}\n\n\\bibliography{math77}\n\\bibliographystyle{math77}\n\n\\subsection{Error Procedures and Restrictions}\n\nThese subprograms issue an error message and terminate execution if\n\n\\hspace{.4in}(a) \\ \\ X $\\leq -1$ or X $\\geq 1$ for SERFI, or\n\n\\hspace{.4in}(b) \\ \\ X $\\leq 0$ or X $\\geq 2$ for SERFCI.\n\nThe error message is issued by way of the error message processor at level 0,\nand the returned function value is 0.0. If it is desired, in the event an error\noccurs, to issue the error message and then terminate program execution\n\n\\hspace{.4in}CALL ERMSET (2)\n\nbefore invoking SERFI or SERFCI.\n\nSee Chapter~19.2 for further description of the error message processor.\n\n\\subsection{Supporting Information}\n\nThe source language is ANSI Fortran~77.\n\nSubprograms designed and developed by W. V. Snyder, JPL, 1986.\n\n\n\\begin{tabular}{@{\\bf}l@{\\hspace{5pt}}l}\n\\bf Entry & \\hspace{.35in} {\\bf Required Files}\\vspace{2pt} \\\\\nDERFCI & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, DERFI, DERM1, DERV1, ERFIN, ERMSG\\rule[-5pt]{0pt}{8pt}}\\\\DERFI & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, DERFI, DERM1, DERV1, ERFIN, ERMSG\\rule[-5pt]{0pt}{8pt}}\\\\SERFCI & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, ERFIN, ERMSG, SERFI, SERM1, SERV1\\rule[-5pt]{0pt}{8pt}}\\\\SERFI & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, ERFIN, ERMSG, SERFI, SERM1, SERV1}\\\\\\end{tabular}\n\n\\begcode\n\n\\bigskip\n\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRDERFI}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{derfi}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODDERFI}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{derfi}}\n\n\\end{document}\n", "meta": {"hexsha": "21397b171eb7b06c5e40cfa45abd8a306ec1bdc5", "size": 5541, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doctex/ch02-13.tex", "max_stars_repo_name": "jacobwilliams/math77", "max_stars_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2016-01-04T03:17:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-25T19:17:42.000Z", "max_issues_repo_path": "doc/doctex/ch02-13.tex", "max_issues_repo_name": "jacobwilliams/math77", "max_issues_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2019-01-17T02:48:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T16:04:58.000Z", "max_forks_repo_path": "doc/doctex/ch02-13.tex", "max_forks_repo_name": "jacobwilliams/math77", "max_forks_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-01-07T09:26:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-11-25T05:32:54.000Z", "avg_line_length": 34.4161490683, "max_line_length": 123, "alphanum_fraction": 0.7016783974, "num_tokens": 1982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.828938825225204, "lm_q2_score": 0.774583389368527, "lm_q1q2_score": 0.6420822448221035}} {"text": "\n\\section{The \\minmaxelement algorithm}\n\\Label{sec:minmaxelement}\n\nThe \\minmaxelement algorithm in the\n\\cxx Standard Library \\cite[\\S 28.7.8]{cxx-17-draft} searches\n\\emph{both} the minimum \\emph{and} the maximum in a sequence.\nThe signature of our version of \\minelement reads:\n\n\\begin{lstlisting}[style = acsl-block]\n\n size_type_pair minmax_element(const value_type* a, size_type n);\n\\end{lstlisting}\n\nNote that \\minmaxelement returns a \\emph{pair} of indices (see \\S\\ref{sec:makepair}).\nThis pair contains the \\emph{first} position where the minimum occurs in\nthe sequence \\inl{a[0..n-1]} and the \\emph{last} position where maximum occurs.\n\nThe properties of the index for the minimum value are the same as the properties\nof \\specref{minelement}.\nHowever, the properties of the index that marks the maximum\nelement, are slightly different from the properties of \\specref{maxelement}.\nThe \\maxelement algorithm returns the position of the \\emph{first} occurrence of\nthe maximum element if it occurs multiple times in the sequence. The\n\\minmaxelement algorithm returns the position of the last occurrence of the\nmaximum element.\n\n%\\clearpage\n\n\\subsection{Formal specification of \\minmaxelement}\n\nThe following listing shows the acsl specification of \\specref{minmaxelement}.\nNote that we use the predicates \\logicref{StrictLowerBound} and\n\\logicref{StrictUpperBound} in order to express that the algorithm returns\nthe positions of both the \\emph{first minimum} and the \\emph{last maximum}.\nWe also use the predicates \\logicref{MinElement} and \\logicref{MaxElement}.\nThus reflects of course the use of this predicates for the algorithms\n\\specref{minelement} and \\specref{maxelement}.\n\n\\input{Listings/minmax_element.h.tex}\n\n\nThe specification is similar to the specifications of \\minelement and\n\\maxelement. The only difference lies in the postcondition \\inl{last}. Here the\npostcondition states that after the position of the maximum element there is no\nvalue greater or equal the maximum element. This differs from the specification\nof \\maxelement, where the first occurrence of the maximum value has to be\nreturned.\n\n%\\clearpage\n\n\\subsection{Implementation of \\minmaxelement}\n\nThe implementation of \\implref{minmaxelement} uses the auxiliary\nfunction \\specref{makepair} to construct a pair of indices.\nWe will focus on the loop invariant \\inl{last}, because it is the\nonly loop invariant that differs from the implementations of \\implref{minelement} and\n\\implref{maxelement}.\n\n\\input{Listings/minmax_element.c.tex}\n\nAs already mentioned we had to alter the range for the predicate\n\\logicref{StrictUpperBound} to fit into the property of returning\nthe last maximum position that occurred.\n\n", "meta": {"hexsha": "23f9c6d418e25da9134cec6e231d92b1c16903a5", "size": 2698, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Informal/maxmin/minmax_element.tex", "max_stars_repo_name": "fraunhoferfokus/acsl-by-example", "max_stars_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 90, "max_stars_repo_stars_event_min_datetime": "2017-06-14T04:17:53.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-07T06:07:36.000Z", "max_issues_repo_path": "Informal/maxmin/minmax_element.tex", "max_issues_repo_name": "fraunhoferfokus/acsl-by-example", "max_issues_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 22, "max_issues_repo_issues_event_min_datetime": "2017-10-18T13:30:41.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-17T07:10:16.000Z", "max_forks_repo_path": "Informal/maxmin/minmax_element.tex", "max_forks_repo_name": "fraunhoferfokus/acsl-by-example", "max_forks_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 19, "max_forks_repo_forks_event_min_datetime": "2017-06-21T13:49:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T16:27:06.000Z", "avg_line_length": 40.8787878788, "max_line_length": 85, "alphanum_fraction": 0.7987398073, "num_tokens": 663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833945721305, "lm_q2_score": 0.8289388083214156, "lm_q1q2_score": 0.6420822360421787}} {"text": "\\documentclass{article}\n\n\\usepackage{mathrsfs,amssymb,amsmath}\n\n\\begin{document}\n\n\\section{Prove that for a finite topological spaces, discrete is equivalent to Hausdorf}\n\nAssume a finite topological space X is discrete. Then given two points $x_1$ and $x_2$ there are the open sets $\\{x_1\\}$ and $\\{x_2\\}$ that are non-intersecting, thus X is Hausdorf. Note the assumption that the space was finite was not needed.\n\nAssume a finite topological space X is Hausdorf. Pick an arbitrary point $x$. For every $x_i \\in X$ where $1 \\le i \\le n=|X|$, there exists an open set containing $x$ that does not contain $x_i$, call it $U_i$. Take $U=\\bigcap_{i=0}^n U_i$. $U = \\{x\\}$ since it is in every set $U_i$ but every other point in $X$ is not in at least one $U_i$. $U$ is open, since it is the finite intersection of open sets.\n\nSince $x$ was chosen arbitrarily, the singleton set around every point is open. Thus since the union of open sets is open, any subset of $X$ is open, thus $X$ is discrete.\n\n\\end{document}\n", "meta": {"hexsha": "c385f1c2eb0a3ccc880be3572cfbc74472a3f52d", "size": 1012, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "analysis/8_C_star_Algebras/extra.tex", "max_stars_repo_name": "lukemassa/math-exercises", "max_stars_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "analysis/8_C_star_Algebras/extra.tex", "max_issues_repo_name": "lukemassa/math-exercises", "max_issues_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "analysis/8_C_star_Algebras/extra.tex", "max_forks_repo_name": "lukemassa/math-exercises", "max_forks_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.25, "max_line_length": 405, "alphanum_fraction": 0.7302371542, "num_tokens": 299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649232, "lm_q2_score": 0.8289388125473629, "lm_q1q2_score": 0.6420822306885893}} {"text": "\\documentclass[twoside]{MATH77}\n\\usepackage{multicol}\n\\usepackage[fleqn,reqno,centertags]{amsmath}\n\\begin{document}\n\\begmath 5.3 Eigenvalues of an Unsymmetric Matrix\n\n\\silentfootnote{$^\\copyright$1997 Calif. Inst. of Technology, \\thisyear \\ Math \\`a la Carte, Inc.}\n\n\\subsection{Purpose}\n\nCompute all eigenvalues of a real N $\\times $ N unsymmetric matrix $A$. Some or all\nof the eigenvalues may be complex.\n\n\\subsection{Usage}\n\n\\subsubsection{Program Prototype, Single Precision}\n\n\\begin{description}\n\\item[REAL] \\ {\\bf A}(LDA, $\\geq $ N) \\ [LDA$\\geq $N]{\\bf , VR}($\\geq $ N)%\n{\\bf , VI}($\\geq $N)\n\n\\item[INTEGER] \\ {\\bf LDA, N, IFLAG}($\\geq $ N)\n\\end{description}\n\nAssign values to A(,), LDA, and N.\n$$\n\\fbox{{\\bf CALL SEVUN(A, LDA, N, VR, VI, IFLAG)}}\n$$\nResults are returned in VR(), VI(), IFLAG(1). The contents of A(,) will be\nmodified.\n\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\\item[A(,), LDA, N] \\ A(,) is [inout], LDA and N are [in]. On entry A(,) must\ncontain the N $\\times $ N matrix $A$ whose eigenvalues are to be computed. The\ninteger LDA is the dimension of the first subscript of the array A(,).\nRequire LDA $\\geq $ N. On return the contents of A(,) will be modified.\n\n\\item[VR(), VI()] \\ [out] The subroutine will store the $J^{th}$ eigenvalue\nin VR($J$) and VI($J$), $J =1$, ..., N. The real part is stored in\nVR($J$), and\nthe imaginary part in VI($J$). If the $J^{th}$ eigenvalue is real VI($J$) will\nbe zero. The eigenvalues will be sorted so that VR(1) $\\leq $ VR(2) $\\leq $\n... $\\leq $ VR(N), and if VR($J$) = VR($J$+1) for some $J$ then $|$VI$(J)|\n\\leq |$VI$(J+1)|.$\n\nComplex eigenvalues will occur in conjugate pairs. Such pairs will be stored\nin adjacent locations with the eigenvalue having positive imaginary part\npreceding its conjugate partner.\n\n\\item[IFLAG()] \\ [out, scratch] The N-array IFLAG() will be used as INTEGER\nworking space. In addition, the first location, IFLAG(1), will be used to\npass information back to the user as follows:\n\n\\begin{itemize}\n\\item[= 1] If successful and all eigenvalues are real.\n\n\\item[= 2] If successful and some eigenvalues are complex.\n\\end{itemize}\n\nSee Section E for use of IFLAG(1) in error conditions.\n\\end{description}\n\n\\subsubsection{Modifications for Double Precision}\n\nChange SEVUN to DEVUN, and the REAL type statement to DOUBLE PRECISION.\n\n\\subsection{Examples and Remarks}\n\n{\\bf Example~1:} Define\n\\begin{equation*}\nA=\\left[\n\\begin{array}{rrr}\n-954 & -464 & -2088 \\\\\n792 & 387 & 1728 \\\\\n264 & 128 & 579\n\\end{array}\n\\right] .\n\\end{equation*}\nThe eigenvalues of $A$ are $\\lambda _1 = 3$, $\\lambda _2 = 3$, $\\lambda _3\n= 6$. This example was constructed by using the matrix called $A^{-1}$ in\nExample~3.1, page~29, of \\cite{Gregory:1969:ACM} as the matrix of\neigenvectors. It illustrates the case of a double eigenvalue having a\nfull set of eigenvectors.\n\n{\\bf Example~2:} Define\n\\begin{equation*}\nB=\\left[\n\\begin{array}{ccc}\n4 & 1 & 1 \\\\\n2 & 4 & 1 \\\\\n0 & 1 & 4\n\\end{array}\n\\right] .\n\\end{equation*}\nThis is Example~5.2 of page~82 of~\\cite{Gregory:1969:ACM}. The\neigenvalues are $\\lambda _1 = 3 $, $\\lambda _2 = 3$, and $\\lambda _3 = 6$.\nThe matrix is defective in that there is only a one-dimensional space of\neigenvectors associated with the eigenvalue~3.\n\n{\\bf Example 3:} Define\n\\begin{equation*}\nC=\\left[\n\\begin{array}{rrr}\n8 & -1 & -5 \\\\\n-4 & 4 & -2 \\\\\n18 & -5 & -7\n\\end{array}\n\\right] .\n\\end{equation*}\nThis is Example~5.4 of page~84 of \\cite{Gregory:1969:ACM}. The\neigenvalues are $\\lambda _1 = 1 $, $\\lambda _2 = 2 + 4i$, and $\\lambda _3\n= 2 - 4i.$\n\nThe demonstration program DRSEVUN below applies SEVUN to compute eigenvalues\nfor the above three matrices. Results are in the file ODSEVUN.\n\n\\subsection{Functional Description}\n\nGiven an N $\\times $ N real unsymmetric matrix $A$ there exists an N $\\times $ N\nnonsingular matrix $C$ such that the matrix\n\\begin{equation*}\nU=C^{-1}AC\n\\end{equation*}\nis N $\\times $ N upper triangular. The matrices $C$ and $U$ may be complex. The\ndiagonal elements of $U$ are called the eigenvalues of $A$. This set of N\nnumbers is uniquely determined by $A$ although $C$ and $U$ are not unique.\nNote that $\\lambda $ is an eigenvalue of $A$ if and only if $A - \\lambda I$\nis singular.\n\nThis subroutine SEVUN was developed using the subroutines BALANC, ELMHES,\nand HQR from the EISPACK package of eigenvalue-eigenvector subroutines,\n\\cite{Smith:1974:MER}. The Fortran subroutines in EISPACK\nare based directly on the earlier set of Algol procedures described in\n\\cite{Wilkinson:1971:HAC}.\n\nSubroutine SEVUN first calls SEVBH which consists of the two EISPACK\nsubroutines, BALANC and ELMHES.\n\nBALANC applies similarity permutations to isolate eigenvalues available by\ninspection, if any. It then applies diagonal similarity scaling to balance\nthe size of the matrix elements%\n\\begin{equation*}\nB=D^{-1}P^TAPD.\n\\end{equation*}\nELMHES reduces $B$ to upper Hessenberg form using stabilized elementary\ntransformations%\n\\begin{equation*}\nH=G^{-1}BG.\n\\end{equation*}\nThe remainder of SEVUN consists of a minor modification of HQR\nwhich applies the QR algorithm to $H$. This is an iterative process that\nreduces $H$ to a real nearly-upper-triangular matrix $R$%\n\\begin{equation*}\nR=Q^THQ.\n\\end{equation*}\nThe matrix $R$ has a mixture of single elements and $2 \\times 2$ blocks on its\ndiagonal and is otherwise upper triangular.\n\nThe eigenvalues of $A$ are the single diagonal elements of $R$ along with the\neigenvalues of the $2 \\times 2$ blocks on the diagonal of $R$. These latter\neigenvalues are computed by direct formulas.\n\nSEVUN reorders the eigenvalues to achieve the ordering described\npreviously in Section~B.\n\n\\bibliography{math77}\n\\bibliographystyle{math77}\n\n\\subsection{Error Procedures and Restrictions}\n\nIf N $\\leq 0$ or if there is convergence failure in the QR algorithm the\nerror processing subroutine ERMSG of Chapter 19.2 will be called with an\nerror level 0 to print an error message. Upon return, IFLAG(1) = 3~or~4 to\nindicate N $\\leq 0$ or convergence failure, respectively.\n\nIn these error conditions all computed eigenvalues should be regarded as\ninvalid.\n\n\\subsection{Supporting Information}\n\nThe source language is ANSI Fortran~77.\n\n\\begin{tabular}{@{\\bf}l@{\\hspace{5pt}}l}\n\\bf Entry & \\hspace{.35in} {\\bf Required Files}\\vspace{2pt} \\\\\nDEVUN & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, DEVBH, DEVUN, ERFIN, ERMSG\\rule[-5pt]{0pt}{8pt}}\\\\\nSEVUN & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\nAMACH, ERFIN, ERMSG, SEVBH, SEVUN}\\\\\n\\end{tabular}\n\nThe EISPACK package of Fortran subroutines was acquired at JPL from Argonne\nNational Laboratories where it was developed with financial support from the\nAEC and the NSF. The subroutine SEVUN was written by F. T. Krogh, JPL,\nOctober~1991.\n\n\n\\begcode\n\n\\medskip\\\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRSEVUN}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{sevun}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODSEVUN}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{sevun}}\n\\end{document}\n", "meta": {"hexsha": "5fa4a799b8f9499a875ad6f44e1c7513ff93ffde", "size": 7077, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doctex/ch05-03.tex", "max_stars_repo_name": "jacobwilliams/math77", "max_stars_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2016-01-04T03:17:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-25T19:17:42.000Z", "max_issues_repo_path": "doc/doctex/ch05-03.tex", "max_issues_repo_name": "jacobwilliams/math77", "max_issues_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2019-01-17T02:48:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T16:04:58.000Z", "max_forks_repo_path": "doc/doctex/ch05-03.tex", "max_forks_repo_name": "jacobwilliams/math77", "max_forks_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-01-07T09:26:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-11-25T05:32:54.000Z", "avg_line_length": 33.3820754717, "max_line_length": 98, "alphanum_fraction": 0.725024728, "num_tokens": 2250, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.8289387998695209, "lm_q1q2_score": 0.6420822208685436}} {"text": "% This is a model template for the solutions in computational science. You can find a very useful documentation for LaTeX in Finnish at ftp://ftp.funet.fi/pub/TeX/CTAN/info/lshort/finnish/ or in English at ftp://ftp.funet.fi/pub/TeX/CTAN/info/lshort/english/. The section List of mathematical symbols in Chapter 3 is especially useful for the typesetting of mathematical formulas.\n\n% Compile the document to PDF by command 'pdflatex model.tex' in the terminal. The command must be run twice for the references in the text to be correct.\n\n\\documentclass[a4paper,11pt]{article}\n\\usepackage[utf8]{inputenc}\n% This includes letters such as � and �\n\\usepackage[T1]{fontenc}\n% Use here 'Finnish' for Finnish hyphenation. You may have to compile the code twice after the change. \n\\usepackage[english]{babel}\n\\usepackage{graphicx}\n% Some math stuff\n\\usepackage{amsmath,amsfonts,amssymb,amsbsy,commath,booktabs,hyperref} \n% This is just to include the urls\n\\usepackage{hyperref,subcaption}\n\\usepackage[margin=2cm]{geometry}\n\n\\setlength{\\parindent}{0mm}\n\\setlength{\\parskip}{1.0\\baselineskip}\n\n\\usepackage{listings}\n\\usepackage{color}\n\\usepackage{pdfpages}\n\n\\definecolor{dkgreen}{rgb}{0,0.6,0}\n\\definecolor{gray}{rgb}{0.5,0.5,0.5}\n\\definecolor{mauve}{rgb}{0.58,0,0.82}\n\n\\lstset{frame=tb,\n\tlanguage=Python,\n\taboveskip=3mm,\n\tbelowskip=3mm,\n\tshowstringspaces=false,\n\tcolumns=flexible,\n\tbasicstyle={\\tiny\\ttfamily},\n\tnumbers=none,\n\tnumberstyle=\\tiny\\color{gray},\n\tkeywordstyle=\\color{blue},\n\tcommentstyle=\\color{dkgreen},\n\tstringstyle=\\color{mauve},\n\tbreaklines=true,\n\tbreakatwhitespace=true,\n\ttabsize=4\n}\n\n\\begin{document}\n\n\\title{Becs-114.1100 Computational Science -- exercise round 8} % Replace the exercise round number\n\\author{Kunal Ghosh, 546247} % Replace with your name and student number\n\\maketitle\n\\section{Solution to Question 3}\n\\subsection{Using Box-Muller algorithm}\\label{prob3a}\nIn this exercise we use the Box-Muller algorithm to generate two gaussing distributed random variables. We generate $10^5$ random number pairs and plot them separately in two histograms shown below.\\\\\nHere we have used the standard python random number generator which uses \\textbf{Mersenel Primes} to generate a random number in the range of [0,1]\n\\begin{figure}[ht]\n \\begin{subfigure}{.5 \\textwidth}\n\t\\centering\n \\includegraphics[scale=0.40]{figure3_y1s.png}\n \\caption{$Y_1$ as per $\\sqrt(-2\\log(x1))*\\cos(2*\\pi*x2)$}\n\t\\label{fig:Y1}\n\\end{subfigure}\n \\begin{subfigure}{.5 \\textwidth}\n\t\\centering\n \\includegraphics[scale=0.40]{figure3_y2s.png}\n \\caption{$Y_1$ as per $\\sqrt(-2\\log(x1))*\\cos(2*\\pi*x2)$}\n\t\\label{fig:Y1}\n\\end{subfigure}\n \\caption{This plot shows Y1 and Y2, two random variables generated from two uniformly distributed random variables(x1 and x2) and transformed as showing under the respective graphs.}\n\\end{figure}\n\nThese are two zero mean and unit variance gaussian distributed random numbers.\n\nThe corresponding python code can be found at \\ref{code:problem3a}\n\\section{Solution 4a}\\label{prob4a}\n\nIn this section we compare the 1st, 2nd and 3rd moments of two random number generators. LCG(128, 0, 509) and Mersene Twister. We compare these to the true moments defined by:\n\\begin{equation}\n \\langle x^{k} \\rangle = \\frac{1}{(k+1)}\n\\end{equation}\nIn the above equation k is the moment. i.e. k = 1 gives us the 1st moment, k = 2 gives us the second moment and so on.\nWe calculate the moments for different random number generators by generating a sequence of $N$ random numbers from the generator and using the formula below. \n\\begin{equation}\n \\langle x^k \\rangle = \\frac{1}{N}\\sum_{i=1}^{N}x_{i}^{k}\n\\end{equation}\n\nIn this experiment we are trying to see whether the the two random number generators are truly uniform. If they are then their moments must be close to the true moments of the uniform distribution.\n\n\\begin{table}[ht]\n\\centering\n\\label{1moment}\n\\begin{tabular}{|c|c|c|c|}\n\\hline\n\\textbf{N} & \\textbf{LCG(128,0,509)}&\\textbf{Mersene Twister}&\\textbf{True Moment} \\\\ \\hline\n10 & 0.0264746457362 & 0.518080782167 & 0.5 \\\\\n100 & 0.00240679565587 & 0.474926738729 & 0.5 \\\\\n1000 & 0.000238511281212 & 0.494459184012 & 0.5 \\\\\n10000 & 2.38296599591e-05 & 0.493928278685 & 0.5 \\\\\n100000 & 2.38275152683e-06 & 0.500197518937 & 0.5 \\\\\n1000000 & 2.38273008204e-07 & 0.499728476939 & 0.5 \\\\\n10000000 & 2.38272793759e-08 & 0.500054377307 & 0.5 \\\\\n\\hline\n\\end{tabular}\n\\caption{Values for the 1st Moment}\n\\end{table}\n\n\\begin{table}[ht]\n\\centering\n\\label{1moment}\n\\begin{tabular}{|c|c|c|c|}\n\\hline\n\\textbf{N} & \\textbf{LCG(128,0,509)}&\\textbf{Mersene Twister}&\\textbf{True Moment} \\\\ \\hline\n10 & 0.00377304392308 & 0.310275753489 & 0.333333333333 \\\\\n100 & 0.000343003993013 & 0.311212396032 & 0.333333333333 \\\\\n1000 & 3.3991386695e-05 & 0.328312667816 & 0.333333333333 \\\\\n10000 & 3.39607913874e-06 & 0.326612059681 & 0.333333333333 \\\\\n100000 & 3.39577348856e-07 & 0.333317261858 & 0.333333333333 \\\\\n1000000 & 3.39574292657e-08 & 0.333169017655 & 0.333333333333 \\\\\n10000000 & 3.3957398704e-09 & 0.3333760948 & 0.333333333333 \\\\\n\\hline\n\\end{tabular}\n\\caption{Values for the 2nd Moment}\n\\end{table}\n\n\\begin{table}[ht]\n\\centering\n\\label{1moment}\n\\begin{tabular}{|c|c|c|c|}\n\\hline\n\\textbf{N} & \\textbf{LCG(128,0,509)}&\\textbf{Mersene Twister}&\\textbf{True Moment} \\\\ \\hline\n10 & 0.000640566707312 & 0.199266027703 & 0.25 \\\\\n100 & 5.82333370283e-05 & 0.228229711742 & 0.25 \\\\\n1000 & 5.77087123704e-06 & 0.246069547559 & 0.25 \\\\\n10000 & 5.7656769335e-07 & 0.243561537242 & 0.25 \\\\\n100000 & 5.76515801739e-08 & 0.249764456768 & 0.25 \\\\\n1000000 & 5.76510613091e-09 & 0.249906635608 & 0.25 \\\\\n10000000 & 5.76510094232e-10 & 0.250034371239 & 0.25 \\\\\n\\hline\n\\end{tabular}\n\\caption{Values for the 3rd Moment}\n\\end{table}\nWe can see that in all the 3 cases above, the LCG deviates from each of the moments quite a bit as we take more samples from the random number generator. However, the Mersent Twister stays quite close to the true moment even when we take a lot of samples from the generator which means that the random number it generates are close to uniformly distributed.\n% \\begin{figure}[ht]\n% \t\\center\n% \\includegraphics[scale=0.75]{matlabS_dash.png}\n% \\caption{Plot showing the first derivative of the natural cubic interpoland S'(X) as calculated using Matlab. Values of $S'(X_i)$ for $X_i$ in the range of (min(knot values), max(knot values)) are shown as red circles.} \n% \t\\label{fig:sdash_matlab}\n% \\end{figure}\n% The corresponding plots created by matlab are much more smoother because matlab implements an additional constraint that the third derivatives of the piecewise polynomials are also equal at the second and the last knots. This are apparently called the \"Not-A-Knot\" end conditions. This is only done when the length of \\textbf{t} and \\textbf{y} are same.\n% \n% Natural splines are a good choice only when the functions have 0 second derivative at the end points. I read it up from here. \\url{http://www.mathworks.com/matlabcentral/newsreader/view_thread/172988} would really appreciate some more information about why using the Not-A-Knot condition is better. \n% \\\\\n% The corresponding matlab code can be found at \\ref{code:problem2b}\n\nThe corresponding python code can be found at \\ref{code:problem4a}\n\n\\section{Solution 4b}\\label{prob4b}\nIn this solution we are trying to validate the central limit theorem. \\textit{An average of measured quantities, taken from the same distribution (which can be any distribution), will asymptotically follow the normal distribution.}\nHere we draw random numbers from the uniform distribution and since the generator generates values close to the uniform distribution the mean of the values as we draw more samples tends to the normal distribution.\n\\begin{figure}[ht]\n \\begin{subfigure}{.5 \\textwidth}\n\t\\centering\n \\includegraphics[scale=0.45]{fig4b.png}\n \\caption{N = 100, m = 10000}\n\t\\label{fig:clt_10000}\n \\end{subfigure}\n \\begin{subfigure}{.5 \\textwidth}\n\t\\centering\n \\includegraphics[scale=0.45]{fig4b_100000.png}\n \\caption{N = 100, m = 100000}\n\t\\label{fig:clt_100000}\n \\end{subfigure}\n \\caption{Figure validating the central limit theorem in the figure to the left we have plotted 10000 means of $\\langle x^2 \\rangle$100 random numbers and can see that the figure is almost like the normal distribution. This is confirmed as the number of means are increased to 100,000 of 100 random numbers and the graph looks very much like the normal distribution. Bins = 100 interval = [0.2,0.5] this interval is chosen because the mean of the values is around 0.33333}\n\\end{figure}\n\n\nThe corresponding python code can be found at \\ref{code:problem4b}\n\\clearpage\n\\section{Appendix A}\\label{code:problem3a}\nPython source code for \\ref{prob3a}.\n{\\footnotesize\n\\begin{lstlisting}\nfrom __future__ import division\nimport random\nimport math\nimport pylab as pl\n\npi = math.pi\nlog = math.log\ncos = math.cos\nsin = math.sin\nsqrt = math.sqrt\nfloor = math.floor\n\ndef get_gauss_random():\n x1 = random.random()\n x2 = random.random()\n\n y1 = sqrt(-2*log(x1))*cos(2*pi*x2)\n y2 = sqrt(-2*log(x1))*sin(2*pi*x2)\n\n # We don't multiply by sigma\n # implying the value is unit variance\n\n # Since are random numbers are uniform\n # between 0 and 1, we have zero mean.\n\n # Hence our distributionis zero mean and \n # unit variance.\n return y1,y2\n\n\ndef get_bin(yi, ymin, ymax, Nbin):\n val = ((yi - ymin) / (ymax - ymin)) * Nbin\n return floor(val)\n\nif __name__ == '__main__':\n seed = 7777\n random.seed(seed)\n Nbin = 100\n bins = [0] * Nbin\n ymin, ymax = -5, 5\n binVals = []\n ybins = []\n for _ in xrange(100000):\n y = get_gauss_random()\n ybins.append(y)\n #binVals.append(get_bin(yi, ymin, ymax, Nbin))\n # binNum = get_bin(yi, ymin, ymax, Nbin)\n # bins[binNum]+=1\n y1s,y2s = zip(*ybins)\n pl.figure()\n pl.grid()\n pl.hist(y1s,bins=Nbin,range=(ymin,ymax))\n #pl.hist(binVals, bins=Nbin)\n pl.xlabel(\"Bins\")\n pl.title(\"Y1s\")\n pl.savefig(\"figure3_y1s.png\")\n pl.figure()\n pl.grid()\n pl.hist(y2s,bins=Nbin,range=(ymin,ymax))\n #pl.hist(binVals, bins=Nbin)\n pl.xlabel(\"Bins\")\n pl.title(\"Y2s\")\n pl.savefig(\"figure3_y2s.png\")\n pl.show()\n\\end{lstlisting}\n}\n\\clearpage\n\\section{Appendix B}\\label{code:problem4a}\nPython source code \\ref{prob4a}.\n{\\footnotesize\n\\begin{lstlisting}\n\nfrom __future__ import division\nimport random\nimport math\nimport pylab as pl\n\npi = math.pi\nlog = math.log\ncos = math.cos\nsin = math.sin\nsqrt = math.sqrt\nfloor = math.floor\n\ndef lcg(x):\n # Implementing lcg(a,b,m)\n # a = 128 b = 0 m = 509\n a = 128\n b = 0\n m = 509\n randVal = (a*x + b)%m\n return randVal / m # scale the rand num to be in 0,1 range\n\ndef get_moment(rand_nums, m):\n return sum([x**m for x in rand_nums])/len(rand_nums)\n\nif __name__ == '__main__':\n seed = 7777\n random.seed(seed)\n randLcg = lcg(seed) \n lcgs = [randLcg]\n srgl = []\n N = 10**7\n moment = [[],[],[]]\n for _ in xrange(N):\n randLcg = lcg(randLcg)\n lcgs.append(randLcg)\n srgl.append(random.random())\n for m in [1,2,3]:\n trueMoment = 1/(m+1)\n print \"M = {}\".format(m)\n for N in [10, 100, 1000, 10000, 100000, 1000000, 10000000]:\n lcgMoment = get_moment(lcgs[1:N], m)\n srglMoment = get_moment(srgl[1:N], m)\n moment[m-1].append((lcgMoment, srglMoment, trueMoment))\n print(\"{} & {} & {} & {} \\\\\\\\\".format(N,lcgMoment,srglMoment,trueMoment))\n\n \n\\end{lstlisting}\n}\n\\section{Appendix C}\\label{code:problem4b}\nPython source code for \\ref{prob4b}.\n{\\footnotesize\n\\begin{lstlisting}\nfrom __future__ import division\nimport random\nimport math\nimport pylab as pl\n\npi = math.pi\nlog = math.log\ncos = math.cos\nsin = math.sin\nsqrt = math.sqrt\nfloor = math.floor\n\ndef get_moment(rand_nums, m):\n return sum([x**m for x in rand_nums])/len(rand_nums)\n\nif __name__ == '__main__':\n seed = 7777\n random.seed(seed)\n N = 100\n m = 10000\n means = []\n for _ in xrange(m):\n temp = 0\n for _ in xrange(N):\n temp += random.random() ** 2\n means.append(temp/N)\n pl.hist(means, bins=100, range=(0.2,0.5))\n pl.savefig(\"fig4b.png\")\n pl.show()\n\\end{lstlisting}\n}\n\\end{document}\n\n \n\n", "meta": {"hexsha": "bf2b17f9c4d11506ff53c42f00b4066942046d5f", "size": 12332, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercise8/report.tex", "max_stars_repo_name": "kunalghosh/BECS-114.1100-Computational-Science", "max_stars_repo_head_hexsha": "ca91ac59cb5276d213c1aec50ae7786efe72ae43", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "exercise8/report.tex", "max_issues_repo_name": "kunalghosh/BECS-114.1100-Computational-Science", "max_issues_repo_head_hexsha": "ca91ac59cb5276d213c1aec50ae7786efe72ae43", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "exercise8/report.tex", "max_forks_repo_name": "kunalghosh/BECS-114.1100-Computational-Science", "max_forks_repo_head_hexsha": "ca91ac59cb5276d213c1aec50ae7786efe72ae43", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.1642228739, "max_line_length": 475, "alphanum_fraction": 0.706373662, "num_tokens": 3960, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085909370422, "lm_q2_score": 0.8175744695262775, "lm_q1q2_score": 0.6420482546497808}} {"text": "\\documentclass[11pt]{article}\n\n\n\\usepackage[margin=1in]{geometry}\n\n\\usepackage{mathpazo}\n\n\\usepackage{amsmath}\n\n\\newcommand{\\numpy}{{\\sffamily NumPy}}\n\n\\usepackage[defaultsans]{cantarell} %% Use option ``defaultsans'' to use cantarell as sans serif only\n\\usepackage[T1]{fontenc}\n\n% this needs to be done before the fncychap style, since that will \n% redo the chapter stuff\n\\usepackage{sectsty}\n\\allsectionsfont{\\sffamily}\n\n\\usepackage{graphicx}\n\n% prefer PDF to PNG\n\\DeclareGraphicsExtensions{% \n .pdf}\n\n\n\\begin{document}\n\n\\begin{center}\n{\\bfseries \\sffamily \\LARGE Making Sense of Power Spectra} \\\\\nMichael Zingale\n\\end{center}\n\nWe are interested in computing the numerical power spectrum of a\ndiscretely-sampled three-dimensional dataset. These notes serve\nto verify the procedure by starting with a real-space\nfunction with a known analytic power spectrum and computing the\npower spectrum of this test function.\n\nConsider a function that is a superposition of sines with different\nwavelengths. The power will be encoded in the amplitude of the sines.\nNow imagine this function discretely sampled at $N$ points. For a domain of\nlength $L$ the highest possible wavenumber is:\n\\begin{equation}\nk_\\mathrm{max} = \\frac{1}{\\Delta x}\n\\end{equation}\nwith $\\Delta x = L/N$. However, for real-valued data, Nyquist sampling\nmeans that $k_\\mathrm{max}/2$ is the meaningful maximum.\nThe lowest wavenumber is:\n\\begin{equation}\nk_\\mathrm{min} = \\frac{1}{L}\n\\end{equation}\nFigure~\\ref{fig:wavenumbers} shows two different sine modes.\n\n\\suppressfloats\n\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=0.75\\linewidth]{wavenumber}\n\\caption{\\label{fig:wavenumbers} The $k = 1/L$ (top) and $k = 4/L$ (bottom)\nmodes.}\n\\end{figure}\n\n\nThe definition of the Fourier transform is:\n\\begin{equation}\n\\mathcal{F}(k) = \\int_{-\\infty}^{\\infty} f(x) e^{-2\\pi i x k} dx\n\\end{equation}\nwith the inverse transform as\n\\begin{equation}\nf(x) = \\int_{-\\infty}^{\\infty} \\mathcal{F}(x) e^{2\\pi i x k} dk\n\\end{equation}\nHowever, we need to work with discretely-sampled data, so we will need\nthe {\\em discrete Fourier transform}, which is efficiently implemented by\nthe FFT algorithm.\n\n\\section*{\\numpy\\ details}\n\nThe \\numpy\\ FFTs have the discrete form:\n\\begin{equation}\n\\mathcal{F}_k = \\sum_{m=0}^{N-1} f_m \\exp \\left \\{ -2\\pi i \\frac{mk}{N} \\right \\}\n\\end{equation}\nand the inverse is\n\\begin{equation}\nf_m = \\frac{1}{N} \\sum_{k=0}^{N-1} \\mathcal{F}_k \\exp \\left \\{ 2\\pi i \\frac{mk}{N} \\right \\}\n\\end{equation}\nWe see from the inverse that if we want to know the amplitude of a single\nmode at a given spatial location (denoted by index $m$), that $\\mathcal{F}/N$\nis the normalized amplitude, i.e., we can think of this expansion as:\n\\begin{equation}\nf_m = \\sum_{k=0}^{N-1} \\left (\\frac{\\mathcal{F}_k}{N} \\right )\n \\exp \\left \\{ 2\\pi i \\frac{mk}{N} \\right \\}\n\\end{equation}\n\n\nThe {\\tt rfftfreq()} function returns the wavenumbers as\n\\begin{equation}\nk = \\left \\{ 0, \\frac{1}{N}, \\frac{2}{N}, \\ldots, \\frac{N-2}{2N}, \\frac{1}{2} \\right \\}\n\\end{equation}\n(for even $n$, and in the mode where we do not specify the sample spacing).\nWe convert these into a physical quantity (with units of cm$^{-1}$) by\nmultiplying by $1/\\Delta x$, giving the set:\n\\begin{equation}\nk = \\left \\{ 0, \\frac{1}{L}, \\frac{2}{L}, \\ldots, \\frac{1}{2\\Delta x} \\right \\}\n\\end{equation}\n\n\\numpy\\ does not have a real-valued FFT function for 3-d data, so we\nneed to use the {\\tt fftn} function. Because $N$ real numbers results\nin $2N$ complex numbers, only 1/2 of the transformed data (in each\ndimension) will be unique. In 3-d this means we consider only an\noctant of data. In practice, this means keeping the positive\nfrequencies and dropping the negative ones (this is what the {\\tt\n rfftfreq} function does for us).\n\n\n\n\n\\section*{Power}\n\nFor a continuous function, $f(x)$, the ``energy'' is defined as\n\\begin{equation}\nE = \\int_{-\\infty}^\\infty |f(x)|^2 dx\n\\end{equation}\nUsing Parseval's theorem, we have\n\\begin{equation}\nE = \\int_{-\\infty}^\\infty |\\mathcal{F}(k)|^2 dk\n\\end{equation}\n%\nSo when we compute the power spectrum, we will be looking at\n$|\\mathcal{F}(k)|^2$. \\\\\n\n\\noindent [see ``spectral density'' Wikipedia] \\\\\n\nThe discrete version of Parseval's theorem in 1-d is\n\\begin{equation}\n\\sum_{n=0}^{N-1} | f(n) |^2 = \\frac{1}{N} \\sum_{k=0}^{N-1} | \\mathcal{F}(k) |^2\n\\end{equation}\n\n\n\\section*{One-dimensional Example}\n\nThe idea of the power spectrum is very straightforward in 1-d. To illustrate\nthis we will create a function in real space that we expect to have a\npower spectrum scaling as:\n\\begin{equation}\nP \\sim k^{-\\eta}\n\\end{equation}\nThis is accomplished by setting the amplitude of each mode to yield\nthis power. Our function is:\n\\begin{equation}\nf(x) = \\sum_{m=1}^{N_\\mathrm{modes}} a_m \\sin \\left ( \\frac{2\\pi m x}{L}\\right )\n\\end{equation}\nwhere $L$ is the width of the physical domain, and \n\\begin{equation}\na_m = \\left (A_0 k^{-\\eta} \\right )^{1/2}\n\\end{equation}\nThis function is shown in Figure~\\ref{fig:phi1d}.\n\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=0.5\\linewidth]{phi_1d}\n\\begin{minipage}[b]{0.45\\linewidth}\n\\caption{\\label{fig:phi1d} The real-space, one-dimensional function $f(x)$,\nfor the case $N_\\mathrm{modes} = 16$, with $A_0 = 1000$ and $\\eta = -4$.\\newline}\n\\end{minipage}\n\\end{figure}\n\nWe then transform this function, denoting the transform as $\\mathcal{F}(k)$. Plotting\n$|\\mathcal{F}(k)|^2$ vs.\\ $k$ gives the power spectrum. Figure~\\ref{fig:ps1d} shows\nthe result, together with the expected power-law scaling, $P$, shown as the \ndashed line.\n\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=0.5\\linewidth]{ps1d}\n\\begin{minipage}[b]{0.45\\linewidth}\n\\caption{\\label{fig:ps1d} The power spectrum of $f(x)$, compared to the\nexpected result (dashed line).\\newline}\n\\end{minipage}\n\\end{figure}\n\n\n\nSee {\\tt ps1d.py} for the implementation of this test.\n\n\n\n\\section*{Three-dimensional Case}\n\nIn 3-d, things are more complex---traditionally we express the power\nspectrum by integrating over spherical shells with a radius $k$. The\npower spectrum is defined as\n\\begin{equation}\nE(k) = \\frac{1}{\\Omega} \\int_{S(k)} \\mathcal{F}(k) \\mathcal{F}^\\star(k) dS\n\\end{equation}\nwhere $\\Omega$ is the volume in physical space, and the integral is done\nover the surface $S(k)$, defined by $|\\vec{k}| = k$. We can\nwrite $dS = 4\\pi k^2 dk$ and integrate along radial $|\\vec{k}|$.\n\nIn discrete form, we bin the $k$ in our transformed space into a\nnumber of uniformly spaced bins, $\\kappa_l$, where the index\n$l$ denote the bin boundaries, with $\\kappa_0$ the leftmost bin boundary.\nWe also denote the bin centers as $\\bar{\\kappa}_l$ such that\n\\begin{equation}\n\\bar{\\kappa}_l = \\frac{1}{2} ( \\kappa_l + \\kappa_{l+1} )\n\\end{equation}\nOur power spectrum becomes\n\\begin{equation}\nE(\\bar{\\kappa}_l) = \\sum_{\\kappa_{l} \\le |\\vec{k}| < \\kappa_{l+1}}\n \\mathcal{F}(k) \\mathcal{F}^\\star(k)\n\\end{equation}\nwhere the sum is over all the zones in the $k_x \\times k_y \\times k_z$\ncube whose wavenumber magnitude, $k$, falls into our bin,\n$[\\kappa_l,\\kappa_{l+1})$. As we go to higher wavenumbers, the number\n of zones will increase as $k^2$---this captures the $dS$ scaling in\n the continuous-form of the power spectrum.\n\n\nAgain, to test the algorithm, we define a test function and set the power at\ndifferent wavenumbers {\\em in physical space}, and then do the transform\nand power spectrum to see if we get the expected result. We will design\nour function to have the same scaling as the 1-d case, $P \\sim k^{-\\eta}$.\nOur test function\nis:\n\\begin{equation}\nf(x,y,z) = \\sum_{m=1}^{N_\\mathrm{modes}} \n \\sum_{n=1}^{N_\\mathrm{modes}} \n \\sum_{p=1}^{N_\\mathrm{modes}} \n a_{m,n,p} \\sin \\left ( \\frac{2\\pi m x}{L_x} + \n \\frac{2\\pi n y}{L_y} + \n \\frac{2\\pi p z}{L_z} \\right )\n\\end{equation}\nwhere we can identify the wavenumbers in each direction as\n\\begin{equation}\nk_m = \\frac{m}{L_x} \\, ; \\quad \nk_n = \\frac{n}{L_y} \\, ; \\quad \nk_p = \\frac{p}{L_z} \\,\n\\end{equation}\nThis is equivalent to defining a vector wavenumber\n\\begin{equation}\n\\vec{k} = k_m \\hat{x} + k_n \\hat{y} + k_p \\hat{z}\n\\end{equation}\nand expressing a single component as\n\\begin{equation}\nf_{m,n,p} = a_{m,n,p} \\sin(2\\pi \\vec{k}\\cdot \\vec{x})\n\\end{equation}\n\n\n\nWe can set the amplitude of component based on our desired\npower-law scaling. Defining $k = |k|$, and $A_0$ as a\nreference amplitude of the power, we take\n\\begin{equation}\na_{m,n,p}^2 = \\frac{A_0 k^{-\\eta}}{W(k)} \\frac{1}{\\Delta S}\n\\end{equation}\nwhere $W(k)$ is the weight that accounts for the fact that multiple\ncombinations of $m, n, p$ can result in the same $k$. Using Iverson bracket\nnotation, defined as\n\\begin{equation}\n[P] = \\begin{cases}\n 1 & \\text{if P is true} \\\\\n 0 & \\text{otherwise}\n \\end{cases}\n\\end{equation}\nwe find $W(k)$ as\n\\begin{equation}\nW(k) = \\sum_{m=1}^{N_\\mathrm{modes}}\n \\sum_{n=1}^{N_\\mathrm{modes}} \n \\sum_{p=1}^{N_\\mathrm{modes}} [ k_m^2 + k_n^2 + k_p^2 = k^2 ]\n\\end{equation}\n\nThe $\\Delta S$ in the weighting reflects the fact that the power \nin mode $k$ resides in a spherical shell---we need to divide this\narea out so when we integrate over $dS$, we get the correct power\nin the shell. For $|k| \\gg \\Delta \\kappa$, we can take $\\Delta S \\sim\n4\\pi k^2$. But we can do better by recognizing that eventually we\nwill bin the data up into the bins given by $\\{\\kappa_l\\}$. If we\ndenote the bin $\\xi$ such that $\\kappa_\\xi \\le k < \\kappa_{\\xi+1}$,\nthen we can take\n\\begin{equation}\n\\Delta S = 4 \\pi \\bar{\\kappa}_\\xi^2\n\\end{equation}\n\nThis normalization should give an amplitude in the power spectrum at the lowest\n$k$ that agrees with the corresponding amplitude in real-space for that same $k$. \nFigure~\\ref{fig:phireal} shows the function in three orthogonal slice planes\nthrough the center of the domain.\n\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=\\linewidth]{phi}\n\\caption{\\label{fig:phireal} $f(x,y,z)$ shown in three orthogonal slice\nplanes through the center of the domain.}\n\\end{figure}\n\nSince the power spectrum integrated over shells, we don't include the $4\\pi$\nin the real space amplitude when comparing our power spectrum to the $P \\sim k^{-\\eta}$\nexpected result. Figure~\\ref{fig:ps} shows the power spectrum of our function.\n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=0.5\\linewidth]{ps}\n\\begin{minipage}[b]{0.45\\linewidth}\n\\caption{\\label{fig:ps} The power spectrum of our function $f(x,y,z)$ compared to\nthe expected result (dashed line).\\newline}\n\\end{minipage}\n\\end{figure}\n\n\nAs a final check on the overall normalization, we can use Parseval's theorem\nto see if we have the correct amount of energy in the spectrum. In 3-d,\nParseval's theorem says:\n\\begin{equation}\nE = \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} |f(x,y,z)|^2 dx dy dz = \n \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} |\\mathcal{F}(k_x,k_y,k_z)|^2 dk_x dk_y dk_z\n\\end{equation}\nor in discrete form:\n\\begin{equation}\n\\sum_{j=0}^{N_x} \\sum_{k=0}^{N_y} \\sum_{l=0}^{N_z}\n | f(x_j, y_k, z_l) |^2 = \\frac{1}{N_x N_y N_z} \n \\sum_{k_x=0}^{N_x-1} \\sum_{k_y=0}^{N_y-1} \\sum_{k_z=0}^{N_z-1} \n | \\mathcal{F}(k_x, k_y, k_z) |^2\n\\end{equation}\nWe can express the integral over $d^3k$ in terms of the spherical $|k|$ as\n\\begin{equation}\nE = 4\\pi \\int_{-\\infty}^{\\infty} |\\mathcal{F}(k)|^2 k^2 dk = \\int_{-\\infty}^{\\infty} E(k) dk\n\\end{equation}\nwhere we substitute in the power spectrum, $E(k)$. This means that if we integrate up our\npower spectrum, we should get the same energy as the integral\nof $|f(x,y,z)|^2$ over physical space. We need to keep track of the normalizations---in\ncreating the power spectrum, we used $\\mathcal{F}/N^3$, so our discrete sum over the energy\nspectrum will have a $N^{-6}$ normalization, while the discrete form of Parseval's theorem\nhas an $N^{-3}$ normalization.\n\n\n\\end{document}\n", "meta": {"hexsha": "c8795dc1b9ed9a5e9ae0d5212fa8afdb69017aa2", "size": 11914, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "overview.tex", "max_stars_repo_name": "zingale/powerspectrum_test", "max_stars_repo_head_hexsha": "140f4a70b43edf3fb68fc8ab1d56818f7a6c7a3b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-03-17T14:53:13.000Z", "max_stars_repo_stars_event_max_datetime": "2018-05-02T21:54:17.000Z", "max_issues_repo_path": "overview.tex", "max_issues_repo_name": "zingale/powerspectrum_test", "max_issues_repo_head_hexsha": "140f4a70b43edf3fb68fc8ab1d56818f7a6c7a3b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "overview.tex", "max_forks_repo_name": "zingale/powerspectrum_test", "max_forks_repo_head_hexsha": "140f4a70b43edf3fb68fc8ab1d56818f7a6c7a3b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.353115727, "max_line_length": 118, "alphanum_fraction": 0.6978344804, "num_tokens": 3975, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.8175744739711883, "lm_q1q2_score": 0.6420482499243692}} {"text": "\\section{Orthogonal projections and Fourier series}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Compute the orthogonal projection of a vector onto a subspace.\n \\item Find the least squares approximation of a function by a\n polynomial of a given degree. \n \\item Calculate the generalized Fourier series of a vector.\n \\item Use Fourier series to approximate a function by polynomials or\n trigonometric functions.\n \\end{enumerate}\n\\end{outcome}\n\nWe now reconsider a problem that we briefly encountered, only in the\ncontext of $\\R^3$, in Section~\\ref{sec:planes}: how to find the\nshortest distance between a point and a subspace. The method we used\nin Section~\\ref{sec:planes} (see\nExample~\\ref{exa:shortest-distance-plane}) relies on the existence of\nnormal vectors and does not generalize beyond $\\R^3$. The following\nproposition gives a much better method for solving this problem,\nprovided that we have an orthogonal basis of the subspace.\n\n\\begin{proposition}{Orthogonal projection onto a subspace}{projection-subspace}\n Let $V$ be an inner product space, and let $W$ be a subspace of\\/\n $V$. Assume $\\set{\\vect{u}_1,\\ldots,\\vect{u}_k}$ is an orthogonal\n basis of\\/ $W$, and $\\vect{v}\\in V$ is any vector. Then the\n following vector $\\vect{v}'$ is the element of $W$ that is closest\n to $\\vect{v}$, i.e., such that $\\norm{\\vect{v}-\\vect{v}'}$ is as\n small as possible.\n \\begin{equation*}\n \\vect{v}' =\n \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n + \\ldots\n + \\frac{\\iprod{\\vect{u}_k,\\vect{v}}}{\\iprod{\\vect{u}_k,\\vect{u}_k}}\\,\\vect{u}_k.\n \\end{equation*}\n Moreover, the vector $\\vect{v}-\\vect{v}'$ is orthogonal to $W$.\n \\begin{center}\n \\begin{tikzpicture}[x={(1cm,-0.1cm)},y={(0.4cm,0.5cm)},z={(0cm,1cm)}]\n \\filldraw[draw=red!80,fill=red!10](-4,-4,0) -- (4,-4,0) -- (4,4,0) -- (-4,4,0) -- cycle;\n \\path[red] (4.5,0,0) node[right] {$W$};\n \\draw[thin,red!50] (-2,-4,0) -- (-2,4,0);\n \\draw[thin,red!50] (0,-4,0) -- (0,4,0);\n \\draw[thin,red!50] (2,-4,0) -- (2,4,0);\n \\draw[thin,red!50] (-4,-2,0) -- (4,-2,0);\n \\draw[thin,red!50] (-4,0,0) -- (4,0,0);\n \\draw[thin,red!50] (-4,2,0) -- (4,2,0);\n \\draw[->,thick,blue!80!black](0,0,0) -- node[below, pos=0.45]{$\\vect{u}_1$} (2,0,0);\n \\draw[->,thick,blue!80!black](0,0,0) -- node[left, pos=0.55]{$\\vect{u}_2$} (0,2,0);\n \\draw[->,thick,green!50!black](0,0,0) -- node[above, pos=0.7] {$\\vect{v}$} (3,2.5,2);\n \\draw[->,thick,green!50!black](0,0,0) -- node[above, pos=0.7] {$\\vect{v}'$} (3,2.5,0);\n \\draw[dashed](3,2.5,0) -- (3,2.5,2);\n \\fill (0,0,0) circle [radius=2.2pt] node [left=3pt] {$\\vect{0}$};\n \\fill (3,2.5,2) circle [radius=2.2pt];\n \\fill (3,2.5,0) circle [radius=2.2pt];\n \\end{tikzpicture}\n \\end{center}\n The vector $\\vect{v}'$ is called the \\textbf{orthogonal projection\n of $\\vect{v}$ onto $W$}%\n \\index{orthogonal projection!onto subspace}%\n \\index{projection!onto subspace}. We also say that $\\vect{v}'$ is\n the \\textbf{best approximation}%\n \\index{approximation} of $\\vect{v}$ in $W$.\n\\end{proposition}\n\n\\begin{proof}\n In case $\\vect{v}\\in W$, then by\n Proposition~\\ref{prop:fourier-coefficients}, we have\n \\begin{equation*}\n \\vect{v} =\n \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n + \\ldots\n + \\frac{\\iprod{\\vect{u}_k,\\vect{v}}}{\\iprod{\\vect{u}_k,\\vect{u}_k}}\\,\\vect{u}_k,\n \\end{equation*}\n and therefore $\\vect{v}'=\\vect{v}$. In that case\n $\\norm{\\vect{v}-\\vect{v}'}=0$, which is clearly as small as\n possible, and $\\vect{v}-\\vect{v}'=\\vect{0}$ is orthogonal to $W$, so\n we are done.\n\n Now assume that $\\vect{v}\\not\\in W$. Let\n $W'=\\sspan\\set{\\vect{u}_1,\\ldots,\\vect{u}_k,\\vect{v}}$. Apply the\n Gram-Schmidt procedure to the $k+1$ vectors\n $\\vect{u}_1,\\ldots,\\vect{u}_k,\\vect{v}$ to obtain an orthogonal\n basis of $W'$. Since the first $k$ vectors are already orthogonal,\n the Gram-Schmidt procedure does not change them, and we therefore\n obtain an orthogonal basis of $W'$ of the form\n $\\vect{u}_1,\\ldots,\\vect{u}_k,\\vect{u}_{k+1}$. Since $\\vect{v}\\in\n W'$, we can write $\\vect{v}$ as a linear combination of these basis\n vectors, i.e., $\\vect{v}=a_1\\vect{u}_1 + \\ldots + a_k\\vect{u}_k +\n a_{k+1}\\vect{u}_{k+1}$. Moreover, by\n Proposition~\\ref{prop:fourier-coefficients}, we know that each $a_j$\n is the corresponding Fourier coefficient of $\\vect{v}$, i.e., $a_j =\n \\frac{\\iprod{\\vect{u}_j,\\vect{v}}}{\\iprod{\\vect{u}_j,\\vect{u}_j}}$.\n\n Now consider any $\\vect{w}\\in W$. Since\n $\\set{\\vect{u}_1,\\ldots,\\vect{u}_k}$ is a basis of $W$, we can write\n $\\vect{w} = x_1\\vect{u}_1 + \\ldots + x_k\\vect{u}_k$. Using\n Proposition~\\ref{prop:norm-orthogonal}, we have\n \\begin{eqnarray*}\n \\norm{\\vect{v}-\\vect{w}}^2\n &=& \\norm{(a_1-x_1)\\vect{u}_1 + \\ldots + (a_k-x_k)\\vect{u}_k + a_{k+1}\\vect{u}_{k+1}}^2 \\\\\n &=& (a_1-x_1)^2\\norm{\\vect{u}_1}^2 ~+~ \\ldots ~+~ (a_k-x_k)^2\\norm{\\vect{u}_k}^2 ~+~ a_{k+1}^2\\norm{\\vect{u}_{k+1}}^2.\n \\end{eqnarray*}\n Here, $a_1,\\ldots,a_{k+1}$ are the fixed coordinates of $\\vect{v}$,\n and $x_1,\\ldots,x_k$ depend on $\\vect{w}$. Therefore,\n $\\norm{\\vect{v}-\\vect{w}}$ takes its smallest value when $x_j=a_j$,\n for $j=1,\\ldots,k$. So the minimum value of\n $\\norm{\\vect{v}-\\vect{w}}$ occurs when\n \\begin{equation*}\n \\vect{w}\n ~~=~~ a_1\\vect{u}_1 + \\ldots + a_k\\vect{u}_k \n ~~=~~ \n \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n + \\ldots\n + \\frac{\\iprod{\\vect{u}_k,\\vect{v}}}{\\iprod{\\vect{u}_k,\\vect{u}_k}}\\,\\vect{u}_k\n ~~=~~ \\vect{v}'.\n \\end{equation*}\n This is what had to be shown. Finally, to show that\n $\\vect{v}-\\vect{v}'$ is orthogonal to $W$, note that\n $\\vect{v}-\\vect{v}' = a_{k+1}\\vect{u}_{k+1}$, which is orthogonal to\n each $\\vect{u}_1,\\ldots,\\vect{u}_k$, and therefore to $W$.\n\\end{proof}\n\n\\begin{example}{Orthogonal projection onto a subspace}{projection-subspace}\n Consider $\\R^4$ with the usual dot product. Let\n $W=\\sspan\\set{\\vect{u}_1,\\vect{u}_2}$, where\n \\begin{equation*}\n \\vect{u}_1 = \\begin{mymatrix}{r} 1 \\\\ 1 \\\\ 2 \\\\ 0 \\end{mymatrix}, \\quad\n \\vect{u}_2 = \\begin{mymatrix}{r} -1 \\\\ -1 \\\\ 1 \\\\ 3 \\end{mymatrix}.\n \\end{equation*}\n Note that $\\vect{u}_1$ and $\\vect{u}_2$ are orthogonal. Let\n \\begin{equation*}\n \\vect{v} = \\begin{mymatrix}{r} 1 \\\\ 2 \\\\ 3 \\\\ 4 \\end{mymatrix}.\n \\end{equation*}\n Find the best approximation of $\\vect{v}$ in $W$.\n\\end{example}\n\n\\begin{solution}\n We calculate $\\iprod{\\vect{u}_1,\\vect{v}}=9$,\n $\\iprod{\\vect{u}_1,\\vect{u}_1}=6$, $\\iprod{\\vect{u}_2,\\vect{v}}=12$,\n and $\\iprod{\\vect{u}_2,\\vect{u}_2}=12$. Therefore, by\n Proposition~\\ref{prop:projection-subspace}, the desired vector is\n \\begin{equation*}\n \\vect{v}'\n ~=~ \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n ~=~ \\frac{9}{6} \\begin{mymatrix}{r} 1 \\\\ 1 \\\\ 2 \\\\ 0 \\end{mymatrix}\n + \\frac{12}{12} \\begin{mymatrix}{r} -1 \\\\ -1 \\\\ 1 \\\\ 3 \\end{mymatrix}\n ~=~ \\begin{mymatrix}{c} 1/2 \\\\ 1/2 \\\\ 4 \\\\ 3 \\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\nNote how straightforward the calculations in this example are. All we\nhad to do is calculate a few inner products. It is not even necessary\nto solve a system of equations. Such is the power of orthogonal bases.\n\nThe next example shows that we can use exactly the same method to find\napproximations%\n\\index{approximation!of functions} of functions.\n\n\\begin{example}{Approximating a function by a polynomial}{projection-subspace-polynomial}\n Let $V=C[-1,1]$ be the vector space of continuous functions, with the\n inner product given by\n \\begin{equation*}\n \\iprod{f,g} = \\int_{-1}^{1} f(x)g(x)\\,dx.\n \\end{equation*}\n Consider the function $f\\in V$ given by\n \\begin{equation*}\n f(x)\n ~=~ 1-\\abs{x}\n ~=~ \\begin{cases}\n 1+x & \\text{if $x<0$,} \\\\\n 1-x & \\text{if $x\\geq 0$.}\n \\end{cases}\n \\end{equation*}\n Find the closest approximation to $f$ by a polynomial of degree at\n most 2. Graph both $f$ and the approximating polynomial.\n\\end{example}\n\n\\begin{solution}\n Let $W=\\sspan\\set{1,x,x^2}$ be the subspace of $V$ consisting of\n polynomials of degree at most 2. What we are looking for is an\n element $g\\in W$ such that $\\norm{f-g}$ is as small as possible. We\n can solve this problem using\n Proposition~\\ref{prop:projection-subspace}.\n\n First we need an orthogonal basis for $W$. We found such an\n orthogonal basis in Example~\\ref{exa:legendre-polynomials}, namely\n the Legendre polynomials $p_0(x) = 1$, $p_1(x) = x$, and\n $p_2(x) = x^2-\\frac{1}{3}$. By\n Proposition~\\ref{prop:projection-subspace}, the desired\n approximation $g\\in W$ is given by:\n \\begin{equation*}\n g\n ~=~\n \\frac{\\iprod{p_0,f}}{\\iprod{p_0,p_0}}\\,p_0\n + \\frac{\\iprod{p_1,f}}{\\iprod{p_1,p_1}}\\,p_1\n + \\frac{\\iprod{p_2,f}}{\\iprod{p_2,p_2}}\\,p_2.\n \\end{equation*}\n We already computed the inner products $\\iprod{p_0,p_0}=2$,\n $\\iprod{p_1,p_1}=\\frac{2}{3}$, and $\\iprod{p_2,p_2}=\\frac{8}{45}$ in\n Example~\\ref{exa:legendre-polynomials}. We calculate the remaining\n inner products:\n \\begin{eqnarray*}\n \\iprod{p_0,f}\n &=& \\int_{-1}^{1} 1\\cdot f(x)\\,dx\n ~=~ \\int_{-1}^{0} 1\\cdot (1+x)\\,dx\n + \\int_{0}^{1} 1\\cdot (1-x)\\,dx\n ~=~ \\frac{1}{2} + \\frac{1}{2}\n ~=~ 1, \\\\\n \\iprod{p_1,f}\n &=& \\int_{-1}^{1} x\\cdot f(x)\\,\\,dx\n ~=~ \\int_{-1}^{0} x\\cdot (1+x)\\,\\,dx\n + \\int_{0}^{1} x\\cdot (1-x)\\,\\,dx\n ~=~ -\\frac{1}{6} + \\frac{1}{6}\n ~=~ 0, \\\\\n \\iprod{p_2,f}\n &=& \\int_{-1}^{1} (x^2-\\frac{1}{3})\\cdot f(x)\\,dx \\\\\n &=& \\int_{-1}^{0} (x^2-\\frac{1}{3})\\cdot (1+x)\\,dx\n + \\int_{0}^{1} (x^2-\\frac{1}{3})\\cdot (1-x)\\,dx\n ~=~ - \\frac{1}{12} - \\frac{1}{12}\n ~=~ -\\frac{1}{6}.\n \\end{eqnarray*}\n Therefore,\n \\begin{eqnarray*}\n g\n &=&\n \\frac{\\iprod{p_0,f}}{\\iprod{p_0,p_0}}\\,p_0\n + \\frac{\\iprod{p_1,f}}{\\iprod{p_1,p_1}}\\,p_1\n + \\frac{\\iprod{p_2,f}}{\\iprod{p_2,p_2}}\\,p_2\n \\\\\n &=& \\frac{1}{2}\\,p_0\n + \\frac{0}{2/3}\\,p_1\n - \\frac{1/6}{8/45}\\,p_2\n \\\\\n &=& \\frac{1}{2}\n - \\frac{15}{16}(x^2-\\frac{1}{3})\n \\\\\n &=& \\frac{13}{16} - \\frac{15}{16}x^2.\n \\end{eqnarray*}\n The following graph shows the function $f(x)$ as well as the\n polynomial $g(x)$:\n \\begin{center}\n \\begin{tikzpicture}[domain=-1:1, scale=3, samples=25]\n \\def\\fA#1{1}\n \\def\\fC#1{(abs((#1)^2)-1/3)}\n \\draw[thick,color=fourier0] (-1,0) -- (0,1) -- (1,0);\n \\draw[thick,color=fourier2] plot (\\x,{0.5*\\fA{\\x} - 0.9375*\\fC{\\x}});\n \\path[color=fourier2t] (0.6,0.6) node[right]{$g(x)= \\frac{13}{16} - \\frac{15}{16}x^2$};\n \\path[color=fourier0t] (0.2,0.9) node[right]{$f(x) = 1-\\abs{x}$};\n \\draw[->] (-1.5,0) -- (1.5,0) node[right] {$x$};\n \\draw[->] (0,-0.3) -- (0,1.1);\n \\draw (1,0) -- (1,-0.05) node[below] {$1$};\n \\draw (-1,0) -- (-1,-0.05) node[below] {$-1$};\n \\end{tikzpicture}\n \\end{center}\n\\end{solution}\n\nWhen approximating a function $f$ by a polynomial, as in the last\nexample, there is no need to stop with polynomials of degree 2. We can\nalso ask what is the best approximation%\n\\index{approximation!of functions} of $f$ by a polynomial of\ndegree 3, of degree 4, of degree 5, and so on. By increasing the\ndegree of the polynomials, we get better and better approximations to\n$f$. This leads us to the concept of a generalized Fourier series.\n\n\\begin{definition}{Generalized Fourier series}{generalized-fourier-series}\n Let $V$ be an inner product space and let\n $\\set{\\vect{u}_1,\\vect{u}_2,\\vect{u}_3,\\ldots}$ be an infinite\n orthogonal set of vectors. Let $\\vect{v}\\in V$. The\n \\textbf{generalized Fourier series}%\n \\index{generalized Fourier series}%\n \\index{Fourier series!generalized}%\n \\index{series!Fourier series} of $\\vect{v}$ (with respect to\n $\\vect{u}_1,\\vect{u}_2,\\vect{u}_3,\\ldots$) consists of the following\n sequence of vectors:\n \\begin{eqnarray*}\n \\vect{v}_1\n &=& \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1, \\\\\n \\vect{v}_2\n &=& \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2, \\\\\n \\vect{v}_3\n &=& \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n + \\frac{\\iprod{\\vect{u}_3,\\vect{v}}}{\\iprod{\\vect{u}_3,\\vect{u}_3}}\\,\\vect{u}_3, \\\\\n \\vect{v}_4\n &=& \\frac{\\iprod{\\vect{u}_1,\\vect{v}}}{\\iprod{\\vect{u}_1,\\vect{u}_1}}\\,\\vect{u}_1\n + \\frac{\\iprod{\\vect{u}_2,\\vect{v}}}{\\iprod{\\vect{u}_2,\\vect{u}_2}}\\,\\vect{u}_2\n + \\frac{\\iprod{\\vect{u}_3,\\vect{v}}}{\\iprod{\\vect{u}_3,\\vect{u}_3}}\\,\\vect{u}_3\n + \\frac{\\iprod{\\vect{u}_4,\\vect{v}}}{\\iprod{\\vect{u}_4,\\vect{u}_4}}\\,\\vect{u}_4, \\\\\n &\\vdots &\n \\end{eqnarray*}\n The vectors $\\vect{v}_i$ are also called \\textbf{generalized Fourier\n approximations}%\n \\index{Fourier approximation}%\n \\index{approximation!Fourier} of $\\vect{v}$.\n\\end{definition}\n\nBy Proposition~\\ref{prop:projection-subspace}, we know that each\n$\\vect{v}_i$ is the best approximation of $\\vect{v}$ in the subspace\n$\\sspan\\set{\\vect{u}_1,\\ldots,\\vect{u}_i}$. In particular,\n$\\norm{\\vect{v}-\\vect{v}_{i+1}} \\leq \\norm{\\vect{v}-\\vect{v}_i}$, so\neach $\\vect{v}_i$ is potentially a better approximation of $\\vect{v}$\nthan the previous one. In a course on analysis%\n\\footnote{``Algebra'' is the name for those subjects where all sums\n are finite. This includes, for example, linear algebra and abstract\n algebra. ``Analysis'' is the name for those subjects where sums are\n potentially infinite (and may or may not converge). This includes,\n for example, calculus, complex analysis, and functional analysis.},\nyou will learn that in many situations, the sequence\n$\\vect{v}_1,\\vect{v}_2,\\vect{v}_3,\\ldots$ can be shown to converge to\n$\\vect{v}$.\n\n\\begin{example}{Generalized Fourier series}{generalized-fourier-series}\n Proceeding as in Example~\\ref{exa:projection-subspace-polynomial},\n find the generalized Fourier approximations of $f(x)=1-\\abs{x}$ up\n to degree 8.\n\\end{example}\n\n\\begin{solution}\n We use the Legendre polynomials $p_0,\\ldots,p_8$ from\n Section~\\ref{sec:gram-schmidt} and calculate the relevant inner\n products, each of which requires solving an integral. As these\n integrals get a bit complicated, it is best to use a computer\n algebra system to compute them.\n \\begin{equation*}\n \\def\\arraystretch{1.4}\n \\begin{array}{rcl}\n \\iprod{p_0,f} &=& 1, \\\\\n \\iprod{p_1,f} &=& 0, \\\\\n \\iprod{p_2,f} &=& -\\frac{1}{6}, \\\\\n \\iprod{p_3,f} &=& 0, \\\\\n \\iprod{p_4,f} &=& \\frac{1}{105}, \\\\\n \\iprod{p_5,f} &=& 0, \\\\\n \\iprod{p_6,f} &=& -\\frac{1}{924}, \\\\\n \\iprod{p_7,f} &=& 0, \\\\\n \\iprod{p_8,f} &=& \\frac{1}{6435}, \\\\\n \\end{array}\n \\quad\n \\begin{array}{rcl}\n \\iprod{p_0,p_0} &=& 2, \\\\\n \\iprod{p_1,p_1} &=& \\frac{2}{3}, \\\\\n \\iprod{p_2,p_2} &=& \\frac{8}{45}, \\\\\n \\iprod{p_3,p_3} &=& \\frac{8}{175}, \\\\\n \\iprod{p_4,p_4} &=& \\frac{128}{11025}, \\\\\n \\iprod{p_5,p_5} &=& \\frac{128}{43659}, \\\\\n \\iprod{p_6,p_6} &=& \\frac{512}{693693}, \\\\\n \\iprod{p_7,p_7} &=& \\frac{512}{2760615}, \\\\\n \\iprod{p_8,p_8} &=& \\frac{32768}{703956825}. \\\\\n \\end{array}\n \\end{equation*}\n We therefore have the following approximations:\n \\begin{eqnarray*}\n f_0 &=& f_1 ~~=~~\n \\frac{1}{2}\\,p_0,\n \\\\\n f_2 &=& f_3 ~~=~~\n \\frac{1}{2}\\,p_0\n - \\frac{15}{16}\\,p_2,\n \\\\\n f_4 &=& f_5 ~~=~~\n \\frac{1}{2}\\,p_0\n - \\frac{15}{16}\\,p_2\n + \\frac{105}{128}\\,p_4,\n \\\\\n f_6 &=& f_7 ~~=~~\n \\frac{1}{2}\\,p_0\n - \\frac{15}{16}\\,p_2\n + \\frac{105}{128}\\,p_4\n - \\frac{3003}{2048}\\,p_6,\n \\\\\n f_8 &=& f_9 ~~=~~\n \\frac{1}{2}\\,p_0\n - \\frac{15}{16}\\,p_2\n + \\frac{105}{128}\\,p_4\n - \\frac{3003}{2048}\\,p_6\n + \\frac{109395}{32768}\\,p_8.\n \\end{eqnarray*}\n The following graph shows the function $f$ as well as its\n approximations $f_0$, $f_2$, $f_4$, $f_6$, and $f_8$. It can be seen\n that each successive approximation is closer to the function $f$\n than the previous one.\n \\begin{center}\n \\begin{tikzpicture}[domain=-1:1, scale=4, samples=25]\n \\def\\fA#1{1}\n \\def\\fB#1{(#1)}\n \\def\\fC#1{(abs((#1)^2)-1/3)}\n \\def\\fD#1{((#1)^3-3/5*(#1))}\n \\def\\fE#1{(abs((#1)^4)-6/7*abs((#1)^2)+3/35)}\n \\def\\fF#1{((#1)^5 - 10/9*(#1)^3+5/21*(#1))}\n \\def\\fG#1{(abs((#1)^6)-15/11*abs((#1)^4)+5/11*abs((#1)^2)-5/231)}\n \\def\\fH#1{((429*(#1)^7 - 693*(#1)^5 + 315*(#1)^3 - 35*(#1))/429)}\n \\def\\fI#1{((6435*abs((#1)^8)-12012*abs((#1)^6)+6930*abs((#1)^4)-1260*abs((#1)^2)+35)/6435)}\n \\draw[thick,color=fourier0] (-1,0) -- (0,1) -- (1,0);\n \\path[color=fourier0t] (0,1) node[right=2ex] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,0.5*\\fA{\\x}) (1,0.5) node[right=1ex,color=fourier1t] {$f_0$};\n \\draw[thick,color=fourier2] plot (\\x,{0.5*\\fA{\\x} - 0.9375*\\fC{\\x}}) (1,-0.18) node[right=1ex,color=fourier2t] {$f_2$};\n \\draw[thick,color=fourier4,samples=50] plot (\\x,{0.5*\\fA{\\x} - 0.9375*\\fC{\\x} + 0.8203125*\\fE{\\x}}) (1,0.18) node[right=1ex,color=fourier4t] {$f_4$};\n \\draw[thick,color=fourier5,samples=50] plot (\\x,{0.5*\\fA{\\x} - 0.9375*\\fC{\\x} + 0.8203125*\\fE{\\x} -1.46630859375*\\fG{\\x}}) (1,-0.06) node[right=1ex,color=fourier5t] {$f_6$};\n \\draw[thick,color=fourier6,samples=75] plot (\\x,{0.5*\\fA{\\x} - 0.9375*\\fC{\\x} + 0.8203125*\\fE{\\x} -1.46630859375*\\fG{\\x}+3.338470458984375*\\fI{\\x}}) (1,0.06) node[right=1ex,color=fourier6t] {$f_8$};\n \\draw[->] (-1.5,0) -- (1.5,0) node[right] {$x$};\n \\draw[->] (0,-0.3) -- (0,1.1);\n \\draw (1,0) -- (1,-0.05) node[below] {$1$};\n \\draw (-1,0) -- (-1,-0.05) node[below] {$-1$};\n \\end{tikzpicture}\n \\end{center}\n\\end{solution}\n\n\\begin{example}{Generalized Fourier series}{generalized-fourier-series2}\n The following graph shows the generalized Fourier approximations of\n the function\n \\begin{equation*}\n f(x) ~=~ \\begin{cases}\n 0 & \\mbox{if $x<0$,} \\\\\n 1 & \\mbox{if $x\\geq 0$}\n \\end{cases}\n \\end{equation*}\n by polynomials up to degree 7. (Note: the function $f$ is not\n continuous, so not technically an element of $C[-1,1]$, but we\n ignore this here. Instead of the vector space of continuous\n functions, we can work in the vector space of piecewise continuous\n functions).\n\\end{example}\n\n\\begin{center}\n \\begin{tikzpicture}[domain=-1:1, scale=4, samples=25]\n \\def\\fA#1{1}\n \\def\\fB#1{(#1)}\n \\def\\fC#1{(abs((#1)^2)-1/3)}\n \\def\\fD#1{((#1)^3-3/5*(#1))}\n \\def\\fE#1{(abs((#1)^4)-6/7*abs((#1)^2)+3/35)}\n \\def\\fF#1{((#1)^5 - 10/9*(#1)^3+5/21*(#1))}\n \\def\\fG#1{(abs((#1)^6)-15/11*abs((#1)^4)+5/11*abs((#1)^2)-5/231)}\n \\def\\fH#1{((429*(#1)^7 - 693*(#1)^5 + 315*(#1)^3 - 35*(#1))/429)}\n \\def\\fI#1{((6435*abs((#1)^8)-12012*abs((#1)^6)+6930*abs((#1)^4)-1260*abs((#1)^2)+35)/6435)}\n \\draw[thick,color=fourier0] (-1,0) -- (0,0) -- (0,1) -- (1,1);\n \\path[color=fourier0t] (0.1,1) node[above] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,0.5*\\fA{\\x}) (1,0.5) node[right=1ex,color=fourier1t] {$f_0$};\n \\draw[thick,color=fourier2] plot (\\x,{0.5*\\fA{\\x} + 0.75*\\fB{\\x}}) (1,1.24) node[right=1ex,color=fourier2t] {$f_1$};\n \\draw[thick,color=fourier4,samples=50] plot (\\x,{0.5*\\fA{\\x} + 0.75*\\fB{\\x} - 1.09375*\\fD{\\x}}) (1,0.82) node[right=1ex,color=fourier4t] {$f_3$};\n \\draw[thick,color=fourier5,samples=50] plot (\\x,{0.5*\\fA{\\x} + 0.75*\\fB{\\x} - 1.09375*\\fD{\\x} + 2.70703125*\\fF{\\x}}) (1,1.10) node[right=1ex,color=fourier5t] {$f_5$};\n \\draw[thick,color=fourier6,samples=75] plot (\\x,{0.5*\\fA{\\x} + 0.75*\\fB{\\x} - 1.09375*\\fD{\\x} + 2.70703125*\\fF{\\x} - 7.855224609375*\\fH{\\x}}) (1,0.96) node[right=1ex,color=fourier6t] {$f_7$};\n \\draw[->] (-1.5,0) -- (1.5,0) node[right] {$x$};\n \\draw[->] (0,-0.3) -- (0,1.1);\n \\draw (1,0) -- (1,-0.05) node[below] {$1$};\n \\draw (-1,0) -- (-1,-0.05) node[below] {$-1$};\n \\end{tikzpicture}\n\\end{center}\n\nComparing the last two examples, we see that the function in\nExample~\\ref{exa:generalized-fourier-series2} is much harder to\napproximate well by polynomials. This is due to the discontinuity in\nthe latter function. Nevertheless, the sequence of approximations\neventually converges to $f$.\n\n\\begin{example}{Generalized Fourier series}{generalized-fourier-series3}\n The following graph shows the generalized Fourier approximations of\n the function\n \\begin{equation*}\n f(x) ~=~ \\begin{cases}\n 0 & \\mbox{if $x<-\\frac{1}{2}$,} \\\\\n x+\\frac{1}{2} & \\mbox{if $-\\frac{1}{2}\\leq x<\\frac{1}{2}$,} \\\\\n 1 & \\mbox{if $x\\geq \\frac{1}{2}$} \\\\\n \\end{cases}\n \\end{equation*}\n by polynomials up to degree 7.\n\\end{example}\n\n\\begin{center}\n \\begin{tikzpicture}[domain=-1:1, scale=4, samples=25]\n \\def\\fA#1{1}\n \\def\\fB#1{(#1)}\n \\def\\fC#1{(abs((#1)^2)-1/3)}\n \\def\\fD#1{((#1)^3-3/5*(#1))}\n \\def\\fE#1{(abs((#1)^4)-6/7*abs((#1)^2)+3/35)}\n \\def\\fF#1{((#1)^5 - 10/9*(#1)^3+5/21*(#1))}\n \\def\\fG#1{(abs((#1)^6)-15/11*abs((#1)^4)+5/11*abs((#1)^2)-5/231)}\n \\def\\fH#1{((429*(#1)^7 - 693*(#1)^5 + 315*(#1)^3 - 35*(#1))/429)}\n \\def\\fI#1{((6435*abs((#1)^8)-12012*abs((#1)^6)+6930*abs((#1)^4)-1260*abs((#1)^2)+35)/6435)}\n \\draw[thick,color=fourier0] (-1,0) -- (-0.5,0) -- (0.5,1) -- (1,1);\n \\path[color=fourier0t] (0.5,1) node[above] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,0.5*\\fA{\\x}) (1,0.5) node[right=1ex,color=fourier1t] {$f_0$};\n \\draw[thick,color=fourier3] plot (\\x,{0.5*\\fA{\\x} + 0.6875*\\fB{\\x}}) (1,1.2) node[right=1ex,color=fourier3t] {$f_1$};\n \\draw[thick,color=fourier4,samples=50] plot (\\x,{0.5*\\fA{\\x} + 0.6875*\\fB{\\x} - 0.615234375*\\fD{\\x}}) (1,0.9) node[right=1ex,color=fourier4t] {$f_3$};\n \\draw[thick,color=fourier5] plot (\\x,{0.5*\\fA{\\x} + 0.6875*\\fB{\\x} - 0.615234375*\\fD{\\x} + 0.38067626953125*\\fF{\\x}}) (1,1.0) node[right=1ex,color=fourier5t] {$f_5$};\n \\draw[thick,color=fourier6,samples=50] plot (\\x,{0.5*\\fA{\\x} + 0.6875*\\fB{\\x} - 0.615234375*\\fD{\\x} + 0.38067626953125*\\fF{\\x} + 1.0494089126587303*\\fH{\\x}}) (1,1.1) node[right=1ex,color=fourier6t] {$f_7$};\n \\draw[->] (-1.5,0) -- (1.5,0) node[right] {$x$};\n \\draw[->] (0,-0.3) -- (0,1.1);\n \\draw (1,0) -- (1,-0.05) node[below] {$1$};\n \\draw (-1,0) -- (-1,-0.05) node[below] {$-1$};\n \\end{tikzpicture}\n\\end{center}\n\nSo far, we have worked with Legendre polynomials, but there are of\ncourse other examples of orthogonal sets of functions. An important\nsuch set is given by sine and cosine waves. We will see that every\nperiodic function can be decomposed into sine and cosine waves of\nvarying frequencies. This was Fourier's original discovery, and the\ncorresponding series are just known as \\textbf{Fourier series}%\n\\index{Fourier series}%\n\\index{series!Fourier series} (i.e., not ``generalized'').\n\n\\begin{example}{Fourier series}{fourier-series}\n Consider the inner product space $C[0,2\\pi]$. Recall from\n Example~\\ref{exa:orthogonal-set-sin-cos} that the following\n functions form an orthogonal set:\n \\begin{eqnarray*}\n \\vect{u}_0 &=& 1 \\\\\n \\vect{u}_1 &=& \\sin x \\\\\n \\vect{u}_2 &=& \\cos x \\\\\n \\vect{u}_3 &=& \\sin 2x \\\\\n \\vect{u}_4 &=& \\cos 2x \\\\\n \\vect{u}_5 &=& \\sin 3x \\\\\n \\vect{u}_6 &=& \\cos 3x \\\\\n &\\vdots&\n \\end{eqnarray*}\n Consider the function $f(x) = x - \\pi$, where $x\\in[0,2\\pi]$. Find\n its Fourier series.\n\\end{example}\n\n\\begin{solution}\n Following Definition~\\ref{def:generalized-fourier-series}, we must\n calculate a number of inner products. We have\n $\\iprod{\\vect{u}_0,\\vect{u}_0}=2\\pi$. Also, for all $i\\geq 1$, we\n have $\\iprod{\\vect{u}_i,\\vect{u}_i}=\\pi$. We note the following\n antiderivatives, for $k\\geq 1$:\n \\begin{equation*}\n \\int x\\sin kx\\,dx ~=~ -\\frac{x}{k}\\cos kx + \\frac{1}{k^2}\\sin kx, \\qquad\n \\int x\\cos kx\\,dx ~=~ \\frac{x}{k}\\sin kx + \\frac{1}{k^2}\\cos kx.\n \\end{equation*}\n Using these formulas, we can compute the following inner products\n quite easily:\n \\begin{eqnarray*}\n \\iprod{\\vect{u}_0, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\n ~~=~~ 0, \\\\\n \\iprod{\\vect{u}_1, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\sin x\n \\,~~=~~ -2\\pi, \\\\\n \\iprod{\\vect{u}_2, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\cos x\n ~~=~~ 0, \\\\\n \\iprod{\\vect{u}_3, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\sin 2x\n \\,~~=~~ -\\frac{2\\pi}{2}, \\\\\n \\iprod{\\vect{u}_4, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\cos 2x\n ~~=~~ 0, \\\\\n \\iprod{\\vect{u}_5, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\sin 3x\n \\,~~=~~ -\\frac{2\\pi}{3}, \\\\\n \\iprod{\\vect{u}_6, f}\n &=& \\int_{0}^{2\\pi} (x-\\pi)\\cos 3x\n ~~=~~ 0,\n \\end{eqnarray*}\n and so on. We therefore find the following Fourier series:\n \\begin{eqnarray*}\n f_1 &=& -2 \\sin x, \\\\\n f_3 &=& -2 \\sin x - \\frac{2}{2}\\sin 2x, \\\\\n f_5 &=& -2 \\sin x - \\frac{2}{2}\\sin 2x - \\frac{2}{3}\\sin 3x, \\\\\n f_7 &=& -2 \\sin x - \\frac{2}{2}\\sin 2x - \\frac{2}{3}\\sin 3x - \\frac{2}{4}\\sin 4x, \\\\\n f_9 &=& -2 \\sin x - \\frac{2}{2}\\sin 2x - \\frac{2}{3}\\sin 3x - \\frac{2}{4}\\sin 4x - \\frac{2}{5}\\sin 5x,\n \\end{eqnarray*}\n and so on.\n\\end{solution}\n\nThe following graph illustrates the successive approximations of this\nFourier series.\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.8, samples=100]\n \\def\\fA#1{1}\n \\def\\fB#1{(sin((#1)/pi*180))}\n \\def\\fC#1{(cos((#1)/pi*180))}\n \\def\\fD#1{(sin(2*(#1)/pi*180))}\n \\def\\fE#1{(cos(2*(#1)/pi*180))}\n \\def\\fF#1{(sin(3*(#1)/pi*180))}\n \\def\\fG#1{(cos(3*(#1)/pi*180))}\n \\def\\fH#1{(sin(4*(#1)/pi*180))}\n \\def\\fI#1{(cos(4*(#1)/pi*180))}\n \\draw[thick,color=fourier0] (0,0) -- (0,-pi) -- (2*pi,pi) -- (2*pi,-pi) -- (4*pi,pi) -- (4*pi,-pi) -- (6*pi, pi) -- (6*pi, 0) (2*pi, pi) node[right,color=fourier0t] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,{((-2)*\\fB{\\x})}) (pi,0) +(0.6*pi,0.6*pi) node[left=4ex,color=fourier1t] {$f_1$};\n \\draw[thick,color=fourier4,samples=200] plot (\\x,{((-2)*\\fB{\\x})+((-1)*\\fD{\\x})}) (pi,0) +(0.75*pi,0.75*pi) node[left=4ex,color=fourier4t] {$f_3$};\n \\draw[thick,color=fourier5,samples=200] plot (\\x,{((-2)*\\fB{\\x})+((-1)*\\fD{\\x})+((-2/3)*\\fF{\\x})}) (pi,0) +(0.9*pi,0.9*pi) node[left=4ex,color=fourier5t] {$f_5$};\n \\draw[thick,color=fourier6,samples=300] plot (\\x,{((-2)*\\fB{\\x})+((-1)*\\fD{\\x})+((-2/3)*\\fF{\\x})+((-1/2)*\\fH{\\x})}) (pi,0) +(1.05*pi,1.05*pi) node[left=4ex,color=fourier6t] {$f_7$};\n \\draw[->] (-0.8,0) -- (6*pi+0.8,0) node[right] {$x$};\n \\draw[->] (0,-3.5) -- (0,3.5);\n \\draw (pi,0) -- (pi,-0.05) node[below] {$\\pi$};\n \\draw (2*pi,0) -- (2*pi,-0.05) node[below] {$2\\pi$};\n \\draw (3*pi,0) -- (3*pi,-0.05) node[below] {$3\\pi$};\n \\draw (4*pi,0) -- (4*pi,-0.05) node[below] {$4\\pi$};\n \\draw (5*pi,0) -- (5*pi,-0.05) node[below] {$5\\pi$};\n \\draw (6*pi,0) -- (6*pi,-0.05) node[below] {$6\\pi$};\n \\end{tikzpicture}\n\\end{center}\nNote that, although the function $f$ is defined on the interval\n$[0,2\\pi]$, we have extended it periodically for $[2\\pi,4\\pi]$,\n$[4\\pi,6\\pi]$, and so on. This makes sense because all of the\northogonal functions $1$, $\\sin x$, $\\cos x$, $\\sin 2x$, and so on,\nhave period $2\\pi$.\n\nWe can also illustrate the same information differently, by showing\nthe individual sine waves making up the wave form of the function\n$f(x)$. In the context of audio signals, these sine waves are also\ncalled the \\textbf{harmonics}%\n\\index{harmonic} of the signal.\n\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.3, samples=100]\n \\def\\fA#1{1}\n \\def\\fB#1{(sin((#1)/pi*180))}\n \\def\\fC#1{(cos((#1)/pi*180))}\n \\def\\fD#1{(sin(2*(#1)/pi*180))}\n \\def\\fE#1{(cos(2*(#1)/pi*180))}\n \\def\\fF#1{(sin(3*(#1)/pi*180))}\n \\def\\fG#1{(cos(3*(#1)/pi*180))}\n \\def\\fH#1{(sin(4*(#1)/pi*180))}\n \\def\\fI#1{(cos(4*(#1)/pi*180))}\n \\begin{scope}[yshift=-12cm,xshift=-29cm]\n \\draw[thick,color=fourier0] (0,0) -- (0,-pi) -- (2*pi,pi) -- (2*pi,-pi) --\n (4*pi,pi) -- (4*pi,-pi) -- (6*pi, pi) -- (6*pi, 0);\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0) node[right]{$~~=~~\\left\\{\\rule{0mm}{3cm}\\right.$};\n \\end{scope}\n \\begin{scope}[yshift=-4cm]\n \\draw[thick,color=fourier1] plot (\\x,{((-2)*\\fB{\\x})}) node[right=2ex,color=fourier1t] {$-\\frac{2}{1}\\sin(x)$};\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-8cm]\n \\draw[thick,color=fourier4] plot (\\x,{((-1)*\\fD{\\x})}) node[right=2ex,color=fourier4t] {$-\\frac{2}{2}\\sin(2x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-12cm]\n \\draw[thick,color=fourier5,samples=200] plot (\\x,{((-2/3)*\\fF{\\x})}) node[right=2ex,color=fourier5t] {$-\\frac{2}{3}\\sin(3x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-16cm]\n \\draw[thick,color=fourier6,samples=200] plot (\\x,{((-1/2)*\\fH{\\x})}) node[right=2ex,color=fourier6t] {$-\\frac{2}{4}\\sin(4x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-20cm]\n \\path (-1.5,0) node[left]{$+$} node[right]{$\\cdots$};\n \\end{scope}\n \\end{tikzpicture}\n\\end{center}\n\n\\begin{example}{Fourier series}{fourier-series2}\n The following graph shows the first few Fourier approximations of the function\n \\begin{equation*}\n f(x) ~=~ \\begin{cases}\n x - \\frac{\\pi}{2} & \\mbox{if $0\\leq x<\\pi$,} \\\\\n \\frac{3\\pi}{2} - x & \\mbox{if $\\pi\\leq x\\leq 2\\pi$.}\n \\end{cases}\n \\end{equation*}\n\\end{example}\n\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.8, samples=100]\n \\def\\fC#1{(cos((#1)/pi*180))}\n \\def\\fG#1{(cos(3*(#1)/pi*180))}\n \\def\\fK#1{(cos(5*(#1)/pi*180))}\n \\def\\fO#1{(cos(7*(#1)/pi*180))}\n \\draw[thick,color=fourier0] (0,-pi/2) -- (pi,pi/2) -- (2*pi,-pi/2) -- (3*pi,pi/2) -- (4*pi,-pi/2) -- (5*pi,pi/2) -- (6*pi, -pi/2) (pi,pi/2) node [above,color=fourier0t] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,{((-1.2732395447351623)*\\fC{\\x})}) (0.6*pi,-pi/2+0.6*pi) node[left=2ex,color=fourier1t] {$f_2$};\n \\draw[thick,color=fourier4,samples=200] plot (\\x,{((-1.2732395447351623)*\\fC{\\x})+((-0.14147106052612904)*\\fG{\\x})}) (0.75*pi,-pi/2+0.75*pi) node[left=2ex,color=fourier4t] {$f_4$};\n \\draw[thick,color=fourier5,samples=200] plot (\\x,{((-1.2732395447351623)*\\fC{\\x})+((-0.14147106052612904)*\\fG{\\x})+((-0.05092958178940641)*\\fK{\\x})}) (0.9*pi,-pi/2+0.9*pi) node[left=2ex,color=fourier5t] {$f_6$};\n \\draw[thick,color=fourier6,samples=200] plot (\\x,{((-1.2732395447351623)*\\fC{\\x})+((-0.14147106052612904)*\\fG{\\x})+((-0.05092958178940641)*\\fK{\\x})+((-0.025984480504798825)*\\fO{\\x})}) (1.05*pi,-pi/2+1.05*pi) node[left=2ex,color=fourier6t] {$f_8$};\n \\draw[->] (-0.8,0) -- (6*pi+0.8,0) node[right] {$x$};\n \\draw[->] (0,-2) -- (0,2);\n \\draw (pi,0) -- (pi,-0.05) node[below] {$\\pi$};\n \\draw (2*pi,0) -- (2*pi,-0.05) node[below] {$2\\pi$};\n \\draw (3*pi,0) -- (3*pi,-0.05) node[below] {$3\\pi$};\n \\draw (4*pi,0) -- (4*pi,-0.05) node[below] {$4\\pi$};\n \\draw (5*pi,0) -- (5*pi,-0.05) node[below] {$5\\pi$};\n \\draw (6*pi,0) -- (6*pi,-0.05) node[below] {$6\\pi$};\n \\end{tikzpicture}\n\\end{center}\n\n\\noindent Note how rapidly this Fourier series converges to $f$. The\nfollowing image shows the function $f$ (extended periodically outside\nthe interval $[0,2\\pi]$) as a sum of its harmonics. Compared to\nExample~\\ref{exa:fourier-series}, we see that the higher harmonics\nhave much smaller amplitudes, which explains the rapid convergence.\n\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.3, samples=100]\n \\def\\fC#1{(cos((#1)/pi*180))}\n \\def\\fG#1{(cos(3*(#1)/pi*180))}\n \\def\\fK#1{(cos(5*(#1)/pi*180))}\n \\def\\fO#1{(cos(7*(#1)/pi*180))}\n \\begin{scope}[yshift=-12cm,xshift=-29cm]\n \\draw[thick,color=fourier0] (0,-pi/2) -- (pi,pi/2) -- (2*pi,-pi/2) -- (3*pi,pi/2) --\n (4*pi,-pi/2) -- (5*pi,pi/2) -- (6*pi, -pi/2);\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0) node[right]{$~~=~~\\left\\{\\rule{0mm}{3cm}\\right.$};\n \\end{scope}\n \\begin{scope}[yshift=-4cm]\n \\draw[thick,color=fourier1] plot (\\x,{((-1.2732395447351623)*\\fC{\\x})}) (6*pi,0) node[right=2ex,color=fourier1t] {$-\\frac{4}{\\pi}cos(x)$};\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-8cm]\n \\draw[thick,color=fourier4] plot (\\x,{((-0.14147106052612904)*\\fG{\\x})}) (6*pi,0) node[right=2ex,color=fourier4t] {$-\\frac{4}{9\\pi}\\cos(3x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-12cm]\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\draw[thick,color=fourier5] plot (\\x,{((-0.05092958178940641)*\\fK{\\x})}) (6*pi,0) node[right=2ex,color=fourier5t] {$-\\frac{4}{25\\pi}\\cos(5x)$};\n \\end{scope}\n \\begin{scope}[yshift=-16cm]\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\draw[thick,color=fourier6] plot (\\x,{((-0.025984480504798825)*\\fO{\\x})}) (6*pi,0) node[right=2ex,color=fourier6t] {$-\\frac{4}{49\\pi}\\cos(7x)$};\n \\end{scope}\n \\begin{scope}[yshift=-20cm]\n \\path (-1.5,0) node[left]{$+$} node[right]{$\\cdots$};\n \\end{scope}\n \\end{tikzpicture}\n\\end{center}\n\n\\begin{example}{Fourier series}{fourier-series3}\n The following graph shows the first few Fourier approximations of the function\n \\begin{equation*}\n f(x) ~=~ \\begin{cases}\n 1 & \\mbox{if $0\\leq x<\\pi$,} \\\\\n -1 & \\mbox{if $\\pi\\leq x\\leq 2\\pi$.}\n \\end{cases}\n \\end{equation*}\n\\end{example}\n\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.8, samples=100]\n \\def\\fB#1{(sin((#1)/pi*180))}\n \\def\\fF#1{(sin(3*(#1)/pi*180))}\n \\def\\fJ#1{(sin(5*(#1)/pi*180))}\n \\def\\fN#1{(sin(7*(#1)/pi*180))}\n \\draw[thick,color=fourier0] (0,0) -- (0,1) -- (pi,1) -- (pi,-1) --\n (2*pi,-1) -- (2*pi,1) --\n (3*pi,1) -- (3*pi,-1) --\n (4*pi,-1) -- (4*pi,1) --\n (5*pi,1) -- (5*pi,-1) --\n (6*pi,-1) -- (6*pi,0) (pi, 1) node[right,color=fourier0t] {$f$};\n \\draw[thick,color=fourier1] plot (\\x,{((1.2732395447351623)*\\fB{\\x})}) (0.4*pi, 1.2) node[above,color=fourier1t] {$f_1$};\n \\draw[thick,color=fourier4,samples=200] plot (\\x,{((1.2732395447351623)*\\fB{\\x})+((0.4244131815783881)*\\fF{\\x})}) (0.6*pi, 1.2) node[above,color=fourier4t] {$f_3$};\n \\draw[thick,color=fourier5,samples=200] plot (\\x,{((1.2732395447351623)*\\fB{\\x})+((0.4244131815783881)*\\fF{\\x})+((0.25464790894703326)*\\fJ{\\x})}) (0.8*pi, 1.2) node[above,color=fourier5t] {$f_5$};\n \\draw[thick,color=fourier6,samples=300] plot (\\x,{((1.2732395447351623)*\\fB{\\x})+((0.4244131815783881)*\\fF{\\x})+((0.25464790894703326)*\\fJ{\\x})+((0.18189136353359486)*\\fN{\\x})}) (1.0*pi, 1.2) node[above,color=fourier6t] {$f_7$};\n \\draw[->] (-0.8,0) -- (6*pi+0.8,0) node[right] {$x$};\n \\draw[->] (0,-2) -- (0,2);\n \\draw (pi,0) -- (pi,-0.05) node[below] {$\\pi$};\n \\draw (2*pi,0) -- (2*pi,-0.05) node[below] {$2\\pi$};\n \\draw (3*pi,0) -- (3*pi,-0.05) node[below] {$3\\pi$};\n \\draw (4*pi,0) -- (4*pi,-0.05) node[below] {$4\\pi$};\n \\draw (5*pi,0) -- (5*pi,-0.05) node[below] {$5\\pi$};\n \\draw (6*pi,0) -- (6*pi,-0.05) node[below] {$6\\pi$};\n \\end{tikzpicture}\n\\end{center}\n\n\\noindent Once again, here is an image showing the function $f$ as a\nsum of its harmonics:\n\n\\begin{center}\n \\begin{tikzpicture}[domain=0:6*pi, scale=0.3, samples=100]\n \\def\\fB#1{(sin((#1)/pi*180))}\n \\def\\fF#1{(sin(3*(#1)/pi*180))}\n \\def\\fJ#1{(sin(5*(#1)/pi*180))}\n \\def\\fN#1{(sin(7*(#1)/pi*180))}\n \\begin{scope}[yshift=-12cm,xshift=-29cm]\n \\draw[thick,color=fourier0] (0,0) -- (0,1) -- (pi,1) -- (pi,-1) --\n (2*pi,-1) -- (2*pi,1) --\n (3*pi,1) -- (3*pi,-1) --\n (4*pi,-1) -- (4*pi,1) --\n (5*pi,1) -- (5*pi,-1) --\n (6*pi,-1) -- (6*pi,0);\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0)\n node[right]{$~~=~~\\left\\{\\rule{0mm}{3cm}\\right.$};\n \\end{scope}\n \\begin{scope}[yshift=-4cm]\n \\draw[thick,color=fourier1] plot (\\x,{((1.2732395447351623)*\\fB{\\x})}) node[right=2ex,color=fourier1t] {$\\frac{4}{\\pi}\\sin(x)$};\n \\draw[->] (-1.5,0) -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-8cm]\n \\draw[thick,color=fourier4] plot (\\x,{((0.4244131815783881)*\\fF{\\x})}) node[right=2ex,color=fourier4t] {$\\frac{4}{3\\pi}\\sin(3x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-12cm]\n \\draw[thick,color=fourier5,samples=150] plot (\\x,{((0.25464790894703326)*\\fJ{\\x})}) node[right=2ex,color=fourier5t] {$\\frac{4}{5\\pi}\\sin(5x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-16cm]\n \\draw[thick,color=fourier6,samples=200] plot (\\x,{((0.18189136353359486)*\\fN{\\x})}) node[right=2ex,color=fourier6t] {$\\frac{4}{7\\pi}\\sin(7x)$};\n \\draw[->] (-1.5,0) node[left]{$+$} -- (6*pi+1.5,0);\n \\end{scope}\n \\begin{scope}[yshift=-20cm]\n \\path (-1.5,0) node[left]{$+$} node[right]{$\\cdots$};\n \\end{scope}\n \\end{tikzpicture}\n\\end{center}\n\n\\noindent\nWatch the video at \\url{http://y2u.be/3IAMpH4xF9Q} for a demonstration\nof what the functions from\nExamples~\\ref{exa:fourier-series}--\\ref{exa:fourier-series3}, and\ntheir harmonics, sound like as audio signals.\n", "meta": {"hexsha": "0ae6f361f108715e31d7733464efb18f32d4997a", "size": 37750, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "baseText/content/InnerProductSpaces-Fourier.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "baseText/content/InnerProductSpaces-Fourier.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "baseText/content/InnerProductSpaces-Fourier.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", 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YES\n2. YES", "lm_q1_score": 0.7248702761768248, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.6419679277353068}} {"text": "%---------------------------Condition-----------------------------\n\\section{Condition}\n\n\\[\nq = \\frac{1}{2} \\max \\left\\{ \\frac {\\normvec{L_0}^2 + \\normvec{L_3}^2 } { \\alpha_0 },\n \\frac {\\normvec{L_1}^2 + \\normvec{L_0}^2 } { \\alpha_1 },\n \\frac {\\normvec{L_2}^2 + \\normvec{L_1}^2 } { \\alpha_2 },\n \\frac {\\normvec{L_3}^2 + \\normvec{L_2}^2 } { \\alpha_3 } \n \\right\\}\n\\]\n\nNote that if $\\alpha_i< DBL\\_MIN$, we set $q = DBL\\_MAX$.\n\n\\quadmetrictable{condition}%\n{$1$}% Dimension\n{$[1,4]$}% Acceptable range\n{$[1,DBL\\_MAX]$}% Normal range\n{$[1,DBL\\_MAX]$}% Full range\n{$1$}% Unit square\n{\\cite{knu:00}}% Citation\n{v\\_quad\\_condition}% Verdict function name\n\n", "meta": {"hexsha": "0f10b3d1ff3a896a1be57f4d2a13b8f627b8fd5a", "size": 941, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadCondition.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadCondition.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadCondition.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 40.9130434783, "max_line_length": 86, "alphanum_fraction": 0.38575983, "num_tokens": 270, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8856314738181875, "lm_q2_score": 0.7248702702332475, "lm_q1q2_score": 0.6419679257536588}} {"text": "\\documentclass[letterpaper]{jpconf}\n\n\\usepackage{amsfonts}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{graphicx,wrapfig}\n\\usepackage{hyperref}\n\\hypersetup{\n bookmarksnumbered = true,\n bookmarksopen=false,\n pdfborder=0 0 0, % make all links invisible, so the pdf looks good when printed\n pdffitwindow=true, % window fit to page when opened\n pdfnewwindow=true, % links in new window\n colorlinks=true, % false: boxed links; true: colored links\n linkcolor=blue, % color of internal links\n citecolor=magenta, % color of links to bibliography\n filecolor=magenta, % color of file links\n urlcolor=cyan % color of external links\n}\n\n\\newcommand{\\deriv}[2]{\\frac{d #1}{d #2}}\n\\newcommand{\\pderiv}[2]{\\frac{\\partial #1}{\\partial #2}}\n\\newcommand{\\vect}[1]{\\boldsymbol{#1}}\n\\newcommand{\\f}[2]{\\frac{#1}{#2}}\n\\newcommand{\\dx}{\\Delta x}\n\\newcommand{\\pd}[2]{\\partial_{#2}{#1}}\n\\newcommand{\\bx}{\\vect{x}}\n\\newcommand{\\bK}{\\vect{K}}\n\\newcommand{\\cT}{\\mathcal{T}}\n\\newcommand{\\xL}{x_{\\textnormal{\\tiny\\textsc{L}}}}\n\\newcommand{\\xH}{x_{\\textnormal{\\tiny\\textsc{H}}}}\n\\newcommand{\\xMin}{x_{\\textnormal{\\tiny m}}}\n\\newcommand{\\xMax}{x_{\\textnormal{\\tiny M}}}\n\\newcommand{\\trans}{\\textnormal{\\tiny\\textsc{T}}}\n\\newcommand{\\leftState}{\\textnormal{\\tiny\\textsc{L}}}\n\\newcommand{\\rightState}{\\textnormal{\\tiny\\textsc{R}}}\n\\newcommand{\\TCI}{\\textnormal{\\tiny\\textsc{TCI}}}\n\\newcommand{\\shock}{\\textnormal{\\tiny\\textsc{Sh}}}\n\\newcommand{\\Pos}{\\textnormal{\\tiny\\textsc{Pos}}}\n\\newcommand{\\TVD}{\\textnormal{\\tiny\\textsc{Tvd}}}\n\\newcommand{\\thornado}{\\texttt{thornado}}\n\\newcommand{\\chimera}{{\\sc Chimera}}\n\n\\newcommand{\\ee}[1]{{\\color{red} EE:~#1}}\n\n\\begin{document}\n\\title{\\thornado-hydro: towards discontinuous Galerkin methods for supernova hydrodynamics\\footnote{This manuscript has been authored by UT-Battelle, LLC under Contract No. DE-AC05-00OR22725 with the U.S. Department of Energy. The United States Government retains and the publisher, by accepting the article for publication, acknowledges that the United States Government retains a non-exclusive, paid-up, irrevocable, world-wide license to publish or reproduce the published form of this manuscript, or allow others to do so, for United States Government purposes. The Department of Energy will provide public access to these results of federally sponsored research in accordance with the DOE Public Access Plan (http://energy.gov/downloads/doe-public-access-plan).}}\n\n\\author{Eirik Endeve$^{1,2,6}$, Jesse Buffaloe$^{2}$, Samuel J Dunham$^{3}$, \\\\ Nick Roberts$^{2}$, Kristopher Andrew$^{4,1}$, Brandon Barker$^{2}$, \\\\ David Pochik$^{2}$, Juliana Pulsinelli$^{5,7}$ and Anthony Mezzacappa$^{2,6}$}\n\\address{$^{1}$Computer Science and Mathematics Division, Oak Ridge National Laboratory, TN 37831}\n\\address{$^{2}$Department of Physics and Astronomy, University of Tennessee Knoxville, TN 37996}\n\\address{$^{3}$Department of Astronomy, Vanderbilt University, TN 37212}\n\\address{$^{4}$Department of Physics and Astronomy, University of Kentucky, Lexington, KY 40506}\n\\address{$^{5}$Princeton University, NJ 08544}\n\\address{$^{6}$Joint Institute for Computational Sciences, Oak Ridge National Laboratory, TN 37831}\n\\address{$^{7}$Physics Division, Oak Ridge National Laboratory, TN 37831}\n\\ead{endevee@ornl.gov}\n\n\\begin{abstract}\nThe {\\bf t}oolkit for {\\bf h}igh-{\\bf or}der {\\bf n}eutrino-r{\\bf ad}iation hydr{\\bf o}dynamics (\\thornado) is being developed for simulations of core-collapse supernovae (CCSNe) and related problems. \nCurrent capabilities in \\thornado\\ include solvers for the Euler equations --- in non-relativistic and special relativistic limits --- and the two-moment model of neutrino transport. \nThe spatial discretization in \\thornado\\ is based on the discontinuous Galerkin (DG) method, which is receiving increased attention from the computational astrophysics community. \nIn this paper, we provide an overview of the numerical methods for the Euler equations in \\thornado, and present some encouraging preliminary numerical results from a set of basic tests in one and two spatial dimensions. \n\\end{abstract}\n\n\\section{Introduction}\n\nAlong with solvers for neutrino transport and gravity, the hydrodynamics solver constitutes a major component of core-collapse supernova (CCSN) simulations. \nThe dynamics of the bounce shock and associated fluid instabilities, which drive turbulent flows and help shape the explosion, must be faithfully captured (see, e.g., \\cite{muller_2016}, for a recent review). \nEssentially, the stellar interior is modeled as a perfect fluid (thermal conduction and viscosity are not explicitly included) using the Euler equations, but the models are extended to accommodate a nuclear equation of state. \n\nThe discontinuous Galerkin (DG) method \\cite{cockburnShu_2001,hesthavenWarburton_2008} appears as an appealing choice to model fluid flows in CCSNe. \nDG methods combine elements of spectral and finite volume methods, and achieve high-order accuracy on a compact stencil. \nData is only communicated with nearest neighbors, regardless of the formal order of accuracy, which leads to a high computation to communication ratio, and favorable parallel scalability on heterogeneous architectures has been demonstrated \\cite{klockner_etal_2009}. \nThey can be used in combination with $hp$-adaptivity \\cite{remacle_etal_2003}, where, in addition to grid refinement with AMR, the local polynomial degree can be chosen differently, and independently, in different cells. \nDG methods can easily be applied to problems involving curvilinear coordinates, which is beneficial in numerical relativity \\cite{teukolsky_2016}. \nCurrently, most hydrodynamics solvers in CCSN codes are based on finite volume methods. \nExploring the utility and performance of DG methods to model CCSNe seems like a worthy exercise by itself. \n\nThe {\\bf t}oolkit for {\\bf h}igh-{\\bf or}der {\\bf n}eutrino-r{\\bf ad}iation hydr{\\bf o}dynamics (\\thornado) is being developed to simulate neutrino-radiation hydrodynamics in CCSNe and related applications in nuclear astrophysics. \nThe spatial discretization of solvers for hyperbolic partial differential equations in \\thornado\\ is based on the DG method, while we foresee using a combination of spectral and continuous finite element methods for solving elliptic equations; e.g., for Newtonian gravity or relativistic gravity employing the conformal flatness approximation \\cite{wilson_etal_1996}. \nWith these approaches, we employ high-order discretization techniques based on a common mathematical framework in all the major model components. \nWhether the high-order approach will improve accuracy and efficiency of CCSN models remains to be demonstrated. \nHowever, in this paper, we provide an initial description and encouraging demonstration of the DG method implemented in \\thornado\\ to solve the Euler equations. \nWe focus on basic tests in one and two spatial dimensions and focus, to some extent, on the implementation of limiters in \\thornado. \nWe note that others (e.g., \\cite{radiceRezzolla_2011,schaal_etal_2015,zanotti_etal_2015,dumbser_etal_2018}) have also investigated the DG method in the context of astrophysical applications and reported encouraging results. \nThe DG limiting techniques implemented in \\thornado\\ are similar to those used by \\cite{schaal_etal_2015}, but is supplemented with a troubled-cell indicator \\cite{fuShu_2017} to prevent excessive limiting (e.g., around smooth extrema), and differs from more recent approaches in the literature (e.g., \\cite{huerta_etal_2012,dumbser_etal_2014,zanotti_etal_2015,sonntagMunz_2017}), which combine the sub-cell resolution of the DG method with high-order finite-volume techniques to improve the performance of the DG limiting procedure. \nWhile the results with current limiters are encouraging, these latter methods appear as attractive options for future implementation in \\thornado. \n\n\\section{The discontinuous Galerkin method}\n\nExcellent books and review articles on the DG method are available (see, e.g., \\cite{cockburnShu_1998,cockburnShu_2001,hesthavenWarburton_2008,shu_2016} and references therein), and we will not go into too much detail here. \nHowever, we review some key concepts to introduce notation, and emphasize specific choices for our implementation in \\thornado. \nTo this end, we consider a system of conservation laws with sources of the form\n\\begin{equation}\n \\pd{}{t}\\big(\\,\\sqrt{\\gamma}\\,\\vect{U}\\,\\big)\n +\\sum_{i=1}^{d}\\pd{}{i}\\Big(\\,\\sqrt{\\gamma}\\,\\vect{F}^{i}(\\vect{U})\\,\\Big)\n =\\sqrt{\\gamma}\\,\\vect{S}(\\vect{U}),\n \\label{eq:extendedEulerCompact}\n\\end{equation}\nwhere $\\vect{U}$ is the evolved state vector, $\\vect{F}^{i}$ are the fluxes, and $\\vect{S}$ is the source vector. \nWe use a formulation of the equations sufficiently general to accommodate curvilinear spatial coordinates encoded in the metric $\\gamma_{ij}$, giving the squared line element $ds^{2}=\\gamma_{ij}dx^{i}dx^{j}$, whose determinant is $\\gamma$. \nHenceforth, we will assume that the spatial metric is time-independent. \nTo solve Eq.~\\eqref{eq:extendedEulerCompact}, the computational domain $D\\subset\\mathbb{R}^{d}$ is divided into a disjoint union $\\cT$ of open elements $\\bK$, so that $D = \\cup_{\\bK \\in \\cT}\\bK$. \nWe require that each element is a box in the logical coordinates; i.e.,\n\\begin{equation}\n \\bK=\\{\\,\\vect{x} : x^{i} \\in K^{i} := (\\xL^{i},\\xH^{i}),\\,i=1,\\ldots,d\\,\\}, \n\\end{equation}\nwith the surface elements denoted $\\partial\\bK^{i}=\\times_{j\\ne i}K^{j}$. \nWe let $V_{\\bK}$ denote the proper element volume\n\\begin{equation}\n V_{\\bK} = \\int_{\\bK}dV, \\quad\\text{where}\\quad dV = \\sqrt{\\gamma}\\,\\prod_{i=1}^{d}dx^{i}. \n\\end{equation}\nWe also define as a set $\\bx=\\{\\tilde{\\bx}^{i},x^{i}\\}$ and $\\dx^{i}=\\xH^{i}-\\xL^{i}$. \n\nWe let the approximation space for the DG method, $\\mathbb{V}^{k}$, be constructed from the tensor product of one-dimensional polynomials of maximal degree $k$. \nNote that functions in $\\mathbb{V}^{k}$ can be discontinuous across element interfaces. \nThe semi-discrete DG problem is to find $\\vect{U}_{h}\\in\\mathbb{V}^{k}$, which approximates $\\vect{U}$ in Eq.~\\eqref{eq:extendedEulerCompact}, such that for all $v\\in\\mathbb{V}^{k}$ and all $\\bK\\in\\mathcal{T}$\n\\begin{align}\n &\\partial_{t}\\int_{\\bK}\\vect{U}_{h}\\,v\\,dV\n +\\sum_{i=1}^{d}\\int_{\\partial\\bK^{i}}\\big(\\,\\sqrt{\\gamma}\\,\\widehat{\\vect{F}}^{i}(\\vect{U}_{h})\\,v\\big|_{\\xH^{i}}-\\sqrt{\\gamma}\\,\\widehat{\\vect{F}}^{i}(\\vect{U}_{h})\\,v\\big|_{\\xL^{i}}\\,\\big)\\,d\\tilde{\\bx}^{i} \\nonumber \\\\\n &\\hspace{24pt}\n -\\sum_{i=1}^{d}\\int_{\\bK}\\vect{F}^{i}(\\vect{U}_{h})\\,\\pd{v}{i}\\,dV\n =\\int_{\\bK}\\vect{S}(\\vect{U}_{h})\\,v\\,dV. \n \\label{eq:semidiscreteDG}\n\\end{align}\nIn Eq.~\\eqref{eq:semidiscreteDG}, $\\widehat{\\vect{F}}^{i}(\\vect{U}_{h})$ is a numerical flux approximating the flux on the $i$th surface of $\\bK$. \nThe numerical flux function $\\vect{f}^{i}$ is evaluated using values from both sides of an element interface; i.e.,\n\\begin{equation}\n \\widehat{\\vect{F}}^{i}(\\vect{U}_{h})=\\vect{f}^{i}(\\vect{U}_{h}(x^{i,-},\\tilde{\\bx}^{i}),\\vect{U}_{h}(x^{i,+},\\tilde{\\bx}^{i})),\n\\end{equation}\nwhere superscripts $-/+$, e.g. in the arguments of $\\vect{U}_{h}$, indicate that the function is evaluated to the immediate left/right of $x^{i}$. \nWe use the Harten-Lax-van Leer (HLL) flux \\cite{harten_etal_1983} or the HLLC flux \\cite{toro_etal_1994,mignoneBodo_2005} for all the numerical experiments presented in Section~\\ref{sec:numerical}. \n\nWe provide further details on the DG method to arrive at the equations that are actually evolved in \\thornado. \nWe start by introducing some notation, defining the polynomial expansion for $\\vect{U}_{h}$ and the quadrature rules used to evaluate the integrals in Eq.~\\eqref{eq:semidiscreteDG}. \nThen we provide explicit expressions for each of the terms in Eq.~\\eqref{eq:semidiscreteDG}. \nFor simplicity we consider one spatial dimension ($d=1$) and drop indices denoting the spatial dimension. \nIn each element $K$, we use a nodal representation of the conserved variables $\\vect{U}$; i.e.,\n\\begin{equation}\n \\vect{U}(x,t)\\approx\n \\vect{U}_{h}(x,t)=\\sum_{i=1}^{N}\\vect{U}_{i}(t)\\,\\ell_{i}(x),\n \\quad\\text{where}\\quad\n \\ell_{i}(\\eta)=\n \\prod_{\\substack{j=1\\\\j\\ne i}}^{N}\\f{\\eta-\\eta_{j}}{\\eta_{i}-\\eta_{j}}\n \\label{eq:conservedNodalExpansion}\n\\end{equation}\nare Lagrange polynomials defined on the reference element $I = \\{ \\eta : \\eta \\in (-0.5,0.5) \\}$, and constructed to interpolate the node set $S_{N}=\\{\\eta_{i}\\}_{i=1}^{N}\\subset I$. \nThe spatial coordinate $x$ and the reference coordinate $\\eta$ are related by the mapping $x(\\eta)=\\xL+(0.5+\\eta)\\,\\dx$. \nThen, for any $\\eta_{j}\\in S_{N}$, $\\ell_{i}(\\eta_{j})=\\delta_{ij}$, so that $\\vect{U}_{h}(x(\\eta_{j}),t)=\\vect{U}_{j}(t)$. \nWe introduce numerical quadratures to evaluate the integrals in Eq.~\\eqref{eq:semidiscreteDG}. \nFirst we define the $M$-point quadrature $Q_{M}:C^{0}(I)\\to\\mathbb{R}$ with abscissas $\\hat{S}_{M}=\\{\\eta_{q}\\}_{q=1}^{M}$ and weights $\\{w_{q}\\}_{q=1}^{M}$, normalized such that $\\sum_{q=1}^{M}w_{q}=1$. \nThe $M$-point Legendre-Gauss quadrature, which we use, integrates polynomials of degree $\\le 2M-1$ exactly. \nThen, if $P_{h}(x)$ is such a polynomial, we have\n\\begin{equation}\n \\f{1}{\\dx}\\int_{K}P_{h}(x)\\,dx=\\int_{I}P_{h}(\\eta)\\,d\\eta=Q_{M}\\big[P_{h}\\big]\\equiv\\sum_{q=1}^{M}w_{q}\\,P_{h}(\\eta_{q}). \n\\end{equation}\n\nNote that the interpolation points $S_{N}$ and the quadrature points $\\hat{S}_{M}$ generally do not coincide. \nHowever, for the sake of computational efficiency we let $M=N$ and $S_{N}=\\hat{S}_{N}$, which is a spectral-type nodal collocation DG approximation \\cite{bassi_etal_2013}. \nInserting Eq.~\\eqref{eq:conservedNodalExpansion} into Eq.~\\eqref{eq:semidiscreteDG}, letting $v(x)=\\ell_{k}(x)$, and using the quadratures defined above, we obtain\n\\begin{align}\n \\pd{}{t}\\int_{K}\\vect{U}_{h}\\,v\\,dV\n &\\approx w_{k}\\,\\sqrt{\\gamma}_{k}\\,\\pd{}{t}\\vect{U}_{k}\\,\\dx\n \\label{eq:timeDerivativeTerm}\n\\end{align}\nfor the time derivative term. \nFor a general metric, the integral in Eq.~\\eqref{eq:timeDerivativeTerm} is approximate since we use the Legendre-Gauss quadrature rule with the nodal points given by the expansion in Eq.~\\eqref{eq:conservedNodalExpansion}. \nHowever, since $\\ell_{i}(\\eta_{k})=\\delta_{ik}$, this leads to a diagonal mass matrix, and simplifies the implementation. \nSimilarly, we obtain\n\\begin{align}\n \\int_{K}\\vect{S}(\\vect{U}_{h})\\,v\\,dV\n &\\approx w_{k}\\,\\sqrt{\\gamma}_{k}\\,\\vect{S}(\\vect{U}_{k})\\,\\dx.\n \\label{eq:sourceTerm}\n\\end{align}\nfor the source term. \nFinally, the volume term (last term on the left-hand side of Eq.~\\eqref{eq:semidiscreteDG}) becomes\n\\begin{equation}\n \\int_{K}\\vect{F}(\\vect{U}_{h})\\,\\pderiv{v}{x}\\,dV\n \\approx \\sum_{q=1}^{N}w_{q}\\,\\sqrt{\\gamma}_{q}\\,\\vect{F}(\\vect{U}_{q})\\,\\pderiv{\\ell_{k}}{\\eta}(\\eta_{q}).\n \\label{eq:volumeTerm}\n\\end{equation}\nUsing Eqs.~\\eqref{eq:timeDerivativeTerm}-\\eqref{eq:volumeTerm} in Eq.~\\eqref{eq:semidiscreteDG} results in the semi-discrete form\n\\begin{align}\n \\deriv{\\vect{U}_{k}}{t}\n =-\\f{1}{w_{k}\\sqrt{\\gamma}_{k}\\dx}\n \\Big\\{\n \\Big[\\,\n \\sqrt{\\gamma}\\widehat{\\vect{F}}\\ell_{k}\\big|_{\\xH}\n -\\sqrt{\\gamma}\\widehat{\\vect{F}}\\ell_{k}\\big|_{\\xL}\n \\,\\Big]\n -\\sum_{q=1}^{N}w_{q}\\,\\sqrt{\\gamma}_{q}\\,\\vect{F}(\\vect{U}_{q})\\,\\pderiv{\\ell_{k}}{\\eta}(\\eta_{q})\n \\Big\\} + \\vect{S}(\\vect{U}_{k}).\n \\label{eq:semidiscreteDiscretized}\n\\end{align}\nEq.~\\eqref{eq:semidiscreteDiscretized} comprises a system of ordinary differential equations (ODEs), which are integrated in time with an ODE solver. \nIn Section~\\ref{sec:numerical} we use the optimal third-order strong stability-preserving Runge-Kutta (SSP-RK3) method from \\cite{shuOsher_1988}. \nWe assume that the source term $\\vect{S}(\\vect{U}_{k})$ is non-stiff so that the equations can be integrated efficiently with explicit methods. \nIn future applications of the DG method to model CCSNe we plan to use a combination of implicit and explicit methods (e.g., IMEX methods \\cite{ascher_etal_1997}) to stably integrate stiff sources due to neutrino-matter interactions. \n\nLimiting of the polynomial representation $\\vect{U}_{h}$, to reduce oscillations and prevent unphysical states, is a critical step in the DG algorithm. \nWe use the TVD-type slope limiter based on limiting characteristic variables, discussed in \\cite{cockburnShu_1998}. \nTo prevent excessive limiting (e.g., at smooth extrema), we use the troubled-cell indicator (TCI) of \\cite{fuShu_2017}:\n\\begin{equation}\n I_{\\bK}(G) = \\f{\\sum_{j}|G_{\\bK}-G_{\\bK}^{(j)}|}{\\max_{j}|G_{\\bK^{(j)}}^{(j)}|},\n \\label{eq:indicator}\n\\end{equation}\nwhere $G\\in\\vect{G}\\subseteq\\vect{U}$. \n(Here, $\\vect{G}$ consists of the first and last element of $\\vect{U}$.) \nIn Eq.~\\eqref{eq:indicator}, the sum in the numerator and the $\\max$ in the denominator extend over the neighboring elements sharing a face with the target element $\\bK$. \n$G_{\\bK}$ is the cell average in $\\bK$, $G_{\\bK}^{(j)}$ is the cell average computed by extrapolating the polynomial representation from the neighbor element $\\bK^{(j)}$ into $\\bK$, and $G_{\\bK^{(j)}}^{(j)}$ is the cell average native to $\\bK^{(j)}$. \nAn element is flagged for limiting if $I_{\\bK}(G)>C_{\\TCI}$ for any $G\\in\\vect{G}$. \nTo prevent negative mass density and pressure (and superluminal velocity for the relativistic Euler equations), we follow the approaches in \\cite{zhangShu_2010,wuTang_2015,qin_etal_2016}. \n\nTo summarize the full RK-DG algorithm we let $\\bar{\\vect{U}}$ denote the global solution vector containing all the unknowns in $D$ and $\\bar{\\vect{F}}$ denote the corresponding DG discretization from the right-hand side of Eq.~\\eqref{eq:semidiscreteDiscretized}, so that the global ODE system can be written as the $d_{t}\\bar{\\vect{U}}=\\bar{\\vect{F}}(\\bar{\\vect{U}})$. \nThe TVD limiter and the `positivity' limiter preventing unphysical states are denoted by $\\Lambda^{\\TVD}\\{\\}$ and $ \\Lambda^{\\Pos}\\{\\}$, respectively (e.g., application of the TVD limiter is $\\bar{\\vect{U}} := \\Lambda^{\\TVD}\\{\\bar{\\vect{U}}\\}$). \nThe general $s$-stage Runge-Kutta time stepping algorithm, which includes application of limiters, can then be written as \\cite{cockburnShu_2001}:\n\\begin{itemize}\n \\item[1.] Set $\\bar{\\vect{U}}^{(0)} = \\bar{\\vect{U}}^{n}$,\n \\item[2.] For $i=1,\\ldots,s$ compute:\n \\begin{equation}\n \\bar{\\vect{U}}^{(i)} \n = \\Lambda^{\\Pos}\\Big\\{\\,\\Lambda^{\\TVD}\\Big\\{\\,\\sum_{j=0}^{i-1}\\Big(\\,\\alpha_{ij}\\,\\bar{\\vect{U}}^{(j)}+\\beta_{ij}\\,\\Delta t\\,\\bar{\\vect{F}}\\big(\\bar{\\vect{U}}^{(j)}\\big)\\,\\Big)\\,\\Big\\}\\,\\Big\\},\n \\label{eq:rkStages}\n \\end{equation}\n \\item[3.] Set $\\bar{\\vect{U}}^{n+1}=\\bar{\\vect{U}}^{(s)}$. \n\\end{itemize}\nThe coefficients $\\alpha_{ij}$ and $\\beta_{ij}$ for SSP-RK3 are given in Table~2.1 in \\cite{cockburnShu_2001}. \nIn Eq.~\\eqref{eq:rkStages}, for each stage, the TVD limiter is applied first (for elements where $I_{\\bK}>C_{\\TCI}$), followed by the positivity limiter. \nThe algorithm is subject to two time step restrictions, the stability condition $\\Delta t_{\\TVD}$ and the positivity condition $\\Delta t_{\\Pos}$, and we take $\\Delta t=\\min_{D}(\\Delta t_{\\TVD},\\Delta t_{\\Pos})$. \nLet $(h/|\\lambda|)=\\min_{i\\in\\{1,\\ldots,d\\}}(\\Delta x^{i}/|\\lambda^{i}|)$, where $|\\lambda^{i}|$ is the largest absolute eigenvalue of the flux Jacobian. \nThen, assuming Cartesian coordinates (e.g., \\cite{cockburnShu_2001,zhangShu_2010}):\n\\begin{equation}\n \\Delta t_{\\TVD} \\le \\f{1}{d}\\,\\f{(h/|\\lambda|)}{(2k+1)} \\quad\\text{and}\\quad\n \\Delta t_{\\Pos} \\le \\f{\\hat{w}_{1}(h/|\\lambda|)}{d},\n \\label{eq:timestepRestrictions}\n\\end{equation}\nwhere $\\hat{w}_{1}$ is the first quadrature weight of the Gauss-Lobatto quadrature that integrates the cell average of $\\vect{U}_{h}$ exactly. \nFor $d=k=2$, the effective Courant-Friedrichs-Lewy (CFL) factor is $0.1$ for $\\Delta t_{\\TVD}$, while it is $1/12$ for $\\Delta t_{\\Pos}$. \nThe positivity limiter checks for unphysical states in all the quadrature points involved in the DG discretization. \nIf any point in an element is flagged as unphysical, the polynomial representation $\\vect{U}_{h}$ is damped towards its cell average $\\vect{U}_{\\bK}$, which is guaranteed to be physical by the $\\Delta t_{\\Pos}$ restriction. \nNote that $\\Delta t_{\\Pos}$ is a sufficient condition. \nIn practice, the time step can often be set larger than $\\Delta t_{\\Pos}$. \n\n\\section{Preliminary numerical results}\n\\label{sec:numerical}\n\nIn this section we show preliminary numerical results obtained with the DG method implemented in \\thornado\\ to solve the non-relativistic and special relativistic Euler equations. \nWe use the ideal gas equation of state where the pressure $p$ is related to the internal energy density $e$ by $p=(\\Gamma-1)e$, and where $\\Gamma$ is the (constant) adiabatic index. \nLet the $d$-dimensional computational domain be $D=\\times_{i=1}^{d}[\\xMin^{i},\\xMax^{i}]$, where $\\xMin^{i}$ and $\\xMax^{i}$ denote the coordinates of the inner and outer boundary in the $i$th coordinate direction, respectively. \nTo limit the scope of this initial study, we use polynomials of degree $k=2$ (except in Section~\\ref{sec:isentropicVortex}, where we also use $k=1$ and $k=3$) combined with SSP-RK3 time stepping, and, unless stated otherwise, run all the tests with a CFL factor of $C=0.1$. \nWe use the HLL Riemann solver when solving the non-relativistic Euler equations and the HLLC Riemann solver when solving the special relativistic Euler equations. \nThe purpose of the tests is to gauge the performance of the DG implementation on a set of benchmarks as an initial measure of its suitability for future CCSN simulations. \n\n\\subsection{Non-relativistic (NR) hydrodynamics}\n\nFor the NR Euler equations, the state, flux, and source vectors in Eq.~\\eqref{eq:extendedEulerCompact} are\n\\begin{equation}\n \\vect{U}=\\big(\\rho,\\rho u_{j},E\\big)^{\\trans},~\n \\vect{F}^{i}=\\big(\\rho u^{i},\\Pi^{i}_{~j},(E+p)u^{i}\\big)^{\\trans},~\n \\text{and}~\n \\vect{S}=\\big(0,\\f{1}{2}\\Pi^{ik}\\pd{\\gamma_{ik}}{j}-\\rho\\pd{\\Phi}{j},-\\rho u^{i}\\pd{\\Phi}{i}\\big)^{\\trans},\n\\end{equation}\nwhere $\\rho$ and $u^{i}$ are the mass density and components of the fluid three-velocity, respectively. \nThe stress tensor is $\\Pi^{i}_{~j}=\\rho u^{i} u_{j}+p\\delta^{i}_{~j}$, and $E=e+\\f{1}{2}\\rho u_{i} u^{i}$ is the fluid (internal plus kinetic) energy density. \nThe sources are due to curvilinear coordinates and Newtonian gravity. \n\n\\subsubsection{Sod shock tube}\n\nThe first test is the classic Riemann problem due to Sod \\cite{sod_1978}, computed using Cartesian coordinates. \nThe one-dimensional computational domain is $D=[0,1]$, and a discontinuity is initially located at $x^{1}=0.5$, separating the left and right states\n\\begin{equation*}\n \\vect{U}_{\\leftState}=(1.0,\\vect{0.0},2.5)^{\\trans}\\quad\\text{and}\\quad\n \\vect{U}_{\\rightState}=(0.125,\\vect{0.0},0.25)^{\\trans},\n\\end{equation*}\nwhere the adiabatic index is $\\Gamma=1.4$. \nThe test is run with $100$ elements until $t=0.2$, using $C_{\\TCI}=0.03$. \nResults are plotted in the left panels of Figure~\\ref{fig:SodSedov}. \nThe DG method captures the main features of the solution well without introducing noticeable oscillations near the discontinuities (upper left panel). \nIn the lower left panel we plot locations in the $xt$-plane of elements flagged for limiting. \nLimiting is mainly triggered by the contact discontinuity and the shock. \n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Sod_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Sedov_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Sod_TCI_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Sedov_TCI_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:SodSedov}{\\it Upper left panel:} numerical solution of Sod's problem using $100$ elements (plusses) compared with the exact solution (solid lines; from \\cite{toro_1999}). {\\it Upper right panel:} numerical solution of the Sedov-Taylor blast wave using $256$ elements (plusses) against the exact solution (solid lines; from \\cite{kammTimmes_2007}). In the lower panels we plot locations in the $xt$-plane of elements flagged for limiting ($I_{\\bK}>C_{\\TCI}$) by the Sod (left) and Sedov (right) problems.}\n\\end{figure}\n\n\\subsubsection{Sedov-Taylor blast wave}\n\nThis test, detailed in \\cite{sedov_1959} (see also {\\S}99 in \\cite{landauLifshitz_1979}), is computed in spherical polar coordinates with the assumption of spherical symmetry.\\footnote{In this paper, all tests in spherical polar coordinates are computed with the assumption of spherical symmetry.}\nThe computational domain is $D=[0,1.2]$, and the initial condition consists of a fluid at rest with density $\\rho=1$. \nThe adiabatic index is $\\Gamma=1.4$, and an amount of thermal energy equal to $1$ is released in the innermost element. \nWe use $256$ elements and run until $t=1$, using $C_{\\TCI}=0.03$. \nAgain, the DG method captures the characteristic of the exact solution (cf. upper right panel of Figure~\\ref{fig:SodSedov}). \nThe maximum density in the shock is somewhat lower for the numerical solution than the exact value (about $5$ versus $6$). \nIn the lower right panel we map locations of elements flagged for limiting in the $xt$-plane. \nThe troubled-cell indicator tracks the shock well (the analytic shock trajectory is given by the dashed cyan line), but also flags elements at smaller radii with low density. \n\n\\subsubsection{Shu-Osher shock tube}\n\nThis test from \\cite{shuOsher_1989} is suitable for measuring the amount of dissipation in a numerical scheme. \nThe one-dimensional computational domain is $D=[-5,5]$, and a discontinuity is placed at $x^{1}=-4$, separating the states\n\\begin{equation*}\n \\vect{U}_{\\leftState}=\\big( 3.857143, 10.14185, 0.0, 0.0, 39.16667 \\big)^{\\trans}\\quad\\text{and}\\quad\n \\vect{U}_{\\rightState}=\\big( 1+0.2\\times\\sin(5 x^{1}), \\vect{0.0}, 2.5 \\big)^{\\trans},\n\\end{equation*}\nwhere the adiabatic index is $\\Gamma=1.4$. \nThe density variations ahead of the Mach=3 shock are compressed through the shock and result in higher-frequency variations downstream, which can be difficult to capture with an excessively dissipative scheme. \n\nWe run this test to $t=1.8$ using $200$ elements. \nIn the two upper panels of Figure~\\ref{fig:ShuOsher} we plot the density for various values of the troubled-cell indicator threshold $C_{\\TCI}$; $0.0$ (full limiting, green), $0.03$ (magenta), $0.3$ (red), and $3.0$ (blue). \nLarger $C_{\\TCI}$ implies less limiting. \nThe results are compared with a reference solution computed with $2048$ elements. \nWith full limiting, the scheme is unable to capture the density variations behind the shock (cf. upper right panel in Figure~\\ref{fig:ShuOsher}). \nThis is expected since the TVD-type limiter clips extrema. \nThe results improve when the troubled-cell indicator is used. \nWhen $C_{\\TCI}=0.03$, the variations are better resolved, but the amplitudes are much reduced relative to the reference solution. \nWith $C_{\\TCI}=3.0$, the variations are well resolved and the amplitudes are comparable to the reference. \n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/ShuOsher_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/ShuOsher_Inset_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{12pc}\n \\includegraphics[width=12pc]{./Figures/ShuOsher_TCI_0015_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{12pc}\n \\includegraphics[width=12pc]{./Figures/ShuOsher_TCI_0150_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}\n \\begin{minipage}{12pc}\n \\includegraphics[width=12pc]{./Figures/ShuOsher_TCI_1500_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:ShuOsher}Results for the Shu-Osher shock tube problem computed with $200$ elements. The upper panels show the mass density at $t=1.8$ for various values of the troubled-cell indicator threshold $C_{\\TCI}$, compared with a high-resolution (2048 elements) reference solution. In the lower panels we show elements flagged for limiting in the $xt$-plane for various values of $C_{\\TCI}$.}\n\\end{figure}\nIn the lower panels of Figure~\\ref{fig:ShuOsher} we plot locations in the $xt$-plane of elements flagged for limiting by the troubled-cell indicator: $C_{\\TCI}=0.03$ (left), $C_{\\TCI}=0.3$ (middle), and $C_{\\TCI}=3.0$ (right). \nWith $C_{\\TCI}=0.03$, limiting is triggered in a wide region around the shock and around the peaks of the three leftmost density variations in the upper left panel in Figure~\\ref{fig:ShuOsher}. \nWith $C_{\\TCI}=3.0$, limiting is only triggered in the shock. \n\n\\subsubsection{Isentropic vortex}\n\\label{sec:isentropicVortex}\n\nThis test from \\cite{shu_1997} is included to verify the accuracy of the DG method implemented in \\thornado. \nThe periodic computational domain is given by $D=[-5,5]\\times[-5,5]$. \nThe initial density and pressure are given by $\\rho_{0}=\\big(\\,1-\\f{(\\Gamma-1)\\,\\beta^{2}}{(8\\pi\\Gamma)}\\exp\\big(1-r^{2}\\big)\\,\\big)^{1/(\\Gamma-1)}$ and $P_{0}=\\rho^{\\Gamma}$, where $r=\\sqrt{(x^{1})^{2}+(x^{2})^{2}}$, while the velocity components are $v_{0}^{1}=1-\\f{x^{2}\\beta}{2\\pi}\\,\\exp\\big((1-r^{2})/2\\big)$ and $v_{0}^{2}=1+\\f{x^{1}\\beta}{2\\pi}\\,\\exp\\big((1-r^{2})/2\\big)$. \nThe strength of the vortex is set to $\\beta=5$, and we use $\\Gamma=1.4$. \nWe evolve this test until $t=t_{f}=10$, when the vortex has returned to its initial position. \n\n\\begin{wrapfigure}{L}{2.825in}\n \\includegraphics[width=18pc]{./Figures/IsentropicVortex_Astronum_2018}\n \\caption{\\label{fig:IsentropicVortex}Convergence results for the Isentropic Vortex: $L^{\\infty}$ error norm versus number of elements in the $x^{1}$-dimension.}\n\\end{wrapfigure}\n\nIn Figure~\\ref{fig:IsentropicVortex} we plot the density error measured in the infinity norm $L^{\\infty}=\\max_{D}|\\rho(\\vect{x},t_{f})-\\rho_{0}(\\vect{x})|$ versus number of cells the $x^{1}$-dimension for various configurations. \nWe ran with spatial resolutions of $25\\times25$, $50\\times50$, $100\\times100$, and $200\\times200$. \nIn the first set of runs, we ran without any limiting and varied the polynomial degree $k$: $k=1$ (solid red), $k=2$ (solid black), and $k=3$ (solid green). \nWe observe the expected order of accuracy for $k=1$ and $k=2$ (second- and third-order accuracy, respectively; cf. dotted reference lines). \nFor $k=3$, the order of accuracy is close to fourth-order for the lower resolutions, but tends to third-order as the resolution increases, presumably due to the use of third-order time stepping. \n(Note that we ran the $k=3$ case with a CFL factor of $1/14$; cf. Eq.~\\eqref{eq:timestepRestrictions}.)\nIn the second set of runs, we kept $k=2$ and ran with limiting turned on. \nWe ran one set with $C_{\\TCI}=0.03$ (dashed magenta) and one set with $C_{\\TCI}=0.1$ (dashed cyan). \nWhen $C_{\\TCI}=0.03$, the limiter is activated for the lowest resolution run and significantly reduces the accuracy of the method. \nFor all the other runs, the limiter is not activated, and the results are identical to the runs with the limiter turned off. \n\n\\subsubsection{Kelvin-Helmholtz (KH) instability}\nThis hydrodynamic instability is of significant astrophysical relevance --- including during the explosion phase of CCSNe --- and may occur in the interface separating fluids in relative motion. \nWe compute this problem on the unit square $D=[0,1]\\times[0,1]$, imposing periodic boundary conditions, and with initial conditions and single-mode perturbations taken from \\cite{mcnally_etal_2012}. \nWe use $256^{2}$ elements and evolve until $t=3.0$, well into the nonlinear regime. \nResults are displayed in Figure~\\ref{fig:KelvinHelmholtz}. \nWe have computed two models, one with $C_{\\TCI}=0.06$ (left panels) and one without limiting ($C_{\\TCI}\\to\\infty$; right panels). \n\n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_15_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_15_noLim_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_24_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_24_noLim_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_30_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_30_noLim_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:KelvinHelmholtz}Mass density from the KH instability at various times, computed with $256^{2}$ elements: $t=1.5$ (top panels), $t=2.4$ (middle panels), and $t=3.0$ (bottom panels). Results for $C_{\\TCI}=0.06$ and $C_{\\TCI}\\to\\infty$ (no limiting) are shown in the left and right panels, respectively.}\n\\end{figure}\n\nEarly on ($t=1.5$), the results from the two runs are practically indistinguishable, show no sign of developing secondary billows, and agree visually with results presented in \\cite{mcnally_etal_2012}. \nWhen $t=2.4$, the run with $C_{\\TCI}=0.06$ has started to develop secondary billows within the coiled-up interface separating dense (white) and less dense (black) fluids, while the model with no limiting has not. \nSee Figure~\\ref{fig:KelvinHelmholtz_TCI} for the spatial distribution of elements flagged for limiting at $t=2.4$. \nWhen $t=3.0$, these secondary instabilities appear to have largely destroyed the coil structure for the model with limiting, while the coils remain intact in the model without limiting (although secondary billows have started to form in this model at this time). \nAs discussed in more detail by \\cite{mcnally_etal_2012}, the development of the secondary billows may be an artifact of numerical perturbations and diffusion, which is delayed when the resolution is increased. \nOur results are consistent with this in the sense that the less diffusive model develops secondary billows later. \n\n\\begin{wrapfigure}{R}{2.825in}\n \\centering\n \\includegraphics[width=18pc]{./Figures/KelvinHelmholtz_TCI_24_Astronum_2018}\n \\caption{\\label{fig:KelvinHelmholtz_TCI}Distribution of elements flagged for limiting in the KH test at $t=2.4$.}\n\\end{wrapfigure}\n\n\\subsubsection{Liska-Wendroff implosion}\n\nThis test from \\cite{liskaWendroff_2003} is computed on a 2D domain $D=[0,0.3]\\times[0,0.3]$ with reflecting boundary conditions using $256^{2}$ elements. \nBelow the line $x^{2}=0.15-x^{1}$, we initially set $\\vect{U}=\\big(0.125,\\vect{0.0},0.35\\big)^{\\trans}$, while $\\vect{U}=\\big(1.0,\\vect{0.0},2.5\\big)^{\\trans}$ elsewhere. \n\n\\begin{figure}[b]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Implosion_01_HighTCI_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Implosion_02_HighTCI_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:Implosion}Plots for the Liska-Wendroff implosion problem at $t=0.045$ (left) and $t=2.5$ (right), intended to match Figures~4.10 and 4.11 in \\cite{liskaWendroff_2003}. The color map shows the pressure distribution, black contours show the mass density, while arrows indicate the velocity.}\n\\end{figure}\n\nResults for $t=0.045$ and $t=2.5$, with $C_{\\TCI}=0.6$, are displayed in Figure~\\ref{fig:Implosion} (left and right panels, respectively), which agree qualitatively with \\cite{liskaWendroff_2003}, who compared the results of eight different schemes for this problem. \nWhen $t=0.045$, a shock wave propagates towards the origin of $D$, while a contact discontinuity is trailing the shock (cf. density contours), and a rarefaction wave is spreading in the opposite directions. \nWhen $t=2.5$, a complex flow pattern with multiple shocks has emerged due to multiple reflections off the boundaries. \nThe initial symmetry about the $x^{1}=x^{2}$ diagonal is maintained, and the results from \\thornado\\ agree qualitatively with the unsplit schemes in \\cite{liskaWendroff_2003}. \nIn the right panel of Figure~\\ref{fig:Implosion}, the chatacteristic ``jet'' has formed and propagates along the diagonal, similar to that displayed by CLAW and WENO in \\cite{liskaWendroff_2003}. \n\n\\subsubsection{Liska-Wendroff explosion}\n\nThis test, also from \\cite{liskaWendroff_2003}, is computed on the domain $D=[0,1.5]\\times[0,1.5]$ with reflecting inner boundaries and outflow outer boundaries using $256^{2}$ elements. \nInside the radius $r=\\sqrt{(x^{1})^{2}+(x^{2})^{2}}=0.4$, we initially set $\\vect{U}=\\big(1.0,\\vect{0.0},2.5\\big)^{\\trans}$, while $\\vect{U}=\\big(0.125,\\vect{0.0},0.25\\big)^{\\trans}$ elsewhere. \nWe did not apply any smoothing to the initial discontinuity. \n\nResults for $t=3.2$, obtained with $C_{\\TCI}=0.1$, are displayed in Figure~\\ref{fig:Explosion}, which agree qualitatively with results in \\cite{liskaWendroff_2003}. \nInitially, this test is similar to the Sod shock tube in cylindrical geometry: a shock and a contact discontinuity propagate in the positive radial direction, while a rarefaction wave propagates toward the origin. \nLater, the initial shock leaves the computational domain, while an instability (resembling Richtmeyr-Meshkov) develops in the contact discontinuity. \nA second shock emerges from the origin of the computational domain and runs through the contact discontinuity, which has developed a complex structure (see density contours in Figure~\\ref{fig:Explosion}). \nThe second shock is located around $r=1.2$ at $t=3.2$. \nIn the right panel we plot the distribution of elements flagged for limiting by the troubled-cell indicator at $t=3.2$. \nLimiting is mainly focused on elements around the shock and the contact discontinuity. \n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Explosion_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/Explosion_TCI_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:Explosion}Plots for the Liska-Wendroff Explosion problem at $t=3.2$. In the left panel, intended to match panels in Figure~4.12 in \\cite{liskaWendroff_2003}, the color map shows the pressure distribution, black contours show the mass density, while arrows indicate the velocity. In the right panel we plot the distribution of elements flagged for limiting $t=3.2$.}\n\\end{figure}\n\n\\subsubsection{Standing accretion shock (SAS)}\n\nThis test is of immediate relevance to simulation of CCSNe. \nThe initial conditions, described in detail in \\cite{blondin_etal_2003}, have been used by many to study the standing accretion shock instability (SASI) and CCSN explosion dynamics. \nWe follow closely the description in \\cite{blondin_etal_2003} to initialize this test. \nWe use spherical polar coordinates, let $D=[0.2,2.0]$, and $\\Gamma=4/3$. \nThe gravitational potential is given by the point-mass formula with $GM=0.5$. \nA stationary shock is placed at a radius $R_{\\shock}=1$. \nAhead of the shock, the flow is essentially in free-fall towards the shock with a constant Mach number of $100$. \nThe mass accretion rate is held fixed at the outer boundary to $\\dot{M}=4\\pi$. \nBelow the shock, $r\\in[0.2,1]$, the settling solution is obtained by solving Bernoulli's equation. \nMatter flows through the inner boundary, and the density and pressure in boundary elements are set by extrapolating from $D$ assuming $\\rho\\propto r^{-3}$ and $E\\propto r^{-4}$, while the momentum components are held fixed to their initial values. \n\n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/SAS_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/SAS_TCI_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{36pc}\n \\includegraphics[width=36pc]{./Figures/SAS_Perturbed_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:SAS}Results from the SAS test using $256$ elements. In the upper left panel we plot the mass density (blue) and velocity (red) versus radius for the initial condition (dashed lines) and for $t=100$ (solid lines) from an unperturbed run. In the upper right panel we show elements flagged for limiting in the $rt$-plane. In the bottom panel we show results from a perturbed model, where we plot the relative deviation of the pressure below the shock; cf. Eq.~\\eqref{eq:pressureDeviation}.}\n\\end{figure}\n\nWe use $256$ elements and run the tests until $t=100$ ($>5$ dynamical times \\cite{blondin_etal_2003}). \nResults are shown in Figure~\\ref{fig:SAS}. \nIn the first test we show results from an unperturbed run to gauge the ability of the DG method to maintain the initial state. \nIn the upper left panel we compare the initial state (dashed lines) with the solution at $t=100$. \nExcept for a slight shift ($\\sim1\\%$) in the position of the shock, the initial and final states are indistinguishable on the plot. \nAlso, there are no oscillations visible in the numerical results. \nIn the upper right panel we show elements flagged for limiting in the $rt$-plane (we set $C_{\\TCI}=0.03$ in this run), which illustrates how limiting is confined to the shock. \nIn the second test we placed a shell with thickness $0.2$ and density three times higher than the ambient density ahead of the shock to induce a strong radial perturbation (similar to \\cite{blondin_etal_2003}; see their Figure~4). \nThe initial condition is stable against radial perturbations. \nIn the lower panel in Figure~\\ref{fig:SAS} we plot the pressure deviation below the shock, defined as\n\\begin{equation}\n \\delta p = (p-\\bar{p})/\\bar{p},\n \\quad\\text{where}\\quad\n \\bar{p}=p_{0}(r_{0}/r)^{4},\n \\label{eq:pressureDeviation}\n\\end{equation}\nand $p_{0}$ and $r_{0}$ are the pressure and radius at the inner boundary. \nAs the shell falls through the shock, the strong perturbation results in an interesting pattern of waves propagating inside the shocked cavity, and oscillations of the shock position. \nThe configuration eventually settles down to a configuration that is close to the initial condition (although the position of the shock is slightly larger than that of the initial state.) \n\n\\subsubsection{Yahil-Lattimer collapse}\n\nThe final test of the NR hydrodynamics in \\thornado\\ involves self-gravity and is due to \\cite{yahilLattimer_1982,yahil_1983}. \nIt models the self-similar collapse of a polytropic star; i.e., $p=\\kappa\\rho^{\\Gamma}$, where $\\kappa$ is the polytropic constant. \nThis test is also of immediate relevance to CCSN simulations. \nIn \\cite{yahilLattimer_1982,yahil_1983}, self-similar solutions to the gravitational collapse problem were constructed for $6/5\\le\\Gamma<4/3$. \nThe solutions smoothly connect an homologously collapsing subsonic inner core (velocity proportional to radius) to a supersonically collapsing outer core in near free-fall ($u\\propto r^{-1/2}$). \n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/YahilLattimerCollapse_MassDensity_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/YahilLattimerCollapse_Velocity_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:YahilLattimer}Results from the Yahil-Lattimer collapse test using $256$ elements. Results obtained with \\thornado\\ (dashed black) are compared with the reference solution from Yahil \\cite{yahil_1983} (solid red). The mass density (left panel) and velocity (right panel) are plotted versus radius. We compare the solutions at select central densities during collapse, approximately $[10^{10},10^{11},10^{12},10^{13},10^{14}]$~g~cm$^{-3}$, which correspond to $(-t)=[51.0,15.0,5.0,1.5,0.5]$~ms. }\n\\end{figure}\nWith two dimensional parameters in the model (the gravitational constant $G$ and the polytropic constant $\\kappa$), the dimensionless similarity variable is\n\\begin{equation}\n X = \\kappa^{-1/2} \\, G^{(\\Gamma-1)/2} \\, r \\, (-t)^{\\Gamma-2},\n\\end{equation}\nwhere the origin of time is the moment of infinite central density. \nAll the hydrodynamic variables can be expressed as a function of $X$, and the time-dependent Euler equations can be recast as a system of ODEs (see \\cite{yahil_1983} for details). \nTo construct a reference to compare with our numerical results we have solved the ODEs given in \\cite{yahil_1983} to obtain these self-similar solutions. \n\nWe show results for a model with $\\Gamma=1.30$. \nWe use spherical polar coordinates, and let $D=[0,1\\times10^{5}]$~km, which is covered with $256$ elements. \nThe Newtonian gravitational potential is obtained by solving Poisson's equation using a third-order accurate continuous finite element method (e.g., \\cite{brennerScott_2008}). \nWe use a geometric grid to resolve the mass distribution as the star collapses and the central density increases from about $10^{9}$~g~cm$^{-3}$ to about $10^{14}$~g~cm$^{-3}$. \nThe size of the innermost element is set to $1$~km while the size of the last element is about $3\\times10^{3}$~km. \nWe specify the polytropic constant by setting a reference pressure $p=6\\times10^{27}$~erg~cm$^{-3}$ for $\\rho=7\\times10^{9}$~g~cm$^{-3}$ (reasonable values for a massive star in the pre-collapse stage). \nWe also set the collapse time to $(-t)=150$~ms. \nResults comparing the mass density and velocity of Yahil's reference solution to those obtained with \\thornado\\ using $C_{\\TCI}=0.03$ are plotted in Figure~\\ref{fig:YahilLattimer}. \nThe agreement of the mass density and velocity profiles is excellent throughout collapse. \n\n\\subsection{Special relativistic (SR) hydrodynamics}\n\nHere we show results from solving the SR Euler equations in Cartesian coordinates with \\thornado.\nIn this case the state, flux, and source vectors in Eq.~\\eqref{eq:extendedEulerCompact} are\n\\begin{equation}\n \\vect{U}=\\big(\\rho W, \\rho h W^{2} u_{j},\\tau\\big)^{\\trans},~\n \\vect{F}^{i}=\\big(\\rho W u^{i},\\Pi^{i}_{~j},\\rho(hW-1)W u^{i}\\big)^{\\trans},~\n \\text{and}~\\vect{S}=0,\n\\end{equation}\nwhere $W$ is the Lorentz factor, $h=1+(e+p)/\\rho$ is the specific enthalpy, $\\tau=\\rho W(hW-1)-p$, and $\\Pi^{i}_{~j}=\\rho h W^{2} u^{i} u_{j}+p\\delta^{i}_{~j}$. \nTo recover primitive from conserved variables we have implemented the procedure for analytic equations of state detailed in Appendix~D in \\cite{rezzollaZanotti_2013}. \nWe solve two Riemann problems and compare the numerical results with exact solutions obtained with the solver from \\cite{martiMuller_2003}. \nLet the primitive state vector be $\\vect{V}=\\big(\\rho,\\vect{u},p\\big)^{\\trans}$. \nIn the first problem (Riemann problem~1), taken from \\cite{mignoneBodo_2005}, we use $\\Gamma=4/3$ and\n\\begin{equation*}\n \\vect{V}_{\\leftState}=\\big(1.0,0.9,0.0,0.0,1.0\\big)^{\\trans}\\quad\\text{and}\\quad\n \\vect{V}_{\\rightState}=\\big(1.0,0.0,0.0,0.0,10.0\\big)^{\\trans},\n\\end{equation*}\nwhile in the second problem (Riemann problem~2), also from \\cite{mignoneBodo_2005}, we use $\\Gamma=5/3$, and let\n\\begin{equation*}\n \\vect{V}_{\\leftState}=\\big(1.0,\\vect{0.0},10^{3}\\big)^{\\trans}\\quad\\text{and}\\quad\n \\vect{V}_{\\rightState}=\\big(1.0,\\vect{0.0},10^{-2}\\big)^{\\trans}. \n\\end{equation*}\nIn both problems, the computational domain is $D=[0,1]$, the discontinuity is initially located at $x^{1}=0.5$, and the solutions are integrated to $t=0.4$. \nResults are plotted in Figure~\\ref{fig:RelativisticHydro}. \n\n\\begin{figure}[h]\n \\centering\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/MB2005_01_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/MB2005_04_Astronum_2018}\n \\end{minipage} \\\\\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/MB2005_01_TCI_Astronum_2018}\n \\end{minipage}\\hspace{0.5pc}%\n \\begin{minipage}{18pc}\n \\includegraphics[width=18pc]{./Figures/MB2005_04_TCI_Astronum_2018}\n \\end{minipage}\n \\caption{\\label{fig:RelativisticHydro}Results from solving the SR Euler equations with \\thornado. Riemann problem~1 (left panels) was solved using $100$ elements, while Riemann problem~2 (right panels) was solved using $400$ elements. The numerical and exact solutions are plotted with plusses and solid lines, respectively. In the lower panels we plot $xt$-plane locations of elements flagged for limiting}\n\\end{figure}\n\nAgain, the main features of the exact solution are captured with the DG method implemented in \\thornado. \nFor Riemann problem~1, we observe some oscillations in the density profile between the leftmost shock and the contact discontinuity located around $x^{1}=0.6$. \nFor Riemann problem~2, the maximum density in the thin density shell between the shock and the contact discontinuity is significantly lower for the numerical solution than the exact solution. \nThis is due to a combination of limited spatial resolution and the action of the slope limiter. \nWe used $C_{\\TCI}=0.03$ in both tests in this section, and we plot locations in the $xt$-plane of elements flagged for limiting in Figure~\\ref{fig:RelativisticHydro}. \nThe troubled-cell indicator tracks the discontinuities well for both Riemann problems. \nWe have found the agreement with the exact solution to improve for larger values of $C_{\\TCI}$, but at the expense of somewhat more oscillatory results. \n\n\\section{Summary and outlook}\n\nWe have presented preliminary algorithm details and numerical results for solvers of the non-relativistic and special relativistic Euler equations of gas dynamics as implemented in the {\\bf t}oolkit for {\\bf h}igh-{\\bf or}der {\\bf n}eutrino-r{\\bf ad}iation hydr{\\bf o}dynamics (\\thornado). \nThe spatial discretization is based on the DG method and the ODEs resulting from this discretization are integrated in time using SSP-RK methods. \nWe employ a spectral-type nodal collocation DG approximation based on Legendre-Gauss points for interpolation and numerical quadrature evaluation \\cite{bassi_etal_2013}. \nThis choice simplifies expressions for the semi-discretized equations, especially when applied to problems involving curvilinear coordinates encoded in a metric (e.g., in numerical relativity). \nResults from a suite of tests in one and two spatial dimensions --- including problems with strong shocks --- demonstrate reliable performance of the implementation. \nThe accuracy of \\thornado\\ on three shock tests (Sod shock tube \\cite{sod_1978}, Sedov-Taylor blast wave \\cite{sedov_1959}, and Shu-Osher shock tube \\cite{shuOsher_1989}) is qualitatively similar to that of the well-established CCSN simulation code \\chimera\\ \\cite{bruenn_etal_2018}, which is based on the PPM finite volume method \\cite{colellaWoodward_1984}. \nTwo tests in spherical symmetry (standing accretion shock \\cite{blondin_etal_2003} and Yahil-Lattimer collapse \\cite{yahilLattimer_1982,yahil_1983}) demonstrate the DG method's ability to handle conditions relevant to CCSN simulations. \nWe will present a more in-depth analysis of the performance of \\thornado\\ on these (and related) problems in a future study. \n\nThe performance of the DG algorithm is sensitive to limiting of the polynomial representation. \nThe combination of a TVD-type limiter (e.g., \\cite{cockburnShu_1998}) and the troubled-cell indicator of \\cite{fuShu_2017} seems to give satisfactory results for a range of problems. \nHowever, in our experience, the optimal value for the indicator threshold $C_{\\TCI}$ seems to vary with the specific problem, and further investigation is needed to determine if there exists an optimal value suitable for CCSN simulations. \nAlong these lines of investigation, it would also be interesting to explore and compare the use of the {\\it a posteriori} subcell limiting approach in \\cite{dumbser_etal_2014,fambri_etal_2018} with our current approach. \n\nOngoing and planned near-future work within \\thornado\\ include coupling to solvers for two-moment neutrino transport, extensions to accommodate nuclear equations of state, general relativity within the conformal flatness approximation, and deployment within an adaptive mesh refinement framework. \nWe hope to report on progress in these directions in the near future. \n\n\\ack Eirik Endeve, Anthony Mezzacappa, Nick Roberts, and Samuel J. 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D} {\\bf 54} 1317\n \\bibitem{wuTang_2015} Wu K and Tang H 2015 {\\it JCP} {\\bf 298} 539\n \\bibitem{yahilLattimer_1982} Yahil A and Lattimer J M 1982 {\\it Supernovae: A Survey of Current Research} Reidel\n \\bibitem{yahil_1983} Yahil A 1983 {\\it ApJ} {\\bf 265} 1047\n \\bibitem{zanotti_etal_2015} Zanotti O, Fambri F and Dumser M 2015 {\\it MNRAS} {\\bf 452} 3010\n \\bibitem{zhangShu_2010} Zhang X and Shu C-W 2010 {\\it JCP} {\\bf 229} 8918\n\\end{thebibliography}\n\n\\end{document}\n\n\n", "meta": {"hexsha": "127bde6f15600c1682ac88d848ef65a207465036", "size": 57108, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documents/Euler/Astronum_2018/ms_endeve.tex", "max_stars_repo_name": "srichers/thornado", "max_stars_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-12-08T16:16:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-24T19:31:21.000Z", "max_issues_repo_path": "Documents/Euler/Astronum_2018/ms_endeve.tex", "max_issues_repo_name": "srichers/thornado", "max_issues_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2019-07-10T20:13:15.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-11T13:21:00.000Z", "max_forks_repo_path": "Documents/Euler/Astronum_2018/ms_endeve.tex", "max_forks_repo_name": "srichers/thornado", "max_forks_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2018-11-14T01:13:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-24T02:08:20.000Z", "avg_line_length": 84.8558692422, "max_line_length": 768, "alphanum_fraction": 0.7385305036, "num_tokens": 17799, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8933093975331751, "lm_q2_score": 0.7185943985973772, "lm_q1q2_score": 0.6419271292817372}} {"text": "\\section{Introduction}\n\\label{sec:Introduction}\n\n\n\\subsection{Model formulation}\n\\label{subsec:Model formulation}\n\nLet us consider a two-stage stochastic program as follows. Let $\\Omega$ represent the set of scenarios, and $p(\\omega)$ be the probability distribution function for the random parameters $\\omega$ with support $\\Omega$. Let $x \\in \\mathbb{R}^{n_1}$ be the first-stage (i.e., here-and-now) decision variables, and let $y(\\omega) \\in \\mathbb{R}^{n_2}$ be the second-stage (i.e., recourse or wait-and-see) decision variables representing recourse decisions corresponding to scenario $\\omega$. Moreover, let the first $i$ variables of $x$ be integer valued. A two-stage stochastic programming formulation is given by\n\\begin{subequations} \\label{eqn:master problem}\n\\begin{align}\n \\min \\quad & c^T x + \\cQ(x) \\\\\n \\text{s.t.} \\quad\n & b^L \\leq A x \\leq b^U \\\\\n & x \\in \\Z^{i}_{+} \\times \\R^{n_1 - i}_{+},\n\\end{align}\n\\end{subequations}\nwith the second stage recourse function $\\cQ(x)\\coloneqq \\E \\lmp Q(x, \\omega) \\rmp$,\n\\begin{subequations} \\label{eqn:subproblem}\n\\begin{align}\n Q(x, \\omega):= \\min \\quad & q(\\omega)^T y(\\omega)) \\\\\n \\text{s.t.} \\quad\n & h^L(\\omega) \\leq W(\\omega) y(\\omega) \\leq h^U(\\omega) \\quad \\forall \\omega \\in \\Omega - T(\\omega) x\\\\\n & y(\\omega) \\in \\R^{n_2}_{+},\n\\end{align}\n\\end{subequations}\nwhere $A \\in \\R^{m_1 \\times n_1}, b \\in \\R^{m_1}, c \\in \\R^{n_1}$ and $T(\\omega) \\in \\R^{m_2 \\times n_1}$, $W(\\omega) \\in \\R^{m_2 \\times n_2}$, $h^L(\\omega), h^U(\\omega) \\in \\R^{m_2}$, $q(\\omega) \\in \\R^{n_2}$ for each $\\omega \\in \\Omega$.\nA reformulation of problem (\\ref{eqn:model formulation}) is the extensive form deterministic equivalent model (DE)\n\\begin{subequations} \\label{eqn:model formulation}\n\\begin{align}\n \\min \\quad & c^T x + \\int_{\\omega \\in \\Omega} p(\\omega) q(\\omega)^T y(\\omega) d\\omega \\\\\n \\text{s.t.} \\quad\n & b^L \\leq A x \\leq b^U \\\\\n & h^L(\\omega) \\leq T(\\omega) x + W(\\omega) y(\\omega) \\geq h^U(\\omega) \\quad \\forall \\omega \\in \\Omega \\\\\n & x \\in \\Z^{i}_{+} \\times \\R^{n_1 - i}_{+}, y(\\omega) \\in \\R^{n_2}_{+}.\n\\end{align}\n\\end{subequations}\n In the stochastic programming literature, $T(\\omega)$ is called a technology matrix and $W(\\omega)$ is called a recourse matrix.\n\n\\subsubsection{Deterministic equivalent format}\n\\label{subsubsec: Deterministic equivalent format}\n\nWhen $\\Omega$ represents a finite set of scenarios, the expectation in (\\ref{eqn:master problem}) can be replaced by a probability-weighted sum. The DE reformulation is the most straight forward approach to solve a two-stage stochastic mixed integer program. Let $S$ be the number of scenarios in $\\Omega$, and $x \\in \\Z^{i}_{+} \\times \\R^{n_1 - i}_{+}, y_1,\\ldots ,y_S \\in \\R^{n_2}_{+}$. The DE formulation of problem (\\ref{eqn:model formulation}) is presented in the following form:\n\\begin{align}\n\\begin{array}{rllllllllll}\n {\\min} & c_f &+ & c^T x & +& p(1) q_1^T y_1 & \\ldots & + &p(S) q_S^T y_S & & \\\\\n & & & & & & & & & \\\\\n \\text{s.t.} & b^L &\\leq& Ax & & & & & &\\leq & b^U, \\\\\n & h^L_1 &\\leq& T_1 x & + &W_1 y_1 & & & & \\leq & h^U_1, \\\\\n & & \\vdots& & & & \\ddots & & & \\vdots & \\\\\n & h^L_S & \\leq &T_S & & & & +& W_S y_S &\\leq & h^U_S,\n\\end{array}\n\\end{align}\nwith the bounds\n\\begin{align}\n\\begin{array}{lllll}\n x^L & \\leq & x & \\leq & x^U, \\\\\n y^L_1 & \\leq & y_1 & \\leq & y^U_1, \\\\\n & \\vdots & & \\vdots & \\\\\n y^L_S & \\leq & y_S & \\leq & y^U_S.\n\\end{array}\n\\end{align}\n\n\\subsubsection{C API}\n\\label{subsubsec:C API}\n\n\\IOP~provides a C API which gives programmatic access to the solver functionality. An online iOptimize C-API user manual can be found in the following website:\n\\[\n %\\mbox{TODO.}\n \\mbox{\\tt http://www.voptimize.org}.\n\\]\n\n\\subsubsection{SMPS Format}\n\\label{subsubsec:SMPS Format}\n\n\\IOP~also supports a subset of SMPS~\\cite{Birge.Louveaux1997IntroductionToStochastic}, which is an early format for stochastic programming problems. In this format three separate data files are provided:\n\n\\begin{itemize}\n \\item Core file: a generic presentation of the variables and constraints in MPS (MPSX) layout. Data values need not feature any particular scenario, but every random data element must be represented.\n \\item Time file: specifying the subdivisions of the core file that belong to each stage.\n \\item Stoch file: specifying every scenario, the values of random data, and probabilities. However, we note that at this stage, only discrete forms of random distributions are implemented in \\IOP.\n\\end{itemize}\n\n\\subsection{External Solvers}\n\\label{subsec:External Solvers}\nThe current version of \\IOP~uses \\Cplex~\\cite{IBMILogCplex} as an external solver. The following versions of \\Cplex~are supported:\n\\begin{itemize}\n \\item 12.5.0.0, 12.5.1.0\n \\item 12.6.0.0, 12.6.1.0, 12.6.2.0, 12.6.3.0\n\\end{itemize}\nA trail or academic version of \\Cplex~can be downloaded from the following website:\n\\[\n \\mbox{\\tt http://goo.gl/gh1SoG}.\n\\]\n\n\\subsection{\\IOP-\\SOS \\ Installation}\n\\label{subsec:Installation}\n\\IOP~supports Linux 64-bit compatible operating system (GLIBC version at least 2.14). After downloading the package, installation requires simply moving the package to the preferred directory. Contents of the package are\n\\begin{itemize}\n \\item ./bin: a folder containing the iOptimize dynamic library.\n \\item ./include: a folder containing the iOptimize header files.\n \\item ./examples: a folder containing examples of using the iOptimize package.\n\\end{itemize}\n\nA gcc makefile is included in the folder examples. To compile the code, use command\n\\[\n {\\tt make~all}.\n\\]\n\nOnce completed, two executable files, namely {\\tt IopTssLpEx1} and {\\tt IopTssLpEx2} shall be generated. To run the examples, use command\n\\[\n {\\tt ./IopTssLpEx1} \\mbox{~~or~~} {\\tt ./IopTssLpEx2}.\n\\]\n\n{\\bf A path to the dynamic library of the external solver \\Cplex~must be specified before running the code}. The path is specified in the parameter file {\\tt TssLp.par} located in folder ./example by modifying the following two parameters:\n\\begin{itemize}\n \\item {\\tt IOP\\_SPA\\_SOS\\_EXTSOL\\_DLL\\_PATH}: A path to the directory where the \\Cplex~dynamic library is currently located. For example, {\\tt\"/PATH/TO/ibm/ILOG/}{\\tt CPLEX\\_Studio1261/}{\\tt cplex/bin/x86-64\\_linux/\"}.\n \\item {\\tt IOP\\_SPA\\_SOS\\_EXTSOL\\_DLL\\_NAME}: A filename of the \\Cplex~dynamic library. For example, {\\tt \"libcplex1261.so\"}.\n\\end{itemize}\nThe editor such as {\\tt vim} may be used to edit the parameter file {\\tt TssLp.par}.\n\n", "meta": {"hexsha": "dbe4b39d0ae6f427299bbff6f7d42256c439a842", "size": 6897, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/ConeOpt_Intro.tex", "max_stars_repo_name": "jupiters1117/ConeOpt", "max_stars_repo_head_hexsha": "b380e4f4f6c1b7cbe1cf02f25b1d404eabe68325", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/ConeOpt_Intro.tex", "max_issues_repo_name": "jupiters1117/ConeOpt", "max_issues_repo_head_hexsha": "b380e4f4f6c1b7cbe1cf02f25b1d404eabe68325", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/ConeOpt_Intro.tex", "max_forks_repo_name": "jupiters1117/ConeOpt", "max_forks_repo_head_hexsha": "b380e4f4f6c1b7cbe1cf02f25b1d404eabe68325", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.0, "max_line_length": 611, "alphanum_fraction": 0.6486878353, "num_tokens": 2199, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951025545426, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.6419093651281262}} {"text": "\\chapter{Spaces}\n\\label{chp:manifold}\n\nA space is a set of points.%\n\\footnote{\\url{https://en.wikipedia.org/wiki/Space_(mathematics)}}%\n\\footnote{\\url{https://en.wikipedia.org/wiki/Topological_space\\#Definition}}\n\n\\enquote{Space} is another word for \\enquote{set}.\n\n\\enquote{Point} is another word for \\enquote{element}.\n\nBecause spaces are sets,\nwe can use set operations such as intersection and union.\nFor example, if \\(A_1\\) is the set of all points on the line \\(L_1\\),\nand \\(A_2\\) is the set of all points on the line \\(L_2\\),\nthen the set intersection \\(A_1 \\cap A_2\\)\nis the set of all points where \\(L_1\\) and \\(L_2\\) intersect.\nA point in \\(A_1 \\cap A_2\\) must be both on \\(L_1\\) and \\(L_2\\) simultaneously.\nIf those lines are parallel (and don't coincide),\nthen \\(A_1 \\cap A_2\\) is empty,\nbecause those lines don't intersect.\n\nThe dimension of a space is a natural number.\n\nA point is an example of a zero-dimensional space.\n\nA line is an example of a one-dimensional space.\n\nA sheet is an example of a two-dimensional space.\n\nA cube is an example of a three-dimensional space.\n", "meta": {"hexsha": "ee49bc7e6644e95c7841312de66559d62996c982", "size": 1089, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "research/physics/space.tex", "max_stars_repo_name": "edom/work", "max_stars_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_stars_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "research/physics/space.tex", "max_issues_repo_name": "edom/work", "max_issues_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_issues_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-12-02T18:37:37.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-16T00:55:32.000Z", "max_forks_repo_path": "research/physics/space.tex", "max_forks_repo_name": "edom/work", "max_forks_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_forks_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-10-02T15:20:22.000Z", "max_forks_repo_forks_event_max_datetime": "2018-10-02T15:20:22.000Z", "avg_line_length": 34.03125, "max_line_length": 79, "alphanum_fraction": 0.7318640955, "num_tokens": 308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438950947024555, "lm_q2_score": 0.7606506472514406, "lm_q1q2_score": 0.6419093499977385}} {"text": "\\chapter{Special relativity}\nThe study of high energy physics involves particles travelling at speeds close to $c$ and often with energies much greater than their mass. Therefore, much of the notation and calculations used stem from a proper treatment of special relativity (SR).\n\n\\section{4-vectors}\nSR places space and time on an equal footing and they adopt the same units in the system of natural units. We combine the information in time and space as components in a 4-vector, for example,\n\\begin{align*}\nx^\\mu = \\mqty(t \\\\ \\vec{r}), \\quad p^\\mu = \\mqty(E \\\\ \\vec{p}).\n\\end{align*}\nNote that in contravariant form (with the index up), 4-vectors are written as column vectors.\n\nThe metric tensor is a symmetric matrix that tells us how to link space and time. In the case of special relativity, it is simply a diagonal matrix\\footnote{The $(+,-,-,-)$ signature is most often used in high energy physics since most 4-vector quantities we deal with will be timelike. By contrast, general relativity usually uses the $(-,+,+,+)$ signature. Using either signature is valid, but it is important to not confuse them.},\n\\begin{equation}\ng_{\\mu\\nu} = \\mqty(\\dmat{1,-1,-1,-1}) = g^{\\mu\\nu}.\n\\end{equation}\nThe metric tensor may be used to perform index gymnastics to raise or lower an index. For example, we can find the covariant forms of the above 4-vectors,\n\\begin{align*}\nx_{\\mu} = g_{\\mu\\nu}x^\\nu = (t,\\, -\\vec{r}), \\quad p_{\\mu} = g_{\\mu\\nu}p^\\nu = (E,\\, -\\vec{p}).\n\\end{align*}\n\nThe scalar product of two 4-vectors is given by\n\\begin{equation}\na_{\\mu} b^{\\mu} = g_{\\mu\\nu}a^{\\nu}b^{\\mu} = b_{\\mu} a^{\\mu}.\n\\end{equation}\n\nNote the scalar product is invariant to which vector is covariant. Often, when only dealing with scalar products, we use the notation of a capital letter with no index for a 4-vector: $X$, $P$, $A$, etc. Then a scalar product may be written, for instance, as $XX = X^2$.\n\nFinally, introduce the 4-vector derivative in covariant form,\n\\begin{equation}\n\\partial_\\mu = \\left( \\pdv{t},\\, \\vec{\\nabla} \\right).\n\\end{equation}\nthen its square length is\n\\begin{equation}\n\\partial^2 = \\partial_\\mu \\partial^\\nu = \\left(\\pdv{t}\\right)^2 - \\nabla^2.\\label{eq:nablaSquared}\n\\end{equation}\n\n\\section{Lorentz transformation}\nThe Lorentz transformation is given by $x^\\mu \\rightarrow (x^\\prime)^\\mu$ where\n\\begin{equation}\n\\mqty(t^\\prime \\\\ x^\\prime \\\\ y^\\prime \\\\ z^\\prime) = \\mqty(\\dmat{\\gamma&{-\\beta\\gamma}\\\\{-\\beta\\gamma}&\\gamma, 1, 1}) \\mqty(t\\\\ x\\\\ y\\\\ z)\n\\end{equation}\nfor a 4-vector boosted by speed $\\beta$ along the $x$-axis. For completeness, we define $\\gamma$ here,\n\\begin{equation}\n\\gamma = \\frac{1}{\\sqrt{1-\\beta^2}}.\n\\end{equation}\n\nThe scalar product $a_\\mu b^\\mu$, and in fact any fully-contracted quantity, is invariant under Lorentz transformations, since they do not depend on the coordinates.\n\n\\section{The light cone}\nConsider two events with 4-coordinates $X$ and $Y$. The square of their spacetime separation is given by\n\\begin{equation}\ns^2 = (X-Y)^2 = (t_x - t_y)^2 - (\\vec{x} - \\vec{y})^2.\n\\end{equation}\nAlong the surface of the light cone, $s=0$. This condition gives\n\\begin{equation}\nt_x - t_y = \\abs{\\vec{x} - \\vec{y}}\n\\end{equation}\ni.e.~if a flash of light is emitted at $X$, it reaches $\\vec{y}$ at $t_y$. If $s^2 > 0$, then $t_x - t_y > \\abs{\\vec{x} - \\vec{y}}$ and a light pulse emitted at $X$ reaches $\\vec{y}$ before $t_y$. Therefore, the events are causally connected and their separation is said to be timelike. For $s^2<0$, the events are have spacelike separation and are causally disjoint.\n\n\\section{Relativistic kinematics}\nIn particle physics, there are two processes we could consider. Firstly, the decay of one particle into daughter particles,\n\\begin{equation}\nA \\to B + C + \\ldots\n\\end{equation}\nwhere the simplest case is that of two-body decay, \\HepProcess{A\\to B+C}. Here the centre of mass (CM) energy is simply the mass of the parent particle, $m_A$. Secondly, a scattering process,\n\\begin{equation}\nA + B \\to C + D + \\ldots\n\\end{equation}\nwhere the CM energy is given by $E_{CM}^2 = (P_A + P_B)^2 = (P_C + P_D + \\ldots) ^2$ (recall that $P_i$ is a 4-momentum).\n\n\\subsection{Fixed target particle production}\nConsider the process\n\\begin{equation}\nA + B \\to C\n\\end{equation}\nwhere the target $B$ is at rest in the laboratory frame. Therefore, $P_B = (m_B, \\vec{0})$.\n\nThe centre of mass energy is given by\n\\begin{align}\nE_{CM}^2 &= (P_A + P_B)^2 \\\\\n&= E_A^2 + 2E_A m_B + m_B^2 - \\abs{\\vec{p_A}}^2\n\\end{align}\nUse that $E_A^2 - \\abs{\\vec{p_A}}^2 = m_A^2$,\n\\begin{equation}\nE_{CM}^2 = m_A^2 + m_B^2 + 2E_A m_V\n\\end{equation}\nA typical beam has energy much greater than either of the masses involved, so we may write\n\\begin{equation}\n\\boxed{\nE_{CM} \\approx \\sqrt{2 E_A m_B}\n}\n\\end{equation}\n\nAt the production threshold, $E_{CM} = E_C$. This means that for a proton target ($m_B \\sim \\SI{1}{\\giga\\electronvolt}$), the incident particle energy required to produce a Higgs boson ($m_C \\sim \\SI{100}{\\giga\\electronvolt}$) is about \\SI{5}{\\tera\\electronvolt}. For a hadronic beam about one tenth of the beam energy goes into collisions so this would require a \\SI{50}{\\tera\\electronvolt} beam. In fact, fixed-target collisions rarely even produce \\Pbottom quarks. This means fixed-target experiemnts are not suitable for modern high-energy discoveries with present accelerator technology, and we must instead look towards beam-beam colliders.\n\n\\subsection{Beam-beam collisions}\nIn this type of collider, two beams are collided with opposite momenta:\n\\begin{align*}\nP_A = \\mqty(E_A \\\\ \\vec{p}), \\quad P_B = \\mqty(E_B \\\\ -\\vec{p})\n\\end{align*}\nso the CM frame is the laboratory frame:\n\\begin{equation}\nP_{CM} = P_A + P_B = \\mqty(E_{CM} \\\\ \\vec{0}) = \\mqty(E_A + E_B \\\\ \\vec{0}).\n\\end{equation}\nWe want to find the required energy for one of the beams, $E_A$, as a function of $E_{CM}$. Start by considering the product\n\\begin{align}\nP_A P_{CM} &= \\left( E_A,\\, \\vec{p} \\right) \\mqty(E_{CM} \\\\ \\vec{0}) = E_A E_{CM} \\\\\n\\text{also,}\\quad P_A P_{CM} &= P_A (P_A + P_B) = P_A^2 + P_A P_B \\label{eq:P_AP_CM}\n\\end{align}\nso, using $P_A^2 = m_A^2$, we have\n\\begin{equation}\nE_A E_{CM} = m_A^2 + P_A P_B. \\label{eq:E_AE_CM}\n\\end{equation}\nTo determine $P_A P_B$ consider\n\\begin{align}\nE_{CM}^2 &= P_{CM}^2 \\\\\n&= (P_A + P_B)^2 \\\\\n&= m_A^2 + m_B^2 + 2P_A P_B.\n\\end{align}\nSubstituting this into \\eqref{eq:E_AE_CM},\n\\begin{align}\nE_A E_{CM} &= m_A^2 + \\frac{1}{2}(E_{CM}^2 - m_A^2 - m_B^2) \\nonumber \\\\\n&= \\frac{E_{CM}^2 + m_A^2 - m_B^2}{2}\n\\end{align}\ntherefore,\n\\begin{equation}\\boxed{\nE_A = \\frac{E_{CM}^2 + m_A^2 - m_B^2}{2E_{CM}}.\n}\n\\end{equation}\n\nIn the case where $m_A = m_B$ this becomes $E_A = E_{CM}/2$.\n\n\\section{Mandelstam variables}\n\\begin{figure}[ht]\n\\centering\n\\include{figures/process}\n\\caption{Some process \\HepProcess{A + B \\to C + D}.\\label{fig:process}}\n\\end{figure}\nConsider some process \\HepProcess{A + B \\to C + D}. The three Mandelstam variables, $s$, $t$, and $u$ are combinations of the incoming and outgoing 4-momenta. For different diagrams, they are the momentum of the propagator:\n\\include{figures/channelTable}\n\nThe sum of $s$, $t$, and $u$ gives the sum of the squares of the mass of the four particles,\n\\begin{align}\ns +t + u &= (P_A + P_B)^2 + (P_A - P_C)^2 + (P_A - P_D)^2 \\\\\n&= 3P_A^2 + P_B^2 + P_C^2 + P_D^2 + 2P_A(P_B - P_C - P_D)\n\\end{align}\nBy conservation of 4-momentum, we have $P_A + P_B$ = $P_C + P_D$, therefore $P_B - P_C - P_D = -P_A$, so\n\\begin{align}\ns + t + u &= 3P_A^2 + P_B^2 + P_D^2 + 2P_A(-P_A) \\\\\n&= P_A^2 + P_B^2 + P_C^2 + P_D^2 \\\\\n&= m_A^2 + m_B^2 + m_C^2 + m_D^2. \\label{eq:madelstamSum}\n\\end{align}\n", "meta": {"hexsha": "1fdb490ac2c1a69cb0b4497d1c5a545b9d77a15f", "size": 7605, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/5_Special_Relativity.tex", "max_stars_repo_name": "adambozson/Standard-Model-I", "max_stars_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/5_Special_Relativity.tex", "max_issues_repo_name": "adambozson/Standard-Model-I", "max_issues_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/5_Special_Relativity.tex", "max_forks_repo_name": "adambozson/Standard-Model-I", "max_forks_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.0328947368, "max_line_length": 646, "alphanum_fraction": 0.691913215, "num_tokens": 2588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9136765187126079, "lm_q2_score": 0.7025300511670689, "lm_q1q2_score": 0.6418852114413178}} {"text": "\\chapter{Lifting Machines}\n\\label{chap:lifting}\n\nWe observe that whenever we want to combine machines, the number of tapes and alphabets of all sub-machines have to agree. For example, we have the\nfollowing typing rule for the sequential composition of labelled Turing machines:\n\\[\n \\inferrule{M_1 : \\TM_\\Sigma^n(L_1) \\and M_2 : \\TM_\\Sigma^n(L_2)}{M_1 \\Seq M_2 : \\TM_\\Sigma^n(L_2)}\n\\]\nAssume that we have a one-tape machine $M_{aux} : \\TM_\\Sigma^1$ that moves the head to the right of the tape. If we need a two-tape machine $M$ that\nmoves both tapes to the right, we would like to use sequential composition and move one tape after the other tape to the right. But according to the\ntyping rule above, we would need two auxiliary two-tape machines $M_1$ and $M_2$, where $M_1$ moves the first tape to the right and $M_2$ the second.\n\nThere are multiple ways to solve this problem. Maybe the most obvious solution is to define a class of $n$-tape machines $M_i : \\TM_\\Sigma^n$\nparametrised over the number of tapes and the tape-index $i$ of the tape to move. The machine $M_i$ moves the $i$th tape to the right. All other\ntapes are ``inactive'' and remain unchanged. This approach of parametrising machines over the tape-indices of ``active'' tapes, however, becomes\nunhandy for machines with a lot of active tapes.\n\nWe choose another approach, because it is in general easier to define and verify machines with a fixed number of tapes. We lift the one-tape machine\n$M_{aux}$ to $n$ different $n$-tape machines, for every $n\\ge1$. Asperti and Ricciotti~\\cite{asperti2015} implement such an operator that translates\na one-tape machine into $n$-tape machines. We implement a generalised operator that takes an $m$-tape machine and a mapping from $\\Fin_m$ to\n$\\Fin_n$, and yields an $n$-tape machine, for~$m \\le n$.\n\nThe second part of this problem is how to combine machines with different alphabets. For example, if we have a machine $\\MS{Add}$ that adds\n(encodings of) natural numbers, we could want to build a machine $\\MS{Sum}$ that computes the sum of a list of numbers. Consider an alphabet\n$\\Sigma_\\Nat$ where we could encode natural numbers on and an alphabet $\\Sigma_{\\List(\\Nat)}$ to encode lists of natural numbers. If the alphabet\n$\\Sigma_\\Nat$ is included in $\\Sigma_{\\List(\\Nat)}$, we would like to lift $\\MS{Add}$ to the alphabet $\\Sigma_{\\List(\\Nat)}$, to define $\\MS{Sum}$.\n\nAsperti and Ricciotti~\\cite{asperti2015} avoid the problem of agreement of alphabets. They consider a fixed alphabet to encode all needed data on and\nimplement a universal Turing machine. However, this approach does not scale when we need to encode many different data types. Whenever the alphabet\nhas to be changed, they also must change the definitions of all auxiliary machines. That is why we introduce another operator that lifts a machine to\na bigger alphabet.\n\nBoth lifting operators are easy to define and verify using the lemmas of the previous chapter (see Section~\\ref{sec:match-proofs}).\n\n\n\\section{Tape-Lift}\n\\label{sec:n-Lift}\n\\setCoqFilename{ProgrammingTuringMachines.TM.Lifting.LiftTapes}%\n\nThe tape-lift takes a machine $M:\\TM_\\Sigma^m(L)$ and a duplicate-free vector $I: \\Fin_n^m$, and yields a machine\n$\\LiftTapes{M}{I} : \\TM_\\Sigma^n(L)$. The tape of $\\LiftTapes{M}{I}$ with the index $i = I[j]$ (with $i:\\Fin_n$, $j:\\Fin_m$) behaves exactly as the\ntape $j$ of $M$. All other tapes of $\\LiftTapes{M}{I}$ that are not in~$I$ are inactive and do not change.\n\n\nThe transition function of $\\LiftTapes{M}{I}$ gets the $n$ read symbols and selects the $m$ relevant symbols. Then it applies the transition function\n$\\delta_M$ with the selected symbols and the current state $q$. $\\delta_M$ yields an $m$-vector $act:\\Act^m$ and the continuation state $q'$. It\nfills ``nop''-actions into $act$, to get an action vector $act':\\Act^n$.\n\n\\begin{definition}[Vector selecting][select]\n \\label{def:select}\n Let $X:\\Type$, $m,n:\\Nat$, $I : \\Fin_n^m$, and $V : X^m$. Then $select~I~V : X^n$ is defined by\n $select~I~V := \\map{\\bigl(\\lambda (j:\\Fin_m).~V[j] \\bigr)}{I}$\n\\end{definition}\n\\begin{lemma}[Correctness of $select$][select_nth]\n By definition, for $j:\\Fin_m$, we have\n \\[\n (select~I~V)[j]=V\\bigl[I[j]\\bigr]\n \\]\n\\end{lemma}\n\n\\begin{definition}[Vector filling][fill]\n Let $X:\\Type$, $m,n:\\Nat$, $I:\\Fin_n^m$, $init:X^n$, and $V:X^m$. Then $fill~I~init~V : X^n$ is defined per recursion:\n \\begin{alignat*}{3}\n & fill~(\\nil )&& ~init~V &~:=~& init \\\\\n & fill~(i \\cons I')&& ~init~V &~:=~& replace~(fill~I'~(\\tl V))~i~(\\hd V)\n \\end{alignat*}\n Where $replace : X^n \\to \\Fin_n \\to X \\to X^n$ replaces the $i$th element of a vector.\n\\end{definition}\n\\begin{lemma}[Correctness of $fill$]\n If $I$ is duplicate-free, then:\n \\begin{enumerate}\n \\coqitem[fill_correct_nth] \\label{lem:fill_correct_nth}\n If $I[j]=i$, then $(fill~I~init~V)[i] = V[j]$.\n \\coqitem[fill_not_index] \\label{lem:fill_not_index}\n If $i \\notin I$, then $(fill~I~init~V)[i] = init[i]$.\n \\end{enumerate}\n\\end{lemma}\n\\begin{proof}\n By induction on $I : \\Fin_n^m$.\n\\end{proof}\n\n\\begin{definition}[$\\LiftTapes{M}{I}$][LiftTapes]\n \\label{def:LiftTapes}\n Let $M:\\TM_\\Sigma^m(L)$ and $I : \\Fin_n^m$. We define $\\LiftTapes{M}{I} : \\TM_\\Sigma^n(L)$. All components are the same as in $M$, except:\n \\begin{alignat*}{2}\n &\\delta(q,s) &~:=~& \\Let{(q', act) := \\delta_M(q, select~I~s)}{\\\\%\n & &~ ~& (q', fill~I~(\\None,N)^n~act)}\n \\end{alignat*}\n\\end{definition}\n\n\\begin{lemma}[Correctness of $\\LiftTapes{M}{I}$][LiftTapes_Realise]\n \\label{lem:LiftTapes_Realise}\n Let $M : \\TM_\\Sigma^m(L)$ and $I : \\Fin_n^m$ duplicate-free. If $M \\Realise R$, then $\\LiftTapes{M}{I} \\Realise \\LiftTapes{R}{I}$ with\n \\[\n \\LiftTapes{R}{I} := \\lambda t~(l,t').~ R~(select~I~t)~(l, select~I~t') \\land \\forall i:\\Fin_n.~i \\notin I \\rightarrow t'[i]=t[i]\n \\]\n\\end{lemma}\n\n\\begin{lemma}[Running time of $\\LiftTapes{M}{I}$][LiftTapes_Terminates]\n \\label{lem:LiftTapes_TerminatesIn}\n Let $M : \\TM_\\Sigma^m(L)$ and $I : \\Fin_n^m$ duplicate-free. If $M \\TerminatesIn T$, then $\\LiftTapes{M}{R} \\TerminatesIn \\LiftTapes{T}{I}$ with\n $ \\LiftTapes{T}{I} := \\lambda t~k.~ T~(select~I~t)~k. $\n\\end{lemma}\n\nThe proofs are similar to the former proofs, i.e.\\ using Lemma~\\ref{lem:loop_lift} and Lemma~\\ref{lem:loop_unlift} with the following configuration\nlifting function:\n\\[\n selectConf(q,t) := (q, select~I~t)\n\\]\n\nHowever, for the second part of the correctness, i.e.\\ tapes that are not in $I$ do not change, we need another lemma about $\\Loop$:\n\n\\begin{lemma}[Mapping loops][loop_map]\n \\label{lem:loop_map}\n Let $A:\\Type$, $f : A \\to A$, $h:A \\to \\Bool$. Let $B:\\Type$ and $g : A \\to B$. If $g(f~a)=g(a)$ for all $a:A$, then\n \\begin{alignat*}{1}\n & \\forall (k:\\Nat)~(a_1~a_2 : A). \\\\\n & \\quad \\Loop~f~h~k~a_1 = \\Some{a_2} \\rightarrow \\\\\n & \\quad g(a_1) = g(a_2)\n \\end{alignat*}\n\\end{lemma}\n\\begin{proof}\n By induction on $k:\\Nat$.\n\\end{proof}\n\nWe apply this lemma in the proof of Lemma~\\ref{lem:LiftTapes_Realise} with $g := \\lambda ((q,t):\\Conf).~t[i]$.\n\n\n\n\\section{Alphabet-Lift}\n\\label{sec:sigma-Lift}\n\\setCoqFilename{ProgrammingTuringMachines.TM.Lifting.LiftAlphabet}%\n\nLet $M : \\TM_\\Sigma^n(L)$ be a machine over the alphabet $\\Sigma$, and $f : \\Sigma \\hookrightarrow \\Tau$ a retraction on another alphabet $\\Tau$.\nNote that then $\\Tau$ has at least as many symbols as $\\Sigma$. We need a default symbol $def:\\Sigma$. In contrast to the default label we needed in\nthe definition of $\\While$, the choice of $def$ is semantically \\textit{relevant}. This means that $def$ should be a symbol that $M$ does not expect\nto read. The alphabet-lift $\\LiftAlphabet{M}{(f,def)} : \\TM_\\Tau^n(L)$ is a machine over the bigger alphabet $\\Tau$.\n\nThe transition function of the lifted machine $\\LiftAlphabet{M}{(f,def)}$ reads the $n$ optional symbols $s : \\Option(\\Tau)^n$ and tries to translate\nthem to $s': \\Option(\\Sigma)^n$ using the partial inversion function $f^{-1} : \\Tau \\to \\Option(\\Sigma)$. If the symbol $\\tau:\\Tau$ has no\ncorresponding symbol in $\\Sigma$, it must be translated to $def$. The transition function $\\delta_M$ of $M$ yields the successor state $q'$ and a\nvector of actions $act: (\\Option(\\Sigma) \\times \\MS{Move})^n$, which is translated using $f:\\Sigma\\to\\Tau$ to\n$act' : (\\Option(\\Tau) \\times \\MS{Move})^n$.\n\n\\begin{definition}[$\\LiftAlphabet{M}{(f,def)}$][LiftAlphabet]\n \\label{def:LiftAlphabet}\n Let $f : \\Sigma \\hookrightarrow \\Tau$, $def:\\Sigma$, and $M : \\TM_\\Sigma^n(L)$. We define the machine %\n $\\LiftAlphabet{M}{(f,def)} : \\TM_\\Tau^n(L)$ with the same components as $M$, except:\n \\begin{alignat*}{2}\n & \\delta(q,s) &~:=~& \\Let{(q', act) := \\delta_M(q, \\map{(mapOpt~(surject~f~def))}{s})}{\\\\%\n & &~ ~& (q', \\map{(mapAct~f)}{act})} \\\\\n \\end{alignat*}\n With $surject~f~def : \\Tau \\to \\Sigma$:\n \\[\n \\coqlink[surjectTape]{surject}~f~def~\\tau :=\n \\begin{cases}\n \\sigma & f^{-1}(\\tau) = \\Some{\\sigma} \\\\\n def & f^{-1}(\\tau) = \\None\n \\end{cases}\n \\]\n and with the canonical functions $mapOpt: \\forall (X~Y:\\Type).~ (X \\to Y) \\to \\Option(X) \\to \\Option(Y)$ and\n $mapAct : \\forall (\\Sigma~\\Tau:\\Type).~(\\Sigma\\to\\Tau) \\to \\Act_\\Sigma \\to \\Act_\\Tau$.\n\\end{definition}\n\n\\setCoqFilename{ProgrammingTuringMachines.TM.TM}%\nFor the correctness and running time lemmas, we also need the canonical function\n\\[\n \\coqlink[mapTape]{mapTape}:\\forall (\\Tau~\\Sigma:\\Type).~(\\Tau\\to\\Sigma)\\to\\Tape_\\Tau\\to\\Tape_\\Sigma\n\\]\nthat maps every symbol on a tape. We write $\\coqlink[mapTapes]{mapTapes}$ for the respective tape-vector function.\n\\setCoqFilename{ProgrammingTuringMachines.TM.Lifting.LiftAlphabet}%\n\n\\begin{lemma}[Correctness of $\\LiftAlphabet{M}{(f,def)}$][LiftAlphabet_Realise]\n \\label{lem:LiftAlphabet_Realise}\n If $M \\Realise R$, then $\\LiftAlphabet{M}{(f,def)} \\Realise \\LiftAlphabet{R}{(f,def)}$ with\n \\small\n \\[\n \\LiftAlphabet{R}{(f,def)} := \\lambda t~(l, t').~R~(mapTapes~(surject~f~def)~t)~(l, mapTapes~(surject~f~def)~t')\n \\]\n\\end{lemma}\n\n\\begin{lemma}[Running time of $\\LiftAlphabet{M}{(f,def)}$][LiftAlphabet_TerminatesIn]\n \\label{lem:LiftAlphabet_TerminatesIn}\n If $M \\TerminatesIn T$, then $\\LiftAlphabet{M}{(f,def)} \\TerminatesIn \\LiftAlphabet{T}{(f,def)}$ with\n \\[\n \\LiftAlphabet{T}{(f,def)} := \\lambda t~k.~T~(mapTapes~(surject~f~def)~t)~k\n \\]\n\\end{lemma}\n\nThe proofs are analogous to the former proofs. The configuration lifting is:\n\\[\n \\coqlink[surjectConf]{surjectConf} (q,t) := (q, mapTapes~(surject~f~def)~t)\n\\]\n\n\n\n%%% Local Variables:\n%%% TeX-master: \"thesis\"\n%%% End:", "meta": {"hexsha": "eab3f1612f2126aae50387d48ffa22059f1f3e10", "size": 10609, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/thesis/Lifting.tex", "max_stars_repo_name": "mwuttke97/CoqTM", "max_stars_repo_head_hexsha": "f4d2aab2008e2158e2c7ca88ebb53b42808a0778", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2018-08-30T14:58:32.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-27T15:44:28.000Z", "max_issues_repo_path": "tex/thesis/Lifting.tex", "max_issues_repo_name": "mwuttke97/CoqTM", "max_issues_repo_head_hexsha": "f4d2aab2008e2158e2c7ca88ebb53b42808a0778", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-04-10T09:16:49.000Z", "max_issues_repo_issues_event_max_datetime": "2019-04-10T09:16:49.000Z", "max_forks_repo_path": "tex/thesis/Lifting.tex", "max_forks_repo_name": "mwuttke97/CoqTM", "max_forks_repo_head_hexsha": "f4d2aab2008e2158e2c7ca88ebb53b42808a0778", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-04-09T19:01:41.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-29T15:39:53.000Z", "avg_line_length": 53.3115577889, "max_line_length": 150, "alphanum_fraction": 0.6795173909, "num_tokens": 3627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199552262967, "lm_q2_score": 0.7662936430859597, "lm_q1q2_score": 0.6418628470118574}} {"text": "\\SecDef{intro}{Introduction}\n\nS-Boxes are used to provide non-linearity in SPN-based block ciphers. They provide basic resistance against linear and differential cryptanalysis, and the rest of the structure ensures that many S-Boxes are activated in a linear or differential trail. The resistance of an S-Box can be quantified. The lower are the \\emph{linearity} and the \\emph{differential uniformity} of an S-Box, the more resistant it is. Everything else being equal, a stronger S-Box allows to use less rounds in the block cipher using it for the same security level.\n\nThe differential uniformity is always even and at least equal to 2. When this bound is achieved, the S-Box is called \\emph{Almost Perfect Non-linear (APN)}. The finite field cube function is APN in all field dimensions~\\cite{Nyb94}. However, it is a permutation only in odd dimensions. This is a problem, since in most cases (e.g. an SPN block cipher) the S-Boxes are required to be bijective. For efficiency reasons, even-dimensional S-Boxes are preferable, especially powers of 2. And this is exactly the case, where the existence of APN functions is not established: bijective S-Boxes in even dimensions with differential uniformity 2, i.e. APN permutations of $\\field{n}$ for $n$ even. For $n = 4$ there exist no APN permutations of $\\field{n}$. For $n = 6$ this question was a long standing problem until Dillon~\\etal{} presented a 6-bit APN permutation~\\cite{DillonAPN,DillonPres} in 2009. Since then, no answers were obtained for even $n \\ge 8$, despite many attempts~\\cite{SubspaceProperty,Uniform4}. This remains a big open problem in the field of Boolean functions.\n\nThe 6-bit APN permutation is found by a computer search, by transforming the 6-bit APN function, called the Kim mapping $\\kim\\colon \\fielde{6} \\to \\fielde{6}$:\n$$\n\\kim(x) \\eqdef \\VV x^{24} + x^{10} + x^3,\n$$\nwhere $v$ is a primitive element of $\\fielde{6}$. Even though the Kim mapping is a trinomial function, the resulting APN permutation is an object without clear structure. For example, its polynomial form contains 52 monomials.\n\nUsing the methods of S-Box reverse-engineering described in previous chapters (developed in~\\cite{OurKuz1}), I and my coauthors managed to find a simple algebraic structure of the Dillon's APN permutation. We call this structure a ``Butterfly'' because of its graphical representation the way it changes by particular transformations. The decomposition is established in Theorem~\\Ref{thm:main}, restated here:\n\n\\textbf{Main theorem (A Family of 6-bit APN Permutations)}.\nThe 6-bit permutation described by Dillon \\etal{} in~\\cite{DillonAPN} is affine equivalent to any involution built using the structure described in \\FigRef{intro-decomp}, where $\\fmult$ denotes multiplication in the finite field $\\fielde{3}$, $\\alpha \\neq 0$ is such that $\\tr(\\alpha) = 0$ and $\\mathcal{A}$ denotes any 3-bit APN permutation.\n\\FigTex{intro-decomp.tex}\n\n\\subsection{Notations}\nFor any $f\\colon \\field{n} \\to \\field{n}$ let $\\parf{f}\\colon \\field{2n} \\to \\field{2n}$ be the parallel application of $f$ given by\n$$\n\\parf{f}(x,y) = (f(x), f(y)).\n$$\nFor any $a,b \\in \\field{n}$ let $\\xorf{f}\\colon \\field{2n} \\to \\field{2n}$ be parallel xor with constants $a,b$:\n$$\n\\xorf{f}(x,y) = (x \\oplus a, y \\oplus b).\n$$\nThe finite field trace function is denoted by $\\Tr\\colon \\fielde{n} \\to \\field{}$, it is given by\n$$\n\\Tr(x) \\eqdef \\sum_{e=0}^{n-1} x^e.\n$$\n\n\\subsection{Outline}\n\\SecRef{decomposition} explains the decomposition process of the APN permutation. In~\\SecRef{properties} I describe new properties of the APN permutation that follow from the discovered structure. \\SecRef{components} studies the flexibility of the structure, i.e. how can we modify the structure while preserving the APN property?\nIn \\SecRef{relations} I show new relations between the APN permutation, the Kim mapping, monomial functions and 3-round Feistel Network structure. Finally, I briefly conclude the chapter in \\SecRef{conclusions}.\n", "meta": {"hexsha": "5f27b501a2cd339abd324c5ef3c26d9f80ac42f4", "size": 3984, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis-source/9strAPN/0intro.tex", "max_stars_repo_name": "hellman/thesis", "max_stars_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 19, "max_stars_repo_stars_event_min_datetime": "2019-05-16T19:55:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-31T15:36:12.000Z", "max_issues_repo_path": "thesis-source/9strAPN/0intro.tex", "max_issues_repo_name": "hellman/thesis", "max_issues_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-08-09T11:26:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-09T11:26:45.000Z", "max_forks_repo_path": "thesis-source/9strAPN/0intro.tex", "max_forks_repo_name": "hellman/thesis", "max_forks_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-05T19:40:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-05T19:40:16.000Z", "avg_line_length": 110.6666666667, "max_line_length": 1075, "alphanum_fraction": 0.7597891566, "num_tokens": 1078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357632379241, "lm_q2_score": 0.740174367770488, "lm_q1q2_score": 0.6417576478890329}} {"text": "\\section{Reachable nonterminals}\n\n$A$ is reachable iif $S \\Rightarrow^* \\alpha A \\beta$\n\nFor finding the reachable nonterminals build the \\textbf{produce} graph and visit it from the axiom.\n\n\\section{Defined nonterminals}\n$A$ is defined iff $L_A(G) \\ne \\emptyset$.\n\nFor finding the defined nonterminals apply the following relations:\n\\begin{align*}\n D &:= \\{ A | (A \\rarr u) \\in P, u \\in \\Sigma^*\\} \\\\\n D &:= D \\cup \\{ A | (A \\rarr B_1\\ldots B_n) \\in P, \\land \\forall B_i: B_i \\in (D \\cup \\Sigma) \\}\n\\end{align*}\n\n\\section{Circular Derivations}\n\nDerivations like $A \\Rightarrow^+ A$, are not essential and introduce ambiguity.\n\n\\section{Clean Grammars}\nA grammar $G$ is clean iif every nonterminal is \\textbf{reachable} and every nonterminal is \\textbf{defined}.\n", "meta": {"hexsha": "c949618df4a2d49298bd3a3209ddd479c193a195", "size": 769, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "grammars/erroneous-grammars.tex", "max_stars_repo_name": "TiberioG/FLC-cheatsheet", "max_stars_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-01-13T14:36:20.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-18T16:22:18.000Z", "max_issues_repo_path": "grammars/erroneous-grammars.tex", "max_issues_repo_name": "TiberioG/FLC-cheatsheet", "max_issues_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "grammars/erroneous-grammars.tex", "max_forks_repo_name": "TiberioG/FLC-cheatsheet", "max_forks_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-05-21T11:05:53.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-17T14:59:50.000Z", "avg_line_length": 34.9545454545, "max_line_length": 109, "alphanum_fraction": 0.7035110533, "num_tokens": 242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357598021707, "lm_q2_score": 0.7401743563075446, "lm_q1q2_score": 0.6417576354071945}} {"text": "% chapters/searching.tex\n\n\\chapter{Searching}\t\\label{chapter:searching}\n\n% deterministic search\nAlgorithm \\ref{alg:deterministic-search} (see Problem $5-2$ of \\cite{CLRS09})\nsearches for a value $x$ in an unsorted array $A$ consisting of $n$ elements \nby checking $A[1], A[2], \\cdots, A[n]$ in order until either it finds $A[i] = x$\nor it reaches the end of the array.\n\n\\input{algs/searching/deterministic-search}\n\n% binary stride\n\\begin{figure}\n \\centering\n \\includegraphics[width = 0.70\\textwidth]{figs/binary-stride}\n \\caption{Binary stride.}\n \\label{fig:binary-stride}\n\\end{figure}\n\n\\input{algs/searching/binary-search-iterative}\n\\input{algs/searching/binary-search-recursive}\n", "meta": {"hexsha": "477201f5528956479e5032e20c7566013a1b70fa", "size": 685, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/searching.tex", "max_stars_repo_name": "hengxin/algorithms-pseudocode", "max_stars_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-11-06T08:52:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-27T13:01:13.000Z", "max_issues_repo_path": "chapters/searching.tex", "max_issues_repo_name": "hengxin/algorithms-pseudocode", "max_issues_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/searching.tex", "max_forks_repo_name": "hengxin/algorithms-pseudocode", "max_forks_repo_head_hexsha": "5c8265b6368f851337ca9c0dd1476c07b6e29f83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.7826086957, "max_line_length": 80, "alphanum_fraction": 0.7445255474, "num_tokens": 211, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.782662489091802, "lm_q2_score": 0.8198933403143929, "lm_q1q2_score": 0.6416997625202546}} {"text": "\\label{App:appendixB}\n\\begin{quotation}\n\t\"The simplest quantum mechanical system, and the system which we will be most concerned with, is the \\emph{qubit}. A qubit has a two-dimensional state space. [...] \n\tThe way a qubit differs from a bit is that superpositions of these two states, of the form $a\\ket{0} + b\\ket{1}$, can also exist, in which it is not possible to say that the qubit is definitely in the state $\\ket0$, or definitely in the state $\\ket1$.\"\n\t\\cite{NC10}\n\\end{quotation}\n\n%\\large{\\textbf{The three postulates}}\\cite{NC10}\n\\subsection{The three postulates}\\footcite{NC10}\n\t\\begin{quote}\n\t\t\\textbf{Postulate 1}: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the \\emph{state space} of the system. \n\t\tThe system is completely described by its \\emph{state vector}, which is a unit vector in the system's state space. \n\t\\end{quote}\n\t\n\t\\begin{quote}\n\t\t\\textbf{Postulate 2}: The evolution of a \\emph{closed} quantum system is described by a \\emph{unitary transformation}. That is, the state $\\ket{\\psi}$ of the system at time $t_1$ is related to the state $\\ket{\\psi'}$ of the system at time $t_2$ by a unitary operator $U$ which depends only on times $t_1$ and $t_2$,\n\t\t$$ \\ket{\\psi'} = U\\ket{\\psi} $$\n\t\\end{quote}\n\t\n\t\\begin{quote}\n\t\t\\textbf{Postulate 3}: Quantum measurements are described by a collection $\\{M_m\\}$ of \\emph{measurements operators}. \n\t\tThese are operators acting on the state space of the system being measured. \n\t\tThe index $m$ refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $\\ket{\\psi}$ immediately before the measurement then the probability that result $m$ occur is given by \n\t\t$$ p(m) = \\bra{\\psi}M_m^{\\dagger}M_m\\ket{\\psi} \\: ,$$\n\t\tand the state of the system after the measurement is \n\t\t$$ \\frac{M_m\\ket{\\psi}}{\\sqrt{\\bra{\\psi}M_m^{\\dagger}M_m\\ket{\\psi}}} \\: . $$\n\t\tThe measurement operators satisfy the \\emph{completeness equation},\n\t\t$$\\sum_m M_m^{\\dagger}M_m = I \\: .$$\n\t\tThe completeness equation expresses the fact that probabilities sum to one:\n\t\t$$ 1 = \\sum_m p(m) = \\sum_m \\bra{\\psi}M_m^{\\dagger}M_m\\ket{\\psi} \\: .$$ \n\t\\end{quote}\n\t\n\tFor our purposes it is enough for us to only consider the quantum system called \\emph{qubit} and its rules of computation following from the tensor product algebra. \n\n\\section{Dirac's bra-ket notation}\n\t%\t\\begin{figure}[h!]\n%\t\t\\centering\n%\t\t\\includegraphics[scale=0.2]{images/sketch1.png} \n%\t\t\\caption{how a physical state is represented}\n%\t\\end{figure}\n\tEvery pure quantum state can be represented as vector in a vector space with inner product, i.e. a \\emph{Hilbert space}. \n\t A complex Hilbert space $\\H$ of dimension $n$ is isomorphic to $ \\mathbb{C}^n $ with the standard inner product. \n\t In $\\mathbb{C}^n $ one can choose a basis and then represent vectors with coordinates with respect to this basis.\n\tThe bra-ket notation is a handy notation introduced by physicist Paul Dirac to deal with such vector representation of quantum states. \n\tFirst of all we note that a state $\\varphi '\\in\\H$ corresponds via the isomorphism to $ \\varphi \\in \\mathbb{C}^n $. It can be represented as a vector with respect of some basis as follows\n\t$$\\ket{\\varphi} = \\begin{pmatrix} \\varphi_1 \\\\ \\varphi_2 \\\\ \\vdots \\end{pmatrix}\t \\text{ is a coloumn \"ket\" vector over } \\H $$\n\t$$\\bra{\\varphi} = \\begin{pmatrix} \\varphi_1 & \\varphi_2 & \\hdots \\end{pmatrix} \\text{ is a row \"bra\" vector over } \\H $$\n\tTo be representative of a quantum state the vector has to have unitary length, $\\|\\varphi\\|= 1$.\n\tFurthermore the conjugate transpose of a \\emph{bra} vector is the corresponding \\emph{ket} vector, and vice versa.\n\t$$ \\bra{\\varphi}^{\\dagger} = \\ket{\\varphi} \\text{, } \\ket{\\varphi}^{\\dagger} = \\bra{\\varphi}$$\n\tMore specifically, for a complex vector space as $\\H$, the components of $\\bra{\\varphi}$ are each the complex conjugate of the components of $\\ket{\\varphi}$.\n\tIt is worth noting that in quantum information we will consider only vectors of finite dimensions, and more often than not, the standard basis for qubits represented by\n\t$$\\ket0 = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}\n\t\\text{ and }\n\t\\ket1 = \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}$$\n\twhich are recognizable as the equivalent of $\\vec{e_1}$ and $\\vec{e_2}$ in $\\mathbb{C}^2$.\n\t\n\tTo summarize then, $\\ket{\\varphi}$ represents a column vector on a complex vector space with inner product equivalent to $\\mathbb{C}^n$ in some basis, and $\\bra{\\varphi}$ is its complex conjugate.\n\t\n\t\n\tSo if we define the matrix\\footnote{The fact that the result of $ \\ketbra{w}{v} $ is indeed a matrix can be seen more directly if we remember that this is nothing less than a column-row vectors multiplication.} $A = \\ketbra{w}{v} $ we observe that\n\t$$ \\ket{w}\\braket{v}{v'} = \\braket{v}{v'}\\ket{w} $$\t\n\twhich is a convenient way of visualizing the action of matrix $A$. In particular if we divide it like $(\\ketbra{w}{v}) (\\ket{v'}) $ it is easy to interpret it as \\textit{matrix $A$ acting on vector $\\ket{v'}$}. The other equivalent form $(\\braket{v}{v'})(\\ket{w})$ can also be seen as multiplying vector $\\ket{w}$ by a value $\\braket{v}{v'}$.\n\t%this part may be too similar to book, page 67...\n\t\n\tThe intuition of this is that $\\ketbra{w}{v}$ can indeed be defined as a (linear) operator from the vector space of $\\ket{v}$ to the vector space of $\\ket{w}$. \n\n\\section{Measurements on a basis} \\label{measurements}\n\n\tTo get any information out of a state one has to \\textit{measure} it. \n\tMeasurement is, mathematically, a projection onto some chosen computational basis. \n\tThe result for each base vector projection is then interpreted as a \\emph{probability}. \n\tThe state then changes after measurement, meaning for example that it will not retain its value as superposition any more.\\\\\n\n\tIf Alice has the state $\\ket{\\psi_i}$ out of $i=1..n$ and all states are orthonormal, then Bob can --- measuring with the same basis --- find out what the choice of $i$ was.\n\tIf the states are not orthonormal there is no quantum measurement capable of distinguishing the states. \n\tIf the states $\\ket{\\psi_1}$ and $\\ket{\\psi_2}$ are not orthogonal, then $\\ket{\\psi_2}$ has a component orthogonal to $\\ket{\\psi_1}$, and also a component parallel to it.\n\tThis will lead to a non-zero probability of measuring a value on $\\ket{\\psi_1}$ for the state $\\ket{\\psi_2}$.\n\n\t\\begin{xmpl}\\cite{NC10}\n\t\t\\begin{equation*} \n\t\t\tZ = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix} \\quad P_{+1} = \\proj{0} , \\; P_{-1} = \\proj{1}\n\t\t\\end{equation*}\n\t\tMeasurement on qubit $ \\ket{\\psi} = \\frac{\\ket0 + \\ket1}{\\sqrt{2}} $ has probability $p_{+1} = \\bra{\\psi}P_{+1}\\ket{\\psi} = \\bk{\\psi}{0}\\bk{0}{\\psi} = \\frac{1}{2}$ and similarly $p_{-1} = \\frac{1}{2}$\n\t\t\t\n\t\\end{xmpl}\n \n\t\\subsubsection*{Linear operators}\n\tA linear operator between two vector spaces is defined as \n\t$$ \\mathbf{A}: V\\longrightarrow W \\text{ , }\\ket{v_i}\\mapsto A\\ket{v_i}$$\n\t$$ \\text{ linear in all inputs, i.e. } A\\left( \\sum_i a_i\\ket{v_i}\\right) = \\sum_i a_i A\\ket{v_i} \\text{ for all } i $$ \n\tLooking back at the definition of the matrix $ A = \\ketbra{w}{v}$ we can now refer to it as a linear operator from now on.\n\tSome well-known linear operators acting on single qubits that we will use later on are the \\textit{Pauli Matrices}\n\t$$ I = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\t \\quad X = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$$\n\t$$ Y= \\begin{bmatrix} 0 & -i \\\\ i & 0 \\end{bmatrix}\t \\quad Z = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix}$$\n\t\n\tIn particular it is safe to say that, unless stated otherwise, the operators that will be presented all have a set of properties and are called Hermitian operators, or \\emph{self-adjoint operators}.\n\t$$ A = A^{\\dagger} \\quad \\Longrightarrow (A\\ket{v})^{\\dagger} = \\bra{v}A^{\\dagger} $$ \n\tOperators have also to be positive, this means that it holds, for every $\\ket{v}$, $\\bra{v}A\\ket{v}$ is real non-negative. \n \n\\section{Mixed states}\n\tAll pure states in QM are normalized vectors in $\\H$.\n\t$$ \\ket{\\psi} \\text{ is a state vector } \\Rightarrow \\ket{\\psi}\\in\\H \\text{ and } \\vert\\bk{\\psi}{\\psi}\\vert = 1$$\n\tThis is instrumental in seeing them as probability vectors. \n\tEvery linear operator has then to be unitary to maintain this property.\n\tA statistical mixture of states corresponds to a \\emph{density matrix}, which is itself a new state. \n\tIt is important to note that a mixture of probability of states is not the same thing as superposition of states. In the latter we don't have a measure of uncertainty of the state, meaning also that in theory we are always able to find a measurement basis that will always output the same result for that state. \n\tIn the former, however, this is not possible because of the intrinsic uncertainty of the state.\n\tDensity matrices have then the properties:\n\t$$ M = \\rho = \\sum_i p_i \\ketbra{\\psi_i}{\\psi_i} = \\sum_i p_i P_{\\ket{\\psi_i}} \\text{ , where state }\\ket{\\psi_i}\\text{ has probability } p_i $$ \n\t$\\rho$ is a positive, trace-1 operator meaning that $\\Tr(\\rho) = 1$ and all eigenvalues of $\\rho$ are positive. \n\tMoreover $\\rho$ is a linear combination of projectors $\\proj{\\psi_i}$ which makes $\\rho\\in\\mathbb{P}(\\H)$ a projector itself on the the Hilbert space.\n\n%\\section{Quantum entanglement}\n%\n%\t\\begin{quotation}\n%\t\tThere exist vectors in $V\\otimes W$ that can not be represented by a single tensor product:\n%\t\tGiven $v_1,v_2\\in V \\; w_1,w_2\\in W$ linear independent:\n% \\begin{equation*}\n%\t\tv_1\\otimes w_1 + v_2\\otimes w_2 = v_1w_1 + v_2w_2 \\in V\\otimes W\n% \\end{equation*}\n% is \\emph{not} separable.\n%\t\tthis may be strange because on physical level tensor product is combination(merging) of quantum systems\n%\t\t\\end{quotation}\\cite{Han13}", "meta": {"hexsha": "bd5d12d704c67d92b6719d6569adb3275f7b9e59", "size": 9779, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "writings/chapters/appendixB.tex", "max_stars_repo_name": "CrashingBrain/BSc_Project", "max_stars_repo_head_hexsha": "44b91601341ff3a59acbad7abbf28389aa99f89d", "max_stars_repo_licenses": ["FSFAP"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "writings/chapters/appendixB.tex", "max_issues_repo_name": "CrashingBrain/BSc_Project", "max_issues_repo_head_hexsha": "44b91601341ff3a59acbad7abbf28389aa99f89d", "max_issues_repo_licenses": ["FSFAP"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "writings/chapters/appendixB.tex", "max_forks_repo_name": "CrashingBrain/BSc_Project", "max_forks_repo_head_hexsha": "44b91601341ff3a59acbad7abbf28389aa99f89d", "max_forks_repo_licenses": ["FSFAP"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 78.8629032258, "max_line_length": 343, "alphanum_fraction": 0.7049800593, "num_tokens": 3024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.8198933403143929, "lm_q1q2_score": 0.6416997542074672}} {"text": "\\documentclass{homework}\n\\course{Math 5522H}\n\\author{Jim Fowler}\n\\input{preamble}\n\n\\begin{document}\n\\maketitle\n\n\\begin{inspiration}\nNature laughs at the difficulties of integration.\n\\byline{Pierre-Simon Laplace} % what's the original source for this quotation?\n\\end{inspiration}\n\n\\section{Terminology}\n\n\\begin{problem}\n What does it mean to say that $\\Omega \\subset \\C$ is connected? Is path-connected? \n \\end{problem}\n \\begin{solution}\n $\\Omega\\subset \\C$ is said to be connected if, for any two disjoint open sets $U,V\\in \\C$ containing $\\Omega$ in their union, either $U\\cap \\Omega$ or $V\\cap \\Omega$ is empty.\n \\end{solution}\n \\begin{problem}\n What is a piecewise-smooth \\textbf{curve}? When are two curves ``the same''? \n \\end{problem}\n \\begin{solution}\n A smooth curve is a smooth function $f$ from an interval $[t_0, t_1]: t_0, t_1\\in\\R$ to the complex numbers $\\gamma(t)\\in \\C$ such that $\\gamma'(t)\\neq 0$ at any point $t$.A piecewise smooth curve is the function defined on the interval $[t_0, t_n]$ by $n$ smooth curves $$\\gamma_i:[t_{i-1}, t_i]\\mapsto \\C$$ satisfying $\\gamma_{i-1}(t_{i}) = \\gamma_{i}(t_{i})$ for $i\\in [1, n]$.\n\n Two curves $\\gamma_a:[a_0, a_1] \\mapsto \\C, \\gamma_b:[b_0, b_1]\\mapsto \\C$ are said to be the same if there is a differentiable function $f: [a_0, a_1] \\mapsto [b_0, b_1]$ such that $\\forall t\\in [a_0, a_1], \\gamma_a(t) = \\gamma_b(f(t))$.\n \\end{solution}\n \\begin{problem}\n Define $\\int_\\gamma f(z) \\, dz$ and define $\\int_\\gamma f(z) \\, dx$ and define $\\int_\\gamma f(z) \\, dy$.\n \\end{problem}\n \\begin{solution}\n First, some notation for $\\gamma$. Define $\\gamma:[a,b]\\to \\{x + iy:x, y \\in \\R\\}$. Then\n \\begin{align*}\n \\int_\\gamma f(z) \\, dz &:= \\int_{a}^b \\gamma'(t)f(\\gamma(t)) \\, dt\\quad \\color{purple}\\text{ complex derivative }\\\\\n \\int_\\gamma f(z) \\, dx &:= \\int_{a}^b \\gamma_x(t)f(\\gamma(t)) \\, dt\\quad\\color{purple}\\text{ first coordinate derivative }\\\\\n \\int_\\gamma f(z) \\, dy &:= i\\int_{a}^b \\gamma_y(t)f(\\gamma(t)) \\,dt\\quad \\color{purple}\\text{ second coordinate derivative }\n \\end{align*}\n \\end{solution}\n \\begin{problem}\n Define $\\int_\\gamma f(z) \\, d\\conj{z}$.\n \\end{problem}\n \\begin{solution}\n \\begin{align*}\n \\int_\\gamma f(z) \\, \\conj{dz} &:= \\conj{\\int_{a}^b \\gamma'(t)\\conj{f(\\gamma(t))} \\, dt}\\\\\n \\end{align*}\n\n \\end{solution}\n \\begin{problem}\n Define $\\int_\\gamma f(z) \\, \\abs{dz}$.\n \\end{problem}\n \\begin{solution}\n \\begin{align*}\n \\int_\\gamma f(z) \\, \\abs{dz} &:= \\int_{a}^b \\abs{\\gamma'(t)}f(\\gamma(t)) \\,dt \\\\\n \\end{align*}\n \\end{solution}\n\n \\begin{problem}\n What does it mean to say that a $1$-form is \\textbf{exact}?\n \\end{problem}\n \\begin{solution}\n $f(z)\\,dz$ is \\textbf{exact} if there it has a \\textbf{primitive} - a complex differentiable function $F(z)$ such that $\\frac{d}{dz} F(z) = f(z)$.\n \\end{solution}\n \\begin{problem}\n What are the \\textbf{poles} and \\textbf{zeros} of a rational\n function $p(z)/q(z)$?\n \\end{problem}\n \\begin{solution}\n Assuming that $p(z)$ and $q(z)$ share no zeros as part of the definition of a rational function:\n The \\textbf{poles} of $p(z)/q(z)$ are the points at which $q(z)=0$. The $\\textbf{zeros}$ are the points at which $p(z)=0$.\n \\end{solution}\n\n \\section{Numericals}\n\n \\begin{problem}\n Consider a piecewise smooth curve $\\gamma$ tracing the boundary of the square\n $$S = \\{ z = x+iy\\in \\C : \\abs{x} \\mbox{ and } \\abs{y} \\leq 1 \\}.$$\n Compute $\\displaystyle\\int_\\gamma \\frac{1}{z} \\, dz$ by hand.\n \\end{problem}\n \\begin{solution}\n Going counter-clockwise, we recognize $\\gamma = \\gamma_1 + \\gamma_2 + \\gamma_3 + \\gamma_4$ with $\\gamma_i$ defined on the interval $[-1, 1]$ satisfying\n \\begin{align*}\n \\gamma_1 = 1 + it &\\qquad \\gamma_3 = -1 - it \\\\\n \\gamma_2 = -t + i &\\qquad \\gamma_4 = t - i\n \\end{align*}\n Now we can compute the integral:\n \\begin{align*}\n \\int_{\\gamma} \\frac{1}{z} \\, dz \n &= \\int_{-1}^1 \\frac{\\gamma_1'}{1 + it} \\, dt +\n \\int_{-1}^1 \\frac{\\gamma_2'}{-t + i} \\, dt +\n \\int_{-1}^1 \\frac{\\gamma_3'}{-1 - it} \\, dt +\n \\int_{-1}^1 \\frac{\\gamma_4'}{t - i} \\, dt\\\\\n &= \\int_{-1}^1 \\frac{i}{1 + it} \\, dt +\n \\int_{-1}^1 \\frac{-1}{-t + i} \\, dt +\n \\int_{-1}^1 \\frac{-i}{-1 - it} \\, dt +\n \\int_{-1}^1 \\frac{1}{t - i} \\, dt\\\\\n &= \\int_{-1}^1 \\frac{4}{t - i} \\, dt\\\\\n &= 4\\Log(t - i)|_{t=-1}^1\\color{purple}\\text{ Log is analytic in an open set containing [-1, 1]} \\\\\n &= 4(\\Log(1 - i) - \\Log(-1 - i))\\\\\n &= 4((\\sqrt{2} + i7\\pi/4) - (\\sqrt{2} + i5\\pi/4))\\\\\n &= 4(i\\pi/2) = 2\\pi i\n \\end{align*}\n\n \\end{solution}\n \\begin{problem}\\label{integral-powers-of-z}Consider the curve $\\gamma : [0,2\\pi] \\to \\C$ given by $\\gamma(\\theta) = e^{i\\theta}$. For an integer $n \\in \\Z$, compute $\\displaystyle\\int_\\gamma z^n \\, dz$ and $\\displaystyle\\int_\\gamma \\conj{z}^n \\, dz$.\n \\end{problem}\n \\begin{solution}\n \\begin{align*}\n \\int_\\gamma z^ndz &= \\int_{0}^{2\\pi} \\gamma'(\\theta) e^{i\\theta n} d\\theta\\\\\n &= i\\int_{0}^{2\\pi}e^{i(1+n)\\theta} d\\theta\\\\\n &= \\begin{cases}i\\int_{0}^{2\\pi} e^{i(1+n)\\theta} d\\theta & n \\neq -1\\\\\n i\\int_{0}^{2\\pi} 1 d\\theta & n = -1 \\end{cases}\\\\\n &= \\begin{cases} 0 & n \\neq -1\\\\\n 2\\pi i & n = -1 \\end{cases}\n \\end{align*}\n For the second integral,\n \\begin{align*}\n \\int_\\gamma \\conj{z}^ndz &= \\int_{0}^{2\\pi} \\gamma'(\\theta) e^{-i\\theta n} d\\theta\\\\\n &= i\\int_{0}^{2\\pi}e^{i(1-n)\\theta} d\\theta\\\\\n &= \\begin{cases}i\\int_{0}^{2\\pi} e^{i(1-n)\\theta} d\\theta & n \\neq 1\\\\\n i\\int_{0}^{2\\pi} 1 d\\theta & n = 1 \\end{cases}\\\\\n &= \\begin{cases} 0 & n \\neq 1\\\\\n 2\\pi i & n = 1 \\end{cases}\n \\end{align*}\n \\end{solution}\n\n \\begin{problem}\\label{one-over-z-around-circle}Let $\\gamma:[a, b]\\mapsto \\C$ be a (positively oriented) parametrization of a circle\n in the plane, and suppose the image of $\\gamma$ does not include the\n origin. Compute $\\displaystyle\\int_\\gamma \\frac{1}{z} \\, dz$.\n \\end{problem}\n \\begin{solution}\n Since $\\gamma$ is a positively oriented parameterization of a circle, we can find a differentiable map $f:[a,b]\\mapsto [0, 2\\pi]$ such that $r(e^{if(t)} + z)= \\gamma(t)$, where $r\\in\\R$ is the radius of and $rz$ is the center of the circle parameterized by $\\gamma$.\n \\begin{align*}\n \\int_\\gamma \\frac{1}{z} \\, dz &= \\int_a^{b} \\frac{\\frac{d}{dt}\\gamma(t)}{\\gamma(t)}dt\\\\\n &= \\int_a^{b} \\frac{\\frac{d}{dt}(re^{if(t)}+ rz)}{re^{if(t)} + rz}dt\\\\\n &= \\int_a^{b} \\frac{if'(t)e^{if(t)}}{e^{if(t)} + z}dt\\\\\n &= \\int_0^{2\\pi} \\frac{ie^{iu}}{e^{iu} + z}du \\quad \\color{purple} u=f(t), du = f'(t) dt\\\\\n &= \\log(e^{ui} + z)\\big|_{u=0}^{2\\pi}\n \\end{align*}\n Though we may be a bit suspcious about which log we are using. However, since the interior of $\\gamma$ does not contain the origin, we can choose a branch of the log such that the angle which has a discontinuity is disjoint from the set of $\\theta$ that satisfy $re^{i\\theta}= e^{ui} +z$ for some $r, u$. Then the value of the above integral is 0 since $e^0 = e^{2\\pi i}$.\n\n On the other hand, if the image does include the origin, we can imagine that the argument of the parameter passed to log will continuously increase by $2\\pi$, so if we switch branches of the log at the place that the derivative is continuous, the difference between the start and end point will be exactly $2\\pi i$.\n\n \\end{solution}\n \\begin{problem}\\label{lacunary-series}What is radius $R$ of convergence of\n $\\displaystyle\\sum_{n=1} x^{(n!)}$? (This is a \\textbf{lacunary\n series} with large gaps between nonzero terms.)\n \\end{problem}\n \\begin{solution}\n We use the following fact: the radius of convergence of a power series $R$ satisfies\n \\[\\frac{1}{R} = \\limsup_{k\\to \\infty} \\sqrt[k]{|x_k|}\\]\n So we need to compute \n \\[\\limsup_{n\\to \\infty} \\sqrt[n!]{1} = 1\\]\n Therefore, the radius of convergence is 1.\n \\end{solution}\n\n \\section{Exploration}\n\n \\begin{problem}For the series in \\ref{lacunary-series}, find a dense\n subset of the circle $\\{ z \\in \\C : \\abs{z} = 1 \\}$ where the series\n diverges.\n \\end{problem}\n \\begin{solution}\n Let $z=e^{2i\\pi \\theta}$ with $\\theta$ some rational number $n/m$, this is a dense subset of the circle. Note that for $x>=m$, \n \\[z^x = (e^{2i\\pi n/m})^{x!} = z^x = e^{(2i\\pi n)*(x!/m)} = 1\\],\n so we are taking a sum of a 1 infinitely many times, which does not seem like a good sign for convergence.\n \\end{solution} \n\n \\begin{problem}\n For which $z \\in \\C$ with $|z|=1$ does the series\n $\\displaystyle\\sum_{n=0}^{\\infty} \\frac{z^n}{n}$ converge? Diverge?\n \\end{problem}\n \\begin{solution}\n If $z=1$ then it's the harmonic series so it diverges. Otherwise it goes in a big spiral that obviously converges. But how to prove... Let's check by letting $z=e^{i\\theta}, \\theta\\neq 0$ and using the integral test (checking if the integral from some lower bound to infinity is finite). It suffices to show that the real and imaginary part converge. For the imaginary part:\n \\begin{align*}\n \\int_{\\pi/\\theta}^\\infty \\frac{\\sin(x\\theta)}{x} dx &=\n \\sum_{k=1}^\\infty (\\int_{(2k + 1)\\pi /\\theta}^{(2k+2)\\pi/\\theta} \\frac{\\sin(x\\theta)}{x} dx +\n \\int_{2k\\pi/\\theta}^{(2k + 1)\\pi/\\theta} \\frac{\\sin(x\\theta)}{x} dx)\n \\end{align*}\n Setting $u=x\\theta, du = \\theta dx$:\n \\begin{align*}\n \\int_{\\pi/\\theta}^\\infty \\frac{\\sin(x\\theta)}{x} dx &= \\sum_{k=1}^\\infty (\\int_{(2k + 1)\\pi}^{(2k+2)\\pi} \\frac{\\sin(u)}{u} du +\n \\int_{2k\\pi}^{(2k + 1)\\pi} \\frac{\\sin(u)}{u} du)\\\\\n &= \\sum_{k=1}^\\infty \\int_{(2k + 1)\\pi}^{(2k+2)\\pi} \\sin(u)(\\frac{1}{u} - \\frac{1}{u-\\pi}) du \\color{purple} \\text{ note it's negative}\\\\\n &\\geq -\\sum_{k=1}^\\infty \\int_{(2k + 1)\\pi}^{(2k+2)\\pi} (\\frac{1}{(2k+2)\\pi} - \\frac{1}{2k\\pi}) du \\\\\n &= \\sum_{k=1}^\\infty \\frac{1}{(2k+2)} - \\frac{1}{2k} du = -\\frac{1}{2}\n \\end{align*}\n This integral is finite so the imaginary part of the series converges. An analogous argument with cosine shows that the real part of the series converges, so the series must converge in the complex plane.\n \\end{solution}\n\n\n \\begin{problem}\n Suppose $f : \\C \\to \\C$ is a rational function which\n sends the unit circle to the real line, i.e., for $z \\in \\C$ with\n $|z| = 1$ we have $f(z) \\in \\R$. Inspired\n by \\ref{schwarz-reflection-principle}, compute\n $\\overline{f(1/\\conj{z})}$ and then discuss the relationship between\n the poles and zeros of $f$.\n \\end{problem}\n \\begin{solution}\n If $|z|=1$, then $f(z)=\\conj{f(z)}$ and $1/\\conj{z} = z$. Thus\n $\\conj{f(1/\\conj{z})} = f(z)$ on the unit circle. Since these are rational functions that agree on an infinite set of points, they must be equal everywhere.\n\n Thus for any zero or pole $z_0$ of $f$, there is another zero or pole respectively at $\\frac{1}{\\conj{z_0}}= \\frac{z_0}{|z_0|}.$\n \\end{solution}\n \\begin{problem}\n In lecture, we briefly saw an example (the\n \\textbf{topologist's sine curve}) of a subset of $\\mathbb{C}$ which\n is connected but not path-connected. For open subsets of\n $\\mathbb{C}$, what is the relationship between connectedness and\n path-connectedness?\n \\end{problem}\n \\begin{solution}\n If a open subset of $\\C$ is path-connected, it's also connected: For contradiction, suppose there was a disconnection formed from disjoint open sets $U$, $V$. Choosing points $P_U \\in U, P_V \\in V$, there is a path $f:[0, 1]\\mapsto \\C$ from $P_U$ to $P_V$. Let $s = \\sup x\\in [0,1]: f(x) \\in U$. s is not 1 since $f$ is continuous and there is a ball around $f(1)$ contained in $V$. Hence by the defintion of sup, we can find an infinite sequence $u_i \\in [0, 1]$ with $f(u_i) \\in U$ that converges to $s$ and any sequence of points $v_i \\in [0, 1]$ with $v_i < s$ converging to $s$ has $f(v_i) \\in V$, so $f(s)$ is a boundary point of both $U$ and $V$ contained in either $U$ or $V$, a contradiction.\n\n Next we show that if a open subset of $S\\in \\C$ is connected, it's also path-connected. For each point $p \\in S$, define $P$ to be the set of points that we can reach from $p$. $P$ is open since if we can reach any point by a path, we can also reach any point from $p$, then we can also reach any point in the (convex) open ball surrounding that point.\n\n Now consider the complement $S/P$. This set is also open - supposing that we can't reach a point $s\\in S/P$, then we can't reach any point in the open ball surrounding that point. So $P=(S/P)\\cup S$ is a union of two disjoint open sets. As $P$ is connected, this implies that one of the $S/P$ and $S$ is empty, and as $s\\in S$, this implies $S/P = \\empty$ and $S=P$. Recalling the definition of $P$, we can reach any point in $S$ by a path starting at $p$, so we can reach any point from any other point by going first to $p$ and then to the second point.\n \\end{solution}\n \\begin{problem}\\label{argument-principle-numerical}Consider $\\gamma : [0,2\\pi] \\to \\C$ given by $\\gamma(\\theta) = e^{i\\theta}$. For an integer $n \\in \\Z$ and $f(z) = z^n$, compute\n \\[\n \\frac{1}{2\\pi i} \\displaystyle\\int_\\gamma \\frac{f'(z)}{f(z)} \\, dz\n \\]\n in two different ways. First, evaluate $f'(z)/f(z)$ and invoke\n \\ref{integral-powers-of-z}. Second, describe a curve $\\gamma_n$ and\n interpret $\\int_\\gamma \\frac{f'(z)}{f(z)} \\, dz$ as\n $\\int_{\\gamma_n} dz/z$ for that different curve $\\gamma_n$. (This\n is our first glimpse of the \\textbf{argument principle}.)\n \\end{problem}\n \\begin{solution}\n \\[\\frac{f'(z)}{f(z)} = \\frac{nz^{n-1}}{z^n} = \\frac{n}{z}\\]\n Then using \\ref{integral-powers-of-z}, the integral's value is $n$.\n\n Next, consider going along the curve $\\gamma_n = f\\circ \\gamma$, it's the path that $f(z)$ traces out as we go along the curve. Algebraically, let $u = f(z)$ so $du = f'(z) dz.$\n Then\n \\[\n \\int_{\\gamma} \\frac{f'(z)}{f(z)}dz = \\int_{\\gamma_n} \\frac{1}{u} du = \\int \\frac{\\gamma_n'(t)}{\\gamma_n(t)}dt = \\int_0^{2\\pi} \\frac{ine^{int}}{e^{int}}dt = 2in\\pi\n \\]\n Dividing out by $2in\\pi$, the value of the integral we were given is $n$. \n \\end{solution}\n\n \\begin{problem}\\label{one-over-z-w-around-circle}\n Consider the curve $\\gamma : [0,2\\pi] \\to \\C$ given by $\\gamma(\\theta) = e^{i\\theta}$. For $w \\in \\C$ with $|w| \\neq 1$, compute\n \\[\n \\frac{1}{2\\pi i} \\displaystyle\\int_\\gamma \\frac{1}{z-w} \\, dz\n \\]\n perhaps by invoking \\ref{one-over-z-around-circle}.\n \\end{problem}\n \\begin{solution}\n Let $v = z - w,$ so $dv = dz$. Defining the function $\\gamma_2(t) := \\gamma(t) - w$, we see that \n \\[\\int_\\gamma \\frac{1}{z-w}dz = \\int_{\\gamma_2} \\frac{1}{v}dv\\]\n This is exactly the integral of a circle, and \\ref{one-over-z-around-circle} shows that if the circle doesn't include the origin, then the integral is 0 and otherwise, (the specific proof I did shows), it is $\\frac{1}{2\\pi i}$. Thus\n \\[\n \\frac{1}{2\\pi i} \\displaystyle\\int_\\gamma \\frac{1}{z-w} \\, dz = \\begin{cases}1 & |w| < 1\\\\ 0 & |w| > 1\\end{cases}.\n \\]\n \\end{solution}\n\n \\begin{problem}\n Yet again consider the curve $\\gamma : [0,2\\pi] \\to \\C$ given by $\\gamma(\\theta) = e^{i\\theta}$. For distinct $w_1, w_2 \\in \\C$ with $\\abs{w_1} \\neq 1$ and $\\abs{w_2} \\neq 1$, let $f(z) = (z - w_1) (z - w_2)$ and compute\n \\[\n \\frac{1}{2\\pi i} \\displaystyle\\int_\\gamma \\frac{f'(z)}{f(z)} \\, dz.\n \\]\n \\end{problem}\n \\begin{solution}\n Let $\\gamma' = f\\circ \\gamma = (e^{i\\theta} - w_1)(e^{i\\theta} - w_2)$, and make the subsitution $w=f(z)$ so $dw = f'(z)dz$.\n Then\n \\begin{align*}\n \\frac{1}{2\\pi i} \\int_\\gamma \\frac{f'(z)}{f(z)} \\, dz\n &= \\frac{1}{2\\pi i} \\int_{\\gamma_1}\\frac{1}{w} \\, dw\\\\\n &= \\frac{1}{2\\pi i} \\int_0^{2\\pi} \\frac{\\gamma_1'(t)}{\\gamma_1(t)} \\, dt\\\\\n &= \\frac{1}{2\\pi i} \\int_0^{2\\pi} \\frac{ie^{i\\theta} - w_1}{e^{i\\theta}} + \\frac{ie^{i\\theta}}{e^{i\\theta} - w_1} \\, dt\\\\\n &= \\frac{1}{2\\pi i} (\\int_0^{2\\pi} \\frac{ie^{i\\theta}}{e^{i\\theta} - w_1} dt + \\int_0^{2\\pi} \\frac{ie^{i\\theta}}{e^{i\\theta} - w_1}dt )\\\\\n &= \\frac{1}{2\\pi i} (\\int_{\\gamma} \\frac{1}{z - w_1} dt + \\int_{\\gamma} \\frac{1}{z - w_1}dt )\n \\end{align*}\n Invoking \\ref{one-over-z-w-around-circle}, this is\n \\[\\left(\\begin{cases}1 & |w_1| < 1\\\\ 0 & |w_1| > 1\\end{cases} \\right) +\n \\left(\\begin{cases}1 & |w_2| < 1\\\\ 0 & |w_2| > 1\\end{cases}\\right)\\]\n \\end{solution}\n\n \\section{Prove or Disprove and Salvage if Possible}\n\n \\begin{problem}\n If $e^z = e^w$, then $z = w$.\n \\end{problem}\n \\begin{solution}\n False, $e^0 = e^{2\\pi i}= 1$. However, since $e^{x + iy} = e^x(\\cos(y) + i\\sin(y))$, $\\exists \\, k \\, s.t. \\, (e^z = e^w \\implies z = w + 2\\pi ik)$\n \\end{solution}\n \\begin{problem}\n If $f \\, dz$ is exact, then $\\displaystyle\\int_\\gamma f \\, dz = 0$.\n \\end{problem}\n \\begin{solution}\n Additionally we require that $\\gamma$ is a closed curve starting and ending at the same point $p$. In this case, letting $F$ be the antiderivative of $f$,\n \\begin{align*}\n \\int_\\gamma f \\, dz \n &= \\int \\gamma'(t)f(\\gamma(t)) \\, dt \\\\\n &= \\int_\\gamma \\frac{d}{dt}(F\\circ\\gamma)(t) \\, dt\\\\\n &= (F\\circ\\gamma)(\\gamma(p) - \\gamma(p)) = 0\n \\end{align*}\n \\end{solution}\n\n \\begin{problem}% orientation issues\n Suppose $\\gamma : [0,1] \\to \\C$ is a smooth curve, and $p : [0,1] \\to [0,1]$ is a smooth bijection.\n Then\n \\[\n \\int_\\gamma f(z) \\, dz = \\int_{\\gamma \\circ p} f(z) \\, dz.\n \\]\n \\end{problem}\n \\begin{solution}\n If $p(0)=1$, then we will switch the direction of the curve which will make the answer negative. \n\n Otherwise, we can prove it as follows.\n \\begin{align*}\n \\int_{\\gamma \\circ p} f(z) \\, dz &= \\int_0^1 (\\frac{d}{dt}(\\gamma\\circ p) \\times (f \\circ \\gamma \\circ p))t \\,dt\\\\\n &= \\int_{0}^1 \\gamma'(p(t))p'(t) (f \\circ \\gamma \\circ p)t\\, dt \\\\\n &= \\int_{0}^1 \\gamma'(u) (f \\circ \\gamma)u\\, du \\quad \\color{purple} u=p(t), du = p'(t)dt\\\\\n &= \\int_{\\gamma} f(z) \\, dz\n \\end{align*}\n \\end{solution}\n\n\n\n \\begin{problem}\\label{identity-theorem}If $f \\in \\C[z]$ is a polynomial with infinitely many zeros, then\n $f \\equiv 0$.\n \\end{problem}\n \\begin{solution}\n By the fundamental theorem of arithmetic, $f$ splits completely in $\\C$ as $\\prod_{i=0}^n (z - r_i)$ where $r_i$ is a root. At any number that's not a root, this product is clearly nonzero unless it is the empty product, so we must have that $f\\equiv 0$.\n \\end{solution} \n \\end{document}\n\n", "meta": {"hexsha": "50c1c4aade17a405ef6761d2665c4a85baadb669", "size": 31676, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problem-solutions/sol3.tex", "max_stars_repo_name": "Alex7Li/math5522h", "max_stars_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "problem-solutions/sol3.tex", "max_issues_repo_name": "Alex7Li/math5522h", "max_issues_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problem-solutions/sol3.tex", "max_forks_repo_name": "Alex7Li/math5522h", "max_forks_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 95.6978851964, "max_line_length": 759, "alphanum_fraction": 0.3548112135, "num_tokens": 6892, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933315126791, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6416997514750897}} {"text": "\\documentclass{notes}\n\n \\title{Collatz Turing Machine}\n \\author{ian.mcloughlin@gmit.ie}\n \\date{\\today}\n\n\\begin{document}\n\n \n\n \n\\section*{Language}\n \\begin{align*}\n A &= \\{ 0, 1 \\} \\\\\n T &= A \\ \\cup \\ \\{ , \\} \\ \\cup \\ \\{ \\blm, X \\} \\\\\n L &= \\{1, 10, 11, 100, 101, 110, 111, 1000, 1001, \\ldots \\} \\\\\n f(s) &= 1s,s \\\\\n \\end{align*}\n\n\\section*{Duplicate and append 1}\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Append a comma. \\\\\n \\( q_1 \\) & Move left to start. \\\\\n \\( q_2 \\) & Check if 0 or 1, mark X. \\\\\n \\( q_3 \\) & Move right to end, append 0. \\\\\n \\( q_4 \\) & Move right to end, append 1. \\\\\n \\( q_5 \\) & Move left to X, overwrite 0, back to \\(q_2\\). \\\\\n \\( q_6 \\) & Move left to X, overwrite 1, back to \\(q_2\\). \\\\\n \\( q_7 \\) & Move to start, prepend 1. \\\\\n \\( q_8 \\) & Back to start, end. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % Right to Write a comma at end on input.\n \\(q_0\\) & \\bl & , & L & \\(q_1\\) \\\\\n \\(q_0\\) & 0 & 0 & R & \\(q_0\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_0\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Left to start.\n \\(q_1\\) & \\bl & \\bl & R & \\(q_2\\) \\\\\n \\(q_1\\) & 0 & 0 & L & \\(q_1\\) \\\\\n \\(q_1\\) & 1 & 1 & L & \\(q_1\\) \\\\\n \\(q_1\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_1\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Overwrite first symbol with X.\n \\(q_2\\) & \\bl & \\bl & R & \\(q_f\\) \\\\\n \\(q_2\\) & 0 & X & R & \\(q_3\\) \\\\\n \\(q_2\\) & 1 & X & R & \\(q_4\\) \\\\\n \\(q_2\\) & , & , & R & \\(q_7\\) \\\\\n \\(q_2\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Left to write a 0 at the end.\n \\(q_3\\) & \\bl & 0 & L & \\(q_5\\) \\\\\n \\(q_3\\) & 0 & 0 & R & \\(q_3\\) \\\\\n \\(q_3\\) & 1 & 1 & R & \\(q_3\\) \\\\\n \\(q_3\\) & , & , & R & \\(q_3\\) \\\\\n \\(q_3\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Left to write a 1 at the end.\n \\(q_4\\) & \\bl & 1 & L & \\(q_6\\) \\\\\n \\(q_4\\) & 0 & 0 & R & \\(q_4\\) \\\\\n \\(q_4\\) & 1 & 1 & R & \\(q_4\\) \\\\\n \\(q_4\\) & , & , & R & \\(q_4\\) \\\\\n \\(q_4\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Left to the X, overwrite with 0\n \\(q_5\\) & \\bl & \\bl & L & \\(q_f\\) \\\\\n \\(q_5\\) & 0 & 0 & L & \\(q_5\\) \\\\\n \\(q_5\\) & 1 & 1 & L & \\(q_5\\) \\\\\n \\(q_5\\) & , & , & L & \\(q_5\\) \\\\\n \\(q_5\\) & X & 0 & R & \\(q_2\\) \\\\\n \\midrule\n % Left to the X, overwrite with 1\n \\(q_6\\) & \\bl & \\bl & L & \\(q_f\\) \\\\\n \\(q_6\\) & 0 & 0 & L & \\(q_6\\) \\\\\n \\(q_6\\) & 1 & 1 & L & \\(q_6\\) \\\\\n \\(q_6\\) & , & , & L & \\(q_6\\) \\\\\n \\(q_6\\) & X & 1 & R & \\(q_2\\) \\\\\n \\midrule\n % Left to start, prepend 1.\n \\(q_7\\) & \\bl & 1 & L & \\(q_8\\) \\\\\n \\(q_7\\) & 0 & 0 & L & \\(q_7\\) \\\\\n \\(q_7\\) & 1 & 1 & L & \\(q_7\\) \\\\\n \\(q_7\\) & , & , & L & \\(q_7\\) \\\\\n \\(q_7\\) & X & X & L & \\(q_7\\) \\\\\n \\midrule\n % Back to start, accept.\n \\(q_8\\) & \\bl & \\bl & R & \\(q_a\\) \\\\\n \\(q_8\\) & 0 & 0 & R & \\(q_f\\) \\\\\n \\(q_8\\) & 1 & 1 & R & \\(q_f\\) \\\\\n \\(q_8\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_8\\) & X & X & R & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n \n\\section*{Add}\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Right to comma. \\\\\n \\( q_1 \\) & Decrease by 1, go to final states if no 1. \\\\\n \\( q_2 \\) & Left to start. \\\\\n \\( q_3 \\) & Increase by 1, use X as comma if overflow. \\\\\n \\( q_4 \\) & Delete to comma/X, replace X with 1. \\\\\n \\( q_5 \\) & Left to start. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % Right to comma/X.\n \\(q_0\\) & \\bl & \\bl & L & \\(q_f\\) \\\\\n \\(q_0\\) & 0 & 0 & R & \\(q_0\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_0\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_1\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_1\\) \\\\\n \\midrule\n % Decrease by 1, go to final states if no 1.\n \\(q_1\\) & \\bl & \\bl & L & \\(q_4\\) \\\\\n \\(q_1\\) & 0 & 1 & R & \\(q_1\\) \\\\\n \\(q_1\\) & 1 & 0 & L & \\(q_2\\) \\\\\n \\(q_1\\) & , & , & L & \\(q_f\\) \\\\\n \\(q_1\\) & X & X & L & \\(q_f\\) \\\\\n % Left to start.\n \\(q_2\\) & \\bl & \\bl & R & \\(q_3\\) \\\\\n \\(q_2\\) & 0 & 0 & L & \\(q_2\\) \\\\\n \\(q_2\\) & 1 & 1 & L & \\(q_2\\) \\\\\n \\(q_2\\) & , & , & L & \\(q_2\\) \\\\\n \\(q_2\\) & X & X & L & \\(q_2\\) \\\\\n \\midrule\n % Increase by 1 - use X in place of comma if overflow.\n \\(q_3\\) & \\bl & \\bl & R & \\(q_f\\) \\\\\n \\(q_3\\) & 0 & 1 & R & \\(q_0\\) \\\\\n \\(q_3\\) & 1 & 0 & R & \\(q_2\\) \\\\\n \\(q_3\\) & , & X & L & \\(q_0\\) \\\\\n \\(q_3\\) & X & X & L & \\(q_f\\) \\\\\n \\midrule\n % Delete to and including comma/X - replace X with 1.\n \\(q_4\\) & \\bl & \\bl & L & \\(q_f\\) \\\\\n \\(q_4\\) & 0 & \\bl & L & \\(q_4\\) \\\\\n \\(q_4\\) & 1 & 1 & L & \\(q_f\\) \\\\\n \\(q_4\\) & , & \\bl & L & \\(q_5\\) \\\\\n \\(q_4\\) & X & 1 & L & \\(q_5\\) \\\\\n \\midrule\n % Left to start.\n \\(q_5\\) & \\bl & \\bl & R & \\(q_a\\) \\\\\n \\(q_5\\) & 0 & 0 & L & \\(q_5\\) \\\\\n \\(q_5\\) & 1 & 1 & L & \\(q_5\\) \\\\\n \\(q_5\\) & , & , & L & \\(q_f\\) \\\\\n \\(q_5\\) & X & X & L & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n\n \\section*{Divide by two}\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Delete 0, move right. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % Delete 0, move right.\n \\(q_0\\) & \\bl & \\bl & L & \\(q_f\\) \\\\\n \\(q_0\\) & 0 & \\bl & R & \\(q_0\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_f\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n \n \\section*{Remove leading zeros}\n Fails on string meaning zero.\n\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Move right to end. \\\\\n \\( q_1 \\) & Delete zeros. \\\\\n \\( q_2 \\) & Left to end. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % Right to end.\n \\(q_0\\) & \\bl & \\bl & L & \\(q_1\\) \\\\\n \\(q_0\\) & 0 & 0 & R & \\(q_0\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_0\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Delete zeros.\n \\(q_1\\) & \\bl & \\bl & R & \\(q_f\\) \\\\\n \\(q_1\\) & 0 & \\bl & L & \\(q_0\\) \\\\\n \\(q_1\\) & 1 & 1 & L & \\(q_2\\) \\\\\n \\(q_1\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_1\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Left to start.\n \\(q_2\\) & \\bl & \\bl & R & \\(q_a\\) \\\\\n \\(q_2\\) & 0 & 0 & R & \\(q_2\\) \\\\\n \\(q_2\\) & 1 & 1 & R & \\(q_2\\) \\\\\n \\(q_2\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_2\\) & X & X & R & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n\n \\section*{Check if 1}\n\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Check if first bit 1. \\\\\n \\( q_1 \\) & Check if second bit blank. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % First bit 1.\n \\(q_0\\) & \\bl & \\bl & R & \\(q_f\\) \\\\\n \\(q_0\\) & 0 & 0 & R & \\(q_f\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_1\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_f\\) \\\\\n \\midrule\n % Check if blank.\n \\(q_1\\) & \\bl & \\bl & L & \\(q_a\\) \\\\\n \\(q_1\\) & 0 & 0 & L & \\(q_f\\) \\\\\n \\(q_1\\) & 1 & 1 & L & \\(q_f\\) \\\\\n \\(q_1\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_1\\) & X & X & R & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n\n \\section*{Check if even}\n\n \\begin{tabular}{rl}\n \\textbf{State} & \\textbf{Description} \\\\\n \\midrule\n \\( q_0 \\) & Check if first bit 0. \\\\\n \\end{tabular}\n\n \\begin{center}\n \\begin{tabular}{x{1cm}x{1cm}x{1cm}x{1cm}x{1cm}}\n \\toprule\n State & Input & Write & Move & Next \\\\\n \\midrule\n % Check 0.\n \\(q_0\\) & \\bl & \\bl & R & \\(q_f\\) \\\\\n \\(q_0\\) & 0 & 0 & R & \\(q_a\\) \\\\\n \\(q_0\\) & 1 & 1 & R & \\(q_f\\) \\\\\n \\(q_0\\) & , & , & R & \\(q_f\\) \\\\\n \\(q_0\\) & X & X & R & \\(q_f\\) \\\\\n \\bottomrule\n \\end{tabular}\n \\end{center}\n\n\n %\\bibliography{bibliography}\n\\end{document}", 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YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7826624789529376, "lm_q1q2_score": 0.6416997507630817}} {"text": "\\section{Termination}\\label{sec:termination}\n\nProgram divergence is, more often than not,\na bug rather than a feature. \n%\nTo account for the common cases, \nby default, \\toolname proves termination of each recursive function.\n%\nFortunately, refinements make this onerous task quite straightforward. \nWe need simply associate a \\emph{well-founded termination metric} % $\\mu$\non the function's parameters, and then use refinement typing to check \nthat the metric strictly decreases at each recursive call. In practice,\ndue to a careful choice of defaults, this amounts to about a line \nof termination-related hints per hundred lines of source. \n%\nIn Chapter~\\ref{chapter:refinedhaskell}\nwe prove soundness of our refinement type based termination checker \nand also we explain how soundness of \\toolname crucially depends on \nthe termination checker. \n%\nHere, we provide an overview on how one can use \\toolname \nto prove termination.\n\n\\mypara{Simple Metrics}\nAs a starting example, consider the @fac@ function\n%\n\\begin{code}\n fac :: n:Nat -> Nat / [n]\n fac 0 = 1 \n fac n = n * fac (n-1)\n\\end{code}\n%\nThe termination metric is simply the parameter @n@; \nas @n@ is non-negative and decreases at the recursive \ncall, \\toolname verifies that @fac@ will terminate.\n%\nWe specify the termination metric in the type signature \nwith the @/[n]@.\n\nTermination checking is performed at the same \ntime as regular type checking, as it can be \nreduced to refinement type checking with a \nspecial terminating fixpoint combinator~\\ref{chapter:refinedhaskell}.\nThus, if \\toolname fails to prove that a given \ntermination metric is well-formed and decreasing, \nit will report a @Termination Check@ @Error@. \n%% \\RJ{Seems untrue -- I just get a plain old liquid type error?\n%% NV: IF you provide termination metrics, you DO get termination check error}.\nAt this point, the user can either debug \nthe specification, or mark the function \nas non-terminating.\n\n\n%%\\mypara{Refinements Enable Termination} \n%%Consider Euclid's GCD:\n%%%\n%%\\begin{code}\n%% gcd :: a:Nat -> {v:Nat | v < a} -> Nat \n%% gcd a 0 = a\n%% gcd a b = gcd b (a `mod` b)\n%%\\end{code}\n%%%\n%%Here, the termination metric is the first parameter @a@.\n%%To prove that @a@ is decreasing requires\n%%the fact that the second parameter is smaller than the first \n%%and that @mod@ returns results smaller than its second \n%%parameter. Both facts are easily expressed as refinements, \n%%but elude non-extensible checkers~\\cite{Giesl11}.\n%%\n%%\\mypara{Explicit Termination Metrics}\n%%The termination metric can be some parameter \\emph{other} than the first \n%%argument.\n%%For example, consider: % As an example, consider the tail-recursive factorial:\n%%%\n%%\\begin{code}\n%% tfac :: Nat -> n:Nat -> Nat / [n] \n%% tfac x 0 = if n == 0 then x\n%% else tfac (n*x) (n-1)\n%%\\end{code}\n%%%\n%%%\n%%It can be checked that @n@, \\ie, the second argument is decreasing at each recursive call.\n%%\n\\mypara{Termination Expressions} \nSometimes, no single parameter decreases across recursive calls,\nbut there is some \\emph{expression} that forms the decreasing \nmetric.\n%\nFor example recall @range lo hi@ (from \\S~\\ref{sec:tool:verification}) \nwhich returns the list of @Int@s from @lo@ to @hi@:\n%\n\\begin{code}\n range lo hi \n | lo < hi = lo : range (lo+1) hi\n | otherwise = [] \n\\end{code}\n%\nHere, neither parameter is decreasing (indeed, the first \none is increasing) but @hi-lo@ decreases across each call. \nTo account for such cases, we can specify as the termination\nmetric a (refinement logic) expression over the function\nparameters. Thus, to prove termination, we could type @range@ as:\n\\begin{code}\n lo:Int -> hi:Int -> [(Btwn lo hi)] / [hi-lo]\n\\end{code}\n\n\\mypara{Lexicographic Termination}\nThe Ackermann function\n%\n\\begin{code}\n ack m n \n | m == 0 = n + 1\n | n == 0 = ack (m-1) 1 \n | otherwise = ack (m-1) (ack m (n-1))\n\\end{code}\n%\nis curious as there exists no simple, natural-valued, \ntermination metric that decreases at each recursive call.\n%\nHowever @ack@ terminates because at each call \\emph{either}\n@m@ decreases \\emph{or} @m@ remains the same and @n@ decreases. \n%\nIn other words, the pair @(m,n)@ strictly decreases according to a\n\\emph{lexicographic} ordering. \n%\nThus \\toolname supports termination metrics that are a \n\\emph{sequence of} termination expressions. For example, \nwe can type @ack@ as:\n%\n\\begin{code}\n ack :: m:Nat -> n:Nat -> Nat / [m, n]\n\\end{code}\n%\nAt each recursive call \\toolname uses a lexicographic \nordering to check that the sequence of termination \nexpressions is decreasing (and well-founded in each component).\n\n\\mypara{Mutual Recursion}\n%\nThe lexicographic mechanism lets us check termination of\nmutually recursive functions, \\eg @isEven@ and @isOdd@\n%\n\\begin{code}\n isEven 0 = True\n isEven n = isOdd $ n-1\n \n isOdd n = not $ isEven n \n\\end{code}\n%\nEach call terminates as either @isEven@ calls @isOdd@ with a \ndecreasing parameter, \\emph{or} @isOdd@ calls @isEven@ with \nthe same parameter, expecting the latter to do the decreasing.\n%\nFor termination, we type:\n%\n\\begin{code}\n isEven :: n:Nat -> Bool / [n, 0]\n isOdd :: n:Nat -> Bool / [n, 1]\n\\end{code}\n%\nTo check termination, \\toolname verifies that at each recursive \ncall the metric of the caller is less than the metric of the \ncallee.\n%\nWhen @isEven@ calls @isOdd@, it proves that the caller's \nmetric, namely @[n,0]@ is greater than the callee's @[n-1,1]@.\nWhen \\hbox{@isOdd@} calls @isEven@, it proves that the \ncaller's metric @[n,1]@ is greater than the callee's @[n,0]@,\nthereby proving the mutual recursion always terminates.\n\n\\mypara{Recursion over Data Types}\nThe above strategies generalize easily to functions that recurse\nover (finite) data structures like arrays, lists, and trees.\nIn these cases, we simply use \\emph{measures} to project the \nstructure onto @Nat@, thereby reducing the verification to \nthe previously seen cases. \nFor example, we can prove that @map@ \n%\n\\begin{code}\n map f (x:xs) = f x : map f xs\n map f [] = []\n\\end{code}\n%\nterminates, by typing @map@ as \n%\n\\begin{code}\n (a -> b) -> xs:[a] -> [b] / [len xs]\n\\end{code}\n%\n\\ie, by using the measure @len xs@, from \\S~\\ref{sec:tool:measures}, \nas the metric.\n\n%%% %\n%%% \\begin{code}\n%%% data L [sz] a = N | C a (L a)\n%%% \\end{code}\n%%% %\n%%% we can define a \\emph{measure}\n%%% %\n%%% \\begin{code}\n%%% measure sz :: L a -> Nat\n%%% sz (C x xs) = 1 + (sz xs)\n%%% sz N = 0\n%%% \\end{code}\n%%% %\n%%% We prove that @map@ terminates using the type:\n%%% %\n%%% \\begin{code}\n%%% map :: (a -> b) -> xs:L a -> L b / [sz xs]\n%%% map f (C x xs) = C (f x) (map f xs)\n%%% map f N = N\n%%% \\end{code}\n%%% %\n%%% That is, by simply using @(sz xs)@ as the \n%%% decreasing metric.\n\n\\mypara{Generalized Metrics Over Datatypes}\nIn many functions there is no single argument \nwhose measure provably decreases. Consider\n%\n\\begin{code}\n merge (x:xs) (y:ys)\n | x < y = x : merge xs (y:ys)\n | otherwise = y : merge (x:xs) ys\n\\end{code}\n%\nfrom the homonymous sorting routine. Here, neither\nparameter decreases, but the \\emph{sum} of their \nsizes does. To prove termination, we can type @merge@ as:\n%\n\\begin{code}\n xs:[a] -> ys:[a] -> [a] / [len xs + len ys]\n\\end{code}\n\n%%%% \\begin{figure*}[!t]\n%%%% \t\\begin{code}\n%%%% \ttype OL a = [a]<{\\fld v -> (v >= fld)}>\n%%%% \n%%%% \tqsort :: (Ord a) => xs:[a] -> OL a / [(len xs), 0]\n%%%% \tqsort [] = []\n%%%% \tqsort (x:xs) = qpart x xs [] []\n%%%% \n%%%% \tqpart :: (Ord a) => x:a -> q:[a] -> r: [{v:a|v p: [{v:a|v>=x}] -> OL a \n%%%% \t / [((len q) + (len r) + (len p)), ((len q) + 1)]\n%%%% \tqpart x [] rlt rge = app x (qsort rlt) (x:qsort rge)\n%%%% \tqpart x (y:ys) rlt rge | x > y = qpart x ys (y:rlt) rge\n%%%% \t | otherwise = qpart x ys rlt (y:rge)\n%%%% \n%%%% \tapp k [] ys = ys\n%%%% \tapp k (x:xs) ys = x : (app k xs ys)\n%%%% \t\\end{code}\n%%%% \\caption{Mutual-recursive qsort}\n%%%% \\label{fig:code:qsort}\n%%%% \\end{figure*}\n\n\n\\mypara{Putting it all Together}\nThe above techniques can be combined to prove \ntermination of the mutually recursive quick-sort (from~\\citep{XiTerminationLICS01})% \\RJ{from where?}\n%\n\\begin{code}\n qsort (x:xs) = qpart x xs [] []\n qsort [] = []\n\n qpart x (y:ys) l r \n | x > y = qpart x ys (y:l) r \n | otherwise = qpart x ys l (y:r)\n qpart x [] l r = app x (qsort l) (qsort r) \n\n app k [] z = k : z\n app k (x:xs) z = x : app k xs z\n\\end{code}\n%\n@qsort (x:xs)@ calls @qpart x xs@ to partition @xs@ \ninto two lists @l@ and @r@ that have elements less \nand greater or equal than the pivot @x@, respectively.\n%\nWhen @qpart@ finishes partitioning it mutually recursively\ncalls @qsort@ to sort the two list and appends the results \nwith @app@. \n%\n\\toolname proves sortedness as well~\\cite{vazou13} but let us \nfocus here on termination. To this end, we type the functions\nas:\n%\n\\begin{code}\n qsort :: xs:_ -> _ \n / [len xs, 0]\n \n qpart :: _ -> ys:_ -> l:_ -> r:_ -> _ \n / [len ys + len l + len r, 1 + len ys]\n\\end{code}\n%\nAs before, \\toolname checks that at each recursive call \nthe caller's metric is less than the callee's. \n%\nWhen @qsort@ calls @qpart@ the length of the unsorted \nlist @len (x:xs)@ exceeds the \\hbox{@len xs + len [] + len []@}.\n%\nWhen @qpart@ recursively calls itself the first component\nof the metric is the same, but the length of the unpartitioned \nlist decreases, \\ie @1 + len y:ys@ exceeds \\hbox{@1 + len ys@}.\n%\nFinally, when @qpart@ calls @qsort@ we have \\hbox{@len ys + len l + len r@}\nexceeds both @len l@ and @len r@, thereby ensuring termination.\n\n\n%%% Before we dive into proving termination, note that the \n%%% type alias @OL a@ uses Abstract Refinements~\\citep{vazou13} to describe \n%%% Ordered Lists. \n%%% Thus, when \\toolname decides that the @qsort@ is SAFE, \n%%% it proves both termination and sortedness.\n%%%%Note that classical appending @rlt ++ rge@ of the two sorted lists will lose \n%%%%the crucial for sorting information that every element of @rlt@ is less than each element of @rge@.\n%%%%%\n%%%%Thus we defined a new version of list appending @app@ that uses the pivot element @k@\n%%%%as a ghost-parameter.\n%%%%%\n%%%%Good news is that \\toolname will automatically infer the appropriate type of @app@!\n\n%% Let \\mus{xs} and \\mup{q}{r}{p} be the (well-founded) termination pairs\n%% for @qsort xs@ and @qpart x q r p@ respectively, as annotated in the type signatures.\n%% %\n%%%$\\mu_s(x:xs) = (1 + len xs, 0) > (len xs + 0 + 0, len xs + 1) = \\mu_p(xs, [], [])$\n%%%$\\mu_p(y:ys, rlt, rge) = ((1 + (len ys)) + (len rlt) + (len rge), (1 + len ys) + 1) > \n%%%(len ys + (1 + (len rlt)) + (len rge), ((len ys) + 1)) = \\mu_p(ys, y:rlt, rge) $\n%%%$\\mu_p([], rlt, rge) = (0 + len rlt + len rge, 1) > (len rlt, 0) = \\mu_s(rlt)$ \n\n%% Existing techniques~\\citep{CookPR11} could be used to \n%% come up with termination metrics.\n%% We leave embedding these techniques into \\toolname as a future work, and instead\n%% we use some defaults to automate termination proving \n%% on functions with trivial metrics.\n\n\\mypara{Automation: Default Size Measures}\n%\nThe @qsort@ example illustrates that while \\toolname is \nvery expressive, devising appropriate termination metrics \ncan be tricky.\n%\nFortunately, such patterns are very uncommon, and the vast\nmajority of cases in real world programs are just structural \nrecursion on a datatype.\n%\n\\toolname automates termination proofs for this common case,\nby allowing users to specify a \\emph{default size measure} \nfor each data type, \\eg @len@ for @[a]@.\n%\nNow, if no explicit termination metric is given, by default \n\\toolname assumes that the \\emph{first} argument whose type\nhas an associated size measure decreases.\n%\nThus, in the above, we need not specify metrics for @fac@ \nor @map@ as the size measure is automatically \nused to prove termination. \n%\nThis heuristic suffices to \\emph{automatically}\nprove 67\\% of recursive functions terminating.\n\n\\mypara{Disabling Termination Checking}\nIn \\texttt{Haskell}'s lazy setting not all functions are terminating.\n% \n\\toolname provides two mechanisms the disable termination proving.\n%\nA user can disable checking a single function by marking \nthat function as lazy. For example, specifying @lazy repeat@ \ntells the tool to not prove @repeat@ terminates.\n%\nOptionally, a user can disable termination checking for a whole\nmodule by using the command line argument \\cmdnotermination\nfor the entire file.\n", "meta": {"hexsha": "cba9faec28313cbdd4e365080733928ad90df6a6", "size": 12480, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "text/realworldhaskell/termination.tex", "max_stars_repo_name": "nikivazou/thesis", "max_stars_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 11, "max_stars_repo_stars_event_min_datetime": "2016-12-02T00:46:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-02-20T07:04:01.000Z", "max_issues_repo_path": "text/realworldhaskell/termination.tex", "max_issues_repo_name": "nikivazou/thesis", "max_issues_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "text/realworldhaskell/termination.tex", "max_forks_repo_name": "nikivazou/thesis", "max_forks_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2016-12-02T00:46:51.000Z", "max_forks_repo_forks_event_max_datetime": "2016-12-02T00:46:51.000Z", "avg_line_length": 33.5483870968, "max_line_length": 103, "alphanum_fraction": 0.6717948718, "num_tokens": 3707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7826624840223698, "lm_q2_score": 0.8198933271118222, "lm_q1q2_score": 0.6416997480307042}} {"text": "\\section{Image Processing}\n\n\\begin{definition}[Aliasing]\n Images are constructed and then reconstructed one or more times - \n each of these reconstructions are referred to as aliases.\n\\end{definition}\n\n\\begin{definition}[Anisotropy]\\label{anisotropy}\n Directionally dependent. Opposite of~\\nameref{isotropy}.\n Example is to apply a filter to a set of images - consider\n what would happen if one image was tilted.\n\\end{definition}\n\n\\begin{definition}[Anisotropic diffusion]\n also called Perona-Malik diffusion, is a technique aiming at reducing image\n noise without removing significant parts of the image content, typically\n edges, lines or other details that are important for the interpretation of\n the image.\n\n\\end{definition}\n\n\\begin{definition}[Aperture]\n A hole which light goes through.\n\\end{definition}\n\n\\begin{definition}[Convolution]\\label{convolution}\n In image-processing, this can in general be thought of as applying a mask\n that takes the local neighbourhood for a pixel x and calculates its\n gradient.\n\n Noteworthingly, convolution as an operation to an image is in general\n $O(n^{2})$.\n\\end{definition}\n\n\\begin{definition}[Curve representation]\n How to represent a curve: computational cost is crucial.\n\n \\begin{description}\n \\item[Explicit] $y = f(x)$ - only lines but really easy to genereate.\n \\item[Implicit] $f(x,y) = 0$. \\dots imagine a circle: to know if a point\n is on the line, we can use pythagoras's $x^{2} + y^{2} - r^{2} = 0$. Costs a lot.\n \\item[Parametric] $(x,y) = ( f(u), g(u))$. Oh yeah..\n \\end{description}\n \n\\end{definition}\n\n\\begin{definition}[Dilation (of functions)]\n To skew a function, e.g.\\ for $f(x) = x$, you skew to $\\hat{f}(x) = x + 2$.\n\\end{definition}\n\n\\begin{definition}[Dirichet boundary condtions]\n In image processing, you can say that this boundary condition is to assume\n for a set of differential equations, the unknowns at the borders are \n 0: $u(0) = u(1) = 0$.\n\n\\end{definition}\n\n\\begin{definition}[extrapolation]\n The process of estimating, beyond the original observation range, the value\n of a variable on the basis of its relationship with another variable.\n\n Creating a tangent line at the end of the known data and extending it\n beyond that limit.\n\n\\end{definition}\n\n\\begin{definition}[Filter]\n Applying a mask or operation to an image to extract or distort values.\n\n \\begin{description}\n \\item[High-pass] takes only the high frequency into play, lower values\n are not affected that strongly (if at all)\n \\item[Low-pass] \\dots\n \\end{description}\n\\end{definition}\n\n\\begin{definition}[Gaussian blur]\n Use a gaussian function on an image to reduce image noise and detail\n\\end{definition}\n\n\\begin{definition}[Gaussian filter]\n Gaussian filters have the properties of having no overshoot to a step\n function input while minimizing the rise and fall time.\n\n\\end{definition}\n\n\\begin{definition}[Gaussian pyramid]\n A stack of images, where for each iteration, you take neighboring pixels\n and make them into one. That is, you reduce the intensity in the picture.\n This compresses the image. \n\n It is a pyramid because for each iteration you take neighboring pixels\n from the previous pyramid to generate the current.\n\\end{definition}\n\n\n\\begin{definition}[Gradient flow]\n $$\n V = \\nabla{f} = \\left(\\\n \\frac{\\partial{f}}{\\partial{x_{1}}},\n \\frac{\\partial{f}}{\\partial{x_{2}}},\n \\dots,\n \\frac{\\partial{f}}{\\partial{x_{n}}}\n \\right)\n $$\n\n Note that for an oridnary image, the gradient at a given position is \n given by:\n $$\n \\nabla{f} = \n \\frac{\\partial{f}}{\\partial{x}}\\hat{x} +\n \\frac{\\partial{f}}{\\partial{y}}\\hat{y}\n $$\n Where the first (x) term is the gradient in direction x, and y is the\n gradient in direction y.\n\n One can also calculate the direction of the gradient using\n $\\theta = a\\tan^{2}{\\frac{\\partial{f}}{\\partial{y}}, \\frac{\\partial{f}}{\\partial{x}}}$\n\\end{definition}\n\n\\begin{definition}[Grating]\n A collection of identical pararell objects, placed next to each other with\n equal spacing.\n \\end{definition}\n\\begin{definition}[Grayscale]\n an image in which the value of each pixel is a single sample, that is, it\n carries only intensity information\n\n Grayscale images are distinct from one-bit bi-tonal black-and-white images,\n which in the context of computer imaging are images with only the two\n colors, black, and white (also called bilevel or binary images). Grayscale\n images have many shades of gray in between.\n\n Grayscale images are often the result of measuring the intensity of light\n at each pixel in a single band of the electromagnetic spectrum \n\n\\end{definition}\n\n\\begin{definition}[Image gradient]\n Gradual blend of color\n\\end{definition}\n\n\\begin{definition}[Image noise]\n Unwanted signal, electrical signals not wanted in an image\n\\end{definition}\n\n\\begin{definition}[Image masking]\n Apply an image over another. This can be used for e.g.\\ putting a square image\n with only contents in the middle over another.\n\\end{definition}\n\n\\begin{definition}[Interpolation]\n A way to estimate (pixel) values when resizing to a larger image.\n\n An example from calculus is: given temps at noon and midnight, estimate\n the temp at 5 PM as the mean of noon and midnight.\n\n Interpolation can also (in signal processing) be referred to as \n \\textbf{upsampling}.\n\n Interpolation can be \\textit{linear}(1D), \\textit{bilinear}(2D) and so on\\dots\n\n\\end{definition}\n\n\\begin{definition}[Isotropy]\\label{isotropy}\n Directionally independent. Opposite of~\\nameref{anisotropy}.\n\\end{definition}\n\n\\begin{definition}[Laplacian]\n Highlight regions of rapid intensity change and is therefore often used for\n edge detection.\n\n Often applied to an image that has first been smoothed with something\n approxiating in order to reduce its sensitivity to noise.\n\n\\end{definition}\n\n\\begin{definition}[Spatial frequency]\n The number of changes in color values that occur per space.\n\n Images with high spatial frequency are detailed, images with low spatial \n frequency will appear blurred.\n\n I.e.\\ sharp transitions from low/high to high/low intensities are said\n to have a large spatial frequency,\n\n\\end{definition}\n\n\n\\begin{definition}[SRGB]\n sRGB is a standard RGB color space created cooperatively by HP and\n Microsoft in 1996 for use on monitors, printers and the Internet.\n\n\\end{definition}\n\n\n\\begin{definition}[Poisson image editing]\n Blend two images together and make them seem alike.\n\\end{definition}\n\n\\begin{definition}[Raster order]\n Begin at top left, proceed to right, then at leftmost pixel in next line.\n\\end{definition}\n\n\\begin{definition}[Raster image]\n An image with pixels.\n\\end{definition}\n\n\\begin{definition}[Ringing]\n The size of the oscillations after a peak frequency. E.g.\\ if you have a \n sudden overshoot (bright color) the amplitude of the next wave is\n the ringing. (???)\n\\end{definition}\n\n\\begin{definition}[Scalar field]\\label{scalarfield}\n Associate a value to every point in a space. E.g.\\ for an image, you\n could assign each pixel a color value.\n\\end{definition}\n\n\\begin{definition}[Seperable filter]\n A filter that can be written as the product of two or more small filters.\n This is usually done to reduce the computational costs.\n E.g.~\\nameref{convolution} is usually faster in 1d than iterating $nD$, for\n $n \\ge 2$.\n\\end{definition}\n\n\\begin{definition}[Spline]\n A line that goes through a series of points.\n\n The section between the lines is interpolated. This can be done linearly,\n quadratic, \\dots so on.\n\\end{definition}\n\n\\begin{definition}[Stencil]\n A figure that connects dots on a grid. Usually a cross-like figure \n (one center dots, and one edge out in 4 directions from the center to\n other dots).\n\\end{definition}\n", "meta": {"hexsha": "2235e6d74ad42788c3a16a9435726bcb0db71bcb", "size": 8017, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/def/image.tex", "max_stars_repo_name": "andsild/NotusVitae", "max_stars_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/def/image.tex", "max_issues_repo_name": "andsild/NotusVitae", "max_issues_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/def/image.tex", "max_forks_repo_name": "andsild/NotusVitae", "max_forks_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.5439330544, "max_line_length": 93, "alphanum_fraction": 0.7169764251, "num_tokens": 2000, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7826624738835052, "lm_q1q2_score": 0.6416997466066878}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{wrapfig}\n\\usepackage{float}\n\\usepackage{graphicx}\n\\usepackage[margin=1in]{geometry} \n\\usepackage{comment}\n\\usepackage{amsmath,amsthm,amssymb}\n% \\usepackage[vlined,linesnumbered,ruled,resetcount]{algorithm2e}\n\n\\DeclareMathOperator*{\\argmax}{arg\\,max}\n\\DeclareMathOperator*{\\argmin}{arg\\,min}\n\\DeclareMathOperator*{\\val}{val}\n\\DeclareMathOperator*{\\best}{best}\n\\DeclareMathOperator*{\\worst}{worst}\n\n\\usepackage{algorithm}\n% \\usepackage{algorithmicx}\n\\usepackage[noend]{algpseudocode}\n\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\Rgz}{\\mathbb{R}_{\\ge 0}}\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\n\\newcommand{\\Es}[2]{\\mathbb{E}_{#1}\\left[{#2}\\right]}\n\\newcommand{\\E}[1]{\\mathbb{E}\\left[{#1}\\right]}\n\\newcommand{\\ip}[2]{\\left\\langle{#1} , {#2}\\right\\rangle}\n\n\n\\newtheorem{definition}{Definition}[section]\n\\newtheorem{lemma}[definition]{Lemma}\n\\newtheorem{corollary}[definition]{Corollary}\n\\newtheorem{theorem}[definition]{Theorem}\n\\newtheorem{claim}[definition]{Claim}\n\n\\begin{document}\n\n% \\renewcommand{\\qedsymbol}{\\filledbox}\n \n\\title{Mechanism Design Notes}\n\\author{Clay Thomas\\\\\nclaytont@princeton.edu}\n\\maketitle\n\n\\clearpage\n\\section{Regret Minimization and Correlated Equilibrium Concepts}\n\nConsider the following online learning problem:\nThere is a set of actions $A$, $|A| = n$. For each time step $t=1,\\ldots,T$,\nthe decision maker picks a distribution on actions $p^t$.\nAfterwards, a cost vector $c^t : A\\to [0,1]$ is selected\n(in any way - even by an adversary who can see $p^t$ before it picks $c^t$)\nand the decision maker ``incurs cost'' $\\Es{a\\sim p^t}{c^t(a)} = \\ip{p^t}{c^t}$.\nAdditionally, the decision maker learns the vector $c^t$.\nThe goal is to minimize the sum of the costs $\\Es{a\\sim p^t}{c^t(a)}$.\n\n\\[ \\sum_{t=1}^T \\min_a c^t(a) \\]\nCan we hope to achieve something approximating this?\nNot quite: an adversary who can first look at $p^t$ can, for example,\nmake all costs very high, except the one with smallest probability in $p^t$.\nA more realistic goal is to compare our cost to the following:\n\\[ \\min_a \\sum_{t=1}^T c^t(a) \\]\nIntuitively, ``the cost of the best action which, in hindsight, we could've\nalways taken''.\n\nThis leads us to define the \\emph{regret} of our learning algorithm as\n\\[ \\Es{a^t\\sim p^t, t=1,\\ldots,T}{\\sum_{t=1}^T c^t(a)} - \\min_a \\sum_{t=1}^T c^t(a) \n = {\\sum_{t=1}^T \\ip{p^t}{c^t}} - \\min_a \\sum_{t=1}^T c^t(a) \n\\]\n\nIt turns out there is a simple algorithm that achieves low regret.\n\\begin{algorithm}\n \\begin{algorithmic}[0]\n \\State \\textbf{Parameters:} $\\epsilon > 0$, $T$\n \\State Initialize $w^1(a) = 1$ for each $a\\in A$\n \\For {$t=1,\\ldots,T$}\n \\State Choose action $a\\in A$ according to distribution $p^t$ proportional to $w^t$\n \\State \\qquad i.e. let $p^t(a) = w^t(a) / \\Phi^t$ where\n $\\Phi^t = \\sum_{a\\in A} w^t(a)$\n \\State Observe costs $c^t$ and update weights as follows for each $a\\in A$:\n \\State \\qquad $w^{t+1}(a) = w^t(a) (1 - \\epsilon)^{c^t(a)}$.\n \\EndFor\n\\end{algorithmic}\n\\end{algorithm}\n\n\\begin{theorem}\n Let $OPT = \\min_a \\sum_{t=1}^T c^t(a)$.\n If we set $\\epsilon = \\sqrt{\\log n/T}$, then after $T$ rounds the total realized\n cost of the multiplicative weights algorithm is $OPT + O(\\sqrt{T\\log n})$.\n\\end{theorem}\n\\begin{proof}\n The proof follows two observations.\n First, if the decision maker makes a bad move\n (i.e. the weight of costly moves is high) then lots of total weight is lost.\n Second, the best action in hindsight contributes has weight\n $(1-\\epsilon)^{OPT}$ at time $T$.\n Intuitively, this says that the only way for the decision maker to have many \n bad moves is if every action was very bad.\n\n For the formal proof, start by bounding $\\Phi^{t+1}$ in terms of $\\Phi^t$.\n In what follows, almost all of the inequalities are standard approximations,\n which can be proved using Taylor series or other estimations.\n This is somewhat interesting, and basically just happens because $\\epsilon$ is\n so small we can go back and forth between exponentiation and addition really\n easily. In a sense, this was exactly the property we needed for our\n multiplicative weights to produce a low additive cost measure.\n Denote by $\\ell^t = \\Es{a\\sim p^t}{c^t(a)}$ be the realized cost at time $t$.\n \\begin{align*}\n \\Phi^{t+1}\n & = \\sum_a w^t(a)(1 - \\epsilon)^{c^t(a)} \\\\\n & \\le \\sum_a w^t(a)(1 - \\epsilon c^t(a)) \\\\\n & = \\sum_a w^t(a) - \\Phi^t \\epsilon \\sum_a \\frac{w^t(a)}{\\Phi^t} c^t(a) \\\\\n & = \\Phi^t \\left( 1 -\\epsilon \\sum_a p^t(a) c^t(a) \\right) \\\\\n & = \\Phi^t \\left( 1 -\\epsilon \\ell^t \\right) \\\\\n \\end{align*}\n Now we can relate $\\Phi^T$ to the total loss $L = \\sum_{t=1}^T \\ell^t$:\n \\begin{align*}\n \\Phi^T\n & = \\Phi^0 \\prod_{t=1}^T \\left( 1 -\\epsilon \\ell^t \\right) \\\\\n & \\le n \\prod_{t=1}^T \\exp\\left(-\\epsilon \\ell^t \\right) \\\\\n & = n \\exp\\left(-\\epsilon \\sum_{t=1}^T \\ell^t \\right) \\\\\n & = n \\exp\\left(-\\epsilon L\\right) \\\\\n \\end{align*}\n\n On the other hand, for any action $a\\in A$ we have\n \\[ (1-\\epsilon)^{OPT} \\le w^T(a) < \\Phi^T \\le n\\exp(-\\epsilon L) \\]\n \\[ OPT(-\\epsilon - \\epsilon^2/2) \\le OPT\\log(1 - \\epsilon) \n \\le \\log n - \\epsilon L \\]\n \\begin{align*} \n L & \\le OPT + ({\\epsilon}/{2}) OPT + \\frac{\\log n}{\\epsilon} \\\\\n & \\le OPT + (\\sqrt{\\log n/T}/{2}) T + \\frac{\\log n}{\\sqrt{\\log n/T}} \\\\\n & = OPT + \\frac{3}{2} \\sqrt{T\\log n}\n \\end{align*}\n\n\\end{proof}\n\nFollowing experts online sounds great and all, but how is this actually useful\nfor proving theorems? Often time we are interested in the \\emph{time averaged}\ncost incurred, typically because that allows us to combine all of the\ndistributions $p^t$ into a distribution which is ``close to optimal'' in some\nsense. The following restatement tells you ``how long you need to go'' to get\nadditive error $\\epsilon$.\n\n\\begin{corollary}\n For any $\\epsilon > 0$, if we run for $T = O(\\log n / \\epsilon^2)$ steps,\n multiplicative weights will have average regret $\\epsilon$.\n\\end{corollary}\n\n\n\n\n\n\n\n\n \\subsection{Swap regret}\n\n The first idea is to run several independent instances of low regret\n algorithm. We'll kind of see that each instance will correspond to a different\n action. However, perhaps the best way to see why this definition works comes\n out in the analysis: we need to align the sum of the costs incurred to the\n individual instances with the swap regret.\n\n Here's a description of the algorithm:\n Run independent instances of the multiplicative weights algorithm (or any\n low-regret online learning algorithm) $M_j$ for $j=1,\\ldots,n = |A|$.\n At each time $t$, the algorithm:\n \\begin{itemize}\n \\item asks each learner for a distribution $q_j^t$ over actions\n \\item calculates $p^t = P(q_1^t,\\ldots,q_n^t)$ using some subroutine $P$,\n and acts according to $p^t$\n \\item observes ``real'' cost $c^t$, but tells each learner $M_j$ that the\n cost incurred to $M_j$ is $p^t(j)c^t$\n \\end{itemize}\n\n Individually, it is a bit difficult to interpret what each $M_j$ is trying to\n minimize: it's like $M_j$ gets punished, but only if the action he is\n associated with is likely played. Collectively, all the $M_j$s are trying to\n minimize $\\sum_j p^t(j) c^t = c^t$, so at least that aligns with what we want.\n\n More formally, the algorithm looks like this:\n \\begin{algorithm}\n \\begin{algorithmic}[0]\n \\State \\textbf{Parameters:} $\\epsilon > 0$, $T$\n \\State Initialize $w^1_j(a) = 1$ for each $a\\in A$ and $j=1,\\ldots,n$\n \\For {$t=1,\\ldots,T$}\n \\For {$j=1,\\ldots,n$}\n \\State Define a distribution $q_j^t$ proportional to $w_j^t$\n \\EndFor\n \\State Let $p^t = P(q^t_1,\\ldots,q^t_n)$, and \n draw an action according to $p^t$\n % \\State Define $p^t$ as follows: $p^t$ is any stationary distribution of\n % the Mark chain whose states \n % \\State \\qquad are actions $j\\in A$ and whose transition\n % probabilities from $j$ to each $i\\in A$ are $q_j^t(i)$\n \\State Observe costs $c^t$\n \\For {$j=1,\\ldots,n$}\n \\State Update weights $w^{t+1}_j(a) = w^t_j(a) (1 - \\epsilon)^{p^t(j) c^t(a)}$\n \\EndFor\n \\EndFor\n \\end{algorithmic}\n \\end{algorithm}\n\n Let's dive into the analysis, during which we'll finally specify $P$:\n\n", "meta": {"hexsha": "09387c44f0d2454f8dd58a47d182c49cb3e96085", "size": 8257, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Tex/Regret.tex", "max_stars_repo_name": "ClathomasPrime/CompetitiveStableMatching", "max_stars_repo_head_hexsha": "8a0b1edadbffb8ea7dde8fac8d7696f91747668f", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Tex/Regret.tex", "max_issues_repo_name": "ClathomasPrime/CompetitiveStableMatching", "max_issues_repo_head_hexsha": "8a0b1edadbffb8ea7dde8fac8d7696f91747668f", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Tex/Regret.tex", "max_forks_repo_name": "ClathomasPrime/CompetitiveStableMatching", "max_forks_repo_head_hexsha": "8a0b1edadbffb8ea7dde8fac8d7696f91747668f", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.8888888889, "max_line_length": 87, "alphanum_fraction": 0.6659803803, "num_tokens": 2751, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933227109649, "lm_q2_score": 0.7826624688140726, "lm_q1q2_score": 0.641699732117137}} {"text": "\\hypertarget{convex-hull}{%\n\\section{Convex Hull}\\label{convex-hull}}\n\n\\hypertarget{introduction}{%\n\\subsection{Introduction}\\label{introduction}}\n\nGiven a set of points P, the Convex Hull of P, denoted conv(P) is the\nsmallest closed simple polygon which contains all the points in P. It\ncan be visualized as a tightly snapped rubber band over the given set of\npoints.\n\nWe would like to compute the convex hull of given set of points by using\na divide and conquer technique. Later, we will also show that the\ncomputational complexity of our algorithm is also O(n log n). Note that\nwe cannot do better than is for computing convex hull\n\n\\hypertarget{how-to-run}{%\n\\subsection{How to Run}\\label{how-to-run}}\n\nThe src folder contains the source code for the convex hull program.\n\\texttt{g++} from the GNU compiler suite is required to compile the\nprogram to a executable.\n\nSteps to Compile:\n\n\\begin{enumerate}\n\\def\\labelenumi{\\arabic{enumi})}\n\\tightlist\n\\item\n \\texttt{cd} into the src directory\n\\item\n Run \\texttt{g++\\ main.cpp} which generates an executable called\n \\texttt{a.out} in the same directory\n\\item\n Run the executable using \\texttt{./a.out} (on linux)\n\n \\begin{enumerate}\n \\def\\labelenumii{\\arabic{enumii})}\n \\tightlist\n \\item\n The executable takes a dataset from command line argument. For\n example, to use an existing dataset, run\n \\texttt{./a.out\\ ../datasets/edge.txt}\n \\item\n If no command-line argument is given, it takes input from the shell\n directly (stdin)\n \\end{enumerate}\n\\end{enumerate}\n\n\\hypertarget{input}{%\n\\subsection{Input}\\label{input}}\n\nThe required file format for the algorithm to work correctly is:\n\n\\begin{itemize}\n\\tightlist\n\\item\n First line must contain the no of Points to be taken as input by the\n program.\n\\item\n Each of next line must contain 2 integers, space seperated denoting\n the (x, y) coordinates of each point.\n\\item\n Each coordinate must be of integer type in the range -10\\^{}6 to\n 10\\^{}6.\n\\item\n No of coordinates must be less than 1 Billion.\n\\item\n \\textbf{Note}: while using floating point datasets, the output\n coordinates might slightly differ as given coordinates are stored as\n floats.\n\\end{itemize}\n\n\\hypertarget{output}{%\n\\subsection{Output}\\label{output}}\n\nEach line of output contains the coordinates of points present on the\nconvex hull in clockwise order.\\\\\nThe last two lines of output show the time taken to take input in\nmicroseconds and the time taken by the algorithm to compute convex hull\n(also in microseconds).\n\nExample output:\n\n\\includegraphics[width=8cm,height=8cm]{img/CH10.png}\\\\\n\n\\hypertarget{documentation-and-report}{%\n\\subsection{Documentation and Report}\\label{documentation-and-report}}\n\nDocumentation of this algorithm, functions and classes can be found in\nthe \\texttt{docs} folder in the current directory. Open the\n\\href{../ConvexHull/docs/html/index.html}{index.html} file from the docs\ndirectory with your preferred browser to go through the documentation\n\n\\hypertarget{performance-analysis}{%\n\\subsection{Performance Analysis}\\label{performance-analysis}}\n\nAnalysis is performed with a system running:\n\n\\begin{itemize}\n\\tightlist\n\\item\n OS: Arch Linux (64Bit) running Linux Kernel version 5.7.2\n\\item\n Processor: Intel Core i7 7700HQ\n\\item\n RAM: 8GB\n\\item\n Compiler: GNU G++ (GCC) 10.1.0\n\\end{itemize}\n\n\\textbf{The following observations are recorded:}\n\n\\begin{longtable}[]{@{}lcccc@{}}\n\\toprule\nFilename & Input Dimentions & Output Diemntions & File read time &\nAlgorithm runtime\\tabularnewline\n\\midrule\n\\endhead\nedge.txt & 5 & 4 & 451 microsec & 8 microsec\\tabularnewline\nsmall.txt & 7 & 7 & 495 microsec & 16 microsec\\tabularnewline\ntest.txt & 34 & 7 & 328 microsec & 56 microsec\\tabularnewline\nradial.txt & 3100 & 9 & 1348 microsec & 6.15 millisec\\tabularnewline\n\\bottomrule\n\\end{longtable}\n\nRandomly generated points (using python):\n\n\\begin{longtable}[]{@{}lcccc@{}}\n\\toprule\nFilename & Input Dimentions & Output Diemntions & File read time &\nAlgorithm runtime\\tabularnewline\n\\midrule\n\\endhead\n10.txt & 10 & 5 & 314 microsec & 16 microsec\\tabularnewline\n20.txt & 20 & 7 & 495 microsec & 31 microsec\\tabularnewline\n50.txt & 50 & 9 & 427 microsec & 108 microsec\\tabularnewline\n100.txt & 100 & 13 & 348 microsec & 165 microsec\\tabularnewline\n500.txt & 500 & 13 & 572 microsec & 935 microsec\\tabularnewline\n1000.txt & 1000 (1K) & 18 & 510 microsec & 2.1 millisec\\tabularnewline\n3000.txt & 3000 (3K) & 24 & 2.49 millisec & 5.5 millisec\\tabularnewline\n5000.txt & 5000 (5K) & 22 & 2.12 millisec & 8.0 millisec\\tabularnewline\n10000.txt & 10000 (10K) & 23 & 3.26 millisec & 17.6\nmillisec\\tabularnewline\n50000.txt & 50000 (50K) & 28 & 17.4 millisec & 80.5\nmillisec\\tabularnewline\n100000.txt & 100000 (1L) & 31 & 39.3 millisec & 164.1\nmillisec\\tabularnewline\n250000.txt & 250000 (2.5L) & 31 & 78.1 millisec & 448.3\nmillisec\\tabularnewline\n\\bottomrule\n\\end{longtable}\n\nUsing real world datasets:\n\n\\begin{longtable}[]{@{}lcccc@{}}\n\\toprule\nFilename & Input Points & Output Diemntions & File read time & Algorithm\nruntime\\tabularnewline\n\\midrule\n\\endhead\nsubway-entrance-ny.txt & 1929 & 16 & 2.85 millisec & 3.81\nmillisec\\tabularnewline\nparking\\_meter.txt & 15191 & 16 & 14.15 millisec & 27.04\nmillisec\\tabularnewline\n\\bottomrule\n\\end{longtable}\n\nSources of datasets:\n\n\\begin{itemize}\n\\tightlist\n\\item\n \\href{https://data.world/city-of-ny/5jsj-cq4s}{Parking meter dataset}\n\\item\n \\href{https://data.world/new-york-city/subway-entrances}{New York\n Subway Entrance dataset}\n\\end{itemize}\n\nIt is highly difficult to calculate the time taken exactly up to the\nmicrosecond, hence the value might vary on different executions. Also,\nplease note that the time calculated here may vary based on the system\nload and other background applications.\n\n\\hypertarget{algorithm-approach}{%\n\\subsection{Algorithm Approach}\\label{algorithm-approach}}\n\nFor more detailed information regarding the algorithm, refer to David\nMount Lecture notes from the course\n\\href{https://www.cs.umd.edu/class/spring2020/cmsc754/lectures.html}{CMSC\n754 - Computational Geometry}\n\nGiven set of points P, we would like to first order them according to\nthe increasing x coordinate. For the sake of simplicity, let us assume\nthat no two points will have same x or y coordinate and no 3 points are\nco-linear.\n\nFirst, we will recursively compute the upper hull and then similarly\ncompute the lower hull of the given set of points. Finally, we will join\nthe upper hull with the lower hull to complete the convex hull and\nreturn/print the clockwise ordering of the points present on the convex\nhull\n\n\\textbf{Upper Hull Computation}\n\nWe divide the input set of points into 2 equal halves and recursively\ncompute the upper hull for both the half. Then, we compute the upper\ntangent to both the hulls by traversing from the right on left hull and\ntraversing from left on the right hull. We perform orientation tests to\ndetermine when the orientation changes on each hull respectively and\nmove upwards a point. We perform this operations until we reach the\nupper part of both hulls.\n\nAll the traversed points on both hulls except the one which lead to\ntangent are deleted (because all of them lie under tangent, hence lie\ninside the polygon) and the resulting points of both hulls are catenated\nto get the upper hull.\n\n\\hypertarget{results}{%\n\\subsection{Results}\\label{results}}\n\nThe following is the computed convex hull of the\n\\href{./datasets/parking_meter.txt}{parking\\_meter.txt} dataset:\n\n\\includegraphics[width=8cm,height=8cm]{img/CHparkingmeter.png}\\\\\n\nThe red boundary in the above image represents the convex hull given as\nthe output by the program. The blue points are the inputs given to the\nprogram.\n\nBy analysis of the above divide and conquer algorithm, It can be\nconcluded that the algorithm runs in O(n log n) time complexity. From\nthe above results of testing of various datasets, we can see that the\ntime increase is proportional to the no of points.\n\n\\includegraphics[width=8cm,height=8cm]{img/CHradial.png}\\\\\n\nThe dataset \\href{./datasets/radial.txt}{radial.txt} is a hand-crafted\ndataset (3K points) which has high concentration of points\n(\\textasciitilde2.1K) in the center, whereas other randomly generated\ndatasets (for example: \\href{./datasets/3000.txt}{3000.txt})have a\nuniform distribution of points (the python random number generator uses\nan underlying uniform distribution). From this we can assume that\nindependant of positioning of points, our algorithm takes approximately\nthe same time to compute the convex hull.\n\n\\hypertarget{conclusion}{%\n\\subsection{Conclusion}\\label{conclusion}}\n\nFor the last dataset which contains 2.5L points, its worth noting that\nthe algorithm runs in just under 500ms which might appear to be fast.\nBut for applications like graphics, rendering and animations, this poses\na huge overhead in terms of repeated computation of convex hull for\nvarious objects.\n\nOne observation we can make here is that the number of points which are\npresent on the convex hull is usually relatively lower most of the\ntimes. So, using an output sensitive algorithms (Ex: Jarvis's march)\nmight help us if our application has fewer vertices on the convex hull.\n\nIt's worth noting that convex hull computation cannot be done under O(n\nlog n) which has been proved. Therefore, we can only hope to reduce the\nconstants of complexity. In our algorithm, we repeatedly recurse, hence\nmake many function calls which is computationally expensive. Hence by\nusing algorithms like Chan's or Grahm's, we can do better\n", "meta": {"hexsha": "32f6d0c21f36780824ea113c9608277dd0088057", "size": 9525, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Latex/ConvexHull.tex", "max_stars_repo_name": "RikilG/Geometry-Algorithms", "max_stars_repo_head_hexsha": "7bdf25e425b93dc6955331a48980a4b4d8051a6d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Latex/ConvexHull.tex", "max_issues_repo_name": "RikilG/Geometry-Algorithms", "max_issues_repo_head_hexsha": "7bdf25e425b93dc6955331a48980a4b4d8051a6d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Latex/ConvexHull.tex", "max_forks_repo_name": "RikilG/Geometry-Algorithms", "max_forks_repo_head_hexsha": "7bdf25e425b93dc6955331a48980a4b4d8051a6d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.8082706767, "max_line_length": 73, "alphanum_fraction": 0.7717585302, "num_tokens": 2598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195385342971, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.6416771741144867}} {"text": "\\subsection{Input Circuit}\nIn Europe the mains voltage in houses is 230 volts, all of the input circuit before the transformer needs to be designed to that voltages. \nAs this part of the input circuit we want to besure that the noise on mains volts do not propegate to our output volts, \nwe are going to use a main emi filter to filter out some of the noise on the mains system. \nThis will also double as filter for any noise we are going to send out to the mains voltage. \n\nAs a means for input protection we are also going to use fuse (see formula \\ref{eq:transfer-transformer} on page \\pageref{eq:transfer-transformer}).\n\n\\subsubsection{Transformer}\nThe transfer formula for the transformer \n\\begin{align}\\label{eq:transfer-transformer}\nm &= \\frac{I_s}{I_p} = \\frac{V_p}{V_s}\n\\end{align}\nFor the input supply I am going to use a transformer 2x18V at 1.5A, that I already have. And I am going to connect the 18V's in series to get 36V at 1.5A. But for finding the transfer value we can use formula \\ref{eq:transfer-transformer}.\n\\begin{align} \nm = \\frac{V_p}{V_s} = \\frac{230V}{18V} = 12.78 \\nonumber\n\\end{align}\n To find out which size of fuse that we are going to be using, we need to find out how much current that are going to be used on the main side of the transformer.\nWe are going rearrange in formula \\ref{eq:transfer-transformer} as we know what $I_s$ is and we have calculated $m$.\n\\begin{align}\nm = \\frac{I_s}{I_p} \\nonumber \\\\\n\\Rightarrow \\nonumber \\\\\nI_p \\cdot m = I_p \\cdot \\frac{I_s}{I_p} = I_s \\nonumber \\\\\n\\Rightarrow \\nonumber \\\\\n\\frac{ I_p \\cdot m}{m} = \\frac{I_s}{m} \\nonumber \\\\\n\\Rightarrow \\nonumber \\\\\nI_p = \\frac{I_s}{m} \\label{eq:found-Ip}\n\\end{align}\nNow we can calculate $I_p$ with $I_s$ and $m$ from \\ref{eq:found-Ip}.\n\\begin{align}\nI_p = \\frac{I_s}{m} = \\frac{1.5A}{12.78} = 0.117A = \\underline{\\underline{117mA}} \\nonumber \n\\end{align}\nSo the primary side of the transformer needs a fuse of 117mA, \nbut as I can not find that size, the one that is going to be used is a 125mA.\nOn the secondary side, I am going to use a 1,5A fuse.\n\n\\subsubsection{DC voltage}\nNow we should calculate if we want to use 2x18V or put them in serial and get 36V. \nOne thing to keep in mind, is that when using a bridge rectifier the input frequency is doubled.\n\\paragraph{36 \\vac}\nThe 36 volt ac are going to be converted to dc, via fullwave diode bridge. \nSinces the ac voltage specified on the transformer is RMS voltage, and not the peak voltage. \nWe can calculate what the dc voltage will become assuming no losses.\n\\begin{align} \\label{eq:calc_vdc}\n\\text{V}_{ \\text{DC} } &= \\frac{ 2 \\cdot \\text{V}_{ \\text{max} } }{ \\pi } \\\\\n &= 0.637 \\cdot \\text{V}_{ \\text{max} } \\\\\n &= 0.9 \\cdot \\text{V}_{ \\text{RMS} } \\\\\n &= 0.9 \\cdot 36 \\text{V}_{\\text{AC}}\\nonumber \\\\\n &= 32.4 \\text{V}_{ \\text{DC} } \\nonumber\n\\end{align}\nBut since we are going to use a bridge rectifier there is two times diode voltage drop that gives total drop of \\(1.4V\\). \nWhich meaning that we are properly only going to see around \\( 31V \\) if measure at the diode bridge output. \n\\paragraph{18 \\vac}\n\\begin{align} \\label{eq:calc_18vdc}\n \\text{V}_{ \\text{DC} } &= \\frac{ 2 \\cdot \\text{V}_{ \\text{max} } }{ \\pi } \\\\\n &= 0.637 \\cdot \\text{V}_{ \\text{max} } \\\\\n &= 0.9 \\cdot \\text{V}_{ \\text{RMS} } \\\\\n &= 0.9 \\cdot 18 \\text{V}_{ \\text{AC} } \\nonumber \\\\\n &= 16.2 \\text{V}_{ \\text{DC} } \\nonumber\n\\end{align}\nAnd again we need to take into account the diode drop, which gives 14.8V after the diode bridge.\n\n\\subsection{Ripple voltage}\nThe diode bridge does not create a dc voltage, but rather flips over the negativ half of the sinus wave to the positive. So there is a need for smoothering out the sinus wave to get a dc voltage. \nWe need to figure out how much ripple voltage, that we can get away with at this point. We are going to use formula \\ref{eq:calc_vripple} to calculate our $V_{ripple}$.\n\\begin{align}\\label{eq:calc_vripple}\n\\text{V}_{ \\text{PP}_{\\text{ripple} }} &= \\frac{ \\text{I}_{ \\text{load} } }{ 2 \\cdot f \\cdot C } [V]\n\\end{align}\nIf we choose a big capacitor, then the time for it going to be discharge from fully charged is going to be long. As long there is voltage on the capacitor the rest of the circuit is working. At table \\ref{tab:choose-c} I have made some calculation of a couple of capacitor that I can buy.\n\\begin{table}[ht]\n\\centering\n\\caption{Choosing smooting capacitor} \\label{tab:choose-c}\n\\begin{tabular}{|c|r|r|r|r|} \n\\hline\nC & $V_{ripple}$ \\@ 1.5A & $V_{ripple}$ \\@ 1A & $V_{ripple}$ \\@ 500mA & $V_{ripple}$ \\@ 10mA \\\\ \\hline \\hline\n 47 uF & 3.525 mV & 2.350 mV & 1.175 mV & 0.024 mV \\\\ \\hline\n100 uF & 7.500 mV & 5.000 mV & 2.500 mV & 0.050 mV \\\\ \\hline\n120 uF & 9.000 mV & 6.000 mV & 3.000 mV & 0.060 mV \\\\ \\hline\n150 uF & 11.250 mV\t& 7.500 mV & 3.750 mV & 0.075 mV \\\\ \\hline\n180 uF & 13.500 mV & 9.000 mV & 4.500 mV & 0.090 mV \\\\ \\hline\n220 uF & 16.500 mV & 11.000 mV & 5.500 mV & 0.110 mV \\\\ \\hline\n270 uF & 20.250 mV & 13.500 mV & 6.750 mV & 0.135 mV \\\\ \\hline\n330 uF & 24.750 mV & 16.500 mV & 8.250 mV & 0.165 mV \\\\ \\hline\n390 uF & 29.250 mV & 19.500 mV & 9.750 mV & 0.195 mV \\\\ \\hline\n470 uF & 35.250 mV & 23.500 mV & 11.750 mV\t& 0.235 mV \\\\ \\hline\n560 uF & 42.000 mV & 28.000 mV\t& 14.000 mV & 0.280 mV \\\\ \\hline\n680 uF & 51.000 mV & 34.000 mV & 17.000 mV & 0.340 mV \\\\ \\hline\n820 uF & 61.500 mV & 41.000 mV & 20.500 mV & 0.410 mV \\\\ \\hline\n\\end{tabular}\n\\end{table}", "meta": {"hexsha": "8d595ea3c1f2e31bccd47c45433c8fb91db15e20", "size": 5511, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/inputcircuit.tex", "max_stars_repo_name": "mtkaalund/psu", "max_stars_repo_head_hexsha": "98bf7e367becc1e3480d916a514af6f1dee106ca", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/inputcircuit.tex", "max_issues_repo_name": "mtkaalund/psu", "max_issues_repo_head_hexsha": "98bf7e367becc1e3480d916a514af6f1dee106ca", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/inputcircuit.tex", "max_forks_repo_name": "mtkaalund/psu", "max_forks_repo_head_hexsha": "98bf7e367becc1e3480d916a514af6f1dee106ca", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.2333333333, "max_line_length": 288, "alphanum_fraction": 0.6741063328, "num_tokens": 1962, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676284, "lm_q2_score": 0.8499711794579723, "lm_q1q2_score": 0.6416559817783778}} {"text": "\\documentclass[a4paper]{article}\n\n\\usepackage[english]{babel}\n\\usepackage[utf8x]{inputenc}\n\\usepackage{amsmath}\n\\newcommand\\numberthis{\\addtocounter{equation}{1}\\tag{\\theequation}}\n\n\\usepackage{graphicx}\n\\usepackage{fullpage}\n\\title{ISE 538 Homework 1}\n\\author{Saket Choudhary\\\\skchoudh@usc.edu}\n\n\\begin{document}\n\\maketitle\n\n\n\n\n\\section*{Problem 1.31}\nLet $D_i$ represent the score on $i^{th}$ dice.\nThen the possible configurations for getting a sum 7 are: $(D_1, D_2): (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\n\nThus $P(D_1=6|D_1+D_2=7) = \\frac{1}{6}$\n\\section*{Problem 1.32}\n\nLet \n\\begin{align*}\nI_i &= \\begin{cases}\n1 & i^{th}\\text{ person gets his hat back}\\\\\n0 & \\text{otherwise}\n\\end{cases}\\\\\nP(\\text{no person gets his hat back}) &= 1 - P(I_1 \\cup I_2 \\dots \\cup I_n)\\\\\nP(I_i=1) &= \\frac{1}{n}\\\\\nP(I_i=1, I_j=1) &= \\frac{1}{n}\\frac{1}{n-1} = \\frac{1}{n(n-1)} = \\frac{(n-2)!}{n!}\\\\\nP(I_i=1, I_j=1) &= \\frac{1}{n}\\frac{1}{n-1}\\frac{1}{n-2} = \\frac{(n-3)!}{n!}\\\\\nP(I_1 \\cup I_2 \\dots \\cup I_n) &= \\sum_{i=1}^{n}P(I_i) - \\sum_{i0$, there exists a bound $h_b>0$, so that $r=|E/\\varepsilon(h)|0$, there exists a bound $k_b>0$, so that $r=|E/\\varepsilon(h)|k_b$.\n \nNote that we divide $E$ by $\\varepsilon(h)$ to get $r$ in the above definition for some range of nonzero values of $h$. So, for any expression to be $o(\\varepsilon(h))$, $\\varepsilon(h)$ must be defined and nonzero for some range of nonzero values of $h$. We therefore restrict $\\varepsilon(h)$ to such admissible functions: \n\n\\begin{quote}\nFor $\\varepsilon(h)$ to be admissible in $o(\\varepsilon(h))$ notation, there must exist some $h_0 > 0$ so that $\\varepsilon(h)$ is defined (finite) and nonzero for $0 < |h| < h_0$. This way we can form the ratio $r=|E/\\varepsilon(h)|$ without dividing by zero for at least some range of $h$.\n\\end{quote}\n \nWe only use admissible $\\varepsilon(h)$ in these notes. In particular, $\\varepsilon(h) = {|h|}^p$ is admissible for any value of $p$ , and $\\varepsilon(h) = h^p$ is admissible for any integer value of $p$. \n\n\\section{Summary}\n\\begin{itemize}\n\\item\nLittle-$o()$ notation is a way of describing an expression as small in comparison to some other value: \n\\begin{quote}\n $E$ is $o(\\varepsilon(h))$ as $h \\rightarrow 0$ means the ratio $r = |E/\\varepsilon(h)|$ can be made as small as desired provided the magnitude of $h$ is small enough (but not zero).\n\\end{quote}\n\n\\item \nGraphically, expressions that differ by $o(h^0)$ need to touch at $h=0$, but expressions that differ by $o(h^1)$ also need to be tangent at $h=0$.\n\n\\item\nSpecifically, $E$ is $o(\\varepsilon(h))$ as $h \\rightarrow 0$ means: \n\\begin{quote} \nFor any $r_b > 0$, there exists $h_b > 0$, so that:\n\\begin{equation*}\n\\text{if\\ } 0 < | h | < h_b\\,,\\text{\\ then\\ }r=\\left|\\frac{E}{\\varepsilon(h)}\\right| 0$, there exists $k_b > 0$, so that:\n\\begin{equation*}\n\\text{if\\ } |k| > k_b\\,,\\text{\\ then\\ }r=\\left|\\frac{E}{\\varepsilon(h)}\\right| 0$ so that $\\varepsilon(h)$ is defined (finite) and nonzero for $0 < | h | < h_0$.\n\\end{quote}\n\n\\item We only use admissible $\\varepsilon(h)$ in these notes. In particular, $\\varepsilon(h) = {|h|}^p$ is admissible for any value of $p$, and $\\varepsilon(h) = h^p$ is admissible for any integer value of $p$.\n\\end{itemize}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=4in]{graphics/model_rail.png}\n\\end{center}\n\\caption[xkcd 878: model rail]{\\url{http://xkcd.com/878} Math is not limited by physics, however.}\n\\label{fig:xkcd878}\n\\end{figure}\n", "meta": {"hexsha": "a6f9327cb019e52bbcb2c3b6eeb3174acf51b78b", "size": 12334, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "basics.tex", "max_stars_repo_name": "wmacevoy/littleo", "max_stars_repo_head_hexsha": "8caebe404f2ddf94ef05acbc4f3109159b964c1a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2016-05-23T08:32:04.000Z", "max_stars_repo_stars_event_max_datetime": "2016-05-23T08:32:04.000Z", "max_issues_repo_path": "basics.tex", "max_issues_repo_name": "wmacevoy/littleo", "max_issues_repo_head_hexsha": "8caebe404f2ddf94ef05acbc4f3109159b964c1a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "basics.tex", "max_forks_repo_name": "wmacevoy/littleo", "max_forks_repo_head_hexsha": "8caebe404f2ddf94ef05acbc4f3109159b964c1a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.5585585586, "max_line_length": 468, "alphanum_fraction": 0.6717204475, "num_tokens": 4434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124812, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.641645020824643}} {"text": "\\chapter*{Add Noise Plugin}\n\\setcounter{chapter}{1}\n\\emph{Add some noise}\n\\section{Introduction}\nThe purpose of the \\emph{AddNoise} plugin is to introduce random noise to \\emph{Tellurium} data.\n\nNoise generation uses the fact that a Rayleigh-distributed random variable $R$, with\nthe probability distribution $F(R) = 0$ if $R < 0$ and $F(R) = 1 - exp(-R^2/2*\\sigma^2)$ if $R >= 0, $\nis related to a pair of Gaussian variables $C$ and $D$ through the transformation $C = R * cos(\\theta)$ and\n$D = R * sin(\\theta)$, where $\\theta$ is a uniformly distributed variable\nin the interval $(0, 2*\\pi())$ \\footnote{From Contemporary Communication Systems\nUSING MATLAB(R), by John G. Proakis and Masoud Salehi, published by\nPWS Publishing Company, 1998, pp 49-50.}\n\nCurrently only Gaussian noise is implemented.\n\n\\section{Plugin Parameters}\nTable \\ref{table:AddNoisePluginParameters} lists available plugin property names, along with their data type and purpose.\n\n\n\\begin{table}[ht]\n\\centering % used for centering table\n\\begin{tabular}{l l p{7.5cm}} % centered columns (4 columns)\n\nParameter Name & Data Type & Purpose \\\\ [0.5ex] % inserts table\n%heading\n\\hline % inserts single horizontal line\nInputData \t\t& \tTelluriumData & Data on which noise will be applied to. \\\\\nSigma,($\\sigma$) \t& \tdouble & Size of applied noise. Noise is generated for each single data value, with a probability corresponding to a Gaussian distribution, centered around the value, and with a variance equal to $\\sigma^2$ .\\\\\nNoiseType \t& \tint & Type of noise applied on data. Only Gaussian noise is currently supported. \\\\\nProgress \t& \tdouble & The progress property communicates the progress (in percent) of Noise application. \\\\\n\n\\hline %inserts single line\n\\end{tabular}\n\\caption{Add noise Plugin Parameters}\n\\label{table:AddNoisePluginParameters}\n\\end{table}\n\n\\section{Plugin Events}\nThe AddNoiseplugin uses all of the available plugin events, i.e. the \\emph{PluginStarted}, \\emph{PluginProgress} and the \\emph{PluginFinished} events.\n\nThe available data variables for each event are internally treated as \\emph{pass trough} variables, so any data, for any of the events, assigned prior to\nthe plugin's execute function (in the assignOn() family of functions), can be retrieved \\emph{unmodified} in the corresponding event function.\n\\begin{table}[ht]\n\\centering % used for centering table\n\\begin{tabular}{l l p{7.5cm}} % centered columns (4 columns)\n\nEvent & Arguments & Purpose \\\\ [0.5ex] % inserts table\n%heading\n\\hline % inserts single horizontal line\n\\hline % inserts single horizontal line\nPluginStarted \t& \tvoid*, void* & Signals to application that the plugin has started applying noise on data. Both parameters are \\emph{pass through} parameters and are unused internally by the plugin.\\\\[0.5ex]\nPluginProgress\t& \tvoid*, void* & Communicates progress of noise generation. Both parameters are \\emph{pass through} parameters and are unused internally by the plugin. \\\\[0.5ex]\nPluginFinished\t& \tvoid*, void* & Signals to application that execution of the plugin has finished. Both parameters are \\emph{pass through} parameters and are unused internally by the plugin.\\\\\n\n\\hline %inserts single line\n\\end{tabular}\n\\caption{AddNoise Plugin Events}\n\\label{table:AddNoisePluginEvents}\n\\end{table}\n\n\\section{The \\texttt{execute(bool inThread)} function}\nThe \\verb|execute()| function will apply noise to all rows and columns of the assigned data, with one exception. Data not affected are data in the first column, and if, and only if, its column header equals \"time\" (case insensitive).\n\nThe \\verb|execute(bool inThread)| method supports a boolean argument indicating if the execution of the plugin work will be done in a thread, or not. Threading is fully implemented in the AddNoise plugin.\n\nThe inThread argument defaults to \\textbf{false}.\n\n\\section{Python examples}\n\n\\subsection{Add noise to data acquired from RoadRunner}\nThe python script below shows how to acquire simulation data from RoadRunner and pass it to the noise plugin. The format of this data, which is obtained from the \\verb|simulate()| function (line 8), is not directly compatible with the Noise plugin's InputData property. This incompatibility is handled by an intermediate data structure in Python, called DataSeries (line 14).\n\nThe plugin's properties, InputData and Sigma, are assigned on line 17 and 20 respectively.\n\nLine 23 denotes the execution of the noise plugin, and after that has finished, data can be visualized by using the \\verb|plot| function (line 26). The output is shown below the script.\n\n\\begin{singlespace}\n\\lstinputlisting[label=plugin_addNoise_header,caption={Add noise example.},language=Python]{Examples/telNoisePluginEx1.py}\n\\end{singlespace}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\textwidth]{AddNoise.png}\n\\caption{Output for the AddNoise python example script discussed above}\n\\label{fig:addNoiseFig1}\n\\end{figure}\n\n\\subsection{Visualization of the noise distribution used in the AddNoise plugin}\nThe Python script below demonstrates how to obtain and visualize the actual distribution (Gaussian) of noise that is applied on data.\n\n\\begin{singlespace}\n\\lstinputlisting[label=plugin_addNoise_header,caption={Noise distribution example.},language=Python]{Examples/telNoisePluginEx2.py}\n\\end{singlespace}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\textwidth]{AddNoise2.png}\n\\caption{Output for the AddNoise python example script discussed above}\n\\label{fig:addNoiseFig2}\n\\end{figure}\n", "meta": {"hexsha": "1c5386fbafbc9fd6f015c879939fee0a83739704", "size": 5515, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "plugins/released/add_noise/docs/add_noise.tex", "max_stars_repo_name": "sys-bio/rrplugins", "max_stars_repo_head_hexsha": "03af6ea70d73462ad88103f1e446dc0c5f3f971c", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "plugins/released/add_noise/docs/add_noise.tex", "max_issues_repo_name": "sys-bio/rrplugins", "max_issues_repo_head_hexsha": "03af6ea70d73462ad88103f1e446dc0c5f3f971c", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2015-12-02T18:20:43.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-20T17:13:34.000Z", "max_forks_repo_path": "plugins/released/add_noise/docs/add_noise.tex", "max_forks_repo_name": "sys-bio/telPlugins", "max_forks_repo_head_hexsha": "03af6ea70d73462ad88103f1e446dc0c5f3f971c", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2015-01-27T18:53:45.000Z", "max_forks_repo_forks_event_max_datetime": "2015-07-13T17:07:50.000Z", "avg_line_length": 54.068627451, "max_line_length": 375, "alphanum_fraction": 0.7718948323, "num_tokens": 1377, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916029436189, "lm_q2_score": 0.7431680029241321, "lm_q1q2_score": 0.6416450133010745}} {"text": "\\subsection{Group presentations}\\label{subsec:group_presentations}\n\n\\begin{definition}\\label{def:free_monoid}\n Let \\( \\mscrS \\) be an arbitrary set. We associate with \\( \\mscrS \\) its \\term{free monoid} \\( F(\\mscrS) \\coloneqq (\\mscrS^{\\ast}, \\cdot) \\), where \\( \\mscrS^{\\ast} \\) is the \\hyperref[def:formal_language/kleene_star]{Kleene star} and \\( \\cdot \\) is \\hyperref[def:formal_language/concatenation]{concatenation}. It is a monoid due to \\fullref{thm:kleene_star_is_monoid}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:free_monoid_is_free_functor}\n The functor \\( F: \\cat{Set} \\to \\cat{Mon} \\), defined pointwise in \\fullref{def:free_monoid}, is \\hyperref[def:category_adjunction]{free}.\n\\end{proposition}\n\\begin{proof}\n Let \\( U: \\cat{Mon} \\to \\cat{Set} \\) be the corresponding forgetful functor and let \\( \\mscrM \\in \\cat{Mon} \\), \\( \\mscrS \\in \\cat{Set} \\). We will first show that\n \\begin{equation*}\n \\cat{Mon}(F(\\mscrS), \\mscrM) = \\cat{Set}(\\mscrS, U(\\mscrM)),\n \\end{equation*}\n where equality means that all the underlying functions are equal.\n\n Every monoid homomorphism is a function, so obviously\n \\begin{equation*}\n \\cat{Mon}(F(\\mscrS), \\mscrM) \\subseteq \\cat{Set}(\\mscrS, U(\\mscrM)).\n \\end{equation*}\n\n Now consider a function \\( f: \\mscrS \\to U(\\mscrM) \\). Define the function\n \\begin{balign*}\n &\\varphi: F(\\mscrS) \\to \\mscrM \\\\\n &\\varphi\\left( \\{ x_k \\}_{k \\in \\mscrK} \\right) \\coloneqq \\prod_{k \\in \\mscrK} x_k.\n \\end{balign*}\n\n Obviously \\( \\varphi \\) is a homomorphism from \\( F(\\mscrS) \\) to \\( \\mscrM \\). Hence,\n \\begin{equation*}\n \\cat{Set}(\\mscrS, U(\\mscrM)) \\subseteq \\cat{Mon}(F(\\mscrS), \\mscrM).\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:free_group}\n Let \\( \\mscrS \\) be an arbitrary set. We will now construct the \\term{free group} \\( F(\\mscrS) \\) of \\( \\mscrS \\). The construction is similar to that of \\hyperref[def:free_monoid]{free monoids}, but it is much more complicated because of special reduction rules for \\hyperref[def:unital_magma_inverse_element]{inverse elements}. Refer to \\cite{code:free_group_grammar_verification} for a, software implementation of the construction.\n\n Let \\( \\star \\) be a \\hyperref[def:formal_language/symbol]{symbol} not in \\( \\mscrS \\). Our goal is, for each \\( a \\in S \\), to make the word \\( a{\\star} \\) behave like the inverse of \\( a \\) in a group. Rather than considering the \\hyperref[def:formal_language/kleene_star]{Kleene star} \\( (S \\cup \\{ \\star \\})^* \\) and removing elements via \\enquote{reductions} as in \\cite{code:free_group_reduction_verification} and \\cite[306]{Knapp2016BasicAlgebra}, we directly build a language of \\term{reduced words} using the mutually recursive \\hyperref[def:formal_grammar]{grammar}\n \\begin{alignedeq}\\label{eq:def:free_group/grammar}\n &I \\to \\varepsilon, && && \\text{\\( I \\) is the initial state} \\\\\n &I \\to S_a \\mid D_a, && a \\in S && \\\\\n &S_a \\to a \\mid a S_a, && a \\in S && S_a \\text{ does not produce words beginning with } a\\star \\\\\n &S_a \\to a D_b, && a, b \\in S, a \\neq b && \\\\\n &D_a \\to a\\star S_b, && a, b \\in S, a \\neq b && D_a \\text{ does not produce words beginning with } a \\\\\n &D_a \\to a\\star \\mid a\\star D_a, && a \\in S && \\\\\n \\end{alignedeq}\n\n The \\term{free group} \\( F(\\mscrS) \\) is defined to be the language of \\eqref{eq:def:free_group/grammar} equipped with the inductively defined operation\n \\begin{equation}\\label{eq:def:free_group/operation}\n w_1 \\odot w_2 \\coloneqq \\begin{cases}\n p \\odot s, &w_1 = p a \\T{and} w_2 = a\\star s \\text{ for some } a \\in S, \\\\\n p \\odot s, &w_1 = p a\\star \\T{and} w_2 = as \\T{and} s \\neq \\star t \\text{ for some } a \\in S, \\\\\n ps, &\\text{otherwise}.\n \\end{cases}\n \\end{equation}\n\n The inverse of the word \\( w = a_1 \\ldots a_n \\) is \\( w^{-1} \\coloneqq b_1 \\ldots b_n \\), where\n \\begin{equation}\\label{eq:def:free_group/inverse}\n b_{n-k+1} \\coloneqq \\begin{cases}\n \\varnothing, &a_k = {\\star} \\\\\n a_k{\\star}, &a_k \\neq {\\star} \\T{and} k = n \\\\\n a_k{\\star}, &a_k \\neq {\\star} \\T{and} a_{k+1} \\neq {\\star} \\\\\n a_k, &a_k \\neq {\\star} \\T{and} k \\neq n \\T{and} a_{k+1} = {\\star} \\\\\n \\end{cases}\n \\end{equation}\n for \\( k = 1, \\ldots, n \\).\n\n The group \\( (F(\\mscrS), \\odot) \\) is called the \\term{free group} generated by \\( \\mscrS \\).\n\\end{definition}\n\\begin{proof}\n The proof of the well-definedness of the group structure of \\( F(\\mscrS) \\) is a straightforward (but tedious) application of induction.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:free_group_is_free_functor}\n The functor \\( F: \\cat{Set} \\to \\cat{Grp} \\), defined pointwise in \\fullref{def:free_group}, is \\hyperref[def:category_adjunction]{free}.\n\\end{proposition}\n\\begin{proof}\n The outline of the proof is similar to the proof of \\fullref{thm:free_monoid_is_free_functor}.\n\\end{proof}\n\n\\begin{definition}\\label{def:group_presentation}\\mcite[314]{Knapp2016BasicAlgebra}\n Let \\( \\mscrS \\) be a set, \\( F(\\mscrS) \\) be the \\hyperref[def:free_group]{free group} and \\( \\mscrR \\subseteq F(\\mscrS) \\) be a subset. Denote by \\( \\mscrN(\\mscrR) \\) the smallest normal subgroup of \\( F(\\mscrS) \\) that includes \\( \\mscrR \\) as a subset.\n\n We define the group\n \\begin{equation}\\label{eq:def:group_presentation/presentation}\n \\mscrG = \\braket{ \\mscrS \\mid \\mscrR} \\coloneqq F(\\mscrS) / \\mscrN(\\mscrR)\n \\end{equation}\n called the group with \\term{generators} \\( \\mscrS \\) and \\term{relators} \\( \\mscrR \\). The expression \\eqref{eq:def:group_presentation/presentation} is called a \\term{presentation} of \\( \\mscrG \\).\n\n If there exists a presentation for \\( \\mscrG \\) such that \\( \\mscrS \\) is finite, it is called a \\term{finitely generated} group. If there exists a presentation such that both \\( \\mscrS \\) and \\( \\mscrR \\) are finite, it is called \\term{finitely presented}.\n\n If \\( \\mscrR = \\varnothing \\), there are no restrictions and we use the notation\n \\begin{equation}\\label{eq:def:group_presentation/free}\n \\mscrG = \\braket{ \\mscrS } \\coloneqq F(\\mscrS)\n \\end{equation}\n for the free group.\n\\end{definition}\n\n\\begin{theorem}\\label{thm:every_group_is_representable}\\mcite[prop. 7.7]{Knapp2016BasicAlgebra}\n Every group \\( \\mscrG \\) has at least one \\hyperref[def:group_presentation]{presentation}.\n\\end{theorem}\n\\begin{proof}\n Let \\( \\mscrG \\) be an arbitrary group and let \\( \\mscrS \\coloneqq U(\\mscrG) \\) be the underlying set. Let \\( F(\\mscrS) \\) be the corresponding free group with \\( \\iota: \\mscrS \\to F(\\mscrS) \\) sending elements of \\( \\mscrS \\) to singleton words in \\( F(\\mscrS) \\). By \\fullref{thm:free_group_is_free_functor}, there exists a unique homomorphism \\( \\varphi: F(\\mscrS) \\to \\mscrG \\) such that\n \\begin{equation*}\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n beginfig(1);\n input metapost/graphs;\n\n v1 := thelabel(\"$\\mscrS$\", origin);\n v2 := thelabel(\"$U(F(\\mscrS))$\", (-1, -1) scaled u);\n v3 := thelabel(\"$U(G)$\", (1, -1) scaled u);\n\n a1 := straight_arc(v1, v2);\n a2 := straight_arc(v1, v3);\n\n d1 := straight_arc(v2, v3);\n\n draw_vertices(v);\n draw_arcs(a);\n\n drawarrow d1 dotted;\n\n label.ulft(\"$\\iota$\", straight_arc_midpoint of a1);\n label.urt(\"$\\id$\", straight_arc_midpoint of a2);\n label.top(\"$U(\\varphi)$\", straight_arc_midpoint of d1);\n endfig;\n \\end{mplibcode}\\fi\n \\end{equation*}\n that is, \\( U(\\varphi) \\circ \\iota = \\id \\). Thus, \\( G = \\mscrS \\subseteq \\ker \\varphi \\). Define \\( \\mscrR \\coloneqq \\ker \\varphi \\). By \\fullref{def:normal_subgroup}, \\( \\mscrR \\) is a normal subgroup of \\( F(\\mscrS) \\), thus\n \\begin{equation*}\n G = \\varphi(F(\\mscrS)) \\cong F(\\mscrS) / \\ker \\varphi = \\braket{ \\mscrS \\mid \\mscrR }.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:cyclic_group}\n For a singleton alphabet \\( \\set{ a } \\), we define the \\term{infinite cyclic group}\n \\begin{equation*}\n C \\coloneqq \\braket{a}\n \\end{equation*}\n and, for positive integers \\( n \\), the \\term{finite cyclic group} of \\term{order} \\( n \\) as\n \\begin{equation*}\n C_n \\coloneqq \\braket{a \\given a^n}.\n \\end{equation*}\n\n We use the same notation independent of \\( a \\) because all cyclic groups of the same order are obviously \\hyperref[def:group/homomorphism]{isomorphic}.\n\n See \\fullref{thm:cyclic_group_isomorphic_to_integers_modulo_n}.\n\\end{definition}\n\n\\begin{definition}\\label{def:group_free_product}\\mcite[323]{Knapp2016BasicAlgebra}\n The \\term{free product} of a nonempty family of groups \\( \\seq{ \\mscrX_k }_{k \\in \\mscrK} \\) with presentations \\( \\braket{\\mscrS_k \\mid \\mscrR_k}, k \\in \\mscrK \\) is the group\n \\begin{equation*}\n \\Ast_{k \\in \\mscrK} \\mscrX_k \\coloneqq \\braket*{ \\coprod_{k \\in \\mscrK} \\mscrS_k \\given* \\coprod_{k \\in \\mscrK} \\mscrR_k },\n \\end{equation*}\n where \\( \\coprod \\) is the \\hyperref[def:disjoint_union]{disjoint union}.\n\\end{definition}\n\n\\begin{definition}\\label{def:free_abelian_group}\n A \\term{free abelian group} is a \\hyperref[def:free_left_module]{free} \\hyperref[thm:abelian_group_iff_z_module]{\\( \\BbbZ \\)-module}. This definition of a free abelian group is different from the definition of a \\hyperref[def:free_group]{free group}.\n\\end{definition}\n", "meta": {"hexsha": "bc42b108a47457e81aa153dcbc3a6841893f1eff", "size": 9342, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/group_presentations.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/group_presentations.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/group_presentations.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.3875, "max_line_length": 577, "alphanum_fraction": 0.660458146, "num_tokens": 3219, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916064587, "lm_q2_score": 0.743167997235783, "lm_q1q2_score": 0.6416450110020974}} {"text": "\\documentclass[11pt,twoside]{article}\n\n\\usepackage[headings]{fullpage}\n\\usepackage[utopia]{mathdesign}\n\n\\pagestyle{myheadings}\n\\markboth{Chemotaxis}{Chemotaxis}\n\n\\input{../../fncextra}\n\n\\begin{document}\n \n\\begin{center}\n \\bf You've been slimed\n\\end{center}\n\nSlime mold cells are able to spontaneously aggregate by secreting a chemical attractant in a process called \\emph{chemotaxis}. A nondimensional 1D model for the process is\n\\begin{equation}\n\\label{eq:1}\n \\begin{split}\n u_t &= u_{xx} - \\chi (uv_x)_x, \\qquad -1\\le x \\le 1, \\quad t\\ge0, \\\\\n v_t &= \\epsilon v_{xx} + u - v,\n \\end{split}\n\\end{equation}\nwhere $u(x,t)$ represents a scaled cell density function, $v(x,t)$\nrepresents the attractant concentration, and $\\chi$ and $\\epsilon$ are nonnegative constants. We will use periodic end conditions. \n\nIn order to apply the method of lines to~\\eqref{eq:1}, first discretize in space using finite differences to get\n\\begin{equation}\n\\label{eq:mol}\n \\begin{split}\n \\bfu' &= \\bff\\,(\\bfu,\\bfv) = \\mD_{xx} \\bfu - \\chi \\mD_x \\bigl[\\bfu \\odot \\bigl(\\mD_{x} \\bfv \\bigr)\\bigr] \n \\\\\n \\bfv' &= \\bfg\\,(\\bfu,\\bfv) = \\epsilon \\mD_{xx} \\bfv + \\bfu - \\bfv,\n \\end{split}\n\\end{equation}\nwhere $\\odot$ is the elementwise product \\verb|.*| in MATLAB. Choosing the Euler time stepping method, for example, leads to this iteration in the time step $j$:\n\\begin{equation}\n\\label{eq:euler}\n\\begin{split}\n\\bfu_{j+1} &= \\bfu_j + \\tau \\bff\\,(\\bfu_j,\\bfv_j) \\\\\n\\bfv_{j+1} &= \\bfv_j + \\tau \\bfg\\,(\\bfu_j,\\bfv_j).\n\\end{split}\n\\end{equation}\n\nTo apply a built-in IVP solver, equation~\\eqref{eq:mol} has to be expressed as a single first-order vector equation. The two ``variables'' of the system are actually the two $m$-vectors $\\bfu$ and $\\bfv$, which are combined into a column vector $\\bfz=\\bigl[ \\bfu\\;;\\; \\bfv \\bigr]$, leading to a system in the standard form $\\bfz'=\\bfF\\,(t,\\bfz)$. \n\n\\subsection*{Goals}\n\nYou will simulate the chemotaxis of slime mold and observe concentration of cells around initially small increases in attractant. You will use both the Euler and \\texttt{ode15s} timestepping methods.\n\n\\subsection*{Preparation}\n\nRead section 11.4. \n\n\\subsection*{Procedure}\n\nThroughout all of these steps, let $\\chi=6$ and $\\epsilon=0.1$ in all the equations above. \n\n\\begin{enumerate}\n\\item Let $m=50$, $u(x,0)=1$, and $v(x,0)=1+0.1e^{-100x^2}$, representing a small drop of extra attractant. (Note: the initial $v$ is not truly periodic, but the exponential is so small at $x=\\pm1$ that is practically so.) Use the Euler timestepping method as shown in~\\eqref{eq:euler} to solve over $0\\le t \\le0.05$ with $n=100$ time steps. Make a waterfall plot of the solution.\n \n\\item Now you will repeat step~1 with $m=200$ and for $0\\le t \\le 0.6$. Due to absolute stability restrictions, there is a minimum acceptable value of $n$. Start with $n=2000$ and increment $n$ repeatedly by 500 until the solution at the final time is finite. (Use the syntax \\texttt{all(isfinite(w))} to determine if a vector is finite.) Output the minimum stable $n$ and plot $u(x,0.6)$. \n \n\\item The next goal is to use the built-in solver \\texttt{ode15s} that works well for stiff problems. To do this, write a function \\texttt{timederiv.m} with headline\n \\begin{verbatim}\n function dzdt = timederiv(u,v,chi,Dx,Dxx)\n \\end{verbatim}\t\n and returning $\\bfz'=\\bigl[ \\bfu'\\;;\\; \\bfv' \\bigr]$ for the given inputs.\n \n\\item Now solve (with the same $m$ and initial conditions as step~2) using \\texttt{ode15s} for $0\\le t \\le 15$. You will need to write an anonymous function of $t$ and $\\bfz$ only that encapsulates a call to \\texttt{timederiv}. Print out the number of time steps that were taken and make a waterfall plot of the solution.\n \n\\item You would expect that if there were two symmetrically placed drops of attractant, the solution would remain symmetric for all time. However, that solution is unstable in time and one of the drops will eventually ``win.'' Repeat step~4 using $$v(x,0)=1+0.1e^{-100(x-0.5)^2}+0.1e^{-100(x+0.5)^2}.$$\n \n\\end{enumerate}\n\n% \\subsection*{Extras}\n% \\begin{enumerate}\n% \\item[E1.] Observe that\n% $$\\frac{d}{dt} \\int_{-1}^{1} u\\, dx = \\int_{-1}^{1} u_t \\,dx = \\int_{-1}^{1} \\bigl[u_x - \\chi(uv_x)\\bigr]_x \\,dx = 0,$$\n% due to the periodicity in $x$. Thus the total ``mass'' $M(t)=\\int u\\,dx$ is conserved in the exact solution. Use the trapezoidal quadrature formula to make a plot of $M(t)$ for the numerical solution of step~5 and check the conservation property. Make sure you apply the formula correctly over the entire interval and account for the periodicity of $u$.\n% \\end{enumerate}\n\n\n\n\\end{document}\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: t\n%%% End: \n\n\n", "meta": {"hexsha": "8d3fe806176f8eee9aa2bf1e83118058aa0dcddc", "size": 4722, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "labs/chapter11/Chemotaxis/Chemotaxis.tex", "max_stars_repo_name": "snowdj/fnc-extras", "max_stars_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 109, "max_stars_repo_stars_event_min_datetime": "2018-04-21T09:02:50.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-20T19:03:54.000Z", "max_issues_repo_path": "labs/chapter11/Chemotaxis/Chemotaxis.tex", "max_issues_repo_name": "snowdj/fnc-extras", "max_issues_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2018-12-04T22:17:44.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-03T21:04:47.000Z", "max_forks_repo_path": "labs/chapter11/Chemotaxis/Chemotaxis.tex", "max_forks_repo_name": "snowdj/fnc-extras", "max_forks_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 49, "max_forks_repo_forks_event_min_datetime": "2017-04-02T17:21:33.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-29T17:19:57.000Z", "avg_line_length": 50.2340425532, "max_line_length": 390, "alphanum_fraction": 0.6994917408, "num_tokens": 1519, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.8311430499496096, "lm_q1q2_score": 0.6415123588777435}} {"text": "%!TEX root = TTK4150-Summary.tex\n\\section{Feedback linearisation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Input-output linearisation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\begin{align}\n\t\\dot{x} &= f(x) + g(x)u \\label{eq:siso-x} \\\\\n\ty &= h(x) \\label{eq:siso-y}\n\\end{align}\n\n\\paragraph{Definition 13.2 (relative degree)}\nThe relative degree $\\rho$ of \\eqref{eq:siso-x}--\\eqref{eq:siso-y} is equal to the number of times $y$ must be differentiated until $u$ appears.\n\n\\paragraph{Theorem 13.1}\nThe system \\eqref{eq:siso-x} can be input-output linearised if the relative degree is well defined in the region of interest $\\mathbb{D}_0$.\n\n\\paragraph{Minimum phase}\nThe system \\eqref{eq:siso-x}--\\eqref{eq:siso-y} is \\emph{minimum phase} if, for the zero-dynamics $\\dot{\\eta} = f_0(\\eta,0)$, $\\eta = 0$ is AS.\n\n\\paragraph{Method}\nThe system is given by \\eqref{eq:siso-x}--\\eqref{eq:siso-y}. First determine relative degree $\\rho$. Define\n\\begin{equation}\n\t\\xi =\n\t\\begin{bmatrix}\n\t\t\\xi_1 \\\\ \\vdots \\\\ \\xi_\\rho\n\t\\end{bmatrix}\n\t=\n\t\\begin{bmatrix}\n\t\ty \\\\ \\vdots \\\\ y^{(\\rho-1)}\n\t\\end{bmatrix}\n\\end{equation}\nso that\n\\begin{equation}\n\t\\dot{\\xi} =\n\t\\begin{bmatrix}\n\t\t\\dot{\\xi}_1 \\\\ \\vdots \\\\ \\dot{\\xi}_\\rho\n\t\\end{bmatrix}\n\t=\n\t\\begin{bmatrix}\n\t\t\\xi_2 \\\\ \\vdots \\\\ L_f^\\rho h + L_g L_f^{\\rho-1} h \\cdot u\n\t\\end{bmatrix}\n\t.\n\\end{equation}\nThen choose $n-\\rho$ coordinates\n\\begin{equation}\n\t\\eta =\n\t\\begin{bmatrix}\n\t\t\\eta_1 \\\\ \\vdots \\\\ \\eta_{n-\\rho}\n\t\\end{bmatrix}\n\\end{equation}\nwith the coordinate transformation\n\\begin{equation}\n\tz = T(x) =\n\t\\begin{bmatrix}\n\t\t\\eta_1 \\\\ \\vdots \\\\ \\eta_{n-\\rho} \\\\ \\xi_1 \\\\ \\vdots \\\\ \\xi_\\rho\n\t\\end{bmatrix}\n\\end{equation}\nand choose $\\eta$ such that\n\\begin{itemize}\n\t\\item $T$ is a diffeomorphism,\n\t\\item $L_g \\eta_i = 0$,\n\t\\item $\\eta_i(0) = 0$.\n\\end{itemize}\nWe have then that\n\\begin{equation}\n\t\\dot{\\eta}_j = \\pd{\\eta_j}{x} \\dot{x} = L_f \\eta_j + \\underbrace{L_g \\eta_j}_{= 0} \\cdot u = f_{0_j} (\\eta_i, \\xi_i)\n\\end{equation}\nand we can write the system on normal form\n\\begin{equation}\n\t\\begin{bmatrix}\n\t\t\\dot{\\eta}_1 \\\\\n\t\t\\vdots \\\\\n\t\t\\dot{\\eta}_{n-\\rho} \\\\\n\t\t\\dot{\\xi}_1 \\\\\n\t\t\\vdots \\\\\n\t\t\\dot{\\xi}_\\rho\n\t\\end{bmatrix}\n\t=\n\t\\begin{bmatrix}\n\t\tf_{0_1}(\\eta_i, \\xi_i) \\\\\n\t\t\\vdots \\\\\n\t\tf_{0_{n-\\rho}}(\\eta_i, \\xi_i) \\\\\n\t\t\\xi_2 \\\\\n\t\t\\vdots \\\\\n\t\tL_f^\\rho h + L_g L_f^{\\rho-1} h \\cdot u\n\t\\end{bmatrix}\n\t.\n\\end{equation}\nThen choose $u$ to cancel the nonlinearities:\n\\begin{equation}\n\tu = \\frac{1}{L_g L_f^{\\rho-1} h} (-L_f^\\rho h + v)\n\\end{equation}\nThis leads to\n\\begin{gather}\n\t\\dot{\\eta} = f_0(\\eta,\\xi) \\\\\n\t\\dot{\\xi} = \n\t\\begin{bmatrix}\n\t\t\\xi_2 \\\\ \\vdots \\\\ v\n\t\\end{bmatrix}\n\\end{gather}\nThen analyse the zero-dynamics---the internal dynamics when the output is kept zero by the input:\n\\begin{equation}\n\t\\dot{\\eta} = f_0(\\eta,0)\n\\end{equation}", "meta": {"hexsha": "9a8999cf611ae1faf3a1ee9eaae2f6322af1f82f", "size": 2891, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-lin.tex", "max_stars_repo_name": "jakoblover/ntnu-course-summaries", "max_stars_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-05-30T09:19:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-11T02:42:40.000Z", "max_issues_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-lin.tex", "max_issues_repo_name": "jakoblover/ntnu-course-summaries", "max_issues_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-lin.tex", "max_forks_repo_name": "jakoblover/ntnu-course-summaries", "max_forks_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.5333333333, "max_line_length": 144, "alphanum_fraction": 0.6015219647, "num_tokens": 1100, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.641512355169013}} {"text": "\n\\subsection{The Lorentz group}\n\nThe Lorentz group is the \\(O(1,3)\\) group.\n\\subsubsection{Symmetries of the Lorentz group}\n\nWe can do the usual \\(3\\) rotations, however there are additional \\(3\\) symmetries, making the Lorzentz group \\(6\\)-dimensional.\n\nThese are the Lorentz boosts.\n\nA symmetry has:\n\n\\(t'^2 - x'2 - y'^2 - z'^2 = t^2 - x^2 - y^2 - z^2\\)\n\nWe consider the case where we just boost on \\(x\\), so \\(y = y'\\) and \\(z = z'\\).\n\n\\(t'^2 - x'2 = t^2 - x^2\\)\n\nOr with \\(c\\):\n\n\\(c^2t'^2 - x'2 = ct^2 - x^2\\)\n\n\n\\subsubsection{New}\n\\(s^2 = t^2 - x^2 - y^z - z^2\\)\n\n\\(s'^2 = t'^2 - x'^2 - y'^2 - z'^2\\)\n\n\\(ds^2 = s'^2 - s^2\\)\n\n\\(ds^2 = (t'^2 - x'^2 - y'^2 - z'^2) - (t^2 - x^2 - y^z - z^2)\\)\n\n\\(ds^2 = (t'^2 - t^2) - (x'^2 - x^2) - (y'^2 - y^2) - (z'^2 - z^2)\\)\n\n\\(ds^2 = dt^2 - dx^2 - dy^2 - dz^2\\)\n\n\nboost:\n\\(s^2 = c^2t^2 - x^2 - y^z - z^2\\)\n\nwe want new t and x where distance is same\n\\(c^2t'^2 - x'^2 - y^z - z^2 = c^2t^2 - x^2 - y^z -z^2\\)\n\\(c^2t'^2 - x'^2 = c^2t^2 - x^2\\)\n\nWe know that both transformations are linear [WHY??], therefore\n\\(x' = Ax + Bt\\)\n\\(t' = Cx+Dt\\)\n\nwe transform to \\(x' = 0\\). so\n\\(Ax + Bt = 0\\)\n\nWe define\n\\(v = \\dfrac{x}{t}\\)\n\nSo:\n\\(x = vt\\)\n\nWe can plug these in:\n\\(Avt + Bt = 0\\)\n\\(Av + B = 0\\)\n\\(\\dfrac{A}{B} = -v\\)\n\n", "meta": {"hexsha": "fee0a2bd8b94b0874a2338d42191e667106e2406", "size": 1253, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/geometry/forms/02-07-lorentz.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/geometry/forms/02-07-lorentz.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/geometry/forms/02-07-lorentz.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 19.8888888889, "max_line_length": 128, "alphanum_fraction": 0.4996009577, "num_tokens": 586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8774767746654976, "lm_q2_score": 0.7310585727705127, "lm_q1q2_score": 0.6414869185262314}} {"text": "\\lab{The Arnoldi Iteration}{The Arnoldi Iteration}\n\\objective{The Arnoldi Iteration is an efficient method for finding the eigenvalues of extremely large matrices.\nInstead of using standard methods, the iteration uses Krylov subspaces to approximate how a linear operator acts on vectors.\nWith this approach, the Arnoldi Iteration facilitates the computation of eigenvalues for enormous matrices without needing to physically create the matrix in memory.\nWe will explore this subject by implementing the Arnoldi iteration algorithm, using our implementation for eigenvalue computation, and then graphically representing the accuracy of our approximated eigenvalues.}\n\n\\section*{Krylov Subspaces} % =================================================\n\nOne of the biggest difficulties in numerical linear algebra is the amount of memory needed to store a large matrix and the amount of time needed to read its entries.\nMethods using Krylov subspaces avoid this difficulty by studying how a matrix acts on vectors, making it unnecessary in many cases to create the matrix itself.\n\nThe \\emph{Arnoldi Iteration} is an algorithm for finding an orthonormal basis of a Krylov subspace.\nOne of its strengths is that it can run on any linear operator without knowing the operator's underlying matrix representation.\nThe outputs of the Arnoldi algorithm can then be used to approximate the eigenvalues of the matrix of the linear operator.\n\nThe order-$n$ Krylov subspace of $A$ generated by $\\x$ is\n\\[\n\\mathcal{K}_n(A, \\x) =\\text{span} \\{\\x, A\\x, A^2\\x, \\ldots, A^{n-1}\\x\\}.\n\\]\nIf the vectors $\\{\\x, A\\x, A^2\\x, \\ldots, A^{n-1}\\x\\}$ are linearly independent, then they form a basis for $\\mathcal{K}_n(A,\\x)$.\nHowever, $A^n x$ frequently converges to a dominant eigenvector of $A$ as $n$ gets large, which fills the basis with many almost parallel vectors.\nThis yields a basis prone to ill-conditioned computations and numerical instability.\n\n\\section*{The Arnoldi Iteration Algorithm} % ==================================\n\nThe Arnoldi iteration focuses on efficiently creating an orthonormal basis for $\\mathcal{K}_n(A,\\x)$ by integrating the creation of $\\{\\x, A\\x, A^2\\x, \\ldots, A^{n-1}\\x\\}$ with the modified Gram-Schmidt algorithm.\nThis process yields an orthonormal basis for $\\mathcal{K}_n(A,\\x)$ that can be used for further computations.\n%A brief walkthrough of this method and a psuedo code representation of it are contained below.\n\n\\begin{comment}\nBefore discussing the specific uses of the Krylov subspace in linear systems and eigenvalue problems, let us address a very practical concern: how can we best compute a basis for the Krylov subspace?\nThe obvious answer is simply to calculate the vectors $x, ax, a^2x, \\ldots, a^{n-1} x$, which we can accomplish using only matrix-vector multiplication.\nStraightforward though this may be, there is a major problem: $a^n x$ tends to converge to a dominant eigenvector of $a$ as $n$ gets large, and consequently these vectors become nearly parallel.\nThus, the basis $\\{x, ax, a^2x, \\ldots, a^{n-1} x\\}$ is far from orthogonal, and matrix computations associated with this basis will likely be ill-conditioned and prone to numerical instability.\nTo redress this problem, we may think to apply the Gram-Schmidt orthogonalization process to the basis, obtaining an orthonormal basis for the krylov subspace that enjoys much better numerical properties.\nThis turns out to be a useful thought, and is the basis for the Arnoldi iteration.\n\nYou may recall from lab \\ref{lab:qrdecomp} that the modified Gram-Schmidt algorithm allows us to find an increasing number of orthogonal vectors but does not require that we run the algorithm to its completion to find a full basis.\nOur goal is to find an orthonormal set of vectors $q_1,\\ldots,q_n$ having the same span as $x, ax, a^2x, \\ldots, a^{n-1} x$.\nWe start things off by setting\n\\[\nq_1 = \\frac{x}{\\|x\\|_2}.\n\\]\nnow, assuming we have obtained $q_1,\\ldots,q_n$, we obtain $q_{n+1}$ by projecting $a q_n$ onto the previous vectors, subtracting out these projections, and then normalizing.\nTo make this more precise, let $h_{i,n} = \\langle q_i, a q_n\\rangle$ for $i = 1,\\ldots, n$. subtract out these projections by calculating\n\\[\np_{n+1} = a^n x - \\sum_{i=1}^n h_{i,n}q_i.\n\\]\ndefine $h_{n+1,n} = \\|p_{n+1}\\|_2$, and normalize $p_{n+1}$ by calculating\n\\[\nq_{n+1} = \\frac{p_{n+1}}{h_{n+1,n}},\n\\]\nour next basis vector.\nThis procedure is outlined (in slightly more python-friendly notation) in algorithm \\ref{alg:arnoldi_iteration}.\n\nPerhaps you noticed a slight discrepancy between the Arnoldi iteration as described above, and the usual Gram-Schmidt procedure.\nSpecifically, you might have expected to compute $q_{n+1}$ by projecting $a^n x$, rather than $a q_n$, onto the previous vectors.\nThankfully, it is straightforward to show that our algorithm produces a valid orthonormal basis for the Krylov subspace, despite this difference, and because of this detail, we do not need to compute and store the original Krylov basis $x, ax, \\ldots, a^{n-1}x$.\nAdditionally, each iteration only requires one matrix-vector calculation, and the individual entries in $a$ are never referenced or modified.\nThus, even if the matrix $a$ is very large in theory, as long as we have a reasonably efficient subroutine to calculate $ax$ for any vector $x$, the Arnoldi iteration is computationally tractable.\n\nThis algorithm produces an orthonormal basis $q_1,\\ldots,q_n$ for the order-$n$ Krylov subspace generated by $a$ and $x$, as well as a collection of numbers $h_{i,j}$.\nIf we define a matrix $h_n$ whose $i,j$'th entry is $h_{i,j}$ for $i \\leq j+1$ and is $0$ otherwise, we now have an upper Hessenberg matrix.\nRecall that an upper Hessenberg matrix has the property that all entries below the first subdiagonal are equal to zero.\nAny square matrix is unitarily similar to an upper Hessenberg matrix.\nDealing with a Hessenberg matrix is often more convenient than dealing with a general matrix, especially when it comes to finding eigenvalues or solving systems of equations, since efficient algorithms designed for these types of matrices exist.\nIt turns out that there is a Hessenberg factorization of $a$, given by\n\\[\na = qhq^*,\n\\]\nWhere $q$ is a unitary matrix and $h$ is upper Hessenberg such that the first $n$ columns of $q$ are $q_1,\\ldots,q_n$, and the upper left $n \\times n$ submatrix of $h$ is equal to $h_n$.\nHence, the Arnoldi iteration provides a connection between the Krylov subspace and the Hessenberg factorization of a matrix.\nEach step in the Arnoldi iteration can be thought of as computing another step in the Hessenberg reduction of $a$.\nEach $h_n$ is really just the $n \\times n + 1$ upper-left block of $h$.\nSolving eigenvalue problems or systems of equations for a general square matrix can thus be reduced, via Arnoldi iteration, to solving these problems for a Hessenberg matrix, a much easier task.\n\nAt this point, we can view the Arnoldi iteration as a means to compute an orthonormal basis for a Krylov subspace, or alternatively, to compute a partial Hessenberg factorization of a matrix.\nBut in lab \\ref{lab:canonical_transformations}, we discussed how orthogonal transformations can be used to transform a matrix to upper Hessenberg form.\nWe were able to find the eigenvalues of such upper Hessenberg matrices in lab \\ref{lab:eigsolve}.\nSo what have we really gained by this new approach?\nThese previous approaches were based on matrix-matrix multiplication and required us to manipulate individual entries of the matrix.\nThe Arnoldi iteration avoids this and relies only on our ability to calculate matrix-vector multiplication.\nFurther, our present approach will allow us to compute only a partial Hessenberg factorization.\nThis is advantageous when, as is often the case, the behavior and properties of a matrix can be well-approximated by only a small portion of its Hessenberg form.\n\\end{comment}\n\nThe algorithm begins by initializing a matrix $H$ which will be an upper Hessenberg matrix and a matrix $Q$ which will be filled with the basis vectors of our Krylov subspace.\nIt also requires an initial vector $\\b \\neq 0$ which is normalized to get $\\q_1 = \\b/\\norm{\\b}$.\nThis represents the basis for the initial Krylov subspace, $\\mathcal{K}_1(A,\\b)$.\n\nFor the $k$th iteration, compute the next basis vector $\\q_{k+1}$ by using the modified Gram-Schmidt process to make $A\\q_k$ orthonormal to $\\q_k$.\nThis entails making each column of $Q$ orthogonal to $\\q_k$ before proceeding to the next iteration.\nThe vectors $\\{\\q_i\\}_{i=1}^k$ are then a basis for $\\mathcal{K}_k(A,\\b)$.\nIf $\\norm{\\q_{k+1}}$ is below a certain tolerance, stop and return $H$ and $Q$.\nOtherwise, normalize the new basis vector new $\\q_{k+1}$ and continue to the next iteration.\n% This process continues until we have an orthonormal basis of an order $k+1$ Krylov subspace.\n\n%This algorithm is described in algorithm \\ref{alg:arnoldi_iteration} for an order $k+1$ Krylov subspace, where $k$ is the number of times we multiply by $A$.\n\n\\begin{algorithm}\n\\begin{algorithmic}[1]\n\\Procedure{Arnoldi}{$\\b$, $A$, $k$, \\li{tol}}\n\t\\State $Q \\gets \\allocate{\\size{\\b}}{k+1}$\t\t\t\\Comment{Some initialization steps}\n\t\\State $H \\gets \\zeros{ k+1}{ k}$\n\t\\State $Q_{:,0} \\gets \\b/\\norm{\\b}_2$\n\t\\For{$j=0\\ldots k-1$}\t\t\t\t\t\t\t\\Comment{Perform the actual iteration.}\n\t\t\\State $Q_{:,j+1} \\gets A(Q_{:,j})$\n\t\t\\For{$i=0\\ldots j$}\t\t\t\t\t\\Comment{Modified Gram-Schmidt.}\n\t\t\t\\State $H_{i,j} \\gets Q_{:,i}\\hrm Q_{:,j+1}$\n\t\t\t\\State $Q_{:,j+1} \\gets Q_{:,j+1} - H_{i,j} Q_{:,i}$\n\t\t\\EndFor\n\t\t\\State $H_{j+1,j} \\gets \\norm{Q_{:,j+1}}_2$\t\t\t\\Comment{Set subdiagonal element of $H$.}\n \\If{$|H_{j+1,j}|<$ \\li{tol}}\t\t\t\t\t\\Comment{Stop if $\\norm{Q_{:,j+1}}_2$ is small enough.}\n\t\t\t\\State \\pseudoli{return} $H_{:j+1,:j+1}$, $Q_{:,:j+1}$\n\t\t\\EndIf\n\t\t\\State $Q_{:,j+1} \\gets Q_{:,j+1}/H_{j+1,j}$\t\t\t\t\\Comment{Normalize $\\q_{j+1}$.}\n\t\\EndFor\n\t\\State \\pseudoli{return} $H_{:-1, :}$, $Q$\t\t\t\\Comment{Return $H_k$ and $Q$.}\n\\EndProcedure\n\\end{algorithmic}\n\\caption{The Arnoldi iteration.\nThis algorithm accepts a square matrix $A$ and a starting vector $\\b$.\nIt iterates $k$ times or until the norm of the next vector in the iteration is less than \\li{tol}.\nThe algorithm returns an upper Hessenberg $H$ and an orthonormal $Q$ such that $H = Q^{\\mathsf{H}}AQ$.}\n\\label{alg:arnoldi_iteration}\n\\end{algorithm}\n\n\\begin{warn}\nIf the starting vector $\\x$ is an eigenvector of $A$ with corresponding eigenvalue $\\lambda$, then by definition $\\mathcal{K}_k(A, \\x)=\\text{span}\\{\\x, \\lambda \\x, \\lambda^2\\x, \\ldots, \\lambda^k \\x\\}$, which is equal to the span of $\\x$.\nSo, when $\\x$ is normalized with $\\q_1 = \\x/\\|\\x\\|$, $\\q_2 = A\\q_1 = \\lambda \\q_1$.\n\nThe vector $\\q_2$ is supposed to be the next vector in the orthonormal basis for $\\mathcal{K}_k(A, \\x)$, but it is not linearly independent of $\\q_1$.\nIn fact, $\\q_1$ already spans $\\mathcal{K}_k(A, \\x)$.\nHence, the Gram-Schmidt process fails and results in a \\li{ZeroDivisionError} or an extremely early termination of the algorithm.\nA similar phenomenon may occur if the starting vector $\\x$ is contained in a proper invariant subspace of $A$.\n\\end{warn}\n\n\\section*{Arnoldi Iteration on Linear Operators} % ============================\n\nA major strength of the Arnoldi iteration is that it can run on a linear operator, even without knowing the matrix representation of the operator.\nIf $L$ is some linear function, then we can modify the pseudocode above by replacing $AQ_{:,j}$ with $A_{mul}(Q_{:,j})$.\nThis makes it possible to find the eigenvalues of an arbitrary linear transformation.\n\n\\begin{problem}\\label{prob:arnoldi}\nWrite a function that accepts a starting vector $\\b$ for the Arnoldi Iteration, a function handle $L$ that describes a linear operator, the number of times $n$ to perform the iteration, and a tolerance \\li{tol} that defaults to $10^{-8}$.\nUse Algorithm \\ref{alg:arnoldi_iteration} to implement the Arnoldi Iteration with these parameters.\nReturn the upper Hessenberg matrix $H$ and the orthonormal matrix $Q$ from the iteration.\n\nConsider the following implementation details.\n\\begin{enumerate}\n\\item Since $H$ and $Q$ will eventually hold complex numbers, initialize them as complex arrays (e.g., \\li{A = np.empty((3,3), dtype=np.complex128)}).\n\\item This function can be tested on a matrix A by passing in \\li{A.dot} for a linear operator.\n\\item Remember to use complex inner products. Here is an example of how to evaluate $A^HA$:\n\\begin{lstlisting}\nb = A.conj() @ B\n\\end{lstlisting}\n\\end{enumerate}\nTest your function by comparing the resulting $H$ with $Q\\hrm A Q$.\n\\end{problem}\n\n\\section*{Finding Eigenvalues Using the Arnoldi Iteration} % ==================\n\nLet $A$ be an $n \\times n$ matrix.\nLet $Q_k$ be the matrix whose columns $\\q_1, \\ldots, \\q_k$ are the orthonormal basis for $\\mathcal{K}_m(A, \\x)$ generated by the Arnoldi algorithm, and let $H_k$ be the $k\\times k$ upper Hessenburg matrix defined at the $k$th stage of the algorithm.\nThen these matrices satisfy\n\\begin{equation}\\label{eq:arnoldi-hqa}\nH_k = Q_k^{\\mathsf H} A Q_k.\n\\end{equation}\nIf $k>> from scipy.sparse import linalg as spla\n>>> B = np.random.random((100,100))\n>>> spla.eigs(B, k=5, return_eigenvectors=false)\narray([ -1.15577072-2.59438308j, -2.63675878-1.09571889j,\n -2.63675878+1.09571889j, -3.00915592+0.j , 50.14472893+0.j ])\n\\end{lstlisting}\n\n\\subsection*{Convergence} % ---------------------------------------------------\nAs more iterations of the Arnoldi method are performed, our approximations are of higher rank.\nConsequently, the Ritz values become more accurate approximations to the eigenvalues of the linear operator.\n\nThis technique converges quickly to eigenvalues whose magnitude is distinctly larger than the rest.\nFor example, matrices with random entries tend to have one eigenvalue of distinctly greatest magnitude.\nConvergence of the Ritz values for such a matrix is plotted in Figure \\ref{fig:arnoldi_random_val_conv}.\n\nHowever, Ritz values converge more slowly for matrices with random eigenvalues.\nFigure \\ref{fig:arnoldi_random_eig_conv} plots convergence of the Ritz values for a matrix with eigenvalues uniformly distributed in $[0,1)$.\n\n\\begin{figure}\n\\centering\n\\begin{subfigure}[b]{.49\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{figures/rand_vals_conv.pdf}\n \\caption{}\n \\label{fig:arnoldi_random_val_conv}\n\\end{subfigure}\n\\begin{subfigure}[b]{.49\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{figures/rand_eigs_conv.pdf}\n \\caption{}\n \\label{fig:arnoldi_random_eig_conv}\n\\end{subfigure}\n\\caption{These plots show the relative error of the ritz values as approximations to the eigenvalues of a matrix.\nThe figure on the left plots the largest 15 Ritz values for a $500\\times 500$ matrix with random entries and demonstrates that the largest eigenvalue (the blue line) converges after 20 iterations.\nThe figure at right plots the largest 15 Ritz values for a $500\\times 500$ matrix with uniformly distributed eigenvalues in $[0,1)$ and demonstrates that all the eigenvalues take from 150 to 250 iterations to converge.}\n\\end{figure}\n\n% todo: this problem may be too difficult but we have added a great deal of supplementary code to make it more doable.\n\\begin{problem}\nWrite a function that accepts a linear operator $A$, the number of Ritz values to plot $n$, and the the number of times to perform the Arnoldi iteration \\li{iters}.\nUse these parameters to create a plot of the absolute error between the largest Ritz values of $A$ and the largest eigenvalues of $A$.\n\\begin{enumerate}\n \\item Find $n$ eigenvalues of $A$ of largest magnitude. Store these in order.\n \\item Create an empty array to store the relative errors for every $k=0,1,\\ldots,$ \\li{iters}.\n \\begin{enumerate}\n \t\\item Use your Ritz function to find the $n$ largest Ritz values of the operator. Note that for small $k$, the matrix $H_k$ may not have this many eigenvalues. Due to this, the graphs of some eigenvalues have to begin after a few iterations.\n \\item Store the absolute error between the eigenvalues of A and the Ritz values of H. Make sure that the errors are stored in the correct order.\n \\end{enumerate}\n %\\item Use array broadcasting to compute the absolute error.\n \\item Iteratively plot the errors for each eigenvalue with the range of the iterations.\n\\end{enumerate}\nHints:\nIf $\\tilde{\\x}$ is an an approximation to $\\x$, then the \\emph{absolute error} in the approximation is $\\|\\x - \\tilde{\\x}\\|$.\n\nSort your eigenvalues from greatest to least.\nAn example of how to do this is included:\n\\begin{lstlisting}\n# Evaluate the eigenvalues\neigvalues = la.eig(A)[0]\n# Sort them from greatest to least (use np.abs to account for complex parts)\neigvalues = eigvalues[np.sort(np.<>(eigvalues))[::-1]]\n\\end{lstlisting}\nIn addition, remember that certain eigenvalues of $H$ will not appear until we are computing enough iterations in the Arnoldi algorithm.\nAs a result, we will have to begin the graphs of several eigenvalues after we are computing sufficient iterations of the algorithm.\n%To be able to keep our graph readable, use masking such as the following to remove any extreme errors that arise in the first iterations.\n%\\begin{lstlisting}\n%errors[errors > 10] = 10.\n%\\end{lstlisting}\n\nRun your function on these examples.\nThe plots should be fairly similar to Figures \\ref{fig:arnoldi_random_eig_conv} and \\ref{fig:arnoldi_random_val_conv}.\n\n\\begin{lstlisting}\n>>> A = np.random.rand(300, 300)\n>>> plot_ritz(a, 10, 175)\n\n>>> # A matrix with uniformly distributed eigenvalues\n>>> d = np.diag(np.random.rand(300))\n>>> B = A @ d @ la.inv(A)\n>>> plot_ritz(B, 10, 175)\n\\end{lstlisting}\n% These may take some time to run.\n\\end{problem}\n\n% once the eigenvalue lab has been rewritten, have them use their own solver to find the eigenvalues of the $h_k$.\n\n% this problem is very ill-conditioned.\n\\begin{comment}\n\\begin{problem}\nfinding the roots of a polynomial can be represented as an eigenvalue problem.\nfinding the roots of a monic polynomial (a polynomial with leading coefficient 1) $p = c_0 + c_1 x + \\dots + c_{n-1} x^{n-1} + x^n$ is equivalent to finding the eigenvalues of the matrix\n\\[c = \\begin{bmatrix}\n0 & 0 & \\dots & 0 & -c_0 \\\\\n1 & 0 & \\dots & 0 & -c_1 \\\\\n0 & 1 & \\dots & 0 & -c_2 \\\\\n\\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\\n0 & 0 & \\dots & 1 & -c_{n-1} \\end{bmatrix}\\]\nthis matrix is called the companion matrix of the polynomial $p$.\nas it happens, every matrix is similar to the companion matrix of its characteristic polynomial, but we won't use that fact here.\n\nThe following is a function that, given an array containing the coefficients $c_0, c_1, \\dots, c_{n-1}$ for a monic polynomial $p$, performs matrix multiplication by the corresponding companion matrix.\n\n\\begin{lstlisting}\ndef companion_multiply(c, u):\n v = np.empty_like(u)\n v[0] = - c[0] * u[-1]\n v[1:] = u[:-1] - c[1:] * u[-1]\n return v\n\\end{lstlisting}\n\nUse the Arnoldi iteration to estimate the five zeros of largest norm of a degree $1000$ monic polynomial with randomly chosen coefficients (the leading coefficient still needs to be 1).\nRun $50$ steps of the Arnoldi iteration.\nCompare your results with the roots of the polynomial computed using NumPy's \\li{poly1d} class.\nThis computation can be done like this (where \\li{c} is the array of random coefficients for the polynomial)\n\n\\begin{lstlisting}\np = np.poly1d([1] + list(c[::-1]))\nroots = p.roots\n# Now sort by absolute value from largest to smallest\nroots = roots[np.<>(roots).argsort()][::-1]\n\\end{lstlisting}\n\nHow close are the first few zeros of largest norm?\n\\end{problem}\n\\end{comment}\n\n\\newpage\n\n\\section*{Additional Material} % ==============================================\n\n\\subsection*{The Lanczos Iteration} % -----------------------------------------\n\nThe Lanczos iteration is a version of the Arnoldi iteration that is optimized to operate on symmetric matrices.\nIf A is symmetric, then \\eqref{eq:arnoldi-hqa} shows that $H_k$ is symmetric and hence tridiagonal.\nThis leads to two simplifications of the Arnoldi algorithm.\n\nFirst, we have $0=H_{k, n}=\\langle \\q_k, A\\q_n \\rangle$ for $k \\leq n-2$; i.e., $A\\q_n$ is orthogonal to $\\q_1, \\ldots, \\q_{n-2}$.\nThus, if the goal is only to compute $H_k$ (say to find the Ritz values), then we only need to store the two most recently computed columns of $Q$.\nSecond, the data of $H_k$ can also be stored in two vectors, one containing the main diagonal and one containing the first subdiagonal of $H_k$\n(by symmetry, the first superdiagonal equals the first subdiagonal of $H_k$).\n\n\\begin{algorithm}[H]\n\\begin{algorithmic}[1]\n\\Procedure{Lanczos}{$\\b, A, k, tol$}\n\t\\State $\\q_0 \\gets \\zeros{\\size{\\b}}$\t\t\t\t\t\t\t\t\\Comment{Some initialization}\n\t\\State $\\q_1 \\gets \\b/\\norm{\\b}_2$\n\t\\State $\\x \\gets \\allocate{k}$\n\t\\State $\\y \\gets \\allocate{k}$\n\t\\For{$i=0\\ldots k-1$}\t\t\t\t\t\t\t\t\t\\Comment{Perform the iteration.}\n\t\t\\State $\\z \\gets A\\q_1$\t\t\t\t\t\\Comment{$\\z$ is a temporary vector to store $\\q_{i+1}$.}\n\t\t\\State $\\x[i] \\gets \\q_1\\trp \\z$\t\t\t\t\\Comment{$\\q_1$ is used to store the previous $\\q_i$.}\n\t\t\\State $\\z \\gets \\z - \\x[i] \\q_1 + \\y[i-1] \\q_0$\t\t\t\t\\Comment{$\\q_0$ is used to store $\\q_{i-1}$.}\n\t\t\\State $\\y[i] = \\norm{\\z}_2$\t\t\t\t\t\t\\Comment{Initialize $\\y[i]$.}\n\t\t\\If{$\\y[i]0,a=qb+r$ such that either $r=0$ or $\\phi(r)<\\phi(b)$.\r\n\\end{definition}\r\n\\begin{proposition}\r\n If $R$ is an Euclidean domain, then it is a PID.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Let $\\phi$ be the Euclidean function and $0\\neq I\\lhd R$ an ideal.\r\n Choose $b\\in I$ such that it is nonzero and $\\phi(b)$ is minimal.\r\n We shall show that $I=(b)$.\r\n Note that we immediately have $(b)\\subset I$.\r\n For the other way, choose $0\\neq a\\in I$, we write $a=qb+r$ with $q,r\\in R$ and either $r=0$ or $\\phi(r)<\\phi(b)$.\r\n If $r=0$ then $a\\in (b)$.\r\n Otherwise, note that $r=a-qb\\in I$, but $\\phi(r)<\\phi(b)$, contradicting the minimality of $b$.\r\n So we must have $a\\in (b)$, hence $I=(b)$.\r\n\\end{proof}\r\n\\begin{remark}\r\n Note that we did not use the first criterion to define the Euclidean function in the above proof.\r\n The reason for us to include that in the definition of a ED is that it allows us to describe the units in $R$ in the way that $\\phi(a)=\\phi(1)$ iff $a$ is unit.\r\n\\end{remark}\r\n\\begin{example}\r\n 1. $\\mathbb Z$ is a ED since we can take $\\phi(n)=|n|$.\\\\\r\n 2. For a field $\\mathbb F$, $\\mathbb F[X]$ is a ED by taking $\\phi(P)=\\deg P$ since we can do division with remainder in the polynomial ring of a field.\\\\\r\n 3. Consider the Gaussian integer $R=\\mathbb Z[i]\\le C$ by taking $\\phi(a+ib)=a^2+b^2$, so $\\phi$ is multiplicative therefore we get the first criterion.\r\n For the second, let $z_1,z_2\\in R$.\r\n Consider $z_1/z_2\\in\\mathbb C$, which has distance strictly less than $1$ from the nearest Gaussian integer, so $z_1/z_2=q+\\epsilon$ where $q\\in R$ and $|\\epsilon|<1$, so $z_1=qz_2+r$ where $r=\\epsilon z_2=z_1-qz_2\\in R$ and $\\phi(r)=|\\epsilon z_2|^2=|\\epsilon|^2\\phi(z_2)<\\phi(z_2)$.\r\n\\end{example}\r\nThe above examples are all also PIDs by the preceding proposition.\r\n\\begin{example}\r\n 1. Let $A$ be an $n\\times n$ matrix in the field $\\mathbb F$, and let $I$ be the set of all $f\\in \\mathbb F[X]$ such that $f(A)=0$.\r\n Now $I$ is trivially an ideal.\r\n But this ideal is principal since $\\mathbb F[X]$ is a ED hence PID.\r\n Suppose $I=(f)$, then for any $g\\in\\mathbb F[X]$ such that $g(A)=0$, we have $f|g$.\r\n So $f$ is the minimal polynomial of $A$.\\\\\r\n 2. Consider $\\mathbb F_2=\\mathbb Z/2\\mathbb Z$ and let $f(X)=X^3+X+1\\in\\mathbb F_2[X]$.\r\n We want to show that $f$ is irreducible.\r\n Suppose $f(X)=g(X)h(X)$ with $\\deg g,\\deg h>0$.\r\n So one of $\\deg g$ and $\\deg f$ must be $1$ since $f$ has degree $3$, so $f$ has a root, but it does not, contradiction.\\\\\r\n But $\\mathbb F_2[X]$ is a ED hence PID, therefore $(f)$ is maximal, so we get to construct a field $\\mathbb F_2[X]/(X^3+X+1)$.\r\n The quotient looks like $\\{aX^2+bX+c+(X^3+X+1):a,b,c\\in\\mathbb F_2\\}$, but $a,b,c$ uniquely determines the ideal, so this is a field of order $8$.\\\\\r\n 3. (non-example) The ring $\\mathbb Z[X]$ is not a PID.\r\n Indeed, consider the ideal $I=(2,X)$, then $I=\\{2f_1(X)+Xf_2(X):f_1,f_2\\in\\mathbb Z[X]\\}=\\{f(X)\\in\\mathbb Z[X]:f(0)\\text{ is even}\\}$.\r\n Suppose $I$ is generated by some polynomial $f\\in\\mathbb Z[X]$, so $2=fg$ for some $g\\in\\mathbb Z[X]$.\r\n But degrees add when polynomials multiply, hence $f,g$ has to be constant.\r\n Therefore $I$ can only be $\\mathbb Z[X]$ or $2\\mathbb Z[X]$, but both are impossible since $1\\notin I$ (so $I$ is not the entire ring) and $X\\in I$ (but $X\\notin 2\\mathbb Z[X]$).\r\n\\end{example}\r\n\\subsection{Unique Factorization Domains}\r\n\\begin{definition}\r\n An integral domain $R$ is called a unique factorization domain (UFD) if\\\\\r\n 1. Every nonzero and nonunit $r\\in R$ is a product of irreducibles.\\\\\r\n 2. If $p_1\\cdots p_m=q_1\\cdots q_n$ where $p_i,q_i$ are irreducibles, then $m=n$ and $\\exists\\sigma\\in S_n$ such that $p_i$ is an associate with $q_{\\sigma(i)}$.\r\n\\end{definition}\r\n\\begin{proposition}\\label{irred_prime_uniq}\r\n Let $R$ be an integral domain having the first property stated above, then $R$ is a UFD iff every irreducible is prime.\r\n\\end{proposition}\r\n\\begin{proof}\r\n If $R$ is a UFD and $r\\in R$ is irreducible and $p|ab$, then $pc=ab$ for some $c\\in R$.\r\n By the first condition, we can write $a,b,c$ as products of irreducibles, so an associate of $p$ must appear in the factorization of $a$ or $b$ by the second condition, so $p|a$ or $p|b$, hence $p$ is prime.\\\\\r\n Suppose every irreducible is prime.\r\n If $p_1\\cdots p_m=q_1\\cdots q_n$.\r\n Now since $p_1$ is prime, then $p_1|q_i$ for some $i$.\r\n But $q_i,p_1$ are both irreducible, so they are associates.\r\n By reordering, we can write $i=1$, and by cancellation law, $p_2\\cdots p_m=q_2\\cdots q_n$.\r\n The proof is finished by descent (or equivalently, induction).\r\n\\end{proof}\r\n\\begin{lemma}\r\n Let $R$ be a PID and there is a nested sequence of ideals $I_1\\subset I_2\\subset\\cdots$, then there exists some $N\\in\\mathbb N$ such that $I_n=I_N$ for every $n\\ge N$.\r\n\\end{lemma}\r\n\\begin{remark}\r\n This condition is one of the formulations of the definition of a Noetherian ring.\r\n\\end{remark}\r\n\\begin{proof}\r\n Consider the union $I=I_1\\cup I_2\\cup\\cdots$.\r\n $I$ is obviously an ideal, so $I=(a)$ for some $a\\in R$.\r\n But $a\\in I_N$ for some $N\\in\\mathbb N$, so for any $n\\ge N$, we have $(a)\\subset I_n\\subset I=(a)$, hence $I_n=(a)=I_N$.\r\n\\end{proof}\r\n\\begin{theorem}\r\n Every PID is a UFD.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Let $R$ be a PID.\r\n By Proposition \\ref{irred_prime_uniq}, since every irreducible in a PID is prime, it suffices to show the first condition of a UFD.\r\n Let $x\\in R$ be nonzero and nonunit.\r\n Suppose $x$ cannot be written as a product of irreducibles, so in particular $x$ is not irreducible.\r\n Therefore we can write $x=x_1y_1$ for nonunit $x_1,y_1$.\r\n But not both of $x_1,y_1$ can be written as a product of irreducibles.\r\n WLOG $x_1$ is not a product of irreducibles, also we have $(x)\\subsetneq (x_1)$ since $y_1$ is not a unit.\r\n Continue the process over and again gives a sequence of strictly nested ideals $(x)\\subsetneq (x_1)\\subsetneq\\cdots$, but this is a sequence of nested ideals that does not terminate, hence contradiction to the preceding lemma.\r\n\\end{proof}\r\n\\begin{example}\r\n We know that ED implies PID implies UFD implies integral, so we have the following table of examples:\r\n \\begin{center}\r\n \\begin{tabular}{c|c|c|c|c|l}\r\n &ED&PID&UFD&Integral&\\\\\r\n $\\mathbb Z/4\\mathbb Z$&&&&No&\\\\\r\n $\\mathbb Z[\\sqrt{5}]$&&&No&Yes&See later.\\\\\r\n $\\mathbb Z[X]$&&No&Yes&&See later.\\\\\r\n $\\mathbb Z\\left[\\frac{1+\\sqrt{-19}}{2}\\right]$&No&Yes&&&See in Number Fields.\\\\\r\n $\\mathbb Z[i]$&Yes&&&&\r\n \\end{tabular}\r\n \\end{center}\r\n\\end{example}\r\n\\subsection{Greatest Common Factors and Least Common Multiples}\r\n\\begin{definition}\r\n Let $R$ be a integral domain.\r\n We say $d$ is a greatest common divisor of $a_1,\\ldots,a_n\\in R$ if $d|a_i$ for each $i$ and if $\\forall i,d'|a_i$, then $d'|d$.\r\n We say $m$ is a least common multiple of $a_1,\\ldots,a_n\\in R$ if $a_i|d$ for each $i$ and if $\\forall i,a_i|d'$, then $d|d'$.\r\n\\end{definition}\r\nBoth GCDs and LCMs, when they exists, they are unique up to associates.\r\n\\begin{proposition}\r\n In a UFD, both LCMs and GCDs exist.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Obvious.\r\n\\end{proof}", "meta": {"hexsha": "0c62b8c89a422423eda6fd2cccfb37cfded23db3", "size": 12484, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "10/ufd.tex", "max_stars_repo_name": "david-bai-notes/IB-Groups-Rings-and-Modules", 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0.6413008651, "num_tokens": 4287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7122321842389469, "lm_q2_score": 0.9005297941266013, "lm_q1q2_score": 0.6413863022430384}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PROBLEM 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{Problem 2}\n\nConsider an infinite slab of material described by diffusion coefficient $D$ and macroscopic cross section $\\Sigma_a$. \nThe material extends infinitely in two dimensions, but has vacuum boundaries at $x=\\pm a$.\nInside the slab is a uniformly distributed source with strength $s'''$ [neutrons per volume]. \nFind the flux in this geometry. \n(You may use the substitution $L = \\sqrt{\\frac{D}{\\Sigma_a}}$.)\n\n", "meta": {"hexsha": "a35041dd7dfc3e578e7c620353e230080001f4ba", "size": 506, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises/drafts/disc10/disc10_exercise02.tex", "max_stars_repo_name": "mitchnegus/NE150-discussion", "max_stars_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "exercises/drafts/disc10/disc10_exercise02.tex", "max_issues_repo_name": "mitchnegus/NE150-discussion", "max_issues_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "exercises/drafts/disc10/disc10_exercise02.tex", "max_forks_repo_name": "mitchnegus/NE150-discussion", "max_forks_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.6, "max_line_length": 119, "alphanum_fraction": 0.6501976285, "num_tokens": 112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9304582516374121, "lm_q2_score": 0.6893056104028799, "lm_q1q2_score": 0.6413700930993228}} {"text": "Disagreements between the obtained and the \\emph{aRTist} simulated projections can be found by subtracting one projection from the other. Any values too big in magnitude can be considered as a defect. However, in the previous chapters, it was found that x-ray photons behave randomly and differences in the comparison can be due to chance. Thus, the comparisons should be done in the face of uncertainty.\n\nA pixel by pixel hypothesis test was proposed to do defect detection. To illustrate this method of inference, the projections from the dataset \\texttt{AbsFilter} at \\ang{120} was used. To recap, 20 replicate projections of a test sample, with purposefully manufactured voids, were obtained. \\emph{aRTist} was used to create a simulation of that projection but as if the voids were not there. Thus, the method should pick these voids up.\n\nBeforehand, linear shading correction was applied to the projections. The \\emph{aRTist} projections were shading corrected using the simulated greyscale projections. The 20 replicate projections were split into two. 19 randomly selected projections were used for the variance-mean model to fit onto. The remaining projection was compared with the \\emph{aRTist} projection. This remaining projection and the \\emph{aRTist} projections are shown in Figures \\ref{fig:inference_inferenceIntro_inference_scan} and \\ref{fig:inference_inferenceIntro_inference_artist} respectively.\n\nA gamma GLM, with a basic linear relationship, was used for the variance-mean model, as described in Chapter \\ref{chapter4}. The variance was predicted using the grey value in the \\emph{aRTist} simulation as the predictor variable.\n\nThe test statistic, for the pixel located at $(x,y)$, is\n\\begin{equation}\n Z_{x,y} =\n \\dfrac{\n \\text{projection}_{x,y} - \\emph{aRTist}_{x,y}\n }\n {\n \\sqrt{\\widehat{y}\\left[\\emph{aRTist}_{x,y}\\right]}\n }\n\\end{equation}\nwhere $\\widehat{y}\\left[\\emph{aRTist}_{x,y}\\right]$ is the predicted grey value variance. The test statistic was calculated for each pixel in the ROI, that is, pixels which represent the test sample. The ROI was created by manually segmenting the test sample from the projection.\n\nIt is important to identify which quantities are random and which are not. For high photon rates, it was shown in the previous chapters that the grey values in the projection can be modelled using a Normal distribution. Thus, $\\text{projection}_{x,y}$ is a random quantity. The simulation $\\emph{aRTist}_{x,y}$ is not random because this was obtained through computer software. Given the 19 selected projections used for training the variance-mean model, the variance prediction $\\widehat{y}\\left[\\emph{aRTist}_{x,y}\\right]$ is not random because the variance-mean model was fitted before encountering the remaining projection to be compared with \\emph{aRTist}. The variance-mean model can be made random if a different set of 19 projections was used to train the model each time this inference was conducted, but this shall not be considered here.\n\nIf there are no significant differences between the obtained and \\emph{aRTist} projections, then the randomness of the test statistics can be approximately quantified as\n\\begin{equation}\nZ_{x,y}\\sim \\normal(0,1) \\ .\n\\end{equation}\nAs with usual statistical convention, an upper case $Z$ denotes a random variable. A lower case $z$ denotes a realisation or an observation of that random variable.\n\nFor each pixel, a test statistic was calculated, which forms a $z$ image. A test statistic too large in magnitude, relative to the anticipated variance, will classify that pixel as a positive result. In two-tailed hypothesis testing \\citep{pearson1900on, neyman1933on, fisher1970statistical}, $p$-values can be used to represent the test statistics in a different way\n\\begin{equation}\n p_{x,y} = 2(1-\\Phi(|z_{x,y}|))\n\\end{equation}\nwhich can takes values $0\\leqslant p_{x,y} \\leqslant 1$. A $p$-value too small is considered a positive result. The $p$-values are shown in Figure \\ref{fig:inference_inferenceIntro_inference_pvalue}.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_scan.eps}\n \\caption{Obtained projection (\\SI{}{\\adu})}\n \\label{fig:inference_inferenceIntro_inference_scan}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_artist.eps}\n \\caption{\\emph{aRTist} projection (\\SI{}{\\adu})}\n \\label{fig:inference_inferenceIntro_inference_artist}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_logp.eps}\n \\caption{$-\\log p\\text{-values}$}\n \\label{fig:inference_inferenceIntro_inference_pvalue}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_positivePixels.eps}\n \\caption{Positive pixels overlaid on a)}\n \\label{fig:inference_inferenceIntro_inference_positivePixels}\n \\end{subfigure}\n }\n \\caption{The obtained projection of the test sample (a), from the \\texttt{AbsFilter} at \\ang{120} dataset, was compared to the \\emph{aRTist} projection (b) to detect purposefully manufactured voids. The $p$-values (c) obtained were used for hypotheses testing. Pixels detected as positive are shown in red in d) using the \\cite{benjamini1995controlling} procedure at the 5\\% false discovery rate level.}\n \\label{fig:inference_inferenceIntro_inference}\n\\end{figure}\n\nThe resulting $p$-values are concerning. This is because the $p$-values are not very smooth on the surfaces of the sample. It should be expected that small $p$-values are in areas of the defects. Pixels were tested positive or considered to be evidence of a defect, when $|z_{x,y}|>\\inputNumber{../figures/inference/inferenceIntro_critical.txt}$ to 2 decimal places. This value was chosen by controlling the false discovery rate at 5\\% using the \\cite{benjamini1995controlling} (BH) procedure. The positive pixels are shown in Figure \\ref{fig:inference_inferenceIntro_inference_positivePixels}.\n\nThis proposed method for defect detection failed because too many false positives were detected. These false positives appeared to have some structure, for example, clustering in the corners or on surfaces. Also, false negatives were detected because not all of the defects were detected.\n\nModel misspecification appeared to be the main source of error. The test statistics and $p$-values were inspected in Figure \\ref{fig:inference_inferenceIntro_histogram}. It can be seen that the test statistics were not compatible with the standard Normal distribution. Also, the majority of the $p$-values did not look uniformly distributed. This seems to suggest that the assumption of $Z_{x,y}\\sim \\normal(0,1)$ is incorrect.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_histogram.eps}\n \\caption{Histogram}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceIntro_pValue.eps}\n \\caption{$p$-values}\n \\label{fig:inference_inferenceIntro_pvalue}\n \\end{subfigure}\n}\n \\caption{a) The histogram of the test statistics, from the \\texttt{AbsFilter} projection at \\ang{120}, is compared with the standard Normal distribution. The standard Normal distribution is known as the null distribution in this scenario. b) The $p$-values are ordered and plotted. The critical region corresponds to controlling the false discovery rate at the 5\\% level using the \\cite{benjamini1995controlling} procedure. The dotted line shows the result if the $p$-values were uniformly distributed.}\n \\label{fig:inference_inferenceIntro_histogram}\n\\end{figure}\n\nThis chapter recaps hypothesis testing, for a single test and then for multiple tests, treating each pixel as a test. Assumptions used in the hypotheses testing can be relaxed by using the empirical null \\citep{efron2004large}, which is reviewed here. The empirical null is then extended to an image filter, called the empirical null filter. This filter adjusts each test statistic according to its neighbours, ironing out false positive results. Simulations and results are shown towards the end of the chapter.\n\n\\section{Literature Review}\n\nConsider a test statistic $Z$ from a pixel. If there are no defects, then the statistic is null and has the null distribution $Z|H_0\\sim\\normal(0,1)$. This can be described by specifying the random variable as $Z\\sim\\normal(\\mu,1)$ and the null hypothesis as $H_0:\\mu=0$. A hypothesis test can judge how much $Z$ deviates from 0 by defining the alternative hypothesis $H_1:\\mu\\neq 0$. A statistic which does not has the null distribution is known as non-null.\n\nIn two-tailed hypothesis testing \\citep{pearson1900on, neyman1933on, fisher1970statistical}, the $p$-value can be compared with the user-defined size of the test $\\alpha$, also known as the significance level in this specific example. A positive result is declared when $p<\\alpha$, else it is a negative result. Correct and incorrect testing of the statistics can occur. For example, testing a null statistic as positive is known as a false positive. It can be shown that $\\alpha$ controls the false positive rate. $\\alpha=5.0\\%$ is a typical choice \\citep{wasserstein2019moving} and is used throughout this thesis.\n\nMultiple hypotheses testing occurs when there are more than one hypotheses to test. For example, $N$ pixels can be tested where the test statistics are $Z_1, Z_2, \\dotsc, Z_N$ and $Z_i\\sim\\normal(\\mu_i,1)$ for $i=1,2,\\dotsc,N$. The null hypotheses are $H_{0,i}:\\mu_i=0$ and are tested against the alternative $H_{1,i}:\\mu_i\\neq 0$ for $i=1,2,\\dotsc,N$. Let the corresponding $p$-values be $p_1, p_2, \\dotsc, p_N$.\n\nThe uncorrected test classify any $p_i<\\alpha$ as positive. This is flawed because the possibility of obtaining at least one false positive increases as the number of tests increases \\citep{shaffer1995multiple}. This method does control the per-comparison error rate (PCER) \\citep{benjamini1995controlling}. PCER is the proportion of false positives out of all tests. The notation for the number of true/false positive/negatives obtained are defined in Table \\ref{table:inference_randomvariables}. Using the notation, the PCER is defined as\n\\begin{equation}\n \\text{PCER}=\n \\dfrac{1}{N}\n \\expectation[V]\n \\ .\n\\end{equation}\nIt can be seen that if the uncorrected test controls the false positive rate such that $\\expectation[V]/N_0 = \\alpha$, then it controls the PCER such that\n\\begin{equation}\n \\text{PCER}\\leqslant\\alpha \\ .\n\\end{equation}\n\n\\begin{table}\n \\centering\n \\begin{tabular}{l|cc|c}\n &Negative&Positive&Total\\\\\\hline\n Null & $U$ & $V$ & $N_0$\\\\\n Non-null & $T$ & $S$ & $N-N_0$\\\\\\hline\n &$N-R$&$R$&$N$\n \\end{tabular}\n \\caption{Random variable definitions for the number of true/false positives/negatives made in multiple hypotheses testing}\n \\label{table:inference_randomvariables}\n\\end{table}\n\nThe Bonferroni correction \\citep{shaffer1995multiple, bland1995multiple, perneger1998what} controls the family-wise error rate (FWER) \\citep{shaffer1995multiple} where\n\\begin{equation}\n \\text{FWER} = \\prob(V\\geqslant1) \\ .\n \\label{eq:inference_fwer}\n\\end{equation}\nThis is done by adjusting the size of the test to be $\\alpha/N$. By using the adjusted size, then $\\text{FWER}=1-\\left[(1-\\alpha/N)^N\\right]$. Using the approximation $(1-\\alpha/N)^N\\approx 1-\\alpha$ then $\\text{FWER}\\approx \\alpha$. This shows that the Bonferroni correction controls the family-wise error rate such that\n\\begin{equation}\n \\text{FWER} \\leqslant \\alpha \\ .\n\\end{equation}\nIn practice, the Bonferroni correction is not very powerful \\citep{perneger1998what}, meaning it gives too many false negatives. This is because the correction traded too many false positives for false negatives.\n\nThe \\cite{benjamini1995controlling} (BH) procedure controls the false discovery rate (FDR) \\citep{benjamini2010discovering} rather than the PCER or FWER. The FDR is the proportion of false positives out of all positive results, that is\n\\begin{equation}\n \\text{FDR} = \\expectation\\left[\\dfrac{V}{R}\\right]\n \\ .\n\\end{equation}\nIt is defined that $V/R=0$ when $R=0$.\n\nThe BH procedure adjusts the size of the test between the Bonferroni correction and the uncorrected test. It adapts to the data and chooses different sizes for different data. The procedure is as follows, the $p$-values are ordered such that $p_{(1)}\\leqslant p_{(2)}\\leqslant \\dotsc \\leqslant p_{(N)}$. Suppose a size $\\alpha$ is provided beforehand. The size is adjusted to\n\\begin{equation}\n \\alpha_{\\text{BH}} = \\frac{\\alpha k}{N}\n\\end{equation}\nwhere\n\\begin{equation}\n k\\text{ is the largest }i\\text{ for which }p_{(i)}\\leqslant\\frac{i}{N}\\alpha\n \\ .\n\\end{equation}\nThis test the statistics with $p$-values $p_{(1)},p_{(2)},\\dotsc,p_{(k)}$ as positive. For the case where $p_{(1)}>\\alpha/N$ then $k=1$ so that the Bonferroni correction is used when there are no positive results, this is only for illustration purposes. For example in Figure \\ref{fig:inference_inferenceIntro_pvalue}, the $p$-values were plotted against their order so that the critical boundary was shown by a linear curve with gradient $\\alpha/N$.\n\nThe BH procedure comes from the fact that if all the statistics are null and independent, then the $p$-values are uniformly distributed \\citep{simes1986improved}. However, it can be shown that the BH procedure works for many scenarios of dependencies \\citep{benjamini2001control}. It can be shown that the BH procedure controls the FDR such that\n\\begin{equation}\n \\text{FDR}\\leqslant\\alpha\n\\end{equation}\n\\citep{benjamini1995controlling}.\n\nIn summary, the uncorrected, Bonferroni and BH correction controls for different error rates according to the threshold $\\alpha$, summarised in Table \\ref{table:inference_corrections}.\n\n\\begin{table}\n \\centering\n \\begin{tabular}{l|l}\n Correction&Controls for\\\\\\hline\n No correction&$\\text{PCER}\\leqslant\\alpha$\\\\\n Bonferroni&$\\text{FWER}\\leqslant\\alpha$\\\\\n BH&$\\text{FDR}\\leqslant\\alpha$\n \\end{tabular}\n \\caption{Different types of corrections for multiple hypotheses testing are listed here, along with what they control for.}\n \\label{table:inference_corrections}\n\\end{table}\n\nConsider a small example where a $\\SI{200}{\\pixel}\\times\\SI{200}{\\pixel}$ section of the $z$ image was investigated, as shown Figure \\ref{fig:inference_inferenceSubsample1_histogram}. The distribution of the test statistics appeared Normal but not centred at zero. By using the BH procedure to obtain a critical region of $|Z|>\\input{../figures/inference/inferenceSubsample1_criticalBoundary.txt}$ to 2 decimal places, almost half of the pixels were tested positive at the 5\\% FDR level. It is questionable whether such many positive results are sensible, in particular, in an area where defects were not expected.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample1_zImage.eps}\n \\caption{Test statistics}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample1_histogram.eps}\n \\caption{Histogram}\n \\end{subfigure}\n }\n \\caption{The resulting test statistics for the \\texttt{AbsFilter} projection at \\ang{120} are shown in a). A $\\SI{200}{\\pixel}\\times\\SI{200}{\\pixel}$ sub-image was taken shown by the dashed lines. A histogram of the test statistics in the sub-image is shown in b). The critical region corresponds to the 5\\% FDR level.}\n \\label{fig:inference_inferenceSubsample1_histogram}\n\\end{figure}\n\nThis problem commonly occurs in large scale multiple hypotheses testing \\citep{efron2004large} such as in microarrays \\citep{hedenfalk2001gene, efron2002empirical, efron2003robbins}. It appeared that the null distribution was misspecified as $Z_i|H_{0,i}\\sim\\normal(0,1)$ when in reality they are distributed as $Z_i|H_{0,i}\\sim\\normal(\\mu_0,\\sigma_0^2)$, where $\\mu_0$ and $\\sigma_0$ are the null mean and null standard deviation respectively.\n\nIn practice, $\\mu_0$ and $\\sigma_0$ are unknown. The empirical null \\citep{efron2004large} replaces the parameters of the null distribution with its estimate. Suppose $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$ are the estimated null mean and null standard deviation respectively. The statistics are specified by $Z_i\\sim\\normal(\\mu_i, \\widehat{\\sigma}_0^2)$ so that the following null hypotheses $H_{0,i}:\\mu_i=\\widehat{\\mu}_0$ are tested against $H_{1,i}:\\mu_i\\neq\\widehat{\\mu}_0$ for $i=1,2,\\dotsc,N$.\n\nIn \\cite{efron2004large}, a few assumptions were made in order to obtain the estimators $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$. Let the test statistic $Z$ have the p.d.f.\n\\begin{equation}\n p_Z(z) =\n \\pi_0 p_{Z|H_0}(z) + \\pi_1 p_{Z|H_1}(z)\n\\end{equation}\nwhere $0\\leqslant\\pi_0\\leqslant 1$ and $\\pi_1 = 1-\\pi_0$. In addition, the null distribution is\n\\begin{equation}\n p_{Z|H_0}(z) =\n \\dfrac{1}{\\sqrt{2\\pi}\\sigma_0}\n \\exp\\left[\n -\\dfrac{1}{2}\n \\left(\n \\dfrac{z-{\\mu}_0}{\\sigma_0}\n \\right)^2\n \\right]\n\\end{equation}\nas it was assumed to be Normal. The non-null distribution $p_{Z|H_1}(z)$ does not need to be specified. Assume that the majority of the data are null and non-null test statistics are rare, say $\\pi_0>0.9$ \\citep{efron2004large}, then around the mode, the probability density function would be dominated by the null distribution. This implies that\n\\begin{equation}\n p_{Z}(z) \\approx \\pi_0 p_{Z|H_0}(z)\n\\end{equation}\nfor values of $z$ around the mode. Finding the mode for $p_{Z}(z)$ and $p_{Z|H_0}(z)$ should yield the same solution. This justify the use of the mode for the empirical null mean \\citep{efron2004large}\n\\begin{equation}\n \\widehat{\\mu}_0 = \\argmax\\widehat{p}_Z(z)\n\\end{equation}\nwhere $\\widehat{p}_Z(z)$ is the density estimation of $p_Z(z)$, for example it could be a smoothing spline fitted onto the histogram \\citep{efron2004large}.\n\nThe null standard deviation is estimated from the log density \\citep{efron2004large}. For values of $z$ at and around the mode\n\\begin{equation}\n \\ln p_{Z}(z) =\n \\ln\\left[\n \\dfrac{\\pi_0}{\\sqrt{2\\pi}{\\sigma}_0}\n \\right]\n -\\dfrac{1}{2}\n \\left(\n \\dfrac{z-{\\mu}_0}{{\\sigma}_0}\n \\right)^2\n \\ .\n\\end{equation}\nTaking derivatives\n\\begin{align}\n \\dfrac{\\partial}{\\partial z} \\ln p_{Z}(z) &=\n -\\left(\n \\dfrac{\n z-{\\mu}_0\n }\n {\n {\\sigma}_0^2\n }\n \\right)\n \\\\\n \\dfrac{\\partial^2}{\\partial z^2} \\ln p_{Z}(z) &=\n -\\dfrac{\n 1\n }\n {\n {\\sigma}_0^2\n }\n\\end{align}\nwhich motivates the estimator \\citep{efron2004large}\n\\begin{equation}\n \\widehat{\\sigma}_0 = \\left[\n \\left.\n -\\dfrac{\\partial^2}{\\partial z^2}\\ln\\widehat{p}(z)\n \\right|_{z=\\widehat{\\mu}_0}\n \\right]^{-1/2}\n \\ .\n \\label{eq:inference_nullStdEstimator}\n\\end{equation}\nThe evaluation of $z=\\widehat{\\mu}_0$ is used because at the mode, it was assumed that the null test statistics would dominate and non-null test statistics would not contribute much to the density estimate.\n\nThis method does not need estimations of $\\pi_0$ which makes it quite convenient. Estimations of $\\pi_0$ are discussed in literature such as \\cite{benjamini2000adaptive, pounds2003estimating, storey2003statistical, pounds2004improving, langaas2005estimating, durnez2014posthoc}. \\cite{efron2004large} motivated the use of the empirical null by investigating various types of false discovery rates \\citep{storey2002direct, storey2003positive, efron2002empirical, efron2007size} which will not be discussed here.\n\nThe empirical null is used widely, including in neuroimaging \\citep{schwartzman2008false, schwartzman2009empirical} and has been extended to include non-Normal null distributions \\citep{schwartzman2008false, schwartzman2008empirical}. There exist various methods for the empirical null, for example, \\cite{schwartzman2008empirical} used Poisson regression on the histogram counts in regions where the null statistics dominate. The empirical characteristic function was used in \\cite{jin2007estimating}.\n\n\\section{Empirical Null}\n\nReturning to the sub-image example in Figure \\ref{fig:inference_inferenceSubsample1_histogram}. The histogram was smoothed using a kernel density estimate \\citep{parzen1962on, friedman2001elements} as shown in Figure \\ref{fig:inference_inferenceSubsample1_densityEstimate}. The kernel density estimate \\citep{parzen1962on} is\n\\begin{equation}\n \\widehat{p}_Z(z)=\n \\frac{1}{nh}\n \\sum_{i=1}^n\\phi\\left(\n \\dfrac{z_i-z}{h}\n \\right)\n \\label{eq:inference_kernel_density_estimate}\n\\end{equation}\nwhere $h$ is the bandwidth, $\\phi(x)$ is the standard Normal density and $n=N$ is number of terms in the summation. The bandwidth was chosen such that\n\\begin{equation}\n h = (0.9n^{-1/5}\\ + 0.16) \\times \\text{min}\\left(s_z,\\text{IQR}_z/1.34\\right)\n \\label{eq:inference_ourruleofthumb}\n\\end{equation}\nwhere $s_z$ and $\\text{IQR}_z$ are the sample standard deviation and sample interquartile range. This bandwidth is justified later on in this section.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\subSize]{../figures/inference/inferenceSubsample1_densityEstimate.eps}\n \\caption{The density of the test statistics, from Figure \\ref{fig:inference_inferenceSubsample1_histogram}, was estimated using a kernel density estimate. The dot-dashed line shows the empirical null distribution multiplied by some constant, this is to illustrate that the two densities have the same curvature at the mode.}\n \\label{fig:inference_inferenceSubsample1_densityEstimate}\n\\end{figure}\n\nThe null mean was estimated by numerically finding the mode of the density estimate. The null standard deviation was estimated using Equation \\eqref{eq:inference_nullStdEstimator}. This resulted in the empirical null density to have the same curvature as the density estimate at the mode, this is illustrated in Figure \\ref{fig:inference_inferenceSubsample1_densityEstimate}. In this particular example, it was found that $\\widehat{\\mu}_0=\\input{../figures/inference/inferenceSubsample1_nullMean.txt}$ and $\\widehat{\\sigma}_0=\\input{../figures/inference/inferenceSubsample1_nullStd.txt}$ to 2 decimal places.\n\nThe use of the kernel density estimate has its advantages. The only tuning parameter is the bandwidth $h$ and the density estimate is simple enough to do calculus on it. Speed may be an issue as an evaluation of the density estimate requires the sum over the $n$ test statistics.\n\nThe test statistics were normalised to $T_1,T_2,\\dotsc,T_N$ by using\n\\begin{equation}\n T_i = \\dfrac{\n Z_i - \\widehat{\\mu}_0\n }\n {\n \\widehat{\\sigma}_0\n }\n \\label{eq:inference_empiricalNullNormalise}\n\\end{equation}\nfor $i=1,2,\\dotsc,N$. The normalised $p$-values were obtained by using\n\\begin{equation}\n p_i = 2(1-\\Phi(|t_i|))\n\\end{equation}\nand were used in the BH procedure. The critical region was found to be $|t|\\geqslant\\inputNumber{../figures/inference/inferenceSubsample1_nullNormalisedCritical.txt}$ at the 5\\% FDR level. In terms of the original units, the critical region was $z \\leqslant \\inputNumber{../figures/inference/inferenceSubsample1_nullCritical1.txt}$ and $\\inputNumber{../figures/inference/inferenceSubsample1_nullCritical2.txt}\\leqslant z$, all to 2 decimal places. The critical region and the normalised $p$-values are shown in Figure \\ref{fig:inference_inferenceSubsample1_nullResults}.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample1_nullHistogram.eps}\n \\caption{Histogram}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample1_nullPValues.eps}\n \\caption{$p$-values}\n \\end{subfigure}\n }\n \\caption{Histogram and $p$-values of the test statistics from the sub-image in Figure \\ref{fig:inference_inferenceSubsample1_histogram}. The critical regions were adjusted using the empirical null and correspond to the 5\\% FDR level.}\n \\label{fig:inference_inferenceSubsample1_nullResults}\n\\end{figure}\n\nNo positive results were found and the $p$-values were sensible as they resemble the uniform distribution. This demonstrated that the empirical null adjusted the parameters of the null distribution to fit onto the majority of the data to make a sensible inference.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample2_zImage.eps}\n \\caption{Test statistics}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample2_subimagePositive.eps}\n \\caption{Sub-image}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample2_nullHistogram.eps}\n \\caption{Histogram}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsample2_nullPValues.eps}\n \\caption{$p$-values}\n \\end{subfigure}\n }\n \\caption{The resulting test statistics for the \\texttt{AbsFilter} projection at \\ang{120} are shown in a). A $\\SI{200}{\\pixel}\\times\\SI{200}{\\pixel}$ sub-image was taken shown by the dashed lines. Positive pixels are shown in red in b). The histogram of the test statistics in the sub-image are shown in c) along with the critical region. The $p$-values are shown in d). The critical regions were adjusted using the empirical null and correspond to the 5\\% FDR level.}\n \\label{fig:inference_inferenceSubsample2}\n\\end{figure}\n\nAnother example is shown in Figure \\ref{fig:inference_inferenceSubsample2} where a $\\SI{200}{\\pixel}\\times\\SI{200}{\\pixel}$ sub-image containing a defect was used. It can be seen that the null distribution was not centred at zero which can be taken into account by using the empirical null. The estimation of the empirical null parameters was robust as it depends on the density estimate at the mode only, it should not be affected by non-null test statistics.\n\nThe BH procedure was conducted using the normalised $p$-values. $\\inputNumber{../figures/inference/inferenceSubsample2_nPositive.txt}$ pixels were tested positive at the 5\\% FDR level. The majority of the positive pixels were found to be clustered together which highlighted the defect. The entire area of the defects was not tested positive but a good portion of them are. This should be enough to raise suspicion in that particular area. Only a few pixels were falsely tested as positive, however, they are typically isolated single pixels. Isolated positive pixels should be discarded as they are more than likely to be tested positive by random chance.\n\nThis section describes the empirical null implementation in detail, in particular, the numerical and computational aspects. The empirical properties of the sampling distribution of the null parameter estimators were studied.\n\n\\subsection{Mode Finding}\n\nThe mode was found by solving $\\widehat{\\mu}_0 = \\argmax\\widehat{p}_Z(z)$ numerically. This was done by using the Newton-Raphson method to solve\n\\begin{equation}\n \\dfrac{\n \\partial\n }\n {\n \\partial z\n }\n \\ln\\widehat{p}_Z(z)\n = 0\n\\end{equation}\nfor $z$ which finds stationary points. The method is an iterative algorithm and requires an initial value $z^{(0)}$. The iterative step is\n\\begin{equation}\n z^{(r+1)} =\n z^{(r)}\n -\\dfrac{\n \\left.\n \\dfrac{\n \\partial\n }\n {\n \\partial z\n }\n \\ln\\widehat{p}_Z(z)\n \\right|_{z = z^{(r)}}\n }\n {\n \\left.\n \\dfrac{\n \\partial^2\n }\n {\n \\partial z^2\n }\n \\ln\\widehat{p}_Z(z)\n \\right|_{z = z^{(r)}}\n }\n\\end{equation}\nfor $r=0,1,2,3,\\dotsc$ until some convergence condition is met. The derivatives for the log density were obtained via the following. Recall the kernel density estimate in Equation \\eqref{eq:inference_kernel_density_estimate}, the log density is\n\\begin{equation}\n \\ln\\widehat{p}_Z(z)=\n \\ln\\left(\n \\dfrac{1}{nh}\n \\right)\n +\n \\ln\\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n \\right]\n \\ .\n\\end{equation}\nTaking the first-order derivative\n\\begin{equation*}\n \\dfrac{\n \\partial\n }\n {\n \\partial z\n }\n \\ln\\widehat{p}_Z(z)\n =\n \\dfrac{\n 1\n }\n {\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n }\n \\times\n \\sum_{i=1}^n\n \\phi'\\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n \\left(\n -\\dfrac{\n 1\n }\n {\n h\n }\n \\right)\n \\ .\n\\end{equation*}\nUsing the fact that $\\phi(z)=(2\\pi)^{-1/2}\\exp(-z^2/2)$, then $\\phi'(z)=-z\\phi(z)$. This is used to simplify the equation to be\n\\begin{equation}\n \\dfrac{\n \\partial\n }\n {\n \\partial z\n }\n \\ln\\widehat{p}_Z(z)\n =\n \\dfrac{\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n \\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n }\n {\n h\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i - z\n }\n {\n h\n }\n \\right)\n }\n \\ .\n\\end{equation}\nTaking the derivative again\n\\begin{multline*}\n \\dfrac{\n \\partial^2\n }\n {\n \\partial z^2\n }\n \\ln\\widehat{p}_Z(z)\n =\n \\left[\n h\\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]^{-2}\n \\times\n \\left\\{\n h\\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]\n \\right.\n \\\\\n \\left.\n \\times\n \\sum_{i=1}^n\\left[\n \\phi'\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\left(\n -\\dfrac{\n 1\n }\n {\n h\n }\n \\right)\n \\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n +\\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\left(\n -\\dfrac{\n 1\n }\n {\n h\n }\n \\right)\n \\right]\n \\right.\n \\\\\n \\left.\n -\n \\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]\n \\left[\n h\\sum_{i=1}^n\n \\phi'\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\left(\n -\\dfrac{\n 1\n }\n {\n h\n }\n \\right)\n \\right]\n \\right\\}\n \\ .\n\\end{multline*}\nUsing the fact that $\\phi'(z)=-z\\phi(z)$, then it is simplified to\n\\begin{multline}\n \\dfrac{\n \\partial^2\n }\n {\n \\partial z^2\n }\n \\ln\\widehat{p}_Z(z)\n =\n \\left[\n h\\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]^{-2}\n \\times\n \\left\\{\n \\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]\n \\right.\n \\\\\n \\left.\n \\times\n \\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\left(\n \\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)^2\n -1\n \\right)\n \\right]\n -\n \\left[\n \\sum_{i=1}^n\n \\phi\\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right.\n \\right.\n \\\\\n \\left.\n \\left.\n \\left(\n \\dfrac{\n z_i-z\n }\n {\n h\n }\n \\right)\n \\right]^2\n \\right\\}\n \\ .\n\\end{multline}\n\nThe convergence criteria were met when either 10 update steps were taken or when\n\\begin{equation}\n \\log\\left[\\left|\n \\left.\n \\dfrac{\n \\partial\n }\n {\n \\partial z\n }\n \\ln\\widehat{p}_Z(z)\n \\right|_{z=z^{(r)}}\n \\right|\\right]\n <-5\n\\end{equation}\nat the current step. This was chosen arbitrary to speed up the algorithm without losing too much accuracy. At the end of the algorithm, for a successful convergence it was required, in addition, that\n\\begin{equation}\n \\left.\n \\dfrac{\n \\partial^2\n }\n {\n \\partial z^2\n }\n \\ln\\widehat{p}_Z(z)\n \\right|_{z=z^{(r)}}\n < 0 \\ .\n\\end{equation}\nFollowing a successful convergence, the estimator $\\widehat{\\sigma}_0$ was calculated straight away.\n\nThe algorithm does depend on the initial value so using different initial values were used. Further initial values were generated by sampling from $\\normal(z^{(0)}, s_z^2)$. This was done multiple times until three valid solutions were obtained. The best solution, the one with the largest $\\ln\\widehat{p}_Z\\left(\\widehat{\\mu}_0\\right)$, out of all the different initial values was used as the final answer. The Newton-Raphson method may converge to a local maximum.\n\nIt can be noted that this mode finding method can be applied to other distributions in general.\n\n\\subsection{Bandwidth Tuning}\n\nThe bandwidth $h$ controls how smooth the kernel density estimator is, higher values produce smoother curves \\citep{friedman2001elements}. Cross-validation methods to select $h$ do exist \\citep{bowman1984alternative, sheather2004density} but they can be computationally expensive. Rules of thumb \\citep{silverman1986density, sheather2004density} can be used instead and are usually of the form\n\\begin{equation}\n h = bn^{-1/5}\\times\\text{min}\\left(s_z,\\text{IQR}_z/1.34\\right)\n\\end{equation}\nwhere $b=0.9$ \\citep{silverman1986density}. This rule of thumb was developed with consideration of bimodal Normal distributions \\citep{silverman1986density}. Other options include $b=1.06$ and $b=1.144$ to produce smoother curves \\citep{silverman1986density, sheather2004density}.\n\nThe kernel density estimator was only used to obtain values of $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$. Thus, a good bandwidth, in the context of the empirical null, is one which has good properties of $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$ rather than the density estimate. Exact properties of these estimators based on the kernel density estimators can be rather complicated. For example, numerous approximations were needed to show that $\\widehat{\\sigma}_0^2$ is an unbiased estimator of $\\sigma_0^2$ to the first-order and higher-order terms are functions of $(\\widehat{\\mu}_0 - \\mu_0)^2$ and $h$. This is shown in Appendix \\ref{chapter:appendix_expectationNullStdEstimator}.\n\nAn experiment was conducted to investigate how the estimators, $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$, varied with $h$ and $n$. This was done on a simulated dataset of $n$ standard Normal random variables. For a given $h$ and $n$, 100 values of $\\widehat{\\mu}_0$ and $\\widehat{\\sigma}_0$ were obtained by repeating the simulation of the dataset. The median squared error was obtained by taking the median over all the squared errors from the 100 estimates. 300 values of $n$ from 10 to $10^6$ and 30 values of $h$ from 0.09 to 1.5 were investigated. Figure \\ref{fig:inference_BandwidthSelection_error} shows the resulting median squared errors with the rules of thumb.\n\n$\\widehat{\\mu}_0$ has a low median squared error for large bandwidths. For low $n$, large bandwidths are particularly useful because smoother curves prevent any false bimodal features appearing, making it easier to find the mode.\n\nA smooth valley can be seen for the median squared error for $\\widehat{\\sigma}_0$. It appeared for a given $n$, there exist a bandwidth which minimises the median squared error. The rules of thumb did not optimise for $\\widehat{\\sigma}_0$ and undershot it. Figure \\ref{fig:inference_BandwidthSelection_error3} shows the median of the estimates of $\\sigma_0$. The true value in this simulation is $\\sigma_0=1$ which is represented as a horizontal plane. Bandwidths too small underestimated $\\sigma_0$, bandwidths too big overestimated $\\sigma_0$.\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection_error1.eps}\n \\caption{$\\mu_0$}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection_error2.eps}\n \\caption{$\\sigma_0$}\n \\end{subfigure}\n \\caption{The median squared error of the estimates of the empirical null parameters over 100 repeats of $n$ simulated standard Normal data. Lines represent the rule of thumb for various values of $b$.}\n \\label{fig:inference_BandwidthSelection_error}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection_error3.eps}\n \\caption{Heatmap}\n \\end{subfigure}\n \\caption{Median value of $\\widehat{\\sigma}_0$ over 100 repeats of $n$ simulated standard Normal data. The true value is $\\sigma_0=1$ which is represented by a horizontal plane.}\n \\label{fig:inference_BandwidthSelection_error3}\n\\end{figure}\n\nOverestimates and underestimates are both dangerous and it can affect the false positive and false negative rate. This is because $\\widehat{\\sigma}_0$ was used to rescale the test statistics accordingly. Recall that the normalised test statistics are $t_i = \\dfrac{z_i - \\widehat{\\mu}_0}{\\widehat{\\sigma}_0}$. Low values of $\\widehat{\\sigma}_0$ produce large values, in magnitude, of $t_i$, potentially testing more of these test statistics as positive. High values of $\\widehat{\\sigma}_0$ do the opposite. A large number of positives can lead to more false positives and vice versa.\n\nThe optimal bandwidth for a given $n$ was found numerically. Figure \\ref{fig:inference_BandwidthSelection2_plot} shows the log squared error against the bandwidth for various $h$ given $n$. A smoothing spline \\citep{friedman2001elements} was fitted and this was optimised to find the bandwidth which minimises the log squared error for a given $n$. \\emph{MATLAB}'s \\texttt{fit(x,y,'smoothingspline')} function was used to fit the smoothing spline which has inbuilt cross validation procedures. The optimisation was done using \\emph{MATLAB}'s gradient free \\texttt{fminsearch} function, using $h=0.9n^{-1/5}$ as the initial value.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot1.eps}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot51.eps}\n \\end{subfigure}\n }\n \\medskip\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot100.eps}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot150.eps}\n \\end{subfigure}\n }\n \\medskip\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot199.eps}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/BandwidthSelection2_plot249.eps}\n \\end{subfigure}\n }\n \\caption{A smoothing spline was fitted on the log squared error, for estimating $\\sigma_0$ on $n$ simulated standard Normal random variables, versus bandwidth. The experiment was repeated by simulating the standard Normal random variables again. The boxplots represent the 100 repeats. The red dashed line shows the bandwidth which minimised the fitted spline.}\n \\label{fig:inference_BandwidthSelection2_plot}\n\\end{figure}\n\nA relationship between the optimal bandwidth and $n$ was attempted to be found. It was assumed such a relationship has the linear form\n\\begin{equation}\n h_{\\text{optimal}} = b n^{-1/5} + a\n\\end{equation}\nwhere $b$ and $a$ are parameters to be fitted. This was chosen to establish a simple relationship with the rules of thumb. Figure \\ref{fig:inference_BandwidthSelection2_ruleOfThumb} show the fitting of an identity link gamma GLM in order to estimate $b$ and $a$. The estimated parameters are shown in Table \\ref{table:inference_BandwidthSelection2_ruleOfThumb}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\mainSize]{../figures/inference/BandwidthSelection2_ruleOfThumb.eps}\n \\caption{An identity link gamma GLM was fitted onto the relationship between the optimal bandwidth and $n^{-1/5}$ from the experiment. The dashed lines show the 68\\% prediction interval.}\n \\label{fig:inference_BandwidthSelection2_ruleOfThumb}\n\\end{figure}\n\n\\begin{table}\n \\centering\n \\begin{tabular}{l|c}\n &Estimate \\\\ \\hline\n Intercept&\\input{../figures/inference/BandwidthSelection2_coeff1.txt}\\\\\n Gradient &\\input{../figures/inference/BandwidthSelection2_coeff2.txt}\n \\end{tabular}\n \\caption{The estimated and standard error of the gradient and intercept from the linear relationship in Figure \\ref{fig:inference_BandwidthSelection2_ruleOfThumb}.}\n \\label{table:inference_BandwidthSelection2_ruleOfThumb}\n\\end{table}\n\nThe rules of thumb can be improved if a small bias is added to it. This is because an intercept was found when fitting a linear relationship between the optimal bandwidth and $n^{-1/5}$. Adding this bias should improve the performance of the estimator $\\widehat{\\sigma}_0$. The gradient is similar to the rules of thumb in the literature \\citep{sheather2004density}.\n\nThe GLM fit in Figure \\ref{fig:inference_BandwidthSelection2_ruleOfThumb} appeared to underfit, which suggest there may be a more complicated relationship between the optimal bandwidth and $n$, for example, a quadratic term can be added to the GLM fit. The method used in fitting the smoothing spline could contribute to the underfitting. For example, if the smoothing spline was not smooth enough so that multiple local minima appeared, it would add variance to the optimal bandwidth if the optimisation algorithm converges to a local minimum. A basic linear relationship was chosen to avoid overfitting and to keep the rule of thumb simple. It would also choose a large bandwidth for small $n$ which would improve the performance $\\widehat{\\mu}_0$ by working on a smoother density estimate for the price of a less optimal $\\widehat{\\sigma}_0$.\n\nIn conclusion, a small bias added to the rule of thumb improved the performance of the estimator $\\widehat{\\sigma}_0$. \\cite{silverman1986density} pointed out that a smaller bandwidth should be used when, in particular, the distribution is bimodal. This is why \\cite{silverman1986density} suggested the use of $b=0.9$. Following from this, it was chosen that the intercept was added to Silverman's rule of thumb for the use in the empirical null $h = \\left(0.9n^{-1/5}+ 0.16\\right) \\times\\text{min}\\left(s_z,\\text{IQR}_z/1.34\\right)$.\n\n\\afterpage{\\clearpage}\n\\subsection{Comparison with Other Robust Estimators}\n\nThe empirical null mean is effectively the mode and the empirical null standard deviation is a measure of dispersion. It was investigated how they compare with various estimates of central tendency and dispersion for varying sample sizes $n$. The empirical sampling distribution of these estimators was observed to look for properties such as bias, variance and robustness.\n\nThe central tendency estimates looked at were the median, the mean and Tukey's biweighted mean \\citep{beaton1974fitting}. In summary, the biweighted mean weighs each data point using a smooth function of its residual and ignores any data points with residuals greater than a certain value. \\emph{MATLAB}'s \\texttt{robustfit} function was used for the biweighted mean and selects any tuning parameters.\n\nThe dispersion estimates looked at were the interquartile range (IQR) $\\div 1.349$, the standard deviation and the median absolute deviation around the mode (MADA-mode) $\\times 1.483$. The normalisation constants were used so that MADA-mode and IQR can be compared directly to the standard deviation. MADA-mode was considered to investigate how estimators, which depends on the mode, perform.\n\nIn the next section, these estimators are used on a circular kernel with radius $r$ containing $\\pi r^2$ pixels. To make results comparable, the sample size $n$ may be represented using $r=\\sqrt{n/\\pi}$.\n\nA sample of standard Normal random variables was simulated to obtain estimates of central tendency and dispersion. This was repeated 100 times to obtain a sampling distribution. The results are shown in Figures \\ref{fig:inference_NullIidMean} and \\ref{fig:inference_NullIidStd}. The mode and empirical null standard deviation both suffered from high variance, much higher than typical robust estimators.\n\nThe empirical null standard deviation had some positive bias and it was found that this bias was quite sensitive to the bandwidth. Further fine-tuning on the bandwidth would be required for the empirical null standard deviation to be unbiased. For example, it appeared there was a consistent positive bias for all $r$ investigated, perhaps suggesting the bandwidth parameter $a$ should be decreased ever so slightly. The source of bias could come from the fact that the bandwidth was tuned to minimise the log squared error rather than the squared bias.\n\nIt is interesting to compare the empirical null standard deviation with MADA-mode, in Figure \\ref{fig:inference_NullIidStd}, because they both depend on the mode. The mode suffered from high variance which may affect dispersion estimators using the mode. The figure shows that MADA-mode had less variance than the empirical null standard deviation. This suggests that using the curvature of the density estimate at the mode was the main source of variance for the empirical null standard deviation.\n\nTo test for robustness, a sample of mixture of Gaussian random variables were simulated with the following distribution: $\\normal(0,1)$ with probability $90\\%$, $\\normal(3,1)$ with probability $10\\%$. The $\\normal(3,1)$ component acts as non-null statistics and robust estimators should be unaffected by it. The results are shown in Figures \\ref{fig:inference_NullIidMixtureMean} and \\ref{fig:inference_NullIidMixtureStd}. The empirical null estimators are robust to the non-null component but still suffers from high variance. The rest of the estimators were affected by the non-null statistics, making them unsuitable to estimate parameters of the null parameters.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullMean1.eps}\n \\caption{Mode}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullMean2.eps}\n \\caption{Bisquare weighted mean}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullMean4.eps}\n \\caption{Median}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullMean3.eps}\n \\caption{Mean}\n \\end{subfigure}\n }\n \\caption{Different types of central tendency estimators were used on $\\pi r^2$ standard Normal random variables. The boxplots represent the sampling distributions over 100 estimates. The dashed lines show the 95\\% confidence interval for the sample mean.}\n \\label{fig:inference_NullIidMean}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullStd1.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullStd2.eps}\n \\caption{MADA-mode}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullStd4.eps}\n \\caption{IQR}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIid_nullStd3.eps}\n \\caption{Standard deviation}\n \\end{subfigure}\n }\n \\caption{Different types of statistical dispersion estimators were used on $\\pi r^2$ standard Normal random variables. The boxplots represent the sampling distributions over 100 estimates. The dashed lines show the 95\\% confidence interval for the sample variance.}\n \\label{fig:inference_NullIidStd}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullMean1.eps}\n \\caption{Mode}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullMean2.eps}\n \\caption{Bisquared weighted mean}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullMean4.eps}\n \\caption{Median}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullMean3.eps}\n \\caption{Mean}\n \\end{subfigure}\n }\n \\caption{Different types of central tendency estimators were used on $\\pi r^2$ mixture of Gaussian random variables. The random variable has distribution $\\normal(0,1)$ with probability $0.9$ and $\\normal(3,1)$ with probability $0.1$. The boxplots represent the sampling distributions over 100 estimates. The dashed lines show the 95\\% confidence interval for the sample mean.}\n \\label{fig:inference_NullIidMixtureMean}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullStd1.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullStd2.eps}\n \\caption{MADA-mode}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullStd4.eps}\n \\caption{IQR}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/NullIidMixture_nullStd3.eps}\n \\caption{Standard deviation}\n \\end{subfigure}\n }\n \\caption{Different types of statistical dispersion estimators were used on $\\pi r^2$ mixture of Gaussian random variables. The random variable has distribution $\\normal(0,1)$ with probability $0.9$ and $\\normal(3,1)$ with probability $0.1$. The boxplots represent the sampling distributions over 100 estimates. The dashed lines show the 95\\% confidence interval for the sample variance.}\n \\label{fig:inference_NullIidMixtureStd}\n\\end{figure}\n\n\\afterpage{\\clearpage}\n\\section{Empirical Null Filter}\n\nReturning to the example at the start of the chapter, the empirical null could be used on the histogram shown in Figure \\ref{fig:inference_inferenceIntro_histogram}. The resulting empirical null distribution is shown in Figure \\ref{fig:inference_inferenceSubsampleAll}. The critical boundary, adjusted using the empirical null, which corresponds to the 5\\% FDR level was found to be $z\\leqslant\\inputNumber{../figures/inference/inferenceSubsampleAll_nullCritical1.txt}$ and $\\inputNumber{../figures/inference/inferenceSubsampleAll_nullCritical2.txt}\\geqslant z$ to 2 decimal places. This would test all pixels as negative which is incorrect.\n\nThe problem was that the empirical distribution of all of the test statistics looked like it did not come from a Normal distribution. It appeared that the test statistics varied spatially, suggesting that the empirical null varied spatially as well. To account for the spatial variability of the null parameters, one possible extension to the empirical null is to spilt the $z$ image into a grid and conduct inference in each section separately. There are problems with this though. The first problem is that how the grid is overlaid can be arbitrary, for example, the grid can be translated to produce different sections. Secondly, it is not clear how to combine the results from each section together \\citep{efron2008simultaneous}.\n\nThe empirical null filter extends the empirical null to images. The parameters of the null distribution varies spatially, slowly and smoothly. Let the test statistic of the pixel at position $(x,y)$ be\n\\begin{equation}\nZ_{x,y}\\sim\\normal(\\mu_{0,x,y},\\widehat{\\sigma}_{0,x,y}^2)\n\\end{equation}\nfor $x=1,2,\\dotsc,W$ and $y=1,2,\\dotsc,H$. $\\mu_{0,x,y}$ and $\\sigma_{0,x,y}$ are the null mean and null standard deviation at position $(x,y)$ respectively. Define the null hypotheses to be\n\\begin{equation}\n H_{0,x,y}:\\mu_{0,x,y}=\\widehat{\\mu}_{0,x,y}\n\\end{equation}\nwhich are tested against\n\\begin{equation}\n H_{1,x,y}:\\mu_{0,x,y}\\neq\\widehat{\\mu}_{0,x,y} \\ .\n\\end{equation}\n The empirical null filter aims to estimate the parameters $\\mu_{0,x,y}$ and $\\sigma_{0,x,y}$ for all $x$ and $y$.\n\n \\begin{figure}[t]\n \\centering\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsampleAll_subimagePositive.eps}\n \\caption{Test statistics}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/inferenceSubsampleAll_densityEstimate.eps}\n \\caption{Distribution of test statistics}\n \\end{subfigure}\n \\caption{The resulting test statistics for the \\texttt{AbsFilter} projection at \\ang{120} are shown in a). The density estimate of the statistics is shown in b). The empirical null distribution multiplied by some constant is also plotted in b); this is to illustrate the curvature is the same as the density estimate at the mode.}\n \\label{fig:inference_inferenceSubsampleAll}\n\\end{figure}\n\nThe empirical null filter uses the empirical null on neighbouring pixels to estimate the null parameters. To estimate $\\mu_{0,x,y}$ and $\\sigma_{0,x,y}$, a circular kernel $C_r(x,y)$ of radius $r$ was centred at $(x,y)$. All the pixels captured by the circular kernel and the ROI were used for the empirical null to obtain $\\widehat{\\mu}_{0,x,y}$ and $\\widehat{\\sigma}_{0,x,y}$, in other words\n\\begin{equation}\n \\widehat{\\mu}_{0,x,y} = \\argmax\\widehat{p}_{Z_{x,y}}(z)\n\\end{equation}\n\\begin{equation}\n \\widehat{\\sigma}_{0,x,y} = \\left[\n \\left.\n -\\dfrac{\\partial^2}{\\partial z^2}\\ln\\widehat{p}_{Z_{x,y}}(z)\n \\right|_{z=\\widehat{\\mu}_{0,x,y}}\n \\right]^{-1/2}\n\\end{equation}\nwhere\n\\begin{equation}\n\\widehat{p}_{Z_{x,y}}(z) =\n\\frac{1}{nh}\n \\sum_{i,j\\in K_{x,y}}\\phi\\left(\n \\dfrac{z_{i,j}-z}{h}\n \\right) \\ ,\n\\end{equation}\n$K_{x,y} = C_r(x,y) \\cap \\text{ROI}$ and $n=\\left|K_{x,y}\\right|$, not to be confused with $N$. A bandwidth of $h = (0.9n^{-1/5}\\ + 0.16) \\times \\text{min}\\left(s_{x,y},\\text{IQR}_{x,y}/1.34\\right)$ was used where $s_{x,y}$ and $\\text{IQR}_{x,y}$ are the sample standard deviation and sample interquartile range of the test statistics in $K_{x,y}$.\n\nAfter obtaining the estimates of the empirical null parameters, the test statistics were normalised using\n\\begin{equation}\n T_{x,y} =\n \\dfrac{\n Z_{x,y}-\\widehat{\\mu}_{0,x,y}\n }\n {\n \\widehat{\\sigma}_{0,x,y}\n }\n\\end{equation}\nand only the normalised test statistics were used in hypotheses testing. It should be noted that because the kernels can overlap, correlation between $T_{x,y}$ for different $(x,y)$ is introduced.\n\nThe parameter of choice is the kernel radius $r$. A radius too small will treat defects as the null and will not detect them, too big can also cause problems as well. The empirical null filter assumes that the null parameters vary smoothly and slowly so that the null test statistics captured by the kernel is somewhat Normal. If the radius is too big, the kernel will capture distanced test statistics with very different null distributions. If all of these test statistics are pulled together into a histogram, the null distribution may not look Normal at all which can cause problems when fitting a Normal null distribution. Furthermore, if $r\\rightarrow \\infty$, then this is just using the empirical null on all pixels.\n\nThe radius controls the proportion of pixels which are null captured by the kernel, denoted as $\\pi_0$. It should be selected so that $\\pi_0$ is of sensible value, for example, $>90\\%$. In practice, the defects are unknown so various kernel radiuses should be tried out. The kernel radius should be much larger than the size of the defect.\n\nThere exist similar filters, for example the local normalisation filter \\citep{sage2003teaching, sage2018local} uses a Gaussian filter to obtain $\\widehat{\\mu}_{0,x,y}$. $\\widehat{\\sigma}_{0,x,y}^2$ was obtained by applying another Gaussian filter on the $\\left(z_{x,y}-\\widehat{\\mu}_{0,x,y}\\right)^2$ image. The purpose of this filter was to normalise images with uneven illumination. The disadvantage of using this filter, for estimating the null parameters, is that it uses a Gaussian filter. This is a type of weighted mean which is not robust to non-null statistics. Also, the two Gaussian filters can have different radiuses, it is not clear how to set them.\n\n\\subsection{Design and Implementation}\n\nThe empirical null filter was implemented using open-source software called \\emph{ImageJ} \\citep{abramoff2004image, schneider2012nih, perez2013image} by modifying the existing class \\texttt{RankFilters}, available on \\emph{GitHub} \\citep{imagej2018rank}. This class implemented filters, such as the mean filter and median filter, using a circular kernel, making this a suitable framework for the empirical null filter. Figure \\ref{fig:inference_fijiGui} shows the graphical user interface of the empirical null filter in \\emph{Fiji} \\citep{schindelin2012fiji}, a distribution of \\emph{ImageJ}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/gui.png}\n \\caption{The graphical user interface of the empirical null filter in \\emph{Fiji}. The user can adjust the kernel radius as well as other advanced options relating to the Newton-Raphson method and the kernel density estimate. By-products such as the empirical null parameters can be shown after the filtering as well.}\n \\label{fig:inference_fijiGui}\n\\end{figure}\n\nAs a test, the empirical null filter with $r=\\SI{5}{\\pixel}$ was used on a \\emph{ImageJ} sample image \\texttt{bridge.gif}, as shown in Figure \\ref{fig:inference_fijiBridgeFilter}. It is meaningless to use the empirical null filter on an arbitrary image, however, it did verify that the filter coped with it with some computational considerations. The empirical null images $\\widehat{\\mu}_{0,x,y}$ and $\\widehat{\\sigma}_{0,x,y}$ could be of interest, in particular, the empirical null mean image could be interpreted as the result of a mode filter \\citep{griffin2000mean}. Figure \\ref{fig:inference_fijiMean} compares the empirical null mean with other averaging filters. The empirical null mean has an impasto effect and preserved edges which are similar to the result in \\cite{griffin2000mean}. Dispersion filters such as the standard deviation filter can be used to detect edges. Figure \\ref{fig:inference_fijiStd} compares the standard deviation filter with the empirical null standard deviation. The resulting images were similar but it was notable that the edges in the empirical null standard deviation were sharper, suggesting that the empirical null standard deviation is a more robust measure of dispersion compared to the standard deviation.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/bridge.png}\n \\caption{Before filtering}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/bridgeFilter.png}\n \\caption{After filtering}\n \\end{subfigure}\n }\n \\caption{The empirical null filter, with kernel radius $r=\\SI{5}{\\pixel}$, was used on the sample image \\texttt{bridge.gif}.}\n \\label{fig:inference_fijiBridgeFilter}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/bridge.png}\n \\caption{No filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/mean.png}\n \\caption{Mean filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/median.png}\n \\caption{Median filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/nullMean.png}\n \\caption{Empirical null mean}\n \\end{subfigure}\n }\n \\caption{Various averaging filters, with kernel radius $r=\\SI{5}{\\pixel}$, were used on the sample image \\texttt{bridge.gif}.}\n \\label{fig:inference_fijiMean}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/std.png}\n \\caption{Standard deviation filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/fiji/nullStd.png}\n \\caption{Empirical null standard deviation}\n \\end{subfigure}\n }\n \\caption{Various dispersion filters, with kernel radius $r=\\SI{5}{\\pixel}$, were used on the sample image \\texttt{bridge.gif}.}\n \\label{fig:inference_fijiStd}\n\\end{figure}\n\nWhen using the empirical null filter in \\emph{ImageJ} or \\emph{Fiji}, the user is presented with a menu, as shown in Figure \\ref{fig:inference_fijiGui}. The user can input the kernel radius, the filter then modifies the currently selected image $z_{x,y}$ to $t_{x,y}$ for all $x$ and $y$ using a circular kernel with the specified radius. The menu also has options for by-product images to be shown: the empirical null mean $\\widehat{\\mu}_{0,x,y}$, the empirical null standard deviation $\\widehat{\\sigma}_{0,x,y}$, the standard deviation filter, quartile filters. Advanced options are available to the user, for example, the number of initial values set the number of valid solutions to be found when using the Newton-Raphson method to find the mode of the density estimate. The user can also set the maximum number of steps and tolerance for the Newton-Raphson method. The bandwidth parameters for the kernel density estimate can also be adjusted here.\n\nLine filtering was done from left to right. At the start on the far left, an initial value of the median over the pixels in the kernel was used. Also, various initial values were randomly tried out until a requested number of valid solutions were found. For the following pixel to the right, the empirical null mean of the neighbouring left pixel was used as the initial value. This was chosen as it was assumed the empirical null mean would vary slowly and smoothly spatially. The implementation uses multiple threads, each thread filters a row in parallel.\n\nWhen the kernel captures pixels outside the boundary of the image or ROI, \\texttt{RankFilters} uses nearest pixel padding which fills in pixels outside the ROI with values to the nearest pixels. Figure \\ref{fig:inference_padding_nearest} shows an example of nearest pixel padding on the top left corner of a rectangle ROI. This is not suitable for the empirical null filter as this will cause bias in the density estimation. The empirical null filter uses \\texttt{NaN} padding which fills pixels outside the ROI with \\texttt{NaN}. This indicates that these pixels are outside the ROI and are ignored. The number of non-\\texttt{NaN} captured by the kernel was kept track for calculations such as the bandwidth.\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{\\subSize}\n \\begin{tabular}{lllllll}\n $\\ddots$ & $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\udots$ \\\\\n $\\dotsc$ & $z_{1,1}$ & $z_{1,1}$ & $z_{1,1}$ & $z_{1,2}$ & $\\dotsc$ \\\\\n $\\dotsc$ & $z_{1,1}$ & $z_{1,1}$ & $z_{1,1}$ & $z_{1,2}$ & $\\dotsc$ \\\\ \\cline{4-6}\n $\\dotsc$ & $z_{1,1}$ & \\multicolumn{1}{l|}{$z_{1,1}$} & $z_{1,1}$ & $z_{1,2}$ & $\\dotsc$ \\\\\n $\\dotsc$ & $z_{2,1}$ & \\multicolumn{1}{l|}{$z_{2,1}$} & $z_{2,1}$ & $z_{2,2}$ & $\\dotsc$ \\\\\n $\\udots$ & $\\vdots$ & \\multicolumn{1}{l|}{$\\vdots$} & $\\vdots$ & $\\vdots$ & $\\ddots$\n \\end{tabular}\n \\caption{Nearest pixel padding}\n \\label{fig:inference_padding_nearest}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\begin{tabular}{lllllll}\n $\\ddots$ & $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\udots$ \\\\\n $\\dotsc$ & \\texttt{NaN} & \\texttt{NaN} & \\texttt{NaN} & \\texttt{NaN} & $\\dotsc$ \\\\\n $\\dotsc$ & \\texttt{NaN} & \\texttt{NaN} & \\texttt{NaN} & \\texttt{NaN} & $\\dotsc$ \\\\ \\cline{4-6}\n $\\dotsc$ & \\texttt{NaN} & \\multicolumn{1}{l|}{\\texttt{NaN}} & $z_{1,1}$ & $z_{1,2}$ & $\\dotsc$ \\\\\n $\\dotsc$ & \\texttt{NaN} & \\multicolumn{1}{l|}{\\texttt{NaN}} & $z_{2,1}$ & $z_{2,2}$ & $\\dotsc$ \\\\\n $\\udots$ & $\\vdots$ & \\multicolumn{1}{l|}{$\\vdots$} & $\\vdots$ & $\\vdots$ & $\\ddots$\n \\end{tabular}\n \\caption{\\texttt{NaN} padding}\n \\label{fig:inference_padding_nan}\n \\end{subfigure}\n \\caption{When a kernel contains pixels outside the ROI, as shown by the solid line, the pixels can either be extrapolated using the nearest pixel or completely ignored by filling in the missing pixels with \\texttt{NaN}.}\n \\label{fig:inference_padding}\n\\end{figure}\n\nWhen filtering arbitrary images such as \\texttt{bridge.gif}, a few computational considerations were needed. \\texttt{bridge.gif} is an 8-bit image so values were represented as integers. This opened up opportunities for the interquartile range and the standard deviation to be zero. This caused problems as this would set the bandwidth to zero. When the standard deviation is zero, it was set to $0.289$ which corresponds to the standard deviation of a uniform random variable. When the inter-quartile range is zero, the inter-quartile range was set to the standard deviation $\\times 1.34$.\n\nWhen the number of initial values is too few, poor solutions to $\\widehat{\\mu}_0$ are possible which then can propagate to the pixel to the right and used as its initial value. Poor initial values could cause the Newton-Raphson method to never converge. If the Newton-Raphson method failed to converge too many times, the median, over pixels in the kernel, is used as the new initial value. Should it fail too, the program gives up and puts a $\\texttt{NaN}$ in that pixel.\n\n\\subsection{Filtering an Image with No Defects}\n\nAn experiment was conducted to demonstrate the effects of the empirical null filter on a simulated $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$ Gaussian image, where all the pixels have value distributed as standard Normal. The purpose is to simulate test statistics which are null and few positive results, subject to the FDR, should be detected. The test statistics after filtering should preserve its original distribution, or at least close to it. The statistical moments before and after filtering should be the same and were investigated. Because the empirical null mean and empirical null standard deviation are random variables, the filtered pixels will never be Normal and any normality tests would be too strict for this experiment.\n\nContamination is defined as some smooth and slowly varying function added and/or multiplied to an image. Conducting hypotheses testing on a contaminated Gaussian image may cause errors if the null distributions were incorrectly specified as standard Normal. In this experiment, the contamination was a gradient such that the pixel at position $(x,y)$ has a value distributed as $Z_{x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_{0,x,y}^2)$ where\n\\begin{equation}\n \\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0) \\ ,\n\\end{equation}\n$(x_0,y_0)$ is the centre of the image and $\\sigma_{0,x,y}=2$ for all $(x,y)$.\n\nA caveat is that some bias to the null standard deviation estimation would be introduced because of the contamination. This is because sources of variance captured by the kernel are from $\\sigma_{0,x,y}$ and also from the variability of $\\mu_{0,x,y}$. This can be shown with an example. Let $Z_{x,y}$ be the value of the pixels captured by the circular kernel, centred at the origin, for integer values of $(x,y)$ such that $x^2+y^2\\leqslant r^2$. Suppose that $Z_{x,y}\\sim\\normal(ax+by,\\sigma^2)$. The quantities of interest are the expectation and variance of all the values contained in the kernel because this is the quantity the estimators are estimating.\n\nThe calculations of the expectation and variance of $Z_{x,y}$ can be approximated by treating $x$ and $y$ as uniformly distributed within a circle centred at the origin with radius $r$. The expectation is\n\\begin{equation}\n\\expectation[Z_{X,Y}] = \\expectation\\expectation[Z_{X,Y}|X,Y]=0 \\ .\n\\end{equation}\n\nThe variance is given as $\\variance[Z_{X,Y}] = \\expectation\\variance[Z_{X,Y}|X,Y] + \\variance\\expectation[Z_{X,Y}|X,Y]$. From the distribution of $Z_{X,Y}$, $\\expectation\\variance[Z_{X,Y}|X,Y] = \\sigma^2$ and $\\variance\\expectation[Z_{X,Y}|X,Y]=\\variance[aX+bY]$. This is worked out to be\n\\begin{align}\n\\variance[aX+bY] &= \\dfrac{1}{\\pi r^2}\\int_{\\rho=0}^{\\rho=r}\\int_{\\theta=0}^{\\theta=2\\pi}\n\\rho^3(a\\cos\\theta + b\\sin\\theta)^2 \\diff \\theta \\diff \\rho\n\\nonumber\\\\\n& = \\dfrac{1}{4}(a^2+b^2)r^2\n\\end{align}\nthus\n\\begin{equation}\n\\variance[Z_{X,Y}] = \\dfrac{1}{4}(a^2+b^2)r^2 + \\sigma^2 \\ .\n\\end{equation}\nThis shows that when estimating the null variance on a contaminated Gaussian image, there may be a $(a^2+b^2)r^2/4$ bias. If the contamination is slowly varying, that is $a$ and $b$ are small, then the bias should be small.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullGaussianScript_beforeFilter.eps}\n \\caption{Unfiltered}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullGaussianScript_afterFilter.eps}\n \\caption{Filtered}\n \\label{fig:inference_allNullGaussianScript_afterFilter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullGaussianScript_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullGaussianScript_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel}\\times \\SI{256}{\\pixel}$ Gaussian image before and after filtering with kernel radius \\SI{20}{\\pixel}. Also shown are the empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_allNullGaussianScript}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullPlaneScript_beforeFilter.eps}\n \\caption{Unfiltered}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullPlaneScript_afterFilter.eps}\n \\caption{Filtered}\n \\label{fig:inference_allNullPlaneScript_afterFilter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullPlaneScript_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullPlaneScript_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel}\\times \\SI{256}{\\pixel}$ contaminated Gaussian image before and after filtering with kernel radius \\SI{20}{\\pixel}. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. Also shown are the empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_allNullPlaneScript}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullGaussianScript_pValue.eps}\n \\caption{No contamination}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/allNullPlaneScript_pValue.eps}\n \\caption{With contamination}\n \\end{subfigure}\n \\caption{$p$-values from a filtered $\\SI{256}{\\pixel}\\times \\SI{256}{\\pixel}$ Gaussian image with kernel radius \\SI{20}{\\pixel}. In b), the Gaussian image was contaminated before filtering. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The $p$-values were obtained from the images in Figures \\ref{fig:inference_allNullGaussianScript_afterFilter} and \\ref{fig:inference_allNullPlaneScript_afterFilter}. The critical region corresponds to the 5\\% FDR level. The dotted lines shows the $p$-values if they were uniformly distributed.}\n \\label{fig:inference_allNull_pValues}\n\\end{figure}\n\nAn example of a Gaussian image with/without contamination before/after filtering are shown in Figures \\ref{fig:inference_allNullGaussianScript} and \\ref{fig:inference_allNullPlaneScript}. In the contaminated example, the filter managed to estimate the null mean, picking up the gradient. This enabled the filtered image to look flat and removed the contamination. The $p$-values after filtering are shown in Figure \\ref{fig:inference_allNull_pValues}. A quick inspection suggests that the filtered pixels appeared reasonably Normal for this particular example.\n\nFor a given kernel radius $r$, 100 different Gaussian images were filtered to investigate the within image mean, standard deviation and kurtosis of the filtered statistics. Various other filters for normalisation were used as well such as the MADA-mode null filter, median IQR null filter and the mean variance null filter. They use different estimators to estimate the null parameters as their name suggests, Table \\ref{table:inference_nullFilters} clarifies them.\n\n\\begin{table}\n \\centering\n \\begin{tabular}{l|l|l}\n Filters&Null mean&Null standard deviation\\\\\\hline\n MADA-mode null&Mode&MADA-mode $\\times1.483$\\\\\n Median IQR null&Median&IQR $\\div1.349$\\\\\n Mean variance null&Mean&Standard deviation\n \\end{tabular}\n \\caption{Various filters using different estimators to estimate the null parameters are described here.}\n \\label{table:inference_nullFilters}\n\\end{table}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianEmpirical_mean.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMadMode_mean.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMedianIqr_mean.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMeanVar_mean.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample mean of a filtered Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullGaussian_mean}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianEmpirical_variance.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMadMode_variance.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMedianIqr_variance.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMeanVar_variance.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample standard deviation of a filtered Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullGaussian_var}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianEmpirical_kurtosis.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMadMode_kurtosis.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMedianIqr_kurtosis.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullGaussianMeanVar_kurtosis.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample kurtosis of a filtered Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullGaussian_kurtosis}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneEmpirical_mean.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMadMode_mean.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMedianIqr_mean.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMeanVar_mean.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample mean of a filtered contaminated Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullPlane_mean}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneEmpirical_variance.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMadMode_variance.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMedianIqr_variance.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMeanVar_variance.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample standard deviation of a filtered contaminated Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullPlane_var}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneEmpirical_kurtosis.eps}\n \\caption{Empirical null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMadMode_kurtosis.eps}\n \\caption{MADA-mode null filter}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMedianIqr_kurtosis.eps}\n \\caption{Median IQR null filter}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/AllNullPlaneMeanVar_kurtosis.eps}\n \\caption{Mean var null filter}\n \\end{subfigure}\n }\n \\caption{The within image sample kurtosis of a filtered contaminated Gaussian image of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The boxplots summarise the 100 repeated simulations of the image. The dashed lines show the 95\\% confidence interval using standard tests and assuming independence.}\n \\label{fig:inference_AllNullPlane_kurtosis}\n\\end{figure}\n\nThe results for the filtered images are shown in Figures \\ref{fig:inference_AllNullGaussian_mean}, \\ref{fig:inference_AllNullGaussian_var} and \\ref{fig:inference_AllNullGaussian_kurtosis}. For the filtered contaminated images, they are in Figures \\ref{fig:inference_AllNullPlane_mean}, \\ref{fig:inference_AllNullPlane_var} and \\ref{fig:inference_AllNullPlane_kurtosis}. The filtered test statistic means did agree with zero as expected, showing that the filters centred the statistics when normalising. When using the empirical null filter, there was some negative bias with the standard deviation of the filtered test statistics. The negative bias comes from a positive bias when estimating the null standard deviation.\n\nWhen filtering the contaminated image, all estimators suffered from negative bias in the standard deviation of the filtered statistics, in particular when the kernel radius increased. This is shown in Figure \\ref{fig:inference_AllNullPlane_var}. As discussed before, the kernel captured the variation due to the contamination and this added bias to the estimation of the null standard deviation.\n\nThe kurtosis in Figures \\ref{fig:inference_AllNullGaussian_kurtosis} and \\ref{fig:inference_AllNullPlane_kurtosis} showed that the empirical null filter caused the filtered statistics to have heavy tails, in particular, for small kernel radiuses. This could cause problems in hypotheses testing because heavy tails could be misinterpreted as a contribution from non-null statistics. The source of the kurtosis inflation is from the estimation of the null standard deviation. This can be seen by comparing the empirical null filter with the MADA-mode null filter because the only difference between the two filters is in the estimation of the null standard deviation.\n\n\\afterpage{\\clearpage}\n\\subsection{Detection of Simulated Defects}\n\nThe empirical null filter was tested to see if it can assist in detecting simulated defects from an image with/without contamination. A defect assigns pixels to have a value not distributed under the null distribution, but instead, under an alternative, or non-null, distribution. For example, suppose $Z_{x,y}|H_{0,x,y}\\sim\\normal(0,1)$, then a defect assign certain pixels to have a value distributed as $Z_{x,y}|H_{1,x,y}\\sim\\normal(\\mu_{1},1)$ where $\\mu_1\\neq0$.\n\nTo recap, contamination is the result of a linear transform of the test statistics $Z_{x,y}$. For example, in this experiment, the image was multiplied by 2 and a gradient was added to it. The resulting null and alternative distributions are\n\\begin{align}\n Z_{x,y}|H_{0,x,y}&\\sim\\normal(\\mu_{0,x,y},2^2)\n \\\\\n Z_{x,y}|H_{1,x,y}&\\sim\\normal(2\\mu_1+\\mu_{0,x,y},2^2)\n\\end{align}\nrespectively where $\\mu_{0,x,y}=0.01(x-x_0)+0.01(y-y_0)$ and $(x_0,y_0)$ is the centre of the image. This was used in this experiment to simulate a contaminated image with defects.\n\nThe empirical null filter aims to estimate the null distribution parameters from $Z_{x,y}$ to normalise it to form $T_{x,y}$. By normalising it, $T_{x,y}|H_{0,x,y}$ should be approximately standard Normal and hypotheses testing can be used to detect defects.\n\nVarious defects were investigated. Speckle defect with density $\\pi_1$ assign all test statistics to be non-null with probability $\\pi_1$ and are null otherwise. This was chosen because the proportion of null statistics captured by the kernel should be the same for all kernel radiuses. A line defect assigns columns of pixels to be non-null, a kernel would only capture a section of the defect. A square defect was also investigated and a kernel can capture the entire defect only if its radius is large enough.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_imageClean.eps}\n \\caption{Uncontaminated}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_imageContaminated.eps}\n \\caption{Contaminated}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_EmpiricalNullFilterimageFiltered.eps}\n \\caption{Filtered}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_EmpiricalNullFilternullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$ Gaussian image with speckle defect is shown in a). The image was contaminated in b) and then filtered with kernel radius \\SI{20}{\\pixel} in c). Non-null pixels have the distribution $\\normal(3,1)$. The speckle defect has density $\\pi_1 = 0.1$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. In a), c) and d), highlighted in red are pixels tested as positive at the 5\\% FDR level.}\n \\label{fig:inference_defectDustExample}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_imageClean.eps}\n \\caption{Uncontaminated}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_imageContaminated.eps}\n \\caption{Contaminated}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_EmpiricalNullFilterimageFiltered.eps}\n \\caption{Filtered}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_EmpiricalNullFilternullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$ Gaussian image with a line defect is shown in a). The image was contaminated in b) and then filtered with kernel radius \\SI{20}{\\pixel} in c). Non-null pixels have the distribution $\\normal(3,1)$. The line defect is \\SI{5}{\\pixel} thick. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. In a), c) and d), highlighted in red are pixels tested as positive at the 5\\% FDR level.}\n \\label{fig:inference_defectLineExample}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_imageClean.eps}\n \\caption{Uncontaminated}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_imageContaminated.eps}\n \\caption{Contaminated}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_EmpiricalNullFilterimageFiltered.eps}\n \\caption{Filtered}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_EmpiricalNullFilternullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$ Gaussian image with a square defect is shown in a). The image was contaminated in b) and then filtered with kernel radius \\SI{20}{\\pixel} in c). Non-null pixels have the distribution $\\normal(3,1)$. The square defect is $\\SI{30}{\\pixel}\\times \\SI{30}{\\pixel}$ in size. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. In a) and c), highlighted in red are pixels tested as positive at the 5\\% FDR level.}\n \\label{fig:inference_defectSquareExample}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_imageClean.eps}\n \\caption{Uncontaminated}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_imageContaminated.eps}\n \\caption{Contaminated}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_EmpiricalNullFilterimageFiltered.eps}\n \\caption{Filtered}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_EmpiricalNullFilternullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$ Gaussian image with a square defect is shown in a). The image was contaminated in b) and then filtered with kernel radius \\SI{40}{\\pixel} in c). Non-null pixels have the distribution $\\normal(3,1)$. The square defect is $\\SI{30}{\\pixel}\\times \\SI{30}{\\pixel}$ in size. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. In a) and c), highlighted in red are pixels tested as positive at the 5\\% FDR level.}\n \\label{fig:inference_defectSquare2Example}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_EmpiricalNullFilterpValueFiltered.eps}\n \\caption{Speckle defect, $r=\\SI{20}{\\pixel}$}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_EmpiricalNullFilterpValueFiltered.eps}\n \\caption{Line defect, $r=\\SI{20}{\\pixel}$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_EmpiricalNullFilterpValueFiltered}\n \\caption{Square defect, $r=\\SI{20}{\\pixel}$}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_EmpiricalNullFilterpValueFiltered}\n \\caption{Square defect, $r=\\SI{40}{\\pixel}$}\n \\end{subfigure}\n }\n \\caption{A $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$ contamined defected Gaussian image was filtered using the empirical null filter with kernel radius $r$. Shown are the $p$-values converted from the filtered images. Non-null pixels have the distribution $\\normal(3,1)$. The speckle defect has density $\\pi_1=0.1$. The line defect is \\SI{5}{\\pixel} thick. The square defect is $\\SI{30}{\\pixel}\\times \\SI{30}{\\pixel}$ in size. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The critical region corresponds to the 5\\% FDR level. The dotted lines shows the $p$-values if they were uniformly distributed.}\n \\label{fig:inference_defectExamplePValue}\n\\end{figure}\n\nAn example of a speckle defected image is shown in Figure \\ref{fig:inference_defectDustExample}, the figure also shows the image contaminated and then filtered. Without filtering, the top-left and bottom-right of the contaminated image have large, in magnitude, test statistics. In these areas, a lot of pixels would be tested as falsely positive. By using the empirical null filter, the resulting filtered image reassemble the uncontaminated image and appropriate inference can be done on the filtered image. However, some statistical power was lost when conducting hypotheses testing on the filtered image because, in areas of high empirical null standard deviation, the normalised statistics became too small which then decreased the detection power. This cannot be avoided as areas of high empirical null standard deviation was due to the randomness in sampling. Similar comments can be made for the line defect as shown in Figure \\ref{fig:inference_defectLineExample}.\n\nThe square defected images, including the contaminated ones, are shown in Figures \\ref{fig:inference_defectSquareExample} and \\ref{fig:inference_defectSquare2Example}. Figure \\ref{fig:inference_defectSquareExample} filtered the contaminated image using a kernel radius of $r=\\SI{20}{\\pixel}$, while Figure \\ref{fig:inference_defectSquare2Example} used a kernel radius of $r=\\SI{40}{\\pixel}$. With $r=\\SI{20}{\\pixel}$, the kernel is smaller than the square defect, therefore, the defect was treated as the null. This is evident because the empirical null mean captured the defect. This resulted in difficulty detecting the defect. With $r=\\SI{40}{\\pixel}$, the kernel is bigger than the defect and treated the defect as non-null. In this scenario, the empirical null filter recovered the gradient in the empirical null which was then used for normalisation.\n\nThe empirical null mean in Figure \\ref{fig:inference_defectSquareExample} demonstrated the multi-thread nature of the implementation of the empirical null filter. It can be seen that there were horizontal streaks where the defect is. This is the result of each thread filtering a row and jumping from one mode, the null, to the other, the non-null, at different times. The horizontal streaks can be removed by using more initial values so that the Newton-Raphson method can pinpoint which mode is greater. However, this would be at a computational cost.\n\nFigure \\ref{fig:inference_defectExamplePValue} shows the $p$-values after filtering a contaminated defected image. The FDR can be estimated in these figures by dividing the number of null statistics in the critical region by the number of statistics in the critical region. With a sensible kernel radius, some of the defects can be detected. False negatives are common but this is inevitable to control for the FDR. The figure shows that using a kernel radius of $r=\\SI{20}{\\pixel}$, for the square defect, failed to capture the defect. This, again, emphasise the importance of a good kernel radius.\n\nThe receiver operating characteristic (ROC) curves \\citep{green1966signal, metz1978basic, hanley1982meaning, friedman2001elements, cook2007use} are shown in Figure \\ref{fig:inference_defectSimulationRoc} for the various defects and filters. The ROC curve is a parametric plot, plotting the true positive rate (sensitivity) against the false positive rate ($1-\\text{specificity}$) for varying thresholds. The area under the ROC curve (AUC) is a commonly used statistic to quantify the performance of the test \\citep{friedman2001elements}. Interpretations of the area do exist \\citep{metz1978basic,hanley1982meaning} and discussed thoroughly in \\cite{cook2007use}.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleDust_roc.eps}\n \\caption{Speckle defect, $r=\\SI{20}{\\pixel}$}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleLine_roc.eps}\n \\caption{Line defect, $r=\\SI{20}{\\pixel}$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare20_roc.eps}\n \\caption{Square defect, $r=\\SI{20}{\\pixel}$}\n \\label{fig:inference_defectSimulationRoc_square20}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectExampleSquare40_roc.eps}\n \\caption{Square defect, $r=\\SI{40}{\\pixel}$}\n \\end{subfigure}\n }\n \\caption{ROC curves for various defected $\\SI{256}{\\pixel}\\times \\SI{256}{\\pixel}$ Gaussian images. The upper/lower dot-dashed lines show the resulting ROC curve when testing on an image without/with contamination respectively. The different curves are the resulting ROC curves after filtering a contaminated image. The speckle defect has density $\\pi_1=0.1$. The line defect is \\SI{5}{\\pixel} thick. The square defect is $\\SI{30}{\\pixel}\\times \\SI{30}{\\pixel}$ in size. Defected pixels have the distribution $\\normal(3,1)$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$.}\n \\label{fig:inference_defectSimulationRoc}\n\\end{figure}\n\nIn terms of the AUC, the empirical null filter improved the performance of hypotheses testing compared with using the unfiltered contaminated image. The performance before contamination cannot be recovered but it was an improvement. For kernel radiuses too small, such as Figure \\ref{fig:inference_defectSimulationRoc_square20}, the empirical null filter deteriorated the performance and it would be better off using the contaminated image.\n\n\\subsection{Speckle Defect Experiment}\n\nThe ROC curves consider all thresholds or specificities used in the hypotheses testing. In the previous experiment, it was found that the variance of the test statistics was different before and after filtering. As a result, for a given threshold, such as the 5\\% FDR level, the specificity may change ever so slightly after filtering.\n\nAn experiment was conducted to investigate how filtering affects hypotheses testing. A $\\SI{256}{\\pixel}\\times \\SI{256}{\\pixel}$ Gaussian image with speckle defect, with density $\\pi_1=0.1$, and various values of $\\mu_1$ were investigated. The AUC, type 1 error, type 2 error and FDR were measured when testing on an image with a defect before and after contamination and then after filtering with the contamination. The AUC was obtained by integrating the ROC curve. A kernel radius of $r=\\SI{20}{\\pixel}$ was used and was repeated 100 times by simulating another image.\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectAltDustEmpirical_roc.eps}\n \\caption{AUC}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectAltDustEmpirical_type1.eps}\n \\caption{Type 1 error}\n \\label{fig:inference_DefectSimulationDust_type1}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectAltDustEmpirical_type2.eps}\n \\caption{Type 2 error}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectAltDustEmpirical_fdr.eps}\n \\caption{FDR}\n \\label{fig:inference_DefectSimulationDust_fdr}\n \\end{subfigure}\n }\n \\caption{AUC and various errors obtained when conducting hypotheses testing on an empirical null filtered contaminated speckle defected Gaussian image of size $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$. In a), the upper/lower dot-dashed lines shows the resulting mean AUC when testing on a defected image before/after contamination respectively. In b), c) and d), the dashed lines show the 95\\% empirical confidence interval of the resulting error when testing the defected image before contamination. The filter used a kernel radius of \\SI{20}{\\pixel}. Defected pixels have the distribution $\\normal(\\mu_1,1)$ where $\\mu_1$ was varied in this experiment. The speckle defect has density $\\pi_1 = 0.1$. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The test was done at the 5\\% FDR level. The boxplots summarise the 100 repeated simulations of the image.}\n \\label{fig:inference_DefectSimulationDust}\n\\end{figure}\n\nThe results are shown in Figure \\ref{fig:inference_DefectSimulationDust}. The AUC quantified showed that using the filter improved the performance of the test from the contamination. The type 1 error, or specificity, did decrease after filtering. However, the results illustrated that the FDR is controlled quite well at around 5\\% after filtering. This meant that in this particular example, FDR control is consistent after filtering. At situations with low detection power, the FDR can fluctuate between 0 and 1 and can take only so many values. For example, if 3 positive pixels were detected, then the FDR can only take values of multiples of $1/3$ between and including 0 and 1.\n\n\\subsection{Optimal Kernel Radius Experiment}\n\nThere is the question of what kernel radius to choose for a given defect. Literature, such as \\cite{efron2004large} and \\cite{schwartzman2008empirical}, suggest that, as a rule of thumb, the proportion of non-null statistics should not be larger than $\\pi_1=0.1$ to satisfy some assumptions made for the empirical null. Given the size of the defect, one could work out the minimum kernel radius by setting a threshold for the maximum proportion of the area of the kernel which contains a defect to 10\\%. In other words, for a kernel with radius $r$\n\\begin{equation}\n0.1 > \\dfrac{\\text{maximum area of defect captured by the kernel}}{\\text{area of kernel}} \\ .\n\\end{equation}\nFor a $d \\times d$ square defect, this is $r > d \\sqrt{10/\\pi}$.\nIn the example of $d=\\SI{30}{\\pixel}$, this is about $r>\\SI{54}{\\pixel}$. For the line defect with thickness $\\SI{5}{\\pixel}$, the radius is about $r>\\SI{32}{\\pixel}$. These are rules of thumb because the kernel is not perfectly circular when used on a grid of pixels. It should be pointed out that the implementation can accept non-integer radiuses if desired.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusLine_roc.eps}\n \\caption{AUC}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusLine_type1.eps}\n \\caption{Type 1 error}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusLine_type2.eps}\n \\caption{Type 2 error}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusLine_fdr.eps}\n \\caption{FDR}\n \\end{subfigure}\n }\n \\caption{AUC and various errors obtained when conducting hypotheses testing on a filtered contaminated line defected Gaussian image, of size $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$, using various kernel radiuses. The dashed lines show the resulting 95\\% empirical confidence interval when the hypotheses testing was done on the uncontaminated defected image. Defected pixels have the distribution $\\normal(3,1)$. The line defect is \\SI{5}{\\pixel} thick. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The test was done at the 5\\% FDR level. The boxplots summarise the 100 repeated simulations of the image.}\n \\label{fig:inference_Experiment_DefectRadiusLine}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusSquare_roc.eps}\n \\caption{AUC}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusSquare_type1.eps}\n \\caption{Type 1 error}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusSquare_type2.eps}\n \\caption{Type 2 error}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectRadiusSquare_fdr.eps}\n \\caption{FDR}\n \\end{subfigure}\n }\n \\caption{AUC and various errors obtained when conducting hypotheses testing on a filtered contaminated square defected Gaussian image, of size $\\SI{256}{\\pixel} \\times \\SI{256}{\\pixel}$, using various kernel radiuses. The dashed lines show the resulting 95\\% empirical confidence interval when the hypotheses testing was done on the uncontaminated defected image. Defected pixels have the distribution $\\normal(3,1)$. The square defect is $\\SI{30}{\\pixel}\\times \\SI{30}{\\pixel}$ pixels in size. The contamination was such that the null distribution is $Z_{x,y}|H_{0,x,y}\\sim\\normal(\\mu_{0,x,y},\\sigma_0^2)$ where $\\mu_{0,x,y} = 0.01 (x-x_0) + 0.01 (y-y_0)$ and $\\sigma_0=2$. The test was done at the 5\\% FDR level. The boxplots summarise the 100 repeated simulations of the image.}\n \\label{fig:inference_Experiment_DefectRadiusSquare}\n\\end{figure}\n\nAn experiment was conducted where a contaminated defected Gaussian image, of size $\\SI{256}{\\pixel}\\times\\SI{256}{\\pixel}$, was filtered using various kernel radiuses for a fixed alternative distribution $\\normal(3,1)$. Hypotheses testing was done on the uncontaminated defected and filtered contaminated defected images. The AUC and various errors were recorded. This was repeated 100 times by simulating the image again. The AUC was obtained by integrating the ROC curve. The results of the uncontaminated defected image were all pooled together to obtain the empirical distribution of the AUC and errors without contamination. Results for the line and square defect are shown in Figures \\ref{fig:inference_Experiment_DefectRadiusLine} and \\ref{fig:inference_Experiment_DefectRadiusSquare} respectively.\n\nThe results showed that the AUC increased with kernel radius. This highlighted that a kernel with a good radius can perform almost as though there was no contamination. It also appeared that the optimal AUC was achieved when using the kernel radius from the rule of thumb, making these results consistent with the literature.\n\nFor large enough kernel radius, the FDR was controlled at around 5\\%. It appeared that a large kernel radius helped preserve FDR control after filtering in these examples. It was observed that there exist a kernel radius which minimised the type 2 error. So a large kernel radius is required for FDR control but too large can lose statistical power. It was noticed that filtering using large kernel radiuses is computationally slower.\n\n\\afterpage{\\clearpage}\n\\subsection{Application to Real Projections}\n\nThe empirical null filter was applied to the \\texttt{AbsFilter} projection at \\ang{120}. Figures \\ref{fig:inference_AbsFilterDeg120_inference_radius1} and \\ref{fig:inference_AbsFilterDeg120_inference_radius4} shows the resulting inference when using a kernel radius of $r=\\addNumber{../figures/inference/DefectDetectAbsFilterDeg120_radius1.txt}\\,\\SI{}{\\pixel}$ and $r=\\addNumber{../figures/inference/DefectDetectAbsFilterDeg120_radius4.txt}\\,\\SI{}{\\pixel}$ respectively. With a small radius, the empirical null treated the voids as the null which is shown by the empirical null mean capturing the features of the voids. When a large enough radius was used, the larger voids were highlighted by the hypotheses test and the empirical null mean became smoother. This is evidence that this method ironed out false positives observed at the start of the chapter.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius1_sig.eps}\n \\caption{Obtained projection (\\SI{}{\\adu})}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius1_logp.eps}\n \\caption{$-\\log p$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius1_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius1_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{Resulting inference when using the empirical null filter on the \\texttt{AbsFilter} projection at \\ang{120}. A kernel radius of $r=\\addNumber{../figures/inference/DefectDetectAbsFilterDeg120_radius1.txt}\\,\\SI{}{\\pixel}$ was used. a) Highlighted in red are positive pixels at the 5\\% FDR level. b) $p$-values on the log scale. c) and d) Empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_AbsFilterDeg120_inference_radius1}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius4_sig.eps}\n \\caption{{Obtained projection (\\SI{}{\\adu})}}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius4_logp.eps}\n \\caption{$-\\log p$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius4_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectAbsFilterDeg120_radius4_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{Resulting inference when using the empirical null filter on the \\texttt{AbsFilter} projection at \\ang{120}. A kernel radius of $r=\\addNumber{../figures/inference/DefectDetectAbsFilterDeg120_radius4.txt}\\,\\SI{}{\\pixel}$ was used. a) Highlighted in red are positive pixels at the 5\\% FDR level. b) $p$-values on the log scale. c) and d) Empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_AbsFilterDeg120_inference_radius4}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/cornerSelect.eps}\n \\caption{$z$ image with section highlighted}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/cornerSelectHist.eps}\n \\caption{Histogram of $z$ in the kernel}\n \\end{subfigure}\n }\n \\caption{A section of the bottom right of an unfiltered $z$ image is shown in a). The $z$ statistics in it are shown as a histogram in b) and indicates a bimodal distribution.}\n \\label{fig:inference_cornerSelect}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\subSize]{../figures/inference/segment.eps}\n \\caption{The $z$ image was segmented further into 7 segments shown by the dotted red lines.}\n \\label{fig:inference_segmentFurther}\n\\end{figure}\n\nFalse positives were detected on the bottom right of the test sample. When placing the kernel on the bottom right corner, the null distribution is bimodal because each face has a different distribution. Figure \\ref{fig:inference_cornerSelect} illustrates this. The geometry of the faces and the kernel used was such that it resulted in one of the faces to be treated as non-null and then tested positive, falsely so.\n\nThe false positives were tackled by manually segmenting the image further as shown in Figure \\ref{fig:inference_segmentFurther}. It was segmented using the edges of the $z$ image. The empirical null filter was then used on each segment, or ROI, independently, ignoring any pixels outside the ROI the filter is working on. The resulting filtered segments were stitched together to form the resulting filtered image. By using this method, the resulting inference is shown in Figure \\ref{fig:inference_subroiAbsFilterDeg120_inference_radius4}. It can be seen that the empirical null mean did change face to face, resulting in a clear boundary set by the edges. Most of the false positives around the corners and edges have been eliminated.\n\nThe same procedure was done on the projection at \\ang{30}, as shown in Figure \\ref{fig:inference_subroiAbsFilterDeg30_inference_radius1}, with similar results.\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg120_radius4_sig.eps}\n \\caption{{Obtained projection (\\SI{}{\\adu})}}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg120_radius4_logp.eps}\n \\caption{$-\\log p$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg120_radius4_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg120_radius4_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{Resulting inference when using the empirical null filter on the \\texttt{AbsFilter} projection at \\ang{120} and on each segment independently. A kernel radius of $r=\\addNumber{../figures/inference/DefectDetectSubRoiAbsFilterDeg120_radius4.txt}\\,\\SI{}{\\pixel}$ was used. a) Highlighted in red are positive pixels at the 5\\% FDR level. The test statistics from all segments were combined in the BH procedure. b) $p$-values on the log scale. c) and d) Empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_subroiAbsFilterDeg120_inference_radius4}\n\\end{figure}\n\n\\begin{figure}[p]\n \\centering\n \\centerline{\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg30_radius1_sig.eps}\n \\caption{{Obtained projection (\\SI{}{\\adu})}}\n \\end{subfigure}\n \\begin{subfigure}[b]{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg30_radius1_logp.eps}\n \\caption{$-\\log p$}\n \\end{subfigure}\n }\n \\centerline{\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg30_radius1_nullMean.eps}\n \\caption{Empirical null mean}\n \\end{subfigure}\n \\begin{subfigure}{\\subSize}\n \\includegraphics[width=\\textwidth]{../figures/inference/DefectDetectSubRoiAbsFilterDeg30_radius1_nullStd.eps}\n \\caption{Empirical null std}\n \\end{subfigure}\n }\n \\caption{Resulting inference when using the empirical null filter on the \\texttt{AbsFilter} projection at \\ang{30} and on each segment independently. A kernel radius of $r=\\addNumber{../figures/inference/DefectDetectSubRoiAbsFilterDeg30_radius1.txt}\\,\\SI{}{\\pixel}$ was used. a) Highlighted in red are positive pixels at the 5\\% FDR level. The test statistics from all segments were combined in the BH procedure. b) $p$-values on the log scale. c) and d) Empirical null mean and empirical null standard deviation.}\n \\label{fig:inference_subroiAbsFilterDeg30_inference_radius1}\n\\end{figure}\n\n\\section{Conclusion}\n\nThe empirical null filter demonstrated that it can adjust the null hypotheses according to the data to make a sensible inference. In the simulations, it was found that the FDR level is preserved after filtering. Also, the empirical null filter outperformed other methods based on quantiles, as suggested by \\cite{efron2004large}. A sensible kernel radius is required to avoid treating defects as the null.\n\nIn the experiment, the larger voids of diameter \\SI{2.4}{\\milli\\metre} were detected in the test sample. False positives do occur but this is unavoidable because tests were done at the 5\\% FDR level. Typically in the experiments, false positives were isolated single pixels and probably occurred due to random chance. Clusters of positive pixels should raise suspicion so it would be a good idea to borrow strength from neighbouring pixels. For example, one could create a binary image, assigning a Boolean value whether that pixel was tested positive or not. A binary image filter, such as erode followed by a dilate, can be used to remove isolated positive pixels to emphasise the cluster of positive pixels.\n\nThe assessment of the inference could be improved if the location of the defects were known in the projections. This would allow identifying which positives are true and false positives so that the analysis can be quantified using a ROC curve.\n\nFor good results, the faces of the test sample were segmented. This was necessary because the empirical null filter assumes that the null parameters varied spatially smoothly and slowly. Manual segmentation was easy to do because a cuboid has 6 faces. However, segmentation of faces cannot be generalised well to AM samples with complicated geometry and curved surfaces. Automatic segmentation may be possible with geometrical information from the CAD model.\n\nThe main issue with implementing this method onto the production line was how slow the filter was. The main bottleneck is the evaluation of the density estimate. Each step in the Newton-Raphson required the evaluation of $\\pi r^2$ data points. As a result, increasing the kernel radius slows down the filter. There may be faster methods for density estimation such as fitting a smoothing spline on the histogram \\citep{efron2004large}, however, that would require tuning the histogram bins as well as the tuning parameters for the spline. On the other hand, the method in \\cite{schwartzman2008empirical}, which uses the histogram count, found that the estimation of the null parameters was insensitive to the histogram binning.\n\nThere are a few strategies to accelerate the filter. The empirical null filter may be accelerated using GPUs \\citep{yang2008parallel, hwu2011gpu, eklund2013medical}, however, efforts to implement the filter in \\emph{CUDA} and \\emph{C++} is fruitless if there may exist a faster method. Instead, accuracy may be sacrificed for speed by only estimating the null parameters for several regularly spaced pixels, the remaining pixels are interpolated. This requires the null parameters to be slowly varying, otherwise, the interpolation may underfit. This may cause problems because of the face to face transition observed in the experiments. It was found that if a bad solution was found for a point, then that solution would spread its bad solution to neighbouring pixels due to the interpolation.\n\nEstimation of the null parameters is essentially robust statistics, estimating the parameters of the null distribution without being affected by non-null statistics. Potential faster methods compared to the empirical null could exist in the literature for robust statistics such as \\cite{hampel1986robust, rousseeuw1987robust, maronna2006robust, huber2009robust, jewson2018principles}. The use of density estimation for robust estimation is also featured when using loss functions derived from the Hellinger-divergence \\citep{beran1977minimum, jewson2018principles} or the beta-divergence \\citep{basu1998robust, jewson2018principles} which bears similarities to the empirical null. However, numerical methods are still required to find the mode. Exact estimation of the mode using Bayesian methods is impossible \\citep{heinrich2013the} which suggest avoiding the use of numerical methods can be difficult.\n\nThe EM algorithm \\citep{dempster1977maximum} could be used to fit a mixture of Gaussians to identify the null distribution and estimate its parameters. However, the power of hypothesis testing and the empirical null comes from the fact that the alternative distribution does not need to be specified.\n", "meta": {"hexsha": "a7c4cb38fec176fa1a89a5ca0c1a11a996e43491", "size": 123639, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "reports/thesis/chapters/5_inference.tex", "max_stars_repo_name": "shermanlo77/oxwasp_phd", "max_stars_repo_head_hexsha": "2b2d337356d35aa3db77dca4ef2a9bd3b6a45c12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "reports/thesis/chapters/5_inference.tex", "max_issues_repo_name": "shermanlo77/oxwasp_phd", "max_issues_repo_head_hexsha": "2b2d337356d35aa3db77dca4ef2a9bd3b6a45c12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "reports/thesis/chapters/5_inference.tex", "max_forks_repo_name": "shermanlo77/oxwasp_phd", "max_forks_repo_head_hexsha": "2b2d337356d35aa3db77dca4ef2a9bd3b6a45c12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 65.3828662084, "max_line_length": 1251, "alphanum_fraction": 0.7384805765, "num_tokens": 33238, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189134878876, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6413016920764267}} {"text": "\\subsection{Functors}\\label{subsec:functors}\n\n\\begin{definition}\\label{def:functor}\\mcite[def. 1.2.1 \\\\ def. 1.2.10]{Leinster2016Basic}\n Fix some \\hyperref[def:category]{categories} \\( \\cat{C} \\) and \\( \\cat{D} \\). A \\term{functor} \\( F: \\cat{C} \\to \\cat{D} \\) is a \\hyperref[eq:def:category_of_small_quivers/homomorphism]{quiver homomorphism} between the underlying quivers that is compatible with composition and identities.\n\n Explicitly, a functor is a family of functions\n \\begin{equation}\\label{eq:def:functor_as_family_of_function}\n \\begin{aligned}\n F_{\\obj}: &\\obj(\\cat{C}) \\to \\obj(\\cat{D}) \\\\\n F_{\\hom(A, B)}: &\\cat{C}(A, B) \\to \\cat{D}(F_{\\obj}(A), F_{\\obj}(B)),\n \\end{aligned}\n \\end{equation}\n where \\( F_{\\hom(A, B)} \\) is a distinct function for every pair of objects \\( A \\) and \\( B \\).\n\n In practice, we usually define the functor as the set\n \\begin{equation}\\label{eq:def:functor_as_single_function}\n F \\coloneqq F_{\\obj} \\cup \\bigcup\\set{ F_{\\hom(A, B)} \\given A, B \\in \\obj(\\cat{C}) }\n \\end{equation}\n\n Since the domains of all constituent functions are disjoint, \\( F \\) is again a total single-valued function. This allows us to justify the notation \\( F(A) \\) for objects and \\( F(f) \\) for morphisms.\n\n \\begin{thmenum}[resume=def:functor]\n \\thmitem{def:functor/domain_and_codomain} We say that the category \\( \\cat{C} \\) is the \\term{domain} and \\( \\cat{D} \\) --- the \\term{codomain} of the functor \\( F \\). These are technically not the domain and codomain of \\( F \\) when regarded as a function, however it is consistent with \\fullref{def:category_of_small_categories}.\n\n \\thmitem{def:functor/endofunctor} Similarly to \\fullref{def:multi_valued_function/endofunction} for functions, if the domain \\( \\cat{C} \\) and codomain \\( \\cat{D} \\) of a functor coincide, we say that it is an \\term{endofunctor}.\n \\end{thmenum}\n\n The definition of a functor additionally requires the following compatibility conditions to hold:\n \\begin{thmenum}[series=def:functor]\n \\thmitem[def:functor/CF1]{CF1} Functors must preserve identities, meaning that for any object \\( A \\in \\cat{C} \\) the following equality must hold:\n \\begin{equation}\\label{eq:def:functor/CF1}\\tag{\\logic{CF1}}\n F(\\id_A) = \\id_{F(A)}.\n \\end{equation}\n\n \\thmitem[def:functor/CF2]{CF2} Functors must preserve composition, meaning that for any pair of morphism \\( f: A \\to B \\) and \\( g: B \\to C \\) in \\( \\cat{C} \\),\n \\begin{equation}\\label{eq:def:functor/CF2}\\tag{\\logic{CF2}}\n F(g \\bincirc f) = F(g) \\bincirc F(f).\n \\end{equation}\n \\end{thmenum}\n\\end{definition}\n\\begin{defproof}\n The definition \\eqref{eq:def:functor_as_family_of_function} ensures that the quiver homomorphism conditions \\eqref{eq:def:category_of_small_quivers/homomorphism/head} and \\eqref{eq:def:category_of_small_quivers/homomorphism/tail} hold.\n\n Indeed, for any morphism \\( f: A \\to B \\) in \\( \\cat{C} \\) we have\n \\begin{equation*}\n F(\\dom(f)) = F(A) = \\dom(F(f)),\n \\end{equation*}\n which implies \\eqref{eq:def:category_of_small_quivers/homomorphism/head}. We also have\n \\begin{equation*}\n F(\\co\\dom(f)) = F(B) = \\co\\dom(F(f)),\n \\end{equation*}\n which implies \\eqref{eq:def:category_of_small_quivers/homomorphism/tail}.\n\\end{defproof}\n\n\\begin{remark}\\label{rem:functor_size}\n It is possible that \\( \\cat{C} \\) is \\( \\mscrU \\)-small in the sense of \\fullref{def:category_size}, but the \\hyperref[def:functor]{functor} \\( F \\), as the set \\eqref{eq:def:functor_as_single_function}, is not \\( \\mscrU \\)-small in the sense of \\fullref{def:large_and_small_sets}. Without using universes, we cannot prove the existence of any functor from the category of smalls sets to itself, for example.\n\\end{remark}\n\n\\begin{example}\\label{ex:unary_functors_in_set}\n In \\fullref{def:basic_set_operations}, we defined some operations on the category \\hyperref[def:category_of_small_sets]{\\( \\cat{Set} \\)} of small sets.\n\n \\begin{thmenum}\n \\thmitem{ex:unary_functors_in_set/power} The \\hyperref[def:basic_set_operations/power_set]{power set} \\( \\pow: \\cat{Set} \\to \\cat{Set} \\) is a canonical example of an \\hyperref[def:functor/endofunctor]{endofunctor}. Explicitly:\n \\begin{equation*}\n \\begin{aligned}\n &\\pow: \\cat{Set} \\to \\cat{Set}, \\\\\n &\\pow(A) \\coloneqq \\set{ S \\given S \\subseteq A }, \\\\\n &\\pow(f: A \\to B) \\coloneqq (S \\mapsto f[S]). \\\\\n \\end{aligned}\n \\end{equation*}\n\n We must verify that it is indeed a functor. \\ref{def:functor/CF2} is satisfied because\n \\begin{equation*}\n \\pow(g) \\bincirc \\pow(f) = (S \\mapsto g[f[S]]) = \\pow(g \\bincirc f).\n \\end{equation*}\n\n The condition \\ref{def:functor/CF2} is also obviously satisfied.\n\n The nuance here is that we send every function \\( f: A \\to B \\) to its \\hyperref[def:multi_valued_function/set_value]{set value} \\( f[S] \\) of some subset \\( S \\) of \\( A \\).\n\n \\thmitem{ex:unary_functors_in_set/union} The \\hyperref[def:basic_set_operations/union]{union} \\( \\bigcup \\) and \\hyperref[def:basic_set_operations/intersection]{intersection} \\( \\bigcap \\) may seem to be good examples of endofunctors in \\( \\cat{Set} \\). Unfortunately, there is no natural way to extend a morphism (function) \\( f: A \\to B \\) to a morphism from \\( \\bigcup A \\) to \\( \\bigcup B \\) or \\( \\bigcap A \\) to \\( \\bigcap B \\).\n \\end{thmenum}\n\\end{example}\n\n\\begin{definition}\\label{def:subcategory}\\mcite[def. 1.2.18]{Leinster2016Basic}\n We call the category \\( \\cat{D} \\) a \\term{subcategory} of \\( \\cat{C} \\) if the following hold:\n \\begin{itemize}\n \\item The underlying quiver \\( U(\\cat{D}) \\) is a \\hyperref[def:quiver/submodel]{subquiver} of \\( U(\\cat{C}) \\). That is, every object in \\( \\cat{D} \\) is an object in \\( \\cat{C} \\) and every morphism in \\( \\cat{D} \\) is a morphism in \\( \\cat{C} \\).\n \\item Composition and identity in \\( \\cat{D} \\) are \\hyperref[def:multi_valued_function/restriction]{restrictions} of composition and identity in \\( \\cat{C} \\).\n \\end{itemize}\n\n \\begin{thmenum}\n \\thmitem{def:subcategory/inclusion} For every subcategory there exists an \\term{inclusion functor} \\( \\Iota: \\cat{D} \\to \\cat{C} \\), which sends every object and morphism of \\( \\cat{D} \\) to itself in \\( \\cat{C} \\).\n\n \\thmitem{def:subcategory/full} We say that \\( \\cat{D} \\) is a \\term{full subcategory} if the underlying quiver \\( U(\\cat{C}) \\) is a \\hyperref[def:quiver/submodel]{full subquiver}. That is, in case \\( \\cat{D}(A, B) = \\cat{C}(A, B) \\) for every pair of objects \\( A \\) and \\( B \\) of \\( \\cat{D} \\).\n\n By \\fullref{thm:def:functor_invertibility/properties/full_subcategory}, this is equivalent to the inclusion functor being \\hyperref[def:functor_invertibility/full]{full}.\n\n \\thmitem{def:subcategory/induced} Every \\hyperref[rem:family_of_sets]{family} \\( \\mscrD \\) of objects in \\( \\cat{C} \\) induces a full subcategory \\( \\cat{D} \\) of \\( \\cat{C} \\), whose objects are those of \\( \\mscrD \\) and whose morphisms are restricted to those whose domain and codomain are both in \\( \\mscrD \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{remark}\\label{rem:contravariant_functor}\\mcite[def. 1.2.10]{Leinster2016Basic}\n We can invert the order of composition in \\ref{def:functor/CF2} in the definition of a functor given in \\fullref{def:functor}.\n \\begin{thmenum}\n \\thmitem[rem:contravariant_functor/CF2]{CF2\\Textprime} We can replace \\ref{def:functor/CF2} with\n \\begin{equation}\\label{eq:rem:contravariant_functor/CF2}\\tag{\\logic{CF2\\Textprime}}\n F(g \\bincirc f) = F(f) \\bincirc F(g).\n \\end{equation}\n \\end{thmenum}\n\n This also requires some other straightforward modifications to the definition of a functor.\n\n A functor that satisfies \\ref{rem:contravariant_functor/CF2} rather than \\ref{def:functor/CF2} is called \\term{contravariant}. In this context, a functor satisfying \\ref{def:functor/CF2} is called \\term{covariant}.\n\n Fortunately, a contravariant functor from \\( \\cat{C}^{\\opcat} \\) to \\( \\cat{D} \\) is identical to a covariant functor from \\( \\cat{C} \\) to \\( \\cat{D} \\). Therefore, there is no formal difference between the two concepts.\n\n The usage of the terms are entirely dictated by context. Unless necessary, we will avoid speaking about contravariant functors to avoid confusion. Some examples where this terminology may be useful are \\fullref{def:dual_functor}, \\fullref{def:hom_functor/unary} and \\fullref{ex:dual_space_contravariant_functor}.\n\\end{remark}\n\n\\begin{example}\\label{ex:dual_space_contravariant_functor}\n We can try to na\\\"ively define a functor that assigns to a \\hyperref[def:vector_space]{vector space} its \\hyperref[def:dual_vector_space]{algebraic dual}:\n \\begin{equation*}\n \\begin{aligned}\n &F: \\cat{Vect_\\BbbK} \\to \\cat{Vect_\\BbbK}, \\\\\n &F(V) \\coloneqq V^*, \\\\\n &F(f: V \\to W) \\coloneqq (\\varphi: W \\to \\BbbK \\mapsto \\varphi \\bincirc f).\n \\end{aligned}\n \\end{equation*}\n\n Unfortunately, \\( F(f) \\) is supposed to be a morphism from \\( V^* \\) to \\( W^* \\), but is actually a morphism from \\( W^* \\) to \\( V^* \\). This makes \\( F \\) a \\hyperref[rem:contravariant_functor]{contravariant functor} or, equivalently, a functor from \\( \\cat{Vect_\\BbbK}^{\\opcat} \\) to \\( \\cat{Vect_\\BbbK} \\).\n\\end{example}\n\n\\begin{definition}\\label{def:discrete_category}\n A \\term{discrete category} is a category with no morphisms except for the identities. Clearly to any set there corresponds exactly one discrete category and vice versa.\n\\end{definition}\n\n\\begin{example}\\label{ex:discrete_category_adjunction}\n Denote by\n \\begin{equation*}\n U: \\cat{Cat} \\to \\cat{Set}\n \\end{equation*}\n the forgetful functor that for any small category \\( \\cat{C} \\) gives us its set of objects \\( \\obj(\\cat{C}) \\). There is also a functor\n \\begin{equation*}\n D: \\cat{Set} \\to \\cat{Cat}\n \\end{equation*}\n that for any small set \\( A \\) gives us the \\hyperref[def:discrete_category]{discrete category} whose set of objects is \\( A \\).\n\n This is actually an \\hyperref[def:category_adjunction]{adjunction} --- see \\fullref{ex:def:category_adjunction/set_cat}.\n\\end{example}\n\n\\begin{proposition}\\label{thm:def:functor/properties}\n \\hyperref[def:functor]{Functors} have the following basic properties:\n \\begin{thmenum}\n \\thmitem{thm:def:functor/properties/half_inverses} Functors preserve inverses. For every functor \\( F: \\cat{C} \\to \\cat{D} \\) and every morphism \\( f: A \\to B \\) in \\( \\cat{C} \\) with a right inverse \\( g: B \\to A \\), \\( F(f) \\) is a right inverse of \\( F(g) \\). Similarly, if \\( g \\) is a left inverse of \\( f \\), then \\( F(g) \\) is a left inverse of \\( F(f) \\).\n\n \\thmitem{thm:def:functor/properties/inverses} For every functor \\( F: \\cat{C} \\to \\cat{D} \\) and every isomorphism \\( f: A \\to B \\) in \\( \\cat{C} \\),\n \\begin{equation}\\label{eq:thm:def:functor/properties/inverses}\n [F(f)]^{-1} = F(f^{-1}).\n \\end{equation}\n\n \\thmitem{thm:def:functor/properties/isomorphisms} Functors preserve \\hyperref[def:morphism_invertibility/isomorphism]{isomorphisms}. That is, for every functor \\( F: \\cat{C} \\to \\cat{D} \\), if \\( f: A \\to B \\) is an isomorphism in \\( \\cat{C} \\), \\( F(f) \\) is an isomorphism in \\( \\cat{D} \\).\n\n Consequently, for every pair of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), from \\( A \\cong B \\) it follows that \\( F(A) \\cong F(B) \\).\n\n The converse sometimes also holds --- see \\fullref{thm:def:functor_invertibility/properties/fully_faithful_reflects_isomorphisms}.\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:functor/properties/half_inverses} Let \\( f: A \\to B \\) be a right inverse of \\( g: B \\to A \\) in \\( \\cat{C} \\). Then\n \\begin{equation*}\n F(g) \\bincirc F(f)\n \\reloset {\\eqref{eq:def:functor/CF2}} =\n F(g \\bincirc f)\n =\n F(\\id_A)\n \\reloset {\\eqref{eq:def:functor/CF1}} =\n \\id_{F(A)}.\n \\end{equation*}\n\n Thus, \\( F(f) \\) is a right inverse of \\( F(g) \\). Since \\( g \\) is a left inverse of \\( f \\), automatically \\( F(g) \\) is a left inverse of \\( F(f) \\).\n\n \\SubProofOf{thm:def:functor/properties/inverses} If \\( f^{-1} \\) is a left inverse of \\( f \\), by \\fullref{thm:def:functor/properties/half_inverses} we have that \\( F(f^{-1}) \\) is a left inverse of \\( F(f) \\). But \\( F(f^{-1}) \\) is also a right inverse, and again by \\fullref{thm:def:functor/properties/half_inverses} \\( F(g) \\) is a right inverse of \\( F(f^{-1}) \\).\n\n Therefore, \\( F(f^{-1}) \\) is a two-sided inverse of \\( F(f) \\). By \\fullref{thm:def:morphism_invertibility/properties/at_most_one_inverse}, it is the only two-sided inverse, hence\n \\begin{equation*}\n [F(f)]^{-1} = F(f^{-1}).\n \\end{equation*}\n\n \\SubProofOf{thm:def:functor/properties/isomorphisms} Follows from \\fullref{thm:def:functor/properties/inverses}.\n\\end{proof}\n\n\\begin{definition}\\label{def:category_of_small_categories}\n Suppose that we are given a \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\), which is safe to assume to be the smallest suitable one as explained in \\fullref{def:large_and_small_sets}.\n\n We denote the \\hyperref[def:category]{category} of \\( \\mscrU \\)-small \\hyperref[def:category]{categories} by \\( \\ucat{Cat} \\) or, if the universe is clear from the context, simply by \\( \\cat{Cat} \\). See \\fullref{def:category_size} for a further discussion of universes and categories.\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(\\cat{Cat}) \\) is the set of all \\( \\mscrU \\)-small categories.\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{Cat}(A, B) \\) from \\( A \\) to \\( B \\) is the set of all \\hyperref[def:functor]{functors} from \\( A \\) to \\( B \\).\n\n \\item The \\hyperref[def:category/composition]{composition of morphisms} is the \\hyperref[def:multi_valued_function/composition]{function composition} of the functors regarded as the functions \\eqref{eq:def:functor_as_single_function}. That is, the composition of \\( F: \\cat{C} \\to \\cat{D} \\) and \\( G: \\cat{D} \\to \\cat{E} \\) is the functor\n \\begin{equation}\\label{eq:def:category_of_small_categories/composition}\n \\begin{aligned}\n &[G \\bincirc F]: \\cat{C} \\to \\cat{E}, \\\\\n &[G \\bincirc F](A) \\coloneqq G(F(A)), \\\\\n &[G \\bincirc F](f) \\coloneqq G(F(f)).\n \\end{aligned}\n \\end{equation}\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the category \\( \\cat{C} \\) is the \\term{identity functor}\n \\begin{equation}\\label{eq:def:category_of_small_categories/identity}\n \\begin{aligned}\n &\\id_{\\cat{C}}: \\cat{C} \\to \\cat{C}, \\\\\n &\\id_{\\cat{C}}(A) \\coloneqq A, \\\\\n &\\id_{\\cat{C}}(f) \\coloneqq f.\n \\end{aligned}\n \\end{equation}\n \\end{itemize}\n\\end{definition}\n\\begin{defproof}\n To see that \\( \\ucat{Cat} \\) is indeed a category, we verify the conditions \\ref{def:category/C1} and \\ref{def:category/C2}.\n\n \\SubProofOf{def:category/C1} For every two \\( \\mscrU \\)-small categories \\( \\cat{C} \\) and \\( \\cat{D} \\) and every functor \\( F: \\cat{C} \\to \\cat{D} \\), for every object \\( A \\in \\cat{C} \\) we have\n \\begin{equation*}\n [\\id_{\\cat{D}} \\bincirc F](A)\n =\n \\id_{\\cat{D}}(F(A))\n =\n F(A)\n =\n F(\\id_{\\cat{C}}(A))\n =\n [F \\bincirc \\id_{\\cat{C}}](A)\n \\end{equation*}\n and analogously for morphisms.\n\n Therefore, \\( \\id_{\\cat{C}} \\) and \\( \\id_{\\cat{D}} \\) satisfy \\eqref{eq:def:category/C1}.\n\n \\SubProofOf{def:category/C2} Associativity of functor composition is inherited from the associativity of function composition.\n\\end{defproof}\n\n\\begin{definition}\\label{def:universal_categories}\n For any \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\), the \\hyperref[def:category_of_small_categories]{category of \\( \\mscrU \\)-small categories} \\( \\ucat{Cat} \\) has an initial and a terminal object.\n\n Similarly to how we use the \\hyperref[def:ordinal]{ordinals} \\( 0 \\) and \\( 1 \\) to denote the initial and terminal object in the category of sets (see \\fullref{ex:def:universal_objects/set}), we denote the initial category by \\( \\cat{0} \\) and the final category by \\( \\cat{1} \\). Note that the final category is only unique up to an isomorphism. They are identical, however, for all universes \\( \\mscrU \\).\n\n These categories are precisely the \\hyperref[def:discrete_category]{discrete categories} induced by the ordinals \\( 0 \\) and \\( 1 \\) as described in \\fullref{thm:order_category_isomorphism}.\n\\end{definition}\n\n\\begin{definition}\\label{def:dual_functor}\\mcite{nLab:opposite_category}\n The \\term{opposite} or \\term{dual} functor of \\( F: \\cat{C} \\to \\cat{D} \\) is the functor\n \\begin{equation*}\n \\begin{aligned}\n &F^{\\opcat}: \\cat{C}^{\\opcat} \\to \\cat{D}^{\\opcat} \\\\\n &F^{\\opcat}(A) \\coloneqq A \\\\\n &F^{\\opcat}(f^{\\opcat}: B \\to A) \\coloneqq [F(f: A \\to B)]^{\\opcat}.\n \\end{aligned}\n \\end{equation*}\n\n For the composition of functors, we then have\n \\begin{equation}\\label{eq:def:dual_functor/composition}\n [G \\bincirc F]^{\\opcat} = G^{\\opcat} \\bincirc F^{\\opcat}.\n \\end{equation}\n\n This is somewhat in contrast to the general practice of inverting morphisms when taking duals. Thus, we define, for any \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\), the \\term{oppositization functor}\n \\begin{equation*}\n (\\anon*)^{\\opcat}: \\ucat{Cat}^{\\opcat} \\to \\ucat{Cat}.\n \\end{equation*}\n\n As an \\hyperref[def:multi_valued_function/endofunction]{endofunction} on \\( \\obj(\\ucat{Cat}) \\), the oppositization functor is clearly an \\hyperref[def:set_with_involution]{involution}.\n\n Dual functors also arise naturally in \\fullref{thm:dual_functor_category}.\n\\end{definition}\n\n\\begin{definition}\\label{def:functor_image}\n The \\term{image} of a functor \\( F: \\cat{C} \\to \\cat{D} \\) is the \\hyperref[def:quiver]{quiver} whose vertex set is\n \\begin{equation*}\n V \\coloneqq \\set{ F(A) \\given A \\in \\cat{C} }\n \\end{equation*}\n and whose arc set is\n \\begin{equation*}\n A \\coloneqq \\set{ F(f) \\given A, B \\in \\cat{C} \\T{and} f \\in \\cat{C}(A, B) }.\n \\end{equation*}\n\n This quiver has no categorical structure --- it is merely a directed multigraph. As shown in \\fullref{ex:functor_image_not_a_category}, imposing a categorical structure na\\\"ively may fail.\n\\end{definition}\n\n\\begin{example}\\label{ex:functor_image_not_a_category}\\mcite{MathSE:image_of_functor_is_not_a_category}\n \\begin{figure}\n \\hfill\n \\includegraphics[page=1]{output/ex__functor_image_not_a_category.pdf}\n \\hfill\n \\hfill\n \\caption{A functor whose image is not a category.}\\label{fig:ex:functor_image_not_a_category}\n \\end{figure}\n\n Consider the functor \\( F: \\cat{C} \\to \\cat{D} \\) from \\cref{fig:ex:functor_image_not_a_category}.\n\n \\begin{itemize}\n \\item The solid arrows are the morphisms in \\( \\cat{C} \\) and their images in \\( F(\\cat{C}) \\).\n \\item The dashed arrows denote the action of the functor \\( F \\).\n \\item The dotted arrow exists in \\( \\cat{D} \\) as the composition of the other two arrows, however it is missing in the image \\( F(\\cat{C}) \\). Thus, composition is not fully defined in \\( F(\\cat{C}) \\), and \\( F(\\cat{C}) \\) fails to be a category.\n \\end{itemize}\n\\end{example}\n\n\\begin{definition}\\label{def:categorical_diagram}\n Fix a category \\( \\cat{I} \\), called an \\term{index category}. A \\term{diagram} in \\( \\cat{C} \\) of shape \\( \\cat{I} \\) is simply a functor \\( D: \\cat{I} \\to \\cat{C} \\), whose domain is \\( \\cat{I} \\). We sometimes identify a diagram functor with its image \\( D(\\cat{I}) \\).\n\n It is often convenient to draw graphically the \\hyperref[def:quiver_geometric_realization]{geometric realizations} of the \\hyperref[def:quiver]{quiver} \\( D(\\cat{I}) \\). An established convention is to allow multiple vertices representing the same object, which can be achieved formally by actually adjoining new vertices to the quiver and labeling them as in \\fullref{def:weighted_set}. Other established conventions for drawing diagrams include not drawing identity morphisms and adding various visual aids. In this regard, categorical diagrams correspond to the everyday sense of the word \\enquote{diagram}.\n\n We say that the diagram \\( D \\) over \\( \\cat{C} \\) \\term{commutes} if, whenever \\( p = (f_1, \\ldots, f_n) \\) and \\( q = (g_1, \\ldots, g_m) \\) are two \\hyperref[def:quiver_path/directed]{directed paths} in \\( D(\\cat{I}) \\) with identical endpoints and either \\( n > 1 \\) or \\( m > 1 \\), then\n \\begin{equation*}\n f_n \\bincirc f_{n-1} \\bincirc \\cdots \\bincirc f_2 \\bincirc f_1\n =\n g_m \\bincirc g_{m-1} \\bincirc \\cdots \\bincirc g_2 \\bincirc g_1.\n \\end{equation*}\n\n We do not really care about how the objects and morphisms in \\( \\cat{I} \\) are labeled, hence we often use placeholder dots like in \\eqref{eq:ex:quivers_as_functors/index/dots}.\n\n The requirement that the one of the paths is nontrivial, however, is crucial in \\fullref{def:equalizers}.\n\\end{definition}\n\n\\begin{remark}\\label{rem:inverting_isomorphisms_may_preserve_commutativity}\n Inverting isomorphisms in a \\hyperref[def:categorical_diagram]{commutative diagram} may or may not preserve commutativity.\n\n If \\( p = (f_1, \\ldots, f_n) \\) and \\( q = (g_1, \\ldots, g_m) \\) are two paths in a commutative diagram, and if \\( f_1 \\) is invertible, then obviously\n \\begin{equation*}\n f_n \\bincirc \\cdots \\bincirc f_1 = g_m \\bincirc \\cdots \\bincirc g_1\n \\end{equation*}\n if and only if\n \\begin{equation*}\n f_n \\bincirc \\cdots \\bincirc f_2 = g_m \\bincirc \\cdots \\bincirc g_1 \\bincirc f_1\n \\end{equation*}\n and similarly if \\( f_n \\) is invertible.\n\n On the other hand, consider \\hyperref[def:ordinal]{ordinals} in \\( \\cat{Set} \\). Denote by \\( \\iota \\) the inclusion maps and by \\( f: \\omega^2 \\to \\omega \\) the bijective map from \\fullref{thm:omega_equinumerous_with_omega_squared}. Then the following diagram commutes:\n \\begin{equation}\\label{eq:rem:inverting_isomorphisms_may_preserve_commutativity/ordinals_commuting}\n \\begin{aligned}\n \\includegraphics[page=1]{output/rem__inverting_isomorphisms_may_preserve_commutativity.pdf}\n \\end{aligned}\n \\end{equation}\n but the following does not:\n \\begin{equation}\\label{eq:rem:inverting_isomorphisms_may_preserve_commutativity/ordinals_not_commuting}\n \\begin{aligned}\n \\includegraphics[page=2]{output/rem__inverting_isomorphisms_may_preserve_commutativity.pdf}\n \\end{aligned}\n \\end{equation}\n\\end{remark}\n\n\\begin{definition}\\label{def:functor_invertibility}\n In connection with \\fullref{def:morphism_invertibility} and \\fullref{def:function_invertibility}, we introduce the following terminology:\n \\begin{thmenum}\n \\thmitem{def:functor_invertibility/injective_on_objects} The \\hyperref[def:functor]{functor} \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{injective on objects} if the \\hyperref[def:multi_valued_function/restriction]{restriction}\n \\begin{equation*}\n F\\restr_{\\obj(C)}: \\obj(C) \\to \\obj(D)\n \\end{equation*}\n is \\hyperref[def:function_invertibility/injective]{injective}.\n\n That is, for every pair of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), from \\( F(A) = F(B) \\) it follows that \\( A = B \\).\n\n If, instead, from \\( F(A) \\cong F(B) \\) it follows that \\( A \\cong B \\), we say that \\( F \\) if \\term{essentially injective on objects}.\n\n \\thmitem{def:functor_invertibility/injective_on_morphisms} The \\hyperref[def:functor]{functor} \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{injective on morphisms} if its restriction to the set\n \\begin{equation*}\n \\bigcup\\set{ \\cat{C}(A, B) \\given A, B \\in \\obj(\\cat{C}) }\n \\end{equation*}\n of all morphisms is injective.\n\n That is, for every pair of morphisms \\( f \\) and \\( g \\) in \\( \\cat{C} \\), from \\( F(f) = F(g) \\) it follows that \\( f = g \\). Note that if the morphisms are not parallel, we assume that they are not equal.\n\n \\thmitem{def:functor_invertibility/faithful}\\mcite[def. 1.2.16]{Leinster2016Basic} The functor \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{faithful} if it is \\hyperref[def:function_invertibility/injective]{injective} on \\( \\hom \\)-sets, i.e. for all pairs of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), the restriction of \\( F \\) to \\( \\cat{C}(A, B) \\) is an injective function.\n\n That is, for every pair of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\) and every pair of morphisms \\( f \\) and \\( g \\) in \\( \\cat{C}(A, B) \\), from \\( F(f) = F(g) \\) it follows that \\( f = g \\).\n\n See \\fullref{thm:def:functor_invertibility/properties/injective} for how faithful functors relate to functors injective on objects or on morphisms.\n\n \\thmitem{def:functor_invertibility/surjective_on_objects}\\mcite[def. 1.3.17]{Leinster2016Basic} The functor \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{surjective on objects} if the restriction\n \\begin{equation*}\n F\\restr_{\\obj(C)}: \\obj(C) \\to \\obj(D)\n \\end{equation*}\n is \\hyperref[def:function_invertibility/surjective]{surjective}.\n\n That is, for every object \\( B \\) in \\( \\cat{D} \\), there exists at least one object \\( A \\) in \\( \\cat{C} \\) such that \\( F(A) = B \\).\n\n If, instead, there exists at least one object \\( A \\in \\cat{C} \\) such that \\( F(A) \\cong B \\), we say that \\( F \\) is \\term{essentially surjective on objects}.\n\n \\thmitem{def:functor_invertibility/surjective_on_morphisms} Similarly, \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{surjective on morphisms} if its restriction to the set of all morphisms is surjective.\n\n That is, for every morphism \\( g \\) in \\( \\cat{D} \\), there exists at least one morphism \\( f \\) in \\( \\cat{C} \\) such that \\( F(f) = g \\).\n\n \\thmitem{def:functor_invertibility/full}\\mcite[def. 1.2.16]{Leinster2016Basic} The functor \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{full} if it is surjective on \\( \\hom \\)-sets, i.e. for all pairs of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), the restriction of \\( F \\) to \\( \\cat{C}(A, B) \\) is a surjective function.\n\n That is, for every pair of objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\) and every morphism \\( g: F(A) \\to F(B) \\) in \\( \\cat{D} \\), there exists at least one morphism in \\( f: A \\to B \\) in \\( \\cat{C} \\) such that \\( F(f) = g \\).\n\n \\thmitem{def:functor_invertibility/fully_faithful} Finally, \\( F: \\cat{C} \\to \\cat{D} \\) is \\term{fully faithful} if it is both full and faithful.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:def:functor_invertibility/properties}\n \\hyperref[def:functor]{Functors} have the following basic properties regarding their \\hyperref[def:functor_invertibility]{invertibility}:\n\n \\begin{thmenum}\n \\thmitem{thm:def:functor_invertibility/properties/injective} A functor is \\hyperref[def:functor_invertibility/injective_on_morphisms]{injective on morphisms} if and only if it is both \\hyperref[def:functor_invertibility/injective_on_objects]{injective on objects} and \\hyperref[def:functor_invertibility/faithful]{faithful}.\n\n \\thmitem{thm:def:functor_invertibility/properties/surjective} A functor is \\hyperref[def:functor_invertibility/surjective_on_morphisms]{surjective on morphisms} if and only if it is both \\hyperref[def:functor_invertibility/surjective_on_objects]{surjective on objects} and \\hyperref[def:functor_invertibility/full]{full}.\n\n \\thmitem{thm:def:functor_invertibility/properties/full_subcategory} A \\hyperref[def:subcategory]{subcategory} \\( \\cat{D} \\) of \\( \\cat{C} \\) is full in the sense of \\fullref{def:subcategory} if and only if the \\hyperref[def:subcategory]{inclusion functor} \\( \\Iota: \\cat{D} \\to \\cat{C} \\) is full in the sense of \\fullref{def:functor_invertibility/full}.\n\n \\thmitem{thm:def:functor_invertibility/properties/faithful_reflects_composition} \\hyperref[def:functor_invertibility/faithful]{Faithful} functors reflect composition. That is, for every functor \\( F: \\cat{C} \\to \\cat{D} \\), if the following diagram commutes:\n \\begin{equation}\\label{eq:thm:def:functor_invertibility/properties/faithful_reflects_composition/image}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__def__functor_invertibility__properties.pdf}\n \\end{aligned}\n \\end{equation}\n then the following diagram are identities:\n \\begin{equation}\\label{eq:thm:def:functor_invertibility/properties/faithful_reflects_composition/source}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__def__functor_invertibility__properties.pdf}\n \\end{aligned}\n \\end{equation}\n\n \\thmitem{thm:def:functor_invertibility/properties/faithful_reflects_cancellative} A \\hyperref[def:functor_invertibility/faithful]{faithful} functor reflects monomorphisms and epimorphisms. For every functor \\( F: \\cat{C} \\to \\cat{D} \\) and morphism \\( f: A \\to B \\) in \\( \\cat{C} \\), if \\( F(f) \\) is a monomorphism (resp. epimorphism), so is \\( f \\).\n\n \\thmitem{thm:def:functor_invertibility/properties/fully_faithful_reflects_identities} A \\hyperref[def:functor_invertibility/fully_faithful]{fully faithful} functor reflects identities. For every functor \\( F: \\cat{C} \\to \\cat{D} \\) and endomorphism \\( f: A \\to A \\) in \\( \\cat{C} \\), if \\( F(f) = \\id_{F(A)} \\), then \\( f = \\id_A \\).\n\n \\thmitem{thm:def:functor_invertibility/properties/fully_faithful_reflects_isomorphisms} A \\hyperref[def:functor_invertibility/fully_faithful]{fully faithful} functor reflects split monomorphisms and split epimorphisms, and hence also isomorphisms.\n\n That is, for every functor \\( F: \\cat{C} \\to \\cat{D} \\) and morphism \\( f: A \\to B \\) in \\( \\cat{C} \\), if \\( F(f) \\) is a split monomorphism (resp. split epimorphism or isomorphism), so is \\( f \\).\n\n \\thmitem{thm:def:functor_invertibility/properties/isomorphism} A functor between \\( \\mscrU \\)-small categories that is both injective and surjective on morphisms is itself an isomorphism in \\( \\ucat{Cat} \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:functor_invertibility/properties/injective}\n \\SufficiencySubProof* Let \\( F: \\cat{C} \\to \\cat{D} \\) be injective on morphisms. It is trivially faithful since faithfulness is a more restrictive condition.\n\n To see that \\( F \\) is injective on objects, let \\( A, B \\in \\cat{C} \\) and suppose that \\( F(A) = F(B) \\). Then \\( \\id_{F(A)} = \\id_{F(B)} \\) and\n \\begin{equation*}\n F(\\id_A)\n \\reloset {\\eqref{eq:def:functor/CF2}} =\n \\id_{F(A)}\n =\n \\id_{F(B)}\n \\reloset {\\eqref{eq:def:functor/CF2}} =\n F(\\id_B).\n \\end{equation*}\n\n Since \\( F \\) is injective on morphisms, it follows that \\( \\id_A = \\id_B \\), hence \\( A = B \\). Thus, \\( F \\) is injective on objects.\n\n \\NecessitySubProof* Let \\( F: \\cat{C} \\to \\cat{D} \\) be faithful and injective on objects. Let \\( f: A \\to B \\) and \\( g: C \\to D \\) be morphisms in \\( \\cat{C} \\) such that \\( F(f) = F(g) \\).\n\n Then both \\( F(f) \\) and \\( F(g) \\) have the same domain \\( F(A) = F(C) \\) and codomain \\( F(B) = F(D) \\). Hence, since \\( F \\) is injective on objects, we have \\( A = C \\) and \\( B = D \\).\n\n Thus, \\( f \\) and \\( g \\) are both morphisms from \\( A \\) to \\( B \\). Since \\( F \\) is also faithful, from \\( F(f) = F(g) \\) it follows that \\( f = g \\).\n\n Therefore, \\( F \\) is injective on morphisms.\n\n \\SubProofOf{thm:def:functor_invertibility/properties/surjective}\n \\SufficiencySubProof* Let \\( F: \\cat{C} \\to \\cat{D} \\) be surjective on morphisms. It is trivially full since fullness is a more restrictive condition.\n\n To see that \\( F \\) is surjective on objects, let \\( C \\in \\cat{D} \\). Then there exists some morphism \\( f: A \\to B \\) in \\( \\cat{C} \\) such that \\( F(f) = \\id_Z \\). We thus necessarily have \\( F(A) = C \\) and \\( F(B) = C \\).\n\n \\NecessitySubProof* Let \\( F: \\cat{C} \\to \\cat{D} \\) be full and injective on objects. Let \\( g: C \\to D \\) be a morphism in \\( \\cat{D} \\).\n\n Since \\( F \\) is surjective on objects, there exists preimages \\( A \\) of \\( C \\) and \\( B \\) of \\( D \\) under \\( F \\). Thus, \\( g \\in \\cat{D}(F(A), F(B)) \\).\n\n Since \\( F \\) is also full, there exists some morphism \\( f: A \\to B \\) such that \\( F(f) = g \\).\n\n Therefore, \\( F \\) is surjective on morphisms.\n\n \\SubProofOf{thm:def:functor_invertibility/properties/full_subcategory} Trivial.\n\n \\SubProofOf{thm:def:functor_invertibility/properties/faithful_reflects_composition} Suppose that \\eqref{eq:thm:def:functor_invertibility/properties/faithful_reflects_composition/image} commutes. Then, since \\( F \\) is faithful and thus injective on the morphism set \\( \\cat{C}(A, C) \\), the equality \\( F(g \\bincirc f) = F(g) \\bincirc F(f) = F(h) \\) implies that \\( g \\bincirc f = h \\). Hence, \\eqref{eq:thm:def:functor_invertibility/properties/faithful_reflects_composition/source} also commutes.\n\n \\SubProofOf{thm:def:functor_invertibility/properties/faithful_reflects_cancellative} Let \\( F(g) \\) be a monomorphism and let \\( f_1, f_2: A \\to B \\) be parallel morphisms such that\n \\begin{equation*}\n g \\bincirc f_1 = g \\bincirc f_2.\n \\end{equation*}\n\n Then, since \\( F(g) \\) is a monomorphism, we have that \\( F(f_1) = F(f_2) \\). Since \\( F \\) is faithful, the restriction \\( F\\restr_{C(A, B)} \\) is injective, and \\( f_1 = f_2 \\).\n\n The proof when \\( F(g) \\) is an epimorphism is analogous.\n\n \\SubProofOf{thm:def:functor_invertibility/properties/fully_faithful_reflects_identities} If \\( F: \\cat{C} \\to \\cat{D} \\) is fully faithful, for every object \\( A \\) in \\( \\cat{C} \\), the identity morphism \\( \\id_{F(A)} \\) has a unique preimage under \\( F \\). By \\ref{def:functor/CF1}, this preimage can only be \\( \\id_A \\).\n\n \\SubProofOf{thm:def:functor_invertibility/properties/fully_faithful_reflects_isomorphisms} Let \\( q \\) be a left inverse of \\( F(f) \\). Since \\( F \\) is fully faithful, there exists a unique morphism \\( g: B \\to A \\) such that \\( F(g) = q \\).\n\n Since\n \\begin{equation*}\n F(g) \\bincirc F(f) = \\id_{F(A)},\n \\end{equation*}\n by \\fullref{thm:def:functor_invertibility/properties/fully_faithful_reflects_identities} we have\n \\begin{equation*}\n g \\bincirc f = \\id_A.\n \\end{equation*}\n\n Therefore, \\( g \\) is a left inverse of \\( F(f) \\).\n\n The proof for right inverses follows from \\fullref{thm:def:morphism_invertibility/properties/inverse_interchanges}.\n\n From \\fullref{thm:def:morphism_invertibility/properties/left_and_right} it follows that if \\( F(f) \\) is an isomorphism, so is \\( f \\).\n\n \\SubProofOf{thm:def:functor_invertibility/properties/isomorphism} If \\( F \\) is both injective and surjective on morphisms, it is also injective and surjective on objects and hence, as a function, is bijective. Therefore, it is both left and right invertible as a consequence of \\fullref{thm:function_invertibility_categorical/fully_invertible}.\n\\end{proof}\n\n\\begin{example}\\label{ex:def:functor_invertibility}\n \\hfill\n \\begin{thmenum}\n \\thmitem{ex:def:functor_invertibility/power} The power set functor described in \\fullref{ex:unary_functors_in_set} is clearly \\hyperref[def:functor_invertibility/injective_on_morphisms]{injective on morphisms}, hence by \\fullref{thm:def:functor_invertibility/properties/injective}, it is also \\hyperref[def:functor_invertibility/injective_on_objects]{injective on objects} and \\hyperref[def:functor_invertibility/faithful]{faithful}.\n\n It is not full, nor surjective on objects.\n\n \\thmitem{ex:def:functor_invertibility/cat_to_set} The forgetful functor \\( D: \\ucat{Cat} \\to \\ucat{Set} \\) discussed in \\fullref{def:discrete_category} is \\hyperref[def:functor_invertibility/surjective_on_morphisms]{surjective on morphisms}, hence by \\fullref{thm:def:functor_invertibility/properties/surjective}, it is also \\hyperref[def:functor_invertibility/surjective_on_objects]{surjective on objects} and \\hyperref[def:functor_invertibility/full]{full}.\n\n It is not faithful, nor injective on objects.\n \\end{thmenum}\n\\end{example}\n\n\\begin{proposition}\\label{thm:commutative_diagrams_preserved_and_reflected}\n Functors preserve commutative diagrams and faithful functors also reflect commutative diagrams.\n\n More precisely, let \\( \\cat{C} \\) be an arbitrary category, let \\( D \\) be a diagram in \\( \\cat{C} \\), and let \\( p = (f_1, \\ldots, f_n) \\) and \\( q = (g_1, \\ldots, g_m) \\) be two \\hyperref[def:quiver_path/directed]{directed paths} with the same endpoints in \\( D \\).\n\n For any functor \\( F: \\cat{C} \\to \\cat{D} \\), if\n \\begin{equation}\\label{eq:thm:commutative_diagrams_preserved_and_reflected/source}\n f_n \\bincirc \\cdots \\bincirc \\bincirc f_1 = g_m \\bincirc \\cdots \\bincirc \\bincirc g_1,\n \\end{equation}\n then\n \\begin{equation}\\label{eq:thm:commutative_diagrams_preserved_and_reflected/image}\n F(f_n) \\bincirc \\cdots \\bincirc \\bincirc F(f_1) = F(g_m) \\bincirc \\cdots \\bincirc \\bincirc F(g_1),\n \\end{equation}\n\n Conversely, if \\( F \\) is faithful, then \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/image} implies \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/source}.\n\\end{proposition}\n\\begin{proof}\n Functors preserve composition by \\ref{eq:def:functor/CF2}, hence \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/image} follows from \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/source} directly.\n\n Now suppose that \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/image} holds for a faithful functor \\( F \\). \\ref{eq:def:functor/CF2} allows us to reduce \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/image} to\n \\begin{equation*}\n F(f_n \\bincirc \\cdots \\bincirc \\bincirc f_1) = F(g_m \\bincirc \\cdots \\bincirc \\bincirc g_1).\n \\end{equation*}\n\n Then, by injectivity of \\( F \\) on the morphism set \\( \\cat{C}(\\dom(f_1), \\co\\dom(f_1)) \\), \\eqref{eq:thm:commutative_diagrams_preserved_and_reflected/source} holds.\n\\end{proof}\n\n\\begin{definition}\\label{def:natural_transformation}\\mcite[def. 1.3.1]{Leinster2016Basic}\n Let \\( F \\) and \\( G \\) be parallel \\hyperref[def:functor]{functors} from the category \\( \\cat{C} \\) to \\( \\cat{D} \\).\n\n A \\term{natural transformation} \\( \\alpha \\) from \\( F \\) to \\( G \\) is an \\hyperref[def:cartesian_product/indexed_family]{indexed family} of\n \\begin{equation}\\label{eq:def:natural_transformation/family}\n \\seq{ \\alpha_A: F(A) \\to G(A) }_{A \\in \\cat{C}}\n \\end{equation}\n of morphisms in \\( \\cat{D} \\) such that, for every morphism \\( f: A \\to B \\) in \\( \\cat{C} \\), the following \\hyperref[def:categorical_diagram]{diagram commutes}:\n \\begin{equation}\\label{eq:def:natural_transformation/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__natural_transformation.pdf}\n \\end{aligned}\n \\end{equation}\n\n The morphisms \\( \\alpha_A \\) are called the components of \\( \\alpha \\). We denote natural transformations by \\( \\alpha: F \\Rightarrow G \\) and, when used in diagrams, by\n \\begin{equation}\\label{eq:def:natural_transformation/notation}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__natural_transformation.pdf}\n \\end{aligned}\n \\end{equation}\n\\end{definition}\n\n\\begin{example}\\label{ex:quivers_as_functors}\\mcite[exmpl. 1.3.46]{Perrone2019}\n In \\fullref{def:quiver}, we have defined a quiver as a set \\( V \\) of vertices, a set \\( A \\) of arcs and two functions --- the head \\( h: A \\to V \\)and tail \\( t: A \\to V \\) of an arc.\n\n Now consider the following \\hyperref[def:categorical_diagram]{index category} \\( \\cat{I}: \\)\n \\begin{equation}\\label{eq:ex:quivers_as_functors/index/dots}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__quivers_as_functors.pdf}\n \\end{aligned}\n \\end{equation}\n\n For the sake of readability, we will give the following explicit labels in this category:\n \\begin{equation}\\label{eq:ex:quivers_as_functors/index/annotated}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__quivers_as_functors.pdf}\n \\end{aligned}\n \\end{equation}\n\n A quiver can then be defined as a functor \\( Q: \\cat{I} \\to \\ucat{Set} \\) to the category \\hyperref[def:category_of_small_sets]{\\( \\ucat{Set} \\)} of \\( \\mscrU \\)-small sets (for a \\hyperref[def:category_size]{fixed Grothendieck universe} \\( \\mscrU \\)).\n\n A \\hyperref[def:natural_transformation]{natural transformation} from the quiver \\( Q: \\cat{I} \\to \\ucat{Set} \\) to \\( R: \\cat{I} \\to \\ucat{Set} \\) is then a pair of functions \\( f_V: Q(V) \\to R(V) \\) and \\( f_A: Q(A) \\to R(A) \\) such that the following \\hyperref[def:categorical_diagram]{diagrams commute}:\n \\begin{equation}\\label{eq:ex:quivers_as_functors/index/diagram}\n \\begin{aligned}\n \\includegraphics[page=3]{output/ex__quivers_as_functors.pdf}\n \\quad\\quad\\quad\\quad\n \\includegraphics[page=4]{output/ex__quivers_as_functors.pdf}\n \\end{aligned}\n \\end{equation}\n\n See \\fullref{ex:isomorphism_of_quiver_categories} for how these functors related to quivers as defined in \\fullref{def:quiver}.\n\\end{example}\n\n\\begin{remark}\\label{rem:natural_transformations_into_set}\n Let \\( \\cat{C} \\) be an arbitrary \\( \\mscrU \\)-small category. A \\hyperref[def:natural_transformation]{natural transformation} \\( \\alpha \\) from \\( F: \\cat{C} \\to \\ucat{Set} \\) to \\( G: \\cat{C} \\to \\ucat{Set} \\) is then a family of functions\n \\begin{equation*}\n \\seq{ \\alpha_A: F(A) \\to G(A) }_{A \\in \\cat{C}}.\n \\end{equation*}\n\n Suppose that for every two objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), the functions \\( \\alpha_A \\) and \\( \\alpha_B \\) agree on \\( F(A) \\cap F(B) \\). This is automatically satisfied in \\( F(A) \\) and \\( F(B) \\) are disjoint whenever \\( A \\neq B \\).\n\n We can then take the set-theoretic union of \\( \\alpha \\) to obtain the function\n \\begin{equation*}\n \\bigcup_{A \\in \\cat{C}} \\alpha_A: \\bigcup\\set{ F(A) \\given A \\in \\cat{C} } \\to \\bigcup\\set{ G(A) \\given A \\in \\cat{C} }.\n \\end{equation*}\n\n Both the domain and codomain are sets as a consequence of \\ref{def:grothendieck_universe/union}, therefore the function is well-defined in the universe \\( \\mscrU \\). Denote it on \\( \\Alpha \\) for brevity.\n\n An advantage of this is that we can define a natural transformation to be a function on a general enough set and then prove that its restrictions satisfy \\eqref{eq:def:natural_transformation/diagram}.\n\n For example, consider the power set functor \\( \\pow: \\ucat{Set} \\to \\ucat{Set} \\) discussed in \\fullref{ex:unary_functors_in_set}. The \\hyperref[def:multi_valued_function/identity]{identity function} \\( \\id_\\mscrU \\) is then a natural transformation from the identity functor \\( \\id_{\\ucat{Set}} \\) to \\( \\pow \\).\n\n Another natural transformation between the same functors is the singleton set operation \\( \\Sigma \\) on sets defined as \\( A \\mapsto \\set{ A } \\). Note that, in this context, \\( \\Sigma \\) operates not on the sets \\( \\id_{\\ucat{Set}}(A) \\) and \\( \\pow(A) \\), but on their members. The diagram \\eqref{eq:def:natural_transformation/diagram} becomes\n \\begin{equation}\\label{eq:rem:natural_transformations_into_set}\n \\begin{aligned}\n \\includegraphics[page=1]{output/rem__natural_transformations_into_set.pdf}\n \\end{aligned}\n \\end{equation}\n\n This diagram commutes because, for every function \\( f: A \\to B \\) and every \\( x \\in A \\), we have\n \\begin{equation*}\n f[\\set{ x }] = \\set{ f(x) }.\n \\end{equation*}\n\\end{remark}\n\n\\begin{definition}\\label{def:functor_category}\n Let \\( \\cat{C} \\) and \\( \\cat{D} \\) be arbitrary \\hyperref[def:category]{categories}. The \\term{functor category} \\( [\\cat{C}, \\cat{D}] \\), also denoted as \\( \\cat{D}^{\\cat{C}} \\), is defined as follows:\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj([\\cat{C}, \\cat{D}]) \\) is the set of all functors from \\( \\cat{C} \\) to \\( \\cat{D} \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( [\\cat{C}, \\cat{D}](F, G) \\) from \\( F \\) to \\( G \\) is the set of all \\hyperref[def:natural_transformation]{natural transformations} from \\( F \\) to \\( G \\).\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms} \\( \\alpha: F \\Rightarrow G \\) and \\( \\beta: G \\Rightarrow H \\) is the natural transformation \\( \\beta \\bincirc \\alpha: F \\Rightarrow H \\) defined in terms of componentwise morphism composition, i.e.\n \\begin{equation}\\label{eq:def:functor_category/composition}\n (\\beta \\bincirc \\alpha)_A \\coloneqq \\beta_A \\bincirc \\alpha_A.\n \\end{equation}\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the functor \\( F: \\cat{C} \\to \\cat{D} \\) is the \\term{identity natural transformation} \\( \\id_F: F \\Rightarrow F \\) with components\n \\begin{equation}\\label{eq:def:functor_category/identity}\n (\\id_F)_A \\coloneqq \\underbrace{\\id_{F(A)}}_{F(\\id_A)}\n \\end{equation}\n \\end{itemize}\n\\end{definition}\n\\begin{defproof}\n Just to verify that the composition \\( \\beta \\bincirc \\alpha \\) defined in \\eqref{eq:def:functor_category/composition} is indeed a natural transformation from \\( F \\) to \\( H \\), note that the following diagram trivially commutes:\n \\begin{equation}\\label{def:functor_category/composition}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__functor_category.pdf}\n \\end{aligned}\n \\end{equation}\n\n Now, to see that \\( [\\cat{C}, \\cat{D}] \\) is indeed a category, we verify the conditions \\ref{def:category/C1} and \\ref{def:category/C2}, which are in turn inherited from the same conditions on the categories \\( \\cat{C} \\) and \\( \\cat{D} \\).\n\n \\SubProofOf{def:category/C1} For every two functors \\( F, G: \\cat{C} \\to \\cat{D} \\) and natural transformation \\( \\alpha: F \\Rightarrow G \\), for every object \\( A \\in \\cat{C} \\) we have\n \\begin{equation*}\n \\id_{G(A)} \\bincirc \\alpha_A\n \\reloset{\\eqref{def:category/C1}} =\n \\alpha_A\n \\reloset{\\eqref{def:category/C1}} =\n \\alpha_A \\bincirc \\id_{F(A)}\n \\end{equation*}\n\n Therefore,\n \\begin{equation*}\n \\id_G \\bincirc \\alpha = \\alpha = \\alpha \\bincirc \\id_F\n \\end{equation*}\n and, after generalizing, we obtain that \\eqref{eq:def:category/C1} holds in \\( [\\cat{C}, \\cat{D}] \\).\n\n \\SubProofOf{def:category/C2} For any quadruple \\( F \\), \\( G \\), \\( H \\) and \\( T \\) of functors from \\( \\cat{C} \\) to \\( \\cat{D} \\) and every combination of natural transformations \\( \\alpha: F \\Rightarrow G \\), \\( \\beta: G \\Rightarrow H \\) and \\( \\gamma: H \\Rightarrow T \\), for every object \\( A \\in \\cat{C} \\) we have\n \\begin{equation*}\n (\\gamma_A \\bincirc \\beta_A) \\bincirc \\alpha_A\n \\reloset{\\eqref{def:category/C2}} =\n \\gamma_A \\bincirc (\\beta_A \\bincirc \\alpha_A).\n \\end{equation*}\n\n Therefore, after generalizing, we obtain that \\eqref{eq:def:category/C2} holds in \\( [\\cat{C}, \\cat{D}] \\).\n\\end{defproof}\n\n\\begin{remark}\\label{rem:functor_category_size}\n If \\( \\cat{C} \\) and \\( \\cat{D} \\) are \\( \\mscrU \\)-large categories in the sense of \\fullref{def:category_size}, we cannot construct the \\hyperref[def:functor_category]{functor category} \\( [\\cat{C}, \\cat{D}] \\). This is the main motivation for the \\hyperref[def:axiom_of_universes]{axiom of universes}, which is discussed in \\fullref{def:large_and_small_sets} and, in relation to category theory, in \\fullref{def:category_size}.\n\\end{remark}\n\n\\begin{example}\\label{ex:isomorphism_of_quiver_categories}\n In \\fullref{ex:quivers_as_functors}, we defined \\hyperref[def:quiver]{quivers} as functors from a certain index category \\( \\cat{I} \\) to \\( \\ucat{Set} \\) (for a \\hyperref[def:category_size]{fixed Grothendieck universe} \\( \\mscrU \\)).\n\n There is then an obvious correspondence between quivers as objects of \\hyperref[def:category_of_small_quivers]{\\( \\ucat{Quiv} \\)}, defined in \\fullref{def:quiver}, and quivers as objects in the \\hyperref[def:functor_category]{functor category} \\( [\\cat{I}, \\ucat{Set}] \\), defined in \\fullref{ex:quivers_as_functors}. Indeed, given any functor \\( Q: \\cat{I} \\to \\ucat{Set} \\), the quadruple\n \\begin{equation*}\n \\parens[\\Big]{ Q(V), Q(A), Q(h), Q(t) }\n \\end{equation*}\n is a quiver in the sense of \\fullref{def:quiver}.\n\n No object in \\( \\ucat{Quiv} \\) is formally equal to any object in \\( [\\cat{I}, \\ucat{Set}] \\) in the sense of \\hyperref[def:zfc]{\\logic{ZFC}}. They are, however, equivalent, as shown above, and this can be formalized by stating that the two categories are isomorphic, in the sense of \\fullref{def:morphism_invertibility/isomorphism}, as objects of the category \\( \\ucat[\\mscrV]{Cat} \\), where \\( \\mscrV \\) is a Grothendieck universe that strictly contains \\( \\mscrU \\). We have already defined this isomorphism explicitly.\n\n This is an example of \\term{isomorphism of categories}. In practice, if two categories are not so obviously identical, we are usually better served by \\term{equivalences of categories} defined in \\fullref{def:category_equivalence}.\n\\end{example}\n\n\\begin{definition}\\label{def:dual_natural_transformation}\\mcite{nLab:opposite_category}\n The \\term{opposite} or \\term{dual} natural transformation of \\( \\alpha: F \\Rightarrow G \\), where \\( F \\) and \\( G \\) are functors from \\( \\cat{C} \\) to \\( \\cat{D} \\), is the natural transformation \\( \\alpha^{\\opcat}: G^{\\opcat} \\Rightarrow F^{\\opcat} \\), in which we take the opposite of each component in \\( \\alpha \\).\n\n Dual natural transformation arise naturally in \\fullref{thm:dual_functor_category}.\n\\end{definition}\n\\begin{defproof}\n The naturality diagram \\eqref{eq:def:natural_transformation/diagram} commutes for \\( \\alpha^{\\opcat} \\) because all morphisms are simply reversed.\n\\end{defproof}\n\n\\begin{proposition}\\label{thm:dual_functor_category}\n For the \\hyperref[def:dual_category]{dual} of the \\hyperref[def:functor_category]{functor category} \\( [\\cat{C}, \\cat{D}] \\) we have\n \\begin{equation*}\n [\\cat{C}, \\cat{D}]^{\\opcat} = [\\cat{C}^{\\opcat}, \\cat{D}^{\\opcat}].\n \\end{equation*}\n\n This is part of the duality principles listed in \\fullref{thm:categorical_principle_of_duality}.\n\\end{proposition}\n\\begin{proof}\n In \\fullref{def:dual_functor}, we have defined the opposite functor \\( F^{\\opcat}: \\cat{C}^{\\opcat} \\to \\cat{D}^{\\opcat} \\) of \\( F: \\cat{C} \\to \\cat{D} \\) in a way that allows us to regard it as an object of \\( [\\cat{C}^{\\opcat}, \\cat{D}^{\\opcat}] \\).\n\n In \\fullref{def:dual_natural_transformation}, we have defined the opposite natural transformation \\( \\alpha^{\\opcat}: G^{\\opcat} \\to F^{\\opcat} \\) of \\( \\alpha: F \\Rightarrow G \\) in a way that allows us to regard it as a morphism of \\( [\\cat{C}^{\\opcat}, \\cat{D}^{\\opcat}] \\).\n\n Furthermore, \\( \\alpha^{\\opcat}: G^{\\opcat} \\to F^{\\opcat} \\) reverses the direction of its morphisms, and hence it is the dual to \\( \\alpha \\) in the category \\( [\\cat{C}, \\cat{D}]^{\\opcat} \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:diagonal_functor}\\mcite[143]{Leinster2016Basic}\n Given an \\term{index category} \\( \\cat{I} \\) and an arbitrary category \\( \\cat{C} \\), for any object \\( A \\) in \\( \\cat{C} \\), we can define the \\term{constant functor}\n \\begin{equation*}\n \\begin{aligned}\n &\\Delta_A^{\\cat{I}}: \\cat{I} \\to \\cat{C}, \\\\\n &\\Delta_A^{\\cat{I}}(X) \\coloneqq X, \\\\\n &\\Delta_A^{\\cat{I}}(g: X \\to Y) \\coloneqq \\id_A.\n \\end{aligned}\n \\end{equation*}\n\n Given two objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\), a natural transformation \\( \\alpha: \\Delta_A^{\\cat{I}} \\Rightarrow \\Delta_B^{\\cat{I}} \\) is an \\hyperref[def:cartesian_product/indexed_family]{indexed family} that gives the same morphism for every object of the index category \\( \\cat{I} \\).\n\n Indeed, the diagram \\eqref{eq:def:natural_transformation/diagram} in this case becomes\n \\begin{equation}\\label{eq:def:diagonal_functor/nat}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__diagonal_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n This diagram implies that \\( \\alpha_A = \\alpha_B \\) for any two objects \\( A \\) and \\( B \\) in \\( \\cat{I} \\). Therefore, all components of \\( \\alpha \\) are equal to some morphism in \\( \\cat{C}(A, B) \\).\n\n We can now define the \\( \\cat{I} \\)-shaped \\term{diagonal functor} on \\( \\cat{C} \\)\n \\begin{equation*}\n \\begin{aligned}\n &\\Delta^{\\cat{I}}: \\cat{C} \\to [\\cat{I}, \\cat{C}], \\\\\n &\\Delta^{\\cat{I}}(A) \\coloneqq \\Delta_A^{\\cat{I}}, \\\\\n &\\Delta^{\\cat{I}}(f: A \\to B) \\coloneqq \\seq{ f: A \\to B }_{k \\in \\cat{I}}.\n \\end{aligned}\n \\end{equation*}\n\n It is called a diagonal functor because, if \\( \\cat{I} \\) is a discrete category of two objects, then \\( \\Delta_{\\cat{I}} \\) gives the diagonal of the \\hyperref[def:product_category]{product category} \\( \\cat{C}^2 \\) by providing, for each object \\( A \\) of \\( \\cat{C} \\), the ordered pair \\( (A, A) \\) (and similarly for morphisms).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:natural_isomorphism}\\mcite{math3ma:natural_transformations}\n Let \\( F \\) and \\( G \\) be parallel \\hyperref[def:functor]{functors} from the category \\( \\cat{C} \\) to \\( \\cat{D} \\). The family \\eqref{eq:def:natural_transformation/family} is an isomorphism in the corresponding \\hyperref[def:functor_category]{functor category} \\( [\\cat{C}, \\cat{D}] \\) if and only if all of its components are isomorphisms and, for any morphism \\( f: A \\to B \\) in \\( \\cat{C} \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:natural_isomorphism/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__natural_isomorphism.pdf}\n \\end{aligned}\n \\end{equation}\n\n We say that \\( \\alpha \\) is a \\term{natural isomorphism}.\n\\end{proposition}\n\\begin{proof}\n If all components of \\( \\alpha \\) are isomorphisms, the condition\n \\begin{equation*}\n \\alpha_B \\bincirc F(f) = G(f) \\bincirc \\alpha_A\n \\end{equation*}\n is equivalent to\n \\begin{equation*}\n F(f) = \\alpha_B^{-1} \\bincirc G(f) \\bincirc \\alpha_A.\n \\end{equation*}\n\n We must now show that, if \\( \\alpha \\) is an isomorphism in \\( [\\cat{C}, \\cat{D}] \\), all of its components are isomorphisms.\n\n If \\( \\alpha: F \\Rightarrow G \\) is an isomorphism in \\( [\\cat{C}, \\cat{D}] \\). Then there exists some natural transformation \\( \\beta: G \\Rightarrow F \\) such that\n \\begin{equation*}\n \\beta \\bincirc \\alpha = \\id_F \\quad\\T{and}\\quad \\alpha \\bincirc \\beta = \\id_G.\n \\end{equation*}\n\n For every object \\( A \\) in \\( \\cat{C} \\), the morphism \\( \\alpha_A: F(A) \\to G(A) \\) composed with \\( \\beta_A: G(A) \\to F(A) \\) is\n \\begin{equation*}\n \\beta_A \\bincirc \\alpha_A = \\id_{F(A)}.\n \\end{equation*}\n\n Therefore, \\( \\alpha_A \\) is left-invertible. Analogously,\n \\begin{equation*}\n \\alpha_A \\bincirc \\beta_A = \\id_{F(A)}\n \\end{equation*}\n and hence \\( \\alpha_A \\) is right-invertible.\n\n Therefore, for every object \\( A \\) in \\( \\cat{C} \\), the morphism \\( \\alpha_A \\) is fully invertible, i.e. an isomorphism.\n\\end{proof}\n\n\\begin{definition}\\label{def:product_category}\\mcite[const. 1.1.11]{Leinster2016Basic}\n We define the \\term{product category} \\( \\cat{C} \\times \\cat{D} \\) of \\( \\cat{C} \\) and \\( \\cat{D} \\) as follows:\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} is the \\hyperref[def:cartesian_product]{Cartesian product}\n \\begin{equation}\\label{eq:def:product_category/objects}\n \\obj(\\cat{C} \\times \\cat{D}) \\coloneqq \\obj(\\cat{C}) \\times \\obj(\\cat{D}).\n \\end{equation}\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} from the pair of objects \\( (A, X) \\) to \\( (B, Y) \\) is the product\n \\begin{equation}\\label{eq:def:product_category/morphisms}\n (\\cat{C} \\times \\cat{D})\\parens[\\Big]{ (A, X), (B, Y) } \\coloneqq \\cat{C}(A, B) \\times \\cat{D}(X, Y).\n \\end{equation}\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms}\n \\begin{align*}\n (f, r)&: (A, X) \\to (B, Y) \\\\\n (g, s)&: (B, Y) \\to (C, Z)\n \\end{align*}\n is the pairwise composition\n \\begin{equation}\\label{eq:def:product_category/composition}\n (g, s) \\bincirc (f, r) \\coloneqq \\underbrace{(g \\bincirc f, s \\bincirc r)}_{(A, X) \\to (C, Z)}.\n \\end{equation}\n\n \\item The \\hyperref[def:category/identity]{identity morphism} of the pair \\( (A, X) \\) is simply the pair of identity morphisms \\( (\\id_A, \\id_X) \\).\n \\end{itemize}\n\\end{definition}\n\n\\begin{definition}\\label{def:hom_functor}\n Let \\( \\cat{C} \\) be a \\hyperref[def:category_size]{locally \\( \\mscrU \\)-small} category. We can regard the morphism sets \\( \\cat{C}(A, B) \\) as a functor parameterized by objects of \\( \\cat{C} \\).\n\n \\begin{thmenum}\n \\thmitem{def:hom_functor/binary} For any pair of morphisms \\( f: B \\to A \\) and \\( g: X \\to Y \\) in \\( \\cat{C} \\), define the operator\n \\begin{equation}\\label{eq:def:hom_functor/t}\n \\begin{aligned}\n &T_{f,g}: \\cat{C}(A, X) \\to \\cat{C}(B, Y) \\\\\n &T_{f,g}(s) \\mapsto g \\bincirc s \\bincirc f.\n \\end{aligned}\n \\end{equation}\n\n The action of \\( T_{f,g} \\) can be expressed graphically as\n \\begin{equation}\\label{eq:def:hom_functor/t_diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__hom_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n We can now define the following \\term{binary hom-functor}:\n \\begin{equation}\\label{eq:def:hom_functor/binary}\n \\begin{aligned}\n &\\cat{C}(\\anon*, \\anon*): \\cat{C}^{\\opcat} \\times \\cat{C} \\to \\ucat{Set} \\\\\n &\\cat{C}(A, X) \\coloneqq \\set{ s: A \\to X } \\\\\n &\\cat{C}(f, g) \\coloneqq T_{f,g}\n \\end{aligned}\n \\end{equation}\n\n \\thmitem{def:hom_functor/unary} Fixing the first argument \\( A \\) in \\eqref{eq:def:hom_functor/binary}, we instead obtain a covariant unary hom-functor:\n \\begin{equation}\\label{eq:def:hom_functor/unary/covariant}\n \\cat{C}(A, \\anon*): \\cat{C} \\to \\ucat{Set}\n \\end{equation}\n\n Analogously, fixing the second argument \\( X \\), we obtain a \\hyperref[def:hom_functor/unary]{contravariant} unary hom-functor:\n \\begin{equation}\\label{eq:def:hom_functor/unary/contravariant}\n \\cat{C}(\\anon*, X): \\cat{C}^{\\opcat} \\to \\ucat{Set}\n \\end{equation}\n \\end{thmenum}\n\\end{definition}\n\\begin{defproof}\n It is sufficient to verify that \\eqref{eq:def:hom_functor/binary} defines a functor. \\ref{def:functor/CF2} can be seen to hold by inspecting the diagram:\n \\begin{equation}\\label{eq:def:hom_functor/inv_composition}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__hom_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n The other functor condition \\ref{def:functor/CF1} is straightforward to prove.\n\\end{defproof}\n\n\\begin{proposition}\\label{thm:currying_is_natural_isomorphism}\n \\hyperref[def:function/currying]{Function currying} is a natural isomorphism between the functors\n \\begin{align*}\n &\\cat{Set}(A \\times B, C)\n &\\cat{Set}(A, \\cat{Set}(B, C))\n \\end{align*}\n\n More concretely, consider the following functors, which are loosely based in \\fullref{def:hom_functor}:\n \\begin{equation*}\n \\begin{aligned}\n &V: \\cat{Set}^3 \\to \\cat{Set} \\\\\n &V(A, B, C) \\coloneqq \\cat{Set}(A \\times B, C) \\\\\n \\Big[& V(f: X \\to A, g: Y \\to B, h: C \\to Z) \\Big](s: A \\times B \\to C) \\coloneqq \\smash{ \\overbrace{ (x, y) \\mapsto h\\parens[\\Bigg]{ s\\parens[\\Big]{ \\underbrace{ f(x), g(y) }_{A \\times B} } } }^{X \\times Y \\to Z} } \\\\\n \\end{aligned}\n \\end{equation*}\n and\n \\begin{equation*}\n \\begin{aligned}\n &W: \\cat{Set}^3 \\to \\cat{Set} \\\\\n &W(A, B, C) \\coloneqq \\cat{Set}(A, \\cat{Set}(B, C)) \\\\\n \\Big[& W(f: X \\to A, g: Y \\to B, h: C \\to Z) \\Big](t: A \\to \\cat{Set}(B, C)) \\coloneqq \\smash{ \\underbrace{ x \\mapsto \\overbrace{y \\mapsto h\\parens[\\Bigg]{ \\overbrace{t(f(x))}^{B \\to C}\\parens[\\Big]{ g(y) } } }^{Y \\to Z} }_{X \\to \\cat{Set}(Y, Z)} } \\\\\n \\end{aligned}\n \\end{equation*}\n\n Then the family of functions\n \\begin{equation*}\n \\begin{aligned}\n &\\alpha: V \\Rightarrow W \\\\\n &\\alpha_{A,B,C}(s: A \\times B \\to C) \\coloneqq a \\mapsto b \\mapsto s(a, b)\n \\end{aligned}\n \\end{equation*}\n is a \\hyperref[thm:natural_isomorphism]{natural isomorphism}.\n\\end{proposition}\n\\begin{proof}\n The function \\( \\varphi \\) is clearly invertible. Fix a triple of functions \\( f: X \\to A \\), \\( g: Y \\to B \\) and \\( h: C \\to Z \\). For every \\( s: A \\times B \\to C \\) we have\n \\begin{align*}\n [W(f, g, h)](\\alpha_{A,B,C}(s))\n &=\n x \\mapsto y \\mapsto \\parens[\\Big]{ h\\parens[\\Big]{ [a \\mapsto b \\mapsto s(a, b)](f(x))(g(y)) } }\n = \\\\ &=\n x \\mapsto y \\mapsto h\\parens[\\Big]{ s(f(x), g(y)) }\n = \\\\ &=\n \\alpha_{X,Y,Z}(V(f, g, h)),\n \\end{align*}\n which proves that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:currying_is_natural_isomorphism/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__currying_is_natural_isomorphism.pdf}\n \\end{aligned}\n \\end{equation}\n\\end{proof}\n", "meta": {"hexsha": "123c186ade3b4ded007339ad444fa45febebd78c", "size": 60210, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/functors.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/functors.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/functors.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.5337620579, "max_line_length": 612, "alphanum_fraction": 0.6822620827, "num_tokens": 20098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891261650248, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6413016852796846}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n%%% \\documentclass{article}\n%%% \\usepackage{graphicx}\n%%% \\usepackage{color}\n\n%%% \\sloppy\n%%% \\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n%%% \\begin{document}\n\n \n \n\\subsection*{Allan Overlapping Deviation}\n\n\\begin{par}\nExample for algorithm OADEV.\n\\end{par} \\vspace{1em}\n\\begin{par}\nOADEV is an algorithm to compute the overlapping Allan deviation for a set of time-domain frequency data.\n\\end{par} \\vspace{1em}\n\\begin{par}\nSee also W. J. Riley, \"The Calculation of Time Domain Frequency Stability\". Implementation: M. A. Hopcroft, \\verb\"mhopeng@gmail.com\", Matlab Central.'\n\\end{par} \\vspace{1em}\n\n\\subsubsection*{Contents}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item Generate sample data\n \\item Call algorithm\n \\item Display results\n\\end{itemize}\n\n\n\\subsubsection*{Generate sample data}\n\n\\begin{par}\nA random numbers with normal probability distribution function will be generated into input data \\lstinline{DI.y.v}. Next a drift will be added.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDI = [];\nDI.y.v = 1.5 + 3.*randn(1, 1e3);\nDI.y.v = DI.y.v + [1:1:1e3]./100;\n\\end{lstlisting}\n\\begin{par}\nLets suppose a sampling frequency is 1 Hz. The algorithm will generate all possible tau values automatically.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDI.fs.v = 1;\n\\end{lstlisting}\n\n\n\\subsubsection*{Call algorithm}\n\n\\begin{par}\nUse QWTB to apply algorithm \\lstinline{OADEV} to data \\lstinline{DI}.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDO = qwtb('OADEV', DI);\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\nQWTB: no uncertainty calculation\n\\end{lstlisting} \\color{black}\n \n\n\\subsubsection*{Display results}\n\n\\begin{par}\nLog log figure is the best to see allan deviation results:\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nfigure; hold on\nloglog(DO.tau.v, DO.oadev.v, '-b')\nloglog(DO.tau.v, DO.oadev.v + DO.oadev.u, '-k')\nloglog(DO.tau.v, DO.oadev.v - DO.oadev.u, '-k')\nxlabel('\\tau (sec)');\nylabel('\\sigma_y(\\tau)');\ntitle(['period = ' num2str(DI.fs.v)]);\ngrid('on'); hold off\n\\end{lstlisting}\n\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{algs_examples_published/OADEV_alg_example_01.pdf}\n\\end{center}\n\n\n\n%%% \\end{document}\n \n", "meta": {"hexsha": "161292e88230875a4c4c876718ed73284af6a9e8", "size": 2368, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/algs_examples_published/doc_OADEV.tex", "max_stars_repo_name": "qwtb/qwtb", "max_stars_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-12-09T13:18:54.000Z", "max_stars_repo_stars_event_max_datetime": "2015-12-09T13:18:54.000Z", "max_issues_repo_path": "doc/algs_examples_published/doc_OADEV.tex", "max_issues_repo_name": "qwtb/qwtb", "max_issues_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 18, "max_issues_repo_issues_event_min_datetime": "2015-12-09T13:08:38.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-13T11:33:41.000Z", "max_forks_repo_path": "doc/algs_examples_published/doc_OADEV.tex", "max_forks_repo_name": "qwtb/qwtb", "max_forks_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-11-11T02:12:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-09-17T12:59:18.000Z", "avg_line_length": 24.9263157895, "max_line_length": 150, "alphanum_fraction": 0.714527027, "num_tokens": 741, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189134878876, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6413016836667522}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage{latexsym}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{graphicx}\n\\usepackage{wrapfig}\n\\usepackage{pseudocode}\n\\usepackage{url}\n\\usepackage{float}\n\\usepackage{subcaption}\n\\usepackage[backref, colorlinks=true, citecolor=red, urlcolor=blue, pdfauthor={Jyh-Ming Lien}]{hyperref}\n\n\n\\newcommand{\\handout}[5]{\n \\noindent\n \\begin{center}\n \\framebox{\n \\vbox{\n \\hbox to 5.78in { {\\bf } \\hfill #2 }\n \\vspace{4mm}\n \\hbox to 5.78in { {\\Large \\hfill #5 \\hfill} }\n \\vspace{2mm}\n \\hbox to 5.78in { {\\em #3 \\hfill #4} }\n }\n }\n \\end{center}\n \\vspace*{4mm}\n}\n\n\\newcommand{\\lecture}[4]{\\handout{#1}{#2}{#3}{#4}{#1}}\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{corollary}[theorem]{Corollary}\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{observation}[theorem]{Observation}\n\\newtheorem{proposition}[theorem]{Proposition}\n\\newtheorem{definition}[theorem]{Definition}\n\\newtheorem{claim}[theorem]{Claim}\n\\newtheorem{fact}[theorem]{Fact}\n\\newtheorem{assumption}[theorem]{Assumption}\n\n% 1-inch margins, from fullpage.sty by H.Partl, Version 2, Dec. 15, 1988.\n\\topmargin 0pt\n\\advance \\topmargin by -\\headheight\n\\advance \\topmargin by -\\headsep\n\\textheight 8.9in\n\\oddsidemargin 0pt\n\\evensidemargin \\oddsidemargin\n\\marginparwidth 0.5in\n\\textwidth 6.5in\n\n\\parindent 0in\n\\parskip 1.5ex\n%\\renewcommand{\\baselinestretch}{1.25}\n\n\\begin{document}\n\n\\lecture{Project 2: Hedcut}{Fall 2019}{William Austin}{CS 633 Computational Geometry}\n\n\\section{Summary of the hedcut method}\n\nThe hedcut method that is implemented in this project is closely related to the idea of stippling, which uses a combination of dots to represent an image. Originally, this style of drawing was used in newspapers because they are typically black-and-white, can be easily printed on a page, and retained the important details of the image throughout this process. However, these works are also produced manually by a skilled artist and take several hours to complete. Therefore, this project explores how this style of art can be generated in an automated way, based off of an input image.\n\nThe provided functionality uses the concepts described in the paper ``Weighted Voronoi Stippling'', which is based on the concept of centroidal Voronoi diagrams. In order for a Voronoi diagram to be centroidal, the generating points must also be the centroid, which can be interpreted as the center of mass. For images, we can calculate the intensity at each pixel: ((R + G + B) / 3), and interpret it as a density function. Then, for each Voronoi cell, we can compute the centroid, based on the ``weight'' and location of each pixel in the cell.\n\nThe implementation is an iterative algorithm that starts off with random samples from the image. Then, we repeatedly:\n\\begin{enumerate}\n\t\\item Compute the Voronoi diagram for the given point locations, and\n\t\\item For each cell, find the centroid, and use it as the generating point for the next iteration.\n\\end{enumerate}\n\nWe must introduce some stopping criteria so that the algorithm eventually halts. In our case, we can specify an upper limit to the number of iterations performed or the distance the centroid moves in a step (total or average.) \n\n\\section{Improvement of hedcuter method}\n\nI've made several improvements to the provided functionality, listed below. In each section, I will give some examples, and intuition for the new features. A quick summary of the changes is given here, in the expanded options list for the hedcuter application:\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[scale=0.3]{HeducterCommandOptions}\n\t\\caption{Hedcuter Application Command Options}\n\\end{figure}\n\n\n\\subsection{Creating Voronoi Diagrams using OpenGL}\n\nThe first large improvement to the application was to use OpenGL to compute the Voronoi diagrams at each step in the algorithm. The idea is to construct a scene with a cone centered on the X-Y plane at each generating point. By fixing the camera to look at the all of the cones in the direction of the Z-axis, using a 2-D projection, the resulting scene is the Voronoi diagram.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[scale=0.35]{VoronoiDiagram}\n\t\\caption{Voronoi Diagram, as computed by OpenGL}\n\t\\label{fig:voronoiDiagram}\n\\end{figure} \n\nOne of the problems that arises during the implementation of this feature is trying to map the regions in the rendered diagram back to the point that generated them. We do this by converting the index of the generating point $[0 \\dots n - 1]$ into the RGB color-space by transforming the index into a point in the range $[0 \\dots 256^3]$, and then representing it as $256^2 \\cdot R + 256 \\cdot G + B$. This tells Open GL what colors to use for the cones, and then we do the reverse transform to map the Voronoi regions of a certain color back to the correct generating point.\n\nSome other challenges that arise include:\n\\begin{itemize}\n\t\\item Model space in OpenGL is represented in X-Y coordinates, so (0, 0) is in the bottom-left of the screen, whereas OpenCV uses row-column format, with the pixel at (0, 0) in the upper-left of the screen. Therefore, when converting between the representations, we must flip the image, and reverse the coordinate order.\n\t\\item OpenGL initialization requires another library to control the window system, such as GLUT. Offline rendering does not seem to be especially well supported.\n\t\\item Internally, the OpenCV library uses a BGR representation for colors instead of the more traditional RGB. Therefore, we need to take care to convert correctly and make sure that the code is accessing the correct channel for the data structure.\n\\end{itemize} \n\nMost of my code for this feature is in the \\verb|vorgpu.hpp| and \\verb|vorgpu.cpp| files. A few changes to other files were required to integrated the command line option and call the correct functionality. This is in the \\verb|wcvt.cpp| file, in the \\verb|compute_weighted_cvt()| function. The functionality is enabled by specifying the \\verb|-useOpenGL| flag, with no options.\n\nI was able to get good stippling results using the OpenGL option. The example below shows the original input on the left, and the generated hedcut with 4000 disks.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t \t\\includegraphics[width=\\linewidth]{Snorlax}\n\t \t\\caption{Original Image, Snorlax}\n\t \t\\label{fig:snorlax}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Snorlax-OpenGL-4000}\n\t\t\\caption{Hedcut Image using OpenGL with 4000 Disks, Snorlax}\n\t\t\\label{fig:snorlaxOpenGL}\n\t\\end{subfigure}\n\t\\caption{OpenGL Example with Snorlax}\n\t\\label{fig:snorlaxExample}\n\\end{figure}\n\nDespite the complete implementation and reasonable output, I did not notice a large speedup when using the OpenGL functionality instead of the built-in method. In most cases, the total time taken is comparable. For example, the output below shows that OpenGL based method was slightly slower for 750 disks.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[scale=0.35]{OpenGL-Performance}\n\t\\caption{Relative OpenGL Performance}\n\t\\label{fig:OpenGLPerformance}\n\\end{figure} \n\n\\subsection{Colored Disks, Grayscale Disks, and Background Color}\n\nFor the next feature to be added into the Hedcuter code, I added a few different ways to specify color in the output image. This includes:\n\n\\begin{itemize}\n\t\\item Use the \\verb|-diskColor R G B| option to specify a constant disk color. The default is black.\n\t\\item Use the \\verb|-bgColor R G B| option to specify a background color for the output. The default is no background (transparent).\n\t\\item Use the \\verb|-useGrayscaleColor| flag to create an image with disks based on the average grayscale intensity for each Voronoi cell to be represented.\n\t\\item Use the \\verb|-useAvgDiskColor| flag to create an image with disks based on the average color for each Voronoi cell to be represented.\n\\end{itemize} \n\nSome sample results are shown below.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Mount-Rushmore}\n\t\t\\caption{Original Image, Mount Rushmore}\n\t\t\\label{fig:mountRushmore}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Mount-Rushmore-Red-15000}\n\t\t\\caption{Red Hedcut Image with 15000 Disks}\n\t\t\\label{fig:mountRusmoreRed}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Mount-Rushmore-Blue-15000}\n\t\t\\caption{Blue Hedcut Image with 15000 Disks}\n\t\t\\label{fig:mountRusmoreBlue}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Mount-Rushmore-Inverted-15000}\n\t\t\\caption{Inverted Hedcut Image with 15000 Disks}\n\t\t\\label{fig:mountRusmoreInverted}\n\t\\end{subfigure}\n\t\\caption{Color Example with Mount Rushmore}\n\t\\label{fig:mountRushmoreEx}\n\\end{figure}\n\nA more dynamic coloring approach can be seen by using the grayscale color and average color options. This is accomplished by simply averaging all of the cell colors together to choose an approximation of the input color when we create the disks for the hedcut.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.52\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Jordan-Pond}\n\t\t\\caption{Original Image, Jordan Pond}\n\t\t\\label{fig:jordanPond}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Jordan-Pond-Grayscale-10000}\n\t\t\\caption{Jordan Pond, Grayscale, 10000 Disks}\n\t\t\\label{fig:jordanPondGray}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Jordan-Pond-Color-10000}\n\t\t\\caption{Jordan Pond, Colored, 10000 Disks}\n\t\t\\label{fig:jordanPondColor}\n\t\\end{subfigure}\n\t\\caption{Colore Example with Jordan Pond}\n\t\\label{fig:jordanPondExample}\n\\end{figure}\n\n\n\\subsection{Disk Sizing based on Cell Area and Intensity, with Regularization}\n\nI have also implemented dynamic disk sizing to produced different visual effects in the final hedcut image. The first way of doing this is to scale the size of each disk by the size of the containing Voronoi cell. This is done by specifying \\verb|-areaScaling scaleFactor|. In the example with Roger Federer, you can see that the disks on the handle of the racket and his grip are much larger than the rest of the disks in the second example, but much smaller in second example because we specified a negative value to invert the relationship.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.52\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Roger-Federer}\n\t\t\\caption{Original Image, Roger Federer}\n\t\t\\label{fig:rf}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Roger-Federer-AreaScale-12000}\n\t\t\\caption{Roger Federer, Area scaled, 12000 Disks}\n\t\t\\label{fig:rf2}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Roger-Federer-ReverseAreaScaling-12000}\n\t\t\\caption{Roger Federer, Reverse area scaled, 12000 Disks}\n\t\t\\label{fig:rf3}\n\t\\end{subfigure}\n\t\\caption{Area Scaling Example with Roger Federer}\n\t\\label{fig:rf4}\n\\end{figure}\n\nWe can perform the same scaling operation, but use the average intensity of the cell as the input, instead of the cell area. This is shown below when we specify the \\verb|-areaScaling scaleFactor| option. As expected, in the first image, the darker areas produce larger disks, and the lighter areas produce smaller disks. The relationship is reversed for the second example when we pass a negative argument.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.52\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Emilee-And-Lola}\n\t\t\\caption{Original Image, Girl and Dog}\n\t\t\\label{fig:gff}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Emilee-And-Lola-IntensityScaled2-10000}\n\t\t\\caption{Intensity scaled, 10000 Disks}\n\t\t\\label{fig:rff2}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Emilee-And-Lola-IntensityScaled-10000}\n\t\t\\caption{Reverse Intensity scaled, 10000 Disks}\n\t\t\\label{fig:rff3}\n\t\\end{subfigure}\n\t\\caption{Intensity Scaling Example with Girl and Dog}\n\t\\label{fig:rff4}\n\\end{figure}\n\nWe can also apply a regularization process to artificially introduce some variation into the stippling. For example, in the area scaling example, most of the cells are about the same size with a few outliers. In the regularization process, we push the distribution of disk sizes to be more uniform, based on a regularization factor from 0 to 1. The images below show an image of train with negative area scaling and regularization values of 0, 0.3 and 0.8. This effect is enabled by using the \\verb|-regularize factor| command. \n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Train-Canada}\n\t\t\\caption{Original Image, Train}\n\t\t\\label{fig:train}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Train-Canada-8000-1}\n\t\t\\caption{Reverse area scaling, R = 0}\n\t\t\\label{fig:train1}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Train-Canada-8000-2}\n\t\t\\caption{Reverse area scaling, R = 0.3}\n\t\t\\label{fig:train2}\n\t\\end{subfigure}\n\t\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Train-Canada-8000-3}\n\t\t\\caption{Reverse area scaling, R = 0.8}\n\t\t\\label{fig:train3}\n\t\\end{subfigure}\n\t\\caption{Regularization Example with Train}\n\t\\label{fig:mountRusddhmoreEx}\n\\end{figure}\n\n\\subsection{More Dynamic Initial Sampling}\n\nWe have also implemented a tweak to allow greater control over the sampling process. This is enabled with the \\verb|-samplingWeight weight| command option. The effect of this is that we are more selective when keeping or throwing away points during the initial sampling. A larger value means that lighter pixels are less likely to get chosen and darker pixels are more likely to be included. In the example below, we see the results for a sampling weight of 0 (uniform distribution), and a sampling weight of 3. Note that we have implemented this as a linear cutoff and large values may lead to an infinite loop because not enough points are eligible for inclusion, so care must be taken.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.52\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Chessboard}\n\t\t\\caption{Original Image, Chessboard}\n\t\t\\label{fig:gfff}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Chessboard-Points-1}\n\t\t\\caption{Uniform sample points, 6000 Disks}\n\t\t\\label{fig:rfff2}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Chessboard-6000-1}\n\t\t\\caption{Stippled Chessboard 1, 600 Disks}\n\t\t\\label{fig:rfff3}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Chessboard-Points-2}\n\t\t\\caption{Weighted sample points, 6000 Disks}\n\t\t\\label{fig:rffdf2}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Chessboard-6000-2}\n\t\t\\caption{Stippled Chessboard 2, 600 Disks}\n\t\t\\label{fig:rfffd3}\n\t\\end{subfigure}\n\t\\caption{Initial Sampling Example with Chessboard}\n\t\\label{fig:rfff4}\n\\end{figure}\n\n\\subsection{Edge Detection}\n\nThe last feature that we have included in the application is the ability to incorporate information about edges in the input image into the stipple pattern. This enabled by using the \\verb|-detectEdges alpha| option. If this is specified, we use OpenCV to detect edges in the input image, using the Sobel operator. Then, we create a linear combination of this output with the grayscale input image, to be used when computing the cell centroids at each step. This helps drag cell centers toward edges in the input image, creating a more well-defined output hedcut, as shown below.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Portrait}\n\t\t\\caption{Original Image, Portrait}\n\t\t\\label{fig:trafin}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Portrait-Sobel}\n\t\t\\caption{Portrait after Sobel Operator (k=3)}\n\t\t\\label{fig:trxain1}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Portrait-7000-1}\n\t\t\\caption{Portrait Stippling Default}\n\t\t\\label{fig:traixn2}\n\t\\end{subfigure}\n\t\\begin{subfigure}[b]{.48\\linewidth}\n\t\t\\includegraphics[width=\\linewidth]{Portrait-7000-2}\n\t\t\\caption{Portrait Stippling with Edge Detection}\n\t\t\\label{fig:traxin3}\n\t\\end{subfigure}\n\t\\caption{Edge Detection Example with Portrait}\n\t\\label{fig:mountRusdxdhmoreEx}\n\\end{figure}\n\n\\section{Conclusion}\n\nThis project was a good example of how computer vision and computation geometry algorithms can be applied to interesting real world problems. All of the code that I implemented is completed, and the submitted application runs without any errors. However, there are several things that could be improved or expanded upon. Some examples are:\n\n\\begin{itemize}\n\t\\item When using the OpenGL Voronoi diagram option, cells seem to converge somewhat more slowly than using the original method. All of my code is a drop-in replacement for the \\verb|vor()| method, so I was a bit confused by this because the \\verb|move_sites()| code is exactly the same, regardless of whether the option is on or not.\n\t\\item There are several cases in the code where I have commented out print statements, or dead code. Throughout the course of this project, I spent considerable time experimenting with different ways of doing things and did not have time to clean up everything.\n\t\\item An idea for improvement is to support alpha channels for the disk and background coloring functionality.\n\t\\item Another improvement would be to have support for color scaling, so that stipples could be different shades of the same color (variable intensity). This would look like a grayscale image, but with a different hue.\n\t\\item There are also performance improvements that could be made. For example, I did not implement Fortune's algorithm. In addition, my focus for this project was to enhance the visual aspects of the produced hedcut images, so it is likely that portions of my code could be implemented more cleanly and efficiently.\n\\end{itemize}\n\n\\bibliographystyle{plain}\n\\bibliography{report}\n\n\\end{document}\n\n\n", "meta": {"hexsha": "5a64ec2c32406d2b53e6c2dcf4fb39dfde582e17", "size": 18380, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/report.tex", "max_stars_repo_name": "william-r-austin/hedcut", "max_stars_repo_head_hexsha": "58c6e452439a4abaf2bf09e09393e9a267fe2fe4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/report.tex", "max_issues_repo_name": "william-r-austin/hedcut", "max_issues_repo_head_hexsha": "58c6e452439a4abaf2bf09e09393e9a267fe2fe4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/report.tex", "max_forks_repo_name": "william-r-austin/hedcut", "max_forks_repo_head_hexsha": "58c6e452439a4abaf2bf09e09393e9a267fe2fe4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.3407821229, "max_line_length": 689, "alphanum_fraction": 0.7722524483, "num_tokens": 5035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891305219504, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6413016802683813}} {"text": "\\chapter{补充内容}\n\\section{$L^{p}$ Spaces for $00 \\mu-a . e .$, define $\\|g\\|_{L^{r}}:=\\left\\|g^{-1}\\right\\|_{L^{|r|}}^{-1}$. Then for $f \\geq 0$ and $g>0 \\mu-$ a.e., we have that\n \\[\n \\|f g\\|_{L^{1}} \\geq\\|f\\|_{L^{q}}\\|g\\|_{L^{q^{\\prime}}}\n \\]\n where $\\frac{1}{q}+\\frac{1}{q^{\\prime}}=1$\n\\end{proposition}\n\\begin{proof}\n If $f g \\notin L^{1}(X, \\mu)$ (i.e. $\\|f g\\|_{L^{1}}=\\infty$ ) or $g^{-1} \\notin L^{q^{\\prime}}(X, \\mu)$, then the inequality is trivial. So assume otherwise. Since $q \\in(0,1)$ and $1=\\frac{1}{q}+\\frac{1}{q^{\\prime}}$, we have that $q^{\\prime}<0$ and\n \\[\n \\frac{1}{q}=\\frac{1}{1}+\\frac{1}{\\left|q^{\\prime}\\right|}\n \\]\n By Hölder's inequality applied to $f g$ and $g^{-1} \\in L^{\\left|q^{\\prime}\\right|}$,\n \\[\n \\|f\\|_{L^{q}}=\\left\\|f g g^{-1}\\right\\|_{L^{q}} \\leq\\|f g\\|_{L^{1}}\\left\\|g^{-1}\\right\\|_{L^{\\prime q^{\\prime}} \\mid} \\Rightarrow\\|f\\|_{L^{q}}\\|g\\|_{L^{q^{\\prime}}}=\\|f\\|_{L^{q}}\\left\\|g^{-1}\\right\\|_{L^{\\left|q^{\\prime}\\right|}} \\leq\\|f g\\|_{L^{1}}\\qedhere\n \\]\n\\end{proof}\n\n\\begin{proposition}[Reverse Minkowski's]\n Let $f_{1}, \\cdots, f_{N} \\in L^{p}(X, \\mu)$, where $01$, by the preceding lemma,\n \\[\n \\left\\|\\sum_{j=1}^{N} f_{j}\\right\\|_{L^{p}}=\\left(\\int_{X}\\left|\\sum_{j=1}^{N} f_{j}\\right|^{p} d x\\right)^{\\frac{1}{p}} \\leq\\left(\\sum_{j=1}^{N} \\int_{X}\\left|f_{j}\\right|^{p} d x\\right)^{\\frac{1}{p}} \\leq N^{\\frac{1}{p}-1} \\sum_{j=1}^{N}\\left(\\int_{X}\\left|f_{j}\\right|^{p} d x\\right)^{\\frac{1}{p}}=N^{\\frac{1-p}{p}} \\sum_{j=1}^{N}\\left\\|f_{j}\\right\\|_{L^{p}}\n \\]\n To see that $N^{\\frac{1-p}{p}}$ is the best possible constant, let $E$ be a measurable set such that $\\mu(E)=\\alpha<\\infty$, and set $E_{j}:=E$ and $f_{j}:=\\mathbf{1}_{E}$ for $1 \\leq j \\leq N$. Then\n \\[\n \\left\\|\\sum_{j=1}^{N} f_{j}\\right\\|_{L^{p}}=\\left(\\sum_{j=1}^{N} \\mu\\left(E_{j}\\right)\\right)^{\\frac{1}{p}}=(N \\alpha)^{\\frac{1}{p}}=N^{\\frac{1-p}{p}}\\left(N \\alpha^{\\frac{1}{p}}\\right)=N^{\\frac{1-p}{p}} \\sum_{j=1}^{N} \\mu\\left(E_{j}\\right)^{\\frac{1}{p}}=N^{\\frac{1-p}{p}} \\sum_{j=1}^{N}\\left\\|f_{j}\\right\\|_{L^{p}}\n \\]\n\\end{proof}\n\\subsection{Day's theorem}\n\\begin{lemma}\n Let $(X, \\mathcal{A}, \\mu)$ be a measure space with the property that given any $f \\in L^{p}(X, \\mu)$ for $p \\in(0,1)$, the functional\n \\[\n \\mathcal{A} \\rightarrow \\mathbb{R}, E \\mapsto \\int_{E}|f|^{p} d \\mu\n \\]\n assumes all values between 0 and $\\|f\\|_{L^{p}}^{p}$. Then $L^{p}(X, \\mu)$, with $00$ such that $\\|f\\|_{L^{p}}<\\delta \\Rightarrow\\|f\\| \\leq 1$. Then $\\forall f \\in L^{p}([0,1]), \\frac{\\alpha \\delta f}{\\|f\\|_{L^{p}}} \\in B_{\\delta}$, where $0<|\\alpha|<1 .$ Hence,\n \\[\n \\left\\|\\frac{\\alpha \\delta f}{\\|f\\|_{L^{p}}}\\right\\| \\leq 1 \\Rightarrow\\|f\\| \\leq \\frac{1}{\\alpha \\delta}\\|f\\|_{L^{p}}\n \\]\n Letting $\\alpha \\rightarrow 1$, we see that the inequality holds for $C=\\frac{1}{\\delta} .$ Choose $C=\\inf \\left\\{K:\\|f\\| \\leq K\\|f\\|_{L^{p}} \\forall f \\in L^{p}([0,1])\\right\\}$ (Note that we do not exclude the possibility that $C=0$ ). By the intermediate value theorem, there exists $c \\in(0,1)$ such that\n \\[\n \\int_{0}^{c}|f|^{p} d \\lambda=\\int_{c}^{1}|f|^{p} d \\lambda=\\frac{1}{2} \\int_{0}^{1}|f|^{p} d \\lambda\n \\]\n Set $g=f \\chi_{[0, c]}$ and $h=f \\chi_{(c, 1]} .$ Then $f=g+h$ and $\\|g\\|_{L^{p}}=\\|h\\|_{L^{p}}=2^{-\\frac{1}{p}}\\|f\\|_{L^{p}} .$ By the triangle inequality,\n \\[\n \\|f\\| \\leq\\|g\\|+\\|h\\| \\leq C\\left(\\|g\\|_{L^{p}}+\\|h\\|_{L^{p}}\\right)=\\frac{C}{2^{\\frac{1}{p}-1}}\\|f\\|_{L^{p}}\n \\]\n Since $p \\in(0,1), \\frac{C}{2^{\\frac{1}{p}-1}} \\leq C \\Rightarrow C=0 \\Rightarrow\\|f\\|=0 ,\\forall f \\in L^{p}(X, \\mu)$, which contradicts that $\\|\\cdot\\|$ is a norm.\n\\end{proof}\n\n\\begin{remark}\n In fact, the non-normability of $L^{p}([0,1])$, when $0 0.5$ current rewards begin to matter more than past values.\n\nThis general method for controlling recent values carries on to more complex models, and longer time horizons. Use it as needed.\n\n\\subsection*{The discount}\nOften values or summed rewards will be discounted as time passes. A common form is:\n\n\\begin{eqnarray}\n \\gamma \\leftarrow \\gamma ^ t\n\\end{eqnarray}\n\nWhere $0 < \\gamma \\leq 1$ and $t \\in \\mathbb{N}_1$ is an integer code for the elapsed discrete time.\n\nAnother form is exponential decay \\cite{Francois-Lavet2015}:\n\\begin{eqnarray}\n \\gamma \\leftarrow \\gamma - \\gamma \\tau\n\\end{eqnarray}\n\nWhere $\\tau \\in \\mathbb{R}_+$. Typically $\\tau \\approx 0.02$.\n\n\\textbf{Note}: anywhere $\\gamma$ is used here assume a discounted form can--and often is--substituted in.\n\n\n% \\vfill\\null\n% \\columnbreak\n\\subsection*{The temporal difference}\nIf the reward is delayed, the minimal model can't give credit to past states. \n\nOne solution is to take the difference between the current $V$ next, $V'$, as in the \\emph{SARSA} rule:\n\n\\begin{eqnarray}\n V \\leftarrow V + \\alpha (r + \\gamma V' - V) \n\\end{eqnarray}\n\nWhere $(0 < \\gamma \\leq 1)$, $(0 < \\alpha \\leq 1)$, and $V_0 = 0$. Unlike the minimal model here $\\gamma$ wieghs the influence of the future, not the past.\n\n\n\\subsection*{The maximum difference}\n\\emph{Q learning} takes the difference between the current value $V$ and the maximum value, $\\argmin_A \\textbf{V}$.\n\n\\begin{eqnarray}\n V \\leftarrow V + \\alpha (r + \\gamma \\argmin_A \\textbf{V} - V) \n\\end{eqnarray}\n\nWhere $(0 < \\gamma \\leq 1)$, $(0 < \\alpha \\leq 1)$, and $V_0 = 0$.\n\n\\textbf{Note:} as policies are define by their target action $a$, \\emph{Q learning} is often called an \\emph{off-policy} learning rule, while the temporal difference is \\emph{on-policy}.\n\n\\subsection*{The average difference}\n\\emph{Avantage learning} works with relative changes in value, comparing $V$ to its average $\\bar{V}$.\n\\begin{eqnarray}\n V \\leftarrow V + \\alpha (r + \\gamma \\bar{V} - V) \\\\\n \\bar{V} = \\frac{1}{k} \\sum_\\textbf{A}{V} \n\\end{eqnarray}\n\nIn practice $\\bar{V}$ is approximated online, often with a discounting scheme.\n\n\\textbf{Note}: advantages play an important role in policy gradients.\n\n\\subsection*{The regretable difference}\n\nTODO....\n\n\n\\vfill\\null\n\\columnbreak\n\\subsection*{The policy gradient}\nPolicy gradients learn to map states to directly to actions. They are most useful when problems are continious in action spaace and time, or when promises of local optima are needed. \n\nPolicy gradients tend however to be sample inefficient, and by definition they are also \\emph{on policy} methods, which can also slow learning down.\n\nA policy is the probability of taking an action $a$, given a state $s$ and some generic parameters $\\theta$.\n\n\\begin{eqnarray}\n\\pi = P(a, s, \\theta)\n\\end{eqnarray}\n\nTo learn a good policy $\\pi$, the parameter gradient should follow the reward gradient. \n\n\\begin{eqnarray}\n\\theta \\leftarrow \\theta + \\alpha \\frac{\\partial \\rho}{\\partial \\theta}\n\\end{eqnarray}\n\nAs with other reinforcement leaning methods, finding a suitable way to estimate $\\rho$ becomes a major concern, and our problem becomes:\n\n\\begin{eqnarray}\n\\theta \\leftarrow \\theta + \\alpha \\frac{\\partial V}{\\partial \\theta}\n\\label{eq:grad}\n\\end{eqnarray}\n\n\\subsection*{The average (again)}\nIn a gradient setting \\emph{avantage learning} increases in $V$ are by definition better than average, $\\bar{V}$.\n\n\\begin{eqnarray}\n V \\leftarrow V + \\alpha (r + \\gamma \\bar{V} - V) \\\\\n \\bar{V} = \\frac{1}{k} \\sum_\\textbf{A}{V} \n\\end{eqnarray}\n\n Gradients driven by advatage updates ensure learning is always better than average \\cite{Sutton}. By eq. \\ref{eq:grad}, purturbations to $\\theta \\leftarrow \\theta + \\delta \\theta$ which increase $V$ must, by definition, be better than average. \n\n\\subsection*{The generalized advantage}\nDiscounting values can introduce a bias to the final estimate. The generalized advantage scheme takes this into account \\cite{Schulman2015a}. \n\n\\subsection*{Actor-critic}\nValue learning and policy can be joined into a single \\emph{actor-critic architecture}.\n\nTODO\\ldots\n\n\n% ------------------------------------------------------------------------\n\\vfill\\null\n\\columnbreak\n\\section*{The planning models}\nPlanning means the learner has, or creates, an \\emph{explicit map} of the state-space. This map can come from outside the system, from some useful oracle, or can be learned.\n\n\\subsection*{The DYNA}\n\n\\subsection*{The prioritized sweep}\n\n\\subsection*{The successor}\n\n\\end{multicols}\n\n\\newpage\n\\bibliography{library}\n\\end{document}", "meta": {"hexsha": "5d23ac3a0adf35fa8831137ed919b3b63b82079c", "size": 8569, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/rl.tex", "max_stars_repo_name": "CoAxLab/azad", "max_stars_repo_head_hexsha": "d1498069dd8856e93ae077b34dd7c9f1c7ce80e6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2018-09-11T21:06:12.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-28T17:36:52.000Z", "max_issues_repo_path": "notes/rl.tex", "max_issues_repo_name": "CoAxLab/azad", "max_issues_repo_head_hexsha": "d1498069dd8856e93ae077b34dd7c9f1c7ce80e6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/rl.tex", "max_forks_repo_name": "CoAxLab/azad", "max_forks_repo_head_hexsha": "d1498069dd8856e93ae077b34dd7c9f1c7ce80e6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2018-09-12T00:40:52.000Z", "max_forks_repo_forks_event_max_datetime": "2018-10-29T15:45:54.000Z", "avg_line_length": 35.5560165975, "max_line_length": 251, "alphanum_fraction": 0.6879449177, "num_tokens": 2416, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8807970904940926, "lm_q2_score": 0.7279754430043072, "lm_q1q2_score": 0.6411986521493419}} {"text": "\\newpage \\subsection{Recurrent Sequences}\t\n\t\n\t\n\t\\begin{myitemize}\n\t\t\\item \\href{https://ufile.io/ywbniil2}{WOOT 2010-11 Recursion}\n\t\\end{myitemize}\n\t\n\t\n\t\\begin{BoxedTheorem}{Sum of Geometric Sequences}{}\\label{theorem:Sum of Geometric Sequences}\n\t\tEvery recurrent sequence can be written as a sum of some geometric sequences. Given a recurrent sequence, \n\t\t\\[x_n = a_1x_{n-1} + a_2x_{n-2}\t+\\dots + a_kx_{n-k}\\]\n\t\tThen $ x_n $ can be written as \n\t\t\\[x_n = c_1r_1^n + c_2r_2^n +\\dots + c_lr_l^n\\]\n\t\tFor all $ c_i $ if $ r_i $ are the roots of the \\emph{characteristic polynomial} of the recursion. Which is:\n\t\t\\[\\tcboxmath[colback=white, colframe=white]{x^k - a_1x^{k-1} - a_2x^{k-2} \\dots - a_k = 0}\\]\n\t\tIf there are double roots, say $ r_1 = r_2 = r_3 $, then we instead have,\n\t\t\\[x_n = \\tcboxmath[colback=white, colframe=white]{c_1r_1^n + c_2n\\ r_2^n + c_3n^2\\ r^n} \\dots + c_lr_l^n\\]\n\t\t \n\t\tReversely, we can say that a sequence defined by a sum of geometric recurrent series is a recursion. \n\t\\end{BoxedTheorem}\n\n\t\\lem{}{Let $ F_n $ be the $ n $th Fibonacci number. Then the following holds:\n\t\t\\[F_{n}^2 + F_{n+1}^2 = F_{2n+1}\\]}\n\t\n\t\\proof{Expanding the general form of the terms, and showing that $ a_n = F_{n}^2 + F_{n+1}^2 - F_{2n+1} $ is a recursion by \\autoref{theorem:Sum of Geometric Sequences}.}\n\t\n\t\\begin{BoxedTheorem}[title=Repertoire Method]{}{}\n\t\tGiven a recurrent function defined by \n\t\t\\[f(n) = A(n)\\a + B(n)\\beta + C(n)\\gamma\\]\n\t\tWe plug in different values for $ f(n) $, for example, $ f(n) = 1, n, 2n $ etc. for which the values are known from the recursion, and then solve for $ A, B, C $.\n\t\\end{BoxedTheorem}", "meta": {"hexsha": "9fcfb585b78efaac5045f8344312e901f8ff958c", "size": 1629, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "combi/sec8_1_Recursion.tex", "max_stars_repo_name": "M-Ahsan-Al-Mahir/BCS_Question_Bank", "max_stars_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 48, "max_stars_repo_stars_event_min_datetime": "2020-10-14T17:15:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-13T19:47:04.000Z", "max_issues_repo_path": "combi/sec8_1_Recursion.tex", "max_issues_repo_name": "AnglyPascal/BCS_Question_Bank", "max_issues_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "combi/sec8_1_Recursion.tex", "max_forks_repo_name": "AnglyPascal/BCS_Question_Bank", "max_forks_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-10-15T08:59:33.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-27T15:19:26.000Z", "avg_line_length": 52.5483870968, "max_line_length": 171, "alphanum_fraction": 0.671577655, "num_tokens": 617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339676722393, "lm_q2_score": 0.7634837743174788, "lm_q1q2_score": 0.641123259060993}} {"text": "\\chapter{Generative Convolutional Neural Networks}\n\nIn this chapter we develop the idea of a generative convolutional neural\nnetwork. Doing so will allow us to use a natural generative pre-training\nalgorithm so that we do not need the same quantity of labeled data\nthat deep neural networks normally require.\n\n\\section{Fixed-length speech samples}\n\nWe begin our discussion with fixed-length vectors of speech signals\n$\\mathbf{X}\\in\\mathbb{R}^{N\\times D}$ which denotes $N$ vectors\nof dimension $D$. We use Numpy-style slicing syntax so that\n$\\mathbf{X}(n_1:n_2,d_1:d_2)\\in \\mathbb{R}^{(n_2-n_1)\\times(d_2-d_1)}$\nrefers to the matrix\n\\begin{equation}\n\\begin{bmatrix}\n \\mathbf{X}(n_1,d_1) & \\mathbf{X}(n_1,d_1+1) & \\cdots & \\mathbf{X}(n_1,d_2-1)\\\\\n \\mathbf{X}(n_1+1,d_1) & \\mathbf{X}(n_1+1,d_1+1) & \\cdots & \\mathbf{X}(n_1+1,d_2-1)\\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n \\mathbf{X}(n_2-1,d_1) & \\mathbf{X}(n_2-1,d_1+1) & \\cdots & \\mathbf{X}(n_2-1,d_2-1)\n\\end{bmatrix}\n\\end{equation}\nand we denote the index set for a vector of length $D$ as $[D]=\\{0,1,\\ldots,D-1\\}$. We will consider a sequence of models for these data. Initially we will begin with \na mixture model and progressively we will develop towards the convolutional neural network model.\n\n\\subsection{Circulant Gaussian Models}\n\nWe may model these vectors using circulant Gaussians $\\mathcal{N}_D(\\mathbf{0},\\mathbf{\\Sigma})$ where\n$\\tilde{\\boldsymbol{\\Sigma}}\\in\\Circ_D$ where $\\Circ_D$ is the set of circulant matrices over $\\mathbb{R}^{D\\times D}$.\nSince $\\boldsymbol{\\Sigma}$ is circulant we may write it as $F_D^*\\diag(\\tilde{\\boldsymbol{\\sigma}})F_D$ for some \nreal vector $\\tilde{\\boldsymbol{\\sigma}}\\in\\mathbb{R}^D$. Since $\\boldsymbol{\\Sigma}$ is a real symmetric\n positive definite circulant matrix we may write it as $\\boldsymbol{\\Sigma}=\\Circ(\\boldsymbol{\\sigma})$\nwhere $\\boldsymbol{\\sigma}$ is the first column of $\\boldsymbol{\\Sigma}$. Due to the symmetry and circularity of\n$\\boldsymbol{\\Sigma}$ this means that $\\boldsymbol{\\sigma}$ is symmetric so that for $1 < d < D$ we have\n$\\boldsymbol{\\sigma}(d)=\\boldsymbol{\\sigma}(D-d)$ and the same holds for $\\tilde{\\boldsymbol{\\sigma}}=\\mathbf{F}\\boldsymbol{\\sigma}$.\n\nThe circulant model is given by a single vector specifying the spectral distribution\n", "meta": {"hexsha": "433340d435e6ea60e8c988ef3807c614e59c07df", "size": 2284, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/ch5_generative_cnn.tex", "max_stars_repo_name": "markstoehr/researchnotes", "max_stars_repo_head_hexsha": "07197f3234a8993d8d4c416fb2878aab0618344d", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-06-06T09:37:00.000Z", "max_stars_repo_stars_event_max_datetime": "2015-06-06T09:37:00.000Z", "max_issues_repo_path": "chapters/ch5_generative_cnn.tex", "max_issues_repo_name": "markstoehr/researchnotes", "max_issues_repo_head_hexsha": "07197f3234a8993d8d4c416fb2878aab0618344d", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/ch5_generative_cnn.tex", "max_forks_repo_name": "markstoehr/researchnotes", "max_forks_repo_head_hexsha": "07197f3234a8993d8d4c416fb2878aab0618344d", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.1052631579, "max_line_length": 168, "alphanum_fraction": 0.7272329247, "num_tokens": 747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339797047029, "lm_q2_score": 0.7634837581726991, "lm_q1q2_score": 0.6411232546902635}} {"text": "\\documentclass[11pt,twoside,a4paper]{article}\n\\usepackage{amsmath}\n\\usepackage{commath}\n\\usepackage{graphicx}\n\\usepackage{tikz}\n\\usetikzlibrary{bayesnet}\n\n\\begin{document}\n\\section{Uivariate Gaussian distribution}\n Let $\\{x_i\\}= \\{x_1, x_2, \\cdots,x_n\\}$ denote $n$ random variates drawn from\n a univariate Normal distribution parametrised by the unknown variance $\\omega$\n and known mean $\\mu = 0$. Let us assume that the a Gaussian noise with known\n variannce $\\zeta$ has been added to the variates yeilding new variates\n $\\{y_i\\}= \\{y_1, y_2, \\cdots,y_n\\}$. Using statistical notations we can write\n \\begin{align}\n \\label{eqn_posteriors_signal_and_noise}\n \\Pr \\left(\\{x_i\\} | \\omega \\right) &= \\prod_{i=1}^n\n \\frac{1}{\\sqrt{2 \\pi \\omega}} \\exp \\left(-\\frac{x_i^2}{2 \\omega} \\right)\\\\\n \\Pr \\left(\\{x_i\\} | \\{y_i\\}, \\zeta \\right) &= \\prod_{i=1}^n\n \\frac{1}{\\sqrt{2 \\pi \\zeta}} \\exp\n \\left(-\\frac{\\left(y_i - x_i \\right)^2}{2 \\zeta} \\right)\n \\end{align}\n Using Bayes theorem the inference about the variance $\\omega$ can be written\n as\n \\begin{align}\n \\label{eqn_posterior_omega_given_noisy_random_variates}\n \\Pr \\left(\\omega | \\{y_i\\} \\right) \\propto\n \\Pr \\left(\\{y_i\\} | \\omega \\right) \\Pr(\\omega).\n \\end{align}\n Assuming a prior distribution $\\Pr(\\omega) = 1$, we can transfom the problem\n to the computation of the conditional distribution which is given by\n \\begin{align}\n \\Pr \\left(\\{y_i\\} | \\omega \\right) &= \\prod_{i=1}^n\n \\frac{1}{\\sqrt{2 \\pi (\\zeta + \\omega)}} \\exp\n \\left(-\\frac{y_i^2}{2(\\zeta + \\omega)} \\right), \\nonumber \\\\\n &= (2 \\pi)^{-\\frac{n}{2}} \\left(\\zeta + \\omega \\right)^{-\\frac{n}{2}}\n \\exp \\left(-\\frac{\\sum_{i=1}^n y_i^2}{2 (\\zeta + \\omega)} \\right) .\n \\end{align}\n This equation represents an inverse-Gamma distribution:\n \\begin{align}\n f(x | \\alpha, \\beta) = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} x^{\\alpha - 1}\n \\exp \\left(-\\frac{\\beta}{x} \\right).\n \\end{align}\n The statistical properties of inverse-Gamma distribution is well known and\n are shown in the table below.\n \\begin{table}\n \\caption{Statistical properties of inverse-Gamma distribution}\n \\label{tab_statistical_properties_of_inverse_gamma}\n \\begin{center}\n \\begin{tabular}{|c|c|c|}\n \\hline\n mode & mean $(\\alpha > 1)$ & variance $(\\alpha > 2)$ \\\\\n \\hline\n $\\frac{\\beta}{\\alpha + 1}$ & $\\frac{\\beta}{\\alpha - 1}$ &\n $\\frac{\\beta^2}{(\\alpha - 1)^2 (\\alpha - 2)}$ \\\\\n \\hline\n \\end{tabular}\n \\end{center}\n \\end{table}\n We can find the values of $n$ for which the variance of the distribution is\n defined as shown below:\n \\begin{align}\n \\alpha > 2 \\\\\n \\frac{n}{2} + 1 > 2 \\\\\n \\frac{n}{2} > 1 \\\\\n \\boxed{n > 2}\n \\end{align}\n This implies that we need at least 2 samples to determine the variance of the\n estimated and at least one sample to determine the mean.\n\n The logarithm of the posterior distribution can be written as\n \\begin{align}\n \\psi &= \\log \\left(\\Pr \\left(\\{y_i\\} | \\omega \\right) \\right) \\\\\n &= -\\frac{n}{2}\\log(2 \\pi) - \\frac{n}{2}\\log(\\zeta + \\omega)\n -\\frac{1}{2} \\frac{\\sum_{i=1}^n y_i^2}{(\\zeta + \\omega)}.\n \\end{align}\n The gradient of the log-posterior with respect to $\\omega$ is given by\n \\begin{align}\n \\frac{\\partial \\psi}{\\partial \\omega} = -\\frac{n}{2 (\\zeta + \\omega)}\n - \\frac{\\sum_{i=1}^n y_i^2}{2 (\\zeta + \\omega)^2}.\n \\end{align}\n The mode of the distribution can be found by setting gradients equal to zero\n and solving the resulting equation:\n \\begin{align}\n \\frac{n}{2 (\\zeta + \\omega)}\n + \\frac{\\sum_{i=1}^n y_i^2}{2 (\\zeta + \\omega)^2} = 0, \\\\\n \\frac{n}{2} + \\frac{\\sum_{i=1}^n y_i^2}{2 (\\zeta + \\omega)} = 0 \\\\\n n + \\frac{\\sum_{i=1}^n y_i^2}{(\\zeta + \\omega)} = 0 \\\\\n n (\\zeta + \\omega) + \\sum_{i=1}^n y_i^2 = 0\n %\\zeta + \\omega + \\frac{\\sum_{i=1}^n y_i^2}{n} = 0 \\\\\n %\\omega = \\frac{\\sum_{i=1}^n y_i^2}{n} - \\zeta \\\\\n %\\omega = \\bar{y} - \\zeta\n \\end{align}\n Thus mode of the distribution is given by\n \\begin{align}\n \\boxed{\\bar{\\omega} = \\bar{y} - \\zeta.}\n \\end{align}\n As we can see, the mode may have chance of being negative or zero which\n corresponds to non-physical variance. Hence we cannot treat this as point\n estimate of the distribution. The question then is what suffices as a\n ``good'' point estimate?\n\n From a Bayesian prespective, we haven't really assed the situaton carefully.\n By choosing uniform prior for $\\omega$ we have allowed non-positive values\n for the variance. However, a sensible prior for the variance should disregard\n any values less than or equal to zero. According to Jeffery's a reasonable\n prior distribution for a variable taking only positive values should be\n a uniform prior on the logarithm of the variable. Let $\\eta = \\log(\\omega)$\n and hence we can write,\n \\begin{align}\n \\Pr(\\eta) = 1.\n \\end{align}\n The rule of chang of variables says that\n \\begin{align}\n \\Pr(\\omega) \\dif \\omega = \\Pr(\\eta) \\dif \\eta.\n \\end{align}\n Therefore the Jeffreys prior on $\\omega$ can be derived as\n \\begin{align}\n \\Pr(\\omega) &= \\Pr(\\eta) \\frac {\\dif \\eta}{\\dif \\omega}, \\\\\n &= \\frac{1}{\\omega}.\n \\end{align}\n\n Applying the Jeffreys prior, the logarithm of the posterior distribution\n becomes\n \\begin{align}\n \\psi = - \\frac{n}{2} \\log (2 \\pi) - \\frac{n}{2} \\log (\\zeta + \\omega)\n - \\frac{1}{2} \\frac{\\sum_{i=1}^n y_i^2}{(\\zeta + \\omega)} - \\log(\\omega).\n \\end{align}\n We compare the two posterior distributions in\n Figure~\\ref{fig_log_post_uniform_jeffreys}\n \\begin{figure}\n \\includegraphics[scale=0.8]{images/log_post_uniform_jeffreys.pdf}\n \\caption{Comparing the log-posterior distributions of $\\omega$ with uniform and Jeffreys prior.}\n \\label{fig_log_post_uniform_jeffreys}\n \\end{figure}\n As we can see the resultng posterior distribution has spike near zero. Can\n there be a probabilisitc explanation for this characterstic? Since we are\n estimating the variance in the presence of noice, we have have to think in\n terms of a hierarchical way.\n\n\\subsection{Hierarchical modelling}\n In hierarchical model, one considers a prior of the form\n \\begin{align}\n \\Pr(\\zeta + \\omega) = \\frac{1}{\\zeta + \\omega}.\n \\end{align}\n The logaritm of the posterior now has the form\n \\begin{align}\n \\psi = - \\frac{n}{2} \\log (2 \\pi) - \\frac{n}{2} \\log (\\zeta + \\omega)\n - \\frac{1}{2} \\frac{\\sum_{i=1}^n y_i^2}{(\\zeta + \\omega)}\n - \\log(\\zeta + \\omega).\n \\end{align}\n\n\\end{document}\n", "meta": {"hexsha": "5d86cdba1134ef5a4feb14af7c0ebb3b4f96b03f", "size": 6532, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/univariate_gaussian/univariate_gaussian_estimation.tex", "max_stars_repo_name": "tbs1980/research-covariance-matrix-estimation", "max_stars_repo_head_hexsha": "33d73b2fc3f31dbfa5dc62eb94229b733be6ac98", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "notes/univariate_gaussian/univariate_gaussian_estimation.tex", "max_issues_repo_name": "tbs1980/research-covariance-matrix-estimation", "max_issues_repo_head_hexsha": "33d73b2fc3f31dbfa5dc62eb94229b733be6ac98", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/univariate_gaussian/univariate_gaussian_estimation.tex", "max_forks_repo_name": "tbs1980/research-covariance-matrix-estimation", "max_forks_repo_head_hexsha": "33d73b2fc3f31dbfa5dc62eb94229b733be6ac98", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-09-14T15:06:48.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-14T15:06:48.000Z", "avg_line_length": 42.4155844156, "max_line_length": 100, "alphanum_fraction": 0.640385793, "num_tokens": 2210, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339596505965, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6411232438983857}} {"text": "\\chapter{Equation Conventions}\n\\label{ch:conventions}\n\n\\textit{In this manual we will} be adhering to the following conventions\n\n\\section{List of Symbols}\n\n\\begin{itemize}\n\t\\item {\\boldmath{$i$}} Receiver identification or {\\boldmath{$r$}}\n\t\\item {\\boldmath{$j$}} Satellite identification or {\\boldmath{$s$}}\n\t\\item {\\boldmath{$k$}} Epoch number or {\\boldmath{$t$}}\n\t\\item {\\boldmath{$q$}} GNSS type (GPS,GALILEO,GLONASS,QZSS)\n\t\\item {\\boldmath{$c$}} Speed of light [m/s]\n\t\\item {\\boldmath{$x$}} Vector of parameters to be estimated\n\t\\item {\\boldmath{$y$}} Vector of observations\n\t\\item {\\boldmath{$v$}} Vector of residuals\n\t\\item {\\boldmath{$H$}} Design matrix\n\t\\item {\\boldmath{$\\sigma$}} Standard deviation of observable\n\t\\item {\\boldmath{$\\Delta$}} Increment to a priori values [m]\n\t\\item {\\boldmath{$\\lambda$}} Wavelength or {\\boldmath{$\\lambda_1,\\lambda_2,\\lambda_5$}}\n\t\\item {\\boldmath{$f_1,f_2,f_5$}} frequency\n\t\\item {\\boldmath{$\\alpha$}} Ambguity or {\\boldmath{$N$}} Real valued ambguity and {\\boldmath{$\\bar{N}$}} Integer part of real valued ambguity\n\t\\item {\\boldmath{$\\alpha$}} level of significance\n\t\n\t\\item {\\boldmath{$\\beta$}} Biases\n\t\\item {\\boldmath{$\\zeta$}} Clock offsets\n\t\\item {\\boldmath{$\\delta t$}} Clock error [s]\n\t\\item {\\boldmath{$\\kappa$}} Correction - relativity\n\t\\item {\\boldmath{$\\iota$}} Ionosphere or {\\boldmath{$I$}}\n\t\\item {\\boldmath{$\\tau$}} Troposphere or {\\boldmath{$T,T_h,T_w$}}\n\t\\item {\\boldmath{$M$}} elevation dependent mapping function for the troposphere wet delay\n\t\\item {\\boldmath{$\\xi$}} Phase wind-up error\n\t\\item {\\boldmath{$\\epsilon$}} Error in observations and unmodelled effects [m]\n\t\\item {\\boldmath{$\\phi_i^j$}} Carrier phase observable (times c) [m]\n\t\\item {\\boldmath{$P_i^j$}} Pseudo range observable [m]\n\t\t\n\\end{itemize}\n\nLets try this for example:\nFor an undifference, uncombined float solution, the linearized observation equations for pseudorange and phase observations from satellite $s$ to receiver $r$ can be described as:\n\n\\begin{math}\n\\Delta P_{r,f}^{q,s} = u_r^{q,s} . \\Delta x + c . (\\delta t_r^q - \\delta t^{q,s}) + M_r^{q,s} . T_r + \\gamma_f^q . I_{r,1}^{q,s} + d_{r,f}^q - d_f^{q,s} + \\epsilon_{P,f}^q\n\\end{math}\\\\\n\\begin{math}\n\\Delta\\phi_{r,f}^{q,s} = u_r^{q,s} . \\Delta x + c . (\\delta t_r^q - \\delta t^{q,s}) + M_r^{q,s} . T_r - \\gamma_f^q . I_{r,1}^{q,s} + \\lambda_f^q . N_{r,f}^{q,s} + b_{r,f}^q - b_f^{q,s} + \\epsilon_{L,f}^q \n\\end{math}\n\nwhere$\\Delta P_{r,f}^{q,s}$ and $\\Delta\\phi_{r,f}^{q,s}$ are the respective pseudorange and phase measurements on the frequency $f$(f=1,2), from which the computed values are removed;\n$u_r^{q,s}$ is the receiver-to-satellite unit vector;\n$\\Delta x$ is the vector of the receiver position corrections to its preliminary position; \n$\\delta t_r^q$ and $\\delta t^{q,s}$ are the receiver and satellite clock errors respectively;\n$c$ is the speed of light in a vaccum\n$M_r^{q,s}$ is the elevation dependent mapping function for the troposphere wet delay from the corresponding zenith one $T_r$;\n$I_{r,1}^{q,s}$ is the ionosphere delay along the line-of-sight from a receiver to a satellite at the first frequency and $\\gamma_f^q = (\\lambda_f^q / \\lambda_1^q)^2$;\n$\\lambda_f^q$ is the wavelength for the frequency $f$ of a GNSS $q$;\n$N_{r,f}^{q,s}$is the phase ambguity \n$d_{r,f}^q$ and $b_{r,f}^q$ are the receiver hardware delays of code and phase observations respectively;\n$d_f^{q,s}$ and $b_f^{q,s}$ are the satellite hardware delays of code and phase observations, respectively;\n$\\epsilon_{P,f}$ and $\\epsilon_{L,f}$ are the code and phase measurement noises respectively. \n\n", "meta": {"hexsha": "f08b412183042d054097a0f7db29cbad3362f7f0", "size": 3590, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/manual/conventions.tex", "max_stars_repo_name": "RodrigoNaves/ginan-bitbucket-update-tests", "max_stars_repo_head_hexsha": "4bd5cc0a9dd0e94b1c2d8b35385e128404009b0c", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 73, "max_stars_repo_stars_event_min_datetime": "2021-07-08T23:35:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T15:17:58.000Z", "max_issues_repo_path": "docs/manual/conventions.tex", "max_issues_repo_name": "RodrigoNaves/ginan-bitbucket-update-tests", "max_issues_repo_head_hexsha": "4bd5cc0a9dd0e94b1c2d8b35385e128404009b0c", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2021-09-27T14:27:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-21T23:50:02.000Z", "max_forks_repo_path": "docs/manual/conventions.tex", "max_forks_repo_name": "RodrigoNaves/ginan-bitbucket-update-tests", "max_forks_repo_head_hexsha": "4bd5cc0a9dd0e94b1c2d8b35385e128404009b0c", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 39, "max_forks_repo_forks_event_min_datetime": "2021-07-12T05:42:51.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T15:15:34.000Z", "avg_line_length": 57.9032258065, "max_line_length": 204, "alphanum_fraction": 0.6896935933, "num_tokens": 1169, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339556397749, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6411232408361885}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\\markright{momftfr}\n\\section*{\\hspace*{-1.6cm} momftfr}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nFrequency moments (order 1 and 2) of a time-frequency representation.\n\\end{minipage}\n\\vspace*{.35cm}\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[tm,T2] = momftfr(tfr)\n[tm,T2] = momftfr(tfr,tmin)\n[tm,T2] = momftfr(tfr,tmin,tmax)\n[tm,T2] = momftfr(tfr,tmin,tmax,time)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.35cm}\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty momftfr} computes the frequeny moments of order 1 and 2 of a\n time-frequency representation\\,:\n\\[t_m(f) = \\frac{1}{E}\\ \\int_{-\\infty}^{+\\infty} t\\ \\mbox{tfr}(t,f)\\ dt\\ \\\n;\\ \\ T^2(f) = \\frac{1}{E}\\ \\int_{-\\infty}^{+\\infty} t^2\\ \\mbox{tfr}(t,f)\\\ndt - t_m(f)^2.\\]\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8.5cm} c} Name &\nDescription & Default value\\\\ \\hline {\\ty tfr} & time-frequency\nrepresentation (size {\\ty (N,M)}). \\\\ {\\ty tmin} & smallest column\nelement of {\\ty tfr} taken into account & {\\ty 1} \\\\ {\\ty tmax} & highest\ncolumn element of {\\ty tfr} taken into account & {\\ty M}\\\\ {\\ty time} &\ntrue time instants & {\\ty (1:M)}\\\\ \\hline {\\ty tm} & averaged time (first order\nmoment)\\\\ {\\ty T2} & squared time duration (second order moment)\\\\\n\\hline\n\\end{tabular*}\n\n\\end{minipage}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf Example}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n sig=fmlin(200,0.1,0.4); [tfr,t,f]=tfrwv(sig); \n [tm,T2]=momftfr(tfr); \n subplot(211); plot(f,tm); subplot(212); plot(f,T2); \n\\end{verbatim}\nThe first order moment represents an estimation of the group delay, and the\nsecond order moment the variance of this estimator. We can see that the\nestimation is better around the time center position than at the edges of\nthe observation interval.\n\\end{minipage}\n\\vspace*{.35cm}\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\nmomttfr, margtfr.\n\\end{verbatim}\n\\end{minipage}\n\n", "meta": {"hexsha": "16ba30fe1e0bec9a8db9b667dac08c5bcf112d82", "size": 2294, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/momftfr.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/momftfr.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/momftfr.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 29.7922077922, "max_line_length": 79, "alphanum_fraction": 0.6591107236, "num_tokens": 856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339516289535, "lm_q2_score": 0.7634837635542925, "lm_q1q2_score": 0.6411232377739916}} {"text": "\\section{The \\variant{\\childrentransitions} problem}\n\\label{sec:simplenodes}\n\nIn this section, we provide the formal problem definition of the \n{\\variant{\\childrentransitions}}\nand describe a greedy polynomial-time algorithm for solving it.\n\n\\begin{problem}[{\\variant{\\nodeitems}}]\n\\label{problem:simplenodes-variant}\nGiven $G=(V,E)$, transition matrix $\\transition$, \ninitial distribution of items to nodes\n$\\initial$ and integer $k$, find\n$S\\subseteq V$ such that $|S|=k$ such that \n$\\objective_{_\\shortchildrentransitions}\\left(S\\right)$ is minimized.\n\\label{problem:simple-nodes-variant}\n\\end{problem}\n\nAs we can see from the definition of \n$\\objective_\\shortchildrentransitions\\left(S\\right)$ \nin \\eqref{eq:shortchildrentransitions},\nthe contribution of each node $u\\in {V-S}$\nto the objective function is equal to\n\\begin{equation}\n\\label{eq:parent-contribution}\n\\objective_{_\\shortchildrentransitions}(\\{u\\}) = \n\t\\initial(u)\\sum_{v\\in V}\\transition(u,v)\\left(1-\\transition(u,v)\\right)\n\\end{equation}\nand independent of any other node in $V-S$.\nIt directly follows, then, that Problem~\\ref{problem:simplenodes-variant}\nis solved by a simple greedy algorithm\nthat places in $S$ the {\\bf top k} nodes $u$ in terms\nof contribution $\\objective_{_\\shortchildrentransitions}(u)$.\n", "meta": {"hexsha": "aabf17376e19d15e4dcfbf77afe5e7b3a04174e9", "size": 1279, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/SDM-2018/simplenodes.tex", "max_stars_repo_name": "chdhr-harshal/MCMonitor", "max_stars_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-11-04T20:35:18.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-05T09:06:43.000Z", "max_issues_repo_path": "paper/SDM-2018/simplenodes.tex", "max_issues_repo_name": "chdhr-harshal/MCMonitor", "max_issues_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/SDM-2018/simplenodes.tex", "max_forks_repo_name": "chdhr-harshal/MCMonitor", "max_forks_repo_head_hexsha": "330fc1a8f8cf83620fd6b0e503707c91e97af16d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-09-05T09:10:41.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-05T09:10:41.000Z", "avg_line_length": 38.7575757576, "max_line_length": 73, "alphanum_fraction": 0.7630961689, "num_tokens": 350, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339516289534, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6411232377739914}} {"text": "\\chapter{Graphical Models}\n\\label{chap:Graphical Models}\n\n\\section{Introduction}\n\n\\section{Directed Graph}\n\n\\section{Undirected Graph}\n\n\\section{Inference in Graphical Models}\n\\subsection{Factor graphs}\nThe joint distribution over a set of variables in the form of a product of factors\n\\begin{align}\np(\\vec{x}) = \\prod_s f_s(\\vec{x}_s)\n\\end{align}\nwhere $\\vec{x}_s$ denotes a set of the variables.We shall denote individual variable by $x_i$,which can comprise groups of variables(such as vectors and matrices).Each factor $f_s$ is a function of a corresponding set of variable $\\vec{x}_s$.\nDirected graphs have factors $f_s(\\vec{x}_s)$ as local conditional distributions.Undirected graphs' factors are potential function over the maximal cliques.\n\n\\subsection{The Sum-product algorithm}\n\\textbf{sum-product} algorithm solves the problem of evaluating the \\textbf{local marginals} over nodes or subsets of nodes.And \\textbf{max-sum} algorithm find the \\textbf{most probable state},evaluating the maximal joint distribution.\n\nThe marginal is obtained by\n\\begin{align}\np(x) = \\sum_{\\vec{x}\\backslash x}{p(\\vec{x})}\n\\end{align}\nwhere $\\backslash x$ denotes the set of variables in $\\vec{x}$ omitting $x$.\n\nJoin distribution \n\\begin{align}\np(\\vec{x}) =\\prod_{s\\in \\text{ne}(x)}{F_s(x,\\vec{X}_s)}\n\\end{align}\n\nThen interchanging the sums and products,we obtain\n\\begin{align}\n\\text{objective} &= p(x)=\\sum_{\\vec{x}\\backslash x}{p(\\vec{x})} \\\\\n&= \\prod_{s\\in\\text{ne}(x)}{\\left[\\sum_{X_s}F_s(x,X_s)\\right]} \\\\\n&= \\prod_{s\\in\\text{ne}(x)}\\mu_{f_s\\rightarrow x}(x)\n\\end{align}\nView \\textbf{messages} from the factor nodes $f_s$ to the variable node $x$ as\n\\begin{align}\n\\mu_{f_s\\rightarrow x}(x) &\\equiv \\sum_{X_s}F_s(x,X_s) \\\\\nF_s(x,X_s) &= f_s(x,x_1,\\ldots,x_M)G_1(x_1,X_{s1})\\ldots G_M(x_M,X_{sM})\n\\end{align}\n\nRecursive inference in the sub-graph and interchanging sums and products leads to \n\\begin{align}\n\\mu_{f_s\\rightarrow x}(x) &= \\sum_{x_1}\\ldots\\sum_{x_M} f_s(x,x_1,\\ldots,x_M)\\prod_{m\\in \\text{ne}(f_s)\\backslash x} \\mu_{x_m\\rightarrow f_s}(x_m) \n\\end{align}\nwhere $\\text{ne}(x)$ denotes the set of neighbor variables.\nAnd define \\textbf{messages} from variable nodes to factor nodes\n\\begin{align}\n\\mu_{x_m\\rightarrow f_s}(x_m) &\\equiv \\sum_{X_{sm}}G_m(x_m,X_{sm})\n\\end{align}\n\nAgain,making use of (sub)-graph factorization,we have\n\\begin{align}\n\\mu_{x_m\\rightarrow f_s}(x_m) &= \\prod_{l\\in \\text{ne}(x_m)\\backslash f_s}\\left[\\sum_{X_{ml}}F_l(x_m,X_{ml}) \\right] \\\\\n&=\\prod_{l\\in \\text{ne}(x_m)\\backslash f_s}\\mu_{f_l\\rightarrow x_m}(x_m)\n\\end{align}\n\n\n\\subsection{Max-sum algorithm}\nThe \\textbf{sum-product} algorithm allows us to take a joint distribution $p(\\vec(x))$ expressed as a factor graph and efficiently find marginal component variables.\\textbf{Max-sum} find a setting of variables that has the largest probability,which is an application of \\textbf{dynamic programming}.\nMaximize the joint distribution\n\\begin{align}\n\\vec{x}^{max} = \\arg\\max_{\\vec{x}}p(\\vec{x})\n\\end{align}\nJoint distribution\n\\begin{align}\np(\\vec{x}) &= \\prod_{s\\in \\text{ne}(x)}F_s(x,X_s)\n\\end{align}\nMaking use of \\textbf{distributive law} for multiplication with max operator,similarly to add operator\n\\begin{align}\n\\text{objective} &= \\max\\ln p(\\vec{x}) =\\max\\sum\\ln F_s(x,X_s) \\\\\n&=\\sum\\max\\ln F_s(x,X_s)\n\\end{align}\nMaking use of recursion,factorize $F_s(x,X_s)$ with sub-graph,we can obtain\n\\begin{align}\n\\mu_{f\\rightarrow x} &= \\max_{x_1,\\ldots,x_M}\\left[\\ln f(x,x_1,\\ldots,x_M) + \\sum_{m\\in \\text{ne}(f_s)\\backslash x} {\\mu_{x_m\\rightarrow f}(x_m)} \\right] \\\\\n\\mu_{x\\rightarrow f}(x) &= \\sum_{l\\in \\text{ne}(x)\\backslash f}{\\mu_{f_l\\rightarrow x}}(x).\n\\end{align}\nwhere $\\mu_{f\\rightarrow x} \\equiv \\max\\ln F_s(x,X_s)$\n\n\n\n\n\n\n\n", "meta": {"hexsha": "212b488b78d51c509412732068a3b2891e744659", "size": 3718, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "prml/Graphical Models.tex", "max_stars_repo_name": "Alexoner/Statistical-formula", "max_stars_repo_head_hexsha": "114a6c2424f206cb57715c7de29ae3d26abcb081", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 24, "max_stars_repo_stars_event_min_datetime": "2015-02-15T17:00:48.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-24T13:46:00.000Z", "max_issues_repo_path": "prml/Graphical Models.tex", "max_issues_repo_name": "Alexoner/Statistical-formula", "max_issues_repo_head_hexsha": "114a6c2424f206cb57715c7de29ae3d26abcb081", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "prml/Graphical Models.tex", "max_forks_repo_name": "Alexoner/Statistical-formula", "max_forks_repo_head_hexsha": "114a6c2424f206cb57715c7de29ae3d26abcb081", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2017-02-25T15:40:39.000Z", "max_forks_repo_forks_event_max_datetime": "2019-09-25T04:26:03.000Z", "avg_line_length": 41.3111111111, "max_line_length": 299, "alphanum_fraction": 0.7227003765, "num_tokens": 1219, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339516289534, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6411232377739914}} {"text": "\\section{FourierTransform: Fast Fourier Transforms}\n[Warren Smith prepared the analysis and modules.]\n\nThis is an {\\em in place} FFT routine with array length: $N = 2^n$.\n\\begin{equation}\n a_{output,k} = \\sum_{m=0}^{N-1} a_{input,m} \n e^{(2 \\pi i \\mbox{direction} m k/N)}\n\\end{equation}\n\nwhere direction $= \\pm 1$.\nYou have to do any scaling (by $1/N$ or $1/\\sqrt{N}$) or zeroing of a[] by\nyourself. \n\nUses $2N \\ln N + 2 N + O(\\ln N)$ [real] multiplications, \nand $ 3N \\ln N + 2 N + O(\\ln N)$ [real] additions.\n(And hopefully the compiler optimizes the subscripting in the inner loop so\nthat only $2 N \\ln N$ total subscripting ops are needed.)\n\nThe test routine shown yields\nerror after 1 forward and one backward use\nat least 3 decimal places better than\nslow FT has after only one use!\n\nThe basic FFT idea (for N a power of 2) is from J.Cooley \\& J.Tukey:\nAn algorithm for the Machine Calculation of Complex Fourier Series,\nMOC 19 (1965) 297-301. (Knuth, Henrici, and Dahlquist/Bjorck also discuss\nthe FFT.) The basic idea is that to calculate the DFT\n\\begin{eqnarray}\n A[k] & = & \\sum_{j=0}^{N-1}a_j*W^{kj} \\, \\mbox{for k=0..N-1}\\\\\n N & = & \\mbox{a power of 2,}\\, N>1\\\\\n W & = & \\mbox{a principal Nth root of unity}\\\\\n \\mbox{e.g., } W & = & e^{2 i \\pi/N}\n\\end{eqnarray}\nWe may write\n\\begin{eqnarray}\n A_k & = & g_k + h_k * W^k \\, \\mbox{ where}\\\\\n g_k & = & \\sum_{0<=2j4/(M-1)$. On PDP-11/44 C, however, rough timing\nhas shown that $M=1.08$ (but with considerable standard deviation.\nIt does\nabout $5*10^4$ additions/sec.) so this idea is not worth it unless N is\nenormous.\n\nSome similar, but worse, ideas have been suggested by Buneman:\n\\begin{verbatim}\n If c=cos(m),s=sin(m), then precompute \n t1 = (1-c)/s = s/(1+c)= tan(m/2) and\n t2 = (1+s)/c = c/(1-s). \n\\end{verbatim}\nThen $X+iY = (a+bi)*(c+is)$ may be found in 3*,3+ by:\n\\begin{verbatim}\n if(|t1|<|t2|) then\n X = a-t1*b; Y = b+s*X; X -= t1*Y;\n else\n X = b+t2*a; Y = c*X-a; X -= t2*Y;\n\\end{verbatim}\ntans may be updated by \n$\\tan(x+y)-\\tan(x-y) = 2*\\tan(y)/(1-(\\tan(x)*\\tan(y))^2)$,\nbut the extra overhead seems not to be worth it].\n\nThere is also a symmetric 3*,5+ (4+ with precomputation) formula for \n$E+iF = (a+bi)*(c+di)$:\n\\begin{eqnarray}\n E & = & a*c-b*d\\\\\n F & = & (a+b)*(c+d)-a*c-b*d\n\\end{eqnarray}\n\nOther FFT algorithms: Winograd has shown how to design FFTs with N prime\n(as opposed to the Cooley-Tukey approach which works for N highly composite)\nthat run in $O(N \\ln N)$ time and even with only O(N) multiplications;\nthe latter figure is optimal. [S.Winograd: Math. of Comput. 32 (1978) 175-179;\nAdvs in Math 32 (1979) 83-117].\n\nWinograd's approach is based on a theorem that allows him (by a permutation\nof the original and transformed variables) to express FFTs for N prime\nin terms of a circular convolution of N-1 elements, plus some additions.\nHe then shows how circular convolutions of k elements (for certain small k)\nmay be computed in a small number of arithmetic operations, (for k=2..6,\nthe number of multiplications Winograd uses is 2,4,5,10,8; for k prime,\nWinograd shows that a (2k-2)* scheme for CC(k) always exists) and further, how\nCC(n1) and CC(n2) algorithms may be composed to make a CC(n1*n2) algorithm, IF\nn1 and n2 are relatively prime, that uses mult(n1)*mult(n2) multiplications.\nAlso, he shows how FFT(n1*n2) may be computed via FFT(n1) and FFT(n2)\nin mult(n1)*mult(n2) multiplications, IF n1,n2 relatively prime, and\nalso gives methods for FFT(prime power). He gives two appendices\ncontaining optimized CC(2..6) and FFT(2..9) algorithms. Winograd's methods\ndon't appear suitable for general N, but if N is specified in advance, they\nmake it possible to do considerable fine tuning at the expense of\nlarge algorithm complexity.\n\nMeanwhile, Nussbaumer [H.J.Nussbaumer: Fast Polynomial Transform algorithms\nfor digital convolutions, IEEE Transactions on Audio, Speech, Signal Processing\n28,2 (April 1980) 205-215 (this article has many references to other FFT\nschemes); see also Knuth 2: 503, 652-653] has found another way to do circular\nconvolutions of arrays of (N=a power of 2) reals without any NTTs, FFTs, trig,\nor complex numbers. His approach is based on viewing circular convolutions\nas polynomial multiplications modulo certain simple polynomials, factoring the\nmodular polynomials, divide and conquer, chinese remainder thm.\nHis approach uses roughly NlgN *,\n$N \\ln N \\ln \\ln N$ +, is fairly complicated to program, and requires extra space.\n\n[Nussbaumer \\& Quandalle: IBM JResDev 22 (1978) 134-144] show how some\nparticularly efficient CC and FFT schemes for N in the range 10-3000\nmay be constructed; their approach is based on some novel ways to combine\nefficient small CC schemes that is rather like the NTT (Number theoretic\ntransform) only in rings of polynomials rather than in the integers.\n\nHowever even the best known arithmetic op count methods only improve on my\nmethod by perhaps 30\\%, and at the cost of considerable complexity.\nC.H.Papadimitriou [Optimality of the FFT, JACM 26 (1979) 95-102 and its refs]\nhas shown that in some models of computation $O(N\\ln N)$ is\noptimal for the FFT, while Patterson et al have shown a lower\nbound of $O(N\\ln N/\\ln \\ln N)$ for integer multiplication on multitape\nTuring machines, see Knuth 2.\n\nFFTs in a finite field (if the ring ZmodK, called \"number theoretic\ntransforms\") are discussed in Aho,Hopcoft,Ullman: The Design and Analysis\nof Computer algorithms, Addison-Wesley 1974. They recommend using W=2,\nN=a power of 2, do all arithmetic in the ring of integers \nmodulo $2^{N/2}+1$\n(in which W is an Nth root of unity, and in which the convolution theorem\n\\begin{equation}\n C_i=A_iB_i \\Longleftrightarrow \n c_i = \\sum_{j=0}{N-1} a_i*b_{i-j \\bmod N}\n\\end{equation}\n[Which makes possible the calculation of discrete convolutions in $N\\ln N$ time]\nstill holds). [See also R.Agrawal\\&C.Burrus: NTTs to implement fast digital\nconvolutions, ProcIEEE 63 (1975) 550; articles by H.Nussbaumer on \"Fermatrix\"\nand \"Mersenne\" transforms, IBMJR\\&D 21 (1976) 282 and 498.]\n\nThese FFFFTs are of use in applications where it is desirable\nto completely eliminate roundoff error and floating point operations,\ne.g. all-integer convolutions. However, as you can see, NTTs have severe\nword length and transform length limitations; the need for high precision\nmodular arithmetic can be a major stumbling block. On the other hand,\nmultiplications by W=2 are easy, while modular arithmetic modulo a\nFermatrix number is not that hard. Thus using N=16, modulo 65537 arithmetic,\nW=2 [left shift and modulo], Winverse=32769 [right shift; modular\naddition correction if inexact], and all numbers in 0..89 allows\ncomputation of CC(16) in 16*, many bit shifts and additions.\n\nRabiner,Schafer,Rader: The Chirp-Z transform and its Applications,\nBSTJ 48,3 (1969) 1249-1292 show how DFTs (for any N) may be calculated in\n$N\\ln N$ time by using fast convolutions; the method also works for an extension\nof FFTs (to W=any complex number, not just the principal Nth root of unity):\n\\begin{equation}\n A_k = \\sum_{j=0}^{N-1}a_j*W^{k*j} \\mbox{ for k=0..N-1}\n\\end{equation}\nmay be calculated in $N\\ln N$ time by a fast convolution by the \"Chirp-Z\ntransform\" identity\n\\begin{equation}\n A_k = W^{k*k/2} \\sum_{j=0}^{N-1} W^{-((j-k)^2)/2} W^{j*j/2}b_j\n\\end{equation}\n\nAho,Stieglitz,Ullman: Evaluating Polynomials at fixed sets of points,\nSIAMJComp 4,4 (Dec 1975) 533-539, demonstrate that a polynomial and all its\nderivatives at one point (or equivalently, an origin shift of an Nth\ndegree polynomial) may be calculated in $N\\ln N$ time by a fast convolution\nvia the (binomial theorem) identity\n\\begin{eqnarray}\n \\sum_{j=0}^{N-1} c_j*(x+q)^j & = & \\sum_{r=0}^{N-1} x^r * d_r/r!\n \\mbox { where}\\\\\n d_r & = & \\sum_{j=r}^{N-1} c_j*j! * q^{j-r}/(j-r)!\n\\end{eqnarray}\n\nSome other applications of FFTs are:\n\nFast multiplication and division of N digit integers may be done in\n(roughly) $N\\ln N$ time by using fast convolutions followed by a carry step.\n(See Aho-Hopcoft-Ullman: Design and Analysis of Computer algorithms, for\nfurther discussion.)\n\nBase conversion of an N digit number may be done in $N(\\ln N)^2$ time by\ndivide and conquer (convert the left and right half of the number recursively,\nthen do a fast multiplication and addition to combine them).\n\nFast polynomial multiplication and division by fast convolutions in $N\\ln N$\ntime are also discussed in AHU. (This may also be done for Chebyshev\nseries...) Given the N roots of a polynomial, you can find its\ncoefficients (as Chebyshev or as regular) in $N(\\ln N)^2$ time by fast\npolynomial multiplications on a binary tree. On the other hand,\nyou can perform a \"root squaring\" transformation on a polynomial\nin (Chebyshev or power form)\n\\begin{equation}\n P(y) = -P(x)*P(-x) \\mbox{ where }y = x^2\n\\end{equation}\nin $N\\ln N$ time by a fast multiplication, or alternatively can implement\na Henrici-Gargantini or Korsak-Pease (or other) simultaneous all root iteration\nstep, in $N(\\ln N)^2$ time by a fast multipoint evaluator, see below.\n\nAll shifted correlations of vectors (and/or autocorrelations) may be\ncalculated in $N\\ln N$ time by fast convolutions; this has application in\nsignal processing, 1D pattern recognition, Electrical engineering.\n\nFast polynomial multiplication/division/remaindering and a divide and\nconquering of the Lagrange interpolation formula may be used to do fast\nNth degree polynomial interpolation and N-point evaluation, as was\nshown by Borodin\\&Moenck [JCompSystSci 1974]. I have extended B\\&M's\nresults to Chebyshev polynomials and less successfully to other\npolynomials.\n\nFast algorithms exist for power-series to continued fraction interconversion;\nthese may also be generalized to Chebyshev series.\n\nFast polynomial evaluation/interpolation at special point sets\n(e.g. $Z^k$ for some Z) may be accomplished in $N\\ln N$ time by the Chirp-Z and\nFF transforms; this also carries over to Chebyshev. Thus fast Taylor\nand Chebyshev series calculations.\n\nFast composition of Taylor series -\n$O((N\\ln N)^{3/2})$ is also possible, as was discovered by Brent\\&Kung,\nvia a \"block Horner\" approach. This may also be extended to Chebyshev.\n\nFast Elliptic linear PDE solvers (by finite differneces or spectrally):\nthere are many schemes based on FFTs that run in $N\\ln N$ time, N=size of\noutput.\n\nA complete list of FFT applications is far too huge to discuss here...\n\n", "meta": {"hexsha": "717bef647b432b4a8c42724eb2f9fa4b50c03eca", "size": 15440, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "m3-libs/arithmetic/doc/fft.tex", "max_stars_repo_name": "jaykrell/cm3", "max_stars_repo_head_hexsha": "2aae7d9342b8e26680f6419f9296450fae8cbd4b", "max_stars_repo_licenses": ["BSD-4-Clause-UC", "BSD-4-Clause", "BSD-3-Clause"], "max_stars_count": 105, "max_stars_repo_stars_event_min_datetime": "2015-03-02T16:58:34.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-28T07:17:49.000Z", "max_issues_repo_path": "m3-libs/arithmetic/doc/fft.tex", "max_issues_repo_name": "jaykrell/cm3", "max_issues_repo_head_hexsha": "2aae7d9342b8e26680f6419f9296450fae8cbd4b", "max_issues_repo_licenses": ["BSD-4-Clause-UC", "BSD-4-Clause", "BSD-3-Clause"], "max_issues_count": 145, "max_issues_repo_issues_event_min_datetime": "2015-03-18T10:08:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T01:27:08.000Z", "max_forks_repo_path": "m3-libs/arithmetic/doc/fft.tex", "max_forks_repo_name": "jaykrell/cm3", "max_forks_repo_head_hexsha": "2aae7d9342b8e26680f6419f9296450fae8cbd4b", "max_forks_repo_licenses": ["BSD-4-Clause-UC", "BSD-4-Clause", "BSD-3-Clause"], "max_forks_count": 26, "max_forks_repo_forks_event_min_datetime": "2015-10-10T09:37:44.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-23T02:02:05.000Z", "avg_line_length": 46.9300911854, "max_line_length": 82, "alphanum_fraction": 0.7267487047, "num_tokens": 4726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511543206819, "lm_q2_score": 0.7490872187162396, "lm_q1q2_score": 0.6411071608251628}} {"text": "\\section{Modelization of a DC-DC converter}\nIn the system we want to simulate, as in many real-life systems, is present a DC-DC converter, which will have its non-idealities too, resulting in energy losses.\n\nThis means, while in an ideal DC-DC converter the input and output power will be equal ($P_{in}=P_{out}$), in a real one the output power will be scaled by a factor $\\eta$, and thus it will be $\\eta P_{in}=P_{out}$. $\\eta$ will be called \\emph{efficiency} of the DC-DC converter. in order to simulate the blocks into SimuLink, we wanted to find a good expression approximating the power loss due to the non-unary efficiency ($P_{loss}$). The chosen analytical expression to fit on the $P_{loss}$ data has been a quadratic one:\n$$\nP_{loss}=c_1I_{out}^2+c_2I_{out}+c_3\n$$\n\nAs in the previous part, the first work has consisted in extrapolating data from a graph present into the datasheet of the converter (\\emph{LTC3789}), in order to obtain a current-efficiency curve. It has been important to notice that the data were in logarithmic scale on the X axis, and so have had to be interpolated as exponents and transformed into linear values post-audit.\n\nAfter extracting data from the graph and linearly interpolating it, in order to fit the analytical onto the $P_{loss}$ data it has been needed to use the relationship (which is valid by definition of $\\eta$) between $\\eta$ and $P_{loss}$:\n\\begin{gather*}\n\\eta=\\frac{P_{out}}{P_{in}} \\\\\n\\Downarrow \\\\\n\\eta=\\frac{P_{out}}{P_{out}+P_{loss}}\n\\end{gather*}\n\nInstead of reverting this equation and doing a two-step process to fit the curve, again the \\emph{fmins} has been used to evaluate fitting between the available $\\eta$ data and the new $P_{loss}$ data. The handle passed to \\emph{fmins} calls a function which evaluates the expected $\\eta$ vector starting from a sample $P_{loss}$ vector, compares it with the actual $\\eta$ vector and returns the squared difference (i.e. the error) between the two. Minimizing this cost function is equivalent to fitting the $P_{loss}$ curve to the desired one.\n\nAgain, the result has been satisfactory enough.\n\n\\begin{center}\n\\begin{tabular}{|c| c|}\n\n \\hline\n Parameter name & Parameter value \\\\\n \\hline\n $c_{1}$ & 0.0309 \\\\\n $c_{2}$ & 0.1166 \\\\\n $c_{3}$ & 0.4535 \\\\\n \\hline\n\\end{tabular}\n\n\\begin{figure}[h]\n \\centering\n\\includegraphics[width=2.5in]{fitted_eta}\n\\caption{Current-Efficiency curve as computed after fitting $P_{loss}$}\n\\end{figure}\n\\begin{figure}[h]\n \\centering\n\\includegraphics[width=2.5in]{ploss}\n\\caption{Obtained Current-Power loss curve for the converter}\n\\end{figure}\n\n\\end{center}\n", "meta": {"hexsha": "71be8174e7bc07cf85877f0fbf365ad77399952a", "size": 2623, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/ass3.tex", "max_stars_repo_name": "conte91/ENOPT_Lab4", "max_stars_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/ass3.tex", "max_issues_repo_name": "conte91/ENOPT_Lab4", "max_issues_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/ass3.tex", "max_forks_repo_name": "conte91/ENOPT_Lab4", "max_forks_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.0217391304, "max_line_length": 544, "alphanum_fraction": 0.7392298894, "num_tokens": 729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031738057795402, "lm_q2_score": 0.7981867825403176, "lm_q1q2_score": 0.6410827158558332}} {"text": "% $Id$\n%\n% Author: David Fournier\n% Copyright (c) 2008 Regents of the University of California\n%\n\n\\section{The necessity for adjoint code}\n\nWhen you write code for variable objects in \\ADM, all the\nderivatives are calculated for you in a transparent manner. \nTo accomplish this, \\ADM\\ must save certain information for\nlater use. We shall refer to this as derivative information.\nEach arithmetic operation generates about 32~bytes of\nderivative information. If you have some simple function that\nhas 20 arithmetic operations, it will therefore\ngenerate 640 bytes of derivative information every time\nit is called. The purpose of writing adjoint code is to\nreduce the amount of derivative information that must\nbe calculated. For a function that is called many times,\nthis can greatly reduce the amount of derivative information that\nmust be stored.\n\nIn this chapter, we investigate how to write and debug\nadjoint code. To begin, we investigate how to write adjoint code for a\nsimple function that takes 1--4 independent variables and returns 1~dependent variable. \nThe adjoint code for such functions is\nsimpler to write than that for a general function---such as the\nsingular value decomposition of a matrix, which we will consider later.\n\n\n\\section{Writing adjoint code: a simple case}\n\nConsider a simple function $f$ which takes 1~independent variable $x$\nand returns a dependent variable $y$, i.e.,\n$$y=f(x)$$\nwhere $f(x)=\\exp(-x^2/2)$. The code for this example can be written like\n\\begin{lstlisting}\n dvariable errf(const prevariable& x)\n {\n return exp(-0.5*square(x));\n }\n\\end{lstlisting}\nThere are three arithmetic operations here: square, multiplication, and exponentiation,\nso 96 bytes of derivative information will be generated. (Actually, the \\texttt{return} operation\nalso generates 32 bytes of derivative information, but we will ignore that for now.)\nA less efficient way to write the code (but more useful for \nshowing adjoint code for this simple example) would be:\n\\begin{lstlisting}\n dvariable errf(const prevariable& x)\n {\n dvariable y;\n y=exp(-0.5*square(x));\n return y;\n }\n\\end{lstlisting}\nHere is the same code with the derivative calculated by the one line of adjoint\ncode \\texttt{double dfx=-value(x)*value(y);}:\n\\begin{lstlisting}\n dvariable errf(const prevariable& x)\n {\n dvariable y;\n value(y)=exp(-0.5*square(value(x)));\n double dfx=-value(x)*value(y);\n AD_SET_DERIVATIVES1(y,x,dfx); // 1 dependent variable \n return y;\n }\n\\end{lstlisting}\nSo what is going on here? Consider the line\n\\begin{lstlisting}\n value(y)=exp(-0.5*square(value(x))); \n\\end{lstlisting}\nThe \\texttt{value} function returns a constant type, that is, a double, \nthat has the same value as the corresponding \\texttt{dvariable} or prevariable.\nIn fact, it is the {\\it same} object. That is, it shares the same address,\nbut the type has been changed to double. So, the above line of\ncode assigns the value \\texttt{exp(-0.5*square(value(x)))} to \\texttt{y},\nbut without generating any derivative code. Similarly, since\nthe calculations are made on \\texttt{value(x)}, these calculations\nwill not generate any derivative code. So, it is the responsibility of the\nprogrammer to calculate the derivative code and store it where it\ncan be used later. The line\n\\begin{lstlisting}\n double dfx=-value(x)*value(y);\n\\end{lstlisting}\ncalculates the derivative $f^\\prime(x)$\nof $y$ with respect to $x$ and stores it with the line of code\n\\begin{lstlisting}\n AD_SET_DERIVATIVES1(y,x,dfx);\n\\end{lstlisting}\nThis code will only generate 32 bytes of derivative information.\n\n\n\\section{Debugging adjoint code: a simple case}\n\nThe simplest way to debug the adjoint code is to put your new\nfunction into an \\ADM\\ template file and use the \\texttt{-dd 1}\ncommand line option to call the derivative checker.\n\\XX{\\fontindexentry{tt}{-dd}}{command line argument}\n\\XX{command line arguments}{\\fontindexentry{tt}{-dd}}\n\\begin{lstlisting}\nDATA_SECTION\nPARAMETER_SECTION\n init_number x\n !! x=2;\n objective_function_value f;\nPROCEDURE_SECTION\n f=square(errf(x));\n \nGLOBALS_SECTION\n #include \n dvariable errf(const prevariable& x)\n {\n dvariable y;\n value(y)=exp(-0.5*square(value(x)));\n double dfx=-value(x)*value(y);\n AD_SET_DERIVATIVES1(y,x,dfx); // 1 dependent variable \n return y;\n }\n\\end{lstlisting}\n\n\n\\section{Adjoint code for more than\\br one independent variable}\n\nThe following code shows how to write the adjoint code for a function with two\nindependent variables:\n\\begin{lstlisting}\nDATA_SECTION\n vector lengths(1,10)\n vector ages(1,10)\n !! lengths.fill_seqadd(1,1);\n !! ages.fill_seqadd(1,1);\n !! lengths=sqrt(lengths);\n\nPARAMETER_SECTION\n init_bounded_number linf(0,10) \n init_bounded_number rho(0,1) \n objective_function_value f;\nPROCEDURE_SECTION\n for (int i=1;i<=10;i++)\n f+=square(lengths(i)-vb_growth(linf,rho,ages(i)));\n \nGLOBALS_SECTION\n #include \n\n dvariable vb_growth(const prevariable& linf, const prevariable& rho, \n double t)\n {\n double clinf=value(linf);\n double crho=value(rho);\n dvariable len;\n value(len)=clinf*(1-pow(crho,t));\n double dflinf=1-pow(crho,t);\n double dfrho=-clinf*t*pow(crho,t-1);\n AD_SET_DERIVATIVES2(len,rho,dfrho,linf,dflinf); // 2 dependent variable \n return len;\n }\n\\end{lstlisting}\nThis approach to writing adjoint code has been implemented for functions of\nup to four independent variables.\n\\begin{lstlisting}\nDATA_SECTION\n vector lengths(1,10)\n vector ages(1,10)\n !! lengths.fill_seqadd(1,1);\n !! ages.fill_seqadd(1,1);\n !! lengths=sqrt(lengths);\n\nPARAMETER_SECTION\n init_bounded_number linf(0,10) \n init_bounded_number rho(0,1) \n init_number t0 \n init_bounded_number gamma(.1,1.9) \n objective_function_value f;\nPROCEDURE_SECTION\n for (int i=1;i<=10;i++)\n f+=square(lengths(i)-vb_growth(linf,rho,t0,gamma,ages(i)));\n \nGLOBALS_SECTION\n #include \n\n dvariable vb_growth(const prevariable& linf, const prevariable& rho, \n const prevariable& t0, const prevariable gamma,double t)\n {\n double clinf=value(linf);\n double ct0=value(t0);\n double crho=value(rho);\n double cgamma=value(gamma);\n dvariable len;\n value(len)=pow(clinf*(1-pow(crho,t-ct0)),cgamma);\n double tmp=cgamma*pow(clinf*(1-pow(crho,t-ct0)),cgamma-1);\n double dflinf=tmp*(1-pow(crho,t-ct0));\n double dft0=tmp*(clinf*log(crho)*pow(crho,t-ct0));\n double dfrho=-tmp*clinf*(t-ct0)*pow(crho,t-ct0-1);\n double dfgamma=value(len)*log(clinf*(1-pow(crho,t-ct0)));\n AD_SET_DERIVATIVES4(len,t0,dft0,rho,dfrho,linf,dflinf,gamma,dfgamma); \n // 4 dependent variable \n return len;\n }\n\\end{lstlisting}\n\n\n\\section{Structured calculation of derivatives in adjoint code}\n\nUntil now, we have deliberately calculated the derivatives with respect to the independent\nvariables in an {\\it ad-hoc} fashion. While this approach works for simple functions,\nit rapidly becomes untenable when the function is more complicated. In the following example,\nwe have calculated the derivatives in a more structured fashion. Notice that to calculate the\nderivatives, every line of code in the function is repeated in the opposite\norder (commented out, of course) and the corresponding derivatives are calculated.\n\\begin{lstlisting}\nDATA_SECTION\n vector lengths(1,10)\n vector ages(1,10)\n !! lengths.fill_seqadd(1,1);\n !! ages.fill_seqadd(1,1);\n !! lengths=sqrt(lengths);\n\nPARAMETER_SECTION\n init_bounded_number linf(0,10) \n init_bounded_number rho(0,1) \n init_number t0 \n init_bounded_number gamma(.1,1.9) \n objective_function_value f;\nPROCEDURE_SECTION\n for (int i=1;i<=10;i++)\n f+=square(lengths(i)-vb_growth(linf,rho,t0,gamma,ages(i)));\n \nGLOBALS_SECTION\n #include \n\n dvariable vb_growth(const prevariable& linf, const prevariable& rho, \n const prevariable& t0, const prevariable gamma,double t)\n {\n double clinf=value(linf);\n double ct0=value(t0);\n double crho=value(rho);\n double cgamma=value(gamma);\n dvariable len;\n double u1=pow(crho,t-ct0);\n double u2=clinf*(1-u1); \n value(len)=pow(u2,cgamma);\n double dflen=1.0;\n //value(len)=pow(u2,cgamma);\n double dfu2=dflen*cgamma*pow(u2,cgamma-1.0);\n double dfgamma=dflen*value(len)*log(u2);\n //double u2=clinf*(1-u1); \n double dflinf=dfu2*(1-u1);\n double dfu1=-dfu2*clinf;\n //double u1=pow(crho,t-ct0);\n double dfrho=dfu1*(t-ct0)*pow(crho,t-ct0-1.0);\n double dft0=-dfu1*u1*log(crho);\n\n AD_SET_DERIVATIVES4(len,t0,dft0,rho,dfrho,linf,dflinf,gamma,dfgamma); \n // 4 dependent variable \n return len;\n }\n\\end{lstlisting}\n\n\n\\section{General adjoint code}\n\nSo far, the adjoint code has been for a simple function which has from~1 to~4\nindependent variables and returns 1~dependent variable. \nNow we consider the general case where the function can take\nany number of dependent variables and return any number of \ndependent variables, and these variables can be in the form of numbers, \nvectors, or matrices.\n\\begin{lstlisting}\n dvar_vector operator * (_CONST dvar_matrix& m,_CONST dvar_vector& x )\n {\n if (x.indexmin() != m.colmin() || x.indexmax() != m.colmax())\n {\n cerr << \" Incompatible array bounds in dvar_vector operator * \"\n << \"(_CONST dvar_matrix& m,_CONST dvar_vector& x)\\n\";\n ad_exit(21);\n }\n\n dvar_vector tmp(m.rowmin(),m.rowmax());\n double sum;\n\n for (int i=m.rowmin(); i<=m.rowmax(); i++)\n {\n sum=0.0;\n for (int j=x.indexmin(); j<=x.indexmax(); j++)\n {\n sum+=(m.elem(i)).elem_value(j)*x.elem_value(j);\n }\n tmp.elem_value(i)=sum;\n }\n save_identifier_string(\"PLACE4\");\n x.save_dvar_vector_value();\n x.save_dvar_vector_position();\n save_identifier_string(\"PLACE3\");\n m.save_dvar_matrix_value();\n m.save_dvar_matrix_position();\n save_identifier_string(\"PLACE2\");\n tmp.save_dvar_vector_position();\n save_identifier_string(\"PLACE1\");\n \n ADJOINT_CODE(dmdv_prod);\n return(tmp);\n }\n\\end{lstlisting}\nTo calculate the adjoint code, it will be necessary to have the\nvalues of the matrix \\texttt{m} and the vector \\texttt{x}.\nThis is accomplished with the instructions\n\\begin{lstlisting}\n x.save_dvar_vector_value();\n m.save_dvar_matrix_value();\n\\end{lstlisting}\nAlso, to calculate the derivatives, it will be necessary to know where the\nderivatives with respect to the independent and dependent variables are \nlocated. This information is saved with the instructions\n\\begin{lstlisting}\n x.save_dvar_vector_position();\n m.save_dvar_matrix_position();\n tmp.save_dvar_vector_position();\n\\end{lstlisting}\nFinally, we need to save the name of the routine that calculates\nthe adjoint code, so that it can be called at the appropriate time. To write the\ncode for the adjoint calculations, keep in mind that everything must be recovered\nfrom the stack in the reverse order from which it was put on the stack.\nThis process can be a bit confusing and if you don't do it properly, the \nstack will become corrupted and nothing will work. To help diagnose\nproblems, function \\texttt{save\\_identifier\\_string} can be\nused to put a string on the stack. This string value can be checked in the\nadjoint code with the \\texttt{verify\\_identifier\\_string} function. At least two of\nthese functions should be left in any adjoint code, so that stack\nintegrity can be monitored if problems show up later. For optimized code,\nthey are not used, and so contribute almost nothing to the overhead.\n\nThe adjoint code begins by reading the information that was saved on the stack.\nAn object of type \\texttt{dvar\\_vector\\_position} contains both the size and\naddress information associated with a \\texttt{dvar\\_vector}---which are needed to\nrecover or store derivative values, or to build a \\texttt{dvector} with the\nsame shape as the {\\t dvar\\_vector}.\nThe function \\texttt{restore\\_dvar\\_vector\\_der\\-ivatives} gets the values of the\nderivatives with respect to the dependent variables, so they can be used in the\nadjoint code. The functions\n\\begin{lstlisting}\n dfx.save_dvector_derivatives(x_pos);\n dfm.save_dmatrix_derivatives(m_pos);\n\\end{lstlisting}\nuse the position information to save the derivatives with respect to the\nindependent variables in the appropriate places.\n\\begin{lstlisting}\nvoid dmdv_prod(void)\n{\n verify_identifier_string(\"PLACE1\");\n dvar_vector_position tmp_pos=restore_dvar_vector_position();\n verify_identifier_string(\"PLACE2\");\n dvar_matrix_position m_pos=restore_dvar_matrix_position();\n dmatrix m=restore_dvar_matrix_value(m_pos);\n verify_identifier_string(\"PLACE3\");\n dvar_vector_position x_pos=restore_dvar_vector_position();\n dvector x=restore_dvar_vector_value(x_pos);\n verify_identifier_string(\"PLACE4\");\n dvector dftmp=restore_dvar_vector_derivatives(tmp_pos);\n \n dmatrix dfm(m_pos);\n dvector dfx(x_pos.indexmin(),x_pos.indexmax());\n dfm.initialize();\n dfx.initialize();\n \n double dfsum;\n for (int i=m.rowmax(); i>=m.rowmin(); i--)\n {\n // tmp.elem_value(i)=sum;\n dfsum=dftmp.elem(i);\n for (int j=x.indexmax(); j>=x.indexmin(); j--)\n {\n //sum+=(m.elem(i)).elem_value(j)*x.elem_value(j);\n dfm.elem(i,j)+=dfsum*x.elem(j);\n dfx.elem(j)+=dfsum*m.elem(i,j);\n }\n //sum=0.0;\n dfsum=0.0;\n }\n dfx.save_dvector_derivatives(x_pos);\n dfm.save_dmatrix_derivatives(m_pos);\n}\n\\end{lstlisting}\n", "meta": {"hexsha": "89acabdc64f5fc3f25d3b3f0c7c815cfa48f1ec3", "size": 13338, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/manuals/admb/adjoint.tex", "max_stars_repo_name": "wStockhausen/admb", "max_stars_repo_head_hexsha": "876ec704ae974d0ed3bcc329f243dbc401ad0e6d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 79, "max_stars_repo_stars_event_min_datetime": "2015-01-16T14:14:22.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-24T06:28:15.000Z", "max_issues_repo_path": "docs/manuals/admb/adjoint.tex", "max_issues_repo_name": "wStockhausen/admb", "max_issues_repo_head_hexsha": "876ec704ae974d0ed3bcc329f243dbc401ad0e6d", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 172, "max_issues_repo_issues_event_min_datetime": "2015-01-21T01:53:57.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-29T19:57:31.000Z", "max_forks_repo_path": "docs/manuals/admb/adjoint.tex", "max_forks_repo_name": "wStockhausen/admb", "max_forks_repo_head_hexsha": "876ec704ae974d0ed3bcc329f243dbc401ad0e6d", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 22, "max_forks_repo_forks_event_min_datetime": "2015-01-15T18:11:54.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-11T21:47:51.000Z", "avg_line_length": 34.8250652742, "max_line_length": 97, "alphanum_fraction": 0.7358674464, "num_tokens": 3675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.798186787341014, "lm_q2_score": 0.8031737940012418, "lm_q1q2_score": 0.6410827103103446}} {"text": "\\section{Extension to Higher Dimensions}\\label{sec:mud-higher-dimensions}\nWe now show that when the dimension of $\\pspace$ is higher, the benefit from maximizing the dimension of the QoI map is even more considerable than what we observed in the two-dimensional cases, as long as we aggregate data into components of the QoI map in a purposeful manner.\nIn the first example, we demonstrate the impact of increasing the volume of the parameter space while fixing all other attributes of the inverse problem.\nIn the second example, we illustrate how we could utilize the solution to the SIP for the two-dimensional case from \\ref{subsec:pde-example} to improve the MUD estimate in five dimensions.\n\n\\subsection{A SIP with a Naive Initial Density}\nConsider a naive implementation in five dimensions, where the same prior knowledge on the bounds of $g$ inform the choice of a parameter space.\nSuppose $\\pspace$ is given by $[-4, 0]^5$, induced by five regularly-spaced knot points in the interior of $x_2 \\in (0,1)$ for which a uniform density is assumed in each component.\nIn Figure~\\ref{fig:pde-highd-initial-5d}, we show what one-thousand such initial functions look like, and note that many of them appear to exhibit fluctuating behavior which may require a finer mesh than the one being used to solve the problem ($36\\times36$), in order to reduce numerical errors.\nThis choice of initial density induces a lot of ``conceptual'' noise into the SIP, since ``common sense'' from a modeler's perspective could rule out functions which zig-zag excessively.\n\n\\begin{figure}\n\\centering\n \\includegraphics[width=0.475\\linewidth]{figures/pde-highd/pde-highd_init_D5.png}\n\\caption{\nOne thousand initial parameter samples (our model evaluation ``bugdet'') were used to estimate $g$, constructed by taking independent uniform samples from $[-4, 0]$ for each direction are shown in purple.\n}\n\\label{fig:pde-highd-initial-5d}\n\\end{figure}\n\nWe solve the SIP for both scalar-- and vector--valued QoI maps.\nThe latter is constructed with horizontal bands\\---which correspond to the partitioning of $\\Omega$ into components of the QoI map\\---shown in the left of Figure \\ref{fig:pde-highd-5d-example}.\nWe refer to this map as $\\qoi_\\text{5D}$, and $\\qoi_\\text{1D}$ will again represent the scalar--valued map.\nIn the right half of Figure~\\ref{fig:pde-highd-5d-example}, the MUD solutions for $\\qoi_\\text{5D}$ and $\\qoi_\\text{1D}$ are shown in parameter space for a representative SIP.\nThe scalar-valued MUD misidentifies the location of $g$'s minimum value, but the vector-valued QoI is able to resolve the general qualitative behavior.\n\n\\begin{figure}\n\\centering\n \\includegraphics[width=0.45\\linewidth]{figures/pde-highd/pde-highd_sensors_D5}\n \\includegraphics[width=0.45\\linewidth]{figures/pde-highd/pde-highd_comp_exmud_D5_m100}\n\\caption{\n(Left): Layout for 5-D vector--valued map and comparison of the two MUD solutions in parameter space.\n(Right): Example MUD solutions for $\\qoi_\\text{5D}$ and $\\qoi_\\text{1D}$.\n}\n\\label{fig:pde-highd-5d-example}\n\\end{figure}\n\nIn Figure~\\ref{fig:pde-highd-5d-mud} we plot the results from twenty repeated trials (perturbations of noise) when using all $100$ measurements.\nWe observe, as before, the same difference in going from scalar-- to vector--valued solutions that we saw in two dimensions in Fig.~\\ref{fig:pde-MUD}.\nIn the left half of Figure~\\ref{fig:pde-highd-5d-mud}, the scalar-valued QoI is unable to differentiate between resolving residual discrepancies in different locations in $\\Omega$.\nBy contrast, the vector-valued QoI shown to the right is constructed with respect to the flow of information in the system, and so many more of the twenty trials land closer to the true minimum value of $g$.\nThe solutions for the vector-valued approach instead explore the available knots (at $x_2=1/6$ and $1/3$), nearest the actual minimum value of $2/7$ instead, a much more valuable area of $\\Lambda$ to explore.\n\n\\begin{figure}\n\\centering\n \\includegraphics[width=0.45\\linewidth]{figures/pde-highd/pde-highd_pair_D5-1_m100}\n \\includegraphics[width=0.45\\linewidth]{figures/pde-highd/pde-highd_pair_D5-5_m100}\n\\caption{ SIP solutions using $\\qoi_\\text{5D}$ and $\\qoi_\\text{1D}$ for twenty realizations of noise polluting the hundred measurements used to construct the map.\n(Left): Scalar-valued solutions.\n(Right): Vector-valued solutions.\n}\n\\label{fig:pde-highd-5d-mud}\n\\end{figure}\n\n\nRecall from \\ref{subsec:pde-example} that we previously solved a two-dimensional version of this problem.\nWe leverage the effort involved in solving this first problem in order to better refine the approximation of $g$ in \\ref{sec:mud-pde-sequence} below.\nBy using the former SIP solution results to define a much smaller region of $\\pspace$ to explore, the results from the second SIP can improve considerably.\n", "meta": {"hexsha": "6165612bb2ddafcf890fdc86e4dcdfd88f304a80", "size": 4832, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "extensions/mud_pde_5d.tex", "max_stars_repo_name": "mathematicalmichael/thesis", "max_stars_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-04-24T08:05:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-28T20:34:29.000Z", "max_issues_repo_path": "extensions/mud_pde_5d.tex", "max_issues_repo_name": "mathematicalmichael/thesis", "max_issues_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 59, "max_issues_repo_issues_event_min_datetime": "2019-12-27T23:15:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-24T17:52:57.000Z", "max_forks_repo_path": "extensions/mud_pde_5d.tex", "max_forks_repo_name": "mathematicalmichael/thesis", "max_forks_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 81.8983050847, "max_line_length": 296, "alphanum_fraction": 0.7800082781, "num_tokens": 1247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.803173791645582, "lm_q2_score": 0.7981867729389246, "lm_q1q2_score": 0.6410826968627072}} {"text": "\\section{Definition of a battery model}\n\nIn order to define a good model for Li-Po battery, comprehending also a good approximation of all its non-idealities, a two-branches model has been used.\nIn the first part, current SoC is computed, by taking as a reference electrical model a current generator and a capacitor. The discharge current $I_{b}$ charges the capacitor, of capacity $C_{m}$ (representing the total capacity of the battery), producing an output voltage $V_{SOC}$ which represents the SoC. \n\n\nThe output equation of the circuit is thus $V_{SOC} = V_{SOC}(0)+\\int\\!\\frac{I_{b}}{C_{m}}dt$, which is equivalent to the SoC expression for the battery.\n\nThe second part of the model is composed by a voltage generator and a series resistance, as in a usual models of voltage sources. The big difference from standard models is that both the voltage and the series resistance of the battery depend on the current state of charge, and are thus functions of $V_{SOC}$.\n\nThe output of this second circuit is the true battery voltage as seen by the load.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=3in]{circuit}\n \\caption{Equivalent circuit of the Li-Po battery}\n\\end{figure}\n\nWhile the first part of the circuit is well-defined by the initial and total capacity of the battery, in the second part the two functions $V(SoC)$ and $R(SoC)$ has to be extracted from available battery data. In this case, the available source of information contained into the battery's datasheet has been the graph showing output voltage ($V_{B}$) as a function of both discharge capacity - which is, after fixing a maximum capacity value, easily convertible to SoC - and load current. The latter dependance is due to the series resistance of the battery.\n\nPoints from these curves have been extracted using a figure digitizer tool into MatLab and have been converted to different Soc-voltage tables. In order to find a good value for R two curves have been necessary (as R affects only the $V_{B}$-$I_{load}$ relationship). To minimize sampling errors, the highest and lowest currents' curves have been digitalized. Finally, the saved data have been interpolated in a linear fashion to obtain a continous sequence of equally-spaced points on the graph.\n\nFrom these data, $V(Soc)$ and $R(Soc)$ values can be easily obtained by solving the second circuit. Although at this point a model of the two circuits has been made, having the values stored as a lookup table is not the best choice, both because of the space occupation which would not scale for bigger graphs, and because of the missing smoothness of the obtained relationship, which is typical of a real-life system. In order to obtain a smoother and simpler relationship, the data have been fitted using MatLab $fmins$ function, in order to have an analytical representation for them:\n\\begin{align*}\nV(SoC)&=b_1e^{(b_2SoC)}+b_3SoC^4+b_4SoC^3+b_5SoC^2+b_6SoC+b_7 \\\\\nR(SoC)&=b_8e^{(b_9SoC)}+b_{10}\n\\end{align*}\n\nThe curve fitting produced a satisfactory result:\n\n\\begin{center}\n\\begin{tabular}{|c| c|}\n\n \\hline\n Parameter name & Parameter value \\\\\n \\hline\n $b_{1}$ & 1.3518 \\\\\n $b_{2}$ & -0.7752 \\\\\n $b_{3}$ & 0.4669 \\\\\n $b_{4}$ & -1.0723 \\\\\n $b_{5}$ & 0.8729 \\\\\n $b_{6}$ & -0.0031 \\\\\n $b_{7}$ & 2.8434 \\\\\n $b_{8}$ & 0.0000 \\\\\n $b_{9}$ & 9.7875 \\\\\n $b_{1}$ & 0.1909 \\\\\n \\hline\n\\end{tabular}\n\n\n\\begin{figure}[h]\n \\centering\n\\includegraphics[width=3in]{fitted_r_curve}\n\\caption{$SoC-R$ curve after fitting}\n\\end{figure}\n\n\n\\begin{figure}[h]\n \\centering\n\\includegraphics[width=4in]{fitted_vsoc_curve} \n\\caption{$SoC-V$ curve after fitting}\n\\end{figure}\n\n\\end{center}\n\n\n", "meta": {"hexsha": "4bcebf77084738d7de42809ee5ad9cab85e9498e", "size": 3664, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/ass1.tex", "max_stars_repo_name": "conte91/ENOPT_Lab4", "max_stars_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/ass1.tex", "max_issues_repo_name": "conte91/ENOPT_Lab4", "max_issues_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/ass1.tex", "max_forks_repo_name": "conte91/ENOPT_Lab4", "max_forks_repo_head_hexsha": "2bd11699bc9274c9f0fe3c36b1f39b66f0a9fd2c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 54.6865671642, "max_line_length": 587, "alphanum_fraction": 0.7399017467, "num_tokens": 991, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569014, "lm_q2_score": 0.798186768138228, "lm_q1q2_score": 0.6410826967674265}} {"text": "\\chapter{R Background}\nThis chapter describes the R programming techniques that we will need in order\nto do Bayesian statistics both within R and also by using JAGS.\n\n\\section{Vectors}\nWe will use vectors frequently throughout the course. A vector can be thought\nof as a list of numbers. In R, you can create a vector using the function\n{\\tt c}, which stands for ``concatenate''.\n\\begin{verbatim}\nmy_vector = c(12.4, -6.2, 4.04)\n\\end{verbatim}\nYou can then examine the contents of the vector by typing its name.\n\\begin{verbatim}\n> my_vector\n[1] 12.40 -6.20 4.04\n\\end{verbatim}\nThere are various helpful things you can do with vectors. You can get the length\nof a vector (the number of elements in it) like so:\n\\begin{verbatim}\n> length(my_vector)\n[1] 3\n\\end{verbatim}\nYou can do arithmetic\nwith vectors too, if they're the same length. For example:\n\\begin{verbatim}\n> c(1,2,3)*c(2,5,3)\n[1] 2 10 9\n\\end{verbatim}\nIf you wanted, you could assign the output to a new variable:\n\\begin{verbatim}\n> x = c(1,2,3)\n> y = c(2,5,3)\n> z = x*y\n> z\n[1] 2 10 9\n\\end{verbatim}\nIt is important to understand how to access subsets (or ``slices'') of vectors\nbased on which elements satisfy a certain condition. Here is an example:\n\\begin{verbatim}\n> x = c(1, 2, 3, 0, 10)\n> test = x > 3\n> test\n[1] FALSE FALSE FALSE FALSE TRUE\n> y = x[x <= 2]\n> y\n[1] 1 2 0\n> z = sum(x[x <= 2])\n> z\n[1] 3\n\\end{verbatim}\nThis kind of thing will be used frequently in the computational parts of the\ncourse.\n\n\n\n\n\\section{Lists}\nLists are a bit like vectors, in that they can contain a lot of information in\na single variable. But instead of just being numbers, lists can contain all\nsorts of things. For example, suppose I want a variable/object in an R program\nto represent a person. A person can have a name and an age. Here is how to make\na list:\n\\begin{verbatim}\na_person = list(name=\"Nicole\", age=21)\n\\end{verbatim}\nIf you needed to extract certain elements from a list, you can do it using the\n\\$ operator in R. For example suppose I wanted to extract the {\\tt name} variable\nfrom within {\\tt a\\_person}. I could do this:\n\\begin{verbatim}\n> a_person$name\n[1] \"Nicole\"\n\\end{verbatim}\nand voila. When we use JAGS, a data set will be represented using a list.\nSo will our JAGS output.\n\nIf you have ever learned C, C++, or Matlab, a list in R is basically the same thing as\na ``struct'' in these languages.\nIf you have learned C++, Java, or Python, a list is\nlike a ``class'' but with just variables and no functions.\nIn Python, ``dictionaries''\nare also very similar to R lists.\n\n\\section{Functions}\nIn this course you'll need to be able to read and understand simple R\nfunctions, and perhaps write a few. A function is like a machine that takes\nan input, does something, and returns an output. You have probably used many\nbuilt-in R functions already, like {\\tt sum()}.\n\n\\begin{verbatim}\n# Defining a function called my_function\nmy_function = function(x)\n{\n # Do some stuff\n result = 3*x + 0.5\n return(result)\n}\n\\end{verbatim}\n\n\n\\section{For Loops}\nFor loops are mostly used to repeat an action many times. Here is an example.\n\\begin{verbatim}\nN = 100\nfor(i in 1:N)\n{\n print(i)\n}\n\\end{verbatim}\n\n\\section{Useful Probability Distributions}\nSince R is a statistics program, it knows about a lot of probability\ndistributions already. So, if I wanted to use the probability density function\nof a normal distribution, instead of having to code something like this:\n\\begin{verbatim}\nf = exp(-0.5*((x - mu)/sigma)**2)/(sigma*sqrt(2*pi))\n\\end{verbatim}\nI can just use the built-in function {\\tt dnorm}.\n\\begin{verbatim}\nf = dnorm(x, mean=mu, sd=sigma)\n\\end{verbatim}\nMuch easier! If, instead of wanting to evaluate the PDF, I wanted to generate\nrandom samples from a normal distribution, I could use {\\tt rnorm}.\n\\begin{verbatim}\n# Generate 1000 samples\nsamples = rnorm(1000, mean=50., sd=10.)\n\\end{verbatim}\n\n", "meta": {"hexsha": "6377a09edc15d2c6cdde573778aa6f97fa524e62", "size": 3876, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "r_background.tex", "max_stars_repo_name": "xulinpan/stat331", "max_stars_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 55, "max_stars_repo_stars_event_min_datetime": "2015-03-09T18:03:23.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-25T03:36:54.000Z", "max_issues_repo_path": "r_background.tex", "max_issues_repo_name": "xulinpan/stat331", "max_issues_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2015-07-07T05:00:32.000Z", "max_issues_repo_issues_event_max_datetime": "2015-07-10T08:48:27.000Z", "max_forks_repo_path": "r_background.tex", "max_forks_repo_name": "xulinpan/stat331", "max_forks_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2015-07-29T14:34:51.000Z", "max_forks_repo_forks_event_max_datetime": "2020-06-04T20:04:47.000Z", "avg_line_length": 30.28125, "max_line_length": 86, "alphanum_fraction": 0.7293601651, "num_tokens": 1114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737869342623, "lm_q2_score": 0.7981867705385762, "lm_q1q2_score": 0.6410826911742974}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage{epstopdf}\n\\usepackage{inputenc}\n\\usepackage{geometry} \n\\usepackage{cancel}\n\\geometry{left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm}\n\n\\begin{document}\n\t\n\t\\title{MITx 15.455x Mathematical Methods for Quantitative Finance\n\t \\\\\n\t\\begin{large} \n\t\tRecitation 3\n\t\\end{large} }\n\n\t\n\t\\maketitle\n\t\n\t\\section*{Module 3: Forecasting}\n\t\n\nLet's talk about forecasting.\nWe saw in lecture that forecasting\nis an application of conditional probability to time series\nmodels.\nSo we think of the time series as generating the data.\nAnd we ask at a given point in time,\ngiven all the observations that exist up\nthrough that point in time, what can we say about the future?\nWe might want to know what happens one time step ahead,\ntwo time steps ahead, or in limit\nas the number of times that goes to infinity,\nbut the key thing is the break between past and future.\nSo we take the present time to be just after we've\nmade our final observation.\nAnd we're first forecast period is one step ahead.\nAnd, as I said, the key tool is in addition to\nour previous ones with expectations,\nlinearity, algebraic substitution, recursion.\nWe had a new one, which is conditional probability.\nIn this setting, conditional probability simply\nmeans that for any variable that was\na random variable in the past that has been realized,\nit's no longer a random variable.\nNow it's a scalar.\nNow it's a number.\nSo those things that show up in the recursion\nthat we previously thought of as random variables,\nin this setting, when we take conditional expectations\nof our defining equations or equations of evolution,\nthen they take a different character,\nbecause they're already known.\nThey're no longer random variables.\nSo we looked at an example in lecture.\nLet's just do a couple of examples together.\nIt's the same structure.\nThey're just one or two things it might be worth\npaying attention to as you look at the blind forecast\nof your own.\nSome of the things that might have look special for the case\nof the AR1 are quite general.\nSo let's take a look at a model.\n\n\\subsection*{ARMA(1,1)}\n\nLet's look at, say, an ARMA(1,1) which is this form:\n\n$$ x_t = c_o + c_1 x_{t-1} \\sigma z_t + \\phi_1 z_{t-1} $$\n\nSo I'm going to keep the notation $\\sigma z_t$ for our shock for innovation in the period.\nYou could give it another name for its coefficient.\nRemember that $\\sigma$, in this case,\nis not the standard deviation of $x$.\nIt's just a scaling parameter for the random shock $z$.\n\nSo the first thing we'd like to do\nis compute the mean value for $x$ and simplify our expression.\nSo we use our usual trick.\n$\\mu$ is the name will give the expectation of $x_t$.\nAnd then we take expectations on the right hand side\nand apply linearity.\nSo we have this is going to be:\n\n\n$$ \\mu = E[x_t] = c_0 + c_1 E[x_{t-1}] + 0 + 0 $$\n\n$$ \\mu = \\frac{c_0}{1-c_1} $$\n\n\nAnd you should recognize that expression from our AR 1 model.\nIt's exactly the same expression.\nAnd therefore, we can substitute and rewrite our equation\nas a bunch of things that are grouped together in such a way\nthat they all have 0 mean.\nThat is to say, if I substitute for $c_0$\nand say it's equal to $\\mu (1 - c_1)$,\nthen I can write this as:\n\n$$ x_t = \\mu (1 - c1) + c_1 x_{t-1} + \\sigma z_t + \\phi_1 z_{t-1} $$\n$$ x_t - \\mu = c_1 (x_{t-1} - \\mu)+ \\sigma z_t + \\phi_1 z_{t-1} $$\n\n\nNow the next thing we want to do is put things\nin forecasting form.\nSo what I'm going to do is I'm going to shift $t$ to $t+1$.\nEverywhere where I see a $t$, I'm going to write $t+1$.\nAnd that puts the future values on the left\nand the present and past values or the known\nvalues on the right hand side.\nSo for one-time step ahead, this makes exactly the split\nthat we want to have:\n\n$$ x_{t+1} = \\mu + c_1 (x_{t} - \\mu)+ \\sigma z_{t+1} + \\phi_1 z_{t} $$\n\n\nSo what's our forecast?\nSo our definition for our forecast\nis going to be the conditional probability.\nSo our first forecast is going to be:\n\n$$f_{t,1} = E[x_{t+1}|x_t, z_t, \\dots] $$\n\nThat is the forecast made at time t for one-time step\nahead in the future.\nAnd there are different notations for this.\nSo what you really should rely on is not the notation,\nbut reduce everything to expectations.\nSo this is the expectation of $t+1$ given everything\nup through that point, which includes $x_t$, $z_t$,\nand any previous failures, but those\ndon't show up in this equation on the right-hand side.\nSo let's compute the expectation.\nAnd what we noticed is that the only random variable\non the right hand side is $z_{t+1}$.\nThat's the only thing that's random.\nSo $x_t$ is known.\nThis $z_t$ is known.\nThere's only one random variable, i.e. $z_{t+1}$ .\nAnd we take expectations, the expectation of $z_{t+1}$\nis going to vanish.\nSo what do we get?\nWe're going to get that the forecast is equal to:\n\n$$f_{t,1} =\\mu + c_1 (x_{t} - \\mu) + \\phi_1 z_{t} $$\n\nSo we just do the calculation.\nIt's exactly the same as the expression\nabove, except for this $\\sigma z_{t+1}$, this random term\nthat dropped out, because it has zero expectation.\nSo let's take a look at the forecast error.\nSo the first forecast error for one-time step\nahead is going to be:\n\n$$e_{t+1} = x_{t+1} - f_{t,1} $$\n\nThat is the forecast error as defined\nas the difference between what we predicted\nand what actually happened.\nWhat we'd like to do and the way in which\nwe have this definition for the forecast being optimal\nis we're going to minimize the mean squared forecast error.\nThat is going to minimize $e^2$\nin expectation of all possible things that would happen.\nNow we could use other kinds of loss functions.\nThose will depend on the settings,\ndepending on the economic values.\nSo it's just a question of doing and minimization,\nbut in this case, we have this basic result\nthat this will be our expectation,\nbut let's compute the forecast error.\nSo the idea is that what we really want, in general, if you\nwant to drive this and check, what we really want\nis we want a predictor for $f$, that some function of all\nthe previous observations, in this case, just $x_t$ and $z_t$,\nwe'd like to be a linear function.\nAnd when we find the linear function that\nminimizes the mean squared error,\nwe find that it's this conditional expectation.\nSo let's compute this quantity.\nAnd we'll see that there's a really nice structure that\nshows up.\n$x_{t+1} - f_{t,1}$, you\nnotice that most of the terms are common.\n\nAnd therefore, the mean squared forecast error\nis:\n$$E[e_{t+1}^2] = E[ x_{t+1} - f_{t,1} ] = E[(\\sigma z_{t+1})^2] =\\sigma^2$$\n\n\nNow what about looking at multiple forecasts?\nSo we'd like to go for future horizons.\nOne-step was easy.\nSo the general rule is that when we\nwant to do it two-step ahead forecast,\nwe're going to shift everything.\nWe're going to add one more.\nSo we write an expression for $x_{t+2}$.\n\nSo here's the general procedure.\nKeep doing the recursion.\nKeep substituting in back until all of the $X$'s on the right\nhand side of the equation have a time index of $t$ or earlier.\nThe $z$'s can be later.\nThat's OK, because we're going to take their expectations,\nand they're going to vanish.\nSo we might have some possibly unknown $z$'s.\nBut the basic idea is once we go beyond one-time step,\nhow do you know what to do?\nIt's easy.\nYou're doing $x_t+h$ for horizon $h$\non the left hand side.\nAnd on the right hand side, you do probably $h$ or $h-1$\nrecursive substitutions of the defining equation\nuntil we can express $x_t$ plus $h$ in terms of $x$, $x_{t-1}$\nminus 1, and so on.\nThose are all known quantities.\nThen we take conditional expectations.\nWe have our forecast.\nWe compute the forecast errors.\nWe take expectations of their squares.\nThat's easy as well, because it's just the same quantities.\nNow we have the known and the unknown ones\nclearly delineated.\nAnd we get our results for our mean squared forecast error.\nAnd there are two really important properties\nto keep in mind for our solution.\nOne of them is the forecast error has zero expectation.\nAnd the forecast error is orthogonal\nto the other variables, to the other to the other predictor variables.\n\n\\subsection*{ARMA(2,2)}\n\nSo let's take a look for another example\nLet's take a look at an ARMA(2,2), shall we?\nSo now when you're given numbers for the forecasts\nor particular horizons, you can either plug them in right away\nor you can leave the parameters general.\nThat generally makes it easier to check your math\nand find any sign errors that might be there,\nbut either way, whether you substitute indefinite numbers\nfor the parameters before or after shouldn't\nmake any difference.\nWhen you're doing forecast, you do want to pay attention though\nto the initial conditions if you have to bootstrap your process,\nbecause if you're asked for a forecast one or three\nor seven steps ahead, you'll need enough data to get\nthe process started before you can generate the recursion.\nThe recursive techniques that we have\nthat I'm writing down here for the forecast\nare technically appropriate for the case, where we go\ninfinitely far into the past.\nEven if the series did exist an infinitely long time,\nwe wouldn't have an infinite amount of data.\nSo we do need to make sure we have the initial conditions set\nto get numerical answers, but the basic rules are compute\nconditional expectations after writing our defining\nequations with variable to be forecast on the left hand side\nand only known observations on the right hand side.\n\n\nSo let's do this for ARMA(2,2).\nit's little bit more complicated at least initially.\nSo it looks like:\n\n$$x_t = c_0+c_1 x_{t-1} + c_2 x_{t-2} + \\sigma z_t + \\phi_1 z_{t-1} + \\phi_2 z_{t-2} $$\n\nAnd you can see how it would go for a general ARMA(p,q).\nI'm going to write it in concrete form\nso we don't have summations running around.\nAnd when we shift, it gets a little messy\nkeeping track of the indices, but you can do it.\nAnd you can take a look at the literature as well for that.\nSo what's our expectation in this case?\n\n$$ \\mu = E[x_t] = c_0+c_1 \\mu + c_2 x_{t-2} + \\mu $$\n$$ \\mu = \\frac{c_0}{1-c_1 - c_2} $$\n\n\nAnd you can guess how this generalizes\nfor the general ARMA(p,q) model, we\nget the generalization of this expression.\nAnd then we can write our equation\nas:\n\n$$ x_t - \\mu = c_1(x_{t-1}-\\mu) + c_2(x_{t-2}-\\mu) + \\sigma z_t + \\phi_1 z_{t-1} + \\phi_2 z_{t-2} $$\n\nLet's do some forecasting, shall we?\n\nYou should notice that each of the terms that I've\nwritten down is a way that it has 0 mean.\nThat just makes it a bit easier to see what's going on,\nto see what the dynamics are, and to do some calculations.\nYou can also, if you want to clean it up,\nyou can find a new variable $y$ to be $x_t - \\mu$ and shift things\nback.\nSo you can do any rescaling things you like.\nThe end results are going to be the same if they're just\nredefinitions of the parameters and rescalings\nof the variables.\n\nSo let's take a look for time $t+1$:\n\n$$ x_{t+1} - \\mu = c_1(x_{t}-\\mu) + c_2(x_{t-1}-\\mu) + \\sigma z_{t+1} + \\phi_1 z_{t} + \\phi_2 z_{t-1} $$\n\n\nSo that one is pretty easy.\nIt looks like the expression that we just\ndid before for the ARMA(1,1).\nThat is everything on the right hand side has an index of \nall of the $x$'s are taken at time $t$ or $t-1$.\nAnd the $z$'s are mostly in the past.\nWe have these two with coefficients\nof $\\phi$'s both in the past.\nAnd this is the only one that's still a random variable,\nbut at time 2, things change.\nSo let's look one more time step ahead.\nLet's contrast a little bit.\nSo if we write this as:\n\n$$ x_{t+2} - \\mu = c_1(x_{t+1}-\\mu) + c_2(x_{t}-\\mu) + \\sigma z_{t+2} + \\phi_1 z_{t+1} + \\phi_2 z_{t} $$\n\n\nSo all I've done is I shifted $t$ to $t+1$ again.\n\n\nSo I have two of the terms on the right hand side\ninvolve the present or the past.\nAnd that's OK.\nThose are known quantities, but I have these other expressions.\nSo I can't just directly compute the expectation\non the left hand side.\nNow I could take the previous work I did before\nand substitute that would be fine.\nBut if I want to see the general approach for doing it\nat a particular time step, what I'm going to say\nis that the expressions $\\sigma z_{t+2}$ and $\\phi_1 z_{t+1}$ , I can leave alone,\nbecause those are both fine.\nThese just involves $z$'s.\nAnd I know their expectations, but I\ndon't want the recursive structure for $x_t+1$, because remember this is saying that the value two\ndays from now depends on the value tomorrow, which\nI haven't yet observed.\nSo the value today, the value yesterday, those are all known,\nbut rather than expressing it this way,\nI'd like to do my recursions, get it out.\nAnd then I can take clean ordinary simple expectations\nand get an answer.\nSo let's do that.\nAnd we just need to substitute one more time and put that in.\nSo which term are we going to substitute--\nwe're going to substitute in?\nWe'll have :\n\n$$ x_{t+2} = \\mu + c_1\\Big[ c_1(x_{t}-\\mu) + c_2(x_{t-1}-\\mu) + \\sigma z_{t+1} + \\phi_1 z_{t} + \\phi_2 z_{t-1} \\Big] + c_2(x_{t}-\\mu) + \\sigma z_{t+2} + \\phi_1 z_{t+1} + \\phi_2 z_{t} $$\n\n\nSo what have we got?\nSo I now have an expression for $x_{t+2}$, where on the right hand side,\neverything depends on known quantities.\nSo we put it in.\nWe turn the crank.\nWe compute the expectation two-time steps ahead.\nThat will give us our forecast today for two-time steps ahead.\nWhen tomorrow is realized, that forecast\nfor what will then be one more day ahead is going to change.\nAnd how will it change?\nIt will be updated by the new observation.\nSo if today is Monday and I'd like\nto know about Wednesday's weather,\nI would make a forecast today.\nTomorrow I'll have a new forecast for Wednesday.\nAnd that will change.\nIt will change by my knowledge of what happened on Tuesday.\nBut from a mathematical point of view,\nwe separated things into the form that we want.\nWe can first take the expectation\nwith the appropriate conditions.\nAnd second, we can compute forecast error in advance.\nAnd we can compute the mean squared forecast\ntwo steps ahead.\nAnd so that's looking ahead for what\nour expected forecast error.\nThe reason that that's important is\nthat we want to think not only about distributions.\nWhat's the exact value?\nBecause that exact value probably won't be realized.\nIt gives us characteristics of the full distribution\nof the actual outcome.\nAnd it lets us have a sense as to how we\nshould evaluate our forecasts.\nSo that's an expectation.\nOnce Wednesday is realized, then we'd like to go back.\nAnd then our Xt plus 2 will be a known quantity.\nWe can compare it with our forecast.\nAnd, of course, this will differ by some amount,\nbut if we'd like to improve our forecasting techniques,\nour forecasting methodology, and our forecasting quality,\nwhat we do is we look over time.\nWe take a large collection of forecasts and outcomes.\nAnd we study the statistics of the forecast errors.\nSo we use forecast errors both in expectation going forward\nand to assess forecast quality and look\nfor improvements going backward once we've\ncollected the relevant data.\n\t\n\\end{document}", "meta": {"hexsha": "73f666bf7defec4ae7b836f5de740c3278afc793", "size": 14996, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "15.455x/recitation_03.tex", "max_stars_repo_name": "j053g/cheatsheets", "max_stars_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-12-14T08:49:07.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-07T17:26:15.000Z", "max_issues_repo_path": "15.455x/recitation_03.tex", "max_issues_repo_name": "j053g/cheatsheets", "max_issues_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "15.455x/recitation_03.tex", "max_forks_repo_name": "j053g/cheatsheets", "max_forks_repo_head_hexsha": "22f7a84879c04d44de40467ddcc0f6e551b812c7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.6650366748, "max_line_length": 186, "alphanum_fraction": 0.7444651907, "num_tokens": 4083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8459424334245618, "lm_q1q2_score": 0.641050423726081}} {"text": "\\chapter{Combinatorial}\n\n\\section{Permutations}\n\t\\subsection{Factorial}\n\t\t\\import{factorial.tex}\n\t\t\\kactlimport{IntPerm.h}\n\n\t\\subsection{Cycles}\n\t\tLet $g_S(n)$ be the number of $n$-permutations whose cycle lengths all belong to the set $S$. Then\n\t\t$$\\sum_{n=0} ^\\infty g_S(n) \\frac{x^n}{n!} = \\exp\\left(\\sum_{n\\in S} \\frac{x^n} {n} \\right)$$\n\n\t\\subsection{Derangements}\n\t\tPermutations of a set such that none of the elements appear in their original position.\n\t\t\\[ \\mkern-2mu D(n) = (n-1)(D(n-1)+D(n-2)) = n D(n-1)+(-1)^n = \\left\\lfloor\\frac{n!}{e}\\right\\rceil \\]\n\n\t\\subsection{Burnside's lemma}\n\t\tGiven a group $G$ of symmetries and a set $X$, the number of elements of $X$ \\emph{up to symmetry} equals\n\t\t \\[ {\\frac {1}{|G|}}\\sum _{{g\\in G}}|X^{g}|, \\]\n\t\t where $X^{g}$ are the elements fixed by $g$ ($g.x = x$).\n\n\t\t If $f(n)$ counts ``configurations'' (of some sort) of length $n$, we can ignore rotational symmetry using $G = \\mathbb Z_n$ to get\n\t\t \\[ g(n) = \\frac 1 n \\sum_{k=0}^{n-1}{f(\\text{gcd}(n, k))} = \\frac 1 n \\sum_{k|n}{f(k)\\phi(n/k)}. \\]\n\n\\section{Partitions and subsets}\n\t\\subsection{Partition function}\n\t\tNumber of ways of writing $n$ as a sum of positive integers, disregarding the order of the summands.\n\t\t\\[ p(0) = 1,\\ p(n) = \\sum_{k \\in \\mathbb Z \\setminus \\{0\\}}{(-1)^{k+1} p(n - k(3k-1) / 2)} \\]\n\t\t\\[ p(n) \\sim 0.145 / n \\cdot \\exp(2.56 \\sqrt{n}) \\]\n\n\t\t\\begin{center}\n\t\t\\begin{tabular}{c|c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c}\n\t\t\t$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 20 & 50 & 100 \\\\ \\hline\n\t\t\t$p(n)$ & 1 & 1 & 2 & 3 & 5 & 7 & 11 & 15 & 22 & 30 & 627 & $\\mathtt{\\sim}$2e5 & $\\mathtt{\\sim}$2e8 \\\\\n\t\t\\end{tabular}\n\t\t\\end{center}\n\n\t\\subsection{Lucas' Theorem}\n\t\tLet $n,m$ be non-negative integers and $p$ a prime. Write $n=n_kp^k+...+n_1p+n_0$ and $m=m_kp^k+...+m_1p+m_0$. Then $\\binom{n}{m} \\equiv \\prod_{i=0}^k\\binom{n_i}{m_i} \\pmod{p}$.\n\n\t\\subsection{Binomials}\n\t\t\\kactlimport{multinomial.h}\n\n\\section{General purpose numbers}\n\t\\subsection{Bernoulli numbers}\n\t\tEGF of Bernoulli numbers is $B(t)=\\frac{t}{e^t-1}$ (FFT-able).\n\t\t$B[0,\\ldots] = [1, -\\frac{1}{2}, \\frac{1}{6}, 0, -\\frac{1}{30}, 0, \\frac{1}{42}, \\ldots]$\n\n\t\tSums of powers:\n\t\t\\small\n\t\t\\[ \\sum_{i=1}^n n^m = \\frac{1}{m+1} \\sum_{k=0}^m \\binom{m+1}{k} B_k \\cdot (n+1)^{m+1-k} \\]\n\t\t\\normalsize\n\n\t\tEuler-Maclaurin formula for infinite sums:\n\t\t\\small\n\t\t\\[ \\sum_{i=m}^{\\infty} f(i) = \\int_m^\\infty f(x) dx - \\sum_{k=1}^\\infty \\frac{B_k}{k!}f^{(k-1)}(m) \\]\n\t\t\\[ \\approx \\int_{m}^\\infty f(x)dx + \\frac{f(m)}{2} - \\frac{f'(m)}{12} + \\frac{f'''(m)}{720} + O(f^{(5)}(m)) \\]\n\t\t\\normalsize\n\n\t\\subsection{Stirling numbers of the first kind}\n\t\tNumber of permutations on $n$ items with $k$ cycles.\n\t\t\\begin{align*}\n\t\t\t&c(n,k) = c(n-1,k-1) + (n-1) c(n-1,k),\\ c(0,0) = 1 \\\\\n\t\t\t&\\textstyle \\sum_{k=0}^n c(n,k)x^k = x(x+1) \\dots (x+n-1)\n\t\t\\end{align*}\n\t\t$c(8,k) = 8, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1$ \\\\\n\t\t$c(n,2) = 0, 0, 1, 3, 11, 50, 274, 1764, 13068, 109584, \\dots$\n\n\t\\subsection{Eulerian numbers}\n\t\tNumber of permutations $\\pi \\in S_n$ in which exactly $k$ elements are greater than the previous element. $k$ $j$:s s.t. $\\pi(j)>\\pi(j+1)$, $k+1$ $j$:s s.t. $\\pi(j)\\geq j$, $k$ $j$:s s.t. $\\pi(j)>j$.\n\t\t$$E(n,k) = (n-k)E(n-1,k-1) + (k+1)E(n-1,k)$$\n\t\t$$E(n,0) = E(n,n-1) = 1$$\n\t\t$$E(n,k) = \\sum_{j=0}^k(-1)^j\\binom{n+1}{j}(k+1-j)^n$$\n\n\t\\subsection{Stirling numbers of the second kind}\n\t\tPartitions of $n$ distinct elements into exactly $k$ groups.\n\t\t$$S(n,k) = S(n-1,k-1) + k S(n-1,k)$$\n\t\t$$S(n,1) = S(n,n) = 1$$\n\t\t$$S(n,k) = \\frac{1}{k!}\\sum_{j=0}^k (-1)^{k-j}\\binom{k}{j}j^n$$\n\n\t\\subsection{Bell numbers}\n\t\tTotal number of partitions of $n$ distinct elements. $B(n) =$\n\t\t$1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, \\dots$. For $p$ prime,\n\t\t\\[ B(p^m+n)\\equiv mB(n)+B(n+1) \\pmod{p} \\]\n\n\t\\subsection{Labeled unrooted trees}\n\t\t\\# on $n$ vertices: $n^{n-2}$ \\\\\n\t\t\\# on $k$ existing trees of size $n_i$: $n_1n_2\\cdots n_k n^{k-2}$ \\\\\n\t\t\\# with degrees $d_i$: $(n-2)! / ((d_1-1)! \\cdots (d_n-1)!)$\n\n\t\\subsection{Catalan numbers}\n\t\t\\[ C_n=\\frac{1}{n+1}\\binom{2n}{n}= \\binom{2n}{n}-\\binom{2n}{n+1} = \\frac{(2n)!}{(n+1)!n!} \\]\n\t\t\\[ C_0=1,\\ C_{n+1} = \\frac{2(2n+1)}{n+2}C_n,\\ C_{n+1}=\\sum C_iC_{n-i} \\]\n\t\t${C_n = 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, \\dots}$\n\t\t\\begin{itemize}[noitemsep]\n\t\t\t\\item sub-diagonal monotone paths in an $n\\times n$ grid.\n\t\t\t\\item strings with $n$ pairs of parenthesis, correctly nested.\n\t\t\t\\item binary trees with with $n+1$ leaves (0 or 2 children).\n\t\t\t\\item ordered trees with $n+1$ vertices.\n\t\t\t\\item ways a convex polygon with $n+2$ sides can be cut into triangles by connecting vertices with straight lines.\n\t\t\t\\item permutations of $[n]$ with no 3-term increasing subseq.\n\t\t\\end{itemize}\n", "meta": {"hexsha": "686a1194d9f96f8b047769dafbd6c362ddecfe7f", "size": 4761, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/combinatorial/chapter.tex", "max_stars_repo_name": "sarafanshul/KACTL", "max_stars_repo_head_hexsha": "fa14ed34e93cd32d8625ed3729ba2eee55838340", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2021-01-25T12:07:17.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-26T17:20:31.000Z", "max_issues_repo_path": "content/combinatorial/chapter.tex", "max_issues_repo_name": "sarafanshul/KACTL", "max_issues_repo_head_hexsha": "fa14ed34e93cd32d8625ed3729ba2eee55838340", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/combinatorial/chapter.tex", "max_forks_repo_name": "sarafanshul/KACTL", "max_forks_repo_head_hexsha": "fa14ed34e93cd32d8625ed3729ba2eee55838340", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2021-02-28T11:13:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-20T12:56:20.000Z", "avg_line_length": 46.6764705882, "max_line_length": 201, "alphanum_fraction": 0.5782398656, "num_tokens": 2185, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424256566559, "lm_q2_score": 0.7577943712746406, "lm_q1q2_score": 0.6410504085850299}} {"text": "%% theorem/proof/lemma/corollary/example blocks\n\\section{Prime Numbers}\n\\subsection{Definition}\n\\begin{frame}{What Are Prime Numbers?}\n \\begin{definition}\n A \\alert{prime number} is a number that has exactly two divisors.\n \\end{definition}\n \\begin{example}\n \\begin{itemize}\n \\item 2 is prime (two divisors: 1 and 2).\n \\item 3 is prime (two divisors: 1 and 3).\n \\item 4 is not prime (\\alert{three} divisors: 1, 2, and 4).\n \\end{itemize}\n \\end{example}\n\\end{frame}\n\n\\lipsum[1]\n\n", "meta": {"hexsha": "b407407db8fead1f69dff91f3d6b8530129841a4", "size": 504, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "beamer-lecture-notes/example1.tex", "max_stars_repo_name": "edgemaster/latex-experiments", "max_stars_repo_head_hexsha": "4b3b3d0f64fad84e23242780c2a0778d1927a7d8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "beamer-lecture-notes/example1.tex", "max_issues_repo_name": "edgemaster/latex-experiments", "max_issues_repo_head_hexsha": "4b3b3d0f64fad84e23242780c2a0778d1927a7d8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2015-02-17T15:45:36.000Z", "max_issues_repo_issues_event_max_datetime": "2015-02-17T15:46:39.000Z", "max_forks_repo_path": "beamer-lecture-notes/example1.tex", "max_forks_repo_name": "edgemaster/beamer-lecture-notes", "max_forks_repo_head_hexsha": "4b3b3d0f64fad84e23242780c2a0778d1927a7d8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.5263157895, "max_line_length": 69, "alphanum_fraction": 0.6746031746, "num_tokens": 164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812554, "lm_q2_score": 0.7956581073313275, "lm_q1q2_score": 0.64100778269002}} {"text": "\\documentclass[main]{subfiles}\n\\begin{document}\n\\section{Poisson Regression}\n\\label{sec:poisson_regression}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsection{}\n\nTransform the poisson distribution function\n\\begin{align*}\n\tp(y;\\lambda)\n\t\t&=\n\t\t\te^{-\\lambda}\n\t\t\t\\frac {(\\lambda)^y} {y!} \t\t\t\\\\\n\t\t&=\n\t\t\t\\frac {1} {y!}\n\t\t\t\\exp (\n\t\t\t\ty \\log \\lambda - \\lambda\n\t\t\t)\n\\end{align*}\n\nCompare it to the general equation that defines the exponential family\n\\begin{center}\n\\begin{math}\n\tp(y;\\lambda) =\n\t\tb(y) \\exp ( \\eta \\, y - a(\\eta) )\n\\end{math}\n\\end{center}\n\nWe get\n\\begin{align*}\n\ta(\\eta) \t&=\n\t\t\t\\lambda = e^{\\eta} \t\t\t\t\t\\\\\n\tb(y)\t\t&=\n\t\t\t\\frac {1} {y!}\t\t\t\t\t\t\\\\\n\t\\eta \t\t&=\n\t\t\t\\log \\lambda\t\t\t\t\t\t\\\\\n\tT(y)\t\t&=\n\t\t\ty\n\\end{align*}\n\n\\subsection{}\n\nThe mean value of a random variable that abides by the Poisson distribution is\n$\\lambda$. Thus the \\textbf{canonical response function} is\n\\begin{center}\n\\begin{math}\n\tg(\\eta) = E[y;\\lambda] = \\lambda = e^{\\eta}\n\\end{math}\n\\end{center}\n\n\\subsection{}\nskipped\n\n\\subsection{}\nGiven $T(y) = y$, the general equation of exponential distributions is\nsimplified to\n\\begin{align*}\n\tp(y|\\eta) =\n\t\tb(y) \\exp ( \\eta \\, y - a(\\eta) )\n\\end{align*}\n\nWith the course assumption $\\eta = \\v{\\theta}\\cdot\\v{x}$, the log-likelihood\nfunction can be written as\n\\begin{align*}\n\tl(\\theta)\n\t\t&=\n\t\t\t\\log p(y^{(i)}|x^{(i)};\\theta) \t\t\\\\\n\t\t&=\n\t\t\t\\log b(y) +\n\t\t\t(\\v{\\theta}\\cdot\\v{x}^{(i)}) y - a(\\v{\\theta}\\cdot\\v{x}^{(i)})\n\\end{align*}\n\nGiven a data point $(x^{(i)}, y^{(i)})$, the gradient of the log-likelihood with\nrespect to $\\theta$ is\n\\begin{align*}\n\t\\pd{l(\\theta)}{\\theta_j}\n\t\t=\n\t\t\tx^{(i)}_j (\n\t\t\t\ty - a^{\\p}(\\eta)\n\t\t\t)\n\\end{align*}\n\nBy observation, it concludes this proof if\n\\begin{align*}\n\ta^{\\p}(\\eta) = g(\\eta)\n\\end{align*}\n\nIt can be proved. First let's define a function of $\\eta$\n\\begin{align*}\n\tI(\\eta)\n\t\t&=\n\t\t\t\\int\n\t\t\tp(y;\\eta)\n\t\t\tdy \t\t\t\t\t\t\t\t\t\\\\\n\t\t&=\n\t\t\t\\int\n\t\t\tb(y) \\exp ( \\eta \\, y - a(\\eta) )\n\t\t\tdy\n\\end{align*}\n\n$I(\\eta)$ is just the sum of the probabilities of all possible states, which\nmust be one, i.e.\n\\begin{center}\n\\begin{math}\n\tI(\\eta) = 1\n\\end{math}\n\\end{center}\n\nTake the derivative with respect to $\\eta$\n\\begin{align*}\n\tI^{\\p}(\\eta)\n\t\t&=\n\t\t\t\\int\n\t\t\t\\pd {} {\\eta}\n\t\t\tb(y) \\exp ( \\eta \\, y - a(\\eta) )\n\t\t\tdy \t\t\t\t\t\t\t\t\t\\\\\n\t\t&=\n\t\t\t\\int\n\t\t\ty \\, b(y) \\exp ( \\eta \\, y - a(\\eta) )\n\t\t\tdy \t\t\t\t\t\t\t\t\t\\\\\n\t\t&-\n\t\t\ta^{\\p}(\\eta)\n\t\t\t\\int\n\t\t\tb(y) \\exp ( \\eta \\, y - a(\\eta) )\n\t\t\tdy \t\t\t\t\t\t\t\t\t\\\\\n\t\t&=\n\t\t\tE[y]\n\t\t\t- a^{\\p}(\\eta) I(\\eta)\t\t\t\t\\\\\n\t\t&=\n\t\t\tg(\\eta)\n\t\t\t- a^{\\p}(\\eta) \t\t\t\t\t\t\\\\\n\t\t&=\n\t\t\t0\n\\end{align*}\n\nThe last step used the fact that the derivative of a constant number is zero.\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\end{document}\n", "meta": {"hexsha": "c7c8b128ed55485fe745c00a059c8d8e7b9e3db3", "size": 2739, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assignments/1/prob2.tex", "max_stars_repo_name": "qzmfranklin/cs299", "max_stars_repo_head_hexsha": "95dff2eaa41c80559b38870b18683fee678e0544", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-03-20T00:40:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-03-20T00:40:43.000Z", "max_issues_repo_path": "assignments/1/prob2.tex", "max_issues_repo_name": "qzmfranklin/cs299", "max_issues_repo_head_hexsha": "95dff2eaa41c80559b38870b18683fee678e0544", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assignments/1/prob2.tex", "max_forks_repo_name": "qzmfranklin/cs299", "max_forks_repo_head_hexsha": "95dff2eaa41c80559b38870b18683fee678e0544", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 19.2887323944, "max_line_length": 80, "alphanum_fraction": 0.5213581599, "num_tokens": 1009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581049086031, "lm_q2_score": 0.8056321866478979, "lm_q1q2_score": 0.6410077788816405}} {"text": "\\documentclass[english]{../thermomemo/thermomemo}\n% NOTE: You must pass either norsk or english as an option!\n\\usepackage[utf8]{inputenc}\n\n\\title{Introduction to thermodynamics}\n\\author{Morten Hammer, Geir Skaugen and Eskil Aursand}\n\n\\usepackage[normalem]{ulem}\n\n\\usepackage{hyperref}\n\\usepackage{color}\n\\usepackage{amsfonts}\n\n\\definecolor{midnightblue}{RGB}{35,35,132}\n\\definecolor{urlblue}{RGB}{70,130,180}\n\n\\definecolor{shadecolor}{gray}{0.9}\n\n\\hypersetup{\n colorlinks=true,\n linkcolor=midnightblue,\n urlcolor=urlblue,\n citecolor=midnightblue,\n linktoc=page\n}\n\n\\usepackage{amsmath}\n\\usepackage{hyperref}\n\\usepackage{framed}\n\\usepackage{siunitx,mhchem,todonotes}\n\\newcommand{\\pone}[3]{\\frac{\\partial #1}{\\partial #2}\\bigg|_{#3}}% partial\n % derivative with information of\n % constant variables\n\\newcommand*{\\vektor}[1]{\\boldsymbol{#1}}%\n\\newcommand{\\dd}[1]{\\mathrm{d}{#1}}\n\n\\DeclareMathOperator*{\\argmin}{arg\\,min }\n\n\n\n\\usepackage[activate={true,nocompatibility},final,kerning=true,tracking=true,spacing=true,stretch=10,shrink=10]{microtype}\n\\microtypecontext{spacing=nonfrench}\n\\SetExtraKerning[unit=space]\n {encoding={*}, family={bch}, series={*}, size={footnotesize,small,normalsize}}\n {\\textendash={400,400}, % en-dash, add more space around it\n \"28={ ,150}, % left bracket, add space from right\n \"29={150, }, % right bracket, add space from left\n \\textquotedblleft={ ,150}, % left quotation mark, space from right\n \\textquotedblright={150, }} % right quotation mark, space from left\n\\SetTracking{encoding={*}, shape=sc}{0}\n\n\\begin{document}\n\\frontmatter\n\n\\tableofcontents\n\n\\section{Introduction}\nThis document is ment as an introduction to thermodynamics for users\nand programmers in thermopack.\n\nIn addition to this memo, there exsist two memos written to document\nthermopack. These can be generated running pdflatex in the thermopack\ndocumentation folder ./doc/memo/.\n\nOtherwise the thermodynamics book by Callen \\cite{callen85} and by\nMichelsen \\cite{michelsen07} is recommended. Much of the work done on\nflash algorithms are based on the work by Michelsen\n\\cite{michelsen82a,michelsen82b,michelsen99}.\n\n\\section{Thermodynamic potentials}\n\\subsection{Internal energy}\nAs a starting point, one assumes a fundamental equation for the internal energy ($U$), with the extensive variables entropy ($S$), volume ($V$) \nand molar numbers ($\\vektor{n})$ as independent variables:\n\\begin{equation}\n U = U(S,V,\\vektor{n})\n \\label{}\n\\end{equation}\nUnder the assumption that $U$ is a \\textit{first order homogeneous function} of the extensive variables, one may use \n\\textit{Euler's theroem of homogeneous functions} to show that:\n\\begin{align}\n U(S,V,\\vektor{n}) &= S \\pone{U}{S}{V,\\vektor{n}}\n +V \\pone{U}{V}{S,\\vektor{n}}\n +\\sum_i n_i \\pone{U}{n_i}{S,V,n_j} \\nonumber \\\\\n &= ST - VP + \\sum_i n_i \\mu_i\n \\label{eq:U_euler}\n\\end{align}\nusing the definitions \n\\begin{equation}\n T \\equiv \\pone{U}{S}{V,\\vektor{n}}, \\quad P \\equiv \\pone{U}{V}{S,\\vektor{n}}, \\quad \\mu_i \\equiv \\pone{U}{n_i}{S,V,n_j}.\n \\label{}\n\\end{equation}\nThe differential of $U$ given its homogeneous property \\eqref{eq:U_euler} is \n\\begin{equation}\n \\dd{U} = S\\dd{T} + T\\dd{S} - V\\dd{P} - P\\dd{V} + \\sum_i n_i \\dd{\\mu_i} + \\sum_i \\mu_i \\dd{n_i},\n \\label{}\n\\end{equation}\nwhile generally the total (exact) differential of \\eqref{eq:U_euler}, given its independent variables, is\n\\begin{equation}\n \\dd{U} = T\\dd{S} - P\\dd{V} + \\sum_i \\mu_i \\dd{n_i},\n \\label{eq:U_totaldiff}\n\\end{equation}\nwhich combine into the \\textit{Gibbs-Duhem} relation for the intensive properties:\n\\begin{equation}\n S\\dd{T} - V\\dd{P} + \\sum_i n_i \\dd{\\mu_i} = 0.\n \\label{eq:gibbsduhem}\n\\end{equation}\n\n\\subsection{Helmholtz energy}\nThe Helmholtz energy is defined as\n\\begin{equation}\n A(T,V,\\vektor{n}) \\equiv U - TS,\n \\label{eq:helmholtz_def}\n\\end{equation}\nand by using \\eqref{eq:U_totaldiff}, one may show that\n\\begin{align}\n \\dd{A} = -S\\dd{T} - P\\dd{V} + \\sum_i \\mu_i \\dd{n_i},\n \\label{}\n\\end{align}\nand one may then see that\n\\begin{equation}\n S = -\\pone{A}{T}{V,\\vektor{n}}, \\quad P = -\\pone{A}{V}{T,\\vektor{n}}, \\quad \\mu_i = \\pone{A}{n_i}{T,V,n_j}.\n \\label{eq:A_differentials}\n\\end{equation}\n\n\n\\subsection{Gibbs energy}\nThe Gibbs energy is defined as \n\\begin{equation}\n G(T,P,\\vektor{n}) \\equiv U + PV - TS,\n \\label{eq:gibbs_def}\n\\end{equation}\nand by using \\eqref{eq:U_totaldiff}, one may show that\n\\begin{align}\n \\dd{G} = -S\\dd{T} + V\\dd{P} + \\sum_i \\mu_i \\dd{n_i}.\n \\label{eq:dG}\n\\end{align}\nand one may then see that\n\\begin{equation}\n S = -\\pone{G}{T}{P,\\vektor{n}}, \\quad V = \\pone{G}{P}{T,\\vektor{n}}, \\quad \\mu_i = \\pone{G}{n_i}{T,P,n_j}.\n \\label{}\n\\end{equation}\n\nCombining \\eqref{eq:gibbsduhem} with \\eqref{eq:dG} leads to \n\\begin{align}\n \\dd{G} = \\sum_i \\mu_i \\dd{n_i} + \\sum_i n_i \\dd{\\mu_i},\n \\label{}\n\\end{align}\nwhich can be recognized as a total differential for the alternative expression for G:\n\\begin{equation}\n G = \\sum_i \\mu_i n_i.\n \\label{eq:G_alt}\n\\end{equation}\n\n\n\\subsection{Enthalpy}\nThe Enthalpy is defined as \n\\begin{equation}\n H(S,P,\\vektor{n}) \\equiv U + PV,\n \\label{}\n\\end{equation}\nand by using \\eqref{eq:U_totaldiff}, one may show that\n\\begin{align}\n \\dd{H} = T\\dd{S} + V\\dd{P} + \\sum_i \\mu_i \\dd{n_i}.\n \\label{eq:dH}\n\\end{align}\nand one may then see that\n\\begin{equation}\n T = \\pone{H}{S}{P,\\vektor{n}}, \\quad V = \\pone{H}{P}{S,\\vektor{n}}, \\quad \\mu_i = \\pone{H}{n_i}{S,P,n_j}.\n \\label{}\n\\end{equation}\n\n\\section{Equilibrium in multi-phase systems with given (T,P)}\n\\subsection{Systems with given (T,P)}\n\\label{sec:eq_givenTP}\nTo characterize the equilibrium state, let us image that a small perturbation from equilibrium happens, and then focus on \nthe spontaneous process ($\\delta$) leading back to the equilibrium state. \nFor an isolated system, the second law of thermodynamics states that all spontaneous changes towards the equilibrium \nstate involve an increase in entropy:\n\\begin{equation}\n \\delta S_\\text{tot} \\geq 0\n \\label{eq:secondlaw}\n\\end{equation}\nwhere $\\delta$ indicates that it is not a change between different equilibrium states, as $\\mathrm{d}$ is used to indicate. \nWe are however not interested in an isolated system, but rather a system with given temperature and pressure. This may be though of as a sub-system of a total \nsystem, the latter satisfying \\eqref{eq:secondlaw}. The other part of the total system is the \\textit{reservoir}, which holds the given temperature and \npressure, and is large enough to force our system to keep the same values. Energy and volume (but not particles) \nmay be freely exchanged between the system (no subscript) and \nthe reservour (subscript r), in order to find the equilibrium of the total system. As entropy is an extensive quantity, the total entropy (which must increase) is \na sum of the entropy of the system and en entropy of the reservoir:\n\\begin{align}\n \\delta S + \\delta S_\\text{r} = \\delta S_\\text{tot} \\geq 0.\n \\label{eq:secondlaw_composite}\n\\end{align}\nDefining $\\delta Q$ as the heat transferred into the system, the change of entropy in the reservoir is given by\n\\begin{align}\n \\delta S_\\text{r} = - \\frac{\\delta Q}{T} = - \\left( \\frac{1}{T}\\delta U + \\frac{P}{T} \\delta V \\right),\n \\label{}\n\\end{align}\nby using the first and second laws of thermodynamics. The implication of \\eqref{eq:secondlaw_composite} is then that\n\\begin{align}\n T \\delta S \\geq \\delta U + P \\delta V.\n \\label{eq:TdS}\n\\end{align}\nThe change in Gibbs energy at constant temperature and pressure is \n\\begin{align}\n \\delta G &= \\delta \\left( U + PV - TS \\right) \\nonumber\\\\ \n &= \\delta U + P \\delta V + V \\delta P - T \\delta S - S \\delta T \\nonumber\\\\\n &= \\delta U + P \\delta V - T \\delta S,\n \\label{}\n\\end{align}\nwhich combined with \\eqref{eq:TdS} shows that\n\\begin{equation}\n \\delta G \\leq 0.\n \\label{}\n\\end{equation}\nIn words, this shows that a process which increases the entropy of the total system must decrease the Gibbs energy of a sub-system, given that the rest of the \ntotal system behaves like a $T,P$ reservoir. Given that this process characterizes returns from perturbations around the equilibrium state, the \nconclusion must be:\n\\begin{framed}\n \\noindent\n For a closed system of given temperature and pressure, the equilibrium condition is that its Gibbs energy ($G$) is at a minimum. \n\\end{framed}\n\n\\subsection{Multi-phase systems}\n\\label{sec:eq_multiphase}\nSo far the internal workings of the system have been ignored. Now we want to take into account that the system may consist of multiple phases, \nwhich essentially are sub-systems which are free to exchange energy, volume and particles.\nAs shown in Sec.~\\ref{sec:eq_givenTP}, the condition for equilibrium is that the Gibbs energy for the entire system is at an extremum with respect to perturbation. \nLet us assume that the perturbation is sufficiently small that the chemical potentials to not change, \\textit{i.e.} $\\delta \\mu_i \\approx 0$. This may also be seen as \nassuming that the perturbation is so small that $\\delta G \\approx \\dd{G}$, where the latter is given by \\eqref{eq:dG}.\nFor constant $T$ and $P$ this means that \n\\begin{align}\n \\delta G = \\sum_j \\delta G_j = \\sum_j \\sum_i \\mu_i^{(j)} \\delta n_i^{(j)} = 0,\n \\label{eq:dG_multiphase}\n\\end{align}\nwhere $j$ indicates the phase, and $i$ indicates the component. The conservation of matter, given no chemical reactions, \ngives that any change in the mole number of a component in an arbitrary phase $\\alpha$, must come from a net opposite change in the same \ncomponent in the other phases:\n\\begin{equation}\n \\delta n_i^{(\\alpha)} = - \\sum_{j\\neq \\alpha} \\delta n_i^{(j)}.\n \\label{eq:matter_cons}\n\\end{equation}\nCombining \\eqref{eq:dG_multiphase} and \\eqref{eq:matter_cons} leads to \n\\begin{equation}\n \\sum_{j\\neq \\alpha} \\sum_i \\left( \\mu_i^{(j)} - \\mu_i^{(\\alpha)} \\right) \\delta n_i^{(j)} = 0.\n \\label{}\n\\end{equation}\nGiven that the above must be true for any variation of the independent mole numbers, this means that \n\\begin{equation}\n \\mu_i^{(\\alpha)} = \\mu_i^{(j)} \\quad \\forall \\quad i,j\n \\label{}\n\\end{equation}\nwhich in words is stated as:\n\\begin{framed}\n \\noindent\n For a multi-phase system of given temperature and pressure, the equilibrium condition is that for each component, the chemical potential \n is equal in all phases.\n\\end{framed}\n\nAdditionally, there is a stability criterion for each phase, stated as:\n\\begin{framed}\n \\noindent\n For a multi-phase system of given temperature and pressure, the condition for phase stability is that the change in total Gibbs energy associated with \n the formation of a new infinitesimal phase within an existing phase is non-negative, for \\textit{any} composition of the new phase. \n This is called the \\textit{tangent plane condition}.\n\\end{framed}\nThe change in Gibbs energy with the formation of a new phase of $\\delta n$ moles and composition $w_i$, from an existing phase of composition $z_i$, is\n\\begin{align}\n \\delta G \n &= \\sum_i \\mu_i(\\vektor{z})\\left( - w_i \\delta n \\right) + \\sum_i \\mu_i(\\vektor{w})\\left( + w_i \\delta n \\right) \\nonumber\\\\\n &= \\delta n \\sum_i w_i \\left( \\mu_i(\\vektor{w}) - \\mu_i(\\vektor{z}) \\right), \n \\label{}\n\\end{align}\nwhich must be non-negative for any composition $\\vektor{w}$, for the existing phase composition $\\vektor{z}$:\n\\begin{align}\n \\sum_i w_i \\left( \\mu_i(\\vektor{w}) - \\mu_i(\\vektor{z}) \\right) \\geq 0.\n \\label{eq:tangentplane}\n\\end{align}\nWhen checking the stability of equilibrium phases, the chemical potential of each component is equal between the phases, and thus the left hand side of \n\\eqref{eq:tangentplane} is identical for each phase.\n\n\n\n\\section{Equilibrium in terms of an equation of state}\n\\subsection{The equation of state}\nAn \\textit{equation of state} is often given as an explicit pressure equation, such as \n\\begin{equation}\n P = P (T,V,\\vektor{n}).\n \\label{}\n\\end{equation}\nHowever, many equations of state is expressed as a Helmholtz energy\nfunction, \n\\begin{equation}\n A = A(T,V,\\vektor{n}).\n \\label{}\n\\end{equation}\nwhich is useful because all thermodynamical properties\nmay be derived by differentiating it.\n\nThe Helmholtz function of new multi--parameter GERG equations of state\n\\cite{kunz07,span96} and the older cubic equations of state is represented as a sum of \nan ideal contribution, and a departure (residual) contribution:\n\\begin{equation}\n\\label{eq:helmholtz}\nA(T,V,\\vektor{n}) = A^{\\text{ig}}(T,V,\\vektor{n}) + A^{\\text{r}}(T,V,\\vektor{n})\n\\end{equation}\n\nOften a reduced (dimensionless) form of the residual Helmholtz energy is used,\n\\begin{equation}\n\\label{eq:reduced_helmholtz}\nF(T,V,\\vektor{n}) = a \\equiv \\frac{A^\\text{r}(T,V,\\vektor{n})}{nRT}\n\\end{equation}\n\n\\subsection{The residual Helmholtz energy}\n\nThe relationship between pressure and the Helmholtz energy two forms is:\n\\begin{equation}\n P(T,V,\\vektor{n}) = -\\pone{A}{V}{T,\\vektor{n}} \n = \\frac{nRT}{V} - \\pone{A^{\\text{r}}(T,V,\\vektor{n})}{V}{T,\\vektor{n}}.\n \\label{}\n\\end{equation}\nAn expression for $A^\\text{r}$ may be found by integrating to the limit of infinite volume at constant temperature and mole numbers:\n\\begin{equation}\n A^\\text{r}(T,V,\\vektor{n}) \n = A^\\text{r}(T,\\infty,\\vektor{n}) + \\int_\\infty^V \\left[ P^\\text{ig}(T,V^\\prime,\\vektor{n}) - P(T,V^\\prime,\\vektor{n}) \\right]\\dd{V^\\prime},\n \\label{}\n\\end{equation}\nwhich when using that real fluids behave as an ideal gas in the limit of zero pressure ($V \\to \\infty$) leads to the expression:\n\\begin{framed}\n\\begin{align}\n A^\\text{r}(T,V,\\vektor{n}) \n = \\int^\\infty_V \\left[ P(T,V^\\prime,\\vektor{n}) - \\frac{nRT}{V^\\prime} \\right]\\dd{V^\\prime}\n \\label{eq:helmholtz_int}\n\\end{align}\n\\end{framed}\n\nBe aware that residual quantities will depend on which set of variables they are represented with. As an example, let's show \nhow this works out for the residual Helmholtz energy:\nFrom the state $(T,V,P,\\vektor{n})$ by definition being a real state, we know that \n\\begin{equation}\n A(T,V,\\vektor{n}) = A(T,P,\\vektor{n}).\n \\label{eq:Aequal}\n\\end{equation}\nWhile the value of $A$ is independent of representation, the distribution of it between the ideal \ncontribution and the residual is not. This is so because \nthe state $(T,V,P,\\vektor{n})$ is not in \ngeneral a valid state of an ideal gas, so \ngenerally we have that\n\\begin{equation}\n A^\\text{ig}(T,V,\\vektor{n}) \\neq A^\\text{ig}(T,P,\\vektor{n}).\n \\label{}\n\\end{equation}\n\nUsing an arbitrary valid ideal gas state at temperature $T$, $(T,V_0,P_0,\\vektor{n})$, we \nhave by definition of this state that \n\\begin{equation}\n A^\\text{ig}(T,P_0,\\vektor{n}) = A^\\text{ig}(T,V_0,\\vektor{n})\n \\label{eq:Aequal_ideal}\n\\end{equation}\n\nCombining \\eqref{eq:helmholtz}, \\eqref{eq:Aequal}, \\eqref{eq:Aequal_ideal} and the ideal gas law \nleads to the following calculation \nfor the difference in residuals between variable representations:\n\\begin{align}\n A^\\text{r}(T,V,\\vektor{n}) - A^\\text{r}(T,P,\\vektor{n}) \n &= A^\\text{ig}(T,P,\\vektor{n}) - A^\\text{ig}(T,V,\\vektor{n})\n \\nonumber \\\\\n &=\\left[A^\\text{ig}(T,P,\\vektor{n}) - A^\\text{ig}(T,P_0,\\vektor{n})\\right]\n -\n \\left[A^\\text{ig}(T,V,\\vektor{n}) - A^\\text{ig}(T,V_0,\\vektor{n})\\right]\n \\nonumber\\\\\n &=\\int_{P_0}^P \n \\pone{A^\\text{ig}}{P}{T,\\vektor{n}}\n \\dd{P}\n -\n \\int_{V_0}^V \n \\pone{A^\\text{ig}}{V}{T,\\vektor{n}}\n \\dd{V}\n \\nonumber\\\\\n &= nRT \\ln\\left( \\frac{PV}{P_0 V_0} \\right) \n = nRT \\ln\\left( \\frac{PV}{nRT} \\right) \n = nRT \\ln\\left( Z \\right),\n \\label{}\n\\end{align}\nwhere $Z$ is the compressibility factor, defined by\n\\begin{equation}\n Z \\equiv \\frac{PV}{nRT} = 1 - \\frac{V}{nRT} \\pone{A^\\text{r}(T,V,\\vektor{n})}{V}{T,\\vektor{n}} .\n \\label{eq:Z_def}\n\\end{equation}\n\nThus we have the following expression for the residual Helmholtz free energy expressed through the \nvariables $(T,P,\\vektor{n})$:\n\\begin{equation}\n\\label{eq:helmholtz_int_TPN}\nA^{\\text{r}}(T,P,\\vektor{n}) = \\int_V^\\infty \\left(P(T,V^\\prime,\\vektor{n}) - \\frac{nRT}{V^\\prime}\\right) \\dd{V^\\prime} \n-nRT \\ln\\left( Z \\right)\n\\end{equation}\n\n\n\n\n\\subsection{Fugacity}\nFor a pure ideal gas, the change in chemical potential corresponding to a change in pressure, at constant temperature, is give by \n\\begin{equation}\n \\frac{P}{P_0} = \\exp\\left( \\frac{\\mu^\\text{ig}(T,P) - \\mu^\\text{ig}(T,P_0) }{RT} \\right).\n \\label{}\n\\end{equation}\nReal fluids do not obey this relation, but one may define an ``effective pressure'' called \n\\textit{fugacity} ($f$) such that the following relation holds:\n\\begin{equation}\n \\frac{f(T,P)}{P_0} = \\exp\\left( \\frac{\\mu(T,P) - \\mu^\\text{ig}(T,P_0) }{RT} \\right).\n \\label{}\n\\end{equation}\nThe reference pressure is arbitrary, so one may set $P_0=P$ to make an expression only in terms of the deviation from ideal gas:\n\\begin{equation}\n \\frac{f(T,P)}{P} = \\exp\\left( \\frac{\\mu(T,P) - \\mu^\\text{ig}(T,P) }{RT} \\right),\n \\label{}\n\\end{equation}\nwhich shows that for ideal gases, $f(T,P)=P$.\n\n\n\n\\subsubsection{Ideal gas mixtures}\nAn ideal gas mixture is defined as a mixture which has the following expression for the Helmholtz free energy:\n\\begin{equation}\n A^\\text{ig}(T,V,\\vektor{n}) \\equiv \\sum_i n_i \\left( \n \\mu_i^\\text{ig}(T,P_0) + RT\\ln \\frac{n_i RT}{P_0 V} - RT\n \\right)\n \\label{eq:def_idealmix}\n\\end{equation}\nwhere $n_i$ is the mole number of component $i$, and\n$\\mu^\\text{ig}_i(T,P)$ is the chemical potential of component $i$ in its pure form \nat the given temperature and pressure. This definition is consistent with the ideal gas law for the mixture as a whole, \n$PV=nRT$, where $n=\\sum n_i$.\nFrom \\eqref{eq:def_idealmix} one may find the Gibbs free energy of the \nideal gas mixture as\n\\begin{align}\n G^\\text{ig}(T,P,\\vektor{n}) \n &= A^\\text{ig}(T,V,\\vektor{n}) + \\left( PV \\right)^\\text{ig} = A^\\text{ig}(T,V,\\vektor{n}) + nRT \\nonumber\\\\\n &= \\sum_i n_i \\left( \n \\mu_i^\\text{ig}(T,P_0) + RT\\ln \\frac{x_i P}{P_0}\n \\right)\n \\label{eq:G_idealmix}\n\\end{align}\nwhere $x_i \\equiv n_i/n$ is the \\textit{mole fraction} of component $i$.\n\nFrom \\eqref{eq:def_idealmix} and \\eqref{eq:G_idealmix} \none may find the chemical potential, in terms of $V$ or $P$, of component $i$ in the ideal mixture as\n\\begin{align}\n \\mu_i^\\text{ig}(T,V,\\vektor{n}) &\\equiv \\pone{A^\\text{ig}}{n_i}{T,V} = \\mu_i^\\text{ig}(T,P_0) + RT\\ln \\frac{n_i RT}{P_0 V} \\nonumber \\\\\n \\mu_i^\\text{ig}(T,P,\\vektor{n}) &\\equiv \\pone{G^\\text{ig}}{n_i}{T,P} = \\mu_i^\\text{ig}(T,P_0) + RT\\ln \\frac{x_i P}{P_0},\n \\label{eq:mu_ig}\n\\end{align}\nwhich are equal if the state $(T,V,P,\\vektor{n})$ is a valid ideal gas state ($PV=nRT$), as expected. Combining the two \nequations in \\eqref{eq:mu_ig}, while setting $P_0=P$, leads the the following useful expression:\n\\begin{equation}\n \\mu_i^\\text{ig}(T,V,\\vektor{n}) - \\mu_i^\\text{ig}(T,P,\\vektor{n})\n =-RT\\ln \\left( Z \\right)\n \\label{eq:mu_ig_PVdiff}\n\\end{equation}\nwhere if $(T,V,P,\\vektor{n})$ is a valid state for an ideal gas, \n$Z=1$, and the right hand side reduces to zero as expected.\n\nThe last equation in \\eqref{eq:mu_ig} may be rearranged to form the inspiration for the definition of \nreal mixture fugacity in the next section:\n\\begin{equation}\n \\frac{P_i}{P_0} = \\exp\\left( \\frac{\\mu^\\text{ig}_i(T,P,\\vektor{n}) - \\mu^\\text{ig}_i(T,P_0)}{RT} \\right)\n \\label{eq:p_mu_ideal}\n\\end{equation}\nwhere $P_i=x_iP$ is the \\textit{partial pressure} of component $i$. For $P_0 = P$ this reduces to \n\\begin{equation}\n \\mu^\\text{ig}_i(T,P,\\vektor{n}) - \\mu^\\text{ig}_i(T,P) = RT\\ln x_i\n \\label{eq:mu_and_xi_idealmix}\n\\end{equation}\n\n\\subsubsection{Real mixtures}\n\\label{sec:realmixtures}\nSimilar to the case of pure components, one may define an ``effective partial pressure'' of a component in a mixture, the \ncomponent fugacity ($f_i$), such that the following relation similar to \\eqref{eq:p_mu_ideal} holds:\n\\begin{equation}\n \\frac{f_i(T,P,\\vektor{n})}{P_0} \\equiv \\exp\\left( \\frac{\\mu_i(T,P,\\vektor{n}) - \\mu_i^\\text{ig}(T,P_0)}{RT} \\right),\n \\label{eq:fug_def_mix}\n\\end{equation}\nwhere one may again set $P_0=P$ to state it in terms of deviations from a pure ideal gas:\n\\begin{equation}\n \\frac{f_i(T,P,\\vektor{n})}{P} \\equiv \\exp\\left( \\frac{\\mu_i(T,P,\\vektor{n}) - \\mu_i^\\text{ig}(T,P)}{RT} \\right),\n \\label{eq:fug_def_mix_P0eqP}\n\\end{equation}\n\nThe chemical potential of a component in a real mixture may be decomposed into an ideal gas part and a residual part:\n\\begin{align}\n \\mu_i(T,V,\\vektor{n}) &= \\mu_i^\\text{ig}(T,V,\\vektor{n}) + \\mu_i^\\text{r}(T,V,\\vektor{n}) \\nonumber\\\\\n \\mu_i(T,P,\\vektor{n}) &= \\mu_i^\\text{ig}(T,P,\\vektor{n}) + \\mu_i^\\text{r}(T,P,\\vektor{n})\n \\label{}\n\\end{align}\nwhich may be combined with \\eqref{eq:mu_ig_PVdiff} to show that \n\\begin{equation}\n \\mu_i^\\text{r}(T,V,\\vektor{n}) - \\mu_i^\\text{r}(T,P,\\vektor{n})\n =RT\\ln \\left( Z \\right)\n \\label{eq:mu_r_PVdiff}\n\\end{equation}\n\nCombining \\eqref{eq:fug_def_mix_P0eqP} and \\eqref{eq:mu_and_xi_idealmix}, one finds that\n\\begin{equation}\n \\frac{f_i(T,P,\\vektor{n})}{x_i P} \\equiv\n \\exp\\left( \\frac{\\mu_i(T,P,\\vektor{n}) - \\mu_i^\\text{ig}(T,P,\\vektor{n})}{RT} \n \\right)\n \\label{eq:fugcoeff_vs_mu}\n\\end{equation}\n\nThe \\textit{fugacity coefficient} may now be defined, calculated using \\eqref{eq:fugcoeff_vs_mu} \nand \\eqref{eq:mu_r_PVdiff} as follows:\n\\begin{align}\n \\phi_i(T,P,\\vektor{n}) \\equiv \\frac{f_i(T,P,\\vektor{n})}{x_i P} &= \n \\exp\\left( \\frac{\\mu_i(T,P,\\vektor{n}) - \\mu_i^\\text{ig}(T,P,\\vektor{n})}{RT}\\right)\n \\nonumber\\\\\n &= \\exp\\left( \\frac{\\mu_i^\\text{r}(T,P,\\vektor{n})}{RT} \\right)\n \\nonumber\\\\\n &= \\exp\\left( \\frac{\\mu_i^\\text{r}(T,V,\\vektor{n})}{RT} - \\ln\\left( Z \\right) \\right)\n \\nonumber\\\\\n &= \\exp\\left( \\frac{1}{RT}\\pone{A^\\text{r}(T,V,\\vektor{n})}{n_i}{T,V,n_j} - \\ln\\left( Z \\right) \\right)\n \\label{eq:fugcoeff}\n\\end{align}\nwhere $A^\\text{r}(T,V,\\vektor{n})$ may be found from \\eqref{eq:helmholtz_int}.\n\nAs shown in Sec.~\\ref{sec:eq_multiphase}, the condition of phase equilibrium is that for each component, \nthe chemical potential must be equal in all phases. Using \\eqref{eq:fug_def_mix_P0eqP}, and the fact that \n$\\mu_i^\\text{ig}(T,P)$ is independent of phase, the condition may be restated in the more useful form:\n\\begin{framed}\n \\noindent\n For a multi-phase system of given temperature and pressure, the equilibrium condition is that for each component $i$, the quantity \n $x_i \\phi_i$ is equal in all phases. The fugacity coefficient may be calculated from the equation of state through \n \\eqref{eq:fugcoeff} and \\eqref{eq:Z_def}.\n\\end{framed}\n\n\\section{TP Flash}\n\n\\subsection{General equations of phase equilibrium and stability}\nAs shown in Sec.~\\ref{sec:eq_multiphase} and~\\ref{sec:realmixtures}, the criterion for equilibrium in a multi-phase system at given temperature \nand pressure is that for each component, the fugacity is equal in all phases.\n\\begin{equation}\n f_i^{(\\alpha)} = f_i^{(j)} \\quad \\forall \\quad i,j,\\alpha\n \\label{}\n\\end{equation}\n\nAdditionally, there is a phase stability criterion, stated as the \\textit{tangent plane condition} in Sec.~\\ref{sec:eq_multiphase}.\nLet us define a \\textit{tangent plane distance function} for the composition $\\vektor{z}$, a function of a trial composition $\\vektor{w}$:\n\\begin{equation}\n \\mathit{TPD}_z(\\vektor{w}) \\equiv \\sum_{i} w_i \\left( \\mu_i(\\vektor{w}) - \\mu_i(\\vektor{z}) \\right).\n \\label{}\n\\end{equation}\nIt is often conventient to use a \n\\textit{reduced tangent plane distance function}\n\\begin{align}\n \\mathit{tpd}_z(\\vektor{w}) &\\equiv \\frac{\\mathit{TPD}(\\vektor{w})}{RT} \n \\nonumber\\\\\n &= \\sum_{i} w_i \\left[ \\ln w_i + \\ln \\phi_i(\\vektor{w}) - \\ln z_i - \\ln \\phi_i(\\vektor{z}) \\right]\n \\nonumber\\\\\n &= \\sum_{i} w_i \\left[ \\ln w_i + \\ln \\phi_i(\\vektor{w}) - d_i(\\vektor{z}) \\right],\n \\label{}\n\\end{align}\nwhere $d_i \\equiv \\ln z_i + \\ln \\phi_i(\\vektor{z}) $\n\nThe phase stability criterion is then\n\\begin{equation}\n \\mathit{tpd}_z(\\vektor{w}) \\geq 0 \\quad \\text{for any valid composition $\\vektor{w}$}\n \\label{}\n\\end{equation}\nand this may be shown to be both a necessary and sufficient condition for phase stability \\cite{michelsen07}. \nIn practice, ensuring that $\\mathit{tpd}_z(\\vektor{w})$ is non-negative everywhere in compositional space is done by finding its \nminima, and then checking if it is negative at these locations. Finding the minima of $\\mathit{tpd}_z(\\vektor{w})$, subject to the constraint of \n$\\sum w_i = 1$, may be solved using the \\textit{method of Lagrange multipliers}.\n\nModified (unconstrained) formulation:\n\\begin{align}\n \\mathit{tm}_z(\\vektor{W}) &= 1 + \\sum_{i} W_i \\left[ \\ln W_i + \\ln \\phi_i(\\vektor{W}) - d_i(\\vektor{z}) -1\\right]\n \\nonumber\\\\\n &= \\left( 1 - W_T + W_T \\ln W_T \\right) + W_T \\mathit{tpd}_z(\\vektor{w})\n \\label{eq:tm}\n\\end{align}\nwhere $W_i = w_i W_T$ ($W_T \\equiv \\sum W_i $) are mole numbers, not composition, and are thus not constrained to sum to one.\nAt stationary points of $\\mathit{tm}_z(\\vektor{W})$ with respect to $W_i$:\n\\begin{equation}\n \\ln W_i^\\text{SP} + \\ln \\phi_i(\\vektor{W}^\\text{SP}) - d_i(\\vektor{z}) = 0 \\quad \\forall \\quad i\n \\label{eq:tm_stationary}\n\\end{equation}\n\\begin{equation}\n \\mathit{tm}_z(\\vektor{W}^\\text{SP}) = 1 - W_T^\\text{SP} \n \\label{}\n\\end{equation}\n\nIt may be shown~\\cite[Ch.~9]{michelsen07} that the stationary points of\n$\\mathit{tm}_z(\\vektor{W})$ are also valid stationary points for the\nconstrained function $\\mathit{tpd}_z(\\vektor{w})$, and\n$\\mathit{tpd}_z(\\vektor{w}^\\text{SP})$ is positive at these minima if and only\nif $W_T^\\text{SP} \\leq 1$, \\textit{i.e.} if\n$\\mathit{tm}_z(\\vektor{W}^\\text{SP}) \\geq 0$. The procedure to check stability\ncan then be formulated as:\n\\begin{framed}\n\\begin{enumerate}\n \\item Find the stationary points of $\\mathit{tm}_z(\\vektor{W})$ with respect to $\\vektor{W}$, \\textit{i.e} solutions of \\eqref{eq:tm_stationary}.\n \\item Check if $W_T^\\text{SP} \\leq 1$ at all the stationary points found. If not, the mixture $\\vektor{z}$ is unstable.\n\\end{enumerate}\n\\end{framed}\nEq.~\\eqref{eq:tm_stationary} may be solved by \\textit{successive substitution}\nor \\textit{Newton's method}. As discussed by Michelsen \\cite{michelsen07},\nsuccessive substitution will converge in most case but may suffer from a low\nrate of convergence, giving merit to the use of Newton's method. \n\nA composition $\\vektor{z}$ is metastable as long as the Hessian matrix \n$H(\\vektor{z})$ \nof $\\mathit{tm}_z(\\vektor{W})$ is positive definite. \nIf not, the state is \\textit{intrinsically unstable}. \nThe limit between the two cases is the \n\\textit{stability limit / spinodal line}.\n\nHow to uncover all negative minima of $\\mathit{tm}_z(\\vektor{W})$ (if any)?\nThis cannot be done with total certainty, but one may use some clever initial\nestimates to perform a finite set of searches which are likely to uncover\nthem. \n\n\\subsection{Tangent plane stability analysis}\n\nTangent plane stability analysis serves two purposes:\n\\begin{itemize}\n \\item Verifies that a single phase state is stable.\n \\item If the single phase state is not stable, gives an improved K-factor guess.\n\\end{itemize}\nAs shown, a phase of composition $\\vektor{z}$ is stable at the specified temperature and pressure \\textit{if and only if}\n$tpd_z(\\vektor{w}) \\geq 0$ for any trial phase composition $\\vektor{w}$. This could checked by inspecting the minima of \n$\\mathit{tm}_z(\\vektor{W})$ \\eqref{eq:tm} and checking the criterion\n$W_T^\\text{SP} \\leq 1$. \n\nThe original composition is \\textit{intrinsically unstable} if the Hessian \n\\begin{equation}\n H_{ij} \\equiv \\frac{\\partial^2 \\mathit{tm}_z}{\\partial W_i \\partial W_j} \n = \\frac{\\delta_{ij}}{W_i} + \\frac{\\partial \\ln \\phi_i}{\\partial W_j}\n \\label{}\n\\end{equation}\nhas any negative eigenvalues at the trivial stationary point at $\\vektor{W}=\\vektor{z}$. If it has no negative eigenvalues at \nthis point, the mixture is either stable ($\\mathit{tm}_z(\\vektor{W}) > 0$ at all its minima) or \n\\textit{metastable} ($\\mathit{tm}_z(\\vektor{W}) < 0$ at some minimum where $\\vektor{W}\\neq \\vektor{z}$).\n\nBy using \\eqref{eq:tm_stationary}, minima of $\\mathit{tm}_z(\\vektor{W})$ may be found by successive substitution:\n\\begin{equation}\n \\ln W_i^{k+1} = d_i(\\vektor{z}) - \\ln \\phi_i \\left( \\vektor{W}^k \\right),\n \\label{}\n\\end{equation}\nwhich may be shown to only converge to minima, not other stationary points. Slow convergence may be remedied by higher order \nmethods.\nOutcomes of stability analysis:\n\\begin{itemize}\n \\item A composition $\\vektor{W}$ where $\\mathit{tm}_z(\\vektor{W}) < 0$ is found. Continue the flash calculation with $\\vektor{w}$ \n as a new phase.\n \\item All attempts converge to either the trivial solution of a positive minimum. The composition $\\vektor{z}$ is thus stable.\n\\end{itemize}\n\n\\subsection{Mixture reduced molar Gibbs energy}\nBy combining \\eqref{eq:G_alt}, \\eqref{eq:fugcoeff_vs_mu} and \\eqref{eq:mu_and_xi_idealmix}, one may show that \nthe Gibbs energy of a multi-phase mixture is given by \n\\begin{align}\n G &= \\sum_j \\sum_i n_i^{(j)} \\mu_i^{(j)} \\nonumber\\\\\n &= \\sum_j\\sum_i n_i^{(j)} \\left[ \\mu_i^\\text{ig}(T,P,\\vektor{n}) + RT\\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right] \\nonumber\\\\\n &= \\sum_j\\sum_i n_i^{(j)} \\left[ \\mu_i^\\text{ig}(T,P) + RT\\ln x_i^{(j)} + RT\\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right] \\nonumber\\\\\n &= \\sum_j\\sum_i n_i^{(j)} \\mu_i^\\text{ig}(T,P) + \\sum_j \\sum_i n_i^{(j)} \\left[ RT\\ln x_i^{(j)} + RT\\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right] \\nonumber\\\\\n &= \\sum_i \\left( \\sum_j n_i^{(j)} \\right) \\mu_i^\\text{ig}(T,P) + \n \\sum_j \\sum_i n_i^{(j)} \\left[ RT\\ln x_i^{(j)} + RT\\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right] \\nonumber\\\\\n &= \\sum_i n_i^{(\\text{tot})} \\mu_i^\\text{ig}(T,P) + \n \\sum_j \\sum_i n_i^{(j)} \\left[ RT\\ln x_i^{(j)} + RT\\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right] \n \\label{eq:G_mix}\n\\end{align}\n\nWhen wanting to compare the total Gibbs energy of different phase configurations possible from the same overall composition, \nthe first term in \\eqref{eq:G_mix} will cancel due to being identical in all configurations. It is thus not necessary to consider \nit further for these purposes. Denoting this term by $G_0 \\equiv \\sum_i n_i^{(\\text{tot})} \\mu_i^\\text{ig}(T,P)$, one may \ndefine a reduced molar Gibbs energy useful for comparing different phase configurations from the same feed by \n\\begin{align}\n g &\\equiv \\frac{G - G_0 }{nRT} \\nonumber\\\\\n &= \\sum_i \\sum_j \\beta_j x_i^{(j)} \\left[ \\ln x_i^{(j)} + \\ln \\phi_i^{(j)}(T,P,\\vektor{n})\\right],\n \\label{eq:g_func}\n\\end{align}\nwhere $\\beta_j$ is the fraction of moles in phase $j$.\n\n\n\\subsection{Two-phase flash}\nConsider a mixture of $C$ components, of overall composition $\\vektor{z}$, and define the liquid and vapor compositions as \n$\\vektor{x}$ and $\\vektor{y}$, respectively. When indices are used instead of vectors, the index $i$ indicates the component.\nIt is assumed that we have a thermodynamic model for the mixture, capable of calculating the fugacity of any component, given a set of \ntemperature, pressure and composition.\n\nThe equilibrium condition is \n\\begin{equation}\n f_i^l = f_i^v \\quad \\forall \\quad i\n \\label{eq:eq_twophase}\n\\end{equation}\n\nDefine $\\beta$ as the overall mole fraction of the vapor phase, giving the material balance:\n\\begin{equation}\n \\beta y_i + (1-\\beta)x_i = z_i \\quad \\forall \\quad i\n \\label{eq:materialbalance}\n\\end{equation}\n\nThe mole fractions must sum to unity, a condition which may be transformed to the convenient form:\n\\begin{equation}\n \\sum_{i=1}^C \\left( x_i - y_i \\right) = 0.\n \\label{eq:mole_frac_sum}\n\\end{equation}\nThe above equations yield $2C + 1$ relations, while the number of variables ($\\vektor{x}$, $\\vektor{y}$, $T$, $P$, $\\beta$) are $2C + 3$. \nThis shows that two additional pieces of information are needed, such as specifying $(T,P)$.\n\n\n\\textit{Equilibrium factors}:\n\\begin{equation}\n K_i \\equiv \\frac{y_i}{x_i}\n \\label{eq:K_factors}\n\\end{equation}\n\n\nCombining \\eqref{eq:K_factors} and \\eqref{eq:materialbalance}\n\\begin{equation}\n x_i = \\frac{z_i}{1 - \\beta + \\beta K_i}, \\quad\\quad y_i = K_i x_i = \\frac{K_i z_i}{1 - \\beta + \\beta K_i} \n \\label{eq:xy_and_Kbeta}\n\\end{equation}\n\nAt equilibrium, where \\eqref{eq:eq_twophase} is satisfied, we have that \n\\begin{equation}\n K_i \\equiv \\frac{y_i}{x_i} = \\frac{\\phi_i^l}{\\phi_i^v}\n \\label{eq:K_eq}\n\\end{equation}\n\nCombining \\eqref{eq:mole_frac_sum} with \\eqref{eq:xy_and_Kbeta} leads to the \\textit{Rachford-Rice} equation:\n\\begin{equation}\n g(\\beta) \\equiv \\sum_{i=1}^C \\left( x_i - y_i \\right) = \\sum_{i=1}^C \\frac{z_i \\left( K_i -1 \\right)}{1 -\\beta + \\beta K_i} = 0\n \\label{eq:rachford_rice}\n\\end{equation}\n\\begin{framed}\nTwo-phase TP flash by successive substitution: \n\\begin{enumerate}\n \\item Use current (or initial) $\\vektor{K}$ in the Rachford-Rice equation \\eqref{eq:rachford_rice}.\n \\item Solve the Rachford-Rice equation for a value of $\\beta$ \\textit{e.g.} by Newton's method.\n \\item Find compositions $x_i$ and $y_i$ corresponding to the above values of $K_i$ and $\\beta$, through \\eqref{eq:xy_and_Kbeta}.\n \\item Find fugacity coefficients from the equation of state given the above compositions, using \\textit{e.g.} \\eqref{eq:fugcoeff}.\n \\item Find a new equilibrium factor $\\vektor{K}$ using the above fugacity coefficients and the equilibrium assunption \\eqref{eq:K_eq}.\n \\item Go to 1 if $\\vektor{K}$ has not converged yet.\n\\end{enumerate}\n\\end{framed}\nEqual fugacities may also be satisfied at ``false solutions'', which are critical points of the Gibbs energy, but not minima.\nFortunately, it may be shown that successive substitution can only converge to minima of the Gibbs energy~\\cite{michelsen07}.\nConvergence will be slow at high pressures or close to the mixture critical point.\n\nInitial values for $\\vektor{K}$:\nWilson's approximation:\n\\begin{equation}\n \\ln K_i = \\ln \\left( \\frac{P_{c_i}}{P} \\right) + 5.373(1+\\gamma_i) \\left( 1-\\frac{T_{c_i}}{T} \\right),\n \\label{eq:wilson_K}\n\\end{equation}\nwhere $P_{c_i}$, $T_{c_i}$ and $\\gamma_{i}$ is the critical temperature, critical pressure and acentric factor, respectively, \nof component $i$.\n\nSingle phase solution?\\\\\nConvergence at $\\beta <0$ or $\\beta>0$, or convergence to the trivial solution $\\vektor{x} = \\vektor{y}$.\\\\\n\nFlash strategy:\n\\begin{enumerate}\n \\item Find Wilson K-factors \\eqref{eq:wilson_K} to use as initial estimates.\n \\item Do three steps of successive substitution. Check for conditions:\n \\begin{itemize}\n \\item Total Gibbs energy of the resulting liquid-vapor system is lower than in the feed (assuming single phase). This verifies the presence of \n multiple phases. Continue iteration, perhaps with higher order methods.\n \\item If not the above, check if $\\mathit{tpd}_z(\\vektor{x})$ or $\\mathit{tpd}_z(\\vektor{y})$ is negative. If the former, set \n $K_i = \\phi_i^x / \\phi_i^z$. If the latter, set $K_i = \\phi_i^z / \\phi_i^y$. Continue iteration.\n \\end{itemize}\n \\item If neither of the above conditions were met after three iterations, do tangent plane stability analysis.\n\\end{enumerate}\nIf at any time $\\beta$ exceeds its bounds, it is very likely that the specification corresponds to a single phase state. \nTo be certain, perform tangent plane stability analysis.\n\n\nInitializing $\\vektor{W}$ for tangent plane analysis: \nInvestigate both ``liquid-like'' and ``vapor-like'' trial compositions, through \nWilson K-factors \\eqref{eq:wilson_K}. If the feed itself may with confidence be said to be ``liquid-like'' or ``vapor-like'', \none may to save time only investigate the trial composition of the opposite kind.\n\n\n\n\\subsubsection{Solution by minimization}\nWhen formulating an equilibrium calculation as a minimization problem, it \nmay be useful to use \\textit{molar amounts} as the independent variables, \ndefined by \n\\begin{align}\n v_i &\\equiv \\beta y_i\\nonumber\\\\\n l_i &\\equiv (1-\\beta) x_i,\n \\label{}\n\\end{align}\nsuch that\n\\begin{equation}\n v_i + l_i = z_i,\n \\label{}\n\\end{equation}\nwhere $\\vektor{z}$ is now molar amounts, but numerically identical to the feed composition.\n\n\\begin{equation}\n y_i = \\frac{v_i}{\\sum v_i}, \\quad x_i = \\frac{l_i}{\\sum l_i}\n \\label{}\n\\end{equation}\n\n\\begin{equation}\n \\beta_L \\equiv 1-\\beta\n \\label{}\n\\end{equation}\n\n\\begin{equation}\n \\beta = \\sum v_i\n \\label{}\n\\end{equation}\n\n\\begin{equation}\n \\beta_L = \\sum l_i\n \\label{}\n\\end{equation}\n\nThe minimization problem to be solved is then \n\\begin{equation}\n \\vektor{v} = \\underset{\\vektor{v} \\in \\mathbb{R}^C}{\\argmin}\\left\\{g(\\vektor{v},\\vektor{l}(\\vektor{v}))\\right\\}, \\quad\\text{with } \n \\vektor{l} = \\vektor{z}-\\vektor{v}\n \\label{}\n\\end{equation}\n\\textit{e.g.}\\ by Newton's method of optimization (ref appendix).\n\n\nBy using \\eqref{eq:g_func} for the case of two-phase systems:\n\\begin{equation}\n g = \\sum_i \\left( v_i \\left[ \\ln y_i + \\ln \\phi_i^v \\right] + l_i \\left[ \\ln x_i + \\ln \\phi_i^l \\right] \\right)\n \\label{eq:g_twophase}\n\\end{equation}\n\n\\begin{align}\n \\frac{\\partial g}{\\partial v_j}\n =& \\ln y_j - \\ln x_j + \\ln \\phi_j^v - \\ln \\phi_j^l \\nonumber\\\\\n &+ 1 - \\underbrace{\\sum_i y_i}_{=1} + \\underbrace{\\sum_i v_i \\frac{\\partial \\ln \\phi_i^v}{\\partial v_j}}_{=0}\\nonumber\\\\\n &- 1 + \\underbrace{\\sum_i x_i}_{=1} - \\underbrace{\\sum_i l_i \\frac{\\partial \\ln \\phi_i^l}{\\partial l_j}}_{=0} \\nonumber\\\\\n =& \\ln y_j - \\ln x_j + \\underbrace{\\ln \\phi_j^v - \\ln \\phi_j^l}_{\\text{From EoS}},\n \\label{}\n\\end{align}\nwhere the sums involving derivatives are zero due to \\textit{Euler's theroem} for homogeneous functions, as explained in\n1.8 of~\\cite{michelsen07}.\n\n\\begin{align}\n \\frac{\\partial^2 g}{\\partial v_i \\partial v_j} \n &= \\frac{\\partial \\ln \\phi_j^v}{\\partial v_i} - \\frac{\\partial \\ln \\phi_j^l}{\\partial v_i} \n - \\left( \\frac{\\beta + \\beta_L}{\\beta \\beta_L} \\right)\n + \\delta_{i,j} \\left( \\frac{z_i}{x_i y_i \\beta \\beta_L} \\right) \\nonumber\\\\\n &= \n \\frac{1}{\\beta \\beta_L} \\left[\n \\beta_L \\underbrace{\\left( \\sum_i v_i\\right) \\frac{\\partial \\ln \\phi_j^v}{\\partial v_i} }_\\text{From EoS}\n + \\beta \\underbrace{\\left( \\sum_i l_i\\right) \\frac{\\partial \\ln \\phi_j^l}{\\partial l_i}}_\\text{From EoS}\n - 1\n + \\delta_{i,j} \\left( \\frac{z_i}{x_i y_i} \\right) \n \\right]\\nonumber\\\\\n \\label{}\n\\end{align}\n\nThe function $g$ in \\eqref{eq:g_twophase} should at the located minimum be smaller than the smallest possible value \nfor the a single phase of feed composition:\n\\begin{equation}\n g_\\text{feed} = \\min\\left\\{ \\sum_i z_i \\left[ \\ln z_i + \\ln \\phi_i(\\vektor{z}) \\right] \\right\\},\n \\label{}\n\\end{equation}\nwhich is either a liquid or a vapor state, depending on which has the lowest Gibbs energy.\n\n\n\\section{Cubic equations of state}\nThe general cubic equation of state has the form\n\\begin{equation}\nP = \\frac{RT}{v -b}- \\frac{\\alpha a}{(v-m_1 b)(v-m_2 b)},\n\\label{eq:gencubic}\n\\end{equation}\nwhere $m_1$ and $m_2$ are dimensionless constants defining the SRK, PR and Van der\nWaals equation of state, $v=V/n~(\\SI{}{\\meter^3\\per\\mol})$ is the specific volume. The\nparameters $a~(\\SI{}{\\meter^3\\joule\\per \\mol^2})$ and $b~(\\SI{}{\\meter^3\\per\\mol})$ typically \ndepend on the composition and the dimensionless quantity \n$\\alpha$ is a function of temperature. \n\n\n\n\\begin{shaded}\n \\noindent\n\\textit{Exercise: Find the reduced residual Helmholtz energy function for a general\ncubic equation of state \\eqref{eq:gencubic}.}\n\n\\noindent\n\\textit{Solution}:\n\\begin{equation}\n A^\\text{r}(T,V,N) = n \\left[ RT\\ln \\left( \\frac{v}{v-b} \\right) + \n \\frac{\\alpha a}{\\left( m_1-m_2 \\right)b} \\ln\\left( \\frac{v-m_1 b}{v-m_2 b} \\right) \\right]\n \\label{}\n\\end{equation}\n\n\\end{shaded}\n\n\\subsection{Mixing rules}\nMixing rules are various ways of letting $a$ and $b$ in \\eqref{eq:gencubic} \ndepend on composition. \n\n\\subsubsection{Classic mixing rule}\nParameters: $a_{ii}$, $k_{ij}$ and $b_i$.\n\\begin{equation}\n a = \\sum_i x_i \\sum_j x_j a_{ij}\n \\label{}\n\\end{equation}\n\n\\begin{equation}\n a_{ij} = a_{ji} = \\sqrt{a_{ii} a_{jj}} \\left( 1-k_{ij} \\right)\n \\label{}\n\\end{equation}\n\n\\begin{equation}\n b = \\sum_i x_i b_i\n \\label{}\n\\end{equation}\n\n\n\\subsubsection{Huron-Vidal mixing rule}\n\n\n\n\\section{Explicit Helmholtz equations of state}\nBased on an explicit expression for the reduced dimensionless Helmholtz energy \\eqref{eq:reduced_helmholtz}. This includes \nequations such as GERG2004/2008 and the Span-Wagner equation of state for \\ce{CO2}.\n\n\\begin{equation}\n a(\\delta,\\tau,\\vektor{x}) = a^\\text{ig}(\\rho,T,\\vektor{x}) + a^\\text{r}(\\delta,\\tau,\\vektor{x}),\n \\label{}\n\\end{equation}\nwhere $\\delta$ is the \\textit{reduced mixture density} and $\\tau$ is the \\textit{reduced mixture temperature}, which are both generally \ndependent on the composition:\n\\begin{equation}\n \\delta = \\frac{\\rho}{\\rho_r(\\vektor{x})}, \\quad \\quad \\tau = \\frac{T}{T_r(\\vektor{x})}.\n \\label{}\n\\end{equation}\n\n\n\\section{Ideal gas Helmholtz energy}\nThe equation of state does not provide the energies.\nIn order to find the total Helmholtz energy, and not just the residual, \nthe ideal gas contribution $A^\\text{ig}$ must be found.\n\nAn ideal gas is defined by two equations. One the equation of state\n\\begin{equation}\n P^\\text{ig} = \\frac{nRT}{V}\n \\label{eq:idealgaslaw}\n\\end{equation}\nand the second is the equation for the internal energy\n\\begin{equation}\n U^\\text{ig}(T,n) = n c_v(T) T\n \\label{eq:U_ideal}\n\\end{equation}\nwhich simply states that the internal energy depends on the temperature and the number of particles only. The function $c_v$\nis the \\textit{heat capacity at constant volume}: \n\\begin{equation}\n C_v = n c_v \\equiv \\pone{Q}{T}{V} = \\pone{U}{T}{V}.\n \\label{}\n\\end{equation}\nAdditionally there is the \\textit{heat capacity at constant pressure}:\n\\begin{equation}\n C_p = n c_p \\equiv \\pone{Q}{T}{P} = \\pone{H}{T}{P},\n \\label{}\n\\end{equation}\nwhere it was used that $\\dd{H} = \\delta Q + V\\dd{P}$.\nFor an ideal gas, the enthalpy is \n\\begin{align}\n H^\\text{ig}(T,n) &\\equiv U^\\text{ig} + p^\\text{ig}V \n \\nonumber\\\\\n &= U^\\text{ig}(T,n) + nRT\n \\nonumber\\\\\n &= n\\left( c_v + R \\right)T\n \\label{}\n\\end{align}\nwhich shows that for an ideal gas\n\\begin{equation}\n c_p = c_v + R.\n \\label{}\n\\end{equation}\n\nChange in ideal gas enthalpy at a constant number of particles:\n\\begin{align}\n H^\\text{ig}(T,n) - H^\\text{ig}_0 &= \\int_{T_0}^T \\pone{H}{T}{P} \\dd{T}\n \\nonumber\\\\\n &= n \\int_{T_0}^T c_p \\dd{T}\n \\label{}\n\\end{align}\n\nChange in ideal gas entropy at a constant number of particles:\n\\begin{align}\n S^\\text{ig}(T,V,n) - S^\\text{ig}_0 \n &= \\int_{T_0}^T \\pone{S}{T}{V} \\dd{T} + \\int_{V_0}^V \\pone{S}{V}{T} \\dd{V}\n \\nonumber\\\\\n &=\\int_{T_0}^T \\frac{n c_v(T)}{T} \\dd{T}\n + \\int_{V_0}^V \\pone{P}{T}{V} \\dd{V}\n \\nonumber\\\\\n &=n\\int_{T_0}^T \\frac{c_p(T) - R}{T} \\dd{T}\n + nR\\int_{V_0}^V \\frac{1}{V} \\dd{V}\n \\nonumber\\\\\n &=n\\int_{T_0}^T \\frac{c_p(T) - R}{T} \\dd{T}\n + nR\\ln \\left( \\frac{V}{V_0} \\right),\n \\label{}\n\\end{align}\nusing \n\\begin{equation}\n \\pone{S}{T}{V} = \\pone{S}{U}{V} \\pone{U}{T}{V} = \\frac{1}{T} \\pone{U}{T}{V} = \\frac{n c_v}{T} \n \\label{}\n\\end{equation}\nand the \\textit{Maxwell relation}\n\\begin{equation}\n\\label{eq:dsrel}\n\\pone{S}{V}{T,\\vektor{n}} = \\pone{P}{T}{V,\\vektor{n}}.\n\\end{equation}\n\nThe total Helmholtz energy is not provided through the cubic equation of state, only the residual part.\nThe ideal gas part is usually specified as \nalgebraic temperature functions for the heat capacity at constant pressure, $c_p(T)$, provided\nfor each component in the mixture. The total Helmholtz energy may be written as \n\\begin{align}\n A &\\equiv U - TS \\nonumber\\\\\n &=U-TS+PV-PV \\nonumber\\\\\n &= H - TS - PV,\n \\label{}\n\\end{align}\nwhich means that the Helmholtz energy of a pure ideal gas may be expressed as:\n\\begin{align}\n A^\\text{ig}_i(T,V,n_i) &= H^\\text{ig} - TS^\\text{ig} - nRT\n \\nonumber\\\\\n &= \\Delta H_i^\\text{ig} + H_{0,i}^\\text{ig} \n - T\\left( \\Delta S_i^\\text{ig} + S_{0,i}^\\text{ig} \\right) - n_i RT\n \\nonumber\\\\\n &= n_i \\int_{T_0}^T c_{p,i}^\\text{ig}(T) \\dd{T} + H_{0,i}^\\text{ig}\n - n_i T \\left[ \\int_{T_0}^T \\frac{c_{p,i}^\\text{ig}(T) - R}{T} \\dd{T} \n + R \\ln \\left( \\frac{V}{V_0} \\right) \\right]\n - T S_{0,i}^\\text{ig} - nRT.\n \\label{}\n\\end{align}\n\nMixing carries with it another entropy contribution, even in the case of ideal mixtures, \\textit{i.e.} with no interactions. The entropy change \nfrom ideal mixing at constant pressure and temperature is:\n\\begin{align}\n \\Delta S^\\text{ig}_\\text{mix} &= -nR \\sum_i x_i \\ln x_i \n \\nonumber\\\\\n &= -R \\sum_i n_i \\ln \\left( \\frac{n_i}{n} \\right),\n \\label{}\n\\end{align}\nwhich means that the total Helmholtz energy of an ideal mixture is given by:\n\\begin{align}\n A^\\text{ig}(T,V,\\vektor{n}) &= \\sum_i A_i^\\text{ig}(T,V,n_i) - T \\Delta S^\\text{ig}_\\text{mix}\n \\label{}\n\\end{align}\nIn terms of the reduced Helmholtz energy, this becomes:\n\\begin{align}\n a^\\text{ig}(T,V,\\vektor{x}) &= \\sum_i x_i a_i^\\text{ig}(T,V) + \\sum_i x_i \\ln x_i, \n \\label{}\n\\end{align}\nwhere \n\\begin{equation}\n a_i^\\text{ig}(T,V) \\equiv \\frac{A^\\text{ig}_i(T,V,n_i)}{n_i RT}.\n \\label{}\n\\end{equation}\n\n\n\\appendix\n\n\\section{Newton's method for root finding}\n\\section{Newton's method for optimization (Modified Newton)}\n\\section{Accelerated direct substitution (Dominant Eigenvalue Method)}\n\n\n\\begin{thebibliography}{9}\n\n\\bibitem{callen85}\n Herbert B. Callen,\n \\emph{Thermodynamics and an Introduction to Thermostatistics}.\n second edition,\n John Wiley \\& Sons Inc,\n 1985.\n\n\\bibitem{kunz07}\n O. Kunz, R. Klimeck, W. Wagner, M. Jaeschke,\n \\emph{The GERG-2004: Wide-Range Equation of State for Natural Gases and Other Mixtures}.\n GERG TM15 Report (http://www.gerg.eu/publications/tm.htm),\n 2007.\n\n\\bibitem{michelsen82a}\n Michael L. Michelsen,\n \\emph{The isothermal flash problem. Part I. stability}.\n Fluid Phase Equilibria,\n 9(1)\n pp. 1-19,\n 1982.\n\n\\bibitem{michelsen82b}\n Michael L. Michelsen,\n \\emph{The isothermal flash problem. Part II. phase-split calculation}.\n Fluid Phase Equilibria,\n 9(1)\n pp. 21-40,\n 1982.\n\n\\bibitem{michelsen99}\n Michael L. Michelsen and J{\\o}rgen M. Mollerup,\n \\emph{State function based flash specifications}.\n Fluid Phase Equilibria,\n 158-160\n pp. 617-626,\n 1999.\n\n\\bibitem{michelsen07}\n Michael L. Michelsen and J{\\o}rgen M. Mollerup,\n \\emph{Thermodynamic models: Fundamentals \\& computational aspects}.\n second edition\n Tie-Line Publication,\n 2007.\n\n\\bibitem{span96}\n Roland Span and Wolfgang Wagner,\n \\emph{A New Equation of State for Carbon Dioxide Covering the Fluid Region\n from the Triple-Point Temperature to 1100 {K} at Pressures up to 800 {MPa}}.\n J. Phys. Chem. Ref. Data,\n 25(6)\n pp. 1509-1596\n 1996.\n\n\\end{thebibliography}\n\n\\end{document}\n", "meta": {"hexsha": "212fbe7b0120962041c93777eff424de8c96e835", "size": 46443, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/memo/intro/intro.tex", "max_stars_repo_name": "SINTEF/Thermopack", "max_stars_repo_head_hexsha": "63c0dc82fe6f88dd5612c53a35f7fbf405b4f3f6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 28, "max_stars_repo_stars_event_min_datetime": "2020-10-14T07:51:21.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T04:59:23.000Z", "max_issues_repo_path": "doc/memo/intro/intro.tex", "max_issues_repo_name": "SINTEF/Thermopack", "max_issues_repo_head_hexsha": "63c0dc82fe6f88dd5612c53a35f7fbf405b4f3f6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 20, "max_issues_repo_issues_event_min_datetime": "2020-10-26T11:43:43.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-30T22:06:30.000Z", "max_forks_repo_path": "doc/memo/intro/intro.tex", "max_forks_repo_name": "SINTEF/Thermopack", "max_forks_repo_head_hexsha": "63c0dc82fe6f88dd5612c53a35f7fbf405b4f3f6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 13, "max_forks_repo_forks_event_min_datetime": "2020-10-27T13:04:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-21T04:59:24.000Z", "avg_line_length": 40.9911738747, "max_line_length": 167, "alphanum_fraction": 0.6945072454, "num_tokens": 15755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581049086031, "lm_q2_score": 0.8056321866478979, "lm_q1q2_score": 0.6410077788816405}} {"text": "\\documentclass[../main.tex]{}\n\\begin{document}\n\n\\subsection{Clarke Wright Saving Algorithm}\nThe Clarke and Wright algorithm \\cite{clarke_wright} is one of the earliest, and most popular heuristic algorithm for the VRP due to its speed, simplicity, and ease of adjustment to handle various constraints in real-life applications. Clarke Wright saving algorithm works equally well for both directed and undirected problems. The algorithm is built on a basic simple idea: maximize saving cost within a route, hence minimize total cost. Consider a depot D and n demand points. Suppose that initially the solution to the VRP consists of using n vehides and dispatching one vehicle to each one of the n demand points. The total cost of this solution is:\n\\[ C = 2\\sum\\limits_{i=1}^n d(D, i)\\]\nwhere d(i, j) is a function calculate distance between point i and , j, and C is total cost. To get a better solution, we now can combine 2 customers served by single vehicle on a single trip, the total distance we can save after combined is:\n\\begin{eqnarray}\ns(i, j) & = & 2d(D, i) + 2d(D, j) \\nonumber \\\\\n&& -\\: [d(D, i) + d(D, j) + d(i, j)] \\nonumber \\\\\n& = & d(D, i) + d(D, j) - d(i, j)\n\\end{eqnarray}•\n\\begin{algorithm}\n\\caption{Clarke Wright saving alorithm}\\label{alg:clarke_wright}\n\\textbf{Step 1: Savings computation}\n\\flushleft\n\\begin{itemize}\n\\item Compute $s_{ij} \\forall(i, j) \\subset E$.\n\\item Sort $s_{ij}$\n\\end{itemize}\n\\textbf{Step 2: Route Extension}\n\\flushleft\n\\begin{itemize}\n\\item for each route $(0, i, ..., j, 0)$\n\\item Determine the first saving ${s_{ki}}$ or ${s_{jl}}$ that can feasibly be used to merge the current route with another route ending with ${(k,0)}$ or starting with ${(0,l)}$.\n\\item Implement the merge and repeat this operation to the current route.\n\\item If not feasible merge exists, consider the next route and reapply the same operations.\n\\item Stop when not route merge is feasible.\n\\end{itemize}\n\\end{algorithm}\n\n\\subsection{Ruin and Recreate Algorithms}\n\nRuin and recreate framework is introduced in \\cite{ref:rr_break}. By performing ruin and recreation on current solutions frequently enough, the better solutions will be obtained. A set of random vehicle routes are created and their total vehicle distance is calculated. The Ruin and Recreate approach is applied frequently to obtain another set of vehicle routes that have the least of total vehicle distance.\n\nThis approach starts with creating a set of initial vehicle routes: a random customer is selected, and combined with the depot location to create a base vector. The base vector starts to rotate clockwise with the depot location at the center and customers on the way are added to the route. If the route is over the vehicle capacity, a new route is created, and so on. Total vehicle distance is calculated for the initial routes.\nNext, some customers from each initial route are removed by different ruin strategies and then added back to the same route in different order or different route using greedy insertion recreation strategy by trying all possible locations to pick the least total vehicle distance.\n There are three ruin strategies are used to remove customers from existing solution: random ruin, sequential ruin, and least customer ruin. Random strategy removes customers randomly from all customers of all routes, which creates the effect of diversifying the removing process. Sequential strategy removes customers sequentially in a selected route. The least customer strategy removes all customers in the route that has the least customers.\nGreedy insertion strategy is used for the recreation of the solution by adding the removed customers back. This approach is to insert each removed customer to the cheapest possible route.\n\n\\begin{algorithm}\n\\caption{Ruin and recreate framework}\\label{alg:RR}\n\\flushleft\n\\begin{itemize}\n\\item create intial solution\n\\item do {\n\\flushleft\n\\begin{itemize}\n\\item choose ruin and recreate strategy\n\\item ruin current solution\n\\item recreate ruined solution\n\\item if new solution better than current solution\n\\item then accept new solution\n\\item else accept new solution with probaility $e^{\\frac{\\Delta C}{T}}$\n\\end{itemize}\n\\item } while stop condition is not meet\n\\item return best solution\n\\end{itemize}\n\\end{algorithm}\n\n\\end{document}\n", "meta": {"hexsha": "48778449eeff0464a5d6c7657ddd8dc6dd80174e", "size": 4290, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/sections/methods.tex", "max_stars_repo_name": "dinhvan5481/ee382v-adv-algorithms-vehicle-routing-problem", "max_stars_repo_head_hexsha": "2aff876c4f6440fe4fb8551f1647ab14085dfa92", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/sections/methods.tex", "max_issues_repo_name": "dinhvan5481/ee382v-adv-algorithms-vehicle-routing-problem", "max_issues_repo_head_hexsha": "2aff876c4f6440fe4fb8551f1647ab14085dfa92", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/sections/methods.tex", "max_forks_repo_name": "dinhvan5481/ee382v-adv-algorithms-vehicle-routing-problem", "max_forks_repo_head_hexsha": "2aff876c4f6440fe4fb8551f1647ab14085dfa92", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.1935483871, "max_line_length": 654, "alphanum_fraction": 0.7771561772, "num_tokens": 1015, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7956581049086031, "lm_q1q2_score": 0.6410077733119761}} {"text": "%intro\n\n\\section{Introduction}\nWe are presenting a brief summory of basic concepts of modern algebraic geometry. In particular, we will discuss standard constructions and expand our learnings to a more abstract scheme theoretic approach.\n\\subsection{Requirements}\nThe reader only needs a minimum understanding in linear and abstract algebra. However, although some definitions will be missed we try to include as many as possible.\\\\\n\\subsection{Basics}\n\\indent We denote by $R$ an arbitrary ring with one (unital), unless mentioned otherwise, always commutative. A field is denoted by $k$ and its algebraic closure with $\\overline{k}$, special fields and rings are denoted following \\cite{Bour}. The polynomial ring for a given ring $R$ is denoted by $R[X] = R[X_1\\ldots X_n]$ and the affine module $\\mathbb{A}^n_R = (R^n, P(R^n), \\phi)$.\n\\subsubsection{Zero sets and the Zariski topology}\nGiven a field $k$ and its polynomial ring $k[X] = k[X_1\\ldots X_n]$ and a nonempty set of polynomials $F \\subset R[X]$, the zero set of $F$, $Z(F)$, is defined:\n$$\\left\\{x \\in \\kaffn : f(x) = 0 \\ \\forall f \\in F\\right\\}.$$\nWe may rewrite:\n$$\\begin{array}{rcl}\nZ(F) &=& \\{x : f(x) = 0 \\forall f \\in F\\}\\\\&&\\\\\n&=& \\bigcap_{f \\in F} \\{x : f(x) = 0\\}.\\\\\n\\end{array}$$\nIn other words, $Z(F)$ is simply the intersection of common zeros of each polynomial in $F$. Therefore, we could have simply defined zero sets for polynomials and then show that zero sets of $F$ are the intersections of the zeros of its elements.\\\\\n\\indent We remark that $Z(F)$ may be empty if no common zero is present (e.g. $f = X, g = X - 1 \\in k[X]$). The zero set of the constant polynomial $1 \\in k[X]$ is therefore empty and the zero set of $0 \\in k[X]$ is the whole affine space.\\\\\n\\paragraph{Zeros and their topology}\nBy the above definition, we get that the set of all zeros is closed under arbitrary intersection. Furthermore, the empty set and the whole affine space are present in $\\cldk := \\{X \\subset \\kaffn : X \\ \\mathrm{zero~set}\\}$. Now, let us check for two polynomials $f, g \\in k[X]$:\n$$\\begin{array}{rcl}\nZ(f) \\cup Z(g) &=& \\{x : f(x) = 0\\} \\cup \\{x : g(x) = 0\\}\\\\&&\\\\\n&=& \\{x : f(x) = 0 \\vee g(x) = 0\\}\\\\&&\\\\\n&=& \\{x : f(x) \\cdot g(x) = 0\\}\\\\&&\\\\\n&=& Z(f g)\\\\\n\\end{array}$$\nThis shows $\\cldk$ is closed under finite union and making it a closed topology (i.e. generated by closed sets) over $\\kaffn$.\\\\\n\\paragraph{Principal open sets and their topology}\n\\indent Now, for a given $f \\in k[X]$ we call the set\n$$D(f) := \\{x \\in \\kaffn : f(x) \\neq 0\\}$$\na principal open set of $\\kaffn$. It is obviously the complement of a zero set. Now, for any nonempty set $F$ of polynomials we define the open set\n$$D(F) := \\bigcup_{f \\in F} D(f).$$\nAgain, we remark that $D(F)$ may be empty or the full affine space:\n$$D(F) = \\kaffn \\Leftrightarrow F \\subset k^\\times,\\ D(F) = \\emptyset \\Leftrightarrow F = \\{0\\}.$$\nAs above, we need to check for closeness under intersection. However, as we defined our subset system via the complements of a topology generated by closed sets we get a topology generated by open sets. Therefore:\n$$\\opnk := \\left\\{U \\subset \\kaffn : U^c := \\kaffn\\backslash U \\in \\cldk\\right\\}$$\nis an open topology on $\\kaffn$.\n\\begin{defi}\nThe topology $\\tau \\subset \\mathcal{P}(\\kaffn)$ generated by both $\\cldk$ and $\\opnk$ is called the Zariski topology on $\\kaffn$.\n\\end{defi}\n\\subsubsection{Affine schemes and the Zariski topology}\nGiven a unital commutative $k$ algebra $R$ and the class of all $R$-submodules of $R$, $\\mathrm{Mod}_R(R)$, (i.e. its ideals)\n$$\\specr := \\{\\prm \\in \\mathrm{Mod}_R(R) : \\prm \\ \\mathrm{prime}\\},$$\nthe set of all prime ideals, is called the spectrum of $R$.\n\\paragraph{Pairing of points and ideals}\nGiven a $P = (P_1,\\ldots,P_n) \\in \\kaffn$, the ideal\n$$\\mxx_P = \\left$$\nis maximal in every polynomial ring $k[X]$ over some field $k$. As every maximal ideal in a unital ring is prime we get a unique subset of $\\specr$:\n$$\\max (R) = \\{\\mxx \\in \\specr : \\mxx \\ \\mathrm{maximal}\\},$$\ni.e. the set of maximal ideals in $R$.\n\\begin{prop}\nLet $k = \\overline{k}$ be a field and $R = k[X]$. The pairing\n$$\\kaffn \\longrightarrow \\max(R),\\ P \\longmapsto \\mxx_P$$\nis a bijection.\n\\end{prop}\n\\bmk This proposition is a rephrasing of Hilberts Nullstellensatz. Thus, the interested reader will find a proof in the standard literature as \\cite{Hart}. In analogy to the Zariski topology on $\\kaffn$ we define for some family of polynomials $F$:\n$$V(F) = \\{\\prm \\in \\speck : f \\in \\prm,\\ \\forall f \\in F\\} = \\bigcap_{f \\in F} \\{\\prm : f \\in \\prm\\}.$$\nAccordingly, we call for some polynomial $f \\in k[X]$ the set $U(f) = \\{\\prm : f \\notin \\prm\\} = V(f)^c$ a prinicipal open set in $\\speck$. Again, we get for some family $F$ of polynomials:\n$$\n\\begin{array}{rclcl}\nU(F) &:=& V(F)^c &=& \\speck \\backslash V(F)\\\\&&&&\\\\ &=& \\speck \\backslash \\left(\\bigcap_{f \\in F} V(f)\\right) &=& \\bigcup_{f \\in F} \\left(\\speck \\backslash V(f)\\right)\\\\&&&&\\\\ &=& \\bigcup_{f \\in F} V(f)^c &=& \\bigcup_{f \\in F} U(f).\\\\\n\\end{array}\n$$\nWith this equivalences in place, we get the following proposition\n\\begin{prop}\nThe sets $\\kaffn$ and $\\X := \\speck$ are isomorphic in the category of topological spaces with corresponding topologies $\\cldk$ and $\\opnk$, as well as\n$$\\mathcal{F}(\\X) = \\{V(F) : F \\subset k[X]\\}\\ \\mathrm{and}\\ \\mathcal{O}(\\X) = \\{V(F)^c : F \\subset k[X]\\},$$\nrespectively. Again, $V(F)^c := \\speck \\backslash V(F)$.\n\\end{prop}\n\\bws We simply use our bijective pairing above:\n$$\\kaffn \\ni P = (P_1,\\ldots,P_n) \\longmapsto \\mxx_P = \\left \\in X.$$\nFurthermore, we expand\n$$\\begin{array}{rclcl}\n\\cldk \\ni Z(F) &=& \\bigcap_{f \\in F} Z(f) \\longmapsto V(F) &=& \\bigcap_{f \\in F} V(f) \\in \\mathcal{F}(\\X)\\\\&&&&\\\\\n\\opnk \\ni D(F) &=& \\bigcup_{f \\in F} D(f) \\longmapsto U(F) &=& \\bigcup_{f \\in F} U(f) \\in \\mathcal{O}(\\X)\\\\\n\\end{array}$$", "meta": {"hexsha": "6d293c61f0e35a7923c30f7ae11c60d7f03eca0e", "size": 5894, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg_geo/intro.tex", "max_stars_repo_name": "gmuel/texlib", "max_stars_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "alg_geo/intro.tex", "max_issues_repo_name": "gmuel/texlib", "max_issues_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg_geo/intro.tex", "max_forks_repo_name": "gmuel/texlib", "max_forks_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 77.5526315789, "max_line_length": 385, "alphanum_fraction": 0.6669494401, "num_tokens": 2025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7956581049086031, "lm_q1q2_score": 0.6410077733119761}} {"text": "\\documentclass[]{article}\n\\usepackage{amsmath, amssymb, graphicx, float, dsfont}\n\n%opening\n\\title{Notes for \"Statistics Informed Decisions using Data\"}\n\\author{Delbert Yip}\n\n\\begin{document}\n\n\\maketitle\n\n\\begin{abstract}\n\tThese notes covers several chapters from \"Statistics Informed Decisions using Data,\" focusing on practical applications. \n\\end{abstract}\n\n\\section{Probability Distributions}\n\\textbf{Poisson distribution}\nA random variable $X$, the number of successes in a fixed interval, follows a Poisson process if:\n\\begin{itemize}\n\t\\item The probability of 2 or more successes in a sufficiently small subinterval, is 0.\n\t\\item The probability of success is the same for any two intervals of equal length. \n\t\\item The number of successes in any interval is independent of the number of successes in any other intervals, as long as said intervals do not overlap. \n\\end{itemize}\n\nA Poisson process is described by:\n\\begin{equation}\nP(x) = \\frac{(\\lambda t)^x}{x!} \\exp(-\\lambda t), \\quad x=0,1,...n\n\\end{equation}\n\nWhere $\\lambda$ is the average number of occurrences of the event in an interval of length 1. \n\n\\section{Inference}\n\\subsection{Hypothesis Tests}\nWe try to find support for the \\textbf{alternative hypothesis}, $H_1$. The \\textbf{null hypothesis}, $H_0$, is assumed true until refuted by evidence, and assumes no change/difference/etc. Three ways to set up null and alternative hypotheses:\n\\begin{enumerate}\n\t\\item Two-tailed test: equal versus not equal\n\t\\begin{align*}\n\t\tH_0 &= x \\\\\n\t\tH_1 &\\neq x\n\t\\end{align*}\n\t\n\t\\item Left-tailed test: equal versus less than \n\t\\begin{align*}\n\tH_0 &= x \\\\\n\tH_1 &< x\n\t\\end{align*}\n\t\n\t\\item Right-tailed test: equal versus greater than \n\t\\begin{align*}\n\tH_0 &= x \\\\\n\tH_1 &> x\n\t\\end{align*}\n\\end{enumerate}\n\n$H_0$ is always a statement of equality. Left- and right-tailed tests are a.k.a one-tailed tests. There are two types of errors in hypothesis testing: Type I (false negative, $H_0$ is rejected when $H_0$ is actually true) and Type II (false positive, $H_0$ is not rejected while $H_1$ is true.) Importantly, the \\textbf{level of significance, $\\alpha$} is the \\textit{probability of making a Type I error/false negative}.The choice of $\\alpha$ depends on the consequences of making a Type I error, or false negative. On the other hand, $\\beta$ is the probability of making a Type II error/false positive. There is an inverse relation between $\\alpha$ and $\\beta$: when $\\alpha$ is small, $\\beta$ is large. \n\n\\end{document}\n", "meta": {"hexsha": "85226ae667745ce0c9759114cc38b79717240efa", "size": 2486, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "_posts/SIDUD_notes.tex", "max_stars_repo_name": "haganenoneko/website.github.io", "max_stars_repo_head_hexsha": "fcb72e2ba797f22f262d54d82a0b0b4b4d9b4a69", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "_posts/SIDUD_notes.tex", "max_issues_repo_name": "haganenoneko/website.github.io", "max_issues_repo_head_hexsha": "fcb72e2ba797f22f262d54d82a0b0b4b4d9b4a69", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "_posts/SIDUD_notes.tex", "max_forks_repo_name": "haganenoneko/website.github.io", "max_forks_repo_head_hexsha": "fcb72e2ba797f22f262d54d82a0b0b4b4d9b4a69", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.8620689655, "max_line_length": 707, "alphanum_fraction": 0.7437650845, "num_tokens": 691, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7956580976404296, "lm_q1q2_score": 0.6410077674565017}} {"text": "\\documentclass[english]{../spicker}\r\n\r\n\\usepackage{amsmath}\r\n\r\n\\usepackage{graphicx}\r\n\\usepackage{tabularx, multirow}\r\n\r\n\\usepackage[style=numeric, backend=bibtex]{biblatex}\r\n\\addbibresource{or.bib}\r\n\r\n\\title{Operations Research}\r\n\\author{Patrick Gustav Blaneck}\r\n\r\n\\begin{document}\r\n\\maketitle\r\n\\tableofcontents\r\n\\newpage\r\n\r\n%\\setcounter{section}{1}\r\n\r\n\\section{Linear Programming}\r\n\r\n\\begin{defi}{Linear Programming}\r\n \\emph{Linear Programming} is the problem of optimizing (maximizing or minimizing) a \\emph{linear objective function} subject to a set of \\emph{linear functional constraints}.\r\n\r\n \\textbf{Given:} $A \\in \\R^{m\\times n}, b \\in \\R^m, c\\in R^n$\r\n\r\n \\textbf{Find:} $x^* \\in \\R^n$ where $x^* = \\arg\\max\\{c^Tx \\mid Ax \\leq b\\}$\r\n\\end{defi}\r\n\r\n\\begin{bonus}{Linear Programming Solvers}\r\n Software that solves linear programs - \\emph{linear programming solvers} - also generate lots of important auxiliary information (as well as the optimum):\r\n \\begin{itemize}\r\n \\item sensitivity analysis\r\n \\item shadow prices\r\n \\item alternative optima\r\n \\item \\ldots\r\n \\end{itemize}\r\n\\end{bonus}\r\n\r\n\\begin{theo}{Ellipsoid Method}\r\n A LP of dimension $n$ can be solved in $\\bigo(L^2 \\cdot n^6)$ time \\cite{khachiyan1979}, where $L =$ \\# bits in the input.\r\n\\end{theo}\r\n\r\n\\begin{theo}{Interior Point Method}\r\n A LP of dimension $n$ can be solved in a \\emph{numerically stable} way in $\\bigo(L^2 \\cdot n^{3.5})$ time \\cite{karmarkar1984}.\r\n\\end{theo}\r\n\r\n\\begin{defi}{Integer Linear Programs (ILP)}\r\n \\textbf{Given:} $A \\in \\R^{m\\times n}, b \\in \\R^m, c\\in R^n$\r\n\r\n \\textbf{Find:} $\\underline{x^*\\in\\Z^n}$ where $x^* = \\arg\\max\\{c^Tx \\mid Ax \\leq b\\}$\r\n\\end{defi}\r\n\r\n\\begin{example}{Integer Linear Program for \\textsc{Vertex Cover}}\r\n \\fbox{\r\n \\parbox{0.95\\textwidth}{\r\n \\underline{\\textsc{Vertex Cover}}\r\n\r\n\r\n \\textbf{Given:} Graph $G = (V, E)$\\\\\r\n \\textbf{Find:} \\textsc{Vertex Cover}, i.e. $V' \\subseteq V$ such that every edge has at least one endpoint in $V'$.\r\n }\r\n }\r\n\r\n \\textbf{Integer Linear Program:}\r\n\r\n For $v\\in V$, let $x_v \\in \\{0, 1\\}$.\r\n\r\n Goal: minimize $\\sum_{v\\in V}x_v$.\r\n\r\n Constraints: for every edge $uv \\in E$, we require $x_u + x_v \\geq 1$.\r\n\\end{example}\r\n\r\n\\printbibliography\r\n\\end{document}\r\n", "meta": {"hexsha": "8e368e85df3e369f0c458843f7e6b7fea9bd148b", "size": 2339, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "or/or.tex", "max_stars_repo_name": "pblan/matse-spicker", "max_stars_repo_head_hexsha": "50109dee722338bb79c70d922cc1adc04c35132d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2021-06-29T08:48:38.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-04T08:55:59.000Z", "max_issues_repo_path": "or/or.tex", "max_issues_repo_name": "pblan/matse-spicker", "max_issues_repo_head_hexsha": "50109dee722338bb79c70d922cc1adc04c35132d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2021-12-13T08:32:00.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-12T12:35:50.000Z", "max_forks_repo_path": "or/or.tex", "max_forks_repo_name": "pblan/matse-spicker", "max_forks_repo_head_hexsha": "50109dee722338bb79c70d922cc1adc04c35132d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2022-03-14T21:11:25.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-14T22:20:59.000Z", "avg_line_length": 30.3766233766, "max_line_length": 179, "alphanum_fraction": 0.6387345019, "num_tokens": 756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245994514084, "lm_q2_score": 0.7690802423634961, "lm_q1q2_score": 0.6408934849135525}} {"text": "\\section{Skipping}\n\\label{sec:skipping}\n\nHere we envisage how to copy a stack while ignoring a given\nitem. [\\emph{The progression is very slow on purpose.}]\n\n\\paragraph{First occurrence}\n\\label{def:linear_search}\n\\index{stack!skipping an item!first occurrence}\n\nLet us suppose that \\(\\fun{sfst}(s,x)\\)\\index{sfst@\\fun{sfst/2}|(}\n(\\emph{skip the first occurrence}) evaluates in a stack identical\nto~\\(s\\) but without the first occurrence of~\\(x\\), starting from the\ntop. In particular, if~\\(x\\) is absent in~\\(s\\), then the value of the\ncall is identical to~\\(s\\). This is our\n\\emph{specification}.\\index{specification} For instance, we expect the\nfollowing evaluations:\\label{sfst_ex}\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,3) & \\twoheadrightarrow & \\el;\\\\\n\\fun{sfst}(\\el,\\el) & \\twoheadrightarrow & \\el;\\\\\n\\fun{sfst}([3,\\el],[5,2]) & \\twoheadrightarrow & [3,\\el];\\\\\n\\fun{sfst}([\\el,[1,2],4,\\el,4],4) & \\twoheadrightarrow &\n [\\el,[1,2],\\el,4];\\\\\n\\fun{sfst}([4,[1,2],\\el,\\el,4],\\el)\n & \\twoheadrightarrow & [4,[1,2],\\el,4].\n\\end{array}\n\\end{equation*}\n\n\\paragraph{First attempt}\n\nLet us try a direct approach. In particular, at this point, it is\nimportant \\emph{not} to seek a definition in tail form. Tail form must\nbe considered as an optimisation and early optimisation is opening\nPandora's jar. The first idea that may come to mind is to define an\nauxiliary function \\fun{mem/2} such that the call \\(\\fun{mem}(s,x)\\)\nchecks whether a given item \\(x\\)~is in a given stack~\\(s\\), because\nthat notion of membership is implicit in the wording of the\nspecification. But two problems then arise. Firstly, what would be the\nresult of such a function? Secondly, what would be the additional cost\nfor using it? For the sake of the argument, let us follow this track\nand find out how where it leads. A stack can either be empty or not,\nso let us make two rules:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nNote that we introduced a variable~\\(y\\), distinct from\nvariable~\\(x\\). \\emph{Two different variables may or may not denote\n the same value, but two occurrences of the same variable always\n denote the same value.} Had we written instead\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{mem}(\\cons{\\underline{x}}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\none case would be missing, namely when the top of the stack is not the\nitem sought for, for instance, \\(\\fun{mem}(3,[4])\\) would fail due to\na match failure. Now, what is the first right\\hyp{}hand side? The\nfirst pattern matches only if the stack is empty. In particular, this\nmeans that the item is not in the stack, since, by definition, an\nempty stack is a stack containing no item. How do we express that?\nSince the original problem is silent on the matter, it is said to be\n\\emph{underspecified}.\\index{underspecification} We may think that\nzero would be a token of choice to denote the absence of the item in\nthe stack:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & 0;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nBut this would be a mistake because there is no natural and necessary\nrelationship between the concept of emptiness and the number\nzero. Zero is best understood algebraically as the number noted~\\(0\\)\nsuch that \\(0 + n = n + 0 = n\\), for any number~\\(n\\). Then, let us\ntry the empty stack:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nThe next step is to find a way to actually compare the value of~\\(x\\)\nto the value of~\\(y\\). We can use the rule above about variables: two\noccurrences of the same variable mean that they hold the same\nvalue. Therefore\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nwas not so bad, after all? Indeed, but we know now that a case is\nmissing, so let us add it at the end, where \\(x \\neq y\\):\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\rightarrow & \\fbcode{CCCCC};\\\\\n\\underline{\\fun{mem}(\\cons{y}{s},x)} & \\rightarrow & \\fbcode{CCCCC}.\n\\end{array}\n\\end{equation*}\nNow, what is the second right\\hyp{}hand side? It is evaluated if the\nitem we were looking for is present at the top of the stack. How do we\nexpress that? We may think of ending with the item itself, the\nrationale being that if the result is the empty stack, then the item\nis not in the input stack, otherwise the result is the item itself:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\rightarrow & x;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nThe last right\\hyp{}hand side is easier to guess since it deals with\nthe case where the top of the stack (\\(y\\)) is not the item we seek\n(\\(x\\)), so a recursive call which ignores~\\(y\\) should come to mind:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\rightarrow & x;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\rightarrow & \\fun{mem}(s,x).\n\\end{array}\n\\end{equation*}\n\nSome tests would increase the confidence that this definition is\ncorrect\\index{soundness} and complete\\index{completeness} with respect\nto the specification. Let us label the rules first:\n\\begin{equation*}\n\\begin{array}{r@{\\;}c@{\\;}l}\n\\fun{mem}(\\el,x) & \\smashedrightarrow{\\zeta} & \\el;\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\smashedrightarrow{\\eta} & x;\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\smashedrightarrow{\\theta} &\n\\fun{mem}(s,x).\n\\end{array}\n\\end{equation*}\nThen we could try the following cases:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l@{\\;}l@{\\;}l}\n\\fun{mem}(\\el,3) & \\smashedrightarrow{\\zeta} & \\el,\\\\\n\\fun{mem}([1],3) & \\smashedrightarrow{\\theta} & \\fun{mem}(\\el,3)\n& \\smashedrightarrow{\\zeta} & \\el,\\\\\n\\fun{mem}([1,3,2],3) & \\smashedrightarrow{\\theta} & \\fun{mem}([3,2],3)\n& \\smashedrightarrow{\\eta} & 3.\n\\end{array}\n\\end{equation*}\nThe code seems to work: item~\\(x\\) is in stack~\\(s\\) if the result\nis~\\(x\\), otherwise it is~\\(\\el\\). However, this function is not\ncorrect.\\index{soundness} The hidden and flawed assumption is ``items\ncan not be stacks'', in spite of the counter\\hyp{}examples given at\nthe beginning for illustrating the expected behaviour of\n\\fun{sfst/2}. In particular, an item can be the empty stack and this\nsituation leads to an ambiguity with our definition of \\fun{mem/2}:\n\\begin{equation*}\n \\fun{mem}(\\el,\\el) \\smashedrightarrow{\\zeta} \\el \\xleftarrow{\\smash{\\eta}} \\fun{mem}([\\el],\\el).\n\\end{equation*}\nIt is impossible to discriminate the two cases, the first one meaning\nabsence of the item and the second presence, because they both end\nwith the empty stack. In fact, we should have distinguished two data\nconstructors to denote the outcomes ``the item was found'' and ``the\nitem was not found''. For example,\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{mem}(\\el,x) & \\smashedrightarrow{\\zeta} & \\fun{false}();\\\\\n\\fun{mem}(\\cons{x}{s},x) & \\smashedrightarrow{\\eta} & \\fun{true}();\\\\\n\\fun{mem}(\\cons{y}{s},x) & \\smashedrightarrow{\\theta} & \\fun{mem}(s,x).\n\\end{array}\n\\end{equation*}\nBut, for now, let us backtrack and ask ourselves again whether using\n\\fun{mem/2} is really a good idea.\n\n\\paragraph{Better approach}\n\nLet us suppose that the input stack contains the item at the\nbottom. Using \\fun{mem/2} to find it leads to a complete traversal of\nthe input stack. Then another traversal from the beginning (the top of\nthe stack) is needed to copy the stack without its last item, so, in\ntotal, two complete traversals are performed.\n\nA better idea consists in \\emph{interleaving} these two passes into\none because the problem stems from the fact that \\fun{mem/2} forgets\nabout the items which are not the item of interest, thus, when it is\nfound or known to be absent, there is no way to build a copy to make\nthe result. By interleaving, we mean that during the traversal, the\nconcepts of membership and of copying are combined, instead of being\nused sequentially as two function calls. A similar situation was\nencountered in the design of a function reversing a stack:\n\\fun{rev\\(_0\\)/1}, which calls \\fun{cat/2}, is much slower than\n\\fun{rev/1}, which uses an auxiliary stack.\n\nHere, our algorithm consists in memorising all visited items and, if\nthe item is not found, the resulting stack is rebuilt from them; if\nfound, the result is built from them \\emph{and} the remaining,\nunvisited, items. There are usually two ways to keep visited items:\neither in an accumulative parameter, called \\emph{accumulator}, or in\nthe context of recursive calls. At this point, it is important to\nrecall a cardinal guideline: Do not try first to design a definition\nin tail form, but opt instead for a direct approach. In some simple\ncases, a direct approach may actually be in tail form, but the point\nis methodological: a priori ignore all concerns about tail\nforms. Accordingly, let us use the context of a recursive call to\nrecord the visited items. A stack being either empty or not, it is\nnatural to start with\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nThen, just as we tried with \\fun{mem/2}, we must distinguish the\ncase when \\(x\\)~is the same as~\\(y\\):\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\underline{\\fun{sfst}(\\cons{x}{s},x)} & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\nThis method is a \\emph{linear search}\\index{linear search}: the items\nin the stack are compared one by one to~\\(x\\), starting from the top,\nuntil the bottom or an item equal to~\\(x\\) is reached. Since we know\nthat the last rule deals with the case when \\(x \\neq y\\), we must\nmemorise~\\(y\\) and go on comparing~\\(x\\) with the other items in~\\(s\\)\n(if any). This is where the recursive call with a context, discussed\nabove, is set as \\(\\cons{y}{\\text{\\textvisiblespace}}\\):\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{sfst}(\\cons{x}{s},x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\cons{y}{\\fun{sfst}(s,x)}.\n\\end{array}\n\\end{equation*}\nVery importantly, let us remark that the position of~\\(y\\) in the\nresult is the same as in the input (the top). The second rule\ncorresponds to the case where the item we are looking for,\nnamely~\\(x\\), is found to be the top of the current stack, which is a\nsubstack of the original input. A stack made of successive items from\nthe beginning of a given stack is called a \\emph{prefix} of the\nlatter. When a stack is a substack of another, that is, it is made of\nsuccessive items including the last, it is called a \\emph{suffix}. We\nknow that the~\\(x\\) in \\(\\cons{x}{s}\\) is the first occurrence\nof~\\(x\\) in the original stack (the one in the first call), because we\nwouldn't be dealing with this case \\emph{again}: the specification\nstates that this first occurrence must be absent from the resulting\nstack; since it is now at the top of a suffix, we just need to end\nwith~\\(s\\), \\emph{which we do not visit}:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{sfst}(\\cons{x}{s},x) & \\rightarrow & \\underline{s}\\,;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\cons{y}{\\fun{sfst}(s,x)}\\,.\n\\end{array}\n\\end{equation*}\nThe first rule handles the case where we traversed the whole original\nstack (up to \\(\\el\\)) without finding~\\(x\\). Thus the result is simply\nthe empty stack because the empty stack without~\\(x\\) is the empty\nstack:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\rightarrow & \\el;\\\\\n\\fun{sfst}(\\cons{x}{s},x) & \\rightarrow & s;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\cons{y}{\\fun{sfst}(s,x)}.\n\\end{array}\n\\end{equation*}\n\nLet us run some tests now and, in order avoid mistakes, it is handy to\nlabel the rules with some Greek letters:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el,x) & \\smashedrightarrow{\\theta} & \\el;\\\\\n\\fun{sfst}(\\cons{x}{s},x) & \\smashedrightarrow{\\iota} & s;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\smashedrightarrow{\\kappa} &\n\\cons{y}{\\fun{sfst}(s,x)}.\n\\end{array}\n\\end{equation*}\nThe item is absent in~\\(\\theta\\), the item is found in~\\(\\iota\\) and\nthe search continues with~\\(\\kappa\\). Note that equality is implicitly\nmeant in non\\hyp{}linear rules like~\\(\\iota\\); in other words, the\ncost of such equality test is~\\(0\\) in our model. Also remark how\nimportant is for rule~\\(\\iota\\) to be written before~\\(\\kappa\\),\notherwise \\(\\iota\\)~would be useless (so\\hyp{}called \\emph{dead\n code}\\index{functional language!dead code}). Here is an\nexample\\index{stack!skipping an item!example} of a successful search:\n\\begin{equation*}\n\\fun{sfst}([3,0,1,2],1) \\xrightarrow{\\smash{\\kappa}}\n\\cons{3}{\\fun{sfst}([0,1,2],1)} \\xrightarrow{\\smash{\\kappa}}\n\\cons{3,0}{\\fun{sfst}([1,2],1)} \\xrightarrow{\\smash{\\iota}} [3,0,2].\n\\end{equation*}\nNow an example of an unsuccessful\\index{stack!skipping an\n item!example} search:\n\\begin{equation*}\n\\fun{sfst}([3,0],4) \\xrightarrow{\\smash{\\kappa}}\n\\cons{3}{\\fun{sfst}([0],4)} \\xrightarrow{\\smash{\\kappa}}\n\\cons{3,0}{\\fun{sfst}(\\el,4)} \\xrightarrow{\\smash{\\theta}}\n[3,0].\n\\end{equation*}\nMore complicated examples, given \\vpageref{sfst_ex}, yield\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}([4,[1,2],\\el,\\el,4],\\el)\n& \\smashedrightarrow{\\kappa} &\n [4|\\fun{sfst}([[1,2],\\el,\\el,4],\\el)]\\\\\n& \\smashedrightarrow{\\kappa} &\n [4|[[1,2]|\\fun{sfst}([\\el,\\el,4],\\el)]]\\\\\n& = & [4,[1,2]|\\fun{sfst}([\\el,\\el,4],\\el)]\\\\\n& \\smashedrightarrow{\\iota} & [4,[1,2]|[\\el,4]]\\\\\n& = & [4,[1,2],\\el,4].\n\\end{array}\n\\end{equation*}\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}([3,\\el],[5,2])\n& \\smashedrightarrow{\\kappa} & [3|\\fun{sfst}([\\el],[5,2])]\\\\\n& \\smashedrightarrow{\\kappa} & [3|[\\el|\\fun{sfst}(\\el,[5,2])]]\\\\\n& = & [3,\\el|\\fun{sfst}(\\el,[5,2])]\\\\\n& \\smashedrightarrow{\\theta} & [3,\\el|\\el]\\\\\n& = & [3,\\el].\n\\end{array}\n\\end{equation*}\nOnce we are convinced that our definition is correct\\index{soundness}\nand complete with respect to the specification, there is a little\nextra worth testing: we can check \\emph{what happens for inputs which\n are not expected by the specification.} Our specification says at\none point that the second argument of \\fun{sfst/2} is a stack. What\nhappens if we supply an integer instead? For example, we have\n\\(\\fun{sfst}(3,\\el) \\nrightarrow\\). We have a match failure, that is,\nthe rewrites are stuck, so that our definition is not\n\\emph{robust},\\index{robustness} in other words, it fails abruptly on\nunspecified inputs.\n\nWhen programming in the small, as we do here, robustness is usually\nnot a concern because we want to focus on learning a language by\nexpressing simple algorithms, but when developing large applications,\nwe must take care of making the code robust by catching and signalling\nerrors. Notice that a program can be complete\\index{completeness} but\nnot robust,\\index{robustness} because completeness is relative to\nwhat is specified behaviour (all valid inputs must be accepted and not\nlead to an error), whereas robustness is relative to what is left\nunspecified.\n\nThese considerations are germane to discussing the merits and\nweaknesses of scripting languages, whose semantics try hard to ignore\nerrors by defaulting on special values (like the empty string) to keep\nrunning. In the setting of our abstract functional language, we can\nuse a data constructor, that is, a function without evaluation rules,\nlike \\(\\fun{error}()\\), to signal an error or notify some piece of\ninformation about the arguments. For instance, here is the definition\nof a function which distinguishes between stacks and non\\hyp{}stack\narguments:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{is\\_a\\_stack}(\\el) & \\rightarrow & \\fun{yes}();\\\\\n\\fun{is\\_a\\_stack}(\\cons{x}{s}) & \\rightarrow & \\fun{yes}();\\\\\n\\fun{is\\_a\\_stack}(s) & \\rightarrow & \\fun{no}().\n\\end{array}\n\\end{equation*}\nData constructors come handy in signalling errors because they are\nlike unique identifiers, therefore they cannot be confused with any\nother kind of data the function computes and so can be detected easily\nby the caller. Consider this robust\\index{robustness} version of\n\\fun{sfst/2} which discriminates errors:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}(\\el, x) & \\rightarrow & \\el;\\\\\n\\fun{sfst}(\\cons{x}{s},x) & \\rightarrow & s;\\\\\n\\fun{sfst}(\\cons{y}{s},x) & \\rightarrow & \\cons{y}{\\fun{sfst}(s,x)};\\\\\n\\fun{sfst}(s,x) & \\rightarrow & \\underline{\\fun{error}()}.\n\\end{array}\n\\end{equation*}\nThen a function calling \\fun{sfst/2} can make the difference between a\nnormal rewrite and an error by using a data constructor in a pattern:\n\\begin{equation*}\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{caller}(s,x) & \\rightarrow & \\fun{check}(\\fun{sfst}(s,x)).\\\\\n\\fun{check}(\\fun{error}()) & \\rightarrow & \\fbcode{CCCCC}\\,;\\\\\n\\fun{check}(r) & \\rightarrow & \\fbcode{CCCCC}\\,.\n\\end{array}\n\\end{equation*}\n%%% END\n\n\\paragraph{Cost}\n\nIn general, the cost\n\\(\\C{\\fun{sfst}}{n}\\)\\index{sfst@$\\C{\\fun{sfst}}{n}$|(} of\n\\(\\fun{sfst}(s,x)\\), where \\(n\\)~is the length of~\\(s\\), depends\non~\\(x\\) being present in~\\(s\\) or not. In the latter case, the trace\nis \\(\\kappa^n\\theta\\), so \\(\\C{\\fun{sfst}}{n} = \\len{\\kappa^n\\theta} =\nn + 1\\). If the former, the cost depends on the position of~\\(x\\)\nin~\\(s\\). Let us set that the top of~\\(s\\) is at position~\\(0\\) and\n\\(x\\)~occurs at position~\\(j\\). We then have \\(\\C{\\fun{sfst}}{n,j} =\n\\len{\\kappa^j\\iota} = j + 1\\)\\index{sfst@$\\C{\\fun{sfst}}{n}$|)}. If we\ndecide that position~\\(n\\) (or greater) means absence, we can actually\nretain the last formula for both cases.\n\nThe minimum cost \\(\\B{\\fun{sfst}}{n}\\)\\index{sfst@$\\B{\\fun{sfst}}{n}$}\nis then the minimum\\index{stack!skipping an item!minimum cost} value\nof \\(\\C{\\fun{sfst}}{n,j}\\), for \\(j\\)~ranging from \\(0\\)~to~\\(n\\),\ntherefore \\(\\B{\\fun{sfst}}{n} = \\C{\\fun{sfst}}{n,0} = 1\\), that is,\nwhen the item occurs at the top, and, dually, the maximum\ncost\\index{stack!skipping an item!maximum cost}\nis\\index{sfst@$\\W{\\fun{sfst}}{n}$} \\(\\W{\\fun{sfst}}{n} =\n\\C{\\fun{sfst}}{n,n} = n + 1\\), that is, when the item is absent. The\naverage cost\n\\(\\M{\\fun{sfst}}{n}\\)\\index{sfst@$\\M{\\fun{sfst}}{n}$}\\index{stack!skipping\n an item!average cost} of a successful search assumes that \\(j\\)~can\ntake all the positions in the stack:\n\\begin{equation*}\n \\M{\\fun{sfst}}{n} = \\frac{1}{n}\\sum_{j=0}^{n-1}\\C{\\fun{sfst}}{n,j} =\n \\frac{1}{n}\\sum_{j=0}^{n-1}(j+1) = \\frac{1}{n}\\sum_{j=0}^{n}j\n = \\frac{n+1}{2} \\sim \\frac{n}{2},\n\\end{equation*}\nby equation~\\eqref{eq:sum_k} \\vpageref{eq:sum_k}.\n\nNotice that rule~\\(\\kappa\\) implies the creation of a\n(\\texttt{|})\\hyp{}node, which we call\n\\emph{cons\\hyp{}node}\\index{cons-node}, as shown in\n\\fig~\\vref{fig:sfst_dag}.\n\\begin{figure}\n\\centering\n\\includegraphics[bb=71 664 345 721]{sfst_dag}\n\\caption{Directed acyclic graphs for \\fun{sfst/2}\n\\label{fig:sfst_dag}}\n\\end{figure}\nHence, whilst the contents of the new stack is shared with the\noriginal stack, \\(j\\)~nodes are newly allocated if~\\(x\\) occurs at\nposition~\\(j\\) in~\\(s\\). The worst case happens when \\(x\\)~is absent\nso the memory\\index{stack!skipping an item!memory} needed amounts to\n\\(n\\)~nodes, all of which being useless because in this case\n\\(\\fun{sfst}(s,x) \\equiv s\\). If we want to avoid this situation,\nanother definition of \\fun{sfst/2} has to be devised, one that\ndiscards all constructed nodes and allows us to reference the input\nwhen~\\(x\\) is missing.\n\nThe crux of the matter is embodied in the construct\n\\(\\cons{y}{\\texttt{\\textvisiblespace}}\\) of rule~\\(\\kappa\\), called\nthe \\emph{context}\\index{functional language!context of a call} of the\ncall \\(\\fun{sfst}(s,x)\\)\\index{sfst@\\fun{sfst/2}|)}, which we want\nif~\\(x\\) is present, but not otherwise. To resolve this conflicting\nrequirement, we opt for removing the context and store the information\nit contains (\\(y\\)) into an accumulator\\index{functional\n language!accumulator}, in a new rule~\\(\\xi\\) derived\nfrom~\\(\\kappa\\). We use the accumulator in a new rule~\\(\\nu\\) derived\nfrom~\\(\\iota\\). The new \\fun{sfst/2} is called\n\\fun{sfst\\(_0\\)/2}\\index{sfst0@\\fun{sfst\\(_0\\)/2}} and is shown in\n\\fig~\\ref{fig:sfst0}.\\index{sfst@\\fun{sfst/4}}\n\\begin{figure}[b]\n\\begin{equation*}\n\\boxed{%\n\\begin{array}{@{}r@{\\;}l@{\\;}lr@{\\;}l@{\\;}l@{}}\n\\fun{rcat}(\\el,t) & \\xrightarrow{\\smash{\\zeta}} & t; &\n\\fun{sfst}(\\el,x,t,u) & \\xrightarrow{\\smash{\\mu}} & u;\\\\\n\\fun{rcat}(\\cons{x}{s},t) & \\xrightarrow{\\smash{\\eta}}\n & \\fun{rcat}(s,\\cons{x}{t}). &\n\\fun{sfst}(\\cons{x}{s},x,t,u) & \\xrightarrow{\\smash{\\nu}}\n & \\fun{rcat}(t,s);\\\\\n\\fun{sfst}_0(s,x) & \\xrightarrow{\\smash{\\lambda}}\n & \\fun{sfst}(s,x,\\el,s). &\n\\fun{sfst}(\\cons{y}{s},x,t,u) & \\xrightarrow{\\smash{\\xi}}\n & \\fun{sfst}(s,x,\\cons{y}{t},u).\n\\end{array}}\n\\end{equation*}\n\\caption{Skipping the first occurrence with maximum sharing\n\\label{fig:sfst0}}\n\\end{figure}\nOf course, whilst in~\\(\\iota\\) we just referenced~\\(s\\), the\nconstruction corresponding to the now missing context of~\\(\\kappa\\)\nmust be performed by~\\(\\nu\\). Also, we must add another argument which\nrefers to the original stack, so we can use it in a new rule~\\(\\mu\\),\nthe pendant of~\\(\\theta\\). Note the shapes of the right\\hyp{}hand\nsides: they are either a value~(\\(\\zeta\\) and~\\(\\mu\\)) or a function\ncall whose arguments contain values. In other words, no function call\nhas a context.\n\nThis syntactic property of a definition is named \\emph{tail\n form}\\index{functional language!tail form}. Intuitively, the\npractical consequence of such a form is that terminating calls unfold\nuntil a value is reached and nothing else is left to be done:\n\\emph{the value of the last call is the value of the first call}. This\nkind of definition enables the sharing in rule~\\clause{\\mu}, where\n\\(u\\) (the reference to the original stack) becomes the value, instead\nof \\(\\fun{rev}(t)\\).\n\nImplementations of functional language often use this property to\noptimise the evaluation, as we shall see in the last part of this\nbook. The downside of \\fun{sfst\\(_0\\)/2} with respect to \\fun{sfst/2}\nis the additional cost incurred by having to reverse~\\(t\\) in\nrule~\\clause{\\nu}, that is, the call \\(\\fun{rcat}(t,s)\\). More\nprecisely, there are two complementary cases: either \\(x\\)~is missing\nin~\\(s\\) or \\(x\\)~occurs in~\\(s\\). Let us assume that \\(s\\)~contains\n\\(n\\)~items and \\(x\\)~is absent in~\\(s\\). The evaluation trace\n\\index{functional language!evaluation!trace} of the call\n\\(\\fun{sfst}_0(s,x)\\) is \\(\\lambda\\xi^n\\mu\\), so\n\\begin{equation*}\n \\C{\\fun{sfst}_0}{n}\n= \\len{\\lambda\\xi^n\\mu} = \\len{\\lambda} + n \\len{\\xi} + \\len{\\mu} = n\n+ 2.\n\\end{equation*}\nLet us now assume that \\(x\\)~occurs at position \\(k\\) in~\\(s\\),\nwith the first item having position~\\(0\\). The evaluation trace is\nthen \\(\\lambda \\xi^{k}\\nu\\eta^k\\zeta\\), hence\n\\begin{equation*}\n \\C{\\fun{sfst}_0}{n,k} = \\len{\\lambda \\xi^{k}\\nu\\eta^k\\zeta} = 2k + 3.\n\\end{equation*}\nClearly now,\n\\begin{align*}\n\\B{\\fun{sfst}_0}{0} & = 2,\\\\\n\\B{\\fun{sfst}_0}{n} & = \\min_{0 \\leqslant k < n}\\{\\C{\\fun{sfst}_0}{n}, \\C{\\fun{sfst}_0}{n,k}\\}\n = \\min_{0 \\leqslant k < n}\\{n+2,2k+3\\} = 3,\\\\\n\\W{\\fun{sfst}_0}{n} & = \\max_{0 \\leqslant k < n}\\{n+2,2k+3\\} = 2n+1,\n\\end{align*}\nwhere the minimum cost occurs when the item is the top of the stack;\nthe maximum cost happens when the sought item is last in the stack (at\nthe bottom).\n\nSince calling \\fun{rcat/2}\\index{rcat@\\fun{rcat/2}} to reverse the\nvisited items is the source of the extra cost, we might try to\nmaintain the order of these items in \\fig~\\vref{fig:sfst1}\\index{sfst1@\\fun{sfst\\(_1\\)/1}}\n\\begin{figure}[b]\n\\begin{equation*}\n\\boxed{%\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{sfst}_1(s,x) & \\rightarrow\n & \\fun{sfst}_2(s,x,\\el,s).\\\\\n\\fun{sfst}_2(\\el,x,t,u) & \\rightarrow & u;\\\\\n\\fun{sfst}_2(\\cons{x}{s},x,t,u) & \\rightarrow & \\fun{cat}(t,s);\\\\\n\\fun{sfst}_2(\\cons{y}{s},x,t,u) & \\rightarrow\n & \\fun{sfst}_2(s,x,\\fun{cat}(t,[y]),u).\n\\end{array}}\n\\end{equation*}\n\\caption{Skipping the first occurrence (bad design)\n\\label{fig:sfst1}}\n\\end{figure}\nbut using stack catenation instead of pushing. The problem is that the\nlast rule of \\fun{sfst\\(_2\\)/4}\\index{sfst@\\fun{sfst\\(_2\\)/4}}\n\\index{cat@\\fun{cat/2}} yields\n\\(\\fun{cat}(\\dots\\fun{cat}(\\fun{cat}(\\el,[x_1]),[x_2])\\dots)\\), whose\ncost we know to be quadratic \\index{cost!quadratic $\\sim$} as in the\nrewrite~\\eqref{eq:rev0}\n\\vpageref{eq:rev0}\\index{rev0@\\fun{rev\\(_0\\)/1}}, from which we can\nquickly conclude that \\(\\W{\\fun{sfst}_1}{n} \\sim \\tfrac{1}{2}n^2\\).\n\n\\paragraph{Last occurrence}\n\\index{stack!skipping an item!last occurrence}\n\nLet us suppose that \\(\\fun{slst}(s,x)\\)\\index{slst@\\fun{slst/2}}\n(\\emph{skip the last occurrence}) evaluates in a stack identical\nto~\\(s\\) but without the last occurrence of~\\(x\\). In particular,\nif~\\(x\\) is absent in~\\(s\\), then the value of the call is identical\nto~\\(s\\). The first design that may come to mind is to see this\nproblem as the dual problem of ignoring the first occurrence:\n\\begin{equation}\n\\fun{slst}_0(s,x) \\xrightarrow{\\smash{\\pi}}\n \\fun{rev}(\\fun{sfst}(\\fun{rev}(s),x)).\n\\label{eq:slst0}\n\\end{equation}\nIf~\\(x\\) is missing in~\\(s\\), we\nhave\\index{slst0@$\\C{\\fun{slst}_0}{n}$}\\index{rev@$\\C{\\fun{rev}}{n}$}\n\\(\\C{\\fun{slst}_0}{n} = 1 + \\C{\\fun{rev}}{n} + \\W{\\fun{sfst}}{n} +\n\\C{\\fun{rev}}{n} = 3n+6\\). If~\\(x\\) occurs in~\\(s\\) at position~\\(k\\),\n\\(\\C{\\fun{slst}_0}{n,k} = 1 + \\C{\\fun{rev}}{n} +\n\\C{\\fun{sfst}}{n,n-k-1} + \\C{\\fun{rev}}{n-1} = 3n - k +\n4\\). Therefore, we can derive the minimum and\nmaximum\\index{stack!skipping an item!minimum\n cost}\\index{stack!skipping an item!maximum cost} costs:\n\\begin{equation*}\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\B{\\fun{slst}_0}{n} = \\min_{k < n}\\{3n+6,3n-k+4\\} = 2n + 5,\n\\end{equation*}\nwhen\\index{slst@$\\B{\\fun{slst}_0}{n}$} \\(x\\)~is last in~\\(s\\), and\n\\begin{equation*}\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\W{\\fun{slst}_0}{n} = \\max_{k < n}\\{3n+6,3n-k+4\\} = 3n+6,\n\\end{equation*}\nwhen\\index{slst@$\\W{\\fun{slst}_0}{n}$} \\(x\\)~is missing in~\\(s\\). The\nmean cost\\index{stack!skipping an item!average\n cost}\\index{slst@$\\M{\\fun{slst}_0}{n}$} when \\(x\\)~is present is\n\\begin{equation*}\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\M{\\fun{slst}_0}{n} = \\frac{1}{n}\\sum_{k=0}^{n-1}\\C{\\fun{slst}_0}{n,k}\n = \\frac{1}{n}\\sum_{k=0}^{n-1}(3n - k + 4)\n = \\frac{5n+9}{2} \\sim \\frac{5}{2}n.\n\\end{equation*}\nWhen \\(x\\)~is present, the worst case is when it is the top of the\nstack: \\(\\W{\\fun{slst}_0}{n} = \\max_{k0}\\). We start with the equality \\((k+1)^3 =\nk^3 + 3k^2 + 3k + 1\\), hence \\((k+1)^3 - k^3 = 3k^2 + 3k + 1\\). Then\nwe can sum these differences, whose terms cancel out, leaving the\nfirst and the last:\n\\begin{align}\n && (1+1)^3 - \\boxed{1^3} &= 3 \\cdot 1^2 + 3 \\cdot 1 + 1\\notag\\\\\n+ && (2+1)^3 - 2^3 &= 3 \\cdot 2^2 + 3 \\cdot 2 + 1\\notag\\\\\n+ &&&\\;\\;\\vdots\\notag\\\\\n+ && \\boxed{(n+1)^3} - n^3 &= 3n^2 + 3n + 1\\notag\n\\intertext{\\rule{\\linewidth}{0.4pt}}\n\\Rightarrow\n &&\\boxed{(n+1)^3} - \\boxed{1^3}\n &= 3 \\sum_{k=1}^{n}{k^2} + 3 \\sum_{k=1}^{n}{k} + n\\notag\\\\\n && n^3 + 3n^2 + 3n\n &= 3 \\sum_{k=1}^{n}{k^2} + 3 \\cdot \\frac{n(n+1)}{2} + n\\notag\\\\\n\\Leftrightarrow && \\sum_{k=1}^{n}{k^2} &= \\frac{n(n+1)(2n+1)}{6}.\n\\label{eq:sum_of_squares}\n\\end{align}\n\n\\paragraph{Exercises}\n\\begin{enumerate}\n\n \\item Prove that \\(\\fun{sfst/2} = \\fun{sfst\\(_0\\)/2}\\).\n \\index{sfst@\\fun{sfst/2}}\\index{sfst0@\\fun{sfst\\(_0\\)/2}}\n\n \\item Show that \\(\\B{\\fun{sfst}_0}{n} = 3\\), \\(\\W{\\fun{sfst}_0}{n} =\n 2n+1\\) and \\(\\M{\\fun{sfst}_0}{n} = n+2\\) (successful\n search).\n\n \\item Prove \\fun{slst/2} = \\fun{slst\\(_0\\)/2}.\n\n \\item Show that, in a worst case to be identified,\n \\(\\fun{slst}_0(s,x)\\) creates \\(3n\\)~useless nodes if~\\(s\\)\n contains \\(n\\)~items. Compare the memory usage of\n \\fun{slst\\(_0\\)/2}\\index{slst@\\fun{slst\\(_0\\)/2}} with that of\n \\fun{slst/2}\\index{slst@\\fun{slst/2}}.\n\n\\end{enumerate}\n", "meta": {"hexsha": "b1cf4e5309fe31d5b5afc6285e5476824e879b12", "size": 34421, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "skipping.tex", "max_stars_repo_name": "rinderknecht/Book", "max_stars_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "skipping.tex", "max_issues_repo_name": "rinderknecht/Book", "max_issues_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "skipping.tex", "max_forks_repo_name": "rinderknecht/Book", "max_forks_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.0173796791, "max_line_length": 98, "alphanum_fraction": 0.657912321, "num_tokens": 12378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707283, "lm_q2_score": 0.8333245953120233, "lm_q1q2_score": 0.6408934773194396}} {"text": "\\def\\mA{\\ensuremath{{\\bf A}}}\n\\def\\mB{\\ensuremath{{\\bf B}}}\n\\def\\mC{\\ensuremath{{\\bf C}}}\n\\def\\mD{\\ensuremath{{\\bf D}}}\n\\def\\mu{\\ensuremath{{\\bf u}}}\n\\def\\mx{\\ensuremath{{\\bf x}}}\n\\def\\my{\\ensuremath{{\\bf y}}}\n\n\\documentclass[12pt]{article}\n\\usepackage{siunitx}\n\\begin{document}\n\\title{Drive kinematics and simulation}\n\\author{sss}\n\\maketitle\n\nA useful reference: http://fbsbook.org.\n\n\\section{Motor constants}\n\nA DC motor typically consists of a permanent magnet rotating relative\nto a conducting coil. As the magnet rotates, it induces an EMF in the\ncoil. This `back-EMF' is proportional to the angular velocity:\n\\begin{equation}\n E = K_e \\omega .\n\\end{equation}\n\nA DC motor can be modeled as a resistance $R$ in series with the back EMF.\nTherefore, given the applied voltage $V$ and current $I$:\n\\begin{equation}\n V = IR + E = IR + K_e \\omega.\n\\end{equation}\n\nIn other words, when a voltage is applied to the motor, it increases in speed\nuntil the back-EMF (plus resistive losses) balance the applied voltage.\n\nIf the current with no load is $I_0$, then the maximum speed of the motor\nis\n\\begin{equation}\n \\omega_{\\max} = {V-I_0R \\over K_e}.\n\\end{equation}\n\nAlso note that when the motor is stalled, $E=0$, so the resistance\ncan be found using the current drawn at the stall:\n\\begin{equation}\n R = V / I_{\\mathrm{stall}}.\n\\end{equation}\n\nIt is common to write $K_V = 1/K_e$. Then the motor speed for a given applied\nvoltage is\n\\begin{equation}\n \\omega = K_V (V-IR).\n\\end{equation}\nIn practice, the $IR$ term is usually small compared to $V$ and can be\nneglected, giving\n\\begin{equation}\n \\omega = K_V V.\n\\end{equation}\n$K_V$ is sometimes called the motor velocity constant, and has SI units\nof $1/Vs$ (though it is sometimes given as $\\textrm{RPM}/V$).\n\nThe torque of the motor is related to the current via the motor\ntorque constant $K_T$:\n\\begin{equation}\n \\tau = K_T I.\n\\end{equation}\nThis constant can be calculated from the torque required to stall the motor\nand the current draw at the stall:\n\\begin{equation}\n K_T = \\tau_{\\textrm{stall}} / I_{\\textrm{stall}}.\n\\end{equation}\n\n$K_T$ and $K_V$ are not independent; they are inversely related.\nThe electrical power input to the motor is\n\\begin{equation}\n P_{\\textrm{in}} = VI = (IR + K_e\\omega)I = I^2R + IK_e\\omega\n\\end{equation}\nwhile the mechanical power out is\n\\begin{equation}\n P_{\\textrm{out}} = \\tau\\omega = IK_T\\omega,\n\\end{equation}\nwhile losses are\n\\begin{equation}\n P_{\\textrm{loss}} = I^2R.\n\\end{equation}.\nBut\n\\begin{equation}\n P_{\\textrm{in}} = P_{\\textrm{out}} + P_{\\textrm{loss}},\n\\end{equation}\nwhich implies that\n\\begin{equation}\n K_T = K_e = 1/K_V.\n\\end{equation}\n\nTo take a specific example, the Falcon 500 motor operates at $\\SI{12}{\\V}$,\nwith $I_0 = \\SI{1.5}{\\A}$, $I_{\\textrm{stall}} = \\SI{257}{\\A}$,\n$\\tau_{\\textrm{stall}} = \\SI{4.69}{\\N\\m}$,\nand $\\omega_{\\max} = \\SI{6380}{RPM} = \\SI{668}{/\\s}$.\nThis then gives $R=V/I_{\\textrm{stall}}=\\SI{0.047}{\\ohm}$,\n$K_V = \\omega_{\\max} / (V-I_0R) = \\SI{60.0}{/\\V\\s}$ and\n$K_T = \\tau_{\\textrm{stall}} / I_{\\textrm{stall}} = \\SI{0.018}{\\N\\m/\\A}$.\nAnd $1/60 = 0.017$, so these are nearly inverse of each other.\n(Note that since power is voltage times current,\n$\\textrm{VA} = \\textrm{Nm}/\\textrm{s}$ or $\\textrm{Vs} = \\textrm{NM}/\\textrm{A}$.)\n\n\n\\section{A unicycle}\n\nLet's now apply this to a putative vehicle with a single wheel,\nconstrained to move in one dimension. In addition to\n$R$, $K_V$, and $K_T$ of the motor, let $m$ be the mass,\n$r_w$ be the wheel radius, and $g$ be the gear ratio between\nthe motor and the wheel.\n\nLet $v$ be the (one-dimensional) velocity and $V$ be the voltage\napplied to the motor. Then we want to write:\n\\begin{equation}\n m \\dot v = c_1 v + c_2 V.\n\\end{equation}\n(Using the notation $\\dot v = dv/dt$.)\nSo the right-hand side represents the force acting on the vehicle.\n\nConsider the $c_2$ term first. If the motor voltage is $V$, then the\ncurrent is $I = V/R$, the motor torque is $\\tau_m = IK_T = K_T V/R$,\nthe wheel torque is $\\tau_w = gK_T V/R$, and the force on the vehicle\nis $F = \\tau_w / r_w$, which implies\n\\begin{equation}\n c_2 = {g K_T \\over R r_w}.\n\\end{equation}\n\nNow, consider the case where the voltage on the motor is zero.\nA rotating motor generates an EMF, so if the voltage is zero, then\nsomething must be absorbing the power produced by the motor; hence,\nthe motor will act as a brake. We can thus write\n\\begin{equation}\n m \\dot v = - { g K_T \\over R r_w } E,\n\\end{equation}\nwhere $E$ is the back-EMF. But $E = K_e\\omega_m = g \\omega_w / K_V =\ng v / r_w K_V$. Putting this together, we get\n\\begin{equation}\n c_1 = - { g^2 K_T \\over K_V R r_w^2}.\n\\end{equation}\n\nTo specify the state of the vehicle, we also need its position $s$.\nThe equation for this is trivial:\n\\begin{equation}\n \\dot s = v.\n\\end{equation}\n\n\\section{State space representation}\n\nThe dynamical variables describing the state of a system make up the\n\\emph{state space}. These variables contain sufficient information\nto find the future state of the system.\n\nFormally, we write the state variables as an $n$-element vector $\\mx$.\nThe system can also have control inputs, which we represent\nas a $p$-element vector $\\mu$. We can also make measurements\non the system, which we represent as an $q$-element vector $\\my$.\nThe system can then be described by\n\\begin{eqnarray}\n {\\bf\\dot x} &=& f(\\mx, \\mu)\\\\\n \\my &=& h(\\mx, \\mu)\n\\end{eqnarray}\n\nIf $f$ and $h$ are time-invariant and linear, then we can write this as\n\\begin{eqnarray}\n {\\bf\\dot x} &=& \\mA\\mx + \\mB\\mu \\\\\n \\my &=& \\mC\\mx + \\mD\\mu.\n\\end{eqnarray}\nThe matrix $\\mA$ is called the `system' or `dynamics' or `state' matrix;\n$\\mB$ is called the `input' or `control' matrix;\n$\\mC$ is called the `output' or `sensor' matrix; and\n$\\mD$ is called the `feedthrough' or `feedforward' or `direct' matrix.\n\n\n\\section{Differential drive}\n\nHere, the system consists of two identical motors with wheels separated\nby distance $2r_b$, and the system can move in two dimensions.\nThe mass is again $m$, and the moment of inertia is $I$.\n\nThe state variables here are the $x$ and $y$ positions, $s_x$ and $s_y$,\nthe heading $\\theta$, the positions of the two wheels\n$d_L$ and $d_R$, and their velocities $\\dot d_L$, $\\dot d_R$.\nWe can write this as a vector:\n$\\mx = [s_x, s_y, \\theta, \\dot d_L, \\dot d_R, d_L, d_R]^T$.\nThe control inputs are the motor voltages: $\\mu = [V_L, V_R]^T$.\n\nUsing the results from the previous section, the forces on the\nvehicle produced by the left and right motors are\n\\begin{equation}\n F_L = c_1 \\dot d_L + c_2 V_L.\n\\end{equation}\nand similarly for the other motor. Given $F_L$ and $F_R$,\nthe total force is $F_{\\textrm{tot}} = F_L + F_R$ and the torque is\n$\\tau = (F_L -F_R) r_b$. Therefore, the acceleration for one\nof the wheels is\n\\begin{eqnarray}\n \\ddot d_L &=& {1\\over m}F_{\\textrm{tot}} + {r_b\\over I}\\tau\\\\\n &=& {1\\over m}(F_L + F_R) + {r_b^2\\over I}(F_L - F_R)\\\\\n &=& \\left({1\\over m} + {r_b^2\\over I}\\right) c_1 \\dot d_L +\n \\left({1\\over m} - {r_b^2\\over I}\\right) c_1 \\dot d_R +\\\\\n & & \\left({1\\over m} + {r_b^2\\over I}\\right) c_2 V_L +\n \\left({1\\over m} - {r_b^2\\over I}\\right) c_2 V_R \\nonumber\\\\\n &=& c_3 c_1 \\dot d_L + c_4 c_1 \\dot d_R + c_3 c_2 V_L + c_4 c_2 V_R ,\n\\end{eqnarray}\nwhere\n\\begin{eqnarray}\n c_3 &=& \\left({1\\over m} + {r_b^2\\over I}\\right) \\\\\n c_4 &=& \\left({1\\over m} - {r_b^2\\over I}\\right).\n\\end{eqnarray}\nSimilarly,\n\\begin{equation}\n \\ddot d_R = c_4 c_1 \\dot d_L + c_3 c_1 \\dot d_R + c_4 c_2 V_L + c_3 c_2 V_R.\n\\end{equation}\n\nIn addition, if we call the forward velocity\n$v = {1\\over 2}(\\dot d_L + \\dot d_R)$, then\n\\begin{eqnarray}\n \\dot s_x &=& v \\cos \\theta \\\\\n \\dot s_y &=& v \\sin \\theta \\\\\n \\dot \\theta &=& (\\dot d_R - \\dot d_L) / 2r_b.\n\\end{eqnarray}\n\nFor the subset of state variables\n$\\mx = [\\dot d_L, \\dot d_R, d_L, d_R]^T$, with $\\mu = [V_L, V_R]^T$,\nthe dynamics can then be written in the notation of the previous\nsection as\n\\begin{equation}\n \\dot\\mx = \\mA\\mx + \\mB \\mu\n\\end{equation}\nwith\n\\begin{equation}\n \\mA = \\left[ \\begin{array}{cccc}\n c_1c_3 & c_1c_4 & 0 & 0 \\\\\n c_1c_4 & c_1c_3 & 0 & 0 \\\\\n 1 & 0 & 0 & 0 \\\\\n 0 & 1 & 0 & 0 \\\\\n\\end{array} \\right]\n\\end{equation}\nand\n\\begin{equation}\n \\mB = \\left[ \\begin{array}{cc}\n c_2c_3 & c_2c_4 \\\\\n c_2c_4 & c_2c_3 \\\\\n 0 & 0 \\\\\n 0 & 0 \\\\\n\\end{array} \\right].\n\\end{equation}\n\n\n\\section{Runge-Kutta integration}\n\nGiven\n\\begin{equation}\n {dy \\over dt} = f'(y, t)\n\\end{equation}\nwith an initial value of $y$ at some $t$, one can use Euler's method\nto integrate this in steps of $t$ of size $h$:\n\\begin{equation}\n y_{n+1} = y_n + hf'(y_n, t_n).\n\\end{equation}\n\nIn practice, however, this is not a good method to use.\nThe error per step is $O(h^2)$, as can be seen from a series expansion:\n\\begin{equation}\n y(t_n+h) = y(t_n) + {dy\\over dt}(t_n) h + O(h^2).\n\\end{equation}\nSo over the $\\sim 1/h$ steps needed to cover a fixed-size interval,\nthe error is $O(h)$. This is not very accurate.\n\nWe can do better by first taking a trial step to the midpoint\nof the interval and then using the derivatives calculated there\nfor entire step. That is,\n\\begin{eqnarray}\n k_1 &=& h f'(t_n, y_n) \\\\\n k_2 &=& h f'(t_n + {h\\over2}, y_n + {k_1\\over2}) \\\\\n y_{n+1} &=& y_n + k_2.\n\\end{eqnarray}\nThis ends up canceling some the first order error terms, giving\na step error of $O(h^3)$.\n\nOne can go further in canceling error terms. Fourth-order Runge-Kutta\nis defined by \n\\begin{eqnarray}\n k_1 &=& h f'(t_n, y_n) \\\\\n k_2 &=& h f'(t_n + {h\\over2}, y_n + {k_1\\over2}) \\\\\n k_3 &=& h f'(t_n + {h\\over2}, y_n + {k_2\\over2}) \\\\\n k_4 &=& h f'(t_n + h, y_n + k_3) \\\\\n y_{n+1} &=& y_n + {k_1\\over6} + {k_2\\over3} + {k_3\\over3} + {k_4\\over6}\n\\end{eqnarray}\nand has a step error of $O(h^5)$. One can extend this further,\nbut that is usually not found to be worthwhile.\n\n\\end{document}\n", "meta": {"hexsha": "ca8b614d7cd294f78ea8546a98b56b63fe99aa50", "size": 9886, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "mechanum-note/mechanum.tex", "max_stars_repo_name": "FRC-Team-496/FRC2022", "max_stars_repo_head_hexsha": "37bda9cbc0c82fd1ad059ca84184b05dd7b23a5a", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "mechanum-note/mechanum.tex", "max_issues_repo_name": "FRC-Team-496/FRC2022", "max_issues_repo_head_hexsha": "37bda9cbc0c82fd1ad059ca84184b05dd7b23a5a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "mechanum-note/mechanum.tex", "max_forks_repo_name": "FRC-Team-496/FRC2022", "max_forks_repo_head_hexsha": "37bda9cbc0c82fd1ad059ca84184b05dd7b23a5a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.3986486486, "max_line_length": 82, "alphanum_fraction": 0.6670038438, "num_tokens": 3580, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672181749421, "lm_q2_score": 0.7931059585194573, "lm_q1q2_score": 0.6408829256187889}} {"text": "\\documentclass[12pt,onecolumn,a4paper]{report}\n\n\\usepackage{mathtools}\n\\usepackage[top=60pt,bottom=60pt,left=50pt, right=50pt]{geometry}\n\\usepackage{graphicx}\n\\usepackage{wrapfig}\n\\usepackage{caption}\n\\usepackage{subcaption}\n\\usepackage{listings}\n\n\\graphicspath{{../plots/}}\n\\renewcommand*\\thesection{\\arabic{section}}\n\n\\newfont{\\pica}{cmpica scaled 800}\n\n\\begin{document}\n\\title{COMP3032 Assignment: Image Contour Extraction}\n\\author{ \\textbf{ Dionisio Perez-Mavrogenis}\\\\\n\t\tSchool of Electronics and Computer Science \\\\\n\t\tFaculty of Physical and Applied Sciences \\\\\n\t\tUniversity of Southampton}\n\\maketitle\n\n\\tableofcontents\n\\pagebreak\n\n\\section{Introduction}\n\\subsection{Equations used}\n\nThe equations used for extracting the optimal contour from the provided image were the following :\n\\begin{equation}\nE(v) = \\sum^{N-2}_{i=1} E_i (v_{i-1},v_i,v_{i+1})\n\\end{equation}\n\n\\begin{equation}\nE_i (v_{i-1},v_i,v_{i+1}) = \\lambda \\frac{|v_{i+1} - 2v_i + v_{i+1}|^2}{|v_{i+1} - v_{i-1}|^2} + (1-\\lambda)I(v_i)\n\\end{equation}\n\n\\begin{equation}\nS_i(v_{i+1},v_i) = \\min_{v_{i-1}}[S_{i-1}(v_i,v_{i-1}) + E_i (v_{i-1},v_i,v_{i+1})]\n\\end{equation}\n\n\\begin{equation}\nE(v_i) = \\lambda |v_i - v_{i-1}| + (1-\\lambda)I(v_i)\n\\end{equation}\n\n\\begin{equation}\nS_i(v_i) = \\min_{v_{i-1}}[S_{i-1}(v_{i-1})+E_i(v_{i-1},v_i)]\n\\end{equation}\n\nwhere $\\lambda \\in [0,1]$ is the regularization parameter and $S_0(v_1,v_0) = 0$.\n\\section{Approach}\nThe coursework was coded in Matlab. Although the approach I have chosen is not exactly the same as the one in the specification(with differences highlighted as encountered), my implementation works fairly well.\nThe image used was the file \"tongue.png\" provided with the specification, but the images displayed in the report are zoomed-in in the area enclosed be the contours because it is easier to spot details.\n\\subsection{Constructing the search space}\nThe construction of the search space is done by the function {\\pica get\\_search\\_space}. Given a number $M$, the search space created will have $M*N$ points, where $M$ includes both points on the starting contours(only $M-2$ points are created). \n\nThe search space is constructed in the following manner : \nfor each point in the first contour, calculate the $x$ and $y$ difference with the corresponding point in the second contour ($ xdif_i = abs[v_{i,1}(x) - v_{i,2}(x)]/(M-1),ydif_i = abs[v_{i,1}(y) - v_{i,2}(y)]/(M-1) $) and then to get each new point $v'(i)$ between points $v_{i,1},v_{i,2}$ do :\n\n\\begin{displaymath}\nv'_i = [xdif_i*(J-1)+v_{i,1}(x),ydif_i*(J-1)+v_{i,1}(y)], \\forall J \\in [1,M]\n\\end{displaymath}\n\nwhere $v_{i,1}$ is point $i$ in contour $1$ and $dif_i$ is the absolute difference between points $(v_{i,1},v_{i,2})$(with $x$ and $y$ denoting the corresponding coordinate).I add $dif*(J-1)$ to $v_{i,1}$ because contour one is assumed to be \"bellow\" contour two, that is, it has $x_1 \\leq x_2$ and $y_1 \\leq y_2$.\nThe {\\pica get\\_search\\_space} function is given as inputs an M, the vector of contour1, the vector of contour2 and the matrix of intensities of the image and returns the intensity matrix for the search space points and matrices $X$ and $Y$, which contain the corresponding $x$ and $y$ coordinates of the search space points.\nBecause the results of calculating the search space point coordinates might not always yield an integer, Matlab's inbuilt function {\\pica round} function was used.\n\nThe code for {\\pica get\\_search\\_space} is given in Appendix A 0.1, Listing A.2, and samples of the generated search space are shown in Appendix B 0.3, Figure B.1 . \n\n\\subsection{Construction of Energy and Position matrices}\n\nThe function {\\pica get\\_matrices} is responsible for calculating the energy and position matrices. It is supplied with arguments : $\\lambda$ (the regularization parameter), the intensity matrix calculated by {\\pica get\\_search\\_space}, and the $X$ and $Y$ matrices(also calculated by {\\pica get\\_search\\_space}), and returns the energy matrix (an $M\\times N\\times M$ matrix) and a vector describing which points belong to the optimal contour (the backtracking is also done in this function), which then is handled by the plotting functions who locate the actual points on the image later on.\n\nFor a given point $v_i$ , the entry in the energy matrix is calculated according to equation(3), with resulting value being inserted in $en(y,x,z)$ and $z$ being equal to the $y$ index of $v_{i+1}$. The energy up to point $v_i$ is taken to be the energy between points $v_{i-1}$ and $v_i$, as this is assumed to be the optimal between these two and thus the optimal between the triplet.\n\nSimultaneously, the position matrix is being built. At each position, $posm(y,x,z)= z' $, where $z'$ is the $y$-index of the optimal $v_{i-1}$ point and $z$ represents the $y$-index of the $v_{i+1}$ point for which this calculation is taking place.\n\nThe first and last columns of the position and energy matrices present a special case, since we need points $v_0$ and $v_{N+1}$ to calculate results for points in columns 1 and N, and such points do not exist. Therefore, column 1 is initialized to 0, and the energy for points in column N is calculated by equations (4) and (5).\n\nBacktracking involved searching through the N-1 column of the energy matrix for the row that contained the minimum value , record the position of the row($y$-index) and the position of the minimum value in the row($z$-index). The rest of the procedure involves setting $y=z$ and reading $posm(y,x,z)$ into $z$, then repeating the process for every $x \\in [2,N-1]$.\n\nThe code for this section is given in Appendix A, Listing A.3. The plotting functions are called in the invoking script, which is given in Appendix A, Listing A.1.\n\n\\section{Evaluation of the algorithm}\nThe evaluation following is not general and refers only to my implementation. Also, timings were performed using Matlab's {\\pica tic toc} function, on machines in the Zepler labs.\n\n\\subsection{Robustness}\n\\paragraph{Robustness to different initialisations }\n\n\n\\begin{figure}[Ht]\n \\centering\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcap-100-0.05000}.png}\n \\caption{Initialized to 0.}\n \\end{subfigure}%\n ~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc. \n %(or a blank line to force the subfigure onto a new line)\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcapnz-100-0.05000}.png}\n \\caption{Initialized to intensity.}\n \\end{subfigure}\n \\caption{Optimal contour for different initialization of the energy matrix(M=100,$\\lambda=0.05$.)}\n\\end{figure}\n\nTwo different initialization methods were tried to test the algorithms robustness. The first method set the energies of the first column to zero, as the coursework suggested, whereas the second set each point's energy to be equal to its intensity.\n\nFigure 1 displays the resulting optimal contour for different initializations. The algorithm is robust enough so that the only difference is not on the actual contour being traced, but on the black space to the left of the tongue. Once the algorithm finds the optimal contour, the two contours are effectively identical.\n\n\n\n\\paragraph{Robustness for varying values of $\\lambda$}\nWith an initial value of $\\lambda=0.5$ the results were not satisfactory,as the result of tracing deviated a lot from the desired result, whereas when varying $\\lambda$ it was found that acceptable results are obtained when $\\lambda < 0.2$, and improved as $\\lambda$ got smaller. \n\n\\begin{figure}[Ht]\n \\centering\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcap-100-0.05000}.png}\n \\caption{$\\lambda=0.05$}\n \\end{subfigure}%\n ~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc. \n %(or a blank line to force the subfigure onto a new line)\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcap-100-0.50000}.png}\n \\caption{$\\lambda=0.5$}\n \\end{subfigure}\n \\caption{Optimal contour for different values of $\\lambda$(M=100)}\n\\end{figure}\n\nFigure 2 shows a comparison for different $\\lambda$ values and Appendix B 0.4 shows resulting traces for for different values of $\\lambda$\n\n\n\\subsection{Performance}\n\n\\begin{figure}[Htb]\n\\includegraphics[scale=0.5]{3D/cw_3d_benchmark.png}\n\\centering\n\\caption{Benchmark of 3D matrix implementation for various values of M.}\n\\end{figure}\n\nThis algorithm performs for $N-1$ columns(where $N$ is the number of points in the initial contours), for $M$ points, $M\\times M$ operations and thus a good approximation for the running time would be $N\\times M^3$, and one hypothesis suggests that the algorithms has $O(M^3)$ running time, since N could be considered a constant. My implementation uses four {\\pica for}-loops to perform the calculations and all calculations have been moved to the outermost loop possible .\n\nFigure 3 is plot of the value of $M$ and the corresponding running times. The shape of the curve suggests an exponential function for the total running time. It has been observed that changing the value of $\\lambda$ has no impact on the running time for any given $M$.\n\n\\subsection{Application to other images}\n\n\\begin{figure}[Ht]\n \\centering\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{curve.png}\n \\caption{The random curve}\n \\end{subfigure}%\n ~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc. \n %(or a blank line to force the subfigure onto a new line)\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{curvez-150-0.05000}.png}\n \\caption{The traced contour.}\n \\end{subfigure}\n \\caption{Application to a random curve(M=150)}\n\\end{figure}\nThe algorithm was applied to the image of a curve that I created, and the results are shown in Figure 4.\nThe result shown in Figure 4(b) is for M=150,$\\lambda$=0.05 and a zero initialization of the energy matrix's first column. \nThere is a small part in the beginning of the curve that is not tracked (missing points from the initial contours), and the two initial points of the optimal contour that are completely off the curve, as well as the last point of the tracing. These might be attributed to the initialization of the energy matrix's first column and different calculation for the last column's energy. However, when the algorithm \"finds\" the black curve it tracks the curve correctly, confirming that the initial and last points pose a special case and could perhaps be ignored, or be dealt with in a different fashion.\n\n\\subsection{Comparison with a different algorithm}\n\n\\begin{figure}[Ht]\n\\includegraphics[scale=0.5]{2D/cw_2d_benchmark.png}\n\\centering\n\\caption{Benchmark of 2D matrix implementation for various values of M.}\n\\end{figure}\n\n\\begin{figure}[Ht]\n \\centering\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcap-100-0.05000}.png}\n \\caption{3D matrix, M=100,$\\lambda=0.05$}\n \\end{subfigure}%\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{2D/{2d-clcap-100-0.05000}.png}\n \\caption{2D matrix, M=100,$\\lambda=0.05$}\n \\end{subfigure}\n \n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{clcap-300-0.05000}.png}\n \\caption{3D matrix, M=300,$\\lambda=0.05$.}\n \\end{subfigure}\n \n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{2D/{2d-clcap-300-0.05000}.png}\n \\caption{2D matrix, M=300,$\\lambda=0.05$.}\n \\end{subfigure}\n \\caption{Comparison of the two algorithms}\n\\end{figure}\n\n\n\\begin{figure}[Ht]\n\\includegraphics[scale=1]{2D/{2d-curve-150-0.00500}.png}\n\\centering\n\\caption{Tracing of curve by 2D algorithm ($M=150,\\lambda=0.005$), with portions of the contour being completely straight, illustrating this algorithm's linear sense of continuity.}\n\\end{figure}\n\n\nA second algorithm was implemented in order to extract the contour of the tongue. This is the algorithm presented in the \"Dynamic Programming\" lecture, and uses Equations (4) and (5) to extract the optimal contour. The algorithm performs a total of $N\\times M\\times M$ operations; a benchmark of this algorithm's performance is given in Figure 5, and its time complexity could be characterized as $O(M^2)$, if we treat $N$ as a constant term.\n\nFigure 6 shows a comparison of the contours extracted from the two algorithms. Both algorithms extracted the optimal contour fairly well, the 2D implementation being simpler and faster to implement and seems to extract a smoother contour. However, the 2D algorithm has got a sense of continuity that approximates a straight line because it only considers two points, something that is illustrated in Figure 7.\n\nCode for this algorithm is given in Appendix A.2.\n\n\\section{Conclusion}\nThis implementation of the algorithm might not be the optimal, as I am not as proficient in Matlab, however it illustrates that the algorithm works and allows comparison with other algorithms.\nAlso, even though the background of the image containing the tongue is the original one, the algorithms run on the negated version of that image (so that the contour we extract has the minimum energy) and the result is plotted on the original image.\n\n%\n\n%\t\tAPPENDICES \n\n%\n\n\n\\pagebreak\n\\appendix\n\\chapter{Code listings}\n\\subsection{Code for the 3D matrix algorithm}\n\n\\lstset{ %\n language=Matlab, % the language of the code\n numbers=left, \n caption={Code used to initialize, run and plot the results for the 3D algorithm.},\n}\n\\lstinputlisting{../start_coursework_3D.m}\n\n\\lstset{ %\n language=Matlab, % the language of the code\n numbers=left, \n caption={Function generating the search space for the 3D algorithm.},\n}\n\\lstinputlisting{../get_search_space.m}\n\n\\lstset{\n language=Matlab,\n numbers=left, \n caption={Function generating the position and energy matrices and doing the backtracking.},\n}\n\\lstinputlisting{../get_matrices_3D.m}\n\n\\lstset{\n language=Matlab,\n numbers=left, \n caption={Function that gets the optimal points returned from {\\pica get\\_matrices} and retrieves their (x,y) coordinates, and adds them to an $N\\times 2$ vector for the plotting functions.},\n}\n\\lstinputlisting{../get_optimals_3D.m}\n\n\\subsection{Energy extraction function of 2D matrix.}\nSince the search space extraction and plotting functions are identical to the 3D implementation, only the energy calculation and back-tracking code for the 2D implementations is shown.\n\\lstset{\n language=Matlab,\n numbers=left, \n caption={Function that generates the position and energy matrices for the 2D implementation of the algorithm.},\n}\n\\lstinputlisting{../2D_implementation/fill_energy_matrix.m}\n\n\\lstset{\n language=Matlab,\n numbers=left, \n caption={Function that does the calculation for the transition between points $v_{i-1},v_i$.},\n}\n\\lstinputlisting{../2D_implementation/point_energy.m}\n\n\\chapter{Some of the images generated}\n\n\\subsection{Images of the search space generated}\nThese are two sample images generated by the {\\pica get\\_search\\_space} function. The search points are not plotted together with the optimal contour because the resulting image would get too cluttered to draw any results.\n\\begin{figure}[ht]\n \\centering\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{sspace-50}.png}\n \\caption{Search space generated for M=50}\n \\end{subfigure}%\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{sspace-50}.png}\n \\caption{Search space generated for M=100}\n \\end{subfigure}\n \\caption{Plotting of two sample search spaces.}\n\\end{figure}\n\n\\subsection{Traced contours for different values of $\\lambda$}\n\\begin{figure}[H]\n\\centering\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{100-0.05000}.png}\n \\caption{$\\lambda=0.05$}\n \\end{subfigure}%\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{100-0.15000}.png}\n \\caption{$\\lambda=0.15$}\n \\end{subfigure}\n \\begin{subfigure}[b]{0.5\\textwidth}\n\t \t\\centering\n \t\t \\includegraphics[width=\\textwidth]{3D/{100-0.30000}.png}\n\t \t\\caption{$\\lambda=0.3$}\n \\end{subfigure}\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{3D/{100-0.50000}.png}\n \\caption{$\\lambda=0.5$}\n \\end{subfigure}\n \\caption{Plotting of two sample search spaces, with M=100.}\n\\end{figure}\n\nThese tracings are generated by the 3D energy matrix algorithm and have $M=100$.\n\\end{document}", "meta": {"hexsha": "080042005c5d666adfcfadc45f298e2d5d18c397", "size": 17565, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "year_4/intelligent_algorithms/report/intalgreport.tex", "max_stars_repo_name": "dperezmavro/courseworks_uni", "max_stars_repo_head_hexsha": "d65e76576ac22c2cec6c1995e462be2a20717090", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "year_4/intelligent_algorithms/report/intalgreport.tex", "max_issues_repo_name": "dperezmavro/courseworks_uni", "max_issues_repo_head_hexsha": "d65e76576ac22c2cec6c1995e462be2a20717090", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "year_4/intelligent_algorithms/report/intalgreport.tex", "max_forks_repo_name": "dperezmavro/courseworks_uni", "max_forks_repo_head_hexsha": "d65e76576ac22c2cec6c1995e462be2a20717090", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.7155963303, "max_line_length": 600, "alphanum_fraction": 0.7070310276, "num_tokens": 4699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059511841119, "lm_q2_score": 0.8080672066194946, "lm_q1q2_score": 0.6408829105266425}} {"text": "\\subsection{What is Pressure?}\nWhile the Zeroth law of thermodynamics allows for the notion of \"temperature\" to be well defined, it isn't a particularly useful definition. While it can tell us that, macroscopically, \"temperature is what you read from a thermometer\", it is of interest to see how the phenomenon of temperature arises microscopically. For this, we will first introduce the notion of pressure\\footnote{{\\tt https://www.youtube.com/watch?v=a01QQZyl-\\_I}}.\n\nWe will here define the pressure $P$ as:\n\n\\begin{equation}\n P = \\frac{F}{A}\n\\end{equation}\n\nWhere $F$ is the force exerted over a certain area $A$. A good way to think about this might be if I have a gas of pressure $P$ contained in a box, that would tell you that if one of the walls of the box has area $A$, there is a perpendicular force of $PA$ being applied.\\\\\n\nJust how does this help us define temperature? In order for us to connect the pieces, it will be helpful to figure out how the pressure of a gas arises from the microscopic properties of the gas. For this, let us consider a very simple scenario of a single molecule of gas bouncing back and forth in a 1-dimensional tube. This might seem like a strange place to start, but we will shortly see that this simple start leads to valuable information about more complicated scenarios! Note that the below derivation is mainly for your own understanding, and is not especially crucial for your success in Thermodynamics.\n\n\\begin{center}\n \\begin{tikzpicture}[scale=4]\n \\draw[draw=black] (0,0) rectangle(2,0.25);\n \\filldraw (1,0.125) circle (0.75pt);\n \\draw[red, ->, thick] (1.025,0.125) -- (1.5,0.125);\n \\node[below] at (1,0.125) {$m$};\n \\node[below] at (1.25,0.125) {$v$};\n \\draw[<->] (0,-0.05) -- (2,-0.05);\n \\draw[<->] (2.05,0) -- (2.05,0.25);\n \\node[below] at (1,-0.05) {$L$};\n \\node[right] at (2.05,0.125) {$A$};\n \\end{tikzpicture}\n\\end{center}\n\nPictured above is our scenario; we have a single molecule of gas of mass $m$ and speed $v$ moving back and forth in a tube of length $L$, as it bounces off the ends of the tube that have area $A$. We assume that the molecule collides elastically off the walls, and hence it goes back and forth with velocities $\\vec{v}$ and $-\\vec{v}$. From our knowledge of kinematics, we obtain that the time it takes for the particle to go back and forth once is:\n\\begin{align*}\n \\Delta t = \\frac{2L}{v}\n\\end{align*}\nNow, we want to figure out the average pressure that this particle applies on the wall; to obtain this, we, we return to how we have defined pressure above:\n\\begin{align*}\n P_{av} = \\frac{F_{av}}{A}\n\\end{align*}\n$A$ is just a constant here, so we then just need to figure out $F_{av}$! For this, we think back to the kinematics unit and the definition of force as the time derivative of momentum\\footnote{Of course you know by know that force and momentum are vector and not scalar quantities, but since our problem is one dimensional, we can just consider the scalar version of the equation here!}:\n\\begin{align*}\n F = \\frac{dp}{dt} = \\frac{d(mv)}{dt}\n\\end{align*}\nSince we are just concerned with the average force, we can just consider the change in momentum $\\Delta p$ of the particle from hitting one of the ends over one cycle over the time it takes for one cycle (which is just $\\Delta t$, above!). We consider that when the molecule hits the end of the tube with speed $v$, it reflects back elastically and starts going in the other direction with the same speed $v$, so its change in velocity is therefore $-2\\Delta v$, and therefore the change in momentum is:\n\\[ \\Delta p_{molecule} = -2m\\Delta v\\]\nAnd therefore the average force applied to the wall is:\n\\[ F_{av} = \\frac{-\\Delta p_{molecule}}{\\Delta t} = \\frac{2mv}{\\frac{2L}{v}} = \\frac{mv^2}{L} \\]\nThen, getting the average pressure (from the one molecule) is simple, using the definition of pressure:\n\\[P_{av} = \\frac{\\frac{mv^2}{L}}{A} = \\frac{mv^2}{V} \\]\nWhere I have used the fact that the volume of the tube is given by $V = AL$. Now, in a real gas, we obviously don't just have one molecule, but a bunch of different molecules, all with different speeds! If we want the total pressure from a whole bunch of (say, $N$) molecules, we can just add up the average pressure of $N$ molecules:\n\\[P_{total} = \\sum_{i=1}^N \\frac{mv_i^2}{V}\\]\nAs a reminder, the greek sigma $\\Sigma$ above is telling us to take the \"sum of N\" things (in this case, molecules). It's clear that the volume can be taken out of the summation (because the volume of the tube is the same no matter which molecule we look at), and if we further assume that all of the masses of the molecules are the same (which is perfectly reasonable if we have a gas made of one element), then we obtain:\n\\[P_{total} = \\frac{m}{V} \\sum_{i=1}^N v_i^2 \\]\nNow I'm going to do a bit of a hack; I'm going to multiply this expression by $1 = \\frac{N}{N}$:\n\\[P_{total} = N\\frac{m}{V} \\left(\\frac{1}{N}\\sum_{i=1}^N v_i^2 \\right)\\]\nYou might recognize from your physics labs that the quantity in brackets is just the \\textit{average} of the squared velocities of all the molecules\\footnote{To be clear, this is not the square of the averages!}; let us denote this by $v^2_{av}$. Now, let us multiply both sides by the volume $V$ of the box:\n\\[PV = Nmv^2_{av} \\]\nNow, if we multiply both sides by a half, we get something that looks familiar on the right hand side:\n\\[\\frac{1}{2}PV = N\\frac{1}{2}mv^2_{av} \\]\nYou can see that we have exactly the average kinetic energy of our molecules (times the number of molecules in our system)! Now, at the beginning I said we were considering a one-dimensional system; but the direction we could have been talking about could have been the x direction, the y direction, or the z direction, and we would have got the same result! Therefore, we have that:\n\\[N\\frac{1}{2}mv^2_{xav} = N\\frac{1}{2}mv^2_{yav} = N\\frac{1}{2}mv^2_{zav} = \\frac{1}{2}PV \\]\nTherefore, for an ideal gas in three dimensions, all we have to do is add the three terms (the three kinetic energy contributions from each of the directions) together:\n\\begin{equation}\n \\label{eqn:(2)}\n \\frac{3}{2}PV = N\\epsilon_{kavg}\n\\end{equation}\nWhere $\\epsilon_{kavg}$ is the average kinetic energy of the molecules in three dimensions, defined as:\n\\begin{equation}\n \\epsilon_{kavg} = \\frac{1}{2}mv^2_{xav} + \\frac{1}{2}mv^2_{yav} + \\frac{1}{2}mv^2_{zav} = \\frac{1}{2}m\\vec{v}_{av} \\cdot \\vec{v}_{av}\n\\end{equation}\nNow at this point, you might be holding your head in your hands because I spent three pages on a derivation\\footnote{I sincerely apologize for your suffering.} and it seems like we're no closer to defining temperature than when we started. However, in the next section we will introduce the ideal gas law, and combining that with equation \\ref{eqn:(2)} we just derived, we will see that we obtain exactly what we want to find. ", "meta": {"hexsha": "556b3124dad5f59276d8a7765918d571cb0c66dd", "size": 6887, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Temperature/pressure.tex", "max_stars_repo_name": "RioWeil/SCIE001-thermo-notes", "max_stars_repo_head_hexsha": "8578248f8f79f5704319dc6cd4ec679ce12b949c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Temperature/pressure.tex", "max_issues_repo_name": "RioWeil/SCIE001-thermo-notes", "max_issues_repo_head_hexsha": "8578248f8f79f5704319dc6cd4ec679ce12b949c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Temperature/pressure.tex", "max_forks_repo_name": "RioWeil/SCIE001-thermo-notes", "max_forks_repo_head_hexsha": "8578248f8f79f5704319dc6cd4ec679ce12b949c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-30T05:36:50.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-30T05:36:50.000Z", "avg_line_length": 102.7910447761, "max_line_length": 614, "alphanum_fraction": 0.7229562945, "num_tokens": 1971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080671950640465, "lm_q2_score": 0.7931059462938815, "lm_q1q2_score": 0.6408828974103131}} {"text": "\\section{\\module{cmath} ---\n Mathematical functions for complex numbers}\n\n\\declaremodule{builtin}{cmath}\n\\modulesynopsis{Mathematical functions for complex numbers.}\n\nThis module is always available. It provides access to mathematical\nfunctions for complex numbers. The functions are:\n\n\\begin{funcdesc}{acos}{x}\nReturn the arc cosine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{acosh}{x}\nReturn the hyperbolic arc cosine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{asin}{x}\nReturn the arc sine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{asinh}{x}\nReturn the hyperbolic arc sine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{atan}{x}\nReturn the arc tangent of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{atanh}{x}\nReturn the hyperbolic arc tangent of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{cos}{x}\nReturn the cosine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{cosh}{x}\nReturn the hyperbolic cosine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{exp}{x}\nReturn the exponential value \\code{e**\\var{x}}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{log}{x}\nReturn the natural logarithm of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{log10}{x}\nReturn the base-10 logarithm of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{sin}{x}\nReturn the sine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{sinh}{x}\nReturn the hyperbolic sine of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{sqrt}{x}\nReturn the square root of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{tan}{x}\nReturn the tangent of \\var{x}.\n\\end{funcdesc}\n\n\\begin{funcdesc}{tanh}{x}\nReturn the hyperbolic tangent of \\var{x}.\n\\end{funcdesc}\n\nThe module also defines two mathematical constants:\n\n\\begin{datadesc}{pi}\nThe mathematical constant \\emph{pi}, as a real.\n\\end{datadesc}\n\n\\begin{datadesc}{e}\nThe mathematical constant \\emph{e}, as a real.\n\\end{datadesc}\n\nNote that the selection of functions is similar, but not identical, to\nthat in module \\refmodule{math}\\refbimodindex{math}. The reason for having\ntwo modules is that some users aren't interested in complex numbers,\nand perhaps don't even know what they are. They would rather have\n\\code{math.sqrt(-1)} raise an exception than return a complex number.\nAlso note that the functions defined in \\module{cmath} always return a\ncomplex number, even if the answer can be expressed as a real number\n(in which case the complex number has an imaginary part of zero).\n", "meta": {"hexsha": "3a11870d5e25efe588e376ead6e955095440ee7b", "size": 2358, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Doc/lib/libcmath.tex", "max_stars_repo_name": "marcosptf/cpython-2.0.1", "max_stars_repo_head_hexsha": "73c739a764e8b1dc84640e73b880bc66e1916bca", "max_stars_repo_licenses": ["PSF-2.0"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2022-03-26T21:53:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-30T21:47:20.000Z", "max_issues_repo_path": "Doc/lib/libcmath.tex", "max_issues_repo_name": "marcosptf/cpython-2.0.1", "max_issues_repo_head_hexsha": "73c739a764e8b1dc84640e73b880bc66e1916bca", "max_issues_repo_licenses": ["PSF-2.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2020-11-18T15:48:14.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-03T21:20:50.000Z", "max_forks_repo_path": "Doc/lib/libcmath.tex", "max_forks_repo_name": "marcosptf/cpython-2.0.1", "max_forks_repo_head_hexsha": "73c739a764e8b1dc84640e73b880bc66e1916bca", "max_forks_repo_licenses": ["PSF-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-07-16T08:14:13.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-27T01:55:17.000Z", "avg_line_length": 25.6304347826, "max_line_length": 75, "alphanum_fraction": 0.7408821035, "num_tokens": 681, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619393159451, "lm_q2_score": 0.7772998663336158, "lm_q1q2_score": 0.6408541552274378}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{subfigure}\n\\usepackage{float}\n\\usepackage{ulem}\n\\usepackage{bm}\n\\usepackage{anysize}\n\\usepackage{pythonhighlight}\n\n\\marginsize{2cm}{2cm}{0.9cm}{1.8cm}\n\n\\title{EECE 5639 Computer Vision\\\\ [2ex] \\begin{large} Homework \\#4 \\end{large} }\n\\author{Jiyu Tian}\n\\date{}\n\n\\begin{document}\n\\maketitle\n\\pagestyle{empty}\n%%---------------------------------------------------------------\n%% Question 1\n%%---------------------------------------------------------------\n\\section{Solution:}\n(a) Since $Z=0$, it is a similarity transformation:\n\\begin{equation*}\n\\left[ \\begin{array}{c}\nu\\\\\nv\\\\\n1\n\\end{array} \\right] \\sim \\left[ \\begin{array}{ccc}\nfcos\\theta & -fsin\\theta & ft_x\\\\\nfsin\\theta & fcos\\theta & ft_y\\\\\n0 & 0 & t_z\n\\end{array} \\right]\\left[ \\begin{array}{c}\nX\\\\\nY\\\\\n1\n\\end{array} \\right]\n\\end{equation*}\n(b) It has 4 degree of freedom.\\\\\n(c) 4 points are needed since it is a 2D to 2D transformation.\\\\\n(d) Yes, more correspondences will be helpful for a more robust solution:\n\\begin{itemize}\n \\item 4 correspondences may be collinear\n \\item points may be inaccurate because of noise\n\\end{itemize}\n(e) Cross ratio of collinear points.\\\\\n(f) Parallellism.\\\\\n(g) The building plane is parallel to the image plane if and only if:\n\\begin{itemize}\n \\item lines of building parallel to $X$ axes are also parallel to each other on the image\n \\item lines of building parallel to $Y$ axes are also parallel to each other on the image\n\\end{itemize}\n\n%%---------------------------------------------------------------\n%% Question 2\n%%---------------------------------------------------------------\n\\section{Solution:}\n(a) SSD\n\\begin{equation*}\nSSD = \\left[ \\begin{array}{cccccccc}\n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times \\\\\n\\times & 24 & 28 & 24 & 12 & 8 & 8 & \\times \\\\\n\\times & 16 & 12 & 16 & 8 & 12 & 8 & \\times \\\\\n\\times & 8 & 8 & 12 & 8 & 16 & 8 & \\times \\\\\n\\times & 10 & 7 & 9 & 12 & 28 & 20 & \\times \\\\\n\\times & 9 & 12 & 9 & 12 & 24 & 20 & \\times \\\\\n\\times & 0 & 7 & 7 & 8 & 12 & 8 & \\times \\\\\n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times \n\\end{array} \\right]\n\\end{equation*}\n(b) Cross-Correlation\n\\begin{equation*}\nC_fg = \\left[ \\begin{array}{cccccccc}\n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times \\\\\n\\times & 4 & 4 & 2 & 0 & 0 & 0 & \\times \\\\\n\\times & 10 & 14 & 8 & 2 & 0 & 2 & \\times \\\\\n\\times & 4 & 4 & 0 & 0 & 0 & 4 & \\times \\\\\n\\times & 4 & 5 & 2 & 0 & 4 & 10 & \\times \\\\\n\\times & 3 & 1 & 0 & 0 & 4 & 8 & \\times \\\\\n\\times & 8 & 4 & 1 & 2 & 8 & 12 & \\times \\\\ \n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times\n\\end{array} \\right]\n\\end{equation*}\n(c) Normalized Cross-Correlation\n\\begin{equation*}\nN_fg = \\left[ \\begin{array}{cccccccc}\n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times \\\\\n\\times & 0.0208 & 0.0179 & 0.0125 & 0 & 0 & 0 & \\times \\\\\n\\times & 0.0446& 0.0547 & 0.0417 & 0.0625& 0 & 0.0625& \\times \\\\\n\\times & 0.0625 &0.0625 & 0 & 0 & 0 & 0.0625 &\\times \\\\\n\\times & 0.05 & 0.0694 & 0.05 & 0 & 0.0179 & 0.0391& \\times \\\\\n\\times & 0.0536 & 0.0208 & 0 & 0 & 0.0208 & 0.0357& \\times \\\\\n\\times & 0.125 & 0.0714 & 0.125 & 0.0625& 0.05 & 0.0625 &\\times \\\\\n\\times & \\times& \\times& \\times& \\times &\\times &\\times & \\times \n\\end{array} \\right]\n\\end{equation*}\n\n\\end{document}\n", "meta": {"hexsha": "1b295b9f83ff8446a78c0e8071dd5baf5b2733d9", "size": 3385, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "EECE5639-Computer-Vision/Homework-4/main.tex", 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YES\n2. YES", "lm_q1_score": 0.7772998714925403, "lm_q2_score": 0.8244619263765707, "lm_q1q2_score": 0.6408541494230007}} {"text": "% !TEX root = index.tex\n\n\\section{The Cocomplex World of Cochain Complexes}\n\\epigraph{The introduction of the digit 0 or the group concept was general nonsense too, and mathematics was more or less stagnating for thousands of years because nobody was around to take such childish steps.}{Grothendieck}\n\\begin{align*}\n \\xymatrix@R-2pc{\n \\U \\ar@{~>}[r] & \\L^\\bullet(\\U)\\\\\n \\mbox{Good cover of } X & \\mbox{Cech Complex of } \\U }\n\\end{align*}\nTo be able to use algebraic techniques, we need a way to convert topological information into algebraic information. We'll do this using locally constant functions and of course, linear algebra. All our vector spaces will be over the base field $ \\F = \\{0, 1\\}$ i.e. all the scalars are either 0 or 1. This is mainly because $-1 = 1$ in $ \\F$\nand hence we do not have to worry about signs.\\\\\n\n\\noindent \\textbf{Notation:} $V = \\F \\langle v_1, \\dots, v_n \\rangle = \\F \\langle \\mathcal{B} \\rangle$ stands for ``$V$ is a vector space over $\\F$ with basis $ \\mathcal{B} = \\{ v_1, \\dots, v_n\\}$''.\n \\begin{ques}\n Show that the elements of $ \\F \\langle \\mathcal{B} \\rangle$ can be identified with subsets of $\\mathcal{B}$ and hence as a set $V$ has size $2^{\\lvert\\mathcal{B}\\rvert}$, where $\\lvert \\mathcal{B} \\rvert$ denotes the size of $\\mathcal{B}$.\n \\end{ques}\n\n\\subsection{Locally Constant Functions}\n\\begin{definition}\n\tFor a topological space $X$, define the vector space of \\textbf{locally constant functions}, denoted $\\L(X)$, to be the space of continuous maps from $X$ to $\\F$.\n\t\\begin{align*}\n\t\t\\L(X) := \\{ \\: f: X \\rightarrow \\F \\: \\mbox{ continuous } \\}\n\t\\end{align*}\n Here we are thinking of $\\F$ as a topological space with 2 points.\n\\end{definition}\n\\begin{ques}\n Show that $\\L(X)$ is naturally a vector space over $\\F$.\n\\end{ques}\n\\begin{ques}\n Show that if $X$ is connected then every continuous function $f: X \\rightarrow \\F$ is a constant function and hence $\\L(X) \\cong \\F$ as a vector space.\n\\end{ques}\n\\begin{ques}\n\tWhat is $\\L(\\phi)$ where $\\phi$ is the empty set (which is a legit topological space)?\n\\end{ques}\n\\noindent Let $X^1, X^2, \\dots, X^k$ be the connected components of $X$. Define $k$ functions $\\delta^1, \\delta^2, \\dots, \\delta^k : X \\rightarrow \\F$ as\n\\begin{align*}\n \\delta^i(x) = \\begin{cases}\n 1 & \\mbox{ if $x \\in X^i$} \\\\\n 0 & \\mbox{ otherwise}\n\\end{cases}\n\\end{align*}\nfor $1 \\le i \\le k$.\\\\\n\\begin{ques}\n\tShow that\n\t\\begin{align*}\n\t\t\\L(X) = \\F \\langle \\delta^1, \\delta^2, \\dots, \\delta^k \\rangle\n\t\\end{align*}\n\tand hence $\\dim \\L(X) = k = \\pi_0(X)$.\n\\end{ques}\n\\begin{ques}*\n Show that the above statement is false if $X$ has infinitely many connected components, for example, if $X$ is the set of integers.\n\\end{ques}\n\\begin{definition}\n\tWith the notation as above, we call $\\delta^1, \\delta^2, \\dots, \\delta^k$ the \\textbf{canonical basis} for $\\L(X)$.\n\\end{definition}\n\nThe following exercise is extremely important, make sure you understand it well.\n\\begin{ques} \\label{q:restrictions}\n For an inclusion of topological spaces $X \\subseteq Y$,\n\t\\begin{enumerate}\n\t\t\\item We can restrict a function $f: Y \\rightarrow \\F$ to the subspace $X$ and get a function $f|_{X} : X \\rightarrow \\F$. Show that this induces a linear transformation\n \\begin{align*}\n\t\t\t\\res_{Y \\rightarrow X}: \\L(Y) \\rightarrow \\L(X)\n\t\t\\end{align*}\n \\item Explicitly compute the matrix for $\\res_{Y \\rightarrow X}$ in the canonical bases when $Y = \\R$ and $X = \\{ -1, 1\\}$.\n \\item Explicitly compute the matrix for $\\res_{Y \\rightarrow X}$ in the canonical bases when $Y = \\R^2 $ minus the $y$-axis and $X = \\{ (-1,0), (1,0), (2,0)\\}$.\n \\item* More generally, show that if $X^i$ and $Y^j$ are the connected components of $X$ and $Y$ respectively, for $1 \\le i \\le k$ and $1 \\le j \\le l$, then in the canonical basis $\\res_{Y \\rightarrow X}$ is a $k \\times l$ matrix whose entry in the $i^{th}$ row and $j^{th}$ column is given by\n \\begin{align*}\n [\\res_{Y \\rightarrow X}]_{i,j} &= \\begin{cases}\n 1 & \\mbox{ if } X^i \\subseteq Y^j \\\\\n 0 & \\mbox{ if } X^i \\not \\subseteq Y^j\n \\end{cases}\n \\end{align*}\n\t\t% \\item Show that $\\res_{U \\rightarrow U}$ is the identity transformation.\n\t\t% \\item Show that $\\res_{V \\rightarrow U} \\circ \\res_{W \\rightarrow V} = \\res_{W \\rightarrow U}$.\n \\end{enumerate}\n\\end{ques}\n\n\n\n\n\n\n\n\\newpage\n\\subsection{Cochain Complexes}\nThe various vector spaces $\\L(U_I)$ naturally assemble themselves into a cochain complex, we'll get to that in the next section. First we need to understand what cochain complexes are.\n\\begin{definition}\n A \\textbf{cochain complex} $ \\V^\\bullet$ consists of the following data:\n \\begin{align*}\n &\\xymatrix{0 \\ar[r] & \\V^0 \\ar[r] & \\cdots \\ar[r] & \\V^{i-1} \\ar^-{d^{i-1}}[r] & \\V^{i} \\ar^{d^{i}}[r]& \\V^{i+1} \\ar[r] & \\cdots \\ar[r] & \\V^n \\ar[r] &0}\n \\end{align*}\n \\begin{enumerate}\n \\item A vector space $ \\V^i$ for each $ i \\in \\Z$, with $ \\V^i \\neq 0$ only if $ 0 \\le i \\le n$ for some positive integer $ n$.\n \\item For each $ i \\in \\Z$ a linear transformation $ d^i: \\V^i \\rightarrow \\V^{i+1}$ that satisfies\n \\begin{align*}\n d^{i} \\circ d^{i-1} = 0\n &&\n \\xymatrix{\n \\V^{i-1} \\ar@/_1pc/_{d^{i} \\circ d^{i-1} = 0}[rrrr] \\ar^{d^{i-1}}[rr] && \\V^i \\ar^{d^{i}}[rr] && \\V^{i+1}\n }\n \\end{align*}\n \\end{enumerate}\n\\end{definition}\n\n\\begin{ques}\n Show that $ \\im d^{i-1} \\subseteq \\ker d^{i}$.\n\\end{ques}\n\n\\begin{definition}\n The $ i^{th}$ \\textbf{cohomology} of $ \\V^\\bullet$ is the quotient vector space\n \\begin{align*}\n H^i(\\V) := \\ker d^{i} / \\im d^{i-1}\n \\end{align*}\n\\end{definition}\n\\begin{remark}\n This is well-defined because of the previous exercise. We'll only be interested in the dimensions\n \\begin{align*}\n \\dim H^i(\\V) = \\dim \\ker d^{i} - \\dim \\im d^{i-1}\n \\end{align*}\n\\end{remark}\n\\noindent \\textbf{Convention:} Even though in a cochain complex there is a vector space $ \\V^i$ for all integers $ i$ it is a common convention to explicitly define $ \\V^i$ only where it is non-zero, it is understood that the rest of the $ \\V^i$ are all 0. The first non-zero vector space in the cochain complex is understood to be $ \\V^0$ unless otherwise specified.\n\\begin{ques}\n A vector space $A$ can be thought of as a cochain complex as\n \\begin{align*}\n \\xymatrix{ \\V^\\bullet= 0 \\ar[r] & A \\ar[r] & 0}\n \\end{align*}\n What are the cohomologies $ H^i(\\V)$, for $ i \\in \\Z$?\n\\end{ques}\n\n\\begin{ques}\n A linear transformation $ f:A \\rightarrow B$ can be thought of as a cochain complex as\n \\begin{align*}\n \\xymatrix{ \\V^\\bullet= 0 \\ar[r] & A \\ar^f[r] & B \\ar[r] & 0}\n \\end{align*}\n What are the cohomologies $ H^i(\\V)$, for $ i \\in \\Z$?\n\\end{ques}\nFor the following exercises it'll be useful to invoke the {\\bf Rank-Nullity / Dimension Theorem}.\n\\begin{ques} Verify that the following are cochain complexes and compute their cohomologies.\n \\begin{enumerate}\n \\item $ \\xymatrix{0 \\ar[r] & \\F \\ar^{0}[r] & \\F \\ar^{0}[r] & \\F \\ar[r] & 0 } $\n \\item $ \\xymatrix{0 \\ar[r] & \\F \\ar^{1}[r] & \\F \\ar^{0}[r] & \\F \\ar[r] & 0 } $\n \\item $ \\xymatrix{0 \\ar[r] & \\F \\ar^{0}[r] & \\F \\ar^{1}[r] & \\F \\ar[r] & 0 } $\n \\item $ \\xymatrix{0 \\ar[r] & \\F \\ar^{\\begin{bmatrix}1 \\\\ 1 \\end{bmatrix}}[rr] && \\F^2 \\ar^{\\begin{bmatrix}1 & 1 \\end{bmatrix}}[rr] && \\F \\ar[r] & 0 } $\n \\end{enumerate}\n\\end{ques}\n\n\\begin{ques} Compute the cohomologies of the following cochain complexes.\n \\begin{enumerate}\n \\item $ \\xymatrix{0 \\ar[r] & \\F^2 \\ar^{\\begin{bmatrix}1 & 1 \\\\ 1 & 1 \\\\ 1 & 1 \\end{bmatrix}}[rr] && \\F^3 \\ar[r] & 0 } \\qquad $ (this is computing $\\check H^* (S^1 \\vee S^1)$).\n \\item $ \\xymatrix{0 \\ar[r] & \\F^3 \\ar^{\\begin{bmatrix}1 & 1 & 0\\\\ 1 & 0 & 1 \\\\ 0 & 1 & 1 \\end{bmatrix}}[rr] && \\F^3 \\ar^{\\begin{bmatrix}1 & 1 & 1\\\\ 1 & 1 & 1 \\end{bmatrix}}[rr] && \\F^2 \\ar[r] & 0 } \\qquad $ (this is computing $\\check H^* (S^2)$).\n \\end{enumerate}\n\\end{ques}\n\n\\begin{ques}\n Given $\\xymatrix{ \\V^\\bullet= 0 \\ar[r] & A \\ar^f[r] & B \\ar^g[r] & C \\ar[r] & 0}$\n \\begin{enumerate}\n \\item Under what conditions on $ f,g$ is $ \\V^\\bullet$ a cochain complex.\n \\item Under what conditions on $ f,g$ is $ H^i(\\V) = 0$ for all $ i$. In this case, we say that $ \\V^\\bullet$ is a \\textbf{short exact sequence}.\n \\end{enumerate}\n\\end{ques}\n\n\\begin{definition}\n More generally, a cochain complex $ \\V^\\bullet$ is said to be \\textbf{exact} (or \\textbf{long exact}) if $H^i(\\V) = 0$ for all $ i$.\n\\end{definition}\n\\newpage\n\n\\subsection*{Optional Problems}\n\\begin{definition}\n Define the \\textbf{Euler characteristic} of a cochain complex to be\n \\begin{align*}\n \\chi(\\V)\n &= \\sum_{i \\in \\Z} (-1)^i \\dim H^i(\\V) \\\\\n &= \\dim H^0(\\V) - \\dim H^1(\\V) + \\dim H^2(\\V) \\pm \\cdots + (-1)^n \\dim H^n(\\V)\n \\end{align*}\n\\end{definition}\n% \\begin{ques} $ $\n% \\begin{enumerate}\n% \\item Show that if $ \\V^\\bullet$ is exact then $ \\chi(\\V) = 0$.\n% \\item Find an example of a cochain complex $ \\V^\\bullet$ which is not exact but for which $ \\chi(\\V) = 0$.\n% \\end{enumerate}\n% \\end{ques}\n\n\\begin{ques}$ $\n \\begin{enumerate}\n \\item Express the Euler characteristic in terms of the $ \\dim \\ker(d^i)$ and $ \\dim \\im(d^i)$ for $i \\in \\Z$.\n \\item Show that\n \\begin{align*}\n \\chi(V) = \\sum_{i \\in \\Z} (-1)^i \\dim \\V^i\n \\end{align*}\n \\end{enumerate}\n \\end{ques}\n \\noindent Thus the Euler characteristic can be computed using the dimensions of the vector spaces of the original cochain complex, but it is really an invariant of the underlying cohomology!!! \\\\\\\\\n\\begin{ques}*\n The direct sum of cochain complexes $(\\V_1 \\oplus \\V_2)^\\bullet$ is defined as\n \\begin{align*}\n (\\V_1 \\oplus \\V_2)^i := \\V_1^i \\oplus \\V_2^i\n \\end{align*}\n and the differential is defined as\n \\begin{align*}\n d^i := d_1^i \\oplus d_2^i\n \\end{align*}\n % \\begin{align*}\n % &\\xymatrix{ \\cdots \\ar[r] & \\V_1^{i-1} \\oplus \\V_2^{i-1} \\ar^-{d_1^{i-1} \\oplus d_2^{i-1}}[rr] && \\V_1^{i} \\oplus \\V_2^{i} \\ar^{d_1^{i} \\oplus d_2^{i}}[rr]&& \\V_1^{i+1} \\oplus \\V_2^{i+1} \\ar[r] & \\cdots}\n % \\end{align*}\n Find $H^*(\\V_1 \\oplus \\V_2)$ in terms of $H^*(\\V_1)$ and $H^*(\\V_2)$.\n\\end{ques}\n\n\\begin{ques}***\n The tensor product $(\\V_1 \\otimes \\V_2)^\\bullet$ of two cochain complexes $\\V_1^\\bullet$, $\\V_2^\\bullet$ is defined as\n \\begin{align*}\n (\\V_1 \\otimes \\V_2)^k := \\bigoplus \\limits_{i + j = k} \\V_1^i \\otimes \\V_2^j\n \\end{align*}\n and the differential is defined as\n \\begin{align*}\n d^k := \\bigoplus \\limits_{i + j = k} d_1^i \\otimes d_2^j\n \\end{align*}\n Find $H^*(\\V_1 \\otimes \\V_2)$ in terms of $H^*(\\V_1)$ and $H^*(\\V_2)$.\n\\end{ques}\n\\newpage\n\\begin{ques}***\n A \\textbf{morphism of cochain complexes} $\\phi: \\V_1^\\bullet \\rightarrow \\V_2^\\bullet$ is a collection of maps $ \\phi: \\V_1^i \\rightarrow \\V_2^i$ for each $ i \\in \\Z$ such that the following diagram commutes\n \\begin{align*}\n \\xymatrix{\n \\V_2^i \\ar^-{d}[r] & \\V_2^{i+1} \\\\\n \\V_1^i \\ar^-{d}[r] \\ar_{\\phi}[u] & \\V_1^{i+1} \\ar^{\\phi}[u]\n }\n \\end{align*}\n \\begin{enumerate}\n \\item Show that a morphism $\\phi: \\V_1^\\bullet \\rightarrow \\V_2^\\bullet$ between cochain complexes naturally induces a map between cohomologies $\\phi^*: H^i(\\V_1) \\rightarrow H^i(\\V_2)$, for all $ i \\in Z$.\n \\item Given two morphisms $\\phi_1: \\V_1^\\bullet \\rightarrow \\V_2^\\bullet$ and $\\phi_2: \\V_2^\\bullet \\rightarrow \\V_3^\\bullet$, show that their composition $\\phi_2 \\circ \\phi_1: \\V_1^\\bullet \\rightarrow \\V_3^\\bullet$ is also a morphism of cochain complexes. Further show that $ (\\phi_2 \\circ \\phi_1)^* = \\phi_2^* \\circ \\phi_1^*$.\n \\end{enumerate}\n\\end{ques}\n\n\\begin{ques}**\n A cochain complex of cochain complexes (i.e. each $\\V^i$ is itself a cochain complex and the differentials $d^i$ are morphisms of cochain complexes) is called a \\textbf{double complex}. Unravel this description of a double complex and describe it more explicitly as a grid of vector spaces.\n\\end{ques}\n\n\\iffalse\n\\begin{ques}[Snake Lemma]**\n Show that a morphism $ \\phi$ between short exact sequences\n \\begin{align}\n \\label{eq:double_complex}\n \\xymatrix{\n 0 \\ar[r] & \\V_1^0 \\ar^{\\phi^0}[d] \\ar[r] & \\V_1^1 \\ar^{\\phi^1}[d] \\ar[r] & \\ar^{\\phi^2}[d] \\V_1^2 \\ar[r] & 0 \\\\\n 0 \\ar[r] & \\V_2^0 \\ar[r] & \\V_2^1 \\ar[r] & \\V_2^2 \\ar[r] & 0\n }\n \\end{align}\n induces a natural exact sequence (not a full cochain complex)\n \\begin{align}\n \\label{eq:snake_lemma}\n \\xymatrix{\\ker \\phi^0 \\ar[r] & \\ker \\phi^1 \\ar[r] & \\ker \\phi^2 \\ar[r] & \\coker \\phi^0 \\ar[r] &\\coker \\phi^1 \\ar[r] &\\coker \\phi^2 }\n \\end{align}\n (Recall the cokernel of a map $ f:A \\rightarrow B$ is $ B/(\\im f)$.) This sequence does not naturally extend to a full cochain complex.\n\\end{ques}\n\\begin{remark}\n Note that \\eqref{eq:double_complex} is a double complex which gives rise to the exact sequence \\eqref{eq:snake_lemma}. Larger double complexes give rise to more complicated structures called \\textbf{spectral sequences}.\n\\end{remark}\n\\fi\n", "meta": {"hexsha": "72ec000d89102c75f863b0371a2cd07998378b7c", "size": 12946, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "02.tex", "max_stars_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_stars_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "02.tex", "max_issues_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_issues_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "02.tex", "max_forks_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_forks_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.7923076923, "max_line_length": 367, "alphanum_fraction": 0.6295380813, "num_tokens": 4809, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619350028204, "lm_q2_score": 0.7772998611746912, "lm_q1q2_score": 0.6408541476215096}} {"text": "\\documentclass{article}\n\n\\usepackage{amsmath}\n\\usepackage{numprint}\n\n\\author{Daniel Fernandes Martins (danielfmt)}\n\\title{Question \\#9 Solution}\n\n\\begin{document}\n\n\\maketitle\n\n\\textbf{Disclaimer.} This is the reasoning I used to solve the problem; it\nmay be wrong though. This is intended just as food for thought.\n\n\\section{Parameters Of Neural Network}\n\nGiven a neural network with 10 input units (including the constant $x_0^{(0)}$\nunit), one output unit, and 36 hidden units (including the necessary number of\nconstant units for a fully-connected network), this question asks what is the\nminimum possible number of weights that such a network can have.\n\n\\subsection{Finding The Minimum}\n\nThe network must be fully connected, and each hidden layer must include a\nconstant (bias) unit. Since the constant units don't get any input from the\nprevious layer, one way to minimize the number of required weights is by\nmaximizing the number of constant units in the network.\n\nThis can be done by creating a network with 18 hidden layers, with 2 units each.\nThe number of weights in this network is given by:\n\n\\begin{equation*}\n(1 \\cdot 10) + (1 \\cdot 2 \\cdot 17) + (2) = 46\n\\end{equation*}\n\n\\end{document}\n", "meta": {"hexsha": "583f67a882d0610d450af1fd7ab15cf12187ca17", "size": 1202, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week-06/math/q09.tex", "max_stars_repo_name": "danielfm/edx-learning-from-data", "max_stars_repo_head_hexsha": "1675e14c20fc1b7ad54d2704b9c8a941e043cbcb", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 98, "max_stars_repo_stars_event_min_datetime": "2015-04-27T06:55:49.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-05T06:09:19.000Z", "max_issues_repo_path": "week-06/math/q09.tex", "max_issues_repo_name": "danielfm/edx-learning-from-data", "max_issues_repo_head_hexsha": "1675e14c20fc1b7ad54d2704b9c8a941e043cbcb", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2016-05-14T19:33:33.000Z", "max_issues_repo_issues_event_max_datetime": "2017-08-12T13:07:41.000Z", "max_forks_repo_path": "week-06/math/q09.tex", "max_forks_repo_name": "danielfm/edx-learning-from-data", "max_forks_repo_head_hexsha": "1675e14c20fc1b7ad54d2704b9c8a941e043cbcb", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 56, "max_forks_repo_forks_event_min_datetime": "2015-01-10T08:18:21.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-28T08:46:22.000Z", "avg_line_length": 31.6315789474, "max_line_length": 80, "alphanum_fraction": 0.768718802, "num_tokens": 306, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619177503206, "lm_q2_score": 0.7772998611746912, "lm_q1q2_score": 0.6408541342111439}} {"text": "% conclusions and future work\n\\section{Discussions and Conclusions}\n\\label{sec:conclusions}\n\nWe presented an effective approach to mine approximate frequent subgraph\npatterns from a single large graph database in the presence of\na label cost matrix.\n\nThere are two main parameters in our method: $K$, the number of random\nwalks, and $\\alpha$ the cost threshold. The value of $K$ is directly\nproportional to the number of maximal approximate patterns we desire,\nand is relatively easy to set.\nOn the other hand, choosing an appropriate value of\n$\\alpha$ is very important as it affects the quality of\npatterns mined. Depending on the application domain\nand the purpose of the graph mining, let $t$ be the number of\nvertices in the pattern for which we allow label mismatches\nin the subgraph isomorphism. One reasonable value of\n$\\alpha$ is $t \\times IMQ$ where $IMQ$ is the inter-quartile mean\ni.e., the mean of the entries between the first quartile ( $25^{th}$\npercentile) and the third quartile ( $75^{th}$ percentile)\nof the entries in the cost matrix arranged in sorted order. \n$t$ can be chosen by first\nenumerating maximal patterns with $\\alpha = 0$ and computing the \naverage size $m$ of the maximal patterns mined from the graph.\nThe value of $t$ then is a fraction of the average size $m$.\nCare has to be taken not to choose a very large $\\alpha$ as it leads to\npatterns of poor quality and also increases the run time of the\nalgorithm significantly as can be seen in Table \\ref{tab:scop_alpha}.\n\nIn terms of future work, we plan to increase the efficiency of our\nmethod by exploiting parallelism. Obviously different walks can be\ncarried out in parallel. However, more interesting is the\nparallelization of the approximate isomorphism generation and\nlabel-based pruning steps, including verification. We also\nwant to explore the idea of label based pruning for more\ngeneral definitions of approximate isomorphism including \nedge mismatches.\n\n", "meta": {"hexsha": "ad6ae761e7e329a376fdbb54baef8ca87998e1e6", "size": 1952, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "finalversion/sigkdd13/conclusions.tex", "max_stars_repo_name": "PranayAnchuri/approx-graph-mining-with-label-costs", "max_stars_repo_head_hexsha": "4bb1d78b52175add3955de47281c3ee0073c7943", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "finalversion/sigkdd13/conclusions.tex", "max_issues_repo_name": "PranayAnchuri/approx-graph-mining-with-label-costs", "max_issues_repo_head_hexsha": "4bb1d78b52175add3955de47281c3ee0073c7943", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "finalversion/sigkdd13/conclusions.tex", "max_forks_repo_name": "PranayAnchuri/approx-graph-mining-with-label-costs", "max_forks_repo_head_hexsha": "4bb1d78b52175add3955de47281c3ee0073c7943", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-05-08T11:17:33.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-08T11:17:33.000Z", "avg_line_length": 48.8, "max_line_length": 72, "alphanum_fraction": 0.7899590164, "num_tokens": 449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619091240701, "lm_q2_score": 0.7772998663336157, "lm_q1q2_score": 0.6408541317592973}} {"text": "\\onecolumn\r\n\\chapter{Measurement Models}\r\n\r\n\r\n\\begin{equation}\r\n \\mathcal{R}_1^{(j)} = \\rho^{(m,j)}_1(t_i) +c\\left(b_r^{(m)}(t_i) - b_t^{(j)}(t_t)\\right) + \\Delta \\rho^{(j)}_{iono}(t_i) + \\Delta \\rho^{(j)}_{tropo}(t_i) + \\sigma_{SA}^{(j)} + \\sigma_n^{(j)} + b_m\r\n\\end{equation}\r\n\r\n\\begin{tabbing}\r\n12345678912345 \\= Reynolds number based on length $s$ \\kill\r\n$\\mathcal{R}_1^{(j)}$ \\> One-way pseudorange measurement from transmitting \\\\\r\n$$ \\> antenna $k$ on satellite $j$ to receiving antenna $m$ on satellite $n$. \\\\\r\n$\\rho^{(m,j)}_1(t_i)$ \\> Geometric distance between transmitting and receiveing antenntas.\\\\\r\n$b_r^{(m)}(t_i)$ \\> Clock bias for receiver \\\\\r\n$b_t^{(j)}(t_t)$ \\> Clock bias for transmitter\\\\\r\n$\\Delta \\rho^{(j)}_{iono}(t_i)$ \\> Correction for ionspheric delay \\\\\r\n$ \\Delta \\rho^{(j)}_{tropo}(t_i) $ \\> Correction for Tropospheric delay\\\\\r\n$\\sigma_{SA}^{(j)}$ \\> Error due to selective availability \\\\\r\n$\\sigma_n^{(j)}$ \\> Measurement noise \\\\\r\n$b_m$ \\> Measurement bias \\\\\r\n$t_t$ \\> Trasmission time \\\\\r\n$t_r$ \\> Receive time\r\n\\end{tabbing}\r\n\r\n\r\nThe transmission time $t_t$ and the geometric range are determined\r\niteratively by using fixed point iteration on the following equation\r\n%\r\n\\begin{equation}\r\n t_t = t_i - \\frac{\\rho^{(m,j)}_1(t_i) + \\Delta \\rho^{(j)}_{iono}(t_i) + \\Delta \\rho^{(j)}_{tropo}(t_i)}{c}\r\n\\end{equation}\r\n\r\n\\section{Range}\r\n\r\nGMAT supports two primary models for spacecraft range: geometric and\r\nradiometric. In both cases the range is a measure of the distance\r\nbetween an observer and a vehicle.\r\n The geometric range is calculated using vector geometry and ignores signal propagation\r\n times and error sources. Most, if not all, ground trackers provide the user with\r\n the round trip signal propagation time. The radiometric range model uses the best\r\n estimate spacecraft state to determine an expected value for the round trip signal\r\n propagation time from observer to spacecraft and back to the observer. Hence,\r\n the raw radiometric value is a measure of round trip range.\r\n\r\nThe general form of the measurement model used by GMAT is\r\n%\r\n\\begin{equation}\r\n \\mathcal{O}_c = \\mathbf{f}_k\\left(\\mathbf{r}_v(t + \\delta t), \\dot{\\mathbf{r}}_v(t + \\delta t),\r\n \\mathbf{r}_o, \\dot{\\mathbf{r}}_o\\right) + b + \\delta_a + \\delta_r\r\n\\end{equation}\r\n%\r\n%\r\n\\begin{tabbing}[htbp!]\r\n12345678 \\= dummy line \\kill\r\n$\\mathbf{f}_k$ \\> The kinematic model specific to a measurement type\\\\\r\n$\\mathbf{r}_v$ \\> Vehicle position\\\\\r\n$\\dot{\\mathbf{r}}_v$ \\> Vehicle velocity\\\\\r\n$\\mathbf{r}_o$ \\> Observer position\\\\\r\n$\\dot{\\mathbf{r}}_o$ \\> Observer velocity\\\\\r\n$t$ \\> Measurement time tag\\\\\r\n$b$ \\> Measurement bias\\\\\r\n$\\delta a$ \\> Atmospheric correction\\\\\r\n$\\delta r$ \\> Relativistic correction\\\\\r\n\\end{tabbing}\r\n\r\nThe kinematic model for geometric range is simply\r\n%\r\n\\begin{equation}\r\n \\rho_c = \\| \\mathbf{r}_v(t) - \\mathbf{r}_g(t)\\|\r\n\\end{equation}\r\n%\r\n\r\nThe kinematic model for radiometric two way range is\r\n%\r\n\\begin{equation}\r\n \\rho_c= \\frac{1}{2}\\left(\\| \\mathbf{r}_v(t_{v}) - \\mathbf{r}_g(t_{gt}) \\| +\r\n \\| \\mathbf{r}_v(t_{v}) - \\mathbf{r}_g(t_{gt}) \\|\\right) \\label{Eq:ExpectedTwoWayRange}\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{tabbing}[htbp!]\r\n12345678 \\= dummy line \\kill\r\n$t_{gt}$ \\> Time the uplink signal is transmitted from ground station\\\\\r\n$t_{vr}$ \\> Time the uplink signal is received at vehicle\\\\\r\n$t_{vt}$ \\> Time the downlink signal is transmitted from vehicle\\\\\r\n$t_{gr}$ \\> Time the downlink signal is received at ground station\\\\\r\n$t_u$ \\> Uplink propagation time, ( $t_{vr}$ - $t_{gt}$ )\\\\\r\n$t_d$ \\> Downlink propagation time, ( $t_{vt}$ - $t_{gr}$ )\\\\\r\n$\\rho_u$ \\> Uplink distance\\\\\r\n$\\rho_d$ \\> Downlink distance\\\\\r\n$d_T$ \\> Distance traveled by vehicle during transponder delay\\\\\r\n$\\mathbf{r}_v(t)$ \\> Position of vehicle at time $t$\\\\\r\n$\\mathbf{r}_t(t)$ \\> Position of transmitter at time $t$\\\\\r\n$\\mathbf{r}_r(t)$ \\> Position of receiver at time $t$\\\\\r\n$t$ \\> time of geometric range measurement \\\\\r\n$\\delta T$ \\> Vehicle's transponder delay time\\\\\r\n$\\delta t_a$ \\> Atmospheric delays\\\\\r\n$\\delta t_r$ \\> Relativistic effects\\\\\r\n\\end{tabbing}\r\n%\r\n\r\nThe radiometric model is derived from the measurement geometry shown\r\nin Fig.~\\ref{Fig:RangeMeasurement}. We see that the total signal\r\npropagation time is the sum of three times, the uplink\r\n signal propagation time, $t_u$, the transponder delay time, $\\delta T$, and\r\n the downlink propagation time, $t_d$. Hence, the observed value for signal propagation time, $t_o$, is\r\n%\r\n\\begin{equation}\r\n \\Delta t_o = t_{gr} - t_{gt} = t_u + \\delta T + t_d\r\n\\end{equation}\r\n%\r\n\\begin{figure}[htbp!]\r\n \\begin{center}\r\n \\begin{picture}(270,215)\r\n \\special{psfile= RangeMeasurement.eps\r\n hscale= 75 vscale= 75 hoffset = -85 voffset = -270}\r\n \\makebox(175,295){ $\\mathbf{r}_v(t_{vr})$}\r\n \\makebox(-25,290){ $\\mathbf{r}_v(t_{vt})$}\r\n \\makebox(85,290){ $\\rho_d$}\r\n \\makebox(-100,90){ $\\mathbf{r}_g(t_{gr})$}\r\n \\makebox(-460,90){ $\\mathbf{r}_g(t_{gt})$}\r\n \\makebox(-485,290){ $\\rho_u$}\r\n \\makebox(-305,390){ $d_T$}\r\n \\end{picture}\r\n \\end{center}\r\n \\vspace{0.2 in}\r\n \\label{Fig:RangeMeasurement}\r\n \\caption{ Geometry of Radiometric Range Measurement}\r\n\\end{figure}\r\n%\r\nThe elapsed time is converted to a measure of the average range\r\nusing\r\n%\r\n\\begin{equation}\r\n \\rho_o = \\frac{c}{2}\\Delta t_o\r\n\\end{equation}\r\n%\r\nThe computed elapsed time is rigorously expressed as\r\n%\r\n\\begin{equation}\r\n \\Delta t_c = \\frac{1}{c}\\| \\mathbf{r}_v(t_{vr}) - \\mathbf{r}_g(t_{gt}) \\| +\r\n \\frac{1}{c}\\| \\mathbf{r}_v(t_{vt}) - \\mathbf{r}_g(t_{gr}) \\| + \\delta T\r\n\\end{equation}\r\n%\r\nAssuming that the transponder delay is modelled as a measurement\r\nbias we can write $ t_{vr} = t_{vt} = t_v$ and convert the computed\r\nround trip time to average range:\r\n%\r\n\\begin{equation}\r\n \\rho_c= \\frac{1}{2}\\left(\\| \\mathbf{r}_v(t_v) - \\mathbf{r}_g(t_{gt}) \\| +\r\n \\| \\mathbf{r}_v(t_v) - \\mathbf{r}_g(t_{gr}) \\|\\right) \\label{Eq:MeasuredTwoWayRange}\r\n\\end{equation}\r\n%\r\nTo solve Eq.~\\ref{Eq:MeasuredTwoWayRange}, we must know $t_{v}$.\r\nFor applications\r\n that do not require high accuracy we can approximate $t_v$ as we describe below.\r\n For higher fidelity applications we must solve for the uplink and downlink\r\n propagation times using two iterative processes. The downlink signal propagation\r\n time is calculated using the following fixed point iteration on the following equation:\r\n%\r\n\\begin{equation}\r\n \\delta t_d^{i+1} = \\frac{1}{c}\\| \\mathbf{r}_v( t - t_d^{i}) - \\mathbf{r}_g(t) \\|\r\n\\end{equation}\r\n%\r\nThe uplink propagation time is calculated using fixed point\r\niteration on\r\n%\r\n\\begin{equation}\r\n \\delta t_u^{i+1} = \\frac{1}{c}\\| \\mathbf{r}_v( t - t_d) - \\mathbf{r}_g(t - t_d - t_u^{i} ) \\|\r\n\\end{equation}\r\n%\r\n\r\n\\begin{eqnarray}\r\n \\frac{\\partial \\rho_c (t)}{\\partial \\mathbf{r}_v(t_v)} &=& \\frac{1}{2\\rho_u\\rho_d}\r\n \\left( \\rho_d(\\mathbf{r}_v^T(t_v) - \\mathbf{r}_T^T(t_{gt}) + \\rho_u(\\mathbf{r}_v^T(t_v) - \\mathbf{r}_R^T(t_{gr} ) \\right)\\\\\r\n %\r\n \\frac{\\partial \\rho_c (t)}{\\partial \\dot{\\mathbf{r}}_v(t_v)} &=& \\mathbf{0}_{1x3}\r\n\\end{eqnarray}\r\n", "meta": {"hexsha": "c412cd421e70f9d1593c4afe44dbc75fbe4848ad", "size": 7438, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/SystemDocs/MathematicalSpecification/MeasurementModels.tex", "max_stars_repo_name": "Randl/GMAT", "max_stars_repo_head_hexsha": "d6a5b1fed68c33b0c4b1cfbd1e25a71cdfb8f8f5", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-01-01T13:14:57.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-09T07:05:07.000Z", "max_issues_repo_path": "doc/SystemDocs/MathematicalSpecification/MeasurementModels.tex", "max_issues_repo_name": "ddj116/gmat", "max_issues_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_issues_repo_licenses": ["NASA-1.3"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2018-03-15T08:58:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-20T20:11:26.000Z", "max_forks_repo_path": "doc/SystemDocs/MathematicalSpecification/MeasurementModels.tex", "max_forks_repo_name": "ddj116/gmat", "max_forks_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_forks_repo_licenses": ["NASA-1.3"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-13T10:26:49.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-09T07:06:55.000Z", "avg_line_length": 41.3222222222, "max_line_length": 201, "alphanum_fraction": 0.628260285, "num_tokens": 2425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9252299612154571, "lm_q2_score": 0.6926419894793246, "lm_q1q2_score": 0.6408531210621525}} {"text": "\\chapter{Beginning Combinatorics}\n\nDiscrete probabilities problems often include some counting. For\nexample, we figured out that there were 36 different ways the two dice\ncould come up, but all of them summed to some number 2 through 12. How\nmany different ways could 3 8-sided dice come up? We would need to\ncount them, right? As the numbers get big we will need some tricks so\nwe don't need to write them all down and count them one-by-one.\n\nThe branch of mathematics that focuses on tricks for counting is\ncalled \\textit{combinatorics}.\\index{combinatorics}\n\nHow can be sure that there were 36 different configurations for the\ntwo 6-sided dice? The first die could have come up as any one of six\nnumbers. For each of those, the second could have come up with any one\nof six numbers. Thus, the number of possibilities is $ 6 \\times 6 =\n36.$\n\nHow many different configurations for 3 8-sided dice? $8 \\times 8\n\\times 8 = 8^3 = 512$.\n\nWhat about 7 dice, each with 20 sides? $20^7=1,280,000,000$\nconfigurations. See, aren't you glad we don't need to write them all\ndown?\n\nNow, let's say that six people (Anne, Brock, Carl, Dev, Edgar, and Fred) are\ngoing to run a race. You have to make a plaque that says who won first\nplace, who won second place, and who won third. If you want to get all\nthe possible plaques created beforehand, and just pull the right one\nout as soon as the race ends, how many plaques would you need to get\nengraved?\n\nIn this case, once someone has been given first place, they can't win\nsecond or third place. Thus, any of the 6 people can come in first,\nbut once you have engraved that person's name on the plaque, there are\nonly 5 people whose name can appear in second place. Once you have\nengraved that name, there are only 4 people whose name can appear in\nthird place. Thus, you would get $6 \\times 5 \\times 4 = 120$ plaques\nengraved.\n\nWhat if the plaque includes all 6 places? Then you would need $6 \\times 5\n\\times 4 \\times 3 \\times 2 \\times 1 = 720$ plaques engraved. We use\nthis process often enough that we gave it a name. We say ``I need 6\nfactorial plaques engraved.'' When we write a factorial, we use an\nexclamation point:\\index{factorial}\n\n$$6! = 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = 720$$\n\nThe rule, then, is $n$ items can be ordered in $n!$ ways. We use the\nword ``permutation'' to mean a particular ordering. Thus\nmathematicians actually say ``If you have a list of $n$ items then I\ncan generate $n!$ different permutations of those items''.\n\nIn Python, there is a \\pyfunction{factorial} function in the math library:\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n> \\textbf{python3} \n>>> \\textbf{import math}\n>>> \\textbf{math.factorial(6)}\n720\n\\end{Verbatim}\n\nHandy, right? Now you don't need to write loop to calculate factorials.\n\nRemember when we only wanted the first three names on the plaque? We can do that problem using factorials:\n\n$$6 \\times 5 \\times 4 = \\frac{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}{3 \\times 2 \\times 1} = \\frac{6!}{3!}$$\n\nThis formulation makes it easy to figure out on any calculator with a ``!'' button.\n\nThe rule on this is to fill $m$ positions from $n$ items, it can be done this many ways:\n\n$$\\frac{n!}{(n-m)!}$$\n\n\\subsection{Choose}\n\nLet's say that there are 12 kids in a classroom, and you need a team\nof 4 to wipe down the desks. How many different possible teams are\nthere? You know that if you were giving out four different positions\n(Like the race gave out 1st, 2nd, and 3rd), the answer would be $12\n\\times 11 \\times 10 \\time 9$ or $12! / (12 - 4)!$.\n\nHowever, once we pick the 4 people, we don't care what order they are\nin, right? In this problem, the team ``Anne, Brad, Carl, and Don'' is\nthe same as the team ``Carl, Don, Brad, and Anne''.\n\nThus, the quantity $12! / (12 - 4)!$ is many times too large because\nit counts each permutation separately. To get the right number, we\njust divide this by the number of possible permuations for a group of\nfour people: $4!$\n\nThat gets us our answer: How many different teams of four can be chosen from 12 people?\n\n$$\\frac{12!}{(12-4)! 4!}= 495$$\n\nIn combinatorics, we use this quantity a lot, so we have given it a name: \\textit{choose}\\index{choose function}\n\nWe have also given it a notation. ``12 choose 4'' is written like this:\n\n$${12 \\choose 4}$$\n\nPython has the \\pyfunction{math.comb} function:\n\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n> \\textbf{python3}\n>>> \\textbf{import math}\n>>> \\textbf{comb(12, 4)}\n495 \n\\end{Verbatim}\n\n", "meta": {"hexsha": "0ed5c6802021cc469ac9f715a56ca4941eb11f52", "size": 4487, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Modules/DiscreteProbability/combinatorics-en_US.tex", "max_stars_repo_name": "rajivjhoomuck/sequence", "max_stars_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Modules/DiscreteProbability/combinatorics-en_US.tex", "max_issues_repo_name": "rajivjhoomuck/sequence", "max_issues_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Modules/DiscreteProbability/combinatorics-en_US.tex", "max_forks_repo_name": "rajivjhoomuck/sequence", "max_forks_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.7909090909, "max_line_length": 116, "alphanum_fraction": 0.7298863383, "num_tokens": 1308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7185943925708561, "lm_q2_score": 0.8918110454379297, "lm_q1q2_score": 0.6408504164844492}} {"text": "\\documentclass{article}\n\\input{commonheader}\n\\usepackage{comment}\n\\usepackage{enumitem}\n\\usepackage{algpseudocode}\n\n\\title{CS 5050: Homework 1} \n\\author{Andrew Pound}\n\\date{\\today}\n\n\\begin{document}\n\\maketitle\n\nI worked on this assignement with Chad Cummings.\n\\section{}\n\\begin{enumerate}[label=(\\alph*)]\n\\item Suppose we are comparing implementations of insertion sort and \nmerge sort on the same machine. For inputs of size n, insertion sort\nruns in $8n^2$ steps, while merge sort runs in $64n \\log n$ steps. For\nwhich values of $n$ does insertion sort beats merge sort? {\\bf (5\n points)} \n\\item[]\nThere are a few different ways to do this. The first way that we\nattacked it was that we solved the inequality equation to easiest way\nto solve this is to tabulate the answers for each side of the equatoin\nand find the point that they cross. (This would actually get a bit\nlong for this particular problem). So instead a plot of the two\nfunctions can be used to see where it crosses.\n\nEquations:\n\\begin{equation}\\label{eq:1a}\n \\begin{split}\n 8n^2 &\\le 64n\\log n\\\\\n \\frac{1}{8} &\\le \\frac{1}{n}\\log n\\\\\n 2^{\\frac{1}{8}} &\\le n^{\\frac{1}{n}}\n \\end{split}\n\\end{equation}\n\nPlotting both sides of this, produces the graph below.\n\\begin{figure}[h!t]\n \\centering\n \\includegraphics[width=.7\\linewidth]{prob1a}\n \\caption{Plot of the inequality in equation \\ref{eq:1a}.}\n \\label{fig:prob1a}\n\\end{figure}\n\nFrom this we can see that the cross over point is around 43.\nTabulating the values of $n^{1/n}$ in Table \\ref{tab:1a}, the\nneighborhood of 43 and \n\\begin{table}[h!t]\n \\centering\n \\caption{Tabulation of values around $n=43$.}\n \\label{tab:1a}\n \\begin{tabular}{cc}\n $n$ & $n^{1/n}$ \\\\\n \\input{prob1a.tex}\n \\end{tabular}\n\\end{table}\ncomparing them with the value of $2^{1/8} = 1.0905$, we can see that\nany number greater than 43 satisfies the inequality. \n\n\n\\item What is the smallest value of n such that an algorithm whose\nrunning time is $100n^2$ runs faster than an algorithm whose running\ntime is $2^n$ on the same machine? {\\bf (5 points) }\n\\item[] This can be solved in much the same manner as the previous\n problem. First, Let's examine the inequality.\n \\begin{equation}\\label{eq:1b}\n \\begin{split}\n 100 n^2 & \\le 2^n\\\\\n \\log_2(100n^2) & \\le log_2(2^n) = n\\\\\n \\frac{2}{n} \\log(10n) &\\le 1\\\\\n \\frac{1}{n} \\log(10n) &\\le \\frac1 2\\\\\n \\log\\left(n^{\\frac{1}{n}}\\right) &\\le \\frac{1}{2}\\\\\n \\left(10n\\right)^{\\frac{1}{n}} &\\le \\sqrt{2}\n \\end{split}\n \\end{equation}\n\\end{enumerate} \n\nThe plot of this function is seen in Figure \\ref{fig:1b}, and the\ntabulation around the cross over is in Table \\ref{tab:1b}. This time\nwe ar comparing with the value $\\sqrt{2} = 1.4142$, and we can see\nthat the cross over happens at $n=15$.\n\\begin{figure}[h!t]\n \\centering\n \\includegraphics[width=.7\\linewidth]{prob1b}\n \\caption{Plot of the inequality in equation \\ref{eq:1b}.}\n \\label{fig:1b}\n\\end{figure}\n\n\\begin{table}[h!t]\n \\centering\n \\caption{Tabulation of values around $n=15$.}\n \\label{tab:1b}\n \\begin{tabular}[h]{cc}\n $n$ & $(10n)^{1/n}$ \\\\\n \\input{prob1b.tex}\n \\end{tabular}\n\\end{table}\n\n\\vspace{3em}\n\\section{}\nWe have discussed the insertion sort algorithm in class and we showed\nthat in the worst case the algorithm runs in $\\Theta(n^2)$ time. You\nmay wonder whether the algorithm can run asymptotically faster for\ncertain ``average'' cases. In this exercise, we will examine the running\ntime of the insertion sort algorithm on some average cases. Let $A$ be \nthe input array of n elements.\n\nNote: For each of the questions, please give your answers in the\nbig-Theta notation and also briefly explain how you obtain your answer.\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item Consider the case where the first $n/2$ elements of $A$ have\nalready been sorted \\emph{increasingly} and the last $n/2$ elements\nhave been sorted \\emph{decreasingly}. For example, $A = \\{1, 3, 5, 7,\n8, 6, 4, 2\\}$. Suppose we want to sort all elements of $A$ in\nincreasing order by using the insertion sort algorithm. What would be\nthe running time of the algorithm? What if the first $n/2$ elements of\n$A$ have already been sorted decreasingly and the last $n/2$ elements\nhave been sorted increasingly? For example, $A = \\{8, 6, 4, 2, 1, 3,\n5, 7\\}$. What would be the running time in this case? {\\bf (10 points)}\n\n\\item[] The insertion sort algorithm that we discussed in class is\n given as\n \\begin{algorithmic}\n \\For{$i=2\\dots n$}\n \\State $x\\gets A[i]$\n \\State $j \\gets i-1$\n \\While{$A[j-1] < x \\And j \\ge 1$}\n \\State $A[j+1]\\gets A[j]$\n \\State $j \\gets j-1$\n \\EndWhile\n \\State $A[j+1]\\gets x$\n \\EndFor\n \\end{algorithmic}\nThe outer for loop consists of $4c$ operations, and the inner loop\n(when run) consists of $2c$ operations.\nNow suppose that the elements of our array are ordered in such a way\nthat the first half of the array are already in ascending order, and\nthe second half is the worst case (descending order) and need to be\nmerged into the first half. If that is the case, then the first half\nof the algorithm is a best case scenario, and only uses\n$4c\\frac{n}{2}$ operations. The second half, though will utilize the\ninner for-loop to move some of the elements in the first half of the\narray in order to insert in the correct place. \nBy considering the first elements in the second half, we can determine\na relation between the number of times in the inner loop and the\niteration index. If the array is in the form specified, then the\nfirst element of the second half will not require any moving to place\nit. The second will require 2 moves (the first element of the 2nd\nhalf and the last element of the first half). The third element in\nthe 2nd half will require 4 moves. We see that for $i > \\frac{n}{2}$,\nthe number of times in the inner loop will be $2(i-\\frac{n}{2})$.\nThus, we can stick this into an expression for the total number of\noperations\n\\begin{equation}\n \\underbrace{4cn}_{\\text{outer loop}} + \\underbrace{\\sum_{i =\n \\frac{n}{2}}^n2\\left(i-\\frac{n}{2}\\right)\\left(2c\\right).\n}_{\\text{inner loop}} \n\\end{equation}\nNow expanding this, we get a total of \n\\begin{equation}\n\\begin{split}\n (4cn) + 4c\\sum_{i=0}^{n/2}i &= 4cn +\n 4c\\frac{\\frac{n}{2}\\left(\\frac{n}{2} + 1\\right)}{2}\\\\\n &= 4cn + 2c\\left(\\frac{1}{4}n^2 + \\frac{1}{2}n\\right)\\\\\n &= 4cn + \\frac{1}{2}n^2c + cn\\\\\n &= \\frac{1}{2}cn^2 + 5cn\\\\\n &= O(n^2)\n\\end{split}\n\\end{equation}\n\n%The second ordering \n\n\n\n\\item Consider the case where the elements of $A$ with odd indices have\nalready been sorted increasingly and elements with even indices have\nbeen sorted decreasingly (we assume the index of $A$ starts from\n1). For example, $A = \\{1, 8, 2, 7, 3, 6, 4, 5\\}$. Suppose we want to\nsort all elements of $A$ in increasing order by using the insertion\nsort algorithm. What would be the running time of the algorithm?\n\nWhat if the elements of $A$ with odd indices have already been sorted\n\\emph{decreasingly} and elements with even indices have been sorted \n\\emph{increasingly}. For example, $A = \\{8, 1, 7, 2, 6, 3, 5,\n4\\}$. What would be the running time in this case? {\\bf (10 points)}\n\n\\item[] \nOk, we'll begin by looking at how the algorithm works on the example\nsequence. The results of the steps of the algorithm are shown below.\nThe bolded numbers identify values that needed to be moved (i.e. the\nvalues for which the inner loop was used). The first line is the\noriginal sequence.\n\\begin{equation}\n \\begin{matrix}\n 1 & 8 & 2 & 7& 3 &6 & 4 & 5\\\\\n 1 &&&&&&&\\\\\n 1 & 8 &&&&&&\\\\\n 1 & 2 & {\\bf 8}&&&&&\\\\ \n 1 & 2 & 7 & {\\bf 8}&&&&\\\\\n 1 & 2 & 3 & {\\bf 7}&{\\bf 8} &&&\\\\\n 1 & 2 & 3 & 6& {\\bf 7}&{\\bf 8} &&\\\\\n 1 & 2 & 3 & 4& {\\bf 6}& {\\bf 7}&{\\bf 8} &\\\\\n 1 & 2 & 3 & 4& 5 &{\\bf 6} & {\\bf 7}&{\\bf 8} \\\\\n \\end{matrix}\n\\end{equation}\nFrom this we can see that for each pair of iterations, the number of\nmoves ends up being predictable. In order to count the time, we then\nstart with the equation\n\\begin{equation}\n \\underbrace{4cn}_{\\text{outer loop}} + \\underbrace{2 \\sum_{i=1}^{n/2}\n (i-1)(2c)}_{\\text{inner loop}}.\n\\end{equation}\nThe outer loop has $4c$ operations and happens for each element in the\narray, thus we get $4cn$. Each pair of iterations is hitting the\ninner loop for the same number of times, namely $i-1$, for the $i$th\npair. The inner loop consists of $2c$ operations, thus we get the\nsummation over all the pairs, and multiply by 2 to get the full effect\nof the inner loop. This is\n\\begin{equation}\n \\begin{split}\n 2 \\sum_{i=1}^{n/2}\n (i-1)(2c) &= 4c \\sum_{i=1}^{n/2}(i - 1)\\\\\n &= 4c \\sum_{i=1}^{n/2}i - 4c\\frac{n}{2}\\\\\n &= 4c \\frac{\\frac{n}{2}\\left(\\frac{n}{2} + 1\\right)}{2} -\n 4c\\frac{n}{2}\\\\ \n &= 2c\\left(\\frac{n^2}{4} + \\frac{n}{2} - n\\right)\\\\\n &= 2c\\left(\\frac{1}{4}n^2 - \\frac{1}{2}n\\right)\n \\end{split}\n\\end{equation}\nThe entire operations count together then would be\n\\begin{equation}\n \\begin{split}\n 4cn + 2c\\left(\\frac{1}{4}n^2 - \\frac{1}{2}n\\right)&= \n 4cn + \\frac{1}{2}cn^2 - cn\\\\\n &= \\frac{1}{2}cn^2 + 3cn\\\\\n &= O(n^2)\n \\end{split}\n\\end{equation}\nNow for the other example, let's see what it looks like:\n\\begin{equation}\n \\begin{matrix}\n 8& 1& 7& 2& 6& 3& 5 &4 \\\\\n 8 &&&&&&&\\\\\n 1 & {\\bf 8} &&&&&&\\\\\n 1 & 7 & {\\bf 8}&&&&&\\\\ \n 1 & 2 & {\\bf 7} & {\\bf 8}&&&&\\\\\n 1 & 2 & 6 & {\\bf 7}&{\\bf 8} &&&\\\\\n 1 & 2 & 3 & {\\bf 6}& {\\bf 7}&{\\bf 8} &&\\\\\n 1 & 2 & 3 & 5& {\\bf 6}& {\\bf 7}&{\\bf 8} &\\\\\n 1 & 2 & 3 & 4& {\\bf 5} &{\\bf 6} & {\\bf 7}&{\\bf 8} \\\\\n \\end{matrix}\n\\end{equation}\nEssentially, the idea is the same, in that the iterations can be split\ninto pairs such that a pair will use the inner loop (to move larger\nnumbers) the same number of times. Thus the analysis is (almost)\nexactly the same: $ O(n^2)$.\n\n\n\\end{enumerate}\n\n\n\\section{}\nFor each of the following pairs of functions, indicate whether it is\none of the three cases: $f (n) = O(g(n))$, $f (n) = \\Omega(g(n))$, or\n\n$f (n) = \\Theta(g(n))$. (30 points)\n\\begin{enumerate}[label=(\\alph*)]\n\\item $f (n) = 100n + \\log n$ and $g(n) = 6n + \\log^2 n$.\n\\item[] $f(n) = \\Omega(g(n))$. My reasoning is this:\n \\begin{equation}\n \\begin{split}\n f(n) &\\: ?\\: g(n)\\\\\n 100n + \\log n &\\: ? \\: 6n+]log^2 n\\\\\n 94n &\\: ?\\: \\log^2 n - \\log n \n \\end{split}\n \\end{equation}\nNow, $\\log^2 n - \\log n < \\log^2 n$, and comparing this as above, we\nsee that \n\\begin{equation}\n \\begin{split}\n 94n > \\log^2 n > \\log^2 n -\\log n.\n \\end{split}\n\\end{equation}\nThus, we can say that $f(n) = \\Omega(g(n))$.\n\\item $f (n) = 20 \\log n + 4$ and $g(n) = \\log n^2 - 100$.\n\\item[] $f(n) = \\Theta(g(n))$. Playing with the equations a bit, we\n can see\n \\begin{equation}\n \\begin{split}\n f(n) & \\: ? \\: g(n)\\\\\n 20\\log n + 4 &\\: ? \\: \\log n^2 - 100\\\\\n 20 \\log n + 4 &\\: ? \\: 2\\log n - 100.\n \\end{split}\n \\end{equation}\nNow, if we choose $c_1 = 20$ and $c_2 = 10$, then we get an asymptotic\nupper and lower bound, respectively, by forming the equations $c_i\ng(n)$. Thus, we know that $f(n) = \\Theta(g(n))$.\n\\item $f (n) =\\frac{n^2}{\\log n}$ and $g(n) = n \\log^2 n$.\n\\item[] $f(n) = \\Omega(g(n))$. Let's play with the equations:\n \\begin{equation}\n \\begin{split}\n \\frac{n^2}{\\log n} &\\: ? \\: n \\log^2 n\\\\\n n &\\: ? \\: \\log^3 n.\n \\end{split}\n \\end{equation}\nThen using a nifty formula:\n \\begin{equation}\n \\log^k n = O(n^d) \\quad \\forall\\: k>0, \\: d > 0,\n \\end{equation}\nwe can see that $d = 1$, and $k = 3$, and thus, $g(n) = O(f(n))$. \nOr in other words, $f(n) = \\Omega(g(n))$.\n\\item $f (n) =\\sqrt{n}$ and $g(n) = \\log^5 n$.\n\\item[] $f(n) = \\Omega(f(n))$. Using the same nifty equation from\n above, we can see that $g(n) = O(f(n))$, as $k = 5$, and $d = 1/2$.\n Thus flipping this around, we see that $f(n) = \\Omega(f(n))$.\n\\item $f (n) = n2^n$ and $g(n) = 3^n$ .\n\\item[] $f(n) = O(g(n))$. We will make use of our other nifty\n formula \n \\begin{equation}\n n^d = o(e^n), \\quad \\forall\\: e > 1.\n \\end{equation}\nFirst, let's play a bit with our equation\n\\begin{equation}\n\\begin{split}\n n2^n &\\: ?\\: 3^n\\\\\n n &\\: ? \\: \\left(\\frac{3}{2}\\right)^n.\n\\end{split}\n\\end{equation}\nThis helps us to see that when $d = 1$ and $ e = 3/2$, then \nwe know that $f(n) = o(g(n))$ and therfore, $f(n) = O(g(n))$.\n\\item $f (n) = 4n \\log n$ and $g(n) = n \\log_3 n$.\n\\item[] $f(n) = \\Theta(g(n))$. Playing with the equation, we get\n \\begin{equation}\n \\begin{split}\n 4n \\log n &\\: ? \\: n \\log_3 n\\\\\n 4 \\log n &\\: ? \\: \\frac{\\log n}{\\log 3}\\\\\n 4 \\log n &\\: ? \\: \\frac{1}{\\log 3} \\log n\\\\\n 4 &\\: ? \\: \\frac{1}{\\log 3}\n \\end{split}\n \\end{equation}\nNow we see that $f(n) = \\Theta(g(n))$.\n\\end{enumerate}\n\n\nNote: For each question, you only need to give your answer and the\nproof is not required. \n\n\n\\section{}\nThe \\emph{knapsack problem} is defined as follows: Given as input a\nknapsack of size $K$ and $n$ items whose sizes are $k_1 , k_2 ,\n\\dots , kn$ , where $K$ and $k_1 , k_2 , \\dots , k_n$ are all real\nnumbers, find a full ``packing'' of the knapsack (i.e., choose a\nsubset of the n items such that the total sum of the sizes of the\nitems in the chosen subset is exactly $K$). \n\nIt is well known that the knapsack problem is NP-complete, which\nimplies that it is very likely that efficient algorithms (i.e., those\nwith a polynomial running time) for this problem do not exist. Thus,\npeople tend to look for good {\\bf approximation algorithms} for\nsolving this problem. In this exercise, we relax the constraint of the\nknapsack problem as follows. We still seek a packing of the knapsack,\nbut we need not look for a ``full'' packing of the knapsack; instead,\nwe look for a packing of the knapsack (i.e., a subset of the n input\nitems) such that the total sum of the sizes of the items in the chosen\nsubset is at least $K/2$ (but no more than $K$). This is called a\n\\emph{factor of 2 approximation solution} for the knapsack problem. To \nsimplify the problem, we assume that a factor of 2 approximation\nsolution for the knapsack problem always exists, i.e, there always\nexists a subset of items whose total size is at least $K/2$ and at\nmost $K$. \n\nDesign a polynomial time algorithm for computing a factor of 2\napproximation solution for this problem, and analyze the running time\nof your algorithm (in the big-O notation). If your algorithm runs in\n$O(n)$ time and is correct, then you get {\\bf 5 extra points}. {\\bf(20\n points)} \n\nNote: You are required to clearly describe the main idea of your\nalgorithm. Although the pseudo-code is not required, you may also give\nthe pseudo-code if you feel it is helpful for you to describe your\nalgorithm. (The reason I want to see the algorithm description instead\nof only the code or pseudo-code is that it would be difficult to\nunderstand another person's code without any explanation.) You also\nneed to briefly explain why your algorithm works, i.e., why your\nalgorithm can produce a factor of 2 approximation solution. Finally,\nplease analyze the running time of your algorithm. \n\n\\paragraph{My algorithm}\n\n\n\n\n\nIn order to find a factor of 2 approximation, we will make use of the\nassumptions that there definitely exists such an approximation, and\nthat all of the $k_i$ sizes are positive.\n\nThe algorithm is as follows:\n\nIterate through the list by pairs, finding the largest of each pair\nand moving it to the lower position in the pair.\nThen iterate and add the largest of each pair (unless it is too large\n(i.e. makes the knapsack over full).\nThen, if at the end of this loop, that knapsack is not at least 1/2\nfull, then iterate through the smaller of the pairs, adding them,\nunless they overfill the knapsack.\n\nBecause there exists a factor of 2 approximation, this algorithm will\nterminate when either the solution is found or the entire list is\nadded during the 3rd loop, which will then constitute a solution\n(because one has to exist).\n\nPseudocode for the algorithm can be laid out as \n{\\singlespacing\n\\begin{algorithmic}\n \\State $cap \\gets$ Knapsack size/capacity\n \\For{$i = 1,3,\\dots,n$} \n \\If{$k[i] \\le k[i+1]$}\n \\State Swap($k[i],k[i+1]$)\n \\EndIf\n \\EndFor\n \\State $Sz = 0$\n \\State // Let's fill in the larger sizes first\n \\For{$i = 1,3,\\dots,n$}\n \\If{ $Sz + k[i] \\le cap$}\n \\State $Sz \\gets Sz + k[i]$\n \\EndIf\n \\If{ $Sz \\ge \\frac{cap}{2}$}\n \\State return\n \\EndIf\n \\EndFor\n \\State // Now, start filling in the smaller sizes\n \\For{$i = 2,4,\\dots,n$}\n \\If{ $Sz + k[i] \\le cap$}\n \\State $Sz \\gets Sz + k[i]$\n \\EndIf\n \\If{ $Sz \\ge \\frac{cap}{2}$}\n \\State return\n \\EndIf\n \\EndFor\n\\end{algorithmic}\n}\nThe runtime for this algorithm is $O(\\frac{3}{2}n)$, because at the\nmost it will utilize all three for-loops which are each over only half\nthe length of the array.\n\\end{document}", "meta": {"hexsha": "f22ddaec7d15c0ee7fef2750b0afb15c4e2f7d01", "size": 16768, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw/hw1/hw1.tex", "max_stars_repo_name": "aepound/advAlgo", "max_stars_repo_head_hexsha": "03d612282a86a8957df6b9197af036c6fc548b0b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw/hw1/hw1.tex", "max_issues_repo_name": "aepound/advAlgo", "max_issues_repo_head_hexsha": "03d612282a86a8957df6b9197af036c6fc548b0b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw/hw1/hw1.tex", "max_forks_repo_name": "aepound/advAlgo", "max_forks_repo_head_hexsha": "03d612282a86a8957df6b9197af036c6fc548b0b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.2622222222, "max_line_length": 72, "alphanum_fraction": 0.6572638359, "num_tokens": 5775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7185943925708561, "lm_q2_score": 0.8918110353738529, "lm_q1q2_score": 0.6408504092524601}} {"text": "%As explained in Section \\ref{sec:background}, comparing elements over $\\mathbb{F}_{p^d}$ boils down to their comparison over $\\mathbb{F}_p$. \nIn this section, we study the structure of basic comparison circuits over $\\mathbb{F}_p$ used in Section~\\ref{sec:background} to compare integers, namely $\\LT_{\\S}$ and $\\EQ_{\\S}$ for some $\\S \\subseteq [0, p-1]$. \n\n%Let $\\S$ be a subset of $[0,p-1]$ such that $\\S$ can be embedded canonically embedded into $\\F_p$.\nFor any choice of $\\S$, the corresponding equality function over $\\F_p$ is equal to\n\\begin{align*}\n \\EQ_{\\S}(x,y) = 1 - (x-y)^{p-1}.\n\\end{align*}\nUnfortunately, $\\LT_{\\S}$ is not that simple and universal and we have to rely on Lagrange interpolation (Lemma~\\ref{lem:interpolation}) to compute it. \nYet, there are two different ways to evaluate it.\nThe first method (as done in \\cite{TLWRK20}) uses $\\S = [0,p-1]$ and directly interpolates $\\LT_{\\F_p}(x,y)$ as a bivariate polynomial over $\\F_{p}$.\nThe second approach (as done in \\cite{NGEG17} and \\cite{PoPETS:SFR20}) has $\\S = [0, (p-1)/2]$ and interpolates a univariate polynomial of $\\LT_{\\S}(z, 0)$ with $z = x - y$.\n%In this case, $x,y$ must belong to the set $[0, (p-1)/2]$. \n%However, these works evaluated $\\LT_{\\F_p}$ without exploiting its structure. \nIn this section, we show how to exploit the structure of these polynomials to speed-up their evaluation.\n\n\\subsection{Bivariate interpolation of $\\LT_\\S$.}\n\n Let $\\S = [0,p-1]$.\n The less-than function can be interpolated using Lemma~\\ref{lem:interpolation} and the following truth table.\n \\begin{table}[h]\n \\centering\n \\begin{tabular}{c|ccccc}\n $<$ & 0 & 1 & 2 & $\\cdots$ & $p-1$ \\\\\n \\hline\n 0 & 0 & 1 & 1 & $\\cdots$ & 1 \\\\\n 1 & 0 & 0 & 1 & $\\cdots$ & 1 \\\\\n 2 & 0 & 0 & 0 & $\\cdots$ & 1 \\\\\n $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\vdots$ & $\\ddots$ & $\\vdots$ \\\\\n $p-1$ & 0 & 0 & 0 & $\\cdots$ & 0 \\\\\n \\end{tabular}\n \\end{table}\n \n In particular, the interpolation polynomial of $\\LT_\\S$ over $\\F_p$ is equal to\n \\begin{align*}%\\label{eq:less_than_function}\n P_{\\LT_\\S}(X,Y) &= \\sum_{a = 0}^{p-2} \\EQ_\\S(X,a)\\sum_{b = a+1}^{p-1} \\EQ_\\S(Y,b) \\nonumber\\\\\n = \\sum_{a = 0}^{p-2}& \\left(1-\\left(X - a\\right)^{p-1}\\right) \\sum_{b = a+1}^{p-1} \\left(1-\\left(Y - b\\right)^{p-1}\\right).\n \\end{align*}\n Surprisingly, the total degree of $P_{\\LT_\\S}(X,Y)$ is only $p$ and its coefficients can be described by the following theorem.\n \\begin{theorem} \\label{thm:less_than_total_degree}\n Let $p>2$ be a prime number and $\\S = [0,p-1]$, then the interpolation polynomial of $\\LT_\\S$ over $\\F_p$ has the following form\n \\begin{align*}\n P_{\\LT_\\S}(X,Y) = Y^{p-1} - \\frac{p-1}{2} (XY)^{\\frac{p-1}{2}} + \\sum_{\\substack{i,j>0, \\\\ i \\ne j, \\\\ i+j \\le p}} a_{ij} X^i Y^j \n \\end{align*}\n where $a_{ij} = \\sum_{a=0}^{p-2} \\sum_{b=a+1}^{p-1} a^{p-1-i} b^{p-1-j} \\in \\F_p$.\n The total degree of $P_{\\LT_\\S}(X,Y)$ is $p$.\n \\end{theorem}\n \\begin{proof}\n See Appendix \\ref{app:proof_thm_less_than_total_degree}.\n \\end{proof}\n From the definition of $\\LT_\\S$, one can easily prove the following facts about $P_{\\LT_\\S}$:\n \\begin{itemize}[label=--]\n \\item $P_{\\LT_\\S} (X, 0) = 0$, thus $Y$ divides $P_{\\LT_\\S} (X, Y)$;\n \\item $P_{\\LT_\\S} (X, X) = P_{\\LT_\\S} (Y,Y) = 0$, thus $(X - Y)$ divides $P_{\\LT_\\S} (X, Y)$;\n \\item $P_{\\LT_\\S} (p-1, Y)=0$ thus $X + 1$ divides $P_{\\LT_\\S} (X, Y)$.\n \\end{itemize}\n Hence, there exists a bivariate polynomial $f(X,Y)$ of total degree $p - 3$ over $\\F_p$ such that:\n \\begin{equation}\n \\label{eq:decomposition-LT}\n P_{\\LT_\\S} (X, Y) = Y(X - Y)(X + 1)f(X, Y).\n \\end{equation}\n The following theorem describes the structure of $f(X,Y)$.\n \\begin{theorem}\\label{thm:decomposition-f}\n Let $p$ be an odd prime and $\\S = [0,p-1]$.\n Let $P_{\\LT_\\S} (X, Y)$ be the interpolation polynomial of $\\LT_\\S$ over $\\F_p$ and $P_{\\LT_\\S} (X, Y) = Y(X - Y)(X + 1)f(X, Y)$.\n Then, for any $z \\in \\F_p$ we have\n \\begin{equation}\n \\label{eq:3}\n f(z,z) = f(z,0) = f(p-1,z).\n \\end{equation}\n As a consequence, \n %there exists a bivariate polynomials $f_1(X,Y)$ of total degree $p-5$ over $\\F_p$ such that:\n %\\begin{align}\n % \\label{eq:dec-f}\n % f(X,Y) &= f(X,0) + Y(X-Y)f_1(X,Y) \\nonumber \\\\\n % &= f_0(X) + Y(X-Y)f_1(X,Y).\n %\\end{align} \n %This decomposition can be applied recursively to $f_1(X,Y)$, so that \n there exists $(p-1)/2$ polynomials $f_i(X)$ over $\\F_p$, $0\\leq i \\leq (p-3)/2$, such that:\n \\begin{align}\\label{eq:decomposition-f-final}\n f(X,Y) = \\sum_{i=0}^{(p-3)/2} f_i(X)Z^i,\n \\end{align}\n with $Z=Y(X-Y)$ and $\\deg (f_{i}(X)) = p-3 - 2i$.\n\n % or equivalently a bivariate polynomials $f'_1$ of total degree $p-5$ over $\\F_p$ such that:\n % \\begin{equation}\n % \\label{eq:dec-f'}\n % f(x,y) = f(p-1,y) + (x+1)(x-y)f'_1(x,y)\n % \\end{equation} \n \\end{theorem} \n\n Since our proof of Theorem~\\ref{thm:decomposition-f} is quite long and with no real interest for the purpose of this work, we defer it to an extended version of this paper. \n In our experiments, we used the decompositions~(\\ref{eq:decomposition-f-final}) of $f(X,Y)$ only for small $p$ (between $3$ and $7$), which you can find in Appendix~\\ref{app:decomposition-f}.\\newline\n \n% In order to show Eq. (\\ref{eq:3}) we will need the following lemma:\n\n% \\begin{lemma}\\label{lem:structure-f}\n% Let $p>3$ be an odd prime, $i\\in[0,p-2]$ and $f$ be the polynomial defined in Eq. (\\ref{eq:decomposition-LT}), then for any $z\\in[0,p-1]$ we have:\n\n% $$ f(i,z) = f(p-1-z,p-1-i) = -\\displaystyle\\sum_{k=i+1}^{p-1}\\frac{1-(z-k)^{p-1}}{(i+1)z(z-i)} \\bmod p.$$\n% \\end{lemma}\n\n% \\begin{proof}\n% We know that for any $z\\in[0,p-1]$ we have:\n% $$\\LT_\\S(i,z) = \\left\\{\n% \\begin{array}{l}\n% 0 ~\\text{ if }~ 0\\leq z\\leq i \\\\\n% 1 ~\\text{ otherwise}~\n% \\end{array}\n% \\right.$$\n% Note that since $i0$ we know from Lemma \\ref{lem:structure-f} that:\n% $$ f(i,z) = -\\displaystyle\\sum_{k=i+1}^{p-1}\\frac{1-(z-k)^{p-1}}{(i+1)z(z-i)} \\bmod p, $$\n% which can be rewritten modulo $p$ as:\n% \\begin{align*}\n% f(i,z) & = -\\frac{p-1-i - \\displaystyle\\sum_{k=i+1}^{p-1}(z-k)^{p-1}}{(i+1)z(z-i)} = -\\frac{p-1-i - \\displaystyle\\sum_{k=i+1}^{p-1}\\displaystyle\\sum_{j=0}^{p-1}z^j k^{p-1-j}}{(i+1)z(z-i)} \\\\\n% & = -\\frac{p-1-i - \\displaystyle\\sum_{j=0}^{p-1}\\left(\\displaystyle\\sum_{k=i+1}^{p-1} k^{p-1-j}\\right)z^j}{(i+1)z(z-i)} = \\frac{\\displaystyle\\sum_{j=1}^{p-1}\\left(\\displaystyle\\sum_{k=i+1}^{p-1} k^{p-1-j}\\right)z^{j-1}}{(i+1)(z-i)}.\n% \\end{align*}\n% Similarly:\n% \\begin{align*}\n% f(i,z) & = -\\frac{p-1-i - \\displaystyle\\sum_{k=1}^{p-1-i}(z-i-k)^{p-1}}{(i+1)z(z-i)} = -\\frac{p-1-i - \\displaystyle\\sum_{k=1}^{p-1-i}\\displaystyle\\sum_{j=0}^{p-1}(z-i)^j k^{p-1-j}}{(i+1)z(z-i)} \\\\\n% & = -\\frac{p-1-i - \\displaystyle\\sum_{j=0}^{p-1}\\left(\\displaystyle\\sum_{k=1}^{p-1-i} k^{p-1-j}\\right)(z-i)^j}{(i+1)z(z-i)} = \\frac{\\displaystyle\\sum_{j=1}^{p-1}\\left(\\displaystyle\\sum_{k=1}^{p-1-i} k^{p-1-j}\\right)(z-i)^{j-1}}{(i+1)z}\n% \\end{align*}\n\n% So for any $0 < i < p-1$, from the first expression we obtain:\n% $$f(i,0) = -\\frac{\\sum_{k=i+1}^{p-1} k^{p-2}}{i(i+1)},$$\n% from the second expression we get:\n% \\begin{align*}\n% f(i,i) &= \\frac{\\sum_{k=1}^{p-1-i} k^{p-2}}{i(i+1)} = \\frac{\\sum_{k'=i+1}^{p-1} (p-k')^{p-2}}{i(i+1)} = \\frac{\\sum_{k'=i+1}^{p-1} (-k')^{p-2}}{i(i+1)} \\bmod p \\\\\n% &= -\\frac{\\sum_{k'=i+1}^{p-1} (k')^{p-2}}{i(i+1)} = f(i,0).\n% \\end{align*}\n\n% At this point we have shown that for $0\\leq i i ~~ x_j = y_j\\,.\n% \\end{align*}\n% This order induces a function $\\LT_{\\F_\\fieldcard}(\\vx, \\vy)$ that returns $1$ if $\\vx < \\vy$ and $0$ otherwise. \n% As done in~\\cite{TLWRK20}, we can employ $\\EQ_{\\F_\\fieldcard}$ and $\\LT_{\\F_\\fieldcard}$ to compute $\\LT_{\\F_\\fieldcard}(\\vx, \\vy)$ as follows\n% \\begin{align*}\n% \\LT_{\\F_\\fieldcard}(\\vx, \\vy) = \\sum_{i=0}^{\\ell-2} \\LT_{\\F_\\fieldcard}(x_{i}, y_{i}) \\prod_{j=i+1}^{\\ell-1} \\EQ_{\\F_\\fieldcard}(x_{j}, y_{j}) + \\LT_{\\F_\\fieldcard}(x_{\\ell-1}, y_{\\ell-1}).\n% \\end{align*}\n% Notice that the multiplicative depth of this function solely depends on the products of the equality functions.\n% In fact, these products compare subvectors $\\vx_i = (x_i, x_{i+1},\\dots,x_{\\ell-1})$ and $\\vy_i = (y_i, y_{i+1},\\dots,y_{\\ell-1})$ for $i \\in [1,\\ell-1]$.\n% Thus, we can rewrite $\\LT_{\\F_\\fieldcard}$ as\n% \\begin{align}\\label{eq:general_lex_order}\n% \\LT_{\\F_\\fieldcard}(\\vx, \\vy) = \\sum_{i=0}^{\\ell-2} \\LT_{\\F_\\fieldcard}(x_{i}, y_{i}) \\EQ_{\\F_\\fieldcard}(\\vx_{i+1}, \\vy_{i+1}) + \\LT_{\\F_\\fieldcard}(x_{\\ell-1}, y_{\\ell-1})\n% \\end{align}\n% with the equality function $\\EQ_{\\F_\\fieldcard}(\\vx_{i+1}, \\vy_{i+1})$ that returns $1$ if $\\vx_{i+1} = \\vy_{i+1}$ and $0$ otherwise.\n% As shown in~\\todo{cite our work}, this function can be realized by a constant-depth circuit in the following way.\n% \\begin{align}\\label{eq:rand_eq_circuit}\n% \\EQ_{\\F_\\fieldcard, e}(\\vx_{i+1},\\vy_{i+1}) = 1 - \\princhar_{\\fieldcard^e}\\left(\\sum_{j=i+1}^{\\ell-1} r_j (x_j - y_j)\\right)\n% \\end{align}\n% where $r_j$ are uniformly random elements from $\\F_{\\fieldcard^e}$.\n% This circuit is false-biased with error probability $\\fieldcard^{-e}$.\n% We can compute all $\\EQ_{\\F_\\fieldcard, e}(\\vx_{i+1},\\vy_{i+1})$ using the same number of multiplications as for the single equality using Algorithm~\\ref{alg:vector_equalities_circuit}.\\todo{What is the complexity?}\n% \\begin{algorithm}[t]\n% \\KwIn{\n% $\\ct_\\vx$ -- a ciphertext encrypting $\\vx \\in \\F^\\ell_\\fieldcard$,\n% $\\ct_\\vy$ -- a ciphertext encrypting $\\vy \\in \\F^\\ell_\\fieldcard$.\n% }\n% \\KwOut{$\\ct$ -- a ciphertext containing the output of $\\EQ_{\\F_\\fieldcard, e}(\\vx_{i+1},\\vy_{i+1})$ in the $i$th SIMD slot}\n% $\\ct_1 \\leftarrow \\Shift(\\ct_\\vx, 1)$ // removes the value $x_0$ and shifts $\\vx$ to the left\\\\\n% $\\ct_2 \\leftarrow \\Shift(\\ct_\\vy, 1)$ // removes the value $y_0$ and shifts $\\vy$ to the left\\\\\n% $\\ct \\leftarrow \\Sub(\\ct_1, \\ct_2)$ // $(x_i - y_i), i \\in [1,\\ell-1]$\\\\\n% $r_0,r_1,\\dots,r_{\\ell-1} \\leftarrow \\udist(\\F_{\\fieldcard^e})$ //\\todo{define uniform distribution}\\\\ \n% $\\pt_r \\leftarrow \\pt(r_0,r_1,\\dots,r_{\\ell-1})$\\\\\n% $\\ct \\leftarrow \\ct \\cdot \\pt_r$ // $r_i(x_i - y_i), i \\in [1,\\ell-1]$\\\\\n% $k \\leftarrow 1$// $\\sum_{j=i}^{\\ell-1} r_i(x_i - y_i), i \\in [1,\\ell-1]$\\\\\n% \\While{$k < \\ell-1$}{\n% $\\ct_{tmp} \\leftarrow \\Shift(\\ct, k)$\\\\\n% $\\ct \\leftarrow \\ct + \\ct_{tmp}$\\\\\n% $k \\leftarrow 2k$\\\\\n% }\n% $\\ct \\leftarrow \\Power(\\ct, \\fieldcard^e-1)$ //$\\princhar_{\\fieldcard^e}\\left(\\sum_{j=i}^{\\ell-1} r_i(x_i - y_i)\\right), i \\in [1,\\ell-1]$ \\\\\n% $\\ct \\leftarrow 1 - \\ct$\\\\\n% \\textbf{Return} $\\ct$.\n% \\caption{Homomorphic circuit computing $\\EQ_{\\F_\\fieldcard, e}(\\vx_{i+1},\\vy_{i+1})$ in parallel.}\\label{alg:vector_equalities_circuit}\n% \\end{algorithm}\n \n% Let us go back to the lexicographic order in equation~(\\ref{eq:general_lex_order}).\n% Assume that we have a ciphertext $\\ct_{\\LT}$ containing $\\LT_{\\F_\\fieldcard}(x_i, y_i)$ in the $i$th slot (\\todo{elaborate on that}) and the output $\\ct_{\\EQ}$ of Algorithm~\\ref{alg:vector_equalities_circuit}.\n% We set the $\\ell-1$ slot of $\\ct_{\\EQ}$ to $1$ and multiply it by $\\ct_{\\LT}$.\n% The resulting ciphertext contains all the products $\\LT_{\\F_\\fieldcard}(x_{i}, y_{i}) \\EQ_{\\F_\\fieldcard}(\\vx_{i+1}, \\vy_{i+1})$ for any $i \\in [0,\\ell-2]$ and $\\LT_{\\F_\\fieldcard}(x_{\\ell-1}, y_{\\ell-1})$, which can be summed in the same way as in the while circuit of Algorithm~\\ref{alg:vector_equalities_circuit}.\n% As a result, we obtain a ciphertext with the output of $\\LT_{\\F_\\fieldcard}(\\vx, \\vy)$ in the first SIMD slot.\n\n% \\subsubsection{Comparing large integers.}\n% Since any integer $z$ can be represented in base $\\fieldcard$ as $z = \\sum_{i=0}^{\\ell-1} z_i q^i$, we can encode $z$ into SIMD slots as a vector $(z_0,z_1,\\dots,z_{\\ell-1}) \\in \\F_{\\fieldcard}^\\ell$.\n% As shown above, we can compare such vectors using the lexicographic order function in~(\\ref{eq:general_lex_order}) and thus compare integers larger than $\\fieldcard$.\n\n% \\todo{Incorporate the following into the complexity analysis.} \n% We can obtain ciphertexts encrypting $(x,x,\\ldots, x)$ and $(y,y,\\ldots, y)$ with $2\\log_2(q-1)\\texttt{Rot}$ and $2\\log_2(q-1)\\texttt{Add}$. \n\n% From there we can obtain encryptions of $(x,x-1,\\ldots, x-q-2)$ and $(y-1,y-2,\\ldots, y-q-1)$ with $2\\texttt{Add}$. We can apply $f$ to these vectors in parallel with a depth of $\\log (p-1) + \\log d$ for a cost of $2(\\log (p-1) + wt(p-1) + d - 2) \\texttt{Mult}$. This can be minimized by choosing $p = 2^d + 1$ for some $d$.\n\n% With $2$ more $\\texttt{Sub}$ we obtain encryptions of $(1-f(x), 1-f(x-1), \\ldots, 1-f(x-q-2))$ and $(1-f(y-1), 1-f(y-2), \\ldots, 1-f(y-q-1))$. From there we can get an encryption of the different partial sums in $\\log_2(q-1)\\texttt{Rot}$, $\\log_2(q-1)\\texttt{Select}$ and $\\log_2(q-1)\\texttt{Add}$. The algorithm works as follow:\n\n% \\begin{center}\n% \\begin{tikzpicture}[scale=0.8]\n% \\newcommand\\y{-4};\n \n% \\draw (0,2) -- (16,2);\n% \\draw (0,0) -- (16,0); \n% \\foreach \\x in {0,...,8}{\n% \\draw (2*\\x,0) -- (2*\\x,2);\n% }\n% \\foreach \\x in {0,...,7}{\n% \\node at (2*\\x+1,1) {$x_\\x$};\n% }\n\n% \\draw (0,2+\\y) -- (16,2+\\y);\n% \\draw (0,\\y) -- (16,\\y); \n% \\foreach \\x in {0,...,8}{\n% \\draw (2*\\x,0+\\y) -- (2*\\x,2+\\y);\n% }\n\n% \\node at (1,1+\\y) {$x_0 + x_1$};\n% \\node at (3,1+\\y) {$x_1 + x_2$};\n% \\node at (5,1+\\y) {$x_2 + x_3$};\n% \\node at (7,1+\\y) {$x_3 + x_4$};\n% \\node at (9,1+\\y) {$x_4 + x_5$};\n% \\node at (11,1+\\y) {$x_5 + x_6$};\n% \\node at (13,1+\\y) {$x_6 + x_7$};\n% \\node at (15,1+\\y) {$x_7$};\n\n% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n \n% \\draw (0,2+2*\\y) -- (16,2+2*\\y);\n% \\draw (0,0+2*\\y) -- (16,0+2*\\y); \n% \\foreach \\x in {0,...,8}{\n% \\draw (2*\\x,0+2*\\y) -- (2*\\x,2+2*\\y);\n% }\n\n% \\node at (1,1+2*\\y) {$\\begin{array}{c}\n% x_0 + x_1 \\\\\n% + x_2 + x_3\n% \\end{array}$};\n% \\node at (3,1+2*\\y) {$\\begin{array}{c}\n% x_1 + x_2 \\\\\n% + x_3 + x_4\n% \\end{array}$};\n% \\node at (5,1+2*\\y) {$\\begin{array}{c}\n% x_2 + x_3 \\\\\n% + x_4 + x_5\n% \\end{array}$};\n% \\node at (7,1+2*\\y) {$\\begin{array}{c}\n% x_3 + x_4 \\\\\n% + x_5 + x_6\n% \\end{array}$};\n% \\node at (9,1+2*\\y) {$\\begin{array}{c}\n% x_4 + x_5 \\\\\n% +x_6 + x_7\n% \\end{array}$};\n% \\node at (11,1+2*\\y) {$\\begin{array}{c}\n% x_5 + x_6 \\\\\n% + x_7\n% \\end{array}$};\n% \\node at (13,1+2*\\y) {$x_6 + x_7$};\n% \\node at (15,1+2*\\y) {$x_7$};\n\n% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n \n% \\draw (0,2+3*\\y) -- (16,2+3*\\y);\n% \\draw (0,0+3*\\y) -- (16,0+3*\\y); \n% \\foreach \\x in {0,...,8}{\n% \\draw (2*\\x,0+3*\\y) -- (2*\\x,2+3*\\y);\n% }\n\n% \\node at (1,1+3*\\y) {$\\begin{array}{c}\n% x_0 + x_1 \\\\\n% + x_2 + x_3 \\\\\n% + x_4 + x_5 \\\\\n% + x_6 + x_7\n% \\end{array}$};\n% \\node at (3,1+3*\\y) {$\\begin{array}{c}\n% x_1 + x_2 \\\\\n% + x_3 + x_4 \\\\\n% + x_5 + x_6 \\\\\n% + x_7 \\\\\n% \\end{array}$};\n% \\node at (5,1+3*\\y) {$\\begin{array}{c}\n% x_2 + x_3 \\\\\n% + x_4 + x_5 \\\\\n% + x_6 + x_7\n% \\end{array}$};\n% \\node at (7,1+3*\\y) {$\\begin{array}{c}\n% x_3 + x_4 \\\\\n% + x_5 + x_6 \\\\\n% +x_7\n% \\end{array}$};\n% \\node at (9,1+3*\\y) {$\\begin{array}{c}\n% x_4 + x_5 \\\\\n% +x_6 + x_7\n% \\end{array}$};\n% \\node at (11,1+3*\\y) {$\\begin{array}{c}\n% x_5 + x_6 \\\\\n% + x_7\n% \\end{array}$};\n% \\node at (13,1+3*\\y) {$x_6 + x_7$};\n% \\node at (15,1+3*\\y) {$x_7$};\n\n% \\foreach \\x in {2,...,8}\n% \\draw[->,thick = 2pt] (2*\\x-0.1-1,2) to[bend right] (2*\\x-3+0.1,2);\n\n% \\foreach \\x in {3,...,8}\n% \\draw[->,thick = 2pt] (2*\\x-0.1-1,2+\\y) to[bend right] (2*\\x-5+0.1,2+\\y);\n\n% \\foreach \\x in {5,...,8}\n% \\draw[->,thick = 2pt] (2*\\x-0.1-1,2+2*\\y) to[bend right] (2*\\x-9+0.1,2+2*\\y);\n\n% \\end{tikzpicture}\n% \\end{center}\n\n% From there we just have to compute the scalar product of the two vectors with $1\\texttt{Mult} + \\log_2(q-1)\\texttt{Rot} + \\log_2(q-1)\\texttt{Add}$. \\newline\n\n% So overall we can compute $<$ over $\\mathbb{F}_q$ for $q = p^d$ in:\n\n% \\begin{itemize}\n% \\item $4\\log_2 (q-1) \\texttt{Rot}$\n% \\item $(4\\log_2 (q-1) + 4) \\texttt{Add}$\n% \\item $(2(\\log (p-1) + wt(p-1) + d - 2) + 1) \\texttt{Mult}$\n% \\item $(\\log_2 (q-1) + 1) \\texttt{Select}$\n% \\end{itemize}\n\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"main_pets\"\n%%% End:\n", "meta": {"hexsha": "77738a01193a1051d7997add7856fb6d35b61c3a", "size": 36900, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/comparison-circuit.tex", "max_stars_repo_name": "iliailia/comparison-circuit-over-fq", "max_stars_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2021-03-24T07:58:48.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-18T00:41:08.000Z", "max_issues_repo_path": "paper/comparison-circuit.tex", "max_issues_repo_name": "iliailia/comparison-circuit-over-fq", "max_issues_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2021-03-24T03:03:11.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-01T09:23:59.000Z", "max_forks_repo_path": "paper/comparison-circuit.tex", "max_forks_repo_name": "iliailia/comparison-circuit-over-fq", "max_forks_repo_head_hexsha": "bc48a9101278997f0847b6ace59c8f3b83884dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-19T16:28:37.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-19T16:28:37.000Z", "avg_line_length": 63.0769230769, "max_line_length": 728, "alphanum_fraction": 0.5776151762, "num_tokens": 14408, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.810478926981208, "lm_q2_score": 0.79053032607222, "lm_q1q2_score": 0.6407081704211174}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{setspace}\n\n\\usepackage{amsmath, amsfonts, amssymb, graphicx, color, fancyhdr, mathrsfs, tikz-cd, mdframed, enumitem, framed, adjustbox, bbm, upgreek, xcolor, hyperref}\n\\usepackage[framed,thmmarks]{ntheorem}\n\\usepackage[style=alphabetic]{biblatex}\n%Set the bibliography file\n\\bibliography{sources}\n\n%Replacement for the old geometry package\n\\usepackage{fullpage}\n\n%Input my definitions\n\\input{./mydefs.tex}\n\n%Shade definitions\n\\theoremindent0cm\n\\theoremheaderfont{\\normalfont\\bfseries} \n\\def\\theoremframecommand{\\colorbox[rgb]{0.9,1,.8}}\n\\newshadedtheorem{defn}[thm]{Definition}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%% Customize Below %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n%header stuff\n\\setlength{\\headsep}{24pt} % space between header and text\n\\pagestyle{fancy} % set pagestyle for document\n\\lhead{Notes on the Grassmannian} % put text in header (left side)\n\\rhead{Nico Courts} % put text in header (right side)\n\\cfoot{\\itshape p. \\thepage}\n\\setlength{\\headheight}{15pt}\n\\allowdisplaybreaks\n\n% Document-Specific Macros\n\\DeclareMathOperator{\\Gr}{Gr}\n\\DeclareMathOperator{\\Flag}{Flag}\n\\DeclareMathOperator{\\GL}{GL}\n\\newcommand{\\Alg}{\\mathbf{Alg}}\n\\newcommand{\\CRing}{\\mathbf{CRing}}\n\\DeclareMathOperator{\\inv}{inv}\n\\newcommand{\\detq}{\\operatorname{det}_q}\n\\let\\k\\relax\n\\newcommand{\\k}{\\mathbbm{k}}\n\n\\begin{document}\n%make the title page\n\\title{Notes on the Grassmannian \\vspace{-1ex}}\n\\author{Nico Courts}\n\\date{Summer 2019}\n\\maketitle\n\n\\begin{abstract}\n\tThese notes are my summary of the realization of the Grassmanian $\\Gr(n,k)$ as a quotient of \n\tthe Lie group $\\GL(n,\\k)$. In particular the focus will be on $\\k=\\bbR$ or $\\bbC$. I also talk about quantum \n\tdeformations of $\\GL(n,\\k)$ and $\\Gr(n,k)$.\n\\end{abstract}\n\n\\section*{Vista: Where We're Headed}\nThe idea that $\\GL_n$ acts on vector subspaces of a Euclidean space should be unsurprising and natural, but \nthe upshot to considering this viewpoint is that is allows us to consider this object from the perspective of smooth \nmanifold theory. This gives us some great machinery to grasp onto to prove some nice properties \nabout $\\Gr(n,k)$.\n\nEventually I will lead us slightly astray by investigating a ``quantum analog'' of $\\Gr(n,k)$. This $q$-deformation\nis related closely to the kinds of structures I have been thinking about recently: specifically quantum groups. One can think of \nthese as ``deformations'' of groups in some controlled way.\n\n\\section{The Grassmannian as a Lie group quotient}\nIn the following, let $\\k=\\bbR$ or $\\bbC$ and let $V=\\k^n$. For a great introduction to this theory see \\cite{LeeISM},\n(specifically chapters 1, 7, and 21), which serves as the primary source for this section.\n\\subsection{Smooth Manifolds}\nRecall that a smooth manifold is a topological manifold equipped with a \\textbf{smooth structure}: that is, an atlas comprised \nof smoothly-compatable charts (ones where the transition functions are smooth as maps of Euclidean spaces).\n\\subsection{The general linear Lie group} \nLet $G=\\GL(V)$, initially considered just as a group. In fact, one can apply to $G$ a smooth structure very naturally\nby recognizing $\\GL(V)\\subseteq M_{n\\times n}(\\k)$, the latter of which is identified with the Euclidean space $\\k^{n^2}$,\nwhose smooth structure is the obvious one (the identity map).\n\nThat $G$ itself is a smooth manifold comes from the fact that $\\GL(V)=\\det^{-1}(\\k\\setminus\\{0\\})$, which is \nitself an open set as the inverse image of an open set under a continuous map. So then $G$ is a (codimension zero) open (smooth) submanifold\nof $M_{n\\times n}(k)$ whose smooth structure is inherited from the ambient space.\n\nIn fact, the algebraic and geometric structure of $G$ come together in a very nice way to define what is called \na \\textbf{Lie group:}\n\\begin{defn}\n\tLet $G$ be a smooth manifold. Then a \\textbf{Lie group} is $G$ along with a group structure given by maps $m:G\\times G\\to G$ and $i:G\\to G$\n\t(representing multiplication and inverse maps) such that $m$ and $i$ are smooth maps (smooth in charts).\n\\end{defn}\n\n\\begin{prop}\n\t$G$ is a Lie group.\n\\end{prop}\n\\begin{prf}\n\tSince we have already defined the smooth and group structures on $G$, it suffices to show that \n\t$m$ and $i$ (matrix multiplication and the inverse map) are smooth maps.\n\n\tMultiplication is smooth since if we consider the $i^{th}$ component function of $m$, where $i=an+b$ for $1\\le a,b\\le n$, we get (for $\\mathbf x$ and $\\mathbf y$ in $\\k^{n^2}$)\n\t\\[m_i(\\mathbf x,\\mathbf y)=\\sum_{k=1}^{n}\\mathbf x_{an+k}\\mathbf y_{kn+b}\\]\n\twhich is a polynomial in the $x_i$ and $y_i$ and is thus (clearly) smooth.\n\n\tThat the inverse map is also smooth follows via similar reasoning: since $\\det$ is smooth (again a polynomial in the entries),\n\tcomputing the adjugate matrix is a smooth operation and so via Cramer's rule (since $\\det A\\ne 0$)\n\t\\[i(A)=\\frac{1}{\\det A}\\operatorname{adj}A\\]\n\tis also a smooth operation.\n\\end{prf}\n\n\\subsection{The action on \\texorpdfstring{$\\Gr(n,k)$}{Gr(n,k)}}\nThe first thing to notice is that there is a natural action \nof $G$ on the set of $k$-dimensional subspaces of $V$ in the following way: let $A\\in\\GL(V)$ and let $W\\le V$ be a $k$-dimensional \nsubspace. Then $A(W)$ is also $k$-dimensional since $A$ is invertible\\footnote{If you are not convinced, remember that $A$ being invertible means all minors are nonsingular,\nso by extending a basis for $W$ to $V$ and looking at $A$ in this basis, the top-left $k\\times k$ matrix defines an isomorphism \nof $W$ onto some other subspace of $V$.}.\n\\begin{prop}\n\tThe action defined above gives a group action of $\\GL(V)$ on $\\Gr(n,k)$.\n\\end{prop}\n\\begin{prf}\n\tThis is clear since the identity matrix $I_n$ fixes subspaces and since matrix multiplication is associative.\n\\end{prf}\n\n\\subsection{Defining a smooth structure on the Grassmanian}\nAs we are working in the context of smooth manifolds, we need to make some sense of how this action fits into the theory of smooth group actions.\nNormally,\n\\begin{defn}\n\tIf $G$ is a Lie group and $M$ is a smooth manifold, we say $G$ acts smoothly on $M$ if $G$ acts on $M$ as a set and the map \n\t\\[\\rho:G\\times M\\to M\\]\n\tgiven by \n\t\\[\\rho(g,m)=g.m\\]\n\tis a smooth map.\n\\end{defn}\nIn this case, we don't have a smooth structure (yet!) on $\\Gr(n,k)$, so we will define one:\n\n\\subsubsection{Stabilizers}\nOur first step is to show that the stabilizer of an arbitrary $W\\in\\Gr(n,k)$, denoted $G_W$, is a closed Lie subgroup (that is a Lie group and closed smooth submanifold). \nTo this end:\n\\begin{thm}[Closed Subgroup Theorem]\\label{thm-closed-subgroup}\n\tSuppose $G$ is a Lie group and $H\\subseteq G$ is a subgroup that is also a closed subset of $G$. Then $H$ is an embedded \n\tLie subgroup.\n\\end{thm}\n\\begin{rmk}\n\tThe proof of this theorem is two pages of Lie theory and is a little involved for this talk. Jack does \n\ta great job (as always) in his proof (cf. \\cite[Thm. 20.12]{LeeISM})\n\\end{rmk}\n\n\\noindent Using this theorem we start by proving something about a particular subset:\n\\begin{prop}\\label{prop-GW}\n\tFix some basis $\\calB$ for $V$. Let $W\\le V$ be the subspace of $\\k^n$ by setting the last $(n-k)$ coordinates in this basis to zero (which is isomorphic to $\\k^k$). Then \n\tthe stabilizer $G_W$ of $W$ is \n\t\\[G_W=\\left\\{\\begin{pmatrix}\n\t\tA & B\\\\ 0 & D\n\t\\end{pmatrix}:A\\in\\GL_k(\\k),\\ D\\in\\GL_{n-k}(\\k),\\ B\\in M_{k\\times n-k}(\\k)\\right\\}\\]\n\tand this set is a closed Lie subgroup of $\\GL(V)$.\n\\end{prop}\n\\begin{prf}\n\tWe begin by noticing that $G_W$ as above is precisely the stabilizer of $W$. But because we picked \n\t$W$ in the way we did, we know immediately that $\\calA\\in\\GL(V)$ fixes $W$ if the $i^{th}$ coordinate\n\tof $\\calA(\\mathbf x)$ is zero for all $\\mathbf x=(x_1,\\dots,x_k,0,\\dots,0)$. Since that $x_i$ are arbitrary,\n\twe get that $\\calA$ is block upper triangular as above. That $A$ and $D$ are invertible comes from the fact that \n\t\\[\\det\\calA=\\det A\\det D\\ne 0.\\]\n\tTherefore the stabilizer of $W$ is contained in $G_W$. The reverse inclusion is clear.\n\n\tThat $G_W$ is a subgroup of $\\GL(V)$ then follows from group theory. In light of thm.~\\ref{thm-closed-subgroup},\n\tproving that $G_W$ is a Lie subgroup amounts to showing that $G_W$ forms a closed subset of $\\GL(V)$. But this is clear because \n\tthis is inverse image of all of $\\GL_k(\\k)$ under $F:\\GL(V)\\to\\GL_k(\\k)$ that maps a matrix to the $k\\times k$ minor in the upper left corner.\n\n\tEquivalently, $G_W$ is defined by the closed condition $\\calA_{ij}=0$ if $i\\ge k_1$ and $j\\le k$. That every such matrix has the form of $G_W$ is \n\tclear as argued above.\n\n\tThus $G_W$ is closed, so is a closed Lie subgroup of $G$.\n\\end{prf}\n\nSo we have shown that a particular stabilizer $G_{\\k^k}$ is a closed Lie subgroup, but in fact this holds for \\textit{any $W\\in\\Gr(n,k)$:}\n\\begin{cor}\n\tFor any $W\\in\\Gr(n,k)$, $G_W$ is a closed Lie subgroup of $G$.\n\\end{cor}\n\\begin{prf}\n\tSimply choose a new basis for $V$ such that all vectors in $W$ are of the form \n\t\\[(v_1,\\dots,v_k,0,\\dots,0)\\]\n\tand apply prop.~\\ref{prop-GW}.\n\\end{prf}\n\n\\subsubsection{The Quotient Manifold Theorem}\nRecall from point-set topology that a proper map is one that pulls back compact sets to compact sets.\n\\begin{defn}\n\tAn action of a Lie group $G$ on a smooth manifold $M$ is \\textbf{proper} if the map \n\t\\[G\\times M\\to M\\times M\\qquad \\text{via}\\qquad (g,p)\\mapsto(g.p,p)\\]\n\tis a proper map.\n\\end{defn}\n\\begin{rmk}\n\tJack (\\cite[Ex. 21.3]{LeeISM}) says that this is a \\textit{strictly weaker} condition than saying\n\t$(g,p)\\mapsto g.p$ is proper.\n\\end{rmk}\n\\begin{thm}[Quotient Manifold Theorem]\\label{thm-QMT}\n\tLet $G$ be a Lie group acting \\textit{smoothly, freely, and properly} on a smooth manifold $M$. Then \n\tthe orbit space $M/G$ is a topological manifold of dimension $\\dim M-\\dim G$ and has a unique smooth structure \n\tmaking the quotient map $\\pi:M\\to M/G$ a smooth submersion.\n\\end{thm}\n\\begin{prf}\n\tAs before, see Jack's proof \\cite[thm. 21.10]{LeeISM}.\n\\end{prf}\n\\begin{rmk}\n\tIn case this wasn't obvious, this is a rather high-powered result and is the primary facilitator of what follows.\n\\end{rmk}\n\n\\subsubsection{Homogeneous Spaces}\n\\begin{defn}\n\tA smooth manifold $M$ endowed with a \\textit{transitive, smooth} action by a Lie group $G$ is called\n\ta \\textbf{homogeneous space.}\n\\end{defn}\nThen we use the quotient manifold theorem to construct a homogeneous space for ourselves (following part of the proof of \\cite[thm. 21.17]{LeeISM}):\n\\begin{prop}\n\tLet $W$ be an arbitrary element in $\\Gr(n,k)$. Then if $G=\\GL(V)$, $G/G_W$ is a homogeneous space given by the action\n\t\\[g\\cdot (g'G_W)=(gg')G_W\\]\n\\end{prop}\n\\begin{prf}\n\tConsider the action of $G_W$ on $G$ by left multiplication. This action is free since if $hg=g$, we get $h=e$. The action is a restriction of \n\t$G$ on itself by left multiplication, which is smooth by virtue of being a Lie group, so the action is smooth.\n\n\tTo see the action is proper, we use the fact (proved in \\cite[prop. 21.5(a,b)]{LeeISM}) that the action is proper if and only if $(h_i)$ is a sequence in $G_W$ and \n\t$(g_i)$ are a sequence in $G$, if both $(h_i)$ and $(h_ig_i)$ converge then the $(g_i)$ converge.\n\n\tLet $(g_i)$ be a convergent sequence in $G$ and $(h_i)$ a sequence in $G_W$ such that $(h_ig_i)$ converges. Then by the continuity of \n\tthe $G$ action on itself, $h_i=(h_ig_i)g_i^{-1}$ converges to a point in $G$. Since $G_W$ is closed in $G$, this limit converges in $G_W$.\n\n\tThus we can apply thm~\\ref{thm-QMT} to say that $G/G_W$ has a unique smooth manifold structure making the quotient map $\\pi:G\\to G/G_W$ a smooth submersion. The action \n\t\\[g\\cdot g'G_W=(gg')G_W\\]\n\tis transitive and smooth, giving us our homogeneous space.\n\\end{prf}\n\\subsubsection{Bringing it home}\nNow that we have that $G/G_W$ is a smooth manifold, consider the map\n\\[F:G/G_W\\to \\Gr(n,k)\\]\nbe defined by \n\\[F(A\\cdot G_W)=A(W).\\]\n\n\\begin{lem}\n\t$F$ defined as above is a $G$-equivariant bijection.\n\\end{lem}\n\\begin{prf}\n\tThis is a well-defined map since if $gG_W=g'G_W$, we get that $g=g'h$ for some $h\\in G_W$, thus\n\t\\[F(gG_W)=g\\cdot W=g'hW=g'W=F(g'G_W).\\]\n\n\tThen that this map is surjective follows since the action of $G$ on $\\Gr(n,k)$ has a single orbit. That \n\tthe map is injective follows since if the images of the same, this means precisely that $gW=g'W$, so $g^{-1}g'\\in G_W$,\n\tproving $gG_W=g'G_W$.\n\n\t$G$-equivariance is a simple consequence of the fact \n\t\\[g\\cdot F(g'G_W)=g\\cdot (g'W)=(gg')W=F(gg'G_W)=F(g(g'G_W)).\\]\n\\end{prf}\n\n\\noindent Now we can give $\\Gr(n,k)$ a smooth structure such that $F$ is a diffeomorphism\\footnote{Open sets are precisely the images of open sets of $G/G_V$ and the smooth \nstructure is the one pulled back through $F$.}. Notice now that the action of $G$ on $\\Gr(n,k)$ is smooth since we can write \n\\[g\\cdot V=F(g\\cdot F^{-1}(V))\\]\nsince $F$ and the action of $G$ on $G/G_W$ is smooth.\n\nWe still need to establish that such a smooth structure is, in some sense, \\textit{the} structure that makes \nthe action provided into a smooth action. This will be the smooth structure we give the Grassmannian.\n\nWe need one more theorem from smooth manifolds to finish this off:\n\\begin{thm}[Equivariant Rank Theorem]\n\tLet $M$ and $N$ be smooth manifolds and $G$ be a Lie group. If $F:M\\to N$ is a smooth, bijective, $G$-equivariant map with respect to \n\ta smooth, transitive action of $G$ on $M$ and any smooth action of $G$ on $N$, then $F$ is a diffeomorphism.\n\\end{thm}\n\\begin{prf}\n\tSee \\cite[thm 7.25]{LeeISM}.\n\\end{prf}\n\\begin{lem}\n\tThere is only one smooth structure on $X=\\Gr(n,k)$ such that the action of $G$ on $X$ is smooth.\n\\end{lem}\n\\begin{prf}\n\tLet $\\tilde X$ be $X$ with any other smooth structure. Then if $G$ acts smoothly on $\\tilde X$, the (smooth) action $\\theta^{(W)}:G\\to \\tilde X$ given by $\\theta^{(W)}(g)=g\\cdot W$ \n\tdescends to the quotient to give us a smooth map $\\calF:G/G_W\\to \\tilde X$, which as before can be shown to be a $G$-equivariant bijection.\n\n\tBut then by the equivariant rank theorem since the action of $G$ on $G/G_W$ is smooth and transitive, and the action of $G$ on $\\tilde X$ is smooth,\n\tthe map $\\calF$ is a diffeomorphism. But then in particular since $F$ gave us a diffeomorphism from $G/G_W$ to $X$, \n\twe have a diffeomorphism $X\\cong \\tilde X$, so the smooth structure on $X$ is unique.\n\\end{prf}\n\\subsection{Denouement: \\texorpdfstring{$\\Gr(n,k)$}{Gr(n,k)} is compact.}\nI found a neat proof of this online \\href{https://math.stackexchange.com/questions/148088/compactness-of-the-grassmannian}{here}.\n\\begin{prop}\n\t$\\Gr(n,k)$ is compact.\n\\end{prop}\n\\begin{prf}\n\tLet $S=\\{v\\in V=\\k^n: \\|v\\|=1\\}$. Notice that $S\\cong \\bbS^{n-1}$, the $n-1$ sphere in $\\k^n$, which is compact. Then $S^d=S\\times \\cdots\\times S$ is compact as well.\n\n\tConsidering $S^d$ as some choice of $d$ unit vectors in $\\k^n$, we can consider the map \n\t\\[S^d\\to\\k^{\\binom{d}{2}}\\qquad\\text{via}\\qquad (v_1\\dots,v_d)\\mapsto (v_1\\cdot v_2,v_1\\cdot v_3,\\dots,v_1\\cdot v_n,v_2\\cdot v_3,\\dots,v_{n-1}\\cdot v_n)\\]\n\tand notice that pulling back $\\mathbf 0$ in this map gets us a (closed whence) compact subspace $C$ of $S^d$ corresponding to the \n\tsets of $d$ orthonormal vectors in $\\k^n$.\n\n\tFinally define the equivalence relation on $C$ where $c_1\\sim c_2$ if $\\operatorname{span} c_1=\\operatorname{span} c_2$. Then the quotient \n\t$C/\\sim$ is precisely $\\Gr(n,k)$ which, as a quotient of a compact space, is compact.\n\\end{prf}\n\n\\section{Aside: Quantized Grassmanians}\nIn this section I will take a little time to develop some of the things I have been learning about quickly,\nspecifically developing non-commutative algebraic geometry by relying on intuitions from traditional algebraic geometry \nwhile ``deforming'' things in a controlled way so that we can use our intuition to develop analogous ideas.\n\n\\subsection{Algebra vs. Geometry}\nThe discussion that follows is a relatively constructive method for $q$-deforming the Grassmannian and studying the action it \ninherits from $\\GL_q(n)$. It gives a good flavor for the field, but is very strongly rooted in algebra.\n\nThat is, the methods we describe take the coordinate ring of the Grassmannian and deform it purely in the realm of algebra, with the understanding \nthat there is some parallel geometric structure that is being analogously twisted.\n\nLate in my research, I found a paper by Aaron Lauve \\cite{lauve} takes the results of Taft and Towber and reimagines them in a context that \npreserves a certain level of geometry. I haven't had time to read it in its entirety but it may be a good place to look for those interested in the geometry.\n\n\\subsection{Hopf Algebras}\nThe leading idea here is that Hopf algebras lie at the heart of a lot of modern algebra---especially representation theory. Sometimes you are working with \na Hopf algebra without knowing it!\n\nLet's begin with a definition:\n\\begin{defn}\n\tA \\textbf{Hopf algebra} is a $k$-vector space $A$ equipped with maps \n\t\\begin{gather*}\n\t\t\\mu:A\\otimes A\\to A\\\\\n\t\t\\eta:k\\to A\\\\\n\t\t\\Delta:A\\to A\\otimes A\\\\\n\t\t\\varepsilon:A\\to k\\\\\n\t\tS:A\\to A\n\t\\end{gather*}\n\tsuch that $\\mu$ and $\\eta$ form an associative algebra structure on $A$, $\\Delta$ and $\\varepsilon$ form a coassociative coalgebra structure on $A$,\n\t$\\mu$ is a coalgebra morphism, $\\Delta$ is an algebra morphism (so far we have defined a bialgebra) and furthermore $S:A\\to A$ is an antihomomorphism:\n\t\\[S(ab)=S(b)S(a)\\]\n\tand such that all the maps satisfy the following:\n\t\\[\\mu\\circ S\\otimes\\id\\circ \\Delta(a)=S(a_1)a_2=a_2S(a_2)=\\mu\\circ \\id\\otimes S\\circ\\Delta(x)=\\eta\\circ\\varepsilon(x)\\]\n\\end{defn}\n\\begin{rmk}\n\tWe used Sweedler notation above to simplify things although it looks deceptively simple to the uninitiated.\n\\end{rmk}\n\\begin{rmk}\n\tIn shorter terms, \n\\end{rmk}\n\n\\subsubsection{Examples of Hopf algebras}\nThese algebras pop up relatively frequently! For instance, the tensor, symmetric, and exterior algebras on a vector space are all \nreadily made into Hopf algebras.\n\nI am a representation theorist, so my primary examples tend to revolve around \nthe group algebra $k[G]$ of a group. It is a Hopf algebra where you define the Hopf algebra structure by making all \nelements of $G$ grouplike\\footnote{$\\Delta(x)=x\\otimes x$, $\\varepsilon(x)=1$ and $S(x)=x^{-1}$} and extending $k$-linearly.\n\nYou can also dualize things somewhat and consider the algebra of regular functions on $G$, denoted $kG$ or $k(G)$. When $G$ is finite,\nthese are isomorphic but when $G$ is infinite these can be quite different. In either of these contexts one can interchange ``group''\nwith ``group scheme'' or ``algebraic group''. A quite nice treatment of the affine case (that is when the Hopf algebra is finite dimensional over a field)\ncan be found in \\cite{waterhouse}. If you are interested in a highly representation-theoretic treatment of algebraic groups, you can can check out \\cite{jantzen-algebraic}\nor \\cite{Milne}.\n\nFinally the last way these arise in nature is in the universal enveloping algebra $\\calU(\\frakg)$ of a Lie algebra $\\frakg$. One can find the theory here developed in Jantzen's \nother book \\cite{jantzen-quantum}.\n\n\\subsection{Quantum deformations of Hopf algebras}\nIn all of the above examples, the Hopf algebra is either commutative or cocommutative and in fact you get some nice equivalences of categories \nbetween, for instance, group schemes over a field $k$ and commutative Hopf algebras over $k$. The ``quantum perspective'' is the tendency to \nask the question: \\textit{what about the non-commutative, non-cocommutative Hopf algebras?}\n\nIn some way, the study of quantum groups is the attempt to consider the Hopf algebras that fall in the gaps between the more ``natural'' ones.\n\n\\subsection{The idea}\nThe primary source for this section is Taft and Towber's \\textit{Quantum deformation of flag schemes and Grassmann schemes} \\cite{quantum-flag}.\nOur tact here is to build up an algebra (thought of as the ring of functions on the flag or Grassmannian scheme) and through deformation of this algebra,\n$q$-deform the space.\n\n\\begin{rmk}\n\tGenerally when working with vector spaces one can more-or-less forget about which basis is chosen when working in coordinates. Unfortunately here\n\tthe commutativity (or lack thereof) of a quantum deformation is highly dependent on the choice of \\textit{ordering} of any basis. Thus we will be considering commutative rings $\\k$ and \n\t$\\k$-algebras freely generated by some fixed \\textbf{ordered} basis $\\calB=\\{v_1,\\dots,v_n\\}$. \n\n\tIn what follows I will suppress the basis $\\calB$ (e.g. write $\\Flag^n(\\calB)$ as $\\Flag^n$) but remember that it is vital to the construction.\n\\end{rmk}\n\n\\subsection{The affine flag scheme \\texorpdfstring{$\\Flag^n(\\calB)$}{Flag-n(B)}}\n\\begin{rmk}\n\tI will admit that here is the most difficult part for me: defining these objects\\footnote{As well as seeing why these objects are what their names imply they should be} requires some geometry \n\tthat is beyond my understanding at the moment. I will do my best, but the geometers among you may notice some \n\tholes in my justification. Corrections are warmly welcomed.\n\\end{rmk}\nThe idea here is to define the scheme $\\Flag^n:=\\Flag^n(\\calB):\\CRing\\to\\Set$ which assigns to any commutative ring $\\k$ the (commutative, associative, unital) $\\k$-algebra\n$\\Flag^n(\\k)$ generated over $\\k$ by the elements of $\\Lambda^\\ast(\\calB)$\n\\[v_{i_1}\\wedge\\cdots\\wedge v_{i_r}\\qquad (1\\le r\\le n,\\ \\forall j,\\ 1\\le i_j\\le n)\\]\nsubject to the relations: $v_{i_1}\\wedge\\cdots\\wedge v_{i_s}=0$ if two $i_k$ coincide and if $\\sigma\\in\\frakS_n$,\n\\[v_{i_{\\sigma 1}}\\wedge\\cdots\\wedge v_{i_{\\sigma s}}=(\\operatorname{sgn}\\sigma)\\ v_{i_1}\\wedge\\cdots\\wedge v_{i_s}\\]\ntogether with the relation \n\\[\\sum_{1\\le \\lambda_1\\le\\cdots\\le\\lambda_r\\le t+r}(-1)^{\\sum \\lambda_i}\\bigwedge_{k\\ne \\lambda_i \\forall i}v_{k}\\times_{\\Lambda^+(\\calB)}(v_{\\lambda_1}\\wedge\\cdots\\wedge v_{\\lambda_r}\\wedge v_{j_1}\\wedge\\cdots\\wedge v_{j_{s-r}})=0\\]\nfor all $1\\le r\\le s\\le t\\le n$ and all choices of $i_1,\\dots,i_{t+r}$ and $j_1,\\dots,j_{s-r}$.\n\n\\subsubsection{The grading}\nThis algebra admits a $P(n)$ grading where $P(n)$ is the semigroup (here no inverses) of nonincreasing sequences of positive integers\n\\[(\\alpha_1,\\dots,\\alpha_k)\\]\nwith $n\\ge \\alpha_1\\ge\\cdots\\ge \\alpha_1\\ge 1$ where multiplication is defined by (disjoint) union and sorting the multiset. The $P(n)$ grading determines the ``\\textbf{shape}'' of a monomial in \n$\\Flag^n(\\k)$ in the following way: we can write each monomial $\\omega\\in\\Flag^n(\\k)$ as $\\omega_1\\cdots\\omega_r$ where $\\omega_i\\in\\Lambda^{\\alpha_i}(\\calB)$\nand such that $n\\ge\\alpha_1\\ge\\cdots\\ge\\alpha_r\\ge 1$. For notation, if $\\alpha=(\\alpha_1,\\dots,\\alpha_r)$, write \n\\[\\Flag^\\alpha(\\k)=\\Lambda^\\alpha(\\calB,\\k)=\\Lambda^\\alpha(\\calB)\\]\nto be the $\\k$-span of all products of shape $\\alpha$.\n\nThis gives us a graded algebra structure (called the \\textbf{shape algebra})\n\\[\\Lambda^+(\\calB)=\\bigoplus_{\\alpha\\in P(n)}\\Lambda^\\alpha(\\calB)\\]\nof $\\Flag^n(\\k)$.\n\nThe upshot here is that each $\\Lambda^\\alpha$ inherits a natural $\\GL(n,\\k)$ action by acting on each $v_i$ independently.\n\nThen defining a deformation of this algebra amounts to deforming $\\GL(n,\\k)$.\n\n\\subsection{The affine Grassmann scheme \\texorpdfstring{$\\Gr^{t,n}(\\calB)$}{Gr(t,n)(B)}}\nWe define the \\textbf{affine Grassmann scheme} $\\Gr^{t,n}:=\\Gr^{t,n}(\\calB)$ to be a closed subscheme of $\\Flag^n$ such that \n$\\Gr^{t,n}(\\k)$ is the $\\k$-subalgebra of $\\Flag^n(\\k)$ generated by the $t$-fold wedges in $\\calB$. Notice this also inherits a grading, \nbut this time we can view it as a $\\bbZ$-grading since all monomials in $\\Gr^{t,n}(\\k)$ are of shape $(t,t,\\dots,t)$.\n\n\\subsection{Quantized \\texorpdfstring{$\\GL(n)$}{GL(n)}}\nLet $q$ be an indeterminate (often thought of as an element in the base ring [e.g an odd root of unity] and always as a \\textit{deformation parameter})\nand let $\\k_q:=\\k[q,q^{-1}]$, the Laurent polynomials in $q$. To proceed we need the \\textit{(strict) inversion statistic} seen in the combinatorics of $\\frakS_n$.\n\\begin{defn}\n\tIf $\\sigma\\in\\frakS_n$, let \n\t\\[\\inv\\sigma=\\#\\{(i,j)|1\\le i\\le j\\le n,\\ \\sigma(i)>\\sigma(j)\\}\\]\n\\end{defn}\n\nThen we can go on to define $M_q(n,\\k)$ to be the $\\k_q$-algebra generated by $n^2$ elements $X_i^j$ for $1\\le i,j\\le n$\nsubject to the $q$-commutativity relations\n\\begin{equation} \n\\begin{gathered}\n\tX_{k}^jX_i^j=qX_i^jX_{k}^j,\\qquad \\text{if } i 0 ) {\n node = node.left;\n } else {\n node = node.right;\n }\n }\n return node.bin.get(); // if null pqueue is empty\n }\n\\end{lstlisting}\n\\hfill\n\\par\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Experiment Description}\n\\par\nThree test cases were provided to exercise this code:\n\\begin{itemize}\n\\item testAdd. This test adds elements with random priority to the priority\nqueue one by one. At the end, it extracts each element and it must be the case\nthat the previously removed element had a lower priority than the current\nelement. \n\\item testParallelAdd. This test spawns 8 threads. Each of the test adds 8\nelements to the queue concurrently. At the end, it performs the same checking as\nin the previous test case. Notice that each thread is in charge of adding\nelements from a fixed range in ascending order.\n\\item testParallelBoth. Does the same as the previous test case but also removes\nelements in parallel. Each remover thread will perform the same checking. Since\nadders add elements in ascending order, it is guaranteed that any removal from\nany thread will retrieve an element with a lower priority than the next one.\n\\end{itemize}\n\\par\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Sample Results}\n\\par\nHere is the result of the execution of the test cases:\n\\par\n\\hfill\n\\begin{verbatim}\n[oraadm@gdlaa008 ch15]$ junit priority.SimpleTreeTest\n.sequential test\nOK.\n.testParallelBoth\nOK.\n.testParallelAdd\nOK.\n\nTime: 0.01\n\nOK (3 tests)\n\\end{verbatim}\n\\hfill\n\\par\nThe tests passed every time\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% --------------------------- %\n% SimpleTreeTest End\n% --------------------------- %\n", "meta": {"hexsha": "3524bc1ab205bd5b9877ff79bf3f392ea5fd7f0a", "size": 3917, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "zava/Multiprocessor/SimpleTreeTest.tex", "max_stars_repo_name": "rzavalet/multiprocessor", "max_stars_repo_head_hexsha": "f27d4dd6f44172bb6c910552e50107838d653f2f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "zava/Multiprocessor/SimpleTreeTest.tex", "max_issues_repo_name": "rzavalet/multiprocessor", "max_issues_repo_head_hexsha": "f27d4dd6f44172bb6c910552e50107838d653f2f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "zava/Multiprocessor/SimpleTreeTest.tex", "max_forks_repo_name": "rzavalet/multiprocessor", "max_forks_repo_head_hexsha": "f27d4dd6f44172bb6c910552e50107838d653f2f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.1949152542, "max_line_length": 81, "alphanum_fraction": 0.6867500638, "num_tokens": 926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8519527982093666, "lm_q1q2_score": 0.640679216503287}} {"text": "\\section{Hydrodynamic Equations in Comoving Coordinates}\n\n\\subsection{Conservative Form}\n\nWe solve the equations of gas dynamics in a coordinate system that is comoving \nwith the expanding universe, with expansion factor, $a,$ related to the redshift, $z$, by $a = 1 / (1 + z).$\nThe continuity equation is written,\n\\begin{equation}\n\\label{eq:dens}\n\\frac{\\partial \\rho_b}{\\partial t} = - \\frac{1}{a} \\nabla \\cdot (\\rho_b \\Ub) \\enskip , \\\\\n\\end{equation}\nwhere $\\rho_b$ is the comoving baryonic density, related to the proper density by $\\rho_b = a^3 \\rho_{proper},$ \nand $\\Ub$ is the proper peculiar baryonic velocity.\n\nThe momentum evolution equation can be expressed as\n\\begin{eqnarray}\n\\frac{\\partial (\\rho_b \\Ub)}{\\partial t} &=& \\frac{1}{a} \\left(\n- \\nabla \\cdot (\\rho_b \\Ub \\Ub) \n- \\nabla p \n+ \\rho_b \\gb \n+ {\\bf S}_{\\rho \\Ub}\n- \\dot{a} \\rho_b \\Ub \\right) \\enskip , \n\\end{eqnarray}\nor equivalently,\n\\begin{eqnarray}\n\\label{eq:momt}\n\\frac{\\partial (a \\rho_b \\Ub)}{\\partial t} &=& \n- \\nabla \\cdot (\\rho_b \\Ub \\Ub) \n- \\nabla p \n+ \\rho_b \\gb \n+ {\\bf S}_{\\rho \\Ub} \\enskip , \n\\end{eqnarray}\nwhere the pressure, $p$, that appears in the \nevolution equations is related to the proper pressure, $p_{proper},$ by $p = a^3 p_{proper}.$ \nHere $\\gb = - \\nabla \\phi$ is the gravitational acceleration vector, and \n${\\bf S}_{\\rho \\Ub}$ represents any external forcing terms.\n\nThe energy equation can be written,\n\\begin{eqnarray}\n\\frac{\\partial (\\rho_b E)}{\\partial t} &=& \\frac{1}{a} \\left[\n- \\nabla \\cdot (\\rho_b \\Ub E + p \\Ub)\n+ ( \\rho_b \\Ub \\cdot \\gb + S_{\\rho E} ) \n- \\dot{a} ( 3 (\\gamma - 1) \\rho_b e + \\rho_b ( \\Ub \\cdot \\Ub) ) \\right] \\enskip . \n\\end{eqnarray}\nor equivalently,\n\\begin{eqnarray}\n\\label{eq:energy}\n\\frac{\\partial (a^2 \\rho_b E)}{\\partial t} &=& a \\left[\n- \\nabla \\cdot (\\rho_b \\Ub E + p \\Ub)\n+ \\rho_b \\Ub \\cdot \\gb \n+ S_{\\rho E} \n+ \\dot{a} ( \\; ( 2 - 3 (\\gamma - 1) ) \\; \\rho_b e ) \\right] \\enskip . \n\\end{eqnarray}\nHere $E = e + \\Ub \\cdot \\Ub / 2$ is the total energy per unit mass, \nwhere $e$ is the specific internal energy. \n$S_{\\rho E} = S_{\\rho e} + \\Ub \\cdot {\\bf S}_{\\rho \\Ub}$\nwhere $S_{\\rho e} = \\Lambda^H - \\Lambda^C$ represents the heating and cooling terms, respectively.\nWe can write the evolution equation for internal energy as\n\\begin{eqnarray}\n\\frac{\\partial (\\rho_b e)}{\\partial t} &=& \\frac{1}{a} \\left[\n- \\nabla \\cdot (\\rho_b \\Ub e)\n- p \\nabla \\cdot \\Ub\n- \\dot{a} ( 3 (\\gamma - 1) \\rho_b e )\n+ S_{\\rho e} \\right] \\enskip . \n\\end{eqnarray}\nor equivalently,\n\\begin{eqnarray}\n\\frac{\\partial (a^2 \\rho_b e)}{\\partial t} &=& a \\left[\n- \\nabla \\cdot (\\rho_b \\Ub e)\n- p \\nabla \\cdot \\Ub\n+ S_{\\rho e} \n+ \\dot{a} ( \\; ( 2 - 3 (\\gamma - 1) ) \\; \\rho_b e ) \\right] \\enskip . \n\\end{eqnarray}\n\nNote that for a gamma-law gas with $\\gamma = 5/3,$ we can write\n\\begin{eqnarray}\n\\frac{\\partial (a^2 \\rho_b E)}{\\partial t} &=& a \\left[\n -\\nabla \\cdot (\\rho_b \\Ub E + p \\Ub)\n+ \\rho_b \\Ub \\cdot \\gb \n+ S_{\\rho e} \\right] \\enskip . \n\\end{eqnarray}\nand\n\\begin{eqnarray}\n\\frac{\\partial (a^2 \\rho_b e)}{\\partial t} &=& a \\left[ \n- \\nabla \\cdot (\\rho_b \\Ub e)\n- p \\nabla \\cdot \\Ub\n+ S_{\\rho e} \\right] \\enskip . \n\\end{eqnarray}\n\n\\subsection{Tracing}\n\nIn order to compute the fluxes on faces, we trace $\\rho, \\Ub, \\rho e$ and $p$ to the faces.\n\n\\noindent Thus we must convert the momentum evolution equation into a velocity evolution equation:\n\\begin{eqnarray}\n\\frac{\\partial\\Ub}{\\partial t} &=& \\frac{1}{\\rho_b} \\left(\n\\frac{\\partial (\\rho_b \\Ub)}{\\partial t} - \\Ub \\frac{\\partial \\rho_b}{\\partial t} \\right) \\\\\n&=& \\frac{1}{a \\rho_b} \\left(\n- \\nabla \\cdot (\\rho_b \\Ub \\Ub)\n- \\nabla p\n+ \\rho_b \\gb\n+ S_{\\rho \\Ub}\n- \\dot{a} \\rho_b \\Ub \\right) \n+ \\frac{1}{a} \\Ub \\; \\nabla \\cdot (\\rho_b \\Ub) \\\\\n&=& \\frac{1}{a} \\left(\n- \\Ub \\cdot \\nabla \\Ub\n- \\frac{1}{\\rho_b} \\nabla p\n+ \\gb\n+ \\frac{1}{\\rho_b} {\\bf S}_{\\rho \\Ub}\n- \\dot{a} \\Ub \\right) \\enskip .\n\\end{eqnarray}\n\n\\section{Subgrid Scale Model in Comoving Coordinates}\n\nThe fundamental modification to the standard compressible equations is the addition\nof a SGS turbulence energy variable, $K$ and associated source terms in the equations\nfor the evolution of velocity, total energy, and $K$ \\cite{schumann:1975,sagaut,schmidt:2006}.\nThe set of conservation equations in comoving coordinates~(\\ref{eq:dens})--(\\ref{eq:energy}) then becomes\n\\cite{maier:2009}:\n\\begin{align}\n\\frac{\\partial \\rho_b}{\\partial t} =& - \\frac{1}{a} \\nabla \\cdot (\\rho_b \\Ub) \\enskip , \\\\\n\\label{eq:momt_les}\n\\frac{\\partial (a \\rho_b \\Ub)}{\\partial t} =& \n- \\nabla \\cdot (\\rho_b \\Ub \\Ub) \n- \\nabla p\n+ \\nabla \\cdot \\boldsymbol{\\tau}\n+ \\rho_b \\gb \\enskip , \\\\\n\\label{eq:energy_les}\n\\frac{\\partial (a^2 \\rho_b E)}{\\partial t} =& - a \\nabla \\cdot (\\rho_b \\Ub E + p \\Ub)\n+ a \\rho_b \\Ub \\cdot \\gb \n+ a \\nabla \\cdot (\\Ub\\cdot\\boldsymbol{\\tau}) - a^2(\\Sigma - \\rho_b \\varepsilon) \\enskip \\\\\n\\nonumber\n&+ a \\dot{a} \\left( ( 2 - 3 (\\gamma - 1) ) \\rho_b e \\right) \n+ a^2 ( \\Lambda^H - \\Lambda^C ) \\enskip , \\\\\n\\label{eq:k_les}\n\\frac{\\partial (a^2\\rho_b K)}{\\partial t} =&\n- a\\nabla \\cdot \\left(\\rho_b \\Ub K\\right) \n+ a\\nabla \\cdot \\left(\\rho_b \\kappa_{\\rm sgs}\\nabla K\\right) \n+ a^2(\\Sigma - \\rho_b \\varepsilon)\\enskip .\n\\end{align}\nThe interaction between resolved and unresolved turbulent eddies is described by\nthe SGS turbulence stress tensor $\\boldsymbol{\\tau}$. Since inertial-range dynamics of \nturbulence is scale-invariant, we conjecture that $\\boldsymbol{\\tau}$ in comoving coordinates \nhas the same form as for non-expanding fluids. For compressible turbulence, \nthe following closure is proposed in \\cite{schmidt:2011}:\n\\begin{equation}\n \\label{eq:tau_nonlin}\n \\tau_{ij}= 2C_{1}\\Delta\\rho_b(2 K_{\\mathrm{sgs}})^{1/2}S_{\\! ij}^{\\ast}\n -4C_{2}\\rho_bK\\frac{U_{i,k}U_{j,k}}{|\\nabla\\Ub|^{2}}\n -\\frac{2}{3}(1-C_{2})\\rho_bK\\delta_{ij}.\n\\end{equation}\nwhere $|\\nabla\\Ub|:=(2U_{i,k}U_{i,k})^{1/2}$ is the norm of the resolved velocity derivative,\n\\begin{equation}\nS_{ij}^{\\ast} = S_{ij} - \\frac{1}{3}\\delta_{ij}d =\n\\frac{1}{2} (U_{i,j} + U_{j,i}) - \\frac{1}{3}\\delta_{ij}U_{k,k}\n\\end{equation}\nis the trace-free rate-of strain, and $\\Delta=(\\vartriangle\\!x\\,\\vartriangle\\!y\\,\\vartriangle\\!z)^{1/3}$\nis the grid scale in comoving coordinates. The production and dissipation terms in equation~(\\ref{eq:k_les}) \nare defined as follows:\n\\begin{eqnarray}\n\\Sigma &=& \\frac{1}{a}\\tau_{ij} S_{ij}, \\\\\n\\varepsilon &=& \\frac{C_\\varepsilon K^{3/2}}{a\\Delta},\n\\end{eqnarray}\nand $\\kappa_{\\rm sgs} = C_{\\kappa}\\Delta K^{1/2}$ is the SGS diffusivity.\nHere we assume that the Reynolds number of turbulence is high such that the damping of turbulent eddies by the microscopic\nviscosity of the fluid occurs entirely on the subgrid scales. Because of the numerical viscosity of PPM,\nhowever, part of the numerically resolved kinetic energy will be dissipated directly into internal energy.\n", "meta": {"hexsha": "b323be93aec8ad8109e7b8fe209bd7d5ae67278a", "size": 6935, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "UsersGuide/ComovingHydro/Equations.tex", "max_stars_repo_name": "Gosenca/axionyx_1.0", "max_stars_repo_head_hexsha": "7e2a723e00e6287717d6d81b23db32bcf6c3521a", "max_stars_repo_licenses": ["BSD-3-Clause-LBNL"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2021-02-18T09:13:17.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T21:27:46.000Z", "max_issues_repo_path": "UsersGuide/ComovingHydro/Equations.tex", "max_issues_repo_name": "Gosenca/axionyx_1.0", "max_issues_repo_head_hexsha": "7e2a723e00e6287717d6d81b23db32bcf6c3521a", "max_issues_repo_licenses": ["BSD-3-Clause-LBNL"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-10-12T08:54:31.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-12T08:54:31.000Z", "max_forks_repo_path": "UsersGuide/ComovingHydro/Equations.tex", "max_forks_repo_name": "Gosenca/axionyx_1.0", "max_forks_repo_head_hexsha": "7e2a723e00e6287717d6d81b23db32bcf6c3521a", "max_forks_repo_licenses": ["BSD-3-Clause-LBNL"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-09-04T10:26:25.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-14T23:51:51.000Z", "avg_line_length": 41.2797619048, "max_line_length": 122, "alphanum_fraction": 0.6454217736, "num_tokens": 2555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9161096204605946, "lm_q2_score": 0.6992544335934766, "lm_q1q2_score": 0.640593713764708}} {"text": "\\graphicspath{{Pics/}}\n\n\\newpage\\subsection{Projective Constructions}\n\n\\begin{minipage}{.6\\linewidth}\n \\begin{construction}[Second Intersection of Line with Conic]\n Given four points $ A, B, C, D $, no three collinear, and a point $ P\n $ on a line $ l $ passing through at most one of the four points,\n construct the point $ P'\\in l $ such that $ A, B, C, D, P, P' $ line\n on the same conic.\n\n \\solu{\n Let $ AP\\cap BC = X,\\ l\\cap CD = Y,\\ XY\\cap AD=Z $. Then by\n Pascal's Hexagrummum Mysticum Theorem, we have, $\\ \\ P' = BZ\\cap\n l $\n }\n \\end{construction}\n\\end{minipage}\\hfill%\n\\begin{minipage}{.35\\linewidth}\n \\figdf{}{constructing_second_intersection_with_conic}{}\n\\end{minipage}\n\n\\begin{minipage}{.6\\linewidth}\n \\begin{construction}[Conic touching conic]\n Given a conic $ \\mathcal{C} $, and two points $ A, B $ on it, and $ C\n $ inside of it. Construct the conic $ \\mathcal{H} $ that is tangent to\n $ \\mathcal{C} $ at $ A, B $ and passes through $ C $.\n \\end{construction}\n \\solu{\n Draw the two tangest at $ A,B $ which meet at $ X $. Take an arbitrary\n line passing through $ X $ that intersects $ AC, BC $ at $ Y, Z $.\n Take $ D=BY\\cap AZ $. Then $ D $ lies on $ \\mathcal{H} $ by Pascal.\n Construct another point $ E $ similarly and draw the conic.\n }\n\\end{minipage}\\hfill%\n\\begin{minipage}{.35\\linewidth}\n \\figdf{.9}{conic_touching_conic}{}\n\\end{minipage}\n\n\\begin{construction}[Inconic of a quadrilateral]\n Given a convex quadrilateral $ ABCD $. $ P=AC\\cap BD $, $ S\\in AD, T\\in BC\n $ such that $ S, P, T $ are collinear. Construct the conic that touches $\n AB, CD $, and also touches $ AD, BC $ at $ S, T $ respectively.\n\\end{construction}\n\n\\solu{[the\\textunderscore Construction] \n Draw the polar line $ l $ of $ P $ wrt to the quadrilateral. Let $\n Z=BC\\cap l $. Let $ ZS\\cap AB = U $, $ ZT\\cap CD = V $. Then $ SSUUTTVV $\n is our desired conic.\n}\t\n\n\\begin{minipage}{.5\\linewidth}\n \\proof{\n If $ U, V\\in CD, AB $ such that $ UV $ passes through $ P $, and if\n the conic passing through $ U, V $ and tangent to $ AD, BC $ at $ S, T\n $ intersects $ CD $ at $ U' $ again, then $ SV, U'T, DB $ are\n concurrent. So to show our construction works, we just need to prove\n that $ U, V, P $ are collinear.\\\\\n\n Since Pascal's theorem works on $ SVBTUD $, we know $ S, V, B, T, U, D $\n lie on a conic $ \\mathcal{H} $ and $ l $ is the pole of $ P $ wrt $\n \\mathcal{H} $. Now, applying Pascal's theorem on $ TDVUBS $, and quadrilateral\n theorem on $ BTUD $ and $ BVSD $, we have, $ ST\\cap UV\\in AC $, which is $ P\n $. So we are done.\n }\t\n\\end{minipage}\\hfill%\n\\begin{minipage}{.45\\linewidth}\n \\figdf{}{conic_touching_quad}{}\n\\end{minipage}\n\n\n\n\\begin{minipage}{.5\\linewidth}\n \\begin{construction}[\\href{http://www.geometry.ru/persons/beluhov/selected.pdf}{Sharygin\n Olympiad 2010}]\n A conic $ \\mathcal{C} $ passing through the vertices of $ \\triangle\n ABC $ is drawn, and three points $ A', B', C' $ on its sides $ BC, CA,\n AB $ are chosen. Then the original triangle is erased. Prove that the\n original triangle can be constructed iff $ AA', BB', CC' $ are\n concurrent.\t\t\n \\end{construction}\n\n \\solu{[the\\textunderscore Construction] \n Draw $ B'C' $. It intersects the circle at $ X_1, X_2 $. Draw the\n conic $ \\mathcal{H} $ that is tangent to $ \\mathcal{C} $ at $ X_1, X_2\n $ and passes through $ A' $. Then $ BC $ is tangent to $ \\mathcal{H} $\n at $ A' $.\n }\n\n\\end{minipage}\\hfill%\n\\begin{minipage}{.45\\linewidth}\n \\figdf{1}{reconstruct_ABC}{}\n\\end{minipage}\n\n\n\n\n\\proof{\n The only if part is easy to prove. Becase if $ AA', BB', CC' $ aren't\n concurrent, then we can get multiple triangles $ ABC $. So suppose that\n they are concurrent. \\\\\n\n Now we define some intersetion points.\n \\begin{center}\n \\begin{tabular}{ccccc}\n $W_1$\t&= &$BB'$\t&$\\cap$ &$\\mathcal{C}$\\\\\n $S$\t&=\t&$X_1X_1$\t&$\\cap$ &$AW_1$\\\\\n $T$\t&=\t&$X_1B$\t&$\\cap$ &$AX_2$\\\\\n $U$\t&=\t&$X_1X_1$\t&$\\cap$ &$BC$\\\\\n $V$\t&=\t&$X_2X_2$\t&$\\cap$ &$BC$\\\\\n $R$\t&=\t&$X_2X_2$\t&$\\cap$ &$AW_1$\\\\\n $Y_1$\t&=\t&$A'B'$\t&$\\cap$ &$SR$\n \\end{tabular}\n \\end{center}\n\n $T, S, B'$ are collinear by Pascal's theorem on $ BX_1X_1X_2AW_1 $. $ T,\n B', V $ are similarly collinear for $ AX_2X_2X_1BC $. And similarly $ R,\n B', U $ are collinear.\\\\\n\n We will prove that $ \\mathcal{H} $ is an inconic of $ SRVU $ that goes\n through $ A', X_1, X_2 $. \\\\\n\n For a point $ X $ on $ UV $, define $ f:UV\\to UV $ such that $ f(X) $ is\n the second intersection of the conic $ X_1X_1X_2X_2X $ ($ X_1X_1 = SU,\n X_2X_2=RV $) with $ UV $. \n $ f $ is an involution by \\autoref{theorem:Three Conic Law}. \\\\\n\n Suppose $A_1$ is the intersection with the inconic of $ SRUV $ through $\n X_1, X_2 $ and $ UV $. Let $ A_2 = X_1X_2\\cap UV $. Then $ f(A_1) =A_1,\n f(A_2)=A_2, f(B)=C $. \\\\\n\n Which means, $A(B, C; A_1, A_2) = -1$. Which means $ A_1=A' $. So, $\n X_1X_2A'X_2X_2 $ is an inconic of $ SRVU $, just as we wanted.\t\n\n \\figdf{.8}{reconstruct_ABC_1}{}\n}\n\n\\newpage\n\\begin{construction}[Focus and Directrix of a Parabola]\n First draw two parallel segments on the parabola, join their midpoints to\n get the line parallel to the axis. Then draw the main axis and find out\n the tip of the parabola. Then draw $f(x) = \\frac{x}{2}$ line through $P$.\n And find the foot of the intersection of it with the parabola. It is the\n focus.\n \\figdf{.5}{parabola_foci_construction}{}\n\\end{construction}\n", "meta": {"hexsha": "2f1bfb8dcdc5d6550feb922d14c95df9b1bfac68", "size": 5796, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "geo/sec8_5_projective_construction.tex", "max_stars_repo_name": "M-Ahsan-Al-Mahir/BCS_Question_Bank", "max_stars_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 48, "max_stars_repo_stars_event_min_datetime": "2020-10-14T17:15:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-13T19:47:04.000Z", "max_issues_repo_path": "geo/sec8_5_projective_construction.tex", "max_issues_repo_name": "AnglyPascal/BCS_Question_Bank", "max_issues_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "geo/sec8_5_projective_construction.tex", "max_forks_repo_name": "AnglyPascal/BCS_Question_Bank", "max_forks_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-10-15T08:59:33.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-27T15:19:26.000Z", "avg_line_length": 39.4285714286, "max_line_length": 92, "alphanum_fraction": 0.5916149068, "num_tokens": 1995, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042767, "lm_q2_score": 0.7879312056025699, "lm_q1q2_score": 0.640483532882625}} {"text": "\\section{Sparse Coding}\n\n\\subsection*{Orthogonal Basis}\nFor $\\mathbf{x}$ and o.n.b. $\\mathbf{U}$ compute $\\mathbf{z} = \\mathbf{U}^\\top \\mathbf{x} $. Approx $ \\mathbf{\\hat{x}} = \\mathbf{U\\hat{z}}$, $\\hat{z}_i = z_i$ if $ \\lvert z_i \\rvert > \\epsilon$ else 0.\nReconstruction Error $\\|\\mathbf{x}-\\mathbf{\\hat{x}}\\|^2 = \\sum_{d\\notin\\sigma}\\langle\\mathbf{x},\\mathbf{u}_d\\rangle ^2$.\nChoice of base depends on signal. Fourier for global, wavelet for local support. PCA basis optimal for given $\\Sigma$. (Haar Wavelet) include the scaling function in the family.\nChange of base complexity = $O(D^2)$ but fast transforms = $O(DlogD)$\n\\subsection*{Overcomplete Basis}\n$\\mathbf{U} \\in \\mathbb{R}^{D \\times L}$ for \\# atoms $ = L > D = \\mathsf{dim}\\text{(data)}$. Decoding involved $\\rightarrow$ add constraint $\\mathbf{z}^\\star \\in \\argmin_\\mathbf{z} \\lVert \\mathbf{z} \\rVert_0$ s.t. $\\mathbf{x} = \\mathbf{Uz}$. NP-hard $\\rightarrow$ approximate with 1-norm (convex) or with MP.\n\n\\textbf{Coherence}\n\\begin{inparaitem}[\\color{red}\\textbullet]\n\t\\item $m(\\mathbf{U}) = \\max_{i,j:\\, i \\neq j} | \\mathbf{u}_i^\\top \\mathbf{u}_j |$\n\t\\item $m(\\mathbf{B}) = 0$ if $\\mathbf{B}$ orthogonal matrix\n\t\\item $m([\\mathbf{B}, \\mathbf{u}]) \\geq \\frac{1}{\\sqrt{D}}$ if atom $\\mathbf{u}$ is added to orthogonal basis $\\mathbf{B}$ (o.n.b. = orthonormal base)\n\\end{inparaitem}\n\\textbf{Matching Pursuit (MP)}\napproximation of $\\mathbf{x}$ onto $\\mathbf{U}$, using $K$ entries.\nObjective: $\\mathbf{z}^\\star \\in \\argmin_{\\mathbf{z}} \\|\\mathbf{x} - \\mathbf{Uz} \\|_2$, s.t. $\\|\\mathbf{z}\\|_0 \\leq K$\n\\begin{inparaenum}[\\color{red}1.]\n\t\\item init: $z \\leftarrow 0, r \\leftarrow x$\n\t\\item while $\\|\\mathbf{z}\\|_0 < K$ do\n\t\\item select atom with smallest angle $i^\\star = \\argmax_i |\\langle \\mathbf{u}_i, \\mathbf{r} \\rangle|$\n\t\\item update coefficients: $z_{i^\\star} \\leftarrow z_{i^\\star} + \\langle \\mathbf{u}_{i^\\star}, \\mathbf{r} \\rangle$\n\t\\item update residual: $\\mathbf{r} \\leftarrow \\mathbf{r} - \\langle \\mathbf{u}_{i^\\star}, \\mathbf{r} \\rangle \\mathbf{u}_{i^\\star}$.\n\\end{inparaenum}\n\\\\\\textbf{Exact recovery} when: $K<1/2( 1+1/m(\\mathbf{U}))$\n\n\\textbf{Compressive Sensing}: Compress data while gathering:\n\\begin{inparaitem}[\\color{red}\\textbullet]\n \\item $\\mathbf{x} \\in \\mathbb{R}^D$, $K$-sparse in o.n.b. $\\mathbf{U}$. $\\mathbf{y} \\in \\mathbb{R}^M$ with $y_i = \\langle \\mathbf{w}_i, \\mathbf{x}\\rangle $: $M$ lin. combinations of signal; $\\mathbf{y} = \\mathbf{Wx} = \\mathbf{WUz} = \\mathbf{\\theta z}$, $\\theta \\in \\mathbb{R}^{M \\times D}$\n \\item Reconstruct $\\mathbf{x} \\in \\mathbb{R}^D$ from $\\mathbf{y}$; find $\\mathbf{z}^\\star \\in \\argmin_{\\mathbf{z}}\\|\\mathbf{z}\\|_0$, s.t. $\\mathbf{y} = \\mathbf{\\theta z}$ (e.g. with MP). Given $\\mathbf{z}$, reconstruct $\\mathbf{x}$ via $\\mathbf{x} = \\mathbf{Uz}$\n\\end{inparaitem}\n\\\\Sufficient conditions:\n\\begin{inparaitem}[\\color{red}\\textbullet]\n \\item $\\mathbf{W} = $ Gaussian random projection, i.e. $w_{ij}\\sim\\mathcal{N}(0, \\frac{1}{D})$\n \\item M $\\geq cK log(\\frac{D}{K})$, where $c$ is some constant\n\\end{inparaitem}\n\n\\subsection*{Dictionary Learning}\nAdapt the dictionary to signal characteristics. Objective: $(\\mathbf{U}^\\star, \\mathbf{Z}^\\star) \\in \\argmin_\\mathbf{U,Z} \\| \\mathbf{X} - \\mathbf{U} \\cdot \\mathbf{Z} \\|_F^2$ not jointly convex but convex in 1 argument.\n\n\\textbf{Matrix Factorization by Iter Greedy Minimization}\n\\begin{inparaenum}[\\color{red} 1.]\n \\item Coding step: $\\mathbf{Z}^{t+1} \\in \\argmin_\\mathbf{Z} \\| \\mathbf{X} - \\mathbf{U}^t \\mathbf{Z} \\|_F^2$ subject to $\\mathbf{Z}$ being sparse ($\\mathbf{z}_n^{t+1}\\in \\argmin_\\mathbf{z}\\|\\mathbf{z}\\|_0$ s.t.$\\|\\mathbf{x}_n - \\mathbf{U}^t\\mathbf{z}\\|_2 \\le \\sigma \\|\\mathbf{x}_n\\|_2$)\n \\item Dict update step: $\\mathbf{U}^{t+1} \\in \\argmin_\\mathbf{U} \\| \\mathbf{X} - \\mathbf{UZ}^{t+1} \\|_F^2$, subj to $\\forall l\\in [L]:\\|\\mathbf{u}_l\\|_2 = 1$. (set $\\mathbf{U} = [\\mathbf{u}_1^t\\cdots \\mathbf{u}_l\\cdots \\mathbf{u}_L^t],~ \\min_{u_l}\\|\\mathbf{X} - \\mathbf{U}\\mathbf{Z}^{t+1}\\|_F^2 = \\min_{u_l}\\|\\mathbf{R}_l^t - \\mathbf{u}_l(\\mathbf{z}_l^{t+1})^\\top\\|_F^2$ with $\\mathbf{R}_l^t = \\tilde{\\mathbf{U}}\\Sigma\\tilde{\\mathbf{V}}^\\top$ by $\\mathbf{u}^*_l=\\tilde{\\mathbf{u}}_1$)\n\\end{inparaenum}\n", "meta": {"hexsha": "cbd6efe7c337dc31facdd9c38913661dce54b270", "size": 4179, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "SparseCoding.tex", "max_stars_repo_name": "Emilien-P/eth-cil-exam-summary", "max_stars_repo_head_hexsha": "ebda1cb98b2e3d17e055b05cb0fd1e7decce507f", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-17T18:13:31.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-17T18:13:31.000Z", "max_issues_repo_path": "SparseCoding.tex", "max_issues_repo_name": "Emilien-P/eth-cil-exam-summary", "max_issues_repo_head_hexsha": "ebda1cb98b2e3d17e055b05cb0fd1e7decce507f", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "SparseCoding.tex", "max_forks_repo_name": "Emilien-P/eth-cil-exam-summary", "max_forks_repo_head_hexsha": "ebda1cb98b2e3d17e055b05cb0fd1e7decce507f", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 87.0625, "max_line_length": 486, "alphanum_fraction": 0.6489590811, "num_tokens": 1716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673087708699, "lm_q2_score": 0.787931190663057, "lm_q1q2_score": 0.6404835064509062}} {"text": "\n\\subsection{Independent Contour Method}\n\\begin{frame}{Definition}\n\t\\begin{block}{Formula}\n\t\tFor a closed mechanism, the number of independent contours $N$ is calculated as follows:\n\t\t\\[N=c-n\\]\n\t\twhere\\\\\n\t\t$c$ is the number of joints\\\\\n\t\t$n$ is the number of links, excluding ground\n\t\\end{block}\n\tThe number of independent contours depends on the number of joint connections to ground. Its graphical representation can be referred to as \\textit{connectivity table}, \\textit{structural diagram} or \\textit{\\textbf{contour diagram}}.\n\\end{frame}\n\\begin{frame}{Velocity Analysis}\n\t\\begin{block}{Formulas}\n\t\tFor a closed kinematic chain:\n\t\t\\begin{itemize}\n\t\t\t\\item $\\displaystyle\\sum_{i}\\vb{\\omega}{i-1,i}=\\vb{0}{}$\n\t\t\t\\item $\\displaystyle\\sum_{i}\\vb{r}{A_i}\\times\\vb{\\omega}{i-1,i} + \\sum_{i}\\vb{v}{A_{i-1,i}}=\\vb{0}{}$\n\t\t\\end{itemize}\n\t\twhere:\\\\\n\t\t$\\vb{\\omega}{i-1,i}$ is the angular velocity of link $i$ relative to link $i-1$\\\\\n\t\t$\\vb{r}{A_i}$ is the position vector of joint $A_i$\\\\\n\t\t$\\vb{v}{A_{i-1,i}}$ is the velocity vector of joint $A_i$ on link $i$ relative to joint $A_i$ on link $A_{i-1}$\\\\\n\t\\end{block}\n\\end{frame}\n\\begin{frame}{Acceleration Analysis}\n\\begin{block}{Formulas}\n\tFor a closed kinematic chain:\n\t\\begin{itemize}\n\t\t\\item $\\displaystyle\\sum_{i}\\vb{\\alpha}{i-1,i}+\\sum_{i}\\vb{\\omega}{i}\\times\\vb{\\omega}{i-1,i}=\\vb{0}{}$\n\t\t\\item $\\displaystyle\\sum_{i}\\vb{r}{A_i}\\times(\\vb{\\alpha}{i-1,i}+\\vb{\\omega}{i}\\times\\vb{\\omega}{i-1,i}) + \\sum_{i}\\vb{a}{A_{i-1,i}} + 2 \\sum_i\\vb{\\omega}{i-1}\\times \\vb{v}{A_{i-1,i}}+ \\sum_{i}\\vb{\\omega}{i}\\times(\\vb{\\omega}{i}\\times \\vb{r}{A_{i+1}A_i})=\\vb{0}{}$\n\t\\end{itemize}\n\twhere:\\\\\n\t$\\vb{\\omega}{i}$, $\\vb{\\omega}{i-1}$ are the angular velocities of link $i$ and $i-1$ relative to ground\\\\\n\t$\\vb{\\alpha}{i-1,i}$ is the angular acceleration of joint $A_i$ relative to joint $A_{i-1}$\\\\\n\t$\\vb{a}{A_{i-1,i}}$ is the acceleration vector of joint $A_i$ on link $i$ relative to joint $A_i$ on link $A_{i-1}$\\\\\n\\end{block}\n\\end{frame}\n\\begin{frame}\n\tIn planar motions, the above equations for acceleration analysis can be reduced to:\n\t\\begin{itemize}\n\t\t\\item$\\displaystyle\\sum_{i}\\vb{\\alpha}{i-1,i}=\\vb{0}{}$\n\t\t\\item $\\displaystyle\\sum_{i}\\vb{r}{A_i}\\times\\vb{\\alpha}{i-1,i} + \\sum_{i}\\vb{a}{A_{i-1,i}} + 2 \\sum_i\\vb{\\omega}{i-1}\\times \\vb{v}{A_{i-1,i}}- \\sum_{i}\\vb{\\omega}{i}^2\\vb{r}{A_{i+1}A_i})=\\vb{0}{}$\n\t\\end{itemize}\\vskip3mm\n\tIn general, the steps of this method are:\n\t\\begin{enumerate}\n\t\t\\item Perform position analysis.\n\t\t\\item Draw the contour diagram of the mechanism.\n\t\t\\item Use the formulas above to obtain the results.\n\t\t\\item Calculate $\\vb{\\omega}{i}$ and $\\vb{a}{i}$ with respect to ground.\n\t\t\\item Compute relevant velocities and accelerations.\n\t\\end{enumerate}\n\\end{frame}\n", "meta": {"hexsha": "d009e51927ebcad1edc44b5cb385b4eb88b262da", "size": 2739, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Finished/velocity_acceleration_analysis_pdf/Sections/Introduction/Independent_contour_method.tex", "max_stars_repo_name": "HungNguyenDang/literate-meme", "max_stars_repo_head_hexsha": "ed3383576918b6ca1480e45c6ed689c87636fc41", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Finished/velocity_acceleration_analysis_pdf/Sections/Introduction/Independent_contour_method.tex", "max_issues_repo_name": "HungNguyenDang/literate-meme", "max_issues_repo_head_hexsha": "ed3383576918b6ca1480e45c6ed689c87636fc41", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Finished/velocity_acceleration_analysis_pdf/Sections/Introduction/Independent_contour_method.tex", "max_forks_repo_name": "HungNguyenDang/literate-meme", "max_forks_repo_head_hexsha": "ed3383576918b6ca1480e45c6ed689c87636fc41", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.7222222222, "max_line_length": 267, "alphanum_fraction": 0.6666666667, "num_tokens": 1012, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673087708699, "lm_q2_score": 0.7879311906630568, "lm_q1q2_score": 0.6404835064509061}} {"text": "\n\\documentclass{article}\n\n\\newcommand{\\steem}{\\texttt{STEEM}}\n\\newcommand{\\mytoken}{\\texttt{MYTOKEN}}\n\n\\begin{document}\n\n\\section{Automated Market Makers for SMTs}\n\n\\section{Setup}\n\n\\subsection{Basic definitions}\n\nIn this article, we'll let $s$ represent a quantity of STEEM, let $t$\nrepresent a quantity of some token (SMT), and let $p$ represent a price,\nsuch that $pt$ is STEEM-valued (i.e. if MYTOKEN is trading at\n$p = 0.05$ STEEM / MYTOKEN then $t = 120$ MYTOKEN has a value of\n$pt = (0.05\\ \\mbox{STEEM / MYTOKEN}) \\cdot (120\\ \\mbox{MYTOKEN}) = 6\\ \\mbox{STEEM}$.\n\nSuppose we have a market maker (or any economic agent) with a two-asset\n``portfolio'' (inventory) of $s$ STEEM and $t$ tokens. If the price of\ntokens is $t$, then we may measure of the value of this portfolio, in\nunits of STEEM, as $v(p, s, t) = s + pt$.\n\nOne common portfolio management policy is to require that STEEM\nshould be some constant fraction $r$ of the portfolio, i.e.\n$s = r v(p, s, t)$ where $0 < r < 1$. We call this policy the\n\\textit{constant portfolio ratio} or CPR policy, and the equation\n$s = r v(p, s, t)$ is the \\textit{CPR invariant}.\n\nA different portfolio management policy, discussed by the Bancor\nwhitepaper, is called CRR or \\textit{constant reserve ratio}. To discuss CRR,\nlet us notate the total number of tokens in existence as $T$. The\n\\textit{CRR invariant} is then defined as $s = r v(p, 0, T-t)$.\n\n\\subsection{Note on conventions}\n\nWe must discuss where our convention varies from the Bancor whitepaper.\nAt some times, when some user Alice interacts with the market maker,\nAlice will remove some tokens from her balance to get STEEM from\nthe market maker's balance. On the other hand, Bob may add some tokens\nto his balance in exchange for sending STEEM to the market maker's balance.\n\nBancor takes the convention that in this example, the market maker \\textit{destroys}\ntokens in its interaction with Alice, and \\textit{creates} tokens in its interaction\nwith Bob. The Bancor convention suggests the market maker is not\nan ordinary actor, but needs system-level ``special powers'' -- specifically,\nthe privilege to operate the token printing press -- in order to function.\n\nIn this paper, we adopt the convention that the tokens sent by Alice to the\nmarket maker are not destroyed, but are instead added to the inventory (balance)\nof the market maker. Likewise, the tokens sent to Bob by the market maker\nare not created out of thin air; they already exist and are merely\ntransferred from the inventory of the market maker to Bob. Thus, we show\nthat the market maker is essentially an ordinary economic agent acting according\nto a deterministic algorithm -- it doesn't actually need ``special powers''!\n\n\\section{Finite trades}\n\n\\subsection{Basic definitions}\n\nA \\textit{trade} is a change in the market maker's balance from\n$(s, t) \\to (s+\\Delta s, t+\\Delta t)$. The \\textit{price} at which the\ntrade occurs is defined as $p = {- \\Delta s \\over \\Delta t}$. We restrict\nourselves to \\textit{well-formed trades} where either $\\Delta s = \\Delta t = 0$,\nor $\\Delta s$ and $\\Delta t$ are both nonzero and have opposite sign.\n\nTheorem: A trade at price $p$ conserves value at price $p$. More rigorously,\nif $\\Delta s, \\Delta t$ represent a trade at price $p$, then\n$v(p, s, t) = v(p, s + \\Delta s, t + \\Delta t)$.\n\nLet us more rigorously define the market maker's \\textit{state} as a tuple\n$M = (s, t, T, r)$. Given some price $p$, we may define the\n\\textit{restoring trade} at $p$ (also called a \\textit{relaxing trade} or\na \\textit{relaxation}) to be a trade which occurs at price $p$ and results in a\nstate that satisfies the CRR invariant.\n\n\\subsection{Computing the restoring trade}\n\nThe restoring trade consists of functions $\\Delta s(M, p)$ and $\\Delta t(M, p)$.\nWe may actually compute these functions from the definition of price and the\nCRR invariant:\n\n\\begin{eqnarray*}\n\\Delta s & = & -p \\Delta t \\\\\ns + \\Delta s & = & r v(p, 0, T-(t+\\Delta t)) \\\\\n\\Rightarrow s - p \\Delta t & = & r v(p, 0, T-t-\\Delta t) \\\\\n & = & r p(T-t-\\Delta t) \\\\\n & = & rpT - rpt - rp \\Delta t \\\\\n\\Rightarrow r p \\Delta t - p \\Delta t & = & r p (T-t) - s \\\\\n\\Rightarrow \\Delta t & = & {r p (T - t) - s \\over rp - p } \\\\\n & = & \\left( {1 \\over 1-r} \\right) \\left( {s \\over p} - r (T-t) \\right) \\\\\n\\Rightarrow \\Delta s & = & - p \\Delta t \\\\\n & = & \\left( {1 \\over 1-r} \\right) \\left( r p (T - t) - s \\right)\n\\end{eqnarray*}\n\n\\subsection{Computing the equilibrium price}\n\nGiven a state $M$, there exists some price $p_{eq}(M)$ for which the restoring\ntrade is zero; call this price the \\textit{equilibrium price}. We may compute\nthe equilibrium price by setting $\\Delta s = 0$:\n\n\\begin{eqnarray*}\n\\Delta s & = & 0 \\\\\n& = & \\left( {1 \\over 1-r} \\right) \\left( r p_{eq} (T - t) - s \\right) \\\\\n\\Rightarrow r p_{eq} (T - t) - s & = & 0 \\\\\n\\Rightarrow r p_{eq} (T - t) & = & s \\\\\n\\Rightarrow p_{eq} & = & {s \\over r (T-t)}\n\\end{eqnarray*}\n\nTheorem: Relaxation is idempotent. That is, after relaxing at price $p$,\nthe equilibrium price of the resulting state is $p$, and a second relaxation\nat price $p$ will be a zero trade.\n\n\\subsection{Example}\n\nExample: Suppose $M = (1200, 3600, 12000, 0.25)$ and $p = 0.5$. Then of the\n$T = 12000$ TOKEN in existence, $t = 3600$ TOKEN is held by the MM, so\n$T-t = 12000 - 3600 = 8400$ TOKEN are ``circulating'' (i.e. exist in balances\noutside the MM). These circulating tokens are worth $p(T-t) = 4200$ STEEM\ntotal, so they ``should be'' backed by a target reserve level of\n$rp(T-t) = 4200 * 0.25 = 1050$ STEEM.\n\nIn this example, there is ``too much'' STEEM in the reserve, so relaxation\nwill buy tokens in the market. This sale will cause two effects: It will\ndecrease the reserve STEEM, and also decrease circulating tokens. The decrease\nin circulating tokens, in turn, causes the target reserve level to decline. For\nevery 1 STEEM used to buy tokens, the target reserve level declines by $r$ STEEM;\nsince $r < 1$ eventually the declining reserve will ``catch up'' to its more slowly\ndeclining target level.\n\nThe above algebra shows that we will catch up at\n$\\Delta s = \\left( {1 \\over 1-r} \\right) \\left( r p (T - t) - s \\right)$ and\n$\\Delta t = \\left( {1 \\over 1-r} \\right) \\left( {s \\over p} - r (T - t) \\right)$.\nRunning the calculations with the numbers defined in this example gives\n$\\Delta s = -200$ STEEM and $\\Delta t = 400$ TOKEN.\n\nLet's check that these computed values $\\Delta s = -200, \\Delta t = 400$\n(a) represent a trade with price $0.5$, and (b) that the CRR invariant holds\nfor the new state $M_{new} = (s+\\Delta s, t+\\Delta t, T, r)$. Calculating\n$p = {- \\Delta s \\over \\Delta t}$ we indeed get $p = 0.5$. After this trade executes,\nthe market maker has $s_{new} = s + \\Delta s = 1200 - 200 = 1000$ STEEM, and\n$t_{new} = t + \\Delta t = 3600 + 400 = 4000$ tokens.\n\nTo check condition (b), that the CRR invariant holds, we effectively repeat the\nanalysis in the initial paragraph of this example with the new numbers. We know\n$M_{new} = (1000, 4000, 12000, 0.25)$ and $p = 0.5$. Then of the $T = 12000$\nTOKEN in existence, $t_{new} = 4000$ TOKEN is now held by the MM, so\n$T-t_{new} = 12000 - 4000 = 8000$ TOKEN are now circulating. These circulating\ntokens are worth $p(T-t_{new}) = 4000$ STEEM total, so they ``should be'' backed\nby a target reserve level of $rp(T-t) = 4000 * 0.25 = 1000$ STEEM. Since the\ntarget reserve level indeed exactly matches the actual reserve level of\n$s_{new} = 1000$ STEEM, we conclude that the CRR invariant is satisfied after\nthis relaxing trade.\n\n\\section{Infinitesimal trades}\n\nThis section is fairly technical; the reader will need a good grasp of\ncalculus and differential equations to follow the results.\n\n\\subsection{Setting up the problem}\n\nSuppose we satisfy the invariant condition at some price\n$p = p_eq$; by the CRR invariant $s = r v( p, 0, T-t) = r p (T-t)$.\nSuppose the price then increases to $p + \\Delta p$ and a relaxing trade\n$\\Delta s, \\Delta t$ occurs at this new price.\n\nIn this section we consider the limiting situation where $\\Delta p$ is\ninfintesimally small, so we will use Leibniz notation ($dp$ for a small change\nin $p$, $ds$ for a small change in $s$, $dt$ for a small change in $t$).\n\n\\subsection{Solving the DE's}\n\nBy applying the substitution $p \\gets p + dp$ to the expression for $\\Delta s$\ncomputed in the previous section, we obtain an expression which simplifies\nto a separable DE which can be solved:\n\n\\begin{eqnarray*}\nds & = & {1 \\over 1-r} \\left( r (p + dp) (T - t) - s \\right) \\\\\n & = & {1 \\over 1-r} \\left( r p (T - t) + r dp (T - t) - r p (T - t) \\right) \\\\\n & = & {1 \\over 1-r} r (T - t) dp \\\\\n & = & {1 \\over 1-r} \\left( {s \\over p} \\right) dp \\\\\n\\Rightarrow {1 \\over p} dp & = & (1-r) \\left( {1 \\over s} ds \\right) \\\\\n\\Rightarrow \\int {1 \\over p} dp & = & (1-r) \\int {1 \\over s} ds \\\\\n\\Rightarrow \\ln(p) & = & (1-r) \\ln(s) + C_0 \\\\\n\\Rightarrow p & = & k_0 s^{1-r}\n\\end{eqnarray*}\n\nSimilarly for $t$, we can start from $dt = -ds / p$ and again obtain and solve\na separable DE:\n\n\\begin{eqnarray*}\ndt & = & -ds / p \\\\\n & = & -{r \\over 1-r} (T - t) dp / p \\\\\n\\Rightarrow {1 \\over T-t} dt & = & -{r \\over 1-r} \\left( {1 \\over p} \\right) dp \\\\\n\\Rightarrow {1 \\over p} dp & = & {1-r \\over r} \\left( {1 \\over t-T} \\right) dt \\\\\n\\Rightarrow \\int {1 \\over p} dp & = & {1-r \\over r} \\int {1 \\over t-T} dt \\\\\n\\Rightarrow \\ln(p) & = & {1-r \\over r} \\ln | t-T | + C_1 \\\\\n\\Rightarrow p & = & k_1 (T-t)^{1-r \\over r}\n\\end{eqnarray*}\n\n\\section{Qualitative discussion}\n\nIn a CRR market maker, where does the ``backing'' for newly emitted tokens come from?\n\nOne option is to lower the reserve ratio $r$. This option results in no immediate market\nactivity, but will weaken the response of the market maker to any future price changes.\nThis is called the ``pay later'' option.\n\nAnother option is to change the dynamical system's initial conditions, i.e. edit the\nconstants of integration. This option will cause the equilibrium price $p_{eq}$ to drop,\nmeaning the market maker will more aggressively sell tokens to replenish the reserve.\nIf order books are deep compared to the amount of emission, and there are adequate buyers\nfor the tokens, then the sales will be able to replenish the reserve and keep the equilibrium\nprice near its old value; the deep order books provide resistance to the price change being\ndriven by the market maker. If order books are thin compared to the amount of emission,\nand there are few/no buyers for the tokens, then the equilibrium price will fall, breaking\nthrough the thin orders and lowering the market price. Even though few/no many tokens were\nsold, so even though the \\textit{absolute} amount of STEEM in the reserve is still\nnearly/exactly the same as before, the reserve's value \\textit{relative} to the now-lower\nmarket cap of the token has increased to the reserve ratio. This option is the ``pay now''\noption.\n\n\n\n\n\\section{FAQ}\n\nFAQ\n\nQ: Where are the pretty pictures and graphs?\n\nA: They're being worked on right now, and will exist in a future version of this paper.\n\nQ: What is the relevance of constant portfolio ratio policy?\n\nA: It may become a supported market maker policy in the future.\n\nQ: Can the reserve ratio go over 100 percent?\n\nA: No.\n\nQ: Can the reserve ratio be exactly 100 percent?\n\nA: Not with the system described in this paper. It might be possible to code as a special case.\n\nQ: In a CRR market maker, where does the ``backing'' for newly emitted tokens come from?\n\nA: As blockchain designers, we have two options for sourcing the ``backing''. One option\nis to lower the reserve ratio $r$. This option results in no immediate market\nactivity, but will weaken the response of the market maker to any future price changes.\nThis is called the ``pay later'' option.\n\nAnother option is to change the dynamical system's initial conditions, i.e. edit the\nconstants of integration. This option will cause the equilibrium price $p_{eq}$ to drop,\nmeaning the market maker will more aggressively sell tokens to replenish the reserve.\nIf order books are deep compared to the amount of emission, and there are adequate buyers\nfor the tokens, then the sales will be able to replenish the reserve to its target level while\nkeeping the equilibrium price near its old value. The deep order books provide resistance to\nthe price change being driven by the market maker.\n\nIf order books are thin compared to the amount of emission, and there are few/no buyers for the\ntokens, then the equilibrium price will fall, breaking through the thin orders and lowering\nthe market price. Even though few/no many tokens were sold, so even though the\n\\textit{absolute} amount of STEEM in the reserve is still nearly/exactly the same as before,\nthe reserve's value \\textit{relative} to the now-lower market cap of the token has increased\nto the reserve ratio. This option is the ``pay now'' option.\n\nQ: Where's the ``don't pay'' option?\n\nA: You have to come up with some answer to where the ``backing'' for newly emitted\ntokens will come from. Unless there's no emission. Or unless there's no ``backing'' for\nany tokens. So the ``don't pay'' option would be to have an SMT with either no\nemission, or no market maker.\n\nQ: Don't fractional exponents require floating point to implement?\n\nA: Only if you need fairly high precision (we don't), don't care about bit-for-bit\nreproducibility across compilers, OS's, CPU's, etc (we do), and need to do massive\nnumbers of calculations quickly (we don't). A fast, approximate, all-integer\nimplementation is possible.\n\nQ: Does this market maker interact with the order book through the existing limit order\nsystem, or is it a separate set of operations?\n\nA: In theory, it could be implemented either way. However, the likely outcome is that the market maker will be implemented outside of order-book markets to allow its code to be modularized. In practice, if implemented as a completely separate subsystem, people will run arbitrage bots which will trade away any price differences\nbetween the reserve system and the existing market system.\n\nQ: Where do the market maker's initial token balances come from?\n\nA: ICO units can specify the market maker as a destination. An ICO creator\nmay direct a percentage of ICO's STEEM contributions to the MM by specifying\nthe market maker similarly to specifying a founder. Or may use the soft cap\nsystem to specify all STEEM above a pre-determined amount goes to the ICO. Likewise,\na fixed or percentage amount of tokens can be added in the ICO to increase the MM's\ntoken balance.\n\nQ: Can someone send STEEM or tokens to the market maker?\n\nA: Yes.\n\nQ: What are the side effects of sending STEEM or tokens to the market maker?\n\nA: The constants of integration are re-initialized, meaning the equilibrium price will\nchange. The market maker will become more aggressive about selling the asset.\n\nQ: Can't this cause manipulation or appropriating the market maker's inventory\nto private profit?\n\nA: Sending assets to the market maker does cause it to engage in trading activity\nwhich affects the price. However, dumping an identical amount on the market will\nresult in a larger amount of trading activity and a larger effect on the price. If\nEve is willing to spend her tokens/STEEM to manipulate prices, she would prefer\nthe strategy of simply dumping tokens/STEEM on the market, as that strategy is more\ncost-effective for her.\n\nQ: Does the market maker's activity generate profits (losses)?\n\nA: It depends on how you measure ``profits.'' If you measure the value of STEEM\nand tokens in some external third currency such as US dollars or bitcoins, the\nmarket maker's inventory, valued in that currency, can definitely increase or\ndecrease. If people voluntarily send STEEM or tokens to the market maker,\nsuch activity definitely increases the value of the market maker regardless of\nyour measurement.\n\nAnother way to define profits is by the constants of integration. If both\nof the constants of integration increase, or one increases while the other\nremains the same, a tiny increase occurs with each trade when the market\nmaker is in ``taker'' mode.\n\nQ: What is ``taker'' mode? How can a market maker be set to operate\nin ``taker'' mode?\n\nA: When orders execute, the order used to set the price is called the maker;\nthe maker's counterparty is the taker. In the STEEM on-chain market (and on\nalmost all trading platforms) the older order is always the maker.\n\nWhen the market maker is in taker mode, its actions are always considered to be\ntaker orders, which execute at the price specified by the user acting as its\ncounterparty -- this price is always at least a little bit more favorable than\nthe market maker is willing to accept. When the market maker is not in taker\nmode, its actions are always considered to be maker orders, which don't\ngenerate changes in the constants of integration.\n\nTaker mode is a runtime parameter that can be set by the SMT's control account.\n\nQ: Who benefits from the profits of a market maker in taker mode?\n\nA: Maybe nobody, or maybe everybody. It's decentralized.\n\nQ: OK, if my SMT reaches a steady price, the STEEM in the reserve is basically\nlocked up forever. That seems not cool. How do I set it up so that this STEEM\ncan be unlocked for the benefits of my SMT users?\n\nA: Set the DRR (decaying reserve ratio) setup parameter. If you set DRR, then the\nreserve ratio will slowly drop over time to a pre-set value, using its excess STEEM\nreserves to buy excess tokens. Setting DRR is an excellent, fair, decentralized\nway to return excess capitalization to contributors in a more-popular-than-anticipated\nICO that raises more than the sponsor can effectively spend.\n\nQ: If the reserve ratio can change over time due to pay-later emissions or DRR,\nit's not really a constant reserve ratio, is it?\n\nA: No, they're not. The reserve ratio's called ``constant'' because it's constant\nover the short-term, in normal conditions, or in the conditions in Bancor which is\nwhere it was named. But the name could be regarded as slightly misleading.\n\nQ: If the constants of integration can change over time, they're not really\nconstants either, are they?\n\nA: No, they're not. They're called constants of integration because that's their\nmathematical role in the calculation that introduces them. Maybe they'll be\ndifferently named in a future version of this paper.\n\nQ: Can I specify a contribution to a DRR to be a pay-later contribution, that\n\\textit{increases} its reserve ratio, the increase to be eventually negated over\ntime by future decay? Why would I want to?\n\nA: Yes. This is effectively contributing to the market maker,\nsubject to the condition that it's not allowed to immediately dump a portion\nof the contribution. It's useful if you want to make a large contribution\nto a market maker without causing it to create a disturbance by immediately\ndumping a significant fraction of your contribution onto the market.\n\nQ: Can I specify a DRR with emission to use pay-later for emissions when the\nRR is decaying?\n\nA: Yes.\n\nQ: Is the market maker specified here equivalent to a Bancor token changer?\n\nA: No. A Bancor token changer has multiple reserve ratios that must sum to\none hundred percent, and involves a third token that effectively represents\nequity in the token changer. This paper's market maker has none of these\nfeatures.\n\nQ: I want to have an initial ``price discovery'' period where people trade\nwithout action from the market maker, then have tokens and STEEM from\nthe ICO gradually flow in over time to the market maker so it has a delayed,\nslow start from zero to full power. Can I do it?\n\nA: This is called ``gradual seeding'' and it may be supported.\n\nQ: What about numerical stability?\n\nA: A market maker will be restricted to only operate when its\nbalances exceed a certain minimum for both assets. Also, reserve\nratios will be restricted to a certain range, all the mechanisms\nthat can set / increase / decrease a reserve ratio will be restricted\nto not allow it to move outside the range. Tentative numerical\nexperiments suggests these limits should be about 10,000 satoshis\nof both assets, 5 percent and 50 percent, respectively. These\nvalues are subject to change based on future experimentation,\nworst-case analysis, and testing.\n\n\\end{document}\n", "meta": {"hexsha": "ae020c84588eafccf22ae102b3ea9ad7c1f9b482", "size": 20376, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "smt-manual/market-maker.tex", "max_stars_repo_name": "blocktradesdevs/smt-whitepaper", "max_stars_repo_head_hexsha": "cc10fc34e20b31ccbacdda436ddd2ee004896488", "max_stars_repo_licenses": ["FSFAP"], "max_stars_count": 35, "max_stars_repo_stars_event_min_datetime": "2017-09-28T19:01:45.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-21T08:07:17.000Z", "max_issues_repo_path": "smt-manual/market-maker.tex", "max_issues_repo_name": "blocktradesdevs/smt-whitepaper", "max_issues_repo_head_hexsha": "cc10fc34e20b31ccbacdda436ddd2ee004896488", "max_issues_repo_licenses": ["FSFAP"], "max_issues_count": 36, "max_issues_repo_issues_event_min_datetime": "2017-09-29T15:49:26.000Z", "max_issues_repo_issues_event_max_datetime": "2019-11-20T18:22:39.000Z", "max_forks_repo_path": "smt-manual/market-maker.tex", "max_forks_repo_name": "blocktradesdevs/smt-whitepaper", "max_forks_repo_head_hexsha": "cc10fc34e20b31ccbacdda436ddd2ee004896488", "max_forks_repo_licenses": ["FSFAP"], "max_forks_count": 28, "max_forks_repo_forks_event_min_datetime": "2017-10-16T23:44:24.000Z", "max_forks_repo_forks_event_max_datetime": "2020-02-15T14:56:19.000Z", "avg_line_length": 48.7464114833, "max_line_length": 331, "alphanum_fraction": 0.7311542992, "num_tokens": 5499, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267864276108, "lm_q2_score": 0.7371581568543043, "lm_q1q2_score": 0.6404627525086258}} {"text": "% Preamble.\n\\documentclass[12pt]{article}\n\\usepackage[margin=1.25in]{geometry}\n\\usepackage[fleqn]{amsmath}\n\\usepackage{textcomp}\n\\usepackage{gensymb}\n\\usepackage{amsfonts}\n\\usepackage{enumitem}\n\\usepackage{siunitx}\n%\\usepackage{subfiles} % Include for subfiles.\n%\\usepackage{tikz} % Include for handmade figures.\n\\usepackage{graphicx} % Include for external figures.\n\n%% Title macros.\n\\newcommand{\\HOMEWORKNUM}{1}\n\\newcommand{\\NAME}{D. Choi}\n\\newcommand{\\DATE}{2020-08-28}\n\n\\title{\\vspace{-2\\baselineskip}PHYS 225 - Written Homework \\#\\HOMEWORKNUM}\n\\author{\\NAME}\n\\date{\\DATE}\n\n%% Formatting options.\n%\\pagenumbering{gobble} % Include for single-page document.\n\n\n% Document.\n\\begin{document}\n\\maketitle\n\n\n\\section*{1.}\n\\textit{Consider the vectors $\\vec{A} = -2\\hat{\\imath} - 2\\hat{\\jmath}$ and\n$\\vec{B} = \\hat{\\imath} + 3\\hat{\\jmath}$.}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item \\textit{Find the unit vectors\n\t$\\hat{A} = \\frac{\\vec{A}}{\\|\\vec{A}\\|}$ and\n\t$\\hat{B} = \\frac{\\vec{B}}{\\|\\vec{B}\\|}$.}\n\t\\begin{gather*}\n\t\t\\hat{A}\n\t\t= \\frac{\\vec{A}}{\\|\\vec{A}\\|}\n\t\t= \\frac{-2\\hat{\\imath} - 2\\hat{\\jmath}}{2\\sqrt{2}}\n\t\t= \\boxed{\n\t\t\t-\\frac{1}{\\sqrt{2}} \\hat{\\imath}\n\t\t\t- \\frac{1}{\\sqrt{2}} \\hat{\\jmath}\n\t\t}, \\\\\n\t\t\\hat{B}\n\t\t= \\frac{\\vec{B}}{\\|\\vec{B}\\|}\n\t\t= \\frac{\\hat{\\imath} + 3\\hat{\\jmath}}{\\sqrt{10}}\n\t\t= \\boxed{\n\t\t\t\\frac{1}{\\sqrt{10}} \\hat{\\imath}\n\t\t\t+ \\frac{3}{\\sqrt{10}} \\hat{\\jmath}\n\t\t}.\n\t\\end{gather*}\n\t\\item \\textit{Express the vector $\\vec{C} = 5\\hat{\\imath} - 3\\hat{\\jmath}$\n\tin terms of $\\hat{A}$ and $\\hat{B}$.}\n\t\\begin{gather*}\n\t\t\\vec{C} = \\hat{A}x + \\hat{B}y, \\\\\n\t\t\\begin{pmatrix} 5 \\\\ -3 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t-\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{10}} \\\\\n\t\t\t-\\frac{1}{\\sqrt{2}} & \\frac{3}{\\sqrt{10}}\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix} x \\\\ y \\end{pmatrix}.\n\t\\end{gather*}\n\tSolving for $x$ and $y$,\n\t\\begin{equation*}\n\t\t\\begin{pmatrix} x \\\\ y \\end{pmatrix}\n\t\t= \\begin{pmatrix} -9\\sqrt{2} \\\\ -4\\sqrt{10} \\end{pmatrix}.\n\t\\end{equation*}.\n\tThus,\n\t\\begin{equation*}\n\t\t\\vec{C} = \\boxed{\n\t\t\t-9\\sqrt{2}\\hat{A} - 4\\sqrt{10}\\hat{B}\n\t\t}.\n\t\\end{equation*}\n\\end{enumerate}\n\n\\section*{2.}\n\\textit{A particle moves in one dimension with a velocity\n$v(t) = v_0 + v_1 t^3$, where $v_0 = 2.00\\si{\\frac{m}{s}}$ and\n$v_1 = 1.00\\si{\\frac{m}{s^4}}$.}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item \\textit{Assuming that at $t_0 = 0$, the particle is at position\n\t$x_0 = 4.00\\si{m}$ measured from the origin, derive a formula for the\n\tparticle's position as a function of time.}\n\t\\begin{gather*}\n\t\tv(t) = v_1 t^3 + v_0, \\\\\n\t\tx(t)\n\t\t= \\int v(t) dt\n\t\t= \\boxed{\n\t\t\t\\frac{v_1}{4}t^4 + v_0 t + x_0\n\t\t}.\n\t\\end{gather*}\n\t\\item \\textit{Obtain a formula for the acceleration of the particle.}\n\t\\begin{equation*}\n\t\ta(t)\n\t\t= \\frac{d}{dt} v(t)\n\t\t= \\boxed{\n\t\t\t3v_1 t^2\n\t\t}.\n\t\\end{equation*}\n\t\\item \\textit{Make quantitative plots of the particle's position,\n\tvelocity, and acceleration as a function of time during the interval\n\t$t_0 = 0$ and $t_1 = 1.00\\si{s}$.}\n\t\\begin{center}\n\t\t\\includegraphics[width=0.6\\textwidth]{fig.pdf}\n\t\\end{center}\n\t\\newpage\n\t\\item \\textit{What is the particle's average velocity and acceleration in\n\tthe time interval $t_0 = 0$ to $t_1 = 1.00\\si{s}$?} \\\\[\\baselineskip]\n\tFor the interval $[0, 1.00\\si{s}]$, let $v_{avg}$ and $a_{avg}$ be the\n\tparticle's average velocity and average acceleration over the interval\n\trespectively.\n\t\\begin{gather*}\n\t\tv_{avg}\n\t\t= \\frac{x(1.00\\si{s}) - x(0)}{1.00\\si{s} - 0}\n\t\t= \\frac{6.25\\si{m} - 4.00\\si{m}}{1.00\\si{s} - 0}\n\t\t= \\boxed{2.25\\si{\\frac{m}{s}}}, \\\\\n\t\ta_{avg}\n\t\t= \\frac{v(1.00\\si{s}) - v(0)}{1.00\\si{s} - 0}\n\t\t= \\frac{3.00\\si{\\frac{m}{s}} - 2.00\\si{\\frac{m}{s}}}{1.00\\si{s} - 0}\n\t\t= \\boxed{1.00\\si{\\frac{m}{s^2}}}.\n\t\\end{gather*}\n\t\\item \\textit{What is the particle's position at the end of this time\n\tinterval (at $t_1 = 1.00\\si{s}$)?}\n\t\\begin{align*}\n\t\tx(1.00\\si{s})\n\t\t&= \\frac{v_1}{4}(1.00\\si{s})^4 + v_0 (1.00\\si{s}) + x_0 \\\\\n\t\t&= {\n\t\t\t\\frac{1.00\\si{\\frac{m}{s^4}}}{4}(1.00\\si{s})^4\n\t\t\t+ 2.00\\si{\\frac{m}{s}} (1.00\\si{s})\n\t\t\t+ 4.00\\si{m}\n\t\t} \\\\\n\t\t&= \\boxed{6.25\\si{m}}\n\t\t.\n\t\\end{align*}\n\\end{enumerate}\n\n\n\\end{document}", "meta": {"hexsha": "e81f8d8f17e78ae2c70317e25d992ae365c9071a", "size": 4079, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "usc-20203-phys-151-50380/hw01/main.tex", "max_stars_repo_name": "Floozutter/coursework", "max_stars_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "usc-20203-phys-151-50380/hw01/main.tex", "max_issues_repo_name": "Floozutter/coursework", "max_issues_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "usc-20203-phys-151-50380/hw01/main.tex", "max_forks_repo_name": "Floozutter/coursework", "max_forks_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.7253521127, "max_line_length": 75, "alphanum_fraction": 0.6038244668, "num_tokens": 1791, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581510799253, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.6404627349729926}} {"text": "\\section{Dijkstra's Algorithm as Framework}\n\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Dijkstra's algorithm}\n \\[\n\t\\dist(v) = \\min_{u \\in N(v)} \\set{\\dist(u) + l(u,v)}\n \\]\n \n \\begin{columns}\n\t\\column{0.35\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\ForAll{$v \\in V$}\n\t\t \\State $\\dist[v] \\gets \\infty$\n\t\t\\EndFor\n\n\t\t\\State $\\dist[s] \\gets 0$\n\t\t\\Statex\n\t\t\\State $Q \\gets \\text{MinPQ}(V)$\n\t \\end{algorithmic}\n\t\\column{0.60\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\While{$Q \\neq \\emptyset$}\n\t\t \\State $u \\gets \\text{deleteMin}(Q)$\n\t\t \\ForAll{$(u,v) \\in E \\land v \\in Q$}\n\t\t\t\\If{$\\dist[v] > \\dist[u] + l(u,v)$}\n\t\t\t \\State $\\dist[v] \\gets \\dist[u] + l(u,v)$\n\t\t\t \\State $\\text{decreaseKey(Q, v)}$\n\t\t\t\\EndIf\n\t\t \\EndFor\n\t\t\\EndWhile\n\t \\end{algorithmic}\n \\end{columns}\n\n \\vspace{0.50cm}\n \\[\n\tO(n + (n+m) \\log n) \\implies O((n + m) \\log n) \\implies O(m \\log n)\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Dijkstra's algorithm}\n \\begin{columns}\n\t\\column{0.45\\textwidth}\n\t Prim's algorithm for MST:\n\t \\begin{algorithmic}\n\t\t\\If{$\\text{cost}[u] > w(u,v)$}\n\t\t \\State $\\text{cost}[u] \\gets w(u,v)$\n\t\t\\EndIf\n\t \\end{algorithmic}\n\t\\column{0.45\\textwidth}\n\t BFS:\n\t \\begin{algorithmic}\n\t\t\\State $Q \\gets \\text{FIFO-Q}(s)$\n\t\t\\Statex\n\t\t\\If{$\\dist[v] = \\infty$}\n\t\t \\State $\\dist[v] \\gets \\dist[u] + 1$\n\t\t\\EndIf\n\t \\end{algorithmic}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Unique shortest paths}\n \\begin{exampleblock}{Unique shortest paths (Problem 6.18)}\n\t Undirected graph $G = (V, E), w(e) > 0, s \\in V$:\n\t \\[\n\t\t\\text{usp}[v] = T \\iff \\exists! s \\leadsto^{\\SP} v \n\t \\]\n \\end{exampleblock}\n\n \\begin{columns}[t]\n\t\\column{0.30\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\ForAll{$v \\in V$}\n\t\t \\State $\\text{usp}[v] \\gets F$\n\t\t\\EndFor\n\t\t\\State $\\text{usp}[s] \\gets T$\n\t \\end{algorithmic}\n\t\\column{0.60\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\If{$\\dist[v] > \\dist[u] + l(u,v)$}\n\t\t \\State $\\dist[v] \\gets \\dist[u] + l(u,v)$\n\t\t \\State $\\text{usp}[v] \\gets \\text{usp}[u]$\n\t\t\\ElsIf{$\\dist[v] = \\dist[u] + l(u,v)$}\n\t\t \\State $\\text{usp}[v] \\gets F$\n\t\t\\EndIf\n\t \\end{algorithmic}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Number of shortest paths}\n \\begin{exampleblock}{Number of shortest paths (Problem 6.31, 5.26)}\n\t\\[\n\t \\# s \\leadsto^{\\SP} v\n\t\\]\n\n\t\\[\n\t O(n + m) \\text{\\emph{ if }} w(e) = 1\n\t\\]\n \\end{exampleblock}\n\n \\[\n\t\\text{Num}[s] = 1 \\text{\\emph{ vs. }} \\text{Num}[s] = 0\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Min-max path problem}\n \\begin{exampleblock}{Min-max path problem (Problem 6.23)}\n\t\\begin{itemize}\n\t \\item $G = (V, E)$: network of highways\n\t \\item $l_e$: road length; $L$: tank capacity\n\t \\item (1) Given $L$, $\\exists? s \\leadsto t$.\n\t \\item (2) Given $G$, compute $\\min L$.\n\t\\end{itemize}\n \\end{exampleblock}\n\n \\[\n\tL[v] = \\min_{u \\in N(v)} \\max \\set{L[u], l(u,v)}\n \\]\n\n \\begin{columns}\n\t\\column{0.35\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\ForAll{$v \\in V$}\n\t\t \\State $L[v] \\gets \\infty$\n\t\t\\EndFor\n\t\t\\State $L[s] \\gets 0$\n\t \\end{algorithmic}\n\t\\column{0.65\\textwidth}\n\t \\begin{algorithmic}\n\t\t\\If{$L[v] > \\max(L[u], l(u,v))$}\n\t\t \\State $L[v] \\gets \\max(L[u], l(u,v))$\n\t\t\\EndIf\n\t \\end{algorithmic}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Min-max path problem}\n \\begin{exampleblock}{Min-max path problem (Problem 6.23)}\n\t\\begin{itemize}\n\t \\item $G = (V, E)$: network of highways\n\t \\item $l_e$: road length; $L$: tank capacity\n\t \\item (1) Given $L$, $\\exists? s \\leadsto t$.\n\t \\item (2) Given $G$, compute $\\min L$.\n\t\\end{itemize}\n \\end{exampleblock}\n\n \\vspace{0.50cm}\n \\centerline{$O(\\log m)$ binary searches for $\\min L$}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Max-min path problem}\n \\begin{exampleblock}{Max-min path problem (Problem 6.26)}\n\t\\begin{itemize}\n\t \\item $G = (V, E)$: network of oil pipelines\n\t \\item $c(u,v)$: capacity of $(u,v)$\n\t \\item (1) Given $s$, compute $\\text{cap}(s, v)$.\n\t \\item (2) Compute all-pair $\\text{cap}(u,v)$.\n\t\\end{itemize}\n \\end{exampleblock}\n\n \\[\n\t\\text{cap}[v] = \\max_{u \\in N(v)} \\min (\\text{cap}[u], c(u,v))\n \\]\n\n \\[\n\tQ \\gets \\text{MaxPQ}(V)\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "21ac29efa5b7d75e95a5ad4e98ca91897931bfe5", "size": 4133, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-paths-20170605/sections/dijkstra-framework.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-paths-20170605/sections/dijkstra-framework.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-paths-20170605/sections/dijkstra-framework.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 24.4556213018, "max_line_length": 69, "alphanum_fraction": 0.5584321316, "num_tokens": 1641, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.872347368040789, "lm_q2_score": 0.7341195269001831, "lm_q1q2_score": 0.6404072371187239}} {"text": "%!TEX root = ../CombinatoricsNotes.tex\n\n\\section{Convexity}\nWe will consider finite collections of points in Euclidean space, $\\R^d$. Let's recall some definitions.\n\\begin{itemize}[]\n\\item[Linear space:] $L\\subset \\R^d$ is a \\defn{linear space}[linear!space] if it is closed under addition and multiplication by scalars.\n\\item[Linear dependence:] We say $v_1,v_2,\\dotsc,v_n$ are \\defn{linearly dependent}[linear!dependence] if there exist $\\alpha_1,\\alpha_2,\\dotsc,\\alpha_n\\in \\R$ not all zero such that\n\\[\n\\alpha_1 v_1 + \\alpha_2v_2 + \\dotsc + \\alpha_n v_n = 0.\n\\]\n\\item[Linear hull:] The \\defn{linear hull}[linear!hull] $\\braket{S}$ of $S\\subset \\R^d$ is the smallest linear subspace containing $S$, i.e. $\\braket{S}=\\bigcap_{L\\supset S} L$ where the intersection is taken over linear subspaces $L$. Equivalently,\n\\begin{align*}\t\n\\braket{S} = \\{ \\alpha_1 v_1 + \\alpha_2 v_2 + \\dotsc + \\alpha_n v_n: \\forall n,\\, v_1,\\dotsc,v_n\\in S,\\, \\alpha_1,\\dotsc,\\alpha_n\\in \\R \\}.\n\\end{align*}\nNote that we may take $n = d$.\n% , because any $d+1$ vectors in $S\\subset \\R^d$ will be linearly dependent.}\n\n\\item[Affine subspace:] An \\defn{affine subspace}[affine!subspace] is a subset of $\\R^d$ of the form $v+ L$ where $L$ is a linear subspace and $v\\in \\R^d$.\n\n\\item[Affine hull:] The \\defn{affine hull}[affine!hull] $\\aff(S)$ of $S\\subset \\R^d$ is the intersection of all affine subspaces containing $S$. Equivalently,\n\\[\n\\aff(S) = \\{\\alpha_1 v_1 + \\alpha_2 v_2 + \\dotsb + \\alpha_n v_n : \\sum_i \\alpha_i= 1,\\, v_1,\\dotsc,v_n\\in S\\}.\n\\]\nNote: if $w\\in v+L$, then $w-v\\in L$, so $w+L = v+L$. \n\n\\item[Affine dependence:]A set of vectors $\\{v_1,\\dotsc,v_n\\}$ are \\defn{affinely dependent}[affine!dependence] if \n\\[\n\\alpha_1 v_1 + \\dotsb + \\alpha_n v_n = 0\n\\]\nfor some $\\sum_{i=1}^n\\alpha_i=0$ with some $\\alpha_i \\neq 0$. This is equivalent to one of the vectors belongs to the affine hull of the other vectors.\n\\begin{remark}\nThe maximum number of affinely independent vectors in $\\R^d$ is $d+1$. Why? Any $d+2$ vectors in $\\R^d$ are affinely dependent. Suppose we have $v_1,\\dotsc,v_{d+2}$. Then WLOG $v_1,\\dotsc,v_d$ form a basis of $\\R^d$. Then $\\alpha_1 v_1 + \\dotsb + \\alpha_d v_d = v_{d+1}$ for some choice of $\\alpha_i$'s. Likewise, $\\beta_1 v_1 +\\dotsb + \\beta_d v_d = v_{d+2}$. Let $\\alpha = \\sum_i \\alpha_i$ and $\\beta = \\sum_i \\beta_i$. So\n\\begin{align*}\t\n\\alpha_1 v_1 + \\dotsb + \\alpha_d v_d - v_{d+1} &= 0\\\\\n\\beta_1 v_1 + \\dotsb + \\beta_d v_d - v_{d+2} &= 0.\n\\end{align*}If $\\alpha=1$ or $\\beta=1$, then we are done. Otherwise,\n\\begin{align*}\t\n(\\beta-1)(\\alpha_1 v_1 + \\dotsb + \\alpha_d v_d - v_{d+1}) - (\\alpha-1)(\\beta_1 v_1 + \\dotsb + \\beta_d v_d - v_{d+2} ) &= 0\n\\end{align*}\nwhich demonstrates affine dependence, as the sum of coefficients is $(\\beta-1)(\\alpha-1) - (\\alpha-1)(\\beta-1)=0$.\n\\end{remark}\n\n\n\\item[Convex:] A set $C\\subset \\R^d$ is \\defn{convex} if for any two points $v_1,v_2\\in C$, the interval joining these two points lies in $C$. That is, if $v_1,v_2\\in C$, then $\\{\\alpha v_1 + (1- \\alpha)v_2: 0\\leq \\alpha \\leq 1\\} \\subset C$.\n\\item[Convex hull:] For $X\\subset \\R^d$, the \\defn{convex hull}[convex!hull] $\\conv(X)$ is the intersection of all convex sets containing $X$. Then\n\\[\n\\conv(X) = \\{\\alpha_1 v_1 + \\dotsc + \\alpha_n v_n:n\\in \\N,\\, v_1,\\dotsc,v_n\\in X, \\sum_i \\alpha_i =1, \\alpha_i\\geq 0 \\}\n\\]\nNote that $\\alpha_1 v_1 + \\dotsc + \\alpha_n v_n$ where $\\alpha_i\\geq 0$ and $\\sum_i \\alpha_i=1$ is called a \\defn{convex combination}[convex!combination] of $v_1,\\dotsc,v_n$.\n\nTo prove this formula, we note by induction on $n$ we see that any convex combination of $n$ vectors in $X$ is in $\\conv(X)$: wlog $\\alpha_n\\neq 0$, $\\alpha_n\\neq 1$. Then \n\\[\n\\alpha_1 v_1 + \\dotsc + \\alpha_n v_n = (1- \\alpha_n) \\underbrace{\\left[ \\frac{\\alpha_1}{1- \\alpha_n}v_1 + \\dotsc + \\frac{\\alpha_{n-1}}{1 - \\alpha_n} v_{n-1} \\right]}_{\\in \\conv(X)} + \\alpha_n \\underbrace{ v_n.}_{\\in \\conv(X)}\n\\]\nSo this set of convex combinations is a subset of $\\conv(X)$. But since the set of convex combinations is itself convex, as it is easy to check, we have equality.\n\n\\item[Convex dependence:]We say $X\\subset \\R^d$ is \\defn{convex dependent}[convex!dependence] if for some $x\\in X$ we have $x\\in \\conv(X\\setminus \\{x\\})$. \\marginnote{That is, one vector in $X$ is a convex combination of some of the others.}\n\nWe say $X$ is \\defn{convex independent}[convex!independence] if it is not convex dependent.\n\n\\begin{remark}\nNo point on the boundary of a circle is a convex combination of the other points, so there is no bound on the size of convex independent sets.\n\\end{remark}\n\n\\end{itemize}\n\n\n\\begin{theorem}[\\cite{Radon1921}] \nLet $A\\subset \\R^d$ be finite, with $|A| \\geq d+2$. Then there exist disjoint $A_1,A_2\\subset A$ such that $\\conv(A_1)\\cap \\conv(A_2)\\neq 0$.\n\n\\end{theorem}\n\\begin{remark}\nLet's first consider the case $d=2$, $|A| =4$. If three points do not fall on a line (a degenerate case), then there are two cases: \n\\begin{figure}\n\\hfill\n\\begin{tikzpicture}[scale=1.5]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (1.5,0);\n\\coordinate (c) at (.75,1);\n\n\\coordinate (d) at (0.5,.3);\n% \\draw[green] (a) -- (d) --;\n% \\draw[blue] (b) -- (c);\n\\filldraw[LightBlue] (a) -- (b) -- (c) -- cycle;\n\\filldraw[black] (a) circle (1pt);\n\\filldraw[black] (b) circle (1pt);\n\\filldraw[black] (c) circle (1pt);\n\\filldraw[black] (d) circle (1pt);\n\n\\end{tikzpicture}\n\\hfill\n\\begin{tikzpicture}[scale=1.5]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (0,1);\n\\coordinate (c) at (1.25,0);\n\\coordinate (d) at (1.25,1);\n\\draw[green] (a) -- (d);\n\\draw[blue] (b) -- (c);\n\n\\filldraw[black] (a) circle (1pt);\n\\filldraw[black] (b) circle (1pt);\n\\filldraw[black] (c) circle (1pt);\n\\filldraw[black] (d) circle (1pt);\n\n% \\node at (a) {\\textbullet};\n\n\\end{tikzpicture}\n\\hfill\n\\caption{Four points in $\\R^2$, when no three form a line. Either one is in the convex hull of the other three (left), or we may divide into pairs so that the convex hulls intersect (right).}\\label{fig:Radon_example_d2}\n\\end{figure}\nIn either case, we may verify Radon's theorem.\n% \\missingfigure{BLue Triangle with dot in middle}\n\\end{remark}\n\n\\begin{proof}\t\nWLOG, we may take $|A|=d+2$; any surplus points we could put in $A_1,A_2$, or neither, without changing the result. Set $A= \\{v_1,v_2,\\dotsc,v_{d+2}\\}$. Since $v_1,v_2,\\dotsc,v_{d+2}$ are affinely dependent\\sidenote{since there are more than $d+1$ of them}, there exist $\\alpha_1,\\dotsc,\\alpha_{d+2}$ with $\\sum_i \\alpha_i=0$ not all zero such that\n\\[\n\\alpha_1v_1 + \\dotsb + \\alpha_{d+2} v_{d+2} = 0.\n\\]\nLet $A_1 = \\{v_i : \\alpha_i > 0\\}$ and $A_2 = \\{ v_i: \\alpha_i < 0\\}$. WLOG, $\\alpha_1,\\dotsc,\\alpha_k > 0$ and $\\alpha_{k+2},\\dotsc,\\alpha_{d+2} < 0$. Let $s = \\alpha_1 + \\dotsb + \\alpha_k = - \\alpha_{k+1} - \\dotsb - \\alpha_{d+2}$.\nThen\n\\begin{align*}\t\n\\left( \\frac{\\alpha_1}{s} \\right) v_1 + \\dotsb + \\left( \\frac{\\alpha_k}{s} \\right) v_{k+1} = \\left( - \\frac{\\alpha_{k+1}}{s} \\right)v_{k+1} + \\dotsb + \\left( - \\frac{\\alpha_{d+2}}{s} \\right)v_{d+2}.\n\\end{align*}\nBut as shown by the LHS, this quantity lies in $\\conv(A_1)$, and as shown by the RHS, the quantity lies in $\\conv(A_2)$.\n\\end{proof}\n\n\\begin{theorem}[\\cite{Caratheodory1911}]\nEvery point in $\\conv(X)$ for $X\\subset \\R^d$ is a convex combination of at most $d+1$ points in $X$.\n\\end{theorem}\n\\begin{proof}\t\nLet $x\\in \\conv(X)$ be written $x = \\alpha_1 v_1 + \\dotsb + \\alpha_n v_n$ for $v_1,\\dotsc,v_n\\in X$ and $\\sum_i \\alpha_i=0$ with $\\alpha_i\\geq 0$ not all zero. Suppose is $n$ is chosen minimally such that a convex combination exists. Suppose $n\\geq d+2$ for the sake of contradiction. Then by Radon's theorem, wlog\n\\[\n\\conv(\\{v_1,\\dotsc,v_k\\})\\cap \\conv(\\{v_{k+1},\\dotsc, v_n \\}).\n\\]\nSo, for some $\\beta_1,\\dotsc,\\beta_n\\geq 0$,\n\\[\n\\beta_1 v_1 + \\dotsb \\beta_k v_k = \\beta_{k+1} v_{k+1} + \\dotsb + \\beta_n v_n\n\\]\nwith $\\sum_{i=1}^k \\beta_i = \\sum_{i=k+1}^n \\beta_i = 1$. Then\n\\vspace{-\\baselineskip}\n\\begin{fullwidth}\n\\begin{align}\t\nx &= (\\alpha_1 + \\epsilon \\beta_1) v_1 + \\dotsb + (\\alpha_k + \\epsilon \\beta_k)v_k + (\\alpha_{k+1} - \\epsilon \\beta_{k+1})v_{k+1}+ \\dotsb + (\\alpha_n - \\epsilon \\beta_n)v_n \\tag{$\\star$} \\label{eq:Cara_proof}\n\\end{align}\n\\end{fullwidth}\nhas the sum of coefficients one for every $\\epsilon$. If $\\epsilon>0$ is minimally such that $\\alpha_i - \\epsilon \\beta_i=0$ for some $i$, then the expression \\eqref{eq:Cara_proof} is a convex combination of $\\{v_1,\\dotsc,v_n\\}\\setminus \\{v_i\\}$, giving a contradiction.\n\\end{proof}\n\\begin{theorem}[\\cite{Helly1923}]\nLet $C_1,C_2\\dotsc,C_n$ be a collection of convex sets in $\\R^d$. If $\\bigcap_{i\\in I} C_i \\neq \\emptyset$ for every $I\\subset [n]$ with $|I| = d+1$, then $\\bigcap_{i=1}^n C_i \\neq \\emptyset$.\n\\end{theorem}\n\\begin{remark}\nFor $\\R^1$, if we consider \\marginnote{Convex sets in $\\R^1$ may be open or closed intervals or rays, but we will take closed intervals for simplicity.}$C_i = [a_i,b_i]$, then if each $C_i\\cap C_j\\neq \\emptyset$, then we need $a_i \\leq b_j$ for each $i,j$. Then $[\\max_i a_i, \\min_i b_i] \\subset C_k$ for all $k$.\n\\end{remark}\n\\begin{proof}[Proof by induction on $n$.] \nLet us postpone the base case, $n=d+2$.\nFor the induction step, let $C_1' = C_1 \\cap C_n$, \\ldots, $C_{n-1}' = C_{n-1} \\cap C_n$. Then by the base case, any $(d+1)$ sets in $\\{C_1',\\dotsc,C_{n-1}'\\}$ have a non-empty intersection. Therefore they all intersect by the induction hypothesis, hence\n\\[\n\\bigcap_{i=1}^n C_i = \\bigcap_{i=1}^{n-1}C_i' \\neq \\emptyset.\n\\]\nSo let us show the base case. Let $v_i \\in \\bigcap_{j\\neq i} C_j$; by the assumption. Let $A = \\{v_1,\\dotsc,v_{d+2}\\}$. Then by Radon's theorem, there exist disjoint $A_1,A_2\\subset A$ such that $\\conv(A_1)\\cap \\conv(A_2)\\neq \\emptyset$. Consider $v\\in \\conv(A_1)\\cap \\conv(A_2)$. We will show that $v\\in C_i$ for every $i$.\n\nSuppose wlog that $v_i \\in A_2$. Then $A_1 \\subset C_i$. Since $C_i$ is convex, then $\\conv(A_1) \\subset C_i$. So $v\\in C_i$, as desired. See \\cref{fig:Helly_proof} for an example of this process in $\\R^2$. \\qedhere\n% \\missingfigure{Draw four disks in $\\R^2$, $C_1,\\dotsc,C_4$. Draw the $v_i$ in all the intersections but $v_i$. Connect $v_1,v_3$ in red $v_2,v_4$ in blue. Then let $v$ be the intersection. $v$ is in all four; it is in $C_1$ because it's in the interval between $v_2,v_4 \\in C_1$. Etc.}\n\\begin{figure}\n\\begin{center}\n\\begin{tikzpicture}[circ/.style={circle,minimum size=\\circdiam cm,draw,opacity=.7}]\n\n\n\\def\\circposrad{1}\n\\def\\circdiam{3}\n\n\\node[circ,label={right:$C_1$}] at (0:\\circposrad cm) {};\n\\node[circ,label={above:$C_2$}] at (90:\\circposrad cm) {};\n\\node[circ,label={left:$C_3$}] at (180:\\circposrad cm) {};\n\\node[circ,label={below:$C_4$}] at (270:\\circposrad cm) {};\n\n\\coordinate (v3) at (0:.7cm);\n\\coordinate (v4) at (90:.7cm);\n\\coordinate (v1) at (180:.7cm);\n\\coordinate (v2) at (270:.7cm);\n\n\\draw[red] (-.7,0) -- (.7,0);\n\\draw[blue] (v2) -- (v4);\n\n\\filldraw[black] (v1) circle (1.65pt) node[label=$v_1$]{};\n\\filldraw[black] (v2) circle (1.65pt) node[label=right:$v_2$]{};\n\\filldraw[black] (v3) circle (1.65pt) node[label=$v_3$]{};\n\\filldraw[black] (v4) circle (1.65pt) node[label=above:$v_4$]{};\n\n\\draw (0,0) circle (1.75pt);% node[label={[label distance=.02cm]north east:$v$}]{};\n\\end{tikzpicture}\n\\end{center}\n\\caption{Helly's theorem in $\\R^2$. Given four convex sets $C_1,\\dots,C_4$ such that every triple has a non-empty intersection $v_i \\in \\bigcap_{j\\neq i} C_j$, we can use Radon's theorem to divide $\\{v_1,v_2,v_3,v_4\\}$ into two disjoint sets with non-empty convex hull. Here, $A_1 = \\{v_1,v_3\\}$ and $A_2 = \\{v_2,v_4\\}$. The point in the intersection $\\conv(A_1)\\cap \\conv(A_2)$ is labelled here with an open circle.} \\label{fig:Helly_proof}\n\\end{figure}\n\\end{proof}\n\n\\begin{remark}\nHelly's theorem may not apply to infinite collections of sets; consider $\\{[n,+\\infty):n\\in \\N\\}$.\n\\end{remark}\n\nFor any set of $n$ points on the line, we can find a point such that there are at least $n/2$ points above it and below it. This is the median. How do we find an analog for $\\R^d$? We'd like to find a point that in any direction away from this point, there are still many points in our set.\n\n\nLet us generalize to $\\R^d$: Given $X\\subset \\R^d$, $|X|=n$, the \\defn{centerpoint} of $x$ is a point $m\\in \\R^d$ such that for every closed halfspace $H\\subset \\R^d$ such that $m\\in H$, then $H$ contains at least $\\frac{n}{d+1}$ points in $X$.\n\n\\begin{theorem}\nFor every finite set $X\\subset \\R^d$, there exists a centerpoint.\n\\end{theorem}\n\\begin{proof}\t\nNote the following are all equivalent:\n\\begin{itemize}\n\\item $m$ is a centerpoint\n\t\\item For every closed halfspace $H\\subset \\R^d$, if $m\\in H$ then $|H\\cap X| \\geq \\frac{n}{d+1}$.\n\t\\item For every closed halfspace $H\\subset \\R^d$, if $|X\\cap H| < \\frac{n}{d+1}$, then $m\\not \\in H$.\n\t\\item For every closed halfspace $H\\subset \\R^d$, if $|X\\cap H^c| > n - \\frac{n}{d+1} = \\frac{dn}{d+1}$, then $m\\in H^c$.\n\t\\item For every open halfspace $H\\subset \\R^d$, if $|X\\cap H| > \\frac{dn}{d+1}$, then $m\\in H$.\n\\end{itemize}\n% Alternatively, every open half space which does not contain $m$ must contain at most $$\nConsider the family \n\\begin{align*}\t\n\\H = \\{H: H \\text{ is an open halfspace of }\\R^d, |H\\cap X| > \\frac{dn}{d+1}\\}.\n\\end{align*}\nWe will show for some $m \\in \\R^d$, we have $m\\in H$ for every $H\\in \\H$, proving that $m$ is a centerpoint. \nFirst, if $H_1,H_2,\\dotsc,H_{d+1}\\in \\H$, then\n\\begin{align*}\t\nH_1\\cap H_2\\cap \\dotsm \\cap H_{d+1}\\neq \\emptyset.\n\\end{align*}\nOtherwise, every point of $X$ belongs to $\\leq d$ out of $d+1$ of these half spaces, and thus\n\\begin{align*}\t\ndn = \\sum_{i=1}^{d+1} \\frac{dn}{d+1} < \\sum_{i=1}^{d+1} |H_i \\cap X| \\leq d n\n\\end{align*}\nwhich is a contradiction.\n\nWe would like to now apply Helly's theorem, but we have an infinite collection $\\H$ instead of a finite one.\n\nLet us instead set $\\H' = \\{ \\conv(Y): Y\\subset X, |Y| > \\frac{dn}{d+1}\\}$. By the previous argument and Helly's theorem, there exists $m\\in \\R^d$ such that $m\\in C$ for every $C\\in \\H'$. But this suffices. For any $H\\in \\H$ there is $Y\\subset X$ with $|Y| > \\frac{dn}{d+1}$ and $Y\\subset H$. Then $\\conv(Y) \\subset H$ since $H$ is convex. But since $m\\in \\conv(Y)$, we have $m\\in H$, as desired.%if $m\\not \\in H$, then $m\\not \\in \\conv(Y)$ which is a contradiction.\n% If $m\\not \\in H$, then we want that $H$ contains\n\\end{proof}\n\n\n\n\\begin{theorem}[Birch]\nLet $X\\subset \\R^2$ be a collection of $3n$ points. Then $X$ can be partitioned into $n$ triples such that the corresponding triangles all share a point in common.\n\\end{theorem}\n\n\\begin{figure}\n\\begin{center}\n\\begin{tikzpicture}\n\\begin{scope}[scale=.6]\n\\coordinate (v1) at (0:3cm);\n\\coordinate (v9) at (20:2cm);\n\\coordinate (v6) at (130:1cm);\n\\coordinate (v3) at (240:2.3cm);\n\\coordinate (v5) at (160:3.3cm);\n\\coordinate (v4) at (290:.3cm);\n\\coordinate (v7) at (80:1.3cm);\n\\coordinate (v2) at (340:1.23cm);\n\\coordinate (v8) at (45:3cm);\n\\foreach \\n in {1,2,3,4,5,6,7,8,9}\n{\n\t\\filldraw (v\\n) circle (1.65pt);\n}\n\\end{scope}\n\\node at (3cm,0){$\\leadsto$};\n\\begin{scope}[scale=.6, xshift=10cm]\n\\coordinate (v1) at (0:3cm);\n\\coordinate (v9) at (20:2cm);\n\\coordinate (v6) at (130:1cm);\n\\coordinate (v3) at (240:2.3cm);\n\\coordinate (v5) at (160:3.3cm);\n\\coordinate (v4) at (290:.3cm);\n\\coordinate (v7) at (80:1.3cm);\n\\coordinate (v2) at (340:1.23cm);\n\\coordinate (v8) at (45:3cm);\n\\foreach \\n in {1,2,3,4,5,6,7,8,9}\n{\n\t\\filldraw (v\\n) circle (1.65pt);\n}\n\n\\filldraw[blue,opacity=.4] (v7) -- (v4) -- (v1) -- cycle;\n\n\\filldraw[green,opacity=.4] (v2) -- (v5) -- (v8) -- cycle;\n\n\\filldraw[red,opacity=.4] (v3) -- (v6) -- (v9) -- cycle;\n\\end{scope}\n\n\n\\end{tikzpicture}\n\\end{center}\n\\caption{An example of Birch's theorem with nine points. On the left, the nine points are depicted; on the right, intersecting triangles chosen.}\\label{fig:birch_ex}\n\\end{figure}\n% \\missingfigure{Draw 9 points and three triangles in different colors so that there is a common area. Label a point of intersection.}\n\\begin{remark}\nSee \\cref{fig:birch_ex} for an example.\n\\end{remark}\\begin{proof}\n% \\missingfigure{Draw rays from centerpoint to each vertex}\n\nLet $m$ be a centerpoint of $X$. Label the points of $X$ by $x_1$, $x_2$, \\ldots, $x_{3n}$ so that the rays $mx_1$, $mx_2$, \\ldots, $mx_{3n}$ are arranged around $m$ in clockwise order, as shown in \\cref{fig:birch_proof1}.\n\\begin{figure}[ht]\n\\begin{center}\n\\begin{tikzpicture}\n\\begin{scope}[scale=1]\n\\coordinate (m) at (.7cm,.3cm);\n\\coordinate (v1) at (0:3cm);\n\\coordinate (v9) at (20:2cm);\n\\coordinate (v6) at (130:1cm);\n\\coordinate (v3) at (240:2.3cm);\n\\coordinate (v5) at (160:3.3cm);\n\\coordinate (v4) at (290:.3cm);\n\\coordinate (v7) at (80:1.3cm);\n\\coordinate (v2) at (340:1.23cm);\n\\coordinate (v8) at (45:3cm);\n\\foreach \\n in {6,7}\n{\n\t\\filldraw (v\\n) circle (1.65pt) node[above]{$x_\\n$};\n}\n\\foreach \\n in {4,5}\n{\n\t\\filldraw (v\\n) circle (1.65pt) node[left]{$x_\\n$};\n}\n\\foreach \\n in {1,8,9}\n{\n\t\\filldraw (v\\n) circle (1.65pt) node[right]{$x_\\n$};\n}\n\\foreach \\n in {2,3}\n{\n\t\\filldraw (v\\n) circle (1.65pt) node[below]{$x_\\n$};\n}\n\n% \\filldraw[blue,opacity=.4] (v5) -- (v4) -- (v1) -- cycle;\n\n% \\filldraw[green,opacity=.4] (v3) -- (v9) -- (v8) -- cycle;\n\n% \\filldraw[red,opacity=.4] (v6) -- (v7) -- (v2) -- cycle;\n\n\t\\foreach \\n in {1,2,...,9}\n{\n\t\\draw[opacity=.7] (m) -- (v\\n);\n}\n\t\\filldraw[red] (m) circle (1.75pt) node[black,right]{$m$};\n\n\\end{scope}\n\\end{tikzpicture}\n\\end{center}\n\\caption{Nine points $\\{x_1,\\dotsc,x_9\\}$ in $\\R^2$ and their centerpoint $m$, in red. Continuing the example from \\cref{fig:birch_ex}, the points are labelled clockwise in order from the centerpoint. }\\label{fig:birch_proof1}\n\\end{figure}\nLet $X_i = \\{ x_i, x_{i+n}, x_{i+2n}\\}$ for $i=1,\\dotsc, n$. Then $m\\in \\conv(X_i)$ for each $i$.\n\nAssume $m\\not \\in \\conv(X_i)$. Then\\sidenote{We are appealing to a separating hyperplane theorem, such as \\cite[Theorem 1.24]{matouvsek2002lectures}, although the result seems clear geometrically in this case.} some closed halfplane containing $m$ separates $m$ from the three points of $X_i$; see \\cref{fig:birch_proof2}. But then the $2n$ rays between $x_i$ and $x_{i+2n}$ fall in the other half of the halfplane, so there are at least $2n+1$ points in $X$ on the other side, contradicting that $m$ is a centerpoint. \\qedhere\n\\begin{marginfigure}\n\\begin{center}\n\\begin{tikzpicture}\n\\coordinate (v3) at (240:2.3cm);\n\\coordinate (v2) at (100:1.5cm);\n\\coordinate (v1) at (130:2cm);\n\\coordinate (m) at (.7cm,.3cm);\n\\foreach \\n in {1,2,3}\n{\n\t\\draw[opacity=.7] (m) -- (v\\n);\n}\n\t\\filldraw (v3) circle (1.65pt) node[left]{$x_i$};\n\t\\filldraw (v2) circle (1.65pt) node[above]{$x_{i+2n}$};\n\t\\filldraw (v1) circle (1.65pt) node[above]{$x_{i+n}$};\n\n\n\\filldraw[blue,opacity=.4] (v1) -- (v2) -- (v3) -- cycle;\n\t\\filldraw[red] (m) circle (1.75pt) node[black,right]{$m$};\n\n\\end{tikzpicture}\n\n\\end{center}\n\\caption{The case where the centerpoint $m$, in red, is not in the convex hull of $X_i$, shown in blue. Then the $2n$ points $\\{x_i,x_{i+1},\\dotsc,x_{i+2n}$ all lie between $x_i$ and $x_{i+2n}$ (angularly) and may be separated from $m$ by a closed hyperplane, contradicting the definition of centerpoint.} \\label{fig:birch_proof2}\n\\end{marginfigure}\n% \\missingfigure{$m$, then line, and three points $x_i$, $x_{i+n}$, $x_{i+2n}$. Draw outline convex hull in red around those three. They are all on the other side of a halfplane from $m$.}\n\n\\end{proof}\n\n\\begin{theorem}[Colorful Carath\\'eodory theorem, \\cite{barany1982generalization}]\nLet $S_1,S_2,\\dotsc,S_{d+1}\\subset \\R^d$ and suppose that $x\\in \\bigcap_{i=1}^{d+1} \\conv(S_i)$. Then there exist $x_1\\in S_1$, $x_2\\in S_2$, \\ldots, $x_{d+1}\\in S_{d+1}$ such that $x\\in \\conv(\\{x_1,x_2,\\dotsc,x_{d+1}\\})$.\n\\end{theorem}\n\\begin{remark}\nThe different sets correspond to different colors. Then $x$ belongs to a convex combination of points each a different color. Taking each set $S_i \\equiv X$, we recover Carath\\'eodory's theorem.\n\\end{remark}\n\\begin{proof}\t\nWe will say $C$ is a \\defn{colorful simplex} if $C = \\conv(\\{ x_1,x_2,\\dotsc,x_{d+1} \\})$ for $x_i\\in S_i$ for $i\\in[d+1]$. Suppose that no colorful simplex contains $x$.\n\nChoose a colorful simplex $C$ such that $\\dist(C,x) = \\min_{c\\in C} \\|c-x\\|$ is minimal.\\marginnote{$\\|c-x\\| = \\dist(c,x)$ is the 2-norm (Euclidean norm).}\n\nLet $z\\in C$ be the closest point to $x$: $\\dist(z,x) = \\dist(C,x)$.\n\n\nLet $H$ be a hyperplane orthogonal to $zx$ through $z$. Let $H^+$ and $H^-$ be closed halfspaces with respect to $H$ such that $x\\in H^-$.\n\n\\begin{claim}\n$C\\subset H^+$.\n\\end{claim}\n\\begin{subproof}\t\nSuppose not: $z' \\in C\\cap (H^- \\setminus H^+)$\\marginnote{So $z'\\not \\in H^+$}. Then angle $z' z x$ is acute and points on $zz'$ near $z$ are closer to $x$ than $z$, a contradiction.\n\\end{subproof}\n\n\\begin{claim}\n$z\\in \\conv(\\{x_1,\\dotsc,x_{d+1}\\}\\cap H)$.\n\\end{claim}\n\\begin{subproof}\t\n$z = \\sum_{i=1}^{d+1} \\lambda_i x_i$ with $\\lambda_i\\geq 0$, $\\sum_i \\lambda_i =1$, since $z\\in C$. Let $f: \\R^d \\to \\R$ linear such that if $f(p)> 0$ for $p\\in H^+\\setminus H$, and $f(p) = 0$ for $p\\in H$. \n\nThen $0 = f(z) = \\sum_{x_i \\in H^+ \\setminus H} \\lambda_i \\underbrace{f(x_i)}_{>0} + \\sum_{x_i \\in H} \\lambda_i \\underbrace{f(x_i)}_{=0}$. So we must have $\\lambda_i=0$ when $x_i \\in H^+ \\setminus H$.\n\\end{subproof}\n\nBy Caratheordy theorem applied to $z$ and $\\{x_i,\\dotsc,x_{d+1}\\}\\cap H$, there exists $j$ such that \\begin{align*}\t\nz\\in \\conv((\\{x_1,x_2,\\dotsc,x_{d+1}\\}\\setminus \\{j\\})\\cap H).\n\\end{align*}\nWe know $x\\in \\conv(S_j)$, so there exists $x'_j \\in S_j \\cap (H^-\\setminus H^+)$. Let $C' = \\conv( \\{x_1,\\dotsc,x_{j-1},x_j',x_{j+1},\\dotsc,x_{d+1}\\})$.\n\n\\begin{claim}\nThen $\\dist(C',x) < \\dist(C,x)$.\n\\end{claim}\n\\begin{subproof}\t\nWe know $z\\in C'$. Then, as before, points on $x_j' z\\subset C'$ are closer to $x$ than $z$. \n\\end{subproof}\nThis last claim yields a contradiction to minimality of $C$.\n\nNote: we assumed the $S_i$ were finite when assuming there exists a minimal colorful simplex $C$. But in fact, we only need to consider $\\conv(S_i)$, which only depends on $d+1$ points, by (the original) Carath\\'eodory points. Thus, we could take $|S_i|=d+1$ for each $i$.\n\\end{proof}\n\\begin{remark}\nThis proof yields an algorithm to improve our convex set one point at a time to the optimal one. Yet, it is unknown if this algorithm or any other can find the optimal $\\conv(\\{x_1,\\dotsc,x_{d+1}\\})$ in polynomial time.\n\\end{remark}\n\n\\begin{theorem}[\\cite{Tverberg1966}]\nLet $A\\subset \\R^d$ with $|A| \\geq (r-1)(d+1)+1$. Then there exist $A_1,A_2,\\dotsc,A_r\\subset A$ pairwise disjoint such that $\\bigcap_{i=1}^r \\conv(A_i)\\neq \\emptyset$.\n\\end{theorem}\n\\begin{remark}\nRadon's theorem is the case $r=2$. If $d=1$, then we have $2r-1$ points in $\\R$ which we wish to write as $r$ groups with intersecting convex hulls.\n\nFor $d=2$, $3r-2$ points in $\\R^2$ can be partitioned into $r$ groups with intersecting convex hulls. This in fact implies Birch's theorem.\n\\end{remark}\n\nIf $u\\in \\R^n$ and $v\\in \\R^m$, then $u\\otimes v \\in \\R^{nm}$. Suppose $u = (u_1,\\dotsc,u_n)$ and $v = (v_1,\\dotsc,v_m)$; i.e., we have chosen bases for $\\R^n$ and $\\R^m$. Then $u\\otimes v $ can be thought of as a $m\\times n$ matrix with components $(u\\otimes v)_{ij} = u_i v_j$.\n\\begin{align*}\t\nu = \\begin{pmatrix}\nu_1\\\\ u_2\\\\ u_3\n\\end{pmatrix}, \\qquad v = \\begin{pmatrix}\nv_1 \\\\ v_2 \\\\ v_3\n\\end{pmatrix}, \\qquad u\\otimes v = \\begin{pmatrix}\nu_1 v_1 & u_1 v_2 & u_1 v_3\\\\\nu_2 v_1 & u_2 v_2 & u_2 v_3 \\\\\nu_3 v_1 & u_3 v_2 & u_3 v_3\n\\end{pmatrix}.\n\\end{align*}\n\n\\begin{proposition}\n~\\begin{enumerate}\n\\item $(\\alpha_1 u_1 + \\alpha_2 u_2) \\otimes v = \\alpha_1 (u_1\\otimes v) + \\alpha_2 (u_2\\otimes v)$\n\\item Suppose $v_1,\\dotsc,v_k$ are linearly independent and $u_1\\otimes v_1 + u_2\\otimes v_2 + \\dotsm + u_k \\otimes v_k = 0$. Then $u_1 = u_2 = \\dotsm = u_k = 0$.\n\\end{enumerate}\n\\end{proposition}\nWe will leave the proof of the proposition as an exercise, and proceed to the proof of Tverberg's theorem.\n\\lect{3}{23}\n% \\marginnote{Lecture X: March 23, 2016.}\n\n\\begin{proof}[Proof (\\cite{sarkaria1992tverberg}).]\nLet $m = (r-1)(d+1)+1$.\n\n\\begin{enumerate}\n\t\\item Instead of the orginal setting, consider $A = \\{v_1,v_2,\\dotsc,v_m\\} \\subset \\R^{d+1}$ such that the vectors of $A$ lie in an affine hyperplane not passing through the origin. I.e., there exists $f: \\R^{d+1}\\to \\R$ linear such that $f(v_i) = 1$ for every $i$. \n\t\\item Let $w_1,w_2,\\dotsc,w_r$ be vectors in $\\R^{r-1}$ such that $w_1+w_2+\\dotsc+w_r=0$ and this is essentially the only relation between these vectors. For instance, we could choose $w_i = (\\underbrace{0,\\dotsc,0}_{i-1},1,0,\\dotsc,0)$ for $i=1,\\dotsc,r-1$ and $w_r = (-1,-1,\\dotsc,-1)$.\n\t\\item Consider $\\{v_i \\otimes w_j: 1\\leq i \\leq m, 1\\leq j \\leq r\\}\\subset \\R^{d+1}\\otimes \\R^{r-1}\\cong \\R^{m-1}$. For each $i$, we may think of $S_i:=\\{v_i\\otimes w_j:1\\leq j\\leq r\\}$ as a copy of the $w_j$ embedded in a hyperplane. Note $0\\in \\conv(S_i)$ for each $i$, since\n\t\\begin{align*}\t\n\t0 &= \\frac{1}{r}v_i \\otimes (w_1+w_2+\\dotsm+ w_r) = \\frac{1}{r} v_i \\otimes w_1 + \\frac{1}{r}v_i\\otimes w_2 +\\dotsm + \\frac{1}{r}v_i\\otimes w_r.\n\t\\end{align*}\n\t\\item Apply the Colorful Carath\\'eodory theorem to $S_1,\\dotsc,S_m$ and the point $0$. We get that for $1\\leq i\\leq m$ there exist $j_i$ and $\\lambda_i\\geq 0$ such that\n\t\\begin{align}\t\n\t\\sum_{i=1}^m \\lambda_i (v_i\\otimes w_{j_i})=0 \\label{eq:Tv_conv_comb}\n\t\\end{align}\n\twith $\\sum_i \\lambda_i=1$.\n\n\tLet $A_j = \\{ v_i : j_i = j \\}$ for $j=1,\\dotsc,r$. Let us rewrite \\cref{eq:Tv_conv_comb} as follows.\n\t\\begin{align*}\t\n\t\\sum_{j=1}^r \\left( \\sum_{v_i \\in A_j} \\lambda_i v_i \\right)\\otimes w_j = 0.\n\t\\end{align*}\n\tLet us write $u_j = \\sum_{v_i \\in A_j} \\lambda_i v_i$. Then we have\n\t\\begin{align*}\t\n\tu_1\\otimes w_1 + u_2\\otimes w_2 + \\dotsm + u_r\\otimes w_r &= 0\\\\\n\t(u_1- u_r)\\otimes w_1 + (u_2-u_r)\\otimes w_2 + \\dotsm + (u_{r-1} - u_r)\\otimes w_{r-1} &= 0\n\t\\end{align*}\n\tusing $w_r = -w_1 - w_2 - \\dotsm - w_{r-1}$. But since $\\{w_1,\\dotsc,w_{r-1}\\}$ are linearly independent, we must have $u_1-u_r = u_2 - u_r = \\dotsm = u_{r-1}- u_r = 0$. That is, \n\t\\[\n\tu_1 = u_2 = \\dotsm = u_r.\n\t\\]\n\tSubstituting the definition of $u_j$,\n\t\\begin{align}\t\n\t\\sum_{v_i \\in A_1} \\lambda_i v_i = \\sum_{v_i\\in A_2} \\lambda_i v_i = \\dotsm = \\sum_{v_i\\in A_r}\\lambda_i v_i. \\label{eq:Tv_lambda_vs_equal}\n\t\\end{align}\n\tSuppose the sum of coefficients $\\lambda_i$ in each expression is the same and is equal to $s$.\\marginnote{In fact, $s= \\frac{1}{r}$.} Then\n\t\\begin{align*}\t\n\t\\sum_{v_i \\in A_1} \\frac{\\lambda_i}{s}v_i = \\dotsm = \\sum_{v_i\\in A_r} \\frac{\\lambda_i}{s}v_i = p\n\t\\end{align*}\n\tand hence $p\\in \\conv(A_k)$.\n\n\tBut if we apply $f$ to \\cref{eq:Tv_lambda_vs_equal}, by linearity we obtain\n\t\\begin{align*}\t\n\t\\sum_{v_i \\in A_1} \\lambda_i f(v_i) = \\sum_{v_i\\in A_2} \\lambda_i f(v_i) = \\dotsm = \\sum_{v_i\\in A_r}\\lambda_i f(v_i). \n\t\\end{align*}\n\tSince $f(v_i)=1$ for all $i$, we find $\\sum_{v_i\\in A_j} \\lambda_i \\equiv s$. \\qedhere\n\\end{enumerate}\n\\end{proof}\n\\flavor{A magical proof. 30 years after the original.}\n\n\\begin{remark}\nIn Helly's theorem, we consider $C_1,\\dotsc,C_n\\subset \\R^d$ convex such that every $(d+1)$-tuple of $C$'s has non-empty intersection. Then we obtain that there is a common point in all of the sets. Suppose instead $\\sim \\frac{1}{2}$ of the $(d+1)$-tuples have a non-empty intersection. What can we say about large intersections?\n\\end{remark}\n\\begin{theorem}[Fractional Helly's theorem]\nFor every $d\\in \\N$ and $0< \\alpha \\leq 1$ there exists a $\\beta = \\beta(\\alpha,d)$ such that the following holds. Let $C_1, C_2,\\dotsc,C_n\\subset \\R^d$ be convex and suppose $\\bigcap_{i\\in I} C_i\\neq \\emptyset$ for at least $\\alpha {n\\choose d+1}$ sets $I\\subset[n]$ with $|I| = d+1$. Then there exists $X \\subset [n]$ with $|X|\\geq \\beta n$ such that $\\bigcap_{i\\in X} C_i \\neq \\emptyset$.\n\n\\end{theorem}\n\\begin{proof}\t\nWe need to assume that $C_1,C_2,\\dotsc,C_n$ are compact; we may do this as follows. For each set $I$ as in the statment, select $p_I \\in \\bigcap_{i\\in I} C_i$. Then replace $C_i$ by $\\conv(\\{p_I: I \\text{ s.t. } I\\ni i\\})$.\n\nLet $F_I = \\bigcap_{i\\in I} C_i$. Let $<$ be a linear lexicographic order on $\\R^d$. Let $p_I= \\min (F_I)$ in this order, if $F_I\\neq \\emptyset$. Since $F_I$ is compact, $p_I$ exists and is unique.\n\n\\begin{claim}\nFor every $I\\subset[n]$ with $|I|=d+1$, s.t. $F_I\\neq \\emptyset$, there exists $J\\subset I$, $|J| = d$ such that $p_I = p_J$.\n\\end{claim}\n\\begin{remark}\nFor any $J\\subset I$, we have $p_J \\leq P_I$.\n\\end{remark}\n\\begin{subproof}[Proof of claim.]\nLet $C = \\{q \\in \\R^d: q< p_I\\}$. Then $C$ is convex. Then $C\\cap \\left( \\bigcap_{i\\in I} C_i \\right) = C\\cap F_I = \\emptyset$. Since this $(d+2)$-tuple intersection is empty, then by the contrapositive of Helly's theorem, not all $(d+1)$-tuples can have empty intersection. But since $F_I\\neq \\emptyset$, the $(d+1)$-tuple with empty intersection must include $C$. So there exists $J\\subset I$ with $|J| = d$ such that\n\\begin{align*}\t\nC \\cap \\left( \\bigcap_{i\\in J}C_i \\right) = \\emptyset.\n\\end{align*}\nThen $p\\geq p_I$ for every $p\\in F_J$.\n\\end{subproof}\n\nNow for every $I\\subset[n]$ with $|I| = d+1$, $F_I\\neq \\emptyset$, select $J\\subset I$ with $|J| = d$ such that $P_J = P_I$. Then there are $\\alpha{n \\choose d+1}$ sets $I$ which we consider, and ${n\\choose d}$ sets $J$ of size $d$. So some set $J_0$ is associated with at least $\\frac{\\alpha {n\\choose d+1}}{{n\\choose d}}$ sets $I$. Such sets are of the form $J_0 \\cup \\{ i\\}$ for some $i$, and $p_{J_0} \\in C_i$ for such $i$. So $p_{J_0}$ belongs to at least \n\\begin{align*}\t\n\\frac{\\alpha{ n\\choose d+1}}{{n\\choose d}} + d = \\frac{\\alpha(n-d)}{d+1}+d \\geq \\frac{\\alpha}{d+1}n\n\\end{align*}\nsets $C_i$ (counting the $d$ sets in $J_0$).\nTherefore, we may take $\\beta = \\frac{\\alpha}{d+1}$.\n\\end{proof}\n\\begin{remark}\nThis constant is not optimal; for instance, when $\\alpha=1$, we would like $\\beta=1$ to recover Helly's theorem. The optimal constant is known, however.\n\\end{remark}\n\n\nFor every $k$, we'd like to prove existence (and estimate the value) of the number $g(k)$ such that in every set of $g(k)$ points in $\\R^2$, with no three lying on a line\\sidenote[][-1cm]{``in general position''} there exist $k$ points which are convexly independent\\sidenote[][-1cm]{``in convex position,'' or are the verticies of a convex polygon}.\n\nLet us consider small $k$. Then $g(1)=1$, $g(2)=2$, $g(3)=3$. But $g(4) > 4$, as shown in \\cref{fig:g4}.\n\\begin{marginfigure}[-1cm]\n\\begin{center}\n\\begin{tikzpicture}[scale=1.5]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (1.5,0);\n\\coordinate (c) at (.75,1);\n\n\\coordinate (d) at (0.5,.3);\n% \\draw[green] (a) -- (d) --;\n% \\draw[blue] (b) -- (c);\n\\filldraw[LightBlue] (a) -- (b) -- (c) -- cycle;\n\\filldraw[black] (a) circle (1pt);\n\\filldraw[black] (b) circle (1pt);\n\\filldraw[black] (c) circle (1pt);\n\\filldraw[black] (d) circle (1pt);\n\n\\end{tikzpicture}\n\\end{center}\n\\caption{An illustration of the fact that $g(4) > 4$. We have four points with no three lying on a line such that not all four are convexly independent: one point is in the convex hull of the other three.} \\label{fig:g4}\n\\end{marginfigure}\n\n\\begin{lemma} \\label{lem:g4_is_5}\n$g(4)=5$.\n\\end{lemma}\n\\begin{proof}\t\nLet $X\\subset \\R^2$ be a set of five points in general position. Consider $\\conv(X)$: this is a polygon with 3, 4, or 5 vertices. If there are 4 or 5 vertices we are done, so, assume $\\conv(X)$ is a triangle with vertices $\\{a,b,c\\}$. Let $\\{d,e\\}$ be the remaining two points of $X$, as shown in \\cref{fig:lem89convX}.\n\\begin{figure}\n\\begin{center}\n\\begin{tikzpicture}[scale=3]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (1.5,0);\n\\coordinate (c) at (.75,1);\n\n\\coordinate (e) at (0.9,.2);\n\\coordinate (d) at (0.7,.4);\n\n% \\draw[green] (a) -- (d) --;\n% \\draw[blue] (b) -- (c);\n\\filldraw[LightBlue] (a) -- (b) -- (c) -- cycle;\n\t\\filldraw[black] (a) circle (1pt) node[label=left:$a$]{};\n\\filldraw[black] (b) circle (1pt)node[label=right:$b$]{};\n\\filldraw[black] (c) circle (1pt)node[label=above:$c$]{};\n\\filldraw[black] (d) circle (1pt)node[label=below:$d$]{};\n\\filldraw[black] (e) circle (1pt)node[label=below:$e$]{};\n\n\\end{tikzpicture}\n\\end{center}\n\\caption{We assume $\\conv(X) = \\conv(\\{a,b,c\\})$; then $\\{d,e\\}:= X\\setminus \\{a,b,c\\}$ lie inside the triangle formed by $\\{a,b,c\\}$.} \\label{fig:lem89convX}\n\\end{figure}\n\nNow assume the line $de$ intersects $ab$ and $ac$, as shown in \\cref{fig:lem89de_line}.\n\\begin{marginfigure}\n\\begin{center}\n\\usetkzobj{all}\n\\begin{tikzpicture}[scale=1.5]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (1.5,0);\n\\coordinate (c) at (.75,1);\n\n\\coordinate (e) at (0.9,.2);\n\\coordinate (d) at (0.6,.4);\n\n% \\draw[green] (a) -- (d) --;\n% \\draw[blue] (b) -- (c);\n\\filldraw[LightBlue] (a) -- (b) -- (c) -- cycle;\n\t\\filldraw[black] (a) circle (1pt) node[label=left:$a$]{};\n\\filldraw[black] (b) circle (1pt)node[label=right:$b$]{};\n\\filldraw[black] (c) circle (1pt)node[label=above:$c$]{};\n\\filldraw[black] (d) circle (1pt)node[label=below:$d$]{};\n\\filldraw[black] (e) circle (1pt)node[label=below:$e$]{};\n\n\n\n % \\tkzDefLine[through d](d,e)\n \\tkzDrawLine[add = 2 and 2](d,e)\n\\end{tikzpicture}\n\\end{center}\n\\caption{We assume the line passing through $d$ and $e$ intersects the sides $ac$ and $ab$ of the triangle. Since no three points in $X$ are colinear, the line passing through $d$ and $e$ must intersect two of the sides of the triangle, so we can always relabel the verticies so that this is the case.} \\label{fig:lem89de_line}\n\\end{marginfigure} Then $bcde$ form a convex quadrangle. We may see this by taking the convex hull of any three of $\\{b,c,d,e\\}$ to form a triangle $T$, and calling the omitted point $z$. Then there are two boundary line segments of the quadrilateral $bcde$ which end in $z$, but each of these line segments only intersects $T$ at the opposite end from $z$. If $z\\in T$, then the whole line between $z$ and those endpoints would have to be included in $T$. Thus, $z\\not \\in T$, and we truly have a convex quadrilateteral $bcde$. This is illustrated in \\cref{fig:lem89T}.\n\\begin{figure}\n\\begin{center}\n\\begin{tikzpicture}[scale=3]\n\\coordinate (a) at (0,0);\n\\coordinate (b) at (1.5,0);\n\\coordinate (c) at (.75,1);\n\n\\coordinate (e) at (0.9,.2);\n\\coordinate (d) at (0.7,.4);\n\n% \\draw[green] (a) -- (d) --;\n% \\draw[blue] (b) -- (c);\n\\filldraw[LightBlue] (a) -- (b) -- (c) -- cycle;\n\t\\filldraw[black] (a) circle (1pt) node[label=left:$a$]{};\n\\filldraw[black] (b) circle (1pt)node[label=right:$b$]{};\n\\filldraw[black] (c) circle (1pt)node[label=above:$c$]{};\n\\filldraw[black] (d) circle (1pt)node[label=below:$d$]{};\n\\filldraw[black] (e) circle (1pt)node[label=below:$e$]{};\n\n\\filldraw[pattern=north west lines] (c) -- (d) -- (b) -- cycle;\n\n\\draw[dashed] (d)--(e);\n\\draw[dashed] (e)--(b);\n\n\\end{tikzpicture}\n\\end{center}\n\\caption{We will take $T= \\conv(\\{c,d,b\\})$, the crosshatched region. Then the two dashed lines only intersect $T$ at the verticies $d,b$. We see if $e$ were moved up into $T$, then the line passing through $d$ and $e$ would no longer intersect the segment $ab$.} \\label{fig:lem89T}\n\\end{figure}\n\\end{proof}\n\\begin{theorem}[\\cite{erdosszekeres1935combinatorial}]\n$g(k)$ exists for all $k$.\n\\end{theorem}\n\\begin{proof}\t\nLet $n=g(k)$ be such that in every coloring of $[n]^{(4)}$ in colors red and blue, one can find either a set of $5$ with all quadruples in red, or a set of size $k$ with all quadruples blue. This exists by the Hypergraph Ramsey theorem\\sidenote{For all positive integers $r,k_1$, and $k_2$ there exists a positive integer $n= R^{(r)}(k_1,k_2)$ so that the following holds. If elements of $[n]^{(r)}$ are colored in colors red and blue then there is a set $Z\\subset [n]$ such that either $|Z| = k_1$ and all elements of $Z^{(r)}$ are red, or $|Z| = k_2$ and all elements of $Z^{(r)}$ are blue.}.\n\nColor quadruples of points in red color if it is not in convex position, and otherwise in blue. By \\cref{lem:g4_is_5}, there exists a set of $k$ points such that every four of them are in convex position.\n\nThen these $k$ points are in convex position by Carath\\'eodory's theorem.\n\\end{proof}\n\\begin{remark}\nWe may obtain much better bounds by longer proofs.\n\\end{remark}\n\n", "meta": {"hexsha": "0dcbc27fcdd0fa7717a3ddc02d831d7f9ef22a0f", "size": 36085, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/ch8_convexity.tex", "max_stars_repo_name": "ericphanson/CombinatoricsNotes", "max_stars_repo_head_hexsha": "6b369a77b77cf6f0281b59f227aaa31e6903079c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2018-04-24T06:43:31.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-20T04:27:41.000Z", "max_issues_repo_path": "chapters/ch8_convexity.tex", "max_issues_repo_name": "ericphanson/CombinatoricsNotes", "max_issues_repo_head_hexsha": "6b369a77b77cf6f0281b59f227aaa31e6903079c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/ch8_convexity.tex", "max_forks_repo_name": "ericphanson/CombinatoricsNotes", "max_forks_repo_head_hexsha": "6b369a77b77cf6f0281b59f227aaa31e6903079c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-01-04T19:38:24.000Z", "max_forks_repo_forks_event_max_datetime": "2020-01-04T19:38:24.000Z", "avg_line_length": 53.618127786, "max_line_length": 594, "alphanum_fraction": 0.6613551337, "num_tokens": 13875, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8418256452674008, "lm_q1q2_score": 0.6403352265131278}} {"text": "% !TEX root = Main.tex\n\\section{Non-Negative Matrix Factorization}\n$\\mathbf{X} \\in \\mathbb{Z}^{N \\times M}_{\\geq 0}$, NMF: $\\mathbf{X} \\approx \\mathbf{U^\\top V}, x_{ij}=\\sum_z{u_{zi}v_{zj}}=\\langle\\mathbf{u}_i \\mathbf{v}_j\\rangle$\nDecompose object into features: topics, face parts, etc.. $\\mathbf{u}$ weights on parts, $\\mathbf{v}$ parts (bases). More interpretable (PCA: holistic repre.).\n\n\\subsection*{EM for MLE for pLSA (NO global opt guarantee)}\n\\textbf{Context Model:} $p(w | d) = \\sum_{z=1}^K p(w | z) p(z | d)$\\\\\n\\textbf{Conditional independence assumption ($*$):}\\\\\n$p(w|d) = \\sum_z p(w,z|d) = \\sum_z p(w|d,z)p(z|d) \\stackrel{*}{=} \\sum_z p(w|z)p(z|d)$\\\\\n\\textbf{Symmetric parameterization:}\\\\\n$p(w, d) = \\sum_z p(z)p(w | z) p(d | z)$ \\\\\nLog-Likelihood: $L(\\mathbf{U}, \\mathbf{V}) = \\sum_{i,j} x_{i,j}\\log p(w_j|d_i) \\\\\n= \\sum_{(i,j) \\in X} \\log \\sum_{z=1}^K p(w_j|z)p(z|d_i)$ \\\\ \n$ p(w_j|z) = v_{zj}$, $p(z|d_i) = u_{zi}$, $\\sum_j^N v_{zj} = \\sum_z^K u_{zi} = 1$\\\\\nE-Step (optimal q: posterior of z over $(d_i, w_j)$):\\\\\n$q_{zij} = \\frac{p(w_j|z)p(z|d_i)}{\\sum_{k=1}^K p(w_j|k)p(k|d_i)} := \\frac{v_{zj}u_{zi}}{\\sum_{k=1}^K v_{kj}u_{ki}}$, $\\sum_z q_{zij}=1$\\\\\nM-Steps:\\\\\n$p(z|d_i) = \\frac{\\sum_j x_{ij}q_{zij}}{\\sum_j x_{ij}}, p(w_j|z) = \\frac{\\sum_i x_{ij}q_{zij}}{\\sum_{i,l}x_{il}q_{zil}}$\\\\\n\n\\subsection*{Latent Dirichlet Allocation}\nTo sample a new document, we need to extend $X$ and $U^T$ with a new row, s.t. $X=U^T V$. (While pLSA fixes both dimensions)\\\\\nFor each $d_i$ sample topic weights $\\mathbf{u}_i$\\textasciitilde Dirichlet($\\alpha$): $p(u_i|\\alpha) = \\prod_{z=1}^K u_{zi}^{\\alpha_k-1}$, then topic $z^t$\\textasciitilde Multi($u_i$), word $w^t$\\textasciitilde Multi($v_{z^t}$)\\\\\nMultinom. obsv. model on wc vec: $p(\\mathbf{x}|V,u) = \\frac{l!}{\\prod_j \\mathbf{x}_j!}\\prod_j \\pi_j^{\\mathbf{x}_j}$ \nwhere $\\pi_j=\\sum_z v_{zj} u_z$, $l=\\sum_j x_j$ \\\\\nBayesian averaging over $\\mathbf{u}$: $p(\\mathbf{x}|\\mathbf{V},\\alpha)=\\int p(\\mathbf{x}|\\mathbf{V},\\mathbf{u})p(\\mathbf{u}|\\alpha)d\\mathbf{u}$\n\n\\subsection*{NMF Algorithm for quadratic cost function}\n\n$\\min_{\\mathbf{U}, \\mathbf{V}} J(\\mathbf{U}, \\mathbf{V}) = \\frac{1}{2} \\|\\mathbf{X} - \\mathbf{U}^\\top\\mathbf{V}\\|_F^2$ (non-negativity)\ns.t. $\\forall i,j,z:u_{zi},v_{zj} \\geq 0 $ \\\\\nComparison with pLSA:\\\\\n1. sampling model: Gaussian vs multinomial\n2. objective: quadratic vs KL divergence\n3. constraints: not normalized \\\\\nAlternating least squares:\\\\\n1. init: $\\mathbf{U}, \\mathbf{V} = rand()$\\\\ 2. repeat 3\\textasciitilde4 for $\\mathit{maxIters}$:\\\\\n3. upd. $(\\mathbf{VV}^\\top)\\mathbf{U} = \\mathbf{VX}^\\top$, proj. $u_{zi} = \\max \\{ 0, u_{zi} \\}$\\\\\n4. update $(\\mathbf{UU}^\\top)\\mathbf{V} = \\mathbf{UX}$, proj. $v_{zj} = \\max \\{ 0, v_{zj} \\}$\n", "meta": {"hexsha": "083833fa0e797ee3270f8390a9dc09da5cb627ef", "size": 2722, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "NMF.tex", "max_stars_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_stars_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-09-24T20:33:35.000Z", "max_stars_repo_stars_event_max_datetime": "2019-02-21T15:52:46.000Z", "max_issues_repo_path": "NMF.tex", "max_issues_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_issues_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "NMF.tex", "max_forks_repo_name": "vscherer/eth-cil-exam-cheatsheet", "max_forks_repo_head_hexsha": "9ae156bcf5e2797e65b5495ff520649b43860cdd", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2019-01-14T16:21:22.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-18T17:17:10.000Z", "avg_line_length": 69.7948717949, "max_line_length": 230, "alphanum_fraction": 0.6223365173, "num_tokens": 1189, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8872045937171068, "lm_q2_score": 0.7217431943271998, "lm_q1q2_score": 0.6403338774911501}} {"text": "\\section{Homology Calculations and Applications}\r\n\\begin{example}\r\n If $S^n\\simeq S^m$, then $m=n$.\r\n This follows immediately from our calculation of the homology groups of simplices which are the triangulations of the spheres.\r\n\\end{example}\r\n\\begin{theorem}\r\n If $\\mathbb R^n\\cong\\mathbb R^m$, then $m=n$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Suppose we are given a homeomorphism $\\phi:\\mathbb R^m\\to\\mathbb R^n$, then by a translation we can assume WLOG that $\\phi(0)=0$.\r\n Then we can restrict $\\phi$ so that $\\mathbb R^n\\setminus\\{0\\}$ and $\\mathbb R^m\\setminus\\{0\\}$, which are homotopic to $S^{n-1}$ and $S^{m-1}$ respectively, consequently $S^{n-1}\\simeq S^{m-1}$, so $n=m$.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Any map $\\phi:D^n\\to D^n$ has a fixed point.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Repeat the argument as in the $D^2$ case and finish the argument by homology.\r\n\\end{proof}\r\n\\subsection{Mayer-Vietoris Theorem}\r\nThe Mayor-Vietoris theorem is an analog of Seifert-van Kampen theorem as in both deals with the algebraic invariants of spaces that are constructed by gluing.\r\n\\begin{definition}\r\n A sequence of homomorphisms of abelian groups\r\n \\[\r\n \\begin{tikzcd}\r\n \\cdots\\arrow{r}&A_{i+1}\\arrow{r}{f_i}&A_i\\arrow{r}{f_{i-1}}&A_{i-1}\\arrow{r}&\\cdots\r\n \\end{tikzcd}\r\n \\]\r\n is exact at $A_i$ if $\\operatorname{Im}f_i=\\ker f_{i-1}$.\r\n We say the sequence is exact if it is exact at every $A_i$.\r\n\\end{definition}\r\nConsequenly, any exact sequence is a chain complex with all homology groups trivial.\r\n\\begin{definition}\r\n A short exact sequence is an exact sequence in the form\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&A\\arrow{r}&B\\arrow{r}&C\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n\\end{definition}\r\nSo necessarily $A\\to B$ is injective, $B\\to C$ is surjective and $C\\cong B/A$.\r\nAn exact sequence that is not short is of course called a long exact sequence.\\\\\r\nSuppose $K$ is a simplicial complex with $K=L\\cup M$ where $L,M$ are subcomplexes of $K$.\r\nThen $N=L\\cap M$ is also a subcomplex.\r\nWe usually write $K=L\\cup_NM$.\r\nAnalogous to what we did in Seifert-van Kampen theorem, we want to relate the homology groups of $L,M,N$ to the homology group of $K$.\r\nOf course we immediately get the natual inclusion maps\r\n$$i:N\\to L,j:N\\to M,l:L\\to K,m:M\\to K$$\r\n\\begin{theorem}[Mayer-Vietoris]\\label{mayer-vietoris}\r\n There exists a map (known as the connecting homomorphism) $\\delta_\\ast:H_n(K)\\to H_{n-1}(N)$ for each $n$ such that the sequence\r\n \\[\r\n \\begin{tikzcd}\r\n \\cdots\\arrow{r}{\\delta_\\ast}&H_n(N)\\arrow{r}{i_\\ast\\oplus j_\\ast}&H_n(L)\\oplus H_n(M)\\arrow{r}{l_\\ast-m_\\ast}&H_n(K)\\arrow[swap,overlay,out=0,in=180]{dll}{\\delta_\\ast}&\\\\\r\n &H_{n-1}(N)\\arrow[swap]{r}{i_\\ast\\oplus j_\\ast}&H_{n-1}(L)\\oplus H_{n-1}(M)\\arrow[swap]{r}{l_\\ast-m_\\ast}&H_{n-1}(K)\\arrow[swap]{r}{\\delta_\\ast}&\\cdots\r\n \\end{tikzcd}\r\n \\]\r\n is exact.\r\n\\end{theorem}\r\nWe will prove this after further development on homological algebra.\r\n\\begin{definition}\r\n We say a sequence of chain complexes\r\n \\[\r\n \\begin{tikzcd}\r\n \\cdots\\arrow{r}&A_\\bullet\\arrow{r}{f_\\bullet}&B_\\bullet\\arrow{r}{g_\\bullet}&C_\\bullet\\arrow{r}&\\cdots\r\n \\end{tikzcd}\r\n \\]\r\n is exact at $B_\\bullet$ if\r\n \\[\r\n \\begin{tikzcd}\r\n \\cdots\\arrow{r}&A_n\\arrow{r}{f_n}&B_n\\arrow{r}{g_n}&C_n\\arrow{r}&\\cdots\r\n \\end{tikzcd}\r\n \\]\r\n is exact for every $n$.\r\n\\end{definition}\r\nSo we can analogously define (short) exact sequences of chain complexes.\r\n\\begin{lemma}[Snake Lemma]\r\n Let\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&A_\\bullet\\arrow{r}{f_\\bullet}&B_\\bullet\\arrow{r}{g_\\bullet}&C_\\bullet\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n be a short exact sequence of chain complexes.\r\n Then for any $n$ there is a homomorphism $\\delta_\\ast:H_{n+1}(C_\\bullet)\\to H_n(A_\\bullet)$ such that\r\n \\[\r\n \\begin{tikzcd}\r\n \\cdots\\arrow{r}{\\delta_\\ast}&H_{n+1}(A_\\bullet)\\arrow{r}{f_\\ast}&H_{n+1}(B_\\bullet)\\arrow{r}{g_\\ast}&H_{n+1}(C_\\bullet)\\arrow[swap,overlay,out=0,in=180]{dll}{\\delta_\\ast}&\\\\\r\n &H_n(A_\\bullet)\\arrow[swap]{r}{f_\\ast}&H_n(B_\\bullet)\\arrow[swap]{r}{g_\\ast}&H_n(C_\\bullet)\\arrow[swap]{r}{\\delta_\\ast}&\\cdots\r\n \\end{tikzcd}\r\n \\]\r\n is exact.\r\n\\end{lemma}\r\n\\begin{proof}\r\n We already have this huge commutative diagram with exact rows\r\n \\[\r\n \\begin{tikzcd}\r\n &\\vdots\\arrow{d}&\\vdots\\arrow{d}&\\vdots\\arrow{d}&\\\\\r\n 0\\arrow{r}&A_{n+1}\\arrow{r}{f_{n+1}}\\arrow{d}{\\partial_{n+1}}&B_{n+1}\\arrow{r}{f_{n+1}}\\arrow{d}{\\partial_{n+1}}&C_{n+1}\\arrow{r}\\arrow{d}{\\partial_{n+1}}&0\\\\\r\n 0\\arrow{r}&A_n\\arrow{r}{f_n}\\arrow{d}{\\partial_n}&B_n\\arrow{r}{f_n}\\arrow{d}{\\partial_n}&C_n\\arrow{r}\\arrow{d}{\\partial_n}&0\\\\\r\n 0\\arrow{r}&A_{n-1}\\arrow{r}{f_{n-1}}\\arrow{d}&B_{n-1}\\arrow{r}{g_{n-1}}\\arrow{d}&C_{n-1}\\arrow{r}\\arrow{d}&0\\\\\r\n &\\vdots&\\vdots&\\vdots&\r\n \\end{tikzcd}\r\n \\]\r\n We now construct $\\delta_\\ast:H_{n+1}(C_\\bullet)\\to H_n(A_\\bullet)$.\r\n Take $[x]\\in H_{n+1}(C_\\bullet)$ for $x\\in Z_{n+1}(C_\\bullet)$.\r\n As $g_{n+1}$ is surjective, there is $x\\in B_{n+1}$ such that $g_{n+1}(y)=x$.\r\n Now $g_n\\circ\\partial_{n+1}(y)=\\partial_{n+1}\\circ g_{n+1}(y)=\\partial_{n+1}(x)=0$.\r\n So by exactness, there exists $z\\in A_n$ such that $f_n(z)=\\partial_{n+1}(y)$.\r\n Note that $f_{n-1}\\circ\\partial_n(z)=\\partial_n\\circ f_n(z)=\\partial_n\\circ\\partial_{n+1}(y)=0$, so $\\partial_n(z)=0$ since $f_{n-1}$ is injective.\r\n This means that $z\\in Z_n(A_\\bullet)$, so we define $\\delta_\\ast([x])=[z]$.\r\n To see it is well-defined, we have two issues to deal with:\\\\\r\n First, if we replace $x$ by $x+\\partial_{n+2}(x')$ and $g_{n+2}(y')=x'$ , then replace $y$ with $y+\\partial_{n+2}(y')$.\r\n Then $g_{n+1}(y+\\partial_{n+2}(y'))=g_{n+1}(y)+\\partial_{n+2}\\circ g_{n+2}(y')=x+\\partial_{n+2}(x')$.\r\n But $\\partial_{n+1}(y+\\partial_{n+2}(y'))=\\partial_{n+1}(y)$, so $z$ does not change.\\\\\r\n Secondly, if we have made up our mind on $x$ and $y,y'$ are chosen with $g_{n+1}(y')=g_{n+1}(y)=x$, then $g_{n+1}(y'-y)=0$, therefore by exactness there is some $z'$ such that $f_{n+1}(z')=y'-y$, thus $y'=y+f_{n+1}(z')$.\r\n So $\\partial_{n+1}(y')=\\partial_{n+1}(y)+\\partial_{n+1}\\circ f_{n+1}(z')=\\partial_{n+1}(y)+f_n\\circ\\partial_{n+1}(z')$, which means $f_n(z+\\partial_{n+1}(z'))=\\partial_{n+1}(y)+f_n\\circ\\partial_{n+1}(z')=\\partial_{n+1}(y')$.\r\n This is saying that replacing $y$ by $y'$ modifies $z$ by adding $\\partial_{n+1}(z')$.\r\n But $[z]=[z+\\partial_{n+1}(z')]$, so $\\delta_\\ast$ is well-defined.\\\\\r\n The proof of the sequence being exact is tedious routine work.\\\\\r\n At $H_n(B_\\bullet)$, if $[a]\\in H_n(A_\\bullet)$, then by definition $g_\\ast\\circ f_\\ast([a])=[g_n\\circ f_n(a)]=0$ by the exactness of our original short exact sequence.\r\n So $\\operatorname{Im}f_\\ast\\subset \\ker g_\\ast$.\r\n Now if $g_\\ast([b])=0$, then $[g_n(b)]=g_\\ast([b])=0$, thus there is some $c\\in C_{n+1}$ with $g_n(b)=\\partial_{n+1}(c)$.\r\n But then choosing $b'\\in B_{n+1}$ with $g_{n+1}(b')=c$ by exactness gives $g_n(b-\\partial_{n+1}(b'))=0$, hence $b-\\partial_{n+1}(b')\\in\\ker g_n=\\operatorname{Im}f_n$, so there is some $a\\in A_n$ such that $f_n(a)=b-\\partial_{n+1}(b')$.\r\n In particular, $f_\\ast([a])=[f_n(a)]=[b-\\partial_{n+1}(b')]=[b]$, therefore $\\operatorname{Im}f_\\ast\\supset\\ker g_\\ast$.\r\n Combining these gives $\\operatorname{Im}f_\\ast=\\ker g_\\ast$ which implies the exactness at $H_n(B_\\bullet)$.\\\\\r\n At $H_n(A_\\bullet)$, suppose $[z]=\\delta_\\ast([x])$, then $f_\\ast([z])=[f_n(z)]=[\\partial_{n+1}(y)]=0$ where $y$ is as in the construction of $\\delta_\\ast$, so $\\operatorname{Im}\\delta_\\ast\\subset\\ker f_\\ast$.\r\n Suppose now that $f_\\ast([z])=0$, then $f_n(z)=\\partial_{n+1}(y)$ for some $y\\in B_{n+1}$.\r\n Take $x=g_{n+1}(y)$, then $\\partial_{n+1}(x)=\\partial_{n+1}\\circ g_{n+1}(y)=g_n\\circ\\partial_{n+1}(y)=g_n\\circ f_n(z)=0$, so $x$ is a cycle, so $[x]$ is a homology class.\r\n But $\\delta_\\ast([x])=[z]$ by construction, hence $\\operatorname{Im}\\delta_\\ast\\supset\\ker f_\\ast$, therefore $\\operatorname{Im}\\delta_\\ast=\\ker f_\\ast$ which gives the exactness at $H_n(A_\\bullet)$.\\\\\r\n At $H_n(C_\\bullet)$, if $[x]\\in\\operatorname{Im}g_\\ast$, then (ater replacing $x$ with another representative if necessary) WLOG there exists a cycle $y\\in B_n$ such that $g_n(y)=x$.\r\n But then immediately $\\delta_\\ast([x])=0$, so $\\operatorname{Im}g_\\ast\\subset\\ker\\delta_\\ast$.\r\n Conversely, suppose $\\delta_\\ast([x])=0$, then let $y,z$ be the corresponding element in the construction of $\\delta_\\ast$.\r\n By hypothesis there exists $a\\in A_n$ such that $\\partial_n(a)=z$.\r\n Then $\\partial_n\\circ f_n(a)=f_{n-1}\\circ\\partial_n(a)=f_{n-1}(z)=\\partial_n(y)$.\r\n Thus $\\partial_n(y-f_n(a))=0$, so $y-f_n(a)$ is a cycle.\r\n Also $g_n(y-f_n(a))=x$, so $[x]=g_\\ast([y-f_n(a)])$.\r\n Consequently $\\operatorname{Im}g_\\ast\\supset\\ker\\delta_\\ast$, so $\\operatorname{Im}g_\\ast=\\ker\\delta_\\ast$.\r\n This finished the proof.\r\n\\end{proof}\r\n\\begin{proof}[Proof of Theorem \\ref{mayer-vietoris}]\r\n Easy to check that $C_\\bullet\\oplus D_\\bullet$ is a chain complex with all data obtained from direct sums of data of $C_\\bullet,D_\\bullet$.\r\n Also $H_n(C_\\bullet\\oplus D_\\bullet)=H_n(C_\\bullet)\\oplus H_n(D_\\bullet)$.\r\n It then suffices to check that\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&C_\\bullet(N)\\arrow{r}{i_\\bullet\\oplus j_\\bullet}&C_\\bullet(L)\\oplus C_\\bullet(M)\\arrow{r}{l_\\bullet-m_\\bullet}&C_\\bullet(K)\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n is exact.\r\n The proof is then done by snake lemma.\\\\\r\n Note that $C_\\bullet(N)$ naturally embeds into $C_\\bullet (L)$ and $C_\\bullet(M)$ via $i_\\bullet\\oplus j_\\bullet$, so in particular $i_\\bullet\\oplus j_\\bullet$ is a injective.\r\n Also since $K=L\\cup M$, for any $c\\in C_n(K)$ we can write $c=c_L+c_M$ where $c_L$ is a linear combination of simplices in $L$ and $c_M$ is that in $M$.\r\n Let $b_L,b_M$ be the respective copies of $c_L,c_M$ in $C_n(L),C_n(M)$ respectively, then $l_\\bullet(b_L)=c_L,m_\\bullet(b_M)=c_M$, and hence $c=(l_\\bullet-m_\\bullet)(b_L,-b_M)$, so $l_\\bullet-m_\\bullet$ is surjective.\\\\\r\n It remains to show the exactness in the middle.\r\n For $(b_L,b_M)\\in C_n(L)\\oplus C_n(M)$, note that $l_\\bullet(b_L)-m_\\bullet(b_M)=0$ iff every simplex which occurs in $b_L$ also occurs in $b_M$ with the same coefficient.\r\n This is just saying that $b_L,b_M$ are linear combinations of simplices in $L\\cap M=N$.\r\n Therefore $\\operatorname{Im}(i_\\bullet\\oplus j_\\bullet)=\\ker (l_\\bullet-m_\\bullet)$, hence the sequence is indeed exact.\r\n\\end{proof}\r\n\\begin{lemma}[Five Lemma]\r\n Given a commutative diagram\r\n \\[\r\n \\begin{tikzcd}\r\n A\\arrow{r}\\arrow{d}{\\alpha}&B\\arrow{r}\\arrow{d}{\\beta}&C\\arrow{r}\\arrow{d}{\\gamma}&D\\arrow{r}\\arrow{d}{\\delta}&E\\arrow{d}{\\epsilon}\\\\\r\n A'\\arrow{r}&B'\\arrow{r}&C'\\arrow{r}&D'\\arrow{r}&E'\r\n \\end{tikzcd}\r\n \\]\r\n If the rows are exact and $\\alpha,\\beta,\\delta,\\epsilon$ are isomorphisms, then $\\gamma$ is also an isomorphism.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Exercise.\r\n\\end{proof}\r\nWe can now prove Proposition \\ref{barycentric_iso_homol}.\r\nRecall that we want to show that $s_\\ast:H_n(K')\\to H_n(K)$ is an isomorphism where $s:K'\\to K$ is a simplicial approximation to the identity map on $|K|=|K'|$ which sends each $\\hat\\sigma$ to a vertex of $\\sigma$.\r\n\\begin{proof}[Proof of Proposition \\ref{barycentric_iso_homol}]\r\n Induction on the number of simplices of $K$.\r\n If $K$ has only one simplex, then it is just a vertex, so the proposition is trivial.\r\n For the induction step, let $\\sigma\\in K$ be of maximal dimension, then $L=K\\setminus \\{\\sigma\\}$ is also a simplicial complex.\r\n Let $M$ be the simplicial complex consists of $\\sigma$ and its faces.\r\n Then $N=M\\cap L$ is just the proper faces of $\\sigma$.\r\n By construction of $s$ we can restrict $s$ to $L',M',N'$ to get the maps\r\n $$s_\\ast:H_n(L')\\to H_n(L),s_\\ast:H_n(M')\\to H_n(M),s_\\ast:H_n(N')\\to H_n(N)$$\r\n By the induction hypothesis, these are all isomorphisms.\r\n Then by Mayer-Vietoris, we have the diagram\r\n \\[\r\n \\begin{tikzcd}[column sep=0.34em]\r\n H_n(N')\\arrow{r}\\arrow{d}{s_\\ast}&H_n(L')\\oplus H_n(M')\\arrow{r}\\arrow{d}{s_\\ast\\oplus s_\\ast}&H_n(K')\\arrow{r}\\arrow{d}{s_\\ast}&H_{n-1}(N')\\arrow{r}\\arrow{d}{s_\\ast}&H_{n-1}(L')\\oplus H_{n-1}(M')\\arrow{d}{s_\\ast\\oplus s_\\ast}\\\\\r\n H_n(N)\\arrow{r}&H_n(L)\\oplus H_n(M)\\arrow{r}&H_n(K)\\arrow{r}&H_{n-1}(N)\\arrow{r}&H_{n-1}(L)\\oplus H_{n-1}(M)\r\n \\end{tikzcd}\r\n \\]\r\n with exact rows.\r\n Easy to check that it commutes.\r\n Also, by the induction hypothesis, all vertical arrows except $s_\\ast:H_n(K')\\to H_n(K)$ are isomorphisms.\r\n Hence $s_\\ast:H_n(K')\\to H_n(K)$ is an isomorphism by five lemma.\r\n\\end{proof}\r\n\\subsection{Homology of Compact Surfaces}\r\nRecall that we constructed the oriented surfaces of genus $g$ as $\\Sigma_g=\\Gamma_{2g}\\cup_{\\rho_g}D^2$ where $\\Gamma_{2g}$ is a bouquet of $2g$ copies of $S^1$ with generators $\\alpha_1,\\ldots,\\alpha_g,\\beta_1,\\ldots,\\beta_g$ and $\\rho_g=\\alpha_1\\beta_1\\alpha_1^{-1}\\beta_1^{-1}\\cdots \\alpha_g\\beta_g\\alpha_g^{-1}\\beta_g^{-1}$.\r\n\\begin{example}\r\n $\\Gamma_r$ can be triangulated in the obvious way by taking it as $r$ hollow triangles joined by a common vertex.\r\n We know that $H_n(\\Gamma_1)$ is $\\mathbb Z$ when $n=0,1$ and $0$ otherwise.\r\n We claim that\r\n $$H_i(\\Gamma_r)=\\begin{cases}\r\n \\mathbb Z\\text{, if $i=0$}\\\\\r\n \\mathbb Z^r\\text{, if $i=1$}\\\\\r\n 0\\text{, otherwise}\r\n \\end{cases}$$\r\n which we shall show inductively.\r\n The cases except for $i=1$ are all trivial, so it suffices to show that.\r\n Suppose we have shown the case for $r-1$, then take $K=\\Gamma_r$, $L$ be a natural copy of $\\Gamma_{r-1}$ in $\\Gamma_r$ and $M$ be the remaining triangle.\r\n Then $N=L\\cap M=\\{\\ast\\}$ and hence Mayer-Vietoris gives the exact sequence\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&H_1(N)\\arrow{r}&H_1(\\Gamma_{r-1})\\oplus H_1(S^1)\\arrow{r}& H_1(\\Gamma_r)\\arrow[overlay, out=0, in=180]{dll}&\\\\\r\n &H_0(N)\\arrow{r}&H_0(\\Gamma_{r-1})\\oplus H_0(S^1)\\arrow{r}& H_0(\\Gamma_r)\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n which, after putting in everything we already know,\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&0\\arrow{r}&\\mathbb Z^{r-1}\\oplus \\mathbb Z\\arrow{r}& H_1(\\Gamma_r)\\arrow[overlay, out=0, in=180]{dll}&\\\\\r\n &\\mathbb Z\\arrow{r}&\\mathbb Z\\oplus \\mathbb Z\\arrow{r}& \\mathbb Z\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n By exactness of the bottom row, the map $\\mathbb Z\\to\\mathbb Z\\oplus\\mathbb Z$ is injective (alternatively this can also be easily seen from the definition of that map), hence the connecting homomophism is the zero map.\r\n Therefore the top row reduces to the exact sequence\r\n \\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&\\mathbb Z^{r-1}\\oplus \\mathbb Z\\arrow{r}& H_1(\\Gamma_r)\\arrow{r}&0\r\n \\end{tikzcd}\r\n \\]\r\n which precisely means that $H_1(\\Gamma_r)\\cong\\mathbb Z^{r-1}\\oplus \\mathbb Z\\cong\\mathbb Z^r$.\r\n\\end{example}\r\nWorth noting that $H_1(\\Gamma_r)$ is the free abelian group on $r$ letters while $\\pi_1(\\Gamma_r)$ is the free group on $r$ letters.\r\nAlso, if $\\alpha_1,\\ldots,\\alpha_r$ are the generating paths of the circles, then $H_1(\\Gamma_r)$ is generated by $\\alpha_1,\\ldots,\\alpha_r$.\r\n\\begin{remark}\r\n Like in the preceding exacmple, whenever $N$ is connected, the map $H_0(N)\\to H_0(L)\\oplus H_0(M)$ is injective, and hence $H_1(L)\\oplus H_1(M)\\to H_1(K)$ is surjective.\r\n\\end{remark}\r\nNow we want to attach our $2$-cell to get $\\Sigma_g$.\r\nWe'll do this in two steps.\\\\\r\nFirst, we attach a cylinder $S^1\\times I$ via the same $\\rho_g$ but on $S^1\\times\\{0\\}$.\r\nWrite $\\Sigma_g^\\star=\\Gamma_{2g}\\cup_{\\rho_g} (S^1\\times I)$\r\nBy shrinking $I$ to a point we can obtain the deformation retraction of $\\Sigma_g^\\star$ onto $\\Gamma_{2g}$, therefore they have the same homology groups.\\\\\r\nNow $\\Sigma_g=\\Sigma_g^\\star\\cup_\\alpha D^2$ where $\\alpha:\\partial D^2\\to S^1\\times \\{1\\}$ is the natural identity.\r\nWe can choose triangulations of $\\Sigma_g^\\star$ and $D^2$ so that they are compatible under this gluing (so we are essentially removing a $D^2$ from $\\Sigma_g$ to get $\\Sigma_g^\\star$).\r\n\\footnote{There are of course some technical details regarding why we can do this, but we are not going into this part of details here. It is not too hard though.}\r\nLet $L$ be the triangulation of $\\Sigma_g^\\star$ and $M$ be the triangulation of $D^2$ such that this holds.\r\nThen $N=L\\cap M$ is a triangulation of $S^1\\times\\{1\\}\\cong S^1$.\r\nMajer-Vietoris then gives the exact sequence\r\n\\[\r\n \\begin{tikzcd}\r\n &H_2(L)\\oplus H_2(M)\\arrow{r}&H_2(\\Sigma_g)\\arrow[overlay, out=0,in=180]{dll}&\\\\\r\n H_1(N)\\arrow[swap]{r}{i_\\ast}&H_1(L)\\oplus H_1(M)\\arrow{r}&H_1(\\Sigma_g)\\arrow{r}&0\r\n \\end{tikzcd}\r\n\\]\r\nNote that the zero at the end is due to the preceding remark.\r\nAgain putting everything we already know into it gives\r\n\\[\r\n \\begin{tikzcd}\r\n 0\\arrow{r}&H_2(\\Sigma_g)\\arrow{r}&\\mathbb Z\\arrow{r}{i_\\ast}&\\mathbb Z^{2g}\\arrow{r}&H_1(\\Sigma_g)\\arrow{r}&0\r\n \\end{tikzcd}\r\n\\]\r\nThe exactness means that $H_2(\\Sigma_g)\\cong\\ker i_\\ast$ and $H_1(\\Sigma_g)\\cong\\operatorname{coker}i_\\ast=\\mathbb Z^{2g}/\\operatorname{Im}i_\\ast$.\r\nSo we just need to understand $i_\\ast:\\mathbb Z\\to\\mathbb Z^{2g}$, or in other words $i_\\ast:H_1(S^1)\\to H_1(\\Sigma_g^\\star)$.\r\nNote that the generator of $H_1(S^1)$ is the cycle runs through it (which is basically the one corresponding to the generator of $\\pi_1(S^1)$).\r\nThe deformation retract of $\\Sigma_g^\\star$ to $\\Gamma_{2g}$ identifies $S^1$ with the image of the cycle under $\\rho_g$.\r\nTherefore $(\\rho_g)_\\ast$ maps the cycle to $[\\alpha_1]+[\\beta_1]-[\\alpha_1]-[\\beta_1]+\\cdots+[\\alpha_g]+[\\beta_g]-[\\alpha_g]-[\\beta_g]=0$.\r\nTherefore $i_\\ast=0$ and hence $H_2(\\Sigma_g)\\cong\\mathbb Z$ and $H_1(\\Sigma_g)\\cong\\mathbb Z^{2g}$.\r\nTo conclude,\r\n$$H_n(\\Sigma_g)\\cong\\begin{cases}\r\n \\mathbb Z\\text{, if $n=0,2$}\\\\\r\n \\mathbb Z^{2g}\\text{, if $n=1$}\\\\\r\n 0\\text{, otherwise}\r\n\\end{cases}$$\r\nwhich in particular allows us to know $g$ given the homology group of some $\\Sigma_g$.\\\\\r\nWe also have the non-orientable surfaces $S_g$ which is given by $\\Gamma_{g+1}\\cup_\\alpha D^2$ where $\\alpha=\\alpha_0^2\\alpha_1^2\\cdots\\alpha_g^2$ where $\\alpha_i$ is the generator of the $i^{th}$ circle in the bouquet.\r\nRepeating the same process gives the exact sequence\r\n\\[\r\n \\begin{tikzcd}[row sep=tiny]\r\n 0\\arrow{r}&H_2(\\Sigma_g)\\arrow{r}&H_1(S^1)\\arrow{r}{i_\\ast}&H_1(\\Gamma_{g+1})\\arrow{r}&H_1(\\Sigma_g)\\arrow{r}&0\\\\\r\n &&\\mathbb Z\\arrow[equal]{u}&\\mathbb Z^{g+1}\\arrow[equal]{u}&\r\n \\end{tikzcd}\r\n\\]\r\nSo again $H_2(S_g)\\cong\\ker i_\\ast$ and $H_1(S_g)\\cong\\operatorname{coker}i_\\ast$.\r\nNote that $i_\\ast$ maps the generator of $H_1(S^1)$ to the nonzero element $2[\\alpha_0]+\\cdots+2[\\alpha_g]$.\r\nHence $\\ker i_\\ast=0$ and $\\operatorname{coker}i_\\ast\\cong\\mathbb Z^{g+1}/(2,\\ldots,2)\\mathbb Z\\cong\\mathbb Z^g\\oplus\\mathbb Z/2\\mathbb Z$.\r\nIn conclusion\r\n$$H_n(S_g)\\cong\\begin{cases}\r\n \\mathbb Z\\text{, if $n=0$}\\\\\r\n \\mathbb Z^g\\oplus\\mathbb Z/2\\mathbb Z\\text{, if $n=1$}\\\\\r\n 0\\text{, otherwise}\r\n\\end{cases}$$\r\nIn particular we can tell apart $S_g$ and $\\Sigma_g$ as well.", "meta": {"hexsha": "2c4c7cd452ddbec522b1dd36a0782d7e9403c43b", "size": 19542, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6/homol.tex", "max_stars_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_stars_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-03T17:38:28.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-03T17:38:28.000Z", "max_issues_repo_path": "6/homol.tex", "max_issues_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_issues_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6/homol.tex", "max_forks_repo_name": "david-bai-notes/II-Algebraic-Topology", "max_forks_repo_head_hexsha": "05767a26daaddb170e563151393371d8213ee741", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.3862068966, "max_line_length": 329, "alphanum_fraction": 0.6425135605, "num_tokens": 7124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.6402676394095056}} {"text": "\\section{Diatomic gas}\nThe heat capacity is\n\\begin{align}\n c_v &= (6N + N)\\frac{1}{2}k_B = \\frac{7}{2}k_B N\n\\end{align}\nHarmonic oscillator.\nFor $k_B T \\gg \\hbar\\omega$, $\\bar{E} = k_B T$.\n\nBack to solids.\nDebye model.\n\\begin{itemize}\n \\item Normal modes have $\\omega_k=v k$ for all 3 modes.\n 2 transverse and 1 longitudinal.\n \\item Brilluoin zone is a sphere\n \\begin{align}\n 3N = 3V \\int^{k_D} \\frac{d^3 k}{(2\\pi)^3}\n =\n \\frac{V}{2\\pi^2} k_D^3\n \\end{align}\n i.e. number of normal modes $3N$ equals number of fourier modes.\n\\end{itemize}\nWe used that\n\\begin{align}\n \\sum_{\\vec{k}} f(\\vec{k}) \\approx\n V \\int \\frac{d^3k}{(2\\pi)^3} f(k)\n\\end{align}\nIf you don't know this,\nwe'll discuss it in detail soon.\n\nThe partition function is\n\\begin{align}\n Z &= \\Tr e^{-\\beta H}\\\\\n &=\n \\prod_{k,\\alpha}\\sum_{n_k=0}^{\\infty} e^{-\\beta\\hbar\\omega\\left( n_k +\n \\frac{1}{2}\n \\right)}\n\\end{align}\nwhere each term is partition function for one oscillator for one oscillator of\nfrequency $\\omega_k$.\nHere $\\alpha = 1,2,3$\nare the polarizations.\nSo\n\\begin{align}\n Z &=\n \\prod_{k,\\alpha}\n \\frac{e^{-\\beta\\hbar\\omega/2}}{1 - e^{-\\beta\\hbar\\omega_k}}\n\\end{align}\nThe free energy is\n\\begin{align}\n F &= -k_B T\\ln Z\\\\\n &=\n k_B T\n \\sum_{\\vec{k},\\alpha}\n \\left[\n \\ln\\left( 1 - e^{-\\beta\\hbar\\omega_k} \\right)\n +\n \\underbrace{\\frac{\\beta\\hbar\\omega_k}{2}}_{\\textrm{negelect zero-point}}\n \\right]\\\\\n &\\approx\n 3 k_B T V \\int^{k_D} \\frac{d^3k}{{(2\\pi)}^3}\n \\ln\\left( 1 - e^{-\\beta\\hbar vk} \\right)\n\\end{align}\nWe neglect the zero-point energy because it's independent of $T$.\nThe factor of 3 comes from the 3 polarizations.\n\nThe entropy is then\n\\begin{align}\n S &= \\left.\\frac{-\\partial F}{\\partial T}\\right|_{V}\\\\\n &=\n -3k_B V \\int^{k_D} d^3k\\, \\ln(\\cdots)\n +\n \\frac{3k_B T}{k_B T^2} V\n \\int^{k_D} d^3k\n \\frac{\\hbar v k e^{-\\beta\\hbar v k}}{1 - e^{-\\beta\\hbar v k}}\n\\end{align}\n\nAnd finally the energy is\n\\begin{align}\n E &= F + TS\\\\\n &=\n 3V \\int^{k_D} \\frac{d^3k}{(2\\pi)^3}\n \\frac{\\hbar vk}{e^{\\beta\\hbar vk} - 1}.\n\\end{align}\nThen the heat capacity is\n\\begin{align}\n C_v &=\n \\left.\\frac{1}{N} \\frac{\\partial E}{\\partial T}\\right|_{V,N}\\\\\n &=\n \\frac{1}{N} V\n \\frac{\\partial}{\\partial T}\n \\int_{0}^{k_D} \\frac{d^3k}{(2\\pi)^3}\n \\frac{3\\hbar\\omega_k}{e^{\\beta\\hbar\\omega_k} - 1}\\\\\n &=\n \\frac{3V}{N} \\frac{1}{k_B T^2}\n \\int_{0}^{k_D} \\frac{d^3k}{(2\\pi)^3}\n \\frac{(\\hbar\\omega_k)^2}{e^{\\beta\\hbar\\omega_k} - 1}\n e^{\\beta\\hbar\\omega_k}\\\\\n &=\n \\frac{3V}{N} \\frac{1}{2\\pi^2}\n \\frac{(\\hbar v)^2}{k_B T^2}\n \\left( \\frac{k_B T}{\\hbar v} \\right)^5\n \\int_{0}^{\\beta\\hbar vk} dx\\,\n \\frac{x^4}{\\left( e^x - 1 \\right)^2} e^x\n\\end{align}\nwhich for $k_B T \\ll \\hbar v k_D$\ngives\n\\begin{align}\n C_V &\\approx\n 3 \\frac{V}{N}\n \\left( \\frac{k_B T^}{\\hbar v} \\right)^3\n \\frac{1}{2\\pi^2}\n \\int_{0}^{\\infty} dx\\,\n \\underbrace{\\frac{x^4 e^{x}}{\\left( e^x - 1 \\right)^2}}_{4\\pi^4/15}\\\\\n &=\n \\frac{V}{N}\n k_B\n \\frac{2\\pi^2}{5}\n \\left( \\frac{k_B T}{\\hbar v} \\right)^3\\\\\n &\\sim T^3\n\\end{align}\nwhich is the $C_v \\sim T^3$ law for low temperatures.\n", "meta": {"hexsha": "c9fa4a30a108bd0fd556dccb9ecf7367bb3dd266", "size": 3249, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "phys612/lecture21.tex", "max_stars_repo_name": "ehua7365/umdphysnotes", "max_stars_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-12-11T12:53:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-11T12:53:46.000Z", "max_issues_repo_path": "phys612/lecture21.tex", "max_issues_repo_name": "ehua7365/umdphysnotes", "max_issues_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "phys612/lecture21.tex", "max_forks_repo_name": "ehua7365/umdphysnotes", "max_forks_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.631147541, "max_line_length": 78, "alphanum_fraction": 0.5641735919, "num_tokens": 1361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382094310355, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6402676354464254}} {"text": "\\section{Introduction}\n\\label{sec:intro}\n\nWe consider a rectangular area of dimension $W \\times L$ subdivided into countries of dimension $w \\times l$ each. We consider a cartesian reference system with $x$ pointing to the East and $y$ to the North as shown in figure~\\ref{fig:area_subdivision}. The areas are indexed in left-to-right and bottom-up order.\nFor simplicity, we assume that these dimensions are given as integers, that $l$ divides $L$ and $w$ divides $W$ (the input will be checked).\nWe use MPI as the implementation technology, so we decide to assign each country to a separate MPI process to improve the regularity of the program, while still not preventing multiple processes to be assigned to the same node.\n\\begin{figure}[h]\n \\centering\n \\begin{subfigure}[c]{0.7\\textwidth}\n \\begin{tikzpicture}\n % grid\n \\draw[xstep=2, ystep=1] (0,0) grid (8,4);\n % axes\n \\draw[->] (0,0) -- (8.25,0) node[right] {$x$};\n \\draw[->] (0,0) -- (0,4.25) node[above] {$y$};\n \\node[below left] at (0,0) {$0$};\n % country dimensions\n \\draw[<->] (0,-.25) -- node[below] {$w$} (2,-.25);\n \\draw[<->] (-.25,0) -- node[left] {$l$} (-.25, 1);\n % world dimensions\n \\draw[<->] (0,-.75) -- node[below] {$W$} (8,-.75);\n \\draw[<->] (-.75,0) -- node[left] {$L$} (-.75, 4);\n % country indices\n \\matrix[matrix of nodes, inner sep=0, anchor=south west,\n nodes={minimum width=2cm, minimum height=1cm, inner sep=0, align=center}]{\n 12 & 13 & 14 & 15 \\\\\n 8 & 9 & 10 & 11 \\\\\n 4 & 5 & 6 & 7 \\\\\n 0 & 1 & 2 & 3 \\\\\n };\n \\end{tikzpicture}\n \\end{subfigure}\n \\begin{subfigure}[c]{0.29\\textwidth}\n \\begin{tikzpicture}\n \\draw[step=2] (-.25,-.25) grid (2.25,2.25);\n \\node at (1,1) {$c$}; % center\n \\node[above] at (1,2) {$c + \\left\\lfloor \\frac{W}{w} \\right\\rfloor$}; % north\n \\node[right] at (2,1) {$c + 1$}; % east\n \\node[below] at (1,0) {$c - \\left\\lfloor \\frac{W}{w} \\right\\rfloor$}; % south\n \\node[left] at (0,1) {$c - 1$}; % west\n \\end{tikzpicture}\n \\end{subfigure}\n \\caption{World subdivision into countries.}\n \\label{fig:area_subdivision}\n\\end{figure}\n\n\\noindent\nOur program then considers a population $N$ and an initial number of infected individuals $I$ which are uniformly distributed among the countries.\n\n\\paragraph{Motion}\nThe individuals will follow a linear motion with velocity $v$ (given in \\si{m/s}), whose direction is randomly generated for each individual at the beginning of the simulation, and is kept also when moving from one country to another. When an individual is detected reaching (or surpassing) the world boundary it “bounces” onto it: the components of its speed perpendicular to the boundary will be inverted, as in figure~\\ref{fig:boundary_bounce}.\n\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[\n dot/.style={circle,fill,inner sep=1.25pt},\n >=stealth\n ]\n \\node[dot, label=below:$p_s$] (p1) at (210:2) {};\n \\node[dot, ] (p2) at (30:0) {};\n \\node[dot, ] (p3) at (30:1) {};\n \\node[dot, label=below:$p_d$] (p4) at (-30:1) {};\n \\draw[-] (p1) -- (p2);\n \\draw[->, dashed] (p2) -- (p3);\n \\draw[->] (p2) -- (p4);\n \\draw[-, very thick] (-2.5,0) -- (2.5,0);\n \\draw[draw=none, pattern=north east lines] (-2.5,0) rectangle (2.5,.1);\n \\end{tikzpicture}\n \\caption{An individual bouncing on the world boundary}\n \\label{fig:boundary_bounce}\n\\end{figure}\n\n\\paragraph{Infection and recovery}\nIf a susceptible individual $i$ spends at least $t_{infection} = \\SI{10}{min}$ continuously at a distance $d_{ij}\\leq d$ from an infected person $j$, it gets \\emph{infected} too.\nAfter being infected for $t_{recovery} = \\SI{10}{day}$ a person will become \\emph{immune}, so it can't be infected or infect others. It will become susceptible again after further $t_{immunity} = \\SI{90}{days}$.\n\nWe only consider individuals in the same country when computing infections, in order to reduce the communication between processes. This means that if two individuals are closer than $d$ but in different countries, they won't be able to infect each other. This approximation should be acceptable in a realistic scenario where there are physical borders between countries and where $d$ is much smaller than the size of a country. However, the accumulated exposure time of an individual is not reset when passing the border.\n\n\\paragraph{Simulation step}\nThe position and status of each individual is computed with a granularity of $t_{step}$ seconds, with no interpolation in between.\nWe compute periodic logs of the status of the population as soon as the end of each day has been surpassed, therefore we assume that $t_{step} \\ll \\SI{1}{day}$. Also, in a realistic scenario we should have $t_{step} \\cdot v \\ll w,l$ so that an individual cannot span a whole country in few steps. However, we only enforce $t_{step} \\cdot v \\le w,l$, so that each country only needs to send individuals out to its neighbors.\nThe simulation ends when\n\\begin{enumerate*}[label=(\\roman*)]\n \\item $t \\geq t_{target}$, or\n \\item there are no infected people left\n\\end{enumerate*}.", "meta": {"hexsha": "379cbd1091efbf7decd68e3505b9a5f22ae1c33a", "size": 5336, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/sections_report/introduction.tex", "max_stars_repo_name": "fuljo/my-population-infection", "max_stars_repo_head_hexsha": "30d2c487afe73516e918eb733920a344b6510905", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/sections_report/introduction.tex", "max_issues_repo_name": "fuljo/my-population-infection", "max_issues_repo_head_hexsha": "30d2c487afe73516e918eb733920a344b6510905", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/sections_report/introduction.tex", "max_forks_repo_name": "fuljo/my-population-infection", "max_forks_repo_head_hexsha": "30d2c487afe73516e918eb733920a344b6510905", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.0465116279, "max_line_length": 522, "alphanum_fraction": 0.6435532234, "num_tokens": 1587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382094310357, "lm_q2_score": 0.7431680029241321, "lm_q1q2_score": 0.6402676305456955}} {"text": "\\documentclass[12pt, leqno]{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n%american mathematical society - AMS \n\n\\usepackage{enumerate}\n\n\\usepackage{wrapfig}\n\\usepackage{subfig}\n\n\\oddsidemargin = 0in\n\\textwidth=6.5in\n\n\\numberwithin{equation}{section} \n\n\\numberwithin{figure}{section}\n\n\\begin{document}\n\n%\\baselineskip=20pt \n\nHello world! This is the beginning of \\LaTeX! \\\\\n\nWe can start a new paragraph at any time by leaving a blank line. \nNot that there are many special characters in \\LaTeX or just \\TeX. \nIn particular \\&, \\%, \\#, \\_, \\{, \\}, carets and tildes. and \\textbackslash\nare all special characters -- if \\LaTeX sees any of these without a preceding backslash, \nit thinks they are commands. \\\\\n\nA macro is just a markup command. \n\n\\section*{Typesetting Mathematics!}\n\nDonnie Knuth is the creator of \\TeX. To typeset mathematics and it is very good at it. \nWe could type a vector as: \n\n$ (v_1, v_2, v_3)^T $.\n\nWe can create display math in a number of ways. \n\n$$ %display math mode.\nf(x) = x^n + 500x^n-1 + \\cdots + 501x\n$$ \nWe can type many things in \\TeX that were not possible in HTML, for example \n\n\\begin{equation}\ny = x \n\\end{equation} \n \n\\begin{displaymath}\ng(x) = \\int_0^\\infty G(t,x) \\, dx % \\, adds just a little bit of extra space -- only in math mode. \n\\end{displaymath} \n\nWe can now do more complicated constructions. For example, we say that $\\lambda$ is an eigenvalue of a matrix $A$ corresponding to a vector $\\vec{u}$ or \\(u\\) if $Au = \\lambda u $ For example \n\n%put some bullshit right here. \n%Remember that you CANNOT HAVE BLANK SPACE AT END OF EQUATION!!!\n\nso $-1$ is an eigenvalue corresponding to eigenvector $[1,0]^T$\n\n\\LaTeX\\ can do more sophisticated mathematical typesetting. We can make piecewise defined functions as follows: \n\n\\[\nf(x) = \\begin{cases}\nx^2 & x\\ge 0 \\\\ \n0 & \\text{otherwise}. \n\\end{cases}\n\\] \n\nHere is an example involving some Greek letters and special functions: \n$$ \n\\Lambda_N = \n\\sum_{n=0}^{N}\\left(\\frac{1}{\\cos\\lambda^n}\\right).\n$$ \n\nWe could write the key part of the proof that the harmonic series diverges as \n\n% &= means align on this thing\n\n\\begin{align}\n\\sum_{i=1}^{2^N-1}\n\\frac{1}{i} &= 1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{2^N-1}\\\\\n&> \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{4} + \n\\overbrace{\\frac{1}{8} + \\cdots + \\frac{1}{8}}^{\\text{4 times}} +\n\\cdots + \\frac{1}{2^N-1} \\quad \\text{for} N > 3\n\\end{align}\n\n\n\\section*{Counting} \n%\\label{sec:math}\n%\\nonumber will give the equation no number. \nAs we saw in Section %\\ref{sec:math} \n\\LaTeX counts many things, such as chapters, sections and other document body parts, equations, \ncitations, theorems and so on. We can manipulate those counters by using the setcounter and addtocounter macros. \n\nTo make our own counter, we just use the \\texttt{setcounter} macro: \n\\newcounter{numberthes}\n\\setcounter{numberthes}{42}\nfor instance, the number of times the word \"the\" has occurred so far is \n\\arabic{numberthes} but now, the number of times ``the\" has occurred. is \n\\addtocounter{numberthes}{2}\n\\arabic{numberthes} \n% these are back quotes by unix geeks `````````\n\n% there are two environments 'align' starts a bunch of stuff(math mode, number equations, align equal signs) \n% there is 'aligned' that does NOT start math mode. It only starts in math mode. \n%ampersands are for alignment \n\n\\subsection*{Theorems \\& Stuff}\n\\label{subsec:theorems}\nIn this subsection, numbered, \\ref{subsec:theorems}, we will discuss the numbering scheme for theorems, definitions corollaries, and such. \n\n\\newtheorem{theorem}{Theorem}[section] %the counter is at the end. \n\\newtheorem{definition}[theorem]{Definition}\n\n\\newtheorem{lemma}[theorem]{Lemma} %counter in the middle uses an exsisting counter\n\n\n\\begin{definition}\nThe definition is where we put the meanings of the terms we use in our theorems. \n\\end{definition}\n\nFirst, we prove a technical result. \n\\begin{lemma}\nhere is where we prove some little technical result taht we will use in the proof of the theorem.\n\\end{lemma}\n\n\\begin{theorem}\nTheorems are where we prove things about the terms we defined in the definitions. \n\\end{theorem}\n\n\\section*{Lists}\nThere are three kinds of lists in every markup language: \nordered lists, unordered lists, and description lists. \n\n\\begin{enumerate}[A)]\n\\item The first item \n\\item The second item \n\\item The ultimate item \n\\end{enumerate}\n\nHere is an unordered list. \n%\\noindent \n\\begin{itemize}\n\\item one \n\\item two \n\\item three\n\\end{itemize}\n\nHere is a description list. \n\\begin{description}\n\\item[The first definition -] the first def. \n\\item[second -] the second. etc etc \n\\end{description}\n\n\n\\section*{Page layout and spacing}\nPage layout in \\TeX\\ is not as transparent as it is in writing in WYSiWYG word processor \ninstead of setting margins, as we woul in a WYSiWYG program, \\LaTeX\\ knows only where the \nupper left corner of hte text area is relative to the page, and how wide \nand tall the text area is. \nFor this purpose it sets many variables that control the positions and sizes of the text area. \nWe can change the values of these variables. \n\nThere are two ways to change the values of these variables: We can do it directly (set the variable equal to exactly what we want.); or we can use the \\texttt{newdimension} command. \nHere is a list(possibly incomplete) of the variables of what we can change and control. \n\n\\begin{description}\n\\item[oddsidemargin-]This is distance of the text area from the left edge of the page on \\textbf{odd} numbered pages. \nThis is useful when you are writing a book! In the article documentclass, this controls both odd and even numbered pages. Note that the distance we specify here is relative to default 1 inch margin. Not that the documentclass also imposes a separate default. \n\\item[evensidemargin-]this is the distance of the text area from the left edge of the page on even-numbered pages This only applies in some documentclasse, such as in book. \n\\item[topmargin-] This is the top of the page. \n\\item[textwidth-]the width of the text area. \n\\item[textheight-] the height of the text area. \n\\item[headheight-] The height of the header. \n\\item[headsep-]The height of the separation of the header from the text area. \n\\item[footskip-] The distance from th text area to the footer. \n\\item[baselineskip-] The distance from the bottom of on line to the next. \n\\item[baselinestrech] This is actually not a dimension; it is a macro that controls how much the space between\nline is stretched. For example, to get double-spaced text you could type\\\\\n%\\texttt{\\textbackslash renewcommand\\{\\textbackslash baselinestretch\\{2\\}\\}}\n\\item[parskip-] The extra distance to place between paragraphs. \n\\item[parindent-] the amount o indentation at the beginning of every paragraph. Except the first paragraph in article class. \n\n\\subsection{Spacing}\n\nBefore we can really discuss spacing, we have to talk about horizontal and vertical mode. \\LaTeX\\ \ntakes characters not separated by spaces and pastes them together into a word. \nIt takes other white space and replaces it by so-called \\textit{glue}.\nThis is illustrated below: \n%\\begin{description}\n\n%\\begin{center}\n%\\fboxsep=0pt\n%\\framebox(stuff){M}\n%\\end{center}\n\nAll of this word and line formation takes place in horizontal mode. \nPutting lines one after another requires going into vertical mode -- where \\LaTeX\\ figures out how to fit the various lines on the page. \n\nwe can make horizontal spaces \\hspace{10pt} any time we are in horizontal mode. \n\\vspace{0.5in} \\\\\nSpace can be either positive or negative. Note that \\LaTeX cannot put in vertical space except when it is in vertical mode. \n\n%\\end{description}\n\nthere are some special spacing commands that allow us to put in a lot of glue at once. \nIn particular in horizontal mode, the \\texttt{hfil} and \\texttt{hfill} commands put in as much space as we need. \nFor example, look below. \n\\begin{center}\nLeft \\hfill Right \\\\\nLeft \\hfil Right \\\\\nLeft \\hfill middle \\hfill right\n\\end{center}\n\nLikewise, there are commands called \\texttt{vfil} and \\texttt{vfill} that give as much vertical space as we need. \nThere are a couple of similar macros that put dots or rules in place of these spaces. These are called \\texttt{dotfill} and \\texttt{hrulefill}, respectively. \n\\begin{center}\nChapter 12 \\dotfill 121 \n\\end{center}\n\nAnother aspect of spacing is the boxes \\TeX\\ constantly uses form lines and so on. \nFor example, we already saw that we could put a box around any text using \\texttt{framebox} \\framebox{framebox}. \nWe can make a a plain box any time using \\makebox[3in]{makebox} \n\n\n\\parbox{2in}{To get a paragraph of a certain size, use \\texttt{parbox}. Not that the width of this is an obligatory argument.}\n\n\n\n\\end{description}\n\n\n\\section{Graphics}\n\nBefore we start, we should probably mention one or two ways of storing graphics that are not traditional \nimage formats. In particular, the EPS(Encapsulated PostScript) format is often used for storing images used in \\TeX. \nThis is an old-fashioned way to do \n \nto incorporate images into your document, you must first load the \\texttt{graphicx}, After that we can use the famous \\texttt{includegraphics} command to put in your image: viz\\\\\n\\begin{figure}[ht] % 'h' means put it here and 'ht' means that put it at top the and here. 'ht!' put it here now! \n\\centering\n%\\includegraphics[width=0.5\\textwidth]{/path/to/picture.jpg}\n%\\caption{Who brought the cat?!}\n\\end{figure}\n\nNote that from now on we will never refer to figures using any formulations such as ``in the figure below''. Instead, we always give the figure a name (number), and then refer to it using that. \n\nthere are many more packages to allow us to place our figures cleverly, attractively, etc., but we must load each \none separately. Note that some of the these packages are not compatible with features of other packages -- never load packages you do not need. For example, if we want to place a figure to one side of the page with text flowing around it, we use the \\texttt{wrapfig} package. \n\n%\\begin{wrapfigure}{l}{2.54in}\n%\\includegraphics[width=2.5in]{/path/to/filname/cat.jpg}\n%caption{Tijgertje's like to sleep}\n%\\label{fig:hanger}\n%\\end{wrapfigure}\n\nNote that if we want to have figure numbers in the form section.figure, we can again use the \\texttt{numberwithin} macro. We can also refer to %\\ref{fig:hanger} \nusing a \\texttt{ref} command. Also note that the figure number is not incremented until we use the \\texttt{caption}\nmacro -- do not use your \\texttt{label} macro until after that happens. \n\nAnother common requirement is to have two figures side by side. We could actually do that in an ordinary \\texttt{figure} environment, but we would not get captions on each individual figure. Figure environment isn't about figures, but about floating elements. If we want captions then we must load another package: the \\texttt{subfig} package. \n%\\begin{figure}[ht!]\n%\\begin{subfloat}[How it is not done.]\n%{\n%\\includegraphics[.45\\textwidth]{/path/to/the/image/cat.jpg}\n%\\label{fig:done}\n%}\n%\\end{subfloat}\n%\\end{figure}\n%\n%\\begin{figure}\n%\\begin{subfloat}[How it is done.]\n%{\n%\\includegraphics[.45\\textwidth]{/path/to/the/image/cat.jpg}\n%}\n%\\end{subfloat}\n%\\end{figure}\n\nNow we can refer to Figure \\ref{fig:done} using the \\texttt{label-ref} system as usual. \n\n\\section*{Tables}\n\nThere is another environment that works exactly like the \\texttt{figure} environment, except that the object in it that the objects in it that have captions are labeled ``Table''. That environment is, not surprisingly, called the \\texttt{table} environment. In order to make an actual table we must invoke the \\texttt{tabular} environment. \nTable \\ref{tab:pets} shows an example of some of the simplest features of the \\texttt{tabular} environment. \n\n\\begin{table}[ht!] %difference between a table and a figure? Nothing other than the name. \n\\caption{A summary of pets}\n\\label{tab:pets}\n\\begin{tabular}{| l | c | p{2in} |}\nAnimal & Sociability & something and feeding \\\\ \n\\hline\nDog & high & blah blah bblah \\\\\n\\hline\nCat & low & blah blah blah \\\\ \n\\cline{2-3}\npolitician & \\multicolumn{2}{c|}{Avoid this pet at all costs} \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\nNote that the wrapfig package supplies a \\texttt{wraptable} environment that lets tables float to one side or the other. \n\n\\section{References}\n\nWe often (always) need to make references to other sources in our documents. One way of handling those is to use \nthe \\texttt{thebibliography} environment. The key macros associated with this are called \\texttt{bibitem} \nand \\texttt{cite} command to actually refer to it. For example, this document contains some bogus references to Donald Knuth's work \\cite{ref:knuth79}.\n\nNote that most scholars create a database of references for the work they cite often. For this they use software called Bib\\TeX. We will say no more about this. \n\n\\begin{thebibliography}{XX}\n\n\\bibitem{ref:knuth79} %when we refer to it by name we use this name. \nDonald Knuth, \\textbf{The \\TeX book} Addison Wesley, New York, 1979. \n\n\\bibitem{ref:lamport69}\nLeslie Lamport, \\textbf{The \\LaTeX\\ Book}, Addison Wesley, New York, 1969. \n\\end{thebibliography}\n\n\\section{Programming}\n\n\\newcommand{\\wsu}{Washington State University}\n\nWe can let \\TeX\\ do a lot of our work for us. For example, if there is some text that appears very often in our \npaper, we can define our own macro to make that text. \nWe can use \\texttt{newcommand} to create a new command that\nwill save us a bunch of time. \n\n\\newcommand{\\tc}[1]{\\textbackslash\\texttt{#1}}\n\nSometimes we need to define commands that operate on some argument. For example we have needed to typeset the \\TeX\\ commands we were talking about, but we were too lazy to type all the stuff every time. Above we defined a \\TeX\\ command called \\tc{tc} that does all that for us. This way we can talk about any command we like, such as \\tc{numberwithin}, and get the output while only having to type the argument. Here is another. \n\n\\newcommand{\\ltx}[2]{L\\raisebox{#1}{a}\\raisebox{#2}{}$\\chi$} \n\nThis one makes a \\LaTeX\\ command that raises the a as much as we want: viz. \\ltx{2pt}{-2pt}.\n\nThere is also a \\tc{newenvironment} command that defines an environment that way we want it. \n\n\\section{Odds 'n ends}\n\n\\begin{enumerate}\n\\item We have seen that tilde is a special character. It is actually a non-breaking space. The hyphenation program is not allowed to hyphenate there. \n\\item If you need to give \\TeX\\ a hint about where a convenient place to hyphenate is, you cna use the \\textbackslash- command. \n\\item The \\texttt{verbatim} environment allows us to typeset text obeying all spaces, line breaks and so on. This is used particularly in typesetting computer code. \n\\end{enumerate}\n\n\\begin{verbatim}\nfunction f(x)\nf = 5 \nreturn f\n\\end{verbatim}\n\n\n\\end{document}", "meta": {"hexsha": "9ac17928a1abeadb33b2019bfc8918d44ed14681", "size": 14786, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "math300/sept21.tex", "max_stars_repo_name": "johnnydevriese/wsu_courses", "max_stars_repo_head_hexsha": "b55efd501c2d8f0651891f422a486e32533f5aa0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "math300/sept21.tex", "max_issues_repo_name": "johnnydevriese/wsu_courses", "max_issues_repo_head_hexsha": "b55efd501c2d8f0651891f422a486e32533f5aa0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "math300/sept21.tex", "max_forks_repo_name": "johnnydevriese/wsu_courses", "max_forks_repo_head_hexsha": "b55efd501c2d8f0651891f422a486e32533f5aa0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.5337078652, "max_line_length": 429, "alphanum_fraction": 0.7473961856, "num_tokens": 4062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.6402676301623184}} {"text": "%---------------------------Aspect Ratio----------------------------------\n\\section{Aspect Ratio\\label{s:tet-aspect-ratio}}\n\nThe aspect ratio of a tetrahedron $K$ is: \n\\[\n\\frac{L_{\\max}}{2\\sqrt{6}r}.\n\\]\n\n\\tetmetrictable{aspect ratio}%\n{$1$}% Dimension\n{$[1,3]$}% Acceptable range\n{$[1,DBL\\_MAX]$}% Normal range\n{$[1,DBL\\_MAX]$}% Full range\n{$1$}% Equilateral tet\n{\\cite{frey:00}}% Citation\n{v\\_tet\\_aspect\\_ratio}% Verdict function name\n", "meta": {"hexsha": "44602b32e6c714d3a25fcf21a1ed54db38b04b55", "size": 530, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetAspectRatio.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetAspectRatio.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TetAspectRatio.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 31.1764705882, "max_line_length": 74, "alphanum_fraction": 0.479245283, "num_tokens": 149, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8615381952105442, "lm_q2_score": 0.7431680029241321, "lm_q1q2_score": 0.6402676199774812}} {"text": "\\section{Generalization of Wright-Fisher model dynamics to non-fixed population size where the population size fluctuates randomly}\n\\subsection{Motivation}\nTodo: Describe motivation - maybe add few sentences to motivate the research\n\\subsection{Results}\nLet me introduce the specific model with three main assumptions:\n\\begin{enumerate}\n\\item\nnon-fixed population size,\n\\item\nnon-overlapping generations,\n\\item\nrandom selection.\n\\end{enumerate}\nThe model is inspired directly by Wright-Fisher model without a fixed population size assumption, hence it can be considered as a generalization of the latter. Random selection feature corresponds to random mating assumption in Wright-Fisher model and means that in order to form a new generation each individual copies the state of a randomly selected parent, so no a priori defined fitness plays role except for the obvious rule that the rich gets richer.\nIt is difficult to observe non-overlapping generations in the nature, however this assumption plays a crucial role in the model so cannot be relaxed. One can think of a model without this limitation but deriving analytics results will be more challenging in such a setup. On the other hand the model should be not more difficult than it has to be and it might be that after appropriate scaling this feature has none or little effect on the quantitative output however may influence the qualitative outcome.\nFurther necessary assumptions are:\n\\begin{enumerate}\n\\item\nStart with fixed number of $n_0$ individuals - each possess a different opinion\n\\item\nExpected population of each generation stays positive i.e. \n\\begin{equation}\n\\mathbb{E}(\\sum_{i=1}X_t(i))>0\n\\end{equation}\n\\item \nPopulation of each generation is independent:\n\\begin{equation}\nn_t\\equiv\\sum_{i=1}X_t(i)\n\\end{equation}\n\\end{enumerate}\n\nIn terms of graph theory language we start with a fixed setup - initial generation $0$ consists of $n_0$ nodes. Each node $x_i$ has a different colour $i$. Every successive generation $t$ consists of $n_t$ nodes. Each node inherits a colour of its parent - a uniformly selected node from the generation $t-1$. \nNow one initial node $i$ (from generation $0$) is selected and a following martingale is considered \n\\begin{equation}\n\\{Y_t(i)=\\frac{X_t(i)}{n_t}\\}_{t\\geq0}\n\\end{equation}\nwhere $X_t(i)$ is a number of offsprings of node $i$ in generation $t$. For a given fixed $T>0$ we consider a following stopping time:\n\\begin{equation}\n\\tau=\\begin{cases} t_0 &\\qquad \\mbox{ if } X_{t_0}(i)=0 \\mbox{ for some time } t_0 \\leq T \\mbox{ (process dies until T) }\\\\ T &\\qquad \\mbox{ if } X_{T}(i)>0\\end{cases}\n\\end{equation}\nNext by applying Optional Stopping Theorem for martingale we have:\n\\begin{equation}\n\\mathbb{E}(Y_{\\tau}(i))=\\frac{1}{n_0}=\\mbox{Pr}\\{Y_T(i)>0\\}\\cdot\\mathbb{E}(Y_{T}(i) | X_{T}(i)>0)\n\\end{equation}\n$Pr\\{Y_T(i)>0\\}$ is a probability that the process will not die until time $T$. Hence, the number of colours at time $T$ is given by the formula:\n\\begin{equation}\n\\mathbb{E}(\\mbox{\\# colours at time } T)=\\frac{1}{\\mathbb{E}(Y_T(i) | X_T(i) >0)}\n\\end{equation}\n\nIn order to go further let’s switch a perspective.\nThe model can be alternatively viewed as a (possibly infinite) series of urns (non-overlapping generations) where the content of each urn (population of the generation) depends only on the content of the previous urn and the capacity of the current urn. Zeroth urn is filled with $n_0$ different balls. Then the process of populating the $t$-th urn is constrained with $n_t$ drawings with replacement from the previous $t-1$-th urn filled with $n_{t-1}$ labelled balls where each ball has exactly single label and some labels could be sticked to many balls. Thus multiplicity of labels/colours/languages can be approximated using mixed Poisson distribution in the following way:\n\nMultiplicity of colours to observe in first generation $X_1$ is a discrete random variable which can be approximated by poisson distribution with $\\lambda_1=\\frac{n_1}{n_0}$, so $Pr\\{X_1(i)=0\\}=\\exp(-\\lambda_1)$ meaning that expected number of $\\exp(-\\frac{n_1}{n_0})$ colours will disappear from the system at the first time step.\nIn general multiplicity of colurs in the first urn is fully described by the Poisson distribution:\n\\begin{equation}\nP(X_1(i)=k)=\\mbox{exp}(-\\lambda_1)\\frac{\\lambda_1^k}{k!}\n\\end{equation}\nLet’s proceed to the content of the next urn. We start from stating an exact formula for conditional probability of $X_2(i)$ under condition that we know the value of $X_1(i)$ (i.e. multiplicity of $i$-th colour in the first urn is known apriori):\n\\begin{equation}\nP\\left(X_2(i)=l|X_1(i)=k\\right)=\\mbox{exp}\\left(-k\\lambda_2\\right)\\frac{\\left(k\\lambda_2\\right)^l}{l!}\n\\end{equation}\nwhere $\\lambda_i\\equiv\\frac{n_i}{n_{i-1}}$ is a factor which corresponds to scaling of the population in the $i$-th step. The above formula is well defined only for $X_1(i)>0$. Let’s further assume that $P(X_t(i)=0|X_{t-1}(i)=0)\\equiv1$ (so if a colour/a label is lost at some point, then it cannot be pushed back in the system in the future). Now we can calculate unconditional probability of $X_2(i)$ using the law of total probability:\n\\begin{equation*}\n\\begin{split}\nP\\left(X_2(i)=l\\right) & =\\sum_{k}P\\left(X_2(i)=l|X_1(i)=k\\right)P\\left(X_1(i)=k\\right) \\\\ & =\\sum_{k=1}\\mbox{exp}\\left(-k\\lambda_2\\right)\\frac{\\left(k\\lambda_2\\right)^l}{l!}\\mbox{exp}\\left(-\\lambda_1\\right)\\frac{\\left(\\lambda_1\\right)^k}{k!} + \\delta_{l0}\\cdot\\mbox{exp}\\left(-\\lambda_1\\right)\n\\end{split}\n\\end{equation*}\n\nThis can be simplified by observing a fact that under the sum we have a sum of probability mass function of another Poisson distribution (except for mass of point $0$ and normalization factor). The computation is lengthy but straightforward and will be given in details only for $X_2(i)$ as further time steps involves the same method:\n\\begin{equation*}\n\\begin{split}\nP\\left(X_2(i)=0\\right) & =\\sum_{k}P\\left(X_2(i)=0|X_1(i)=k\\right)P\\left(X_1(i)=k\\right) \\\\ & =\\sum_{k=1}\\mbox{exp}\\left(-k\\lambda_2\\right)\\mbox{exp}\\left(-\\lambda_1\\right)\\frac{\\left(\\lambda_1\\right)^k}{k!} + \\mbox{exp}\\left(-\\lambda_1\\right) \\\\ &\n=\\mbox{exp}\\left(-\\lambda_1\\right)\\left(\\sum_{k=1}\\frac{\\left(\\lambda_1\\mbox{exp}\\left(-\\lambda_2\\right)\\right)^k}{k!} + 1 \\right) \\\\ &\n=\\mbox{exp}\\left(-\\lambda_1\\right)\\left(\\left(1-\\mbox{exp}\n\\left(-\\lambda_1\\mbox{exp}\\left(-\\lambda_2\\right)\\right)\\right)\\mbox{exp}\\left(\\lambda_1\n\\mbox{exp}\\left(-\\lambda_2\\right)\\right)+1\\right)\n\\end{split}\n\\end{equation*}\n\nFinally, the calculation above gives the elegant exact formula for probability of extinction until $t=2$:\n\\begin{equation}\nP(X_2(i)=0)=\\mbox{exp}\\left(-\\lambda_1\\left(1-\\mbox{exp}\\left(-\\lambda_2\\right)\\right)\\right)\n\\end{equation}\nTODO: Now the missing part - please compute explicit the value of $P(X_2(i)=l>0)$.\nWithout that computation, all remaining results are not valid.\nPlease check \\url{https://en.wikipedia.org/wiki/Poisson_distribution#Higher_moments} for further reference.\n\n\nIn fact the following exact formula is valid for any $t>1$:\n\\begin{equation}\nP(X_t(i)=0)=\\mbox{exp}\\left(- \\lambda_1 \\left(\\cdots\\left(1-\\mbox{exp}\\left(- \\lambda_{t-1}\\left(1 - \\mbox{exp}\\left(-\\lambda_{n_t}\\right)\\right)\\right)\\right)\\cdots\\right)\\right)\n\\end{equation}\n\nProof of this formula is based on induction and very useful property explained above on the recurrent use of sum of probability mass function except mass point $0$ - that for any $a$, $b$ and $\\left(x_k\\right)_{k\\in \\mathbb{N}}$ being a collection of positive iterators, we have: \n\\begin{equation}\n\\sum_{x_k=1}\\exp(a)^{x_k}\\frac{(x_{k-1}b)^{x_k}}{x_k!}=\n\\exp(x_{k-1}b\\exp(a))-1\n\\end{equation}\n\n\\begin{equation*}\n\\begin{split}\nP\\left(X_{t}(i)=0\\right) & =\\sum_{x_{t-1}}P\\left(X_{t}(i)=0|X_{t-1}(i)=x_{t-1}\\right)P\\left(X_{t-1}(i)=x_{t-1}\\right) \\\\ & \n=\\sum_{x_{t-1}}P\\left(X_{t}(i)=0|X_{t-1}(i)=x_{t-1}\\right)\\cdots \\sum_{x_{1}}P\\left(X_{2}(i)=x_{2}|X_{1}(i)=x_{1}\\right)P\\left(X_{1}(i)=x_1\\right)\n\\end{split}\n\\end{equation*}\n\n\\begin{equation*}\n\\begin{split}\n& P\\left(X_{t}(i)=0\\right) \\\\ & = \\sum_{x_{t-1}=1} \\exp(-\\lambda_{t})^{x_{t-1}}\n\\sum_{x_{t-2}=1} \\exp(- \\lambda_{t-1})^{x_{t-2}}\\frac{\\left(x_{t-2}\\lambda_{t-1}\\right)^{x_{t-1}}}{x_{t-1}!} \\cdots \\sum_{x_{1}=1} \\exp( - \\lambda_1)\\frac{\\lambda_1^{x_1}}{x_1!} + P\\left(X_{t-1}(i)=0\\right)\n\\end{split}\n\\end{equation*}\n\n", "meta": {"hexsha": "3e274441d954c26b28188cf669f6cf99ce88ed78", "size": 8325, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "LaTeX/Analytic_WFExtended.tex", "max_stars_repo_name": "eMaerthin/agent-based-models-for-language-evolution-phd-thesis", "max_stars_repo_head_hexsha": "d61ad554f7a3bf5093aeffb5d83d56a8055ad9b8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "LaTeX/Analytic_WFExtended.tex", "max_issues_repo_name": "eMaerthin/agent-based-models-for-language-evolution-phd-thesis", "max_issues_repo_head_hexsha": "d61ad554f7a3bf5093aeffb5d83d56a8055ad9b8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "LaTeX/Analytic_WFExtended.tex", "max_forks_repo_name": "eMaerthin/agent-based-models-for-language-evolution-phd-thesis", "max_forks_repo_head_hexsha": "d61ad554f7a3bf5093aeffb5d83d56a8055ad9b8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.0263157895, "max_line_length": 678, "alphanum_fraction": 0.732972973, "num_tokens": 2668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615381987656672, "lm_q2_score": 0.743167997235783, "lm_q1q2_score": 0.6402676177188048}} {"text": "\\chapter{Deep Learning and Large Scale Computing}\n\\section{Deep Learning and Large Scale Computing}\n\t\\subsection{What is a learning algorithm?}\n\t\t\n\t\tA \\textbf{machine learning algorithm} is an algorithm that is able to learn form data. But\twhat do we mean by learning? Mitchell (1997) provides the definition \"A computer program is said to learn from experience $E$\twith respect to some class of tasks\t$T$\tand performance measure\t$P$, if its performance at tasks in\t$T$, as measured by $P$, improves with experience $E$.\"Learning is not the task, but the means of attaining the ability to perform the task. The result of the learning is a model that is used to perform tasks.\n\t\t\n\t\tThe machine learning algorithm grew out of work in \\textbf{Artificial Intelligence (AI)} and take advantage of new capability of computing resources. \\textbf{Deep learning} is a subset of machine learning.\n\n\t\\subsection{Types of learning algorithm}\n\t\t\n\t\t Machine learning algorithms can be broadly categorized as \\textbf{unsupervised} or \\textbf{supervised} or \\textbf{Reinforcement learning algorithms\n\t\t }\tby what kind of experience they are allowed to have during the learning process.\n\t\\begin{itemize}\t\t\n\t\t\\item Unsupervised learning algorithms learn from a dataset containing many features, then learn useful properties of the structure of this dataset. E.g. K-means clustering\t\n\t\t\\item Supervised learning algorithms learn from a dataset containing features, but each example is also associated with a label or target. E.g. classification , regression method to predict housing price\t\n\t\t\\item Reinforcement learning algorithms interact with an environment, so there is feedback loop between the learning system and its experience. But it's out of the scope of this note.\t\t\n\t\\end{itemize}\t\n\tHere we offer a simple example to illustrate the difference between the supervised and unsupervised learning algorithm.\n\t\n\t Suppose we have a set of points in $\\mathbb{R}^2: \\{(x_1,y_1),(x_2,y_2),\\dots,(x_n,y_n)\\}$.\n\t \\begin{itemize}\n\t \t\\item Ther's no more information about the points except their coordinates. We want to split them into several subsets (which is 'clustering') based on their coordinates. That's an unsupervised algorithm.\n\t \t\\item The points are painted red or blue. We want to learn from the data about the reason why the point is red or blue. If given a new point, we can label it as red or blue according the points before. That's an example of classification, which is one kind of supervised learning algorithm.\n\t \\end{itemize}\n\t\\section{Maximum Likelihood Estimation}\n\tAs the process of machine learning is to estimate a set of parameters $\\theta$ based on observations of examples X, \\textbf{maximum likelihood} is often considered the preferred estimator to use for machine learning.\n\t\n\t\\begin{itemize}\n\t\t\\item Consider a set of $m$ examples $\\mathbb X=\\{\\bm x^{(1)},...,\\bm x^{(m)}\\}$ drawn independently from the true but unknown data generating distribution $p_{data}(\\bm x)$.\n\t\t\\item Let $p_{model}(\\bm x;\\bm \\theta)$ be a parametric family of probability distributions over the same space indexed by $\\bm \\theta$.\n\t\t\\item The maximum likelihood estimator for $\\bm \\theta$ is then defined as\n\t\t\\begin{equation*}\n\t\t\\begin{split}\n\t\t\\bm \\theta_{ML} &= \\arg \\max_{\\bm \\theta}p_{model}(\\mathbb X;\\bm \\theta) \\\\\n\t\t&=\\arg \\max_{\\bm \\theta}\\prod_{\\bm x \\in \\mathbb X} p_{model}(\\bm x;\\bm \\theta)\\\\\n\t\t&=\\arg \\max_{\\bm \\theta}\\sum_{\\bm x \\in \\mathbb X} \\log p_{model}(\\bm x;\\bm \\theta)\\\\\n\t\t\\end{split}\n\t\t\\end{equation*}\n\t\\end{itemize}\n\n\t\\section{Input data}\n\tPreviously observed data in the experience can be split to three categories:\n\t\\begin{itemize}\n\t\t\\item Training data: used to train the model\n\t\t\\item Validation data: used to adjust the model\n\t\t\\item Test data: used to evaluate the generalization of the model\n\t\\end{itemize}\n\tTraining data and validation data are usually split from a large observed data set. The ratio between training data and validation data is commonly 4:1 or 9:1.\n\t\\subsection{Capacity and Performance}\t\n\n\tThe capacity of the learning algorithms can be underfitting, appropriate capacity and overfitting. Underfitting occurs when the model is not able to obtain a sufficiently low error value on the training set. Overfitting occurs when the gap between the training error and test error is too large.\n\t\n\tThe factors determining the performance of the machine learning algorithms are:\n\t\\begin{itemize}\n\t\t\\item Test error (final error of the generated model) The error is calculated by a cost function based on the test data set\n\t\t\\item The gap between training and validation error\n\t\\end{itemize}\n\tThe central challenge in machine learning is that we must perform well on new, previously unseen inputs, not just those on which our model was trained. So Machine learning algorithms indirectly minimize the test error. Regularization is the modification of machine learning algorithms to reduce validation error but not the training error in order to avoid overfitting.\n\t\n\\section{Neural Network}\nAccording to Maureen Caudill{\\color{red} LZ:???} neural network is a computing system made up of a number of simple, highly interconnected processing elements, which process information by their dynamic state response to external inputs.\n\n\n{\\bf Activation function}. In a neural network, an activation function of a node determines the output of that node for a given input. Inspired by biological neurons, an activation function basically works as a switch that can set the neuron on \"ON\" or \"OFF\".\n\nThe sigmoid function is a common choice for activation function. It is defined by\n\\begin{equation}\n\\sigma(t) = \\frac{1}{1+e^{-t}}.\n\\end{equation}\nThe sigmoid function is monotonically increasing over $\\mathbb{R}$ and is infinitely-differentiable. It has limit 0 and 1 as $t$ approaches $-\\infty$ and $+\\infty$, respectively.\n%The graph of $\\sigma(t)$ is shown in Figure \\ref{fsig}.\n\nA useful property of $\\sigma(t)$ is that its derivative can be expressed by the function itself.\n{\\color{red} LZ:Add graph of sigmoid}\n\\begin{lemma}\n\\label{lsig}\n$\\sigma'(t) = \\sigma(t)(1-\\sigma(t))$.\n\\end{lemma}\n\n\nBesides the sigmoid function, commonly used activation functions are:\n\n\\begin{enumerate}\n\\item\nBinary step function: $f(t) = \\mathbbm{1}_{t\\ge 0}$.\n\\item\nHyperbolic tangent function: $f(t) = \\tanh(t)=\\frac{2}{1+e^{2t}}-1$.\n\\item\nArctan function: $f(t) = \\tan^{-1}(t)$.\n\\item\nRectified linear unit (ReLU) $f(t) = \\max\\{0, t\\}$.\n\\end{enumerate}\n\nA \\textit{neural network} is a network model with non-linear unites on its hidden layers.\n\n\n\n\n{\\bf Hyper-parameters.} Most machine learning algorithms involve hyper-parameters which are variables set before actually optimizing the model's parameters. For a neural network, hyper-parameters include the depth of the network $d$, the width of each layer $n$, the regularization parameter $\\lambda$, the learning rate for gradient descent $r$, and the initial values for weight matrices $(\\bm{W_1}(0),...,\\bm{W_d}(0))$ and biases $(\\bm{b_1}(0),...,\\bm{b_d}(0))$. {\\color{red} LZ:???}There is no theorem stating how to choose the best hyper-parameters, we now choose them merely by experience and luck.\n\n\\section{Back Propagation}\n\n{\\bf Training a neural network model}. Suppose we have $m$ labeled training examples\n$$\\{(\\bm{x^1, y^1}),...,(\\bm{x^m, y^m})\\}.$$ Let $\\sigma_k$, $\\bm{W_k}$, $\\bm{b_k}$, be the activation function, weight matrix, bias vector, resp. for the $k_{th}$ hidden layer, $k=1,2,...,d-1$. Let $\\sigma_d$, $\\bm{W_d}$ and $\\bm{b_d}$ be the activation function, weight matrix and bias vector for the output layer. \n\nThe neural network model can be defined recursively as follows:\n\\begin{equation}\n\\label{enn}\n\\aligned\n\\bm{z_1}=&\\bm{W_1^Tx}+\\bm{b_1}\\\\\n\\bm{a_1}=&\\sigma_1(\\bm{z_1})\\\\\n\\bm{z_k}=&\\bm{W_k^T}\\bm{a_{k-1}}+\\bm{b_k}\\\\\n\\bm{a_k}=&\\sigma_k(\\bm{z_k}), k=2,...,d.\n\\endaligned\n\\end{equation}\nHere we use the convention $\\sigma(x_1,...,x_n) = (\\sigma(x_1),...,\\sigma(x_n))$ and call $\\bm{z_k}$ weighted input and $\\bm{a_k}$ weighted output hence $\\bm{a_d}(\\bm{x})$ is the output from the neural network given $\\bm{x}$ as the input data.\n\n{\\bf Cost function}.\nThe error of a neural network with respect to a sample $(\\bm{x^j, y^j})$ is defined by cost (loss) function. One way is to use the quadratic loss function\n\\begin{equation}\nL(\\bm{x^j, y^j}) =\\|\\bm{y^j} - \\bm{a_d}(\\bm{x^j})\\|^2+\\lambda \\sum_{i=1}^d ||\\bm{W_i}||_F^2,\n\\end{equation}\nwhere the second term is the regularization term to avoid overfitting and $\\lambda$ is the regularization parameter.{\\color{red} LZ: remove regularization?}\n\nAn alternative way is to use the cross-entropy loss function\n\\begin{equation}\nL(\\bm{x^j, y^j}) = -[\\bm{y^j} \\ln (\\bm{a_d}(\\bm{x^j}))+(1-\\bm{y^j}) \\ln (1-\\bm{a_d}(\\bm{x^j}))]+\\lambda \\sum_{i=1}^d ||\\bm{W_i}||_F^2,\n\\end{equation}\n{\\color{red} LZ: remove regularization?}\n\nThe goal of training a neural network model is to find parameters $(\\bm{W_1},...,\\bm{W_d},\\bm{b_1},..,\\bm{b_d})$ such that the total loss\n\\begin{equation}\n\\label{etotalloss}\nL_{total} = \\frac{1}{m} \\sum_{j=1}^m L(\\bm{x^j, y^j})\n\\end{equation}\nis as small as possible.\n\n{\\bf Back propagation}.\nThe parameters are adjusted using the following process called \\textit{back-propagation} w.r.t. $L_{total}$.\n\nThere is an intuition to introduce the error $\\delta_l^j$.\nSuppose there is a very small change $\\Delta z_l^j$ added to the weighted input to the $j_{th}$ neuron in the $l_{th}$ layer, after the forward propagation, the final cost function changes by $\\frac{\\partial}{\\partial \\bm{z_l^j}}L_{total} \\Delta z_l^j$. If the partial derivative $\\frac{\\partial}{\\partial \\bm{z_l^j}}L_{total}$ is large, we can choose $\\Delta z_l^j$ with the different sign to decrease the cost function. Otherwise if the partial derivative is near zero, it's difficult to change the cost function. Therefore we use $\\delta_l^j = \\frac{\\partial}{\\partial \\bm{z_l^j}}L_{total}$ to denote the error of the $j_{th}$ neuron in the $l_{th}$ layer.\nFirst we compute the error in the last layer.\n{\\color{red}LZ: bold sigma}\n\n\\begin{equation}\n\\aligned\n\\bm{\\delta_d^j} =& \\frac{\\partial}{\\partial z_d^j}L_{total}\\\\\n=& \\sum_{k} \\frac{\\partial L_{total}}{\\partial a_k^j}\\frac{\\partial a_k^j}{\\partial z_d^j}\\\\\n=& \\frac{\\partial L_{total}}{\\partial a_d^j} \\sigma'(z_d^j),\\\\\n\\bm{\\delta_d} =& \\nabla_{a_d} L_{total}\\odot \\sigma'(z_d).\n\\endaligned\n\\end{equation}\nHere $\\odot$ is the Hadamard product meaning component-wise multiplication of two vectors. In the third equality results from that $a_d^k$ relies only on the weighted input $z_d^k$.\n\nThen we compute the error in the $l_{th}$ layer, $l=1,2,...,d-1$.\n\\begin{equation}\n\\aligned\n\\delta_l^j =& \\frac{\\partial}{\\partial {z_l^j}}L_{total}\\\\\n=& \\sum_{k} \\frac{\\partial L_{total}}{\\partial z_{l+1}^k}\\frac{\\partial z_{l+1}^k}{\\partial z_l^j}\\\\\n=& \\sum_{k} \\frac{\\partial z_{l+1}^k}{\\partial z_l^j} \\bm\\delta_{l+1}^k\\\\\n=& \\sum_{k} W_{l+1}^{kj} \\delta_{l+1}^k \\sigma'(z_l^j),\\\\\n\\bm{\\delta_l} =& ((W_{l+1})^T \\bm\\delta_{l+1} \\odot \\sigma'(z_l).\n\\endaligned\n\\end{equation}\n\n\nThe computation of $\\bm{\\delta^k}$ depends on the knowledge of $\\bm{\\delta^{k+1}}$, and here comes the name back-propagation. The gradient of the loss function with respect to each parameter can be computed as follows: for each $k=1,...,d$ we have\n\\begin{equation}\n\\label{ebpparam}\n\\aligned\n\\frac{\\partial}{\\partial W_l^{jk}}L_{total}\n=& \\sum_{i} \\frac{\\partial L_{total}}{\\partial z_l^i}\\frac{\\partial z_l^i}{\\partial W_l^{jk}}\\\\\n=& \\frac{\\partial L_{total}}{\\partial z_l^j}\\frac{\\partial z_l^j}{\\partial W_l^{jk}}\\\\\n=& \\delta_l^j a_{l-1}^k,\\\\\n\\frac{\\partial}{\\partial b_l^{j}}L_{total}\n=& \\sum_{i} \\frac{\\partial L_{total}}{\\partial z_l^i}\\frac{\\partial z_l^i}{\\partial b_l^{j}}\\\\\n=& \\frac{\\partial L_{total}}{\\partial z_l^j}\\frac{\\partial z_l^j}{\\partial b_l^{j}}\\\\\n=& \\delta_l^j.\n\\endaligned\n\\end{equation}\n{\\color{red}LZ: matrix form:\n\\begin{align*}\n \\frac{\\partial L_{total}}{\\partial \\bm W_l} =& \\bm{\\delta_l}\\bm a_{l-1}\\\\\n \\frac{\\partial L_{total}}{\\partial \\bm b_l} =& \\bm{\\delta_l}\n\\end{align*}}\n\nWith the gradients computed, we use the following update rule to adjust the parameters:\n\\begin{equation}\n\\label{ebpup}\n\\aligned\nW_l^{jk} =& W_l^{jk} - r\\frac{\\partial L_{total}}{\\partial W_l^{jk}}\\\\\nb_l^{j} =& b_l^{j} - r\\frac{\\partial L_{total}}{\\partial b_l^{j}},\n\\endaligned\n\\end{equation}\nwhere r is the learning rate, for $l=1,...,d$. The learning rate r and the regularization parameter $\\lambda$ are hyper parameters, we still don't know why and how to choose them.{\\color{red}LZ: $\\lambda$? overstatement?} Therefore we normally choose them by experience and manually parameter tuning to decide the optimal hyper parameters.\n\n\n", "meta": {"hexsha": "b6dca246759211380b069469e293901097b4e2d7", "size": 12688, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/DeepLearningandLargeScaleComputingJiLiu.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/DeepLearningandLargeScaleComputingJiLiu.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/DeepLearningandLargeScaleComputingJiLiu.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.7587939698, "max_line_length": 659, "alphanum_fraction": 0.7239123581, "num_tokens": 3735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835493924954, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6402257372615309}} {"text": "\\subsection{Rings}\\label{subsec:rings}\n\n\\begin{definition}\\label{def:ring}\n A \\term{ring} is a \\hyperref[def:semiring]{semiring} with additive inverses. More precisely, this means that the additive monoid is a group.\n\n As for semirings, rings can also be nonunital, with \\hyperref[def:semiring_ideal]{ring ideals} being the main example.\n\n Rings have the following metamathematical properties:\n \\begin{thmenum}\n \\thmitem{def:ring/theory} We can construct a \\hyperref[def:first_order_theory]{first-order theory} for rings by adding a unary functional symbol \\( - \\) and the involution axiom \\eqref{eq:def:group/theory/inverse_axiom} to the \\hyperref[def:semiring/theory]{theory of semirings}.\n\n \\thmitem{def:ring/homomorphism} A \\hyperref[def:first_order_homomorphism]{first-order homomorphism} between the rings \\( R \\) and \\( T \\) is a \\hyperref[def:semiring/homomorphism]{semiring homomorphism} \\( \\varphi: R \\to T \\) that additionally preserves additive inverses.\n\n As shown in \\fullref{thm:group_homomorphism_single_condition}, this condition is not only redundant, but the structure of a ring rather than semiring also automatically implies that \\( \\varphi(0_R) = 0_S \\).\n\n \\thmitem{def:ring/submodel} The set \\( A \\subseteq R \\) is a \\hyperref[thm:substructure_is_model]{submodel} of \\( R \\) if it is both a \\hyperref[def:semiring]{sub-semiring} of \\( R \\) and an additive submonoid of \\( R \\).\n\n As a consequence of \\fullref{thm:positive_formulas_preserved_under_homomorphism}, the image of a ring homomorphism is a subring of its range.\n\n \\thmitem{def:ring/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} ring is the \\hyperref[def:pointed_set/trivial]{trivial pointed set} \\( \\set{ 0 } \\).\n\n \\thmitem{def:ring/commutative} If multiplication is commutative, we call the ring itself \\term{commutative}. Unless multiplication corresponds to function composition, most rings we will encounter will be commutative.\n\n \\thmitem{def:ring/category} The corresponding \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} \\( \\ucat{Ring} \\) is \\hyperref[def:concrete_category]{concrete} over \\hyperref[def:monoid]{\\( \\ucat{SRing} \\)}. We denote the category of commutative rings by \\( \\cat{CRing} \\).\n\n Unlike the category \\hyperref[def:group/category]{\\( \\cat{Grp} \\)} of groups, \\( \\cat{Ring} \\) is not as well-behaved. Nevertheless, kernels and quotients of rings are commonly established concepts.\n\n The category of unital rings does not have a zero object, but the category of nonunital rings does, and we will sometimes consider nonunital ring homomorphisms between unital rings. That is, ring homomorphisms that may not preserve the multiplicative identity.\n\n \\thmitem{def:ring/kernel} The \\term{kernel} of a ring homomorphism \\( \\varphi: R \\to T \\) is simply its \\hyperref[def:zero_locus]{zero locus} \\( \\varphi^{-1}(0_S) \\). This is precisely the kernel of the additive group in the sense of \\fullref{def:group/kernel}, and the \\hyperref[def:zero_morphisms/cokernel]{categorical kernel} in the category of nonunital rings.\n\n Furthermore, \\( \\ker \\varphi \\) is an ideal of \\( R \\) because, if \\( x \\in \\ker \\varphi \\),\n \\begin{equation*}\n \\varphi(xy)\n =\n \\varphi(x) \\varphi(y)\n =\n 0_S \\varphi(y)\n =\n 0_S,\n \\end{equation*}\n and thus \\( xy \\in \\ker \\varphi \\).\n\n Despite being categorical kernels only in the category of nonunital rings, the kernel is defined and used mainly for unital ring homomorphism.\n\n \\thmitem{def:ring/quotient} The \\hyperref[def:zero_morphisms/cokernel]{categorical cokernel} of homomorphism \\( \\varphi: R \\to T \\) in the category of nonunital rings is, similarly to the case for groups in \\fullref{def:group/quotient}, a partition of \\( T \\) induced by the image of \\( \\varphi \\).\n\n This is not merely the cokernel \\( T / \\img \\varphi \\) of the additive group, however. Multiplication induces an additional restriction on congruences: \\( x \\cong x' \\) and \\( y \\cong y' \\) together imply \\( x y \\cong x' y' \\). Hence, \\( [x][y] = [xy] \\). Denote the coset \\( [0_S] \\) by \\( I \\). We have \\( I[x] = [0x] = I \\), therefore the cokernel inherits absorption from \\( T \\).\n\n Additive subgroups of \\( T \\) that absorb multiplication are precisely the \\hyperref[def:semiring_ideal]{two-sided ideals} of \\( T \\). Hence, \\( I \\) is the ideal generated by \\( \\img \\varphi \\). From the general case for groups it follows that quotient ring cosets have the form \\( x + I \\).\n\n Finally, given an ideal \\( I \\) of an arbitrary ring \\( R \\), we can define the \\term{quotient ring} \\( R / I \\) as the cokernel of the inclusion \\( \\iota: I \\to R \\). That is, \\( R / I \\) consists of the cosets \\( x + I \\) for \\( x \\in R \\). In practice, quotients are conveniently characterized by \\fullref{thm:quotient_algebra_universal_property}.\n\n Somewhat similarly to \\fullref{thm:def:group/kernel_cokernel_compatibility} for groups, the kernel \\( \\ker \\pi \\) of the canonical projection \\( \\pi(x) \\coloneqq x + I \\) is the ideal \\( I \\) itself.\n\n Despite being \\hyperref[def:zero_morphisms/cokernel]{categorical cokernel} only in the category of nonunital rings, the quotient \\( R / I \\) is defined and used mainly for unital rings.\n\n Fortunately, for unital ring \\( R \\), the quotient \\( R / I \\) is also unital. The projection morphism is an epimorphism by \\fullref{thm:equalizer_invertibility}, and hence \\( R / I \\) is a \\hyperref[def:subobject_and_quotient]{categorical quotient object}.\n\n \\thmitem{def:ring/simple} Analogously to \\hyperref[def:group/simple]{simple groups}, if the only proper \\hyperref[def:semiring_ideal]{ideal} of \\( R \\) is the \\hyperref[def:ring/trivial]{trivial ideal} \\( \\set{ 0_R } \\), we say that \\( R \\) is a \\term{simple ring}.\n\n The trivial ring itself is not simple, because it has no proper ideals.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:quotient_equality_via_difference}\n Given an \\hyperref[def:semiring_ideal]{ideal} \\( I \\) in a \\hyperref[def:ring]{ring} \\( R \\), we have \\( x + I = y + I \\) if and only if \\( x - y \\in I \\).\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ring_of_integers_modulo}\n For a positive integer \\( n > 1 \\), we extend the group \\hyperref[thm:group_of_integers_modulo]{\\( \\BbbZ_n \\)} of integers modulo \\( n \\) with the operation\n \\begin{equation*}\n x \\odot y \\coloneqq \\rem(xy, n).\n \\end{equation*}\n\n Then \\( \\BbbZ_n \\) is a \\hyperref[def:ring/commutative]{commutative ring} called the \\term{ring of integers modulo} \\( n \\).\n\\end{proposition}\n\\begin{proof}\n Note that\n \\begin{balign*}\n &\\phantom{{}\\cong{}} \\rem(x, n) \\rem(y, n)\n &\\cong \\pmod n \\\\ &\\cong\n (x - n \\quot(x, n)) (y - n \\quot(y, n))\n &\\cong \\pmod n \\\\ &\\cong\n xy - n \\quot(x, n) - n \\quot(y, n) + n^2 \\quot(x, n) \\quot(y, n)\n &\\cong \\pmod n \\\\ &\\cong\n xy.\n \\end{balign*}\n\n The proof that multiplication in \\( \\BbbZ_n \\) is associative, unital and commutative becomes trivial.\n\n We will prove that multiplication distributes over addition. Fix \\( x, y, z \\in \\BbbZ_n \\). We have\n \\begin{balign*}\n (x \\oplus y) \\odot z\n & =\n \\rem((x \\oplus y) z, n)\n = \\\\ &=\n \\rem(\\rem(x + y, n) z, n)\n = \\\\ &=\n \\rem((x + y - n \\quot(x + y, n)) z, n)\n = \\\\ &=\n \\rem((x + y)z, n).\n \\end{balign*}\n and\n \\begin{balign*}\n (x \\odot z) \\oplus (y \\odot z)\n & =\n \\rem([(x \\odot z) + (y \\odot z)], n)\n = \\\\ &=\n \\rem([xz - n \\quot(xz, n) + yz - n \\quot(yz, n)], n)\n = \\\\ &=\n \\rem(xz + yz, n)\n = \\\\ &=\n \\rem((x + y)z, n).\n \\end{balign*}\n\n Hence,\n \\begin{equation*}\n (x \\oplus y) \\odot z = (x \\odot z) \\oplus (y \\odot z).\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ring_characteristic_homomorphism}\n Similarly to how \\( \\BbbN \\) is an \\hyperref[def:universal_objects/initial]{initial object} in the category \\hyperref[def:semiring/category]{\\( \\cat{SRing} \\)} of semirings, \\( \\BbbZ \\) is an initial object in the category \\hyperref[def:ring/category]{\\( \\cat{Ring} \\)} of rings.\n\\end{proposition}\n\\begin{proof}\n Follows from \\fullref{thm:semiring_characteristic_homomorphism} with the addition of \\( (-n)x = -nx \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:ring_characteristic}\\mimprovised\n We define the \\term{characteristic} \\( \\op{char}(R) \\) of a ring \\( R \\) via the following equivalent definitions:\n \\begin{thmenum}\n \\thmitem{def:ring_characteristic/embedding} \\( \\op{char}(R) \\) is the unique nonnegative integer \\( n \\) for which \\( \\BbbZ_n \\) can be embedded into \\( R \\). That is,\n \\begin{equation*}\n \\BbbZ_n \\cong \\BbbZ / \\ker \\iota,\n \\end{equation*}\n where \\( \\iota \\) is the homomorphism from the integers defined via \\eqref{eq:thm:ring_characteristic_homomorphism}.\n\n We use here that \\( \\BbbZ_0 \\) is the \\hyperref[def:ring/trivial]{trivial ring}.\n\n \\thmitem{def:ring_characteristic/direct} \\( \\op{char}(R) \\) is the \\hyperref[def:group_order]{order} of the additive group of \\( R \\) and, if it exists, and \\( 0 \\) otherwise.\n\n That is, \\( \\op{char}(R) \\) is the smallest positive integer \\( n \\) such that \\( n \\cdot 1_R = 0_R \\) and \\( \\op{char}(R) = 0 \\) if \\( 0_R \\) cannot be obtained in this way.\n \\end{thmenum}\n\\end{definition}\n\\begin{defproof}\n \\EquivalenceSubProof{def:ring_characteristic/homomorphism}{def:ring_characteristic/direct} Let \\( n \\) be such that\n \\begin{equation*}\n n \\BbbZ = \\ker\\iota.\n \\end{equation*}\n\n In particular, \\( \\iota(0) = \\iota(n) \\).\n\n If \\( n = 0 \\), \\( \\ker\\iota \\) is a trivial group and \\( \\iota \\) is an embedding. Then there cannot exist a positive integer \\( n \\) such that\n \\begin{equation*}\n n \\cdot 1_R = 0_R.\n \\end{equation*}\n\n Otherwise, \\( n \\) is the smallest positive integer such that\n \\begin{equation*}\n n \\cdot 1_R = 0 \\cdot 1_R = 0_R.\n \\end{equation*}\n\\end{defproof}\n\n\\begin{proposition}\\label{thm:ring_embedding_preserves_characterstic}\n If \\( \\varphi: R \\to S \\) is a \\hyperref[def:ring/homomorphism]{ring embedding}, then \\( S \\) inherits its \\hyperref[def:semiring_characteristic]{characteristics} from \\( R \\).\n\\end{proposition}\n\\begin{proof}\n First suppose that \\( R \\) has positive characteristic \\( n \\). Then \\( n \\cdot 1_R = 0_R \\), which implies \\( n \\cdot \\varphi(1_R) = \\varphi(0_R) \\), hence \\( \\op{char}(T) \\leq n \\). But \\( \\varphi \\) is an embedding, hence if \\( k \\cdot 1_R \\neq 0_R \\), then\n \\begin{equation*}\n k \\cdot \\varphi(1_R) \\neq \\varphi(0_R).\n \\end{equation*}\n\n This implies that \\( \\op{char}(S) \\geq \\op{char}(R) \\), which in turn shows that \\( \\op{char}(S) = \\op{char}(R) \\).\n\n If \\( R \\) has characteristic zero, then \\( \\iota: \\BbbN \\to R \\) is an embedding and thus \\( \\varphi \\bincirc \\iota: \\BbbN \\to S \\) is also an embedding. It is unique as shown in \\fullref{thm:ring_characteristic_homomorphism}. Therefore, \\( S \\) also has characteristic zero.\n\\end{proof}\n\n\\begin{example}\\label{ex:def:ring_characteristic}\n The following are examples of \\hyperref[def:ring_characteristic]{ring characteristics}:\n \\begin{thmenum}\n \\thmitem{ex:def:ring_characteristic/natural_numbers} The \\hyperref[def:set_of_integers]{integers} \\( \\BbbZ \\) have characteristic \\( \\op{char}(\\BbbZ) = 0 \\) because \\( \\iota \\) is an isomorphism. Consequently, by \\fullref{thm:ring_embedding_preserves_characterstic}, any superring of \\( \\BbbZ \\) has characteristic zero, most notably the fields \\( \\BbbQ \\), \\( \\BbbR \\) and \\( \\BbbC \\).\n\n \\thmitem{ex:def:ring_characteristic/integers_modulo} The ring \\hyperref[thm:ring_of_integers_modulo]{\\( \\BbbZ_n \\)} of integers modulo \\( n \\) has characteristic \\( \\op{char}(\\BbbZ_n) = n \\) because of \\fullref{thm:integers_modulo_isomorphic_to_quotient_group}.\n\n \\thmitem{ex:def:ring_characteristic/polynomial_ring} An \\hyperref[def:algebra_over_semiring]{algebra} \\( M \\) over a nontrivial commutative unital ring \\( R \\) has the same characteristic as \\( R \\) because of the canonical embedding of \\( R \\) in \\( M \\). In particular, the \\hyperref[def:polynomial_algebra]{polynomial ring} \\( R[X] \\) has the same characteristic as their ring.\n \\end{thmenum}\n\\end{example}\n\n\\begin{proposition}\\label{thm:grothendieck_semiring_completion}\\mcite[80]{OpenLogicFull}\n The \\hyperref[def:monoid_grothendieck_completion]{Grothendieck completion} \\( \\overline{R} \\) of the additive monoid of a \\hyperref[def:semiring]{semiring} \\( R \\) becomes a \\hyperref[def:ring]{ring} with the operation\n \\begin{equation*}\n [(a, b)] \\odot [(c, d)] \\coloneqq [(ac + bd, ad + bc)].\n \\end{equation*}\n\n This definition is motivated in the proof of \\fullref{thm:grothendieck_semiring_completion_universal_property}.\n\\end{proposition}\n\\begin{proof}\n Multiplication on \\( R \\) does not depend on the representative of the equivalence class. Indeed, let \\( (a, b) \\sim (a', b') \\) and \\( (c, d) \\sim (c', d') \\). Then there exist \\( u \\) and \\( v \\) such that\n \\begin{align*}\n a + b' + u &= a' + b + u, \\\\\n c + d' + v &= c' + d + v.\n \\end{align*}\n\n Then\n \\begin{align*}\n &\\phantom{{}={}}\n \\hi{ac} + b'c + uc + a'd + \\hi{bd} + ud + a'c + \\hi{a'd'} + a'v + \\hi{b'c'} + b'd + b'v\n = \\\\ &=\n (a + b' + u)c + (a' + b + u)d + a'(c + d' + v) + b'(c' + d + v)\n = \\\\ &=\n (a' + b + u)c + (a + b' + u)d + a'(c' + d + v) + b'(c + d' + v)\n = \\\\ &=\n a'c + \\hi{bc} + uc + \\hi{ad} + b'd + ud + \\hi{a'c'} + a'd + a'v + b'c + \\hi{b'd'} + b'v.\n \\end{align*}\n\n Therefore,\n \\begin{equation*}\n (a \\cdot c + b \\cdot d, a \\cdot d + b \\cdot c) \\sim (a' \\cdot c' + b' \\cdot d', a' \\cdot d' + b' \\cdot c').\n \\end{equation*}\n\n Associativity and distributivity in \\( \\overline{R} \\) are inherited from \\( R \\).\n\\end{proof}\n\n\\begin{theorem}[Grothendieck semiring completion universal property]\\label{thm:grothendieck_semiring_completion_universal_property}\n The \\hyperref[thm:grothendieck_semiring_completion]{Grothendieck completion} \\( \\overline{R} \\) of a semiring \\( R \\) satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every ring \\( T \\) and every semiring homomorphism \\( \\varphi: R \\to T \\), there exists a unique ring homomorphism \\( \\widetilde{\\varphi}: \\overline{R} \\to T \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:grothendieck_semiring_completion_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__semiring_grothendieck_completion_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Via \\fullref{rem:universal_mapping_property}, \\( \\overline{\\anon} \\) becomes \\hyperref[def:category_adjunction]{left adjoint} to the \\hyperref[def:concrete_category]{forgetful functor}\n \\begin{equation*}\n U: \\cat{CRing} \\to \\cat{CSRing}.\n \\end{equation*}\n\n Compare this result to \\fullref{thm:grothendieck_monoid_completion_universal_property}.\n\\end{theorem}\n\\begin{proof}\n \\Fullref{thm:grothendieck_monoid_completion_universal_property} suggests the definition\n \\begin{equation*}\n \\overline{\\varphi}([(a, b)]) \\coloneqq \\varphi(a) - \\varphi(b).\n \\end{equation*}\n\n We must only show that \\( \\overline{\\varphi} \\) is a ring homomorphism. Clearly\n \\begin{equation*}\n \\overline{\\varphi}([(1, 0)]) = \\varphi(1) - \\varphi(0),\n \\end{equation*}\n which implies that \\( \\varphi \\) preserves multiplicative identities. Also,\n \\begin{balign*}\n \\overline{\\varphi}\\parens[\\Big]{ [(a, b)] \\odot [(c, d)] }\n &=\n \\overline{\\varphi}\\parens[\\Big]{ [(a \\cdot b + c \\cdot d, a \\cdot d + b \\cdot c)] }\n = \\\\ &=\n \\varphi(a \\cdot b + c \\cdot d) - \\varphi(a \\cdot d + b \\cdot c)\n = \\\\ &=\n \\varphi(c) \\parens[\\Big]{ \\varphi(d) - \\varphi(b) } - \\varphi(a) \\parens[\\Big]{ \\varphi(d) - \\varphi(b) }\n = \\\\ &=\n \\parens[\\Big]{ \\varphi(c) - \\varphi(a) } \\parens[\\Big]{ \\varphi(d) - \\varphi(b) }\n = \\\\ &=\n \\overline{\\varphi}\\parens[\\Big]{ [(a, c)] } \\overline{\\varphi}\\parens[\\Big]{ [(b, d)] }.\n \\end{balign*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:grothendieck_semiring_completion}\n The \\hyperref[thm:grothendieck_semiring_completion]{Grothendieck completion} \\( \\overline{R} \\) of a semiring \\( R \\) satisfies the following basic properties:\n \\begin{thmenum}\n \\thmitem{thm:def:grothendieck_semiring_completion/commutative} If \\( R \\) is commutative, so is \\( \\overline{R} \\).\n \\thmitem{thm:def:grothendieck_semiring_completion/entire} If \\( R \\) is \\hyperref[def:divisibility/zero]{entire}, so is \\( \\overline{R} \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:grothendieck_semiring_completion/commutative} This is clear from the definition of multiplication.\n \\SubProofOf{thm:def:grothendieck_semiring_completion/entire} Suppose that\n \\begin{equation*}\n \\underbrace{[(a, b)] \\cdot [(c, d)]}_{[(ac + bd, ad + bc)]} = [(0, 0)]\n \\end{equation*}\n\n Then there exists an element \\( u \\) in \\( R \\) such that\n \\begin{equation*}\n (ac + bd) + 0 + u = 0 + (ad + bc) + u.\n \\end{equation*}\n\n Suppose that \\( d = c + e \\). Then\n \\begin{equation*}\n ac + b(c + e) = a(c + e) + bc\n \\end{equation*}\n and\n \\begin{equation*}\n (ac + bc) + be = (ac + bc) + ae.\n \\end{equation*}\n\n Cancelling \\( e \\), we obtain that \\( a = b \\). But \\( [(a, b)] = [(0, 0)] \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:ring_commutator}\n Let \\( R \\) be an arbitrary ring. We define the \\term{commutator} of the elements \\( x \\) and \\( y \\) as\n \\begin{equation*}\n [x, y] \\coloneqq xy - yx.\n \\end{equation*}\n\n The \\term{commutator ideal} \\( [R, R] \\) of \\( R \\) is the two-sided ideal \\hyperref[def:generated_ring_ideal]{generated} by all the commutators in \\( G \\).\n\\end{definition}\n\n\\begin{theorem}[Ring abelianization universal property]\\label{thm:ring_abelianization_universal_property}\\mcite[prop. 7.4]{Knapp2016BasicAlgebra}\n The quotient \\( R / [R, R] \\) of a ring \\( R \\) by its commutator ideal \\( [R, R] \\) is a commutative ring, which we call the \\term{abelianization} of \\( R \\), and satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every commutative ring \\( T \\) and every ring homomorphism \\( \\varphi: R \\to T \\), \\( \\varphi \\) \\hyperref[def:factors_through]{uniquely factors through} \\( R / [R, R] \\). That is, there exists a unique ring homomorphism \\( \\widetilde{\\varphi}: R / [R, R] \\to T \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:ring_abelianization_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__ring_abelianization_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Via \\fullref{rem:universal_mapping_property}, the abelianization functor becomes \\hyperref[def:category_adjunction]{left adjoint} to the \\hyperref[def:concrete_category]{forgetful functor}\n \\begin{equation*}\n U: \\cat{CRing} \\to \\cat{Ring}.\n \\end{equation*}\n\n Compare this result to \\fullref{thm:group_abelianization_universal_property}.\n\\end{theorem}\n\\begin{proof}\n This is a refinement of \\fullref{thm:group_abelianization_universal_property}, and we only need to show that \\( R / [R, R] \\) is a commutative ring. For \\( x \\) and \\( y \\) in \\( R \\), since \\( yx - xy \\in I \\), we have\n \\begin{equation*}\n (x + I) (y + I)\n =\n (xy + I)\n =\n (xy + yx - xy + I)\n =\n (yx + I)\n =\n (y + I) (x + I).\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:multiplicative_set_in_semiring}\\mcite[428]{Knapp2016BasicAlgebra}\n We call the subset of the semiring \\( R \\) a \\term{multiplicative set} if it contains \\( 1_R \\) and, furthermore, it is closed under multiplication.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:complement_of_prime_ideal}\n The \\hyperref[def:semiring_ideal]{ideal} \\( P \\) in the \\hyperref[def:semiring/commutative]{commutative semiring} \\( R \\) is \\hyperref[def:semiring_ideal/prime]{prime} if and only if \\( R \\setminus P \\) is a \\hyperref[def:multiplicative_set_in_semiring]{multiplicative set}.\n\n Not all multiplicative sets are obtained as complements of prime ideals --- see \\fullref{ex:def:ring_localization/powers_of_two}.\n\\end{proposition}\n\\begin{proof}\n By \\fullref{thm:def:semiring_ideal/ideal_containing_unit}, \\( P \\) is a proper ideal if and only if \\( 1_R \\in R \\setminus P \\).\n\n By \\fullref{thm:def:semiring_ideal/prime_pointwise}, \\( P \\) is prime if and only if \\( x, y \\in R \\setminus P \\) implies \\( xy \\in R \\setminus P \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:ring_localization}\\mcite[428]{Knapp2016BasicAlgebra}\n Let \\( R \\) be a \\hyperref[def:ring/commutative]{commutative ring} and let \\( S \\subseteq R \\) be a \\hyperref[def:multiplicative_set_in_semiring]{multiplicative set}.\n\n Define the equivalence relation \\( (r, s) \\sim (r', s') \\) on \\( R \\times S \\) to hold if and only if there exists some \\( u \\in S \\) such that \\( u r s' = u r' s \\).\n\n Consider the set\n \\begin{equation*}\n S^{-1} R \\coloneqq R \\times S / \\sim,\n \\end{equation*}\n whose cosets we will denote by \\( \\ifrac r s \\) rather than \\( [(r, s)] \\).\n\n Define on \\( S^{-1} R \\) the operations\n \\begin{align*}\n \\frac a b + \\frac c d &\\coloneqq \\frac {a d + b c} {b d}, \\\\\n \\frac a b \\cdot \\frac c d &\\coloneqq \\frac {a c} {b d},\n \\end{align*}\n and the canonical inclusion\n \\begin{equation*}\n \\begin{aligned}\n &\\iota: R \\to S^{-1} R \\\\\n &\\iota(r) \\coloneqq \\frac r {1_R}.\n \\end{aligned}\n \\end{equation*}\n\n This ring is called the \\term{localization} of \\( R \\) with respect to \\( A \\); we denote it by \\( S^{-1} R \\). In case \\( S \\) is the \\hyperref[thm:boolean_algebra_of_subsets/complement]{complement} of a \\hyperref[def:semiring_ideal/prime]{prime ideal}, we may denote the localization by \\( R_P \\) (or \\( R_p \\) if \\( P = \\braket{ p } \\)).\n\n The image under \\( \\iota \\) of every element \\( s \\) of \\( S \\) is invertible in \\( S^{-1} R \\), and we call the inverse \\( \\ifrac {1_R} s \\) the \\term{reciprocal} of \\( s \\).\n\n This construction is very similar to the \\hyperref[def:monoid_grothendieck_completion]{Grothendieck completion} of a monoid or semiring, although with notable differences --- the set \\( S \\) may be a strict subset of \\( R \\), and addition in the Grothendieck completion corresponds to multiplication in the localization, while addition in the completion has no analogy.\n\\end{definition}\n\\begin{defproof}\n The proof that \\( {\\sim} \\) is an equivalence relation is the same as in \\fullref{def:monoid_grothendieck_completion}. The result is then a ring if the operations are well-defined.\n\n We will show that both operations are well-defined. Let \\( u ab' = u a'b \\), meaning that \\( (a, b) \\sim (a', b') \\) and hence \\( \\ifrac a b = \\ifrac {a'} {b'} \\), and let \\( v cd' = v c'd \\).\n\n For addition, we have\n \\begin{align*}\n u v (ad + bc) b' d'\n &=\n v dd' (u ab') + u bb' (v cd')\n = \\\\ &=\n v dd' (u a'b) + u bb' (v c'd)\n = \\\\ &=\n u v (a'd' + b'c') b d,\n \\end{align*}\n hence \\( (ad + bc, bd) \\sim (a'd' + b'c', b'd') \\).\n\n The proof for correctness of multiplication is the same as the proof of correctness of addition in \\fullref{def:monoid_grothendieck_completion}.\n\\end{defproof}\n\n\\begin{theorem}[Ring localization universal property]\\label{thm:ring_localization_universal_property}\\mcite[431]{Knapp2016BasicAlgebra}\n The \\hyperref[def:ring_localization]{localization} of \\( R \\) by \\( S \\) satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every commutative ring \\( T \\) and every ring homomorphism \\( \\varphi: R \\to T \\) such that \\( \\varphi(s) \\) is invertible in \\( T \\) for every \\( s \\in S \\), \\( \\varphi \\) \\hyperref[def:factors_through]{uniquely factors through} \\( S^{-1} R \\). That is, there exists a unique ring homomorphism \\( \\widetilde{\\varphi}: S^{-1} R \\to T \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:ring_localization_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__ring_localization_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\\end{theorem}\n\\begin{proof}\n The condition suggests the definition\n \\begin{equation*}\n \\widetilde{\\varphi}\\parens*{ \\frac r s } \\coloneqq \\varphi(r) \\varphi(s)^{-1}.\n \\end{equation*}\n\\end{proof}\n\n\\begin{example}\\label{ex:def:ring_localization}\n We list several examples of \\hyperref[def:ring/commutative]{commutative ring} \\hyperref[def:ring_localization]{localization}.\n\n \\begin{thmenum}\n \\thmitem{ex:def:ring_localization/zero} If \\( S \\) contains \\( 0_R \\), then \\( S^{-1} R \\) is the trivial ring.\n\n \\thmitem{ex:def:ring_localization/powers_of_two} The localization \\( S^{-1} \\BbbZ \\) by the set \\( S \\coloneqq \\set{ 2^n \\given n \\geq 0 } \\) is (a ring isomorphic to) the rational numbers with denominators that are powers of two. This is an example of a multiplicative set that is not the complement of a prime ideal.\n\n This ring is isomorphic to the ring \\( \\BbbZ[\\ifrac 1 2] \\) obtained by \\hyperref[thm:adjoining_elements_to_semiring]{adjoining} the rational number \\( \\ifrac 1 2 \\) to \\( \\BbbZ \\).\n\n \\thmitem{ex:def:ring_localization/prime_number} Let \\( p \\) be a \\hyperref[def:prime_number]{prime number}. The localization \\( S^{-1} \\BbbZ \\) by \\( S \\coloneqq \\BbbZ \\setminus \\braket{ p } \\) is (a ring isomorphic to) the rational numbers with denominators coprime to \\( p \\).\n\n For \\( p = 2 \\), this localization consists of rational numbers whose denominator is an odd number.\n \\end{thmenum}\n\\end{example}\n\n\\begin{proposition}\\label{thm:def:ring_localization}\n \\hyperref[def:ring_localization]{Ring localization} has the following basic properties:\n\n \\begin{thmenum}\n \\thmitem{thm:def:ring_localization/image_of_ideal}\\mcite[432]{Knapp2016BasicAlgebra} Localization preserves \\hyperref[def:semiring_ideal]{ideals}. More precisely, given a commutative ring \\( R \\), a multiplicative set \\( S \\) and an ideal \\( I \\), the set\n \\begin{equation*}\n S^{-1} I \\coloneqq \\set*{ \\frac r s \\given* r \\in I \\T{and} s \\in S }\n \\end{equation*}\n is an ideal of the localization \\( S^{-1} R \\).\n\n \\thmitem{thm:def:ring_localization/prime_ideals}\\mcite[exer. 4.3]{КоцевСидеров2016} The map \\( I \\mapsto S^{-1} I \\) is a \\hyperref[def:partially_ordered_set/homomorphism]{strict order isomorphism} between the set of \\hyperref[def:ring/submodel]{prime ideals} of \\( R \\) not intersecting \\( S \\) and the set of all prime ideals of \\( S^{-1} R \\).\n\n \\thmitem{thm:def:ring_localization/by_prime_ideal}\\mcite[exer. 4.2a)]{КоцевСидеров2016} The localization \\( R_P \\) by a \\hyperref[def:semiring_ideal/prime]{prime ideal} \\( P \\) has a unique maximal ideal \\( S^{-1} P \\) (here \\( S \\coloneqq R \\setminus P \\)).\n\n \\thmitem{thm:def:ring_localization/injective_inclusion} The canonical inclusion \\( \\iota: R \\to S^{-1} R \\) is injective if and only if \\( S \\) contains no zero divisors.\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:ring_localization/image_of_ideal} Trivial since \\( S \\) is closed under multiplication.\n\n \\SubProofOf{thm:def:ring_localization/prime_ideals} Let \\( P \\) be a prime ideal in \\( R \\) disjoint from \\( S \\). By \\fullref{thm:def:ring_localization/image_of_ideal}, \\( S^{-1} P \\) is an ideal of \\( S^{-1} R \\). If the product \\( \\ifrac {ac} {bd} \\) belong to \\( S^{-1} P \\), then \\( ac \\in P \\) and \\( bd \\in S \\). Since \\( P \\) is prime, \\( a \\in P \\) or \\( c \\in P \\). If \\( a \\in P \\), then \\( ba \\in P \\) and \\( \\ifrac a d = \\ifrac {ba} {bd} \\in S^{-1} P \\); if \\( c \\in P \\), we proceed analogously. Thus, \\( S^{-1} P \\) is a prime ideal, i.e. the image under \\( I \\mapsto S^{-1} I \\) of a prime ideal is a prime ideal.\n\n \\SubProofOf[def:function_invertibility/injective/equality]{injectivity} Let \\( S^{-1} P = S^{-1} Q \\) for prime ideals \\( P \\) and \\( Q \\) disjoint from \\( S \\). Suppose that \\( P \\setminus Q \\) contains at least one element, say \\( p \\). Then \\( \\iota(p) = \\ifrac p 1 \\) belongs to both \\( S^{-1} P \\) and \\( S^{-1} Q \\); hence, \\( Q \\) contains an element \\( q \\) such that, for some \\( s \\in S \\) and \\( u \\in S \\),\n \\begin{equation*}\n p \\cdot s \\cdot u = 1 \\cdot q \\cdot u.\n \\end{equation*}\n\n Since \\( Q \\) is an ideal, \\( qu \\in Q \\), and hence \\( psu \\in Q \\). But neither \\( p \\), \\( s \\) nor \\( u \\) belong to \\( Q \\), which contradicts the assumption that \\( Q \\) is prime. Therefore, \\( P \\setminus Q \\) is empty. Generalizing, we obtain that \\( I \\mapsto S^{-1} I \\) is injective on prime ideals.\n\n \\SubProofOf[def:function_invertibility/surjective/existence]{surjectivity} Fix a prime ideal \\( T \\) in \\( S^{-1} R \\) and let \\( P \\) be the set of numerators in \\( T \\), i.e. if \\( \\ifrac p s \\in T \\), then \\( p \\in P \\). We will show that \\( P \\) is a prime ideal; clearly \\( T = S^{-1} P \\).\n\n Clearly \\( 0_R \\in P \\). Let \\( a, c \\in P \\). Then there exist \\( b, d \\in S \\) such that \\( \\ifrac a b \\) and \\( \\ifrac c d \\) belong to \\( T \\). But \\( T \\) is closed under multiplication with members of \\( R \\), hence \\( \\ifrac a {1_R} = b (\\ifrac a b) \\) and \\( \\ifrac c {1_R} = d (\\ifrac c d) \\) also belong to \\( T \\). Then their sum \\( \\ifrac {(a + c)} 1 \\) belongs to \\( T \\), and hence also to \\( P \\). Thus, \\( P \\) is closed under addition. We analogously obtain that it is closed under multiplication.\n\n We have shown that \\( P \\) is an ideal in \\( R \\). We must show that it is a prime ideal. Let \\( ac \\in P \\). Then\n \\begin{equation*}\n \\frac a b \\cdot \\frac c d \\in T\n \\end{equation*}\n for some \\( b, d \\in S \\). Hence, \\( \\ifrac a b \\) or \\( \\ifrac c d \\) belongs to \\( T \\), implying that \\( a \\in P \\) or \\( c \\in P \\).\n\n \\SubProofOf[def:partially_ordered_set/homomorphism]{monotonicity} Follows from \\fullref{thm:order_embedding_is_strict}.\n\n \\SubProofOf{thm:def:ring_localization/by_prime_ideal} In the localization \\( R_P \\) be a prime ideal, all members of \\( P \\) become invertible. Hence, a maximal ideal cannot contain members of \\( P \\). By \\fullref{thm:def:ring_localization/image_of_ideal}, \\( S^{-1} P \\) is an ideal, therefore it must be the largest proper ideal.\n\n \\SubProofOf{thm:def:ring_localization/injective_inclusion} Let \\( sr = 0 \\) for \\( s \\in S \\). Then \\( \\iota(s) = \\ifrac s {1_R} \\) is invertible in \\( S^{-1} R \\) and hence\n \\begin{equation*}\n \\frac {0_R} {1_R}\n =\n \\frac {sr} {1_R}\n =\n \\frac {1_R} s \\cdot \\frac {sr} {1_R}\n =\n \\frac r {1_R}.\n \\end{equation*}\n\n Hence, \\( \\iota(r) = \\iota(0_R) \\).\n\n It follows that \\( \\iota \\) is injective if and only if \\( S \\) contains no zero divisors.\n\\end{proof}\n\n\\begin{definition}\\label{def:division_ring}\\mcite[144]{Knapp2016BasicAlgebra}\n If every nonzero element of a ring is \\hyperref[def:divisibility/unit]{invertible}, we call it a \\term{division ring}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:division_ring_is_entire}\n A nontrivial \\hyperref[def:division_ring]{division ring} is \\hyperref[def:entire_semiring]{entire}.\n\\end{proposition}\n\\begin{proof}\n Let \\( xy = 0 \\). If \\( x \\) is nonzero, multiplying both sides by \\( x^{-1} \\), we obtain \\( y = 0 \\). Analogously, \\( y \\neq 0 \\) implies that \\( x = 0 \\). In all cases, either \\( x \\) or \\( y \\) is necessarily zero.\n\n Therefore, the ring has no nontrivial zero divisors.\n\\end{proof}\n\n\\begin{definition}\\label{def:field}\n We will call the \\hyperref[def:ring/trivial]{nontrivial} \\hyperref[def:ring]{ring} \\( \\BbbK \\) a \\term{field} if any of the following equivalent conditions hold:\n \\begin{thmenum}\n \\thmitem{def:field/simple} \\( \\BbbK \\) is \\hyperref[def:ring/commutative]{commutative} and \\hyperref[def:ring/simple]{simple}.\n \\thmitem{def:field/division_ring} \\( \\BbbK \\) is a \\hyperref[def:ring/commutative]{commutative} \\hyperref[def:division_ring]{division ring}.\n \\end{thmenum}\n\n Fields have the following metamathematical properties:\n \\begin{thmenum}\n \\thmitem{def:field/theory} We can construct a \\hyperref[def:first_order_theory]{first-order theory} for fields by adding to the \\hyperref[def:semiring/theory]{theory of rings} the axioms \\( \\neg (0 \\doteq 1) \\) and\n \\begin{equation}\\label{eq:def:field/theory/invertibility}\n (\\xi \\doteq 0) \\vee \\qexists \\eta (\\xi \\cdot \\eta \\doteq 1).\n \\end{equation}\n\n These axioms are not \\hyperref[def:positive_formula]{positive formulas}, hence fields automatically get worse metamathematical properties than rings, for example.\n\n \\thmitem{def:field/homomorphism}\\mcite[453]{Knapp2016BasicAlgebra} A \\hyperref[def:first_order_homomorphism]{first-order homomorphism} between fields is simply a \\hyperref[def:ring/homomorphism]{unital ring homomorphism}.\n\n \\thmitem{def:field/submodel} If for two fields \\( \\Bbbk \\) and \\( \\BbbK \\) are have \\( \\Bbbk \\subseteq \\BbbK \\), we say that \\( \\BbbK \\) is a \\term{field extension} of \\( \\Bbbk \\) and that \\( \\Bbbk \\) is a \\term{subfield} of \\( \\BbbK \\). In particular, if \\( \\BbbK = \\Bbbk \\), we say that the extension is trivial.\n\n \\thmitem{def:field/category} The category of \\hyperref[def:large_and_small_sets]{\\( \\mscrU \\)-small} fields \\( \\ucat{Field} \\) is a full subcategory of \\hyperref[def:ring/category]{\\( \\ucat{CRing} \\)} with objects restricted to fields.\n \\end{thmenum}\n\\end{definition}\n\\begin{defproof}\n The equivalence of definitions follows from \\fullref{thm:def:semiring_ideal/units}.\n\\end{defproof}\n\n\\begin{proposition}\\label{thm:field_of_fractions}\n Let \\( D \\) be an \\hyperref[def:integral_domain]{integral domain}. The \\hyperref[def:ring_localization]{localization} of \\( D \\) at the zero ideal \\( \\set{ 0_R } \\) is a \\hyperref[def:field]{field}, which we call the \\term{field of fractions} of \\( D \\).\n\\end{proposition}\n\\begin{proof}\n By \\fullref{thm:def:ring_localization/prime_ideal}, the localization by the prime ideal \\( \\set{ 0_R } \\) has whose only maximal ideal is \\( S^{-1} \\set{ 0_R } \\). Since \\( 0_R \\) is absorbing, \\( S^{-1} \\set{ 0_R } \\) is again the zero ideal. Therefore, it is the only proper ideal of the localization \\( S^{-1} D \\), and hence the localization is a \\hyperref[def:ring/simple]{simple ring}.\n\n Since \\( D \\) is an integral domain, by \\fullref{thm:def:ring_localization/injective_inclusion}, \\( S^{-1} D \\) is a superring of \\( D \\). It is therefore a nontrivial commutative simple ring, and thus it satisfies \\fullref{def:field/simple}.\n\\end{proof}\n\n\\begin{theorem}[Field of fractions universal property]\\label{thm:field_of_fractions_universal_property}\n The \\hyperref[def:field_of_fractions]{field of fractions} \\( \\BbbK \\) of the integral domain \\( D \\) satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every field \\( \\BbbL \\) and every ring homomorphism \\( \\varphi: D \\to \\BbbL \\), \\( \\varphi \\) \\hyperref[def:factors_through]{uniquely factors through} \\( \\BbbK \\). That is, there exists a unique field homomorphism \\( \\widetilde{\\varphi}: \\BbbK \\to \\BbbL \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:field_of_fractions_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__field_of_fractions_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\\end{theorem}\n\\begin{proof}\n This is simply a special case of \\fullref{thm:ring_localization_universal_property}.\n\\end{proof}\n\n\\begin{definition}\\label{def:rational_function_field}\n The field of \\term{rational algebraic functions} \\( D(\\mscrX) \\) for the set of indeterminates \\( \\mscrX \\) over the \\hyperref[def:integral_domain]{integral domain} \\( D \\) is the \\hyperref[thm:field_of_fractions]{field of fractions} of the \\hyperref[def:polynomial_algebra]{polynomial ring} \\( D[\\mscrX] \\).\n\n Despite the name, elements of the field of fractions are not actually functions, but merely formal expressions. In particular, an analog of \\fullref{thm:polynomial_algebra_universal_property} does not really make sense.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:adjoining_elements_to_field}\n Let \\( \\Bbbk \\subseteq \\BbbK \\) be \\hyperref[def:field]{fields} and let \\( A \\) be an arbitrary subset of \\( \\BbbK \\).\n\n Let \\( \\Bbbk[A] \\) be the ring obtained by adjoining the elements of \\( A \\) to \\( \\Bbbk \\) as described in \\fullref{thm:adjoining_elements_to_semiring}. The \\hyperref[def:field_of_fractions]{field of fractions} of \\( \\Bbbk[A] \\) is the smallest field extension of \\( \\Bbbk \\) containing \\( A \\).\n\n We denote this extension by \\( \\Bbbk(A) \\). It should not be confused with the image of the evaluation homomorphism on the \\hyperref[def:rational_function_field]{field of rational functions}; the rational functions do not actually have an evaluation homomorphism.\n\\end{proposition}\n\\begin{proof}\n It follows from \\fullref{thm:adjoining_elements_to_semiring} that \\( \\Bbbk[A] \\) is the smallest superring of \\( \\Bbbk \\) containing \\( A \\). By \\fullref{thm:field_of_fractions_universal_property}, \\( \\Bbbk(A) \\) is the smallest field containing \\( \\Bbbk[A] \\).\n\\end{proof}\n", "meta": {"hexsha": "c7aa70bfecb68a9d9e11f95187d299ca523cac65", "size": 37240, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/rings.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/rings.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/rings.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.8765008576, "max_line_length": 631, "alphanum_fraction": 0.6757518797, "num_tokens": 12200, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835452961425, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6402257341225217}} {"text": "\\documentclass[a4paper]{article}\n\n\\def\\npart{IB}\n\n\\def\\ntitle{Optimisation}\n\\def\\nlecturer{M.\\ Tehranchi}\n\n\\def\\nterm{Easter}\n\\def\\nyear{2017}\n\n\\input{header}\n\n\\begin{document}\n\n\\input{titlepage}\n\n\\tableofcontents\n\n\\section{Introduction}\n\nThe typical problem is of the form\n\n\\begin{center}\n minimise $f(x)$ subject to $g(x) = b,\\: x \\in X$,\n\\end{center}\n\nwhere\n\n\\begin{itemize}\n\\item $f : \\mathbb{R}^n → \\mathbb{R}$ is the \\emph{objective function},\n\\item $X \\subseteq \\mathbb{R}^n$ defines a \\emph{regional constraint},\n\\item $g: \\mathbb{R}^n → \\mathbb{R}^m$ defines $m$ \\emph{functional constraints},\n\\item $b \\in \\mathbb{R}^m$ is the \\emph{right-hand side}.\n\\end{itemize}\n\nWe will also use the terminology\n\n\\begin{itemize}\n\\item a \\emph{feasible solution} is any $x \\in X$ s.t. $g(x) = b$,\n\\item an \\emph{optimal solution} is a feasible solution $x^*$ s.t. $f(x^*) ≤ f(x)$ for all feasible $x$.\n\n\\end{itemize}\n\nA problem with inequality constraints can be put into equality form by introducing \\emph{slack variables}:\n\n\\begin{center}\n minimise $f(x)$ subject to $g(x) ≤ b,\\: x \\in X$\n\n minmise $f(x)$ subject to $g(x) + z = b,\\: x \\in X$.\n\\end{center}\n\n\\section{Lagrangian Methods}\n\nConsider the problem\n\n\\begin{center}\n minimise $f(x)$ subject to $g(x) = b,\\: x \\in X$.\n\\end{center}\n\nIntroduce a new function $L : \\mathbb{R}^n × \\mathbb{R}^m → \\mathbb{R},\\: L(x, \\lambda) = f(x) + \\lambda^T (b - g(x))$, the \\emph{Lagrangian} of the problem. The components $\\lambda_i$ is called the \\emph{Lagrangian multiplier} for the $i$-th functional constraint.\n\n\\subsection{Lagrangian Sufficiency}\n\n\\begin{theorem}[Lagrangian Sufficiency Theorem]\n Let $x^*$ be feasible. Suppose there exists $\\lambda^* \\in \\mathbb{R}^m$ s.t.\n \\[\n L(x^*, \\lambda^*) ≤ L(x, \\lambda^*) \\: \\text{for all } x \\in X\n \\]\n\n then $x^*$ is optimal.\n\n\\end{theorem}\n\n\\begin{proof}\n For any feasible $x$ and any $\\lambda$ we have\n \\[\n L(x, \\lambda) = f(x) + \\lambda^T (b - g(x)) = f(x)\n \\]\n\n so\n \\begin{align*}\n f(x^*) &= L(x^*, \\lambda^*) \\\\\n &≤ L(x, \\lambda^*) \\: \\text{for all} x \\in X \\: \\text{by assumption } \\\\\n &= f(x) \\: \\text{for all feasible } x \\in X\n \\end{align*}\n\\end{proof}\n\n\\subsection{Steps}\n\nConsider the problem to\n\n\\begin{center}\n minimise $f(x)$ subject to $g(x) = b,\\: x \\in X$\n\\end{center}\n\n\\begin{itemize}\n\\item[Step 1] $\\Lambda := \\{\\displaystyle \\lambda \\in \\mathbb{R}^n: \\inf_{x \\in X} L(x, \\lambda) > -∞\\}$\n\\item[Step 2] For each $\\lambda \\in \\Lambda$ find the optimal solution to the unconstrained problem\n\n\\begin{center}\n minimise $L(x, \\lambda)$ subject to $x \\in X$\n\\end{center}\n\nLet $x(\\lambda)$ be the minimiser\n\\item[Step 3] Find a $\\lambda^* \\in \\Lambda$ s.t. $x^* = x(\\lambda^*)$ is feasible for the original problem, i.e. $g(x^*) = b$.\n\\end{itemize}\n\nNote that $x^*$ is optimal by Lagrangian sufficiency theorem.\n\n\\subsection{Complementary Slackness}\n\nGiven an inequality constraint, we introduce the equivalent equality constraint problem with slack variables. The Lagrangian is now\n\\[\n L(x, z, \\lambda) = f(x) + \\lambda^T (b - g(x)) - \\lambda^T z\n\\]\n\nNote that for the infimum to exist, $\\lambda ≤ 0$. The inequality constraint $g(x) ≤ b $ for the variable $x$ introduces a \\emph{sign constraint} $\\lambda ≤ 0$ for the Lagrange multiplier $\\lambda$.\n\nIn addition, for $\\lambda ≤ 0$ we have $\\displaystyle \\inf_{z ≥ 0} (-\\lambda^T z) = 0$. Thus for each $\\lambda \\in \\Lambda$, the optimal $z = z(\\lambda)$ satisfis the \\emph{compementary slacknes} condition $\\lambda^T z = 0$. Hence $\\lambda_i z_i = 0$ for all $i$.\n\n\\subsection{Lagrangian Necessity}\n\nNote that for any Lagrange multiplier $\\lambda$ we have\n\n\\begin{align*}\n \\inf_{x \\in X,\\: g(x) = b} f(x) &= \\inf_{x \\in X,\\: g(x) = b} (f(x) + \\lambda^T (b - g(x))) \\\\\n &= \\inf_{x \\in X,\\: g(x) = b} L(x, \\lambda) \\\\\n &≥ \\inf_{x \\in X} L(x, \\lambda)\n\\end{align*}\n\nsince $\\{x \\in X: g(x) = b\\} ⊆ X$. We say the Lagrangian method works if there exists a Lagrange multiplier $\\lambda^*$ s.t. there is an equality. To characterise when the Lagrangian method works, we need to define some terms.\n\n\\begin{definition}\n $\\psi: \\mathbb{R}^m → \\mathbb{R}$ has a \\emph{supporting hyperplane} at a point $b \\in \\mathbb{R}^m$ if there exist a $\\lambda$ s.t.\n \\[\n \\psi(c) ≥ \\psi(b) + \\lambda^T (c-b)\n \\]\n for all $c \\in \\mathbb{R}^m$.\n\\end{definition}\n\n\\begin{definition}\n The \\emph{value function} $φ$ on $\\mathbb{R}^n$ is defined by $\\displaystyle\\phi(c) = \\inf_{x \\in X,\\: g(x) = c} f(x)$.\n\\end{definition}\n\n\\begin{theorem}[Lagrangian Necessity Theorem]\n The Lagrangian method works for the problem iff the value function has a supporting hyperplane at $b$.\n\\end{theorem}\n\n\\begin{proof}\n The Lagrangian method works iff there exists a $\\lambda$ s.t.\n \\[\n \\phi(b) = \\inf_{x \\in X} (f(x) + \\lambda^T (b - g(x)))\n \\]\n\n The value function has a supporting hyperplane at $b$ iff there exists a $\\lambda$ s.t.\n \\[\n \\phi(b) = \\inf_{c \\in \\mathbb{R}^m} (\\phi(c) + \\lambda^T (b - c))\n \\]\n\n Thus the equivalence of the two hypotheses is proven by noting the equality\n \\begin{align*}\n \\inf_{x \\in X} (f(x) + \\lambda^T (b - g(x))) &= \\inf_{c \\in \\mathbb{R}^m} \\underbrace{\\inf_{x \\in X,\\: g(x) = c} (f(x)}_{\\phi(c)} + \\lambda^T \\underbrace{(c - g(x))}_{= 0} + \\lambda^T (b - c)) \\\\\n &= \\inf_{c \\in \\mathbb{R}^m} (\\phi(c) + \\lambda^T (b - c))\n \\end{align*}\n\\end{proof}\n\nTo check whether a function has a supporting hyperplane, we have to define a few terms.\n\n\\begin{definition}\n A subset $C \\subseteq \\mathbb{R}^n$ is \\emph{convex} if\n \\[\n x, y \\in C \\: \\text{implies } \\theta x + (1 - \\theta)y \\in C \\: \\text{for all } 0 \\subseteq \\theta \\subseteq 1.\n \\]\n\\end{definition}\n\n\\begin{definition}\n A function $\\psi: \\mathbb{R}^m \\rightarrow \\mathbb{R}^m$ is \\emph{convex} if\n \\[\n \\psi(\\theta x + (1 - \\theta) y \\leq \\theta\\psi(x) + (1-\\theta) \\psi(y) \\: \\text{for all } x, y \\in \\mathbb{R}^m \\: \\text{and } 0 \\leq \\theta \\leq 1.\n \\]\n\\end{definition}\n\n\\begin{corollary}\n A fucntion $\\psi : \\mathbb{R}^m \\rightarrow \\mathbb{R}$ is convex iff the set\n \\[\n C := \\{(x, y) : \\psi(x) \\leq y \\} \\subseteq \\mathbb{R}^{m + 1}\n \\]\n is convex\\footnote{The set $C$ defined above is call the \\emph{epigraph} of $\\psi$.}.\n\\end{corollary}\n\n\\begin{theorem}[Non-examinable]\n A function is convex iff it has a supporting hyperplane at each point.\n\\end{theorem}\n\n\\begin{proposition}\n If\n \\begin{enumerate}\n \\item the set $X$ is convex,\n \\item the objective function $f$ is convex, and\n \\item the functional constraint is\n \\begin{itemize}\n \\item either $g(x) = b$ and $g$ is linear, or\n \\item $g(x) \\leq b$ and $g$ is convex.\n \\end{itemize}\n \\end{enumerate}\n\n then $\\psi$ is convex.\n\\end{proposition}\n\n\\section{Dual Problem}\n\nConsider the \\emph{primal} problem\n\n\\begin{center}\n $P:$ minimise $f(x)$ subject to $g(x) = b,\\: x \\in X$.\n\\end{center}\n\nAs before, introduce the Lagrangian $L$ and the set of Lagrange multipliers $\\Lambda$. Now define the \\emph{dual objective function} $h: \\Lambda \\rightarrow \\mathbb{R}$ by\n\\[\n h(\\lambda) = \\inf_{x \\in X} L(x, \\lambda).\n\\]\n\nThe \\emph{dual} problem is defined to be\n\n\\begin{center}\n $D:$ maximise $h(\\lambda)$ subject to $\\lambda \\in \\Lambda$.\n\\end{center}\n\nThe set $\\Lambda$ is the set of \\emph{feasible solutions to the dual problem}.\n\n\\begin{theorem}[Weak Duality]\n Let $x$ be feasible for $P$ and let $\\lambda$ be feasible for $D$. Then\n \\[\n h(\\lambda) \\leq f(x)\n \\]\n\n and in particular\n \\[\n \\sup_{\\lambda \\in \\Lambda} \\leq \\inf_{x \\in X,\\: g(x) = b} f(x).\n \\]\n\\end{theorem}\n\n\\begin{proof}\n Let $x$ and $f$ be feasible for their respective problems, then\n \\begin{align*}\n h(\\lambda) &= \\inf\\{L(x', \\lambda): x' \\in X \\} \\\\\n &\\leq L(x, \\lambda) \\: \\text{for all } x \\in X \\\\\n &= f(x) \\: \\text{for all feasible } x\n \\end{align*}\n\\end{proof}\n\nThe difference\n\\[\n \\inf_{x \\in X,\\: g(x) = b} f(x) - \\sup_{\\lambda \\in \\Lambda} h(\\lambda)\n\\]\n\nis called the \\emph{duality gap}. Weak duality says that the duality gap is non-negative while in the case where the conditions of Lagrangian necessity are met, the duality gap is zero. This is called strong duality.\n\n\\begin{eg}[Linear Programming]\n Consider the primal problem\n \\begin{center}\n $P:$ maximise $c^T x$ subject to $Ax \\leq b,\\: x \\geq 0$\n \\end{center}\n where $A$ is a $m \\times n$ matrix, $b \\in \\mathbb{R}^m$, and $c \\in \\mathbb{R}^n$.\n\n The dual problem is found as follows:\n \\begin{enumerate}\n \\item Introduce slack variables\n \\begin{center}\n $P:$ maximise $c^T x$ subject to $Ax + z = b, \\: x \\geq 0, \\: z \\geq 0$\n \\end{center}\n \\item The Lagrangian is\n \\[\n L(x, z, \\lambda) = b^T \\lambda + (c - A^T \\lambda)^T x - \\lambda ^T z.\n \\]\n \\item The set of feasible solutions to the dual problem is\n \\begin{align*}\n \\Lambda &= \\{\\lambda \\in \\mathbb{R}^m: \\sup_{x \\geq 0,\\: z \\geq 0} L(x, z, \\lambda) < \\infty \\} \\\\\n &= \\{\\lambda \\in \\mathbb{R}^m: A^T \\lambda \\geq c,\\: \\lambda \\geq 0\\}.\n \\end{align*}\n \\item The dual objective funtion\n \\[\n \\sup_{x \\geq 0, z \\geq 0} L(x, z, \\lambda) = b^T \\lambda \\: \\text{for } \\lambda \\in \\Lambda.\n \\]\n\n The dual problem is then\n\n \\begin{center}\n $D:$ minimise $b^T \\lambda$ subject to $A^T \\lambda \\geq c,\\: \\lambda \\geq 0$.\n \\end{center}\n \\end{enumerate}\n\\end{eg}\n\nWe can verify that the dual of the dual is the original problem.\n\n\\begin{theorem}[Fundamental Theorem of Linear Programming]\n Consider the problem\n \n \\begin{center}\n $P:$ maximise $c^T x$ subject to $Ax \\leq b,\\: x \\geq 0$.\n \\end{center}\n \n A vector $x^* \\in \\mathbb{R}^n$ is optimal for $P$ iff there exists a vector $\\lambda^* \\in \\mathbb{R}^m$ s.t.\n\n \\begin{itemize}\n \\item $A x^* \\leq b,\\: x^* \\geq 0$ \\: (primal feasibility)\n \\item $A^T \\lambda^* \\geq c,\\: \\lambda^* \\geq 0$ \\: (dual feasibility)\n \\item $(\\lambda^*)^T (b - A x^*) = 0 = (x^*)^T (c - A^T \\lambda ^*)$ \\: (complementary slackness)\n \\end{itemize}\n\\end{theorem}\n\nIn this case, the value of the primal problem $c^T x^* = b^T \\lambda^*$ equals the value of the dual problem.\n\n\\subsection{Extreme Points and Basic Feasible Solutions}\n\nSuppose $\\psi$ is a convex function, then give $x, y \\in X$ and $0 \\leq \\theta \\leq 1$,\n\n\\begin{align*}\n \\psi(\\theta x + (1 - \\theta) y) & \\leq \\theta \\psi(x) + (1 - \\theta)\\psi(y) \\\\\n &\\leq \\max\\{\\psi(x), \\psi(y)\\}\n\\end{align*}\n\nThat is to say the maximimum of $\\psi$ on any segment occurs at one of the end points. Thus to find the maximum of $\\psi$ over $X$ we over have to consider points of $X$ that do not lie on a line segment contained in $X$.\n\n\\begin{definition}\n Let $C \\subseteq \\mathbb{R}^n$ be a convex set. A point $x \\in C$ is an \\emph{extreme point} if\n\n \\[\n x = \\theta y + (1 - \\theta) z\n \\]\n for $y, z \\in X$ and $0 \\leq \\theta \\leq 1$ imples $x = y = z$.\n\\end{definition}\n\n\\begin{definition}\n The \\emph{standard form} of a linear programme is\n\n \\begin{center}\n maximise $c^T x$ subject to $Ax = b,\\: x\\geq 0$\n \\end{center}\n where $c \\in \\mathbb{R}^n,\\: b \\in \\mathbb{R}^m$ and $A$ is a $m \\times n$ matrix. The set $C := \\{x \\in \\mathbb{R}^n: Ax = b\\}$ is the set of feasible solutions to the problem.\n\\end{definition}\n\n\\begin{proposition}\n The set $C$ is convex.\n\\end{proposition}\n\n\\begin{proof}\n Suppose $x, y \\in C$. Then $Ax = b,\\: x \\geq 0$, and $Ay = b,\\: y \\geq 0$. Fix $\\theta \\in [0, 1]$ and let $z = \\theta x + (1 - \\theta)y$. Then\n\n \\begin{align*}\n Az &= \\theta Ax + (1 - \\theta) Ay = b \\\\\n z_i &= \\theta x_i + (1 - \\theta) y_i \\geq 0 \\: \\text{for all } i\n \\end{align*}\n\n Hence $z \\in C$.\n\\end{proof}\n\nA solution $x \\in \\mathbb{R}^n$ of the equation $Ax = b$ is called \\emph{basic} if at least $n - m$ entries of $x$ are zero. If $x$ is a basic solution and $x \\geq 0$ then $x$ is called a \\emph{basic feasible solution}, abbreviated \\emph{b.f.s}.\n\n\\begin{theorem}\n Let $x$ be a point in $C$ with the property that have at least $m+1$ indices $i$ s.t. $x_i>0$. Then $x$ is not a extreme point of $C$.\n\\end{theorem}\n\n\\begin{proof}\n \\texttt{to be filled in}\n\\end{proof}\n\n\\begin{theorem}\n Suppose that every set of $m$ columns of $A$ is linearly independent. Let $x$ be a point in $C$ with the property that at most $m$ indices $i$ are s.t. $z_i>0$. Then $x$ is an extreme point of $C$.\n\\end{theorem}\n\n\\begin{proof}\n \\texttt{to be filled in}\n\\end{proof}\n\n\\subsection{Simplex Algorithm}\n\nWe assume $A$ is a $m\\times n$ matrix and $n > m$ and that every set of $m$ columns of $A$ is linearly independent. From discussion above it suffices to consider the extreme points of $C$. Fix $B \\subset \\{1, \\ldots ,n\\}$ with $|B|=m$, and let $N=\\{1,\\ldots, n\\} \\setminus B$. If $B=\\{i_1,\\ldots,i_m\\}$, let $A_B=(A_{i_1}\\:\\ldots\\:A_{i_m})$ bethe $m\\times m$ matrix formed by takning the colums of $A$ indexed by $i \\in B$. By assumption $A_B$ is invertible. Define $x_B$ and $c_B$ similarly. Similar for the set $N$.\n\nUsing this notation, the equation is\n\\[\n A_B x_B + A_N x_N = b\n\\]\n\nSetting $x_N=0$ yields $x_B=A_B^{-1}b$ so by rearranging the coordinates we may write the basic point as $x=\\binom{x_B}{x_N}\\binom{A_B^{-1}b}{0}$. To check this $x$ is feasible, we need $A_B^{-1}b \\geq 0$. If it is, compute the objective function $c^Tx=c_B^Tx_B$.\n\nTo check the optimality, we use the fundamental theorem. For the b.f.s. $x=\\binom{A_B^{-1}b}{0}$, we associate to it a Lagrange multiplier $\\lambda$ by complementary slackness\n\n\\begin{align*}\n 0 &= (c - A^T\\lambda)^Tx \\\\\n &= (c_B - A_B^T\\lambda)^Tx_B\n\\end{align*}\n\nAssuming \\emph{non-degeneracy}, we take $\\lambda=(A_B^T)^{-1}c_B$. By construction $x$ satisfies primal feasibility and $\\lambda$ satisfies complementary slackness. Thus $x$ is optimal iff $\\lambda$ satisfies dual feasibility. Thus if $A^T\\lambda\\geq c$ then we know we have found an optimal solution.\n\n\\section{Two-person Zero-sum Game}\n\nSuppose Player I and Player II are competing. I has $m$ choices of strategies, labelled $i \\in \\{1,\\ldots,m\\}$ while II has $n$, labelled $j \\in \\{1,\\ldots,n\\}$. Zero-sum means that if I chooses strategy $i$ and II chooses strategy $j$ then\n\n\\begin{itemize}\n\\item I is paid $a_{i,j}$\n\\item II is paid $-a_{i,j}$\n\\end{itemize}\n\nSo the net payment is zero. The matrix $A=(a_{i,j})_{i,j}$ is called the \\emph{payoff} matrix of the game\n\\end{document}\n", "meta": {"hexsha": "960cd5ad3f1ba3ec1770fac6e34c50cfb0be547d", "size": 14449, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "IB/optimisation.tex", "max_stars_repo_name": "geniusKuang/tripos", "max_stars_repo_head_hexsha": "127e9fccea5732677ef237213d73a98fdb8d0ca0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 27, "max_stars_repo_stars_event_min_datetime": "2018-01-15T05:02:27.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-10T15:48:31.000Z", "max_issues_repo_path": "IB/optimisation.tex", "max_issues_repo_name": "b-mehta/tripos", "max_issues_repo_head_hexsha": "8d3037ede28fed3a3cdb82a88dd3a005bf94b310", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-10-11T20:43:21.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-14T21:29:15.000Z", "max_forks_repo_path": "IB/optimisation.tex", "max_forks_repo_name": "b-mehta/tripos", "max_forks_repo_head_hexsha": "8d3037ede28fed3a3cdb82a88dd3a005bf94b310", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2017-11-08T16:16:20.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-25T17:20:19.000Z", "avg_line_length": 35.5012285012, "max_line_length": 517, "alphanum_fraction": 0.6264101322, "num_tokens": 5173, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8354835350552604, "lm_q1q2_score": 0.6402257218158316}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage{enumitem}\n\\usepackage{physics}\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{1em}\n\\relpenalty=10000\n\\binoppenalty=10000\n\n\\begin{document}\n\n\\section{Vectors}\n\\subsection{Rules}\n\\begin{enumerate}\n\\item\nThe angle between vectors $\\mathbf{a}$ and $\\mathbf{b}$ is the angle one should rotate vector $\\mathbf{a}$ to align it with $\\mathbf{b}$.\nThen:\n\\begin{enumerate}\n\\item \t\n$\\angle (\\mathbf{a},\\mathbf{b}) = - \\angle (\\mathbf{b},\\mathbf{a})$\n\\item \n$\\angle (\\mathbf{a},\\mathbf{b}) + \\angle (\\mathbf{b},\\mathbf{c}) = \\angle (\\mathbf{a},\\mathbf{c})$\n\\item\n$\\angle (\\mathbf{a},\\mathbf{b}) = \\angle (\\mathbf{-a},\\mathbf{b}) - 180 ^\\circ$\n\\end{enumerate}\n\n\\item \nThe dot product (also inner product or scalar product) of vectors $\\mathbf{a}$ and $\\mathbf{b}$ is: \\\\\n$$\\mathbf{a} \\cdot \\mathbf{b} \\equiv (\\mathbf{a},\\mathbf{b}) = \\abs{\\mathbf{a}} \\abs{\\mathbf{b}} \\cos \\angle (\\mathbf{a},\\mathbf{b}) $$\nThen:\n\\begin{enumerate}\n\\item \n$\\mathbf{a} \\cdot \\mathbf{b} = \\mathbf{b} \\cdot \\mathbf{a}$\n\\item \n$\\abs{\\mathbf{a} \\cdot \\mathbf{b}} \\leq \\abs{\\mathbf{a}} \\abs{\\mathbf{b}} $ \n\\item\n$(\\lambda \\mathbf{a} + \\mu \\mathbf{b}) \\cdot \\mathbf{c} = \\lambda \\mathbf{a} \\cdot \\mathbf{c} + \\mu \\mathbf{b} \\cdot \\mathbf{c}$\n\\item \nIf $\\abs{\\mathbf{a}},\\abs{\\mathbf{b}} \\ne 0$,then $\\mathbf{a} \\cdot \\mathbf{b}=0$ if and only if when $\\mathbf{a} \\perp \\mathbf{b}$.\n\\end{enumerate}\n\\item \nThe cross product (also vector product) of vectors $\\mathbf{a}$ and $\\mathbf{b}$ has direction for which $\\mathbf{c} \\perp \\mathbf{a}$ and $\\mathbf{c} \\perp \\mathbf{b}$ and magnitude of: \\\\\n$$\\abs{\\mathbf{a} \\cross \\mathbf{b}} = \\abs{\\mathbf{a}} \\abs{\\mathbf{b}} \\sin \\angle (\\mathbf{a},\\mathbf{b})$$ \\\\\nAlternatively:\n$$ \\mathbf{a} \\cross \\mathbf{b} = \\det\n\\left(\n\\begin{matrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\na_1 & a_2 & a_3 \\\\\nb_1 & b_2 & b_3 \n\\end{matrix}\n\\right)\n$$\nwhere $\\mathbf{i},\\mathbf{j},\\mathbf{k}$ are unit vectors. Then:\n\\begin{enumerate}\n\\item \n$\\mathbf{a} \\cross \\mathbf{b} = - \\mathbf{b} \\cross \\mathbf{a} $\n\\item \n$(\\lambda \\mathbf{a} + \\mu \\mathbf{b}) \\times \\mathbf{c} = \\lambda \\mathbf{a} \\times \\mathbf{c} + \\mu \\mathbf{b} \\times \\mathbf{c}$\n\n\\item \nIf $\\abs{\\mathbf{a}},\\abs{\\mathbf{b}} \\ne 0$,then $\\mathbf{a} \\cross \\mathbf{b}=0$ if and only if when $\\mathbf{a} \\parallel \\mathbf{b}$.\n\\end{enumerate}\n\\end{enumerate}\n\n\\newpage\n\\subsection{Problems}\n\\begin{enumerate}\n\\item\nProve that $\\abs{\\mathbf{a} + \\mathbf{b}}^2 + \\abs{\\mathbf{a} - \\mathbf{b}}^2 = 2(\\abs{\\mathbf{a}}^2 + \\abs{\\mathbf{b}}^2)$.\n\n\\item\nProve that if $(\\mathbf{a}+\\mathbf{b}) \\perp (\\mathbf{a}-\\mathbf{b})$, then $\\abs{\\mathbf{a}} = \\abs{\\mathbf{b}}$.\n\n\\item \nLet $\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC} = 0$ and $\\abs{OA}=\\abs{OB}=\\abs{OC}$. Prove that $ABC$ is an equilateral triangle\n\n\\item %Prasolov 13.1\nProve that it is possible to make another triangle $KLM$ from the medians $AA_1$,$BB_1$,$CC_1$ of a triangle $ABC$. Similarly, triangle $XYZ$ is formed from the medians of $KLM$. Prove that $\\triangle ABC \\sim \\triangle XYZ$ and find their similarity coefficient. \n\n\\item %Prasolov 13.4\nFrom a point inside a convex $n$-gon, the rays are drawn perpendicular to the sides and intersecting the sides (or their continuations). On these rays the vectors $\\mathbf{a_1}, \\dots , \\mathbf{a_n}$ whose lengths are equal to the lengths of the corresponding sides are drawn. Prove that\n$\\mathbf{a_1} + \\dots + \\mathbf{a_n} = 0$.\n\n\n\\item % Prasolov 13.7\nConsider n pairwise non-codirected vectors ($n \\geq 3$) whose sum is equal to zero. Prove that there exists a convex $n$-gon such that the set of vectors formed by its sides coincides with the given set of vectors.\n\n\\item % Napoleonic triangles\n(\\emph{Napoleon's theorem})Three equilateral triangles $ABD$, $BCE$, $CAF$ are constructed outside the triangle $ABC$. Prove that the centres of the constructed triangles form an equilateral triangle. \n\n\\item %Prasolov 13.12\n\\begin{enumerate}\n\\item Let A, B, C and D be arbitrary points on a plane. Prove that\n$$ \\overrightarrow{AB} \\cdot \\overrightarrow{CD} +\n\\overrightarrow{BC} \\cdot \\overrightarrow{AD} +\n\\overrightarrow{CA} \\cdot \\overrightarrow{BD} = 0\n $$\n\\item Prove that the heights of a triangle intersect at one point.\t\n\\end{enumerate}\n\n\\item %Prasolov 13.13\nLet $O$ be the centre of the circumcircle of triangle $ABC$ and let point $H$\nsatisfy $\\overrightarrow{OH} = \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}$. Prove that $H$ is the intersection point of heights of triangle.\n\n\\item %Prasolov 13.19\nGiven points $A$, $B$, $C$ and $D$. Prove that $AB^2 + BC^2 + CD^2 + DA^2 \\geq AC^2 + BD^2$,\nwhere the equality is attained only if $ABCD$ is a parallelogram.\n\n\\item % Prasolov13.22.\nPoints $A_1,\\dots, A_n$ lie on a circle with center $O$ and $\\overrightarrow{OA_1} +\\dots+ \\overrightarrow{OA_n}=0$. Prove\nthat for any point $X$ we have $XA_1 + \\dots + XA_n \\geq nR$, where $R$ is the radius of the circle.\n\n\n\n\\item \nProve Ceva's theorem: $X$, $Y$ and $Z$ are points on the sides $BC$, $CA$ and $AB$ of a triangle $ABC$ respectively. Then, lines $AX$, $BY$,$CZ$ intersect at a single point if and only if\n$$\\frac{\\overrightarrow{BX}}{\\overrightarrow{XC}} \\cdot \\frac{\\overrightarrow{CY}}{\\overrightarrow{YA}} \\cdot \\frac{\\overrightarrow{AZ}}{\\overrightarrow{ZB}} = 1$$\n\n\\item \nProve Menelaus's theorem: $X$, $Y$ and $Z$ are points on the sides (or their elongations) $BC$, $CA$ and $AB$ of a triangle $ABC$ respectively. Then, $X$,$Y$ and $Z$ lie on the same line if and only if\n$$\\frac{\\overrightarrow{BX}}{\\overrightarrow{XC}} \\cdot \\frac{\\overrightarrow{CY}}{\\overrightarrow{YA}} \\cdot \\frac{\\overrightarrow{AZ}}{\\overrightarrow{ZB}} = -1$$\n\n\n\\item % IMO-2013-4\nLet $ABC$ be an acute-angled triangle with orthocentre $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\\omega_1$ the circumcircle of $BWN$, and let $X$ be the point on $\\omega_1$ such that $WX$ is a diameter of $\\omega_1$. Analogously, denote by $\\omega_2$ the circumcircle of $CWM$, and let $Y$ be the point on $\\omega_2$ such that $WY$ is a diameter of $\\omega_2$. Prove that $X$, $Y$ and $H$ are collinear.\n\n\\item \nIs it possible to construct a triangle for which both coordinates of each vertex are integers?\n\n\\item %http://www.math.olympiaadid.ut.ee/eng/archive/prob1314.pdf p12\nThe angles of a triangle are $22.5^\\circ$, $45^\\circ$ and $112.5^\\circ$. Prove that inside this triangle there exists a point that is located on the median through one vertex, the angle bisector through another vertex and the altitude through the third vertex.\n\n\n\n\\end{enumerate}\n\n\\end{document}", "meta": {"hexsha": "65a738756f5c793fe7b6aef686f33876d1e89546", "size": 6830, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "08_vectors.tex", "max_stars_repo_name": "ZhaoWanLong/maths-olympiad", "max_stars_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-21T21:57:43.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-21T21:57:43.000Z", "max_issues_repo_path": "08_vectors.tex", "max_issues_repo_name": "ZhaoWanLong/maths-olympiad", "max_issues_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "08_vectors.tex", "max_forks_repo_name": "ZhaoWanLong/maths-olympiad", "max_forks_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-08T07:04:43.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-08T07:04:43.000Z", "avg_line_length": 48.4397163121, "max_line_length": 539, "alphanum_fraction": 0.6872620791, "num_tokens": 2408, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152325073083131, "lm_q2_score": 0.7853085834000791, "lm_q1q2_score": 0.640209085455986}} {"text": "% !TeX root = ../libro.tex\n% !TeX encoding = utf8\n%\n%*******************************************************\n% Summary\n%*******************************************************\n\n\\selectlanguage{english}\n\\chapter*{Summary}\\label{ch:summary}\n%\\addcontentsline{toc}{chapter}{Summar}\n\nNowadays experimental research in Neural Networks is more advanced than theoretical\nresults. \nFrom this we aim to establish a solid mathematical theory so as to optimize the current neural network models. \n\n\nAs a result of our study, we have proposed a novel neural \nnetwork model, and adapted and optimized\nevaluation and learning methods to it. \nMoreover, we have discovered some theorems that prove the \nequivalence among some activation functions, and hence we propose a new\n algorithm to initialize the weights of neural networks. Thanks to the\nfirst result, we obtain a criterion to choose the most \nsuitable activation function to maintain accuracy and reduce computational costs.\n Thanks to the second one, we might accelerate \nlearning convergence methods.\n\nIn addition, the models, methods and algorithms have been \nimplemented in Julia, resulting in the \\textit{OptimizedNeuralNetwork.jl} library. \n\nAll the theory development, designs, decisions and results are \nwritten in this memory, which has the following structure: \n\\begin{itemize}\n \\item \\textbf{Chapter \\ref{ch00:methodology}: Description of the methodology followed.} We have organized our project according to an agile philosophy based on personas methodology, user stories, milestones and tests. This method has conducted and linked mathematical and technical results and implementations, giving them coherence and validation methods. \n\n \\item \\textbf{Chapter \\ref{chapter:Introduction-neuronal-networks}: Description of the learning problem.} We defined the characteristic and type of machine learning problems. We will focus on supervised learning ones. \n\n \\item \\textbf{Chapter \\ref{ch03:teoria-aproximar}: Approximation theory.} In order to establish a solid theory, we will start our work trying to solve machine learning problems by traditional approximation methods. The main result we prove is the Stone Weierstrass's theorem. As a conclusion of this chapter we will achieve knowledge of the virtues and faults of traditional methods and understanding the necessity of new methods and structures such as neural networks. \n\n \\item \\textbf{Chapter \\ref{chapter4:redes-neuronales-aproximador-universal}: Neural networks are universal approximators.} In this chapter we introduce our neural \n network model and compare it with the conventional ones. In order to show it is well \n defined, we will prove the universal convergence of our model to any measurable \n function. In addition, we will give some results about how our model solves \n classification and regression problems as its number of neurons rises. Finally, we \n will argue if all of those math results can actually solve real life problems. The \n idea behind the debate is the computability representation of real numbers. \n\n \\item \\textbf{Chapter \\ref{chapter:construir-redes-neuronales}: The design and implementation of neural networks.} We will carefully describe the design and \n implementation of our model of neural network. Thanks to that we will obtain some \n mathematical results about bias and classification function. This will be useful to \n compare our model with the conventional ones and justify \nour selection. Moreover, we will explain, justify and design learning and evaluation \nmethods to our model. These methods are optimized versions of Forward Propagation and \nBackpropagation. \n\n\\item \\textbf{Chapter \\ref{funciones-activacion-democraticas-mas-demoscraticas}: Democratization of activation functions.} \nWe will explain in this chapter if there are better activation functions. In this \ndirection we will prove two original results which show that there are families of \nactivation functions that under the same conditions will solve problems with the same \naccuracy. As a result, if we compare the computational cost of the members of those \nfamilies and choose the faster one, we will obtain a method to optimize evaluation and \nlearning of neural networks without loss of accuracy. We have used the Wilcoxon \nsigned-rank test as a statistical hypothesis test so as to give a rigorous study of \nour criteria. \n\n\\item \\textbf{Chapter \\ref{section:inicializar_pesos}: Weight initializing algorithm.} \nSince Backpropagation and other iterative methods are sensitive to the initial \nvalue of a neural network, we will show an original method to initialize its weights \nfrom training data. This process not only will produce a better initial step but also \nhas lower computational cost than Backpropagation. To test the potential of this \nmethod we will use the Wilcoxon signed-rank test again and also, from the experiment's \nrequirements we will design and create our OptimizedNeuralNetwork.jl library. In this chapter we \nwill also explain every decision done during the design and implementation of the \nlibrary in order to be as efficient as possible.\t\n\n\\item \\textbf{Chapter \\ref{ch08:genetic-selection}: Use of genetic algorithm in the selection of activation function.} \nIn this chapter we will explain a future work. Given a fixed number of neurons, the \nselection of its activation function may be crucial to reduce the train and test \nerror. However, adding more free params to the search space increases its complexity \nand at same time the cost of finding a solution. Nevertheless, the result obtained in \nchapter \\ref{funciones-activacion-democraticas-mas-demoscraticas} and a property of our neural model will reduce the space complexity.\n\n\\item \\textbf{Chapter \\ref{ch09:conclusion}: Conclusions.}\n\\end{itemize} \n\n\\paragraph{KEYWORDS:}\n\\begin{itemize*}[label=,itemsep=1em,itemjoin=\\hspace{1em}]\n \\item neural networks\n \\item optimization\n \\item activation functions\n \\item weights initializing\n \\item machine learning library\n\\end{itemize*}\n\n% Al finalizar el resumen en inglés, volvemos a seleccionar el idioma español para el documento\n\\selectlanguage{spanish}\n\\endinput\n", "meta": {"hexsha": "295f70313e2b47525ca92eebdb4280575e4a6e93", "size": 6148, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Memoria/preliminares/summary.tex", "max_stars_repo_name": "BlancaCC/TFG-Estudio-de-las-redes-neuronales", "max_stars_repo_head_hexsha": "e16d039ba972c6f2fb4eeed899b3abfa6e121d07", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2022-01-09T18:56:07.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-26T17:05:44.000Z", "max_issues_repo_path": "Memoria/preliminares/summary.tex", "max_issues_repo_name": "BlancaCC/TFG-Estudio-de-las-redes-neuronales", "max_issues_repo_head_hexsha": "e16d039ba972c6f2fb4eeed899b3abfa6e121d07", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 73, "max_issues_repo_issues_event_min_datetime": "2021-11-27T06:47:40.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-21T17:38:01.000Z", "max_forks_repo_path": "Memoria/preliminares/summary.tex", "max_forks_repo_name": "BlancaCC/TFG-Estudio-de-las-redes-neuronales", "max_forks_repo_head_hexsha": "e16d039ba972c6f2fb4eeed899b3abfa6e121d07", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.7346938776, "max_line_length": 473, "alphanum_fraction": 0.7895250488, "num_tokens": 1308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324938410783, "lm_q2_score": 0.7853085859124002, "lm_q1q2_score": 0.6402090769281766}} {"text": "\\documentclass{homework}\n\\course{Math 5522H}\n\\author{Jim Fowler}\n\\input{preamble}\n\\DeclareMathOperator{\\Res}{Res}\n\n\\begin{document}\n\\maketitle\n\n\\begin{inspiration} % https://frinkiac.com/caption/S01E04/240324\n As far as anybody knows, we're a nice normal family.\n\\byline{Homer Simpson in S01E04}\n\\end{inspiration}\n\n\\section{Terminology}\n\n\\begin{problem}\n What is a \\textbf{normal family} of holomorphic functions on an open set $U$?\n\\end{problem}\n\n\\begin{problem}\\label{terminology-infinite-product}What does it mean to say that the infinite product\n \\(\n \\prod_{n=1}^\\infty \\left( 1 + a_n \\right)\n \\)\n converges?\n\\end{problem}\n\n\\section{Numericals}\n\n\\begin{problem}\n For which $z$ does the series\n \\[\n \\sum_{n=0}^\\infty \\frac{\\cos \\left( nz \\right)}{n!}\n \\]\n converge?\n\n Does it converge to a holomorphic function you recognize?\n\\end{problem}\n\n\\begin{problem}\n For which $z$ does the series\n \\[\n \\sum_{0 \\neq n \\in \\Z} \\left( \\frac{1}{z+n} - \\frac{1}{n} \\right)\n \\]\n converge?\n\n Does it converge to a meromorphic function you recognize?\n \\textit{Hint:} \\ref{residues-all-one}.\n\\end{problem}\n\n\\section{Exploration}\n\n\\begin{problem}\\label{poisson-summation}Describe conditions on $f$ so\n that, for a suitable $\\gamma_1$ and $\\gamma_2$,\n \\[\n \\sum_{n=-\\infty}^\\infty f(n) = \\int_{\\gamma_1} \\frac{f(z)}{e^{2\\pi i z} - 1} \\, dz - \\int_{\\gamma_2} \\frac{f(z)}{e^{2\\pi i z} - 1} \\, dz.\n \\]\n Then expand $1/(e^{2\\pi i z} - 1)$ as a geometric series to deduce the \\textbf{Poisson summation formula}\n \\[\n \\sum_{n=-\\infty}^\\infty f(n) = \\sum_{n=-\\infty}^\\infty \\hat{f}(n) \n \\]\n where $\\hat{f}$ is the Fourier transform, i.e.,\n \\[\n {\\hat {f}}(\\xi ) := \\int _{-\\infty }^{\\infty} f(x) \\, e^{-2\\pi ix \\xi} \\,dx.\n \\]\n\\end{problem}\n\n\n\\begin{problem}\\label{modularity}For $a > 0$ define\n $\\vartheta(a) = \\sum_{n=-\\infty}^\\infty e^{-a \\pi z^2}$. Recalling\n \\ref{fourier-transform-itself} shows that if $f(z) = e^{-a \\pi z^2}$\n we can compute $\\hat{f}(z)$. Use $\\hat{f}(z)$ and\n \\ref{poisson-summation} to verify\n \\[\n \\vartheta(a) = \\vartheta(1/a) / \\sqrt{a}.\n \\]\n\\end{problem}\n\n\\begin{problem}\n Having just celebrated $\\pi$-day, some computer calculations revealed\n\\begin{align*}\n \\vartheta(1/(4\\pi)) = \\sum_{n=-\\infty}^\\infty e^{-n^2/4} &=\n3.544907701811032\\textbf{10533931955126186}\\ldots \\\\\n 2\\sqrt{\\pi} &=\n3.544907701811032\\textbf{05459633496668229}\\ldots\n\\end{align*}\nIs my computer broken? (For more, see \\texttt{https://arxiv.org/abs/1809.10907}.)\n\\end{problem}\n\n\\begin{problem}\n Define\n \\[\n \\wp(z) = \\frac{1}{z^2} + \\sum_{0 \\neq \\lambda \\in \\Z[i]} \\left( \\frac{1}{(z - \\lambda)^2} - \\frac{1}{\\lambda^2} \\right)\n \\]\n which is \\textbf{Weierstrass' elliptic function} on the square lattice. Use \\ref{sum-one-over-gaussian-integers} to verify that $\\wp(z)$ converges.\n\\end{problem}\n\n\\begin{problem}\\label{elliptic-derivative-periodic}Compute $\\wp'(z)$ by differentiating term-by-term to show that $\\wp'(z + \\lambda) = \\wp'(z)$ for $\\lambda \\in \\Z[i]$.\n\\end{problem}\n\n\\begin{problem}\n Compare $\\wp(z)$ and $\\wp(-z)$. Use this to relate $\\wp(1/2)$ and $\\wp(-1/2)$ and to relate $\\wp(i/2)$ and $\\wp(-i/2)$ and with \\ref{elliptic-derivative-periodic}, conclude that $\\wp(z + \\lambda) = \\wp(z)$ for $\\lambda \\in \\Z[i]$.\n\\end{problem}\n\n\\section{Prove or Disprove and Salvage if Possible}\n\n\\begin{problem}\\label{differentiating-taylor-series}If\n $f(z) = \\sum_{n=0}^\\infty a_n z^n$ has radius of convergence $r$,\n then the series\n \\[\n g(z) = \\sum_{n=1}^\\infty n a_n z^{n-1} \n \\]\n has radius of convergence $r$ and if $|z| < r$ then $f'(z) = g(z)$.\n\\end{problem}\n\n\\begin{problem}\\label{sum-one-over-gaussian-integers}For $n > 2$, the\n series \\( \\displaystyle\\sum_{0 \\neq \\lambda \\in \\Z[i]} \\frac{1}{|\\lambda|^n} \\).\n converges.\n\\end{problem}\n\n\\begin{problem}\\label{elliptic-more-than-two-poles}There is no meromorphic function $f : \\C \\to \\hat{\\C}$ having simple poles at $a+bi \\in \\Z[i]$ and satisfying $f(z) = f(z+a+bi)$ for all $z \\in \\C$ and $a+bi \\in \\Z[i]$.\n\\end{problem}\n\n\\begin{problem}\\label{infinite-product-sine}For all $z \\in \\C$, we have \n \\(\n \\sin \\left( \\pi z \\right) = \\pi \\displaystyle\\prod_{n=1}^\\infty \\left( 1 - \\frac{z^2}{n^2} \\right)\n \\). % missing factor of z\n\\end{problem}\n\n\\begin{problem}\\label{normal-family-example}Define\n $f_w : B_1(0) \\to \\C$ by $f_w(z) = z/(z-w)$. The family of\n functions $\\mathcal{F} := \\{ f_w \\mid w \\in \\C \\}$ is a normal\n family.\n\\end{problem}\n\n\\begin{problem}\\label{derivatives-normal-then-not-normal}Suppose $\\mathcal{F}$ is a family of functions defined on the domain $B_1(0)$. If $\\mathcal{F}' := \\{ f' \\mid f \\in \\mathcal{F} \\}$, the\n family consisting of derivatives of functions in the family\n $\\mathcal{F}$, is normal, then the original family $\\mathcal{F}$ is\n normal. % assume that \\{ f(0) \\mid f \\in \\mathcal{F} \\} is bounded\n\\end{problem}\n\n\\end{document}\n", "meta": {"hexsha": "1e38fd4df53f9670bf2dfbcd392c133f4a2c1e17", "size": 4884, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problem-sets/set10.tex", "max_stars_repo_name": "kisonecat/math5522h", "max_stars_repo_head_hexsha": "c9fc5eb915c6d29d91a864dfe066878b75305c42", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-13T03:38:29.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-13T03:38:29.000Z", "max_issues_repo_path": "problem-sets/set10.tex", "max_issues_repo_name": "kisonecat/math5522h", "max_issues_repo_head_hexsha": "c9fc5eb915c6d29d91a864dfe066878b75305c42", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problem-sets/set10.tex", "max_forks_repo_name": "kisonecat/math5522h", "max_forks_repo_head_hexsha": "c9fc5eb915c6d29d91a864dfe066878b75305c42", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-01-11T18:43:51.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-11T18:43:51.000Z", "avg_line_length": 34.1538461538, "max_line_length": 233, "alphanum_fraction": 0.6457821458, "num_tokens": 1822, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.8152324871074607, "lm_q1q2_score": 0.6402090675439575}} {"text": "\\input{Setup.tex}\n\\begin{document}\n\n\\mainmatter\n\n\\section{Hebbian Learning}\n\\vspace{1mm}\n\\begin{quote}\n\\textsl{When an axon of cell A is near enough to excite a cell B and repeatedly or persistently takes part in firing it, some growth process or metabolic change takes places in one or both cells such that A's efficiency, as one of the cells firing B, is increased.}\\cite{Hebb1949}\n\\end{quote}\n\n\\section{Spike Timing Dependant Plasticity}\nWrite about STDP, see the 'Postulates.pdf'. \\\\\n\nWith spike timing dependant plasticity we want to study how two neurons change their synaptic strength based on the time delay between spikes. When $\\theta(t)_j > \\pi$ we say that the neuron $\\theta_j$ spikes at time $t$. \nLet us say that $\\theta_i$ spikes at time $t_i$ and $\\theta_j$ spikes at $t_j$. Taking the time difference $\\Delta t_{ij}$ as $t_j - t_i$, we can say that when $\\Delta t_{ij} > 0$ the spikes are correlated (there exists a temporally causal relation), and we can model an increase in synaptic strength of the connection $K_{ij}$ from $\\theta_i$ to $\\theta_j$. In the same fashion we can decrease $K_{ji}$ when $\\Delta t_{ij} < 0$ as there is no causal relation. \\\\\nThe functions $W(t)$ that relate $\\Delta t_{ij}$ to $\\Delta K_{ij}$ are called \\textsl{learning windows}, as they define a range in which $K_{ij}$ is able to adapt. When signals between neurons show a very large time difference (negative or positive) we do not expect them to be correlated. Because the learning windows are generally not symmetrical we can also expect an asymmetrical adjacency matrix.\\\\\nAnother characteristic is the integral over the learning window. A window with a negative integral directs synaptic strengths mostly towards inhibitory behaviour, and vice versa with a positive integral. An integral of zero would mean that both inhibitory and excitatory synapses are stimulated equally.\\\\\n\nNow that we have a feeling of how \\STDP works, we need to formulate the behaviour exactly. One by one, we will denote our ideas into mathematics.\n\n\n\\subsection{Biphasic learning windows}\n\n\\subsubsection{Kempter1999}\nFollowing the notation in \\cite{Kempter1999}, we will denote the spike train coming from each neuron $\\theta_i$ as $S_i^{\\rm out}(t) = \\sum_{n} \\delta (t-t_{i}^{n})$, where $t_{i}^{n}$ is the time that $\\theta_i$ has fired. Similarly, we will denote the spike train coming into each neuron $\\theta_i$ as $S_i^{\\rm in}(t) = \\sum_{f} \\delta (t-t_{i}^{f})$. Now we can say that the synaptic strengths are adjusted as:\n\\begin{align}\n\\Delta K_{ij} &= \\int_{t}^{t+\\mathcal{T}} w^{\\rm{out}} S_i^{\\rm out}(\\tau) + w^{\\rm{in}} S_{j}^{\\rm {in}}(\\tau) \\mathrm{d}\\tau\n+ \\iint_{t}^{t+\\mathcal{T}} W( \\tau^\\prime - \\tau) S_{i}^{\\rm out}(\\tau) S_{j}^{\\rm in}( \\tau^\\prime) \\mathrm{d} \\tau \\mathrm{d} \\tau^\\prime\n\\label{eq:KempterSTDPFormulation1} \\\\\n&= \\sum_{t_i^{n}\\in \\mathcal{T}} w^{\\mathrm{out}} + \\sum_{t_{j}^{f} \\in \\mathcal{T}} w^{\\mathrm{in}} + \\sum_{t_{j}^{f}, t_i^{n} \\in \\mathcal{T}} W (t_{j}^{f}-t_i^{n} ) \\label{eq:KempterSTDPFormulation2}\n\\end{align}\nwhere \\cite{Kempter1999} proposes the following learning window:\n\\begin{align}\nW(t)_K = \\alpha\n\\begin{cases}\n\\left[A_{p}\\left(1-\\frac{t}{\\tilde{\\tau}_{p}}\\right)+A_{n}\\left(1-\\frac{t}{\\tilde{\\tau}_{n}}\\right)\\right] \\cdot \\exp \\left( \\frac{t}{\\tau_{\\rm syn}} \\right) & \\text{for } t \\leq 0 \\\\\nA_{p} \\cdot \\exp \\left(-\\frac{t}{\\tau_{p}}\\right) + A_{n} \\cdot \\exp \\left(-\\frac{t}{\\tau_{n}} \\right) & \\text{for } t > 0\n\\end{cases} \\label{eq:learningwindowKempter1999}\n\\end{align}\nHere $t$ is the delay between presynaptic spike arrival and postsynaptic firing, $\\alpha$ is a small learning parameter and all $\\tau$ are time constants. Numerical values are usually $\\alpha = 0.05$, $\\tau_{\\rm syn} = 5$ ms, $\\tau_{p} = 1$ ms, $\\tau_{n} = 20$ ms and $A_p = 1$ and $A_{n} = -1$. $\\tilde{\\tau}_{p} \\equiv \\tau_{\\rm syn} \\tau_{p} / (\\tau_{\\rm syn} + \\tau_{p})$ and $\\tilde{\\tau}_{n} \\equiv \\tau_{\\rm syn} \\tau_{n} / (\\tau_{\\rm syn} + \\tau_{n})$. \\\\\n$\\int W(s)_K \\mathrm{d}s = 2.56 \\times 10^{-4}$.\n\n\n\\subsubsection{Song 2000}\nThe first formulation of \\STDP as a mathematical model was in \\cite{Song2000}. It is postulated without being concerned about the biological aspect too much. The synaptic strengths are simply updated as:\n\\begin{align}\n\\Delta K_{ij} &= K_{ij} \\sum_{t_{j}^{f}, t_i^{n} \\in \\mathcal{T}} W (t_{j}^{f}-t_i^{n} ) \\label{eq:SongSTDPFormulation}\n\\end{align}\nThe learning window is defined as a discontinuous function:\n\\begin{align}\nW(t)_S =\n\\begin{cases}\nA_{p} \\cdot \\exp \\left(\\frac{-t}{\\tau_p}\\right) & \\text{for } s > 0 \\\\\nA_{n} \\cdot \\exp \\left(\\frac{t}{\\tau_n}\\right) & \\text{for } s \\leq 0\n\\end{cases} \\label{eq:learningwindowKempter1999}\n\\end{align}\nwhere we will use $A_p = 0.005$, $A_n = -0.00525$ and $\\tau_p = \\tau_n = 20$ ms. $\\int W(s)_S \\mathrm{d}s = -3.70 \\times 10^{-4}$ so we expect the weights to be suppressed towards a negative value.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width = \\textwidth]{../Figures/LearningWindowsBiphasic.png}\n\\caption{Two different biphasic learning windows. We can see how in $W(t)_s$ a larger weight is put on the anti-Hebbian learning.}\n\\label{fig:LearningWindowsBiphasic}\n\\end{figure}\n\n\n\\subsection{Triphasic Learning windows}\nBased off of neuronal actiity in the hippocmpus, biologically inspired triphasic windows have been developed. The papers describing the learning process are not specifying exactly how the weights are updated, as we have seen before there are a few options. We assume the change is simply:\n\\begin{align}\n\\Delta K_{ij} = \\sum_{t_{j}^{f}, t_i^{n} \\in \\mathcal{T}} W (t_{j}^{f}-t_i^{n} ) \\label{eq:WaddingtonSTDPFormulation}\n\\end{align}\nas there is no mention of other parameters.\n\n\\subsubsection{Chrol-Cannon 2012}\n\\begin{align}\nW(t)_C = A_{p} \\cdot \\exp \\left(\\frac{-\\left(t - 15 \\right)^{2}}{ \\tau_{p}}\\right) - A_{n} \\cdot \\exp \\left(\\frac{-\\left(t - 20\\right)^{2}}{ \\tau_{n}}\\right) \\label{eq:learningwindowChrolCannon2012}\n\\end{align}\nwhere $A_{p}=0.23$, $A_{n}=0.15$, $\\tau_{p}=200$ and $\\tau_n = 2000$. $\\int W(s)_C \\mathrm{d}s = -60.0 \\times 10^{-4}$.\n\n\\subsubsection{Waddington 2014}\nThe learning window is then defined as:\n\\begin{align}\nW(t)_W = A \\left[1-\\frac{\\left(t-\\alpha\\right)^{2}}{\\alpha^{2}}\\right] \\cdot \\exp \\left(\\frac{-\\left|t - \\alpha\\right|}{\\alpha}\\right) \\label{eq:learningwindowWaddington2014}\n\\end{align}\nWe will use $A = 0.1$ and $\\alpha = 4.0$ ms . $\\int W(s)_W \\mathrm{d}s = -8.0 \\times 10^{-4}$.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width = \\textwidth]{../Figures/LearningWindowsTriphasic.png}\n\\caption{The mean degree $\\kmean$ evolves over time as a function of the spiking behaviour, just like we expect.}\n\\label{fig:LearningWindowsTriphasic}\n\\end{figure}\n\n\n\\subsection{Interpretation}\nThe learning windows generally have $W(t^{\\ast}) = 0$ for $t^{\\ast} > 0$. This means that no learning will take place when the delay between neuron spikes is exactly $t^{\\ast} > 0$. The triphasic windows show two of those points.\n\n\\subsection{Results}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width = \\textwidth, trim={60 0 60 0}]{../Figures/LearningWithoutScaling.png}\n\\caption{Two different triphasic learning windows. We can see how in $W(t)_C$ a much higher penalty is given to signals that arrive too early or too late. The moments at which signals should peak also differ by about 10 ms.}\n\\label{fig:LearningWithoutScaling}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width = 0.2\\textheight]{../Figures/STDPbeforeIP.png}\n\\caption{Results from \\STDP.}\n\\label{fig:STDPbeforeIP}\n\\end{figure}\n\n\n\n\\subsection{Synaptic scaling}\nThere is no upper or lower bound on the synapse strength. Also generally, connection strengths are nonzero. This means that the notion of a \\textsl{network} is lost.\\\\\nOne technique we can apply to keep the strengths within a definitive range is to scale homeostatically - a method where increases in synaptic strength will balance out any decreases by scaling:\n\\begin{align}\nK_{ij}^s = K_{ij} \\frac{\\frac{1}{N} \\sum_{i,j} K_{ij}}{\\sum_{i} K_{ij}}\n\\end{align}\nIn this way, the out-degrees will remain constant. Using this approach, something has to remain constant, whether that is $\\kmean$, or $\\kmean^2$ or any other property of the adjacency matrix.\n\n\n\\subsection{Intrinsic plasticity}\nInstead of scaling the weights to preserve a certain quantity in the network, we can allow the neurons to adjust the sensitivity to incoming signals. So when some synaptic strengths are increased, we can reduce the excitability. In \\cite{Song2017} such a method is introduced in detail. We can simply update $\\eta_{i} \\rightarrow \\eta_{i} + \\eta_{\\max } \\cdot \\phi_{i}$ where \n\\begin{align}\n\\phi_{i} (t) =\\left\\{\\begin{array}{l}-\\alpha \\cdot \\exp \\left(\\frac{T_{\\min }-t}{T_{\\rm min }}\\right) \\text { if } tT_{\\rm max } \\\\ 0, \\text { otherwise }\\end{array}\\right.\n\\end{align}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width = 0.5\\textwidth]{../Figures/IPlearningFunction.png}\n\\caption{The learning behaviour for the intrinsic plastiity.}\n\\label{fig:IPlearningFunction}\n\\end{figure}\n\n\n\n\\subsection{Custom learning windows}\nSome of the behaviour we want to observe is currently not accounted for: synaptic strengths are unbounded and are generally nonzero. How can we model the behaviour where synaptic strengths can also settle on being zero? Perhaps a better idea than updating the synaptic strength by adding a new value, we can scale it.\n\\begin{itemize}\n\\item When $\\Delta t_{ij}$ is very large (both positive and negative) we expect no change in the synaptic strength: $W(-\\infty) = W(-\\infty) = 0$. \n\\item We expect a specific positive time delay to yield the most amount of synaptic strengthening: $\\Delta t_{\\rm best} = \\underset{x}{\\arg \\max} W(t)$\n\\item We expect small connections that have been reducing in size to quickly become zero, \n\\item A better scaling system would be that not the out- or in-degree is constant, but \n\\end{itemize}\n\n\n\n\\bibliographystyle{utphys}\n\\small{\\bibliography{references}}\n\n\\label{LastPage}~\n\n\\end{document}\n", "meta": {"hexsha": "fd802d84ab9a056eaea51d3ee67cc908fc004106", "size": 10173, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Writing/Notes/SynapticPlasticity.tex", "max_stars_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_stars_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Writing/Notes/SynapticPlasticity.tex", "max_issues_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_issues_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Writing/Notes/SynapticPlasticity.tex", "max_forks_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_forks_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.7961783439, "max_line_length": 464, "alphanum_fraction": 0.7169959697, "num_tokens": 3181, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8152324803738429, "lm_q2_score": 0.7853085859124003, "lm_q1q2_score": 0.6402090663522412}} {"text": "% Written by Anders Sjoqvist and Ulf Lundstrom, 2009\n% The main sources are: tinyKACTL, Beta and Wikipedia\n\n\\chapter{Mathematics}\n\n\\section{Recurrences}\nIf $a_n = c_1 a_{n-1} + \\dots + c_k a_{n-k}$, and $r_1, \\dots, r_k$ are distinct roots of $x^k + c_1 x^{k-1} + \\dots + c_k$, there are $d_1, \\dots, d_k$ s.t.\n\\[a_n = d_1r_1^n + \\dots + d_kr_k^n. \\]\nNon-distinct roots $r$ become polynomial factors, e.g. $a_n = (d_1n + d_2)r^n$.\n\n\\section{Trigonometry}\n\\begin{align*}\n\\sin(v+w)&{}=\\sin v\\cos w+\\cos v\\sin w\\\\\ncos(v+w)&{}=\\cos v\\cos w-\\sin v\\sin w\\\\\n\\end{align*}\n\\begin{align*}\n\\tan(v+w)&{}=\\dfrac{\\tan v+\\tan w}{1-\\tan v\\tan w}\\\\\n\\sin v+\\sin w&{}=2\\sin\\dfrac{v+w}{2}\\cos\\dfrac{v-w}{2}\\\\\n\\cos v+\\cos w&{}=2\\cos\\dfrac{v+w}{2}\\cos\\dfrac{v-w}{2}\n\\end{align*}\n\\[ (V+W)\\tan(v-w)/2{}=(V-W)\\tan(v+w)/2 \\]\nwhere $V, W$ are lengths of sides opposite angles $v, w$.\n\\begin{align*}\n\ta\\cos x+b\\sin x&=r\\cos(x-\\phi)\\\\\n\ta\\sin x+b\\cos x&=r\\sin(x+\\phi)\n\\end{align*}\nwhere $r=\\sqrt{a^2+b^2}, \\phi=\\operatorname{atan2}(b,a)$.\n\n\\section{Geometry}\n\n\\subsection{Triangles}\nSide lengths: $a,b,c$\\\\\nSemiperimeter: $p=\\dfrac{a+b+c}{2}$\\\\\nArea: $A=\\sqrt{p(p-a)(p-b)(p-c)}$\\\\\nCircumradius: $R=\\dfrac{abc}{4A}$\\\\\nInradius: $r=\\dfrac{A}{p}$\\\\\nLength of median (divides triangle into two equal-area triangles): $m_a=\\tfrac{1}{2}\\sqrt{2b^2+2c^2-a^2}$\\\\\nLength of bisector (divides angles in two): $s_a=\\sqrt{bc\\left[1-\\left(\\dfrac{a}{b+c}\\right)^2\\right]}$\\\\\nLaw of sines: $\\dfrac{\\sin\\alpha}{a}=\\dfrac{\\sin\\beta}{b}=\\dfrac{\\sin\\gamma}{c}=\\dfrac{1}{2R}$\\\\\nLaw of cosines: $a^2=b^2+c^2-2bc\\cos\\alpha$\\\\\nLaw of tangents: $\\dfrac{a+b}{a-b}=\\dfrac{\\tan\\dfrac{\\alpha+\\beta}{2}}{\\tan\\dfrac{\\alpha-\\beta}{2}}$\\\\\n\n\\subsection{Spherical coordinates}\n\\centerline{\\includegraphics[width=25mm]{../content/math/sphericalCoordinates}}\n\\[\\begin{array}{cc}\nx = r\\sin\\theta\\cos\\phi & r = \\sqrt{x^2+y^2+z^2}\\\\\ny = r\\sin\\theta\\sin\\phi & \\theta = \\textrm{acos}(z/\\sqrt{x^2+y^2+z^2})\\\\\nz = r\\cos\\theta & \\phi = \\textrm{atan2}(y,x)\n\\end{array}\\]\n\n\\section{Derivatives/Integrals}\n\\begin{align*}\n\t\\dfrac{d}{dx}\\arcsin x = \\dfrac{1}{\\sqrt{1-x^2}} &&& \\dfrac{d}{dx}\\arccos x = -\\dfrac{1}{\\sqrt{1-x^2}} \\\\\n\t\\dfrac{d}{dx}\\tan x = 1+\\tan^2 x &&& \\dfrac{d}{dx}\\arctan x = \\dfrac{1}{1+x^2} \\\\\n\t\\int\\tan ax = -\\dfrac{\\ln|\\cos ax|}{a} &&& \\int x\\sin ax = \\dfrac{\\sin ax-ax \\cos ax}{a^2} \\\\\n\t\\int e^{-x^2} = \\frac{\\sqrt \\pi}{2} \\text{erf}(x) &&& \\int xe^{ax}dx = \\frac{e^{ax}}{a^2}(ax-1)\n\\end{align*}\n\nIntegration by parts:\n\\[\\int_a^bf(x)g(x)dx = [F(x)g(x)]_a^b-\\int_a^bF(x)g'(x)dx\\]\n\n\\section{Sums}\n\\[ c^a + c^{a+1} + \\dots + c^{b} = \\frac{c^{b+1} - c^a}{c-1}, c \\neq 1 \\]\n\\begin{align*}\n\t1 + 2 + 3 + \\dots + n &= \\frac{n(n+1)}{2} \\\\\n\t1^2 + 2^2 + 3^2 + \\dots + n^2 &= \\frac{n(2n+1)(n+1)}{6} \\\\\n\t1^3 + 2^3 + 3^3 + \\dots + n^3 &= \\frac{n^2(n+1)^2}{4} \\\\\n\t1^4 + 2^4 + 3^4 + \\dots + n^4 &= \\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \\\\\n\\end{align*}\n\n\\section{Series} \n$$e^x = 1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\dots,\\,(-\\infty = \n\t\t\\frac{1}{|G|}\n\t\t\\sum_{g \\in G} f_1(g) \\ol{f_2(g)}.\n\t\\]\n\\end{definition}\nThis is the same ``dot product'' that we mentioned at the beginning,\nwhen we looked at the character table of $S_3$.\nWe now aim to prove the following orthogonality theorem,\nwhich will imply (III) and (IV) from earlier.\n\\begin{theorem}[Orthogonality]\n\tFor any finite-dimensional complex representations $V$ and $W$\n\tof $G$ we have\n\t\\[ \\left< \\chi_V, \\chi_W \\right> = \\dim \\Homrep(W, V). \\]\n\tIn particular, if $V$ and $W$ are irreps then\n\t\\[ \\left< \\chi_V, \\chi_W \\right> \n\t\t=\n\t\t\\begin{cases}\n\t\t\t1 & V \\cong W \\\\\n\t\t\t0 & \\text{otherwise}.\n\t\t\\end{cases}\n\t\\]\n\\end{theorem}\n\\begin{corollary}[Irreps give an orthonormal basis]\n\tThe characters associated to irreps\n\tform an \\emph{orthonormal} basis of $\\FunCl(G)$.\n\\end{corollary}\n\nIn order to prove this theorem, we have to define\nthe dual representation and the tensor representation,\nwhich give a natural way to deal with the quantity $\\chi_V(g)\\ol{\\chi_W(g)}$.\n\\begin{definition}\n\tLet $V = (V, \\rho)$ be a representation of $G$.\n\tThe \\vocab{dual representation} $V^\\vee$ is the representation on $V^\\vee$\n\twith the action of $G$ given as follows: for each $\\xi \\in V^\\vee$,\n\tthe action of $g$ gives a $g \\cdot \\xi \\in V^\\vee$ specified by\n\t\\[ v \\xmapsto{g \\cdot \\xi} \\xi\\left( \\rho(g\\inv)(v) \\right). \\]\n\\end{definition}\n\\begin{definition}\n\tLet $V = (V, \\rho_V)$ and $W = (W, \\rho_W)$\n\tbe \\emph{group} representations of $G$.\n\tThe \\vocab{tensor product} of $V$ and $W$ is the group representation\n\ton $V \\otimes W$ with the action of $G$ given on pure tensors by\n\t\\[\n\t\tg \\cdot (v \\otimes w)\n\t\t= \n\t\t(\\rho_V(g)(v)) \\otimes (\\rho_W(g)(w)) \\]\n\twhich extends linearly to define the action of $G$ on all of $V \\otimes W$.\n\\end{definition}\n\\begin{remark}\n\tWarning: the definition for tensors does \\emph{not} extend to algebras.\n\tWe might hope that $a \\cdot (v \\otimes w) = (a \\cdot v) \\otimes (a \\cdot w)$\n\twould work, but this is not even linear in $a \\in A$\n\t(what happens if we take $a=2$, for example?).\n\\end{remark}\n\n\\begin{theorem}\n\t[Character traces]\n\tIf $V$ and $W$ are finite-dimensional representations of $G$,\n\tthen for any $g \\in G$.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii $\\chi_{V \\oplus W}(g) = \\chi_V(g) + \\chi_W(g)$.\n\t\t\\ii $\\chi_{V \\otimes W}(g) = \\chi_V(g) \\cdot \\chi_W(g)$.\n\t\t\\ii $\\chi_{V^\\vee}(g) = \\ol{\\chi_V(g)}$.\n\t\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\n\tParts (a) and (b) follow from the identities\n\t$\\Tr(S \\oplus T) = \\Tr(S) + \\Tr(T)$\n\tand $\\Tr(S \\otimes T) = \\Tr(S) \\Tr(T)$.\n\tHowever, part (c) is trickier.\n\tAs $(\\rho(g))^{|G|} = \\rho(g^{|G|}) = \\rho(1_G) = \\id_V$\n\tby Lagrange's theorem, we can diagonalize $\\rho(g)$,\n\tsay with eigenvalues $\\lambda_1$, \\dots, $\\lambda_n$\n\twhich are $|G|$th roots of unity,\n\tcorresponding to eigenvectors $e_1$, \\dots, $e_n$.\n\tThen we see that in the basis $e_1^\\vee$, \\dots, $e_n^\\vee$,\n\tthe action of $g$ on $V^\\vee$ has eigenvalues\n\t$\\lambda_1\\inv$, $\\lambda_2\\inv$, \\dots, $\\lambda_n\\inv$.\n\tSo\n\t\\[\n\t\t\\chi_V(g) = \\sum_{i=1}^n \\lambda_i \\quad\\text{and}\\quad\n\t\t\\chi_{V^\\vee}(g) = \\sum_{i=1}^n \\lambda_i\\inv = \\sum_{i=1}^n \\ol{\\lambda_i}\n\t\\]\n\twhere the last step follows from the identity $|z|=1 \\iff z\\inv = \\ol z$.\n\\end{proof}\n\\begin{remark}\n\t[Warning]\n\tThe identities (b) and (c) do not extend linearly to $\\CC[G]$,\n\ti.e.\\ it is not true for example that $\\chi_V(a) = \\ol{\\chi_V(a)}$\n\tif we think of $\\chi_V$ as a map $\\CC[G] \\to \\CC$.\n\\end{remark}\n\\begin{proof}\n\t[Proof of orthogonality relation]\n\tThe key point is that we can now reduce\n\tthe sums of products to just a single character by\n\t\\[ \\chi_V(g) \\ol{\\chi_W(g)} = \\chi_{V \\otimes W^\\vee} (g). \\]\n\tSo we can rewrite the sum in question as just\n\t\\[\n\t\t\\left< \\chi_V, \\chi_W \\right>\n\t\t= \\frac{1}{|G|} \\sum_{g \\in G} \\chi_{V \\otimes W^\\vee} (g)\n\t\t= \\chi_{V \\otimes W^\\vee}\n\t\t\\left( \\frac{1}{|G|} \\sum_{g \\in G} g \\right).\n\t\\]\n\tLet $P : V \\otimes W^\\vee \\to V \\otimes W^\\vee$ be the\n\taction of $\\frac{1}{|G|} \\sum_{g \\in G} g$,\n\tso we wish to find $\\Tr P$.\n\t\\begin{exercise}\n\t\tShow that $P$ is idempotent.\n\t\t(Compute $P \\circ P$ directly.)\n\t\\end{exercise}\n\tHence $V \\otimes W^\\vee = \\ker P \\oplus \\img P$ (by \\Cref{prob:idempotent})\n\tand $\\img P$ is the subspace of elements which are fixed under $G$.\n\tFrom this we deduce that\n\t\\[ \\Tr P = \\dim \\img P =\n\t\t\\dim \\left\\{ x \\in V \\otimes W^\\vee\n\t\t\\mid g \\cdot x = x \\; \\forall g \\in G \\right\\}.\n\t\t\\]\n\tNow, consider the natural isomorphism $V \\otimes W^\\vee \\to \\Hom(W, V)$.\n\t\\begin{exercise}\n\t\tLet $g \\in G$.\n\t\tShow that under this isomorphism, $T \\in \\Hom(W, V)$\n\t\tsatisfies $g \\cdot T = T$ if and only if\n\t\t$T(g \\cdot w) = g \\cdot T(w)$ for each $w \\in W$.\n\t\t(This is just unwinding three or four definitions.)\n\t\\end{exercise}\n\tConsequently, $\\chi_{V \\otimes W^\\vee}(P) = \\Tr P = \\dim \\Homrep(W,V)$\n\tas desired.\n\\end{proof}\n\nThe orthogonality relation gives us a fast and mechanical way to check\nwhether a finite-dimensional representation $V$ is irreducible.\nNamely, compute the traces $\\chi_V(g)$ for each $g \\in G$,\nand then check whether $\\left< \\chi_V, \\chi_V \\right> = 1$.\nSo, for example, we could have seen the three representations of\n$S_3$ that we found were irreps directly from the character table.\nThus, we can now efficiently verify any time we have\na complete set of irreps.\n\n\\section{Examples of character tables}\n\\begin{example}\n\t[Dihedral group on $10$ elements]\n\tLet $D_{10} = \\left< r,s \\mid r^5 = s^2 = 1, rs = sr\\inv \\right>$.\n\tLet $\\omega = \\exp(\\frac{2\\pi i}{5})$.\n\tWe write four representations of $D_{10}$:\n\t\\begin{itemize}\n\t\t\\ii $\\Ctriv$, all elements of $D_{10}$ act as the identity.\n\t\t\\ii $\\Csign$, $r$ acts as the identity while $s$ acts by negation.\n\t\t\\ii $V_1$, which is two-dimensional and given by\n\t\t$r \\mapsto \\begin{bmatrix} \\omega & 0 \\\\ 0 & \\omega^4 \\end{bmatrix}$\n\t\tand $s \\mapsto \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$.\n\t\t\\ii $V_2$, which is two-dimensional and given by\n\t\t$r \\mapsto \\begin{bmatrix} \\omega^2 & 0 \\\\ 0 & \\omega^3 \\end{bmatrix}$\n\t\tand $s \\mapsto \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$.\n\t\\end{itemize}\n\tWe claim that these four representations are irreducible\n\tand pairwise non-isomorphic.\n\tWe do so by writing the character table:\n\t\\[\n\t\t\\begin{array}{|c|rccr|}\n\t\t\t\\hline\n\t\t\tD_{10} & 1 & r, r^4 & r^2, r^3 & sr^k \\\\ \\hline\n\t\t\t\\Ctriv & 1 & 1 & 1 & 1 \\\\\n\t\t\t\\Csign & 1 & 1 & 1 & -1 \\\\\n\t\t\tV_1 & 2 & \\omega+\\omega^4 & \\omega^2+\\omega^3 & 0 \\\\\n\t\t\tV_2 & 2 & \\omega^2+\\omega^3 & \\omega+\\omega^4 & 0 \\\\ \\hline\n\t\t\\end{array}\n\t\\]\n\tThen a direct computation shows the orthogonality relations,\n\thence we indeed have an orthonormal basis.\n\tFor example, $\\left< \\Ctriv, \\Csign \\right> = 1 + 2 \\cdot 1 + 2 \\cdot 1 + 5 \\cdot (-1) = 0$.\n\\end{example}\n\n\\begin{example}\n\t[Character table of $S_4$]\n\tWe now have enough machinery to to compute the character\n\ttable of $S_4$, which has five conjugacy classes\n\t(corresponding to cycle types $\\id$, $2$, $3$, $4$ and $2+2$).\n\tFirst of all, we note that it has two one-dimensional representations,\n\t$\\Ctriv$ and $\\Csign$, and these are the only ones\n\t(because there are only two homomorphisms $S_4 \\to \\CC^\\times$).\n\tSo thus far we have the table\n\t\\[\n\t\t\\begin{array}{|c|rrrrr|}\n\t\t\t\\hline\n\t\t\tS_4 & 1 & (\\bullet\\;\\bullet) & (\\bullet\\;\\bullet\\;\\bullet)\n\t\t\t\t& (\\bullet\\;\\bullet\\;\\bullet\\;\\bullet)\n\t\t\t\t& (\\bullet\\;\\bullet)(\\bullet\\;\\bullet)\n\t\t\t\t\\\\ \\hline\n\t\t\t\\Ctriv & 1 & 1 & 1 & 1 & 1 \\\\\n\t\t\t\\Csign & 1 & -1 & 1 & -1 & 1 \\\\ \n\t\t\t\\vdots & \\multicolumn{5}{|c|}{\\vdots}\n\t\t\\end{array}\n\t\\]\n\tNote the columns represent $1+6+8+6+3=24$ elements.\n\n\tNow, the remaining three representations have dimensions\n\t$d_1$, $d_2$, $d_3$ with\n\t\\[ d_1^2 + d_2^2 + d_3^2 = 4! - 2 = 22 \\]\n\twhich has only $(d_1, d_2, d_3) = (2,3,3)$ and permutations.\n\tNow, we can take the $\\refl_0$ representation\n\t\\[ \\left\\{ (w,x,y,z) \\mid w+x+y+z=0 \\right\\} \\]\n\twith basis $(1,0,0,-1)$, $(0,1,0,-1)$ and $(0,0,1,-1)$.\n\tThis can be geometrically checked to be irreducible,\n\tbut we can also do this numerically by computing the\n\tcharacter directly (this is tedious):\n\tit comes out to have $3$, $1$, $0$, $-1$, $-1$\n\twhich indeed gives norm\n\t\\[\n\t\t\\left< \\chi_{\\refl_0}, \\chi_{\\refl_0} \\right>\n\t\t=\n\t\t\\frac{1}{4!}\n\t\t\\left( \n\t\t\t\\underbrace{3^2}_{\\id}\n\t\t\t+ \\underbrace{6\\cdot(1)^2}_{(\\bullet\\;\\bullet)}\n\t\t\t+ \\underbrace{8\\cdot(0)^2}_{(\\bullet\\;\\bullet\\;\\bullet)}\n\t\t\t+ \\underbrace{6\\cdot(-1)^2}_{(\\bullet\\;\\bullet\\;\\bullet\\;\\bullet)}\n\t\t\t+ \\underbrace{3\\cdot(-1)^2}_{(\\bullet\\;\\bullet)(\\bullet\\;\\bullet)}\n\t\t\\right)\n\t\t= 1.\n\t\\]\n\tNote that we can also tensor this with the sign representation,\n\tto get another irreducible representation\n\t(since $\\Csign$ has all traces $\\pm 1$, the norm doesn't change).\n\tFinally, we recover the final row using orthogonality\n\t(which we name $\\CC^2$, for lack of a better name);\n\thence the completed table is as follows.\n\t\\[\n\t\t\\begin{array}{|c|rrrrr|}\n\t\t\t\\hline\n\t\t\tS_4 & 1 & (\\bullet\\;\\bullet) & (\\bullet\\;\\bullet\\;\\bullet)\n\t\t\t\t& (\\bullet\\;\\bullet\\;\\bullet\\;\\bullet)\n\t\t\t\t& (\\bullet\\;\\bullet)(\\bullet\\;\\bullet)\n\t\t\t\t\\\\ \\hline\n\t\t\t\\Ctriv & 1 & 1 & 1 & 1 & 1 \\\\\n\t\t\t\\Csign & 1 & -1 & 1 & -1 & 1 \\\\ \n\t\t\t\\CC^2 & 2 & 0 & -1 & 0 & 2 \\\\ \n\t\t\t\\refl_0 & 3 & 1 & 0 & -1 & -1 \\\\\n\t\t\t\\refl_0 \\otimes \\Csign & 3 & -1 & 0 & 1 & -1 \\\\\\hline\n\t\t\\end{array}\n\t\\]\n\\end{example}\n\n\\section\\problemhead\n\n\\begin{dproblem}\n\t[Reading decompositions from characters]\n\tLet $W$ be a complex representation of a finite group $G$.\n\tLet $V_1$, \\dots, $V_r$ be the complex irreps of $G$\n\tand set $n_i = \\left< \\chi_W, \\chi_{V_i} \\right>$.\n\tProve that each $n_i$ is a non-negative integer and\n\t\\[ W = \\bigoplus_{i=1}^r V_i^{\\oplus n_i}. \\]\n\t\\begin{hint}\n\t\tObvious.\n\t\tLet $W = \\bigoplus V_i^{m_i}$ (possible since $\\CC[G]$ semisimple)\n\t\tthus $\\chi_W = \\sum_i m_i \\chi_{V_i}$.\n\t\\end{hint}\n\\end{dproblem}\n\n\\begin{problem}\n\tConsider complex representations of $G = S_4$.\n\tThe representation $\\refl_0 \\otimes \\refl_0$\n\tis $9$-dimensional, so it is clearly reducible.\n\tCompute its decomposition in terms of the five\n\tirreducible representations.\n\t\\begin{hint}\n\t\tUse the previous problem, with $\\chi_W = \\chi_{\\refl_0}^2$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\t$\\Csign \\oplus \\CC^2 \\oplus \\refl_0 \\oplus (\\refl_0\\otimes\\Csign)$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t[Tensoring by one-dimensional irreps]\n\tLet $V$ and $W$ be irreps of $G$, with $\\dim W = 1$.\n\tShow that $V \\otimes W$ is irreducible.\n\t\\begin{hint}\n\t\tCharacters. Note that $|\\chi_W| = 1$ everywhere.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tFirst, observe that $|\\chi_W(g)|=1$ for all $g \\in G$.\n\t\t\\begin{align*}\n\t\t\t\\left< \\chi_{V \\otimes W}, \\chi_{V \\otimes W} \\right>\n\t\t\t&= \\left< \\chi_V \\chi_W, \\chi_V \\chi_W \\right> \\\\\n\t\t\t&= \\frac{1}{|G|} \\sum_{g \\in G} \n\t\t\t\\left\\lvert \\chi_V(g) \\right\\rvert^2\n\t\t\t\\left\\lvert \\chi_W(g) \\right\\rvert^2 \\\\\n\t\t\t&= \\frac{1}{|G|} \\sum_{g \\in G} \n\t\t\t\\left\\lvert \\chi_V(g) \\right\\rvert^2 \\\\\n\t\t\t&= \\left< \\chi_V, \\chi_V \\right> = 1.\n\t\t\\end{align*}\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t[Quaternions]\n\tCompute the character table of the quaternion group $Q_8$.\n\t\\begin{hint}\n\t\tThere are five conjugacy classes, $1$, $-1$\n\t\tand $\\pm i$, $\\pm j$, $\\pm k$.\n\t\tGiven four of the representations, orthogonality\n\t\tcan give you the fifth one.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tThe table is given by\n\t\t\\[\n\t\t\t\\begin{array}{|c|rrrrr|}\n\t\t\t\t\\hline\n\t\t\t\tQ_8 & 1 & -1 & \\pm i & \\pm j & \\pm k \\\\ \\hline\n\t\t\t\t\\Ctriv & 1 & 1 & 1 & 1 & 1 \\\\\n\t\t\t\t\\CC_i & 1 & 1 & 1 & -1 & -1 \\\\\n\t\t\t\t\\CC_j & 1 & 1 & -1 & 1 & -1 \\\\\n\t\t\t\t\\CC_k & 1 & 1 & -1 & -1 & 1 \\\\\n\t\t\t\t\\CC^2 & 2 & -2 & 0 & 0 & 0 \\\\\\hline\n\t\t\t\\end{array}\n\t\t\\]\n\t\tThe one-dimensional representations (first four rows)\n\t\tfollows by considering the homomorphism $Q_8 \\to \\CC^\\times$.\n\t\tThe last row is two-dimensional and can be recovered\n\t\tby using the orthogonality formula.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{sproblem}\n\t[Second orthogonality formula]\n\t\\label{prob:second_orthog}\n\t\\gim\n\tLet $g$ and $h$ be elements of a finite group $G$,\n\tand let $V_1$, \\dots, $V_r$ be the irreps of $G$.\n\tProve that\n\t\\[\n\t\t\\sum_{i = 1}^r \\chi_{V_i}(g) \\ol{\\chi_{V_i}(h)}\n\t\t=\n\t\t\\begin{cases}\n\t\t\t|C_G(g)| & \\text{if $g$ and $h$ are conjugates} \\\\\n\t\t\t0 & \\text{otherwise}.\n\t\t\\end{cases}\n\t\\]\n\tHere, $C_G(g) = \\left\\{ x \\in G : xg = gx \\right\\}$\n\tis the centralizer of $g$.\n\t\\begin{hint}\n\t\tWrite as \n\t\t\\[ \\sum_{i=1}^r \\chi_{V_i \\otimes V_i^\\vee} (gh\\inv)\n\t\t\t= \\chi_{\\bigoplus_i V_i \\otimes V_i^\\vee}(gh\\inv)\n\t\t\t= \\chi_{\\CC[G]}(gh\\inv).\n\t\t\\]\n\t\tNow look at the usual basis for $\\CC[G]$.\n\t\\end{hint}\n\\end{sproblem}\n", "meta": {"hexsha": "c3a4e316c46fec9601bda061fee2efc2b9703081", "size": 18897, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/rep-theory/characters.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/rep-theory/characters.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/rep-theory/characters.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.7645914397, "max_line_length": 93, "alphanum_fraction": 0.6415304017, "num_tokens": 7009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8479677545357569, "lm_q1q2_score": 0.6401435662795332}} {"text": "\\documentclass[10pt]{article}\n\\usepackage{amsmath}\n\n\\title{Some Informal Notes for Radiative Transfer Code}\n\\author{Mark Baum}\n\n\\begin{document}\n\n\\small\n\n\\maketitle\n\n\\section*{Pressure Coordinates}\n\nGet pressure coordinates by using the hydrostatic relation,\n\\begin{equation}\n\t\\frac{dP}{dz} = -\\rho g \\, ,\n\\end{equation}\nto replace $dz$ wherever it appears,\n\\begin{equation}\n\tdz = -\\frac{dP}{\\rho g} \\, .\n\\end{equation}\nConsider optical depth\n\\begin{equation}\n\t\\tau\t = \\int \\sigma N dz \\, ,\n\\end{equation}\nwhere $\\sigma$ is the absorption cross-section (m$^2$/molecule) and $N$ is the number density of the absorber (molecules/m$^3$). Switching to pressure coordinates yields\n\\begin{equation}\n\t\\tau\t = \\int \\sigma N \\frac{1}{\\rho g} dP \\, .\n\t\\label{tau}\n\\end{equation}\nThen the ideal gas law can be used to substitute for both $N$ and $\\rho$.\n\\begin{align}\n\tN &= \\frac{P}{k_B T} \\\\[1ex]\n\t\\rho &= \\frac{P \\mu}{k_B T N_A}\n\\end{align}\nwhere $\\mu$ is the molar mass in kg/mole and $N_A$ is Avogadro's number. Because we have $N/\\rho$ above in equation (\\ref{tau}) for optical depth, lots of things cancel.\n\\begin{equation}\n\t\\frac{N}{\\rho} = \\frac{P}{k_B T} \\frac{k_B T N_A}{P \\mu} = \\frac{N_A}{\\mu}\n\\end{equation}\nPutting this into equation (\\ref{tau}),\n\\begin{equation}\n\t\\tau = \\int \\sigma \\frac{N_A}{\\mu g} dP \\, .\n\\end{equation}\nThe negative sign in the original $dz=-dP/\\rho g$ has been dropped for convenience. Obviously, the optical depth must be positive, so the coordinate switch probably brings in another negative sign, perhaps when integral limits are switched because low pressure is high altitude and vice versa.\n\nThe same steps apply to the closely related Schwarzschild equation,\n\\begin{equation}\n\tdI = \\sigma N [B_{\\nu}(T) - I] dz \\, ,\n\\end{equation}\nexpressed here in differential form. We have the same $N dz$ factor that, through the hydrostatic relation and ideal gas law, gets replaced by $N_A/\\mu g$.\n\\begin{equation}\n\tdI = \\sigma\t\\frac{N_A}{\\mu g} [B_{\\nu}(T) - I] dP \\, .\n\\end{equation}\nImportant to remember that the molar mass here has units of kg/mole.\n\n\\section*{log Pressure Coordinates}\n\nSometimes there is a case for very high resolution in the upper atmosphere, per unit mass. To get that, log pressure coordinates can be used, which is essentially the same as using altitude directly but lets you keep the equations in terms of $P$. The same steps from above are applied, with one more. Simply use\n\\begin{equation}\n\t\\frac{1}{P} dP = d\\ln P\n\\end{equation}\nor, rearranged\n\\begin{equation}\n\tdP = P d\\ln P \\, .\n\\end{equation}\nPutting this into the optical depth and Schwarzschild equations,\n\\begin{align}\n\t\\tau\t &= \\int \\sigma N \\frac{1}{\\rho g} P d\\ln P \\\\[1em]\n\tdI &= \\sigma\t\\frac{N_A}{\\mu g} [B_{\\nu}(T) - I] P d\\ln P \\, .\n\\end{align}\nIt's the same as pressure coordinates, but multiplied by $P$. Then, of course, these equations must be integrated with the appropriate limits, $\\ln(P_{\\textrm{s}})$ and $\\ln(P_{\\textrm{toa}})$. Its important to consistently use the natural log, not log$_{10}$.\n\n\\section*{Integrating Multiple Streams}\n\nIn the absence of scattering, the total flux up or down in the atmosphere must account for irradiance at all angles in the hemisphere pointing up/down.\n\\begin{equation}\n\tF = \\int_0^{2\\pi} \\int_0^{\\pi/2} I(\\theta\t,\\phi) \\cos(\\theta) \\sin(\\theta) d\\theta d\\phi \\, ,\n\\end{equation}\nwhere $\\theta$ is the azimuthal angle (zero is straight up/down), $\\phi$ is the ``latitude\" angle going around the whole hemisphere. The cosine factor accounts for the angle of $I$ with respect to the horizontal planar surface that the flux is passing through. The sine factor simply comes out of the spherical integration, accounting for area attenuation near the very top of the hemisphere.\n\nAssuming streams are identical with respect to the latitude angle $\\phi$, that dimension is trivial to integrate.\n\\begin{equation}\n\tF = 2\\pi\t \\int_0^{\\pi/2} I(\\theta) \\cos(\\theta) \\sin(\\theta) d\\theta\n\\end{equation}\nThe azimuthal dimension generally must be integrated numerically. $I$ should be quite smooth with respect to $\\theta$, so gaussian integration is a good way to do it. The flux is then\n\\begin{equation}\n\tF = 2\\pi \\sum_{i=0}^N I(\\theta_i) w_i \\cos(\\theta_i) \\sin(\\theta_i)\n\\end{equation}\nwhere $\\theta_i$ are the gaussian quadrature nodes mapped to $[0,\\pi/2]$ and $w_i$ are the appropriately scaled weights. For a given number of streams $N$, one can precompute the factors $2\\pi w_i \\cos(\\theta_i) \\sin(\\theta_i)$.\n\n\\end{document}", "meta": {"hexsha": "60f9a48701c44ee61f7405931b84e78498624815", "size": 4484, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/notes.tex", "max_stars_repo_name": "markmbaum/ClearSky.jl", "max_stars_repo_head_hexsha": "84e8aec24a648dd7ebbef4c8d1f743bb2669a47d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-31T18:12:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-31T18:12:56.000Z", "max_issues_repo_path": "notes/notes.tex", "max_issues_repo_name": "wordsworthgroup/ClearSky.jl", "max_issues_repo_head_hexsha": "84e8aec24a648dd7ebbef4c8d1f743bb2669a47d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-07-23T20:42:43.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-23T20:42:44.000Z", "max_forks_repo_path": "notes/notes.tex", "max_forks_repo_name": "wordsworthgroup/ClearSky.jl", "max_forks_repo_head_hexsha": "84e8aec24a648dd7ebbef4c8d1f743bb2669a47d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.7391304348, "max_line_length": 392, "alphanum_fraction": 0.721677074, "num_tokens": 1352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677506936879, "lm_q2_score": 0.7549149868676284, "lm_q1q2_score": 0.6401435633790977}} {"text": "\\section{Jinchao's inputs}\n\n\\subsection{Classification versus regression}\nClassification is a special form of regression. Basically, it tries to find the (continuous) boundary of domains (that contain different clusters of data)\n\n\\subsection{Overfitting}\n\nIt seems to be related to the Runge phenomenon in polynomial interpolation for non-analytic function. It would be interesting to study, theoretically, whether deep neural network also has Runge phenomenon. \n\n\\subsection{CNN versus multigrid}\n$$\nAu=f\n$$\n\n\\paragraph{Multigrid}\n\nWe know $f$, then try to find $u$.\n\nWe have multilevel\n\nWe also use average, namely convolution, which is smoother:\n\n\\begin{itemize}\n\\item filter out the high frequencies\n\\end{itemize}\n\n\\paragraph{CNN}\nWe know $u$, then try to find $f$. In order to find $f$, we have to find $A$ first. \n\nWe use ``finite difference'', namely convolution, which is ``rougher''.\n\\begin{itemize}\n\\item filter out the low frequencies\n\\end{itemize}\n\n\n\\subsection{Role of probability theory and tool}\n\nThe role of probability is important for sure, but it is not essential .... \n\nIn the idealistic situation, namely all the data are good, no noise etc, we can approach deep learning problem without relying on probability theory and tools.\n\nTo understand the essence of deep learning, can we first skip probability tool?\n\nIf we skip probability, how much do we miss? We would not miss very much for the understanding how and why DNN works. \n\n\\begin{itemize}\n\\item High school probability is enough.\n\\end{itemize}\n\n", "meta": {"hexsha": "878f86a3271dc21f2be3fde53fe9ae2a6b7b5a55", "size": 1531, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/jinchao.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/jinchao.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/jinchao.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.62, "max_line_length": 208, "alphanum_fraction": 0.7668190725, "num_tokens": 360, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677506936878, "lm_q2_score": 0.7549149758396752, "lm_q1q2_score": 0.640143554027749}} {"text": "\\documentclass{notes}\n\n \\title{Computability}\n \\author{ian.mcloughlin@gmit.ie}\n \\date{\\today}\n\n\\begin{document}\n\n \n\n \n\n \\section*{Natural numbers}\n \\begin{align*}\n \\mathbb{N} &= \\{ 1, 2, 3, 4, \\ldots \\} \\\\\n \\mathbb{N}_0 &= \\{ 0, 1, 2, 3, \\ldots \\} \\\\\n 2\\mathbb{N}_0 &= \\{ 0, 2, 4, 6, \\ldots \\} \\\\\n 3\\mathbb{N}_0 &= \\{ 0, 3, 6, 9, \\ldots \\} \\\\\n \\mathbb{Z} &= \\{ \\ldots, -2, -1, 0, 1, 2, \\ldots \\} \\\\\n \\end{align*}\n \n \\section*{Real numbers}\n \n \\begin{center}\n \\begin{tikzpicture}\n % https://tex.stackexchange.com/a/148253\n \\draw[latex-latex] (-1.5,0) -- (3.5,0) ; %edit here for the axis\n \n \\draw[shift={(0,0)},color=black] (0pt,3pt) -- (0pt,-3pt);\n \\draw[shift={(1,0)},color=black] (0pt,3pt) -- (0pt,-3pt);\n \\draw[shift={(1.41,0)},color=black] (0pt,3pt) -- (0pt,-3pt);\n \\draw[shift={(3.12,0)},color=black] (0pt,3pt) -- (0pt,-3pt);\n \n \\draw[shift={(0,0)},color=black] (0pt,-3pt) -- (0pt,0pt) node[above] {$0$};\n \\draw[shift={(1,0)},color=black] (0pt,-3pt) -- (0pt,0pt) node[above] {$1$};\n \\draw[shift={(1.41,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below] {$\\sqrt{2}$};\n \\draw[shift={(3.14,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below] {$\\pi$};\n %\\draw[very thick] (,0) -- (1.92,0);\n \\end{tikzpicture}\n \\end{center}\n\n \\noindent Real numbers are difficult to define:\n \\begin{enumerate}\n \\item Draw an infinite\n straight line.\n \\item Mark two distinct points.\n \\item Call the left-most 0, the right-most 1.\n \\item One unit is the length between 0 and 1.\n \\item Mark any point on the line and measure its length $x$ in units from 0.\n \\end{enumerate}\n Then $x$ is a real number.\n The set $\\mathbb{R}$ is the set of all such $x$'s.\n \n \\section*{Bijections}\n \n \\[ f: \\mathbb{N}_0 \\rightarrow 2\\mathbb{N}_0 : n \\rightarrow 2n \\]\n \n \\begin{align*}\n f(0) &= 0 \\\\\n f(1) &= 2 \\\\\n f(2) &= 4 \\\\\n f(3) &= 6 \\\\\n \\vdots \\\\\n \\end{align*}\n \n \\noindent\n Every element of $\\mathbb{N}_0$ is mapped to a different element of $2\\mathbb{N}_0$.\n For instance, $1$ is the only number mapped to $2$.\n \n Also, every element of $2 \\mathbb{N}_0$ is mapped to from some element of $\\mathbb{N}_0$.\n There's no element of $2 \\mathbb{N}_0$ that isn't the image of some element of $\\mathbb{N}_0$.\n \n There are lots of functions like $f$ to and from each of the sets $\\mathbb{N}, \\mathbb{N}_0, \\mathbb{Z}$ and so on, but there is none to $\\mathbb{R}$.\n \n \\section*{Diagonilsation}\n \n Suppose there was a map $f: \\mathbb{N}_0 \\rightarrow \\mathbb{R}$.\n List out each element $n$ of $\\mathbb{N}_0$ with $f(n)$ beside it with all of the decimal places displayed.\n For example, suppose the first few are as follows.\n \n \\begin{alignat*}{2}\n 0 & \\quad \\rightarrow \\quad & 0&.16346234234234 \\ldots \\\\\n 1 & \\quad \\rightarrow \\quad & 10&.56775344747474 \\ldots \\\\\n 2 & \\quad \\rightarrow \\quad & 214&.99999999999999 \\ldots \\\\\n 3 & \\quad \\rightarrow \\quad & -1&.33333333333333 \\ldots \\\\\n 4 & \\quad \\rightarrow \\quad & 0&.00000000000000 \\ldots \\\\\n & \\quad \\;\\; \\vdots \\quad & & \\\\\n \\end{alignat*}\n\n While the list contains all of the natural numbers $\\mathbb{N}_0$, we can show that it can't possibly contain all of the real numbers $\\mathbb{R}$.\n We'll use Cantor's diagonal argument to construct a real number not in the list.\n \n Start by constructing a new number $0.$ and add decimal places as follows.\n Take the digit in the first decimal place of the first real number in the list.\n The first real number is $0.1634\\ldots$ so the first decimal place contains a $1$.\n Add $1$ to this number, giving $2$, and use this as the digit in the first decimal place of our new number: $0.2$.\n \n Next add 1 to the digit in the second decimal place of the second real number.\n That gives 6, so the digit in the second decimal place of our new number is $6$, giving $0.26$ so far.\n Keep going down the list, adding one to each successive decimal place in each successive number.\n Any time you encounter $9$, adding $1$ would give $10$, so replace the $9$ with $0$ instead.\n That happens for the third real number in the list, so our new number becomes $0.260$ at that point.\n \n The new number differs from the first real number in the list in the first decimal place, the second real number in the second decimal place and so on.\n Therefore, the new number is not in the list --- so the list can't contain all real numbers.\n We have to conclude that the real numbers were not and can not be paired up with the natural numbers.\n \n \\section*{Computable numbers}\n Cantor's argument has profound implications for what a computer can do.\n Each file on a computer is stored in binary -- a string over the alphabet $\\{ 0, 1 \\}$.\n The possible files are then $\\{ 0 , 1 \\}^* = \\{ \\epsilon, 0, 1, 00, 01, 10, 11, 000, \\ldots \\}$.\n We can pair $\\mathbb{N}_0$ up with $\\{ 0 , 1 \\}^*$ using that listing: $0 \\rightarrow \\epsilon$, $1 \\rightarrow 0$, $2 \\rightarrow 1$, $3 \\rightarrow 00$, and so on.\n \n Executables are just files.\n Consider an executable that prints a real number to the screen.\n For instance, we could write a program to print $0.0000\\ldots$ to the screen where it just keeps printing $0$'s.\n It's printing the real number $0$.\n For any given real number, is there a program to print it out?\n There must be real numbers for which no such program exists -- because there are less possible programs than real numbers.\n \n Computers can't even print all numbers out.\n \n\n \\bibliography{bibliography}\n\\end{document}\n", "meta": {"hexsha": "8adff9b7d259311a46d87b3b418b9ba780152eb7", "size": 5879, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "computability.tex", "max_stars_repo_name": "ianmcloughlin/latex-notes", "max_stars_repo_head_hexsha": "2ce8e4de828f7ff916d8e21d46ad610f62653acb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "computability.tex", "max_issues_repo_name": "ianmcloughlin/latex-notes", "max_issues_repo_head_hexsha": "2ce8e4de828f7ff916d8e21d46ad610f62653acb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "computability.tex", "max_forks_repo_name": "ianmcloughlin/latex-notes", "max_forks_repo_head_hexsha": "2ce8e4de828f7ff916d8e21d46ad610f62653acb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.6587301587, "max_line_length": 169, "alphanum_fraction": 0.612178942, "num_tokens": 1897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.782662489091802, "lm_q2_score": 0.8175744761936438, "lm_q1q2_score": 0.6398848745556435}} {"text": "\\documentclass{article}\n\n\\usepackage{mathrsfs,amsmath}\n\\usepackage{xcolor}\n\\usepackage{titlesec}\n\\usepackage{listings}\n\\usepackage{syntax}\n\\usepackage{pythonhighlighting}\n\\usepackage{fancyvrb}\n\n\\usepackage{graphicx}\n\n\\graphicspath{ {./assets/} }\n\n\\usepackage[margin=1.4in]{geometry}\n\n\\title{Homework \\#3 | Fall 2021} \n\\author{Jared Dyreson\\\\ \n California State University, Fullerton}\n\n\\DeclareRobustCommand{\\bowtie}{%\n \\mathrel\\triangleright\\joinrel\\mathrel\\triangleleft}\n\n\n\\usepackage [english]{babel}\n\\usepackage [autostyle, english = american]{csquotes}\n\\MakeOuterQuote{\"}\n\n\\titlespacing*{\\section}\n{0pt}{5.5ex plus 1ex minus .2ex}{4.3ex plus .2ex}\n\\titlespacing*{\\subsection}\n{0pt}{5.5ex plus 1ex minus .2ex}{4.3ex plus .2ex}\n\n\\usepackage{hyperref}\n\\hypersetup{\n colorlinks,\n citecolor=black,\n filecolor=black,\n linkcolor=black,\n urlcolor=black\n}\n\n\\begin{document}\n\n\\maketitle\n\\tableofcontents\n\n\\newpage\n\n\n\\section{Questions}\n\n\\subsection{Offline Ski Rental}\n\n\\subsubsection{Problem Description}\n\n\\begin{table}[!h]\n\\begin{tabular}{|l|l|}\n\\hline\n\\textbf{input} & a daily ski rental price $r > 0$, purchase price $p > 0$, and number of days $d > 0$ \\\\ \\hline\n\\textbf{output} & \\begin{tabular}[c]{@{}l@{}}True if it is cheaper to rent skis for d days at r dollars per day, or False if it is\\\\ cheaper to buy skis for p dollars\\end{tabular} \\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\subsubsection{Pseudocode}\n\n\\begin{verbatim}\nfunction ski(int rate, int purchasePrice, int days):\n rental_total = rate * days\n return rental_total < purchasePrice\n\\end{verbatim}\n\n\\subsubsection{Time Complexity}\n\nThis function is quite simple, as we are given all the information up front.\nEach operation we have will complete in a time complexity of $O(1)$.\nTherefore, the best/worst case of this is $O(1)$.\n\n\\subsubsection{Implementation}\n\n\\begin{python}\n\ndef solution(rental_price: int,\n purchase_price: int, days: int) -> bool:\n\n return (rental_price * days) < purchase_price\n\\end{python}\n\n\\newpage\n\n\\subsection{List Reversal}\n\n\\subsubsection{Problem Description}\n\n\\begin{table}[!h]\n\\begin{tabular}{|l|l|}\n\\hline\n\\textbf{input} & a list L of n elements \\\\ \\hline\n\\textbf{output} & a list containing the elements of L but in reversed order \\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\subsubsection{Pseudocode}\n\n\\begin{verbatim}\nfunction reverse(container):\n last_index = container.length # O(1)\n for element in container: # O(n)\n swap(element, container[last_index]) # O(1)\n last_index = last_index - 1 # O(1)\n return container # O(1)\n\\end{verbatim}\n\n\\subsubsection{Time Complexity}\n\nThis function will modify the container in place and does not create a copy.\nOther variants can leave the original container intact and return a reversed copy.\nThe benefit of this would be data integrity, however, that is at the expense of having to have duplicate data.\nWe need to iterate over the entire container of \"n\" elements and we assume the swap function is of constant time.\nTherefore, time complexity of this function is linear, or $O(n)$.\n\\subsubsection{Implementation}\n\n\\begin{python}\ndef solution(container: list) -> list:\n end = len(container) - 1\n for x, _ in enumerate(container):\n if x < end:\n container[x], container[end] = container[end], container[x]\n end -= 1\n return container\n\\end{python}\n\n\\newpage\n\n\\subsection{Pythagorean Triple}\n\n\\subsubsection{Problem Description}\n\\begin{table}[!h]\n\\begin{tabular}{|l|l|}\n\\hline\n\\textbf{input} & two positive integers; $a$ and $b$ with $a < b$ \\\\ \\hline\n\\textbf{output} & \\begin{tabular}[c]{@{}l@{}}a Pythagorean triple (x, y, z), such that $x$, $y$, and $z$ are positive integers,\\\\ $a \\le x \\le y \\le z \\le b$ and $x^{2} + y^{2} = z^{2}$, or None if no such triples exist\\end{tabular}\\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\subsubsection{Pseudocode}\n\n\\begin{verbatim}\nfunction validator(tuple[int, int, int] triplet):\n x, y, z = triplet # O(3), assignment and unpacking for three elements\n return z == square_root(x ** 2 + y ** 2) and x <= y <= z\n # O(n) <- for the squaring of x and y. This is building off the previous\n # assumption made in Project 2 for the Fibonacci sequence\n\nfunction pythagorean_triples(int a /* lower bound */, int b /* upper bound */):\n candidates = [] # O(1)\n for x in range(a, b + 1): # get range of values, bounded by a and b # O(n)\n candidates.append(x) # O(1)\n # create all permutations\n # description does not specifically instruct to create an algorithm\n # to create these permutations, therefore we use the best case\n # of O(n^r), where r represents the length of the tuple -> O(n^3)\n permutations = generate_permutations(iterable=candidates, r=3) # O(n^3)\n # this is the function signature of itertools.combinations\n valid_triplets = [] # O(1)\n\n # this for loop block runs in O(n^2)\n for permuation in permutations: # O(n)\n if(validator(permuation)): # O(n)\n valid_triplets.append(permutation) # O(1)\n return valid_triplets # O(1)\n\\end{verbatim}\n\n\\subsubsection{Time Complexity}\n\nThe main portion of this algorithm comes solely from creating all possible combinations the range presents.\nHere, we are tasked with creating triplets, given a range of numbers $a$ and $b$.\nThe creation of all possible combinations will take $O(n^{r})$, where $r$ is the length of the container you're producing.\nTherefore in our case it is $O(n^{3})$ and more information regarding this can be found in \\href{https://stackoverflow.com/a/20765011}{\\underline{this StackOverflow comment thread}}.\nAll of these triplets need to then be processed by our validator function, which was defined in the problem description.\nThese validations will occur in $O(n)$ time, because we are assuming raising a number to a given power will take that time complexity.\nIt is disputed in implementations in Python, either $O(1)$ or $O(n)$ and since a previous homework assignment uses the latter, we will go with this assumption.\nOverall, the time complexity will be of $O(n^{3})$, as it the leading term in the polynomial $O(n^{3}) + O(n^{2})$.\n\n\n\\newpage\n\n\\subsubsection{Implementation}\n\n\\begin{python}\nimport itertools\nimport math\n\ndef validator(container: tuple[int, int, int]) -> bool:\n x, y, z = container # raise ValueError if container len != 3\n # x ^ 2 + y ^ 2 = z ^ 2\n # where x, y and z are perfect squares\n return z == math.sqrt(x ** 2 + y ** 2) and x <= y <= z\n\ndef solution(integers: tuple[int, int]) -> list[tuple[int, int, int]]:\n a, b = integers # a is lower bound and b is upper bound\n if not(isinstance(a, int)\n and isinstance(b, int)\n and a < b):\n raise ValueError(\n f'[ERROR] Expected integers for `a` ({a}) and `b` ({b}), where a < b')\n candidates = [_ for _ in range(a, b+1)]\n \"\"\"\n Given a = 1 and b = 4\n - (1, 2, 3)\n - (2, 3, 4)\n are all valid combinations\n \"\"\"\n\n return [candidate for candidate in\n itertools.combinations(candidates, 3) if validator(candidate)]\n\\end{python}\n\n\n\\newpage\n\n\\subsection{Missing Integer}\n\n\\subsubsection{Problem Description}\n\n\\begin{table}[!h]\n\\begin{tabular}{|l|l|}\n\\hline\n\\textbf{input} & \\begin{tabular}[c]{@{}l@{}}an array $A$ of size $N-1$ that contains all integers in the range of $1$ to $N$,\\\\ except one missing integer. This list contains no duplicates.\\end{tabular} \\\\ \\hline\n\\textbf{output} & the missing number from the list \\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\subsubsection{Pseudocode}\n\n\\begin{verbatim}\nfunction missing_element(container: list[int]):\n N = len(container) + 1\n intended_sum = N * (N + 1) / 2 # Gaussian Sum\n return intended_sum - sum(container) \n # the missing element is the difference between each container\n\\end{verbatim}\n\n\n\\subsubsection{Time Complexity}\n\nThis function is quite efficient in finding the missing integer, as it only needs to loop through the container once.\nAlso, the algorithm does not require the list to be in sorted order and can be extended to tuples, sets and other Python data containers.\nFirst, we need to compute the Gaussian Sum of the container as if it was intact.\nThe formula for this is:\n$$\\frac{N \\times (N + 1)}{2}$$\nand assumes the sequence is monotonically increasing, and contains all integers from $1$ to $N$ inclusive.\nHere, we have access to a key characteristic of the container without having to reconstruct it.\nThen, we translate the current container in terms of an integer and we apply the sum function $O(n)$.\nSubtracting the difference between the two will give you the missing integer.\nThis will run in $O(n)$ time, as the only linear aspect is to find the sum of the current container.\n\\subsubsection{Implementation}\n\n\\begin{python}\ndef missing_integer(container: typing.List[int], dialation: int = 1) -> int:\n \"\"\"\n Find the missing integer in a monotonically increasing sequence\n \"\"\"\n\n intended_sum = gaussian_sum(len(container) + 1)\n\n return intended_sum - sum(container)\n\n\ndef gaussian_sum(N: int, dialation: int = 1) -> int:\n \"\"\"\n This is the sum of a contiguous sequence of integers\n Dialation is applied when a constant is applied to all elements in the sequence\n \"\"\"\n\n return int(N * (N + 1) / 2) * dialation\n\\end{python}\n\n\n\\newpage\n\n\\section{Extra Credit}\n\n\\subsection{Street Map}\n\n\\begin{figure}[!h]\n\\centering\n\\includegraphics[width=10cm]{StreetMap}\n\\caption{Street View of Troy High School backroads}\n\\end{figure}\n\nEach node can contain the following information, and applying this to node \\#1 we get:\n\n\\begin{itemize}\n\\item Lateral intersection | Dorthy Lane (East/West)\n\\item Medial intersection | Acacia Avenue (South/North)\n\\item Longitude | SomeValue\n\\item Latitude | SomeValue\n\\item Dead end | False\n\\end{itemize}\n\nEach edge on the graph represents the street connecting each intersection and each edge is bidirectional (forgot to make it so in the diagram).\n\n\\newpage\n\n\\subsection{Course Requirements}\n\n\\begin{figure}[!h]\n\\centering\n\\includegraphics[width=10cm]{CourseRequirements}\n\\caption{Example set of courses to be taken}\n\\end{figure}\n\nEach class should be taken in sequential order and must have it's prerequisites met before.\nThis is called a topological graph and is applied in different areas such as package managers and neuroscience.\nAs shown here in this example, some sections or \"forests\" can exist independently of one another.\nThese do not have interlaced dependencies and can be completed in their own continuum.\nOthers are connected on a basis where the course must be taken in conjunction with the another class.\nThis can be seen in CS 121 and Math 150A, however this can also be applied to lab courses and their associated lecture (PHYS 225 and PHYS 225L).\n\n\\newpage\n\n\\subsection{Social Media Network}\n\n\\begin{figure}[!h]\n\\centering\n\\includegraphics[width=13cm]{SocialMediaNetwork}\n\\caption{Example Social Media Network}\n\\end{figure}\n\nGiven the list of people and their associated friends, we can construct a graph showing their relationships.\n\\begin{itemize}\n\\item Sussy $\\rightarrow$ Alex, Jo, Albert\n\\item Alex $\\rightarrow$ Jacob, Sussy, Jon\n\\item Bobby $\\rightarrow$ Nick, Nicole, Sheldon\n\\end{itemize}\n\nSingly directed edges can be read as \"Person 1 is following Person 2 but Person 2 is not following Person 1\".\nBidirectional edges are when both parties are following each other.\nEach \"forest\" is a collection of nodes where they are all connected in some fashion.\nThe separation occurs when there are not mutual connections between every element in the network\n\n\\end{document}\n", "meta": {"hexsha": "a7630f19393efe88168ce2f7dfae053e462c6566", "size": 11591, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "3/Homework_3.tex", "max_stars_repo_name": "JaredsAlgorithms/Assignments", "max_stars_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "3/Homework_3.tex", "max_issues_repo_name": "JaredsAlgorithms/Assignments", "max_issues_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "3/Homework_3.tex", "max_forks_repo_name": "JaredsAlgorithms/Assignments", "max_forks_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.6947674419, "max_line_length": 241, "alphanum_fraction": 0.7186610301, "num_tokens": 3110, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355187, "lm_q2_score": 0.8289388125473629, "lm_q1q2_score": 0.6398110412720355}} {"text": "\\documentclass[]{article}\n\n% Define page\n\\usepackage[top=1in, bottom=1in, left=1in, right=1in]{geometry}\n\n% Include packages for text modification\n\\usepackage{amsmath,amsfonts,amssymb}\n\\usepackage{bm}\n\n% Include packages for citing\n\\usepackage[noadjust]{cite}\n\n% Include packages for graphics\n\\usepackage{graphicx,psfrag}\n\\usepackage{subfigure}\n\n% Include packages for tables\n\\usepackage{array,multirow}\n\\usepackage{ctable}\n\n% Define new commands\n\\newcommand{\\etal}{\\emph{et al.}}\n\\newcommand{\\vT}{\\bm{T}}\n\\newcommand{\\vmu}{\\bm{\\mu}}\n\\newcommand{\\vTmu}{\\bm{T_{\\mu}}}\n\\newcommand{\\vTx}[1][]{\\bm{T}_{#1}(\\bm{x})}\n\\newcommand{\\vTy}[1][]{\\bm{T}_{#1}(\\bm{y})}\n\\newcommand{\\vx}[1][]{\\bm{x}_{#1}}\n\\newcommand{\\vxt}[1][]{\\bm{\\widetilde x}_{#1}}\n\\newcommand{\\vy}[1][]{\\bm{y}_{#1}}\n\\newcommand{\\D}[2]{\\frac{\\partial #1}{\\partial #2}}\n\\newcommand{\\Dd}[3]{\\frac{\\partial^2 #1}{\\partial #2 \\partial #3}}\n\\newcommand{\\elastix}{\\texttt{elastix}}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{document}\n\n\\title{Spatial derivatives and penalty terms in ITK and \\elastix}\n\n\\author{Marius Staring and Stefan Klein}\n\\date{}\n\\maketitle\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% main text\n\n\\section{Introduction}\n\nImage registration is the process of aligning images, and can be\ndefined as an optimisation problem:\n\\begin{align}\n\\hat \\vmu &= \\arg \\min_{\\vmu} \\mathcal{C}(I_F, I_M;\n\\vmu),\\label{eq:reg}\n\\end{align}\nwith $I_F$ and $I_M$ the $d$-dimensional fixed and moving image,\nrespectively, and $\\vmu$ the vector of parameters of size $N$ that\nmodel the transformation $\\vT$. The cost function $\\mathcal{C}$\nconsists of a similarity measure $\\mathcal{S}(I_F, I_M; \\vmu)$ that\ndefines the quality of alignment. Examples are the mean square\ndifference, normalised correlation, and mutual information measure.\nIn order to regularise the transformation $\\vT_{\\vmu}$ often a\npenalty term $\\mathcal{P}(\\vmu)$ is added to the cost function, so\nit becomes:\n\\begin{align}\n\\mathcal{C} &= \\alpha_1 \\mathcal{S} + \\alpha_2 \\mathcal{P},\n\\end{align}\nwhere $\\alpha_1, \\alpha_2$ user-defined constants that weigh\nsimilarity against regularity.\n\nPenalty term are often based on the first or second order spatial\nderivatives of the transformation. For example the bending energy of\nthe transformation, which is arguably the most common penalty term,\nis defined in 2D as:\n\\begin{align}\n\\mathcal{P}_{\\mathrm{BE}}(\\vmu) &= \\frac{1}{P} \\sum_{\\vxt[i]}\n\\left\\| \\frac{\\partial^2 \\vT}{\\partial \\vx \\partial \\vx^T}(\\vxt[i])\n\\right\\|_F^2 \\\\\n&= \\frac{1}{P} \\sum_{\\vxt[i]} \\sum_{j = 1}^2 \\left( \\frac{\\partial^2\nT_j}{\\partial x_1^2}(\\vxt[i]) \\right)^2 + 2 \\left( \\frac{\\partial^2\nT_j}{\\partial x_1 \\partial x_2}(\\vxt[i]) \\right)^2 + \\left(\n\\frac{\\partial^2 T_j}{\\partial x_2^2}(\\vxt[i]) \\right)^2,\n\\end{align}\nwhere $P$ is the number of points $\\vxt[i]$, and the tilde denotes\nthe difference between a variable and a given point over which a\nterm is evaluated.\n\nThe optimisation problem (\\ref{eq:reg}) is frequently solved using\nan iterative gradient descent routine:\n\\begin{align}\n\\vmu_{k+1} &= \\vmu_k - a_k \\frac{\\partial \\mathcal{C}}{\\partial\n\\vmu} = \\vmu_k - a_k \\left( \\alpha_1 \\frac{\\partial\n\\mathcal{S}}{\\partial \\vmu} + \\alpha_2 \\frac{\\partial\n\\mathcal{P}}{\\partial \\vmu} \\right),\\label{eq:opt}\n\\end{align}\nwith $a_k$ a user-defined declining function that defines the step\nsize.\n\nThe derivative of the similarity measure usually involves\ncomputation of the spatial derivative of the moving image:\n$\\D{I_M}{\\vx}$, and the derivative of the transformation to its\nparameters: $\\D{\\vT}{\\vmu}$. In the ITK the last derivative is\nimplemented using $\\texttt{transform->GetJacobian()}$, i.e. the\nderivative to the transformation parameters $\\vmu$ is referred to as\n`Jacobian'.\n\nPenalty terms usually consist of the first and second order\n\\emph{spatial} derivatives of the transformation, i.e.\n$\\D{\\vT}{\\vx}$ and $\\Dd{\\vT}{\\vx}{\\vx^T}$. We will refer to these\nderivatives as the `SpatialJacobian' and the `SpatialHessian' to\nclearly distinguish between these derivatives and the `Jacobian'. In\norder to apply the gradient descent optimisation routine\n(\\ref{eq:opt}), we additionally need the derivatives $\\D{}{\\vmu}\n\\D{\\vT}{\\vx}$ and $\\D{}{\\vmu} \\Dd{\\vT}{\\vx}{\\vx^T}$. These we call\nthe `JacobianOfSpatialJacobian' and `JacobianOfSpatialHessian',\nrespectively. See Table \\ref{tab:notation} for details.\n\n\\begin{table}[h]\n\\centering\n\\begin{tabular}{llll}\n\\toprule \\toprule\nName & definition & matrix size & written out in 2D \\\\\n\\midrule Transformation & $\\vT = \\vT_{\\vmu}(\\widetilde \\vx)$ & $d\n\\times 1$ & $\\begin{bmatrix} T_1(\\vx) \\\\ T_2(\\vx) \\end{bmatrix}$ \\\\[3ex]\nJacobian & $\\frac{\\partial \\vT_{\\vmu}}{\\partial \\vmu}(\\widetilde\n\\vx)$ & $d \\times N$ & $\\begin{bmatrix}\n \\frac{\\partial T_1}{\\partial \\mu_1}(\\widetilde \\vx) & \\cdots & \\frac{\\partial T_1}{\\partial \\mu_N}(\\widetilde \\vx) \\\\\n \\frac{\\partial T_2}{\\partial \\mu_1}(\\widetilde \\vx) & \\cdots & \\frac{\\partial T_2}{\\partial \\mu_N}(\\widetilde \\vx) \\\\\n\\end{bmatrix}$ \\\\[3ex]\nSpatialJacobian & $\\frac{\\partial \\vT_{\\vmu}}{\\partial\n\\vx}(\\widetilde \\vx)$ & $d \\times d$ & $\\begin{bmatrix}\n \\frac{\\partial T_1}{\\partial x_1}(\\widetilde \\vx) \\frac{\\partial T_1}{\\partial x_2}(\\widetilde \\vx) \\\\\n \\frac{\\partial T_2}{\\partial x_1}(\\widetilde \\vx) \\frac{\\partial T_2}{\\partial x_2}(\\widetilde \\vx) \\\\\n\\end{bmatrix}$ \\\\[3ex]\nJacobianOfSpatialJacobian & $\\frac{\\partial}{\\partial \\vmu}\n\\frac{\\partial \\vT_{\\vmu}}{\\partial \\vx}(\\widetilde \\vx)$ & $d\n\\times d \\times N$ & $\\begin{bmatrix}\n \\frac{\\partial}{\\partial \\mu_1} \\frac{\\partial \\vT_{\\vmu}}{\\partial\n\\vx}(\\widetilde \\vx) & \\cdots & \\frac{\\partial}{\\partial \\mu_N}\n\\frac{\\partial \\vT_{\\vmu}}{\\partial \\vx}(\\widetilde \\vx)\n\\end{bmatrix}$ \\\\[3ex]\nSpatialHessian & $\\frac{\\partial^2 \\vT_{\\vmu}}{\\partial \\vx\n\\partial \\vx^T}$ & $d \\times d \\times d$ & $\\left\\{ \\begin{bmatrix}\n \\frac{\\partial^2 T_1}{\\partial x_1 \\partial x_1}(\\widetilde \\vx) \\frac{\\partial^2 T_1}{\\partial x_2 \\partial x_1}(\\widetilde \\vx) \\\\\n \\frac{\\partial^2 T_1}{\\partial x_2 \\partial x_1}(\\widetilde \\vx) \\frac{\\partial^2 T_1}{\\partial x_2 \\partial x_2}(\\widetilde \\vx) \\\\\n\\end{bmatrix}, \\begin{bmatrix}\n \\frac{\\partial^2 T_2}{\\partial x_1 \\partial x_1}(\\widetilde \\vx) \\frac{\\partial^2 T_2}{\\partial x_2 \\partial x_1}(\\widetilde \\vx) \\\\\n \\frac{\\partial^2 T_2}{\\partial x_2 \\partial x_1}(\\widetilde \\vx) \\frac{\\partial^2 T_2}{\\partial x_2 \\partial x_2}(\\widetilde \\vx) \\\\\n\\end{bmatrix} \\right\\}$ \\\\[3ex]\nJacobianOfSpatialHessian & $\\frac{\\partial}{\\partial \\vmu}\n\\frac{\\partial^2 \\vT_{\\vmu}}{\\partial \\vx \\partial \\vx^T}$ & $d\n\\times d \\times d \\times N$ & $\\begin{bmatrix}\n \\frac{\\partial}{\\partial \\mu_1} \\frac{\\partial^2 \\vT_{\\vmu}}{\\partial \\vx\n\\partial \\vx^T} & \\cdots & \\frac{\\partial}{\\partial \\mu_N} \\frac{\\partial^2 \\vT_{\\vmu}}{\\partial \\vx\n\\partial \\vx^T}\n\\end{bmatrix}$ \\\\\n\\bottomrule \\bottomrule\n\\end{tabular}\n\\caption{Naming conventions and definitions for the transformation\nand its derivatives used in this paper.}\\label{tab:notation}\n\\end{table}\n\n\\section{Support for penalty terms in the ITK}\n\nThe derivative of the penalty term is not supported in the ITK. We\npropose to add the following functions in the\n\\texttt{itk::Transform} class:\n\\begin{quote}\n\\begin{verbatim}\nvirtual void GetSpatialJacobian(\n const InputPointType &,\n SpatialJacobianType & ) const;\n\nvirtual void GetSpatialHessian(\n const InputPointType &,\n SpatialHessianType & ) const;\n\nvirtual void GetJacobianOfSpatialJacobian(\n const InputPointType &,\n JacobianOfSpatialJacobianType &,\n NonZeroJacobianIndicesType & ) const;\n\nvirtual void GetJacobianOfSpatialHessian(\n const InputPointType &,\n JacobianOfSpatialHessianType &,\n NonZeroJacobianIndicesType & ) const;\n\\end{verbatim}\n\\end{quote}\nand additionally a function to implement a sparse version of the\nJacobian:\n\\begin{quote}\n\\begin{verbatim}\nvirtual void GetJacobian(\n const InputPointType &,\n JacobianType &,\n NonZeroJacobianIndicesType & ) const;\n\\end{verbatim}\n\\end{quote}\n\nThe ITK structures that were used to store the data are given in\nTable \\ref{tab:datastructures}. The Jacobian is of size $d \\times\nN$, and since the number of transformation parameters is flexible\nfor some transformations, the data structure used for storing the\nJacobian is an \\texttt{itk::Array2D} object, which inherits from the\n\\texttt{vnl\\_matrix}. This was already chosen previously in the ITK.\nThe SpatialJacobian is of fixed size $d \\times d$, and therefore we\nchoose to use the \\texttt{itk::Matrix} to store the SpatialJacobian,\nwhich inherits from the \\texttt{vnl\\_matrix\\_fixed}. For derivatives\nto $\\vmu$ we choose to use the \\texttt{std::vector}. The\nSpatialHessian gives us some problems, since we really need a 3D\nmatrix, but currently no such thing exists in the ITK. Therefore, we\nopt for an \\texttt{itk::FixedArray} of \\texttt{itk::Matrix}.\n\n\\begin{table}[h]\n\\centering\n\\begin{tabular}{ll}\n\\toprule \\toprule\nName & ITK structure \\\\\n\\midrule Transformation & \\\\\nJacobian & \\texttt{Array2D = vnl\\_matrix} \\\\\nSpatialJacobian & \\texttt{Matrix = vnl\\_matrix\\_fixed} \\\\\nJacobianOfSpatialJacobian & \\texttt{std::vector< Matrix >} \\\\\nSpatialHessian & \\texttt{FixedArray< Matrix >}\\footnotemark \\\\\nJacobianOfSpatialHessian & \\texttt{std::vector< FixedArray< Matrix > >} \\\\\nNonZeroJacobianIndices & \\texttt{std::vector< unsigned long >} \\\\\n\\bottomrule \\bottomrule\n\\end{tabular}\n\\caption{The ITK structures that store the\ndata.}\\label{tab:datastructures}\n\\end{table}\n\\footnotetext{by lack of a good 3D matrix structure}\n\nFrom the function definitions above, notice that we chose to pass\nthe variables by reference, and as function arguments. The\n\\texttt{GetJacobian} in the \\texttt{itk::Transform} is defined as:\n\\begin{quote}\n\\begin{verbatim}\nvirtual const JacobianType & GetJacobian( const InputPointType & ) const;\n\\end{verbatim}\n\\end{quote}\nwhich returns a reference to the member variable\n\\texttt{m\\_Jacobian}. It is possible, however, that this member is\nonly valid for the input point, namely for transformations with a\nderivative dependent of the spatial position. Therefore, we think\nthat it should not be possible to access this parameter at a later\ntime, when the input point has possibly changed.\n\nAlso notice the \\texttt{NonZeroJacobianIndicesType} in the function\ndefinitions. These are meant for the support of sparse Jacobians,\nJacobianOfSpatialJacobians, etc.\n\n\\section{Affine transformation}\n\nFor the affine transformation, the derivatives evaluate to the\nfollowing in 2D:\n\\begin{align}\n\\vTmu(\\vxt) &= A \\vx + \\bm{t} = \\begin{bmatrix} a_{11} & a_{12}\n\\\\ a_{21} & a_{22} \\end{bmatrix} \\begin{bmatrix} \\widetilde x_1 \\\\\n\\widetilde x_2 \\end{bmatrix} + \\begin{bmatrix} t_1 \\\\ t_2\n\\end{bmatrix} = \\begin{bmatrix} \\mu_0 & \\mu_1\n\\\\ \\mu_2 & \\mu_3 \\end{bmatrix} \\begin{bmatrix} \\widetilde x_1 \\\\\n\\widetilde x_2 \\end{bmatrix} + \\begin{bmatrix} \\mu_4 \\\\ \\mu_5\n\\end{bmatrix}, \\\\\n\\D{\\vT}{\\vmu}(\\vxt) &= \\begin{bmatrix} \\widetilde x_1 & \\widetilde\nx_2 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & \\widetilde x_1 & \\widetilde x_2 & 0 &\n1 \\end{bmatrix}, \\\\\n\\D{\\vT}{\\vx}(\\vxt) &= \\begin{bmatrix} \\mu_0 & \\mu_1 \\\\ \\mu_2 & \\mu_3 \\end{bmatrix}, \\\\\n\\Dd{\\vT}{\\vx}{\\vx^T}(\\vxt) &= O_{d \\times d \\times d}, \\\\\n\\D{}{\\vmu} \\D{\\vT}{\\vx}(\\vxt) &= \\left\\{ \\begin{bmatrix} 1 & 0 \\\\\n0 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 1 \\\\ 0 & 0\n\\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 1 & 0\n\\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1\n\\end{bmatrix} \\right\\}, \\\\\n\\D{}{\\vmu} \\Dd{\\vT}{\\vx}{\\vx^T}(\\vxt) &= O_{d \\times d \\times d\n\\times N},\n\\end{align}\nwhere $O_s$ is a zero matrix of size $s$.\n\n%centre of rotation!!!\n\nThe \\texttt{GetJacobianOfSpatialJacobian()} returns\n\\texttt{nonZeroJacobianIndices} = $[ 0, 1, 2, 3 ]$.\n\n%For the SpatialHessian and the JacobianOfSpatialHessian we choose to\n%not return matrices filled with zeros, but instead zero sized\n%matrices. The implementation of the penalty term is assumed to check\n%for the size. In case of an affine transform the size is zero, and\n%the penalty term can simply return zero. This is a performance\n%benefit compared to walking over the zero matrix, and adding and\n%multiplying everything, which in the end also gives zero.\n\nThese derivatives are implemented in the\n\\texttt{itk::AdvancedMatrixOffsetTransformBase} class.\n\n\\section{B-spline transformation}\n\nA transformation parameterised by third order B-splines can be\nwritten in 2D as follows:\n\\begin{align}\n\\vTmu(\\vxt) &= \\begin{bmatrix} T_1(\\vxt;\\vmu) \\\\ T_2(\\vxt;\\vmu)\n\\end{bmatrix} = \\begin{bmatrix} \\widetilde{x}_1 \\\\ \\widetilde{x}_2 \\end{bmatrix}\n + \\begin{bmatrix} \\sum_{i} \\mu_i \\beta^3 \\left(\n\\frac{ \\widetilde{x}_1 - x_{1}^i}{\\sigma_1} \\right)\n\\beta^3 \\left( \\frac{\\widetilde{x}_2 - x_{2}^i}{\\sigma_2}\\right) \\\\\n \\sum_{i} \\mu_{i+16} \\beta^3 \\left( \\frac{ \\widetilde{x}_1 - x_{1}^i}\n {\\sigma_1} \\right) \\beta^3 \\left( \\frac{\\widetilde{x}_2 - x_{2}^i}{\\sigma_2} \\right)\n\\end{bmatrix},\n\\end{align}\nwith $\\vx^i$ the control points within the support of the B-spline\nbasis functions $\\beta^3(\\cdot)$, and $\\sigma$ the B-spline grid\nspacing. Here, we labelled the $x$-direction of the control points\nwith $\\mu_0, \\ldots, \\mu_{15}$ and the $y$-direction with $\\mu_{16},\n\\ldots, \\mu_{31}$. (Since the support of a cubic B-spline is $4^d =\n16$.)\n\nFor short notation, define:\n\\begin{align}\nb_{33}^{i} &= \\beta^3\\left( (\\widetilde{x}_1 - x_{1}^i) / \\sigma_1\n\\right) \\cdot \\beta^3 \\left( (\\widetilde{x}_2 - x_{2}^i) / \\sigma_2\n\\right), \\\\\nb_{23}^{i} &= \\left[ \\beta^2 \\left( (\\widetilde{x}_1 -\nx_{1}^i)/\\sigma_1 + \\tfrac{1}{2} \\right) - \\beta^2 \\left(\n(\\widetilde{x}_1 - x_{1}^i)/\\sigma_1 - \\tfrac{1}{2} \\right) \\right]\n\\cdot \\beta^3 \\left( (\\widetilde{x}_2 - x_{2}^i)/\\sigma_2 \\right) /\n\\sigma_1, \\\\\nb_{32}^{i} &= \\beta^3 \\left( (\\widetilde{x}_1 - x_{1}^i)/\\sigma_1\n\\right) \\cdot \\left[ \\beta^2 \\left( (\\widetilde{x}_2 -\nx_{2}^i)/\\sigma_2 + \\tfrac{1}{2} \\right) - \\beta^2 \\left( (\n\\widetilde{x}_2 - x_{2}^i)/\\sigma_2 - \\tfrac{1}{2} \\right) \\right] /\n\\sigma_2, \\\\\n\\begin{split}\nb_{22}^{i} &= \\left[ \\beta^2 \\left( ( \\widetilde{x}_1 -\nx_{1}^i)/\\sigma_1 + \\tfrac{1}{2} \\right) - \\beta^2 \\left( (\n\\widetilde{x}_1 - x_{1}^i)/\\sigma_1 - \\tfrac{1}{2} \\right) \\right] \\\\\n& \\qquad \\qquad \\cdot \\left[ \\beta^2 \\left( ( \\widetilde{x}_2 -\nx_{2}^i)/\\sigma_2 + \\tfrac{1}{2} \\right) - \\beta^2 \\left( (\n\\widetilde{x}_2 - x_{2}^i)/\\sigma_2 - \\tfrac{1}{2} \\right) \\right] /\n(\\sigma_1 \\sigma_2),\n\\end{split} \\\\\nb_{13}^{i} &= \\left[ \\beta^1 \\left( (\\widetilde{x}_1 -\nx_{1}^i)/\\sigma_1 + 1 \\right) - 2 \\beta^1 \\left( ( \\widetilde{x}_1 -\nx_{1}^i)/\\sigma_1 \\right) + \\beta^1 \\left( (\\widetilde{x}_1 -\nx_{1}^i)/\\sigma_1 - 1 \\right) \\right] \\cdot \\beta^3 \\left(\n(\\widetilde{x}_2 - x_{2}^i)/\\sigma_2 \\right) / \\sigma_1^2, \\\\\nb_{31}^{i} &= \\beta^3 \\left( (\\widetilde{x}_1 - x_{1}^i)/\\sigma_1\n\\right) \\cdot \\left[ \\beta^1 \\left( (\\widetilde{x}_2 -\nx_{2}^i)/\\sigma_2 + 1 \\right) - 2 \\beta^1 \\left( ( \\widetilde{x}_2 -\nx_{2}^i)/\\sigma_2 \\right) + \\beta^1 \\left( (\\widetilde{x}_2 -\nx_{2}^i)/\\sigma_2 - 1 \\right) \\right] / \\sigma_2^2.\n\\end{align}\nFrom these equations we derive:\n\\begin{align}\n\\D{\\vT}{\\vmu}(\\vxt) &= \\begin{bmatrix} b_{33}^{0} & \\cdots &\nb_{33}^{15} & 0 & \\cdots & 0 \\\\ 0 & \\cdots & 0 & b_{33}^{0} &\n\\cdots & b_{33}^{15} \\end{bmatrix}, \\\\\n\\D{\\vT}{\\vx}(\\vxt) &= \\begin{bmatrix} 1 + \\sum_i \\mu_i b_{23}^i &\n\\sum_i \\mu_i b_{32}^i \\\\ \\sum_i \\mu_{i+16} b_{23}^{i} & 1 + \\sum_i\n\\mu_{i+16} b_{32}^{i} \\end{bmatrix}, \\\\\n\\Dd{\\vT}{\\vx}{\\vx^T}(\\vxt) &= \\left\\{ \\begin{bmatrix} \\sum_i \\mu_i\nb_{13}^i & \\sum_i \\mu_i b_{22}^i \\\\ \\sum_i \\mu_{i} b_{22}^{i} &\n\\sum_i \\mu_{i} b_{31}^{i} \\end{bmatrix}, \\begin{bmatrix} \\sum_i\n\\mu_{i+16} b_{13}^{i} & \\sum_i \\mu_{i+16} b_{22}^{i} \\\\ \\sum_i\n\\mu_{i+16} b_{22}^{i} & \\sum_i \\mu_{i+16} b_{31}^{i}\n\\end{bmatrix} \\right\\} \\\\\n\\D{}{\\vmu} \\D{\\vT}{\\vx}(\\vxt) &= \\left\\{ \\begin{bmatrix} b_{23}^{0}\n& b_{32}^{0} \\\\ 0 & 0 \\end{bmatrix}, \\cdots, \\begin{bmatrix}\nb_{23}^{15} & b_{32}^{15} \\\\ 0 & 0 \\end{bmatrix}, \\begin{bmatrix} 0\n& 0 \\\\ b_{23}^{0} & b_{32}^{0} \\end{bmatrix}, \\cdots,\n\\begin{bmatrix} 0 & 0 \\\\ b_{23}^{15} & b_{32}^{15} \\end{bmatrix} \\right\\}\n\\\\\n\\D{}{\\vmu} \\Dd{\\vT}{\\vx}{\\vx^T}(\\vxt) &= \\left\\{ \\left\\{\n\\begin{bmatrix} b_{13}^0 & b_{22}^0 \\\\ b_{22}^0 & b_{31}^0 \\end{bmatrix},\nO_{d \\times d} \\right\\}, \\cdots, \\left\\{ \\begin{bmatrix} b_{13}^{15}\n& b_{22}^{15} \\\\ b_{22}^{15} & b_{31}^{15} \\end{bmatrix}, O_{d\n\\times d} \\right\\}, \\right. \\\\\n& \\qquad \\left. \\left\\{ O_{d \\times d},\n\\begin{bmatrix} b_{13}^{0} & b_{22}^{0} \\\\ b_{22}^{0} &\nb_{31}^{0} \\end{bmatrix} \\right\\}, \\cdots, \\left\\{ O_{d \\times d},\n\\begin{bmatrix} b_{13}^{15} & b_{22}^{15} \\\\ b_{22}^{15} &\nb_{31}^{15} \\end{bmatrix} \\right\\} \\right\\}.\n\\end{align}\n\nThese derivatives are implemented in the\n\\texttt{itk::AdvancedBSplineDeformableTransform} class.\n\n\\section{Combining transformations}\n\n\\texttt{elastix} supports combining multiple transformations by\naddition or composition. Adding transformations is done via:\n\\begin{align}\n\\vTx &= \\vTx[0] + \\vTx[1] - \\vx,\n\\end{align}\nwhere $\\vTx[0]$ is the initial transformation and $\\vTx[1]$ the\ncurrent transformation. Only the current transformation is optimised\nduring the registration (as a choice). Composition of\ntransformations is defined via:\n\\begin{align}\n\\vTx &= \\vT_1(\\vTx[0]).\n\\end{align}\n\nFor these combined transformations we need to derive the relations\nfor the derivatives. Define $\\bm{y} = \\vT_0(\\vxt)$.\n\n\\begin{align}\n\\intertext{Jacobian:}\n\\D{\\vTmu}{\\vmu}(\\vxt) &= \\D{}{\\vmu} \\left( \\vT_0(\\vxt) + \\vT_1(\\vxt)\n- \\vxt \\right) = \\D{}{\\vmu} \\vT_1(\\vxt), \\\\\n\\D{\\vTmu}{\\vmu}(\\vxt) &= \\D{}{\\vmu} \\left( \\vT_1(\\vT_0(\\vxt))\n\\right) = \\D{}{\\vmu} \\vT_1(\\bm{y}).\n\\intertext{SpatialJacobian:}\n\\D{\\vTmu}{\\vx}(\\vxt) &= \\D{}{\\vx} \\left( \\vT_0(\\vxt) + \\vT_1(\\vxt) -\n\\vxt \\right) = \\D{}{\\vx} \\vT_0(\\vxt) + \\D{}{\\vx} \\vT_1(\\vxt) - \\bm{I}, \\\\\n%\\D{\\vTmu}{\\vx}(\\vxt) &= \\D{}{\\vx} \\vT_1(\\vT_0(\\vxt)) = \\D{}{\\vx}\n%\\vT_1(\\bm{y}) \\cdot \\D{}{\\vx} \\vT_0(\\vxt). \\\\\n\\D{T_{\\vmu,k}}{x_i}(\\vxt) &= \\left( \\D{T_{1,k}}{\\vx}(\\vy) \\right)^T\n\\D{\\vT_0}{x_i}(\\vxt) = \\left(\\D{\\vT_0}{x_i}(\\vxt) \\right)^T\n\\D{T_{1,k}}{\\vx}(\\vy)\n\\intertext{JacobianOfSpatialJacobian:}\n\\D{}{\\vmu} \\D{\\vTmu}{\\vx}(\\vxt) &= \\D{}{\\vmu} \\D{}{\\vx} \\left(\n\\vT_0(\\vxt) + \\vT_1(\\vxt) - \\vxt \\right) = \\D{}{\\vmu} \\D{}{\\vx}\n\\vT_1(\\vxt), \\\\\n\\D{}{\\vmu} \\D{\\vTmu}{\\vx}(\\vxt) &= \\D{}{\\vmu} \\D{}{\\vx} \\left(\n\\vT_1(\\vT_0(\\vxt)) \\right) = \\D{}{\\vmu} \\D{}{\\vx} \\vT_1(\\bm{y}) \\cdot\n\\D{}{\\vx} \\vT_0(\\vxt).\n\\intertext{SpatialHessian:}\n\\Dd{\\vTmu}{x_i}{x_j} &= \\Dd{}{x_i}{x_j} \\left( \\vT_0(\\vxt) +\n\\vT_1(\\vxt) \\right) = \\Dd{}{x_i}{x_j} \\vT_0(\\vxt) +\n\\Dd{}{x_i}{x_j} \\vT_1(\\vxt), \\\\\n%\\Dd{\\vTmu}{x_i}{x_j} &= \\Dd{}{x_i}{x_j} \\left( \\vT_1(\\vT_0(\\vxt))\n%\\right) = \\D{}{x_i} \\vT_0(\\vxt) \\Dd{}{x_i}{x_j} \\vT_1(\\bm{y})\n%\\D{}{x_j} \\vT_0(\\vxt) + \\Dd{}{x_i}{x_j} \\vT_0(\\vxt) \\D{}{x_j}\n%\\vT_1(\\bm{y}). \\\\\n\\Dd{T_{\\vmu,k}}{x_i}{x_j}(\\vxt) &= \\left( \\D{T_{1,k}}{\\vx}(\\vy)\n\\right)^T \\Dd{\\vT_0}{x_i}{x_j}(\\vxt) + \\left(\\D{\\vT_0}{x_i}(\\vxt)\n\\right)^T \\Dd{T_{1,k}}{\\vx}{\\vx^T}(\\bm{y}) \\D{\\vT_0}{x_j}(\\vxt)\n\\intertext{JacobianOfSpatialHessian:}\n\\D{}{\\vmu} \\Dd{\\vTmu}{x_i}{x_j} &= \\D{}{\\vmu} \\Dd{}{x_i}{x_j} \\left(\n\\vT_0(\\vxt) + \\vT_1(\\vxt) - \\vxt \\right) = \\D{}{\\vmu}\n\\Dd{}{x_i}{x_j} \\vT_1(\\vxt), \\\\\n%\\D{}{\\vmu} \\Dd{\\vTmu}{\\vx}{\\vx^T} &= \\D{}{\\vmu} \\Dd{}{x_i}{x_j}\n%\\left( \\vT_1(\\vT_0(\\vxt)) \\right) \\\\\n% &= \\D{}{x_i} \\vT_0(\\vxt) \\D{}{\\vmu} \\Dd{}{x_i}{x_j} \\vT_1(\\bm{y})\n%\\D{}{x_j} \\vT_0(\\vxt) + \\Dd{}{x_i}{x_j} \\vT_0(\\vxt) \\D{}{\\vmu}\n%\\D{}{x_j} \\vT_1(\\bm{y}). \\\\\n\\D{}{\\vmu} \\Dd{T_{mu,k}}{x_i}{x_j}(\\vxt) &=\n\\Dd{T_{1,k}}{\\vmu}{\\vx^T}(\\vy) \\Dd{\\vT_0}{x_i}{x_j}(\\vxt) +\n\\left(\\D{\\vT_0}{x_i}(\\vxt) \\right)^T \\left(\\D{}{\\vmu}\n\\Dd{T_{1,k}}{\\vx}{\\vx^T}(\\bm{y})\\right) \\D{\\vT_0}{x_j}(\\vxt)\n\\end{align}\n\nThese derivatives are implemented in the\n\\texttt{itk::AdvancedCombinationTransform} class.\n\n\\section{Bending energy penalty term}\n\nThe bending energy was defined as:\n\\begin{align}\n\\mathcal{P}_{\\mathrm{BE}}(\\vmu) &= \\frac{1}{P} \\sum_{\\vxt[i]}\n\\sum_{k,l,m = 1}^2 \\left( \\Dd{T_k}{x_l}{x_m}(\\vxt[i]) \\right)^2.\n\\end{align}\n\nWe constructed an \\texttt{itk::TransformBendingEnergyPenaltyTerm}\nwhich inherits from the \\texttt{itk::ImageToImage\\-Metric}. The\n\\texttt{GetValue()}-method is implemented like:\n\\begin{quote}\n\\begin{verbatim}\nSpatialHessianType spatialHessian;\nfor all samples\n this->GetTransform()->GetSpatialHessian( sample, spatialHessian );\n for all k, l, m\n measure += spatialHessian[ k ][ l ][ m ]^2;\n end\nend\nmeasure /= this->m_NumberOfPixelsCounted;\n\\end{verbatim}\n\\end{quote}\n\nFor the \\texttt{GetValueAndDerivative()} we have:\n\\begin{align}\n\\D{}{\\vmu} \\mathcal{P}_{\\mathrm{BE}}(\\vmu) &= \\frac{1}{P}\n\\sum_{\\vxt[i]} \\sum_{k,l,m = 1}^2 2 \\Dd{T_k}{x_l}{x_m} (\\vxt[i])\n\\D{}{\\vmu} \\Dd{T_k}{x_l}{x_m}(\\vxt[i]),\n\\end{align}\nwhich is implemented like:\n\\begin{quote}\n\\begin{verbatim}\nSpatialHessianType spatialHessian;\nJacobianOfSpatialHessianType jacobianOfSpatialHessian;\nfor all samples\n this->GetTransform()->GetSpatialHessian( sample, spatialHessian );\n this->GetTransform()->GetJacobianOfSpatialHessian( sample,\n jacobianOfSpatialHessian, nonZeroJacobianIndices );\n for all nonZeroJacobianIndices, k, l, m\n derivative[ nonZeroJacobianIndices[ mu ] ] += 2.0\n * spatialHessian[ k ][ i ][ j ]\n * jacobianOfSpatialHessian[ mu ][ k ][ i ][ j ];\n end\nend\nderivative /= this->m_NumberOfPixelsCounted;\n\\end{verbatim}\n\\end{quote}\n\nThis penalty term is implemented in the\n\\texttt{itk::TransformBendingEnergyPenaltyTerm} class.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Discussion and Conclusion}\n\nThis document describes the use and implementation of spatial\nderivatives of coordinate transformations. These spatial derivatives\nwere exploited by the bending energy penalty term. Their usage is,\nhowever, not limited to that penalty term, and many more penalty\nterms can be implemented using the new functionality.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n%\\bibliographystyle{plainnat}\n%\\bibliography{mi}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\end{document}\n", "meta": {"hexsha": "7ffe4429590f87bcd0e1ef2f6385905bd691d6b4", "size": 22075, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "dox/misc/penalty.tex", "max_stars_repo_name": "eliseemond/elastix", "max_stars_repo_head_hexsha": "0e8572f4a315e0a8f08b07d5947b4f3ac160b575", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 318, "max_stars_repo_stars_event_min_datetime": "2017-05-22T11:39:46.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-27T04:40:13.000Z", "max_issues_repo_path": "dox/misc/penalty.tex", "max_issues_repo_name": "eliseemond/elastix", "max_issues_repo_head_hexsha": "0e8572f4a315e0a8f08b07d5947b4f3ac160b575", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 358, "max_issues_repo_issues_event_min_datetime": "2017-05-22T11:36:05.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-18T15:49:10.000Z", "max_forks_repo_path": "dox/misc/penalty.tex", "max_forks_repo_name": "eliseemond/elastix", "max_forks_repo_head_hexsha": "0e8572f4a315e0a8f08b07d5947b4f3ac160b575", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 102, "max_forks_repo_forks_event_min_datetime": "2017-05-22T11:38:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-23T20:27:51.000Z", "avg_line_length": 42.0476190476, "max_line_length": 134, "alphanum_fraction": 0.6573046433, "num_tokens": 8387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388167733099, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.6398110401833503}} {"text": "\\chapter{Multi-Rotor Development}\n\\label{chap:MR}\n\\section{Range \\& Endurance}\nFor a multirotor, range can be estimated by how far it can be controlled by the controller before losing its connection. The range of a multirotor is dependent on certain factors like weight, motor power, weather conditions, etc. Based on the application requirement, they can be customized for longer ranges and better performances. In this project, the forward flight momentum theory was used to estimate the range and endurance of the aircraft for a given set of forward speeds. A velocity range of 0-20 m/s was chosen to perform this calculation and the estimated values of range, endurance, and forward speeds of the multi-rotor drone were calculated using the following steps as follows: \\\\\\\n\n\\subsection{Forward Flight Momentum Theory}\n\nThe forward flight expression can be written as:\n\n\\begin{equation}\n\t\\label{eq:range_multirotor}\n\tv = \\frac{T}{2} \n\\end{equation}\n\nAt a given forward speed, V we can then solve for $ α_D $, v, T, $ P_{ind} $, $ P_{tot} $ as follows: \\\\\\\n\n\\begin{enumerate}\n\t\\item Calculate the quadrotor drag $ D = \\frac{\\rho C_D V^2}{2} $ at $ α_D = 0 $ Assume $ α_D $ does not affect the drag, otherwise need to iterate to find solution\n\t\\item Solve $ α_D = tan^-1(\\frac{D}{W}) $\n\t\\item Square both sides of \\ref{eq:range_multirotor} replace $ T_2 $ by $ W_2 $ + $ D_2 $ and re-arrange to get, $ v_4 + (2 V sin(α_D)) v_3 + V 2 v_2 −(W2 + D2)/(2 ρ A)2 = 0 $\n\t\\item Positive real root of equation gives v, then $ P_{ind} = T_v; P_{tot} = T (v + V sin(α_D)) $\n\t\\item Notice that this total power does not include profile drag, swirl, or additional\n\tlosses due to non-uniform induced velocity.\n\t\\item Solve for a range of speeds and plot results versus V\n\t\t\n\\end{enumerate}\n\nThe maximum range of the quadcopter is calculated using the formula: \\\\\n\nMax. Range = $ (\\frac{E_b * m_e * esc_e }{Minimum Total Power/Velocity}) $ \\\\\n\nWhere, $ E_b $ = Energy of the battery; $ M_e $ = Motor efficiency; $ Esc_e $ = ESC efficiency \\\\\n\nThe maximum endurance of the quadcopter is calculated using the formula: \\\\\n\nMax. Endurance = $ (E_b * m_e * esc_e) / (Minimum Total Power) $ \\\\\nWhere, $ E_b $ = Energy of the battery; $ M_e $ = Motor efficiency; $ Esc_e $ = ESC efficiency \\\\\n\nBased on the above-mentioned procedures, the results of the multi-rotor drone (quadcopter) are as follows: \\\\\n\\begin{itemize}\n\t\\item Max. Range = 34 kms\n\t\\item Max. Endurance = 68.1 minutes\n\t\\item Forward Speed for range = 9.5 m/s\n\t\\item Forward Speed for endurance = 7 m/s\n\\end{itemize}\n\n\\section{Quadrotor Dynamics Model}\n\nThe development of dynamics model of quadrotor was done by using the following state space matrices given for each control channel.\n\n\\textbf{Given State-Space Matrices} : \\\\\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.8\\textwidth]{Images/ss_1}\n\t\\includegraphics[width=0.8\\textwidth]{Images/ss_2}\n\t\\caption{State-Space Matrices}\n\t\\label{fig:ss}\n\\end{figure}\n\nThe output from position controller was fed into the dynamics model. The dynamics model consists of each control channel mentioned above put in a separate state-space block in Simulink. For roll control the input was the desired roll angle and the output from the state-space block was the actual roll angle. This actual roll angle was then fed into the roll to v control channel as input, to get the actual velocity v about y-axis.\n\nFor pitch control the input was desired pitch angle and the output was the actual pitch angle. This actual pitch angle was then fed into the pitch to u control channel which gives the actual velocity about x-axis. \n\nYaw control channel used desired yaw rate, also called as desired heading rate, for input while the output of this control channel is actual yaw angle. The last control channel used for the dynamic model of this quadrotor is altitude control, for this the input was desired vertical velocity w while the output was orientation along z- axis, which is also called altitude. \n\nThe Simulink design of Dynamic Model is shown in the figure below:\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.8\\textwidth]{Images/position_control}\n\t\\caption{Simulink Design Dynamic Model}\n\t\\label{fig:pc}\n\\end{figure}\n\n\\section{Development of Position/Orientation control system:}\n\nPosition/Orientation Control System contains two types of state input, one is the target x, y, and z coordinate input and the other is actual x, y and z coordinates of the quadrotor. A reference yaw command is also given into the position controller to ensure that the quadrotor maintains the commanded yaw while operating for the target position.\n\nThe target x coordinate is given into a summation block while the actual x orientation of the quadrotor is given to the summation block as feedback. This command is fed into the PD controller, which gives corresponding derivative $\\dot{x}$ (i.e., speed along x axis). This derived velocity u is given to the summation block whereas the actual velocity $ u_a $ is given as feedback in the same block. A PD controller is used again which finally gives the derivative $\\ddot{x}$ . \n\nSimilarly, target y position and actual y position of the quadrotor is also given into the summation block and then into the PD controller. The output from this PD controller is further given into a summation block where the actual velocity v is given as feedback. This output is then given into the PD controller to get acceleration along y-axis. \n\nActual heading angle psi, along with $\\ddot{x}$ and $\\ddot{y}$ is used to calculate the desired pitch angle and desired roll angle. The following equations [reference no?] are used for this: \\\\\n\n\\left[\\begin{array}{c}\n\t\\phi_{d} \\\\\n\t\\theta_{d}\n\\end{array}\\right]=\\left[\\begin{array}{cc}\n\t-\\sin \\psi & -\\cos \\psi \\\\\n\t\\cos \\psi & -\\sin \\psi\n\\end{array}\\right]^{-1} \\frac{m}{U_{1}}\\left[\\begin{array}{l}\n\t\\ddot{x}_{d} \\\\\n\t\\ddot{y}_{d}\n\\end{array}\\right] \\\\\n\n\nSince the derivation of these equations in (reference no) uses small angle approximation, we must ensure that the desired angles $\\theta_{d}$ and $\\phi_{d}$ are within the limit of –20° and 20°. Therefore, a saturation block is placed at the output of MATLAB function block. \n\nWe also need the desired vertical velocity w and the desired heading rate $\\dot{psi$ to be given into the dynamics model of the quadrotor. For this, desired z coordinate and actual z coordinate are given into a summation block and PD controller is used to obtain the desired w velocity. Similarly, actual yaw angle and desired yaw angle along with PD controller are used to obtain heading rate $\\dot{psi}$. \n\t\n\\section{Development of Position Estimation:}\n\nFor estimating the position of the quadrotor, the output states from the dynamic model i.e., psi, psi, theta, h, u and v, are used. The dynamic model provides information about the actual value of states of the quadrotor. To acquire data regarding the actual x and y coordinated of the quadrotor we integrate the u and v velocities, respectively. The Simulink model of position estimation subsystem is shown in \\ref{fig:pe}.\n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.8\\textwidth]{Images/pe}\n\t\\caption{Position Estimation}\n\t\\label{fig:pe}\n\\end{figure}\n \n \\section{Development of linear simulation model:}\n \n For the development of the linear simulation model, a reference subsystem was created which was used for giving the desired coordinate information to the position controller. The desired pitch and roll angles $ \\theta_{d} $ and $\\phi_{d}$ along with the heading rate $\\dot{psi}$ and desired vertical velocity wd goes into a summation block followed by a PID controller. To this summation block feedback is given of actual orientation of the respective states. \n \n Further, these values are given into the dynamics model subsystem which is also explained in the previous section. The actual values obtained from the dynamic model are used in the position estimation subsystem to get information about Euler angles, position, and velocities of the quadrotor. \n \n The reference position subsystem developed in for this project is an If-Else condition block which is used with Action block to activate a particular command for an If-Else condition. The reference command subsystem is shown in the \\ref{fig:rp}.\n \\begin{figure}\n \t\\centering\n \t\\includegraphics[width=0.8\\textwidth]{Images/rp}\n \t\\caption{Reference Position}\n \t\\label{fig:rp}\n \\end{figure}\n\nThe complete linear simulation system that was developed in Simulink is shown in the \\ref{fig:cm} \n\nPID Tuning methods used for this project were Inbuilt PID tuners provided by Simulink. For tuning PD controllers, the following steps were taken:\n\\begin{enumerate}\n\t\\item Proportional gain Kp was increased until steady oscillations were obtained\n\t\\item Derivate gain Kd was increased until the oscillations were critically damped\n\\end{enumerate} \\\\\n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=1.1\\textwidth]{Images/multirotor}\n\t\\caption{Complete Multirotor Dynamic System}\n\t\\label{fig:cm}\n\\end{figure}\n\nTotal simulation time was taken to be 1500 seconds for the following four maneuvers: \\\\\n\\begin{enumerate}\n\t\\item Take off and hover at 2 meters above origin\n\t\\item Fly to the first target (x = 5 m, y = 6 m, h = 4 m) and hover\n\t\\item Fly to the second target (x = -5 m, y = -6 m, h = 4 m) and hover\n\t\\item Return to 2 meters above origin and land\n\\end{enumerate}\n\n\\section{Result}\n\nThe quadcopter was successfully able to complete all the given maneuvers for this project. The simulation result, which shows the trajectory of x, y and z coordinates of the quadcopter for 1500 seconds is shown in . \n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=1.1\\textwidth]{Images/multirotor_sol}\n\t\\caption{Quadrotor Simulation Solution}\n\t\\label{fig:m_sol}\n\\end{figure}\\\\\n\nIn the figure the quadcopter is initially at x=0, y=0 and z=0 position, in the first 300 seconds it takes off and hover at x=0 y=0 and z=2 m. Next, the quadcopter reaches to x=5, y=6 and z=4.For the third maneuver it goes from its current position to the target position, x=-5, y=-6 and z=4 and then for the last maneuver is goes from current position to target position of x=0, y=0 and z= 2 and then back to the origin. ", "meta": {"hexsha": "5cc7a67803516e08e31f944315d987a141ca61d7", "size": 10220, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "project/report/Chapters/Multirotor.tex", "max_stars_repo_name": "Rutvikrj26/AER1216", "max_stars_repo_head_hexsha": "c809846ba49d086415827829ec57ef42ae37ba9c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "project/report/Chapters/Multirotor.tex", "max_issues_repo_name": "Rutvikrj26/AER1216", "max_issues_repo_head_hexsha": "c809846ba49d086415827829ec57ef42ae37ba9c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "project/report/Chapters/Multirotor.tex", "max_forks_repo_name": "Rutvikrj26/AER1216", "max_forks_repo_head_hexsha": "c809846ba49d086415827829ec57ef42ae37ba9c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.2767295597, "max_line_length": 697, "alphanum_fraction": 0.7610567515, "num_tokens": 2607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8198933359135361, "lm_q1q2_score": 0.6395109809782137}} {"text": "\\chapter{Multi-scale Analysis}\n\\section{Introduction}\n\nIn this chapter, the tools for multi-scale and multi-resoltuion\nprocessing (analysis, synthesis and fusion) will be presented. Most of\nthe algorithms are based on pyramidal approaches. These approaches\nwere first used for image compression and they are based on the fact\nthat, once an image has been low-pass filtered it does not have\ndetails beyond the cut-off frequency of the low-pass\nfilter any more. Therefore, the image can be subsampled -- decimated -- without\nany loss of information.\n\nA pyramidal decomposition is thus performed applying the following 3\nsteps in an iterative way:\n\\begin{enumerate}\n \\item Low pas filter the image $I_{n}$ in order to produce $F(I_n)$;\n \\item Compute the difference $D_n = I_n - F(I_n)$ which corresponds\n to the details at level $n$;\n \\item Subsample $F_(I_n)$ in order to obtain $I_{n+1}$.\n\\end{enumerate}\n\nThe result is a series of decrasing resolution images $I_k$ and a\nseries of decreasing resolution details $D_k$.\n\n\n\n\\section{Morphological Pyramid}\\label{secMorphoPyr}\n\nIf the smoothing filter used in the pyramidal analysis is a\nmorphological filter, one cannot safely subsample the filtered image\nwithout loss of information. However, by keeping the details possibly\nlost in the down-sampling operation, such a decomposition can be used.\n\nThe Morphological Pyramid is an approach to such a\ndecomposition. Its computation process is an iterative analysis\ninvolving smoothing by the morphological filter, computing the\ndetails lost in the smoothing, down-sampling the current image, and\ncomputing the details lost in the down-sampling.\n\n\\input{MorphologicalPyramidAnalysisFilterExample.tex}\n\n\\input{MorphologicalPyramidSynthesisFilterExample.tex}\n\n\n\\subsection{Morphological Pyramid Exploitation}\nOne of the possible uses of the morphological pyramid is the\nsegmentation of objects -- regions -- of a particular scale.\n\n\\input{MorphologicalPyramidSegmenterExample.tex}\n\nThis same approach can be applied to all the levels of the\nmorphological pyramid analysis.\n\n\\input{MorphologicalPyramidSegmentationExample.tex}\n", "meta": {"hexsha": "39d1531f81ad8c8d05b1eb5c480d0b5cb34a1bdf", "size": 2123, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documentation/SoftwareGuide/Latex/MultiScaleAnalysis.tex", "max_stars_repo_name": "xcorail/OTB", "max_stars_repo_head_hexsha": "092a93654c3b5d009e420f450fe9b675f737cdca", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-03-30T18:05:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-08-28T01:03:49.000Z", "max_issues_repo_path": "Documentation/SoftwareGuide/Latex/MultiScaleAnalysis.tex", "max_issues_repo_name": "xcorail/OTB", "max_issues_repo_head_hexsha": "092a93654c3b5d009e420f450fe9b675f737cdca", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2015-10-14T10:11:38.000Z", "max_issues_repo_issues_event_max_datetime": "2015-10-15T08:26:23.000Z", "max_forks_repo_path": "Documentation/SoftwareGuide/Latex/MultiScaleAnalysis.tex", "max_forks_repo_name": "xcorail/OTB", "max_forks_repo_head_hexsha": "092a93654c3b5d009e420f450fe9b675f737cdca", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-10-08T12:04:06.000Z", "max_forks_repo_forks_event_max_datetime": "2018-06-19T08:00:47.000Z", "avg_line_length": 38.6, "max_line_length": 79, "alphanum_fraction": 0.8035798398, "num_tokens": 492, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199552262966, "lm_q2_score": 0.7634837743174788, "lm_q1q2_score": 0.6395092448598105}} {"text": "%\\pagestyle{plain}\n\n\\chapter{Event Detection}\nIn this chapter, we will introduce the mechanism of event detection in this project. \nWe first apply some noise reduction or audio enhancement technique on the training data for our events. \nThen we extract MFCC features from the training clips. \nAfter getting features, a Gaussian Mixture Model is built on those features. \n\n\\section{Audio Enhancement}\nBecause the audios downloaded from SSEs are uploaded by other users. \nThose clips are recorded by various types of devices and in different environments. \nHence we need to apply a noise reduction step to enhance the audio clips. \n\nAmong algorithms for speech enhancement, spectral subtraction was one of the first algorithms proposed. \nAlthough it was originally proposed for speech enhancement, we use it in our project for noise is pretty much the same in speech or in more general audios. \nThe basic principles is that we assume the clean signal add some additive noise becomes the audio we get. \nThen if we subtract the noise spectrum from the audio, we may get signal close to the original clean one. \n\nMathematically speaking, assume the input signal is $ y(t) = x(t) + n(t)$, where the signal $x(t)$ is the signal of interest and $n(t)$ is the additive noise. \nThe audio enhancement process takes on the following steps:\n\\begin{enumerate}\n\\item The input signal $y(t)$ is resampled to 16kHz and multiple channels are averaged to one channel. \n\\item Using short time Fourier transform (STFT), we obtain $Y_i(k) = X_i(k) + N_i(k)$, where $i$ is the frame indices and $k$ is the frequency bin indices of the spectrum. \n\\item The Minimum Statistics (MS) estimator \\cite{martin2001noise} is used to the noise power spectral density from $y(t)$. \n\\item Spectral subtraction is used to remove the estimated noice spectrum from the input spectrum $Y_i(k)$. \nThen we could get an estimated clean signal spectrum: $\\hat{X_i}(k)$. \n\\item Applying inverse short time Fourier transform (ISTFT) on $\\hat{X_i}(k)$ to get the estimated clean signal $\\hat{x}(t)$. \n\\end{enumerate}\n\nA speech processing toolbox for Matlab called VoiceBox is used for spectral subtraction\\footnote{\\url{http://www.ee.ic.ac.uk/hp/staff/dmb/voicebox/voicebox.html}}.\n% noise subtraction example here \n\n\\section{Feature Extraction}\nSince the audio data is downloaded from a sound website where clips are crowdsourced, the clips are in various format. \nWe converted all audio clips to WAV format and averaged the channels into one if it has multiple channels. \nMoreover, clips are all downsampled to 16khz sample rate for feature extraction. \n\nFor the audio data, mel-frequency cepstrum coefficients (MFCCs) features are a widely-used feature. \nIt was brought up by Davis and Mermelstein in the 1980s. \nMFCC features extract the spectral envelope from spectrums of audio frames.\nBasically, the process of extracting MFCCs start by applying Fast Fourier transform (FFT) on framed audio. \nFFT transform audio data from Amplitude-Time domain (frame) into Amplitude-Frequency domain (spectrum). \nAs each audio frame is transformed into a spectrum, the spectral envelope is extracted by Inverse FFT (IFFT).\n\nActually, one more filtering is performed before applying IFFT on spectrums. \nBecause MFCC features are based on human perceptual experiment, and human's ear is like a natural filter, where low-frequency area has more filter and high-frequency area has less. So a nonlinear function is applied to original spectrum to transform the frequency axis into the mel scale. \nA popular formula to convert $f$ herts into $m$ mel is: \n\\begin{equation}\n\tm = 2595 \\times \\log_{10}(1+\\frac{1}{700})\n\\end{equation} \n\n% melscale.eps on 5566\n\\begin{figure}[htb]\n\\centering\n\\input{figure/eventdetect/melscale}\n\\caption{Mel scale versus hertz scale}\n\\label{fig:melscale}\n\\end{figure}\n\nFigure \\ref{fig:melscale} shows the non-linear transformation of hertz scale to mel scale. \nWe could see that the slope of the curve are decreasing as the pitch goes up in hertz scale. \nBecause of this feature of suppressing the higher frequency bands, MFCC enables a focus of the more useful range of bands in lower scale \\cite{davis1980comparison}.\n\n% Paraphrase !!!\n\n\\section{Model Selection}\nFrom the previous feature extraction process, we could a matrix representation for an audio clip. \nThe column number stands for the dimension of the features and each row corresponds to one observation. \nFrom these data, we need to build a model that can capture the overall feature distribution and also are convenient to use for testing data.\n\nFor this goal, Gaussian Mixture Model (GMM) is used. \nFirst, a multivariate gaussian distribution has the following probability density:\n\n% multivariate gaussian \n\\begin{equation}\n \\mathcal{N}(\\mathbf{x}| \\mathbf{\\mu}, \\Sigma) = \n\\frac{1}{\\sqrt{(2\\pi)^D|\\Sigma|}}e^{-\\frac{1}{2}(\\mathbf{x}-\\mathbf{\\mu})^T \\Sigma^{-1} (\\mathbf{x}-\\mathbf{\\mu})}\n\\end{equation}\n\nIn this equation, $\\mathbf{\\mu}$ and $\\mathbf{\\Sigma}$ are all vectors in $D$ dimensions, for discribing high dimensional data. \nA gaussian mixture model (or density) is a weighted sum of N gaussian densities. \nTypically, these gaussian densities have the same dimension, say $D$, and each gaussian are called a component. \nPut many gaussian distributions together, we could get the density function for GMM: \n\n% GMM density function\n\\begin{equation}\nP(\\mathbf{x}|\\mathbf{\\pi},\\mathbf{\\mu},\\Sigma) = \\sum_{k = 1}^{M} \\pi_k\n\\mathcal{N}(\\mathbf{x}|\\mathbf{\\mu}_k, \\Sigma_k),\n\\end{equation} \n\nThe intuition of using GMM to model the audio events is that the individual component densities of a multi-model density may model some underlysing set of acoustic classes \\cite{reynolds1995robust}. \nLike in speech, a words may consist of some vowels and consonants. \nAn audio event is naturally more complicated than speech, and may therefore contain more characteristics. \nTherefore, we use different gaussian densities with different $\\mathbf{\\mu}$ and $\\mathbf\\sigma$, where the $\\mathbf\\mu$ may capture the overall value for one class and the $\\mathbf\\sigma$ shows the variation of that class. \nWhen these densities are added together, forming a GMM, they can have a good depiction of some unusual distributions. \n\n%Before training GMMs for event training data, we first apply K-means algorithm on the data. \n%K-means is also an unsupervised clustering method. \n%It uses a Expectation-Maximization step to update the cluster centroid. \n%Because K-means has a faster speed for iterating than GMMs, so we first run K-means and use its centroid result as initials for further iterating of GMMs. \n%This part use the gaussian toolbox provided by Matlab\\footnote{\\url{http://cn.mathworks.com/help/stats/gmdistribution-class.html}} . \n\n\n\\section{Training Data Selection}\nAs previously mentioned, many of the clips in those Sound Search Engines are crowdsourced. \nHence, the quality of those clips are not guaranteed to be suitable as our event training data. \nWe need to select out those data which are similar, and discard those outliers. \nBecause outliers are very likely to be a noise, thus adding no knowledge for our event models, perhaps even bring bad effects. \n\nTherefore, in our project, we added a step for comparing the similarity of the audio clips, and cluster similar clips together, while leaving others clips out. \nYet signal data of the audio clips are in different volume, and different duration, it is hard to directly compare their similarity. \nSo we resort to the distance between their corresponding GMMs. \nBecause the trained GMM are in the same dimension for different audio clips, and GMM represent the overall features, it is reasonable to cluster similar GMMs together. \n\nIn order to measure the distance between GMMs, we use Kullback-Leibler (KL) divergence. \nKL divergence is a measure of the difference between two probability distributions. \nMore precisely, KL divergence of $Q$ from $P$, denoted by $KL(P||Q)$, is a measure of the information loss when we use $Q$ to approximate $P$.\n\nFor continuous random variables $\\mathbf{x}$ and two distributions $P$ and $Q$, the KL divergence of $Q$ from $P$ is defined as:\n\\begin{equation}\nKL(P||Q) = \\int_{-\\infty}^{+\\infty}\\ln(\\frac{p(\\mathbf{x})}{q(\\mathbf{x})})p(\\mathbf{x})\\mathrm{d}x,\n\\label{eq:kl}\n\\end{equation}\nwhere $p(\\mathbf{x})$ and $q(\\mathbf{x})$ are the density functions of $P$ and $Q$.\nThe divergence satisifies three properties: \n\\begin{enumerate}\n\\item{Self similarity: $KL(P||P) = 0$}. \n\\item{Self identification: $KL(P||Q) = 0$, only if $P = Q$}. \n\\item{Non-Negativity: $KL(P||Q) >= 0$ for any $P, Q$}. \n\\end{enumerate}\n\nBecause of these properties, KL divergence is used in many aspects of speech recognition \\cite{olsen2003efficient}. \nBut we need to take notice that KL divergence is not a distance because it is not symmetric: $KL(P||Q) \\neq KL(Q||P)$. \nFor two gaussian distributions, the KL divergence has a closed formed expression. \nBut for two gaussian mixture models, there is no such closed form expression. \nTo tackle this issue, we use an approximation method to the KL divergence proposed in \\cite{hershey2007approximating}\nIt uses ``variational approximation'' for calculating KL divergence. \nWe use the \\textit{gaussmixk} function in VoiceBox toolbox for implementing the approximated KL divergence\\footnote{\\url{http://www.ee.ic.ac.uk/hp/staff/dmb/voicebox/doc/voicebox/gaussmixk.html}}. \n\n\\section{Summary}\nThe method we propose for audible event detection are introduced in this chapter. \nWe first apply a noise reduction technique on the downloaded audio clips. \nThis audio enhancement process uses Minimum Statistics estimator to estimate the noise power spectral density. \nThen this estimated noise are subtracted from the input signal to get a more cleaned signal for event models trainging. \nFeature extraction are also reviewed in this chapter. \nThe features we use in this task are MFCC features. \nThey are notable for good capture of audio features because of the non-linear mel scale could suppress higher frequency bands and put more focus on the lower part. \nFrom the extracted features, we build Gaussian Mixture Models (GMMs) on them. \nThe reason that GMMs are chosen against other models is that they comprised of multiple gaussian distributions, which are called components. \nThese components could be used to model the underlying sound generating process and help us model more complicated sound events. \nIn the end, we touch on the problem of removing some low-quality sound clips. \nThis removal process is carried out by clustering audio clips. \nWe use an approximation for Kullback-Leibler (KL) divergence to represent the distance between different GMMs, and thus, can also be viewed as the distance between audio clips. \nClip features which have similar GMMs are clustered together, and we stop the clustering process when the divergence is large enough. \nThrough this way, we could remove some noisy clips for their feature data is too far away from others. \n", "meta": {"hexsha": "79aa086d23b3ffaeace12fc7a3478b45dd38dd2c", "size": 11031, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/eventdetect.tex", "max_stars_repo_name": "findslowly/thesis", "max_stars_repo_head_hexsha": "177115f287b00d81434a13b00dd449ed9944607d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/eventdetect.tex", "max_issues_repo_name": "findslowly/thesis", "max_issues_repo_head_hexsha": "177115f287b00d81434a13b00dd449ed9944607d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/eventdetect.tex", "max_forks_repo_name": "findslowly/thesis", "max_forks_repo_head_hexsha": "177115f287b00d81434a13b00dd449ed9944607d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 70.7115384615, "max_line_length": 289, "alphanum_fraction": 0.7769921131, "num_tokens": 2641, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438950868503682, "lm_q2_score": 0.7577943658046608, "lm_q1q2_score": 0.6394989421454439}} {"text": "\n\\subsection{Testing for stationarity with Dickey-Fuller (DF) and Augmented Dicky-Fuller (ADF)}\n\n\\subsubsection{Stationarity}\n\n\\subsubsection{Unit roots}\n\n\n\\subsubsection{Integration order}\n\n\\subsubsection{Dickey-Fuller}\n\nThe Dickey-Fuller test tests if there is a unit root.\n\nThe AR(\\(1\\)) model is:\n\n\\(y_t=\\alpha + \\beta y_{t-1}+\\epsilon_t\\)\n\nWe can rewrite this as:\n\n\\(\\Delta y_t=\\alpha + (\\beta -1)y_{t-1}+\\epsilon_t\\)\n\nWe test if \\(\\beta -1)=0\\).\n\nIf the coefficient on the last term is \\(1\\) we have a random walk, and the process is non-stationary.\n\nIf the last term is \\(<1\\) then we have a stationary process.\n\n\\subsubsection{Variation: Removing the drift}\n\nIf our model has no intercept it is:\n\n\\(y_t=\\beta y_{t-1}+\\epsilon_t\\)\n\n\\(\\Delta y_t=(\\beta -1)y_{t-1}+\\epsilon_t\\)\n\n\\subsubsection{Variation: Adding a deterministic trend}\n\nIf our model has a time trend it is:\n\n\\(y_t=\\alpha \\beta y_{t-1}+\\gamma t + \\epsilon_t\\)\n\n\\(\\Delta y_t=\\alpha + (\\beta -1)y_{t-1}+\\gamma t+\\epsilon_t\\)\n\n\\subsubsection{Augmented Dickey-Fuller}\n\nWe include more lagged variables.\n\n\\(y_t=\\alpha + \\beta t + \\sum_i^p \\theta_i y_{t-i}+\\epsilon_t\\)\n\nIf no unit root, can do normal OLS?\n\n\n", "meta": {"hexsha": "ac5592671f367548fce0c888eb00a428efa2a197", "size": 1173, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/probability/stochasticWold/01-02-stationarity.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/probability/stochasticWold/01-02-stationarity.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/forecastingUni/02-02-stationarity.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.7222222222, "max_line_length": 102, "alphanum_fraction": 0.7007672634, "num_tokens": 371, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430645886583, "lm_q2_score": 0.7690802317779601, "lm_q1q2_score": 0.6392157007544894}} {"text": "% !TEX root = manual.tex\n\\section{PCA Analysis}\\label{sec:pca}\n\nThe emulator is dependent on the principal component analysis (PCA). PCA also provides insight of its own. The principal components show which linear combinations of observables are likely to be well constrained, and which combinations are uninteresting. Further, one can also estimate which parameters are driving specific principal components, at least within a linear picture.\n\nPCA is performed by the following command,\n\n\\commandline{madai\\_pca\\_decompose stat1}\n\nWhen running, \\path{madai_pca_decompose} will read the parameters file \\path{stat1/settings.dat}. The PCA analysis uses the following parameters:\n\n\\begin{quote}\n{\\tt MODEL\\_OUTPUT\\_DIR \"./model\\_output\"}\\\\\n{\\tt EXPERIMENTAL\\_RESULTS\\_FILE \"./experimental\\_results.dat\"}\\\\\n$\\vdots$\n\\end{quote}\nThe PCA analysis uses the experimental information only in calculating the error for comparing a model value to the experimental value, each of which has its own error.\n\n\\path{madai_pca_decompose} produces a file \\path{stat1/PCADecomposition.dat}. This file holds the eigenvalues of $\\Xi$ (the strength of the principal components) and the eigenvectors of $\\Xi$ (which define the rotation of $\\tilde{y}$ into the principal components, $z$. If $M$ experimental values are written down, the PCA analysis will determine and record the information recorded to rotate the vector $y_1\\cdots y_M$ to $z_1\\cdots z_M$. The emulator will in turn use only those components $z_1\\cdots z_{M_Z}$ that correspond to the largest eigenvalues, $\\lambda_i$, of the matrix $\\Xi$. The number of components retained, $M_Z$, can be chosen during the emulator tuning or initialization.\n\nThe \\path{madai_pca_decompose} file has the following format:\n\\begin{quote}\n{\\tt OUTPUT\\_MEANS}\\\\\n$M~~\\leftarrow$ Number of observables.\\\\\n$y_1$~~~$\\leftarrow$ Means for observables averaged over the $N$ model runs.\\\\\n$y_2$~~~$\\swarrow$\\\\\n$\\vdots$\\\\\n$y_M$\\\\\n{\\tt OUTPUT\\_UNCERTAINTY\\_SCALES}\\\\\n$M~~\\leftarrow$ Number of observables.\\\\\n$\\sigma_1$~~~$\\leftarrow$ Uncertainties that incorporate both experimental and model uncertainties.\\\\\n$\\sigma_2$~~~$\\leftarrow$ If different model runs have different $\\sigma_i$, model error is averaged over runs.\\\\\n$\\sigma_3$~~~$\\swarrow$\\\\\n$\\vdots$\\\\\n$\\sigma_M$\\\\\n{\\tt OUTPUT\\_PCA\\_EIGENVALUES}\\\\\n$M$~~$\\leftarrow$ Number of observables\\\\\n$\\lambda^{(M)}$~$\\leftarrow$ Smallest eigenvalue of $\\Xi_{ab}$ in Eq. \\eqref{eq:Xidef}\\\\\n$\\lambda^{(M-1)}$~$\\swarrow$\\\\\n$\\vdots$\\\\\n$\\lambda^{(1)}$~~~~$\\leftarrow$ Largest eigenvalue\\\\\n{\\tt OUTPUT\\_PCA\\_EIGENVECTORS}\\\\\n$M~~~~M$\\\\\n\\vspace*{-10pt}\n\\[\n\\hspace*{-130pt}\n\\begin{array}{cccc}\n~~~~\\xi_1^{\\rm(M)}&\\xi_1^{\\rm(M-1)}&\\cdots&\\xi_1^{(1)}\\\\\n~~~~\\xi_2^{\\rm(M)}&\\xi_2^{\\rm(M-1)}&\\cdots&\\xi_2^{(1)}\\\\\n~~~~\\vdots&\\vdots&\\ddots&\\vdots\\\\\n~~~~\\xi_M^{\\rm(M)}&\\xi_M^{\\rm(2)}&\\cdots&\\xi_M^{(1)}\\\\\n\\end{array} ~~~\\leftarrow{\\rm These~are~the~normalized~eigenvectors}\n\\]\n{\\tt END\\_OF\\_FILE}\n\\end{quote}\n\nBefore building an emulator, the experimental observables and the corresponding model values are rotated into a new basis. The rotation can be stated as,\n\\[\nz_a=U_{ab}y_b,~~y_a=U^{-1}_{ab}z_b.\n\\]\nAs described in Eq. (\\ref{eq:pcarotation}), the rotation matrix is comprised of eigenvectors of the matrix $\\Xi_{ab}=\\langle \\tilde{y}_a\\tilde{y}_b\\rangle$, where $\\tilde{y}_a\\equiv (y_a-\\langle y_a\\rangle)/\\sigma_a$, with $\\langle y_a\\rangle$ denoting $y_a({\\bf x})$ averaged over model values calculated from the parameters taken from the prior distribution.\n", "meta": {"hexsha": "9f71a84aaa1dd74c6ba3cf02f19edf2450de1201", "size": 3515, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/manual/pca.tex", "max_stars_repo_name": "scottedwardpratt/MADAI", "max_stars_repo_head_hexsha": "9f9ee0dac704d77492d9905b4d90a57746201912", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2015-04-02T17:37:35.000Z", "max_stars_repo_stars_event_max_datetime": "2017-03-28T20:14:23.000Z", "max_issues_repo_path": "doc/manual/pca.tex", "max_issues_repo_name": "scottedwardpratt/MADAI", "max_issues_repo_head_hexsha": "9f9ee0dac704d77492d9905b4d90a57746201912", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/manual/pca.tex", "max_forks_repo_name": "scottedwardpratt/MADAI", "max_forks_repo_head_hexsha": "9f9ee0dac704d77492d9905b4d90a57746201912", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2015-08-20T14:07:41.000Z", "max_forks_repo_forks_event_max_datetime": "2017-03-28T20:15:23.000Z", "avg_line_length": 56.6935483871, "max_line_length": 691, "alphanum_fraction": 0.7305832148, "num_tokens": 1068, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430520409023, "lm_q2_score": 0.76908023177796, "lm_q1q2_score": 0.6392156911042582}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%\n% Author: David Oniani\n%\n% _ _____ __ __\n% | | __ |_ _|__\\ \\/ /\n% | | / _` || |/ _ \\\\ /\n% | |__| (_| || | __// \\\n% |_____\\__,_||_|\\___/_/\\_\\\n%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Document Definition\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass{article}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Packages and Related Settings\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Global, document-wide settings\n\\usepackage[english]{babel}\n\\usepackage[margin=1in]{geometry}\n\\usepackage[utf8]{inputenc}\n\n% Other packages\n\\usepackage{caption}\n\\usepackage{hyperref}\n\\usepackage{mathtools}\n\\usepackage{minted}\n\\usepackage{sectsty}\n\\usepackage{xcolor}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Setup\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Black-blue color\n\\colorlet{bb}{black!50!blue}\n\n% Orange color\n\\definecolor{orange}{HTML}{f0360e}\n\n% PDF information and nice-looking urls\n\\hypersetup{%\n pdfauthor={David Oniani},\n pdftitle={Bresenham's Circle Drawing Algorithm},\n pdfsubject={bresenham, circle, geometry},\n pdfkeywords={bresenham, circle, geometry},\n pdflang={English},\n colorlinks=true,\n linkcolor={bb},\n citecolor={bb},\n urlcolor={bb}\n}\n\n% Change section color\n\\sectionfont{\\color{orange}}\n\n% Remove indentations from paragraphs\n\\setlength{\\parindent}{0pt}\n\n% Setting style for the minted package\n\\setminted{\n fontsize=\\small,\n frame=lines,\n linenos,\n}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Author(s), Title, and Date\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Author(s)\n\\author{David Oniani\\\\\n \\href{mailto:onianidavid@gmail.com}{onianidavid@gmail.com}}\n\n% Title\n\\title{\\textcolor{orange}{\\textit{Bresenham's Circle Drawing Algorithm}}}\n\n% Date\n\\date{\\today}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Beginning of Document\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{document}\n\\maketitle\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Derivation\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Derivation}\n\nLet the radius of the circle centered at \\((0, 0)\\) be given and equal \\(r\\). Any point \\((x, y)\\)\non the circle satisfies the following:\n\\begin{equation}\n x^2 + y^2 = r^2\n\\end{equation}\n\nConsider the upper half of the first quadrant. The next point could then be placed either to the\neast \\((x + 1, y)\\) or southeast \\((x + 1, y - 1)\\). The errors for both scenarios are:\n\\begin{align}\n &e(x + 1, y) = {(x + 1)}^2 + y^2 - r^2\\\\\n &e(x + 1, y - 1) = {(x + 1)}^2 + {(y - 1)}^2 - r^2\n\\end{align}\n\nThe decision parameter \\(d\\) representing the total error is defined as follows:\n\\begin{equation}\n d = e(x + 1, y) + e(x + 1, y - 1)\n = 2{(x + 1)}^2 + y^2 + {(y - 1)}^2 - 2r^2\n\\end{equation}\n\nSince the \\(x\\) coordinate always increases, the next point is some \\((x + 1, y_n)\\) and we get:\n\\begin{equation}\n d_n = e(x + 2, y_n) + e(x + 2, y_n - 1)\n = 2{(x + 2)}^2 + {y_n}^2 + {(y_n - 1)}^2 - 2r^2\n\\end{equation}\n\nThe difference is equal to the following:\n\\begin{equation}\n d_n - d = 2(2x + 3) + \\big({y_n}^2 - y^2\\big) + \\big({(y_n - 1)}^2 - {(y - 1)}^2\\big)\n\\end{equation}\n\nFinally, we have:\n\\begin{align}\n &d_n - d = 2(2x + 3) &&\\implies d_n\n = d + 4x + 6 &&\\text{ if } d \\leq 0\\ (y_n = y)\\\\\n &d_n - d = 2(2x + 3) - 4y + 4 &&\\implies d_n\n = d + 4(x - y) + 10 &&\\text{ if } d > 0\\ (y_n = y - 1)\\\\\n &d_0 = 2{(0 + 1)}^2 + r^2 + {(r - 1)}^2 - 2r^2 &&\\implies d_0\n = 3 - 2r &&\\text{ at initial point } (0, r)\n\\end{align}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Implementation\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\clearpage\n\\section{Implementation}\n\n\\begin{itemize}\n \\item Time Complexity: \\(O(r)\\) where \\(r\\) is the radius of the circle\n \\item Space Complexity: \\(O(1)\\)\n\\end{itemize}\n\n\\begin{figure}[H]\n \\inputminted{python}{bresenham.py}\n \\caption*{A Python Implementation of Bresenham's Circle Drawing Algorithm}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% End of Document\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\end{document}\n", "meta": {"hexsha": "4cfd4d3ef0170b8f5f941ac066658fc7271f53e0", "size": 5200, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "bresenham.tex", "max_stars_repo_name": "oniani/bresenham", "max_stars_repo_head_hexsha": "a01bd090a578372dbfacbcce9b730b446a5c4cb6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "bresenham.tex", "max_issues_repo_name": "oniani/bresenham", "max_issues_repo_head_hexsha": "a01bd090a578372dbfacbcce9b730b446a5c4cb6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "bresenham.tex", "max_forks_repo_name": "oniani/bresenham", "max_forks_repo_head_hexsha": "a01bd090a578372dbfacbcce9b730b446a5c4cb6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.298136646, "max_line_length": 100, "alphanum_fraction": 0.3996153846, "num_tokens": 1292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125848754471, "lm_q2_score": 0.8499711775577735, "lm_q1q2_score": 0.6391890223048489}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Written By Michael Brodskiy\n% Class: Analytic Geometry & Calculus III (Math-292)\n% Professor: V. Cherkassky\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass[12pt]{article} \n\\usepackage{alphalph}\n\\usepackage[utf8]{inputenc}\n\\usepackage[russian,english]{babel}\n\\usepackage{titling}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{enumitem}\n\\usepackage{amssymb}\n\\usepackage[super]{nth}\n\\usepackage{everysel}\n\\usepackage{ragged2e}\n\\usepackage{geometry}\n\\usepackage{fancyhdr}\n\\geometry{top=1.0in,bottom=1.0in,left=1.0in,right=1.0in}\n\\newcommand{\\subtitle}[1]{%\n \\posttitle{%\n \\par\\end{center}\n \\begin{center}\\large#1\\end{center}\n \\vskip0.5em}%\n\n}\n\\usepackage{hyperref}\n\\hypersetup{\ncolorlinks=true,\nlinkcolor=blue,\nfilecolor=magenta, \nurlcolor=blue,\ncitecolor=blue,\n}\n\n\\urlstyle{same}\n\n\n\\title{Lecture XIV Notes}\n\\date{\\today}\n\\author{Michael Brodskiy\\\\ \\small Professor: V. Cherkassky}\n\n% Mathematical Operations:\n\n% Sum: $$\\sum_{n=a}^{b} f(x) $$\n% Integral: $$\\int_{lower}^{upper} f(x) dx$$\n% Limit: $$\\lim_{x\\to\\infty} f(x)$$\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Multiple Integrals $-$ 15.1}\n\nIn calculus I, integrals were defined as the area, $A$, under a curve, where:\n\n$$A=\\lim_{x\\to\\infty}\\sum_{i=1}^n f(x_i^*) \\Delta x$$\n\nThis formula was simplified to what is known as the integral:\n\n$$\\int_a^b f(x)\\,dx$$\n\nThe volume of a multivariable function, then, is given by:\n\n$$V \\approx \\sum_{i=1}^m \\sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \\Delta A$$\n$$V=\\lim_{m,n\\to\\infty}\\sum_{i=1}^m \\sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \\Delta A $$\n$$V=\\iint_R f(x,y)\\,dA$$\n\n\\subsection{Double Integral Properties}\n\nDouble integrals have many of the same properties as single integrals:\n\n$$1. \\iint_R [f(x,y)+g(x,y)]\\,dA=\\iint_R f(x,y)\\,dA + \\iint_R g(x,y)\\,dA$$\n\nIf $c$ is a constant:\n\n$$\\iint_R cf(x,y)\\,dA=c\\iint_R f(x,y)\\,dA$$\n\nIf $f(x,y)\\geq g(x,y)$ for all $(x,y)$, then:\n\n$$\\iint_R f(x,y)\\,dA \\geq \\iint_R g(x,y)\\,dA$$\n\nIf $g(x)$ is a function of $x$ and $h(y)$ is a function of $y$:\n\n$$\\iint_R g(x)h(y) dA=\\int_a^b g(x)\\,dx\\int_c^d h(y)\\,dy$$\n\n\\subsection{Iterated Integrals}\n\nThe double integral, $\\iint_R f(x,y)\\,dA$, may be broken up in order to be calculated with ease:\n\n$$\\iint_R f(x,y) dA=\\int_a^b \\int_c^d f(x,y)\\,dy\\,dx$$\n\nThis may be solved using the method of partial integrals, where, much like partial derivatives, only one of the variables is treated as a variable at once. As a result of this, just like with partial derivatives, the order that the operation is done in yields the same result no matter what:\n\n$$\\int_a^b \\int_c^d f(x,y)\\,dy\\,dx=\\int_c^d\\int_a^b f(x,y)\\,dx\\,dy$$\n\n\\subsection{Fubini's Theorem}\n\nIf $f(x,y)$ is continuous on $R=\\{(x,y)|a\\leq x\\leq b, c\\leq y\\leq d\\}$\n\n$$\\iint_R f(x,y) dA=\\int_a^b \\int_c^d f(x,y)\\,dy\\,dx=\\int_c^d\\int_a^b f(x,y)\\,dx\\,dy$$\n\n\\end{document}\n", "meta": {"hexsha": "d1642c8201421e0e3019963029b2f5fceda95353", "size": 3157, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture Notes/Lecture14.tex", "max_stars_repo_name": "MDBrodskiy/Vector_Calculus", "max_stars_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T15:51:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T15:51:52.000Z", "max_issues_repo_path": "Lecture Notes/Lecture14.tex", "max_issues_repo_name": "MDBrodskiy/Vector_Calculus", "max_issues_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture Notes/Lecture14.tex", "max_forks_repo_name": "MDBrodskiy/Vector_Calculus", "max_forks_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.7830188679, "max_line_length": 291, "alphanum_fraction": 0.5929680076, "num_tokens": 1073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125848754472, "lm_q2_score": 0.8499711699569787, "lm_q1q2_score": 0.6391890165889557}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{exercises}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 2.2 Covariant derivative of $v_{ab}$}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n \\nabla{#}::Derivative.\n \\partial{#}::PartialDerivative.\n\n # template for covariant derivative of a vector\n\n derivU := \\nabla_{a}{A?^{b}} -> \\partial_{a}{A?^{b}} + \\Gamma^{b}_{c a} A?^{c}.\n derivD := \\nabla_{a}{A?_{b}} -> \\partial_{a}{A?_{b}} - \\Gamma^{c}_{b a} A?_{c}.\n\n vab := v_{a b} -> A_{a} B_{b}.\n iab := A_{a} B_{b} -> v_{a b}.\n\n pab := \\partial_{a}{A_{b}} B_{c} -> \\partial_{a}{A_{b} B_{c}} - A_{b} \\partial_{a}{B_{c}}.\n\n # create an object\n\n Dvab := \\nabla_{a}{v_{b c}}. # cdb (ex-0202.101,Dvab)\n\n # apply the rule, then simplify\n\n substitute (Dvab,vab) # cdb (ex-0202.102,Dvab)\n product_rule (Dvab) # cdb (ex-0202.103,Dvab)\n substitute (Dvab,derivD) # cdb (ex-0202.104,Dvab)\n substitute (Dvab,derivU) # cdb (ex-0202.105,Dvab)\n distribute (Dvab) # cdb (ex-0202.106,Dvab)\n substitute (Dvab,pab) # cdb (ex-0202.107,Dvab)\n canonicalise (Dvab) # cdb (ex-0202.108,Dvab)\n substitute (Dvab,iab) # cdb (ex-0202.109,Dvab)\n sort_product (Dvab) # cdb (ex-0202.110,Dvab)\n\\end{cadabra}\n\n\\begin{align}\n \\cdb{ex-0202.101} &= \\Cdb{ex-0202.102}\\\\\n &= \\Cdb{ex-0202.103}\\\\\n &= \\Cdb{ex-0202.104}\\\\\n &= \\Cdb{ex-0202.105}\\\\\n &= \\Cdb{ex-0202.106}\\\\\n &= \\Cdb{ex-0202.107}\\\\\n &= \\Cdb{ex-0202.108}\\\\\n &= \\Cdb{ex-0202.109}\\\\\n &= \\Cdb{ex-0202.110}\n\\end{align}\n\n\\end{document}\n", "meta": {"hexsha": "45938b399f6a94007664b248efaa214756e6fe5a", "size": 1931, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/exercises/ex-0202.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/exercises/ex-0202.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/exercises/ex-0202.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 33.8771929825, "max_line_length": 94, "alphanum_fraction": 0.4899016054, "num_tokens": 722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711794579723, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6391890142858403}} {"text": "\\section{Lines}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Find the vector, parametric, and symmetric equations of a line.\n \\item Determine whether a point is on a given line.\n \\item Determine whether two lines intersect.\n \\item Find the angle between two lines.\n \\item Find the projection of a point onto a line.\n \\end{enumerate}\n\\end{outcome}\n\nWe can use the concept of vectors and points to find equations for\nlines in $\\R^n$. Consider a straight line $L$ that passes through a\npoint $P$ in the direction given by a non-zero vector $\\vect{d}$.\n\\begin{center}\n \\begin{tikzpicture}[rotate=10]\n % Note: I deliberately made the red a bit lighter and the blue a bit\n % darker, so that it will also look okay in black-and-white.\n \\draw[red!80](-4,0) -- node [above=3pt, pos=0.875] {$L$} (9,0);\n \\draw[->,thick,blue!80!black](0,0) -- node[above left] {$\\vect{d}$} (2,0);\n \\fill (0,0) circle [radius=2.2pt] node [above=3pt] {$P$};\n \\fill (5,0) circle [radius=2.2pt] node [above=3pt] {$Q$};\n \\end{tikzpicture}\n\\end{center}\nThe line $L$ is infinitely long in both directions, although the\npicture only shows a finite part of it. To find an equation for this\nline, first suppose that $Q$ is an arbitrary point on $L$. Then the\nvector $\\longvect{PQ}$ is parallel to $\\vect{d}$. In other words,\nthere exists some real number $t$ such that\n\\begin{equation*}\n \\longvect{PQ} = t\\,\\vect{d}.\n\\end{equation*}\nIf $\\vect{p}$ is the position vector of $P$ and $\\vect{q}$ is the\nposition vector of $Q$, we can write\n\\begin{equation*}\n \\longvect{PQ} = \\vect{q}-\\vect{p}.\n\\end{equation*}\nPutting together the last two equations, we get $\\vect{q}-\\vect{p} =\nt\\,\\vect{d}$, which we can write as\n\\begin{equation*}\n \\vect{q} = \\vect{p} + t\\,\\vect{d}.\n\\end{equation*}\nThis is called the \\textbf{vector equation}%\n\\index{vector equation!of a line}%\n\\index{line!vector equation} of the line $L$. The vector $\\vect{d}$ is\ncalled the \\textbf{direction vector}%\n\\index{direction vector}%\n\\index{vector!direction vector}, and $t$ is called a\n\\textbf{parameter}%\n\\index{parameter}. The parameter $t$ can be any real number; each time\nwe plug in a different number for $t$, we get a different point $Q$ on\nthe line. The following picture shows the effect of the parameter:\n\\begin{center}\n \\begin{tikzpicture}[rotate=10]\n % Note: I deliberately made the red a bit lighter and the blue a bit\n % darker, so that it will also look okay in black-and-white.\n \\draw[red!80](-4,0) -- node [above=3pt, pos=0.92] {$L$} (10,0);\n \\draw[->,thick,blue!80!black](0,0) -- node[above=3pt] {$\\vect{d}$} (2,0);\n \\draw[->,thick,blue!80!black](3,-2) -- node[below=3pt] {$\\vect{p}$} (0,0);\n \\fill (0,0) circle [radius=2.2pt] node [above=3pt] {$P$};\n \\fill (-2,0) circle [radius=2.2pt] node [below=9pt] {$t=-1$};\n \\fill (0,0) circle [radius=2.2pt] node [below=10pt] {$t=0$};\n \\fill (2,0) circle [radius=2.2pt] node [below=10pt] {$t=1$};\n \\fill (4,0) circle [radius=2.2pt] node [below=10pt] {$t=2$};\n \\fill (6,0) circle [radius=2.2pt] node [below=10pt] {$t=3$};\n \\fill (8,0) circle [radius=2.2pt] node [below=10pt] {$t=3$};\n \\fill (3,-2) circle [radius=2.2pt] node [right=3pt] {$0$};\n \\end{tikzpicture}\n\\end{center}\nThe following definition summarizes the above.\n\n\\begin{definition}{Vector equation of a line}{vector-equation-of-line}\n Let $\\vect{p}$ be a vector and $\\vect{d}$ a non-zero vector. Then\n \\begin{equation*}\n \\vect{q} = \\vect{p} + t\\,\\vect{d}\n \\end{equation*}\n is the \\textbf{vector equation}%\n \\index{vector equation!of a line}%\n \\index{line!vector equation} of a straight line $L$. Specifically,\n as the parameter $t$ ranges over the real numbers, $\\vect{q}$ ranges\n over the position vectors of all the points $Q$ on the line $L$.\n The vector $\\vect{d}$ is called the \\textbf{direction vector}%\n \\index{direction vector}%\n \\index{vector!direction vector} of the line.\n\\end{definition}\n\n\\begin{example}{A line from a point and a direction vector}{line-point-and-direction-vector}\n Find a vector equation for the line which contains the point\n $P = (2,0,3)$ and has direction vector\n $\\vect{d} = \\mat{1,2,1}^T$.\n\\end{example}\n\n\\begin{solution}\n The position vector of the point $P$ is\n $\\vect{p}=\\mat{2,0,3}^T$. The equation of the line is\n $\\vect{q}= \\vect{p} + t\\,\\vect{d}$, which we can write as\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\n\\begin{example}{A line from two points}{line-from-two-points}\n Find a vector equation for the line through the points\n $P = (1,2,0,1)$ and $R = (2,-4,6,3)$.\n\\end{example}\n\n\\begin{solution}\n We can use $P$ as the base point; its position vector is\n $\\vect{p}=\\mat{1,2,0,1}^T$. We can use\n $\\vect{d}=\\longvect{PR}=\\mat{1,-6,6,2}$ as the direction vector. Then\n a vector equation of the line is\n $\\vect{q} = \\vect{p} + t\\,\\vect{d}$,\n which we can also write as\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\\\ w \\end{mymatrix}\n = \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 0 \\\\ 1\\end{mymatrix}\n + t \\begin{mymatrix}{r} 1 \\\\ -6 \\\\ 6 \\\\ 2\\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\nWhen we write a vector equation in the form\n\\begin{equation*}\n \\begin{mymatrix}{c} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{mymatrix}\n = \\begin{mymatrix}{c} p_1 \\\\ p_2 \\\\ \\vdots \\\\ p_n \\end{mymatrix}\n + t \\begin{mymatrix}{r} d_1 \\\\ d_2 \\\\ \\vdots \\\\ d_n \\end{mymatrix},\n\\end{equation*}\nit is also called the \\textbf{component form}%\n\\index{vector equation!of a line!component form}%\n\\index{component form!line}%\n\\index{line!component form} of the vector equation.\n\nNotice that the vector equation of a line is not unique. In fact,\nthere are infinitely many vector equations for the same line. For\nexample, we can replace the parameter $t$ with another parameter, say\n$3s$ or $1-r$.\n\n\\begin{example}{Change of parameter}{change-of-parameter}\n Consider the vector equation from\n Example~\\ref{exa:line-point-and-direction-vector},\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}.\n \\end{equation*}\n Find two other equations for the same line, by changing the\n parameter%\n \\index{line!change of parameter}%\n \\index{change of parameters!line} to $3s$ and to $1-r$.\n\\end{example}\n\n\\begin{solution}\n If we let $t=3s$, we get\n \\begin{equation*}\n \\begin{array}{rcl}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n &=& \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + 3s \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\\\\\\\\[-1ex]\n &=& \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + s \\begin{mymatrix}{c} 3 \\\\ 6 \\\\ 3 \\end{mymatrix}.\n \\end{array}\n \\end{equation*}\n If we let $t=1-r$, we get\n \\begin{equation*}\n \\begin{array}{rcl}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n &=& \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + (1-r) \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\\\\\\\\[-1ex]\n &=& \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 3 \\end{mymatrix}\n + \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\n - r \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\\\\\\\\[-1ex]\n &=& \\begin{mymatrix}{c} 3 \\\\ 2 \\\\ 4 \\end{mymatrix}\n + r \\begin{mymatrix}{c} -1 \\\\ -2 \\\\ -1 \\end{mymatrix}.\n \\end{array}\n \\end{equation*}\n\\end{solution}\n\n\\begin{definition}{Parametric equations of a line}{parametric-equations}\n A line with vector equation\n \\begin{equation*}\n \\begin{mymatrix}{c} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{mymatrix}\n = \\begin{mymatrix}{c} p_1 \\\\ p_2 \\\\ \\vdots \\\\ p_n \\end{mymatrix}\n + t \\begin{mymatrix}{r} d_1 \\\\ d_2 \\\\ \\vdots \\\\ d_n \\end{mymatrix}\n \\end{equation*}\n can also be written as a set of $n$ scalar equations:\n \\begin{equation*}\n \\begin{array}{c@{~}c@{~}c}\n x_1 &=& p_1 + t\\,d_1, \\\\\n x_2 &=& p_2 + t\\,d_2, \\\\\n &\\vdots& \\\\\n x_n &=& p_n + t\\,d_n,\n \\end{array}\n \\end{equation*}\n When written in this form, they are called the \\textbf{parametric\n equations}%\n \\index{parametric equations!of a line}%\n \\index{line!parametric equations} of the line.\n\\end{definition}\n\n\\begin{example}{Parametric equations}{parametric-equation}\n Find parametric equations for the line through the points\n $P = (1,2,0,1)$ and $R = (2,-4,6,3)$.\n\\end{example}\n\n\\begin{solution}\n This is a same line as in Example~\\ref{exa:line-from-two-points}. We\n can easily convert the vector equation\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\\\ w \\end{mymatrix}\n = \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 0 \\\\ 1\\end{mymatrix}\n + t \\begin{mymatrix}{r} 1 \\\\ -6 \\\\ 6 \\\\ 2\\end{mymatrix}\n \\end{equation*}\n to a set of parametric equations:\n \\begin{equation*}\n \\begin{array}{c@{~}c@{~}l}\n x &=& 1 + t, \\\\\n y &=& 2 - 6t, \\\\\n z &=& 6t, \\\\\n w &=& 1 + 2t.\n \\end{array}\n \\end{equation*}\n\\end{solution}\n\n\\begin{example}{Determine whether a point is on a line}{point-on-line}\n Determine whether the point $P=(5,8,4)$ is on the line $L$ given by\n the vector equation\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 2 \\\\ 3 \\\\ 1 \\end{mymatrix}.\n \\end{equation*}\n\\end{example}\n\n\\begin{solution}\n The point $P$ is on the line $L$ if and only if there exists some\n $t\\in\\R$ such that\n \\begin{equation*}\n \\begin{mymatrix}{c} 1 \\\\ 2 \\\\ 1 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 2 \\\\ 3 \\\\ 1 \\end{mymatrix}\n = \\begin{mymatrix}{c} 5 \\\\ 8 \\\\ 4 \\end{mymatrix}.\n \\end{equation*}\n Subtracting $\\mat{1,2,1}^T$ from both sides of the equation, this is\n equivalent to\n \\begin{equation*}\n t \\begin{mymatrix}{c} 2 \\\\ 3 \\\\ 1 \\end{mymatrix}\n = \\begin{mymatrix}{c} 4 \\\\ 6 \\\\ 3 \\end{mymatrix}.\n \\end{equation*}\n We can write this as a set of parametric equations:\n \\begin{equation*}\n \\begin{array}{c@{~}c@{~}l}\n 2t &=& 4, \\\\\n 3t &=& 6, \\\\\n t &=& 3.\n \\end{array}\n \\end{equation*}\n This is a system of three linear equations in one variable, and we\n quickly see that it is inconsistent. Therefore, the point $P$ does\n not lie on the line $L$.\n\\end{solution}\n\n\\begin{example}{Determine whether two lines intersect}{lines-intersect}\n Determine whether the lines\n \\begin{equation*}\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{c} 3 \\\\ 1 \\\\ 0 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 1 \\end{mymatrix}\n \\quad\\mbox{and}\\quad\n \\begin{mymatrix}{c} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{c} 1 \\\\ -1 \\\\ 5 \\end{mymatrix}\n + s \\begin{mymatrix}{c} 2 \\\\ 1 \\\\ -2 \\end{mymatrix}\n \\end{equation*}\n intersect. If yes, find the point of intersection.%\n \\index{intersection!of two lines}\n\\end{example}\n\n\\begin{solution}\n The two lines intersect if and only if there exist $t,s\\in\\R$ such\n that\n \\begin{equation*}\n \\begin{mymatrix}{c} 3 \\\\ 1 \\\\ 0 \\end{mymatrix}\n + t \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 1 \\end{mymatrix}\n = \\begin{mymatrix}{c} 1 \\\\ -1 \\\\ 5 \\end{mymatrix}\n + s \\begin{mymatrix}{c} 2 \\\\ 1 \\\\ -2 \\end{mymatrix}.\n \\end{equation*}\n Bringing $s$ to the left-hand side, and subtracting $\\mat{3,1,0}^T$\n from both sides of the equation, this is equivalent to\n \\begin{equation*}\n t \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 1 \\end{mymatrix}\n - s \\begin{mymatrix}{c} 2 \\\\ 1 \\\\ -2 \\end{mymatrix}\n = \\begin{mymatrix}{c} -2 \\\\ -2 \\\\ 5 \\end{mymatrix}.\n \\end{equation*}\n If we write this vector equation as a set of three parametric\n equations, it is a system of 3 linear equations in 2 variables. The\n augmented matrix of the system is\n \\begin{equation*}\n \\begin{mymatrix}{cc|c}\n 2 & -2 & -2 \\\\\n 0 & -1 & -2 \\\\\n 1 & 2 & 5 \\\\\n \\end{mymatrix}.\n \\end{equation*}\n This system has {\\rref}\n \\begin{equation*}\n \\begin{mymatrix}{cc|c}\n 1 & 0 & 1 \\\\\n 0 & 1 & 2 \\\\\n 0 & 0 & 0 \\\\\n \\end{mymatrix},\n \\end{equation*}\n and has the unique solution $t=1$ and $s=2$. Therefore, the lines\n intersect. (Other possible cases are: If the system is inconsistent,\n the lines do not intersect. If the system has more than one\n solution, the lines are identical). We find the point of\n intersection by plugging the parameter $t=1$ into the equation of\n the first line (or equivalently, but plugging $s=2$ into the\n equation of the second line - doing it both ways is a good way to\n double-check your answer). Therefore, the point of intersection is\n \\begin{equation*}\n \\begin{mymatrix}{c} 3 \\\\ 1 \\\\ 0 \\end{mymatrix}\n + 1 \\begin{mymatrix}{c} 2 \\\\ 0 \\\\ 1 \\end{mymatrix}\n = \\begin{mymatrix}{c} 5 \\\\ 1 \\\\ 1 \\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\nThere is one other form for a line which is useful, which is the\n\\textbf{symmetric form}. Consider the line given by\n\\begin{equation*}\n \\begin{array}{c@{~}c@{~}l}\n x &=& 1 + 2t, \\\\\n y &=& 1 - t, \\\\\n z &=& 3 + 2t. \\\\\n \\end{array}\n\\end{equation*}\nWe can solve each equation for $t$:\n\\begin{equation*}\n \\begin{array}{c@{~}c@{~}l}\n t &=& \\frac{x-1}{2}, \\\\\\\\[-2ex]\n t &=& \\frac{y-1}{-1}, \\\\\\\\[-2ex]\n t &=& \\frac{z-3}{2}.\n \\end{array}\n\\end{equation*}\nFinally, we can eliminate $t$ from the equations by setting all three\nequations equal to one another:\n\\begin{equation*}\n \\frac{x-1}{2} = \\frac{y-1}{-1} = \\frac{z-3}{2}.\n\\end{equation*}\nThe latter is really a system of 2 equations in 3 variables. This is\nthe \\textbf{symmetric form}%\n\\index{line!symmetric form equation}%\n\\index{symmetric form} of the equation of the line. In the following\nexample, we look at how to convert the equation of a line from\nsymmetric form to parametric form.\n\n\\begin{example}{Change symmetric form to parametric form}{symmetric-to-parametric}\n Consider the line whose equations are given in \\textbf{symmetric form} as\n \\begin{equation*}\n \\frac{x-2}{3}=\\frac{y-1}{2}=\\frac{z+3}{1}.\n \\end{equation*}\n Find parametric and vector equations for this line.\n\\end{example}\n\n\\begin{solution}\n We set all three quantities equal to $t$:\n \\begin{equation*}\n t=\\frac{x-2}{3}, \\quad\n t=\\frac{y-1}{2}, \\quad\n t=\\frac{z+3}{1}.\n \\end{equation*}\n Solving these equations for $x,y,z$ yields\n \\begin{equation*}\n \\begin{array}{c}\n x = 2 + 3t, \\\\\n y = 1 + 2t, \\\\\n z = -3 + t.\n \\end{array}\n \\end{equation*}\n These are the parametric equations for the line. The vector equation\n is\n \\begin{equation*}\n \\begin{mymatrix}{c}\n x \\\\\n y \\\\\n z\n \\end{mymatrix} =\n \\begin{mymatrix}{r}\n 2 \\\\\n 1 \\\\\n -3\n \\end{mymatrix}\n +\n t\n \\begin{mymatrix}{r}\n 3 \\\\\n 2 \\\\\n 1\n \\end{mymatrix}.\n \\end{equation*}\n\\end{solution}\n\nWe can use the dot product to find the angle between two intersecting\nlines. This is simply the smallest angle between (any of) their\ndirection vectors. The only subtlety here is that if $\\vect{u}$ is a\ndirection vector for a line, then so is $-\\vect{u}$, and thus we will\nfind pairs of complementary angles. We will take the smaller of the\ntwo angles.\n\\begin{center} \\begin{tikzpicture}\n \\filldraw[fill=green!20,draw=green!50!black] (0,0) -- (-25:8mm)\n arc (-25:25:8mm) -- cycle;\n \\filldraw[fill=yellow!20,draw=yellow!50!black] (0,0) -- (25:8mm)\n arc (25:155:8mm) -- cycle;\n \\node at (0:5mm) {$\\theta$};\n \\node at (90:4.2mm) {$\\pi-\\theta$};\n \\draw[red!80](25:-2.5) -- (25:2.5);\n \\draw[red!80](-25:-2.5) -- (-25:2.5);\n \\draw[->,thick,red!80!black](0,0) -- node[above,pos=0.6]{$\\vect{v}$} (25:2);\n \\draw[->,thick,blue!80!black](0,0) -- node[below,pos=0.3]{$\\vect{u}$} (-25:1.5);\n \\draw[->,thick,blue!80!black](0,0) -- node[below,pos=0.6]{$-\\vect{u}$} (155:1.5);\n \\end{tikzpicture}\n\\end{center}\n\n\\begin{example}{Find the angle between two lines}{angle-between-two-lines}\n Find the angle between the two lines%\n \\index{line!angle between}%\n \\index{angle!between two lines}\n \\begin{equation*}\n L_1: \\;\n \\begin{mymatrix}{r} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{r} 1 \\\\ 2 \\\\ 0 \\end{mymatrix}\n + t \\begin{mymatrix}{r} -1 \\\\ 1 \\\\ 2 \\end{mymatrix}\n \\end{equation*}\n and\n \\begin{equation*}\n L_2: \\;\n \\begin{mymatrix}{r} x \\\\ y \\\\ z \\end{mymatrix}\n = \\begin{mymatrix}{r} 0 \\\\ 3 \\\\ 2 \\end{mymatrix}\n + s \\begin{mymatrix}{r} 2 \\\\ 1 \\\\ -1 \\end{mymatrix}.\n \\end{equation*}\n\\end{example}\n\n\\begin{solution}\n The direction vectors are\n \\begin{equation*}\n \\vect{u}=\\begin{mymatrix}{r} -1 \\\\ 1 \\\\ 2 \\end{mymatrix}\n \\quad\\mbox{and}\\quad\n \\vect{v}=\\begin{mymatrix}{r} 2 \\\\ 1 \\\\ -1 \\end{mymatrix}.\n \\end{equation*}\n The answer is given by\n \\begin{equation*}\n \\cos\\theta =\n \\frac{\\vect{u}\\dotprod\\vect{v}}{\\norm{\\vect{u}}\\norm{\\vect{v}}} = -\\frac{1}{2},\n \\end{equation*}\n which gives $\\theta=\\frac{2\\pi}{3}$. Now the angles between any two\n direction vectors for these lines will either be $\\frac{2 \\pi}{3}$ or\n its complement $\\phi = \\pi - \\frac{2 \\pi}{3} = \\frac{\\pi}{3}$. We\n choose the smaller angle, and therefore conclude that the angle\n between the two lines is $\\frac{\\pi}{3}$.\n\\end{solution}\n\nFinally, we will show how to use projections to find the shortest\ndistance from a point to a line.\n\n\\begin{example}{Shortest distance from a point to a line}{shortest-point-line}\n Let $L$ be the line which goes through the point $P = (0,4,-2)$ with\n direction vector\n $\\vect{d} = \\begin{mysmallmatrix}{r} 2 \\\\ 1 \\\\\n 2 \\end{mysmallmatrix}$, and let $Q=(1,3,5)$. Find the shortest\n distance%\n \\index{distance!point to line} from $Q$ to the line $L$, and find\n the point $R$ on $L$ that is closest to $Q$.%\n \\index{projection!point to line}\n \\begin{center}\n \\begin{tikzpicture}[rotate=-20]\n \\draw[red!80](-1.5,0) -- node[above, pos=0.9]{$L$} (4.5,0);\n \\draw[->,thick,blue!80!black](0,0) -- node[above right, pos=0.4]{$\\vect{d}$} (2,0);\n \\draw[dashed] (3,1.5) -- (3,0);\n \\draw[fill](0,0) circle [radius=2.25pt] node[below=1ex]{$P$};\n \\draw[fill](3,1.5) circle [radius=2.25pt] node[right=0.5ex]{$Q$};\n \\draw[fill](3,0) circle [radius=2.25pt] node[below=1ex]{$R$};\n \\end{tikzpicture}\n \\end{center}\n\\end{example}\n\n\\begin{solution}\n In order to determine the shortest distance from $Q$ to $L$, we will\n first find the vector $\\longvect{PQ}$ and then find the projection\n of this vector onto $L$. The vector $\\longvect{PQ}$ is given by\n \\begin{equation*}\n \\longvect{PQ}=\n \\begin{mymatrix}{r} 1 \\\\ 3 \\\\ 5 \\end{mymatrix}\n - \\begin{mymatrix}{r} 0 \\\\ 4 \\\\ -2 \\end{mymatrix}\n = \\begin{mymatrix}{r} 1 \\\\ -1 \\\\ 7 \\end{mymatrix}.\n \\end{equation*}\n Then, if $R$ is the point on $L$ closest to $Q$, it follows that\n \\begin{equation*}\n \\longvect{PR}\n ~=~ \\proj_{\\vect{d}}\\longvect{PQ} \\\\\n ~=~ \\frac{\\vect{d}\\dotprod\\longvect{PQ}}{\\norm{\\vect{d}}^2}\\,\\vect{d} \\\\\n ~=~ \\frac{15}{9} \\begin{mymatrix}{r} 2 \\\\ 1 \\\\ 2 \\end{mymatrix} \\\\\n ~=~ \\frac{5}{3} \\begin{mymatrix}{r} 2 \\\\ 1 \\\\ 2 \\end{mymatrix}.\n \\end{equation*}\n Now, the distance from $Q$ to $L$ is given by\n \\begin{equation*}\n \\norm{\\longvect{RQ}} = \\norm{\\longvect{PQ} - \\longvect{PR}}\n = \\sqrt{26}.\n \\end{equation*}\n The point $R$ is found by adding the vector $\\longvect{PR}$ to the\n position vector $\\longvect{0P}$ for $P$ as follows\n \\begin{equation*}\n \\begin{mymatrix}{r} 0 \\\\ 4 \\\\ -2 \\end{mymatrix}\n + \\frac{5}{3} \\begin{mymatrix}{r} 2 \\\\ 1 \\\\ 2 \\end{mymatrix}\n ~=~\n \\begin{mymatrix}{c}\n 10/3 \\\\\n 17/3 \\\\\n 4/3\n \\end{mymatrix}.\n \\end{equation*}\n Therefore, $R = (\\frac{10}{3}, \\frac{17}{3}, \\frac{4}{3})$.\n\\end{solution}\n", "meta": {"hexsha": "c44e2429111efa5d459612521819e650a0e75e34", "size": 19655, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "baseText/content/LinesAndPlanes-Lines.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "baseText/content/LinesAndPlanes-Lines.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "baseText/content/LinesAndPlanes-Lines.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 37.653256705, "max_line_length": 92, "alphanum_fraction": 0.62284406, "num_tokens": 7385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.734119526900183, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.6391224560369261}} {"text": "\\chapter{Probability Models}\nWrite something here.\n\\newpage\n\\section{Multivariable Gaussian Distributions}\n\\subsection{Product of Experts}\n\\[ p(x) = \\frac{1}{Z}\\prod_{m=1}^M p_m(x) \\]\nIs (informally) equivalent to the intersection of Gaussian distributions.\n\\begin{remark}\n Product of Experts is beyond the scope of this course.\n\\end{remark}\n\\subsection{Mixture of Experts}\nWeighted sum of multiple distributions.\n\\[ p(x) = \\sum_{m=1}^M c_m p_m(x) \\]\n\\begin{remark}\n Weighted sum of Gaussians is not necessarily a Gaussian. So mixture of Gaussians can be used to ``fit'' different distributions.\n\\end{remark}\n\n\\section{Gaussian Mixture Model}\n\\subsection{Sampling from a GMM}\n\\begin{enumerate}\n \\item Sample $c_m$ from $c_1,c_2,\\dots,c_M$\n \\item Sample data from $\\mathcal{N}(x|\\mu_m,\\Sigma_m)$\n\\end{enumerate}\n\n\\section{Some Information Theory}\n\\textbf{Information} contains \\textbf{uncertainty}. The more uncertain something is, the more information is needed to eliminate its uncertainty.\n\n\\begin{definition}\n The information in discrete variable is\n \\[ I(x) = -\\log_2 P(x) \\]\n\\end{definition}\n\\begin{definition}[Entropy]\n The \n \\[ H = \\mathbb{E}[-\\log_2 P(x)] = -\\sum_{x\\in X}P(x)\\log_2 P(x) \\]\n \\[ H = \\mathbb{E}[I(x) = -\\int_{-\\infty}^{\\infty}p(x)\\ln(p(x))\\mathrm{d}x] \\]\n\\end{definition}\n\\begin{remark}\n \\begin{itemize}\n \\item Entropy is a function of distribution $p(x)$.\n \\item Is the expectation of information of random variables.\n \\end{itemize}\n\\end{remark}\n\n\\subsection{Entropy of common distriutions}\n\n\\subsection{Conditional Entropy}\n\\begin{definition}\n \\[ H[y|x] = \\sum_{x'}P(x)H(y|x=x') = -\\sum_{x}P(x)\\sum_yP(y|x)\\ln P(TooFast) \\]\n\\end{definition}", "meta": {"hexsha": "962e7b233250b16a7173293473e56294e8bf6327", "size": 1715, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Speech Recognition/Probability.tex", "max_stars_repo_name": "YBRua/CourseNotes", "max_stars_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-03-20T10:40:58.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-25T08:15:15.000Z", "max_issues_repo_path": "Speech Recognition/Probability.tex", "max_issues_repo_name": "YBRua/CourseNotes", "max_issues_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Speech Recognition/Probability.tex", "max_forks_repo_name": "YBRua/CourseNotes", "max_forks_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T11:31:00.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T11:31:00.000Z", "avg_line_length": 35.0, "max_line_length": 145, "alphanum_fraction": 0.6979591837, "num_tokens": 531, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972549785203, "lm_q2_score": 0.7341195327172402, "lm_q1q2_score": 0.6391224500097433}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\n\\usepackage{fullpage}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\n\\title{Numerical Methods for Solving Navier-Stokes Equations}\n\\author{Nikolas Tselepidis}\n\\date{}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Finite Pr Equations}\n\n\\paragraph{Conservation of momentum (dimensionless)}\n\n\\begin{equation}\n\\frac{1}{Pr} \\left( \\frac{\\partial \\vec{v}}{\\partial t} + \\vec{v} \\cdot \\nabla \\vec{v} \\right) =\n - \\nabla P + \\nabla^2 \\vec{v} + Ra T \\hat{y}\n\\end{equation}\n\nTaking the curl of equation (1), and substituting the vorticity $\\vec{\\omega} = \\nabla \\times \\vec{v}$, it follows that:\n\n\\begin{equation}\n \\frac{1}{Pr} \\left( \\frac{\\partial \\omega}{\\partial t} + v_x \\frac{\\partial \\omega}{\\partial x} + v_y \\frac{\\partial \\omega}{\\partial y} \\right) =\n \\nabla^2 \\omega - Ra \\frac{\\partial T}{\\partial x}\n\\end{equation}\n\nIn 2D only one component of vorticity is needed, i.e. the one perpedicular to the 2D plane: $\\nabla^2 \\psi = \\omega_z$.\n\n\\paragraph{Conservation of mass (continuity)}\n\\begin{equation}\n \\frac{\\partial T}{\\partial t} + \\vec{v} \\cdot \\nabla T = \\kappa \\nabla^2 T\n\\end{equation}\n\n\\paragraph{Conservation of energy}\n\\begin{equation}\n \\nabla \\cdot \\vec{v} = 0\n\\end{equation}\n\n\n\\section{Solve}\n\n\\begin{equation}\n \\nabla^2 \\psi = \\omega\n\\end{equation}\n\n\\begin{equation}\n \\left( v_x, v_y \\right) = \\left( \\frac{\\partial \\psi}{\\partial y}, -\\frac{\\partial \\psi}{\\partial x} \\right)\n\\end{equation}\n\n\\begin{equation}\n \\frac{\\partial T}{\\partial t} = -v_x \\frac{\\partial T}{\\partial x} -v_y \\frac{\\partial T}{\\partial y} + \\kappa \\nabla^2 T\n\\end{equation}\n\n\\begin{equation}\n \\frac{\\partial \\omega}{\\partial t} = -v_x \\frac{\\partial \\omega}{\\partial x} -v_y \\frac{\\partial \\omega}{\\partial y} + Pr \\nabla^2 \\omega - Ra Pr \\frac{\\partial T}{\\partial x}\n\\end{equation}\n\n\\end{document}\n", "meta": {"hexsha": "09dd9c39af7c7379d5311eb7f72db1963b939324", "size": 1862, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doc.tex", "max_stars_repo_name": "ntselepidis/FDM-Navier-Stokes", "max_stars_repo_head_hexsha": "fbb9a1741b1d8e3e550db0c2e756d25780c48ed6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-21T15:16:04.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-21T15:16:04.000Z", "max_issues_repo_path": "doc/doc.tex", "max_issues_repo_name": "ntselepidis/FDM-Navier-Stokes", "max_issues_repo_head_hexsha": "fbb9a1741b1d8e3e550db0c2e756d25780c48ed6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/doc.tex", "max_forks_repo_name": "ntselepidis/FDM-Navier-Stokes", "max_forks_repo_head_hexsha": "fbb9a1741b1d8e3e550db0c2e756d25780c48ed6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.09375, "max_line_length": 179, "alphanum_fraction": 0.6761546724, "num_tokens": 672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972583359805, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6391224423458924}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{examples}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% ============================================================================================\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n import cdblib\n checkpoint_file = 'tests/semantic/output/example-11.json'\n cdblib.create (checkpoint_file)\n checkpoint = []\n\\end{cadabra}\n\\egroup\n\n\\clearpage\n\n% ============================================================================================\n\\section*{Example 11 The RNC connection.}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,u#}::Indices(position=independent).\n\n D{#}::PartialDerivative.\n \\nabla{#}::Derivative.\n\n g_{a b}::Metric.\n g^{a b}::InverseMetric.\n g^{a b}::Weight(label=gnum,value=1).\n\n \\delta{#}::KroneckerDelta.\n\n R_{a b c d}::RiemannTensor.\n R_{a b c d}::Depends(\\nabla{#}).\n\n x^{a}::Depends(D{#}).\n x^{a}::Weight(label=xnum,value=1).\n\n Dx := D_{a}{x^{b}} -> \\delta^{b}_{a}. # cdb (ex-11.000,Dx)\n\n gab := g_{a b} -> g_{a b}\n - (1/3) x^{c} x^{d} R_{a c b d}\n - (1/6) x^{c} x^{d} x^{e} \\nabla_{c}{R_{a d b e}}\n + (1/180) x^{c} x^{d} x^{e} x^{f} ( 8 g^{g h} R_{a c d g} R_{b e f h}\n -9 \\nabla_{c d}{R_{a e b f}} ). # cdb (ex-11.001,gab)\n\n iab := g^{a b} -> g^{a b}\n + (1/3) x^{c} x^{d} g^{a e} g^{b f} R_{c e d f}\n + (1/6) x^{c} x^{d} x^{e} g^{a f} g^{b g} \\nabla_{c}{R_{d f e g}}\n + (1/60) x^{c} x^{d} x^{e} x^{f} g^{a g} g^{b h}\n ( 4 g^{i j} R_{c g d i} R_{e h f j}\n +3 \\nabla_{c d}{R_{e g f h}} ). # cdb(ex-11.002,iab)\n\n distribute (gab)\n distribute (iab)\n\n ChrSym := \\Gamma^{a}_{b c} -> 1/2 g^{a d} ( D_{b}{g_{d c}}\n + D_{c}{g_{b d}}\n - D_{d}{g_{b c}} ). # cdb (ex-11.003,ChrSym)\n\n Gamma := \\Gamma^{a}_{b c}. # cdb (ex-11.100,Gamma)\n\n substitute (Gamma,ChrSym) # cdb (ex-11.101,Gamma)\n substitute (Gamma,gab) # cdb (ex-11.102,Gamma)\n substitute (Gamma,iab) # cdb (ex-11.103,Gamma)\n distribute (Gamma) # cdb (ex-11.104,Gamma)\n unwrap (Gamma) # cdb (ex-11.105,Gamma)\n product_rule (Gamma) # cdb (ex-11.106,Gamma)\n distribute (Gamma) # cdb (ex-11.107,Gamma)\n substitute (Gamma,Dx) # cdb (ex-11.108,Gamma)\n eliminate_kronecker (Gamma) # cdb (ex-11.109,Gamma)\n\n def truncate (obj,n):\n\n ans = Ex(\"0\") # create a Cadabra object with value zero\n\n for i in range (0,n+1):\n foo := @(obj).\n bah = Ex(\"xnum = \" + str(i))\n distribute (foo)\n keep_weight (foo, bah)\n ans = ans + foo\n\n return ans\n\n checkpoint.append (Gamma)\n\n # sort_product (Gamma) # 52.3 sec, 49 Mbyte\n # rename_dummies (Gamma) # 58.6 sec, 51 Mbyte\n # canonicalise (Gamma) # killed after 20 mins and over 500 Mbyte\n\n Gamma = truncate (Gamma,3) # cdb (ex-11.110,Gamma) # allow up to 3rd order in x^a\n\n sort_product (Gamma)\n rename_dummies (Gamma)\n canonicalise (Gamma)\n\n checkpoint.append (Gamma)\n\n # ==========================================================================\n # the remaining code is just for pretty printing\n\n def product_sort (obj):\n substitute (obj,$ g^{a b} -> A001^{a b} $)\n substitute (obj,$ x^{a} -> A002^{a} $)\n substitute (obj,$ z^{a} -> A003^{a} $)\n substitute (obj,$ R_{a b c d} -> A004_{a b c d} $)\n substitute (obj,$ \\nabla_{e}{R_{a b c d}} -> A005_{a b c d e} $)\n substitute (obj,$ \\nabla_{e f}{R_{a b c d}} -> A006_{a b c d e f} $)\n sort_sum (obj)\n sort_product (obj)\n rename_dummies (obj)\n substitute (obj,$ A001^{a b} -> g^{a b} $)\n substitute (obj,$ A002^{a} -> x^{a} $)\n substitute (obj,$ A003^{a} -> z^{a} $)\n substitute (obj,$ A004_{a b c d} -> R_{a b c d} $)\n substitute (obj,$ A005_{a b c d e} -> \\nabla_{e}{R_{a b c d}} $)\n substitute (obj,$ A006_{a b c d e f} -> \\nabla_{e f}{R_{a b c d}} $)\n\n def get_xterm (obj,n):\n\n foo := @(obj).\n bah = Ex(\"xnum = \" + str(n))\n distribute (foo)\n keep_weight (foo, bah)\n\n return foo\n\n def get_gterm (obj,n):\n\n foo := @(obj).\n bah = Ex(\"gnum = \" + str(n))\n distribute (foo)\n keep_weight (foo, bah)\n\n return foo\n\n def reformat (obj,scale):\n\n foo = Ex(str(scale))\n bah := @(foo) @(obj).\n\n distribute (bah)\n product_sort (bah)\n rename_dummies (bah)\n canonicalise (bah)\n factor_out (bah,$x^{a?},g^{b? c?}$)\n ans := @(bah) / @(foo).\n\n return ans\n\n gam1 = get_xterm (Gamma, 1) # cdb (ex-11.200,gam1)\n gam2 = get_xterm (Gamma, 2) # cdb (ex-11.201,gam2)\n gam3 = get_xterm (Gamma, 3) # cdb (ex-11.202,gam3)\n\n gam31 = get_gterm (gam3, 1) # cdb (ex-11.210,gam31)\n gam32 = get_gterm (gam3, 2) # cdb (ex-11.211,gam31)\n\n gam1 = reformat (gam1, 3) # cdb (ex-11.220,gam1)\n gam2 = reformat (gam2, 12) # cdb (ex-11.221,gam2)\n\n gam31 = reformat (gam31, 40) # cdb (ex-11.222,gam31)\n gam32 = reformat (gam32, 45) # cdb (ex-11.223,gam32)\n\n Gamma := @(gam1) + @(gam2) + @(gam31) + @(gam32). # cdb (ex-11.230,Gamma)\n Scaled := 360 @(Gamma). # cdb (ex-11.231,Scaled)\n\n checkpoint.append (Gamma)\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\Dmath*{\\cdb{ex-11.100}(x) = \\Cdb*[\\hskip 3.0cm]{ex-11.230}}\n\\end{dgroup*}\n\n\\begin{dgroup*}\n \\Dmath*{360 \\cdb{ex-11.100}(x) = \\Cdb*[\\hskip 3.0cm]{ex-11.231}}\n\\end{dgroup*}\n\n\\clearpage\n\n% ============================================================================================\n% export to json format\n\n% --------------------------------------------------------------------------------------------\n\\subsection*{Save $\\Gamma^{a}{}_{bc}$ for later use in Example 12.}\n\n\\begin{cadabra}\n jsonfile = 'example-11.json'\n cdblib.create (jsonfile)\n cdblib.put ('Gamma',Gamma,jsonfile)\n\\end{cadabra}\n\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n for i in range( len(checkpoint) ):\n cdblib.put ('check{:03d}'.format(i),checkpoint[i],checkpoint_file)\n\\end{cadabra}\n\\egroup\n\n\\end{document}\n", "meta": {"hexsha": "6e2da1dc9f7b5e4b8fdbdf98f6a1b7b598baebef", "size": 7066, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/example-11.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/example-11.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/example-11.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 33.9711538462, "max_line_length": 112, "alphanum_fraction": 0.4363147467, "num_tokens": 2249, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972583359806, "lm_q2_score": 0.7341195152660688, "lm_q1q2_score": 0.6391224372815785}} {"text": "\n\n\n\nThe chapter presents two ways of doing hypothesis tests on whole\nmodels: (1) permutation tests where the connection is severed between\nthe explanatory and response variables, (2) tests such as \\ANOVA\\\nwhere the sampling distribution is calculated from first principles.\nIn practice, first-principle tests are used most of the time.\nStill, the permutation test is useful for developing intuition about\nhypothesis testing --- our main purpose here --- \nand for those not-so-rare occasions where the assumptions\nbehind the first-principle tests are dubious.\n\n\\subsection{The Permutation Test}\n\nThe idea of a permutation test is to enforce the null hypothesis that\nthere is no connection between the response variables and the\nexplanatory variables. An effective way to do this is to randomize\nthe response variable in a way that is consistent with sampling\nvariability. When constructing confidence intervals, the\n\\function{resampling} function was used. Re-sampling will typically\nrepeat some cases and omit others. Here, the \\function{shuffle}\nfunction will be used instead, to scramble the order of one or more\nvariables while leaving the others in their original state.\n\n\nTo illustrate, consider a model for exploring whether \\VN{sex} and\n\\VN{mother}'s height are related to the height of the child: \\datasetGalton\n\\begin{Schunk}\n\\begin{Sinput}\n> galton = fetchData(\"galton.csv\")\n> mod = lm(height ~ sex + mother, data=galton)\n> coefficients(mod)\n\\end{Sinput}\n\\begin{Soutput}\n(Intercept) sexM mother \n 41.450 5.177 0.353 \n\\end{Soutput}\n\\end{Schunk}\nThe coefficients indicate that typical males are taller than typical \nfemales by about 5 inches and that for each inch taller the mother is,\na child will typically be taller by 0.35 inches. \nA reasonable test statistic to summarize the whole model is $R^2$:\n\\begin{Schunk}\n\\begin{Sinput}\n> rsquared(mod)\n\\end{Sinput}\n\\begin{Soutput}\n[1] 0.562\n\\end{Soutput}\n\\end{Schunk}\n\nFor confidence intervals, re-sampling was applied to the entire data\nframe. This selects random cases, but each selected case is an\nauthentic one that matches exactly the original values for that case.\nThe point of re-sampling is to get an idea of the variability\nintroduced by random sampling of authentic cases.\n\\begin{Schunk}\n\\begin{Sinput}\n> do(5) * lm(height ~ sex + mother, data=resample(galton))\n\\end{Sinput}\n\\begin{Soutput}\n Intercept sexM mother sigma r-squared\n1 42.3 5.37 0.337 2.29 0.593\n2 42.1 4.93 0.344 2.34 0.550\n3 43.5 5.23 0.321 2.36 0.568\n4 37.9 5.26 0.409 2.34 0.579\n5 39.9 5.14 0.378 2.29 0.570\n\\end{Soutput}\n\\end{Schunk}\nThe \\indicatorVar{sex}{M} coefficients are tightly grouped near 5\ninches, the \\VN{mother} coefficients are around 0.3 to 0.4.\n\n\\index{P}{resample@\\texttt{resample}*}\n\\index{P}{shuffle@\\texttt{shuffle}*}\n\\index{P}{Hypothesis Testing!resample@\\texttt{resample}*}\n\\index{P}{Hypothesis Testing!shuffle@\\texttt{shuffle}*}\n\n\\index{P}{Resampling!resample@\\texttt{resample}*}\n\\index{P}{Resampling!shuffle@\\texttt{shuffle}*}\n\\index{C}{permutation test!computing}\n\nIn order to carry out a permutation test, do not randomize the\nwhole data frame. Instead, shuffle just the response variable:\n\\begin{Schunk}\n\\begin{Sinput}\n> do(5) * lm(shuffle(height) ~ sex + mother, data=galton)\n\\end{Sinput}\n\\begin{Soutput}\n Intercept sexM mother sigma r-squared\n1 63.7 -0.30663 0.0507 3.58 0.00300\n2 72.3 0.00676 -0.0859 3.58 0.00306\n3 70.8 0.00563 -0.0629 3.58 0.00164\n4 70.2 0.02282 -0.0533 3.58 0.00119\n5 68.3 -0.55719 -0.0192 3.58 0.00613\n\\end{Soutput}\n\\end{Schunk}\nNow the \\indicatorVar{sex}{M} and \\VN{mother} \ncoefficients are close to zero, as would be expected when\nthere is no relationship between the response variable and the\nexplanatory variables.\n\nIn constructing the sampling distribution under the null hypothesis,\nyou should do hundreds of trials of fitting the model to the scrambled\ndata, calculating the test statistic ($R^2$ here) for each trial. Note that each\ntrial itself involves all of the cases in your sample, but those cases\nhave been changed so that the shuffled variable almost certainly\ntakes on a different value in every case than in the original data.\n\\begin{Schunk}\n\\begin{Sinput}\n> nulltrials = do(500) * rsquared(\n lm(shuffle(height) ~ sex + mother, data=galton))\n\\end{Sinput}\n\\end{Schunk}\nNotice that the \\function{rsquared} operator has been used to calculate\nthe test statistic $R^2$ from the model. The output of \\function{do}\nis a data frame: \n\\begin{Schunk}\n\\begin{Sinput}\n> nulltrials\n\\end{Sinput}\n\\end{Schunk}\n\\begin{Schunk}\n\\begin{Soutput}\n var\n1 0.000920\n2 0.001854\n3 0.000533\n... for 500 cases altogether ...\n\\end{Soutput}\n\\end{Schunk}\nNaturally, all of the $R^2$ values for the trials are close to zero.\nAfter all, there is no\nrelation between the response variable (after randomization with\n\\function{shuffle}) and the explanatory variables.\n\nThe p-value can be calculated directly from the trials, by comparison\nto the observed value in the actual data: $R^2$ was 0.5618. \n\\begin{Schunk}\n\\begin{Sinput}\n> count(var > 0.5618, nulltrials)\n\\end{Sinput}\n\\begin{Soutput}\ncount.FALSE \n 500 \n\\end{Soutput}\n\\end{Schunk}\nNone of the 500 trials were greater than the value of the test statistic, 0.5618.\nIt wouldn't be fair to claim that $p=0$, since we only did 500 trials,\nbut it is reasonable to say that the permutation test shows the \np-value is $p \\leq 1/500$.\n\n\n\\subsection{First-Principle Tests}\n\n\\index{C}{regression report!hypothesis testing}\n\\index{C}{hypothesis testing!in regression report}\nOn modern computers, the permutation test is entirely practical. But\na few decades ago, it was not. Great creativity was applied to\nfinding test statistics where the sampling distribution could be\nestimated without extensive calculation. One of these is the F\nstatistic. This is still very useful today and is a standard part of\nthe regression report in many statistical packages. \n\n\\index{P}{summary@\\texttt{summary}!for linear models}\n\\index{P}{Hypothesis Testing!summary@\\texttt{summary}!for linear models}\n\nHere is the regression report from the\n\\model{\\VN{height}}{\\VN{sex}+\\VN{mother}}:\n\\begin{Schunk}\n\\begin{Sinput}\n> mod = lm( height ~ sex + mother, data=galton)\n> summary(mod)\n\\end{Sinput}\n\\begin{Soutput}\n...\n Estimate Std. Error t value Pr(>|t|) \n(Intercept) 41.4495 2.2095 18.8 <2e-16 ***\nsexM 5.1767 0.1587 32.6 <2e-16 ***\nmother 0.3531 0.0344 10.3 <2e-16 ***\n---\nSignif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 \n\nResidual standard error: 2.37 on 895 degrees of freedom\nMultiple R-squared: 0.562,\tAdjusted R-squared: 0.561 \nF-statistic: 574 on 2 and 895 DF, p-value: <2e-16 \n\\end{Soutput}\n\\end{Schunk}\nThe last line of the report\nshows an F statistic of 574 based on an $R^2$ of 0.562 and\ntranslates this to a p-value that is practically zero: \\code{<2e-16}.\n\nBy way of showing that the regression report is rooted in the same\napproach shown in the chapter, you can confirm the calculations.\nThere are $m=3$ coefficients and $n=898$ cases,\nproducing $n-m=895$ degrees of freedom in the denominator and $m-1=2$\ndegrees of freedom in the numerator. The calculation of the F\nstatistic from $R^2$ and the degrees of freedom follows the formula\ngiven in the chapter. \n$$ F = \\frac{R^2}{m-1} \\Biggm/ \\frac{1-R^2}{n-m} $$\nPlugging the values into the formula\n\\begin{Schunk}\n\\begin{Sinput}\n> (0.562/2) / ((1-.562)/895)\n\\end{Sinput}\n\\begin{Soutput}\n[1] 574\n\\end{Soutput}\n\\end{Schunk}\nF is the test statistic. To convert it to a p-value, you need to\ncalculate how extreme the value of F$=574.2$ is with reference to the\nF distribution with 895 and 2 degrees of freedom.\n\\begin{Schunk}\n\\begin{Sinput}\n> 1 - pf( 574.2, 2, 895)\n\\end{Sinput}\n\\begin{Soutput}\n[1] 0\n\\end{Soutput}\n\\end{Schunk}\n\\index{P}{pf@\\texttt{pf}}\n\\index{P}{Hypothesis Testing!pf@\\texttt{pf}}\n\\index{C}{p value!from F statistic}\nThe calculation of p-values from F always follows this form. In the\ncontext of the F distribution, ``extreme'' always means ``bigger\nthan.'' So, calculate the area under the F distribution to the right\nof the observed value.\n\n", "meta": {"hexsha": "b1e87f1529f03561582a5b03a3bd8375456808f6", "size": 8258, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ComputationalTechnique-Orig/WholeModels/computer-wholemodels.tex", "max_stars_repo_name": "dtkaplan/SM3", "max_stars_repo_head_hexsha": "56fef8d4368e7afa7ccce006d8f4acc6cf6c1fd1", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-04-01T01:28:07.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-01T01:28:07.000Z", "max_issues_repo_path": "ComputationalTechnique-Orig/WholeModels/computer-wholemodels.tex", "max_issues_repo_name": "BriannaBarry/SM3", "max_issues_repo_head_hexsha": "56fef8d4368e7afa7ccce006d8f4acc6cf6c1fd1", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ComputationalTechnique-Orig/WholeModels/computer-wholemodels.tex", "max_forks_repo_name": "BriannaBarry/SM3", "max_forks_repo_head_hexsha": "56fef8d4368e7afa7ccce006d8f4acc6cf6c1fd1", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-02-14T05:22:29.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-08T12:42:15.000Z", "avg_line_length": 35.9043478261, "max_line_length": 81, "alphanum_fraction": 0.7308064907, "num_tokens": 2538, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920068519378, "lm_q2_score": 0.7981867849406659, "lm_q1q2_score": 0.6391017786768378}} {"text": "\\documentclass{article}\n\\input{../../FrequentlyUsed/latex/mypackages}\n\\input{../../FrequentlyUsed/latex/mydefs}\n\n\\usepackage[makeroom]{cancel}\n\n\\title{Math for Taehoon}\n\\author{Sunghee Yun}\n\n\\begin{document}\n\\maketitle\n\n\\tableofcontents\n\n\\newpage\n\n\\section{27-SEP-2019}\n\\begin{itemize}\n\\item What is the smallest possible value of $n$ that makes $7! \\times n$ a perfect square?\n\n{\\bf Answer}: $35$\n\n\\begin{proof}\nSince\n\\begin{equation}\n7! = 2 \\times 3 \\times (2 \\times 2 ) \\times 5 \\times (2\\times 3) \\times 7\n= 2^4 \\times 3^2 \\times 5 \\times 7\n= (2^2 \\times 3)^2 \\times 5 \\times 7,\n\\end{equation}\nwe need at least one $5$ and one $7$ to make $7!\\times n$ a perfect square.\nTherefore the smallest possible value of $n$ that makes $7! \\times n$ a perfect square\nis $5\\times 7 = 35$.\nNote that $7! \\times 35 = (2^2 \\times 3 \\times 5 \\times 7)^2$.\n\n\\end{proof}\n\n\\item If there are 18 dragons with either $2$ or $3$ heads,\nand the total number of heads is $42$,\nhow many $2$-headed dragons are there? \n\n{\\bf Answer}: $12$\n\n\\begin{proof}\nLet $x$ be the number of $2$-headed dragons\nand\n$y$ be the number of $3$-headed dragons.\nSince there are $18$ dragons, we have the following equation:\n\\begin{equation}\n\\label{eq:dfihj}\nx + y = 18\n\\end{equation}\nAlso, the total number of heads is $42$, thus we have\n\\begin{equation}\n2 \\times x + 3 \\times y = 42\n\\end{equation}\n\nNow if we multiply $3$ to both the left-hand-side (LHS) and right-hand-side (RHS) of (\\ref{eq:dfihj}),\nwe obtain the following two equations:\n\\begin{eqnarray}\n3\\times x + 3\\times y &=& 3\\times(x+y) = 3\\times 18\n\\label{eq:oges}\n\\\\\n2 \\times x + 3 \\times y &=& 42\n\\label{eq:uydh}\n\\end{eqnarray}\n\nNow we subtract (\\ref{eq:uydh}) from (\\ref{eq:oges}), we get\n\\begin{equation}\nx = (3\\times x + 3\\times y) - (2 \\times x + 3 \\times) = 3 \\times 18 - 42 = 54 - 42 = 12.\n\\end{equation}\nSince we initially assumed that $x$ is the number of $2$-headed dragons,\nthe answer is $12$.\n\\end{proof}\n\n\n\\item Brenda and Sally run in opposite directions on a circular track,\nstarting at diametrically opposite points.\nThey first meet after Brenda has run $100$ meters.\nThey next meet after Sally has run $150$ meters past their first meeting point.\nEach girl runs at a constant speed.\nWhat is the length of the track in meters?\n\n{\\bf Answer}: $350$ meters.\n\n\\begin{proof}\n\\figurename~\\ref{fig:race} shows the track Brenda and Sally run.\n$X$ refers to the distance that Sally runs until she meets Brenda\nand $Y$ refers to the distance that Brenda runs past the first meeting point until she meets Sally again.\nNow since they run at constant speed,\nthe ratio of $100$ to $X$ and that of $Y$ to $150$ are the same,\n\\ie,\n\\begin{equation}\n\\label{eq:bysio}\n100 : X = Y : 150\n\\Leftrightarrow\n\\frac{100}{X} = \\frac{Y}{150}\n\\Leftrightarrow\nX \\times Y = 100 \\times 150.\n\\end{equation}\nNow we know that the sum of the distance that Brenda runs and the distance that Sally runs until they first meet\nis half the circumference of the track.\nWe also know that the sum of the distance that Brenda runs and the distance that Sally runs\nuntil they meets second time is the circumference of the track.\nIf we express this as an equation, we have\n\\begin{equation}\nY + 150 = 2 \\times (X + 100).\n\\end{equation}\nNow if we multiply $2$ to both side of this equation and use (\\ref{eq:bysio}),\nwe have\n\\begin{eqnarray*}\n&& X \\times Y + 150 \\times X = 2 \\times (X \\times X + 100 \\times X)\n\\\\\n&\\Leftrightarrow&\n100 \\times 150 + 150 \\times X = 2 \\times (X \\times X + 100 \\times X)\n\\\\\n&\\Leftrightarrow&\nX\\times X + 25 \\times X - 100 \\times 75 = 0\n\\\\\n&\\Leftrightarrow&\n(X - 75) \\times (X + 100) = 0.\n\\end{eqnarray*}\nTherefore $X=75$ and the circumference (or the length) of the track\nis $2\\times (X+100) = 350$.\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=.5\\linewidth]{figures/sally_and_brenda}\n\\caption{Race of Brenda and Sally. B and S stand for Brenda and Sally respectively.}\n\\label{fig:race}\n\\end{center}\n\\end{figure}\n\n\\end{proof}\n\n\\end{itemize}\n\n\n\n\\section{02-NOV-2019}\n\n\\begin{enumerate}\n\n\\item In this addition problme, \\ldots\\\\\n\n\\begin{tabular}{cccc}\n&$T$\t&$W$\t&$O$\n\\\\\n+\t&$T$\t&$W$\t&$O$\n\\\\\n\\hline\n$F$\t&$O$\t&$U$\t&$R$\n\\end{tabular}\n\n\\begin{solution}\n\nSince $O$ is an even number, the previous addition, \\ie, $W+W$ should not generate overflow.\nTherefore, $W$ should be less than $5$.\nThen $O=4$ and $F=1$ since $7+7=14$.\nSince $O=4$, $R=8$.\nNow the above equation becomes\\\\\n\n\\begin{tabular}{cccc}\n&$7$\t&$W$\t&$4$\n\\\\\n+\t&$7$\t&$W$\t&$4$\n\\\\\n\\hline\n$1$\t&$4$\t&$U$\t&$8$\n\\end{tabular}\n\n\\ \\\\\n\n\n\nNow since $O=4$ and $F=1$, the only possible values for $W$ are $2$ and $3$.\nBut if $W=2$, then $U=4$ and $U=O$. Therefore $W=3$.\nIn summary,\\\\\n\n\\begin{tabular}{cccc}\n&$7$\t&$3$\t&$4$\n\\\\\n+\t&$7$\t&$3$\t&$4$\n\\\\\n\\hline\n$1$\t&$4$\t&$6$\t&$8$\n\\end{tabular}\n\n\\end{solution}\n\n\\item Mr. Harman needs ...\n\n\\begin{description}\n\\item[Answer:] $(122+125+127)/2 = 187$.\n\\end{description}\n\n\\begin{solution}\nLet $X$, $Y$, and $Z$ be the weights of the three boxes. Then we have\n\\begin{eqnarray*}\nX + Y &=& 122\n\\\\\nY + Z &=& 125\n\\\\\nX + Z &=& 127\n\\end{eqnarray*}\n\nSince the sum of all three quantities in left-hand-side (LHS)\nis the same as the sum of all three quantities in right-hand-side (RHS),\nwe have\n\\begin{equation}\n2 \\times ( X + Y + Z ) = 122+125+127,\n\\end{equation}\nthus\n\\begin{equation}\nX + Y + Z = (122+125+127)/2 = 187.\n\\end{equation}\n\n\\end{solution}\n\n\n\\item The students in Mrs. Reed's English class are reading \\ldots\n\n\\begin{description}\n\\item[Answer:] $760 \\times \\frac{1.5}{1 + 1.5} = 760 \\times \\frac{3}{5}= 456.$\n\\end{description}\n\n\\begin{solution}\nNote that Chandra's reading speed is $1.5$ times bigger than that of Bob.\nIf we let $x$ be the number of pages that Bob should read,\nthen $1.5 \\times x$ is the number of pages that Chandra should read.\nThus we have $x + 1.5 \\times x = 2.5 \\times x = 760$ and\n\\begin{equation}\nx = \\frac{760}{2.5} = 760 \\times \\frac{2}{5}.\n\\end{equation}\nThus the number of pages Chandra should read is\n\\begin{equation}\nx \\times 1.5 = 760 \\times \\frac{\\cancel{2}}{5} \\times \\frac{3}{\\cancel{2}}\n= 152 \\times 3 = 456.\n\\end{equation}\n\\end{solution}\n\n\n\\item Landy drove \\ldots\\\\\n\n\\begin{description}\n\\item[Answer:]\n\\begin{equation}\n\\frac{10+20+30}{10/30 + 20/20 + 30/10} = 13.8 \\mbox{ miles/hour}\n\\end{equation}\n\\end{description}\n\n\n\\item The Incredible \\ldots\\\\\n\n\\begin{description}\n\\item[Answer:] The $11$th jump!\n\\end{description}\n\n\\begin{solution}\n\\begin{eqnarray*}\n1\\mbox{st jump} &=& 1\n\\\\\n2\\mbox{nd jump} &=& 2\n\\\\\n3\\mbox{rd jump} &=& 4\n\\\\\n4\\mbox{th jump} &=& 8\n\\\\\n5\\mbox{th jump} &=& 16\n\\\\\n6\\mbox{th jump} &=& 32\n\\\\\n7\\mbox{th jump} &=& 64\n\\\\\n8\\mbox{th jump} &=& 128\n\\\\\n9\\mbox{th jump} &=& 256\n\\\\\n10\\mbox{th jump} &=& 512\n\\\\\n11\\mbox{th jump} &=& 1024\n\\end{eqnarray*}\n\\end{solution}\n\n\n\\item The Amaco \\ldots\\\\\n\n\\begin{description}\n\\item[Answer:] $15 - 11 = 4$\n\\end{description}\n\n\\begin{solution}\nLet $X$ and $Y$ be the number of seventh graders and that of sixth graders\nrespectively who bought pencils.\nLet $P$ be the price of one pencil in cents.\nThen we have\n\\begin{eqnarray*}\nX \\times P &=& 143 = 11 \\times 13\n\\\\\nY \\times P &=& 195 = 15 \\times 13\n\\end{eqnarray*}\nThus $P$ is a common divisor of $143$ and $195$.\nThus $P$ is $1$ or $13$. However, if $P=1$, then $Y=195$. But $Y\\leq 30$, hence $P=13$.\nTherefore, $X=11$ and $Y=15$.\nThus $Y-X = 4$.\n\\end{solution}\n\n\\end{enumerate}\n\n\\begin{description}\n\n\\item[Q1] $88+86+91+92+87+90+89+93+92+88$?\n\n\\begin{description}\n\\item[Answer:] \n\\begin{eqnarray*}\n\\lefteqn{\n88+86+91+92+87+90+89+93+92+88}\n\\\\\n&=&\n(90 - 2)\n+ (90 - 4)\n+ (90 + 1)\n+ (90 + 2)\n+ (90 - 3)\n\\\\\n&&\n+ (90 + 0)\n+ (90 - 1)\n+ (90 + 3)\n+ (90 + 2)\n+ (90 - 2)\n\\\\\n&=&\n90 \\times 10\n+(- 2 - 4 + 1 + 2 - 3 + 0 - 1 + 3 + 2 - 2)\n\\\\\n&=&\n90 \\times 10 - 4 = 896.\n\\end{eqnarray*}\n\\end{description}\n\n\n\\item[Q2]\n\\begin{description}\n\\item[Answer:] $90$ inches.\n\\end{description}\n\n\n\\item[Q3]\n\\begin{description}\n\\item[Answer:] $4$.\n\\end{description}\n\n\n\\item[Q4]\n\\begin{description}\n\\item[Answer:] $3 \\times 8 \\times 2 = 48$.\n\\end{description}\n\n\n\\item[Q5]\n\\begin{description}\n\\item[Answer:] $1249$\n\\end{description}\n\n\n\\item[Q6]\n\\begin{description}\n\\item[Answer:] $56.25$\n\\begin{equation}\nx \\times \\frac{4}{5} \\times \\frac{4}{5} = 36\n\\Rightarrow x = 36 \\times \\frac{5}{4} \\times \\frac{5}{4} = 56.25\n\\end{equation}\n\\end{description}\n\n\n\\item[Q7]\n\\begin{description}\n\\item[Answer:] $2 \\times 2 / 2 = 2$\n\\end{description}\n\n\n\\item[Q8]\n\\begin{description}\n\\item[Answer:] Captain Hook found $30$ diamonds.\n\\begin{eqnarray*}\nH + P &=& 80\n\\\\\nP + S &=& 70\n\\\\\nH + S &=& 50\n\\end{eqnarray*}\nthus\n\\[\n2 \\times H = (H+P) + (H+S) - (P+S) = 80+50-70 = 60\n\\Rightarrow H = 30.\n\\]\n\\end{description}\n\n\n\\item[Q9]\n\\begin{description}\n\\item[Answer:] $15 + 10 \\times 2 = 35$.\n\\end{description}\n\n\n\\item[Q10]\n\\begin{description}\n\\item[Answer:] $25$ hours.\n\nSince the 1st pipe can fill the swimming pool $1.5$ times faster than the 2nd pipe,\nusing both pipes will be $2.5$ $(=1 + 1.5)$ times fater than using the 2nd pipe alone.\nThus using only 2nd pipe will be $2.5$ slower than using both pipes, hence it'd take $10 \\times 2.5 = 25$ hours to fill the swimming pool.\n\\end{description}\n\n\n\\item[Q11]\n\\begin{description}\n\\item[Answer:] $12 \\times 12 - 4 \\times 4 - 4 \\times (4\\times 4)/2 = 96$.\n\\end{description}\n\n\n\\item[Q12]\n\\begin{description}\n\\item[Answer:] $(30+90+50)/(1+2+1) = 42.5$ miles/hour.\n\\end{description}\n\n\n\\item[Q13]\n\\begin{description}\n\\item[Answer:] $4$ months\n\\[\n1\\ \\cancel{2}\\ \\cancel{3}\\ \\cancel{4}\\ 5\\ \\cancel{6}\\ 7\\ \\cancel{8}\\ \\cancel{9}\\ \\cancel{10}\\ 11\\ \\cancel{12}\n\\]\n\\end{description}\n\n\n\\item[Q17]\n\\begin{description}\n\\item[Answer:] $12$ kilograms\n\n\nAssume that $x$ kilograms of raisins and $y$ kilograms of nuts were used for making the snakcs.\nThen\n\\begin{eqnarray*}\nx + y &=& 20\n\\\\\n3.5 \\times x + 4.75 \\times y &=& 20 \\times 4\n\\end{eqnarray*}\n\nIf we multiply $4.75$ to both sides of the first equation,\nwe have\n\\begin{eqnarray*}\n4.75 \\times x + 4.75 \\times y &=& 20 \\times 4.75\n\\\\\n3.5 \\times x + 4.75 \\times y &=& 20 \\times 4\n\\end{eqnarray*}\n\nIf we subtract the second equation from the first one,\nwe have\n\\begin{eqnarray*}\n&&(4.75 - 3.5) \\times x = 20 \\times (4.75 - 4)\n\\\\\n&\\Leftrightarrow&\n1.25 \\times x = 20 \\times 0.75\n\\\\\n&\\Leftrightarrow&\nx = 20 \\times 0.75 / 1.25 = 20 \\times \\frac{3}{5} = 12\n\\end{eqnarray*}\n\n\n\n\n\n\n\\end{description}\n\n\n\\end{description}\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "05a5d74b4b48071c3a6294f0e880e10814b4c00b", "size": 10300, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problem_sovling/math4taehoon_Main.tex", "max_stars_repo_name": "sungheeyun/MathFoTaeHoon", "max_stars_repo_head_hexsha": "7271948c2a9653320dee2e5e46203c1d914f41fb", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-01-18T22:59:17.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-18T22:59:17.000Z", "max_issues_repo_path": "problem_sovling/math4taehoon_Main.tex", "max_issues_repo_name": "sungheeyun/MathForFamily", "max_issues_repo_head_hexsha": "7271948c2a9653320dee2e5e46203c1d914f41fb", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problem_sovling/math4taehoon_Main.tex", "max_forks_repo_name": "sungheeyun/MathForFamily", "max_forks_repo_head_hexsha": "7271948c2a9653320dee2e5e46203c1d914f41fb", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.2371134021, "max_line_length": 138, "alphanum_fraction": 0.6548543689, "num_tokens": 3945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.8006920116079209, "lm_q1q2_score": 0.6391017767071818}} {"text": "\\documentclass[paper.tex]{subfiles}\n\n\\usepackage{tikz}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{tabularx}\n\\usepackage{multicol}\n\\usepackage{algpseudocode}\n\\usepackage{algorithm}\n\n% Add vertical spacing to tables\n\\renewcommand{\\arraystretch}{1.4}\n\n% Begin Document\n\\begin{document}\n\n\\section{Brute Force Approach}\n\nCalculating the Minimum Domating Set of a graph is a difficult problem.\nThe size and complexity of a graph quickly grows with the number of nodes, $n$, and the number of edges, $e$, that make up the graph.\nAs such, the only known way to guarantee finding the minimum dominating set is through a brute force approach.\n\nThis entails checking every possible combination of nodes in the graph.\nFor each combination, the algorithm would check if it dominates every node in the graph, and if it is then checks if it's smaller than the currently known smallest dominating set.\nAfter all combinations are checked, the smallest known set is then guaranteed to be the Minimum Dominating Set.\n\nTo put this concept into pseudocode:\n\n\\begin{algorithm}[H]\n\n \\caption{Minimum Dominating Set brute force algorithm}\n\n \\begin{algorithmic}[1]\n \\Procedure{Minimum Dominating Set}{}\n \\State $i \\gets 1$\n \\While{$i < 2^n$}\n \\State{Convert $i$ into a binary array, $binArray$}\n \\State{Create an array of $0s$ for the dominated nodes, $domArray$}\n \\For{each $1 \\in binArray$}\n \\State{Get the node at that index from the graph}\n \\For{each neighbor of the node}\n \\State{Set it to dominated in $domArray$}\n \\EndFor\n \\EndFor\n \\If{all nodes dominated}\n \\If{Current Solution Size $<$ Best Solution Size}\n \\State{$bestSolution \\gets i$}\n \\EndIf\n \\EndIf\n \\State{Increment $i$}\n \\EndWhile\n \\EndProcedure\n\n \\end{algorithmic}\n\n\\end{algorithm}\n\nAs this is a brute force algorithm, the amount of iterations needed is strictly dependent on the size of the graph.\nWe can represent each solution as a set of bits, such that any index with a $1$ indicates that node is a part of the dominating set.\nThe number of possible solutions is therefore $2^n$, resulting in a runtime complexity of $O(2^n)$.\nThe amount of iterations quickly scales up, crossing the one billion mark for a graph of size 30.\nThis indicates that any sufficiently large graphs are essentially outside the limits of brute force computing.\nWe later cover experimental data that shows this.\n\n\\end{document}", "meta": {"hexsha": "2bf7f849f70badaa331df25d1ff79028248c5718", "size": 2662, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignments/Minimum Dominating Set/docs/tex/bruteforce.tex", "max_stars_repo_name": "Bkrenz/calu-csc360", "max_stars_repo_head_hexsha": "8600fb644e145cca27e10b084e9ddf62fbc84f4e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignments/Minimum Dominating Set/docs/tex/bruteforce.tex", "max_issues_repo_name": "Bkrenz/calu-csc360", "max_issues_repo_head_hexsha": "8600fb644e145cca27e10b084e9ddf62fbc84f4e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignments/Minimum Dominating Set/docs/tex/bruteforce.tex", "max_forks_repo_name": "Bkrenz/calu-csc360", "max_forks_repo_head_hexsha": "8600fb644e145cca27e10b084e9ddf62fbc84f4e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.9538461538, "max_line_length": 179, "alphanum_fraction": 0.681066867, "num_tokens": 607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006919925839874, "lm_q2_score": 0.7981867873410141, "lm_q1q2_score": 0.6391017692102879}} {"text": "\\documentclass[hidelinks, 11pt, openleft]{scrartcl}\n\n\\input{config/bendell-notes-config}\n\n\\title{Midterm Vocabularly List}\n\\author{Cody Mazza-Anthony \\\\ 260405012}\n\\date{March 2014}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Vocabularly from Chapters 1-3}\n\\subsection*{Dot Product in $\\mathbb{R}^2$ (1.4 pg. 4)}\n\n\\begin{itemize}\n \\item The $u \\cdot v$ is obtained by multiplying corresponding components and adding the resulting products. The vectors $u$ and $v$ are said to be \\textit{orthogonal} (or \\textit{perpendicular}) if their dot product is zero --that is, if $u \\cdot v = 0$.\n \\item The \\textit{dot product} or \\textit{inner product} of $u$ and $v$ is defined by:\n \\begin{center}\n $ u \\cdot v = a_{1}b_{1} + a_{2}b_{2} + \\dots + a_{n}b_{n}$\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Norm (Length) of a Vector (1.4 pg. 5)}\n\n\\begin{itemize}\n \\item The \\textit{norm} and \\textit{length} of a vector $u$ in $\\mathbb{R}^n$, denoted by $\\|u\\|$, is defined to be the nonnegative square root of $u \\cdot u$. \n \\item If $u = (a_{1}, \\dots, a_{n})$, then\n \\begin{center}\n $\\|u\\| = \\sqrt{u \\cdot u} = \\sqrt{a_1^2 + \\dots + a_n^2}$ \n \\end{center}\n \\item $\\|u\\|$ is the square root of the sum of squares of each of its components of $u$. Thus, $\\|u\\| \\ge 0$, and $\\|u\\|=0$ if and only if $u=0$.\n \\item A vector $u$ is called a unit vector if $\\|u\\|=1$ or, equivalently, if $u \\cdot u = 1$. For any nonzero vector $v$ in $\\mathbb{R}^n$, the vector\n \\begin{center}\n $\\hat{v} = \\frac{v}{\\|v\\|}$\n \\end{center}\n is the unique unit vector in the same direction as $v$. The process of finding $\\hat{v}$ from $v$ is called \\textit{normalizing} $v$.\n\\end{itemize}\n\n\\subsection*{Distance and Angle Between Vectors (1.4 pg. 6)}\n\\begin{itemize}\n \\item The \\textit{distance} between vectors $u=(a_{1}, \\dots. a_{n})$ and $v=(b_{1}, \\dots , b_{n})$ in $\\mathbb{R}^n$ is defined by:\n \\begin{center}\n $d(u,v) = \\|u-v\\| = \\sqrt{(a_1-b_1)^2 + \\dots + (a_n-b_n)^2}$ \n \\end{center}\n \\item The angle $\\theta$ between nonzero vectors $u,v$ in $\\mathbb{R}^n$ is defined by\n \\begin{center}\n ${cos\\theta} = {u \\cdot v \\over \\|u\\|\\|v\\| }$\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Complex Numbers (1.7 pg. 11)}\n\\begin{itemize}\n \\item Formally, a complex number is an ordered pair (a,b) of real numbers.\n \\item The complex number (0,1) is denoted by $i$. Important Property: $i^2 = -1$\n \\item A complex number can be written $z=(a,b)= a + bi$\n\\end{itemize}\n\n\\subsection*{Complex Conjugation (1.7 pg. 12)}\n\\begin{itemize}\n \\item Consider the complex number $z=a+bi$. The \\textit{conjugate} of $z$ is denoted and defined by: $\\overline{z} = \\overline{a+bi} = a-bi$.\n\\end{itemize}\n\n\\subsection*{Absolute Value of a Complex Number (1.7 pg. 12)}\n\\begin{itemize}\n \\item The \\textit{absolute value} of $z$, denoted by $|z|$, is defined to be the nonnegative square root of $z\\overline{z}$. Namely, $|z|= \\sqrt{z\\overline{z}}= \\sqrt{a^2+b^2}$\n\\end{itemize}\n\n\\subsection*{Dot Product in $C^n$ (1.8 pg. 13)}\n\\begin{itemize}\n \\item The dot or inner product of $u$ and $v$ is denoted by\n \\begin{align*}\n u \\cdot v = u_1\\overline{v_1} + \\cdots + u_n\\overline{v_n}\n \\end{align*}\n\\end{itemize}\n\n\\subsection*{Matrix Multiplication (2.5 pg. 31)}\n\n \\begin{align*}\n c_{ij} &= a_{i1}b_{1j} + \\cdots + a_{ip}b_{pj} \\\\\n &= \\sum_{k=1}^{p} a_{ik}b_{kj}\n \\end{align*}\n\n\\subsection*{Polynomials in Matrices ()}\n\\vspace{10 cm}\n\n\\subsection*{Invertible Matrices (2.9 pg. 34)}\n\\begin{itemize}\n \\item A square matrix $A$ is said to be \\textit{invertible} or \\textit{nonsingular} if there exists a matrix $B$ such that: \n \\begin{align*}\n AB=BA=I\n \\end{align*}\n where $I$ is the identity matrix. Such a matrix $B$ is unique. That is, if $AB_1=B_1A = I$ and $AB_2=B_2A=I$, then\n \\begin{align*}\n B_1=B_1I=B_1(AB_2) = (B_1A)B_2=IB_2=B_2\n \\end{align*}\n \\item We call such a matrix B the inverse of A and dneote it by $A^-1$. Observe that the above relatin is symmetric; that is, if $B$ is the inverse $A$, then $A$ is the inverse of $B$.\n\\end{itemize}\n\n\\subsection*{Equivalent Systems of Linear Equations (3.3 pg. 60)}\n\\begin{itemize}\n \\item Two systems of linear equations have the same solutoins if and only if each equation in each system is alinear combination of the equations in the other system.\n \\item Two systems of linear equations are said to be equivalent if they have the same solutions. \n\\end{itemize}\n\n\\subsection*{Augmented Matrix of a Linear System (3.2 pg. 59)}\n\\begin{itemize}\n \\item Is the matrix of coefficients plus the last column which is made up of constants, sometimes written $M=[A,B]$.\n\\end{itemize}\n\n\\subsection*{Coefficient Matrix of a Linear System (3.3 pg. 59)}\n\\begin{itemize}\n \\item Is the matrix of coefficients which is associated with a general system of $m$ equations and $n$ unknowns.\n\\end{itemize}\n\n\\subsection*{Matrix Equation of a system of linear Equations (3.2 pg. 58)}\n\\begin{itemize}\n \\item Is a list of linear equations with the same unknowns -- that is, a system of $m$ linear equations and $n$ unknowns.\n \\item The coefficients, constants, and equations can be represented in a matrix as the two above definitions state(augmented or coefficient matrices).\n\\end{itemize}\n\n\\subsection*{Elementary Row Operations(ERO) (3.3 pg. 61) }\n\\begin{itemize}\n \\item The following operations on a system of linear equations:\n \\begin{enumerate}\n \\item Interchange two of the equations.\n \\item Replace an equation by a nonzero multiple of itself.\n \\item Replace an equation by the sum of a multiple of another equation and itself.\n \\end{enumerate}\n\\end{itemize}\n\n\\subsection*{Elementary Matrices (3.12 pg. 84)}\n\\begin{itemize}\n \\item Let $e$ denote an elementary row operation and let $e(A)$ denote the results of applying the operation $e$ to a matrix $A$. Now let $E$ be the matrix obtained by applying $e$ to the identity matrix $I$; that is,\n \\begin{align*}\n E=e(I)\n \\end{align*}\n \\item Then $E$ is called the \\textit{elementary matrix} corresponding to the elementary row operation $e$. Note that $E$ is always a square matrix\n\\end{itemize}\n\n\\subsection*{Row Equivalence (3.7 pg. 72)}\n\\begin{itemize}\n \\item A matrix $A$ is said to be \\textit{row equivalent} to a matrix $B$, written $A~B$\n \\item If $B$ can be obtained from $A$ by a sequence of elementary row operations. In the case that $B$ is also an echelon matrix, $B$ is called an \\textit{echelon form} of $A$.\n\\end{itemize}\n\n\\subsection*{Echelon Form (3.7 pg. 70)}\n\\begin{itemize}\n \\item A matrix $A$ is called an \\textit{echelon matrix}, or is said to be in \\textit{echelon form},if the following two conditions hold (where a leading nonzero element of a row A is the first nonzero element in the row):\n \\begin{enumerate}\n \\item All zero rows, if any, are at the bottom of the matrix\n \\item Each leading nonzero entry in a row is to the right of the leading nonzero entry in the preceding row. \n \\end{enumerate}\n\\end{itemize}\n\n\\subsection*{Row Canonical Form (3.7 pg. 71)}\n\\begin{itemize}\n \\item A matrix $A$ is said to be in \\textit{row canonical form} if it is an echelon matrix -- that is, if it satisfies the above properties (1) and (2), and if it satisfies the following additional two properties \n \\item The major difference between an echelon matrix and a matrix in row canonical form is that in an echelon matrix there must be zeros below the pivots, but in a matrix in row canonical form, each pivot must also equal 1 and there must also be zeros above the pivots. \n \\item The zero matrix of any size and the identity matrix $I$ of any size are important special examples of matrices in row canonical form. \n\\end{itemize}\n\n\\subsection*{Free Variables and Pivot Variables (3.5 pg. 65)}\n\\begin{itemize}\n \\item The leading unknowns in a system are called \\textit{pivot} variables, and the other unknowns are called \\textit{free variables}.\n\\end{itemize}\n\n\\subsection*{Gaussian Elimination (3.8 pg. 73)}\n\\begin{itemize}\n \\item Algorithms to convert matrices to echelon form and row canonical form by using elementary row operations, are simply restatements of the Gaussian elimination as applied to matrices rather than linear equations.\n\\end{itemize}\n\n\\subsection*{Homogeneous Systems of Linear Equations associated to a Matrix (3.11 pg. 81)}\n\\begin{itemize}\n \\item A system of linear equations is said to be \\textit{homogeneous} if all the constant terms are zero. Thus, a homogeneous system has the form $AX=0$\n \\item Such a system has the zero vector as a solution. We are usually interested in whether or not the system has a nonzero solution.\n \\item Here $r$ denotes the number of equations in echelon form and $n$ denotes the number of unknowns. Thus, the echelon system has $n-r$ free variables.\n \\begin{enumerate}\n \\item $r=n$. The system has only the zero solution. \n \\item $r < n$. The system has a nonzero solution. \n \\end{enumerate}\n\\end{itemize}\n\n\n\\section{Vocabularly from Chapter 4}\n\n \\subsection*{Vector Space (4.2 pg. 112)}\n \\begin{itemize}\n \\item Let $V$ be a nonempty set with two operations:\n \\begin{enumerate}\n \\item Vector Addition: This assigns to any $u, v \\in V$ a sum $u+v$ in $V$.\n \\item Scalar Multiplication: This assigns to any $u \\in V, k \\in K$ a product $ku \\in V$.\n \\end{enumerate}\n \\item There are additional axioms that need to hold located on pg. 113 of the textbook. They can be summarized by saying $V$ is a \\textit{commutative group} under addition. Also, \\textit{subtraction} in $V$ is defined by $u-v = u + (-v)$, where $-v$ is a unique negative of $v$.\n \\end{itemize}\n \n \\subsection*{Subspace (4.5 pg. 117)}\n \\begin{itemize}\n \\item Let $V$ be a vector space over field $K$ and let $W$ be a subset of $V$. Then $W$ is a \\textit{subspace} of $V$ if $W$ is itself a vector space over $K$ with respect to the operations of vector addition and scalar multiplication on $V$.\n \\item The way you would show that any set $W$ is a vector space is to show that $W$ satisfies the eight axioms of a vector space. However, if $W$ is a subset of a vector space $V$, then some of the axioms automatically hold in $W$, because they already hold in $V$. \n \\item To identify a vector space suppose $W$ is a subset of a vector space $V$. Then $W$ is a subspace of $V$ if the following two conditions hold:\n \\begin{enumerate}\n \\item The zero vector 0 belongs to $W$\n \\item For ever $u, v \\in W, k \\in K$: (i) The sume $u+v \\in W$ (ii) The multiple $ku \\in W$.\n \\end{enumerate}\n \\end{itemize}\n \n \\subsection*{Linear Combination (4.4 pg. 115)}\n \\begin{itemize}\n \\item A vector $v$ in $V$ is a linear combination of vectors $u_{1}, u_{2}, \\dots , u_{m}$ in V if there exists scalars $a_{1}, a_{2}, \\dots , a_{m}$ in K such that\n \\begin{center}\n $v = a_{1}u_{1} + a_{2}u_{2} + \\dots + a_{m}u_{m}$\n \\end{center}\n \\item Alternatively, $v$ is a linear combination of $u_{1}, u_{2}, \\dots , u_{m}$ if there is a solution to the vector equation\n \\begin{center}\n $v = x_{1}u_{1} + x_{2}u_{2} + \\dots + x_{m}u_{m}$\n \\end{center}\n where $x_{1},x_{2}, \\dots , x_{m}$ are unknown scalars.\n \\item A system can have a unique solution, many solutions, or no solution. No solution means that the system cannot be written as a linear combination.\n \\end{itemize}\n \n \\subsection*{Spanning Set (4.4 pg. 116)}\n \\begin{itemize}\n \\item Let $V$ be a vector space over $K$. Vectors $u_{1}, u_{2}, \\dots , u_{m}$ in $V$ are said to span V or to form a spanning set of $V$ if every $v$ in $V$ is a linear combination of the vectors $u_{1}, u_{2}, \\dots , u_{m}$ - that is, if there exists scalars $a_{1}, a_{2}, \\dots , a_{m}$ in $K$ such that \n \\begin{center}\n $v = a_{1}u_{1} + a_{2}u_{2} + \\dots + a_{m}u_{m}$\n \\end{center}\n \\end{itemize}\n\n \\subsection*{Span of a Set of Vectors ()}\n \\vspace{10 cm}\n \n \\subsection*{Row Space of a Matrix ()}\n \\vspace{10 cm}\n \n \\subsection*{Column Space of a Matrix ()}\n \\vspace{10 cm}\n\n \\subsection*{Linear Dependence (4.7 pg. 121)}\n \\begin{itemize}\n \\item We say that vectors $v_1, \\cdots ,v_m$ in $V$ are \\textit{linearly depedent} if there exists scalars $a_1, \\cdots, a_m$ in $K$, not all of them 0, such that\n \\begin{align*}\n a_1v_1 + \\cdots + a_mv_m = 0\n \\end{align*}\n \\item Otherwise, we say that the vectors are \\textit{linearly independent}.\n \\end{itemize}\n \n \\subsection*{Linear Independence (4.7 pg. 121}\n \\begin{itemize}\n \\item Consider the vector equation\n \\begin{align*}\n x_1v_1 + \\cdots + x_mv_m = 0\n \\end{align*}\n \\item where the x's are unknown scalars in $K$. This equation always has the \\textit{zero solution} $x_1=0, \\cdots,x_m=0$. Suppose this is the only solution; that is, suppose we can show:\n \\begin{align*}\n x_1v_1 + \\cdots + x_mv_m = 0\n \\end{align*}\n implies\n \\begin{align*}\n x_1=0, \\cdots ,x_m=0\n \\end{align*}\n \\item Then the vectors $v_1, \\cdots ,v_m$ are \\textit{linearly independent}\n \\end{itemize}\n \n \\subsection*{Basis (4.8 pg. 124)}\n \\begin{itemize}\n \\item Definition 1: A set $S = \\{u_1, \\cdots, u_n\\}$ of vectors is a \\textit{basis} of $V$ if it has the following two properties: (1) S is linearly independent (2) $S$ spans $V$.\n \\item Definition 2: A set $S = \\{u_1, \\cdots, u_n\\}$ of vectors is a \\textit{basis} of $V$ if every $v \\in V$ can be written uniquely as a linear combination of the basis vectors. \n \\end{itemize}\n \n \\subsection*{Dimension ()}\n \\vspace{10 cm}\n \n \\subsection*{Rank of Matrix (4.9 pg. 126)}\n \\begin{itemize}\n \\item The \\textit{rank} of a matrix $A$, written rank(A), is equal to the maximum number of linearly independent rows of $A$ or, equivalently, the dimension of the row space of $A$.\n \\item The rank can be equal to the number of pivots in echelon form.\n \\item The rank can correspond to the nonzero rows in echelon form, which form a basis for the row space of the original vector.\n \\end{itemize}\n \n \\subsection*{Sum of two Subspaces (4.10 pg. 129)}\n \\begin{itemize}\n \\item Suppose $U$ and $W$ are subspaces of $V$. Then one can shows that $U+W$ is a subspace of $V$. Recall that $U \\cap W$ is also a subspace of $V$. \n \\item Suppose $U$ and $W$ are finite-dimensional subspaces of a vector space $V$. Then $U+W$ has finite dimension and\n \\begin{align*}\n dim(U+W) = dimU + dimW - dim(U \\cap W)\n \\end{align*}\n \\end{itemize}\n \n \\subsection*{Direct Sum (4.10 pg. 129)}\n \\begin{itemize}\n \\item The vector space $V$ is said to be the \\textit{direct sum} of its subspaces $U$ and $W$, denoted by\n \\begin{align*}\n V=U \\oplus W\n \\end{align*}\n \\item if every $v in V$ can be written in one and only one way as $v=u+w$ where $u \\in U$ and $w \\in W$.\n \\end{itemize}\n \n \\subsection*{Coordinates (4.11 pg. 130)}\n \\begin{itemize}\n \\item Let $V$ be an n-dimensional vector space over $K$ with basis $S=\\{u_1, \\cdots , u_n\\}$. Then any vector $v \\in V$ can be expressed uniquely as a linear combination of the basis vectors ini $S$, say\n \\begin{align*}\n v = a_1u_1 + \\cdots + a_nu_n\n \\end{align*}\n \\item Then $n$ scalars $a_1, \\cdots , a_n$ are called the \\textit{coordinates} of $v$ relative to the basis $S$, and they form a vector $[a_1, \\cdots , a_n]$ in $K^n$ called the \\textit{coordinate vector} of $v$ relative to $S$. \n \\item We denote this vector by $[v]_S$, or simple $[v]$, when $S$ is understood. Thus,\n \\begin{align*}\n [v]_S = [a_1, \\cdots , a_n]\n \\end{align*}\n \\end{itemize}\n \n \\subsection*{Isomorphism of vector spaces (4.11 pg. 130)}\n \\vspace{10 cm}\n \n \n\\section{Vocabularly from Chapters 5 and 6}\n\n\\subsection*{Function/Mapping (5.2 pg. 164)}\n\\begin{itemize}\n \\item Let $A$ and $B$ be arbitrary sets. Suppose to each element in $a \\in A$ there is assigned a unique element of B; called the \\textit{image} of $a$. \n \\item The collection of $f$ of such assignments is called a \\textit{mapping} from $A$ to $B$, and is denoted by\n \\begin{center}\n $f: A \\to B$\n \\end{center}\n \\item The set $A$ is called the \\textit{domain} of the mapping, and $B$ is called the \\textit{target set}.\n \\item One may also view a mapping $f: A \\to B$ as a computer that, for each input value $a \\in A$, produces a unique output $f(a) \\in B$.\n\\end{itemize}\n\n\\subsection*{Range/Image (5.2 pg. 164)}\n\\begin{itemize}\n \\item Is a subset of the \\textit{target set} that the function actually maps to. \n \\item Sometimes the \"barred\" arrow $\\mapsto$ is used to denote the range of an arbitrary element $x \\in A$ under a mapping $f: A \\to B$ by writing $x \\mapsto f(x)$.\n\\end{itemize}\n\n\n\\subsection*{One-to-One Mapping (5.2 pg. 166)}\n\\begin{itemize}\n \\item A mapping $f: A \\to B$ is said to be \\textit{one-to-one} (or 1-1 or injective) if different elements of A have distinct images; that is,\n \\begin{center}\n If $f(a) = f(a')$, then $a=a'$\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Onto Mapping (5.2 pg. 166)}\n\\begin{itemize}\n \\item A mapping $f: A \\to B$ is said to be \\textit{onto} (or $f$ maps $A$ onto $B$ or \\textit{surjective}) if ever $b \\in B$ is the range of at least one $a \\in A$.\n\\end{itemize}\n\n\\subsection*{Linear Transformation (5.3 pg. 167)}\n\\begin{itemize}\n \\item Let $V$ and $U$ be vector spaces over the same field $K$. A mapping $F: V \\to U$ is called a \\textit{linear mapping} or \\textit{linear transformation} if it satisfies the following two conditions:\n \\begin{enumerate}\n \\item For any vectors $v, w \\in V, F(v+w) = F(v) + F(w)$.\n \\item For any scalar $k$ and vector $v \\in V, F(kv) = kF(v)$.\n \\end{enumerate}\n \\item Basically, $F: V \\to U$ is linear if it \"preserves\" the two basic operations of a vector space, that of vector addition and that of scalar multiplication.\n\\end{itemize}\n\n\\subsection*{Kernal (5.4 pg. 169)}\n\n\\begin{itemize}\n \\item Let $F: V \\to U$ be a linear transformation/mapping. The \\textit{kernel} of $F$, written Ker F, is the set of elements in $V$ that map into the zero vector 0 in $U$; that is,\n \\begin{center}\n Ker $F = \\{v \\in V: F(v) = 0\\}$\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Image (5.4 pg. 169)}\n\n\\begin{itemize}\n \\item The \\textit{image} of $F$, written Im $F$, is the set of image points in $U$; that is,\n \\begin{center}\n Im $F = \\{u \\in U: there exists v \\in V for which F(v) = u\\}$\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Rank (5.4 pg. 171)}\n\\begin{itemize}\n \\item Let $F: V \\to U$ be a linear mapping. The \\textit{rank} of $F$ is defined to be the dimension of its image.\n \\begin{center}\n rank(F) = dim(Im F)\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Nullity (5.4 pg. 171)}\n\\begin{itemize}\n \\item The \\textit{nullity} of $F$ is defined to be the dimension of its kernel\n \\begin{center}\n nullity(F) = dim(Ker F)\n \\end{center}\n\\end{itemize}\n\n\\subsection*{Linear Transformation associated to a Matrix (Matrix Mapping) () }\n\\vspace{10 cm}\n\n\\subsection*{Singular Linear Transformation (5.5 pg. 172)}\n\\begin{itemize}\n \\item Let $F: V \\to U$ be a linear mapping. Recall that $F(0) = 0$. $F$ is said to be $singular$ if the image of some nonzero vector $v$ is 0 -- that is, if there exists $v \\ne 0$ such that $F(v) = 0$.\n\\end{itemize}\n\n\\subsection*{Matrix Representation of Linear Transformation ()}\n\\vspace{10 cm}\n\n\n\\end{document}\n", "meta": {"hexsha": "e3c0f052f74caef65a179775790a28a581147fa7", "size": 20243, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Vocab.tex", "max_stars_repo_name": "sorinmuchi/math-223-mcgill-notes", "max_stars_repo_head_hexsha": "31f6cea755bc29fade21bdfe9ed7118aa6affcfa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-11-03T23:09:18.000Z", "max_stars_repo_stars_event_max_datetime": "2015-11-03T23:09:18.000Z", "max_issues_repo_path": "Vocab.tex", "max_issues_repo_name": "sorinmuchi/math-223-mcgill-notes", "max_issues_repo_head_hexsha": "31f6cea755bc29fade21bdfe9ed7118aa6affcfa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Vocab.tex", "max_forks_repo_name": "sorinmuchi/math-223-mcgill-notes", "max_forks_repo_head_hexsha": "31f6cea755bc29fade21bdfe9ed7118aa6affcfa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.8596059113, "max_line_length": 322, "alphanum_fraction": 0.6437780961, "num_tokens": 6387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.8006920020959545, "lm_q1q2_score": 0.6391017691148559}} {"text": "\\documentclass[a4paper]{article}\n\n\\input{temp}\n\n\\setcounter{section}{-1}\n\n\\begin{document}\n\n\\title{Logic and Set Theory}\n\n\\maketitle\n\n\\newpage\n\n\\tableofcontents\n\n\\newpage\n\n\\section{Miscellaneous}\n\nSome introductory speech\n\n\\newpage\n\n\\section{Propositional logic}\nLet $P$ denote a set of \\emph{primitive proposition}, unless otherwise stated, $P=\\{p_1,p_2,...\\}$.\n\n\\begin{defi}\nThe \\emph{language} or \\emph{set of propositions} $L=L(P)$ is defined inductively by:\\\\\n(1) $p \\in L$ $\\forall p \\in P$;\\\\\n(2) $\\perp \\in L$, where $\\perp$ is read as 'false';\\\\\n(3) If $p,q \\in L$, then $(p \\implies q) \\in L$. For example, $(p_1 \\implies L)$, $((p_1 \\implies p_2) \\implies (p_1 \\implies p_3))$.\n\\end{defi}\n\nNote that at this point, each proposition is only a finite string of symbols from the alphabet $(,),\\implies,\\perp,p_1,p_2,...$ and do not really mean anything (until we define so).\\\\\nBy \\emph{inductively define}, we mean more precisely that we set $L_1 = P \\cup \\{\\perp\\}$, and $L_{n+1} = L_n \\cup \\{(p \\implies q):p,q \\in L_n\\}$, and then put $L = L_1 \\cup L_2 \\cup ...$.\n\nEach proposition is built up \\emph{uniquely} from 1) and 2) using 3). For example, $((p_1 \\implies p_2) \\implies (p_1 \\implies p_3))$ came from $(p_1 \\implies p_2)$ and $(p_1 \\implies p_3)$. We often omit outer brackets or use different brackets for clarity.\n\nNow we can define some useful things:\\\\\n$\\bullet$ $\\neg p$ (not $p$), as an abbreviation for $p \\implies \\perp$;\\\\\n$\\bullet$ $p \\vee q$ ($p$ or $q$), as an abbreviation for $(\\neg p) \\implies q$;\\\\\n$\\bullet$ $p \\wedge q$ ($p$ and $q$), as an abbreviation for $\\neg (p \\implies (\\neg q))$.\n\nThese definitions 'make sense' in the way that we expect them to.\n\n\\begin{defi}\nA \\emph{valuation} is a function $v:L \\to \\{0,1\\}$ s.t.\\\\\n(1) $v(\\perp) = 0$;\n(2) \n\\begin{equation*}\n\\begin{aligned}\nv(p \\implies q) = \\left\\{ \\begin{array}{ll}\n0 & v(p) =1, v(q) = 0\\\\\n1 & else\n\\end{array}\n\\right. \\forall p,q \\in L\n\\end{aligned}\n\\end{equation*}\n\\end{defi}\n\n\\begin{rem}\nOn $\\{0,1\\}$, we could define a constant $\\perp$ by $\\perp = 0$, and an operation $\\implies$ by $a \\implies b = 0$ if $a=1, b=0$ and $1$ otherwise. Then a valuation is a function $L \\to \\{0,1\\}$ that preserves the structure ($\\perp$ and $\\implies$), i.e. a homomorphism.\n\\end{rem}\n\n\\begin{prop}\n(1) If $v,v'$ are valuations with $v(p) = v'(p)$ $\\forall p \\in P$, then $v=v'$ (on $L$).\\\\\n(2) For any $w:P \\to \\{0,1\\}$, there exists a valuation $v$ with $v(p) = w(p)$ $\\forall p \\in P$.\\\\\nIn short, a valuation is defined by its value on $p$, and any values will do.\n\\begin{proof}\n(1) We have $v(p)=v'(p)$ $\\forall p \\in L_1$. However, if $v(p) = v'(p)$ and $v(q) = v'(q)$ then $v(p \\implies q) = v'(p \\implies q)$, so $v=v'$ on $L_2$. Continue inductively we have $v=v'$ on $L_n \\forall n$.\\\\\n(2) Set $v(p) = w(p)$ $\\forall p \\in P$ and $v(\\perp) = 0$: this defines $v$ on $L_1$. Having defined $v$ on $L_n$, use the rules for valuation to inductively define $v$ on $L_{n+1}$ so we can extend $v$ to $L$.\n\\end{proof}\n\\end{prop}\n\n\\begin{defi}\nWe say $p$ is a \\emph{tautology}, written $\\vDash p$, if $v(p) = 1$ $\\forall$ valuations $v$. Some examples:\\\\\n(1) $p \\implies (q \\implies p)$: a true statement is implies by anything. We can verify this by:\n\\begin{equation*}\n\\begin{aligned}\n\\begin{matrix}\nv(p) & v(q) & v(q \\implies p) & v(p \\implies (q \\implies p))\\\\\n1 & 1 & 1 & 1\\\\\n1 & 0 & 1 & 1\\\\\n0 & 1 & 0 & 1\\\\\n0 & 0 & 1 & 1\n\\end{matrix}\n\\end{aligned}\n\\end{equation*}\n\nSo we see that this is indeed a tautology;\\\\\n(2) $(\\neg\\neg p) \\implies p$, i.e. $((p \\implies \\perp) \\implies \\perp) \\implies p$, called the \"law of excluded middle\";\n\n(3) $[p \\implies (q \\implies r)] \\implies [(p \\implies q) \\implies (p \\implies r)]$.\\\\\nIndeed, if not then we have some $v$ with $v(p\\implies(q \\implies r)) = 1$, $v(\\implies (p \\implies q) \\implies (p \\implies r)) = 0$. So $v(p\\implies q) = 1$, $v(p \\implies r) =0$. This happens when $v(p) = 1$, $v(r) = 0$, so also $v(q) = 1$. But then $v(q \\implies r)=0$, so $v(p \\implies (q \\implies r)) = 0$.\n\\end{defi}\n\n\\begin{defi}\nFor $S \\subset L$, $t \\in L$, say $S$ \\emph{entails} or \\emph{semantically implies} $t$, written $S \\vDash t$ if $v(s) = 1 \\forall s \\in S \\implies v(t) = 1$, for each valuation $v$.\\\\\n(\"Whenever all of $S$ is true, $t$ is true as well.\")\n\nFor example, $\\{p \\implies q, q \\implies r\\} \\vDash (p \\implies r)$. To prove this, suppose not: so we have $v$ with $v(p\\implies q) = v(q\\implies r) = 1$ but $v(p\\implies r) = 0$. So $v(p) = 1$, $v(r) = 0$, so $v(q) = 0$, but then $v(p \\implies q)$ = 0.\n\nIf $v(t) = 1$ we say $t$ is true in $v$ or that $v$ is a model of $t$.\n\nFor $S \\subset L$, $v$ is a model of $S$ if $v(s) = 1$ $\\forall s \\in S$. So $S \\vDash t$ says that every model of $S$ is a model of $t$. For example, in fact $\\vDash t$ is the same as $\\phi \\vDash t$.\n\\end{defi}\n\n\\newpage\n\n\\section{Syntactic implication}\n\nFor a notion of 'proof', we will need axioms and deduction rules. As axioms, we'll take:\\\\\n1. $p \\implies (q \\implies p)$ $\\forall p,q \\in L$;\\\\\n2. $[p\\implies (q \\implies r)] \\implies [(p \\implies q) \\implies (p \\implies r)]$ $\\forall p,q,r \\in L$;\\\\\n3. $(\\neg\\neg p) \\implies p$ $\\forall p \\in L$.\n\nNote: these are all tautologies. Sometimes we say they are 3 axiom-schemes, as all of these are infinite sets of axioms.\n\nAs deduction rules, we'll take just \\emph{modus ponens}: from $p$, and $p\\implies q$, we can deduce $q$.\n\nFor $S \\subset L$, $t \\in L$, a \\emph{proof} of $t$ from $S$ cosists of a finite sequence $t_1,...,t_n$ of propositions, with $t_n = t$, s.t. $\\forall i$ the proposition $t_i$ is an axiom, or a member of $S$, or there exists $j,k < i$ with $t_j=(t_k \\implies t_i)$.\n\nWe say $S$ is the \\emph{hypotheses} or \\emph{premises} and $t$ is the \\emph{conclusion}.\n\nIf there exists a proof of $t$ from $S$, we say $S$ \\emph{proves} or \\emph{syntactically implies} $t$, written $S \\vdash t$.\n\nIf $\\phi \\vdash t$, we say $t$ is a \\emph{theorem}, written $\\vdash t$.\n\n\\begin{eg}\n$\\{p \\implies q, q \\implies r\\} \\vdash p \\implies r$.\\\\\nwe deduce by the following:\\\\\n(1) $[p \\implies (q \\implies r)] \\implies [(p \\implies q) \\implies (p \\implies r)]$; (axiom 2)\\\\\n(2) $q \\implies r$; (hypothesis)\\\\\n(3) $(q \\implies r) \\implies (p \\implies (q \\implies r))$; (axiom 1)\\\\\n(4) $p \\implies (q \\implies r)$; (mp on 2,3)\\\\\n(5) $(p \\implies q) \\implies (p \\implies r)$ (mp on 1,4);\\\\\n(6) $p \\implies q$; (hypothesis)\\\\\n(7) $p \\implies r$. (mp on 5,6)\n\\end{eg}\n\n\\begin{eg}\nLet's now try to prove $\\vdash p \\implies p$. Axiom 1 and 3 probably don't help so look at axiom 2; if we make $(p \\implies q)$ and $p\\implies (q \\implies r)$ something that's a theorem, and make $p \\implies r$ to be $p \\implies p$ then we are done. So we need to take $p=p,q=(p \\implies p), r=p$. Now:\\\\\n(1) $[p \\implies ((p \\implies p) \\implies p)] \\implies [( p \\implies (p \\implies p)) \\implies (p \\implies p)]$; (axiom 2)\\\\\n(2) $p \\implies ((p \\implies p) \\implies p)$; (axiom 1)\\\\\n(3) $(p \\implies (p \\implies p)) \\implies (p \\implies p)$; (mp on 1,2)\\\\\n(4) $p \\implies (p \\implies p)$; (axiom 1)\\\\\n(5) $p \\implies p$. (mp on 3,4)\n\\end{eg}\n\nProofs are made easier by:\n\\begin{prop} (2, deduction theorem)\\\\\nLet $S \\subset L$, $p,q \\in L$. Then $S \\vdash (p \\implies q)$ if and only if $(S \\cup \\{p\\}) \\vdash q$.\n\\begin{proof}\nForward: given a proof of $p \\implies q$ from $S$, add the lines $p$ (hypothesis), $q$ (mp) to optaion a proof of $q$ from $S\\cup\\{p\\}$.\\\\\nBackward: if we have proof $t_1,...,t_n=q$ of $q$ from $S \\cup \\{p\\}$. We'll show that $S \\vdash (p \\implies t_i) \\forall i$, so $p \\implies t_n=q$.\\\\\nIf $t_i$ is an axiom, then we have $\\vdash t_i \\implies (p \\implies t_i)$, so $\\vdash p \\implies t_i$;\\\\\nIf $t_i \\in S$, write down $t_i$, $t_i \\implies (p \\implies t_i)$, $p \\implies t_i$ we get a proof of $p \\implies t_i$ from $S$;\\\\\nIf $t_i = p$: we know $\\vdash (p \\implies p)$, so done;\\\\\nIf $t_i$ obtained by mp: in that case we have some earlier lines $t_j$ and $t_j \\implies t_i$. By induction, we may assume $S \\vdash (p \\implies t_j)$ and $S \\vdash (p \\implies (t_j \\implies t_i))$. Now we can write down $[p \\implies (t_j \\implies t_i)] \\implies [(p \\implies t_j) \\implies (t_i)]$ by axiom 2, $p \\implies (t_j \\implies t_i)$, $p \\implies t_j) \\implies (p \\implies t_i)$ (mp), $p \\implies t_j$, $p \\implies t_i$ (mp) to obtain $S \\vdash (p \\implies t_i)$.\n\nThese are all of the cases. So $S \\vdash (p \\implies q)$.\n\\end{proof}\nThis is why we chose axiom 2 as we did -- to make this proof work.\n\\end{prop}\n\n\\begin{eg}\nTo show $\\{p \\implies q, q \\implies r\\} \\vdash (p \\implies r)$, it's enough to show that $\\{p \\implies q, q \\implies r, p\\} \\vdash r$, which is trivial by mp.\n\\end{eg}\n\nNow, how are $\\vdash$ and $\\vDash$ related? We are going to prove the \\emph{completeness theorem}: $S \\vdash t \\iff S \\vDash t$.\n\nThis ensures that our proofs are sound, in the sense that everything it can prove is not absurd ($S \\vdash t$ then $S \\vDash t$), and are adequate, i.e. our axioms are powerful enough to define every semantic consequence of $S$, which is not obvious ($S \\vDash t$ then $S \\vdash t$).\n\n\\begin{prop} (3)\\\\\nLet $S \\subset L$, $t \\in L$. Then $S \\vdash t \\implies S \\vDash t$.\n\\begin{proof}\nGiven a valuation $v$ with $v(s) =1$ $\\forall s \\in S$, we want $v(t) = 1$.\\\\\nWe have $v(p) = 1$ $\\forall p$ axiom as our axioms are all tautologies (proven earier); $v(p)=1$ $\\forall p \\in S$ by definition of $v$; also if $v(p) = 1$ and $v(p \\implies q) = 1$, then also $v(q) = 1$ (by definition of $\\implies$). So $v(p) = 1$ for each line $p$ of our proof of $t$ from $S$.\n\\end{proof}\n\\end{prop}\n\nWe say $S \\subset L$ consistent if $S \\not\\vdash \\perp$. One special case of adequacy is: $S \\vDash \\perp \\implies S \\vdash \\perp$, i.e. if $S$ has no model then $S$ inconsistent, i.e. if $S$ is consistent then $S$ has a model. This implies adequacy: given $S \\vDash t$, we have $S \\cup \\{\\neg t\\} \\vDash \\perp$, so by our special case we have $S \\cup \\{\\neg t\\} \\vdash \\perp$, i.e. $S \\vdash ((\\neg t) \\implies t)$ by deduction theorem, so $S \\vdash \\neg\\neg t$. But $S \\vdash ((\\neg\\neg t) \\implies t)$ by axiom 3, so $S \\vdash t$ (mp).\n\n\\begin{thm} (4)\\\\\nLet $S \\subset L$ be consistent, then $S$ has a model.\\\\\nThe idea is that we would like to define valuation $v$ by $v(p) = 1 \\iff p \\in S$, or more sensibly, $v(p)=1 \\iff S \\vdash p$.\\\\\nBut maybe $S \\not \\vdash p_3, S \\not\\vdash \\neg p_3$, but a valuation maps half of $L$ to 1, so we want to 'grow' $S$ to contain one of $p$ or $\\neg p$ for each $p \\in L$, while keeping consistency.\n\\begin{proof}\nClaim: for any consistent $S \\subset L$, $p \\in L$, $S \\cup \\{p\\}$ or $S \\cup \\{\\neg p\\}$ consistent.\n\\emph{Proof of claim.} If not, then $S \\cup \\{p\\} \\vdash \\perp$ and $S \\cup \\{\\neg p\\} \\vdash \\perp$, then $S \\vdash (p \\implies \\perp)$ (deduction theorem), i.e. $S \\vdash \\not p$, so $S \\vdash \\perp$ contradiction.\n\nNow $L$ is countable as each $L_n$ is countable, so we can list $L$ as $t_1,t_2,...$. Put $S_0 = S$; set $S_1 = s_0 \\cup \\{t_1\\}$ or $s_0 \\cup (\\neg t_1\\}$ so that $S_1$ is consistent. Then set $S_2 = S_1 \\cup \\{t_2\\}$ or $S_1 \\cup \\{\\neg t_2\\}$ so that $S_2$ is consistent, and continue likewise. Set $\\bar{S} = S_0 \\cup S_1 \\cup S_2 \\cup ...$ Then $\\bar{S} \\supset S$, and $\\bar{S}$ is consistent (as each $S_n$ is, and each proof is finite). $\\forall p \\in L$, we have either $p \\in S$ or $(\\neg p) \\in S$. Also, $\\bar{S}$ is \\emph{deductively closed}, meaning that is $\\bar{S} \\vdash p$ then $p \\in \\bar{S}$: if $p \\not\\in \\bar{S}$ then $(\\neg p) \\in \\bar{S}$, so $\\bar{S} \\vdash p$, $\\bar{S} \\vdash (\\not p)$ so $\\bar{S} \\vdash \\perp$ contradiction.\\\\\nDefine $v:L \\to \\{0,1\\}$ by $p \\to 1$ if $p \\in \\bar{S}$, $0$ otherwise. Then $v$ is a valuation: $v(\\perp) = 0$ as $\\perp \\not\\in \\bar{S}$; for $v(p \\implies q)$:\\\\\nIf $v(p) = 1$, $v(q) = 0$: We have $p \\in \\bar{S}$, $q \\not\\in \\bar{S}$, and want $v(p \\implies q) = 0$, i.e. $(p \\implies q \\not\\in \\bar{S}$. But if $9p \\implies q) \\in \\bar{S}$ then $\\bar{S} \\vdash q$ contradiction;\\\\\nIf $v(q) = 1$: have $q \\in \\bar{S}$, and want $v(p \\implies q) =1$, i.e. $(p \\implies q) \\int \\bar{S}$. But $\\vdash q \\implies (p \\implies q)$ so $\\bar{S} \\vdash (p \\implies q)$;\\\\\nIf $v(p) = 0$: have $p \\not\\in \\bar{S}$, i.e. $(\\neg p) \\in \\bar{S}$ and want $(p \\implies q) \\in \\bar{S}$. So we need $(p \\implies \\perp) \\vdash (p \\implies q)$, i.e. $p \\implies \\perp,p \\vdash q$ (deduction theorem). Thus it's enough to show that $\\perp \\vdash q$. But $(\\neg\\neg q) \\implies q$, and $\\vdash (\\perp \\implies (\\neg\\neg q))$ (axiom 3 and 1 -- to see the second one, write $\\neg$ explicitly using $\\implies$ and $\\perp$), so $\\vdash(\\perp \\implies q)$, i.e. $\\perp\\vdash q$.\n\\end{proof}\n\\end{thm}\n\n\\begin{rem}\nSometimes this is called 'completeness theorem'. The proof used $P$ being countable to get $L$ countable; in fact, result still holds if $P$ is uncountable (see chapter 3).\n\\end{rem}\n\nBy remark before theorem 4, we have\n\\begin{coro} (5, adequacy)\\\\\nLet $S \\subset L$, $t \\in L$. Then if $S \\vDash t$ then $S \\vdash t$.\n\\end{coro}\n\nAnd hence,\n\\begin{thm} (6, completeness theorem)\\\\\nLet $S \\subset L$, $t \\in L$. Then $S \\vdash t \\iff S \\vDash t$.\n\\end{thm}\n\nSome consequences:\n\\begin{coro} (7, compactness theorem)\\\\\nLet $S \\subset L$, $t \\in L$ with $S \\vDash t$. Then $\\exists$ finite $S' \\subset S$ with $S' \\vDash t$.\\\\\nThis is trivial if we replace $\\vDash$ by $\\vdash$ (as proofs are finite).\n\\end{coro}\n\nSpecial case for $t=\\perp$: If $S$ has no model then some finite $S' \\subset S$ has no model. Equivalently,\n\\begin{coro} (7', compactness theorem, equivalent form)\\\\\nLet $S \\subset L$. If every finite subset of $S$ has a model then $S$ has a model.\\\\\nThis \\emph{isi} equivalent to corollary 7 because $S \\vDash t$ $\\iff S \\cup \\{\\neg t\\}$ has no model and $S' \\vDash t \\iff S' \\cup (\\neg t)$ has no model.\n\\end{coro}\n\n\\begin{coro} (8, decidability theorem)\\\\\nThere is an algorithm to determine (in finite time) whether or not, for a given finite $S \\subset L$ and $t \\in L$, we have $S \\vdash t$.\\\\\nThis is highly non-obviuos; however it's trivial to decide if $S \\vDash t$ just by drawing a truth table, and $\\vDash \\iff \\vdash$.\n\\end{coro}\n\n\\newpage\n\n\\section{Well-Orderings and Ordinals}\n\n\\begin{defi}\nA \\emph{total order} or \\emph{linear order} on a set $X$ is a relation $<$ on $X$, such that\\\\\n(1) Irreflexive: Not $xx$ if $xx_2>x_3>...$ in $X$.\\\\\nClearly if there is such a sequence then $S=\\{x_1,x_2,...\\}$ has no least element. Conversely, if $S \\subset X$ has no least element, then for each element $x \\in S$ there exists a $x' \\in S$ with $x' < x$, so we can just pick $x,x',...$ inductively.\n\\end{rem}\n\n\\begin{defi}\nWe say total orders $X,Y$ are \\emph{isomorphic} if there exists a bijection $f:X \\to Y$ that is order-preserving, i.e. $x a$, then some $x'>x$ has $f(x') = a$ by surjectivity, contradiction. The same shows $g(x) = $least element of $Y\\setminus \\{g(y):yy$ for all $y \\in X$. This is a well-ordering on $X \\cup \\{x\\}$: written $X^+$. Clearly $X < X^+$.\n\nPut some together:\\\\\nLet $(X,<_X)$ and $(Y,<_Y)$ be well-orderings. Say $Y$ extends $X$ if $X \\subset Y$, and $<_X$, $<_Y$ agree on $X$, and $X$ an initial segment of $(Y,<_Y)$.\\\\\nWell-orderings $(X_i: i \\in I)$ are nested if $\\forall i,j \\in I:$ $X_i$ extends $X_j$ or $X_j$ extends $X_i$.\n\n\\begin{prop} (7)\\\\\nLet $(X_i:i \\in I)$ be a nested family of well-orderings. Then there exist well-ordering $X$ with $X \\geq X_i$ $\\forall i$.\n\\begin{proof}\nLet $X = \\cup_{i \\in I} X_i$, with $x \\alpha$. Also, if $\\{\\alpha_i:i \\in I\\}$ is a set of ordinals, then there exists $\\alpha$ with $\\alpha \\geq \\alpha_i \\forall i$ (by applying prop 7 to the nested family of $I_{\\alpha_i};i \\in I$).\n\nIn fact, there is therefore a least upper bound for $\\{\\alpha_i: i \\in I\\}$ by applying prop 9 to the set $\\{\\beta \\leq \\alpha: \\beta$ an upper bound for the $\\alpha_i\\}$. This is written $\\sup\\{\\alpha_i:i \\in I\\}$, e.g. $\\sup\\{2,4,6,8,...\\} = \\omega$.\n\nSome ordinals: $0,1,2,...,\\omega,\\omega+1$(officially $\\omega^+$),$\\omega+2$,...,\\\\\n$\\omega+\\omega =\\omega2 = \\sup\\{\\omega+1,\\omega+2,...,\\}$, $\\omega^2+1,\\omega^2+2,...$,\\\\\n$\\omega3,...,\\omega4,...,...,\\omega\\omega = \\omega^2 = \\sup\\{\\omega,\\omega2,\\omega3,...\\}$,\\\\\n$\\omega^2+1,...,\\omega^2+\\omega,\\omega^2+\\omega+1,...\\omega^2+\\omega2,...,\\omega^2+\\omega^2=\\omega^2 2,...,\\omega^2 3,..., \\omega^2 4,..., \\omega^2 5,..., \\omega^2 \\omega = \\omega^3, ...\\omega^3 2,...,\\omega^4,...,\\omega^\\omega = \\sup\\{\\omega,\\omega^2,\\omega^3,...\\}$,\\\\\n$\\omega^\\omega+1,...,\\omega^\\omega 2,...,\\omega^\\omega \\omega = \\omega^{\\omega+1}$,\\\\\n$\\omega^{\\omega+2},...,\\omega^{\\omega+3},...,\\omega^{\\omega^2},...,\\omega^{\\omega^3},...,\\omega^{\\omega^\\omega}$,...\\\\\nAnd as expected we have $\\omega^{\\omega^{\\omega^{\\omega^{...}}}} = \\sup\\{\\omega,\\omega^2,\\omega^3,...\\} := \\varepsilon_0$, and then $\\varepsilon_0+1,...$, and then the whole thing again until $\\varepsilon_1 = \\varepsilon_0^{\\varepsilon_0^{...}}$.\n\nHowever, although this thing looks quite magnificent, they are all just countable (as we have just done it). Is there an uncoutnable ordinal? In other words, is there an uncountable well-ordered set?\n\n\\begin{thm} (11)\\\\\nThere is an uncountable ordinal.\n\\begin{proof}\n\\[IDEA: take sup of all countable ordinals. However, this might not be a set.\\]\\\\\nLet $R=\\{A \\in \\mathcal{P}(\\N \\times \\N)\\}$ s.t. $A$ is a well-ordering of a subset of $\\N$. Let $S$ be image of $R$ under 'order-type', i.e. $S$ is the set of all order-types of well-orderings of some subset of $\\N$. Then $S$ is the set of all countable ordinals. Let $\\omega_1$ be $\\sup S$. Then $\\omega_1$ is uncountable: otherwise, then $\\omega_1 \\in S$, so $\\omega_1$ would be the greatest member of $S$. But then $\\omega_1+1$ is also in $S$.\n\\end{proof}\n\\end{thm}\n\nNote that, by contradiction, $\\omega_1$ is the \\emph{least} uncountable ordinal. $\\omega_1$ has some strange properties, e.g.\\\\\n1. $\\omega_1$ is uncountable, but for any $\\alpha < \\omega_1$, we have $\\{\\beta:\\beta < \\alpha\\}$ countable.\\\\\n2. If $\\alpha_1,\\alpha_2,... < \\omega_1$ is any sequence, then it is bounded in $\\omega_1$: $\\sup\\{\\alpha_1,...,\\alpha_2\\}$ is countable, so is less than $\\omega_1$.\n\nSimilarly we have\n\n\\begin{thm} (11', Hartogs' lemma)\\\\\nFor any set $X$, there is an ordinal that does not inject into $X$.\\\\\nTo see that, just replace $\\mathcal{P}(\\N\\times\\N)$ by $\\mathcal{P}(X \\times X)$ in the previous proof.\\\\\n\\end{thm}\n\nWrite $\\gamma(X)$ for the least such ordinal -- e.g. $\\gamma(\\omega) = \\omega_1$.\n\n\\newpage\n\n\\subsection{Successors and limits}\nGiven ordinal $\\alpha$, does $\\alpha$ (any set of order-type $\\alpha$, e.g. $I_\\alpha$) have a greatest element?\\\\\nIf yes: say $\\beta$ is that greatest element. Then $\\gamma<\\beta$ or $\\gamma =\\beta$ $\\implies$ $\\gamma<\\alpha$, and $\\gamma<\\alpha \\implies \\gamma < \\beta$ or $\\gamma = \\beta$ (as we can't have $\\gamma > \\beta$). In other words, $\\alpha = \\beta^+$. In that case, we call $\\alpha$ a \\emph{successor};\\\\\nIf not: then $\\forall \\beta < \\alpha$, $\\exists \\gamma < \\alpha$ s.t. $\\gamma > \\beta$. So $\\alpha = \\sup\\{\\beta:\\beta < \\alpha\\}$. (this is false in general, e.g. $\\omega+5$). We call $\\alpha$ a \\emph{limit}.\n\nFor example, $5$ is a successor, $\\omega+5$ is a successor, $\\omega$ is a limit, $\\omega+\\omega$ is a limit. ($0$ is a limit as well).\n\nFor ordinals $\\alpha,\\beta$, define $\\alpha+\\beta$ by recursion on $\\beta$ ($\\alpha$ fixed) by: $\\alpha+0=\\alpha$, $\\alpha+\\beta^+ = (\\alpha+\\beta)^+$, $\\alpha+\\lambda = \\sup \\{\\alpha+\\gamma: \\gamma < \\lambda\\}$ for $\\lambda$ a non-zero limit.\n\nFor example, $\\omega+1 = (\\omega+0)^+ = \\omega^+$, $\\omega+2 = \\omega^{++}$, $1+\\omega = \\sup \\{1+\\gamma:\\gamma < \\omega\\} =\\omega$ -- so addition is not commutative.\n\nOfficially, by 'recursion on the ordinals', we mean: define $\\alpha+\\gamma$ on $\\{\\gamma:\\gamma \\leq \\beta\\}$ (a set) recursively, plus uniqueness. Similarly for induction: if know $p(\\beta) \\forall \\beta < \\alpha \\implies p(\\alpha)$ (for each $\\alpha$), then must have $p(\\alpha) \\forall \\alpha$. If not, say $p(\\alpha)$ false: then look at $\\{\\beta \\leq \\alpha : p(\\beta)$ false $\\}$.\n\nNote that $\\beta \\leq \\gamma \\implies \\alpha + \\beta \\leq \\alpha +\\gamma$ (induction on $\\gamma$). Also, $\\beta <\\gamma \\implies \\alpha+\\beta<\\alpha+\\gamma$. Indeed, $\\gamma \\geq \\beta^+$, so $\\alpha+\\gamma\\geq \\alpha+\\beta^+ = (\\alpha+\\beta)^+ > \\alpha+\\beta$. However, $1<2$, but $1+\\omega = 2+\\omega$.\n\n\\begin{prop} (12)\\\\\n$\\alpha+(\\beta+\\gamma) = (\\alpha+\\beta)+\\gamma \\forall \\alpha,\\beta,\\gamma$ ordinals.\n\\begin{proof}\nInduction on $\\gamma$:\\\\\n$0$: $\\alpha+(\\beta+0) = \\alpha+\\beta = (\\alpha+\\beta)+0$.\\\\\nSuccessors: $(\\alpha+\\beta)+\\gamma^+ = ((\\alpha+\\beta)+\\gamma)^+ = (\\alpha+(\\beta+\\gamma))^+ = \\alpha+(\\beta+\\gamma)^+ = \\alpha+(\\beta+\\gamma^+)$.\\\\\n$\\lambda$ a non-zero limit: $(\\alpha+\\beta)+\\lambda = \\sup\\{(\\alpha+\\beta)+\\gamma: \\gamma<\\lambda\\} = \\sup\\{\\alpha+(\\beta+\\gamma):\\gamma<\\lambda\\}$.\n\nClaim: $\\beta+\\lambda$ is a limit.\\\\\nProof of claim: We have $\\beta + \\gamma = \\sup\\{\\beta+\\gamma:\\gamma<\\lambda\\}$. But $\\gamma<\\lambda \\implies \\exists \\gamma' <\\lambda$ with $\\gamma <\\gamma'$ $\\implies $ $\\beta+\\gamma < \\beta+\\gamma'$. So $\\{\\beta+\\gamma : \\gamma<\\lambda\\}$ does not have a greatest element.\n\nBack to the main proof, now $\\alpha+(\\beta+\\gamma) = \\sup\\{\\alpha+\\delta:\\delta<\\beta+\\lambda\\}$. So want $\\sup\\{\\alpha+(\\beta+\\gamma):\\gamma<\\lambda\\{ = \\sup\\{\\alpha+\\delta:\\delta <\\beta+\\lambda\\}$. \\\\\n$\\leq$: $\\gamma < \\lambda \\implies \\beta+\\gamma<\\beta+\\lambda$, so LHS $\\subset$ RHS;\\\\\n$\\geq$: $\\delta < \\beta+\\lambda \\implies \\delta<\\beta+\\gamma$, some $\\gamma<\\lambda$ (definition of $\\beta+\\lambda$). So $\\alpha+\\delta \\leq \\alpha+(\\beta+\\gamma)$.\n\\end{proof}\n\\end{prop}\n\nAlternative viewpoint:\\\\\nAbove is the 'inductive' definition of $+$. Thereis also a synthetic definition: $\\alpha+\\beta$ is the order-type of $\\alpha \\sqcup \\beta$ ($\\alpha$ disjoint union $\\beta$), with all of $\\alpha$ coming before all of $\\beta$.\n\nClearly we have $\\alpha+(\\beta+\\gamma) = (\\alpha+\\beta)+\\gamma$ with this definition (same order-type). We need:\n\n\\begin{prop} (13)\\\\\nThe synthetic and inductive definition of $+$ coincide.\n\\begin{proof}\nWrite $\\alpha+\\beta$ for inductive, $\\alpha +' \\beta$ for synthetic. Do induction on $\\beta$ ($\\alpha$ fixed).\\\\\n$0$: $\\alpha+0=\\alpha=\\alpha+'0$:\\\\\nSuccessors: $\\alpha+'\\beta^+ = (\\alpha+'\\beta)^+ = (\\alpha+\\beta)^+=\\alpha+\\beta^+$;\\\\\n$\\lambda$ a non-zero limit: $\\alpha+'\\gamma$ = order-type of $\\alpha \\sqcup \\lambda$ = $\\sup$ of order-type of $\\alpha\\sqcup \\gamma$, $\\gamma <\\lambda$ (nest union, so order-type of union = $\\sup$ -- this was proved before) = $\\sup(\\alpha+'\\gamma:\\gamma<\\lambda) = \\sup(\\alpha+\\gamma:\\gamma<\\lambda) = \\alpha+\\lambda$.\n\\end{proof}\n\\end{prop}\n\nNormally we prefer to use synthetic than inductive, \\emph{if} we do have a synthetic definition available.\n\nOrdinal multiplication:\\\\\nDefine $\\alpha \\beta$ recursively by:\\\\\n$\\alpha 0 = 0$, $\\alpha(\\beta^+) = \\alpha\\beta + \\alpha$, $\\alpha\\lambda = \\sup\\{\\alpha\\gamma:\\gamma <\\lambda \\}$ for $\\lambda$ a non-zero limit. e.g:\\\\\n$\\omega 1 = \\omega 0 + \\omega = 0 + \\omega = \\omega$;\\\\\n$\\omega 2 = \\omega 1 + \\omega = \\omega+\\omega$;\\\\\n$\\omega\\omega = \\sup\\{0,\\omega,\\omega+\\omega,\\omega+\\omega+\\omega,...\\}$ (as in our big picture)\\\\\n$2\\omega = \\sup\\{2 \\gamma:\\gamma < \\omega\\} = \\omega$, so multiplication is not commutative.\n\nSimilarly, this also has a synthetic definition: $\\alpha\\beta$ is the order-type of $\\alpha\\times\\beta$, with $(x,y) < (z,t)$ if either $yx$, but no $z$ that $y>z>x$.\n\nHasse diagrams can be useful to visualize a poset (e.g. $\\N$, usual order), or useless (e.g. $\\Q$, usual order).\n\nIn a poset $X$, a \\emph{chain} is a set $S \\subset X$ that is totally ordered ($\\forall x,y\\in S: x \\leq y$ or $y \\leq x$). \n\nNote: chains can be uncountable, e.g. in $(\\R,\\leq)$ take $\\R$.\n\nWe say $S \\subset X$ is an antichain if no two elmeent are related.\n\nFor $S \\subset X$, an \\emph{upper bound} for $S$ is an $x \\in X$ s.t. $x \\geq y$ $\\forall y \\in S$.\n\nSay $X$ is a \\emph{least upper bound}, or \\emph{supremum} for $S$, if $x$ is an upper bound for $S$, and $x \\leq y$ for every upper bound $y$ of $S$.\n\nWrite $x = \\sup S$ or $x = \\vee S$.\n\ne.g. In $\\R$, $\\{x:x^2<2\\}$ has 7 as least upper bound, and $\\sup = \\sqrt{2}$ (so $\\sup S$ need not be in $S$). In $\\R$, $\\Z$ has no upper bound. In $\\Q$, $\\{x:x^2 < 2\\}$ has 7 as an upper bound, but no least upper bound.\n\nWe say a poset is \\emph{complete} if every subset has a sup.\n\ne.g. $(\\R,\\leq)$ is not complete: $\\Z$ has no sup (so different to notion of 'completeness' from analysis);\\\\\n$[0,1]$ is complete; $(0,1)$ is not complete: itself has no sup;\\\\\n$\\P(S)$ is always complete: $\\{A_i: i \\in I\\}$ has sup $\\cup_{i \\in I} A_i$.\n\nA function $f:X \\to X$, where $X$ is any poset, is order-preserving if $f(x) \\leq f(y)$ $\\forall x \\leq y$.\n\ne.g. on $\\N$ : $f(x) = x+1$; on $[0,1]: f(x) = \\frac{1+x}{2}$ (halve the distance to 1); on $\\P(S)$: $f(A) = A \\cup \\{i\\}$ for some fixed $i \\in S$.\n\nnot every order-preserving $f$ has a fixed point ($f(x) = x$), e.g. $f(x) = x+1$ on $\\N$.\n\n\\begin{thm} (1, Knaster-Tarski fixed point theorem):\\\\\nLet $X$ be a complete poset. Then every order-preserving function $f:X \\to X$ has a fixed point.\n\\begin{proof}\nLet $E = \\{x \\in X: x \\leq f(x)\\}$, and put $s = \\sup E$. To show $f(s) = s$, we'll show that $s \\leq f(s)$ and $s \\geq f(s)$.\\\\\n$s \\leq f(S)$: Enough to show $f(s)$ is an upper bound for $E$ (as $s$ the \\emph{least} upper bound). But $x \\in E$ $\\implies x \\leq s$ $\\implies f(x) \\leq f(s) \\implies x \\leq f(x) \\leq f(s)$.\\\\\n$s \\geq f(s)$: Enough to show $f(s) \\in E$ (as $s$ an upper bound). We know $s \\leq f(s)$, and want $f(s) \\leq f(f(s))$. But that's true because $f$ is order preserving.\n\\end{proof}\n\\end{thm}\n\nNote: in any complete poset $X$, we have a greatest element ($x s.t. x \\geq y \\forall y$), namely $\\sup X$. A typical application of knaster-tarski:\n\n\\begin{thm} (2, schr$\\ddot{o}$der-bernstein theorem)\\\\\nLet $a,B$ be sets s.t. there exists injection $f:A \\to B$ and an injection $g:B \\to A$. Then there exists an bijection from $A$ to $B$.\n\\begin{proof}\nSeek partition $A = P \\sqcup Q,B = R \\sqcup S$ s.t. $f(P) = R$ and $g(S) = Q$. Then we are done: set $h$ to be $f$ on $P$, $y^{-1}$ on $Q$, then $h:A \\to B$ is a bijection.\\\\\ni.e. we seek $P \\subset A$ s.t. $A \\setminus g(B\\setminus f(P)) = P$. Define $\\theta: \\mathcal{P}(A) \\to \\mathcal{P}(A)$ via $P \\to A \\setminus g(B\\setminus f(P))$. Then since $\\mathcal{P}(A)$ is complete, $\\theta$ order-preserving, there is a fixed point by K-T theorem.\n\\end{proof}\n\\end{thm}\n\n\\subsection{Zorn's Lemma}\nAn element $x$ in poset $X$ is \\emph{Maximal} if no $y \\in X$ has $y >x$.\n\nPosets need not have a maximal element, for example $\\Z,\\Q,\\R$.\n\n\\begin{thm} (3, Zorn's lemma)\\\\\nLet $X$ be a non-empty poset in which every chain has an u.b.. Then $X$ has a maximal element.\n\\begin{proof}\nSuppose not. Then for each $x \\in X$ there is some $x' \\in X$ with $x' > x$. Also, for any chain $C$ we have an upper bound $u(C)$. Pick $x \\in X$. Define $x_\\alpha \\in X$, each $\\alpha < \\gamma(x)$ ($\\gamma(x)$ is the u.b.?) recursively by: $x_0 = x$, $x_{\\alpha+1} = x'_\\alpha$, $x_\\lambda = u(\\{x_\\alpha: \\alpha < \\lambda\\})$ for $\\lambda$ a non-zero limit (this is a chain by induction). Then $\\alpha \\to x_\\alpha$ is an injection from $\\gamma(X) to X$.\n\\end{proof}\n\\end{thm}\n\nA typical application of Zorn: does every vecotr space have a basis? Recall that a basis is a LI spanning set.\n\ne.g. $V=$ space of all real polynomials. We can take $1,x,x^2,...$\\\\\nLet $V$ now be all real sequences. But $l_1 = (1,0,0,0,...)$, $l_2 = (0,1,0,0,...)$, then $l_1,l_2$ LI but not spanning! (recall span must be a finite linear combination!) It's easy to check that there is no countable basis. Also, it turns out that there is no \\emph{explicit} basis.\\\\\n$\\R$ as a vector space over $\\Q$. Basis is called a Hamel basis.\n\n\\begin{thm} (4)\nEvery vector space $V$ has a basis.\n\\begin{proof}\nLet $X = \\{A\\subset V: A$ is LI$\\}$, ordered by $\\subset$. We seek a maximal element $M$ of $X$ (then we are done: if $M$ does not span then choose $x \\not\\in \\bra M\\ket$, and now $M \\cup \\{x\\}$ is LI, contradiction.\\\\\nWe have $X\\neq\\phi$, as $\\phi \\in X$.\\\\\nGiven a chain $\\{A_i: i \\in I\\}$ in $X$, put $A = \\cup_{i \\in I} A_i$, then $A > A_i$ $\\forall i$, so just need $A \\in X$, i.e. $A$ LI. Suppose $A$ is not LI, hten $\\sum_{i=1}^n \\lambda_i x_i = 0$ for some $x_1,...,x_n \\in A$, and $\\lambda_i$ scalars not all zero. We have $x_i \\in A_{i_1},...,x_n \\in A_{i_n}$ for some $i_1,...,i_n \\in I$. But $A_{i_1},...,A_{i_n} \\in A_{i_k}$, some $k$ (as they are nested), contradicting $A_{i_k}$ being LI.\n\\end{proof}\n\\end{thm}\n\nNote: the only actualy maths (i.e. linear alebra) in the proof was the 'then done' part.\n\nAnother application: completeness theorem when proposition language uncountable.\n\n\\begin{thm} (5)\\\\\nLet $S \\subset L(P)$, where $P$ is any set. Then $S$ consistent implies that $S$ has a model.\n\\begin{proof}\nWe seek a maximal consistent $\\bar{S} \\supset S$. Then done: for each $t \\in L(p)$ we have $\\bar{S} \\cup \\{t\\}$ or $\\bar{S} \\cup \\{\\neg t\\}$ consistent (see chapter 1), hence $t \\in \\bar{S}$ or $\\neg t \\in \\bar{S}$ by maximality of $\\bar{S}$. Now define $v(t) = 1$ if $t \\in \\bar{S}$, $0$ otherwise (as in chapter 1). Let $X$ be the set of all consistent subsets of $L(P)$, ordered by $\\subset$. Then $X \\neq \\phi$, as $S \\in X$. Given a non-empty chain $(T_i:i \\in I)$ in $X$, put $T = \\cup_{i \\in I} T_i$. Then $T \\supset T_i$ for each $i$, so we just need $T \\in X$. We have $S \\subset T$ as $T \\neq \\phi$. Also $T$ is consistent: if $T \\vdash \\perp$, then $\\{t_1,...,t_n\\} \\vdash \\perp$ for some $t_1,...,t_n \\in T$. We have $t_1 \\in T_{i_1},...,t_n \\in T_{i_n}$ for some $i_1,...,i_n \\in I$. But $T_{i_1},...,T_{i_n} \\subset T_{i_k}$ for some $k$ (nested), contradicting $T_{i_k}$ being consistent.\n\\end{proof}\n\\end{thm}\n\nOne more:\n\n\\begin{thm} (6, well-ordering principle)\\\\\nEvery set $S$ can be well-ordered.\\\\\nNote that this is very surprising for e.g $S=\\R$.\n\\begin{proof}\nLet $X = \\{(A,R):A \\subset S$ and $R$ is a well-ordering of $A\\}$. We order this by: $(A,R) \\leq (A',R')$ if $(A',R')$ extends $(A,R)$. Then $X \\neq \\phi$, as $(\\phi,\\phi) \\in X$. Given a chain $((A_i,R_i) : i \\in I)$, we have $(\\cup_{i \\in I} A_i, \\cup_{i \\in I} R_i) \\in X$, and extends each $(A_i,R_i)$ from chapter 2. So by Zorn's lemma, $X$ has a maximal element $(A,R)$. We must have $A = S$: otherwise choose $x \\in S\\setminus A$ and take 'successor': well-order $A\\cup\\{x\\}$ by putting $x > a$ $\\forall a \\in A$, contradicting maximality of $(A,R)$.\n\\end{proof}\n\\end{thm}\n\n\\begin{rem}\nProof of zorn was easy, but we used a lot of machinery there (ordinals, recursion, hartog's lemma).\n\\end{rem}\n\n\\subsection{Zorn's lemma and the axiom of choice}\nIn proof of Zorn's kemma, we chose, for each $x \\in X$, and $x' \\supset x$, i.e. we made infinitely many arbitrary choices, even by time we get to $x_\\omega$. We did the same in part IA, to prove that a countable union of countable sets is countable. This is appealing to the axiom of choice, saying that we may choose an element of each set in a family of non-empty sets.\n\nMore precisely, the axiom of choice states that, if $(A_i : i \\in I)$ is a family of sets, we have a choice function, meaning a function $f:I \\to \\cup_{i \\in I} A_i$ s.t. $f(i) \\in A_i$ $\\forall i$. This is of a different characterto the other set-building rules in that the object whose existence is asserted is not uniquely specified by its properties (unlike ,e.g., $A \\cup B$).\\\\\nSo often one points out when one has used axiom of choice.\n\nNote that AC is trivial $|I| = 1$ ($A \\neq \\phi$ means $\\exists x \\in A$). Similarly for $I$ finite by induction. However, there is no derivation of AC from the other set-building rules for general $I$.\n\nAlso, we cannot prove ZL without AC because we can deduce AC from ZL:\\\\\nGiven family $(A_i:i \\in I)$ of non-empty sets, a partial choice function is an $f:J \\to \\cup_{i \\in I} A_i$ for some $J \\subset I$, s.t. $f(j) \\in A_j \\forall j \\in J$. Put $(J,f) \\leq (J',f')$ if $J \\subset J'$ and $f'|J = f$. This poset is not empty. Also, given a chain we have an upper bound being the union of them. So by ZL, there is a maximal of such. We must have $J=I$ in that case, as if not we can choose (???) $i \\in I \\setminus J$, $x \\in A_i$ and put $J' = J \\cup \\{i\\}$, $f' = f \\cup \\{(i,x)\\}$. Contradiction.\n\nConclusion: ZL $\\iff$ AC (in presence of the other set-building rules).\n\nAlso, we had $ZL \\implies WO$, and $WO \\implies AC$ trivially (well order $\\cup i \\in I A_i$ and let $f(i)$ be the least element of $A_i$). So we get $ZL \\iff AC \\iff WO$.\n\n\\subsection{The Bourbaki-Witt theorem}\nPoset $X$ is \\emph{chain-complete} if $X \\neq \\phi$ and every non-empty chain has a sup.\\\\\nFor example, any complete poset is chain-complete; any finite poset is chain-complete; and $\\{A \\subset V: A$ is LI$\\}$, for a vector space $V$ is also.\n\nWe say $f:X \\to X$ is \\emph{inflationary} if $f(x) \\geq x$ $\\forall x$.\n\n\\begin{thm} (Bourbaki-Witt)\\\\\n$X$ chain-complete, $f:X \\to X$ inflationary. Then $f$ has a fixed point.\\\\\nNote that BW follows instantly from ZL: take maximal $x$, and now $f(x) \\geq x$ $\\implies f(x) = x$.\\\\\nHowever, we can prove BW without AC: we pick some $x_0 \\in X$, then let $x_1 = f(x_0)$, $x_2 = f(x_1)$, ..., and let $x_\\omega$ be the sup of them.\n\nIn chapter 2, we did not use AC, except in remark that well-ordering $\\iff$ no decreasing sequence, and that $\\omega_1$ is not a countable sup.\n\\end{thm}\n\nIn fact, it's easy to deduce ZL from BW (using AC). So we can view BW as the choice-free version of ZL.\n\n\\newpage\n\\section{Predicate Logic}\nRecall that a group is a set equipped with functions:\\\\\n$M:A^2 \\to A$ ('arity' (slots) 2) and inverse $iA \\to A$ ('arity' 1), and a constant $e \\in A$ (kind of 'arity' 0), s.t.\n\\begin{equation*}\n\\begin{aligned}\n& (\\forall x,y,z \\in A) (M(x,M(y,z)) = M(M(x,y),z)),\\\\\n(\\forall x \\in A) (M(x,e) = x \\wedge M(e,x) = x),\\\\\n(\\forall x \\in A) (M(x,i(x)) = e \\wedge M(i(x),x) = e)\n\\end{aligned}\n\\end{equation*}\n\nAnd a poset is a set $A$ equipped with a predicate (relation) $\\leq$ (arity 2) $\\subset A^2$ s.t\\\\\n\\begin{equation*}\n\\begin{aligned}\n(\\forall x \\in A) (x \\leq x),\\\\\n(\\forall x,y,z \\in A) ((x \\leq y) \\wedge (y \\leq z) \\implies x \\leq z),\\\\\n(\\forall x,y \\in A) ((x \\leq y \\wedge y \\leq x) \\implies x = y)\n\\end{aligned}\n\\end{equation*}\n\nWe try to establish these correspondence between propositional logic and predicate logic:\nLanguage $\\to$ e.g. language of groups (thinks like the definitions above);\\\\\nValuation $\\to$ structure (set equipped with functions and relations of given arities);\\\\\nModel of $S$ (valuation making each $s \\in S$ true) $\\to$ model of $S$ (structure in which each $s \\in S$ holds);\\\\\n$S \\vDash t$ $\\to $ same (e.g. In language of groups, should have the above 3 definitions $\\vDash M(e,e) = e$ etc);\\\\\n$S \\vdash t$ $\\to $ same (but a bit more complicated).\n\nLet $\\Omega$ (function symbols) and $\\Pi$(relation symbols) be disjoint sets, and $\\alpha$ (arity) : $\\Omega \\cup \\Pi \\to \\N$. The \\emph{language} $L=L(\\Omega,\\Pi,\\alpha)$ is the set of \\emph{formulae}, defined by:\\\\\n$\\bullet$ variables: $x_1,x_2,x_3,...$ (can use $x,y$, etc);\\\\\n$\\bullet$ terms: defined inductively by:\\\\\n(i) each variable is a term;\\\\\n(ii) If $f \\in \\Omega$, $\\alpha(f) = n$, and $t_1,...,t_n$ are terms, then $ft_1...t_n$ is a term (and as always, we can add brackets, commas, etc).\nFor example, in the language of groups: $\\Omega = \\{m,i,e\\}$ of arities $2,1,0$, $\\Pi = \\phi$. Some terms: $x_1,m(x_1,x_2),e,m(e,e),m(x_1,i(x_1))$, etc.\\\\\n$\\bullet$ Atomic formulae, consists of:\\\\\n(i) $\\perp$;\\\\\n(ii) $(s=t)$, any terms $s,t$;\\\\\n(iii) $\\phi(t_1,...,t_n)$, any $\\phi \\in \\Pi$, $\\alpha(\\phi) = n$, and terms $t_1,...,t_n$.\\\\\nAgain use the language of groups as example: $m(x,y) = m(y,x)$, $m(x,i(x)) = e$;\\\\\nIn language of posets: $\\Omega = \\phi$, $\\Pi = \\{\\leq\\}$ of arity 2. We could take $x=y,x\\leq y,x \\leq x$.\\\\\n$\\bullet$ Formulae: defined inductively by:\\\\\n(i) Each atomic formula is a formula;\\\\\n(ii) If $p,q$ are formulae, then so is $(p \\implies q)$;\\\\\n(iii) If $p$ is a formulae, $x$ is a variable, then $(\\forall x) p$ is a formula.\\\\\ne.g. in language of groupsL $(\\forall x)(m(x,x) = e)$, $(\\forall x)((m(x,x) = e) \\implies (\\exists y)(m(y,y) = x))$ (note that we have not talked about $\\exists$ yet; we'll do that later).\\\\\nIn language of posets: $(\\forall x)(x \\leq x)$.\\\\\n\nNotes:\\\\\n1. A formula is just a string of symbols.\\\\\n2. We can now write $\\neg p$ for $p \\implies \\perp$, and similarly for $p \\wedge q$, $p \\vee q$ etc, and $(\\exists x) p$ for $\\neg(\\forall x)(\\neg p)$.\n\nA term is \\emph{closed} if it contains no variables. For example, $e,m(e,e),m(e,m(e,e))$. However, $m(x,i(x))$ is \\emph{not} closed.\\\\\nAn occurrence of variable $x$ in formular $p$ is \\emph{bound} if it is inside the brackets of '$\\forall x$' quantifier. Otherwise, it is \\emph{free}.\\\\\nFor example, in $m(x,x) = e \\implies (\\exists y) (m(y,y) = x)$, each $x$ is free and each $y$ is bound.\\\\\nNote that in some cases we can make a variable both free and bound: $(m(x,x) = e) \\implies (\\forall x)(\\forall y) (m(x,y) = m(y,x))$. We see that $x$ in LHS is free, but in RHS is bound (although it's not a very helpful expression).\n\nA \\emph{sentence} is a formula without free variables: e.g., $(\\forall x) (m(x,e) = x)$. For formula $p$, variable $x$, term $t$, the \\emph{substitution} $p[t/x]$ is obtained by replacing each free occurence of $x$ with $t$.\\\\\nFor example, if $p$ is $(\\exists y) (m(y,y) = x)$, then $p[e/x]$ is $(\\exists y) (m(y,y) = e)$.\n\n\\emph{Semantic entailment}:\nAn \\emph{$L$-structure} consists of a non-empty (see later wfor why) set $A$ equipped with, for each $f \\in \\Omega$ with $\\alpha(f) = m$, a function $f_A:A^m \\to A$, and for each $\\phi \\in \\Pi$, with $\\alpha(\\phi) = n$, a relation $\\phi_A \\subset A^n$.\n\nFor example, let $L$ be the language of groups: an $L$-structure is a set $A$ with functions $m_A:A^2 \\to A$, $i_A: A \\to A$, $e_A$ an element of $A$ (need not be a group! These have no 'meaning' yet).\\\\\nAnother example: $L$ be the language of posets: an $L$-structure is a set $A$ with a relation $\\leq_A \\subset A^2$.\n\nWe want to define the \\emph{interpretation} $p_A \\in \\{0,1\\}$ of a sentence $p$ in structure $A$, e.g. $(\\forall x)(m(x,x) = e)$ shold be 'true in $A$' if $\\forall a \\in A: m_A(a,a) = e_A$.\\\\\nSo: 'insert $\\in A$ subsubscript $A$ and say it aloud'.\n\n\\emph{Formal bit}:\nFor $L$-structure $A$, define \\emph{interpretation} of a closed term $t$ to be $t_A \\in A$, defined inductively by:\\\\\n$(ft_1...t_n)_A = f_A({t_1}_A,...,{t_n}_A)$ for any $f \\in \\Omega$, $\\alpha(f) = n$, closed terms $t_1,...,t_n$.\\\\\ne.g. $m(e,i(e))_A =m_A (e_A, i_A (e_A))$ (and $e_A$ already defined).\\\\\n\nAtomic formulae: define $p_A \\in \\{9,1\\}$ for $p$ atomic by:\\\\\n(i) $\\perp_A = 0$;\\\\\n(ii) \n\\begin{equation*}\n\\begin{aligned}\n(s=t)_A = \\left\\{\\begin{array}{ll}\n1 & s_A = t_A\\\\\n0 & else\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nfor $s,t$ closed terms;\\\\\n(iii) \n\\begin{equation*}\n\\begin{aligned}\n\\phi(t_1...t_n)_A = \\left\\{\\begin{array}{ll}\n1 & ({t_1}_A,...,{t_n}_A) \\in \\phi_A\\\\\n0 & else\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nfor $\\phi \\in \\Pi$, $\\alpha(\\phi) = n$, closed terms $t_1,...,t_n$.\n\nSentences: $p_A$ defined inductively by:\\\\\n(i)\n\\begin{equation*}\n\\begin{aligned}\n(p \\implies q)_A =\\left\\{\\begin{array}{ll}\n0 & p_A=1,q_A=0\\\\\n1 & else\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\n(ii)\n\\begin{equation*}\n\\begin{aligned}\n((\\forall i)_p)_A =\\left\\{\\begin{array}{ll}\n1 & p[\\bar{a}/x]_A = 1 \\text{ for all } a \\in A\\\\\n0 & else\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nwhere, for any $a \\in A$, add constant symbol $\\bar{a}$ to $L$, obtaining $L'$, and make $A$ an $L'$-structure by setting $\\bar{a}_A = a$.\n\nIf $p$ has free variables, we can define $p_A \\subset A^{\\text{number of free variables of } p}$.\\\\\ne.g. if $p$ is $(\\exists y)(m(y,y) = x)$, then $p_A = \\{a \\in A: \\exists b \\in A$ with $m_A(b,b) = a\\}$.\n\nIf $p_A = 1$, say $p$ \\emph{true} in $A$, or $p$ holds in $A$, or $A$ is a \\emph{model} of $p$. For $T$ a theoy (set of sentences), say $T$ semantically entails $p$, written $T \\vDash p$, if every model of $T$ is a model of $p$.\n\n$p$ is a \\emph{tautology} if $\\phi \\vDash p$ (or just $\\vDash p$), i.e. $p$ holds in every $L$-structure. For example, $\\vDash (\\forall x)(x=x)$.\n\nExamples: theory of groups: $\\Omega = (m,i,e)$, $\\Pi = \\phi$. Let\n\\begin{equation*}\n\\begin{aligned}\nT = \\{(\\forall x)(\\forall y)(\\forall z)(m(x,m(y,z)) = m(m(x,y),z),(\\forall x)(m(x,e) = x \\wedge m(e,x) = x),(\\forall x)(m(x,i(x)) = e \\wedge m(i(x),x) = e)\\}\n\\end{aligned}\n\\end{equation*}\nThen an $L$-structure is a model of $T$ $\\iff$ it is a group.\n\nSay $T$ 'axiomatises' the class of groups or 'axiomatises the theory of groups'.\n\nSometimes call the elements of $T$ the 'axioms' of $T$.\n\nTheory of fields: $\\Omega = \\{+,\\times,-,0,1\\}$. $T$ is: abelian group under $(+,-,0)$; $X$ is commutative, associative, distributive under $+$; $(\\forall x)(1x=x)$, $\\neg(1=0)$, $(\\forall x)((\\neg(x=0)) \\implies (\\exists y) (xy = 1))$. Then $T$ axiomatises the class of fields. E.g., $T\\vDash$ inverses are unique: $(\\forall x) ((\\neg(x \\neq 0)) \\implies ((\\forall y)(\\forall x)((yx = 1 \\wedge zx=1) \\implies y=z))$.\n\nTheory of posets: $\\Omega = \\phi, \\Pi = \\{\\leq\\}$.\n\n$T$ is: $(\\forall x)(x \\leq x)$, $(\\forall x)(\\forall y)(\\forall z)((x \\leq y \\wedge y \\leq z) \\implies x \\leq z)$, $(\\forall x)(\\forall y)((x \\leq y \\wedge y \\leq x) \\implies x=y)$.\n\nTheory of graphs: $\\Omega = \\phi$, $\\Pi =\\{a\\}$ ('is adjacent to').\n\n$T$ is $(\\forall x)(\\neg a(x,x))$, $(\\forall x)(\\forall y)(a(x,y) \\implies a(y,x))$.\n\nProofs:\n\nLogical axioms:\\\\\n(1) $p \\implies (q \\implies p)$ (any formulae $p,q$);\\\\\n(2) $p \\implies (q \\implies r)) \\implies ((p \\implies q) \\implies (p \\implies r))$ (any formulae $p,q,r$);\\\\\n(3) $(\\neg\\neg p) \\implies p$ (any formula $p$);\\\\\n(4) $(\\forall x) (x=x)$; (any variable $x$);\\\\\n(5) $(\\forall x)(\\forall y)(x=y) \\implies (p\\implies p[y/x]))$ (any variables $x,y$, formula $p$ where $y$ is a bound);\\\\\n(6) $((\\forall x) p) \\implies p [t/x]$ (any variable $x$, term $t$, formula $p$ with no variable in $t$ occuring bound in $p$)\\\\\n(7) $((\\forall x)(p \\implies q)) \\implies (p \\implies (\\forall x) q)$ (any variable $x$, formulae $p,q$ with $x$ not occurring free in $p$).\n\nAs rules of deduction, we take:\\\\\n\\emph{Modus Ponens}: From $p,p \\implies q$ can deduce $q$;\\\\\n\\emph{Generalisation}: From $p$ can deduce $(\\forall x)p$, if $x$ does not occur free in any premise used to prove $p$.\n\nFor $S \\subset L$, $p \\in L$, a proof of $p$ from $S$ is a finite sequence of formulae, ending with $p$, s.t. each line is a logical axiom, or a member of $S$, or follows from earlier lines by MP or GEN. Write $S\\vdash p$ ('$S$ proves $P$') if there exists a proof of $p$ from $S$.\n\nExample: $\\{x=y,x=z\\} \\vdash\\{y=z\\}$ (use axiom 5, with $p$ being $'x=z')$.\n\n1. $(\\forall x)(\\forall y) (x=y \\implies (x=z \\implies y=z))$ (axiom 5);\\\\\n2. $(\\forall x)(\\forall y)(x=y \\implies (x=z \\implies y=z)) \\implies (\\forall y)(x=y \\implies (x=z \\implies y=z))$ (axiom 6, $t='x'$);\\\\\n3. $(\\forall y)(x=y \\implies (x=z \\implies y=z))$ (MP on 1,2);\\\\\n4. $(\\forall y) (x=y \\implies (x=z \\implies y=z)) \\implies (x=y \\implies (x=z \\implies y=z))$ (axiom 6);\\\\\n5. $x=y \\implies (x=z \\implies y=z)$ (MP on 3,4);\\\\\n6. $x=y$ (hypothesis)\\\\\n7. $x=y \\implies y=z$ (mp on 5,6)\\\\\n8. $x \\implies z$ (hypothesis)\\\\\n9. $y=z$ (mp on 7,8).\n\nAim: $T \\vdash p$ $\\iff$ $T \\vDash p$.\n\ne.g. if $p$ holds in every group then $p$ can be proved from the three group axioms (completely obvious).\n\n\\begin{prop} (1, deduction theorem)\\\\\nLet $S \\subset L$, $p,q \\in L$. Then $S \\vdash (p\\implies q) \\iff S \\cup \\{p\\} \\vdash q$.\n\\begin{proof}\nForward: as for propositional logic, from $p \\implies q$ write down $p$ and apply MP to obtain $S \\cup \\{p\\} \\vdash q$;\\\\\nBackward: as for propositional logic: the only new case is 'generalisation'. So in proof of $q$ from $S \\cup \\{p\\}$ we have something like $r$ then $(\\forall x) r$ (Gen), and have a proof of $p \\implies r$ from $S$ (induction), and we want $S \\vdash p \\implies (\\forall x) r$. In proof of $r$ from $S \\cup \\{p\\}$, no premise had $x$ free. So in proof of $p \\implies r$ from $S$, no premise had $x$ free. Hence $S \\vdash (\\forall x)(p \\implies r$ (gen).\\\\\n$\\bullet$ If $x$ does not occur free in $p$: we have $S \\vdash p \\implies (\\forall x) r$ by axiom 6 and MP;\\\\\n$\\bullet$ If $x$ does occur free in $p$: proof of $r$ from $S \\cup \\{p\\}$ cannot have used $p$. So in fact $S \\vdash (\\forall x) r$ whence $S \\vdash (p \\implies (\\forall x) r)$ by axiom 1.\n\\end{proof}\n\\end{prop}\n\n\\begin{prop} (2, soundness)\\\\\nLet $S$ be a set of sentences, $p$ a sentence. Then if $S \\vdash p$ then $S \\vDash p$.\n\\begin{proof}\nWe have proof of $p$ from $S$, and a model $A$ of $S$, and we want $p_x = 1$. This is an induction down the lines of the proof.\n\\end{proof}\n\\end{prop}\n\nFor adequacy, we want if $S \\vDash p$, i.e. that if $S \\cup \\{\\neg p\\} \\vDash \\perp$, then $S \\cup \\{\\neg p\\} \\vdash \\perp$.\n\n\\begin{thm} (3, model existence lemma, or completeness theorem)\\\\\nLet $S \\subset L$ be a set of setences. Then $S$ consistent implies that $S$ have a model.\\\\\nIdeas:\\\\\n$\\bullet$ 1. Build model out of language: let $A$ be the set of closed terms of $L$, with operation line $(1+1) +_A (1+1) = (1+1) +(1+1)$;\\\\\n$\\bullet$ 2. Say for $S$ be the theory of fields: $(1+1)+1 \\neq 1+(1+1)$, but $S \\vdash (1+1)+1 = 1+(1+1)$. So quotient out by $s \\sim t$ if $S \\vdash s = t$;\\\\\n$\\bullet$ 3. Suppose $s$ is the fields of characteristic $2$ or $3$, i.e. field axioms, and the statement $1+1=0 \\vee 1+1+1=0$. Then $S \\not\\vdash 1+1=0$. So $[1+1] \\neq [0]$, where $[\\cdot]$ denotes the equivalent class unrder $\\sim$. Also, $S \\not\\vdash 1+1+1=0$, so $[1+1+1] \\neq [0]$.\n\nSo our structure does not satisfy $1+1=0 \\vee 1+1+1 = 0$. Then we need to extend $S$ to maximal consistent.\n\n$\\bullet$ 4. If $S$ is 'fields with a sqaure root of $2$': field axioms + $(\\exists x) (xx=1+1)$. Maybe no closed term $t$ has $[tt] = [1+1]$. So $s$ lacks 'witnesses'.\\\\\nSolution: for each $(\\exists x|p$ in $S$, add new constant $c$ to language, and add $p[c/x]$ to $S$. (e.g. $cc=1+1$).\\\\\nNow no longer maximal consistent, so go back to step 3.\\\\\nProblem: this might not terminate.\n\\begin{proof}\nWe have consistent $S$ in language $L_0 = L(\\Omega,\\Pi)$. Extend to maximal consistent $S_1$ (zorn), so for each sentence $p \\in L$, we have $p \\in S_1$, or $(\\neg p) \\in S_1$. Thus $S_1$ is complete (for every $p$, $S_1 \\vdash p$ or $S_1 \\vdash (\\neg p)$). Add witnesses: for each $(\\exists x) p$ in $S_1$, add new constant $c$ and axiom $p[c/x]$. We obtain $T_1$ in language $L_1 = L(\\Omega \\cup C_1,\\Pi)$ that has \\emph{witnesses} for $S_1$ (if $(\\exists x) p \\in S$, then some closed term $t$ has $p[t/x] \\in T_1$). It's easy to check $T_1$ consistent. Now extend $T_1$ to maximal consistent $S_2$ (in $L$). Add witnesses, obtaining $T_2$ in language $L_2 =L(\\Omega \\cup C_1 \\cup C_2, \\Pi)$.\\\\\nContinue inductively.\\\\\nPut $\\bar{S} = S_1 \\cup S_2 \\cup ...$. In language $\\bar{L} = L(\\Omega \\cup C_1\\cup C_2 \\cup ...)$.\\\\\n$\\bullet$ $\\bar{S}$ is consistent: If $\\bar{S} \\vdash \\perp$, then some $S_n \\vdash \\perp$ (as proofs are finite), contradiction;\\\\\n$\\bullet$ $\\bar{S}$ is complete: given sentence $p \\in \\bar{L}$, we have $p \\in L_n$ for some $n$ (as $p$ mentions only finitely many constants), so $S_{n+1} \\vdash p$ or $S_{n+1} \\vdash (\\neg p)$ (choice of $S_{n+1}$).\\\\\n$\\bullet$ $\\bar{S}$ has witnesses (for itself): given $(\\exists x)p \\in \\bar{S}$, we have $(\\exists x)p \\in S_n$ for some $n$. So $p[t/x] \\in T_n$ for some closed term $t$ (choice of $T_n$), whence $p[t/x] \\in \\bar{S}$.\n\\end{proof}\n\\end{thm}\n\nOn set of closed terms of $\\bar{L}$, define $s \\sim t$ if $\\bar{S} \\vdash (s=t)$.\n\nThis is clearly an equivalent relationship. let $A$ be the set of equivalent clases. Make $A$ into an $\\bar{L}$-structure by setting $f_A([t_1],...,[t_2]) = [ft_1...t_n]$ (each $f \\in \\bar{\\Omega},\\alpha(f) = n$, closed terms $t_1...t_n$), $\\varphi_A = \\{([t_1],...,[t_n]): \\bar{S} \\vdash \\phi(t_1,...,t_n)\\}$ (each $\\phi \\in \\Pi$, $\\alpha(\\phi) = n$, closed terms $t_1...t_n$).\n\nClaim: $\\phi_A = 1 \\iff \\bar{S} \\vdash p$ for each setnence $p \\in \\bar{L}$. (Then done: $A$ is a model of $\\bar{S}$, so $A$ is a model of $S$.\n\\begin{proof}\nAn easy induction:\\\\\n\\emph{Atomic sentences}:\\\\\n$\\perp$: $\\perp_A = 0$ and $\\bar{S} \\not\\vdash \\perp$.\\\\\n$s=t$: \n\\begin{equation*}\n\\begin{aligned}\n\\bar{S} \\vdash (s=t) &\\iff [s] = [t]\\\\\n&\\iff s_A = t_A\\\\\n&\\iff (s=t)_A = 1\n\\end{aligned}\n\\end{equation*}\n$\\phi(t_1...t_n)$: same.\n\n\\emph{Induction step}:\\\\\n$p \\implies q$: \n\\begin{equation*}\n\\begin{aligned}\n\\bar{S} \\vdash (p \\implies q) &\\iff \\bar{S} \\vdash (\\neg p)\\text{ or } \\bar{S} \\vdash q\\\\\n&\\iff p_A = 0\\text{ or } q_A = 1 (induction)\\\\\n&\\iff (p \\implies q)_A =1 \n\\end{aligned}\n\\end{equation*}\nwhere the second step is because, say if the forward direction doesn't hold, then $\\bar{S} \\vdash p$, $\\bar{S} \\vdash (\\neg q)$ (since $\\bar{S}$ is complete), but then $\\bar{S} \\vdash \\neg(p \\implies q)$, contradiction).\n\n$(\\exists x) p$:\n\\begin{equation*}\n\\begin{aligned}\n\\bar{S} \\vdash (\\exists x) p &\\iff \\bar{S} \\vdash p[t/x]\\\\\n&\\iff p[t/x]_A = 1\\\\\n&\\iff ((\\exists x)p)_A = 1\n\\end{aligned}\n\\end{equation*}\nfor some closed term $t$. The last line is because $A$ is the set of equivalent classes of closed terms.\n\\end{proof}\n\nBy remark before theorem 3 we have\n\n\\begin{coro} (4,adequacy)\\\\\nIf $S \\vDash p$, then $S \\vdash o$.\n\\end{coro}\n\nHence:\n\\begin{thm} (5, G$\\ddot{o}$del's completeness theorem for first-order logic)\\\\\nLet $S$ be a set of sentences and $p$ a sentence (in language $L$). Then $S \\vDash p \\iff S \\vdash p$.\\\\\nThe proof is just soundness + adequacy.\n\\end{thm}\n\nNote:\\\\\n$\\bullet$ If $L$ is countable (i.e .$\\Omega,\\Pi$ countable), then we don't need Zorn's lemma;\\\\\n$\\bullet$ 'First-order' means variables range over elements of our structure (not, e.g., subsets).\n\n\\begin{thm} (6, compactness)\\\\\nLet $S \\subset L$ be a set of sentences. Then if every finite subset of $S$ has a model, then $S$ has a model.\n\\begin{proof}\nThis is trivial if we replace $\\vDash$ with $\\vdash$ (as proofs are finite).\n\\end{proof}\n\\end{thm}\n\nNote: we have no decidability theorem -- how to check if $S \\vDash t$?\n\nSome consequences of completeness/compactness:\\\\\nCan we axiomatise the class of finite groups? In other words, we want some sentences $S$ (in language of groups) s.t. a structure is a model for $S$ $\\iff$ it is a finite group.\n\nHowever, this is not possible.\n\n\\begin{coro} (7)\\\\\nthe class of finite groups cannot be axiomatised (in language of groups).\n\\begin{proof}\nSuppose $S$ axiomatises finite groups. We add to $S$ the sentences:\n\\begin{equation*}\n\\begin{aligned}\n(\\exists x_1) (\\exists x_2) (\\neg (x_1 =x_2))\\\\\n(\\exists x_1) (\\exists x_2) (\\exists x_3) (\\neg (x_1 = x_2) \\wedge \\neg(x_1 = x_3) \\wedge \\neg(x_2 = x_3))\\\\\n...\n\\end{aligned}\n\\end{equation*}\nwhich stands for $|G| \\geq 2$, $|G| \\geq 3$, etc.\\\\\nThen ever finite subset has a model (e.g. $\\Z_n$, $n$ large). However, the set itself has no model -- contradicting compactness.\n\\end{proof}\n\\end{coro}\n\nSimilarly,\n\\begin{coro} (7')\\\\\nLet $S$ be a theory in a language $L$. Then if $S$ has arbitrarily large finite models, then it has an infinite model.\n\\begin{proof}\nAdd sentences as in corollary 7, and apply compactness theorem.\n\\end{proof}\n\\end{coro}\n\nSo we know \\emph{finiteness is not a first-order property}.\n\n\\begin{coro} (8, upward L$\\ddot{o}$wenheim-Skolem theorem)\\\\\nIf a theory $S$ has an infinite model, then it has an uncoutnable model.\n\\begin{proof}\nAdd uncoutnably many consttants $\\{c_i:i \\in I\\}$ to the language, and add to $S$ the set of sentences $c_i \\neq c_j$ (for each distinct $i,j \\in I$). Then any finite subset has a model. So the whole set has a model by compactness.\n\\end{proof}\n\\end{coro}\n\nSimilarly, we could find a model into which $P(P(R))$ injects (choose $I = P(P(R))$). E.g., there exists an infinite field ($\\Q$), so there exists field as big as $P(P(R))$.\n\n\\begin{coro} (9, downward L$\\ddot{o}$wenheim-Skolem theorem):\\\\\nLet $S$ be a theory in countable language $L$. If $S$ has a model, then it has a countable model.\n\\begin{proof}\nThe model constructed in theorem 3 is countable.\n\\end{proof}\n\\end{coro}\n\n\\subsection{Peano Arithmetic}\nWe try to make the usual axioms for $\\N$ into a first-order theory.\\\\\n$L : \\Omega = \\{0,s,+,\\times\\}$, $\\Pi = \\phi$, axioms:\\\\\n1. $(\\forall x) (\\neg s(x) = 0)$;\\\\\n2. $(\\forall x)(\\forall y) (s(x) = s(y) \\implies x=y)$;\\\\\n3. $(\\forall y_1)...(\\forall y_n) [(p[0/x] \\cap (\\forall x) (p \\implies p[s(x)/x])) \\implies (\\forall x) p]$.\\\\\n($y_i$ in 3 are parameters).\\\\\n4. $(\\forall x) (x+0=x)$;\\\\\n5. $(\\forall x)(\\forall y) (x+s(y) = s(x+y))$;\\\\\n6. $(\\forall x) (x+0=0)$;\\\\\n7. $(\\forall x)(\\forall y) (x\\times(y) = (x+y)+x)$.\n\nThese axioms are called Peano Arithmetic or Formal Number Theory.\n\nNote on axiom 3: first guess shold have been \n\\begin{equation*}\n\\begin{aligned}\n(p[0/x] \\cap (\\forall x|(p \\implies p[s(x)/x])) \\implies (\\forall x) p\n\\end{aligned}\n\\end{equation*}\nBut then missing properties like $x \\geq y$ ($y$ chosen earlier).\n\nThen PA has an infinite model, so by upward L-S, PA has an uncountable model that is not isomorphic to $\\N$ trivially. Doesn't this contradict the fact that the usual axioms characterise $\\N$ uniquely?\n\nAnswer: axiom 3 is only 'first-order induction' -- even in $\\N$ itself, it refers to only countably many subsets (as opposed to true induction).\n\nA subset $S \\subset \\N$ is called \\emph{definable} if there exists $p \\in L$, free variable $x$, s.t. $\\forall m \\in \\N$ we have: $m \\in S \\iff p[m/x]$ holds in $\\N$ (where by $m$ we mean $1+1+...+1$ ($m$ times)).\n\ne.g. set of squares: $p(x)$ is $(\\exists y) (yy=x)$;\\\\\nset of primes: $p(x)$ is: $\\neg(x=0) \\cap \\neg(x=1) \\neg (\\forall y) (y |x) \\implies ((y=1) \\vee (y=x))$, where $y|x$ is a short hand for $(\\exists z) (yz = x)$, and by $1$ we mean $s(0)$.\\\\\nPowers of $2$: $p(x)$ is $(\\forall y) ((y|x \\wedge y\\ prime) \\implies (y=2))$.\n\nExercise: powers of 4; challenge: powers of 6.\n\nIs PA complete? in other words, for each sentence $p$, PA $\\vdash p$ or PA $\\vdash \\neg p$?\n\n\\begin{thm} (G$\\ddot{o}$del's incompleteness theorem)\\\\\nPA is not complete.\\\\\nTake $p$ with PA $\\not\\vdash p$, $PA \\not\\vdash \\neg p$. We have $p$ holding in $\\N$ or $(\\neg p)$ holding in $\\N$. Conclution: $\\exists$ sentence $p$ s.t. $p$ is true in $\\N$, but $PA \\not\\vdash p$.\n\\end{thm}\n\nThis does not contradict completeness; it shows that if $p$ true in all models of PA, then PA $\\vdash p$.\n\n\\newpage\n\\section{Set Theory}\nAim: what does 'the universe of sets' look like?\n\nKey starting point: view set theory as 'just another finite-order theory'.\n\n\\subsection{Zermelo-Fraenkel set theory}\nWe have $L$: $\\Omega = \\phi$, $\\Pi = \\{\\varepsilon\\}$, $\\alpha(\\epsilon) = 2$.\n\nWe'll have the ZF axioms: 2 to get started, 4 to build things, and 3 you might not think of at first.\n\nThen a 'universe of sets' will mean a model ($V,\\epsilon$) of the ZF axioms.\n\n1. \\emph{Axiom of extension}:\\\\\nIf two sets have the same mebmers, then they are equal:\\\\\n$(\\forall x)(\\forall y) ((\\forall z) (z \\in x \\iff z \\in y) \\implies (x=y))$.\n\nNote: converse is an instance of a logical axiom.\n\n2. \\emph{Axiom of separtion}:\\\\\nWe can form a subset of a set, or precisely, given set $x$ and property $p(z)$, we can form the set of all $z \\in x$ such that $p(z)$ holds:\\\\\n$(\\forall t_1) ... (\\forall t_n) (\\forall x)(\\exists y) (\\forall z) (z \\in y \\iff (z \\in x \\wedge p))$\\\\\nThis is actually an axiom scheme: for each formula $p$ and free variables $t_i$.\n\nNote: we do want parameters, e.g. to have $\\{z \\in x: t \\in z\\}$, $t$ chosen earlier.\n\n3. \\emph{Axiom of empty-set}:\\\\\nThere is a set with no members.\\\\\n$(\\exists x) (\\forall y) (\\neg y \\in x)$.\n\nWe write $\\phi$ for the unique (by extension axiom) such set $x$. This is just an abbreviation: so $p(\\phi)$ means $(\\exists x) ((\\forall y) (\\neg y \\in x) \\wedge p(x))$.\n\nSimilarly, write $\\{z \\in x: p (z)\\}$ for the set guaranteed by separation.\n\n4. \\emph{Axiom of pair-set}:\\\\\nWe can form $\\{x,y\\}$.\\\\\n$(\\forall x)(\\forall y)(\\exists z) (\\forall t) (t \\in z \\iff t = x \\vee t = y)$.\n\nWe write $\\{x,y\\}$ for this set, and $\\{x\\}$ for $\\{x,x\\}$.\\\\\nWe can now define the 'ordered pair' $(x,y)$ to be $\\{\\{x\\},\\{x,y\\}\\}$.\\\\\nIt's easy to check that $(x,y) = (t,u) \\implies x=t \\wedge y=u$ (follows from axiom so far).\\\\\nSay $x$ is an ordered pair if $(\\exists y) (\\exists z) (x=(y,z))$, and we say $f$ is a function to mean $(\\forall x) (x \\in f \\implies x$ is an ordered pair) $\\wedge (\\forall x)(\\forall y)(\\forall z)((x,y) \\in f \\wedge (x,z) \\in f \\implies y=z)$.\n\nCan now define the domain of a function as follows: write $x = Dom f$ if $(f$ is a function) $\\wedge (\\forall z)(z \\in x \\iff (\\exists t)((z,t) \\in f)))$.\n\nAnd write $f:x \\to y$ for $(f$ is a function) $\\wedge (x=Dom f|\\wedge (\\forall z)((\\exists t)((t,z) \\in f) \\implies z \\in y))$.\n\n5. \\emph{Axiom of union}:\\\\\nWe can form unions.\\\\\n$(\\forall x)(\\exists y)(\\forall z)(z \\in y \\iff (\\exists t) (z \\in t \\wedge t \\in x))$.\n\n6. \\emph{Axiom of power-set}:\\\\\nWe can form power-sets.\\\\\n$(\\forall x)(\\exists y) (\\forall z) (z \\in y \\iff z \\subset x)$.\\\\\nHere by $z \\subset x$ we mean $(\\forall t)(t \\in z \\implies t \\in x)$.\n\nNotes:\\\\\n1. write $\\cup x$ and $\\mathcal{P}(x)$ for these two sets. We can write $x \\cup y$, etc.\\\\\n2. No extra axiom needed for interseionts: we can form $\\cap x$ ($x \\neq \\phi$) as a subset of $y$ any $y \\in x$. So ok by separation.\\\\\n3. We can now form $x \\times y$ as a suitable subset of $\\mathcal{P}\\mathcal{P}(x\\cup y)$ -- since if $t \\in x,u \\in y$, then $(t,u) = \\{\\{t\\},\\{t,u\\}\\}\\in\\mathcal{P}\\mathcal{P}(x \\cup y)$. And then we can form the set of all functions from $x$ to $y$, as a subset of $\\mathcal{P}(x \\times y)$.\n\nThe next three are more subtle:\n\n7. \\emph{Axiom of infinity}:\\\\\nSo far, $V$ (the branch symbol) must be inifinite. For example, write $x^+ = x \\cup \\{x\\}$, then easy to check that $\\phi,\\phi^+,\\phi^{++},...$ are all distinct. We often write $0$ for $\\phi$, $1$ for $\\phi^+$,$2$ for $\\phi^{++}$, etc. So $1 = \\{0\\}, 2 = \\{0,1\\},3=\\{0,1,2\\}$,etc. But does the structure $(V,\\epsilon)$ have an infinite set -- e.g. $x$ with $\\phi \\in x,\\phi^+ \\in x$, ...?\n\nWe say $x$ is a successor set if $(\\phi \\in x) \\wedge (\\forall y) (y \\in x \\implies y^+ \\in x)$.\n\nNow let's state the axiom:\\\\\nThere is an infintie set/there is a successor set.\\\\\n$(\\exists x) (x$ is a successor set$)$.\n\nNote that any intersection of successor sets is a successor set, so there exists a least one, called $\\omega$. This will be our version, in $V$, of the natural numbers.\n\nThus $(\\forall x)(x \\in \\omega \\iff (\\forall y)(y$ a successor set $\\implies x \\in y))$.\n\nNote that if $x \\subset \\omega$ is a successor set then $x=\\omega$ by definition:\\\\\n$(\\forall x)(x \\subset \\omega \\wedge \\phi \\in x \\wedge (\\forall y)(y \\in x \\implies y^+ \\in x)) \\implies x=\\omega)$. This is induction: genuine induction, over all $x \\subset \\omega$ (as opposed to in PA).\n\nAlso, it's easy to check $(\\forall x \\in \\omega) (\\neg x^+ = \\phi)$, and $(\\forall x \\in \\omega)(\\forall y \\in \\omega)(x^+ = y^+ \\implies x=y)$.\n\nThus: $\\omega$ satisfies (in $V$) all the usual axioms for the natural numbers.\n\nSay $x$ is finite if $(\\exists y) (y \\in \\omega \\wedge x$ bijects with $y$).\n\nAnd then $x$ is countable if $x$ is finite or $x$ bijects with $y$.\n\n8. \\emph{Axiom of Foundation}:\\\\\n\"Sets are build up from simpler sets\". We want to disallow $x \\in x$: note that $\\{x\\}$ has no $\\varepsilon$-minimal member; and also disallow $x \\in y \\in x$: note $\\{x,y\\}$ has no $\\varepsilon$-minimal element, etc. And we also want to disallow the infinite sequence $x_1 \\in x_0$, $x_2 \\in x_1$, $x_3 \\in x_2$,..., in which case $\\{x_0,x_1,...\\}$ has no $\\varepsilon$-minimal element.\n\nThe axiom: every (non-empty) set has an $\\varepsilon$-minimal element.\\\\\n$(\\forall x) (x \\neq \\phi \\implies (\\exists y) (y \\in x \\wedge (\\forall z) (z \\in x \\implies z \\not\\in y))$.\n\nBonus lecture on next Wednesday 1pm (proof of incompleteness theorem, consistency of ZF)\n\n9. \\emph{Axiom of Replacement}:\\\\\nWe often say \"for each $i \\in I$ have $A_i$ -- take $\\{A_i: i \\in I\\}$. However, how do we know they form a set? Alternatively, how do we know that $i \\to A_i$ is a function?\\\\\nWe want to say \"the image of a set under something that looks like a function is a set\".\n\nA digression on classes:\\\\\nIdea: $x \\to \\{x\\}$ (for all $x$). This looks like a function, but it isn't: e.g. every function has a domain as functions are sets of ordered pairs, and the domain is just the left element of all those pairs. However, the 'domain' of $x \\to \\{x\\}$ is not a set (the universal 'set').\n\nFor an $L$-structure $V$, a collection $C$ of elements of $V$ is called a \\emph{class} if there is a formula $p$, free variables $x$ (and maybe more) s.t. $x \\in C$ $\\iff p(x)$ holds in $V$. E.g. $V$ is a class: take $p(x)$ to be $x=x$.\n\nFor any $t$, $\\{x: t \\in x\\}$ is a class: take $p(x)$ to be $t \\in x$.\\\\\nNote that every set $y$ is a class: take $p(x)$ to be $x \\in y$.\n\nIf $C$ is not a set (in $V$), i.e. $\\not (\\exists y)(\\forall x)(x \\in y \\iff p(x))$, say $C$ is a proper class. E.g., $V$ is a proper class, as is $\\{x:x$ infinite$\\}$, where by infinite we mean not finite.\n\nSimilarly, a function-class is a collection $F$ of ordered pairs from $V$, s.t. for some formula $p$, free variables $x,y$ (and maybe more), have $(x,y) \\in F \\iff p(x,y)$, and if $(x,y) \\in F, (x,z) \\in F$, then $y=z$.\\\\\nFor example, $x \\to \\{X\\}$ is a function class: take $p(x,y)$ to be $y=\\{x\\}$.\n\n---End of digression---\n\nLet's now state the axiom of replacement: \"the image of a set under a function-class is a set.\\\\\n$(\\forall t_1) ... (\\forall t_n) ([(\\forall x)(\\forall y)(\\forall z) ((p \\wedge p[z/y] ) \\implies y=z)] \\implies [(\\forall x)(\\exists y) (\\forall z) (z \\in y \\iff (\\exists t) (t \\in x \\wedge p[t/x,z/y])])$\\\\\nFor each formula $p$, free variables $x,y,t_1,...,t_n$, i.e., the image of $x$ under $p$ is a set.\n\nEg. for any set $x$, we can form $\\{\\{t\\}:t \\in x\\}$ using function class $t \\to \\{t\\}$.\n\nThis is a 'bad' example, as it didn't need replacement -- see later for 'good' examples.\n\nThose are the ZF axioms.\n\nNote:\\\\\n1: Sometimes separation is called 'comprehension', and sometimes fundation is called 'regularity'.\\\\\n2. ZF axioms do not include AC: ZF + AC is called ZFC, where axiom of choice is: \"every family of (non-empty) sets has a choice function\" -- $ (\\forall f) (f$ is a function $\\wedge (\\forall x) (x \\in Dom f \\implies f(x) \\neq \\phi)) \\implies (\\exists y) (y$ is a function $ \\wedge Dom y = Dom f \\wedge (\\forall x)(x \\in Dom f \\implies g(x) \\in f(x))))$.\n\nGoal: what does a model $(V,\\epsilon)$ of ZF look like?\n\nRemark: we haven't proved ZF consistent (i.e. $\\exists$ model of ZF). Sadly, ZF $\\not\\vdash$ \"ZF has a model\", i.e. it cannot be proved in ordinary maths (ZF or ZFC).\n\nSay $x$ is transitive if every member of $x$ is itself a member of $x$: $(\\forall y)((\\exists z)(y \\in z \\wedge z \\in x) \\implies (y \\in x)$, i.e. $\\cup x \\subset x$.\n\nE.g. $2=\\{\\phi,\\{\\phi\\}\\}$ is transitive; $\\omega$ is transitive as $n=\\{0,1,...,n-1\\}$ $\\forall n \\in \\omega$.\n\nLemma 1: every set $x$ is contained in a transitive set.\n\nRemarks: 1. Officially, let $(V,\\epsilon)$ be a model of ZF. Then in $V$, ... holds, or equivalently, $ZF \\vdash ...$.\\\\\n2. Any $\\cap$ of transitive sets is transitive, so we'll then know that there exists a least transitive set containing $x$, called the transitive closure of $x$, written $TC(x)$.\n\\begin{proof}\nWe'll take $x \\cup (\\cup x) \\cup(\\cup\\cup x) \\cup)\\cup\\cup\\cup x) \\cup ...$ which is a set by union axiom, which is a set by replacement (a good example of replacement): $0 \\to x, 1 \\to \\cup x$, etc. But why is this a function class?\\\\\nTo show that, define $f$ is a an attempt to mean (recall we've done similar things before in chapter 2) $(f$ is a function $) \\cap (Dom f \\in \\omega) \\cap (Dom f \\neq \\phi) \\cap (f (0) = x) \\cap (\\forall n) (n \\in Dom f \\cap n \\neq 0 \\implies f(n) = \\cup f(n-1))$. Then $(\\forall n\\in\\omega) (\\forall f) (\\forall f') ((f,f'$ attempts $\\wedge n \\in Dom f') \\implies f(n) = f'(n))$ (by $\\omega$-induction). And $(\\forall n \\in \\omega) (\\exists f) (f$ an attempt $\\cap n \\in Dom f)$ (again, by $\\omega$-induction). So take $p(y,z)$ to be $(\\exists f) (f$ an attempt $\\cap$ $y \\in Dom f \\cap f(y) = z)$.\n\\end{proof}\n\nWe want foundation to be saying 'sets are built out of simpler sets'. If so, we would want: suppose $p(y) \\forall y \\in x$ implies $p(x)$, then $p(x) \\forall x$.\n\n\\begin{thm} (2, principle of $\\epsilon$-induction): let $p$ be a formula with free variables $t_1,...,t_n,x$. Then $(\\forall t_1)...(\\forall t_n) ((\\forall x) ((\\forall y)(y \\in x \\implies p(y) \\implies p(x)) \\implies (\\forall x) p(x))$. Note that formally, $p(y)$ should be $p[y/x]$, and $p(x)$ should just be $p$.\n\\begin{proof}\nGiven $t_1,...,t_n$, have $p(y) \\forall y \\in x \\implies p(x)$, and suppose $(\\forall x) p(x)$ not true. So $(\\exists x) (\\neg p(x))$. We want ot say 'choose $\\epsilon$-minimal member of $\\{x: \\neg p(x))$, then contradiction'; however, this might not be a set -- e.g. if $p(x)$ is $x \\neq x$.\n\nLet $t = TC(\\{x\\})$. So $x \\in t$, and $\\neg p(x)$. Let $u =\\{y \\in t: \\neg p(y))$, and let $y$ be an $epsilon$-minimal element of $u$. Then $\\neg p(y)$. But $(\\forall z \\in y) p(z)$ (as $z \\in y \\implies z \\in t$ and $y$ is $\\epsilon$-minimal in $u$).\n\\end{proof}\n\\end{thm}\n\nRemarks: 1. we used existence of transitive closures (i.e. lemma 1).\\\\\n2. In fact, $\\epsilon$-induction equivalent to foundatoin: as can deduce foundation from $\\epsilon$-induction (in the presence of the other ZF axioms): say $x$ is regular if $(\\forall y) (x \\in y \\implies y$ has an $\\epsilon$-minimal element$)$. Foundation says every set is regular. To prove this by $\\epsilon$ induction, given $y$ regular $\\forall y \\in x$, we want to prove $x$ is regular. For $x \\in z$, if $x$ minimal then done. Otherwise, some $y \\in x$ has $y \\in z$. But $y$ is regular. So $z$ has a minimal element.\n\nHow about recursion? we want '$f(x)$ defined in terms of the $f(y)$, $y \\in x$'.\n\n\\begin{thm} (3, $\\epsilon$-recursion theorem)\\\\\nLet $G$ be a function-class ($(x,y) \\in G \\iff p(x,y)$ for some formula $p$), everywhere defined. Then there is a function-class $F$ ($(x,y) \\in F \\iff q(x,y)$, for some formula $q$) s.t. $(\\forall x) (F(x) = G(F|x))$. Moreover, $F$ is unique.\\\\\nNote: $F|x = \\{(z,f(z)): z \\in x\\}$ is a set, by replacement.\n\\begin{proof}\nSay $f$ is an attempt if: $(f$ is a function $) \\wedge (Dom f$ transitive $) \\wedge (\\forall x) (x \\in Dom f \\implies f(x) = G(f|x))$ ($f|x$ is defined, as $Dom f$ is transitive).\\\\\nThen $(\\forall x) (f,f'$ attempts defined at $x \\implies f(x) = f'(x))$ by $\\epsilon$-induction.\\\\\nSince, if $f,f'$ agree at all $y \\in x$, then they agree at $x$.\\\\\nAlso, $(\\forall x) (\\exists$ attempt $f$ defined at $x)$ by $\\epsilon$-induction.\\\\\nIndeed, suppose $|forall y \\in x$ $\\exists$ attempt defined at $y$. So $\\forall y \\in x$ $\\exists$ unique attempt $f_y$ defined on $TC(\\{y\\})$. Put $f = \\cup_{y \\in x} f_y$, and now put $f' = f \\cup \\{(x,G(f|x)\\}$. So done: take $q(x,y)$ to be $(\\exists f) (f$ an attempt $\\wedge x \\in Dom f \\wedge f(x) = y)$.\n\\end{proof}\n\\end{thm}\n\nNote: $\\epsilon$-induction and $\\epsilon$-recursion proofs look very similar to induction and recursion from chapter 2.\n\nWhat properties of the 'relation-class' $\\epsilon$ (i.e. the formula $p(x,y) = x\\epsilon y$) have we used?\n\n1. $p$ is well-founded: every non-empty set has a $p$-minimal element;\\\\\n2. $p$ is local: $(y:p(y,x))$ is a set, for each $x$.\n\nSo in fact we have $p$-induction and $p$-recursion for any $p(x,y)$ that is well-founde and local.\n\nFor a relation $r$ on a set $a$, trivially $r$ is local (as $a$ is a set). So to have $r$-induction and $r$-recursion, just need $r$ to be well-founded.\n\nThus induction and recursion from chapter 2 are special cases of this.\n\nCan we 'model' a relation by $\\varepsilon$?\n\nE.g. let $a = \\{a_1,a_2,a_3\\}$ and $r=\\{(a_1,a_2),(a_2,a_3)\\}$.\n\nPut $b = \\{b_1,b_2,b_3\\}$, where $b_1 = \\phi$, $b_2 = \\{\\phi\\}$, $b_3 = \\{\\{\\phi\\}\\}$. Then $a_i r a_j \\iff b_i r b_j \\forall i,j$. Moreover, $b$ transitive.\n\nSay relation $r$ on set $a$ is extensional if $(\\forall x,y \\in a) ((\\forall z \\in a)(zrx \\iff zry) \\implies x=y)$, e.g. above relation on above $a$, or relation $\\epsilon$ on any transitive set.\n\nAnalogue of subset collapse is:\n\n\\begin{thm} (4, Mostowski's collapse theorem):\\\\\nLet $r$ be a relation on a set $a$ that is well-founded and extensional. Then $\\exists$ transitive $b$ and bijection $f:a \\to b$ s.t. $(\\forall x,y \\in a)(x \\vee y \\iff f(x) \\in f(y))$. Moreover, $b$ and $f$ are unique.\n\\begin{proof}\nDefine $f(x)=\\{f(y):yrx\\}$ a definition by $r$-recursion on the set $a$. ($f$ is a function, not just a function-class, as it is an image of the set $a$).\n\nLet $b = \\{f(x):x \\in a\\}$ (a set, by replacement).\n\nThen $b$ transitive (definition of $f$), and $f$ surjective (definition of $b$). We need $f$ injective, then also have $xry \\iff f(x) \\in f(y)$.\n\nWe'll show that $(\\forall y) (f(y) = f(x) \\implies y=x)$ holds $\\forall x \\in a$, by $r$-induction on $x$.\n\nSo given $y$ with $f(y) =f(x)$, we want $y=x$, and may assume that $(\\forall t)(\\forall n) ((t,n \\in a \\wedge trx \\wedge f(y)=f(t) ) \\implies n=t)$.\n\nFrom $f(y)=f(x)$, we have $\\{f(n):nry\\} = \\{f(t):trx\\}$, whence $\\{n:nry\\} = \\{t:trx\\}$.\n\nThus $x=y$ as $r$ extensional.\n\nExistence: if $f,f'$ suitable then $(\\forall x \\in a) (f(x) = f''(x))$ by $r$-induction.\n\\end{proof}\n\\end{thm}\n\nAn ordimal or Von Neumann ordinal is a transitive set that is well-orderd by $\\epsilon$. (or 'totally ordered, thanks to foundation)\n\ne.g. $\\phi,\\{\\phi\\}$, any $n \\in \\omega$ (as $n=\\{0,1,2,...,\\{n-1\\}$), $\\omega$ itself.\n\nSo mostowski tells us: any well-ordered $X$ is order-isomorphic to a unique ordinal $\\alpha$. Say $X$ has order-type $\\alpha$. (this was owed from chapter 2).\n\nRemark (irrelevant): we know that for any ordinal $\\alpha$, have $\\{\\beta:\\beta<\\alpha\\}$ is a well-ordered set of order-type $\\alpha$.\n\nHence, by definition of $f$ in theorem 4, we have: $\\alpha<\\beta \\iff \\alpha \\in \\beta$.\n\nSo $\\alpha = \\{\\beta:\\beta<\\alpha\\}$.\n\nSo e.g. $\\alpha^+ = \\alpha \\cup \\{\\alpha\\}$, and $\\sup\\{\\alpha_i:i \\in I\\} = \\cup\\{\\alpha_i:i\\in I\\}$.\n\nPicture of the universe:\n\n\"start with $\\phi$, and take $\\P$ (power sets) many times. Define sets $V_\\alpha$ for each ordinal $\\alpha$ be recursion: $V_0 = \\phi$, $V_{\\alpha+1} = \\P(V_\\alpha)$, $V_\\lambda = \\cup_{\\alpha < \\lambda} V_\\alpha$ for $\\lambda$ a non-zero limit. We want every set $x$ to belong to some $V_\\alpha$.\n\n\\begin{lemma} (5)\\\\\nEach $V_\\alpha$ is transitive.\n\\begin{proof}\nInduction on $\\alpha$: 0 is trivial.\\\\\nSuccessors: given $x \\in y \\in V_{\\alpha+1}$, we have $y \\in \\P(V_\\alpha)$, so $x \\in V_\\alpha$. So $x \\subset V_\\alpha$ (as $V_\\alpha$ transitive), i.e. $x \\in \\P(V_\\alpha) = V_{\\alpha+1}$.\\\\\nLimits: any union of transitive sets is transitive.\n\\end{proof}\n\\end{lemma}\n\n\\begin{lemma} (6)\\\\\nWe have $V_\\alpha \\subset V_\\beta$ whenever $\\alpha \\leq \\beta$.\n\\begin{proof}\nInduction on $\\beta$ ($\\alpha$ fixed):\\\\\n$\\beta=\\alpha$ is trivial.\\\\\nSuccessors ($\\beta$): given $V_\\alpha \\subset V_\\beta$, we want $V_\\alpha \\subset \\P(V_\\beta)$. But $V_\\beta \\subset \\P(V_\\beta)$, $x \\in V_\\beta \\implies x \\subset V_\\beta$ as that is transitive.\\\\\nLimits are trivial as well as it is the union of all $V_\\alpha$ below.\n\\end{proof}\n\\end{lemma}\n\n\\begin{thm} (7)\\\\\n$(\\forall x) (\\exists \\alpha) (x \\in V_\\alpha)$.\\\\\nSlogan: $V = \\cup_{\\alpha \\in ON} V_\\alpha$, where $ON$ is the class of ordinals. However that's not allowed, as we cannot take union over a class.\\\\\nNotes: 1. $x \\subset V_\\alpha \\iff x \\in V_{\\alpha+1}$.\\\\\n2. If $x \\subset V_\\alpha$, then there exists least such $\\alpha$ -- called the rank of $x$. For example, $rank(\\phi) = 0$, $rank(\\{\\phi\\}) = 1$, $rank(\\omega) = \\omega$, and $rank(\\alpha) = \\alpha$ for all ordinals $\\alpha$ (by induction).\n\\begin{proof}\nWe'll show $(\\forall x) (\\exists \\alpha) (x \\subset V_\\alpha)$ by $\\epsilon$-induction. Given $x$, have $y \\subset V_\\alpha$ for some $\\alpha$ (for each $y \\in x)$, so $y \\subset V_{rank(y)}$, i.e. $y \\in V_{rank(y)+1}$ for each $y \\in x$.\\\\\nLet $\\alpha =\\sup \\{ rank(y) + 1: y \\in x\\}$. Then $x \\subset V_\\alpha$.\n\\end{proof}\n\\end{thm}\n\n\\begin{rem}\n1. The $V_\\alpha$ are called the Von-Neumann Hierarchy.\\\\\n2. Proof gives: $rank(x) = \\sup\\{rank(y)+1: y \\in x\\}$ (this is the right way to think about rank). For example, what's the rank of $\\{6\\}$? For each ordinal the rank is itself, so $rank(6) = 6$. So $rank\\{6\\} =\\sup\\{rank(6)+1\\} = 7$.\\\\\n3. (useless comment) Most of maths takes place in $V_{\\omega+10}$, apart from order-types, etc. i.e. in this course.\n\\end{rem}\n\n\\section{Cardinals}\nLet's look at 'sizes' of sets. Work in ZFC.\n\nWe want to define $Card(x)$ so that $Card(x) = Card(y) \\iff x \\leftrightarrow y$, which is a short hand for 'there is a bijection from $x$ to $y$'.\n\n(Note: We cannot take $Card(x) = \\{y:y \\leftrightarrow x\\}$, as this may not be a set.)\n\nWe do know $x \\leftrightarrow \\alpha$ for some ordinal $\\alpha$, so can define $Card(x)$ to be the least such $\\alpha$. Thus $Card(x) = Card(y) \\iff x \\leftrightarrow y$. \n\n(In just ZF, use Scott trick: define the essential rank of $x$ to be $essrank(x)=$least rank of any $y \\leftrightarrow x$, and then define $Card(x) = \\{y \\subset V_{essrank(x)}:y \\leftrightarrow x\\}$)\n\nSay $m$ is a cardinal or a cardinality if $m=Card(x)$ for some $x$.\n\nFor cardinals $m,n$, say $m \\leq n$ if $M$ injects into $N$ for some $M,N$ with $Card(M) = m$, $Card(N) = n$ (does not depend on choice of $M$ and $N$).\n\nWrite $m (intrinsic,homography)}\n\n\\begin{description}\n\\cvarg{line}{the main object axis direction (vector (dx,dy,dz))}\n\\cvarg{center}{object center ((cx,cy,cz))}\n\\cvarg{intrinsic}{intrinsic camera parameters (3x3 matrix)}\n\\cvarg{homography}{output homography matrix (3x3)}\n\\end{description}\n\nThe function calculates the homography\nmatrix for the initial image transformation from image plane to the\nplane, defined by a 3D oblong object line (See \\_\\_Figure 6-10\\_\\_\nin the OpenCV Guide 3D Reconstruction Chapter).\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{CalibrateCamera2}\n\\else\n\\cvCppFunc{calibrateCamera}\n\\fi\nFinds the camera intrinsic and extrinsic parameters from several views of a calibration pattern.\n\n\\cvdefC{double cvCalibrateCamera2( \\par const CvMat* objectPoints,\\par const CvMat* imagePoints,\\par const CvMat* pointCounts,\\par CvSize imageSize,\\par CvMat* cameraMatrix,\\par CvMat* distCoeffs,\\par CvMat* rvecs=NULL,\\par CvMat* tvecs=NULL,\\par int flags=0 );}\n\\cvdefPy{CalibrateCamera2(objectPoints,imagePoints,pointCounts,imageSize,cameraMatrix,distCoeffs,rvecs,tvecs,flags=0)-> None}\n\\cvdefCpp{double calibrateCamera( const vector >\\& objectPoints,\\par\n const vector >\\& imagePoints,\\par\n Size imageSize,\\par\n Mat\\& cameraMatrix, Mat\\& distCoeffs,\\par\n vector\\& rvecs, vector\\& tvecs,\\par\n int flags=0 );}\n\\begin{description}\n\\ifCPy\n\\cvarg{objectPoints}{The joint matrix of object points - calibration pattern features in the model coordinate space. It is floating-point 3xN or Nx3 1-channel, or 1xN or Nx1 3-channel array, where N is the total number of points in all views.}\n\\cvarg{imagePoints}{The joint matrix of object points projections in the camera views. It is floating-point 2xN or Nx2 1-channel, or 1xN or Nx1 2-channel array, where N is the total number of points in all views}\n\\cvarg{pointCounts}{Integer 1xM or Mx1 vector (where M is the number of calibration pattern views) containing the number of points in each particular view. The sum of vector elements must match the size of \\texttt{objectPoints} and \\texttt{imagePoints} (=N).}\n\\fi\n\\ifCpp\n\\cvarg{objectPoints}{The vector of vectors of points on the calibration pattern in its coordinate system, one vector per view. If the same calibration pattern is shown in each view and it's fully visible then all the vectors will be the same, although it is possible to use partially occluded patterns, or even different patterns in different views - then the vectors will be different. The points are 3D, but since they are in the pattern coordinate system, then if the rig is planar, it may have sense to put the model to the XY coordinate plane, so that Z-coordinate of each input object point is 0}\n\\cvarg{imagePoints}{The vector of vectors of the object point projections on the calibration pattern views, one vector per a view. The projections must be in the same order as the corresponding object points.}\n\\fi\n\\cvarg{imageSize}{Size of the image, used only to initialize the intrinsic camera matrix}\n\\cvarg{cameraMatrix}{The output 3x3 floating-point camera matrix $A = \\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{1}$.\\newline\nIf \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS} and/or \\texttt{CV\\_CALIB\\_FIX\\_ASPECT\\_RATIO} are specified, some or all of \\texttt{fx, fy, cx, cy} must be initialized before calling the function}\n\\cvarg{distCoeffs}{The output vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements}\n\\cvarg{rvecs}{The output \\cvCPy{3x\\emph{M} or \\emph{M}x3 1-channel, or 1x\\emph{M} or \\emph{M}x1 3-channel array}\\cvCpp{vector} of rotation vectors (see \\cvCross{Rodrigues2}{Rodrigues}), estimated for each pattern view. That is, each k-th rotation vector together with the corresponding k-th translation vector (see the next output parameter description) brings the calibration pattern from the model coordinate space (in which object points are specified) to the world coordinate space, i.e. real position of the calibration pattern in the k-th pattern view (k=0..\\emph{M}-1)}\n\\cvarg{tvecs}{The output \\cvCPy{3x\\emph{M} or \\emph{M}x3 1-channel, or 1x\\emph{M} or \\emph{M}x1 3-channel array}\\cvCpp{vector} of translation vectors, estimated for each pattern view.}\n\\cvarg{flags}{Different flags, may be 0 or combination of the following values:\n\\begin{description}\n\\cvarg{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS}{\\texttt{cameraMatrix} contains the valid initial values of \\texttt{fx, fy, cx, cy} that are optimized further. Otherwise, \\texttt{(cx, cy)} is initially set to the image center (\\texttt{imageSize} is used here), and focal distances are computed in some least-squares fashion. Note, that if intrinsic parameters are known, there is no need to use this function just to estimate the extrinsic parameters. Use \\cvCross{FindExtrinsicCameraParams2}{solvePnP} instead.}\n\\cvarg{CV\\_CALIB\\_FIX\\_PRINCIPAL\\_POINT}{The principal point is not changed during the global optimization, it stays at the center or at the other location specified when \\newline \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS} is set too.}\n\\cvarg{CV\\_CALIB\\_FIX\\_ASPECT\\_RATIO}{The functions considers only \\texttt{fy} as a free parameter, the ratio \\texttt{fx/fy} stays the same as in the input \\texttt{cameraMatrix}. \\newline When \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS} is not set, the actual input values of \\texttt{fx} and \\texttt{fy} are ignored, only their ratio is computed and used further.}\n\\cvarg{CV\\_CALIB\\_ZERO\\_TANGENT\\_DIST}{Tangential distortion coefficients $(p_1, p_2)$ will be set to zeros and stay zero.}}\n\\cvarg{CV\\_CALIB\\_FIX\\_K1,...,CV\\_CALIB\\_FIX\\_K6}{Do not change the corresponding radial distortion coefficient during the optimization. If \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS} is set, the coefficient from the supplied \\texttt{distCoeffs} matrix is used, otherwise it is set to 0.}\n\\cvarg{CV\\_CALIB\\_RATIONAL\\_MODEL}{Enable coefficients k4, k5 and k6. To provide the backward compatibility, this extra flag should be explicitly specified to make the calibration function use the rational model and return 8 coefficients. If the flag is not set, the function will compute \\cvCpp{and return} only 5 distortion coefficients.}\n\\end{description}\n\\end{description}\n\nThe function estimates the intrinsic camera\nparameters and extrinsic parameters for each of the views. The\ncoordinates of 3D object points and their correspondent 2D projections\nin each view must be specified. That may be achieved by using an\nobject with known geometry and easily detectable feature points.\nSuch an object is called a calibration rig or calibration pattern,\nand OpenCV has built-in support for a chessboard as a calibration\nrig (see \\cvCross{FindChessboardCorners}{findChessboardCorners}). Currently, initialization\nof intrinsic parameters (when \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS}\nis not set) is only implemented for planar calibration patterns\n(where z-coordinates of the object points must be all 0's). 3D\ncalibration rigs can also be used as long as initial \\texttt{cameraMatrix}\nis provided.\n\nThe algorithm does the following:\n\\begin{enumerate}\n \\item First, it computes the initial intrinsic parameters (the option only available for planar calibration patterns) or reads them from the input parameters. The distortion coefficients are all set to zeros initially (unless some of \\texttt{CV\\_CALIB\\_FIX\\_K?} are specified).\n \\item The initial camera pose is estimated as if the intrinsic parameters have been already known. This is done using \\cvCross{FindExtrinsicCameraParams2}{solvePnP}\n \\item After that the global Levenberg-Marquardt optimization algorithm is run to minimize the reprojection error, i.e. the total sum of squared distances between the observed feature points \\texttt{imagePoints} and the projected (using the current estimates for camera parameters and the poses) object points \\texttt{objectPoints}; see \\cvCross{ProjectPoints2}{projectPoints}.\n\\end{enumerate}\n\n\\ifPy\n\\else\nThe function returns the final re-projection error.\n\\fi\n\nNote: if you're using a non-square (=non-NxN) grid and\n\\cvCppCross{findChessboardCorners} for calibration, and \\texttt{calibrateCamera} returns\nbad values (i.e. zero distortion coefficients, an image center very far from\n$(w/2-0.5,h/2-0.5)$, and / or large differences between $f_x$ and $f_y$ (ratios of\n10:1 or more)), then you've probably used \\texttt{patternSize=cvSize(rows,cols)},\nbut should use \\texttt{patternSize=cvSize(cols,rows)} in \\cvCross{FindChessboardCorners}{findChessboardCorners}.\n\nSee also: \\cvCross{FindChessboardCorners}{findChessboardCorners}, \\cvCross{FindExtrinsicCameraParams2}{solvePnP}, \\cvCppCross{initCameraMatrix2D}, \\cvCross{StereoCalibrate}{stereoCalibrate}, \\cvCross{Undistort2}{undistort}\n\n\\ifCpp\n\n\\cvCppFunc{calibrationMatrixValues}\nComputes some useful camera characteristics from the camera matrix\n\n\\cvdefCpp{void calibrationMatrixValues( const Mat\\& cameraMatrix,\\par\n Size imageSize,\\par\n double apertureWidth,\\par\n double apertureHeight,\\par\n double\\& fovx,\\par\n double\\& fovy,\\par\n double\\& focalLength,\\par\n Point2d\\& principalPoint,\\par\n double\\& aspectRatio );}\n\\begin{description}\n\\cvarg{cameraMatrix}{The input camera matrix that can be estimated by \\cvCppCross{calibrateCamera} or \\cvCppCross{stereoCalibrate}}\n\\cvarg{imageSize}{The input image size in pixels}\n\\cvarg{apertureWidth}{Physical width of the sensor}\n\\cvarg{apertureHeight}{Physical height of the sensor}\n\\cvarg{fovx}{The output field of view in degrees along the horizontal sensor axis}\n\\cvarg{fovy}{The output field of view in degrees along the vertical sensor axis}\n\\cvarg{focalLength}{The focal length of the lens in mm}\n\\cvarg{principalPoint}{The principal point in pixels}\n\\cvarg{aspectRatio}{$f_y/f_x$}\n\\end{description}\n\nThe function computes various useful camera characteristics from the previously estimated camera matrix.\n\n\\cvCppFunc{composeRT}\nCombines two rotation-and-shift transformations\n\n\\cvdefCpp{void composeRT( const Mat\\& rvec1, const Mat\\& tvec1,\\par\n const Mat\\& rvec2, const Mat\\& tvec2,\\par\n Mat\\& rvec3, Mat\\& tvec3 );\\newline\nvoid composeRT( const Mat\\& rvec1, const Mat\\& tvec1,\\par\n const Mat\\& rvec2, const Mat\\& tvec2,\\par\n Mat\\& rvec3, Mat\\& tvec3,\\par\n Mat\\& dr3dr1, Mat\\& dr3dt1,\\par\n Mat\\& dr3dr2, Mat\\& dr3dt2,\\par\n Mat\\& dt3dr1, Mat\\& dt3dt1,\\par\n Mat\\& dt3dr2, Mat\\& dt3dt2 );}\n\\begin{description}\n\\cvarg{rvec1}{The first rotation vector}\n\\cvarg{tvec1}{The first translation vector}\n\\cvarg{rvec2}{The second rotation vector}\n\\cvarg{tvec2}{The second translation vector}\n\\cvarg{rvec3}{The output rotation vector of the superposition}\n\\cvarg{tvec3}{The output translation vector of the superposition}\n\\cvarg{d??d??}{The optional output derivatives of \\texttt{rvec3} or \\texttt{tvec3} w.r.t. \\texttt{rvec?} or \\texttt{tvec?}}\n\\end{description}\n\nThe functions compute:\n\n\\[ \\begin{array}{l}\n\\texttt{rvec3} = \\mathrm{rodrigues}^{-1}\\left(\\mathrm{rodrigues}(\\texttt{rvec2}) \\cdot\n\\mathrm{rodrigues}(\\texttt{rvec1})\\right) \\\\\n\\texttt{tvec3} = \\mathrm{rodrigues}(\\texttt{rvec2}) \\cdot \\texttt{tvec1} + \\texttt{tvec2}\n\\end{array}, \\]\n\nwhere $\\mathrm{rodrigues}$ denotes a rotation vector to rotation matrix transformation, and $\\mathrm{rodrigues}^{-1}$ denotes the inverse transformation, see \\cvCppCross{Rodrigues}.\n\nAlso, the functions can compute the derivatives of the output vectors w.r.t the input vectors (see \\cvCppCross{matMulDeriv}).\nThe functions are used inside \\cvCppCross{stereoCalibrate} but can also be used in your own code where Levenberg-Marquardt or another gradient-based solver is used to optimize a function that contains matrix multiplication.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{ComputeCorrespondEpilines}\n\\else\n\\cvCppFunc{computeCorrespondEpilines}\n\\fi\nFor points in one image of a stereo pair, computes the corresponding epilines in the other image.\n\n\\cvdefC{void cvComputeCorrespondEpilines( \\par const CvMat* points,\\par int whichImage,\\par const CvMat* F, \\par CvMat* lines);}\n\\cvdefPy{ComputeCorrespondEpilines(points, whichImage, F, lines) -> None}\n\\cvdefCpp{void computeCorrespondEpilines( const Mat\\& points,\\par\n int whichImage, const Mat\\& F,\\par\n vector\\& lines );}\n\\begin{description}\n\\cvCPy{\\cvarg{points}{The input points. \\texttt{2xN, Nx2, 3xN} or \\texttt{Nx3} array (where \\texttt{N} number of points). Multi-channel \\texttt{1xN} or \\texttt{Nx1} array is also acceptable}}\n\\cvCpp{\\cvarg{points}{The input points. $N \\times 1$ or $1 \\times N$ matrix of type \\texttt{CV\\_32FC2} or \\texttt{vector}}}\n\\cvarg{whichImage}{Index of the image (1 or 2) that contains the \\texttt{points}}\n\\cvarg{F}{The fundamental matrix that can be estimated using \\cvCross{FindFundamentalMat}{findFundamentalMat}\nor \\cvCross{StereoRectify}{stereoRectify}.}\n\\cvarg{lines}{\\cvCPy{The output epilines, a \\texttt{3xN} or \\texttt{Nx3} array.}\\cvCpp{The output vector of the corresponding to the points epipolar lines in the other image.} Each line $ax + by + c=0$ is encoded by 3 numbers $(a, b, c)$}\n\\end{description}\n\nFor every point in one of the two images of a stereo-pair the function finds the equation of the\ncorresponding epipolar line in the other image.\n\nFrom the fundamental matrix definition (see \\cvCross{FindFundamentalMat}{findFundamentalMat}),\nline $l^{(2)}_i$ in the second image for the point $p^{(1)}_i$ in the first image (i.e. when \\texttt{whichImage=1}) is computed as:\n\n\\[ l^{(2)}_i = F p^{(1)}_i \\]\n\nand, vice versa, when \\texttt{whichImage=2}, $l^{(1)}_i$ is computed from $p^{(2)}_i$ as:\n\n\\[ l^{(1)}_i = F^T p^{(2)}_i \\]\n\nLine coefficients are defined up to a scale. They are normalized, such that $a_i^2+b_i^2=1$.\n\n\\ifCPy\n\\cvCPyFunc{ConvertPointsHomogeneous}\n\\else\n\\cvCppFunc{convertPointsHomogeneous}\n\\fi\nConvert points to/from homogeneous coordinates.\n\n\\cvdefC{void cvConvertPointsHomogeneous( \\par const CvMat* src,\\par CvMat* dst );}\n\n\\cvdefPy{ConvertPointsHomogeneous( src, dst ) -> None}\n\n\\cvdefCpp{void convertPointsHomogeneous( const Mat\\& src, vector\\& dst );\\newline\nvoid convertPointsHomogeneous( const Mat\\& src, vector\\& dst );}\n\n\\begin{description}\n\\ifC\n\\cvarg{src}{The input point array, \\texttt{2xN, Nx2, 3xN, Nx3, 4xN or Nx4 (where \\texttt{N} is the number of points)}. Multi-channel \\texttt{1xN} or \\texttt{Nx1} array is also acceptable}\n\\cvarg{dst}{The output point array, must contain the same number of points as the input; The dimensionality must be the same, 1 less or 1 more than the input, and also within 2 to 4}\n\\else\n\\cvarg{src}{The input array or vector of 2D, 3D or 4D points}\n\\cvarg{dst}{The output vector of 2D or 2D points}\n\\fi\n\\end{description}\n\nThe \\cvC{function converts}\\cvCpp{functions convert} 2D or 3D points from/to homogeneous coordinates, or simply \\cvC{copies or transposes}\\cvCpp{copy or transpose} the array. If the input array dimensionality is larger than the output, each coordinate is divided by the last coordinate:\n\n\\[\n\\begin{array}{l}\n(x,y[,z],w) -> (x',y'[,z'])\\\\\n\\text{where} \\\\\nx' = x/w \\\\\ny' = y/w \\\\\nz' = z/w \\quad \\text{(if output is 3D)}\n\\end{array}\n\\]\n\nIf the output array dimensionality is larger, an extra 1 is appended to each point. Otherwise, the input array is simply copied (with optional transposition) to the output.\n\n\\cvC{\\textbf{Note} because the function accepts a large variety of array layouts, it may report an error when input/output array dimensionality is ambiguous. It is always safe to use the function with number of points $\\texttt{N} \\ge 5$, or to use multi-channel \\texttt{Nx1} or \\texttt{1xN} arrays.}\n\n\\ifCPy\n\n\\cvCPyFunc{CreatePOSITObject}\nInitializes a structure containing object information.\n\n\\cvdefC{\nCvPOSITObject* cvCreatePOSITObject( \\par CvPoint3D32f* points,\\par int point\\_count );\n}\\cvdefPy{CreatePOSITObject(points)-> POSITObject}\n\n\\begin{description}\n\\ifC\n\\cvarg{points}{Pointer to the points of the 3D object model}\n\\cvarg{point\\_count}{Number of object points}\n\\else\n\\cvarg{points}{List of 3D points}\n\\fi\n\\end{description}\n\nThe function allocates memory for the object structure and computes the object inverse matrix.\n\nThe preprocessed object data is stored in the structure \\cvCPyCross{CvPOSITObject}, internal for OpenCV, which means that the user cannot directly access the structure data. The user may only create this structure and pass its pointer to the function.\n\nAn object is defined as a set of points given in a coordinate system. The function \\cvCPyCross{POSIT} computes a vector that begins at a camera-related coordinate system center and ends at the \\texttt{points[0]} of the object.\n\nOnce the work with a given object is finished, the function \\cvCPyCross{ReleasePOSITObject} must be called to free memory.\n\n\\cvCPyFunc{CreateStereoBMState}\nCreates block matching stereo correspondence structure.\n\n\\cvdefC{\n\nCvStereoBMState* cvCreateStereoBMState( int preset=CV\\_STEREO\\_BM\\_BASIC,\n int numberOfDisparities=0 );\n\n}\n\\cvdefPy{CreateStereoBMState(preset=CV\\_STEREO\\_BM\\_BASIC,numberOfDisparities=0)-> StereoBMState}\n\n\\begin{description}\n\\cvarg{preset}{ID of one of the pre-defined parameter sets. Any of the parameters can be overridden after creating the structure. Values are\n\\begin{description}\n\\cvarg{CV\\_STEREO\\_BM\\_BASIC}{Parameters suitable for general cameras}\n\\cvarg{CV\\_STEREO\\_BM\\_FISH\\_EYE}{Parameters suitable for wide-angle cameras}\n\\cvarg{CV\\_STEREO\\_BM\\_NARROW}{Parameters suitable for narrow-angle cameras}\n\\end{description}\n}\n\\cvarg{numberOfDisparities}{The number of disparities. If the parameter is 0, it is taken from the preset, otherwise the supplied value overrides the one from preset.}\n\\end{description}\n\nThe function creates the stereo correspondence structure and initializes\nit. It is possible to override any of the parameters at any time between\nthe calls to \\cvCPyCross{FindStereoCorrespondenceBM}.\n\n\\cvCPyFunc{CreateStereoGCState}\nCreates the state of graph cut-based stereo correspondence algorithm.\n\n\\cvdefC{\n\nCvStereoGCState* cvCreateStereoGCState( int numberOfDisparities,\n int maxIters );\n\n}\n\\cvdefPy{CreateStereoGCState(numberOfDisparities,maxIters)-> StereoGCState}\n\n\\begin{description}\n\\cvarg{numberOfDisparities}{The number of disparities. The disparity search range will be $\\texttt{state->minDisparity} \\le disparity < \\texttt{state->minDisparity} + \\texttt{state->numberOfDisparities}$}\n\\cvarg{maxIters}{Maximum number of iterations. On each iteration all possible (or reasonable) alpha-expansions are tried. The algorithm may terminate earlier if it could not find an alpha-expansion that decreases the overall cost function value. See \\cite{Kolmogorov03} for details. }\n\\end{description}\n\nThe function creates the stereo correspondence structure and initializes it. It is possible to override any of the parameters at any time between the calls to \\cvCPyCross{FindStereoCorrespondenceGC}.\n\n\\cvclass{CvStereoBMState}\nThe structure for block matching stereo correspondence algorithm.\n\n\\ifC\n\\begin{lstlisting}\ntypedef struct CvStereoBMState\n{\n //pre filters (normalize input images):\n int preFilterType; // 0 for now\n int preFilterSize; // ~5x5..21x21\n int preFilterCap; // up to ~31\n //correspondence using Sum of Absolute Difference (SAD):\n int SADWindowSize; // Could be 5x5..21x21\n int minDisparity; // minimum disparity (=0)\n int numberOfDisparities; // maximum disparity - minimum disparity\n //post filters (knock out bad matches):\n int textureThreshold; // areas with no texture are ignored\n int uniquenessRatio;// invalidate disparity at pixels where there are other close matches\n // with different disparity\n int speckleWindowSize; // the maximum area of speckles to remove\n // (set to 0 to disable speckle filtering)\n int speckleRange; // acceptable range of disparity variation in each connected component\n \n int trySmallerWindows; // not used \n CvRect roi1, roi2; // clipping ROIs\n \n int disp12MaxDiff; // maximum allowed disparity difference in the left-right check \n \n // internal data\n ...\n}\nCvStereoBMState;\n\\end{lstlisting}\n\\fi\n\\begin{description}\n\\cvarg{preFilterType}{type of the prefilter, \\texttt{CV\\_STEREO\\_BM\\_NORMALIZED\\_RESPONSE} or the default and the recommended \\texttt{CV\\_STEREO\\_BM\\_XSOBEL}, int}\n\\cvarg{preFilterSize}{~5x5..21x21, int}\n\\cvarg{preFilterCap}{up to ~31, int}\n\\cvarg{SADWindowSize}{Could be 5x5..21x21 or higher, but with 21x21 or smaller windows the processing speed is much higher, int}\n\\cvarg{minDisparity}{minimum disparity (=0), int}\n\\cvarg{numberOfDisparities}{maximum disparity - minimum disparity, int}\n\\cvarg{textureThreshold}{the textureness threshold. That is, if the sum of absolute values of x-derivatives computed over \\texttt{SADWindowSize} by \\texttt{SADWindowSize} pixel neighborhood is smaller than the parameter, no disparity is computed at the pixel, int}\n\\cvarg{uniquenessRatio}{the minimum margin in percents between the best (minimum) cost function value and the second best value to accept the computed disparity, int}\n\\cvarg{speckleWindowSize}{the maximum area of speckles to remove (set to 0 to disable speckle filtering), int}\n\\cvarg{speckleRange}{acceptable range of disparity variation in each connected component, int}\n\\cvarg{trySmallerWindows}{not used currently (0), int}\n\\cvarg{roi1, roi2}{These are the clipping ROIs for the left and the right images. The function \\cvCPyCross{StereoRectify} returns the largest rectangles in the left and right images where after the rectification all the pixels are valid. If you copy those rectangles to the \\texttt{CvStereoBMState} structure, the stereo correspondence function will automatically clear out the pixels outside of the \"valid\" disparity rectangle computed by \\cvCPyCross{GetValidDisparityROI}. Thus you will get more \"invalid disparity\" pixels than usual, but the remaining pixels are more probable to be valid.}\n\\cvarg{disp12MaxDiff}{The maximum allowed difference between the explicitly computed left-to-right disparity map and the implicitly (by \\cvCPyCross{ValidateDisparity}) computed right-to-left disparity. If for some pixel the difference is larger than the specified threshold, the disparity at the pixel is invalidated. By default this parameter is set to (-1), which means that the left-right check is not performed.}\n\\end{description}\n\n\nThe block matching stereo correspondence algorithm, by Kurt Konolige, is very fast single-pass stereo matching algorithm that uses sliding sums of absolute differences between pixels in the left image and the pixels in the right image, shifted by some varying amount of pixels (from \\texttt{minDisparity} to \\texttt{minDisparity+numberOfDisparities}). On a pair of images WxH the algorithm computes disparity in \\texttt{O(W*H*numberOfDisparities)} time. In order to improve quality and readability of the disparity map, the algorithm includes pre-filtering and post-filtering procedures.\n\nNote that the algorithm searches for the corresponding blocks in x direction only. It means that the supplied stereo pair should be rectified. Vertical stereo layout is not directly supported, but in such a case the images could be transposed by user.\n\n\\cvclass{CvStereoGCState}\nThe structure for graph cuts-based stereo correspondence algorithm\n\n\\ifC\n\\begin{lstlisting}\ntypedef struct CvStereoGCState\n{\n int Ithreshold; // threshold for piece-wise linear data cost function (5 by default)\n int interactionRadius; // radius for smoothness cost function (1 by default; means Potts model)\n float K, lambda, lambda1, lambda2; // parameters for the cost function\n // (usually computed adaptively from the input data)\n int occlusionCost; // 10000 by default\n int minDisparity; // 0 by default; see CvStereoBMState\n int numberOfDisparities; // defined by user; see CvStereoBMState\n int maxIters; // number of iterations; defined by user.\n\n // internal buffers\n CvMat* left;\n CvMat* right;\n CvMat* dispLeft;\n CvMat* dispRight;\n CvMat* ptrLeft;\n CvMat* ptrRight;\n CvMat* vtxBuf;\n CvMat* edgeBuf;\n}\nCvStereoGCState;\n\\end{lstlisting}\n\\else\n\\begin{description}\n\\cvarg{Ithreshold}{threshold for piece-wise linear data cost function (5 by default)}\n\\cvarg{interactionRadius}{radius for smoothness cost function (1 by default; means Potts model)}\n\\cvarg{K, lambda, lambda1, lambda2}{parameters for the cost function (usually computed adaptively from the input data)}\n\\cvarg{occlusionCost}{10000 by default}\n\\cvarg{minDisparity}{0 by default; see \\cross{CvStereoBMState}}\n\\cvarg{numberOfDisparities}{defined by user; see \\cross{CvStereoBMState}}\n\\cvarg{maxIters}{number of iterations; defined by user.}\n\\end{description}\n\\fi\n\nThe graph cuts stereo correspondence algorithm, described in \\cite{Kolmogorov03} (as \\textbf{KZ1}), is non-realtime stereo correspondence algorithm that usually gives very accurate depth map with well-defined object boundaries. The algorithm represents stereo problem as a sequence of binary optimization problems, each of those is solved using maximum graph flow algorithm. The state structure above should not be allocated and initialized manually; instead, use \\cvCPyCross{CreateStereoGCState} and then override necessary parameters if needed.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{DecomposeProjectionMatrix}\n\\else\n\\cvCppFunc{decomposeProjectionMatrix}\n\\fi\nDecomposes the projection matrix into a rotation matrix and a camera matrix.\n\n\\cvdefC{\nvoid cvDecomposeProjectionMatrix( \\par const CvMat *projMatrix,\\par CvMat *cameraMatrix,\\par CvMat *rotMatrix,\\par CvMat *transVect,\\par CvMat *rotMatrX=NULL,\\par CvMat *rotMatrY=NULL,\\par CvMat *rotMatrZ=NULL,\\par CvPoint3D64f *eulerAngles=NULL);\n}\n\n\\cvdefPy{DecomposeProjectionMatrix(projMatrix, cameraMatrix, rotMatrix, transVect, rotMatrX = None, rotMatrY = None, rotMatrZ = None) -> eulerAngles}\n\n\\cvdefCpp{void decomposeProjectionMatrix( const Mat\\& projMatrix,\\par\n Mat\\& cameraMatrix,\\par\n Mat\\& rotMatrix, Mat\\& transVect );\\newline\nvoid decomposeProjectionMatrix( const Mat\\& projMatrix, \\par\n Mat\\& cameraMatrix,\\par\n Mat\\& rotMatrix, Mat\\& transVect,\\par\n Mat\\& rotMatrixX, Mat\\& rotMatrixY,\\par\n Mat\\& rotMatrixZ, Vec3d\\& eulerAngles );}\n\\begin{description}\n\\cvarg{projMatrix}{The 3x4 input projection matrix P}\n\\cvarg{cameraMatrix}{The output 3x3 camera matrix K}\n\\cvarg{rotMatrix}{The output 3x3 external rotation matrix R}\n\\cvarg{transVect}{The output 4x1 translation vector T}\n\\cvarg{rotMatrX}{Optional 3x3 rotation matrix around x-axis}\n\\cvarg{rotMatrY}{Optional 3x3 rotation matrix around y-axis}\n\\cvarg{rotMatrZ}{Optional 3x3 rotation matrix around z-axis}\n\\cvarg{eulerAngles}{Optional 3 points containing the three Euler angles of rotation}\n\\end{description}\n\nThe function computes a decomposition of a projection matrix into a calibration and a rotation matrix and the position of the camera.\n\nIt optionally returns three rotation matrices, one for each axis, and the three Euler angles that could be used in OpenGL.\n\nThe function is based on \\cvCross{RQDecomp3x3}{RQDecomp3x3}.\n\n\\ifCPy\n\\cvCPyFunc{DrawChessboardCorners}\n\\else\n\\cvCppFunc{drawChessboardCorners}\n\\fi\nRenders the detected chessboard corners.\n\n\\cvdefC{\nvoid cvDrawChessboardCorners( \\par CvArr* image,\\par CvSize patternSize,\\par CvPoint2D32f* corners,\\par int count,\\par int patternWasFound );\n}\\cvdefPy{DrawChessboardCorners(image,patternSize,corners,patternWasFound)-> None}\n\\cvdefCpp{void drawChessboardCorners( Mat\\& image, Size patternSize,\\par\n const Mat\\& corners,\\par\n bool patternWasFound );}\n\\begin{description}\n\\cvarg{image}{The destination image; it must be an 8-bit color image}\n\\cvarg{patternSize}{The number of inner corners per chessboard row and column. (patternSize = cv::Size(points\\_per\\_row,points\\_per\\_column) = cv::Size(rows,columns) )}\n\\cvarg{corners}{The array of corners detected, this should be the output from findChessboardCorners wrapped in a cv::Mat().}\n\\cvC{\\cvarg{count}{The number of corners}}\n\\cvarg{patternWasFound}{Indicates whether the complete board was found \\cvCPy{$(\\ne 0)$} or not \\cvCPy{$(=0)$}. One may just pass the return value \\cvCPyCross{FindChessboardCorners}{findChessboardCorners} here}\n\\end{description}\n\nThe function draws the individual chessboard corners detected as red circles if the board was not found or as colored corners connected with lines if the board was found.\n\n\\ifCPy\n\\cvCPyFunc{FindChessboardCorners}\n\\else\n\\cvCppFunc{findChessboardCorners}\n\\fi\nFinds the positions of the internal corners of the chessboard.\n\n\\cvdefC{int cvFindChessboardCorners( \\par const void* image,\\par CvSize patternSize,\\par CvPoint2D32f* corners,\\par int* cornerCount=NULL,\\par int flags=CV\\_CALIB\\_CB\\_ADAPTIVE\\_THRESH );}\n\\cvdefPy{FindChessboardCorners(image, patternSize, flags=CV\\_CALIB\\_CB\\_ADAPTIVE\\_THRESH) -> corners}\n\\cvdefCpp{bool findChessboardCorners( const Mat\\& image, Size patternSize,\\par\n vector\\& corners,\\par\n int flags=CV\\_CALIB\\_CB\\_ADAPTIVE\\_THRESH+\\par\n CV\\_CALIB\\_CB\\_NORMALIZE\\_IMAGE );}\n\\begin{description}\n\\cvarg{image}{Source chessboard view; it must be an 8-bit grayscale or color image}\n\\cvarg{patternSize}{The number of inner corners per chessboard row and column\n( patternSize = cvSize(points\\_per\\_row,points\\_per\\_colum) = cvSize(columns,rows) )}\n\\cvarg{corners}{The output array of corners detected}\n\\cvC{\\cvarg{cornerCount}{The output corner counter. If it is not NULL, it stores the number of corners found}}\n\\cvarg{flags}{Various operation flags, can be 0 or a combination of the following values:\n\\begin{description}\n \\cvarg{CV\\_CALIB\\_CB\\_ADAPTIVE\\_THRESH}{use adaptive thresholding to convert the image to black and white, rather than a fixed threshold level (computed from the average image brightness).}\n \\cvarg{CV\\_CALIB\\_CB\\_NORMALIZE\\_IMAGE}{normalize the image gamma with \\cvCross{EqualizeHist}{equalizeHist} before applying fixed or adaptive thresholding.}\n \\cvarg{CV\\_CALIB\\_CB\\_FILTER\\_QUADS}{use additional criteria (like contour area, perimeter, square-like shape) to filter out false quads that are extracted at the contour retrieval stage.}\n \\cvarg{CALIB\\_CB\\_FAST\\_CHECK}{Runs a fast check on the image that looks for chessboard corners, and shortcuts the call if none are found. This can drastically speed up the call in the degenerate condition when\n no chessboard is observed.}\n\\end{description}}\n\\end{description}\n\nThe function attempts to determine\nwhether the input image is a view of the chessboard pattern and\nlocate the internal chessboard corners. The function returns a non-zero\nvalue if all of the corners have been found and they have been placed\nin a certain order (row by row, left to right in every row),\notherwise, if the function fails to find all the corners or reorder\nthem, it returns 0. For example, a regular chessboard has 8 x 8\nsquares and 7 x 7 internal corners, that is, points, where the black\nsquares touch each other. The coordinates detected are approximate,\nand to determine their position more accurately, the user may use\nthe function \\cvCross{FindCornerSubPix}{cornerSubPix}.\n\nSample usage of detecting and drawing chessboard corners:\n\\begin{lstlisting}\nSize patternsize(8,6); //interior number of corners\nMat gray = ....; //source image\nvector corners; //this will be filled by the detected corners\n\n//CALIB_CB_FAST_CHECK saves a lot of time on images \n//that don't contain any chessboard corners\nbool patternfound = findChessboardCorners(gray, patternsize, corners, \n\tCALIB_CB_ADAPTIVE_THRESH + CALIB_CB_NORMALIZE_IMAGE \n\t+ CALIB_CB_FAST_CHECK);\n\nif(patternfound)\n cornerSubPix(gray, corners, Size(11, 11), Size(-1, -1), \n TermCriteria(CV_TERMCRIT_EPS + CV_TERMCRIT_ITER, 30, 0.1));\n \ndrawChessboardCorners(img, patternsize, Mat(corners), patternfound);\n\\end{lstlisting}\n\n\\textbf{Note:} the function requires some white space (like a square-thick border, the wider the better) around the board to make the detection more robust in various environment (otherwise if there is no border and the background is dark, the outer black squares could not be segmented properly and so the square grouping and ordering algorithm will fail).\n\n\\ifCPy\n\\cvCPyFunc{FindExtrinsicCameraParams2}\n\\else\n\\cvCppFunc{solvePnP}\n\\fi\nFinds the object pose from the 3D-2D point correspondences\n\n\\cvdefC{void cvFindExtrinsicCameraParams2( \\par const CvMat* objectPoints,\\par const CvMat* imagePoints,\\par const CvMat* cameraMatrix,\\par const CvMat* distCoeffs,\\par CvMat* rvec,\\par CvMat* tvec, \\par int useExtrinsicGuess=0);}\n\\cvdefPy{FindExtrinsicCameraParams2(objectPoints,imagePoints,cameraMatrix,distCoeffs,rvec,tvec,useExtrinsicGuess=0)-> None}\n\\cvdefCpp{void solvePnP( const Mat\\& objectPoints,\\par\n const Mat\\& imagePoints,\\par\n const Mat\\& cameraMatrix,\\par\n const Mat\\& distCoeffs,\\par\n Mat\\& rvec, Mat\\& tvec,\\par\n bool useExtrinsicGuess=false );}\n\\begin{description}\n\\cvarg{objectPoints}{The array of object points in the object coordinate space, 3xN or Nx3 1-channel, or 1xN or Nx1 3-channel, where N is the number of points. \\cvCpp{Can also pass \\texttt{vector} here.}}\n\\cvarg{imagePoints}{The array of corresponding image points, 2xN or Nx2 1-channel or 1xN or Nx1 2-channel, where N is the number of points. \\cvCpp{Can also pass \\texttt{vector} here.}}\n\\cvarg{cameraMatrix}{The input camera matrix $A = \\vecthreethree{fx}{0}{cx}{0}{fy}{cy}{0}{0}{1} $}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{rvec}{The output rotation vector (see \\cvCross{Rodrigues2}{Rodrigues}) that (together with \\texttt{tvec}) brings points from the model coordinate system to the camera coordinate system}\n\\cvarg{tvec}{The output translation vector}\n\\cvarg{useExtrinsicGuess}{If true (1), the function will use the provided \\texttt{rvec} and \\texttt{tvec} as the initial approximations of the rotation and translation vectors, respectively, and will further optimize them.}\n\\end{description}\n\nThe function estimates the object pose given a set of object points, their corresponding image projections, as well as the camera matrix and the distortion coefficients. This function finds such a pose that minimizes reprojection error, i.e. the sum of squared distances between the observed projections \\texttt{imagePoints} and the projected (using \\cvCross{ProjectPoints2}{projectPoints}) \\texttt{objectPoints}.\n\n\n\\ifCPy\nThe function's counterpart in the C++ API is \\hyperref[cppfunc.solvePnP]{cv::solvePnP}\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{FindFundamentalMat}\n\\else\n\\cvCppFunc{findFundamentalMat}\n\\fi\n\nCalculates the fundamental matrix from the corresponding points in two images.\n\n\\cvdefC{\nint cvFindFundamentalMat( \\par const CvMat* points1,\\par const CvMat* points2,\\par CvMat* fundamentalMatrix,\\par int method=CV\\_FM\\_RANSAC,\\par double param1=1.,\\par double param2=0.99,\\par CvMat* status=NULL);\n}\n\\cvdefPy{FindFundamentalMat(points1, points2, fundamentalMatrix, method=CV\\_FM\\_RANSAC, param1=1., param2=0.99, status = None) -> None}\n\\cvdefCpp{Mat findFundamentalMat( const Mat\\& points1, const Mat\\& points2,\\par\n vector\\& status, int method=FM\\_RANSAC,\\par\n double param1=3., double param2=0.99 );\\newline\nMat findFundamentalMat( const Mat\\& points1, const Mat\\& points2,\\par\n int method=FM\\_RANSAC,\\par\n double param1=3., double param2=0.99 );}\n\\begin{description}\n\\cvarg{points1}{Array of \\texttt{N} points from the first image.\\cvCPy{It can be \\texttt{2xN, Nx2, 3xN} or \\texttt{Nx3} 1-channel array or \\texttt{1xN} or \\texttt{Nx1} 2- or 3-channel array}. The point coordinates should be floating-point (single or double precision)}\n\\cvarg{points2}{Array of the second image points of the same size and format as \\texttt{points1}}\n\\cvCPy{\\cvarg{fundamentalMatrix}{The output fundamental matrix or matrices. The size should be 3x3 or 9x3 (7-point method may return up to 3 matrices)}}\n\\cvarg{method}{Method for computing the fundamental matrix\n\\begin{description}\n \\cvarg{CV\\_FM\\_7POINT}{for a 7-point algorithm. $N = 7$}\n \\cvarg{CV\\_FM\\_8POINT}{for an 8-point algorithm. $N \\ge 8$}\n \\cvarg{CV\\_FM\\_RANSAC}{for the RANSAC algorithm. $N \\ge 8$}\n \\cvarg{CV\\_FM\\_LMEDS}{for the LMedS algorithm. $N \\ge 8$}\n\\end{description}}\n\\cvarg{param1}{The parameter is used for RANSAC. It is the maximum distance from point to epipolar line in pixels, beyond which the point is considered an outlier and is not used for computing the final fundamental matrix. It can be set to something like 1-3, depending on the accuracy of the point localization, image resolution and the image noise}\n\\cvarg{param2}{The parameter is used for RANSAC or LMedS methods only. It specifies the desirable level of confidence (probability) that the estimated matrix is correct}\n\\cvarg{status}{The \\cvCPy{optional} output array of N elements, every element of which is set to 0 for outliers and to 1 for the other points. The array is computed only in RANSAC and LMedS methods. For other methods it is set to all 1's}\n\\end{description}\n\nThe epipolar geometry is described by the following equation:\n\n\\[ [p_2; 1]^T F [p_1; 1] = 0 \\]\n\nwhere $F$ is fundamental matrix, $p_1$ and $p_2$ are corresponding points in the first and the second images, respectively.\n\nThe function calculates the fundamental matrix using one of four methods listed above and returns \\cvCpp{the found fundamental matrix}\\cvCPy{the number of fundamental matrices found (1 or 3) and 0, if no matrix is found}. Normally just 1 matrix is found, but in the case of 7-point algorithm the function may return up to 3 solutions ($9 \\times 3$ matrix that stores all 3 matrices sequentially).\n\nThe calculated fundamental matrix may be passed further to\n\\cvCross{ComputeCorrespondEpilines}{computeCorrespondEpilines} that finds the epipolar lines\ncorresponding to the specified points. It can also be passed to \\cvCross{StereoRectifyUncalibrated}{stereoRectifyUncalibrated} to compute the rectification transformation.\n\n\\ifC\n% Example. Estimation of fundamental matrix using RANSAC algorithm\n\\begin{lstlisting}\nint point_count = 100;\nCvMat* points1;\nCvMat* points2;\nCvMat* status;\nCvMat* fundamental_matrix;\n\npoints1 = cvCreateMat(1,point_count,CV_32FC2);\npoints2 = cvCreateMat(1,point_count,CV_32FC2);\nstatus = cvCreateMat(1,point_count,CV_8UC1);\n\n/* Fill the points here ... */\nfor( i = 0; i < point_count; i++ )\n{\n points1->data.fl[i*2] = ;\n points1->data.fl[i*2+1] = ;\n points2->data.fl[i*2] = ;\n points2->data.fl[i*2+1] = ;\n}\n\nfundamental_matrix = cvCreateMat(3,3,CV_32FC1);\nint fm_count = cvFindFundamentalMat( points1,points2,fundamental_matrix,\n CV_FM_RANSAC,1.0,0.99,status );\n\\end{lstlisting}\n\\fi\n\\ifCpp\n\\begin{lstlisting}\n// Example. Estimation of fundamental matrix using RANSAC algorithm\nint point_count = 100;\nvector points1(point_count);\nvector points2(point_count);\n\n// initialize the points here ... */\nfor( int i = 0; i < point_count; i++ )\n{\n points1[i] = ...;\n points2[i] = ...;\n}\n\nMat fundamental_matrix =\n findFundamentalMat(points1, points2, FM_RANSAC, 3, 0.99);\n\\end{lstlisting}\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{FindHomography}\n\\else\n\\cvCppFunc{findHomography}\n\\fi\nFinds the perspective transformation between two planes.\n\n\\cvdefC{void cvFindHomography( \\par const CvMat* srcPoints,\\par const CvMat* dstPoints,\\par CvMat* H \\par\nint method=0, \\par double ransacReprojThreshold=3, \\par CvMat* status=NULL);}\n\\cvdefPy{FindHomography(srcPoints,dstPoints,H,method,ransacReprojThreshold=3.0, status=None)-> None}\n\\cvdefCpp{Mat findHomography( const Mat\\& srcPoints, const Mat\\& dstPoints,\\par\n Mat\\& status, int method=0,\\par\n double ransacReprojThreshold=3 );\\newline\nMat findHomography( const Mat\\& srcPoints, const Mat\\& dstPoints,\\par\n vector\\& status, int method=0,\\par\n double ransacReprojThreshold=3 );\\newline\nMat findHomography( const Mat\\& srcPoints, const Mat\\& dstPoints,\\par\n int method=0, double ransacReprojThreshold=3 );}\n\\begin{description}\n\n\\cvCPy{\\cvarg{srcPoints}{Coordinates of the points in the original plane, 2xN, Nx2, 3xN or Nx3 1-channel array (the latter two are for representation in homogeneous coordinates), where N is the number of points. 1xN or Nx1 2- or 3-channel array can also be passed.}\n\\cvarg{dstPoints}{Point coordinates in the destination plane, 2xN, Nx2, 3xN or Nx3 1-channel, or 1xN or Nx1 2- or 3-channel array.}}\n\\cvCpp{\\cvarg{srcPoints}{Coordinates of the points in the original plane, a matrix of type \\texttt{CV\\_32FC2} or a \\texttt{vector}.}\n\\cvarg{dstPoints}{Coordinates of the points in the target plane, a matrix of type \\texttt{CV\\_32FC2} or a \\texttt{vector}.}}\n\n\\cvCPy{\\cvarg{H}{The output 3x3 homography matrix}}\n\\cvarg{method}{ The method used to computed homography matrix; one of the following:\n\\begin{description}\n\\cvarg{0}{a regular method using all the points}\n\\cvarg{CV\\_RANSAC}{RANSAC-based robust method}\n\\cvarg{CV\\_LMEDS}{Least-Median robust method}\n\\end{description}}\n\\cvarg{ransacReprojThreshold}{The maximum allowed reprojection error to treat a point pair as an inlier (used in the RANSAC method only). That is, if\n\\[\\|\\texttt{dstPoints}_i - \\texttt{convertPointsHomogeneous}(\\texttt{H} \\texttt{srcPoints}_i)\\| > \\texttt{ransacReprojThreshold}\\]\nthen the point $i$ is considered an outlier. If \\texttt{srcPoints} and \\texttt{dstPoints} are measured in pixels, it usually makes sense to set this parameter somewhere in the range 1 to 10.}\n\\cvarg{status}{The optional output mask set by a robust method (\\texttt{CV\\_RANSAC} or \\texttt{CV\\_LMEDS}). \\emph{Note that the input mask values are ignored.}}\n\\end{description}\n\nThe \\cvCPy{function finds}\\cvCpp{functions find and return} the perspective transformation $H$ between the source and the destination planes:\n\n\\[\ns_i \\vecthree{x'_i}{y'_i}{1} \\sim H \\vecthree{x_i}{y_i}{1}\n\\]\n\nSo that the back-projection error\n\n\\[\n\\sum_i\n\\left( x'_i-\\frac{h_{11} x_i + h_{12} y_i + h_{13}}{h_{31} x_i + h_{32} y_i + h_{33}} \\right)^2+\n\\left( y'_i-\\frac{h_{21} x_i + h_{22} y_i + h_{23}}{h_{31} x_i + h_{32} y_i + h_{33}} \\right)^2\n\\]\n\nis minimized. If the parameter \\texttt{method} is set to the default value 0, the function\nuses all the point pairs to compute the initial homography estimate with a simple least-squares scheme.\n\nHowever, if not all of the point pairs ($srcPoints_i$,\n$dstPoints_i$) fit the rigid perspective transformation (i.e. there\nare some outliers), this initial estimate will be poor.\nIn this case one can use one of the 2 robust methods. Both methods,\n\\texttt{RANSAC} and \\texttt{LMeDS}, try many different random subsets\nof the corresponding point pairs (of 4 pairs each), estimate\nthe homography matrix using this subset and a simple least-square\nalgorithm and then compute the quality/goodness of the computed homography\n(which is the number of inliers for RANSAC or the median re-projection\nerror for LMeDs). The best subset is then used to produce the initial\nestimate of the homography matrix and the mask of inliers/outliers.\n\nRegardless of the method, robust or not, the computed homography\nmatrix is refined further (using inliers only in the case of a robust\nmethod) with the Levenberg-Marquardt method in order to reduce the\nre-projection error even more.\n\nThe method \\texttt{RANSAC} can handle practically any ratio of outliers,\nbut it needs the threshold to distinguish inliers from outliers.\nThe method \\texttt{LMeDS} does not need any threshold, but it works\ncorrectly only when there are more than 50\\% of inliers. Finally,\nif you are sure in the computed features, where can be only some\nsmall noise present, but no outliers, the default method could be the best\nchoice.\n\nThe function is used to find initial intrinsic and extrinsic matrices.\nHomography matrix is determined up to a scale, thus it is normalized so that\n$h_{33}=1$.\n\nSee also: \\cvCross{GetAffineTransform}{getAffineTransform}, \\cvCross{GetPerspectiveTransform}{getPerspectiveTransform}, \\cvCross{EstimateRigidMotion}{estimateRigidMotion}, \\cvCross{WarpPerspective}{warpPerspective}, \\cvCross{PerspectiveTransform}{perspectiveTransform}\n\n\\ifCPy\n\n\\cvCPyFunc{FindStereoCorrespondenceBM}\nComputes the disparity map using block matching algorithm.\n\n\\cvdefC{\n\nvoid cvFindStereoCorrespondenceBM( \\par const CvArr* left, \\par const CvArr* right,\n \\par CvArr* disparity, \\par CvStereoBMState* state );\n\n}\\cvdefPy{FindStereoCorrespondenceBM(left,right,disparity,state)-> None}\n\n\\begin{description}\n\\cvarg{left}{The left single-channel, 8-bit image.}\n\\cvarg{right}{The right image of the same size and the same type.}\n\\cvarg{disparity}{The output single-channel 16-bit signed, or 32-bit floating-point disparity map of the same size as input images. In the first case the computed disparities are represented as fixed-point numbers with 4 fractional bits (i.e. the computed disparity values are multiplied by 16 and rounded to integers).}\n\\cvarg{state}{Stereo correspondence structure.}\n\\end{description}\n\nThe function cvFindStereoCorrespondenceBM computes disparity map for the input rectified stereo pair. Invalid pixels (for which disparity can not be computed) are set to \\texttt{state->minDisparity - 1} (or to \\texttt{(state->minDisparity-1)*16} in the case of 16-bit fixed-point disparity map)\n\n\\cvCPyFunc{FindStereoCorrespondenceGC}\nComputes the disparity map using graph cut-based algorithm.\n\n\\cvdefC{\n\nvoid cvFindStereoCorrespondenceGC( \\par const CvArr* left, \\par const CvArr* right,\n \\par CvArr* dispLeft, \\par CvArr* dispRight,\n \\par CvStereoGCState* state,\n \\par int useDisparityGuess = CV\\_DEFAULT(0) );\n\n}\n\\cvdefPy{FindStereoCorrespondenceGC( left, right, dispLeft, dispRight, state, useDisparityGuess=(0))-> None}\n\n\\begin{description}\n\\cvarg{left}{The left single-channel, 8-bit image.}\n\\cvarg{right}{The right image of the same size and the same type.}\n\\cvarg{dispLeft}{The optional output single-channel 16-bit signed left disparity map of the same size as input images.}\n\\cvarg{dispRight}{The optional output single-channel 16-bit signed right disparity map of the same size as input images.}\n\\cvarg{state}{Stereo correspondence structure.}\n\\cvarg{useDisparityGuess}{If the parameter is not zero, the algorithm will start with pre-defined disparity maps. Both dispLeft and dispRight should be valid disparity maps. Otherwise, the function starts with blank disparity maps (all pixels are marked as occlusions).}\n\\end{description}\n\nThe function computes disparity maps for the input rectified stereo pair. Note that the left disparity image will contain values in the following range: \n\n\\[\n-\\texttt{state->numberOfDisparities}-\\texttt{state->minDisparity}\n< dispLeft(x,y) \\le -\\texttt{state->minDisparity},\n\\]\n\nor\n\\[\ndispLeft(x,y) == \\texttt{CV\\_STEREO\\_GC\\_OCCLUSION}\n\\]\n\nand for the right disparity image the following will be true: \n\n\\[\n\\texttt{state->minDisparity} \\le dispRight(x,y) \n< \\texttt{state->minDisparity} + \\texttt{state->numberOfDisparities}\n\\]\n\nor\n\n\\[\ndispRight(x,y) == \\texttt{CV\\_STEREO\\_GC\\_OCCLUSION}\n\\]\n\nthat is, the range for the left disparity image will be inversed,\nand the pixels for which no good match has been found, will be marked\nas occlusions.\n\nHere is how the function can be used:\n\n\\ifC\n\\begin{lstlisting}\n// image_left and image_right are the input 8-bit single-channel images\n// from the left and the right cameras, respectively\nCvSize size = cvGetSize(image_left);\nCvMat* disparity_left = cvCreateMat( size.height, size.width, CV_16S );\nCvMat* disparity_right = cvCreateMat( size.height, size.width, CV_16S );\nCvStereoGCState* state = cvCreateStereoGCState( 16, 2 );\ncvFindStereoCorrespondenceGC( image_left, image_right,\n disparity_left, disparity_right, state, 0 );\ncvReleaseStereoGCState( &state );\n// now process the computed disparity images as you want ...\n\\end{lstlisting}\n\nand this is the output left disparity image computed from the well-known\nTsukuba stereo pair and multiplied by -16 (because the values in the\nleft disparity images are usually negative):\n\n\\begin{lstlisting}\nCvMat* disparity_left_visual = cvCreateMat( size.height, size.width, CV_8U );\ncvConvertScale( disparity_left, disparity_left_visual, -16 );\ncvSave( \"disparity.pgm\", disparity_left_visual );\n\\end{lstlisting}\n\n\\includegraphics{pics/disparity.png}\n\n\\else\n\n\\lstinputlisting{python_fragments/findstereocorrespondence.py}\n\nand this is the output left disparity image computed from the well-known\nTsukuba stereo pair and multiplied by -16 (because the values in the\nleft disparity images are usually negative):\n\n\\includegraphics{pics/disparity.png}\n\n\\fi\n\n\\fi\n\n\\ifCpp\n\\cvCppFunc{getDefaultNewCameraMatrix}\nReturns the default new camera matrix\n\n\\cvdefCpp{Mat getDefaultNewCameraMatrix(\\par\n const Mat\\& cameraMatrix,\\par\n Size imgSize=Size(),\\par\n bool centerPrincipalPoint=false );}\n\\begin{description}\n\\cvarg{cameraMatrix}{The input camera matrix}\n\\cvarg{imageSize}{The camera view image size in pixels}\n\\cvarg{centerPrincipalPoint}{Indicates whether in the new camera matrix the principal point should be at the image center or not}\n\\end{description}\n\nThe function returns the camera matrix that is either an exact copy of the input \\texttt{cameraMatrix} (when \\texttt{centerPrinicipalPoint=false}), or the modified one (when \\texttt{centerPrincipalPoint}=true).\n\nIn the latter case the new camera matrix will be:\n\n\\[\\begin{bmatrix}\nf_x && 0 && (\\texttt{imgSize.width}-1)*0.5 \\\\\n0 && f_y && (\\texttt{imgSize.height}-1)*0.5 \\\\\n0 && 0 && 1\n\\end{bmatrix},\\]\n\nwhere $f_x$ and $f_y$ are $(0,0)$ and $(1,1)$ elements of \\texttt{cameraMatrix}, respectively.\n\nBy default, the undistortion functions in OpenCV (see \\texttt{initUndistortRectifyMap}, \\texttt{undistort}) do not move the principal point. However, when you work with stereo, it's important to move the principal points in both views to the same y-coordinate (which is required by most of stereo correspondence algorithms), and maybe to the same x-coordinate too. So you can form the new camera matrix for each view, where the principal points will be at the center.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{GetOptimalNewCameraMatrix}\n\\else\n\\cvCppFunc{getOptimalNewCameraMatrix}\n\\fi\nReturns the new camera matrix based on the free scaling parameter\n\n\\cvdefC{void cvGetOptimalNewCameraMatrix(\n \\par const CvMat* cameraMatrix, const CvMat* distCoeffs,\n \\par CvSize imageSize, double alpha,\n \\par CvMat* newCameraMatrix,\n \\par CvSize newImageSize=cvSize(0,0),\n \\par CvRect* validPixROI=0 );}\n\\cvdefPy{GetOptimalNewCameraMatrix(cameraMatrix, distCoeffs, imageSize, alpha, newCameraMatrix, newImageSize=(0,0), validPixROI=0) -> None}\n\\cvdefCpp{Mat getOptimalNewCameraMatrix(\n \\par const Mat\\& cameraMatrix, const Mat\\& distCoeffs,\n \\par Size imageSize, double alpha, Size newImageSize=Size(),\n \\par Rect* validPixROI=0);}\n \n\\begin{description}\n\\cvarg{cameraMatrix}{The input camera matrix}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{imageSize}{The original image size}\n\\cvarg{alpha}{The free scaling parameter between 0 (when all the pixels in the undistorted image will be valid) and 1 (when all the source image pixels will be retained in the undistorted image); see \\cvCross{StereoRectify}{stereoRectify}}\n\\cvarg{newCameraMatrix}{The output new camera matrix.}\n\\cvarg{newImageSize}{The image size after rectification. By default it will be set to \\texttt{imageSize}.}\n\\cvarg{validPixROI}{The optional output rectangle that will outline all-good-pixels region in the undistorted image. See \\texttt{roi1, roi2} description in \\cvCross{StereoRectify}{stereoRectify}}\n\\end{description}\n\nThe function computes \\cvCpp{and returns} the optimal new camera matrix based on the free scaling parameter. By varying this parameter the user may retrieve only sensible pixels \\texttt{alpha=0}, keep all the original image pixels if there is valuable information in the corners \\texttt{alpha=1}, or get something in between. When \\texttt{alpha>0}, the undistortion result will likely have some black pixels corresponding to \"virtual\" pixels outside of the captured distorted image. The original camera matrix, distortion coefficients, the computed new camera matrix and the \\texttt{newImageSize} should be passed to \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap} to produce the maps for \\cvCross{Remap}{remap}.\n\n\\ifCPy\n\\cvCPyFunc{InitIntrinsicParams2D}\n\\else\n\\cvCppFunc{initCameraMatrix2D}\n\\fi\n\nFinds the initial camera matrix from the 3D-2D point correspondences\n\n\\cvdefC{void cvInitIntrinsicParams2D(\\par const CvMat* objectPoints,\n \\par const CvMat* imagePoints,\n \\par const CvMat* npoints, CvSize imageSize,\n \\par CvMat* cameraMatrix,\n \\par double aspectRatio=1.);} \n\\cvdefPy{InitIntrinsicParams2D(objectPoints, imagePoints, npoints, imageSize, cameraMatrix, aspectRatio=1.) -> None}\n\\cvdefCpp{Mat initCameraMatrix2D( const vector >\\& objectPoints,\\par\n const vector >\\& imagePoints,\\par\n Size imageSize, double aspectRatio=1.);}\n\\begin{description}\n\\ifCPy\n\\cvarg{objectPoints}{The joint array of object points; see \\cvCross{CalibrateCamera2}{calibrateCamera}}\n\\cvarg{imagePoints}{The joint array of object point projections; see \\cvCross{CalibrateCamera2}{calibrateCamera}}\n\\cvarg{npoints}{The array of point counts; see \\cvCross{CalibrateCamera2}{calibrateCamera}}\n\\fi \n\\ifCpp\n\\cvarg{objectPoints}{The vector of vectors of the object points. See \\cvCppCross{calibrateCamera}}\n\\cvarg{imagePoints}{The vector of vectors of the corresponding image points. See \\cvCppCross{calibrateCamera}}\n\\fi\n\\cvarg{imageSize}{The image size in pixels; used to initialize the principal point}\n\\cvCPy{\\cvarg{cameraMatrix}{The output camera matrix $\\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{1}$}}\n\\cvarg{aspectRatio}{If it is zero or negative, both $f_x$ and $f_y$ are estimated independently. Otherwise $f_x = f_y * \\texttt{aspectRatio}$}\n\\end{description}\n\nThe function estimates and returns the initial camera matrix for camera calibration process.\nCurrently, the function only supports planar calibration patterns, i.e. patterns where each object point has z-coordinate =0.\n\n\\ifCPy\n\\cvCPyFunc{InitUndistortMap}\nComputes an undistortion map.\n\n\\cvdefC{void cvInitUndistortMap( \\par const CvMat* cameraMatrix,\\par const CvMat* distCoeffs,\\par CvArr* map1,\\par CvArr* map2 );}\n\\cvdefPy{InitUndistortMap(cameraMatrix,distCoeffs,map1,map2)-> None}\n\n\\begin{description}\n\\cvarg{cameraMatrix}{The input camera matrix $A = \\vecthreethree{fx}{0}{cx}{0}{fy}{cy}{0}{0}{1} $}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{map1}{The first output map \\cvCPy{of type \\texttt{CV\\_32FC1} or \\texttt{CV\\_16SC2} - the second variant is more efficient}}\n\\cvarg{map2}{The second output map \\cvCPy{of type \\texttt{CV\\_32FC1} or \\texttt{CV\\_16UC1} - the second variant is more efficient}}\n\\end{description}\n\nThe function is a simplified variant of \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap} where the rectification transformation \\texttt{R} is identity matrix and \\texttt{newCameraMatrix=cameraMatrix}.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{InitUndistortRectifyMap}\n\\else\n\\cvCppFunc{initUndistortRectifyMap}\n\\fi\nComputes the undistortion and rectification transformation map.\n\n\\cvdefC{void cvInitUndistortRectifyMap( \\par const CvMat* cameraMatrix,\n \\par const CvMat* distCoeffs,\n \\par const CvMat* R,\n \\par const CvMat* newCameraMatrix,\n \\par CvArr* map1, \\par CvArr* map2 );}\n\\cvdefPy{InitUndistortRectifyMap(cameraMatrix,distCoeffs,R,newCameraMatrix,map1,map2)-> None}\n\\cvdefCpp{void initUndistortRectifyMap( const Mat\\& cameraMatrix,\\par\n const Mat\\& distCoeffs, const Mat\\& R,\\par\n const Mat\\& newCameraMatrix,\\par\n Size size, int m1type,\\par\n Mat\\& map1, Mat\\& map2 );}\n\\begin{description}\n\\cvarg{cameraMatrix}{The input camera matrix $A=\\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{1}$}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{R}{The optional rectification transformation in object space (3x3 matrix). \\texttt{R1} or \\texttt{R2}, computed by \\cvCross{StereoRectify}{stereoRectify} can be passed here. If the matrix is \\cvCPy{NULL}\\cvCpp{empty}, the identity transformation is assumed}\n\\cvarg{newCameraMatrix}{The new camera matrix $A'=\\vecthreethree{f_x'}{0}{c_x'}{0}{f_y'}{c_y'}{0}{0}{1}$}\n\\cvCpp{\\cvarg{size}{The undistorted image size}\n\\cvarg{m1type}{The type of the first output map, can be \\texttt{CV\\_32FC1} or \\texttt{CV\\_16SC2}. See \\cvCppCross{convertMaps}}}\n\\cvarg{map1}{The first output map \\cvCPy{of type \\texttt{CV\\_32FC1} or \\texttt{CV\\_16SC2} - the second variant is more efficient}}\n\\cvarg{map2}{The second output map \\cvCPy{of type \\texttt{CV\\_32FC1} or \\texttt{CV\\_16UC1} - the second variant is more efficient}}\n\\end{description}\n\nThe function computes the joint undistortion+rectification transformation and represents the result in the form of maps for \\cvCross{Remap}{remap}. The undistorted image will look like the original, as if it was captured with a camera with camera matrix \\texttt{=newCameraMatrix} and zero distortion. In the case of monocular camera \\texttt{newCameraMatrix} is usually equal to \\texttt{cameraMatrix}, or it can be computed by \\cvCross{GetOptimalNewCameraMatrix}{getOptimalNewCameraMatrix} for a better control over scaling. In the case of stereo camera \\texttt{newCameraMatrix} is normally set to \\texttt{P1} or \\texttt{P2} computed by \\cvCross{StereoRectify}{stereoRectify}.\n\nAlso, this new camera will be oriented differently in the coordinate space, according to \\texttt{R}. That, for example, helps to align two heads of a stereo camera so that the epipolar lines on both images become horizontal and have the same y- coordinate (in the case of horizontally aligned stereo camera).\n\nThe function actually builds the maps for the inverse mapping algorithm that is used by \\cvCross{Remap}{remap}. That is, for each pixel $(u, v)$ in the destination (corrected and rectified) image the function computes the corresponding coordinates in the source image (i.e. in the original image from camera). The process is the following:\n\n\\[\n\\begin{array}{l}\nx \\leftarrow (u - {c'}_x)/{f'}_x \\\\\ny \\leftarrow (v - {c'}_y)/{f'}_y \\\\\n{[X\\,Y\\,W]}^T \\leftarrow R^{-1}*[x\\,y\\,1]^T \\\\\nx' \\leftarrow X/W \\\\\ny' \\leftarrow Y/W \\\\\nx\" \\leftarrow x' (1 + k_1 r^2 + k_2 r^4 + k_3 r^6) + 2p_1 x' y' + p_2(r^2 + 2 x'^2) \\\\\ny\" \\leftarrow y' (1 + k_1 r^2 + k_2 r^4 + k_3 r^6) + p_1 (r^2 + 2 y'^2) + 2 p_2 x' y' \\\\\nmap_x(u,v) \\leftarrow x\" f_x + c_x \\\\\nmap_y(u,v) \\leftarrow y\" f_y + c_y\n\\end{array}\n\\]\nwhere $(k_1, k_2, p_1, p_2[, k_3])$ are the distortion coefficients. \n \nIn the case of a stereo camera this function is called twice, once for each camera head, after \\cvCross{StereoRectify}{stereoRectify}, which in its turn is called after \\cvCross{StereoCalibrate}{stereoCalibrate}. But if the stereo camera was not calibrated, it is still possible to compute the rectification transformations directly from the fundamental matrix using \\cvCross{StereoRectifyUncalibrated}{stereoRectifyUncalibrated}. For each camera the function computes homography \\texttt{H} as the rectification transformation in pixel domain, not a rotation matrix \\texttt{R} in 3D space. The \\texttt{R} can be computed from \\texttt{H} as \n\n\\[ \\texttt{R} = \\texttt{cameraMatrix}^{-1} \\cdot \\texttt{H} \\cdot \\texttt{cameraMatrix} \\]\n\nwhere the \\texttt{cameraMatrix} can be chosen arbitrarily.\n\n\\ifCpp\n\n\\cvCppFunc{matMulDeriv}\nComputes partial derivatives of the matrix product w.r.t each multiplied matrix\n\n\\cvdefCpp{void matMulDeriv( const Mat\\& A, const Mat\\& B, Mat\\& dABdA, Mat\\& dABdB );}\n\\begin{description}\n\\cvarg{A}{The first multiplied matrix}\n\\cvarg{B}{The second multiplied matrix}\n\\cvarg{dABdA}{The first output derivative matrix \\texttt{d(A*B)/dA} of size $\\texttt{A.rows*B.cols} \\times {A.rows*A.cols}$}\n\\cvarg{dABdA}{The second output derivative matrix \\texttt{d(A*B)/dB} of size $\\texttt{A.rows*B.cols} \\times {B.rows*B.cols}$}\n\\end{description}\n\nThe function computes the partial derivatives of the elements of the matrix product $A*B$ w.r.t. the elements of each of the two input matrices. The function is used to compute Jacobian matrices in \\cvCppCross{stereoCalibrate}, but can also be used in any other similar optimization function.\n\n\\fi\n\n\\ifCPy\n\n\\cvCPyFunc{POSIT}\nImplements the POSIT algorithm.\n\n\\cvdefC{\nvoid cvPOSIT( \\par CvPOSITObject* posit\\_object,\\par CvPoint2D32f* imagePoints,\\par double focal\\_length,\\par CvTermCriteria criteria,\\par CvMatr32f rotationMatrix,\\par CvVect32f translation\\_vector );\n}\n\\cvdefPy{POSIT(posit\\_object,imagePoints,focal\\_length,criteria)-> (rotationMatrix,translation\\_vector)}\n\n\\begin{description}\n\\cvarg{posit\\_object}{Pointer to the object structure}\n\\cvarg{imagePoints}{Pointer to the object points projections on the 2D image plane}\n\\cvarg{focal\\_length}{Focal length of the camera used}\n\\cvarg{criteria}{Termination criteria of the iterative POSIT algorithm}\n\\cvarg{rotationMatrix}{Matrix of rotations}\n\\cvarg{translation\\_vector}{Translation vector}\n\\end{description}\n\nThe function implements the POSIT algorithm. Image coordinates are given in a camera-related coordinate system. The focal length may be retrieved using the camera calibration functions. At every iteration of the algorithm a new perspective projection of the estimated pose is computed.\n\nDifference norm between two projections is the maximal distance between corresponding points. The parameter \\texttt{criteria.epsilon} serves to stop the algorithm if the difference is small.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{ProjectPoints2}\n\\else\n\\cvCppFunc{projectPoints}\n\\fi\nProject 3D points on to an image plane.\n\n\\cvdefC{void cvProjectPoints2( \\par const CvMat* objectPoints,\\par const CvMat* rvec,\\par const CvMat* tvec,\\par const CvMat* cameraMatrix,\\par const CvMat* distCoeffs,\\par CvMat* imagePoints,\\par CvMat* dpdrot=NULL,\\par CvMat* dpdt=NULL,\\par CvMat* dpdf=NULL,\\par CvMat* dpdc=NULL,\\par CvMat* dpddist=NULL );}\n\n\\cvdefPy{ProjectPoints2(objectPoints,rvec,tvec,cameraMatrix,distCoeffs, imagePoints,dpdrot=NULL,dpdt=NULL,dpdf=NULL,dpdc=NULL,dpddist=NULL)-> None}\n\n\n\\cvdefCpp{void projectPoints( const Mat\\& objectPoints,\\par\n const Mat\\& rvec, const Mat\\& tvec,\\par\n const Mat\\& cameraMatrix,\\par\n const Mat\\& distCoeffs,\\par\n vector\\& imagePoints );\\newline\nvoid projectPoints( const Mat\\& objectPoints,\\par\n const Mat\\& rvec, const Mat\\& tvec,\\par\n const Mat\\& cameraMatrix,\\par\n const Mat\\& distCoeffs,\\par\n vector\\& imagePoints,\\par\n Mat\\& dpdrot, Mat\\& dpdt, Mat\\& dpdf,\\par\n Mat\\& dpdc, Mat\\& dpddist,\\par\n double aspectRatio=0 );}\n\n\\begin{description}\n\\cvarg{objectPoints}{The array of object points, 3xN or Nx3 1-channel or 1xN or Nx1 3-channel \\cvCpp{(or \\texttt{vector})}, where N is the number of points in the view}\n\\cvarg{rvec}{The rotation vector, see \\cvCross{Rodrigues2}{Rodrigues}}\n\\cvarg{tvec}{The translation vector}\n\\cvarg{cameraMatrix}{The camera matrix $A = \\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{_1} $}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{imagePoints}{The output array of image points, 2xN or Nx2 1-channel or 1xN or Nx1 2-channel \\cvCpp{(or \\texttt{vector})}}\n\\cvarg{dpdrot}{Optional 2Nx3 matrix of derivatives of image points with respect to components of the rotation vector}\n\\cvarg{dpdt}{Optional 2Nx3 matrix of derivatives of image points with respect to components of the translation vector}\n\\cvarg{dpdf}{Optional 2Nx2 matrix of derivatives of image points with respect to $f_x$ and $f_y$}\n\\cvarg{dpdc}{Optional 2Nx2 matrix of derivatives of image points with respect to $c_x$ and $c_y$}\n\\cvarg{dpddist}{Optional 2Nx4 matrix of derivatives of image points with respect to distortion coefficients}\n\\end{description}\n\nThe function computes projections of 3D\npoints to the image plane given intrinsic and extrinsic camera\nparameters. Optionally, the function computes jacobians - matrices\nof partial derivatives of image points coordinates (as functions of all the\ninput parameters) with respect to the particular parameters, intrinsic and/or\nextrinsic. The jacobians are used during the global optimization\nin \\cvCross{CalibrateCamera2}{calibrateCamera},\n\\cvCross{FindExtrinsicCameraParams2}{solvePnP} and \\cvCross{StereoCalibrate}{stereoCalibrate}. The\nfunction itself can also used to compute re-projection error given the\ncurrent intrinsic and extrinsic parameters.\n\nNote, that by setting \\texttt{rvec=tvec=(0,0,0)}, or by setting \\texttt{cameraMatrix} to 3x3 identity matrix, or by passing zero distortion coefficients, you can get various useful partial cases of the function, i.e. you can compute the distorted coordinates for a sparse set of points, or apply a perspective transformation (and also compute the derivatives) in the ideal zero-distortion setup etc.\n\n\n\\ifCPy\n\\cvCPyFunc{ReprojectImageTo3D}\n\\else\n\\cvCppFunc{reprojectImageTo3D}\n\\fi\nReprojects disparity image to 3D space.\n\n\\cvdefC{void cvReprojectImageTo3D( const CvArr* disparity,\\par\n CvArr* \\_3dImage, const CvMat* Q,\\par\n int handleMissingValues=0);}\n\n\\cvdefPy{ReprojectImageTo3D(disparity, \\_3dImage, Q, handleMissingValues=0) -> None}\n\n\\cvdefCpp{void reprojectImageTo3D( const Mat\\& disparity,\\par\n Mat\\& \\_3dImage, const Mat\\& Q,\\par\n bool handleMissingValues=false );}\n\\begin{description}\n\\cvarg{disparity}{The input single-channel 16-bit signed or 32-bit floating-point disparity image}\n\\cvarg{\\_3dImage}{The output 3-channel floating-point image of the same size as \\texttt{disparity}.\n Each element of \\texttt{\\_3dImage(x,y)} will contain the 3D coordinates of the point \\texttt{(x,y)}, computed from the disparity map.}\n\\cvarg{Q}{The $4 \\times 4$ perspective transformation matrix that can be obtained with \\cvCross{StereoRectify}{stereoRectify}}\n\\cvarg{handleMissingValues}{If true, when the pixels with the minimal disparity (that corresponds to the outliers; see \\cvCross{FindStereoCorrespondenceBM}{StereoBM}) will be transformed to 3D points with some very large Z value (currently set to 10000)}\n\\end{description}\n \nThe function transforms 1-channel disparity map to 3-channel image representing a 3D surface. That is, for each pixel \\texttt{(x,y)} and the corresponding disparity \\texttt{d=disparity(x,y)} it computes: \n\n\\[\\begin{array}{l}\n[X\\; Y\\; Z\\; W]^T = \\texttt{Q}*[x\\; y\\; \\texttt{disparity}(x,y)\\; 1]^T \\\\\n\\texttt{\\_3dImage}(x,y) = (X/W,\\; Y/W,\\; Z/W)\n\\end{array}\\]\n\nThe matrix \\texttt{Q} can be arbitrary $4 \\times 4$ matrix, e.g. the one computed by \\cvCross{StereoRectify}{stereoRectify}. To reproject a sparse set of points {(x,y,d),...} to 3D space, use \\cvCross{PerspectiveTransform}{perspectiveTransform}.\n\n\\ifCPy\n\\cvCPyFunc{RQDecomp3x3}\n\\else\n\\cvCppFunc{RQDecomp3x3}\n\\fi\nComputes the 'RQ' decomposition of 3x3 matrices.\n\n\\cvdefC{\nvoid cvRQDecomp3x3( \\par const CvMat *M,\\par CvMat *R,\\par CvMat *Q,\\par CvMat *Qx=NULL,\\par CvMat *Qy=NULL,\\par CvMat *Qz=NULL,\\par CvPoint3D64f *eulerAngles=NULL);\n}\n\\cvdefPy{RQDecomp3x3(M, R, Q, Qx = None, Qy = None, Qz = None) -> eulerAngles}\n\\cvdefCpp{void RQDecomp3x3( const Mat\\& M, Mat\\& R, Mat\\& Q );\\newline\nVec3d RQDecomp3x3( const Mat\\& M, Mat\\& R, Mat\\& Q,\\par\n Mat\\& Qx, Mat\\& Qy, Mat\\& Qz );}\n\n\\begin{description}\n\\cvarg{M}{The 3x3 input matrix}\n\\cvarg{R}{The output 3x3 upper-triangular matrix}\n\\cvarg{Q}{The output 3x3 orthogonal matrix}\n\\cvarg{Qx}{Optional 3x3 rotation matrix around x-axis}\n\\cvarg{Qy}{Optional 3x3 rotation matrix around y-axis}\n\\cvarg{Qz}{Optional 3x3 rotation matrix around z-axis}\n\\cvCPy{\\cvarg{eulerAngles}{Optional three Euler angles of rotation}}\n\\end{description}\n\nThe function computes a RQ decomposition using the given rotations. This function is used in \\cvCross{DecomposeProjectionMatrix}{decomposeProjectionMatrix} to decompose the left 3x3 submatrix of a projection matrix into a camera and a rotation matrix.\n\nIt optionally returns three rotation matrices, one for each axis, and the three Euler angles \\cvCpp{(as the return value)} that could be used in OpenGL.\n\n\\ifC\n\n\\cvCPyFunc{ReleasePOSITObject}\nDeallocates a 3D object structure.\n\n\\cvdefC{\nvoid cvReleasePOSITObject( \\par CvPOSITObject** posit\\_object );\n}\n\n\\begin{description}\n\\cvarg{posit\\_object}{Double pointer to \\texttt{CvPOSIT} structure}\n\\end{description}\n\nThe function releases memory previously allocated by the function \\cvCPyCross{CreatePOSITObject}.\n\n\\fi\n\n\\ifC\n\n\\cvCPyFunc{ReleaseStereoBMState}\nReleases block matching stereo correspondence structure.\n\n\\cvdefC{void cvReleaseStereoBMState( CvStereoBMState** state );}\n\\cvdefPy{ReleaseStereoBMState(state)-> None}\n\n\\begin{description}\n\\cvarg{state}{Double pointer to the released structure.}\n\\end{description}\n\nThe function releases the stereo correspondence structure and all the associated internal buffers. \n\n\\cvCPyFunc{ReleaseStereoGCState}\nReleases the state structure of the graph cut-based stereo correspondence algorithm.\n\n\\cvdefC{void cvReleaseStereoGCState( CvStereoGCState** state );}\n\\cvdefPy{ReleaseStereoGCState(state)-> None}\n\n\\begin{description}\n\\cvarg{state}{Double pointer to the released structure.}\n\\end{description}\n\nThe function releases the stereo correspondence structure and all the associated internal buffers. \n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{Rodrigues2}\n\\else\n\\cvCppFunc{Rodrigues}\n\\fi\nConverts a rotation matrix to a rotation vector or vice versa.\n\n\\cvdefC{int cvRodrigues2( \\par const CvMat* src,\\par CvMat* dst,\\par CvMat* jacobian=0 );}\n\\cvdefPy{Rodrigues2(src,dst,jacobian=0)-> None}\n\n\\cvdefCpp{void Rodrigues(const Mat\\& src, Mat\\& dst);\\newline\nvoid Rodrigues(const Mat\\& src, Mat\\& dst, Mat\\& jacobian);}\n\n\\begin{description}\n\\cvarg{src}{The input rotation vector (3x1 or 1x3) or rotation matrix (3x3)}\n\\cvarg{dst}{The output rotation matrix (3x3) or rotation vector (3x1 or 1x3), respectively}\n\\cvarg{jacobian}{Optional output Jacobian matrix, 3x9 or 9x3 - partial derivatives of the output array components with respect to the input array components}\n\\end{description}\n\n\\[\n\\begin{array}{l}\n\\theta \\leftarrow norm(r)\\\\\nr \\leftarrow r/\\theta\\\\\nR = \\cos{\\theta} I + (1-\\cos{\\theta}) r r^T + \\sin{\\theta}\n\\vecthreethree\n{0}{-r_z}{r_y}\n{r_z}{0}{-r_x}\n{-r_y}{r_x}{0}\n\\end{array}\n\\]\n\nInverse transformation can also be done easily, since\n\n\\[\n\\sin(\\theta)\n\\vecthreethree\n{0}{-r_z}{r_y}\n{r_z}{0}{-r_x}\n{-r_y}{r_x}{0}\n=\n\\frac{R - R^T}{2}\n\\]\n\nA rotation vector is a convenient and most-compact representation of a rotation matrix\n(since any rotation matrix has just 3 degrees of freedom). The representation is\nused in the global 3D geometry optimization procedures like \\cvCross{CalibrateCamera2}{calibrateCamera},\n\\cvCross{StereoCalibrate}{stereoCalibrate} or \\cvCross{FindExtrinsicCameraParams2}{solvePnP}.\n\n\n\\ifCpp\n\n\\cvclass{StereoBM}\nThe class for computing stereo correspondence using block matching algorithm.\n\n\\begin{lstlisting}\n// Block matching stereo correspondence algorithm\\par\nclass StereoBM\n{\n enum { NORMALIZED_RESPONSE = CV_STEREO_BM_NORMALIZED_RESPONSE,\n BASIC_PRESET=CV_STEREO_BM_BASIC,\n FISH_EYE_PRESET=CV_STEREO_BM_FISH_EYE,\n NARROW_PRESET=CV_STEREO_BM_NARROW };\n\n StereoBM();\n // the preset is one of ..._PRESET above.\n // ndisparities is the size of disparity range,\n // in which the optimal disparity at each pixel is searched for.\n // SADWindowSize is the size of averaging window used to match pixel blocks\n // (larger values mean better robustness to noise, but yield blurry disparity maps)\n StereoBM(int preset, int ndisparities=0, int SADWindowSize=21);\n // separate initialization function\n void init(int preset, int ndisparities=0, int SADWindowSize=21);\n // computes the disparity for the two rectified 8-bit single-channel images.\n // the disparity will be 16-bit signed (fixed-point) or 32-bit floating-point image of the same size as left.\n void operator()( const Mat& left, const Mat& right, Mat& disparity, int disptype=CV_16S );\n\n Ptr state;\n};\n\\end{lstlisting}\n\nThe class is a C++ wrapper for \\hyperref[CvStereoBMState]{cvStereoBMState} and the associated functions. In particular, \\texttt{StereoBM::operator ()} is the wrapper for \\cvCPyCross{FindStereoCorrespondceBM}. See the respective descriptions.\n\n\n\\cvclass{StereoSGBM}\nThe class for computing stereo correspondence using semi-global block matching algorithm.\n\n\\begin{lstlisting}\nclass StereoSGBM\n{\n StereoSGBM();\n StereoSGBM(int minDisparity, int numDisparities, int SADWindowSize,\n int P1=0, int P2=0, int disp12MaxDiff=0,\n int preFilterCap=0, int uniquenessRatio=0,\n int speckleWindowSize=0, int speckleRange=0,\n bool fullDP=false);\n virtual ~StereoSGBM();\n \n virtual void operator()(const Mat& left, const Mat& right, Mat& disp);\n \n int minDisparity;\n int numberOfDisparities;\n int SADWindowSize;\n int preFilterCap;\n int uniquenessRatio;\n int P1, P2;\n int speckleWindowSize;\n int speckleRange;\n int disp12MaxDiff;\n bool fullDP;\n \n ...\n};\n\\end{lstlisting}\n\nThe class implements modified H. Hirschmuller algorithm \\cite{HH08}. The main differences between the implemented algorithm and the original one are:\n\n\\begin{itemize}\n \\item by default the algorithm is single-pass, i.e. instead of 8 directions we only consider 5. Set \\texttt{fullDP=true} to run the full variant of the algorithm (which could consume \\emph{a lot} of memory)\n \\item the algorithm matches blocks, not individual pixels (though, by setting \\texttt{SADWindowSize=1} the blocks are reduced to single pixels)\n \\item mutual information cost function is not implemented. Instead, we use a simpler Birchfield-Tomasi sub-pixel metric from \\cite{BT96}, though the color images are supported as well.\n \\item we include some pre- and post- processing steps from K. Konolige algorithm \\cvCPyCross{FindStereoCorrespondceBM}, such as pre-filtering (\\texttt{CV\\_STEREO\\_BM\\_XSOBEL} type) and post-filtering (uniqueness check, quadratic interpolation and speckle filtering)\n\\end{itemize}\n\n\\cvCppFunc{StereoSGBM::StereoSGBM}\nStereoSGBM constructors\n\n\\cvdefCpp{\nStereoSGBM::StereoSGBM();\\newline\nStereoSGBM::StereoSGBM(\n \\par int minDisparity, int numDisparities, int SADWindowSize,\n \\par int P1=0, int P2=0, int disp12MaxDiff=0,\n \\par int preFilterCap=0, int uniquenessRatio=0,\n \\par int speckleWindowSize=0, int speckleRange=0,\n \\par bool fullDP=false);\n}\n\\begin{description}\n\\cvarg{minDisparity}{The minimum possible disparity value. Normally it is 0, but sometimes rectification algorithms can shift images, so this parameter needs to be adjusted accordingly}\n\\cvarg{numDisparities}{This is maximum disparity minus minimum disparity. Always greater than 0. In the current implementation this parameter must be divisible by 16.}\n\\cvarg{SADWindowSize}{The matched block size. Must be an odd number \\texttt{>=1}. Normally, it should be somewhere in \\texttt{3..11} range}.\n\\cvarg{P1, P2}{Parameters that control disparity smoothness. The larger the values, the smoother the disparity. \\texttt{P1} is the penalty on the disparity change by plus or minus 1 between neighbor pixels. \\texttt{P2} is the penalty on the disparity change by more than 1 between neighbor pixels. The algorithm requires \\texttt{P2 > P1}. See \\texttt{stereo\\_match.cpp} sample where some reasonably good \\texttt{P1} and \\texttt{P2} values are shown (like \\texttt{8*number\\_of\\_image\\_channels*SADWindowSize*SADWindowSize} and \\texttt{32*number\\_of\\_image\\_channels*SADWindowSize*SADWindowSize}, respectively).}\n\\cvarg{disp12MaxDiff}{Maximum allowed difference (in integer pixel units) in the left-right disparity check. Set it to non-positive value to disable the check.}\n\\cvarg{preFilterCap}{Truncation value for the prefiltered image pixels. The algorithm first computes x-derivative at each pixel and clips its value by \\texttt{[-preFilterCap, preFilterCap]} interval. The result values are passed to the Birchfield-Tomasi pixel cost function.}\n\\cvarg{uniquenessRatio}{The margin in percents by which the best (minimum) computed cost function value should \"win\" the second best value to consider the found match correct. Normally, some value within 5-15 range is good enough}\n\\cvarg{speckleWindowSize}{Maximum size of smooth disparity regions to consider them noise speckles and invdalidate. Set it to 0 to disable speckle filtering. Otherwise, set it somewhere in 50-200 range.}\n\\cvarg{speckleRange}{Maximum disparity variation within each connected component. If you do speckle filtering, set it to some positive value, multiple of 16. Normally, 16 or 32 is good enough.}\n\\cvarg{fullDP}{Set it to \\texttt{true} to run full-scale 2-pass dynamic programming algorithm. It will consume O(W*H*numDisparities) bytes, which is large for 640x480 stereo and huge for HD-size pictures. By default this is \\texttt{false}}\n\\end{description}\n\nThe first constructor initializes \\texttt{StereoSGBM} with all the default parameters (so actually one will only have to set \\texttt{StereoSGBM::numberOfDisparities} at minimum). The second constructor allows you to set each parameter to a custom value.\n\n\\cvCppFunc{StereoSGBM::operator ()}\nComputes disparity using SGBM algorithm for a rectified stereo pair\n\n\\cvdefCpp{\nvoid SGBM::operator()(const Mat\\& left, const Mat\\& right, Mat\\& disp);\n}\n\\begin{description}\n\\cvarg{left}{The left image, 8-bit single-channel or 3-channel.}\n\\cvarg{right}{The right image of the same size and the same type as the left one.}\n\\cvarg{disp}{The output disparity map. It will be 16-bit signed single-channel image of the same size as the input images. It will contain scaled by 16 disparity values, so that to get the floating-point disparity map, you will need to divide each \\texttt{disp} element by 16.}\n\\end{description}\n\nThe method executes SGBM algorithm on a rectified stereo pair. See \\texttt{stereo\\_match.cpp} OpenCV sample on how to prepare the images and call the method. Note that the method is not constant, thus you should not use the same \\texttt{StereoSGBM} instance from within different threads simultaneously.\n\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{StereoCalibrate}\n\\else\n\\cvCppFunc{stereoCalibrate}\n\\fi\nCalibrates stereo camera.\n\n\\cvdefC{double cvStereoCalibrate( \\par const CvMat* objectPoints, \\par const CvMat* imagePoints1,\n \\par const CvMat* imagePoints2, \\par const CvMat* pointCounts,\n \\par CvMat* cameraMatrix1, \\par CvMat* distCoeffs1,\n \\par CvMat* cameraMatrix2, \\par CvMat* distCoeffs2,\n \\par CvSize imageSize, \\par CvMat* R, \\par CvMat* T,\n \\par CvMat* E=0, \\par CvMat* F=0,\n \\par CvTermCriteria term\\_crit=cvTermCriteria(\n \\par CV\\_TERMCRIT\\_ITER+CV\\_TERMCRIT\\_EPS,30,1e-6),\n \\par int flags=CV\\_CALIB\\_FIX\\_INTRINSIC );}\n\n\\cvdefPy{StereoCalibrate( objectPoints, imagePoints1, imagePoints2, pointCounts, cameraMatrix1, distCoeffs1, cameraMatrix2, distCoeffs2, imageSize, R, T, E=NULL, F=NULL, term\\_crit=(CV\\_TERMCRIT\\_ITER+CV\\_TERMCRIT\\_EPS,30,1e-6), flags=CV\\_CALIB\\_FIX\\_INTRINSIC)-> None}\n\n\\cvdefCpp{double stereoCalibrate( const vector >\\& objectPoints,\\par\n const vector >\\& imagePoints1,\\par\n const vector >\\& imagePoints2,\\par\n Mat\\& cameraMatrix1, Mat\\& distCoeffs1,\\par\n Mat\\& cameraMatrix2, Mat\\& distCoeffs2,\\par\n Size imageSize, Mat\\& R, Mat\\& T,\\par\n Mat\\& E, Mat\\& F,\\par\n TermCriteria term\\_crit = TermCriteria(TermCriteria::COUNT+\\par\n TermCriteria::EPS, 30, 1e-6),\\par\n int flags=CALIB\\_FIX\\_INTRINSIC );}\n\n\\begin{description}\n\\ifCPy\n \\cvarg{objectPoints}{The joint matrix of object points - calibration pattern features in the model coordinate space. It is floating-point 3xN or Nx3 1-channel, or 1xN or Nx1 3-channel array, where N is the total number of points in all views.}\n \\cvarg{imagePoints1}{The joint matrix of object points projections in the first camera views. It is floating-point 2xN or Nx2 1-channel, or 1xN or Nx1 2-channel array, where N is the total number of points in all views}\n \\cvarg{imagePoints2}{The joint matrix of object points projections in the second camera views. It is floating-point 2xN or Nx2 1-channel, or 1xN or Nx1 2-channel array, where N is the total number of points in all views}\n \\cvarg{pointCounts}{Integer 1xM or Mx1 vector (where M is the number of calibration pattern views) containing the number of points in each particular view. The sum of vector elements must match the size of \\texttt{objectPoints} and \\texttt{imagePoints*} (=N).}\n\\fi\n\\ifCpp\n \\cvarg{objectPoints}{The vector of vectors of points on the calibration pattern in its coordinate system, one vector per view. If the same calibration pattern is shown in each view and it's fully visible then all the vectors will be the same, although it is possible to use partially occluded patterns, or even different patterns in different views - then the vectors will be different. The points are 3D, but since they are in the pattern coordinate system, then if the rig is planar, it may have sense to put the model to the XY coordinate plane, so that Z-coordinate of each input object point is 0}\n \\cvarg{imagePoints1}{The vector of vectors of the object point projections on the calibration pattern views from the 1st camera, one vector per a view. The projections must be in the same order as the corresponding object points.}\n \\cvarg{imagePoints2}{The vector of vectors of the object point projections on the calibration pattern views from the 2nd camera, one vector per a view. The projections must be in the same order as the corresponding object points.}\n\\fi\n \\cvarg{cameraMatrix1}{The input/output first camera matrix: $ \\vecthreethree{f_x^{(j)}}{0}{c_x^{(j)}}{0}{f_y^{(j)}}{c_y^{(j)}}{0}{0}{1}$, $j = 0,\\, 1$. If any of \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS}, \\newline \\texttt{CV\\_CALIB\\_FIX\\_ASPECT\\_RATIO}, \\texttt{CV\\_CALIB\\_FIX\\_INTRINSIC} or \\texttt{CV\\_CALIB\\_FIX\\_FOCAL\\_LENGTH} are specified, some or all of the matrices' components must be initialized; see the flags description}\n \\cvarg{distCoeffs}{The input/output vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. \\cvCpp{On output vector length depends on the flags.}}\n \\cvarg{cameraMatrix2}{The input/output second camera matrix, as cameraMatrix1.}\n \\cvarg{distCoeffs2}{The input/output lens distortion coefficients for the second camera, as \\texttt{distCoeffs1}.}\n\\cvarg{imageSize}{Size of the image, used only to initialize intrinsic camera matrix.} \n\\cvarg{R}{The output rotation matrix between the 1st and the 2nd cameras' coordinate systems.}\n\\cvarg{T}{The output translation vector between the cameras' coordinate systems.}\n\\cvarg{E}{The \\cvCPy{optional} output essential matrix.}\n\\cvarg{F}{The \\cvCPy{optional} output fundamental matrix.}\n\\cvarg{term\\_crit}{The termination criteria for the iterative optimization algorithm.}\n\\cvarg{flags}{Different flags, may be 0 or combination of the following values:\n\\begin{description}\n\\cvarg{CV\\_CALIB\\_FIX\\_INTRINSIC}{If it is set, \\texttt{cameraMatrix?}, as well as \\texttt{distCoeffs?} are fixed, so that only \\texttt{R, T, E} and \\texttt{F} are estimated.}\n\\cvarg{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS}{The flag allows the function to optimize some or all of the intrinsic parameters, depending on the other flags, but the initial values are provided by the user.}\n\\cvarg{CV\\_CALIB\\_FIX\\_PRINCIPAL\\_POINT}{The principal points are fixed during the optimization.}\n\\cvarg{CV\\_CALIB\\_FIX\\_FOCAL\\_LENGTH}{$f^{(j)}_x$ and $f^{(j)}_y$ are fixed.}\n\\cvarg{CV\\_CALIB\\_FIX\\_ASPECT\\_RATIO}{$f^{(j)}_y$ is optimized, but the ratio $f^{(j)}_x/f^{(j)}_y$ is fixed.}\n\\cvarg{CV\\_CALIB\\_SAME\\_FOCAL\\_LENGTH}{Enforces $f^{(0)}_x=f^{(1)}_x$ and $f^{(0)}_y=f^{(1)}_y$} \\cvarg{CV\\_CALIB\\_ZERO\\_TANGENT\\_DIST}{Tangential distortion coefficients for each camera are set to zeros and fixed there.}\n\\cvarg{CV\\_CALIB\\_FIX\\_K1,...,CV\\_CALIB\\_FIX\\_K6}{Do not change the corresponding radial distortion coefficient during the optimization. If \\texttt{CV\\_CALIB\\_USE\\_INTRINSIC\\_GUESS} is set, the coefficient from the supplied \\texttt{distCoeffs} matrix is used, otherwise it is set to 0.}\n\\cvarg{CV\\_CALIB\\_RATIONAL\\_MODEL}{Enable coefficients k4, k5 and k6. To provide the backward compatibility, this extra flag should be explicitly specified to make the calibration function use the rational model and return 8 coefficients. If the flag is not set, the function will compute \\cvCpp{and return} only 5 distortion coefficients.}\n\\end{description}}\n\\end{description}\n\nThe function estimates transformation between the 2 cameras making a stereo pair. If we have a stereo camera, where the relative position and orientation of the 2 cameras is fixed, and if we computed poses of an object relative to the fist camera and to the second camera, (R1, T1) and (R2, T2), respectively (that can be done with \\cvCross{FindExtrinsicCameraParams2}{solvePnP}), obviously, those poses will relate to each other, i.e. given ($R_1$, $T_1$) it should be possible to compute ($R_2$, $T_2$) - we only need to know the position and orientation of the 2nd camera relative to the 1st camera. That's what the described function does. It computes ($R$, $T$) such that:\n\n\\[\nR_2=R*R_1\nT_2=R*T_1 + T,\n\\]\n\nOptionally, it computes the essential matrix E:\n\n\\[\nE=\n\\vecthreethree\n{0}{-T_2}{T_1}\n{T_2}{0}{-T_0}\n{-T_1}{T_0}{0}\n*R\n\\]\n\nwhere $T_i$ are components of the translation vector $T$: $T=[T_0, T_1, T_2]^T$. And also the function can compute the fundamental matrix F:\n\n\\[F = cameraMatrix2^{-T} E cameraMatrix1^{-1}\\]\n\nBesides the stereo-related information, the function can also perform full calibration of each of the 2 cameras. However, because of the high dimensionality of the parameter space and noise in the input data the function can diverge from the correct solution. Thus, if intrinsic parameters can be estimated with high accuracy for each of the cameras individually (e.g. using \\cvCross{CalibrateCamera2}{calibrateCamera}), it is recommended to do so and then pass \\texttt{CV\\_CALIB\\_FIX\\_INTRINSIC} flag to the function along with the computed intrinsic parameters. Otherwise, if all the parameters are estimated at once, it makes sense to restrict some parameters, e.g. pass \\texttt{CV\\_CALIB\\_SAME\\_FOCAL\\_LENGTH} and \\texttt{CV\\_CALIB\\_ZERO\\_TANGENT\\_DIST} flags, which are usually reasonable assumptions.\n\nSimilarly to \\cvCross{CalibrateCamera2}{calibrateCamera}, the function minimizes the total re-projection error for all the points in all the available views from both cameras.\n\\ifPy\n\\else\nThe function returns the final value of the re-projection error.\n\\fi\n\n\\ifCPy\n\\cvCPyFunc{StereoRectify}\n\\else\n\\cvCppFunc{stereoRectify}\n\\fi\nComputes rectification transforms for each head of a calibrated stereo camera.\n\n\\cvdefC{void cvStereoRectify( \\par const CvMat* cameraMatrix1, const CvMat* cameraMatrix2,\n \\par const CvMat* distCoeffs1, const CvMat* distCoeffs2,\n \\par CvSize imageSize, const CvMat* R, const CvMat* T,\n \\par CvMat* R1, CvMat* R2, CvMat* P1, CvMat* P2,\n \\par CvMat* Q=0, int flags=CV\\_CALIB\\_ZERO\\_DISPARITY,\n \\par double alpha=-1, CvSize newImageSize=cvSize(0,0),\n \\par CvRect* roi1=0, CvRect* roi2=0);}\n\\cvdefPy{StereoRectify( cameraMatrix1, cameraMatrix2, distCoeffs1, distCoeffs2, imageSize, R, T, R1, R2, P1, P2, Q=NULL, flags=CV\\_CALIB\\_ZERO\\_DISPARITY, alpha=-1, newImageSize=(0,0))-> (roi1, roi2)}\n\n\\cvdefCpp{void stereoRectify( const Mat\\& cameraMatrix1, const Mat\\& distCoeffs1,\\par\n const Mat\\& cameraMatrix2, const Mat\\& distCoeffs2,\\par\n Size imageSize, const Mat\\& R, const Mat\\& T,\\par\n Mat\\& R1, Mat\\& R2, Mat\\& P1, Mat\\& P2, Mat\\& Q,\\par\n int flags=CALIB\\_ZERO\\_DISPARITY );\\newline\nvoid stereoRectify( const Mat\\& cameraMatrix1, const Mat\\& distCoeffs1,\\par\n const Mat\\& cameraMatrix2, const Mat\\& distCoeffs2,\\par\n Size imageSize, const Mat\\& R, const Mat\\& T,\\par\n Mat\\& R1, Mat\\& R2, Mat\\& P1, Mat\\& P2, Mat\\& Q,\\par\n double alpha, Size newImageSize=Size(),\\par\n Rect* roi1=0, Rect* roi2=0,\\par\n int flags=CALIB\\_ZERO\\_DISPARITY );}\n\\begin{description}\n\\cvarg{cameraMatrix1, cameraMatrix2}{The camera matrices $\\vecthreethree{f_x^{(j)}}{0}{c_x^{(j)}}{0}{f_y^{(j)}}{c_y^{(j)}}{0}{0}{1}$.}\n\\cvarg{distCoeffs1, distCoeffs2}\\cvarg{distCoeffs}{The input vectors of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements each. If the vectors are NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{imageSize}{Size of the image used for stereo calibration.}\n\\cvarg{R}{The rotation matrix between the 1st and the 2nd cameras' coordinate systems.}\n\\cvarg{T}{The translation vector between the cameras' coordinate systems.}\n\\cvarg{R1, R2}{The output $3 \\times 3$ rectification transforms (rotation matrices) for the first and the second cameras, respectively.}\n\\cvarg{P1, P2}{The output $3 \\times 4$ projection matrices in the new (rectified) coordinate systems.}\n\\cvarg{Q}{The output $4 \\times 4$ disparity-to-depth mapping matrix, see \\cvCppCross{reprojectImageTo3D}.}\n\\cvarg{flags}{The operation flags; may be 0 or \\texttt{CV\\_CALIB\\_ZERO\\_DISPARITY}. If the flag is set, the function makes the principal points of each camera have the same pixel coordinates in the rectified views. And if the flag is not set, the function may still shift the images in horizontal or vertical direction (depending on the orientation of epipolar lines) in order to maximize the useful image area.}\n\\cvarg{alpha}{The free scaling parameter. If it is -1\\cvCpp{ or absent}, the functions performs some default scaling. Otherwise the parameter should be between 0 and 1. \\texttt{alpha=0} means that the rectified images will be zoomed and shifted so that only valid pixels are visible (i.e. there will be no black areas after rectification). \\texttt{alpha=1} means that the rectified image will be decimated and shifted so that all the pixels from the original images from the cameras are retained in the rectified images, i.e. no source image pixels are lost. Obviously, any intermediate value yields some intermediate result between those two extreme cases.}\n\\cvarg{newImageSize}{The new image resolution after rectification. The same size should be passed to \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap}, see the \\texttt{stereo\\_calib.cpp} sample in OpenCV samples directory. By default, i.e. when (0,0) is passed, it is set to the original \\texttt{imageSize}. Setting it to larger value can help you to preserve details in the original image, especially when there is big radial distortion.}\n\\cvarg{roi1, roi2}{The optional output rectangles inside the rectified images where all the pixels are valid. If \\texttt{alpha=0}, the ROIs will cover the whole images, otherwise they likely be smaller, see the picture below}\n\\end{description}\n\nThe function computes the rotation matrices for each camera that (virtually) make both camera image planes the same plane. Consequently, that makes all the epipolar lines parallel and thus simplifies the dense stereo correspondence problem. On input the function takes the matrices computed by \\cvCppCross{stereoCalibrate} and on output it gives 2 rotation matrices and also 2 projection matrices in the new coordinates. The 2 cases are distinguished by the function are: \n\n\\begin{enumerate}\n\\item Horizontal stereo, when 1st and 2nd camera views are shifted relative to each other mainly along the x axis (with possible small vertical shift). Then in the rectified images the corresponding epipolar lines in left and right cameras will be horizontal and have the same y-coordinate. P1 and P2 will look as: \n\n\\[\\texttt{P1}=\n\\begin{bmatrix}\nf & 0 & cx_1 & 0\\\\\n0 & f & cy & 0\\\\\n0 & 0 & 1 & 0\n\\end{bmatrix}\n\\]\n\\[\\texttt{P2}=\n\\begin{bmatrix}\nf & 0 & cx_2 & T_x*f\\\\\n0 & f & cy & 0\\\\\n0 & 0 & 1 & 0\n\\end{bmatrix}\n,\n\\]\n\nwhere $T_x$ is horizontal shift between the cameras and $cx_1=cx_2$ if \\texttt{CV\\_CALIB\\_ZERO\\_DISPARITY} is set.\n\\item Vertical stereo, when 1st and 2nd camera views are shifted relative to each other mainly in vertical direction (and probably a bit in the horizontal direction too). Then the epipolar lines in the rectified images will be vertical and have the same x coordinate. P2 and P2 will look as:\n\n\\[\n\\texttt{P1}=\n\\begin{bmatrix}\nf & 0 & cx & 0\\\\\n0 & f & cy_1 & 0\\\\\n0 & 0 & 1 & 0\n\\end{bmatrix}\n\\]\n\\[\n\\texttt{P2}=\n\\begin{bmatrix}\nf & 0 & cx & 0\\\\\n0 & f & cy_2 & T_y*f\\\\\n0 & 0 & 1 & 0\n\\end{bmatrix}\n,\n\\]\n\nwhere $T_y$ is vertical shift between the cameras and $cy_1=cy_2$ if \\texttt{CALIB\\_ZERO\\_DISPARITY} is set.\n\\end{enumerate} \n\nAs you can see, the first 3 columns of \\texttt{P1} and \\texttt{P2} will effectively be the new \"rectified\" camera matrices. \nThe matrices, together with \\texttt{R1} and \\texttt{R2}, can then be passed to \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap} to initialize the rectification map for each camera.\n\nBelow is the screenshot from \\texttt{stereo\\_calib.cpp} sample. Some red horizontal lines, as you can see, pass through the corresponding image regions, i.e. the images are well rectified (which is what most stereo correspondence algorithms rely on). The green rectangles are \\texttt{roi1} and \\texttt{roi2} - indeed, their interior are all valid pixels.\n\n\\includegraphics[width=0.8\\textwidth]{pics/stereo_undistort.jpg}\n\n\\ifCPy\n\\cvCPyFunc{StereoRectifyUncalibrated}\n\\else\n\\cvCppFunc{stereoRectifyUncalibrated}\n\\fi\nComputes rectification transform for uncalibrated stereo camera.\n\n\\cvdefC{void cvStereoRectifyUncalibrated( \\par const CvMat* points1, \\par const CvMat* points2,\n \\par const CvMat* F, \\par CvSize imageSize,\n \\par CvMat* H1, \\par CvMat* H2,\n \\par double threshold=5 );}\n\\cvdefPy{StereoRectifyUncalibrated(points1,points2,F,imageSize,H1,H2,threshold=5)-> None}\n\\cvdefCpp{bool stereoRectifyUncalibrated( const Mat\\& points1,\\par\n const Mat\\& points2,\\par\n const Mat\\& F, Size imgSize,\\par\n Mat\\& H1, Mat\\& H2,\\par\n double threshold=5 );}\n\\begin{description}\n\\cvarg{points1, points2}{The 2 arrays of corresponding 2D points. The same formats as in \\cvCross{FindFundamentalMat}{findFundamentalMat} are supported}\n\\cvarg{F}{The input fundamental matrix. It can be computed from the same set of point pairs using \\cvCross{FindFundamentalMat}{findFundamentalMat}.}\n\\cvarg{imageSize}{Size of the image.}\n\\cvarg{H1, H2}{The output rectification homography matrices for the first and for the second images.}\n\\cvarg{threshold}{The optional threshold used to filter out the outliers. If the parameter is greater than zero, then all the point pairs that do not comply the epipolar geometry well enough (that is, the points for which $|\\texttt{points2[i]}^T*\\texttt{F}*\\texttt{points1[i]}|>\\texttt{threshold}$) are rejected prior to computing the homographies.\nOtherwise all the points are considered inliers.}\n\\end{description}\n\nThe function computes the rectification transformations without knowing intrinsic parameters of the cameras and their relative position in space, hence the suffix \"Uncalibrated\". Another related difference from \\cvCross{StereoRectify}{stereoRectify} is that the function outputs not the rectification transformations in the object (3D) space, but the planar perspective transformations, encoded by the homography matrices \\texttt{H1} and \\texttt{H2}. The function implements the algorithm \\cite{Hartley99}. \n\nNote that while the algorithm does not need to know the intrinsic parameters of the cameras, it heavily depends on the epipolar geometry. Therefore, if the camera lenses have significant distortion, it would better be corrected before computing the fundamental matrix and calling this function. For example, distortion coefficients can be estimated for each head of stereo camera separately by using \\cvCross{CalibrateCamera2}{calibrateCamera} and then the images can be corrected using \\cvCross{Undistort2}{undistort}, or just the point coordinates can be corrected with \\cvCross{UndistortPoints}{undistortPoints}.\n\n\n\\ifCPy\n\\cvCPyFunc{Undistort2}\n\\else\n\\cvCppFunc{undistort}\n\\fi\nTransforms an image to compensate for lens distortion.\n\n\\cvdefC{void cvUndistort2( \\par const CvArr* src,\\par CvArr* dst,\\par const CvMat* cameraMatrix,\n \\par const CvMat* distCoeffs, \\par const CvMat* newCameraMatrix=0 );}\n\\cvdefPy{Undistort2(src,dst,cameraMatrix,distCoeffs)-> None}\n\n\\cvdefCpp{void undistort( const Mat\\& src, Mat\\& dst, const Mat\\& cameraMatrix,\\par\n const Mat\\& distCoeffs, const Mat\\& newCameraMatrix=Mat() );}\n\\begin{description}\n\\cvarg{src}{The input (distorted) image}\n\\cvarg{dst}{The output (corrected) image; will have the same size and the same type as \\texttt{src}}\n\\cvarg{cameraMatrix}{The input camera matrix $A = \\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{1} $}\n\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvCpp{\\cvarg{newCameraMatrix}{Camera matrix of the distorted image. By default it is the same as \\texttt{cameraMatrix}, but you may additionally scale and shift the result by using some different matrix}}\n\\end{description}\n\nThe function transforms the image to compensate radial and tangential lens distortion.\n\nThe function is simply a combination of \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap} (with unity \\texttt{R}) and \\cvCross{Remap}{remap} (with bilinear interpolation). See the former function for details of the transformation being performed.\n\nThose pixels in the destination image, for which there is no correspondent pixels in the source image, are filled with 0's (black color).\n\nThe particular subset of the source image that will be visible in the corrected image can be regulated by \\texttt{newCameraMatrix}. You can use \\cvCross{GetOptimalNewCameraMatrix}{getOptimalNewCameraMatrix} to compute the appropriate \\texttt{newCameraMatrix}, depending on your requirements.\n\nThe camera matrix and the distortion parameters can be determined using\n\\cvCross{CalibrateCamera2}{calibrateCamera}. If the resolution of images is different from the used at the calibration stage, $f_x, f_y, c_x$ and $c_y$ need to be scaled accordingly, while the distortion coefficients remain the same.\n\n\n\\ifCPy\n\\cvCPyFunc{UndistortPoints}\n\\else\n\\cvCppFunc{undistortPoints}\n\\fi\nComputes the ideal point coordinates from the observed point coordinates.\n\n\\cvdefC{void cvUndistortPoints( \\par const CvMat* src, \\par CvMat* dst,\n \\par const CvMat* cameraMatrix,\n \\par const CvMat* distCoeffs,\n \\par const CvMat* R=NULL,\n \\par const CvMat* P=NULL);}\n\\cvdefPy{UndistortPoints(src,dst,cameraMatrix,distCoeffs,R=NULL,P=NULL)-> None}\n\n\\cvdefCpp{void undistortPoints( const Mat\\& src, vector\\& dst,\\par\n const Mat\\& cameraMatrix, const Mat\\& distCoeffs,\\par\n const Mat\\& R=Mat(), const Mat\\& P=Mat());\\newline\nvoid undistortPoints( const Mat\\& src, Mat\\& dst,\\par\n const Mat\\& cameraMatrix, const Mat\\& distCoeffs,\\par\n const Mat\\& R=Mat(), const Mat\\& P=Mat());}\n\n\\begin{description}\n\n\\cvarg{src}{The observed point coordinates, 1xN or Nx1 2-channel (CV\\_32FC2 or CV\\_64FC2).} \n\\cvarg{dst}{The output ideal point coordinates, after undistortion and reverse perspective transformation\\cvCPy{, same format as \\texttt{src}}.}\n\\cvarg{cameraMatrix}{The camera matrix $\\vecthreethree{f_x}{0}{c_x}{0}{f_y}{c_y}{0}{0}{1}$}\n\\cvarg{distCoeffs}\\cvarg{distCoeffs}{The input vector of distortion coefficients $(k_1, k_2, p_1, p_2[, k_3[, k_4, k_5, k_6]])$ of 4, 5 or 8 elements. If the vector is NULL/empty, the zero distortion coefficients are assumed.}\n\\cvarg{R}{The rectification transformation in object space (3x3 matrix). \\texttt{R1} or \\texttt{R2}, computed by \\cvCppCross{StereoRectify} can be passed here. If the matrix is empty, the identity transformation is used}\n\\cvarg{P}{The new camera matrix (3x3) or the new projection matrix (3x4). \\texttt{P1} or \\texttt{P2}, computed by \\cvCppCross{StereoRectify} can be passed here. If the matrix is empty, the identity new camera matrix is used}\n\\end{description}\n\nThe function is similar to \\cvCross{Undistort2}{undistort} and \\cvCross{InitUndistortRectifyMap}{initUndistortRectifyMap}, but it operates on a sparse set of points instead of a raster image. Also the function does some kind of reverse transformation to \\cvCross{ProjectPoints2}{projectPoints} (in the case of 3D object it will not reconstruct its 3D coordinates, of course; but for a planar object it will, up to a translation vector, if the proper \\texttt{R} is specified).\n\n\\begin{lstlisting}\n// (u,v) is the input point, (u', v') is the output point\n// camera_matrix=[fx 0 cx; 0 fy cy; 0 0 1]\n// P=[fx' 0 cx' tx; 0 fy' cy' ty; 0 0 1 tz]\nx\" = (u - cx)/fx\ny\" = (v - cy)/fy\n(x',y') = undistort(x\",y\",dist_coeffs)\n[X,Y,W]T = R*[x' y' 1]T\nx = X/W, y = Y/W\nu' = x*fx' + cx'\nv' = y*fy' + cy',\n\\end{lstlisting}\n\nwhere undistort() is approximate iterative algorithm that estimates the normalized original point coordinates out of the normalized distorted point coordinates (\"normalized\" means that the coordinates do not depend on the camera matrix).\n\nThe function can be used both for a stereo camera head or for monocular camera (when R is \\cvC{NULL}\\cvPy{None}\\cvCpp{empty}).\n", "meta": {"hexsha": "96edb875968f27cea20e696243761f356f6a5ff0", "size": 109140, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "to/lang/OpenCV-2.2.0/doc/calib3d.tex", "max_stars_repo_name": "eirTony/INDI1", "max_stars_repo_head_hexsha": "42642d8c632da53f60f2610b056547137793021b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "to/lang/OpenCV-2.2.0/doc/calib3d.tex", "max_issues_repo_name": "eirTony/INDI1", "max_issues_repo_head_hexsha": "42642d8c632da53f60f2610b056547137793021b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 14, "max_issues_repo_issues_event_min_datetime": "2016-11-24T10:46:39.000Z", "max_issues_repo_issues_event_max_datetime": "2016-12-10T07:24:15.000Z", "max_forks_repo_path": "to/lang/OpenCV-2.2.0/doc/calib3d.tex", "max_forks_repo_name": "eirTony/INDI1", "max_forks_repo_head_hexsha": "42642d8c632da53f60f2610b056547137793021b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.9720670391, "max_line_length": 806, "alphanum_fraction": 0.7446948873, "num_tokens": 30498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891305219504, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.6390874885308283}} {"text": "% -*- root: 00-main.tex -*-\n\n\\renewcommand{\\theequation}{A.\\arabic{equation}}\n\\renewcommand{\\thesubsection}{Appendix \\arabic{subsection}}\n\n\\section*{Appendix}\n\n\\subsection{Simplifying the regularization term}\\label{app:reg_term}\nThe exponentials of the Thikonov regularization prior \\eqref{eq:regseg-thikonov} have the general form\n $\\vec{v}^T \\mathbf{M} \\vec{v}$.\nIf $\\mathbf{M}$ is a $n \\times n$ diagonal matrix such that $\\mathbf{M} = \\vec{m} \\, \\mathbf{I}_n$,\n then:\n\n\\begin{equation*}\n\\vec{v}^T \\mathbf{M} \\vec{v} = \\vec{m} \\cdot (\\vec{v}^T \\mathbf{I}_n \\vec{v}) = \\vec{m} \\cdot \\vec{v}^{\\circ2},\n\\end{equation*}\n where we have introduced the Hadamard power notation\\footnote{The Hadamard power of a matrix or a vector\n is the power of its elements $\\mathbf{M}^{\\circ p} = ({m_{ij}}^{p})$}.\n\nIn general, the anisotropy of the distortion field is aligned with the \n voxel coordinate system, so\n $\\mathbf{A}$ and $\\mathbf{B}$ of \\eqref{eq:regseg-energy} can be simplified to diagonal matrices\n to regularize the registration process, such that\n $\\mathbf{A}= \\, \\boldsymbol{\\alpha}\\,\\vec{I}_n$ and\n $\\mathbf{B}= \\, \\boldsymbol{\\beta}\\,\\vec{I}_n$.\nBy substituting into equation \\eqref{eq:regseg-energy}, we obtain:\n\n \\begin{align}\n E(\\vec{u}) &= \\const + \\, \\underset{l}{\\sum} \\int_{\\Omega_l} \\mdist{f'}{l} \\,d\\vec{r} \\,\n \\ifthenelse{\\boolean{review}}{+}{+ \\notag\\\\ &+}\n \\int_{\\Omega} \\frac12 \\left[ \\boldsymbol{\\alpha} \\cdot \\vec{u}^{\\circ2} + \\boldsymbol{\\beta} \\cdot (\\nabla \\vec{u})^{\\circ2} \\right] \\,d\\vec{r}.\n \\label{eq:regseg-app_energy}\n \\end{align}\n\n\\subsection{Application of the shape-gradients}\\label{app:shape_gradients}\nThe computation of gradients at the locations of the active contours in the\n instant $t$ is based on the work of \\cite{herbulot_segmentation_2006}.\nLet $F(\\vec{r})$ be an ``arbitrary'' function over the image domain\n $\\Omega = \\Omega_l \\cup \\Omega_m$ split in two regions $l$ and\n $m$, and $\\Gamma_{l,m}$ a closed boundary between them.\nWe now derive the domain integral w.r.t. $t$:\n\n \\begin{equation}\n \\frac{\\partial}{\\partial t} \\int_\\Omega F(\\vec{r}) d\\vec{r} =\n \\int_\\Omega \\frac{\\partial}{\\partial t}F(\\vec{r}) d\\vec{r}\n - \\int_{\\Gamma_{l,m}} F(\\vec{r}) \\left\\langle \\frac{\\partial \\Gamma_{l,m} }{\\partial t},\n N_{\\Gamma_{l,m}}\\right\\rangle d\\vec{r},\n \\end{equation}\n%\n where $\\left\\langle\\frac{\\partial\\Gamma_{l,m}}{\\partial t}, N_{\\Gamma_{l,m}}\\right\\rangle$ is\n the projection of the boundary movement on the unit inward normal $N_{\\Gamma_{l,m}}$.\nAssuming that the region descriptors $\\{\\boldsymbol{\\mu}_l, \\boldsymbol{\\Sigma}_l\\}$ vary slowly enough, we can consider\n that $\\frac{\\partial}{\\partial t} F(\\vec{r}) = 0$ and thus:\n\n \\begin{equation}\n \\frac{\\partial}{\\partial t} \\int_\\Omega F(\\vec{r}) d\\vec{r} =\n - \\int_{\\Gamma_{l,m}} F(\\vec{r}) \\left\\langle \\frac{\\partial \\Gamma_{l,m} }{\\partial t},\n N_{\\Gamma_{l,m}}\\right\\rangle d\\vec{r}.\n \\label{eq:regseg-shape_gradients}\n \\end{equation}\n\nThe equation \\eqref{eq:regseg-shape_gradients} is discretized as follows.\nFirst, the surface between limiting regions $l$ and $m$ ($\\Gamma_{l,m}$) is explicitly represented by\n a discrete set of vertices $\\vec{v}_i$, with $i \\in \\{0, \\ldots, N_p -1 \\}$.\nConsequently, the inwards normal of the surface $N_{\\Gamma_{l,m}}$ is represented by the discrete\n set of normals $\\hat{\\vec{n}}_i$ at each vertex of the mesh.\nThe resulting summation is, therefore, discrete and the integral operator is replaced by the sum:\n\n \\begin{align}\n \\frac{\\partial}{\\partial t} \\int_\\Omega F(\\vec{r}) d\\vec{r} &=\n \\underbracket{\\cancel{\\int_\\Omega \\frac{\\partial}{\\partial t}F(\\vec{r}) d\\vec{r} }}_{\\text{Functional's evolution}}\n - \\underbracket{\\int_{\\Gamma_{l,m}} F(\\vec{r}) \\left\\langle \\frac{\\partial \\Gamma_{l,m}}{\\partial t},\n N_{\\Gamma_{l,m}}\\right\\rangle d\\vec{r}}_{\\text{Shape's evolution}} \\notag \\\\\n & = - \\underset{p}{\\sum} \\frac{1}{A_p} \\underset{i}{\\sum} \\, a_i \\, F(\\vec{v}_i) \\left\\langle \\underbracket{\\frac{\\partial \\vec{v}_i}{\\partial t}}_{\\text{speed of }\\vec{v}_i},\n \\hat{\\vec{n}}_{i}\\right\\rangle.\n \\label{eq:regseg-shape_gradient_orig}\n \\end{align}\nwhere $a_i$ is the area corresponding to vertex $\\vec{v}_i$, and $A_p = \\sum_i a_i$ is the total area of the surface $p$.\nIn the following, we will refer as $w_{p,i} = a_i / A_p $ to the area contribution of $\\vec{v}_i$ to the\n total area of the surface it belongs to.\nFor simplicity, the sum over $p$ can be also removed, as the vertices belong to only one of the total $P$ contours.\n\nThen, the speed of $\\vec{v}_i$ is discretized using the artificial time-step parameter $\\delta$, as the displacement\n $\\frac{\\partial \\vec{v}_i}{\\partial t} = \\vec{v}_i(\\delta = t+1) - \\vec{v}_i(\\delta = t)$:\n\n \\begin{equation}\n \\frac{\\partial}{\\partial t} \\int_\\Omega F(\\vec{r}) \\, d\\vec{r} =\n - \\underset{i}{\\sum} w_{p,i} F(\\vec{v}_i) \\frac{\\partial \\vec{v}_i}{\\partial t} \\cdot \\hat{\\vec{n}}_i.\n \\label{eq:regseg-shape_gradient_disc1}\n \\end{equation}\n\nSince the energy functional is defined over competing regions, the displacement of $\\vec{v}_i$ will cause\n an energy exchange between the limiting regions, and therefore $F(\\vec{r})$ must be split in\n two terms, $F_{in}(\\vec{r})$ corresponding to the interior region and $F_{out}(\\vec{r})$ to the exterior:\n\n \\begin{equation}\n \\frac{\\partial}{\\partial t} \\int_\\Omega F(\\vec{r}) \\, d\\vec{r} =\n - \\underset{i}{\\sum} \\, \\frac{\\partial \\vec{v}_i}{\\partial t} \\cdot\n \\underbracket{w_{p,i} \\, \\Big[ F_{out}(\\vec{v}_i) - F_{in}(\\vec{v}_i) \\Big] \\hat{\\vec{n}}_i}_{\\bar{s}_i \\text{ in Figure 1}}.\n \\label{eq:regseg-shape_gradient_disc2}\n \\end{equation}\n\nThen, we identify the shape gradient contribution $\\vec{g}_k$ on the coefficients $\\vec{u}_k$ of the B-spline grid \n\\revcomment[R\\#1-C4]{to obtain the definition of $\\vec{g}_k$ given in \\eqref{eq:regseg-final_gradient}}:\n\n \\begin{equation}\n \\label{eq:regseg-gradient_wshape}\n \\begin{split}\n \\vec{g}_k &= - \\underset{i}{\\sum} \\left\\langle \\frac{\\partial \\vec{v}_i'}{\\partial \\vec{u}_k}, \\bar{s}_i'\\right\\rangle \\\\\n \\text{with }\n \\bar{s}_i' &= w_i \\left[ \\mdist{f_i'}{out} - \\mdist{f_i'}{in} \\right] \\, \\hat{\\vec{n}}_i, \\\\\n \\text{and }\n \\frac{\\partial \\vec{v}_i'}{\\partial \\vec{u}_k} &=\n \\frac{\\partial}{\\partial \\vec{u}_k} \\left\\{ \\vec{v}_i + \\sum_k \\psi_k(\\vec{v}_i) \\vec{u}_k \\right\\} = \\psi_k(\\vec{v}_i)\\, \\hat{\\vec{e}},\n \\end{split}\n \\end{equation}%\n\\revcomment[R\\#1-C4]{where $\\psi_k$ and $\\vec{u}_k$ define our B-spline deformation model (\\autoref{eq:regseg-nodes_tfm})} and\n $\\hat{\\vec{e}}$ is the coordinates system's unit vector.\n\n\n", "meta": {"hexsha": "e1eaf3d012332a0099a9629083dd2a45e3f90a6f", "size": 6589, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2015-NeuroImage/08-appendix-00-shapepriors.tex", "max_stars_repo_name": "oesteban/RegSeg-NeuroImage2016", "max_stars_repo_head_hexsha": "434aba23a032a373b287fe72939cbfe4a6caedca", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2015-NeuroImage/08-appendix-00-shapepriors.tex", "max_issues_repo_name": "oesteban/RegSeg-NeuroImage2016", "max_issues_repo_head_hexsha": "434aba23a032a373b287fe72939cbfe4a6caedca", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2015-NeuroImage/08-appendix-00-shapepriors.tex", "max_forks_repo_name": "oesteban/RegSeg-NeuroImage2016", "max_forks_repo_head_hexsha": "434aba23a032a373b287fe72939cbfe4a6caedca", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 54.4545454545, "max_line_length": 177, "alphanum_fraction": 0.6709667628, "num_tokens": 2356, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916240341031, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6390603445092725}} {"text": "\\section{Features}\n\\label{sec:Features}\nEach sample is split into windowed and overlapping blocks. Different features are calculated on each block. The blocks have a length of 20~ms, 50~\\% overlap and are Hann windowed.\\\\\nFeatures used are the RMS, peak, crest factor, spectral centroid, logarithmic spectral centroid, spectral variance, spectral skewness, spectral flatness, spectral brightness and mean spectral absolute slope. They are described in detail as follows.\\\\\nFor every feature, the block is given as $x$, $x[k]$ is a frame within the block, $K$ is the number of frames within one block. The FFT of the block is $X$, $X[n]$ is an FFT bin, $2 N$ is the FFT-Length. The features are calculated as follows:\n\\begin{description}\n \\item[RMS:]\n The RMS is a measure for the energy of the block.\n \\[\n \\mathrm{RMS} = \\sqrt{\\frac{1}{K}\\sum_{k=0}^{K-1}{x[k]^2}}\n \\]\n \\item[Peak:]\n The peak is simply the maximum of all absolute values within the block.\n \\[\n \\mathrm{PK} = \\max{|x[k]|}\n \\]\n \\item[Crest Factor:]\n The ratio of the peak to the RMS value is the crest factor.\n \\[\n \\mathrm{CF} = \\frac{\\mathrm{PK}}{\\mathrm{RMS}}\n \\]\n \\item[Spectral Centroid:]\n The spectral centroid is the normalized frequency weighted mean of the absolute spectrum. The frequency is normalized to the sampling frequency.\n \\[\n \\mathrm{SC} = \\sum_{n=0}^{N}{|X[n]| \\cdot \\frac{n}{2N}}\n \\]\n\n \\item[Logarithmic Spectral Centroid:]\n To account for human frequency perception, the spectral centroid is calculated relative to a logarithmic frequency axis as well.\n \\[\n \\mathrm{LSC} = \\sum_{n=0}^{N}{|X[n]| \\cdot \\log{\\frac{n}{2 N}+1}}\n \\]\n \\item[Spectral Variance:]\n The spectral variance gives a value of the change of the absolute spectrum over all frequencies.\n \\[\n \\mathrm{SV} = \\sum_{n=0}^{\\frac{K}{2}}{(|X[n]|-\\overline{|X[n]|})^2}\n \\]\n with the spectral mean\n \\[\n \\overline{|X[n]|} = \\frac{1}{N+1}\\sum_{n=0}^{N}{|X[n]|}.\n \\]\n \\item[Spectral Skewness:]\n \t\tThe skewness of the absolute spectrum shows how much its distribution leans towards high or low frequencies.\n \\[\n \\mathrm{SS} = \\sum_{n=0}^{\\frac{K}{2}}{(|X[n]|-\\overline{|X[n]|})^3}.\n \\]\n \\item[Spectral Flatness:]\n Given the spectral mean $\\overline{|X[n]|}$ and the geometric spectral mean\n \\[\n \\overline{|X[n]|}_\\mathrm{geom} = \\sqrt[N+1]{\\prod_{n=0}^{N}{|X[n]|}},\n \\]\n the spectral flatness is calculated as the ratio of them.\n \\[\n \t\t\\frac{\\overline{|X[n]|}_\\mathrm{geom}}{\\overline{|X[n]|}}\n \\]\n \\item[Spectral Brightness:]\n The spectral brightness is the ratio of high to the low frequency energy. To get the energy of low and high frequency, the weighted sum of the absolute spectrum is taken. The weighting of the high frequency bins is given as\n \\begin{eqnarray*}\n w_\\mathrm{high} &=& 0.5 - \\frac{\\cos(\\omega)}{2}\\\\\n w_\\mathrm{low} &=& \\frac{\\cos(\\omega)}{2} + 0.5\n \\end{eqnarray*}\n with $\\omega$ as $N$ logarithmic ordered values in the range $[0;~\\pi]$. The point of intersection between the low and high frequency weighting was set to approximately 2~kHz.\n \\[\n \\mathrm{SB} = \\frac{\\sum_{n=0}^{\\frac{K}{2}}{|X[n]|\\cdot w_\\mathrm{high}}}{\\sum_{n=0}^{\\frac{K}{2}}{|X[n]|\\cdot w_\\mathrm{low} }}\n \\]\n \\item[Mean Absolute Spectral Slope:]\n The mean of the absolute spectral slope gives the amount of change in the spectrum over the frequency. It is calculated by\n \\[\n \\mathrm{MASS} = \\frac{1}{N}\\sum_{n=1}^{N}{||X[n]|-|X[n-1]||}\n \\]\n\\end{description}\nThe features of all blocks of all samples are saved within a Pandas DataFrame\\footnote{http://pandas.pydata.org/}. This makes them easily searchable and saveable.", "meta": {"hexsha": "724c6fe5392389fdd683e6ee8a4daffd6fe81ece", "size": 4004, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/Features.tex", "max_stars_repo_name": "bastibe/MusicTagger", "max_stars_repo_head_hexsha": "ed4d6a642f6d624325b3d4fb4bdc022671e24777", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-12-02T19:16:29.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-02T19:16:29.000Z", "max_issues_repo_path": "Report/Features.tex", "max_issues_repo_name": "bastibe/MusicTagger", "max_issues_repo_head_hexsha": "ed4d6a642f6d624325b3d4fb4bdc022671e24777", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/Features.tex", "max_forks_repo_name": "bastibe/MusicTagger", "max_forks_repo_head_hexsha": "ed4d6a642f6d624325b3d4fb4bdc022671e24777", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.6111111111, "max_line_length": 250, "alphanum_fraction": 0.6136363636, "num_tokens": 1130, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.863391617003942, "lm_q2_score": 0.7401743620390162, "lm_q1q2_score": 0.6390603393057274}} {"text": "%-------------------------------------------------------------------------------\n\\section{Working Example}\n%-------------------------------------------------------------------------------\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{0.48\\columnwidth}\n \\includegraphics[width=\\linewidth]{figs/sosp-working_ex3}\n \\caption{\\label{fig:working_ex1}Proximal cost function at $\\bar{x} = 7$.}\n \\end{subfigure}\n \\hspace{0.1cm}\n \\begin{subfigure}[b]{0.48\\columnwidth}\n \\includegraphics[width=\\linewidth]{figs/sosp-working_ex2}\n \\caption{\\label{fig:working_ex2}Evaluated Proximal Gradient for $\\bar{x} = 7$.}\n \\end{subfigure}\n \\vspace{-15pt}\n \\caption{\\label{fig:working_ex}Example of Proximal Gradient Evaluation.}\n \\vspace{-15pt}\n\\end{figure}\n\nTo give a better intuition for how proximal gradients work in practice, we provide a working example. Consider the function $f(x) = x \\texttt{\\&} 4$ shown in figure \\ref{fig:ex_funcs}, and suppose we are evaluating it with $\\bar{x} = 7$. This type of bitwise operation is common in programs that parse a file for input, since they frequently need to check if flag bits are set in a header or section header of the file. However, discrete functions like bitwise operations are nondifferentiable so we instead evaluate the proximal gradient.\n\nTo evaluate the proximal gradient on $f\\left(x\\right)$, the sampling region must first be defined. \\tc{x \\& 4} has a Lipschitz Constant of $K=4$ (any bitwise operation with a constant has a Lipschitz Constant with the value of the most significant bit in the constant), so assuming a proximal scaling factor of $\\lambda = 1$, the sampling distance bound is $2K = 8$. Therefore, a total of 16 samples must be taken, for $x = [-1, 15]$, and the cost function of the proximal operator, $f(x) + ||x - \\bar{x}||_2^2$ evaluated for each possible $x$. Figure \\ref{fig:working_ex1} shows the value of the cost function near $x = 7$, there is a clear minimum at $x=8$, where $f\\left(8\\right) + ||8-7||_2^2 = 1$, so the proximal operator evaluates to $x=8$. The proximal gradient is then $prox_{\\nabla f}\\left(7\\right) = \\frac{f\\left(7\\right) - f\\left(8\\right)}{7 - 8} = -4$.\n\n\n", "meta": {"hexsha": "c2d3a4bed242e89643e7fb1b58439980af53215e", "size": 2172, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "submissions/sosp2019/sections/working_example.tex", "max_stars_repo_name": "sillywalk/grazz", "max_stars_repo_head_hexsha": "a0adb1a90d41ff9006d8c1476546263f728b3c83", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "submissions/sosp2019/sections/working_example.tex", "max_issues_repo_name": "sillywalk/grazz", "max_issues_repo_head_hexsha": "a0adb1a90d41ff9006d8c1476546263f728b3c83", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "submissions/sosp2019/sections/working_example.tex", "max_forks_repo_name": "sillywalk/grazz", "max_forks_repo_head_hexsha": "a0adb1a90d41ff9006d8c1476546263f728b3c83", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 83.5384615385, "max_line_length": 865, "alphanum_fraction": 0.6731123389, "num_tokens": 607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.8031738010682209, "lm_q1q2_score": 0.6390517405784424}} {"text": "\\section{Euclidean Structure}\n\n\\subsection{Exercise 1}\nApplying Cauchy-Schwarz gives that $(x, y) \\leq \\norm{x}\\norm{y} = \\norm{x}$, which yields\nthe desired result.\n\n\\subsection{Exercise 2}\nBoth of these just follow from the fact that a linear space with Euclidean structure is\nisomorphic to $R^k$, which has the desired properties. The following feel like cop-out\nsolutions, but I feel they're fair given this isn't supposed to be a real analysis text.\n\n(i) Let $x_k = \\sum_i a_i^{(k)} x^{(i)}$ and $x_j = \\sum_i a_i^{(j)} x^{(i)}$. Then we have that\n$\\norm{x_k - x_j} \\to 0 \\implies \\abs{a_i^{(k)} - a_i^{(j)}} \\to 0$. Since $\\mathbb{R}$ is\ncomplete, $a_i^{(n)} \\to a_i$ for some $a_i \\in \\mathbb{R}$. Thus $x_k \\to x = \\sum a_i x^{(i)}$.\n\n(ii) Same logic as (i): the individual $a_i^{(n)}$ have convergent subsequences.\n\n\\subsection{Exercise 3}\nWe also need to assume $X$ is finite dimensional (I think).\n\n(i) Since $X$ is finite dimensional, we have that $x = \\sum a_i e^{(i)}$ for some basis $e^{(i)}$. Let\n$v$ be the vector whose components are $v_i = \\norm{Ae^{(i)}}$. Then\n\\begin{align*}\n \\norm{Ax} &= \\norm{\\sum a_i Ae^{(i)}} \\\\\n &= \\norm{(x, v)} \\\\\n &\\leq \\norm{x} \\norm{v}\n\\end{align*}\nSince $\\norm{v}$ is a constant, $\\norm{Ax}$ is bounded on the unit sphere.\n\n(ii) We have that $\\norm{A} = \\max_x \\frac{\\norm{Ax}}{\\norm{x}} = \\max_x \\norm{Ax}$.\n The result then follows from $(Ax, y) \\leq \\norm{Ax}$.\n\n(iii) Let $v$ be as in (i). Then $\\norm{Ax_i - Ax_j} \\leq \\norm{A} \\norm{x_i - x_j}$. Since\n$\\norm{A} \\leq \\norm{v}$ from (i), we are done (we can make $ \\norm{x_i - x_j}$ as small\nas we'd like).\n\n\\subsection{Exercise 4}\nFollows immediately from $(Ax, y) = (x, A^{*}y)$ and Exercise 3 (ii).\n\n\\subsection{Exercise 5}\nLet $x = y_1 + y_1^{\\perp}$ and $z = y_2 + y_2^{\\perp}$. Then\n\\begin{align*}\n (P_Y x, z) &= (y_1, y_2 + y_2^{\\perp}) \\\\\n &= (y_1, y_2) + 0 \\\\\n &= (y_1, y_2) + (y_1^{\\perp}, y_2) \\\\\n &= (x, P_Y z)\n\\end{align*}\n\n\\subsection{Exercise 6}\nReflection across $x_3 = 0$ sends $(x_1, x_2, x_3) \\to (x_1, x_2, -x_3)$. Hence it can be represented as\n\\begin{align*}\n A =\n \\begin{pmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & -1\n \\end{pmatrix}\n\\end{align*}\nWhich has determinant -1 (product of diagonal terms).\n\n\\subsection{Exercise 7}\n(a) If $A$ is orthogonal, then $A^{*}A = I$. Since $A^{*} = A^{\\top}$ in this case, we immediately get\nthat the column vectors of $A$ are pairwise orthogonal unit vectors. Similarly, a matrix $A$ with\npairwise orthogonal unit vectors satisfies $A^{*} A = I$, implying that it is orthogonal.\n\n(b) $A^{*} A = I \\implies A^{*} = A^{-1} \\implies A A^{*} = I$ so $A$ orthogonal implies $A^{*}$ orthogonal.\nThe result then follows from plugging $A^{*}$ into (a).\n\n\\subsection{Exercise 8}\nThe proof is almost identical to the non-complex case.\n\\begin{align*}\n (x + ty, x + ty) &= (x, x) + (ty, x) + (x, ty) + \\norm{t}^2 (y, y) \\\\\n &= (x, x) + t(y, x) + \\bar{t}(x, y) + \\norm{t}^2 (y, y)\n\\end{align*}\nPlugging in $t = \\frac{(x, y)}{(y, y)}$ and using the fact that $(x + ty, x + ty) \\geq 0$ for all\ncomplex $t$ gives the desired inequality.\n\n\\subsection{Exercise 9}\n\\subsubsection{Theorem 4}\nThe proof of Theorem 4 is the same, except we use $y = \\sum \\bar{b_k} x^{(k)}$ instead.\n\n\\subsubsection{Theorem 5}\nNo changes need to be made to the proof of Theorem 5.\n\n\\subsubsection{Theorem 6}\nAgain, no changes need to be made.\n\n\\subsubsection{Theorem 7}\nNo changes need to be made here either, since $(y, y^{\\perp}) = 0 \\implies \\overline{(y, y^{\\perp})} = 0$.\n\n\\subsection{Exercise 10}\n\\subsubsection{Theorem 8}\nParts (i)-(iii) remain the same. For part (iv), we see that \n\\begin{align*}\n (Ax, y) = (x, A^{*}y) = \\overline{(A^{*}y, x)} = \\overline{(y, A^{**}x)} = (A^{**}x, y)\n\\end{align*}\n\n\\subsubsection{Theorem 9}\n(i) $\\norm{kA} = \\max_x \\frac{\\sqrt{(kAx, kAx)}}{\\norm{x}} = \\sqrt{k\\bar{k}} \\norm{A} = \\norm{k} \\norm{A}$.\n\n(ii) Applying triangle inequality to the definition of norm gives the result.\n\n(iii) Comes immediately from $\\norm{A(Bx)} \\leq \\norm{A} \\norm{Bx}$.\n\n\\subsection{Exercise 11}\nFollows from $\\norm{M(x) - M(y)} = (\\bar{x} - \\bar{y}) (x - y) = \\norm{x - y}$.\n\n\\subsection{Exercise 12}\nSame idea as proof of Theorem 10.\n\n\\subsection{Exercise 13}\nAs in Exercise 7, $M^{*} M = I \\implies M^{*} = M^{-1} \\implies M M^{*} = I$. Similarly,\n$M^{*} = M^{-1} \\implies (M^{*})^{-1} M^{-1} = I$.\n\n\\subsection{Exercise 14}\nAssociativity follows from associativity of composition. From Exercise 13, if $M$ is unitary, so is $M^{-1}$.\nFinally, $I$ is also unitary, so the unitary maps are a group with unit $I$.\n\n\\subsection{Exercise 15}\nAgain, same idea as in Theorem 10: $\\det M^{*} \\det M = 1$.\n\n\\subsection{Exercise 16}\n\\begin{align*}\n (Mf, Mg) = \\int_{-1}^{1} m^2(s) f(s) \\bar{g}(s) = \\int_{-1}^{1} f(s) \\bar{g}(s)\n\\end{align*}\n\n", "meta": {"hexsha": "97abaf00b66e8c2024b26ab8c723f8f50bd7f033", "size": 4992, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Linear_Algebra_Lax/chapter_7.tex", "max_stars_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_stars_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-09-19T07:33:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-19T07:33:25.000Z", "max_issues_repo_path": "Linear_Algebra_Lax/chapter_7.tex", "max_issues_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_issues_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Linear_Algebra_Lax/chapter_7.tex", "max_forks_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_forks_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.3070866142, "max_line_length": 109, "alphanum_fraction": 0.5971554487, "num_tokens": 1825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569016, "lm_q2_score": 0.7956581024858786, "lm_q1q2_score": 0.6390517387757118}} {"text": "\n\n\\subsection{Looking from a \\emph{modular} perspective}\n\nGiven a Riordan array $\\mathcal{M}$, we are interested to look at it\ndoing a transformation of the matrix $\\lbrace m_{nk}\\rbrace_{n,k\\in\\mathbb{N}}$,\nbuilding a new one defined by $\\lbrace m_{nk}\\mod\\,p\\rbrace_{n,k\\in\\mathbb{N}}$,\nwhere $p\\in\\mathbb{N}$ is a module, usually a \\emph{prime} number.\n\nThis transformation allows us to represent any array $\\mathcal{M}$ pictorially,\nassociating a colour to each \\emph{remainder class} of $\\equiv_{p}$ relation:\ninspecting the produced coloured triangle it is possible to recognize some\npatterns in the colouring and we will see that for same classes of Riordan\narrays, these colourings are pretty interesting, while for other classes they\nseems to be quite random.\n\nWe will attempt to formally prove what can be see in the pictures, in\nparticular we will discuss the Pascal array $\\mathcal{P}$ and the Catalan array\n$\\mathcal{C}$. We believe that this kind of study is important in order to\nunderstand and to make conjectures about the combinatorial structure underlying\nthose arrays and, why not, any array $\\mathcal{M}$ in general.\n\n\\input{Chapters/back-to-the-basics/main-idea-modular-characterization/assembling-include-figures}\n\nIn \\autoref{fig:main:idea:modular:characterization} is shown this kind of study,\nand it will be addressed in \\autoref{ch:modular:characterization}.\n\n\\subsection{Looking for characterizations, again}\n\nAltough pretty interesting and fairly general characterizations exists for\na coefficient $m_{nk}$ in a Riordan array $\\mathcal{M}$, we would like to\nrework some of them, looking from different points of view.\n\nIn particular we provide a new characterization, called $h$-\\emph{characterization},\nwhich allows us to rewrite a Riordan array $\\mathcal{M}=(d(t),h(t))$ only\nusing function $h$, namely $\\mathcal{M}_{h(t)}(\\alpha(h(t)),h(t))$, for some\nfunction $\\alpha$ in the ``variable'' $h(t)$. If the given array $\\mathcal{M}$\nis in the \\emph{Renewal} subgroup, then the $h$-characterization shows the $A$-sequence's\n\\ac{gf} in the first component of array $\\mathcal{M}_{h(t)}$, automagically.\n\nMoreover, we provide another derivation of a concept similar to $A$-matrix characterization,\nfound by reasoning from a \\emph{column}'s \\ac{gf} point of view, instead of\nfocusing on the combination for a generic element $m_{n+1,k+1}$. This study will be\naddressed in \\autoref{ch:h:characterization}.\n\n\\subsection{Python implementation}\n\nIn order to support the present work, we have coded a bunch of classes in the\n\\emph{Python} programming language, that implement a subset of \\emph{Riordan\ngroup} theory. We rest on \\emph{Sage} \\cite{sage} mathematical framework in\norder to do hard math computation and our implementation aims to be truly\nobject oriented and very minimal, in order to be extended easily. \n\nNonetheless, it is possible to play with Riordan arrays, inverting them, doing\nraw matrix expansions and building a modular representation producing\n\\LaTeX\\,ready code. All pictures in this document are generated by compilation\nof \\emph{TikZ} statements, produced by our implementation.\n\n", "meta": {"hexsha": "c9d6f613959e1be4e3f0ec861cff40b69f838dc2", "size": 3130, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "classicthesis/Chapters/introduction/targets.tex", "max_stars_repo_name": "massimo-nocentini/master-thesis", "max_stars_repo_head_hexsha": "0d82bfcc82c92512d0795f286256a19f39b9b1f9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "classicthesis/Chapters/introduction/targets.tex", "max_issues_repo_name": "massimo-nocentini/master-thesis", "max_issues_repo_head_hexsha": "0d82bfcc82c92512d0795f286256a19f39b9b1f9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "classicthesis/Chapters/introduction/targets.tex", "max_forks_repo_name": "massimo-nocentini/master-thesis", "max_forks_repo_head_hexsha": "0d82bfcc82c92512d0795f286256a19f39b9b1f9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.0508474576, "max_line_length": 97, "alphanum_fraction": 0.7734824281, "num_tokens": 816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.803173791645582, "lm_q2_score": 0.7956581024858786, "lm_q1q2_score": 0.6390517350271122}} {"text": "\\chapter{Game Theory}\n\\label{chap:game-theory}\n\nAs a warm-up, we include the following result.\n\n\\begin{lemma}\n\tFor every game $G$, there exists at least one individually rational strategy profile.\n\tIn other words, $\\mathcal{S}_1(G) \\ne \\emptyset$.\n\\end{lemma}\n\n\\begin{proof}\n\tFor every player $i \\in P$ we denote $m_i \\in S_i$ the strategy that lower-bounds utility of player $i$:\n\t\\[\n\t\tm_i = \\text{arg} \\max_{t_i \\in S_i} \\min_{\\vect_{-i} \\in S_{-i}} u_i(t_i, \\vect_{-i}).\n\t\\]\n\tWe claim that the strategy profile $\\mm = (m_1, \\dots, m_n) \\in S$ is individually rational.\n\tIf $\\mm$ was not individually rational, there would have to be some player $j \\in P$ and a strategy $s_j^* \\in S_j$ that would ensure a better worst-case payoff: $\\min_{\\vecs_{-j} \\in S_{-j}}u_j(s_j^*, \\vecs_{-j}) > u_j(\\mm)$ which would contradict the choice of~$m_j$.\n\\end{proof}\n\nThe main goal of this work is to create a gaming platform of human vs. computer, which trains the human to play towards a PTE.\nThis \\enquote{training} should be conducted by the human playing many games and trying to figure out how to maximize their payoffs.\nThe primary means of affecting this process is through choosing an appropriate strategy chosen by the computer in each play.\nHow precisely this response should be chosen is not clear and needs to be determined by experiments.\n\nThis chapter introduces three new equilibria which might be useful for the cause described in the previous paragraph: the perfectly transparent best response equilibrium and the perfectly transparent optimal profile equilibrium.\nThen, we discuss some relations between these two equilibria and between the other equilibria introduced in \\autoref{chap:background}.\n\nWe start by defining the most natural way to translate the preemption process for determining a PTE to a notion of best response.\n\n\\begin{definition}[Perfectly transparent best response]\n\tIn a game (possibly with ties) $G = (P, S, \\uu)$ we say a strategy $s_i^* \\in S_i$ is a perfectly transparent best response to a strategy profile $\\vecs_{-i} \\in S_{-i}$ of the opponent players if it is the Nashian best response to $\\vecs_{-i}$ across all profiles that survive the maximum possible number of preemption rounds.\n\tThe weak/strict notion is inherited from the Nashian best response.\n\tFormally $s_i^*$ is a (\\textit{weak}) perfectly transparent best response if either $(s_i^*, \\vecs_{-i})$ is a PTE or $\\exists k \\in \\N: (s_i^*, \\vecs_{-i}) \\in \\mathcal{S}_k(G), (s_i, \\vecs_{-i}) \\notin \\mathcal{S}_{k+1}(G) \\forall s_i \\in S_i$, and\n\t\\[\n\t\t\\forall (s_i', \\vecs_{-i}) \\in \\mathcal{S}_k(G): u_i(s_i^*, \\vecs_{-i}) \\ge u_i(s_i', \\vecs_{-i}).\n\t\\]\n\tIt is a \\textit{strict} perfectly transparent best response if either $(s_i^*, \\vecs_{-i})$ is a PTE or $\\exists k \\in \\N: (s_i^*, \\vecs_{-i}) \\in \\mathcal{S}_k(G), (s_i, \\vecs_{-i}) \\notin \\mathcal{S}_{k+1}(G) \\forall s_i \\in S_i$, and\n\t\\[\n\t\t\\forall (s_i', \\vecs_{-i}) \\in \\mathcal{S}_k(G): s_i' \\ne s_i^* \\implies u_i(s_i^*, \\vecs_{-i}) > u_i(s_i', \\vecs_{-i}).\n\t\\]\n\\end{definition}\n\nA best response naturally induces an equilibrium.\n\n\\begin{definition}[Perfectly transparent best response equilibrium]\n\tWe say a strategy profile $\\vecs = (s_1, \\dots, s_n) \\in S$ is a (\\textit{weak}) perfectly transparent best response equilibrium (PTBRE) if $s_i$ is a (weak) perfectly transparent best response to $\\vecs_{-i}$ for all players $i \\in P$.\n\t\n\tSimilarly, $\\vecs$ is a \\textit{strict} PTBRE if $s_i$ is a strict perfectly transparent best response to $\\vecs_{-i}$ for all players $i \\in P$.\n\\end{definition}\n\nIt is easy to see that for any game, if a strategy profile is a PTE, then it is also a PTBRE.\nOur initial hope was that the other direction also holds, so PTE and PTBRE are equivalent.\nUnfortunately, this is not true, as we can see in the following sections.\n\nNext, we can consider the following situation: we are playing a game of 2 players, and we know that our opponent's strategy will always be a perfectly transparent best response to our strategy.\nIn this case, what strategy should we choose to maximize our payoff?\nThis is captured in the following definition.\n\n\\begin{definition}[Perfectly transparent $i$-best profile]\n\tFor a game $G = (P, S, \\uu)$ of two players with $P = \\{i, j\\}$\n\twe say a strategy profile $\\vecs = (s_1, s_2) \\in S$ is a (\\textit{weak}) perfectly transparent $i$-best profile if\n\t$j$'s strategy $s_j$ is a (weak) perfectly transparent best response to $s_i$, and\n\t\\[\n\t\t\\forall s_i' \\in S_i: u_i(\\vecs) \\ge u_i\\left(s_i', b(s_i')\\right),\n\t\\]\n\twhere $b(s_i')$ is some weak perfectly transparent best response of player $j$ to $s_i'$ (note that there may be multiple best responses to $s_i'$ but they must all have the same utility for player $i$).\n\n\tIt is a \\textit{strict} perfectly transparent $i$-best profile if $s_j$ is a strict perfectly transparent best response to $s_i$ and\n\t\\[\n\t\t\\forall s_i' \\in S_i: s_i' \\ne s_i \\implies u_i(\\vecs) > u_i\\left(s_i', b(s_i')\\right).\n\t\\]\n\tNote that we still consider $b(s_i')$ to be a weak best response even in the strict definition.\n\tThis is to ensure that every strict best profile is also a weak best profile.\n\\end{definition}\n\nAgain, this notion naturally induces an equilibrium.\n\n\\begin{definition}[Perfectly transparent best profile equilibrium]\n\tFor a game of two players, we say that a strategy profile $\\vecs \\in S$ is a (\\textit{weak}) perfectly transparent best profile equilibrium (PTBPE) if $\\vecs$ is a (weak) perfectly transparent $i$-best profile for both players $i \\in P$.\n\n\tSimilarly, $\\vecs$ is a \\textit{strict} PTBPE if it is a strict perfectly transparent $i$-best profile for both $i \\in P$.\n\\end{definition}\n\nAnother situation that we consider is the following: we are playing a game of two players, and the strategy that we are going to choose is a perfectly transparent best response to the opponent's strategy.\nWhat is the best payoff that we can get---that is, what payoff would we get if the opponent knew we would choose a perfectly transparent best response and were optimizing for \\textit{our} payoff instead of theirs?\nThis is captured by the following definition.\n\n\\begin{definition}[Perfectly transparent $i$-optimal profile]\n\tFor a given game $G = (P, S, \\uu)$ and player $i \\in P$ we say a strategy profile $\\vecs = (s_1, \\dots, s_n) \\in S$ is a (\\textit{weak}) perfectly transparent $i$-optimal profile if\n\t$s_i$ is a (weak) perfectly transparent best response to $\\vecs_{-i}$ and\n\t\\[\n\t\t\\forall \\vecs'_{-i} \\in S_{-i}: u_i(\\vecs) \\ge u_i\\left(b(\\vecs'_{-i}), \\vecs'_{-i}\\right),\n\t\\]\n\twhere $b(\\vecs'_{-i})$ is some weak perfectly transparent best response to $\\vecs'_{-i}$ (note that there may be multiple best responses to $\\vecs'_{-i}$ but they must all have the same utility for player $i$).\n\n\tIt is a \\textit{strict} perfectly transparent $i$-optimal profile if $s_i$ is a strict perfectly transparent best response to $\\vecs_{-i}$ and\n\t\\[\n\t\t\\forall \\vecs_{-i}' \\in S_{-i}: \\vecs_{-i}' \\ne s_{-i} \\implies u_i(\\vecs) > u_i\\left(b(\\vecs_{-i}'), \\vecs_{-i}'\\right).\n\t\\]\n\tNote that even in the strict definition, we still consider $b(\\vecs_{-1})$ to be a weak best response.\n\tThis is to ensure that every strict optimal profile is also a weak optimal profile.\n\\end{definition}\n\nThis definition also induces an equilibrium.\n\n\\begin{definition}[Perfectly transparent optimal profile equilibrium]\n\tWe say that a strategy profile $\\vecs \\in S$ is a (\\textit{weak}) perfectly transparent optimal profile equilibrium (PTOPE) if $\\vecs$ is a (weak) perfectly transparent $i$-optimal profile for all $i \\in P$.\n\n\tSimilarly, $\\vecs$ is a \\textit{strict} PTOPE if it is a strict perfectly transparent $i$-optimal profile for all $i \\in P$.\n\\end{definition}\n\nWhile the other equilibria model selfish behavior, PTOPE models altruistic behavior.\nIt can be seen as the non-Nashian counterpart to the Berge equilibrium \\cite{berge1957theorie}.\n\nRecall that for any game $G = (P, S, \\uu)$, a strategy profile $\\vecs \\in S$ is individually rational if and only if it survives the first preemption round: $\\vecs \\in \\mathcal{S}_1(G)$.\nThis means that every PTE must be individually rational.\nIf we think of these equilibria as sets of strategy profiles in any possible game, we can express this as PTE $\\subset$ IR.\nIt is not hard to see that the inclusion is indeed strict as there are games in which $\\mathcal{S}_2 \\subsetneq \\mathcal{S}_1$, so not all individually rational profiles are PTEs.\n\nIn the rest of this chapter, we present several such inclusion results that help to understand the relations between individual rationality, perfectly transparent best response equilibria, perfectly transparent equilibria, perfectly transparent optimal profile equilibria, and minimax rationalizability.\n\nThese results were obtained by first computing equilibria for all games in large datasets \\cite{bigDataset}, and analyzing the results using Apache Spark.\nMost of the counterexamples found in the Appendix come from these datasets.\nMoreover, not finding counterexamples for certain inclusions initiated many hypotheses that we proved at a later point.`'\nAn overview of all inclusions that we prove or disprove can be found in \\autoref{sec:inclusions-overview} of the Appendix.\n\n\\section{Games with Ties}\n\\label{sec:games-with-ties}\n\nWe start by presenting results about the most general setup: games with ties.\nAll the proofs presented in this section naturally hold also for the more specific cases (symmetric games, games without ties).\nThe results are summarized by a Venn diagram in \\autoref{fig:venn-with-ties}.\n\nWe consider the most interesting result of this section to be that both PTBPE and PTOPE are subsets of PTE.\nThis means that both selfishness and altruism can achieve a perfectly transparent equilibrium under the non-Nashian assumptions.\n\n\\begin{figure}[b]\n\t\\centering\n\t\\begin{tikzpicture}[line width=0.25pt]\n\t\t\\draw (0,0) ellipse (1 and 0.5);\n\t\t\\draw (-0.2,0.2) ellipse (1.5 and 1.1);\n\t\t\\draw (1.75,0) ellipse (1 and 0.5);\n\t\t\\draw (2,0.2) ellipse (1.5 and 1.1);\n\t\t\\draw (0.9,0.3) ellipse (2 and 2.5);\n\t\t\\draw (0.3,0.4) ellipse (3.3 and 3.3);\n\t\t\\draw (1.7,-0.2) ellipse (3.5 and 2.5);\n\t\t\\draw[rounded corners=2ex] (-3.2,4) rectangle (5.5,-3.2);\n\t\t\\draw (4.1,-0.3) ellipse (3.7 and 3);\n\t\t\\node at (-0.25,0) {sPTOPE};\n\t\t\\node at (-0.3,0.8) {PTOPE};\n\t\t\\node at (1.95,0) {sPTBPE};\n\t\t\\node at (1.85,0.8) {PTBPE};\n\t\t\\node at (0.4,1.6) {sPTBRE};\n\t\t\\node at (-1.5,2.0) {PTBRE};\n\t\t\\node at (4.3,0) {PTE};\n\t\t\\node at (-2.5,3) {IR};\n\t\t\\node at (6.5,0) {MR};\n\t\\end{tikzpicture}\n\t\\caption{A Venn diagram depicting the inclusion of different equilibria in games with ties.}\n\t\\label{fig:venn-with-ties}\n\\end{figure}\n\n\\begin{observation}\n\t\\label{th:strict-sub-weak}\n\tWe can see from our definition of PTOPE, PTBPE, and PTBRE that any strategy profile satisfying the strict definition of each equilibrium also satisfies the weak definition of the respective equilibrium.\n\tThus, we have sPTOPE $\\subset$ PTOPE, sPTBPE $\\subset$ PTBPE, and sPTBRE $\\subset$ PTBRE.\n\\end{observation}\n\n\\begin{observation}\n\t\\label{th:ptope-subset-ptbre}\n\tFor any game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTOPE, then it is a PTBRE.\n\tSimilarly, if $\\vecs$ is a strict PTOPE, then it is a strict PTBRE.\n\tThus, we have PTOPE $\\subset$ PTBRE, and sPTOPE $\\subset$ sPTBRE.\n\\end{observation}\n\n\\begin{proof}\n\tIf $\\vecs$ is a PTOPE, it is by definition a perfectly transparent $i$-optimal profile for every player $i \\in P$.\n\tIf $\\vecs$ is a perfectly transparent $i$-optimal profile, then, again by definition, $s_i$ is a perfectly transparent best response to $\\vecs_{-i}$.\n\tThis being true for all $i \\in P$ is precisely the definition of a PTBRE.\n\tThe same argument holds for strict equilibria.\n\\end{proof}\n\n\\begin{lemma}\n\t\\label{th:ptope-subset-pte}\n\tIf a game $G = (P, S, \\uu)$ has a PTOPE $\\vecs \\in S$, then $\\vecs$ is also a PTE, i.e., we have the inclusion PTOPE $\\subset$ PTE.\n\\end{lemma}\n\n\\begin{proof}\n\tFor the sake of deriving a contradiction, suppose that there is a game $G = (P, S, \\uu)$ with a strategy profile $\\vecs = (s_1, \\dots, s_n) \\in S$ such that $\\vecs$ is a PTOPE but not a PTE and let $k$ be the last preemption round in which $\\vecs$ is not eliminated (i.e., $\\vecs \\in \\mathcal{S}_k(G)$ but $\\vecs \\notin \\mathcal{S}_{k+1}(G)$).\n\tBecause $\\vecs$ is eliminated in the $(k+1)$\\textsuperscript{st} round, there must be some player $i \\in P$ with a strategy $s_i^* \\ne s_i$ which causes the elimination:\n\t\\[\n\t\t\\forall \\vecs_{-i}' \\in S_{-i}: (s_i^*, \\vecs_{-i}') \\in \\mathcal{S}_k(G) \\implies u_i(s_i^*, \\vecs_{-i}') > u_i(\\vecs).\n\t\\]\n\tLet $\\vecs_{-i}^* \\in S_{-i}$ be the opponents' profile that minimizes $u_i(s_i^*, \\vecs_{-i}^*)$, subject to $(s_i^*, \\vecs_{-i}^*)\\in \\mathcal{S}_k(G)$ (there must be at least one such profile to cause $\\vecs \\notin \\mathcal{S}_{k+1}(G)$).\n\tLet $b \\in S_i$ be the perfectly transparent best response to $\\vecs_{-i}^*$.\n\tBy the choice of $\\vecs_{-i}^*$, we must have $u_i(b, \\vecs_{-i}^*) \\ge u_i(s_i^*, \\vecs_{-i}^*) > u_i(\\vecs)$.\n\tThis means that $\\vecs$ is not a perfectly transparent $i$-optimal profile, contradicting $\\vecs$ being a PTOPE.\n\\end{proof}\n\n\\begin{lemma}\n\t\\label{th:ptbpe-subset-pte}\n\tIf a game $G = (P, S, \\uu)$ of two players has a PTBPE $\\vecs \\in S$, then $\\vecs$ is a PTE; we have PTBPE $\\subset$ PTE.\n\\end{lemma}\n\n\\begin{proof}\n\tThe proof is very similar to the proof of \\autoref{th:ptope-subset-pte}.\n\tWe suppose for the sake of contradiction that there is a game $G = (P, S, \\uu)$ with $P = \\{1, 2\\}$, and a strategy profile $\\vecs = (s_1, s_2) \\in S$ such that $\\vecs$ is a PTBPE but not a PTE and let $k$ be the last preemption round in which $\\vecs$ is not eliminated (i.e., $\\vecs \\in \\mathcal{S}_k(G)$ but $\\vecs \\notin \\mathcal{S}_{k+1}(G)$).\n\tBecause $\\vecs$ is eliminated in the $(k+1)$\\textsuperscript{st} round, there must be some player $i \\in P$ with a strategy $s_i^* \\ne s_i$ which causes the elimination:\n\t\\[\n\t\t\\forall s_j' \\in S_j: (s_i^*, s_j') \\in \\mathcal{S}_k(G) \\implies u_i(s_i^*, s_j') > u_i(\\vecs),\n\t\\]\n\twhere $j = 3 - i$ is the opponent of $i$ and there is at least one $s_j' \\in S_j$ which satisfies $(s_i^*, s_j') \\in \\mathcal{S}_k(G)$.\n\tLet $b \\in S_j$ be the perfectly transparent best response of player $j$ to $s_i^*$.\n\tWe must have $(s_i^*, b) \\in \\mathcal{S}_k$ and so $u_i(s_i^*, b) > u_i(\\vecs)$.\n\tThis means that $\\vecs$ is not a perfectly transparent $i$-best profile, contradicting $\\vecs$ being a PTBPE.\n\\end{proof}\n\n\\begin{lemma}\n\t\\label{th:ptbpe-subset-ptbre}\n\tIf a game $G = (P, S, \\uu)$ of two players has a PTBPE $\\vecs \\in S$, then $\\vecs$ is a PTBRE.\n\tSimilarly, if $\\vecs$ is a strict PTBPE, then it is a strict PTBRE.\n\tWe have PTBPE $\\subset$ PTBRE, and sPTBPE $\\subset$ sPTBRE.\n\\end{lemma}\n\n\\begin{proof}\n\tSuppose that a game $G = (P, S, \\uu)$ with $P = \\{i, j\\}$ has a PTBPE $\\vecs = (s_i, s_j) \\in S$.\n\tBy definition, this means that $\\vecs$ is a perfectly transparent $i$-best profile, hence $s_j$ is a perfectly transparent best response to $s_i$.\n\tMoreover, $\\vecs$ is a perfectly transparent $j$-best profile, hence $s_i$ is a perfectly transparent best response to $s_j$.\n\tThis implies that $\\vecs$ is a PTBRE.\n\tThe same argument holds for strict equilibria.\n\\end{proof}\n\n\\begin{observation}\n\t\\label{th:pte-subset-ir}\n\tIn any game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTE, then it is individually rational.\n\tWe have PTE $\\subset$ IR.\n\\end{observation}\n\n\\begin{proof}\n\tBy definition, a profile is IR if it survives the first preemption round, i.e., $\\vecs \\in \\mathcal{S}_1$.\n\tIf $\\vecs$ is a PTE, it survives all preemption rounds.\n\tThus, it must satisfy $\\vecs \\in \\mathcal{S}_1$.\n\\end{proof}\n\n\\begin{lemma}\n\t\\label{th:ptbre-subset-ir}\n\tFor any game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTBRE, then it is individually rational.\n\tThus, we have PTBRE $\\subset$ IR.\n\\end{lemma}\n\n\\begin{proof}\n\tLet us suppose that there is a game $G = (P, S, \\uu)$ and a strategy profile $\\vecs = (s_1, \\dots, s_n) \\in S$ such that $\\vecs$ is a PTBRE but not individually rational.\n\tThis is equivalent to $s_i$ being a perfectly transparent best response to $\\vecs_{-i}$ for all $i \\in P$, and $\\vecs \\notin \\mathcal{S}_1(G)$.\n\tNote that we must also have $(s_i', \\vecs_{-i}) \\notin \\mathcal{S}_1(G)\\ \\forall s_i' \\in S_i,\\ \\forall i \\in P$ because otherwise $s_i$ would not be a perfectly transparent best response to $\\vecs_{-i}$ for some player $i$.\n\tThis means the perfectly transparent best response of every player $i \\in P$ to $\\vecs_{-i}$ is simply a Nashian best response (since all considered strategy profiles are eliminated in the same preemption round).\n\tThus, $\\vecs$ is a Nash equilibrium, which is always individually rational.\n\\end{proof}\n\n\\begin{lemma}\n\t\\label{th:ptbpe-subset-mr}\n\tIn any game $G = (P, S, \\uu)$ of two players with $P = \\{i, j\\}$, if a strategy profile $\\vecs \\in S$ is a PTBPE, then it is minimax rationalizable.\n\tWe have PTBPE $\\subset$ MR.\n\\end{lemma}\n\n\\begin{proof}\n\tSuppose for the sake of contradiction that there is a strategy profile $\\vecs \\in S$, which is a PTBPE but not MR.\n\tThis means there is some $k \\in \\N$ such that $\\vecs \\in \\mathcal{R}_k$ and $\\vecs \\notin \\mathcal{R}_{k+1}$.\n\tSuppose WLOG that player $i$ caused the elimination of $\\vecs$ in the $k+1$\\textsuperscript{st} round.\n\tThis means, player $i$ has some strategy $s_i' \\in S_i'$ such that\n\t\\[\n\t\t\\forall (s_i', s_j') \\in \\mathcal{R}_k: u_i(s_i', s_j') > u_i(\\vecs),\n\t\\]\n\tand there is at least one $s_j' \\in S_j$ such that $(s_i', s_j') \\in \\mathcal{R}_k$.\n\tNow, we can observe that $(s_i', s_j) \\in \\mathcal{R}_k$; otherwise either the whole row or column corresponding to $(s_i', s_j)$ would have to be eliminated in some previous round but we know that $\\vecs \\in \\mathcal{R}_k$ and $(s_i', s_j') \\in \\mathcal{R}_k$ for some $s_j' \\in S_j$.\n\tSince $u_i(s_i', s_j) > u_i(\\vecs)$, it cannot be that $s_i$ is a perfectly transparent best response to $s_j$. Hence, $\\vecs$ cannot be a perfectly transparent $j$-best profile, a contradiction.\n\\end{proof}\n\nAll the described inclusions are strict.\nIn \\autoref{tab:overview-nosym-dup} an overview of all inclusions can be found.\nFor each pair of equilibria $A, B$, it refers to either a proof or a counterexample of the claim \\enquote{$A \\subseteq B$} (note that $A \\subseteq B \\land B \\not\\subseteq A$ implies strict inclusion: $A \\subset B$).\n\nMinimax rationalizability intersects with all the other equilibria, as shown in \\autoref{tab:PTOPE-eq-minimax}.\n\n\n\\section{Symmetric Games with Ties}\nEvery proof from the previous section holds even for symmetric games with ties.\nHowever, some counterexamples that we gave are non-symmetric.\nIn particular, for the following claims we do not have any counterexamples that would prove them:\n\\begin{enumerate}\n\t\\item sPTOPE $\\not\\subseteq$ MR. For non-symmetric games proved by \\autoref{tab:PTOPE-ne-minimax}.\n\t\\item PTOPE $\\not\\subseteq$ MR. For non-symmetric games holds transitively because sPTOPE~$\\subset$~PTOPE (\\autoref{th:strict-sub-weak}), and sPTOPE~$\\not\\subseteq$~MR (point 1).\n\t\\item PTE $\\not\\subseteq$ MR. For non-symmetric games holds because PTOPE $\\subset$ PTE (\\autoref{th:ptope-subset-pte}), and PTOPE $\\not\\subset$ MR (point 2).\n\\end{enumerate}\nSince we have not found any counterexamples in our datasets, we believe the opposite statements to be true.\n\n\\begin{conjecture}\n\tFor any game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTOPE, then it is minimax rationalizable.\n\tWe have PTOPE $\\subset$ MR.\n\\end{conjecture}\n\n\\begin{conjecture}\n\tFor any game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTE, then it is minimax rationalizable.\n\tWe have PTE $\\subset$ MR.\n\\end{conjecture}\n\nDue to limited time constraints, we have not been able to prove these conjectures.\n\nAll other inclusions are the same as in \\autoref{sec:games-with-ties}.\nAn overview of all inclusion proofs and counterexamples can be found in \\autoref{tab:overview-sym-dup}.\n\n\\section{Games without Ties}\nWe can observe that in games without ties, similar to Nash equilibria, weak and strict equilibria are also equivalent in the case of PTBRE and PTOPE.\nThus, in this section, we do not distinguish between them.\nMoreover, all the \\enquote{is subset of} relations are inherited from \\autoref{sec:games-with-ties}.\nThe only difference is that PTE is not intersecting with PTBRE anymore (the example in \\autoref{tab:ties-pte-not-sub-ptbre} indeed has ties).\nAll inclusions for games without ties are depicted by a Venn diagram in \\autoref{fig:venn-general-without-ties}.\n\n\\begin{observation}\n\t\\label{th:pte-subset-ptbre}\n\tIf a strategy profile is a PTE, then it must be a PTBRE because there is only one strategy profile to choose from in the last preemption round.\n\tWe have PTE $\\subset$ PTBRE.\n\\end{observation}\n\n\\begin{figure}[t]\n\t\\centering\n\t\\begin{tikzpicture}[line width=0.25pt]\n\t\t\\draw (-0.9,0) ellipse (1.1 and 1.1);\n\t\t\\draw (0.9,0) ellipse (1.1 and 1.1);\n\t\t\\draw (0,0) circle (2.2);\n\t\t\\draw (0,0) circle (3.3);\n\t\t\\draw[rounded corners=2ex] (-3.5,3.5) rectangle (3.5,-3.5);\n\t\t\\draw (2.8,0) ellipse (3 and 2.8);\n\t\t\\node at (-1,0) {PTOPE};\n\t\t\\node at (1,0) {PTBPE};\n\t\t\\node at (-0.5,1.45) {PTE};\n\t\t\\node at (-1.3,2.45) {PTBRE};\n\t\t\\node at (-2.5,3.0) {IR};\n\t\t\\node at (4.3,0) {MR};\n\t\\end{tikzpicture}\n\t\\caption{A Venn diagram depicting the inclusion of different equilibria in games without ties.}\n\t\\label{fig:venn-general-without-ties}\n\\end{figure}\n\nAn overview of all inclusion proofs and counterexamples can be found in \\autoref{tab:overview-nosym-nodup}.\n\n\\section{Symmetric Games without Ties}\nIn this section, we restrict ourselves to symmetric games without ties.\nAll the \\enquote{subset} relations from the previous section still hold since symmetric games are a special case of general games.\nMoreover, none of the equilibria collapses (i.e., they are still strict subsets, even when restricted to symmetric games); this because all the provided counterexamples are symmetric games.\nThe only new relation that emerges is between minimax rationalizability and~PTE.\nAll the inclusions are depicted in \\autoref{fig:venn-sym-without-ties}.\nFurthermore, we present some other interesting observations about symmetric games.\n\n\\begin{figure}[t]\n\t\\centering\n\t\\begin{tikzpicture}[line width=0.25pt]\n\t\t\\draw (-0.9,0) ellipse (1.1 and 1.1);\n\t\t\\draw (0.9,0) ellipse (1.1 and 1.1);\n\t\t\\draw (0,0) circle (2.2);\n\t\t\\draw (0,0) circle (3.3);\n\t\t\\draw[rounded corners=2ex] (-3.5,3.5) rectangle (3.5,-3.5);\n\t\t\\draw (2.8,0) ellipse (5 and 3.2);\n\t\t\\node at (-1,0) {PTOPE};\n\t\t\\node at (1,0) {PTBPE};\n\t\t\\node at (-0,1.35) {PTE};\n\t\t\\node at (-1.3,2.45) {PTBRE};\n\t\t\\node at (-2.5,3.0) {IR};\n\t\t\\node at (5,0) {MR};\n\t\\end{tikzpicture}\n\t\\caption{A Venn diagram depicting the inclusion of equilibria in symmetric games without ties.}\n\t\\label{fig:venn-sym-without-ties}\n\\end{figure}\n\n\\begin{observation}\n\t In a symmetric game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTE, then it lies on the diagonal, i.e., we have $\\vecs = (a, a, \\dots, a)$ for some strategy $a$.\n\\end{observation}\n\n\\begin{proof}\n\tIf there was a PTE somewhere else than on the diagonal, we would necessarily have multiple PTEs by symmetry, contradicting the uniqueness of PTE.\n\\end{proof}\n\n\\begin{corollary}\n\tIf a symmetric game $G = (P, S, \\uu)$ has a PTE $\\vecs \\in S$ and $\\vecs$ is the perfectly transparent $i$-optimal profile for some player $i \\in P$, then $\\vecs$ is a PTOPE. \n\\end{corollary}\n\n\\begin{proof}\n\tSince $\\vecs$ is on the diagonal and it is the perfectly transparent $i$-optimal profile for some $i \\in P$, by symmetry, it must actually be the perfectly transparent $j$-optimal profile for all $j \\in P$, which is the definition of PTOPE.\n\\end{proof}\n\n\\begin{lemma}[Fourny \\cite{Fourny20}]\n\t\\label{th:pte-sym-subset-mr}\n\tIn a symmetric game $G = (P, S, \\uu)$, if a strategy profile $\\vecs \\in S$ is a PTE, then it is minimax rationalizable.\n\tThus, we have PTE\\textsuperscript{sym} $\\subset$ MR\\textsuperscript{sym}.\n\\end{lemma}\n\n\\begin{proof}\n\tSuppose, for the sake of deriving a contradiction, that there is a game $G = (P, S, \\uu)$ and a strategy profile $\\vecs \\in S$ that is a PTE but not minimax rationalizable.\n\tBy definition, this means that there is some player $i \\in P$ such that $(s_i, \\vecs_{-i}')$ is not minimax rationalizable for any $\\vecs' \\in S$.\n\tBy symmetry, if this statement holds for one player, then it must hold for all players.\n\tLet $s_j$ be the strategy that minimax-dominates $s_i$ (i.e. causes that $s_i$ is not minimax rationalizable).\n\tWe denote $\\vecs^* = (s_j, s_j, \\dots, s_j)$.\n\tIn the round where strategy $s_i$ is eliminated, strategy profile $\\vecs^*$ must survive---otherwise, strategy $s_j$ would be eliminated as well, but we supposed that $s_j$ is the strategy that caused the elimination of $s_i$.\n\tMoreover, we must have $u_i(\\vecs^*) > u_i(\\vecs)\\ \\forall i \\in P$ because $s_j$ minimax-dominates $s_i$.\n\tThis means $\\vecs$ is not Pareto-optimal which contradicts $\\vecs$ being a PTE.\n\\end{proof}\n\nAn overview of all inclusion proofs and counterexamples can be found in \\autoref{tab:overview-sym-nodup}.\n", "meta": {"hexsha": "b1337f5f4bed1c7584fb113881362f5da54e945a", "size": 25188, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/chapters/game-theory.tex", "max_stars_repo_name": "KuceraMartin/perfect-prediction-game", "max_stars_repo_head_hexsha": "61296cdae8fd33f902d73a819d39535c7945501a", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/chapters/game-theory.tex", "max_issues_repo_name": "KuceraMartin/perfect-prediction-game", "max_issues_repo_head_hexsha": "61296cdae8fd33f902d73a819d39535c7945501a", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/chapters/game-theory.tex", "max_forks_repo_name": "KuceraMartin/perfect-prediction-game", "max_forks_repo_head_hexsha": "61296cdae8fd33f902d73a819d39535c7945501a", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.7352941176, "max_line_length": 348, "alphanum_fraction": 0.7134746705, "num_tokens": 7907, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.80563219364797, "lm_q2_score": 0.7931059560743422, "lm_q1q2_score": 0.6389516911874429}} {"text": "\\documentclass[12pt]{article}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{color}\n\\usepackage{geometry}\n \\geometry{\n a4paper,\n total={210mm,297mm},\n left=20mm,\n right=20mm,\n top=20mm,\n bottom=20mm,\n}\n\\usepackage[parfill]{parskip}\n\\usepackage[colorlinks=true]{hyperref}\n\\hypersetup{urlcolor=blue, citecolor=red}\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{definition}{Definition}\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{rem}[theorem]{Remark}\n\\newtheorem{pro}[theorem]{Proposition}\n\\newtheorem{con}[theorem]{Conjecture}\n\n\\renewcommand{\\vec}[1]{\\ensuremath{\\overrightarrow{#1}}}\n\n\\begin{document}\n\\begin{center}\n{\\huge \\bf Idealized 2 links} \\\\\n\\end{center}\n\\vspace{1ex}\n\\begin{flushright}\nFerenc A. Bartha\\\\\nfab2@rice.edu\n\\end{flushright}\n\n\\subsection*{Generalized Coordinates}\nWe use the following coordinates and notation:\n\\begin{itemize}\n \\item The middle mass point is given in polar coordinates w.r. to the origin and the fixed $x$-, $y$-axes.\n \\begin{itemize}\n \\item the connection to the origin (fixed) is a weightless rod,\n \\item the weight of the mass point is $m_1$,\n \\item radius$\\, = d_1$, constant,\n \\item angle from $x$-axis to the direction $\\vec{e_{r_1}} = \\theta_1$.\n \\end{itemize}\n \\item The upper mass point is given in polar coordinates w.r. to middle point\n \\begin{itemize}\n \\item the connection to the middle point is a weightless rod,\n \\item the weight of the mass point is $m_2$,\n \\item radius$\\, = d_2$, constant,\n \\item angle from the direction $\\vec{e_{r_1}}$ to the direction $\\vec{e_{r_2}} = \\theta_2$.\n \\end{itemize}\n \\item Consequently, we describe our system by $\\theta_1$ and $\\theta_2$ and their derivatives.\n\\end{itemize}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[scale=0.8]{01_Idealized_2links.png}\n \\caption{The coordinates}\n\\end{figure}\n\nNote that none of these coordinate systems (describing the middle or the upper point) is an inertial coordinate system. Both are rotating (thus accelerating) compared to the one fixed in the origin with axis $x$ and $y$.\n\n\\newpage\n\n\\subsection*{The Lagrangian}\nWe assume that no external forces act on the system, so we have conservation of energy, that is in our case conservation of kinetic energy.\nThe Lagrangian is\n$$\\mathcal{L}(\\theta_1, \\dot \\theta_1, \\theta_2, \\dot \\theta_2) = \\frac{1}{2} m_1 |\\vec{v_1}|^2 + \\frac{1}{2} m_2 |\\vec{v_2}|^2.$$\n\\vspace{1ex}\n\\noindent\nWe have\n\\begin{itemize}\\itemsep 0.7ex\n \\item conservation of energy: $\\mathcal{L} = $ constant,\n \\item the position vector of the middle mass from the origin is $d_1 \\vec{e_{r_1}}$,\n \\item the velocity vector is the derivative $\\vec{v_1} = \\dot{d_1} \\vec{e_{r_1}} + d_1 \\dot{\\vec{e_{r_1}}} = d_1 \\dot{\\theta_1} \\vec{e_{\\theta_1}}$\\\\\n as $\\dot d_1 = 0$ and $\\dot{\\vec{e_{r_1}}} = \\dot{\\theta_1} \\vec{e_{\\theta_1}}$ (see {\\bf Appendix}),\n \\item thus, the speed$^2$ is $|\\vec{v_1}|^2 = (d_1 \\dot{\\theta_1})^2$,\n \\item the position vector of the upper mass from the origin is $d_1 \\vec{e_{r_1}} + d_2 \\vec{e_{r_2}}$,\n \\item the velocity vector is $\\vec{v_2} = \\vec{v_1} + \\dot{d_2} \\vec{e_{r_2}} + d_2 \\dot{\\vec{e_{r_2}}} = d_1 \\dot{\\theta_1} \\vec{e_{\\theta_1}} + d_2 (\\dot{\\theta_1} + \\dot{\\theta_2}) \\vec{e_{\\theta_2}}$\\\\\n as $\\dot d_2 = 0$ and $\\dot{\\vec{e_{r_2}}} = (\\dot{\\theta_1} + \\dot{\\theta_2}) \\vec{e_{\\theta_2}}$ (see {\\bf Appendix}),\n \\item thus, the speed$^2$ is $|\\vec{v_2}|^2 = (d_1 \\dot{\\theta_1})^2 + (d_2 (\\dot{\\theta_1} + \\dot{\\theta_2}))^2 + 2 d_1 d_2 \\dot{\\theta_1} (\\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{cos}(\\mathrm{angle}(\\vec{e_{\\theta_1}}, \\vec{e_{\\theta_2}}))$\\\\\n that is $|\\vec{v_2}|^2 = (d_1 \\dot{\\theta_1})^2 + (d_2 (\\dot{\\theta_1} + \\dot{\\theta_2}))^2 + 2 d_1 d_2 \\dot{\\theta_1} (\\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{cos}\\theta_2$.\n\\end{itemize}\n\n\\vspace{1ex}\nThus, \n\\begin{equation*}\n\\begin{split}\n\\mathcal{L}(\\theta_1, \\dot \\theta_1, \\theta_2, \\dot \\theta_2) = &\\,\\frac{1}{2} (m_1 + m_2) d_1^2 \\dot{\\theta_1}^2 \n + \\frac{1}{2} m_2 d_2^2 (\\dot{\\theta_1} + \\dot{\\theta_2})^2\n + m_2 d_1 d_2 \\dot{\\theta_1} (\\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{cos}\\theta_2.\n\\end{split}\n\\end{equation*}\n\n\\subsection*{Equations of Motion}\nThe equations of motion are derived by taking\\\\ {\\it(Lagrange equations of the $2nd$ kind a.k.a Euler--Lagrange equations)}\n$$d/dt [ \\partial\\mathcal{L}/\\partial \\dot q_j ] = \\partial\\mathcal{L}/\\partial q_j$$\nfor all generalized coordinates $q_j \\in \\{\\theta_1, \\theta_2 \\}$.\n\nThis gives us\n$$\\frac{d}{dt}\\left[(m_1 + m_2) d_1^2 \\dot{\\theta_1} \n + m_2 d_2^2 (\\dot{\\theta_1} + \\dot{\\theta_2})\n + m_2 d_1 d_2 (2 \\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{cos}\\theta_2\\right] = 0 \\quad \\text{ for } q_j = \\theta_1$$\nand\n\\begin{equation*}\n \\frac{d}{dt}\\left[m_2 d_2^2 (\\dot{\\theta_1} + \\dot{\\theta_2})\n + m_2 d_1 d_2 \\dot{\\theta_1} \\mathrm{cos}\\theta_2 \\right] = \n - m_2 d_1 d_2 \\dot{\\theta_1} (\\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{sin}(\\theta_2) \\quad \\text{ for } q_j = \\theta_2.\n\\end{equation*}\n\nWriting in the time derivatives we get\n\\begin{equation*}\n\\begin{split}\n (m_1 + m_2) d_1^2 \\ddot{\\theta_1} \n + m_2 d_2^2 (\\ddot{\\theta_1} + \\ddot{\\theta_2})\n + m_2 d_1 d_2 [(2 \\ddot{\\theta_1} + \\ddot{\\theta_2}) \\mathrm{cos}\\theta_2 - (2 \\dot{\\theta_1} + \\dot{\\theta_2})\\dot{\\theta_2} \\mathrm{sin}\\theta_2] = 0\n\\end{split}\n\\end{equation*}\nthat results in \n\\begin{equation}\\label{eq-theta1}\n\\begin{split}\n \\left( m_2 d_2^2 + m_2 d_1 d_2 \\mathrm{cos}\\theta_2 \\right) \\ddot{\\theta_2} = \n &- \\left( (m_1 + m_2) d_1^2 + m_2 d_2^2 + 2 m_2 d_1 d_2 \\mathrm{cos}\\theta_2 \\right) \\ddot{\\theta_1}\\\\ \n &+ m_2 d_1 d_2 (2 \\dot{\\theta_1} + \\dot{\\theta_2}) \\dot{\\theta_2} \\mathrm{sin}\\theta_2 \n \\qquad \\qquad \\qquad \\text{ for } q_j = \\theta_1\n\\end{split}\n\\end{equation}\nand\n\\begin{equation*}\n m_2 d_2^2 (\\ddot{\\theta_1} + \\ddot{\\theta_2})\n + m_2 d_1 d_2 [\\ddot{\\theta_1} \\mathrm{cos}\\theta_2 - \\dot{\\theta_1} \\dot{\\theta_2} \\mathrm{sin}\\theta_2] =\n- m_2 d_1 d_2 \\dot{\\theta_1} (\\dot{\\theta_1} + \\dot{\\theta_2}) \\mathrm{sin}(\\theta_2)\n\\end{equation*}\nthat is transformed to\n$$m_2 d_2^2 \\ddot{\\theta_2} =\n- m_2 d_2^2 \\ddot{\\theta_1} - m_2 d_1 d_2 \\ddot{\\theta_1} \\mathrm{cos}\\theta_2 \n- m_2 d_1 d_2 \\dot{\\theta_1}^2 \\mathrm{sin}(\\theta_2)\n$$\nand simplified to \n\\begin{equation}\\label{eq-theta2}\n \\ddot{\\theta_2} = - \\left( \\left[ 1 + \\frac{d_1}{d_2} \\mathrm{cos}\\theta_2 \\right] \\ddot{\\theta_1} + \n \\frac{d_1}{d_2} \\dot{\\theta_1}^2 \\mathrm{sin}(\\theta_2) \\right)\n \\qquad \\qquad \\qquad \\qquad \\text{ for } q_j = \\theta_2\n\\end{equation}\n\n\\subsection*{Further Transformation of the Equations of Motion}\nWe want to solve the equations \\eqref{eq-theta1} and \\eqref{eq-theta2} for \n$\\ddot{\\theta_1}$ and $\\ddot{\\theta_1}$ \n(some tools can model them as they readily are, but we need now a more standard form).\n\nSubstituting \\eqref{eq-theta2} into \\eqref{eq-theta1} yields\n\\begin{equation*}\n\\begin{split}\n - \\left( m_2 d_2^2 + m_2 d_1 d_2 \\mathrm{cos}\\theta_2 \\right) &\\left( \\left[ 1 + \\frac{d_1}{d_2} \\mathrm{cos}\\theta_2 \\right] \\ddot{\\theta_1} + \\frac{d_1}{d_2} \\dot{\\theta_1}^2 \\mathrm{sin}(\\theta_2) \\right) = \\\\\n &- \\left( (m_1 + m_2) d_1^2 + m_2 d_2^2 + 2 m_2 d_1 d_2 \\mathrm{cos}\\theta_2 \\right) \\ddot{\\theta_1}\\\\ \n &+ m_2 d_1 d_2 (2 \\dot{\\theta_1} + \\dot{\\theta_2}) \\dot{\\theta_2} \\mathrm{sin}\\theta_2 \n\\end{split}\n\\end{equation*}\nthat is transformed to\n\\begin{equation*}\n\\begin{split}\n \\big( m_1 d_1^2 + m_2 d_1^2 -& m_2 d_1^2 \\mathrm{cos}^2\\theta_2 \\big) \\ddot{\\theta_1} =\\\\\n & m_2 d_1 d_2 (2 \\dot{\\theta_1} + \\dot{\\theta_2}) \\dot{\\theta_2} \\mathrm{sin}\\theta_2 \n + \\left( m_2 d_2^2 + m_2 d_1 d_2 \\mathrm{cos}\\theta_2 \\right) \\frac{d_1}{d_2} \\dot{\\theta_1}^2 \\mathrm{sin}(\\theta_2)\n\\end{split}\n\\end{equation*}\nand simplified to\n\\begin{equation}\\label{eq-theta1b}\n \\left( \\tfrac{m_1}{m_2} + \\mathrm{sin}^2\\theta_2 \\right) \\ddot{\\theta_1} =\n \\tfrac{d_2}{d_1} (2 \\dot{\\theta_1} + \\dot{\\theta_2}) \\dot{\\theta_2} \\mathrm{sin}\\theta_2 \n + \\left( \\tfrac{d_2}{d_1} + \\mathrm{cos}\\theta_2 \\right) \\dot{\\theta_1}^2 \\mathrm{sin}(\\theta_2)\n\\end{equation}\n\nSubstitute $\\ddot{\\theta_1}$ from \\eqref{eq-theta1b} into \\eqref{eq-theta2} and we reach the desired form.\n\n\\subsection*{Deriving the Initial Conditions}\nOur experiment is the following. The initial alignment is $\\theta_1(0) = \\frac{\\pi}{2}$ and $\\theta_2(0) = 0$. A momentary impulse gives velocity to the middle point and does not affect the upper point.\n\\begin{itemize}\n \\item $\\vec{v_1}_{, init} = \\vec{v_1}(0) = d_1 \\dot{\\theta_1}(0) \\vec{e_{\\theta_1}}(0)$ \\\\\n \\hspace*{0.5cm} $\\Rightarrow$ we obtain the initial value $\\dot{\\theta_1}(0)$.\n \n \\item The upper point is not affected that is $\\vec{0} = \\vec{v_2}(0) = d_1 \\dot{\\theta_1}(0) \\vec{e_{\\theta_1}}(0) + d_2 (\\dot{\\theta_1}(0) + \\dot{\\theta_2}(0)) \\vec{e_{\\theta_2}}(0)$\\\\\n \\hspace*{0.5cm} $\\Rightarrow$ we obtain the initial value $\\dot{\\theta_2}(0) = - \\frac{d_1}{d_2} \\dot{\\theta_1}(0) - \\dot{\\theta_1}(0)$\\\\\n \\hspace*{1.1cm} as $\\vec{e_{\\theta_1}}(0) = \\vec{e_{\\theta_2}}(0)$ due to the initial alignment.\n\\end{itemize}\n\nRecall that $\\dot\\theta_2$ is a relative coordinate with respect to the middle point that is in motion at time $0$. \nThus, as the upper point is not in motion viewing from the fixed coordinate system centered at the origin, it is moving \nwhen viewed from the middle point, therefore, it has nonzero {\\bf relative} velocity $d_2 (\\dot{\\theta_1}(0) + \\dot{\\theta_2}(0)) \\vec{e_{\\theta_2}}(0) = - d_1 \\dot{\\theta_1}(0) \\vec{e_{\\theta_2}}(0) = - d_1 \\dot{\\theta_1}(0) \\vec{e_{\\theta_1}}(0) = - \\vec{v_1}(0)$.\n\n\\newpage\n\\subsection*{Appendix}\nThe derivation for $\\dot{\\vec{e_{r_1}}}(t) = \\dot{\\theta_1}(t) \\vec{e_{\\theta_1}}(t)$ is pretty straightforward. Centered at the origin, \n$\\vec{e_{r_1}}(t)$ and $\\vec{e_{\\theta_1}}(t)$ form a coordinate system that rotates with speed $\\dot{\\theta_1}(t)$ around the origin \n(thus, rotates with respect to the inertial coordinate system). This rotation is actually around the $z$-axis (lifting it into 3D).\n\nGiven a position vector $\\vec{p}_{relative}(t)$ of a moving point in this rotating coordinate system, its velocity (within this rotating system) is $\\dot{\\vec{p}}_{relative}(t)$. Looking from the origin in the inertial coordinate system, the effect of the rotation needs to be taken into account:\n$$\\dot{\\vec{p}}_{global}(t) = \\dot{\\vec{p}}_{relative}(t) + \\dot\\theta_1(t) \\vec{k} \\times \\vec{p}_{global}(t).$$\nHere $\\vec{k}$ is the axis of rotation, the direction of the $z$-axis.\n\nApplying this for $\\vec{p}_{global}(t) = \\vec{e_{r_1}}(t)$, we get \n$$\\dot{\\vec{e_{r_1}}}(t) = \\dot{(1,0)} + \\dot\\theta_1(t) \\vec{k} \\times \\vec{e_{r_1}}(t) = \\dot\\theta_1(t) \\vec{k} \\times \\vec{e_{r_1}}(t).$$\nNow as $\\vec{e_{\\theta_1}}(t) = \\vec{k} \\times \\vec{e_{r_1}}(t)$ by definition, we have shown the sought relation.\n\n\\vspace{1cm}\nSimilar argument works for $\\dot{\\vec{e_{r_2}}}(t) = (\\dot{\\theta_1}(t) + \\dot{\\theta_2}(t)) \\vec{e_{\\theta_2}}(t)$ as well. \nTranslate $\\dot{\\vec{e_{r_2}}}(t)$ to the origin and observe that it actually rotates with speed \n$(\\dot{\\theta_1}(t) + \\dot{\\theta_2}(t))$ around the $z$-axis.\n\n\\subsection*{Further reading}\n\\begin{itemize}\n \\item \\href{http://en.wikipedia.org/wiki/Lagrangian_mechanics#Euler.E2.80.93Lagrange_equations}{Wikipedia - Lagrangian Mechanics}\n \\item \\href{http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec08.pdf}{MIT OpenCourseware - Dynamics - Relative Motion using Rotating Axis}\n \\item \\href{http://en.wikibooks.org/wiki/Kinematics/2D_Coordinate_Systems}{Wikibooks - 2D Coordinate Systems}\n \\item \\href{http://en.wikipedia.org/wiki/Double_pendulum}{Wikipedia - Double Pendulum}\n\\end{itemize}\n\\end{document}\n\n", "meta": {"hexsha": "239d4b9a756cfa0ad3db35921028336905505c7f", "size": 11704, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "examples/03_Projects/06_Ferenc_Bartha/01_Idealized_2links.tex", "max_stars_repo_name": "janduracz/acumen-dev", "max_stars_repo_head_hexsha": "8fdb5c72d4f3617c709bf354b83347faf7a57b60", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "examples/03_Projects/06_Ferenc_Bartha/01_Idealized_2links.tex", "max_issues_repo_name": "janduracz/acumen-dev", "max_issues_repo_head_hexsha": "8fdb5c72d4f3617c709bf354b83347faf7a57b60", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "examples/03_Projects/06_Ferenc_Bartha/01_Idealized_2links.tex", "max_forks_repo_name": "janduracz/acumen-dev", "max_forks_repo_head_hexsha": "8fdb5c72d4f3617c709bf354b83347faf7a57b60", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-07-30T08:20:29.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-13T12:10:32.000Z", "avg_line_length": 52.7207207207, "max_line_length": 296, "alphanum_fraction": 0.6756664388, "num_tokens": 4627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059609645723, "lm_q2_score": 0.8056321843145404, "lm_q1q2_score": 0.6389516877247711}} {"text": "% LaTeX file for a 1 page document\n\\documentclass[12pt]{article}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{physymb}\n\\usepackage{braket}\n\\title{EP 307 Assignment 2}\n\n\\author{Manish Goregaokar\\\\120260006}\n\n\\begin{document}\n\\maketitle\n\n\\section*{Problem 1}\n\n\\begin{align*}\n\\Braket{e_1|e_1}=\\Braket{e_2|e_2} &= 2\\\\\n \\Braket{e_3|e_3} &=4\\\\\n \\Braket{e_1|e_2} &=i\\sqrt{2}\\\\\n \\Braket{e_1|e_3} &=1+i\\\\\n \\Braket{e_2|e_3} &=2\n\\end{align*}\n\nFirst, we normalize the states to get:\n\n\\begin{align*}\n\\Braket{a_1|a_1}=\\Braket{a_2|a_2} &= 1\\\\\n \\Braket{e_3|a_3} &=1\\\\\n \\Braket{a_1|a_2} &=i\\frac1{\\sqrt{2}}\\\\\n \\Braket{a_1|a_3} &=\\frac{1+i}{2\\sqrt{2}}\\\\\n \\Braket{a_2|a_3} &=\\frac1{\\sqrt{2}}\n\\end{align*}\n\n\\newcommand{\\proj}[1]{\\hat{p}_{a_{#1}}}\nTaking $\\proj i = \\Ket{a_i}\\Bra{a_i}$ to be the projection operator with respect to $a_i$, by Gram-Schmidt orthogonalization procedure,\n\n\\begin{align*}\n&\\Ket{\\phi_1}=\\Ket{a_1}\\\\\n&\\Ket{\\phi_2}=\\Ket{a_2}-\\proj1\\Ket{a_2}\\\\\n&\\Ket{\\phi_3}=\\Ket{a_3}-\\proj1\\Ket{a_3}-\\proj2\\Ket{a_3}\\\\\n\\end{align*}\n\\renewcommand{\\proj}[2]{\\Ket{a_{#1}}\\Braket{a_{#1}|#2}}\nThus\n\\begin{align*}\n&\\Ket{\\phi_1}=\\Ket{a_1}\\\\\n&\\Ket{\\phi_2}=\\Ket{a_2}-\\proj1{a_2}\\\\\n&\\Ket{\\phi_3}=\\Ket{a_3}-\\proj1{a_3}-\\proj2{a_3}\\\\\n\\end{align*}\nand\n\\begin{align*}\n&\\Ket{\\phi_1}=\\Ket{a_1}\\\\\n&\\Ket{\\phi_2}=\\Ket{a_2}+\\frac{1}{\\sqrt{2}i}\\Ket{a_1}\\\\\n&\\Ket{\\phi_3}=\\Ket{a_3}-\\frac{1+i}{2\\sqrt{2}}\\Ket{a_1}-\\frac{1}{\\sqrt{2}}\\Ket{a_2}\\\\\n\\end{align*}\n\nWe can calculate \\begin{align*}&\\Braket{\\phi_1|\\phi_1}=1\\\\\n&\\Braket{\\phi_2|\\phi_2}=1-\\frac{1}{2}=\\frac12\\\\ &\\Braket{\\phi_3|\\phi_3}=1+\\frac{(1-i)(1+i)}{8}+\\frac12=\\frac74\n\\end{align*}\nand from here we get an orthonormal basis\n\n\\begin{align*}\n&\\Ket{\\psi_1}=\\Ket{a_1}\\\\\n&\\Ket{\\psi_2}=\\sqrt{2}\\left(\\Ket{a_2}+\\frac{1}{\\sqrt{2}i}\\Ket{a_1}\\right)\\\\\n&\\Ket{\\psi_3}=\\frac{2}{\\sqrt{7}}\\left(\\Ket{a_3}-\\frac{1+i}{2\\sqrt{2}}\\Ket{a_1}-\\frac{1}{\\sqrt{2}}\\Ket{a_2}\\right)\\\\\n\\end{align*}\n\nRewriting in terms of the original vectors\n\n\\begin{align*}\n&\\Ket{\\psi_1}=\\frac{1}{\\sqrt{2}}\\Ket{e_1}\\\\\n&\\Ket{\\psi_2}=\\Ket{e_2}+\\frac{1}{\\sqrt{2}i}\\Ket{e_1}\\\\\n&\\Ket{\\psi_3}=\\frac{4}{\\sqrt{7}}\\left(\\Ket{e_3}-\\frac{1+i}{2\\sqrt{2}}\\Ket{e_1}-\\frac{1}{\\sqrt{2}}\\Ket{e_2}\\right)\\\\\n\\end{align*}\n\n$\\Ket{\\psi_1},\\Ket{\\psi_2},\\Ket{\\psi_3}$ are our orthonormal basis vectors.\n\n\\section*{Problem 2}\nFirst, we normalize them by taking integrals in $[-1,1]$, and get $c_0=\\frac1{\\sqrt{2}},c_1=\\sqrt{\\frac32},c_2=\\sqrt{\\frac52}$\n\nApplying the orthogonalization process,\n\n\\begin{align*}\n\\Ket{\\phi_1}&=\\Ket{\\psi_1}\\\\\n\\Ket{\\phi_2}&=\\Ket{\\psi_2}-\\Ket{\\psi_1}\\Braket{\\psi_1|\\psi_2}\\\\\n\\Ket{\\phi_3}&=\\Ket{\\psi_3}-\\Ket{\\psi_1}\\Braket{\\psi_1|\\psi_3}-\\Ket{\\psi_2}\\Braket{\\psi_2|\\psi_3}\n\\end{align*}\n\ncalculating the inner products, we get\n\n\\begin{align*}\n\\Ket{\\phi_1}&=\\Ket{\\psi_1}\\\\\n\\Ket{\\phi_2}&=\\Ket{\\psi_2}-0\\Ket{\\psi_1}\\\\\n\\Ket{\\phi_3}&=\\Ket{\\psi_3}-\\frac{\\sqrt5}{2}\\frac23\\Ket{\\psi_1}-0\\Ket{\\psi_2}\n\\end{align*}\n\nwhich evaluates to\n\n\\begin{align*}\n\\phi_1&=\\frac1{\\sqrt{2}}\\\\\n\\phi_2&=\\sqrt{\\frac32}x\\\\\n\\phi_3&=\\sqrt{\\frac52}\\left(x^2-\\frac13\\right)\n\\end{align*}\n\\section*{Problem 3}\n\n$$\\hat Bg(x)=g(-x)$$\nThe operator is linear, since $\\hat B(\\alpha f(x)+\\beta g(x))=\\alpha f(-x)+\\beta g(-x)=\\alpha \\hat Bf(x)+\\beta\\hat B g(x)$\n\nIt also is Hermitian, as \\begin{align*}\\Bra{f}(\\hat B\\Ket{g})&=\\int_{-\\infty}^\\infty \\bar{f}(x)\\hat B g(x) dx\\\\\n&= \\int_{-\\infty}^\\infty \\bar{f}(x) g(-x) dx\\\\\n&= \\int_{-\\infty}^\\infty \\bar{f}(-x)g(x)dx\\\\\n&= \\int_{-\\infty}^\\infty \\bar{f}(-x)g(x)dx\\\\\n&=\\int_{-\\infty}^\\infty \\overline{\\hat B f(x)}g(x)\\\\\n&=\\Braket{\\overline{\\hat B} f|g}\\\\\n\\therefore \\Braket{f|\\hat B g}&=\\Braket{\\overline{\\hat B} f|g}\n\\end{align*}\n\nTo find eigenvalues, $\\hat B g(x)=b g(x)=g(-x)$\n\nSince $b g(x)=g(-x)$ and $b g(-x)=g(x)$, $b=\\pm 1$ (neglecting the trivial $b=0$ solution).\n\nThus, we have eigenvectors:\n\\begin{align*}\n&\\sum_n a_n x^{2n} & \\text{(even function)}\\qquad &\\text{for eigenvalue $b=1$}\\\\\n&\\sum_n b_n x^{2n+1}& \\text{(odd function)}\\qquad &\\text{for eigenvalue $b=-1$}\n\\end{align*} \nTo calculate the commutator,\n\\begin{align*}\n [B,\\hat{x}^n]f&=\\hat B \\hat x^n f(x)-\\hat xn \\hat B f(x)\\\\\n &=\\hat B x^n f(x) - \\hat x^n f(-x)\\\\\n &= (-x)^{n}f(-x)-x^n f(-x)\\\\\n &= f(-x)((-x)^{n}-x^n)\n\\end{align*}\n\nFor this to be zero for all states $f$, $n$ must be even. Thus the commutator is zero only for even $n$\n\\section*{Problem 4}\n\n$$\\hat \\Omega = \\Ket{\\psi}\\Bra{\\phi}$$\n\nFor it to be Hermitian, $\\hat \\Omega^\\dagger =\\Omega$. Applying it to the state $\\Ket{\\Psi}=\\Ket{\\psi}+\\Ket{\\phi}$,\n\n\\begin{align*}\n\\Ket{\\psi}\\Braket{\\phi|\\Psi}&=\\Ket{\\phi}\\Braket{\\psi|\\Psi}\\\\\n\\Ket{\\psi}\\Braket{\\phi|\\psi}+\\Ket{\\phi}\\Braket{\\psi|\\psi} &= \\Ket{\\psi}\\Braket{\\phi|\\phi}+\\Ket{\\phi}\\Braket{\\psi|\\phi}\\\\\n\\left(\\Braket{\\phi|\\psi}-\\Braket{\\phi|\\phi}\\right)\\Ket{\\psi} &= \\left(\\Braket{\\psi|\\phi}-\\Braket{\\psi|\\psi}\\right)\\Ket{\\phi}\n\\end{align*}\n\nFor this to hold, either $\\Ket{\\psi}=c\\ket{\\phi}$ or $\\left(\\Braket{\\phi|\\psi}-\\Braket{\\phi|\\phi}\\right)=0=\\left(\\Braket{\\psi|\\phi}-\\Braket{\\psi|\\psi}\\right)$\n\nIn the latter case, we get $\\Braket{\\psi|\\psi}=\\Braket{\\phi|\\psi}=\\Braket{\\phi|\\phi}$, which can only happen if the two states are equivalent.\n\nThus condition for the operator to be hermitian is that $\\Ket{\\psi}=c\\Ket{\\phi}$.\n\nFor it to be a projection operator, $c=\\frac1{\\Braket{\\phi|\\phi}}$\n\n\\section{Problem 5}\n\n$$\\hat B \\psi(x)=\\int_{-\\infty}^x x' \\psi(x')dx'$$\n\nThe eigenvalue problem $\\hat B \\psi =\\lambda\\psi$ thus becomes $\\int_{-\\infty}^x x' \\psi(x')dx' = \\lambda \\psi(x)$.\n\nDifferentiating with respect to $x$, this is \\begin{align*}\nx\\psi(x)&=\\lambda\\psi'(x)\\\\\n\\implies \\frac{\\psi'(x)}{\\psi(x)}&=\\frac{x}{\\lambda}\\\\\n\\implies \\int_{\\psi(0)}^\\psi\\frac{\\psi'(x)}{\\psi(x)}&=\\int_0^x\\frac{x}{\\lambda}\\\\\n\\implies \\log\\frac{\\psi(x)}{\\psi(0)} &= \\frac1{\\lambda} \\frac{x^2}2\\\\\n\\implies \\psi(x)=\\psi(0)e^{\\frac{x^2}{2\\lambda}}\n\\end{align*}\n\nThis is an acceptable (square-integrable) wavefunction only when $\\lambda$ is negative.\n\\end{document}\n\n\\documentclass[12pt]{}", "meta": {"hexsha": "29a3381369139d3bd8bb7b7b491bbab7aeacfb4c", "size": 5962, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 2/assign2.tex", "max_stars_repo_name": "CourseResources/CourseResources", "max_stars_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-05-28T05:59:31.000Z", "max_stars_repo_stars_event_max_datetime": "2015-05-28T05:59:31.000Z", "max_issues_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 2/assign2.tex", "max_issues_repo_name": "CourseResources/CourseResources", "max_issues_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 2/assign2.tex", "max_forks_repo_name": "CourseResources/CourseResources", "max_forks_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.4624277457, "max_line_length": 159, "alphanum_fraction": 0.6278094599, "num_tokens": 2733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812553, "lm_q2_score": 0.7931059536292271, "lm_q1q2_score": 0.6389516855163802}} {"text": "\r\n\\input{SingleAssignmentSetup.tex}\r\n\\input{../WeekTitles.tex}\r\n\\begin{document}\r\n\r\n\r\n\\begin{center}\r\n\\subsection*{MNTC P01 - Week \\#3 - \\WeekTitleThree}\r\n\\end{center}\r\n\r\n\\subsection*{Taylor Polynomials}\r\n\r\nFor reference, the general formula for the Taylor polynomial centered at x = a is:\r\n\\[P(x) = f(a) + \\frac{f'(a)}{1!} (x-a) + \\frac{f''(a)}{2!} (x-a)^2 + ... + \\frac{f^n(a)}{n!} (x-a)^n\\]\r\nwhere $n!$ means ``n factorial'', or $n \\times (n-1) \\times (n-2) \\times ... 2 \\times 1$. E.g. $4! = 4 \\cdot 3 \\cdot 2 \\cdot 1 = 24$.\r\n\r\n\\begin{enumerate}[1.]\r\n\\begin{multicols}{2}\r\n\r\n\\item\r\n\\begin{Question}\r\n Suppose \\(g\\) is a function which has continuous derivatives, and that\r\n\\(g(5)=-3, g'(5)=-2\\), \\(g''(5)=1\\),\r\n\\(g'''(5)=-5\\).\r\n\\begin{enumerate}[(a)]\r\n\\item What is the Taylor polynomial of degree 2 for \\(g\\) near \r\n\\(5\\)?\r\n\\item What is the Taylor polynomial of degree 3 for \\(g\\) near \r\n\\(5\\)? \\\\\r\n\\item Use the two polynomials that you found in parts (a) and (b) to\r\napproximate \\(g(5.1)\\).\r\n\\end{enumerate}\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nWe have\r\n\\[g(x) = g(5) + g'(5)(x-5) + \\frac{g''(5)}{2!}(x-5)^2\r\n+\\frac{g'''(5)}{3!}(x-5)^3+\\cdots\\]\r\nSubstituting gives\r\n\\[g(x) = -3 - 2(x-5) + \\frac{1}{2!}(x-5)^2\r\n +\\frac{-5}{3!}(x-5)^3+\\cdots\\]\r\n \\begin{enumerate}[(a)]\r\n\\item \r\nThe degree 2 Taylor polynomial, \\(P_2 (x)\\), is obtained by truncating\r\nafter the \\((x-5)^2\\) term:\r\n\\[P_2(x) = -3+\\left(-2\\right)\\!\\left(x-5\\right)+\\frac{1}{2!}\\cdot 1\\!\\left(x-5\\right)^{2}.\\]\r\n\\par \r\n\\item \r\nThe degree 3 Taylor polynomial, \\(P_3 (x)\\), is obtained by truncating\r\nafter the\r\n\\((x-5)^3\\) term:\r\n\\[P_3 (x) = -3+\\left(-2\\right)\\!\\left(x-5\\right)+\\frac{1}{2!}\\cdot 1\\!\\left(x-5\\right)^{2}+\\frac{1}{3!}\\!\\left(-5\\right)\\!\\left(x-5\\right)^{3}\\]\r\n\\par \r\n\\item \r\nSubstitute \\(x=5.1\\) into the Taylor polynomial of degree 2:\r\n\\[P_2(5.1) = -3+\\left(-2\\right)\\!\\left(5.1-5\\right)+\\frac{1}{2!}\\cdot 1\\!\\left(5.1-5\\right)^{2} = -3.195.\\]\r\nFrom the Taylor polynomial of degree 3, we obtain\r\n\\begin{align*}\r\nP_3(5.1) =& -3+\\left(-2\\right)\\!\\left(5.1-5\\right)+\\frac{1}{2!}\\cdot 1\\!\\left(5.1-5\\right)^{2}+ \\\\\r\n& \\frac{1}{3!}\\!\\left(-5\\right)\\!\\left(5.1-5\\right)^{3} \\\\\r\n =& -3.19583.\r\n\\end{align*}\r\n \\end{enumerate}\r\n\\par\\end{Solution}\r\n\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the Taylor polynomial of degree \\(n = 4\\) for \\(x\\) near the\r\n point \\(a = \\frac{\\pi }{4}\\) for the function\r\n \\(\\cos\\!\\left(4x\\right)\\).\\\\\r\n Use MATLAB to graph both the function and the Taylor polynomial on a\r\n reasonable interval around $x=\\pi/4$.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nLet \\(f(x)= \\cos\\!\\left(4x\\right)\\). Then \\(f(\\frac{\\pi }{4}) = -1\\), and\r\n\\par\\smallskip\\begin{center}\\begin{tabular}{|c|c|} \\hline\r\n\\(f'(x) = -4\\sin\\!\\left(4x\\right)\\) &\\(f'(\\frac{\\pi }{4}) = 0\\) \\\\ \\hline \r\n\\(f''(x) = -16\\cos\\!\\left(4x\\right)\\) &\\(f'''(\\frac{\\pi }{4}) = 16\\) \\\\ \\hline \r\n\\(f'''(x) = 64\\sin\\!\\left(4x\\right)\\) &\\(f'''(\\frac{\\pi }{4}) = 0\\) \\\\ \\hline \r\n\\(f''''(x) = 256\\cos\\!\\left(4x\\right)\\) &\\(f''''(\\frac{\\pi }{4}) = -256\\) \\\\ \\hline \r\n\\end {tabular}\\end{center}\\par\\smallskip\r\nSo,\r\n\\(P_4(x) = -1+\\frac{16}{2}\\!\\left(x-\\frac{\\pi }{4}\\right)^{2}+\\frac{-256}{24}\\!\\left(x-\\frac{\\pi }{4}\\right)^{4}\\).\r\n\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorTrig1.m}{W03TaylorTrig1.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the fifth order Taylor polynomial for\r\n \\(\\sin\\!\\left(x\\right)\\) near $x = \\pi/4$. \\\\\r\n Use MATLAB to graph both the function and the Taylor polynomial on\r\n a reasonable interval around $x = \\pi/4$.\r\n \\end{Question}\r\n\r\n\\begin{Solution}\r\nThe fifth order Taylor polynomial is based on the first five derivatives of $\\sin(x)$.\r\n\\begin{align*}\r\nf(x) &= \\sin(x) & f(\\pi/4) = \\frac{1}{\\sqrt{2}} \\approx 0.7071\\\\\r\nf'(x) &= \\cos(x) & f'(\\pi/4) = \\frac{1}{\\sqrt{2}} \\approx 0.7071\\\\\r\nf''(x) &= -\\sin(x) & f''(\\pi/4) \\approx -0.7071\\\\\r\nf'''(x) &= - \\cos(x)& f'''(\\pi/4) \\approx -0.7071\\\\\r\nf^{(4)}(x) &= \\sin(x)& f^{(4)}(\\pi/4)\\approx 0.7071\\\\\r\nf^{(5)}(x) &= \\cos(x)& f^{(5)}(\\pi/4) \\approx 0.7071\\\\\r\n\\end{align*}\r\nUsing the Taylor polynomial formula, \r\n\\begin{align*}\r\nP_5(x) = &\r\n\\frac{1}{\\sqrt{2}} + \r\n\\frac{1}{\\sqrt{2}} (x-\\pi/4) - \r\n\\frac{1}{2}\\frac{1}{\\sqrt{2}} (x-\\pi/4)^2 \\\\\r\n& - \r\n\\frac{1}{3!}\\frac{1}{\\sqrt{2}} (x-\\pi/4)^3 + \r\n\\frac{1}{4!}\\frac{1}{\\sqrt{2}} (x-\\pi/4)^4 \\\\\r\n& + \r\n\\frac{1}{5!}\\frac{1}{\\sqrt{2}} (x-\\pi/4)^5 \r\n\\end{align*}\r\n\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorTrig2.m}{W03TaylorTrig2.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the Taylor polynomial of degree 3 around the point \\(x = -4\\) of\r\n\\(f(x) = \\sqrt{5+x}\\). \\\\\r\n Use MATLAB to graph both the function and the Taylor polynomial on\r\n a reasonable interval around the reference point.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nLet \\(f(x) = \\sqrt{5+x} = (5 +x)^{1/2}\\).\r\n\\par \r\nThen\r\n\\[f'(x) = \\frac{1}{2\\sqrt{5 + x}},\\]\r\n\\[f''(x) = -\\frac{1}{4(5 + x)^{3/2}},\\]\r\nand\r\n\\[f'''(x) = \\frac{3}{8(5 + x)^{5/2}}.\\]\r\nThe Taylor polynomial of degree three about \\(x =-4\\) is thus\r\n\\begin{align*}\r\nP_3(x) & = \\sqrt{5 - 4} + \\frac{1}{2\\sqrt{5 - 4}} (x + 4 ) + (\\frac{1}{2!})\\,\\left(-\\frac{1}{4(5 - 4)^{3/2}}\\right)\r\n (x + 4)^2 + \\\\\r\n&(\\frac{1}{3!})\\left(\\frac{3}{8(5 - 4)^{5/2}}\\right) (x + 4)^3. \\\\\r\n& = 1 + \\frac{1}{2} (x+4) + \\frac{1}{2!} \\frac{-1}{4} (x+4)^2\r\n+ \\frac{1}{3!} \\frac{3}{8} (x+4)^3\r\n\\end{align*}\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorRoot1.m}{W03TaylorRoot1.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \r\n Calculate the Taylor polynomials \\(P_2(x)\\) and \\(P_3(x)\\) centered\r\n at \\(x = 1\\) for \\(f(x) = \\ln\\!\\left(x+1\\right)\\). \\\\\r\n Use MATLAB to graph the function and both the Taylor polynomials on\r\n a reasonable interval around the reference point.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nRecall the general formula for the Taylor polynomial centered at \\(x\\) = a:\r\n\\[P(x) = f(a) + \\frac{f'(a)}{1!} (x-a) + \\frac{f''(a)}{2!} (x-a)^2 + ... + \\frac{f^n(a)}{n!} (x-a)^n\\]\r\n\\[f(x) = \\ln\\!\\left(x+1\\right)\\]\r\n\\[f'(x) = \\frac{1}{x+1}\\]\r\n\\[f''(x) = \\frac{-1}{\\left(x+1\\right)^{2}}\\]\r\n\\[f'''(x) = \\frac{2}{\\left(x+1\\right)^{3}}\\]\r\nSo, \\(f(x) = \\ln\\!\\left(x+1\\right)\\) and \r\n\\(P_2(x) = A + B(x-1) + C(x-1)^2\\) \\\\\r\nwhere \\\\\r\n\\(A=f(1)=\\ln (2)=0.693147\\) \\\\\r\n\\(B=f'(1)=\\frac{1}{(2)}=0.5\\) \\\\\r\n\\(C=\\frac{f''(1)}{2!}=-\\frac{1}{(8)}=-0.125\\) \\\\\r\nThus, \r\n\\[P_2(x) = \\ln (2) + \\frac{1}{(2)} (x - 1) - \\frac{1}{(8)} (x - 1)^2\\] \\\\\r\nor,\r\n\\[P_2(x) = 0.693147+0.5\\!\\left(x-1\\right)-0.125\\!\\left(x-1\\right)^{2}\\] \\\\\r\nSimilarly, \\\\\r\n\\(P_3(x) = D + E(x-1) + F(x-1)^2 + G(x-1)^3\\) \\\\\r\nwhere \\\\\r\n\\(D=f(1)=\\ln (2)=0.693147\\) \\\\\r\n\\(E=f'(1)=\\frac{1}{(2)}=0.5\\) \\\\\r\n\\(F=\\frac{f''(1)}{2!}=-\\frac{1}{(8)}=-0.125\\) \\\\\r\n\\(G=\\frac{f'''(1)}{3!}=\\frac{1}{(24)}=0.0416667\\) \\\\\r\nThus, \r\n\\[P_3(x) = \\ln (2) + \\frac{1}{(2)} (x - 1) - \\frac{1}{(8)} (x - 1)^2 + \\frac{1}{(24)} (x - 1)^3\\] \\\\\r\nor,\r\n\\[P_3(x) = P_2(x) + 0.0417 (x-1)^3\\] \r\n\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorLn1.m}{W03TaylorLn1.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \r\n Calculate the Taylor polynomials \\(P_2(x)\\) and \\(P_3(x)\\) centered\r\n at \\(x = \\frac{\\pi }{6}\\) for \\(f(x) = \\sin\\!\\left(x\\right)\\). \\\\\r\nUse MATLAB to graph the function and both the Taylor polynomials on a\r\nreasonable interval around the reference point.\r\n\\end{Question}\r\n\\begin{Solution}\r\nThe 2nd and 3rd order Taylor polynomials are based on the first three derivatives of $\\sin(x)$.\r\n\\begin{align*}\r\nf(x) &= \\sin(x) & f(\\pi/6) = \\frac{1}{2} \\\\\r\nf'(x) &= \\cos(x) & f'(\\pi/6) = \\frac{\\sqrt{3}}{2} \\approx 0.866\\\\\r\nf''(x) &= -\\sin(x) & f''(\\pi/6) = \\frac{-1}{2} \\\\ \r\nf'''(x) &= - \\cos(x)& f'''(\\pi/6) = \\frac{-\\sqrt{3}}{2} \\approx -0.866\\\\\r\n\\end{align*}\r\nUsing the Taylor polynomial formula, \r\n\\begin{align*}\r\nP_2(x) = &\r\n\\frac{1}{2} + \r\n\\frac{\\sqrt{3}}{2} (x-\\pi/6) - \r\n\\frac{1}{2!}\\frac{1}{2} (x-\\pi/6)^2 \r\n\\end{align*}\r\nand $P_3$ will be the same, just with the cubic term added at the end.\r\n\\begin{align*}\r\nP_3(x) = &\r\n\\frac{1}{2} + \r\n\\frac{\\sqrt{3}}{2} (x-\\pi/6) - \r\n\\frac{1}{2!}\\frac{1}{2} (x-\\pi/6)^2 - \r\n\\frac{1}{3!}\\frac{\\sqrt{3}}{2} (x-\\pi/6)^3 \r\n\\end{align*}\r\n\r\n\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorTrig3.m}{W03TaylorTrig3.m}\r\n \r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \r\n Calculate the Taylor polynomials \\(P_2(x)\\) and \\(P_3(x)\\) centered\r\n at \\(x = 7\\) for \\(f(x) = \\frac{1}{1+x}\\). \\\\\r\n Use MATLAB to graph both the function and the Taylor polynomial on a\r\n reasonable interval around the reference point.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\n\\[f(x) = \\frac{1}{1+x}\\]\r\n\\[f'(x) = \\frac{-1}{(1+x)^2}\\]\r\n\\[f''(x) = \\frac{2}{(1+x)^3}\\]\r\n\\[f'''(x) = \\frac{-6}{(1+x)^4}\\]\r\nand \r\n\\(P_2(x) = A + B(x-7) + C(x-7)^2\\) \\\\\r\nwhere \\\\\r\n\\(A=f(7)=\\frac{1}{(8)}=0.125\\) \\\\\r\n\\(B=f'(7)=-\\frac{1}{(64)}\\approx-0.0156\\) \\\\\r\n\\(C=\\frac{f''(7)}{2!}=\\frac{1}{(512)}=0.00195\\) \\\\\r\nThus, \r\n\\[P_2(x) = \\frac{1}{(8)} - \\frac{1}{(64)} (x - 7) + \\frac{1}{(512)} (x - 7)^2\\] \\\\\r\nor,\r\n\\[P_2(x) = 0.125-0.0156\\!\\left(x-7\\right)+0.00195\\!\\left(x-7\\right)^{2}\\] \\\\\r\nSimilarly, \\\\\r\n\\(P_3(x) = D + E(x-7) + F(x-7)^2 + G(x-7)^3\\) \\\\\r\nwhere \\\\\r\n\\(D=f(7)=\\frac{1}{(8)}=0.125\\) \\\\\r\n\\(E=f'(7)=-\\frac{1}{(64)}=-0.0156\\) \\\\\r\n\\(F=\\frac{f''(7)}{2!}=\\frac{1}{(512)}=0.00195\\) \\\\\r\n\\(G=\\frac{f'''(7)}{3!}=-\\frac{1}{(4096)}=-0.000244\\) \\\\\r\nThus, \r\n\\[P_3(x) = \\frac{1}{(8)} - \\frac{1}{(64)} (x - 7) + \\frac{1}{(512)} (x - 7)^2 - \\frac{1}{(4096)} (x - 7)^3\\] \\\\\r\nor,\r\n\\[P_3(x) = P_2(x) - 0.000244 (x-7)^3\\] \\\\\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorRecip1.m}{W03TaylorRecip1.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Calculate the Taylor polynomials \\(P_2(x)\\) and \\(P_3(x)\\) centered\r\n at \\(x = 2\\) for \\(f(x) = e^{-x}+e^{-2x}\\). \\\\\r\n Use MATLAB to graph the function and both the Taylor polynomials on a\r\n reasonable interval around the reference point.\r\n\\end{Question}\r\n\\begin{Solution}\r\n \r\n\\[f(x) = e^{-x}+e^{-2x}\\]\r\n\\[f'(x) = -e^{-x}-2e^{-2x}\\]\r\n\\[f''(x) = e^{-x}+4e^{-2x}\\]\r\n\\[f'''(x) = -e^{-x}-8e^{-2x}\\]\r\nand \r\n\\(P_2(x) = A + B(x-2) + C(x-2)^2\\) \\\\\r\nwhere \\\\\r\n\\(A=f(2)=e^{-2}+e^{-4}\\) \\\\\r\n\\(B=f'(2)=-\\left(e^{-2}+2e^{-4}\\right)\\) \\\\\r\n\\(C=\\frac{f''(2)}{2!}=\\frac{e^{-2}+4e^{-4}}{2}\\) \\\\\r\nThus, \r\n\\[P_2(x) = (e^{-2}+e^{-4})-\\left(e^{-2}+2e^{-4}\\right)\\!\\left(x-2\\right)+\\frac{e^{-2}+4e^{-4}}{2}\\!\\left(x-2\\right)^{2}\\] \\\\\r\nSimilarly, \\\\\r\n\\(P_3(x) = D + E(x-2) + F(x-2)^2 + G(x-2)^3\\) \\\\\r\nwhere \\\\\r\n\\(D=f(2)=e^{-2}+e^{-4}\\) \\\\\r\n\\(E=f'(2)=-\\left(e^{-2}+2e^{-4}\\right)\\) \\\\\r\n\\(F=\\frac{f''(2)}{2!}=\\frac{e^{-2}+4e^{-4}}{2}\\) \\\\\r\n\\(G=\\frac{f'''(2)}{3!}=-\\frac{e^{-2}+8e^{-4}}{6}\\) \\\\\r\nThus, \\(P_3(x) =\\)\r\n\\[\\begin{array}{lcl} \r\n& = & e^{-2}+e^{-4}-\\left(e^{-2}+2e^{-4}\\right)\\!\\left(x-2\\right)+\\frac{e^{-2}+4e^{-4}}{2}\\!\\left(x-2\\right)^{2} \\cr\\cr \r\n& & - {\\frac{e^{-2}+8e^{-4}}{6}} (x-2)^3 \r\n\\end{array}\\] \r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorExp1.m}{W03TaylorExp1.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \r\n Calculate the Taylor polynomials \\(P_2(x)\\) and \\(P_3(x)\\) centered\r\n at \\(x = \\frac{\\pi }{4}\\) for \\(f(x) = \\tan\\!\\left(x\\right)\\).\r\n\r\n Use MATLAB to graph the function and both the Taylor polynomials on\r\n a reasonable interval around the reference point.\r\n\\end{Question}\r\n\\begin{Solution}\r\n \r\n\\[f(x) = \\tan\\!\\left(x\\right)\\]\r\n\\[\\frac{d}{dx}\\tan\\!\\left(x\\right) = \\frac{1}{\\cos^{2}\\!\\left(x\\right)}\\]\r\n\\[\\frac{d^2}{dx^2}\\tan\\!\\left(x\\right) = \\frac{2\\sin\\!\\left(x\\right)}{\\cos^{3}\\!\\left(x\\right)}\\]\r\n\\[\\frac{d^3}{dx^3}\\tan\\!\\left(x\\right) = \\frac{2\\!\\left(2\\sin^{2}\\!\\left(x\\right)+1\\right)}{\\cos^{4}\\!\\left(x\\right)}\\]\r\nSo, in this case, \r\n\\[P_2(x) = 1+2\\!\\left(x-\\frac{\\pi }{4}\\right)+2\\!\\left(x-\\frac{\\pi }{4}\\right)^{2}\\]\r\nand\r\n\\[P_3(x) = 1+2\\!\\left(x-\\frac{\\pi }{4}\\right)+2\\!\\left(x-\\frac{\\pi }{4}\\right)^{2} + \\frac{8}{3} (x-\\frac{\\pi }{4})^3\\]\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorTrig4.m}{W03TaylorTrig4.m}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the second-degree Taylor polynomial \\(P_2(x)\\) for the function\r\n \\(f(x) = \\sqrt{15+x^2}\\) at the number \\(x=1\\).\r\n\r\n Use MATLAB to graph both the function and the Taylor polynomial on a\r\n reasonable interval around $x = \\pi/4$.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n$P_2(x)= 4+(1/4)*(x-1)+(15/(2(16)4))(x-1)^2$\r\n\r\nMATLAB graphing: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03TaylorRoot2.m}{W03TaylorRoot2.m}\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the second-degree Taylor polynomial for\r\n\\(f(x)=2 x^2 - 3 x + 8\\) about\r\n\\(x=0\\).\r\nWhat do you notice about your polynomial?\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nWe note that \\(f(0) = 8\\);\r\n\\(f'(x) = 4 x - 3\\), so that \\(f'(0) = -3\\); and\r\n\\(f''(x) = 4\\), so that \\(f''(0) = 4\\).\r\n\\par \r\nThus\r\n\\[P_2(x) = 8 - 3 x + {4\\over 2!} x^2\r\n = 8 - 3 x + 2 x^2.\\]\r\nWe notice that \\(f(x) = P_2(x)\\) in this case, which makes\r\nsense because \\(f(x)\\) is a polynomial.\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Suppose that \\(P_2(x)=a+bx+cx^2\\) is the\r\nsecond degree Taylor polynomial for the function \\(f\\) about \\(x=0\\).\r\nWhat can you say about the signs of\r\n\\(a\\), \\(b\\), \\(c\\) if \\(f\\) has the graph given below? \\\\\r\nNote that the central lines are the $x$ and $y$ axes.\r\n\\begin{center} \r\n\\includegraphics[width=0.25\\linewidth]{graphics/Week03_TaylorPolynomials/aableson-2281-setUnit_06_TaylorPolynomialsprob16image1}\r\n\\end{center} \r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nSince \\(P_2(x)\\) is the second degree Taylor polynomial for \\(f(x)\\)\r\nabout \\(x=0\\), \\(P_2(0)=f(0)\\), which says \\(a=f(0)\\). Since\r\n\\(\\frac{d}{dx}P_2(x) \\big|_{x=0} = f'(0)\\),\r\n\\(b=f'(0)\\); and since\r\n\\(\\frac{d^2}{dx^2} P_2(x) \\big|_{x=0} = f''(0)\\),\r\n\\(2c=f''(0)\\). In other words, \\(a\\) is the \\(y\\)-intercept of\r\n\\(f(x)\\), \\(b\\) is the slope of the tangent line to \\(f(x)\\) at\r\n\\(x=0\\) and \\(c\\) tells us the concavity of \\(f(x)\\) near \\(x=0\\).\r\n\\par \r\nThus\r\n\\(a < 0\\); \\(b > 0\\); and \\(c > 0\\)\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n The function \\(f(x)\\) is approximated near \\(x=0\\) by the second\r\ndegree Taylor polynomial \\(P_2(x) = 3x-3+8x^{2}\\).\r\n\\par \r\nGive values the values of $f(0)$, $f'(0)$, and $f''(0)$.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nUsing the fact that\r\n\\[f(x)\\approx\r\nP_2(x)=f(0)+f'(0)x+\\frac{f''(0)}{2!}x^2\\]\r\nand identifying coefficients with those given for \\(P_2(x)\\), we obtain:\r\n\\par \r\n\\(f(0)=\\) the constant term, which equals \\(-3\\), so \\(f(0)=-3\\).\r\n\\(f'(0)=\\) the coefficient of \\(x\\), which equals \\(3\\), so\r\n\\(f'(0)=3\\).\r\n\\(\\frac{f''(0)}{2!}=\\) the coefficient of \\(x^2\\), which equals \\(8\\),\r\nso \\(f''(0)=16\\).\r\n\\par\\end{Solution}\r\n\\end{multicols} \r\n\r\n\\hrulefill\r\n\r\n\\subsection*{Optimization Introduction}\r\n\\begin{multicols}{2}\r\n\r\n\\item\r\n\\begin{Question}\r\nLet \\(f(x)=x^{2}-10x+13\\), and consider the interval \\([0, 10]\\). \r\n\\begin{enumerate}[(a)]\r\n\\item Find the critical point \\(c\\) of \\(f(x)\\) and compute \\(f(c)\\).\r\n\\item Compute the value of \\(f(x)\\) at the endpoints of the interval \\([0,10]\\).\r\n\\item Determine the global min and max of \\(f(x)\\) on \\([0,10]\\).\r\n\\item Find the global min and max of \\(f(x)\\) on \\([0,1]\\). (Note: not the same interval as before)\r\n\\end{enumerate}\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item The critical point of \\(f(x)\\) is the solution to \\(f'(x)=0\\). The\r\n derivative is\\(f'(x)=2x-10\\). Setting this equal to zero and solving\r\n for \\(x\\) gives \\(x=5\\). Evaluating \\(f(5)\\) yields $-12$.\r\n\\item Evaluating \\(f(0)\\) and \\(f(10)\\), we find that each is equal to 13.\r\n\\item The global min and max values must occur at critical points or\r\n at the endpoints of the interval. Since the value at the critical\r\n point is smaller than the value at the endpoints, the value of\r\n \\(f(5)\\) is a minimum, and the value of \\(f(0)\\) (or \\(f(10)\\) since\r\n they are equal) is a maximum.\r\n\\item Since there are no critical points in the interval \\([0,1]\\), so\r\n the global min and max values lie at the endpoints of the\r\n interval. Computation yields \\(f(0)=13\\),\\(f(1)=4\\), so the minimum\r\n is 4 and the maximum is 13.\r\n \\end{enumerate}\r\n\\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\nFind the maximum and minimum values of the function \\(\\ds f(x)=\\frac{\\ln\\!\\left(x\\right)}{x}\\) on the interval [1,3]. \r\n\\end{Question}\r\n\\begin{Solution}\r\n First we check for critical points. The critical point of \\(f(x)\\)\r\n is the solution to \\(f'(x)=0\\). \\\\\r\n The derivative\r\n is\\(\\ds f'(x)=x^{-2}-\\frac{\\ln\\!\\left(x\\right)}{x^{2}}\\). \\\\\r\n Setting this\r\n equal to zero and solving for \\(x\\) gives \\(x=e^1=e\\).\\\\\r\n Evaluating \\(f(e)\\) yields the value \r\n \\(\\frac{1}{e}=0.3679\\). \\\\\r\n The values of the function at the endpoints of the interval are\r\n \\(f(1)=0\\), \\(f(3)=0.3662\\), so the minimum value is 0, and the\r\n maximum value is \\(\\frac{1}{e}=0.3679\\).\r\n\\par \r\n\\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n Find the minimum and maximum values of \\(y = \\sqrt{10}\\theta-\\sqrt{5}\\sec\\theta\\) on the interval \\([ 0, \\frac {\\pi}{3}]\\).\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\n\\par \r\nLet \\(f(\\theta) = \\sqrt{10} \\theta - \\sqrt{5}\\sec \\theta\\). \\leavevmode\\\\\\relax \r\nOn the interval \\([ 0, \\frac {\\pi}{3}]\\) , \\(f'(\\theta)= \\sqrt{10} - \\sqrt{5}\\sec \\theta \\tan \\theta = 0\\) at \\(\\theta = \\frac{\\pi}{4}\\). \\\\\r\n(This can be found either by inspiration, guessing that the answer is a `nice' angle like $\\pi/3$, $\\pi/4$ or $\\pi/6$, or more mechanically by writing the whole equation in terms of $\\sin(\\theta)$, using $\\cos^2(\\theta) = 1 - \\sin^2(\\theta)$ and then solving a quadratic equation.) \\\\\r\nThe minimum value of \\(f\\) on this interval is at \\leavevmode\\\\\\relax \r\n the endpoint \\(\\theta=0\\), where \\(f(0) = -2.2361\\), \\leavevmode\\\\\\relax \r\nwhereas the maximum value over this interval is \\leavevmode\\\\\\relax \r\n\\(f(\\frac{\\pi}{4})=\\sqrt{10}(\\frac{\\pi}{4} - 1) = -0.6786\\). \\leavevmode\\\\\\relax \r\nAt the second endpoint \\(\\theta=\\frac{\\pi}{3}\\), \\leavevmode\\\\\\relax \r\n \\(f(\\frac{\\pi}{3})= \\sqrt{10} \\frac{\\pi}{3} - 2 \\sqrt{5}= -1.1606\\). \r\n\\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n \r\n\\par \r\nFind the maximum and minimum values of the function \\(\\ds f(x)=x-\\frac{125x}{x+5}\\) on the interval [0,21]. \r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nFirst we check for critical points. The critical point of \\(f(x)\\) is\r\nthe solution to \\(f'(x)=0\\). \\\\\r\nThe derivative\r\nis\\(f'(x)=1-\\frac{625}{\\left(x+5\\right)^{2}}\\). \\\\\r\nSetting this equal to\r\nzero and solving for \\(x\\) gives \\(x=-5\\pm 25\\), and of these two\r\ncritical points only -5+25=20 lies in our interval.\\\\\r\n Evaluating\r\n\\(f(-5+25)=f(20)\\) yields the value -80.\\\\\r\n The values of the\r\nfunction at the endpoints of the interval are \\(f(0)=0\\),\r\n\\(f(21)=-79.9615\\), so the minimum value is -80, and the maximum value\r\nis 0.\r\n\\par \r\n\\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n The function \\(f(x) = -2 x^3 + 21 x^2 - 36 x + 10\\)\r\nhas one local minimum and one local maximum. Find their $(x,y)$ locations.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n To identify any local extrema, we start by identifying critical\r\n points. We note that $f(x)$ is a polynomial, so its derivative is\r\n defined everywhere, so only points where $f'(x) = 0$ will be\r\n critical points.\r\n \\begin{align*}\r\n f'(x) & = -6x^2 + 42 x - 36 \\\\\r\n\\mbox{ Setting $f'(x) = 0$, }~~~~~~~ 0 & = -6 x^2 + 42 x - 36 \\\\\r\n\\mbox{ Factoring, }~~~~~~~ 0 & = -6 (x^2 - 7 x + 6) \\\\\r\n0 & = -6 (x-1)(x-6) \r\n \\end{align*}\r\n The two critical points are at $x=1$ and $x=6$. Subbing those $x$\r\n values back into the original function $f(x)$ gives us the points\r\n (1, -7) and $(6, 118)$.\r\n\r\nUsing test points and the first derivative test, or taking another derivative and using the second derivative test, you can find that: \\\\\r\nthere is a local minimum at (1,-7), and \\\\\r\nthere is a local maximum at (6, 118).\r\n\r\n\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n A University of Rochester student decided to depart from Earth after his\r\ngraduation to find work on Mars. Before building a shuttle, he conducted careful\r\ncalculations. A model for the velocity of the shuttle, from liftoff at t = 0\r\ns until the solid rocket boosters were jettisoned at t = 80 s, is given by\r\n\\[v(t) = 0.001094333 t^3- 0.08215 t^2 + 28.6 t - 4.3\\] (in feet per\r\nsecond). Using this model, estimate the global\r\nmaximum value and global minimum value of the {\\bf acceleration } of\r\nthe shuttle between liftoff and the jettisoning of the boosters.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n\r\nFor simplicity of presentation, let $c_3 = 0.001094333$ and $c_2 = 0.08215$, so\r\n$$v(t) = c_3 t^3 - c_2 t^2 + 28.6 t - 4.3$$\r\nDifferentiating once gives the acceleration\r\n$$a(t) = 3c_3 t^2 - 2c_2 t + 28.6 $$\r\nTo find the critical points of the acceleration, we need to know when its rate of change is\r\nzero: \r\n$$a'(t) = 6 c_3 t - 2c_2 $$\r\nThis will have a zero value when \r\n\\begin{align*}\r\n 0 & = 6 c_3 t - 2c_2 \\\\\r\n t & = \\frac{2c_2}{6 c_3} = \\frac{1}{3} \\frac{0.08215}{0.001094333} \\approx 25.022\r\n\\end{align*}\r\nThus $t \\approx 25.022$ is the only critical point. \r\n\r\nWe compute the acceleration at the end points of the interval ($t=0$\r\nand 80), and at the critical point: \\\\\r\n$a(0) = 28.6 \\mbox{ ft/s}^2$ \\\\\r\n$a(25.022) = 26.5444 \\mbox{ ft/s}^2$ \\\\\r\n$a(80) = 36.4672 \\mbox{ ft/s}^2$ \\\\\r\n\r\nThe global maximum acceleration is $36.4672$ ft/s$^2$ and occurs at\r\n$t=80$, while the global minimum acceleration is 26.5444 ft/s$^2$,\r\nand it occurs at $t=25.022$ s.\r\n\r\n\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n Use the given graph of the function on the interval \\((0,8]\\) to answer the following questions.\r\n\\par \r\n\\begin{center}\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week03_OptimizationIntro/crit_point_graph}\r\n\\end{center} \\par \r\n\\begin{enumerate}[(a)]\r\n\\item Where does the function \\(f\\) have a local maximum?\r\n\\item Where does the function \\(f\\) have a local minimum?\r\n\\item What is the global maximum of \\(f\\)?\r\n\\item What is the global minimum of \\(f\\)?\r\n\\end{enumerate}\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n\\item $x=$3. ($x=8$ is an end-point, and so is {\\bf not} considered a local max or min using our definitions.)\r\n\\item $x=$2, 5 \r\n\\item none: at the left end, the interval is open, so the maximum is never reached.\r\n\\item The global minimum of the function occurs at $x=5$, and the value there is $f(5) = 0$.\r\n \\end{enumerate}\r\n\r\n\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n \\par \r\nFind the global maximum and minimum values of the following function on the given interval. \r\n\\par \r\n\\(f(t)= 4 t \\sqrt{4-t^2}, \\ [-1,2]\\)\r\n\\par\\end{Question}\r\n\\begin{Solution} \r\nThe global maximum occurs at $x= 1.4142$ and $y = 8$. \r\n\r\nThe global minimum occurs at $x=-1$, and $y = -6.9282$\r\n\r\n\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n \\par \r\n An object with weight W is dragged along a horizontal plane by a\r\n force acting along a rope attached to the object. If the rope makes\r\n an angle \\(\\theta\\) with the plane, then the magnitude of the force\r\n is\r\n\\par \r\n$$F = \\dfrac {\\mu W} {\\mu \\sin ( \\theta ) + \\cos ( \\theta ) }$$\r\n\\par \r\nwhere \\(\\mu\\) is a positive constant called the coefficient of\r\nfriction and where \\(0 \\leq \\theta \\leq \\pi/2\\). Find the value for\r\n\\(\\tan \\theta\\) which minimizes the force. Your answer may depend on W\r\nand \\(\\mu\\).\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\nTo minimize $F$, we differentiate with respect to $\\theta$:\r\n\\begin{align*}\r\n && F(\\theta) & = \\mu W (\\mu \\sin(\\theta) + \\cos(\\theta))^{-1} \\\\\r\n\\mbox{ so } && F'(\\theta) & = - \\frac{\\mu W}{(\\mu \\sin(\\theta) + \\cos(\\theta))^2}(\\mu \\cos(\\theta) - \\sin(\\theta)) \\\\\r\n\\end{align*}\r\nSetting the derivative equal to zero to identify critical points, \r\n\\begin{align*}\r\n0 & = - \\frac{\\mu W}{(\\mu \\sin(\\theta) + \\cos(\\theta))^2}(\\mu \\cos(\\theta) - \\sin(\\theta)) \\\\\r\n\\mbox{ requires } 0 & = (\\mu \\cos(\\theta) - \\sin(\\theta)) \\\\\r\n\\sin(\\theta) & = \\mu \\cos(\\theta) \\\\\r\n\\frac{\\sin(\\theta)}{\\cos(\\theta)} & = \\mu \\\\\r\n\\tan(\\theta) & = \\mu \\\\\r\n\\end{align*}\r\nThe question asked for the value of $\\tan(\\theta)$, so we have that now as $\\mu$.\r\n\r\nThe greater the coefficient of friction, $\\mu$, the more of our force\r\nshould be directed upwards rather than forwards, to help minimize the\r\nfriction effect.\r\n\r\n\\end{Solution}\r\n\r\n\\item\r\n\\begin{Question}\r\n Find the exact global maximum and minimum values of the\r\nfunction \\(f(t) = \\frac{3 t}{8 + t^2}\\) if its domain is all real\r\nnumbers.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nDifferentiating using the quotient rule gives\r\n\\[f'(t)=\\frac{3(8 + t^2) - 3 t(2 8 t)}{(8+t^2)^2} = \r\n \\frac{3(8 - t^2)}{(8+t^2)^2}.\\]\r\nThe critical points are the solutions to\r\n\\(\\frac{3(8 - t^2)}{(8+t^2)^2} = 0\\), which are \r\n\\(t = \\pm\\sqrt{8}\\).\r\n\\par \r\nSince \\(f'(t)>0\\) for \\(-\\sqrt{8} 0$: concave up; $x=0$ is a local min. \\\\\r\n$g''(5) = -3000 < 0$: concave down; $x=5$ is a local max. \\\\\r\n\r\nThe values of $g(-5) = 9375$ and $g(5) = 9375$ are the slopes of the\r\noriginal function at $x=-5$ and $x=5$. They are equal, so they are\r\nboth the common global maximum slope of $9375$.\r\n\\end{Solution}\r\n\r\n\\end{multicols}\r\n\\hrulefill\r\n\r\n\\begin{multicols}{2}\r\n\r\n\\subsection*{Optimization With MATLAB}\r\n\r\nFor Questions \\ref{qoptmatlab1}-\\ref{qoptmatlab2}, use MATLAB to:\r\n\\begin{itemize}\r\n\\item generate a graph of the given function on the domain shown, and\r\n\\item use the \\texttt{fminbnd} function to find the global maximum and\r\n global minimum of the function on that interval.\r\n\\end{itemize}\r\n\r\n% ******************************\r\n\\item \\label{qoptmatlab1}\r\n\\begin{Question}\r\n$f(x)= 7 e^{7 x^3 - 7 x} , \\mbox{ on } -1 \\leq x \\leq 0$\r\n\\end{Question}\r\n\\begin{Solution}\r\n\r\nAll the examples will be solved with the same basic architecture. \r\n\r\nIn the MATLAB plots, \r\n\\begin{itemize}\r\n\\item the global {\\bf min} will be shown as a {\\bf red} dot, and\r\n\\item the global {\\bf max} will be shown as a {\\bf green} dot.\r\n\\end{itemize}\r\n\r\nIf there is anything special that is required in the solving, it will\r\nbe mentioned in text, as well as a commetn in the MATLAB code.\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptExp1.m}{W03OptExp1.m}\r\n\r\n\\lstinputlisting{MATLAB/W03OptExp1.m}\r\n\r\nFinal answer:\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n -0.0001 7.0032\r\n\r\nGlobal max (x, y) \r\n -0.5774 103.5662\r\n\\end{verbatim}\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n$f(x)= 7 x - 21 \\ln (x),\\mbox{ on } \\ [1,4]$\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n\r\n For all the following problems, we will just provide a link to the\r\n solution MATLAB file, and the final answer, rather than including it\r\n in the PDF.\r\n\r\n Looking at the graph, this function has a clear global max at the\r\n left boundary of the interval, and a global min at the critical\r\n point at $x=1$.\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptLn1.m}{W03OptLn1.m}\r\n\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n 3.0000 -2.0709\r\n\r\nGlobal max (x, y) \r\n 1.0000 6.9994\r\n\\end{verbatim}\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \\par \r\n$f(x)= 4 e^{-x} - 4 e^{-2 x}, \\mbox{ on }\\ [0,1]$\r\n\\end{Question}\r\n\\begin{Solution}\r\n Looking at the graph, this function has a clear global max at the\r\n left boundary of the interval, and a global min at the critical\r\n point at $x\\approx 0.6931$ (exact value is $x = \\ln(2)$).\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptExp2.m}{W03OptExp2.m}\r\n\r\nNotice that MATLAB doesn't return {\\em exactly} $x=0$ for the global\r\nminimum, but just the very small\r\n$x = 0.0661\\times 10^{-3} = 0.0000661$. This kind of slight deviation\r\nfrom the exact value is common with numerical methods.\r\n\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n 1.0e-03 *\r\n\r\n 0.0661 0.2644\r\n\r\nGlobal max (x, y) \r\n 0.6931 1.0000\r\n\\end{verbatim}\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n$f(x)= 7 x - 14 \\cos (x), \\mbox{ on } [-\\pi,\\pi]$\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n Looking at the graph, this function has \r\n \\begin{itemize}\r\n \\item a clear global max at the\r\n right boundary of the interval, and \r\n\\item a global min at the critical point at $x\\approx -0.5236$.\r\n \\end{itemize}\r\n\r\n There is a local max around $x=-2.8$ that isn't found by\r\n \\texttt{fminbnd} when the interval is $[-\\pi, pi]$.\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptTrig1.m}{W03OptTrig1.m}\r\n\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n -0.5236 -15.7895\r\n\r\nGlobal max (x, y) \r\n 3.1415 35.9907\r\n\\end{verbatim}\r\n\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n \\par \r\n$f(x)= ( 3 \\cos x ) / (20 + 10 \\sin x ), \\ 0 \\leq x \\leq 2 \\pi$\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n\r\n This example requires some care, as the default search with\r\n \\texttt{fminbnd} does {\\bf not} find the correct global maximum.\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptTrig2.m}{W03OptTrig2.m}\r\n\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n 3.6652 -0.1732\r\n\r\nEndpoint (x, y) \r\n 0.0001 0.1500\r\n\r\nGlobal max (x, y) \r\n 5.7596 0.1732\r\n\\end{verbatim}\r\n\\end{Solution}\r\n% ******************************\r\n\\item\\label{qoptmatlab2}\r\n\\begin{Question}\r\n \\par \r\n \\(f(t)=\\frac{10}{t}+4, 0 < t \\leq 1\\) \\\\\r\n Note the open end due to the $0 < t$ instead of $0 \\le t$.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n\r\n {\\bf Variables}: note that you can choose to change the $x$\r\n variables in your script to \\texttt{t}, or just note that the\r\n function $\\ds f(x) = \\frac{10}{x} + 4$ has all the same properties\r\n as $\\ds f(t) = \\frac{10}{t} + 4$.\r\n\r\n {\\bf Open Interval: }Even though we are working on an open interval,\r\n we can still try to use \\texttt{fminbnd}. We just need to be\r\n careful when running it, and interpreting its results.\r\n\r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W03OptRecip1.m}{W03OptRecip1.m}\r\n\r\n If you run your \\texttt{fminbnd} with $t=0$ as a boundary, you will get an error:\r\n\\begin{verbatim}\r\nError using fminbnd (line 219)\r\nUser supplied objective function must return a scalar value.\r\n\\end{verbatim}\r\n This points to the fact that $f(0)$ is actually undefined (dividing\r\n by zero). As a result, the best you can do is try to use a small\r\n but non-zero left limit, but then check your results against the\r\n graph to be sure.\r\n\r\n\\begin{verbatim}\r\nGlobal min (x, y) \r\n 0.9999 14.0005\r\n\r\nGlobal max (x, y) \r\n 1.0e+04 *\r\n\r\n 0.0000 6.0209\r\n\\end{verbatim}\r\n {\\bf However}, the global max is actually incorrect. This function won't have a global max, \r\n because there is a vertical asymptote at $t=0$, so there is no single highest point for this function.\r\n\r\n\\end{Solution}\r\n\r\n\\end{multicols}\r\n\\hrulefill\r\n\\subsection*{Optimization Word Problems}\r\n\r\n\\begin{multicols}{2}\r\n\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Some airlines have restrictions on the size of items of luggage that\r\npassengers are allowed to take with them. Suppose that one has a rule\r\nthat the sum of the length, width and height of any piece of luggage\r\nmust be less than or equal to 192 cm. A passenger wants to take a\r\nbox of the maximum allowable volume. \r\n\\begin{enumerate}[(a)]\r\n\\item If the length and width are to\r\nbe equal, what should the dimensions be?\r\n\\item In this case, what is the volume? \r\n\\item If the length is be twice the width, what should the dimensions be? \r\n\\item In this case, what is the volume? \r\n\\end{enumerate}\r\nInclude units in all your answers.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\n Let the length, width and height of the box be \\(L\\), \\(w\\) and\r\n \\(h\\), respectively. Then the volume of the box is \\(V = Lwh\\).\r\n The sum \\(L + w + h = 192\\), and, for the first part, we know that\r\n \\(L = w\\). Thus \\(2w + h = 192\\), so \\(h = 192 - 2w\\), and the\r\n volume equation becomes \\(V = Lwh = w\\cdot w\\cdot(192 - 2w) =\r\n 192\\cdot w^2 - 2w^3\\). Since we need $h \\ge 0$ and $h = 192 - 2w$,\r\n the domain for \\(w\\) is \\(0\\le w\\le 96\\).\r\n\\par \r\nCritical points are where \r\n\\(\\frac{dV}{dw} = 2\\cdot 192\\cdot w - 6\\cdot w^2 = 0\\), so \r\n\\(w = 0\\) or \\(w = 64\\). The global maximum must occur either at\r\nthis point or at the end points. \\(V(64) > 0\\) while \r\n\\(V(0) = V(96) = 0\\), so the global maximum is at \\(L=w=64\\), \r\nin which case \\(h=64\\) as well. The volume is then \r\n\\(V = 64^3 = 262144 {\\rm cm}^3\\).\r\n\\par \r\nIf \\(L = 2w\\), \\(2w + w + h = 192\\), so \r\n\\(V = (2w)(w)(192 - 3w) = 384 w^2 - 6 w^3\\). Proceeding as\r\nbefore, we find \\(w = \\frac{128}{3}\\), \\(L = \\frac{256}{3}\\) and \\(h = 64\\), so that \r\n\\(V = \\frac{2097152}{9}\\).\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n A wire 3 meters long is cut into two pieces. One piece is bent\r\ninto a square for a frame for a stained glass ornament, while the\r\nother piece is bent into a circle for a TV antenna. \r\n\\begin{enumerate}[(a)]\r\n\\item To reduce storage\r\nspace, where should the wire be cut to {\\bf minimize} the total area of both\r\nfigures? \r\n\\item Where should the wire be cut to {\\bf maximize} the total area? \r\n\\end{enumerate}\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Note that we are interested in {\\it the total area enclosed by\r\nthe two figures} . Our first task is therefore to find an\r\nequation for this area, which will be the sum of the areas of the two\r\nfigures.\r\n\\par \r\nSuppose we cut \\(x\\) meters of wire to make the circular antenna.\r\nThen there are \\(3 - x\\) meters left for the square. To find the\r\narea of the circle we need its radius. The circumference of a circle\r\nof radius \\(r\\) is \\(C = 2\\pi r\\), so the radius of the circle is\r\ngiven by \\(2\\pi r = x\\), and so \\(r = \\frac{x}{2\\pi}\\). The area\r\nof the circular antenna is therefore \r\n\\(A_c = \\pi r^2 = \\frac{1}{4\\pi} x^2\\).\r\n\\par \r\nThen the perimeter of a square with side length \\(s\\) is \r\n\\(P = 4 s = 3 - x\\), so the side length is \r\n\\(s = \\frac{1}{4}(3 - x)\\). Then the area of a square is \r\n\\(A_s = s^2\\), so the area of the square is \r\n\\(A_s = (\\frac{1}{16})(3 - x)^2\\).\r\n\\par \r\nThe total area is therefore\r\n\\[A = \\frac{1}{4\\pi} x^2 + (\\frac{1}{16})(3 - x)^2\r\n = \\frac{1}{4\\pi} x^2 + \\frac{1}{ (16)}(9 - 6 x + x^2).\\]\r\nThe domain for \\(x\\) is \\(0\\le x\\le 3\\). \r\n\\par \r\nThe maximum and minimum of \\(A\\) will occur at critical or end\r\npoints. Critical points are where \\(dA/dx = 0\\), or, where\r\n\\[\\frac{1}{2\\pi} x + \\frac{1}{ (16)}(2x - 6) = 0.\\]\r\nCollecting all terms in \\(x\\) we have\r\n\\[\\left( \\frac{1}{2\\pi} + \\frac{2 }{ (16)}\\right) x = \r\n \\frac{6 }{16},\\]\r\nso, after simplifying, \r\n\\[x = \\frac{3 \\pi}{4 + \\pi}.\\]\r\n\\par \r\nTo determine if this is a local maximum or minimum, we use the second\r\nderivative test. \r\n\\[A'' = \\left( \\frac{1}{2\\pi} + \\frac{2 }{ (16)}\\right) > 0,\\]\r\nso the function is concave up everywhere and this is a local minimum.\r\nAlso, because this is the {\\em only} critical point, this is also a {\\em global}\r\nminimum for the area.\r\n\r\nThus to minimize area we use \\(\\frac{3 \\pi}{4 + \\pi}\\) meters of wire\r\nfor the circle and \\(3 - \\frac{3 \\pi}{4 + \\pi}\\) meters for the\r\nsquare.\r\n\\item To {\\em maximize} the area, we can't use our critical point,\r\n which was a minimum; instead we must use the endpoints. The areas\r\n at the endpoints are\r\n\\[A(0) = \\frac{9}{16} \\approx 0.56\\qquad{\\rm and}\\qquad\r\n A(3) = \\frac{9}{4\\pi} \\approx 0.72,\\]\r\nthe larger of which is \\(A(3)\\), so the maximum area occurs when\r\nall of the wire is used for the circle and none for the square.\r\n \\end{enumerate}\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n A printed poster is to have a total area of 799 square inches with top and\r\nbottom margins of 6 inches and side margins of 4 inches. What should be the\r\ndimensions of the poster so that the printed area be as large as possible?\r\nLet \\(x\\) denote the width of the poster and let \\(y\\)\r\ndenote the length. \r\n\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week03_OptimizationWordProblems/Poster}\r\n\r\n\\begin{enumerate}[(a)]\r\n\\item Write the function of $x$ and $y$ that you need to maximize.\r\n\\item Express that function in terms of $x$ alone. \r\n\\item Find the critical points of the function. \r\n\\item Use the second derivative test to verify that \\(f(x)\\) has a maximum at this critical point \r\n\\item Find the optimal dimensions of the poster, and the resulting area. Include units.\r\n\\end{enumerate}\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n\r\n\\item $\\mbox{Area} = A = (x-2\\cdot 4 )(y - 2\\cdot 6)= (x - 8)(y - 12)$\r\n\\item By using the requirement that $799 = x y$, we get $A = (x - 8)(799/x - 12)$\r\n\\item Differentiating and setting the derivative equal to zero, we obtain $x = 23.08$.\r\n\\item The second derivative of $A$ will be negative at $x = 23.08$, so\r\n $A$ is concave down there, indicating $x=23.08$ is a local maximum for the printed area.\r\n\\item The dimensions of the poster with the largest printed area will be\r\n$23.08 \\times 34.62$, with a net printed area of 341.09 cm.\r\n \\end{enumerate}\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n A box with an open top has vertical sides, a square bottom, and a volume of\r\n32 cubic meters. If the box has the least possible surface area, find its\r\ndimensions.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\nThis is the same question studied in the course videos. Let the dimensions\r\nof the box be $w$ and $h$; the bottom is square so $w$ can represent the length of both \r\nsides of the bottom. These combine to produce\r\n$$A = w^2 + 2(wh) + 2(wh), ~~~~ V = w^2h = 32$$\r\nSolving for $h$ in the $V$ equation, $\\ds h = 32/w^2$, we can write\r\n$A$ as just a function of $w$:\r\n\\begin{align*}\r\nA & = w^2 + 4w(32/w^2) \\\\\r\nA & = w^2 + 128/w \\\\\r\n\\end{align*}\r\nDifferentiating and setting $A' = 0$, you will find $w = 4$, and consequently $h = 32/w^2 = 2$.\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the dimensions of the rectangle of largest area that can be\r\ninscribed in an equilateral triangle with sides of length 2 if \r\none side of the rectangle lies on the base of the triangle.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\nThe optimal rectangle will be 1 unit on the base, and have height $\\ds \\frac{\\sqrt{3}}{2}$.\r\n\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n Find the minimum distance from the parabola \r\n \\[x - y^2 = 0\\] to the point {(0,3)}.\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n\r\n\r\nThe point of closest approach will occur at $y=1$, and that will give a distance\r\nof $\\sqrt{ 1^2 + (3-1)^2} = \\sqrt{5}$.\r\n\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\n I have enough pure silver to coat \\(2\\) square meters of surface\r\n area. I plan to coat a sphere and a cube.\r\n \\begin{enumerate}[(a)]\r\n \\item Allowing for the possibility of all the silver going onto one\r\n of the solids, what dimensions should they be if the total volume\r\n of the silvered solids is to be a maximum? \r\n \\item Now allowing for the possibility of all the silver going onto\r\n one of the solids, what dimensions should they be if the total\r\n volume of the silvered solids is to be a minimum? \r\n \\end{enumerate}\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\n Let \\(s\\) be the length of the cube, \\(A\\) the total area, and \\(V\\)\r\n the total volume. Then\r\n\\[A=4\\pi r^2+6s^2\\]\r\nand\r\n\\[V= \\frac{4}{3}\\pi r^3 + s^3.\\]\r\nObviously, the volume is maximized if we put all of our stock in the\r\nsphere. In that case, \r\n\\[r = \\sqrt{\\frac{A}{4\\pi}} \\approx 0.3989\\hbox{~meters}\\]\r\n(and \\(s=0\\) meters).\r\nTo minimize the volume we eliminate one of the variables and find a\r\nstationary point as usual.\r\nSolving the area equation for \\(s\\) gives\r\n\\[s= \\sqrt{\\frac{A-4\\pi r^2}{6}}.\\] Substituting this value in the\r\nvolume equation gives\r\n\\[V= \\frac{4}{3}\\pi r^3 + \\left(\\frac{A-4 \\pi\r\nr^2}{6}\\right)^{\\frac{3}{2}}.\\]\r\nDifferentiating with respect to \\(r\\) and setting to zero gives:\r\n\\[V'=4\\pi r^2 - \\frac{3}{2}\\times\\frac{8\\pi r}{6} \\left(\\frac{A-4 \\pi\r\nr^2}{6}\\right)^{\\frac{1}{2}} = 0.\\]\r\nThis simplifies to\r\n\\[2r = \\sqrt{\\frac{A-4 \\pi\r\nr^2}{6}}.\\]\r\nSquaring gives\r\n\\[4r^2 = \\frac{A-4 \\pi\r\nr^2}{6}\\]\r\nwhich gives\r\n\\[r = \\sqrt{\\frac{A}{24+4\\pi}} \\approx 0.2339\\hbox{~meters.}\\]\r\nThe corresponding value of \\(s\\) is\r\n\\[s = 2r = \\sqrt{\\frac{A}{6+\\pi}} \\approx 0.4677\\hbox{~meters.}\\]\r\n\\par\\end{Solution}\r\n% ******************************\r\n\\item\r\n\\begin{Question}\r\nSuppose that 241 ft of fencing are used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle as the following figure:\\leavevmode\\\\\\relax \r\n\\includegraphics[width=0.3\\linewidth]{graphics/Week03_OptimizationWordProblems/WindowShape}\r\n\r\nFind the dimensions of the corral with maximum area. \r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\n\\par \r\nFrom the picture, we see that \\(x\\) is the width of the corral, and therefore the diameter of the semicircle, \\leavevmode\\\\\\relax and that \\(y\\) is the height of the rectangular section. Thus the perimeter of the corral can be expressed \\leavevmode\\\\\\relax \r\nby the equation \\(2y+ x+ \\frac{\\pi}{2}x= 2y + (1+ \\frac{\\pi}{2})x= 241\\) ft or equivalently, \\leavevmode\\\\\\relax \r\n\\(y=\\frac{1}{2}(241 - (1+ \\frac{\\pi}{2})x)\\). Since \\(x\\) and \\(y\\) must both be non-negative, it follows that \\(x\\) must \\leavevmode\\\\\\relax \r\nbe restricted to the interval \\([0, \\frac{241}{1+ \\pi/2}]\\). The area of the corral is the sum of the area of the \\leavevmode\\\\\\relax \r\nrectangle and semicircle, \\(A= xy + \\frac{\\pi}{8}x^2\\). Making the substitution for \\(y\\) from the \\leavevmode\\\\\\relax \r\nconstraint equation, \\par \r\n\\(A(x)=\\frac{1}{2}x(241 - (1+ \\frac{\\pi}{2})x) + \\frac{\\pi}{8}x^2 = 120.5 x - \\frac{1}{2}(1+ \\frac{\\pi}{2})x^2 + \\frac{\\pi}{8}x^2\\). \\leavevmode\\\\\\relax \r\nNow, \\(A'(x) = 120.5 - (1 + \\frac{\\pi}{2})x + \\frac{\\pi}{4}x = 0\\) implies \\(x= \\frac{120.5}{(1+ \\frac{\\pi}{4})} \\approx 67.4919\\). \\leavevmode\\\\\\relax \r\nWith \\(A(0)=0\\), \\par \r\n\\(A(\\frac{120.5}{(1+ \\frac{\\pi}{4})}) \\approx 4066.39 \\qquad\\) and \\(\\qquad A(\\frac{241}{1+ \\frac{\\pi}{2}}) \\approx 3451.11\\), \\leavevmode\\\\\\relax \r\nit follows that the corral of maximum area has dimensions \\par \r\n\\(x= \\frac{120.5}{1+ \\frac{\\pi}{4}} \\qquad\\) and \\\\\r\n\\(\\qquad y= \\frac{1}{2}(241 - (1+ \\frac{\\pi}{2}) \\frac{120.5}{1+ \\frac{\\pi}{4}}) \\approx 33.746\\).\r\n\\par\\end{Solution}\r\n% \\item\r\n% \\begin{Question}\r\n \r\n% \\par \r\n% Find the maximum area of a triangle formed in the first quadrant by the \\(x\\)-axis, \\(y\\)-axis and a tangent line to the graph of \\(f=(x + 2)^{-2}\\).\r\n% \\par \\end{Question}\r\n% \\begin{Solution}\r\n \r\n% Let \\(P\\left(t,\\frac{1}{(t+2)^2}\\right)\\) be a point on the graph of the curve \\(y=\\frac{1}{(x+2)^2}\\) in the first quadrant. The tangent line to the curve at \\(P\\) is\r\n% \\[L(x)=\\frac{1}{(t+2)^2}-\\frac{2(x-t)}{(t+2)^3},\\]\r\n% which has \\(x\\)-intercept \\(a=\\frac{3t+2}{2}\\) and \\(y\\)-intercept \\(b=\\frac{3t+2}{(t+2)^3}\\). The area of the triangle in question is\r\n% \\[A(t) = \\frac{1}{2}ab = \\frac{(3t+2)^2}{4(t+2)^3}.\\]\r\n% Solve\r\n% \\[A'(t) = \\frac{(3t+2)(3\\cdot 2-3t)}{4(t+2)^4} = 0\\]\r\n% for \\(0 \\le t\\) to obtain \\(t = 2\\). Because \\(A(0) = \\frac{1}{4 \\cdot 2}\\), \\(A(2) = \\frac{1}{2 \\cdot 2}\\) and \\(A(t) \\to 0\\) as \\(t \\to \\infty\\), it follows that the maximum area is \\(A(2) = 0.25\\).\r\n% \\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n \r\n\\par \r\nA box is constructed out of two different types of metal. The metal for\r\nthe top and bottom, which are both square, costs \\$4 per square foot\r\nand the metal for the sides costs \\$6 per square foot. Find the\r\ndimensions that minimize cost if the box has a volume of 35 cubic\r\nfeet.\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nLet \\(x>0\\) be the length of a side of the square base and \\(z>0\\) the height of the box. With volume \\(x^{2} z=35\\), we have \\(z=35/x^{2}\\) and cost\r\n\\[C(x)=4\\cdot 2\\cdot x^{2}+6\\cdot 4\\cdot xz=8x^{2}+840\\frac{1}{x}.\\]\r\nSolve \\(C'(x)=8\\cdot 2x-840x^{-2}=0\\) to obtain \\(x= \\left(\\frac{35 \\cdot 6}{4} \\right)^{1/3}\\). \r\nSince \\(C(x) \\to \\infty\\) as \\(x \\to 0+\\) and as \\(x \\to \\infty\\), the minimum cost is \r\n\\(C\\left( (\\frac{35 \\cdot 6}{4})^{1/3} \\right)\r\n\\approx \\$ 336.499\\) \r\nwhen \r\n\\(x\\approx 3.74444\\textrm{ ft}\\) \r\nand \r\n\\(z\\approx 2.49629\\textrm{ ft}\\).\r\n\\par\\end{Solution}\r\n%\\item\r\n%\\begin{Question}\r\n% A manufacturer has been selling 1550 television sets a week at \\$540\r\n% each. A market survey indicates that for each \\$23 rebate offered to\r\n% a buyer, the number of sets sold will increase by 230 per week. \r\n% \\begin{enumerate}[(a)]\r\n% \\item Find the price, $p(x)$, as a function of $x$, where \\(x\\) is\r\n% the number of the television sets sold per week.\r\n% \\item How large rebate should the company offer to a buyer, in order\r\n% to maximize its revenue?\r\n% \\item If the weekly cost function is \\(139500 + 180 x\\), how should\r\n% it set the size of the rebate to maximize its profit?\r\n% \\end{enumerate}\r\n%\\par\\end{Question}\r\n%\\begin{Solution}\r\n% \\begin{enumerate}[(a)]\r\n%\\item $p(x) = (1550-x)/10 +540$.\r\n%\\item Revenue = price $\\times$ demand = $(540 - 23\\cdot x) (1550-x)/10 +540$.\r\n%Finding the critical point of the revenue function, you will find $x = 192.5$.\r\n%\\item Build profit as revenue - cost. Compute the critical points, and\r\n% you will find $x= 102.5$ produces a maximum profit.\r\n% \\end{enumerate}\r\n%\r\n%\\end{Solution}\r\n% \\item\r\n% \\begin{Question}\r\n% A baseball team plays in he stadium that holds 72000\r\n% spectators. With the ticket price at 12 the average attendence has\r\n% been 30000. When the price dropped to 11, the averege attendence\r\n% rose to 36000. \\leavevmode\\\\\\relax \r\n% \\begin{enumerate}[(a)]\r\n% \\item Find the demand function \\(p(x)\\), where \\(x\\) is the number\r\n% of the spectators. (assume \\(p(x)\\) is linear.)\r\n% \\item How should be set a ticket price to maximize revenue?\r\n% \\end{enumerate}\r\n% \\par\\end{Question}\r\n% \\begin{Solution}\r\n% {\r\n% \\vspace{-\\parskip}\\begin{itemize}\r\n% \\item $(30000-x)*1/(6*1000) +12$\r\n% \\item 8.5 \r\n% \\end{itemize}}\\par\r\n%\\end{Solution}\r\n%\\item\r\n%\\begin{Question}\r\n% For the cost function \\(C(x) = 40000+800 x + x^2\\) find: \r\n% \\begin{enumerate}[(a)]\r\n%\\item The cost at the production level 1650.\r\n%\\item The average cost at the production level 1650.\r\n%\\item The marginal cost at the production level 1650 (this is the in.\r\n%\\item The production level that will minimize the average cost.\r\n%\\item The minimal average cost. \r\n% \\end{enumerate}\r\n%\\par\\end{Question}\r\n%\\begin{Solution}\r\n% \\begin{enumerate}[(a)]\r\n%\\item 4082500 \r\n%\\item 2474.24242424242 \r\n%\\item 4100 \r\n%\\item 200 \r\n%\\item 1200 \r\n% \\end{enumerate}\r\n%\r\n%\\end{Solution}\r\n% \\item\r\n% \\begin{Question}\r\n% A rectangle is inscribed with its base on the \\(x\\) axis and its\r\n% upper corners on the parabola \\(y= 12 - x^2\\). What are the\r\n% dimensions of such a rectangle with the greatest possible area?\r\n% \\par \\end{Question}\r\n% \\begin{Solution}\r\n \r\n% To solve this problem, we need to find an expression for the area of the \r\n% rectangle in terms of one of its dimensions, and then use derivatives to\r\n% maximize this area. First, however, we can simply things quite a bit by \r\n% noting that the parabola given by \\(y = 12 - x^2\\) is symmetric about \r\n% the \\(y\\)-axis. Therefore, the inscribed rectangle will also be symmetric \r\n% about the \\(y\\)-axis. So it is enough to find the dimensions of half of this \r\n% rectangle and double the x value.\r\n% \\par \r\n% Our rectangle will therefore have the bottom left corner \\((0,0)\\) and the top\r\n% right corner \\((x,12-x^2)\\) where \\(x\\) is the width of the rectangle, \r\n% and \\(12 - x^2\\) is its height. Thus, the area of the rectangle is\r\n% given by \\(f(x) = x(12-x^2) = {12}x-x^3\\) where \\(x\\) is the width \r\n% of the rectangle. We now find the derivative of this and solve for zero \r\n% to find critical points.\r\n% \\par \r\n% The derivative is \\(f'(x) = 12 - 3x^2\\). Setting this equal to 0 and \r\n% recalling that we are talking about widths, so that all \\(x\\) values \r\n% should be positive, we get:\r\n% \\par \r\n% \\[\\begin{aligned}\r\n% f'(x) & = 0 \\\\\r\n% 12 - 3x^2 & = 0 \\\\\r\n% 3x^2 & = 12 \\\\\r\n% x^2 & = \\frac{12}{3} = 4 \\\\\r\n% x & = \\sqrt{4} = 2\r\n% \\end{aligned}\\]\r\n% \\par \r\n% It is easy to check that the second derivative of \\(f(x)\\) is negative everywhere, \r\n% so this is a maximum of \\(f(x)\\). Therefore, this is the width of the rectangle with the\r\n% maximum area. Actually, it is the width of half of that rectangle, since we were\r\n% ignoring the half on the left side of the y-axis. So the width of the whole rectangle is \\(2\\cdot 2 = 4\\). The height is given by \r\n% plugging \\(x=2\\) into the formula for the parabola, giving\r\n% \\(12 - (2)^2 = 12 - 4 = 8 \\).\r\n% \\par\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n Centerville is the headquarters of Greedy Cablevision Inc. The\r\ncable company is about to expand service to two nearby towns,\r\nSpringfield and Shelbyville. There needs to be cable connecting\r\nCenterville to both towns. The idea is to save on the cost of\r\ncable by arranging the cable in a Y-shaped configuration.\r\nCenterville is located at\r\n\\((8,0)\\) in the \\(xy\\)-plane, Springfield is at \\((0,5)\\), and\r\nShelbyville is at \\((0,- 5)\\). The cable runs from Centerville\r\nto some point \\((x,0)\\) on the \\(x\\)-axis where it splits into two branches going to\r\nSpringfield and Shelbyville. Find the location \\((x,0)\\)\r\nthat will minimize the amount of cable between the 3 towns and\r\ncompute the amount of cable needed. Justify your answer.\r\n\\par \r\n\\begin{enumerate}[(a)]\r\n\\item What function of $x$ needs to be minimized to solve this problem?\r\n\\item Find the critical points of $f(x)$.\r\n\\item Use the second derivative test to verify that \\(f(x)\\) has a minimum at this critical point.\r\n\\item Compute the minimum amount of wire needed.\r\n\\end{enumerate}\r\n\\par\\end{Question}\r\n\\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n\\item Draw a sketch. \\\\\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week03_OptimizationWordProblems/Cablevision}\r\n\r\nWith $x$ being the horizontal component of the diagonal lines, the\r\ntotal length of the cable will be $L(x) = 2 \\sqrt{x^2+5^2}+(8 -x)$.\r\n\\item Taking the derivative and finding critical points of $L(x)$ yields $x = 2.89$.\r\n\\item The second derivative of $L(x)$ will be positive at $x = 2.89$, indicating that\r\nthe critical point is a local minimum for the length of cable. \r\n\\item $L(2.89) = 2 \\sqrt{(2.89)^2 + 25} + (8-2.89) = 16.66$ units of cable.\r\n \\end{enumerate}\r\n\r\n\\end{Solution}\r\n\\item\r\n\\begin{Question}\r\n A cylinder is inscribed in a right circular cone of height 4 m and\r\n radius (at the base) equal to 3.5 m. What are the dimensions of\r\n such a cylinder which has maximum volume?\r\n\\par \\end{Question}\r\n\\begin{Solution}\r\n \r\nAs we are attempting to maximize the volume of the inscribed cylinder,\r\nwe must first come up with a formula for the volume of this cylinder.\r\nLet \\(x\\) be the radius of the cylinder, \\(v(x)\\) the volume. We \r\nknow from basic geometry that the formula for volume is give by \r\n\\(\\pi x^2h\\) where \\(x\\) is the radius and \\(h\\) is the height\r\nof the cylinder. So in order to come up with a formula for volume\r\nin terms of \\(x\\) only, we need to relate \\(x\\) to \\(h\\).\r\n\\par \r\nThis is where the information about the cone comes in handy. The cone is \r\na right circular cone. Thus, inscribing the cylinder will fill up\r\nsome of the base of the cone, and just touch the slanted side, leaving\r\na similar right circular cone at the top. This new cone will have a\r\nradius of \\(x\\) and a height of \\(4 - h\\) where \\(x\\) and \\(h\\) are as in the formula for the volume of our cylinder. As this cone is\r\nsimilar to the original, we can use ratios to get:\r\n\\par \r\n\\[\\frac{x}{3.5}=\\frac{4 - h}{4}\\]\r\n\\par \r\nSimplifying this, we get \\(h = 4 - \\frac{4}{3.5}x\\). \r\nTherefore, our formula for volume in terms of \\(x\\) becomes \\(v(x) = \\pi \r\nx^2(4 - \\frac{4}{3.5}x) = (4\\pi)x^2 - \r\n(\\frac{4}{3.5}\\pi)x^3\\)\r\n\\par \r\nNow, we want to maximize this. So we will first take the derivative. \r\nUsing the rules for differentiation of polynomials, the derivative is \\(v'(x) = 2(4\\pi)x - 3(\\frac{4}{3.5}\\pi)x^2\\). Solving\r\nfor zero, we get, as we don't want \\(x = 0\\), the following.\r\n\\par \r\n\\[\\begin{aligned}\r\n v'(x) & = 0 \\\\\r\n 2(4\\pi)x - 3(\\frac{4}{3.5}\\pi)x^2 & = 0 \\\\\r\n \\pi x(2(4) - 3\\frac{4}{3.5}x) & = 0 \\\\\r\n 2(4) - 3\\frac{4}{3.5}x & = 0 \\\\\r\n 3\\frac{4}{3.5}x & = 2(4) \\\\\r\n x & = \\frac{2}{3}(3.5) = 2.333 \\mbox{ m}\\\\\r\n \\end{aligned}\\]\r\nThen, using the formula for height we came up with before, the height\r\ncan be determined by:\r\n\\[h = 4 - \\frac{4}{3.5}(2.333) = 1.333\\mbox{ m}\\]\r\n\\par\\end{Solution}\r\n% \\item\r\n% \\begin{Question}\r\n% A small island is 3 miles from the nearest point P on the straight\r\n% shoreline of a large lake. If a woman on the island can row a boat\r\n% 2 miles per hour and can walk 3 miles per hour, where should\r\n% the boat be landed in order to arrive at a town 8 miles down the\r\n% shore from P in the least time? Let \\(x\\) be the distance (in miles) between\r\n% point P and where the boat lands on the lakeshore.\r\n% \\begin{enumerate}[(a)]\r\n% \\item Enter a function \\(T(x)\\) that describes the total amount of\r\n% time the trip takes as a function of the distance \\(x\\).\r\n% \\item What is the distance \\(x = c\\) that minimizes the travel time?\r\n% \\item What is the least travel time? \r\n% \\end{enumerate}\r\n% \\end{Question}\r\n% \\begin{Solution}\r\n% See the similar example in the course notes.\r\n% \\begin{enumerate}[(a)]\r\n% \\item \\begin{align*}\r\n% T(x)& = \r\n% \\frac{\\mbox{water dist}}{\\mbox{water speed}} + \r\n% \\frac{\\mbox{land dist}}{\\mbox{land speed}} \\\\\r\n% & = \\frac{\\sqrt{9+x^2}}{2}+\\frac{(8-x)}{3}\r\n% \\end{align*}\r\n% \\item The minimum of $T(x)$ will occur when $x = 2.68$ miles.\r\n% \\item For that landing point, the travel time will be $T(2.68) = 3.78$\r\n% hours or about 3 hours and 45 minutes.\r\n% \\end{enumerate}\r\n%\r\n%\\end{Solution}\r\n\r\n\r\n\\end{multicols}\r\n\r\n\\end{enumerate}\r\n\\end{document}\r\n", "meta": {"hexsha": "1b39bd73e5352df1e62b73dc8586427eae8d3fe3", "size": 58400, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "PracticeProblems/Week03.tex", "max_stars_repo_name": "aableson/MNTCP01", "max_stars_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-11-27T16:10:35.000Z", "max_stars_repo_stars_event_max_datetime": "2015-11-27T16:10:35.000Z", "max_issues_repo_path": "PracticeProblems/Week03.tex", "max_issues_repo_name": "aableson/MNTCP01", "max_issues_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "PracticeProblems/Week03.tex", "max_forks_repo_name": "aableson/MNTCP01", "max_forks_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.4861392833, "max_line_length": 285, "alphanum_fraction": 0.6128424658, "num_tokens": 20765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.793105951184112, "lm_q2_score": 0.8056321889812553, "lm_q1q2_score": 0.6389516835465168}} {"text": "\\chapter{Interpolation of the data}\n\\label{Sec:interpolate}\nThe data in \\xendl\\ representing the probability density\n$\\pi(E', \\mu \\mid E ) = \\pi_r(E', \\mu \\mid E )$ in the integrals (\\ref{Inum}) and~(\\ref{Ien})\nare given in various forms. In the case of tabulated data, intermediate\nvalues must be obtained via some sort of interpolation. Interpolation\nwith respect to one independent variable is described first, followed by a discussion of\nthe 2-dimensional case. In \\xendl\\ full 3-dimensional interpolation\nof $\\pi(E', \\mu \\mid E )$ data is reduced to a sequence of 2-dimensional\ninterpolations.\n\n\\section{Interpolation methods for a single variable}\\label{Sec:1d-interp}\nFor the sake of having a specific application,\nthe discussion here is given in terms of tables of data $\\{E_i, f(E_i)\\}$,\nwith values $E_i$ of the energy of the outgoing particle as independent variable.\nThese ideas are applicable to one dimension for any tabular data.\nThe types of interpolation method used in \\xendl\\ \nfor such tables are: histogram, linear-linear, \nlog-linear, linear-log, and log-log. The algorithms for interpolation\nof $F(E)$ on an interval $E_0 < E < E_1$ with given $f(E_0)$ and~$f(E_1)$\nare as follows. In these definitions it is assumed that the argument of a\nlogarithm is positive.\n\n\\subsection{Histograms}\nFor histogram interpolation set\n$$\n f(E) = f(E_0) \\quad \\text{for $E_0 \\le E < E_1$}.\n$$\n\n\\subsection{Linear-linear}\n For linear-linear interpolation set\n\\begin{equation}\n \\alpha = \\frac{E - E_0}{E_1 - E_0}\n \\label{def-alpha}\n\\end{equation}\nand take\n$$\n f(E) = (1 - \\alpha)f(E_0) + \\alpha f(E_1)\n \\quad \\text{for $E_0 \\le E \\le E_1$}.\n$$\n\n\\subsection{Log-linear}\nFor log-linear interpolation take $\\alpha$ as in Eq.~(\\ref{def-alpha}),\nand set\n$$\n \\log f(E) = (1 - \\alpha)\\log f(E_0) + \\alpha \\log f(E_1)\n \\quad \\text{for $E_0 \\le E \\le E_1$}.\n$$\nThis relation may also be written as\n\\begin{equation}\n f(E) = f(E_0)^{1 - \\alpha} f(E_1)^\\alpha.\n \\label{loglin-interp}\n\\end{equation}\n\n\\subsection{Linear-log}\n For linear-log interpolation set\n\\begin{equation}\n \\alpha' = \\frac{\\log(E / E_0)}{\\log( E_1 / E_0 )}\n \\label{def-alpha-prime}\n\\end{equation}\nand take\n$$\n f(E) = (1 - \\alpha')f(E_0) + \\alpha' f(E_1)\n \\quad \\text{for $E_0 \\le E \\le E_1$}.\n$$\n\n\\subsection{Log-log}\nFor log-log interpolation take $\\alpha'$ as in Eq.~(\\ref{def-alpha-prime}),\nand set\n$$\n \\log f(E) = (1 - \\alpha')\\log f(E_0) + \\alpha' \\log f(E_1)\n \\quad \\text{for $E_0 \\le E \\le E_1$}.\n$$\nThis is equivalent to\n\\begin{equation}\n f(E) = f(E_0)^{1 - \\alpha'} f(E_1)^{\\alpha'}.\n \\label{loglog-interp}\n\\end{equation}\n\n\\textbf{Remark.}\nWith log-linear interpolation written in the form of Eq.~(\\ref{loglin-interp})\nand log-log interpolation written as Eq.~(\\ref{loglog-interp}), it is permitted\nthat $f(E_0) = 0$ or $f(E_1) = 0$. These cases all lead to the result that\n$f(E) = 0$ for $E_0 < E < E_1$, however.\n\n\\section{Interpolation methods for probability densities}\\label{Sec:2d-interp}\nIn order to explain the methods for interpolation of probability densities, it suffices to\nconsider a table of values $\\pi( E' \\mid E )$\n\\begin{equation}\n \\{ E'_{j,k}, \\pi(E'_{j,k} \\mid E_k )\\} \\quad \\text{for $j = 0$, 1, $\\ldots\\,$, $J_k$}\n \\label{EPtable}\n\\end{equation}\ngiven at values of the incident energy~$E_k$, for\n$k = 0$,~1, $\\ldots\\,$,~$K$.\nIn Eq.~(\\ref{EPtable}) it is required that the outgoing energies\nbe ordered\n\\begin{equation}\n E'_{0,k} < E'_{1,k} \\le E'_{2,k} \\le \\cdots \\le E'_{J_k-1,k} < E'_{J_k,k}.\n \\label{Eout-order}\n\\end{equation}\n\nThe condition Eq.~(\\ref{Eout-order}) permits the data of\nEq.~(\\ref{EPtable}) to have equal consecutive intermediate\noutgoing energies $E'_{j-1,k} = E'_{j,k}$, so that the probability\ndensity~$\\pi(E' \\mid E_k)$ may have a jump discontinuity there.\nJump discontinuities are not allowed at the end points $E' = E'_{0,k}$\nand~$E' = E'_{J_k,k}$. In Eq.~(\\ref{EPtable}) the possibility \nof three or more consecutive equal outgoing energies may be ruled out,\nbecause all but the first and last would be redundant. \nThe convention adopted here is that the value of $\\pi(E' \\mid E_k)$\nat a discontinuity is the second data value\n$$\n \\pi(E' \\mid E_k) = \\pi(E'_{j,k} \\mid E_k) \\quad\n \\text{if $E' = E'_{j-1,k} = E'_{j,k}$.}\n$$\n\nFor fixed incident energy $E_k$, the rules for interpolation of \n$\\pi(E' \\mid E_k )$ in outgoing energy~$E'$ are as given\nin Section~\\ref{Sec:1d-interp}. The following types\nof interpolation with respect to $E$ are discussed in subsequent subsections:\n\\begin{enumerate}\n \\item direct interpolation.\n \\item unit-base interpolation,\n \\item interpolation using cumulative points.\n\\end{enumerate}\n\nThe method referred to here as ``interpolation by cumulative points''\nis closely related to ``interpolation by corresponding energies'' as described\nin the \\ENDF\\ manual~\\cite{ENDFB}. \nFor a more-detailed discussion of 2-dimensional interpolation methods,\nsee the reference~\\cite{interpolation}.\n\nFor a discussion of interpolation of data Eq.~(\\ref{EPtable}), it suffices\nto consider interpolation between incident energies $E_0$ and~$E_1$.\nThus, it is desired to interpolate to incident energy~$E$ with\n$E_0 < E < E_1$ the data\n\\begin{equation}\n\\begin{split}\n \\{ E'_{j,0}, \\pi(E'_{j,0} \\mid E_0 )\\} &\\quad \\text{for $j = 0$, 1, $\\ldots\\,$, $J_0$}, \\\\\n \\{ E'_{j,1}, \\pi(E'_{j,1} \\mid E_1 )\\} &\\quad \\text{for $j = 0$, 1, $\\ldots\\,$, $J_1$}.\n \\label{EPtables}\n\\end{split}\n\\end{equation}\nThe ideas presented apply equally well to interpolation of data\nin Eq.~(\\ref{EPtable}) between any consecutive pair of\nincident energies $E_{k-1} < E_k$.\n\nThe methods of 2-dimensional interpolation are described in turn.\nFor each of these methods the interpolated probability density $\\pi(E' \\mid E)$\nfor $E_0 < E < E_1$ has the proper norm\n\\begin{equation}\n \\int\n \\pi(E' \\mid E) \\ dE' = 1\n \\label{piNorm}\n\\end{equation}\nwhen the interpolation rule with respect to incident energy~$E$ for the\ndata Eq.~\\eqref{EPtables} is linear-linear, histogram, or linear-log.\nThe norm condition Eq.~\\eqref{piNorm} is not usually satisfied when\nlog-linear or log-log interpolation is used for the incident energy.\n\n\\subsection{Direct interpolation}\\label{Sec:direct-interp}\nIt is common to do direct interpolation for interpolating tables of angular probability\ndensity~$\\pi( \\mu \\mid E)$ with respect to incident energy~$E$, because the\nrange of direction cosines is usually $-1 \\le \\mu \\le 1$. For example, in order\nto determine the value of $\\pi( \\mu \\mid E)$ for $E_0 < E < E_1$ from data\nEq.~(\\ref{EPtables}), one first interpolates in $\\mu$\nat fixed incident energies to obtain $\\pi( \\mu \\mid E_0)$ and~$\\pi( \\mu \\mid E_1)$.\nOne then obtains the value of $\\pi( \\mu \\mid E)$ by interpolating between\n$\\pi( \\mu \\mid E_0)$ and~$\\pi( \\mu \\mid E_1)$.\n\nThe trouble with the application of direct interpolation to tables of \nenergy distributions\n is that the range of outgoing energy~$E'$\nusually depends on the incident energy~$E$.\nThus, for the data in Eq.~(\\ref{EPtables}), the ranges of outgoing energies\nare given by\n\\begin{equation}\n\\begin{split}\n \\Eminzero' = E'_{0,0} \\quad \\text{and} \\quad\n \\Emaxzero' = E'_{J_0,0} \\quad \\text{for $E = E_0$}, \\\\\n \\Eminone' = E'_{0,1} \\quad \\text{and} \\quad\n \\Emaxone' = E'_{J_1,1} \\quad \\text{for $E = E_1$}.\n \\label{Eout-ranges}\n\\end{split}\n\\end{equation}\n\n\\textbf{Remark.}\nIn the definition of the range of outgoing energies Eq.~(\\ref{Eout-ranges}),\nit is natural to expect that the data in Eq.~(\\ref{EPtables}) are such that\nfor each incident energy $E_k$ with $k = 0,$~1, $\\ldots\\,$,~$K$, the probability\ndensity $\\pi(E' \\mid E_k )$ is not equal to zero on the entire lowest outgoing\nenergy range $E'_{0,k} < E' < E'_{1,k}$ or highest outgoing energy range\n$E'_{J_k - 1,k} < E' < E'_{J_k,k}$. That is, \nEq.~(\\ref{Eout-ranges}) ought to give the\nactual range of outgoing energies.\nSome nuclear data libraries, e.~g.,\n\\ENDFdata~\\cite{ENDFdata}, have\ndata of the form Eq.~(\\ref{EPtable}) which imply that\n$\\pi(E' \\mid E_k ) = 0$ on the lowest or highest outgoing energy ranges.\nThe sample input data given in\nSection~\\ref{Sec:isotropic-table-lab} illustrates the problem.\n\nIt is convenient to describe the process of direct interpolation using\nnotation of set theory, with the sets\n\\begin{equation}\n\\begin{split}\n \\calA_0 = \\{ E': \\Eminzero' \\le E' \\le\n \\Emaxzero' \\}, \\\\\n \\calA_1 = \\{ E': \\Eminone' \\le E' \\le\n \\Emaxone' \\}.\n \\label{def-calA01}\n\\end{split}\n\\end{equation}\nThe union of these two sets is denoted by\n\\begin{equation}\n \\calA_X = \\calA_0 \\cup \\calA_1,\n \\label{def-calA-X}\n\\end{equation}\nand the intersection is denoted by\n\\begin{equation}\n \\calA_T = \\calA_0 \\cap \\calA_1,\n \\label{def-calA-T}\n\\end{equation}\n\nThere are two obvious interpretations of direct interpolation of\nthe data in Eq.~(\\ref{EPtables}) when the outgoing energy ranges\ndiffer, $\\calA_0 \\ne \\calA_1$. One may do \\textit{direct interpolation\nwith extrapolation} or \\textit{direct interpolation\nwith truncation}. Linear-linear versions of these methods are\ndescribed here. \n\nFor direct interpolation with extrapolation the probability densities\n$\\pi(E' \\mid E_0 )$ and $\\pi(E' \\mid E_1 )$ constructed from the\ntables in Eq.~(\\ref{EPtables}) are extrapolated to\n\\begin{equation}\n \\pi_X(E' \\mid E_0 ) = \\begin{cases}\n \\pi(E' \\mid E_0 ) &\\quad \\text{for $E'$ in $\\calA_0$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_X \\setminus \\calA_0$,}\n \\end{cases}\n \\label{def-pi-X0}\n\\end{equation}\nand\n\\begin{equation}\n \\pi_X(E' \\mid E_1 ) = \\begin{cases}\n \\pi(E' \\mid E_1 ) &\\quad \\text{for $E'$ in $\\calA_1$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_X \\setminus \\calA_1$.}\n \\end{cases}\n \\label{def-pi-X1}\n\\end{equation}\nFor direct interpolation\nto incident energy $E$ with $E_0 < E < E_1$,\nthe proportionality factor $q$ is defined as\n\\begin{equation}\n q = \\frac{ E - E_0}{ E_1 - E_0 }.\n \\label{def-q}\n\\end{equation}\nIn linear-linear direct interpolation with extrapolation,\nthe interpolant is taken to be\n\\begin{equation}\n \\pi_X(E' \\mid E ) = ( 1 - q )\\, \\pi_X(E' \\mid E_0 ) + q\\, \\pi_X(E' \\mid E_1 )\n \\label{def-pi-X}\n\\end{equation}\nfor $E'$ in the set~$\\calA_X$.\n\nThe method of direct interpolation with truncation differs from\nthat using extrapolation, in that this method uses the truncated\nprobability densities\n\\begin{equation}\n \\begin{split}\n \\pi_T(E' \\mid E_0 )& = C_0 \\pi(E' \\mid E_0 ), \\\\\n \\pi_T(E' \\mid E_1 )& = C_1 \\pi(E' \\mid E_1 )\n \\end{split}\n \\label{def-pi-T01}\n\\end{equation}\nfor outgoing energy $E'$ in the set~$\\calA_T$. Here, $C_0$ \nand~$C_1$ are normalization constants such that\n$$\n \\int_{\\calA_T} dE' \\, \\pi_T(E' \\mid E_0 ) = 1\n \\quad \\text{and} \\quad\n \\int_{\\calA_T} dE' \\, \\pi_T(E' \\mid E_1 ) = 1.\n$$\nFor linear-linear direct interpolation with truncation of the data in Eq.~(\\ref{EPtables})\nto incident energy $E$ with $E_0 < E < E_1$, the factor~$q$\nis chosen as in Eq.~(\\ref{def-q}), and the interpolant is\n$$\n \\pi_T(E' \\mid E ) = ( 1 - q )\\, \\pi_T(E' \\mid E_0 ) + q\\, \\pi_T(E' \\mid E_1 )\n$$\nfor $E'$ in the set~$\\calA_T$.\n\n\\textbf{Remarks.}\nThe \\ENDFdata\\ data~\\cite{ENDFdata} contains many instances in which\nlinear-linear direct interpolation is specified, but the \\ENDF\\ manual~\\cite{ENDFB}\nsays nothing about how to deal with differences in range of outgoing\nenergies. Both versions can be expected to produce violation of\nenergy conservation. The {\\gettransfer} code\ncurrently uses direct interpolation with extrapolation.\n\n\\subsection{Unit-base interpolation}\\label{Sec:unitBase}\nOnly the linear-linear version of unit-base interpolation is discussed here.\nThe first step in unit-base interpolation is the construction of the\nrange of energies of the outgoing particle. \nThe minimum and maximum\noutgoing energies for the data in Eq.~(\\ref{EPtables})\nare given by Eq.~(\\ref{Eout-ranges}).\nFor incident energy~$E$ with $E_0 < E < E_1$, the factor~$q$\nis taken as in Eq.~(\\ref{def-q}), and\n the minimum and maximum outgoing energies\nare given by\n\\begin{equation}\n \\begin{split}\n \\Emin'& = ( 1 - q )\\Eminzero' + q \\Eminone', \\\\\n \\Emax' & = ( 1 - q )\\Emaxzero' + q \\Emaxone'.\n \\label{EoutRange}\n \\end{split}\n\\end{equation}\n\nThe interpolated probability density $\\pi(E' \\mid E)$ must\nsatisfy the normalization condition\n\\begin{equation}\n \\int_{\\Emin'}^{\\Emax'} dE' \\, \\pi(E' \\mid E) = 1.\n \\label{probabilityNorm}\n\\end{equation}\nOne way to ensure this is to first map the outgoing energy ranges\nEq.~(\\ref{Eout-ranges}) to unit base $0 \\le \\Ehat' \\le 1$\nand to scale the probability densities Eq.~(\\ref{EPtables})\naccordingly. Thus, for the data in Eq.~(\\ref{EPtables}) at\nincident energy~$E_0$, set\n\\begin{equation}\n \\Ehat' = \\frac{ E' - \\Eminzero'}{ \\Emaxzero' - \\Eminzero'}\n \\label{unit-base-map}\n\\end{equation}\nand scale the probability density\n\\begin{equation}\n \\pihat(\\Ehat' \\mid E_0) = ( \\Emaxzero' - \\Eminzero' )\\pi(E' \\mid E_0).\n \\label{unitbaseMap}\n\\end{equation}\nFor incident energy $E_1$, the outgoing energy is scaled\nas\n\\begin{equation}\n \\Ehat' = \\frac{ E' - \\Eminone'}{ \\Emaxone' - \\Eminone'},\n \\label{unit-base-map1}\n\\end{equation}\nand the probability density\nis scaled to define the unit-base probability\ndensity \n\\begin{equation}\n \\pihat(\\Ehat' \\mid E_1) = ( \\Emaxone' - \\Eminone' )\\pi(E' \\mid E_1)\n \\label{unitbaseMap1}\n\\end{equation}\nfor $0 \\le \\Ehat' \\le 1$.\n\nIf linear-linear interpolation with respect to incident energy is\ndesired, the proportionality factor $q$ defined in\nEq.~(\\ref{def-q}) is used to linearly interpolate between\n$\\pihat(\\Ehat' \\mid E_0)$ and $\\pihat(\\Ehat' \\mid E_1)$\nby setting\n\\begin{equation}\n \\pihat(\\Ehat' \\mid E) = (1 - q)\\,\\pihat(\\Ehat' \\mid E_0) +\n q \\,\\pihat(\\Ehat' \\mid E_1)\n \\label{unitbaseInterp}\n\\end{equation}\nfor $0 \\le \\Ehat' \\le 1$. \n\nFinally, in order to define the interpolated probability\ndensity $\\pi(E' \\mid E)$, invert the mappings Eq.~(\\ref{unit-base-map})\nand~Eq.~(\\ref{unitbaseMap}). Specifically, with $\\Emin'$ and\n$\\Emax'$ as in Eq.~(\\ref{EoutRange}), set\n\\begin{equation}\n E' = \\Emin' + ( \\Emax' - \\Emin')\\Ehat'\n \\label{range-inv}\n\\end{equation}\nand take\n\\begin{equation}\n \\pi(E' \\mid E) = \\frac{\\pihat(\\Ehat' \\mid E)}{ \\Emax' - \\Emin' }.\n \\label{unitbaseInvert}\n\\end{equation}\n\nUnit-base interpolation is ordinarily not used with tables of\nangular probability densities~$\\pi( \\mu \\mid E)$, because the\nrange of direction cosines is usually~$-1 \\le \\mu \\le 1$. One\nmay want to use it for a table with forward emission given in\nthe laboratory frame, however.\n\n\\subsection{Interpolation by cumulative points}\\label{Sec:cumProb}\nThe method of interpolation by cumulative points that is\nused in the code \\gettransfer\\ is proposed in~\\cite{interpolation},\nand it is a modification of interpolation by corresponding energies as described\nin the \\ENDF\\ manual~\\cite{ENDFB}. \nInterpolation by corresponding energies requires the selection of\n$N$ equiprobable energy bins, so the result depends on the value of~$N$.\nIt is shown in~\\cite{interpolation} that for data Eq.~(\\ref{EPtable}) which are\nhistograms with respect to outgoing energy $E'$, interpolation by cumulative \npoints is equivalent to interpolation by corresponding energies with~$N = \\infty$.\nThe \\gettransfer\\ code therefore uses interpolation by cumulative points\nwhenever the data specify interpolation by corresponding energies.\n\nOne objection to unit-base interpolation is that the mapping\n(\\ref{unitbaseMap}) depends only\non the range of outgoing energies.\nOne can often get a better approximation to the physics if the\ninterpolation method incorporates\nknowledge of the local behavior of each $\\pi( E' \\mid E_k)$\nin Eq.~(\\ref{EPtable}).\nOne method of doing so is based on the cumulative\nprobability function\n\\begin{equation}\n \\Pi( E' \\mid E_k) = \\int_{E'_{k,\\text{min}}}^{E'} dx\\, \\pi( x \\mid E_k)\n \\label{cumProb}\n\\end{equation}\nfor $k = 0$,~1, $\\ldots\\,$,~$K$.\n\n\n\n\nBecause the data $\\pi( E' \\mid E_k)$ consist of probability densities,\nit follows that $\\pi( E' \\mid E_k) \\ge 0$. Hence, $\\pi( E' \\mid E_k)$ \ns a non-decreasing function of~$E'$. There are\nenergy distributions in the data library\n\\xendl~\\cite{GND} for which $\\pi( E' \\mid E_k ) = 0$\non an interval $E_{j - 1,k}' < E' < E_{j,k}$ for $k = 0$~or~1\nand for some values of $j = 1$, 2, $\\ldots\\,$, $J_k$.\nThe corresponding cumulative probability is constant\non such intervals. The method of cumulative points\ndepends on solutions of the equation\n$$\n \\Pi(E' \\mid E_k ) = Y\n$$\nfor given value of $0 \\le Y \\le 1$. Denote the\nlargest solution by\n$$\n S = \\max \\left \\{ E': \\Pi(E' \\mid E_k ) = Y \\right \\}\n$$\nand the smallest solution by\n$$\n T = \\min \\left \\{ E': \\Pi(E' \\mid E_k ) = Y \\right \\}.\n$$\nIf $\\pi( E' \\mid E_k ) = 0$ only at discrete\npoints, then $\\Pi(E' \\mid E_k )$ is strictly increasing\nand $S = T$.\n\nThe cumulative points method is defined as\nfollows.\n\\begin{enumerate}\n\\item For probability density data Eqs.~\\eqref{EPtables}\nat incident energy~$E_k$ with $k = 0$ and~1, compute the cumulative\nprobabilities $\\Pi( E' \\mid E_k )$ in Eqs.~\\eqref{cumProb}\nat the data points~$E_{j,k}'$ for $j = 0$, 1, $\\ldots\\,$, $J_k$. \nDenote the result as\n\\begin{equation}\n y_{j,k} = \\Pi( E_{j,k}' \\mid E_k ).\n \\label{def-phijk}\n\\end{equation}\n\\item Form the union of these two sets\n\\begin{equation}\n \\{Y_\\ell\\} = \\{y_{j,0}\\} \\cup \\{y_{j,1}\\},\n \\label{def-Phi-cup}\n\\end{equation}\nremove duplicates, and arrange the remaining values\nin increasing order,\n\\begin{equation}\n 0 = Y_0 < Y_1 < \\cdots < Y_L = 1.\n \\label{Phi-ordered}\n\\end{equation}\n\\item For $k = 0$ and~1 and for each $Y = Y_\\ell$ with\n$\\ell = 1$, 2, $\\ldots\\,$, $L$,\ncompute the outgoing energies\n$S_{\\ell,k}'$ and~$T_{\\ell,k}'$ by solving the equations\n\\begin{equation}\n \\begin{split}\n S_{\\ell,k}' &= \\max \\left \\{ E': \\Pi(E' \\mid E_k ) = Y_{\\ell-1} \\right \\}, \\\\\n T_{\\ell,k}' &= \\min \\left \\{ E': \\Pi(E' \\mid E_k ) = Y_\\ell \\right \\}.\n \\end{split}\n \\label{T-cum-prob-inv}\n\\end{equation}\n\\item For $k = 0$ and~1 form the intervals $\\calB_\\ell(E_k)$\n\\begin{equation}\n \\begin{split}\n \\calB_\\ell(E_k) &= \\{E' : S_{\\ell,k}' \\le E' < T_{\\ell,k}' \\} \n \\quad \\text{for $\\ell = 1$,~2, $\\ldots\\,$,~$L-1$,} \\\\\n \\calB_L(E_k) &= \\{E' : S_{L,k}' \\le E' \\le T_{L,k}' \\}.\n \\end{split}\n \\label{def-B-ell}\n\\end{equation}\nThe reason for omitting right-hand endpoints of\nintervals $\\calB_\\ell(E_k)$ for $\\ell < L$ in Eqs.~\\eqref{def-B-ell}\nis that they may be points of discontinuity\nof the original energy distributions in Eq.~\\eqref{EPtables}.\n\\item For incident energy $E$ with $E_0 < E < E_1$ do unit-base\ninterpolation of $\\pi( E' \\mid E_0 )$ on $\\calB_\\ell(E_0)$ with\n$\\pi( E' \\mid E_1 )$ on $\\calB_\\ell(E_1)$\nfor $\\ell = 1$,~2, $\\ldots\\,$,~$L$. The interpolation with\nrespect to~$E$ may be histogram, linear-linear, or linear-log.\nThe linear-linear version is as follows. \nFor $k = 0$ and~1 use\n\\begin{equation}\n \\Ehat' = \\frac{ E' - S_{\\ell,k}'}{ T_{\\ell,k}' - S_{\\ell,k}'}\n \\label{unit-base-map-cum}\n\\end{equation}\nto map the interval $\\calB_\\ell(E_k)$\nto $0 \\le \\Ehat' < 1$ for $\\ell = 1$,~2, $\\ldots\\,$,~$L-1$ and to\n$0 \\le \\Ehat' \\le 1$ for $\\ell = L$. Scale the probability density\naccordingly,\n\\begin{equation}\n \\pihat(\\Ehat' \\mid E_k) = (T_{\\ell,k}' -S_{\\ell,k}' )\\pi(E' \\mid E_0).\n \\label{unit-base-scale}\n\\end{equation}\nDefine the proportionality factor\n\\begin{equation}\n \\alpha = \\frac{ E - E_0 }{ E_1 - E_0 }\n \\label{def-alpha-cum}\n\\end{equation}\nand interpolate to obtain\n\\begin{equation}\n \\pihat(\\Ehat' \\mid E) = (1 - \\alpha)\\pihat(\\Ehat' \\mid E_0) +\n \\alpha \\pihat(\\Ehat' \\mid E_1)\n \\label{unitbaseInterpCum}\n\\end{equation}\nfor $0 \\le \\Ehat' < 1$ and $\\ell < L$ and for $0 \\le \\Ehat' \\le 1$ and $\\ell = L$.\nFor $\\ell = 1$,~2, $\\ldots\\,$,~$L$\ninterpolate the end points of the intervals $\\calB_\\ell(E_k)$ to\nobtain\n\\begin{equation}\n \\begin{split}\n S_{\\ell,\\alpha}' &= (1 - \\alpha)S_{\\ell,0}' + \\alpha S_{\\ell,1}', \\\\\n T_{\\ell,\\alpha}' &= (1 - \\alpha)T_{\\ell,0}' + \\alpha T_{\\ell,1}'.\n \\end{split}\n \\label{energy-range}\n\\end{equation}\nAt incident energy~$E$ \nthe range of outgoing energies~$E'$ as taken as\n\\begin{equation}\n \\begin{split}\n \\calB_\\ell(E) &= \\{E' : S_{\\ell,\\alpha}' \\le E' < T_{\\ell,\\alpha}' \\}\n \\quad \\text{for $\\ell = 1$,~2, $\\ldots\\,$,~$L-1$,} \\\\\n \\calB_L(E) &= \\{E' : S_{L,\\alpha}' \\le E' \\le T_{L,\\alpha}' \\}.\n \\end{split}\n \\label{def-B-ell-alpha}\n\\end{equation}\nFinally, the interpolated probability\ndensity $\\pi( E' \\mid E )$ is obtained by inversion of the mapping\nEq.~\\eqref{unit-base-scale}, giving\n\\begin{equation}\n \\pi(E' \\mid E) = \\frac{\\pihat(\\Ehat' \\mid E)}{ T_{\\ell,\\alpha}' - S_{\\ell,\\alpha}'}\n \\label{map-inv}\n\\end{equation}\nfor $E'$ in $\\calB_\\ell(E)$.\nSet $\\pi(E' \\mid E) = 0$ for all outgoing energies~$E'$ which \nare not in any of the sets $\\calB_\\ell(E)$ for \n$\\ell = 1$,~2, $\\ldots\\,$,~$L$.\n\\end{enumerate}\n\nFor tables of angular probability densities~$\\pi( \\mu \\mid E)$, direct interpolation\nis often used for interpolating with respect to incident energy~$E$,\nbecause the direction cosines usually range over $-1 \\le \\mu \\le 1$.\nFor angular distributions with strong local features, interpolation by cumulative points \nmay be preferable.\n\n\n\\subsubsection{Practical considerations: intervals of zero length} \n\\label{Sec:cum-points-trivial}\nExperience with interpolation\nby cumulative points as described above shows that it must be\nmodified to deal with the inaccuracy of\ncomputer arithmetic. In particular, it may happen that \nfor $k=0$ or~1 and for some $\\ell = \\widehat \\ell$\nthe values of $S'_{ \\ell,k}$ and~$T'_{ \\ell,k}$\nas obtained from Eq.~(\\ref{T-cum-prob-inv}) are computed to be equal.\nThe mapping Eq.~(\\ref{unit-base-map-cum}) to unit base is then undefined.\n\nThis phenomenon may be understood in terms of the computer's\nmachine epsilon, which is the smallest number $\\epsilon_{\\text{mach}}$\nsuch that $1 + \\epsilon_{\\text{mach}} > 1$ in the computer's arithmetic.\nIn double precision arithmetic a common value is\n$$\n \\epsilon_{\\text{mach}} = 2^{-52} \\approx 2.22 \\times 10^{-16}.\n$$\nIn particular, if the exact values of $S_{\\widehat \\ell,k}' $ and $T_{\\widehat \\ell,k}' $\nare such that $S_{\\widehat \\ell,k}' > 0$ and\n$$\n T_{\\widehat \\ell,k}' - S_{\\widehat \\ell,k}' < \\epsilon_{\\text{mach}} S_{\\widehat \\ell,k}',\n$$\nthen the computer will say that $S_{\\widehat \\ell,k}' = T_{\\widehat \\ell,k}'$.\n\nConsider first the case that the probability density $\\pi(E' \\mid E_k )$ is\ngiven as a histogram. Then in exact arithmetic the cumulative points algorithm \nensures the existence of $S_{\\widehat \\ell,k}' $ and $T_{\\widehat \\ell,k}' $\nsuch that\n\\begin{equation}\n \\Pi(S_{\\widehat \\ell,k}' \\mid E_k ) = Y_{\\widehat \\ell-1}\n \\quad \\text{and} \\quad\n \\Pi(T_{\\widehat \\ell,k}' \\mid E_k ) = Y_{\\widehat \\ell}.\n \\label{get-S-T}\n\\end{equation}\nFurthermore, it is assured that $S_{\\widehat \\ell,k}' < T_{\\widehat \\ell,k}' $\nand that\n$$\n \\pi(E' \\mid E_k ) = \\pi_{\\widehat \\ell,k}\n \\quad \\text{for} \\quad\n S_{\\widehat \\ell,k}' \\le E' < T_{\\widehat \\ell,k}',\n$$\nwhere $\\pi_{\\widehat \\ell,k}$ is a positive constant. Thus, if it happens\nthat\n\\begin{equation}\n Y_{\\widehat \\ell} - Y_{\\widehat \\ell-1} <\n \\epsilon_{\\text{mach}} \\pi_{\\widehat \\ell,k} S_{\\widehat \\ell,k}' ,\n \\label{bad-cum-prob}\n\\end{equation}\nthe computer will conclude that $S_{\\widehat \\ell,k}' = T_{\\widehat \\ell,k}'$.\nBecause of the factor $\\pi_{\\widehat \\ell,k}$ appearing on the right-hand\nside of Eq.~{\\ref{bad-cum-prob}), this phenomenon is usually associated\nwith narrow resonances in the data. The size of $ \\pi_{\\widehat \\ell,k}$ depends\non the units used for energy, but the product $ \\pi_{\\widehat \\ell,k} S_{\\widehat \\ell,k}' $\nis dimensionless, and it is typically large at a resonance. For example, \nfor $S_{\\widehat \\ell,k}' = 1\\,\\text{MeV}$ the value $ \\pi_{\\widehat \\ell,k} \n= 1.0 \\times 10^8\\,\\text{MeV}^{-1}$ is reasonable, and Eq.~(\\ref{bad-cum-prob})\ncould easily be satisfied.\n\nThe situation for a piecewise linear probability density $\\pi(E' \\mid E_k )$ is\na little different. For one thing, the value $ \\pi_{\\widehat \\ell,k}$ in\nEq.~(\\ref{bad-cum-prob}) is replaced by the average\n$$\n \\frac{1}{2} \\left(\n \\pi( S_{\\widehat \\ell,k}' \\mid E_k ) + \\pi( T_{\\widehat \\ell,k}' \\mid E_k ) \n \\right).\n$$\nAnother difference from the histogram case is that the determination of\n$S_{\\widehat \\ell,k}' $ and $T_{\\widehat \\ell,k}' $ in Eq.~(\\ref{get-S-T}) \nrequire the solution of quadratic equations, with additional inaccuracy\nintroduced by the computer arithmetic.\n\nThis raises the question of what to do when the solution of\nEq.~(\\ref{T-cum-prob-inv}) gives values of $S_{ \\widehat \\ell,k}'$ and \n$T_{\\widehat \\ell,k}'$\nwhich are nearly equal for some $\\widehat \\ell$ and~$k$.\nThe first requirement is a definition of equality for computer arithmetic.\nIn the \\gettransfer\\ code $S'_{ \\widehat \\ell,k}$ and~$T'_{ \\widehat \\ell,k}$\nare considered to be essentially equal if\n\\begin{equation}\n \\left| \n S'_{ \\widehat \\ell,k} - T'_{ \\widehat \\ell,k}\n \\right| <\n \\frac{ \\delta_r} {2} \\left(\n S'_{ \\widehat \\ell,k} + T'_{ \\widehat \\ell,k}\n \\right),\n \\label{bad-unit-base}\n\\end{equation}\nwhere $\\delta_r$ is the parameter \\texttt{tight\\_tol}\ndescribed in Section~\\ref{Sec:floatingPoint}. \n\n The decision on what to do in \\gettransfer\\ about such near equality of\n$S'_{ \\widehat \\ell,k}$ and~$T'_{ \\widehat \\ell,k}$ is based on the size of\n$Y_{\\widehat \\ell} - Y_{\\widehat \\ell-1}$. Let $\\delta_c$\nbe the parameter \\texttt{cum\\_prob\\_skip} described in\nSection~\\ref{Sec:cum-prob-skip}. \nIf Eq~(\\ref{bad-unit-base})\nis satisfied and\n\\begin{equation}\n Y_{\\widehat \\ell} - Y_{\\widehat \\ell-1} < \\delta_c,\n \\label{drop-B-ell}\n\\end{equation}\nthen the sets $\\calB_{\\widehat \\ell}(E_k)$ with $k = 0$,~1\nin Eq.~(\\ref{def-B-ell}) are omitted from the computation of the\ntransfer matrix. This omission introduces an error of at most one part\nin~$1/\\delta_c$ in the transfer matrix.\n\nOn the other hand, if Eq~(\\ref{bad-unit-base})\nis satisfied and Eq~(\\ref{drop-B-ell}) is violated for some\n$\\widehat \\ell$ and~$k$, then in \\gettransfer\\ the interval\n$\\calB_{\\widehat \\ell}(E_k)$ is reduced to the point\n\\begin{equation}\n \\calB_{\\widehat \\ell}(E_k) = \\{E' : E' = \\widetilde S_{\\widehat \\ell,k}' \\} \n \\label{def-B-ell-point}\n\\end{equation}\nwith\n\\begin{equation}\n \\widetilde S_{\\widehat \\ell,k}' =\n \\frac{ 1} {2} \\left(\n S'_{ \\widehat \\ell,k} + T'_{ \\widehat \\ell,k}\n \\right),\n \\label{def-S-ell-point}\n\\end{equation}\nand $\\pi( E' \\mid E_k )$ on $\\calB_{\\widehat \\ell}(E_k)$ is taken\nto be a delta function\n\\begin{equation}\n \\pi( E' \\mid E_k ) = \\left(\n Y_{\\widehat \\ell} - Y_{\\widehat \\ell - 1}\n \\right)\n \\delta( E' - \\widetilde S_{\\widehat \\ell,k}' ).\n \\label{delta-data}\n\\end{equation}\nNote that in exact arithmetic the intervals $\\calB_{\\ell}( E_k )$ defined\nin Eq.~(\\ref{def-B-ell}) are disjoint. For the construction here,\nhowever, there is no guarantee that the point $\\calB_{\\widehat \\ell}(E_k)$\nobtained from Eqs.~(\\ref{def-B-ell-point}) and~(\\ref{def-S-ell-point})\nis not contained in $\\calB_{\\widehat \\ell - 1}( E_k )$ or~$\\calB_{\\widehat \\ell + 1}( E_k )$.\n\nIf it happens that the delta function Eq.~(\\ref{delta-data}) is used\nfor both $k = 0$ and $k = 1$, then for incident energy~$E$ with\n$E_0 < E < E_1$, the set $\\calB_{\\widehat \\ell}(E)$ in\nEq.~(\\ref{def-B-ell-alpha}) is defined as the single point\n$$\n \\widetilde S_{\\widehat \\ell, \\alpha}' = ( 1 - \\alpha ) \\widetilde S_{\\widehat \\ell, 0}' +\n \\alpha \\widetilde S_{\\widehat \\ell,1}'\n$$\nwith $\\alpha$ given by Eq.~(\\ref{def-alpha-cum}). The\nprobability density on $\\calB_{\\widehat \\ell}(E)$ is taken to be\nthe delta function\n$$\n \\pi( E' \\mid E ) = \\left(\n Y_{\\widehat \\ell} - Y_{\\widehat \\ell - 1}\n \\right)\n \\delta( E' - \\widetilde S_{\\widehat \\ell, \\alpha}' ).\n$$\n\nIt remains to consider the case when Eq~(\\ref{bad-unit-base})\nis satisfied and Eq~(\\ref{drop-B-ell}) is false for only one value\nof~$k$, say $k = 0$. In that case the set $ \\calB_{\\widehat \\ell}(E_0)$\nreduces to a point as in Eqs.~(\\ref{def-B-ell-point}) and~(\\ref{def-S-ell-point}), and\n$\\pi( E' \\mid E_0 )$ on $ \\calB_{\\widehat \\ell}(E_0)$ is defined to be as\nin Eq.~(\\ref{delta-data}) with $k = 0$. For the mapping of this\n$\\pi( E' \\mid E_0 )$ on $ \\calB_{\\widehat \\ell}(E_0)$ to unit base,\nEq.~(\\ref{unit-base-scale}) is replaced by in \\gettransfer\\ by\n$$\n \\pihat(\\Ehat' \\mid E_0) = Y_{\\widehat \\ell} - Y_{\\widehat \\ell - 1}\n \\quad \\text{for} \\quad\n 0 \\le \\Ehat' \\le 1.\n$$\nAt incident energy~$E$ with $E_0 < E < E_1$ the interpolation\nEq.~(\\ref{unitbaseInterpCum}) is performed using $\\alpha$\ndefined by Eq.~(\\ref{def-B-ell-alpha}). For inversion of the\nunit base map at incident energy~$E$, the range of outgoing\nenergies Eq.~(\\ref{energy-range}) is replaced by\n\\begin{equation*}\n \\begin{split}\n S_{\\ell,\\alpha}' &= (1 - \\alpha) \\widetilde S_{\\widehat \\ell, 0}' + \\alpha S_{\\ell,1}', \\\\\n T_{\\ell,\\alpha}' &= (1 - \\alpha) \\widetilde S_{\\widehat \\ell, 0}' + \\alpha T_{\\ell,1}'.\n \\end{split}\n %\\label\n\\end{equation*}\nThe probability density $\\pi( E' \\mid E )$ for $S_{\\ell,\\alpha}' \\le E' <\nT_{\\ell,\\alpha}'$ is then calculated using the scaling Eq.~(\\ref{map-inv}).\n\n\n\\section{Unscaled interpolation of Kalbach-Mann data}\\label{Sec:Kalbach-r-interp}\nThe above discussion pertains to the interpolation of tables of probability densities,\nfor which maintenance of the norm condition Eq.~(\\ref{probabilityNorm})\nis essential. The parameter~$r(\\Ecm', E)$ in Eq.~(\\ref{Kalbach-eta}) for the\nKalbach-Mann model of double-differential data is given as tables\ndepending on the energy $E$ of the incident particle and the energy $\\Ecm'$ of\nthe outgoing particle in the center-of-mass frame, and it has the different\nconstraint,\n\\begin{equation}\n 0 \\le r \\le 1.\n \\label{Kalbach-r-constraint}\n\\end{equation}\n\nAgain, it suffices to describe interpolation between Kalbach-Mann $r$\ndata between tables at incident energies $E_0$ and~$E_1$ with\n$E_0 < E_1$. As in Eq.~(\\ref{def-calA01}),\nconsider the sets $\\calA_0$ of outgoing energies\nat $E = E_0$ and $\\calA_1$ at $E = E_1$. For unscaled direct interpolation\nwith extrapolation, take $\\calA_X = \\calA_0 \\cup \\calA_1$ as in\nEq.~(\\ref{def-calA-X}), so that the extrapolated $r$ parameter is\n\\begin{equation*}\n r_X(E', E_0 ) = \\begin{cases}\n r(E', E_0 ) &\\quad \\text{for $E'$ in $\\calA_0$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_X \\setminus \\calA_0$,}\n \\end{cases}\n% \\label{def-r-X0}\n\\end{equation*}\nand\n\\begin{equation*}\n r_X(E', E_1 ) = \\begin{cases}\n r(E', E_1 ) &\\quad \\text{for $E'$ in $\\calA_1$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_X \\setminus \\calA_1$.}\n \\end{cases}\n% \\label{def-r-X1}\n\\end{equation*}\nThen for $E_0 < E < E_1$, for $q$ as in Eq.~(\\ref{def-q}) and for\n$E'$ in the set~$\\calA_X$, the linear-linear form of unscaled direct interpolation\nwith extrapolation becomes as in Eq.~(\\ref{def-pi-X}),\n\\begin{equation}\n r_X(E', E ) = ( 1 - q )\\, r_X(E', E_0 ) + q\\, r_X(E', E_1 ).\n \\label{r-direct-extrapolation}\n\\end{equation}\nThe extrapolation version of direct interpolation of the Kallbach-Mann\n$r$ parameter as in Eq.~(\\ref{r-direct-extrapolation}) is implemented in\nthe \\gettransfer\\ code.\n\nFor unscaled direct interpolation of the Kalbach-Mann $r$ parameter with\ntruncation, the outgoing energy $E'$ is restricted to the common \ndomain $\\calA_T = \\calA_0 \\cap \\calA_1$, and there is no change\nof scale analogous to that used for probability densities in\nEq.~(\\ref{def-pi-T01}). Thus, the truncated Kalbach-Mann $r$ parameters\nfor incident energies $E_0$ and~$E_1$ are\n\\begin{equation*}\n r_T(E', E_0 ) = \\begin{cases}\n r(E', E_0 ) &\\quad \\text{for $E'$ in $\\calA_T$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_0 \\setminus \\calA_T$,}\n \\end{cases}\n% \\label{def-r-T0}\n\\end{equation*}\nand\n\\begin{equation*}\n r_X(E', E_1 ) = \\begin{cases}\n r(E', E_1 ) &\\quad \\text{for $E'$ in $\\calA_T$,} \\\\\n 0 &\\quad \\text{for $E'$ in $\\calA_1 \\setminus \\calA_T$.}\n \\end{cases}\n% \\label{def-r-T1}\n\\end{equation*}\nThe linear-linear version of unscaled direct\ninterpolation with truncation is\n\\begin{equation}\n r_T(E', E ) = ( 1 - q )\\, r_T(E', E_0 ) + q\\, r_T(E', E_1 )\n \\label{r-direct-truncation}\n\\end{equation}\nwith $E'$ restricted to~$\\calA_T$.\nThe \\gettransfer\\ code does not currently implement unscaled direct\ninterpolation with truncation of the Kalbach-Mann $r$ parameter\ngiven in Eq.~(\\ref{r-direct-truncation}).\n\nThere is also an unscaled version of unit-base interpolation with\nEqs.~(\\ref{unitbaseMap}) and~(\\ref{unitbaseMap1}) replaced by\n\\begin{equation*}\n \\begin{split}\n \\widehat r(\\Ehat', E_0) &= r(E', E_0), \\\\\n \\widehat r(\\Ehat', E_1) &= r(E', E_1), \\\\\n \\end{split}\n\\end{equation*}\nfor $0 \\le \\Ehat' \\le 1$ with $\\Ehat'$ as in Eq.~(\\ref{unit-base-map-cum})\nfor $E = E_0$ and as in Eq.~(\\ref{unit-base-map1})\nfor $E = E_1$. For linear-linear unscaled unit-base interpolation\nto incident energy $E$ with $E_0 < E < E_1$, interpolate the\nminimal and maximal outgoing energies as in Eq.~(\\ref{EoutRange}),\ninterpolate $\\widehat r$ using\n$$\n \\widehat r(E', E ) = ( 1 - q )\\, \\widehat r(E', E_0 ) + q\\, \\widehat r(E', E_1 ),\n$$\nand invert the unit-base map using Eq.~(\\ref{range-inv})\nand \n$$\n r(E', E) = \\widehat r(\\Ehat', E)\n$$\nfor $\\Emin' \\le E' \\le \\Emax'$.\n\nWhen the energy probability density $\\pi_E( E' \\mid E )$ in Eq.~(\\ref{Kalbach-prob})\nis interpolated using the method of cumulative points, the interpolated values of\n$r(E', E)$ in Eq.~(\\ref{Kalbach-eta}) are obtained using the method of\nunscaled cumulative points defined as follows. Because the data for\n $r( E' , E )$ and $\\pi_E( E' \\mid E )$ are given at the same energy points\n $E$ and~$E'$, it is natural to use for $r( E' , E_0 )$ and $r( E' , E_1 )$ the\n intervals $\\calB_{ \\ell}(E_k )$, k = 0,~1, constructed for $\\pi_E( E' \\mid E_k )$\nin Eq.~(\\ref{def-B-ell}). The \\gettransfer\\ code uses unscaled unit-base interpolation \nbetween $r( E', E_0)$ on $\\calB_{\\ell}(E_0 )$ and $r( E', E_0)$ on $\\calB_{ \\ell}(E_1 )$\nin sequence for $\\ell = 1,$ 2, $\\ldots\\,$, $L$. This method is applicable\neven when a set $\\calB_{ \\ell}(E_k )$ reduces to a point as discussed in\nSection~\\ref{Sec:cum-points-trivial}, as here is no problem of possible division by zero.\n\n\n", "meta": {"hexsha": "adf3d1e208a9194100a66aaab8ea693cec6f537b", "size": 34259, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Merced/Doc/interpolate.tex", "max_stars_repo_name": "brown170/fudge", "max_stars_repo_head_hexsha": "4f818b0e0b0de52bc127dd77285b20ce3568c97a", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 14, "max_stars_repo_stars_event_min_datetime": "2019-08-29T23:46:24.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T10:16:25.000Z", "max_issues_repo_path": "Merced/Doc/interpolate.tex", "max_issues_repo_name": "brown170/fudge", "max_issues_repo_head_hexsha": "4f818b0e0b0de52bc127dd77285b20ce3568c97a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-08-04T16:14:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-01T01:54:34.000Z", "max_forks_repo_path": "Merced/Doc/interpolate.tex", "max_forks_repo_name": "brown170/fudge", "max_forks_repo_head_hexsha": "4f818b0e0b0de52bc127dd77285b20ce3568c97a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2022-03-03T22:41:41.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-03T22:54:43.000Z", "avg_line_length": 39.8360465116, "max_line_length": 94, "alphanum_fraction": 0.6746840246, "num_tokens": 11922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059414036511, "lm_q2_score": 0.8056321796478255, "lm_q1q2_score": 0.638951668264664}} {"text": "\\documentclass{amsart}\n\\usepackage{style/preamble}\n\\usepackage{parskip}\n\\begin{document}\n \\section{Complex analysis theorems}\n Let $U$ be an open subset of $\\bbc$.\n Let $f: U \\rightarrow \\bbc$ be a continuous function.\n\n $f$ is called (once) \\emph{complex differentiable} at $a \\in U$ if the limit\n \\begin{align*}\n f'(z) := \\lim_{h \\rightarrow 0} \\dfrac{f(z + h) - f(z)}{h}\n \\end{align*}\n exists for all $z$ in some neighborhood of $a$.\n\n \\begin{theorem}[Taylor series]\n If $f$ is once differentiable at $a$, then $f$ is infinitely differentiable at $a$ and $f$ has a Taylor series converging to it in a neighborhood of $a$.\n \\begin{align*}\n f(z) = f(a) + f'(a){(z-a)} + f''(a)\\dfrac{(z-a)^2}{2!} + \\dots + f^{(n)}(a)\\dfrac{(z-a)^n}{n!} + \\dots\n \\end{align*}\n \\end{theorem}\n\n \\begin{theorem}[Open mapping theorem]\n If $f$ is complex differentiable then for any open set $V \\subseteq U$, the set $f(V)$ is an open subset of $\\bbc$.\n \\end{theorem}\n\n \\begin{theorem}[Isolated zeroes]\n If $f$ is complex differentiable then the set of zeroes of $f$ are isolated i.e. if $f(a) = 0$ for some $a \\in U$ then there exists a neighborhood $V$ of $a$ such that $a$ is the only zero of $f$ in $V$.\n \\end{theorem}\n\n\\vspace{1cm}\n\n A complex differentiable function $g: \\bbc \\rightarrow \\bbc$ is called an \\emph{entire} function.\n \\begin{theorem}[Liouville's theorem]\n Let $g$ be an entire function.\n If there exists a real number $M$ such that $|g(z)| < M$ for all $z \\in \\bbc$ then $g(z)$ is a constant function.\n \\end{theorem}\n\n \\begin{theorem}[Little Picard's theorem]\n Let $g$ be an entire function.\n If $g$ is not a constant function then the image of $g$ is either the whole complex plane or the plane minus a single point.\n \\end{theorem}\n\\end{document}\n", "meta": {"hexsha": "e71423eb74d08e0c89d8f652d8b21c9593029261", "size": 1801, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ComplexAnalysisTheorems.tex", "max_stars_repo_name": "apurvnakade/mc2019-Riemann-surfaces", "max_stars_repo_head_hexsha": "edebdd1c81027b9cedb264eba1eeaa09dd4e311f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ComplexAnalysisTheorems.tex", "max_issues_repo_name": "apurvnakade/mc2019-Riemann-surfaces", "max_issues_repo_head_hexsha": "edebdd1c81027b9cedb264eba1eeaa09dd4e311f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ComplexAnalysisTheorems.tex", "max_forks_repo_name": "apurvnakade/mc2019-Riemann-surfaces", "max_forks_repo_head_hexsha": "edebdd1c81027b9cedb264eba1eeaa09dd4e311f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.8837209302, "max_line_length": 207, "alphanum_fraction": 0.6563020544, "num_tokens": 594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303285397349, "lm_q2_score": 0.8080672135527631, "lm_q1q2_score": 0.638801639812054}} {"text": "\\chapter{Governing Equations}\n\\label{cha:governing:equations}\n\nWe present here a brief derivation of the equations for both quasi-static\nand dynamic computations. Since the general equations are the same\n(except for the absence of inertial terms in the quasi-static case),\nwe first derive these equations. We then present solution methods\nfor each specific case. In all of our derivations, we use the notation\ndescribed in Table \\vref{tab:notation} for both index\nand vector notation. When using index notation, we use the common\nconvention that repeated indices indicate summation over the range\nof the index.\n\n\\begin{table}[htbp]\n \\caption{Mathematical notation}\n \\label{tab:notation}\n \\begin{tabular}{ccp{3in}}\n \\multicolumn{2}{c}{{\\bf Symbol}} & {\\bf Description} \\\\\n {\\bf Index notation} & {\\bf Vector Notation} & \\\\\n \\hline \n $a_{i}$ & \\raisebox{12pt}{}$\\overrightarrow{a}$ & Vector field a \\\\\n $a_{ij}$ & $\\underline{a}$ & Second order tensor field a \\\\\n $u_{i}$ & $\\overrightarrow{u}$ & Displacement vector field \\\\\n $d_{i}$ & $\\vec{{d}}$ & Fault slip vector field \\\\\n $f_{i}$ & $\\overrightarrow{f}$ & Body force vector field \\\\\n $T_{i}$ & $\\overrightarrow{T}$ & Traction vector field \\\\\n $\\sigma_{ij}$ & $\\underline{\\sigma}$ & Stress tensor field \\\\\n $n_{i}$ & $\\overrightarrow{n}$ & Normal vector field \\\\\n $\\rho$ & $\\rho$ & Mass density scalar field \\\\\n \\hline \n \\end{tabular}\n\\end{table}\n\n\n\\section{Derivation of Elasticity Equation}\n\n\\subsection{Index Notation}\n\nConsider volume $V$ bounded by surface $S$. Applying a Lagrangian\ndescription of the conservation of momentum gives\n\\begin{equation}\n\\label{eqn:momentum:index}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial u_{i}}{\\partial t}\\, dV=\\int_{V}f_{i}\\, dV+\\int_{S}T_{i}\\, dS.\n\\end{equation}\nThe traction vector field is related to the stress tensor through\n\\begin{equation}\nT_{i}=\\sigma_{ij}n_{j},\n\\end{equation}\nwhere $n_{j}$ is the vector normal to $S$. Substituting into equation\n\\vref{eqn:momentum:index} yields\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial u_{i}}{\\partial t}\\, dV=\\int_{V}f_{i}\\, dV+\\int_{S}\\sigma_{ij}n_{j}\\, dS.\n\\end{equation}\nApplying the divergence theorem,\n\\begin{equation}\n\\int_{V}a_{i,j}\\: dV=\\int_{S}a_{j}n_{j}\\: dS,\n\\end{equation}\nto the surface integral results in\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial u_{i}}{\\partial t}\\, dV=\\int_{V}f_{i}\\, dV+\\int_{V}\\sigma_{ij,j}\\, dV,\n\\end{equation}\nwhich we can rewrite as\n\\begin{equation}\n\\int_{V}\\left(\\rho\\frac{\\partial^{2}u_{i}}{\\partial t^{2}}-f_{i}-\\sigma_{ij,j}\\right)\\, dV=0.\n\\end{equation}\nBecause the volume $V$ is arbitrary, the integrand must be zero at\nevery location in the volume, so that we end up with\n\\begin{gather}\n\\rho\\frac{\\partial^{2}u_{i}}{\\partial t^{2}}-f_{i}-\\sigma_{ij,j}=0\\text{ in }V,\\\\\n\\sigma_{ij}n_{j}=T_{i}\\text{ on }S_{T}\\text{,}\\\\\nu_{i}=u_{i}^{o}\\text{ on }S_{u}\\text{, and}\\\\\nR_{ki}(u_{i}^{+}-u_{i}^{-})=d_{k}\\text{ on }S_{f}.\n\\end{gather}\nWe specify tractions, $T_{i}$, on surface $S_{f}$, displacements,\n$u_{i}^{o}$, on surface $S_{u}$, and slip, $d_{k}$, on fault surface\n$S_{f}$ (we will consider the case of fault constitutive models in\nSection \\vref{sec:fault}). The rotation matrix $R_{ki}$ transforms\nvectors from the global coordinate system to the fault coordinate\nsystem. Note that since both $T_{i}$ and $u_{i}$ are vector quantities,\nthere can be some spatial overlap of the surfaces $S_{T}$ and $S_{u}$;\nhowever, the same degree of freedom cannot simultaneously have both\ntypes of boundary conditions.\n\n\n\\subsection{Vector Notation}\n\nConsider volume $V$ bounded by surface $S$. Applying a Lagrangian\ndescription of the conservation of momentum gives\n\\begin{equation}\n\\label{eqn:momentum:vec}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial\\vec{u}}{\\partial t}\\, dV=\\int_{V}\\overrightarrow{f}\\, dV+\\int_{S}\\overrightarrow{T}\\, dS.\n\\end{equation}\nThe traction vector field is related to the stress tensor through\n\\begin{equation}\n\\overrightarrow{T}=\\underline{\\sigma}\\cdot\\overrightarrow{n},\n\\end{equation}\nwhere $\\overrightarrow{n}$ is the vector normal to $S$. Substituting\ninto equation \\vref{eqn:momentum:vec} yields\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial\\overrightarrow{u}}{\\partial t}\\, dV=\\int_{V}\\overrightarrow{f}\\, dV+\\int_{S}\\underline{\\sigma}\\cdot\\overrightarrow{n}\\, dS.\n\\end{equation}\nApplying the divergence theorem,\n\\begin{equation}\n\\int_{V}\\nabla\\cdot\\overrightarrow{a}\\: dV=\\int_{S}\\overrightarrow{a}\\cdot\\overrightarrow{n}\\: dS,\n\\end{equation}\nto the surface integral results in\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\int_{V}\\rho\\frac{\\partial\\overrightarrow{u}}{\\partial t}\\, dV=\\int_{V}\\overrightarrow{f}\\, dV+\\int_{V}\\nabla\\cdot\\underline{\\sigma}\\, dV,\n\\end{equation}\nwhich we can rewrite as\n\\begin{equation}\n\\int_{V}\\left(\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}-\\overrightarrow{f}-\\nabla\\cdot\\overrightarrow{\\sigma}\\right)\\, dV=\\vec{0}.\n\\end{equation}\nBecause the volume $V$ is arbitrary, the integrand must be the zero\nvector at every location in the volume, so that we end up with\n\\begin{gather}\n\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}-\\overrightarrow{f}-\\nabla\\cdot\\overrightarrow{\\sigma}=\\vec{0}\\text{ in }V,\\\\\n\\underline{\\sigma}\\cdot\\overrightarrow{n}=\\overrightarrow{T}\\text{ on }S_{T}\\text{,}\\\\\n\\overrightarrow{u}=\\overrightarrow{u^{o}}\\text{ on }S_{u},\\text{ and}\\\\\n\\underbar{R}\\cdot(\\vec{u^{+}}-\\vec{u^{-}})=\\vec{d}\\text{ on }S_{f}.\n\\end{gather}\nWe specify tractions, $\\vec{T}$, on surface $S_{f}$, displacements,\n$\\overrightarrow{u^{o}}$, on surface $S_{u}$, and slip, $\\vec{d}$,\non fault surface $S_{f}$ (we will consider the case of fault constitutive\nmodels in Section \\vref{sec:fault}). The rotation matrix $\\underline{R}$\ntransforms vectors from the global coordinate system to the fault\ncoordinate system. Note that since both $\\overrightarrow{T}$ and\n$\\overrightarrow{u}$ are vector quantities, there can be some spatial\noverlap of the surfaces $S_{T}$ and $S_{u}$; however, the same degree\nof freedom cannot simultaneously have both types of boundary conditions.\n\n\\section{Finite-Element Formulation of Elasticity Equation}\n\nWe formulate a set of algebraic equations using Galerkin's method.\nWe consider (1) a trial solution, $\\vec{u}$, that is a piecewise\ndifferentiable vector field and satisfies the Dirichlet boundary conditions\non $S_{u}$, and (2) a weighting function, $\\vec{\\phi}$, that is\na piecewise differentiable vector field and is zero on $S_{u}$.\n\n\\subsection{Index Notation}\n\nWe start with the wave equation (strong form),\n\\begin{gather}\n\\sigma_{ij,j}+f_{i}=\\rho\\ddot{u_{i}}\\text{ in }V,\\\\\n\\sigma_{ij}n_{j}=T_{i}\\text{ on }S_{T},\\\\\nu_{i}=u_{i}^{o}\\text{ on }S_{u},\\\\\nR_{ki}(u_{i}^{+}-u_{i}^{-})=d_{k}\\text{ on }S_{f},\\text{ and}\\\\\n\\sigma_{ij}=\\sigma_{ji}\\text{ (symmetric).}\n\\end{gather}\nWe construct the weak form by computing the dot product of the wave\nequation and weighting function and setting the integral over the\ndomain to zero:\n\\begin{gather}\n\\int_{V}\\left(\\sigma_{ij,j}+f_{i}-\\rho\\ddot{u}_{i}\\right)\\phi_{i}\\, dV=0\\text{, or }\\\\\n\\int_{V}\\sigma_{ij,j}\\phi_{i}\\: dV+\\int_{V}f_{i}\\phi_{i}\\: dV-\\int_{V}\\rho\\ddot{u}_{i}\\phi_{i}\\: dV=0.\n\\end{gather}\n Consider the divergence theorem applied to the dot product of the\nstress tensor and the weighting function, $\\sigma_{ij}\\phi_{i}$,\n\\begin{equation}\n\\int_{V}(\\sigma_{ij}\\phi_{i})_{,j}\\, dV=\\int_{S}(\\sigma_{ij}\\phi_{i})n_{i}\\, dS.\n\\end{equation}\nExpanding the left-hand side yields\n\\begin{gather}\n\\int_{V}\\sigma_{ij,j}\\phi_{i}\\: dV+\\int_{V}\\sigma_{ij}\\phi_{i,j}\\: dV=\\int_{S}\\sigma_{ij}\\phi_{i}n_{i}\\: dS,\\text{ or}\\\\\n\\int_{V}\\sigma_{ij,j}\\phi_{i}\\: dV=-\\int_{V}\\sigma_{ij}\\phi_{i,j}\\, dV+\\int_{S}\\sigma_{ij}\\phi_{i}n_{i}\\, dS.\n\\end{gather}\nSubstituting into the weak form gives\n\\begin{equation}\n-\\int_{V}\\sigma_{ij}\\phi_{i,j}\\, dV+\\int_{S}\\sigma_{ij}\\phi_{i}n_{i}\\, dS+\\int_{V}f_{i}\\phi_{i}\\, dV-\\int_{V}\\rho\\ddot{u}_{i}\\phi_{i}\\, dV=0.\n\\end{equation}\nTurning our attention to the second term, we separate the integration\nover $S$ into integration over $S_{T}$ and $S_{u}$ (we will consider\ntractions over the fault surface, $S_{f}$, associated with the fault\nconstitutive model in Section \\vref{sec:fault}),\n\\begin{equation}\n-\\int_{V}\\sigma_{ij}\\phi_{i,j}\\, dV+\\int_{S_{T}}\\sigma_{ij}\\phi_{i}n_{i}\\, dS+\\int_{S_{u}}\\sigma_{ij}\\phi_{i}n_{i}\\, dS+\\int_{V}f_{i}\\phi_{i}\\, dV-\\int_{V}\\rho\\ddot{u}_{i}\\phi_{i}\\, dV=0,\n\\end{equation}\nand recognize that\n\\begin{gather}\n\\sigma_{ij}n_{i}=T_{i}\\text{ on }S_{T}\\text{ and}\\\\\n\\phi_{i}=0\\text{ on }S_{u},\n\\end{gather}\nso that the equation reduces to\n\\begin{equation}\n\\label{eq:elasticity:integral}\n-\\int_{V}\\sigma_{ij}\\phi_{i,j}\\: dV+\\int_{S_{T}}T_{i}\\phi_{i}\\, dS+\\int_{V}f_{i}\\phi_{i}\\, dV-\\int_{V}\\rho\\ddot{u}_{i}\\phi_{i}\\, dV=0.\n\\end{equation}\nWe express the trial solution and weighting function as linear combinations\nof basis functions,\n\\begin{gather}\nu_{i}=\\sum_{m}a_{i}^{m}N^{m},\\\\\n\\phi_{i}=\\sum_{n}c_{i}^{n}N^{n}.\n\\end{gather}\nNote that because the trial solution satisfies the Dirichlet boundary\ncondition, the number of basis functions for $u$ is generally greater\nthan the number of basis functions for $\\phi$, i.e., $m>n$. Substituting\nin the expressions for the trial solution and weighting function yields\n\\begin{gather}\n-\\int_{V}\\sigma_{ij}\\sum_{n}c_{i}^{n}N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}\\sum_{n}c_{i}^{n}N^{n}\\, dS+\\int_{V}f_{i}\\sum_{n}c_{i}^{n}N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\ddot{a}_{i}^{m}N^{m}\\sum_{n}c_{i}^{n}N^{n}\\ dV=0,\\text{ or}\\\\\n\\sum_{n}c_{i}^{n}(-\\int_{V}\\sigma_{ij}N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}N^{n}\\, dS+\\int_{V}f_{i}N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\ddot{a}_{i}^{m}N^{m}N^{n}\\ dV)=0.\n\\end{gather}\n Because the weighting function is arbitrary, this equation must hold\nfor all $c_{i}^{n}$, so that the quantity in parenthesis is zero\nfor each $c_{i}^{n}$\n\\begin{equation}\n\\label{eq:elasticity:integral:discretized}\n-\\int_{V}\\sigma_{ij}N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}N^{n}\\, dS+\\int_{V}f_{i}N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\ddot{a}_{i}^{m}N^{m}N^{n}\\ dV=\\vec{0}.\n\\end{equation}\nWe want to solve this equation for the unknown coefficients $a_{i}^{m}$\nsubject to\n\\begin{gather}\nu_{i}=u_{i}^{o}\\text{ on }S_{u},\\text{ and}\\\\\nR_{ki}(u_{i}^{+}-u_{i}^{-})=d_{k}\\text{ on }S_{f},\n\\end{gather}\n\n\n\\subsection{Vector Notation}\n\nWe start with the wave equation (strong form),\n\\begin{gather}\n\\nabla\\cdot\\underline{\\sigma}+\\overrightarrow{f}=\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\text{ in }V,\\\\\n\\underline{\\sigma}\\cdot\\overrightarrow{n}=\\overrightarrow{T}\\text{ on }S_{T},\\\\\n\\overrightarrow{u}=\\overrightarrow{u^{o}}\\text{ on }S_{u},\\\\\n\\underbar{R}\\cdot(\\overrightarrow{u^{+}}-\\overrightarrow{u^{-}})=\\vec{d}\\text{ on }S_{f}\\\\\n\\underline{\\sigma}=\\underline{\\sigma}^{T}\\text{ (symmetric).}\n\\end{gather}\nWe construct the weak form by multiplying the wave equation by a weighting\nfunction and setting the integral over the domain to zero. The weighting\nfunction is a piecewise differential vector field, $\\overrightarrow{\\phi}$,\nwhere $\\overrightarrow{\\phi}=0$ on $S_{u}.$ Hence our weak form\nis\n\\begin{gather}\n\\int_{V}\\left(\\nabla\\cdot\\underline{\\sigma}+\\overrightarrow{f}-\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\right)\\cdot\\overrightarrow{\\phi}\\, dV=0\\text{, or }\\\\\n\\int_{V}(\\nabla\\cdot\\underline{\\sigma})\\cdot\\overrightarrow{\\phi}\\: dV+\\int_{V}\\overrightarrow{f}\\cdot\\overrightarrow{\\phi}\\: dV-\\int_{V}\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\: dV=0.\n\\end{gather}\n Consider the divergence theorem applied to the dot product of the\nstress tensor and the trial function, $\\underline{\\sigma}\\cdot\\overrightarrow{\\phi}$,\n\\begin{equation}\n\\int_{V}\\nabla\\cdot(\\underline{\\sigma}\\cdot\\overrightarrow{\\phi})\\, dV=\\int_{S}(\\underline{\\sigma}\\cdot\\overrightarrow{\\phi})\\cdot\\overrightarrow{n}\\, dS.\n\\end{equation}\nExpanding the left-hand side yields\n\\begin{equation}\n\\int_{V}(\\nabla\\cdot\\underline{\\sigma})\\cdot\\overrightarrow{\\phi}\\: dV+\\int_{V}\\underline{\\sigma}:\\nabla\\overrightarrow{\\phi}\\: dV=\\int_{S}(\\underline{\\sigma}\\cdot\\overrightarrow{\\phi})\\cdot\\overrightarrow{n}\\: dS,\\text{ or}\n\\end{equation}\n\\begin{equation}\n\\int_{V}(\\nabla\\cdot\\underline{\\sigma})\\cdot\\overrightarrow{\\phi}\\: dV=-\\int_{V}\\underline{\\sigma}:\\nabla\\overrightarrow{\\phi}\\, dV+\\int_{S}\\underline{\\sigma}\\cdot\\overrightarrow{n}\\cdot\\overrightarrow{\\phi}\\, dS.\n\\end{equation}\nSubstituting into the weak form gives\n\\begin{equation}\n-\\int_{V}\\underline{\\sigma}:\\nabla\\overrightarrow{\\phi}\\, dV+\\int_{S}\\underline{\\sigma}\\cdot\\overrightarrow{n}\\cdot\\overrightarrow{\\phi}\\, dS+\\int_{V}\\overrightarrow{f}\\cdot\\overrightarrow{\\phi}\\, dV-\\int_{V}\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\, dV=0.\n\\end{equation}\nWe separate the integration over $S$ into integration over $S_{T}$\nand $S_{u}$,\n\\begin{multline}\n-\\int_{V}\\underline{\\sigma}:\\nabla\\overrightarrow{\\phi}\\, dV+\\int_{S_{T}}\\underline{\\sigma}\\cdot\\overrightarrow{n}\\cdot\\overrightarrow{\\phi}\\, dS+\\int_{S_{u}}\\underline{\\sigma}\\cdot\\overrightarrow{n}\\cdot\\overrightarrow{\\phi}\\, dS+\\int_{V}\\overrightarrow{f}\\cdot\\overrightarrow{\\phi}\\, dV-\\int_{V}\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\, dV=0,\n\\end{multline}\nand recognize that\n\\begin{gather}\n\\underline{\\sigma}\\cdot\\overrightarrow{n}=\\overrightarrow{T}\\text{ on }S_{T}\\text{ and}\\\\\n\\overrightarrow{\\phi}=0\\text{ on }S_{u},\n\\end{gather}\nso that the equation reduces to\n\\begin{equation}\n-\\int_{V}\\underline{\\sigma}:\\nabla\\overrightarrow{\\phi}\\: dV+\\int_{S_{T}}\\overrightarrow{T}\\cdot\\overrightarrow{\\phi}\\, dS+\\int_{V}\\overrightarrow{f}\\cdot\\overrightarrow{\\phi}\\, dV-\\int_{V}\\rho\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\, dV=0.\n\\end{equation}\nWe express the trial solution and weighting function as linear combinations\nof basis functions,\n\\begin{gather}\n\\vec{u}=\\sum_{m}\\overrightarrow{a^{m}}N^{m},\\\\\n\\vec{\\phi}=\\sum_{n}\\overrightarrow{c^{n}}N^{n}.\n\\end{gather}\nNote that because the weighting function is zero on $S_{u}$, the\nnumber of basis functions for $\\vec{u}$ is generally greater than\nthe number of basis functions for $\\vec{\\phi}$, i.e., $m>n$. Substituting\nin the expressions for the trial solution and weighting function yields\n\\begin{multline}\n-\\int_{V}\\underline{\\sigma}:\\sum_{n}\\overrightarrow{c^{n}}\\nabla N_{,}^{n}\\, dV+\\int_{S_{T}}\\vec{T}\\cdot\\sum_{n}\\overrightarrow{c^{n}}N^{n}\\, dS+\\int_{V}\\vec{f}\\cdot\\sum_{n}\\overrightarrow{c^{n}}N^{n}\\, dV\\\\\n-\\int_{V}\\rho\\sum_{m}\\frac{\\partial^{2}\\overrightarrow{a^{m}}}{\\partial t^{2}}N^{m}\\cdot\\sum_{n}\\overrightarrow{c^{n}}N^{n}\\ dV=0.\n\\end{multline}\n Because the weighting function is arbitrary, this equation must hold\nfor all $\\overrightarrow{c^{n}}$, so that\n\\begin{equation}\n-\\int_{V}\\underline{\\sigma}:\\nabla N^{n}\\, dV+\\int_{S_{T}}\\vec{T}N^{n}\\, dS+\\int_{V}\\vec{f}N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\frac{\\partial^{2}\\overrightarrow{a^{m}}}{\\partial t^{2}}N^{m}N^{n}\\, dV=\\vec{0}.\n\\end{equation}\nWe want to solve this equation for the unknown coefficients $\\overrightarrow{a^{m}}$\nsubject to\n\\begin{gather}\n\\vec{u}=u^{o}\\overrightarrow{}\\text{ on }S_{u},\\text{ and}\\\\\n\\underline{R}(\\overrightarrow{u^{+}}-\\overrightarrow{u^{-}})=\\vec{d}\\text{ on }S_{f},\n\\end{gather}\n\n\\section{Solution Method for Quasi-Static Problems}\n\nFor brevity we outline the solution method for quasi-static problems\nusing only index notation. In quasi-static problems we neglect the\ninertial terms, so equation \\eqref{eq:elasticity:integral:discretized}\nreduces to\n\\begin{equation}\n-\\int_{V}\\sigma_{ij}N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}N^{n}\\, dS+\\int_{V}f_{i}N^{n}\\, dV=\\vec{0}.\n\\end{equation}\nAs a result, time-dependence only enters through the constitutive\nrelationships and the loading conditions. We consider the deformation\nat time $t+\\Delta t$,\n\\begin{equation}\n\\label{eq:elasticity:integral:quasistatic}\n-\\int_{V}\\sigma_{ij}(t+\\Delta t)N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}(t+\\Delta t)N^{n}\\, dS+\\int_{V}f_{i}(t+\\Delta t)N^{n}\\, dV=\\vec{0}.\n\\end{equation}\nWe solve this equation through formulation of a linear algebraic system\nof equations ($Au=b$), involving the residual ($r=b-Au$) and Jacobian\n($A$). The residual is simply\n\\begin{equation}\nr_{i}^{n}=-\\int_{V}\\sigma_{ij}(t+\\Delta t)N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}(t+\\Delta t)N^{n}\\, dS+\\int_{V}f_{i}(t+\\Delta t)N^{n}\\, dV.\n\\end{equation}\nWe employ numerical quadrature in the finite-element discretization\nand replace the integrals with sums over the cells and quadrature\npoints,\n\\begin{multline}\nr_{i}^{n}=-\\sum_{\\text{vol cells}}\\sum_{\\text{quad pts}}\\sigma_{ij}(x_{q},t+\\Delta t)N_{,j}^{n}(x_{q})\\: w_{q}|J_{cell}(x_{q})|+\\sum_{\\text{vol cells}}\\sum_{\\text{quad pt}s}f_{i}(x_{q},t+\\Delta t)N^{n}(x_{q})\\, w_{q}|J_{cell}(x_{q})|\\\\\n+\\sum_{\\text{tract cells}}\\sum_{\\text{quad pts}}T_{i}(x_{q},t+\\Delta t)N^{n}(x_{q})\\, w_{q}|J_{cell}(x_{q})|,\n\\end{multline}\nwhere $r_{i}^{n}$ is an $nd$ vector ($d$ is the dimension of the\nvector space) and $i$ is a vector space component, $x_{q}$ are the\ncoordinates of the quadrature points, $w_{q}$ are the weights of\nthe quadrature points, and $|J_{cell}(x_{q})|$ is the determinant\nof the Jacobian matrix evaluated at the quadrature points associated\nwith mapping the reference cell to the actual cell. The quadrature\nscheme for the integral over the tractions is one dimension lower\nthan the one used in integrating the terms for the volume cells.\n\nIn order to find the Jacobian of the system, we let\n\\begin{equation}\n\\sigma_{ij}(t+\\Delta t)=\\sigma_{ij}(t)+d\\sigma_{ij}(t).\n\\end{equation}\nIsolating the term associated with the increment in stresses yields\n\n\\begin{equation}\n\\int_{V}d\\sigma_{ij}(t)N_{j}^{n}\\ dV=-\\int_{V}\\sigma_{ij}(t)N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}(t+\\Delta t)N^{n}\\, dS+\\int_{V}f_{i}(t+\\Delta t)N^{n}\\, dV\n\\end{equation}\nWe associate the term on the left-hand-side with the action of the\nsystem Jacobian on the increment of the displacement field. We approximate\nthe increment in stresses using linear elasticity and infinitesimal\nstrains,\n\n\\begin{gather}\nd\\sigma_{ij}(t)=C_{ijkl}(t)d\\varepsilon_{kl}(t)\\\\\nd\\sigma_{ij}(t)=\\frac{1}{2}C_{ijkl}(t)(du_{k.l}(t)+du_{l,k}(t))\\\\\nd\\sigma_{ij}(t)=\\frac{1}{2}C_{ijkl}(t)(\\sum_{m}da_{k,l}^{m}(t)N^{m}+\\sum_{m}da_{l,k}^{m}(t)N^{m})\n\\end{gather}\nNow, $d\\sigma_{ij}\\phi_{i,j}$ is a scalar, so it is symmetric,\n\\begin{equation}\nd\\sigma_{ij}\\phi_{i,j}=d\\sigma_{ji}\\phi_{j,i},\n\\end{equation}\nand we know that $d\\sigma_{ij}$ is symmetric, so\n\\begin{equation}\nd\\sigma_{ij}\\phi_{i,j}=d\\sigma_{ij}\\phi_{j,i},\n\\end{equation}\nwhich means\n\\begin{equation}\n\\phi_{i,j}=\\phi_{j,i},\n\\end{equation}\nwhich we can write as\n\\begin{equation}\n\\phi_{i,j}=\\frac{1}{2}(\\phi_{i,j}+\\phi_{j,i}).\n\\end{equation}\nIn terms of the basis functions, we have\n\n\\begin{equation}\n\\sum_{n}c_{i}^{n}N_{,j}^{n}=\\frac{1}{2}(\\sum_{n}c_{i}^{n}N_{,j}^{n}+\\sum_{n}c_{j}^{n}N_{,i}^{n}).\n\\end{equation}\nCombining these expressions for the increment in stresses and making\nuse of the symmetry of the weighting functions, we find the system\nJacobian is\n\n\\begin{equation}\nA_{ij}^{nm}=\\int_{V}\\frac{1}{4}C_{ijkl}(N_{,l}^{m}+N_{,k}^{m})(N_{,j}^{n}+N_{,i}^{n})\\ dV.\n\\end{equation}\nWe employ numerical quadrature in the finite-element discretization\nand replace the integral with a sum over the cells and quadrature\npoints,\n\\begin{equation}\nA_{ij}^{nm}=\\sum_{\\text{vol cells}}\\sum_{\\text{quad pts}}\\frac{1}{4}C_{ijkl}(N_{,l}^{m}(x_{q})+N_{,k}^{m}(x_{q}))(N_{,j}^{n}(x_{q})+N_{,i}^{n}(x_{q}))w_{q}|J_{cell}(x_{q}).\n\\end{equation}\n\n\\section{Solution Method for Dynamic Problems}\n\nFor brevity we outline the solution method for dynamic problems using\nonly index notation. Time-dependence enters through the constitutive\nrelationships, loading conditions, and the inertial terms. We consider\nthe deformation at time $t$,\n\\begin{equation}\n\\label{eq:elasticity:integral:dynamic:t}\n-\\int_{V}\\sigma_{ij}(t)N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}(t)N^{n}\\, dS+\\int_{V}f_{i}(t)N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\ddot{a}_{i}^{m}(t)N^{m}N^{n}\\ dV=\\vec{0}.\n\\end{equation}\nWe solve this equation through formulation of a linear algebraic system\nof equations ($Au=b$), involving the residual ($r=b-Au$) and Jacobian\n($A$). The residual is simply\n\\begin{equation}\nr_{i}^{n}=-\\int_{V}\\sigma_{ij}(t)N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}(t)N^{n}\\, dS+\\int_{V}f_{i}(t)N^{n}\\, dV-\\int_{V}\\rho\\sum_{m}\\ddot{a}_{i}^{m}(t)N^{m}N^{n}\\ dV.\n\\end{equation}\nWe employ numerical quadrature in the finite-element discretization\nand replace the integrals with sums over the cells and quadrature\npoints,\n\\begin{multline}\nr_{i}^{n}=-\\sum_{\\text{vol cells}}\\sum_{\\text{quad pts}}\\sigma_{ij}(x_{q},t)N^{n}(x_{q})\\: w_{q}|J_{cell}(x_{q})|+\\sum_{\\text{vol cells}}\\sum_{\\text{quad pt}s}f_{i}(x_{q},t)N^{n}(x_{q})\\, w_{q}|J_{cell}(x_{q})|\\\\\n+\\sum_{\\text{tract cells}}\\sum_{\\text{quad pts}}T_{i}(x_{q},t)N^{n}(x_{q})\\, w_{q}|J_{cell}(x_{q})|-\\sum_{\\text{vol cells}}\\sum_{\\text{quad pts}}\\rho\\sum_{m}\\ddot{a}_{i}^{m}(t)N^{m}N^{n}\\ w_{q|J_{cell}(x_{q})},\n\\end{multline}\nwhere $x_{q}$ are the coordinates of the quadrature points, $w_{q}$\nare the weights of the quadrature points, and $|J_{cell}(x_{q})|$\nis the determinant of the Jacobian matrix evaluated at the quadrature\npoints associated with mapping the reference cell to the actual cell.\nThe quadrature scheme for the integral over the tractions is one dimension\nlower than the one used in integrating the terms for the volume cells. \n\nWe find the system Jacobian matrix by making use of the temporal discretization\nand isolating the term for the increment in the displacement field\nat time $t$. Using the central difference method to approximate the\nacceleration (and velocity),\n\\begin{gather}\n\\ddot{u}_{i}(t)=\\frac{1}{\\Delta t^{2}}\\left(u_{i}(t+\\Delta t)-2u_{i}(t)+u_{i}(t-\\Delta t)\\right)\\\\\n\\dot{u}_{i}(t)=\\frac{1}{2\\Delta t}\\left(u_{i}(t+\\Delta t)-u_{i}(t-\\Delta t)\\right)\n\\end{gather}\nand writing the displacement at time $t+\\Delta t$ in terms of the\ndisplacement at $t$ (for consistency with the displacement increment\nquasi-static formulation),\n\\begin{gather}\nu_{i}(t+\\Delta t)=u_{i}(t)+du_{i}(t),\\\\\n\\ddot{u}_{i}(t)=\\frac{1}{\\Delta t^{2}}\\left(du_{i}(t)-u_{i}(t)+u_{i}(t-\\Delta t)\\right),\\\\\n\\dot{u}_{i}(t)=\\frac{1}{2\\Delta t}\\left(du_{i}(t)+u_{i}(t)-u_{i}(t-\\Delta t)\\right).\n\\end{gather}\nSubstituting into equation \\eqref{eq:elasticity:integral:dynamic:t}\nyields\n\\begin{multline}\n\\frac{1}{\\Delta t^{2}}\\int_{V}\\rho\\sum_{m}da_{i}^{m}(t)N^{m}N^{n}\\ dV=-\\int_{V}\\sigma_{ij}N_{,j}^{n}\\: dV+\\int_{S_{T}}T_{i}N^{n}\\, dS+\\int_{V}f_{i}N^{n}\\, dV\\\\\n-\\frac{1}{\\Delta t^{2}}\\int_{V}\\rho\\sum_{m}(a_{i}^{m}(t)-a_{i}^{m}(t-\\Delta t))N^{m}N^{n}\\ dV.\n\\end{multline}\nThus, the Jacobian for the system is\n\\begin{equation}\nA_{ij}^{nm}=\\delta_{ij}\\frac{1}{\\Delta t^{2}}\\int_{V}\\rho N^{m}N^{n}\\ dV,\n\\end{equation}\nand using numerical quadrature in the finite-element discretization\nto replace the integrals with sums over the cells and quadrature points,\n\n\\begin{equation}\nA_{ij}^{nm}=\\delta_{ij}\\frac{1}{\\Delta t^{2}}\\sum_{\\text{vol cells}}\\sum_{\\text{quad pts}}\\rho(x_{q})N^{m}(x_{q})N^{n}(x_{q}),\n\\end{equation}\nwhere $A_{ij}^{mn}$ is a $nd$ by $md$ matrix ($d$ is the dimension\nof the vector space), $m$ and $n$ refer to the basis functions and\n$i$ and $j$ are vector space components. We consider the contributions\nassociated with the fault in section \\vref{sec:fault} and with absorbing\nboundaries is section \\vref{sec:absorbing:boundaries}.\n\n\n\\section{Small (Finite) Strain Formulation}\n\\label{sec:small:strain:formulation}\n\nIn some crustal deformation problems sufficient deformation may occur\nthat the assumptions associated with infinitesimal strains no longer\nhold. This is often the case for problems when one wants to include\nthe effects of gravitational body forces on vertical deformation.\nIn such cases we want to account for both rigid body motion and small\nstrains. We use a total Lagrangian formulation (quantities are associated\nwith the undeformed configuration) based on the one presented by Bathe\n\\cite{Bathe:1995}.\n\nStarting from the governing equation, written for the deformed configuration\n(denoted by the subscript $t$), we have\n\\begin{equation}\n\\label{eq:governing:equation:deformed}\n\\int(\\nabla_{t}\\cdot\\underline{\\sigma})\\cdot\\vec{\\phi}\\: dV_{t}+\\int_{Vt}\\overrightarrow{f_{t}}\\cdot\\overrightarrow{\\phi}\\, dV_{t}-\\int_{Vt}\\rho_{t}\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\, dV_{t}=0.\n\\end{equation}\nFor the total Lagrangian formulation we want to transform these integrals\nover the deformed configuration to integrals over the undeformed configuration.\nWe require that the deformed and undeformed configurations use the\nsame coordinate system (origin and orientation). Conservation of mass\nrequires that $\\rho\\, dV_{t}=\\rho_{0}\\, dV_{0}$. We define the body\nforce as a force per unit volume that does not depend on the configuration,\nwhich leads to $\\vec{f}_{t}\\, dV_{t}=\\vec{f}_{0}\\, dV_{0}$.\n\nThe Green-Lagrange strain provides a measure of the strain relative\nto the original, undeformed configuration.\n\\begin{gather}\n\\underline{\\epsilon}=\\frac{1}{2}(\\nabla u_{i,j}+(\\nabla u)^{T}+u_{k,i}u_{k,j}),\\text{ or}\\\\\n\\underline{\\epsilon}=\\underline{X}_{0}^{T}\\underline{\\, X}_{0}-\\underline{I},\\text{ where}\\\\\n\\underline{X_{0}}=\\frac{\\partial}{\\partial x_{j}}(\\vec{x}(0)+\\vec{u}(t)),\n\\end{gather}\nand $\\underline{X}$ is the deformation gradient tensor. The second\nPiola-Kirchhoff stress tensor, $\\underline{S}$, is the work conjugate\nof the Green-Lagrange strain tensor. As a result, they are related\nthrough the elasticity constants,\n\n\\begin{equation}\n\\underline{S}=\\underline{C\\,}\\underline{\\varepsilon},\n\\end{equation}\nin the same manner as the Cauchy stress is related to the infinitesimal\nstrain. The Cauchy stress is related to the second Piola-Kirchoff\nstress through the deformation gradient tensor,\n\\begin{equation}\n\\underline{\\sigma}=\\frac{1}{|\\underline{X}_{0}|}\\underline{\\, X}_{0}\\,\\underline{S\\,}\\underline{X}_{0}^{T},\n\\end{equation}\n where $\\mathit{det}(\\underline{X}_{0})=|\\underline{X}_{0}|$. Additionally,\nthe first Piola-Kirhoff stress is define to be \n\\begin{equation}\n\\underline{P}=\\underline{S\\,}\\underline{X}_{0}^{T}.\n\\end{equation}\n\n\nApplying the divergence theorem, making use of the fact that $dV_{t}=|\\underline{X}_{0}|\\, dV_{0}$,\nand recognizing that the gradient in the deformed configuration is\nrelated to the gradient in the undeformed configuration through the\ndeformation gradient tensor, we can show that\n\\begin{equation}\n\\int_{V_{t}}\\nabla_{t}\\cdot\\underline{\\sigma}\\cdot\\vec{\\phi}\\: dV_{t}=-\\int_{V_{0}}\\underline{P}:\\nabla\\overrightarrow{\\phi}\\, dV_{0}+\\int_{S_{0}}\\overrightarrow{T_{0}}\\cdot\\overrightarrow{\\phi}\\, dS_{0},\n\\end{equation}\nwhere we assume the the tractions on the boundary do not depend on\nthe configuration. That is, the normal and share traction components\nare defined in terms of the undeformed configuration. Incorporating\nthe other relationships between the underformed and deformed configurations\nallows us to rewrite Equation \\vref{eq:governing:equation:deformed}\nin the undeformed configuration,\n\\begin{equation}\n-\\int_{V_{0}}\\underline{P}:\\nabla\\overrightarrow{\\phi}\\, dV_{0}+\\int_{S_{0}}\\overrightarrow{T_{0}}\\cdot\\overrightarrow{\\phi}\\, dS_{0}+\\int_{V_{0}}\\overrightarrow{f_{0}}\\cdot\\overrightarrow{\\phi}\\, dV_{0}-\\int_{V_{0}}\\rho_{0}\\frac{\\partial^{2}\\overrightarrow{u}}{\\partial t^{2}}\\cdot\\overrightarrow{\\phi}\\, dV_{0}=0.\n\\end{equation}\n\n\n\\subsection{Quasi-static Problems}\n\nThe system Jacobian for quasi-static problems includes terms associated\nwith elasticity. For the small strain formulation, we write the elasticity\nterm at time $t+\\Delta t$ and consider the first terms of the Taylor\nseries expansion,\n\\begin{equation}\n\\int_{v}S_{ij}(t+\\Delta t)\\delta\\varepsilon_{ij}(t+\\Delta t)\\: dV=\\int_{V}(S_{ij}(t)\\delta\\varepsilon_{ij}(t)+dS_{ij}(t)\\delta\\varepsilon_{ij}(t)+S_{ij}(t)d\\delta\\varepsilon_{ij}(t))\\: dV.\n\\end{equation}\nWe approximate the increment in the stress tensor using the elastic\nconstants,\n\\begin{equation}\ndS_{ij}=C_{ijkl}d\\varepsilon_{kl},\n\\end{equation}\nand the increment in the ``virtual'' strain via\n\\begin{equation}\nd\\delta\\varepsilon_{ij}=\\frac{1}{2}(du_{k,i}\\delta u_{k,j}+du_{k,j}\\delta u_{k,i}).\n\\end{equation}\nWe associate the system Jacobian with the terms involving the increment\nin displacements. After substituting in the expressions for the increment\nin the stresses and the increment in the ``virtual'' strains, we\nhave\n\\begin{equation}\nA_{ij}^{nm}=\\int_{V}\\frac{1}{4}C_{ijkl}(N_{,k}^{m}+(\\sum_{r}a_{p}^{r}N_{,l}^{r})N_{,k}^{m})(N_{,i}^{n}+(\\sum_{r}a_{p}^{r}N_{,j}^{r})N_{,i}^{n})+\\frac{1}{2}S_{kl}N_{,l}^{m}N_{,l}^{n}\\delta_{ij}\\: dV.\n\\end{equation}\nThe small strain formulation produces additional terms associated\nwith the elastic constants and a new term associated with the stress\ntensor.\n\n\n\\subsection{Dynamic Problems}\n\nThe system Jacobian matrix in dynamic problems does not include any\nterms associated with elasticity, so the system Jacobian matrix in\nthe small strain formulation matches the one used in the infinitesimal\nstrain formulation.\n", "meta": {"hexsha": "c07e1a2eef5b7b33fa68e3cbfa99047eda1dcfe2", "size": 29173, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/userguide/governingeqns/governingeqns.tex", "max_stars_repo_name": "joegeisz/pylith", "max_stars_repo_head_hexsha": "f74060b7b19d7e90abf8597bbe9250c96593c0ad", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-20T17:18:28.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-20T17:18:28.000Z", "max_issues_repo_path": "doc/userguide/governingeqns/governingeqns.tex", "max_issues_repo_name": "joegeisz/pylith", "max_issues_repo_head_hexsha": "f74060b7b19d7e90abf8597bbe9250c96593c0ad", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/userguide/governingeqns/governingeqns.tex", "max_forks_repo_name": "joegeisz/pylith", "max_forks_repo_head_hexsha": "f74060b7b19d7e90abf8597bbe9250c96593c0ad", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.0017825312, "max_line_length": 388, "alphanum_fraction": 0.7055153738, "num_tokens": 10156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7905303211371898, "lm_q1q2_score": 0.6388016338303001}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[margin=0.5in]{geometry}\n\\usepackage{hyperref}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\n\\setlength{\\parindent}{0em}\n\n\\title{\\textbf{COMP90056 Assignment A}}\n\\author{Tingsheng (Tinson) Lai (731319)}\n\\date{\\today}\n\n\\begin{document}\n \\maketitle\n \\section{Part 1}\n Let $S$ denotes the union, i.e. $$S = S_1 \\cup S_2$$\n From the inequality, we can claim that the inequality inherently holds for $\\forall x \\in S \\backslash S_2$ and $\\forall x \\in S \\backslash S_1$. So the only task is to prove that the inequality holds for the third circumstance where $$\\forall x \\in S_1 \\cap S_2$$\n Use $\\hat{S}$ as the intersection $$\\hat{S} = S_1 \\cap S_2$$\n First prove the $RHS$ where $\\forall x \\in \\hat{Z}$ $\\hat{c} \\left( x \\right) \\leq f_{x}$. As $x \\in S_1 \\wedge x \\in S_2$, we know that $\\hat{c}_1 \\left( x \\right) \\leq f_{1,x}$ and $\\hat{c}_2 \\left( x \\right) \\leq f_{2,x}$. So $$\\hat{c} \\left( x \\right) = \\hat{c}_1 \\left( x \\right) + \\hat{c}_2 \\left( x \\right) \\leq f_{1,x} + f_{2,x} = f_{x}$$\n Even if the subtraction to shrink the size happens, the $RHS$ still holds as $\\forall i \\in \\mathbb{U}$, $f_{i}$ will not be changed throughout the whole process. For the $LHS$, using the same trick we can get $$f_{1,x} - \\frac{m_1 - \\tau_1}{k_1} + f_{2,x} - \\frac{m_2 - \\tau_2}{k_2} = f_x - \\frac{m - \\tau}{x}$$ So $$\\hat{c} \\left( x \\right) \\geq f_x - \\frac{m - r}{k} = LHS$$ Subtracting $a$ from each $x \\in S$ will decrease the value of $\\tau$ by at least $k \\cdot a$. Denoting $\\Delta \\hat{c}_x$ = -a as the change in the counter and $\\Delta \\tau$ as the change in $\\tau \\leq -a \\cdot k$. It is clear that $f_x$ and $m$ and $k$ are fixed, so $\\Delta LHS \\leq -a$. So the decrease in the $LHS$ are even greater than $\\hat{c} \\left( x \\right)$, so the inequality still holds.\n\\end{document}\n", "meta": {"hexsha": "e5a9e20646c5a3abfd1c1854d05a1abb33e490e3", "size": 1852, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment/B/part1.tex", "max_stars_repo_name": "laitingsheng/2019S2-COMP90056", "max_stars_repo_head_hexsha": "adc65917942ce0057cd51602f700c8a7e09cfaea", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment/B/part1.tex", "max_issues_repo_name": "laitingsheng/2019S2-COMP90056", "max_issues_repo_head_hexsha": "adc65917942ce0057cd51602f700c8a7e09cfaea", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment/B/part1.tex", "max_forks_repo_name": "laitingsheng/2019S2-COMP90056", "max_forks_repo_head_hexsha": "adc65917942ce0057cd51602f700c8a7e09cfaea", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 84.1818181818, "max_line_length": 782, "alphanum_fraction": 0.6517278618, "num_tokens": 689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996143, "lm_q2_score": 0.8080672204860317, "lm_q1q2_score": 0.6388016293416686}} {"text": "\\section*{Ex.34.5-6}\n\\subsection*{Hamiltonian path NP complete}\n\nTo show that HAM--PATH is NP complete we show\n\\begin{enumerate}[]\n\t\\item HAM--PATH is NP\n\t\\item HAM--CYCLE $\\leq_P$ HAM--PATH\n\\end{enumerate}\nThat HAM--PATH is NP, is shown in EX.34.2-6.\n\nTo see that HAM--CYCLE $\\leq_P$ HAM--PATH:\n\nGiven a graph $G$ where there is an instance of HAM--CYCLE, pick a node $u$ and create a new graph $G'$ where $u$ is replaced with two new nodes $u'$ and $u''$. All the edges that goes in to $u$ are also going into $u''$ and all of the outgoing edges from $u$ are also outgoing edges from $u'$. No edges are going from $u'$ to $u''$.\n\\\\\nThen it holds that\n\\\\\n$G$ has HAM--CYCLE $\\Leftrightarrow$ $G'$ has HAM--PATH\n\nBefore we prove it, note that a HAM--PATH must start in $u'$ and end in $u''$\n\nWe show $\\Rightarrow$:\n\nAssume $G$ has a HAM-CYCLE, then the same nodes creates a HAM--CYCLE in $G'$ if we split $u$ into $u'$ and $u''$.\n\nWe show $\\Leftarrow$:\n\nAssume $G'$ has HAM-PATH, then combining $u'$ and $u''$ into $u$ will create a HAM-CYCLE in $G$.\n\n\nThis show item 2. Hence we have shown that HAM-PATH is NP complete.\n", "meta": {"hexsha": "c7ecd97e02d27ccd8acfdece4a5e07e7d8fcb11b", "size": 1121, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Uge4/Ex.34.5-6.tex", "max_stars_repo_name": "pdebesc/AADS", "max_stars_repo_head_hexsha": "a26e24d18adee973d3ce88bdfd96d857ec472fdf", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Uge4/Ex.34.5-6.tex", "max_issues_repo_name": "pdebesc/AADS", "max_issues_repo_head_hexsha": "a26e24d18adee973d3ce88bdfd96d857ec472fdf", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Uge4/Ex.34.5-6.tex", "max_forks_repo_name": "pdebesc/AADS", "max_forks_repo_head_hexsha": "a26e24d18adee973d3ce88bdfd96d857ec472fdf", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.1612903226, "max_line_length": 333, "alphanum_fraction": 0.6743978591, "num_tokens": 384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996143, "lm_q2_score": 0.8080672066194945, "lm_q1q2_score": 0.6388016183797507}} {"text": "\\paragraph{Def} $x_i \\in [l_i, u_i]$, $a_i^\\le \\le x_i \\le a_i^\\ge$, $a_i^\\cdot = \\sum_j w_j x_j + v$\n\\paragraph{Complexity} Affine: $\\mathcal{O}(\\left(\\text{\\#layers} \\right) \\left(\\text{max \\#neurons}\\right)^2)$, ReLU: $\\mathcal{O}(1)$\n\\paragraph{ReLU}\n\\begin{tabular}[t]{c|c|c|c|c}\n Condition & $a_j^\\le$ & $a_j^\\ge$ & $l_j$ & $u_j$ \\\\\\midrule\n $u_i \\le 0$ & 0 & 0 & 0 & 0 \\\\\n $l_i \\ge 0$ & $x_i$ & $x_i$ & $l_i$ & $u_i$ \\\\\n $l_i < 0 < u_i$ & $\\lambda x_i$ & $u_i \\dfrac{x_i-l_i}{u_i-l_i}$ & $\\lambda l_i$ & $u_i$ \n\\end{tabular}\nFor any $\\lambda \\in [0,1]$\n\\paragraph{Area heuristic} Choose $\\lambda=0$ if $u\\le-l$, else $\\lambda=1$\n\n\\paragraph{Backsub} After affine layer, repeatedly backsubstitute (for $l$, choose $a^\\le$ for $w \\ge 0$, $a^\\ge$ otherwise)\n\\paragraph{Verification} Backsubstitute $x_\\text{label} - x_i$ to find lower bound, verifies iff $> 0$", "meta": {"hexsha": "c7b811ec710b1e9550f5a76b966c4cf04f342c7d", "size": 876, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "deeppoly.tex", "max_stars_repo_name": "cknabs/RIAI-summary-HS2020", "max_stars_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-01-20T21:27:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-24T20:28:56.000Z", "max_issues_repo_path": "deeppoly.tex", "max_issues_repo_name": "cknabs/RIAI-summary-HS2020", "max_issues_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-01-25T09:29:16.000Z", "max_issues_repo_issues_event_max_datetime": "2021-01-25T10:50:09.000Z", "max_forks_repo_path": "deeppoly.tex", "max_forks_repo_name": "cknabs/RIAI-summary-HS2020", "max_forks_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.5714285714, "max_line_length": 135, "alphanum_fraction": 0.6050228311, "num_tokens": 370, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8774767810736692, "lm_q2_score": 0.7279754548076477, "lm_q1q2_score": 0.6387815587852551}} {"text": "In Chapter~\\ref{chapter:bdp}, we assumed that an event organizer is only allowed to inspect a group of columns (as a batch) by polling all invitees at once.\nIn this chapter, we relax this restriction, and assume that the organizer can inspect each entry of a random matrix at unit cost (by querying a certain invitee about a certain date/time option). \nConsequently, \\Time and \\Inconvenience are always equal in this setting, and the goal is to find an optimal inspection sequence which is a permutation of entries of a given random matrix.\nAs in \\BDP, we would like to optimize the expected cost, which is the expected number of inspections until a feasible column is found or all columns are deemed infeasible -- we call this problem the Probabilistic Matrix Inspection Problem (\\PMIP). \n\n\nThe main difference between the Batched Doodle Problem (\\BDP) and the Probabilistic Matrix Inspection Problem (\\PMIP) lies in the solution domain (a unit operation on a group of columns versus entries) and the cost model (a function of Time and Inconvenience versus the number of inspections). \nThe Batched Doodle generalizes Doodle by allowing columns-by-columns inspections, and we further generalize it by allowing entry-by-entry inspections. \n\n%TODO: To re-write intro?\n\n\\section{Notation and Definitions} \\label{PMIP:sec:notation}\n\n We use the same definition of feasibility as in previous chapter, which is copied below. \n\n \\begin{definition}[Feasibility]\n Let $A$ be a matrix whose entries are from $\\{0,1\\}$, and refer to the entry of $A$ at row $r$ and column $c$ as $a_{r,c}$.\n We say that a column $c$ of $A$ is {\\em feasible} if it consists only of 1's (otherwise it is {\\em infeasible}). \n We say that $A$ is {\\em feasible} if it contains at least one feasible column (otherwise it is {\\em infeasible}).\n \\end{definition}\n\n In the Probabilistic Matrix Inspection Problem, we do not know of the values of the entries of $A$, as they are Bernoulli random variables, but we know of probability distribution of each entry.\n An input to the problem is this probability distribution. \n \\begin{definition}[Input instance]\n An input instance of the Probabilistic Matrix Inspection problem is a pair of matrices $(A, P_A)$ of size $n$ by $m$\n where $A$ is a matrix of Bernoulli random variables \n and $P_A$ is the probability matrix associated with it.\n We denote an entry of $A$ as $a_{r,c}$ and of $P_A$ as $p_{r,c}$.\n Each entry $a_{r,c}$ of $A$ is a Bernoulli random variable, and $p_{r,c}$ is the probability of success for $a_{r,c}$ (i.e., $p_{r,c} = \\mathbf{P}[a_{r,c} = 1]$).\n We define $s_c = \\prod_{r=1}^{n} p_{r,c}$ to denote the probability that column $c$ is feasible (probability of success for column $c$).\n \\end{definition}\n Throughout this work we will assume that the set of random variables $\\{a_{r,c}\\}$ are mutually independent. This assumption is crucial to our technical results because it allows to compute (in polynomial time) the probability of a specific realization of $A$ conditioning on the event in which some entries of $A$ have already been realized. \n Without this assumption, it is unclear how one can compute such probabilities without having an access to the joint probability distribution over all realizations of $A$ (whose size is exponential in the size of $A$).\n\n We define an ``inspection'' as an operation that can be performed on $A$. \n One can inspect an arbitrary entry $a_{r,c}$ of $A$ at unit cost, so as to know of the realization of the random variable. In group scheduling, an inspection corresponds to querying an agent about her availability for a certain outcome. \n The objective of the problem is to determine whether $A$ is feasible or not, with minimum (expected) number of inspections possible. The expectation is with respect to the probability distribution specified by $P_A$.\n\n Because we are interested in determining feasibility of $A$ with minimum number of inspections, there are certain ``unnecessary inspections'' that an optimal strategy must avoid.\n For instance, if a certain entry $a_{r,c}$ is found to be unsuccessful (i.e., $a_{r,c} = 0$ is realized), then there is no need to inspect any other entry from the same column because the column is already known to be infeasible. Similarly, if a certain column is found to be feasible (which implies $A$ is feasible) or if all columns are found to be infeasible (which implies $A$ is infeasible), then there is no need to inspect any other entries of the matrix. Lastly, if $p_{r,c} = 0$ or $p_{r,c} = 1$, then there is no need to inspect the entry $a_{r,c}$ because we already know its realization with probability $1$.\n Therefore, without loss of generality, we will assume that $p_{r,c} \\in (0,1)$ (i.e., $p_{r,c}\\neq 0,1$) in this work. \n\n Let us define what constitutes a solution to the problem. \n \\begin{definition}[Inspection policy]\n \tGiven an input instance $(A,P_{A})$, a solution is any permutation of the entries of $A$, and we call it an ``inspection policy'' (or simply, a ``policy'').\n \\end{definition}\n The interpretation of a permutation is as follows. The entries of $A$ will be inspected in order specified by the permutation. After each inspection, if $A$ is found to be feasible or infeasible, the inspection process ends. Otherwise, it continues inspecting the entries as specified, but it will not make any unnecessary inspections as mentioned earlier. \n\n If $\\pi$ is a permutation of the entries of $A$, we write $C(\\pi)$ to denote the number of inspections performed by $\\pi$. $C(\\pi)$ is a random variable whose probability distribution is determined by $P_A$. We are interested in finding an optimal policy which minimizes the expected number of inspections, $\\mathbf{E}[C(\\pi)]$.\n Note that there are $(nm)!$ permutations of the entries of $A$, and therefore exhaustive search for an optimal permutation will not produce an efficient algorithm. \n\n \\subsection{Example} \\label{sec:example}\n\n Consider a 2-by-2 matrix $A$ of Bernoulli random variables whose probability of success is given by $P_A$ as follows. In group scheduling, this corresponds to two agents and a set of two date/time options being considered.\n \\begin{equation*}\n A = \n \t\\begin{bmatrix}\n \t\ta_{1,1} & a_{1, 2} \\\\\n \t\ta_{2,1} & a_{2, 2}\n \t\\end{bmatrix},\n P_A = \n \t\\begin{bmatrix}\n \t\t0.6 & 0.7 \\\\\n \t\t0.9 & 0.8 \n \t\\end{bmatrix}\n \\end{equation*}\n Let us consider an inspection policy $\\pi$ which inspects the entries of $A$ column-by-column while inspecting them from top to bottom within a column:\n \\begin{equation*}\n \t\\pi = \n \t\\begin{pmatrix} \n \t1 & 2 & 3 & 4 \\\\\n \ta_{1,1} & a_{2,1} & a_{1,2} & a_{2,2} \n \t\\end{pmatrix}.\n \\end{equation*}\t\n Suppose that the realization of $A$ happens to be the identity matrix of size $2$ (i.e., $a_{1,1} = a_{2,2} = 1$ and $a_{2,1} = a_{1,2} = 0$). If we use $\\pi$, it will first inspect $a_{1,1}$ and learn its realization. Since $a_{1,1} = 1$, it will inspect $a_{2,1}$ next only to find that column $1$ is infeasible after all. It will then inspect $a_{1,2}$ and learn that column $2$ is also infeasible, which implies that $A$ is infeasible. At this point, the inspection process terminates without inspecting $a_{2,2}$.\n This specific realization of $A$ happens with probability $p_{1,1}(1 - p_{2,1})(1-p_{1,2})p_{2,2}$, and yields $C(\\pi) = 3$ because 3 inspections would occur. \n If the realization of $A$ happens to be the null matrix (i.e., all entries are $0$'s), then $\\pi$ would only inspect $a_{1,1}$ and $a_{1,2}$ but skip $a_{2,1}$ and $a_{2,2}$. In this manner one can consider all $2^{2\\cdot 2} = 16$ possible realizations of $A$, and compute $\\mathbf{E}[C(\\pi)]$ in this example. \n\n Another way to compute $\\mathbf{E}[C(\\pi)]$ is by de-coupling $C(\\pi)$ into two random variables $N(\\pi, c)$ with $c\\in\\{1,2\\}$ where $N(\\pi,c)$ denotes the number of inspections (on column $c$) performed by $\\pi$ conditioning on the event that (at least one element of) column $c$ is inspected.\n We can efficiently compute these: $\\mathbf{E}[N(\\pi, c)] = 1 \\cdot \\mathbf{P}[a_{1,c} = 0] + 2 \\cdot \\mathbf{P}[a_{1,c} = 1]$ for $c\\in \\{1, 2\\}$.\n To express $\\mathbf{E}[C(\\pi)]$ in $N(\\pi,c)$'s, we need to take conditional probability into account, as column 2 is inspected only if column 1 is infeasible: \n $\\mathbf{E}[C(\\pi)] = \\mathbf{E}[N(\\pi,1)] + \\mathbf{E}[N(\\pi,2)] (1-s_1) = 2.382$. Recall that $s_c$ is the probability of success for column $c$.\n\n\n\\section{Technical Results and Algorithm} \\label{PMIP:sec:results}\n\n\n We first consider two special cases (1-row or 1-column matrices) of the Probabilistic Matrix Inspection problem, which admit intuitive, greedy algorithms. We then discuss a couple of interesting properties of an optimal inspection policy, which leads to our main result and algorithm.\n\n \\subsection{1-Row Matrix and 1-Column Matrix}\n Let us first consider the case where an input matrix $A$ has only one row (i.e., $n = 1$).\n In this case it is natural to inspect entries with largest probability first because we can stop as soon as we find an entry whose value is $1$ -- which makes its column and $A$ feasible. This intuition is exactly what an optimal policy should do in the single row case.\n\n \\begin{lemma}[1-Row Matrix]\\label{lemma:single_row_opt}\n When $n = 1$, an inspection policy $\\pi$ is optimal if and only if it inspects the entries in non-increasing order of their associated probabilities.\n \\end{lemma}\n \\begin{proof}\n \tWithout loss of generality let us assume that a policy $\\pi$ inspects the entries in increasing order of their column index; that is, $\\pi(i) = a_{1,i}$. \n \tRecall that $C(\\pi)$ is a random variable that denotes the number of inspections $\\pi$ incurs. \n \tWe can express the expectation of $C(\\pi)$ in terms of $p_{1,c}$'s as follows:\n \t\\small\n \t\\begin{equation} \\label{eqn:exp_cost_pi_1row}\n \t\t\\mathbf{E}\\left[C(\\pi)\\right] = \n \t\tm \\left(\\prod_{k=1}^{m-1}(1 - p_{1,k})\\right) + \n \t\t\\sum_{j=1}^{m-1} j \\left(p_j \\prod_{k=1}^{j-1} (1-p_{1,k}) \\right).\n \t\\end{equation}\n \t\\normalsize\n\t\n \tSuppose that there exists some $c^*$ such that $p_{1,c^*} < p_{1,c^*+1}$ (if no such $c^*$ exists, then $\\pi$ is an inspection policy that inspects the entries in non-increasing order of probabilities).\n \tLet $\\pi'$ be the same policy as $\\pi$ except we swap the order of $a_{1,c^*}$ and $a_{1,c^*+1}$. \n \tThat is, $\\pi'$ is defined as follows.\n \t\\begin{equation*}\n \t\t\\pi'(j) = \n \t\t\\begin{cases}\n \t\t\t\\pi'(j) = \\pi(c^*+1)& \\mbox{if~} j = c^* \\\\\n \t\t\t\\pi'(j) = \\pi(c^*) & \\mbox{if~} j = c^*+1 \\\\\n \t\t\t\\pi'(j) = \\pi(j) & \\mbox{if~} j \\neq c^* \\land j \\neq c^*+1\n \t\t\\end{cases}\n \t\\end{equation*}\t\n \tAfter expressing $\\mathbf{E}[C(\\pi)]$ and $\\mathbf{E}[C(\\pi')]$ as in Equation~\\ref{eqn:exp_cost_pi_1row}, one can re-arrange the terms to obtain the following:\n \t\\begin{equation} \\label{eqn:proof_pi_1row}\n \t\\mathbf{E}\\left[C(\\pi)\\right] - \\mathbf{E}\\left[C(\\pi')\\right] = \n \t\\left(\\prod_{j=1}^{c^*-1} (1-p_{1,j})\\right) (p_{1,c^*+1} - p_{1,c^*}).\n \t\\end{equation}\n \tThis quantity is positive if $p_{1,c^*+1} > p_{1,c^*}$ (recall that $p_{1,j} \\in (0,1)$ for all $j$ as mentioned in Section~\\ref{PMIP:sec:notation}). \n\t\n \tThis proves the lemma because any policy that inspects an entry with smaller probability before another entry with higher probability is suboptimal, and therefore an optimal policy must inspect entries in non-increasing order of their associated probabilities. \n \\end{proof}\n Although the proof of Lemma~\\ref{lemma:single_row_opt} is simple, it confirms correctness of our intuition. Equation~\\ref{eqn:proof_pi_1row} illustrates this intuition; conditioning on the event that the first $c^*-1$ inspections fail (whose probability is the product term in Equation~\\ref{eqn:proof_pi_1row}), the difference $\\mathbf{E}[C(\\pi)] - \\mathbf{E}[C(\\pi')]$ depends on the difference in the probabilities of success between the next-entry-to-be-inspected by $\\pi$ and $\\pi'$.\n\n We can also consider the case where an input matrix $A$ has only one column.\n Intuitively, if we wish to minimize the expected number of inspections, we must inspect entries with smallest probability first because we can stop as soon as we determine that $A$ is infeasible.\n Lemma~\\ref{lemma:single_column_opt} formally states this intuition about optimal policy, and we omit a proof of it as it can be easily done by following the proof of Lemma~\\ref{lemma:single_row_opt}.\n \\begin{lemma}[1-Column Matrix] \\label{lemma:single_column_opt}\n When $m = 1$, an inspection policy $\\pi$ is optimal if and only if it inspects the entries in non-decreasing order of their associated probabilities.\n \\end{lemma}\n\n\n \\subsection{Inspection of Entire Column}\n Another interesting property of an optimal inspection policy is that once it inspects the first entry of a column, then it must commit to it and continue inspecting the remaining entries of the column until feasibility of the column is determined. Otherwise, if the policy switches to another column too soon, then it is not optimal. \n \\begin{theorem}[Optimality of inspecting entire column] \\label{theorem:col_by_col_opt}\n Consider any inspection policy $\\pi$.\n Without loss of generality, let us assume that for each column $c$, $\\pi$ inspects $a_{n,c}$ the last among $n$ entries of the column. \n Let $b_c$ be the index of $\\pi$ such that $\\pi(b_c) = a_{n,c}$. \n Without loss of generality, assume $b_1 < b_2 < \\cdots < b_m$ (we can do this by re-labeling the columns of $A$). \n If there is some column $c^*$ such that $b_{c^*} > n\\cdot c^*$, then $\\pi$ is not optimal.\n \\end{theorem}\n \\begin{proof}\n First, note that $b_c \\geq n\\cdot c$ for all $c$ because we assumed $b_1 < b_2 < \\cdots < b_m$, and therefore the entries of previous columns must appear before the last entry of each column.\n\n Let $\\pi$ be an inspection policy being considered in the theorem for which there exists some $c$ with $b_{c} > n \\cdot c$.\n Let us construct a different inspection policy $\\pi'$.\n First, $\\pi'$ inspects all entries of column $1$ in the same order $\\pi$ does. \n Then, $\\pi'$ inspects all entries of column $2$ in the same order $\\pi$ does, and so on. \n In particular, $\\pi'$ inspects all entries of a column before inspecting another column, while preserving the original ordering of the entries within each column that is given by $\\pi$. \n We will show that $\\mathbf{E}[C(\\pi')] < \\mathbf{E}[C(\\pi)]$.\n\n Let us define a set of new random variables which can be used to express $C(\\cdot)$, as we did in Section~\\ref{sec:example} when analyzing an example.\n Recall that $s_c = \\prod_{r=1}^{n} a_{r,c}$ is the probability of success for column $c$.\n Let $N(\\pi, c)$ ($N(\\pi', c)$, respectively) be a random variable that denotes the number of entries of column $c$ that is inspected by $\\pi$ (by $\\pi'$, respectively), conditioning on the event that column $c$ is inspected (i.e., when the previous $c-1$ columns are infeasible).\n We can then express $\\mathbf{E}[C(\\pi)]$ and $\\mathbf{E}[C(\\pi')]$ as follows:\n \\begin{equation} \\label{eqn:exp_c_pi_decoupled}\n \t\\mathbf{E}[C(\\pi)] = \\sum_{c=1}^{m} \\mathbf{E}[N(\\pi, c)]\\left( \\prod_{k=1}^{c-1} (1 - s_c) \\right)\n \\end{equation}\n and\n \\begin{equation} \\label{eqn:exp_c_pi2_decoupled}\n \t\\mathbf{E}[C(\\pi')] = \\sum_{c=1}^{m} \\mathbf{E}[N(\\pi', c)]\\left( \\prod_{k=1}^{c-1} (1 - s_c) \\right).\n \\end{equation}\n\n\n To prove the theorem we will first show that for any realization of $A$, $N(\\pi, c) \\geq N(\\pi', c)$ holds for all $c$; this immediately implies $\\mathbf{E}[C(\\pi)] \\geq \\mathbf{E}[C(\\pi')]$.\n We will then show that there exists at least one realization of $A$ such that for some column $c'$ the strict inequality $N(\\pi, c') > N(\\pi', c')$ holds. These two statements together imply that $\\mathbf{E}[C(\\pi)] > \\mathbf{E}[C(\\pi')]$.\n\n\n Consider any realization of $A$ with the condition that the first $m-1$ columns are infeasible (recall that $m$ is the number of columns of $A$).\n Then $N(\\pi, c) = N(\\pi', c)$ for all $c$ regardless of feasibility of column $m$.\n To see why, both $\\pi$ and $\\pi'$ would inspect the same set of entries in each of the first $m-1$ columns in the same order until the column is determined to be infeasible, and therefore $N(\\pi, c) = N(\\pi', c)$ if $c c'$; therefore we have $N(\\pi, c) \\geq N(\\pi', c)$ for all $c>c'$. In summary $N(\\pi, c) \\geq N(\\pi', c)$ holds for all $c$ in this case as well. \n\n So far we proved the first claim we stated earlier: for all realizations of $A$, we have $N(\\pi, c) \\geq N(\\pi', c)$ for all $c$.\n Let us now prove the second claim. Let $c^*$ be the smallest index $c$ of columns such that $b_c > nc$ (note that $c^* < m$ because $b_m = nm$ by definition). \n Consider any realization of $A$ with the condition that the first $c^*-1$ columns are infeasible and column $c^*$ is feasible (feasibility of other columns do not matter). \n Using the same arguments we used earlier, we can show that $N(\\pi, c) = N(\\pi', c)$ for all $c < c^*$, that $N(\\pi, c^*) = N(\\pi', c^*) = n$, and that $N(\\pi', c) = 0$ for all $c> c^*$. \n However, because $b_{c^*} > n \\cdot c^*$, there is at least one entry $a_{r',c'}$ with $c' > c^*$ which appears before $b_{n, c^*}$ in $\\pi$ (otherwise, if no such entry exists, then $b_{c^*}$ would be equal to $n \\cdot c^*$). This implies that there exists some $c'$ with $c' > c^*$ such that $N(\\pi, c') > 0$. \n This proves the second claim that for some realization of $A$, there is some column $c'$ for which $N(\\pi, c') > N(\\pi', c')$, and together with the first claim we proved earlier, this implies that $\\mathbf{E}[C(\\pi)] > \\mathbf{E}[C(\\pi')]$. \n\n This proves the theorem: Any policy that does not inspect all entries of a column consecutively is suboptimal.\n \\end{proof}\n By Theorem~\\ref{theorem:col_by_col_opt}, when seeking an optimal policy, it is sufficient to consider the set of policies that inspect an entire column before committing to another column. Lemma~\\ref{lemma:single_column_opt} hints that one should inspect the entries of each column in increasing order of probabilities, and this is what we prove next.\n\n\n \\subsection{Optimal Ordering within Column} \n\n Lemma~\\ref{lemma:single_column_opt} states that\n an optimal policy must inspect the entries in increasing order of their probability of success, if $A$ is a 1-column matrix.\n This argument can be generalized to the case where there is more than one column: If an optimal policy is to inspect an entry of some column $c$, it must inspect the entry with smallest probability of success first. \n \\begin{theorem}[Optimal ordering within column] \\label{theorem:within_column_opt}\n \tConsider any inspection policy $\\pi$.\n \tIf there exist two entries $a_{r_1,c}$ and $a_{r_2,c}$ from the same column such that $a_{r_1,c}$ appears before $a_{r_2,c}$ in $\\pi$ and $p_{r_1,c} > p_{r_2,c}$, then $\\pi$ is not optimal. \n \tIn other words, when restricted to each column, an optimal policy must inspect the entries of the column in non-decreasing order of probabilities.\n \\end{theorem}\n \\begin{proof}\n \tLet $\\pi$ be an inspection policy being considered in the theorem.\n \tBecause of Theorem~\\ref{theorem:col_by_col_opt} we can assume, without loss of generality, that $\\pi$ inspects all entries of column 1, followed by column 2, and so on.\n \tFurther let us assume that $\\pi$ inspects the entries of each column in increasing order of their row index (we can do so by re-labeling the indices of entries). Precisely, $\\pi(r + n (c-1)) = a_{r,c}$ defines $\\pi$. Let $p_{r,c^*}$ and $p_{r+1,c^*}$ be the entries with $p_{r,c^*} > p_{r+1, c^*}$.\n \tLet us consider a different inspection policy $\\pi'$ that is the same as $\\pi$ except that $\\pi'$ inspects $p_{r+1,c^*}$ before $p_{r,c^*}$, by swapping the ordering of them.\n \t\\begin{equation*}\n \t\t\\pi'(j) = \n \t\t\\begin{cases}\n \t\t\t\\pi'(j) = a_{r+1,c^*} & \\mbox{if~} \\pi(j) = a_{r,c^*} \\\\\n \t\t\t\\pi'(j) = a_{r,c^*} & \\mbox{if~} \\pi(j) = a_{r+1,c^*} \\\\\n \t\t\t\\pi'(j) = \\pi(j) & \\mbox{otherwise} \n \t\t\\end{cases}\n \t\\end{equation*}\t\n\t\n \tWe claim that $\\mathbf{E}[C(\\pi')] < \\mathbf{E}[C(\\pi)]$, which implies that $\\pi$ is not optimal.\n\t\n \tLet us define new random variables $N(\\pi, c)$ and $N(\\pi', c)$ as we did in our proof of Theorem~\\ref{theorem:col_by_col_opt} (i.e., the number of inspections performed by the respective policy on column $c$, conditioning on the event that the column is inspected).\n \tThen we can express $\\mathbf{E}[C(\\pi)]$ and $\\mathbf{E}[C(\\pi')]$ in terms of the new random variables and $s_c$'s as we did in Equations~\\ref{eqn:exp_c_pi_decoupled} and \\ref{eqn:exp_c_pi2_decoupled}.\n\t\n \tObserve that $N(\\pi, c) = N(\\pi', c)$ for any realization of $A$ if $c \\neq c^*$.\n \tTo see this, first note that column $c$ would not be inspected by $\\pi$ or by $\\pi'$ if any of the previous columns (that is, columns $1$ through $c-1$) is found to be feasible, in which case $N(\\pi, c) = N(\\pi', c) = 0$. Otherwise, if column $c$ is inspected, both policies would inspect the entries of $c$ in the very same order, so $N(\\pi, c) = N(\\pi', c)$ must hold. Therefore we conclude that $\\mathbf{E}[N(\\pi, c)] = \\mathbf{E}[N(\\pi', c)]$ when $c \\neq c^*$. \n\t\n \tWe will now show that $\\mathbf{E}[N(\\pi, c^*)] > \\mathbf{E}[N(\\pi', c^*)]$ holds. \n \tThis immediately implies $\\mathbf{E}[C(\\pi)] > \\mathbf{E}[C(\\pi')]$ due to Equations~\\ref{eqn:exp_c_pi_decoupled} and \\ref{eqn:exp_c_pi2_decoupled}.\n \tLet us express $\\mathbf{E}[N(\\pi, c^*)]$ in terms of $p_{r,c^*}$'s.\n \t\\begin{equation*}\n \t\t\\mathbf{E}[N(\\pi, c^*)] = n\\left( \\prod_{k=1}^{n-1} p_{k,c^*}\\right) + \n \t\t\\sum_{j=1}^{n-1} j (1-p_{j,c^*}) \\left( \\prod_{k=1}^{j-1} p_{k,c^*} \\right)\n \t\\end{equation*}\n \tNote that the event $N(\\pi, c^*) = j$ occurs if the first $j-1$ entries are feasible while the $j$-th entry is not feasible when $j < n$, and $N(\\pi, c^*) = n$ occurs if the first $n-1$ entries are feasible (but the $n$-th entry's feasibility does not matter).\n\t\n \tWe can express $\\mathbf{E}[N(\\pi', c^*)]$ in a similar manner, and simplify $\\mathbf{E}[N(\\pi, c^*)] - \\mathbf{E}[N(\\pi', c^*)]$ as follows:\n \t\\small\n \t\\begin{equation*}\n \t\t\\mathbf{E}[N(\\pi, c^*)] - \\mathbf{E}[N(\\pi', c^*)] = \n \t\tr \\left( \\prod_{k=1}^{r-1} p_{k,c^*} \\right) (p_{r,c^*} - p_{r+1,c^*}).\n \t\\end{equation*}\n \t\\normalsize\n \tThe quantity above is positive if $p_{r,c^*} > p_{r+1,c^*}$, which is the assumption we began with.\n \tThis proves the theorem.\n \\end{proof}\n\n\n Theorems~\\ref{theorem:col_by_col_opt} and \\ref{theorem:within_column_opt} together tell us that \n in order to find an optimal policy we only need to decide the ordering of the columns. \n There are still $m!$ orderings of columns, and an exhaustive search algorithm would not be efficient.\n As we were able to generalize Lemma~\\ref{lemma:single_column_opt} to Theorem~\\ref{theorem:within_column_opt} by generalizing the optimal solution for 1-column case, it would be natural to consider generalizing Lemma~\\ref{lemma:single_row_opt} in a similar manner.\n\n This idea leads to the following greedy algorithm: First we sort columns by their probability of success ($s_c = \\prod_{r=1}^{n} p_{r,c}$) in decreasing order, and inspect the entries of each column in increasing order of their associated probabilities. However, as the following example shows, this algorithm is suboptimal. \n \\begin{equation*}\n A = \n \t\\begin{bmatrix}\n \t\ta_{1,1} & a_{1, 2} \\\\\n \t\ta_{2,1} & a_{2, 2}\n \t\\end{bmatrix},\n \tP_A =\n \t\\begin{bmatrix}\n \t\t0.4459 & 0.2262 \\\\\n \t\t0.4459 & 0.8114\n \t\\end{bmatrix}\n \\end{equation*}\n Here we have $s_1 ~= 0.199$ and $s_2 ~= 0.184$, and the greedy algorithm would produce $\\pi = \\begin{pmatrix} a_{1,1} & a_{2,1} & a_{1,2} & a_{2,2} \\end{pmatrix}$. Its expected cost, $\\mathbf{E}[C(\\pi)]$, is $2.428$, but if we inspect the second column first, then the expected cost is $2.407$ which is optimal in this example.\n One can consider another greedy algorithm which inspects the columns in increasing order of their expected number of inspections (within column), but this algorithm turns out to be suboptimal as well. \n\n\n \\subsection{Main Result and Algorithm}\n Let us present the main result that leads to an efficient algorithm for finding an optimal inspection policy. \n\n \\begin{theorem} \\label{thm:main_result}\n \tLet $s_c$ be the probability of success for column $c$ as before. \n \tLet $\\mu_c$ be the expected number of inspections that column $c$ incurs if its entries are inspected in increasing order of their probability of success, conditioning on the event that column $c$ is inspected and infeasible.\n \tAn optimal policy must be a column-by-column policy (due to Theorem~\\ref{theorem:col_by_col_opt}), must inspect the entries of each column in non-decreasing order of probabilities (due to Theorem~\\ref{theorem:within_column_opt}), and must inspect the columns in non-decreasing order of $\\mu_c(1 - s_c)/s_c$.\n \\end{theorem}\n \\begin{proof}\n \tConsider a column-by-column inspection policy $\\pi$ which inspects the column 1 through $m$ in increasing order of their index (we can assume this without loss of generality by re-labeling columns).\n\t\n \tAs before, let $N(\\pi,c)$ be a random variable that denotes the number of inspections performed by $\\pi$ on column $c$, conditioning on the event that column $c$ is inspected.\n \tThen we can express $\\mathbf{E}[N(\\pi, c)]$ in terms of $s_c$ and $\\mu_c$ as follows.\n \t\\begin{equation}\\label{eqn:exp_cost_column_by_condition}\n \t\t\\mathbf{E}[N(\\pi, c)] = s_c \\cdot n + (1-s_c) \\cdot \\mu_c\n \t\\end{equation}\n \tThis equation holds because if the column is feasible (with probability $s_c$), it would require $n$ inspections, but if it is not (with probability $1-s_c$), it would require $\\mu_c$ inspections in expectation. The equation above simply considers these two events, and calculates the expected value of $N(\\pi, c)$. \n\t\n \tSuppose that there is some column $c^*$ such that $\\mu_{c^*}(1 - s_{c^*}) / s_{c^*} > \\mu_{c^*+1} (1 - s_{c^*+1}) / s_{c^*+1}$. Because $\\pi$ inspects column $c^*$ before column $c^*+1$, it would not be inspecting the columns in increasing order of $\\mu_c(1-s_c)/s_c$.\n \tConsider a different inspection policy $\\pi'$ which inspects the columns in the same order as $\\pi$ except that $\\pi'$ inspects column $c^*+1$ before $c^*$ by swapping the inspection ordering of the two.\n \tWe can relate $N(\\pi, \\cdot)$ to $N(\\pi', \\cdot)$ as follows.\n \t\\begin{equation*}\n \t\tN(\\pi', c) = \n \t\t\\begin{cases}\n \t\t\tN(\\pi, c^*+1) & \\mbox{if~} c = c^* \\\\\n \t\t\tN(\\pi, c^*) & \\mbox{if~} c = c^* + 1 \\\\\n \t\t\tN(\\pi, c) & \\mbox{otherwise}\n \t\t\\end{cases}\n \t\\end{equation*}\n\t\n \tAs we did in proofs of Theorems~\\ref{theorem:col_by_col_opt} and \\ref{theorem:within_column_opt}, we can use Equations~\\ref{eqn:exp_c_pi_decoupled} and \\ref{eqn:exp_c_pi2_decoupled}, and simplify $\\mathbf{E}[C(\\pi)] - \\mathbf{E}[C(\\pi')]$ as follows.\n\t\\small\n \t\\begin{equation} \\label{eqn:main_result_diff}\n \t\t\\begin{aligned}\n \t\t\\mathbf{E}[C(\\pi)] - \\mathbf{E}[C(\\pi')] ={} & \\left(\\frac{\\mathbf{E}[N(\\pi, c^*)]}{s_{c^*}} - \\frac{\\mathbf{E}[N(\\pi, c^*+1)]}{s_{c^*+1}} \\right) \\\\\n \t\t& \\cdot \\left( \\prod_{j=1}^{c^*-1} (1 - s_j)\\right) s_{c^*} s_{c^*+1}\n \t\t \\end{aligned}\n \t\\end{equation}\t\n\t\\normalsize\n \tThe quantity in Equation~\\ref{eqn:main_result_diff} is positive if the difference of the weighted expected values (in the first parentheses) are positive. Using Equation~\\ref{eqn:exp_cost_column_by_condition} we obtain the following inequality.\n \t\\begin{equation*}\n \t\t\\begin{aligned}\n \t\t\t{} &\n \t\t\\frac{\\mathbf{E}[N(\\pi, c^*)]}{s_{c^*}} > \\frac{\\mathbf{E}[N(\\pi, c^*+1)]}{s_{c^*+1}} \\\\\n \t\t\\Leftrightarrow &\n \t\t\\mu_{c^*} (1 - s_{c^*}) / s_{c^*} > \\mu_{c^*+1}(1 - s_{c^*+1}) / s_{c^*+1} \n \t\t\\end{aligned}\n \t\\end{equation*}\n \tBy definition of $c^*$, the second inequality above holds, which implies $\\mathbf{E}[C(\\pi)] > \\mathbf{E}[C(\\pi')]$.\n \tTherefore, an optimal inspection policy must inspect the columns in non-decreasing order of $\\mu_c(1 - s_c)/s_c$. \n \\end{proof}\n Because there is a unique ordering of columns if we sort them by $\\mu_c(1-s_c)/s_c$ (up to ties), Theorem~\\ref{thm:main_result} leads to the following algorithm: We inspect the columns in increasing order of $\\mu_c(1-s_c)/s_c$, and in each column, we inspect the entries of it in increasing order of probabilities. \n This algorithm can easily be implemented to run in polynomial time. \n\n\n\\section{Discussion and Future Work} \\label{PMIP:sec:discussion}\nIn this work we defined the Probabilistic Matrix Inspection problem motivated by group scheduling and Doodle. \nWe first considered two special cases, and discovered interesting properties of an optimal inspection policy which agree with our intuition. We then generalized our findings to design an efficient algorithm to solve the general case, and along the way we showed that two natural greedy algorithms fail to find an optimal solution. While we believe that our technical results make a great starting point for studying and optimizing a group scheduling process, there remain several open problems and future work to be done.\n\nAs we discussed in previous chapter, our model and algorithm rely on the assumption that probability estimates on availability of agents are available. We suggested several ideas motivated by previous work in the literature, but it will be important to deploy such ideas into a system, and integrate it with our algorithm. From a theoretical perspective, there remain several open problems. While we assumed that an inspection can be performed on a single entry at unit cost, one can generalize the cost model by allowing an inspection of any subset of entries whose cost depends on, for example, the number of entries being inspected. In the context of group scheduling, an inspection on many entries means querying multiple agents at the same time for one or many outcomes (but an inspection is not limited to the entries from the same column or row). This generalization is particularly useful when scheduling takes place in a hierarchical setting such as corporates. For instance the event organizer may feel that the cost of querying a supervisor is significantly different from that of querying a colleague.\n\nLastly, although we focused on relating our model to group scheduling, the Probabilistic Matrix Inspection problem has other applications. Finding the right childcare facility, for example, involves extensive inquiries as parents wish to gather more information about how they would handle certain situations, what benefits and environments they provide, and so on. Through advertisements or brochures parents may even be able to gauge the likelihood of a certain facility satisfying their needs. Yet they still need to inquire facilities for precise information, which can be modeled by our probabilistic matrix model where columns correspond to facilities and rows correspond to the needs of parents. \n\n", "meta": {"hexsha": "0ab6225f9d066722cf859f4ffddd7b3917bede38", "size": 31727, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "prior to 2016-09-18/30Matrix.tex", "max_stars_repo_name": "ltdtl/thesis", "max_stars_repo_head_hexsha": "b1585aa3e57e06b4368fb51540bbbf1c64c491df", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2016-09-18T17:20:45.000Z", "max_stars_repo_stars_event_max_datetime": "2016-09-18T17:20:45.000Z", "max_issues_repo_path": "prior to 2016-09-18/30Matrix.tex", "max_issues_repo_name": "ltdtl/thesis", "max_issues_repo_head_hexsha": "b1585aa3e57e06b4368fb51540bbbf1c64c491df", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-01-29T07:21:01.000Z", "max_issues_repo_issues_event_max_datetime": "2017-01-29T07:21:01.000Z", "max_forks_repo_path": "prior to 2016-09-18/30Matrix.tex", "max_forks_repo_name": "ltdtl/thesis", "max_forks_repo_head_hexsha": "b1585aa3e57e06b4368fb51540bbbf1c64c491df", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 94.9910179641, "max_line_length": 1113, "alphanum_fraction": 0.7077883191, "num_tokens": 9474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.839733963661418, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.6387441920906125}} {"text": "%---------------------------Shape & Size-----------------------------\n\\section{Shape and Size\\label{s:tri-shape-and-size}}\n\nLet $R$ be the relative size squared as defined in \\S\\ref{s:tri-rel-size-squared}\nand $S$ be the shape as defined in \\S\\ref{s:tri-shape}.\nThen the ``shape and size'' metric is \n\\[\n q = RS\n\\]\n\n\\trimetrictable{shape and size}%\n{$1$}% Dimension\n{$[0.25,1]$}% Acceptable range\n{$[0,1]$}% Normal range\n{$[0,1]$}% Full range\n{Dependent on $\\overline{A}$}% Unit equilateral triangle value\n{\\cite{knu:03}}% Reference(s) \n{v\\_tri\\_shape\\_and\\_size}% Verdict function name\n\n", "meta": {"hexsha": "aabb1beb2e50054cc9cd8932ceac9cb60cf5bff4", "size": 867, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TriShapeAndSize.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TriShapeAndSize.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/TriShapeAndSize.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 43.35, "max_line_length": 85, "alphanum_fraction": 0.4186851211, "num_tokens": 189, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8887587993853655, "lm_q2_score": 0.7185943805178139, "lm_q1q2_score": 0.6386570788740827}} {"text": "\\newpage\n\\part{Introduction}\n%\\section{Introduction}\n\n\\section{Terminologies \\& Intros }\n\n\\subsection{Categories of Business Analytics}\n\\begin{itemize}\n\t\\item descriptive analytics\n\t\t\\begin{itemize}\n\t\t\t\\item data engineering(organizing data, queries) \\& statistics (mean, trend, standard deviation, test hypotheses)\n\t\t\\end{itemize}\n\t\n\t \\item predictive analytics\n\t \t\\begin{itemize}\n\t \t\t\\item machine learning \\& econometrics\n\t \t\t\\item learn the pattern of data\n\t \t\\end{itemize}\n \t\n \t\\item prescriptive analytics\n \t\\begin{itemize}\n \t\t\\item algorithm \\& optimization models\n \t\\end{itemize}\n\t\n\\end{itemize}\n\n\\subsection{From Data to Information}\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.8\\textwidth]{datatoinfo.png}\n\\end{figure}\n\n\\subsection{Types of Analytic Exercises}\n\\begin{itemize}\n\t\\item Numeric Prediction\n\t\\begin{itemize}\n\t\t\\item \\textbf{supervised} learning\n\t\t\\item input: a collection of data input + \\textbf{known} output\n\t\t\n\t\t\t job: create a function $f(\\text{input}_{0}) = \\text{output}_0$\n\t\t\t \n\t\t\t output: \\textbf{prediction} values to a \\textbf{new} collection of data: $f(\\text{input}_1) = ?$ \n\t\t\\item example: linear regression\n\t\\end{itemize}\n\n\t\\item Classification\n\t\\begin{itemize}\n\t\t\\item \\textbf{supervised} learning\n\t\t\\item input: a collection of data input + \\textbf{known} label\n\t\t\t\n\t\t\t job: create a classifier $f(\\text{input}_0) = \\text{label}_0$\n\t\t\t \n\t\t\t output: \\textbf{prediction} label to a \\textbf{new} collection of data: $f(\\text{input}_1) = ?$\n\t\\end{itemize}\n\n\t\\item Clustering\n\t\\begin{itemize}\n\t\t\\item \\textbf{unsupervised} learning\n\t\t\\item input: a collection of data input\n\t\t\t \n\t\t\t job: identify \"natural\" grouping in data\n\t\t\t \n\t\t\t output: clustered data\n\t\\end{itemize}\n\n\t\\item Association Rule Analysis\n\t\\begin{itemize}\n\t\t\\item \\textbf{unsupervised} learning\n\t\t\\item input: a collection of data list\n\t\t\t \n\t\t\t job: identify \\textbf{relationships} in data from \\textbf{co-occuring items}\n\t\t\\item grocery store purchases analysis\n\t\\end{itemize}\n\\end{itemize}\n\n\\subsection{Machine Learning Terminology in categorizing analytic exercises}\n\\begin{itemize}\n\t\\item \\textbf{supervised learning}\n\t\\begin{itemize}\n\t\t\\item a training set is given\n\t\t\\item find relationship between input \\& target attributes\t\t\n\t\t\\item examples: numeric prediction, classification\n\t\\end{itemize}\n\n\t\\item \\textbf{unsupervised learning}\n\t\\begin{itemize}\n\t\t\\item only input data available, no training set\n\t\t\\item find regularities, irregularities, relationships, similarities among data points\n\t\t\\item examples: clustering, association rule analysis\n\t\\end{itemize}\n\\end{itemize}\n\n\\subsection{Model}\nexamples: \n\\begin{itemize}\n\t\\item linear model\n\t\\item decision tree\n\t\\item neural network\n\\end{itemize}\n\n\\section{Statistics Recap}\n\\subsection{Categories}\n\\begin{itemize}\n\t\\item \\hl{descriptive statistics}: \\textbf{summary} of data\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item examples: mean, standard deviation\n\t\\end{itemize}\n\t\n\t\\item \\hl{inferential statistics}: model \\textbf{patterns} of data, accounting for \\textbf{randomness} and drawing \\textbf{inferences about a larger population}\n\t\\begin{itemize}\n\t\t\\item estimation\n\t\t\\item hypothesis testing\n\t\t\\item forecasting\n\t\t\\item correlation\n\t\t\\item regression\n\t\\end{itemize}\n\t\n\\end{itemize}\n\n\\subsection{Random Variables}\n$X$ is a \\textbf{random variable} if:\n\\begin{itemize}\n\t\\item it represents a \\textbf{random draw} from a population\n\t\\item it's associated with a \\textbf{probability distribution}\n\t\\item either \\textbf{discrete} or \\textbf{continuous}\n\t\\item example: a random variable that follows a \\textbf{normal distribution} $N(\\mu, \\sigma^2)$ has a probability density function of \n\t$$f(x) = \\frac{1}{\\sigma \\sqrt{2\\pi}}e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}}$$\n\\end{itemize}\n\n\\subsection{Normal Distribution \\& Standard Normal Distribution}\nA random variable $X$ follows a normal distribution $N(\\mu, \\sigma^2)$\n\nTo \\textbf{''standardize''} a random variable $x$ to \\textbf{standard normal distribution }\n$N(0,1)$: $$Z = \\frac{X - \\mu}{\\sigma}$$\n\n\\subsection{Probability density function \\& Cumulative density function}\n\\textbf{Probability density function (pdf): f(x)} $$f(x) = P(X = x)$$\n\n\\textbf{Cumulative distribution fuction (cdf): F(x)} $$F(x) = P(X \\leq x)$$ $$P(X \\geq x) = 1 - F(x)$$ $$P(x_1 \\leq X \\leq x_2) = F(x_2) - F(x_1)$$\n\n\\subsection{Statistical Estimation}\n\\textbf{Statistical estimation}: \n\nThe parameters of a population are \\textbf{unknown}. However, we can \\textbf{estimate} the parameters by \\textbf{drawing a random sample} out of the population, analyzing this random sample and getting the statistics. The statistics should infer the parameters.\n\\\\ \\ \\\\\nRequirements of random sampling: \n\\begin{itemize}\n\t\\item each variable $X$ from the population is a random variable. \n\t\\item each combination of $n$ sample points has an \\textbf{equal} chance of being selected. \n\t\n\\end{itemize}\n\t$\\longrightarrow$ a random sample is a set of \\textbf{independent, identically distributed (i.i.d)} random variables.\n\\\\ \\ \\\\\n\\textbf{Categories:}\n\\begin{itemize}\n\t\\item Point Estimate\n\t\\begin{itemize}\n\t\t\\item sample mean\n\t\t\\item sample proportion\n\t\\end{itemize}\n\n\t\\item Interval Estimate\n\t\\begin{itemize}\n\t\t\\item \\textbf{confidence interval} for sample mean\n\t\t\\item \\textbf{confidence interval} for sample proportion\n\t\\end{itemize}\n\\end{itemize}\nPoint estimate is always within the interval estimate.\n\n\\subsubsection{Point Estimate: Population Mean and Its Estimation}\n\\begin{itemize}\n\t\\item \\textbf{Expected Value of X and the Population Mean}\n\t\n\tThe expected value of a probability \\textbf{weighted average} of $X$, $E(X)$, is the mean/expected value of the probability distribution of $X$. \n\t\n\t$f(x_i)$ is the (discrete) probability that $X = x_i$.\n\t$$\\mu_x = E(X) = \\Sigma_{i=1}^{n} x_i f(x_i)$$ or $$\\mu_x = E(X) = \\int_{-\\infty}^{+\\infty} xf(x) dx$$ \n\t\n\tIf this random sample is a set of i.i.d random variables, the \\textbf{expected value of X} is the \\textbf{population mean}(unknown).\n\t\n\t\\item \\textbf{Estimation of the Population Mean by Sample Mean}\n\t\n\t\\textbf{Sample Mean $\\bar{X}$:} the \\textbf{random variable} for the arithmetic mean of the sample. $\\bar{x}$ is the mean of a \\textbf{particular realization} of a sample.\n\t\n\t \n\t$$\\bar{X} = \\frac{\\Sigma X_i}{n} $$\n\t\n\tThis is a random variable, because a lot of samples are drawn repeatedly, the arithmetic mean of the sample has also a probability distribution. \\textbf{The mean(center) of this distribution should estimate the mean of the whole population. }\n\t\n\t\n\t\n\t\\item Requirements for an estimator: \\textbf{unbiased}\n\t\n\tExample: $$E(\\bar{X}) = \\mu_x$$\n\t\n\t\\item \\textbf{Standard Error of the Sample Mean}: \n\t\n\t\\begin{itemize}\n\t\t\\item \\hl{\\textbf{standard error of the sample mean $SE_{\\bar{X}}$}} : an estimate of how far the \\textbf{sample mean} is likely to be \\textbf{from the population mean}. \n\t\t\n\t\tWhen $n \\rightarrow \\infty$, $SE_{\\bar{X}} \\rightarrow \\sigma_{\\bar{X}}$ (\\hl{\\textbf{true standard deviation of sample mean}}).\n\t\t\n\t\t$$SE_{\\bar{X}} = \\frac{s}{\\sqrt{n}}$$\n\t\t$$\\sigma_{\\bar{X}} = SD(\\bar{X}) = \\sqrt{Var(\\bar{X})} = \\frac{\\sigma}{\\sqrt{n}}$$\n\t\t\n\t\t\\item\\hl{\\textbf{sample standard deviation $s$}} : the degree to which \\textbf{individuals within the sample} differ from the \\textbf{sample mean}\n\t\t\n\t\t$$s = \\sqrt{\\frac{1}{n-1} \\Sigma_{i=1}^{n} (X_i - \\bar{X})^2}$$\n\t\\end{itemize}\n\t\n\t\\item \\textbf{Law of Large Numbers}\n\t\n\tIf $n \\rightarrow \\infty, \\bar{X}_n \\rightarrow \\mu$. \n\t\n\tIf the size of the random sample is large enough, then the arithmetic mean converge to the real population mean. \n\t$$\\lim\\limits_{n\\rightarrow \\infty} P(|\\bar{X}_n - \\mu| > \\epsilon) = 0$$\n\t\n\t\\item \\textbf{Central Limit Theorem}\n\t\n\tIf \\textbf{$n \\rightarrow \\infty$}, the \\textbf{average $\\bar{X}$} of any population of \\textbf{i.i.d.} random variables $X_i$ with the population mean $\\mu_X$ and population variance $\\sigma^2$ \\textbf{follows asymptotically a normal distribution} $\\bar{X} \\sim N(\\mu_X, \\frac{\\sigma^2}{n})$. \n\t$$\\bar{X} = \\frac{X_1 + X_2 + \\dots + X_n}{n}$$\n\tThe \\textbf{standardize average} $Z \\sim N(0,1)$: $$Z = \\frac{\\bar{X} - \\mu_X}{\\frac{\\sigma}{\\sqrt{n}}}$$\n\\end{itemize}\n\n\\subsubsection{Interval Estimate: }\n\\begin{itemize}\n\t\\item \\textbf{Confidence Interval of Sample Mean:}\n\t\\begin{itemize}\n\t\t\\item Assumption: samples drawn from a population that follows a normal distribution $N(\\mu_X, \\sigma^2)$. \n\t\t\n\t\teg: The sample mean $\\bar{X}$ follows asymptotically a normal distribution $N(\\mu_X, \\frac{\\sigma^2}{n})$\n\t\t\\item a level of confidence($1 - \\alpha$) is given.\n\t\t\\item \\textbf{two-sided} $\\rightarrow z_{(1 - \\frac{\\alpha}{2})} $ / $z_{(1 + \\frac{\\alpha}{2})}$\n\t\t\\item z value: \\textbf{standardized}. \\textbf{Find z value from cdf-table given an $\\alpha$}.\n\t\t\\item if population standard deviation $\\sigma$ is given, \n\t\t$$CI = \\left[ \\bar{X} - z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{\\sigma}{\\sqrt{n}} , \\bar{X} + z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{\\sigma}{\\sqrt{n}} \\right] $$\n\t\t$$Pr(\\bar{X} - z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{\\sigma}{\\sqrt{n}} < \\mu_X < \\bar{X} + z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{\\sigma}{\\sqrt{n}} ) = 1 - \\alpha$$\n\t\t\n\t\t\\item if population standard deviation $\\sigma$ is unknown, \n\t\t\\begin{itemize}\n\t\t\t\\item n is \\textbf{small}: use \\textbf{sample standard deviation $s$} and \\textbf{t-distribution}\n\t\t\t$$CI = \\left[ \\bar{X} - t_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{s}{\\sqrt{n}} , \\bar{X} + t_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{s}{\\sqrt{n}} \\right] $$\n\t\t\t\\item n is \\textbf{large}: use \\textbf{sample standard deviation $s$} and \\textbf{normal distribution}\n\t\t\t$$CI = \\left[ \\bar{X} - z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{s}{\\sqrt{n}} , \\bar{X} + z_{(1 - \\frac{\\alpha}{2})}\\cdot \\frac{s}{\\sqrt{n}} \\right] $$\n\t\t\\end{itemize}\n\t\tIf $n \\rightarrow \\infty$, the T-Distribution converges to a Normal Distribution.\n\t\t\n\t\\end{itemize}\n\n\t\\item Effects on Confidence Intervals:\n\t\\begin{itemize}\n\t\t\\item \\textbf{sample size} n: n $\\uparrow$, interval size $\\downarrow$ \n\t\t\n\t\t(the larger the size, more precise is the estimation)\n\t\t\\item \\textbf{confidence level} ($1 - \\alpha$): confidence level $\\uparrow$, interval size $\\uparrow$ \n\t\t\n\t\t(given the same sample size, the higher the confidence level, the more values need to be included.)\n\t\t\\item \\textbf{population standard deviation} $\\sigma$: $\\sigma \\uparrow$, interval size $\\uparrow$ \n\t\t\n\t\t(the more spreaded the population, the more values need to be included to achieve same confidence level.)\n\t\\end{itemize}\n\\end{itemize}\n\n\\subsection{Statistical Tests}\n\\subsubsection{Process} \n\\begin{itemize}\n\t\\item number of samples\n\t\\begin{itemize}\n\t\t\\item 1 sample:\n\t\t\\begin{itemize}\n\t\t\t\\item $\\sigma$ known: \\textbf{Z-Test}\n\t\t\t\\item $\\sigma$ unknown: \\textbf{T-Test}\n\t\t\\end{itemize}\n\t\t\\item 2 samples:\n\t\t\\begin{itemize}\n\t\t\t\\item dependent:\\textbf{ Paired T-Test} \n\t\t\t\\item independent: \\textbf{Welch-Test}\n\t\t\\end{itemize}\n\t\\end{itemize}\n\t\\item Formulate \\textbf{null and alternative hypothesis} ($H_0$ and $H_1$), $H_1$ is the hypothesis we want to test.\n\t\\item Choose an \\textbf{$\\alpha$ level}. (type I error: probability of falsely rejecting $H_0$)\n\t\\item Find corresponding \\textbf{distribution}, calculate \\textbf{test statistic}, \n\t\\item Find the \\textbf{critical value} in the table and corresponding \\textbf{p-value}\n\t\\item Conclusion: \n\t\\begin{itemize}\n\t\t\\item $p \\leq \\alpha$, reject $H_0$\n\t\t\\item $p > \\alpha$, reject $H_1$ \n\t\\end{itemize}\n\tInterpretation of p-value: the probability of having the other mean $\\bar{x}$, given that $H_0$ is true.\n\\end{itemize}\n\n\\subsubsection{Two-sided or One-sided Test}\nThree possible alternative hypotheses $H_1$:\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.5\\textwidth]{H1.png}\n\\end{figure}\nCritical Value:\n\\begin{itemize}\n\t\\item two-sided test: $z_{\\frac{\\alpha}{2}}, z_{1-\\frac{\\alpha}{2}}$ / $t_{\\frac{\\alpha}{2}}, t_{1-\\frac{\\alpha}{2}}$\n\t\\item one-sided test: $z_{1-\\alpha}$ / $t_{1-\\alpha}$\n\\end{itemize}\n\n\n\n\\subsubsection{Z-Test}\n\\begin{itemize}\n\t\\item Requirements: \\textbf{1 sample, $\\mu$, $\\sigma$ known} (population mean and population standard deviation) \n\t\\item Distribution: \\textbf{standardized normal distribution}\n\t\\item test statistic: \\large{$$ z = \\frac{\\bar{X} - \\mu_0}{\\frac{\\sigma}{\\sqrt{n}}}$$}\n\t\\item critical value: $z^c_{1-\\frac{\\alpha}{2}}$ / $z^c_{1-\\alpha}$\n\t\\item $H_0$ Rejection region/ $H_1$ Acceptance region :\n\t\n\t\\begin{table}[H]\n\t\t\\begin{center}\n\t\t\t\\begin{tabular}{|c|c|c|}\n\t\t\t\t\\hline\n\t\t\t\t$H_1$ \t\t\t\t& Rejection Region\t\t\t\t\t& test variant \\\\ \\hline\n\t\t\t\t$\\mu\\mu \\neq \\mu_0$\t& $|z| \\geq z^c_{1-\\frac{\\alpha}{2}}$\t& two-sided \\\\ \\hline\n\t\t\t\t$\\mu\\mu > \\mu_0$\t& $ z \\geq z^c_{1-\\alpha}$\t\t\t& one-sided \\\\ \\hline\n\t\t\t\t$\\mu\\mu < \\mu_0$\t& $ z \\leq -z^c_{1-\\alpha} $\t\t\t& one-sided \\\\ \\hline\n\t\t\t\\end{tabular}\n\t\t\\end{center}\n\t\\end{table} \t\n\\end{itemize}\n\n\\subsubsection{Single Sample T-test}\n\\begin{itemize}\n\t\\item Requirements: \\textbf{1 sample, $\\sigma$ unknown}\n\t\\item Distribution: Student \\textbf{t-Distribution}\n\t\\item Degree of Freedom($df$): determines how spread the distribution is. \n\t\\large{$$df = n - 1$$}\n\t\\item test statistic: \n\t\n\t$s$ is the standard error(empirical)\n\t\\large{$$t(df) = \\frac{\\bar{X} - \\mu_0}{\\frac{s}{\\sqrt{n}}}$$}\n\t\\item critical value: $t^c_{1-\\frac{\\alpha}{2}, df}$ / $t^c_{1-\\alpha, df}$\n\t\\item $H_0$ Rejection region/ $H_1$ Acceptance region :\n\t\n\t\\begin{table}[H]\n\t\t\\begin{center}\n\t\t\t\\begin{tabular}{|c|c|c|}\n\t\t\t\t\\hline\n\t\t\t\t$H_1$ \t\t\t\t& Rejection Region\t\t\t\t\t& test variant \\\\ \\hline\n\t\t\t\t$\\mu\\mu \\neq \\mu_0$\t& $|t| \\geq t^c_{1-\\frac{\\alpha}{2}, df}$\t& two-sided \\\\ \\hline\n\t\t\t\t$\\mu\\mu > \\mu_0$\t& $ t \\geq t^c_{1-\\alpha, df}$\t\t\t& one-sided \\\\ \\hline\n\t\t\t\t$\\mu\\mu < \\mu_0$\t& $ t \\leq -t^c_{1-\\alpha,df} $\t\t\t& one-sided \\\\ \\hline\n\t\t\t\\end{tabular}\n\t\t\\end{center}\n\t\\end{table} \t\t\n\\end{itemize}\n\n\\subsubsection{Paired Sample T-test}\n\\begin{itemize}\n\t\\item Requirements: \\textbf{2 samples, $\\sigma$ unknown, dependent}\n\t\n\teg: means obtained in 2 conditions(time,places,etc.) by a single group of participants\n\t\\item Distribution: \\textbf{T-Distribution}\n\t\\item Test of relationship between 2 linked samples (eg: Difference)\n\t\\item Degree of Freedom($df$): $$df = n -1 $$\n\t\\item Example null hypothesis: $ H_0: \\mu_d = \\mu_1 - \\mu_2 = \\Delta_0$\n\t\\item test statistic: \n\t\\large{$$t = \\frac{\\bar{d} - \\Delta_0}{\\frac{s}{\\sqrt{n}}}$$}\n\t\\item critical value: $t^c_{1-\\frac{\\alpha}{2}, df}$ / $t^c_{1-\\alpha, df}$\n\t\\item $H_0$ Rejection Region/ $H_1$ Acceptance Region:\n\t\\begin{table}[H]\n\t\t\\begin{center}\n\t\t\t\\begin{tabular}{|c|c|c|}\n\t\t\t\t\\hline\n\t\t\t\t$H_1$ \t\t\t\t& Rejection Region\t\t\t\t\t& test variant \\\\ \\hline\n\t\t\t\t$\\mu_d \\neq \\Delta_0$\t& $|t| \\geq t^c_{1-\\frac{\\alpha}{2}, df}$\t& two-sided \\\\ \\hline\n\t\t\t\t$\\mu_d > \\Delta_0$\t& $ t \\geq t^c_{1-\\alpha, df}$\t\t\t& one-sided \\\\ \\hline\n\t\t\t\t$\\mu_d < \\Delta_0$\t& $ t \\leq -t^c_{1-\\alpha,df} $\t\t\t& one-sided \\\\ \\hline\n\t\t\t\\end{tabular}\n\t\t\\end{center}\n\t\\end{table} \t\n\\end{itemize}\n\n\\subsubsection{Independent T-Test/ Welch-Test}\n\\begin{itemize}\n\t\\item Requirements: \\textbf{2 samples, $\\sigma$ unknown, independent}\n\t\n\teg: credit card debt difference between urban and rural households/ shopping expenses between male and female\n\t\\item Distribution: \\textbf{T-Distribution}\n\t\\item Test of relationship between 2 independent samples (doesn't need to be same size)\n\t\\item Degree of Freedom($df$): round to \\textbf{nearest integer} number\n\t\\large{$$df = \\dfrac{(\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2})^2}{\\frac{(\\frac{s_1^2}{n_1})^2}{n_1 -1} + \\frac{(\\frac{s_2^2}{n_2})^2}{n_2 -1}}$$} \n\t\\item test statistic:\n\t\\large{$$t = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\mu_0}{s_{\\bar{x}_1 -\\bar{x}_2}}$$}\n\t\\large{$$s_{\\bar{x}_1 -\\bar{x}_2} = \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}$$}\n\t\\item $H_0$ Rejection Region/ $H_1$ Acceptance Region:\n\t\\begin{table}[H]\n\t\t\\begin{center}\n\t\t\t\\begin{tabular}{|c|c|c|}\n\t\t\t\t\\hline\n\t\t\t\t$H_1$ \t\t\t\t& Rejection Region\t\t\t\t\t& test variant \\\\ \\hline\n\t\t\t\t$\\mu_d \\neq \\Delta_0$\t& $|t| \\geq t^c_{1-\\frac{\\alpha}{2}, df}$\t& two-sided \\\\ \\hline\n\t\t\t\t$\\mu_d > \\Delta_0$\t& $ t \\geq t^c_{1-\\alpha, df}$\t\t\t& one-sided \\\\ \\hline\n\t\t\t\t$\\mu_d < \\Delta_0$\t& $ t \\leq -t^c_{1-\\alpha,df} $\t\t\t& one-sided \\\\ \\hline\n\t\t\t\\end{tabular}\n\t\t\\end{center}\n\t\\end{table} \n\t\\item Confidence Interval of both samples: if both confidence intervals overlap each other $\\rightarrow$ \\textbf{cannot reject} $H_0$\n\\end{itemize}\n\\subsubsection{Using Confidence Intervals in Significance Tests}\nFind confidence intervals for $\\mu_x$, which -- under $H_0$ -- contains the true value $\\mu_x$ with a probability of at least $1 - \\alpha$: \n\\begin{itemize}\n\t\\item $\\sigma$ known: \\textbf{Normal Distribution}\n\t$$CI = [\\bar{x} - z^c_{(1-\\frac{\\alpha}{2})} \\cdot \\frac{\\sigma}{\\sqrt{n}}, \\bar{x} + z^c_{(1-\\frac{\\alpha}{2})} \\cdot \\frac{\\sigma}{\\sqrt{n}}]$$\n\t\\item $\\sigma$ unknown: \\textbf{t-Distribution}\n\t$$CI = [\\bar{x} - t^c_{(1-\\frac{\\alpha}{2}, n-1)} \\cdot \\frac{s}{\\sqrt{n}}, \\bar{x} + t^c_{(1-\\frac{\\alpha}{2}, n-1)} \\cdot \\frac{s}{\\sqrt{n}}]$$\n\\end{itemize}\t\nConclusion:\n\\begin{itemize}\n\t\\item Accept $H_0$: if $\\mu_0$ lies \\textbf{within} the confidence interval\n\t\\item Rejection $H_0$: if $\\mu_0$ lies \\textbf{outside} the confidence interval\n\\end{itemize}\n\n\\subsubsection{Other Tests}\n\\begin{itemize}\n\t\\item Parametric Tests: eg: T-tests, F-test(comparing variance of 2 samples)\n\t\\item Non-parametric Tests: eg: Wilcoxon signed-rank test(2 paired i.i.d. samples), Mann-Whitney-U test(2 independent i.i.d. samples)\n\t\\item Test of Probability Distribution: Kolmogorov-Smirnov test, Chi-square test\n\\end{itemize}\n\n\\section{Description of a Dataset}\n\\begin{itemize}\n\t\\item \\textbf{Dependent} and \\textbf{independent} variables\n\t\\item Scales of measurement of the variables \n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[width=0.8\\textwidth]{scale_measurement.jpg}\n\t\\end{figure}\n\t\\begin{itemize}\n\t\t\\item \\textbf{Nominal}: \\textbf{categorical} variable scale. A scale used for labeling variables into distinct classifications, \\textbf{doesn't involve quantitative value or order. }\n\t\t\\begin{itemize}\n\t\t\t\\item eg: Gender, City, Nationality, Jobs\n\t\t\\end{itemize}\n\t\t\\item \\textbf{Ordinal}: scale to depict \\textbf{order} of the variables for non-mathematical ideas. It maintains a \\textbf{descriptional} quality and the \\textbf{difference between variables can't be calculated}. \n\t\t\\begin{itemize}\n\t\t\t\\item eg: Frequency(high, medium, low), Happiness, Satisfaction, Pain level, Cloth size (S, M, L), rank of Unis\n\t\t\\end{itemize}\n\t\t\\item \\textbf{Interval}: numeric scale where \\textbf{order} of the variables as well as the \\textbf{difference between variables are known}. There exists \\textbf{no true 0}. Variables can be added and subtracted, but \\textbf{not multiplied or divided}. \n\t\t\\begin{itemize}\n\t\t\t\\item eg: GPA, GRE, Celsius, Fahrenheit (20$^\\circ$C is 10$^\\circ$C higher than 10$^\\circ$C, it doesn't mean 2 times warmer.)\n\t\t\\end{itemize}\n\t\t\\item \\textbf{Ratio}: numeric scale that's ordered, difference between variables known, and there \\textbf{exists true 0}. Variables can be added, subtracted, multiplied and divided.\n\t\t\\begin{itemize}\n\t\t\t\\item eg: weight, height, time, Kelvin temperature, money\n\t\t\\end{itemize}\n\t\\end{itemize}\n\t\\item \\textbf{Cross-sectional, time series, panel data} (see 7.1.5)\n\\end{itemize}\n\n\n%\\section{Summary of Possible Exam Questions}\n%\\begin{itemize}\n%\t\\item Calculation of arithmetic weighted mean of sample $\\bar{X}$, variance $\\sigma^2$ of the sample, standard deviation $\\sigma$ of the sample.\n%\t\\item Calculation of variance/standard deviation of the sample mean $\\sigma_{\\bar{X}}^2$/$\\sigma_{\\bar{X}}$\n%\t\\item Standardization of a normal distribution \n%\t\\item Calculation of probability, find values from pdf or cdf table \n%\t\\item independent, identically distributed sample points (i.i.d.)\n%\t\\item Performance of Significance Test according to requirements\n%\t\\item p-value: Interpretation when to reject $H_0$ ($p < \\alpha$)\n%\t\\item description of dataset\n%\\end{itemize}\n", "meta": {"hexsha": "15d2c9f88ca33b8619bdc2d76fe6b5ca39ca654b", "size": 20086, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Business Analytics/lectures/introduction.tex", "max_stars_repo_name": "YourPsychiatrist/TUM", "max_stars_repo_head_hexsha": "12e60881c225408d057b8637594c37fa54c3bcfa", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 225, "max_stars_repo_stars_event_min_datetime": "2019-10-02T10:49:41.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-29T22:25:38.000Z", "max_issues_repo_path": "Business Analytics/lectures/introduction.tex", "max_issues_repo_name": "YourPsychiatrist/TUM", "max_issues_repo_head_hexsha": "12e60881c225408d057b8637594c37fa54c3bcfa", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-02-16T12:22:43.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-31T19:35:57.000Z", "max_forks_repo_path": "Business Analytics/lectures/introduction.tex", "max_forks_repo_name": "YourPsychiatrist/TUM", "max_forks_repo_head_hexsha": "12e60881c225408d057b8637594c37fa54c3bcfa", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 69, "max_forks_repo_forks_event_min_datetime": "2019-10-02T21:46:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-17T19:27:50.000Z", "avg_line_length": 42.8272921109, "max_line_length": 295, "alphanum_fraction": 0.6787314547, "num_tokens": 6677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8459424353665381, "lm_q1q2_score": 0.6386146224854996}} {"text": "\\documentclass{ximera}\n\\input{../preamble}\n\\title{Exercises: Partial Fractions}\n%%%%%\\author{Philip T. Gressman}\n\n\\begin{document}\n\\begin{abstract}\nVarious exercises relating to partial fractions and integration.\n\\end{abstract}\n\\maketitle\n\n\nCompute the indefinite integrals below. Since there are many possible answers (which differ by constant values), use the given instructions if needed to choose which possible answer to use. Do not forget absolute value signs inside logarithms when they are needed.\n\n\\begin{exercise}%[APEX0604PF07]\n\\[ \\frac{7x+7}{x^2+3x-10} = \\frac{\\answer{3}}{x-2} + \\frac{\\answer{4}}{x+5} \\]\n\\[ \\int \\frac{7x+7}{x^2+3x-10}\\ dx = \\answer{3 \\ln |x-2|+4 \\ln |x+5|}+C\\]\n(Do not include any constant terms in your antiderivative.)\n%%\n%%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0604PF08]\n\\[ \\int \\frac{7x-2}{x^2+x}\\ dx = \\answer{9 \\ln |x+1|-2 \\ln |x|}+C\\]\n(Do not include any constant terms in your antiderivative.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0604PF11]\n\\[ \\frac{x+7}{(x+5)^2} = \\frac{\\answer{1}}{x+5} + \\frac{\\answer{2}}{(x+5)^2} \\]\n\\[ \\int \\frac{x+7}{(x+5)^2}\\ dx = \\answer{\\ln |x+5|-\\frac{2}{x+5}}+C\\]\n(Do not include any constant terms in your antiderivative.)\n%\n%\n\\end{exercise}\n\n\n\\begin{exercise}%[APEX0604PF13]\n\\[ \\int \\frac{9x^2+11x+7}{x(x+1)^2}\\ dx = \\answer{\\frac{5}{x+1}+7 \\ln |x|+2 \\ln |x+1|}+C\\]\n(Do not include any constant terms in your answer.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0604PF16]\n\\[ \\int \\frac{x^2+x+1}{x^2+x-2}\\ dx = \\answer{x+\\ln |x-1|-\\ln |x+2|}+C \\]\n(Do not include any constant terms in your answer.)\n\\begin{hint}\nDon't forget polynomial long division; it is needed in this case because the degree of the numerator is at least as large as the degree of the denominator.\n\\end{hint}\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0604PF20]\n\\[ \\int \\frac{x^2+x+5}{x^2+4x+10}\\ dx = \\answer{-\\frac{3}{2} \\ln \\left|x^2+4 x+10\\right|+x+\\frac{\\arctan \\left(\\frac{x+2}{\\sqrt{6}}\\right)}{\\sqrt{6}}}+C \\]\n(Do not include any constant terms in your answer.)\n\\begin{hint}\n\\[ \\frac{x^2+x+5}{x^2+4x+10} = \\answer{1} + \\frac{\\answer{-3} x + \\answer{-5}}{x^{2} + 4 x + 10} \\]\nSince the derivative of the denominator is $2x + 4$, we should rewrite the numerator of the big fraction to have $x+2$'s if possible:\n\\[ \\frac{x^2+x+5}{x^2+4x+10} = \\answer{1} + \\frac{\\answer{-3}(x+2) + \\answer{1}}{x^2 + 4x + 10}. \\]\nFor expressions like\n\\[ \\int \\frac{x+2}{x^2 +4x+10} dx \\]\nwe should do a substitution. For terms like\n\\[ \\int \\frac{1}{x^2 + 4x + 10} dx \\]\nwe should first complete the square: $x^2 + 4x + 10 = (x + 2)^2 + 6$ and then make the substitution $x + 2 = u \\sqrt{6}$.\n\\end{hint}\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0604PF23]\n\\[ \\int \\frac{2x^2+x+1}{(x+1)(x^2+9)}\\ dx = \\answer{\\frac{9}{10} \\ln \\left|x^2+9\\right|+\\frac{1}{5} \\ln |x+1|-\\frac{4}{15} \\arctan \\left(\\frac{x}{3}\\right)} +C\\]\n(Do not include any constant terms in your answer.)\n%\n%\n\\end{exercise}\n\n\\section*{Sample Quiz Questions}\n\n\\begin{question}%%%%%[PartialFracHeav001v2]\n\nCompute the integral\n\\[\\int_{3}^{4}\\frac{2x-3}{x^2-3x+2}~dx.\\]\n(Hints won't be revealed until after you choose a response.)\n\\begin{multiplechoice}\n\\choice{\\(\\displaystyle \\ln 2\\)}\n\\choice[correct]{\\(\\displaystyle \\ln 3\\)}\n\\choice{\\(\\displaystyle \\ln 4\\)}\n\\choice{\\(\\displaystyle \\ln 5\\)}\n\\choice{\\(\\displaystyle \\ln 6\\)}\n\\choice{\\(\\displaystyle \\ln 7\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nFirst factor the denominator of the integrand: \\(x^2-3x+2 = (x-1)(x-2)\\). Since the roots are distinct, it is possible to use the Heaviside cover-up method. \\begin{hint} The partial fractions expansion will take the form \\[\\frac{A}{x-1} + \\frac{B}{x-2}, \\] where the coefficient \\(A\\) can be computed by cancelling the factor of \\(x-1\\) in the denominator and evaluating the result at \\(x = 1\\), i.e., \\[A = \\frac{2(1)-3}{(1)-2} = 1. \\] Similarly, \\[B = \\frac{2(2)-3}{(2)-1} = 1,\\] which gives that \\[\\frac{2x-3}{(x-1)(x-2)} = \\frac{1}{x-1} + \\frac{1}{x-2}. \\] \\begin{hint} Therefore\n\\[ \\begin{aligned} \n\\int_{3}^{4}\\frac{2x-3}{x^2-3x+2}~dx & = \\int_{3}^{4}\\left(\\frac{1}{x-1} + \\frac{1}{x-2}\\right)~dx \\\\\n & = \\left(\\ln |4-1| + \\ln |4-2| \\right) - \\left(\\ln |3-1| + \\ln |3-2| \\right)\\\\\n & = \\ln 3 + \\ln 2 + \\ln \\frac{1}{2} + 0 = \\ln 3.\\end{aligned}\\] \\end{hint} \\end{hint}\n\\end{feedback}\n\n\\end{question}\n\n\\section*{Sample Exam Questions}\n\n\\begin{question}%%%%%[2019WrittenByClass01]\nCompute the volume of the solid of revolution obtained by revolving around the $y$-axis the region below the graph\n\\[ y = \\frac{1}{(x-1)^2}, \\]\nabove $y=0$, and between $x=2$ and $x=3$.\n(Hints won't be revealed until after you choose a response.)\n\\begin{multiplechoice}\n\\choice{\\(\\displaystyle \\pi\\)}\n\\choice{\\(\\displaystyle \\pi ( \\ln 2 + 3)\\)}\n\\choice[correct]{\\(\\displaystyle \\pi (2 \\ln 2 + 1)\\)}\n\\choice{\\(\\displaystyle \\pi (2 \\ln 3 + 1)\\)}\n\\choice{\\(\\displaystyle \\pi (3 \\ln 2 + 1)\\)}\n\\choice{\\(\\displaystyle \\pi (3 \\ln 3 + 1)\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nChoosing $x$ as the variable of integration, slices will be parallel to the $y$-axis, indicating that the shell method should be used. The radius of a shell is $x$ (because the axis lies to the left of the region) and the height will be $(x-1)^{-2}$, so \n\\[ V = \\int_2^3 \\frac{2 \\pi x}{(x-1)^2} dx = 2 \\pi \\int_2^3 \\frac{x}{(x-1)^2} dx. \\] \\begin{hint}\nThe integral can be computed by partial fractions; the expansion has the form\n\\[ \\frac{x}{(x-1)^2} = \\frac{A}{x-1} + \\frac{B}{(x-1)^2}. \\]\nThe coefficients $A$ and $B$ can be found by usual methods (but note that the Heaviside cover up method will \\textit{not} work in this case), but it is also possible to find them directly by carefully rewriting the numerator of the fraction in terms of $x-1$:\n\\[ \\frac{x}{(x-1)^2} = \\frac{(x-1) + 1}{(x-1)^2} = \\frac{(x-1)}{(x-1)^2} + \\frac{1}{(x-1)^2} = \\frac{1}{x-1} + \\frac{1}{(x-1)^2}. \\]\nTherefore\n\\[ \\begin{aligned} V & = 2 \\pi \\int_2^3 \\left[ \\frac{1}{x-1} + \\frac{1}{(x-1)^2} \\right] dx = 2 \\pi \\left. \\left[ \\ln |x-1| - \\frac{1}{x-1} \\right] \\right|_{2}^3 \\\\\n& = 2 \\pi \\left( \\ln 2 - \\frac{1}{2} \\right) - 2 \\pi \\left( 0 - 1 \\right) = \\pi(2 \\ln 2 + 1). \\end{aligned}\\] \\end{hint}\n\\end{feedback}\n\n\\end{question}\n\n\\begin{question}%%%%%[2016C.04]\n\nCompute the constants \\(A\\) and \\(B\\) in the partial fractions expansion indicated below. \\offline{To receive full credit, it is not necessary to compute \\(C, D,\\) or \\(E\\).}\n\\[ \\frac{x^4 + 16}{x^4 - 16} =A + \\frac{B}{x-2} + \\frac{C}{x+2} + \\frac{Dx + E}{x^2 + 4} \\]\n(Hints won't be revealed until after you choose a response.)\n\\begin{multiplechoice}\n\\choice{\\(A=-1, B=1\\)}\n\\choice{\\(A = 0, B = 1\\)}\n\\choice[correct]{\\(A = 1, B = 1\\)} \n\\choice{\\(A=-1, B=-1\\)}\n\\choice{\\(A = 0, B = -1\\)}\n\\choice{\\(A = 1, B = -1\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nYou'll need to do polynomial long division first. To compute $B$, you can use Heaviside cover-up.\n\\begin{hint}\n\\[ \\frac{x^4 + 16}{x^4 - 16} = 1 + \\frac{1}{x-2} - \\frac{1}{x+2} - \\frac{4}{x^2+4} \\]\n\\end{hint}\n\\end{feedback}\n\n\n\\end{question}\n\n\\begin{question}%%%%%[2017C.04]\n\nEvaluate \\(\\displaystyle \\int_1^2 \\frac{x^2+x+1}{x^2+x} dx\\).\n\\begin{multiplechoice}\n\\choice{\\(0\\)}\n\\choice{\\(1\\)}\n\\choice[correct]{\\(\\displaystyle 1 + \\ln \\left(\\frac{4}{3}\\right)\\)}\n\\choice{\\(2\\)}\n\\choice{\\(\\displaystyle 2 + \\ln \\left(\\frac{8}{3}\\right)\\)}\n\\choice{none of these}\n\\end{multiplechoice}\n\n\\end{question}\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "cd22679067f189c06c76d3b25006023eabc08b19", "size": 7360, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "techniques/12partialfractionspractice.tex", "max_stars_repo_name": "ptgressman/math104", "max_stars_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "techniques/12partialfractionspractice.tex", "max_issues_repo_name": "ptgressman/math104", "max_issues_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "techniques/12partialfractionspractice.tex", "max_forks_repo_name": "ptgressman/math104", "max_forks_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.0571428571, "max_line_length": 584, "alphanum_fraction": 0.6315217391, "num_tokens": 2862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619350028204, "lm_q2_score": 0.7745833893685269, "lm_q1q2_score": 0.6386145200198188}} {"text": "\\subsection{Modules}\\label{subsec:modules}\n\n\\begin{definition}\\label{def:left_module}\n Let \\( R \\) be a \\hyperref[def:semiring/dioid]{dioid} and \\( M \\) be an abelian group. Analogously to \\fullref{def:left_monoid_action}, we say that \\( M \\) is a left \\( R \\)-\\term{module} if it has some additional structure, which can be defined equivalently as\n \\begin{thmenum}\n \\thmitem{def:left_module/homomorphism} a dioid homomorphism from \\( R \\) to the endomorphism \\hyperref[def:endomorphism_dioid]{ring} \\( \\End(M) \\)\n\n \\thmitem{def:left_module/multiplication}\\cite[374]{Knapp2016BasicAlgebra} an \\hyperref[def:magma/associative]{associative} and \\hyperref[def:unital_magma]{unital} \\hyperref[def:left_monoid_action/operation]{operation} \\( \\cdot: R \\times M \\to M \\), written using juxtaposition.\n\n We require that \\( \\cdot \\) is associative, distributive over \\( + \\), and compatible with the identity in \\( R \\). Explicitly, the following are satisfied for \\( x, y \\in M \\) and \\( s, t \\in R \\):\n \\begin{thmenum}\n \\thmitem{def:left_module/associativity}(associativity) \\( s \\cdot (t \\cdot x) = (s t) \\cdot x \\).\n \\thmitem{def:left_module/scalar_distributivity}(scalar distributivity) \\( (s + t) \\cdot x = s \\cdot x + t \\cdot x \\).\n \\thmitem{def:left_module/vector_distributivity}(vector distributivity) \\( t \\cdot (x + y) = t \\cdot x + t \\cdot y \\).\n \\thmitem{def:left_module/identity}(identity) \\( 1_R \\cdot x = x \\).\n \\end{thmenum}\n \\end{thmenum}\n\n In analogy with \\hyperref[def:vector_space]{vector spaces}, we call elements of \\( R \\) scalars and elements of \\( M \\) vectors. See \\fullref{def:vector_space}.\n\n We denote the category of modules over \\( R \\) by \\( \\cat{Mod}_R \\).\n\\end{definition}\n\\begin{proof}\n \\ImplicationSubProof{def:left_module/homomorphism}{def:left_module/multiplication} Let \\( \\tau: R \\to \\End(M) \\) be a ring homomorphism. Define the operation\n \\begin{balign*}\n & \\cdot: R \\times M \\to M \\\\\n & \\cdot(r, m) \\coloneqq \\tau(r)(m).\n \\end{balign*}\n\n Let \\( x, y \\in M \\) and \\( s, t \\in R \\). From the fact that \\( \\tau \\) is a ring homomorphism, we have\n \\SubProofOf{def:left_module/associativity}\n \\begin{equation*}\n s \\cdot (t \\cdot x)\n =\n \\tau(s)(\\tau(t)(x))\n =\n (\\tau(s) \\circ \\tau(t))(x)\n =\n \\tau(st)(x).\n \\end{equation*}\n\n \\SubProofOf{def:left_module/scalar_distributivity}\n \\begin{equation*}\n (s + t) \\cdot x\n =\n \\tau(s + t)(x)\n =\n \\tau(s)(x) + \\tau(t)(x)\n =\n s \\cdot x + t \\cdot x.\n \\end{equation*}\n\n \\SubProofOf{def:left_module/vector_distributivity}\n \\begin{equation*}\n t \\cdot (x + y)\n =\n \\tau(t)(x + y).\n \\end{equation*}\n\n Now since \\( \\tau(t) \\) is a group endomorphism, we have \\( \\tau(t)(x + y) = \\tau(t)(x) + \\tau(t)(y) \\). Thus,\n \\begin{equation*}\n t \\cdot (x + y)\n =\n t \\cdot x + t \\cdot y.\n \\end{equation*}\n\n \\SubProofOf{def:left_module/identity}\n \\begin{equation*}\n 1_R \\cdot x\n =\n \\tau(1_R)(x)\n =\n \\id(x)\n =\n x.\n \\end{equation*}\n\n \\ImplicationSubProof{def:left_module/multiplication}{def:left_module/homomorphism} Let \\( \\cdot: R \\times M \\to M \\) be a left scalar multiplication operation. We define\n \\begin{balign*}\n & \\tau: R \\to \\End(M) \\\\\n & \\tau(t) \\coloneqq (x \\mapsto t \\cdot x).\n \\end{balign*}\n\n This function is well-defined since for each \\( t \\in R \\), the function \\( \\tau(t) \\) is an abelian group homomorphism (due to \\fullref{def:left_module/associativity}).\n\n \\( \\tau \\) is a ring homomorphism because\n \\begin{itemize}\n \\item it preserves addition:\n \\begin{equation*}\n \\tau(s + t)\n =\n (x \\mapsto (s + t) \\cdot x)\n =\n (x \\mapsto s \\cdot x + t \\cdot x)\n =\n \\tau(s) + \\tau(t).\n \\end{equation*}\n\n \\item it preserves multiplication:\n \\begin{equation*}\n \\tau(st)\n =\n (x \\mapsto (st) \\cdot x)\n =\n (x \\mapsto (s \\cdot (t \\cdot x)))\n =\n \\tau(s) \\circ \\tau(t).\n \\end{equation*}\n\n \\item it preserves identities:\n \\begin{equation*}\n \\tau(1_R)\n =\n \\id,\n \\end{equation*}\n which is the multiplicative unit in \\( \\End(M) \\).\n \\end{itemize}\n\\end{proof}\n\n\\begin{definition}\\label{def:right_module}\n We say that \\( \\tau: R \\to \\End(A) \\) is a \\term{right \\( R \\)-module} if the same function is a \\hyperref[def:left_module]{left module} on the opposite \\hyperref[def:opposite_ring]{ring} \\( R^{-1} \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:bimodule}\n An abelian group \\( M \\) that is both a left \\( L \\)-module and right \\( R \\)-module is called an \\( L, R \\)-\\term{bimodule} if \\( l \\in L, r \\in R \\) and \\( x \\in M \\) implies\n \\begin{equation*}\n (lx)r = l(xr).\n \\end{equation*}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:def:left_module/properties}\n Any left \\( R \\)-module \\( M \\) has the following basic properties:\n \\begin{thmenum}\n \\thmitem{thm:def:left_module/properties/ring_zero_is_absorbing} Multiplication by \\( 0_R \\) is \\term{absorbing}, that is, \\( 0_R x = 0_M \\) for any \\( x \\in M \\).\n\n \\thmitem{thm:def:left_module/properties/module_zero_is_absorbing} Multiplication by \\( 0_M \\) is absorbing, that is, \\( t 0_M = 0_M \\) for any \\( t \\in R \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:left_module/properties/ring_zero_is_absorbing} For any \\( x \\in M \\) we have that \\( 0_R x = (0_R + 0_R)x = 0_R x + 0_R x \\), thus \\( 0_R x \\) is an additive identity and \\( 0_R x = 0_M \\).\n\n \\SubProofOf{thm:def:left_module/properties/module_zero_is_absorbing} For any \\( t \\in R \\) we have that \\( t 0_M = t (0_M + 0_M) = t 0_M + t 0_M \\), thus \\( t 0_M \\) is an additive identity and \\( t 0_M = 0_M \\).\n\\end{proof}\n\n\\begin{example}\\label{ex:module/ideal_of_ring}\n Every unital ring \\( R \\) is a module over itself. Every ideal \\( I \\subseteq R \\) is an \\( R \\)-module since it is closed under multiplication with \\enquote{scalars} from \\( R \\).\n\\end{example}\n\n\\begin{definition}\\label{def:left_module_kernel}\n The \\term{kernel} \\( \\ker(f) \\) of a left \\( R \\)-module homomorphism \\( f: M \\to N \\) is the \\hyperref[def:zero_locus]{zero locus} of \\( f \\), that is, \\hyperref[thm:def:function/properties/preimage]{preimage} \\( f^{-1}(0_N) \\).\n\n It is an instance of \\fullref{def:zero_morphisms/kernel}.\n\\end{definition}\n\n\\begin{definition}\\label{def:quotient_left_module}\n Let \\( M \\) be a left module and \\( N \\) be a submodule of \\( M \\). Define the \\term{quotient module}\n \\begin{equation*}\n M / N \\coloneqq \\{ x + N \\colon x \\in M \\}\n \\end{equation*}\n with the operations\n \\begin{balign*}\n x + N \\oplus y + N \\coloneqq x + y + N.\n t \\odot x + N \\coloneqq tx + N.\n \\end{balign*}\n\n Define the canonical projection homomorphism\n \\begin{balign*}\n & \\pi: G \\to G / N \\\\\n & \\pi(x) \\coloneqq x + N.\n \\end{balign*}\n\n The kernel of \\( \\pi \\) is precisely \\( N \\).\n\\end{definition}\n\\begin{proof}\n The proof of correctness is similar to \\fullref{def:quotient_group}.\n\\end{proof}\n\n\\begin{theorem}\\label{thm:homomorphism_theorem_for_left_modules}\n Fix a ring \\( R \\). Let \\( \\varphi: M \\to K \\) be a homomorphism of left \\( R \\)-modules. We have the isomorphism\n \\begin{equation*}\n M / \\ker \\varphi \\cong \\img \\varphi.\n \\end{equation*}\n\\end{theorem}\n\\begin{proof}\n Analogous to \\fullref{thm:homomorphism_theorem_for_groups}.\n\\end{proof}\n\n\\begin{definition}\\label{def:linear_operator}\n Let \\( M \\) and \\( N \\) be two left \\( R \\)-modules. We say that the function \\( f: M \\to N \\) is \\term{linear} or a \\term{linear operator} if it satisfies the conditions\n \\begin{thmenum}\n \\thmitem{def:linear_operator/additivity}(additivity) \\( f(x + y) = f(x) + f(y) \\) for any \\( x, y \\in M \\).\n \\thmitem{def:linear_operator/homogeneity}(homogeneity) \\( f(tx) = t f(x) \\) for any \\( t \\in R \\) and \\( x \\in M \\) (see \\fullref{def:homogenous_function}).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:map_is_linear_iff_homomorphism}\n A function \\( f: M \\to N \\) between left \\( R \\)-modules is \\hyperref[def:linear_operator]{linear} if and only if it is a module homomorphism in the sense of \\fullref{def:first_order_homomorphism}.\n\\end{proposition}\n\\begin{proof}\n Since \\( (M, +) \\) and \\( (N, +) \\) are groups, it follows from \\fullref{thm:group_homomorphism_single_condition} that \\fullref{def:linear_operator/additivity} is equivalent to the requirement that \\( f \\) is a homomorphism between \\( (M, +) \\) and \\( (N, +) \\).\n\n Now fix \\( t \\in R \\). For \\( f \\) to be a homomorphism, it must satisfy\n \\begin{equation*}\n f(\\cdot_M^{(t)}(x)) = \\cdot_t^{(N)}(f(x)),\n \\end{equation*}\n which is just a more formal way to write \\fullref{def:linear_operator/homogeneity}.\n\\end{proof}\n\n\\begin{definition}\\label{def:affine_operator}\n Let \\( M \\) and \\( N \\) be two left \\( R \\)-modules. We say that the function \\( f: M \\to N \\) is \\term{affine} if it is a translation of a \\hyperref[def:linear_operator]{linear function}, that is, if there exists a linear function \\( l: M \\to N \\) and a constant \\( a \\in N \\) such that \\( f(x) = l(x) + a \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:multilinear_function}\n Generalizing \\fullref{def:linear_operator}, if \\( M_1, \\ldots, M_k \\) and \\( N \\) are \\( R \\)-modules, we say that the function\n \\begin{equation*}\n f: M_1 \\times \\ldots \\times M_k \\to N\n \\end{equation*}\n is \\term{multilinear} or \\term{\\( k \\)-linear} (\\term{bilinear} for \\( k = 2 \\), \\term{trilinear} for \\( k = 3 \\)) if it is linear in each component, that is, for each component \\( i = 1, \\ldots, k \\), and for each tuple not containing elements from \\( M_i \\),\n \\begin{equation*}\n (u_1, \\ldots, u_{i-1}, u_{i+1}, \\ldots, u_k) \\in M_1 \\times \\ldots \\times M_{i-1} \\times M_{i+1} \\times \\ldots \\times M_k \\to N\n \\end{equation*}\n the following function is linear:\n \\begin{balign*}\n & f_i: M_i \\to N \\\\\n & f_i(u_i) \\coloneqq f(u_1, \\ldots, u_{i-1}, u_i, u_{i+1}, \\ldots, u_k).\n \\end{balign*}\n\\end{definition}\n\n\\begin{definition}\\label{def:abelian_group_z_module}\\mcite[375]{Knapp2016BasicAlgebra}\n Let \\( G \\) be an abelian group. Associate with \\( G \\) the \\( \\BbbZ \\)-module \\( M \\) with scalar multiplication\n \\begin{balign*}\n nu \\coloneqq \\begin{cases}\n 0, & n = 0 \\\\\n u + \\ldots + u, & n > 0 \\\\\n -((-n)u), & n < 0.\n \\end{cases}\n \\end{balign*}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:abelian_group_iff_z_module}\\mcite[375]{Knapp2016BasicAlgebra}\n Every abelian group is isomorphic to exactly one \\( \\BbbZ \\)-module.\n\\end{proposition}\n\\begin{proof}\n We already saw in \\fullref{def:abelian_group_z_module} how every abelian group can be regarded as a \\( \\BbbZ \\)-module. Every \\( \\BbbZ \\)-module can then be identified with its additive group.\n\n Scalar multiplication ensures that there is exactly one way to define a \\( \\BbbZ \\)-module structure on an abelian group since \\( na = (n-1)a + a \\) and \\( 0a = 0 \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:left_module_direct_product}\n Let \\( \\{ X_k \\}_{k \\in \\mscrK} \\) be a nonempty family of left \\( R \\)-modules.\n\n Analogously to \\fullref{def:group_direct_product}, we define their \\term{direct product} as the module \\( \\prod_{k \\in \\mscrK} X_k \\), the operations defined componentwise as\n \\begin{balign*}\n & \\{ x_k \\}_{k \\in \\mscrK} + \\{ y_k \\}_{k \\in \\mscrK}\n \\coloneqq\n \\{ x_k + y_k \\}_{k \\in \\mscrK}, \\\\\n & t \\{ x_k \\}_{k \\in \\mscrK}\n \\coloneqq\n t \\{ t x_k \\}_{k \\in \\mscrK}.\n \\end{balign*}\n\n We define their \\term{direct sum} as the submodule of \\( \\prod_{k \\in \\mscrK} X_k \\) (see \\fullref{def:left_module_direct_product}) where only finitely many components of any module element are different from zero.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:module_categorical_limits}\n We are interested in \\hyperref[def:category_of_cones/limit]{categorical limits} and \\hyperref[def:category_of_cones/colimit]{colimits} in the category \\( \\cat{Mod}_R \\) of left-modules over \\( R \\). Fix an indexed family \\( \\{ X_k \\}_{k \\in \\mscrK} \\) of \\( R \\)-modules.\n \\begin{thmenum}\n \\thmitem{thm:module_categorical_limits/product} Their \\hyperref[def:discrete_category_limits]{categorical product} is their direct \\hyperref[def:left_module_direct_product]{product} \\( \\prod_{k \\in \\mscrK} X_k \\), the projection morphisms being inherited from \\fullref{thm:discrete_category_limits_in_set}.\n\n \\thmitem{thm:module_categorical_limits/coproduct} Their \\hyperref[def:discrete_category_limits]{categorical coproduct} is the \\hyperref[def:group_direct_product]{direct sum} \\( \\oplus_{k \\in \\mscrK} X_k \\), the embedding morphisms being inherited from \\fullref{thm:abelian_group_categorical_limits/coproduct}.\n \\end{thmenum}\n\\end{proposition}\n\n\\begin{definition}\\label{def:linear_combination}\n Let \\( M \\) be a left \\( R \\)-module. Like \\hyperref[def:polynomial]{polynomials}, we define linear combinations to be tuples \\( (t_1, t_2, \\ldots, t_n) \\) of scalars from \\( R \\). Unlike with polynomials, we are not interested in defining operations on them, but rather defining the function\n \\begin{equation}\\label{def:linear_combination/function}\n (x_1, \\ldots, x_n) \\mapsto \\sum_{k=1}^n t_k x_k.\n \\end{equation}\n\n The scalars \\( t_1, \\ldots, t_n \\) are called the \\term{coefficients} of the linear combination. A linear combination is said to be \\term{trivial} if all coefficients are equal to \\( 0_R \\).\n\n For convenience, given set of vectors \\( x_1, \\ldots, x_n \\in M \\), we also call the sum \\( \\sum_{k=1}^n t_k x_k \\) a linear combination.\n\n In the special case where \\( R \\) is a superring of \\( \\BbbR \\), we define the following special types of linear combinations:\n \\begin{thmenum}\n \\thmitem{def:linear_combination/affine} an \\term{affine combination} if \\( \\sum_{k=1}^n t_k = 1 \\).\n \\thmitem{def:linear_combination/conic} a \\term{conic combination} if all of the coefficients are nonnegative real numbers.\n \\thmitem{def:linear_combination/convex} a \\term{convex combination} if it is both affine and conic.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:left_module_linear_dependence}\n Let \\( M \\) be a left \\( R \\)-module and let \\( A \\subseteq M \\). We say that the set \\( A \\) is \\term{linearly independent} if for any linear \\hyperref[def:linear_combination/function]{combination}, the equality\n \\begin{equation*}\n \\sum_{i=1}^n t_i x_k = 0_M\n \\end{equation*}\n for \\( x_1, \\ldots, x_n \\in A \\) implies that the combination is trivial.\n\n We say that the vectors \\( x_1, \\ldots, x_n \\) are linearly independent if the corresponding set \\( \\{ x_1, \\ldots, x_n \\} \\) is linearly independent.\n\n We say that \\( A \\) is \\term{linearly dependent} if it is not linearly independent.\n\\end{definition}\n\n\\begin{definition}\\label{def:left_module_hamel_basis}\n A subset \\( B \\) of the left \\( R \\)-module \\( M \\) is called a \\term{Hamel basis} or simply \\term{basis} of \\( M \\) if \\( B \\) is a \\hyperref[def:partially_ordered_set_extremal_points/maximal_and_minimal_element]{minimal} (with respect to set inclusion) linearly independent subset of \\( M \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:free_left_module}\\mcite[377]{Knapp2016BasicAlgebra}\n We say that the left \\( R \\)-module \\( M \\) is a \\term{free left module} if it has a \\hyperref[def:left_module_hamel_basis]{basis}.\n\n Let \\( S \\) be any set. If we regard \\( R \\) as a left module over itself, then the \\hyperref[thm:module_categorical_limits/coproduct]{direct sum}\n \\begin{equation*}\n F(S) \\coloneqq \\oplus_{s \\in S} R\n \\end{equation*}\n with injections \\( \\{ \\iota_s \\}_{s \\in S} \\) is called the free left module \\term{generated by \\( S \\)}. Define the function\n \\begin{balign*}\n & \\varphi: S \\to F(S) \\\\\n & \\varphi(s) \\coloneqq \\iota_s(1_R).\n \\end{balign*}\n\n The image \\( \\varphi(S) \\) is then a basis of \\( F(S) \\).\n\n The cardinality of the basis of a free left module \\( M \\) is called the \\term{rank} \\( \\rank M \\) of \\( M \\). \\Fullref{thm:left_module_basis_cardinality} tells us that this rank is unique for commutative unital rings. If the rank of a module is finite, we say that the module is \\term{finitely generated}.\n\n We also denote \\( F(S) = \\braket S \\), especially in finitely generated modules.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:free_module_is_free_functor}\n The functor \\( F: \\cat{Set} \\to \\cat{Mod}_R \\), defined pointwise in \\fullref{def:free_left_module}, is \\hyperref[def:category_adjunction]{free}.\n\\end{proposition}\n\n\\begin{proposition}\\label{thm:left_module_basis_decomposition}\n Let \\( B \\) be a basis of the free left \\( R \\)-module \\( M \\). Then each element \\( u \\) of \\( M \\) can be uniquely (up to a permutation) represented as a linear \\hyperref[def:linear_combination]{combination} of elements of \\( B \\).\n\\end{proposition}\n\\begin{proof}\n Let\n \\begin{equation*}\n u = \\sum_{i=1}^n t_i v_i\n \\end{equation*}\n and\n \\begin{equation*}\n u = \\sum_{j=1}^m s_i w_i\n \\end{equation*}\n be two representations of \\( u \\) as a linear combination over \\( B \\).\n\n Define the function\n \\begin{balign*}\n & t: M \\to R \\\\\n & t(v) \\coloneqq \\begin{cases}\n t_i, & v = v_i, \\\\\n 0, & v \\not\\in \\{ v_1, \\ldots, v_n \\}.\n \\end{cases}\n \\end{balign*}\n and analogously for \\( s: M \\to R \\). Define the set\n \\begin{equation*}\n B' \\coloneqq \\{ v_1, \\ldots, v_n, w_1, \\ldots, w_m \\}.\n \\end{equation*}\n\n Thus,\n \\begin{equation*}\n u = \\sum_{b \\in B'} t(b) b = \\sum_{b \\in B'} s(b) b\n \\end{equation*}\n and\n \\begin{equation*}\n 0 = u - u = \\sum_{b \\in B'} (t(b) - s(b)) b.\n \\end{equation*}\n\n The set \\( B' \\) is linearly independent as a subset of the basis \\( B \\), hence only a trivial linear combination can be the zero vector. This gives us\n \\begin{equation*}\n t(b) = s(b), b \\in B'.\n \\end{equation*}\n\n In particular, the two decompositions of \\( u \\) along \\( B \\) are identical up to a permutation.\n\\end{proof}\n\n\\begin{definition}\\label{def:left_module_basis_projection}\n Let \\( M \\) be a left \\( R \\)-module and let \\( B \\) be a basis of \\( M \\). For each \\( b \\in B \\), we define the \\text{coordinate projection functional} \\( \\pi_b: M \\to R \\) that gives us the unique coefficient in the basis decomposition. Thus, for every \\( x \\in M \\) we have\n \\begin{equation*}\n x = \\sum_{b \\in B} \\pi_b(x) b.\n \\end{equation*}\n\n The sum is well-defined since only finitely many terms are nonzero.\n\n When the basis \\( B \\) is finite and ordered:\n \\begin{equation*}\n B = \\{ b_1, \\ldots, b_n \\},\n \\end{equation*}\n we also write\n \\begin{equation*}\n x = \\sum_{i=1}^n x_k b_i.\n \\end{equation*}\n\\end{definition}\n\\begin{proof}\n By \\fullref{thm:left_module_basis_decomposition}, this decomposition is unique given a basis \\( B \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:left_module_basis_projections_are_linear}\n The basis projection \\hyperref[def:left_module_basis_projection]{maps} are linear.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{def:linear_operator/homogeneity} Let \\( t \\in R \\) and \\( x \\in M \\). We have the unique decompositions\n \\begin{balign*}\n x & = \\sum_{b \\in B} \\pi_b(x) b, \\\\\n tx & = \\sum_{b \\in B} \\pi_b(tx) b.\n \\end{balign*}\n\n Since both decompositions have only finitely many terms, their difference also has only finitely many nonzero terms. Thus,\n \\begin{equation*}\n 0\n =\n tx - tx\n =\n t \\left( \\sum_{b \\in B} \\pi_b(x) b \\right) - \\sum_{b \\in B} \\pi_b(tx) b\n =\n \\sum_{b \\in B} (t \\pi_b(x) - \\pi_b(tx)) b.\n \\end{equation*}\n\n Since the vectors in \\( B \\) are linearly independent, no nontrivial linear combination can equal the zero vector. Thus, for all \\( b \\in B \\),\n \\begin{equation*}\n t \\pi_b(x) = \\pi_b(tx).\n \\end{equation*}\n\n \\SubProofOf{def:linear_operator/additivity} Analogous.\n\\end{proof}\n\n\\begin{theorem}\\label{thm:linear_map_iff_function_on_basis}\n Let \\( M \\) and \\( N \\) be left \\( R \\)-modules and let \\( B \\) be a basis of \\( M \\). Then there exists a bijection between the \\hyperref[def:function]{functions} from \\( B \\) to \\( N \\) and the module \\hyperref[def:linear_operator]{homomorphisms} from \\( M \\) to \\( N \\), such that each linear map is an \\hyperref[def:multi_valued_function/restriction]{extension} of the corresponding function.\n\\end{theorem}\n\\begin{proof}\n Let \\( \\varphi: M \\to N \\) be a homomorphism. Define the function\n \\begin{balign*}\n & f: B \\to N \\\\\n & f(b) \\coloneqq \\varphi(b).\n \\end{balign*}\n\n Now define the linear map\n \\begin{balign*}\n & \\hat \\varphi: M \\to N \\\\\n & \\hat \\varphi(x) \\coloneqq \\sum_{b \\in B} \\pi_b(x) f(b).\n \\end{balign*}\n\n Since the projections \\( \\pi_b(x) \\) are linear functions by \\fullref{thm:left_module_basis_projections_are_linear} and since we only use the value of \\( f \\) on fixed vectors, it follows that \\( \\hat \\varphi \\) is also linear.\n\n It remains to show that \\( \\varphi = \\hat \\varphi \\). For each \\( x \\in M \\), by linearity of \\( \\varphi \\) we have\n \\begin{equation*}\n \\hat \\varphi(x)\n =\n \\sum_{b \\in B} \\pi_b(x) f(b)\n =\n \\sum_{b \\in B} \\pi_b(x) \\varphi(b)\n =\n \\varphi(x).\n \\end{equation*}\n\\end{proof}\n\n\\begin{remark}\\label{rem:linear_map_iff_function_on_basis}\n \\Fullref{thm:linear_map_iff_function_on_basis} is very powerful in that is allows to study linear maps given their value at only a small subset of vectors. The connection between multilinear \\hyperref[def:multilinear_function]{maps} and \\hyperref[def:left_module_tensor_product]{tensors} is based on this idea.\n\\end{remark}\n\n\\begin{corollary}\\label{thm:linear_maps_agree_on_free_module_if_they_agree_on_basis}\n If two linear maps from the free left module \\( M \\) to \\( N \\) agree on a basis of \\( M \\), they agree on the whole module.\n\\end{corollary}\n\n\\begin{proposition}\\label{thm:left_module_basis_cardinality}\\mcite{ProofWiki:bases_of_free_module_have_same_cardinality}\n All bases in a free left module over a commutative unital ring have the same cardinality.\n\\end{proposition}\n\n\\begin{definition}\\label{def:left_module_tensor_product}\\mcite[574]{Knapp2016BasicAlgebra}\n Let \\( R \\) be a unital ring. Let \\( M \\) be a right \\( R \\)-module and \\( N \\) be a left \\( R \\)-module. Define the free \\hyperref[def:free_abelian_group]{abelian group} \\( G \\) generated by the basis \\( M \\times N \\), that is,\n \\begin{equation*}\n G \\coloneqq \\oplus_{(m,n) \\in M \\times N} \\BbbZ.\n \\end{equation*}\n\n Denote by \\( e_{m,n} \\) the \\( (m,n) \\)-th basis vector and by \\( t_{m,n} \\) the \\( (m,n) \\)-th coordinate of \\( t \\in G \\) (we can have only a finite amount of nonzero coordinates since \\( G \\) is a direct sum).\n\n We can regard \\( G \\) as a left \\( R \\)-module with scalar multiplication given by\n \\begin{equation*}\n (r t)_{(m,n)} \\coloneqq t_{(rm,n)}.\n \\end{equation*}\n\n Let \\( H \\) be the submodule of \\( G \\) generated by\n \\begin{itemize}\n \\item \\( e_{(m_1 - m_2, n)} - e_{(m_1,n)} - e_{(m_2,n)} \\), \\( m_1, m_2, n \\in G \\)\n \\item \\( e_{(m, n_1 - n_2)} - e_{(m,n_1)} - e_{(m,n_2)} \\), \\( m, n_1, n_2 \\in G \\)\n \\item \\( e_{(rm,n)} - e_{(m,rn)} \\), \\( m, n \\in G \\) and \\( r \\in R \\)\n \\end{itemize}\n\n Define the \\term{tensor product} of \\( M \\) and \\( N \\) to be the \\( R \\)-module\n \\begin{equation*}\n M \\otimes N \\coloneqq G / H.\n \\end{equation*}\n\\end{definition}\n\n\\begin{theorem}\\label{thm:tensor_product_universal_property}\\mcite[thm. 10.18]{Knapp2016BasicAlgebra}\n Let \\( R \\) be a unital ring. Let \\( M \\) be a right \\( R \\)-module and \\( N \\) be a left \\( R \\)-module and let \\( M \\otimes N \\) be their \\hyperref[def:left_module_tensor_product]{tensor product} with \\( q: M \\times N \\to M \\otimes N \\) being the corresponding quotient map.\n\n The tensor product \\( M \\otimes N \\) satisfies the following universal mapping property: for every \\( R \\)-module \\( K \\) and any bilinear \\hyperref[def:multilinear_function]{map} \\( f: M \\times N \\to K \\) there exists a unique map \\( \\hat f: M \\otimes N \\to K \\) such that\n \\begin{equation*}\n f = \\hat f \\circ q,\n \\end{equation*}\n that is, the following diagram commutes:\n\n \\begin{alignedeq}\\label{thm:tensor_product_universal_property/diagram}\n \\text{\\todo{Add diagram}}\\iffalse\\begin{mplibcode}\n beginfig(1);\n input metapost/graphs;\n\n v1 := thelabel(\"$M \\times N$\", origin);\n v2 := thelabel(\"$M \\otimes N$\", (2, 0) scaled u);\n v3 := thelabel(\"$K$\", (1, -1) scaled u);\n\n a1 := straight_arc(v1, v2);\n a2 := straight_arc(v2, v3);\n a3 := straight_arc(v1, v3);\n\n draw_vertices(v);\n draw_arcs(a);\n\n label.top(\"$q$\", straight_arc_midpoint of a1);\n label.lrt(\"$\\hat f$\", straight_arc_midpoint of a2);\n label.llft(\"$f$\", straight_arc_midpoint of a3);\n endfig;\n \\end{mplibcode}\\fi\n \\end{alignedeq}\n\\end{theorem}\n\n\\begin{proposition}\\label{thm:tensor_product_with_underlying_ring}\\mcite[677]{Knapp2016BasicAlgebra}\n Let \\( R \\) be a unital ring (regarded as a right module) and \\( B \\) be a left \\( R \\)-module. Their \\hyperref[def:left_module_tensor_product]{tensor product} satisfies\n \\begin{equation*}\n R \\otimes M \\cong M.\n \\end{equation*}\n\\end{proposition}\n\n\\begin{definition}\\label{def:algebra_over_ring}\\mcite[408]{Knapp2016BasicAlgebra}\n Let \\( R \\) be a commutative unital ring. We say that the left \\( R \\)-module \\( A \\) is an \\( R \\)-\\term{algebra} if we define an additional bilinear \\term{vector multiplication} operation\n \\begin{equation*}\n \\odot: A \\times A \\to A\n \\end{equation*}\n such that for \\( x, y \\in M \\) and \\( t \\in R \\)\n \\begin{equation*}\n t \\cdot (x \\odot y) = (t \\cdot x) \\odot y = x \\odot (t \\cdot y).\n \\end{equation*}\n\n Both vector and scalar multiplication are usually denoted by juxtaposition.\n\n If \\( \\odot \\) is associative, commutative, unital or invertible, we add this prefix to \\( A \\), e.g. A is a commutative algebra if \\( \\odot \\) is commutative.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:functions_over_ring_form_algebra}\n Let \\( X \\) be an arbitrary nonempty set and \\( R \\) be a commutative unital ring. Define\n \\begin{equation*}\n A \\coloneqq \\cat{Set}(X, R)\n \\end{equation*}\n to be the set of all functions from \\( X \\) to \\( R \\) (see \\fullref{def:category_of_small_sets}). Then \\( A \\) is an \\( R \\)-algebra with the operations being defined pointwise, that is,\n \\begin{balign*}\n [f + g](x) & \\coloneqq f(x) + g(x) \\\\\n [f \\odot g](x) & \\coloneqq f(x) \\circ g(x) \\\\\n [rf](x) & \\coloneqq r f(x)\n \\end{balign*}\n\n We call the algebra \\( A \\) the \\term{algebra of functions} from \\( X \\) to \\( R \\).\n\n If \\( X \\) itself has a ring structure, we consider the set of ring \\hyperref[thm:ring_homomorphism_simpler_conditions]{homomorphisms}\n \\begin{equation*}\n \\cat{Ring}(X, R),\n \\end{equation*}\n which is usually a strict subset of \\( \\cat{Set}(X, R) \\). This set is usually denoted by \\( \\hom(X, R) \\).\n\n If \\( R \\) is a module, but not necessarily a ring, then \\( \\cat{Set}(X, R) \\) is a only module since we do not necessarily have multiplication. See \\fullref{def:linear_operator}.\n\\end{proposition}\n\n\\begin{definition}\\label{def:homogenous_function}\n Let \\( M \\) and \\( N \\) be left \\( R \\)-modules. We say that the function \\( f: M \\to N \\) is homogeneous with degree \\( n \\) if for all \\( t \\in R \\) and \\( x \\in M \\) we have\n \\begin{equation*}\n f(t x) = t^n f(x).\n \\end{equation*}\n\\end{definition}\n\n\\begin{definition}\\label{def:left_module_annihilator}\\mcite[30]{КоцевСидеров2016}\n Fix a subset \\( S \\subseteq M \\) of a left \\( R \\)-module \\( M \\). We define the \\term{annihilator} of \\( S \\) as the ideal\n \\begin{equation*}\n \\op{ann}(S) \\coloneqq \\{ r \\in R \\colon rS = \\{ 0 \\} \\}.\n \\end{equation*}\n\\end{definition}\n", "meta": {"hexsha": "c5cbc0498c31347d122052a75c169c0fce01a1ed", "size": 28053, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/modules.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/modules.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/modules.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.9538461538, "max_line_length": 398, "alphanum_fraction": 0.6475599758, "num_tokens": 9264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619350028205, "lm_q2_score": 0.7745833789613196, "lm_q1q2_score": 0.6386145114394726}} {"text": "\\chapter[Linear Programming]{Linear Programming}\\label{chp:linear_programming}\n\n\n% Introduction\n\\chapterinitial{F}{inding} the best configuration of some system can be\nchallenging, especially when there is a seemingly endless amount of possible\nsolutions. Optimisation techniques are a way to mathematically derive solutions\nthat maximise or minimise some objective function, subject to a number of\nfeasibility constraints. When all components of the problem can be written in a\nlinear way, then linear programming\\index{linear programming} is one technique\nthat can be used to find the solution.\n\n\\section{Problem}\\label{sec:linear_programming_problem}\n\nA university runs 14 modules over three subjects: Art, Biology, and Chemistry.\nEach subject runs core modules and optional modules.\nTable~\\ref{tab:modules} gives the module numbers for each of these.\n\n\\begin{table}\n\\begin{center}\n\\begin{tabular}{lll}\n\\toprule\n\\textbf{Art Core} & \\textbf{Biology Core} & \\textbf{Chemistry Core} \\\\\n\\midrule\nM00 & M05 & M09 \\\\\nM01 & M06 & M10 \\\\\n\\midrule\n\\textbf{Art Optional} & \\textbf{Biology Optional} & \\textbf{Chemistry Optional}\\\\\n\\midrule\nM02 & M07 & M11 \\\\\nM03 & M08 & M12 \\\\\nM04 & & M13 \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n\\caption{List of modules on offer at the university.}\n\\label{tab:modules}\n\\end{table}\n\nThe university is required to schedule examinations for each of these modules.\nThe university would like the exams to be scheduled using the least amount of\ntime slots. However not all modules can be scheduled at the same time\nas they share some students:\n\n\\begin{itemize}\n \\item All art modules share students;\n \\item all biology modules share students;\n \\item all chemistry modules share students;\n \\item biology students have the option of taking optional modules from\n chemistry, so all biology modules may share students with optional chemistry\n modules;\n \\item chemistry students have the option of taking optional modules from\n biology, so all chemistry modules may share students with optional biology\n modules;\n \\item biology students have the option of taking core art modules, and so all\n biology modules may share students with core art modules.\n\\end{itemize}\n\nHow can every exam be scheduled with no clashes, using the least amount of\ntime slots?\n\n\\section{Theory}\\label{sec:linear_programming_theory}\n\nLinear programming is a method that solves a type of\noptimisation\\index{optimisation} problem of a number of variables by making use\nof some concepts of higher dimensional geometry~\\cite{conforti2014integer}.\nOptimisation here refers to finding the variable that gives either the maximum\nor minimum of some linear function, called the objective function.\n\nLinear programming employs algorithms such as the Simplex\nmethod\\index{Simplex method} to efficiently search some feasible convex\nregion, stopping at the optimum. To do this, an objective function function and\nconstraints need to be defined.\n\nTo illustrate this a classic 2-dimensional example will be used:\n\\(\\pounds 50\\) of profit can be made on each\ntonne of paint A produced, and \\(\\pounds 60\\) profit on each tonne of paint B\nproduced. A tonne of paint A needs 4 tonnes of component X and 5 tonnes of\ncomponent Y. A tonne of paint B needs 6 tonnes of component X and 4 tonnes of\ncomponent Y. Only 24 tonnes of X and 20 tonnes of Y are available per day. How\nmuch of paint A and paint B should be produced to maximise profit?\n\nThis is formulated as a linear objective function, representing total profit,\nthat is to be maximised; and two linear constraints, representing the\navailability of components X and Y. They are written as:\n\n\\begin{align}\n\\text{Maximise: } 50 A + 60 B & \\label{eqn:objective_paint} \\\\\n\\text{Subject to: } & \\nonumber \\\\\n4 A + 6 B &\\leq 24 \\label{eqn:ingredientX} \\\\\n5 A + 4 B &\\leq 20 \\label{eqn:ingredientY}\n\\end{align}\n\nNow this is a linear system in 2-dimensional space with coordinates A and B.\nThese are called the decision variables, what is required are the values of A\nand B that optimises the objective function given by\nexpression~\\ref{eqn:objective_paint}.\n\nInequalities~\\ref{eqn:ingredientX} and~\\ref{eqn:ingredientY} correspond to the\namount of component X and Y available per day. These, along with the additional\nconstraints that a negative amount of paint cannot be produced (\\(A \\geq 0\\) and\n\\(B \\geq 0\\)), form a convex region, shown in Figure~\\ref{fig:paint_lp}.\nThis shaded region shows the pairs of values of \\(A\\) and \\(B\\) which are\nfeasible, that is they satisfy the constraints.\n\n\\begin{figure}\n\\begin{center}\n\\includestandalone[width=\\textwidth]{assets/paint_LP}\n\\end{center}\n \\caption{Visual representation of the paint linear programming\n problem. The feasible convex\n region is shaded in grey; the objective function with arbitrary value is shown\nin a dashed line.}\n\\label{fig:paint_lp}\n\\end{figure}\n\nExpression~\\ref{eqn:objective_paint} corresponds to the total profit, which is\nthe value to be maximised. As a line in 2-dimensional space,\nthis expression fixes its gradient, but its value determines the size of the\n\\(y\\)-intercept. Therefore optimising this function corresponds to pushing a line\nwith that gradient to its furthest extreme within the feasible region,\ndemonstrated in Figure~\\ref{fig:paint_lp}.\nTherefore for this problem the optimum occurs in a particular vertex of the\nfeasible region, at \\(A = \\frac{12}{7}\\) and \\(B = \\frac{20}{7}\\).\n\nThis works well as \\(A\\) and \\(B\\) can take any real value in the feasible region.\nSome problems must be formulated as integer linear\nprogramming\\index{integer linear programming} problems where the decision\nvariables are restricted to integers. There are a number of methods that can\nhelp adapt a real solution to an integer solution. These include cutting\nplanes\\index{cutting planes}, which introduce new constraints around the real\nsolution to force an integer value; and branch and bound\\index{branch and bound}\nmethods, where we iteratively convert decision variables to their closest two\nintegers and remove any infeasible solutions~\\cite{conforti2014integer}.\n\nBoth Python and R have libraries that carry out linear and integer\nprogramming algorithms. When solving these kinds of problems, formulating\nthem as linear systems is the most important challenge.\n\nConsider again the exam scheduling\\index{exam scheduling} problem from\nSection~\\ref{sec:linear_programming_problem} which\nwill now be formulated as an integer linear programming problem.\nDefine \\(M\\) as the set of all modules to be scheduled, and define \\(T\\) as the set\nof possible time slots. At worst each exam is\nscheduled for a different day, thus \\(|T| = |M| = 14\\) in this case.\nLet \\(\\{X_{mt} \\text{ for } m \\in M \\text{ and } t \\in T\\}\\) be a set of binary\ndecision variables, that is \\(X_{mt} = 1\\) if module \\(m\\) is scheduled for time\n\\(t\\), and \\(0\\) otherwise.\n\nThere are six distinct sets of modules in which exams cannot be scheduled\nsimultaneously: \\(A_c\\), \\(A_o\\) representing core and optional art modules\nrespectively; \\(B_c\\), \\(B_o\\) representing core and optional biology modules\nrespectively; and \\(C_c\\), \\(C_o\\) representing core and optional chemistry modules\nrespectively.\nTherefore \\(M = A_c \\cup A_o \\cup B_c \\cup B_o \\cup C_c \\cup C_o\\).\n\nAdditionally there are further clashes between these sets:\n\\begin{itemize}\n \\item No modules in \\(A_c \\cup A_o\\) can be scheduled together as they may\n share students, this is given by the constraint in inequality~\\ref{eqn:clique1}.\n \\item No modules in \\(B_c \\cup B_o \\cup A_c\\), can be scheduled together as\n they may share students, given by inequality~\\ref{eqn:clique2}.\n \\item No modules in \\(B_c \\cup B_o \\cup C_o\\), can be scheduled together as\n they may share students, given by inequality~\\ref{eqn:clique3}.\n \\item No modules in \\(B_o \\cup C_c \\cup C_o\\), can be scheduled together as\n they may share students, given by inequality~\\ref{eqn:clique4}.\n\\end{itemize}\n\nDefine \\(\\{Y_t \\text{ for } t \\in T\\}\\) as a set of auxiliary binary\ndecision variables, where \\(Y_t\\) is 1 if time slot \\(t\\) is being used. This is\nenforced by Inequality~\\ref{eqn:auxiliary}.\n\nEquation~\\ref{eqn:schedule_all_modules},\nensures all modules are scheduled once and once only.\nThus altogether the formulation becomes:\n\n\\begin{align}\n\\text{Minimise: } \\sum_{t \\in T} Y_j & \\label{eqn:objective_modules} \\\\\n\\text{Subject to: } & \\nonumber \\\\\n\\frac{1}{|M|} \\sum_{m \\in M} X_{mt} &\\leq Y_j \\text{ for all } j \\in T \\label{eqn:auxiliary} \\\\\n\\sum_{t \\in T} X_{mt} &= 1 \\text{ for all } m \\in M \\label{eqn:schedule_all_modules} \\\\\n\\sum_{m \\in A_c \\cup A_o} X_{mt} &\\leq 1 \\text{ for all } t \\in T \\label{eqn:clique1} \\\\\n\\sum_{m \\in B_c \\cup B_o \\cup A_c} X_{mt} &\\leq 1 \\text{ for all } t \\in T \\label{eqn:clique2} \\\\\n\\sum_{m \\in B_c \\cup B_o \\cup C_o} X_{mt} &\\leq 1 \\text{ for all } t \\in T \\label{eqn:clique3} \\\\\n\\sum_{m \\in B_o \\cup C_c \\cup C_o} X_{mt} &\\leq 1 \\text{ for all } t \\in T \\label{eqn:clique4}\n\\end{align}\n\nAnother common way to define this linear programming problem is by representing the\ncoefficients of the constraints as a matrix\\index{matrix}.\nThat is:\n\n\\begin{align}\n\\text{Minimise: } c^T Z & \\\\\n\\text{Subject to: } & \\nonumber \\\\\nA Z & \\star b\n\\end{align}\n\nwhere \\(Z\\) is a vector\\index{vector} representing the decision variables, \\(c\\)\nis the coefficients of \\(Z\\) in the objective function, \\(A\\) is the matrix of the\ncoefficients of \\(Z\\) in the constraints, \\(b\\) is the vector of the right hand\nside of the constraints, and \\(\\star\\) represents either \\(\\leq\\), \\(=\\) or\n\\(\\geq\\) as required.\n\nAs \\(Z\\) is a one-dimensional vector of decisions variables,\nthe matrix \\(X\\) and the vector \\(Y\\) can be `flattened' together to form this\nnew variable.\nThis is done by first ordering \\(X\\) then \\(Y\\), within that\nordering by time slot, then within that ordering by module number.\nTherefore:\n\n\\begin{align}\nZ_{|M|t + m} &= X_{mt}\\\\\nZ_{|M|^2 + m} &= Y_m\n\\end{align}\n\nwhere \\(t\\) and \\(m\\) are indices starting at 0.\nFor example \\(Z_{17}\\) would correspond to \\(X_{3, 2}\\), the decision variable\nrepresenting whether module number 4 is scheduled on day 3; \\(Z_{208}\\) would\ncorrespond to \\(Y_{12}\\), the decision variable representing whether there is an\nexam scheduled for day 12.\n\nParameters \\(c\\), \\(A\\), and \\(b\\) can be determined by using this same conversion\nfrom the model in Equations~\\ref{eqn:objective_modules} to \\ref{eqn:clique4}.\nThe vector \\(c\\) would be \\(|M|^2\\) zeroes followed by \\(|M|\\) ones.\nThe vector \\(b\\) would be zeroes for all the rows representing\nEquation~\\ref{eqn:auxiliary}, and ones for all other constraints.\n\n\\section{Solving with Python}\\label{sec:linear_programming_solving-with-python}\n\nIn this book the Python library Pulp\\index{Pulp}~\\cite{mitchell2011pulp} will be\nused to formulate and solve the integer programming problem. First a function to\ncreate the binary problem variables for a given set of times and modules is needed:\n\n\\begin{pyin}\nimport pulp\n\n\ndef get_variables(modules, times):\n \"\"\"Returns the binary variables for a given timetabling\n problem.\n\n Args:\n modules: The complete collection of modules to be\n timetabled.\n times: The collection of available time slots.\n\n Returns:\n A tuple containing the decision variables x and y.\n \"\"\"\n xshape = (modules, times)\n x = pulp.LpVariable.dicts(\"X\", xshape, cat=pulp.LpBinary)\n y = pulp.LpVariable.dicts(\"Y\", times, cat=pulp.LpBinary)\n return x, y\n\\end{pyin}\n\nThe specific modules and times relating to the problem can now be used to\nobtain the corresponding variables:\n\n\\begin{pyin}\nAc = [0, 1]\nAo = [2, 3, 4]\nBc = [5, 6]\nBo = [7, 8]\nCc = [9, 10]\nCo = [11, 12, 13]\nmodules = Ac + Ao + Bc + Bo + Cc + Co\ntimes = range(14)\nx, y = get_variables(modules=modules, times=times)\n\\end{pyin}\n\nNow \\mintinline{python}{y} is a dictionary of binary decision variables, with\nkeys as elements of the list \\mintinline{python}{times}. \\(Y_3\\) corresponds to\nthe third day:\n\n\\begin{pyin}\nprint(y[3])\n\\end{pyin}\n\n\\begin{pyout}\nY_3\n\\end{pyout}\n\nWhile \\mintinline{python}{x} is a dictionary of dictionaries of binary decision\nvariables, with keys as elements of the lists \\mintinline{python}{modules} and\n\\mintinline{python}{times}. \\(X_{2,5}\\) is the variable corresponding\nto module 2 being scheduled on day 5:\n\n\\begin{pyin}\nprint(x[2][5])\n\\end{pyin}\n\n\\begin{pyout}\nX_2_5\n\\end{pyout}\n\n\nThe next step is to create a Pulp object with the corresponding variables,\nobjective function, constraints and solve it. This is done with the following\nfunction:\n\n\\begin{pyin}\ndef get_solution(Ac, Ao, Bc, Bo, Cc, Co, times):\n \"\"\"Returns the binary variables corresponding to the solution\n of given timetabling problem.\n\n Args:\n Ac: The set of core art modules\n Ao: The set of optional art modules\n Bc: The set of core biology modules\n Bo: The set of optional biology modules\n Cc: The set of core chemistry modules\n Co: The set of optional chemistry modules\n times: The collection of available time slots.\n\n Returns:\n A tuple containing the decision variables x and y.\n \"\"\"\n modules = Ac + Ao + Bc + Bo + Cc + Co\n x, y = get_variables(modules=modules, times=times)\n prob = pulp.LpProblem(\"ExamScheduling\", pulp.LpMinimize)\n objective_function = sum([y[day] for day in times])\n prob += objective_function\n\n M = 1 / len(modules)\n for day in times:\n prob += M * sum(x[m][day] for m in modules) <= y[day]\n prob += sum([x[mod][day] for mod in Ac + Ao]) <= 1\n prob += sum([x[mod][day] for mod in Bc + Bo + Co]) <= 1\n prob += sum([x[mod][day] for mod in Bc + Bo + Ac]) <= 1\n prob += sum([x[mod][day] for mod in Cc + Co + Bo]) <= 1\n\n for mod in modules:\n prob += sum(x[mod][day] for day in times) == 1\n\n prob.solve(pulp.apis.PULP_CBC_CMD(msg=False))\n return x, y\n\\end{pyin}\n\nUsing this, the solution \\(x\\) of the original problem can be obtained:\n\n\\begin{pyin}\nx, y = get_solution(\n Ac=Ac, Ao=Ao, Bc=Bc, Bo=Bo, Cc=Cc, Co=Co, times=times\n)\n\\end{pyin}\n\nThese can be\ninspected, for example \\(x_{25}\\) is a boolean variable relating to if module 2\nis scheduled on the 5th day.\n\n\\begin{pyin}\nprint(x[2][5].value())\n\\end{pyin}\n\n\\begin{pyout}\n0.0\n\\end{pyout}\n\nThis was assigned the value 0, and so module 2 was not scheduled for that day.\nHowever, module 2 was scheduled for day 9:\n\n\\begin{pyin}\nprint(x[2][9].value())\n\\end{pyin}\n\n\\begin{pyout}\n1.0\n\\end{pyout}\n\nThis was assigned a value of 1, and so module 2 was scheduled for that day.\n\nThe following function creates a readable schedule:\n\n\\begin{pyin}\ndef get_schedule(x, y, Ac, Ao, Bc, Bo, Cc, Co, times):\n \"\"\"Returns a human readable schedule corresponding to the\n solution of given timetabling problem.\n\n Args:\n Ac: The set of core art modules\n Ao: The set of optional art modules\n Bc: The set of core biology modules\n Bo: The set of optional biology modules\n Cc: The set of core chemistry modules\n Co: The set of optional chemistry modules\n times: The collection of available time slots.\n\n Returns:\n A string with the schedule\n \"\"\"\n modules = Ac + Ao + Bc + Bo + Cc + Co\n\n schedule = \"\"\n for day in times:\n if y[day].value() == 1:\n schedule += f\"\\nDay {day}: \"\n for mod in modules:\n if x[mod][day].value() == 1:\n schedule += f\"{mod}, \"\n return schedule\n\\end{pyin}\n\nThus:\n\n\\begin{pyin}\nschedule = get_schedule(\n x=x,\n y=y,\n times=times,\n Ac=Ac,\n Ao=Ao,\n Bc=Bc,\n Bo=Bo,\n Cc=Cc,\n Co=Co,\n)\nprint(schedule)\n\\end{pyin}\n\ngives:\n\n\\begin{pyout}\n\nDay 0: 1, 12, \nDay 5: 0, 13, \nDay 6: 11, \nDay 7: 4, 6, 10, \nDay 8: 3, 5, 9, \nDay 9: 2, 7, \nDay 13: 8,\n\\end{pyout}\n\nThe order of the days do not matter here, but we 7 days are required in order to\nschedule all exams with no clashes, with at most three exams scheduled each day.\n\n\\section{Solving with R}\\label{sec:linear_programming_solving-with-R}\n\nThe R package ROI\\index{ROI}, the R Optimization\nInfrastructure\\index{R Optimization Infrastructure} will be used here. This is\na library of code that acts as a front end to a number of other solvers that\nneed to be installed externally, allowing a range of optimisation problems to be\nsolved with a number of different solvers. The solver that will be used here is\ncalled the CBC~\\cite{forrest2005cbc} MILP Solver\\index{CBC MILP Solver}, which\nneeds to be installed. The rcbc\\index{rcbc}~\\cite{shumacher2021rcbc} package is\nalso necessary but cannot be installed in the usual way. Installation\ninstructions for both, depend on the operating system and can be found at the\ndocumentation page for the ROI~\\cite{theussl2017roi} package\\footnote{As of the\ntime of writing, this can be found at \\url{https://roi.r-forge.r-project.org/installation.html}}.\n\nThe ROI package requires that the linear programming problem is\nrepresented in its matrix form, with a one-dimensional array of decision\nvariables. Therefore the form of the model described at the end of\nSection~\\ref{sec:linear_programming_theory} will be used.\nFunctions that define the coefficients of the objective function \\(c\\), the coefficient\nmatrix \\(A\\), the vector of the right hand side of the constraints \\(b\\), and the\nvector of equality or inequalities directions \\(\\star\\) are needed.\n\nFirst the objective function:\n\n\\begin{Rin-no-test}\n#' Writes the row of coefficients for the objective function\n#'\n#' @param n_modules: the number of modules to schedule\n#' @param n_days: the maximum number of days to schedule\n#'\n#' @return the objective function row to minimise\nwrite_objective <- function(n_modules, n_days){\n all_days <- rep(0, n_modules * n_days)\n Ys <- rep(1, n_days)\n append(all_days, Ys)\n}\n\\end{Rin-no-test}\n\nFor 3 modules and 3 days:\n\n\\begin{Rin-no-test}\nobjectives <- write_objective(n_modules = 3, n_days = 3)\nprint(objectives)\n\\end{Rin-no-test}\n\nWhich gives the following array, corresponding to the coefficients of the\narray \\(Z\\) for Equation~\\ref{eqn:objective_modules}.\n\n\\begin{Rout-no-test}\n[1] 0 0 0 0 0 0 0 0 0 1 1 1\n\\end{Rout-no-test}\n\nThe following function is used to write one row of that coefficients matrix, for\na given day, for a given set of clashes, corresponding to\nInequalities~\\ref{eqn:clique1} to \\ref{eqn:clique4}:\n\n\\begin{Rin-no-test}\n#' Writes the constraint row dealing with clashes\n#'\n#' @param clashes: a vector of module indices that all cannot\n#' be scheduled at the same time\n#' @param day: an integer representing the day\n#'\n#' @return the constraint row corresponding to that set of\n#' clashes on that day\nwrite_X_clashes <- function(clashes, day, n_days, n_modules){\n today <- rep(0, n_modules)\n today[clashes] = 1\n before_today <- rep(0, n_modules * (day - 1))\n after_today <- rep(0, n_modules * (n_days - day))\n all_days <- c(before_today, today, after_today)\n full_coeffs <- c(all_days, rep(0, n_days))\n full_coeffs\n}\n\\end{Rin-no-test}\n\nwhere \\mintinline{R}{clashes} is an array containing the module numbers of a set\nof modules that may all share students.\n\nThe following function is used to write one row of the coefficients matrix, for\neach module, ensuring that each module is scheduled on one day and one day only,\ncorresponding to Equation~\\ref{eqn:schedule_all_modules}:\n\n\\begin{Rin-no-test}\n#' Writes the constraint row to ensure that every module is\n#' scheduled once and only once\n#'\n#' @param module: an integer representing the module\n#'\n#' @return the constraint row corresponding to scheduling a\n#' module on only one day\nwrite_X_requirements <- function(module, n_days, n_modules){\n today <- rep(0, n_modules)\n today[module] = 1\n all_days <- rep(today, n_days)\n full_coeffs <- c(all_days, rep(0, n_days))\n full_coeffs\n}\n\\end{Rin-no-test}\n\nThe following function is used to write one row of the coefficients matrix\ncorresponding to the auxiliary constraints of Inequality~\\ref{eqn:auxiliary}:\n\n\\begin{Rin-no-test}\n#' Writes the constraint row representing the Y variable,\n#' whether at least one exam is scheduled on that day\n#'\n#' @param day: an integer representing the day\n#'\n#' @return the constraint row corresponding to creating Y\nwrite_Y_constraints <- function(day, n_days, n_modules){\n today <- rep(1, n_modules)\n before_today <- rep(0, n_modules * (day - 1))\n after_today <- rep(0, n_modules * (n_days - day))\n all_days <- c(before_today, today, after_today)\n all_Ys <- rep(0, n_days)\n all_Ys[day] = -n_modules\n full_coeffs <- append(all_days, all_Ys)\n full_coeffs\n}\n\\end{Rin-no-test}\n\nFinally the following function uses all previous functions to assemble a\ncoefficients matrix.\nIt loops though the parameters for each constraint row required, uses the\nappropriate function to create the row of the coefficients matrix, sets the\nappropriate inequality direction (\\(\\leq\\), \\(=\\), \\(\\geq\\)), and the value of\nthe right hand side.\nIt returns all three components:\n\n\\begin{Rin-no-test}\n#' Writes all the constraints as a matrix, column of\n#' inequalities, and right hand side column.\n#'\n#' @param list_clashes: a list of vectors with sets of modules\n# that cannot be scheduled at the same time\n#'\n#' @return f.con the LHS of the constraints as a matrix\n#' @return f.dir the directions of the inequalities\n#' @return f.rhs the values of the RHS of the inequalities\nwrite_constraints <- function(list_clashes, n_days, n_modules){\n all_rows <- c()\n all_dirs <- c()\n all_rhss <- c()\n n_rows <- 0\n\n for (clash in list_clashes){\n for (day in 1:n_days){\n clashes <- write_X_clashes(clash, day, n_days, n_modules)\n all_rows <- append(all_rows, clashes)\n all_dirs <- append(all_dirs, \"<=\")\n all_rhss <- append(all_rhss, 1)\n n_rows <- n_rows + 1\n }\n }\n for (module in 1:n_modules){\n reqs <- write_X_requirements(module, n_days, n_modules)\n all_rows <- append(all_rows, reqs)\n all_dirs <- append(all_dirs, \"==\")\n all_rhss <- append(all_rhss, 1)\n n_rows <- n_rows + 1\n }\n for (day in 1:n_days){\n Yconstraints <- write_Y_constraints(day, n_days, n_modules)\n all_rows <- append(all_rows, Yconstraints)\n all_dirs <- append(all_dirs, \"<=\")\n all_rhss <- append(all_rhss, 0)\n n_rows <- n_rows + 1\n }\n f.con <- matrix(all_rows, nrow = n_rows, byrow = TRUE)\n f.dir <- all_dirs\n f.rhs <- all_rhss\n list(f.con, f.dir, f.rhs)\n}\n\\end{Rin-no-test}\n\nFor demonstration, with 2 modules and 2 possible days, with the single\nconstraint that both modules cannot be scheduled at the same time, then:\n\n\\begin{Rin-no-test}\nwrite_constraints(\n list_clashes = list(c(1, 2)),\n n_days = 2,\n n_modules = 2\n)\n\\end{Rin-no-test}\n\nThis would give 3 components:\n\n\\begin{itemize}\n \\item a coefficient matrix of the left hand side of the constraints, \\(A\\),\n (rows 1 and 2 corresponding to the clash on days 1 and 2, row 3 ensuring\n module 1 is scheduled on one day only, row 4 ensuring module 2 is scheduled on\n one day only, and rows 5 and 6 defining the decision variables \\(Y\\));\n \\item an array of direction of the constraint inequalities, \\(\\star\\);\n \\item and an array of the right hand side values of the constraints, \\(b\\).\n\\end{itemize}\n\n\\begin{Rout-no-test}\n[[1]]\n [,1] [,2] [,3] [,4] [,5] [,6]\n[1,] 1 1 0 0 0 0\n[2,] 0 0 1 1 0 0\n[3,] 1 0 1 0 0 0\n[4,] 0 1 0 1 0 0\n[5,] 1 1 0 0 -2 0\n[6,] 0 0 1 1 0 -2\n\n[[2]]\n[1] \"<=\" \"<=\" \"==\" \"==\" \"<=\" \"<=\"\n\n[[3]]\n[1] 1 1 1 1 0 0\n\\end{Rout-no-test}\n\nNow, the problem will be solved.\nFirst some parameters, including the sets of modules that all share\nstudents, that is the list of clashes are needed:\n\n\\begin{Rin-no-test}\nn_modules = 14\nn_days = 14\nAc <- c(0, 1)\nAo <- c(2, 3, 4)\nBc <- c(5, 6)\nBo <- c(7, 8)\nCc <- c(9, 10)\nCo <- c(11, 12, 13)\nlist_clashes <- list(\n c(Ac, Ao),\n c(Bc, Bo, Co),\n c(Bc, Bo, Ac),\n c(Bo, Cc, Co)\n)\n\\end{Rin-no-test}\n\nThen, the functions defined above are used to create the objective function and\nthe 3 elements of the constraints:\n\n\\begin{Rin-no-test}\nconstraints <- write_constraints(\n list_clashes = list_clashes,\n n_days = n_days,\n n_modules = n_modules\n)\nf.con <- constraints[[1]]\nf.dir <- constraints[[2]]\nf.rhs <- constraints[[3]]\nf.obj <- write_objective(n_modules = n_modules, n_days = n_days)\n\\end{Rin-no-test}\n\nFinally, once these objects are in place, the\nROI library is used to construct an optimisation problem object:\n\n\\begin{Rin-no-test}\nlibrary(ROI)\n\nmilp <- OP(\n objective = L_objective(f.obj),\n constraints = L_constraint(L = f.con, dir = f.dir, rhs = f.rhs),\n types = rep(\"B\", length(f.obj)),\n maximum = FALSE\n)\n\\end{Rin-no-test}\n\nThis creates an \\mintinline{R}{OP} object from our objective row\n\\mintinline{R}{f.obj}, and our constraints which are made up from the three\ncomponents \\mintinline{R}{f.con}, \\mintinline{R}{f.dir} and\n\\mintinline{R}{f.rhs}.\nWhen creating this object the \\mintinline{R}{types} as binary\nvariables are indicated (an array of \\mintinline{R}{\"B\"} for each decision\nvariable).\nThe objective function is to be minimised so\n\\mintinline{R}{maximum = FALSE} is used.\n\nNow to solve:\n\n\\begin{Rin-no-test}\nsol <- ROI_solve(milp)\n\\end{Rin-no-test}\n\nThe solver will output information about the solve process and runtime.\n\n\\begin{Rin-no-test}\nprint(sol$solution)\n\\end{Rin-no-test}\n\n\\begin{Rout-no-test}\n [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0\n [30] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n [59] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0\n [88] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0\n[117] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0\n[146] 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0\n[175] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0\n[204] 1 0 1 1 1 0 1\n\\end{Rout-no-test}\n\nThis gives the values of each of the \\(Z\\) decision variables.\nWe know the structure of this, that is the first 14 variables are the modules\nscheduled for day 1, and so on.\nThe following code prints a readable schedule:\n\n\\begin{Rin-no-test}\n#' Gives a human readable schedule corresponding to the\n#' solution of a given timetable problem.\n#'\n#' @param sol: a solution to the timetabling problem\n#' @param n_modules: the number of modules to schedule\n#' @param n_days: the maximum number of days to schedule\n#'\n#' @return A string with the schedule\nget_schedule <- function(sol, n_days, n_modules){\n schedule = \"\"\n for (day in 1:n_days){\n if (sol$solution[(n_days * n_modules) + day] == 1){\n schedule <- paste(schedule, \"\\n\", \"Day\", day, \":\")\n for (module in 1:n_modules){\n var <- ((day - 1) * n_modules) + module\n if (sol$solution[var] == 1){\n schedule <- paste(schedule, module)\n }\n }\n }\n }\n schedule\n}\n\\end{Rin-no-test}\n\nThus:\n\n\\begin{Rin-no-test}\nschedule <- get_schedule(\n sol = sol,\n n_days = n_days,\n n_modules = n_modules\n)\ncat(schedule)\n\\end{Rin-no-test}\n\ngives:\n\n\\begin{Rout-no-test}\n\n \"Day 2 : 4 11\"\n \"Day 6 : 1 12\"\n \"Day 8 : 7\"\n \"Day 10 : 8\"\n \"Day 11 : 3 13\"\n \"Day 12 : 2 6 9 14\"\n \"Day 14 : 5 10\"\n\\end{Rout-no-test}\n\nThis gives that 7 days are the minimum required to schedule the 14 exams without\nclashes, with either 1, 2 or 4 exams scheduled on each day.\n\n\\section{Wider context}\\label{sec:linear_programming_wider_context}\n\nThe overview given here on linear programming covers a wide breath of the subject\nalthough not much depth. For specific algorithmic approaches to the underlying\nalgorithms and problem types, such as branch and bound\\index{branch and bound} and\ncutting plane\\index{cutting plane} methods as well as some minor extensions\nsee~\\cite{conforti2014integer, sultan2014linear}.\n\nThe efficiency of linear programming as well as the ability to model\nlinear situations imply that it is often used for a variety of applications.\nTheatre scheduling as one such application is given\nin~\\cite{guerriero2011operational}.\nScheduling is indeed a wide ranging sub application of linear programming\nwhich can also be used to schedule sport seasons~\\cite{duran2007scheduling}.\n\nOther applications include the transportation\nproblem\\index{transportation problem}~\\cite{diaz2014survey} which can be used to\nfind a best allocation of a fleet of delivery vehicles; the fire station\nlocation problem\\index{fire station location problem}~\\cite{schreuder1981application}\nused to minimise travel times to at-risk areas; and the bin packing\\index{bin packing}\nproblem~\\cite{hifi2010linear} in which a number of, possibly irregular,\nshapes are packed into the smallest possible number of bins.\n", "meta": {"hexsha": "ea30a1ada5641535cd3bb92a7a7bd3384a13ba43", "size": 28672, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/chapters/08/main.tex", "max_stars_repo_name": "drvinceknight/amwoss", "max_stars_repo_head_hexsha": "8b0bf80f0a06dc5cf9bfeef4b9f9e174ccadf06d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-03-21T21:35:44.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T21:35:44.000Z", "max_issues_repo_path": "src/chapters/08/main.tex", "max_issues_repo_name": "drvinceknight/amwoss", "max_issues_repo_head_hexsha": "8b0bf80f0a06dc5cf9bfeef4b9f9e174ccadf06d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 71, "max_issues_repo_issues_event_min_datetime": "2019-11-18T11:00:25.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-21T22:49:40.000Z", "max_forks_repo_path": "src/chapters/08/main.tex", "max_forks_repo_name": "drvinceknight/amwoss", "max_forks_repo_head_hexsha": "8b0bf80f0a06dc5cf9bfeef4b9f9e174ccadf06d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-01-15T12:00:49.000Z", "max_forks_repo_forks_event_max_datetime": "2020-01-15T12:00:49.000Z", "avg_line_length": 35.2235872236, "max_line_length": 97, "alphanum_fraction": 0.7120884487, "num_tokens": 8450, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.8244619220634456, "lm_q1q2_score": 0.6386145057070209}} {"text": "\\section{Independence Models}\n\n\\begin{frame}{Statistical Models}\n \\begin{block}{Definition}\n A \\emph{statistical model} $\\mc{P}$ is the collection of probability distributions, usually parameterised by a function called a \\emph{parametrisation}\n $$ \\Theta \\ra \\mc{P}, \\quad \\text{given by} \\quad \\theta \\mapsto P_{\\theta},\\quad \\text{so that}\\quad \\mc{P} = \\{ P_{\\theta} : \\theta \\in \\Theta \\}, $$\n where $\\Theta$ is the \\emph{parameter space}. $\\Theta$ is usually a subset of $\\RR^{n}$.\n \\end{block}\n\n \\begin{block}{McCullagh, \\cite{PM2002}}\n This should be defined using category theory.\n \\end{block}\n\n\\end{frame}\n\n\\begin{frame}{Contingency Tables}\n \n \\begin{itemize}\n \\item Classify using two criteria with $r$ and $c$ levels, yielding two random variables $X$ and $Y$. \n \\item Note outcomes as $[r] := \\{1,\\ldots, r\\}$, and $[c] := \\{ 1, \\ldots, c \\}$.\n \\end{itemize}\n \n All information about $X$ and $Y$ is contained in the \\emph{joint probabilities}:\n $$ p_{ij} = P(X = i; Y = j), \\quad i \\in [r],\\ j \\in [c]. $$\n\n \\begin{itemize}\n \\item These in turn determine the \\emph{marginal probabilities}:\n \\end{itemize}\n \\vspace*{-1pt}\n \\begin{equation*}\n \\begin{split}\n p_{i+} &:= \\sum_{j = 1}^{c} p_{ij} = P(X = i), \\quad i \\in [r], \\\\\n p_{+j} &:= \\sum_{i = 1}^{r} p_{ij} = P(Y = j), \\quad j \\in [c].\n \\end{split}\n \\end{equation*}\n\n\\end{frame}\n\n\\begin{frame}\n \\begin{block}{Definition}\n \\begin{itemize}\n \\item Two random variables $X$ and $Y$ are \\emph{independent} if the joint probabilities factor as $p_{ij} = p_{i+}\\cdot p_{+j}$, for all $i \\in [r]$ and $j \\in [c]$. \n \\item Denote independence of $X$ and $Y$ by $X \\indep Y$.\n \\end{itemize}\n \\end{block}\n\n \\begin{block}{Proposition, \\cite{BSSSMD2009}}\n \\emph{Two random variables $X$ and $Y$ are independent if and only if the $(r \\times c)$-matrix, $p = (p_{ij})$, has rank one.}\n \\end{block}\n\n For a $(2 \\times 2)$-table, we thus have:\n\n\\begin{center}\n\\begin{tabular}{lcc}\n & $P(Y=1)$ & $P(Y=2)$\\\\\\hline\n$P(X=1)$ & $p_{11}$ & $p_{12}$\\\\\n$P(X=2)$ & $p_{21}$ & $p_{22}$\\\\\\hline\n\\end{tabular}\n$ \\ \\overset{X \\indep Y}{\\longleadsto{1.25}} \\ p_{11}p_{22} = p_{12}p_{21}. $\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{Some Geometry}\n\n Suppose now we select $n$ cases, giving rise to $n$ independent pairs of discrete random variables:\n $$ \\begin{pmatrix} X^{(1)} \\\\ Y^{(1)} \\end{pmatrix}, \\begin{pmatrix} X^{(2)} \\\\ Y^{(2)} \\end{pmatrix}, \\ldots, \\begin{pmatrix} X^{(n)} \\\\ Y^{(n)} \\end{pmatrix}, $$\n all drawn from the same distribution, i.e.:\n $$ P( X^{(k)} = i; Y^{(k)} = j ) = p_{ij}, \\quad \\text{for all } i \\in [r],\\ j \\in [c],\\ k \\in [n]. $$\n\n \\begin{block}{Probability Simplices}\n Joint probability matrix $p = (p_{ij})$ is an \\emph{unknown} element of the $(rc-1)$-dimensional \\emph{probability simplex}:\n $$ \\Delta_{rc-1} = \\Set{ q \\in \\RR^{r\\times c} | q_{ij} \\geq 0, \\text{ for all } i,j, \\text{ and } \\sum_{i = 1}^{r}\\sum_{j = 1}^{c} q_{ij} = 1 }. $$\n \\end{block}\n\n\\end{frame}\n\n\\begin{frame}\n\n \\begin{block}{Definitions}\n \\begin{itemize}\n \\item A \\emph{statistical model} $\\mc{M}$ is a subset of $\\Delta_{rc-1}$. It represents the set of all candidates for the unknown distribution $p$.\n\n \\item The \\emph{independence model} for $X$ and $Y$ is the set\n $$ \\mc{M}_{X \\indep Y} := \\Set{ p \\in \\Delta_{rc - 1} | \\rank(p) = 1 }. $$\n \\end{itemize}\n \\end{block}\n\n $\\mc{M}_{X \\indep Y}$ is the intersection of $\\Delta_{rc-1}$ and the set of all matrices $p = (p_{ij})$ such that\n $$ p_{ij}p_{kl} - p_{il}p_{jk} = 0, \\quad (1 \\leq i < k \\leq r, \\text{ and } 1 \\leq j < l \\leq c). $$\n\n These are called \\emph{Segre varieties} in algebraic geometry.\n\n\\end{frame}\n", "meta": {"hexsha": "51181e214245bd15d752339e50b4534481d5cf3e", "size": 3851, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/independence.tex", "max_stars_repo_name": "bencwbrown/PGColloquium-06-11-2020", "max_stars_repo_head_hexsha": "853e91be69ff80c3e214ffcd97afe1db2ae458e8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "content/independence.tex", "max_issues_repo_name": "bencwbrown/PGColloquium-06-11-2020", "max_issues_repo_head_hexsha": "853e91be69ff80c3e214ffcd97afe1db2ae458e8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/independence.tex", "max_forks_repo_name": "bencwbrown/PGColloquium-06-11-2020", "max_forks_repo_head_hexsha": "853e91be69ff80c3e214ffcd97afe1db2ae458e8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.9680851064, "max_line_length": 175, "alphanum_fraction": 0.5798493898, "num_tokens": 1417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.78793120560257, "lm_q1q2_score": 0.6386016236240624}} {"text": "\\section{Introduction}\n% a contextual description of the goals of experiment\n\nIn 1921, Michelson and Peace developed the concept of {\\it astronomy interferometry} \\cite{michel}, measuring the {\\it angular diameter} of one the brightness star in the sky, {\\it Betelgeuse}, with an {\\it optical telescope}. Nowadays, Michelson and Peace techniques have broad applications in astronomy to measure stellar diameters \\cite{sbu}.\n\n{\\it Optical} (wavelength ranges of 390 to 750 nanometers) or {\\it Radio interference} (wavelength range of hundreds of meters to millimeters) \\cite{wiki} can be achieved by superimposing signals from more than one telescope, yielding superior {\\it resolutions} than that offered by the telescope alone \\cite{brand}. This is due the fact that the orders of magnitude of the wavelengths in the radio electromagnetic spectra are scalables to the {\\it baselines distances}. Interferometry is a powerful tool in radio astronomy. \n\nA {\\it radio interferometer} measuring an extensive source such as the Sun will show a proportional relation of the dimensions of the experiment (\\ie measurable baseline lengths), to the radio wavelength detected, and to the {\\it sizes of fringe}, in the resulting power spectra. In addition, precise values for the baseline lengths can be inferred by observing the power spectra of a point source (\\eg an artificial geostationary satellite).\n\nIn our experiment, we make use of a two-mirrors radio interferometer, assembled in the Stony Brook University\\cite{sbu}, to measure the angular diameter of the Sun, $\\Phi_{sun}$. We show that the Sun is roughly resolved when viewed in {\\it single-dish mode}, but it is well resolved when {\\it observed interferometrically}. \n\\bigskip\n\n\n\\subsection{The Geometry of Radio Interferometry}\n\nThe principle of {\\it superposition of waves}, figure \\ref{int}, states that when two or more waves are incident on the same point, the {\\it total displacement} at that point is equal to the {\\it vector sum of the displacements of the individual waves}. If a crest of a wave meets a crest of another wave of the same frequency at the same point, the {\\it magnitude of the displacement} is the {\\it sum of the individual magnitudes}, yielding a {\\it constructive interference}. If a crest of one wave meets a trough of another wave, the magnitude of the displacements equals the difference in the individual magnitudes, yielding a {\\it destructive interference} \\cite{brand}.\n\n\\begin{figure}[htb]\n\\begin{center}\n\\includegraphics[scale=0.55]{figs/inter.png}\n\\caption{Interference between two plane wave from a distance source at different angles, from \\cite{brand}. The signals are vectorially added, constructively (a), (c), and destructively (b). }\n \\end{center}\n\\label{int}\n\\end{figure}\n\n\n\nAn interferometer will detect and vectorially add signals from two different positions, figure \\ref{fig:t}-a. The signal arrives on the two {\\it mirrors}, being then mixed in the {\\it antenna receiver}. The separation between the two positions (mirrors) is named {\\it baseline length}, $B$.\n\n\\begin{figure}[htb]\n\\begin{center}\n\n \\includegraphics[scale=0.45]{figs/time.png} \\includegraphics[scale=0.81]{figs/ele.jpg}\n\\caption{(left) a - Signal detection on two mirrors on a radio telescope separated by a baseline length $B$, from \\cite{sbu}. (right) b - Elevation and azimuth angles in the dish.}\n\\label{fig:t}\n \n\\end{center}\n\\end{figure}\n\n\nFrom an arbitrary origin, we define the horizontal variable angle of the telescope pointing, $\\theta$, and the fixed horizontal angle coordinates of the object in the sky, $\\theta_0$, as in figure \\ref{fig:t}. The angle $\\theta_0$ will be later related to the actual azimuth, $A$, of the object. It will be also useful to calculate the {\\it declination}, $\\delta$ of the object (equatorial coordinates), that can obtained from the {\\it elevation}, $E$, the {\\it azimuth}, and the geographical latitude $\\phi$ (horizontal coordinates)\\cite{wiki},\n\\begin{equation}\n\\sin \\delta = \\sin \\phi \\cdot \\sin E + \\cos \\phi \\cdot \\cos E \\cdot \\cos A.\n\\label{dec}\n\\end{equation}\n\n\n\nThe signal wavefronts arrive at the telescope in phases because one of the sources is overhead. The {\\it time delay}, $\\tau$, of the signal at the position 2 (figure \\ref{fig:t}-a), is related to difference between these two angles, the baseline length $B$, and the electromagnetic radiation speed, {\\it c},\n\\begin{equation}\n\\tau = \\frac{B \\sin(\\theta - \\theta_0)}{c} \\approx \\frac{B(\\theta - \\theta_0)}{c},\n\\label{t}\n\\end{equation}\nwhere we use the fact that $\\theta-\\theta_0\\ll1$, such that the Taylor-expansion for small-angle approximation allows $ \\sin \\theta \\approx \\theta $.\n\n\\bigskip\n\n\\subsection{The Physics of Radio Interferometry}\n\nElectromagnetic radiation can be written in terms of their perpendicular {\\it electric fields, E}, and {\\it magnetic fields, B}, where the last can be simply derived from the first with the {\\it Maxwell Equations} \\cite{jackson}.\n\nThe radio signal is electromagnetic radiation. At the frequency $\\nu$, its time-sinusoidal amplitude arriving at position 1 (figure \\ref{fig:t}-a) can be described as a plane wave with the electric field amplitude \\cite{sbu},\n$$\\mathbf E_1(t) = \\mathbf E(\\theta_0) \\cos [2\\pi \\nu t] .$$\n\nThe signal arriving at position 2 at that same time has traveled $B \\sin(\\theta - \\theta_0)$ more than $E_1(t)$, and its electric field amplitude is\n$$\\mathbf E_2(t) = \\mathbf E(\\theta_0) \\cos [2\\pi \\nu (t-\\tau)] .$$\n\n\n\nIn the interferometer receiver, the two signals are vectorially added,\n$$\n\\mathbf E_{tot}(t) = \\mathbf E_1(t) + \\mathbf E_2(t),\n$$\n and the receiver detects the {\\it total power} of the signal, $P(t)$. The radio frequency, $\\nu$, is large compared to a data sampling rate of the receiver, and the total power detected is {\\it time averaged} (or integrated) \\cite{sbu} \\cite {brand} \\cite{jackson},\n\\begin{eqnarray}\nP(t) &=& \\langle\\mathbf E_{tot}(t)^2 \\rangle, \\nonumber \\\\\n& =& E^2(\\theta_0) [1+ \\cos (2\\pi \\nu \\tau)], \\nonumber \\\\ \nP(\\theta)&=& E^2(\\theta_0) [1+ \\cos (2\\pi B_{\\lambda} (\\theta - \\theta_0))],\n\\label{p}\n\\end{eqnarray}\nwhere \n$B_{\\lambda} \\equiv B/\\lambda,$ \nis the {\\it normalized baseline length} to the wavelength $\\lambda = c/\\nu$ and we have applied the results from equation (\\ref{t}).\n\nConsidering that the object has an extension in the sky, we can rewrite equation (\\ref{p}) as a continuum total power law,\n\\begin{eqnarray}\nP(\\theta) = \\int \\varepsilon (\\theta_0) d\\theta_0 \\Bigg [1+ \\cos \\Big (2\\pi B_{\\lambda} (\\theta - \\theta_0)\\Big)\\Bigg ].\n\\label{p22}\n\\end{eqnarray}\n\nAn astronomical object can have any kind of structure, this will result on many kinds of the {\\it energy density distribution}, $\\varepsilon(\\theta_0)$. The telescope interferometer receiver will measure the sinusoidal power intensity response of the object given by equation (\\ref{p22}), where $\\theta$ are the fringes pattern due the interferometry. This quantity will be directly related to the {\\it Fourier transformation} of the energy density distribution of the object.\n\n\\bigskip\n\n\n\n\\subsection{Obtaining the Baseline Lengths from a Point Source }\n\n\n\nFor a punctual source (artificial geostationary satellite), the energy density varying on $\\theta_0$ (as we sweep the telescope) is a $\\delta$-function at the position of the object:\n\\begin{eqnarray}\n \\varepsilon_{sat} (\\theta_0) = \\varepsilon_0 \\delta (\\theta_0). \n\\label{del}\n\\end{eqnarray}\n\nAn expect profile of the total power (amplitude squared) of a point source measured interferometrically can be seen in the figure \\ref{bas}. The pattern of this figure can be thought as resolved as a serie of point sources, slightly displaced, where the interference and the peaks and valleys shrink towards the central value of one. \n\n\nWe insert equation (\\ref{del}) into (\\ref{p22}) and integrate. As we have mentioned in the beginning of this session, measurements of a point source yields the accurate measurement of baseline lengths. The total power is zero in the fringe's minimums (destructive interference, as shown in the figure \\ref{int}), becomes\n\\begin{eqnarray}\nP_{sat}(\\theta) = \\varepsilon_0 \\Bigg [1+ \\cos \\Big (2\\pi B_{\\lambda} (\\theta - \\theta_0) \\Big ) \\Bigg] \\equiv 0,\n\\label{p223}\n\\end{eqnarray}\nmeaning\n\\begin{eqnarray}\n2\\pi B_{\\lambda} (\\theta - \\theta_0) & =& (2n+1) \\pi \\nonumber \\\\\nB_{\\lambda} (\\theta - \\theta_0) & =& \\frac{2n+1}{2}. \\nonumber\n\\end{eqnarray}\n\nTherefore, the separation between adjacent null positions (\\ie the {\\it angular size of the component} we are measuring) gives the value of the baseline lengths,\n\\begin{eqnarray}\nB_{\\lambda} \\Delta \\theta_{12} &=& \\frac{2(n_1 - n_2)}{2},\\nonumber \\\\\nB^{\\Delta \\theta}_{\\lambda} &\\approx& \\frac{1}{\\Delta \\theta}.\n\\end{eqnarray}\n(as we can observe, again in the figure \\ref{bas}). We say it is an approximate value because the rotation of the earth plays a role here, it continuously reorients the telescope, causing the directions of the constructive/destructive interference to scan across the radio source. For instance, the earth rotates $0.5^o$ each every 2 minutes. This oscillation/modulation should be at a much lower frequency than the radio frequency \\cite{brand}.\n\n\\begin{figure}[htb]\n\\centering\n \\includegraphics[scale=0.55]{figs/bas.png}\n\\caption{Example plot of total power as function of telescope pointing, $\\theta$ for a point source such as the satellite, from \\cite{sbu}. }\n\\label{bas}\n\\end{figure}\n\n\nIn addition, with the actual declination of the object, $\\delta$, from equation (\\ref{dec}), the fringe $\\Delta \\theta$ is the {\\it Fourier transform} of the fringe period $\\Delta t$. Considering the angular frequency of earth rotation as $\\omega = 15^o / h$, we can also obtain the baseline lengths writing \n\\begin{equation}\n B^{\\Delta t}_{\\lambda} = \\frac{1}{\\cos(\\delta) \\sin(\\omega \\Delta t)}.\n\\label{BB}\n\\end{equation}\n\nIn the rest of this session, we shall write $B_{\\lambda}$ for simplicity.\n\n\n\n\n\\bigskip\n\n\\subsection{Calculating the Angular Diameter of the Sun} \\label{visb}\n\n\nWe can rewrite the total power given by equation (\\ref{p22}),\n$$ P_{sun}(\\theta) = \\int \\varepsilon (\\theta_0) d\\theta_0 + \\int \\varepsilon (\\theta_0) \\cos (2\\pi B_{\\lambda} (\\theta - \\theta_0)) d\\theta_0.$$\n\nThis is the the Fourier transformation of the object's energy density distribution and we can prove it rewriting the above equation in the same fashion as in \\cite{sbu}:\n$$P_{sun}(\\theta) = S_{sun} [1+V(\\theta,B_{\\lambda})],$$\nwhere\n$$ S_{sun} = \\int \\varepsilon (\\theta_0) d\\theta_0,$$\nis the full spectra of the Sun (obtained as a single dish rather than as interferometer), and\n\\begin{eqnarray}\nV(\\theta, B_{\\lambda})&=& \\frac{1}{S_{sun}} \\int \\varepsilon (\\theta_0) \\cos [2\\pi B_{\\lambda} (\\theta-\\theta_0)] d\\theta_0 ,\\nonumber \\\\\n&=& V_{sun} (B_{\\lambda}) \\cos [2\\pi B_{\\lambda} \\theta],\n\\end{eqnarray}\n where we gauge the {\\it zero-phase position} of the coordinates to eliminate a phase. $V_{sun}$ is the visibility function of the Sun (an amplitude of the Fourier transform a baseline length $B_{\\lambda}$),\n\\begin{equation}\nV_{sun}(B_{\\lambda}) = \\frac{1}{S_{sun}} \\int \\varepsilon (\\theta_0) e^{-i2\\pi B_{\\lambda} \\theta_0}d\\theta_0.\n\\end{equation}\n\nWe can compare this to the observed total power law,\n$$P_{sun}(\\theta) = S_{sun} \\Big [1 + V_0(B_{\\lambda}) \\cos [2\\pi B_{\\lambda} (\\theta - \\Delta \\theta)]\\Big].$$\n\n\\bigskip\n\nTo extract the power function from an astronomical source, $P_{sun}(\\theta)$, we sweep the object by changing the direction of the telescope pointing, $\\theta$. We see a sinusoidal curve plus an {\\it offset} proportional to $\\theta$, as shown in the figure \\ref{psun}, together with maximum and minimum values for the total power function.\n\n\n\n\\begin{figure}[htb]\n\\centering\n \\includegraphics[scale=0.55]{figs/psun.png}\n\\caption{Example plot of total power as function of telescope pointing, $\\theta$ for a extended source, such as the Sun, from \\cite{sbu}. }\n\\label{psun}\n\\end{figure}\n\nIt is straightforward to rewrite out results in terms of $P_{max}$ and $P_{min}$,\n\\begin{equation}\nP_{max} = S_{sun} [1 + V_{sun}(B_{\\lambda})], \n\\end{equation}\n\\begin{equation}\n P_{min} = S_{sun }[1-V_{sun}(B_{\\lambda})].\n\\end{equation}\n\nThe {\\it visibility function} is then\n\\begin{eqnarray}\nV_{sun}(B_{\\lambda}) &=& \\frac{P_{max} - P_{min}}{P_{max} + P_{min}}.\n\\label{vis}\n\\end{eqnarray}\nwhich is a {\\it sinc} function in term of the baseline lengths,\n\\begin{eqnarray}\nV_{sun}(B_{\\lambda}) &=& \\frac{\\sin ( \\pi B_{\\lambda} \\Phi_{sun} )}{ \\pi B_{\\lambda} {\\Phi_{sun}}}, \\nonumber \\\\\n&\\equiv& \\mbox{ sinc } (B_{\\lambda} {\\Phi_{sun}}).\n\\label{vissinc}\n\\end{eqnarray}\n\n\nIn this experiment we apply these properties of the visibility function as a Fourier component of our measurement to obtain the final values of the angular diameter of the sun.\n\n\n", "meta": {"hexsha": "9414fc1297e8960d61d90bb55c22f0f28a84e9dd", "size": 12842, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/intro.tex", "max_stars_repo_name": "bt3gl/Tool-Calculating_the_Diameter_of_Sun", "max_stars_repo_head_hexsha": "f8f7729c2caad2f411e9835a3a28d33a03dff73e", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2022-02-21T17:41:06.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-21T17:48:02.000Z", "max_issues_repo_path": "tex/intro.tex", "max_issues_repo_name": "bt3gl/Calculating_the_Diameter_of_Sun", "max_issues_repo_head_hexsha": "f8f7729c2caad2f411e9835a3a28d33a03dff73e", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/intro.tex", "max_forks_repo_name": "bt3gl/Calculating_the_Diameter_of_Sun", "max_forks_repo_head_hexsha": "f8f7729c2caad2f411e9835a3a28d33a03dff73e", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.4449760766, "max_line_length": 675, "alphanum_fraction": 0.7326740383, "num_tokens": 3590, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7879311906630568, "lm_q1q2_score": 0.6386016115159021}} {"text": "\\documentclass[notitlepage, twocolumn]{article}\n\\usepackage{fullpage}\n\\usepackage[affil-it]{authblk}\n\\usepackage{hyperref}\n\\usepackage{algpseudocode}\n\\usepackage[]{algorithmicx}\n\\usepackage{algorithm}\n\\usepackage{amsfonts}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage[backend=bibtex, style=numeric]{biblatex}\n\\usepackage{caption}\n\\usepackage{pdfpages}\n\n\\captionsetup[ruled]{labelsep=period}\n\n\\makeatletter\n\\@addtoreset{algorithm}{section}% algorithm counter resets every section\n\\makeatother\n\n\\renewcommand{\\thealgorithm}{\\thesection.\\arabic{algorithm}}\n\n\\addbibresource{report.bib}\n\n\\def\\ft{\\mathcal{F}}\n\\algblockdefx[spawn]{Spawn}{EndSpawn}\n[1]{\\textbf{spawn process for} #1 \\textbf{do}}\n{\\textbf{end spawn}}\n\n\\def\\CC{\\mathbb{C}}\t% Complex\n\n\\newcommand{\\set}[1]{\\lbrace #1 \\rbrace}\n\\newcommand{\\paren}[1]{\\left( #1 \\right)}\n\n\\title{\\bf\nGPU Accelerated Fast Fourier Transform\n}\n\n\\date{December 2015}\n\\author{\nMatt McCarthy\n}\n\\affil{Christopher Newport University\\\\\n\\textbf{CPSC 621 Parallel Processing}\\\\\n\\texttt{\\href{mailto:matthew.mccarthy.12@cnu.edu}{matthew.mccarthy.12@cnu.edu}}\n}\n\n\\begin{document}\n\\nocite{*}\n\\maketitle\n\n\\noindent\\textbf{Abstract}\nWe empirically investigate the performance benefits of parallel fast Fourier transform running on the GPU over a sequential version running on the GPU.\n\n\\section{Background}\n\n\\subsection{Discrete Fourier Transform}\n\nThe discrete Fourier transform is a mathematical transformation that takes a set of Complex-valued signals and outputs a set of Complex-valued frequencies.\nFor an $n$-dimensional Complex-valued vector $\\mathbf{X}$, the discrete Fourier transform $\\mathbf{Y}=\\mathcal{F}(X)$ is given by\n\\[\n\tY_j = \\sum_{k=0}^n x_k \\omega^{jk}\n\\]\nwhere $\\omega$ is the $n$-th root of unity, $e^{2\\pi i/n}$.\nSince $\\mathbf{Y}$ is an $n$-dimensional, Complex-valued vector, we can see that the discrete Fourier transform has a complexity of $\\Theta(n^2)$.\n\n\\subsection{Fast Fourier Transform}\n\nFurthermore, we can split the discrete Fourier transform into even and odd sums for $n=2m$, yielding\n\\[\n\tY_j = \\sum_{k=0}^m x_{2k}\\omega^{2jk} + \\omega^j\\sum_{k=0}^m x_{2k+1}\\omega^{2jk}\n\\]\nwhich is two separate discrete Fourier transforms.\nSuppose $n=2^k$.\nIf we iterate this process, we get the following algorithm called the one-dimensional, unordered radix 2, fast Fourier transform in Algorithm 1.1.\n\\begin{algorithm}\n\t\\caption{Recursive FFT}\n\t\\begin{algorithmic}[1]\n\t\t\\Function{R-FFT}{$\\mathbf{X}$,$\\mathbf{Y}$,$n$,$\\omega$}\n\t\t\t\\If{$n$=1}\n\t\t\t\t\\State $y_0=x_0$\n\t\t\t\\Else\n\t\t\t\t\\State Let $\\mathbf{Q}=\\mathbf{0},\\mathbf{T}=\\mathbf{0}\\in\\CC^n$\n\t\t\t\t\\State Let $\\mathbf{X_e}=(x_0, x_2,\\ldots,x_{n-2})$\n\t\t\t\t\\State Let $\\mathbf{X_o}=(x_1, x_3,\\ldots,x_{n-1})$\n\t\t\t\t\\State \\Call{R-FFT}{$\\mathbf{X_e}$,$\\mathbf{Q_e}$,$n/2$,$\\omega^2$}\n\t\t\t\t\\State \\Call{R-FFT}{$\\mathbf{X_o}$,$\\mathbf{T_o}$,$n/2$,$\\omega^2$}\n\t\t\t\t\\ForAll{$j\\in\\set{0,1,\\ldots,n-1}$}\n\t\t\t\t\t\\State $y_j=q_{j\\mod{n/2}}+\\omega^i t_{j\\mod{n/2}}$\n\t\t\t\t\\EndFor\n\t\t\t\\EndIf\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\\subsubsection{Cooley Tukey}\n\nFurthermore, we have an iterative formulation of the prior algorithm, called the Cooley Tukey algorithm for one-dimensional, unordered radix 2, fast Fourier transforms.\n\n\\begin{algorithm}\n\t\\caption{Cooley-Tukey FFT}\n\t\\begin{algorithmic}[1]\n\t\t\\Function{I-FFT}{$\\mathbf{X}$,$\\mathbf{Y}$,$n$}\n\t\t\t\\State $t:=\\lg n$\n\t\t\t\\State $\\mathbf{R}=\\mathbf{X}$\n\t\t\t\\For{$m=0$ to $t-1$}\n\t\t\t\t\\State $\\mathbf{S}=\\mathbf{R}$\n\t\t\t\t\\For{$l=0$ to $n-1$}\n\t\t\t\t\t\\State Let $(b_0b_1\\ldots b_{t-1})$ be the binary expansion of $l$\n\t\t\t\t\t\\State $j:=(b_0\\ldots b_{m-1}0b_{m+1}\\ldots b_{t-1})$\n\t\t\t\t\t\\State $k:=(b_0\\ldots b_{m-1}1b_{m+1}\\ldots b_{t-1})$\n\t\t\t\t\t\\State $r_i:= s_j+s_k\\omega^{(b_mb_{m-1}\\ldots b_0 0\\ldots0)}$\n\t\t\t\t\\EndFor\n\t\t\t\\EndFor\n\t\t\t\\State $\\mathbf{Y}:=\\mathbf{R}$\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\\end{algorithm}\n\nFrom this pseudo-code, we can determine the complexity of fast Fourier transform as described by Algorithm 1.2.\nWe begin by noting that we iterate through the outer loop precisely $\\lg n$ times and the inner loop $n$ times.\nTherefore our complexity is $T_1=\\Theta(n\\lg n)$.\n\n\\subsection{Parallelization}\n\nFor our parallelization, we use a simplified version of the binary exchange algorithm, a parallelization of the Cooley Tukey algorithm designed for use on a hypercube.\nSince our implementation runs on a single GPU, any thread can access any memory location via a pointer.\nHowever, this also complicates the matter by introducing a potential for data races.\nWe solve this by modifying the algorithm to work as follows in Algorithm 1.3.\n\\begin{algorithm}\n\t\\caption{Parallel FFT}\n\t\\begin{algorithmic}[1]\n\t\t\\Function{PAR-FFT}{$\\mathbf{X}$,$\\mathbf{Y}$,$n$}\n\t\t\t\\State $t:=\\lg n$, $BLK:=n/p$\n\t\t\t\\State $\\mathbf{R}=\\mathbf{X}$\n\t\t\t\\State $\\mathbf{S}=\\mathbf{0}$\n\t\t\t\\For{$m=0$ to $t-1$}\n\t\t\t\t\\State Swap pointers $\\mathbf{R}$ and $\\mathbf{S}$\n\t\t\t\t\\Spawn{$l=0$ to $BLK-1$}\n\t\t\t\t\t\\For{$c=l\\cdot BLK$, to $l\\cdot(BLK+1)$}\n\t\t\t\t\t\t\\State Let $(b_0b_1\\ldots b_{t-1})$ be the binary expansion of $c$\n\t\t\t\t\t\t\\State $j:=(b_0\\ldots b_{m-1}0b_{m+1}\\ldots b_{t-1})$\n\t\t\t\t\t\t\\State $k:=(b_0\\ldots b_{m-1}1b_{m+1}\\ldots b_{t-1})$\n\t\t\t\t\t\t\\State $r_i:= s_j+s_k\\omega^{(b_mb_{m-1}\\ldots b_0 0\\ldots0)}$\n\t\t\t\t\t\\EndFor\n\t\t\t\t\\EndSpawn\n\t\t\t\t\\State \\textbf{sync}\n\t\t\t\\EndFor\n\t\t\t\\State $\\mathbf{Y}:=\\mathbf{R}$\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\\end{algorithm}\nIn each iteration, we only write to $\\mathbf{R}$ and only read from $\\mathbf{S}$.\nSince we wait until each thread is complete before moving on to the next iteration, we avoid the potential to use old or incorrect data.\n\nIf we inspect the parallelization, we see that the outer loop runs $\\lg n$ times.\nHowever, the inner loop is ran on $p$ processes, each of which handle $n/p$ iterations.\nTherefore, the computation cost is $\\Theta(n/p\\lg n)$.\nMoreover, for communication we simply do two reads from VRAM on each iteration of the inner loop, however some values will be in cache so we may get better performance than that.\nErgo, the communication cost is $O(n/p\\lg n)$.\nThus, the parallel runtime will be $T_p = \\Theta(n/p \\lg n)$ and is cost optimal if and only if $p\\leq n$.\n\n\\section{Experimental Design}\n\nThe goal of the experiment is to empirically determine the effect of parallelization on the runtime of the Cooley Tukey algorithm.\nTo this end we want to measure the runtime of fast Fourier transform using differing sizes of $n$, the dimension of our Complex valued vector, and $t$, the number of threads.\nLet $\\mathcal{N}$ be the set of vector dimensions we will test and let $\\mathcal{T}$ be the set of thread counts we will test.\nFurthermore, we specify that each $n\\in\\mathcal{N}$ and each $t\\in\\mathcal{T}$ be powers of two.\n\nFor $t=1$, we will run the Cooley Tukey algorithm on random Complex valued vectors of dimension $n$ for each $n\\in\\mathcal{N}$.\nOtherwise, we run our parallelization of the Cooley Tukey algorithm on Complex valued vectors of dimension $n$ for each $n\\in\\mathcal{N}$.\nMoreover, before and after each run, we copy the input to the GPU and the output from the GPU.\nWe take the system time before copying the input to the GPU and again after copying the output to the host.\nWe do not check the output for correctness during the test in order to save time, but instead checked the correctness of the algorithm beforehand.\n\n\\section{Test Environment}\n\n\\subsection{Test System}\n\nThe machine used to run the test has an Intel i7 4770k running at 4.2GHz, 16GB of RAM, and a Nvidia GTX 970 with a core clock of 1342MHz and memory clock of 7000MHz.\nThe computer was running Arch Linux on Linux kernel 4.2.5 with Nvidia driver version 358.16 and CUDA 7.5.\n\n\\subsection{Test Program}\n\nFirstly, all code can be located at \\url{https://github.com/matt-mccarthy/fast-fourier-transform} in the \\verb|fast-fourier| folder.\nWe also include a version of the code in the Appendix.\n\nWe wrote all code in CUDA C++.\nFurthermore all code was compiled with the CUDA toolkit 7.5 and GCC 5.2.0 using the command \\verb|nvcc| \\verb| -std=c++11| \\verb|-rdc=true| \\verb|-arch=compute_52| \\verb|-code=sm_52 -O3|.\nThe library source file is \\verb|src/fast-fourier.cu|, and the sequential and parallel main functions are located in \\verb|sequential-fft.cu| and \\verb|parallel-fft.cu| respectively.\n\nIn both the sequential fast Fourier and parallel fast Fourier, we ran into issues with dynamically allocating memory in a CUDA kernel.\nNamely, if we allocated and freed often enough, we ran out of memory.\nTo avoid this issue entirely, we made all allocations of giant arrays occur in main once and then reused those arrays.\nAs a note, the \\verb|binary_stor| array is where we store the binary representation of our index $l$ for each thread.\nWhile this design is suboptimal (ideally, each thread would create its own so others can not touch it), it works and the performance impact should be neglible for large enough $n$.\n\nMoreover, our GPU code is entirely without any conditional statements except the ones for determining whether or not our for loop is within its bounds.\nAnother note about the code is that at the end of each iteration of the outer for loop, we call \\verb|cudaDeviceSynchronize| which blocks until our \\verb|transformer| kernel completes.\nWe do this in order to ensure syncrhonization across all threads and prevent data races.\nFurthermore, the \\verb|transformer| kernel must always be called from another kernel that runs on exactly one thread.\nWe do this in order to keep all code running on the device and prevent useless communication between the CPU and GPU.\n\nFor our test we took, $\\mathcal{N}=\\set{2^{17},2^{19},2^{21},2^{23},2^{25}}$ and $\\mathcal{T}=\\set{1,4,16,64,256,1024}$, all of which are less than the number of cores on our GTX 970.\nWe chose $\\mathcal{N}$ such that we would get a runtime for 1MB of data on the low end, and 1GB of data on the high end since each complex value is a typedef'd \\verb|float2|.\nMoreover, we ran each combination of $n$ and $t$ 10 times each.\nAll runtimes were measured and recorded in milliseconds, and our result files are included in Appendix B.\n\n\\section{Results}\n\n\\begin{figure}\\label{results}\n\t\\centering\n\n\t\\includegraphics[scale=0.35]{results.png}\n\t\\caption{Runtime vs Array Size}\n\\end{figure}\n\n\\begin{figure}\\label{res-table}\n\t\\centering\n\t\\begin{small}\n\t\\[\n\t\t\\begin{array}{r|c|c|c|c|c}\n\t\t\t& 2^{17} & 2^{19} & 2^{21} & 2^{23} & 2^{25}\\\\\\hline\n\t\t\t2^0& 7.37 & 35.8 & 162 & 773 & 3.42\\cdot10^6\\\\\\hline\n\t\t\t2^2& 1.75 & 8.47 & 38.5 & 183 & 813\\\\\\hline\n\t\t\t2^4& .446 & 2.15 & 9.75 & 46.4 & 205.8\\\\\\hline\n\t\t\t2^6& .117 & .552 & 2.51 & 11.9 & 52.8\\\\\\hline\n\t\t\t2^8& .034 & .149 & .669 & 3.14 & 14.3\\\\\\hline\n\t\t\t2^{10}& 0.011 & .046 & .201 & .975 & 4.65\n\t\t\\end{array}\n\t\\]\n\t\\end{small}\n\t\\caption{The table of runtimes at $(n,t)$ in s}\n\\end{figure}\n\nFigure 2 lists the runtimes for each $t$ (row index) at size $n$ (column index).\nMoreover, the graphs all look like the standard $x\\ln x$ curve.\nBy inspection, we see that $T_{2k}\\approx T_{2k+2}/4$, which is exactly what we are looking for.\nSo naively, we can say that this has a linear speedup.\n\nWe also attempted to compute on 1GB of random complex numbers, or precisely, $2^{27}$ complex numbers.\nThe parallel runtime for $t=1024$ was approximately 22s, however, after many hours of running the sequential version at this $n$ did not complete.\nUpon deeper analysis, we learned that with perfect linear speedup, the sequential version would have a runtime roughly 1000 times greater than $T_{1024}$, namely 6 hours.\nDue to time constraints, we had to abandon this size of $n$ and make our maximum $2^{25}$.\nFurthermore, due to memory constraints, we could not test for any $n\\geq 2^{28}$ since we have to allocate two arrays of 2GB each in order to compute on it and our 970 has 4GB VRAM.\n\n\\section{Future Work}\n\nIn our study, VRAM limitations and time constraints prevented us from harnessing extremely large values of $n$.\nOne way to extend this study to larger values of $n$ is to simply use GPUs that have more than 4GB of VRAM like the Titan X and simply run tests with larger values of $n$.\n\nAnother, more interesting path to explore is to distribute the problem across multiple GPUs.\nIf we split it smartly, we should be able to utilize the extra VRAM without incurring too much communication overhead.\nOf course, the inter-GPU communication is also a factor and we would need to decide the best way to manage it.\nThis latter method can also scale to a GPU farm.\n\n\\section{Conclusion}\n\nOur analysis of the algorithm determined that for $t \\leq n$, we should get perfect linear speedup and the data provides empirical evidence for that, with 1024 threads executing approximately 1000 times faster than the sequential.\n\n\\printbibliography\n\n\\end{document}\n", "meta": {"hexsha": "d7aa7794203e9efb6e297ba016f9d8e40b31e18d", "size": 12771, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/paper.tex", "max_stars_repo_name": "matt-mccarthy/fast-fourier-transform", "max_stars_repo_head_hexsha": "58ce4e63b9e2a00fe7c22b27fab98b66aaf120e5", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-12-10T01:26:51.000Z", "max_stars_repo_stars_event_max_datetime": "2015-12-10T01:26:51.000Z", "max_issues_repo_path": "paper/paper.tex", "max_issues_repo_name": "matt-mccarthy/fast-fourier-transform", "max_issues_repo_head_hexsha": "58ce4e63b9e2a00fe7c22b27fab98b66aaf120e5", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/paper.tex", "max_forks_repo_name": "matt-mccarthy/fast-fourier-transform", "max_forks_repo_head_hexsha": "58ce4e63b9e2a00fe7c22b27fab98b66aaf120e5", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.1924528302, "max_line_length": 230, "alphanum_fraction": 0.7284472633, "num_tokens": 3939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.8104789018037399, "lm_q1q2_score": 0.638601610141561}} {"text": "% !TEX root = main.tex\n\nAn important problem in statistics is to estimate the distribution of a random variable from a set of observations. For example, certain observations may be known (or assumed) to have an exponential distribution and we want to estimate the rate parameter $\\lambda$ of the distribution, or observations may be taken from a normal distribution and we want to estimate the mean $\\mu$ and variance $\\sigma^2$ of the distribution.\n\n%-------------------------------------------------\n\\section{Random samples}\\label{sec:random_samples}\n\nFirst we define random vectors. These are simply vectors of random variables.\n\n\\begin{definition}\\label{def:random_vector}\nLet $X_1,X_2,\\ldots,X_n$ be random variables defined on the same probability space. The vector-valued function\n\\[\n\\begin{array}{rccl}\n\\mathbf{X}:\t& \\Omega \t& \\longrightarrow \t\t& \\R^n \\\\\n\t\t\t& \\omega\t& \\mapsto\t& \\big[X_1(\\omega), X_2(\\omega),\\ldots, X_n(\\omega)\\big]\n\\end{array}\n\\]\nis called a \\emph{random vector} of size $n$. The individual $X_i$ are called the \\emph{component variables} of $\\mathbf{X}$ and a vector $\\mathbf{x} = (x_1,x_2\\ldots,x_n)\\in\\R^n$ is called a \\emph{realisation} of $\\mathbf{X}$, where $x_i$ is the value taken by the corresponding component variable $X_i$.\n\\end{definition}\n\nThe behaviour of a random vector is completely described by the joint CDF of its component variables.\n\\begin{definition}\nLet $\\mathbf{X}=(X_1,X_2,\\ldots,X_n)$ be a random vector.\n\\ben\n\\it \nThe \\emph{joint distribution} of $\\mathbf{X}$ is the function $\\prob_{\\mathbf{X}}(B) = \\prob(\\mathbf{X}\\in B)$ defined on subsets of $\\R^n$.\n\\it\nThe \\emph{joint CDF} of $\\mathbf{X}$ is the function $F_{\\mathbf{X}}:\\R^n\\to[0,1]$ given by\n\\begin{align*}\nF_{\\mathbf{X}}(\\mathbf{x}) \t= \\prob(\\mathbf{X}\\leq\\mathbf{x})\n\t\t\t\t\t\t\t& = \\prob(X_1\\leq x_1, X_2\\leq x_2, \\ldots, X_n\\leq x_n) \\\\\n%\t\t\t\t\t\t\t& = \\prob\\big(\\{\\omega:X_1(\\omega)\\leq x_1, X_2(\\omega)\\leq x_2, \\ldots, X_n(\\omega)\\leq x_n\\}\\big).\n\\end{align*}\n\\een\n\\end{definition}\n\nIn general the component variables $X_i$ might depend on each other and have different distributions. Our analysis is greatly simplified by assuming that the $X_i$ are \\emph{independent} and \\emph{identically distributed}.% random variables.\n\n\\begin{definition}\\label{def:random_sample}\nLet $X$ be a random variable. A \\emph{random sample from the distribution of $X$} is a random vector with the property that the component variables $X_i$ are independent and have the same distribution as $X$.\n\\end{definition}\n\nBy independence the joint CDF of a random sample is just the product of its marginal CDFs and because each $X_i$ has the same distribution as $X$, this can be expressed entirely in terms of the CDF of $X$.\n \n\\begin{lemma}\nLet $\\mathbf{X}=(X_1,X_2,\\ldots,X_n)$ be a random sample from the distribution of $X$. Then the joint CDF and joint PMF/PDF of $\\mathbf{X}$ can be written respectively as\n\\[\nF_{\\mathbf{X}}(\\mathbf{x}) = \\textstyle\\prod_{i=1}^n F_X(x_i) \n\\qquad\\text{and}\\qquad\nf_{\\mathbf{X}}(\\mathbf{x}) = \\textstyle\\prod_{i=1}^n f_X(x_i)\n%\\qquad\\text{respecively},\n\\]\nwhere $F_X$ and $f_X$ are the CDF and PMF/PDF of $X$ respectively, and $\\boldx=(x_1,x_2,\\ldots,x_n)$ is a realisation of the sample.. \n\\end{lemma}\n\\begin{proof}\n\\begin{align*}\nF_{\\mathbf{X}}(\\mathbf{x})\n\t& = \\prob(X_1\\leq x_1, X_2\\leq x_2, \\ldots, X_n\\leq x_n) \\\\\n\t& = \\prob(X_1\\leq x_1)\\prob(X_2\\leq x_2)\\ldots\\prob(X_n\\leq x_n) \\quad\\text{(by independence)} \\\\\n\t& = F_X(x_1)F_X(x_2)\\cdots F_X(x_n) \\quad\\text{(because each $X_i$ has the same distribution as $X$)}\\\\\n\t& = \\textstyle\\prod_{i=1}^n F_X(x_i).\n\\end{align*}\n\\end{proof}\n\n%\\begin{example}\n%Let $\\mathbf{X}=(X_1,X_2,\\ldots,X_n)$ be a random sample from the $\\text{Exponential}(\\lambda)$ distribution, where $\\lambda>0$ is a rate parameter. Then the joint CDF and joint PDF of $\\mathbf{X}$ are respectively\n%\\begin{align*}\n%F_{\\mathbf{X}}(\\mathbf{x}) \n%\t& = \\textstyle\\prod_{i=1}^n\\big[1-\\exp(-\\lambda x_i)\\big] \\\\\n%f_{\\mathbf{X}}(\\mathbf{x}) \n%\t& = \\textstyle\\prod_{i=1}^n\\lambda \\exp(\\lambda x_i) = \\lambda^n\\exp(-\\lambda\\sum_{i=1}^n x_i).\n%\\end{align*}\n%\\end{example}\n%The same relationship holds for PMFs (discrete case) and PDFs (continuous case).\n%\\begin{corollary}\n%If $\\boldX=(X_1,X_2,\\ldots,X_n)$ is a random sample from the distribution of $X$, the joint PMF/PDF of $\\mathbf{X}$ can be written as\n%\\[\n%f_{\\mathbf{X}}(\\mathbf{x}) = \\textstyle\\prod_{i=1}^n f_X(x_i)\n%\\]\n%for all $\\boldx=(x_1,x_2,\\ldots,x_n)\\in\\R^n$, where $f_X$ is the PMF/PDF of $X$.\n%\\end{corollary}\n\n", "meta": {"hexsha": "1e28925b9d38812a09e646e047fdd344423fa3ca", "size": 4521, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/07A_random_samples.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/07A_random_samples.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/07A_random_samples.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 54.4698795181, "max_line_length": 425, "alphanum_fraction": 0.696084937, "num_tokens": 1548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.746138993030751, "lm_q2_score": 0.8558511488056151, "lm_q1q2_score": 0.638583914354033}} {"text": "\\input{settings}\n\n\\begin{document}\n\n\\lhead{Sergio Arnaud}\n\\rhead{ Statistical learning (MT7038) \\\\ Proyect 3 - Neural networks }\n\\cfoot{\\thepage of \\pageref{LastPage}}\n\n\\section{Backpropagation}\n\nThe purpose of this assignment is to implement a complete solution for training a neural network. In practice this is of course done by packages and programs, but it is important to have done it at least once yourselves.\n\nUsing backpropagation and stochastic gradient descent we will fit a nerual network with two hidden layers, and sigmoid activation, to the function $y(x) = x^3$ on the interval $(-2, 2)$. Let $\\theta = \\{(W_i, b_i), i = 1, 2, 3\\}$ be the parameters of the model, then the network function is given by\n\n\\begin{align*}\n f(x,\\theta) = W_3h_2 + b_3 \\quad h_2 = \\sigma(W_2h_1 + b_2) \\quad h_1 = \\sigma(W_1x + b_1)\n\\end{align*}\n\n\\subsection{Dimensions of the Parameters}\n\nAssume that both hidden layers has $n$ hidden units, i.e. $h = (h1, . . . , hn)$ . Give the dimensions of all the\nparameters $(W_i,bi)$, $i=1,2,3$\n\\begin{align*}\n W_1 \\in \\mathbb{R}^{n \\times 1} \\quad b_1 \\in \\mathbb{R}^{n \\times 1}\\\\\n W_2 \\in \\mathbb{R}^{n \\times n} \\quad b_2 \\in \\mathbb{R}^{n \\times 1} \\\\\n W_3 \\in \\mathbb{R}^{1 \\times n} \\quad b_3 \\in \\mathbb{R} \\ \\ \\ \\ \\\n\\end{align*}\n\n\\subsection{Calculating the Derivatives.}\n\nLet $L(x_i, y_i, \\theta) = (y_i - f(x_i; \\theta))^2$. Given the observation $(x,y)$ calculate the derivatives of $L(x,y;\\theta)$.\n\n\\begin{align*}\n \\frac{\\partial L}{\\partial W_{i}^3} &= 2(y-f(x,\\theta))h^2_i\\\\\n \\frac{\\partial L}{\\partial b_{i}^3} &= 2(y-f(x,\\theta))\\\\\n \\frac{\\partial L}{\\partial h_{i}^2} &= \\frac{\\partial L}{\\partial b_{i}^3}W_i^3 \\\\\\\\\n \\frac{\\partial L}{\\partial W_{ij}^2} &= \\frac{\\partial L}{\\partial h_{i}^2}(h_i^2(1-h_i2))h_j^1 \\\\\n \\frac{\\partial L}{\\partial b_{i}^2} &= \\frac{\\partial L}{\\partial h_{i}^2}(h_i^2(1-h_i2)) \\\\\n \\frac{\\partial L}{\\partial h_{i}^1} &= \\frac{\\partial L}{\\partial b_{i}^2}W_{ii}^2 \\\\\\\\\n \\frac{\\partial L}{\\partial W_{i}^3} &= \\frac{\\partial L}{\\partial h_{i}^1}(h_i^1(1-h_i^1))x \\\\\n \\frac{\\partial L}{\\partial b_{i}^3} &= \\frac{\\partial L}{\\partial h_{i}^1}(h_i^1(1-h_i^1)) \\\\\n\\end{align*}\n\n\\subsection{Fitting the Model}\n\nThe following code fites a feed forward neural network with 2 hiddel layers (125 hidden units) to the function $y(x) = x^3$.\n\n\\begin{lstlisting}[language=R]\n\nsigm <- function(x){\n f <- function(y){\n return(1/(1+exp(-y)))\n }\n return( sapply(x,f) )\n}\n\ninitParams <- function(nrHiddenUnits, xDim) {\n W <- list()\n b <- list()\n\n # First hidden layer\n W_1 <- matrix(0.05*rnorm(nrHiddenUnits * xDim), ncol = xDim,\n nrow = nrHiddenUnits, byrow = TRUE)\n b_1 <- matrix(0.05*rnorm(nrHiddenUnits), ncol = 1,\n nrow = nrHiddenUnits, byrow = TRUE)\n\n # Second hidden layer\n W_2 <- matrix(0.05*rnorm(nrHiddenUnits * xDim), ncol = nrHiddenUnits,\n nrow = nrHiddenUnits, byrow = TRUE)\n b_2 <- matrix(0.05*rnorm(nrHiddenUnits), ncol = 1,\n nrow = nrHiddenUnits, byrow = TRUE)\n\n # Output layer\n W_3 <- matrix(0.05*rnorm(nrHiddenUnits), ncol = nrHiddenUnits,\n nrow = 1, byrow = TRUE)\n b_3 <- 0.05*rnorm(1)\n\n W[[1]] <- W_1\n W[[2]] <- W_2\n W[[3]] <- W_3\n\n b[[1]] <- b_1\n b[[2]] <- b_2\n b[[3]] <- b_3\n\n return(list(\"W\" = W,\"b\" = b))\n}\n\n\nforwardPass <- function(x,W,b){\n #First hidden Layer\n h_1 <- sigm(W[[1]] %*% x + b[[1]])\n\n #Second hidden Layer\n h_2 <- sigm(W[[2]] %*% h_1 + b[[2]])\n\n #Output Layer\n y <- W[[3]] %*% h_2 + b[[3]]\n\n return( list(\"output\" = y, \"hiddenLayer1\" = h_1, \"hiddenLayer2\" = h_2 ) )\n}\n\ngradient <- function(x,y,W,b){\n\n net <- forwardPass(x,W,b)\n h_1 <- net$hiddenLayer1\n h_2 <- net$hiddenLayer2\n\n output <- net$output[[1]]\n\n W_3 <- W[[3]]\n W_2 <- W[[2]]\n W_1 <- W[[1]]\n\n b_1 <- b[[1]]\n b_2 <- b[[2]]\n b_3 <- b[[3]]\n\n #dL/dW_3\n dW_3 <- 0*W_3\n for (j in 1:dim(dW_3)[2]) {\n dW_3[1,j] <- -2*(y-output)*h_2[j]\n }\n #dL/db_3\n db_3 <- -2*(y-output)\n #dL/dh_2\n dh_2 <- 0*h_2\n for ( i in 1:length(h_2) ) {\n dh_2[i] <- db_3*W_3[1,i]\n }\n\n\n #dL/dW_2\n dW_2 <- 0*W_2\n for (i in 1:(dim(dW_2)[1])){\n for (j in 1:(dim(dW_2)[2])) {\n dW_2[i,j] <- dh_2[i]*(h_2[i]*(1-h_2[i]))*h_1[j]\n }\n }\n #dL/db_2\n db_2 <- 0*b_2\n for (i in 1:length(db_2)) {\n db_2[i] <- dh_2[i]*(h_2[i]*(1-h_2[i]))\n }\n #dL/dh_1\n dh_1 <- 0*h_1\n for (i in 1:length(dh_1)) {\n dh_1[i] <- db_2[i]*W_2[i,i]\n }\n\n\n #dL/dW_1\n dW_1 <- 0*W_1\n for (i in 1:(dim(dW_1)[1])){\n for (j in 1:(dim(dW_1)[2])) {\n dW_1[i,j] <- dh_1[i]*(h_1[i]*(1-h_1[i]))*x\n }\n }\n #dL/db_1\n db_1 <- 0*b_1\n for (i in 1:length(db_1)) {\n db_1[i] <- dh_1[i]*(h_1[i]*(1-h_1[i]))\n }\n\n return( list(\"dW_1\" = dW_1, \"dW_2\" = dW_2, \"dW_3\" = dW_3,\n \"db_1\" = db_1, \"db_2\" = db_2, \"db_3\" = db_3) )\n}\n\nbackPropagation <- function(){\n learningRate <- 0.01\n params <- initParams(125,1)\n\n N= 100000\n for (iter in 1:N) {\n x <- -2+4*runif(1)\n y <- x^3\n grad <- gradient(x,y,params$W, params$b)\n\n params$W[[1]] <- params$W[[1]]-learningRate*grad$dW_1\n params$W[[2]] <- params$W[[2]]-learningRate*grad$dW_2\n params$W[[3]] <- params$W[[3]]-learningRate*grad$dW_3\n\n params$b[[1]] <- params$b[[1]]-learningRate*grad$db_1\n params$b[[2]] <- params$b[[2]]-learningRate*grad$db_2\n params$b[[3]] <- params$b[[3]]-learningRate*grad$db_3\n\n #if (iter %% 1000 == 0) print(iter)\n }\n return( params )\n}\n\nans = backPropagation()\n\\end{lstlisting}\n\nNow, we plot the predicted values of our model against the true curve $f(x) = x^3$. As we can see, the results resemble the true model really good.\n\n\\begin{lstlisting}[language=R]\ncall_forward_pass <- function(x){\n forwardPass(x,ans$W, ans$b)$output\n}\n\ndf <- tibble(x = seq(-2,2,.05)) %>%\n mutate(y = map(x, ~call_forward_pass(.x)), y = as.numeric(y))\n\nggplot(df, aes(x,y)) +\n geom_point() +\n stat_function(fun=function(x) x^3, col = 'red') +\n scale_colour_manual(\"Lgend title\", values = c(\"red\", \"blue\"))\n\\end{lstlisting}\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[width=0.9\\textwidth]{img/img1.jpg}\n \\caption{}\n \\end{center}\n\\end{figure}\n\n\\section{Learning Keras}\nIn this assignment we will fit a neural network for classifying images. We will use Keras, which is an extensive deep learning package in Python. Keras can be used in R through the keras package. But first we need to install it, see the steps below.\n\nTo get started with Keras follow the tutorial at \\url{https://keras.rstudio.com/articles/getting\\_started.html}.\n\nThe documentationt to the fit function is availible at: \\url{https://keras.rstudio.com/reference/fit.keras.engine}. training.Model.html. Look at the documentation and answer the following questions:\n\n- Explain what a batch size is.\n- Explain what an epoch is.\n- Explain why the shuffle argument is default set to true, i.e. why do you want to shuffle? Can you think of example’s when shuffling is a bad idea?\n\n\\subsection{Batch Size}\nIs the number of samples that will be propagated through the network per gradient update, that is to say in one forward/backward pass. Some of the advantages of using a batch size smaller than the number of all samples are that it requires less memory and that the training process is usually faster because we update the weights after each propagation. A disadvantage of using a batch size smaller than the number of samples is that the smaller the batch the less accurate the estimate of the gradient will be. In general we have\n\\begin{itemize}\n \\item \\textbf{Batch Gradient Descent}. Batch Size = Size of Training Set\n \\item \\textbf{Stochastic Gradient Descent}. Batch Size = 1\n \\item \\textbf{Mini-Batch Gradient Descent}. 1 < Batch Size < Size of Training Set. Popular batch sizes include 32, 64, and 128 samples.\n\\end{itemize}\n\n\\subsection{Epoch}\nIn general, one epoch means that each sample in the training dataset has had an opportunity to update the internal model parameters, an epoch is comprised of one or more batches. A way to understand it is the following: we have a first for-loop over the number of epochs, in each loop we will a nested for-loop that iterates over each batch of sample.\n\nIn keras, the epoch parameter defines the number times that the learning algorithm will work through the entire training dataset.\n\n\\subsection{Shuffle}\nThis argument is a boolean that determines whether to shuffle the trining data before each epoch. The reason to set it by default is that we want the model to understand all inputs equally. For example, if the data is ordered by class, then the training result will be a bad one because the model will have a higher accuracy for the last class it trains than the first one.\n\nAnother reason why shuffle is true by default is that the objective function might have numerous minima, and therefore gradient descent algorithms are susceptible to becoming \"stuck\" in those minima. This is likely to occur if the set of observations is unchanged over all training iterations. Shuffling observations might help the solver to \"bounce\" out of a local minimum.\n\nFor the Batch Gradient Descent process (Batch Size = Size of Training Set) shuffling is unnecessary.\n\n\n\\section{Fitting a Model in keras}\n\nNow we'll fit a feed-forward neural network to the dataset \\textit{dataset\\_fashion\\_mnist()}. For the following part python was used.\n\n\\begin{python}\nimport keras\nfrom keras.models import Sequential\nfrom keras.layers import Dense, Dropout, Flatten, Conv2D, MaxPooling2D\nfrom keras.datasets import fashion_mnist\n\nfrom matplotlib import pyplot as plt\n\n# Loading fashion_mnist dataset\nfashion_mnist = keras.datasets.fashion_mnist\n(train, test = fashion_mnist.load_data()\n(train_images, train_labels) = train\n(test_images, test_labels) = test\n\n# Turning pixel values to values in the interval $[0,1]$\ntrain_images = train_images / 255.0\ntest_images = test_images / 255.0\n\n# Reshaping\ntrain_images = train_images.reshape(-1,28,28,1)\ntrain_images = train_images.astype('float32')\ntest_images = test_images.reshape(-1,28,28,1)\ntest_images = test_images.astype('float32')\n\n# Neural net architecture\nmodel = Sequential()\n\n# Convolutional and max pooling layers since we are dealing with images\nmodel.add(Conv2D(filters=64, kernel_size=2,\n padding='same', activation='relu',\n input_shape=(28,28,1)))\nmodel.add(MaxPooling2D(pool_size=2))\n# Dropout to deal with overfitting\nmodel.add(Dropout(0.25))\n\n# Repeat convolutional - max pooling - dropout\nmodel.add(Conv2D(filters=32, kernel_size=2,\n padding='same', activation='relu'))\nmodel.add(MaxPooling2D(pool_size=2))\nmodel.add(Dropout(0.25))\n\n# Flattening\nmodel.add(Flatten())\n\n# Dense layer with 128 hidden units and relu activation\nmodel.add(Dense(128, activation='relu'))\n\n# Droput to deal with overfitting\nmodel.add(Dropout(0.5))\n\n# Output layer\nmodel.add(Dense(10, activation='softmax'))\n\n# Compiling model\nmodel.compile(loss = 'sparse_categorical_crossentropy',\n optimizer = keras.optimizers.adam(),\n metrics = ['accuracy'])\n\n# Training\nmodel.fit(train_images, train_labels, epochs=10, validation_split=0.2)\n\n# Evaluating\ntest_loss, test_acc = model.evaluate(test_images, test_labels, verbose=2)\n# Test error = 0.914\n\n# Plotting validation curve\nhistory = model.history\nplt.plot(history.history['accuracy'])\nplt.plot(history.history['val_accuracy'])\nplt.title('model accuracy')\nplt.ylabel('accuracy')\nplt.xlabel('epoch')\nplt.legend(['train', 'val'], loc='upper left')\nplt.show()\n\\end{python}\n\nThe test error was $.914$ and the validation curve is given in the following figure:\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[width=0.8\\textwidth]{img/img2.jpg}\n \\caption{Validation curve}\n \\end{center}\n\\end{figure}\n\n\n\\end{document}\n", "meta": {"hexsha": "d0bcf1ef0f27c515c8c62e8b75a27d956bdf5ab0", "size": 11734, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW3/Proyect3_solution.tex", "max_stars_repo_name": "SergioArnaud/Statistical-Learning", "max_stars_repo_head_hexsha": "e4ee81c02fea996794df70a277f1ad51e6fb1a83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-13T21:25:23.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-13T21:25:23.000Z", "max_issues_repo_path": "HW3/Proyect3_solution.tex", "max_issues_repo_name": "SergioArnaud/Statistical-Learning", "max_issues_repo_head_hexsha": "e4ee81c02fea996794df70a277f1ad51e6fb1a83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW3/Proyect3_solution.tex", "max_forks_repo_name": "SergioArnaud/Statistical-Learning", "max_forks_repo_head_hexsha": "e4ee81c02fea996794df70a277f1ad51e6fb1a83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.5117647059, "max_line_length": 530, "alphanum_fraction": 0.6781148798, "num_tokens": 3762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511543206819, "lm_q2_score": 0.7461389873857264, "lm_q1q2_score": 0.6385839136377387}} {"text": "\\chapter{Interpreting a Logistic Regression Model \\label{chapter:logreg}}\n\nThis chapter is similar to Chapter~\\ref{chapter:linreg} but focuses on logistic regression models. As we saw in Chapter~\\ref{chapter:linreg}, linear regression models are used in situations where the outcome of a supervised learning problem, $y$, follows a normal distribution, conditional on the values of the predictors. \\textbf{Logistic regression} models, in contrast, handle situations where the outcome, $y$, is binary: either 0 or 1. We first encountered these models as examples of classification algorithms in Chapter~\\ref{chapter:classification}. Because of their popularity in the clinical domain, it's important to understand how these models are fit and how to interpret the summary output produced by software. \n\nUnfortunately, a full understanding of logistic regression requires knowledge of maximum likelihood estimation. We will, therefore, skip over some of the details until we've had more time to explore this topic. \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{ER Readmissions Example from Chapter~\\ref{chapter:classification} \\label{sect:eragain}}\n\nIn Chapter~\\ref{chapter:classification}, we saw an example where information about two predictors -- a disease severity score ($x_1$) and a social determinants score ($x_2$) -- was used to predict a binary outcome: whether a patient would be readmitted to the ER within 30 days of discharge. We tried three different supervised learning algorithms, one of which was a logistic regression model (Section~\\ref{ssect:logreg}). The output from that model is repeated below.\n\n\\begin{center}\n\\includegraphics[width=0.35\\textwidth]{img/esl-logistic.png}\n\\includegraphics[width=0.64\\textwidth]{img/glm-binomial-example.png}\n\\end{center}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Understanding the Model Summary}\n\nA logistic regression model looks like this (see also Section~\\ref{ssect:logreg}):\n\\begin{equation} \\log \\frac{\\mu}{1-\\mu} = \\beta_0 + \\beta_1 x_1 + \\dots + \\beta_p x_p \\label{eqn:logistic} \\end{equation}\nwhere $\\mu$ is the mean of the Bernoulli distribution (see Section~\\ref{sect:bernoulli}) governing our binary outcome, $y$; in other words, it is the probability that $y = 1$. \n\nYou will note that there is no independent error term here as there was in linear regression. That's because the variance and mean of a Bernoulli distribution are coupled and depend only on $\\mu$ (again, see Section~\\ref{sect:bernoulli}). \n\\vspace{3mm}\n\n\\begin{question}{question:logreglink}\nIn logistic regression, $\\mu$ itself is not equal the sum of the predictors; instead, the \\textbf{logit} of $\\mu$ is their sum. Based on what you know about $\\mu$, why is a logistic regression model not of the form\n$$ \\mu = \\beta_0 + \\beta_1 x_1 + \\dots + \\beta_p x_p \\text{?}$$\nWe will explore this further in Chapter~\\ref{chapter:glms}.\n\\end{question}\n\n\\begin{question}{}\nThe decision boundary in logistic regression (see picture above) occurs where the sum of the linear predictors, $\\beta_0 + \\beta_1 x_1 + \\dots + \\beta_p x_p$, is zero. What value of $\\mu$ does this correspond to? Why does this make sense, intuitively?\n\\end{question}\n\n\\subsection{The Call}\n\nThe first line of the output repeats the call you made to the \\verb|glm| function in R to fit the model. The \\texttt{glm} package fits a variety of different generalized linear models using maximum likelihood estimation (see Chapter~\\ref{chapter:mlebasics}; this will also be discussed in more detail in Chapter~\\ref{chapter:glms}). The \\texttt{family = \"binomial\"} argument tells the function to fit a logistic regression model.\n\n\\subsection{Coefficients and Standard Errors}\n\nLogistic regression models, like other GLMs, are fit using maximum likelihood (see Chapter~\\ref{chapter:mlebasics} for a brief introduction). We will skip most of the details for now, but you can gain intuition by staring at Equation~\\ref{eqn:logistic}. This equation says that the model's predicted value of $\\mu$, the probability that the outcome will be positive ($y=1$), is controlled by the values of the predictors and their coefficients $\\beta_0, \\dots, \\beta_p$. \n\nBy adjusting the values of the $\\beta$s, the model causes $\\mu$ to be high in regions of the feature space where $y=1$ and low where $y=0$. The values of the $\\beta$s that do this the best are called the \\textbf{maximum likelihood estimates}, and they are the coefficients shown in the model output.\n\nAs with linear regression, a full understanding of the standard errors requires matrix multiplication. However, they are related to the same factors that drive the standard errors in linear regression: (1) the spread of the values of the corresponding covariate about its mean (more spread will decrease the standard error) and (2) correlations between that covariate and other covariates in the model (tighter correlations will increase the standard error).\n\\vspace{5mm}\n\n\\begin{question}{}\nLooking at the form of the logistic regression model\n$$ \\log \\frac{\\mu}{1-\\mu} = \\beta_0 + \\beta_1 x_1 + \\dots + \\beta_p x_p $$\nwhat does the value of each of the $\\beta$s mean? What is $\\beta_j$ telling us about how $y$ varies with the predictor $j$, all else being equal?\n\\end{question}\n\n\\begin{question}{}\nThe \\textbf{odds} of something happening are defined as $\\mu/(1-\\mu)$, where $\\mu$ is the probability that the thing occurs. In our example model, we are interested in the odds that $y=1$ (the patient is readmitted). Does a unit increase in $x_1$ (disease severity score) increase or decrease the odds that a patient will be readmitted? What about $x_2$ (social determinants score)? \n\\end{question}\n\n\\begin{question}{}\nWhat are the odds of readmission for a patient with:\n\\begin{enumerate}[label=(\\alph*)] \n\\item $x_1 = 0.1$ and $x_2 = 0.3$?\n\\item $x_1 = 0.1$ and $x_2 = -1.3$?\n\\item $x_1 = 1.1$ and $x_2 = 0.3$?\n\\end{enumerate}\n\\end{question}\n\n\\subsection{Hypothesis Tests of Coefficients}\n\nJust as in linear regression, we can use our coefficients and standard errors to perform a hypothesis test on each regression coefficient, $\\beta_j$, against the null hypothesis that $\\beta_j = 0$ (the predictor $x_j$ has no effect on the outcome). In logistic regression, the quantity $\\hat{\\beta_j}/\\text{se}(\\hat{\\beta}_j)$ will follow a normal distribution under the null.\n\\vspace{3mm}\n\n\\begin{question}{question:nulllogregexample}\nBelow are the null distributions for the hypothesis tests of our three regression coefficients, $\\beta_0$, $\\beta_1$, and $\\beta_2$. In each graph, the maximum likelihood estimate of the coefficient is shown as a vertical dashed line. Based on these graphs, can you tell why the $p$-values for $\\beta_0$ and $\\beta_2$ are low and the one for $\\beta_1$ is high? What is the intuition behind this?\n\\begin{center}\n\\includegraphics[width=0.45\\textwidth]{img/logreg-example-beta-0.png}\n\\includegraphics[width=0.45\\textwidth]{img/logreg-example-beta-1.png}\n\\includegraphics[width=0.45\\textwidth]{img/logreg-example-beta-2.png}\n\\end{center}\n\\end{question}\n\n\\subsection{Deviance and Deviance Residuals}\n\nThe \\textbf{deviance} (called \\textbf{residual deviance} in the model output) plays a role in GLMs akin to that of the residual standard error in linear regression; it is a measure of the residual variation in the outcome not explained by the model. The \\textbf{null deviance} is the deviance for a model that only includes an intercept. Under the null hypothesis that all of the $\\beta$s are zero except the intercept (i.e, a model with no predictors explains the data as well as our model), the difference\n$$ \\text{null deviance} - \\text{residual deviance} $$\nis distributed as $\\chi^2_p$, a chi-squared distribution (see Section~\\ref{sect:chisqdist}) with $p$ degrees of freedom, where $p$ is the number of predictors.\n\\vspace{5mm}\n\n\\begin{question}{}\nThis test is a hypothesis test of the null hypothesis that a model with no predictors fits our data as well as our model, where goodness of fit is measured by the deviance (lower is better). What is this hypothesis test akin to in the linear regression model output?\n\\end{question}\n\n\\begin{question}{}\nThe difference in null and residual deviances in this case is $67.72$. It follows a $\\chi^2_2$ distribution under the null. A plot of the $\\chi^2_2$ distribution and our test statistic is shown below. What do these findings indicate about the $p$-value of this goodness of fit test and what does it mean?\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/logreg-example-goodness-of-fit-test.png}\n\\end{center}\n\\end{question}\n\nIn the GLM context, there are multiple types of residual (more on this later). \\textbf{Deviance residuals} quantify the contributions of the individual samples to the deviance. Unfortunately, the output from \\texttt{glm} is confusing because what \\texttt{glm} calls a deviance residual in the model summary is actually something called a \\textbf{working residual}. We will ignore this part of the output until we understand more about the inner workings of GLMs. \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Example: Low Birthweight Dataset \\label{sect:lowbwt}}\n\nThe goal of this study was to identify risk factors associated with giving birth to a low birth weight baby (a baby weighing less than 2500 grams). Infant mortality rates and birth defect rates are very high for low birth weight babies. A woman's behavior during pregnancy (including diet, smoking habits, and receiving prenatal care) can greatly alter the chances of carrying the baby to term and, consequently, of delivering a baby of normal birth weight.\n\nData were collected on 189 women, 59 of which had low birth weight babies and 130 of which had normal birth weight babies\\footnote{SOURCE: Hosmer and Lemeshow (2000) \\emph{Applied Logistic Regression: Second Edition}. Data were collected at Baystate Medical Center, Springfield, Massachusetts during 1986.}.\n\n\\begin{center}\n\\texttt{ \\small\n\\begin{tabular}{ll}\n\\toprule\nLOW & Low birth weight (0 = birth weight $\\geq$ 2500 g;\\\\\n& 1 = birth weight $< 2500$ g) \\\\\nAGE & Age of mother in years \\\\\nLWT & Mother's weight in pounds at last menstrual period \\\\\nRACE & Race (1 = white, 2 = black, 3 = other) \\\\\nSMOKE & Smoking status during pregnancy (1 = yes, 0 = no) \\\\\nPTL & History of premature labor (0 = none, 1 = one, etc.) \\\\\nHT & History of hypertension (0 = no, 1 = yes) \\\\\nUI & Presence of uterine irritability (0 = no, 1 = yes) \\\\\nFTV & Number of physician visits during the first trimester \\\\\nBWT & Birth weight in grams \\\\\n\\bottomrule\n\\end{tabular}\n}\n\\end{center}\n\nWe will build a model that predicts the value of \\texttt{LOW} based on all of the other covariates except, of course, \\texttt{BWT}. (Why not use \\texttt{BWT}?) \n\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/logreg-low-bwt-example-model.png}\n\\end{center}\n\n\\begin{question}{}\nIn this model, is the effect of one predictor (say, \\verb|AGE|) impacted by the value(s) of any of the other predictor(s)? How does this differ from the other classification algorithms we've seen (KNN and decision trees)? What are the advantages and disadvantages of this choice? \n\\end{question}\n\n\\begin{question}{}\nComment on how the variable \\texttt{RACE} enters into the model here. Does this make sense in light of what that variable means and how it potentially interacts with the other study variables?\n\\end{question}\n\n\\begin{question}{}\nInterpret the values of each of these coefficients. Based on the coefficient values and their standard errors, which predictor(s) do you think have the greatest impact on whether or not a woman has a low birthweight baby? \n\\end{question}\n", "meta": {"hexsha": "cd7a660f69b6ba841dcc04b19a2d220e83c703a6", "size": 11764, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/mcds-logistic-regression.tex", "max_stars_repo_name": "blpercha/mcds-notes", "max_stars_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-12-10T16:51:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-03T01:31:23.000Z", "max_issues_repo_path": "tex/mcds-logistic-regression.tex", "max_issues_repo_name": "blpercha/mcds-notes", "max_issues_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/mcds-logistic-regression.tex", "max_forks_repo_name": "blpercha/mcds-notes", "max_forks_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T17:16:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T17:16:44.000Z", "avg_line_length": 78.4266666667, "max_line_length": 725, "alphanum_fraction": 0.75, "num_tokens": 2939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511579973931, "lm_q2_score": 0.7461389817407017, "lm_q1q2_score": 0.6385839115497753}} {"text": "\\documentclass[12pt]{amsart}\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem*{corollary}{Corollary}\n\\newtheorem*{lemma}{Lemma}\n\\theoremstyle{definition}\n\\newtheorem{problem}{Problem}\n\n\\newcommand{\\C}{\\mathbb{C}}\n\\DeclareMathOperator{\\re}{Re}\n\\renewcommand{\\iint}{\\int\\!\\!\\!\\int} \n\\newcommand{\\Beta}{B}\n\n\\addtolength{\\headheight}{1.5pt} \n\n\\begin{document}\n\n\\title{Lu Qi-Keng's Problem}\n\n\\author{Harold P. Boas}\n\n\\address{Department of Mathematics, Texas A\\&M University,\n College Station, TX 77843--3368, USA}\n\n\\email{boas@math.tamu.edu}\n\n\\thanks{This article is based on a lecture at the third Korean\n several complex variables symposium in December 1998. The\n author thanks GARC, the Global Analysis Research Center at\n Seoul National University, for sponsoring his participation in\n this international conference.} \n\n\\thanks{The author's research was partially supported by grant\n number DMS-9801539 from the National Science Foundation of the\n United States of America.}\n\n\n\\begin{abstract}\n This expository article, intended to be accessible to students,\n surveys results about the presence or absence of zeroes of the\n Bergman kernel function of a bounded domain in~\\(\\C^n\\). Six\n open problems are stated.\n\\end{abstract}\n\n\\subjclass{Primary 32H10}\n\n\\maketitle\n\n\\setcounter{tocdepth}{1}\n\\tableofcontents\n\n\\section{Introduction}\nWhen does a convergent infinite series of holomorphic functions\nhave zeroes? This question is a fundamental, difficult problem\nin mathematics.\n\nFor example, the series \\(\\sum_{n=0}^\\infty z^n/n!\\) is\nzero-free, but how can one tell this without a priori knowledge\nthat the series represents the exponential function? Changing the\ninitial term of this series produces a new series that does have\nzeroes, since the range of the exponential function is all\nnon-zero complex numbers. The problem of determining when a\nseries has zeroes is essentially equivalent to the hard problem\nof determining the range of a holomorphic function that is\npresented as a series.\n\nA famous instance of the problem of locating zeroes of infinite\nseries is the Riemann hypothesis about the zeta function: namely,\nthe conjecture that when \\(0<\\re z<1/2\\), the convergent series\n\\(\\sum_{n=1}^\\infty (-1)^n/n^z\\) has no zeroes. This formulation\nof the Riemann hypothesis is equivalent to the usual statement\nthat the zeroes of \\(\\zeta(z)\\) in the critical strip where\n\\(0<\\re z<1\\) all lie on the line where \\(\\re z=1/2\\). Indeed,\nwhen \\(\\re z>1\\), absolute convergence justifies writing that\n\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{(-1)^n}{n^z} = -\\sum_{n=1}^\\infty\n\\frac{1}{n^z} +2\\sum_{\\text{even \\(n\\)}}\\frac{1}{n^z}\n=\\zeta(z)(2^{1-z}-1).\n\\end{equation*}\nBy the principle of persistence of functional relationships, the\nexpressions on the outer ends of this equality still agree when\n\\(0<\\re z<1\\), so \\(\\zeta(z)\\) and \\(\\sum_{n=1}^\\infty\n(-1)^n/n^z\\) have the same zeroes in the interior of the critical\nstrip. Moreover, the well known functional equation for the zeta\nfunction implies that the zeroes of~\\(\\zeta\\) in the critical\nstrip are symmetric about the point~\\(1/2\\), so it suffices to\nexamine the left half of the critical strip for zeroes.\n\nIn this article, I shall discuss a different instance of the\ngeneral problem of locating zeroes of infinite series. The\nBergman kernel function is most conveniently expressed as the sum\nof a convergent infinite series. In 1966, Lu Qi-Keng \\cite{lu}\nasked: for which domains is the Bergman kernel function\n\\(K(z,w)\\) zero-free? I shall address three general methods for\napproaching this problem, and I shall give examples both of\ndomains whose Bergman kernel functions are zero-free and of\ndomains whose Bergman kernel functions have zeroes.\n\n\\section{The Bergman kernel function}\n\\subsection{Definition}\nThe Bergman kernel function \\(K(z,w)\\) of a domain~\\(D\\) in\n\\(\\C^n\\) is the unique\nsesqui-holomorphic\\footnote{Sesqui-holomorphic means holomorphic\n in the first variable and conjugate holomorphic in the second\n variable.} function satisfying the skew-symmetry property that\n\\(K(z,w)=\\overline {K(w,z)}\\) and the reproducing property that\n\\begin{equation*}\nf(z)=\\int_D K(z,w) f(w) \\,d\\text{Volume}_w \\qquad \\text{for all\n $z$ in $D$}\n\\end{equation*}\nfor every square-integrable holomorphic function~\\(f\\) on~\\(D\\).\nEquivalently, \\(K(z,w)=\\sum_j\n\\varphi_j(z)\\overline{\\varphi_j(w)}\\), where \\(\\{\\varphi_j\\}\\) is\nan orthonormal basis for the Hilbert space of square-integrable\nholomorphic functions on~\\(D\\). To compute the Bergman kernel\nfunction, one typically chooses an orthogonal basis, calculates\nthe normalizing factors, and sums the series.\n\nFor example, the monomials \\(1\\), \\(z\\), \\(z^2\\), \n\\ldots, are orthogonal on the unit disk in the complex plane, and\nthe norm of \\(z^k\\) is \\((\\int_{0}^{2\\pi} \\int_{0}^{1}\nr^{2k+1}\\,dr\\,d\\theta)^{1/2}\\), or \\(\\sqrt{2\\pi/(2k+2)}\\).\nTherefore the Bergman kernel function \\(K(z,w)\\) of the unit disk\nequals\n\\begin{equation}\n \\label{eq:disk}\n \\sum_{k=0}^\\infty \\frac{k+1}{\\pi} \\cdot z^k \\bar w^k \n = \\frac{1}{\\pi}\\cdot \\frac{1}{(1-z\\bar w)^2}.\n\\end{equation}\nThis result is compatible with the Cauchy integral formula\n\\begin{equation*}\nf(z)=\\frac{1}{2\\pi i}\\oint_{|w|=1}\n\\frac{f(w)}{w-z}\\,dw,\n\\end{equation*}\nwhich can be rewritten by Green's formula as\n\\begin{equation*}\nf(z)=\\frac{1}{\\pi}\\iint_{|w|<1} \\frac{f(w)}{(1-z\\bar w)^2}\n\\,d\\text{Area}_w,\n\\end{equation*}\nthus confirming that the kernel function~\\eqref{eq:disk} does\nhave the required reproducing property.\n \n\\subsection{Transformation rule}\nIf \\(f:D_1\\to D_2\\) is a biholomorphic\\footnote{Biholomorphic\n means holomorphic with a holomorphic inverse.} mapping, and if\n\\(K_1\\) and~\\(K_2\\) denote the Bergman kernel functions of the\ndomains \\(D_1\\) and~\\(D_2\\) in~\\(\\C^n\\), then\n\\begin{equation}\n\\label{eq:transform}\nK_1(z,w)=(\\det f'(z)) K_2(f(z), f(w)) (\\,\\overline{\\det f'(w)}\\,).\n\\end{equation}\nThis relationship holds because if \\(\\{\\varphi_j\\}\\) is an\northonormal basis for the square-integrable holomorphic functions\non~\\(D_2\\), then \\(\\{(\\det f') \\,\\varphi_j\\circ f\\}\\) is an\northonormal basis for the square-integrable holomorphic functions\non~\\(D_1\\). For example, scaling~\\eqref{eq:disk} shows that the\nBergman kernel function of the disk of radius~\\(r\\) is equal to\n\\(\\dfrac{1}{\\pi r^2}\\cdot \\dfrac{1}{(1-z\\bar w/r^2)^2}\\).\n\nIt is an observation of Steven~R. Bell \\cite{bell1981, bell1982}\nthat a similar transformation rule holds even when \\(f:D_1\\to\nD_2\\) is a branched \\(m\\)-fold covering (a proper holomorphic\nmapping): namely,\n\\begin{equation*}\n\\sum_{k=1}^m K_1(z, f_k^{-1}(w)) \n(\\,\\overline{\\det (f_k^{-1})'(w)}\\,) =\n(\\det f'(z)) K_2(f(z),w),\n\\end{equation*}\nwhere the \\(f_k^{-1}\\) are the \\(m\\) holomorphic local inverses\nof~\\(f\\). This formula is not valid on the branching set, where\nlocal inverses are not defined.\n\nThe Bergman kernel function \\(K(z,w)\\) of a simply-connected\nplanar domain~\\(D\\) is related to the Riemann mapping\nfunction~\\(f\\) that maps~\\(D\\) onto the unit disk, taking the\npoint~\\(a\\) to~\\(0\\): namely,\n\\begin{equation}\n\\label{eq:riemann}\nf'(z)=K(z,a)\\sqrt{\\frac{\\pi}{K(a,a)}}.\n\\end{equation}\nIndeed, the transformation rule~\\eqref{eq:transform} implies that\n\\begin{equation*}\nK(z,w)=f'(z)\\cdot\\frac{1}{\\pi}\\cdot\n\\frac{1}{(1-f(z)\\overline{f(w)}\\,)^2}\\cdot\\overline{f'(w)}.\n\\end{equation*}\nSince \\(f(a)=0\\), and \\(f'(a)\\) is real and positive,\nsetting \\(w=a\\) implies that\n\\begin{math}\n\\pi K(z,a)=f'(z) f'(a)\n\\end{math},\nand then setting \\(z=a\\) makes it possible to eliminate~\\(f'(a)\\)\nto obtain~\\eqref{eq:riemann}.\n\nSince the Riemann mapping function solves a certain extremal\nproblem, the connection in one dimension with the Bergman kernel\nfunction suggests studying the Bergman kernel function in higher\ndimensions in connection with extremal problems.\n\n\\subsection{Extremal properties}\n\\label{extremal}\nOne can use extremal characterizations of the Bergman kernel\nfunction to help prove theorems about convergence of the Bergman\nkernel functions of a convergent sequence of domains. In the\nfollowing two properties, the point~\\(w\\) is fixed in a\ndomain~\\(D\\) in~\\(\\C^n\\), and \\(\\{\\varphi_j\\}\\) is an orthonormal\nbasis for the Hilbert space \\(A^2(D)\\) of square-integrable\nholomorphic functions on~\\(D\\).\n\n\\begin{enumerate}\n\\item In the class of holomorphic functions~\\(f\\) on~\\(D\\) such\n that \\(\\int_D |f|^2\\le 1\\), the maximal value of \\(|f(w)|^2\\)\n is \\(K(w,w)\\). In other words, \\(K(w,w)\\) is the square of the\n norm of the functional from \\(A^2(D)\\) to~\\(\\C\\) that\n evaluates a function at the point~\\(w\\).\n \n Indeed, if \\(f(z)=\\sum_j c_j \\varphi_j(z)\\), and if \\(\\sum_j\n |c_j|^2\\le1\\), then the Cauchy-Schwarz inequality implies that\n \\(|\\sum_j c_j \\varphi_j(w)|^2\\) is bounded above by \\(\\sum_j\n |\\varphi_j(w)|^2\\), which equals \\(K(w,w)\\); and the upper\n bound is attained if \\(c_j\\) is taken equal to\n \\(\\overline{\\varphi_j(w)}/(\\sum_j |\\varphi_j(w)|^2)^{1/2}\\).\n \n\\item In the class of holomorphic functions~\\(f\\) on~\\(D\\)\n satisfying the nonlinear constraint that \\(f(w)\\ge \\int_D\n |f|^2\\), the function with the maximal value at~\\(w\\) is\n \\(K(\\cdot,w)\\).\n \n Indeed, it is evident that the function \\(K(\\cdot,w)= \\sum_j\n \\overline{\\varphi_j(w)}\\,\\varphi_j\\) is in the class. On the\n other hand, if \\(f=\\sum_j c_j \\varphi_j\\) is an arbitrary\n member of the class, then the preceding extremal property\n implies that \\(f(w)^2\\le K(w,w)\\sum_j |c_j|^2\\); and since the\n defining property of the class implies that \\((\\sum_j\n |c_j|^2)^2 \\le f(w)^2\\), it follows that \\(\\sum_j|c_j|^2\\le\n K(w,w)\\), and hence \\(f(w)\\le K(w,w)\\).\n\\end{enumerate}\n\n\\section{Motivation for Lu Qi-Keng's problem}\nThe Riemann mapping theorem characterizes the planar domains that\nare biholomorphically equivalent to the unit disk. In higher\ndimensions, there is no Riemann mapping theorem,\\footnote{More\n precisely, in order to obtain a generalized Riemann mapping\n theorem, one needs either new hypotheses \\cite{chernji} or new\n definitions \\cite{semmes}.} and two natural problems arise.\n\n\\begin{enumerate}\n\\item Are there canonical representatives of biholomorphic\n equivalence classes of domains?\n \n\\item How can one tell that two particular domains are\nbiholomorphically inequivalent?\n\\end{enumerate}\n\nAs an approach to the first question, Stefan Bergman introduced\nthe notion of a ``representative domain'' to which a given domain\nmay be mapped by ``representative coordinates''. If \\(g_{jk}\\)\ndenotes the Bergman metric\n\\begin{math}\n\\dfrac{\\partial^2}{\\partial z_j\\partial\\bar z_k}\\log K(z,z)\n\\end{math},\nwhere \\(K\\)~is the Bergman kernel function, then the local\nrepresentative coordinates based at the point~\\(a\\) are\n\\begin{equation*}\n\\sum_{k=1}^n\ng_{kj}^{-1}(a) \\left. \\frac{\\partial}{\\partial \\bar w_k} \\log\n\\frac{K(z,w)}{K(w,w)} \\right|_{w=a}, \\qquad j=1, \\dotsc, n.\n\\end{equation*}\nThese coordinates take \\(a\\) to~\\(0\\) and have complex Jacobian\nmatrix at~\\(a\\) equal to the identity.\n\nZeroes of the Bergman kernel function \\(K(z,w)\\) evidently pose\nan obstruction to the global definition of Bergman representative\ncoordinates. This observation was Lu Qi-Keng's motivation for\nasking which domains have zero-free Bergman kernel functions.\n\nOn the other hand, if the Bergman kernel function of a domain\ndoes have zeroes, then the transformation\nrule~\\eqref{eq:transform} shows that the zero set is a\nbiholomorphically invariant object. Therefore zero sets of\nBergman kernel functions could be a tool for addressing the\nsecond question stated above. This idea has not yet been\nexploited in the literature.\n\n\\section{First examples}\n\\label{first}\nThe Bergman kernel function~\\eqref{eq:disk} of the unit disk is\nevidently zero-free. Consequently, the Bergman kernel function\nof every bounded, simply connected, planar domain is zero-free:\napply either the transformation rule~\\eqref{eq:transform} or the\nexplicit formula~\\eqref{eq:riemann} relating the Riemann mapping\nfunction to the Bergman kernel function.\n\nOn the other hand, the Bergman kernel function of every annulus\ndoes have zeroes \\cite{rosenthal,skwar}; more generally, the\nBergman kernel function of every bounded, multiply connected,\nplanar domain with smooth boundary has zeroes \\cite{suitayamada}.\n\nIsolated singularities of square-integrable holomorphic functions\nare removable, and therefore the Bergman kernel function does not\nsee isolated punctures in a domain. For example, the Bergman\nkernel function of a punctured disk is zero-free. On the other\nhand, a finitely connected planar domain with no singleton\nboundary component can be mapped biholomorphically to a smoothly\nbounded domain. Consequently, if a bounded planar domain is\nfinitely connected and has at least two non-singleton boundary\ncomponents, then its Bergman kernel function has zeroes.\n\nI do not know if a corresponding statement holds for infinitely\nconnected planar domains. For example, delete from the open unit\ndisk a countable sequence of pairwise disjoint closed disks that\naccumulate only at the boundary of the unit disk. Does the\nBergman kernel function of the resulting domain have zeroes?\n\n\\begin{problem}\n Give necessary and sufficient conditions on an infinitely\n connected planar domain for its Bergman kernel function to have\n zeroes.\n\\end{problem}\n\nIt is easy to see that in higher dimensions, the Bergman kernel\nfunction of a product domain is the product of the Bergman kernel\nfunctions of the lower dimensional domains. Consequently, the\nBergman kernel function of a polydisc is zero-free, while the\nBergman kernel function of the Cartesian product of a disc with\nan annulus does have zeroes. The Bergman kernel function\n\\(K(z,w)\\) of the unit ball in~\\(\\C^n\\) is the zero-free function\n\\(\\dfrac{n!}{\\pi^n}\\cdot \\dfrac{1}{(1-\\langle\nz,w\\rangle)^{n+1}}\\), where \\(\\langle z,w\\rangle\\) denotes the\nscalar product \\(z_1\\bar w_1 + \\dots + z_n \\bar w_n\\).\n\nEven without knowing this explicit formula, one can see that the\nBergman kernel function of the unit ball is zero-free. Since the\nball is a complete circular domain,\\footnote{A domain is called\n complete circular if whenever it contains a point~\\(z\\), it\n also contains the one-dimensional disk \\(\\{\\,\\lambda z:\n |\\lambda|\\le1\\,\\}\\).} its space of square-integrable\nholomorphic functions has an orthonormal basis whose first\nelement is a constant (namely, the reciprocal of the square root\nof the volume of the domain) and whose other elements are\nfunctions that vanish at the origin. Consequently, the Bergman\nkernel function~\\(K\\) has the property that \\(K(z,0)\\) is a\nnon-zero constant function of~\\(z\\). Since the ball is\nhomogeneous,\\footnote{A domain is called homogeneous if it has a\n transitive automorphism group, that is, if any point of the\n domain can be mapped to any other point by a biholomorphic\n self-mapping of the domain.} the transformation\nrule~\\eqref{eq:transform} implies that the Bergman kernel\nfunction is nowhere zero. The same argument shows that every\nbounded, homogeneous, complete circular domain has a zero-free\nBergman kernel function \\cite{belldavid}.\n\nFor many years it was thought that sufficiently nice,\ntopologically trivial, bounded domains in~\\(\\C^n\\) should have\nzero-free Bergman kernel functions. For example, all strongly\nconvex,\\footnote{The statement is also true for strongly\n pseudoconvex domains, that is, domains that locally can be\n mapped biholomorphically to strongly convex domains.}\nsufficiently small perturbations of the ball have zero-free\nBergman kernel functions if ``small'' is interpreted in the\n\\(C^\\infty\\) topology on domains \\cite{greenekrantz}. It turns\nout, however, to be the generic situation for the Bergman kernel\nfunction of a domain of holomorphy to have zeroes, if ``generic''\nis interpreted in the very flexible Hausdorff topology on domains\n\\cite{boasgeneric}.\n\n\\begin{problem}\n Do there exist arbitrarily small class \\(C^1\\) perturbations of\n the ball whose Bergman kernel functions have zeroes?\n\\end{problem}\n\nI shall now discuss three techniques that can be used to show the\nexistence of interesting domains whose Bergman kernel functions\nhave zeroes.\n\n\\section{Variation of domains}\n\\label{variation-domains}\nIf reasonable domains \\(\\Omega_j\\) converge in a reasonable way\nto a limiting domain \\(\\Omega\\), then the Bergman kernel\nfunctions \\(K_{\\Omega_j}(z,w)\\) converge to \\(K_\\Omega(z,w)\\)\nuniformly on compact subsets of \\(\\Omega\\times\\Omega\\). The word\n``reasonable'' can be made precise \\cite{boasgeneric}, but here I\nwill simply mention two examples of reasonable behavior. The\nfirst example is a fundamental theorem of I.~P. Ramadanov\n\\cite{ramadanov} which started the whole theory.\n\\begin{itemize}\n\\item The \\(\\Omega_j\\) form an increasing sequence whose union\n is~\\(\\Omega\\).\n \n\\item The \\(\\Omega_j\\) are bounded pseudoconvex\\footnote{A domain\nis pseudoconvex if it is the union of an increasing sequence of\nstrongly pseudoconvex domains, as defined in the preceding\nfootnote. According to the solution of the Levi problem (see, for\nexample, \\cite{Krantz}), pseudoconvex domains are the same as\ndomains of holomorphy.} domains whose complements converge in the\nHausdorff metric to the complement of an~\\(\\Omega\\) whose\nboundary is locally a graph.\n\\end{itemize}\nAn example of unreasonable convergence is a sequence of disks\nshrinking down to a disk with a slit \\cite{skwarczynski}.\n\nThe proof of the convergence theorem exploits the extremal\ncharacterization of the Bergman kernel function from\nsection~\\ref{extremal}. The application to Lu Qi-Keng's problem\nis that by Hurwitz's theorem, if the Bergman kernel function of\nthe limiting domain~\\(\\Omega\\) has zeroes, then so does the\nBergman kernel function of the approximating domain~\\(\\Omega_j\\)\nwhen \\(j\\)~is sufficiently large.\n\nConsequently, to construct a nice domain whose Bergman kernel\nfunction has zeroes, it suffices to construct a degenerate domain\nwhose Bergman kernel function has zeroes, and then to approximate\nthe degenerate domain by nice ones. For example, an easy\ncalculation shows that constant functions are not\nsquare-integrable on the domain\n\\begin{equation}\n\\label{eq:counterex}\n \\left\\{\\, (z_1,z_2)\\in\\mathbf{C}^2: |z_2|<\\frac{1}{1+|z_1|}\\,\n\\right\\},\n\\end{equation}\nalthough the domain does support some non-constant\nsquare-integrable holomorphic functions. Therefore the Bergman\nkernel function of this domain has a zero at the origin, but is\nnot identically zero. By approximating this domain from inside,\none sees that there exists a bounded, smooth, logarithmically\nconvex, complete Reinhardt domain\\footnote{A domain is called\n complete Reinhardt if whenever it contains a point\n \\((z_1,\\dotsc,z_n)\\), it also contains the polydisc\n \\(\\{\\,(\\lambda_1z_1, \\dotsc, \\lambda_nz_n): |\\lambda_1|\\le1,\n \\dotsc, |\\lambda_n|\\le1\\,\\}\\). The logarithmically convex\n complete Reinhardt domains are the convergence domains of power\n series.} whose Bergman kernel function has zeroes\n\\cite{boascounterexample}. This example was surprising when it\nwas first discovered; indeed, it contradicts a theorem previously\npublished by two different authors \\cite[Theorem~1]{matsuura},\n\\cite[Corollary]{kakurai} and applied by a third \\cite{kanemaru}.\n\nNguy\\^en Vi\\^et Anh, a student in Marseille, recently showed\n\\cite{anh} how to approximate the domain~\\eqref{eq:counterex}\nfrom inside by \\emph{concrete} domains that are smooth,\nalgebraic, logarithmically convex, complete Reinhardt domains.\nNamely, the domain defined by the inequality\n\\begin{equation}\n\\label{eq:anh}\n |z_2|^{2k}(1+|z_1|)^{2k} + |z_2|^{2k}(1-|z_1|)^{2k} + \\left(\n \\frac{|z_1|^2+|z_2|^2} {k}\\right)^k <1\n\\end{equation}\nhas the indicated properties when \\(k\\)~is a positive integer,\nand so the Bergman kernel function of this domain must have\nzeroes when \\(k\\)~is sufficiently large.\n\n\\begin{problem}\n How large must \\(k\\) be in order for the Bergman kernel\n function of the domain~\\eqref{eq:anh} to have zeroes?\n\\end{problem}\n\nThe technique of variation of domains can be used to prove the\nstatement at the end of section~\\ref{first} that every nice\ndomain can be arbitrarily closely approximated in the Hausdorff\nmetric by a nice domain whose Bergman kernel function has zeroes.\nView the starting domain as the Earth, and place in orbit around\nthe Earth a small copy of one of the bounded domains just\ndiscussed. The Bergman kernel function of this disconnected\nregion has zeroes because the Bergman kernel function of the\nsatellite has zeroes. Now attach the satellite to the Earth by a\nthin tether. If the tether is sufficiently thin, then the\nprinciple of variation of domains implies that the Bergman kernel\nfunction of the joined domain has zeroes. According to the\nbarbell lemma \\cite[Chapter~5, Exercise~21]{Krantz} in a suitable\nformulation \\cite{boasgeneric}, the joined domain can be made\nstrongly pseudoconvex if the Earth is.\n\n\\section{Variation of weights}\nIn the representation of the Bergman kernel function as an\ninfinite series \\(\\sum_j \\varphi_j(z)\\overline{\\varphi_j(w)}\\),\nthe basis elements~\\(\\varphi_j\\) are supposed to be orthonormal\nfor integration with respect to Lebesgue measure. It is natural\nto consider the analogous construction when Lebesgue measure is\nmultiplied by a positive weight function.\n\nFor instance, if \\(G\\)~is a domain in~\\(\\C^n\\), and \\(\\Omega\\)~is\na Hartogs domain in~\\(\\C^{n+1}\\) with base~\\(G\\), which means\nthat \\(\\Omega = \\{\\,(z,z_{n+1})\\in\\C^{n+1}: z\\in G\\) and\n\\(|z_{n+1}|0\\). The square of the norm of the\nmonomial \\(z^k\\) with weight factor \\( (1-|z|)^q\\) is\n\\begin{equation*}\n\\int_0^{2\\pi}\\int_0^1 r^{2k+1} (1-r)^q \\,dr\\,d\\theta = \n2\\pi \\Beta(2k+2,q+1),\n\\end{equation*}\nwhere \\(\\Beta\\) is the Beta function defined in terms of the\nGamma function by \\(\\Beta(a,b)= \\Gamma(a)\n\\Gamma(b)/\\Gamma(a+b)\\). Consequently, the weighted Bergman\nkernel function \\(K_q(z,w)\\) equals\n\\((2\\pi)^{-1}\\sum_{k=0}^\\infty (z\\bar w)^k/\\Beta(2k+2,q+1)\\). A\nclosed form expression for this series is most conveniently\nwritten in terms of the squares of the variables:\n\\begin{equation}\n\\label{eq:q}\nK_q(z^2, w^2)= \\frac{(q+1)}{4\\pi z\\bar w} \\left[ \\frac{1}{(1-z\\bar\nw)^{q+2}} - \\frac{1}{(1+z\\bar w)^{q+2}} \\right].\n\\end{equation}\nThe powers of $(1\\pm z\\bar w)$ are to be understood as principal\nbranches. The validity of this closed form expression can be\nverified by the binomial series expansion.\n\nThe explicit expression~\\eqref{eq:q} implies that the weighted\nBergman kernel function~\\(K_q\\) has zeroes in the interior of the\nunit disk if and only if \\(q>2\\). Indeed, taking limits\nin~\\eqref{eq:q} shows that \\(K_q\\)~is not equal to~\\(0\\) when\neither coordinate is equal to~\\(0\\), so it is only necessary to\ndecide if \\((1-t)^{q+2}=(1+t)^{q+2}\\) for some non-zero~\\(t\\) in\nthe unit disk. Since the mapping \\(t\\mapsto (1+t)/(1-t)\\) takes\nthe unit disk bijectively to the right half-plane, with the\norigin going to the point~\\(1\\), following this mapping by the\nmapping \\(u\\mapsto u^{q+2}\\) produces a composite mapping that\ntakes some non-zero point of the open unit disk to the\npoint~\\(1\\) if and only if \\(q>2\\).\n\nConsequently, the Bergman kernel function of the\ndomain~\\eqref{eq:ex} is guaranteed to have zeroes if\n\\(p_2+\\dots+p_n >2\\). Moreover, this domain is geometrically\nconvex if no~\\(p_j\\) exceeds~\\(2\\). Therefore, one can exhibit\nmany concrete examples of convex domains whose Bergman kernel\nfunctions have zeroes \\cite{boasfustraube}. Here are some:\n\\begin{equation}\n\\label{eq:convex}\n\\begin{gathered}\n\\{\\,z\\in\\C^3: |z_1|+|z_2|+|z_3|<1\\,\\}, \\\\\n\\{\\,z\\in\\C^3: |z_1|+|z_2|+|z_3|^2<1\\,\\}, \\\\\n\\{\\,z\\in\\C^4: |z_1|+|z_2|^2+|z_3|^2+|z_4|^4<1\\,\\}.\n\\end{gathered}\n\\end{equation}\n\nUsing a different method, Peter Pflug and E.~H. Youssfi found\nsome other interesting examples of convex domains whose Bergman\nkernel functions have zeroes. Even though the ``minimal ball''\nin~\\(\\C^n\\) defined by the inequality\n\\begin{equation}\n\\label{eq:minimal}\n |z_1|^2+\\dots+|z_n|^2 + |z_1^2+\\dots+z_n^2| <1\n\\end{equation}\nlacks multi-circular symmetry, its Bergman kernel function is\nknown explicitly \\cite{opy}, and in \\cite{pflugyoussfi} the\nauthors analyzed the explicit formula to see that the Bergman\nkernel function of this domain has zeroes when \\(n\\ge4\\).\n\nAlthough the convex domains defined by \\eqref{eq:convex}\nand~\\eqref{eq:minimal} do not have smooth boundaries, they can be\napproximated from inside by smoothly bounded, strongly convex\ndomains. From the method of variation of domains in\nsection~\\ref{variation-domains}, it follows that when \\(n\\ge3\\),\nthere exist smoothly bounded, strongly convex domains in~\\(\\C^n\\)\nwhose Bergman kernel functions have zeroes. Pflug and Youssfi\neven showed in \\cite{pflugyoussfi} how to write down concrete\nexamples of bounded, smooth, algebraic, strongly convex domains\nthat approximate~\\eqref{eq:minimal} from inside.\n\nUsing the same idea, Nguy\\^en Vi\\^et Anh \\cite{anh} gave concrete\nexamples of bounded, smooth, algebraic, strongly convex,\nReinhardt domains in~\\(\\C^n\\) whose Bergman kernel functions have\nzeroes when \\(n\\ge3\\). For example, when \\(k\\)~is a sufficiently\nlarge positive integer, the inequality\n\\begin{equation}\n\\label{eq:alg}\n(|z_1|^2+|z_2|^2+|z_3|^2)^{2k} + \\sum_{\\substack{ \\pm \\\\ \\text{(8\nterms)}}} (\\pm |z_1| \\pm |z_2| \\pm |z_3|)^{2k} <1\n\\end{equation}\ndefines such a domain in~\\(\\C^3\\). \n\nTo see that~\\eqref{eq:alg} has the required properties, first\nobserve that the \\(\\ell_{2k}\\) norm decreases to the\n\\(\\ell_\\infty\\) norm as \\(k\\to\\infty\\), so these domains are\ninterior approximations to the domain \\(\\{\\,z\\in\\C^3:\n|z_1|+|z_2|+|z_3|<1\\,\\}\\), which is one of the domains\n\\eqref{eq:convex} whose Bergman kernel functions have zeroes.\nThe odd powers of the \\(|z_j|\\) in the expansion\nof~\\eqref{eq:alg} cancel out by symmetry, so the defining\nfunction is equivalent to a polynomial. It would be obvious that\nthe defining function~\\eqref{eq:alg} is convex if it had \\(\\re\nz_j\\) in place of~\\(|z_j|\\), for a convex function of a linear\nfunction is convex; now observe that positive combinations of\neven powers are increasing, and the composite of a convex\nincreasing function with the convex function~\\(|z_j|\\) is convex.\n\nThe two-dimensional domain defined by the inequality\n\\(|z_1|+|z_2|<1\\) is a borderline case for the preceding\nconsiderations. It turns out \\cite{boasfustraube} that the\nBergman kernel function of this domain has no zeroes in the\ninterior of the domain, although it does have zeroes on the\nboundary.\n\n\\begin{problem}\n Exhibit a bounded convex domain in~\\(\\C^2\\) whose Bergman\n kernel function has zeroes in the interior of the domain.\n\\end{problem}\n\n\\section{Conclusion}\nIt is a difficult problem to determine whether the Bergman kernel\nfunction of a specific domain has zeroes or not. If the kernel\nfunction is presented as an infinite series, then locating the\nzeroes may be of the same order of difficulty as proving the\nRiemann hypothesis; and even if the series can be summed in\nclosed form, determining whether or not \\(0\\)~is in the range may\nbe hard.\n\nIn this article, I have emphasized examples in which the Bergman\nkernel function does have zeroes. As the subject developed\nhistorically, such examples were considered surprising. From our\ncurrent perspective, it would be more surprising to find some\nsimple geometric condition guaranteeing that the Bergman kernel\nfunction is zero-free.\n\nStudents planning further investigation of the Bergman kernel\nfunction might consult, in addition to the journal articles I\nhave cited, Bell's book \\cite{bellbook} about the one-dimensional\ntheory, the book of Jarnicki and Pflug \\cite{jarnickipflug}, and\nStefan Bergman's own book \\cite{bergman}. I offer the following\nproblem as an illustration of how much remains to be discovered\nabout the zeroes of the Bergman kernel function.\n\n\\begin{problem}\n Characterize the vectors \\((p_1, p_2, \\dots, p_n)\\) of positive\n numbers for which the Bergman kernel function of the domain\n in~\\(\\C^n\\) defined by the inequality\n\\begin{equation*}\n|z_1|^{2/p_1}+|z_2|^{2/p_2} + \\dots + |z_n|^{2/p_n} <1\n\\end{equation*}\nis zero-free.\n\\end{problem}\n\n\n%\\bibliographystyle{amsplain} \\bibliography{luqikeng}\n%\\providecommand{\\bysame}{\\leavevmode\\hbox to3em{\\hrulefill}\\thinspace}\n\n\\begin{thebibliography}{10}\n\n\\bibitem{anh} Nguy\\^en~Vi\\^et Anh, \\emph{The {L}u {Q}i-{K}eng\nconjecture fails for strongly convex algebraic complete\n{R}einhardt domains in $\\mathbf{C}\\sp n\\ (n\\ge3)$},\nProc. Amer. Math. Soc., to appear.\n\n\\bibitem{belldavid} David Bell, \\emph{Some properties of the\n{B}ergman kernel function}, Compositio Math. \\textbf{21} (1969),\n329--330.\n\n\\bibitem{bell1981} Steven~R. Bell, \\emph{Proper holomorphic\nmappings and the {B}ergman projection}, Duke Math. J. \\textbf{48}\n(1981), no.~1, 167--175.\n\n\\bibitem{bell1982} \\bysame, \\emph{The {B}ergman kernel function\nand proper holomorphic mappings},\nTrans. Amer. Math. Soc. \\textbf{270} (1982), no.~2, 685--691.\n\n\\bibitem{bellbook} \\bysame, \\emph{The {C}auchy transform,\npotential theory, and conformal mapping}, CRC Press, 1992.\n\n\\bibitem{bergman} Stefan Bergman, \\emph{The kernel function and\nconformal mapping}, revised ed., American Mathematical Society,\n1970.\n\n\\bibitem{boascounterexample} Harold~P. Boas, \\emph{Counterexample\nto the {L}u {Q}i-{K}eng conjecture}, Proc.\nAmer. Math. Soc. \\textbf{97} (1986), no.~2, 374--375.\n\n\\bibitem{boasgeneric} \\bysame, \\emph{The {L}u {Q}i-{K}eng\nconjecture fails generically}, Proc. Amer.\nMath. Soc. \\textbf{124} (1996), no.~7, 2021--2027.\n\n\\bibitem{boasfustraube} Harold~P. Boas, Siqi Fu, and\nEmil~J. Straube, \\emph{The {B}ergman kernel function: explicit\nformulas and zeroes}, Proc. Amer. Math. Soc. \\textbf{127} (1999),\nno.~3, 805--811.\n\n\\bibitem{chernji} Shiing-Shen Chern and Shanyu Ji, \\emph{On the\n{R}iemann mapping theorem}, Ann. of Math. (2) \\textbf{144}\n(1996), no.~2, 421--439.\n\n\\bibitem{englis} Miroslav Engli{\\v{s}}, \\emph{Asymptotic\nbehaviour of reproducing kernels of weighted {B}ergman spaces},\nTrans. Amer. Math. Soc. \\textbf{349} (1997), no.~9, 3717--3735.\n\n\\bibitem{greenekrantz} R.~E. Greene and Steven~G. Krantz,\n \\emph{Stability properties of the {B}ergman kernel and\n curvature properties of bounded domains}, Recent developments\n in several complex variables, Princeton Univ. Press, 1981,\n pp.~179--198.\n \n\\bibitem{jarnickipflug} Marek Jarnicki and Peter Pflug,\n \\emph{Invariant distances and metrics in complex analysis},\n Walter de Gruyter \\& Co., 1993.\n\n\\bibitem{kakurai}\nShigeki Kakurai, \\emph{On the Lu Qi-Keng conjecture},\nRep. Fac. Engrg. Kanagawa Univ. no.~17 (1979), 3--4.\n\n\\bibitem{kanemaru}\nTadayoshi Kanemaru, \n\\emph{A remark on the Lu Qi Keng conjecture},\nMem. Fac. Ed. Kumamoto Univ. Natur. Sci. no.~31 (1982), 1--3. \n\n\\bibitem{Krantz} Steven~G. Krantz, \\emph{Function theory of\nseveral complex variables}, second ed., Wadsworth \\& Brooks/Cole,\n1992.\n\n\\bibitem{lu} Lu~Qi-keng, \\emph{On {K}aehler manifolds with\nconstant curvature}, Chinese Math.--Acta \\textbf{8} (1966),\n283--298.\n\n\\bibitem{matsuura} Shozo Matsuura, \\emph{On the {L}u {Q}i-{K}eng\nconjecture and the {B}ergman representative domains}, Pacific\nJ. Math. \\textbf{49} (1973), 407--416.\n\n\\bibitem{opy} K.~Oeljeklaus, P.~Pflug, and E.~H. Youssfi,\n\\emph{The {B}ergman kernel of the minimal ball and applications},\nAnn. Inst. Fourier (Grenoble) \\textbf{47} (1997), no.~3,\n915--928.\n\n\\bibitem{pflugyoussfi} P.~Pflug and E.~H. Youssfi, \\emph{The {L}u\n{Q}i-{K}eng conjecture fails for strongly convex algebraic\ndomains}, Arch. Math. (Basel) \\textbf{71} (1998), no.~3,\n240--245.\n\n\\bibitem{ramadanov} I.~Ramadanov, \\emph{Sur une propri\\'et\\'e de\nla fonction de {B}ergman}, C. R. Acad. Bulgare Sci. \\textbf{20}\n(1967), 759--762.\n\n\\bibitem{rosenthal} Paul Rosenthal, \\emph{On the zeros of the\n{B}ergman function in doubly-connected domains},\nProc. Amer. Math. Soc. \\textbf{21} (1969), 33--35.\n\n\\bibitem{semmes} Stephen Semmes, \\emph{A generalization of\n{R}iemann mappings and geometric structures on a space of domains\nin $\\mathbf{C}\\sp n$}, Mem. Amer. Math. Soc. \\textbf{98} (1992),\nno.~472.\n\n\\bibitem{skwar} M.~Skwarczy{\\'n}ski, \\emph{The invariant distance\nin the theory of pseudoconformal transformations and the {L}u\n{Q}i-keng conjecture}, Proc. Amer. Math. Soc. \\textbf{22}\n(1969), 305--310.\n\n\\bibitem{skwarczynski} Maciej Skwarczy{\\'n}ski,\n\\emph{Biholomorphic invariants related to the {B}ergman\nfunction}, Dissertationes Math. \\textbf{173} (1980).\n\n\\bibitem{suitayamada} Nobuyuki Suita and Akira Yamada, \\emph{On\nthe {L}u {Q}i-keng conjecture}, Proc.\nAmer. Math. Soc. \\textbf{59} (1976), no.~2, 222--224.\n\n\\end{thebibliography}\n\n\n\\end{document}\n", "meta": {"hexsha": "8d6058b66baaa58467cab1af9c8206aaef23e4c1", "size": 37218, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "benchmark/src/test-data/0001/math0001036/math0001036.tex", "max_stars_repo_name": "e-sim/pdf-text-extraction-benchmark", "max_stars_repo_head_hexsha": "42eede9867e5795a6fc040b0a7ce92da3ddd3120", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-08-23T19:07:01.000Z", "max_stars_repo_stars_event_max_datetime": "2018-08-23T19:07:01.000Z", "max_issues_repo_path": "benchmark/src/test-data/0001/math0001036/math0001036.tex", "max_issues_repo_name": "e-sim/pdf-text-extraction-benchmark", "max_issues_repo_head_hexsha": "42eede9867e5795a6fc040b0a7ce92da3ddd3120", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "benchmark/src/test-data/0001/math0001036/math0001036.tex", "max_forks_repo_name": "e-sim/pdf-text-extraction-benchmark", "max_forks_repo_head_hexsha": "42eede9867e5795a6fc040b0a7ce92da3ddd3120", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.9929078014, "max_line_length": 71, "alphanum_fraction": 0.7359610941, "num_tokens": 11430, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333246118695629, "lm_q2_score": 0.7662936537604181, "lm_q1q2_score": 0.6385713615980096}} {"text": "%\\documentclass[notitlepage,aps,pre,onecolumn]{revtex4-1}\n\\documentclass{article}\n\\usepackage{fullpage}\n\n\\usepackage{color}\n\\usepackage{amsmath,amsfonts,graphicx,enumerate,color,hyperref}\n\\usepackage{bm,braket,cancel}\n\\usepackage{tabularx}\n\\usepackage[normalem]{ulem}\n\\newcommand{\\pa}{\\partial}\n\\newcommand{\\abs}[1]{\\left| #1 \\right|}\n\\newcommand{\\avg}[1]{\\left\\langle #1 \\right\\rangle}\n\\newcommand{\\red}[1]{{\\color{red} #1}}\n\\newcommand{\\dd}[2]{\\frac{d^2 #1}{d #2^2}}\n\\newcommand{\\pd}[2]{\\frac{\\partial #1}{\\partial #2}}\n\\newcommand{\\pdd}[2]{\\frac{\\partial^2 #1}{{\\partial #2}^2}}\n\\renewcommand{\\L}{\\mathfrak{L}}\n\\renewcommand{\\O}{\\mathcal{O}\\left(\\epsilon^2\\right)}\n\\newcommand{\\s}[1]{^{(#1)}}\n\\newcommand{\\0}{^{(0)}}\n\n\\usepackage[usenames]{xcolor}\n\\hypersetup{\n colorlinks,\n linkcolor={blue!50!black},%{red!80!black},\n citecolor={blue!50!black},\n urlcolor={blue!80!black}\n}\n\\begin{document}\n\\title{Advection-diffusion of a scalar}\n\\author{YBS}\n\\maketitle\n\n\\section{Governing equations}\nThe time evolution of the concentration $c(\\bm r, t)$ of a chemical which undergoes simultaneous diffusion (with diffusion constant $\\eta$) and advection by a velocity field $\\bm v(\\bm r , t)$ is given by \n\\begin{align}\n\\pd{c}{t}+\\bm v \\cdot \\nabla c=\\eta \\nabla^2 c\n\\label{eq:eom}\n\\end{align}\nTe left-hand-side is the material time derivative (aka advective derivative). The right-hand-side is diffusion. Physically, $\\bm v$ should also satisfy an evolution equation of its own, but for our purposes we'll just assume it's constant in time. However, we will add the (physically valid) constraint that $\\bm v$ is divergence-free, i.e.~$\\nabla\\cdot v=\\sum_i \\pa_i v_i=0$. In 2D, a divergence-free vector field over a periodic domain $[0,2\\pi]\\times[0,2\\pi]$ can be decomposed to Fourier components of the form\n\\begin{align}\nv_x&= \\sum_i A_i m_i \\cos (m_i y+\\beta_i ) \\cos (n_i x+\\alpha_i )\\ , &\nv_y&=\\sum_i A_i n_i \\sin (m_i y+\\beta_i ) \\sin (n_i x+\\alpha_i )\\ ,\n\\end{align}\nwhere $A_i$ is the amplitude, $m_i, n_i$ are integers ($(m_i, n_i)$ is the wave-vector), and $\\alpha_i, \\beta_i$ are phases. For the simulation I simply draw all these parameters from a uniform distribution, using 5 terms for each simulation. \n\n\\section{Numerical implementation}\nFirst, we write the time derivative explicitly as\n\\begin{align}\n\\pd{c}{t}=\\eta \\nabla^2 c - \\bm v\\cdot\\nabla c=\\eta \\nabla^2 c - \\nabla \\cdot (c \\bm v)\\ .\n\\label{eq:3}\n\\end{align}\nThe last transition is valid because $\\nabla\\cdot \\bm v=0$.\n\nThe equations are solved on a square grid, using a staggered grid for $v$. That is, $c$ is calculated on the points $ \\left(2\\pi \\frac{i}{N},2\\pi \\frac{j}{N}\\right)$ where $N$ is the number of grid points in each dimension (currently 200) and $i,j$ are integers between 0 and $N-1$. $v$ is (pre-)computed on the points $ \\left(2\\pi \\frac{i-\\frac{1}{2}}{N},2\\pi \\frac{j-\\frac{1}{2}}{N}\\right)$. This allows implementing the right-hand-side of Eq.~\\eqref{eq:3} in a conservative way, i.e.~the spatial integral the right-hand-side vanishes (or alternatively, the spatial integral of $c$ remains constant in time, as it should).\n \nThe generation of the velocity field (and the shifted spatial mesh) is done by \\verb|generate_v_field|. The script \\verb|run_collect_save| runs many simulations and collects them in a \\verb|hdf5| file. \nThe file produced by \\texttt{run\\_collect\\_save} have this format: \n\n\\begin{itemize}\n\t\\item The name of the file is the value of $\\eta$.\n\t\\item \\verb|t| is the list of times for which the solutions are calculated\t(uniformly spaced).\n\t\\item \\verb|x| and \\verb|y| are matrices of shape $N\\times N$ with the \\texttt{x} and \\texttt{y} coordinates of the\tfinite-differences grid for \\verb|c|.\n\t\\item Each file contains 50 realizations, which are at groups \\texttt{/001},\n\t\\texttt{/002}, \\ldots{}, \\texttt{/050}: \n\t\\item \\texttt{/XXX/c} is an\n\t\\texttt{ndarray} of shape \\texttt{[n,n,len(t)]} that contains the\n\tconcentration field. \\texttt{c[i,j,k]} is the value of the\n\tconcentration field at position \\texttt{x[i,j]} and\n\t\\texttt{y[i,j]} at time \\texttt{t[k]}. In other words,\n\t\\texttt{c[...,i]} is the snapshot of the concentration field at time\n\t\\texttt{t[i]}. \n\t\\item The group \\texttt{/XXX} also contains\n\t\\texttt{/XXX/u},\\texttt{/XXX/v}, which are the \\texttt{x} and \\texttt{y}\n\tcomponents of the velocity field. Note that \\texttt{u} and \\texttt{v}\n\tare not evaluated at the same grid points as c since we use a staggered\n\tscheme to ensure conservation. The functional form of \\texttt{u} and\n\t\\texttt{v} is given in both MATLAB and Mathematica syntax in the\n\tattributes of \\texttt{/XXX}.\n\\end{itemize} \n\n\n\\end{document}", "meta": {"hexsha": "9072629217f9d14c2e240d24d8f304e2c3ac8de6", "size": 4649, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "explanation.tex", "max_stars_repo_name": "yohai/advection-diffusion", "max_stars_repo_head_hexsha": "1e277812b548d010a82605cecd835ec07c79c605", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "explanation.tex", "max_issues_repo_name": "yohai/advection-diffusion", "max_issues_repo_head_hexsha": "1e277812b548d010a82605cecd835ec07c79c605", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "explanation.tex", "max_forks_repo_name": "yohai/advection-diffusion", "max_forks_repo_head_hexsha": "1e277812b548d010a82605cecd835ec07c79c605", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 56.0120481928, "max_line_length": 624, "alphanum_fraction": 0.7143471714, "num_tokens": 1515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245828938677, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.638571334946474}} {"text": "\n\\subsection{Extreme learning machines}\n\nThis is an alternative to backprobagation for training a feedforward neural network.\n\nWe start with random parameters for each layer \\(W_i\\).\n\nWe have:\n\n\\(\\hat y=W_2\\sigma (W_1 x)\\)\n\nEtc.\n\nWe calculate:\n\n\\(W_2=\\sigma(W_1x)^+Y\\)\n\nSo \\(W_1\\) is random and not updated.\n\n\\(W_2\\) is assigned to minimise loss, where \\(W_2\\) has no activation function.\n\n", "meta": {"hexsha": "88096d2e6cbde4da133f09eac185e310e4780e7a", "size": 390, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/neuralNetworksAlternative/01-03-extreme.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/neuralNetworksAlternative/01-03-extreme.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/neuralNetworksAlternative/01-03-extreme.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 17.7272727273, "max_line_length": 84, "alphanum_fraction": 0.7179487179, "num_tokens": 111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9263037363973295, "lm_q2_score": 0.6893056231680122, "lm_q1q2_score": 0.6385063742602194}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{examples}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% ============================================================================================\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n import cdblib\n checkpoint_file = 'tests/semantic/output/example-05.json'\n cdblib.create (checkpoint_file)\n checkpoint = []\n\\end{cadabra}\n\\egroup\n\n\\clearpage\n\n% ============================================================================================\n\\section*{Example 5a Keeping focused}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices.\n\n expr := A_{a} v^{a} + B_{a} v^{a} + C_{a} v^{a}. # cdb (ex-05.100,expr)\n\n zoom (expr,$B_{a} Q??$) # cdb (ex-05.101,expr)\n substitute (expr, $v^{a} -> w^{a}$) # cdb (ex-05.102,expr)\n unzoom (expr) # cdb (ex-05.103,expr)\n\n checkpoint.append (expr)\n\\end{cadabra}\n\n\\begin{align*}\n \\cdb{ex-05.100} &= \\Cdb{ex-05.101}\\\\\n &= \\Cdb{ex-05.102}\\\\\n &= \\Cdb{ex-05.103}\n\\end{align*}\n\n\\clearpage\n\n% ============================================================================================\n\\section*{Example 5b Tags}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n def add_tags (obj,tag):\n n = 0\n ans = Ex('0')\n for i in obj.top().terms():\n foo = obj[i]\n bah = Ex(tag+'_{'+str(n)+'}')\n ans := @(ans) + @(bah) @(foo).\n n = n + 1\n return ans\n\n def clear_tags (obj,tag):\n ans := @(obj).\n foo = Ex(tag+'_{a?} -> 1')\n substitute (ans,foo)\n return ans\n\n expr := 2 V_{p q} - 3 V_{q p}. # cdb (ex-05.200,expr)\n\n expr = add_tags (expr,'\\\\mu') # cdb (ex-05.201,expr)\n\n zoom (expr, $\\mu_{1} Q??$) # cdb (ex-05.202,expr)\n substitute (expr, $V_{a b} -> - V_{b a}$) # cdb (ex-05.203,expr)\n unzoom (expr) # cdb (ex-05.204,expr)\n\n expr = clear_tags (expr,'\\\\mu') # cdb (ex-05.205,expr)\n\n checkpoint.append (expr)\n\\end{cadabra}\n\n\\begin{align*}\n \\cdb{ex-05.200} &= \\Cdb{ex-05.201}\\\\\n &= \\Cdb{ex-05.202}\\\\\n &= \\Cdb{ex-05.203}\\\\\n &= \\Cdb{ex-05.204}\\\\\n &= \\Cdb{ex-05.205}\n\\end{align*}\n\n\\clearpage\n\n% ============================================================================================\n% export to json format\n\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n for i in range( len(checkpoint) ):\n cdblib.put ('check{:03d}'.format(i),checkpoint[i],checkpoint_file)\n\\end{cadabra}\n\\egroup\n\n\\end{document}\n", "meta": {"hexsha": "ac8f193709fac0cd581369d0e292a2f9f08446d0", "size": 2776, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/example-05.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/example-05.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/example-05.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 27.4851485149, "max_line_length": 94, "alphanum_fraction": 0.4347982709, "num_tokens": 837, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673133042217, "lm_q2_score": 0.7853085909370422, "lm_q1q2_score": 0.6383516844297176}} {"text": "\\input{../../UCLHeader.tex}\n\\input{../../UCLCommands.tex}\n\\begin{document}\n\\title{Error Correction - Lecture 1}\n\\author{with Dan Browne}\n\\maketitle\n\\tableofcontents\n\n\\section{Introduction}\nError correction is at the core of realistic quantum computing. If you think of quantum computation, you should also think of error correction. Any vision of quantum computing without error correction is fundamentally flawed. This course will introduce all the various notions of quantum error correction that are common in the field. \n\n\\section{Classical Repetition Code}\nIn this section, we will first have a look at classical repetition code in order to understand how classical error correction is done. \n\nFor the classical bit $x$, which can take on values $ x \\in [0,1]$, there are two possible errors. Either the bit gets flipped, such that $0 \\rightarrow 1 $ or $1 \\rightarrow 0$, or the bit is lost. \n\nWe can decompose most codes in terms of these errors. To protect from these errors, we \\textbf{encode} our bits into several bits, such that \n\\beq\n0 \\rightarrow 000\n\\eeq\n\\beq\n1 \\rightarrow 111\n\\eeq\nThen, an error on the 2nd bit would cause\n\\beq\n000 \\rightarrow 010\n\\eeq\n\\beq\n111 \\rightarrow 101\n\\eeq\nIf we query the values of the bits, we can take a majority vote and then correct the bit accordingly. This would work for the error described above. However, for two errors on two of the qubits, we find\n\\beq\n000 \\xrightarrow{error} 110 \\xrightarrow{correction} 111 \\xrightarrow{decoding} 1\n\\eeq\n\\beq\n111 \\xrightarrow{error} 001 \\xrightarrow{correction} 000 \\xrightarrow{decoding} 0\n\\eeq\nAs we can see, the correction and decoding process are unable to correctly negate the error. \n\nWe call the expression for the encoded bit a \\textbf{codeword}. We also make a distinction between detection and correction of an error. It is always true that correction is easier than correction. Finally, consider\n\\beq\n000 \\xrightarrow{error} 111\n\\eeq\n\\beq\n111 \\xrightarrow{error} 000\n\\eeq\nThis error cannot be corrected because we cannot detect it. Whenever an error maps a codeword into another codeword, we become unable to correct it. It is a very general concept. \n\n\\section{Code distance}\nWe can define the \\textbf{Hamming distance}. Given two $n$ bit strings, $A$ and $B$, the Hamming distance is the number of bits that you need to flip in order to go from $A$ to $B$. As an example, let $A = 010$ and $B = 001$. We require two flips to go from $A$ to $B$. Thus, $D_H = 2$. \n\nIt also follows that the Hamming distance is the number of errors that we cannot correct for. \n\nLet us define another important concept, the \\textbf{Hamming weight}. The Hamming weight is the number of nonzero bits in our codeword. \n\nFinally, the \\textbf{code distance} is the maximum distance of the code, which in turn is the smallest Hamming distance between codewords. \n\n\\emph{Example}: For an $n$ bit repetition code, the distance of the code is $n$. A number of $n-1$ bitflips are detectable. \n\nUltimately, the error correction that we perform is limited by our choice of correction method. Consider the following table of errors on the $000$ codeword:\n\n\\begin{tabular}{c| c}\n\\hline\nNo. of errors &Codeword\\\\ \n0 & 000 \\\\ \n1 & 001 \\\\ \\hline\n2 & 011 \\\\\n3 &111\n\\end{tabular}\nWe have drawn the line at the point where we can no longer correct for errors. We can draw the Hamming distance around a codeword as a sphere. Every error inside the sphere will be mapped onto the codeword at its centre, but once an error takes us out of the sphere, it will be corrected into a codeword different from its original one. \n\nIn other words, we don't want the spheres to overlap. This leaves us with a number of $\\frac{n-1}{2}$ correctable errors for an $n$ bit repetition code. Note that when $n-1$ is odd, we must always round down to the nearest integer. \n\n\n\\section{Computations}\nConsider the encoded NOT gate. It has the following function on the encoded bits:\n\\beq\n0 \\xrightarrow{NOT} 1\n\\eeq\n\\beq\n1 \\xrightarrow{NOT} 0 \n\\eeq\nIn the repetition code, flipping all $n$ bits implements a \\textbf{logical NOT gate}. By logical, we mean that the gate affects the entire codeword that we have stored to mimic the effect of a single bit. \n\n\\section{Going Quantum}\nWould the same concepts carry over without any problem into the quantum regime? Unfortunately not. The repetition code is a prime example where the \\textbf{no-cloning theorem} prevents us from copying arbitrary quantum states into multiples of themselves. \n\nThe following aspects significantly complicate the transition to quantum error correction:\n\\begin{itemize}\n\\item[1] The no-cloning theorem forbids arbitrary qubit states from being copied. \n\\item[2] Measurements change the state, which complicates detection. \n\\item[3] Quantum errors are much more complicated than classical errors - there are more ways in which they can occur. \n\\end{itemize}\n\nThe last point might require some clarification. Any unitary transformation can become an error. In addition, we have the various ways in which a quantum state can decohere. Together, they add up to an infinite amount of errors, compared to just the single classical error!\n\n\\section{The Quantum Repetition Code}\nThis is an example of the most naive way in which we can start considering the quantum correction of errors. However, as we shall see later, even the quantum repetition code is an example of a more powerful formalism. \n\nWe take the basis states, and we clone the computational basis states, such that \n\\beq\n\\ket{0} \\rightarrow \\ket{000} = \\ket{0}_L\n\\eeq\n\\beq\n\\ket{1} \\rightarrow \\ket{111} = \\ket{1}_L\n\\eeq\nHere, $\\ket{0}_L$ and $\\ket{1}_L$ are the logical qubits. We also say that the codewords, $\\ket{000}, \\ket{111}$ live in a \\textbf{codespace}. We say that the codespace is spanned by the codewords. \n\nFor an arbitrary superposition, we obtain\n\\beq\n\\alpha \\ket{0} + \\beta \\ket{1} \\rightarrow \\alpha \\ket{000} + \\beta \\ket{111}\n\\eeq\nNote that since we are only copying the basis states, it is not cloning and therefore perfectly allowed. \n\n\\subsection{The bit-flip error}\nThe bit-flip error takes us\n\\beq\n\\ket{0} \\rightarrow \\ket{1}\n\\eeq\nand vice versa. It is equivalent to the Pauli $X$ matrix\n\\beq\nX = \\bpmat 0 & 1 \\\\ 1 & 0\\epmat\n\\eeq\nSo an error on the 2nd qubit of our three-qubit example would look like\n\\beq\nI \\otimes X \\otimes I = \\alpha \\ket{010} + \\beta \\ket{101}\n\\eeq\nHow can we correct for this error? We are not allowed to measure the qubits individually, since this measurement would tell us about the state and thus collapse the superposition. Instead, we introduce the concept of \\textbf{parity}. The state $\\ket{00}$ has even parity, and $\\ket{01}$ has odd parity. We can then measure without collapsing the superposition, because the states in the above superposition, for example, have the same parity. We can measure a property as long as it belongs to every state in the superposition. \n\nFor the three-qubit code example, we have that $\\ket{000}$, and $\\ket{111}$ are examples of even parity states, and $\\ket{010}, \\ket{110}$ are examples of odd parity states. \n\nSo, we need a measurement that tells us about the parity without distinguishing the states. In order to perform this measurement, we require a CNOT gate and an ancilla qubit. Then, we apply the CNOT gate to each of our two (or three) qubits and the ancilla, and finally we read off the ancilla. Thus, \n\\beq\n\\ket{00} \\rightarrow \\ket{0}_A\n\\eeq\n\\beq\n\\ket{11} \\rightarrow \\ket{0}_A\n\\eeq\n\\beq\n\\ket{01} \\rightarrow \\ket{1}_A\n\\eeq\n\\beq\n\\ket{10} \\rightarrow \\ket{1}_A\n\\eeq\nwhere $\\ket{0}_A$ is the ancilla. However, we can also perform this measurement in a simpler way, that uses fewer gates. For example, the $Z$ gate will measure Pauli $Z$ in the computational basis. This will tell us about the parity of the qubits. \n\nWe must consider the parity of all the pair of the code. For the three-qubit code, the $Z\\otimes Z$ measurement will have outcome $+1$ for even parity, and $-1$ for odd parity when applied to two of the three qubits. \n\nSo, for the three-qubit repetition code, we can think of three measurements that allows us to check the parity. \n\\beq\nZ \\otimes Z \\otimes I\n\\eeq\n\\beq\nZ \\otimes I \\otimes Z\n\\eeq\n\\beq\nI \\otimes Z \\otimes Z\n\\eeq\nBut note that since $ZZ = I$, the last measurement can actually be formed as a product of the other two. This is a sign of an underlying group-theoretical structure which we will explore more thoroughly in upcoming lectures. \n\n\\section{Correcting quantum errors}\nAssume that there is only one error on our arbitrary superposition, \n\\beq\nZ \\otimes I \\otimes I \\left( \\alpha \\ket{000} + \\beta \\ket{111} \\right) = \\alpha \\ket{100} + \\beta \\ket{011} \n\\eeq\nThen, if we apply the parity measurements defined above, we find\n\\beq\nZ \\otimes Z \\otimes I \\rightarrow - 1\n\\eeq\n\\beq\nZ \\otimes I \\otimes Z \\rightarrow - 1 \n\\eeq\n\\beq\nI \\otimes Z \\otimes Z \\rightarrow +1\n\\eeq\nand similarly for every other error. \n\nWe can write down a so-called \\textbf{syndrome table} for all possible one-qubit $X$ errors. \n\n\\begin{tabular}{|c|c|c|}\n$X\\otimes I \\otimes I $& $I \\otimes X \\otimes I$ & $I \\otimes I \\otimes X$ \\\\ \\hline\n-- &-- & + \\\\ \\hline\n-- & + & -- \\\\ \\hline\n+ & -- & -- \\\\ \\hline\n\\end{tabular}\n\nSo, our correction strategy should be:\n\\begin{itemize}\n\\item Identify where the error occurred \n\\item Apply a bit-flip operation where the error occurs\n\\end{itemize}\n\n\\section{More general errors}\nLet us consider a general error\n\\beq\nU = e^{- i \\theta X} = \\cos{\\theta} I - i \\sin{\\theta} X\n\\eeq\nSo that for our general superposition, we have\n\\beq\n\\alpha \\ket{000} + \\beta \\ket{111} = \\cos{\\theta} \\left( \\alpha \\ket{000} + \\beta \\ket{111} \\right) - i \\sin{\\theta} \\left( \\alpha \\ket{100} + \\beta \\ket{011} \\right)\n\\eeq\nBut now, measuring with $Z\\otimes Z \\otimes I$ collapses the superposition to either of the terms. But, if $\\theta \\ll 1$, then detecting the error will correct it!\n\nThis simple example is very powerful. It tells us that our error correction strategy will work in general. \n\n\n\\section{The encoding circuit}\nIn order to encode our qubits into logical qubits, for the case of the repetition code we want a circuit that gives us\n\\beq\n\\alpha \\ket{0} \\rightarrow \\alpha \\ket{000}\n\\eeq\nThis cannot be just a unitary. Instead, we need an \\textbf{isometry}, which is a unitary plus a Hilbert space expansion. The following circuit encodes the qubit:\n\nstuff\n\nAll encoding is based on CNOT gates. In summary, we have that \n\\begin{itemize}\n\\item The encoding circuit encodes the state\n\\item Two measurements detect the errors \n\\item $X$ gates corrects the errors\n\\end{itemize}\n\n\\section{Logical operators}\nConsider the logical $X$ operator. It flips the computational basis state. How can we achieve this on our encoded qubit? One stupid way to do this would be to decode our codewords, such that \n\\beq\n\\alpha \\ket{000} + \\beta \\ket{111} \\rightarrow \\alpha \\ket{0} + \\beta \\ket{1}\n\\eeq\napply a single gate and then re-encode them. This is not a good way, since it requires a lot of gates. \n\nThe good way would be to apply the operator\n\\beq\nX\\otimes X \\otimes X = \\bar{X}\n\\eeq\nwhich we call the logical $X$ operator. Clearly\n\\beq\n\\bar{X} \\ket{0}_L = \\bar{X} \\ket{000} = \\ket{111} = \\ket{1}_L\n\\eeq\nwhich is what we want. \n\nSimilarly for the logical $Z$ operator, we want\n\\beq\n\\alpha \\ket{000} + \\beta \\ket{111} \\rightarrow \\alpha \\ket{000} - \\beta \\ket{111}\n\\eeq\nBut now, there are many different ways in which we can do this. We have $ZII, IZI, IIZ, $ and $ZZZ$ that all achieve the right thing (note that we have dropped the tensor product notation). Each of the first three operators has weight 1. This means that they are indistinguishable from a single $Z$ error. \n\nRecalling what we said before about code distance and the Hamming distance, it means that the code cannot protect against a $Z$ error, also called a phase error. \n\n\\section{Distance of a Quantum Code}\nThe logical operators are equal to the smallest undetectable error. We can define the minimum distance $d$ in a quantum code as the smallest weight of a non-identity logical operator. That is, the smallest weight is the undetectable error. \n\nSo, given what we know about the repetition code, we find that it has distance $d = 1$, which is very bad. \n\n\\emph{Note on notation}: we will use so called ($n,k,d$) notation, where \\\\\n$n$ is the number of physical qubits used to write one logical qubit, \\\\$k$ is the number of encoded qubits \\\\\n$d$ is the quantum distance of the code\n\nIt follows that $k < n$ always. \n\n\\section{Correcting phase errors}\nCan we think of a way to correct a single $Z$ error? We first note that \n\\beq\nZ = HXH\n\\eeq\nand we know that the repetition code can correct for $X$ error. Thus, we try applying $H\\otimes H\\otimes H $ on the repetition code. We find\n\\beq\n\\alpha \\ket{000} + \\beta \\ket{111} \\rightarrow \\alpha \\ket{+++} + \\beta \\ket{---}\n\\eeq\nThen, we measure with the operators $XXI, XIX$ and $IXX$. Again note that the last measurement can be made up of the previous two. We then measure the parity in the $\\{\\ket{+}, \\ket{-}\\}$ basis. We again make use of CNOT and an ancilla to record the measurement. \n\nWe get positive outcomes for even parity, and negative outcomes for odd parity. This works exactly the same as in the $\\{\\ket{0}, \\ket{1}\\}$ basis. \n\nBut just like the previous code failed for phase errors, this code fails for bit-flip errors. So is there a way to combine both approaches? \n\n\\section{The Shor code}\nThe code idea is: encode a bit-flip code in a phase flip code. That is, take\n\\beq\n\\ket{+} \\rightarrow \\ket{+++}\n\\eeq\n\\beq\n\\ket{-} \\rightarrow \\ket{---}\n\\eeq\nTake each qubit and encode them in the bit-flip code. This is called \\textbf{code concatenation}. We encode each codeword into more codewords. \n\nSo we get the following code\n\\beq\n\\ket{0}_L \\rightarrow \\ket{+}_{bitflip} \\ket{+}_{bitflip} \\ket{+}_{bitflip} = \\left( \\ket{000} + \\ket{111} \\right)^{\\otimes 3}\n\\eeq\n\\beq\n\\ket{1}_L \\rightarrow \\ket{-}_{bitflip} \\ket{-}_{bitflip}\\ket{-}_{bitflip} = \\left( \\ket{000}- \\ket{111} \\right)^{\\otimes 3}\n\\eeq\nNote here that we have neglected to write out the normalisation. This is intentional, as it would quickly get very messy. \n\nThe logical operators are then for this code\n\\beq\n\\bar{X} = ZII \\otimes ZII \\otimes ZII = ZIIZIIZII\n\\eeq\n\\beq\n\\bar{Z} = XXXIIIIII\n\\eeq\nThe distance of the Shor code is 3, and the error that we can correct for is\n\\beq\n\\frac{3-1}{2} \\sim 1\n\\eeq\n\\end{document}\n\n", "meta": {"hexsha": "02169e2ab56388b3af26d75b9b12715fbe57f51c", "size": 14426, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Error Correction Lecture 22 Feb 2016.tex", "max_stars_repo_name": "sqvarfort/Quantum-Error-Correction-Notes", "max_stars_repo_head_hexsha": "3628ece1bf999b4ed57ce1badb376bd91b40dc26", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2016-04-01T04:53:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-09T07:01:28.000Z", "max_issues_repo_path": "Error Correction Lecture 22 Feb 2016.tex", "max_issues_repo_name": "sqvarfort/Quantum-Error-Correction-Notes", "max_issues_repo_head_hexsha": "3628ece1bf999b4ed57ce1badb376bd91b40dc26", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Error Correction Lecture 22 Feb 2016.tex", "max_forks_repo_name": "sqvarfort/Quantum-Error-Correction-Notes", "max_forks_repo_head_hexsha": "3628ece1bf999b4ed57ce1badb376bd91b40dc26", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.1437908497, "max_line_length": 528, "alphanum_fraction": 0.7421322612, "num_tokens": 4164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673178375734, "lm_q2_score": 0.7853085733507947, "lm_q1q2_score": 0.6383516736945117}} {"text": "\\documentclass{article}\n\n\\usepackage[letterpaper, margin=1.3cm]{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage{siunitx}\n\\usepackage{mathtools}\n\\usepackage{multicol}\n\n\\title{ECE 203 Problem Set 6}\n\\author{Michael Kwok}\n\\begin{document}\n\n\\maketitle\n\\begin{multicols}{2}\n\\section*{1}\n\\subsection*{a}\n\\begin{align}\n \\left(I_1 + 10^{-5} V_2\\right) 1000 &= V_1\\\\\n 10000 I_2 - 100 V_1 &= V_2\n\\end{align}\n\nExpanding equation (1) then substituting in equation (2):\n\n\\begin{align*}\n &1000 I_1 + 0.01 V_2 = V_1\\\\\n &1000 I_1 + 0.01 \\left(10000 I_2 - 100 V_1\\right) = V_1\\\\\n &1000 I_1 + 100 I_2 = 2 V_1\n\\end{align*}\n\nForming the following equations:\n\n\\begin{align}\n V_1 &= 500 I_1 + 50 I_2\\\\\n V_2 &= -50000 I_1 + 5000 I_2\n\\end{align}\n\n$$\n [Z] = \n \\begin{bmatrix}\n 500 & 50\\\\\n -50000 & 5000\n \\end{bmatrix}\n \\si{\\ohm}\n$$\n\\subsection*{b}\nThe network is not reciprocal. $Z_{12} \\neq Z_{21}$\n\\subsection*{c}\n$$\n\\frac{V_1}{I_1} = R_{in}\n$$\n\\begin{align*}\n V_2 &= -I_2 R_L\\\\\n &= -4000 I_2\n\\end{align*}\n\\begin{align*}\n -4000 I_2 &= -50000 I_1 + 5000 I_2\\\\\n -9000 I_2 &= -50000 I_1\\\\\n I_2 &= \\frac{50}{9}I_1\\\\\n V_1 &= 500 I_1 + \\frac{2500}{9} I_1\\\\\n \\frac{V_1}{I_1} &= \\SI{778}{\\ohm}\\\\\n R_in &= \\SI{778}{\\ohm}\n\\end{align*}\n\\section*{2}\n\\begin{align}\n I_1 &= \\frac{V_1}{100} + \\frac{V_a}{500} + \\frac{V_1}{250}\\\\\n I_2 &= \\frac{V_2}{200} - \\frac{V_1}{250}\\\\\n V_a &= 250 I_2 + V_1\n\\end{align}\n\nEquation (7) into (5)\n\n\\begin{equation}\n I_1 = \\frac{V_1}{100} + \\frac{250 I_2 + V_1}{500} + \\frac{V_1}{250}\n\\end{equation}\n\\begin{align*}\n 500 I_1 &- 250 I_2 = 5 V_1 + V_1 + 2 V_1\\\\\n 500 I_1 &- 250 I_2 = 8 V_1 \\\\\n 500 I_1 &- 250 \\left( \\frac{V_2}{200} - \\frac{V_1}{250} \\right) = 8 V_1\\\\\n 500 I_1 &- \\frac{250}{200} V_2 + V_1 = 8 V_1\\\\\n 7V_1 &= 500 I_1 - \\frac{5}{4}V_2\\\\\n V_1 &= \\frac{500}{7} I_1 - \\frac{5}{28} V_2\n\\end{align*}\n\n\\begin{align*}\n I_2 &= \\frac{V_2}{200} - \\frac{1}{250} \\frac{500}{7} I_1 + \\frac{5}{28}\\frac{1}{250} V_2\\\\\n &= -\\frac{2}{7} I_1 + \\frac{1}{175} V_2\n\\end{align*}\n\n$$\n [h] = \n \\begin{bmatrix}\n \\frac{500}{7} & -\\frac{5}{28}\\\\\n -\\frac{2}{7} & \\frac{1}{175}\n \\end{bmatrix}\n$$\n\n$h_{11}$ has unit of $\\si{\\ohm}$, $h_{22}$ has unit of $\\si{\\per\\ohm}$\n\n\\section*{3}\n$R_{eq}$ can be calculated with $V_s = 0$\n\n$$\n V_1 = -4 I_1\n$$\n\n\\begin{align}\n -4 I_1 &= 2 I_1 + V_2\\\\\n I_2 &= 5 I_1 + 2 V_2\n\\end{align}\n\\begin{align*}\n -6 I_1 &= V_2\\\\\n I_1 &= -\\frac{1}{6} V_2\n\\end{align*}\n\\begin{align*}\n I_2 &= -\\frac{5}{6} V_2 + 2 V_2\\\\\n \\frac{I_2}{V_2} &= \\frac{7}{6}\\\\\n R_{eq} &= \\frac{6}{7}\\si{\\ohm}\n\\end{align*}\n\n\\section*{4}\nFor $y_1$, the CL circuit.\n\\begin{align}\nI_1 = \\frac{V_1}{\\frac{1}{sC}} + \\frac{V_1 - V_2}{sL}\\\\\nI_2 = \\frac{V_2 - V_1}{sL} + \\frac{V_2}{\\frac{1}{sC}}\n\\end{align}\n\\begin{align*}\n I_1 = sV_1 + \\frac{1}{s} V_1 - \\frac{1}{s}V_2\\\\\n I_2 = sV_2 + \\frac{1}{s} V_2 - \\frac{1}{s}V_1\n\\end{align*}\n$$\n [y_1] = \n \\begin{bmatrix}\n \\frac{s^2 +1}{s} & -\\frac{1}{s}\\\\\n -\\frac{1}{s} & \\frac{s^2 +1}{s}\n \\end{bmatrix}\n \\si{\\per\\ohm}\n$$\nFor $y_2$, the RC circuit.\n\\begin{align}\nI_1 = \\frac{V_1-V_2}{\\frac{1}{sC}} + \\frac{V_1}{1}\\\\\nI_2 = \\frac{V_2-V_1}{\\frac{1}{sC}} + \\frac{V_2}{1}\n\\end{align}\n\\begin{align*}\n I_1 = (s+1)V_1 - s V_2\\\\\n I_2 = -sV_1 + (s+1)V_2\n\\end{align*}\n$$\n [y_2] = \n \\begin{bmatrix}\n s+1 & -s\\\\\n -s & s+1\n \\end{bmatrix}\n \\si{\\per\\ohm}\n$$\n$$\n[y] = [y_1] + [y_2] = \n \\begin{bmatrix}\n \\frac{2s^2 + s + 1}{s} & -\\frac{s^2+1}{s}\\\\\n -\\frac{s^2+1}{s} & \\frac{2s^2 + s + 1}{s}\n \\end{bmatrix}\n \\si{\\per\\ohm}\n$$\n\\section*{5}\n\\subsection*{a}\n\\begin{align}\n I_1 &= \\frac{V_1}{4} + \\frac{V_1 - V_2}{2}\\\\\n I_2 &= \\frac{V_2}{3} + \\frac{V_2 - V_1}{2}\n\\end{align}\n\\begin{align}\n 4 I_1 &= V_1 + 2 V_1 - 2 V_2\\\\\n 6 I_2 &= 2 V_2 + 3 V_2 - 3 V_1\n\\end{align}\nUsing Equation (18):\n\\begin{align*}\n 6 I_2 - 5 V_2 &= - 3 V_1\\\\\n V_1 &= \\frac{5}{3} V_2 - 2 I_2\n\\end{align*}\n\nUsing Equation (17):\n\\begin{align*}\n 4 I_1 &= 5 V_2 - 6 I_2 -2 V_2\\\\\n 4 I_1 &= 3 V_2 - 6 I_2\\\\\n I_1 &= \\frac{3}{4} V_2 - \\frac{3}{2} I_2\n\\end{align*}\n\n\\begin{align*}\n3 V_1 &= 4 I_1 + 2 V_2\\\\\n V_1 &= \\frac{4}{3} \\left( \\frac{3}{4} V_2 - \\frac{3}{2} I_2\\right) +\\frac{2}{3} V_2\\\\\n &= V_2 - 2 I_2 + \\frac{2}{3} V_2\\\\\n &= \\frac{5}{3} V_2 - 2I_2\n\\end{align*}\n$$\n[t] = \n \\begin{bmatrix}\n \\frac{5}{3} & 2\\\\\n \\frac{3}{4} & \\frac{3}{2}\n \\end{bmatrix}\n$$\n$t_{12}$ has unit of $\\si{\\ohm}$, $t_{21}$ has unit of $\\si{\\per\\ohm}$\n\\subsection*{b}\n$$\n[t] = [t_1][t_2] = \n \\begin{bmatrix}\n \\frac{5}{3} & 2\\\\\n \\frac{3}{4} & \\frac{3}{2}\n \\end{bmatrix}\n \\begin{bmatrix}\n 1 & 0\\\\\n \\frac{1}{6} & 1\n \\end{bmatrix}\n$$\n$$\n= \n \\begin{bmatrix}\n 2 & 2\\\\\n 1 & \\frac{3}{2}\n \\end{bmatrix}\n$$\n\n\\end{multicols}\n\\end{document}\n", "meta": {"hexsha": "9235435bbefb2cf7fb38b3acff5afd26d3082985", "size": 4739, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignments/ECE203/ECE203Ps6.tex", "max_stars_repo_name": "n30phyte/SchoolDocuments", "max_stars_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignments/ECE203/ECE203Ps6.tex", "max_issues_repo_name": "n30phyte/SchoolDocuments", "max_issues_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignments/ECE203/ECE203Ps6.tex", "max_forks_repo_name": "n30phyte/SchoolDocuments", "max_forks_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.3468468468, "max_line_length": 94, "alphanum_fraction": 0.5376661743, "num_tokens": 2467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.896251378675949, "lm_q2_score": 0.7122321781307374, "lm_q1q2_score": 0.6383390715870474}} {"text": "\\documentclass{article}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n% Main maths packages\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{mathrsfs}\n\n\\usepackage{tabularx}\n\n% First and foremost, never use the environment eqnarray. It is not recommended because spacing is inconsistent.\n\n\\begin{document}\n\n\\section*{System of equations}\n\n\n\\subsection*{Summary of the environments}\n\n\\begin{table}[h]\n\t\\begin{tabularx}{\\textwidth}{| X | X | X |}\n\t\\hline\n\t\\textbf{Environment names} & \\textbf{Description} & \\textbf{Notes} \\\\\n\t\\hline\n\t\\emph{gather} and \\emph{gather*} & Consecutive equations with an alignment to the center. & \\\\\n\t\\hline\n\t\\emph{align}, \\emph{align*}, \\emph{aligned} and \\emph{split} & Consecutive equations with user-defined alignment. & \\\\\n\t\\hline\n\t\\emph{flalign} and \\emph{flalign*} & Similar to \\emph{align}, but left aligns first equation column, and right aligns last column. & \\\\\n\t\\hline\n\t\\emph{alignat} and \\emph{alignat*} & Takes an argument specifying number of columns. Allows control of the horizontal space between equations. & This environment takes one argument, the number of “equation columns”: count the maximum number of \\&s in any row, add 1 and divide by 2. \\\\\n\t\\hline\n\t\\emph{array} & Advanced alignment scenario. Enables to control how to columns are aligned. & \\\\\n\t\\hline\n\t\\emph{multline} and \\emph{multline*} & First line left aligned, last line right aligned. & Equation number aligned vertically with first line and not centered as with other environments. \\\\\n\t\\hline\n\t\\emph{eqnarray} and \\emph{eqnarray*} & Similar to \\emph{align} and \\emph{align*}. & Not recommended because spacing is inconsistent. \\\\\n\t\\hline\n\t\\end{tabularx}\n\\end{table}\n\nThis table is based on the one available on Wikibooks, \\LaTeX Advanced Mathematics.\n\n\\newpage\n\n\n\\subsection*{Grouping and centering equations with \\emph{gather}}\n\n\\begin{gather}\n\tx + 2y - z = 4 \\\\\n\tx + y - 5z = -1 \\\\\n\t2x - z = 10\n\\end{gather}\n\nIn addition, the environment \\emph{gather*} removes the equation numbering. The environment \\emph{gathered} and can be used in another mathematical environment.\n\n\n\\subsection*{Align equations with \\emph{align}}\n\n\\begin{align}\n\tx + 2y - z &= 4 \\\\\n\tx + y - 5z &= -1 \\\\\n\t2x - z &= 10\n\\end{align}\n\nThis environment can also be used to align equations on the same line:\n\n\\begin{align*}\n\tf(x) &= a x^2+b x +c & g(x) &= d x^3 \\\\\n\tf'(x) &= 2 a x +b & g'(x) &= 3 d x^2\n\\end{align*}\n\nIn addition, the environment \\emph{align*} removes the equation numbering. The environment \\emph{aligned} and \\emph{split} are the similar to \\emph{align} and \\emph{align*} but can be used in another mathematical environment.\n\n\\subsubsection*{Align equations with \\emph{align} and \\emph{subequations}}\n\nThe environment \\emph{subequations} changes the way how \\LaTeX counts the equations.\n\n\\begin{subequations}\n\t\\begin{align}\n\t\tx + 2y - z &= 4 \\\\\n\t\tx + y - 5z &= -1 \\\\\n\t\t2x - z &= 10\n\t\\end{align}\n\\end{subequations}\n\n\\subsubsection*{Braces a system of equations}\n\nThe environment \\emph{aligned} is similar to align, to be used inside another mathematics environment.\n\n% The commands \\left and \\right insert a brace. The type of the brace is written just after the command.\n% When \\left is called, \\right is mandatory.\n\\[\n\\left(\n\\begin{aligned}\n\tx + 2y - z &= 4 \\\\\n\tx + y - 5z &= -1 \\\\\n\t2x - z &= 10\n\\end{aligned}\n\\right)\n\\]\n\n% In the case where only one brace is needed, the other brace must be marked with a point (.) to hide it.\n\\[\n\\left\\{\n\\begin{aligned}\nx + 2y - z &= 4 \\\\\nx + y - 5z &= -1 \\\\\n2x - z &= 10\n\\end{aligned}\n\\right.\n\\]\n\n\n\\subsection*{Align equations to the left with \\emph{flalign}}\n\n\\begin{flalign}\na &= b+c &\\\\\n&= 1+1 &\\\\\n&= 2 &\n\\end{flalign}\n\nIn addition, the environment \\emph{flalign*} removes the equation numbering.\n\n\n\\subsection*{Align equations with \\emph{alignat}}\n\nThe environment \\emph{alignat} enables to control of the horizontal space between equations. Indeed, no addition space is added between equation, at the opposite of the environment \\emph{align}.\n\nThe mathematical development below is aligned according the arrow and the equal signs.\n\n\\begin{alignat}{2}\n\t&& \\sqrt{4 x^2} - 1 &= 0 \\\\\n\t&\\Rightarrow & \\sqrt{4} \\sqrt{x^2} &= 1 \\\\\n\t&\\Rightarrow & 2 \\left|x\\right| &= 1 \\\\\n\t&\\Rightarrow & x &= \\pm \\frac{1}{2}\n\\end{alignat}\n\nThis environment takes an argument specifying number of columns. The rule of thumb to determine the argument is to count the maximum number of \\& symbols on one row, add 1 and divide by 2.\n\nIn addition, the environment \\emph{alignat*} removes the equation numbering. The environment \\emph{alignedat} and \\emph{alignedat*} and can be used in another mathematical environment.\n\n\n\\subsection*{Align equations with \\emph{array}}\n\nThe environment \\emph{array} is for more advanced scenario. It is basically the same as \\emph{align}, but the columns and their alignment are explicitly indicated. It must be used in the math mode.\n\n\\[\n\\left\\{\n\\begin{array}{ r c l}\n\t\tx + z &=& y - 4 \\\\\n\t\t5z &=& x + y -1 \\\\\n\t\t2x - y &=& 10 - y\n\\end{array}\n\\right.\n\\]\n\n\n\\subsection*{Displaying long equations}\n\nThe environment \\emph{multline} can be used to display formula on multiple lines.\n\n\\begin{multline*}\n\tf(x) = 60x^{15} + 56x^{14} + 52x^{13} + 48x^{12} + 44x^{11} + 40x^{10} + 36x^{9} + 32x^{8}\\\\\n\t+ 28x^{7} + 24x^{6} + 20x^{5} + 16x^{4} + 12x^{3} + 8x^{2} + 4\n\\end{multline*}\n\n\\end{document}", "meta": {"hexsha": "bf2f47a1f627c203615121b9fcc6a68e025e0be1", "size": 5365, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compendium/mathematics/system-of-equations.tex", "max_stars_repo_name": "ZenLulz/LatexCompendium", "max_stars_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-07-30T21:43:55.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-23T20:16:19.000Z", "max_issues_repo_path": "compendium/mathematics/system-of-equations.tex", "max_issues_repo_name": "ZenLulz/LatexCompendium", "max_issues_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compendium/mathematics/system-of-equations.tex", "max_forks_repo_name": "ZenLulz/LatexCompendium", "max_forks_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.5588235294, "max_line_length": 286, "alphanum_fraction": 0.6943150047, "num_tokens": 1698, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7025300573952054, "lm_q2_score": 0.9086179031191509, "lm_q1q2_score": 0.6383313876286082}} {"text": "\\chapter{Background \\& Related Work}\n\\label{chap:background}\n\nIn this section, constraint optimization and the distributed, as well as dynamic variants are briefly explained and brought into context of the related work. Also, the meeting scheduling problem will be described and different algorithm designs and their advantages and disadvantages are going to be briefly discussed.\n \n\\section{Dynamic Distributed Constraint Optimization}\n\n% -----------------Constraint Optimization --------------------\n \nA constraint optimization problem (COP) contains a set of variables \\(V=\\{V_{1},V_{2}, ..., V_{n}\\}\\). These variables are assigned to a value or state \\(s_{j} \\in S_{j}\\), which is contained in a set of possible values defined by a finite problem domain \\(D=\\{D_{1},D_{2}, ..., D_{n}\\}\\). A constraint \\(C = \\) contains one (unary), two (binary) or multiple (k-ary) variables and their relationship. The constraint defines a rule for the variables that needs to be fulfilled. One of those rules could be that none of the variables should take the same value. This would for example be the case for a meeting scheduling problem where none of the meetings should take place at the same time. \\newline\nA utility function for the constraint \\({c}_{k}\\) on variable state \\(s\\) in the form of \\(u_{{c}_{k}}(s_{{c}_{k}})\\) needs to be formulated that defines a certain cost respectively reward for a given configuration of the involved states. The global utility function \\(u_{g}\\) would then be the summation of all utility functions of all constraints. \n\n\\[u_{g}(s) = u_{c_{1}}(s_{c_{1}}) \\oplus \\cdots \\oplus u_{c_{k}}(s_{c_{k}}) \\oplus \\cdots \\oplus u_{c_{l}}(s_{c_{l}}) \\] \n\nConstraints can be attributed with varying levels of importance through weighting. One can, instead of so-called soft constraints, define hard constraints by multiplying their utility instead of using addition in the global utility function. By defining the utility of a violated hard constraint as 0, the global utility would also go to 0 if this hard constraint is not satisfied \\cite{Chapman2011, Petcu2003}. A problem only containing hard constraints would represent a constraint satisfaction problem (CSP). A formal definition of such a combined utility function including soft constraints (SC) and hard constraints (HC) would look like the following formula, where the product of all hard constraints is multiplied by the sum of all utilities of a state \\(s\\) in the soft constraint utility functions:\n\n\\[ u_{g}(s) = \\prod_{\\substack{hc_{k} \\in HC}} u_{SC_{g}}(s) \\bigg( \\sum_{sc_{k} \\in SC} u_{SC_{g}}(s) \\bigg)\\] \n\n\n% ----------------- Distributed Constraint Optimization --------------------\nThe definition of a distributed constraint optimization problem (DCOP) extends the basic constraint optimization by distributing sets of variables to autonomous agents. These agents all have the goal to maximize the utility of their variables in a private utility function and thereby also contribute to a global utility function. Agent's whose variables are linked to at least one common constraint are called neighbours \\cite{Chapman2011, Farinelli, Petcu2003}.\n\\newline\\newline \n% ----------------- Dynamic Distributed Constraint Optimization --------------------\nThe problem definition in dynamic distribute constraint optimization (DynDCO) is, as a further extension to DCOPs, moved from a static to a dynamic attribute. Constraints can change and therefore change neighbourhoods and the outcome of private and global utility functions. A change of constraints inherently changes the area of satisfying solutions if hard constraints have been included in a problem definition. \\cite{Nguyen2012} state that changing the constraints might lead to the discovery of a better global optima. \\cite{Maillera} define a dynamic distributed constraint satisfaction problem (DCSP) as a sequence of DCSPs \\(\\{P_{0}, P_{1}, ..., P_{n}\\}\\) where every DCSP is a static problem definition. \\(P_{i}\\) is therefore a result of the previous DSCP in the sequence and also a result of the added and removed constraints: \\(P_{i} = P_{i-1} + c_{i^{a}} - c_{i^{r}}\\). This definition should also hold for DynDCOPs. Utility functions could also be dynamically changed. Modifying this property could especially have an impact on real-world problems like meeting scheduling, where it could move the global optima from one disconnected solution space to another \\cite{Nguyen2012}. Furthermore, variables could be added or removed in a dynamic setting and the problem domain \\(D\\) also could be changed during the course of the problem solving process.\n\n\\section{Meeting Scheduling Problem} \n\n%--------------------- Introduction with examples\nScheduling is the problem of allocating tasks to a given set of ressources in an optimal order. The meeting scheduling problem is an exemplary type of this family of problems and is supposedly well-know to all of us. Participants of a meeting have private schedules with preferences when a meeting should be held according to their calendar. The challenge is to identify a time for a meeting that maximizes the preferences of all participants while being valid in then sense that every person is able to attend \\cite{Farinelli}. \\cite{Scheduling} have formally defined a meeting scheduling problem as:\n\n\\begin{itemize}\n\\item \\(P = {p1, p, ... pn}\\) is the set of people where every person has a calendar that holds r slots, \\(S = {s1,s2,s3}\\)\n\\item \\(M = {m1, m2, ..., m3}\\) is a set of k meetings\n\\item \\(At = {at1, at2, ..., atk}\\) defines all attendee's of a meeting\n\\end{itemize}\n\nThe c parameter has been neglected as it is not relevant to this thesis. From the definition of a valid solution, one can derive two important criteria to the problem solving process:\n\\theoremstyle{definition}\n\\newtheorem{hardconstraint1}{Validity Criterium}\n\\begin{hardconstraint1}\nAll participants need to agree on the same time for the meeting.\n\\end{hardconstraint1}\n\\begin{hardconstraint1}\nMeetings need to be scheduled in a way that there are no overlaps of meeting times in the schedules of the participants.\n\\end{hardconstraint1}\n \nThere is further an inherent privacy aspect to the problem. Meeting participants are often not willing to share their schedules with others except to find a time for the specific meeting. It will later be shown that some of the algorithms can guarantee this privacy to a certain degree \\cite{Farinelli} \\cite{Scheduling}. The meeting scheduling problem will be mapped as a distributed constraint optimization problem in the design chapter.\n \n %--------------------- Explanation of the components with formal definition\n\\section{Algorithm Design Approaches}\n\n\\begin{figure}[h]\n\\includegraphics[width=400px]{graphics/overview_algos}\n\\caption{Categorization of DCO algorithms \\cite{Chapman2011}}\n\\label{fig:categorization}\n\\end{figure}\n\n\\cite{Chapman2011} categorize distributed constraint optimization algorithms into local-iterative and distributed complete algorithms. They further divide local-iterative into message-passing algorithms and approximate best-response algorithms (Figure \\ref{fig:categorization}). The following subsections are going to explain the differences between the three categories and introduce the specific algorithms, which have been chosen from these three different approaches for benchmarking. Advantages, as well as disadvantages will be described and which behaviour one can expect of these algorithms under certain parameter configurations.\n\n%Other Approaches: Bee Hive optimization, Genetic Algorithms, .. DynDCOAA, SBDO, Bee Colony algorithm, Ant colony algorithm, adopt, dsa-a, dsa-b stochastic ...\\cite{Likhachev}\n \n\\subsection{Distributed Complete}\n\n%----------------- Basic Concept and Advantages Disadvantes ----------------------\nDistributed complete algorithms always discover a configuration of value assignments for a set of variables that maximizes the global utility function. This completeness guarantee increases the complexity of computation and leads to exponentially growing message numbers or calculations when increasing the amount of variables in a problem. Messages between agents often contain complex structures and constraint problems usually need to be transformed to an extensive graph structure \\cite{Chapman2011}. These types of algorithms therefore are not expected to scale well and quickly find qualitative solutions, but they fit well if one wants to find the maximal utility of a problem. \n %----------------- Chosen algorithm ----------------------\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=250px]{graphics/pseudotree}\n\\caption{Pseudotree in DPOP \\cite{Petcu2003}}\n\\label{fig:pseudotree}\n\\end{figure}\n\nFor this thesis, it was decided to implement the Dynamic Programming OPptimization algorithm (DPOP) proposed by \\cite{Petcu2003} as a comparison to the local-iterative approaches. In this algorithm, constraint optimization problems need to be converted to a pseudotree (Fig. \\ref{fig:pseudotree}), which is a modification of a DFS Tree. The original DCOP graph is transformed in a way that previous neighbours are placed in the same branches of a binary tree. They are connected trough ordinary tree edges and additionally, so-called back-edges between unconnected previous neighbours are established. The leaf nodes propose UTIL messages containing their utility values for each value assignment upwards the tree and the root node sends a VALUE message downwards, containing the best value to choose as a variable state. Nodes in the middle of the tree propagate UTIL and VALUE messages. The message structure is fairly complex as it involves all the utilities of the pseudoparents connected by the back-edges and their context in the graph, which increases the message size exponentially. The number of messages on the other hand is linear \\cite{Petcu2003}.\n\n%----------- Dynamics -----------------------\n\\subsection{Local-Iterative - Best Response}\n\n%----------------- Basic Concept and Advantages Disadvantes ----------------------\nIn a local-iterative best-response algorithm, agents only communicate their current state, e.g. their value assignment and react to these value messages in the best possible way from their perspective. The agents are only connected to their neighbours with whom they share constraints and there exists no complex graph structure controlling the message flow \\cite{Chapman2011}. Through this local property, the types of algorithms should be inherently scalable as the messages and computations do not increase exponentially. Further, this approach is optimal from a privacy perspective as the neighbours only share their current preference and no other details of their schedule \\cite{Chapman2010}. %\\cite{Maheswaran} % Achtung muss das richtig paper finden\n\\newline\\newline\n%----------------- Chosen algorithm: Structure, Messages --------------------- \nFor this thesis, it was decided to implement the Maximum-Gain Messaging algorithm (MGM). In this algorithm, agents calculate the maximal gain in utility they can achieve when assigning to another value and send this value as a message. If they have the highest gain compared to all received gain messages from their neighbours, the local value is changed. Otherwise the local value stays the same. This algorithm fullfills the anytime property, i.e it can provide a solution at every timepoint during calculation and also reaches good solutions quickly \\cite{Chapman2010}. As the decision of an agent depends on a complete set of message of all of its neighbours, this algorithm will supposedly not perform well in asynchronous running mode. This type of algorithm does further not always converge and the deliverance of an optimal solution to a problem is not guaranteed.\n\n\\subsection{Local-Iterative - Message Passing}\n%----------------- Basic Concept and Advantages Disadvantes ----------------------\nThe difference of message-passing to best-response algorithms lays in the fact that the agents send and receive messages containing a specific data structure, which contains the utilities respectively costs that various assignments hold for a local variable. Received messages are used to calculate the next message, which is sent to the connected neighbours. These types of algorithms are - like best-response algorithms - able to provide an acceptable solution in a short period of time, but also share the charasteric to sometimes not converge or not providing an optimal solution \\cite{Chapman2011}.\n%----------------- Choosen algorithm ----------------------\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=170px]{graphics/factorgraph}\n\\caption{Conversion of a general DCOP to a factor graph \\cite{Zivan2012}}\n\\label{fig:factorgraph}\n\\end{figure}\n\nFor this thesis, it was decided to implement the MaxSum algorithm introduced by \\cite{Farinelli2008}. The algorithm has currently gotten a lot of attention from researchers. For this work, the algorithm is especially interesting because of its proposed abilities in dynamic environments. \\cite{Farinelli2008} wrote in their paper:\n\\begin{quote}\n[...] we note that if messages are continuously propagated,\nand the states of the agents are continuously updated, then the algorithm may be applied to dynamic problems where the interactions between agents, or the utilities resulting from these interactions, may change at any time.\n\\end{quote}\n\nIn MaxSum, the original DCOP graph is transformed to a factor graph, which is a form of a bipartite graph and of cyclic nature (Fig. \\ref{fig:factorgraph}). After the transformation, an agent is made up of a variable and a function node, whereas variables are connected to all corresponding function nodes of their previous neighbours. The function nodes are vice versa connected to all previous neighbours of its variable node. \n% ------------- Messages sent --------------------\nThe messages sent from variable nodes differ from the function nodes. A message from variable to function contains for every value \\(d \\in D_{x}\\) the sum of utilities regarding this value, which the node has received from all connected function nodes. It is important to note that this sum does not include the values provided by the message target. The values are normalized at this point to avoid an infinite increase of the sum of the utilities. A message from a function node to a variable node holds for every value \\(d \\in D_{x}\\) the summation of all costs received from all connected variable nodes except the message receiver and the original cost respectively utility of the constraint represented by the function node \\cite{Zivan2012}.\n\n\n \n ", "meta": {"hexsha": "3c1d594a574f0b3a4690eaec8779d07889778966", "size": 14685, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documentation/report/section-relatedwork/relatedwork.tex", "max_stars_repo_name": "danihegglin/DynDCO", "max_stars_repo_head_hexsha": "4186857d1183eeff5721434f12afdca664af9c4b", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "documentation/report/section-relatedwork/relatedwork.tex", "max_issues_repo_name": "danihegglin/DynDCO", "max_issues_repo_head_hexsha": "4186857d1183eeff5721434f12afdca664af9c4b", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "documentation/report/section-relatedwork/relatedwork.tex", "max_forks_repo_name": "danihegglin/DynDCO", "max_forks_repo_head_hexsha": "4186857d1183eeff5721434f12afdca664af9c4b", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 135.9722222222, "max_line_length": 1363, "alphanum_fraction": 0.7721484508, "num_tokens": 3106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519528132451417, "lm_q2_score": 0.7490872075132153, "lm_q1q2_score": 0.6381869538068311}} {"text": "\\documentclass[11pt,letterpaper]{article}\n\\usepackage[T1]{fontenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\n\\usepackage[charter]{mathdesign}\n\\usepackage{fullpage}\n\\pagestyle{empty}\n\n\\input{../fncextra}\n\n\\begin{document}\n\t\n\\begin{center}\n \\bf \n Project: To be continued\n\\end{center}\n\t\n\nRecall the case of a circular elastic membrane pinned at $z=1$ and subjected to an electrostatic force applied on the plane $z=0$. If we exploit the circular symmetry, we can reduce the problem to a BVP in the disk radius $r$:\n\\begin{equation}\n \\label{eq:mems}\n u''(r) + \\frac{1}{r} u'(r) = \\frac{\\lambda}{u^2}, \\qquad u'(0)=0, \\quad u(1)=1.\n\\end{equation}\nThis has been used as a model for an actuator device in a microelectromechanical system (MEMS). Because we would have some headaches with division by zero in this ODE, we multiply through by $r$ to get\n\\begin{equation}\n \\label{eq:rmems}\n r u''(r) + u'(r) = \\frac{r\\lambda}{u^2}.\n\\end{equation}\n\nThe behavior of this system as a function of the field strength $\\lambda$ is quite interesting. Clearly $u\\equiv 1$, $\\lambda=0$ is a trivial solution. As $\\lambda$ increases from zero, the membrane begins to deflect downward toward zero, with the greatest amount of deflection (i.e., smallest value of $u$) at the center point $r=0$. The amount of deflection continues to decrease until, at some critical value $\\lambda^*$, no solution exists. However, for some values $\\lambda < \\lambda^*$ there are at least two valid solutions. It's not at all clear how to figure out good initial guesses that will drive a Newton-type iteration to find the others. \n\nWhen the BVP is discretized, we use $n+1$ unknown values of $u$ and $n+1$ equations to specify them. But if we also think of $\\lambda$ as a parameter, we have $n+1$ equations in $n+2$ variables. This defines a curve or path in $(n+2)$-dimensional space---just like $x^2+y^2=1$ defines a curve in the plane. The situation can be visualized by this graph:\n\\begin{center}\n \\includegraphics{strip_path}\n\\end{center}\n(This picture is \\emph{not} quantitatively accurate for your results!) \n\nWhat we really want to do is follow along the path, a technique known as \\emph{continuation}. There are many ways this can be done, but this particular problem gives us a fairly easy option. The key is to observe in the figure above that $\\lambda$ is a poor choice for parameterizing the path, because it fails the ``vertical line test.'' We are fortunate that in this problem, the value $u(0)$ is monotonic along the path, so it will serve as the parameter. To be specific, we solve a discretized form of~\\eqref{eq:rmems} for the $n+2$ unknowns $\\lambda,u_0,u_1,\\ldots,u_n$. As has been our practice, the ODE is discretized at the $n-1$ interior nodes only. To this are added the \\emph{three} supplemental conditions\n\\begin{equation}\n \\label{eq:threebc}\n u'(0)=0, \\qquad u(1)=1, \\qquad u(0)=1-s,\n\\end{equation}\nwhere $0\\le s < 1$ is a value under your control. You start at $s=0$ with the trivial solution mentioned above, then gradually increase $s$ to get a sequence $s_1,s_2,\\ldots$, corresponding to solutions $(\\lambda_1,\\bfu_1),(\\lambda_2,\\bfu_2)\\ldots$, which are points along the path. When solving the nonlinear equations for $s=s_{k+1}$, the starting guess should be $(\\lambda_k,\\bfu_k)$, which is hopefully near the next point. \n\n\n\\subsection*{Project assignment}\n\\label{sec:project-assignment}\n\nThe heart of the assignment is a function\n\\begin{verbatim}\n[u,lambda] = mems_solve(u0,lam0,s)\n\\end{verbatim}\nthat solves the ODE with the three side conditions in~\\eqref{eq:threebc}. Here \\texttt{lam0} is an initial guess for $\\lambda$, and the $(n+1)$-vector \\texttt{u0} is an initial guess for $\\bfu$. You should model this function after \\texttt{bvp}, but your function does not need to solve a general problem, just the specific one at hand. Like \\texttt{bvp}, your function will call \\texttt{levenberg} on a subfunction named \\texttt{residual}, but your version of \\texttt{residual} evaluates $n+2$ quantities for values of the $n+2$ variables in a given vector $\\bfz=\n\\begin{bmatrix}\n \\lambda \\\\ \\bfu \n\\end{bmatrix}$. \nOnce \\verb!mems_solve! is working, you can call it within a loop to implement the steady increase of $s$. \n\nYour objectives are:\n\\begin{enumerate}\n\\item Find $\\lambda^*$ to at least two significant digits. This means taking small enough steps along the curve as well as giving evidence that the answer has converged as a function of the discretization size $n$.\n\\item Find a second turning point, $\\lambda^\\dagger<\\lambda^*$, to two digits. Past this turning point, $\\lambda$ begins to increase again as a function of $s$. \n\\item Make a plot like the one above for $0\\le s \\le 0.99$. \n\\item On one graph, plot $u(r)$ against $r$ for $s=0.1,0.3,0.5,0.7,0.9,0.95$. \n\\item How accurately can you find $\\lambda^*$? Think about how to get the most out the discretization accuracy and how to use something more sophisticated than just taking super-small steps in $s$. Style points matter here!\n\\end{enumerate}\n\nSubmit a PDF report describing your results and providing any necessary supporting information. Keep it concise and to the point---for instance, don't describe multiple attempts at the last item above, just your best one. Don't forget to label your plots! Also include your \\verb!mems_solve.m! and the code to reproduce the results for items~1--3 above. \n\n\\end{document}\n\n", "meta": {"hexsha": "8e925105bb7c7a0c3b2f38588c6370cc20037bfc", "size": 5444, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "projects/mems/mems.tex", "max_stars_repo_name": "snowdj/fnc-extras", "max_stars_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 109, "max_stars_repo_stars_event_min_datetime": "2018-04-21T09:02:50.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-20T19:03:54.000Z", "max_issues_repo_path": "projects/mems/mems.tex", "max_issues_repo_name": "snowdj/fnc-extras", "max_issues_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2018-12-04T22:17:44.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-03T21:04:47.000Z", "max_forks_repo_path": "projects/mems/mems.tex", "max_forks_repo_name": "snowdj/fnc-extras", "max_forks_repo_head_hexsha": "ef51fada748de1326a4ce645fbcb0c2499cb2b8a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 49, "max_forks_repo_forks_event_min_datetime": "2017-04-02T17:21:33.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-29T17:19:57.000Z", "avg_line_length": 72.5866666667, "max_line_length": 717, "alphanum_fraction": 0.7441219691, "num_tokens": 1536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147276, "lm_q2_score": 0.8519528038477825, "lm_q1q2_score": 0.6381869515396136}} {"text": "\\chapter{Introduction to linear regression}\n\\label{linRegrForTwoVar}\n\n\\index{linear regression|textbf}\n\nLinear regression is a very powerful statistical technique. Many people have some familiarity with regression just from reading the news, where graphs with straight lines are overlaid on scatterplots. Linear models can be used for prediction or to evaluate whether there is a linear relationship between two numerical variables.\n\nFigure~\\ref{perfLinearModel} shows two variables whose relationship can be modeled perfectly with a straight line. The equation for the line is\n\\begin{eqnarray*}\ny = 5 + 57.49x\n\\end{eqnarray*}\nImagine what a perfect linear relationship would mean: you would know the exact value of $y$ just by knowing the value of $x$. This is unrealistic in almost any natural process. For example, if we took family income $x$, this value would provide some useful information about how much financial support $y$ a college may offer a prospective student. However, there would still be variability in financial support, even when comparing students whose families have similar financial backgrounds.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.6\\textwidth]{05/figures/perfLinearModel/perfLinearModel}\n \\caption{Requests from twelve separate buyers were simultaneously placed with a trading company to purchase Target Corporation stock (ticker \\texttt{TGT}, April 26th, 2012), and the total cost of the shares were reported. Because the cost is computed using a linear formula, the linear fit is perfect.}\n \\label{perfLinearModel}\n\\end{figure}\n\nLinear regression assumes that the relationship between two variables, $x$ and $y$, can be modeled by a straight line:\n\\begin{eqnarray}\ny = \\beta_0 + \\beta_1x\n\\label{genLinModelWNoErrorTerm}\n\\end{eqnarray}\n\\marginpar[\\raggedright\\vspace{-10mm}\n\n$\\beta_0, \\beta_1$\\vspace{0.7mm}\\\\\\footnotesize Linear model\\\\ parameters]{\\raggedright\\vspace{-10mm}\n\n$\\beta_0, \\beta_1$\\vspace{0.7mm}\\\\\\footnotesize Linear model\\\\ parameters}where $\\beta_0$ and $\\beta_1$ represent two model parameters\\index{parameter} ($\\beta$ is the Greek letter \\emph{beta}\\index{Greek!beta@beta ($\\beta$)}). These parameters are estimated using data, and we write their point estimates as $b_0$ and $b_1$. When we use $x$ to predict $y$, we usually call $x$ the explanatory or \\textbf{predictor} variable, and we call $y$ the response.\n\nIt is rare for all of the data to fall on a straight line, as seen in the three scatterplots in Figure~\\ref{imperfLinearModel}. In each case, the data fall around a straight line, even if none of the observations fall exactly on the line. The first plot shows a relatively strong downward linear trend, where the remaining variability in the data around the line is minor relative to the strength of the relationship between $x$ and $y$. The second plot shows an upward trend that, while evident, is not as strong as the first. The last plot shows a very weak downward trend in the data, so slight we can hardly notice it. In each of these examples, we will have some uncertainty regarding our estimates of the model parameters, $\\beta_0$ and $\\beta_1$. For instance, we might wonder, should we move the line up or down a little, or should we tilt it more or less? As we move forward in this chapter, we will learn different criteria for line-fitting, and we will also learn about the uncertainty associated with estimates of model parameters.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth]{05/figures/imperfLinearModel/imperfLinearModel}\n \\caption{Three data sets where a linear model may be useful even though the data do not all fall exactly on the line.}\n \\label{imperfLinearModel}\n\\end{figure}\n\nWe will also see examples in this chapter where fitting a straight line to the data, even if there is a clear relationship between the variables, is not helpful. One such case is shown in Figure~\\ref{notGoodAtAllForALinearModel} where there is a very strong relationship between the variables even though the trend is not linear. We will discuss nonlinear trends in this chapter\\MultipleRegressionChapter{ and the next}{}, but the details of fitting nonlinear models are saved for a later course.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.87\\textwidth]{05/figures/notGoodAtAllForALinearModel/notGoodAtAllForALinearModel}\n \\caption{A linear model is not useful in this nonlinear case. These data are from an introductory physics experiment.}\n \\label{notGoodAtAllForALinearModel}\n\\end{figure}\n\n\\textA{\\newpage}\n\n\n%__________________\n\\section{Line fitting, residuals, and correlation}\n\\label{lineFittingResidualsCorrelation}\n\nIt is helpful to think deeply about the line fitting process. In this section, we examine criteria for identifying a linear model and introduce a new statistic, \\emph{correlation}.\n\n\\subsection{Beginning with straight lines}\n\n\\index{data!possum|(}\n\nScatterplots were introduced in Chapter~\\ref{introductionToData} as a graphical technique to present two numerical variables simultaneously. Such plots permit the relationship between the variables to be examined with ease. Figure~\\ref{scattHeadLTotalL} shows a scatterplot for the head length and total length of 104 brushtail possums from Australia. Each point represents a single possum from the data.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.85\\textwidth]{05/figures/scattHeadLTotalL/scattHeadLTotalL}\n \\caption{A scatterplot showing head length against total length for 104 brushtail possums. A point representing a possum with head length 94.1mm and total length 89cm is highlighted.}\n \\label{scattHeadLTotalL}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.8\\textwidth]{05/figures/possumPic/possumPic}\n \\caption{The common brushtail possum of Australia. Photo by wollombi on Flickr}\n \\label{possumPic}\n\\end{figure}\n\nThe head and total length variables are associated. Possums with an above average total length also tend to have above average head lengths. While the relationship is not perfectly linear, it could be helpful to partially explain the connection between these variables with a straight line.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.9\\textwidth]{05/figures/scattHeadLTotalLTube/scattHeadLTotalLTube}\n \\caption{The figure on the left shows head length versus total length, and reveals that many of the points could be captured by a straight band. On the right, we see that a curved band is more appropriate in the scatterplot for \\textbf{weight} and \\textbf{mpgCity} from the \\textbf{cars} data set.}\n \\label{scattHeadLTotalLTube}\n\\end{figure}\n\nStraight lines should only be used when the data appear to have a linear relationship, such as the case shown in the left panel of Figure~\\ref{scattHeadLTotalLTube}. The right panel of Figure~\\ref{scattHeadLTotalLTube} shows a case where a curved line would be more useful in understanding the relationship between the two variables.\n\n\\begin{caution}\n{Watch out for curved trends}\n{We only consider models based on straight lines in this chapter. If data show a nonlinear trend, like that in the right panel of Figure~\\ref{scattHeadLTotalLTube}, more advanced techniques should be used.\\vspace{0.7mm}}\n\\end{caution}\n\n\n\\textA{\\pagebreak}\n\n\\subsection{Fitting a line by eye}\n\nWe want to describe the relationship between the head length and total length variables in the possum data set using a line. In this example, we will use the total length as the predictor variable, $x$, to predict a possum's head length, $y$. We could fit the linear relationship by eye, as in Figure~\\ref{scattHeadLTotalLLine}. The equation for this line is\n\\begin{eqnarray}\n\\hat{y} = 41 + 0.59x\n\\label{headLLinModTotalL}\n\\end{eqnarray}\nWe can use this line to discuss properties of possums. For instance, the equation predicts a possum with a total length of 80 cm will have a head length of\n\\begin{align*}\n\\hat{y} &= 41 + 0.59\\times 80 \\\\\n\t&= 88.2 % mm\n\\end{align*}\nA ``hat'' on $y$ is used to signify that this is an estimate. This estimate may be viewed as an average: the equation predicts that possums with a total length of 80 cm will have an average head length of 88.2 mm. Absent further information about an 80 cm possum, the prediction for head length that uses the average is a reasonable estimate.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.8\\textwidth]{05/figures/scattHeadLTotalLLine/scattHeadLTotalLLine}\n \\caption{A reasonable linear model was fit to represent the relationship between head length and total length.}\n \\label{scattHeadLTotalLLine}\n\\end{figure}\n\n\\subsection{Residuals}\n\n\\index{residual|(}\n\n\\termsub{Residuals}{residual} are the leftover variation in the data after accounting for the model fit:\n\\begin{align*}\n\\text{Data} = \\text{Fit} + \\text{Residual}\n\\end{align*}\nEach observation will have a residual. If an observation is above the regression line, then its residual, the vertical distance from the observation to the line, is positive. Observations below the line have negative residuals. One goal in picking the right linear model is for these residuals to be as small as possible.\n\nThree observations are noted specially in Figure~\\ref{scattHeadLTotalLLine}. The observation marked by an ``$\\times$'' has a small, negative residual of about -1; the observation marked by ``$+$'' has a large residual of about +7; and the observation marked by ``$\\triangle$'' has a moderate residual of about -4. The size of a residual is usually discussed in terms of its absolute value. For example, the residual for ``$\\triangle$'' is larger than that of ``$\\times$'' because $|-4|$ is larger than $|-1|$.\n\n\\begin{termBox}{\\tBoxTitle{Residual: difference between observed and expected}\nThe residual of the $i^{th}$ observation $(x_i, y_i)$ is the difference of the observed response ($y_i$) and the response we would predict based on the model fit ($\\hat{y}_i$):\n\\begin{eqnarray*}\ne_i = y_i - \\hat{y}_i\n\\end{eqnarray*}\nWe typically identify $\\hat{y}_i$ by plugging $x_i$ into the model.}\n\\end{termBox}\n\n\\begin{example}{The linear fit shown in Figure~\\ref{scattHeadLTotalLLine} is given as $\\hat{y} = 41 + 0.59x$. Based on this line, formally compute the residual of the observation $(77.0, 85.3)$. This observation is denoted by ``$\\times$'' on the plot. Check it against the earlier visual estimate,~-1.}\nWe first compute the predicted value of point ``$\\times$'' based on the model:\n\\begin{eqnarray*}\n\\hat{y}_{\\times} = 41+0.59x_{\\times} = 41+0.59\\times 77.0 = 86.4\n\\end{eqnarray*}\nNext we compute the difference of the actual head length and the predicted head length:\n\\begin{eqnarray*}\ne_{\\times} = y_{\\times} - \\hat{y}_{\\times} = 85.3 - 86.4 = -1.1\n\\end{eqnarray*}\nThis is very close to the visual estimate of -1.\n\\end{example}\n\n\\begin{exercise}\nIf a model underestimates an observation, will the residual be positive or negative? What about if it overestimates the observation?\\footnote{If a model underestimates an observation, then the model estimate is below the actual. The residual, which is the actual observation value minus the model estimate, must then be positive. The opposite is true when the model overestimates the observation: the residual is negative.}\n\\end{exercise}\n\n\\begin{exercise}\nCompute the residuals for the observations $(85.0, 98.6)$ (``$+$'' in the figure) and $(95.5, 94.0)$ (``$\\triangle$'') using the linear relationship $\\hat{y} = 41 + 0.59x$. \\footnote{($+$) First compute the predicted value based on the model: $$\\hat{y}_{+} = 41+0.59x_{+} = 41+0.59\\times 85.0 = 91.15$$ Then the residual is given by $$e_{+} = y_{+} - \\hat{y}_{+} = 98.6-91.15=7.45$$This was close to the earlier estimate of 7.\n\n($\\triangle$) $\\hat{y}_{\\triangle} = 41+0.59x_{\\triangle} = 97.3$. $e_{\\triangle} = y_{\\triangle} - \\hat{y}_{\\triangle} = -3.3$, close to the estimate of -4.}\n\\end{exercise}\n\nResiduals are helpful in evaluating how well a linear model fits a data set. We often display them in a \\textbf{residual plot} such as the one shown in Figure~\\ref{scattHeadLTotalLResidualPlot} for the regression line in Figure~\\ref{scattHeadLTotalLLine}. The residuals are plotted at their original horizontal locations but with the vertical coordinate as the residual. For instance, the point $(85.0,98.6)_{+}$ had a residual of 7.45, so in the residual plot it is placed at $(85.0, 7.45)$. Creating a residual plot is sort of like tipping the scatterplot over so the regression line is horizontal. \n\\index{data!possum|)}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.73\\textwidth]{05/figures/scattHeadLTotalLResidualPlot/scattHeadLTotalLResidualPlot}\n \\caption{Residual plot for the model in Figure~\\ref{scattHeadLTotalLLine}.}\n \\label{scattHeadLTotalLResidualPlot}\n\\end{figure}\n\n\\begin{example}{One purpose of residual plots is to identify characteristics or patterns still apparent in data after fitting a model. Figure~\\ref{sampleLinesAndResPlots} shows three scatterplots with linear models in the first row and residual plots in the second row. Can you identify any patterns remaining in the residuals?}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth]{05/figures/sampleLinesAndResPlots/sampleLinesAndResPlots}\n \\caption{Sample data with their best fitting lines (top row) and their corresponding residual plots (bottom row).}\n \\label{sampleLinesAndResPlots}\n\\end{figure}\nIn the first data set (first column), the residuals show no obvious patterns. The residuals appear to be scattered randomly around the dashed line that represents 0.\n\nThe second data set shows a pattern in the residuals. There is some curvature in the scatterplot, which is more obvious in the residual plot. We should not use a straight line to model these data. Instead, a more advanced technique should be used.\n\nThe last plot shows very little upwards trend, and the residuals also show no obvious patterns. It is reasonable to try to fit a linear model to the data. However, it is unclear whether there is statistically significant evidence that the slope parameter is different from zero. The point estimate of the slope parameter, labeled $b_1$, is not zero, but we might wonder if this could just be due to chance. We will address this sort of scenario in Section~\\ref{inferenceForLinearRegression}.\n\\index{residual|)}\n\\end{example}\n\n\\subsection{Describing linear relationships with correlation}\n\n\\index{correlation|(}\n\n\\begin{termBox}{\\tBoxTitle{Correlation: strength of a linear relationship}\n\\termsub{Correlation}{correlation}, which always takes values between -1 and 1, describes the strength of the linear relationship between two variables. We denote the correlation by $R$.}\n\\end{termBox}\\marginpar[\\raggedright\\vspace{-11.5mm}\n\n$R$\\\\\\footnotesize correlation]{\\raggedright\\vspace{-11.5mm}\n\n$R$\\\\\\footnotesize correlation}\n\nWe can compute the correlation using a formula, just as we did with the sample mean and standard deviation. However, this formula is rather complex,\\footnote{Formally, we can compute the correlation for observations $(x_1, y_1)$, $(x_2, y_2)$, ..., $(x_n, y_n)$ using the formula\n\\begin{eqnarray*}\nR = \\frac{1}{n-1}\\sum_{i=1}^{n} \\frac{x_i-\\bar{x}}{s_x}\\frac{y_i-\\bar{y}}{s_y}\n\\end{eqnarray*}\nwhere $\\bar{x}$, $\\bar{y}$, $s_x$, and $s_y$ are the sample means and standard deviations for each variable.} so we generally perform the calculations on a computer or calculator. Figure~\\ref{posNegCorPlots} shows eight plots and their corresponding correlations. Only when the relationship is perfectly linear is the correlation either -1 or 1. If the relationship is strong and positive, the correlation will be near +1. If it is strong and negative, it will be near -1. If there is no apparent linear relationship between the variables, then the correlation will be near zero.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.9\\textwidth]{05/figures/posNegCorPlots/posNegCorPlots}\n \\caption{Sample scatterplots and their correlations. The first row shows variables with a positive relationship, represented by the trend up and to the right. The second row shows variables with a negative trend, where a large value in one variable is associated with a low value in the other.}\n \\label{posNegCorPlots}\n\\end{figure}\n\nThe correlation is intended to quantify the strength of a linear trend. Nonlinear trends, even when strong, sometimes produce correlations that do not reflect the strength of the relationship; see three such examples in Figure~\\ref{corForNonLinearPlots}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.96\\textwidth]{05/figures/posNegCorPlots/corForNonLinearPlots}\n \\caption{Sample scatterplots and their correlations. In each case, there is a strong relationship between the variables. However, the correlation is not very strong, and the relationship is not linear.}\n \\label{corForNonLinearPlots}\n\\end{figure}\n\n\\begin{exercise}\nIt appears no straight line would fit any of the datasets represented in Figure~\\ref{corForNonLinearPlots}. Instead, try drawing nonlinear curves on each plot. Once you create a curve for each, describe what is important in your fit.\\footnote{We'll leave it to you to draw the lines. In general, the lines you draw should be close to most points and reflect overall trends in the data.}\n\\index{correlation|)}\n\\end{exercise}\n\n\n%__________________\n\\section{Fitting a line by least squares regression}\n\\label{fittingALineByLSR}\n\n\\index{least squares regression|(}\n\nFitting linear models by eye is open to criticism since it is based on an individual preference. In this section, we use \\emph{least squares regression} as a more rigorous approach.\n\nThis section considers family income and gift aid data from a random sample of fifty students in the 2011 freshman class of Elmhurst College in Illinois.\\footnote{These data were sampled from a table of data for all freshman from the 2011 class at Elmhurst College that accompanied an article titled \\emph{What Students Really Pay to Go to College} published online by \\emph{The~Chronicle of Higher Education}: \\href{http://chronicle.com/article/What-Students-Really-Pay-to-Go/131435}{chronicle.com/article/What-Students-Really-Pay-to-Go/131435}} Gift aid is financial aid that does not need to be paid back, as opposed to a loan. A scatterplot of the data is shown in Figure~\\ref{elmhurstScatterW2Lines} along with two linear fits. The lines follow a negative trend in the data; students who have higher family incomes tended to have lower gift aid from the university.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{05/figures/elmhurstPlots/elmhurstScatterW2Lines}\n\\caption{Gift aid and family income for a random sample of 50 freshman students from Elmhurst College. Two lines are fit to the data, the solid line being the \\emph{least squares line}.}\n\\label{elmhurstScatterW2Lines}\n\\end{figure}\n\n\\begin{exercise}\nIs the correlation positive or negative in Figure~\\ref{elmhurstScatterW2Lines}?\\footnote{Larger family incomes are associated with lower amounts of aid, so the correlation will be negative. Using a computer, the correlation can be computed: -0.499.}\n\\end{exercise}\n\n\n\\subsection{An objective measure for finding the best line}\n\nWe begin by thinking about what we mean by ``best''. Mathematically, we want a line that has small residuals. Perhaps our criterion could minimize the sum of the residual magnitudes:\\textA{\\vspace{-4mm}}\n\\begin{eqnarray}\n|e_1| + |e_2| + \\dots + |e_n|\n\\label{sumOfAbsoluteValueOfResiduals}\n\\end{eqnarray}\nwhich we could accomplish with a computer program. The resulting dashed line shown in Figure~\\ref{elmhurstScatterW2Lines} demonstrates this fit can be quite reasonable. However, a more common practice is to choose the line that minimizes the sum of the squared residuals:\\textA{\\vspace{-1mm}}\n\\begin{eqnarray}\ne_{1}^2 + e_{2}^2 + \\dots + e_{n}^2\n\\label{sumOfSquaresForResiduals}\n\\end{eqnarray}\nThe line that minimizes this \\textbf{least squares criterion} is represented as the solid line in Figure~\\ref{elmhurstScatterW2Lines}. This is commonly called the \\textbf{least squares line}. The following are three possible reasons to choose Criterion~\\eqref{sumOfSquaresForResiduals} over Criterion~\\eqref{sumOfAbsoluteValueOfResiduals}:\n\\begin{enumerate}\n\\item It is the most commonly used method.\n\\item Computing the line based on Criterion~\\eqref{sumOfSquaresForResiduals} is much easier by hand and in most statistical software.\n\\item In many applications, a residual twice as large as another residual is more than twice as bad. For example, being off by 4 is usually more than twice as bad as being off by 2. Squaring the residuals accounts for this discrepancy.\n\\end{enumerate}\nThe first two reasons are largely for tradition and convenience; the last reason explains why Criterion~\\eqref{sumOfSquaresForResiduals} is typically most helpful.\\footnote{There are applications where Criterion~\\eqref{sumOfAbsoluteValueOfResiduals} may be more useful, and there are plenty of other criteria we might consider. However, this book only applies the least squares criterion.}\n\n\n\\subsection{Finding the least squares line}\n\\label{findingTheLeastSquaresLineSection}\n\nFor the Elmhurst data, we could write the equation of the least squares regression line as\n\\begin{eqnarray*}\n\\widehat{aid} = \\beta_0 + \\beta_{1}\\times family\\_\\hspace{0.3mm}income\n\\end{eqnarray*}\nHere the equation is set up to predict gift aid based on a student's family income, which would be useful to students considering Elmhurst. These two values, $\\beta_0$ and $\\beta_1$, are the \\emph{parameters} of the regression line.\n\nAs in Chapters~4-6, the parameters are estimated using observed data. In practice, this estimation is done using a computer in the same way that other estimates, like a sample mean, can be estimated using a computer or calculator. However, we can also find the parameter estimates by applying two properties of the least squares line:\n\\begin{itemize}\n\\setlength{\\itemsep}{0mm}\n\\item The slope of the least squares line can be estimated by\n\\begin{eqnarray}\nb_1 = \\frac{s_y}{s_x} R\n\\label{slopeOfLSRLine}\n\\end{eqnarray}\nwhere $R$ is the correlation between the two variables, and $s_x$ and $s_y$ are the sample standard deviations of the explanatory variable %(variable on the horizontal axis) \nand response% (variable on the vertical axis)\n, respectively.\n\\item If $\\bar{x}$ is the mean of the horizontal variable (from the data) and $\\bar{y}$ is the mean of the vertical variable, then the point $(\\bar{x}, \\bar{y})$ is on the least squares line.\n\\end{itemize}\n\\marginpar[\\raggedright\\vspace{0.5mm}\n\n$b_0, b_1$\\vspace{0.5mm}\\\\\\footnotesize Sample\\\\estimates\\\\ of $\\beta_0$, $\\beta_1$]{\\raggedright\\vspace{0.5mm}\n\n$b_0, b_1$\\vspace{0.5mm}\\\\\\footnotesize Sample\\\\estimates\\\\ of $\\beta_0$, $\\beta_1$}We use $b_0$ and $b_1$ to represent the point estimates of the parameters $\\beta_0$ and $\\beta_1$.\n\n\\begin{exercise}\nTable~\\ref{summaryStatsOfSATGPAData} shows the sample means for the family income and gift aid as \\$101,800 and \\$19,940, respectively. Plot the point $(101.8, 19.94)$ on Figure~\\vref{elmhurstScatterW2Lines} to verify it falls on the least squares line (the solid line).\\footnote{If you need help finding this location, draw a straight line up from the x-value of 100 (or thereabout). Then draw a horizontal line at 20 (or thereabout). These lines should intersect on the least squares line.}\n\\end{exercise}\n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{l rr}\n\\hline\n\\vspace{-4mm} & & \\\\\n\\vspace{0.4mm}\t&\t\\ \\ family income, in \\$1000s (``$x$'')\t& \\ \\ gift aid, in \\$1000s (``$y$'') \\\\\n\\hline\n \\vspace{-3.9mm} & & \\\\\nmean\t& $\\bar{x} = 101.8$\t\t& $\\bar{y} = 19.94$ \\\\\nsd\t\t& $s_x = 63.2$\t\t& $s_y = 5.46$\\vspace{0.4mm} \\\\\n\\hline\n\\vspace{-4mm}\\ &\\\\\n\t& \\multicolumn{2}{r}{$R=-0.499$} \\\\\n\\hline\n\\end{tabular}\n\\caption{Summary statistics for family income and gift aid.}\n\\label{summaryStatsOfSATGPAData}\n\\end{table}\n\n\\begin{exercise} \\label{findingTheSlopeOfTheLSRLineForIncomeAndAid}\nUsing the summary statistics in Table~\\ref{summaryStatsOfSATGPAData}, compute the slope for the regression line of gift aid against family income.\\footnote{Apply Equation~\\eqref{slopeOfLSRLine} with the summary statistics from Table~\\ref{summaryStatsOfSATGPAData} to compute the slope:\n\\begin{eqnarray*}\nb_1 = \\frac{s_y}{s_x} R = \\frac{5.46}{63.2}(-0.499) = -0.0431\n\\end{eqnarray*}\\vspace{-3mm}}\n\\end{exercise}\n\nYou might recall the \\textbf{point-slope} form of a line from math class (another common form is \\emph{slope-intercept}). Given the slope of a line and a point on the line, $(x_0, y_0)$, the equation for the line can be written as\n\\begin{eqnarray}\ny - y_0 = slope\\times (x - x_0)\n\\label{pointSlopeFormForALine}\n\\end{eqnarray}\nA common exercise to become more familiar with foundations of least squares regression is to use basic summary statistics and point-slope form to produce the least squares line. \n\n\\begin{tipBox}{\\tipBoxTitle{Identifying the least squares line from summary statistics}\nTo identify the least squares line from summary statistics:\\vspace{-1mm}\n\\begin{itemize}\n\\setlength{\\itemsep}{0mm}\n\\item Estimate the slope parameter, $\\beta_1$, by calculating $b_1$ using Equation~\\eqref{slopeOfLSRLine}.\n\\item Noting that the point $(\\bar{x}, \\bar{y})$ is on the least squares line, use $x_0=\\bar{x}$ and $y_0=\\bar{y}$ along with the slope $b_1$ in the point-slope equation:\n$$y - \\bar{y} = b_1 (x - \\bar{x}) $$\n\\item Simplify the equation.\\vspace{0.5mm}\n\\end{itemize}}\n\\end{tipBox}\n\n\\textA{\\pagebreak}\n\n\\begin{example}{Using the point $(101.8, 19.94)$ from the sample means and the slope estimate $b_1 = -0.0431$ from Guided Practice~\\ref{findingTheSlopeOfTheLSRLineForIncomeAndAid}, find the least-squares line for predicting aid based on family income.} \\label{exampleToFindLSRLineOfElmhurstData}\nApply the point-slope equation using $(101.8, 19.94)$ and the slope $b_1 = -0.0431$:\n\\begin{align*}\ny - y_0 &= b_1 (x - x_0) \\\\\ny - 19.94 &= -0.0431(x - 101.8)\n\\end{align*}\nExpanding the right side and then adding 19.94 to each side, the equation simplifies:\n$$ \\widehat{aid} = 24.3 - 0.0431 \\times family\\_\\hspace{0.3mm}income $$\nHere we have replaced $y$ with $\\widehat{aid}$ and $x$ with $family\\_\\hspace{0.3mm}income$ to put the equation in context.\n\\end{example}\n\nWe mentioned earlier that a computer is usually used to compute the least squares line. A summary table based on computer output is shown in Table~\\ref{rOutputForIncomeAidLSRLine} for the Elmhurst data. The first column of numbers provides estimates for ${b}_0$ and ${b}_1$, respectively. Compare these to the result from Example~\\ref{exampleToFindLSRLineOfElmhurstData}.\n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{l rrrr}\n \\hline\n \\vspace{-3.7mm} & & & & \\\\\n & Estimate & Std. Error & t value & Pr($>$$|$t$|$) \\\\ \n \\hline\n \\vspace{-3.6mm} & & & & \\\\\n(Intercept) & 24.3193 & 1.2915 & 18.83 & 0.0000 \\\\ \nfamily\\_\\hspace{0.3mm}income & -0.0431 & 0.0108 & -3.98 & 0.0002 \\\\ \n \\hline\n\\end{tabular}\n\\caption{Summary of least squares fit for the Elmhurst data. Compare the parameter estimates in the first column to the results of Example~\\ref{exampleToFindLSRLineOfElmhurstData}.}\n\\label{rOutputForIncomeAidLSRLine}\n\\end{table}\n\n\\begin{example}{Examine the second, third, and fourth columns in Table~\\ref{rOutputForIncomeAidLSRLine}. Can you guess what they represent?}\nWe'll describe the meaning of the columns using the second row, which corresponds to~$\\beta_1$. The first column provides the point estimate for $\\beta_1$, as we calculated in an earlier example: -0.0431. The second column is a standard error for this point estimate: 0.0108. The third column is a $t$ test statistic for the null hypothesis that $\\beta_1 = 0$: $T=-3.98$. The last column is the p-value for the $t$ test statistic for the null hypothesis $\\beta_1=0$ and a two-sided alternative hypothesis: 0.0002. We will get into more of these details in Section~\\ref{inferenceForLinearRegression}.\n\\end{example}\n\n\\begin{example}{Suppose a high school senior is considering Elmhurst College. Can she simply use the linear equation that we have estimated to calculate her financial aid from the university?}\nShe may use it as an estimate, though some qualifiers on this approach are important. First, the data all come from one freshman class, and the way aid is determined by the university may change from year to year. Second, the equation will provide an imperfect estimate. While the linear equation is good at capturing the trend in the data, no individual student's aid will be perfectly predicted.\n\\index{least squares regression|)}\n\\end{example} \n\n\\subsection{Interpreting regression line parameter estimates}\n\n\\index{least squares regression!interpreting parameters|(}\n\nInterpreting parameters in a regression model is often one of the most important steps in the analysis.\n\n\\begin{example}{The slope and intercept estimates for the Elmhurst data are -0.0431 and 24.3. What do these numbers really mean?}\nInterpreting the slope parameter is helpful in almost any application. For each additional \\$1,000 of family income, we would expect a student to receive a net difference of $\\$\\text{1,000}\\times (-0.0431) = -\\$43.10$ in aid on average, i.e. \\$43.10 \\emph{less}. Note that a higher family income corresponds to less aid because the coefficient of family income is negative in the model. We must be cautious in this interpretation: while there is a real association, we cannot interpret a causal connection between the variables because these data are observational. That is, increasing a student's family income may not cause the student's aid to drop. (It would be reasonable to contact the college and ask if the relationship is causal, i.e. if Elmhurst College's aid decisions are partially based on students' family income.)\n\nThe estimated intercept $b_0=24.3$ (in \\$1000s) describes the average aid if a student's family had no income. The meaning of the intercept is relevant to this application since the family income for some students at Elmhurst is \\$0. In other applications, the intercept may have little or no practical value if there are no observations where $x$ is near zero.\n\\end{example}\n\n\\begin{termBox}{\\tBoxTitle{Interpreting parameters estimated by least squares}\nThe slope describes the estimated difference in the $y$ variable if the explanatory variable $x$ for a case happened to be one unit larger. The intercept describes the average outcome of $y$ if $x=0$ \\emph{and} the linear model is valid all the way to $x=0$, which in many applications is not the case.}\n\\index{least squares regression!interpreting parameters|)}\n\\end{termBox}\n\n\\subsection{Extrapolation is treacherous}\n\n\\index{least squares regression!extrapolation|(}\n\n{\\em\\small When those blizzards hit the East Coast this winter, it proved to my satisfaction that global warming was a fraud. That snow was freezing cold. But in an alarming trend, temperatures this spring have risen. Consider this: On February $6^{th}$ it was 10 degrees. Today it hit almost 80. At this rate, by August it will be 220 degrees. So clearly folks the climate debate rages on.\\vspace{0.5mm}}\n\n\\noindent\\hspace{\\textwidth}\\hspace{-40mm}Stephen Colbert\n\n\\noindent\\hspace{\\textwidth}\\hspace{-40mm}April 6th, 2010 \\footnote{\\urlwofont{http://www.colbertnation.com/the-colbert-report-videos/269929/}} \\\\\n\nLinear models can be used to approximate the relationship between two variables. However, these models have real limitations. Linear regression is simply a modeling framework. The truth is almost always much more complex than our simple line. For example, we do not know how the data outside of our limited window will behave.\n\n\\textA{\\pagebreak}\n\n\\begin{example}{Use the model $\\widehat{aid} = 24.3 - 0.0431\\times family\\_\\hspace{0.3mm}income$ to estimate the aid of another freshman student whose family had income of \\$1 million.}\nRecall that the units of family income are in \\$1000s, so we want to calculate the aid for $family\\_\\hspace{0.3mm}income = 1000$:\n\\begin{align*}\n24.3 - 0.0431\\times family\\_\\hspace{0.3mm}income = 24.3 - 0.0431\\times 1000 = -18.8\n\\end{align*}\nThe model predicts this student will have -\\$18,800 in aid (!). Elmhurst College cannot (or at least does not) require any students to pay extra on top of tuition to attend.\n\\end{example}\n\nApplying a model estimate to values outside of the realm of the original data is called \\textbf{extrapolation}. Generally, a linear model is only an approximation of the real relationship between two variables. If we extrapolate, we are making an unreliable bet that the approximate linear relationship will be valid in places where it has not been explored.\n\n\\index{least squares regression!extrapolation|)}\n\n\\subsection{Using $R^2$ to describe the strength of a fit}\n\n\\index{least squares regression!R-squared ($R^2$)|(}\n\nWe evaluated the strength of the linear relationship between two variables earlier using the correlation, $R$. However, it is more common to explain the strength of a linear fit using $R^2$, called \\termsub{R-squared}{least squares regression!R-squared ($R^2$)}. If provided with a linear model, we might like to describe how closely the data cluster around the linear fit.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{05/figures/elmhurstPlots/elmhurstScatterWLSROnly}\n\\caption{Gift aid and family income for a random sample of 50 freshman students from Elmhurst College, shown with the least squares regression line.}\n\\label{elmhurstScatterWLSROnly}\n\\end{figure}\n\nThe $R^2$ of a linear model describes the amount of variation in the response that is explained by the least squares line. For example, consider the Elmhurst data, shown in Figure~\\ref{elmhurstScatterWLSROnly}. The variance of the response variable, aid received, is $s_{aid}^2=29.8$. However, if we apply our least squares line, then this model reduces our uncertainty in predicting aid using a student's family income. The variability in the residuals describes how much variation remains after using the model: $s_{_{RES}}^2 = 22.4$. In short, there was a reduction of\n$$\\frac{s_{aid}^2 - s_{_{RES}}^2}{s_{_{aid}}^2}\n\t= \\frac{29.8 - 22.4}{29.8} = \\frac{7.5}{29.8}\n\t= 0.25$$\nor about 25\\% in the data's variation by using information about family income for predicting aid using a linear model. This corresponds exactly to the R-squared value:\n\\begin{align*}\nR &= -0.499 &R^2 &= 0.25\n\\end{align*}\n\n\\begin{exercise}\nIf a linear model has a very strong negative relationship with a correlation of -0.97, how much of the variation in the response is explained by the explanatory variable?\\footnote{About $R^2 = (-0.97)^2 = 0.94$ or 94\\% of the variation is explained by the linear model.}\n\\index{least squares regression!R-squared ($R^2$)|)}\n\\end{exercise}\n\n\\subsection{Categorical predictors with two levels}\n\\label{categoricalPredictorsWithTwoLevels}\n\nCategorical variables are also useful in predicting outcomes. Here we consider a categorical predictor with two levels (recall that a \\emph{level} is the same as a \\emph{category}). We'll consider Ebay auctions for a video game, \\emph{Mario Kart} for the Nintendo Wii, where both the total price of the auction and the condition of the game were recorded.\\footnote{These data were collected in Fall 2009 and may be found at \\href{http://www.openintro.org}{openintro.org}.} Here we want to predict total price based on game condition, which takes values \\resp{used} and \\resp{new}. A plot of the auction data is shown in Figure~\\ref{marioKartNewUsed}.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.65\\textwidth]{05/figures/marioKartNewUsed/marioKartNewUsed}\n\\caption{Total auction prices for the video game \\emph{Mario Kart}, divided into used ($x=0$) and new ($x=1$) condition games. The least squares regression line is also shown.}\n\\label{marioKartNewUsed}\n\\end{figure}\n\nTo incorporate the game condition variable into a regression equation, we must convert the categories into a numerical form. We will do so using an \\textbf{indicator variable} called \\textbf{cond\\_\\hspace{0.3mm}new}, which takes value 1 when the game is new and 0 when the game is used. Using this indicator variable, the linear model may be written as\n\\begin{align*}\n\\widehat{price} = \\beta_0 + \\beta_1 \\times \\text{\\textbf{cond\\_\\hspace{0.3mm}new}}\n\\end{align*}\nThe fitted model is summarized in Table~\\ref{marioKartNewUsedRegrSummary}, and the model with its parameter estimates is given as\n\\begin{align*}\n\\widehat{price} = 42.87 + 10.90 \\times \\text{\\textbf{cond\\_\\hspace{0.3mm}new}}\n\\end{align*}\n\n\\begin{table}\n\\centering\n\\begin{tabular}{rrrrr}\n \\hline\n \\vspace{-3.7mm} & & & & \\\\\n & Estimate & Std. Error & t value & Pr($>$$|$t$|$) \\\\ \n \\hline\n \\vspace{-3.6mm} & & & & \\\\\n(Intercept) & 42.87 & 0.81 & 52.67 & 0.0000 \\\\ \n cond\\_\\hspace{0.3mm}new & 10.90 & 1.26 & 8.66 & 0.0000 \\\\ \n \\hline\n\\end{tabular}\n\\caption{Least squares regression summary for the final auction price against the condition of the game.}\n\\label{marioKartNewUsedRegrSummary}\n\\end{table}\n\n\\begin{example}{Interpret the two parameters estimated in the model for the price of \\emph{Mario Kart} in eBay auctions.}\nThe intercept is the estimated price when \\textbf{cond\\_\\hspace{0.3mm}new} takes value 0, i.e. when the game is in used condition. That is, the average selling price of a used version of the game is \\$42.87.\n\nThe slope indicates that, on average, new games sell for about \\$10.90 more than used games.\n\\end{example}\n\n\\begin{tipBox}{\\tipBoxTitle{Interpreting model estimates for categorical predictors.}\nThe estimated intercept is the value of the response variable for the first category (i.e. the category corresponding to an indicator value of 0). The estimated slope is the average change in the response variable between the two categories.}\n\\end{tipBox}\n\n\\MultipleRegression{We'll elaborate further on this Ebay auction data in Chapter~\\ref{multipleRegressionAndANOVA}, where we examine the influence of many predictor variables simultaneously using multiple regression\\index{multiple regression}. In multiple regression, we will consider the association of auction price with regard to each variable while controlling for the influence of other variables. This is especially important since some of the predictors are associated. For example, auctions with games in new condition also often came with more accessories.}{}\n\n\n%__________________\n\\section{Types of outliers in linear regression}\n\\label{typesOfOutliersInLinearRegression}\n\nIn this section, we identify criteria for determining which outliers are important and influential.\n\nOutliers in regression are observations that fall far from the ``cloud'' of points. These points are especially important because they can have a strong influence on the least squares line.\n\n\\textA{\\pagebreak}\n\n\\begin{example}{There are six plots shown in Figure~\\ref{outlierPlots} along with the least squares line and residual plots. For each scatterplot and residual plot pair, identify any obvious outliers and note how they influence the least squares line. Recall that an outlier is any point that doesn't appear to belong with the vast majority of the other points.}\\label{outlierPlotsExample}\n\\begin{itemize}\n\\item[(1)] There is one outlier far from the other points, though it only appears to slightly influence the line.\n\\item[(2)] There is one outlier on the right, though it is quite close to the least squares line, which suggests it wasn't very influential.\n\\item[(3)] There is one point far away from the cloud, and this outlier appears to pull the least squares line up on the right; examine how the line around the primary cloud doesn't appear to fit very well.\n\\item[(4)] There is a primary cloud and then a small secondary cloud of four outliers. The secondary cloud appears to be influencing the line somewhat strongly, making the least square line fit poorly almost everywhere. There might be an interesting explanation for the dual clouds, which is something that could be investigated.\n\\item[(5)] There is no obvious trend in the main cloud of points and the outlier on the right appears to largely control the slope of the least squares line.\n\\item[(6)] There is one outlier far from the cloud, however, it falls quite close to the least squares line and does not appear to be very influential.\n\\end{itemize}\n\\end{example}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{05/figures/outlierPlots/outlierPlots}\n\\caption{Six plots, each with a least squares line and residual plot. All data sets have at least one outlier.}\n\\label{outlierPlots}\n\\end{figure}\n\nExamine the residual plots in Figure~\\ref{outlierPlots}. You will probably find that there is some trend in the main clouds of (3) and (4). In these cases, the outliers influenced the slope of the least squares lines. In (5), data with no clear trend were assigned a line with a large trend simply due to one outlier (!).\n \n \\begin{termBox}{\\tBoxTitle{Leverage}\nPoints that fall horizontally away from the center of the cloud tend to pull harder on the line, so we call them points with \\textbf{high leverage}.}\n\\end{termBox}\n\nPoints that fall horizontally far from the line are points of high leverage; these points can strongly influence the slope of the least squares line. If one of these high leverage points does appear to actually invoke its influence on the slope of the line -- as in cases (3), (4), and (5) of Example~\\ref{outlierPlotsExample} -- then we call it an \\textbf{influential point}. Usually we can say a point is influential if, had we fitted the line without it, the influential point would have been unusually far from the least squares line.\n\nIt is tempting to remove outliers. Don't do this without a very good reason. Models that ignore exceptional (and interesting) cases often perform poorly. For instance, if a financial firm ignored the largest market swings -- the ``outliers'' -- they would soon go bankrupt by making poorly thought-out investments.\n\n\\begin{caution}{Don't ignore outliers when fitting a final model}\n{If there are outliers in the data, they should not be removed or ignored without a~good reason. Whatever final model is fit to the data would not be very helpful if it ignores the most exceptional cases.}\n\\end{caution}\n\n\\begin{caution}{Outliers for a categorical predictor with two levels}{\nBe cautious about using a categorical predictor when one of the levels has very few observations. When this happens, those few observations become influential points.}\n\\end{caution}\n\n\n%__________________\n\\section{Inference for linear regression}\n\\label{inferenceForLinearRegression}\n\nIn this section we discuss uncertainty in the estimates of the slope and y-intercept for a regression line. Just as we identified standard errors for point estimates in previous chapters, we first discuss standard errors for these new estimates. However, in the case of regression, we will identify standard errors using statistical software.\n\n\n\\subsection{Conditions for the least squares line}\n\nWhen performing inference on a least squares line, we generally require the following:\n\\begin{description}\n\\setlength{\\itemsep}{0mm}\n\\item[Linearity.] The data should show a linear trend. If there is a nonlinear trend (e.g. left panel of Figure~\\ref{whatCanGoWrongWithLinearModel}), an advanced regression method from another book or later course should be applied.\n\\item[Nearly normal residuals.] Generally the residuals must be nearly normal.\nWhen this condition is found to be unreasonable, it is usually because of outliers or concerns about influential points, which we will discuss in greater depth in Section~\\ref{typesOfOutliersInLinearRegression}. An example of non-normal residuals is shown in the second panel of Figure~\\ref{whatCanGoWrongWithLinearModel}.\n\\item[Constant variability.] The variability of points around the least squares line remains roughly constant. An example of non-constant variability is shown in the third panel of Figure~\\ref{whatCanGoWrongWithLinearModel}.\n\\item[Independent observations.] Be cautious about applying regression to data collected sequentially in what is called a \\textbf{time series}. Such data may have an underlying structure that should be considered in a model and analysis. An example of a time series where independence is violated is shown in the fourth panel of Figure~\\ref{whatCanGoWrongWithLinearModel}.\n%The observations in the data set must be independent. A common example of data that are not independent are data collected sequentially, called \\textbf{time series data}, where consecutive observations tend to be close to one another. Plotting the data in the order it was collected is often helpful for evaluating this condition.\n\\end{description}\nFor additional information on checking regression conditions, see Section~\\ref{multipleRegressionModelAssumptions}.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{05/figures/whatCanGoWrongWithLinearModel/whatCanGoWrongWithLinearModel}\n\\caption{Four examples showing when the methods in this chapter are insufficient to apply to the data. In the left panel, a straight line does not fit the data. In the second panel, there are outliers; two points on the left are relatively distant from the rest of the data, and one of these points is very far away from the line. In the third panel, the variability of the data around the line increases with larger values of $x$. In the last panel, a time series data set is shown, where successive observations are highly correlated.}\n\\label{whatCanGoWrongWithLinearModel}\n\\end{figure}\n\n\\begin{example}{Should we have concerns about applying inference to the Elmhurst data in Figure~\\ref{elmhurstScatterLSRLine_conditions}?}\nThe trend appears to be linear, the data fall around the line with no obvious outliers, the variance is roughly constant. These are also not time series observations. It would be reasonable to analyze the model using inference.\n\\end{example}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{05/figures/elmhurstPlots/elmhurstScatterLSRLine_conditions}\n\\caption{Gift aid and family income for a random sample of 50 freshman students from Elmhurst College. Two lines are fit to the data, the solid line being the \\emph{least squares line}.}\n\\label{elmhurstScatterLSRLine_conditions}\n\\end{figure}\n\n\n\n\n\\subsection{Midterm elections and unemployment}\n\n\\index{data!midterm elections|(}\n\nElections for members of the United States House of Representatives occur every two years, coinciding every four years with the U.S. Presidential election. The set of House elections occurring during the middle of a Presidential term are called \\indexthis{midterm elections}{midterm election}. In America's two-party system, one political theory suggests the higher the unemployment rate, the worse the President's party will do in the midterm elections.\n\nTo assess the validity of this claim, we can compile historical data and look for a connection. We consider every midterm election from 1898 to 2010, with the exception of those elections during the Great Depression. Figure~\\ref{unemploymentAndChangeInHouse} shows these data and the least-squares regression line: \\vspace{-2mm}\n\\begin{align*}\n&\\text{\\% change in House seats for President's party} \\\\\n&\\qquad\\qquad= -6.71 - 1.00\\times \\text{(unemployment rate)}\n\\end{align*}\nWe consider the percent change in the number of seats of the President's party (e.g. percent change in the number of seats for Democrats in 2010) against the unemployment rate.\n\nExamining the data, there are no clear deviations from linearity, the constant variance condition, or in the normality of residuals (though we don't examine a normal probability plot here). While the data are collected sequentially, a separate analysis was used to check for any apparent correlation between successive observations; no such correlation was found.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.95\\textwidth]{05/figures/unemploymentAndChangeInHouse/unemploymentAndChangeInHouse}\n\\caption{The percent change in House seats for the President's party in each election from 1898 to 2010 plotted against the unemployment rate. The two points for the Great Depression have been removed, and a least squares regression line has been fit to the data.}\n\\label{unemploymentAndChangeInHouse}\n\\end{figure}\n\n\\begin{exercise}\nThe data for the Great Depression (1934 and 1938) were removed because the unemployment rate was 21\\% and 18\\%, respectively. Do you agree that they should be removed for this investigation? Why or why not?\\footnote{We will provide two considerations. Each of these points would have very high leverage on any least-squares regression line, and years with such high unemployment may not help us understand what would happen in other years where the unemployment is only modestly high. On the other hand, these are exceptional cases, and we would be discarding important information if we exclude them from a final analysis.}\n\\end{exercise}\n\nThere is a negative slope in the line shown in Figure~\\ref{unemploymentAndChangeInHouse}. However, this slope (and the y-intercept) are only estimates of the parameter values. We might wonder, is this convincing evidence that the ``true'' linear model has a negative slope? That is, do the data provide strong evidence that the political theory is accurate? We can frame this investigation into a two-sided statistical hypothesis test. We use a two-sided test since a statistically significant result in either direction would be interesting.\n\\begin{itemize}\n\\item[$H_0$:] $\\beta_1 = 0$. The true linear model has slope zero.\n\\item[$H_A$:] $\\beta_1 \\neq 0$. The true linear model has a slope different than zero. The higher the unemployment, the greater the loss for the President's party in the House of Representatives, or vice-versa.\n\\end{itemize}\nWe would reject $H_0$ in favor of $H_A$ if the data provide strong evidence that the true slope parameter is less than zero. To assess the hypotheses, we identify a standard error for the estimate, compute an appropriate test statistic, and identify the p-value.\n\n\n\\subsection{Understanding regression output from software}\n\\label{testStatisticForTheSlope}\n\nJust like other point estimates we have seen before, we can compute a standard error and test statistic for $b_1$. We will generally label the test statistic using a $T$, since it follows the $t$ distribution.\n\nWe will rely on statistical software to compute the standard error and leave the explanation of how this standard error is determined to a second or third statistics course. Table~\\ref{midtermElectionUnemploymentRRegressionOutput} shows software output for the least squares regression line in Figure~\\ref{unemploymentAndChangeInHouse}. The row labeled \\emph{unemp} represents the information for the slope, which is the coefficient of the unemployment variable.\n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{rrrrr}\n \\hline\n \\vspace{-3.7mm} & & & & \\\\\n & Estimate & Std. Error & t value & Pr($>$$|$t$|$) \\\\ \n \\hline\n \\vspace{-3.6mm} & & & & \\\\\n(Intercept) & -6.7142 & 5.4567 & -1.23 & 0.2300 \\\\ \n unemp & -1.0010 & 0.8717 & -1.15 & 0.2617 \\\\ \n \\hline\n \\multicolumn{5}{r}{$df=25$} \\\\\n\\end{tabular}\n\\caption{Output from statistical software for the regression line modeling the midterm election gains and losses for the President's party as a response to unemployment.}\n\\label{midtermElectionUnemploymentRRegressionOutput}\n\\end{table}\n\n\\begin{example}{What do the first and second columns of Table~\\ref{midtermElectionUnemploymentRRegressionOutput} represent?}\nThe entries in the first column represent the least squares estimates, $b_0$ and $b_1$, and the values in the second column correspond to the standard errors of each estimate.\n\\end{example}\n\nWe previously used a $t$ test statistic for hypothesis testing in the context of numerical data. Regression is very similar. In the hypotheses we consider, the null value for the slope is 0, so we can compute the test statistic using the T (or Z) score formula:\n\\begin{align*}\nT = \\frac{\\text{estimate} - \\text{null value}}{\\text{SE}} = \\frac{-1.0010 - 0}{0.8717} = -1.15\n\\end{align*}\nWe can look for the two-tailed p-value -- shown in Figure~\\ref{oneSidedTailForMidtermUnemploymentHT} -- using the probability table for the $t$ distribution in Appendix~\\ref{tDistributionTable} on page~\\pageref{tDistributionTable}.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.82\\textwidth]{05/figures/pValueMidtermUnemploymentHT/pValueMidtermUnemploymentHT}\n\\caption{The distribution shown here is the sampling distribution for $b_1$, if the null hypothesis was true. The shaded tail represents the p-value for the hypothesis test evaluating whether there is convincing evidence that higher unemployment corresponds to a greater loss of House seats for the President's party during a midterm election.}\n\\label{oneSidedTailForMidtermUnemploymentHT}\n\\end{figure}\n\n\\textA{\\pagebreak}\n\n\\begin{example}{Table~\\ref{midtermElectionUnemploymentRRegressionOutput} offers the degrees of freedom for the test statistic $T$: $df=25$. Identify the p-value for the hypothesis test.}\nLooking in the 25 degrees of freedom row in Appendix~\\ref{tDistributionTable}, we see that the absolute value of the test statistic is smaller than any value listed, which means the tail area and therefore also the p-value is larger than 0.200 (two tails!). Because the p-value is so large, we fail to reject the null hypothesis. That is, the data do not provide convincing evidence that unemployment is a good predictor of how well a president's party will do in the midterm elections for the House of Representatives.\n\\index{data!midterm elections|)}\n\\end{example}\n\nWe could have identified the $t$ test statistic from the software output in Table~\\ref{midtermElectionUnemploymentRRegressionOutput}, shown in the second row (unemp) and third column (t value). The entry in the second row and last column in Table~\\ref{midtermElectionUnemploymentRRegressionOutput} represents the p-value for the two-sided hypothesis test where the null value is zero.\n\n\\begin{termBox}{\\tBoxTitle{Inference for regression}\nWe usually rely on statistical software to identify point estimates and standard errors for parameters of a regression line. After verifying conditions hold for fitting a line, we can use the methods learned in Section~\\ref{oneSampleMeansWithTDistribution} for the $t$ distribution to create confidence intervals for regression parameters or to evaluate hypothesis tests.}\n\\end{termBox}\n\n\\begin{caution}{Don't carelessly use the p-value from regression output}{The last column in regression output often lists p-values for one particular hypothesis: a two-sided test where the null value is zero. If a hypothesis test should be one-sided or a comparison is being made to a value other than zero, be cautious about using the software output to obtain the p-value.}\n\\end{caution}\n\n\\textA{\\pagebreak}\n\n\\begin{example}{Examine Figure~\\vref{elmhurstScatterWLSROnly}, which relates the Elmhurst College aid and student family income. How sure are you that the slope is statistically significantly different from zero? That is, do you think a formal hypothesis test would reject the claim that the true slope of the line should be zero?} \\label{overallAidIncomeInformalAssessmentOfRegressionLineSlope}\nWhile the relationship between the variables is not perfect, there is an evident decreasing trend in the data. This suggests the hypothesis test will reject the null claim that the slope is zero.\n\\end{example}\n\n\\begin{exercise}\nTable~\\ref{rOutputForIncomeAidLSRLineInInferenceSection} shows statistical software output from fitting the least squares regression line shown in Figure~\\ref{elmhurstScatterWLSROnly}. Use this output to formally evaluate the following hypotheses. $H_0$: The true coefficient for family income is zero. $H_A$: The true coefficient for family income is not zero.\\footnote{We look in the second row corresponding to the family income variable. We see the point estimate of the slope of the line is -0.0431, the standard error of this estimate is 0.0108, and the $t$ test statistic is -3.98. The p-value corresponds exactly to the two-sided test we are interested in: 0.0002. The p-value is so small that we reject the null hypothesis and conclude that family income and financial aid at Elmhurst College for freshman entering in the year 2011 are negatively correlated and the true slope parameter is indeed less than 0, just as we believed in Example~\\ref{overallAidIncomeInformalAssessmentOfRegressionLineSlope}.}\n\\end{exercise}\n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{rrrrr}\n \\hline\n \\vspace{-3.7mm} & & & & \\\\\n & Estimate & Std. Error & t value & Pr($>$$|$t$|$) \\\\ \n \\hline\n \\vspace{-3.6mm} & & & & \\\\\n(Intercept) & 24.3193 & 1.2915 & 18.83 & 0.0000 \\\\ \nfamily\\_\\hspace{0.3mm}income & -0.0431 & 0.0108 & -3.98 & 0.0002 \\\\ \n \\hline\n \\multicolumn{5}{r}{$df=48$} \\\\\n\\end{tabular}\n\\caption{Summary of least squares fit for the Elmhurst College data.\\textA{\\vspace{-5mm}}}\n\\label{rOutputForIncomeAidLSRLineInInferenceSection}\n\\end{table}\n\n\\begin{tipBox}{\\tipBoxTitle{Always check assumptions}\nIf conditions for fitting the regression line do not hold, then the methods presented here should not be applied. The standard error or distribution assumption of the point estimate -- assumed to be normal when applying the $t$ test statistic -- may not be valid.}\n\\end{tipBox}\n\n\\ANOVASection{\\subsection{An alternative test statistic}\n\nWe considered the $t$ test statistic as a way to evaluate the strength of evidence for a hypothesis test in Section~\\ref{testStatisticForTheSlope}. However, we could focus on $R^2$. Recall that $R^2$ described the proportion of variability in the response variable ($y$) explained by the explanatory variable ($x$). If this proportion is large, then this suggests a linear relationship exists between the variables. If this proportion is small, then the evidence provided by the data may not be convincing.\n\nThis concept -- considering the amount of variability in the response variable explained by the explanatory variable -- is a key component in some statistical techniques. The \\emph{analysis of variance (ANOVA)}\\index{analysis of variance (ANOVA)} technique introduced in Section~\\ref{anovaAndRegrWithCategoricalVariables} uses this general principle. The method states that if enough variability is explained away by the categories, then we would conclude the mean varied between the categories. On the other hand, we might not be convinced if only a little variability is explained. ANOVA can be further employed in advanced regression modeling to evaluate the inclusion of explanatory variables, though we leave these details to a later course.}{}\n\n\n", "meta": {"hexsha": "7ebb1bcca3793349b5713744f33c25c8c5497e7c", "size": 58759, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ISRS1/Tex/05-V01.tex", "max_stars_repo_name": "amthapar/Crump_Stat", "max_stars_repo_head_hexsha": "8baa0a79d4e996c10b30d41af4d331e887bbaf1c", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-03-09T23:52:33.000Z", "max_stars_repo_stars_event_max_datetime": "2020-03-09T23:52:33.000Z", "max_issues_repo_path": "ISRS1/Tex/05-V01.tex", "max_issues_repo_name": "amthapar/Crump_Stat", "max_issues_repo_head_hexsha": "8baa0a79d4e996c10b30d41af4d331e887bbaf1c", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ISRS1/Tex/05-V01.tex", "max_forks_repo_name": "amthapar/Crump_Stat", "max_forks_repo_head_hexsha": "8baa0a79d4e996c10b30d41af4d331e887bbaf1c", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 81.158839779, "max_line_length": 1043, "alphanum_fraction": 0.7791487262, "num_tokens": 15013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434873426302, "lm_q2_score": 0.8267117983401363, "lm_q1q2_score": 0.6380921174581481}} {"text": "\\section{Circuits}\n \\subsection{Capacitants}\n Previously we found that $E=\\frac{\\sigma}{2\\epsilon_0}+\\frac{\\sigma}{2\\epsilon_0}=\\frac{\\sigma}{\\epsilon_0}$ where $\\sigma = \\frac{Q}{A}$. Let's find potential $V$ between the plates. Recall that $v=-\\int\\vec{E}\\cdot d\\vec{s}$.\n \\begin{align*}\n V&=-\\int_0^d\\vec{E}\\cdot d\\vec{s}\\\\\n \\text{for } \\vec{E}&=\\frac{\\sigma}{\\epsilon_0}\\\\\n V&=-\\int_0^d Eds=-\\int_0^d Edx\\\\\n &=-E\\int_0^d dx\\\\\n V&=-Ed \\text{ This is from the pos to negative plate}\\\\\n \\shortintertext{Equivilently from - to + plate:}\\\\\n V&=-int_0^d \\vec{E}\\cdot d\\vec{s}\\\\\n &=-int_0^d(-Eds)=\\int_0^d Edx\\\\\n V&=Ed \\text{ This is from neg to pos plate}\n \\shortintertext{Now we want to relate our two equations to a charge to find the charge within the plate.}\n \\end{align*}\n Consider that if there are more charges on the plates, then the potential (difference) between them will be larger. Difference is in paranthesis because we could have had a $\\Delta V$ instead of just a V if we wanted. The potential goes with the field, and the field is larger if we have more charges in it.\n \\begin{align*}\n Q~V\\\\\n \\shortintertext{Let us introduce capacitance $C$ such that $Q=CV$. In magnitude, we found $V=Ed$, which gives the potential difference accross the plates. And $E=\\frac{\\sigma}{\\epsilon_0}$}\\\\\n V&=\\frac{\\sigma}{\\epsilon_0}d=\\frac{1}{\\epsilon}\\frac{Q}{A}d\\\\\n \\shortintertext{Using our new definition of $V=\\frac{Q}{C}$, we find:}\\\\\n \\frac{1}{\\epsilon_0}\\frac{Q}{A}d&=\\frac{Q}{C}\\\\\n C&=\\epsilon_0 \\frac{A}{d}\\\\\n \\shortintertext{This is for parallel plate capacitance. This only depends on geometry. Due to the equation $Q=CV$, as soon as we know the charge on the plates we can easily determine the potential energy of the plates. $Q=CV$ is true for all capacitors.}\n \\end{align*}\n Let's discuss combining capacitors. For a parallel plate capacitor, we found that it's capacitance is $C=\\epsilon_0 \\frac{A}{d}$. What if we stacked plates together? What would the capacitancebe for this combined system?\n \\begin{align*}\n A_{new}&=A_1+A_2+A_3\\\\\n &=\\frac{d}{\\epsilon_0}C_1+\\frac{d}{\\epsilon_0}C_2+\\frac{d}{\\epsilon_0}C_3\\\\\n \\frac{d}{\\epsilon_0}C_{new}&=\\frac{d}{\\epsilon_0}C_1+\\frac{d}{\\epsilon_0}C_2+\\frac{d}{\\epsilon_0}C_3\\\\\n \\shortintertext{d is the same everywhere, and $\\epsilon_0$ is a constant so they both cancel}\\\\\n C_{new}&=C_1+C_2+C-3\\\\\n C_{new}&= C_{parallel}\\\\\n \\shortintertext{Capacitants increase when area is increased.}\n \\end{align*}\n What if instead we put capacitors one after another (in sequential order)? The area is the same, and the two middle sets of plates are going to have zero charge.\n \\begin{align*}\n d_{new}&=d_1+d_2+d_3\\\\\n \\epsilon_0\\frac{A}{C_{new}}&=\\epsilon_0\\frac{A}{C_1}+\\epsilon_0\\frac{A}{C_2}+\\epsilon_0\\frac{A}{C_3}\\\\\n \\frac{1}{C_{series}}&=\\frac{1}{C_1}+\\frac{1}{C_2}+\\frac{1}{C_3}+...\\\\\n C_{series}&=\\left[\\frac{1}{C_1}+\\frac{1}{C_2}+\\frac{1}{C_3}+...\\right]\\\\\n \\end{align*}\n Capacitance let's us calculate the energy stored inside of the field between its plates. This is similar to throwing a ball from a lower surface to a hgiher one. The kinetic energy is changed into potential energy. \n \\begin{align*}\n \\shortintertext{Recall that $U=VQ$.}\\\\\n \\shortintertext{Consider:}\\\\\n dU&=VdQ\\\\\n \\shortintertext{and $Q=CV$}\\\\\n \\to dU&=\\frac{Q}{C}dQ\\\\\n U&=\\frac{1}{C}\\int QdQ\\\\\n &=\\frac{1}{C}\\frac{1}{2}Q^2\\\\\n &=\\frac{1}{2}\\frac{1}{C}C^2V^2\\\\\n \\alignedbox{U}{=\\frac{1}{2}CV^2}\n \\shortintertext{Or}\\\\\n \\alignedbox{U}{=\\frac{1}{2}\\frac{Q^2}{C}}\\\\\n U&=\\frac{1}{2}Q^2\\frac{V}{Q}\\\\\n \\alignedbox{U}{\\frac{1}{2}QV}\\\\\n \\end{align*}\n Let's discuss energy density.\n \\begin{align*}\n u&=\\frac{U}{volume}\\\\\n \\shortintertext{This is always true, but for a parallel-plate capacitor, the volume is:}\n volume&=Ad\\\\\n \\shortintertext{Then,}\\\\\n U&=\\frac{1}{2}CV^2\\\\\n \\to u&=\\frac{1}{2}CV^2\\frac{1}{Ad}\\\\\n &=\\frac{1}{2}\\left(\\epsilon_0\\frac{A}{d}\\right)V^2\\frac{1}{Ad}\\\\\n u&=\\frac{\\epsilon_0}{2}\\frac{V^2}{d^2}\\\\\n u&=\\frac{\\epsilon_0}{2}\\left(\\frac{V}{d}\\right)^2\\\\\n \\shortintertext{Recall,}\\\\\n V&=Ed\\\\\n \\to E&=\\frac{V}{d}\\\\\n \\shortintertext{We find}\\\\\n \\alignedbox{u}{=\\frac{\\epsilon_0}{2}E^2}\\\\\n \\shortintertext{Energy density relies solely on the energy field between the plates.}\\\\\n \\end{align*}\n\n \\subsubsection{Example 1}\n Let's find the capacitance C of the demo capacitor. The area of the capacitor is A ($A=\\pi r^2$) for $r=6cm$. $A=\\pi 35cm^2$, also d is 2cm. We also need $\\epsilon_0$. This number is a constant. i$\\epsilon_0 = 8.85\\times10^{-12} \\frac{F}{m}$, where F stands for ferads. For parallel plates, $C=\\epsilon_0\\frac{A}{d}$. The units are ferads $(F)$.\n \\begin{align*}\n C&=\\left(8.85\\times10^{-12}\\frac{F}{m}\\right)\\left(\\frac{\\pi 36cm^2}{2cm}\\right)\\left(\\frac{1m}{100cm}\\right)\\\\\n \\alignedbox{C}{=5.00\\times10^{-12}F=5pF}\\\\\n \\shortintertext{C of the other two objects:}\\\\\n C_{blue}&=10\\mu F=10\\times10^{-6}F=10^{-5}F\\\\\n C_{brown}&=0.33\\mu F=0.33\\times10^{-6}F=3.3\\times10^{-7}\\\\\n \\end{align*}\n The capacitance of a capacitor is typically given within tolerances, not as raw numbers.\n Let's find the electric field around the first capacitor. If there is no charge on the plates, then there is no electric field between the plates, thus the capacitor is not storing any energy. An analogy to gravity: if there is no mass for two plates you're trying to calculate gravity for, then there is no gravity between the two plates. By adding electric charge to the plates, a field is created between them in which energy is stored.\n \\begin{align*}\n U&=\\frac{1}{2}CV^2\\\\\n Q&=CV\\\\\n U&=\\frac{1}{2}QV\\\\\n &=\\frac{1}{2}\\frac{Q^2}{C}\\\\\n \\shortintertext{Let's put a 9V battery across our capacitor.}\\\\\n U&=\\frac{1}{2}CV^2\\\\\n &=\\frac{1}{2}(5\\times10^{-12}F)(9V)^2\\\\\n U&=2.03\\times10^{-10}J\\\\\n \\alignedbox{U}{=.203nJ}\\\\\n \\end{align*}\n Consider a microwave oven. Many are powered at around 1200 Watts. We run it for 2 minutes. How much energy does this use? How much time is how much energy used?\n \\begin{align*}\n P&=\\frac{\\Delta E}{\\Delta T}\\\\\n \\Delta E &= P\\Delta T\\\\\n &=\\left(1200\\frac{J}{s}\\right)(120s)\\\\\n \\alignedbox{\\Delta E}{=144000J}\\\\\n &=144kJ\n \\end{align*}\n There are some benefits for something using a low amount of power. Let's recall that $C=5\\times10^{-5}$. Q of d would like instead $10\\times10^{-17}F$. d can combine capacitors. Consider $C=\\epsilon\\frac{A}{d}$. If you want more capacitance, you can increase the area. If you double the area you double the capacitance.\n \\begin{align*}\n C&=C_1+C_2\\\\\n &=2C\\\\\n &=10\\times10^{-12}\\\\\n \\end{align*}\n \\subsubsection{Example 2}\n There is a potential across the plates such that if a charge is placed you can easily calculate the potential energy and then the kinetic energy. Let's find the capacitants of something that is not necessarily parallel plates. We still need 2 plates, but they are not going to be flat plates. We are going to calculate the capacitance of a spherical capacitor.\n \\begin{align*}\n Q&=CV\\\\\n C_{parallelplates}=\\epsilon_0\\frac{A}{d}\\\\\n \\shortintertext{The capacitance of a capacitor is always fully defined.}\\\\\n \\shortintertext{For the spherical case, let's find C. Q is the charge on either plate, NOT BOTH. This is to prevent having zero as your Q value. The easiest way to find the voltage is by using Gauss' law to find $\\vec{E}$ to find the electric field then using that to solve for the volutage.}\\\\\n \\vec{E}&=0 \\text{ }rb\\\\\n \\shortintertext{For $a_=$I_0$] (4,4)\n to[resistor=$R_1$, i>_=$I_0$] (4,0)\n to[resistor=$R_2$, i>_=$I_0$] (0,0)\n ;\n \\end{circuitikz}\n \\end{figure}\n There is one current that describes this entire circuit. Let's let $V_0 = 11V$, $R_0=10k\\Omega$, $R_1=12k\\Omega$, and $R_2=35k\\Omega$. To get $I_0$ we need to know $R_eq$. These resistors are all in series with each other. Because they are all in series,we know that:\n \\begin{align*}\n R_{eq}&=R_0+R_1+R_2\\\\\n &=57k\\Omega\\\\\n \\shortintertext{Then with $V=IR$ we set:}\\\\\n V_0&=I_0R_{eq}\\\\\n \\alignedbox{I_0}{=\\frac{V_0}{R_{eq}}}\\\\\n I_0&=\\frac{11V}{57k\\Omega}=0.193mA\n \\end{align*}\n If you start somewhere in a gravatational field, and you move something down, it has changed in its potential energy. Because electric force is conservative, a charge that travels any path and then comes back to its original location has no net change in potential energy\n \\begin{align*}\n \\shortintertext{For a given loop,}\\\\\n \\sum qV_{loop}&=0\\\\\n \\sum V_{loop}&=0\\\\\n \\end{align*}\n To use this, pick any starting location. Take a complete path that's called a loop and consider the change in voltage across various circuit elements. For a battery or a power supply, if you travel from the negative to the positive, you have an increase in voltage across the battery. Converse also works. For a resistor, if you are traveling \\underline{with} the current, voltage will \\underline{drop} across the resistor. \n \\newline\\newline\n Let's try a clockwise loop from the top left corner.\n \\begin{align*}\n \\sum V_{loop} &=\\\\\n -I_0R_0-I_0R_1-I_0R_2+V_0&=0\\\\\n V_0&=I_0(R_0+R_1+R_2)\\\\\n V_{R_0}&=-I_0R_0=(-0.193mA)(10k\\Omega)\\\\\n V_{R_0}&=-1.93V\\\\\n V_{R_1}&=-2.32V\\\\\n V_{R_2}&=-6.76V\\\\\n \\alignedbox{\\sum V_{loop}}{=-11.0V}\\\\\n P_{R_0}&=IV\\\\\n IV&=0.37mW\\\\\n \\end{align*}\n \\subsubsection{Example 2}\n \\begin{figure}[!h]\n \\centering\n \\begin{circuitikz}\n \\draw\n (4,4) to[battery=$I_0$] (0,4) \n to[short, -*, i=$I_0$] (0,2)\n node[label={[font=\\footnotesize]left:X}] {}\n to[R=$R_1$, i>_=$I_1$] (4,2) \n node[label={[font=\\footnotesize]right:Y}] {} -- (4,4)\n (0,2) -- (0,0)\n to[R=$R_2$, i>_=$I_2$] (4,0) \n to[short, -*] (4,2)\n ;\n \\end{circuitikz}\n \\end{figure}\n \\begin{align*}\n \\sum V_{loop}&=0\\\\\n a&:-V_0+I_1R_1=0\\\\\n b&:-I_1R_1+I_2R_2=0\\\\\n \\alignedbox{\\sum I_{in}}{=\\sum I_{out}}\\\\\n x&:I_0=I_1+I_2\\\\\n y&:I_1+I_2=I_0\\\\\n \\end{align*}\n \\newline\n There are basically two rules when it comes to circuit analysis. $\\sum V_{loop} = 0$ and $\\sum I_{in}=I_{out}$. This is essentially conservation of energy and conservation of charge. Let's consider the following circuit.\n \\begin{figure}[!h]\n \\centering\n \\begin{circuitikz}\n \\draw\n (4,4) to[isource=$I_0$] (0,4) \n to[short, -*, i=$I_0$] (0,2)\n node[label={[font=\\footnotesize]left:X}] {}\n to[R=$R_1$, i>_=$I_1$] (4,2) \n node[label={[font=\\footnotesize]right:Y}] {} -- (4,4)\n (0,2) -- (0,0)\n to[R=$R_2$, i>_=$I_2$] (4,0) \n to[short, -*] (4,2)\n ;\n \\end{circuitikz}\n \\end{figure}\n Let's let $V_0=11V$, $R_1=12k\\Omega$, and $R_2=35k\\Omega$. First let's find the current directly out of the battery.\n \\begin{align*}\n R_{eq}&=\\frac{R_1R_2}{R_1+R_2}=8.94k\\Omega\\\\\n V_0&=I_0R_{eq}\\\\\n \\alignedbox{I_0}{=\\frac{V_0}{R_{eq}}}\\\\\n I_0&=\\frac{11V}{8.94k\\Omega}=1.23mA\\\\\n P_0=I_0V_0&=(1.23mA)(11V)=13.5mW\\\\\n \\shortintertext{Now let's find the current through each resistor. Let's start with going from the batter to $R_1$ and back (a). and then going from $R_2$ through $R_1$. The third loop could go from the battery to $R_2$ and back but it is not important to do so because all of the variables show up in these two equations already.}\\\\\n \\sum V_{loop}&=0\\\\\n a&) V_0-I_1R_1 = 0\\\\\n b&) I_1R_1-I_1R_2 = 0\\\\\n \\sum I_{in}&=\\sum I_{out}\\\\\n x&)I_0=I_1+I_2\\\\\n y&)I_1+I_2=I_0\\\\\n \\shortintertext{We have already done the physics, and now we must solve for the things that we do not know. This is just basic algebra.}\\\\\n \\to V_0&=I_1R_1\\\\\n I_1&=\\frac{V_0}{R_0}\\\\\n &=\\frac{11V}{12k\\Omega}\\\\\n \\alignedbox{I_1}{=0.917mA}\\\\\n \\to I_2&=I_0-I_1\\\\\n &=1.23mA-0.917mA\\\\\n \\alignedbox{I_2}{=0.313mA}\n \\shortintertext{Now let's find the power of these two resistors. Remember $V=IV=I^2R=\\frac{V^2}{R}$}\\\\\n P_1&=I_1^2R_1=(0.917mA)^2(12k\\Omega)\\\\\n P_1&=10.1mW\\\\\n P_2&=I_2^2R_2=(0.313mA)^2(35k\\Omega)\\\\\n P_2&=3.43mW\\\\\n \\shortintertext{Notice that $P_1+P_2=13.5mW$, which is the same amount of power that the batter puts out! Now let's find the voltage drops across $R_1+R_2$ from $IV$}\\\\\n V_1&=\\frac{P_1}{I_1}=\\frac{10.1mW}{0.916mA}\\\\\n V_1&=11.0V\\\\\n V_2&=\\frac{P_2}{I_2}=\\frac{3.43mW}{0.313mA}\\\\\n V_2&=11.0V\\\\\n \\shortintertext{Notice that $V_1=V_2=V_0$. We get the same voltage across all three.}\n \\end{align*}\n The voltage drops across parallel segments of circuit are \\underline{equal}. $V_0$, $R_1$, and $R_2$ are all in parallel, which means they all have the same voltage. We essentially did it with the first loop but we did not. Energy is conserved, but voltage is \\underline{NOT} conservative. To determine how to combine the resistors, you \\underline{must} first determine where the charge will flow.\n \\subsubsection{Example (Draw Later)}\n \\begin{figure}[h!]\n \\centering\n \\begin{circuitikz}\n \\draw (10,0) to[isource=$V_0$] (10,8) \n to[R=$R_0$](5,8) -- (0,8)\n to[R=$R_4$](0,4) \n to[R=$R_5$](0,0) -- (10,0)\n (5,8) to[R=$R_1$] (5,4) -- (7.5, 4) \n to[R=$R_3$](7.5,0)\n (5,4) -- (2.5, 4) \n to[R=$R_2$](2.5, 0);\n \\end{circuitikz} \n \\end{figure}\n There is a short within this circuit, which means it flows without resistance. Even with a short, you still calculate everything the same way. Let's find the current out of the battery and the power output of $R_0$\n \\begin{align*}\n \\sum V_{bat} &=0\\\\\n \\sum I_{in} &= \\sum I_{out}\\\\\n \\shortintertext{A quick way to get $I_0$ is using $V_0=I_0R_{eq}$. Let's find $R_{eq}$}\\\\\n R_{eq}&=R_0\\frac{R_{45}R_{123}}{R_{45}+R{123}}\\\\\n \\text{Also, } R_{45}&=R_4+R_5\\\\\n R_{123}&=R_1+\\frac{R_2R_3}{R_2+R_3}\\\\\n \\alignedbox{R_{eq}}{=R_0+\\frac{\\left(R_4+R_5\\right)\\left(R_1+\\frac{R_2R_3}{R_2+R_3}\\right)}{R_4+R_5+R_1+\\frac{R_2R_3}{R_2+R_3}}}\\\\\n \\end{align*}\n \\subsubsection{Using measuring tools}\n How to measure current using a digital multimeter.\\newline\\newline\n To measure current, we must carefullt add the ammeter to the circuit. To use an ammeter you must break the circuit. So in order to add an ammeter, first be sure the circuit is \\underline{not} currently powered! Plug the hole with the ammeter and \\underline{then} turn on the power supply.\\newline\\newline\n How to measure voltage using a DMM.\\newline\\newline\n Set the DMM to the voltmeter. Set the DMM to be a voltage. We then add the loads across the element whose voltage drop we want to measure. We must not break the circuit or turn off the power supply to measure the voltage.\\newline\\newline\n Ammeters go into circuits and have low internal resistance, while voltmeters go across circuit elements and have high internal resistance.\n \\subsection{Resistors and Capacitors in Circuits}\n \\begin{align*}\n V_R&=IR\\\\\n V_C=\\frac{Q}{C}\\\\\n \\shortintertext{When the swtich is closed:}\\\\\n V_R&=V_C=0\\\\\n \\shortintertext{Let's throw the switch to the position (a):}.\\\\\n \\shortintertext{There are no junctions and there is only one loop:}\\\\\n \\sum V_{loop} = 0\\\\\n V_0-IR-\\frac{Q}{C}&=0\\\\\n \\end{align*}\n How much charge acumulates on the plates? What is Q in terms of capacitance, resistance, and the battery. The amount of charge that accumulates within the circuit is a property of the circuit only, wheras battery voltage, resistance, and capacitance are functions of themselves. Using $I=\\frac{dQ}{dt}$, we get an equation that's a differential. We must get $dQ$ with $Q$ and we must get $dt$ with $t$.\n \\begin{align*}\n V_0-R\\frac{dQ}{dt}-\\frac{Q}{C}&=0\\\\\n V_0-\\frac{Q}{C}&=R\\frac{dQ}{dt}\\\\\n CV_0-Q&=RC\\frac{dQ}{dt}\\\\\n \\frac{1}{RC}&=\\frac{1}{CV_0-Q}\\frac{dQ}{dt}\\\\\n \\frac{dt}{RC}&=\\frac{dQ}{CV_0-Q}\\\\\n \\int_0^t\\frac{dt}{RC}&=\\int_0^Q\\frac{dQ}{CV_0-Q}\\\\\n \\shortintertext{Let $U=CV_0-Q$, and $du=-dQ$:}\\\\\n \\frac{1}{RC}(t-0)&=\\int_{CV_0}^{CV_0-Q}\\frac{-du}{u}\\\\\n \\frac{t}{RC}&=-ln\\left(\\frac{CV_0-Q}{CV_0}\\right)\\\\\n e^{-\\frac{t}{RC}}&=\\frac{CV_0-Q}{CV_0}\\\\\n CV_0e^{-\\frac{t}{RC}}&=CV_0-Q\\\\\n Q&=CV_0-CV_0e^{-\\frac{t}{RC}}\\\\\n \\alignedbox{Q(t)}{=CV_0\\left(1-e^{-\\frac{t}{RC}}\\right)}\\\\\n \\text{Charging} &\\text{ a capacitor}\\\\\n \\shortintertext{At $t=0$}\n Q(0)&=CV_0\\left(1-e^{-\\frac{0}{RC}}\\right)=0\\\\\n \\shortintertext{After a long time, $Q=CV_0$}\\\\\n V_0&=IR-\\frac{CV_0}{C}=0\\\\\n V_0-IR&=V_0\\\\\n \\shortintertext{This is only true if $I=0$. If we take the dervivative of our $Q(t)$ then we can see how current depends on time:}\\\\\n I(t)&=\\frac{d}{dt}Q(t)\\\\\n &=CV_0\\left(0-\\frac{-1}{RC}e^{-\\frac{t}{RC}}\\right)\\\\\n \\alignedbox{I(t)}{=\\frac{V_0}{R}e^{-\\frac{t}{RC}}}\\\\\n \\shortintertext{Charging a capacitor. Notice that RC has dimension of time.}\n \\end{align*}\n \\newpage", "meta": {"hexsha": "e8e0eca7da666934d21290eb16eefda9b1c415ec", "size": 26993, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "physics204/Sections/5Circuits.tex", "max_stars_repo_name": "CameronSWilliamson/GU-MATH", "max_stars_repo_head_hexsha": "a501bcb919b60bc35fa43b99eb6ed2a2630cb100", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-18T00:49:14.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-18T00:49:14.000Z", "max_issues_repo_path": "physics204/Sections/5Circuits.tex", "max_issues_repo_name": "therealkeyisme/Math-Notes", "max_issues_repo_head_hexsha": "a501bcb919b60bc35fa43b99eb6ed2a2630cb100", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "physics204/Sections/5Circuits.tex", "max_forks_repo_name": "therealkeyisme/Math-Notes", "max_forks_repo_head_hexsha": "a501bcb919b60bc35fa43b99eb6ed2a2630cb100", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.8082788671, "max_line_length": 868, "alphanum_fraction": 0.6493164895, "num_tokens": 9845, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434873426302, "lm_q2_score": 0.8267117855317474, "lm_q1q2_score": 0.6380921075720765}} {"text": "% section3\r\n\\chapter {Matrices and Matrix Algebra}\r\n\r\n\\section{Operations on Matrices}\r\n\r\nNo MATLAB problems in this section.\r\n\r\n\r\n\\section{Inverses; Algebraic Properties of Matrices}\r\n\r\n\\begin{exer}\r\nIn this problem, we compute $A^{5} - 3A^{3} + 7A - 4I$ for the matrix $A$, where\r\n$$\r\nA = \\left[\\begin{array}{rrrr} 1&\\hspace{3mm} 2&\\hspace{1mm} -3&\\hspace{2mm} 0\\\\ 1 & 1 & -2 & 1 \\\\ 2 & 1 & 3 & 4 \\\\ -3 & 2 & 2 & -8 \\end{array} \\right].\r\n$$\r\n\\begin{enumerate}\r\n\r\n\\vspace{2mm}\r\n\\item[(a)] Using the syntax $A$\\hspace{1mm}$\\hat{}$\\hspace{1mm}$k$ which produces the $k$-th power of a square matrix and the command \\textit{eye} for the identity matrix, compute the above matrix polynomial.\r\n\r\n\\vspace{1mm}\r\n\\item[(b)] Using the command \\textit{polyvalm}, compute the above matrix polynomial.\r\n\r\n\\vspace{1mm}\r\n\\item[(c)] Tell what happens if you type the syntax $A.$\\hspace{1mm}$\\hat{}$\\hspace{1mm}$k$.\r\n\r\n\\end{enumerate}\r\n\\end{exer}\r\n\r\n\\begin{sol}\r\n\r\n\\begin{verbatim}\r\n\r\n% Construct the matrix A.\r\nA = [1 2 -3 0; 1 1 -2 1; 2 1 3 4; -3 2 2 -8]; \r\n\r\n% (a)\r\nresult_a = A^5 + (-3)*A^3 + 7*A + (-4)*eye(4);\r\n\r\n% Display the matrix polynomial.\r\ndisp('The result of the matrix polynomial is');\r\ndisp(result_a) \r\n\r\n% (b)\r\n% Coefficient of the matrix polynomial.\r\ncoeff_poly = [1 0 -3 0 7 -4]; \r\n\r\n% Evaluate the matrix polynomial of coefficient\r\n% with coeff_poly vector with the input matrix A.\r\nresult_b = polyvalm(coeff_poly, A);\r\n\r\n% Display the matrix polynomial.\r\ndisp('The result of the matrix polynomial is');\r\ndisp(result_b);\r\n\r\n% (c)\r\ndisp('The result of A.^2 is'); disp(A.^2);\r\ndisp('The result of A.^3 is'); disp(A.^3);\r\ndisp('The result of A.^4 is'); disp(A.^4);\r\n\\end{verbatim}\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nThe result of the matrix polynomial is\r\n 874 -1272 -39 3021\r\n 2580 -2306 -723 7536\r\n 5191 -4121 -2444 14563\r\n -16852 12539 5649 -46917\r\n\r\nThe result of the matrix polynomial is\r\n 874 -1272 -39 3021\r\n 2580 -2306 -723 7536\r\n 5191 -4121 -2444 14563\r\n -16852 12539 5649 -46917\r\n\r\nThe result of A.^2 is\r\n 1 4 9 0\r\n 1 1 4 1\r\n 4 1 9 16\r\n 9 4 4 64\r\n\r\nThe result of A.^3 is\r\n 1 8 -27 0\r\n 1 1 -8 1\r\n 8 1 27 64\r\n -27 8 8 -512\r\n\r\nThe result of A.^4 is\r\n 1 16 81 0\r\n 1 1 16 1\r\n 16 1 81 256\r\n 81 16 16 4096\r\n\\end{verbatim}\r\n\\end{outputs}\r\n\r\n\\noindent From the results, we can see that the syntax $A.$\\hspace{1mm}$\\hat{}$\\hspace{1mm}$k$ produces the entrywise $k$-th powers of the matrix $A$.\r\n\r\n\\end{sol}\r\n\r\n\r\n\r\n\\section{Elementary Matrices; A Method for Finding $A^{-1}$}\r\n\r\n\r\n\r\n\\begin{exer}\r\n\r\nIn this problem, we solve the linear system $A \\mathbf{x} = \\mathbf{b}$ by using matrix inversion, where\r\n$$\r\nA = \\left[\\begin{array}{rrrr} 3 &\\hspace{2.5mm} 3 & -4 & -3 \\\\ 0 & 6 & 1 & 1\\\\ 5 & 4 & 2 & 1 \\\\ 2 & 3 & 3 & 2 \\end{array} \\right] \\hspace{2mm} \\mathrm{and} \\hspace{3mm} \\textbf{b} = \\left[\\begin{array}{r} -2 \\\\ 3 \\\\ 5 \\\\ 1 \\end{array} \\right].\r\n$$\r\n\r\n\\begin{enumerate}\r\n%8a\r\n\\item[(a)] Use the MATLAB command \\textit{inv} or the syntax $A$\\hspace{1mm}$\\hat{}$\\hspace{1mm}$(-1)$ to find the inverse of $A$.\r\n\\vspace{1mm}\r\n%8b\r\n\\item[(b)] Display the output matrix as a rational form, NOT decimally. You may use the command \\textit{format}.\r\n\\vspace{1mm}\r\n%8c\r\n\\item[(c)] Using the result of (a), compute the solution of the linear system $A \\mathbf{x} = \\mathbf{b}$ by taking $\\mathbf{x} = A^{-1} \\mathbf{b}$.\r\n\r\n\\end{enumerate}\r\n\r\n\\end{exer}\r\n\r\n\r\n\\begin{sol}\r\n\r\n\\begin{verbatim}\r\n\r\n% Construct the matrix A and the right-hand-side vector b.\r\nA = [3 3 -4 -3; 0 6 1 1; 5 4 2 1; 2 3 3 2]; \r\nb = [-2 3 5 1]'; \r\n\r\n% (a)\r\n% Use the command inv.\r\nInv_A1 = inv(A); \r\n\r\n% Use the syntax A^(-1).\r\nInv_A2 = A^(-1); \r\n\r\n% (b)\r\nformat rat; \r\ndisp('The result of the command inv is'); disp(Inv_A1);\r\ndisp('The result of the syntax A^(-1) is'); disp(Inv_A2);\r\n\r\n% (c)\r\n% Since A is invertible, the solution to Ax=b is x=A^(-1)*b.\r\nx = Inv_A1 * b;\r\ndisp('The solution to Ax=b is x = A^(-1)*b'); disp(x');\r\n\\end{verbatim}\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nThe result of the command inv is\r\n -7 5 12 -19\r\n 3 -2 -5 8\r\n 41 -30 -69 111\r\n-59 43 99 -159\r\n\r\nThe result of the syntax A^(-1) is\r\n -7 5 12 -19\r\n 3 -2 -5 8\r\n 41 -30 -69 111\r\n-59 43 99 -159\r\n\r\nThe solution to Ax=b is x = A^(-1)*b\r\n 70 -29 -406 583\r\n\\end{verbatim}\r\n\\end{outputs}\r\n\\end{sol}\r\n\r\n\r\n\r\n\\section{Subspaces and Linear Independence}\r\n\r\n\r\n\\begin{exer} (\\textit{Sigma notation})\\\\\r\nCompute the linear combination \r\n$$\\mathbf{v}=\\Sigma_{j=1}^{25} c_{j}\\mathbf{v}_{j}$$\r\nfor $c_{j}=1/j$ and $\\mathbf{v}_{j}=(\\sin j, \\cos j).$\r\n\r\n\\end{exer}\r\n\r\n\r\n\r\n\\begin{sol}\r\n\\begin{verbatim}\r\n\r\nv=zeros(1,2);\r\nfor i=1:25\r\n v=v+(1/i)*[sin(i), cos(i)];\r\nend\r\ndisp(v);\r\n\\end{verbatim}\r\n\r\n\r\n\r\n\\begin{outputs}\r\n\\begin{verbatim}\r\n\r\n1.0322 0.0553\r\n\\end{verbatim}\r\n\\end{outputs}\r\n\\end{sol}\r\n\r\n\\vspace{3mm}\r\n\r\n\r\n\\begin{exer} Let $\\mathbf{v_{1}}=(4, 3, 2, 1)$, $\\mathbf{v_{2}}=(5, 1, 2, 4)$, $\\mathbf{v_{3}}=(7, 1, 5, 3)$, $\\mathbf{x}=(16, 5, 9, 8)$, and $\\mathbf{y}=(3, 1, 2, 7)$. Determine whether $\\mathbf{x}$ and $\\mathbf{y}$ lie in $\\textrm{span}\\{\\mathbf{v_{1}}, \\mathbf{v_{2}}, \\mathbf{v_{3}}\\}$.\r\n\r\n\\end{exer}\r\n\r\n\r\n\\begin{sol}\r\n\r\n\\begin{verbatim}\r\n\r\n% Construct v1, v2, v3, x, y\r\nv1=[4 3 2 1]'; v2=[5 1 2 4]'; v3=[7 1 5 3]';\r\nx=[16 5 9 8]'; y=[3 1 2 7]';\r\n\r\n% Augmented matrices [v1|v2|v3|x] and [v1|v2|v3|y]\r\nX=[v1 v2 v3 x];\r\nY=[v1 v2 v3 y];\r\n\r\ndisp('Reduced row echelon form of [v1 v2 v3 x] is');\r\ndisp(rref(X));\r\ndisp('Reduced row echelon form of [v1 v2 v3 y] is');\r\ndisp(rref(Y));\r\n\\end{verbatim}\r\n\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nReduced row echelon form of [v1 v2 v3 x] is\r\n 1 0 0 1 \r\n 0 1 0 1 \r\n 0 0 1 1 \r\n 0 0 0 0 \r\n\r\nReduced row echelon form of [v1 v2 v3 y] is\r\n 1 0 0 0 \r\n 0 1 0 0 \r\n 0 0 1 0 \r\n 0 0 0 1 \r\n\\end{verbatim}\r\n\\end{outputs}\r\n\r\n\\noindent Therefore, $\\mathbf{x}$ lies in $\\textrm{span}\\{\\mathbf{v_{1}}, \\mathbf{v_{2}}, \\mathbf{v_{3}}\\}$ and $\\mathbf{y}$ does not lie in $\\textrm{span}\\{\\mathbf{v_{1}}, \\mathbf{v_{2}}, \\mathbf{v_{3}}\\}$. \r\n\\end{sol}\r\n\r\n\r\n%\\begin{exer} (\\textit{Linear Combinations})\\\\\r\n%Use the MATLAB command \\textit{rref} to express the vector $\\mathbf{b}=(-21, \\hspace{1mm}-60, \\hspace{1mm}-3, \\hspace{1mm}108, \\hspace{1mm}84)$ as a linear combination of $\\mathbf{v_{1}}$, $\\mathbf{v_{2}}$, and $\\mathbf{v_{3}}$ where $\\mathbf{v_{1}}=(1, \\hspace{1mm} -1, \\hspace{1mm}3, \\hspace{1mm}11, \\hspace{1mm}20)$, $\\mathbf{v_{2}}=(10, \\hspace{1mm}5, \\hspace{1mm}15, \\hspace{1mm}20, \\hspace{1mm}11)$, and $\\mathbf{v_{3}}=(3, \\hspace{1mm}3, \\hspace{1mm}4, \\hspace{1mm}4, \\hspace{1mm}9)$.\r\n%\r\n%\\end{exer}\r\n%\r\n%\r\n%\\begin{sol}\r\n%\r\n%\\begin{verbatim}\r\n%\r\n%% Construct b as a column vector.\r\n%b = [-21 -60 -3 108 84]'; \r\n%\r\n%% Set v1, v2, v3 as column vectors.\r\n%v1 = [1 -1 3 11 20]';\r\n%v2 = [10 5 15 20 11]';\r\n%v3 = [3 3 4 4 9]'; \r\n%\r\n%% Set a matrix A with column vectors v1, v2 and v3. \r\n%A = [v1 v2 v3];\r\n%\r\n%% Construct the augmented matrix [A | b].\r\n%augA = [A b]; \r\n%\r\n%% Find the reduced row echelon form of augA.\r\n%rref_augA = rref(augA); \r\n%\r\n%% Extract the solution vector from rref_augA.\r\n%x = rref_augA(1:3, 4); \r\n%\r\n%% From the result of rref_augA, we get rank(A) = rank([A | b]),\r\n%% hence, we can find each coefficient of this linear combination.\r\n%% Otherwise, we cannot find a linear combination of b as v1, v2, and v3.\r\n%\r\n%% Moreover, since rank(A) = rank([A | b]) = the number of columns of A,\r\n%% b is uniquely expressed as a linear combination of v1, v2, and v3.\r\n%\r\n%format rat % Display the result as an integer form.\r\n%disp('b is a linear combination of x(1)*v1+x(2)*v2+x(3)*v3, where');\r\n%disp('x(1) ='); disp(x(1)); \r\n%disp('x(2) ='); disp(x(2));\r\n%disp('x(3) ='); disp(x(3));\r\n%\\end{verbatim}\r\n%\r\n%\r\n%\\begin{outputs}\r\n%\r\n%\\begin{verbatim}\r\n%\r\n%b is a linear combination of x(1)*v1+x(2)*v2+x(3)*v3, where\r\n%x(1) =\r\n% 12\r\n%\r\n%x(2) =\r\n% 3\r\n%\r\n%x(3) =\r\n% -21\r\n%\\end{verbatim}\r\n%\\end{outputs}\r\n%\\end{sol}\r\n\r\n\\section{The Geometry of Linear Systems}\r\n\r\n\r\nNo MATLAB problems in this section.\r\n\r\n%%% Week2\r\n\\newpage\r\n\\section{Matrices with Special Forms}\r\n\r\n\\begin{exer} (\\textit{Inverting $(I-A)$})\r\n\\begin{enumerate}\r\n%7a\r\n\\item[(a)] (\\textit{Inverting $(I-A)$ when $A$ is nilpotent}) Using MATLAB, show that the matrix \r\n$$\r\nA = \\left[\\begin{array}{rrr} 2&\\quad 11&\\quad 3\\\\ -2 & -11 & -3\\\\ 8 & 35 & 9 \\end{array} \\right]\r\n$$\r\nis nilpotent, and then use Theorem~3.6.6 in the text book to compute $(I-A)^{-1}$. Check your answer by computing the inverse directly in MATLAB.\r\n\\vspace{1mm}\r\n\r\n\\item[(b)] (\\textit{Approximating $(I-A)^{-1}$ by a power series}) Using MATLAB, confirm that the matrix\r\n$$\r\nA = \\left[\\displaystyle\\begin{array}{rrr} 0&\\quad \\displaystyle\\frac{1}{4}&\\quad \\displaystyle\\frac{1}{8}\\\\ \\displaystyle\\frac{1}{4} & \\displaystyle\\frac{1}{8} & \\displaystyle\\frac{1}{10}\\\\ \\displaystyle\\frac{1}{8} & \\displaystyle\\frac{1}{10} & \\displaystyle\\frac{1}{10} \\end{array} \\right]\r\n$$\r\nsatisfies the condition in Theorem~3.6.7 of the text book. You may use the command \\textit{sum}. Since $A$ satisfies that condition, $(I-A)$ is invertible and can be expressed by the series in Formula~(18) in Section~3.6 of the text book. Compute the approximation $$(I-A)^{-1}\\approx I+A+A^2+A^3+\\cdots+A^{10},$$ and compare it with the inverse of $I-A$ produced directly by MATLAB. To how many decimal places do the results agree? You may use the command \\textit{format} to display the output with long digits.\r\n\r\n\\end{enumerate}\r\n\\end{exer}\r\n\r\n\r\n\\begin{sol}\r\n\\verb\"\"\r\n\\begin{enumerate}\r\n\\item[(a)]\r\n\\begin{verbatim}\r\n% (a)-i\r\nA = [ 2 11 3 ; -2 -11 -3; 8 35 9]; % Construct the matrix A.\r\n% Compute the A^2, A^3, ... , and display.\r\ndisp('A^2 is'); disp(A^2);\r\ndisp('A^3 is'); disp(A^3);\r\n\r\n% (a)-ii Comparing two result\r\n\r\n% By Theorem 3.6.6, (I-A)^(-1)=I+A+A^2.\r\nresult1=eye(3)+A+A^2; \r\n\r\n% Compute the inverse of (I-A) directly.\r\nresult2=inv(eye(3)-A);\r\ndisp('I+A+A^2 is'); disp(result1);\r\ndisp('(I-A)^(-1) is'); disp(result2);\r\n\r\n% Display as a rational form.\r\nformat rat;\t\r\ndisp('Rational form of (I-A)^(-1) is');disp(result2);\r\n\\end{verbatim}\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nA^2 is\r\n 6 6 0\r\n -6 -6 0\r\n 18 18 0\r\n\r\nA^3 is\r\n 0 0 0\r\n 0 0 0\r\n 0 0 0\r\n\r\nI+A+A^2 is\r\n 9 17 3\r\n -8 -16 -3\r\n 26 53 10\r\n\r\n(I-A)^(-1) is\r\n 9.0000 17.0000 3.0000\r\n -8.0000 -16.0000 -3.0000\r\n 26.0000 53.0000 10.0000\r\n\r\nRational form of (I-A)^(-1) is\r\n 9 17 3 \r\n -8 -16 -3 \r\n 26 53 10 \r\n\\end{verbatim}\r\n\r\n\\end{outputs}\r\n\r\n\\noindent Since $A^{3} = \\mathbf{0}$, $A$ is nilpotent. \r\nBy the Theorem $3.6.6$, since $A^{3} = \\mathbf{0}$, $I-A$ is invertible and $(I-A)^{-1} = I + A + A^{2}.$ To check answer by computing the inverse directly in MATLAB, we implement as in the next page.\r\n\r\n\r\n\\item[(b)]\r\n\\begin{verbatim}\r\n% Construct the matrix A.\r\nA=[0 1/4 1/8; 1/4 1/8 1/10; 1/8 1/10 1/10]; \r\n\r\n% Check that the condition in Theorem 3.6.7 \r\n% of the text book is satisfied for matrix A.\r\ncolumn_sum=sum(abs(A),1); % column-wise sum \r\nrow_sum=sum(abs(A),2); % row-wise sum\r\ndisp('The sum of the absolute values of the entries in each column is');\r\ndisp(column_sum);\r\ndisp('The sum of the absolute values of the entries in each row is');\r\ndisp(row_sum);\r\n\r\nresult3=eye(size(A))+A+A^2+A^3+A^4+A^5+A^6+A^7+A^8+A^9+A^10;\r\nresult4=inv(eye(3)-A);\r\n\r\nformat long;\t% Display the result with long digits\r\ndisp('With format long');\r\ndisp('Approximated inv(I-A) is'); disp(result3);\r\ndisp('Exact inv(I-A) is'); disp(result4);\r\n\\end{verbatim}\r\n\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nThe sum of the absolute values of the entries in each column is\r\n 3/8 19/40 13/40 \r\n\r\nThe sum of the absolute values of the entries in each row is\r\n 3/8 \r\n 19/40 \r\n 13/40 \r\n\r\nWith format long\r\nApproximated inv(I-A) is\r\n 1.108587459181130 0.338615080927493 0.191581699462210\r\n 0.338615080927493 1.260966638806045 0.187122081247432\r\n 0.191581699462210 0.187122081247432 1.158500720998029\r\n\r\nExact inv(I-A) is\r\n 1.108610894508188 0.338643199287067 0.191600757491367\r\n 0.338643199287067 1.261000334187368 0.187144925921800\r\n 0.191600757491367 0.187144925921800 1.158516208087334\r\n\\end{verbatim}\r\n\r\n\\end{outputs}\r\n\r\n\r\n\\noindent The approximation result agrees with the exact result to 2 decimal places.\r\n\\end{enumerate}\r\n\\end{sol}\r\n\r\n\r\n\\section{Matrix Factorizations; $LU$-Decomposition}\r\n\r\n\\begin{exer}(\\textit{LU-decompositions})\r\nIn this problem, we find an $LU$-decomposition of $A$, where $A$ is given in the \\mbox{Example $2$} of the Section $3.7$.\r\n\r\n\\vspace{2mm}\r\n\\begin{enumerate}\r\n\\item[(a)] Find an $LU$-decomposition of $A$ by following the procedure given in the Example $2$.\r\n\\vspace{1mm}\r\n\\item[(b)] Solve the linear system $A \\mathbf{x} = \\mathbf{b}$ by using the $LU$-decomposition of $A$ obtained in (a), where $\\textbf{b} = \\left[\\begin{array}{r} 0 \\\\ -2 \\\\ 1 \\end{array} \\right].$\r\n\\vspace{1mm}\r\n\\item[(c)] Tell what happens if you use the MATLAB command \\textit{lu} of $A$. Explain why this result differs from the result in (a).\r\n\r\n\\end{enumerate}\r\n\r\n\\end{exer}\r\n\r\n\r\n\\begin{sol}\r\n\r\n\\begin{verbatim}\r\n\r\n%(a)\r\nA = [6 -2 0; 9 -1 1; 3 7 5]; % Set the matrix A.\r\n\r\nformat rat; % Display results as a rational form.\r\n\r\n% Initialization of U and L.\r\nU = A; L = eye(3); \r\n\r\n% Multiply the first row by 1/6.\r\nU(1,:)=(1/6)*U(1,:); \r\n% L(1,1) is the inverse of 1/6.\r\nL(1,1)=(1/6)^(-1); \r\n\r\n% Add (-9) times the first to the second.\r\nU(2,:)=((-9)*U(1,:))+U(2,:);\r\n% L(2,1) is the negative of (-9).\r\nL(2,1)=-(-9); \r\n\r\n% Add (-3) times the first to the third.\r\nU(3,:)=((-3)*U(1,:))+U(3,:); \r\n% L(3,1) is the negative of (-3).\r\nL(3,1)=-(-3); \r\n\r\n% Multiply the second row by 1/2.\r\nU(2,:)=(1/2)*U(2,:);\r\n% L(2,2) is the inverse of 1/2.\r\nL(2,2)=(1/2)^(-1); \r\n\r\n% Add (-8) times the second to the third.\r\nU(3,:)=((-8)*U(2,:))+U(3,:); \r\n% L(3,2) is the negative of (-8).\r\nL(3,2)=-(-8); \r\n\r\ndisp('A is'); disp(A);\r\ndisp('The Lower Triangular part L is'); disp(L);\r\ndisp('The Upper Triangular part U is'); disp(U);\r\ndisp('The product L*U is'); disp(L*U);\r\n\r\n%(b)\r\n% Solve the linear system Ax=b \r\n% by using the LU-decomposition obtained in (a).\r\n\r\n% First, let us solve L*y = b by forward substitution.\r\n % Set the right-hand-side vector b.\r\nb = [0 -2 1]';\r\n\r\n % Initialization of the solution vector y.\r\ny = zeros(3, 1);\r\ny(1) = b(1) / L(1, 1);\r\ny(2) = (b(2) - (L(2, 1)*y(1))) / L(2, 2);\r\ny(3) = (b(3) - (L(3, 1)*y(1)) - (L(3, 2)*y(2))) / L(3, 3);\r\n\r\n% Next, let us solve U*x = y by backward substitution.\r\nx = zeros(3, 1); % Initialization of the solution vector x.\r\nx(3) = y(3) / U(3, 3);\r\nx(2) = (y(2) - (U(2, 3)*x(3))) / U(2, 2);\r\nx(1) = (y(1) - (U(1, 3)*x(3)) - (U(1, 2)*x(2))) / U(1, 1);\r\n\r\ndisp('The solution to Ax=b by the LU-decomposition is'); disp(x');\r\n\r\n% (c)\r\nfprintf('Using MATLAB command lu\\n');\r\n% LU decomposition of A with a permutation matrix.\r\n[L U P] = lu(A);\r\n\r\ndisp('Lower triangular part L is'); disp(L);\r\ndisp('Upper triangular part U is'); disp(U);\r\ndisp('The permutation matrix P is'); disp(P);\r\ndisp('PA='); disp(P*A); disp('LU='); disp(L*U);\r\n\r\n\\end{verbatim}\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\nA is\r\n 6 -2 0\r\n 9 -1 1\r\n 3 7 5\r\n\r\nThe Lower Triangular part L is\r\n 6 0 0\r\n 9 2 0\r\n 3 8 1\r\n\r\nThe Upper Triangular part U is\r\n 1 -1/3 0\r\n 0 1 1/2\r\n 0 0 1\r\n\r\nThe product L*U is\r\n 6 -2 0\r\n 9 -1 1\r\n 3 7 5\r\n\r\nThe solution to Ax=b by the LU-decomposition is\r\n -11/6 -11/2 9\r\n\r\nUsing MATLAB command lu\r\n\r\nLower triangular part L is\r\n 1 0 0\r\n 1/3 1 0\r\n 2/3 -2/11 1\r\nUpper triangular part U is\r\n 9 -1 1\r\n 0 22/3 14/3\r\n 0 0 2/11\r\nThe permutation matrix P is\r\n 0 1 0\r\n 0 0 1\r\n 1 0 0\r\nPA=\r\n 9 -1 1\r\n 3 7 5\r\n 6 -2 0\r\nLU=\r\n 9 -1 1\r\n 3 7 5\r\n 6 -2 0\r\n\r\n\\end{verbatim}\r\n\r\n\\end{outputs}\r\n\r\n\\noindent Since the permutation matrix $P$ is not the identity matrix, the MATLAB command \\textit{lu} gave us an $LU$-decomposition after multiplying $A$ by the permutation matrix $P$, hence, this decomposition is a $PLU$-decomposition of $A$ because $PA=LU$. Since at least one row interchange of $A$ occurred in the process of $LU$-decomposition, this result is different from the previous decomposition obtained in $(a)$.\r\n\r\n\\end{sol}\r\n\r\n\r\n\\vspace{3mm}\r\n\\begin{exer} (\\textit{LU-decomposition})\r\n\\begin{enumerate}\r\n\\item[(a)] The MATLAB command \\textit{lu} is used to find the $LU$-decomposition of a matrix $A$. Tell what happens if you use the command $lu$ for $A$, where $A$ is given in the Example 2 of the Section 3.7. Explain why this result differs from the result in the textbook.\r\n\\vspace{1mm}\r\n\\item[(b)] Using MATLAB, observe what happens when you try to find an $LU$-decomposition of a singular matrix.\r\n\r\n\\end{enumerate}\r\n\\end{exer}\r\n\r\n\\begin{sol}\r\n\r\n\\begin{verbatim}\r\n\r\n% (a)\r\n% Construct the matrix A.\r\nA=[6 -2 0; 9 -1 1; 3 7 5]; \r\n\r\n% LU decomposition of A.\r\n[L U P]=lu(A); \r\ndisp('[L U P]=lu(A)');\r\ndisp('L'); disp(L); disp('U'); disp(U); disp('P'); disp(P);\r\n\r\n% (b)\r\n% Construct the some singular matrices.\r\nA1=[1 0 0; -2 0 0; 4 6 1]; \r\nA2=[1 -2 7; -4 8 5; 2 -4 3];\r\nA3=[1 0 0; -2 0 0; 4 6 1]; \r\n\r\n% LU decompositions of them.\r\n[L1 U1 P1]=lu(A1); [L2 U2 P2]=lu(A2); [L3 U3 P3]=lu(A3); \r\ndisp('[L1 U1 P1]=lu(A1)'); disp('L1');disp(L1);disp('U1');disp(U1);\r\ndisp('[L2 U2 P2]=lu(A2)'); disp('L2');disp(L2); disp('U2');disp(U2);\r\ndisp('[L3 U3 P3]=lu(A3)'); disp('L3');disp(L3); disp('U3');disp(U3);\r\n\\end{verbatim}\r\n\r\n\\begin{outputs}\r\n\r\n\\begin{verbatim}\r\n\r\n[L U P]=lu(A)\r\nL\r\n 1.0000 0 0\r\n 0.3333 1.0000 0\r\n 0.6667 -0.1818 1.0000\r\n\r\nU\r\n 9.0000 -1.0000 1.0000\r\n 0 7.3333 4.6667\r\n 0 0 0.1818\r\n\r\nP\r\n 0 1 0\r\n 0 0 1\r\n 1 0 0\r\n\r\n\r\n[L1 U1 P1]=lu(A1)\r\nL1\r\n 1.0000 0 0\r\n -0.5000 1.0000 0\r\n 0.2500 -0.5000 1.0000\r\nU1\r\n 4.0000 6.0000 1.0000\r\n 0 3.0000 0.5000\r\n 0 0 0\r\n \r\n[L2 U2 P2]=lu(A2)\r\nL2\r\n 1.0000 0 0\r\n -0.2500 1.0000 0\r\n -0.5000 0 1.0000\r\nU2\r\n -4.0000 8.0000 5.0000\r\n 0 0 8.2500\r\n 0 0 5.5000\r\n \r\n[L3 U3 P3]=lu(A3)\r\nL3\r\n 1.0000 0 0\r\n -0.5000 1.0000 0\r\n 0.2500 -0.5000 1.0000\r\nU3\r\n 4.0000 6.0000 1.0000\r\n 0 3.0000 0.5000\r\n 0 0 0\r\n \r\n\\end{verbatim}\r\n\\end{outputs}\r\n\r\n\\noindent \\textit{Remark on (a).} Since the permutation matrix $P$ is not the identity matrix, the MATLAB command \\textit{lu} gave us an $LU$-decomposition after multiplying $A$ by the permutation matrix $P$, hence, this decomposition is a $PLU$-decomposition of $A$ because $PA=LU$. Since at least one row interchange of $A$ occurred in the process of $LU$-decomposition, this result is different from the decomposition result in the textbook.\r\n\r\n\\vspace{2mm}\r\n\\noindent \\textit{Remark on (b).} When we try $LU$-decomposition of the sigular matrices using the MATLAB command \\textit{lu}, the resulting upper triangular matrices are singular.\r\n\r\n\\end{sol}", "meta": {"hexsha": "983c54ac603ab403bc5a3784ddf5adf6eecea95b", "size": 20934, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "_codes/intro/Learning MATLAB with Linear Algebra (Jeon, Lee)/section3.tex", "max_stars_repo_name": "mireiffe/mas109_matlab_2021_2", "max_stars_repo_head_hexsha": "f955eb2789b463d8cffbfbbb321bcd057d32933a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "_codes/intro/Learning MATLAB with Linear Algebra (Jeon, Lee)/section3.tex", "max_issues_repo_name": "mireiffe/mas109_matlab_2021_2", "max_issues_repo_head_hexsha": "f955eb2789b463d8cffbfbbb321bcd057d32933a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-09-19T08:29:55.000Z", "max_issues_repo_issues_event_max_datetime": "2021-09-19T08:29:55.000Z", "max_forks_repo_path": "files/intro/Learning MATLAB with Linear Algebra (Jeon, Lee)/section3.tex", "max_forks_repo_name": "mireiffe/mas109_matlab_2021_2", "max_forks_repo_head_hexsha": "f955eb2789b463d8cffbfbbb321bcd057d32933a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.1559888579, "max_line_length": 513, "alphanum_fraction": 0.5418935703, "num_tokens": 7519, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324983301568, "lm_q2_score": 0.7826624738835051, "lm_q1q2_score": 0.6380518839333109}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{graphicx}\n\\usepackage{multicol}\n\\usepackage[top=0.8in, bottom=0.8in, left=0.8in, right=0.8in]{geometry}\n% add other packages here\n\\usepackage[font=small,labelfont=bf]{caption} % Required for specifying captions to tables and figures\n\n\n% put your group number and names in the author field\n\\title{\\bf Exercise 2: A Reactive Agent for the Pickup and Delivery Problem}\n\\author{Group \\textnumero 70: Jelena Banjac, Stratos Triantafyllou}\n\n% the report should not be longer than 3 pages\n\n\\begin{document}\n\\maketitle\n\n\n\\section{Problem Representation}\n\n\\subsection{Representation Description}\n% describe how you design the state representation, the possible actions, the reward table and the probability transition table\n\nOur implementation assumes as state a route from an origin city to a destination city. The next city the vehicle will visit next is modeled as an action. If a vehicle on a route is carrying out a delivery task, no destination city is defined - implemented with the method \\texttt{hasTask()}.\n\nThis flag is used in the creation of the reward table. If the vehicle is carrying out a delivery task next, then the added reward R(s, a) is computed based on the income for the delivery from the destination city in s to the next city as defined in the action a, subtracting the distance between these two cities. If the vehicle is just moving to the next city without delivering a task, we just subtract the distance between the cities.\n\nTo compute transition probabilities, if the vehicle is carrying out a delivery task, we use the probability that the agent moves between the origin and the destination city. If the vehicle is just moving to a neighbor city, each neighbor has the same probability and the transition is random.\n\n\n\\subsection{Implementation Details}\n% describe the implementation details of the representations above and the implementation details of the reinforcement learning algorithm you implemented\nImplementation of the state is in the \\texttt{state.RouteState} with parameters \\texttt{fromCity} and \\texttt{toCity}. The initialization of all the states is done in the method \\texttt{initializeStates}. In total we have ${N}^2$ states. \nThe action is implemented in the class \\texttt{action.RouteAction} with parameter \\texttt{neighborCity}. The method \\texttt{initializeActions} is initializing all the actions. In total we have N actions.\nBefore Reinforcement Learning Algorithm (RLA), we fill in the Reward table - \\texttt{R(s,a)}. Reward is calculated differently depending whther the agent has a tas or not. We also initialize Q-table - \\texttt{Q(s,a)} together with the vector with state values and corresponding action to take in that state - \\texttt{V}. When Q table is initialized, it does not use all the combinations of state and action. The constraints are: the agent cannot stay in the same city and if there is no delivery, the action needs to be a neighbor city.\nThese tables are all located in the \\texttt{tables.Tables}.\nWe compute this vector in method \\texttt{rla()} by value iteration. This value improves with each iteration, and our stopping criteria is until the difference between two successive iteration is smaller than $\\epsilon$=0.001.\n\n\n\\section{Results}\n% in this section, you describe several results from the experiments with your reactive agent\nIn this section, we will describe several results from the experiments with our reactive agent.\n\n\\subsection{Experiment 1: Discount factor}\n% the purpose of this experiment is to understand how the discount factor influences the result\nIn the first experiment, we want to understand how discount factor $\\gamma$ influences the results. The discount factor represents how much we discount the future, i.e. how much an agent is taking care of rewards from the future states. If the $\\gamma$=0, it means that we discard the future totally. Whereas, the higher the value of $\\gamma$, the higher the effect of the future on the value of state value V(S). \n\n\\subsubsection{Setting}\n% you describe how you perform the experiment (you also need to specify the configuration used for the experiment)\nThis experiment was performed with the following settings:\n(1) discount factor for our reactive agent: $\\gamma$ $\\epsilon$ [0.15, 0.50, 0.85, 0.95, 0.99]; (2) cost per km for all vehicles is 5 CHF; (3) capacity of all the vehicles is 30 KG; (4) speed of all the vehicles vehicle is 220 KM/H; (5) topology is Switzerland.\n\n\n\\subsubsection{Observations}\n% you describe the experimental results and the conclusions you inferred from these results\nNumber of iterations of RLA algorithm until good enough is: [6, 9, 11, 9, 7] for the discount factors $\\gamma$ $\\epsilon$ [0.99, 0.95, 0.95, 0.85, 0.50, 0.15], respectively. We run several times (3 times) in order to get a confidence intervals of total profits and reward per kilometer results. Using this information we can find the best value of $\\gamma$. \n\n\n\\begin{multicols}{2}\n\\includegraphics[scale=0.2]{ex1-rpk-full.png}\n\\captionof{figure}{Reward per kilometer - full image}\n\n\\includegraphics[scale=0.2]{ex1-rpk-zoomed.png}\n\\captionof{figure}{Reward per kilometer - zoomed}\n\\end{multicols}\n\n\\begin{multicols}{2}\n\\includegraphics[scale=0.2]{ex1-tp-full.png}\n\\captionof{figure}{Total profit - full image}\n\n\\includegraphics[scale=0.2]{ex1-tp-zoomed.png}\n\\captionof{figure}{Total profit - zoomed}\n\\end{multicols}\n\nFrom the images above, we can see that with our settings the agent that performs the best is the one with the discount factor $\\gamma$ = 0.85. It means that it takes into an account the future pretty highly.\n\n\\subsection{Experiment 2: Comparisons with dummy agents}\n% you compare the results of your agent with two dummy agents: the random agent that was already given in the starter files and another dummy agent that you define and create. You should report the results from the simulations using the topologies given in the starter files and optionally, additional topologies that you create.\nIn this experiment, we compare the results of our agent with two dummy agents: the random agent in the starter files and another dummy agent that has $\\gamma$=0. We perform this experiment on different typologies.\n\n\\subsubsection{Setting}\n% you describe how you perform the experiment and you describe the dummy agent you created (you also need to specify the configuration used for the experiment)\nThe first dummy agent is a random agent does not implement the Reinforcement Learning Algorithm (RLA) and chooses the next path to move randomly. The second dummy agent is the agent that discards the effect of the future states, therefore, it acts only to maximize an immediate reward. The third agent is our final agent that performs the best.\nAdditional settings we have are: (1) cost per km for all vehicles is 5 CHF; (2) capacity of all the vehicles is 30 KG; (3) speed of all the vehicles vehicle is 220 KM/H\n\n\n\\subsubsection{Observations}\n% elaborate on the observed results\n\n\\begin{multicols}{2}\n\\includegraphics[scale=0.18]{ex2-rpk-full.png}\n\\captionof{figure}{Switzerland topology - Reward per kilometer (full image)}\n\n\\includegraphics[scale=0.18]{ex2-rpk-zoomed.png}\n\\captionof{figure}{Switzerland - Reward per kilometer (zoomed)}\n\\end{multicols}\n\n\\begin{multicols}{2}\n\\includegraphics[scale=0.19]{ex2-tp-full.png}\n\\captionof{figure}{Switzerland - Total profit (full image)}\n\n\\includegraphics[scale=0.19]{ex2-tp-zoomed.png}\n\\captionof{figure}{Switzerland - Total profit (zoomed)}\n\\end{multicols}\nFrom the images of the performance of agents in Swiss topology, we can see that random agent (green) performs worse than the ones that implement the RLA, see Fig.5, Fig.6, Fig.7, and Fig.8. Even the agent that is using the discount factor $\\gamma$=0 is performing better than the random one (which is expected since it is maximizing an immediate reward and does not go randomly). The agent which we think was the best was the one with $\\gamma$=0.85 (blue). \n\n\\begin{multicols}{2}\n\\includegraphics[scale=0.19]{fr1.png}\n\\captionof{figure}{France - Total profit (zoomed)}\n\n\\includegraphics[scale=0.19]{fr2.png}\n\\captionof{figure}{France - Reward per kilometer (full)}\n\\end{multicols}\nHowever, when we ran the agents in a different topology (France), we got that the random one still performs the worse, however, the agent with $\\gamma$=0.95 performs better than the one with $\\gamma$=0.85. And the agent that gets the immediate reward has performance with the result in between these agents, see Fig.9 and Fig.10. We did the same for the topology of England. The discount factor that performs the best there was $\\gamma$=0.5. More information can be found in the notebooks.\n\n\\end{document}", "meta": {"hexsha": "5d2e99e6a06068e45c041fdfc882b05d6080790a", "size": 8738, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2-reactive/banjac-triantafyllou-ex2/doc/banjac-triantafyllou-ex2.tex", "max_stars_repo_name": "JelenaBanjac/intelligent-agents", "max_stars_repo_head_hexsha": "c6a7f412c9c60b50a1f312d1ceab71cbe654a5bc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-12-14T13:32:26.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-14T13:32:26.000Z", "max_issues_repo_path": "2-reactive/banjac-triantafyllou-ex2/doc/banjac-triantafyllou-ex2.tex", "max_issues_repo_name": "JelenaBanjac/intelligent-agents", "max_issues_repo_head_hexsha": "c6a7f412c9c60b50a1f312d1ceab71cbe654a5bc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2-reactive/banjac-triantafyllou-ex2/doc/banjac-triantafyllou-ex2.tex", "max_forks_repo_name": "JelenaBanjac/intelligent-agents", "max_forks_repo_head_hexsha": "c6a7f412c9c60b50a1f312d1ceab71cbe654a5bc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.4285714286, "max_line_length": 536, "alphanum_fraction": 0.7823300526, "num_tokens": 2110, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.8152324915965392, "lm_q1q2_score": 0.6380518827959273}} {"text": "\\RequirePackage[l2tabu, orthodox]{nag}\n\\documentclass[landscape, letterpaper, 8pt]{extarticle}\n\n\\usepackage[margin=.25in]{geometry}\n\\usepackage{fontspec}\n\\usepackage{amssymb}\n\\usepackage{mathtools}\n\\usepackage{multicol}\n\\usepackage{enumitem}\n\\usepackage{scrextend}\n\n%\\setmainfont{Lato}\n%\\linespread{0.2}\n\n\\renewcommand{\\complement}[1]{{#1}^\\mathsf{c}}\n\\DeclarePairedDelimiter{\\floor}{\\lfloor}{\\rfloor}\n\n\\begin{document}\n\\begin{multicols}{3}\n \\section*{Logic Properties}\n \\begin{itemize}[noitemsep,nolistsep]\n \\item \\textbf{Generalization: } $p~\\therefore p \\lor q$ and $q~\\therefore p \\lor q$\n \\item \\textbf{Specialization: } $p \\land q~\\therefore p$ or $q$\n \\item \\textbf{Conjunction: } $p,~q~\\therefore p \\land q$\n \\item \\textbf{Eliminiation: } $p \\lor q,~\\neg q~\\therefore p$\n \\item \\textbf{Division into Cases: } $p \\lor q,~p\\implies r,~q\\implies r~\\therefore r$\n \\item \\textbf{Contradiction Rule: } $\\neg p \\implies c~\\therefore p$\n \\end{itemize}\n \\subsection*{Implies}\n \\begin{itemize}[noitemsep,nolistsep]\n \\item \\textbf{Statement: } $p \\implies q$\n \\item \\textbf{Converse: } $q \\implies p$\n \\item \\textbf{Inverse: } $\\neg p \\implies \\neg q$\n \\item \\textbf{Contrapositive: } $\\neg q \\implies \\neg p$\n \\end{itemize}\n \\subsection*{Quantified}\n \\begin{itemize}[noitemsep,nolistsep]\n \\item \\textbf{Universal Instantiation: } $\\forall x,~P(x)~\\therefore P(a)$ also applies to $\\exists$.\n \\item \\textbf{Universal Generalization: }$P(a)~\\therefore \\forall x,~P(x)$ also applies to $\\exists$.\n \\item \\textbf{Negation: }$\\neg(\\forall x, Q(X)) \\equiv \\exists x, \\neg Q(x)$ and $\\neg(\\exists x, Q(X)) \\equiv \\forall x, \\neg Q(x)$.\n \\item \\textbf{Universal Modus Ponens: } $\\forall x, P(x)\\implies Q(x),~P(a)~\\therefore Q(a)$ does not apply to $\\exists$.\n \\item \\textbf{Universal Modus Tolens: } $\\forall x, P(x)\\implies Q(x),~\\neg Q(a) \\therefore \\neg P(a)$ does not apply to $\\exists$.\n \\item \\textbf{Converse Error: }$ \\begin{aligned}\n & \\forall x, P(x) \\implies Q(x) \\\\\n & Q(a) \\\\\n \\therefore & P(a)\n \\end{aligned}$\n \\item \\textbf{Inverse Error: } $\\begin{aligned}\n & \\forall x, P(x) \\implies Q(x) \\\\\n & \\neg P(a) \\\\\n \\therefore & Q(a)\n \\end{aligned}$\n\n \\end{itemize}\n \\section*{Induction}\n \\subsection*{Strong Induction}\n \\textbf{Let} $P(n)$ be the property to be proven, let $a$ and $b$ be fixed integers where $a\\leq b$.\n\n Suppose the following:\n \\begin{enumerate}[noitemsep,nolistsep]\n \\item \\textbf{Basis:} $P(a), P(a+1), \\ldots P(b)$ are true.\n \\item \\textbf{Induction:} For all integers $k\\geq b$, if $P(i)$ is true for all integers from $a$ to $k$ then $P(k+1)$ is true.\n \\item Hence for all integers $n\\geq a, P(n)$ is true.\n \\end{enumerate}\n \\section*{Number Theory}\n \\begin{itemize}[noitemsep,nolistsep]\n \\item \\textbf{Even} $n$ is even $\\iff \\exists k, n = 2k$.\n \\item \\textbf{Odd} $n$ is odd $\\iff \\exists k, n = 2k+1$.\n \\item \\textbf{Prime} $n$ is prime $\\iff \\forall r, s$, if $n = rs$ then $r=1$ and $s = n$ or vice versa.\n \\item \\textbf{Composite} $n$ is composite $\\iff \\exists r, s$, if $n = rs$ and $1 < r < s$ and $1 < s < n$.\n \\item \\textbf{Rational} $r$ is rational $\\iff \\exists a, b$ such that $r = \\frac{a}{b}$ and $b \\neq 0$.\n \\item \\textbf{Divisible} $d|n \\iff \\exists k$ such that $n=dk$.\n \\end{itemize}\n \\section*{Sets}\n Can be proven by using Induction, Algebra or Element method.\n\n Element Method: Usually with equality or subset relations: prove that an element in LHS is an element in RHS.\n \\begin{itemize}[noitemsep,nolistsep]\n \\item Proper Subset: $A \\subseteq B$ \\textbf{AND} there is one element in B that is \\textbf{not} in A\n \\item Power Set: $\\wp(\\{x, y, z\\}) = \\{\\emptyset, \\{x\\}, \\{y\\}, \\{z\\},\\\\\n \\{x, y\\},\\{x, z\\}, \\{y, z\\}, \\{x, y, z\\}$\n \\item Disjoint set:$A\\cap B = \\varnothing$\n \\end{itemize}\n \\subsection*{Identities}\n \\begin{itemize}[noitemsep,nolistsep]\n \\item Commutative Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup B = B \\cup A $ and $A \\cap B = B \\cap A$\n \\end{itemize}\n \\item Associative Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $(A \\cup B) \\cup C = A \\cup (B \\cup C)$ and $(A \\cap B) \\cap C = A \\cap (B \\cap C)$\n \\end{itemize}\n \\item Distributive Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup (B \\cap C) = (A \\cup B) \\cap (A \\cup C)$\n \\item $A \\cap (B \\cup C) = (A \\cap B) \\cup (A \\cap C)$\n \\end{itemize}\n \\item Identity Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup \\varnothing = A$ and $A \\cap U$\n \\end{itemize}\n \\item Idempotent Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup A = A$ and $A\\cap A = A$\n \\end{itemize}\n \\item Universal Bound Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup U = U$ and $A \\cap \\varnothing = \\varnothing$\n \\end{itemize}\n \\item DeMorgan's Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $\\complement{(A \\cup B)} = \\complement{A} \\cap \\complement{B}$ and $\\complement{(A \\cap B)} = \\complement{A} \\cup \\complement{B}$\n \\end{itemize}\n \\item Absorption Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A \\cup (A \\cap B) = A$ and $A \\cap (A \\cup B) = A$\n \\end{itemize}\n \\item Set Difference Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A-B = A \\cap \\complement{B}$\n \\end{itemize}\n \\item Complement Law\n \\begin{itemize}[noitemsep,nolistsep]\n \\item $A\\cup \\complement{A} = U$ and $A \\cap \\complement{A} = \\varnothing$\n \\end{itemize}\n \\end{itemize}\n \\section*{Functions}\n A function $f$ from a set $X$ to a set $Y$ denoted by $f: X \\to Y$ where $X$ is the domain and $Y$ is the co-domain.\n Every element in $X$ relates to some element in $Y$ and each element in $X$ must map to at most one element in $Y$.\n \\subsection*{One-to-One functions}\n $\\forall x_1, x_2 \\in X[(F(x_1) = F(x_2)) \\implies (x_1 = x_2)]$\n \\subsection*{Onto functions}\n $\\forall y \\in Y[ \\exists x \\in X \\text{ where } F(x)=y]$\n \\subsection*{Inverse functions}\n $F^{-1}(y) = x\\text{ where }F(x) = y$.\n $F(x)$ has to be bijective (both one-to-one and onto).\n \\section*{Relations}\n A partition of set $A$ is a collection of nonempty, mutually disjoint subsets whose union is $A$, i.e. $A_1 \\cup A_2 \\cup A_3 \\cup \\ldots \\cup A_i = A$\n \\begin{itemize}[noitemsep,nolistsep]\n \\item Reflexive $\\forall x \\in A[(x, x) \\in \\mathbb{R}]$\n \\item Symmetric $\\forall x, y \\in A[(x, y) \\in \\mathbb{R} \\implies (y, x) \\in \\mathbb{R}]$\n \\item Transitive $\\forall x, y, z \\in A[(x R y \\land y R z)\\implies x R z]$\n \\item Partition-induced relation: $\\forall x, y \\in A[x R y \\iff (x, y) \\in \\text{single partition }A_i]$\n \\item Equivalence Relation: When a relation is relation is reflexive, symmetric and transitive.\n \\item Identity Relation: $I_A = \\{(x, y) \\in A \\times A~|~x=y\\}$\n \\end{itemize}\n \\subsection*{Equivalence Classes}\n The subset of all elements that are related to $a$.\n\n Suppose $A$ is a set and $R$ is an equivalence relation on $A$. For each element $a$ in $A$, the \\textbf{equivalence class of $a$}, $[a]$ or \\textbf{class of $a$} is the set of all elements $x$ in $A$ such that $x$ is related to $a$ by $R$. \\[[a] = \\{x \\in A~|~x~R~a\\}\\]\n \\section*{Partial Order Relations}\n A relation that is reflexive, antisymmetric and transitive.\n \\subsection*{Antisymmetry}\n A relation is antisymmetric if $x R y \\land y R x,~x = y$\n A relation is \\textbf{not} antisymmetric if $x R y \\land y R x\\text{ but }x \\neq y$\n \\section*{Counting}\n $P(E)$ is \\[P(E) = |E|/|S|\\] where S is the set of possible outcomes and E is the set of events.\n\n Number of elements: $|{n, n+1, n+2, \\ldots, m}| = m - n + 1$\n \\subsection*{Permutations and Combinations}\n Number of possible arrangements of n units: $n!$\n \\begin{itemize}[noitemsep,nolistsep]\n \\item With $n_1$ duplicates: $\\frac{n!}{n_1!}$\n \\item With $m$ units in certain order: $(n-(m-1))!$\n \\item In a circle/rotation: $\\frac{n!}{n}$ (Let 1 be fixed, arrange the rest.)\n \\end{itemize}\n\n $P(n, r) = \\frac{n!}{(n-r)!}$\n\n $C(n, r) = \\binom{n}{r} = \\frac{P(n, r)}{r!}$\n\n Number of Combinations with repetitions allowed: $\\binom{r+n-1}{r}$\n\n $\\binom{n}{r} = \\binom{n}{n-r}$\n\n Pascal's Formula: $\\binom{n+1}{r} = \\binom{n}{r-1}+ \\binom{n}{r}$\n\n Binomial Theorem:\\begin{align*}\n (a+b)^n & = \\sum_{k=0}^n\\binom{n}{k} a^{n-k}b^k \\\\\n & = a^n + \\binom{n}{1} a^{n-1} b^1 \\\\\n & + \\binom{n}{2} a^{n-2} b^2 + \\ldots + \\binom{n}{n-1} a^1 b^{n-1} + b^n \\\\\n \\end{align*}\n \\subsection*{Pigeonhole Principle}\n If $n$ pigeons fly into $m$ pigeonholes, and $n > km$ for some $k \\in \\mathbb{Z}^+$ then at least one pigeonhole contains k+1 or more pigeons.\n\n Generalized: \\[k = \\floor*{\\frac{n}{m}}\\]\n\n \"The maximum value has at least the average value, for any non-empty finite bag of real numbers\" -Prof. Djikstra\n \\section*{Examples}\n \\newtheorem{example}{Example}\n \\subsection*{Strong Induction}\n \\begin{example}\n \\begin{equation*}\n \\begin{aligned}\n f(n) = \\begin{cases}\n 1 & \\text{if }n=1 \\\\\n 3 & \\text{if }n=2 \\\\\n f(n-1) + 2f(n-2) +3 & \\text{if } n\\geq 3\n \\end{cases}\n \\end{aligned}\n \\end{equation*}\n Using mathematical induction, prove that for all $n \\geq 3: f(n) \\geq 2^n$\n\n Answer:\n \\begin{equation*}\n \\begin{aligned}\n f(k+1) =f(k)+ 2f(k-1)+3 \\\\\n \\text{Use fact that $f(n) \\geq 2^n$} \\\\\n f(k+1) \\geq 2^k+2\\cdot 2^{k-1} +3 \\\\\n \\text{and } f(k+1) \\geq 2^k + 2^k +3 = 2^{k+1} + 3 \\geq 2^{k+1}\n \\end{aligned}\n \\end{equation*}\n \\end{example}\n \\subsection*{Weak Induction}\n \\begin{example}\n What is the smallest value of k such that any integer postage of k, or\n more, cents can be formed by using only 3-cent and 8-cent stamps?\n\n Use mathematical induction to prove that all integer postages of k, or more, cents can be\n obtained.\n \\begin{equation*}\n \\begin{aligned}\n \\textbf{Basis: } P(n): & n = 3a + 8b \\\\\n \\textbf{Induction: } P(n+1): & n+1 = 3a' + 8b' \\\\\n \\textbf{Case 1: } \\text{if } b \\geq 1, n = 3a + 8b & \\text{, so} \\\\\n n+1 = 3a+1+8b & = 8(b-1) + 9 + 3a \\\\\n n+1 & = 8(b-1) +3(a+3) \\\\\n \\textbf{Hence }a' = a+3 \\text{ and } & b' = b - 1 \\\\\n \\textbf{Case 2: } \\text{if } b = 0, n = 3a & \\\\\n \\text{Since }n \\geq 14,~a \\geq 5 \\\\\n n+1 = 1 + 15 + 3(l-5) & = 16+ 3(l-5) \\\\\n \\textbf{Hence }a' = l-5 \\text{ and } & b' = 2 \\\\\n \\end{aligned}\n \\end{equation*}\n \\end{example}\n \\begin{example}\n Recall that the $n$'th Harmonic number, $H_n$, is defined as:\n \\[\n H_n = 1+ \\frac{1}{2} + \\frac{1}{3} + \\ldots + \\frac{1}{n}\n \\]\n Use mathematical induction to prove that for all positive integers $n$:\n \\[\n H_1 + H_2 + \\ldots + H_n = (n+1)H_n -n\n \\]\n Answer:\n \\begin{equation*}\n \\begin{aligned}\n \\textbf{Basis: } P(n): & H_1 + H_2 + \\ldots + H_n = (n+1) H_n-n \\\\\n P(1): & H_1 = 1 = 2H_1 -1 \\\\\n \\end{aligned}\n \\end{equation*}\n \\begin{equation*}\n \\begin{aligned}\n \\textbf{Induction: } & H_1 + \\ldots + H_k + H_{k+1} = (k+1)H_k - k + H_{k+1} \\\\\n \\text{Using IH for k} \\\\\n = & (k+1)(1+\\frac{1}{2}+\\ldots+\\frac{1}{k})-k +(1+\\ldots + \\frac{1}{k+1}) \\\\\n = & \\ldots \\\\\n % Rest of the fucking owl\n = & (k+2) (1+\\frac{1}{2} + \\ldots + \\frac{1}{k+1}) - (k+1)\n \\end{aligned}\n \\end{equation*}\n \\end{example}\n \\subsection*{Functions}\n Let $f: X \\mapsto Y$ and $g: Y \\mapsto Z$ be functions. Indicate whether each statement below is true or false. Give a proof for each true statement and a counterexample for each false statement.\n\n \\textbf{a) If $g \\circ f$ is onto then $f$ is onto.}\n\n \\textbf{Answer: } False.\n\n $X = {1, 2}, Y = {1, 2, 3}, Z = {1, 2}, f(1) = 1, f(2) = 2, g(1) = 1, g(2) = 2, g(3) = 2$\n\n \\textbf{b) If $g \\circ f$ is onto then $g$ is onto.}\n\n \\textbf{Answer: } True.\n\n Let $z \\in Z$. There exists $x \\in X$ such that $g(f(x)) = z$ since $g \\circ f$ is onto. So $g(y) = z$ for $y = f(x)$. Therefore $g$ is onto.\n\n \\textbf{c) If $g \\circ f$ is one-to-one then $f$ is one-to-one.}\n\n \\textbf{Answer: } True.\n\n Let $x_1, x_2 \\in X$ such that $f(x_1) = f(x_2)$.\n \n Then $g(f(x_1)) = g(f(x_2))$. Now $g(f(x_1)) = g(f(x_2))$ by definition of composition. Therefore $x_1 = x_2$ since $g \\circ f$ is one-to-one. This implies $f$ is one-to-one.\n\n \\textbf{d) If $g \\circ f$ is one-to-one then $g$ is one-to-one.}\n\n \\textbf{Answer: } False.\n\n Use part (a).\n\n \\subsection*{Counting}\n \\begin{example}\n How many three-digit integers are divisible by 5?\n \\begin{equation*}\n \\begin{aligned}\n \\text{Three-digit integers} = {100, 101, \\ldots, 999} \\\\\n \\left\\lfloor \\frac{100}{5} \\right\\rfloor = 20, \\left\\lfloor \\frac{999}{5} \\right\\rfloor = 199 \\\\\n \\text{by number of elements theorem: } 199-20 + 1 = 180\n \\end{aligned}\n \\end{equation*}\n \\end{example}\n \\begin{example}\n Consider a robot moving in an $m \\times n$ grid, where $m\\leq n$. It starts at the coordinate $(1,1)$. At each step, when it is at position $(i,j)$, it either moves one square up to $(i+1, j)$ or one square right to $(i, j+1)$. How many ways can the robot travel from $(1,1)$ to $(m,n)$? Justify your answer.\n\n Answer:\n \\begin{align*}\n m - 1 \\text{ possible steps up.} \\\\\n n - 1 \\text{ possible steps right.} \\\\\n (m - 1) + (n - 1) \\text{ total steps.} \\\\\n \\text{Each step up and right are the same, so} \\\\\n \\frac{(m+n - 2)!}{(m-1)!(n-1)!} \\\\\n \\end{align*}\n \\end{example}\n \\subsection*{Direct proofs}\n \\begin{example}\n $6-7\\sqrt{2}$ is irrational.\n \\begin{align*}\n \\text{Assume } 6-7\\sqrt{2} = \\frac{p}{q} \\text{ where } p, q \\in \\mathbb{Z} \\land q \\not = 0. \\\\\n \\sqrt{2} = \\frac{\\left(\\frac{p}{q}-6\\right)}{-7} = \\frac{p-6q}{-7q} \\\\\n \\end{align*}\n This is rational as both the numerator and denominator are products and differences of integers and thus are integers, and the denominator is not equal to zero.\n \\end{example}\n \\subsection*{Pigeonhole Princple}\n \\begin{example}\n For every 27 word sequence in a paragraph, at least two words will start with the same letter.\n\n \\textbf{Reason: } There are 27 words that can start with the 26 different English letters. By the pigeonhole principle, two of the words must start with the same letter.\n \\end{example}\n \\begin{example}\n If a theatre holds 1300 people, how many seats need to be filled to ensure that two people have the same first and last initials?\n\n \\textbf{Answer: } 26 letters in English language, Fname.Lname = 1 letter on each initial. $26 * 26$ means at least 1 person as each combination of initials. $26 * 26 +1$ ensures at least 1 duplicate.\n \\end{example}\n \\begin{example}\n Let ABC be an equilateral triangle with AB = 1. Show that by selecting 10 points, there are at least two with distance $\\leq \\frac{1}{3}$ apart.\n\n \\textbf{Answer: } If you split up the triangle by $\\frac{1}{3}$ lengths, you will create 9 slots. The 10th will be in the same triangle as another, which is less than $\\frac{1}{3}$.\n \\end{example}\n \\begin{example}\n Prove with the pigeonhole principle that having 100 whole numbers, one can choose 15 of them so that the difference of any 2 is divisible by 7?\n\n \\textbf{Answer: } Use the generalized pigeonhole principle.\n\n Let $S: \\{1\\ldots100\\}.~S' = S \\mod 7 = \\{1 \\mod 7 \\ldots 100 \\mod 7\\}$\n\n Partition $S'$ into $7$ subsets of equal size where each subset has same value for $S \\mod 7$.\n\n By generalized PHP, where $\\lfloor \\frac{units}{holes}\\rfloor$ units in each hole, there are at least $\\lfloor \\frac{100}{7} \\rfloor = 14$ remainders in each subset of $S'$.\n\n $14 * 7 = 98$ which is $2$ less than $100$. We can place those $2$ with the other subsets that contain the same number, increasing the number to at least 15 for one or more of the subsets.\n\n \\end{example}\n\\end{multicols}\n\\end{document}\n", "meta": {"hexsha": "acfb31550d0b959153c4cf4e7245f7e5541cb68a", "size": 18393, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/CMPUT272.tex", "max_stars_repo_name": "n30phyte/SchoolDocuments", "max_stars_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Notes/CMPUT272.tex", "max_issues_repo_name": "n30phyte/SchoolDocuments", "max_issues_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/CMPUT272.tex", "max_forks_repo_name": "n30phyte/SchoolDocuments", "max_forks_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.5210084034, "max_line_length": 316, "alphanum_fraction": 0.5323220791, "num_tokens": 5947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8418256472515684, "lm_q1q2_score": 0.6379307370818696}} {"text": "\\section{$I$-invariance of $\\Bun_G$, and $G$-CW-complexes}\nLet $G$ be a topological group. We need to show that the functor\n$\\Bun_G:\\Top^{op}\\to\\Set$ is $I$-invariant, i.e., the projection\n$X\\times I\\xrightarrow{\\pr}X$ induces an isomorphism\n$\\Bun_G(X)\\xrightarrow{\\simeq}\\Bun_G(X\\times I)$.\nInjectivity is easy: the composite $X\\xrightarrow{\\inc_0} X\\times\nI\\xrightarrow{\\pr}X$ gives you a splitting\n$\\Bun_G(X)\\xrightarrow{\\pr_\\ast}\\Bun_G(X\\times I)\\xrightarrow{\\inc_0}\\Bun_G(X)$\nwhose composite is the identity.\n\nThe rest of this lecture is devoted to proving surjectivity. We will prove this\nwhen $X$ is a CW-complex (Husemoller does the general case; see \\cite[\\S\n4.9]{husemoller}). We begin with a small digression.\n%Consider two principal $G$-bundles $P\\to X$ and $Q\\to Y$, with a map $f:X\\to\n%Y$. We would like to understand maps $P\\to f^\\ast Q$ over $X$, i.e.,\n%$G$-equivariant dotted maps that make the following diagram commute.\n%\\begin{equation*}\n% \\xymatrix{\n%\tP\\ar@{-->}[r]^g\\ar[d] & Q\\ar[d]\\\\\n%\tX\\ar[r]_f & Y\n% }\n%\\end{equation*}\n%Suppose I have $P\\to X\\times I$; then we get $in_0^\\ast P\\to X$. This has to\n%be what you get when you ---. All we have to do is construct a map $P\\to\n%in_0^\\ast P$ like in the diagram above.\n\\subsection{$G$-CW-complexes}\nWe would like to define CW-complexes with an action of the group $G$. The\nna\\\"ive definition (of a space with an action of the group $G$) will not be\nsufficient; rather, we will require that each cell have an action of $G$.\n\nIn other words, we will build $G$-CW-complexes out of ``$G$-cells''. This is\nsupposed to be something of the form $D^n\\times H\\backslash G$, where $H$ is a\nclosed subgroup of $G$. Here, the space $H\\backslash G$ is the orbit space,\nviewed as a right $G$-space. The boundary of the $G$-cell $D^n\\times\nH\\backslash G$ is just $\\partial D^n\\times H\\backslash G$. More precisely:\n\\begin{definition}\n A $G$-CW-complex is a (right) $G$-space $X$ with a filtration\n $0=X_{-1}\\subseteq X_0\\subseteq \\cdots\\subseteq X$ such that for all $n$,\n there exists a pushout square:\n \\begin{equation*}\n\t\\xymatrix{\n\t \\coprod\\partial D^n_\\alpha\\times H_\\alpha\\backslash G\\ar[r]\\ar[d] &\n\t \\coprod D^n_\\alpha\\times H_\\alpha\\backslash G\\ar[d]\\\\\n\t X_{n-1}\\ar[r] & X_n,\n\t }\n \\end{equation*}\n and $X$ has the direct limit topology.\n\\end{definition}\nNotice that a CW-complex is a $G$-CW-complex for the trivial group $G$.\n\\begin{theorem}\n If $G$ is a compact Lie group and $M$ a compact smooth $G$-manifold, then\n $M$ admits a $G$-CW-structure.\n\\end{theorem}\nThis is the analogue of the classical result that a compact smooth manifold is\nhomotopy equivalent to a CW-complex, but it is much harder to prove the\nequivariant statement.\n\nNote that if $G$ acts principally (Definition \\ref{principaldefn}) on $P$, then\nevery $G$-CW-structure on $P$ is ``free'', i.e., $H_\\alpha = 0$.\n\\begin{enumerate}\n \\item If $X$ is a $G$-CW-complex, then $X/G$ inherits a CW-structure whose\n\t$n$-skeleton is given by $(X/G)_n = X_n/G$.\n \\item If $P\\to X$ is a principal $G$-bundle, then a CW-structure on $X$\n\tlifts to a $G$-CW-structure on $P$.\n\\end{enumerate}\n\\subsection{Proof of $I$-invariance}\nRecall that our goal is to prove that every $G$-bundle over $X\\times I$ is\npulled back from some vector bundle over $X$.\n\nAs a baby case of Theorem \\ref{Iinvariance} we will prove that if\n$X$ is contractible, then any principal $G$-bundle over $X$ is trivial, i.e.,\n$P\\simeq X\\times G$ as $G$-bundles.\n\nLet us first prove the following: if $P\\downarrow X$ has a section, then it's\ntrivial. Indeed, suppose we have a section $s:X\\to P$. Since $P$ has an action\nof the group on it, we may extend this to a map $X\\times G\\to P$ by sending\n$(x,g)\\mapsto gs(x)$. As this is a map of $G$-bundles over $X$, it is an\nisomorphism by Theorem \\ref{morphismiso}, as desired.\n\nTo prove the statement about triviality of any principal $G$-bundle over a\ncontractible space, it therefore suffices to construct a section for any\nprincipal $G$-bundle. Consider the constant map $X\\to P$. Then the following\ndiagram commutes up to homotopy, and hence (by Exercise\n\\ref{sectionuptohomotopy}(1)) there is an \\emph{actual} section of $P\\to X$, as\ndesired.\n\\begin{equation*}\n \\xymatrix{\n\t& P\\ar[d]\\\\\n\tX\\ar[ur]^{\\mathrm{const}}\\ar[r] & X\n }\n\\end{equation*}\nFor the general case, we will assume $X$ is a CW-complex. For notational\nconvenience, let us write $Y=X\\times I$. We will use descending induction to\nconstruct the desired principal $G$-bundle over $X$.\n\nTo do this, we will filter $Y$ by subcomplexes. Let $Y_0 = X\\times 0$; in\ngeneral, we define\n$$Y_n = X\\times 0\\cup X_{n-1}\\times I.$$\nIt follows that we may construct $Y_n$ out of $Y_{n-1}$ via a pushout:\n\\begin{equation*}\n \\xymatrix{\n\t\\coprod_{\\alpha\\in\\Sigma_{n-1}}(\\partial D^{n-1}\\times I \\cup\n\tD^{n-1}_\\alpha\\times 0) \\ar[r]\\ar[d]_{\\coprod_{\\alpha\\in\\Sigma_{n-1}}\n\tf_\\alpha\\times 1_I\\cup\\phi_\\alpha\\times 0} &\n\t\\coprod_\\alpha(D^{n-1}_\\alpha\\times I) \\ar[d]\\\\\n\tY_{n-1}\\ar[r] & Y_n,\n }\n\\end{equation*}\nwhere the maps $f_\\alpha$ and $\\phi_\\alpha$ are defined as:\n\\begin{equation*}\n \\xymatrix{\n\t\\partial D^{n-1}_\\alpha\\ar[r]^{f_\\alpha}\\ar[d] & X_{n-2}\\ar[d]\\\\\n\tD^{n-1}_\\alpha\\ar[r]_{\\phi_\\alpha} & X_{n-1}\n }\n\\end{equation*}\nIn other words, the $f_\\alpha$ are the attaching maps and the $\\phi_\\alpha$ are\nthe characteristic maps.\n\nConsider a principal $G$-bundle $P\\xrightarrow{p}Y = X\\times I$. Define $P_n =\np^{-1}(Y_n)$; then we can build $P_n$ from $P_{n-1}$ in a similar way:\n\\begin{equation*}\n \\xymatrix{\n\t\\coprod_\\alpha(\\partial D^{n-1}_\\alpha\\times I\\cup D^{n-1}_\\alpha\\times\n\t0)\\times G\\ar[r]\\ar[d] & \\coprod_\\alpha (D^{n-1}_\\alpha\\times I)\\times\n\tG\\ar[d]\\\\\n\tP_{n-1}\\ar[r] & P_n\n }\n\\end{equation*}\n%This makes sense since $D^n_\\alpha\\times I$ is a contractible space, and same\n%thing for the other factor.\nNote that this isn't \\emph{quite} a $G$-CW-structure. Recall that we are\nattempting to fill in a dotted map:\n\\begin{equation*}\n \\xymatrix{\n\tP\\ar@{-->}[r]\\ar[d] & P_0\\ar[d]\\\\\n\tY\\ar[r]_{\\pr} & Y_0 = X\n }\n\\end{equation*}\n\\todo{finish this...}\nI'm constructing this inductively-- we have $P_{n-1}\\to P_0$. So I want to define $\\coprod_\\alpha(D^{n-1}_\\alpha\\times I)\\times G\\to P_0$ that's equivariant. That's the same thing as a map $\\coprod_\\alpha (D^{n-1}_\\alpha\\times I)\\to P_0$ that's compatible with the map from $\\coprod(\\partial D^{n-1}_\\alpha\\times I\\cup D^{n-1}_\\alpha\\times 0)$. Namely, I want to fill in:\n\\begin{equation}\\label{finally}\n \\xymatrix{\n\t\\coprod_\\alpha (\\partial D^{n-1}_\\alpha\\times I\\cup D^{n-1}_\\alpha\\times 0)\\ar[r]\\ar[d] & \\coprod_\\alpha(D^{n-1}_\\alpha\\times I)\\ar@{-->}[dddr]\\ar[d] & \\\\\n\t\\coprod_\\alpha(\\partial D^{n-1}_\\alpha\\times I\\cup D^{n-1}_\\alpha\\times 0)\\times G\\ar[r]\\ar[d] & \\coprod_\\alpha (D^{n-1}_\\alpha\\times I)\\times G\\ar@{-->}[ddr]\\ar[d] & \\\\\n\tP_{n-1}\\ar[drr]_{\\mathrm{induction}}\\ar[r] & P_n\\ar[dr] & \\\\\n\t& & P_0\\ar[d]\\\\\n\t& & X\n }\n\\end{equation}\nNow, I know that $(D^{n-1}\\times I,\\partial D^{n-1}\\times I\\cup D^{n-1}\\times 0)\\simeq (D^{n-1}\\times I,D^{n-1}\\times 0)$. So what I have is:\n\\begin{equation*}\n \\xymatrix{\n\tD^{n-1}\\times 0\\ar[d]\\ar[r]^{\\mathrm{induction}} & P_0\\ar[d]\\\\\n\tD^{n-1}\\times I\\ar[r]_{\\phi\\circ pr}\\ar@{-->}[ur] & X\n }\n\\end{equation*}\nSo the dotted map exists, since $P_0\\to X$ is a fibration!\n\nOK, so note that I haven't checked that the outer diagram in Equation \\ref{finally} commutes, because otherwise we wouldn't get $P_n\\to P_0$.\n\\begin{exercise}\n Check my question above.\n \n Turns out this is easy, because you have a factorization:\n\\begin{equation*}\n \\xymatrix{\n\tD^{n-1}\\times 0\\ar[d]\\ar[r] & P_{n-1}\\ar[r]^{\\mathrm{induction}} & P_0\\ar[d]\\\\\n\tD^{n-1}\\times I\\ar[rr]_{\\phi\\circ pr}\\ar@{-->}[urr] & & X\n }\n\\end{equation*}\n\\end{exercise}\nOh my god, look what time it is! Oh well, at least we got the proof done.\n", "meta": {"hexsha": "c40c8006b477e4d2dfa7078195a8f383226d38d7", "size": 7884, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "906/lec-54-I-invariance.tex", "max_stars_repo_name": "ichung/algtop-notes", "max_stars_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-04-26T15:00:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T22:47:06.000Z", "max_issues_repo_path": "906/lec-54-I-invariance.tex", "max_issues_repo_name": "ichung/algtop-notes", "max_issues_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-03-13T17:54:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-13T17:59:46.000Z", "max_forks_repo_path": "906/lec-54-I-invariance.tex", "max_forks_repo_name": "ichung/algtop-notes", "max_forks_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-21T18:15:11.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-13T17:38:04.000Z", "avg_line_length": 45.8372093023, "max_line_length": 371, "alphanum_fraction": 0.6752917301, "num_tokens": 2859, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8418256452674008, "lm_q1q2_score": 0.6379307355782785}} {"text": "\\section{Chain Adjacency and Incidence}\n\\label{chain-adjacencies-and-incidensies}\n%=========================================\n\nBoundary decompositions (B-reps) are the typical representations used in solid modeling and computer graphcs. The model boundary is partitioned into vertices (\\texttt{V}: 0-cells), edges (\\texttt{E}: 1-cells), and faces (\\texttt{F}: 2-cells), where faces are often triangles or more general convex cells. \n\n\\subsection{Boundary representation of solid models}\n\\label{boundary-representation-of-solid-models}\n%--------------------------------------------------\n\n\nIn particular, a typical solid modeling representation employs some specialized data structure to efficiently traverse the boundary, moving from some element to the adjacent ones (with same dimension) or to incident ones (with different dimension). Several data structures have been used for this purpose~\\cite{HofShapiro:2017}, taking into account both the efficiency of topological queries and the storage compactness. Three binary adjacency relations and six binary incidence relations can be needed by algorithms, as shown in Table~\\ref{tab:relations}a.\n\nUsing chain operators, or better their matrices, when an ordering of elements has been fixed inside the sets \\texttt{V}, \\texttt{E}, and \\texttt{F} using one-dimensional arrays, such binary relations can be represented and/or computed as shown in Table~\\ref{tab:relations}b. It is easy to show (see Table~\\ref{tab:relations}b), that all linear operators between chain spaces corresponding to binary relations between boundary elements can be derived from $\\partial_1$ and $\\partial_2$ via matrix transposition or product over semirings~\\cite{DBLP:journals/corr/KepnerABBFGHKLM16,GraphBLAS:API:2017}.\n\n\\begin{table}[htp]\n\\caption{(a) The 9 binary relations between boundary elements, and the 16 binary relations between decompositive elements; (b) corresponding linear operators between chain spaces. Remember that $\\partial_p: C_p\\to C_{p-1}$ and that $\\delta_p = \\partial_{p+1}^\\top$. }\n\\begin{center}\n\\begin{tabular}{|c|ccc|c|}\n\\hline\n &\\texttt{V} & \\texttt{E} & \\texttt{F} & \\texttt{C} \\\\ \n\\hline\n\\texttt{V} &\\texttt{VV} & \\texttt{VE} & \\texttt{VF} & \\texttt{VC} \\\\ \n\\texttt{E} &\\texttt{EV} & \\texttt{EE} & \\texttt{EF} & \\texttt{EC} \\\\ \n\\texttt{F} &\\texttt{FV} & \\texttt{FE} & \\texttt{FF} & \\texttt{FC} \\\\ \n\\hline\n\\texttt{C} &\\texttt{CV} & \\texttt{CE} & \\texttt{CF} & \\texttt{CC} \\\\ \n\\hline\n\\end{tabular}\n$\\qquad$\n\\begin{tabular}{|c|ccc|c|}\n\\hline\n &\\texttt{$C_0$} & \\texttt{$C_1$} & \\texttt{$C_2$} & \\texttt{$C_3$} \\\\ \n\\hline\n\\texttt{$C_0$} & $[\\partial_1][\\delta_0]$ & $[\\partial_1]$ & $[\\partial_1][\\partial_2]$ & $[\\partial_1][\\partial_2][\\partial_3]$\\\\ \n\\texttt{$C_1$} & $[\\delta_0]$ & $[\\delta_0][\\partial_1]$ & $[\\partial_2]$ & $[\\partial_2][\\partial_3]$ \\\\ \n\\texttt{$C_2$} & $[\\delta_1][\\delta_0]$ & $[\\delta_1]$ & $[\\delta_1][\\partial_2]$ & $[\\partial_3]$ \\\\ \n\\hline\n\\texttt{$C_3$} & $[\\delta_2][\\delta_1][\\delta_0]$ & $[\\delta_2][\\delta_1]$ & $[\\delta_2]$ & $[\\delta_2][\\partial_3]$ \\\\ \n\\hline\n\\end{tabular}\n\\end{center}\n\\label{tab:relations}\n\\end{table}%\n\n\nNote that we have assumed vectors as column matrices, so that an operator maps the column space to the row space of its matrix. Hence, in order to maintain the consistency between the two representations we should read the relation $\\texttt{AB} \\subset \\texttt{A}\\times \\texttt{B}$ as a map $\\texttt{AB}: \\texttt{B} \\to \\texttt{A}$.\n\n\n\\subsection{Space Decompositions}\\label{space-decompositions}\n%-------------------------------------------------------------\nA similar representation scheme---called \\emph{decompositive} in~\\cite{Requicha:1980:RRS:356827.356833}---is also used, both in solid modeling and in order to represent the domain decomposition in FEM and other discretizations of physical models. In this case a partition of the whole domain is provided, using cellular 3-complexes, given by \\texttt{V}, \\texttt{E}, \\texttt{F}, and \\texttt{C} sets of $0$-, 1-, 2- and 3-cells, where often the \\texttt{C} elements are either tetrahedra or hexahedra. \n\nFor the topology of a graph (cellular 1-complex) just $C_0$ (vertices) and $C_1$ (edges) are needed, so that a complete representation is given by the signed or unsigned matrix $[\\partial_1]$, corresponding to the relation $\\texttt{EV} \\equiv C_1\\to C_0$.\n\nIt is worth noting that operators $\\partial_1$, $\\partial_2$, and $\\partial_3$ are sufficient to represent the complete collection of $4\\times 4$ operators (see Table~\\ref{tab:relations}b), via matrix transposition or product over semirings. In other words, the \\emph{chain complex} $C_\\bullet = (C_p, \\partial_p)$ of boundary matrices, $1\\leq p\\leq 1,2,3$ is a \\emph{complete representation} of the topology of (a) graphs, (b) B-reps, and (c) decompositive representations, respectively. \n\nCellular complexes may be either \\emph{oriented} (signed) or \\emph{non-oriented} (unsigned). In the first case the elements of operator matrices are taken over the domain $D=\\{0,1\\}$; in the second case they belong to the domain $D=\\{-1,0,1\\}$, so that passing from a coefficient $+1$ to a coefficient $-1$ (or viceversa) the element orientation is reversed.\n\n\n\\subsection{Minimal representations}\\label{minimal-reps}\n\nMost of earlier algorithms and procedures~\\cite{\n4055948,\nAla:1992:PAB:616022.617736,\nBaumgart:1972:WEP:891970,\nbowyer1995introducing,\nbowyer1995svlis,\nBraid:1975:SSB:360715.360727,\nBrisson:1989:RGS:73833.73858,\ncadanda,\nDobkin:1987:PMT:41958.41967,\nGomes:1999:MMB:304012.304039,\nGuibas:1985:PMG:282918.282923,\nHoffmannK01,\nKalay:1989:HET:63718.63719,\nLee:2001:PES:376957.376976,\nLienhardt:1991:TMB:115604.115610,\nMantyla:1988:ISM:60949,\nPaoluzzi:1989:BAO:70248.70249,\nPaoluzzi:1993:DMS:169728.169719,\nPaoluzzi:1995:GPP:212332.212349,\nPascucci:1995:DCB:218013.218055,\nPratt94ashape,\nRaghothama:1999:CUD:304012.304019,\nRap97,\nRequichaVoelcker:77,\nRossignac:1991:CNG:115604.115606,\nRossignac:SGC:90,\nShapiro:1991:RSS:124951,\nShapiro:1995:PFS:218013.218029,\nSilva:81,\nWeiler:86,\nWeiler:88,\nWoo:85,\nwozny1990geometric,\nYamaguchi:85,\nyamaguchi1995ntb,\nZhou:2016:MAS:2897824.2925901,\nbieri:95,\nRossignac:89,\nHoffmann:91,\nHoffmann:1989:GSM:74803,\nHoffmann:1987:RSO:866286} \nwork with data\nstructures optimized for selected classes of geometric\nobjects. By contrast, our formulation, representation, and algorithms, cast in\nterms of (co)chain complexes of (co)boundary maps, may be\napplied to very different geometric objects, ranging from solid models\nto engineering meshes, geographical systems, biomedical images.\n\nGiven a set $S=\\{s_j\\}$, the \\emph{characteristic function} $\\chi_A: S\\to\\{0,1\\}$ takes value 1 for all elements of $A\\subseteq S$ and 0 at all elements of $S$ not in $A$. \nWe call \\emph{characteristic matrix} $M$ of a collection of subsets $A_i\\subseteq S$ ($i=1,\\ldots,n$) the binary matrix $M=(m_{ij})$, with $m_{ij} = \\chi_{A_i}(s_j)$. {A matrix $M_p$, whose rows are indexed by unit $p$-chains and columns are indexed by unit $0$-chains, provides a useful representation of a basis for the linear space $C_p$. Permuting (reindexing) either rows or columns provides a different basis.} While chains are mostly presented as formal sums of cells, in the actual implementation their signed coordinate vectors are used as \\emph{sparse} arrays, and in particular as CSC (Compressed Sparse Column) maps : $\\N\\to\\{-1,0,1\\}$.\n\nIt is possible to show~\\cite{} that, when the $d$-cells are convex, the topology of a cellular $d$-complex is fully described by $M_{d-1}$ and by an embedding function $\\mu: \\texttt{V}\\to \\E^d$. When $d$-cells are more complex---say, non-convex or with holes---the triple $(M_{d-1}, M_{d-2}, \\mu)$ is needed to get a full knowledge of the topology of the complex.\n\nWe call \\texttt{LAR} (Linear Algebraic Representation) of subsets of a set \\texttt{V} (vertices), the array, indexed by ordinals (one-to-one with subsets), of arrays of indices to \\texttt{V} elements. This one is a compact representation of the \\emph{characteristic matrix} of the collection of subsets.\n\n\n", "meta": {"hexsha": "ce95741419e85a4b24a33a28109a62c766675058", "size": 8060, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/incidence.tex", "max_stars_repo_name": "cvdlab/Chain-BLAS", "max_stars_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/incidence.tex", "max_issues_repo_name": "cvdlab/Chain-BLAS", "max_issues_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/incidence.tex", "max_forks_repo_name": "cvdlab/Chain-BLAS", "max_forks_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.1666666667, "max_line_length": 652, "alphanum_fraction": 0.7305210918, "num_tokens": 2546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835371034369, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6378781197916169}} {"text": "\\chapter{Graphing Polynomials}\n\nIn using polynomials to solve real-world problems, it is often handy\nto know what the graph of the polynomial looks like. You have many of\nthe tools you need to start to sketch out the the graphs:\n\\begin{itemize}\n\\item To find where the graph crosses the y-axis, you can evaluate the polynomial at $x = 0$.\n\\item To find where the graph crosses the x-axis, you can find the roots of the polynomial.\n\\item To find the level spots on the graph (often the top of a hump or the bottom of a dip), you can take the derivative of the polynomial (which is a polynomial), and find the roots of that.\n\\end{itemize}\\index{polynomial!graphing}\n\n\\textit{FIXME: Diagram of those things}\n\nFor example, if you wanted to graph the polynomial $f(x) = -x^3 -x^2 +\n6x$, you might plug in a few values that are easy to compute:\n\\begin{itemize}\n\\item $f(-2) = -8$\n\\item $f(-1) = -6$\n\\item $f(0) = 0$\n\\item $f(1) = 4$\n\\item $f(2) = 0$ \n\\end{itemize}\n\nSo, right away we know two roots: $x = 0$ and $x = 2$. Are there\nothers? We won't know until we factor the polynomial:\n\\begin{multline*}\n -x^3 -x^2 + 6x \\\\\n = (-1x)(x^2 + x - 6) \\\\\n = (-1x)(x + 3)(x - 2)\n\\end{multline*}\nSo, yes, there is a third root: $x = -3$\n\nWhat about the level spots? $f'(x) = -3x^2 - 2x + 6$. Where is that zero?\n\\begin{multline*}\n -3x^2 -2x + 6 = 0 \\\\\n x^2 + \\frac{2}{3}x - 2 = 0\n\\end{multline*}\nWe have a formula for quadratics like this:\n\\begin{multline*}\n x = -\\frac{b}{2} \\pm \\frac{\\sqrt{b^2 - 4c}}{2} \\\\\n = -\\frac{\\frac{2}{3}}{2} \\pm \\frac{\\sqrt{\\left(\\frac{2}{3}\\right)^2 - 4(-2)}}{2} \\\\\n = -\\frac{1}{3} \\pm \\frac{\\sqrt{\\frac{4}{9} + 8}}{2} \\\\\n = -\\frac{1}{3} \\pm \\frac{\\sqrt{\\frac{85}{9}}}{2} \\\\\n = -\\frac{1}{3} \\pm \\frac{\\sqrt{85}}{6} \\\\\n \\approx 1.20 \\text{ and } -1.87 \n\\end{multline*}\n\nNow you might plug those numbers in:\n\\begin{itemize}\n\\item $f(1.2) \\approx 4.0 $\n\\item $f(-1.87) \\approx -8.2$\n\\end{itemize}\n\n\\section{Leading term in graphing}\n\nThere is one more trick you need before you can draw a good graph of a\npolynomial. As you go father and farther to the left and right, where\ndoes the function go? That is, does the graph go up on both ends\n(like a smile)? Or does it go down on both ends (like a frown)? Or\ndoes the negative end go down (frowny) while the positive end go up\n(smiley)? Or does the negative go up (smiley) and the positive end go\ndown (frowny)?\n\nAssuming the polynomial is not constant, there are only those four\npossibilties. It is determined entirely by the leading term of the\npolynomial. If the degree of the leading term is even, both ends go\nin the same direction (both are smiley or both are frowny). If the\ncoefficient of the leading term is positive, the positive end is\nsmiley.\n\nThe graph we are working on has a leading term of $-1x^3$. The degree is odd, thus the ends go in different directions. The coefficient is negative, so the positive end points down. Now you can draw the graph, which should look something like this:\n\n\\includegraphics[width=\\textwidth]{annotated_graph.png}\n", "meta": {"hexsha": "0cc850046c06972883ca3ac458e16f64950f672e", "size": 3034, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Modules/Polynomials/graphs-en_US.tex", "max_stars_repo_name": "rajivjhoomuck/sequence", "max_stars_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2020-06-13T17:19:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-05T00:43:44.000Z", "max_issues_repo_path": "Modules/Polynomials/graphs-en_US.tex", "max_issues_repo_name": "rajivjhoomuck/sequence", "max_issues_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Modules/Polynomials/graphs-en_US.tex", "max_forks_repo_name": "rajivjhoomuck/sequence", "max_forks_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-05T00:43:58.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-05T00:43:58.000Z", "avg_line_length": 41.0, "max_line_length": 249, "alphanum_fraction": 0.6849044166, "num_tokens": 1005, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837743174788, "lm_q2_score": 0.8354835309589074, "lm_q1q2_score": 0.6378781195966008}} {"text": "\\input{../header_function}\r\n\r\n%---------- start document ---------- %\r\n \\section{gcd -- gcd algorithm}\\linkedzero{gcd}\r\n%\r\n \\subsection{gcd -- the greatest common divisor}\\linkedone{gcd}{gcd}\r\n \\func{gcd}{\\hiki{a}{integer},\\ \\hiki{b}{integer}}{\\out{integer}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the greatest common divisor of two integers \\param{a} and \\param{b}.\\\\\r\n \\spacing\r\n % added document\r\n %\\spacing\r\n % input, output document\r\n \\quad \\param{a},\\ \\param{b} must be int, long or \\linkingone{rational}{Integer}.\r\n Even if one of the arguments is negative, the result is non-negative.\\\\\r\n%\r\n \\subsection{binarygcd -- binary gcd algorithm}\\linkedone{gcd}{binarygcd}\r\n \\func{binarygcd}{\\hiki{a}{integer},\\ \\hiki{b}{integer}}{\\out{integer}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the greatest common divisor of two integers \\param{a} and \\param{b} by binary gcd algorithm.\\\\\r\n \\spacing\r\n % added document\r\n \\quad \\negok This function is an alias of \\linkingone{arygcd}{binarygcd}\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{a},\\ \\param{b} must be int, long, or \\linkingone{rational}{Integer}.\\\\\r\n%\r\n \\subsection{extgcd -- extended gcd algorithm}\\linkedone{gcd}{extgcd}\r\n \\func{extgcd}{\\hiki{a}{integer},\\ \\hiki{b}{integer}}{(\\out{integer},\\ \\out{integer},\\ \\out{integer})}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the greatest common divisor $d$ of two integers \\param{a} and \\param{b} and $u,\\ v$ such that $d = \\param{a}u + \\param{b}v$.\\\\\r\n \\spacing\r\n % added document\r\n %\\spacing\r\n % input, output document\r\n \\quad \\param{a},\\ \\param{b} must be int, long, or \\linkingone{rational}{Integer}.\\\\\r\n The returned value is a tuple (\\param{u},\\ \\param{v},\\ \\param{d}).\\\\\r\n%\r\n \\subsection{lcm -- the least common multiple}\\linkedone{gcd}{lcm}\r\n \\func{lcm}{\\hiki{a}{integer},\\ \\hiki{b}{integer}}{\\out{integer}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the least common multiple of two integers \\param{a} and \\param{b}.\\\\\r\n \\spacing\r\n % added document\r\n \\negok If both \\param{a} and \\param{b} are zero, then it raises an exception.\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{a},\\ \\param{b} must be int, long, or \\linkingone{rational}{Integer}.\\\\\r\n%\r\n \\subsection{gcd\\_of\\_list -- gcd of many integers}\\linkedone{gcd}{gcd\\_of\\_list}\r\n \\func{gcd\\_of\\_list}{\\hiki{integers}{list}}{\\out{list}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return gcd of multiple integers.\\\\\r\n \\spacing\r\n % added document\r\n \\quad For given \\param{integers} $[x_1,\\ldots,x_n]$, return a list $[d,\\ [c_1,\\ldots,c_n]]$ such that $d=c_1 x_1+\\cdots+c_n x_n$, where $d$ is the greatest common divisor of $x_1,\\ldots, x_n$.\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{integers} is a list which elements are int or long\\\\\r\n This function returns $[d,\\ [c_1,\\ldots,c_n]]$, where $d,\\ c_i$ are an integer.\\\\\r\n%\r\n \\subsection{coprime -- coprime check}\\linkedone{gcd}{coprime}\r\n \\func{coprime}{\\hiki{a}{integer},\\ \\hiki{b}{integer}}{\\out{bool}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return True if \\param{a} and \\param{b} are coprime, False otherwise.\\\\\r\n \\spacing\r\n % added document\r\n %\\spacing\r\n % input, output document\r\n \\quad \\param{a},\\ \\param{b} are int, long, or \\linkingone{rational}{Integer}.\\\\\r\n%\r\n \\subsection{pairwise\\_coprime -- coprime check of many integers}\\linkedone{gcd}{pairwise\\_coprime}\r\n \\func{pairwise\\_coprime}{\\hiki{integers}{list}}{\\out{bool}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return True if all integers in \\param{integers} are pairwise coprime, False otherwise.\\\\\r\n \\spacing\r\n % added document\r\n %\\spacing\r\n % input, output document\r\n \\quad \\param{integers} is a list which elements are int, long, or \\linkingone{rational}{Integer}.\\\\\r\n%\r\n\\begin{ex}\r\n>>> gcd.gcd(12, 18)\r\n6\r\n>>> gcd.gcd(12, -18)\r\n6\r\n>>> gcd.gcd(-12, -18)\r\n6\r\n>>> gcd.extgcd(12, -18)\r\n(-1, -1, 6)\r\n>>> gcd.extgcd(-12, -18)\r\n(1, -1, 6)\r\n>>> gcd.extgcd(0, -18)\r\n(0, -1, 18)\r\n>>> gcd.lcm(12, 18)\r\n36\r\n>>> gcd.lcm(12, -18)\r\n-36\r\n>>> gcd.gcd_of_list([60, 90, 210])\r\n[30, [-1, 1, 0]]\r\n\\end{ex}%Don't indent!(indent causes an error.)\r\n\\C\r\n\r\n%---------- end document ---------- %\r\n\r\n\\input{../footer}\r\n", "meta": {"hexsha": "1d49f2501ea39751c688b0122e24bddfc5134210", "size": 4346, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manual/en/gcd.tex", "max_stars_repo_name": "turkeydonkey/nzmath3", "max_stars_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-05-26T19:22:17.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-26T19:22:17.000Z", "max_issues_repo_path": "manual/en/gcd.tex", "max_issues_repo_name": "turkeydonkey/nzmath3", "max_issues_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "manual/en/gcd.tex", "max_forks_repo_name": "turkeydonkey/nzmath3", "max_forks_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.1228070175, "max_line_length": 198, "alphanum_fraction": 0.6352968247, "num_tokens": 1425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835207180243, "lm_q2_score": 0.763483758172699, "lm_q1q2_score": 0.6378780982891552}} {"text": "%!TEX root = ../dokumentation.tex\n\\chapter{Implementation}\n\nThe implementation of ARIMA needs two separate scripts: One for predictions/forecasting and one for training a suitable model.\nNaturally the training script is going to use the prediction script in some way to test the current model and improve it.\n\nBoth scripts need to do three things: \nFirst, loading the time series and parameters p, d, q, P, D, Q, and s. Then differencing the time series according to d and D. And finally they both have to construct the auxiliary matrix $Z$. This matrix is needed to do different calculations within both scripts. The idea behind it, is to provide all the $X_{t-h}$ for any $h \\in [1, max(p, q, Ps, Qs, p + Ps, q + Qs]$ for each $\\hat{X}_t$ to be computed.\nThis means that there is p columns for \\acs{AR}$(p)$, q columns for \\acs{MA}$(q)$, P columns for \\acs{SAR}$(P)$, Q columns for \\acs{SMA}$(Q)$ and when combining non seasonal and seasonal models there is also additional $p \\cdot P$ and $q \\cdot Q$ columns. Each one of those columns corresponds to one of the $X_{t-h}$ terms found in equation for \\eqref{eq:multiplicative_SARIMA}. So essentially Z is constructed in a way that the n-th row of Z contains all the time series values needed to predict $X_n$.\n\n\\begin{figure}[ht]\n\t\\centering\n\t\\scalebox{1}{\\includegraphics[width=1\\textwidth]{images/Zmatrix.png}}\n \\caption{Schema of auxiliary matrix Z}\n\\end{figure}\n\nThe following function is designed to built such a matrix:\n\\begin{lstlisting}[caption=Prediction of \\acs{AR}$(p)$ or \\acs{SAR}$(P)$ ,captionpos=b]\nconstructPredictorMatrix = function (Matrix[Double] X, Integer p, Integer P, Integer q, Integer Q, Integer s)return( Matrix[Double] Z){\n\tZ = matrix(0, nrow(X), p+P+Q+q+p*P+q*Q)\n\n\t# fills Z with values used for non seasonal AR prediction\n\tfor\t(i in seq(1, p, 1)){\n\t\tZ = addShiftedMatrix(Z, X, i, i)\n\t}\n\n\t#prediction values for seasonal AR \n\tfor(i in seq(1, P, 1)){\n\t\tZ = addShiftedMatrix(Z, X, (i * s), p + i)\n\t}\n\n\t#prediction values for combined models of non-seasonal and seasonal AR\n\tZ = addValuesForCombinedModel (Z, X, p, P, s, p + P)\n\n\t#prediction values for non seasonal MA \n\tfor(i in seq(1, q, 1)){\n\t\tZ = addShiftedMatrix(Z, X, i, p + P + p*P + i)\n\t}\n\n\t#prediction values for seasonal MA \n\tfor(i in seq(1,Q, 1)){\n\t\tZ = addShiftedMatrix(Z, X, (i * s), p + P + p*P + q + i)\n\t}\n\n\t#prediction values for combined models of non-seasonal and seasonal MA\n\tZ = addValuesForCombinedModel (Z, X, q, Q, s, p + P + p*P + q + Q )\n}\n\\end{lstlisting}\n\nThe idea behind this function is to add each column step by step. It starts with all the columns for non seasonal and seasonal \\acl{AR} and proceeds to \\acl{MA}.\n\n\n%TODO: Maybe add example of Z matrix here\n\n\\section{Prediction Script}\n\nThe task of the prediction script is simply to calculate the t-dimensional vector $\\hat{X}$ containing all the predictions $\\hat{X}_t$ for each $t \\in [1,T]$ with T being the number of rows of $Z$. When using the prediction script for training purposes this is also equal to the size of the time series $X$ used for training.\n\nIn the simplest case of forecasting \\acs{AR}$(p)$ or \\acs{SAR}$(P)$ the code for the \\lstinline{predict()} function is quite straight forward:\n\n\\begin{lstlisting}[caption=Prediction of \\acs{AR}$(p)$ or \\acs{SAR}$(P)$ ,captionpos=b]\npredict = function(Matrix[Double] weights, Matrix[Double] Z) return (Matrix[Double] approximated_solution){\n\tapproximated_solution = Z%*%weights\n}\n\\end{lstlisting}\n\nFor predicting non seasonal and seasonal \\acs{SAR} at once the combination of the purely non seasonal and purely seasonal part have to be taken into consideration. This is done by constructing another auxiliary matrix $\\omega$ (referred to in the scripts as 'combined\\_weights') that holds all weights for $p$ and $P$ as well as all the combinations of those weights:\n\n\\begin{lstlisting}[caption=Prediction of \\acs{SAR}$(p)(P)$ ,captionpos=b]\ncombined_weights = weights\nif (p>0 & P>0){\n\tcombined_weights = rbind(combined_weights, matrix(weights[1:p,] %*% t(weights[p+1:p+P,]), rows=p*P, cols=1))\n}\napproximated_solution = Z%*%combined_weights\n\\end{lstlisting}\n\nLike already discussed in the chapter \\ref{arimamodel} and described in more detail with the equations \\eqref{eq:MA_q} through \\eqref{eq:example_MA_three_system_2}, calculating $\\hat{X}$ for \\acs{MA}(p) requires to construct a third matrix R to be used to describe the system of linear equations that needs to be solved with:\n\\[A = I + R\\]\n\n\\begin{lstlisting}[caption=Prediction of \\acs{MA}$(q)$ ,captionpos=b]\nb = Z%*%weights\nR = matrix(0, nrow(Z), nrow(Z))\nfor(i in seq(1, q, 1)){\n\td_ns = matrix(as.scalar(weights[P+p+i,1]), nrow(R)-i, 1)\n\tR[1+i:nrow(R),1:ncol(R)-i] = R[1+i:nrow(R),1:ncol(R)-i] + diag(d_ns)\n}\napproximated_solution = numerical_solvers::jacobi (A=A, b=b)\n\\end{lstlisting}\n\nThe equivalent for-loop can be implemented for \\acs{SMA} as well. With the only difference that instead of the term $i$ the term $i\\cdot s$ is to be used.\n\nTo solve the system of linear equations the \\textit{Jacobi} solver had to be implemented in the way it was described in chapter \\ref{jacobi}:\n\\begin{lstlisting}[caption=Prediction of \\acs{MA}$(q)$ ,captionpos=b]\njacobi = function (Matrix[Double] R, Matrix[Double] b, Double tolerance = 0.01, Integer max_iterations = 1000) return (Matrix[Double] x){\n\tx = matrix(0, nrow(R), 1)\n \tk = 0\n\tdiff = tolerance+1\n\twhile(k < max_iterations & diff > tolerance){\n\t\tx_k = b - R%*%x\n\t\tdiff = sum((x_k-x)^2)\n\t\tx = x_k\n\t\tk += 1\n\t}\n}\n\\end{lstlisting}\n\n\\section{Training Script}\n\n\n%Pseudo code/description arima training\n\n%Load Time Series X_t and parameters p, d, q, P, D, Q, and s\n%Difference X_tnon seasonaland seasonal\n%Build auxiliary matrix Z with the series X_t as the first column, the target values and all the previous values from X_t that can be used to predict the target value at time \n%Solve nonlinear optimization problem for ARMA(p,q)〖(P,Q)〗_s to find best values for 〖 ϕ〗_i,〖 Φ〗_j,〖 θ〗_k,〖 Θ〗_l over a) Maximum likelihood estimation b) Nonlinear least squares method \n\n\n\n%arima residuals and css \n%gradient of arima\n%bfgs\n\n%building Z\n\t%addshiftedMatrix\n %addValuesForCombinedModel\n \n%differencing\n\n\\chapter{Result}\n% What was tested\n\n\\section{Correctness Test}\n\n\\section{Performance Test}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "9985918cc8b0fc6a29df6066ab019f168e99d56d", "size": 6308, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/04kapitel.tex", "max_stars_repo_name": "TobiasSchmidtDE/Scalable-Time-Series-Forecasting-with-ARIMA-for-SystemML", "max_stars_repo_head_hexsha": "c7935e293aae2d5dd3d10803537423c9dd24dc8e", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "content/04kapitel.tex", "max_issues_repo_name": "TobiasSchmidtDE/Scalable-Time-Series-Forecasting-with-ARIMA-for-SystemML", "max_issues_repo_head_hexsha": "c7935e293aae2d5dd3d10803537423c9dd24dc8e", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/04kapitel.tex", "max_forks_repo_name": "TobiasSchmidtDE/Scalable-Time-Series-Forecasting-with-ARIMA-for-SystemML", "max_forks_repo_head_hexsha": "c7935e293aae2d5dd3d10803537423c9dd24dc8e", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.2287581699, "max_line_length": 504, "alphanum_fraction": 0.7190868738, "num_tokens": 1861, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267660487573, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6378226893959148}} {"text": "\\section{Table of Laplace Transforms}\r\n\\noindent\r\nBelow is a brief table of Laplace transforms. Although most differential equations textbooks have a longer table, this table covers all Laplace transforms done in this text.\r\n\\renewcommand{\\arraystretch}{1.75}\r\n\\begin{table}[H]\r\n\t\\centering\r\n\t\\begin{tabular}{ l | l }\r\n\t\t$f(t)$ & $F(s) = \\Laplace{f}$\t\\\\\r\n\t\t\\hline\r\n\t\t$f(at)$ & $\\frac{1}{a}F\\left(\\frac{s}{a}\\right)$\t\\\\\r\n\t\t$f'(t)$ & $sF(s) - f(0)$\t\\\\\r\n\t\t$f^{(n)}(t)$ & $s^nF(s) - \\sum_{i=0}^{n-1}{s^{n-i-1}f^{(i)}}(0)$\t\\\\\r\n\t\t$t^nf(t)$ & $(-1)^nF^{(n)}(s)$\t\\\\\r\n\t\t$\\frac{1}{t}f(t)$ & $\\int_s^\\infty{F(u) \\mathrm{d}u}$\t\\\\\r\n\t\t$a$ & $\\frac{a}{s}\\text{, } s > 0$ \\\\\r\n\t\t$e^{at}$ & $\\frac{1}{s-a}\\text{, } s > a$\t\\\\\r\n\t\t$t^n, n = 1,2,\\ldots$ & $\\frac{n!}{s^{n+1}}\\text{, } s < 0$\t\\\\\r\n\t\t$e^{at}t^n, n = 1,2,\\ldots$ & $\\frac{n!}{(s-a)^{n+1}}\\text{, } s > a$\t\\\\\r\n\t\t$\\sin{(bt)}$ & $\\frac{b}{s^2 + b^2}\\text{, } s > 0$ \\\\\r\n\t\t$\\cos{(bt)}$ & $\\frac{s}{s^2 + b^2}\\text{, } s > 0$ \\\\\r\n\t\t$e^{at}\\sin{(bt)}$ & $\\frac{s}{(s-a)^2 + b^2}\\text{, } s > a$ \\\\\r\n\t\t$e^{at}\\cos{(bt)}$ & $\\frac{s-a}{(s-a)^2 + b^2}\\text{, } s > a$ \\\\\r\n\t\\end{tabular}\r\n\\end{table}", "meta": {"hexsha": "815c849cb8f1a9d0170c4a61f4c74a2553968da4", "size": 1132, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/appendix/laplaceTable/laplaceTable.tex", "max_stars_repo_name": "aneziac/Math-Summaries", "max_stars_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2020-03-26T06:20:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T17:38:45.000Z", "max_issues_repo_path": "diffEq/appendix/laplaceTable/laplaceTable.tex", "max_issues_repo_name": "aneziac/Math-Summaries", "max_issues_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 26, "max_issues_repo_issues_event_min_datetime": "2020-03-28T17:44:18.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T04:47:03.000Z", "max_forks_repo_path": "diffEq/appendix/laplaceTable/laplaceTable.tex", "max_forks_repo_name": "aneziac/Math-Summaries", "max_forks_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2020-04-10T05:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-17T15:21:12.000Z", "avg_line_length": 47.1666666667, "max_line_length": 174, "alphanum_fraction": 0.4770318021, "num_tokens": 550, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267694452331, "lm_q2_score": 0.7341195152660688, "lm_q1q2_score": 0.637822686835319}} {"text": "% !TEX root = Main.tex\n\\section{Generative Models}\nDeep Probabilistic Models: $x^{l+1} = F^l(x^l) + W^l z^l$, $z^{l}\\sim \\mathcal{N}(0,1)$. Allows for easy sampling from complex distributions.\n\\subsection{Autoencoders (AE)}\nLearn low dim. representation. \nLinear: $min \\frac{1}{2k} \\sum_{i}^n ||x_i - DCx_i||^2$, C encoder, D decoder. Frobenius norm optimal solution from SVD(X): $C^*=U_d^T, D*=U_d$, $U$\nNon-linear: Use feedforward NN for encoding and decoding $F = H \\circ G$. \\\\\nDenoising: Perturb input $x := x + \\eta$\n\\subsection{Variational AE}\nGet gen. AE by constraining distribution of latent vars to follow a given distrib.\n\\begin{inparaenum}[\\color{red}1.]\n\t\\item Condition on y, compute $q(z|y)$\n\t\\item sample $z^l \\sim q(z^l|y)$\n\t\\item forward $z^l$\n\t\\item backpropagate for generated example.\n\\end{inparaenum}\nMax ELBO $E_q[log p(x^{1:L},y;\\theta) - log q(x^{1:L};y)] \\leq log p (y;\\theta)$ \\\\ \nOr: $log p_\\theta(x^{i}) \\geq E_z[log p_\\theta(x^{i}|z)] - D_{KL}(q_\\theta(z|x^{i}||p_\\theta(z)))$\nAutoregressive Model: Gen. output 1 var at a time. \nPixel CNN: $p(x) = \\prod_i^{n^2} p(x_i|x_1,...,x_{i-1})$. \\\\\nAutoregr. process of order p: $x_t = \\sum_{i=1}^p \\phi_i x_{t-p} + \\delta_t$.\nSolve for $\\phi^*$ by MLE: $p(x, \\phi) = p(x_1)\\prod_{i=2}^n p(x_i|x_{i-1},...,x_1,\\phi)$ or via least sq. solution of the problem $\\phi^* = (X^TX)^{-1}X^Ty$", "meta": {"hexsha": "ecd8a967583d5957756870df40f5d5f8c35bed07", "size": 1356, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "GenerativeModels.tex", "max_stars_repo_name": "phil9987/eth-cil-exam-summary", "max_stars_repo_head_hexsha": "6c4986f399ca40c2270b8fc01150a69f89650db7", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "GenerativeModels.tex", "max_issues_repo_name": "phil9987/eth-cil-exam-summary", "max_issues_repo_head_hexsha": "6c4986f399ca40c2270b8fc01150a69f89650db7", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "GenerativeModels.tex", "max_forks_repo_name": "phil9987/eth-cil-exam-summary", "max_forks_repo_head_hexsha": "6c4986f399ca40c2270b8fc01150a69f89650db7", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.6363636364, "max_line_length": 157, "alphanum_fraction": 0.6482300885, "num_tokens": 527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240964782011, "lm_q2_score": 0.7371581568543044, "lm_q1q2_score": 0.6378070002258016}} {"text": "\n\\section{Kernels\\label{app:kernels}}\nHere we list the elliptic \\pde's investigated in this work along with the associated kernels for their single- and double-layer potentials.\nIn this section, $\\vx$ and $\\vy$ are in $\\mathbb{R}^3$, $\\vx$ is the point of evaluation and $\\vy$ is a point on the boundary and $\\vr = \\vx - \\vy$. \nRecall that $\\vn$ is the outward pointing unit normal at $\\vy$ to the domain boundary $\\Gamma$.\nWe denote the single layer kernel, also known as the \\textit{fundamental solution} or \\textit{Green's function} of the \\pde, by $S$ and the double layer kernel by $D$.\n\\begin{enumerate}\n \\item \\textit{Laplace equation}:\n \\begin{align*}\n &\\Delta u = 0\\\\\n &S(\\vx,\\vy) = \\frac{1}{4\\pi}\\frac{1}{\\|\\vr\\|}, \\quad \n D(\\vx, \\vy) = - \\frac{1}{4\\pi} \\frac{\\vr \\cdot \\mathbf{n}}{\\|\\vr\\|^3}\n \\end{align*}\n \\item \\textit{Stokes equation}:\n \\begin{align*}\n &\\mu\\Delta u - \\nabla p = 0, \\,\\, \\nabla \\cdot u = 0\\\\\n &S(\\vx,\\vy) = \\frac{1}{8\\pi\\mu}\\left( \\frac{1}{\\|\\vr\\|} + \\frac{\\vr \\otimes\\vr}{\\|\\vr\\|^3}\\right), \\quad \n D(\\vx, \\vy) = - \\frac{3}{4\\mu\\pi} \\frac{\\vr \\otimes\\vr}{\\|\\vr\\|^5}(\\vr \\cdot \\mathbf{n})\n \\end{align*}\n \\item \\textit{Elasticity equation}:\n \\begin{align*}\n &\\mu\\Delta u - \\frac{\\mu}{1-2\\nu}\\nabla(\\nabla \\cdot u) = 0\\\\\n &S(\\vx,\\vy) = \\frac{1}{16\\pi\\mu(1-\\nu)}\\left( \\frac{3-4\\nu}{\\|\\vr\\|} + \\frac{\\vr \\otimes\\vr}{\\|\\vr\\|^3}\\right), \\\\\n &D(\\vx, \\vy) = - \\frac{1-2\\nu}{8\\mu(1-\\nu)} \\left(\n \\frac{1}{\\|\\vr\\|^3} \\left(\n \\vr \\otimes \\vn - (\\vr\\cdot\\vn) I - \\vn \\otimes \\vr \n \\right) - \\frac{3}{1-2\\nu}\n \\frac{(\\vr\\cdot\\vn) (\\vr \\otimes \\vr)}{\\|\\vr\\|^5}\n \\right)\n \\end{align*}\n\\end{enumerate}\n", "meta": {"hexsha": "941bdad6553ac70d2bba1247839dfd0ff61d2267", "size": 1708, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hedgehog/kernels.tex", "max_stars_repo_name": "mmorse1217/nyu-thesis-template", "max_stars_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hedgehog/kernels.tex", "max_issues_repo_name": "mmorse1217/nyu-thesis-template", "max_issues_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hedgehog/kernels.tex", "max_forks_repo_name": "mmorse1217/nyu-thesis-template", "max_forks_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.375, "max_line_length": 167, "alphanum_fraction": 0.5690866511, "num_tokens": 684, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240756264639, "lm_q2_score": 0.7371581684030623, "lm_q1q2_score": 0.6378069948470367}} {"text": "\\section*{Problem 2 Solution}\n\n\\begin{enumerate}[a)]\n\n\\item \n\nThe $Q$-value of a reaction is given by:$$Q=[m(\\text{x})+m(\\text{X})-m(\\text{y})-m(\\text{Y})]c^{2}.$$\n\n\\begin{spacing}{1.75}\t\n$^{235}\\text{\\normalfont{U}}+n \\rightarrow \\-\\ ^{135}\\text{\\normalfont{I}}+ \\! ^{98}\\text{\\normalfont{Y}}+3n$\\\\\n\t\\tab $Q = [m(^{235}\\text{\\normalfont{U}})+m(n)-m(^{135}\\text{\\normalfont{I}})-m(^{98}\\text{\\normalfont{Y}})-3m(n)]c^{2}$\\\\\n\t\\tab $Q = [235.043924\\text{u}+1.008665\\text{u}-134.910048\\text{u}-97.922203\\text{u}-3(1.008665\\text{u})]c^{2}$\\\\\n\t\\tab $Q = 0.194343\\text{u} \\cdot c^{2}$\\\\\n\t\\tab $\\boxed{Q = 181.031 \\text{ MeV}}$ \n\\end{spacing}\n\n\n\\item \n\nThe binding energy $B(Z,A)$ for any nuclei can be found approximately from the equation:$$M(Z,A) = Zm(^{1}\\text{H})+(A-Z)m_{n}-B(Z,A)/c^{2},$$ where $M(Z,A),\\; Zm(^{1}\\text{H})$ and $m_{n}$ are experimentally calculated values. Per nucleon, the binding energy can be expressed as: $$B_{A}(Z,A) = [Zm(^{1}\\text{H})+(A-Z)m_{n}-M(Z,A)]c^{2}/A.$$\n\n\\begin{spacing}{1.75}\n\n\\textbf{$^{2}$H}\\\\\n$B_{A}(1,2) = [m(^{1}\\text{H})+(2-1)m_{n}-M(1,2)]c^{2}/2$\\\\\n$B_{A}(1,2) = [(1.007825\\text{u})+(1.008665\\text{u})-(2.014102\\text{u})]c^{2}/2$\\\\\n$B_{A}(1,2) = 0.001194\\text{u} \\cdot c^{2}$\\\\\n$\\boxed{B_{A}(1,2) = 1.112\\text{ MeV}}$\\\\\n\\textbf{$^{56}$Fe}\\\\\n$B_{A}(26,56) = [26m(^{1}\\text{H})+(56-26)m_{n}-M(26,56)]c^{2}/56$\\\\\n$B_{A}(26,56) = [26(1.007825\\text{u})+30(1.008665\\text{u})-(55.934939\\text{u})]c^{2}/56$\\\\\n$B_{A}(26,56) = 0.009437\\text{u} \\cdot c^{2}$\\\\\n$\\boxed{B_{Z}(26,56) = 8.791\\text{ MeV}}$\\\\\n\\textbf{$^{235}$U}\\\\\n$B_{A}(92,235) = [92m(^{1}\\text{H})+(235-92)m_{n}-M(92,235)]c^{2}/235$\\\\\n$B_{A}(92,235) = [92(1.007825\\text{u})+143(1.008665\\text{u})-(235.043924\\text{u})]c^{2}/235$\\\\\n$B_{A}(92,235) = 0.00814924 u\\text{u} \\cdot c^{2}$\\\\\n$\\boxed{B_{Z}(92,235) = 7.591\\text{ MeV}}$\n\\end{spacing}\n\n\\end{enumerate}\n\n", "meta": {"hexsha": "0784a91ada6b3329f6709e36b4d59cf35a0d7f8c", "size": 1854, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises/drafts/disc01/disc01_solution02.tex", "max_stars_repo_name": "mitchnegus/NE150-discussion", "max_stars_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "exercises/drafts/disc01/disc01_solution02.tex", "max_issues_repo_name": "mitchnegus/NE150-discussion", "max_issues_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "exercises/drafts/disc01/disc01_solution02.tex", "max_forks_repo_name": "mitchnegus/NE150-discussion", "max_forks_repo_head_hexsha": "1d2afe0fc4830c3d13d491b9d6ccb7819083c5ad", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.1162790698, "max_line_length": 342, "alphanum_fraction": 0.5609492988, "num_tokens": 923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8723473846343394, "lm_q2_score": 0.7310585903489891, "lm_q1q2_score": 0.6377370493054075}} {"text": "\\graphicspath{{Pics/combi/geo/}}\n\n\\newpage\n\\section{Combinatorial Geometry}\n\n\n\\begin{myitemize}\n\\item \\href{https://blogm4e.files.wordpress.com/2016/08/combinatorial-geometry-maria-monks-mop-2010.pdf}{Combinatorial Geometry - Maria Monk (MOP 2010)}\n\\end{myitemize}\n\n\n\\begin{take_note*}{}\n \\begin{itemize}[wide=0pt]\n \\item Consider the convex hull made up of the points.\n \\item Consider the extreme points: smallest or highest $ x $ or $ y $ coordinate.\n \\item Find the triangle (quadrilateral, pentagon, etc.) with the vertices being the points from your set $ S $, so that the area of the triangle is minimal/maximal.\n \\end{itemize}\n\\end{take_note*}\n\n\\theo{https://en.wikipedia.org/wiki/Helly's_theorem}{Helly's Theorem}{Let $ X_1, ..., X_n $ be a finite collection of convex subsets of $ \\R^d $, with $ n > d $. If the intersection of every $ d + 1 $ of these sets is nonempty, then the whole collection has a nonempty intersection; that is,\n\\[\\bigcap _{j=1}^{n}X_{j}\\neq \\varnothing\\]}\n\n\n\\vspace{1em}\n\n\\begin{myitemize}\n \\item \\href{https://math.berkeley.edu/~moorxu/misc/equiareal.pdf}{Sperner's\n Lemma - Moor Xu}\n\\end{myitemize}\n\n\\theo{https://www.wikiwand.com/en/Sperner's_lemma}\n{Sperner's Lemma}{\n Given a triangle $ABC$, and a triangulation $\\mathcal{T}$ of the triangle,\n the set $S$ of vertices of $\\mathcal{T}$ is colored with three colors in\n such a way that\n \\begin{enumerate}\n \\item $A, B$, and $C$ are colored $1, 2$, and $3$ respectively.\n \\item Each vertex on an edge of $ABC$ is to be colored only with one of the\n two colors of the ends of its edge. For example, each vertex on $AC$\n must have a color either $1$ or $3$.\n \\end{enumerate}\n Then there exists a triangle from $\\mathcal{T}$, whose vertices are colored with the\n three different colors. More precisely, there must be an odd number of\n such triangles. \n}\n\n\n\\begin{prooof}\n Consider a graph $G$ built from the triangulation $\\mathcal{T}$ as\n follows:\\\\\n\n {\\color{solC}The vertices of $G$ are the members of $\\mathcal{T}$ plus the\n area outside the triangle. Two vertices are connected with an edge if\n their corresponding areas share a common border with an edge $1$--$2$.\\\\}\n\n Note that on the interval $AB$ there is an odd number of borders colored\n $1$--$2$. Therefore, the vertex of $G$ corresponding to the outer area has\n an odd degree. \\\\\n\n But since in a finite graph there is an even number of\n vertices with odd degree, in the remaining graph, excluding the\n outer area, there is {\\color{solC}an odd number of vertices with odd\n degree} corresponding to members of $\\mathcal{T}$.\\\\\n\n It can be easily seen that the only possible degree of a triangle from\n $\\mathcal{T}$ is $0, 1$, or $2$, and that the degree $1$ corresponds to a\n triangle colored with the three colors $1, 2$, and $3$.\\\\\n\n Thus we have obtained a slightly stronger conclusion, which says that in a\n triangulation $\\mathcal{T}$ there is an odd number (and at least one) of\n full-colored triangles. \n\\end{prooof}\n\n\n\\theo{https://www.wikiwand.com/en/Brouwer_fixed-point_theorem}\n{Brouwer fixed-point theorem}{\n Let $B^n$ be the $n$th dimensional ball. Then any continuous map\n $f:B^n \\to B^n$ has a fixed point.\n}\n\n\\theo{https://en.wikipedia.org/wiki/Monsky's_theorem}\n{Monsky's theorem}{\n If we triangulate a square with triangles of equal area, then there must\n be an even number of triangles used.\n}\n\n\n\\theo{https://en.wikipedia.org/wiki/Pick's_theorem}\n{Pick's Theorem}{\n A simple polygon $P$ has all of its vertices on the lattice points of $xy$\n grid. If it's area is $A$, the number of lattice points inside the\n polygon is $i$ and the number of lattice points on the boundary is $b$,\n then \\textbf{Pick's theorem} states that\n \\[A = i + \\frac{b}{2} -1\\] \n}\n\n\n\n\n\\newpage\n\\subsection{Problems}\n\n\\prob{https://artofproblemsolving.com/community/c6h535002p3067558}{ARO 2013 P9.4}{E}{$ N $ lines lie on a plane, no two of which are parallel and no three of which are concurrent. Prove that there exists a non-self-intersecting broken line $ A_1A_2A_3\\dots A_N $ with $ N $ parts, such that on each of the $ N $ lines lies exactly one of the $ N $ segments of the line.}\\label{problem:constructive_algo_8}\\label{problem:induction_type1_7}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1424941p8024557}{EGMO 2017 P3}{M}{There are $ 2017 $ lines in the plane such that no three of them go through the same point. Turbo the snail sits on a point on exactly one of the lines and starts sliding along the lines in the following fashion: she moves on a given line until she reaches an intersection of two lines. At the intersection, she follows her journey on the other line turning left or right, alternating her choice at each intersection point she reaches. She can only change direction at an intersection point. Can there exist a line segment through which she passes in both directions during her journey?}\\label{problem:plane_coloring_1}\n\n\\solu{The condition that tells us to go either right or left, seems very non-rigorous. So to rigorize this condition, instead of using right or left condition in the direction, we consider what’s on our right and left. (INTUITION) After some experiment we see (not all of us) that if we color the plane with two colors in a way where every neighboring regions have different colors, we find some interesting stuff. (CREATIVITY) With this we are done. \\hrf{plane_coloring}{Color the Plane}}\n\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h101306p571973}{ISL 2006 C2}{TE}{Let $ P $ be a regular $ 2006 $ -gon. A diagonal is called good if its endpoints divide the boundary of $ P $ into two parts, each composed of an odd number of sides of $ P $. The sides of $ P $ are also called good.\n\nSuppose $ P $ has been dissected into triangles by $ 2003 $ diagonals, no two of which have a common point in the interior of $ P $. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.}\\label{problem:induction_type2_4}\\label{problem:bijection_9}\n\n\n\\solu{The straight way, induction.}\n\n\\solu{The intuitive way, bijection. There are at most $ n $ good triangles, there are $ 2n $ edges, so a mapping that takes a two edges to a single good triangle must exist. Finding it is not that hard.}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h589865p3493114}{ARO 2014 P9.3}{E}{In a convex $ n $ -gon, several diagonals are drawn. Among these diagonals, a diagonal is called good if it intersects exactly one other diagonal drawn (in the interior of the $ n $ -gon). Find the maximum number of good diagonals.}\\label{problem:induction_type2_1}\n\n\\solu{There can be two cases, two good diagonals intersecting each other, and no two good diagonals intersecting each other. In the first case, we just use induction, and in the later, all of the good diagonals create a ``triangulation'' of the polygon, which gives us the numbers.}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1181527p5720110}{ISL 2013 C2, IMO 2013 P2}{E}{A configuration of $ 4027 $ points in the plane is called Colombian if it consists of $ 2013 $ red points and $ 2014 $ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied:\n\n \\begin{enumerate}\n\n \\item No line passes through any point of the configuration.\n \\item No region contains points of both colors.\n\n \\end{enumerate}\n\nFind the least value of $ k $ such that for any Colombian configuration of $ 4027 $ points, there is a good arrangement of $ k $ lines.}\\label{problem:sandwiching_points_1}\n\n\\solu{Obviously a n00b would think about induction. The only problem occurs when the convex hull completely consists of red points. In this case, after some investigation, we should get the sandwiching two points idea.}\n\n\\solu{Another way of inductive approach is like this, as the problem condition says that no region contains points of both colors, which means if we connect any two red and blue points, some line must bisect this segment. Now \\hrf{problem:convex_hull_1}{it is known} that there is non intersecting partition of the points in to red-blue segments. So suppose in such a partition, we draw bisectors of each segments. Now there will be some holes in this proof. We see that to fill these holes, we have to focus on two red points with their respective blue partners, and draw the two bisectors in a way that separates the two red points form the blue points. So to remove further holes, we get the sandwiching idea.}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h366745p2018324}{ILL 1985}{E}{Let $A$ and $B$ be two finite disjoint sets of points in the plane such that no three distinct points in $A \\cup B$ are collinear. Assume that at least one of the sets $A, B$ contains at least five points. Show that there exists a triangle all of whose vertices's are contained in $A$ or in $B$ that does not contain in its interior any point from the other set.}\n\n\\solu{Concentrating on one of the sets five points such that there is no other points of the same set inside the hull of those five points.}\\label{problem:convex_hull_2}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h79789p456611}{APMO 1999 P5}{M}{Let $S$ be a set of $2n+1$ points in the plane such that no three are collinear and no four concyclic. A circle will be called ``Good'' if it has $ 3 $ points of $S$ on its circumference, $n-1$ points in its interior and $n-1$ points in its exterior. Prove that the number of good circles has the same parity as $n$.}\n\n\n\\solu{When thinking about induction, got a feeling that double counting with the number of good circles going through pairs of points might be useful, because a good circle will be counted three times, if we can show that every pair has odd number of good circles, we are done. So, take a pair. Now we need to `sort' the points somehow. See that, we can't sort the points in a trivial way with numbers, so moving to angles. Now setting conditions for a point inside of a circle in terms of angles, we see amazing patter, and an easy way to calculate the number of good circle of that pair of points.}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1113183p5083543}{ISL 2014 C1}{E}{Let $ n $ points be given inside a rectangle $ R $ such that no two of them lie on a line parallel to one of the sides of $ R $. The rectangle $ R $ is to be dissected into smaller rectangles with sides parallel to the sides of $ R $ in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect $ R $ into at least $ n + 1 $ smaller rectangles.}\\label{problem:extreme_object_7}\\label{problem:double_counting_3}\n\n\\solu{Work with the largest continuous segments, and their endpoints.}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h195050p1071295}{ISL 2007 C2}{EM}{A rectangle $ D$ is partitioned in several ($ \\ge2$) rectangles with sides parallel to those of $ D$. Given that any line parallel to one of the sides of $ D$, and having common points with the interior of $ D$, also has common interior points with the interior of at least one rectangle of the partition; prove that there is at least one rectangle of the partition having no common points with $ D$'s boundary.}\\label{problem:extreme_object_17}\n\n\\begin{minipage}{.6\\linewidth}\n \\solu{\n There existing such a rectangle means that there is a rectangular region\n inside of the original rectangle. So what if we walked along the segments,\n and cut a smaller rectangle from the inside of the rectangle? Like the way\n in the game.\n }\n\\end{minipage}\\hfill%\n\\begin{minipage}{.4\\linewidth}\n \\figdf{.8}{2007c2}{}\n\\end{minipage}\n\n\n\\solu{Starting from one corner, and taking the opposite corner of the rectangle containing that corner, we use infinite decent to reach a contradiction.}\n\n\\solu{Using \\hrf{problem:extreme_object_7}{ISL 2014 C1} as a lemma.}\n\n\\solu{Take one side of the square. Take a ``sandwiched'' rectangle touching that side. If no such rectangle exists, then it's just a special case that can be dealt with ease.}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h5753p18977}{ISL 2003 C2}{E}{Let $D_1,D_2\\dots,D_n$ be closed discs in the plane. (A closed disc is the region limited by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most $2003$ discs $D_i$. Prove that there exists a disc $D_k$ which intersects at most $7\\cdot 2003 - 1 = 14020$ other discs $D_i$.}\n\n\\solu{Just go with the natural idea.}\n\n\\prob{https://artofproblemsolving.com/community/c6h5785p19086}{ISL 2003 C3}{E}{Let $n \\geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles.}\\label{problem:double_counting_10}\n\n\\solu{double count}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1389042p7736716}{Tournament of Towns 2015S S4}{A convex$N-$gon with equal sides is located inside a circle. Each side is extended in both directions up to the intersection with the circle so that it contains two new segments outside the polygon. Prove that one can paint some of these new $2N$ segments in red and the rest in blue so that the sum of lengths of all the red segments would be the same as for the blue ones.}\n\n\\solu{Just use what's the most natural, POP, on one vertex point.}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h145844p825495}{USAMO 2007 P2}{E}{A square grid on the Euclidean plane consists of all points $(m,n)$, where $m$ and $n$ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least $5$?}\n\n\n\\begin{minipage}{.5\\linewidth}\n \\prob{https://artofproblemsolving.com/community/c6h1135648p5301617}\n {MEMO 2015 T4}{EM}{\n Let $N$ be a positive integer. In each of the $N^2$ unit squares of an\n $N\\times N$ board, one of the two diagonals is drawn. The drawn\n diagonals divide the $N\\times N$ board into $K$ regions. For each $N$,\n determine the smallest and the largest possible values of $K$.\n }\n\\end{minipage}\\hfill%\n\\begin{minipage}{.4\\linewidth}\n \\figdf{.7}{MEMO2015T4}{}\n\\end{minipage}\n\n\n\\solu{An Algorithmic Approach: Consider each diagonal as $ 0 $ or $ 1 $, prove that the maximum configuration is the one with alternating $ 0, 1 $s and the minimum one is the one with all $ 0 $s.}\n\n\\solu{A Counting Approach: Just count and bound with the minimum areas of the regions.}\n\n\n\\prob{}{}{EM}{In every cells of a $ m\\times n $ grid, one of the two diagonals are drawn. Prove that there exist a path on these diagonals from left to right or from up to bottom of the grid.}\n\n\\solu{First remove the cycles, then take the largest path from left to right, and use induction.}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c132h1546693p9385334}{Math Price for Girls 2017 P4}{E}{A lattice point is a point in the plane whose two coordinates are both integers. A lattice line is a line in the plane that contains at least two lattice points. Is it possible to color every lattice point red or blue such that every lattice line contains exactly 2017 red lattice points? Prove that your answer is correct.}\n\n\\solu{Transfinite induction.}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1217113p6063817}{China TST 2016 T3P2}{E}{In the coordinate plane the points with both coordinates being rational numbers are called rational points. For any positive integer $n$, is there a way to use $n$ colours to colour all rational points, every point is coloured one colour, such that any line segment with both endpoints being rational points contains the rational points of every colour?}\n\n\\solu{Transfinite induction}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1709989p11022201}{IGO 2018 A3}{E}{Find all possible values of integer $n > 3$ such that there is a convex $n$-gon in which, each diagonal is the perpendicular bisector of at least one other diagonal.}\n\n\\solu{Taking maximum terminal triangle.}\n\n\n\\prob{}{Lithuania ??}{E}{Prove that in every polygon there is a diagonal that cuts off a triangle and lies completely within the polygon.}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h202941p1116402}{Romanian TST 2008 T1P4}{E}{Prove that there exists a set $ S$ of $ n - 2$ points inside a convex polygon $ P$ with $ n$ sides, such that any triangle determined by $3$ vertices of $ P$ contains exactly one point from $ S$ inside or on the boundaries.}\n\n\\solu{Checking small cases inductively quickly shows a construction.}\n\n\n\\prob{}{Iran TST ??}{E}{In an isosceles right-angled triangle shaped billiards table, a ball starts moving from one of the vertices adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if we reach to a vertex then it is not the vertex at initial position}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1662915p10561203}{APMO 2018 P4}{EM}{Let $ABC$ be an equilateral triangle. From the vertex $A$ we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle $\\alpha$, it leaves with a directed angle $180^{\\circ}-\\alpha$. After $n$ bounces, the ray returns to $A$ without ever landing on any of the other two vertices. Find all possible values of $n$.}\n\n\n\\solu{Reflect the whole board when just reflecting the ball doesn't seem to be helping. GLOBAL}\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h195043p1071276}{ISL 2007 C5}{EM}{In the Cartesian coordinate plane define the strips $ S_n = \\{(x,y)|n\\le x < n + 1\\}$, $ n\\in\\mathbb{Z}$ and color each strip black or white. Prove that any rectangle which is not a square can be placed in the plane so that its vertices have the same color.}\n\n\\solu{Proceed step by step. See what happens if the parity of $ a, b $ are different. Then the case with two coprimes. In this case, we want to tilt the rectangle to some extent where the desired result is achieved. We just need to show that this is possible. A bit of wishful thinking and a bit of algebra does the rest.}\t\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1662910p10561186}{APMO 2018 P3}{M}{A collection of $n$ squares on the plane is called tri-connected if the following criteria are satisfied:\n\n \\begin{enumerate}\n \\item All the squares are congruent.\n \\item If two squares have a point $P$ in common, then $P$ is a vertex of each of the squares.\n \\item Each square touches exactly three other squares.\n \\end{enumerate}\n\nHow many positive integers $n$ are there with $2018\\leq n \\leq 3018$, such that there exists a collection of $n$ squares that is tri-connected?}\n\n\\solu{Play around to find that $ 6k $ for $ k>4 $ is good. Then play around a little bit more for a different construction. Another construction for $ 6k $ gives rise to a construction for $ 10k $. Which integers can be written as a sum of $ 6k $ and $ 10k $?}\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1082935p4768413}{Iran 2005}{E}{A simple polygon is one where the perimeter of the polygon does not intersect itself (but is not necessarily convex). Prove that a simple polygon $P$ contains a diagonal which is completely inside $P$ such that the diagonal divides the perimeter into two parts both containing at least $\\frac{n}{3} - 1$ vertices. (Do not count the vertices which are endpoints of the diagonal.)}\n\n\\solu{Triangulate.}\t\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h287860p1555907}{ISL 2008 C3}{E}{In the coordinate plane consider the set $ S$ of all points with integer coordinates. For a positive integer $ k$, two distinct points $ a$, $ B\\in S$ will be called $ k$-friends if there is a point $ C\\in S$ such that the area of the triangle $ ABC$ is equal to $ k$. A set $ T\\subset S$ will be called $ k$-clique if every two points in $ T$ are $ k$-friends. Find the least positive integer $ k$ for which there exits a $ k$-clique with more than 200 elements.}\n\n\\solu{When does $ ax+by= c $ have integer solution? Fix one point as origin and check other points friendliness with other points.}\n\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1112753p5079689}{ISL 2015 C2}{E}{We say that a finite set $\\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\\mathcal{S}$, there is a point $C$ in $\\mathcal{S}$ such that $AC=BC$. We say that $\\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\\mathcal{S}$, there is no points $P$ in $\\mathcal{S}$ such that $PA=PB=PC$.\\\\\n\n \\begin{enumerate}\n \\item Show that for all integers $n\\ge 3$, there exists a balanced set consisting of $n$ points.\n \\item Determine all integers $n\\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.\n\\end{enumerate}}\n\n\\solu{Simple, think about circles, then think about ``center-free'' in a graph theoritic manner.}\n\n\n\n\n\n\\newpage\n\\subsection{Chessboard Pieces}\n\n\n\\lem{}{What is the maximum number of knights that can be placed on a chessboard such that no two knights attack each other?}\\label{lemma:maximum_knight_problem}\n\n\\solu{A knight's move always changes the color of the cell.}\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1671290p10632348P}{IMO 2018 P4}{E/H}{A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20.\\\\\n\n Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone.\\\\\n\nFind the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.}\\label{problem:coloring_3}\n\n\n\\solu{Using the \\hrf{lemma:maximum_knight_problem}{maximum knight problem} as a lemma.}\n\n\n\n\\prob{}{}{E}{How many rooks can be placed on an $ n\\times n $ board such that each rook attacks at most one other rook?}\n\n\\solu{Use graphs with one set of degrees being rows, and the other set of degrees being columns.}\n\n\n\\prob{https://en.wikipedia.org/wiki/Eight_queens_puzzle}{Eight queens puzzle}{}{How many queens can be placed on an $ n\\times n $ board such that no queen attacks another queen?}\n\n\\begin{minipage}[t][][b]{0.195\\linewidth}\n \\figdf{1}{queen_88}{}\n\\end{minipage}\\hfill%\n\\begin{minipage}[t][][b]{0.342\\linewidth}\n \\figdf{1}{queen_1414}{}\n\\end{minipage}\\hfill%\n\\begin{minipage}[t][][b]{0.366\\linewidth}\n \\figdf{1}{queen_1515}{}\n\\end{minipage}\\hfill%\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1417932p7979120}{Serbia National D2P2}{HM}{How many queens can be placed on an $ n\\times n $ board such that each queen attacks at most one other queen?}\n\n\n\\prob{}{BdMO 2019 P10}{E}{Define a new chess piece named warrior. it can either go three steps forward and one step to the side, or t2wo steps forward and two steps to the side in some orientation. In a $ 2020\\times 2020 $ chessboard, prove that the mazimum number of warriors so that none of them attack each other is leass htan or equal to $ \\dfrac{2}{5} $ of the number of cells.}\n\n\\solu{Color and partition}\n\n\n\\begin{minipage}{.5\\linewidth}\n \\prob{https://artofproblemsolving.com/community/c6h1790447p11841777}\n {RMM 2019 P4}{EM}{\n Prove that for every positive integer $n$ there exists a (not\n necessarily convex) polygon with no three collinear vertices, which\n admits exactly $n$ diffferent triangulations.\n\n (A triangulation is a dissection of the polygon into triangles by\n interior diagonals which have no common interior points with each\n other nor with the sides of the polygon)\n }\n\\end{minipage}\\hfill%\n\\begin{minipage}{.4\\linewidth}\n \\figdf{.9}{RMM2019P4}{Fixes}\n\\end{minipage}\n\n\n\n\n\n\\prob{https://artofproblemsolving.com/community/c6h1062931p4608924}{China TST 2015 T1D2P1}{EM}{Prove that : For each integer $n \\ge 3$, there exists the positive integers $a_1 a_3 $ while keeping the area rational i.e. keeping the height and base rational.}\n\n\n\n\\prob{https://codeforces.com/problemset/problem/1158/D}{Codeforces 1158D}{E}{You are given $ n $ points on the plane, and a sequence $ S $ of length $ n-2 $ consisting of $ L $ and $ R $. You need to generate a sequence of the points $ a_1, a_2\\dots a_n $ such that \n \\begin{itemize}\n \\item the polyline $ a_1a_2\\dots a_n $ is not self intersecting.\n \\item the directed segment $ a_{i+1}a_{i+2} $ is on the left side of the the directed segment $ a_{i}a_{i+1} $ if $ S_i = L $, and on the right side if $ S_i = R $.\n\\end{itemize}}\n\n\n\n", "meta": {"hexsha": "c4161f3e85a598fe140cb7cabe71044add371201", "size": 25779, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "combi/sec7_combi_geo.tex", "max_stars_repo_name": "M-Ahsan-Al-Mahir/BCS_Question_Bank", "max_stars_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 48, "max_stars_repo_stars_event_min_datetime": "2020-10-14T17:15:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-13T19:47:04.000Z", "max_issues_repo_path": "combi/sec7_combi_geo.tex", "max_issues_repo_name": "AnglyPascal/BCS_Question_Bank", "max_issues_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "combi/sec7_combi_geo.tex", "max_forks_repo_name": "AnglyPascal/BCS_Question_Bank", "max_forks_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-10-15T08:59:33.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-27T15:19:26.000Z", "avg_line_length": 64.9345088161, "max_line_length": 713, "alphanum_fraction": 0.7404088599, "num_tokens": 7006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585786300048, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.6377370354431251}} {"text": "\\chapter{Bayesian Neural Networks}\nA Bayesian neural network (BNN) is a probabilistic model build on the architecture of a neural network, but trained using Bayesian inference. The intent of such a model is to use the approximation capabilities of the neural networks, described in section \\ref{sec:neural_network}, combined with the capability of estimating uncertainty from Bayesian inference. These uncertainty estimating capabilities comes from considering the model parameters $\\boldsymbol{\\theta}$ as a random variable, such that predictions are made by considering the probability distribution of $\\boldsymbol{\\theta}$. BNNs can therefore be viewed as a special case of ensemble learning (\\cite{zhou_ensemble}) where the model ensemble is constructed by considering all possible values for $\\boldsymbol{\\theta}$. In practice we are not able to consider all possible values of $\\boldsymbol{\\theta}$ and sampling methods are used to sample different models as an approximation and there predictions are aggregated to make a single prediction.\\\\\n\\\\\nIn section \\ref{sec:bayesian_stat} we introduce two principal concepts of statistical inference; the frequentist and Bayesian paradigm and the main differences between these two paradigms. In section \\ref{sec:MC_methods} we introduce the basic theory of Monte Carlo methods and in section \\ref{sec:simple_BNN} we illustrate the main ideas behind Bayesian neural networks with a simple example. Since basic Monte Carlo methods are often inefficient when sampling from complex distributions we examine more sophisticated sampling methods based on Markov chains in section \\ref{sec:MCMC}.\n\\\\\n\\\\\nOne of these methods is the Metropolis algorithm, which we examine in section \\ref{sec:Metropolis}. Afterwards we examine the Hamiltonian Monte Carlo in section \\ref{sec:HMC}. This algorithm builds on the principles of Metropolis, but explores the target distribution more efficiently. Lastly in section \\ref{sec:nuts} we cover an extension of Hamiltonian Monte Carlo, that fixes possible inefficient \"U-turns\", called the No-U-Turn sampler. We end this chapter by briefly examining the choice of prior distributions in section \\ref{sec:priors} and how one can choose the parameters for these using a distribution.\n\n\n\n\n\n\\section{Bayesian \\& Frequentist Views of Learning}\\label{sec:bayesian_stat}\nIn this section we will introduce two concepts of statistical inference. These are the Bayesian and the frequentist paradigm.\n\\\\\n\\\\\nThe ideas behind Bayesian statistics goes back to the 18th century and is named after Thomas Bayes (\\cite{stigler1986history}). In Bayesian learning we consider the model parameters $\\boldsymbol{\\theta}$ as random and aim to learn the probability distribution of these. On the other hand the conventional frequentist methodology considers the model parameters $\\boldsymbol{\\theta}$ as fixed but unknown, while the point estimate $\\hat{\\boldsymbol{\\theta}}$ is a random variable, as it is a function of the dataset, which is assumed to be random.\n\\\\\n\\\\\nTo illustrate the difference between these two approaches in more detail, we will consider an example, which involves a simple coin toss. The uncertainty of the coin showing head or tails can be expressed by the coins probability $p$ of showing heads, which is often referred to as the bias of the coin. Since the properties of the coin is not known beforehand, we do not know the exact probability of showing heads. It could be a fair coin and have the probability $p=\\frac{1}{2}$ or it could be an unbalanced coin meaning that $p\\neq \\frac{1}{2}$.\n\\\\\n\\\\\nA Bayesian statistician would express this uncertainty by a probability distribution over possible values of the unknown probability $p$ and would then update the distribution as more observations become known. Frequentists would find the introduction of a distribution over parameter weights as pure nonsense. The frequentist would instead flip the coin a given number of times to form a dataset, and choose some estimator, for the unknown probability $p$ which would be most consistent with the data, an obvious choice would be the relative frequency of heads in the past coin tosses. \n\\\\\n\\\\\nThis example illustrates the main differences between these two paradigms and these differences will be highlighted in a more formal way in the subsequent sections.\n\n\n\n\\subsection{Maximum Likelihood Estimation} \\label{sec:mle}\nLet us now consider a dataset $\\boldsymbol{X}$, with $n$ examples $\\boldsymbol{x}^{(1)}, \\boldsymbol{x}^{(2)},\\ldots \\boldsymbol{x}^{(n)}$ drawn independently from the true but unknown distribution. We let\n$L(\\boldsymbol{\\theta}\\mid \\boldsymbol{X})\\equiv p(\\boldsymbol{X}\\mid \\boldsymbol{\\theta})$ denote the likelihood function, where $p\\lr{\\boldsymbol{X}\\mid\\boldsymbol{\\theta}}$ is a parametric family of probability distributions with parameter $\\boldsymbol{\\theta}$. A frequentist is, as described earlier, trying to find an estimate of the true parameter that have generated the data, we call such an estimate a point estimate and denote it by $\\hat{\\boldsymbol{\\theta}}$ to separate it from the true parameter. A point estimate can be viewed as a function of data $\\hat{\\boldsymbol{\\theta}}=g\\lr{\\boldsymbol{x}^{(1)}, \\boldsymbol{x}^{(2)},\\ldots \\boldsymbol{x}^{(n)}}$ which is drawn from a random process meaning that $\\hat{\\boldsymbol{\\theta}}$ is a random variable.\nIn most cases of frequentist inference a point estimate is found by maximizing the likelihood and is called the maximum likelihood estimate (MLE)\n\\begin{equation*}\n\\begin{split}\n \\hat{\\boldsymbol{\\theta}}_{\\text{MLE}}&=\\argmax_{\\boldsymbol{\\theta}}{L(\\boldsymbol{\\theta}\\mid \\boldsymbol{X})}\\\\\n & = \\argmax_{\\boldsymbol{\\theta}}{\\prod^n_{i=1}p\\lr{\\boldsymbol{x}^{(i)}\\mid \\boldsymbol{\\theta}}}\n\\end{split}\n\\end{equation*}\nIt is often more convenient to maximize a sum instead of a product, not only is it easier to handle sums when differentiating but it also helps stabilize the calculations numerically (\\cite{Goodfellow-et-al-2016}). Thus, we take the logarithm of the likelihood function to obtain the log-likelihood function $\\ell(\\boldsymbol{\\theta}\\mid \\boldsymbol{X})\\equiv \\log{p(\\boldsymbol{X}\\mid \\boldsymbol{\\theta})}$,\n\\begin{equation*}\n\\begin{split}\n \\hat{\\boldsymbol{\\theta}}_{\\text{MLE}}&=\\argmax_{\\boldsymbol{\\theta}}{\\ell(\\boldsymbol{\\theta}\\mid \\boldsymbol{X})}\\\\\n &=\\argmax_{\\boldsymbol{\\theta}}{\\sum^n_{i=1}\\log{p\\lr{\\boldsymbol{x}^{(i)}\\mid\\boldsymbol{\\theta}}}}\n\\end{split}\n\\end{equation*}\nand since the logarithm is a monotonic-increasing function, optimizing the log-likelihood is equivalent to optimizing the likelihood. \n\\\\\n\\\\\nWhen the size of the dataset is small the MLE is often prone to overfitting and regularization methods such as penalized maximum likelihood are applied, see \\cite{hastie01statisticallearning}.\n\\\\\n\\\\\nWe can generalize the maximum likelihood estimator to the case where our goal is to estimate a conditional distribution $p\\lr{\\boldsymbol{y}\\mid \\boldsymbol{X},\\boldsymbol{\\theta}}$, in order to predict $\\boldsymbol{y}$ given $\\boldsymbol{X}$ as described for supervised learning in section \\ref{sec:ml_basic}. In this case the conditional maximum likelihood is given by\n\\begin{equation*}\n \\hat{\\boldsymbol{\\theta}}_{\\text{MLE}}=\\argmax_{\\boldsymbol{\\theta}}L\\lr{\\boldsymbol{\\theta}\\mid \\boldsymbol{y}}=\\argmax_{\\boldsymbol{\\theta}} p(\\boldsymbol{y} \\mid \\boldsymbol{X}, \\boldsymbol{\\theta})\n\\end{equation*}\nand if assume that the targets $\\boldsymbol{y}^{(1)},\\ldots \\boldsymbol{y}^{(n)}$ are i.i.d, we can write\n\\begin{equation*}\n \\hat{\\boldsymbol{\\theta}}_{\\text{MLE}}=\\underset{\\boldsymbol{\\theta}}{\\arg \\max } \\sum_{i=1}^{n} \\log p\\left(\\boldsymbol{y}^{(i)} \\mid \\boldsymbol{x}^{(i)},\\boldsymbol{\\theta}\\right)\n\\end{equation*}\nThe sum of squared errors, obtained by summing over all datapoints in equation \\ref{eq:mse}, can be justified theoretically by the use of maximum likelihood with a Gaussian likelihood.\nTo see this, consider a regression model that outputs $f(\\boldsymbol{X};\\boldsymbol{\\theta})=\\boldsymbol{\\theta}\\boldsymbol{X}$, with real-valued targets $\\boldsymbol{y}$. If we define the likelihood as the conditional distribution of $\\boldsymbol{y}$ as Gaussian with mean given by the regression output $\\hat{y}\\equiv f(\\boldsymbol{X};\\boldsymbol{\\theta})$ and standard deviation $\\sigma$, we can write\n\\begin{equation*}\n L(\\boldsymbol{\\theta}\\mid \\boldsymbol{y})=\\prod_{i=1}^{n} p\\left(y^{(i)} \\mid \\boldsymbol{x}^{(i)} , \\boldsymbol{\\theta}\\right)=\\prod_i^n\\frac{1}{\\sqrt{2\\pi\\sigma}}\\exp\\left(-(\\hat{y}^{(i)}-y^{(i)})^2/2\\sigma^2\\right)\n\\end{equation*}\nNext we take the\nlogarithm of the likelihood function which gives us the log-likelihood function\n\\begin{equation*}\n\\begin{split}\n \\ell(\\boldsymbol{\\theta}\\mid \\boldsymbol{y})&=\\sum_i^n\\log\\frac{1}{\\sqrt{2\\pi\\sigma}}\\exp\\left(-(\\hat{y}^{(i)}-y^{(i)})^2/2\\sigma^2\\right)\\\\\n &=-\\frac{n}{2} \\log \\sigma^{2}-\\frac{n}{2} \\log (2 \\pi)-\\frac{1}{2 \\sigma^{2}} \\sum_{i}^n\\left(\\hat{y}^{(i)}-y^{(i)}\\right)^{2}\n\\end{split}\n\\end{equation*}\nnote that $-\\frac{n}{2} \\log \\sigma^{2}-\\frac{n}{2} \\log (2 \\pi)$ does not depend on the model parameters $\\boldsymbol{\\theta}$ and can therefore be omitted when maximizing. Maximizing the log-likelihood is the same as minimizing the negative log-likehood, so we can write \n\\begin{equation*}\n\\begin{split}\n \\min_{\\boldsymbol{\\theta}}{-\\ell(\\boldsymbol{\\theta}\\mid \\boldsymbol{y})}=\\frac{1}{2 \\sigma^{2}} \\sum_{i}^n\\left(\\hat{y}^{(i)}-y^{(i)}\\right)^{2}\n\\end{split}\n\\end{equation*}\nand since $\\frac{1}{2 \\sigma^{2}}$ does not depend on $\\boldsymbol{\\theta}$ either, we can see that minimizing the negative log-likelihood is the same as minimizing the sum of squared errors. \n\n\n\n\\subsection{Bayesian Learning and Prediction}\nA different approach than the frequentist perspective of the parameter value $\\boldsymbol{\\theta}$ as fixed but unknown and the point estimator $\\hat{\\boldsymbol{\\theta}}$ as a random variable, is taken with Bayesian inference. The Bayesian paradigm considers the dataset as fixed and observable, while the true parameter $\\boldsymbol{\\theta}$ is unknown, and thus considered a random variable. \n\\\\\n\\\\\nIn Bayesian statistics, we begin with defining a prior distribution $p(\\boldsymbol{\\theta})$ over the parameters. This prior distribution expresses our initial view on the parameters, before any data has been observed. When data becomes available, we update our prior distribution to a posterior distribution. The posterior distribution is defined by Bayes' rule\n\\begin{equation*}\n p(\\boldsymbol{\\theta}|\\boldsymbol{X},\\boldsymbol{y})=\\frac{p(\\boldsymbol{\\theta})p(\\boldsymbol{y}|\\boldsymbol{X},\\boldsymbol{\\theta})}{p(\\boldsymbol{y})}\n\\end{equation*}\nand it combines the information about the data, that comes from the likelihood function $p(\\boldsymbol{y}|\\boldsymbol{X},\\boldsymbol{\\theta})$, with the prior distribution. $p(\\boldsymbol{y})$ is called the model evidence and is the distribution of the observed data marginalized over the parameters $p(\\boldsymbol{y})=\\int p(\\boldsymbol{y}\\mid \\boldsymbol{X}, \\boldsymbol{\\theta})p(\\boldsymbol{\\theta})d\\boldsymbol{\\theta}$. The model evidence is often intractable, since it requires integration over all possible values of $\\boldsymbol{\\theta}$, which in many applications requires integration over high-dimensional spaces. As a result, finding analytical solutions of the posterior is not possible for complex models. \nWe often ignore the evidence term, as it does not depend on $\\boldsymbol{\\theta}$ and thus for a fixed $\\boldsymbol{y}$ can be interpreted as a constant, we thus write\n\\begin{equation} \\label{eq:posterior}\n p(\\boldsymbol{\\theta}|\\boldsymbol{X},\\boldsymbol{y})\\propto p(\\boldsymbol{\\theta})p(\\boldsymbol{y}\\mid \\boldsymbol{X},\\boldsymbol{\\theta})\n\\end{equation}\nAn important quality of the Bayesian method, is that it uses a full distribution over the parameters $\\boldsymbol{\\theta}$ to make predictions. Let us for example consider the case where we have observed a sample consisting of $n$ examples $\\boldsymbol{X}=\\boldsymbol{x}^{(1)}, \\boldsymbol{x}^{(2)},\\ldots, \\boldsymbol{x}^{(n)}$. To predict an unobserved label $\\boldsymbol{y}^{(n+1)}$ for a new example $\\boldsymbol{x}^{(n+1)}$ we need the \nposterior predictive distribution, that is to integrate the model predictions by the posterior\n\\begin{equation} \\label{eq:post_pred_distribution}\n \\begin{split}\n &p\\left(\\boldsymbol{y}^{(n+1)} \\mid \\boldsymbol{x}^{(n+1)}, \\lr{\\boldsymbol{x}^{(1)},\\boldsymbol{y}^{(1)}}, \\ldots, \\lr{\\boldsymbol{x}^{(n)}, \\boldsymbol{y}^{(n)}}\\right)\\\\\n &=\\int p\\left(\\boldsymbol{y}^{(n+1)} \\mid \\boldsymbol{x}^{(n+1)}, \\boldsymbol{\\theta} \\right) p \\lr{\\boldsymbol{\\theta} \\mid \\lr{\\boldsymbol{x}^{(1)}, \\boldsymbol{y}^{(1)}}, \\ldots, \\lr{\\boldsymbol{x}^{(n)}, \\boldsymbol{y}^{(n)}}} \\, d \\boldsymbol{\\theta}\n \\end{split}\n\\end{equation}\nThis is quite different from the maximum likelihood method, that uses a point estimate for $\\boldsymbol{\\theta}$ to make predictions on any unobserved data, the Bayesian method takes the uncertainty of estimating $\\boldsymbol{\\theta}$ into account when making predictions, which tends to do well in avoidance of overfitting (\\cite{Goodfellow-et-al-2016}). \n\\\\\n\\\\\nAn important difference between the Bayesian approach and MLE, lies on the contribution of a prior distribution. The prior has the effect of shifting the probability mass towards regions of the parameter space, that are preferred a priori. According to \\cite{Goodfellow-et-al-2016}, the prior often expresses a preference for models that are simpler or more smooth. Critics of the Bayesian method often point their fingers at the prior distribution, and criticize it for being a subjective component that can affect the predictions of the model. According to \\cite{neal2012bayesian} Bayesian methods often do a lot better than a frequentist model, when training data is limited in availability, but suffers from high computational cost when the number of training examples are large. \n\\\\\n\\\\\n\\cite{neal2012bayesian} and \\cite{mackay1991} argues that Bayesian models embodies Occam's Razor, which is the principle that we should prefer simpler models to complex ones. This principle is a component often found in machine learning since a too complex model might overfit the data. This belief is justified when model parameters are estimated by maximum likelihood, but \\cite{neal2012bayesian} argues that one should not limit the complexity of Bayesian neural networks to prevent overfitting. The approach of training by minimizing loss might lead to a choice of model with increasing complexity the more data that is available. With a Bayesian approach adjusting the complexity of the model based on the amount of data available makes no sense, since a correct prior and model for 10 observations must be correct for 10.000 observations as well. One might however switch to a simple model if it seems unlikely that a complex computational expensive model will provide significant benefit. \n\\\\\n\\\\\nAs mentioned earlier, in many practical examples, the posterior distribution is intractable and therefore must be derived in some other way. Often we will use methods such as Monte Carlo to approximate the posterior distribution. \n\n\\subsection{Maximum a Posteriori (MAP) Estimation}\nA way to avoid the computational hurdle of approximating the entire Bayesian posterior, is to use a point estimate as an approximation. Instead of turning completely to frequentist methods and use MLE, one can still benefit of the Bayesian method, by allowing the prior to influence the choice of the point estimate. One way to do this, is to use the maximum a posteriori (MAP) point estimate. The MAP estimate is obtained by maximizing the posterior distribution\n\\begin{equation}\\label{eq: MAP}\n \\begin{split}\n \\hat{\\boldsymbol{\\theta}}_{\\text{MAP}}&=\\argmax_{\\boldsymbol{\\theta}}{p(\\boldsymbol{\\theta}\\mid \\boldsymbol{X},\\boldsymbol{y}})=\\argmax_{\\boldsymbol{\\theta}}{p(\\boldsymbol{y}\\mid \\boldsymbol{X},\\boldsymbol{\\theta}}) p(\\boldsymbol{\\theta})\\\\\n &=\\argmax_{\\boldsymbol{\\theta}}{\\log p(\\boldsymbol{y}\\mid \\boldsymbol{X},\\boldsymbol{\\theta}})+ \\log p(\\boldsymbol{\\theta})\n \\end{split}\n\\end{equation}\nnote that the evidence term has been omitted, since it does not depend on the parameter $\\boldsymbol{\\theta}$ and thus vanishes under maximization anyway. The bottom part of equation \\ref{eq: MAP} can be recognized as an equation consisting of the standard log-likelihood term plus a log-prior term. \\\\\n\\\\\nAs an example, consider a model with a Gaussian prior placed on the regression weights $\\boldsymbol{\\theta}$. If we specifically choose the prior to be given by $\\mathcal{N}\\left(\\boldsymbol{\\theta},0,\\frac{1}{\\alpha}\\boldsymbol{I}^2\\right)$, then the log-prior term in equation \\ref{eq: MAP} is proportional to the L2 norm introduced in \\ref{eq:L2_reg}, plus a term that does not depend on $\\boldsymbol{\\theta}$.\nNote also that the MAP estimate is the same as the MLE, when choosing a uniform prior, since the $p(\\boldsymbol{\\theta})$ becomes a constant function in equation \\ref{eq: MAP} and consequently we can ignore it when maximizing the expression. Compared to MLE, MAP estimation has the advantage that it can benefit from the information in the prior that cannot be found in the dataset. According to \\cite{Goodfellow-et-al-2016} this additional information, gained from the choice of prior, can reduce the variance in the MAP point estimate in direct comparison with MLE estimate, but this advantage has a price of an increased bias.\n\\\\\n\\\\\nMAP is closely related to Bayes optimal estimation, instead of finding the most probable hypothesis (set of parameters), it aims at finding the most probable label for a new example. The Bayes optimal estimation is done by predicting the $\\boldsymbol{y}^{(n+1)}$, which maximizes the posterior predictive distribution in equation \\ref{eq:post_pred_distribution}. We do not pursue the idea of Bayes optimal estimation, as we aim to minimize a loss function which is not always equivalent to predicting the most probable label. In this way we preserve the idea that the loss function dictates how much the algorithm should care about making certain predictions as explained in \\ref{sec:loss_func}.\n\\\\\n\\\\\nAn obvious disadvantage from using MAP estimation, is that it discards the information contained in the distribution. It is estimating distributions that make Bayesian methods attractive, especially if one wants to evaluate the uncertainty of the parameters. As we value access to this distribution higher than faster computational time we will not pursue MAP estimation any further.\n\n\\section{Monte Carlo Methods}\\label{sec:MC_methods}\nOne way of approximating intractable integrals, like the posterior predictive distribution in equation \\ref{eq:post_pred_distribution}, is to use Monte Carlo methods. The idea behind Monte Carlo methods is to view the integral as an expectation of some random variable with respect to a probability distribution $p(\\cdot)$. In the case of BNNs our random variable is the model parameters $\\boldsymbol{\\theta}$, and we can write the posterior predictive distribution as\n\\begin{equation}\n \\begin{split}\n &p\\left(\\boldsymbol{y}^{(n+1)} \\mid \\boldsymbol{x}^{(n+1)}, \\lr{\\boldsymbol{x}^{(1)},\\boldsymbol{y}^{(1)}}, \\ldots, \\lr{\\boldsymbol{x}^{(n)}, \\boldsymbol{y}^{(n)}}\\right)\\\\\n &=\\int p\\left(\\boldsymbol{y}^{(n+1)} \\mid \\boldsymbol{x}^{(n+1)}, \\boldsymbol{\\theta}\\right) p \\lr{\\boldsymbol{\\theta} \\mid \\lr{\\boldsymbol{x}^{(1)}, \\boldsymbol{y}^{(1)}}, \\ldots, \\lr{\\boldsymbol{x}^{(n)}, \\boldsymbol{y}^{(n)}}} \\, d \\boldsymbol{\\theta}\\\\\n &=\\int f\\lr{\\boldsymbol{\\theta}}\\hat{p}\\lr{\\boldsymbol{\\theta}} d\\boldsymbol{\\theta}\n \\end{split}\n\\end{equation}\nwhere we use a simpler notation of $\\hat{p}\\lr{\\boldsymbol{\\theta}}$ to denote the posterior distribution. Such an integral can be interpreted as an expectation taking under the probability distribution $\\hat{p}(\\boldsymbol{\\theta})$ \n\\begin{equation*}\n s=\\int f(\\boldsymbol{\\theta})\\hat{p}(\\boldsymbol{\\theta}) d \\boldsymbol{\\theta}=\\mathbb{E}_{\\hat{p}}[f(\\boldsymbol{\\theta})]\n\\end{equation*}\nNow in order to approximate $s$ we can draw samples from the distribution $\\hat{p}(\\boldsymbol{\\theta}) $ and approximate the expected value by the empirical average. If we draw $n$ samples $\\boldsymbol{\\theta}\\sim \\hat{p}(\\boldsymbol{\\theta})$ we can approximate $s$ by $\\hat{s}_n$\n\\begin{equation}\\label{eq:empirical_mean_MC}\n \\hat{s}_{n}=\\frac{1}{n} \\sum_{i=1}^{n} f\\left(\\boldsymbol{\\theta}^{(i)}\\right)\n\\end{equation}\nThis implies the simplest situation, where it is possible to simulate directly from the probability distribution, which is often not possible.\n\\\\\n\\\\\nWe can justify this approximation theoretically, by noticing that $\\hat{s}_n$ is an unbiased estimator of $s$\n\\begin{equation*}\n \\mathbb{E}_{\\hat{p}}\\left[\\hat{s}_{n}\\right]=\\frac{1}{n} \\sum_{i=1}^{n} \\mathbb{E}_{\\hat{p}}\\left[f\\left(\\boldsymbol{\\theta}^{(i)}\\right)\\right]=\\frac{1}{n} \\sum_{i=1}^{n} s=s\n\\end{equation*}\nadditionally the law of large numbers, states that if the samples $\\boldsymbol{\\theta}^{(i)}$ are independent and identical distributed (i.i.d), the empirical average converges to the true expectation almost surely\n\\begin{equation*}\n \\lim _{n \\rightarrow \\infty} \\hat{s}_{n}=s\n\\end{equation*}\nthis only holds if the variance of the individual terms $\\operatorname{Var}[f\\lr{\\boldsymbol{\\theta}^{(i)}}]$ is bounded. To see this, we note that $\\operatorname{Var}[\\hat{s}_n]$ converges to zero as n goes to infinity, if and only if $\\operatorname{Var}[f\\lr{\\boldsymbol{\\theta}}]<\\infty$\n\\begin{equation*}\n \\begin{split}\n\\operatorname{Var}\\left[\\hat{s}_{n}\\right] &=\\operatorname{Var}\\left[\\frac{1}{n} \\sum_{i=1}^{n} f(\\boldsymbol{\\theta}^{(i)})]\\right] \\\\ &=\\frac{1}{n^{2}} \\sum_{i=1}^{n} \\operatorname{Var}[f(\\boldsymbol{\\theta}^{(i)})]\n=\\frac{\\operatorname{Var}[f(\\boldsymbol{\\theta})]}{n}\n\\end{split}\n\\end{equation*}\nFurther the central limit theorem states that, if $\\mathbb{E}_{\\hat{p}}[f(\\boldsymbol{\\theta})]=s$ and $\\operatorname{Var}[f(\\boldsymbol{\\theta})]<\\infty$ then\n\\begin{equation*}\n \\frac{\\hat{s}_{n}-s}{\\sqrt{\\operatorname{Var}[f(\\boldsymbol{\\theta})] / n}} \\sim \\mathcal{N}(0,1)\n\\end{equation*}\nwhich is equivalent to\n\\begin{equation*}\n \\hat{s}_n\\sim \\mathcal{N}\\left(s,\\frac{\\operatorname{Var}[f(\\boldsymbol{\\theta})]}{n}\\right)\n\\end{equation*}\nWhich gives us a way of estimating confidence intervals around the estimate $\\hat{s}$. \\\\\n\\\\\nWhen it is infeasible or not possible to make simulations of $\\boldsymbol{\\theta}$ directly, Markov chain Monte Carlo methods can be used. such methods simulate from a target distribution by running a Markov chain, that eventually will converge to target distribution. Markov chain Monte Carlo methods will be examined more thoroughly in section \\ref{sec:MCMC}.\n\n\n\n\\section{A Simple Bayesian Neural Network} \\label{sec:simple_BNN}\nA simple example, inspired by \\cite{neal2012bayesian}, will illustrate the general concept of Bayesian learning for neural networks and the inefficiency of brute force methods of sampling. Figure \\ref{fig:simple_BNN} shows six BNNs whose weights and biases were drawn from independent standard normal prior distributions except output weights, which had a standard deviation of $\\frac{1}{\\sqrt{16}}$. The networks performs regression on six data points. \n\\\\\n\\\\\nThe six networks was chosen from a larger pool of $10^5$ networks with weights and biases sampled from identical prior distributions. The likelihood was computed for each of these networks and scaled so that the largest likelihood was 1. The networks were then accepted with the probability of this scaled likelihood for which only six was accepted. This approach resembles rejection sampling and embodies the posterior from equation \\ref{eq:posterior} by making the prior control the generation of candidate networks and the likelihood control which of these candidates are rejected.\nWe follow the suggestion of \\cite{neal2012bayesian} to model regression tasks with a conditional distribution for the real valued targets $\\boldsymbol{y}_k$, from $k$ neural networks with outputs $f_k(\\boldsymbol{x})$, defined by a Gaussian distribution\n\\begin{equation}\\label{eq:regr_taget_distribution}\n p(\\boldsymbol{y} \\mid \\boldsymbol{x}) = \\prod_k \\frac{1}{\\sqrt{2 \\pi} \\sigma_k} \\exp{- \\frac{\\lr{f_k(\\boldsymbol{x}) - y_k}^2}{2 \\sigma_k^2}}\n\\end{equation}\nwith mean $f_k(\\boldsymbol{x})$ and standard deviation $\\sigma_k$ as a hyperparameter, which we choose to be $\\sigma_k = 0.1$ for all $k$.\n\\\\\n\\\\\nAccording to \\cite{neal2012bayesian} the optimal way to predict the target associated with various new examples, assuming we want to minimize the expected squared error, is to predict the mean of the predictive distribution in equation \\ref{eq:post_pred_distribution}. For a regression model defined by equation \\ref{eq:regr_taget_distribution} this is equal to predicting \n\\begin{equation*}\n \\hat{\\boldsymbol{y}}^{(n+1)} = \\int f\\lr{\\boldsymbol{x}^{(n+1)}, \\boldsymbol{\\theta}} p\\lr{\\boldsymbol{\\theta} \\mid \\lr{\\boldsymbol{x}^{(1)}, \\boldsymbol{y}^{(1)}}, \\dots, \\lr{\\boldsymbol{x}^{(n+1)}, \\boldsymbol{y}^{(n+1)}}} \\, d \\boldsymbol{\\theta}\n\\end{equation*}\nAs we do not know this distribution, we resort to a Monte Carlo approximation by averaging over the outputs from the six neural networks, obtained by sampling parameters from the posterior. The average is shown in figure \\ref{fig:simple_BNN} by the solid line. But Bayesian inference can do more than a single-valued guess. By examining the function we can also see the uncertainty of the guesses, for example how rapidly uncertainty increases beyond the region of the data points. \n\\\\\n\\\\\nThis illustrates some of the benefits of using Bayesian inference for neural networks, but what remains is the downside of computational time. Generating $10^5$ samples to get six draws from the posterior is not very efficient and this only becomes more infeasible as the number of data points increase. \n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth,height=\\textheight,keepaspectratio]{pics/figure_simple_BNN.pdf}\n \\caption{Sampled neural networks from a posterior predictive distribution that is based on a Gaussian prior and a Gaussian likelihood on the six data points. The average prediction of the networks are plotted along with a filled area defined by the average plus minus the standard deviation of the network predictions to represent uncertainty. The Python code for implementing this Bayesian neural network and the production of this figure can be seen in appendix \\ref{app:simple_BNN}}\n \\label{fig:simple_BNN}\n\\end{figure}\n\\clearpage\n\\section{Markov Chain Monte Carlo}\\label{sec:MCMC}\n As mentioned in section \\ref{sec:bayesian_stat} the posterior distribution is often intractable and we might need simulation based methods to find a feasible solution. We want to be able to calculate posterior summaries like $\\mathbb{E}_{\\hat{p}}\\left[f(\\boldsymbol{\\theta})\\right]$ where $f(\\boldsymbol{\\theta})$ is some function over the parameters and the expectation is taken under the posterior distribution. Such an expectation is straightforward to approximate using the simple Monte Carlo method, described in section \\ref{sec:MC_methods}, when dealing with nice and low dimensional distributions. Other methods like importance sampling has often been proposed in the litterateur for simple problems, but rarely for high-dimensional problems as we often face with neural networks.\n \\\\\n \\\\\nThe simple example in section \\ref{sec:simple_BNN} illustrated that brute force methods are also not very useful in complex models like BNNs and we are instead motivated to use a collection of more sophisticated methods. This thesis will mainly pursue this motivation by exploring the Markov chain Monte Carlo (MCMC) methods. The main idea is to simulate a Markov chain, which has the posterior distribution as its stationary distribution. We will initially give a concise explanation of the fundamentals of Markov chains in section \\ref{sec:basic_mc} and follow this by exploring the most popular MCMC algorithms for sampling in BNNs.\n \n \n\\subsection{Markov Chains}\\label{sec:basic_mc}\nA stochastic process is a set of random variables that are defined over some probability space $\\left\\{X^{(t)} \\right\\}_{t\\in T}$, where $T\\subseteq \\mathbb{R}$ is the indexation and may be interpreted as a time index. In the context of machine learning, the set $T$ can be interpreted as the iterations of a simulation scheme. In stochastic simulation we typically have that $T=\\mathbb{N}$ and we can write the stochastic process as $\\left\\{X^{(n)}\\right\\}_{n\\geq 0}$, where $n$ reminds us that we have a discrete index set. Throughout this thesis we will only consider discrete time stochastic processes, since this is found to be most relevant when discussing stochastic simulation. \\\\\nA stochastic process is defined over a space of possible values called the state space $\\mathbb{S}$ and will in most applications be either integers or real values. \n\\\\\n\\\\\nWith the objective of modelling BNNs, we want to sample model parameters from the posterior distribution. These samples will then be used to make predictions for unseen data by approximating the posterior predictive distribution from equation \\ref{eq:post_pred_distribution} via Monte Carlo integration. In order to do this using MCMC we will consider the stochastic process $\\{\\boldsymbol{\\theta}^{(n)}\\}_{n\\geq 0}$. To make the description of MCMC methods more general in the next sections, we will denote the distribution we aim to sample from the target distribution and denote it $\\hat{p}(\\cdot)$. The target distribution in the case of BNNs is the posterior of the weights such that $\\hat{p}(\\boldsymbol{\\theta})\\equiv p(\\boldsymbol{\\theta}\\mid \\boldsymbol{X},\\boldsymbol{y})$.\n\\\\\n\\\\\nUsing Markov chain Monte Carlo we aim to sample from a stochastic process satisfying the Markov property. Such a process is defined by being dependant only on the previous state of the process and a set of transitional probabilities, or densities for a infinite state space, and is called a Markov chain. The Markov chain is defined by an initial distribution for the first state of the chain $\\boldsymbol{\\theta}^{(0)}$, and a transition density for the next states in the system. We write the transition density from transitioning from state $\\boldsymbol{\\theta}^{(n-1)}$ to another state $\\boldsymbol{\\theta}^{(n)}$ as $T(\\boldsymbol{\\theta}^{(n)}\\mid \\boldsymbol{\\theta}^{(n-1)})$. \\\\\n\\\\\nWhen sampling from the Markov chain we want to make sure that samples are actually coming from the desired distribution $\\hat{p}(\\boldsymbol{\\theta})$ no matter the initial distribution. To ensure this we must generate a Markov chain that has our desired distribution $\\hat{p}\\lr{\\boldsymbol{\\theta}}$ as a stationary distribution. That is if $\\boldsymbol{\\theta}^{(n-1)}$ has distribution $\\hat{p}(\\cdot)$, then $\\boldsymbol{\\theta}^{(n)}$ will have the same distribution, and this will hold for all future states of the chain. The property of a Markov chain having a stationary distribution is called the invariance property and is defined by\n\\begin{equation*}\n \\pi(\\boldsymbol{\\theta}^{(n)})=\\int T(\\boldsymbol{\\theta}^{(n)}\\mid \\boldsymbol{\\theta}) \\pi(\\boldsymbol{\\theta})d\\boldsymbol{\\theta}\n\\end{equation*}\nA sufficient, but not necessary condition, that ensures that a particular $\\hat{p}(\\boldsymbol{\\theta})$ is stationary distribution is the detailed balance condition. The condition states if we let $T(\\cdot,\\cdot)$ be a transition density, which satisfies the following condition \n$$T(\\boldsymbol{\\theta}^{(n-1)}, \\boldsymbol{\\theta}^{(n)}) \\hat{p}(\\boldsymbol{\\theta}^{(n)})= T(\\boldsymbol{\\theta}^{(n)}, \\boldsymbol{\\theta}^{(n-1)})\\hat{p}(\\boldsymbol{\\theta}^{(n-1)})$$\nthen $\\hat{p}(\\cdot)$ is a stationary distribution of the Markov chain associated with the transition density $T(\\cdot,\\cdot)$. This property however only ensures that $\\hat{p}\\lr{\\boldsymbol{\\theta}}$ is stationary distribution but not that it is the only one, meaning that our Markov chain can end up sampling from the wrong distribution, even though it satisfies the detailed balance condition.\n\\\\\n\\\\\nTo guarantee that we sample from $\\hat{p} \\lr{\\boldsymbol{\\theta}}$ we need to ensure that the Markov chain has only this distribution as a stationary distribution. A Markov chain which has a unique stationary distribution, from which it converges to, from any initial state is called an ergodic Markov chain (see e.g \\cite{turkman2019computational}). For a Markov chain on a finite state space to be ergodic, it has to be irreducible and aperiodic. The same goes for a Markov chain on a infinite state space, but a slightly stronger condition called Harris recurrence is needed, see \\cite{gamerman2006markov}. \n\\\\\n\\\\\nOften we discard or burn some of the initial states, since these may not be representative of the desired distribution, as the chain might not have reached the stationary distribution yet. These discarded steps are part of what is called the burn-in period of the Markov chain. When the chain has reached the stationary distribution, it is possible to draw as many identical distributed samples as we wish for, but one should note that any successive samples will be highly correlated with each other and therefore not necessarily a good representative for the target distribution. \\cite{Goodfellow-et-al-2016} suggest a way to mitigate this problem, by only returning every $n$ successive sample. Because of both the burn-in period and the time required for the chain to return uncorrelated samples MCMCs are often computational expensive. \\\\\n\\\\\nIn order to produce truly independent samples, \\cite{Goodfellow-et-al-2016} suggest to run multiple Markov chains in parallel. They also mention that practitioners often chooses the number of chains to run in parallel, similar to the number of examples in a mini-batch and then draw the samples needed from this set of Markov chains. \n\n\n\\subsection{The Metropolis algorithm} \\label{sec:Metropolis}\nThe Metropolis algorithm is a MCMC method, which is used to sample from a distribution. \nThe Metropolis algorithm was originally introduced by \\cite{Metropolis1953} and was developed to simulate the states for a system of molecules. This was later further developed by \\cite{hastings70}, so that the algorithm could now simulate from a general distribution and not just a symmetric one, as it was previously based on. The Metropolis algorithm is an attractive MCMC method due to its versatility and simplicity.\\\\\n\\\\\nThe algorithm considers a target distribution $\\hat{p}(\\boldsymbol{\\theta})$ and a proposal distribution $q(\\boldsymbol{\\theta})$. The algorithm generates a Markov chain, by starting the chain at some arbitrarily point generated from the proposal distribution $\\boldsymbol{\\theta}^{(0)}\\sim q(\\boldsymbol{\\theta})$ and then proposing a candidate state for the next state in the chain $\\boldsymbol{\\theta}^{\\text{\\text{cand}}}$, where this candidate state is drawn from the conditional distribution on the previous state $\\boldsymbol{\\theta}^{\\text{\\text{cand}}}\\sim q(\\boldsymbol{\\theta}^{(n+1)}\\mid\\boldsymbol{\\theta}^{(n)})$. The next step is to decide whether or not to reject this new state based on the relative density to the old state. If the relative density is larger than one, we accept the new state, if the relative density is less than one, we accept the new state with probability $\\frac{\\hat{p}(\\boldsymbol{\\theta}^{\\text{\\text{cand}}})}{\\hat{p}(\\boldsymbol{\\theta}^{(n)})}$. In this context the Metropolis algorithm imposes the symmetry condition of on the proposal distribution, so that\n\\begin{equation*}\n q(\\boldsymbol{\\theta}^{(n)}\\mid \\boldsymbol{\\theta}^{(n-1)})=q(\\boldsymbol{\\theta}^{(n-1)}\\mid \\boldsymbol{\\theta}^{(n)})\n\\end{equation*}\nA pseudocode version of the Metropolis algorithm can be seen in algorithm \\ref{algo_2}.\n% Metropolis Algorithm\n\\begin{algorithm}\\label{algo_2}\n\n\\SetAlgoLined\n\\KwInput{A proposal distribution $q$}\n\\KwOutput{A set of parameters $\\boldsymbol{\\theta^{(n)}}$ for $n = 1, \\dots, N$}\ninitialize $\\boldsymbol{\\theta}^{(0)}\\sim q(\\boldsymbol{\\theta})$;\n\n\\For{$n=1,2,\\ldots, N$}{\nPropose: $\\boldsymbol{\\theta}^{\\text{\\text{cand}}} \\sim q\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right)$\n\nAcceptance Probability:\n\n$ \\alpha\\left(\\boldsymbol{\\theta}^{\\text{\\text{cand}}} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right)=\\min \\left\\{1, \\frac{\\hat{p}\\left(\\boldsymbol{\\theta}^{\\text{\\text{cand}}}\\right)}{ \\hat{p}\\left(\\boldsymbol{\\theta}^{(n-1)}\\right)}\\right\\} $\n\n$u \\sim \\text { Uniform }(0,1)\n$\n\n \\uIf{$u<\\alpha$}{\n Accept the proposal: $\\boldsymbol{\\theta}^{(n)} \\leftarrow \\boldsymbol{\\theta}^{\\text{\\text{cand}}}$\\;\n }\n \\Else{\n Reject the proposal: $\\boldsymbol{\\theta}^{(n)} \\leftarrow \\boldsymbol{\\theta}^{(n-1)}$ \\;\n }\n }\n\\caption{Metropolis algorithm}\n\\end{algorithm}\nOne apparent problem is that due to the evidence term we can not calculate the posterior exactly, which is our target distribution $\\hat{p}\\lr{\\boldsymbol{\\theta}}$, so we can not directly calculate $\\frac{\\hat{p}(\\boldsymbol{\\theta}^{(n)})}{\\hat{p}\\left(\\boldsymbol{\\theta}^{(n-1)}\\right)}$. \nBut a nice property of Metropolis acceptance probability is that we only need a function that is proportional to the posterior, as any constant of proportionality will cancel out in the calculation of the acceptance probability. As the evidence term can be interpreted as a constant of proportionality, it will cancel out and we can instead use equation \\ref{eq:posterior} and calculate the ratio of the likelihood times the prior, which we are often able to. \n\\\\\n\\\\\nTo show that the Metropolis algorithm is in fact sampling from the target distribution, we must show that the Markov chain converges to a stationary distribution which is our target distribution. If we assume that the Markov Chain is ergodic we can do this by showing that the metropolis satisfies the detailed balanced condition explained in section \\ref{sec:MCMC}.\\\\\n\\\\\nFor $\\boldsymbol{\\theta}^{(n)}\\neq \\boldsymbol{\\theta}^{(n-1)}$ the Metropolis algorithm has the following transitions densities,\n\\begin{equation*}\n T\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right)=q\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right) \\min \\left(1, \\frac{\\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right)}{ \\hat{p}(\\boldsymbol{\\theta}^{(n-1)})}\\right)\n\\end{equation*}\nWe can show that this satisfies the detailed balanced condition by\n\\begin{equation*}\n\\begin{aligned}\nT\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right) \\hat{p}(\\boldsymbol{\\theta}^{(n-1)}) &=q\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right) \\min \\left(1,\\frac{ \\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right) }{ \\hat{p}(\\boldsymbol{\\theta}^{(n-1)})}\\right) \\hat{p}(\\boldsymbol{\\theta}^{(n-1)}) \\\\\n&=q\\left(\\boldsymbol{\\theta}^{(n)} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right) \\min \\left(\\hat{p}(\\boldsymbol{\\theta}^{(n-1)}), \\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right)\\right) \\\\\n&=q\\left(\\boldsymbol{\\theta}^{(n-1)} \\mid \\boldsymbol{\\theta}^{(n)}\\right) \\min \\left(\\hat{p}\\left(\\boldsymbol{\\theta}^{(n-1)}\\right), \\hat{p}(\\boldsymbol{\\theta}^{(n-1)})\\right) \\\\\n&=q\\left(\\boldsymbol{\\theta}^{(n-1)} \\mid \\boldsymbol{\\theta}^{(n)}\\right) \\min \\left(1, \\frac{\\hat{p}(\\boldsymbol{\\theta}^{(n-1)}) }{ \\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right)}\\right) \\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right) \\\\\n&=T\\left(\\boldsymbol{\\theta}^{(n-1)} \\mid \\boldsymbol{\\theta}^{(n)}\\right) \\hat{p}\\left(\\boldsymbol{\\theta}^{(n)}\\right)\n\\end{aligned}\n\\end{equation*}\nThis shows that the transitions proposed by the algorithm leaves the target distribution $\\hat{p}(\\boldsymbol{\\theta})$ invariant and therefore samples produced by this Markov chain all has the same stationary distribution $\\hat{p}(\\boldsymbol{\\theta})$, provided that the Markov chain is ergodic. However according to \\cite{neal2012bayesian} the Metropolis algorithm will not always produce an ergodic Markov chain and that this depends on the details of the target distribution and proposal distribution. If the produced Markov chain is not ergodic we might end up sampling from a stationary distribution that is not our target distribution.\n\\\\\n\\\\\nThere are many possible choices for the proposal distribution. \\cite{neal2012bayesian} mentions that a simple choice could be a Gaussian centered on $\\boldsymbol{\\theta}^{(n)}$ with standard deviation chosen so that the acceptance probability of the candidate state is reasonably high, since a very low acceptance ratio is usually unwanted, as many rejections means that we are inefficiently wasting computation time. He also notes that when sampling from high-dimensional and complex distributions, which is often the case with posteriors in BNNs, the standard deviation of such a proposal distribution will often have to be small compared to the extent of the target distribution, as large changes almost certainly will lead to a region of low probability. This will result in highly dependant states and many steps will be needed to arrive at distant points in the distribution. As suggested in section \\ref{sec:basic_mc} one way of coping with this problem is to throw away some of the samples or run multiple chains in parallel. \\\\\n\\\\\n\\cite{neal2012bayesian} further mentions that this problem with Metropolis is made worse due to its movements taking the inefficient form of a random walk instead of a systematic path, as can be seen in the illustration of a sampling with Metropolis in figure \\ref{fig:MH_sampling}. This inefficient movement yields slower convergence to the target distribution. This drawback, will be even more prominent in higher dimension and for more complex target distributions according to \\cite{gelmanbda04}. \\\\\n\\\\\nA more generalized version of the algorithm is the one introduced by \\cite{hastings70}, which allows for non-symmetric proposal distributions $q(\\boldsymbol{\\theta}^{(n)}\\mid \\boldsymbol{\\theta}^{(n-1)}) \\neq q(\\boldsymbol{\\theta}^{(n-1)}\\mid \\boldsymbol{\\theta}^{(n)})$. To correct for this asymmetry in the proposal distribution, the acceptance ratio is replaced by\n\\begin{equation}\\label{eq: hasti_pasti}\n\\alpha\\left(\\boldsymbol{\\theta}^{c a n d} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right)=\\min \\left\\{1, \\frac{q\\left(\\boldsymbol{\\theta}^{(n-1)} \\mid \\boldsymbol{\\theta}^{c a n d}\\right) \\hat{p}\\left(\\boldsymbol{\\theta}^{c a n d}\\right)}{q\\left(\\boldsymbol{\\theta}^{c a n d} \\mid \\boldsymbol{\\theta}^{(n-1)}\\right) \\hat{p}\\left(\\boldsymbol{\\theta}^{(n-1)}\\right)}\\right\\} \n\\end{equation}\nOne should note that the Metropolis algorithm is an instance of the generalized version, since equation \\ref{eq: hasti_pasti} is identical to the acceptance ratio in the original algorithm when allowance for a symmetric distribution. \\\\\n\\\\\nThe introduction of an asymmetric proposal distribution, is often useful when we want to increase the speed of the random walk generated by the Metropolis algorithm. However, this is often not sufficient for complicated models with high-dimensional target distributions as we face with Bayesian neural networks, see \\cite{gelmanbda04}.\n\\begin{figure}\n \\centering\n \\includegraphics[width=\\textwidth, height=\\textheight, keepaspectratio]{pics/mh_randomWalk_behavior.pdf}\n \\caption{Illustration of the convergence to a target distribution with 300 samples from the Metropolis algorithm. The target distribution is a bivariate Gaussian with mean \n $\\boldsymbol{\\mu}= \\begin{bmatrix}\n 0 & 0\n \\end{bmatrix}$ and covariance matrix\n $\\boldsymbol{\\Sigma}= \n \\begin{bmatrix}\n 1 & 0.6\\\\\n 0.6 & 1 \n \\end{bmatrix}$. The Python code for producing this figure can be found in appendix \\ref{app:MH_code}.\n }\n \\label{fig:MH_sampling}\n\\end{figure}\n\\clearpage\n\n\\section{Hamiltonian Monte Carlo}\\label{sec:HMC}\nAnother way to generate proposals with a higher efficiency, is by updating the parameters by dynamical simulation and then use the Metropolis algorithm to accept or reject these proposals. Such an algorithm suppress the local random walk behavior and allow it to move faster and more rapidly through the target distribution. This method is called Hamiltonian Monte Carlo (HMC), and is commonly used in computational physics and statistics. The algorithm was originally proposed by \\cite{Duane1987216} for calculations used in lattice quantum chromodynamics, but was later introduced to the field of computational statistics when it was used for Bayesian neural networks in \\cite{neal2012bayesian}. This means that the Markov chain from which we sample in BNNs is produced analogous to paths of particles using Hamiltonian dynamics, and we will explain these dynamics using this physical analogy, as it gives a more intuitive idea of HMC.\n\\\\\n\\\\\nIt turns out that the HMC algorithm reduces correlation between successive sampled states, compared to the Metropolis Hastings algorithm in section \\ref{sec:Metropolis}, by proposing moves to distant states of the target distribution, which maintain a high probability of acceptance due to the properties of the simulated Hamiltonian dynamics. The reduced correlation imply that fewer Markov chain samples are needed to approximate integrals with respect to the target probability distribution.\n\n\\subsection{Hamiltonian Dynamics}\nBefore we move to the actual HMC algorithm, we will explain the Hamiltonian dynamics from which we produce the Markov chain for the algorithm. The Hamiltonian dynamics are used to describe how an object move around in a system or space. It is defined by the objects position $\\boldsymbol{q}\\in \\mathbb{R}^d$ and its momentum $\\boldsymbol{\\rho}\\in \\mathbb{R}^d$, which in the field of physics is equivalent to an object's mass times its velocity at some point in time. When performing HMC to generate samples from the target distribution $\\hat{p}(\\boldsymbol{\\theta})$, the position variable plays the role of the parameter vector and we will from now on let $\\boldsymbol{\\theta}\\equiv \\boldsymbol{q}$. The object's position is associated with a potential energy $U(\\boldsymbol{\\theta})$ and the momentum is associated with a kinetic energy $K(\\boldsymbol{\\rho})$. The sum of the potential and kinetic energy is regarded as the total energy of the system often called the Hamiltonian\n\\begin{equation*}\nH(\\boldsymbol{\\theta},\\boldsymbol{\\rho})=U(\\boldsymbol{\\theta})+K(\\boldsymbol{\\rho}) \n\\end{equation*} \nAn important property of the Hamiltonian is that it conserves the sum of the potential and kinetic energy, meaning that it is constant over time. Taking the partial derivative with respect to time of the position and momentum shows us how they evolve over time\n\\begin{equation}\\label{eq:hamilton_equations}\n\\begin{split}\n\\frac{d \\theta_{i}}{d t}=\\frac{\\partial H}{\\partial \\rho_{i}}=\\frac{\\partial K(\\boldsymbol{\\rho})}{\\partial \\rho_{i}} \\\\\n\\frac{d \\rho_{i}}{d t}=-\\frac{\\partial H}{\\partial \\theta_{i}}=-\\frac{\\partial U(\\boldsymbol{\\theta})}{\\partial \\theta_{i}}\n\\end{split}\n\\end{equation}\nfor $i=1,2, \\ldots,d$. These are named Hamiltonian equations and represent a differential equations system. The Hamiltonian equations are useful, since if we are able to evaluate the partial derivatives from equation \\ref{eq:hamilton_equations}, we are able to predict the position and momentum variables of the object at any point in the future $t^\\prime>t$. \\cite{neal2012bayesian} shows that these dynamics, along with the energy conserving property of the Hamiltonian, results in the process being reversible and preserving volume of the state space, which in turn provides a stationary distribution. \n\n\n\\subsection{Discretizing Hamiltonian Equations}\nThe Hamiltonian equations describe how an objective evolve in continuous time, but for simulating Hamiltonian dynamics on a computer we have to approximate the differential equations which is done by discretizing time. We do this by splitting the time interval $dt$ into smaller intervals $\\varepsilon$. \n\\\\\n\\\\\nThe usual discretizing scheme for simulating Hamiltonian equations, is the Leapfrog method. The Leapfrog method takes a half step to update momentum variable then a whole step to update the position value, and finally the last half step to update momentum\n\\begin{equation*}\n\\begin{split}\n\\rho_{i}^{(t+\\varepsilon / 2)}=\\rho_{i}^{(t)}-(\\varepsilon / 2) \\frac{\\partial U}{\\partial \\theta_{i}^{(t)}} \\\\\n\\theta_{i}^{(t+\\varepsilon)}=\\theta_{i}^{(t)}+\\varepsilon \\frac{\\partial K}{\\partial \\rho_{i}^{(t+\\varepsilon / 2)}} \\\\\n\\rho_{i}^{(t+\\varepsilon)}=\\rho_{i}^{(t)}-(\\varepsilon / 2) \\frac{\\partial U}{\\partial \\theta_{i}^{(t+\\varepsilon)}}\n\\end{split}\n\\end{equation*}\nAccording to \\cite{neal2012mcmc} the Leapfrog method preserves the HMC properties of being reversible and preserve volume of the state space, ensuring that we sample from a stationary distribution.\n\n\n\\subsection{The Hamiltonian and Probability Distributions}\nWe have now explained what a Hamiltonian is and how we can simulate its dynamics by using the Leapfrog method. We will now connect this to MCMC theory from the previous sections in order to explain how to use HMC to sample from the posterior of the parameters in a BNN. In order to perform this connection, we need to relate the target distribution and the Hamiltonian, such that we can use the Hamiltonian equations to sample from the target distribution. A way of doing this, proposed by \\cite{neal2012bayesian}, is to use a concept from statistical mechanics known as the canonical (Boltzmann) distribution. We can write a probability distribution on $\\boldsymbol{\\theta}$ under the canonical distribution as\n\\begin{equation*}\n p(\\boldsymbol{\\theta})\\propto \\exp\\left(\\frac{-E(\\boldsymbol{\\theta})}{T}\\right)\n\\end{equation*}\nwhere $E(\\boldsymbol{\\theta})$ can be any energy function defined over $\\boldsymbol{\\theta}$. $T$ is often called the temperature of the system and usually chosen to be equal to 1 as it plays no role in this application, see \\cite{neal2012bayesian}.\nOne should note that any probability distribution that is nowhere zero can be put into this form by letting $E(\\boldsymbol{\\theta})=-\\log p(\\boldsymbol{\\theta})-\\log Z$, for any convenient choice of normalization constant $Z$. Since the Hamiltonian is an energy function for the joint state of both position and momentum, a joint distribution can be defined by\n\\begin{equation*}\np(\\boldsymbol{\\theta},\\boldsymbol{\\rho})\\propto \\exp(-\\H(\\boldsymbol{\\theta},\\boldsymbol{\\rho})) = \\exp(-U(\\boldsymbol{\\theta}))\\exp(-K(\\boldsymbol{\\rho}))\n\\end{equation*}\nAssuming Independence between $\\boldsymbol{\\theta}$ and $\\boldsymbol{\\rho}$ we can by the equation above write $U(\\boldsymbol{\\theta})=-\\log p(\\boldsymbol{\\theta})$ and $K(\\boldsymbol{\\rho})=-\\log p(\\boldsymbol{\\rho})$ meaning that the Hamiltonian can be interpreted as the log joint distribution on $(\\boldsymbol{\\theta},\\boldsymbol{\\rho})$. \n\\\\\n\\\\\nWe now have a joint distribution, in terms of the Hamiltonian function, which we know how to simulate from. But we are in fact only interested in samples of the position variable $\\boldsymbol{\\theta}$, which is samples from our target distribution, and not samples from the momentum variable $\\boldsymbol{\\rho}$, which is only introduced to make the algorithm move faster through the parameter space. This means that we can choose the marginal distribution of the momentum arbitrarily. This is often chosen by practitioners to be Gaussian, $\\boldsymbol{\\rho}\\sim \\mathcal{N}\\left(0, \\boldsymbol{\\Sigma} \\right)$, where $\\boldsymbol{\\Sigma}$ is some symmetric, positive-definite covariance matrix and often chosen to be diagonal, such that $\\boldsymbol{\\rho}$ is $d$-dimensional multivariate Gaussian, with the $d$ variables being independent. We follow the simple approach from \\cite{hoffman2011nouturn} and let $\\boldsymbol{\\Sigma}$ be the identity matrix $\\boldsymbol{I}$. This makes the dynamics of equation \\ref{eq:hamilton_equations} simplify to \n\\begin{equation*}\n\\begin{split}\n\\frac{d \\theta_{i}}{d t}&=\\rho_i \\\\\n\\frac{d \\rho_{i}}{d t}&=\\frac{\\partial \\log p(\\theta_{i})}{\\partial \\theta_i} \n\\end{split}\n\\end{equation*}\nA more rigorous examination of possible choices for the covariance matrix is provided by \\cite{neal2012mcmc}. \n\n\n\n\\subsection{The Hamiltonian Monte Carlo Algorithm}\nWe start the HMC algorithm from an initial state $\\lr{\\boldsymbol{\\theta}^{(0)},\\boldsymbol{\\rho}^{(0)}}$, and then we simulate the Hamiltonian dynamics for $t+\\varepsilon$ using the Leapfrog method. The states generated for the position and momentum variables at the end of the Leapfrog simulation is used as proposals for a new state $(\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}})$. These states needs to be accepted according to a criterion, because the leapfrog-discretization provides an error term in its approximation of the continuous Hamiltonian dynamics. This criterion is the Metropolis acceptance criterion,\n\\begin{equation}\\label{eq:hmc_acceptance}\n\\begin{split}\n \\alpha\\left((\\boldsymbol{\\theta},\\boldsymbol{\\rho}) \\mapsto (\\boldsymbol{\\theta}^{\\text{cand}} , \\boldsymbol{\\rho}^{\\text{cand}} )\\right) &= \\min\\left\\{1, \\frac{p(\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}})}{p(\\boldsymbol{\\theta},\\boldsymbol{\\rho})} \\right\\}\\\\\n &= \\min\\{1,\\exp\\left(\\log p(\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}})- \\log p(\\boldsymbol{\\theta}, \\boldsymbol{\\rho}) \\right)\\\\\n &= \\min \\left\\{1,\\exp\\left(-\\H(\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}}) +\\H(\\boldsymbol{\\theta},\\boldsymbol{\\rho})\\right) \\right\\}\n\\end{split}\n\\end{equation}\nThis means that we follow the same logic as in the Metropolis algorithm, but use distributions provided by the Hamiltonian dynamics. With $\\boldsymbol{\\rho}\\sim \\mathcal{N}\\left(0,\\boldsymbol{I}\\right)$ this is equivalent to\n\\begin{equation*}\n\\alpha\\left((\\boldsymbol{\\theta},\\boldsymbol{\\rho}) \\mapsto (\\boldsymbol{\\theta}^{\\text{cand}} , \\boldsymbol{\\rho}^{\\text{cand}} )\\right) \n=\\min\\left\\{1,\\frac{\\exp\\left(\\mathcal{L}(\\boldsymbol{\\theta}^{\\text{cand}})-\\frac{1}{2}\\boldsymbol{\\rho}^{\\text{(cand)}^\\top}\\boldsymbol{\\rho}^{\\text{(cand)}}\\right)}{\\exp\\left(\\mathcal{L}(\\boldsymbol{\\theta})-\\frac{1}{2}\\boldsymbol{\\rho}^\\top\\boldsymbol{\\rho}\\right)}\\right\\}\n\\end{equation*}\nwhere $\\mathcal{L}\\left(\\boldsymbol{\\theta}\\right)$ is the log-probability distribution on $\\boldsymbol{\\theta}$.\n\\\\ \n\\\\\nWe see from the last part in equation \\ref{eq:hmc_acceptance} that if we could perfectly discretize the Hamiltonian dynamics, the Metropolis acceptance criterion would always be equal to 1 due to the energy conservation property of the Hamiltonian. Since this is usually not possible, the Metropolis acceptance criterion will often be lower than 1. We can see that if we get a small error in the discretization the term $\\H(\\boldsymbol{\\theta},\\boldsymbol{\\rho})-\\H(\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}})$ in the exponent should be small, yielding a high acceptance rate. This way of making proposals is beneficial since it allows the Markov chain to effectively make large and uncorrelated moves in the state space, while keeping a high acceptance probability. HMC is written in pseudocode in algorithm \\ref{alg:HMC}. It takes in an initial value for the parameters $\\boldsymbol{\\theta}^{(0)}$, which is the starting point of the algorithm. The input $\\mathcal{L}$ is the log-probability distribution of $\\boldsymbol{\\theta}$, which is defined to be equal to the negative potential energy function. In BNN this is identical to the log-posterior distribution on the neural network parameters. One need to be able to at least evaluate the posterior distribution and its gradient or alternatively something proportional to it. In our case where the target distribution is the posterior distribution, we know that it is proportional to the likelihood times the prior, which we in most cases are able to evaluate. The $M$ input is the number of total iterations one would like to perform.\n\\\\\n\\\\\nThe algorithm also relies on a stepsize $\\varepsilon$ variable, that defines the size of the leapfrog step. If $\\varepsilon$ is chosen too be too large, the leapfrog simulation of the Hamiltonian will be inaccurate and lead to a very low rate of acceptance making the algorithm ineffectively waste of computational time. On the other hand, if we choose $\\varepsilon$ to be too small, we will waste computational time by taking too small steps. The sampling is also affected by a hyperparameter $L$, that defines how many leapfrog steps the algorithm performs before proposing a new candidate state. A very small value for $L$ will give successive samples that lie close to each other which results in the same undesirable random walk behavior as the Metropolis algorithm from section \\ref{sec:Metropolis}. Too large a value for $L$ might produce trajectories that loop back and retrace their steps, a behavior called U-turns. This behavior results in the algorithm inefficiently wasting time, sampling from the same area of the distribution again and again. Tuning these parameters can be hard and one are often forced to rely on heuristics based preliminary runs, see \\cite{neal2012mcmc}.\n\\\\\n\\\\\nIn figure \\ref{fig:HMC_Example} we have illustrated how HMC propose a candidate sample for a bivariate $\\mathcal{N}\\left(\\boldsymbol{0},\\boldsymbol{I}\\right)$ target distribution. In subfigure (a) and (b) we have chosen a proper value for $\\varepsilon$ and $L$ such that the algorithm generates proposals that are appropriately far from the previous ones. In subfigure (c) and (d) we have chosen larger values for $\\varepsilon$ and $L$, and we can see that the algorithm starts to loop-back, which results in identical proposals for each iteration and a large proportion of the target distribution will therefore never be visited. In the next section we will look into a modification of the HMC algorithm, so that we can avoid this kind of U-turn behavior. \n\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=\\textwidth]{pics/HMC_Example.png}\n \\caption{A simulation example with HMC for a bivariate Gaussian target distribution with $\\boldsymbol{\\mu}= \\begin{bmatrix}\n 0 & 0\n \\end{bmatrix}$ and covariance matrix\n $\\boldsymbol{\\Sigma}= \n \\begin{bmatrix}\n 1 & 0\\\\\n 0 & 1 \n \\end{bmatrix}$. Subfigure (a) and (b) shows a HMC simulation example with a proper choice for $\\varepsilon$ and $L$, where the target distribution is thoroughly explored. Subfigure (c) and (d) is the same example, modified with a poor choice of values for $\\varepsilon$ and $L$, where the target distribution is poorly approximated. This is especially clear on the plots of the marginal distributions where the histograms are far from the plotted correct distribution. The result is the U-Turn effect, which yield an ineffective exploration of the target distribution. The example has been generated with the interactive gallery provided by \\cite{feng} \\textcopyright MIT.}\n \\label{fig:HMC_Example}\n\\end{figure}\n\n\\begin{algorithm}\\label{alg:HMC}\n\\SetAlgoLined\n\\SetKwFunction{Leapfrog}{Leapfrog}\n\\KwInput{An intial parameter $\\boldsymbol{\\theta}^{(0)}$}\n\\KwInput{A log-probability distribution $\\mathcal{L}$}\n\\KwInput{Total number of iterations $M$}\n\\KwInput{Size of leapfrog-steps $\\varepsilon$}\n\\KwInput{Number of leapfrog-steps before generating candidate state, $L$.}\n\\KwOutput{Set of accepted parameters}\n\n$\\text{Given } \\boldsymbol{\\theta}^{(0)}, \\mathcal{L}, M, \\varepsilon, L,$\\\\\n\\For{m=1,2 \\dots, M}{\nSample $\\boldsymbol{\\rho}^{(0)} \\sim \\mathcal{N}(0, \\boldsymbol{\\boldsymbol{I}})$ \\\\\nSet $ \\boldsymbol{\\theta}^{\\text{cand}} \\leftarrow \\boldsymbol{\\theta}^{(m-1)}, \\boldsymbol{\\rho}^{\\text{cand}} \\leftarrow \\boldsymbol{\\rho}^{(0)}$\\\\\n\\For{i = 1 to L}{\nSet $\\boldsymbol{\\theta}^{\\text{cand}}, \\boldsymbol{\\rho}^{\\text{cand}} \\leftarrow \\Leapfrog(\\boldsymbol{\\theta}^{\\text{cand}}, \\boldsymbol{\\rho}^{\\text{cand}}, \\varepsilon)$\n}\nNegate momentum: $\\boldsymbol{\\rho}^{\\text{cand}} \\leftarrow - \\boldsymbol{\\rho}^{\\text{cand}}$\\\\\nCompute acceptance probability:\\\\\n$\\alpha=\\min \\left\\{1, \\frac{\\exp \\left\\{\\mathcal{L}(\\boldsymbol{\\theta}^{\\text{cand}})-\\frac{1}{2} \\boldsymbol{\\rho}^{\\text{(cand)}^\\top} \\boldsymbol{\\rho}^{\\text{cand}}\\right\\}}{\\exp \\left\\{\\mathcal{L}\\left(\\boldsymbol{\\theta}^{(m-1)}\\right)-\\frac{1}{2} \\boldsymbol{\\rho}^{(0)^\\top} \\boldsymbol{\\rho}^{(0)}\\right\\}}\\right\\}$\\\\\nSample $u\\sim \\text{Uniform}(0,1)$\\\\\n \\uIf{$u<\\alpha$}{\n Accept the proposal: $\\boldsymbol{\\theta}^{(m)} \\leftarrow \\boldsymbol{\\theta}^{c a n d}$\\\\\n }\n \\Else{\n Reject the proposal: $\\boldsymbol{\\theta}^{(m)} \\leftarrow \\boldsymbol{\\theta}^{(m-1)}$ \\\\\n }\n}\n\\SetKwProg{Fn}{Function}{:}{\\KwRet{$\\boldsymbol{\\theta}^{\\text{cand}},\\boldsymbol{\\rho}^{\\text{cand}}$}}\n\\Fn{\\Leapfrog{$\\boldsymbol{\\theta}$, $\\boldsymbol{\\rho}$, $\\varepsilon$}}{\n Set $\\boldsymbol{\\rho}^{\\text{cand}} \\leftarrow \\boldsymbol{\\rho}^{\\text{cand}}+(\\varepsilon / 2) \\nabla_{\\theta} \\mathcal{L}(\\theta)$\\\\\nSet $\\boldsymbol{\\theta}^{\\text{cand}} \\leftarrow \\theta+\\varepsilon \\boldsymbol{\\rho}^{\\text{cand}}$\\\\\nSet $\\boldsymbol{\\rho}^{\\text{cand}} \\leftarrow \\boldsymbol{\\rho}\n^{\\text{cand}}+(\\varepsilon / 2) \\nabla_{\\theta} \\mathcal{L}(\\boldsymbol{\\theta}^{\\text{cand}})$\n }\n\\caption{Hamiltonian Monte Carlo}\n\\end{algorithm}\n\n\n\n\n\\clearpage\n\\subsection{No-U-Turn Hamiltonian Monte Carlo}\\label{sec:nuts}\nIn this section we present an algorithm introduced by \\cite{hoffman2011nouturn} and evaluated by \\cite{nishio_arakawa_nouturn}, whose explanation we found more clear and concise. No-U-Turn (NUTS) extends HMC by eliminating the need to specify the trajectory length $L$. This algorithm gets its name by avoiding the possible U-Turning behavior shown in figure \\ref{fig:HMC_Example} (c) and (d). This is done by introducing a criterion that tells us when to stop simulating the dynamics to prevent a possible U-turn. The authors define this criterion to be when performing another leapfrog step will no longer increase the distance between the proposed state $\\boldsymbol{\\theta}^{\\text{cand}}$ and the initial value $\\boldsymbol{\\theta}^{(0)}$. More specifically, they choose a criterion, based on the derivative with respect to time of half the squared distance between the initial parameter $\\boldsymbol{\\theta}^{(0)}$ and the current state $\\boldsymbol{\\theta}^{\\text{cand}}$, meaning that leapfrog steps are performed until \n\\begin{equation*}\n\\begin{split}\n \\frac{d}{d t} \\frac{(\\boldsymbol{\\theta}^{\\text{cand}}-\\boldsymbol{\\theta})^\\top \\cdot(\\boldsymbol{\\theta}^{\\text{cand}}-\\boldsymbol{\\theta})}{2}&=(\\boldsymbol{\\theta}^{\\text{cand}}-\\boldsymbol{\\theta})^\\top \\cdot \\frac{d}{d t}(\\boldsymbol{\\theta}^{\\text{cand}}-\\boldsymbol{\\theta})\\\\\n &=(\\boldsymbol{\\theta}^{\\text{cand}}-\\boldsymbol{\\theta})^\\top \\cdot \\boldsymbol{\\rho}^{\\text{cand}} < 0\n\\end{split}\n\\end{equation*}\nHowever, \\cite{hoffman2011nouturn} notes that by doing this we do not have the guarantee of time reversibility, so the sampling algorithm might not converge to the correct distribution. NUTS overcomes this problem by using slice sampling and applying a double method suggested by \\cite{neal_slice_sampling}. \n\\\\\n\\\\\nNUTS augments the distribution of HMC, $p \\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}} \\propto \\exp{\\left(\\mathcal{L} \\lr{\\boldsymbol{\\theta}} - \\frac{1}{2} \\boldsymbol{\\rho}^\\top \\boldsymbol{\\rho}\\right)}$, to include a slice variable $u$ so that the joint probability of $\\boldsymbol{\\theta}, \\boldsymbol{\\rho}$ and $u$ is \n\\begin{equation*}\n p \\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}, u} \\propto \\mathbf{1} \\lrs{ u \\in \\lrs{0,\\exp{\\left(\\mathcal{L} \\lr{\\boldsymbol{\\theta}} - \\frac{1}{2} \\boldsymbol{\\rho}^\\top \\boldsymbol{\\rho}\\right)}}}\n\\end{equation*}\nmeaning that the un-normalized marginal probability of $\\boldsymbol{\\theta}$ and $\\boldsymbol{\\rho}$ (gained by integrating over $u$) is \n\\begin{equation*}\n p \\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}} \\propto \\exp{\\left(\\mathcal{L} \\lr{\\boldsymbol{\\theta}} - \\frac{1}{2} \\boldsymbol{\\rho}^\\top \\boldsymbol{\\rho}\\right)}\n\\end{equation*}\nThe conditional probabilities $p \\lr{u \\mid \\boldsymbol{\\theta}, \\boldsymbol{\\rho}}$ and $p \\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho} \\mid u}$ are each uniform as long as the condition\n\\begin{equation} \\label{eq:nuts_unif_condition}\n u \\leq \\exp{\\left(\\mathcal{L} \\lr{\\boldsymbol{\\theta}} - \\frac{1}{2} \\boldsymbol{\\rho}^\\top \\boldsymbol{\\rho}\\right)}\n\\end{equation}\nis satisfied. The challenge of slice sampling is to find the bounds of the region for which this condition is satisfied. \\cite{neal_slice_sampling} proposes a doubling method, where the size of the initial segment containing the current value of $\\boldsymbol{\\theta}$ is randomly chosen and afterwards expanded by doubling its size until the samples are outside of the region. The expanding directions are randomly chosen to be leapfrog steps forward or backward in the Markov chain to satisfy reversibility.\n\\\\\n\\\\\nNUTS generates a finite set of all $\\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}}$ by iteratively doubling its size. The doubling process is stopped to satisfy the condition \n\\begin{equation*}\n \\lr{\\boldsymbol{\\theta}^+ - \\boldsymbol{\\theta}^-}^\\top \\boldsymbol{\\rho}^- < 0 \\quad \\text{or} \\quad \\lr{\\boldsymbol{\\theta}^- - \\boldsymbol{\\theta}^+}^\\top \\boldsymbol{\\rho}^+ < 0\n\\end{equation*}\nwhere $\\boldsymbol{\\theta}^+, \\boldsymbol{\\rho}^-$ and $\\boldsymbol{\\theta}^+, \\boldsymbol{\\rho}^-$ is the leftmost and rightmost variables of all $\\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}}$ generated by the doubling process respectively. \n\\\\\n\\\\\nA subset of proposal candidates $\\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}}$, denoted by $\\mathbb{C}$, is selected from the doubling process to satisfy the condition in equation \\ref{eq:nuts_unif_condition}. The new values of $\\lr{\\boldsymbol{\\theta}^*, \\boldsymbol{\\rho}^*}$ are then afterwards sampled uniformly from $\\mathbb{C}$. \n\\\\\n\\\\\nTo further improve this algorithm \\cite{hoffman2011nouturn} used the following transition kernel in each step of doubling\n\\begin{equation*}\n\\resizebox{\\textwidth}{!}{$\nT\\left(\\boldsymbol{\\theta}^{*}, \\boldsymbol{\\rho}^{*} \\mid \\boldsymbol{\\theta},\\boldsymbol{\\rho}, \\mathbb{C}\\right)=\\begin{cases} \\frac{\\mathbf{1}\\left[\\boldsymbol{\\theta}^{*}, \\boldsymbol{\\rho}^{*} \\in \\mathbb{C}^{\\text{new}}\\right]}{\\left|\\mathbb{C}^{\\text{new}}\\right|} & \\text{ when } \\left|\\mathbb{C}^{\\text{new}}\\right|>\\left|\\mathbb{C}^{\\text{old}}\\right|\\\\\n\\frac{\\left|\\mathbb{C}^{\\text{new}}\\right|}{\\left|\\mathbb{C}^{\\text{old}}\\right|} \\frac{\\mathbf{1}\\left[\\boldsymbol{\\theta}^{*}, \\boldsymbol{\\rho}^{*} \\in \\mathbb{C}^{\\text{new}}\\right]}{\\left|\\mathbb{C}^{\\text{new}}\\right|}+\\left(1-\\frac{\\left|\\mathbb{C}^{\\text{new}}\\right|}{\\left|\\mathbb{C}^{\\text{old}}\\right|}\\right) \\mathbf{1}\\left[\\left(\\boldsymbol{\\theta}^{*}, \\boldsymbol{\\rho}^{*}\\right)=(\\boldsymbol{\\theta}, \\boldsymbol{\\rho})\\right] \n&\\text{ when } \\left|\\mathbb{C}^{\\text{new}}\\right|\\leq\\left|\\mathbb{C}^{\\text{old}}\\right|\n\\end{cases}$\n}\n\\end{equation*}\nwhere $\\mathbb{C}^{\\text{new}}$ is the subset of $\\lr{\\boldsymbol{\\theta}, \\boldsymbol{\\rho}}$, added by the last step of the doubling process and $\\mathbb{C}^{\\text{old}}$ is the disjoint subset of $\\mathbb{C}$ so that $\\mathbb{C} = \\mathbb{C}^{\\text{new}} \\cup \\mathbb{C}^{\\text{old}}$. This transition kernel proposes a move from a state in $\\mathbb{C}^{\\text{old}}$ to a random state in $\\mathbb{C}^{\\text{new}}$ and accepts this move with probability $\\frac{\\mathbb{C}^{\\text{new}}}{\\mathbb{C}^{\\text{old}}}$. The authors show that $T$ satisfies the detailed balance condition so that it leaves the uniform distribution over $\\mathbb{C}$ invariant. According to \\cite{nishio_arakawa_nouturn} this transitional kernel permits memory-efficient implementation and produces larger jumps on average than simple uniform sampling. \n\\\\\n\\\\\nNUTS is especially efficient as it can automatically choose a step size that achieves an acceptance probability around a desired level. The stepsize $\\varepsilon$ for the $j$th iteration of a NUTS Markov chain is tuned as follows\n\\begin{equation*}\n \\begin{split}\n \\log\\lr{\\varepsilon_{j+1}} &= \\mu - \\frac{\\sqrt{j}}{\\gamma} \\frac{1}{j + j_0} \\sum_{i=1}^j \\lr{\\delta - \\alpha_i}\\\\\n \\log \\lr{\\bar{\\varepsilon}_{j+1}} &= \\eta_j \\log \\lr{\\varepsilon_{j+1}} + \\lr{1 - \\eta_j} \\log \\lr{\\bar{\\varepsilon}_j} \\\\\n \\varepsilon_{j+1} &= \\bar{\\varepsilon}_{j+1}\n \\end{split}\n\\end{equation*}\nwhere $\\alpha_j$ is an actual acceptance probability for the $j$th iteration, $\\delta$ is the desired average acceptance probability, $\\mu$ is a freely chosen point that the iterated $\\varepsilon_j$ shrinks towards, $\\gamma$ is a free parameter that controls the shrinkage amount towards $\\mu$ and $j_0$ is a free parameter that dampens early exploration. \n\\\\\n\\\\\n\\cite{hoffman2011nouturn} introduces the variables $\\eta_j = j^{-\\kappa}$ with $\\kappa < 1$ to give more recent iterates more weight. They show that this way of adapting stepsize guarantees that $\\alpha \\rightarrow \\delta$. They recommend setting $\\mu = \\log \\lr{10 \\varepsilon_1}$ and $\\delta \\approx 0.6$.\n\\\\\n\\\\\nNUTS tunes $\\varepsilon$ during a predetermined warm-up phase and fixes this value thereafter. Since the algorithm accepts or rejects $\\lr{\\boldsymbol{\\theta}^*, \\boldsymbol{\\rho}^*}$ from multiple candidates, an alternative statistic to the Metropolis acceptance probability must be defined. They do this by defining for each iteration the acceptance probability by\n\\begin{equation*}\n \\alpha_{j}=\\frac{1}{\\left|B_{j}\\right|} \\sum_{\\boldsymbol{\\theta}, \\boldsymbol{\\rho} \\in B_{j}} \\min \\left\\{1, \\frac{p\\left(\\boldsymbol{\\theta}^{j}, \\boldsymbol{\\rho}^{j}\\right)}{p\\left(\\boldsymbol{\\theta}^{j-1}, \\boldsymbol{\\rho}^{j, 0}\\right)}\\right\\}\n\\end{equation*}\nThe pseudocode for NUTS can be seen in algorithm \\ref{alg:nuts}.\n\n\\begin{algorithm} \\label{alg:nuts}\n \\SetAlgoLined\n \\KwInput{Initial parameters $\\boldsymbol{\\theta}^{(0)}$}\n \\KwInput{A log-probability distribution $\\mathcal{L}$}\n \\KwInput{Initial size of leapfrog-steps $\\bar{\\varepsilon_0}$}\n \\KwInput{Desired average of acceptance probability $\\delta$}\n \\KwInput{Aim point for values of the iterated $\\varepsilon_j$ values $\\mu$}\n \\KwInput{Parameter controlling shrinkage towards $\\mu$, $\\gamma$}\n \\KwInput{Parameter that controls dampening of early exploration $j_0$}\n \\KwInput{Parameter that controls how much more weight are given to more recent iterates $\\kappa < 1$}\n \\KwInput{Total number of total number of iterations $J$}\n \\KwInput{Total number of iterations for adapting size of leapfrog steps $J^{\\text{adapt}}$}\n \\KwOutput{Samples from the target distribution}\n \\For{j=0, \\dots, J}{\n Sample momentum $\\boldsymbol{\\rho}^{\\text{init}} \\sim N(0, \\boldsymbol{I})$\\\\\n Sample auxiliary variable $u \\sim \\operatorname{Uniform}\\left(0, \\exp \\left(\\mathcal{L}\\left(\\boldsymbol{\\theta}^{(j)}\\right)-\\frac{1}{2}\\boldsymbol{\\rho}^{\\text{init}^{\\top}} \\boldsymbol{I}^{-1} \\boldsymbol{\\rho}^{\\text{init}}\\right)\\right)$\\\\\n Generate $\\mathbb{C}$ by using the doubling method with transition kernel $T$. \\\\\n Compute acceptance probability: $\\alpha_{j}=\\frac{1}{\\left|B_{j}\\right|} \\sum_{\\theta, \\boldsymbol{\\rho} \\in B_{j}} \\min \\left\\{1, \\frac{p\\left(\\boldsymbol{\\theta}^{j+1}, \\boldsymbol{\\rho}^{j+1}\\right)}{p\\left(\\boldsymbol{\\theta}^{j}, \\boldsymbol{\\rho}^{\\text{init}}\\right)}\\right\\}$\\\\\n Accept the proposal $\\left(\\boldsymbol{\\theta}^{*}, \\boldsymbol{\\rho}^{*}\\right)$ with probability $\\alpha_{j}$. \\\\\n \\uIf{$j\\leq J^{\\text{adapt}}$}{\n $\\log\\lr{\\varepsilon_{j+1}} \\leftarrow \\mu - \\frac{\\sqrt{j}}{\\gamma} \\frac{1}{j + j_0} \\sum_{i=1}^j \\lr{\\delta - \\alpha_i}$\\\\\n $\\log \\lr{\\bar{\\varepsilon}_{j+1}} \\leftarrow j^{-\\kappa}\\log \\lr{\\varepsilon_{j+1}} + \\lr{1 - j^{-\\kappa}} \\log \\lr{\\bar{\\varepsilon}_j}$ \\\\\n $\\varepsilon_{j+1} \\leftarrow \\bar{\\varepsilon}_{j+1}$\n }\n \\Else{\n $\\varepsilon_{j+1}=\\varepsilon_{J^{\\text{adapt}}}$\n } \n }\n \\caption{No-U-Turn Sampler with Dual Averaging. One can easily change this pseudocode to one that runs until a certain number of samples are collected. For a more detailed pseudocode see \\cite{hoffman2011nouturn}.}\n\\end{algorithm}\n\n\n\n\\clearpage\n\\section{Priors}\\label{sec:priors}\nFrom section \\ref{sec:bayesian_stat} and the example given in section \\ref{sec:simple_BNN} it is seen that Bayesian inference starts with a prior for the model parameters, which is supposed to embody ones prior beliefs about the assigned task. The prior is an important component and choosing a bad prior can affect the resulting posterior greatly. That said, the prior does have a diminishing effect on the posterior as the number of samples grow, as the likelihood will concentrate the posterior around a few highly likely parameters. One might think that if one has no qualified prior belief, then a weak prior, like a wide uniform distribution, can be a safe bet, but this can however have various unforeseen consequences as pointed out by \\cite{lemoine2019} and \\cite{sarma_kay2020}. \\\\\n\\\\\nThe prior component is however more than a dangerous element that threatens the quality of the posterior. The prior provides a principled mechanism for researchers to incorporate previous research and knowledge into the model. Priors can also be beneficial in small sample sizes as the prior acts to regularize and reduce the chances of overfitting.\nAlthough in neural networks the relationship between parameters and the problem can be very abstract and not as intuitive as in other machine learning models like linear regression models or support vector machines, so having qualified prior beliefs about the weights might not be so easy. \n\\\\\n\\\\\n\\cite{neal2012bayesian} stresses that even though it can seem like BNNs can be threatened by a lack of a suitable prior this is not the case, as much past work shows useful criteria for selecting suitable priors, even without full understanding of what the prior over the parameters will mean in terms of the output of the network. \\cite{mackay1991} and \\cite{MacKay1992} has produced results, that \\cite{neal2012bayesian} describes as at least reasonable, by giving the parameters Gaussian prior distributions. He lets the standard deviation of these distributions be selected as a hyperparameter, which allows the model to adapt to the data. \n\\\\\n\\\\\nAccording to \\cite{neal2012bayesian}, our prior knowledge will often be too unspecific to fix the parameter values chosen for the prior distribution, even if we have complete insight into their effects on the prior. We may then wish to treat these values as unknown hyperparameters, giving them a higher-level broad prior distribution, which we call a hyper-prior. \\cite{neal2012bayesian} refers to these as Hierarchical models. One benefit of such models is that the appropriate degree of regularization for the task can be determined automatically from the data, see \\cite{mackay1991} and \\cite{MacKay1992}. \n\n\n\n\n\n\n\n", "meta": {"hexsha": "077af2d76da2477004d8fcfd3e081659956d291b", "size": 75630, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "bayesian_neural_networks.tex", "max_stars_repo_name": "mraabo/Dissertation--Bayesian-Neural-Networks", "max_stars_repo_head_hexsha": "629b1c5f4bbdb80ef1d1037b4a0a1b7f95ac710b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "bayesian_neural_networks.tex", "max_issues_repo_name": "mraabo/Dissertation--Bayesian-Neural-Networks", "max_issues_repo_head_hexsha": "629b1c5f4bbdb80ef1d1037b4a0a1b7f95ac710b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "bayesian_neural_networks.tex", "max_forks_repo_name": "mraabo/Dissertation--Bayesian-Neural-Networks", "max_forks_repo_head_hexsha": "629b1c5f4bbdb80ef1d1037b4a0a1b7f95ac710b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 127.5379426644, "max_line_length": 1613, "alphanum_fraction": 0.7477984927, "num_tokens": 20168, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8723473713594992, "lm_q2_score": 0.7310585786300049, "lm_q1q2_score": 0.6377370293776966}} {"text": "%!TEX root = ../gfkg.tex\n\\subsection{Chapter 1}\n\n\\begin{p}%1\n\\end{p}\n\n\\begin{p}%2\n\\end{p}\n\n\\begin{p}%3\n\\end{p}\n\n\\begin{p}%4\n\\end{p}\n\n\n\\begin{p}%{5}\n{Given a Lie group $G$, define its identity component $G_0$ to be the connected component \ncontaining the identity element. Show that the identity component of any Lie group is a subgroup, and a Lie\ngroup in its own right.}\n\\end{p}\n{Suppose we have a path from the identity to $g\\in G_0$. Now map this \npath to a new path by multiplying each element by $h\\in G_0$. This path starts at $h$ and since the mapping is continuous, must remain in $G_0$. (Otherwise, smoothly mapping the group manifold to $\\R^n$ would show a \ndiscontinuity at some point.) Thus $hg\\in G_0$ for all $h,g\\in G_0$.\n\nThere's a certain tension between having a smooth manifold with disconnected pieces --- given a map from the \nthe manifold to itself, one must take care that it does not have a discontinuous action, mapping some points in one component to another component. When this map is an element of the group, this requirement makes\n$G_0$ into a subgroup.}\n\n\n\\begin{p}%{6}\n{Show that every element of O(3) is either a rotation about some axis or a rotation about some axis followed by a reflection through some plane. Show that the former class of elements are all in the identity component of O(3), while the latter are not. Conlude that the identity component of O(3) is SO(3).}\n\\end{p}\n{Rotations preserve the inner product between vectors, so with any rotation $R$ we can transform the standard orthonormal basis $\\{e_1,e_2,e_3\\}$ to another. But $Re_1$ is just the first column of $R$, so this fact means that $R^T R=\\mathbbm{1}$. Conversely, the condition $Q^TQ=\\mathbbm{1}$ means that $Q$ can be thought of as a set of orthonormal basis elements, and the action of $Q$ on the standard basis will produce this new basis. \nThis new basis need not be a rotated version of the standard basis, however, as it could also involve reflections. This corresponds to a negative determinant as the orientation of the standard basis is not preserved. \nAany such $Q$ can be decomposed into $Q=PR$ by using a reflection operator. Let $P'$ be the reflection through\nthe plane defined by the first two standard basis vectors, i.e.~the one flipping the third standard basis vector. Clearly $QP'=R$ for some rotation $R$, since it now has a unit determinant. Then write $QP'=QP'Q^TQ=PQ$,\ndefining $P=QP'Q^T$, which is symmetric: $P^T=QP'^TQ^T=QP'Q^T=P$. Thus $Q=PR$ for some rotation $R$ and reflection $P$. All rotations are connected to the identity, since we can imagine continuously varying the rotation amount. Thus the identity-connected component is SO(3). The reflections could not be the SO(3) since\nthey do not form a subgroup; the product of two reflections is a rotation.}\n\n\\begin{p}%7\n\\end{p}\n\n\\begin{p}%{8}\n{Show that if $\\rho:G\\rightarrow H$ is a homomorphism of groups, then $\\rho(1)=1$ and $\\rho(g^{-1})=\\rho(g)^{-1}$.}\n\\end{p}\n{First, the identity and inverses in a group are unique. Suppose there were\ntwo identity elements, $e$ and $1$. Then $ge=eg=g=g1=1g$ for all $g$ and in particular $e=e1=1$, using the\nidentity properties of first $1$ and then $e$. The uniqueness of the identity implies the uniqueness of inverses,\nfor suppose $g$ had two inverses, $h$ and $k$. Then $gh=gk=hg=hk=1$ and thus $hgh=h=hgk=k$.\nNow if $\\rho$ is to be a homomorphism, then $\\rho(g)=\\rho(g1)=\\rho(g)\\rho(1)$ for all $g$, whence $\\rho(1)$ is\nthe identity. Similarly, $\\rho(1)=\\rho(gg^{-1})=\\rho(g)\\rho(g^{-1})$ for all $g$. So $\\rho(g^{-1})$ must be \nthe inverse of $\\rho(g)$, i.e.~$\\rho(g)^{-1}$.}\n\n\\begin{p}%{9}\n{A $1\\times 1$ matrix is just a number, so show that U(1)$=\\{e^{i\\theta}:\\theta\\in\\R\\}$. In physics, an element of U(1) is a called a phase. Show that U(1) is isomorphic to SO(2), with an isomorphism being given by \n\\[\\rho(e^{i\\theta})=\\left(\\begin{array}{cc}\\cos\\theta & \\sin\\theta\\\\-\\sin\\theta & \\cos\\theta\\end{array}\\right).\\]}\n\\end{p}\n{First we check if the map is a homomorphism, and then if it is bijective. For the homomorphism we need\nto show that $\\rho(e^{i\\theta})\\rho(e^{i\\phi})=\\rho(e^{i(\\theta+\\phi)})$, which follows since\n\\[\\left(\\begin{array}{cc}\\cos\\theta & \\sin\\theta\\\\-\\sin\\theta & \\cos\\theta\\end{array}\\right)\\left(\\begin{array}{cc}\\cos\\phi & \\sin\\phi\\\\-\\sin\\phi & \\cos\\phi\\end{array}\\right)=\\left(\\begin{array}{cc}\\cos(\\theta+\\phi) & \\sin(\\theta+\\phi)\\\\-\\sin(\\theta+\\phi) & \\cos(\\theta+\\phi)\\end{array}\\right).\\]\nThe map is onto, since every element in SO(2) is of the given form, for some $\\theta$. It is one-to-one since\nunless $\\theta=\\phi$, $\\rho(e^{i\\theta})\\neq \\rho(e^{i\\phi})$.}\n\n\\begin{p}%10\n\\end{p}\n\n\\begin{p}%11\n\\end{p}\n\n\\begin{p}%{12}\n{Show that for any bilinear function $f:V\\times V'\\rightarrow W$ from vector spaces $V$ and $V'$ to $W$ there exists a unique linear function $F:V\\otimes V'\\rightarrow W$ such that $f(v,v')=F(v\\otimes v')$.}\n\\end{p}\n{\nHere we're not trying to prove abstractly that the tensor product has this property; this property is the definition\nof tensor product. We want to show this property holds for the tensor product given. For the function $f$ to be\nbilinear, it must obey $f(v,v')=f(v^ie_i,v'^j e'_j)=v^iv'^jf(e_i,e'_j)$. But $v\\otimes v'=v^iv'^je_i\\otimes e'_j$, so we can just define $F$ by the action on the basis: $F(e_i\\otimes e'_j)=f(e_i,e'_j)$. This set is a linearly independent basis, since the parts are linearly independent, and so the function defined this way is unique.} \n\n\\begin{p}%13\n\\end{p}\n\n\\begin{p}%14\n\\end{p}\n\n\\begin{p}%15\n\\end{p}\n\n\\begin{p}%16\n\\end{p}\n\n\\begin{p}%17\n\\end{p}\n\n\\begin{p}%{18}\n{Show that any $2\\times 2$ matrix may be uniquely expressed as a linear combination of Pauli matrixes $\\sigma_0,\\dots,\\sigma_3$ with complex coefficients, and that the matrix is hermitian iff these coefficients are real. Show tha tthe matrix is traceless iff the coefficient of $\\sigma_0$ vanishes.}\n\\end{p}\n{A general\n$2\\times 2$ matrix $A$ has components $a_{jk}$, i.e.~in the ``standard basis'' of matrices. By direct calculation we can also represent $A$ as $\\sum_{k}c_k\\sigma_k$ for $2c_0=a_{00}+a_{11}$, $2c_3=a_{00}-a_{11}$, $2c_1=a_{01}+a_{10}$, and $2c_2=i(a_{01}-a_{10})$. Hermiticity requires\nthat the $c_j$ are real, since the $\\sigma_j$ are, which can also be seen from direct calculation. $c_0$ is half the\ntrace, so the matrix is traceless exactly when $c_0=0$.}\n\n\\begin{p}%19\n\\end{p}\n\n\\begin{p}%{20}\n{Show that the determinant of the $2\\times 2$ matrix $a+bI+cJ+dK$ is $a^2+b^2+c^2+d^2$. Show that if $a,b,c,d$ are real and $a^2+b^2+c^2+d^2=1$, this matrix is unitary. Conclude that SU(2) is the unit sphere in $\\mathbb{H}$.}\n\\end{p}\n{In the usual representation the matrix is just $\\left(\\begin{array}{cc}a-id & -ib-c\\\\ -ib+c& a+id\\end{array}\\right)$, so the determinant follows. Direct calculation reveals $MM^\\dagger=$ det$(M)\\mathbbm{1}$. Regarding the matrices $I,J,K$ as the three quaternionic units, which we can do since they obey the same algebra as the matrices, we can see that any element of SU(2) can be regarded as a point\non the quaternionic unit sphere.}\n\n\\begin{p}%{21}\n{Show that the spin-0 representation of SU(2) is equivalent to the trivial representation in which every element of the group acts on $\\mathbbm{C}$ as the identity.}\n\\end{p}\n{To construct the representations, we \nstart from the homogeneous polynomials $f(x,y)$ on $(x,y)\\in\\mathbbm{C}^2$ of degree $2j$. This is a \nvector space of dimension $2j+1$ since $x^{2j},x^{2j-1}y,\\dots y^{2j}$ forms a basis. \nNow for any $g\\in$SU(2), let $U_j(g)$ be such that $(U_j(g)f)(v)=f(g^{-1}v)$ for $v\\in\\mathbbm{C}^2$. \nIn the case of spin-0, $f(x,y)=c$ for some constant $c$. Thus $f(g^{-1}v)=c=(U_0(g)f)(v)$ \nand we must therefore have $U_0(g)f=f$.}\n\n\\begin{p}%{22}\n{Show that the spin-1/2 represenation of SU(2) is equivalent to the fundamental representation in which every element $g\\in$SU(2) acts on $\\mathbbm{C}^2$ by matrix multiplication.}\n\\end{p}\n{Here the \nhomogeneous polynomials are simply $f(x,y)=ax+by$, i.e.~elements of $\\mathbbm{C}^2$. Then regarding\n$f$ as the column vector $\\bar{f}$, $f(x,y)$ becomes the inner product $\\bar{f}^T\\cdot(x,y)$. \n$f(g^{-1}v)$ is therefore $\\bar{f}^T\\cdot g^{-1}v=(g\\bar{f})^T\\cdot v$, so $U_{1/2}(g)f=gf$.}\n\n\\begin{p}%{23}\n{Show that for any representation $\\rho$ of a group $G$ on a vector space $V$ there is a dual or \ncontragredient representation $\\rho^*$ of $G$ on $V^*$, given by \n\\[(\\rho^*(g)f)(v)=f(\\rho(g^{-1})v)\\] for all $v\\in V, f\\in V^*$. \nShow that all the representations $U_j$ of \nSU(2) are equivalent to their duals.}\n\\end{p}\n{Clearly $\\rho^*(e)=1$, so we need to show that $\\rho^*(gh)=\\rho^*(g)\\rho^*(h)$. We have $(\\rho^*(gh)f)(v)=f(\\rho((gh)^{-1})v)=f(\\rho(h^{-1}g^{-1})v)=f(\\rho(h^{-1})\\rho(g^{-1})v)=(\\rho^*(g)\\rho^*(h)f)(v).$ }\n\n", "meta": {"hexsha": "a3fc870d7a0ce9c19c43dd82605936725bbfc7da", "size": 8801, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/II1.tex", "max_stars_repo_name": "joerenes/Baez-Muniain-solutions", "max_stars_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-04-13T12:10:03.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T18:18:34.000Z", "max_issues_repo_path": "src/II1.tex", "max_issues_repo_name": "joerenes/Baez-Muniain-solutions", "max_issues_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-04-13T12:15:30.000Z", "max_issues_repo_issues_event_max_datetime": "2017-04-13T20:19:44.000Z", "max_forks_repo_path": "src/II1.tex", "max_forks_repo_name": "joerenes/Baez-Muniain-solutions", "max_forks_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.7753623188, "max_line_length": 438, "alphanum_fraction": 0.6928758096, "num_tokens": 2926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585669110202, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.6377370264331853}} {"text": "\\subsection{Continuum extrapolation before infinite volume limit}\n\n\\begin{figure}[!htb]\n\\scalebox{1.05}{\\input{figure/continuum-extrapolation.pgf}}\n\\caption{\n \\label{fig:continuum-extrapolation-spectrum}\n Continuum extrapolation of the discrete finite volume spectrum with $L=1\\,[\\mathrm{fm}]$.\n Each column represents a different implementation of the kinetic operator, rows correspond to eigenvalues of the hamiltonian sorted by value.\n For visualization purposes we present each second eigenvalue starting at $E_2$ ($E_0$ was used to tune the interaction and is thus constant by construction).\n Black dots are the eigenvalues at different lattice spacings, the green band is the model averaged fit function for best parameters and the blue band parallel to the x-axis is the continuum-extrapolated energy.\n The uncertainty is dominated by the fluctuations over models; the propagated numerical uncertainty is negligible in comparison.\n The dashed line corresponds to the expected result obtained by computing the intersection of the zeta function $S^{\\spherical}_3$ with the phase shifts.\n The boundary of each frame corresponds to the poles of the zeta function.\n Different energy extrapolations in the continuum agree with zeros of the \\Luscher zeta within uncertainty.\n For finite discretization implementations ($n_s < \\infty$), the uncertainty drastically increases with the number of excited states ($\\sim 3$ orders of magnitude from $E_2^{(n_s)}$ to $E_{20}^{(n_s)}$).\n}\n\\end{figure}\n\nAfter tuning the contact interaction to the first zero of the spherical zeta function, we compute the spectrum of the hamiltonian.\nNext, we extrapolate the obtained energy eigenvalues to the continuum $\\epsilon \\to 0$ using a polynomial fit\n\\begin{equation}\n E^{(n_s)}_i(\\epsilon) = E_i^{(n_s)} + \\sum\\limits_{n=1}^{n_\\mathrm{max}} e_{i,n}^{(n_s)} \\epsilon^n \\, .\n\\end{equation}\nBecause the contact interaction is expected to scale linear with the momentum cutoff and thus linear in $1/\\epsilon$ (see \\eqref{three-d-counterterm}), one cannot generally expect the fit coefficients $e_{i,n}^{(n_s)}$ to be zero for odd $n$ or $n < n_s$, despite the kinetic improvement~\\eqref{gamma determination}.\nNevertheless, we would expect the small $n$ coefficient for larger $n_s$ to be relatively smaller then small $n$ coefficients for smaller $n_s$: $e_{i,n}^{(n_{s_1})} < e_{i,n}^{(n_{s_2})}$ on average for $n_{s_1} > n_{s_2}$.\n\nWe individually fit each discretization implementation to extract the continuum energies $E_i^{(n_s)}$ using the software provided by \\Ref{peter_lepage_2016_60221}.\nBecause our numerical uncertainties have an estimated relative error at the order $~10^{-13}$, we must in principle fit the energy for relatively high values of $n_\\mathrm{max}$ which would require having many data points over different scales of $\\epsilon$.\nFor this reason we add further lattice spacings\n\\begin{equation}\n\t\\left\\{\n\t\t\\epsilon \\, [\\mathrm{fm}] =\n\t\t\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{10},\n\t\t\\frac{1}{15}, \\frac{1}{20}, \\frac{1}{25},\n\t\t\\frac{1}{ 30}, \\frac{1}{ 35}, \\frac{1}{ 40},\n\t\t\\frac{1}{ 41}, \\frac{1}{ 42}, \\frac{1}{ 43},\n\t\t\\frac{1}{ 44}, \\frac{1}{45}, \\frac{1}{ 46},\n\t\t\\frac{1}{ 47}, \\frac{1}{ 48}, \\frac{1}{ 49}, \\frac{1}{ 50}\n\t\\right\\}\n\t\\, .\n\\end{equation}\nHowever, we still obtain $\\chi^2_{\\mathrm{d.o.f}} \\gg 1$ up to the point where it is computationally not feasible to add new data points for even smaller lattice spacings as the dimension of the hamiltonian scales with $(L/\\epsilon)^3$.\n\nFor this reason, we have decided to fit multiple fit models over the span of $n_\\mathrm{max} = \\{2, 3, 4, 5\\}$ and compare their results to estimate a systematic extrapolation uncertainty (unweighted average and standard deviation of results over models).\nWe repeat this procedure for each discretization and compare different continuum energies to decide wether the fits are consistent.\nThese values are compared to the spectrum predicted by L\\\"uscher's formalism.\n\n\\begin{figure}[H]\n\t\\centering\n \\scalebox{0.8}{\\input{figure/continuum-extrapolation-ere.pgf}}\n \\caption{\n \\label{fig:continuum-extrapolation-ere}\n Phase shifts computed by inserting the continuum-extrapolated spectrum for different discretization implementations $n_s$ and finite volumes $L$ in the zeta function $S^{\\spherical}_3$.\n Data points indicate locations of the eigenvalue.\n We show the propagated error associated with continuum extrapolation as an uncertainty band.\n The black dashed line represents physical phase shifts.\n Bands stop at different $x$ values because we stop presenting results after uncertainties become too large (but are still consistent with the physical phase shifts).\n }\n\\end{figure}\n\nWe present the model average over best fits of the spectrum in \\figref{continuum-extrapolation-spectrum}.\nAlso, we provide access to the raw data and fitting scripts online at \\cite{luescher-nd_201}.\nWe observe that the model average for polynomials of degree 2 up to 5 is consistent over different discretization and agrees with the expected continuum results.\nWe noted that including higher polynomials with $n_\\mathrm{max} > 6$ resulted in overfitting of higher energy levels visible in oscillating fit functions which were generally were more favorable in model selection criteria\\footnote{\n A potential cure for overfitting of higher polynomials would have been the marginalization of higher contributions which would cast the contributions of higher neglected epsilon terms into the uncertainty of the data.\n We eventually settled for an unweighted model average over smaller $n_\\mathrm{max}$ because the continuum-extrapolated spectrum was more consistent over different $n_s$.\n}.\nAs expected, the continuum limit becomes more uncertain for excited states.\nFurthermore, the $n_s = \\infty$ implementation provides the most precise results.\nSurprisingly a few energy levels in the $n_s = 1$ implementation have a more precise continuum limit on average than some improved implementations -- even though non-extrapolated energy values are further apart from the continuum as in the improved cases.\nThis effect is related to the continuum convergence pattern.\nWhile the $n_s = 1$ (and $n_s = \\infty$) energy values seem to converge against the continuum result from below (and respectively from above) for all excited states, the improved derivative eigenvalues change their convergence pattern.\nThe slope of the extrapolation function changes it sign from $E_2 \\to E_4$ for $n_s = 2$ and from $E_6 \\to E_8$ for $n_s = 4$.\nThis suggests that the importance of fit model coefficients $e_{i,n}^{(n_s)}$ changes and thus makes it more difficult to perform the continuum limit.\n\nIn the next step, we use the continuum-extrapolated spectrum to convert it to phase shifts using the spherical zeta function.\nWe present the phase shifts in \\figref{continuum-extrapolation-ere}.\nIndependent of discretization scheme, we observe that the continuum-extrapolated results agree with the constant input phase shifts.\nBecause the zeta function is relatively steep, uncertainties in the continuum limit get drastically enhanced when converting to phase shifts (on average more than an order of magnitude).\nWe observe that for $x > 5$ all discretizations besides the exact-$p^2$ discretization come with significant uncertainties.\n\nWe emphasize that these findings are not unique to the unitary case, we obtain similar results for a non-zero scattering length.\nWe present data for an example non-unitarity scenario with $a_{30} = - 5$~fm in our repository \\cite{luescher-nd_201}.\n\n\\subsection{Using \\Luscher's formula before continuum extrapolation}\nNext we want to discuss what effects finite discretization artifacts have when applying L\\\"uscher's formalism to a spectrum for finite lattice spacings.\nWe insert the energy levels presented in \\figref{continuum-extrapolation-spectrum} before taking the continuum limit and present results in figure \\figref{unimproved spherical}.\n\n\\begin{figure}[H]\n\t\\centering\n \\scalebox{0.8}{\\input{figure/ere-contact-fitted_a-inv_+0.0_zeta_spherical_projector_a1g_n-eigs_200.pgf}}\n \\caption{\n \\label{fig:unimproved spherical}\n We present energy eigenvalues presented in \\figref{continuum-extrapolation-spectrum} directly inserted $S^{\\spherical}_3$--without a continuum limit.\n In the top row we show results for $L=1.0$~fm, in bottom we show $L=2.0$~fm, while in different columns we show different discretization schemes.\n Even though results for $n_s = 2$ seem to be close to the continuum limit result, they start to drastically oscillate for higher energies.\n While more improved discretization schemes seem to oscillate less, they do not lay on top of the continuum result where the difference is related to the lattice spacing.\n }\n\\end{figure}\n\n\\begin{figure}[htb]\n \\scalebox{1.0}{\\input{figure/ere-continuum-extrapolation.pgf}}\n \\caption{\n \\label{fig:iv-continuum}\n Continuum limit of different phase shift points computed by inserting finite lattice spacing eigenstates in $S^{\\spherical}_3$ (see \\figref{unimproved spherical}).\n Each column represents a different kinetic operator and each row tracks a different eigenvalue of the discrete finite-volume hamiltonian.\n Note that both axis have a log scale and thus on these scales, a linear trend for the phase shifts suggests that they extrapolate to zero.\n }\n\\end{figure}\n\nWe note that the phase shifts for $x > 10$ start to oscillate wildly.\nThis is the case because energy values are close to the poles of the zeta function (close to the frame boundaries in \\figref{continuum-extrapolation-spectrum}). With an imperfect kinetic operator, the lattice artifacts in the energy can push energy levels past a pole in the continuum zeta. This leads to multiple interacting energy levels on a single segment of the zeta function.\n\nFurthermore it seems like the small $x$ results for $n_s = 2$ seem to be closer to the expected flat result than other discretization schemes.\nThis behavior can be explained by \\figref{continuum-extrapolation-spectrum}.\nWhile other discretization schemes for $x < 8$ monotonically converge against the continuum limit, $n_s = 2$ data points converge non-monotonically and are therefore closer to the continuum by accident.\nIn this sense it is possible to select a discretization scheme which in principle converges slower against the continuum, but has an accidental good agreement with the continuum even though it is discrete.\n\nFor small energies, better discretization schemes or small lattice spacings, we observe that the phase shifts do not oscillate and monotonically increase in $x$ with no or small curvature.\nThis non-flat $x$-dependence seems to depend less on the employed discretization scheme but certainly on the value of the lattice spacing.\nThis suggests that artifacts of the imperfect kinetic operator are negligible compared to cutoff effects of the lattice spacing itself.\nThe non-zero lattice spacing induces effective-range-like effects.\nAs we will show in the next section, this effect arises from using the continuum $S^{\\spherical}$ rather than the lattice-aware $S^{\\dispersion}$.\n\nWe visualize the continuum limit of phase shift points in \\figref{iv-continuum}.\nSimilar to the case where we first extrapolated the spectrum to the continuum and computed phase shifts afterwards, the best discretization allows to also extrapolate higher excited states to zero--visible by the linear log-log dependence of the phase shifts on epsilon.\nWe note that similar to the case where we first extrapolated the spectrum to the continuum, the extrapolation of the phase shifts seems to work best for the same discretization schemes in the same energy range.\nFor example, while we find a linear log-log scaling region in \\figref{iv-continuum} for $n_s = 2$ and $x<6$, uncertainties of the $n_s =2$ extrapolation also start to increase in \\figref{continuum-extrapolation-spectrum} after $x>6$.\nHowever the $n_s = 4$ implementation seems to be stable longer in \\figref{iv-continuum} which is related to the $x>9$ state having a relatively larger continuum extrapolation uncertainty while also being close to the continuum value.\n", "meta": {"hexsha": "6b4537883830055f7c575b8cbc8b4837dddaa7ae", "size": 12315, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/luescher-nd/section/three-dimensions/regular-luescher.tex", "max_stars_repo_name": "ckoerber/luescher-nd", "max_stars_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-04-12T22:19:38.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-26T14:06:49.000Z", "max_issues_repo_path": "paper/luescher-nd/section/three-dimensions/regular-luescher.tex", "max_issues_repo_name": "ckoerber/luescher-nd", "max_issues_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2019-12-16T19:49:19.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-02T00:50:31.000Z", "max_forks_repo_path": "paper/luescher-nd/section/three-dimensions/regular-luescher.tex", "max_forks_repo_name": "ckoerber/luescher-nd", "max_forks_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 94.7307692308, "max_line_length": 383, "alphanum_fraction": 0.7751522533, "num_tokens": 2937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382200964035, "lm_q2_score": 0.7401743563075447, "lm_q1q2_score": 0.6376884974942032}} {"text": "\\columnbreak\n\\section{Model Validation}\n\n% ===\n\\subsection{Stability-based Validation}\n\n\\emph{Stability:}\\enspace\nSolutions on two data sets drawn from the same source should be similar.\n\n%\\emph{(In-)Stability:} Expected (dis-)agreement of solutions from two data sets from the same source.\n\n% ===\n\\subsection{Information-theoretic Validation}\n\n\\subsubsection{Shannon's Channel Coding Thm.}\n\n\\begin{itemize}\n \\item \\textbf{Channel:}\\enspace\n $(\\mathcal S, \\{p(\\cdot\\mid s)\\}_{s\\in\\mathcal S} )$,\\enspace\n $\\mathcal S$: alphabet\n \\begin{itemize}\n \\item $\\epsilon$-noisy binary channel:\n $p(\\hat s\\mid s) = \\big\\{ \\substack{\n 1-\\epsilon \\:\\:\\text{if}\\:\\: \\hat s=s \\\\\n \\epsilon \\phantom{1-} \\:\\:\\text{if}\\:\\: \\hat s\\neq s}$\n \\end{itemize}\n \\vspace{-3pt}\n \\item \\textbf{Capacity:}\\enspace\n \\highlight{$\\mathrm{cap} = \\max_p I(S;\\hat S)$}\n $\\rightsquigarrow p_S(s)$\n \\item \\textbf{$\\bm{(M,n)}$-code:}\\enspace\n is a pair $(Enc,Dec)$\\hfill\n $\\color{gray} \\leftarrow\\text{scr. p.87}$\\\\\\quad\n where $M$: \\#messages, $n$: code-length\n \\begin{itemize}\n \\item \\textbf{Rate:}\\enspace\n $r = \\frac{\\log_2 M}{n}$\n $\\Leftrightarrow$ $M=\\floor{2^{nr}}$\n \\item \\textbf{Commu. err.:}\\enspace\n $p\\ped{err} \\coloneqq \\max\\limits_{i\\leq M}\n \\P{ Dec(\\widehat{ Enc(i) }) \\neq i }$\n \\end{itemize}\n\\end{itemize}\n\nGoal / \\textbf{Best code:}\\enspace\n\\highlight*{$\\lim\\limits_{n\\to\\infty} \\frac{\\log M}{n}$\\enspace s.t.\\enspace $\\lim\\limits_{n\\to\\infty} p\\ped{err} \\to 0$}\n\n\\emph{Asymptotic equiparition property (AEP):}\n\\vspace{-2pt}\n\\begin{itemize}\n \\item $A_\\epsilon^{(n)}$:\n Typical set of sequences $(s_1,\\ldots,s_n) \\in \\mathcal S^n$\\\\\\qquad\n $\\abs*{-\\frac1n \\log p_{S^n} (\\bm s^n) - H[S]} < \\epsilon$\\hfill\n $\\color{gray} \\leftarrow \\text{scr. p.89}$\n \\item $\\P*{(\\bm S^n, \\bm{\\hat S}^n) \\in A_\\epsilon^{(n)}} \\overset{n\\to\\infty}{\\to} 1$\\hfill\n $\\color{gray} \\leftarrow \\text{scr. p.90}$\n \\item \\highlight{$p\\ped{err} \\leq 2^{-n(\\mathrm{cap} - 3\\epsilon -r)}$}\n $\\overset{n\\to\\infty}{\\to} 0$\\enspace if\\enspace \\highlight{$r<\\mathrm{cap}$}\n\\end{itemize}\n\n\n\\subsubsection{Algorithm Validation}\n\n\\textbf{Assumptions:}\n\\begin{itemize}\n \\item Exponential solution space, i.e. $\\log\\abs{\\mathcal C} = \\mathcal O(n)$\n \\item $\\mathcal A$'s output is probabilistic, i.e. $p(\\cdot \\mid X')$\n\\end{itemize}\n\n\\emph{Ideal variant:}\n\n\\textbf{Messages:}\\enspace\n$\\mathcal M = \\brace{ X'_1, \\ldots, X'_m }$\n\\vspace{-2pt}\n\n\\textbf{Code:}\\enspace\n$X'_i \\xrightarrow{Enc_{\\mathcal A}} p(\\cdot\\mid X'_i)\n\\xrightarrow{\\mathcal{C_A}} p(\\cdot\\mid X''_i)\n\\xrightarrow{Dec_{\\mathcal A}} \\hat X$\n\\iffalse\n \\begin{itemize}\n \\item $Enc_{\\mathcal A}$:\\enspace\n encodes $X'_i \\in \\mathcal M$ as $p(\\cdot\\mid X'_i)$.\n \\item $Dec_{\\mathcal A}$:\\enspace\n selects $\\hat X = \\arg\\max\\limits_X \\kappa(X''_i, X)$\\\\\n whereby $\\kappa(X'', X') \\coloneqq \\sum_c p(c\\mid X'') p(c\\mid X')$\n \\end{itemize}\n\\fi\n\n\\emph{Empirical variant:}\n\n\\textbf{Messages:}\\enspace\n$\\mathcal M = \\brace{ \\tau_1, \\ldots, \\tau_m }$\ndrawn u.a.r. from $\\mathbb T$\n\\begin{itemize}\n \\item Require\\enspace $\\sum_\\tau p(c\\mid \\tau\\circ X') \\approx \\frac{\\abs{\\mathbb T}}{\\abs{\\mathcal C}} \\pm \\rho$\\hfill\n $\\color{gray} \\leftarrow \\text{scr. p.95}$\n\\end{itemize}\n\\vspace{-2pt}\n\n\\textbf{Code:}\\enspace\n$\\tau_i \\xrightarrow{Enc} p(\\cdot\\mid \\tau_i \\circ X')\n\\xrightarrow{\\mathcal{C_A}} p(\\cdot\\mid \\tau_i \\circ X'')\n\\xrightarrow{Dec} \\hat\\tau$\n\\begin{itemize}\n \\item $Enc_{\\mathcal A}$:\\enspace\n encodes $\\tau_i \\in \\mathcal M$ as $p(\\cdot\\mid \\tau_i \\circ X')$\n \\item $Dec_{\\mathcal A}$:\\enspace\n selects $\\hat\\tau = \\arg\\max\\limits_{\\color{red}\\tau} \\kappa(\\tau_i \\circ X'', {\\color{red}\\tau} \\circ X')$\\\\\n whereby \\highlight{$\\kappa(X'', X') \\coloneqq \\sum_c p(c\\mid X'') p(c\\mid X')$}\n\\end{itemize}\n\n\\emph{Asymptotic Equipartition Property (AEP):}\n\n\\textit{AEP fulfilled}\\enspace if\\enspace \\highlight*{$\\log \\kappa(X',X'') \\overset{n\\to\\infty}{\\to} \\mathcal E$}\\\\\n\\enspace whereby $\\mathcal E \\coloneqq \\E[X',X'']{\\log\\kappa(X',X'')}$\n\\begin{itemize}\n \\item $A^{(n)}_\\epsilon$: set of $(\\epsilon,n)$-typical pairs $X', X''$\\\\\\qquad\n $\\abs*{ \\log\\kappa(X',X'') - \\mathcal E } < \\epsilon$\n \\item \\highlight{$p\\ped{err} \\leq P_{(n)} \\color{gray} \\text{ c.f. scr. (6.19)}$}\n $\\overset{n\\to\\infty}{\\to} 0$ \\enspace if\\enspace \\highlight{$\\frac{\\log m}{\\log\\abs{\\mathcal C}} < I$}\\\\\n where $I \\coloneqq \\frac{1}{\\log\\abs{\\mathcal C}} \\E[X',X'']{\\log(\\abs{\\mathcal C} \\kappa(X',X''))}$\n\\end{itemize}\n\n% ===\n\\subsection{Applications of PA}\n\n\\emph{PA:}\\enskip\n\\textit{quantifies the amount of information that algorithms extract from phenomena.}\n$\\to$ quantified by \\textbf{capacity} (max. \\# distinguishable messages that can be communicated)\n\n\\emph{Temperature:}\\enspace\n$T^\\ast = \\arg\\max_T \\kappa(X',X'')$\n\n\\emph{Cost functions:}\\enspace\nGiven $R_1(\\cdot,\\cdot),\\ldots,R_s(\\cdot,\\cdot)$\\\\\n$\\max\\limits_{\\ell\\leq s} \\kappa_\\ell(X',X'') = \\max\\limits_{\\ell\\leq s} \\frac{1}{Z_{X'} Z_{X''}} \\sum\\limits_c \\eu^{-\\frac1T R_\\ell(c,X')} \\eu^{-\\frac1T R_\\ell(c,X'')}$\n\n\\emph{Algorithms:}\\enspace\n%\\textit{here for MST (min. span. tree)}\nMany MST (min. spanning tree) algo's are \\textbf{contractive} ($\\to$ sequence of candidate sol's).\n\\\\\n\\textbf{Approximation Set Coding (ASC):}\\\\\n\\quad $p\\ap{ASC}(c\\mid X') = \\begin{cases}\n 1 / \\abs*{G_\\gamma(X')} & \\text{if}\\enspace c\\in G_\\gamma(X') \\\\[-2pt]\n 0 & \\text{otw.}\n\\end{cases}$\\\\\n\\quad $G_\\gamma(X') \\coloneqq \\brace*{ c\\in\\mathcal C : R(c,X') - \\min\\limits_{c\\in\\mathcal C} R(c,X') \\leq \\gamma }$\n\\begin{enumerate}\n \\item Run $\\mathcal A$ to compute $G_t^{\\mathcal A}(X')$ and $G_t^{\\mathcal A}(X'')$,\\enspace for all $t$\n \\vspace{2pt}\n \\item $t^\\ast = \\arg\\max\\limits_t \\kappa(X',X'')\n = \\arg\\max\\limits_t \\frac{\\abs*{G_t^{\\mathcal A}(X') \\:\\cap\\: G_t^{\\mathcal A}(X'')}}{\\abs*{G_t^{\\mathcal A}(X')} \\:\\cdot\\: \\abs*{G_t^{\\mathcal A}(X'')}}$\n \\item $c^\\ast \\xleftarrow{\\$\\:\\text{sample}} \\mathrm{Unif}\\paren*{G_{t^\\ast}^{\\mathcal A}(X') \\cap G_{t^\\ast}^{\\mathcal A}(X'')}$\n\\end{enumerate}\n\n\n% ===\n", "meta": {"hexsha": "87d6f242018994f29de592cfdd6559619f88cc40", "size": 6379, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/SLT21/sections/09_model_validation.tex", "max_stars_repo_name": "tstreule/eth-cheat-sheets", "max_stars_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-26T23:11:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T23:11:57.000Z", "max_issues_repo_path": "src/SLT21/sections/09_model_validation.tex", "max_issues_repo_name": "tstreule/eth-cheat-sheets", "max_issues_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/SLT21/sections/09_model_validation.tex", "max_forks_repo_name": "tstreule/eth-cheat-sheets", "max_forks_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.1194968553, "max_line_length": 169, "alphanum_fraction": 0.5983696504, "num_tokens": 2425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615381952105442, "lm_q2_score": 0.7401743735019595, "lm_q1q2_score": 0.6376884938879734}} {"text": "\\chapter{Milky Way diffuse background study: denoising and inpainting}\n\\label{ch_inpainting}\n\n% \\markright{Milky Way diffuse background study: denoising and inpainting}\n\nIn order to extract a diffuse emission, we want to remove the point sources from the data. \nAs our HSD algorithm is very close to the MCA algorithm~\\citep{starck:sta04}, an idea is to mask the most intense sources and to modify our algorithm in order to interpolate through the gaps exactly as in the MCA-Inpainting algorithm~\\citep{inpainting:abrial06}. \nThis modified algorithm can be called MS-VSTS-Inpainting algorithm.\n\nThe problem can be reformulated as a convex constrained minimization problem:\n\n\\begin{equation}\n\\label{inp_eq34}\n\\begin{split}\n\\text{Arg} \\min_{\\mathbf{X}} \\| \\mathbf{ \\Phi}^{T}\\mathbf{X}\\|_1,\n\\text{s.t.} \\\\ \\: \\left\\{\\begin{array}{c}\\mathbf{X} \\geqslant 0 , \\\\\\forall (j,k)\\in \\mathcal{M}, (\\mathbf{ \\Phi}^{T}\\Pi \\mathbf{X})_j[k]=(\\mathbf{ \\Phi}^{T} \\mathbf{Y})_j[k] , \\end{array}\\right. \n\\end{split}\n\\end{equation}\nwhere $\\Pi$ is a binary mask ($1$ on valid data and $0$ on invalid data).\n\nThe iterative scheme can be adapted to cope with a binary mask, which gives:\n\\begin{eqnarray}\n\\tilde{\\mathbf{X}} = P_{+}[\\mathbf{ X}^{(n)} + \\mathbf{ \\Phi} P_{\\mathcal{M}} \\mathbf{ \\Phi}^{T} \\Pi (\\mathbf{ Y} - \\mathbf{ X}^{(n)})] , \\\\\n\\mathbf{X}^{(n+1)} = \\mathbf{ \\Phi} \\text{ST}_{\\lambda_n}[\\mathbf{ \\Phi}\\tilde{\\mathbf{X}}] .\n\\end{eqnarray}\n\n\nThe thresholding strategy has to be adapted. Indeed, for the impainting task we need to have a very large initial threshold in order to have a very smooth image in the beginning and to refine the details progressively. We chose an exponentially decreasing threshold:\n\\begin{equation}\n\\label{eq42}\n\\lambda_{n} = \\lambda_{\\max} (2^{(\\frac{N_{\\max} - n}{N_{\\max} - 1})} -1),n=1,2,\\cdots,N_{\\max} ,\n\\end{equation}\nwhere $\\lambda_{\\max} = \\max (\\mathbf{\\Phi}^{T}\\mathbf{X})$.\n\n\\begin{algorithm}\n\\caption{MS-VST + IUWT Denoising + Inpainting}\n\\label{alg2}\n\n\\begin{algorithmic}[1]\n\\REQUIRE $\\quad$ data $a_0:=\\mathbf{Y}$, mask $\\Pi$, number of iterations $N_{\\max}$, threshold $\\kappa$.\\\\\n\\underline{\\emph{\\textbf{Detection}}} \\\\\n\\FOR{$j=1$ to $J$}\n\\STATE Compute $a_j$ and $d_j$ using (\\ref{eq27}).\n\\STATE Hard threshold $|d_j[k]|$ with threshold $\\kappa \\sigma_j$ and update $\\mathcal{M}$.\n\\ENDFOR \\\\\n\\underline{\\emph{\\textbf{Estimation}}} \\\\\n\\STATE Initialize $\\mathbf{X}^{(0)}=0$, $\\lambda_{0} = \\lambda_{\\max}$.\n\\FOR{$n=0$ to $N_{\\max}-1$}\n\\STATE $\\tilde{\\mathbf{X}}= P_{+}[\\mathbf{ X}^{(n)} + \\mathbf{ \\Phi} P_{\\mathcal{M}} \\mathbf{ \\Phi}^{T} \\Pi(\\mathbf{ Y} - \\mathbf{ X}^{(n)})]$.\n\\STATE $\\mathbf{X}^{(n+1)} = \\mathbf{ \\Phi}^\\text{ST}_{\\lambda_n}[\\mathbf{ \\Phi}^{T}\\tilde{\\mathbf{X}}]$.\n\\STATE $\\lambda_{n+1} = \\lambda_{\\max} (2^{(\\frac{N_{\\max} - (n+1)}{N_{\\max} - 1})} -1)$\n\\ENDFOR\n\\STATE Get the estimate $\\hat{\\mathbf{\\Lambda}} = \\mathbf{X}^{(N_{\\max})}$.\n\n\\end{algorithmic}\n\\end{algorithm}\n\n\n\n\\section*{Experiment}\n\nWe applied this method on simulated Fermi data where we masked the most luminous sources.\n\nThe results are on Figure~\\ref{impainting}. The MS-VST + IUWT + Inpainting method (Algorithm~\\ref{alg2}) interpolates the missing data very well. Indeed, the missing part can not be seen anymore in the inpainted map, which shows that the diffuse emission component has been correctly reconstructed.\n\n\n\\begin{figure}[htb]\n\\centering{\n\\includegraphics[width=5.5in]{13822fg22.pdf} \n\\includegraphics[width=5.5in]{13822fg23.pdf} \n}\n\\caption{MS-VSTS - Inpainting.\n\\emph{Top}: Fermi simulated map with Poisson noise and the most luminous sources masked.\n\\emph{Bottomt}: Fermi simulated map denoised and inpainted with wavelets (Algorithm~\\ref{alg2}).\nPictures are in logarithmic scale.\n}\n\\label{impainting}\n\\end{figure}\n", "meta": {"hexsha": "f6e48576b2225a6fa9fdae367eb802fe1ee68b54", "size": 3769, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_isap/msvst_inpainting.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_isap/msvst_inpainting.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_isap/msvst_inpainting.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.7088607595, "max_line_length": 299, "alphanum_fraction": 0.6914300876, "num_tokens": 1297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677506936878, "lm_q2_score": 0.7520125793176222, "lm_q1q2_score": 0.6376824153773226}} {"text": "\n\\documentclass[12pt]{amsart}\n\\usepackage{geometry} % see geometry.pdf on how to lay out the page. There's lots.\n\\usepackage{calculational}\n\\usepackage{unitb}\n\\usepackage{amsmath}\n\\geometry{a4paper} % or letter or a5paper or ... etc\n% \\geometry{landscape} % rotated page geometry\n\n% See the ``Article customise'' template for come common customisations\n\n\\title{A Sample Unit-B Toolkit proof}\n\\author{Simon Hudon}\n\\date{} % delete this line to display the current date\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{lemma}{Lemma}\n\n%%% BEGIN DOCUMENT\n\\begin{document}\n\n\\maketitle\n%\\tableofcontents\n\n%\\section{}\n\n%\\begin{theorem} \n%Any conjunctive function is monotonic\n%\\begin{assumption}\n%f.(x \\land y) ~ \\equiv~ f.x \\land f.y\n%\\end{assumption}\n%\\end{theorem}\n%\\begin{dummy}\n\\begin{calculation}\n\tf.x \\le f.y\n\\hint{=}{ $\\le$ to $=$ and $\\uparrow$ }\n\tf.x \\uparrow f.y \\,=\\, f.y\n\\hint{=}{ $f$ over $\\uparrow$ }\n\tf.(x \\uparrow y) \\,=\\, f.y\n\\hint{\\follows}{ Leibniz }\n\tx \\uparrow y \\,=\\, y\n\\hint{=}{ $\\le$ from $=$ and $\\uparrow$ }\n\tx \\le y\n\\end{calculation}\n\n\\begin{calculation}\n\tf.x\n\\hint{\\le}{ $x \\le x\\uparrow y$ }\n\tf.x \\uparrow f.y\n\\hint{=}{ $f$ over $\\uparrow$ }\n\tf.(x \\uparrow y)\n\\hint{=}{ $x \\le y$ cast as $x\\uparrow y = y$ }\n\tf.y\n\\end{calculation}\n\n%\\subsection{}\n\n\\end{document}", "meta": {"hexsha": "cb93f5f5801346b487b392616be4328578f11935", "size": 1279, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Tests/sample.tex", "max_stars_repo_name": "literate-unitb/literate-unitb", "max_stars_repo_head_hexsha": "0d843456dc103bb09babc5b12855435d2e10f534", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-07-27T11:05:56.000Z", "max_stars_repo_stars_event_max_datetime": "2018-01-20T14:53:33.000Z", "max_issues_repo_path": "Tests/sample.tex", "max_issues_repo_name": "unitb/literate-unitb", "max_issues_repo_head_hexsha": "0d843456dc103bb09babc5b12855435d2e10f534", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 32, "max_issues_repo_issues_event_min_datetime": "2017-06-25T03:53:02.000Z", "max_issues_repo_issues_event_max_datetime": "2017-06-25T04:28:38.000Z", "max_forks_repo_path": "Tests/sample.tex", "max_forks_repo_name": "literate-unitb/literate-unitb", "max_forks_repo_head_hexsha": "0d843456dc103bb09babc5b12855435d2e10f534", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 22.0517241379, "max_line_length": 82, "alphanum_fraction": 0.6653635653, "num_tokens": 442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677506936878, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6376824059516099}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{amsmath}\n\n\\begin{document}\n\n\\section{Geometry definition of \\texttt{pyFAI}}\n\nAll dimentions are in meter and all rotation are in radians.\n\n\\texttt{pyFAI} describes the position and orientation of the detector\nby six variables, collectively called the PONI, for point of normal\nincidence. In addition, a detector calibration is provided in the\nPONI-file to convert pixel coordinates into real-space coordinates. In\nthe simplest case this is a pixel size. \n\n\\texttt{pyFAI} uses a coordinate system where the first axis (1) is \nvertically up (or $y$), the second axis (2) is horizontal ($x$) towards the ring\ncenter (starboard), and the third axis (3) along the beam $z$. Note that\nin this order (1,2,3) is a right-handed coordinate system, which makes\n$xyz$ in the usual order a left-handed coordinate system!\n\nThe PONI parameters are: a distance $L$, the $y$ and $x$ coordinates\nof the point of normal incidence in meters, $\\mathrm{poni}_{1,2}$, and\n3 detector rotations, $\\theta_{1,2,3}$ about the $y$, $x$ and $z$\naxes. The inversion of the $x$ and $y$ axes is due to the arrangement\nof the detector data, with $x$-rows being the ``slow'' axis and\n$y$-columns the ``fast'' axis. Extra care has to be taken with the\nsigns of the rotations when converting form this coordinate system to\nanother.\n\nThe detector rotations are taken about the origin of the coordinate\nsystem (sample position). They are represented by the rotation\nmatrices:\n\n\\begin{align}\n \\mathrm{R}_1(\\theta_1)\n & =\n \\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & \\cos(\\theta_1) & \\sin(\\theta_1) \\\\\n 0 & -\\sin(\\theta_1) & \\cos(\\theta_1)\n \\end{bmatrix}\n \\\\\n \\mathrm{R}_2(\\theta_2)\n & =\n \\begin{bmatrix}\n \\cos(\\theta_2) & 0 & -\\sin(\\theta_2) \\\\\n 0 & 1 & 0 \\\\\n \\sin(\\theta_2) & 0 & \\cos(\\theta_2)\n \\end{bmatrix}\n \\\\\n \\mathrm{R}_3(\\theta_3)\n & =\n \\begin{bmatrix}\n \\cos(\\theta_3) & -\\sin(\\theta_3) & 0\\\\\n \\sin(\\theta_3) & \\cos(\\theta_3) & 0\\\\\n 0 & 0 & 1 \n \\end{bmatrix}\n \\\\\n R(\\theta_1, \\theta_2, \\theta_3)\n & =\n R_3(\\theta_3) \\cdot R_2(\\theta_2) \\cdot R_1(\\theta_1)\n\\end{align}\n\nLet $P$ be a point on the detector at pixel coordinates $(d_1, d_2)$\nin units $\\mathrm{px}$. In the most simple case (in the absence of\ndistortions) these are converted into coordinates in units of meters\nby multiplication with a scaling factor ``pixel size''\n($\\mathrm{pxsize}_{1,2}$m in units of $\\mathrm{m}/\\mathrm{px}$) and\nsubsequent subtraction of an offset ($\\mathrm{poni}_{1,2}$, in units of\nmeters and yielding the coordinates $(p_1, p_2)$ in meters.\n\n\\begin{align}\n p_1 & = \\mathrm{pxsize}_1 d_1 - \\mathrm{poni}_1 \\\\\n p_2 & = \\mathrm{pxsize}_2 d_2 - \\mathrm{poni}_2\n\\end{align}\n\nLet $L$ be the distance from the\norigin/sample/diffractometer center of rotation. In the absence of any\ndetector rotations, $L$ is taken along $z$, $p_1$ along the $x$-axis\n(horizontal) and $p_2$ along the $y$ axis (vertical). Then, in the\n``detector coordinate system''\n\n\\begin{align}\n P & = \\begin{bmatrix} p_1 \\\\ p_2 \\\\ L \\end{bmatrix}\n\\end{align}\n\nThe coordinates in the laboratory system are obtained by\nmultiplication with the rotation matrix $R$\n\n\\begin{align}\n \\begin{bmatrix} t_1 \\\\ t_2 \\\\ t_3 \\end{bmatrix}\n = &\n R_{\\mathtt{pyFAI}}(\\theta_1, \\theta_2, \\theta_3) \\cdot P\n \\\\\n = &\n R_3(\\theta_3) \\cdot R_2(\\theta_2) \\cdot R_1(\\theta_1) \\cdot\n \\begin{bmatrix}\n \\mathrm{pxsize}_1 d_1 - \\mathrm{poni}_1 \\\\\n \\mathrm{pxsize}_2 d_2 - \\mathrm{poni}_2 \\\\\n L\n \\end{bmatrix}\n \\\\\n = &\n R_3(\\theta_3) \\cdot R_2(\\theta_2) \\cdot R_1(\\theta_1)\n \\nonumber \\\\ &\n \\cdot\n \\left(\n \\begin{bmatrix}\n \\mathrm{pxsize}_1 & 0 & 0 \\\\\n 0 & \\mathrm{pxsize}_2 & 0 \\\\\n 0 & 0 & 1\n \\end{bmatrix}\n \\cdot\n \\begin{bmatrix}\n d_1 \\\\\n d_2 \\\\\n 0\n \\end{bmatrix}\n +\n \\begin{bmatrix}\n - \\mathrm{poni}_1 \\\\\n - \\mathrm{poni}_2 \\\\\n L\n \\end{bmatrix}\n \\right)\n \\label{eq-tpyFAI}\n\\end{align}\n\nThe inversion is straight-forward:\n\n\\begin{align}\n R_1(-\\theta_1)\\cdot R_2(-\\theta_2) \\cdot R_3(-\\theta_3) \\cdot\n \\begin{bmatrix} t_1 \\\\ t_2 \\\\ t_3\n \\end{bmatrix}\n & =\n \\begin{bmatrix} p_1 \\\\ p_2 \\\\ L \\end{bmatrix}\n \\label{eq-find-alpha}\n \\\\\n \\begin{bmatrix}\n t_1 \\\\ t_2 \\\\ t_3\n \\end{bmatrix}\n & =\n \\begin{bmatrix}\n A_1 \\\\ A_2 \\\\ A_3 \n \\end{bmatrix}\n + \\alpha\n \\begin{bmatrix}\n k_1 \\\\ k_2 \\\\ k_3\n \\end{bmatrix},\n\\end{align}\nwhere $A_{1,2,3}$ are the coordinates, in the laboratory system, of\nthe scattering center (grain, sub-grain or voxel), and $k_{1,2,3}$ is\nthe direction of the scattered beam. The third line ($\\ldots = L$) of\neq.~\\ref{eq-find-alpha} is then used to determine the free parameter\n$\\alpha$, which in turn is used in the first and second lines to find\n$p_{1,2}$ and thus $d_{1,2}$.\n\nAs the most trivial example we consider the case of no rotations,\n$\\theta_1 = \\theta_2 = \\theta_3 = 0$. Then\n\n\\begin{align}\n A_3 + \\alpha k_3 & = L \\\\\n \\alpha & = \\frac{L-A_3}{k_3} \\\\\n p_1 & = A_1 + (L-A_3) \\frac{k_1}{k_3} \\\\\n p_2 & = A_2 + (L-A_3) \\frac{k_2}{k_3}.\n\\end{align}\n\nWe see also that when all rotations are zero, $(\\mathrm{poni}_1,\n\\mathrm{poni_2})$ are the real space coordinates of the direct beam\n($A_{1,2,3}=k_{1,2}=0$) .\n\n\\section{Geometry definition of \\texttt{ImageD11}}\n\nFor maximum convenience, \\texttt{ImageD11} defines almost everything\ndifferently than \\texttt{pyFAI}.\n\nCoordinate system: \\texttt{ImageD11} uses the ID06 coordinate system\nwith $x$ along the beam, $y$ to port (away from the ring center), and\n$z$ up.\n\nUnits: As the system is somewhat scale-invariant, \\texttt{ImageD11}\nallows a free choice of the unit of length, which we will call $X$\nhere. The same unit has to be used for all translations, and for the\npixel size of the detector. The default used in the code appears to be\n$X = 1\\,\\mathrm{\\mu m}$, but it might as well be Planck lengths,\nmillimeters, inches, meters, tlalcuahuitl, furlongs, nautical miles,\nlight years, kparsec, or whatever else floats your boat. The only\nrequirement is that you can actually measure and express the detector\npixel size and COR-to-detector distance in your units of choice.\n\nParameters: \\texttt{ImageD11} defines the detector geometry via the\nfollowing parameters:\n\n\\begin{itemize}\n\\item Beam center $y_{\\mathrm{center}}$ and $z_{\\mathrm{center}}$:\n Position of the direct beam on the detector, in $\\mathrm{px}$.\n\\item Pixel size $y_{\\mathrm{size}}$ and $z_{\\mathrm{size}}$ in\n ${X}/{\\mathrm{px}}$.\n\\item Detector tilts $t_x$, $t_y$, and $t_z$, in $\\mathrm{rad}$. The\n center of rotation is the point where the direct beam intersects the\n detector.\n\\item Distance $D$, in units $X$, between the origin to the point\n where the direct beam intersects the detector.\n\\item Detector flip matrix $O = \\begin{bmatrix} o_{11} & o_{12}\n \\\\ o_{21} & o_{22} \\end{bmatrix}$. This matrix takes care of\n correcting typical problems with the way pixel coordinates are\n arranged on the detector. If, e.g., the detector is rotated by\n $90^{\\circ}$, then $O=\\begin{bmatrix} 0 & 1 \\\\ -1 &\n 0\\end{bmatrix}$. If left and right (or up and down) are inverted on the\n detector, then $o_{22} = -1$ ($o_{11}=-1$). \n\\end{itemize}\n\nIt appears that these conventions where defined under the assumption\nthat the detector is more or less centered in the direct beam, and\nthat the detector tilts are small.\n\nThe implementation in the code \\texttt{transform.py} is using the\nfollowing equations:\n\n\\begin{align}\n R_{\\mathtt{ImageD11}}(\\theta_x, \\theta_y, \\theta_z)\n & =\n R_1(\\theta_x) \\cdot R_2(\\theta_y) \\cdot R_3(\\theta_z)\n \\\\\n \\begin{bmatrix}\n p_z \\\\ p_y\n \\end{bmatrix}\n & =\n \\begin{bmatrix}\n o_{11} & o_{12}\n \\\\ o_{21} & o_{22}\n \\end{bmatrix}\n \\cdot\n \\begin{bmatrix}\n (d_z - z_{\\mathrm{center}}) z_{\\mathrm{size}} \\\\\n (d_y - y_{\\mathrm{center}}) y_{\\mathrm{size}}\n \\end{bmatrix}\n \\label{eq-p}\n \\\\\n \\begin{bmatrix}\n t_x \\\\ t_y \\\\ t_z\n \\end{bmatrix}\n & =\n R_{\\mathtt{ImageD11}}(\\theta_x, \\theta_y, \\theta_z)\n \\cdot\n \\begin{bmatrix}\n 0 \\\\ p_y \\\\ p_z\n \\end{bmatrix}\n +\n \\begin{bmatrix}\n D \\\\ 0 \\\\ 0\n \\end{bmatrix}\n \\label{eq-t}\n\\end{align}\nNote that the order of $y$ and $z$ is not the same in eqs.~\\ref{eq-p} and \\ref{eq-t}.\n\nThe same could be achieved more easily by\n\n\\begin{align}\n \\begin{bmatrix}\n t_x \\\\ t_y \\\\ t_z\n \\end{bmatrix}\n & =\n R_{\\mathtt{ImageD11}}(\\theta_x, \\theta_y, \\theta_z)\n \\cdot\n \\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & y_{\\mathrm{size}} o_{22} & z_{\\mathrm{size}} o_{21} \\\\\n 0 & y_{\\mathrm{size}} o_{12} & z_{\\mathrm{size}} o_{11}\n \\end{bmatrix}\n \\cdot\n \\begin{bmatrix}\n D \\\\ p_y - y_{\\mathrm{center}} \\\\ p_z - z_{\\mathrm{center}}\n \\end{bmatrix}.\n \\label{eq-tImageD11}\n\\end{align}\n\n\\section{Conversion}\n\nThe transformation between the different coordinate systems is\nachieved by\n\\begin{align}\n t_{\\mathtt{ImageD11}}\n =\n \\begin{bmatrix}\n 0 & 0 & 1 \\\\ 0 & -1 & 0 \\\\ 1 & 0 & 0\n \\end{bmatrix}\n t_{\\mathtt{pyFAI}},\n \\label{eq-coordconv}\n\\end{align}\nwhere $t_{\\mathtt{ImageD11}}$ is given by eq.~\\ref{eq-tImageD11}, and\n$t_{\\mathtt{pyFAI}}$ is given by eq.~\\ref{eq-tpyFAI}. Substituting\nthese equations into eq.~\\ref{eq-coordconv}, one can them attempt to\nconvert \\texttt{pyFAI} parameters into \\texttt{ImageD11} parameters\nand vice versa. As \\texttt{pyFAI} does not allow for detector\nflipping, $o_{11}=o_{22}=1$ and $o_{12}=o_{21}=0$. Furthermore,\n$\\mathrm{pxsize}_1 = z_{\\mathrm{size}}$ and $\\mathrm{pxsize}_2 =\ny_{\\mathrm{size}}$. Because we assume that both equations describe the\nsame pixel, twice in detector coordinates and twice in real-space\ncoordinates, $p_y = p_2$ and $p_z = p_1$.\n\nCHECK IF $p_y = p_2$ or $p_y = - p_2$/$o_{22}=-1$ DUE TO\n\\texttt{pyFAI} LEFT HANDED COORDINATE SYSTEM!!!\n\nFor the translations, the equivalence is tractable by setting $p_1 =\np_2 = p_y = p_z = 0$. One obtains 3 equations that are coupled via the\nrotation matrices --- as the center of rotations are different, there\nshould be a noticeable influence of the distance on the offsets. In\ngeneral, $D \\geq L$ with $D=L$ only for $\\theta_1=\\theta_2=0$\n($\\theta_y = \\theta_z = 0$. Because the rotations are applied in\ndifferent order one can not assume, \\emph{a priori}, that the angles\nare the same! (If at least two tilt angles are small, this may be a\ngood approximation)\n\n\\end{document}\n", "meta": {"hexsha": "9ff0417cb0038f6f27a532e17491c961a3936a6f", "size": 10283, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/source/geometry.tex", "max_stars_repo_name": "yugangzhang/pyFAI", "max_stars_repo_head_hexsha": "e0453b279dac1f165f637e2a2ed1d4ddf57d31ba", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2016-07-16T19:43:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-12T16:53:47.000Z", "max_issues_repo_path": "doc/source/geometry.tex", "max_issues_repo_name": "yugangzhang/pyFAI", "max_issues_repo_head_hexsha": "e0453b279dac1f165f637e2a2ed1d4ddf57d31ba", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1125, "max_issues_repo_issues_event_min_datetime": "2016-06-09T07:47:57.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T20:34:00.000Z", "max_forks_repo_path": "doc/source/geometry.tex", "max_forks_repo_name": "yugangzhang/pyFAI", "max_forks_repo_head_hexsha": "e0453b279dac1f165f637e2a2ed1d4ddf57d31ba", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 52, "max_forks_repo_forks_event_min_datetime": "2016-06-09T07:30:46.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-14T08:25:11.000Z", "avg_line_length": 32.3364779874, "max_line_length": 85, "alphanum_fraction": 0.6752893125, "num_tokens": 3583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9161096181702031, "lm_q2_score": 0.6959583124210896, "lm_q1q2_score": 0.6375741038544633}} {"text": "\\documentclass{article}\n\\usepackage{amsmath,amssymb}\n\\usepackage{hyperref}\n\\usepackage{graphicx}\n\\usepackage{listings}\n\\title{\\bf{Turbine Formulation}}\n\\author{Nicholas Malaya, Roy Stogner, Robert D. Moser \\\\ Institute for\nComputational Engineering and Sciences \\\\ University of Texas at Austin}\n\\date{} \n\n\\begin{document}\n\\maketitle\n\\newpage\n\\section{Turbine Definitions}\n\nThe normal in the blade's velocity direction is, \n\\begin{equation}\nn_B = \\frac{u_B}{||u_B||}. \n\\end{equation}\nWhere $u_B$ is the blade velocity and is specified.\nThe normal in the fan vertical direction is also set, and is typically, \n\\begin{equation}\nn_v = \\left(0,0,1\\right)\n\\end{equation}\ne.g. pointing ``up''. Then the normal in the radial direction must be, \n\\begin{equation}\nn_r = n_B \\times n_v. \n\\end{equation}\n% The fan-wing-plane bit means that we're only looking at the projection\n% of velocity into the plane that's defined by the base velocity and\n% vertical direction. \n\n% (01:03:44 PM) Roy Stogner: The \"local relative velocity\" means that\n% we're taking the velocity not in the reference frame of the domain, but\n% in the reference frame of the wing. So if the base velocity is U_B and\n% the air velocity is U, then the local relative velocity is U - U_B. \n% (01:04:25 PM) Roy Stogner: Note that we simplify that equation a tiny\n% bit by using the fact that U_B and N_R are perpendicular. \n\nThen, the fan-wing-plane component (e.g. the plane perpendicular to the \nradius) of local relative velocity is\n\\begin{equation}\nu_p = u - (u\\cdot n_r)\\cdot n_r - u_B. \n\\end{equation}\nWe can now define the lift and drag normals, where the direction\nopposing drag is, \n\\begin{equation}\nn_{\\text{drag}} = \\frac{u_p}{||u_p||} \n\\end{equation}\nand the direction opposing lift orthogonal to the drag and the radial direction, \n\\begin{equation}\nn_{\\text{lift}}= n_{\\text{drag}} \\times n_r. \n\\end{equation}\nNow the ``forward velocity'' in the reference frame of the turbine is, \n\\begin{equation}\nu_{\\text{fwd}}= -u_p \\cdot n_B\n\\end{equation}\nand the ``upward'' velocity in this frame is, \n\\begin{equation}\nu_{\\text{up}} = u_p \\cdot n_v. \n\\end{equation}\nFinally, we can specify the angle with respect to the fan velocity\ndirection as, \n\\begin{equation}\n \\theta_f = \\text{atan2}\\left(\\frac{u_{\\text{up}}}{u_{\\text{fwd}}}\\right)\n %\\theta_f = \\text{tan}^{-1}\\left(\\frac{u_{\\text{up}}}{u_{\\text{fwd}}}\\right)\n\\end{equation}\nwhile the angle with respect to the chord is this with the addition of\nthe angle of attack of the blade, \n\\begin{equation}\n \\theta = \\theta_f + \\alpha(r).\n\\end{equation}\nNow, we only need the drag polars in order to fully specify the force\non the blades.\n\n\\newpage\n\\section{Drag Polars}\n\nWe have been provided lift and drag as a function of angle of attack for\ntwo cases, the flat plate and semi-circle turbine blades. \nFor the flat plate, the drag polars are specified as,\n\n\\begin{lstlisting}\n theta := ((t+pi/2)%pi)-pi/2; \n lift = 'if(abs(theta) c_2)\\) and therefore,\n\n\\begin{equation}\n\\pi = \\Phi\\left(\\frac{\\mu_1-\\mu_2}{\\sqrt{\\sigma^2_1+\\sigma^2_2}}\\right)\n(\\#eq:c2)\n\\end{equation}\n\n\\textbf{Utilities}\n\nIf \\(\\pi > 1 - \\pi\\) then the optimal action is to purchase asset\n\\(A_1\\), otherwise it is to purchase \\(A_2\\).\n\nWe can assign utilities to each combination of the condition of \\(c_1\\)\nand \\(c_2\\), and each action such that,\n\n\\begin{equation}\n\\begin{aligned}\nu(c_1 > c_2, A_1)&=1\\\\\nu(c_1 < c_2, A_1)&=0\\\\\nu(c_1 > c_2, A_2)&=0\\\\\nu(c_1 < c_2, A_2)&=1\\\\\n\\end{aligned}\n(\\#eq:utilitiesapen)\n\\end{equation}\n\nTherefore the expected values of taking each action are\n\\(\\mathrm{E}[u(A_1)]=\\pi\\) and \\(\\mathrm{E}[u(A_2)]=1-\\pi\\).\n\n\\textbf{Value of perfect information}\n\nIf we could reduce the uncertainty in the prior probabilites of \\(c_1\\)\nand \\(c_2\\) such that \\(\\pi\\) approached either limit, then the expected\nvalue of perfect information is,\n\n\\begin{equation}\n\\mathrm{EVPI}=1-\\max(\\pi, 1-\\pi)=\\min(\\pi, 1-\\pi)\n(\\#eq:evpiaucapen)\n\\end{equation}\n\n\\textbf{Preposterior analysis}\n\nNow let's assume we can sample from some process and learn about \\(c_1\\)\nand \\(c_2\\).\\\\\nLet \\(X_1\\) and \\(X_2\\) be observations from a process described by the\nfollowing sampling distributions,\n\n\\begin{equation}\nX_1\\sim\\mathcal{N}\\left(c_1, 1\\right)\n(\\#eq:x1)\n\\end{equation}\n\nand\n\n\\begin{equation}\nX_2\\sim\\mathcal{N}\\left(c_2, 1\\right)\n(\\#eq:x2)\n\\end{equation}\n\nwhere for simplicity the standard deviation is one.\n\nFurthermore, we can make \\(Mp\\) and \\(M(p-1)\\) observations of \\(X_1\\)\nand \\(X_2\\) respectively. Where \\(M\\) is the total number of samples the\nbudget allows and \\(p\\) and \\(p-1\\) are proportions of that budget\nallocated to each assets. Given the observations of \\(X_1\\) and \\(X_1\\)\nwith sample sizes \\(Mp\\) and \\(M(p-1)\\) we can update the prior beliefs\nin \\(c_1\\) and \\(c_2\\) and arrive at posterior distributions,\n\n\\begin{equation} \n\\begin{aligned}\nc_1^\\prime &\\sim \\mathcal{N}(\\mu^\\prime_1, \\sigma^\\prime_1)\\\\\nc_2^\\prime &\\sim \\mathcal{N}(\\mu^\\prime_2, \\sigma^\\prime_2)\n\\end{aligned}\n(\\#eq:posteriorcs)\n\\end{equation}\n\nwhere,\n\n\\begin{equation}\n\\begin{aligned}\n\\mu^\\prime_1 &= \\frac{\\mu_1 + X_1Mp\\sigma^2_1}{Mp\\sigma^2_1 + 1}\\\\\n\\sigma^\\prime_1 &= \\sqrt{\\frac{\\sigma^2_1}{Mp\\sigma^2_1 + 1}}\\\\\n\\mu^\\prime_2 &= \\frac{\\mu_2 + X_2M(p-1)\\sigma^2_2}{M(p-1)\\sigma^2_2 + 1}\\\\\n\\sigma^\\prime_2 &= \\sqrt{\\frac{\\sigma^2_2}{M(p-1)\\sigma^2_2 + 1}}\n\\end{aligned}\n(\\#eq:posteriormusig)\n\\end{equation}\n\n\\textbf{Value of sample information}\n\nNow we make the simplifying assumptions that\n\\(\\mathrm{E}[X_1]=\\mathrm{E}[c_1]=\\mu_1\\) and\n\\(\\mathrm{E}[X_1]=\\mathrm{E}[c_1]=\\mu_1=0\\).\n\nIn this scenario the prior probability of \\(c_1>c_2\\) is,\n\n\\begin{equation}\n\\pi=\\Phi\\left(\\frac{\\mu_1}{\\sqrt{\\sigma^2_1+\\sigma^2_2}}\\right)\n(\\#eq:priorpi)\n\\end{equation}\n\n, and the posterior is\n\n\\begin{equation}\n\\pi^\\prime=\\Phi\\left(\\frac{\\mu_1}{\\sqrt{\\frac{\\sigma^2_1}{Mp\\sigma^2_1 + 1}+\\frac{\\sigma^2_2}{M(p-1)\\sigma^2_2 + 1}}}\\right)\n(\\#eq:posteriorpi)\n\\end{equation}\n\nWe can now calculate the expected value of sample information,\n\n\\begin{equation}\n\\mathrm{EVSI}=\\max(\\pi^{\\prime}, 1-\\pi^{\\prime})-\\max(\\pi, 1-\\pi)\n(\\#eq:evsiaucapen)\n\\end{equation}\n\n\\textbf{Optimal sampling allocation}\n\nWith equations 10-12, for any given set of \\(\\mu_1\\), \\(\\sigma_1\\),\n\\(\\mu_2\\), and \\(M\\) we can find the optimal allocation to sample for\nasset \\(A_1\\), \\(p\\) to maximise the value of sample information.\n\n\\textbf{Summary}\n\n\\begin{itemize}\n\\tightlist\n\\item\n Optimal allocation insensitive to ratio of \\(\\mu_a\\) to \\(\\mu_b\\).\n\\item\n Optimal allocation sensitive to ratio of \\(\\sigma_1\\) to \\(\\sigma_2\\).\n\\item\n Always preferential to sample asset with greater uncertainty.\n\\item\n Solution is symmetrical.\n\\item\n The \\(M\\) at which you start to allocate sampling to both assets is\n proportional to the ratio of \\(\\sigma_1\\) to \\(\\sigma_2\\).\n\\item\n The \\(M\\) at which you start to allocate sampling to both assets is\n insensitive to whether \\(\\sigma_1 < \\sigma_2\\) or\n \\(\\sigma_1 > \\sigma_2\\).\n\\item\n Optimal allocation always has greater EVSI when\n \\(\\sigma_1 \\neq \\sigma_2\\).\n\\end{itemize}\n\n\\textbf{\\(\\mu_1 > \\mu_2, \\sigma_1 \\gg \\sigma_2\\)}\n\nFirst let's examine the case where our prior belief is that \\(c_1\\) is\nsomewhat greater than \\(c_2\\), where \\(\\mu_1=.2\\) and \\(\\mu_2 = 0\\) but\nthe uncertainty in \\(\\theta_1\\) is far greater than in \\(\\theta_2\\),\n\\(\\sigma_1 = 2\\) and \\(\\sigma_2 = .2\\).\n\n\\includegraphics{figure/unnamed-chunk-2-1.png} \\clearpage\n\nIn this case, until we can take more than a total of 24 samples\n(\\(N=24\\)) the optimal allocation is to allocate all sampling effort to\nasset \\(A_1\\) (a). With a total budget greater than 24 samples, we start\nallocating sampling to asset \\(A_2\\) at a diminishing rate with total\nsample size and asymptoting at \\(p=0.5\\). The expected value of sample\ninformation (EVSI) increases with sample size with diminishing returns\nasymptoting below the EVPI (c). At all sample sizes, the optimal\nallocation has greater expected value than naive assumption of constant\nallocation of \\(p=0.5\\). The greatest benefit of allocating optimally is\nwhen the sample size is \\(N=10\\). As sample size increases the\nadditional benefit of allocating optimally declines asymptotically as\nthe optimal allocation approaches \\(p=0.5\\) (d).\n\n\\textbf{\\(\\mu_1 \\simeq \\mu_2, \\sigma_1 \\gg \\sigma_2\\)}\n\nHolding all the other parameters constant, let's examine a scenario\nwhere the prior expection of \\(c_1\\) is only marginally better than\n\\(c_2\\) (i.e., reduce \\(\\mu_1\\) to \\(.01\\)).\n\n\\includegraphics{figure/unnamed-chunk-3-1.png} \\clearpage\n\nWe still allocate in the same way as before (a). The EVPI approaches its\ntheoretical maxmimum of 0.5. But the value of sample information\nachievable for anything less than 200 samples is reduced to near zero\n(c). The shape of the additional benefit from optimal allocation is the\nsame but the scale has reduced and the optimal value of N has increased\nmeaning more samples must be taken to maxmise the additional gain in\nEVSI by using the optimal allocation vs the naive allocation of\n\\(p=0.5\\) (d).\n\n\\textbf{\\(\\mu_1 \\gg \\mu_2, \\sigma_1 \\gg \\sigma_2\\)}\n\nAll other parameters still the same, but now with \\(\\mu_1=1\\) much\ngreater than \\(\\mu_2\\).\n\n\\includegraphics{figure/unnamed-chunk-4-1.png} \\clearpage\n\nStill same optimal allocation (a). EVPI reduced as we start already more\ncertain that \\(A_1\\) is a better asset than \\(A_2\\). The return on\ninvesting in each additional sample diminishes quicker and sooner as\nEVSI asymptotes at nearer EVPI (c). The additional benefit in EVSI seen\nby sampling optimally has increased by peaks earlier at \\(N=2\\) (d).\n\n\\textbf{\\(\\mu_1 > \\mu_2, \\sigma_1 > \\sigma_2\\)}\n\nResetting \\(\\mu_1\\) now we examine what happens when \\(\\sigma_1\\) is\ndouble \\(\\sigma_2\\).\n\n\\includegraphics{figure/unnamed-chunk-5-1.png} \\clearpage\n\nThe optimal allocation has the same shape as before but now we allocate\nsampling to asset \\(A_2\\) sooner than before. Now if we take more than 3\nsamples we will allocate an increasing amount of them to asset \\(A_2\\)\n(a). The EVPI has been reduced as we have started off more certain that\n\\(A_1\\) is greater \\(A_2\\) (c). There is less advantage to sampling\noptimally rather than the naive allocation but the point at which the\nadditional benefit in EVSI peaks is at higher \\(N\\) (\\(N=2\\)) than when\n\\(\\sigma_1\\) was much more than \\(\\sigma_2\\) (d).\n\n\\textbf{\\(\\mu_1 > \\mu_2, \\sigma_1 = \\sigma_2\\)}\n\nNow increase the precision of \\(c_1\\) so that \\(\\sigma_1 = \\sigma_2\\).\n\n\\includegraphics{figure/unnamed-chunk-6-1.png} \\clearpage\n\nNow the optimal allocation is to just allocate evenly between the assets\n(i.e., the naive allocation is now optimal and there is effectively no\nadvantage to optimize) (a). The allocation is still insentive to the\nvalue of \\(\\mu_1\\) and the value of \\(\\sigma\\)'s. Decreasing the prior\nprecisions increases EVPI, and with it the point at which EVSI vs \\(N\\)\nasymptotes, but without changing the rate EVPI increases with \\(N\\).\nIncreaseing the value of \\(\\mu_1\\) decreases EVPI (but less sensitively\nthan changing the precision) and also increases the rate at which EVPI\nvs \\(N\\) approaches EVPI (c).\n\n\\textbf{\\(\\mu_1 > \\mu_2, \\sigma_1 \\ll \\sigma_2\\)}\n\nReturning to the original paramterisation now reverse the prior\nprecisions so that \\(\\tau^\\prime_a\\) is much greater than\n\\(\\tau^\\prime_b\\).\n\n\\includegraphics{figure/unnamed-chunk-7-1.png} \\clearpage\n\nThe allocation is the same but now reflected so that we still allocate\nto the asset with more uncertainty (a). This applies no matter what the\nother parameter values are (i.e., the problem is symetrical).\n\n\\captionsetup{labelformat=default}\n", "meta": {"hexsha": "ae4803922c4379e205e53fc4a027f3667799fc4c", "size": 9519, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "appendix1.tex", "max_stars_repo_name": "wkmor1/voiConsAuc", "max_stars_repo_head_hexsha": "71b466f70188c7faa85de6a1a677700be1a72295", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "appendix1.tex", "max_issues_repo_name": "wkmor1/voiConsAuc", "max_issues_repo_head_hexsha": "71b466f70188c7faa85de6a1a677700be1a72295", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "appendix1.tex", "max_forks_repo_name": "wkmor1/voiConsAuc", "max_forks_repo_head_hexsha": "71b466f70188c7faa85de6a1a677700be1a72295", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.8681318681, "max_line_length": 142, "alphanum_fraction": 0.7184578212, "num_tokens": 3096, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.8289388040954683, "lm_q1q2_score": 0.6375204519708686}} {"text": "\\documentclass[./\\jobname.tex]{subfiles}\n\\begin{document}\n\n\\chapter{State of the Art}\n\\label{chap:state_of_the_art}\nThis chapter provides an overview of the current state of the art in solving \\gls{pde}. Included are the widely used \\gls{fem} as well as heuristic optimisation methods. Further, an introduction to the \\gls{de} framework is given, which provides the basis for the algorithms described in this thesis. \n\n\\section{Finite Element Method}\nCurrently, the Finite Element Method is the go-to approach to solve partial differential equations. The domain $\\Omega$ on which the \\gls{pde} is posed, is discretised into multiple smaller elements - as the name suggests. Thus, \\gls{fem} counts to the category of meshed methods. The underlying solution function $u(\\mathbf{x})$ to the PDE is then approximated by so called ``basis-functions'' $\\Phi(\\mathbf{x})$ limited to these finite elements. This thesis uses the open-source Netgen/NGSolve \\gls{fem} package (\\cite{schoberl_ngsolvengsolve_2020}). \n\nThe general steps taken to solve a PDE with an FEM solver are: \n\\begin{enumerate}\n\t\\item \\underline{Step: Strong Form} \\\\\n\t\t This is the standard formulation of the linear \\gls{pde}. $\\mathbf{L}$ and $\\mathbf{B}$ are linear differential operators that include the derivatives. \\\\\n\t\t \\begin{equation}\n\t\t \\label{eq: strong form}\n\t\t\t \\begin{split}\n\t\t\t \t\\mathbf{x} \\in \\mathbb{R}^2 \\\\\n\t\t\t \tu(\\mathbf{x}), f(\\mathbf{x}), g(\\mathbf{x}): \\Omega \\rightarrow \\mathbb{R} \\\\\n\t\t\t\t\\mathbf{L} u(\\mathbf{x}) = f(\\mathbf{x}) \\text{ on $\\Omega$} \\\\\n\t\t\t\t\\mathbf{B} u(\\mathbf{x}) = g(\\mathbf{x}) \\text{ on $\\partial \\Omega$}\n\t\t\t \\end{split}\n\t\t \\end{equation}\n\t\t Further, only Dirichlet boundary conditions are considered, thus the boundary operator is always the identity matrix $\\mathbf{B} = \\mathbb{I}$. Therefore, the linear operator on the boundary $\\mathbf{B}$ can be disregarded, resulting in \\\\\n\t\t \\begin{equation}\n\t\t \tu(\\mathbf{x})|_{\\partial \\Omega} = g(\\mathbf{x}) .\n\t\t \\end{equation}\n\t\t \n\t\\item \\underline{Step: Weak Form} \\\\\n\t\t The next step is to reformulate the strong form into a usable weak form. This is equivalent to the strong form but written in an integral notation. In this equation, the $A$, $b$ and $c$ correspond to the constant factors of the derivatives in strong form. For the sake of completeness, this is kept abstract. In the \\gls{pde}s considered in this work, $A = \\mathbb{I}$ and $b=\\mathbf{0}$, $c = 0$. Currently, the so-called test-function $v(\\mathbf{x})$ is an arbitrary function, but it has to be 0 on the boundary $v(\\mathbf{x})|_{\\Omega} = 0$. The choice of the test-function correspond to different \\gls{fem} types (\\cite[p. 6f]{shen_spectral_2011}).\\\\\n\t\t \\begin{equation}\n\t\t \\label{eq: weak form}\n\t\t\t \\begin{split}\n\t\t\t \\underbrace{\\int_{\\Omega} - (\\nabla^T A \\nabla) u(\\mathbf{x}) v(\\mathbf{x}) dV - \\int_{\\Omega} b^T \\nabla u(\\mathbf{x}) v(\\mathbf{x}) dV + \\int_{\\Omega} c u(\\mathbf{x}) v(\\mathbf{x}) dV}_{a(u,v)} \\\\ = \\underbrace{\\int_{\\Omega} f(\\mathbf{x}) v(\\mathbf{x}) dV}_{F(v)}\n\t\t\t \\end{split}\n\t\t \\end{equation} \n\t\\item \\underline{Step: Discretisation of $\\Omega$} \\\\\n\t\t Create a mesh of finite elements that span the whole domain. Usually these are triangles. Thus, this step is sometimes called ``triangulation''.\n\t\\item \\underline{Step: Basis functions} \\\\\n\t\t Choose a basis function $\\Phi(\\mathbf{x})$ that can be used to approximate the solution $u(\\mathbf{x}) \\approx u_{h}(\\mathbf{x}) = \\sum_{i = 1}^{N} u_i \\Phi_i(\\mathbf{x})$. A common choice are Lagrange or Chebyshev polynomials. In the Galerkin type \\gls{fem}, the test-function $v(\\mathbf{x})$ is the same as the trail-function, thus $v(\\mathbf{x}) = \\sum_{j = 1}^{N} v_j \\Phi_j(\\mathbf{x})$. The choice of the basis function $\\Phi(\\mathbf{x})$ largely influences the computational effort. $\\Phi(\\mathbf{x})$ should have a small support, to produce a thinly populated matrix $A$ in the linear system of equations \\eqref{eq:linear_system_of_equations} below.\n\t\\item \\underline{Step: Solution} \\\\\n\t\t In the weak form, as seen in equation \\eqref{eq: weak form}, $a(u,v)$ is a continuous bilinear form and $F(v)$ is a continuous linear functional. Substituting $u$ and $v$ with their corresponding approximation from \\mbox{step 4} results in \n\t\t \\begin{equation}\n\t\t \\sum_{j=1}^{N} v_j \\sum_{i=1}^{N} u_i a(\\Phi_i, \\Phi_j) = \\sum_{j=1}^{N} v_j F(\\Phi_j).\n\t\t \\end{equation} \n\t\t Dividing by the $v_j$ values on both sides results in a linear system of equations, where the constant factors $u_i$ need to be determined. \n\t\t \\begin{equation}\n\t\t \\label{eq:linear_system_of_equations}\n\t\t \\underbrace{\\sum_{i=1}^{N} u_i a(\\Phi_i, \\Phi_j)}_{\\mathbf{A u}} = \\underbrace{F(\\Phi_j)}_{\\mathbf{b}} \\text{ for $j=1,...N$}\n\t\t \\end{equation}\n\\end{enumerate}\n\nModern solvers include more complex and advanced techniques to further improve the solution error and the computation time. Some of the most important concepts that are also available in NGSolve are listed here. \n\n\\begin{itemize}\n\t\\item \\underline{Static Condensation}: \\\\\n\t\t Depending on the number of discrete elements, the $\\mathbf{A}$ matrix can be very large. Inverting large matrices is very time consuming. Static condensation, also called Guyan reduction (\\cite{guyan_reduction_1965}), reduces this dimensionality by exploiting the structure of $\\mathbf{A}$. \n\t\\item \\underline{Preconditioner}: \\\\\n\t\t Instead of solving the $\\mathbf{A}^{-1}$ exactly, this can also be approximated by a matrix that is similar to $\\mathbf{A}^{-1}$. The actual inverse can be iteratively approximated. NGSolve implements multiple different preconditioners and it even allows to create your own method. \n\t\\item \\underline{Adaptive Mesh Refinement}: \\\\\n\t\tThe accuracy of a FEM-approximated solution mainly depends on the density of the mesh. Typically, finer meshes tend to produce more accurate solutions, but the computation time is longer. This trade-off can be overcome by a self-adaptive mesh. NGSolve implements that in an adaptive loop that executes: \n\t\t\\begin{itemize}\n\t\t\t\\item Solve PDE (with coarse mesh)\n\t\t\t\\item Estimate Error (for every element)\n\t\t\t\\item Mark Elements (that have the greatest error)\n\t\t\t\\item Refine Elements (that were previously marked)\n\t\t\t\\item Repeat until degrees of freedom exceed a specified $N$\n\t\t\\end{itemize}\n\\end{itemize}\n\n\n\\section{Computational Intelligence Methods} \n\\label{chap:literature_overview}\n\nThe research community interested in computational intelligence solvers for differential equations has been steadily growing over the past 20 years. This chapter summarises the most important works done in the general field of development and application of such statistical numerical solvers. The following table \\ref{tab:literature_research} gives a brief overview of these papers and sorts them historically. \n\nIn general, all of these papers from the table use the \\gls{wrm}, or some variant of that concept, to transform their differential equation into an optimisation problem. This serves as the fitness function and is necessary to evaluate a possible candidate solution and perform the evolutionary selection. The fitness function is the function to be optimised. It is also called objective function and these terms are used interchangeably in this thesis. In short, the residual $R$ is defined through the differential equation itself and can be calculated by $R(u(\\mathbf{x})) = \\mathbf{L}u(\\mathbf{x}) - f(\\mathbf{x})$. The residual can be thought of as a functional that substitutes $u(\\mathbf{x})$ with an approximate solution $u_{apx}(\\mathbf{x})$ and returns a numerical score. The \\gls{wrm} method is further described in chapter \\ref{chap:opt_problem}.\n\n\\cite{howard_genetic_2001} is one of the first advances in this field. They approximate a subset of the convection-diffusion equations with \\gls{gp} (\\cite{koza_genetic_1992}). Their main idea is to use a polynomial of variable length as the candidate solution that is forced to satisfy the boundary condition. Their fitness value, as seen in equation \\eqref{eq:howard_fitness_2001}, is calculated by squaring the residual $R$ and integrating it over the domain. Since the polynomials are known, and the problems are restricted to a specific differential equation, the integral can be evaluated analytically. \n\n\\begin{equation}\n\\label{eq:howard_fitness_2001}\n\tF(u_{apx}(\\mathbf{x})) = -\\int_{\\Omega} R(u_{apx}(\\mathbf{x}))^2 dx\n\\end{equation}\n\n\\cite{kirstukas_hybrid_2005} proposes a three-step procedure. The first step is time consuming and employs \\gls{gp} techniques to find basis functions that span the solution space. The second step is faster and uses a Gram–Schmidt algorithm to compute the basis function multipliers to develop a complete solution for a given set of boundary conditions. Using linear solver methods, a set of coefficients is found that produces a single function that both satisfies the differential equation and the boundary or initial conditions at distinct points over the domain. These points are further called collocation points. \n\n\\cite{tsoulos_solving_2006} use \\gls{ge} (\\cite{ryan_grammatical_1998}) to find solutions to various differential equations. In contrary to \\gls{gp}, \\gls{ge} uses vectors instead of trees to represent the candidate string. The solution is evaluated as an analytical string, constructed of the functions $sin$, $cos$, $exp$ and $log$, as well as all digits and all four basic arithmetic operations. Because the \\gls{ge} step could result in virtually any function, the fitness integral from equation \\eqref{eq:howard_fitness_2001} can not be calculated analytically. Thus, the integral is approximated by evaluating the residual at collocation points within the domain. This is seen in equation \\eqref{eq:fit_func_tsoulos}. The algorithm was tested on multiple problems of \\gls{ode}, system of ODEs and \\gls{pde}. Only the results for ODEs were promising.\n\n\\begin{equation}\n\\label{eq:fit_func_tsoulos}\nF(u_{apx}(\\mathbf{x})) = \\sum_{i=1}^{n_C} ||R(u_{apx}(\\mathbf{x}_i))||^2\n\\end{equation} \n\n\\cite{mastorakis_unstable_2006} couples a \\gls{ga} (\\cite{holland_outline_1962}) with a \\gls{ds} method (\\cite{nelder_simplex_1965}) for the local solution refinement. The candidates are represented as polynomials of the order 5 where the coefficients are optimised. The boundary condition is directly incorporated into the candidate, thus simplifying the objective function to equation \\eqref{eq:fit_func_tsoulos}. The focus here is on unstable ODEs that can not be solved with finite difference methods. \n\n\\cite{sobester_genetic_2008} tried a radical different approach to incorporate the boundary condition into the solution. They found that using \\gls{gp} for the inner domain is only effective if the algorithm does not have to consider the boundary. They split the solution $u_{apx}(\\mathbf{x})$ into two parts where $u_{GP}(\\mathbf{x})$ represents the solution for the inner domain and $u_{RBF}(\\mathbf{x})$ ensures the boundary condition \n\\begin{equation}\n\\label{eq:solution_sobester}\nu(\\mathbf{x})_{apx} = u_{GP}(\\mathbf{x}) + u(\\mathbf{x})_{RBF}.\n\\end{equation}\nAt first, the \\gls{gp} step produced a trial solution according to the objective function \\eqref{eq:fit_func_tsoulos}. After the \\gls{gp} procedure, a linear combination of radial basis functions $u(\\mathbf{x})_{RBF} = \\sum_{j=1}^{n_B} \\alpha_j \\Phi (||\\mathbf{x}-\\mathbf{x}_{j}||)$ is specifically tailored to $u_{GP}(\\mathbf{x})$ that ensures the boundary condition at all $\\mathbf{x}_{j}$ points on $\\partial \\Omega$. Finding the parameters $\\alpha_j$ can be formulated as a least squares problem. \n\n\\cite{howard_genetic_2011} use a \\gls{gp} scheme to find the solution to a specific set of simplified convection-diffusion equations. They represent a candidate as discrete function value points over the domain. The function between these points is interpolated. The fitness function is similar to equation \\eqref{eq:fit_func_tsoulos} with the exception that the $n_C$ points are not predetermined. These points are sampled randomly in the domain, thus allowing the algorithm to approximate the solution aside from fixed base points. \n\n\\cite{chaquet_solving_2012} use a simple self-adaptive \\gls{es} (as developed by \\cite{schwefel_evolutionsstrategien_1977} and \\cite{rechenberg_evolutionsstrategien_1978}) to evolve the coefficients of a partial Fourier series. The fitness function is expressed in equation \\eqref{eq:fit_func_chaquet}. This is similar to the fitness function \\ref{eq:fit_func_tsoulos}, but it extends the definition of the boundary to also include Neumann conditions by introducing the linear differential operator $\\mathbf{B}$. The limit $n_C$ denotes the number of inner collocation points $\\mathbf{x}_i$ within the domain $\\Omega$, whereas $n_B$ is the number of discrete points $\\mathbf{x}_j$ on the boundary $\\partial \\Omega$. Further, a penalty factor $\\phi$ shifts the focus of the fitness to the boundary. Additionally, this objective function can also represent systems of differential equations, where the number of equations is denoted by $m$. To reduce the search dimension (represented by the number of harmonics), they developed a scheme that only optimises one harmonic at a time and freezes the other coefficients. This scheme is based on the often observed principle that lower frequencies are more important in reconstructing a signal than higher ones. Albeit this concept might not be valid for all possible functions, it worked on all differential equations of their testbed. \n\n\\begin{equation}\n\\label{eq:fit_func_chaquet_2012}\nF(u_{apx}(\\mathbf{x})) = \\frac{\\sum_{i=1}^{n_C} || \\mathbf{L}u_{apx}(\\mathbf{x}_i) - f(\\mathbf{x}_i)||^2 + \\phi \\sum_{j=1}^{n_B} || \\mathbf{B}u_{apx}(\\mathbf{x}_j) - g(\\mathbf{x}_j)||^2}{m (n_C + n_B)} \n\\end{equation}\n\n\\cite{babaei_general_2013} takes a similar approach. They approximate a solution using a partial Fourier series. The optimal parameters for the candidates are found using a \\gls{pso} algorithm (\\cite{kennedy_particle_1995}). The fitness function consists of two parts, one for the inner area (equation \\eqref{eq:inner_WRF}) and one for the boundary (equation \\eqref{eq:boundary_penalty}). These are added together resulting in equation \\eqref{eq:inner_and_boundary_fitness}.\n\nThe weighted residual integral WRF is exactly the formulation of the \\gls{wrm} from chapter \\ref{chap:opt_problem}. $W$ is an arbitrary weighting function. The absolute values of $W$ and $R$ ensure that only positive values count towards the fitness. Instead of using a sum over collocation points, the integral is evaluated using a numerical integration scheme.\n\n\\begin{equation}\n\\label{eq:inner_WRF}\nWRF(u_{apx}(\\mathbf{x})) = \\int_{\\Omega} |W(\\mathbf{x)}| |R(u_{apx}(\\mathbf{x}))| dx\n\\end{equation} \n\nThe boundary condition is incorporated by summing up its normed violations at distinct points $\\mathbf{x}_i$. $K_j$ are penality multipliers that shift the focus to different points of the boundary. The concept of this penalty function originates from \\cite{rajeev_discrete_1992}.\n\n\\begin{equation}\n\\label{eq:boundary_penalty}\nPFV(u_{apx}(\\mathbf{x})) = WRF(u_{apx}(\\mathbf{x})) \\cdot \\sum_{j=1}^{n_B} K_j \\left(\\frac{u_{apx}(\\mathbf{x}_i)}{g(\\mathbf{x}_i)} - 1\\right)\n\\end{equation}\n\n\\begin{equation}\n\\label{eq:inner_and_boundary_fitness}\nF(u_{apx}(\\mathbf{x})) = WRF(u_{apx}(\\mathbf{x})) + PFV(u_{apx}(\\mathbf{x}))\n\\end{equation}\n\n\\cite{panagant_solving_2014} use polynomials as a candidate representation. They do not specify the order or the type of the polynomial. They test five different simple versions of the optimisation algorithm \\gls{de} (\\cite{storn_differential_1997}). Further, they introduce a so called DE-New that increases the population size after every generation. Their proposition is that greater population sizes are better at finding good solutions. \n\n\\cite{sadollah_metaheuristic_2017} compares three different optimisation algorithms to approximate differential equations: \\gls{pso}, \\gls{hs} (\\cite{geem_new_2001}) and \\gls{wca} (\\cite{eskandar_water_2012}). They use the formulation in equation \\eqref{eq:inner_WRF}, where the weighting function is the same as the residual $|W(\\mathbf{x})| = |R(u_{apx}(\\mathbf{x}))| \\rightarrow WRF = \\int_{\\Omega} |R(u_{apx}(\\mathbf{x}))|^2 dx$. The integral is again approximated using a numerical integration scheme. They find that the \\gls{pso} is slightly better at producing low error solutions, however \\gls{wca} is better at satisfying the boundary condition. \n\nIn their paper \\cite{chaquet_using_2019} describe an algorithm that approximates a solution with a linear combination of Gaussian \\gls{rbf} as kernels:\n\\begin{equation}\nu(\\mathbf{x})_{apx} = \\sum_{i=1}^{N} \\omega_i e^{\\gamma_i (\\left||\\mathbf{x} - \\mathbf{c}_i\\right||^2)}\n\\end{equation}\nThe approximated function $u(\\mathbf{x})_{apx}$ can be fully determined by a finite number of parameters: $\\omega_i, \\gamma_i, \\mathbf{c}_i$. These are stacked together into a vector $\\mathbf{p_{apx}}$ and called the decision variables which are optimised by the algorithm. \nThe objective function can be seen in equation \\eqref{eq:fit_func_chaquet}. This is an update of the objective function in \\ref{eq:fit_func_chaquet_2012} where the inner collocation points also get scaled by a weighting function $\\xi(\\mathbf{x}_i)$.\n\\begin{equation}\n\\label{eq:fit_func_chaquet}\nF(u_{apx}(\\mathbf{x})) = \\frac{\\sum_{i=1}^{n_C} \\xi (\\mathbf{x}_i) || \\mathbf{L}u_{apx}(\\mathbf{x}_i) - f(\\mathbf{x}_i)||^2 + \\phi \\sum_{j=1}^{n_B} || \\mathbf{B}u_{apx}(\\mathbf{x}_j) - g(\\mathbf{x}_j)||^2}{m (n_C + n_B)} \n\\end{equation}\nThe multipliers $\\xi(\\mathbf{x}_i)$ and $\\phi$ are weighting factors for either the inner or the boundary term. The whole term is normalised with the number of collocation points. \nThe parameters of the kernels are determined via a \\gls{cma_es} (\\cite{hansen_reducing_2003}). To further improve the solution, the evolutionary algorithm is coupled with a \\gls{ds} method to carry out the local search. The authors show empirically that the local search significantly improves the performance by testing the algorithm on a set of 32 differential equations. \n\n\\cite{fateh_differential_2019} use a simple variant of \\gls{de} where candidates are represented as discrete function value points within the domain. The function values between the grid points are linearly interpolated. This is a radical brute force approach that results in a massive search space dimension. Yet, the main advantage is that the solution is not limited to a decomposition of kernel functions and thus, even non-smooth functions can be approximated. Since this approach does not produce an analytical solution, the differential equation and the boundary condition is incorporated into the fitness function by taking the sum of squared residuals at every grid point, as seen in equation \\eqref{eq:fit_fateh}. The derivatives within the residual are calculated between two neighbouring points by the difference quotient. \n\\begin{equation}\n\\label{eq:fit_fateh}\nF(\\mathbf{x}) = \\sqrt{\\sum_{i=0}^{n} R(\\mathbf{x}_i)^2}\n\\end{equation}\n\n\n\n\\begin{table}[H]\n\t\\centering\n\t\\noindent\\adjustbox{max width=\\linewidth}{\n\t\t\\begin{tabular}{|c|c|c|c|}\n\t\t\t\n\t\t\t\\hline\n\t\t\t\\rowcolor[HTML]{\\farbeTabA}\n\t\t\t\n\t\t\tPaper & Algorithm & Representation & Problems \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{howard_genetic_2001}} & \\multilinecell{\\gls{gp}} & \\multilinecell{polynomial of \\\\ arbitrary length} & \\multilinecell{one-dimensional \\\\ steady-state \\\\ model of \\\\ convection-diffusion \\\\ equation} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{kirstukas_hybrid_2005}} & \\multilinecell{\\gls{gp}} & \\multilinecell{algebraic \\\\ expression} & \\multilinecell{heating of thin rod \\\\ heating by current} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{tsoulos_solving_2006}} & \\multilinecell{\\gls{ge}} & \\multilinecell{algebraic term} & \\multilinecell{set of ODEs \\\\ system of ODEs \\\\ and PDEs} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{mastorakis_unstable_2006}} & \\multilinecell{\\gls{ga}\\\\(global); \\\\ \\gls{ds}\\\\(local)} & \\multilinecell{5th order \\\\ polynomial}& \\multilinecell{unstable \\\\ ODEs} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{sobester_genetic_2008}} & \\multilinecell{\\gls{gp} \\\\ and \\\\ RBF-NN} & \\multilinecell{algebraic term \\\\ for inner; \\\\ RBF for boundary} & elliptic PDEs \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{howard_genetic_2011}} & \\multilinecell{\\gls{gp}} & function value grid & \\multilinecell{convection–diffusion \\\\ equation \\\\ at different \\\\ Peclet numbers } \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{chaquet_solving_2012}} & \\multilinecell{\\gls{es}} & \\multilinecell{partial sum \\\\ of Fourier series} & \\multilinecell{testbench of \\\\ ODEs \\\\ system of ODEs \\\\ and PDEs} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{babaei_general_2013}} & \\multilinecell{\\gls{pso}} & \\multilinecell{partial sum\\\\of Fourier series} & \\multilinecell{integro-differential equation\\\\system of linear ODEs \\\\ Brachistochrone \\\\ nonlinear Bernoulli} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{panagant_solving_2014}} & \\multilinecell{\\gls{de}} & \\multilinecell{polynomial of \\\\ unspecified order} & \\multilinecell{set of 6 \\\\ different PDEs} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{sadollah_metaheuristic_2017}} & \\multilinecell{\\gls{pso}\\\\\\gls{hs}\\\\\\gls{wca}} & \\multilinecell{partial sum\\\\of Fourier series} & \\multilinecell{singular BVP} \\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{chaquet_using_2019}} & \\multilinecell{\\gls{cma_es}\\\\(global); \\\\ \\gls{ds}\\\\(local)} & \\multilinecell{linear combination \\\\ of Gaussian kernels} & \\multilinecell{testbench of \\\\ ODEs \\\\ system of ODEs \\\\ and PDEs}\\\\ \\hline\n\t\t\t\n\t\t\t\\multilinecell{\\cite{fateh_differential_2019}} & \\multilinecell{\\gls{de}} & \\multilinecell{function value\\\\grid} & elliptic PDEs \\\\ \\hline\n\t\t\t\n\t\t\\end{tabular}\n\t}\n\t\\unterschrift{Literature research on the general topic of stochastic solver and their application. The papers are sorted by date of release. }{}{}\n\t\\label{tab:literature_research}\n\\end{table}\n\n\n\\section{Differential Evolution}\n\nThe differential evolution framework was first introduced in \\cite{storn_differential_1997}. Due to its simple and flexible structure, it quickly became one of the most successful evolutionary algorithm. Over the years, several adaptations to the original framework have been proposed and some of them currently count to the best performing algorithms, as the 100-Digit Challenge at GECCO 2019 (\\cite{suganthan_suganthancec2019_2020}) shows. \n\nThe main \\gls{de} framework consists of three necessary steps that continuously update a population of possible solutions. The population can be interpreted as a matrix, where each row-vector $\\mathbf{x}_i$, also called individual, represents a point within the search domain and has a fitness value corresponding to the fitness function $f(\\mathbf{x}_i): \\mathbb{R}^n \\rightarrow \\mathbb{R}$. The goal is to minimise the fitness function. These steps are performed in a loop until a predefined termination condition is reached. Each individual step is controlled by a user-defined parameter: \n\\begin{itemize}\n\t\\item \\underline{Mutation}: \\\\\n\t\t Mutation strength parameter F;\\\\\n\t\t The mutation uses the information from within the population to create a trial vector $v_i$. This is done by scaling the difference between some vectors in the population - hence the name \\textit{differential} evolution. The \\textit{/current-to-pbest/1} mutation operator can be seen in equation \\eqref{eq:mut_rand_1} where $x_i$ is the current individual, $x_{best}^p$ is one random vector of the p\\% top vectors, $x_{r1}$ is a random vector from the population while $\\tilde{x}_{r2}$ is randomly chosen from the population and the archive. $x_{r1}$ and $\\tilde{x}_{r2}$ must not describe the same individual.\n\t\t \\begin{equation}\n\t\t \\label{eq:mut_rand_1}\n\t\t v_i = x_{i} + F_i(x_{best}^p - x_{i}) + F_i(x_{r1} - \\tilde{x}_{r2})\n\t\t \\end{equation}\n\t\\item \\underline{Crossover}: \\\\\n\t\t Crossover probability parameter CR;\\\\\n\t\t The crossover procedure randomly mixes the information between the trial vector $v_i$ and a random candidate from the population $x_{i}$ to create a new trial vector $u_i$. The binomial crossover from equation \\eqref{eq:crs_bin} randomly takes elements from both vectors, where $K$ is a random index to ensure that at least one element from the trial vector $v_i$ is taken.\n\t\t \\begin{equation}\n\t\t \\label{eq:crs_bin}\n\t\t u_{ij}=\\begin{cases}\n\t\t v_{ij}, &\\text{if $j = K \\lor rand[0,1] \\leq CR$}\\\\\n\t\t x_{ij}, &\\text{otherwise}\n\t\t \\end{cases}\n\t\t \\end{equation}\n\t\\item \\underline{Selection}: \\\\\n\t\t Population size N;\\\\\n\t\t The selection replaces the old candidate $x_i$ if the trial candidate $u_i$ is better as measured by the fitness function. This is performed for every individual in the population, then the next generation is started.\n\\end{itemize} \n\nIn modern \\gls{de} variants, these parameters are self-adapted during the evolutionary process. This means that the algorithms can balance out between exploration of the search-space and exploitation of promising locations. \n\nA prominent example of a modern \\gls{de} with self-adaption is JADE, which was developed by \\cite{zhang_jade_2009}. The adaption is performed by taking successful F and CR values of the last generation into account. If a certain setting is successful in generating better candidates, newly selected F and CR gravitate towards that setting. The pseudocode is presented in the appendix \\ref{chap:pscode_jade}. \n\nThis idea was later refined by \\cite{tanabe_success-history_2013}. They propose a similar self-adaptive scheme but extend the ``memory'' for good F and CR parameters over multiple generations. This idea improves the robustness as compared to JADE. The pseudocode in appendix \\ref{chap:pscode_shade} shows the outline of this so called SHADE algorithm. \n\nThe latest iteration of SHADE is called L-SHADE (\\cite{tanabe_improving_2014}), which improves the performance by including a deterministic adaptive concept for the population size. At first, L-SHADE starts with a big population size, and reduces the number of individuals in a linear fashion by deleting bad candidates. This has the effect of reducing the number of unnecessary function evaluations. The code is displayed in the appendix \\ref{chap:pscode_lshade}. \n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "87a1b36b4066164d9f7d060fc2c93b33b5dd240b", "size": 26304, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "master_thesis_doc/tex/State_of_the_Art.tex", "max_stars_repo_name": "nicolai-schwartze/Masterthesis", "max_stars_repo_head_hexsha": "7857af20c6b233901ab3cedc325bd64704111e16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-06-13T10:02:02.000Z", "max_stars_repo_stars_event_max_datetime": "2020-06-13T10:02:02.000Z", "max_issues_repo_path": "master_thesis_doc/tex/State_of_the_Art.tex", "max_issues_repo_name": "nicolai-schwartze/Masterthesis", "max_issues_repo_head_hexsha": "7857af20c6b233901ab3cedc325bd64704111e16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "master_thesis_doc/tex/State_of_the_Art.tex", "max_forks_repo_name": "nicolai-schwartze/Masterthesis", "max_forks_repo_head_hexsha": "7857af20c6b233901ab3cedc325bd64704111e16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 109.6, "max_line_length": 1380, "alphanum_fraction": 0.7553984185, "num_tokens": 7144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289387998695209, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.637520448720776}} {"text": "%% -*- coding:utf-8 -*-\n\\chapter{Objects and morphisms}\n\n\\section{Equality}\n\nThe important question is how can we decide whenever an\nobject/morphism is equal to another object/morphism? The trivial\nanswer is possible if an \\mynameref{def:object} is a\n\\mynameref{def:set}. In the case we can say that 2 objects are equal\nif they contain the equivalent collection of elements. Unfortunately\nwe cannot do the same trick for categorical \\mynameref{def:object}s as\nsoon as they don't have any internal \nstructure but can use a \\mynameref{def:categorical_approach} (see \n\\mynameref{rem:set_vs_category}): if we cannot use ``microscope'' lets\nuse ``telescope'' and define the equality of objects and morphisms of\na category $\\cat{C}$ in the terms of whole $\\cathom{C}$.\n\n\\begin{definition}[Objects equality]\n\\label{def:object_equality}\nTwo \\mynameref{def:object}s $a$ and $b$ in \\mynameref{def:category}\n$\\cat{C}$ are equal if there exists an unique\n\\mynameref{def:isomorphism} $a \\cong_f b$. This also means that\nalso exist unique isomorphism $b \\cong_g a$. These two\n\\mynameref{def:morphism}s ($f$ and $g$) are related each other via \nthe following equations: $f \\circ g = \\idm{a}$ and $g \\circ f\n= \\idm{b}$. \n\\end{definition}\n\nUnlike \\mynameref{def:function}s between \\mynameref{def:set}s we don't\nhave any additional info \n\\footnote{\nfor instance info about sets internals. i.e. which elements of the sets\nare connected by the considered functions\n}\nabout \\mynameref{def:morphism}s except\ncategory theory axioms which the morphisms satisfy \n\\cite{bib:stackexchange:morphism:equality}. This leads us to the\nfollowing definition of morphims equality:\n\\begin{definition}[Morphisms equality]\n\\label{def:morphism_equality}\nTwo \\mynameref{def:morphism}s $f$ and $g$ in \\mynameref{def:category}\n$\\cat{C}$ are equal if the equality can be derived from the base axioms: \n\\begin{itemize}\n\\item \\mynameref{axm:composition}\n\\item \\mynameref{axm:associativity}\n\\item \\mynameref{def:id}: \\eqref{eq:leftid}, \\eqref{eq:rightid}\n\\end{itemize}\nor \\mynameref{def:commutative_diagram}s which postulate the equality.\n\\end{definition}\nAs an example lets proof the following theorem\n\\begin{theorem}[Identity is unique]\n\\label{thm:identity_unique}\nThe \\mynameref{def:id} is unique.\n\\begin{proof}\nConsider an \\mynameref{def:object} $a$ and it's \\mynameref{def:id} \n$\\idm{a}$. Let $\\exists f: a \\to a$ such that $f$ is also\nidentity. In the case \\eqref{eq:leftid} for $f$ as identity gives\n\\[\nf \\circ \\idm{a} = \\idm{a}.\n\\]\nFrom other side \\eqref{eq:rightid} for $\\idm{a}$ satisfied\n\\[\nf \\circ \\idm{a} = f\n\\]\ni.e.\n\\[\nf = f \\circ \\idm{a} = \\idm{a}\n\\] \nor\n$f = \\idm{a}$.\n\\end{proof}\n\\end{theorem}\n\n\\section{Initial and terminal objects}\n\\begin{definition}[Initial object]\n\\label{def:initial_object}\nLet $\\cat{C}$ is a \\mynameref{def:category}, the\n\\mynameref{def:object} $i \\in \\catob{C}$ is called\n\\textit{initial object} if $\\forall x \\in \\catob{C}\n\\exists! f_x: i \\to x \\in \\cathom{C}$.\n\\end{definition}\n\n\\begin{example}[Initial object][\\textbf{Set}]\n\\label{ex:set_initial_object}\nNote that there is only one function from empty set to any other sets\n\\cite{bib:proofwiki:Empty_Mapping_is_Unique} that makes the empty set\nas the \\mynameref{def:initial_object} in \\mynameref{def:setcategory}. \n\\end{example}\n\n\\begin{definition}[Terminal object]\n\\label{def:terminal_object}\nLet $\\cat{C}$ is a \\mynameref{def:category}, the\n\\mynameref{def:object} $t \\in \\catob{C}$ is called\n\\textit{terminal object} if $\\forall x \\in \\catob{C}\n\\exists! g_x: x \\to t \\in \\cathom{C}$.\n\\end{definition}\n\n\\begin{example}[Terminal object][\\textbf{Set}]\n\\label{ex:set_terminal_object}\n\\mynameref{def:terminal_object} in \\mynameref{def:setcategory} is a set\nwith one element i.e \\mynameref{def:singleton_set}. \n\\end{example}\n\nAs you can see the initial and terminal objects are opposite each\nother. I.e. if $i$ is an \\mynameref{def:initial_object} in $\\cat{C}$\nthen it will be \\mynameref{def:terminal_object} in\nthe \\mynameref{def:op_category} $\\cat{C^{op}}$.\n\n\\begin{theorem}[Initial object is unique]\n\\label{thm:initial_object_unique}\nLet $\\cat{C}$ is a category and $i, i' \\in \\catob{C}$ two\n\\mynameref{def:initial_object}s then there exists an unique\n\\mynameref{def:isomorphism} $u: i \\to i'$ (see\n\\mynameref{def:object_equality}) \n\\begin{proof}\nConsider the following \\mynameref{def:commutative_diagram} (see\n\\cref{fig:initial_object_unique}). As soon as $i$ initial object\n$\\exists! \\, u: i \\to i'$. From other side $i'$ is also initial object\nand therefore $\\exists! \\, u^{-1}: i' \\to i$. Combining them together\nvia composition we can get $u^{-1} \\circ u: i \\to i$ and $u \\circ\nu^{-1}: i' \\to i'$. From the fact that $i$ is initial object one can\nget that there exists\nonly one morphism $\\idm{i}: i \\to i$. The same is the truth for $i'$.\nTherefore $u^{-1} \\circ u = \\idm{i}$ and $u \\circ u^{-1} = \\idm{i'}$.\nThese complete the commutative diagram build and finishes the proof. \n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=above:$i$] (i1) at (0,3) {}; \n \\node[ele,label=above:$i'$] (i2) at (3,3) {}; \n \\node[ele,label=below:$i$] (i12) at (3,0) {};\n \\node[ele,label=below:$i'$] (i22) at (6,0) {};\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (i1) to\n node[sloped,above]{$u$} (i2);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (i12) to\n node[sloped,above]{$u$} (i22); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (i1) to\n node[sloped,above]{$\\idm{i}$} (i12); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (i2) to\n node[sloped,above]{$\\idm{i'}$} (i22); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (i2) to\n node[right]{$u^{-1}$} (i12); \n \\end{tikzpicture}\n \\caption{Commutative diagram for initial object uniqueness proof}\n \\label{fig:initial_object_unique}\n\\end{figure}\n\\end{proof}\n\\end{theorem}\n\n\\begin{theorem}[Terminal object is unique]\n\\label{thm:terminal_object_unique}\nLet $\\cat{C}$ is a category and $t, t' \\in \\catob{C}$ two\n\\mynameref{def:terminal_object}s then there exists an unique\n\\mynameref{def:isomorphism} $v: t' \\to t$ (see\n\\mynameref{def:object_equality}) \n\\begin{proof}\nJust got to the \\mynameref{def:op_category} and revert arrows in\n\\cref{fig:initial_object_unique}. The result shown on\n\\cref{fig:terminal_object_unique} and it proofs the theorem statement.\n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=above:$t$] (t1) at (0,3) {}; \n \\node[ele,label=above:$t'$] (t2) at (3,3) {}; \n \\node[ele,label=below:$t$] (t12) at (3,0) {};\n \\node[ele,label=below:$t'$] (t22) at (6,0) {};\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (t2) to\n node[sloped,above]{$v$} (t1);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (t22) to\n node[sloped,above]{$v^{-1}$} (t12); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (t12) to\n node[sloped,above]{$\\idm{t}$} (t1); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (t22) to\n node[sloped,above]{$\\idm{t'}$} (t2); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (t12) to\n node[right]{$v^{-1}$} (t2); \n \\end{tikzpicture}\n \\caption{Commutative diagram for terminal object uniqueness proof}\n \\label{fig:terminal_object_unique}\n\\end{figure}\n\\end{proof}\n\\end{theorem}\n\n\\begin{example}[Toy example]\nIn our toy example \\cref{fig:pl_example} the type String is\n\\mynameref{def:initial_object} and type Bool is the\n\\mynameref{def:terminal_object}.\n\\end{example}\n\n\\section{Product and sum}\nThe pair of 2 objects is defined via the universal property in\nthe following way:\n\\begin{definition}[Product]\n\\label{def:product}\nLet we have a category $\\cat{C}$ and $c_1, c_2 \\in \\catob{C}$ -two\n\\mynameref{def:object}s then the product of the objects $c_1, c_2$ is\nanother object in $\\cat{C}$ $c = c_1 \\times c_2$ with 2\n\\mynameref{def:morphism}s $\\pi_1, \\pi_2$ such that $c_1 = \\pi_1(c),\nc_2 = \\pi(c_2)$ and the \nfollowing universal property is satisfied: $\\forall c' \\in \\catob{C}$\nand morphisms $\\pi'_1: c' \\to c_1, \\pi'_2: c' \\to c_2$, exists unique\nmorphism $h$ such that the following diagram (see \\cref{fig:product})\ncommutes, i.e. $\\pi'_1 = \\pi_1 \\circ h, \\pi'_2 = \\pi_2 \\circ h$.\n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=left:$c_1$] (a) at (0,0) {}; \n \\node[ele,label=right:$c_2$] (b) at (4,0) {}; \n \\node[ele,label=below:$c$] (c) at (2,0) {};\n \\node[ele,label=above:$c'$] (c') at (2,3) {};\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (c) to\n node[sloped,above]{$\\pi_1$} (a);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (c) to\n node[sloped,above]{$\\pi_2$} (b); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (c') to\n node[sloped,above]{$\\pi'_1$} (a); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (c') to\n node[sloped,above]{$\\pi'_2$} (b); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (c') to\n node[right]{$h$} (c); \n \\end{tikzpicture}\n \\caption{Product $c = c_1 \\times c_2$. $\\forall c, \\exists! h \\in\n \\cathom{C}: \\pi'_1 = \\pi_1 \\circ h, \\pi'_2 = \\pi_2 \\circ h$.}\n \\label{fig:product}\n\\end{figure}\nIn other words $h$ factorizes $\\pi'_{1,2}$.\n\\end{definition}\n\n\\begin{example}[Product][\\textbf{Set}]\n\\label{ex:set_product}\nThe \\mynameref{def:product} of two sets $A$ and $B$ in\n\\mynameref{def:setcategory} is defined \nas a \\mynameref{def:cartesian_product}: $A \\times B = \\{(a,b)| a \\in A, b \\in B\\}$. \n\\end{example}\n\nIf we invert arrows in \\mynameref{def:product} we will got another\nobject definition that is called sum\n\n\\begin{definition}[Sum]\n\\label{def:sum}\nLet we have a category $\\cat{C}$ and $c_1, c_2 \\in \\catob{C}$ -two\n\\mynameref{def:object}s then the sum of the objects $c_1, c_2$ is\nanother object in $\\cat{C}$ $c = c_1 \\oplus c_2$ with 2\n\\mynameref{def:morphism}s $i_1, i_2$ such that $c = i_1(c_1), c = i_2\n(c_2)$ and the \nfollowing universal property is satisfied: $\\forall c' \\in \\catob{C}$\nand morphisms $i'_1: c_1 \\to c', i'_2: c_2 \\to c'$, exists unique\nmorphism $h$ such that the following diagram (see \\cref{fig:sum})\ncommutes, i.e. $i'_1 = h \\circ i_1, i'_2 = h \\circ i_2$.\n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=left:$c_1$] (a) at (0,0) {}; \n \\node[ele,label=right:$c_2$] (b) at (4,0) {}; \n \\node[ele,label=below:$c$] (c) at (2,0) {};\n \\node[ele,label=above:$c'$] (c') at (2,3) {};\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (a) to\n node[sloped,above]{$i_1$} (c);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (b) to\n node[sloped,above]{$i_2$} (c); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (a) to\n node[sloped,above]{$i'_1$} (c'); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (b) to\n node[sloped,above]{$i'_2$} (c'); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (c) to\n node[right]{$h$} (c'); \n \\end{tikzpicture}\n \\caption{Sum $c = c_1 \\oplus c_2$. $\\forall c, \\exists! h \\in\n \\cathom{C}: i'_1 = h \\circ i_1, i'_2 = h \\circ i_2 $.}\n \\label{fig:sum}\n\\end{figure}\nIn other words $h$ factorizes $i'_{1,2}$.\n\\end{definition}\n\n\\begin{definition}[Disjoint union]\n \\label{def:disjoint_union}\n Let $\\{A_i: i \\in I\\}$ be a family of sets \n indexed by $I$. The \n \\textit{disjoint union} \\cite{wiki:disjointunion} of this family is\n the set \n \\[\n \\sqcup_{i \\in I} A_i = \\cup_{i \\in I}\\left\\{\n \\left(x, i\\right): x \\in A_i\n \\right\\}.\n \\]\n The elements of the disjoint union are ordered pairs $(x, i)$. Here $i$\n serves as an auxiliary index that indicates which $A_i$ the element $x$\n came from.\n\\end{definition}\n\n\\begin{example}[Sum][\\textbf{Set}]\n \\label{ex:set_sum}\n The \\mynameref{def:sum} of two sets $A$ and $B$ in\n \\mynameref{def:setcategory} is defined as\n \\mynameref{def:disjoint_union}. \n\\end{example}\n\nThe \\mynameref{def:product} of objects will provide also a definition\nfor product of morphisms\n\\begin{definition}[Product of morphisms]\n\\label{def:product_of_morphisms}\nLet $\\cat{C}$ is a category and $a,a' \\in \\catob{C}$ and \n$b, b' \\in \\catob{C}$ are 2 pairs of \\mynameref{def:object}s that admit\n\\cref{def:product}. Consider 2 morphisms that connects the objects: \n$f: a \\to b, f': a' \\to b'$ then we can create a new unique morphism that\nconnects the products: $f \\times f': a \\times a' \\to b \\times b'$ and\nmakes the diagram commute (see \\cref{fig:product_of_morphisms}). \n\n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=above:$a$] (a) at (0,2) {}; \n \\node[ele,label=above:$a \\times a'$] (aa') at (2,2) {}; \n \\node[ele,label=above:$a'$] (a') at (4,2) {}; \n \\node[ele,label=below:$b$] (b) at (0,0) {}; \n \\node[ele,label=below:$b \\times b'$] (bb') at (2,0) {}; \n \\node[ele,label=below:$b'$] (b') at (4,0) {}; \n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (a) to\n node[sloped,above]{$f$} (b);\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (a') to\n node[sloped,above]{$f'$} (b');\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (aa') to\n node[sloped,above]{$f \\times f'$} (bb');\n\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (aa') to\n node[sloped,above]{$\\pi_a$} (a);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (aa') to\n node[sloped,above]{$\\pi_{a'}$} (a'); \n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (bb') to\n node[sloped,above]{$\\pi_b$} (b);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (bb') to\n node[sloped,above]{$\\pi_{b'}$} (b'); \n \\end{tikzpicture}\n \\caption{Product of morphisms.}\n \\label{fig:product_of_morphisms}\n\\end{figure}\n\\end{definition} \n\n\\section{Category as a monoid}\nConsider the following definition from abstract algebra\n\\begin{definition}[Monoid]\n \\label{def:monoid}\n The set of elements $M$ with defined binary operation $\\circ$ we will call\n as a monoid if the following conditions are satisfied.\n \\begin{enumerate}\n \\item Closure: $\\forall a, b \\in M$: $a \\circ b \\in M$\n \\item Associativity: $\\forall a, b, c \\in M$:\n $a \\circ \\left( b \\circ c \\right) =\n \\left( a \\circ b \\right) \\circ c$\n \\item Identity element: $\\exists e \\in M$ such that\n $\\forall a \\in M$: $e \\circ a = a \\circ e = a$\n \\end{enumerate}\n\\end{definition}\n\nWe can consider 2 \\mynameref{def:monoid}s. The first one has\n\\mynameref{def:product} as the binary operation and\n\\mynameref{def:terminal_object} as the identity element. As result we\njust got an analog of multiplication in the category theory. This is why\nthe terminal object is often denoted as $\\mathbf{1}$ and the operation\nis called as the product. \n\nAnother one is additional \\mynameref{def:monoid} that has\n\\mynameref{def:initial_object} as the identity element and the\n\\mynameref{def:sum} as the binary operation. The initial object in\nthat case is often denoted as $\\mathbf{0}$. I.e. we can see a direct\nconnection with addition in algebra.\n\nIf we do such consideration then we can make a step forward and look\nat the distributive law that sum and multiplication satisfy.\n\\begin{definition}[Distributive category]\n\\label{def:distributive_category}\nA category $\\cat{C}$ is \\textit{distributive} if \n\\cite{wiki:distributive_category}\nit has finite\n\\mynameref{def:product}s and \\mynameref{def:sum}s such that \n$\\forall a,b,c \\in \\catob{C}$:\n\\[\na \\times b \\oplus a \\times c \\cong a \\times ( b \\oplus c )\n\\]\nand \n\\[\na \\times 0 \\cong 0\n\\]\nwhere $\\mathbf{0}$ is the \\mynameref{def:initial_object}.\n\\end{definition} \n\n\\begin{example}[Distributive category]\n\\label{ex:distributive_category}\n\\mynameref{def:setcategory} is an example \n\\cite{wiki:distributive_category}\nof \\mynameref{def:distributive_category}\n\nFrom other hand not all categories which have both product and sum are\ndistributive. One of such example is a category of all groups\n\\cite{wiki:distributive_category} \n$\\cat{Grp}$ where groups are considered as objects and group\nhomomorphisms as morphisms. \n\\end{example}\n\n\\section{Exponential}\nWe are going to talk about functions (aka morphisms) as\n\\mynameref{def:object}s. \n\n\\begin{example}[Hom set]\n\\label{ex:homset}\nConsider 2 sets $A$ and $B$ the set of functions between the 2 sets\nform a new set that is called as \\textit{Hom-set} and denoted as $A\n\\to B$. Thus if $A,B \\in \\catob{Set}$ then the Hom-set will also $A\n\\to B \\in \\catob{Set}$. \n\\end{example}\n\nThe construction of \\mynameref{ex:homset} is applied to\n\\mynameref{def:setcategory} but not to an arbitrary category because\nthe Hom-set is a set and therefore the object in the \\textbf{Set}\ncategory. I.e. if $\\cat{C}$ is a category and $a, b \\in \\catob{C}$\nthen the Hom-set $a \\to b \\in \\catob{Set}$ but we now want to\nconstruct something like to the Home-set but that is an object in\n$\\cat{C}$. This will be called as the function object. We will use the\nuniversal construction for the object definition.\n\n\\begin{definition}[Exponential]\n\\label{def:exponential}\n\\begin{figure}\n \\centering\n \\begin{tikzpicture}[ele/.style={fill=black,circle,minimum\n width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner\n sep=-2pt}]\n\n % the texts\n \n \\node[ele,label=above:$z'$] (z1) at (0,3) {}; \n \\node[ele,label=below:$z^y$] (zy) at (0,0) {}; \n \\node[ele,label=above:$z' \\times y$] (z1y) at (3,3) {};\n \\node[ele,label=below:$z^y \\times y$] (zyy) at (3,0) {};\n \\node[ele,label=below:$z$] (z) at (6,0) {};\n\n \\draw[->,thick,shorten <=2pt,shorten >=2pt] (z1) to\n node[left]{$h$} (zy);\n \\draw[->,thick,shorten <=2pt,shorten >=2] (z1y) to\n node[left]{$h \\times \\idm{y}$} (zyy); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (z1y) to\n node[sloped,above]{$e'$} (z); \n \\draw[->,thick,shorten <=2pt,shorten >=2] (zyy) to\n node[sloped,above]{$e$} (z); \n \\end{tikzpicture}\n \\caption{Exponential object}\n \\label{fig:exponential}\n\\end{figure}\nLet $\\cat{C}$ is a category and $z, y \\in \\catob{C}$. We also assume\nthat $\\cat{C}$ allows all \\mynameref{def:product}s with $y$, i.e.\n$\\forall z' \\in \\catob{C}, \\exists z' \\times y$. An object $z^y$\ntogether with a \\mynameref{def:morphism} $e: z^y \\times y \\to z$ is \nan \\textit{exponential object} if $\\forall e' \\in \\cathom{C}$ and\n$\\forall z' \\in \\catob{C}$ exists an\nunique morphism $h: z' \\to z$ such that the\n\\mynameref{def:commutative_diagram} shown in \\cref{fig:exponential}\ncommutes:\n\\[\ne' = e \\circ \\left(h \\times \\idm{y} \\right)\n\\]\n\\end{definition}\n\n\\begin{example}[Exponential][\\textbf{Set}]\n\\label{ex:exponential_set}\nLets look at the \\mynameref{def:exponential} in $\\cat{Set}$. We want\nto show that the object corresponds to the function. Really if we want\nto define a function $f: X \\to Y$ then we should look at the\n\\mynameref{ex:homset} $F = X \\to Y$. $f \\in F$ - is an element of the\nHom-set. For the function application we have to take the argument $x\n\\in X$ and the function we want to apply $f \\in F$. Then we construct\nthe pair $(f,x) \\in F \\times X$. For the function application we have\nto call a \\mynameref{def:morphism} $e: F \\times X \\to Z$.\n\\footnote{$e$ from the word ``eval''}\nI.e. the\napplication $e(f, x)$ gives us $e(f, x) = y \\in Y$ - the function\nvalue. \n\\end{example}\n\n\\begin{definition}[Cartesian closed category]\n\\label{def:cartesian_closed_category}\nIf a category $\\cat{C}$ satisfies the following conditions then it is\ncalled \\textit{Cartesian closed category}\n\\begin{enumerate}\n\\item It has \\mynameref{def:terminal_object}\n\\item $\\forall a,b \\in \\catob{C}$ exists \\mynameref{def:product} $a\n \\times b \\in \\catob{C}$.\n\\item $\\forall a,b \\in \\catob{C}$ exists \\mynameref{def:exponential}\n $a^b \\in \\catob{C}$\n\\end{enumerate}\n\\end{definition}\n\n\\begin{theorem}[Cartesian closed category]\n\\label{thm:ccc}\nIf $\\cat{C}$ is a \\mynameref{def:cartesian_closed_category} with\nfinite \\mynameref{def:sum} then it is a\n\\mynameref{def:distributive_category}.\n\\begin{proof}\nTBD\n\\end{proof}\n\\end{theorem}\n\n\\section{Type algebra and Curry-Howard-Lambek correspondence}\n\\label{sec:curry_howard_lambek}\nThere is an interesting correspondence between computer programs and\nmathematical proofs. First of all consider a category of proofs\n\\subsection{\\textbf{Proof} category}\n\\begin{definition}[Proposition]\n\\label{def:proposition}\n\\textit{Proposition} is a statement that either true or false.\n\\end{definition}\n\nThere are 2 main propositions\n\\begin{definition}[True]\n\\label{def:true}\nA true statement is one that is correct, either in all cases or at\nleast in the sample case \\cite{bib:studycom:truefalse}. \n\\end{definition}\nand\n\\begin{definition}[False]\n\\label{def:false}\nA false statement is one that is not correct \\cite{bib:studycom:truefalse}. \n\\end{definition}\n\n\n\\begin{example}[Proposition]\n\\label{ex:proposition}\nThere is an example of correct (true) proposition\n\\[\n\\forall n \\in \\mathbb{R}: n^2 \\ge 0\n\\]\n\nThere is an example of incorrect (false) proposition\n\\[\n\\forall n \\in \\mathbb{C}: n^2 \\ge 0,\n\\]\nfor instance $i \\in \\mathbb{C}$ gives $i^2 = -1$.\n\n\\end{example}\n\n\\begin{definition}[Implication]\n\\label{def:implication} \\textit{An implication} is a\n\\mynameref{def:proposition} of the form $P \\implies Q$ i.e. if $P$\nthen $Q$ \\cite{bib:whatisaproof}.\n\\end{definition}\n\nThe main logical deduction rule is the following\n\\begin{definition}[Modus ponens]\n\\label{def:modusponens}\nIf $P$ is true and $P \\implies Q$ is true then $Q$ is also true. The\nrule is often written as \\cite{bib:whatisaproof}\n\\[\n\\frac{\n\\begin{array}{c}\nP \\\\\nP \\implies Q\n\\end{array}\n}{Q}\n\\]\nwhere if statements above the line are true then the statement below\nthe line is also true.\n\\end{definition}\n\n\\begin{definition}[Proof]\n\\label{def:proof}\n\\textit{Proof} is a verification \\cite{bib:whatisaproof} of a\n\\mynameref{def:proposition} by a chain of logical deduction from a\nbase set of axioms.\n\\end{definition}\n\nPropositions can be combined into new propositions via the following\nlogical operations\n\\begin{definition}[Conjunction]\n\\label{def:conjunction}\nConjunction or logical AND is the operation with following rules\n \\begin{table}[H]\n \\centering\n \\caption{Conjunction}\n \\label{tab:conjunction}\n \\begin{tabular}{|l|l|l|}\n \\hline\n $a$ & $b$ & $a \\land b$ \\\\ \\hline\n True & True & True \\\\ \\hline\n True & False & False \\\\ \\hline\n False & True & False \\\\ \\hline\n False & False & False \\\\ \\hline\n \\end{tabular}\n \\end{table}\n\\end{definition}\n\\begin{definition}[Disjunction]\n\\label{def:disjunction}\nConjunction or logical OR is the operation with following rules\n \\begin{table}[H]\n \\centering\n \\caption{Disjunction}\n \\label{tab:disjunction}\n \\begin{tabular}{|l|l|l|}\n \\hline\n $a$ & $b$ & $a \\lor b$ \\\\ \\hline\n True & True & True \\\\ \\hline\n True & False & True \\\\ \\hline\n False & True & True \\\\ \\hline\n False & False & False \\\\ \\hline\n \\end{tabular}\n \\end{table}\n\\end{definition}\n\nOperations in Boolean logic follow the distributive low:\n\\begin{equation}\na \\land ( b \\lor c) = (a \\land b) \\lor (a \\land c) \n\\nonumber\n\\end{equation}\ni.e. the operation $\\land$ corresponds to multiplication and $\\lor$ to\nsum. Therefore the $\\cat{Proof}$ can be considered as a\n\\mynameref{def:distributive_category}. \n\n\\begin{definition}[\\textbf{Proof} category]\n\\label{def:proof_category}\nThe $\\cat{Proof}$ category is a category where \\mynameref{def:proposition}s are\n\\mynameref{def:object}s and \\mynameref{def:proof}s are\n\\mynameref{def:morphism}s. I.e. proofs are used as connectors between\ndifferent propositions.\n\\end{definition}\n\nConsider different objects and constructions of the proof (logic)\ntheory from the categorical point of view\n\\begin{example}[Initial object][\\textbf{Proof}]\n\\label{ex:proof_initial_object}\nThe \\textit{false} statement can be considered as the initial object\nbecause for any other statement exists only one proof from the false\nstatement to that one.\n\\end{example}\n\n\\begin{example}[Terminal object][\\textbf{Proof}]\n\\label{ex:proof_terminal_object}\nThe \\textit{true} statement can be considered as the terminal object\n\\end{example}\n\n\\begin{example}[Product][\\textbf{Proof}]\n\\label{ex:proof_product}\n\\mynameref{def:conjunction} can be considered as\n\\mynameref{def:product} in \\mynameref{def:proof_category}.\n\\end{example}\n\n\\begin{example}[Sum][\\textbf{Proof}]\n\\label{ex:proof_sum}\n\\mynameref{def:disjunction} can be considered as\n\\mynameref{def:sum} in \\mynameref{def:proof_category}.\n\\end{example}\n\nThus we can declare the following correspondence (see\n\\cref{tab:curry_howard_lambek}) between logic \nproofs and \\mynameref{def:cartesian_closed_category} and therefore\nalso between programming languages.\n\\begin{table}[H]\n \\centering\n \\caption{Relation between logic proofs and programming languages}\n \\label{tab:curry_howard_lambek}\n \\begin{adjustbox}{width=1\\textwidth}\n \\small\n \\begin{tabular}{l|l|l}\n \\toprule\n \\mynameref{def:proof_category} & Programming language & \n \\mynameref{def:cartesian_closed_category}\\\\\n \\midrule\n \\mynameref{def:proposition}/\\mynameref{def:implication} & Type &\n \\mynameref{def:object} \\\\ \n \\mynameref{def:proof} & Function type & \\mynameref{def:exponential} \\\\\n \\mynameref{def:conjunction} & Product type & \\mynameref{def:product} \\\\\n \\mynameref{def:disjunction} & Sum type & \\mynameref{def:sum} \\\\\n \\mynameref{def:true} & unit type & \\mynameref{def:terminal_object} \\\\\n \\mynameref{def:false} & botom type & \\mynameref{def:initial_object} \\\\\n \\bottomrule\n \\end{tabular}\n \\end{adjustbox}\n\\end{table}\n\n\\subsection{\\textbf{Hask} category}\n\\begin{example}[Initial object][\\textbf{Hask}]\nIf we avoid lazy evaluations in Haskell (see\n\\mynameref{rem:hask_lazy_eval}) then we can found several types\nas candidates for initial and terminal object in Haskell. \n\\label{ex:hask_initial_object}\n\\mynameref{def:initial_object} in \\mynameref{def:haskcategory} is a\ntype without values \n\\begin{minted}{haskell}\ndata Void\n\\end{minted}\ni.e. you cannot construct a object of the type.\n\nThere is only one function from the initial object:\n\\begin{minted}{haskell}\nabsurd :: Void -> a\n\\end{minted}\nThe function is called absurd because it does absurd action. Nobody\ncan proof that it does not exist. For the existence proof the\nfollowing absurd argument can be used: ``Just provide me an object type\n\\mintinline{haskell}{Void} and I will provide you the result of\nevaluation''. \n\nThere is no function in opposite direction because it would had been\nused for the \\mintinline{haskell}{Void} object creation. \n\\end{example}\n\n\\begin{example}[Terminal object][\\textbf{Hask}]\n\\label{ex:hask_terminal_object}\nTerminal object (unit) in \\mynameref{def:haskcategory} keeps only one element\n\\begin{minted}{haskell}\ndata () = ()\n\\end{minted}\ni.e. you can create only one element of the type. You can use the\nfollowing function for the creation:\n\\begin{minted}{haskell}\nunit :: a -> ()\nunit _ = ()\n\\end{minted}\n\\end{example}\n\n\\begin{example}[Product][\\textbf{Hask}]\n\\label{ex:hask_product}\nThe \\mynameref{def:product} in \\mynameref{def:haskcategory} keeps a\npair and the constructor defined as follows\n\\begin{minted}{haskell}\n(,) :: a -> b -> (a, b)\n(,) x y = (x, y)\n\\end{minted}\nThere are 2 projectors: \n\\begin{minted}{haskell}\nfst :: (a, b) -> a\nfst (x, _) = x\nsnd :: (a, b) -> b\nsnd (_, y) = y\n\\end{minted}\n\\end{example}\n\n\\begin{example}[Sum][\\textbf{Hask}]\n\\label{ex:hask_sum}\nThe \\mynameref{def:sum} in \\mynameref{def:haskcategory} defined as\nfollows \n\\begin{minted}{haskell}\ndata Either a b = Left a | Right b\n\\end{minted}\n\nThe typical usage is via pattern matching for instance \n\\begin{minted}{haskell}\nfactor :: (a -> c) -> (b -> c) -> Either a b -> c\nfactor f _ (Left x) = f x\nfactor _ g (Right y) = g y\n\\end{minted}\n\\end{example}\n\n\\begin{example}[Distributive category][\\textbf{Hask}]\nAs soon as $\\cat{Hask}$ is a \\mynameref{def:cartesian_closed_category}\nthen by \\cref{thm:ccc} it is a \\mynameref{def:distributive_category}\ni.e. one can conclude that\n\\begin{minted}{haskell}\n(a,Either b c)\n\\end{minted}\nis the same to\n\\begin{minted}{haskell}\nEither (a, b) (a, c)\n\\end{minted}\n\\end{example}\n\n\\begin{example}[Exponential][\\textbf{Hask}]\n\\label{ex:exponential_hask}\nIt's not surprisingly that the \\mynameref{def:exponential} in\n$\\cat{Hask}$ is a function object i.e. $b^a$ can be written as \n\\mintinline{haskell}{a -> b}.\n\\end{example}\n\n\\begin{example}[Type algebra]\n\\cref{ex:exponential_hask} gives interesting results with types\nmanipulations. For instance the type $a^{b+c}$ can be written as\n\\begin{minted}{haskell}\nEither b c -> a\n\\end{minted}\nfor the function we should have both functions \n\\mintinline{haskell}{b -> a} and \\mintinline{haskell}{b -> c}. I.e.\nthe code is equivalent to the following one\n\\begin{minted}{haskell}\n(b -> a, c -> a)\n\\end{minted}\nThese transformations correspond to the following simple algebraic\nequation \n\\[\na^{b+c} = a^b a^c.\n\\]\nThis is also called as \\textit{type algebra}.\n\\end{example}\n\n\\subsection{\\textbf{C++} category}\n\\begin{example}[Initial object][\\textbf{C++}]\n\\label{ex:cpp_initial_object}\nIn C++ exists a special type that does not hold any values and as\nresult cannot be created: \\mintinline{c++}{void}.\nYou cannot create an object of that type i.e. you will get a compiler\nerror if you try.\n\\end{example}\n\n\\begin{example}[Terminal object][\\textbf{C++}]\n\\label{ex:cpp_terminal_object}\nC++ 17 introduced a special type that keeps only one value -\n\\mintinline{c++}{std::monostate}: \n\\begin{minted}{c++}\nnamespace std {\n struct monostate {};\n}\n\\end{minted}\n\\end{example}\n\n\\begin{example}[Product][\\textbf{C++}]\n\\label{ex:cpp_product}\nThe \\mynameref{def:product} in \\mynameref{def:cppcategory} keeps a\npair and the constructor defined as follows\n\\begin{minted}{c++}\nnamespace std {\n template< class A, class B > struct pair {\n A first;\n B second;\n };\n}\n\\end{minted}\n\nThere is a simple usage example\n\\begin{minted}{c++}\n std::pair p(0, false);\n\n std::cout << \"First projector: \" << p.first << std::endl;\n std::cout << \"Second projector: \" << p.second << std::endl;\n\n\\end{minted}\nReally any \\mintinline{c++}{struct} or \\mintinline{c++}{class} can be\nconsidered as a product. \n\\end{example}\n\n\\begin{example}[Sum][\\textbf{C++}]\n \\label{ex:cpp_sum}\n If we consider \\mynameref{def:object}s as types then\n \\mynameref{def:sum} is an object that can be either one or another\n type. The corresponding C/C++ construction that provides an ability\n to keep one of two types is \\mintinline{c++}{union}. \n\n %% There is an \\mintinline{haskell}{Either} implementation from\n %% \\mynameref{ex:hask_sum} \n %% \\begin{minted}{c++}\n %% template class Either\n %% {\n %% public:\n %% Either(const Either& e) : is_left_(e.is_left_){\n %% if (is_left_){\n %% data.l = e.data.l; \n %% } else {\n %% data.r = e.data.r; \n %% } \n %% }\n %% ~Either(){\n %% if (is_left_){\n %% data.l.~A(); \n %% } else {\n %% data.r.~B(); \n %% } \n %% } \n %% Either(const A& l) : data(l), is_left_(true){\n %% }\n %% Either(const B& r) : data(r), is_left_(false){\n %% }\n %% const A& left() const {\n %% if (!is_left_){\n %% throw std::logic_error(\"no left\"); \n %% } \n %% return data.l; \n %% }\n %% const B& right() const {\n %% if (is_left_){\n %% throw std::logic_error(\"no right\"); \n %% } \n %% return data.r; \n %% }\n %% private:\n %% union Data {\n %% Data() {}\n %% Data( const A& a) : l(a) {}\n %% Data (const B& b) : r(b) {}\n %% ~Data() {}\n %% A l;\n %% B r;\n %% } data; \n %% bool is_left_; \n %% };\n %% \\end{minted}\n %% The usage example:\n %% \\begin{minted}{c++}\n %% template \n %% auto factor(A f, B g, const Either& either) {\n %% try {\n %% return f(either.left()); \n %% }\n %% catch(...) {\n %% return g(either.right()); \n %% }\n %% };\n \n %% auto stringLength = [](std::string s) {\n %% return static_cast(s.size()); };\n %% auto id = [](auto x) { return x; };\n \n %% Either str = std::string(\"abc\");\n %% std::cout << \"String length:\" <<\n %% factor<>(stringLength, id, str) << std::endl;\n %% Either i = 4;\n %% std::cout << \"id(int):\" <<\n %% factor<>(stringLength, id, i) << std::endl;\n %% \\end{minted}\n\n C++17 suggests \\mintinline{c++}{std:variant} as a safe replacement\n for \\mintinline{c++}{union}. The example of the \\mintinline{c++}{factor}\n function is below\n \\begin{minted}{c++}\n template \n auto factor(A f, B g, const std::variant& either) {\n try {\n return f(std::get(either)); \n }\n catch(...) {\n return g(std::get(either)); \n }\n };\n \\end{minted}\n The simple usage as follows:\n \\begin{minted}{c++}\n std::variant var = std::string(\"abc\");\n std::cout << \"String length:\" <<\n factor<>(stringLength, id, var) << std::endl;\n var = 4;\n std::cout << \"id(int):\" <<\n factor<>(stringLength, id, var) << std::endl; \n \\end{minted}\n\n\\end{example}\nTBD\n\\subsection{\\textbf{Scala} category}\n\\begin{example}[Initial object][\\textbf{Scala}]\n\\label{ex:scala_initial_object}\nWe used a same trick as for \\mynameref{ex:hask_initial_object} in\n\\mynameref{def:haskcategory} and define\n\\mynameref{def:initial_object} in \\mynameref{def:scalacategory} as a\ntype without values \n\\begin{minted}{scala}\nsealed trait Void\n\\end{minted}\ni.e. you cannot construct a object of the type.\n\\end{example}\n\n\\begin{example}[Terminal object][\\textbf{Scala}]\n\\label{ex:scala_terminal_object}\nWe used a same trick as for \\mynameref{ex:hask_terminal_object}\nin \\mynameref{def:haskcategory} and define\n\\mynameref{def:terminal_object} in \\mynameref{def:scalacategory} as a\ntype with only one value\n\\begin{minted}{scala}\nabstract final class Unit extends AnyVal\n\\end{minted}\nTBD\ni.e. you can create only one element of the type.\n\\end{example}\n\nTBD\n\n\\section{Quantum mechanics}\n\\begin{example}[Initial object][\\textbf{FdHilb}]\n\\label{ex:quant_initial_object}\nWe will use a Hilber space of dimensional 0 as the\n\\mynameref{def:initial_object}. I.e. the set that does not have any\nstates in it.\n\\end{example}\n\n\\begin{example}[Terminal object][\\textbf{FdHilb}]\n \\label{ex:quant_terminal_object}\nWe will use a Hilber space of dimensional 1 as the\n\\mynameref{def:terminal_object}. I.e. the set of complex numbers\n$\\mathbb{C}$. \n\\end{example}\n\n\\begin{example}[Product][\\textbf{FdHilb}]\n \\label{ex:quant_product}\n The \\mynameref{def:product} in \\mynameref{def:fdhilbcategory} is a\n \\mynameref{def:fdhilb_direct_sum}.\n\\end{example}\n\n\\begin{example}[Sum][\\textbf{FdHilb}]\n \\label{ex:quant_sum}\n The \\mynameref{def:sum} in \\mynameref{def:fdhilbcategory} is a\n \\mynameref{def:fdhilb_direct_sum}.\n\\end{example}\n\nTBD\n\n", "meta": {"hexsha": "ced9944914b23a310e987a008f1c3674e1b99da9", "size": 35577, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cattheory/objectsmorphisms.tex", "max_stars_repo_name": "ivanmurashko/articles", "max_stars_repo_head_hexsha": "522db3ad21e96084490acd39a146a335763e5beb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-09-27T08:59:55.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-27T08:59:55.000Z", "max_issues_repo_path": "cattheory/objectsmorphisms.tex", "max_issues_repo_name": "ivanmurashko/articles", "max_issues_repo_head_hexsha": "522db3ad21e96084490acd39a146a335763e5beb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cattheory/objectsmorphisms.tex", "max_forks_repo_name": "ivanmurashko/articles", "max_forks_repo_head_hexsha": "522db3ad21e96084490acd39a146a335763e5beb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.5743440233, "max_line_length": 84, "alphanum_fraction": 0.67338449, "num_tokens": 11943, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680199891789, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.6374658228443074}} {"text": "\\section{Change of Bases, Canonical Forms and Symmetries}\r\n\\subsection{Change of Bases in General}\r\nRecall that given a linear operator $T:V\\to W$, we can write its matrix form once we have specified the bases $\\{\\underline{e_1},\\underline{e_2},\\ldots,\\underline{e_n}\\}$ and $\\{\\underline{f_1},\\underline{f_2},\\ldots,\\underline{f_m}\\}$ for $V,W$.\r\nSo $T(\\underline{e_i})=A_{ai}\\underline{f_a}$ for some matrix $A$,\r\nFor other bases $\\{\\underline{e_i'}\\}$ and $\\{\\underline{f_i'}\\}$ for $V,W$, we have a matrix $A'$ with $T(\\underline{e_i'})=A'_{ai}\\underline{f_a'}$.\r\nSuppose the bases are related by $\\underline{e_i'}=P_{ji}\\underline{e_j}, \\underline{f_a'}=Q_{ba}\\underline{f_b}$.\r\nSo $P,Q$ are invertible and\r\n\\begin{proposition}\r\n $A'=Q^{-1}AP$.\r\n\\end{proposition}\r\nWe say $P,Q$ constitutes a change of bases.\r\nThe column $i$ of $A$ consists of components of $T(\\underline{e_i})$ with respect to the basis $\\{\\underline{f_a}\\}$, column $i$ of $P$ consists of components of new basis vectors with respect to the old basis, similar for $Q$.\r\n\\begin{example}\r\n Let $\\dim V=n=2$, $\\dim W=m=3$, and consider the linear map $T$ with $T(\\underline{e_1})=\\underline{f_1}+2\\underline{f_2}-\\underline{f_3}$ and $T(\\underline{e_2})=-\\underline{f_1}+2\\underline{f_2}+\\underline{f_3}$, so $T$ has matrix\r\n $$A=\\begin{pmatrix}\r\n 1&-1\\\\\r\n 2&2\\\\\r\n -1&1\r\n \\end{pmatrix}$$\r\n Under these choice of bases.\r\n Now consider the new basis $\\underline{e_i'}$ for $V$ by\r\n $$\\underline{e_1'}=\\underline{e_1}-\\underline{e_2},\\underline{e_2'}=\\underline{e_1}+\\underline{e_2}$$\r\n which yields a changes of basis matrix\r\n $$P=\\begin{pmatrix}\r\n 1&1\\\\\r\n -1&1\r\n \\end{pmatrix}$$\r\n And a new basis $\\underline{f_i'}$ for $W$ is defined by $$\\underline{f_1'}=\\underline{f_1}-\\underline{f_3},\\underline{f_2'}=\\underline{f_2},\\underline{f_3'}=\\underline{f_1}+\\underline{f_3}$$\r\n which has a matrix\r\n $$Q=\\begin{pmatrix}\r\n 1&0&1\\\\\r\n 0&1&0\\\\\r\n -1&0&1\r\n \\end{pmatrix}$$\r\n So the change of basis formula gives\r\n $$A'=Q^{-1}AP=\\begin{pmatrix}\r\n 2&0\\\\\r\n 0&4\\\\\r\n 0&0\r\n \\end{pmatrix}$$\r\n which can be verified by direct calculation.\r\n\\end{example}\r\n\\begin{proof}\r\n For vectors $\\underline{y},\\underline{x}$ with $\\underline{y}=A\\underline{x}$ for some invertible matrix (representing a basis) $A$, we denote the components in the respective components by\r\n $$X=\\begin{pmatrix}\r\n x_1\\\\\r\n x_2\\\\\r\n \\vdots\\\\\r\n x_n\r\n \\end{pmatrix},Y=\\begin{pmatrix}\r\n y_1\\\\\r\n y_2\\\\\r\n \\vdots\\\\\r\n y_n\r\n \\end{pmatrix}$$\r\n to avoid confusion.\r\n So for a change of basis by matrix $P,Q$, the formula $Y=AX$ turns to $Y'=A'X'$, hence, we have $\\underline{x}=x_i\\underline{e_i}=x_j'\\underline{e_j'}=x_j'(\\underline{e_i}P_{ij})=P_{ij}x_j'\\underline{e_i}$.\r\n Therefore $X=PX'$, similarly $Y=QY'$, so $A'X'=Y'=Q^{-1}Y=Q^{-1}AX=(Q^{-1}AP)X'$, so $A'=Q^{-1}AP$\r\n\\end{proof}\r\nIn the special cases where $V=W$ and the bases are the same for $V,W$ before and after the change of basis, then $A$ is transformed to a matrix similar to $A$ by that change-of-basis matrix.\r\nNote that this can justify immediately that $\\operatorname{tr}A'=\\operatorname{tr}A,\\det A'=\\det A,\\chi_{A'}=\\chi_A$ for similar matrices $A,A'$.\\\\\r\nIf $V=W=\\mathbb F^n$ where $\\mathbb F=\\mathbb R$ or $\\mathbb C$.\r\nFor a diagonalizable matrix $M$, then by changing the standard basis into the eigenbasis, $M$ becomes diagonal.\r\n\\begin{proof}[Alternative proof]\r\n Take the linear map $T$, then\r\n $$\\underline{f_a}Q_{ab}A'_{bi}=\\underline{f_b'}A'_{bi}=T(\\underline{e_i'})=T(\\underline{e_j}P_{ji})=P_{ji}T(\\underline{e_j})=\\underline{f_a}A_{aj}P_{ji}$$\r\n So $AP=QA'$ by considering the coefficient of $\\underline{f_a}$.\r\n\\end{proof}\r\n\\subsection{Jordan Canonical/Normal Form}\r\nThis result classifies complex $n\\times n$ matrices up to similarity (i.e. conjugacy classes).\r\n\\begin{proposition}\r\n Any $2\\times 2$ complex matrix $A$ is similar to one of the followings:\r\n $$\r\n \\begin{pmatrix}\r\n \\lambda_1&0\\\\\r\n 0&\\lambda_2\r\n \\end{pmatrix},\\lambda_1\\neq\\lambda_2;\r\n \\begin{pmatrix}\r\n \\lambda&0\\\\\r\n 0&\\lambda\r\n \\end{pmatrix};\r\n \\begin{pmatrix}\r\n \\lambda&1\\\\\r\n 0&\\lambda\r\n \\end{pmatrix}\r\n $$\r\n Where $\\lambda_1,\\lambda_2,\\lambda\\in\\mathbb C$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n If $A$ is diagonalizable (which includes the case where $\\chi_A$ has two distinct roots) then the proposition is immediate.\r\n Otherwise, $\\chi_A(t)=c(t-\\lambda)^2,c\\neq 0$ and $\\operatorname{null}(A-\\lambda I)=1$ and we shall show that $A$ is similar to a matrix of the third form.\r\n Indeed, let $\\underline{v}$ be an eigenvector for $\\lambda$ and $\\underline{w}$ any vector that is independent from $\\underline{v}$.\r\n Then $A\\underline{v}=\\lambda\\underline{v},A\\underline{w}=\\alpha\\underline{v}+\\beta\\underline{w}$, then the matrix of this transformation under the basis $\\underline{v},\\underline{w}$ is\r\n $$\\begin{pmatrix}\r\n \\lambda&\\alpha\\\\\r\n 0&\\beta\r\n \\end{pmatrix}$$\r\n But then $\\beta=\\lambda$ by looking at the eigenvalue of matrices of this form, and $\\alpha\\neq 0$ by assumption.\r\n So we can set $\\underline{u}=\\alpha\\underline{v}$, then under the basis $\\underline{u},\\underline{w}$, we get the matrix of $A$ to be\r\n $$\\begin{pmatrix}\r\n \\lambda&1\\\\\r\n 0&\\lambda\r\n \\end{pmatrix}$$\r\n As claimed.\r\n\\end{proof}\r\n\\begin{proof}[Alternative approach for the third case]\r\n If $A$ has characteristic polynomial of the form $\\chi_A(t)=c(t-\\lambda)^2$ with $c\\neq 0$ and $A\\neq \\lambda I$, then there is some vector $\\underline{w}$ with\r\n $$(A-\\lambda I)\\underline{w}=\\underline{u}\\neq\\underline{0}$$\r\n but by Cayley-Hamilton, we have\r\n $$(A-\\lambda I)\\underline{u}=(A-\\lambda I)^2\\underline{w}=\\underline{0}$$\r\n so $A$ is of the form under the basis $\\underline{u},\\underline{w}$.\r\n\\end{proof}\r\n\\begin{example}\r\n Consider\r\n $$A=\\begin{pmatrix}\r\n 1&4\\\\\r\n -1&5\r\n \\end{pmatrix}$$\r\n Then $\\lambda=3$ and we can have $\\underline{w}=(1,0)^\\top$, $\\underline{u}=(-2,-1)^\\top$, therefore\r\n $$\\begin{pmatrix}\r\n 3&1\\\\\r\n 0&3\r\n \\end{pmatrix}=\r\n \\begin{pmatrix}\r\n -2&1\\\\\r\n -1&0\r\n \\end{pmatrix}^{-1}\r\n \\begin{pmatrix}\r\n 1&4\\\\\r\n -1&5\r\n \\end{pmatrix}\r\n \\begin{pmatrix}\r\n -2&1\\\\\r\n -1&0\r\n \\end{pmatrix}$$\r\n\\end{example}\r\n\\begin{theorem}\r\n Any $n\\times n$ complex matrix is similar to a matrix of the form\r\n $$A'=\r\n \\begin{pmatrix}\r\n J_{n_1}(\\lambda_1)&&\\\\\r\n &\\ddots&\\\\\r\n &&J_{n_r}(\\lambda_r)\r\n \\end{pmatrix}$$\r\n where $\\lambda_1,\\ldots,\\lambda_r$ are the eigenvalues, $n_1+n_2+\\cdots+n_r=n$ and $J_p(\\lambda)$, the Jordan Block, is of the form\r\n $$J_p(\\lambda)=\\begin{pmatrix}\r\n \\lambda&1&&&\\\\\r\n &\\lambda&1&&\\\\\r\n &&\\lambda&&\\\\\r\n &&&\\ddots&1\\\\\r\n &&&&\\lambda\r\n \\end{pmatrix}$$\r\n\\end{theorem}\r\n\\subsection{Quadrics and Conics}\r\nA quadric in $\\mathbb R^n$ is a hypersurface defined by\r\n$$Q(\\underline{x})=\\underline{x}^\\top A\\underline{x}+\\underline{b}^\\top\\underline{x}+c=0$$\r\nFor some real symmetric $n\\times n$ matrix $A$, $\\underline{b}\\in \\mathbb R^n,c\\in\\mathbb R$.\r\nSo $Q(\\underline{x})=A_{ij}x_ix_j+b_ix_i+c$.\\\\\r\nConsider classifying solutions up to geometrical equivalence, i.e. no distinction up to isometries.\r\n\\begin{theorem}\r\n Any isometry in $\\mathbb R^n$ is a composition of translation and orthogonal transformation.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\nIf $A$ in invertible, we can complete the square.\r\nWe set $\\underline{y}=\\underline{x}+A^{-1}\\underline{b}$, so $\\mathcal{F}(\\underline{y})=\\underline{y}^\\top A\\underline{y}=Q(\\underline{x})-k$ for some constant $k$.\r\nSo $Q(\\underline{x})=0\\iff \\mathcal F(\\underline{y})=k$ for some $k$, but the quadratic form is with respect to a new origin.\r\nIn this case, we can diagonalize $\\mathcal F$ and change the basis as appropriate and think of them as new coordinate (bases).\r\nThe case in $\\mathbb R^3$ has been discussed in the previous sections and we see that the quadrics are ellipsoids and hyperboloids (in $2$ directions).\\\\\r\nIf $A$ is singular, that is, if $A$ has one or more $0$ eigenvalue, things change.\\\\\r\nBack in $\\mathbb R^2$, if $A$ is invertible, then by our analysis above, we can write it as a (diagonalized) quadratic form $k=\\lambda_1x_1^2+\\lambda_2x_2^2$.\r\nFor $\\lambda_1,\\lambda_2>0$, $k>0$ gives an ellipse, $k=0$ gives a point and $k<0$ no solution.\r\nFor $\\lambda_1\\lambda_2<0$, so WLOG $\\lambda_1>0>\\lambda_2$, then we get a hyperbola for any $k\\neq 0$, and a pair of lines for $k=0$.\\\\\r\nIf $\\det A=0$, then unless $A$ is zero, we have $\\lambda_1\\neq 0$ and $\\lambda_2=0$, so we diagonaize $A$ in the original formula to get\r\n$$\\lambda_1x_1'^2+b_1'x_1'+b_2'x_2'+c=0\\iff \\lambda_1x_1''^2+b_2'x_2'+c'=0$$\r\nBy a shift in $x_1$.\r\nIf $b_2'=0$, then we again get a pair of lines for $c'<0$, a single line for $c'=0$ and no solution for $c'>0$.\r\nOtherwise $b_2'\\neq 0$, it gives a parabola.\\\\\r\nNote that all these changes of coordinates is prepresented by an isometry.\r\nThey are called conics because these shapes can be obtained by slicing the cone.\r\nThe general form for conics can be expressed in terms of $a,b$ semi-major and semi-minor axes or in terms of length unit $\\ell$ and eccentricity $e$.\\\\\r\nFor ellipses, we have the general form\r\n$$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$$\r\nWLOG $b>a$, then $b^2=(1-e^2)a^2$ for some $0\\le e<1$.\r\nSo we take this $e$ to be the eccentricity.\r\nSo the foci are at $x=\\pm ae$.\\\\\r\nAnd for parabola in the form $y^2=4ax$, the focus is at $x=a$ and $e=1$.\\\\\r\nFor hyperbola, i.e.\r\n$$\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1$$\r\nSo for $b^2=a^2(e^2-1)$ for $e>1$ the eccentricity.\r\nThe foci are $x=\\pm ae$.\r\n\\subsection{Symmetries and Transformation Groups}\r\nRecall from a previous section that $R$ is orthogonal iff $R^\\top R=RR^\\top=I$ iff it preserves dot products.\r\nThe set of such matrices is a group $\\operatorname{O}(n)$.\r\nAlso, given the properties above and the multiplictivity of $\\det$, $\\det R=\\pm 1$.\r\nAnd $\\operatorname{SO}(n)\\le \\operatorname{O}(n)$ is the set of orthogonal matrices with determinant $1$.\r\nWhile orthogonal matrices preserves lengths, angles and volumns (by alternating forms), special orthogonal matrices also preserves orientation.\r\nReflections belongs in $\\operatorname{O}(n)\\setminus\\operatorname{SO}(n)$.\r\nFor a specific $H\\in \\operatorname{O}(n)\\setminus\\operatorname{SO}(n)$, any element in $\\operatorname{O(n)}$ is of the form $R$ or $RH$ for some $R\\in\\operatorname{SO}(n)$.\r\nFor example, if $n$ is odd, then we can take $H=-I$\r\nWe can regard the transformation $x_i'=R_{ij}x_j$ in two ways:\\\\\r\nThe ``active'' way is to say the rotations transform vectors.\r\nThen $\\operatorname{SO}(n)\\star\\underline{x}$ would be a hypersphere.\r\nThe components $x_i'$ are components of new vector.\\\\\r\nBy contrast, the ``passive'' point of view is to think about the basis vectors and the vectors that they are mapped to, so $x_i'$ are the components of the same vector $\\underline{x}$ wrt a new (rotated) orthonormal basis.\r\n\\subsection{2-Dimensional Minkowski Space and Lorentz Transformations}\r\nConsider a new ``inner product'' in $\\mathbb R^2$ given by\r\n$$(\\underline{x},\\underline{y})=x^\\top J\\underline{y},J=\\begin{pmatrix}\r\n 1&0\\\\\r\n 0&-1\r\n\\end{pmatrix}$$\r\nAnd also we write $\\underline{x}=(x_0,x_1)^\\top$ and $\\underline{y}=(y_0,y_1)^\\top$.\r\nNote that all original properties of inner products hold except positive definiteness.\r\nWe can choose basis vectors that are orthonormal, which are the standard basis $\\underline{e_0},\\underline{e_1}$ with $(\\underline{e_0},\\underline{e_0})=1=-(\\underline{e_1},\\underline{e_1}), (\\underline{e_0},\\underline{e_1})=0$.\r\n\\begin{definition}\r\n This ``new inner product'' is called the Minkowski metric, and $\\mathbb R^2$ equipped with it is called Minkowski space.\r\n\\end{definition}\r\nConsider\r\n$$M=\\begin{pmatrix}\r\n M_{00}&M_{01}\\\\\r\n M_{10}&M_{11}\r\n\\end{pmatrix}$$\r\nwhich preserves Minkowski metric if and only if $(M\\underline{x},M\\underline{y})=(\\underline{x},\\underline{y})$ if and only if\r\n$$\\underline{x}^\\top M^\\top JM\\underline{y}=\\underline{x}^\\top J\\underline{y}$$\r\nfor any $\\underline{x},\\underline{y}\\in\\mathbb R^2$, which happens iff $M^\\top JM=J$.\r\nBy taking determinant, $\\det(M)=\\pm 1$.\r\nSo all such $M$ constitutes a group and has a subgroup with $\\det M=1$ and $M_00>0$ is the Lorentz group.\\\\\r\nTo determine the general form of $M$ in Lorentz group, we can find it by requiring $M\\underline{e_0},M\\underline{e_1}$ to be orthonormal in this generalized sense, so $M$ is in the general form\r\n$$M(\\theta)=\\begin{pmatrix}\r\n \\cosh\\theta&\\sinh\\theta\\\\\r\n \\sinh\\theta&\\cosh\\theta\r\n\\end{pmatrix}$$\r\nFixed by $(M\\underline{e_0},M\\underline{e_0})=1$ and $M_{00}>0$.\r\nNote that $M(\\theta)M(\\phi)=M(\\theta+\\phi)$.\\\\\r\nCurves with $(\\underline{x},\\underline{x})=k$ with $k$ constant are simply hyperbolic.\r\nSince lengths are preserve, $\\underline{x'}=M\\underline{x}$ must lie on the same hyperbola.\\\\\r\nPhysically, we want to set\r\n$$M(\\theta)=\\gamma(v)\\begin{pmatrix}\r\n 1&v\\\\\r\n v&1\r\n\\end{pmatrix}$$\r\nwhere $v=\\tanh\\theta$, then $|v|<1$ and $\\gamma(v)=(1-v^2)^{-1/2}$\r\nThen (with $x_0=t$ the time dimension and $x_1=x$ the space parameter)\r\n$$\\begin{cases}\r\n t'=\\gamma(v)(t+vx)\\\\\r\n x'=\\gamma(v)(x+vt)\r\n\\end{cases}$$\r\nThis is the Lorentz Transformation, or boost, relating time and space coordinates for observers moving with relative velocity $v<1$.\r\nThe $\\gamma$ factor here implies time dilation and length contraction.\r\nNote that $M(\\theta_3)=M_{\\theta_1}M_{\\theta_2}$ and $v_i=\\tanh(\\theta_i)$ gives $v_3=\\frac{v_1+v_2}{1+v_1v_2}$ which is consistent with $v<1$.", "meta": {"hexsha": "2c89a26b76abb298b12c83ef917eeb7706b7406b", "size": 13885, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "7/bases.tex", "max_stars_repo_name": "david-bai-notes/IA-Vectors-and-Matrices", "max_stars_repo_head_hexsha": "7fc43486ec5276262d4058c9daaea12affdb6bac", 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YES\n2. YES", "lm_q1_score": 0.743167997235783, "lm_q2_score": 0.8577681122619885, "lm_q1q2_score": 0.6374658100824602}} {"text": "\\section{Theory}\n\nIn the following section, equations and algorithms used to model the behavior of\na symmetric ideal gas centrifuge cascade from individual centrifuge properties\nare described. Details will also be provided to model centrifuge behavior\nwhen fed with a different feed enrichment than designed.\n\n\\subsection{Centrifuge properties}\n\n\\subsubsection{Separative power}\n\n\\paragraph{R\\\"aetz equation}\n\nAs described by Glaser in \\cite{glaser.2008}, the separative power of a single centrifuge can be express as an analytical solution \\cite{raetz.phd} of the\ndifferential equation for the gas centrifuge:\n\n\\begin{eqnarray}\n \\label{eq:raetz}\n \\delta U(L,F,\\theta,Z_{p}) &= &\\frac{1}{2}\n F\\theta(1 - \\theta)\n \\left(\\frac{\\Delta M}{2 RT}v_{a}^{2}\\right)^{2}\n \\left(\\frac{r_{2}}{a}\\right)^{4}\n \\left[1 - \\left(\\frac{r_{1}}{r_{2}}\\right)^{2}\\right]^{2}\n \\label{eq_raetz}\\\\\n &&\n \\left[\n \\left(\\frac{1+L/F}{\\theta}\\right)\n (1- exp[ - A_{P}(L,F,\\theta)Z_{p}]) \\nonumber \\right. \\\\\n &&~~ + \\left.\\left(\\frac{L/F}{1 - \\theta}\\right)\n (1 - exp[ -A_{W}(L,F,\\theta)(Z - Z_{p}])\\right]^{2}, \\nonumber\n \\\\\n \\textrm{with~ ~ ~ ~ ~}\n A_{P} &= &\\frac{2\\pi D\\rho }{ ln(r_{2}/r_{1}) }\n \\frac{ 1 }{ F }\n \\frac{ 1-\\theta }{(1+L/F)(1-\\theta+L/F) }\\\\\n A_{W} &= &\\frac{2\\pi D\\rho }{ ln(r_{2}/r_{1}) }\n \\frac{ 1 }{ F }\n \\frac{ 1-\\theta }{ (L/F)(1-\\theta+L/F) }\n\\end{eqnarray}\n\nIn this equation, the parameters of average gas temperature, $T$, peripheral speed,\n$v_a$, height, $h$, diameter, $d$, pressure ratio, $x$, feed flow rate, $F$,\ncounter-current flow ratio, $L/F$, are intrinsic to the centrifuge design.\nTo match the cascade design describe in \\cite{glaser.2008} and \\cite{walker.2017},\nP1-type centrifuge properties have been chosen (Table \\ref{tab:centrifuges}).\n\n\\begin{table}[htb]\n \\centering\n \\caption{Summary of the centrifuge parameters.}\n \\begin{tabular}{ccccccc}\n \\toprule\n $T$[K] & $v$[m/s] & $h$[m] & $d$[m] & $x$ & $F$[mg/s] & $L/F$ \\\\\n \\midrule\n 320 & $320$ & $1.8$ & $0.105$ & $10^{3}$ & $13$ & 2 \\\\\n \\bottomrule\n \\end{tabular}\n \\label{tab:centrifuges}\n\\end{table}\n\nThe variable $Z_p$ is the rectifier length, or the location of the feed point,\nand has an optimal axial location as defined by \\cite{raetz.phd}:\n\n\\begin{equation}\n Z_p = \\frac{(1-\\theta)(1+L/F)}{1-\\theta+L/F}Z\n\\end{equation}\n\nThis optimizes the recitifer length based on the cut, $\\theta$, which is an expression\nof the fraction of the centrifuge feed that is output as product, and the counter-current\nflow, $L/F$. In practice, this value is a design parameter that is defined in the model\nduring the design of the centrifuge cascade.\n\nThe parameters $r_1$ and $r_2$ are the separation radii of the enriched material\n(here ${}^{235}\\mathrm{U}$ vs. ${}^{238}\\mathrm{U}$), $r_1$ being the withdrawal\nradius for the lighter isotope and $r_2$ for the heavier isotope. R\\\"aetz's\ntwo-shell model looks for optimal values between these two, as described by\nthe hydrodynamic equations. The radii ratio is optimized using the following relationship:\n\n\\begin{equation}\n \\max \\left\\{ \\left[1-\\left(\\frac{r_1}{r_2}\\right)^2 \\right]^2 \\times \\left[ \\ln \\left(\\frac{r_2}{r_1} \\right) \\right]^{-1} \\right\\}\n\\end{equation}\n\nThis ratio can further be constrained by approximating $r_2$ as being very close\nto the inner centrifuge wall with radius $a$:\n\n\\begin{equation}\n \\left(\\frac{r_1}{r_2}\\right) \\approx \\left(\\frac{r_1}{a}\\right) = \\sqrt{1 - \\frac{2RT}{M}(\\ln x)\\frac{1}{v_{a}^{2}}}\n\\end{equation}\n \nHere the gas constants are molar mass, $M$, temperature $T$, universal gas constant, $R$,\nand the pressure ratio, $x$ (typically $1000:1$) \\cite{}. %needs citation?\nThis relationship is valid when $v_a > 380 \\frac{\\mathrm{m}}{\\mathrm{s}}$.\nOtherwise, the relationship can be approximated $\\{\\frac{r_1}{r_2} \\approx 0.534 \\mid v_a \\leq 380 \\frac{\\mathrm{m}}{\\mathrm{s}}\\}$.\nIn order to decompose the ratio into each individual radius, knowledge on one\nof them is required. Glaser \\cite{glaser.2008} states that $r_2$ typically ranges\nfrom $96\\%$ to $99\\%$ the value of the inner centrifuge wall radius $a$. The exact\nbehavior of $r_2$ between these two values in operational conditions is unknown,\nbut a linear approximation can be made. It is assumed that $r_2$ is always at\nthe midpoint of these two extremes (i.e. $r_2 = 0.975 a$). Then, $r_1$ can be\nfound by simply multiplying this value by the ratio.\n\nWith this, all parameters of the separative power equation \\ref{eq:raetz} are\ndefined and a value can be calculated. Separative power, $\\delta U$, has units similar to the feed flow, that of $[mg/s]$.\n\n\\paragraph{First principle}\n\nIt can be shown \\cite{avery} that the separative power of a single centrifuge can be\nderived from first principles and expressed as a function of the feed-to-product enrichment factor, $\\alpha$, the cut, $\\theta$, and the feed rate, $F$:\n%I think we should replicate at least some of that here\n\n\\begin{equation} \\label{eq_alpha_principle}\n \\delta U = \\frac{F}{2}\\frac{\\theta}{1-\\theta}(\\alpha-1)^{2}\n\\end{equation}\n\n\\subsection{Centrifuges basic properties and definition}\n\nThe outputs of a centrifuge relative to its input can be described with ratios\nof the abundance ($R = \\frac{N}{1-N}$) where the feed, product, and tail enrichment $N, N', N''$ respectively. Enrichment factors of $\\alpha$ (feed-to-product),\n$\\beta$ (feed-to-tail), and $\\gamma$ (tail-to-product) can then be defined:\n\n\\begin{subequations}\n \\label{eq_alphabeta}\n \\begin{equation} \\label{eq_alpha_def}\n \\alpha = \\frac{1-N}{N}\\frac{N'}{1-N'}\n \\end{equation}\n \\begin{equation}\\label{eq_beta-def}\n \\beta = \\frac{1-N''}{N''}\\frac{N}{1-N}\n \\end{equation}\n \\begin{equation}\\label{eq_gamma-def}\n \\gamma = \\alpha \\, \\beta\n \\end{equation}\n\\end{subequations}\n\n\n\\subsection{Cascade Design}\n\\subsubsection{Symmetric Cascade}\n\nA symmetric cascade is a cascade where a stage is fed using the tail, $T$, of the next stage and the product, $P$, of the previous one. The the cascade feeding stage, $F_{i}$ flow, with external feed, $F_{ext}$, is given by:\n\n\\begin{equation}\n F_{i} = T_{i+1} + F_{i-1} ~(+ F_{ext})\n\\end{equation}\n\n\\subsubsection{Symmetric Ideal Cascade}\nThis model constructs cascades as symmetrical and ideal, with no losses in\nseparative work. This means that the tail assay of the next stage ($N''_{i+1}$)\nis the product assay of the previous stage ($N'_{i-1}$), which can be\nexpressed as:\n\n\\begin{equation}\n \\forall i~ N_{i} = N'_{i-1} = N''_{i+1}~ ~\\Leftrightarrow~ ~\\forall (i,j)~\n \\alpha_{i} = \\beta_{j}\n\\end{equation}\n\n\n\n\\subsection{Building the cascade}\n\nDesigning a symmetric ideal cascade beings with the feeding\nstage. As all the enrichment factors are equal across all the cascade, the\nfeeding stage is used to determine all subsequent stages.\nThe feed assay of the feeding stage, $N_{0}$, is fixed by the external feed assay\nprovided as an input.\n\nFrom equation \\eqref{eq_raetz} and \\eqref{eq_alpha_principle} it is possible to\nexpress the feed-to-product enrichment factor $\\alpha$ as a function of the feed rate $F$, the separative performance\n$\\delta U$ (a function of $\\theta$), and the cut $\\theta$:\n\n\\begin{equation} \\label{eq_alpha}\n \\alpha = \\sqrt{\\frac{2\\delta U(\\theta)}{F} \\frac{1-\\theta}{\\theta}} + 1\n\\end{equation}\n\n\nFrom equations \\eqref{eq_alphabeta}, the product assay can be expressed as:\n\n\\begin{equation}\\label{eq_product_assay}\n N' = \\frac{\\alpha \\frac{N}{1-N}}{1+\\alpha \\frac{N}{1-N}} = \\frac{\\alpha R}{1 + \\alpha R}\n\\end {equation}\n\n\nThen, from mass conservation, $N = \\theta N' + (1-\\theta)N''$ and equation\n\\eqref{eq_product_assay}, it is possible to express the feed-to-tail enrichment factor $\\beta$ as a function of only the feed abundance, $R$, the cut $\\theta$, and the feed-to-product enrichment factor, $\\alpha$:\n\n\\begin{subequations}\n \\begin{equation}\\label{eq_beta_interim}\n \\beta = \\left( 1 - \\frac{N - \\theta N'}{1-\\theta} \\right)\n \\left( \\frac{R}{\\frac{N - \\theta N'}{1 - \\theta}} \\right)\n \\end{equation}\n \\begin{equation}\\label{eq_beta}\n \\beta = R \\left(\\dfrac{1-\\theta}\n {\\dfrac{R}{R+1}- \\theta \\dfrac{\\alpha R}{1+\\alpha R}} -1\\right)\n \\end{equation}\n\\end{subequations}\n\nFinally from equation \\eqref{eq_alpha} and \\eqref{eq_beta} it is possible to\ndetermine the cut, $\\theta$, required to build an ideal cascade:\n\n\\begin{eqnarray}\n \\theta_{i} = \\dfrac{N_{i} - \\dfrac{1}{1 + \\beta/R_{i}}}{ \\dfrac{\\alpha R_{i}}{1 + \\alpha R_{i}} -\n \\dfrac{1}{1 + \\beta/R_{i}}}\n \\label{eq_theta}\n\\end{eqnarray}\n\n\nTo construct an ideal stage, a cut, $\\theta$, must first be computed.\nThis is found iteratively, searching for the optimal cut value where $\\alpha = \\beta$\nfor given feed and centrifuge parameters. With a separative power calculated,\nan $\\alpha$ and $\\beta$ value can be determined for an initial cut guess. This\nmodel assumes that the ideal cut for a stage should be between 0.1 and 0.9. The\ntwo enrichment factors are compared, and the higher or lower cut value is chosen\nby which pair of factors are closer. A new cut is determined from the chosen\nfactors, a new separative power is computed, and new $\\alpha$ and $\\beta$ values\nare compared. This process continues until, to some precision, the resulting\nenrichment factors are approximately equal. As illustrated in Fig. \\ref{fig_a_m_b},\nfor a given input feed assay, the ideal feeding stage typically has a $\\theta$\nvalue between 0.45 and 0.525 when $\\alpha = \\beta$.\n\n\\begin{figure}[h!] % replace 't' with 'b' to force it to be on the bottom\n \\centering\n \\includegraphics[scale=0.5]{alpha_minus_beta}\n \\caption{Evolution of the difference between $\\alpha$ and $\\beta$ as a\n function of the cut value for different of feed assays, 0.007 (black),\n 0.035 (red), 0.2 (green), 0.7 (black). }\n \\label{fig_a_m_b}\n\\end{figure}\n\nThe number of machines required to construct a stage can then be computed using\nequation \\eqref{eq_alpha_principle} to solve for the centrifuge feed flow:\n\n\\begin{equation}\\label{eq_cent_feed}\n F_c = \\frac{2 \\delta U}{(\\alpha - 1)^2} \\frac{M}{M_{238}} \\frac{1-\\theta}{\\theta}\n\\end{equation}\n\nWhere the molar mass ratio $\\frac{M}{M_{238}}$ accounts for the molar mass\ndifferences between the feed gas, $UF_6$, and the individual uranium isotopes\nbeing separated. The stage feed flow can then be divided by the individual\ncentrifuge feed flow, equation \\eqref{eq_cent_feed}, to find the exact number of\nmachines needed for the ideal stage. In practice, this number is rounded up to\naccount for fractional machines required.\n\n\nIn a cascade, as the stage feed-to-product enrichment factor $\\alpha_{i}$ and stage feed-to-tail enrichment factor $\\beta_{i}$ remain constant, only the value of\nthe cut, $\\theta_{i}$, changes across the different stages of a cascade. This\nalgorithm assumes that the corresponding separative power $\\delta U$ (not\nre-computed) can be achieved with the chosen centrifuge design by tuning other\noperational parameters such as the rotation speed, counter-current flow\nratio, etc. Once $\\theta_{i}$ is determined, it is possible to compute the\nproduct and the tail assays.\n\n\nThe design of the ideal symmetric cascade is performed through 2 steps. First,\nthe configuration and number of stages is determined, adding stages with product assay $N'_i$ calculated using equation \\ref{eq_product_assay}, until the\nproduct assay of the final stage is greater than or equal to the product targeted assay. The stage tails assay $N''_i$ is calculated similarly, until it is less than or equal to the tails desired assay. This determines the number of enriching and stripping stages as well as their enrichment properties ($N_{i}$, $N'_{i}$, $N''_{i}$,$\\theta_{i}$).\n\n\nThe second step determines the relative flows at each stage, solving the linear\nflow equation, \\eqref{eq_flow}.\nThe cascade can then be populated with actual machines until the maximum number\nof available machines is reached.\n\n\\begin{equation}\n\\resizebox{\\linewidth}{!}{%\n$\\displaystyle\n\\setcounter{MaxMatrixCols}{20}\n\\begin{bmatrix}\n -1 & 1-\\theta_{_{S+1}} & 0 & ... & 0 & 0 & 0 & 0 & 0 & ... & 0 & 0 & 0 \\\\\n\\theta_{_S} & -1 & 1-\\theta_{_{S+2}} & ... & 0 & 0 & 0 & 0 & 0 & ... & 0 & 0 & 0 \\\\\n & & & & & & ... & & & & & & \\\\\n 0 & 0 & 0 & ... & \\theta_{_-2} & -1 & 1-\\theta_{_0} & 0 & 0 & ... & 0 & 0 & 0 \\\\\n 0 & 0 & 0 & ... & 0 & \\theta_{_-1} & -1 & 1-\\theta_{_1} & 0 & ... & 0 & 0 & 0 \\\\\n 0 & 0 & 0 & ... & 0 & 0 & \\theta_{_0} & -1 & 1-\\theta_{_2} & ... & 0 & 0 & 0 \\\\\n & & & & & & ... & & & & & & \\\\\n 0 & 0 & 0 & ... & 0 & 0 & 0 & 0 & 0 & ... & \\theta_{_{E-2}} & -1 & 1-\\theta_{_E} \\\\\n 0 & 0 & 0 & ... & 0 & 0 & 0 & 0 & 0 & ... & 0 & \\theta_{_{E-1}} & -1\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n F_{_{S}} \\\\\n F_{_{S+1}} \\\\\n \\cdots \\\\\n F_{_{-1}} \\\\\n F_{_{0} } \\\\\n F_{_{1} } \\\\\n \\cdots \\\\\n F_{_{E-1}} \\\\\n F_{_{E}}\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n \\cdots \\\\\n 0 \\\\\n F \\\\\n 0 \\\\\n \\cdots \\\\\n 0 \\\\\n 0\n\\end{bmatrix}\n$}\n%\\caption{caption needed!}\n\\label{eq_flow}\n\\end{equation}\n\n\n\n\\subsection{Misuse models}\n\nLittle information is available about optimising an existing enrichment cascade\nthat is being fed with a feed enrichment that does not match the design\nenrichment. Here, 3 different methods will be investigated.\n\nThe first method, A, assumes that no changes are being made to the cascade, i.e $\\delta U$, $F$ and $\\theta$ are fixed across all stages. The second method, B, assumes the cut value $\\theta$ is re-tuned at each stage to maintain the ideal state of the cascade, while $\\alpha$ and $\\beta$ remain fixed. The last method, C, described in \\cite{walker.2017} assumes the tails-to-product enrichment factor $\\gamma$ and the cut $\\theta$ remain constants\n(eq. \\eqref{eq_gamma-def}: $\\gamma = \\alpha\\beta$). Model behaviors and\nassumptions are summarized in Tab. \\ref{tab:models}.\n\n\\begin{table}[htb]\n\\centering\n \\caption{Summary of misuse model properties.}\n\\begin{tabular}{l|ccc}\n\\toprule\n\nModel & A & B & C \\\\\n\\midrule\nConstant parameters & $\\alpha_i, \\theta_i$ & $\\alpha_i=\\beta_i$ & $\\gamma_i=\\alpha_i\\beta_i, \\theta_i$ \\\\\nVarying parameters & $\\beta_i$ & $\\theta_i$ & $\\alpha_i, \\beta_i$ \\\\\nAssays determination & blended & ideal & blended \\\\\nFlow & unchanged & reduced & unchanged \\\\\n\n\\bottomrule\n\\end{tabular}\n \\label{tab:models}\n\\end{table}\n\n\\subsubsection{Model A}\n\nThe tuning method A does not re-optimize $\\theta_i$, keeping the same flow as the\nideal configuration. From equation \\eqref{eq_alpha}, maintaining $\\delta U$ and\n$F$ while $\\theta$ is unchanged implies $\\alpha$ remains unchanged as well.\nAccording to equation \\eqref{eq_beta}, when $\\alpha$ and $\\theta$ are fixed, if\nthe feed assay ($N$) changes, $\\beta$ will change accordingly. This breaks the\nideal status of the cascade, i.e. $N_{i} \\neq N'_{i-1} \\neq N''_{i+1}$.\n\n\nIn order to compute the proper product and tails assay at each stage, the tail\nand the product from the next and the previous stage respectively must be\nblended in order to determine the correct stage feed assay. All feed assays are\niteratively updated, blending the proper product and tails, then using the\nupdated feed assay, the new product and tails assays are recomputed. This\nprocess is repeated until the sum of the square difference in assays is smaller\nthan $10^{-8}$. As the cut remains fixed at each stage, the flows do not need\nto be recomputed.\n\nThis model assumes that it is possible to maintain the separative power of a\ncentrifuges, $\\delta U$, for any feed assays $N$ while maintaining its cut\n$\\theta$ and feed flow $F$.\n\n\\subsubsection{Model B}\n\nUsing the second method, the cut value at each stage, $\\theta_i$, is retuned in\norder to maintain the $\\alpha_i$ and $\\beta_i$ at their original values\n(equation \\eqref{eq_theta}). Since the cascade remains ideal, the product and\ntails assay at each stage can easily be determined using equations\n\\eqref{eq_alphabeta}.\n\nAs the cut values change, the relative flow rates between the different stages\nare recomputed using equation \\eqref{eq_flow}. Under this model, the flow at\neach stage of the original ideal cascade is assumed to be the maximum flow\nallowed at that stage. Therefore, all of the recomputed flow rates are scaled\ntogether to ensure that no stage experiences a flow rate larger than that of\nthe original cascade. Some stages may now experience flow rates much lower\nthan the original cascade.\n\n%% The flow rates are determinted as\n%% the largest flow rates allowed by the cascade design, number of centrifuges\n%% limiting the flow at each stage.\n\nThis model assumes that it is possible to tune a centrifuge separative power\n$\\delta U$, for any feed assay $N$, cut $\\theta$ and feed flow $F$, in order to\nmaintain its constant feed to product enrichment factor $\\alpha$.\n\n\n\n\n\\subsubsection{Model C}\nThe last model assumes that the tails to product enrichment factor remains\nconstant regardless to the feed assays. To compute the response of the cascade\none need to determine $\\alpha$ and $\\beta$ such that their product and\n$\\theta$ remain fixed.\nFrom equations \\eqref{eq_alphabeta} and the assay conservation equation $N =\n\\theta N' + (1-\\theta)N''$ it is possible to express the product, $N'$,\ndependent on the feed assay $N$, $\\gamma$, and the cut, $\\theta$, as one\nsolution of the second order equation \\eqref{eq_gamma_p}:\n\n\\begin{equation}\\label{eq_gamma_p}\n \\theta(\\gamma-1)N'^2+((N+\\theta)(\\gamma-1)+1)N'-N\\gamma = 0\n\\end{equation}\n\n\nThe only solution allowing product assay to range between 0 and 1 is the\nfollowing :\n\n\\begin{equation}\\label{eq_model_b_sol}\n N' = \\frac{N+\\theta}{2\\theta} +\n \\frac{1 - \\sqrt{\\gamma^{2}(N-\\theta)^{2}\n + 2\\gamma( N^{2} + N - \\theta^{2} + \\theta)\n + (N + \\theta + 1)^{2}}}\n {2\\theta(\\gamma - 1)}\n\\end{equation}\n\nOnce the product assay is known, one can trivially determine the tails assay,\n$\\alpha$ and $\\beta$, using equations \\eqref{eq_alphabeta} and mass\nconservation.\n\nSimilar to model A, because the cut values remain constant, the flows do not\nneed to be recomputed and the correct assays, $\\alpha$ and $\\beta$, are\ndetermined through iterative blending of the product assays of the previous\nstage and the tails assay of the next stage using equation\n\\eqref{eq_model_b_sol}.\n\nThis model assumes that it is possible to tune the centrifuge separative power\n$\\delta U$ in order to maintain for any feed assay $N$ and its tails to product\nenrichment factor $\\gamma$ while maintaining its cut $\\theta$ and feed flow $F$.\n\n%The speed at which each models reach a threshold of $90\\%$ enrichment is\n%illustrated in Figure \\ref{fig:model_comparison}. The production curve of\n%enrichment for each model is shown. The step functions for each model represent\n%the cascade configuration. Each step is one cascade in a chain of cascades and\n%the shape of the curve for each model gives an indication of how quickly the model\n%can reach the enrichment threshold. The main difference is that Model B reaches\n%the threshold faster, whereas Models A and C reach a higher enrichment value.\n%This highlights some of the immediate differences that would influence the choice\n%of using each model in a breakout scenario.\n\n%\\begin{figure}[ht]\n% \\centering\n% \\includegraphics[scale=0.4]{ModelComparison}\n% \\caption{The curve of resulting product assays for each model is plotted.\n% Since the feed assay of each cascade is fed into the next cascade in the\n% chain, a linear relationship between the product and feed assays is plotted\n% in black. The step function for each model represents the output of the\n% cascade chain (each step is a cascade). While there are four steps for each\n% model, Model B (green) reached the $90\\%$ enrichment threshold before\n% Model A (red) or Model C (blue). However, in the same amount of steps, Models\n% A and C are able to achieve a higher enrichment percent than Model B.}\n% \\label{fig:model_comparison}\n%\\end{figure}\n", "meta": {"hexsha": "412a469a7a2d4076b71fba0dd0cb32b4fa7208fa", "size": 21566, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "theory.tex", "max_stars_repo_name": "nuclearkatie/enrichment-paper", "max_stars_repo_head_hexsha": "223eea892a43948ded8ed611a2e4829ba19fe6ce", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "theory.tex", "max_issues_repo_name": "nuclearkatie/enrichment-paper", "max_issues_repo_head_hexsha": "223eea892a43948ded8ed611a2e4829ba19fe6ce", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "theory.tex", "max_forks_repo_name": "nuclearkatie/enrichment-paper", "max_forks_repo_head_hexsha": "223eea892a43948ded8ed611a2e4829ba19fe6ce", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.2460850112, "max_line_length": 447, "alphanum_fraction": 0.6522767319, "num_tokens": 6062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8577680904463334, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6374658036283322}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% =================================================================================================\n\\section*{Symmetrised derivatives of the connection}\n\nHere we compute, just for a check, the symmetrised derivatives of the connections.\nThese are defined by\n\\begin{align}\n \\Gamma^{d}_{a(b,\\uc)} = \\Gamma^{d}_{a(b,c_1,c_2,\\cdots c_n)}\n\\end{align}\nNote that these are \\emph{not} the generalised connections. The generalised connections\ninvolve $\\Gamma^{d}_{(ab,\\uc)}$ and quadratice combinations of lower order generalised\nconnections (see eq (1) of {\\tt\\small ../genGamma.pdf}). Note that the\ngeneralised connections vanish at the origin (unlike the $\\Gamma^{d}_{a(b,\\uc)}$).\n\nThese results agree with those of Hatzinikitas equation (12) (arXiv:hep-th/0001078).\n\nThis code provides an indirect check on our results for the connection. It does not\nprove that the code {\\tt connection.tex} is correct but it does show that our\nresults are consistent with those of Hatzinikitas.\n\n\\clearpage\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w#}::Indices(position=independent).\n\n D{#}::Derivative.\n \\nabla{#}::Derivative.\n \\partial{#}::PartialDerivative.\n\n g_{a b}::Metric.\n g^{a b}::InverseMetric.\n g_{a}^{b}::KroneckerDelta.\n g^{a}_{b}::KroneckerDelta.\n \\delta^{a}_{b}::KroneckerDelta.\n \\delta_{a}^{b}::KroneckerDelta.\n\n R_{a b c d}::RiemannTensor.\n R^{a}_{b c d}::RiemannTensor.\n R_{a b c}^{d}::RiemannTensor.\n\n \\Gamma^{a}_{b c}::TableauSymmetry(shape={2}, indices={1,2}).\n\n x^{a}::Depends(D{#}).\n\n g_{a b}::Depends(\\partial{#}).\n R_{a b c d}::Depends(\\partial{#}).\n R^{a}_{b c d}::Depends(\\partial{#}).\n \\Gamma^{a}_{b c}::Depends(\\partial{#}).\n\n R_{a b c d}::Depends(\\nabla{#}).\n R^{a}_{b c d}::Depends(\\nabla{#}).\n\n import cdblib\n\n Gamma = cdblib.get ('Gamma','../connection.json')\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{cadabra}\n tmp := D_{p}{@(Gamma)}.\n distribute (tmp)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n eliminate_kronecker (tmp)\n substitute (tmp,$x^{a}->0$)\n sort_product (tmp)\n rename_dummies (tmp)\n canonicalise (tmp)\n\n foo := A^{p} A^{b} @(tmp).\n distribute (foo) # cdb(foo.301,foo)\n sort_product (foo) # cdb(foo.302,foo)\n rename_dummies (foo) # cdb(foo.303,foo)\n canonicalise (foo) # cdb(foo.304,foo)\n\n # save the result\n dGamma1 := @(foo). # cdb (dGamma1.000,dGamma1)\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\begin{dmath*} \\cdb*{foo.301} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.302} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.303} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.304} \\end{dmath*}\n\\end{dgroup*}\n\n\\clearpage\n\n\\begin{cadabra}\n tmp := D_{p q}{@(Gamma)}.\n distribute (tmp)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n eliminate_kronecker (tmp)\n substitute (tmp,$x^{a}->0$)\n sort_product (tmp)\n rename_dummies (tmp)\n canonicalise (tmp)\n\n foo := A^{p} A^{q} A^{b} @(tmp).\n distribute (foo) # cdb(foo.401,foo)\n sort_product (foo) # cdb(foo.402,foo)\n rename_dummies (foo) # cdb(foo.403,foo)\n canonicalise (foo) # cdb(foo.404,foo)\n\n # save the result\n dGamma2 := @(foo). # cdb (dGamma2.000,dGamma2)\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\begin{dmath*} \\cdb*{foo.401} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.402} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.403} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.404} \\end{dmath*}\n\\end{dgroup*}\n\n\\clearpage\n\n\\begin{cadabra}\n tmp := D_{p q r}{@(Gamma)}.\n distribute (tmp)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n eliminate_kronecker (tmp)\n substitute (tmp,$x^{a}->0$)\n sort_product (tmp)\n rename_dummies (tmp)\n canonicalise (tmp)\n\n foo := A^{p} A^{q} A^{r} A^{b} @(tmp).\n distribute (foo) # cdb(foo.501,foo)\n sort_product (foo) # cdb(foo.502,foo)\n rename_dummies (foo) # cdb(foo.503,foo)\n canonicalise (foo) # cdb(foo.504,foo)\n\n # save the result\n dGamma3 := @(foo). # cdb (dGamma3.000,dGamma3)\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n % \\begin{dmath*} \\cdb*{foo.501} \\end{dmath*}\n % \\begin{dmath*} \\cdb*{foo.502} \\end{dmath*}\n % \\begin{dmath*} \\cdb*{foo.503} \\end{dmath*}\n \\begin{dmath*} \\cdb*{foo.504} \\end{dmath*}\n\\end{dgroup*}\n\n\\clearpage\n\n\\begin{cadabra}\n tmp := D_{p q r s}{@(Gamma)}.\n distribute (tmp)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n eliminate_kronecker (tmp)\n substitute (tmp,$x^{a}->0$)\n sort_product (tmp)\n rename_dummies (tmp)\n canonicalise (tmp)\n\n foo := A^{p} A^{q} A^{r} A^{s} A^{b} @(tmp).\n distribute (foo)\n sort_product (foo)\n rename_dummies (foo)\n canonicalise (foo)\n\n # save the result\n dGamma4 := @(foo). # cdb (dGamma4.000,dGamma4)\n\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{cadabra}\n tmp := D_{p q r s t}{@(Gamma)}.\n distribute (tmp)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n unwrap (tmp)\n product_rule (tmp)\n distribute (tmp)\n substitute (tmp,$D_{a}{x^{b}}->\\delta_{a}^{b}$)\n eliminate_kronecker (tmp)\n substitute (tmp,$x^{a}->0$)\n sort_product (tmp)\n rename_dummies (tmp)\n canonicalise (tmp)\n\n foo := A^{p} A^{q} A^{r} A^{s} A^{t} A^{b} @(tmp).\n distribute (foo)\n sort_product (foo)\n rename_dummies (foo)\n canonicalise (foo)\n\n # save the result\n dGamma5 := @(foo). # cdb (dGamma5.000,dGamma5)\n\n\\end{cadabra}\n\n\\clearpage\n\nCompare these results against those of Hatzinikitas equation (12) (arXiv:hep-th/0001078).\nOur final $d\\Gamma$ is zero because our metric was expanded to order $x^5$ so the $\\Gamma$\nonly contain terms to order $x^4$. Hence the 5-th partial derivatices are zero.\n\n\\begin{dgroup*}\n \\begin{dmath*} A^b A^c \\Gamma^{d}{}_{ab,c} = \\Cdb*{dGamma1.000} \\end{dmath*}\n \\begin{dmath*} A^b A^c A^e \\Gamma^{d}{}_{ab,ce} = \\Cdb*{dGamma2.000} \\end{dmath*}\n \\begin{dmath*} A^b A^c A^e A^f \\Gamma^{d}{}_{ab,cef} = \\Cdb*{dGamma3.000} \\end{dmath*}\n \\begin{dmath*} A^b A^c A^e A^f A^g \\Gamma^{d}{}_{ab,cefg} = \\Cdb*{dGamma4.000} \\end{dmath*}\n \\begin{dmath*} A^b A^c A^e A^f A^g A^h \\Gamma^{d}{}_{ab,cefgh} = \\Cdb*{dGamma5.000} \\end{dmath*}\n\\end{dgroup*}\n\n\\end{document}\n", "meta": {"hexsha": "6c9c90418e507c12dba29c6f94f169d4b5cfb4b1", "size": 7781, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/checks/check-pGamma.tex", "max_stars_repo_name": "leo-brewin/riemann-normal-coords", "max_stars_repo_head_hexsha": "4e6546028229b6f43fcef1c0b83660cddc021716", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-12-20T16:15:58.000Z", "max_stars_repo_stars_event_max_datetime": "2019-12-20T16:15:58.000Z", "max_issues_repo_path": "source/cadabra/checks/check-pGamma.tex", "max_issues_repo_name": "leo-brewin/riemann-normal-coords", "max_issues_repo_head_hexsha": "4e6546028229b6f43fcef1c0b83660cddc021716", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/checks/check-pGamma.tex", "max_forks_repo_name": "leo-brewin/riemann-normal-coords", "max_forks_repo_head_hexsha": "4e6546028229b6f43fcef1c0b83660cddc021716", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.1920289855, "max_line_length": 99, "alphanum_fraction": 0.5820588613, "num_tokens": 2846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7772998714925403, "lm_q1q2_score": 0.6373029846431818}} {"text": "%==============================================================================\n\\chapter{Distances in Quantum Theory}\n\\label{chap:distappd}\n%==============================================================================\n\nThis chapter follows mostly \\cite{Wilde2009}. Some conventions follow those in \n\\cite{Petz2008}.\n\n%------------------------------------------------------------------------------\n\\section{Trace Distance}\n\\label{sec:trace-dist}\n%------------------------------------------------------------------------------\n\nTo begin with, define the \\emph{trace norm} or \\emph{$l_1$-norm} of an \nHermitian operator $M$ as\n\\begin{equation}\n\t\\dvbr{M}_1 \\coloneqq \\tr\\sqrt{M^\\dagger M}.\n\\end{equation}\nWhen the spectral decomposition $M = \\sum_i \\mu_i \\Ket{i}\\Bra{i}$ exists, the \ntrace norm reads\n\\begin{equation}\n\t\\dvbr{M}_1 \\equiv \\sum_i\\vbr{\\mu_i},\n\\end{equation}\nso the name $l_1$-norm comes. It is positive definite and homogeneous; the \ntriangle inequality also holds. Thus it can be used to define the \\emph{trace \ndistance} between Hermitian operators ${M}$ and ${N}$ as\n\\begin{equation}\n\\rfun{T}{{M}, {N}} \\coloneqq \\frac{1}{2} \\dvbr{M-N}_1 \\equiv\n\\frac{1}{2} \\tr \\sqrt{\\rbr{{M}-{N}}^\\dagger\\rbr{{M}-{N}}}.\n\\label{eq:def-trace-dist}\n\\end{equation}\nNow consider density operators $\\rho$ and $\\sigma$ only. Since $\\dvbr{\\rho}_1 \n\\equiv 1$, one sees\n\\begin{equation}\n0 \\le \\rfun{T}{{\\rho}, {\\sigma}} \\le 1,\n\\label{eq:def-trace-prop}\n\\end{equation}\nfollowed from positive definiteness and triangle inequality.\n\nThe following lemma helps constructing a physical interpretation of the \ndistance. Let the Hermitian operator $\\Lambda$ be such that all its eigenvalues \nlies within $\\sbr{0,1}$, then\n\\begin{equation}\n\\rfun{T}{{\\rho}, {\\sigma}} = \\max_{0 \\le {\\Lambda} \\le \\Bbbone}\n\\cfun{\\tr}{{\\Lambda}\\rbr{{\\rho}-{\\sigma}}}.\n\\end{equation}\nTo understand this, take $\\Lambda \\equiv \\Ket{\\alpha}\\Bra{\\alpha}$,\nwhere $\\Ket{\\alpha}$ is the eigenket of some observable $\\Alpha$, with \neigenvalue $\\alpha$. Then $\\tr\\cbr{\\Lambda \\rho}$ tells the probability of \nmeasuring $\\Alpha$ which gives the result $\\alpha$. Therefore $\\tr\\cbr{\\Lambda \n\\rbr{\\rho-\\sigma}}$ gives the difference of the probability above for $\\rho$ and \n$\\sigma$, and $\\rfun{T}{{\\rho}, {\\sigma}}$ is the maximal value of the \ndifference above.\n\nThough the trace distance is used in formulating \\cref{chap:quantifing}, it is \nrather formidable to compute due to the operatorial square root in \n\\cref{eq:def-trace-dist}. Therefore the author seeks other ways to evaluate the \nquantity.\n\n%------------------------------------------------------------------------------\n\\section{Fidelity}\n\\label{sec:fidelity}\n%------------------------------------------------------------------------------\n\nFidelity is another means to compare two quantum states. The simplest case \nof fidelity is that of two pure states,\n\\begin{equation}\n\\rfun{F}{\\Ket{\\alpha},\\Ket{\\beta}} \\coloneqq \\vbr{\\Braket{\\alpha | \\beta }}.\n\\end{equation}\nOne sees that it is just the modulus of the transition amplitude, a measure of \n\\emph{faithfulness}. For more general cases, fidelity is defined as\n\\begin{align}\n\\rfun{F}{\\Ket{\\alpha},{\\sigma}} &\\coloneqq \\sqrt{\\Braket{\\alpha | {\\sigma} | \n\\alpha}}, \\\\\n\\label{eq:fidelity-pure-mixed}\n\\rfun{F}{{\\rho},{\\sigma}} &\\coloneqq \\tr\\sqrt{{\\rho}^\\frac{1}{2} \n{\\sigma} {\\rho}^\\frac{1}{2}}.\n\\end{align}\nIn applications in this work, fidelity is solvable because only\n\\cref{eq:fidelity-pure-mixed} is needed.\n\nAn important property of fidelity is that it follows the \\emph{Fuchs--van de \nGraaf inequality} \\cite{Fuchs1999})\n\\begin{equation}\n1 - \\rfun{F}{{\\rho},{\\sigma}} \\le \\rfun{T}{{\\rho},{\\sigma}}\n\\le \\sqrt{1 - \\rfun{F^2}{{\\rho},{\\sigma}}}.\n\\label{eq:ineq-fvdg}\n\\end{equation}\nIn this work, the trace distances are evaluated by computing the fidelities \nexactly and inserting them to \\cref{eq:ineq-fvdg}.\n\nConsider a single harmonic oscillator with intrinsic frequency $\\Omega$, a \nthermalised state of it at temperature $T$ described by a density operator \n$\\rfun{\\rho}{T}$, as well as a generalised Gaussian state $\\Ket{\\omega}$ (see \n\\cref{sec:single-harosc}). In the following the fidelity of them will be \ncalculated.\n\nSince $\\rfun{\\rho}{T}$ is diagonal in the energy-eigenstate basis $\\Ket{n}$, one \nhas\n\\begin{equation}\n\\Braket{\\omega | \\rfun{\\rho}{T} | \\omega} = \\sum_{n=0}^{+\\infty}\n\\Braket{\\omega | n} \\Braket{n | \\rfun{\\rho}{T} | n} \\Braket{n | \\omega},\n\\label{eq:fidelity-1}\n\\end{equation}\nin which\n\\begin{align}\n\\Braket{n | \\omega} &= \n\\begin{cases} \\rbr{\\Omega \\Re\\omega }^{\\frac{1}{4}}\n\\dfrac{2^{\\frac{1}{2}-m} \\sqrt{(2 m)!} }{m!}\n\\dfrac{(\\Omega -\\omega )^m}{(\\omega +\\Omega )^{m+\\frac{1}{2}}}\n& n = 2m, \\\\\n0 & n = 2m+1,\n\\end{cases}\n\\tag{\\ref{eq:n|omega} revisited} \\\\\n\\Braket{n | \\rho | n} &=\n\\frac{1}{Z}\\cfun{\\exp}{-\\rbr{n+\\frac{1}{2}}\\frac{\\Omega}{T}}\n\\equiv \n\\rbr{\\ee^{\\Omega/T}-1} \\cfun{\\exp}{-\\rbr{n+1}\\frac{\\Omega}{T}}.\n\\label{eq:n|rho|n}\n\\end{align}\nInserting \\cref{eq:n|omega,eq:n|rho|n} into \\cref{eq:fidelity-1}, one finds \nthat each term in the summation is of the form $b\\cdot\\binom{2m}{m}a^{2m}$ \nwhere $a$ and $b$ are expressions of $\\Omega$, $\\omega$ and $T$, which can be \ncomputed by using\n\\begin{equation}\n\\arcsin z = \\sum_{n=0}^{+\\infty} \\binom{2n}{n}\\frac{z^{2n+1}}{4^n\\rbr{2n+1}}\n\\end{equation}\nand taking derivative with respect to $z$ on both sides. Therefore the fidelity \nof $\\rho(T)$ and $\\Ket{\\omega}$ is computed to be\n\\begin{equation}\n\\rfun{F}{\\Ket{\\omega}, \\rho(T)} = \\sqrt{2} \\sqrt[4]{\\frac{\\Omega \\Re{\\omega} \n\\left(\\ee^{\\Omega /T}-1\\right)^2}{(\\Omega -\\Re{\\omega})^2+\\Im{\\omega}^2 - \n\\ee^{2 \\Omega/T} \\rbr{(\\Omega +\\Re{\\omega})^2+\\Im{\\omega}^2})}},\n\\label{eq:fidelity-gg-th}\n\\end{equation}\nwhich can be used to set a bound on the trace distance by \\cref{eq:ineq-fvdg}.\n\n", "meta": {"hexsha": "6cf1caf0d4ca9d77ecca50c6784e8ee22eac0720", "size": 5790, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ubonn-thesis-current/mythesis/sections/thesis_distappd.tex", "max_stars_repo_name": "cmp0xff/Masterarbeit", "max_stars_repo_head_hexsha": "b29c84f9a29e4a7c9a3499658a1dfa7f87d64c9c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ubonn-thesis-current/mythesis/sections/thesis_distappd.tex", "max_issues_repo_name": "cmp0xff/Masterarbeit", "max_issues_repo_head_hexsha": "b29c84f9a29e4a7c9a3499658a1dfa7f87d64c9c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ubonn-thesis-current/mythesis/sections/thesis_distappd.tex", "max_forks_repo_name": "cmp0xff/Masterarbeit", "max_forks_repo_head_hexsha": "b29c84f9a29e4a7c9a3499658a1dfa7f87d64c9c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.9565217391, "max_line_length": 81, "alphanum_fraction": 0.6274611399, "num_tokens": 1870, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8198933337131076, "lm_q1q2_score": 0.6373029744732532}} {"text": "%!TEX root = ../main.tex\n% !TEX useTabs\n\n\\subsection{$r, \\theta$}\nApart from arbitrary roads running north-south and east-west, no one would ever walk solely\nin horizontal and vertical lines. When you want to fly your plane somewhere, you turn the\nnose towards where you want to go, and walked the shortest path. For example, you might\nknow that your friend is 3m east of you and 4m north, but you would never walk the 7m such \nludacrous path takes. What is the shortest path ($r$)? What angle would you turn ($\\theta$)?\n\n\\begin{tikzpicture}\n\t\\draw (0,0) -- (3,0) -- (3,4) -- (0,0);\n\t\\draw (1.5,0) node[anchor=north] {3m};\n\t\\draw (3,2) node[anchor=west] {4cm};\n\t\\draw (1.4,2) node[anchor=south] {$r$};\n\t\\draw (0.5,0.4) node {$\\theta$};\n\\end{tikzpicture}\n\nWell, by this point in this class, you are undoubtedly very used to using the Pythagorean Theorem\nto find hypotenuses, and inverse tangent to calculate angle. While you should certainly be able to\ncalculate $r=5$, a calculator or table is required to know when tangent is $1.\\overline{3}$, \napproximately 0.927 radians.\n\nOn the other hand, suppose you knew something was 6 miles away, at an angle of $\\frac{2}{3}\\tau$.\nHow would you determine how far east-west and north-south it is? Just as we learned with vectors,\nwe can use sine and cosine to calculate these values. $\\sin(\\frac{2}{3}\\tau)=\\frac{y}{6}$ and \n$\\cos(\\frac{2}{3}\\tau)=\\frac{x}{6}$. Solving for $x$ and $y$, we see the coordinates of the\ndestination are $(-3,-3\\sqrt{3})$.\n\n\\begin{derivation}{Polar Conversion Formulae}\n\\begin{equation}\n\tr^2=x^2+y^2\n\\end{equation}\n\\begin{equation}\n\t\\tan\\theta=\\frac{y}{x}\n\\end{equation}\n\\begin{equation}\n\tx=r\\cdot\\cos\\theta\n\\end{equation}\n\\begin{equation}\n\ty=r\\cdot\\sin\\theta\n\\end{equation}\n*The tangent equation doesn't work everywhere: what are the limitations on how we can use it?\n\\end{derivation}\n\n\\subsection{Properties}\nWorking with $r$ and $\\theta$ to the exclusion of $x$ and $y$ yields a different way of looking\nat the coordinate plane, called \\emph{polar coordinates}. Locations are still labeled via ordered\npairs, but of the form $(r,\\theta)$. Polar graph paper looks like this:\n\n \\begin{tikzpicture}[scale=0.5]\n %Circles \n \\foreach \\r in {1, 2,...,7}\n \\draw[thick] (0,0) circle (\\r); \n %half radii \n %\\foreach \\r in {0.5, 1.5,...,7}\n %\\draw[thin] (0,0) circle (\\r);\n %1° Rays\n %\\foreach \\a in {0, 1,...,359}\n %\\draw[-] (\\a:7.7) -- (\\a:8);\n %5° Rays\n %\\foreach \\a in {0, 5,...,359}\n %\\draw[-] (\\a:7.5) -- (\\a:8); \n %15° Rays\n \\foreach \\a in {0, 15,...,359}\n \\draw[thick] (\\a:1) -- (\\a:8); \n %30° Rays\n \\foreach \\a in {0, 30,...,359}\n \\draw[thick] (0, 0) -- (\\a:8);\n %Radius labels (background filled white)\n %\\foreach \\r in {1, 2,...,7}\n %\\draw (\\r,0) node[inner sep=1pt,below=3pt,rectangle,fill=white] {$\\r$};\n %Main rays\n \\foreach \\a in {0, 90,...,359}\n \\draw[very thick] (0, 0) -- (\\a:8);\n\\end{tikzpicture}\n\n\\subsection{Negative Radii}\nThe ``spokes'' are every $\\frac{\\tau}{24}$, and the circles are every unit. You might begin by turning\nthe angle first, and then ``walking out'' the radius. On the polar graph, try locating $(3,\\frac{\\tau}{4})$.\nIt shouldn't be too complicated to see this is the same as the rectangular coordinate (0,3). Angles\ngreater than $\\tau$ are not difficult to reduce to simpler terms, but what about negative radii? \nWhere would $(-3,\\frac{3}{4}\\tau)$ take you? We might imagine a person standing at the\npole (the origin on a polar graph), he or she would begin facing to the right, and then turning \ncounter-clockwise. $\\frac{3}{4}\\tau$ is straight down, so our person's back is to the ``north''. A\nnegative radii can be thought of as walking \\emph{backwards}, so $(3,\\frac{\\tau}{4})$ and\n$(-3,\\frac{3}{4}\\tau)$ take you to the same place!\n\n\n\\subsection{Polar Equations}\nJust as Cartesian (or rectangular) equations as $y$ equals something in terms of $x$, so\ntoo polar equations are written as $r$ equals something in terms of $\\theta$. Press the `mode'\nbutton on your calculator, and you can see that there is an option called `POL', short for POLAR.\nAccept that option and see what the `y=' button has become! What has the X button become?\n\nThe exercise or problems will walk you through many different types of $r=$ equations, and they\ncan all be demonstrated in your calculator. However, just as you have to know how to graph\n$x=n$ even though it is not a function in rectangular, so too you should take a moment and reason\nthrough how to graph $\\theta=\\frac{1}{5}\\tau$ and similar equations. \n\nWhen we attempt plot points to determine the make-up of $x=3$, we see that $y$ can be anything.\nIn other words, (3,0); (3,1); (3,2); (3,-2); (3,-10), etc. are all on the graph. When we take this principle\nover to graphing $\\theta=\\frac{5}{6}$, we recognize that any radius is possible, including negatives.\n~\\vfill\n", "meta": {"hexsha": "b0a3fd2c357fefb4bd947fb69855eb79cb148e22", "size": 4937, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ch12/1201.tex", "max_stars_repo_name": "aquatiki/AnalysisTextbook", "max_stars_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-10-08T15:05:17.000Z", "max_stars_repo_stars_event_max_datetime": "2019-07-07T12:32:53.000Z", "max_issues_repo_path": "ch12/1201.tex", "max_issues_repo_name": "aquatiki/AnalysisTextbook", "max_issues_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ch12/1201.tex", "max_forks_repo_name": "aquatiki/AnalysisTextbook", "max_forks_repo_head_hexsha": "011c16427ada1b1e3df8e66c02566a5d5ac8abcf", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.019047619, "max_line_length": 109, "alphanum_fraction": 0.6775369658, "num_tokens": 1561, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998508568416, "lm_q2_score": 0.8198933271118222, "lm_q1q2_score": 0.637302960882539}} {"text": "\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{amsmath} \n\n\\usepackage{pgfplots}\n\\usepackage{tikz}\n\\usepackage{nicefrac}\n\\pgfplotsset{every axis legend/.append style={\nat={(0,0)},\nanchor=north east}}\n\\usetikzlibrary{shapes,positioning,intersections,quotes}\n\n\\definecolor{darkgreen}{rgb}{0.0, 0.6, 0.0}\n\\definecolor{darkred}{rgb}{0.7, 0.0, 0.0}\n\n\\title{Interpolation}\n\\begin{document}\n\\def\\horzbar{\\text{magic}}\n\n \\pagenumbering{gobble}\n \\maketitle\n \\newpage\n \\pagenumbering{arabic}\n\n\\section*{Introduction}\n\nWe have to arrays of numbers $X$ and $Y$. Array $X$ contains independent data points. Array $Y$ contains dependent data points $y_i,i=1,…,m$.\n\nWe want to find a function $\\hat{y}(x)$, which gets the exact same value with given points.\\\\\n\n\n\\section*{Linear Interpolation}\n\nLinear interpolation is achieved by connecting two data points with a straight line.\n\nFor $x_i < x < x_{i+1}$:\n\n$$\\hat{y}(x) = y_i + \\frac{(y_{i+1} - y_{i})(x - x_{i})}{(x_{i+1} - x_{i})}.$$\n\n\\section*{Derivation}\n\n \n\\begin{tikzpicture}\n\n\t\\draw [dashed] (-3, -1.9) -- (3, -1.9);\n\t\\draw [dashed] (3, -5) -- (3, -1.9);\n\t\\draw [-stealth](-3,-5) -- (9,-5);\n\t\\draw [-stealth](-3,-5) -- (-3,1);\n\n \\draw [stealth-stealth, blue](0,-3) -- (8,0);\n \\draw [stealth-stealth, red](8,0) -- (8,-3);\n \\draw [stealth-stealth, green](0,-3) -- (8,-3);\n \\draw [stealth-stealth, black](0,-3.1) -- (3,-3.1);\n\n\t\\node[above right=0pt of {(8, 0)}, outer sep=2pt,fill=none] {$(x_2, y_2)$};\n\t\\node[above right=0pt of {(8, -1.8)}, outer sep=2pt,fill=none, darkred] {$y_2 - y_1$};\n\t\\node[above right=0pt of {(4.5, -3.6)}, outer sep=2pt,fill=none, darkgreen] {$x_2 - x_1$};\n\t\\node[above right=0pt of {(1, -3.6)}, outer sep=2pt,fill=none] {$x - x_1$};\n\t\\node[above right=0pt of {(-1,-3.7)}, outer sep=2pt,fill=none] {$(x_2, y_2)$};\n\t\\node[above right=0pt of {(-1,-1.7)}, outer sep=2pt,fill=none] {$y$};\n\t\\node[above right=0pt of {(3,-4.2)}, outer sep=2pt,fill=none] {$x$};\n\t\\node[above right=0pt of {(3,-2.7)}, outer sep=2pt,fill=none] {$h$};\n\n\\end{tikzpicture}\n\n\n$$\\alpha = \\frac{y_2 - y_1}{x_2 - x_1}$$\n\n$$h = \\alpha \\cdot (x - x_1)$$\n\n$$y = y_1 + h$$\n\n$$y = y_1 + (x - x_1) \\cdot \\frac{y_2 - y_1}{x_2 - x_1}$$\n\n\\section*{Example}\n\nWe are given two points A(-2, 0) and B (2, 2).\\\\\n\n\\begin{tikzpicture}\n \\begin{axis}[\n axis x line=middle,\n axis y line=middle,\n width=8cm,\n height=8cm,\n xmin=-5, % start the diagram at this x-coordinate\n xmax= 5, % end the diagram at this x-coordinate\n ymin=-5, % start the diagram at this y-coordinate\n ymax= 5, % end the diagram at this y-coordinate\n xlabel=$x$,\n ylabel=$y$,\n legend cell align=left,\n legend pos=south east,\n legend style={draw=none},\n tick align=outside,\n enlargelimits=false]\n % plot the function\n \\addplot[domain=-5:5, blue, ultra thick,samples=500] {0.5*x + 1};\n \\fill[red] (700,700) circle (3pt);\n \t \\fill[red] (300, 500) circle (3pt);\n \t \\draw [dashed] (500, 700) -- (680, 700);\n \t \\draw [dashed] (700, 500) -- (700, 680);\n \t \\node[above right=0pt of {(255,510)}, outer sep=2pt,fill=none] {A};\n\t \\node[above right=0pt of {(655,710)}, outer sep=2pt,fill=none] {B};\n\t\\legend{$\\nicefrac{1}{2} \\cdot x$ + 1}\n \\end{axis}\n\\end{tikzpicture}\n\nLet's try to evaluate the value of the function at $x=1$\n\n$$\\hat{y}(x) = y_i + \\frac{(y_{i+1} - y_{i})(x - x_{i})}{(x_{i+1} - x_{i})} = 0 + \\frac{(2 - 0)(1 - (-2))}{(2 - (-2))} = 1.5$$\n\n\\newpage\n\\section*{Cubic Spline}\n\nThe interpolating function in cubic spline interpolation is a set of piecewise cubic functions.\\\\\n\nFor $x_i < x < x_{i+1}$:\\\\\n\nWe have two points $(x_i, y_i)$ and $(x_{i+1}, y_{i+1})$ joined with a cubic polynomial:\n\n$$S_i(x) = a_i x^3 + b_i x^2 + c_i x + d_i$$\n\nFor $n$ points, there are $n-1$ cubic functions to find, and each cubic function requires four coefficients ($a_i, b_i, c_i, d_i$).\n\nThere are $4(n-1)$ unknowns to find.\\\\\n\n\\begin{tikzpicture}\n \\begin{axis}[\n axis x line=middle,\n axis y line=middle,\n width=13cm, height=13cm, % size of the image\n grid = none,\n grid style={dashed, gray!0},\n %xmode=log,log basis x=10,\n %ymode=log,log basis y=10,\n xmin=-2, % start the diagram at this x-coordinate\n xmax= 4, % end the diagram at this x-coordinate\n ymin=-7, % start the diagram at this y-coordinate\n ymax= 7, % end the diagram at this y-coordinate\n %/pgfplots/xtick={0,1,...,60}, % make steps of length 5\n %extra x ticks={23},\n %extra y ticks={0.507297},\n axis background/.style={fill=white},\n ylabel=y,\n xlabel=x,\n %xticklabels={,,},\n %yticklabels={,,},\n tick align=outside,\n tension=0.08]\n % plot the stirling-formulae\n \\addplot[name path global=a, domain=-2:4, blue, thick,samples=500] {-x*x*x + 4*x*x-x-4};\n \\fill[red] (211, 39.3) circle (3pt);\n \\fill[red] (455, 108.8) circle (3pt);\n \\node[above right=0pt of {(211, 39.3)}, outer sep=2pt,fill=none] {$y_1$};\n \t \\node[above right=0pt of {(455, 108.8)}, outer sep=2pt,fill=none] {$y_2$};\n \t \\node[above right=0pt of {(250, 98.8)}, outer sep=2pt,fill=none] {$S_1(x)$};\n\\draw [-stealth](280,98) -- (350,82\n);\n \\end{axis}\n\\end{tikzpicture}\n\n\\section*{Derivation}\n\nWe are trying to find a function $S_i(x) = a_i x^3 + b_i x^2 + c_i x + d_i$ going trough both points: $(x_i, y_i)$ and $x_{i+1}, y_{i+1}$.\n\n\\begin{align}\nS_i(x_i) &= y_i,\\quad i = 1,\\ldots,n-1,\n\\end{align}\n\\begin{align}\nS_i(x_{i+1}) &= y_{i+1},\\quad i = 1,\\ldots,n-1,\n\\end{align}\n\nSmoothness condition:\n\\begin{align}\nS'_i(x_{i+1}) &= S^{\\prime}_{i+1}(x_{i+1}),\\quad i = 1,\\ldots,n-2,\n\\end{align}\n\\begin{align}\nS''_i(x_{i+1}) &= S''_{i+1}(x_{i+1}),\\quad i = 1,\\ldots,n-2,\n\\end{align}\n\nBoundry condition: The curve is a “straight line” at the end points:\n\\begin{align}\nS''_1(x_1) &= 0\n\\end{align}\n\\begin{align}\nS''_{n-1}(x_n) &= 0\n\\end{align}\n\nLet $h_{i}=x_{i}-x_{i-1}$\n\nLet $S_i^{''}(x_i) = S_i^{''}(x_{i+1}) = M_i$\n\n$S_1^{''}(x_1)= M_0 = 0$ and $S_n^{''}(x_n) = M_n = 0$\n\nOther $M_i$ are unknown.\n\nBy Lagrange interpolation, we can interpolate each $S''_{i}$ on $[x_{i-1},x_{i}]$:\n\n$$S''_{i}(x)=M_{i-1}{\\frac {x_{i}-x}{h_{i}}}+M_{i}{\\frac {x-x_{i-1}}{h_{i}}} \\quad for \\quad x\\in [x_{i-1},x_{i}]$$\n\nIntegrating the above equation twice and using the condition that $C_{i}(x_{i-1})=y_{i-1}$ and $ C_{i}(x_{i})=y_{i}$ to determine the constants of integration, we have.\n\n$$ S_{i}(x)=M_{i-1}{\\frac {(x_{i}-x)^{3}}{6h_{i}}}+M_{i}{\\frac {(x-x_{i-1})^{3}}{6h_{i}}}+\\left(y_{i-1}-{\\frac {M_{i-1}h_{i}^{2}}{6}}\\right){\\frac {x_{i}-x}{h_{i}}}+\\left(y_{i}-{\\frac {M_{i}h_{i}^{2}}{6}}\\right){\\frac {x-x_{i-1}}{h_{i}}}$$\n$${\\text{for}}\\quad x\\in [x_{i-1},x_{i}] $$\\\\\n\nThis expression gives us the cubic spline S(x) if $ M_{i},i=0,1,\\cdots ,n$ can be determined.\n$$S'_{i+1}(x)=-M_{i}{\\frac {(x_{i+1}-x)^{2}}{2h_{i+1}}}+M_{i+1}{\\frac {(x-x_{i})^{2}}{2h_{i+1}}}+{\\frac {y_{i+1}-y_{i}}{h_{i+1}}}-{\\frac {M_{i+1}-M_{i}}{6}}h_{i+1}$$\n\n$$S'_{i+1}(x_{i})=-M_{i}{\\frac {h_{i+1}}{2}}+{\\frac {y_{i+1}-y_{i}}{h_{i+1}}}-{\\frac {M_{i+1}-M_{i}}{6}}h_{i+1}$$\n\nSimilarly, when $x\\in [x_{i-1},x_{i}]$, we can shift the index to obtain\n\n\\begin{align}\nS'_{i}(x) &=-M_{i-1}{\\frac {(x_{i}-x)^{2}}{2h_{i}}}+M_{i}{\\frac {(x-x_{i-1})^{2}}{2h_{i}}}+{\\frac {y_{i}-y_{i-1}}{h_{i}}}-{\\frac {M_{i}-M_{i-1}}{6}}h_{i}\n\\end{align}\n\n \n$$ S'_{i}(x_{i})=M_{i}{\\frac {h_{i}}{2}}+{\\frac {y_{i}-y_{i-1}}{h_{i}}}-{\\frac {M_{i}-M_{i-1}}{6}}h_{i}$$\n\nSince $ S'_{i+1}(x_{i})=S'_{i}(x_{i})$, we can derive:\n\n$$\\mu _{i}M_{i-1}+2M_{i}+\\lambda _{i}M_{i+1}=d_{i}\\quad {\\text{for}}\\quad i=1,2,\\cdots ,n-1,$$\n \n$$\\mu _{i}={\\frac {h_{i}}{h_{i}+h_{i+1}}},\\quad \\lambda _{i}=1-\\mu _{i}={\\frac {h_{i+1}}{h_{i}+h_{i+1}}},\\quad {\\text{and}}\\quad d_{i}=6f[x_{i-1},x_{i},x_{i+1}]$$\n\nand $f[x_{i-1},x_{i},x_{i+1}]$ is a divided difference.\\\\\n\nAccording to different boundary conditions, we can solve the system of equations above to obtain the values of $M_{i}$'s.\\\\\n\n$S'_{1}(x_{0})=f'_{0}$ and $S'_{n}(x_{n})=f'_{n}$. According to equation (7), we can obtain:\n\n$$S'_{1}(x_{0})=-M_{0}{\\frac {(x_{1}-x_{0})^{2}}{2h_{1}}}+M_{1}{\\frac {(x_{0}-x_{0})^{2}}{2h_{1}}}+{\\frac {y_{1}-y_{0}}{h_{1}}}-{\\frac {M_{1}-M_{0}}{6}}h_{1}$$\n$$\\Rightarrow f'_{0}=-M_{0}{\\frac {h_{1}}{2}}+f[x_{0},x_{1}]-{\\frac {M_{1}-M_{0}}{6}}h_{1}$$\n$$\\Rightarrow 2M_{0}+M_{1}={\\frac {6}{h_{1}}}(f[x_{0},x_{1}]-f'_{0})=6f[x_{0},x_{0},x_{1}]$$\n\nAnalogously:\n\n$$ S'_{n}(x_{n})=-M_{n-1}{\\frac {(x_{n}-x_{n})^{2}}{2h_{n}}}+M_{n}{\\frac {(x_{n}-x_{n-1})^{2}}{2h_{n}}}+{\\frac {y_{n}-y_{n-1}}{h_{n}}}-{\\frac {M_{n}-M_{n-1}}{6}}h_{n}$$\n\n$$M_{n-1}+2M_{n}={\\frac {6}{h_{n}}}(f'_{n}-f[x_{n-1},x_{n}])=6f[x_{n-1},x_{n},x_{n+1}]$$\n\nLet:\\\\\n$\\lambda _{0}=\\mu _{n}=1,$\\\\\n$d_{0}=6f[x_{0},x_{0},x_{1}]$ and\\\\ \n$d_{n}=6f[x_{n-1},x_{n},x_{n}]$\n\n \n\\begin{equation*}\n \\begin{bmatrix}\n 2 & \\lambda_0 \\\\ \n \\mu_1 & 2 & \\lambda_1 \\\\ \n & \\ddots & \\ddots & \\ddots \\\\\n && \\ddots & \\ddots & \\ddots \\\\\n\t&&& \\ddots & \\ddots & \\ddots \\\\\n\t&&&& \\mu_{n-1} & 2 & \\lambda_{n-1} \\\\ \n\t&&&&& \\mu_{n} & 2 \\\\ \n \\end{bmatrix}\n %\n \\begin{bmatrix}\n M_0 \\\\\n M_1 \\\\\n \\vdots \\\\\n \\vdots \\\\\n \\vdots \\\\\n M_{n-1} \\\\\n M_n \\\\\n \\end{bmatrix} \n =\n %\n \\begin{bmatrix}\n d_0 \\\\\n d_1 \\\\\n \\vdots \\\\\n \\vdots \\\\\n \\vdots \\\\\n d_{n-1} \\\\\n d_n \\\\\n \\end{bmatrix} \n\\end{equation*}\n\n \\newpage\n\\section*{Lagrange Polynomial Interpolation}\n\nLagrange polynomial interpolation gives us a single polynomial that connects all of the data points.\n\nThat polynomial is denoted as $L(x)$. It is true that $L(x_i) = y_i$ for all points $(x_i, y_i)$ .\n\n$$L(x) = \\sum_{i = 1}^n y_i P_i(x).$$\n\nEach polynomial appearing in the sum is called a Lagrange basis polynomials, $P_i(x)$.\n\n$$P_i(x) = \\prod_{j = 1, j\\ne i}^n\\frac{x - x_j}{x_i - x_j},$$\n\n\n\\section*{Example}\n\nWe are given three points A(-1, 1), B(2, 3) and C(3,5).\\\\\n\n$$P_1(x) = \\frac{(x - x_2)(x - x_3)}{(x_1-x_2)(x_1-x_3)} = \\frac{(x - 2)(x - 3)}{(-1-2)(-1-3)} = \\frac{1}{12}(x^2 - 5x + 6)$$\n\n$$P_2(x) = \\frac{(x - x_1)(x - x_3)}{(x_2-x_1)(x_2-x_3)} = \\frac{(x + 1)(x - 3)}{(2 + 1)(2-3)} = -\\frac{1}{3}(x^2 - 2x - 3)$$\n\n$$P_3(x) = \\frac{(x - x_1)(x - x_2)}{(x_3-x_1)(x_3-x_2)} = \\frac{(x + 1)(x - 2)}{(3 + 1)(3-2)} =\\frac{1}{4}(x^2 -x - 2)$$\n\n$$ L(x) = 1 \\cdot P_1(x) + 3 \\cdot P_2(x) + 5 \\cdot P_3(x) $$\n\n$$ L(x) = 1 \\cdot P_1(x) + 3 \\cdot P_2(x) + 5 \\cdot P_3(x) $$\n\n$$ L(x) = \\frac{1}{3} x^2 + \\frac{1}{3} x + 1 $$\n\n\\begin{tikzpicture}\n \\begin{axis}[\n axis x line=middle,\n axis y line=middle,\n width=10cm,\n height=10cm,\n xmin=-5, % start the diagram at this x-coordinate\n xmax= 6, % end the diagram at this x-coordinate\n ymin= -1, % start the diagram at this y-coordinate\n ymax= 8, % end the diagram at this y-coordinate\n xlabel=$x$,\n ylabel=$y$,\n legend cell align=left,\n legend pos=north east,\n legend style={draw=none},\n tick align=outside,\n enlargelimits=false,\n xtick distance=1,\n ytick distance=1]\n \n % plot the function\n \\addplot[domain=-5:10, blue, ultra thick,samples=500] {1/3*(x^2 + x + 3)};\n\n \\fill[red] (400, 20) circle (3pt);\n \t \\fill[red] (700, 40) circle (3pt);\n \t \\fill[red] (800, 60) circle (3pt);\n\n \t \\node[above right=0pt of {(340, 13)}, outer sep=2pt,fill=none] {A};\n\t \\node[above right=0pt of {(630, 40)}, outer sep=2pt,fill=none] {B};\n\t \\node[above right=0pt of {(730, 60)}, outer sep=2pt,fill=none] {C};\n\n\t\\legend{$\\frac{1}{3} x^2 + \\frac{1}{3} x + 1$}\n\n \\end{axis}\n\\end{tikzpicture}\n\n\\end{document}", "meta": {"hexsha": "8756c85492aa79c630898b4ddbf9b530a6d4f597", "size": 11717, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "slides/interpolation.tex", "max_stars_repo_name": "djeada/Numerical-Methodes", "max_stars_repo_head_hexsha": "45a5288f4719568a62a82374efbb3fc06d33ec46", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "slides/interpolation.tex", "max_issues_repo_name": "djeada/Numerical-Methodes", "max_issues_repo_head_hexsha": "45a5288f4719568a62a82374efbb3fc06d33ec46", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "slides/interpolation.tex", "max_forks_repo_name": "djeada/Numerical-Methodes", "max_forks_repo_head_hexsha": "45a5288f4719568a62a82374efbb3fc06d33ec46", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.7665706052, "max_line_length": 239, "alphanum_fraction": 0.5567978151, "num_tokens": 5022, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7122321720225278, "lm_q2_score": 0.8947894675053567, "lm_q1q2_score": 0.6372978459442212}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=0.7in]{geometry}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{hyperref}\n\\hypersetup{colorlinks=true, urlcolor=green}\n\\DeclareMathOperator*{\\argmax}{argmax}\n\\begin{document}\n\n\\title{The Gory Details}\n\\author{Pavel Komarov}\n\\date{March 12, 2018}\n\\maketitle\n\nHere lies a full explanation of how multivariate Projection Pursuit Regression (PPR) and univariate Classification work, to the degree I currently understand. It is as much for me as for you, because I much prefer my own notation, and every time I have to dig in to the bones of the code and faff with the loss function, I end up having to refer to \\href{http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-3824.pdf}{the original paper}'s somewhat ambiguous, derivationless equations. Formerly I was attempting to deposit some of this knowledge in code comments, but they took up too much space while somehow remaining marvelously unreadable.\n\nThis is all in \\LaTeX\\ because \\href{https://github.com/github/markup/issues/897}{native math in markdown is amazingly still not supported by github}, so a \\texttt{.md} would really be no less uncomfortable than code comments. If you happen to want to know how I accomplished all this gorgeous formatting, the source \\texttt{.tex} is in the \\href{https://github.com/pavelkomarov/projection-pursuit/tree/master/doc}{/doc} subdirectory of the \\href{https://github.com/pavelkomarov/projection-pursuit}{repo} and can be compiled with \\texttt{pdflatex}.\n\nI attempt to follow a structure based on the \\href{https://www.youtube.com/watch?v=H6ygbGX8OV8&t=9m19s}{five components of any machine learning algorithm}:\n\n\\begin{enumerate}\n\t\\setlength\\itemsep{-2mm}\n\t\\item A Task/Problem Reduction\n\t\\item A Model\n\t\\item A Loss Function\n\t\\item An Optimization Scheme\n\t\\item Data\n\\end{enumerate}\n\n--or at least the middle three, since by the time the data for your task meets my algorithm you'll have abstracted away concerns about where it came from and what it actually means.\n\n\\section{The Model}\n\nPPR is a statistical model of the form:\n\n$$\\vec{y_i} = \\sum_{j=1}^{r} f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\vec{\\beta_j}^T$$\n\nwhere:\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $i$ iterates examples, the rows of input and output matrices\n\t\\item $j$ iterates the number of terms in the PPR ``additive model\"\n\t\\item $r$ is the total number of projections and functions (terms) in the PPR\n\t\\item $\\vec{y_i}$ is a $d$-dimensional vector, the $i$\\textsuperscript{th} row in an output matrix $\\pmb{Y} \\in \\mathbb{R}^{n \\times d}$\n\t\\item $\\vec{x_i}$ is a $p$-dimensional vector, the $i$\\textsuperscript{th} row of an input matrix $\\pmb{X} \\in \\mathbb{R}^{n \\times p}$\n\t\\item $\\vec{\\alpha_j}$ is the $j$\\textsuperscript{th} projection vector in the mdoel, a $p$-dimensional vector inner-producted with $\\vec{x_i}$\n\t\\item $f_j$ is the $j$\\textsuperscript{th} function in the model, mapping from $\\mathbb{R}^1 \\rightarrow \\mathbb{R}^1$\n\t\\item $\\vec{\\beta_j}^T$ is the transpose of $\\vec{\\beta_j}$, a $d$-dimensional vector outer-producted with the result of $f_j$ to yield a result in the output space\n\t\\item $\\cdot$ is an inner product\n\t\\item $\\otimes$ is an outer product\n\\end{itemize}\n\nThis can be ``vectorized\" as\n\n$$\\pmb{\\hat{Y}} = \\sum_{j=1}^r f_j(\\pmb{X} \\cdot \\vec{\\alpha_j}) \\otimes \\vec{\\beta_j}^T$$\n\nwhere the ``hat\" \\string^ distinguishes the answers predicted, as opposed to known training answers.\\newline\n\nI also term this the ``evaluation function\". It may seem complicated, but the idea is simple:\n\n\\begin{enumerate}\n\t\\setlength\\itemsep{-2mm}\n\t\\item Linearly project the input down to one dimension where it is easier to work with, thereby sidestepping the curse of dimensionality.\n\t\\item Find a sensible mapping from this reduced space to weighted ``residuals\", unexplained variance in the outputs. This is where the nonlinearity happens.\n\t\\item Unpack from the single-dimensional residual space to the output space with a kind of inverse projection.\n\\end{enumerate}\n\n\\subsection{A Word on Additive Models}\n\nIn practice a single projection-mapping-expansion is not descriptive enough to capture the richness of what may be a very complicated underlying relationship between $\\pmb{X}$ and $\\pmb{Y}$, so it is repeated $r$ times, each new ``stage\" only accounting for the variance left unexplained by the stages that have come before. Notice that, as per Taylor's Theorem and the no-doubt familiar universal approximation theorems, for certain classes of functions $f$, as $r$ goes to infinity the evaluation function can approximate any continuous functional relationship between inputs and outputs.\n\n\\section{The Loss Function}\n\nThe (supervised) learning process consists of minimizing a standard quadratic cost, or ``loss\", function:\n\n$$loss = \\sum_{i=1}^n w_i (y_i - \\hat{y_i})^2$$\n\nwhere:\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $i$ iterates all training examples\n\t\\item $n$ is the total number of training examples\n\t\\item $w_i$ is the weight of the $i$\\textsuperscript{th} example\n\t\\item $y_i$ is the known answer for example $i$\n\t\\item $\\hat{y_i}$ (``y-i-hat\") is the answer predicted by the model for example $i$\n\\end{itemize}\n\nIn words: Get as close as you can for all examples, where the penalty for predicting an example incorrectly scales with the square of how wrong you are.\n\nThere are other choices of loss function, including an absolute value loss with no square penalty. This can work better in cases where a nonliner penalty might cause the algorithm to account too strongly for a few outlier examples.\n\n(TODO) It is also possible to add regularization terms, which bias the optimization process to find smaller parameter values. This is good for numerical stability and has the added benefit of reducing the likelihood of overfit.\n\nUnfortunately, modifying the loss function in either of these ways means altering the optimization process, and that's a mathematical bear I just haven't chosen to wrestle with.\\newline\n\nPlugging the evaluation function in to the cost function yields a relationship between model parameters and loss. Because there are multiple dimensions in our vector $\\vec{y_i}$, we introduce a sum over them so the PPR is motivated to make good predictions for all entries of the output:\n\n$$loss = \\sum_{i=1}^n w_i \\sum_{k=1}^d w_k [y_{ik} - \\sum_{j=1}^r f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk}]^2$$\n\nwhere this new fauna:\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $k$ iterates the columns of the output $\\pmb{Y}$\n\t\\item $d$ is the number of outputs, the width of the output matrix $\\pmb{Y}$\n\t\\item $w_k$ is a scalar weight, the relative importance of the $k$\\textsuperscript{th} output dimension\n\t\\item $y_{ik}$ is the scalar $k$\\textsuperscript{th} entry in the vector $\\vec{y_i}$, itself the $i$\\textsuperscript{th} row of $\\pmb{Y}$\n\t\\item $\\beta_{jk}$ is the scalar $k$\\textsuperscript{th} entry of $\\vec{\\beta_j}$ from the evaluation function\n\\end{itemize}\n\nThis can be vectorized as\n\n$$loss = (\\vec{w_I}^T \\cdot (\\pmb{Y} - \\pmb{\\hat{Y}})^2) \\cdot \\vec{w_K}$$\n\nwhere\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $\\vec{w_I}$ is an $n$-vector containing all example weights, $w_i$\n\t\\item $\\vec{w_K}$ is a $d$-vector containing all output dimension weights, $w_k$\n\\end{itemize}\nand the sums disappear inside the inner products.\\newline\n\nThe parameters we need to optimize to make the PPR ``learn\" are $\\vec{\\alpha_j}$, $f_j$, and $\\vec{\\beta_j}$. $w_k$ are hyperparameters chosen by the user, just as $r$ is chosen. If output dimensions are on differing scales and the user wishes for the loss to be relative to those scales, then choosing $w_k$ inversely proportional to the variance of outputs causes the loss to treat outputs more evenly. If some output dimensions are more important than others, then setting their weights higher causes predictions in those dimensions to be closer.\n\n\\section{The Optimization Scheme}\n\nThe macroscopic optimization scheme to solve for so many different parameters is non-obvious but straightforward:\n\\begin{enumerate}\n\t\\setlength\\itemsep{-2mm}\n\t\\item Initialize all $\\vec{\\alpha_j}$, $f_j$ and $\\vec{\\beta_j}$ to something random. Let $j=1$. \n\t\\item Find the ``residual\" variance unexplained by all stages fit so far.\n\t\\item Project the input in to single dimension: $\\pmb{X} \\cdot \\vec{\\alpha_j}$.\n\t\\item Fit $f_j$ to a weighted residual target versus projections.\n\t\\item Use this $f_j$ to find a better setting for $\\vec{\\beta_j}$.\n\t\\item Use a Gauss-Newton scheme to solve for an update to $\\vec{\\alpha_j}$.\n\t\\item Repeat steps 3-6 until $f_j$, $\\vec{\\beta_j}$, and $\\vec{\\alpha_j}$ converge.\n\t\\item (optional) Use the newly converged parameters to retune all previous $f_t$, $\\vec{\\beta_t}$, $\\vec{\\alpha_t}$ where $t \\leq j$. (backfitting)\n\t\\item Increment $j$ and go back to step 2 until $j$ reaches $r$.\n\\end{enumerate}\n\nThe inner loop here is a form of \\textit{alternating optimization}, wherein all parameters except one are held constant, the best setting for that parameter given those constants is found, and the process cycled through all parameters until convergence. The outer loop builds up the additive model.\n\nBut this leaves some details unexplained. How exactly is the residual found? How are parameters found given fixed solutions to the others? How do we test for convergence?\n\n\\subsection{Finding The Residual}\n\nThe residual trick cleverly separates the contribution of the $j$\\textsuperscript{th} stage from the other terms in the additive model. Consider rephrasing the loss function as:\n\n$$loss = \\sum_{i=1}^n w_i \\sum_{k=1}^d w_k [y_{ik} - \\sum_{t \\neq j} f_t(\\vec{x_i} \\cdot \\vec{\\alpha_t}) \\otimes \\beta_{tk} - f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk}]^2$$\n\nNow if we let\n\n$$r_{ijk} = y_{ik} - \\sum_{t \\neq j} f_t(\\vec{x_i} \\cdot \\vec{\\alpha_t}) \\otimes \\beta_{tk}$$\n\nthen\n\n$$loss,\\ L = \\sum_{i=1}^n w_i \\sum_{k=1}^d w_k [r_{ijk} - f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk}]^2$$\n\nIn practice we will wish to find all $r_{ijk}$ for a particular $j$. Call this $\\pmb{R}_j \\in \\mathbb{R}^{n \\times d}$, the same space as the output. It can be found with\n\n$$\\pmb{R}_j = \\pmb{Y} - \\sum_{t \\neq j} f_t(\\pmb{X} \\cdot \\vec{\\alpha_t}) \\otimes \\vec{\\beta_t}$$\n\nwhere $f_t(\\pmb{X} \\cdot \\vec{\\alpha_t})$ is actually an $n$-vector.\n\n\\subsection{Optimizing $\\vec{\\beta_j}$ Given $\\vec{\\alpha_j}$ and $f_j$}\n\n$$loss\\ for\\ the\\ jth\\ term,\\ L_j = \\sum_{i=1}^n w_i \\sum_{k=1}^d w_k [r_{ijk} - f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk}]^2$$\n\nTo optimize with respect to a parameter, use good ol' calculus: Take a derivative, set equal to zero, and solve. Let's select $\\beta_{jk'}$, the $(k=k')$\\textsuperscript{th} entry of $\\vec{\\beta_j}$ as the parameter of interest.\n\n$$\\frac{\\partial L_j}{\\partial \\beta_{jk'}} = \\sum_{i=1}^n w_i w_{k'} [2(r_{ijk'} - f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk'})(-f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j}))] = 0$$\n\nNotice that the sum over $k$ disappears because no term where $k \\neq k'$ will contain our variable $\\beta_{jk'}$, so for the purposes of differentiation they are constant, and the derivative of constants is zero. Only the weight $w_{k'}$ remains.\n\nAlso, I've been using $\\otimes$ for consistency, but since $\\beta_{jk'}$ is scalar, an outer product is nothing special, just an ordinary multiplication. So we can do some algebra:\n\n$$ -2 w_{k'} \\sum_{i=1}^n w_i [r_{ijk'} f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j})] + 2 w_{k'} \\beta_{jk'} \\sum_{i=1}^n w_i [f_j^2(\\vec{x_i} \\cdot \\vec{\\alpha_j})] = 0 $$\n\n$$\\rightarrow \\beta_{jk'} = \\frac{\\sum_{i=1}^n w_i [r_{ijk'} f_j(\\vec{x_i} \\cdot \\vec{\\alpha_j})]}{\\sum_{i=1}^n w_i [f_j^2(\\vec{x_i} \\cdot \\vec{\\alpha_j})]}$$\n\nThis can be vectorized to find all entries of $\\vec{\\beta_j}$ at once:\n\n$$ \\vec{\\beta_j} = \\frac{\\pmb{R}_j^T \\cdot (\\vec{w_I} \\odot f_j(\\pmb{X} \\cdot \\alpha_j))}{f_j(\\pmb{X} \\cdot \\alpha_j) \\cdot (\\vec{w_I} \\odot f_j(\\pmb{X} \\cdot \\alpha_j))}$$\n\nwhere $\\odot$ is a Hadamard product, $\\vec{w_I}$ is an $n$-vector, $\\pmb{R}_j$ is the residuals, and the dot products are taken along the length-$n$ dimensions of these objects.\n\n\\subsection{Optimizing $f_j$ Given $\\vec{\\alpha_j}$ and $\\vec{\\beta_j}$}\n\nNow a similar argument, but this time consider the parameter of interest to be $f_{i'j}$, the $(i=i')$\\textsuperscript{th} entry of the $n$-vector formed by taking the inner product of $\\pmb{X}$ with $\\alpha_j$ and applying $f_j$ to each entry.\n\n\n$$\\frac{\\partial L_j}{\\partial f_{i'j}} = w_{i'} \\sum_{k=1}^d w_k [2(r_{i'jk} - f_{i'j} \\otimes \\beta_{jk})(-\\beta_{jk})] = 0$$\n\nThe sum over $i$ disappears because only the single term where $i=i'$ isn't constant to the derivative.\n\n$$\\rightarrow w_{i'} \\sum_{k=1}^d w_k [-2 r_{i'jk} \\beta_{jk} + 2 f_{i'j} \\beta_{jk}^2 ] = 0$$\n\n$$\\rightarrow 2 f_{i'j} \\sum_{k=1}^d w_k \\beta_{jk}^2 = 2 \\sum_{k=1}^d w_k r_{i'jk} \\beta_{jk}$$\n\n$$\\rightarrow f_{i'j} = \\frac{\\sum_{k=1}^d w_k r_{i'jk} \\beta_{jk}}{\\sum_{k=1}^d w_k \\beta_{jk}^2}$$\n\nAs in the case of $\\vec{\\beta_j}$, this can be vectorized.\n\n$$ f_j(\\pmb{X} \\cdot \\vec{\\alpha_j}) = \\frac{\\pmb{R}_j \\cdot (\\vec{w_K} \\odot \\vec{\\beta_j})}{\\vec{\\beta_j} \\cdot (\\vec{w_K} \\odot \\vec{\\beta_j})}$$\n\nThis provides targets for the funtion $f_j$. The task is to find the function that maps from this input to this output, for which there are numerous solvers (finding a polynomial by reducing to a linear inverse problem, for example). The example weights $w_i$ disappear in the algebra and so do not affect the targets, but they can be passed on to the function-fitter so it considers some examples more important than others.\n\n\\subsection{Optimizing $\\vec{\\alpha_j}$ Given $\\vec{\\beta_j}$ and $f_j$}\n\nThis is by far the toughest set of parameters to optimize, because they are nested inside the function. This time express the loss as:\n\n$$L_j = \\sum_{k=1}^d w_k [\\vec{w_I} \\odot \\vec{g_{jk}}^2]$$\n\nwhere\n\n$$ \\vec{g_{jk}} = \\vec{r_{jk}} - f_j(\\pmb{X} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk}$$\n\nwhere $\\vec{r_{jk}}$ is the $n$-vector formed by stacking $r_{ijk} \\forall i$ together, or equivalently the $k$\\textsuperscript{th} column of $\\pmb{R}_j$.\n\nThe weights can be factored in to the square to yield a form solveable with \\href{https://en.wikipedia.org/wiki/Gauss%E2%80%93Newton_algorithm#Description}{Gauss-Newton}.\n\n$$L_j = \\sum_{k=1}^d \\vec{g_{jkw}}(\\vec{\\alpha_j})^2,$$\n\n$$\\vec{g_{jkw}} = \\sqrt{w_k} \\sqrt{\\vec{w_I}} \\odot (\\vec{r_{jk}} - f_j(\\pmb{X} \\cdot \\vec{\\alpha_j}) \\otimes \\beta_{jk})$$\n\nFind the Jacobian:\n\n$$J_{jk}[u,v] = \\frac{\\partial \\vec{g_{jkw}}(\\vec{\\alpha_j})[u]}{\\partial \\vec{\\alpha_j}[v]} = -\\sqrt{w_k} \\sqrt{w_u} \\dot{f_j}(\\vec{x_u} \\cdot \\vec{\\alpha_j}) \\beta_{jk} x_{uv}$$\n\nwhere $w_u$ is $\\vec{w_I}[u]$, the $u$\\textsuperscript{th} example weight, as distinct from $w_k$, which, remember, are output dimension weights.\n\n\\[\n\\rightarrow J_{jk} = -\\sqrt{w_k} \\beta_{jk}\n\\begin{bmatrix}\n\t\\sqrt{w_0} \\dot{f_j}(\\vec{x_0} \\cdot \\vec{\\alpha_j}) x_{00} & \\sqrt{w_0} \\dot{f_j}(\\vec{x_0} \\cdot \\vec{\\alpha_j}) x_{01} & \\dots & \\sqrt{w_0} \\dot{f_j}(\\vec{x_0} \\cdot \\vec{\\alpha_j}) x_{0p} \\\\\n \\sqrt{w_1} \\dot{f_j}(\\vec{x_1} \\cdot \\vec{\\alpha_j}) x_{10} & \\sqrt{w_1} \\dot{f_j}(\\vec{x_1} \\cdot \\vec{\\alpha_j}) x_{11} & \\dots & \\sqrt{w_1} \\dot{f_j}(\\vec{x_1} \\cdot \\vec{\\alpha_j}) x_{1p}\\\\\n \\vdots & & \\ddots & \\vdots\\\\\n \\sqrt{w_n} \\dot{f_j}(\\vec{x_n} \\cdot \\vec{\\alpha_j}) x_{n0} & \\hdots & \\hdots & \\sqrt{w_n} \\dot{f_j}(\\vec{x_n} \\cdot \\vec{\\alpha_j}) x_{np}\n\\end{bmatrix}\n\\]\n\n$$= -\\sqrt{w_k} \\beta_{jk} (\\sqrt{\\vec{w_I}} \\odot \\dot{f_j}(\\pmb{X} \\cdot \\vec{\\alpha_j})) \\odot \\pmb{X}$$\n\nwhere everything left of the last $\\odot$ simplifies to an $n$-vector, so the Hadamard product is taken with each column of $\\pmb{X}$ individually.\\newline\n\nAs per Gauss-Newton, the update to the parameter $\\vec{\\alpha_j}$ to the function $\\vec{g_{jkw}}$ is given by the solution $\\vec{\\delta}$ to:\n\n$$\\left[ \\sum_{k=1}^d J_{jk}^T J_{jk} \\right] \\vec{\\delta} = \\sum_{k=1}^d J_{jk}^T \\vec{g_{jkw}}$$\n\nOn the left side is a $p \\times p$ matrix, and on the right a $p \\times 1$ vector, so we have an easy-to-solve linear inverse problem.\n\n$$ \\vec{\\alpha_j} = \\vec{\\alpha_j} + \\vec{\\delta} $$\n\nIn practice $\\vec{\\alpha_j}$ is renormalized after this update to avoid numerical drift and keep it from blowing up. Because $\\vec{\\alpha_j}$ is a projection vector only meant to flatten the data along some dimension, its magnitude does not really matter.\n\n\\subsection{Testing for Convergence}\n\nParameters are considered ``converged\" when their values across iterations cause a small-enough change in the value of the loss function.\n\nBecause we only make updates to the parameters governing stage $j$, we can make use the convenient identity\n\n$$\\pmb{Y} - \\pmb{\\hat{Y}} = \\pmb{R}_j - f_j(\\pmb{X} \\cdot \\vec{\\alpha_j}) \\otimes \\vec{\\beta_j}^T$$\n\nto calculate the loss quickly.\n\n\\section{Classification}\n\nLet the miscalssification risk $R$ be\n\n$$R = \\sum_{i=1}^n \\min_{k \\in [1,q]} \\sum_{c=1}^q l_{ck} p(c | \\vec{x_i})$$\n\nwhere\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $i$ iterates over examples\n\t\\item $q$ is the total number of classes in the problem\n\t\\item $\\min_k$ implements the optimal decision rule for each example\n\t\\item $l_{ck}$ is the user-specified loss for predicting $y=k$ when in truth $y=c$\n\t\\item the inner sum is the total loss for predicting $y=k$\n\t\\item $p(c | \\vec{x_i})$ is the true probability $y=c$ given input $\\vec{x_i}$\n\\end{itemize}\n\nThe unknown here is that conditional probability. If we define an indicator variable\n\n$$h_{ci} = 1\\ if\\ y_i=c,\\ 0\\ otherwise$$\n\nthen Friedman says we can rewrite the conditional probability as\n\n$$p(c | \\vec{x_i}) = \\frac{\\pi_c S}{s_c} E[h_{ci} | \\vec{x_i}]$$\n\nwhere\n\\begin{itemize}\n\t\\setlength\\itemsep{-2mm}\n\t\\item $\\pi_c$ is the prior probability that $y_i=c$ ($h_{ci}=1$), calculable from the training set with $\\frac{\\sum_{i=1}^n h_{ci}}{n}$\n\t\\item $s_c = \\sum_{i=1}^n w_i h_{ci}$, the cumulative weight of examples with classification $c$\n\t\\item $S = \\sum_{c=1}^q s_c$, the cumulative weight of everything, a constant\n\t\\item $E$ means the expected value\n\\end{itemize}\n\nThe coefficient can be expanded to\n\n$$\\frac{\\sum_{c=1}^q \\sum_{i=1}^n w_i h_{ci} * \\frac{\\sum_{i=1}^n h_{ci}}{n}}{\\sum_{i=1}^n w_i h_{ci}} \\propto \\frac{\\sum_{i=1}^n h_{ci}}{\\sum_{i=1}^n w_i h_{ci}}$$\n\nThis doesn't make any sense, because it says probabilities are \\textit{inversely} related to example weights, but we really want the probability of predicting a given class to be \\textit{directly} related to the importances of examples in that class. Convinced Friedman accidentally inverted the equation, the code and the following equations reinvert his inversion.\n\nNotice that if the weights $w_i$ in $s_c$ are uniform (so no example is considered any more important than any other), then $\\pi_c = s_c/S$, and all those terms cancel (regardless of the inversion).\n\nAdditionally, $l_{ck}$ is often simplified as\n\n$$l_{ck} = 1\\ if\\ c \\neq k,\\ 0\\ if\\ c=k$$\n\nSo rewrite the risk as\n\n$$R = \\sum_{i=1}^n \\min_{k \\in [1,q]} \\frac{1}{S} \\sum_{c=1}^q \\frac{s_c l_{ck}}{\\pi_c} E[h_{ci} | \\vec{x_i}]$$\n\nAnd with the simplifying assumptions that all examples are equally important and misclassification is equally bad between all class pairs it becomes:\n\n$$R = \\sum_{i=1}^n \\min_{k \\in [1,q]} \\sum_{c \\neq k} E[h_{ci} | \\vec{x_i}]$$\n\nOr equivalently:\n\n$$R = \\sum_{i=1}^n \\max_{k \\in [1,q]} E[h_{ci} | \\vec{x_i}]$$\n\nbecause the sum is minimized by excluding the largest expectation.\\newline\n\nNow, recognize $E[\\vec{h_c} | \\pmb{X}]$ is a vector of the values $E[h_{ci} | \\vec{x_i}]$, and for training data the expectation that $h_{ci}$ has a given value given $\\vec{x_i}$ is known to be either a one or a zero, so $E[\\vec{h_c} | \\pmb{X}] = \\vec{h_c}$, $h_{ci} \\forall i$ stacked together.\n\nFurther, recognize that stacking $\\vec{h_c}$ for all classes $c$ together as columns yields $\\pmb{H}$, a one-hot representation of the true classifications $\\pmb{Y}$. That is:\n\n\\[\n\\pmb{Y} =\n\\begin{bmatrix}\n1\\\\\n0\\\\\n2\\\\\n.\\\\\n.\\\\\n3\\\\\n\\end{bmatrix}\n%%%%%\n\\ \\ \\ \\ \\ \\ \\ \\pmb{H} =\n\\begin{bmatrix}\n0 & 1 & 0 & 0\\\\\n1 & 0 & 0 & 0\\\\\n0 & 0 & 1 & 0\\\\\n& \\hdots & \\hdots\\\\\n& \\hdots & \\hdots\\\\\n0 & 0 & 0 & 1\n\\end{bmatrix}\n\\]\n\nAnd now we can model $\\pmb{H}$ with a multivariate projection pursuit regression model, where we take the predicted class of example $i$ to be:\n\n$$\\hat{y_i} = \\argmax_c \\hat{h_{ci}} $$\n\nThat is: the predicted class is the index of the column where the largest value in the $i$\\textsuperscript{th} row of $\\hat{\\pmb{H}}$, the predicted $\\pmb{H}$, is located. If $\\pmb{Y}$ is filled with generalized categories rather than numbers, then categoricals can be assigned numbers for the construction of $\\pmb{H}$, and argmaxes can be translated back at prediction-time.\n\nAnd just like that we have reduced univariate classification to multivariate regression! There is one caveat: Training the model to make these predictions should ideally involve optimizing the the misclassification risk as the loss function, not the sum-of-squares loss function from section 2 as is done for actual regression. But the $\\max_k$ in the risk equation makes it nonconvex, which means we can no longer employ the methods detailed in section 3 to find model parameters.\n\nFortunately, Friedman assures us (and experiment bears out) that using the quatratic loss function is acceptable, and if we wish to account for examples being of differing importances or specify a funky non-uniform pairwise loss scheme, all we have to do is use weights:\n\n$$w_c = \\frac{s_c}{S \\pi_c} \\sum_{k \\in [1,q]} l_{ck}$$\n\nSince weights are relative, the constant $S$ and the normalization by the constant $n$ in the calculation of $\\pi_c$ can be dropped to yeild:\n\n$$w_c = \\frac{\\sum_{i=1}^n w_i h_{ci}}{\\sum_{i=1}^n h_{ci}} \\sum_{k \\in [1,q]} l_{ck}$$\n\nVectorizing to obtain this weight for all classes at once we get:\n\n$$\\vec{w_C} = \\frac{\\vec{w_I} \\cdot \\pmb{H}}{\\sum_{i=1}^n \\pmb{H}} \\odot \\sum_{k=1}^q \\pmb{L}$$\n\nwhere $\\pmb{L}[c,k] = l_{ck}$, the sum over $k$ amounts to a sum along the horizontal (second) axis, the sum over $n$ of $\\pmb{H}$ amounts to a sum along the vertical (first) axis, and the division is pointwise on the vectors involved.\n\n\\end{document}\n", "meta": {"hexsha": "c25501d74eb12864c7246806019912731565faee", "size": 22286, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/math.tex", "max_stars_repo_name": "pavelkomarov/projection-pursuit-regression", "max_stars_repo_head_hexsha": "d8d4d0a72a0ba7f64d240bc3b14129ccd9c7cf9e", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 29, "max_stars_repo_stars_event_min_datetime": "2018-02-08T21:24:01.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-29T03:05:26.000Z", "max_issues_repo_path": "doc/math.tex", "max_issues_repo_name": "pavelkomarov/projection-pursuit-regression", "max_issues_repo_head_hexsha": "d8d4d0a72a0ba7f64d240bc3b14129ccd9c7cf9e", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-12-20T00:22:05.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-09T19:59:57.000Z", "max_forks_repo_path": "doc/math.tex", "max_forks_repo_name": "pavelkomarov/projection-pursuit-regression", "max_forks_repo_head_hexsha": "d8d4d0a72a0ba7f64d240bc3b14129ccd9c7cf9e", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2019-02-24T00:28:53.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-10T01:48:03.000Z", "avg_line_length": 59.5882352941, "max_line_length": 644, "alphanum_fraction": 0.7099973077, "num_tokens": 7091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127603871312, "lm_q2_score": 0.7461389930307512, "lm_q1q2_score": 0.6371376071713631}} {"text": "\\chapter{Breaking the continuum hypothesis}\nWe now use the technique of forcing to break the\nContinuum Hypothesis by choosing a good poset $\\Po$.\nAs I mentioned earlier, one can also build a model\nwhere the Continuum Hypothesis is true;\nthis is called the \\emph{constructible universe},\n(this model is often called ``$V=L$'').\nHowever, I think it's more fun when things break\\dots\n\n%\\section{Forcing $V \\neq L$ is really easy}\n%As a small aside, to check we're on the right track we show the following result.\n%\n%\\begin{theorem}[$V \\ne L$]\n%\tLet $M$ be a countable transitive model of $\\ZFC$.\n%\tLet $\\Po \\in M$ be \\emph{any} splitting poset,\n%\tand let $G \\subseteq \\Po$ be $M$-generic.\n%\tThen $M[G] \\vDash (V \\neq L)$.\n%\\end{theorem}\n%\\begin{proof}\n%\tSince $L$ has a $\\Sigma_1$ definition,\n%\twe have \\[ L^{M[G]} = L^M \\subseteq M \\subsetneq M[G] \\]\n%\twhere the last part follows from $G \\notin M[G]$.\n%\\end{proof}\n%\n%Thus $M[G] \\vDash \\ZFC + (V \\ne L)$ for any splitting poset $\\Po$,\n%and we are one step closer to breaking $\\CH$.\n\n\\section{Adding in reals}\nStarting with a \\emph{countable} transitive model $M$.\n\nWe want to choose $\\Po \\in M$ such that $(\\aleph_2)^M$ many real numbers appear,\nand then worry about cardinal collapse later.\n\nRecall the earlier situation where we set $\\Po$ to be the infinite complete binary tree; its nodes can be thought of as partial functions $n \\to 2$ where $n < \\omega$.\nThen $G$ itself is a path down this tree; i.e.\\ it can be encoded as a total function $G : \\omega \\to 2$,\nand corresponds to a real number.\n\n\\begin{center}\n\t\\begin{asy}\n\t\tsize(8cm);\n\t\tpair P = Drawing(\"\\varnothing\", (0,4), dir(90), red);\n\t\tpair P0 = Drawing(\"0\", (-5,2), 1.5*dir(90), red);\n\t\tpair P1 = Drawing(\"1\", (5,2), 1.5*dir(90));\n\t\tpair P00 = Drawing(\"00\", (-7,0), 1.4*dir(120));\n\t\tpair P01 = Drawing(\"01\", (-3,0), 1.4*dir(60), red);\n\t\tpair P10 = Drawing(\"10\", (3,0), 1.4*dir(120));\n\t\tpair P11 = Drawing(\"11\", (7,0), 1.4*dir(60));\n\n\t\tpair P000 = Drawing(\"000\", (-8,-3));\n\t\tpair P001 = Drawing(\"001\", (-6,-3));\n\t\tpair P010 = Drawing(\"010\", (-4,-3), red);\n\t\tpair P011 = Drawing(\"011\", (-2,-3));\n\n\t\tpair P100 = Drawing(\"100\", (2,-3));\n\t\tpair P101 = Drawing(\"101\", (4,-3));\n\t\tpair P110 = Drawing(\"110\", (6,-3));\n\t\tpair P111 = Drawing(\"111\", (8,-3));\n\n\t\tdraw(P01--P0--P00);\n\t\tdraw(P11--P1--P10);\n\t\tdraw(P0--P--P1);\n\t\tdraw(P000--P00--P001);\n\t\tdraw(P100--P10--P101);\n\t\tdraw(P010--P01--P011);\n\t\tdraw(P110--P11--P111);\n\n\t\tdraw(P--P0--P01--P010--(P010+2*dir(-90)), red+1.4);\n\t\tMP(\"G\", P010+2*dir(-90), dir(-90), red);\n\t\\end{asy}\n\\end{center}\n\nWe want to do something similar,\nbut with $\\omega_2$ many real numbers instead of just one.\nIn light of this, consider in $M$ the poset\n\\[\n\t\\Po = \\opname{Add} \\left( \\omega_2, \\omega \\right)\n\t\\defeq \\left( \\left\\{ p : \\omega_2 \\times \\omega \\to 2,\n\t\t\\dom(p) \\text{ is finite} \\right\\},\n\t\\supseteq \\right).\n\\]\nThese elements $p$ (conditions) are ``partial functions'':\nwe take some finite subset of $\\omega_2 \\times \\omega$ and map it into $2=\\{0,1\\}$.\n(Here $\\dom(p)$ denotes the domain of $p$,\nwhich is the finite subset of $\\omega_2 \\times \\omega$ mentioned.)\nMoreover, we say $p \\le q$ if $\\dom(p) \\supseteq \\dom(q)$\nand the two functions agree over $\\dom(q)$.\n\n\\begin{ques}\n\tWhat is the maximal element $1_\\Po$ here?\n\\end{ques}\n\n\\begin{exercise}\n\tShow that a generic $G$ can be encoded as a function $\\omega_2 \\times \\omega \\to 2$.\n\\end{exercise}\n\n%Let $G \\subseteq \\opname{Add}(\\omega_2, \\omega)$ be an $M$-generic.\n%We claim that, like in the binary case, $G$ can be encoded as a function $\\omega_2 \\times \\omega \\to 2$.\n%To see this, consider $\\alpha \\in \\omega_2$ and $n \\in \\omega$; we have the dense set\n%\\[ D_{\\alpha, n}\n%\t= \\left\\{ p \\in \\opname{Add}(\\omega_2, \\omega)\n%\t\\mid (\\alpha, n) \\in \\dom(p) \\right\\}\n%\\]\n%(this is obviously dense, given any $p$ add in $(\\alpha, n)$ if it's not in there already).\n%So $G$ hits this dense set, meaning that for every $(\\alpha, n)$ there's a function in $G$ which defines it.\n%Using the fact that $G$ is upwards closed and a filter, we may as before we may interpret $G$ as a function $\\omega_2 \\times \\omega \\to 2$.\n\n\\begin{lemma}[$G$ encodes distinct real numbers]\n\tFor $\\alpha \\in \\omega_2$ define\n\t\\[ G_\\alpha = \\left\\{ n \\mid G\\left( \\alpha,n \\right) = 0 \\right\\} \\in \\PP(\\NN). \\]\n\tThen $G_\\alpha \\neq G_\\beta$ for any $\\alpha \\neq \\beta$.\n\\end{lemma}\n\\begin{proof}\n\tWe claim that the set\n\t\\[ D = \\left\\{ q \\mid \\exists n \\in \\omega :\n\t\tq\\left( \\alpha, n \\right) \\neq q\\left( \\beta, n \\right)\n\t\t\\text{ are both defined}\n\t\\right\\} \\]\n\tis dense.\n\t\\begin{ques}\n\t\tCheck this.\n\t\t(Use the fact that the domains are all finite.)\n\t\\end{ques}\n%\tThis is pretty easy to see.\n%\tConsider $p \\in \\opname{Add}(\\omega_2, \\omega)$.\n%\tThen you can find an $n$ such that\n%\tneither $(\\alpha, n)$ nor $(\\beta, n)$ is defined,\n%\tjust because $\\dom(p)$ is finite.\n%\tThen you make $p'$ as $p$ plus $p'( (\\alpha, n) ) = 1$\n%\tand $p'( (\\beta, n) ) = 0$.\n%\tHence the set is dense.\n\n\tSince $G$ is an $M$-generic it hits this dense set $D$.\n\tHence $G_\\alpha \\neq G_\\beta$.\n\\end{proof}\n\nSince $G \\in M[G]$ and $M[G] \\vDash \\ZFC$,\nit follows that each $G_\\alpha$ is in $M[G]$.\nSo there are at least $\\aleph_2^M$ real numbers in $M[G]$.\nWe are done once we can show there is no cardinal collapse.\n\n\\section{The countable chain condition}\nIt remains to show that with $\\Po = \\opname{Add}(\\omega, \\omega_2)$, we have that\n\\[ \\aleph_2^{M[G]} = \\aleph_2^M. \\]\nIn that case, since $M[G]$ will have $\\aleph_2^M = \\aleph_2^{M[G]}$ many reals, we will be done.\n\nTo do this, we'll rely on a combinatorial property of $\\Po$:\n\\begin{definition}\n\tWe say that $A \\subseteq \\mathcal P$ is a \\vocab{strong antichain}\n\tif for any distinct $p$ and $q$ in $A$, we have $p \\perp q$.\n\\end{definition}\n\\begin{example}[Example of an antichain]\n\tIn the infinite binary tree, \n\tthe set $A = \\{00, 01, 10, 11\\}$ is a strong antichain\n\t(in fact maximal by inclusion).\n\\end{example}\nThis is stronger than the notion of ``antichain'' than you might be used to!\\footnote{%\n\tIn the context of forcing, some authors use ``antichain'' to refer to ``strong antichain''.\n\tI think this is lame.}\nWe don't merely require that every two elements are incomparable,\nbut that they are in fact \\emph{incompatible}.\n\\begin{ques}\n\tDraw a finite poset and an antichain of it which is not strong.\n\\end{ques}\n\n\\begin{definition}\n\tA poset $\\Po$ has the \\vocab{$\\kappa$-chain condition}\n\t(where $\\kappa$ is a cardinal) if all strong antichains\n\tin $\\Po$ have size less than $\\kappa$.\n\tThe special case $\\kappa = \\aleph_1$ is called the \\vocab{countable chain condition},\n\tbecause it implies that every strong antichain is countable.\n\\end{definition}\n\nWe are going to show that if the poset has the $\\kappa$-chain condition\nthen it preserves all cardinals greater than $\\kappa$.\n% or was it > \\kappa?\nIn particular, the countable chain condition will show that $\\Po$ preserves all the cardinals.\nThen, we'll show that $\\opname{Add}(\\omega, \\omega_2)$ does indeed have this property.\nThis will complete the proof.\n\nWe isolate a useful lemma:\n\\begin{lemma}[Possible values argument]\n\tSuppose $M$ is a transitive model of $\\ZFC$ and $\\Po$ is a partial order\n\tsuch that $\\Po$ has the $\\kappa$-chain condition in $M$.\n\tLet $X,Y \\in M$ and let $f: X \\to Y$\n\tbe some function in $M[G]$, but $f \\notin M$.\n\n\tThen there exists a function $F \\in M$, with $F: X \\to \\PP(Y)$ and such that\n\tfor any $x \\in X$,\n\t\\[ f(x) \\in F(x) \\quad\\text{and}\\quad \\left\\lvert F(x) \\right\\rvert^M < \\kappa. \\]\n\\end{lemma}\nWhat this is saying is that if $f$ is some new function that's generated,\n$M$ is still able to pin down the values of $f$ to at most $\\kappa$ many values.\n\n\\begin{proof}\n\tThe idea behind the proof is easy: any possible value of $f$ gives us some condition in\n\tthe poset $\\Po$ which forces it.\n\tSince distinct values must have incompatible conditions,\n\tthe $\\kappa$-chain condition guarantees\n\tthere are at most $\\kappa$ such values.\n\n\tHere are the details.\n\tLet $\\dot f$, $\\check X$, $\\check Y$ be names for $f$, $X$, $Y$.\n\tStart with a condition $p$ such that $p$ forces the sentence\n\t\\[ \\text{``$\\dot f$ is a function from $\\check X$ to $\\check Y$''}. \\]\n\tWe'll work just below here.\n\n\tFor each $x \\in X$, we can consider (using the Axiom of Choice) a maximal strong antichain $A(x)$\n\tof incompatible conditions $q \\le p$ which forces $f(x)$ to equal some value $y \\in Y$.\n\tThen, we let $F(x)$ collect all the resulting $y$-values.\n\tThese are all possible values, and there are less than $\\kappa$ of them.\n\\end{proof}\n\n\\section{Preserving cardinals}\nAs we saw earlier, cardinal collapse can still occur.\nFor the Continuum Hypothesis we want to avoid this possibility,\nso we can add in $\\aleph_2^M$ many real numbers and have $\\aleph_2^{M[G]} = \\aleph_2^M$.\nIt turns out that to verify this, one can check a weaker result.\n\n\\begin{definition}\n\tFor $M$ a transitive model of $\\ZFC$ and $\\Po \\in M$ a poset,\n\twe say $\\Po$ \\vocab{preserves cardinals} if\n\t$\\forall G \\subseteq \\Po$ an $M$-generic,\n\tthe model $M$ and $M[G]$ agree on the sentence ``$\\kappa$ is a cardinal'' for every $\\kappa$.\n\tSimilarly we say $\\Po$ \\vocab{preserves regular cardinals} if $M$ and $M[G]$\n\tagree on the sentence ``$\\kappa$ is a regular cardinal'' for every $\\kappa$.\n\\end{definition}\nIntuition:\nIn a model $M$, it's possible that two ordinals which are in bijection in $V$ are no longer in bijection in $M$.\nSimilarly, it might be the case that some cardinal $\\kappa \\in M$ is regular,\nbut stops being regular in $V$ because some function $f : \\ol\\kappa \\to \\kappa$ is cofinal but happened to only exist in $V$.\nIn still other words, ``$\\kappa$ is a regular cardinal '' turns out to be a $\\Pi_1$ statement too.\n\nFortunately, each implies the other.\nWe quote the following without proof.\n\\begin{proposition}[Preserving cardinals $\\iff$ preserving regular cardinals]\n\tLet $M$ be a transitive model of $\\ZFC$.\n\tLet $\\Po \\in M$ be a poset.\n\tThen for any $\\lambda$,\n\t\t$\\Po$ preserves cardinalities less than or equal to $\\lambda$\n\t\tif and only if $\\Po$ preserves regular cardinals less than or equal to $\\lambda$.\n\tMoreover the same holds if we replace ``less than or equal to''\n\tby ``greater than or equal to''.\n\\end{proposition}\n\nThus, to show that $\\Po$ preserves cardinality and cofinalities\nit suffices to show that $\\Po$ preserves regularity.\nThe following theorem lets us do this:\n\\begin{theorem}[Chain conditions preserve regular cardinals]\n\tLet $M$ be a transitive model of ZFC, and let $\\Po \\in M$ be a poset.\n\tSuppose $M$ satisfies the sentence ``$\\Po$ has the $\\kappa$ chain condition and $\\kappa$ is regular''.\n\tThen $\\Po$ preserves regularity greater than or equal to $\\kappa$.\n\\end{theorem}\n\\begin{proof}\n\tUse the Possible Values Argument.\n\t\\Cref{prob:chain}.\n\\end{proof}\n\nIn particular, if $\\Po$ has the countable chain condition then $\\Po$ preserves \\emph{all} the cardinals (and cofinalities).\nTherefore, it remains to show that $\\opname{Add}(\\omega, \\omega_2)$ satisfies the countable chain condition.\n\n\\section{Infinite combinatorics}\nWe now prove that $\\opname{Add}(\\omega, \\omega_2)$ satisfies the countable chain condition.\nThis is purely combinatorial, and so we work briefly.\n\n\\begin{definition}\n\tSuppose $C$ is an uncountable collection of finite sets.\n\t$C$ is a \\vocab{$\\Delta$-system} if there exists a \\vocab{root} $R$\n\twith the condition that for any distinct $X$ and $Y$\n\tin $C$, we have $X \\cap Y = R$.\n\\end{definition}\n\n\\begin{lemma}\n\t[$\\Delta$-System lemma] Suppose $C$ is an uncountable collection of finite sets.\n\tThen $\\exists \\ol C \\subseteq C$ such that $\\ol C$ is an uncountable $\\Delta$-system.\n\\end{lemma}\n\\begin{proof}\n\tThere exists an integer $n$ such that $C$ has uncountably many guys of length $n$.\n\tSo we can throw away all the other sets, and just assume that all sets in $C$ have size $n$.\n\n\tWe now proceed by induction on $n$.\n\tThe base case $n=1$ is trivial, since we can just take $R = \\varnothing$.\n\tFor the inductive step we consider two cases.\n\n\tFirst, assume there exists an $a \\in C$ contained in uncountably many $F \\in C$.\n\tThrow away all the other guys.\n\tThen we can just delete $a$, and apply the inductive hypothesis.\n\n\tNow assume that for every $a$, only countably many members of $C$ have $a$ in them.\n\tWe claim we can even get a $\\ol C$ with $R = \\varnothing$.\n\tFirst, pick $F_0 \\in C$.\n\tIt's straightforward to construct an $F_1$ such that $F_1 \\cap F_0 = \\varnothing$.\n\tAnd we can just construct $F_2, F_3, \\dots$\n\\end{proof}\n\n\\begin{lemma}\n\tFor all $\\kappa$, $\\opname{Add}(\\omega, \\kappa)$ satisfies the countable chain condition.\n\\end{lemma}\n\\begin{proof}\n\tAssume not. Let\n\t\\[ \\left\\{ p_\\alpha : \\alpha < \\omega_1 \\right\\} \\]\n\tbe a strong antichain. Let\n\t\\[ C = \\left\\{ \\dom(p_\\alpha) : \\alpha < \\omega_1 \\right\\}. \\]\n\tLet $\\ol C \\subseteq C$ be such that $\\ol C$ is uncountable, and $\\ol C$ is a $\\Delta$-system with root $R$.\n\tThen let\n\t\\[ B = \\left\\{ p_\\alpha : \\dom(p_\\alpha) \\in R \\right\\}. \\]\n\tEach $p_\\alpha \\in B$ is a function $p_\\alpha : R \\to \\{0,1\\}$,\n\tso there are two that are the same.\n\\end{proof}\n\nThus, we have proven that the Continuum Hypothesis cannot be proven in $\\ZFC$.\n\n\\section\\problemhead\n\\begin{problem}\n\t\\label{prob:chain}\n\tLet $M$ be a transitive model of ZFC, and let $\\Po \\in M$ be a poset.\n\tSuppose $M$ satisfies the sentence ``$\\Po$ has the $\\kappa$ chain condition and $\\kappa$ is regular''.\n\tShow that $\\Po$ preserves regularity greater than or equal to $\\kappa$.\n\t\\begin{hint}\n\t\tAssume not, and take $\\lambda > \\kappa$ regular in $M$;\n\t\tif $f : \\ol \\lambda \\to \\lambda$,\n\t\tuse the Possible Values Argument on $f$ to generate a function in $M$\n\t\tthat breaks cofinality of $\\lambda$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tIt suffices to show that $\\Po$ preserves regularity greater than or equal to $\\kappa$.\n\t\tConsider $\\lambda > \\kappa$ which is regular in $M$,\n\t\tand suppose for contradiction that $\\lambda$ is not regular in $M[G]$.\n\t\tThat's the same as saying that there is a function $f \\in M[G]$,\n\t\t$f : \\ol \\lambda \\to \\lambda$ cofinal, with $\\ol \\lambda < \\lambda$.\n\t\tThen by the Possible Values Argument,\n\t\tthere exists a function $F \\in M$ from $\\ol \\lambda \\to \\PP(\\lambda)$\n\t\tsuch that $f(\\alpha) \\in F(\\alpha)$ and $\\left\\lvert F(\\alpha) \\right\\rvert^M < \\kappa$\n\t\tfor every $\\alpha$.\n\n\t\tNow we work in $M$ again.\n\t\tNote for each $\\alpha \\in \\ol\\lambda$,\n\t\t$F(\\alpha)$ is bounded in $\\lambda$ since $\\lambda$ is regular in $M$ and\n\t\tgreater than $\\left\\lvert F(\\alpha) \\right\\rvert$.\n\t\tNow look at the function $\\ol \\lambda \\to \\lambda$ in $M$ by just\n\t\t\\[ \\alpha \\mapsto \\cup F(\\alpha) < \\lambda. \\]\n\t\tThis is cofinal in $M$, contradiction.\n\t\\end{sol}\n\\end{problem}\n", "meta": {"hexsha": "bf270ad51cc206776403049efa7da959eff43952", "size": 14670, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/set-theory/CH.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/set-theory/CH.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/set-theory/CH.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.6453488372, "max_line_length": 167, "alphanum_fraction": 0.678868439, "num_tokens": 4781, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.837619979547273, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.6371361886829685}} {"text": "%\n% Teori.tex\n% Part of VertEgg documentation\n%\n\n\\chapter{Theory of Vertical Egg Distributions}\n\n\\section{The Equations}\\label{sec:equations}\n\n\nNeglecting horizontal processes, the evolution of the vertical\ndistribution\nof a class of fish eggs is governed by a simple \\emph{conservation}\nprinciple\\index{conservation principle}\n\n\\begin{equation}\\label{eq:consprin}\n\\begin{split}\n \\text{The change}\\ & \\text{in the number of eggs between two depth levels} \\\\\n & = \\text{the number of eggs entering at the lower level} \\\\\n & - \\text{the number of eggs leaving at the upper level} \\\\\n & + \\text{the number of eggs spawned} \\\\\n & - \\text{the number of eggs that hatched or died}\n\\end{split}\n\\end{equation}\n\nTo put this in a more mathematical formulation, let the variable\n$z$ denote depth and $t$ time. The $z$-axis is chosen to point\n\\emph{up}, that is $z = 0$ at the surface and negative values are\nused in the water column. This choice is opposite from\n\\cite{sund83,sund91} and \\cite{west89}. The present choice is\nmotivated by the most\ncommon axis-convention in 3D hydrodynamic ocean modelling.\n\nTo make the theory more convenient mathematically, a\n\\emphi{continuum} approach is used, where the eggs are\nreplaced by a continuous egg \\emphi{concentration}. Let $\\phi =\n\\phi(z,t)$ denote this egg concentration at depth $z$ and time\n$t$. Let $F = F(z,t)$ be the upwards egg \\emphi{flux}, i.e.\\ the net\nnumber of eggs passing upwards per time unit at depth $z$ and time\n$t$. Similarly, let $Q = Q(z,t)$ denote the \\emphi{source term}, the\nnumber of eggs spawned minus the numbers that hatched or died per\ndepth and time unit at depth $z$ and time $t$. The source term can\nalso be used to take care of other processes, such as loss of eggs by\nhorizontal advection.\n\nThe conservation principle \\ref{eq:consprin} can then be formulated\nin \\emphi{integral form}\n\\begin{equation}\\label{eq:intlaw}\n\\int_{z_1}^{z_2} \\!\\phi(z,t_2)\\,dz - \\int_{z_1}^{z_2}\\!\\phi(z,t_1)\\,dz\n = \\int_{t_1}^{t_2} \\!F(z_1,t)\\,dt\n - \\int_{t_1}^{t_2} \\!F(z_2,t)\\,dt\n + \\int_{t_1}^{t_2}\\!\\int_{z_1}^{z_2} \\!Q(z,t)\\,dzdt\n\\end{equation}\nMathematically this can be reformulated in mixed form, differential in time\nand integral in space\n\\begin{equation}\\label{eq:diffintlaw}\n \\pddt \\left( \\int_{z_1}^{z_2} \\!\\phi(z,t)\\,dz \\right) =\n F({z_1},t) - F({z_2},t) + \\int_{z_1}^{z_2}\\!Q(z,t)\\,dz\n \\quad \\mbox{for} \\ -H \\le {z_1} < {z_2} \\le 0 .\n\\end{equation}\nor in pure \\emph{differential} form\\index{differential form}\n\\begin{equation}\\label{eq:difflaw}\n \\Pd{\\phi}{t} = - \\Pd{F}{z} + Q\n\\end{equation}\n\nThe flux is decomposed in two parts, the \\emphi{convective flux} and the\n\\emphi{diffusive flux}. The convection is due to the terminal buoyant\nvelocity $w = w(z,t)$ of the egg which is computed by the density\ndifference from the surroundings and the egg size as described in section\n\\ref{sec:velocity}. The formulation is simply,\n\\begin{equation}\\label{eq:convflux}\n \\Fconv = w \\phi .\n\\end{equation}\nThe diffusion is caused by turbulent mixing and is modelled by\nFick's law\\index{Fick's law} using the vertical eddy diffusion\n\\index{eddy diffusion} $K = K(z,t)$,\n\\begin{equation}\\label{eq:fick}\n \\Fdiff = - K \\frac{\\partial \\phi}{\\partial z} .\n\\end{equation}\n\nOften the source term $Q$ can be separated in a\n\\emph{spawning}\\index{spawning term} or \\emph{production}\nterm\\index{production term} independent of the egg concentration\n\\begin{equation}\\label{eq:spawn}\n \\Qspawn = P\n\\end{equation}\nand a \\emph{mortality}\\index{mortality term} or \\emph{loss}\nterm\\index{loss term} depending on the concentration\n\\begin{equation}\\label{eq:loss}\n \\Qloss = - \\alpha \\phi\n\\end{equation}\nUsing this the differential conservation law (\\ref{eq:difflaw})\nbecomes\n\\bel{eq:difflawfull}\n\\Pd{\\phi}{t} = - \\pddz (w \\phi) + \\pddz (K \\Pd{\\phi}{z}) + P - \\alpha \\phi\n\\end{equation}\nThis parabolic partial differential equations is\ncalled the \\emph{convection-diffusion}\\index{convection-diffusion equation}\nor \\emph{transport} equation\\index{transport equation}.\n\nThe boundary conditions\\index{boundary conditions} are simply no flux\nacross the surface and bottom,\n\\bel{eq:boundcond}\n F(0,t) = F(-H,t) = 0,\n\\end{equation}\nand the initial condition is given by\n\\bel{eq:initcond}\n \\phi(z,0) = \\phi_0(z), \\quad -H \\le z \\le 0.\n\\end{equation}\n\n\\section{Solutions of the equations}\n\nUnder simplified circumstances the equations in section~\\ref{sec:equations}\ncan be solved analytically. This section examines some of these solutions.\n\n\\subsection{Vertical integrated equation}\\label{sec:vertint}\n\nLet $\\Phi$ denote the total concentration, $\\Phi = \\Vint \\phi dz$.\nTake the vertical integral of equation~(\\ref{eq:difflawfull}) and\nuse the boundary conditions (\\ref{eq:boundcond}) gives\n\\be\n\\Od{\\Phi}{t} = P^{tot} - \\Vint \\!\\alpha \\phi \\, dz\n\\end{equation}\nwhere $P^{tot} = \\Vint P dz$ is the total spawning contribution.\nWithout any source terms the total concentration $\\Phi$ is constant.\nTo reach a steady state solution with spawning, the loss term $\\alpha$\nmust be nonzero.\n\n\nIf $\\alpha$ is a positive\nconstant, the solution to the equation above is\n\\be\n \\Phi(t) = \\e^{-\\alpha t} \\Phi(0)\n + (1 - \\e^{-\\alpha t}) \\frac{P^{tot}}{\\alpha}\n\\end{equation}\nwith steady state solution $\\Phi = P^{tot}/\\alpha$.\nIf $\\alpha = 0$ the solution is simply\n\\be\n \\Phi(t) = \\Phi(0) + t P^{tot} .\n\\end{equation}\n\n\n\\subsection{Stationary solution}\\index{stationary solution}\n\nA stationary solution solves the conservation law\n\\pref{eq:difflaw} without the time derivative\n\\begin{equation}\\label{eq:stateq}\n \\Od{F}{z} = Q\n\\end{equation}\nwith boundary conditions $F(0) = F(-H) = 0$ given by \\pref{eq:boundcond}.\nThe condition for the existence of this solution is that the total\nsource term vanishes,\n\\begin{equation}\n Q^{tot} = \\Vint \\!Q \\, dz= 0\n\\end{equation}\nthe flux function is then given by integrating \\pref{eq:stateq}\n\\begin{equation}\n F(z) = \\int_{-H}^z \\! Q(s) \\, ds = - \\int_z^0 \\! Q(s) \\, ds .\n\\end{equation}\n\nWithout source term the expression above reduces to $F = 0$, i.e.\\ the\nnet flux is zero everywhere. This simply says that in a steady state\nconvection is balanced by diffusion\n\\bel{eq:sstate}\n w \\phi - K \\Pd{\\phi}{z} = 0 .\n\\end{equation}\nThis is an ordinary differential equation for $\\phi(z)$.\nThe boundary conditions \\pref{eq:boundcond} does not contain more information.\nA unique solution can nevertheless be singled out by the\nintegral condition\n\\begin{equation}\\label{eq:intcond}\n \\Vint \\!\\phi(z)\\,dz = \\Phi .\n\\end{equation}\nIn other words, among all the solutions of (\\ref{eq:sstate})\nchoose the one with correct vertically integrated concentration.\n\nTo simplify the notation put $m = w / K$ and let $M(z) = - \\int_z^0 m(s)\nds$. Then the stationary solution is\n\\begin{equation}\\label{eq:sstatesol}\n \\phi(z) = \\frac{\\Phi}{\\int_{-H}^0 \\e^{M(s)} ds} \\e^{M(z)} .\n\\end{equation}\n\n\\subsubsection{Constant coefficients}\n\nThe case with constant coefficients was studied by Sundby\n\\shortcite{sund83}. In this case $M(z) = m z$ and the solution is a\ntruncated exponential distribution,\n\\begin{equation}\\label{eq:eggsact}\n \\phi_m(z) = \\Phi \\frac{m}{1 - \\e^{-mH}} \\e^{mz} .\n\\end{equation}\nIn the toolbox, this solution is computed by the function \\edbi{eggsact}.\nThis solution has the following symmetry between ascending and\ndescending velocity,\n\\begin{equation}\\label{eq:symmetry}\n \\phi_{-m}(z) = \\phi_m(-H-z)\n\\end{equation}\n\nWith large depth and/or high ascending velocity the\neffect of the bottom may be neglected. In this case the distribution\nis well approximated by an exponential distribution \\index{exponential\ndistribution} with parameter $\\lambda = 1 / m$, that is\n\\begin{equation}\n \\phi_m(z) \\approx \\Phi m \\e^{mz}, \\quad \\text{for $mH \\gg 1$} .\n\\end{equation}\n\n\nFor positive values of $m$ the following series development from\n\\citep{sund83} is valid\n\\begin{equation}\n \\phi_m(z) = \\Phi m \\e^{mz} \\sum_{j = 0}^{\\infty} \\e^{-jmH}, \\quad m > 0 .\n\\end{equation}\nThe zero-th term of this expansion is the exponential distribution above.\nFor negative values of $m$ the symmetry relation (\\ref{eq:symmetry})\ncan be used,\n\\begin{equation}\n \\phi_m(z) = - \\Phi m \\e^{mz} \\sum_{j = 1}^{\\infty} \\e^{jmH}, \\quad m < 0 .\n\\end{equation}\n\nTo compute the mean and variance of the distribution the constant\nfactor $\\Phi$ can be dropped. The moment generating function is then\n\\begin{equation}\\label{eq:momgen}\n \\psi(t) = \\int_{-H}^0 \\frac{m}{1 - \\e^{-mH}} \\e^{(m+t)z} dz\n = \\frac{m}{1 - \\e^{-mH}} \\frac{1 - \\e^{-(m+t)H}}{m+t}.\n\\end{equation}\nDifferentiating $\\psi$ gives the moments of the distribution.\nThe mean depth is given by\n\\begin{equation}\\label{eq:eggsactmean}\n \\mu = \\psi'(0) = - \\frac{1}{m} + \\frac{H}{\\e^{mH} - 1}\n\\end{equation}\nThe function $\\mu = \\mu(m)$ with $H = 100\\m$ is plotted in\nfigure~\\ref{fig:mdepth}. The variance is\n\\begin{equation}\\label{eq:eggsactvar}\n \\psi''(0) - \\mu^2\n = \\frac{2 - \\e^{-mH}(m^2 H^2 + 2 m H + 2)}{m^2 ( 1 - \\e^{-mH})} - \\mu^2\n\\end{equation}\n\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[height=6cm]{mdepth}\n\\end{center}\n\\caption{Mean depth $\\mu$ as a function of $m$}\\label{fig:mdepth}\n\\end{figure}\n\n\\subsubsection{Piecewise constant coefficients}\n\nThe solution (\\ref{eq:eggsact}) can easily be extended to a piecewise\nconstant $m$.\nLet $0 = z_0 > \\cdots > z_m = -H$ be a partition of the water column\nwith $m(z) = m_i$ on the interval $I_i = (z_i, z_{i-1})$.\nThen the stationary solution restricted to the interval $I_i$ is\ngiven as\n\\begin{equation}\\label{eq:pwconst}\n \\phi(z) = C_i \\e^{m_i z}, \\quad \\mbox{for} \\ z_i < z < z_{i-1}\n\\end{equation}\nThe $C_i$-s are determined by continuity\nof the solution\n\\begin{equation}\n C_i \\e^{m_i z_i} = C_{i+1} \\e^{m_{i+1} z_i}, \\quad i = 1, \\ldots, m-1\n\\end{equation}\nand the integral condition~(\\ref{eq:intcond})\n\\begin{equation}\n \\sum_{i = 1}^m C_i \\int_{z_i}^{z_{i-1}} \\e^{m_i z} dz\n = \\sum_{i = 1}^m C_i \\frac{\\e^{m_i z_{i-1}} - \\e^{m_i z_i}}{m_i}\n = \\Phi .\n\\end{equation}\nThis method is used in the function \\edbi{sstate} to approximate the\ngeneral steady state solution. More details of the implementation is\ngiven in section~\\ref{sec:statprob}.\n\n\n\\subsubsection{Linear coefficients}\n\nIn Sundby \\shortcite{sund91} the solution is derived for $m$ linear.\nLet $m$ be given as $m(z) = a (z-z_0)$. Then $M(z) =\n\\half a ((z - z_0)^2 - z_0^2)$. Taking the last term into the\nconstant $C$, the solution becomes\n\\begin{equation}\n \\phi(z) = C \\e^{\\half a (z - z_0)^2} .\n\\end{equation}\nBathypelagic eggs are neutral buoyant at $z = z_0$, raising if deeper\nand sinking if higher in the water column. In this case the coefficient\n$a$ is negative and the concentration has a normal distribution about $z =\nz_0$ with variance $\\sigma^2 = {1 / |a|}$.\n\n%In a sample of eggs the diameters and buoyancies differ from egg to egg.\n%A more realistic model of the stationary situation will therefore be to\n%use a distribution of the $m$-values. This task was studied by Sundby (1983).\n%...\n\n\n\\subsubsection{Stationary solution with source terms}\n\nWith source terms, the steady state equation is\n\\begin{equation}\\label{eq:srcsstate}\n 0 = - \\pddz (w \\phi) + \\pddz (K \\Pd{\\phi}{z}) + P - \\alpha \\phi .\n\\end{equation}\nThis is a general second order ordinary\ndifferential equation. With constant coefficients the equation becomes\n\\begin{equation}\\label{eq:srcsacteq}\n K \\phi'' - w \\phi' - \\alpha \\phi = - P\n\\end{equation}\nand the no flux boundary conditions are\n\\begin{equation}\n K \\phi' - w \\phi, \\quad z = 0, z = -H .\n\\end{equation}\nThe solution can be written\n\\bel{eq:srssact}\n \\phi(z) = A \\e^{az} - B \\e^{-bz} + \\frac{P}{\\alpha}\n\\end{equation}\nwith\n\\begin{align}\n a & = \\frac{1}{2K} ( \\sqrt{w^2 + 4 \\alpha K} + w ) \\\\\n b & = \\frac{1}{2K} ( \\sqrt{w^2 + 4 \\alpha K} - w ) \\\\\n A & = \\frac{w P}{\\alpha b K} \\frac{1-\\e^{-b H}}{1-\\e^{-(a+b)H}} \\\\\n B & = \\frac{w P}{\\alpha a K} \\e^{-b H} \\frac{1-\\e^{-a H}}{1-\\e^{-(a+b)H}} .\n\\end{align}\n\nThis function is computed in the toolbox by the function \\edbi{srcsact}.\nThe signs are chosen such that positive velocity $w$ makes all terms\npositive. For negative velocities a symmetry property\nsimilar to equation~(\\ref{eq:symmetry})\ncan be used, if\n$\\phi(z)$ is a solution to equation (\\ref{eq:srcsacteq}) and boundary\nconditions (\\ref{eq:boundcond}) then $\\phi(-H-z)$ is a solution the\nsame equation and boundary conditions with the opposite sign on $w$.\nIn other words,\n\\begin{equation}\n \\phi_{-w}(z) = \\phi_w(-H-z) .\n\\end{equation}\n\n\n\\section{The Terminal Egg Velocity}\\label{sec:velocity}\n\\index{egg velocity}\\index{terminal velocity}\n\nAn egg in sea water will reach its terminal velocity $w$, where the\nbuoyant forcing balances the frictional drag. This velocity is a\nfunction of the difference $\\Delta \\rho = \\rho - \\rho_e$ between the\ndensity of the water and the egg, the egg diameter $d$, the\nacceleration $g$ due to gravity and the molecular\nviscosity\\index{molecular viscosity}. Here the \\emph{dynamic}\nmolecular viscosity\\index{dynamic viscosity} $\\mu$ is used.\n\nThe situation is characterised by the non-dimensional \\emphi{Reynolds\nnumber},\n\\begin{equation}\n Re = \\frac{\\rho d w}{\\mu}\n\\end{equation}\nFor low values, $Re < 0.5$, the terminal velocity is given by\n\\emphi{Stokes' formula}\n\\begin{equation}\\label{eq:stokes}\n w = \\frac{1}{18} \\frac{g d^2 \\Delta \\rho}{\\mu} .\n\\end{equation}\nThis formula was obtained by Stokes \\shortcite{stok1851}. The\nderivation of the formula is given in almost any textbook on\nfluid dynamics, for instance \\citep{yih77}.\n\n\nCombining these equations, one obtains an expression for $D$, the maximum\ndiameter for which Stokes' velocity applies,\n\\begin{equation}\n D^3 = \\frac{9 \\mu^2}{\\rho g \\Delta \\rho}\n\\end{equation}\n\n\nIn the intermediate region $0.5 < Re < 5$, Dallavalle (1948) gave an\nempirical formula\\index{Dallavalle's formula}\n\\begin{equation}\\label{eq:dallavalle}\n w = K_I (d - \\zeta D) \\Delta \\rho^{2/3} \\mu^{-1/3}\n\\end{equation}\nwhere $\\zeta = 0.4$ for a sphere. The coefficient $K_I$ is determined\nby the requirement that both formulas should give the same answer for\n$Re = 0.5$ or equivalently $d = D$.\nThis gives\n\\begin{equation}\n K_I = \\frac{5}{54} 9^{1/3} g^{2/3} \\rho^{-1/3}\n = 0.0875 \\text{kg}^{-1/3} \\,\n \\text{m}^{5/3} \\text{s}^{-4/3} .\n\\end{equation}\n\nThe formulas above are implemented as the function \\edbi{eggvel} in\nthe VertEgg toolbox. The buoyancy of a fish egg is often given as the\nsalinity $S_e$ where the egg is neutrally buoyant. The function\n\\edbi{eggvelst} computes the terminal velocity in this case.\n\n\nTo compute the egg velocity the density, $\\rho$ of water is needed.\nIn the toolbox only density at surface pressure is presently\navailable. This is computed by the function \\edbi{dens0} by the\nUNESCO formula \\citep{UNES81}. The function \\edbi{sw\\_dens} in the\nSEAWATER toolbox\\index{SEAWATER} \\citep{morg94} has implemented the\nfull UNESCO equation of state.\n\nThe dynamic molecular\nviscosity\\index{viscosity!molecular}\\index{viscosity!dynamic} $\\mu$ of\nsea water is tabulated in table \\ref{tab:molvisc} taken from Sverdrup\n\\emph{et al.}\n\\shortcite{sver52}. The values decrease with temperature and increase\nslowly with salinity. The dependence on pressure is insignificant, and is\nneglected here.\n\n\\begin{table}[h]\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n\\hline\\hline\nSalinity & \\multicolumn{7}{c|}{Temperature [\\deg C]} \\\\\n\\cline{2-8}\n[psu] & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\\\\n\\hline\n 0 & 1.79 & 1.52 & 1.31 & 1.14 & 1.01 & 0.89 & 0.80 \\\\\n10 & 1.82 & 1.55 & 1.34 & 1.17 & 1.03 & 0.91 & 0.82 \\\\\n20 & 1.85 & 1.58 & 1.36 & 1.19 & 1.05 & 0.93 & 0.84 \\\\\n30 & 1.88 & 1.60 & 1.38 & 1.21 & 1.07 & 0.95 & 0.86 \\\\\n35 & 1.89 & 1.61 & 1.39 & 1.22 & 1.09 & 0.96 & 0.87 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n \\caption{Dynamic molecular viscosity of sea water\n with unit $10^{-3} \\,\\text{kgm}^{-1}\\text{s}^{-1}$}\n \\label{tab:molvisc}\n\\end{table}\n\n\nUsing linear least squares regression the table is approximated by the\nfollowing function where $T$ is the temperature in \\deg C and $S$ is\nthe salinity in psu.\n\\begin{equation}\\label{eq:molvisc}\n \\mu = 10^{-3} \\, (1.7915 - 0.0538 \\, T + 0.007 \\, T^2 - 0.0023 \\, S )\n \\: \\text{kgm}^{-1}\\text{s}^{-1} ,\n\\end{equation}\nThis reproduces table \\ref{tab:molvisc} with an absolute error less\nthan $2 \\times 10^{-5}\\, \\text{kgm}^{-1}\\text{s}^{-1}$ and a relative\nerror of 1.7 \\%. This is good enough to compute egg velocities where\nthe uncertainty in the other variables are larger. This formula is\nimplemented by the function\n\\edbi{molvisc} in the toolbox. Riley and Skirrow \\shortcite{rile75}\ngive a more precise formula which requires more computational effort.\n", "meta": {"hexsha": "743a25596ba9545eede1c276d872e7716424f39d", "size": 16652, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Rapport/teori.tex", "max_stars_repo_name": "bjornaa/Vertegg", "max_stars_repo_head_hexsha": "836ed0857b2f6bc1a7399a20b28b3e4320810b15", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Rapport/teori.tex", "max_issues_repo_name": "bjornaa/Vertegg", "max_issues_repo_head_hexsha": "836ed0857b2f6bc1a7399a20b28b3e4320810b15", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Rapport/teori.tex", "max_forks_repo_name": "bjornaa/Vertegg", "max_forks_repo_head_hexsha": "836ed0857b2f6bc1a7399a20b28b3e4320810b15", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.0182648402, "max_line_length": 79, "alphanum_fraction": 0.6976939707, "num_tokens": 5597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199633332891, "lm_q2_score": 0.7606506526772884, "lm_q1q2_score": 0.6371361718049927}} {"text": "\\documentclass[10pt]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{xcolor}\n\\usepackage[T1]{fontenc}\n\\usepackage{pagecolor}\n\\usepackage{amssymb}\n\\usepackage{lmodern}\n\\usepackage{mathtools, nccmath}\n\\usepackage{courier}\n\\usepackage[dvipsnames]{xcolor}\n\n\\definecolor{myyellow}{RGB}{225,225,100}\n\\definecolor{myred}{RGB}{220,100,100}\n\\definecolor{mygreen}{RGB}{120,225,120}\n\\definecolor{myblue}{RGB}{100,200,255}\n\\definecolor{mypurple}{RGB}{200,50,255}\n\\color{white}\n\n\\title{Geometry with complex numbers \\#1}\n\\author{@all.about.mathematics}\n\n\\begin{document}\n\n\\maketitle\n\\pagecolor{black}\n\\LARGE\n\\section{Introduction}\nIn this series of posts, we discuss representations of geometrical objects with complex numbers and use them to prove some famous geometry theorems! We'll focus on lines and points in this post!\n\n\\noindent{Enjoy :D}\n\\newpage\n\\Large\n\\section{Lines with complex numbers}\nFirst, let's think about lines segments on the complex plane. Let $z_1, z_2 \\in\\mathbb{C}$. How can we represent the line segment from $z_1$ and $z_2$? \n\n\\noindent{We know that the vector from $z_1$ to $z_2$ is $z_2-z_1$, so we can represent the line segment by the equation}\n$$l=z_1+t(z_2-z_1)=z_1+tv$$\nWhere $t$ is a parameter so that $0\\leq t\\leq 1$ and $v$ is a non-zero vector in the direction of the line. Therefore, the line segment is the collection of all possible points $l$. Besides, if we let $t$ be any real number, then we get the line passing through $z_1$ and $z_2$.\n\n\\noindent\nFor example, let $z_1=3-2i$ and $z_2=4+5i$. Then the equation\n$$L:3-2i+t(1+7i)$$\nWith $0\\leq t\\leq 1$ is the line segment from $z_1$ to $z_2$, and if $t\\in\\mathbb{R}$ then it is the line passing through $z_1$ and $z_2$. \n\n\\newpage\n\\section{Finding points of division}\nLet's say we have 2 points $z_1,z_2\\in\\mathbb{C}$. Suppose we want to find a $z_3$ on the line segment $z_1 z_2$ such that it divides the length of the line segment by the ratio $a:b$, which is equivalent to \n$$|z_1-z_3|:|z_2-z_3|=a:b$$\nConsider the equation \n$$L:=z_1+t(z_2-z_1)\\qquad 0\\leq t \\leq 1$$\nTo find $z_3$, \n$$t=\\frac{a}{a+b}\\implies z_3=z_1+\\frac{a}{a+b}(z_2-z_1)=\\frac{bz_1+az_2}{a+b}$$\n\n\\noindent\nFor example, let $z_1=4+5i\\:,\\: z_2=-4-3i$ and suppose we want to find $z_3$ such that it divides $z_1z_2$ by $3:5$. Then\n$$z_3=\\frac{5(4+5i)+3(-4-3i)}{3+5}=\\frac{8+16i}{8}=1+2i$$\n\n\\newpage\n\\section{Parallel lines}\nGiven 2 lines $L_1\\:,\\: L_2$ and their equations \n$$L_1:z_1+tv_1 \\qquad t\\in\\mathbb{R}$$\n$$L_1:z_2+sv_2 \\qquad s\\in\\mathbb{R}$$\nIt is easy to see that if $v_1$ and $v_2$ are parallel, i.e.\n$$\\exists k\\in\\mathbb{R}:v_1=kv_2 $$\nStay tuned for the next post! We are going to use what we've learned here to easily prove a very useful theorem!\n\\end{document}", "meta": {"hexsha": "aaa7028724079cf8f762420ae1619c44d7d5d188", "size": 2729, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Tig. and Geometry/all.about.mathematics' questions/Complex number and geometry 1.tex", "max_stars_repo_name": "Nanu00/LaTeX", "max_stars_repo_head_hexsha": "0f08a90c4e9ef78af42797670903636059ca0df2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 15, "max_stars_repo_stars_event_min_datetime": "2020-05-29T17:22:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-28T18:47:05.000Z", "max_issues_repo_path": "Tig. and Geometry/all.about.mathematics' questions/Complex number and geometry 1.tex", "max_issues_repo_name": "Nanu00/LaTeX", "max_issues_repo_head_hexsha": "0f08a90c4e9ef78af42797670903636059ca0df2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-06-26T07:33:59.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-11T12:14:49.000Z", "max_forks_repo_path": "Tig. and Geometry/all.about.mathematics' questions/Complex number and geometry 1.tex", "max_forks_repo_name": "Shreenabh664/LaTeX", "max_forks_repo_head_hexsha": "675e03f3ec555456b9a2cc714825ec75317848c2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2020-06-22T07:50:41.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-08T05:11:14.000Z", "avg_line_length": 41.3484848485, "max_line_length": 278, "alphanum_fraction": 0.719311103, "num_tokens": 1027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199552262967, "lm_q2_score": 0.7606506472514406, "lm_q1q2_score": 0.6371361610936053}} {"text": "\\chapter{Separating Hyperplanes, Lagrange Multipliers, and Convex Duality}\n% \\documentclass[draft,11pt]{article}\n%\\documentclass[11pt]{article}\n%\\input{../defs}\n%\\input{../layout-defs}\n%\\usepackage{amsmath}\n%\\usepackage{amssymb}\n%\\usepackage{amsthm}\n%\\usepackage{graphicx}\n%\\usepackage{float}\n%\\usepackage[ruled,vlined]{algorithm2e}\n%\\SetKwBlock{Repeat}{repeat}{}\n%%% for this lecture\n%\\newcommand{\\gap}{\\text{gap}}\n% \\newtheorem{theorem}{Definition}\n%\\interfootnotelinepenalty=10000\n% \\newtheorem{lemma}{Lemma}\n\n%\\usepackage{tikz}\n%\\usetikzlibrary{arrows}\n\n\n%%%% ADD MACROS HERE\n% feel free to add more macros here\n\n%\\begin{document}\n\n\\sloppy\n%\\lecture{12 --- Wednesday, May 13th}\n%{Spring 2020}{Rasmus Kyng, Scribe: Tim Taubner}{\n% Separating Hyperplanes, Lagrange Multipliers, and Convex Duality}\n\n\\section{Overview}\nFirst part of this chapter introduces the concept of a separating hyperplane of two sets followed by a proof that for two closed, convex and disjoint sets a separating hyperplane always exists.\nThis is a variant of the more general \\emph{separating hyperplane theorem}%\n\\footnote{Wikipedia is good on this: \\href{https://en.wikipedia.org/wiki/Hyperplane_separation_theorem}{https://en.wikipedia.org/wiki/Hyperplane\\_separation\\_theorem}} %\ndue to Minkowski.\nThen \\emph{Lagrange multipliers} $\\xx$, $\\ss$ of a convex optimization\nproblem\n\\begin{align*}\n \\min_y{}\\ & \\energy(\\yy) \\\\ \\nonumber\n\\text{s.t.}\\ & \\AA\\yy = \\bb \\\\ \\nonumber\n & c(\\yy) \\leq 0\n\\end{align*}\nare introduced and with that, the \\emph{Lagrangian} \\begin{equation*} \\LL(\\yy, \\xx, \\ss) = \\energy(\\yy) + \\xx^{\\trp}(\\bb - \\AA\\yy) + \\ss^{\\trp}c(\\yy) \\end{equation*}\nis defined.\nFinally, we deal with the dual problem\n\\begin{align*}\n \\max_{\\xx, \\ss, \\ss \\geq 0}{} L(\\xx, \\ss),\n\\end{align*}\nwhere $L(\\xx, \\ss) = \\min_y L(\\yy, \\xx, \\ss)$. We show \\emph{weak\n duality}, i.e.\\ $L(\\yy, \\xx, \\ss) \\leq \\energy(\\yy)$ and that assuming \\emph{Slater's condition} the values of both the primal and dual is equal, which is referred to as \\emph{strong duality}.\n\n\\section{Separating Hyperplane Theorem}\nSuppose we have two convex subsets $A, B \\subseteq \\R^n$ that are disjoint ($A \\cup B = \\emptyset$).\nWe wish to show that there will always be a (hyper-)plane $H$ that separates these two sets, i.e.~ $A$ lies on one side, and $B$ on the other side of $H$.\n\nSo what exactly do we mean by Hyperplane? Let's define it.\n\\begin{definition}[Hyperplane]\nA \\emph{hyperplane} $H$ of dimension $n$ is the subset $H := \\{\\xx \\in \\R^n : \\ip{\\nn, \\xx} = \\mu\\}$.\nWe say $H$ has \\emph{normal} $\\nn \\in \\R^n$ and \\emph{threshold} $\\mu$.\nIt is required that $\\nn \\neq \\veczero$.\n\\end{definition}\n\nEvery hyperplane divides $\\R^n$ into two halfspaces $\\{\\xx : \\ip{\\vv, \\xx} \\geq \\mu\\}$ and $\\{\\xx : \\ip{\\vv, \\xx} \\leq \\mu\\}.$\nIt separates two sets, if they lie in different halfspaces. We formally define separating hyperplane as follows.\n\n\\begin{definition}[Separating Hyperplane]\nWe say a hyperplane $H$ \\emph{separates} two sets $A, B$ iff\n\\begin{align*}\n \\forall \\aa \\in A:& \\ip{\\nn, \\aa} \\geq \\mu \\\\\n \\forall \\bb \\in B:& \\ip{\\nn, \\bb} \\leq \\mu\n\\end{align*}\nIf we replace $\\geq$ with $>$ and $\\leq$ with $<$ we say $H$ \\emph{strictly} separates $A$ and $B$.\n\\end{definition}\n\nIt is easy to see that there exists disjoint non-convex sets that can not be\nseparated by a hyperplane (e.g.\\ a point cannot be separated from a\nring around it). But can two disjoint convex sets always be strictly separated by a hyperplane?\nThe answer is no: consider the two-dimensional case depicted in\n\\autoref{fig:lefthyper} with $A = \\{ (x,y): x \\leq 0\\}$ and $B = \\{\n(x, y) : x> 0 \\text{ and } y \\geq \\frac{1}{x}\\}$. Clearly they are disjoint; however the only separating hyperplane is $H = \\{ (x, y): x = 0\\}$ but it intersects $A$.\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=1\\textwidth]{fig/lec12-non-strict-separator.png}\n \\caption{The sets $A = \\{ (x, y): x \\leq 0\\}$ and $B = \\{ (x, y): x\n > 0 \\text{ and } y \\geq \\frac{1}{x} \\}$ only permit a non-strictly separating hyperplane.}\n \\label{fig:lefthyper}\n\\end{figure}\n\nOne can prove that there exists a non-strictly separating hyperplane for any two disjoint convex sets.\nWe will prove that if we further require $A$,$B$ to be closed and\nbounded, then a strictly separating hyperplane always exists. (Note in\nthe example above how our choice of $B$ is not bounded.)\n\n\\begin{theorem}[Separating Hyperplane Theorem; closed, bounded sets] \\label{th:shtcb}\nFor two closed, bounded, and disjoint convex sets $A, B \\in \\R^n$ there exists a strictly separating hyperplane $H$.\nOne such hyperplane is given by normal $\\nn = \\dd - \\cc$ and threshold $\\mu = \\frac{1}{2}\\left(\\norm{\\dd}_2^2 - \\norm{\\cc}_2^2\\right)$, where $\\cc \\in A$, $\\dd \\in B$ are the minimizers of the distance between $A$ and $B$\n\\begin{equation*} \\text{dist}(A, B) = \\min_{\\aa\\in A, \\bb \\in B}\\norm{\\aa-\\bb}_2 > 0.\\end{equation*}\n% Note that this distance is strictly larger than zero.\n\\end{theorem}\n\\begin{proof}\n We omit the proof that $\\text{dist}(A, B) = \\min_{\\aa\\in A, \\bb \\in\n B}\\norm{\\aa-\\bb}_2 > 0$, which follows from $A,B$ being disjoint,\n closed, and bounded.\nNow, we want to show that $\\ip{\\nn,\\bb} > \\mu$ for all $\\bb \\in B$; then $\\ip{\\nn,\\aa} < \\mu$ for all $\\aa\\in A$ follows by symmetry.\nObserve that\n\\vspace{-1em}\n\\begin{align*}\n \\ip{\\nn, \\dd} - \\mu&= \\ip{\\dd-\\cc,\\dd} - \\frac{1}{2}\\left(\\norm{\\dd}_2^2-\\norm{\\cc}_2^2\\right)\\\\\n &= \\norm{\\dd}_2^2 - \\dd^{\\trp}\\cc - \\frac{1}{2}\\norm{\\dd}_2^2 + \\frac{1}{2}\\norm{\\cc}_2^2 \\\\\n\t\t\t\t\t\t\t\t\t\t&= \\frac{1}{2}\\norm{\\dd-\\cc}_2^2 > 0.\n\\end{align*}\nSo suppose there exists $\\uu \\in B$ such that $\\ip{\\nn, \\uu} - \\mu \\leq 0$.\nWe now look at the line defined by the distance minimizer $\\dd$ and the point on the ``wrong side'' $\\uu$.\nDefine $\\bb(\\lambda) = \\dd + \\lambda(\\uu - \\dd)$, and take the derivative of the distance between $\\bb(\\lambda)$ and $\\cc$.\nEvaluated at $\\lambda=0$ (which is when $\\bb(\\lambda) = \\dd$), this yields\n\\begin{equation*}\n \\left.\\frac{d}{d\\lambda} \\norm{\\bb(\\lambda) - \\cc}_2^2\\right\\rvert_{\\lambda=0} =\n\t\\left.2\\ip{\\dd-\\lambda\\dd+\\lambda\\uu-\\cc,\\uu-\\dd}\\right\\rvert_{\\lambda=0} =\n\t2\\ip{\\dd-\\cc, \\uu-\\dd}.\n\\end{equation*}\n\nHowever, this would imply that the gradient is strictly negative since\n\\begin{align*}\n\\ip{\\nn, \\uu} - \\mu & = \\ip{\\dd-\\cc,\\uu} - \\ip{\\dd-\\cc,\\dd} + \\ip{\\dd-\\cc,\\dd} - \\mu \\\\\n & = \\ip{\\dd-\\cc,\\uu-\\dd} + \\norm{\\dd}_2^2 - \\ip{\\cc,\\dd} - \\frac{1}{2}\\norm{\\dd}_2^2+\\frac{1}{2}\\norm{\\cc}_2^2 \\\\\n & = \\ip{\\dd-\\cc,\\uu-\\dd} + \\frac{1}{2}\\norm{\\dd-\\cc}_2^2 \\leq 0.\n\\end{align*}\nThis contradicts the minimality of $\\dd$ and thus concludes this proof.\n\\end{proof}\n\nA more general separating hyperplane theorem holds even when the sets\nare not closed and bounded:\n\n\\begin{theorem}[Separating Hyperplane Theorem] \\label{th:sht}\nGiven two disjoint convex sets $A, B \\in \\R^n$ there exists a hyperplane $H$\nseparating them.\n\\end{theorem}\n\n\n\\section{Lagrange Multipliers and Duality of Convex Problems}\n\nIn this Section, we'll learn about \\emph{Langrange Multipliers} and\nhow they lead to convex duality.\nBut first, let's see an example to help illustrate where these ideas\ncome from.\n\nImagine you were to prove that for all $\\xx \\in \\R^n$ we have $\\norm{\\xx}_p \\leq n^{\\frac{1}{2} - \\frac{1}{p}}\\norm{\\xx}_2$ for some $1 \\leq p \\leq 2;$.\nWe can look at this as optimizing $\\max_{\\xx} \\norm{\\xx}_p$ subject to $\\norm{\\xx}_2$ being constant, e.g.\\ simply $\\norm{\\xx}_2 = 1$. Then the statement above follows from a scaling argument.\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=.48\\textwidth]{fig/lec12-2p-norms.jpeg}\n \\caption{Looking at fixed $\\norm{\\xx}_p = \\alpha$ and $\\norm{\\xx}_2=1$. (Here, $p = 1.5$.)}\n \\label{fig:2-p-norm}\n\\end{figure}\n\nIf we move from $\\xx$ to $\\xx + \\ddelta$ with $\\ddelta \\perp \\grad_{\\xx}\\norm{\\xx}_2$ and $\\ddelta \\not\\perp \\grad_{\\xx}\\norm{\\xx}_p$ means that for infinitesimally small $\\ddelta$ the 2-norm stays constant but the $p$-norm changes.\nThat means for either $\\xx - \\ddelta$ or $\\xx + \\ddelta$ the $p$-norm increases\nwhile the 2-norm stays constant. Hence at the maximum of $\\norm{\\xx}_p$ the gradients of both norms have to be parallel, i.e.~\\begin{equation*}\\grad_{\\xx}\\left(\\norm{\\xx}_p - \\lambda\\norm{\\xx}_2\\right) = 0.\\end{equation*}\n%Here, $\\lambda$ is the\nThis insight is the core idea of Lagrange multipliers (in this case $\\lambda$).\n\nNote that here the problem is not convex, because $\\{ \\xx: \\norm{\\xx}_2^2 =\n1 \\}$ is not convex and because we are asking to \\emph{maximize} a norm.\nIn the following we will study Lagrange multipliers for general convex problems.\n\n\\subsection{General Convex Problems}\n\nA full formal treament of convex duality would require us to be more careful\nabout using $\\inf$ and $\\sup$ in place of $\\min$ and $\\max$, as well\nas considering problems that have no feasible solutions.\nToday, we'll ignore these concerns.\n\nLet us consider a general convex optimization problem with convex objective, linear equality constraints and convex inequality constraints\n\\begin{align}\n \\label{eq:primalprob}\n \\min_{y \\in S}{} & \\energy(\\yy) \\\\ \\nonumber\n\\text{s.t.}\\ & \\AA\\yy = \\bb \\\\ \\nonumber\n & \\cc(\\yy) \\leq \\veczero,\n\\end{align}\nwhere $\\energy(\\yy): S \\rightarrow \\R$ is defined on a convex subset $S \\subseteq \\R^n$, $\\AA \\in \\R^{m\\times n}$\nand $\\cc(\\yy)$ is a vector of constraints $\\cc(\\yy) = \\left(c_i(\\yy)\\right)_{i \\in [k]}$.\nFor every $i \\in [k]$ the function $c_i: S \\rightarrow \\R$ is convex.\n\n\\begin{definition}[Primal feasibility]\nWe say that $\\yy \\in S$ is \\emph{primal feasible} if all constraints are satisfied, i.e.~$\\AA\\yy = \\bb$ and $\\cc(\\yy) \\leq \\veczero$.\n\\end{definition}\n\nIn the following we will denote by $\\alpha^* = \\energy(\\yy^*)$ the optimal value of the primal program where $\\yy^*$ is an minimizer.\n\n\\begin{definition}\nNext we introduce the \\emph{dual variables} $\\xx \\in \\R^m$, $\\ss \\in \\R^k$ and define the \\emph{Lagrangian} as\n\\begin{equation*} L(\\yy, \\xx, \\ss) = \\energy(\\yy) + \\xx^{\\trp}(\\bb-\\AA\\yy) + \\ss^{\\trp}\\cc(\\yy). \\end{equation*}\nWe also define a Lagrangian only in terms of the dual variables by minimizing over $\\yy$ as\n\\begin{equation*} L(\\xx, \\ss) = \\min_{\\yy}L(\\yy, \\xx, \\ss). \\end{equation*}\n\\end{definition}\n\n\\begin{definition}[Dual feasibility]\nWe say $(\\xx, \\ss)$ is dual feasible if $\\ss \\geq 0$.\nIf additionally $\\yy$ is primal feasible, we say $(\\yy, \\xx, \\ss)$ is primal-dual feasible.\n\\end{definition}\n\n\\begin{definition}[Dual problem]\nWe define the \\emph{dual problem} as\n\\begin{align}\n \\label{eq:dualprob}\n \\max_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}}\n \\min_{\\yy}L(\\yy, \\xx, \\ss)\n =\n \\max_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}} L(\\xx, \\ss)\n \\end{align}\nand denote the optimal dual value by $\\beta^*$.\n\\end{definition}\n\n\nFor each $\\yy$, the Lagrangian $L(\\yy, \\xx,\n\\ss)$ is linear in $(\\xx,\\ss)$ and hence also concave in them.\nHence $L(\\xx, \\ss)$ is a concave function, because it is the pointwise\nminimum (over $\\yy$), of a collection of concave functions in\n$(\\xx,\\ss)$.\n\nThis also means that the dual problem is really a convex optimization\nproblem in disguise, because we can flip the sign of $-L(\\xx, \\ss)$\nto get a convex function and minimizing this is equivalent to\nmaximizing $L(\\xx, \\ss)$.\n\\begin{align*}\n \\max_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}} L(\\xx, \\ss)\n =\n -\n \\min_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}} -L(\\xx, \\ss)\n\\end{align*}\n\n% Of course, we can write the dual problem using the\n% $\\min_{\\yy}L(\\yy, \\xx, \\ss)$ notation as\n% \\[\n% \\max_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}}\n% \\min_{\\yy}L(\\yy, \\xx, \\ss)\n% =\n% \\max_{\\substack{\\xx, \\ss \\\\ \\ss \\geq 0}} L(\\xx, \\ss)\n% \\]\n\n\n\n\\subsection{Weak Duality}\n\\label{sec:weakdual}\nFirst we see that the primal problem can be written in terms of the Lagrangian as\n\\begin{equation}\n \\label{eq:primalminmax}\n \\alpha^* = \\min_{\\yy}\\max_{\\xx; \\ss\\geq\\veczero} L(\\yy, \\xx, \\ss)\n\\end{equation}\nThis is because for a minimizing $\\yy$ all constraints have to be satisfied and the Lagrangian simplifies to $L(\\yy,\\xx,\\ss) = \\energy(\\yy)$.\nIf $\\AA\\xx - \\bb = \\veczero$ was violated, making $\\xx$ large sends $L(\\yy,\\xx,\\ss) \\rightarrow \\infty$.\nAnd if $\\cc(\\yy) \\leq \\veczero$ is violated, we can make $L(\\yy,\\xx,\\ss) \\rightarrow \\infty$ by choosing large $\\ss$.\n\nNote that we require $\\ss \\geq \\veczero$, as we only want to penalize the violation of the inequality constraints in one direction, i.e.\\ when $\\cc(\\yy) > 0$.\n\n%\\begin{theorem}[Weak Duality]\nFor any primal-dual feasible $\\yy,\\xx,\\ss$ we have\n$L(\\yy, \\xx, \\ss) \\leq \\energy(\\yy)$ and hence also $L(\\xx, \\ss) = \\min_y L(\\yy, \\xx, \\ss) \\leq \\energy(\\yy)$.\n\nIn other words $\\max_{\\xx; \\ss \\geq 0} L(\\xx, \\ss) = \\beta^* \\leq \\alpha^*$.\nThis is referred to as \\emph{weak duality}.\n\nUsing the forms in Equations~\\eqref{eq:dualprob} and\n\\eqref{eq:primalminmax}, we can also state this as\n\\begin{equation*}\n \\alpha^* = \\min_{\\yy}\\max_{\\xx; \\ss\\geq\\veczero} L(\\yy, \\xx, \\ss)\n \\geq\n \\max_{\\xx; \\ss\\geq\\veczero} \\min_{\\yy} L(\\yy, \\xx, \\ss)\n =\n \\beta^*\n .\n\\end{equation*}\n\n\\subsection{Strong Duality}\nSo now that we have proved weak duality $\\beta^* \\leq \\alpha^*$,\nwhat is strong duality? $\\beta^* = \\alpha^*$?\nThe answer is yes, but strong duality only holds under some conditions.\n\nOne sufficient condition we look at today is a variant of \\emph{Slater's condition}%\n\n\\begin{definition}[Slater's condition with full domain] \\label{def:slater}\nA (primal) problem as defined in \\eqref{eq:primalprob} with $S = \\R^n$ fulfills\nSlater's condition if there exists a \\emph{strictly feasible} point,\ni.e.\\ there exists $\\yytil$ s.t.\\ $\\AA\\yytil = \\bb$ and $\\cc(\\yytil) <\n\\veczero$.\nThis means that the strictly feasible point $\\yytil$ lies strictly inside the set $\\{\\yy : \\cc(\\yy) \\leq \\veczero\\}$ defined by the inequality constraints.\n\\end{definition}\n\n\\begin{theorem}\n \\label{thm:slaterstrongduality}\nFor a problem satisfying Slater's condition, strong duality holds, i.e.\\ $\\alpha^* = \\beta^*$.\nIn other words, the optimal value of the primal problem $\\alpha^*$ is equal to the optimal value of the dual.\n\\end{theorem}\n\nTo extend Slater's condition to the case when the domain $S$ is a\nstrict subset of $\\R^n$, we need the notion of a ``relative interior''.\n\n% \\begin{definition}[Affine hull] \\label{def:affhull}\n% Given a set $S \\subset \\R^n$, the \\emph{affine hull} of $S$, denoted $\\operatorname{aff}(S)$, is the\n% union of $S$ and all lines through any pair of points in $S$.\n% We can also write this as\n% \\[\n% \\operatorname{aff}(S) = \\setof{\\alpha \\xx + (1-\\alpha) \\yy : \\xx,\n% \\yy \\in S, \\alpha \\in \\R}. \n% \\]\n% \\end{definition}\n\n\\begin{definition}[Relative interior] \\label{def:relint}\n Given a convex set $S \\subset \\R^n$, the \\emph{relative interior} of $S$ is\n \\[\n \\operatorname{relint}(S) = \\setof{x \\in S : \\text{ for all } \\yy\n \\in S \\text{ there exists } \\epsilon > 0 \\text{ such that }\n \\xx - \\epsilon(\\yy - \\xx) \\in S}\n .\n \\]\n\\end{definition}\nIn other words, $\\xx \\in \\operatorname{relint}(S)$ if starting at $\\xx\n\\in S$ we can move\n``away'' from any $\\yy \\in S$ by a little and still be in $\\SS$.\nAs an example, suppose $S = \\setof{(s,t) \\in \\R^2 \\text{ such that } s\n \\geq 0 \\text{ and } t = 0}$.\nThen $(0,0) \\in S$ but $(0,0) \\not\\in \\operatorname{relint}(S)$, while\n$(1,0) \\in \\operatorname{relint}(S)$.\n\nNow, we can state a more general version of Slater's condition.\n\n\\begin{definition}[Slater's condition] \\label{def:slatergeneral}\nA (primal) problem as defined in \\eqref{eq:primalprob} fulfills\nSlater's condition if there exists a \\emph{strictly feasible} point\n$\\yy \\in \\operatorname{relint}(S)$.\nWe require $\\AA\\yytil = \\bb$ and $\\cc(\\yytil) <\n\\veczero$.\nThis means that the strictly feasible point $\\yytil$ lies strictly inside the set $\\{\\yy : \\cc(\\yy) \\leq \\veczero\\}$ defined by the inequality constraints.\n\\end{definition}\n\nTheorem~\\ref{thm:slaterstrongduality} also holds given the general\nSlater's condition of Definition~\\ref{def:slatergeneral}.\n\n\\paragraph{How are we going to prove this?} Before we prove the theorem, let's make a few observations to get us\nwarmed up. If you get bored, skip ahead to the proof.\n\nIt is sufficient to prove that $\\alpha^* \\leq \\beta^*$, as the statement then follows in conjunction with weak duality.\nWe define the set\n\\begin{equation*} G = \\{(\\energy(\\yy), \\AA\\yy - \\bb, \\cc(\\yy)) : \\yy \\in S \\}, \\end{equation*}\nwhere $S \\subseteq \\R^n$ is the domain of $\\energy$.\n\nImmediately, we observe that we can write the optimal primal value as\n\\begin{equation*} \\alpha^* = \\min \\{t:(t,\\vv,\\uu) \\in G, \\vv = \\veczero, \\uu \\leq \\veczero\\}. \\end{equation*}\nSimilarly, we can write the Lagrangian (after minimizing over $\\yy$)\n\\begin{equation*} L(\\xx, \\ss) = \\min_{(t,\\vv,\\uu) \\in G} (1, \\xx, \\ss)^{\\trp} (t, \\vv, \\uu). \\end{equation*}\nThis is equivalent to the inequality, for $(t,\\vv,\\uu) \\in G$,\n\\begin{equation*} (1, \\xx, \\ss)^{\\trp}(t, \\vv, \\uu) \\geq L(\\xx, \\ss). \\end{equation*}\nwhich defines a hyperplane with $\\nn = (1,\\xx,\\ss)$ and $\\mu = L(\\xx,\n\\ss)$ such that $G$ is on one side.\n\nTo establish strong duality, we would like to show the existence of a\nhyperplane such that for $(t,\\vv,\\uu) \\in G$\n\\[\n \\nn^{\\trp}(t, \\vv, \\uu) \\geq \\alpha^* \\text{ and } \\nn = (1,\\xxhat,\\sshat)\n \\text{ with } \\sshat \\geq \\veczero.\n\\]\nThen we would immediately get\n\\begin{equation*}\n \\beta^* \\geq L(\\xxhat,\\sshat)= \\min_{(t,\\vv,\\uu) \\in G} (1, \\xx,\n \\ss)^{\\trp} (t, \\vv, \\uu) \\geq \\alpha^* .\n\\end{equation*}\n\nPerhaps not surprisingly, we will use the Separating\nHyperplane Theorem.\nWhat are the challenges we need to deal with?\n\\begin{itemize}\n\\item We need to replace $G$ with a convex set (which we will call\n $A$) and separate $A$ from some other convex set (which we will call $B$).\n\\item We need to make sure the hyperplane normal $\\nn$ has $1$ in the first\n coordinate and $\\ss \\geq \\veczero$, and the hyperplane threshold is $\\alpha^*$.\n\\end{itemize}\n\n% The proof we are about to see handles these things in one neat\n% stroke: Essentially, we replace $G$ with the convex set that comes from replacing\n% the convex functions $\\energy$ and $\\cc$ with their epigraphs.\n\n% If strong duality indeed holds,\n% and is obtained at some $L(\\yyhat,\\xxhat,\\sshat)$\n% then the reasoning we used to prove weak duality in\n% Section~\\ref{sec:weakdual} seems to suggest that a dual point\n% $(\\xx^*,\\ss^*)$ maximizing $L(\\xx,\\ss)$ will occur at a feasible\n% $\\yy$, since otherwise $L(\\yy,\\xx,\\ss)$ could be reduced further by\n% changing $(\\xx,\\ss)$ to penalize $\\yy$.\n% Of course, it's not clear whether this could then be counter-acted by\n% changing $\\yy$.\n\n\n% \\begin{itemize}\n% \\item and correspond to a feasible $\\yy$ in the sense that\n% $L(\\xx^*,\\ss^*) = (\\yy,\\xx^*,\\ss^*)$\n% \\item be dual feasible,\n% \\end{itemize}\n% If the minimizing $\\yy$ is not feasible, but\n\n\n\n% Then by considering a feasible $\\yyhat$ for the primal problem, we\n% get\n% \\begin{align**}\n% \\beta^*\n% & \\geq L(\\xxhat,\\sshat) \\\\\n% &= (1, \\xxhat,\\sshat)^{\\trp}(\\energy(\\yyhat), \\AA\\yyhat - \\bb, \\cc(\\yyhat))\n% \\geq \\energy(\\yyhat) \\geq \\alpha^*\n% \\end{align**}\n% This sandwiching also tells us that $\\yyhat$ and $(\\xxhat,\\sshat)$\n% must be optimal for the primal and dual problems respectively.\n\n% Clearly, we like to ensure that\n\n\n% We'd like to show that\n% for some dual-feasible $(\\xxhat,\\sshat)$ maximizing $L(\\xx,\n% \\ss)$, then the hyperplane inequality doesn't have any slack, i.e. it\n% holds with equality at some point, and furthermore, equality is\n% obtained at some $(t, \\vv, \\xx) = (\\energy(\\yyhat), \\AA \\yyhat - \\bb,\n% \\cc(\\yyhat))$ where $\\yyhat$ is feasible.\n\n\n\n\n\n% When we proved weak duality in Section~\\ref{sec:weakdual}, our reasoning\n\n\\begin{proof}[Proof of Theorem~\\ref{thm:slaterstrongduality}.]\n For simplicity, our proof will assume that $S = \\R^n$, but only a little\n extra work is required to handle the general case.\n\n Let's move to on finding two convex disjoints sets $A, B$ to enable the use of the separating hyperplane \\autoref{th:sht}.\n\nFirst set we define $A$, roughly speaking, as a multi-dimensional epigraph of\n$G$. More precisely\n\\begin{equation*} A = \\{(t, \\vv, \\uu) : \\exists \\yy \\in S, t \\geq \\energy(\\yy), \\vv = \\AA\\yy - \\bb, \\uu \\geq \\cc(\\yy)\\}. \\end{equation*}\nNote that $A$ is a convex set. The proof is similar to the proof that the epigraph of a convex function is a convex set.\nThe optimal value of the primal program can be now written as\n\\begin{equation*} \\alpha^* = \\min_{(t,\\veczero,\\veczero) \\in A} t. \\end{equation*}\nAnd we define another set $B$ of the same dimensionality as $A$ by\n\\begin{equation*} B := \\{(r \\in \\R, \\veczero \\in \\R^m, \\veczero \\in \\R^k) : r < \\alpha^* \\}. \\end{equation*}\nThis set $B$ is convex, as it is a ray.\nAn example of two such sets $A,B$ is illustrated in \\autoref{fig:ab}.\n\nWe show that $A \\cap B = \\emptyset$ by contradiction.\nSuppose $A, B$ are not disjoint; then there exists $\\yy$ such that\n\\begin{equation*} (\\energy(\\yy), \\AA\\yy-\\bb, \\cc(\\yy)) = (r, \\veczero, \\uu) \\end{equation*}\nwith $\\uu \\leq \\veczero$.\nBut this means that $\\yy$ is feasible and $\\energy(\\yy) = r < \\alpha^*$; contradicting the optimality of $\\alpha^*$.\n\n% Now let's move back to proving that strong duality follows from Slater's condition.\n%As Slater's condition requires the existence of a $\\yy$ such that $\\AA\\yy = \\bb$\nTo make things simpler,\nwe assume that our linear constraint matrix $\\AA \\in \\R^{m\\times n}$,\nhas full row rank and $m < n$ (but very little extra work is required to\ndeal with the remaining cases, which we omit).\n\nAs we just proved, $A$ and $B$ are convex and disjoint sets and hence the separating hyperplane theorem (\\autoref{th:sht}) we introduced earlier in this chapter implies the existence a separating hyperplane.\nThis means there exists a normal\n$\\nn = (\\tilde\\rho, \\xxtil, \\sstil)$ and threshold\n% \\footnote{To prevent confusion, the threshold here is denoted $\\mu$ instead of $alpha$} %\n$\\mu$ and with $A$ on one side, i.e.\\\n%\\begin{equation*} (t, \\vv, \\uu) \\in A \\implies \\begin{pmatrix}t \\vv^{\\trp} \\uu^{\\trp}\\end{pmatrix}\\begin{pmatrix}\\tilde\\rho \\\\ \\xxtil \\\\ \\sstil\\end{pmatrix} \\geq \\alpha \\end{equation*}\n\\begin{equation} \\label{eq:A}\n(t, \\vv, \\uu) \\in A \\implies (t, \\vv, \\uu)^{\\trp}(\\tilde\\rho, \\xxtil, \\sstil) \\geq \\mu\n\\end{equation}\nand the set $B$ on the other side:\n\\begin{equation} \\label{eq:B}\n(t, \\vv, \\uu) \\in B \\implies (t, \\vv, \\uu)^{\\trp}(\\tilde\\rho, \\xxtil, \\sstil) \\leq \\mu.\n\\end{equation}\n\nNow, we claim that $\\sstil \\geq 0$. Suppose $\\sstil(i) < 0$, then for $\\uu(i) \\rightarrow \\infty$ the threshold would grow unbounded, i.e.\\ $\\mu \\rightarrow -\\infty$ contradicting that the threshold $\\mu$ is finite by the separating hyperplane theorem.\nSimilarly we claim $\\tilde\\rho \\geq 0$, as if this were not the case, having $t \\rightarrow \\infty$ implies that $\\mu \\rightarrow -\\infty$ again contradicting the finiteness of $\\mu$.\n\nFrom Equation~\\eqref{eq:B} it follows that $t\\tilde\\rho \\leq \\mu$ for all $t < \\alpha^*$ which implies that $t\\tilde\\rho \\leq \\mu$ for $t = \\alpha^*$ by taking the limit. Hence we have $\\alpha^*\\tilde\\rho \\leq \\mu$.\nFrom $(t, \\vv, \\uu) \\in A$ we get from Equation~\\eqref{eq:A}\n\\begin{equation*}\n(\\tilde\\rho, \\xxtil, \\sstil)^{\\trp}(t, \\vv, \\uu) \\geq \\mu \\geq \\alpha^*\\tilde\\rho\n\\end{equation*}\nand thus\n\\begin{equation}\\label{eq:rhotilde}\n(\\tilde\\rho, \\xxtil, \\sstil)^{\\trp}(\\energy(\\yy), \\AA\\yy-\\bb,\\cc(\\yy)) \\geq \\alpha^*\\tilde\\rho.\n\\end{equation}\n\n\nNow we consider two cases; starting with the ``good'' case where $\\tilde\\rho > 0$.\nDividing Equation~\\eqref{eq:rhotilde} by $\\tilde\\rho$ gives\n\\begin{equation*}\n \\energy(\\yy) + \\frac{\\xxtil^{\\trp}}{\\tilde\\rho}(\\AA\\yy - \\bb) + \\frac{\\sstil^{\\trp}}{\\tilde\\rho}\\cc(\\yy) \\geq \\alpha^*.\n\\end{equation*}\nNoting that the left hand side above is $L(\\yy, \\frac{\\xxtil}{\\tilde\\rho}, \\frac{\\sstil}{\\tilde\\rho})$\nand that the equation holds for arbitrary $\\yy$; therefore also for the minimum we get\n\\begin{equation*} \\min_{\\yy}{L\\left(\\yy, \\frac{\\xxtil}{\\tilde\\rho}, \\frac{\\sstil}{\\tilde\\rho}\\right)}\\geq \\alpha^* \\end{equation*}\nand hence via definition of $\\beta^*$ finally\n\\begin{equation*} \\beta^* \\geq L\\left(\\frac{\\xxtil}{\\tilde\\rho},\\frac{\\sstil}{\\tilde\\rho}\\right) \\geq \\alpha^*. \\end{equation*}\n\nNext consider the ``bad'' case $\\tilde\\rho = 0$.\nAs $\\alpha^*\\tilde\\rho \\leq \\mu$, we have $0 \\leq \\mu$.\nFrom Equation~\\eqref{eq:A} we get\n\\begin{equation*}\n \\cc(\\yy)^{\\trp}\\ss + \\xx^{\\trp}(\\bb-\\AA\\yy) \\geq \\mu \\geq 0.\n\\end{equation*}\nAs Slater's condition holds, there is an interior point $\\yytil$, i.e.\\ it satisfies $\\bb - \\AA\\yytil = \\veczero$ and $\\cc(\\yytil) < \\veczero$. Together with the equation above this yields\n\\begin{equation*} \\cc(\\yytil)^{\\trp}\\sstil + \\xxtil^{\\trp}\\veczero \\geq 0 \\end{equation*}\nwhich implies\n$\\cc(\\yytil)^{\\trp}\\sstil \\geq 0 $\nand as $\\cc(\\yytil) < \\veczero$ this means $\\sstil = \\veczero$.\n\nAs the normal $(\\tilde\\rho,\\sstil,\\xxtil)$ of the hyperplane can not be all zeroes, this means the last ``component'' $\\xxtil$ must contain a non-zero entry, i.e.\\ $\\xxtil \\neq \\veczero$.\nFurthermore $\\xxtil^{\\trp}(\\bb - \\AA\\yytil) = \\veczero$, $\\cc(\\yytil) < \\veczero$ and $\\AA$ has full row rank, hence there exists $\\ddelta$ such that\n\\begin{equation*}\n\\xxtil^{\\trp}(\\bb-\\AA(\\yytil+\\ddelta)) < \\veczero \\text{ and } \\cc(\\yytil + \\ddelta) < 0.\n\\end{equation*}\nThis, however, means that there is a point in $A$ on the wrong side of the hyperplane, as\n\\begin{equation*}\n(\\tilde\\rho, \\xxtil, \\sstil)^{\\trp}(\\energy(\\yytil+\\ddelta), \\bb-\\AA(\\yytil+\\ddelta), \\cc(\\yytil+\\ddelta)) < 0\n\\end{equation*}\nbut the threshold is $\\mu \\geq 0$.\n\\end{proof}\n\n\\begin{remark*}\n Note that our reasoning about why $\\ss \\geq \\veczero$ in the proof\n above is very similar to our reasoning for why the primal program\n can be written as Problem~\\eqref{eq:primalminmax}.\n\\end{remark*}\n\n\\paragraph{Example.}\nAs an example of $A$ and $B$ as they appear in the above proof,\nconsider\n\\[\n\\min_{\\substack{y \\in (0,\\infty) \\\\ 1/y\n -1 \\leq 0}}\ny^2\n\\]\n This leads to $\\alpha^* = 1,\n y^* =1$, and $A = \\setof{ (t,u): y \\in (0,\\infty) \\text{ and } t >y^2\n \\text{ and } u \\geq 1/y-1}$, and $B = \\setof{ (t,0) : t < 1}$ and the separating hyperplane normal is $\\nn = (1,2)$.\nThese two sets $A,B$ are illustrated in \\autoref{fig:ab}.\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.9\\textwidth]{fig/lec12-the-sets-a-b.jpg}\n \\caption{Example of the convex sets $A$ and $B$ we wish to separate by hyperplane.}\n\\label{fig:ab}\n\\end{figure}\n\n\n\n\\subsection{The Gradient Perspective}\nLet's come back to what we said earlier about parallel gradient.\nSuppose $\\yy^*$ is an optimizer of the primal problem and $\\xx^*, \\ss^*$ for the dual.\nWe thus have\n\\[ L(\\yy^*, \\xx^*, \\ss^*) = \\alpha^* = \\beta^*. \\]\nBecause $L(\\yy, \\xx^*, \\ss^*)$ is a convex function in $\\yy$, it also\nfollows that if $\\energy: S \\to \\R$ and $\\cc$ are differentiable and the minimizer\n$\\yy^*$ is not on the boundary of $S$, then we must have that the\ngradient w.r.t. $\\yy$ is zero, i.e.\n\\begin{equation*} \\left.\\grad_{\\yy}L(\\yy, \\xx^*, \\ss^*)\\right|_{\\yy=\\yy^*} = 0 \\end{equation*}\nand plugging in\n\\[ L(\\yy, \\xx^*, \\ss^*) = \\energy(\\yy) + \\xx^{\\trp}(\\bb - \\AA\\yy) + \\ss^{\\trp}\\cc(\\yy) \\]\nyields\n\\begin{equation*} \\grad\\energy(\\yy) + \\xx^{\\trp}\\grad_y(\\bb-\\AA\\yy) + \\ss^{\\trp}\\grad\\cc(\\yy) = \\veczero. \\end{equation*}\nAnd this connects to our point of the parallel gradients from the beginning of this section.\n\n\\subsection{Complementary Slackness}\n\nWe will see more of this next time, but if we look at $\\yy^*$ we see that\n\\begin{equation*}\\energy(\\yy^*) = \\alpha^* = \\energy(\\yy^*) + \\xx^{\\trp}(\\bb - \\AA\\yy^*) + \\ss^{\\trp}\\cc(\\yy^*) = \\energy(\\yy^*) + \\ss^{\\trp}\\cc(\\yy^*) \\end{equation*}\nand hence when the $i$-th convex constraint is not active, i.e.\\ $c_i(\\yy^*) < 0$ the \\emph{slack} must be zero, i.e.\\ $\\ss(i) = 0$.\nConversely if the slack is non-zero, that is $\\ss(i) \\neq 0$ implies that the constraint is active, i.e.\\ $c_i(\\yy^*) = 0$.\n\nA good reference for this is Boyd's free online book ``Convex\noptimization'' (linked to on the course website).\nIt provides a number of different interpretations of duality. One\nparticularly interesting one comes from economics:\neconomists see the slack variables $\\ss$ as prices for violating the constraints.\n\n%\\end{document}\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"agao21_script\"\n%%% End:\n", "meta": {"hexsha": "f921f7b21dcde7bf1663519ee984544718e9cf13", "size": 28431, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "agao21_script/lecture12_mod.tex", "max_stars_repo_name": "rjkyng/agao21_script", "max_stars_repo_head_hexsha": "772f8c17b0802ec43d45e1480f7193dd0eceadb7", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-15T09:04:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-26T05:39:09.000Z", "max_issues_repo_path": "agao21_script/lecture12_mod.tex", "max_issues_repo_name": "rjkyng/agao21_script", "max_issues_repo_head_hexsha": "772f8c17b0802ec43d45e1480f7193dd0eceadb7", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "agao21_script/lecture12_mod.tex", "max_forks_repo_name": "rjkyng/agao21_script", "max_forks_repo_head_hexsha": "772f8c17b0802ec43d45e1480f7193dd0eceadb7", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2021-03-11T12:35:23.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-13T06:04:51.000Z", "avg_line_length": 48.4344122658, "max_line_length": 252, "alphanum_fraction": 0.6675108157, "num_tokens": 9760, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7981867801399695, "lm_q1q2_score": 0.6371021340742857}} {"text": "\\subsection{Group actions}\\label{subsec:group_actions}\n\n\\begin{definition}\\label{def:endomorphism_monoid}\n For every object \\( X \\) in an arbitrary \\hyperref[def:category]{category} \\( \\cat{C} \\), the set \\( \\cat{C}(X) \\) is a \\hyperref[def:monoid]{monoid} with morphism composition as the monoid operation and \\( \\id_X \\) as the monoid identity.\n\n Outside of \\hyperref[sec:category_theory]{category theory}, whenever the category \\( \\cat{C} \\) is clear from the context, we call \\( \\cat{C}(X) \\) the \\term{endomorphism monoid} over \\( X \\) and denote it by \\( \\End(X) \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:monoid_action}\n Let \\( M \\) be a \\hyperref[def:monoid]{monoid} and let \\( X \\) be an object in some \\hyperref[def:category]{category} \\( \\cat{C} \\).\n\n We will define monoid actions of \\( M \\) on \\( X \\), which we will sometimes call \\term{left monoid actions}. There are also \\term{right monoid actions}, which are only briefly mentioned in \\fullref{def:monoid_action/functor}.\n\n A \\term{monoid action} can be defined equivalently as:\n \\begin{thmenum}\n \\thmitem{def:monoid_action/homomorphism} A \\hyperref[def:monoid/homomorphism]{homomorphism} from \\( M \\) to the \\hyperref[def:endomorphism_monoid]{endomorphism monoid} \\( \\End(X) \\).\n\n \\thmitem{def:monoid_action/functor} A \\hyperref[def:functor]{functor} from the \\hyperref[def:monoid_delooping]{delooping} \\( \\cat{B}_M \\) to \\( \\cat{C} \\).\n\n Right actions are \\hyperref[rem:contravariant_functor]{contravariant functors}.\n\n \\thmitem{def:monoid_action/family} An \\hyperref[def:cartesian_product/indexed_family]{indexed family} \\( \\seq{ \\Phi_m }_{m \\in M} \\) of \\hyperref[def:morphism_invertibility/endomorphism]{endomorphisms} of \\( X \\) such that\n \\begin{align}\n &\\Phi_e = \\id_X, \\label{eq:def:monoid_action/family/identity}\\tag{\\logic{MA1}} \\\\\n &\\Phi_{mn} = \\Phi_m \\bincirc \\Phi_n. \\label{eq:def:monoid_action/family/compatibility}\\tag{\\logic{MA2}}\n \\end{align}\n\n This defines a function \\( \\Phi: M \\times A \\to A \\).\n \\end{thmenum}\n\\end{definition}\n\\begin{proof}\n \\ImplicationSubProof{def:monoid_action/homomorphism}{def:monoid_action/functor} Suppose that we have a monoid homomorphism \\( \\Phi: M \\to \\End(X) \\). Define the functor\n \\begin{equation*}\n \\begin{aligned}\n &F: \\cat{B}_M \\to \\cat{C} \\\\\n &F(\\anon) \\coloneqq X \\\\\n &F(m) \\coloneqq \\Phi(m).\n \\end{aligned}\n \\end{equation*}\n\n This is indeed a functor because \\eqref{eq:def:functor/CF1} follows from \\eqref{eq:def:pointed_set/homomorphism} and \\eqref{eq:def:functor/CF2} follows from \\eqref{eq:def:magma/homomorphism}.\n\n \\ImplicationSubProof{def:monoid_action/functor}{def:monoid_action/family} Suppose that we have a functor \\( F: \\cat{B}_M \\to \\cat{C} \\). Let \\( X \\coloneqq F(\\anon) \\) and define the \\( M \\)-indexed family\n \\begin{equation*}\n \\begin{aligned}\n &\\Phi_m: X \\to X \\\\\n &\\Phi_m \\coloneqq F(m).\n \\end{aligned}\n \\end{equation*}\n\n It satisfies the necessary axioms:\n \\begin{itemize}\n \\item \\ref{eq:def:monoid_action/family/identity} holds:\n \\begin{equation*}\n \\Phi_e\n =\n F(e)\n \\reloset {\\eqref{eq:def:functor/CF1}} =\n \\id_A.\n \\end{equation*}\n\n \\item \\ref{eq:def:monoid_action/family/compatibility} holds: for every pair \\( m, n \\in M \\), we have\n \\begin{equation*}\n \\Phi_{mn}\n =\n F(mn)\n \\reloset {\\eqref{eq:def:functor/CF2}} =\n F(m) \\bincirc F(n)\n =\n \\Phi_m \\bincirc \\Phi_n\n \\end{equation*}\n \\end{itemize}\n\n \\ImplicationSubProof{def:monoid_action/family}{def:monoid_action/homomorphism} Suppose that we have an indexed family \\( \\seq{ \\Phi_m }_{m \\in M} \\) of endomorphisms of \\( A \\) that satisfies the axioms for left action. Regard this indexed family as a function \\( \\Phi: M \\to \\End(X) \\).\n\n Then \\( \\Phi \\) is a monoid homomorphism because \\ref{eq:def:monoid_action/family/identity} implies \\( \\Phi(e) = \\id_X \\) and \\eqref{eq:def:monoid_action/family/compatibility} implies\n \\begin{equation*}\n \\Phi(mn) = \\Phi(m) \\bincirc \\Phi(n).\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:monoid_is_action}\n Every \\hyperref[def:monoid]{monoid} \\hyperref[def:monoid_action]{acts} on itself via the family of functions \\( h \\mapsto g \\cdot h \\) indexed by \\( g \\). These functions are not monoid homomorphisms in general.\n\n Compare this result to \\fullref{thm:cayleys_theorem}.\n\\end{proposition}\n\\begin{proof}\n The family satisfies \\fullref{def:monoid_action/family}:\n \\begin{itemize}\n \\item \\ref{eq:def:monoid_action/family/identity} follows from \\eqref{eq:def:monoid/theory/identity}.\n\n \\item \\ref{eq:def:monoid_action/family/compatibility} follows from associativity:\n \\begin{equation*}\n [h \\mapsto g_1 \\cdot h] \\bincirc [h \\mapsto g_2 \\cdot h] = [h \\mapsto g_1 \\cdot (g_2 \\cdot h)] = [h \\mapsto (g_1 \\cdot g_2) \\cdot h].\n \\end{equation*}\n \\end{itemize}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:exponentiation_monoid_action}\n The \\hyperref[def:set_of_natural_numbers]{natural numbers} \\( \\BbbN \\) (\\hyperref[rem:peano_arithmetic_zero]{with zero}) act on any \\hyperref[def:monoid]{monoid} by \\hyperref[def:monoid/exponentiation]{exponentiation} via the family of function \\( g \\mapsto g^n \\) indexed by \\( n \\in \\BbbN \\).\n\n Compare this result to \\fullref{thm:exponentiation_group_action}.\n\\end{proposition}\n\\begin{proof}\n This family satisfies \\fullref{def:monoid_action/family}:\n \\begin{itemize}\n \\item \\ref{eq:def:monoid_action/family/identity} is obvious.\n \\item \\ref{eq:def:monoid_action/family/compatibility} follows from \\fullref{thm:magma_exponentiation_properties/repeated}.\n \\end{itemize}\n\\end{proof}\n\n\\begin{definition}\\label{def:automorphism_group}\n For every object \\( X \\) in a \\hyperref[def:groupoid]{groupoid} \\( \\cat{G} \\), the set \\( \\cat{G}(X) \\) is a \\hyperref[def:group]{group} with morphism composition as the group operation.\n\n Similarly to \\hyperref[def:endomorphism_monoid]{endomorphism monoids}, whenever the groupoid \\( \\cat{G} \\) is clear from the context, we call \\( \\cat{G}(X) \\) the \\term{automorphism group} over \\( X \\) and denote it by \\( \\aut(X) \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:symmetric_group}\n We call the \\hyperref[def:automorphism_group]{automorphism group} of a set \\( A \\) the \\term{symmetric group}\\footnote{The term possibly comes from symmetric functions defined in \\fullref{def:symmetric_function}} on \\( A \\) and denote it by \\( S(A) \\). The group \\( S(A) \\) consists of bijective functions, which we call \\term{permutations}.\n\n Rather than for arbitrary sets, we often consider symmetric group\n \\begin{equation*}\n S_n \\coloneqq S(\\set{ 1, 2, \\ldots, n }).\n \\end{equation*}\n\n It is conventional to call \\( S_n \\) the \\enquote{symmetric group on \\( n \\) letters}.\n\n \\begin{thmenum}\n \\thmitem{def:symmetric_group/permutation} It is common to write a permutation \\( \\sigma \\) in \\( S_n \\) as\n \\begin{equation*}\n \\begin{pmatrix}\n 1 & \\cdots & n \\\\\n \\sigma(1) & \\cdots & \\sigma(n)\n \\end{pmatrix}\n \\end{equation*}\n\n \\thmitem{def:symmetric_group/cycle} If there exists a finite sequence \\( k_1, \\ldots, k_m \\) of distinct numbers such that \\( \\sigma(k_m) = k_1 \\) and \\( \\sigma(k_{i+1}) = \\sigma(k_i) \\) for each \\( i < m \\), we say that the permutation is \\term{cyclic} or a \\term{cycle} of \\term{length} \\( m \\). For brevity, we denote this cycle by \\( \\cycle{k_1, \\cdots, k_m} \\).\n\n We call cycles of length \\( 1 \\) \\term{trivial cycles} or \\term{loops} and cycles of length \\( 2 \\) --- \\term{transpositions}. Every \\hyperref[def:symmetric_group/cycle]{transposition} is an \\hyperref[def:set_with_involution]{involution}, and thus a transposition is equal to its inverse permutation.\n\n \\thmitem{def:symmetric_group/disjoint_cycle} If \\( \\cycle{k_1, \\cdots, k_m} \\) and \\( \\cycle{s_1, \\cdots, s_l} \\) are two cycles and if the sets \\( \\set{ k_1, \\ldots, k_m } \\) and \\( \\set{ s_1, \\ldots, s_l } \\) are disjoint, we say that the cycles themselves are \\term{disjoint}.\n\n Two disjoint cycles commute. That is,\n \\begin{equation*}\n \\cycle{k_1, \\cdots, k_m} \\bincirc \\cycle{s_1, \\cdots, s_l} = \\cycle{s_1, \\cdots, s_l} \\bincirc \\cycle{k_1, \\cdots, k_m}.\n \\end{equation*}\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:cycle_transposition_decomposition}\n Every nontrivial cycle \\( \\cycle{ k_1, \\cdots, k_m } \\) can be decomposed into the product of transpositions\n \\begin{equation*}\n \\cycle{ k_1, \\cdots, k_m } = \\cycle{ k_1, k_m } \\bincirc \\cycle{ k_1, k_{m-1} } \\bincirc \\cdots \\bincirc \\cycle{ k_1, k_2 }.\n \\end{equation*}\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:symmetric_group_cardinality}\n The \\hyperref[def:symmetric_group]{symmetric group} \\( S_n \\) has \\( n! \\) elements.\n\\end{proposition}\n\\begin{proof}\n We use induction on \\( n \\). The case \\( n = 1 \\) is trivial. Suppose that \\( S_{n-1} \\) has \\( (n-1)! \\) elements. Then \\( S_n \\) is obtained by permuting \\( n \\) with each element of \\( S_{n-1} \\). That is,\n \\begin{equation*}\n S_n = \\set{ \\cycle{ k, n } \\bincirc \\sigma \\given \\sigma \\in S_{n-1} \\T{and} 1 \\leq k \\leq n }.\n \\end{equation*}\n\n It follows that\n \\begin{equation*}\n \\card(S_n) = n \\cdot \\card(S_{n-1}) = n (n-1)! = n!.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:group_action}\n Let \\( G \\) be a \\hyperref[def:group]{group} and let \\( X \\) be an object in some \\hyperref[def:concrete_category]{concrete category} \\( \\cat{C} \\).\n\n We will define group actions as a special case of \\hyperref[def:monoid_action]{monoid actions}, with the same remarks regarding left and right group actions.\n\n A \\term{group action} can be defined equivalently as:\n \\begin{thmenum}\n \\thmitem{def:group_action/homomorphism} A \\hyperref[def:group/homomorphism]{homomorphism} from \\( G \\) to the \\hyperref[def:automorphism_group]{automorphism group} \\( \\aut(X) \\).\n\n Right actions are homomorphisms from the \\hyperref[def:monoid/opposite]{opposite} group \\( G^{\\opcat} \\) to \\( \\aut(X) \\).\n\n \\thmitem{def:group_action/functor} A \\hyperref[def:functor]{functor} from the \\hyperref[def:monoid_delooping]{delooping} \\( \\cat{B}_G \\) to \\( \\cat{C} \\).\n\n Right actions are \\hyperref[rem:contravariant_functor]{contravariant functors}.\n\n \\thmitem{def:group_action/family} An \\hyperref[def:cartesian_product/indexed_family]{indexed family} \\( \\seq{ \\Phi_x }_{x \\in G} \\) of \\hyperref[def:morphism_invertibility/isomorphism]{isomorphisms} of \\( X \\) such that, for every pair \\( g, h \\in G \\),\n \\begin{equation}\\label{eq:def:group_action/family/compatibility}\\tag{\\logic{GA}}\n \\Phi_{gh} = \\Phi_g \\bincirc \\Phi_h.\n \\end{equation}\n\n This defines a function \\( \\Phi: G \\times A \\to A \\).\n \\end{thmenum}\n\\end{definition}\n\\begin{proof}\n The proof of equivalence is simple; it is similar to \\fullref{def:monoid_action}.\n\\end{proof}\n\n\\begin{theorem}[Cayley's theorem]\\label{thm:cayleys_theorem}\n Every \\hyperref[def:group]{group} \\hyperref[def:group_action]{acts} on itself via the family of functions \\( y \\mapsto x \\cdot y \\) indexed by \\( x \\). These functions are not group homomorphisms in general.\n\n Compare this result to \\fullref{thm:monoid_is_action}.\n\\end{theorem}\n\\begin{proof}\n Follows directly from \\fullref{thm:monoid_is_action} and \\fullref{thm:group_operation_induces_bijections}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:exponentiation_group_action}\n The \\hyperref[def:set_of_integers]{integers} \\( \\BbbZ \\) act on any \\hyperref[def:group]{group} by \\hyperref[def:monoid/exponentiation]{exponentiation} via the family of function \\( g \\mapsto g^n \\) indexed by \\( n \\in \\BbbZ \\).\n\n Compare this result to \\fullref{thm:exponentiation_monoid_action}.\n\\end{proposition}\n\\begin{proof}\n Follows from \\fullref{thm:exponentiation_monoid_action}.\n\\end{proof}\n\n\\begin{definition}\\label{def:group_action_orbit}\\mcite[163]{Knapp2016BasicAlgebra}\n The \\term{orbit} of \\( x \\) under the \\hyperref[def:group_action]{group action} \\( \\Phi: G \\to \\End(X) \\) is the set\n \\begin{equation*}\n \\set{ \\Phi_g(x) \\given g \\in G }.\n \\end{equation*}\n\n This is the set of all members of \\( X \\) \\enquote{reachable} from \\( x \\) via the action.\n\n The relation \\( g \\sim h \\) on \\( G \\), defined to hold if \\( g \\) and \\( h \\) have the same \\hyperref[def:group_action_orbit]{orbit}, is an \\hyperref[def:equivalence_relation]{equivalence relation}. The quotient set \\( G / {\\sim} \\) is a partition of \\( G \\) into sets called \\term{orbits}.\n\\end{definition}\n\n\\begin{example}\\label{ex:plane_rotation_action_orbits}\n Consider the action of the additive group of \\( \\BbbR \\) on \\( \\BbbR^2 \\) given by the \\hyperref[def:euclidean_transformation/rotation]{rotation} \\hyperref[def:array/matrix]{matrices}\n \\begin{equation*}\n \\Phi_r \\coloneqq \\begin{pmatrix}\n \\cos(r) & \\sin(r) \\\\\n -\\sin(r) & \\cos(r) \\\\\n \\end{pmatrix}\n \\end{equation*}\n\n Fix a nonzero vector \\( (x, y)^T \\) in \\( \\BbbR^2 \\) with norm \\( l \\). Since rotation matrices are orthogonal, they preserve norms. Furthermore, given a vector of norm \\( l \\), with the angle \\( r \\) defined via \\eqref{eq:def:angle/measure}, \\( \\Phi_r^{-1} \\) sends the vector to \\( (x, y)^T \\).\n\n The \\hyperref[def:group_action_orbit]{orbit} of \\( (x, y)^T \\) is thus a \\hyperref[def:quadratic_plane_curve/ellipse]{circle} at the origin with radius \\( l \\).\n\\end{example}\n\n\\begin{proposition}\\label{thm:group_conjugation_action}\\mcite[165]{Knapp2016BasicAlgebra}\n Every \\hyperref[def:group]{group} \\hyperref[def:group_action]{acts} on itself via the \\term{conjugation automorphisms} defined as\n \\begin{equation*}\n \\begin{aligned}\n &\\Phi_g: G \\to G, \\\\\n &\\Phi_g(h) \\coloneqq g h g^{-1}.\n \\end{aligned}\n \\end{equation*}\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{definition}\\label{def:inner_and_outer_automorphisms}\\mcite[exer. 6.12.15]{Knapp2016BasicAlgebra}\n Denote by \\( \\Phi: G \\to \\aut(G) \\) the \\hyperref[thm:group_conjugation_action]{conjugation action} on the group \\( G \\).\n\n The \\term{inner automorphisms group} of \\( G \\) is\n \\begin{equation*}\n \\op{inn}(G) \\coloneqq \\set{ \\Phi_g \\given g \\in G }.\n \\end{equation*}\n\n The \\term{outer automorphism group} is the \\hyperref[def:group/quotient]{quotient group}\n \\begin{equation*}\n \\op{out}(G) \\coloneqq \\aut(G) / \\op{inn}(G).\n \\end{equation*}\n\\end{definition}\n\\begin{defproof}\n We will show that \\( \\op{inn}(G) \\) is a \\hyperref[thm:normal_subgroup_equivalences]{normal subgroup} of \\( \\aut(G) \\).\n\n Fix a member \\( g \\) of \\( G \\) and define the inner automorphism\n \\begin{equation*}\n \\varphi(h) \\coloneqq gh^{-1}g.\n \\end{equation*}\n\n Let \\( \\psi \\) be an arbitrary automorphism of \\( G \\). Then\n \\begin{equation*}\n [\\varphi \\bincirc \\psi \\bincirc \\varphi^{-1}](h)\n =\n \\varphi(g \\varphi^{-1}(h) g^{-1})\n =\n \\varphi(g) h \\varphi(g^{-1}),\n \\end{equation*}\n and thus \\( \\varphi \\bincirc \\psi \\bincirc \\varphi^{-1} \\) is again an inner automorphism.\n\\end{defproof}\n\n\\begin{proposition}\\label{thm:permutation_decomposition_existence}\n Let \\( S_n \\) be a \\hyperref[def:symmetric_group]{symmetric group} and let \\( \\sigma \\) be a permutation in \\( S_n \\). Then there exists a finite sequence \\( c_1, \\ldots, c_m \\) of nontrivial \\hyperref[def:symmetric_group/disjoint_cycle]{disjoint cycles} such that\n \\begin{equation*}\n \\sigma = c_1 \\bincirc \\cdots \\bincirc c_m.\n \\end{equation*}\n\n The case where \\( \\sigma \\) is the identity permutation corresponds to \\( m = 0 \\).\n\n We call this a \\term{cycle decomposition} of \\( \\sigma \\). Via \\fullref{thm:cycle_transposition_decomposition}, this permutation can further be decomposed into a composition of individual transpositions. We call the latter a \\term{transposition decomposition} of \\( \\sigma \\).\n\n Neither decomposition is unique. Nonetheless, existence is sufficient for most practical purposes, including permutation parity defined in \\fullref{def:permutation_parity}.\n\\end{proposition}\n\\begin{proof}\n Suppose that \\( \\sigma \\) is not the identity.\n\n Consider the \\hyperref[def:group_action]{group action}\n \\begin{equation*}\n \\begin{aligned}\n &\\Phi^{(\\sigma)}: \\braket{ \\sigma } \\times \\set{ 1, \\ldots, n } \\to \\set{ 1, \\ldots, n } \\\\\n &\\Phi^{(\\sigma)}_\\sigma(k) \\coloneqq \\sigma(k)\n \\end{aligned}\n \\end{equation*}\n of the \\hyperref[def:first_order_generated_substructure]{generated subgroup} \\( \\braket{ \\sigma } \\).\n\n The \\hyperref[def:group_action_orbit]{orbit} of \\( k \\) under \\( \\Phi^{(\\sigma)} \\) is the set\n \\begin{equation*}\n O_\\sigma(k) \\coloneqq \\set{ \\sigma^m(k) \\given m \\in \\BbbZ }\n \\end{equation*}\n of all numbers reachable from \\( k \\) via iterated application of \\( \\sigma \\) or \\( \\sigma^{-1} \\). The family\n \\begin{equation*}\n \\set{ O_\\sigma(k) \\given k = 1, \\ldots, n }\n \\end{equation*}\n of orbits partitions \\( 1, \\ldots, n \\) into disjoint subsets.\n\n Each orbit \\( O \\) has a smallest element \\( o \\); and \\( O_\\sigma(o) = O \\). This smallest element uniquely identifies a cycle\n \\begin{equation*}\n \\cycle{o, \\sigma(o), \\sigma^2(o), \\ldots, \\sigma^{c-1}(o)},\n \\end{equation*}\n where \\( c \\) is the cardinality of \\( O_\\sigma(o) \\).\n\n Ignoring the trivial cycles, we obtain a unique set of disjoint nontrivial cycles for every permutation \\( \\sigma \\) in \\( S_n \\), which we call the \\term{cycle decomposition} of \\( \\sigma \\). Denote this set by \\( C_\\sigma \\).\n\n We must prove that \\( C_\\sigma \\) is family of disjoint sets. Let \\( O_\\sigma(k_1) \\) and \\( O_\\sigma(k_2) \\) be two orbits.\n\n Suppose that there exists a number \\( k \\in O_\\sigma(k_2) \\cap O_\\sigma(k_1) \\). Then there exist \\( m_1, m_2 \\leq n \\) such that \\( k = \\sigma^{m_1}(k_1) \\) and \\( k = \\sigma^{m_2}(k_2) \\). Then\n \\begin{equation*}\n k_1 = \\sigma^{-m_1}(k) = \\sigma^{-m_1}(\\sigma^{m_2}(k_2)) = \\sigma^{m_2 - m_1}(k_2),\n \\end{equation*}\n and thus \\( O_\\sigma(k_2) \\subseteq O_\\sigma(k_1) \\). We can analogously obtain the converse inclusion.\n\n Therefore, if two orbits have a nonempty intersection, they are equal.\n\n Finally, we will use induction on \\( m \\) to show that these cycles give us \\( \\sigma \\). The case \\( m = 0 \\) is trivial. For the inductive hypothesis, note that \\( c_1^{-1} \\sigma \\) has as cycles \\( c_2, \\ldots, c_m \\) because all the numbers in \\( c_1 \\) are fixed points of \\( c_1^{-1} \\sigma \\). Thus, the inductive hypothesis holds for \\( c_1^{-1} \\sigma \\).\n\n From\n \\begin{equation*}\n c_1^{-1} \\sigma = c_2 \\bincirc \\cdots \\bincirc c_m\n \\end{equation*}\n it follows that\n \\begin{equation*}\n \\sigma = c_1 \\bincirc \\cdots \\bincirc c_m.\n \\end{equation*}\n\\end{proof}\n\n\\begin{lemma}\\label{thm:permutation_parity_correctness}\n If a \\hyperref[def:symmetric_group/permutation]{permutation} \\( \\sigma \\in S_n \\) can be decomposed into \\hyperref[def:symmetric_group/cycle]{transpositions} as both\n \\begin{equation}\\label{eq:thm:permutation_parity_correctness/k}\n \\sigma = \\underbrace{\\cycle{ k_1, k_2 } \\bincirc \\cycle{ k_3, k_4 } \\bincirc \\cdots \\bincirc \\cycle{ k_{2n-1}, k_{2n} }}_{n \\T*{transpositions}}\n \\end{equation}\n and\n \\begin{equation}\\label{eq:thm:permutation_parity_correctness/l}\n \\sigma = \\underbrace{\\cycle{ l_1, l_2 } \\bincirc \\cycle{ l_3, l_4 } \\bincirc \\cdots \\bincirc \\cycle{ l_{2m-1}, l_{2m} }}_{m \\T*{transpositions}},\n \\end{equation}\n then \\( n - m \\) is an even number.\n\\end{lemma}\n\\begin{proof}\n We will use induction on \\( n \\). First consider the base case \\( n = 0 \\). Then \\( \\sigma \\) is the identity. Hence, every transposition in \\eqref{eq:thm:permutation_parity_correctness/l} should be present twice so that its action cancels out. Therefore, \\( m \\) is an even number.\n\n Now suppose that the statement holds for \\( n - 1 \\). Add (compose on the right) the last transposition \\( \\cycle{ k_{2n-1}, k_{2n} } \\) of \\eqref{eq:thm:permutation_parity_correctness/k} to both \\eqref{eq:thm:permutation_parity_correctness/k} and \\eqref{eq:thm:permutation_parity_correctness/l}. The obtained permutations are obviously equal. Furthermore, since \\( \\cycle{ k_{2n-1}, k_{2n} } \\) is its own inverse, we can just as well remove \\( \\cycle{ k_{2n-1}, k_{2n} } \\) from \\eqref{eq:thm:permutation_parity_correctness/k} to obtain a decomposition into \\( n - 1 \\) (rather than \\( n + 1 \\)) transpositions.\n\n By the inductive hypothesis, \\( (n - 1) - (m + 1) = n - m - 2 \\) is an even number. Therefore, \\( n - m \\) is also an even number.\n\\end{proof}\n\n\\begin{definition}\\label{def:permutation_parity}\n We say that a \\hyperref[def:symmetric_group/permutation]{permutation} \\( \\sigma \\in S_n \\) is \\term{even} (resp. \\term{odd}) if a decomposition of \\( \\sigma \\) into \\hyperref[def:symmetric_group/cycle]{transpositions} has an even (resp. odd) number of transpositions.\n\n By \\fullref{thm:permutation_decomposition_existence}, such a decomposition exists. By \\fullref{thm:permutation_parity_correctness}, all such decompositions yield the same parity even if they decompose into a differing number of transpositions.\n\n We correspondingly define the \\term{sign} of a permutation as\n \\begin{equation*}\n \\begin{aligned}\n &\\sgn: S_n \\to \\BbbZ, \\\\\n &\\sgn(\\sigma) \\coloneqq \\begin{cases}\n 1, &\\sigma \\T{is even} \\\\\n -1, &\\sigma \\T{is odd}\n \\end{cases}\n \\end{aligned}\n \\end{equation*}\n\\end{definition}\n\n\\begin{definition}\\label{def:alternating_group}\n The \\term{alternating group} \\( A_n \\) on \\( n \\) letters is the subgroup of all \\hyperref[def:permutation_parity]{even permutation} in the \\hyperref[def:symmetric_group]{symmetric group} \\( S_n \\).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:alternating_group_cardinality}\n The \\hyperref[def:alternating_group]{alternating group} \\( A_n \\) has \\( \\ifrac {n!} 2 \\) elements.\n\\end{proposition}\n\\begin{proof}\n The proof is similar to that of \\fullref{thm:symmetric_group_cardinality}, but is a little different.\n\n We use induction on \\( n \\). The case \\( n = 1 \\) is trivial. Suppose that \\( A_{n-1} \\) has \\( \\ifrac {(n-1)!} 2 \\) elements. Then\n \\begin{equation*}\n A_n = \\set{ \\cycle{ k, n } \\bincirc \\sigma \\given \\sigma \\in S_{n-1} \\setminus A_{n-1} \\T{and} 1 \\leq k \\leq n }.\n \\end{equation*}\n\n We obtain \\( A_n \\) by taking all the odd permutations in \\( S_{n-1} \\) and composing them with one new transposition. It follows that\n \\begin{equation*}\n \\card(A_n) = n \\cdot \\card(S_{n-1} \\setminus A_{n-1}) = n \\frac {(n-1)!} 2 = \\frac {n!} 2.\n \\end{equation*}\n\\end{proof}\n\n\\begin{example}\\label{ex:s3_and_a3}\n The \\hyperref[def:symmetric_group]{symmetric group} \\( S_3 \\) contains the following \\hyperref[def:symmetric_group/permutation]{permutations}:\n \\begin{equation*}\n S_3\n \\coloneqq\n \\set[\\vast]\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 1 & 2 & 3\n \\end{pmatrix},\n \\underbrace\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 2 & 1 & 3\n \\end{pmatrix}\n }_{\n \\cycle{ 1, 2 }\n },\n \\underbrace\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 2 & 3 & 1\n \\end{pmatrix}\n }_{\n \\cycle{ 1, 2, 3 }\n },\n \\underbrace\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 3 & 2 & 1\n \\end{pmatrix}\n }_{\n \\cycle{ 1, 3 }\n },\n \\underbrace\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 3 & 1 & 2\n \\end{pmatrix}\n }_{\n \\cycle{ 1, 3, 2 }\n },\n \\underbrace\n {\n \\begin{pmatrix}\n 1 & 2 & 3 \\\\\n 1 & 3 & 2\n \\end{pmatrix}\n }_{\n \\cycle{ 2, 3 }\n }\n }\n \\end{equation*}\n\n We can observe the following:\n \\begin{itemize}\n \\item The permutations \\( \\cycle{ 1, 2, 3 } \\) and \\( \\cycle{ 1, 3, 2 } \\) are inverses of each other and all other permutations are involutions.\n\n \\item Every conjugation automorphism is unique. This can be verified explicitly. Therefore, the \\hyperref[def:inner_and_outer_automorphisms]{inner automorphism group} \\( \\op{inn}(S_3) \\) is isomorphic to \\( S_3 \\).\n\n \\item The \\hyperref[def:alternating_group]{alternating group} \\( A_3 \\) consists of the identity and the odd-length cycles \\( (1, 2, 3) \\) and \\( (1, 3, 2) \\).\n\n \\item When restricted to \\( A_3 \\), all conjugation automorphisms are trivial. This can be verified explicitly. Therefore, the inner automorphism group \\( \\op{inn}(A_3) \\) is trivial, and hence\n \\begin{equation*}\n \\aut(A_3) \\cong \\op{out}(A_3).\n \\end{equation*}\n\n \\item The map \\( \\sigma \\mapsto \\sigma^{-1} \\), which fixes the identity and exchanges the two other permutations, is an automorphism of \\( A_3 \\). It is distinct from the identity, hence it is an outer automorphism.\n\n This map is given by the restriction of the conjugation \\( \\sigma \\mapsto \\cycle{1, 2} \\sigma \\cycle{1, 2} \\) to \\( A_3 \\). It is an inner automorphism of \\( S_3 \\), but an outer automorphism of \\( A_3 \\).\n \\end{itemize}\n\\end{example}\n\n\\begin{proposition}\\label{thm:group_epimorphisms_are_surjective}\\mcite[exer. I.5.5]{MacLane1994}\n Every \\hyperref[def:morphism_invertibility/right_cancellative]{epimorphism} in \\hyperref[def:group/category]{\\( \\cat{Grp} \\)} is \\hyperref[def:function_invertibility/surjective]{surjective}.\n\\end{proposition}\n\\begin{proof}\n Let \\( \\varphi: G \\to H \\) be an epimorphism and suppose that it is not surjective. Let \\( M \\) be the smallest normal subgroup of \\( H \\) containing \\( \\img \\varphi \\).\n\n If \\( M \\) has index \\( 2 \\) in \\( H \\), consider the quotient map \\( \\pi: H \\to H / M \\) and the constant map \\( c(h) \\coloneqq M \\). Then\n \\begin{equation*}\n \\pi \\bincirc \\varphi = c \\bincirc \\varphi.\n \\end{equation*}\n\n Since \\( \\varphi \\) is an epimorphism, we have \\( \\pi = c \\). But we have deliberately taken \\( \\pi \\) and \\( c \\) so that \\( \\pi \\neq c \\). The obtained contradiction shows that \\( M \\) must have an index greater than \\( 2 \\).\n\n Let \\( M \\), \\( uM \\) and \\( vM \\) be different cosets. Define \\( \\sigma: H \\to H \\) as the \\hyperref[def:symmetric_group/permutation]{permutation} on \\( H \\) that exchanges \\( xu \\) with \\( xv \\) for every \\( x \\in M \\). Define the homomorphism\n \\begin{equation*}\n \\begin{aligned}\n &\\psi: H \\to S(H) \\\\\n &\\psi(h) \\coloneqq (x \\mapsto hx),\n \\end{aligned}\n \\end{equation*}\n where \\( S(H) \\) is the \\hyperref[def:symmetric_group]{symmetric group}.\n\n This is indeed a homomorphism by \\fullref{thm:cayleys_theorem}. By \\fullref{thm:group_conjugation_action}, another homomorphism is\n \\begin{equation*}\n \\begin{aligned}\n &\\theta: H \\to S(H) \\\\\n &\\theta(h) \\coloneqq \\sigma^{-1} \\bincirc \\psi(h) \\bincirc \\sigma.\n \\end{aligned}\n \\end{equation*}\n\n Since \\( \\sigma \\) fixes the members of \\( M \\) in-place, we have \\( \\theta(h)\\restr_M = \\psi(h)\\restr_M \\). Since \\( M \\) contains the image of \\( \\varphi \\), this implies\n \\begin{equation*}\n \\psi \\bincirc \\varphi = \\theta \\bincirc \\varphi.\n \\end{equation*}\n\n Since \\( \\varphi \\) is an epimorphism, we have \\( \\psi = \\theta \\). But we have deliberately constructed \\( \\psi \\) and \\( \\theta \\) such that \\( \\psi \\neq \\theta \\). The obtained contradiction shows that \\( \\img \\varphi \\) cannot be a strict subgroup of \\( G \\). Therefore, \\( \\varphi \\) must be surjective.\n\\end{proof}\n\n\\begin{definition}\\label{def:dynamical_system}\\mimprovised\n Suppose that \\( \\cat{C} \\) is a \\hyperref[def:concrete_category]{concrete category} and let \\( X \\) be an object of \\( \\cat{C} \\).\n\n A \\term{dynamical system} is a \\hyperref[def:monoid_action]{monoid action} \\( \\Phi: T \\times X \\to X \\). We call \\( X \\) the \\term{phase space} of the system. In applications, we interpret the monoid \\( T \\) as \\term{time} and consider it to be \\hyperref[rem:additive_magma]{additive}. We call \\( \\Phi \\) the \\term{evolution function} of the system.\n\n \\begin{thmenum}\n \\thmitem{def:dynamical_system/ifs} If \\( T \\) is either the additive monoid of the \\hyperref[def:set_of_natural_numbers]{natural numbers} or the additive group of the \\hyperref[def:set_of_integers]{integers}, we say that the dynamical system has \\term{discrete time}.\n\n Due to \\eqref{eq:def:monoid_action/family/compatibility}, \\( \\Phi_{n+1} = \\Phi_n \\bincirc \\Phi_1 \\) for any integer \\( n \\). Using \\hyperref[rem:induction/peano_arithmetic]{natural number induction} and \\fullref{thm:def:group/negative_power}, we can show that \\( \\Phi_n = \\Phi_1^n \\) for every integer \\( n \\).\n\n Therefore, the entire evolution function of a discrete-time dynamical system is determined by a single endofunction \\( \\varphi: X \\to X \\). For this reason, we also refer to discrete-time dynamical systems as \\term{iterated function systems}.\n\n \\thmitem{def:dynamical_system/semiflow} If \\( T \\) is the additive monoid of \\hyperref[def:set_of_real_numbers]{real numbers}, with or without \\hyperref[def:extended_real_numbers]{an infinite element}, we say that the system is a \\term{semiflow}.\n\n \\thmitem{def:dynamical_system/flow} If \\( T \\) is the additive group of \\hyperref[def:set_of_real_numbers]{real numbers}, we say that the system is a \\term{flow}.\n \\end{thmenum}\n\n We will call the system \\term{discrete} if \\( T \\) is the monoid of zero-based \\hyperref[def:set_of_natural_numbers]{natural numbers} and \\term{continuous} if \\( T \\) is the monoid of nonnegative \\hyperref[def:set_of_real_numbers]{real numbers}, with or without \\hyperref[def:extended_real_numbers]{an infinite element}.\n\n The monoid \\( T \\) can theoretically be a \\hyperref[def:group]{group}, in which case we consider \\hyperref[def:group_action]{group actions}, however negative time is not as often needed in practice.\n\\end{definition}\n\n\\begin{definition}\\label{def:dynamical_system_trajectory}\n Fix a \\hyperref[def:dynamical_system]{dynamical system} with evolution function \\( \\Phi: T \\times X \\to X \\).\n\n A \\term{trajectory} in a starting at the \\term{initial state} \\( x_0 \\in X \\) is an \\hyperref[def:cartesian_product/indexed_family]{indexed family} \\( \\seq{ x_t }_{t \\in T} \\) obtained as\n \\begin{equation*}\n x_t \\coloneqq \\Phi_t(x_0).\n \\end{equation*}\n\n The condition \\ref{eq:def:monoid_action/family/identity} ensures that \\( \\Phi_0(x_0) = x_0 \\), and \\ref{eq:def:monoid_action/family/compatibility} ensures that\n \\begin{equation*}\n x_{t + s}\n =\n \\Phi_{t + s}(x_0)\n =\n \\Phi_t(x_s).\n \\end{equation*}\n\n For discrete dynamical systems, trajectories are sequences.\n\\end{definition}\n", "meta": {"hexsha": "da75c25e38e7799b981089966b8688b334b8441c", "size": 30565, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/group_actions.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/group_actions.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/group_actions.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.8116197183, "max_line_length": 615, "alphanum_fraction": 0.6729919843, "num_tokens": 10061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867729389246, "lm_q2_score": 0.7981867825403176, "lm_q1q2_score": 0.6371021321583593}} {"text": "\\begin{quote}\n Spectral sequences are one of those things for which anybody who is\n anybody must suffer through. Once you've done that, it's like linear\n algebra. You stop thinking so much about the `inner workings' later.\n\n -- Haynes Miller\n\\end{quote}\n\\section{The spectral sequence of a filtered complex}\nOur goal will be to describe a method for computing the homology of a chain\ncomplex. We will approach this problem by assuming that our chain complex is\nequipped with a filtration; then we will discuss how to compute the associated\ngraded of an induced filtration on the homology, given the homology of the\nassociated graded of the filtration on our chain complex.\n\nWe will start off with a definition.\n\\begin{definition}\n A \\emph{filtered chain complex} is a chain complex $C_\\ast$ along with a\n sequence of subcomplexes $F_s C_\\ast$ such that the group $C_n$ has a\n filtration by\n $$F_0 C_n\\subset F_1 C_n\\subseteq \\cdots,$$\n such that $\\bigcup F_s C_n = C_n$.\n\\end{definition}\nThe differential on $C_\\ast$ begets the structure of a chain\ncomplex on the associated graded $\\gr_s C_n = F_s C_n/F_{s-1} C_n$; in other\nwords, the differential on $C_\\ast$ respects the filtration, hence begets a\ndifferential $d:\\gr_s C_n\\to \\gr_s C_{n-1}$.\n\nThe canonical example of a filtered chain complex to keep in mind is the\nhomology of a filtered space (such as a CW-complex). Let $X$ be a filtered\nspace, i.e., a space equipped with a filtration $X_0\\subseteq X_1\\subseteq\n\\cdots$ such that $\\bigcup X_n = X$. We then have a filtration of the chain\ncomplex $C_\\ast(X)$ by the subcomplexes $C_\\ast(X_n)$.\n\nFor ease of notation, let us write\n$$E^0_{s,t} = \\gr_s C_{s+t} = F_s C_{s+t}/F_{s-1} C_{s+t},$$\nso the differential on $C_\\ast$ gives a differential $d^0:E^0_{s,t}\\to\nE^0_{s,t-1}$. A first approximation to the homology of $C_\\ast$ might therefore\nbe the homology $H_{s+t}(\\gr_s C_\\ast)$. We will denote this group by\n$E^1_{s,t}$. This is the homology of the associated graded of the filtration\n$F_\\ast C_\\ast$.\n\nWe can get an even better approximation to $H_\\ast C_\\ast$ by noticing that\nthere is a differential even on $E^1_{s,t}$. By construction, there is a short\nexact sequence of chain complexes\n$$0\\to F_{s-1} C_\\ast\\to F_s C_\\ast\\to \\gr_s C_\\ast\\to 0,$$\nso we get a long exact sequence in homology. The differential on $E^1_{s,t}$ is\nthe composite of the boundary map in this long exact sequence with the natural\nmap $H_\\ast(F_{s-1} C_\\ast)\\to H_\\ast(\\gr_{s-1} C_\\ast)$; more precisely, it is\nthe composite\n$$d^1:E^1_{s,t} = H_{s+t}(\\gr_s C_\\ast)\\xar{\\partial} H_{s+t-1}(F_{s-1}\nC_\\ast)\\to H_{s+t-1}(\\gr_{s-1}C_\\ast) = E^1_{s-1,t}.$$\nIt is easy to check that $(d^1)^2 = 0$.\n\nThis construction is already familiar from cellular chains: in this case,\n$E^1_{s,t}$ is exactly $H_{s+t}(X_{s},X_{s-1})$, which is exactly the cellular\n$s$-chains when $t=0$ (and is $0$ if $t\\neq 0$). The $d^1$ differential is\nconstructed in exactly the same way as the differential on cellular chains.\n\nIn light of this, we define $E^2_{s,t}$ to be the homology of the chain complex\n$(E^1_{\\ast,\\ast}, d^1)$; explicitly, we let\n$$E^2_{s,t} = \\ker(d^1:E^1_{s,t}\\to E^1_{s-1,t})/\\img(d^1:E^1_{s+1,t}\\to\nE^1_{s,t}).$$\nDoes this also have a differential $d^2$? The answer is yes. We will\ninductively define $E^r_{s,t}$ via a similar formula: if $E^{r-1}_{\\ast,\\ast}$\nand the differential $d^{r-1}:E^{r-1}_{s,t}\\to E^{r-1}_{s-r+1,t+r-2}$ are both\ndefined, we set\n$$E^{r}_{s,t} = \\ker(d^{r-1}:E^{r-1}_{s,t}\\to\nE^{r-1}_{s-r+1,t+r-2})/\\img(d^{r-1}:E^{r-1}_{s+r-1,t-r+2}\\to E^{r-1}_{s,t}).$$\nThe differential $d^{r}:E^{r}_{s,t}\\to E^{r}_{s-r,t+r-1}$ is defined as\nfollows. Let $[x]\\in E^r_{s,t}$ be represented by an element of $x\\in\nE^1_{s,t}$, i.e., an element of $H_{s+t}(\\gr_s C_\\ast)$. As above, the boundary\nmap induces natural maps $\\partial:H_{s+t}(\\gr_s C_\\ast)\\to H_{s+t-1}(F_{s-1}\nC_\\ast)$ and $\\partial:H_{s+t-1}(F_{s-r} C_\\ast)\\to H_{s+t-1}(\\gr_{s-r}\nC_\\ast)$. The element $\\partial x\\in H_{s+t-1}(F_{s-1} C_\\ast)$ in fact lifts\nto an element of $H_{s+t-1}(F_{s-r} C_\\ast)$. The image of this element under\n$\\partial$ inside $H_{s+t-1}(\\gr_{s-r} C_\\ast) = E^1_{s-r,t+r-1}$ begets a\nclass in $E^r_{s-r,t+r-1}$; this is the desired differential.\n\\begin{exercise}\n Fill in the missing details in this construction of $d^r$, and show that\n $(d^r)^2 = 0$.\n\\end{exercise}\nWe have proven most of the statements in the following theorem.\n\\begin{thm-defn}\\label{filtered-sseq}\n Let $F_\\ast C$ be a filtered complex. Then there exist natural\n \\begin{enumerate}\n\t\\item bigraded groups $(E^r_{s,t})_{s\\geq 0, t\\in\\Z}$ for any $r\\geq\n\t 0$, and\n\t\\item differentials $d^r:E^r_{s,t} \\to E^r_{s-r,t+r - 1}$ for any\n\t $r\\geq 0$.\n \\end{enumerate}\n such that $E^{r+1}_{s,t}$ is the homology of $(E^r_{\\ast,\\ast},d^r)$, and\n $(E^0, d^0)$ and $(E^1, d^1)$ are as above. If $F_\\ast C$ is bounded\n below, then this spectral sequence \\emph{converges to $\\gr_\\ast\n H_\\ast(C)$}, in the sense that there is an isomorphism:\n \\begin{equation}\\label{convergence}\n\tE^\\infty_{s,t} \\simeq \\gr_s H_{s+t}(C).\n \\end{equation}\n\\end{thm-defn}\nThis is called a \\emph{homology spectral sequence}. One should think of each\n$E^r_{\\ast,\\ast}$ as a ``page'', with lattice points $E^r_{s,t}$. We still need\nto describe the symbols used in the formula \\eqref{convergence}.\n\nThere is a filtration $F_s H_n(C):=\\img(H_n(F_s C) \\to H_n(C))$, and $\\gr_s\nH_\\ast(C)$ is the associated graded of this filtration. Taking formula\n\\eqref{convergence} literally, we only obtain information about the associated\ngraded of the homology of $C_\\ast$. Over vector spaces, this is sufficient to\ndetermine the homology of $C_\\ast$, but in general, one needs to solve an\nextension problem.\n\nTo define the notation $E^\\infty$ used above, let us assume that the filtration\n$F_\\ast C$ is bounded below (so $F_{-1} C = 0$). It follows that $E^0_{s,t} =\nF_s C_{s+t}/F_{s-1} C_{s+t} = 0$ for $s<0$, so the spectral sequence of\nTheorem-Definition \\ref{filtered-sseq} is a ``right half plane'' spectral\nsequence. It follows that in our example, the differentials from the group in\nposition $(s,t)$ must have vanishing $d^{s+1}$ differential.\n\nIn turn, this implies that there is a surjection $E^{s+1}_{s,t} \\to\nE^{s+2}_{s,t}$. This continues: we get surjections\n$$E^{s+1}_{s,t}\\to E^{s+2}_{s,t}\\to E^{s+3}_{s,t}\\to\\cdots,$$\nand the direct limit of this directed system is defined to be $E^\\infty_{s,t}$.\n\nFor instance, in the case of cellular chains, we argued above that $E^1_{s,t} =\nH_{s+t}(X_s, X_{s-1})$, so that $E^1_{s,t} = 0$ if $t\\neq 0$, and the $d^1$\ndifferential is just the differential in the cellular chain complex. It follows\nthat $E^2_{s,t} = H_s^{cell}(X)$ if $t=0$, and is $0$ if $t\\neq 0$. All higher\ndifferentials are therefore zero (because either the target or the source is\nzero!), so $E^r_{s,t} = E^2_{s,t}$ for every $r\\geq 2$. In particular\n$E^\\infty_{s,t} = H_s^{cell}(X)$ when $t=0$, and is $0$ if $t\\neq 0$. There are\nno extension problems either: the filtration on $X$ is bounded below, so\nTheorem-Definition \\ref{filtered-sseq} implies that $\\gr_s H_{s+0}(X) = H_s(X)\n\\simeq H_s^{cell}(X) = E^\\infty_{s,t}$.\n\n%OK, so for our theorem, we have the following picture of $E^\\infty_{\\ast,\\ast}$:\n%\\begin{equation*}\n% \\xymatrix{\n%\t\\gr_0 H_n(C) \\simeq E^\\infty_{0,n} & & & \\\\\n%\t& \\gr_1 H_n(C) \\simeq E^\\infty_{1,n-1} & & \\\\\n%\t& & \\gr_2 H_n(C) \\simeq E^\\infty_{2,n-2} & & \\\\\n%\t& & & \\ddots\n% }\n%\\end{equation*}\n%Note that $\\gr_0 H_n(C) = F_0 H_n(C) = \\img(H_n(F_0 C) \\to H_n(C))$.\n%\n%Maybe you still don't get the homology of $n$ (like if $F_\\ast$ is $0$).\n%But if $F_\\ast C$ is exhaustive, then $F_\\ast H_\\ast C$ is also exhaustive (I said this on Monday).\n%With this assumption, then we get all the associated graded pieces.\nIn a very precise sense, the datum of the spectral sequence of a filtered\ncomplex $F_\\ast C_\\ast$ determines the homology of $C_\\ast$:\n\\begin{corollary}\n Let $C\\xrightarrow{f} D$ be a map of filtered complexes.\n Assume that the filtration on $C$ and $D$ are bounded below and exhaustive.\n Assume also that $E^r(f)$ is an isomorphism for some $r$.\n Then $f_\\ast : H_\\ast(C) \\to H_\\ast(D)$ is an isomorphism.\n\\end{corollary}\n\\begin{proof}\n The map $E^r(f)$ is an isomorphism which is also also a chain map, i.e., it\n is compatible with the differential $d^r$. It follows that $E^{r+1}(f)$ is\n an isomorphism. By induction, we conclude that $E^\\infty_{s,t}(f)$ is an\n isomorphism for all $s,t$. Theorem-Definitino \\ref{filtered-sseq} implies\n that the map $\\gr_s(f_\\ast):\\gr_s H_\\ast(C) \\to \\gr_s H(D)$ is an\n isomorphism.\n \n We argue by induction using the short exact sequence:\n $$\n 0 \\to F_s H_\\ast(C) \\to F_{s+1} H_\\ast(C) \\to \\gr_{s+1} H_\\ast(C) \\to 0.\n $$\n We have $\\gr_0 H_n(C) = F_0 H_n(C) = \\img(H_n(F_0 C) \\to H_n(C))$, so the\n base case follows from the five lemma. In general, $f$ induces an\n isomorphism an isomorphism on the groups on the left (by the inductive\n hypothesis) and right (by the above discussion), so it follows that $F_s\n f_\\ast$ is an isomorphism by the five lemma. Since the filtration $F_\\ast\n C_\\ast$ was exhaustive, it follows that $f_\\ast$ is an isomorphism.\n\\end{proof}\n\\subsection{Serre spectral sequence}\\label{serre-sseq}\nIn this book, we will give two constructions of the Serre spectral sequence.\nThe second will appear later. Fix a fibration $E\\xrightarrow{p} B$, with $B$ a\nCW-complex. We obtain a filtration on $E$ by taking the preimage of the\n$s$-skeleton of $B$, i.e., $E_s = p^{-1} \\sk_s B$. It follows that there is a\nfiltration on $S_\\ast(E)$ given by\n$$F_s S_\\ast (E) = \\img(S_\\ast(p^{-1} \\sk_s(B)) \\to S_\\ast E).$$\nThis filtration is bounded below and exhaustive. The resulting spectral\nsequence of Theorem-Definition \\ref{filtered-sseq} is the Serre spectral\nsequence.\n%The low terms might depend on the CW-structure, but higher terms won't.\n\nLet us be more explicit. We have a pushout square:\n\\begin{equation*}\n \\xymatrix{\n\tE_{s-1}\\ar[r]\\ar[d] & E_s\\ar[d]\\\\\n\tB_{s-1}\\ar[r] & B_s\\\\\n\t\\coprod_{\\alpha\\in\\Sigma_s}S^{s-1}_\\alpha\\ar[r]\\ar[u] & \\coprod_{\\alpha\\in\\Sigma_s} D^s_\\alpha\\ar[u]\n }\n\\end{equation*}\n%I can pullback $E_s$ via:\n%\\begin{equation*}\n% \\xymatrix{\n%\t\\coprod_{\\alpha\\in \\Sigma_s} D^s_\\alpha\\times F_\\alpha\\ar[r]\\ar[d] & E_s\\ar[d]\\\\\n%\t\\coprod_{\\alpha\\in \\Sigma_s} D^s_\\alpha\\ar[r] & B_s\n% }\n%\\end{equation*}\nLet $F_\\alpha$ be the preimage of the center of $\\alpha$ cell. In particular,\nwe have a pushout:\n\\begin{equation*}\n \\xymatrix{\n\tE_{s-1}\\ar[r] & E_s\\\\\n\t\\coprod_{\\alpha\\in\\Sigma_s} S^{s-1}_\\alpha\\times F_\\alpha\\ar[r]\\ar[u] & \\coprod_{\\alpha\\in\\Sigma_s} D^s_\\alpha \\times F_\\alpha\\ar[u]\n }\n\\end{equation*}\nWe know that\n$$E^1_{s,t} = H_{s+t}(E_s,E_{s-1}) = \\bigoplus_{\\alpha\\in \\Sigma_s}\nH_{s+t}(D^s_\\alpha\\times F_\\alpha, S^{s-1}_\\alpha\\times F_\\alpha).$$\nWe can suggestively view this as $\\bigoplus_{\\alpha\\in\\Sigma_s}\nH_{s+1}((D^s_\\alpha, S^{s-1}_\\alpha)\\times F_\\alpha)$. By the K\\\"unneth\nformula (at least, if our coefficients are in a field), this is exactly\n$\\bigoplus_{\\alpha\\in\\Sigma_s} H_t(F_\\alpha)$. In analogy with our discussion\nabove regarding the spectral sequence coming from the cellular chain complex,\none would like to think of this as ``$C_s(B;H_t(F_\\alpha))$''. Sadly, there are\nmany things wrong with writing this.\n\nFor instance, suppose $B$ isn't connected. The fibers $F_\\alpha$ could have\ncompletely different homotopy types, so the symbol $C_s(B;H_t(F_\\alpha))$\ndoes not make any sense. Even if $B$ was path-connected, there would still be\nno canonical way to identify the fibers over different points. Instead, we\nobtain a functor $H_t(p^{-1}(-)):\\Pi_1(B) \\to \\mathbf{Ab}$, i.e., a ``local\ncoefficient system'' on $B$. So, the right thing to say is ``$E^2_{s,t} =\nH_s(B;\\underline{H_t(\\mathrm{fiber})})$''.\n\nTo define precisely what $H_s(B;\\underline{H_t(\\mathrm{fiber})})$ means, let us\npick a basepoint in $B$, and build the universal cover $\\widetilde{B}\\to B$.\nThis has an action of $\\pi_1(B,\\ast)$, so we obtain an action of\n$\\pi_1(B,\\ast)$ on the chain complex $S_\\ast(\\widetilde{B})$. Said differently,\n$S_\\ast(\\widetilde{B})$ is a chain complex of right modules over\n$\\Z[\\pi_1(B)]$. If $B$ is connected, a local coefficient system on $B$ is the\nsame thing as a (left) action of $\\pi_1(B)$ on $H_t(p^{-1}(\\ast))$. Then, we\ndefine a chain complex:\n$$\nS_\\ast(B;\\underline{H_t(p^{-1}(\\ast))}) = S_\\ast(\\widetilde{B})\n\\otimes_{\\Z[\\pi_1(B)]} H_t(p^{-1}(\\ast));\n$$\nthe differential is induced by the $\\Z[\\pi_1(B)]$-equivariant differential on\n$S_\\ast(\\widetilde{B})$. Our discussion above implies that the homology of this\nchain complex is the $E^2$-page.\n\nWe will always be in the case where that local system is trivial, so that\n$H_\\ast(B;\\underline{H_\\ast(p^{-1}(\\ast))})$ is just\n$H_\\ast(B;H_\\ast(p^{-1}(\\ast)))$. For instance, this is the case if $\\pi_1(B)$\nacts trivially on the fiber. 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YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7981867681382279, "lm_q1q2_score": 0.6371021244946543}} {"text": "\\documentclass[fleqn]{article}\n\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath}\n\\usepackage[colorlinks=true]{hyperref}\n\\usepackage{tikz}\n\\usetikzlibrary{calc,patterns,angles,quotes}\n\n\\begin{document}\n\\begin{center}\n {\\bfseries Solution to assignment \\#2}\\\\\n Introduction to GR, 2020 Fall\\\\\n International Centre for Theoretical Sciences\\\\\n Instructor: Prof. Bala Iyer, Tutor: Md Arif Shaikh\\footnote{\\href{mailto: arif.shaikh@icts.res.in}{arif.shaikh@icts.res.in}}\\\\\n % Due on October 04, 2020 11:59 PM.\n\\end{center}\n\\hrule\n\n\n\\section{Parallel transport along $r=$ constant line on the surface of a cone}\n\n\\begin{center}\n \\begin{tikzpicture}\n \\draw[cyan] (0,3) ellipse (2cm and 0.5cm);\n \\draw[->] (0, 0) -- (0, 4) node[above] {$z$};\n \\draw[->] (0, 0) -- (4, 0) node[right] {$y$};\n \\draw[->] (0, 0) -- (-3, -2) node[left] {$x$};\n \\draw[cyan] (0, 0) -- (2, 3);\n \\draw[cyan] (0, 0) -- (-2, 3);\n \\draw (0, 0.5) arc (90:110:1) node[left] {$\\alpha$} ;\n \\draw[orange] (0, 1.5) ellipse (1cm and 0.25cm);\n \\draw[fill=red] (0.5, 1.28) circle (0.05cm);\n \\draw (0, 1.5) -- (0.5, 1.28) node[right] {$(r, \\phi)$};\n \\draw (0, 1.5) -- (-3, -0.5);\n \\draw (0, 1.5) -- (4, 1.5);\n \\draw[red] (0.5, 1.28) arc (-80:-100:2.3) node[below] {$\\phi$}; \n\\end{tikzpicture}\n\\end{center}\n\nWe have a cone with opening angle $2\\alpha$ embedded in the 3-dimensional flat space. $r$ is the distance measured from the apex. A constant $r$ line on the surface of the cone would be parameterised by the angle $\\phi$. The 3-dimensional flat metric in spherical polar coordinate is given by\n\n\\begin{equation}\n \\label{eq:3-d-flat-metric}\n ds^2_{3d} = dr^2 + r^2d\\theta^2 + r^2\\sin^2\\theta d\\phi^2.\n\\end{equation}\n$\\theta$ is the angle w.r.t to the $z$-axis. On the surface of the cone, $\\theta=\\alpha=$ constant. Thus the metric on the surface of the cone becomes\n\\begin{equation}\n \\label{eq:2-d-metric-on-cone}\n \\boxed{ds^2 = dr^2 + r^2\\sin^2\\alpha d\\phi^2.}\n\\end{equation}\n\n\\noindent\\fbox{\\parbox{\\textwidth}{{\\bfseries What is parallel transport of a tensor?}\n Parallel transport is defined in the following way: A tensor field $T^{\\mu...}_{\\nu...}$ is said to be {\\itshape parallel transported} along a curve $\\gamma:=x^\\alpha(\\lambda)$ if the covariant derivative of the tensor field along the curve vanishes: $D_\\lambda T^{\\mu...}_{\\nu...} = u^\\eta \\nabla_\\eta T^{\\mu...}_{\\nu...}= 0$, where $u^\\eta = dx^\\eta/d\\lambda$.}}\n\n\\vspace{0.5cm}\n\nIn the given problem we are asked to compute the rotation of a vector $V^\\mu$ when it is parallel transported along the $r=$ constant line from $\\phi=0$ to $\\phi=2\\pi$. Thus the curve $\\gamma$ is represented by $x^\\mu(\\lambda) = (r_0, \\phi(\\lambda))$. The condition for parallel transport of $V^\\mu$ along $\\gamma$ then becomes\n\n\\begin{equation}\n \\label{eq:parallel-transport-along-r-constant}\n u^\\mu\\nabla_\\mu V^\\nu = 0 \\to \\dot{\\phi}\\nabla_\\phi V^\\mu = 0\\to \\nabla_\\phi V^\\mu = 0 \\to \\boxed{\\partial_\\phi V^\\mu + \\Gamma^\\mu_{\\phi \\nu} V^\\nu = 0}.\n\\end{equation}\n\nTo find out the Christofell symbols we use the geodesic equations. The geodesic equations could be obtained from the Lagrangian $L = (1/2)g_{\\mu\\nu}\\dot{x}^\\mu\\dot{x}^\\nu$ where $\\dot{x}^\\mu = dx^\\mu/d\\tau$ (we take $\\lambda = \\tau$ the propertime). The geodesic e.o.m is given by\n\\begin{equation}\n \\label{eq:geodesic-eom}\n \\frac{d}{d\\tau}\\left(\\frac{\\partial L}{\\partial \\dot{x^\\mu}}\\right) - \\frac{\\partial L}{\\partial x^\\mu} = 0\n\\end{equation}\n\nWith the Lagrangian given by\n\\begin{equation}\n \\label{eq:Lagrangian}\n L = \\frac{1}{2}(\\dot{r}^2 + r^2\\sin^2\\alpha\\dot{\\phi}^{2})\n\\end{equation}\nThe e.o.m for $r$ becomes\n\\begin{equation}\n \\label{eq:eom-r}\n \\ddot{r} - r\\sin^2\\alpha \\dot{\\phi}^{2} = 0,\n\\end{equation}\nwhich provides $\\Gamma^{r}_{\\phi\\phi} = - r\\sin^2\\alpha$.\nSimilarly e.o.m for $\\phi$ is given by\n\\begin{equation}\n \\label{eq:eom-phi}\n \\ddot{\\phi} + \\frac{2}{r}\\dot{r}\\dot{\\phi} = 0,\n\\end{equation}\nwhich provides $\\Gamma^{\\phi}_{r\\phi} = \\Gamma^{r}_{\\phi r} = \\frac{1}{r}$. Thus we have\n\\begin{equation}\n \\label{eq:Christofell-symbols}\n \\boxed{\\Gamma^{r}_{\\phi\\phi} = - r\\sin^2\\alpha,\\qquad \\Gamma^{\\phi}_{r\\phi} = \\Gamma^{r}_{\\phi r} = \\frac{1}{r}}.\n\\end{equation}\n\nThus Eq. (\\ref{eq:parallel-transport-along-r-constant}) becomes \n\\begin{equation}\n \\label{eq:parallel-transport}\n \\partial_\\phi V^{\\mu} + \\Gamma^{\\mu}_{\\phi\\nu}V^\\nu = 0.\n\\end{equation}\n\nFor $\\mu=r,\\phi$ this gives the following equations\n\\begin{equation}\n \\label{eq:r-phi}\n \\partial_\\phi V^{r} + \\Gamma^{r}_{\\phi\\nu}V^\\nu = 0,\\qquad\n \\partial_\\phi V^{\\phi} + \\Gamma^{\\phi}_{\\phi\\nu}V^\\nu = 0.\n\\end{equation}\nwhich using Eq.~\\eqref{eq:Christofell-symbols} becomes\n\\begin{equation}\n \\label{eq:r-phi-2}\n \t \\partial_\\phi V^{r} + \\Gamma^{r}_{\\phi\\nu}V^\\nu = 0,\\qquad\n \\partial_\\phi V^{\\phi} + \\Gamma^{\\phi}_{\\phi\\nu}V^\\nu = 0.\n\\end{equation}\n\nor after differentiating with respect to $\\phi$,\n\\begin{equation}\n \\label{eq:r-phi-3}\n \\partial_\\phi^2 V^r + \\sin^2\\alpha V^r = 0,\\qquad \\partial_\\phi^2 V^\\phi + \\sin^2\\alpha V^\\phi = 0\n\\end{equation}\nWhich has the following general solutions\n\\begin{equation}\n \\label{eq:r-phi-solution}\n \\boxed{V^r = A^r_1 \\sin(\\sin\\alpha \\phi) + A^r_2 \\cos(\\sin\\alpha \\phi),\\quad V^\\phi = A^\\phi_1 \\sin(\\sin\\alpha \\phi) + A^\\phi_2 \\cos(\\sin\\alpha \\phi)}.\n\\end{equation}\nDenoting the values of the components at $\\phi=0$ with suffix `0' the solutions becomes\n\n\\begin{equation}\n \\label{eq:r-phi-solution-2}\n \\boxed{V^r = A^r_1 \\sin(\\sin\\alpha \\phi) + V^r_0 \\cos(\\sin\\alpha \\phi),\\quad V^\\phi = A^\\phi_1 \\sin(\\sin\\alpha \\phi) + V^\\phi_0 \\cos(\\sin\\alpha \\phi)}.\n\\end{equation}\nNow, the constants are not all independent, there are only two independent constants. Thus the other two components could be written in terms of the $A^r_0$ and $A^\\phi_0$ using equation \\eqref{eq:r-phi-2}. This provides\n\\begin{equation}\n \\label{eq:constants}\n A^r_1 = r \\sin\\alpha V^\\phi_0,\\quad A^\\phi_1 = - V^r_0/r\\sin\\alpha.\n\\end{equation}\nThus finally we have the solutions\n\\begin{equation}\n \\label{eq:r-phi-solution-final}\n \\boxed{V^r = V^\\phi_0 r \\sin\\alpha \\sin(\\phi\\sin\\alpha) + V^r_0 \\cos(\\phi\\sin\\alpha),\\quad V^\\phi = - \\frac{V^r_0}{r\\sin\\alpha} \\sin(\\phi\\sin\\alpha) + V^\\phi_0 \\cos(\\phi\\sin\\alpha)}.\n\\end{equation}\nNow we move to a orthonormal set of basis given by $E_r = \\partial_r, E_\\phi = (1/\\sin\\alpha)\\partial_\\phi$. In this basis, we can rewrite the components of the $V$ as\n\\begin{equation}\n \\label{eq:V-in-orthonormal}\n \\hat{V}^r = V^r,\\quad \\hat{V}^\\phi = r\\sin\\alpha V^\\phi.\n\\end{equation}\nThus in this basis, we have the transformation\n\\begin{equation}\n \\label{eq:transformation-orthonormal}\n \\boxed{\\begin{bmatrix}\n \\hat{V}^r \\\\\n \\hat{V}^\\phi\n \\end{bmatrix} =\n \\begin{bmatrix}\n \\cos(\\phi\\sin\\alpha) & \\sin(\\phi\\sin\\alpha)\\\\\n -\\sin(\\phi\\sin\\alpha) & \\cos(\\phi\\sin\\alpha)\n \\end{bmatrix}\n \\begin{bmatrix}\n \\hat{V}^r_0 \\\\\n \\hat{V}^\\phi_0\n \\end{bmatrix}}\n\\end{equation}\nThus the components are rotated by an angle $\\beta = \\phi\\sin\\alpha$. For $\\alpha = 2\\pi$ this rotation equals $\\beta = 2\\pi\\sin\\alpha$.\n\n\\end{document}", "meta": {"hexsha": "6a69e90296a8be5d3ef67068ce65f70623b1d901", "size": 7108, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sol-assignment-2/Solution_assignment_2_icts_igr_2020_fall.tex", "max_stars_repo_name": "mdarifshaikh/GR-Fall-2020-ICTS", "max_stars_repo_head_hexsha": "4b651c965b58d899a98f6ebe838a6bc30c3c37ef", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sol-assignment-2/Solution_assignment_2_icts_igr_2020_fall.tex", "max_issues_repo_name": "mdarifshaikh/GR-Fall-2020-ICTS", "max_issues_repo_head_hexsha": "4b651c965b58d899a98f6ebe838a6bc30c3c37ef", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sol-assignment-2/Solution_assignment_2_icts_igr_2020_fall.tex", "max_forks_repo_name": "mdarifshaikh/GR-Fall-2020-ICTS", "max_forks_repo_head_hexsha": "4b651c965b58d899a98f6ebe838a6bc30c3c37ef", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.425, "max_line_length": 369, "alphanum_fraction": 0.6551772651, "num_tokens": 2722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867681382279, "lm_q2_score": 0.798186775339273, "lm_q1q2_score": 0.637102122578728}} {"text": "%% LyX 2.3.2-1 created this file. For more info, see http://www.lyx.org/.\n%% Do not edit unless you really know what you are doing.\n\\documentclass[english]{article}\n\\usepackage[T1]{fontenc}\n\\usepackage[latin9]{luainputenc}\n\\usepackage{babel}\n\\usepackage{array}\n\\usepackage{textcomp}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage[unicode=true]\n {hyperref}\n\n\\makeatletter\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.\n%% Because html converters don't know tabularnewline\n\\providecommand{\\tabularnewline}{\\\\}\n\n\\makeatother\n\n\\begin{document}\n\\title{Chapter 7: Chi-square Test}\n\\maketitle\n\n\\section{Example}\n\\begin{enumerate}\n\\item $H_{0}:$ The die is fair $\\left(p=\\frac{1}{6}\\right)$\n\\item $H_{1}:$ The die is unfair $\\left(p\\neq\\frac{1}{6}\\right)$\n\\item At $\\alpha=0.05$, critical value = $\\chi_{\\alpha;m-t-1}^{2}$.\n\\item ~\n\\begin{align*}\n\\chi_{0.05;6-0-1}^{2} & =\\chi_{0.05;5}^{2}\\\\\n & =11.070\n\\end{align*}\n\\item Rejection region: (remember Chi-square distribution does NOT have\na negative region)\n\\[\n\\chi^{2}>11.07\n\\]\n\\item %\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n\\hline \nNumber & 1 & 2 & 3 & 4 & 5 & 6 & Total\\tabularnewline\n\\hline \n\\hline \n$O_{i}$ & 89 & 113 & 98 & 104 & 117 & 79 & 600\\tabularnewline\n\\hline \n$E_{i}$ & 100 & 100 & 100 & 100 & 100 & 100 & 600\\tabularnewline\n\\hline \n$O_{i}-E_{i}$ & -11 & 13 & -2 & -4 & 17 & -21 & \\tabularnewline\n\\hline \n$\\left(O_{i}-E_{i}\\right)^{2}$ & 121 & 169 & 4 & 16 & 289 & 441 & \\tabularnewline\n\\hline \n$\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$ & 1.21 & 1.69 & 0.04 & 0.16 & 2.89 & 4.41 & 10.40\\tabularnewline\n\\hline \n\\end{tabular}\n\\item Since $\\chi^{2}=10.40<11.07$, we failed to reject $H_{0}$ at 5\\%\nsignificance level. Hence, we do not have enough evidence to conclude\nthat the observed frequencies are significantly different from those\nexpected of a fair die.\n\\end{enumerate}\n\n\\section{Example}\n\nSince every city have equal sales potential, therefore, each sale\nshould be $\\frac{\\sum x}{n}$, which is the mean.\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Claim: Each of the seven cities have equal sales potential. $\\left(H_{0}\\right)$\n\\item Opposite: Each of the seven cities do not have equal sales potential.\n$\\left(H_{1}\\right)$\n\\item $H_{0}:A=B=C=D=E=F=G$\n\\item $H_{1}:A\\neq B\\neq C\\neq D\\neq E\\neq F\\neq G$\n\\end{enumerate}\n\\item Find the critical value\n\\begin{enumerate}\n\\item $\\alpha=0.05$\n\\item $\\chi_{0.05;7-1}^{2}=\\chi_{0.05;6}^{2}=12.592$\n\\end{enumerate}\n\\item Find the rejection region\n\\begin{enumerate}\n\\item $\\chi^{2}>12.592$\n\\end{enumerate}\n\\item %\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|}\n\\hline \nCity & A & B & C & D & E & F & G & Total\\tabularnewline\n\\hline \n\\hline \n$O_{i}$ & 120 & 185 & 260 & 190 & 210 & 175 & 260 & \\tabularnewline\n\\hline \n$E_{i}$ & 200 & 200 & 200 & 200 & 200 & 200 & 200 & 1400\\tabularnewline\n\\hline \n$\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$ & $\\frac{\\left(120-200\\right)^{2}}{200}=32$ & $\\frac{\\left(185-200\\right)^{2}}{200}=1.125$ & $\\frac{\\left(260-200\\right)^{2}}{200}=18$ & $\\frac{\\left(190-200\\right)^{2}}{200}=0.5$ & $\\frac{\\left(210-200\\right)^{2}}{200}=0.5$ & $\\frac{\\left(175-200\\right)^{2}}{200}=3.125$ & $\\frac{\\left(260-200\\right)^{2}}{200}=18$ & 73.25\\tabularnewline\n\\hline \n\\end{tabular}\n\\item Since $\\chi^{2}=73.25>12.592$, we reject $H_{0}$and do not have\nenough evidence to conclude that each of the seven cities have equal\nsales potential.\n\\end{enumerate}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Claim: The local hospital follows the national pattern. $\\left(H_{0}\\right)$\n\\item Oppo: The local hospital do not follow the national pattern. $\\left(H_{1}\\right)$\n\\end{enumerate}\n\\item Find the rejection region\n\\begin{enumerate}\n\\item $\\alpha=0.05$\n\\item $\\chi_{0.05;7-1}^{2}=\\chi_{0.05;6}^{2}=12.592$\n\\item Rejection region: $\\chi^{2}>12.592$\n\\end{enumerate}\n\\item Find the Chi-square score\n\\item %\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|}\n\\hline \nNumber of times admitted & 1 & 2 & 3 & 4 & 5 & 6 & 7 & Total\\tabularnewline\n\\hline \n\\hline \n$O_{i}$ & 165 & 79 & 50 & 44 & 32 & 20 & 10 & 400\\tabularnewline\n\\hline \n$E_{i}$ & 400{*}40\\%=160 & 80 & 56 & 40 & 32 & 24 & 8 & \\tabularnewline\n\\hline \n$\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$ & $\\frac{\\left(165-160\\right)^{2}}{160}$ & $\\frac{\\left(79-80\\right)^{2}}{80}$ & $\\frac{\\left(50-56\\right)^{2}}{56}$ & $\\frac{\\left(44-40\\right)^{2}}{40}$ & $\\frac{\\left(32-32\\right)^{2}}{32}$ & $\\frac{\\left(20-24\\right)^{2}}{24}$ & $\\frac{\\left(10-8\\right)^{2}}{8}$ & 2.378\\tabularnewline\n\\hline \n\\end{tabular}\n\\item Since $\\chi^{2}=2.378<12.592$, we failed to reject $H_{0}$. Hence,\nwe conclude that the local hospital follows the national pattern.\n\\end{enumerate}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Claim: The results support the theory $\\left(H_{0}\\right)$\n\\item Opposite: The results do not support the theory $\\left(H_{1}\\right)$\n\\end{enumerate}\n\\item Find the rejection region\n\\begin{enumerate}\n\\item $\\chi_{0.05;3-1}^{2}=5.991$\n\\item $\\chi^{2}>5.991$\n\\end{enumerate}\n\\item Calculate the Chi-score\n\\begin{enumerate}\n\\item %\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline \nColor & $O_{i}$ & $E_{i}$ & $\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$ & \\tabularnewline\n\\hline \n\\hline \nRed & 84 & 83.25 & 0.00676 & \\tabularnewline\n\\hline \nBlue & 92 & 83.25 & 0.91967 & \\tabularnewline\n\\hline \nPurple & 157 & 166.5 & 0.54204 & \\tabularnewline\n\\hline \n & 333 & & 1.46847 & \\tabularnewline\n\\hline \n\\end{tabular}\n\\end{enumerate}\n\\item $\\chi^{2}=1.46847<5.991$. We failed to reject $H_{0}$ at $\\alpha=0.05$\nand conclude that the result support the genetic theory.\n\\end{enumerate}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Claim: The random number generator is working incorrectly. $H_{1}$\n\\item Hypo: The random number generator is working correctly. $H_{0}$\n\\end{enumerate}\n\\item The critical value\n\\begin{enumerate}\n\\item $\\chi_{0.05;6-1}^{2}=11.070$\n\\item Rejection region: $\\chi^{2}>11.070$\n\\end{enumerate}\n\\item Find the test statistic\n\\begin{enumerate}\n\\item %\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline \nInterval & $O_{i}$ & Probability, $P_{i}$ & $E_{i}=500*P_{i}$ & $\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$\\tabularnewline\n\\hline \n\\hline \n$s<4$ & 10 & $P\\left(Z<\\frac{4-6}{1}\\right)=P\\left(Z<\\text{\\textminus}2\\right)=0.02275$ & 11.375 & 0.16621\\tabularnewline\n\\hline \n$4\\le s<5$ & 75 & 0.13595 & 67.975 & 0.74601\\tabularnewline\n\\hline \n$5\\le s<6$ & 163 & 0.34130 & 170.65 & 0.34294\\tabularnewline\n\\hline \n$6\\le s<7$ & 174 & 0.34130 & 170.65 & 0.06576\\tabularnewline\n\\hline \n$7\\le s<8$ & 66 & 0.13595 & 67.975 & 0.05738\\tabularnewline\n\\hline \n$s\\ge8$ & 12 & 0.02285 & 11.375 & 0.3434\\tabularnewline\n\\hline \nTotal & & 1.0000 & 500 & \\textbf{1.39264}\\tabularnewline\n\\hline \n\\end{tabular}\n\\item Note: For the probability, use the Z-score\n\\[\nP\\left(s>\\bar{x}\\right)=P\\left(Z>\\frac{\\bar{x}-\\mu}{\\sigma}\\right)\n\\]\n\\item Use \\href{http://onlinestatbook.com/2/calculators/normal_dist.html}{Normal Distribution Calculator from OnlineStatBook}\nto quickly calculate without converting to normal.\n\\end{enumerate}\n\\item Conclusion\n\\begin{enumerate}\n\\item Since $\\chi_{2}=1.39264<11.070$, we failed to reject $H_{0}$, and\nhence conclude that the random number generator is working correctly.\n\\end{enumerate}\n\\end{enumerate}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Claim: The distribution of X is $B\\left(3,0.25\\right)$ $\\left(H_{0}\\right)$\n\\item Oppo: The distribution of X is not $B\\left(3,0.25\\right)$ $\\left(H_{1}\\right)$\n\\end{enumerate}\n\\item Find the critical value\n\\[\n\\alpha=0.01\n\\]\n\n\\begin{enumerate}\n\\item Critical value\n\\begin{align*}\n\\chi_{\\alpha;m-t-1}^{2} & =\\chi_{0.01;3-0-1}^{2}\\\\\n & =\\chi_{0.01;2}^{2}\\\\\n & =9.210\n\\end{align*}\n\\end{enumerate}\n\\item Rejection region\n\\[\n\\chi^{2}>9.210\n\\]\n\n\\begin{enumerate}\n\\item If this is true, the probability distribution of X is \n\\[\nP\\left(X=x\\right)=^{3}C_{x}\\left(0.25\\right)^{x}0.75^{3-x}\n\\]\n\\end{enumerate}\n\\item %\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline \nX & $O_{i}$ & Probability, $p_{i}$ & $E_{i}=64p_{i}$ & $\\frac{\\left(O_{i}-E_{i}\\right)^{2}}{E_{i}}$ & Total\\tabularnewline\n\\hline \n\\hline \n0 & 21 & $\\frac{21}{21+31+12+0}=0.4219$ & 27 & 1.333 & \\tabularnewline\n\\hline \n1 & 31 & 0.4219 & 27 & 0.5926 & \\tabularnewline\n\\hline \n2,3 & 12 & 0.1563 & 10 & 0.4 & 2.3259\\tabularnewline\n\\hline \n\\end{tabular}\n\\begin{enumerate}\n\\item $p_{0}=^{3}C_{0}*\\left(0.25\\right)^{0}*\\left(0.75\\right)^{3}=0.4219$\n\\item $p_{1}=^{3}C_{1}*\\left(0.25\\right)^{1}*\\left(0.75\\right)^{2}=0.4219$\n\\item $p_{2+3}=^{3}C_{2}*\\left(0.25\\right)^{2}*\\left(0.75\\right)^{1}+{}^{3}C_{3}*\\left(0.25\\right)^{3}=0.1563$\n\\end{enumerate}\n\\item Since $\\chi^{2}=2.3259<9.210.$ We failed to reject $H_{0}$, and\nhence do not have enough evidence to show that the recorded data do\nnot fit $B\\left(3,0.25\\right)$.\n\\end{enumerate}\n\n\\section{Example}\n\nNote: We should not round it from 2.02 to 2.00, oterhwise it will\nbe inaccurate during exams.\n\n\\section{Example}\n\n\\subsection{Extra Notes}\n\nBy following the probability multiplication rule, if they are independent,\nthe expected frequency for a field in a two-way table will be\n\\begin{enumerate}\n\\item Proportion probability of something happening in the row, multiple\nby the proportion probability of something happening in a column (to\nget row {*} column proportion) and then multipled by the sample size.\n\\item Eg:\n\\[\n1000\\left(sample\\right)*\\left(\\frac{100}{1000}\\left(row\\right)*\\frac{100}{1000}\\left(column\\right)\\right)=10\\left(expected\\right)\n\\]\n\\end{enumerate}\n\n\\subsection{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item The number of accidents depend on the visits by the inspector. $\\left(H_{0}\\right)$\n\\item The number of accidents do not depend on the visits by the inspector.\n$\\left(H_{1}\\right)$\n\\end{enumerate}\n\\item Critical value\n\\begin{align*}\n\\chi_{0.05;\\left(4-1\\right)\\left(2-1\\right)}^{2} & =\\chi_{0.05;3}^{2}\\\\\n & =7.815\n\\end{align*}\n\n\\begin{enumerate}\n\\item Note : $D.O.F.=\\left(Row-1\\right)\\left(Column-1\\right)$\n\\item ~\\includegraphics[width=0.8\\paperwidth,height=0.8\\textheight,keepaspectratio]{lyx-img/c7q8expected}\n\\end{enumerate}\n\\item Rejection region\n\\[\n\\chi^{2}>7.815\n\\]\n\\item Find the test-statistic\n\\begin{align*}\n\\chi^{2} & =\\frac{\\left(33-27.778\\right)^{2}}{27.778}+\\frac{\\left(8-13.889\\right)^{2}}{13.889}+\\frac{\\left(5-5.556\\right)^{2}}{5.556}+\\frac{\\left(4-2.778\\right)^{2}}{5.556}+\\frac{\\left(67-72.222\\right)^{2}}{72.222}+\\frac{\\left(42-36.111\\right)^{2}}{36.111}+\\frac{\\left(15-14.444\\right)^{2}}{14.444}+\\frac{\\left(6-7.222\\right)^{2}}{7.222}\\\\\n & =5.3692\n\\end{align*}\n\\item Conclusion\n\\begin{enumerate}\n\\item Since $\\chi^{2}=5.3692<7.815$. We failed to reject $H_{0}$, and\nhence do not have enough evidence to conclude that the number of accidents\ndo not depend on the visits by the inspector.\n\\end{enumerate}\n\\end{enumerate}\n\n\\section{Example}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item Hypo: The coin is fair $\\left(H_{1}\\right)$\n\\item Oppo: The coin is unfair $\\left(H_{0}\\right)$\n\\end{enumerate}\n\\item Critical value at $\\alpha=0.05$\n\\begin{enumerate}\n\\item Critical value: $\\chi_{0.05;2-0-1}^{2}=\\chi_{0.05;1}^{2}=3.841$\n\\end{enumerate}\n\\item Rejection range: $\\chi^{2}>3.841$\n\\item Test-statistics\n\\begin{align*}\n\\chi^{2} & =\\sum_{i=1}^{m}\\frac{\\left(\\left|O_{i}-E_{i}\\right|-0.5\\right)^{2}}{E_{i}}\\\\\n & =\\frac{\\left(\\left|115-100\\right|-0.5\\right)^{2}}{100}+\\frac{\\left(\\left|85-100\\right|-0.5\\right)^{2}}{100}\\\\\n & =4.205\n\\end{align*}\n\\end{enumerate}\n\n\\section{Example}\n\\begin{enumerate}\n\\item Hypothesis\n\\begin{enumerate}\n\\item $\\left(H_{0}\\right)$ There is no association between taking the new\ndrug and attack by the disease\n\\item $\\left(H_{1}\\right)$ There is an association between taking the new\ndrug and attack by the disease\n\\end{enumerate}\n\\item Critical value\n\\begin{enumerate}\n\\item $\\chi_{0.05;\\left(2-1\\right)\\left(2-1\\right)}^{2}=\\chi_{0.05;1}^{2}=3.841$\n\\end{enumerate}\n\\item Rejection range\n\\begin{enumerate}\n\\item $\\chi^{2}>3.841$\n\\end{enumerate}\n\\item Test-statistic (apply Yate's correction because $D.O.F.=1$)\n\\begin{enumerate}\n\\item ~%\n\\begin{tabular}{|>{\\centering}p{0.8\\textwidth}|c|c|c|}\n\\hline \n & Drugged & Drugless & Total row\\tabularnewline\n\\hline \n\\hline \nAttacked & 24 & 32 & 56\\tabularnewline\n\\hline \nExpected & (35.47) & (20.53) & \\tabularnewline\n\\hline \nNot attacked & 52 & 12 & 64\\tabularnewline\n\\hline \nExpected & (40.53) & (23.47) & \\tabularnewline\n\\hline \nTotal column & 76 & 44 & 120\\tabularnewline\n\\hline \n\\end{tabular}\n\\item $\\chi^{2}=\\frac{\\left(\\left|24-35.47\\right|-0.5\\right)^{2}}{35.47}+\\frac{\\left(\\left|32-20.53\\right|-0.5\\right)^{2}}{20.53}+\\frac{\\left(\\left|52-40.53\\right|-0.5\\right)^{2}}{40.53}+\\frac{\\left(\\left|12-23.47\\right|-0.5\\right)^{2}}{23.47}=17.351$\n\\end{enumerate}\n\\item Conclusion\n\\begin{enumerate}\n\\item Since $\\chi^{2}=17.351>3.841$, we reject $H_{0}$. Therefore we have\nenough evidence to conclude that there is an association between taking\nthe new drug and attack by the disease.\n\\end{enumerate}\n\\end{enumerate}\n\n\\end{document}\n", "meta": {"hexsha": "ae7275f6df93e348804e14a259c3728be78f1df9", "size": 13079, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "y2s1/statistics-ii/lectures/c7-chi-square.tex", "max_stars_repo_name": "ouldevloper/university", "max_stars_repo_head_hexsha": "4bb4a7e86912b46aa625900d2985b4d1e436cf4e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2019-06-22T14:29:51.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-22T16:58:12.000Z", "max_issues_repo_path": "y2s1/statistics-ii/lectures/c7-chi-square.tex", "max_issues_repo_name": "ouldevloper/university", "max_issues_repo_head_hexsha": "4bb4a7e86912b46aa625900d2985b4d1e436cf4e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-06-30T06:38:00.000Z", "max_issues_repo_issues_event_max_datetime": "2020-07-08T06:37:57.000Z", "max_forks_repo_path": "y2s1/statistics-ii/lectures/c7-chi-square.tex", "max_forks_repo_name": "ouldevloper/university", "max_forks_repo_head_hexsha": "4bb4a7e86912b46aa625900d2985b4d1e436cf4e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 9, "max_forks_repo_forks_event_min_datetime": "2020-04-04T12:59:50.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-27T22:44:58.000Z", "avg_line_length": 32.9445843829, "max_line_length": 383, "alphanum_fraction": 0.671534521, "num_tokens": 5055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581097540519, "lm_q2_score": 0.8006920020959544, "lm_q1q2_score": 0.6370770848828544}} {"text": "\\documentclass[../PHYS306Notes.tex]{subfiles}\n\n\\begin{document}\n\\subsection{Worksheet - Lagrangians and Coordinate Transformations}\n\\begin{p}\nIf we transform $x, y$ into polar coordinates, what happens to the principle of least action? What do Lagrange’s equations become for a particle in a two-dimensional potential $U(x,y)$, now using polar coordinates. What are the generalized forces and generalized momenta? \n\\end{p}\n\\begin{s}\nSince the Lagrangian is invariant of the choice of coordinates, nothing happens; the principle of least action still holds (the Lagrangian and action integral are equivalent between the two coordinate systems):\n\\[S[r, \\theta] = \\int_{t_1}^{t_2}\\LL(r, \\dot{r}, \\theta, \\dot{\\theta}, t)dt = \\int_{t_1}^{t_2} \\LL(x, \\dot{x}, y, \\dot{y}, t)dt = S[x, y]\\]\nUsing the fact that $\\v{r} = r\\rhat$ and $\\dot{\\v{r}} = \\dot{r}\\rhat + r\\dot{\\phi}\\phihat$ The Lagrangian becomes:\n\\[\\LL = T - U = \\frac{1}{2}m\\dot{\\v{r}}^2 - U(x,y) = \\frac{1}{2}m(\\dot{r}^2 + r^2\\dot{\\phi}^2) - U(r, \\theta)\\]\nSo Lagrange's equations become:\n\\[\\dpd{\\LL}{r} = \\dod{}{t}\\dpd{\\LL}{\\dot{r}} \\implies -\\dpd{U}{r} + mr\\dot{\\phi}^2 = m\\ddot{r}\\]\nSo the generalized force is $-\\dpd{U}{r} + mr\\dot{\\phi}^2$ and the generalized momentum is $m\\dot{r}$.\n\\[\\dpd{\\LL}{\\phi} = \\dod{}{t}\\dpd{\\LL}{\\dot{\\phi}} \\implies -\\dpd{U}{\\phi} = mr^2\\ddot{\\phi} \\]\nSo the generalized force is $-\\dpd{U}{\\phi}$ (which is just the torque!) and the generalized momentum is $mr^2\\dot{\\phi}$ (which is the angular momentum!).\n\\end{s}\n\n\\begin{p}\nWrite the Lagrangian for two particles interacting through a potential $U(\\v{r}_1, \\v{r}_2)$, using the “lab frame” coordinates $\\v{r}_1, \\v{r}_2$. How does the potential simplify if it is translationally-invariant? How does it simplify if it is orientationally-invariant (i.e. central)? \n\\end{p}\n\\begin{s}\nThe Lagrangian is given by:\n\\[\\LL = \\frac{1}{2}m_1\\dot{\\v{r}}_1^2 + \\frac{1}{2}m_2\\dot{\\v{r}}_2^2 - U(x,y)\\]\nIf the potential is translationally invariant, then there is no difference if we move the entire system through space; hence, it can only depend on the difference, i.e. $U(\\v{r}_1 - \\v{r}_2)$. If it is orientationally-invariant (central), then it only depends on the magnitude of the the difference, i.e. $U(\\abs{\\v{r}_1 - \\v{r}_2})$.\n\\end{s}\n\n\\begin{p}\nRewrite the kinetic energy in terms of the centre of mass (CM) and relative coordinates, $\\v{R}, \\v{r}$. What are the Lagrange equations in these coordinates?\n\\begin{center}\n \\includegraphics[scale=0.6]{Lecture-3/W3-img1.png}\n\\end{center}\n\\end{p}\n\\begin{s}\nWe have that the relative coordinate is $\\v{r} = \\v{r}_1 - \\v{r}_2$ and $U = U(\\abs{\\v{r}})$. The CM position is given by:\n\\[\\v{R} = \\frac{m_1\\v{r}_1 + m_2\\v{r}_2}{m_1 + m_2}\\]\nLet us also define the combined mass $M = m_1 + m_2$, we then have that:\n\\[T = \\frac{1}{2}\\left(m_1\\dot{\\v{r}}_1^2 + m_2\\dot{\\v{r}}_2^2\\right) = \\frac{1}{2}\\left(m_1\\left(\\dot{\\v{R}} + \\frac{m_2}{M}\\dot{\\v{r}}\\right)^2 + m_2\\left(\\dot{\\v{R}} - \\frac{m_1}{M}\\dot{\\v{r}}\\right)^2\\right) = \\frac{1}{2}\\left(M\\dot{\\v{R}}^2 + \\frac{m_1m_2}{M}\\dot{\\v{r}}^2\\right)\\]\nSo defining the reduced mass $\\mu = \\frac{m_1m_2}{M}$ we have:\n\\[\\LL = T - U = \\frac{M}{2}\\dot{\\v{R}}^2 + \\left(\\frac{\\mu}{2}\\dot{\\v{r}}^2 - U(r)\\right)\\]\nWe end up with a Lagrangian that has essentially two independent terms; a COM motion term (which is trivial, just a particle of mass $M$) and a relative position term which is equivalent to a particle of mass $\\mu$ subject to potential $U(r)$. We have essentially converted a two particle problem into what is effectively a one particle problem. This makes the two-body problem analytically (somewhat) easy to solve with this method.\n\\end{s}\n\n\\begin{p}\nWrite the expressions for $\\v{r}_1$ and $\\v{r}_2$ in terms of appropriate generalized coordinates, for the double pendulum. \n\\begin{center}\n \\includegraphics[scale=0.6]{Lecture-3/W3-img2.png}\n\\end{center}\n\\end{p}\n\\begin{s}\nTrigonometry to get the position vectors:\n\\[\\v{r}_1 = l_1\\sin\\phi_1\\xhat + l_1\\cos\\phi_1\\yhat\\]\n\\[\\v{r}_2 = \\left(l_1\\sin\\phi_1 + l_2\\sin\\phi_2\\right)\\xhat + \\left(l_1\\cos\\phi_1 + l_2\\cos\\phi_2\\right)\\yhat\\]\nThis problem has four degrees of freedom (motion in the plane) but two constraints that $\\abs{\\v{r}_1} = l_1$ and $\\abs{\\v{r}_2 - \\v{r}_1} = l_2$. This leaves to two generalized coordinates. Note that the constraints $f(\\v{r}_1, \\cdots, \\v{r}_n, t) = 0$ are called \"holonomic\" and are in general nice for solving problems.\n\\end{s}\n\\end{document}", "meta": {"hexsha": "518d92aeb7b8a2a34a291a3613b01d92e1eef66b", "size": 4469, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture-3/Worksheet-3.tex", "max_stars_repo_name": "RioWeil/PHYS306-notes", "max_stars_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Lecture-3/Worksheet-3.tex", "max_issues_repo_name": "RioWeil/PHYS306-notes", "max_issues_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture-3/Worksheet-3.tex", "max_forks_repo_name": "RioWeil/PHYS306-notes", "max_forks_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 78.4035087719, "max_line_length": 433, "alphanum_fraction": 0.6791228463, "num_tokens": 1616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.8006919925839875, "lm_q1q2_score": 0.6370770695551566}} {"text": "\\documentclass{homework}\n\\course{Math 5522H}\n\\author{Jim Fowler}\n\\input{preamble}\n\n\\begin{document}\n\\maketitle\n\n\\begin{inspiration}\nSome of the most important results (e.g. Cauchy's theorem) are so surprising at first sight that nothing short of a proof can make them credible.\n\\byline{Sir Harold Jeffreys} % and where is this quotation from?\n\\end{inspiration}\n\n\\section{Terminology}\n\n\\begin{problem}\n What is a \\textbf{Jordan curve}?\n\\end{problem}\n\n\\begin{problem}\n Define the \\textbf{winding number} $n(\\gamma,z)$ of a piecewise\n smooth closed curve $\\gamma$ around a point $z \\in \\C$.\n\\end{problem}\n\n\\section{Numericals}\n\n\\begin{problem} Compute the \\textbf{Fresnel integrals}\n \\[\n \\int_0^\\infty \\sin \\left( x^2 \\right) \\, dx \\mbox{ and } \\int_0^\\infty \\cos \\left( x^2 \\right) \\, dx.\n \\]\n\\end{problem}\n\n\\begin{problem}\n For an integrable function $f : \\R \\to \\C$, define the\n \\textbf{Fourier transform} of $f$, denoted $\\hat{f}$, by\n \\[\n {\\hat {f}}(\\xi ) := \\int _{-\\infty }^{\\infty} f(x) \\, e^{-2\\pi ix \\xi} \\,dx.\n \\]\n Find the Fourier transform of $f(x) = e^{-\\pi x^2}$.\n\\end{problem}\n\n\\section{Exploration}\n\n\\begin{problem}\n Does the function $f : \\C \\to \\C$ given by $f(z) = \\conj{z}$ have a\n primitive?\n\\end{problem}\n\n\\begin{problem}\n We proved Goursat's theorem for rectangles. Without simply\n recapitalulating the proof (e.g., your argument should not again\n invoke \\ref{nested-subsets-convergence}), deduce a theorem for\n \\textit{triangles} from our result about rectangles.\n\\end{problem}\n\n\\begin{problem}\n Suppose $f, g : \\C \\to \\C$ are holomorphic and agree on the unit\n circle. How do $f$ and $g$ relate? Morally, this problem is\n related to \\ref{identity-theorem}.\n\\end{problem}\n\n \\begin{problem}\n Suppose $f : \\C \\to \\C$ is holomorphic, and consider the circle\n $\\gamma$ with center $z_0$. How does the real part of the average\n value of $f$ on the circle $\\gamma$ relate to $f(z_0)$?\n \\end{problem}\n\n \\begin{problem}\n Recalling \\ref{harmonic-conjugate}, suppose $u, v : \\C \\to \\R$ are\n harmonic functions and $f = u + iv$ is holomorphic, so $u$ and $v$\n are harmonic conjugates. Find the \\textbf{Poisson kernel} for the\n unit disc, i.e., find a function $P_r(\\theta)$ so that\n \\[\n u(re^{i\\theta}) = \\frac {1}{2\\pi} \\int_{-\\pi }^{\\pi } P_{r}(\\theta -t) \\, u(e^{it}) \\, dt\n \\]\n for $r < 1$.\n \\end{problem}\n\n\\section{Prove or Disprove and Salvage if Possible}\n\n\\begin{problem}\n Suppose $D = \\{ z \\in \\C : \\abs{z} < 1 \\}$ and $f : D \\to \\C$ is\n continuous and $\\gamma : [a,b] \\to D$ is a piecewise smooth curve.\n Then $\\displaystyle\\int_\\gamma f\\, dz = 0$.\n\\end{problem}\n \n\\begin{problem}\n Suppose $U$ is an open set and $f : U \\to \\C$ is analytic and\n $\\gamma : [a,b] \\to U$ is a piecewise smooth curve. Then\n $\\displaystyle\\int_\\gamma f \\, dz = 0$.\n\\end{problem}\n\n\\begin{problem}\n Define $\\gamma_r : [0,2\\pi] \\to \\C$ by\n $\\gamma_r(\\theta) = r e^{i \\theta}$, and for real numbers\n $R > r > 0$, define the annulus\n \\[\n A(r,R) := \\{ z \\in \\C : r < |z| < R \\}\n \\]\n and suppose $f : A(r/2,2R) \\to \\C$ is holomorphic. Then\n \\[\n \\int_{\\gamma_r} f \\, dz = \\int_{\\gamma_R} f \\, dz.\n \\]\n\\end{problem}\n\n\\begin{problem}\\label{cauchy-integral-formula}Define $B_R(0) := \\{ z \\in C : \\abs{z} < R \\}$ and suppose\n $f : B_R(0) \\to \\C$ is holomorphic and let $\\gamma$ be the\n positively-oriented circle of positive radius $r < R$. Then\n \\[\n f(a) = \\int_\\gamma \\frac{f(z)}{z-a} \\, dz.\n \\]\n and moreover\n \\[\n f'(a) = \\int_\\gamma \\frac{f(z)}{(z-a)^2} \\, dz.\n \\] \n\\end{problem}\n\n\\begin{problem}\\label{cauchy-inequalities}If $f : U \\to \\C$ is\n holomorphic and $U \\supset B_r(z_0)$, then\n \\[\n \\abs{f(z_0)} \\leq \\sup_{z \\in \\partial B_r(z_0)} \\abs{f(z)}\n \\]\n and\n \\[\n \\abs{f'(z_0)} \\leq \\sup_{z \\in \\partial B_r(z_0)} \\abs{f(z)}. % missing (1/r) factor\n \\]\n \\end{problem}\n\n\\begin{problem}\\label{liouville-theorem}Recall that a function $f : U \\to \\C$ is\n \\textbf{bounded} if there exists $M > 0$ so that for all $z \\in U$\n we have $\\abs{f(z)} \\leq M$. A bounded holomorphic function\n $f : U \\to \\C$ is constant.\n\\end{problem}\n\n\\begin{problem}\n For every polynomial $p \\in \\C[z]$ there is $z \\in \\C$ so that $p(z) = 0$.\n\\end{problem}\n\n\\end{document}\n", "meta": {"hexsha": "62944bfa6c136e77371251cd3214a8bd5a1e83b8", "size": 4254, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problem-sets/set04.tex", "max_stars_repo_name": "Alex7Li/math5522h", "max_stars_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "problem-sets/set04.tex", "max_issues_repo_name": "Alex7Li/math5522h", "max_issues_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problem-sets/set04.tex", "max_forks_repo_name": "Alex7Li/math5522h", "max_forks_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.3857142857, "max_line_length": 145, "alphanum_fraction": 0.6290550071, "num_tokens": 1558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149978955811, "lm_q2_score": 0.8438951005915208, "lm_q1q2_score": 0.6370690680871391}} {"text": "\\section{Working Principle}\n\nFeature Kernel can be implemented in various ways, here we'll describe our implementation. \n\n\\subsection{Usage Ratio} \\label{sec:usage-ratio}\nOne of the first decisions that need to be made is size of the sub model.\nFirst of all, we can define the \\emph{Usage Ratio} (ur) as \n\\begin{align}\n ur := \\frac{\\text{Sub model size}}{\\text{Full model size}} \\label{eq:usage-ratio}\n\\end{align}\n\nOf course a fixed $ur$, very easy to program, is not a good idea. \nIt is completely arbitrary and it may be to much in some case and not enough in other situations. \nA good choice is to having an automatically adjusting $ur$. In the current implementation $ur$ is initially set to the fixed value\\footnote{This value is hard-coded} of $0.01$, so initially\na sub model large one hundredth of the whole model. This value has been chosen because it is small enough, so it is very unlikely that a sub problem so small\nis already feasible, but also it is not too small because it can lead to a slow convergence. \n\n\n$ur$ value is automatically adjusted according to the status of the current sub problem:\n\n\n\\subsubsection*{Sub problem linearly infeasible}\\label{subsec:lin-infeasible-action}\n\n In this case the random sub model is so small that is very unlikely that another random sub model\n with the same size can be feasible. So to find a feasible sub model (at least in the continuous) \n it is necessary to increase $ur$ significantly. This is the function used to update $ur$ in this case. \n \\begin{align}\n &ur_{i+i} = \\sqrt[m]{u_{i}^{n}}\\label{eq:ur-infease-grow} \\\\\n With:&\\nonumber \\\\\n &m > n > 1 \\nonumber\n \\end{align}\n\n This function, or better class of functions, has been chosen for three reasons:\n \\begin{itemize}\n \\item It has a fixed point in 1, so is not possible that such a series produces an $ur$ larger then 1.\n \\item It is pretty fast as the beginning and than it becomes slower as $ur$ gets closer to 1.\n \\item It allows to have a large number of points during the growth.\n \\end{itemize}\n\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\textwidth,keepaspectratio]{plot_initial_increase}\n \\caption{Infeasible $ur$ Growth example}\\label{fig:inf-grow-plot}\n \\end{figure}\n \n In Figure \\ref{fig:inf-grow-plot} are shown four different configuration. This possible to notice that those function\n grows quickly at the beginning, approximately from $ur = 0.01$ to $ur = 0.6 / 0.8$ then it starts settling\n to $ur = 0.95 / 1$.\n\n \n Because the porpoise is to grow $ur$ until the generate model is large enough to be feasible, but not so large \n that the sub model is practically as complex as the whole model. Also we'd like to have $ur$ to grow fast, so the number\n of infeasible models is as limited as possible, but not too fast otherwise we'll get sub model that are too large for their\n purpose. Last but not least we'd like to have an large number of points so the number of iteration to get to one should be high \n but not excessive.\n For those reason the current Feature Kernel implementation uses $n = 4$ and $m = 5$ to define the growth function.\n\n \n\n\n\\subsubsection*{Sub problem is linearly feasible}\n \\begin{align}\n &u_{i+i} = 1.1 u_{i} \\label{eq:ur-lin-fease-grow}\n \\end{align}\n\n Once the model is proven to be feasible in the continuous, Feature Kernel tries to make it also feasible in the \n integer. This can be done by simply trying over and over, and by chance it is possible that some of those sub models\n become integer feasible. Feature Kernel does not rely (too much) on chance and to increase its possibility to get \n a feasible sub model it increase $ur$ by 10\\%. This value has been chosen hoping that when the solution is feasible \n in the continuous it is also close to be feasible in the integer. So this 10\\% increase should be done just a few times\n before getting a feasible integer problem. If that holds true $ur$ should never become too close to 1. \n It also possible that some models requires all the variables to become integer feasible. In this case this 10\\% increase \n can lead to a $ur > 1$, if this happens Feature Kernel simply set $ur = 1$.\n\n\\subsubsection*{Sub problem is integer feasible}\n \\begin{align}\n &u_{i+i} = 0.9 u_{i} \\label{eq:ur-fease-grow}\n \\end{align}\n\n\n If an integer feasible sub model is found means that the current $ur$ is able to provide some other integer feasible solutions. \n The Feature Kernel goal is to find a ranking based on a machine learning technique, so to find a valuable answer Feature Kernel needs\n some sub models that are integer feasible and some other that are feasible only in the continuous. This is why when an integer feasible\n sub model is found Feature Kernel decreases $ur$ by 10\\%. This value has been chosen to mirror the value from Eq \\ref{eq:ur-lin-fease-grow}.\n Of course a 10\\% decrease can lead to a very small $ur$ in few iterations, but consider that once the problem turn again into an integer infeasible \n size the growth function turns into the method described in \\Cref{subsec:lin-infeasible-action}.\n\n\nThose three growth function will try to keep $ur$ switching between integer feasibility, continuous feasibility and infeasibility. This should provide \na meaningful dataset, rich of different statuses sub problem, to fed into the machine learning algorithm. \n\n\n\\subsubsection{Feasibility Threshold}\n\nNow it is possible to define those two indexes:\n\\begin{itemize}\n \\item Continuous Feasibility Threshold (CFT): is the average $ur$ of sub problems that are feasible in the continuous but not in the integer.\n \\item Integer Feasibility Threshold (IFT): is the average $ur$ of sub problems that are integer feasible.\n\\end{itemize}\n\n\nThose values allow the user to get an idea of some properties of the specific instance. They allow to understand if the problem is suitable for Kernel Search or not. \nFor example is it happens that IFT is very close to 1 it is very unlikely that an incremental solution, like the one built by Kernel Search, can provide a good solution in\na shot period of time. On the other side if IFT (and/or CFT) are reasonably low it is possible for this problem to be solved using Kernel Search, in fact in this situation\nan incremental solution is a good idea to solve the problem, potentially to a value vary close to the optimal. \n\n\n\\subsection{Sub Problem Construction}\n\nThe algorithm to build the random sub model is presented in the following pseudocode\n\\input{algorithms/generate_random_sub_model.tex}\n\n\n\n\\subsection{Sub Problem Solution}\nEach sub problem needs to be solved in order to understand if it is integer feasible, continuous feasible or infeasible. \nThis is done in two steps: first prove continuous feasibility, second prove integer feasibility. \n\n\\subsubsection{Continuous Feasibility}\nIn this first step the sub model is relaxed (integer constraints are removed) and then it is solved to the optimal. The optimal solution is needed for a \npractical reason: Random Forrest requires for each variable of each instance to have a value. This value cannot be a random one or simply one or zero. It\nmust be a significant value, so we'll use the value of the variable in the linear solution. Out of base variables are not set to any special value, they are \nsimply kept to zero. Base variables of course are set to their base value. \n\nIf the sub problem happens to be infeasible in the continuous domain then this sub problem is considered as meaningless and it will not be uses in the training of the Random Forrest\n\n\n\\subsubsection{Integer Feasibility}\nIn the second step the model is solved as it is. In this scenario is of course possible to solve the problem at the optimal but is pointless: \n\\begin{itemize}\n \\item it's slow\n \\item there is already a value for the variables\n \\item we are looking for feasibility not for a solution\n\\end{itemize}\nSo in our implementation the solver is set to search for just one solution, so we are able to prove if the problem is feasible or not: if a solution is found then the model is feasible, otherwise\nis infeasible.\n\n\n\\subsubsection{Time Limit}\\label{sec:timeout}\nIn some cases even a sub problem solved once may take to long. This issue can be unacceptable within a method like Feature Kernel where the number of sub problem to solve can be very high: it \nis pointless to use a method for a faster initial kernel generation if it take more then the standard method. To avoid this issue is important to properly set a time limit. Feature Kernel\nsupports two time limits:\n\\begin{itemize}\n \\item Minimum Time: this is the time limit used at the beginning of method and is increased each time a sub model end due to a timeout\n \\item Maximum Time: the upper time limit, when this value is reached the actual time limit value won't increase even if the model ends due to a timeout\n\\end{itemize}\n\n\n\\subsubsection{Possible Sub Model Results}\\label{sec:submodel-result}\nConsidering the various steps and their configuration a sub problem can end with four different status:\n\\begin{enumerate}\n \\item Infeasible: the model is infeasible in the continuous domain\n \\item Linear Feasible: the model is feasible in the continuous but not in the integer\n \\item Integer Feasible: the model is feasible in the integer\n \\item Timeout: the solver reach the configured time limit, nothing could be said about the feasibility. \n\\end{enumerate}\n\n\\input{algorithms/solve_sub_model.tex}\n\n\\subsection{Kernel Construction}\n\nThis is the final step and is made out of two components: variable sorting and variable split. \nVariable sorting is the step involving the Random Forrest, where the solved models are joined into a dataset and then used to train the Model. The second component, namely variable split,\ntakes care for the split of variables between the initial kernel and the bucket variables.\n\n\\subsubsection{Variable Sort}\nWhen all the sub problem have been solved those solutions must be joined into a dataset. In our implementation Feature Kernel build the dataset only from instances \nwith final status of integer or continuous feasibility. Of course infeasible instances cannot be used in the dataset: variables does not have a value. For the same reason\nalso sub problems ended in timeout are to be considered insignificant, also them haven't a value.\n\nIn order to train the Random Forrest is important to modify the collected solution accordingly to the Random Forrest requirements. Once built the dataset it is possible to \nstart training the Random Forrest. In general it is an extremely fast process. We won't show a general pseudocode about this \nstep because it is very reach of implementation details. \nThe only thing that meters is that now we have a variable ranking.\n\n\n\\subsubsection{Variable Split}\nOnce we've got a variable ranking, Feature Kernel sorts the variables (from the highest ranking one). The top variables will be uses as kernel variables, the bottom ones will be\nthe buckets. Now Feature Kernel need to now the point to split kernel and variables. In our implementation this point is computed automatically according to the $ur$. \nAt the current time it is possible to use four different split points:\n\\begin{itemize}\n \\item Minimum Continuous Feasibility: the size of the smallest sub model that happens to be continuous feasible\n \\item Maximum Continuous Feasibility: the size of the largest sub model that happens to be continuous feasible\n \\item Minimum Integer Feasibility: the size of the smallest sub model that happens to be integer feasible\n \\item Maximum Integer Feasibility: the size of the largest sub model that happens to be integer feasible\n\\end{itemize}\n\n\nOnce the variables are split Feature Kernel can build\\footnote{Depending on the actual Kernel Search implementation} the initial Kernel and the buckets. From now on Kernel Search can \nproceed as usual. \n\n\n\\subsubsection{Feature Kernel Pseudocode}\nAt this point, after all this theoretical description, we'll provide a pseudocode that should provide a guide for those who wants to implement Feature Kernel them self.\n\n\\input{algorithms/feature_kernel.tex}\n\n\\input{algorithms/build_solutions.tex}\n\n", "meta": {"hexsha": "df24dd71d56126c52592e94353cd33eab44f80e8", "size": 12344, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "working_principle.tex", "max_stars_repo_name": "Optimization-Algorithms/Feature-Kernel", "max_stars_repo_head_hexsha": "157c3775c5ca239000557d18cf3d6bbb3b350e33", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "working_principle.tex", "max_issues_repo_name": "Optimization-Algorithms/Feature-Kernel", "max_issues_repo_head_hexsha": "157c3775c5ca239000557d18cf3d6bbb3b350e33", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "working_principle.tex", "max_forks_repo_name": "Optimization-Algorithms/Feature-Kernel", "max_forks_repo_head_hexsha": "157c3775c5ca239000557d18cf3d6bbb3b350e33", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.4129353234, "max_line_length": 195, "alphanum_fraction": 0.7669313027, "num_tokens": 2813, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438950986284991, "lm_q2_score": 0.754914997895581, "lm_q1q2_score": 0.6370690666052246}} {"text": "\\chapter{Spinless \\Pelectron + \\Pmuon scattering}\nFor the following calculations we will consider the elastic electromagnetic scattering of scalar electrons and muons. That is, the particles considered are solutions of the Klein-Gordon equation.\n\nThe process has one diagram:\n\\begin{figure}[h]\n\\centering\n\\include{figures/emuscatter}\n\\caption{Diagram for the t-channel process \\HepProcess{\\Pelectron+\\Pmuon\\to\\Pelectron+\\Pmuon} with scalar particles.\\label{fig:emuscatter}}\n\\end{figure}\n\n\\section{Electrodynamics of scalars}\\label{sec:EMdynamics}\nThe equations of motion for a particle may be modified to include the electromagnetic interaction via a 4-potential $A$ via the substitution\n\\begin{equation}\np^\\mu \\rightarrow p^\\mu + eA^\\mu.\n\\end{equation}\nThen the Klein-Gordon equation, \\eqref{eq:KleinGordon}, becomes\n\\begin{equation}\n\\left( i\\partial_\\mu + eA_\\mu \\right)\\left( i\\partial^\\mu + eA^\\mu \\right)\\psi = m^2\\psi.\n\\end{equation}\nExpanding out the left-hand side and collecting the free-particle kinetic and mass terms, we can identify an expression for the electromagnetic potential, $V$,\n\\begin{equation}\n\\left( \\partial_\\mu\\partial^\\mu + m^2 \\right)\\psi = \\left( ie\\partial_\\mu A^\\mu + ieA_\\mu\\partial^\\mu + e^2A_\\mu A^\\mu \\right)\\psi \\equiv -V(x) \\,\\psi.\n\\end{equation}\nUnder the weak field approximation, we may ignore the second-order term $A_\\mu A^\\mu$, so that\n\\begin{equation}\nV(x) \\approx -ie\\left( \\partial_\\mu A^\\mu + A_\\mu\\partial^\\mu \\right)\n\\end{equation}\n\n\\section{Scattering amplitude}\nThe amplitude for a particle with wavefunction $\\phi_i$ to scatter from potential $V$ and subsequently have the wavefunction $\\phi_f$ is given by\n\\begin{align}\nT_{fi} &= -i \\int \\dd[4]{x} \\, \\phi_f^* \\, V \\, \\phi_i \\\\\n&= -i \\int \\dd[4]{x} \\, \\phi_f^* \\, (-ie) \\left( \\partial_\\mu A^\\mu + A_\\mu\\partial^\\mu \\right) \\, \\phi_i\n\\end{align}\nwhere the weak field form of the EM potential has been substituted for $V$. Integrating by parts, the first term in the integral evaluates to\n\\begin{equation}\n\\int \\dd[4]{x} \\, \\phi_f^* \\, \\partial_\\mu A^\\mu \\, \\phi_i = \\left[ \\phi_f^* \\, A^\\mu \\, \\phi_i \\right]_{-\\infty}^{+\\infty} - \\int \\dd[4]{x} \\, \\left( \\partial_\\mu \\phi_f^* \\right) A^\\mu \\, \\phi_i\n\\end{equation}\nand the boundary term in square brackets goes to zero for a vanishing (i.e.~local) potential. Therefore, the scattering amplitude becomes\n\\begin{align}\nT_{fi} &= -i \\int \\dd[4]{x} \\,(-ie)\\left( \\phi_f^* A_\\mu \\partial^\\mu \\phi_i - (\\partial_\\mu \\phi_f^*) A^\\mu \\phi_i \\right) \\nonumber \\\\\n&= -i \\int \\dd[4]{x} \\, A^\\mu \\,(-ie)\\left( \\phi_f^* \\partial_\\mu \\phi_i - (\\partial_\\mu \\phi_f^*) \\phi_i \\right) \\nonumber \\\\\n&= -i \\int \\dd[4]{x} \\, A^\\mu \\, j_\\mu^{fi} \\label{eq:scatterAmp}\n\\end{align}\nwhere we have identified the current density,\n\\begin{equation}\nj_\\mu^{fi} = -ie \\left( \\phi_f^* \\partial_\\mu \\phi_i - (\\partial_\\mu \\phi_f^*) \\phi_i \\right).\n\\end{equation}\nThis is the current-potential formulation, where we identify the scattering process with a current $j_\\mu^{fi}$ and potential $A^\\mu$.\n\nAt the vertex changing A to C, the scalar electron has plane wavefunctions,\n\\begin{equation}\n\\phi_A(x) = N_A \\, e^{-iP_Ax}, \\quad \\phi_C(x) = N_C \\, e^{-iP_Cx},\n\\end{equation}\ngiving the current,\n\\begin{equation}\nj_\\mu^{CA} = -e \\, N_C^* \\, N_A \\, e^{i(P_C-P_A)x} \\, (P_A + P_C)_\\mu. \\label{eq:current}\n\\end{equation}\nThis will be the current in the current-potential formulation. Similarly, the current flowing through the other vertex is given by\n\\begin{equation}\nj_\\mu^{DB} = -e \\, N_D^* \\, N_B \\, e^{i(P_D-P_B)x} \\, (P_B + P_D)_\\mu.\n\\end{equation}\n\nWe wish to identify the potential $A^\\mu$ associated with the current $j_\\mu^{DB}$. For the potential to describe fields that behave according to Maxwell's equations, it must itself obey the Poisson equation (see Section \\ref{sec:maxwell} below),\n\\begin{equation}\n\\partial^2 A^\\mu = j^\\mu.\n\\end{equation}\nIt can be verified that the potential\n\\begin{equation}\nA_{DB}^\\mu = \\frac{- g^{\\mu\\nu} j^{DB}_\\nu}{q^2} \\label{eq:potential}\n\\end{equation}\nis a solution via substitution,\n\\begin{align}\n\\partial^2 \\left( \\frac{-g^{\\mu\\nu} j^{DB}_\\nu}{q^2} \\right) &= -\\frac{1}{q^2} \\left[ i(P_D - P_B) \\right]^2 \\, j_{DB}^\\mu \\nonumber \\\\\n&= j_{DB}^\\mu\n\\end{align}\nwhere $q^2 = (P_D - P_B)^2 = t$ is the 4-momentum of the propagating photon.\n\nInserting the above forms for the current, \\eqref{eq:current}, and potential, \\eqref{eq:potential}, into the expression for the scattering amplitude, \\eqref{eq:scatterAmp},\n\\begin{equation}\nT_{fi} = \\frac{ie^2}{q^2} \\int \\dd[4]{x} \\, N_C^* \\, N_A \\, N_D^* \\, N_B \\, (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\, e^{i(P_C+P_D-P_A-P_B)x}\n\\end{equation}\n\n\\section{Scattering rate}\nTo calculate the cross-section, we would like to know the rate $W_{fi}$ of particles being scattered per unit time and volume,\n\\begin{equation}\\boxed{\nW_{fi} = \\frac{T_{fi}^* T_{fi}}{T \\times V}\n}\\end{equation}\n\n\nAssume the scattering takes place in a volume $V$. We use the covariant normalisation to $2E$ particles in the volume, giving\n\\begin{equation}\nN_i = \\frac{1}{\\sqrt{V}} \\quad \\text{for $i=A,B,C,D$.}\n\\end{equation}\nThen the amplitude becomes\n\\begin{equation}\nT_{fi} = \\frac{ie^2}{q^2 \\, V^2} \\int \\dd[4]{x} \\, (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\, e^{i(P_C+P_D-P_A-P_B)x}\n\\end{equation}\ngiving a probability\n\\begin{align}\nT_{fi}^* T_{fi} &= \\frac{e^4}{q^4 \\, V^4} \\int_{TV} \\dd[4]x \\int_{TV} \\dd[4]x^\\prime \\, \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2 \\, e^{i(P_C+P_D-P_A-P_B)(x-x^\\prime)} \\\\\n&= \\frac{e^4}{q^4 \\, V^4} \\left(\\sqrt{2\\pi}\\right)^4 \\, \\sqrt{TV} \\int_{TV} \\dd[4]{x} \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2 \\, e^{i(P_C+P_D-P_A-P_B)x} \\nonumber \\\\\n&= \\frac{e^4}{q^4 \\, V^4} \\left(\\sqrt{2\\pi}\\right)^8 \\, \\left(\\sqrt{TV}\\right)^2 \\, \\delta^{(4)}(P_C+P_D-P_A-P_B) \\, \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2 \\nonumber \\\\\n&= \\frac{e^4}{q^4 \\, V^4} \\, (2\\pi)^4 \\, \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2\\, \\delta^{(4)}(P_C+P_D-P_A-P_B)\\, TV\n\\end{align}\nwhich gives the rate\n\\begin{equation}\nW_{fi} = \\frac{(2\\pi)^4\\,e^4}{q^4 \\, V^4} \\, \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2\\, \\delta^{(4)}(P_C+P_D-P_A-P_B).\n\\end{equation}\n\n\\section{Density of states}\nThe larger the number of final states per unit volume $N_f$ available to a process, the more likely it is to happen and hence the cross-section is proportional to $N_f$.\n\nIn the scattering problem the number of final states is the number of possible momenta inside $[\\vec{p}, \\vec{p}+\\dd[3]{\\vec{p}}]$ for the propagating photon with a given energy. This is given by\n\\begin{equation}\nN_f = \\frac{1}{2E} \\frac{\\dd[3]{\\vec{p}}}{\\Delta p}\n\\end{equation}\nwhere $\\Delta p$ is the separation in momentum space between adjacent states and the $1/2E$ factor is due to the fact that there are $2E$ particles inside the volume and we want the number of states for one particle.\n\nSince the scattering process is modelled as happening inside a volume $V$, the propagating photon has the wavefunction of a particle in an infinite potential well with the dimensions of $V$ (Dirichlet boundary conditions, $\\psi(0) = \\psi(L) = 0$). This leads to the quantisation of its allowed momentum,\n\\begin{align}\np_x &= \\frac{2\\pi}{L_x}\\, n_x, \\quad\np_y = \\frac{2\\pi}{L_y}\\, n_y, \\quad\np_z = \\frac{2\\pi}{L_z}\\, n_z.\n\\end{align}\nTherefore, the separation between momentum states is $2\\pi/L$ in each dimension. This leads to the density of states,\n\\begin{align}\nN_f &= \\frac{1}{2E} \\frac{L_x}{2\\pi} \\frac{L_x}{2\\pi} \\frac{L_x}{2\\pi} \\dd[3]{\\vec{p}} \\nonumber \\\\\n&= \\frac{1}{2E} \\frac{V}{(2\\pi)^3} \\dd[3]{\\vec{p}}.\n\\end{align}\n\nNow the photon's energy is shared amongst particles $C$ and $D$, so the density of final states becomes\n\\begin{align}\nN_f^{CD} &= \\frac{1}{2E_C} \\frac{V}{(2\\pi)^3} \\dd[3]{\\vec{p}_C} \\, \\frac{1}{2E_D} \\frac{V}{(2\\pi)^3} \\dd[3]{\\vec{p}_D} \\\\\n&= \\frac{1}{4E_CE_D}\\frac{V^2}{(2\\pi)^6} \\dd[3]{\\vec{p}_C} \\, \\dd[3]{\\vec{p}_D}\n\\end{align}\n\n\\section{Incoming flux}\nThe rate $W_{fi}$ scales with the flux of incoming particles, so the in cross-section calculation, it should be divided by a flux factor, $\\rho_A\\rho_Bu_{AB}$, where $\\rho_{AB}$ give the density of initial-state particles and $u_{AB} = u_A - u_B$ is their relative velocity.\n\\begin{align}\n\\rho_A\\rho_Bu_{AB} &= \\frac{2E_A}{V} \\frac{2E_B}{V} \\left( \\frac{p_A}{E_A} - \\frac{p_B}{E_B} \\right) \\\\\n&= \\frac{4}{V^2} \\left( p_A E_B - p_B E_A \\right).\n\\end{align}\nFor a collider in the centre of momentum (CM) frame, we have $p_A = -p_B$,\n\\begin{equation}\n\\rho_A\\rho_Bu_{AB} = \\frac{4p_A}{V^2} (E_B + E_A) = \\frac{4p_A\\sqrt{s}}{V^2}\n\\end{equation}\nwhere $\\sqrt{s} = E_A + E_B$ is the CM energy.\n\n\\section{Cross-section}\nNow we are ready to calculate the differential cross-section,\n\\begin{align}\n\\dd{\\sigma} &= \\frac{W_{fi}}{\\rho_A\\rho_Bu_{AB}} N_f \\\\\n&= \\left\\{ \\frac{(2\\pi)^4\\,e^4}{q^4 \\, V^4} \\, \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2\\, \\delta^{(4)}(P_C+P_D-P_A-P_B) \\right\\} \\frac{V^2}{4p_A\\sqrt{s}} \\frac{1}{4E_CE_D}\\frac{V^2}{(2\\pi)^6} \\dd[3]{\\vec{p}_C} \\, \\dd[3]{\\vec{p}_D} \\nonumber \\\\\n&= \\frac{e^4}{q^4V^2} \\frac{\\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2}{4p_A\\sqrt{s}} \\, \\dd{Q}\n\\end{align}\nwhere $\\dd{Q}$ is the Lorentz-invariant phase space factor\n\\begin{equation}\n\\dd{Q} = \\frac{V}{2E_C} \\frac{\\dd[3]{\\vec{p}_C}}{(2\\pi)^3} \\frac{V}{2E_D} \\frac{\\dd[3]{\\vec{p}_D}}{(2\\pi)^3} (2\\pi)^4 \\, \\delta^{(4)}(P_C+P_D-P_A-P_B)\n\\end{equation}\nIn the CM frame, we have that $\\vec{p}_A = -\\vec{p}_B$, so\n\\begin{equation}\n\\dd{Q} = \\frac{V^2}{(2\\pi)^2} \\, \\delta(\\sqrt{s} - E_C - E_D) \\, \\delta^{(3)}(\\vec{p_C} + \\vec{p_D}) \\, \\frac{\\dd[3]{\\vec{p}_C}}{2E_C} \\frac{\\dd[3]{\\vec{p}_D}}{2E_D}.\n\\end{equation}\nWe now integrate over all momenta for $D$, the unobserved particle. This gives,\n\\begin{align}\n\\dd{Q} &= \\frac{V^2}{(2\\pi)^2} \\frac{1}{4E_C E_D} \\, \\delta(\\sqrt{s} - E_C - E_D) \\, \\dd[3]{\\vec{p}_C} \\\\\n&= \\frac{V^2}{(2\\pi)^2} \\frac{1}{4E_C E_D} \\, \\delta(\\sqrt{s} - E_C - E_D) \\, p_C^2 \\, \\dd{p_C} \\, \\dd{\\Omega}\n\\end{align}\n\nNow from the relation $E^2 = p^2 + m^2$ we have that $E \\dd{E} = p\\dd{p}$, so\n\\begin{equation}\n\\dd{Q} = \\frac{V^2}{(2\\pi)^2} \\frac{p_C}{4 E_D} \\, \\delta(\\sqrt{s} - E_C - E_D) \\, \\dd{E_C} \\, \\dd{\\Omega}\n\\end{equation}\n\nWe next want to integrate over $E_C$. To do so, write the expression in the $\\delta$ function as\n\\begin{align}\nf(E_C) &= \\sqrt{s} - E_C - E_D \\\\\n&= \\sqrt{s} - E_C - \\sqrt{p_D^2 + m_D^2} \\nonumber \\\\\n&= \\sqrt{s} - E_C - \\sqrt{p_C^2 + m_D^2} \\quad \\text{since $\\vec{p}_C = -\\vec{p}_D$} \\nonumber \\\\\n&= \\sqrt{s} - E_C - \\sqrt{E_C^2 - m_C^2 + m_D^2}.\n\\end{align}\nTherefore,\n\\begin{align}\n\\abs{\\dv{f}{E_C}} &= 1 + \\frac{E_C}{\\sqrt{E_C^2 - m_C^2 + m_D^2}} \\nonumber \\\\\n&= \\frac{\\sqrt{s}}{E_D}\n\\end{align}\nso the $\\delta$-function becomes\n\\begin{equation}\n\\delta(\\sqrt{s} - E_C - E_D) = \\delta(E_C) \\, \\frac{E_D}{\\sqrt{s}}\n\\end{equation}\nand the Lorentz-invariant phase space is\n\\begin{equation}\n\\dd{Q} = \\frac{V^2}{(2\\pi)^2} \\frac{p_C}{4\\sqrt{s}} \\, \\delta(E_C)\\, \\dd{E_C} \\, \\dd{\\Omega}.\n\\end{equation}\nIntegrating over the final-state particle's energy, $E_C$,\n\\begin{equation}\n\\dd{Q} = \\frac{V^2}{(2\\pi)^2} \\frac{p_C}{4\\sqrt{s}} \\, \\dd{\\Omega}.\n\\end{equation}\n\nSubstituting this back into the expression for the differential cross section,\n\\begin{equation}\\boxed{\n\\dv{\\sigma}{\\Omega} = \\frac{1}{64\\pi^2}\\frac{e^4}{q^4} \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2 \\frac{1}{s} \\, \\frac{p_C}{p_A}\n}.\\end{equation}\nThe fact that the cross-section is proportional to $1/s$, the inverse square of the SM energy means that we need increasingly higher luminosities to observe high energy scattering processes.\n\n\\section{Application: high-energy scattering}\nConsider the ultra-relativistic case, where $E \\gg m$ for all particles. In the CM frame, all particles now have the same energy $E = s/2$. Then we have that the Mandelstam $t$ variable is\n\\begin{align}\nq^2 = t &= (P_A - P_C)^2 \\\\\n&= \\mqty(E_A - E_C \\\\ \\vec{p}_A - \\vec{p}_C)^2 \\nonumber \\\\\n&= -2E^2 (1-\\cos\\theta)\n\\end{align}\n\nIn this case, the differential cross-section becomes\n\\begin{equation}\n\\dv{\\sigma}{\\Omega} = \\frac{e^4}{64\\pi^2} \\frac{\\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2}{4E^4 (1-\\cos\\theta)^2} \\frac{1}{s}.\n\\end{equation}\n\nNow we have, in the CM frame,\n\\begin{equation*}\nP_A = \\mqty(E \\\\ \\vec{p}) \\, \\quad P_B = \\mqty(E \\\\ -\\vec{p}) \\, \\quad P_C = \\mqty(E \\\\ \\vec{p^\\prime}) \\, \\quad P_D = \\mqty(E \\\\ -\\vec{p^\\prime}),\n\\end{equation*}\nand therefore,\n\\begin{align}\n(P_A + P_C)_\\mu (P_B + P_D)^\\mu &= \\mqty(2E \\\\ \\vec{p} + \\vec{p^\\prime}) \\mqty(2E \\\\ -\\vec{p} - \\vec{p^\\prime} ) \\quad \\text{with $\\abs{\\vec{p}} = \\abs{\\vec{p^\\prime}} = E$} \\nonumber \\\\\n&= (2E)^2 + (\\vec{p} + \\vec{p^\\prime})^2 \\nonumber \\\\\n&= 6E^2 + 2E^2 \\cos\\theta\n\\end{align}\n\\begin{equation}\n\\Rightarrow\\quad \\left[ (P_A + P_C)_\\mu (P_B + P_D)^\\mu \\right]^2 = 4E^4 (3 + \\cos\\theta)^2\n\\end{equation}\n\nSo the final answer is\n\\begin{equation}\n\\dv{\\sigma}{\\Omega} = \\frac{e^4}{64\\pi^2} \\left(\\frac{3+\\cos\\theta}{1-\\cos\\theta}\\right)^2 \\frac{1}{s}.\n\\end{equation}\n\n\\section{$1 \\to 2$ scattering (decay)}\nFor a single body of mass $M$ decaying into two particles, we replace the flux factor with the normalised energy density,\n\\begin{equation}\n\\rho = \\frac{2M}{V}\n\\end{equation}\nso the cross-section is given by\n\\begin{equation}\n\\dd{\\sigma} = \\frac{V}{2M} \\, W_{fi} \\, N_f.\n\\end{equation}\nwhere for a two-body final state, $N_f$ is the same as for above. However, $T_{fi}$ will obviously be different.\n", "meta": {"hexsha": "beed8ec9fafb5a6e7bb1e1390cda06e695f4b2eb", "size": 13434, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/8_Spinless_e_mu_scattering.tex", "max_stars_repo_name": "adambozson/Standard-Model-I", "max_stars_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/8_Spinless_e_mu_scattering.tex", "max_issues_repo_name": "adambozson/Standard-Model-I", "max_issues_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/8_Spinless_e_mu_scattering.tex", "max_forks_repo_name": "adambozson/Standard-Model-I", "max_forks_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.9518072289, "max_line_length": 303, "alphanum_fraction": 0.6540866458, "num_tokens": 5229, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569016, "lm_q2_score": 0.7931059462938815, "lm_q1q2_score": 0.6370019137980897}} {"text": "\\section{Multi-armed Bandits}\nReinforcement learning involves evaluative feedback rather than instructive feedback. We get told whether our actions are good ones or not, rather than what the single best action to take is. This is a key distinction between reinforcement learning and supervised learning.\n\n\\subsection{A $k$-armed Bandit Problem}\nIn the $k$-armed bandit problem there are $k$ possible actions, each of which yields a numerical reward drawn from a stationary probability distribution for that action. We want to maximise the expected total reward, taking an action at each \\emph{time step}. Some notation:\n\n\\begin{itemize}\n \\item Index timesteps by $t$\n \\item Action $A_t$\n \\item Corresponding reward $R_t$\n \\item \\emph{Value} of action $a$ is $q_*(a) = \\mathbb{E}[R_t | A_t = a]$\n \\item Estimate of value of action $a$ at $t$ is denoted $Q_t(a)$\n\\end{itemize}\n\nWe therefore want to choose $\\{a_1, \\dots, a_T\\}$ to maximise $\\sum_{t = 1}^T q_*(a_t)$.\\\\\n\\mbox{}\\\\ \nAt each timestep, the actions with the highest estimated reward are called the \\emph{greedy} actions. If we take this action, we say that we are \\emph{exploiting} our understanding of the values of actions. The other actions are known as \\emph{non-greedy} actions, sometimes we might want to take one of these to improve our estimate of their value. This is called \\emph{exploration}. The balance between exploration and exploitation is a key concept in reinforcement learning.\n\n\n\\subsection{Action-value Methods}\nWe may like to form estimates of the values of possible actions and then choose actions according to these estimates. Methods such as this are known as \\emph{action-value methods}. There are, of course, many ways of generating the estimates $Q_t(a)$. \\\\\n\\mbox{}\\\\\nAn $\\varepsilon$-greedy method is one in which with probability $\\varepsilon$ we take a random draw from all of the actions (choosing each action with equal probability), providing some exploration.\n\n\n\\setcounter{subsection}{4}\n\\subsection{Tracking a Non-stationary Problem}\nIf we decide to implement the sample average method, then at each iteration that we choose the given action we update our estimate by\n\\begin{equation}\n Q_{n+1} = Q_n + \\frac1n [R_n - Q_n]\n\\end{equation}\nNote that this has the (soon to be familiar) form\n\\begin{equation}\n \\mathrm{NewEstimate} \\gets \\mathrm{OldEstimate} + \\mathrm{StepSize}\\times[\\mathrm{Target} - \\mathrm{OldEstimate}].\n\\end{equation}\n\\mbox{}\\\\\nIf the problem was non-stationary, we might like to use a time weighted exponential average for our estimates (\\emph{exponential recency-weighted average}). This corresponds to a constant step-size $\\alpha \\in (0, 1]$ (you can check).\n\\begin{equation}\n Q_{n+1} = Q_n + \\alpha [R_n - Q_n].\n\\end{equation}\n\\mbox{}\\\\\nWe might like to vary the step-size parameter. Write $\\alpha_n(a)$ for the step-size after the $n^{\\mathsf{th}}$ reward from action $a$. Of course, not all choices of $\\alpha_n(a)$ will give convergent estimates of the values of $a$. To converge with probability 1 we must have\n\\begin{equation}\n \\sum_n \\alpha_n(a) = \\infty \\quad\\quad \\mathsf{and} \\quad\\quad \\sum_n \\alpha_n(a)^2 < \\infty.\n\\end{equation}\nMeaning that the coefficients must be large enough to recover from initial fluctuations, but not so large that they don't converge in the long run. Although these conditions are used in theoretical work, they are seldom used in empirical work or applications. (Most reinforcement learning problems have non-stationary rewards, in which case convergence is undesirable.)\n\n\\subsection{Optimistic Initial Values}\nThe exponential recency weighted method is biased by the initial value one gives. If we like, we may set initial value estimates artificially high to encourage exploration in the short run -- this is called \\emph{optimistic initial values}. This is a useful trick for stationary problems, but does not apply so well to non-stationary problems as the added exploration is only temporary.\n\n\n\\subsection{Upper-Confidence Bound Action Selection}\nWe might like to discriminate between potential explorative actions. Note that $\\varepsilon$-greedy does not do this. We define the \\emph{upper-confidence bound} action at $t$ as follows\n\\begin{equation}\n A_t \\doteq \\argmax_{a}\\left[ \\, Q_t(a)+ c \\sqrt{\\frac{\\mathrm{ln}(t)}{N_t(a)}} \\, \\right]\n\\end{equation}\nwhere $Q_t(a)$ is the value estimate for the action $a$ at time $t$, $c > 0$ is a parameter that controls the degree of exploration and $N_t(a)$ is the number of times that $a$ has been selected by time $t$. If $N_t(a) = 0$ then we consider $a$ a maximal action.\\\\\n\nThis approach favours actions with a higher estimated rewards but also favours actions with uncertain estimates (more precisely, actions that have been chosen few times).\n\n\n\\subsection{Gradient Bandit Algorithms}\nSuppose that we choose actions probabilistically based on a preference for each action, $H_t(a)$. Let the action at $t$ be denoted by $A_t$. We then define the probability of choosing action $a$ via the softmax\n\\begin{equation}\n \\pi_t(a) \\doteq \\P{}(A_t = a) = \\frac{e^{H_t(a)}}{\\sum_i e^{H_t(i)}}.\n\\end{equation}\nWe then iteratively perform updates according to \n\\begin{equation}\n H_{t+1}(a) = H_t(a) + (R_t - \\bar{R}_t)(\\mathds{1}_{A_t = a} - \\pi_t(a)),\n\\end{equation}\nwhere $\\bar{R}_t$ is the mean of previous rewards. The box in the notes shows that this is an instance of stochastic gradient ascent since the expected value of the update is equal to the update when doing gradient ascent on the (total) expected reward.", "meta": {"hexsha": "89b63a3199e6023edd6647a8ed514996c164d6cd", "size": 5563, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/chapters/chapter2/chapter2_content.tex", "max_stars_repo_name": "ElliotMunro200/reinforcement_learning_an_introduction", "max_stars_repo_head_hexsha": "c4fccb46a4bb00955549be3505144ec49f0132e5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 234, "max_stars_repo_stars_event_min_datetime": "2018-09-01T00:26:29.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T03:55:50.000Z", "max_issues_repo_path": "notes/chapters/chapter2/chapter2_content.tex", "max_issues_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_issues_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2018-11-29T21:04:36.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-13T17:11:50.000Z", "max_forks_repo_path": "notes/chapters/chapter2/chapter2_content.tex", "max_forks_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_forks_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 63, "max_forks_repo_forks_event_min_datetime": "2018-07-31T04:53:21.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T04:03:43.000Z", "avg_line_length": 78.3521126761, "max_line_length": 477, "alphanum_fraction": 0.7524716879, "num_tokens": 1469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569016, "lm_q2_score": 0.7931059438487663, "lm_q1q2_score": 0.6370019118342373}} {"text": "\\chapter{Audio Frequency Scaling methods}\\label{ch:scaling_methods}\n\\section{Introduction}\nThe human hearing system can detect \nacoustic vibrations and translate \nthose vibrations into sounds.\nThe detectable range of frequencies by the human ear\nis referred to as audio or sonic. This range\nspans over approximately \\(20kHz\\),\nstarting at \\(20Hz\\) to about \\(20kHz\\)\n\\cite{hearthres}.\n\nAs a result of aging, the hearing system's dynamic range \nas to the detectable bandwidth decreases,\nand by middle-age are set at about\n\\(20Hz \\div 14KHz\\)\\cite{Wiley2008ChangesIH}.\nAs such, the maximum hearable frequency\ndeclines with age.\n\nThe human ear's ability to distinguish \nbetween two different frequencies \nis not symmetric. \nFor example, the spectral distance between two \ndifferent frequencies in one distinct region does \nnot equal the spectral distance \nbetween two additional frequencies in other regions.\nDue to that asymmetry, the conventional linear \nspectral mapping is impractical for\nspeech analysis applications.\nThus, a different spectral mapping \nbased on a different scaling system\nthat mimics the human hearing as possible\nis applied instead as an alternative.\n\n\\section{Mel-Scaling}\nThe Mel-scaling method \nis a suggested solution to mapping \nstandard audio frequencies to perceived frequencies.\nThe basic idea that lies underneath it is that for\ndifferent pitches we assign varying bandwidth,\nsuch that they are equal in distance\nfrom each other, as rated by listeners.\nThe reference point has been chosen to be \n\\(1000 Hz = 1000 Mels\\).\n\nMel-scale was first described in \\cite{Volkmann} by Stevens and Volkmann,\nwhere the authors presented different curves for Mel-scaling.\n\nTwo common tables were composed\naccording to the Mel-scaling curves. One table by \nBeranek in 1949 \\cite{beranek1988acoustical} \nand the second by Umesh et al. in 1999 \\cite{UmeshMel}.\n\nThe most popular equation that models the Mel-scale\nis given by \\cite{o1987speech} and is typically referenced as\nthe \"Logarithm based Mel scale\":\n\\begin{equation}\\label{eq:mel_1}\n Mel = \\ln \\left( 1 + \\frac{f}{700} \\right) \\cdot \\frac{1000}{\\ln(1+\\frac{1000}{700})} \n\\end{equation}\n\nEquation\\;[\\ref{eq:mel_1}] can be simplified as follows:\n\\begin{align}\n Mel & = 1127 \\ln \\left( 1 + \\frac{f}{700} \\right) \\nonumber \\\\\n Mel & = 2595 \\log_{10}\\left( 1 + \\frac{f}{700} \\right)\n\\end{align}\n\nThen, the reverse equation, converting Mels back to Hz,\ncan be written as:\n\n\\begin{align}\n f[Hz] & = 700 \\left( 10^{\\frac{Mel}{2595}} -1 \\right)\n\\end{align}\n\n% \\subsection{Logarithm based Mel scale}\n% The most popular equation that models the Mel-scale\n% is given by \\cite{o1987speech}:\n% \\begin{equation}\n% Mel = \\ln \\left( 1 + \\frac{f}{700} \\right) \\cdot \\frac{1000}{\\ln(1+\\frac{1000}{700})} \n% \\end{equation}\n% This can be simplified as follows:\n% \\begin{align}\n% Mel & = 1127 \\ln \\left( 1 + \\frac{f}{700} \\right) \\nonumber \\\\\n% Mel & = 2595 \\log_{10}\\left( 1 + \\frac{f}{700} \\right)\n% \\end{align}\n% Then, the reverse equation, converting Mels back to Hz,\n% can be written as:\n% \\begin{align}\n% f[Hz] & = 700 \\left( 10^{\\frac{Mel}{2595}} -1 \\right)\n% \\end{align}\n% These modeling equations became de-facto the default way to\n% map audio frequencies to Mels.\n\n\\subsection{Mel scale approximations}\nComputing a logarithm for hardware devices, \nwhether it is the natural logarithm or any other base, \nis not very straightforward.\nFor example, this kind of computation might require special\ntechniques or long LUTs (look-up tables),\nwhich are extraordinarily resource hungry.\n\nInstead, other approximations that do not involve\ntrigonometric or logarithms, but only\nsimple arithmetic structures can be applied.\nBy doing so, we benefit from low resource\nutilization while maintaining high accuracy.\n\nMultiple approximation methods were studied in \\cite{fitmelscale}.\nTwo approximations are the most prominent for target HW devices.\n\\begin{align}\n \\label{eq:melapproxa} Mel & = a + b \\cdot f \\\\ \n Mel & = \\frac{f}{a \\cdot f + b} \\label{eq:melapproxb}\n\\end{align}\n\nWhere \\(a\\), \\(b\\) in Equation \\ref{eq:melapproxa} are defined as follows:\n\\begin{align}\n a & = \\begin{cases}\n 127.7 &,\\;f \\leq 1000 \\\\\n 1322 &,\\;f > 1000\n \\end{cases} \\nonumber \\\\\n b & = \\begin{cases}\n 0.9 &,\\;f \\leq 1000 \\\\\n 0.19 &,\\;f > 1000\n \\end{cases}\n\\end{align}\n\nwhile in Equation \\ref{eq:melapproxb} \\(a\\), \\(b\\) are:\n\\begin{align}\n a & = \\begin{cases}\n 0.000244 &,\\;f \\leq 1000 \\\\\n 0.0004 &,\\;f > 1000\n \\end{cases} \\nonumber \\\\\n b & = \\begin{cases}\n 0.741 &,\\;f \\leq 1000 \\\\\n 0.603 &,\\;f > 1000\n \\end{cases}\n\\end{align}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Scaling/images/mel_methods_comp}\n \\caption{Mel scale comparisons with Beranek \\& Umesh tables}\\label{fig:mel_methods_comp}\n\\end{figure}\n\nFigure \\ref{fig:mel_methods_comp} has comparisons of three different \nMel-scale implementations with Beranek \\& Umesh tables.\nThe first column, Log Based Mel, represents\nO'Shaugnessy's famous log-based Mel modeling.\nThe Mel option \\#1, and Mel option \\#2 columns \nfollow the suggested approximations\ngiven in Equations \\ref{eq:melapproxa} \nand \\ref{eq:melapproxb}, respectively.\n\nFrom the last row of graphs, we can deduce that the approximation\nin Equation \\ref{eq:melapproxb} \nis the closest along with the range of audio frequencies\nto the tables provided by Beranek \\& Umesh.\nOn the other hand, the more simplified approximation \nin Equation\\ref{eq:melapproxa} seems \nto yield the highest errors.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Scaling/images/mel_lut}\n \\caption{LUT based Mel FPGA implementation results}\\label{fig:mel_lut}\n\\end{figure}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Scaling/images/mel_approx_no_opt}\n \\caption{Mel approx. \\#1 FPGA implementation results}\\label{fig:mel_approx_no_opt}\n\\end{figure}\n\nFigures \\ref{fig:mel_lut}, \\ref{fig:mel_approx_no_opt} \npresent the results of FPGA implementations\nof O'Shaugnessy's log-based Mel and Mel approximation \\#1.\nNonetheless, high precision quantization settings were chosen\nfor the Mel approximation, U32.22, \nthe shifting error received is higher compared \nto the conventional Log Mel scaling implementation.\nAlthough this error shift is compensated \njust by selecting the approximation method, \nthe straightforward approach \nturned out to be the non-optimized solution \nin terms of HW resources and power consumption,\nwhich utilized four times higher wattage on top of \\(25 \\div 30 \\%\\) \nadditional resources.\n\nInstead, two optimization workarounds were tested.\nThe first is the multiplication of the \\(a, b\\) coefficients\nin \\ref{eq:melapproxb} by 1000. \nThe second optimization is reorganizing the equation\nand storing the result in a sufficient precision\nfor the fractional part but lower resolution for the integer part. \nThese optimizations lead to a reduction in the \nrequired number of bits for the fractional part.\nAs a result, both the frequency shifting error \nand the overall resource utilization are greatly improved.\n\nYet, the split in frequency bands results in two multiplied sets\nof coefficients for each band calculation.\nTherefore, choosing a more generalized set \nof coefficients for the entire audio band can help \nin the reduction of redundant LUTs and other combinational logic, \nsuch as selectors and multi-bus multiplexer cells.\n\nSelection of \\(a=0.24,\\;b=0.741\\), showed better results\nas can be seen in Figure \\ref{fig:mel_approx_logic_opt_generic}.\nThe accuracy estimation for the non-generic implementation\nis shown in Figure\\;\\ref{fig:mel_approx_logic_opt}.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Scaling/images/mel_approx_logic_opt}\n \\caption{Mel approx. \\#1 \\underline{optimized} FPGA implementation results}\\label{fig:mel_approx_logic_opt}\n\\end{figure}\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=\\linewidth]{Scaling/images/mel_approx_logic_opt_generic}\n \\caption{Mel approx. \\#1 \\underline{optimized, generic} FPGA implementation results}\\label{fig:mel_approx_logic_opt_generic}\n\\end{figure}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n % \\centering\n \\hspace*{-1.8cm}\n \\arrayrulecolor{mtblborder}\n\\begin{tabular}{ !{\\color{mtblborder}\\vrule}l!{\\color{mtblborder}\\vrule}rcccc| } \n \\hline\n\n \\hline\n \\rowcolor{mtblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{Latency \\([ns]\\)} \n & \\color{white}\\bf{Quant.} \n & \\color{white}\\bf{Max Err. \\([\\Delta Hz]\\)}\n & \\color{white}\\bf{Mean Err.}\n & \\color{white}\\bf{Std Err.} \\\\\n \\hline\n\n \\hline\n \\rowcolor{mtbl} Log LUT Mel & \\(14\\) (3 C.C) & U16/4 & 1.146 & 0.268 & 0.106 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Mel \\#1 & \\(9\\) (2 C.C) & U32/22 & 1.338 & -0.873 & -0.359 \\\\\n \\hline\n \n \\hline\n \\rowcolor{mtbl} Mel \\#1 Opt & \\(9\\) (2 C.C) & U16/4 & 0.190 & -0.053 & 0.065 \\\\\n \\hline\n\n \\hline\n \\rowcolor{wtbl} Mel \\#1 Generic & \\(9\\) (2 C.C) & U16/4 & \\color{gtblcaption}0.174 & \\color{gtblcaption}-0.035 & \\color{gtblcaption}0.061 \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel-Approx, log-based Mel performance comparison}\n\\label{tbl:mel_scale_performance}\n\\end{table}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{ytblborder}\n\\begin{tabular}{ !{\\color{ytblborder}\\vrule}l!{\\color{ytblborder}\\vrule}rrrrr| } \n \\hline\n\n \\hline\n \\rowcolor{ytblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{FF} \n & \\color{white}\\bf{LUT} \n & \\color{white}\\bf{DSP} \n & \\color{white}\\bf{LUTRAM} \n & \\color{white}\\bf{BRAM} \\\\\n % & \\color{white}\\bf{ORM} \n % & \\color{white}\\bf{Clean} \\\\\n \\hline\n\n \\hline\n \\rowcolor{ytbl} Log LUT Mel & 82(0.04\\%) & 276(0.23\\%) & 1(<1\\%) & 10(0.02\\%) & 1.5(1.04\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Mel \\#1 & 47(0.02\\%) & 530(0.45\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{ytbl} Mel \\#1 Opt & 47(0.02\\%) & 271(0.23\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n\n \\hline\n \\rowcolor{wtbl} Mel \\#1 Generic & \\color{gtblcaption}47(0.02\\%) & \\color{gtblcaption}269(0.19\\%) & \\color{gtblcaption}1(<1\\%) & \\color{gtblcaption}0(0\\%) & \\color{gtblcaption}0(0\\%) \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel scaling methods resource utilization table}\n\\label{tbl:mel_resource_util}\n\\end{table}\n\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{gtblborder}\n\\begin{tabular}{ |l|cccc| } \n \\hline\n\n \\hline\n \\rowcolor{gtblcaption} \\color{white}\\bf{Parameter} \n & \\color{white}\\bf{Log LUT Mel} \n & \\color{white}\\bf{Mel \\#1} \n & \\color{white}\\bf{Mel \\#1 Opt} \n & \\color{white}\\bf{Mel \\#1 Generic} \\\\\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{5}{|c|}{\\bf{Dynamic Power [W]}}\\\\\n \\hline\n \\rowcolor{gtbl} Signals & 4.947 & 5.221 & 3.011 & 2.916 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Logic & 6.50 & 6.792 & 3.751 & 3.070 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} DSP & 0.014 & 0.013 & 0.014 & 0.014 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} I/O & 18.236 & 26.263 & 7.207 & 6.378 \\\\\n \\hline\n \n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{dynmic}}\\) & \\textbf{29.697} & \\textbf{38.290} & \\textbf{13.983} & \\textbf{\\colorbox{Goldenrod!70}{\\color{MidnightBlue}12.379}} \\\\\n \\hline\n\n % \\hline\n % \\rowcolor{gtbl} Bark Scale & 56(0.02\\%) & 218(\\%) & 1(<1\\%) \\\\\n % % \\cline{2-8}\n % % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{5}{|c|}{\\bf{Static Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} PL Static & 2.364 & 2.466 & 0.499 & 0.499 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} PS Static & 0.068 & 0.071 & 0.020 & 0.018 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{static}}\\) & \\textbf{2.432} & \\textbf{2.537} & \\textbf{0.519} & \\textbf{\\colorbox{Goldenrod!70}{\\color{MidnightBlue}0.517}} \\\\\n \\hline\n\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{5}{|c|}{\\bf{Total Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{total}}\\) & \\textbf{32.13} & \\textbf{40.827} & \\textbf{14.502} & \\textbf{\\colorbox{Goldenrod!70}{\\color{MidnightBlue}12.896}} \\\\\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel-Approx, log-based Mel, Bark Scale Power consumption}\n\\label{tbl:mel_scale_pwr_tbl}\n\\end{table}\n\nThe LUT implementation makes use of a \\(log_{2}\\) look-up table.\nHowever, an additional step is \nneeded for the natural logarithm or other log bases. \nSince the logarithm bases are constant, \nthe LUT result is divided by the \\(log_{2}\\) of the base,\nwhether it be the natural base or base ten.\nThis logarithm bases \nconvention is described in Equation\\;\\ref{eq:log_base_conv}.\n\\begin{align}\\label{eq:log_base_conv}\n \\log_{b}(a) & = \\frac{\\log_{x}(a)}{\\log_{x}(b)}\n\\end{align}\n\n\nTable\\;\\ref{tbl:mel_scale_performance} summarizes\nthe performance comparison between the different Mel scaling\nimplementation approaches. \n\nThe HW setup is for the PYNQ-Z1 development board.\nHence, the results are unique to that specific \nHW device and probably change for other \nFPGA devices and development boards.\n\nLatencies were simulated with Xilinx Vivado Suit\nfor an operating clock frequency of \\(225MHz\\).\nIn this operating condition, no timing violations were reported.\n\nTables\\;\\ref{tbl:mel_resource_util} and \\ref{tbl:mel_scale_pwr_tbl}\nshow the synthesis and implementation results \nplus the power estimation reports. \nThese reports were taken from the Xilinx Vivado Suit application.\n\n\\section{Bark-Scaling}\nAnother scaling method is the Bark scale which is based on\nthe same principle of retaining the same perceptual distances.\nThe Bark scale is divided into critical \nbands corresponding to the critical hearing bands. \nEach band has a bandwidth similar \nto the psychoacoustic band of the corresponding \n``filter'' in the human hearing system \nand is ranked with a unique number.\n\n\\subsection{Bark Critical Bands}\nLike the Mel scale, several equations were proposed to model best\nthe Bark scale and its critical bands.\n\nThe first method was introduced in \\cite{1908630}.\nA proposed approximation to the Bark scaling is described in \n\\cite{TraunmullerScale}; this paper also introduces the \ncorrection of the band boundaries to ensure more \ncorrectness with the original Bark scaling.\n\nFour different equations were proposed to model\nthe Bark scale.\n\nThe first by Zwicker is well described in\\cite{1908630} and\nis given by the following equation:\n\\begin{align}% Zwicker Eq\n Bark & = \\tan^{-1}\\left\\{ 0.00073f \\right\\} + 3.5 \\tan^{-1}\\left\\{ \n \\left( \\frac{f}{7500} \\right)^{2}\n \\right\\}\n\\end{align}\n\nAnother proposed equation by Traunmuller \\cite{TraunmullerScale} is:\n\\begin{align}\\label{eq:traunmuller_no_fix}% Traunmuller Eq + Correction\n Bark & = \\frac{26.81f}{1960 + f} - 0.53\n\\end{align}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/Traunmuller_nofix}\n \\caption{Traunmuller no fix}\\label{fig:Traunmuller_nofix}\n\\end{figure}\n\nDenoting Traunmuller's original\nBark scale given in Equation \\ref{eq:traunmuller_no_fix} as \\(Bark'\\),\nthe fixed form for Traunmuller's Bark equation is:\n\\begin{align}\n Bark & = \\begin{cases}\n 0.3 + 0.85\\cdot \\left( Bark' \\right) \n &,\\;Bark' < 2 \\\\\n Bark' + 0.22\\cdot \\left( Bark' - 20.1 \\right) \n &,\\;Bark' > 20.1\n \\end{cases}\n\\end{align}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/Traunmuller_fix}\n \\caption{Traunmuller fix}\\label{fig:Traunmuller_fix}\n\\end{figure}\n\nThe fourth possible modeling equation is proposed by\nSchroeder in \\cite{SchroederScale} and is as follows:\n% Schroeder Eq\n\\begin{align}\n Bark & = 7\\ln \\left( \\frac{f}{650} + \\sqrt{1 + \\frac{f^{2}}{422500} } \\right)\n\\end{align}\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/bark_comparison}\n \\caption{Bark comparisons}\\label{fig:bark_comparison}\n\\end{figure}\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/bark_comparison2}\n \\caption{Bark comparison}\\label{fig:wer_23}\n\\end{figure}\n\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/Zwicker}\n \\caption{Zwicker}\\label{fig:Zwicker}\n\\end{figure}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Experiments/images/Schroeder}\n \\caption{Schroeder}\\label{fig:Schroeder}\n\\end{figure}\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Scaling/images/bark_traunmuller_no_fix}\n \\caption{Schroeder}\\label{fig:bark_traunmuller_no_fix}\n\\end{figure}\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n % \\centering\n \\hspace*{-1.8cm}\n \\arrayrulecolor{mtblborder}\n\\begin{tabular}{ !{\\color{mtblborder}\\vrule}l!{\\color{mtblborder}\\vrule}rcccc| } \n \\hline\n\n \\hline\n \\rowcolor{mtblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{Latency \\([ns]\\)} \n & \\color{white}\\bf{Quant.} \n & \\color{white}\\bf{Max Err. \\([\\Delta Hz]\\)}\n & \\color{white}\\bf{Mean Err.}\n & \\color{white}\\bf{Std Err.} \\\\\n \\hline\n\n \\hline\n \\rowcolor{mtbl} Traunmuller w/ Fix & 14 (3 C.C) & U10/5 & N.A & N.A & N.a \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Traunmuller w/o Fix & 9 (2 C.C) & U9/4 & 0.0443 & 0.0015 & 0.0180 \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Traunmuller's Bark scale implementations performance comparison}\n\\label{tbl:bark_implementations_performance}\n\\end{table}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{ytblborder}\n\\begin{tabular}{ !{\\color{ytblborder}\\vrule}l!{\\color{ytblborder}\\vrule}rrrrr| } \n \\hline\n\n \\hline\n \\rowcolor{ytblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{FF} \n & \\color{white}\\bf{LUT} \n & \\color{white}\\bf{DSP} \n & \\color{white}\\bf{LUTRAM} \n & \\color{white}\\bf{BRAM} \\\\\n % & \\color{white}\\bf{ORM} \n % & \\color{white}\\bf{Clean} \\\\\n \\hline\n\n \\hline\n \\rowcolor{ytbl} Traunmuller w/ Fix & 43(0.02\\%) & 302(0.26\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Traunmuller w/o Fix & 27(0.01\\%) & 284(0.24\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Traunmuller's Bark scale implementations resource utilization table}\n\\label{tbl:Bark_resource_util}\n\\end{table}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{gtblborder}\n\\begin{tabular}{ |l|cc| } \n \\hline\n\n \\hline\n \\rowcolor{gtblcaption} \\color{white}\\bf{Parameter} \n & \\color{white}\\bf{Traunmuller w/ Fix} \n & \\color{white}\\bf{Traunmuller w/o Fix} \\\\\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Dynamic Power [W]}}\\\\\n \\hline\n \\rowcolor{gtbl} Signals & 2.380 & 2.192 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Logic & 3.079 & 2.957 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} DSP & 1.445 & 0.014 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} I/O & 4.022 & 4.022 \\\\\n \\hline\n \n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{dynmic}}\\) & \\textbf{10.923} & \\textbf{\\color{gtblborder}9.186} \\\\\n \\hline\n\n % \\hline\n % \\rowcolor{gtbl} Bark Scale & 56(0.02\\%) & 218(\\%) & 1(<1\\%) \\\\\n % % \\cline{2-8}\n % % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Static Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} PL Static & 0.469 & 0.437 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} PS Static & 0.017 & 0.016 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{static}}\\) & \\textbf{0.486} & \\textbf{\\color{gtblborder}0.453} \\\\\n \\hline\n\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Total Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{total}}\\) & \\textbf{11.412} & \\textbf{\\color{gtblborder}9.639} \\\\\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Traunmuller's Bark scale implementations Power consumption}\n\\label{tbl:bark_scale_pwr_tbl}\n\\end{table}\n\n\n\\section{ERB - Equivalent Rectangular Bandwidth}\nLike the Bark scaling method, the ERB scale aims \nto rescale the audio spectrum in different \nbandwidths corresponding to the human hearing ``filters''.\n\nThis approach slightly differs from the Bark scale\nbecause the filters' model is according\nto a rectangular filter with an equivalent\nbandwidth.\n\n\\begin{align}\\label{eq:erb_eq}\n ERBs = 11.17\\ln (47.065 - \\frac{676170.42}{f + 14678.5}) \n\\end{align}\n\nA proposed approximation is given by:\n\\begin{align}\\label{eq:erb_approx_eq}\n ERBs = 21.4 \\cdot \\log_{10} (1 + 0.00437f) \n\\end{align}\n\n\n% \\subsection{ERB Critical Bands}\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Scaling/images/erb}\n \\caption{LUT based ERB FPGA implementation results}\\label{fig:erb_fpga}\n\\end{figure}\n\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\linewidth]{Scaling/images/erb_approx}\n \\caption{LUT based ERB approx. FPGA implementation results}\\label{fig:erb_approx}\n\\end{figure}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n % \\hspace*{-1.8cm}\n \\arrayrulecolor{mtblborder}\n\\begin{tabular}{ !{\\color{mtblborder}\\vrule}l!{\\color{mtblborder}\\vrule}rcccc| } \n \\hline\n\n \\hline\n \\rowcolor{mtblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{Latency \\([ns]\\)} \n & \\color{white}\\bf{Quant.} \n & \\color{white}\\bf{Max Err. \\([\\Delta C]\\)}\n & \\color{white}\\bf{Mean Err.}\n & \\color{white}\\bf{Std Err.} \\\\\n \\hline\n\n \\hline\n \\rowcolor{mtbl} ERB & 23 (5 C.C) & U16/10 & 0.0153 & 0.0130 & 0.0018 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} ERB Approx & 14 (3 C.C) & U16/10 & 6.210m & -0.408m & 0.657m \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{ERB and ERB approx performance comparison}\n\\label{tbl:erb_scale_performance}\n\\end{table}\n\nTable\\;\\ref{tbl:erb_scale_performance} presents the performance\ncomparison between the implementations of the pure ERB\nand the ERB approximation\ngiven by Equations\\;\\ref{eq:erb_eq} \nand \\ref{eq:erb_approx_eq}, respectively. The error terms\nfor the ERBs are measured in \\(\\Delta C\\), indicating the\nerror in Cams rather than in frequency units (Hz). \nEquation\\;\\ref{eq:cams_hz} can be used to convert Cams to Hz.\n\n\\begin{align}\\label{eq:cams_hz}\n f = \\frac{676170.42}{47.065 - e^{0.0895\\cdot C}} - 14678.5\n\\end{align}\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{ytblborder}\n\\begin{tabular}{ !{\\color{ytblborder}\\vrule}l!{\\color{ytblborder}\\vrule}rrrrr| } \n \\hline\n\n \\hline\n \\rowcolor{ytblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{FF} \n & \\color{white}\\bf{LUT} \n & \\color{white}\\bf{DSP} \n & \\color{white}\\bf{LUTRAM} \n & \\color{white}\\bf{BRAM} \\\\\n % & \\color{white}\\bf{ORM} \n % & \\color{white}\\bf{Clean} \\\\\n \\hline\n\n \\hline\n \\rowcolor{ytbl} ERB & 112(0.05\\%) & 824(0.70\\%) & 1(<1\\%) & 1(0.02\\%) & 1.5(1.04\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} ERB Approx & 103(0.04\\%) & 564(0.48\\%) & 1(<1\\%) & 1(0.01\\%) & 1.5(1.04\\%) \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel scaling methods resource utilization table}\n\\label{tbl:ERB_resource_util}\n\\end{table}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{gtblborder}\n\\begin{tabular}{ |l|cc| } \n \\hline\n\n \\hline\n \\rowcolor{gtblcaption} \\color{white}\\bf{Parameter} \n & \\color{white}\\bf{ERB} \n & \\color{white}\\bf{ERB Approx} \\\\\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Dynamic Power [W]}}\\\\\n \\hline\n \\rowcolor{gtbl} Signals & 12.691 & 5.525 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Logic & 17.576 & 7.478 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} DSP & 0.067 & 0.049 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} I/O & 17.379 & 17.373 \\\\\n \\hline\n \n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{dynmic}}\\) & \\textbf{47.713} & \\textbf{\\color{gtblborder}31.519} \\\\\n \\hline\n\n % \\hline\n % \\rowcolor{gtbl} Bark Scale & 56(0.02\\%) & 218(\\%) & 1(<1\\%) \\\\\n % % \\cline{2-8}\n % % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Static Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} PL Static & 2.971 & 1.510 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} PS Static & 0.084 & 0.046 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{static}}\\) & \\textbf{3.055} & \\textbf{\\color{gtblborder}1.556} \\\\\n \\hline\n\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{3}{|c|}{\\bf{Total Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{total}}\\) & \\textbf{50.768} & \\textbf{\\color{gtblborder}33.075} \\\\\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{ERB vs. ERB approx Power consumption}\n\\label{tbl:erb_scale_pwr_tbl}\n\\end{table}\n\n\n\n\n% \\begin{table}[H]\n% % for more info see: https://www.overleaf.com/learn/latex/tables\n% \\centering\n% \\begin{tabular}{|c|r|r|r|r|r|}\n% \\hline\n% $N$ & Latency ($\\mu s$) & FF & LUT & DSP48 & BRAM \\\\\n% \\hline\n% 3 & 4.80 & 3889 (27.6\\%) & 3901 (54.9\\%) & 90 (25.0\\%) & 0 (0.0\\%) \\\\\n% 4 & 6.34 & 63657 (45.1\\%) & 64149 (90.4\\%) & 144 (40.0\\%) & 0 (0.0\\%) \\\\\n% 8 & 12.50 & 231199 (\\textcolor{red}{164\\%}) & 252446 (\\textcolor{red}{356\\%}) & 272 (75.6\\%) & 0 (0.0\\%) \\\\\n% 16 & 24.82 & 895377 (\\textcolor{red}{635\\%}) & 1040992 (\\textcolor{red}{1466\\%}) & 144 (40.0\\%) & 0 (0.0\\%) \\\\\n% \\hline\n% \\end{tabular}\n% \\caption{\\bf{ IQRD Performance:} Latency, Resource Utilization, and Power for \\(log_{10}\\) LUT on FPGA}\n% \\label{tbl:IQRD_perf}\n% \\end{table}\n\n\\section{Summary}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n \\centering\n \\arrayrulecolor{ytblborder}\n\\begin{tabular}{ !{\\color{ytblborder}\\vrule}l!{\\color{ytblborder}\\vrule}rrrrr| } \n \\hline\n\n \\hline\n \\rowcolor{ytblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{FF} \n & \\color{white}\\bf{LUT} \n & \\color{white}\\bf{DSP} \n & \\color{white}\\bf{LUTRAM} \n & \\color{white}\\bf{BRAM} \\\\\n % & \\color{white}\\bf{ORM} \n % & \\color{white}\\bf{Clean} \\\\\n \\hline\n\n \\hline\n \\rowcolor{ytbl} Log-Based Mel & 82(0.04\\%) & 276(0.23\\%) & 1(<1\\%) & 10(0.02\\%) & 1.5(1.04\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Mel \\#1 Generic & 47(0.02\\%) & 269(0.19\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n \n \\hline\n \\rowcolor{ytbl} Bark Scale w/ fix & 43(0.02\\%) & 302(0.26\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n\n \\hline\n \\rowcolor{wtbl} Bark Scale w/o fix & 27(0.01\\%) & 284(0.24\\%) & 1(<1\\%) & 0(0\\%) & 0(0\\%) \\\\\n \\hline\n\n \\hline\n \\rowcolor{ytbl} Approx. ERB & 103(0.04\\%) & 564(0.48\\%) & 1(<1\\%) & 1(0.01\\%) & 1.5(1.04\\%) \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel-Approx, log-based Mel, Bark Scale}\n\\label{tbl:IQRD_perf}\n\\end{table}\n\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n % \\centering\n \\hspace*{-1cm}\n \\arrayrulecolor{gtblborder}\n\\begin{tabular}{ |l|ccccc| } \n \\hline\n\n \\hline\n \\rowcolor{gtblcaption} \\color{white}\\bf{Parameter} \n & \\color{white}\\bf{Mel LUT}\n & \\color{white}\\bf{Mel Gen}\n & \\color{white}\\bf{Bark w/}\n & \\color{white}\\bf{Bark w/o}\n & \\color{white}\\bf{Approx. ERB} \\\\\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{6}{|c|}{\\bf{Dynamic Power [W]}}\\\\\n \\hline\n \\rowcolor{gtbl} Signals & 4.947 & 2.916 & 2.380 & 2.192 & 5.525 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Logic & 6.50 & 3.070 & 3.079 & 2.957 & 7.478 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} DSP & 0.014 & 0.014 & 1.445 & 0.014 & 0.049 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} I/O & 18.236 & 6.378 & 4.022 & 4.022 & 17.373 \\\\\n \\hline\n \n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{dynmic}}\\) & \\textbf{29.697} & \\textbf{12.379} & \\textbf{10.923} & \\textbf{9.186} & \\textbf{31.519} \\\\\n \\hline\n\n % \\hline\n % \\rowcolor{gtbl} Bark Scale & 56(0.02\\%) & 218(\\%) & 1(<1\\%) \\\\\n % % \\cline{2-8}\n % % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{6}{|c|}{\\bf{Static Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} PL Static & 2.364 & 0.499 & 0.469 & 0.437 & 1.510 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} PS Static & 0.068 & 0.018 & 0.017 & 0.016 & 0.046 \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{static}}\\) & \\textbf{2.432} & \\textbf{0.517} & \\textbf{0.486} & \\textbf{0.453} & \\textbf{1.556} \\\\\n \\hline\n\n \\hline\\hline\n \\rowcolor{wtbl}\\multicolumn{6}{|c|}{\\bf{Total Power [W]}} \\\\\n \\hline\n\n \\hline\n \\rowcolor{gtbl} \\(\\mathbf{P_{total}}\\) & \\textbf{32.13} & \\textbf{12.896} & \\textbf{11.412} & \\textbf{9.639} & \\textbf{33.075} \\\\\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel-Approx, log-based Mel, Bark Scale Power consumption}\n\\label{tbl:sum_scale_pwr_tbl}\n\\end{table}\n\n\n\n\n% \\begin{table}[H]\n% % for more info see: https://www.overleaf.com/learn/latex/tables\n% \\centering\n% \\arrayrulecolor{gtblborder}\n% \\begin{tabular}{ !{\\color{gtblborder}\\vrule}l!{\\color{gtblborder}\\vrule}rrrrrr| } \n% \\hline\n\n% \\hline\n% \\rowcolor{gtblcaption} \\color{white}\\bf{Algorithm} \n% & \\color{white}\\bf{Signals} \n% & \\color{white}\\bf{Logic} \n% & \\color{white}\\bf{DSP} \n% & \\color{white}\\bf{IO} \n% & \\color{white}\\bf{BRAM}\n% & \\color{white}\\bf{Total} \\\\\n% \\hline\n\n% \\hline\n% \\rowcolor{gtbl} Log Based Mel & 82(0.04\\%) & 276(\\%) & 1(<1\\%) & 18 & 1.5 & 0 \\\\\n% \\hline\n \n% \\hline\n% \\rowcolor{wtbl} Approx. Mel & 47(0.02\\%) & 230(\\%) & 1(<1\\%) & 0 & 0 & 0 \\\\\n% \\hline\n \n% \\hline\n% \\rowcolor{gtbl} Bark Scale & 56(0.02\\%) & 218(\\%) & 1(<1\\%) & 0 & 0 & 0 \\\\\n% % \\cline{2-8}\n% % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n% \\hline\n\n% \\hline\n% \\end{tabular}\n% \\arrayrulecolor{black}\n% \\caption{Mel-Approx, log based Mel, Bark Scale Power consumption}\n% \\label{tbl:scale_pwr_tbl}\n% \\end{table}\n\n\n\\begin{table}[H]\n % for more info see: https://www.overleaf.com/learn/latex/tables\n % \\centering\n \\hspace*{-1.8cm}\n \\arrayrulecolor{mtblborder}\n\\begin{tabular}{ !{\\color{mtblborder}\\vrule}l!{\\color{mtblborder}\\vrule}rcccc| } \n \\hline\n\n \\hline\n \\rowcolor{mtblcaption} \\color{white}\\bf{Algorithm} \n & \\color{white}\\bf{Latency \\([ns]\\)} \n & \\color{white}\\bf{Quant.} \n & \\color{white}\\bf{Max Err. \\([\\Delta Hz]\\)}\n & \\color{white}\\bf{Mean Err.}\n & \\color{white}\\bf{Std Err.} \\\\\n \\hline\n\n \\hline\n \\rowcolor{mtbl} Log-Based Mel & 14 (3 C.C) & U16/4 & 1.146 & 0.268 & 0.106 \\\\\n \\hline\n \n \\hline\n \\rowcolor{wtbl} Mel \\#1 Generic & 9 (2 C.C) & U16/4 & 0.174 & -0.035 & 0.061 \\\\\n \\hline\n \n \\hline\n \\rowcolor{mtbl} Bark Scale w/ fix & 14 (3 C.C) & U10/5 & N.A & N.A & N.A \\\\\n % \\cline{2-8}\n % \\multirow{-2}{*}{\\cellcolor{ytbl}\\#1(5)} & 14.03/17.78 & 27.06 & 13.73 & - & - & - & 7.72 \\\\ \n \\hline\n\n \\hline\n \\rowcolor{wtbl} Bark Scale w/o fix & 9 (2 C.C) & U9/4 & 0.0443 & 1.5m & 0.0015 \\\\\n \\hline\n\n \\hline\n \\rowcolor{mtbl} ERB Approx. & 14 (3 C.C) & U16/10 & 0.292 & 0.104 & 0.134 \\\\\n \\hline\n\n \\hline\n\\end{tabular}\n\\arrayrulecolor{black}\n\\caption{Mel-Approx, log-based Mel, Bark Scale performance comparison}\n\\label{tbl:scale_pwr_tbl}\n\\end{table}\n\n\n\n", "meta": {"hexsha": "b9eacd1127bf8cb0d644d74b3405fa28ff05bcf6", "size": 33149, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Dissertation/Scaling/scaling.tex", "max_stars_repo_name": "aviadb/research", "max_stars_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Dissertation/Scaling/scaling.tex", "max_issues_repo_name": "aviadb/research", "max_issues_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Dissertation/Scaling/scaling.tex", "max_forks_repo_name": "aviadb/research", "max_forks_repo_head_hexsha": "d710d21cd18a5f5c01acf65d85c8ef364a3ca219", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.4506641366, "max_line_length": 196, "alphanum_fraction": 0.630456424, "num_tokens": 11851, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059414036511, "lm_q2_score": 0.8031737987125612, "lm_q1q2_score": 0.6370019117386725}} {"text": "\\documentclass{article}\n\n\\usepackage[letterpaper, margin=1.3cm]{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage{siunitx}\n\\usepackage{mathtools}\n\\usepackage{multicol}\n\\usepackage{pgfplots}\n\\usepackage[RPvoltages, american, betterproportions, siunitx]{circuitikz}\n\\pgfplotsset{compat=1.16}\n\\pgfplotsset{scaled x ticks=true}\n\n\\title{ECE 203 Problem Set 5}\n\\author{Michael Kwok}\n\\date{March 2020}\n\\begin{document}\n\n\\maketitle\n\\begin{multicols}{2}\n\\section*{1}\n\\subsection*{a}\n\n\\begin{align*}\n \\omega_0 &= \\frac{1}{\\sqrt{L C}} = \\boxed{\\SI{1390.77}{\\radian\\per\\second}}\\\\\n f_0 &= \\frac{\\omega_0}{2\\pi} = \\boxed{\\SI{221.3}{\\hertz}}\\\\\n Q_0 &= \\omega_0 R C\\\\\n &= 1390.77 \\cdot 1000 \\cdot \\num{47d-6} = \\boxed{65.37}\n\\end{align*}\n\n\\subsection*{b}\n\n\\begin{align*}\n \\omega_0 \\text{ parallel} &= \\omega_0 \\text{ series}\\\\\n f_0 \\text{ parallel} &= f_0 \\text{ series}\\\\\n Q_0 \\text{ series} &= \\frac{1}{Q_0 \\text{ parallel}}\\\\\n\\end{align*}\n\\begin{align*}\n \\omega_0 &= \\boxed{\\SI{1390.77}{\\radian\\per\\second}}\\\\\n f_0 &= \\boxed{\\SI{221.3}{\\hertz}}\\\\\n Q_0 &= \\boxed{0.0153}\n\\end{align*}\n\n\\subsection*{c}\n\nThe parallel configuration is better as the quality factor is higher, which by definition means that the filter is more selective.\n\n\\section*{2}\n\\subsection*{a}\n\n\\begin{align*}\n Q_0 &= \\omega_0 R C\\\\\n \\omega_0 &= 2\\pi f_0 \\\\\n Q_0 &= 2\\pi f_0 R C\\\\\n\\end{align*}\n\n\\begin{align*}\n R &= \\frac{V}{I}\\\\\n &= \\frac{\\num{125e-3}}{2e-6}\\\\\n &= \\boxed{\\SI{62500}{\\ohm}}\\\\\n RC &= \\frac{40}{2\\pi \\cdot 1000000}\\\\\n C &= \\frac{\\num{6.366197724e-6}}{62500}\\\\\n &= \\SI{1.018591636e-10}{\\farad}\\\\\n &= \\boxed{\\SI{102}{\\pico\\farad}}\n\\end{align*}\n\\begin{align*}\n \\omega_0 &= \\frac{1}{\\sqrt{LC}}\\\\\n \\num{2\\pi e-6} &= \\frac{1}{\\sqrt{LC}}\\\\\n LC &= \\left(\\frac{1}{\\num{2\\pi e-6}}\\right)^2\\\\\n L &= \\boxed{\\num{2.5e-4}}\n\\end{align*}\n\n\\subsection*{b}\n\\begin{align*}\n\\text{Using Z at } \\omega = 2000,\\\\\nR= \\boxed{\\SI{8}{\\kilo\\ohm}}\n\\end{align*}\n\\begin{align*}\n Z&=\\left( \\frac{1}{R} + \\frac{1}{j \\omega L} + j \\omega C \\right) ^{-1}\\\\\n \\frac{1}{4000} e^{-j \\theta} &= \\frac{1}{8000} + \\frac{1}{1800 L j} + 1800 C j\\\\\n \\frac{1}{4000} \\cos \\theta &= \\frac{1}{8000}\\\\\n \\cos\\theta &= \\frac{1}{2}\\\\\n \\theta &= \\frac{\\pi}{3}\n\\end{align*}\n\\begin{align*}\n -j\\frac{1}{4000} \\sin\\theta &= -j\\left( \\frac{1}{1800L} - 1800 C\\right)\\\\\n \\frac{1800L}{4000} \\sin\\theta &= -(1800)^2 LC + 1\\\\\n \\omega_0 &= 2000\\\\\n &= \\frac{1}{\\sqrt{LC}}\\\\\n LC&= \\frac{1}{\\num{4e6}}\\\\\n\\end{align*}\n\\begin{align*}\n L\\sin\\theta &= 0.4222\\\\\n L &= \\boxed{\\SI{0.488}{\\henry}}\\\\\n C &= \\boxed{\\SI{5.133e-7}{\\farad}}\n\\end{align*}\n\n\\section*{3}\n\\subsection*{a}\n\\begin{align*}\n Z_{in} &= 10 + j\\omega 3 + \\left( \\frac{1}{75} + \\frac{1}{15 + \\frac{675}{j\\omega}} \\right) ^{-1}\\\\\n &= \\frac{90\\omega + (6 \\omega^2 - 1275)j}{2\\omega - 15j}\\\\\n &= \\frac{\\left[ 90\\omega + (6 \\omega^2 - 1275)j \\right] [2\\omega + 15j]}{4\\omega^2 + 15^2}\\\\\n &= \\frac{90\\omega^2 +19125 + \\left( 12\\omega^3 -1200 \\omega j \\right)}{4\\omega^2 + 15^2}\n\\end{align*}\n\\begin{align*}\n 12\\omega^3-1200\\omega &= 0\\\\\n 12\\omega^3 &= 1200\\omega\\\\\n \\omega^2 &= 100\\\\\n \\omega &= \\boxed{\\SI{10}{\\radian\\per\\second}}\n\\end{align*}\n\n\\subsection*{b}\n\\begin{align*}\n Z_{in} &= 10 + 30j + \\left( \\frac{1}{75} + \\frac{1}{15 + \\frac{675}{j10}} \\right)^{-1}\\\\\n &= 10 + 30j - 30j + 35\\\\\n &= \\boxed{\\SI{45}{\\ohm}}\n\\end{align*}\n\n\\section*{4}\nUse a test voltage $V_{in}$\n\n\\begin{circuitikz}\n\\draw (0,0) to[V=$V_{in}$] (0,2) -- (4,2);\n\\draw (0,0) -- (7,0) to[cV] (7,2);\n\\draw (2,2) to[C, i=$I_C$](2,0);\n\\draw (4,0) to[R, i=$I_R$](4,2);\n\\draw (4,2) to[L] (7,2);\n\\end{circuitikz}\n\n\\begin{align*}\n I_C &= V_{in}\\num{e-8}j\\omega\\\\\n I_R &= -\\frac{V_{in}}{10000}\\\\\n\\end{align*}\n Using KCL on node above resistor,\n\\begin{align*}\n -I_C + I_R &+\\frac{V_{in} - \\num{e5}I_R}{\\num{4.4e-3}j\\omega } = 0\\\\\n -V_{in}\\num{e-8}j\\omega-\\frac{V_{in}}{10000} &+\\frac{V_{in} - \\num{e5}I_R}{\\num{4.4e-3}j\\omega} = 0\\\\\n Y = &\\frac{V_{in}\\num{e-8}j\\omega +\\frac{V_{in}}{10000} +\\frac{V_{in} \\left( 1+\\frac{\\num{e5}}{10000}\\right)}{\\num{4.4e-3}j\\omega}}{V_{in}}\\\\\n = &\\boxed{\\num{e-8}j\\omega +\\frac{1}{10000} -\\frac{2500}{\\omega}j}\n\\end{align*}\nFinding $\\omega_0$ and $Q_0$.\n\n$\\omega_0$ is when coefficient of $j = 0$\n\\begin{align*}\n \\text{Start with: }\\num{e-8}j\\omega&+\\frac{1}{10000} -\\frac{2500}{\\omega}j\\\\\n \\num{e-8}j\\omega &-\\frac{2500}{\\omega}j = 0\\\\\n \\omega^2 &= \\num{2.5e4}\\\\\n \\omega &= \\boxed{\\SI{500}{\\kilo\\radian}}\\\\\n Q_0 &= \\omega_0 R C\\\\\n &\\boxed{= 50}\n\\end{align*}\n\\section*{5}\n\\subsection*{a}\n$H_{dB} = 20 \\log(|\\tilde{H}(j\\omega)|)$\n\\begin{enumerate}\n \\item $20 \\log(0.2) = \\boxed{-13.979}$\n \\item $20 \\log(13) = \\boxed{22.3}$\n \\item $20 \\log(|\\frac{100}{23j +21}|) = \\log(|\\frac{2300 j - 2100}{23^2+21^2}|) = 20 \\log(3.21) = \\boxed{10.13}$\n\\end{enumerate}\n\\subsection*{b}\n$|\\tilde{H}(j\\omega)| = 10^{\\frac{H_{dB}}{20}}$\n\\begin{enumerate}\n \\item $|\\tilde{H}(j\\omega)| = \\boxed{39.8}$\n \\item $|\\tilde{H}(j\\omega)| = \\boxed{0.447}$\n \\item $|\\tilde{H}(j\\omega)| = \\boxed{1.00}$\n\\end{enumerate}\n\\end{multicols}\n\\subsection*{c}\n\\begin{align*}\n \\tilde{H}(j\\omega) &= \\frac{V_{out}}{V_{in}}\\\\\n Z &= 1+ \\frac{1}{j\\omega \\num{100e-6}}\\\\\n \\tilde{H}(j\\omega) &= \\frac{1}{1 + j\\omega\\num{100e-6}}\\\\\n H_{dB} &= 20 \\log(|\\tilde{H}(j\\omega)|) = 20\\log\\left(\\frac{1}{\\num{e-8}\\omega^{2} + 1}\\right)\\\\\n \\angle\\tilde{H}(j\\omega) &= -\\arctan\\left(0.0001\\omega\\right)\n\\end{align*}\n\\begin{figure}\n \\centering\n\\begin{tikzpicture}\n\\begin{axis}[xmode=log,\nxmin=1,\nxmax=1000000,\nymin=-40,\nymax=10,\ngrid=both]\n\\addplot table [smooth, x={omega}, y={result}, col sep=comma, mark=none]{res1.csv};\n\\addplot[gray,ultra thick, opacity=0.6] coordinates {\n(1,0)\n(10000,0)\n(99500, -40)};\n\\end{axis}\n\\end{tikzpicture}\n \\caption{Magnitude bode plot}\n \\label{fig:my_label}\n\\end{figure}\n\\begin{figure}\n \\centering\n\\begin{tikzpicture}\n\\begin{axis}[xmode=log,\nxmin=1,\nxmax=1000000,\nymin=-135,\nymax=45,\nytick={45,0,-45,-90,-135},\ngrid=both]\n\\addplot table [smooth, x={omega}, y={atan}, col sep=comma, mark=none]{res1.csv};\n\\addplot[gray,ultra thick, opacity=0.6] coordinates {\n(1,0)\n(1000,0)\n(100000,-90)\n(1000000, -90)};\n\\end{axis}\n\\end{tikzpicture}\n \\caption{Phase bode plot}\n \\label{fig:label2}\n\\end{figure}\n\\end{document}\n", "meta": {"hexsha": "ad0c0fd93e897790aa2c7348d281c58ca2ef1f61", "size": 6379, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignments/ECE203/ECE203PS5.tex", "max_stars_repo_name": "n30phyte/SchoolDocuments", "max_stars_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignments/ECE203/ECE203PS5.tex", "max_issues_repo_name": "n30phyte/SchoolDocuments", "max_issues_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignments/ECE203/ECE203PS5.tex", "max_forks_repo_name": "n30phyte/SchoolDocuments", "max_forks_repo_head_hexsha": "79652ec7e3345d67e67f0cffe3bea468708622bd", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.3963133641, "max_line_length": 145, "alphanum_fraction": 0.578147045, "num_tokens": 2878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737869342623, "lm_q2_score": 0.7931059438487663, "lm_q1q2_score": 0.6370019043610861}} {"text": "\\section{Algebraic Structures}\n\n\\subsection{COFE}\n\n\\begin{defn}[Chain]\n Given some set $T$ and an indexed family $({\\nequiv{n}} \\subseteq T \\times T)_{n \\in \\mathbb{N}}$ of equivalence relations, a \\emph{chain} is a function $c : \\mathbb{N} \\to T$ such that $\\All n, m. n \\leq m \\Ra c (m) \\nequiv{n} c (n)$.\n\\end{defn}\n\n\\begin{defn}\n A \\emph{complete ordered family of equivalences} (COFE) is a tuple $(T, ({\\nequiv{n}} \\subseteq T \\times T)_{n \\in \\mathbb{N}}, \\lim : \\chain(T) \\to T)$ satisfying\n \\begin{align*}\n \\All n. (\\nequiv{n}) ~& \\text{is an equivalence relation} \\tagH{cofe-equiv} \\\\\n \\All n, m.& n \\geq m \\Ra (\\nequiv{n}) \\subseteq (\\nequiv{m}) \\tagH{cofe-mono} \\\\\n \\All x, y.& x = y \\Lra (\\All n. x \\nequiv{n} y) \\tagH{cofe-limit} \\\\\n \\All n, c.& \\lim(c) \\nequiv{n} c(n+1) \\tagH{cofe-compl}\n \\end{align*}\n\\end{defn}\n\n\\ralf{Copy the explanation from the paper, when that one is more polished.}\n\n\\begin{defn}\n An element $x \\in A$ of a COFE is called \\emph{discrete} if\n \\[ \\All y \\in A. x \\nequiv{0} y \\Ra x = y\\]\n A COFE $A$ is called \\emph{discrete} if all its elements are discrete.\n\\end{defn}\n\n\\begin{defn}\n A function $f : A \\to B$ between two COFEs is \\emph{non-expansive} if\n \\[\\All n, x \\in A, y \\in A. x \\nequiv{n} y \\Ra f(x) \\nequiv{n} f(y) \\]\n It is \\emph{contractive} if\n \\[ \\All n, x \\in A, y \\in A. (\\All m < n. x \\nequiv{m} y) \\Ra f(x) \\nequiv{n} f(x) \\]\n\\end{defn}\n\n\\begin{defn}\n The category $\\COFEs$ consists of COFEs as objects, and non-expansive functions as arrows.\n\\end{defn}\nNote that $\\COFEs$ is cartesian closed.\n\n\\begin{defn}\n A functor $F : \\COFEs \\to COFEs$ is called \\emph{locally non-expansive} if its actions $F_1$ on arrows is itself a non-expansive map.\n Similarly, $F$ is called \\emph{locally contractive} if $F_1$ is a contractive map.\n\\end{defn}\n\n\\subsection{RA}\n\n\\ralf{Define this, including frame-preserving updates.}\n\n\\subsection{CMRA}\n\n\\begin{defn}\n A \\emph{CMRA} is a tuple $(\\monoid, (\\mval_n \\subseteq \\monoid)_{n \\in \\mathbb{N}}, \\mcore{-}: \\monoid \\to \\monoid, (\\mtimes) : \\monoid \\times \\monoid \\to \\monoid, (\\mdiv) : \\monoid \\times \\monoid \\to \\monoid)$ satisfying\n \\begin{align*}\n \\All n, m.& n \\geq m \\Ra V_n \\subseteq V_m \\tagH{cmra-valid-mono} \\\\\n \\All \\melt, \\meltB, \\meltC.& (\\melt \\mtimes \\meltB) \\mtimes \\meltC = \\melt \\mtimes (\\meltB \\mtimes \\meltC) \\tagH{cmra-assoc} \\\\\n \\All \\melt, \\meltB.& \\melt \\mtimes \\meltB = \\meltB \\mtimes \\melt \\tagH{cmra-comm} \\\\\n \\All \\melt.& \\mcore\\melt \\mtimes \\melt = \\melt \\tagH{cmra-core-id} \\\\\n \\All \\melt.& \\mcore{\\mcore\\melt} = \\mcore\\melt \\tagH{cmra-core-idem} \\\\\n \\All \\melt, \\meltB.& \\melt \\leq \\meltB \\Ra \\mcore\\melt \\leq \\mcore\\meltB \\tagH{cmra-core-mono} \\\\\n \\All n, \\melt, \\meltB.& (\\melt \\mtimes \\meltB) \\in \\mval_n \\Ra \\melt \\in \\mval_n \\tagH{cmra-valid-op} \\\\\n \\All \\melt, \\meltB.& \\melt \\leq \\meltB \\Ra \\melt \\mtimes (\\meltB \\mdiv \\melt) = \\meltB \\tagH{cmra-div-op} \\\\\n \\All n, \\melt, \\meltB_1, \\meltB_2.& \\omit\\rlap{$\\melt \\in \\mval_n \\land \\melt \\nequiv{n} \\meltB_1 \\mtimes \\meltB_2 \\Ra {}$} \\\\\n &\\Exists \\meltC_1, \\meltC_2. \\melt = \\meltC_1 \\mtimes \\meltC_2 \\land \\meltC_1 \\nequiv{n} \\meltB_1 \\land \\meltC_2 \\nequiv{n} \\meltB_2 \\tagH{cmra-extend} \\\\\n \\text{where}\\qquad\\qquad\\\\\n \\melt \\leq \\meltB \\eqdef{}& \\Exists \\meltC. \\meltB = \\melt \\mtimes \\meltC \\tagH{cmra-incl}\n \\end{align*}\n\\end{defn}\n\nNote that every RA is a CMRA, by picking the discrete COFE for the equivalence relation.\n\n\\ralf{Copy the rest of the explanation from the paper, when that one is more polished.}\n\n\\paragraph{The division operator $\\mdiv$.}\nOne way to describe $\\mdiv$ is to say that it extracts the witness from the extension order: If $\\melt \\leq \\meltB$, then $\\melt \\mdiv \\meltB$ computes the difference between the two elements (\\ruleref{cmra-div-op}).\nOtherwise, $\\mdiv$ can have arbitrary behavior.\nThis means that, in classical logic, the division operator can be defined for any PCM using the axiom of choice, and it will trivially satisfy \\ruleref{cmra-div-op}.\nHowever, notice that the division operator also has to be \\emph{non-expansive} --- so if the carrier $\\monoid$ is equipped with a non-trivial $\\nequiv{n}$, there is an additional proof obligation here.\nThis is crucial, for the following reason:\nConsidering that the extension order is defined using \\emph{equality}, there is a natural notion of a \\emph{step-indexed extension} order using the step-indexed equivalence of the underlying COFE:\n\\[ \\melt \\mincl{n} \\meltB \\eqdef \\Exists \\meltC. \\meltB \\nequiv{n} \\melt \\mtimes \\meltC \\tagH{cmra-inclM} \\]\nOne of the properties we would expect to hold is the usual correspondence between a step-indexed predicate and its non-step-indexed counterpart:\n\\[ \\All \\melt, \\meltB. \\melt \\leq \\meltB \\Lra (\\All n. \\melt \\mincl{n} \\meltB) \\tagH{cmra-incl-limit} \\]\nThe right-to-left direction here is trick.\nFor every $n$, we obtain a proof that $\\melt \\mincl{n} \\meltB$.\nFrom this, we could extract a sequence of witnesses $(\\meltC_m)_{m}$, and we need to arrive at a single witness $\\meltC$ showing that $\\melt \\leq \\meltB$.\nWithout the division operator, there is no reason to believe that such a witness exists.\nHowever, since we can use the division operator, and since we know that this operator is \\emph{non-expansive}, we can pick $\\meltC \\eqdef \\meltB \\mdiv \\melt$, and then we can prove that this is indeed the desired witness.\n\\ralf{Do we actually need this property anywhere?}\n\n\\paragraph{The extension axiom (\\ruleref{cmra-extend}).}\nNotice that the existential quantification in this axiom is \\emph{constructive}, \\ie it is a sigma type in Coq.\nThe purpose of this axiom is to compute $\\melt_1$, $\\melt_2$ completing the following square:\n\n\\ralf{Needs some magic to fix the baseline of the $\\nequiv{n}$, or so}\n\\begin{center}\n\\begin{tikzpicture}[every edge/.style={draw=none}]\n \\node (a) at (0, 0) {$\\melt$};\n \\node (b) at (1.7, 0) {$\\meltB$};\n \\node (b12) at (1.7, -1) {$\\meltB_1 \\mtimes \\meltB_2$};\n \\node (a12) at (0, -1) {$\\melt_1 \\mtimes \\melt_2$};\n\n \\path (a) edge node {$\\nequiv{n}$} (b);\n \\path (a12) edge node {$\\nequiv{n}$} (b12);\n \\path (a) edge node [rotate=90] {$=$} (a12);\n \\path (b) edge node [rotate=90] {$=$} (b12);\n\\end{tikzpicture}\\end{center}\nwhere the $n$-equivalence at the bottom is meant to apply to the pairs of elements, \\ie we demand $\\melt_1 \\nequiv{n} \\meltB_1$ and $\\melt_2 \\nequiv{n} \\meltB_2$.\nIn other words, extension carries the decomposition of $\\meltB$ into $\\meltB_1$ and $\\meltB_2$ over the $n$-equivalence of $\\melt$ and $\\meltB$, and yields a corresponding decomposition of $\\melt$ into $\\melt_1$ and $\\melt_2$.\nThis operation is needed to prove that $\\later$ commutes with existential quantification and separating conjunction:\n\\begin{mathpar}\n \\axiom{\\later(\\Exists\\var:\\type. \\prop) \\Lra \\Exists\\var:\\type. \\later\\prop}\n \\and\\axiom{\\later (\\prop * \\propB) \\Lra \\later\\prop * \\later\\propB}\n\\end{mathpar}\n(This assumes that the type $\\type$ is non-empty.)\n\n\\begin{defn}\n An element $\\munit$ of a CMRA $\\monoid$ is called the \\emph{unit} of $\\monoid$ if it satisfies the following conditions:\n \\begin{enumerate}[itemsep=0pt]\n \\item $\\munit$ is valid: \\\\ $\\All n, \\munit \\in \\mval_n$\n \\item $\\munit$ is left-identity of the operation: \\\\\n $\\All \\melt \\in M. \\munit \\mtimes \\melt = \\melt$\n \\item $\\munit$ is discrete\n \\end{enumerate}\n\\end{defn}\n\n\\begin{defn}\n It is possible to do a \\emph{frame-preserving update} from $\\melt \\in \\monoid$ to $\\meltsB \\subseteq \\monoid$, written $\\melt \\mupd \\meltsB$, if\n \\[ \\All n, \\melt_f. \\melt \\mtimes \\melt_f \\in \\mval_n \\Ra \\Exists \\meltB \\in \\meltsB. \\meltB \\mtimes \\melt_f \\in \\mval_n \\]\n\n We further define $\\melt \\mupd \\meltB \\eqdef \\melt \\mupd \\set\\meltB$.\n\\end{defn}\nNote that for RAs, this and the RA-based definition of a frame-preserving update coincide.\n\n\\ralf{Describe discrete CMRAs, and how they correspond to RAs.}\n\n\\begin{defn}\n A function $f : M \\to N$ between two CMRAs is \\emph{monotone} if it satisfies the following conditions:\n \\begin{enumerate}[itemsep=0pt]\n \\item $f$ is non-expansive\n \\item $f$ preserves validity: \\\\\n $\\All n, x \\in M. x \\in \\mval_n \\Ra f(x) \\in \\mval_n$\n \\item $f$ preserves CMRA inclusion:\\\\\n $\\All x, y. x \\mincl y \\Ra f(x) \\mincl f(y)$\n \\end{enumerate}\n\\end{defn}\n\n\\begin{defn}\n The category $\\CMRAs$ consists of CMRAs as objects, and monotone functions as arrows.\n\\end{defn}\nNote that $\\CMRAs$ is a subcategory of $\\COFEs$.\nThe notion of a locally non-expansive (or contractive) functor naturally generalizes to functors between these categories.\n\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: \"iris\"\n%%% End: \n", "meta": {"hexsha": "63cb6f8189feb8f5c0f8b216e41b584edb4db197", "size": 8662, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/algebra.tex", "max_stars_repo_name": 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YES\n2. YES", "lm_q1_score": 0.8902942232112239, "lm_q2_score": 0.7154239897159438, "lm_q1q2_score": 0.6369378451908307}} {"text": "\r\n\\input{SingleAssignmentSetup.tex}\r\n\\input{../WeekTitles.tex}\r\n\\begin{document}\r\n\r\n\r\n\\begin{center}\r\n\\subsection*{MNTC P01 - Week \\#4 - \\WeekTitleFour}\r\n\\end{center}\r\n\r\n\\begin{enumerate}[1.]\r\n\\subsection*{Distance And Velocity}\r\n\r\n\\begin{multicols}{2}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The graph below shows the velocity, $v$, of an object (in\r\n meters/sec). Estimate the total distance the object traveled\r\n between $t = 0$ and $t = 6$.\r\n\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week04_DistanceAndVelocity/Velocity_1}\r\n \r\n \\end{Question}\r\n \\begin{Solution}\r\n Just counting the squares (each of which has area representing 10\r\n (m/s)$\\cdot$(s) = 10 m of distance), and allowing for the partial\r\n squares, we can see that the area under the curve from $t=0$ to\r\n $t=6$ is between 140 and 150 units. Therefore the distance\r\n traveled is between 140 and 150 meters.\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The figure below shows the velocity of a particle, in cm/sec,\r\n along the $t$-axis for $-3 \\le t \\le 3$ ($t$ in seconds). \r\n \\begin{enumerate}\r\n \\item Describe the motion in words. Is the particle changing\r\n direction or always moving in the same direction? Is the\r\n particle speeding up or slowing down?\r\n \\item Make over- and underestimates of the distance traveled for \r\n $-3 \\le t \\le 3$.\r\n \\end{enumerate}\r\n\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week04_DistanceAndVelocity/Velocity_2}\r\n \\end{Question}\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item The velocity is always positive, so the particle is moving\r\n in the same direction throughout. However, the particle is\r\n speeding up until shortly before $t=0$, and slowing down\r\n thereafter.\r\n \\item The distance traveled is represented by the area under the\r\n curve. Using whole grid squares, we can overestimate the area\r\n as 3+3+3+3+2+1 = 15 squares, and we can underestimate the area\r\n as 1+2+2+1+0+0= 6 squares. Each square represents 1\r\n (cm/sec)$\\cdot$(s) = 1 cm, so the particle moved in one\r\n direction between 6 and 15 cm.\r\n\r\n \\end{enumerate}\r\n \r\n \\end{Solution}\r\n\r\n\r\n%*****************\r\n \\begin{Question}\r\nFor questions \\ref{q:graph1_start} to \\ref{q:graph1_end}, the graph\r\n shows the velocity, in cm/sec, of a particle moving along the\r\n $x$-axis. Compute the particle's change in position, left (negative)\r\n or right (positive), between times $t = 0$ and $t = 5$ seconds.\r\n\\end{Question}\r\n\r\n\\item ~\\\\\\label{q:graph1_start} \r\n \\begin{Question}\r\n\r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_DistanceAndVelocity/ParticleVelocity1}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n The velocity is constant and negative, so the change in position is $-3 \\cdot 5$ cm, that is 15 cm to the left.\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item ~\\\\\r\n \\begin{Question}\r\n \r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_DistanceAndVelocity/ParticleVelocity2}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n From $t = 0$ to $t = 3$, the velocity is constant and positive, so\r\n the change in position is $2 \\cdot 3$ cm, that is 6 cm to the right. \\\\\r\n From $t = 3$ to $t = 5$, the velocity is negative and constant, so\r\n the\r\n change in position is $3 \\cdot 2$ cm, that is 6 cm to the left. \\\\\r\n Thus the total change in position is 0. The particle moves 6 cm to\r\n the right, followed by 6 cm to the left, and returns to where it\r\n started.\r\n \\end{Solution}\r\n\r\n%******************\r\n\\item ~ \\\\\r\n \\begin{Question}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_DistanceAndVelocity/ParticleVelocity3}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n From $t = 0$ to $t = 5$ the velocity is positive so the change in\r\n position is to the right. The area under the velocity graph gives\r\n the distance traveled. The region is a triangle, and so has area\r\n $(1/2)bh = (1/2)5 \\cdot 10 = 25$. Thus the change in position is 25 cm\r\n to the right.\r\n \\end{Solution}\r\n\r\n%******************\r\n\\item ~ \\\\\\label{q:graph1_end}\r\n \\begin{Question}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_DistanceAndVelocity/ParticleVelocity4}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n From $t = 0$ to $t = 4$ the velocity is positive so the change in\r\n position is to the right. The area under the velocity graph gives\r\n the distance traveled. The region is a triangle, and so has area\r\n $(1/2)bh = (1/2)4\\cdot 8 = 16$. Thus the change in position is 16 cm\r\n to the right for$ t = 0$ to $t = 4$. \r\n\r\n From $t = 4$ to $t = 5$, the velocity is negative so the change in\r\n position is to the left. The distance traveled to the left is\r\n given by the area of the triangle, $(1/2)bh = (1/2)1\\cdot 2 =\r\n 1$. \r\n\r\n Thus the total change in position is $16-1= 15$ cm to the\r\n right.\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n\\begin{Question}\r\n A car going 80 ft/s ( about 90 km/h) brakes to a stop in five seconds. Assume the deceleration is constant. \r\n \\begin{enumerate}\r\n \\item Graph the velocity against time, $t$, for $0 \\le t \\le 5$ seconds. \r\n \\item Represent, as an area on the graph, the total distance traveled\r\n from the time the brakes are applied until the car comes to a stop.\r\n \\item Find this area and hence the distance traveled. \r\n% \\item Now find the total distance traveled using antidifferentiation.\r\n \\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}\r\n \\item ~\\\\\r\n\\includegraphics*[width=0.8\\linewidth]{graphics/Week04_DistanceAndVelocity/graph_car_decel}\r\n\\item The total distance is represented by the shaded region $A$, the area under the graph of $v(t)$.\r\n\\item $A$ is a triangle, so its area is \r\n$$A = \\frac{1}{2} \\mbox{(base)(height)} = \\frac{1}{2} (5 \\mbox{ sec})(80 \\mbox{ ft/s}) = 200 \\mbox{ ft}$$\r\n%\\item \r\n \\end{enumerate}\r\n \r\n \\end{Solution}\r\n\r\n\r\n%******************\r\n\\item\r\n \\begin{Question}\r\n A baseball thrown directly upward at 96 ft/sec has velocity $v(t) = 96 - 32t$ ft/ sec at time $t$ seconds.\r\n \\begin{enumerate}[(a)]\r\n\\item Graph the velocity from $t = 0$ to $t = 6$. \r\n\\item When does the baseball reach the peak of its flight? How high does it go? \r\n\\item How high is the baseball at time $t = 5$?\r\n \\end{enumerate}\r\n \\end{Question}\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item See the figure below.\r\n\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week04_DistanceAndVelocity/ChangeInPosition_2}\r\n\r\n\\item The peak of the flight is when the velocity is 0, namely $t =\r\n 3$. The height at $t = 3$ is given by the area under the graph of\r\n the velocity from $t = 0$ to $t = 3$; see the figure above. The\r\n region is a triangle of base 3 seconds and altitude 96 ft/sec, so\r\n the height is $(1/2)3 \\cdot 96 = 144$ feet.\r\n\\item The velocity is negative from $t = 3$ to $t = 5$, so the motion\r\n is downward then. The distance traveled downward can be calculated\r\n by the area of the triangular region which has base of 2 seconds and\r\n altitude of -64 ft/sec. Thus, the baseball travels $(1/2)2 \\cdot 64\r\n = 64$ feet downward from its peak height of 144 feet at $t =\r\n 3$. Thus, the height at time $t = 5$ is the total change in\r\n position, $144 - 64 = 80$ feet.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n%******************\r\n\\item\r\n \\begin{Question}\r\n Two cars start at the same time and travel in the same direction\r\n along a straight road. The graph below gives the velocity, $v$, of\r\n each car as a function of time, $t$. Which car:\r\n \\begin{enumerate}[(a)]\r\n \\item Attains the larger maximum velocity? \r\n \\item Stops first? \r\n \\item Travels farther?\r\n \\end{enumerate}\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week04_DistanceAndVelocity/CarVelocity1}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Car A has the largest maximum velocity because the peak of\r\n car A's velocity curve is higher than the peak of B's.\r\n \\item Car A stops first because the curve representing its\r\n velocity hits zero (on the $t$-axis) first.\r\n \\item Car B travels farther because the {\\bf area} under car B's\r\n velocity curve is the larger, and area under the velocity graph\r\n reprsents distance.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n%******************\r\n\\item\r\n \\begin{Question}\r\n Two cars travel in the same direction along a straight road. The graph below shows the velocity, $v$, of each car at time $t$. Car B starts 2 hours after car A and car B reaches a maximum velocity of 50 km/ hr. \r\n \\begin{enumerate}[(a)]\r\n \\item For approximately how long does each car travel? \\item\r\n Estimate car A's maximum velocity. \\item Approximately how far\r\n does each car travel?\r\n \\end{enumerate}\r\n\\includegraphics[width=0.7\\linewidth]{graphics/Week04_DistanceAndVelocity/CarVelocity2}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Since car B starts at $t = 2$, the tick marks on the\r\n horizontal axis (which we assume are equally spaced) are 2 hours\r\n apart. Thus car B stops at $t = 6$ and travels for 4 hours. Car\r\n A starts at $t = 0$ and stops at $t = 8$, so it travels for 8\r\n hours.\r\n \\item Car A's maximum velocity is approximately twice that of car B (50 km/h), so A's max velocity is 100 km/hr.\r\n \\item The distance traveled is given by the area of under the\r\n velocity graph. Using the formula for the area of a triangle,\r\n the distances are given by\r\n\\begin{align*}\r\n\\mbox{Car A travels} &=\r\n\\frac{1}{2} \\cdot \\mbox{Base }\\cdot \\mbox{Height} \\\\\r\n& = \\frac{1}{2} \\cdot 8 \\cdot 100 = 400 \\mbox{ km} \\\\\r\n\\mbox{Car B travels} &=\r\n\\frac{1}{2} \\cdot \\mbox{Base }\\cdot \\mbox{Height}\\\\\r\n& = \\frac{1}{2} \\cdot 4 \\cdot 50 = 100 \\mbox{ km}. \\\\\r\n\\end{align*}\r\n \r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n\\end{multicols}\r\n\\hrulefill\r\n\\subsection*{The Definite Integral}\r\n\r\n\\begin{multicols}{2}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The figure below shows a Riemann sum approximation with $n$ subdivisions to $\\ds \\int_a^bf(x) dx$.\r\n \\begin{enumerate}\r\n \\item Is it a left- or right-hand approximation? Would the other one be larger or smaller? \r\n \\item What are $a$, $b$, $n$ and $\\D x$?\r\n \\end{enumerate}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral1}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Left-hand sum. Right-hand sum would be smaller. \r\n \\item We have $a = 0$, $b = 2$, $n = 6$, $\\Delta x = \\frac{2}{6} = \\frac{1}{3}$.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Using the figure below, draw rectangles representing each of the\r\n following Riemann sums for the function $f$ on the interval $0 \\le t \\le\r\n 8$. Calculate the value of each sum. \r\n\r\n\\includegraphics[width=0.5\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral2}\r\n\\begin{enumerate}[(a)]\r\n\\item Left-hand sum with $\\D t = 4.$\r\n\\item Left-hand sum with $\\D t = 2$.\r\n\\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n\\begin{tabular}{cc}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral2_solutions1} & \r\n\\includegraphics[width=0.4\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral2_solutions3} \\\\\r\n(a) & (b) \\\\ \r\n\\end{tabular}\r\n\r\n(a) The left-hand sum with $n=2$ intervals, or $\\D t = 4$: $32 \\cdot 4 + 24 \\cdot 4$ = 224. \\\\\r\n(b) Left-hand sum with $n=4$ intervals, or $\\D t = 2$: $32 \\cdot 2 + 30 \\cdot 2 + 24 \\cdot 2 + 14 \\cdot 2 = 200$. \\\\\r\n\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The graph of a function $f(t)$ is given in the figure below.\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral3}\r\n\r\nWhich of the following four numbers could be an estimate of $\\ds \\int_0^1f(t) ~dt$, accurate to two decimal places? Explain how you chose your answer. \\\\\r\n%\\begin{tabular}{ll}\r\n(a) -98.35 (b) 71.84 \r\n(c) 100.12 (d) 93.47\r\n%\\end{tabular}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n The graph given shows that $f$ is positive for $0 \\le t \\le\r\n 1$, so the integral value must be positive: the answer cannot be -98.35. \r\n\r\n Since the graph is contained within a rectangle of height 100 and\r\n length 1, the answer 100.12 is too large.\r\n\r\n The graph of $f$ is well above the horizontal line y = 80 for\r\n $0\\le t \\le 0.95$, so the integral is likely much higher then\r\n 71.84 $<$ 80, so out of the choices given the best estimate\r\n is 93.47: answer (d).\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n\\begin{enumerate}[(a)]\r\n\\item What is the area between the graph of $f(x)$ shown below and\r\n the $x$-axis, between $x = 0$ and $x = 5$?\r\n\\item What is $\\ds \\int_0^5 f(x)~dx$?\r\n\\end{enumerate}\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral4}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item The total area between $f(x)$ and the $x$-axis is the sum of\r\n the two given areas, so area = 7 + 6 = 13.\r\n\\item To find the integral, we note that from $x = 3$ to $x = 5$, the function lies below the $x$-axis, and hence makes a negative\r\ncontribution to the integral. So\r\n$$\\ds \\int_0^5 f(x)~dx = \\int_0^3 f(x)~dx + \\int_3^5 f(x)dx = 7 - 6 = 1$$\r\n \\end{enumerate}\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n \\begin{enumerate}[(a)]\r\n \\item On a sketch of $y =\\ln(x)$, represent the left Riemann sum\r\n with $n = 2$ approximating $\\ds \\int_1^2 \\ln(x)~dx$. Write out\r\n the terms in the sum, but do not evaluate it.\r\n \\item Will this LEFT(4) estimate be an overestimate or an\r\n underestimate of $\\ds \\int_1^2 \\ln(x)~dx$?\r\n \\end{enumerate}\r\n \\end{Question}\r\n \\begin{enumerate}[(a)]\r\n \\item Below are the left and right sums respectively. \\\\\r\n\\includegraphics[width=0.9\\linewidth]{graphics/Week04_TheDefiniteIntegral/Ln_solutions} \\\\\r\n\\begin{align*}\r\n\\mbox{ Left sum} & = f(1) \\Delta x + f(1.5) \\Delta x \\\\\r\n& = \\underbrace{(\\ln1)}_{=0}0.5 + \\ln(1.5)0.5 = (\\ln 1.5)0.5\r\n\\end{align*}\r\n\\item Because the $y$ values on the left end of each interval are the\r\n smallest on that interval, the sum built on them will be an\r\n underestimate of the exact integral value.\r\n \\end{enumerate}\r\n \\begin{Solution}\r\n \r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Estimate $\\ds \\int_1^2 x^2~ dx$ using left-hand sum with four\r\n subdivisions. \r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n Left-hand sum gives: \\\\$1^2(1/4) + (1.25)^2(1/4) + (1.5)^2(1/4) + (1.75)^2(1/4) = 1.96875.$ \r\n\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Without computation, decide if\r\n $\\ds \\int_0^{2\\pi} e^{-x} \\sin x~dx$ is positive or\r\n negative. [Hint: use MATLAB to draw $e^{-x} \\sin(x)$ over the\r\n given interval.]\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/ExpSin_solutions} \\\\\r\n\r\nLooking at the graph of $y = e^{-x} \\sin x$ shown above for\r\n$0 \\le x \\le 2\\pi$, we see that the area, $A_1$, which contributes a\r\npositive amount to the integral\r\n$\\ds \\int_0^{2 \\pi} e^{-x} \\sin(x)~dx$, is much larger than the area\r\n$A_2$, which contributes a negative amount to the integral.\r\n\r\nSince the positive integral contributions are larger than the\r\nnegative, the overall integral $\\ds \\int_0^{\\pi} e^{-x} \\sin(x)~dx$ will be {\\bf positive}.\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n \\begin{enumerate}[(a)]\r\n \\item Graph $\\ds f(x) = \\begin{cases} 1-x & \\text{ if $0 \\le x\\le 1$} \\\\\r\nx-1 & \\text{ if $1 < x\\le 2$} \\\\\r\n\\end{cases}$\r\n\\item Find the {\\em exact} value of $\\ds \\int_0^2 f(x)~ dx$ (hint: sketch and see what shapes you get). \r\n\\item Calculate the 4-term left Riemann sum approximation to the\r\n definite integral. How does the approximation compare to the exact\r\n value?\r\n \\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item ~\\\\\r\n\\includegraphics[width=0.9\\linewidth]{graphics/Week04_TheDefiniteIntegral/Piecewise_solutions} \\\\ \r\n\\item The area is made up of two triangles, with total area of 2 $\\times$ (1/2)(1)(1) = 1 square unit. \r\n\\item Using $\\D x = 1/2$ in the 4-term Riemann sum shown in right-side\r\n graph above, we have \r\n\\begin{align*}\r\n& \\mbox{Left hand sum} \\\\\r\n&= f(0)\\D x + f(0.5)\\D x + f(1)\\D x + f(1.5)\\D x \\\\\r\n& = 1 \\left(\\frac{1}{2} \\right) + \\frac{1}{2} \\left(\\frac{1}{2} \\right) + 0 \\left(\\frac{1}{2} \\right) + \\frac{1}{2} \\left(\\frac{1}{2} \\right) \\\\\r\n& = 1.\r\n\\end{align*} We notice that in this case the approximation is exactly\r\nequal to the exact value of the integral. This is mostly coincidence\r\ndue to the simple shape of $f(x)$. In general, approximations will\r\nnot work out to be exactly the same as the value of the integral.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\nUsing the figure below, find the values of \r\n\r\n\\begin{tabular}{ll}\r\n(a) $\\ds \\int_a^b f(x)~dx$ & (b) $\\ds \\int_b^c f(x)~dx$ \\\\\r\n(c) $\\ds \\int_a^c f(x)~dx$ & (d) $\\ds \\int_a^c ~|f(x)|~dx$\r\n\\end{tabular}\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral5}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item The area between the graph of $f(x)$ and the $x$-axis\r\n between $x = a$ and $x = b$ is 13, so $\\ds \\int_a^bf(x)~dx = 13$.\r\n \\item Since the graph of $f(x)$ is below the x-axis for $b < x <\r\n c$, $\\ds \\int_b^c f(x)~ dx = -2$.\r\n\\item Since the graph of $f(x)$ is above the $x$-axis for $a < x < b$ and below for $b < x < c$,\r\n$\\ds \\int_a^c f(x)~dx = 13 - 2 = 11$.\r\n\\item The graph of $|f(x)|$ is the same as the graph of $f(x)$, except\r\n that the part {\\em below} the $x$-axis is reflected to be {\\em\r\n above} it (see graph of $|f(x)|$ below). Thus $\\ds \\int_a^c\r\n |f(x)|~ dx = 13 + 2 = 15$.\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/AbsValue_solutions}\r\n \\end{enumerate}\r\n \\end{Solution}\r\n%*****************\r\n\\item \r\n \\begin{Question}\r\n Given the figure below, and the statement that $\\ds \\int_{-2}^0 f(x)~dx = 4$, estimate \r\n\r\n\\begin{tabular}{ll}\r\n(a) $\\ds \\int_0^2 f(x)~dx$ & (b) $\\ds \\int_{-2}^2 f(x)~dx$ \\\\\r\n(c) The total shaded area.\r\n\\end{tabular}\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral6}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n The region shaded between $x = 0$ and $x = 2$ appears to have\r\n approximately the same area as the region shaded between $x = -2$\r\n and $x = 0$, but it lies below the axis. Since $\\ds \\int_{-2}^0 f(x)~dx = 4$, we\r\n have the following results: \r\n \\begin{enumerate}\r\n \\item $\\ds \\int_0^2f(x)~dx \\approx \\mbox{ negative of }\\int_{-2}^0 f(x)~dx =-4$\r\n \\item $\\ds \\int_{-2}^2 f(x)~ dx \\approx 4 - 4 = 0$. \r\n \\item The total area shaded is approximately 4 + 4 = 8.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n \\begin{enumerate}[(a)]\r\n \\item Using the graph below, find $\\ds \\int_{-3}^0 f(x)~dx$. \r\n\\item If the area of the shaded region is $A$, estimate $\\ds \\int_{-3}^4 f(x)~ dx$.\r\n \\end{enumerate}\r\n\\includegraphics[width=0.9\\linewidth]{graphics/Week04_TheDefiniteIntegral/DefiniteIntegral7}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}\r\n \\item $\\ds \\int_{-3}^0 f(x)~dx = -2$: counting squares or\r\n computing areas of the triangles and rectangles in this region.\r\n \\item We break the integral over the interval $x=-3\\ldots 5$ into\r\n pieces, each of which we can find the areas of.\r\n\\begin{align*}\r\n \\int_{-3}^4 f(x)~dx & = \\int_{-3}^0 f(x)~dx + \\int_{0}^3 f(x)~dx + \\int_{3}^4 f(x)~dx \\\\\r\n& = -2 + 2 - \\frac{A}{2} \\\\\r\n&= \\frac{-A}{2}\r\n\\end{align*}\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Find the exact value of $\\ds \\int_0^{2\\pi} \\sin \\theta d\\theta$\r\n without calculation (i.e. from a sketch).\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_TheDefiniteIntegral/Sine_solutions}\r\n\r\nFrom the graph of sine above, and the fact that the integral includes\r\na complete $[0, 2\\pi]$ cycle, it is clear from symmetry that the\r\npositive and negative contributions to the integral must be equal, so\r\n$$\\int_0^{2 \\pi} \\sin(\\theta)~d\\theta = 0$$\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The width, in feet, at various points along the fairway of a hole\r\n on a golf course is given in the figure below. If one pound of\r\n fertilizer covers 200 square feet, use a LEFT estimate the amount\r\n of fertilizer needed to fertilize the fairway.\r\n\r\n\\includegraphics[width=0.9\\linewidth]{graphics/Week04_TheDefiniteIntegral/GolfHole}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n We approximate the area of the playing field by using Riemann\r\n sums. On each 100 m horizontal interval, we use the width printed at the left end of the\r\ninterval as the other dimension.\r\n\\begin{align*}\r\n \\mbox{LEFT}(10) = & (0)(100) + 80(100) + \\\\\r\n&85(100) +95(100) + \\\\\r\n&110(100) + 105(100) + \\\\\r\n&100(100) + 100(100) + \\\\\r\n&105(100) + 110(100) \\\\\r\n= & 89,000 \\mbox{ sq feet} \r\n\\end{align*}\r\n\r\n Thus approximately\r\n $\\ds \\frac{89,000 \\mbox{ sq. ft.}}{200 \\mbox{ sq. ft./lb}} = 445$\r\n lbs. of fertilizer should be necessary.\r\n\r\n \\end{Solution}\r\n\r\n\\end{multicols}\r\n\r\n\\hrulefill\r\n\r\n\\subsection*{Definite Integrals in Modeling}\r\n\r\n\\begin{multicols}{2}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The rate at which the world's oil is being consumed is\r\n continuously increasing. Suppose the rate of oil consumption (in\r\n billions of barrels per year) is given by the function $r = f(t)$,\r\n where $t$ is measured in years and $t = 0$ is the start of 2004. \r\n \\begin{enumerate}[(a)]\r\n \\item Write a definite integral which represents the total\r\n quantity of oil used between the start of 2004 and the start of\r\n 2009.\r\n \\item Suppose $r = 32e^{0.05t}$. Using a left-hand sum with five\r\n subdivisions, find an approximate value for the total quantity of\r\n oil used between the start of 2004 and the start of 2009. \r\n \\item Interpret each of the five terms in the sum from part (b) in\r\n terms of oil consumption.\r\n \\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Quantity used = $\\ds \\int^5_0 f(t)~dt$.\r\n\\item Using a left sum, our approximation is\r\n\\begin{align*}&32e^{0.05(0)}+ 32e^{0.05(1)}+ 32e^{0.05(2)}+ 32e^{0.05(3)}+ 32e^{0.05(4)} \\\\\r\n&= 177.270.\r\n\\end{align*}\r\nSince $f$ is an increasing function, the left endpoint values of $f$\r\nare lower than anywhere else on the subintervals, so this represents\r\nan underestimate.\r\n\\item Each term is a lower estimate of one year's consumption of oil. \r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n As coal deposits are depleted, it becomes necessary to strip-mine\r\n larger areas for each ton of coal. The graph below shows the\r\n number of acres of land per million tons of coal that will be\r\n defaced during strip-mining as a function of the number of million\r\n tons removed, starting from the present day.\r\n \\begin{enumerate}[(a)]\r\n \\item Estimate the total number of acres defaced in extracting the\r\n next 4 million tons of coal (measured from the present\r\n day). Draw four rectangles under the curve, and compute their\r\n area.\r\n \\item Re-estimate the number of acres defaced using rectangles\r\n above the curve.\r\n \\item Combine your answers to parts (a) and (b) to get a better\r\n estimate of the actual number of acres defaced.\r\n \\end{enumerate}\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/Coal}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n\\item ~\\\\\r\n\\includegraphics[width=1.0\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/Coal_solutions} \\\\\r\nUsing rectangles under the curve, we get\r\n$$\\mbox{Acres defaced} \\approx (1)(0.2 + 0.4 + 1 + 2) = 3.6 \\mbox{ acres.}$$\r\n\\item Using rectangles above the curve, we get\r\n$$\\mbox{Acres defaced} \\approx (1)(0.4 + 1 + 2 + 3.5) = 6.9 \\mbox{ acres.}$$\r\n\\item The number of acres defaced is between 3.6 and 6.9, so we\r\n estimate the average, 5.25 acres.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\nCoal gas is produced at a gasworks. Pollutants in the gas are removed by scrubbers, which become less and less efficient as time goes on. The following measurements, made at the start of each month, show the rate at which pollutants are escaping (in tons/ month) in the gas: \r\n\\begin{tabular}{lrrrrrrrr} \\hline\r\nTime (months)&0&1&2&3&4&5&6\\\\ \\hline\r\n Rate pollutants escape&5&7&8&10&13&16&20 \\\\ \r\n(tons/month) & \\\\ \\hline\r\n\\end{tabular} \r\n\\begin{enumerate}[(a)]\r\n\\item Make an overestimate and an underestimate of the total quantity\r\n of pollutants that escape during the {\\bf first month}.\r\n\\item Make an overestimate and an underestimate of the total quantity\r\n of pollutants that escape during the {\\bf six months} shown in the table.\r\n\\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item An overestimate is (7 tons/month)$\\cdot$ (1 month) = 7 tons. An underestimate is (5 tons/month)$\\cdot$(1 month) = 5 tons.\r\n \\item An overestimate is 7+8+10+13+16+20 = 74 tons. An\r\n underestimate is 5+7+8+10+13+16 = 59 tons.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n The graph below shows the rate of change of the quantity of water\r\n in a water tower, in liters per day, during the month of April. If\r\n the tower had 12,000 liters of water in it on April 1, estimate\r\n the quantity of water in the tower on April 30.\r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/WaterTower2}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n Suppose $V(t)$ represents the total quantity of water in the water\r\n tower at time $t$, where $t$ is in days since April 1. Then the\r\n graph shown in the problem is a graph of the volume's rate of\r\n change, or $\\ddt{V}$ or $V'(t)$. By the Fundamental Theorem, the\r\n change in the volume over the 30 days of April is given by:\r\n$$V(30) - V(0) = \\int^{30}_0 V'(t)~dt.$$\r\nThis relationship means we can calculate the change in the volume of\r\nwater by calculating the area under the curve. Each box represents\r\nabout (50 l/d)$\\cdot$(6 d) = 300 liters. \r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/WaterTower2_soln}\r\n\r\nIn the red region, from $t = 0$ to $t = 12$, the graph is below the\r\n$t$ axis, indicating that water is being lost from the tower. There\r\nis a little more than one\r\nbox, but for simplicity we'll call it one square, so roughly $\\approx 300$ liters were lost. \\\\\r\nFrom $t = 12$ to $t = 30$ (blue region), there are between 6 and 7\r\nboxes in total\r\n(let's say 6 for simplicity), or roughly +1800 liters. \\\\\r\nThus, $$\\int^{30}_0 V'(t)~dt \\approx 1800 - 300 = 1500$$ This is the\r\nnet {\\em change} in the volume of water. Solving for $V(30)$, the\r\n{\\em amount} of water on April 30th,\r\n\\begin{align*}\r\nV(30) & = V(0) + \\int^{30}_0 V'(t)~dt \\\\ \r\n& = 12,000 + 1500 = 13,500 \\mbox{ liters.}\r\n\\end{align*}\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n A bicyclist pedals along a straight road with velocity $v$ given\r\n in the graph below. She starts 5 miles from a lake; positive\r\n velocities take her away from the lake and negative velocities\r\n take her toward the lake. When is the cyclist farthest from the\r\n lake, and how far away is she then?\r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/LakeCyclist}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n Notice that the area of a square on the graph represents (10 mph)$\r\n \\cdot$ (1/6) h = (5/3) miles. \r\n\r\n\\includegraphics[width=1.0\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/LakeCyclist_solutions}\r\n\r\nAt $t = 1/3$ hours, $v = 0$. The area between the curve $v$ and the\r\n$t$-axis over the interval $0 \\le t \\le 1/3$ is $\\ds -\\int^{1/3}_0 v\r\n~dt \\approx \\mbox{ one square} = 5/3$ miles. Since $v$ is negative\r\nhere, she is moving toward the lake. Since she starts 5 miles from the\r\nlake at time $t=0$, at $t = 1/3$ she is about 5 - 5/3 = 10/3 miles\r\nfrom the lake (so closer than when she started).\r\n\r\n\r\nFor $1/3 \\le t \\le 1$, $v$ is positive, so she is moving away from the\r\nlake. Her {\\em change} in distance from the lake between the times\r\n$t=0$ and the end point $t=1$ is given by\r\n\\begin{align*}\r\n\\int^1_0 v ~dt & =\r\n\\int^{1/3}_0 v ~dt +\r\n\\int^1_{1/3} v ~dt \\\\\r\n& \\approx -\r\n\\frac{5}{3} + 8 \\cdot\r\n\\frac{5}{3} =\r\n\\frac{35}{ 3} = 11.667 \\mbox{ miles,}\r\n\\end{align*}\r\nRelative to her starting point of 5 miles at $t=0$, the cyclist is\r\nabout 5 + 35/3 = 50/3 = 16.667 miles from the lake at $t=1$. Since,\r\nstarting from the moment $t = 1/3$, she is always moving away from the\r\nlake, the cyclist will be farthest from the lake at the latest\r\npossible time, $t = 1$. The maximal distance at that time equals 50/3\r\n= 16.667 miles.\r\n \\end{Solution}\r\n\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Height velocity graphs are used by endocrinologists (doctors\r\n specializing in the study of hormones) to follow the progress of\r\n children with growth deficiencies. The graph below shows the\r\n height velocity curves of an average boy and an average girl\r\n between ages 3 and 18.\r\n\r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_DefiniteIntegralsInModeling/HeightVelocity}\r\n\\begin{enumerate}[(a)]\r\n\\item Which curve is for girls and which is for boys? Explain how\r\n you can tell.\r\n\\item About how much does the average boy grow between ages 3 and\r\n 10?\r\n \\item The growth spurt associated with adolescence and the onset of\r\n puberty occurs between ages 12 and 15 for the average boy and\r\n between ages 10 and 12.5 for the average girl. Estimate the height\r\n gained by each average child during this growth spurt.\r\n \\item When fully grown, about how much taller is the average man\r\n than the average woman? (The average boy and girl are about the\r\n same height at age 3.)\r\n\\end{enumerate}\r\n\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Since men are generally taller than women, the curve with\r\n the larger area under it is the height velocity of the boys.\r\n The area under each curve represents the change in growth in\r\n centimeters. Therefore, the black curve is for boys, the\r\n lighter grey one for girls.\r\n \\item Each square below the height velocity curve has area (1 cm/yr) $\\cdot$\r\n 1 yr = 1 cm. Counting squares lying below the black curve\r\n gives about 43 cm. Thus, on average, boys grow about 43 cm\r\n between ages 3 and 10.\r\n \\item Counting squares lying below the black curve gives about 23\r\n cm growth for boys during their growth spurt. Counting squares\r\n lying below the colored curve gives about 18 cm for girls during\r\n their growth spurt.\r\n \\item We can measure the difference in growth by counting squares\r\n that lie between the two curves. Between ages 2 and 12.5, the\r\n average girl grows faster than the average boy. Counting squares\r\n yields about 5 cm between the colored and black curves for $2 \\le\r\n x \\le 12.5$. Counting squares between the curves for $12.5 \\le x \\le\r\n 18$ gives about 18 squares. Thus, there is a net increase of\r\n boys over girls by about 18 - 5 = 13 cm.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n\\end{multicols}\r\n\r\n\\hrulefill\r\n\r\n\\subsection*{Properties of Definite Integrals}\r\n\r\n\\begin{multicols}{2}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Without integrating, show that $$2 \\le \\ds \\int_0^2 \\sqrt{1 + x^3}\r\n ~dx \\le 6.$$\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n Notice that $f(x) = \\sqrt{1 + x^3}$ is increasing for $0 \\le x \\le\r\n 2$, since $x^3$ gets bigger as $x$ increases. This means that,\r\n looking at the interval [0, 2] over which we are integrating,\r\n $f(0) \\le f(x) \\le f(2)$. \r\n\r\n\r\nFor this function, $f(0) = \\sqrt{1 +0} = 1$ and $f(2) = \\sqrt{1 + 8} = 3$. Thus, the area under\r\n$f(x)$ lies between the area under the line $y = 1$ and the area under\r\nthe line $y = 3$ on the interval $0 \\le x \\le 2$. See the diagram\r\nbelow. \r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_PropertiesOfDefiniteIntegrals/IntegralBound1_solutions}\r\n\r\nThat is,\r\n$$\\underbrace{1}_h\\underbrace{(2 - 0)}_w \\le \\int^2_0 \\sqrt{ 1 + x^3}~dx \\le \\underbrace{3}_h\\underbrace{(2 - 0)}_w$$\r\n\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Without calculating the integral, explain why the following statements are false.\r\n \\begin{enumerate}[(a)]\r\n \\item $\\ds \\int_{-2}^{-1} e^{x^2} ~dx = -3$ \r\n \\item $\\ds \\int_{-1}^{1} \\left| \\frac{\\cos(x+2)}{1 + \\tan^2x} \\right| ~dx = 0$.\r\n \\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n (a) The integrand is always positive, so the integral (sum of positive $f(x)$ times positive $\\D x$ values) cannot be negative. \\\\\r\n (b) The integrand is always $\\ge 0$ because of the absolute value\r\n sign. If the integral = 0, then with a non-negative integrand, the\r\n integrand must be exactly 0 at every $x$ value on the interval,\r\n which is definitely not the case here.\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n Using the graph of $f(x)$ shown below, arrange the following\r\n quantities in increasing order, from least to greatest.\r\n\r\n\\includegraphics[width=0.6\\linewidth]{graphics/Week04_PropertiesOfDefiniteIntegrals/IntegralRanking}\r\n\r\n\\begin{tabular}{ll}\r\n(i) $\\ds \\int_0^1 f(x)~dx $ & \r\n(ii)$\\ds \\int_1^2 f(x)~dx $ \\\\\r\n(iii)$\\ds \\int_0^2 f (x)~dx $ & \r\n(iv) $\\ds \\int_2^3 f(x)~dx $ \\\\\r\n(v) $-\\ds \\int_1^2 f(x)~dx $ &\r\n(vi) The number 0 \\\\\r\n(vii) The number 20 &\r\n(viii) The number -10\r\n\\end{tabular}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n See the figure below.\r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_PropertiesOfDefiniteIntegrals/IntegralRanking_solutions}\r\n\r\nSince $\\ds \\int^1_0 f(x)~dx = A_1$ and\r\n$\\ds \\int^2_1 f(x)~dx = -A_2$ and\r\n$\\ds \\int^3_2 f(x)~dx = A_3$, we know that\r\n$$0 < \\int^1_0 f(x)~dx < - \\int^2_1 f(x)~dx < \\int^3_2 f(x)~dx.$$\r\nIn addition,\r\n$\\ds \\int^2_0 f(x)~dx = A_1 - A_2$, which is negative, but smaller in magnitude than\r\n$\\ds \\int^2_1 f(x)~dx$. Thus\r\n$$\\int^2_1 f(x)~dx < \\int^2_0 f(x)~dx < 0.$$\r\nThe area $A_3$ lies inside a rectangle of height 20 and base 1, so\r\n$A_3 < 20$. \\\\\r\nThe area $A_2$ lies inside a rectangle below the $x$-axis of\r\nheight 10 and width 1, so $-10 < A_2$. Thus: \r\n\\begin{center}\r\n(viii) $<$ (ii) $<$ (iii) $<$ (vi) $<$ (i) $<$ (v) $<$ (iv) $<$ (vii).\r\n\\end{center}\r\n \\end{Solution}\r\n%*****************\r\n\\item\r\n \\begin{Question}\r\n \\begin{enumerate}[(a)]\r\n \\item Use the graph of $y = x e^{-x^2}$ shown below to explain why $\\ds \\int_{-3}^3 x e^{-x^2} ~dx= 0$.\r\n\\begin{center}\r\n\\includegraphics[width=0.5\\linewidth]{graphics/Week04_PropertiesOfDefiniteIntegrals/xexp_integral}\r\n\\end{center}\r\n\r\n \\item Find the left-hand sum approximation with $n = 3$ to $\\ds\r\n \\int_{0}^3 x e^{-x^2}~dx$. Give your answer to four decimal\r\n places.\r\n \\item Repeat part (b) for $\\ds \\int_{-3}^0 x e^{-x^2}~dx$.\r\n \\item Do your answers to parts (b) and (c) add to 0? Explain.\r\n \\end{enumerate}\r\n \\end{Question}\r\n\r\n \\begin{Solution}\r\n \\begin{enumerate}[(a)]\r\n \\item Since the function is symmetric around the origin, and the\r\n interval used is also symmetric across $x=0$, the areas above\r\n and below the x-axis cancel when we compute the integral. Thus,\r\n$$\\ds\\int^0_{-3} xe^{-x^2} dx = -\r\n\\int^3_0 xe^{-x^2} dx,$$\r\n $$\\mbox{ so } \\ds \\int^3_{-3} xe^{-x^2} dx =\r\n\\underbrace{\\int^0_{-3} xe^{-x^2} dx +\r\n\\int^3_0 xe^{-x^2} dx}_{\\mbox{equal magnitude, opposite signs}} = 0.$$\r\n\\item For $0 \\le x \\le 3$ with $n = 3$, we have $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3$, and $\\D x = 1$. See the figure below. \r\n\r\n\\includegraphics[width=0.8\\linewidth]{graphics/Week04_PropertiesOfDefiniteIntegrals/xexp_integral_solutions}\r\n\r\n\\begin{align*}\r\n\\mbox{Left sum}& = f(x_0)\\D x + f(x_1)\\D x + f(x_2)\\D x \\\\\r\n& = 0e^{-0^2} \\cdot 1 + 1e^{-(1^2)} \\cdot 1 + 2e^{-(2^2)} \\cdot 1 \\\\\r\n& = 0.4045.\r\n\\end{align*}\r\n\\item For $-3 \\le x \\le 0$, with $n = 3$, we have $x_0 = -3, x_1 = -2, x_2 = -1, x_3 = 0$, and $\\D x = 1$. (See the diagram again.)\r\n\\begin{align*}\r\n\\mbox{Left sum} & = f(x_0)\\D x + f(x_1)\\D x + f(x_2)\\D x \\\\\r\n& = -3e^{-(-3)^2}\r\n\\cdot 1 - 2e^{-(-2)^2}\r\n\\cdot 1 - 1e^{-(-1)^2}\r\n\\cdot 1 \\\\\r\n& = -0.4049. \\\\\r\n\\end{align*}\r\n\\item No. The rectangles between $x=-3$ and 0 are not the same size as\r\n those between $x=0$ and 3. See the diagram. There are three\r\n rectangles with nonzero height on [-3, 0] and only two on [0, 3], so\r\n the estimates computed must be different, so we won't get the exact\r\n result of zero using left sums.\r\n \\end{enumerate}\r\n \\end{Solution}\r\n\r\n\\end{multicols}\r\n\\hrulefill\r\n\r\n\\subsection*{Integration By Anti-Derivatives}\r\n\r\n\\begin{Question}\r\n To practice computing anti-derivatives, do as many of the\r\n problems from the following section as you need.\r\n\r\n\r\n These problems are a good start for practicing anti-derivatives or\r\n integration. If you want more though, consult any calculus\r\n textbook.\r\n\\end{Question}\r\n\r\n\\begin{Question}\r\nEvaluate the following integrals.\r\n\\end{Question}\r\n\r\n\\item \\begin{Question}\r\n $\\ds \\int_{-1}^2 (x^3 - 2x) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{-1}^2 (x^3 - 2x) dx = \\left[ \\frac{x^4}{4} - x^2 \r\n \\right]_{-1}^2 = \\left( \\frac{2^4}{4} - 2^2 \\right) - \\left( \\\r\n \\frac{(-1)^4}{4} - (-1)^2 \\right) = (4 - 4) - (\\frac{1}{4} - 1) = 0\r\n - (- \\frac{3}{4}) = \\frac{3}{4}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n $\\ds \\int_{-1}^1 x^{100} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{-1}^1 x^{100} dx = \\left[ \\frac{1}{101} x^{101} \r\n \\right]_{-1}^{1} = \\frac{1}{101} - (- \\frac{1}{101}) = \r\n \\frac{2}{101}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n $\\ds \\int_{1}^4 (5 - 2t +3t^2) dt$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^4 (5 - 2t - 3t^2) dt = [5t - t^2 + t^3]_1^4 = \r\n (20 - 16 + 64) - (5 - 1 + 1) = 68 - 5 = 63$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n $\\ds \\int_0^1 (1 + \\frac{1}{2} u^4 - \\frac{2}{5} u^9) du$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int_0^1 (1 + \\frac{1}{2} u^4 - \\frac{2}{5} u^9) du = \r\n [u + \\frac{1}{10} u^5 - \\frac{1}{25} u^10]_0^1 = \r\n (1 + \\frac{1}{10} - \\frac{1}{25}) - 0 = \\frac{50 + 5 - 2}{50} = \\frac{53}{50}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n $\\ds \\int_1^9 \\sqrt{x} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^9 \\sqrt{x} dx = \\int_1^9 x^{1 / 2} dx = \r\n \\left[ \\frac{x^{3 / 2}}{3 / 2} \\right]_1^9 =\r\n \\frac{2}{3} \\left[ x^{2 / 3} \\right]_1^9 =\r\n \\frac{2}{3} (9^{3 / 2} - 1^{3 / 2}) = \r\n \\frac{2}{3} (27 - 1) = \\frac{52}{3}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^8 x^{-2 / 3} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n \r\n $\\ds \\int_1^8 x^{-2 / 3} dx = \r\n \\left[ \\frac{x^{1 / 3}}{1 / 3} \\right]_1^8 = \r\n 3 \\left[ x^{1 / 3} \\right]_1^8 = \r\n 3(8^{1 / 3} - 1^{1 / 3}) = 3(2 - 1) = 3$\r\n \r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_{\\pi / 6}^{\\pi} \\sin \\theta d \\theta$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{\\pi / 6}^{\\pi} \\sin \\theta d \\theta =\r\n \\left[ - \\cos \\theta \\right]_{\\pi / 6}^{\\pi} =\r\n - \\cos \\pi - ( - \\cos \\frac{\\pi}{6} ) = \r\n -(-1) - (- \\sqrt{3} / 2) = 1 + \\sqrt{3} / 2$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_{-5}^5 e~ dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n The number $e$ is just a constant. \r\n\r\n $\\ds \\int_{-5}^5 e ~dx = [e x]_{-5}^5 = 5e - (-5e) = 10e$\r\n \r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^1 (u + 2)(u - 3) du$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n You need to expand the product before you can integrate. \r\n\r\n $\\ds \\int_0^1 (u + 2)(u - 3) du = \\int_0^1 (u^2 - u - 6) du = \\left[\r\n \\frac{1}{3} u^3 - \\frac{1}{2} u^2 - 6u \\right]_0^1 =\r\n \\left(\\frac{1}{3} - \\frac{1}{2} - 6 \\right) - 0 = - \\frac{37}{6}$\r\n\\end{Solution}\r\n\r\n\r\n \\item \\begin{Question}\r\n $\\ds \\int_0^4 (4 - t) \\sqrt{t} ~dt$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\nTo evaluate this integral, you need to expand the product first; you can't integrate the product $(4-t)\\sqrt{t}$ as it is originally written.\r\n\r\n $\\ds \\int_0^4 (4 - t) \\sqrt{t} dt = \\int_0^4 (4 - t) t^{1 / 2} dt =\r\n \\int_0^4 (4t^{1 / 2} - t^{3 / 2}) dt = \r\n \\left[ \\frac{8}{3} t^{3 / 2} - \\frac{2}{5} t^{5 / 2} \\right]_0^4 =\r\n \\frac{8}{3} (8) - \\frac{2}{5} (32) = \r\n \\frac{320 - 192}{15} = \\frac{128}{15}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^9 \\frac{x - 1}{\\sqrt{x}} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n \r\n $\\ds \\int_1^9 \\frac{x - 1}{\\sqrt{x}} dx =\r\n \\int_1^9 \\left( \\frac{x}{\\sqrt{x}} - \\frac{1}{\\sqrt{x}} \\right) dx\r\n = \\int_1^9 (x^{1 / 2} - x^{-1 / 2}) dx =\r\n \\left[\\frac{2}{3} x^{3 / 2} - 2x^{1 / 2} \\right]_1^9 = \r\n (\\frac{2}{3} \\cdot 27 - 2 \\cdot 3) = (\\frac{2}{3} - 2) = \r\n 12 - (- \\frac{4}{3}) = \\frac{40}{3}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^2 (y - 1)(2y + 1) dy$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n \r\n $\\ds \\int_0^2 (y - 1)(2y + 1) dy = \\int_0^2 (2y^2 - y - 1) dy =\r\n \\left[ \\frac{2}{3} y^3 - \\frac{1}{2} y^2 - y \\right]_0^2 = \r\n ( \\frac{16}{3} - 2 - 2) - 0 = \\frac{4}{3}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^{\\pi / 4} \\sec^2 t ~dt$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n \r\n $\\ds \\int_0^{\\pi / 4} \\sec^2 t ~dt = [\\tan t]_0^{\\pi / 4} =\r\n \\tan \\frac{\\pi}{4} - \\tan 0 = 1 - 0 = 1$\r\n \r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^{\\pi / 4} \\sec \\theta \\tan \\theta d \\theta$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_0^{\\pi / 4} \\sec \\theta \\tan \\theta d \\theta = \r\n [ \\sec \\theta ]_0^{\\pi / 4} = \r\n \\sec \\frac{\\pi}{4} - \\sec 0 = \\sqrt{2} - 1$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^2 (1 + 2y)^2 dy$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^2 (1 + 2y)^2 dy = \r\n \\int_1^2 (1 + 4y + 4y^2) dy =\r\n [y + 2y^2 + \\frac{4}{3} y^3]_1^2 =\r\n (2 + 8 + \\frac{32}{3}) - (1 + 2 + \\frac{4}{3}) = \r\n \\frac{62}{3} - \\frac{13}{3} = \\frac{49}{3}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^3 (2 \\sin x - e^x) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_0^3 (2 \\sin x - e^x) dx = \r\n [-2 \\cos 3 - e^3) - (-2 - 1) =\r\n 3 - 2 \\cos 3 - e^3$\r\n \r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^2 \\frac{v^3 + 3v^6}{v^4} dv$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^2 \\frac{v^3 + 3v^6}{v^4} =\r\n \\int_1^2 \\left( \\frac{1}{v} + 3v^2 \\right) dv =\r\n [ \\ln |v| + v^3]_1^2 = \r\n ( \\ln 2 + 8 ) - ( \\ln 1 + 1) =\r\n \\ln 2 + 7$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^{18} \\sqrt{\\frac{3}{z}} dz$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^{18} \\sqrt{\\frac{3}{z}} dz =\r\n \\int_1^{18} \\sqrt{3} z^{-1/2} dz =\r\n \\sqrt{3} [ 2z^{1/2} ]_1^{18} =\r\n 2 \\sqrt{3} (18^{1/2} - 1^{1/2}) = \r\n 2 \\sqrt{3} (3 \\sqrt{2} - 1)$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_0^1 (x^e + e^x) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n \r\n $\\ds \\int_0^1 (x^e e^x) dx = \r\n \\left[ \\frac{x^{e + 1}}{e + 1} + e^x \\right]_0^1 =\r\n \\left( \\frac{1}{e + 1} + e \\right) - (0 + 1) = \r\n \\frac{1}{e + 1} + e - 1$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_{1 / \\sqrt{3}}^{\\sqrt{3}} \\frac{8}{1 + x^2} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{1 / \\sqrt{3}}^{\\sqrt{3}} \\frac{8}{1 + x^2} dx =\r\n \\left[ 8 \\arctan x \\right]_{1 / \\sqrt{3}}^{\\sqrt{3}} =\r\n 8 \\left( \\frac{\\pi}{3} - \\frac{\\pi}{6} \\right) =\r\n 8 \\left( \\frac{\\pi}{6} \\right) = \\frac{4 \\pi}{3}$\r\n \r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_1^2 \\frac{4 + u^2}{u^3} du$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_1^2 \\frac{4 + u^2}{u^3} du =\r\n \\int_1^2 (4u^{-3} + u^{-1}) du =\r\n \\left[ \\frac{4}{-2} u^{-2} + \\ln |u| \\right]_1^2 =\r\n \\left[ \\frac{-2}{u^2} + \\ln u \\right]_1^2 =\r\n (-\\frac{1}{2} + \\ln 2) - (-2 + \\ln 1) = \r\n \\frac{3}{2} + \\ln 2$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_{-1}^1 e^{u + 1} du$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{-1}^1 e^{u + 1} du = \r\n [e^{u + 1}]_{-1}^1 =\r\n e^2 - e^0 = e^2 - 1$ [or start with $e^{u + 1} = e^u e^1$]\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n $\\ds \\int_{1 / 2}^{1 / \\sqrt{2}} \\frac{4}{\\sqrt{1 - x^2}} dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\int_{1/2}^{1 / \\sqrt{2}} \\frac{4}{\\sqrt{1 - x^2}} dx =\r\n \\left[ 4 \\arcsin x \\right]_{1/2}^{1 / \\sqrt{2}} =\r\n 4 \\left( \\frac{\\pi}{4} - \\frac{\\pi}{6} \\right) =\r\n 4(\\frac{\\pi}{12}) = \\frac{\\pi}{3}$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n$\\ds \\int_0^{\\pi} f(x) dx \\text{ where } f(x) = \r\n \\begin{cases}\r\n \\sin x & \\text{ if } 0 \\leq x < \\pi / 2 \\\\\r\n \\cos x & \\text{ if } \\pi / 2 \\leq x \\leq \\pi \\\\\r\n \\end{cases}\r\n $ \r\n\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\text{If } f(x) = \r\n \\begin{cases}\r\n \\sin x & \\text{ if } 0 \\leq x < \\pi / 2 \\\\\r\n \\cos x & \\text{ if } \\pi / 2 \\leq x \\leq \\pi \\\\\r\n \\end{cases}\r\n $ then\r\n \r\n $\\ds \\int_{0}^{\\pi} f(x) dx = \r\n \\int_0^{\\pi/2} \\sin x dx + \\int_{\\pi/2}^{\\pi} \\cos x dx =\r\n [ - \\cos x]_0^{\\pi/2} + [ \\sin x ]_{\\pi / 2}^{\\pi} =\r\n - \\cos \\frac{\\pi}{2} + \\cos 0 + \\sin \\pi - \\sin \\frac{\\pi}{2} = \r\n -0 + 1 + 0 - 1 = 0$\r\n\\end{Solution}\r\n\r\n\\item \\begin{Question}\r\n \r\n$\\ds \\int_{-2}^2 f(x) dx \\text{ where } f(x) = \r\n \\begin{cases}\r\n 2 & \\text{ if } -2 \\leq x \\leq 0 \\\\\r\n 4 - x^2 & \\text{ if } 0 < x \\leq 2 \\\\\r\n \\end{cases}\r\n $ \r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\ds \\text{If } f(x) =\r\n \\begin{cases}\r\n 2 & \\text{ if } -2 \\leq x \\leq 0 \\\\\r\n 4 - x^2 & \\text{ if } 0 < x \\leq 2 \\\\\r\n \\end{cases}\r\n $ then\r\n \r\n $\\ds \\int_{-2}^{2} f(x) dx =\r\n \\int_{-2}^{0} 2 dx + \\int_0^2 (4 - x^2) dx = \r\n [2x]_{-2}^{0} + [4x - \\frac{1}{3} x^3 ]_0^2 =\r\n [0 - (-4)] + (\\frac{16}{3} - 0) = \\frac{28}{3}$\r\n \r\n\\end{Solution}\r\n\r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n $\\int (x^2 + x^{-2}) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int (x^2 + x^{-2})dx = \\frac{x^3}{3} + \\frac{x^{-1}}{-1} + C =\r\n \\frac{1}{3} x^3 - \\frac{1}{x} + C$\r\n \r\n\\end{Solution}\r\n \r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n$\\int (\\sqrt{x^3} + \\sqrt[3]{x^2}) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int (\\sqrt{x^3} + \\sqrt[3]{x^2}) dx = \r\n \\int (x^{3/2} + x^{2/3})dx =\r\n \\frac{x^{5/2}}{5/2} + \\frac{x^{5/3}}{5/3} + C =\r\n \\frac{2}{5} x^{5/2} + \\frac{3}{5} x^{5/3} + C$\r\n \r\n\\end{Solution}\r\n \r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n $\\int (x^4 - \\frac{1}{2} x^3 + \\frac{1}{4} x - 2) dx$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int (x^4 - \\frac{1}{2} x^3 + \\frac{1}{4} x - 2) dx = \\frac{x^5}{5}\r\n - \\frac{1}{2} \\frac{x^2}{4} + \\frac{1}{4} \\frac{x^2}{2} - 2x + C =\r\n \\frac{1}{5} x^5 - \\frac{1}{8} x^4 + \\frac{1}{8} x^2 - 2x + C$\r\n \r\n\\end{Solution}\r\n \r\n \r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n $\\int (u + 4)(2u + 1) du$\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int (u + 4)(2u + 1) du =\r\n \\int (2u^2 + 9u + 4)du = \r\n 2 \\frac{u^3}{3} + 9 \\frac{u^2}{2} + 4u + C =\r\n \\frac{2}{3} u^3 + \\frac{9}{2} u^2 + 4u + c$\r\n\\end{Solution}\r\n\r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n $\\int v (v^2 + 2)^2 dv$\r\n \r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int v (v^2 + 2)^2 dv = \\int v(v^4 + 4v^2 + 4)dv = \\int (v^5 + 4v^3\r\n + 4v)dv = \\frac{v^6}{6} + 4 \\frac{v^4}{4} + 4 \\frac{v^2}{2} + C =\r\n \\frac{1}{6} v^6 + v^4 + 2 v^2 + C$\r\n \r\n\\end{Solution}\r\n \r\n \r\n%*******************************\r\n\\item\r\n\\begin{Question}\r\n $\\int \\frac{x^3 - 2 \\sqrt{x}}{x} dx$\r\n \r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n $\\int \\frac{x^3 - 2 \\sqrt{x}}{x} dx = \r\n \\int \\left( \\frac{x^3}{x} - \\frac{2x^{1/2}}{x} \\right) dx =\r\n \\int (x^2 - 2x^{-1/2})dx =\r\n \\frac{x^3}{3} - 2 \\frac{x^{1/2}}{1/2} + C =\r\n \\frac{1}{3} x^3 - 4 \\sqrt{x} + C$\r\n\\end{Solution}\r\n\r\n \r\n%*******************************\r\n\\item\r\n \\begin{Question}\r\n $\\int \\left( x^2 + 1 + \\frac{1}{x^2 + 1} \\right) dx$\r\n \\end{Question}\r\n \r\n \\begin{Solution}\r\n $\\int \\left( x^2 + 1 + \\frac{1}{x^2 + 1} \\right) dx =\r\n \\frac{x^3}{3} + x + \\tan^{-1} x + C$\r\n \\end{Solution}\r\n\r\n\r\n\\hrulefill\r\n\r\n\\subsection*{Numerical Integration with MATLAB}\r\n\r\n\\begin{Question}\r\n In Questions \\ref{qmat1}-\\ref{qmat2}, consider the integrals which\r\n were evaluated earlier in the practice problems using anti-derivatives\r\nand the Fundamental Theorem of Calculus.\r\n\r\nFor each integral, in MATLAB, now use the LEFT$(n)$ rule with $n=1000$\r\nto approximate the value of the integral; this does {\\bf not} require\r\nusing anti-derivatives.\r\n\\end{Question}\r\n\\item \\begin{Question} \\label{qmat1}\r\n $\\ds \\int_{-1}^2 (x^3 - 2x) dx$, which equals $\\ds \\frac{3}{4}$.\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\n See the file linked here: \\\\\r\n \\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W04LEFT1.m}{W04LEFT1.m} \\\\\r\n Estimate with $n = 1000$ is $\\approx 0.7455$. Exact value is $\\frac{3}{4} = 0.75$.\r\n\\end{Solution}\r\n% ******************************\r\n\\item \\begin{Question}\r\n $\\ds \\int_{-1}^1 x^{100} dx$, which equals $\\ds \\frac{2}{101}$.\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\nSee the file linked here: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W04LEFT2.m}{W04LEFT2.m} \\\\\r\n Estimate with $n = 1000$ is $\\approx 0.0199$. Exact value is $\\frac{2}{101} = 0.0198$.\r\n \r\n\\end{Solution}\r\n\r\n% ******************************\r\n\\item \\begin{Question}\r\n $\\ds \\int_{1}^4 (5 - 2t +3t^2) dt$, which equals 63.\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\nSee the file linked here: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W04LEFT3.m}{W04LEFT3.m} \\\\\r\n Estimate with $n = 1000$ is $\\approx 62.9415$. Exact value is 63.0000.\r\n \r\n\\end{Solution}\r\n\r\n% ******************************\r\n\\item \\begin{Question} \\label{qmat2}\r\n $\\ds \\int_0^1 (1 + \\frac{1}{2} u^4 - \\frac{2}{5} u^9) du$, which\r\n equals $\\ds \\frac{53}{50}$.\r\n\\end{Question}\r\n\r\n\\begin{Solution}\r\nSee the file linked here: \\\\ \r\n\\href{http://www.mast.queensu.ca/~apsc171/MNTCP01/PracticeProblems/MATLAB/W04LEFT4.m}{W04LEFT4.m} \\\\\r\n Estimate with $n = 1000$ is $\\approx 1.0599$. Exact value is $\\frac{53}{50} = 1.06$.\r\n\\end{Solution}\r\n\r\n\\end{enumerate}\r\n\r\n\\end{document} \r\n", "meta": {"hexsha": "84d74d207118b6d5a2f3943f17e171ad3dcaee38", "size": 51034, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "PracticeProblems/Week04.tex", "max_stars_repo_name": "aableson/MNTCP01", "max_stars_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-11-27T16:10:35.000Z", "max_stars_repo_stars_event_max_datetime": "2015-11-27T16:10:35.000Z", "max_issues_repo_path": "PracticeProblems/Week04.tex", "max_issues_repo_name": "aableson/MNTCP01", "max_issues_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "PracticeProblems/Week04.tex", "max_forks_repo_name": "aableson/MNTCP01", "max_forks_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.1716057891, "max_line_length": 276, "alphanum_fraction": 0.605341537, "num_tokens": 17765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217432062975979, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.6368863198928134}} {"text": "%-------------------------------------------------------------------------------------------------------------------------------------------------\n\\section{Genetic Algorithm}\n\\label{section:GA}\n%-------------------------------------------------------------------------------------------------------------------------------------------------\nOne of the drawbacks of using a Reinforcement Learning strategy is that the training schema can potentially lead to a localized solution space. Such a solution might not have evaluated all possible strategies. This drawback can be countered by using the robustness of methods such as Genetic Algorithms, which carry out more thorough search. We use such a method in order to evolve a Neural Network controller managing the bot in a 1 unit Vs 1 unit fight. This methodology is called \\emph{Neuro Evolution} in general.\n\n%-------------------------------------------------------------------------------------------------------------------------------------------------\n\\subsection{Representation} \nOur objective for the neural network is to be able to make informed decisions by interpreting the game state. Hence, a representation for the game state needs to be formulated. This representation needs to be minimal since the number of inputs to the neural network directly correlate to the size of the chromosome to be evolved. In this approach, we use 8 inputs, 4 for each unit involved in the combat: \n\\begin{enumerate}\n\\item \\emph{Shots to Death} - This heuristics utilizes the fact that all units in the game, at any given time, will be eliminated by a particular opponent unit in certain number of attack commands. This parameter can be used to represent several values at the same time - \\emph{Hit-points}, \\emph{Armor}, \\emph{Opponent's Damage} and \\emph{Damage Multiplier}, combined in equation:\n\\begin{equation}\n\\label{equation:nShots}\n\\text{shots}=\\left \\lceil\\frac{\\text{HP}}{(\\text{damage}\\times\\text{times}-\\text{armor}) \\times\\text{mul}} \\right \\rceil \\text{,}\n\\end{equation}\nThis input is normalized over the maximum value of shots left to death when the game starts (maximum hot-points).\n\\item \\emph{Distance} - This number signifies the distance of the unit from the opponent, which is normalized over its weapon range.\n\\item \\emph{Cooldown Left} - The number of updates left before the unit can shoot again, which is normalized with the largest weapon cooldown time for the two units.\n\\item \\emph{Speed} - The unit's top movement speed possible. This is again normalized by the maximum speed out of the two units.\n\\end{enumerate}\nThe neural network can provide 3 output values, each corresponding to an action's desirability ({\\bf Move Towards Enemy}, {\\bf Move Away from Enemy} or {\\bf Attack}). The action with the maximum desirability is performed.\n\n%-------------------------------------------------------------------------------------------------------------------------------------------------\n\\subsection{Algorithm Parameters}\n\\begin{enumerate}\n\\item \\emph{Fitness}: The fitness of a particular chromosome is an integer value representing a Win ($1$) or a Loss ($-1$). To average out inaccuracy in fitness due to randomness, each chromosome is tested $20$ times and the fitness is averaged out. An alternate method to judge the fitness of a chromosome is to take an average of difference of hitpoints of the two units at the end of a match.\n\\item \\emph{Choosing parents}: The natural selection method used in this project is Elite Selection. In a population of $N$ chromosomes, each reproduction cycle selects $\\frac{N}{3}$ parents for breeding. However the selection method is modified in such a way that the chromosome with the best fitness is always selected (in order to not forget fitness comparison). The rest of the parents selected are uniformly distributed from the top portion of the population and bottom portion. This version is also comparable to the Roulette-Wheel selection methodology, where segments of the total population have certain selection probabilities.\n\\item \\emph{Crossover}: The selected parents reproduce themselves by a simple one point crossover approach. The selection of the split point is normally distributed for each gene.\n\\item \\emph{Mutation}: Each child gets mutated with a predefined mutation probability (standard value used in experiments is $0.2$). It is a recommended strategy to mutate a gene representing neural network connection weight by adding a Gaussian Random number. This enables the weight to be not bound within a particular range and enhance progression in the search space.\n\\item \\emph{Population Refresh}: All newly produced children, two per parent (i.e. $\\frac{N}{3}$ parents generate $\\frac{N}{3}$ children), automatically replace the $\\frac{N}{3}$ chromosomes at the bottom of the population (not replacing if a chromosome which was selected for parenting). The remaining $\\frac{N}{3}$ part of the population is completely mutated, meaning that their each gene mutates with a probability of $1.0$.\n\\item \\emph{Optimization}: Genetic Algorithms need a lot of fitness updates, especially in this case where a match to test fitness of a chromosome takes approximately 7 sec. Therefore one performance optimization made for the project is to preserve the fitness value of chromosomes that have not changed. In general, these are the chromosomes that were selected for parenting.\n\\end{enumerate}\n\n%-------------------------------------------------------------------------------------------------------------------------------------------------\n\\subsection{Results} \n%%%\n\n\\begin{table}\n\\caption{average winning rate}\n\\begin{center}\n% Table generated by Excel2LaTeX from sheet 'Sheet1'\n\\begin{tabular}{|r|r|r|r|}\n\\hline\n & {\\bf mean }& {\\bf stdev} & {\\bf p-value} \\\\\n\\hline\n\t\\emph{8.8.4.3 (ref)} & $ 0.025$ & $ 0.099 $ & --- \\\\\n\\hline\n\t\\emph{ 8.8.3 }& $0.019$ & $ 0.099 $ & $ 0.459 $ \\\\\n\\hline\n \t\\emph{8.1.3 }& $ 0.014$ & $ 0.104 $ & $ 0.094 $ \\\\\n\\hline\n\\end{tabular} \n\\label{table:winningRate}\n\\end{center}\n\\end{table}\n% ---------\n\n\\begin{enumerate}\n\\item Neural Network Structure - While designing the agent, one of the things on our mind was for the neural network to be able to process the inputs efficiently and try to decipher the optimal output values. But there is no way to understand the relation of the input values to the playing strategy for the agent. To test what structure of the neural network can achieve the most efficient results for the bot, we run T-tests with $3$ different structures of the network:\n\\begin{itemize}\n\\item Inputs (8), Hidden Layer (1), Output Layer (3)\n\\item Inputs (8), Hidden Layer (8), Output Layer (3)\n\\item Inputs (8), Hidden Layer (8), Hidden Layer (4), Output Layer (3)\n\\end{itemize}\nThe agent is trained in each case for a certain number of iterations (in this case they were trained for $60$ iterations) and then the best chromosome in each pool plays the game to create a sample space to be tested. The results for the T-test carried out on the three samples are presented in Table \\ref{table:winningRate}. The samples are compared with respect to the $8,8,4,3$ network. While comparing the network with the $8,8,3$ neural network, the relatively high p-value signifies that they are similar. This means that the using the larger network is an overkill as it provides relatively similar output quality as the smaller network. This, however is not the case with the $8,1,3$ network. The low p-value means that the samples are not similar. Then the decision to select a network falls on the which has a higher success-rate (which is already being measured by the average fitness value of the chromosomes).\n\\begin{figure}[htp]\n\\centerline{\\includegraphics[width=1.0\\columnwidth]{fig_GA_AverageFitness}}\n\\caption{Graph showing the progression of Average Fitness of population of size $10$, $20$ and $30$, respectively: NN configuration is $8,8,3$}\n\\label{fig:AverageFitness}\n\\end{figure}\n\\item Population Size - We are also interested in finding out the rate of convergence of the Genetic Algorithm, as well as its dependency on the population size. Generally a larger population size is preferred, since the high number of chromosomes provide a bigger search space for the algorithm to find a fit individual in. But this leads to a lot of overhead on the system to process individuals introduced on each iteration in the larger population. To assess this parameter, we look at a comparative Figure \\ref{fig:AverageFitness} showing the progression of average fitness of the populations of sizes $10$, $20$ and $30$. The population with size 10 seems to be too small and has slow convergence to the other two populations. The other two populations are very close in average performance and hence a smaller and more efficient population size is selected. \n\\begin{figure}[htp]\n\\centerline{\\includegraphics[width=1.0\\columnwidth]{fig_GA_Fitness_MinAvgMax}}\n\\caption{Graph showing the minimum, average and maximum fitness of population of size $10$, NN configuration is $8,8,3$}\n\\label{fig:minAvgMaxFitness}\n\\end{figure}\n\\item Fitness Spread - In a genetic algorithm, the reproduction and discard rules are used to introduce new individuals to the population that can spread the solution in a much broader spectrum. This counters the problems provided by other computation intelligence algorithms to get stuck in local minima. A way to test the spread of the population is by graphing the minimum, average and maximum fitness of a population (size = $10$), as shown in Figure \\ref{fig:minAvgMaxFitness}. A large distance between the minimum and maximum fitness points to the conclusion that the population is taking care of introducing diverse individuals.\n\\end{enumerate}\n\n%%%\n\\begin{figure}[htp]\n\\centerline{\\includegraphics[width=1.0\\columnwidth]{fig_GA_Speciation_Top2}}\n\\caption{Graph of Specie difference between the top two chromosomes}\n\\label{fig:specieTop2}\n\\end{figure}\n\\begin{figure}[htp]\n\\centerline{\\includegraphics[width=1.0\\columnwidth]{fig_GA_Speciation_next2}}\n\\caption{Graph of Specie difference between $2^{nd}$ and $3^{rd}$ ranked chromosomes}\n\\label{fig:specieNext2}\n\\end{figure}\n\n%-------------------------------------------------------------------------------------------------------------------------------------------------\n\\subsection{Discussion}\nAccording to our tests with the network structure and the population size, the best solution for our one versus one approach is a network structure of 8,8,3 and a population size of $20$. Although the bigger network also evolved well, in terms of performance we clearly have to choose the smaller one.\nAccording to the figure \\ref{fig:AverageFitness}, the population size of $20$, roughly scores as good as the population size of $30$. When we take the fact, that a smaller population size will evolve $\\frac{1}{3}$ faster, into account, we have to chose a population size of $20$.\n\nIf we measure the cumulative distance between each gene of the top two candidates in the populations which is depicted in Figure \\ref{fig:specieTop2}, we can see that the larger populations in genetic evolution hamper the evolving chances because of higher preservation of the top candidates. This can be attributes to the elite selection methodology we have employed that gives increasingly higher probability to the $1^{st}$ and $2^{nd}$ chromosomes to be selected for parenting.\nThis pattern becomes much more varied if we calculate the distance between the $2^{nd}$ and $3^{rd}$ chromosomes (Figure \\ref{fig:specieNext2}), which depicts the higher chances for the chromosomes on those positions to be mutated.\n%%%", "meta": {"hexsha": "a58d493e7afb99b0e52285875a2d06ad8fe0f203", "size": 11665, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/_ga.tex", "max_stars_repo_name": "indusninja/sc-bw-ai", "max_stars_repo_head_hexsha": "c00a29ad3bf3421d7fdb1b0514424d69154ae7b9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/_ga.tex", "max_issues_repo_name": "indusninja/sc-bw-ai", "max_issues_repo_head_hexsha": "c00a29ad3bf3421d7fdb1b0514424d69154ae7b9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/_ga.tex", "max_forks_repo_name": "indusninja/sc-bw-ai", "max_forks_repo_head_hexsha": "c00a29ad3bf3421d7fdb1b0514424d69154ae7b9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 117.8282828283, "max_line_length": 922, "alphanum_fraction": 0.7195885126, "num_tokens": 2534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8824278695464501, "lm_q2_score": 0.7217432003123989, "lm_q1q2_score": 0.636886314611307}} {"text": "\\documentclass{article}\n\\usepackage{latexsym}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{graphicx}\n\n\\newtheorem{theorem}{Theorem}[section]\n\\begin{document}\n\\title{Eisenstein Triples: Connecting geometry and number theory}\n\\author{Dave Neary}\n\n\\maketitle\n\n\\section{A number theory problem}\n\nWe start with a number theory puzzle: Can you find positive integer triples $(a,b,c)$ such that:\n\\[ \\frac{1}{a+c} + \\frac{1}{b+c} = \\frac{3}{a+b+c} \\]\n\nTo get started with our exploration, let's clear denominators by multiplying across by\n$(a+c)(b+c)(a+b+c)$ to get:\n\\begin{equation*}\n\\begin{split}\n(b+c)(a+b+c) + (a+c)(a+b+c) & = 3(a+c)(b+c) \\\\\na^2 + b^2 + 2c^2 + 2ab + 3bc + 3ac & = 3ab + 3ac + 3bc + 3c^2 \\\\\na^2 - ab + b^2 &= c^2\n\\end{split}\n\\end{equation*}\n\nOur goal will be to find positive integer solutions to this equation.\n\n\\section{A geometric aside}\n\nThis equation may look familiar to anyone who has come across the cosine rule in geometry.\nFor the sides of a triangle $(a,b,c)$ with opposite angles $(A,B,C)$:\n\n\\[ c^2 = a^2 + b^2 -2ab\\cos(C) \\]\n\nwith similar equations for $a^2, b^2$.\n\nRewriting the equation above in this form, we can see that the problem is equivalent to finding\ntriangles with integer sides $(a,b,c)$ such that $\\cos(C) = \\frac{1}{2}$ - that is,\n$C = \\frac{\\pi}{3}$. In other words, our challenge is to find triangles with one angle measuring\n$60^{\\circ}$ which have integer side lengths.\n\nThis is a well known problem that has been explored by many number theorists in the past, and\ntriples of this form are known as Eisenstein triples, named after the mathematician Eisenstein.\n\nBy considering the problem as one of triangles with a certain constraint, perhaps we can approach\nthis problem the way we would approach finding Pythagorean triples.\n\n\\section{Factoring sums of squares}\n\nOne way to generate Pythagorean triples is to factor $a^2+b^2$ over the Gaussian integers - that is,\nthe subset of the complex numbers $a+ib$ with $a,b\\in \\mathbb{Z}$. In the Gaussian integers, we can \nrewrite $ a^2 + b^2 = a^2 - b^2i^2$ as a difference of two squares - yielding a factorization:\n\\[ c^2 = a^2 + b^2 = (a+ib)(a-ib) \\]\n\nHow does this help? Well, if we take an arbitrary Gaussian integer $z = a+ib$, the length of $z$ is\n$|z| = \\sqrt{a^2+b^2}$ - so if we take the length of $z^2$, we are guaranteed to get an integer\nlength, since $|z|^2 = |z^2|$. We will have:\n\\begin{equation*}\n\t\\begin{split} \n\t\tz^2 &= (a+ib)^2 \\\\\n\t\t& = a^2-b^2 + i(2ab) \\\\\n\t\t|z^2| &= \\sqrt{(a^2-b^2)^2 + (2ab)^2 }\\\\\n\t\t|z|^2 &= a^2 + b^2 \\\\\n\t\t|z^2| &= |z|^2 \\\\\n\t\t\\Rightarrow (a^2-b^2)^2 + (2ab)^2 &= (a^2 + b^2)^2\n\t\\end{split}\n\\end{equation*}\n\nSo if we take \\textbf{any} positive integers $(a,b)$, we can generate a Pythagorean triple \n$(a^2-b^2, 2ab, a^2+b^2)$ from them.\n\nCan we use a similar approach with our puzzle? Indeed we can!\n\n\\section{Factoring the quadratic in $\\mathbb{C}$}\n\nWe can factor $a^2 - ab + b^2$ in complex numbers too by setting it equal to zero and using \nthe quadratic formula:\n\\begin{equation*}\n\t\\begin{split} \n\t\ta^2 - ab + b^2 &= 0 \\\\\n\t\ta &= \\frac{1}{2}\\left(b \\pm \\sqrt{b^2-4b^2}\\right) \\\\\n\t\t&= \\frac{b}{2}\\left(1 \\pm \\sqrt{3}i\\right)\n\t\\end{split}\n\\end{equation*}\n\nThe factorization is:\n\\[ a^2 - ab + b^2 = \n\\left(a + \\frac{b}{2}\\left(-1 - \\sqrt{3}i\\right) \\right)\n\\left(a + \\frac{b}{2}\\left(-1 + \\sqrt{3}i\\right) \\right) \\]\n\nIf you have done any complex analysis these roots might look familiar to you. Given \n$z = \\frac{1}{2}(-1+\\sqrt{3}i)$ then:\n\\begin{equation*}\n \\begin{split} \n\t\tz^2 & = \\frac{1}{4}(-1+\\sqrt{3}i)^2 \\\\\n\t\t& = \\frac{1}{4}(1 - 2\\sqrt{3}i - 3) \\\\\n\t\t& = \\frac{1}{2}(-1 - \\sqrt{3}i) \\\\\n\t\t& = \\bar{z}\n \\end{split}\n\\end{equation*}\n\nand:\n\n\\begin{equation*}\n \\begin{split}\n z^3 & = \\frac{1}{8}(-1+\\sqrt{3}i)^3 \\\\\n\t\t& = \\frac{1}{8}(- 1 + 3\\sqrt{3}i +9 -3\\sqrt{3}i) \\\\\n & = 1\n \\end{split}\n\\end{equation*}\n\nThese are the cube roots of unity, $\\omega = \\frac{1}{2}(-1+\\sqrt{3}i), \\omega^2 = \\frac{1}{2}(-1-\\sqrt{3}i)$.\nOne identity of the cube roots that we will use below is that $1 + \\omega + \\omega^2 = 0$, so \n$\\omega = -1 -\\omega^2$ and $\\omega^2 = -1 -\\omega$.\n\nWe can rewrite our original problem expression as:\n\\[ c^2 = (a+\\omega b)(a + \\omega^2 b) \\]\n\nLet's try the same trick as we used with Pythagorean triples, and see what happens when we start with\n$z = a + \\omega b$ for $a,b\\in \\mathbb{Z}$.\n\n\\begin{equation*}\n \\begin{split}\n\t\t|z| & = \\sqrt{(a+\\frac{b}{2})^2 + (\\frac{\\sqrt{3}b}{2})^2} \\\\ \n\t\t &= \\sqrt{a^2-ab+b^2} \n\t\\end{split}\n\\end{equation*}\n\nIn general, $|a+ \\omega b| = |a + \\omega^2 b| = \\sqrt{a^2-ab+b^2}$.\n\n\\begin{equation*}\n \\begin{split}\n z^2 & = a^2 + 2ab\\omega +\\omega^2 b^2 \\\\\n\t\tz^2 & = a^2 + 2ab\\omega +(-1 - \\omega) b^2 \\\\\n\t\t & = (a^2 -b^2) + (2ab - b^2)\\omega \\\\\n\t\t|z^2| &= \\sqrt{(a^2-b^2)^2 - (a^2-b^2)(2ab-b^2) + (2ab-b^2)^2}\n\t\\end{split}\n\\end{equation*}\n\n\nAnd since $|z^2| = |z|^2$ we get:\n\\[ (a^2-ab+b^2)^2 = (a^2-b^2)^2 - (a^2-b^2)(2ab-b^2) + (2ab-b^2)^2 \\]\n\nSo, given any positive integers $(a,b)$, we can generate triples satisfying the equation \n$(a^2-b^2, 2ab-b^2, a^2-ab+b^2)$.\n\nTrying some values we find:\n\\begin{equation*}\n \\begin{split}\n\t\t(a,b) = (2,1) & \\Rightarrow (a^2-b^2, 2ab-b^2, a^2-ab+b^2) = (3, 3, 3) \\\\\n\t\t(a,b) = (3,1) & \\Rightarrow (a^2-b^2, 2ab-b^2, a^2-ab+b^2) = (8, 5, 7) \\\\\n\t\t(a,b) = (4,3) & \\Rightarrow (a^2-b^2, 2ab-b^2, a^2-ab+b^2) = (7, 15, 13)\n\t\\end{split}\n\\end{equation*}\n\nWe can easily check that $7^2 = 8^2 - (8)(5) + 5^2$ and $13^2 = 7^2 - (7)(15) + 15^2$.\n\nWe have a formula to generate an infinite number of triples satisfying our equation $a^2-ab+b^2=c^2$ - nice!\nLet's check that they also solve our initial puzzle:\n\n\\begin{equation*}\n \\begin{split}\n\t\t\\frac{1}{7+13} + \\frac{1}{15+13} & = \\frac{1}{20} + \\frac{1}{28} \\\\\n\t\t& = \\frac{48}{560} \\\\\n\t\t& = \\frac{3}{35} \\\\\n\t\t& = \\frac{3}{7 + 13 + 15}\n \\end{split}\n\\end{equation*}\n\n\\section{A geometric approach}\n\nDoes this method yield all possible triples? It is hard to tell, but we have other tools that we can\nuse to find that answer. Again, we are going to start with a search for Pythagorean triples to\nexplore this approach.\n\n\\begin{figure}[htb]\n \\center{\\includegraphics[width=0.9 \\linewidth]{Pythagorean_circle.png}}\n \\caption{Finding rational points with a line of rational slope}\n \\label{fig:pyth_circle}\n\\end{figure}\n\nReplacing $x = \\frac{a}{c}, y = \\frac{b}{c}$, we can see that finding integer solutions to $a^2 + b^2 = c^2$\nis equivalent to finding rational solutions to $x^2 + y^2 = 1$. This is the equation of a circle of radius 1,\ncentered on the origin.\n\nIf we can find a way to find all rational points on the perimeter of the circle, then we will have found all\npossible pythagrean triples, up to multiplication by a scalar (since if $a^2 + b^2 = c^2$, then\n$(ka)^2 + (kb)^2 = (kc)^2$ for all $k \\in \\mathbb{Z}$). One way we can do that is to find one rational\npoint on the circle (say, $(0,-1)$, and draw lines through that point with a rational slope. We can show\nthat, since the slope of the line is rational, and the equation of the circle has integer coefficients, then\nif one intersection point is rational (as it is by design), then the other point must also be rational.\n\nLet's take an example to see how it happens. The line of slope $t$ through $(x,y) = (-1,0)$ is:\n\\[ y = tx +t\\]\n\nClearly, $(-1,0)$ is on this line. Now, substituting this value of $y$ into the equation of the circle,\n$x^2+y^2-1 = 0$, we get:\n\\[x^2 + (tx+t)^2 -1 = (1+t^2)x^2 +2t^2x + t^2 -1 = 0 \\]\nAnd solving for $x$ we get $x = -1$ or $x = \\frac{1-t^2}{1+t^2}$. Substituting this value of $x$ back into\nour line, we get $y = \\frac{2t}{1+t^2}$. We can easily verify that for any value of $t$ that this satisfies\nthe equation of the circle, as expected. And since $t$ is rational, $x$ and $y$ are guaranteed to be\nrational too.\n\nIs it possible, though, that we would somehow miss rational points in the circle by this method? \nNo - given any rational point $(x,y)$ on the perimeter of the circle, the slope of the line through\n$(x,y)$ and $(-1,0)$ is $\\frac{y}{x+1}$, which is clearly rational. Since this would be a line with rational\nslope which goes through the point $(-1,0)$, we are guaranteed that by enumerating all of the rational\nnumbers for line slopes, we are guaranteed to hit all of the rational points on the circle.\n\nIf we take the slope to be $t = \\frac{m}{n}$ for $m,n \\in \\mathbb{Z}$ we can multiply across the\ntop and bottom by $n^2$ in the formulas for $x = \\frac{1-t^2}{1+t^2}, y = \\frac{2t}{1+t^2}$ to give\nthe Pythagorean triples $a = n^2-m^2, b = 2mn, c = n^2 + m^2$ for the familiar formula we found above.\n\nTaking this approach with our problem, we substitute $x = \\frac{a}{c}, y = \\frac{b}{c}$ to get the equation:\n\\[ x^2 - xy + y^2 = 1 \\]\n\n\\begin{figure}[htb]\n \\center{\\includegraphics[width=0.9 \\linewidth]{Eisenstein_ellipse.png}}\n \\caption{Finding rational points with a line of rational slope}\n \\label{fig:eisen_ellipse}\n\\end{figure}\n\nThis is an ellipse, centered on the origin, which goes through the points $(0,-1), (0,1), (-1,0), (1,0)$\non the X and Y axis. There are also a few other \"nice\" rational points on the curve at $(-1,-1), (1,1)$.\n\nNow taking the same approach, given any rational number $t$, we can find the intersection points of our\nellipse and the line $y = t(x+1)$ (which again goes through $(-1,0)$):\n\n\\begin{equation*}\n \\begin{split}\n\t\tx^2 - xy + y^2 & = x^2 - tx(x+1)+t^2(x+1)^2 \\\\\n\t\t&= (1-t+t^2)x^2 +(2t^2-t)x + t^2 \\\\\n\t\t&= 1 \n\t\\end{split}\n\\end{equation*}\n\nNow factoring to find the intersection points, we find, as expected, one of the points at $x=-1$, and\nthe other is a rational function of $t$:\n\\begin{equation*}\n\t\\begin{split}\n\t\t(1-t+t^2)x^2 + (2t^2-t)x + (t^2-1) &= 0 \\\\\n\t\t(x + 1)((1-t+t^2)x + (t^2 - 1)) = 0\n \\end{split}\n\\end{equation*}\n\nSo we have $x = -1$ or $x = \\frac{1 - t^2}{1-t+t^2}$ for rational $t$.\n\nSubstituting the valie for $x$ back into the equation for the line:\n\\[ y = t(x+1) = \\frac{2t - t^2}{1-t+t^2} \\]\n\nAnd replacing $t = \\frac{m}{n}$ and multiplying through with $n^2$ above and below the line we get:\n\\[(x,y) = \\left(\\frac{n^2-m^2}{n^2-mn+m^2}, \\frac{2nm - m^2}{n^2-mn+ m^2}\\right) \\]\n\nAnd clearing a common denominator, we get an integer triple\n$(a,b,c) = (n^2 - m^2, 2nm - m^2, m^2 - mn + n^2)$ as before.\n\nA similar argument as before shows that this will find \\textbf{all} rational points on the\nellipse, giving all integer solutions to the original problem (after dividing out any common factors\nof $a, b, c$).\n\nInterestingly, if we take a different rational point on the ellipse, we will get a different, but\nequally valid, formula for generating Eisenstein triples from rational numbers. The equation of\na line through $(-1,-1)$, for example, is $y + 1 = t(x + 1)$. Isolating $y$, and substituting this\ninto the equation of the ellipse, we get:\n\\begin{equation*}\n \\begin{split}\n\t\tx^2 - xy + y^2 - 1 & = x^2 - x(t(x+1)-1) + (t(x+1)-1)^2 - 1\\\\\n\t\t& = x^2 -tx^2 -tx + x + t^2(x^2+2x+1)-2t(x+1)+1 \\\\\n\t\t& = (1 - t + t^2)x^2 + (1 - 3t + 2t^2)x + (t^2-2t) \\\\\n\t\t& = (x+1)((1 - t + t^2)x + (t^2-2t)) = 0\n\t\\end{split}\n\\end{equation*}\n\nGiving the solutions $x = -1$ or $x = \\frac{2t-t^2}{1-t+t^2}$.\nSubstituting this back into the equation for the line, we get:\n\\begin{equation*} \n \\begin{split}\n\t\ty &= t\\left(\\frac{2t-t^2}{1-t+t^2} + 1\\right) - 1 \\\\\n\t\t& = \\frac{t^2 + t}{1-t+t^2} - 1 \\\\\n\t\t& = \\frac{2t - 1}{1 - t + t^2}\n\t\\end{split}\n\\end{equation*}\n\nReplacing $t = \\frac{m}{n}$, we get:\n\\[ x = \\frac{2mn - m^2}{m^2 - mn + n^2}, y = \\frac{2mn - n^2}{m^2 - mn + n^2} \\]\nWhich gives an alternative form for triples of:\n\\[ (a,b,c) = (2mn - m^2, 2mn - n^2, m^2 -mn + n^2) \\]\n\nwhich, surprisingly perhaps, is exactly equivalent to the formula we found earlier.\n\n\\end{document}\n", "meta": {"hexsha": "5d2853ac067c2ce2f0aa000b049a585fcd862614", "size": 11910, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "eisensten_triples.tex", "max_stars_repo_name": "dneary/math", "max_stars_repo_head_hexsha": "129b2093c01b12ddc2e61abd331c95da2177803c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "eisensten_triples.tex", "max_issues_repo_name": "dneary/math", "max_issues_repo_head_hexsha": "129b2093c01b12ddc2e61abd331c95da2177803c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "eisensten_triples.tex", "max_forks_repo_name": "dneary/math", "max_forks_repo_head_hexsha": "129b2093c01b12ddc2e61abd331c95da2177803c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.2364864865, "max_line_length": 110, "alphanum_fraction": 0.6278757347, "num_tokens": 4374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321936479701, "lm_q2_score": 0.7905303260722198, "lm_q1q2_score": 0.6368766807388075}} {"text": "\\chapter{Acceptance Sampling}\n\\chaptermark{Acceptance}\n\n% AQL\n% LTPD\n% RQL\n% LQL\n% Type A and Type B RO curves.\n\nWe can improve quality (read- conformance to specification) by introducing an inspection stage in our process.\nClearly, a full inspection is time consuming. \nIt may also be destructive (you don't want to re-package ice-cream after checking its texture \\dots).\nNo-inspection may be appropriate if you don't particularly care about your brand, or if production has very high capability indices.\nA reasonable, intermediate approach, is a partial random inspection, known as \\emph{acceptance sampling}.\nAs the name suggests, in acceptance sampling, one samples, then checks, then accepts (or not).\n\nAcceptance sampling can be seen as a control chart monitoring that triggers active intervention in the production. As such ,it is a crude type of \\emph{engineering control} (Sec.~\\ref{sec:terminology_statistical}).\nThe intervention is obvious. The monitoring is based on some continuous (variable) or discrete (attribute) of a sample of units from a \\emph{batch}, \\aka, a \\emph{lot}.\nSeen as a feedback control, it is not surprising that when designing an acceptance sampling scheme, we have similar decisions as when designing a control chart:\n\\begin{enumerate}\n\\item What is a batch? Just like choosing the sampling frequency in a Shewart chart. \nWe would like homogenous batches, i.e., with low inner variability. A box, a shipment, a day's production, are typical batches. \n\\item Within batch sampling scheme: just like rational grouping in Shewart chart. Typical approaches include \\emph{single sampling plans}, \\emph{double}, \\emph{multiple}, and \\emph{sequential sampling plans}.\nThis can be seen as the design of an experiment to be performed on each batch.\n\\item How many units? Just like choosing the sample size in a Shewart chart.\n\\item Decision cutoff: Just like setting control limits in a Shewart chart. \n\\end{enumerate}\nWe can readily see that the design of an acceptance sampling scheme is very similar to the design of a control chart. \nWe may construct an full blown economical optimization problem to design the sampling, as we did in Section~\\ref{sec:economical_considerations}. Just like control charts, however, it is more common to design sampling schemes using ``first-order'' power considerations. \nFor this reason, the \\emph{power function} will play a crucial role.\n\n\\section{Acceptance Sampling Terminology}\nAdapted from \\cite{natrella_nist/sematech_2010}.\n\\begin{description}\n\\item [LASP] A \\emph{lot acceptance sampling plan}, ultimately, a statistical test at the end of which we either accept a batch. In this text we typically use the \\emph{batch acceptance sampling scheme} for the same purpose. \n\\item [AQL] The \\emph{acceptable quality level}, or \\emph{acceptable quality limit}, is the highest proportion of defects acceptable to the producer. \n\\item [LTPD] The \\emph{lot tolerance percent defective} is the highest proportion of defects acceptable to the consumer. Clearly, $AQL4$, when $n=21$.\n\\begin{remark}[Approximate Power Calculations]\nThe problem to solve in Eq.(\\ref{eq:power_acceptance}) requires some non trivial iterations because of the discrete nature.\nIt is quite more convenient to replace the exact form of Eq.(\\ref{eq:power_accpet}) with a normal approximation, so that Eq.(\\ref{eq:power_acceptance}) has a closed form solution. \n\\end{remark}\n\n\n\n\\subsubsection{Type-A Power Function}\nIt is quite wired that we would sample with replacement from a batch.\nIt is quite more probable that we used the replacement assumption, only as an approximation because $n$ is small compared to the batch size $N$. \nIf this is not the case, the binomial distribution in Eq.(\\ref{eq:power_accpet}) should be replaced with the Hypergeometric distribution. \nFor all practical purposes, this means using the \\rcode{phyper()} \\R function, instead of \\rcode{pbinom()}.\n\n\n\n\\subsection{Double Sampling Scheme}\nIn a double sampling scheme, we first example $n_1$ units. \nWe may then decide to accept, reject, or sample another $n_2$ units. \nAfter those $n_2$ samples, we can accept or reject. \nThe idea of a power function remains the same, even if calculations are slightly more cumbersome.\nHere is our our policy:\nFor $x_1$ computed on the first $n_1$ samples:\nIf $x_1 < a_1$ then accept the batch;\nIf $x_1 \\geq c_1$ then reject the batch;\nOtherwise, compute $x_2$ with $n_1+n_2$ samples.\nIf $x_2 < a_2$ accept the batch;\nIf $x_2 \\geq c_2$ then reject the batch.\n\nFor brevity, we denote all the design parameters of the scheme by $\\gamma:= (n_1,n_2,c_1,c_2,a_1,a_2)$. \nThe power function of such a scheme would thus be:\n\\begin{align}\n\t\\pi_\\gamma &:= P(\\set{\\x_1 \\geq c_1} \\union \\set{\\x_1 \\in [a_1,c_1], \\x_2 \\geq c_2}) \\\\\n\t&= P(\\x_1 \\geq c_1) + \\sum_{k=a_1}^{c_1} P(\\x_1 =k, \\x_2-\\x_1 \\geq c_2-k) \\\\\n\t\t&= P(\\x_1 \\geq c_1) + \\sum_{k=a_1}^{c_1} P(\\x_1 =k) P(\\x_2-\\x_1 \\geq c_2-k).\n\\end{align}\nWe may now use the fact that $\\x_1 \\sim Binom(n_1,p)$ and that $\\x_2-\\x_1 \\sim Binom(n_2,p)$, and quickly compute the power in \\R.\n\n\\begin{remark}[Redundancy]\nUnlike the single stage LASP, where we have two equations with two variables, in the two-stage case there are many $\\gamma$ configurations that will achieve given consumer and producer risks ($\\alpha_0,\\pi_0$).\nThe choice of the particular configuration should depend on the type of signal we expect. \nFor quick detection of strong signal (large $p$), choose small $n_1$. \nFor sensitive detection of subtle signal, choose large $n_1$. \n\\end{remark}\n\n\\begin{remark}[No Free Lunch]\nWhile it may seem that a two stage LASP is always better than a single stage LASP, this is not the case.\nTo see why, consider a weak signal ($p$ close to AQL). We may need all $n_1+n_2$ samples to get decent power. The first stage then add nothing except logistic complications.\n\\end{remark}\n\n\n\\section{Sequential Scheme}\nAt this point you should be thinking: why only two stages? \nClearly we may reject or accept a sample as each unit comes in.\nThis is exactly what Sequential LASPs are all about.\nWe will not give the details, except the observation that this is merely a type of sequential experiment as described in Section~\\ref{sec:sequantial}.\n\n\n\n\n\\section{Bibliographic Notes}\n[TODO]", "meta": {"hexsha": "c04f13049010a8ffb9b17a06545511390af76934", "size": 8722, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Class_notes/acceptance.tex", "max_stars_repo_name": "johnros/qualityEngineering", "max_stars_repo_head_hexsha": "4a1c0959672fb5c5a6e59829e543c95beb4e5b44", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Class_notes/acceptance.tex", "max_issues_repo_name": "johnros/qualityEngineering", "max_issues_repo_head_hexsha": "4a1c0959672fb5c5a6e59829e543c95beb4e5b44", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Class_notes/acceptance.tex", "max_forks_repo_name": "johnros/qualityEngineering", "max_forks_repo_head_hexsha": "4a1c0959672fb5c5a6e59829e543c95beb4e5b44", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 66.5801526718, "max_line_length": 269, "alphanum_fraction": 0.7575097455, "num_tokens": 2368, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.7905303211371898, "lm_q1q2_score": 0.6368766804521679}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{exercises}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 3.8 Symmetry of $R_{ab}$}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n \\partial{#}::PartialDerivative;\n\n g_{a b}::Metric;\n g^{a b}::InverseMetric;\n\n dgab := \\partial_{c}{g^{a b}} -> - g^{a e} g^{b f} \\partial_{c}{g_{e f}}.\n # cdb (dgab.000,dgab)\n\n Gamma := \\Gamma^{a}_{b c} -> (1/2) g^{a e} ( \\partial_{b}{g_{e c}}\n + \\partial_{c}{g_{b e}}\n - \\partial_{e}{g_{b c}}).\n # cdb (Gamma.000,Gamma)\n\n Rabcd := R^{a}_{b c d} ->\n \\partial_{c}{\\Gamma^{a}_{b d}} + \\Gamma^{a}_{e c} \\Gamma^{e}_{b d}\n - \\partial_{d}{\\Gamma^{a}_{b c}} - \\Gamma^{a}_{e d} \\Gamma^{e}_{b c}.\n # cdb (Rabcd.000,Rabcd)\n\n Rab := R_{a b} -> R^{c}_{a c b}. # cdb (Rab.000,Rab)\n\n expr := 4 (R_{a b} - R_{b a}). # cdb (ex-0308.100,expr)\n\n substitute (expr, Rab) # cdb (ex-0308.101,expr)\n substitute (expr, Rabcd) # cdb (ex-0308.102,expr)\n substitute (expr, Gamma) # cdb (ex-0308.103,expr)\n\n distribute (expr) # cdb (ex-0308.104,expr)\n product_rule (expr) # cdb (ex-0308.105,expr)\n canonicalise (expr) # cdb (ex-0308.106,expr)\n\n substitute (expr, dgab) # cdb (ex-0308.107,expr)\n canonicalise (expr) # cdb (ex-0308.108,expr)\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}[spread={3pt}]\n \\Dmath*{\\cdb{ex-0308.100} = \\Cdb*{ex-0308.101}\n = \\Cdb*{ex-0308.102}\n = \\Cdb*{ex-0308.103}\n = \\Cdb*{ex-0308.104}\n = \\Cdb*[\\hskip2.0cm\\hfill]{ex-0308.105}\n = \\Cdb*{ex-0308.106}\n = \\Cdb*{ex-0308.107}\n = \\Cdb*{ex-0308.108}}\n\\end{dgroup*}\n\n\\clearpage\n\n% --------------------------------------------------------------------------------------------\n\\section*{Exercise 3.8 Symmetry of $R_{ab}$ alternative solution}\n\nThis differs from the previous code by the inclusion of a call to \\verb|canonicalise| immediately\nafter the first two substitutions and a declaration that $\\Gamma^{a}{}_{bc}$ is symmetric in $bc$.\nThis pair of changes produces a more compact set of results than given above. Incidently, this also\nshows that $\\partial_{a}\\Gamma^{c}{}_{bc} = \\partial_{b}\\Gamma^{c}{}_{ac}$.\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u#}::Indices(position=independent).\n\n \\partial{#}::PartialDerivative;\n\n \\Gamma^{a}_{b c}::TableauSymmetry(shape={2}, indices={1,2}).\n\n g_{a b}::Metric;\n g^{a b}::InverseMetric;\n\n dgab := \\partial_{c}{g^{a b}} -> - g^{a e} g^{b f} \\partial_{c}{g_{e f}}.\n # cdb (dgab.000,dgab)\n\n Gamma := \\Gamma^{a}_{b c} -> (1/2) g^{a e} ( \\partial_{b}{g_{e c}}\n + \\partial_{c}{g_{b e}}\n - \\partial_{e}{g_{b c}}).\n # cdb (Gamma.000,Gamma)\n\n Rabcd := R^{a}_{b c d} ->\n \\partial_{c}{\\Gamma^{a}_{b d}} + \\Gamma^{a}_{e c} \\Gamma^{e}_{b d}\n - \\partial_{d}{\\Gamma^{a}_{b c}} - \\Gamma^{a}_{e d} \\Gamma^{e}_{b c}.\n # cdb (Rabcd.000,Rabcd)\n\n Rab := R_{a b} -> R^{c}_{a c b}. # cdb (Rab.000,Rab)\n\n expr := 4 (R_{a b} - R_{b a}). # cdb (ex-0308.200,expr)\n\n substitute (expr, Rab) # cdb (ex-0308.201,expr)\n substitute (expr, Rabcd) # cdb (ex-0308.202,expr)\n canonicalise (expr) # cdb (ex-0308.203,expr)\n substitute (expr, Gamma) # cdb (ex-0308.204,expr)\n\n distribute (expr) # cdb (ex-0308.205,expr)\n product_rule (expr) # cdb (ex-0308.206,expr)\n canonicalise (expr) # cdb (ex-0308.207,expr)\n\n substitute (expr, dgab) # cdb (ex-0308.208,expr)\n canonicalise (expr) # cdb (ex-0308.209,expr)\n\\end{cadabra}\n\n\\begin{dgroup*}[spread={3pt}]\n \\Dmath*{\\cdb{ex-0308.200} = \\Cdb*{ex-0308.201}\n = \\Cdb*{ex-0308.202}\n = \\Cdb*{ex-0308.203}\n = \\Cdb*{ex-0308.204}\n = \\Cdb*{ex-0308.205}\n = \\Cdb*{ex-0308.206}\n = \\Cdb*{ex-0308.207}\n = \\Cdb*{ex-0308.208}\n = \\Cdb*{ex-0308.209}}\n\\end{dgroup*}\n\n\\end{document}\n", "meta": {"hexsha": "59b2e0674fcae7b3862f5e5551e9034c94efb0c4", "size": 5450, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/exercises/ex-0308.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/exercises/ex-0308.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/exercises/ex-0308.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 43.9516129032, "max_line_length": 99, "alphanum_fraction": 0.399266055, "num_tokens": 1573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.7905303162021596, "lm_q1q2_score": 0.6368766764763487}} {"text": "\\section{Ambiguity}\n\nA sentence $x$ of $G$ is ambiguous iff it admits different syntax trees. In that case $G$ is ambiguous. The degree of ambiguity of $x$ is the number of distinct trees of $x$, of $G$ is the maximum over its sentences.\n\n\\paragraph{RegExp ambiguity} given an r.e. f , we number its letters and we obtain a numbered r.e. is ambiguous if the language defined by the numbered r.e. contains two distinct strings x and y that become identical when the numbers are erased.\n\n\n\\paragraph{Bilateral Recursions} $E \\rarr E + E | i$ becomes $E \\rarr i + E | i$.\n\n\\paragraph{Left-Right Recurions in different rules} $A \\rarr aA | Ab | c$. Remedies: generate using different rules or force an order of derivation.\n\n\\paragraph{Union of Languages} assume that the two non-terminal sets are disjoint, otherwise the union grammar would generate\na superlanguage that strictly contains both languages. If $L_1 \\cap L_2 \\ne \\emptyset$ then $L_1 \\cup L_2$ is ambiguous (the intersection has 2 derivations). Remedy: provide disjointed set of rules: $L_1 \\cap L_2$, $L_1 \\setminus L_2$ and $L_2 \\setminus L_1$.\n\n\\paragraph{Concatenation of Languages} if a suffix of a sentence of language one is also a prefix of a sentence of language two $G_1 . G_2$ is ambiguous if $\\exists x_1 \\in L_1, x_2 \\in L_2$ such that $x_1 = uv$ with $u \\in L_1$ and $x_2 = vz$ with $z \\in L_2$: $uvz$ can be $(uv)z$ or $u(vz)$.\n\n\\paragraph{Inherent Ambiguity} A language is inherently ambiguous if all its grammar are ambiguous, e.g. those where the intersection is not CF.\n\n\\paragraph{Lack of Order in Derivations} add a nonterminal to impose that one rule must follw the other $S \\rarr bSc | bbSc | \\epsilon$ becomes $S \\rarr bSc | D$, $D \\rarr bbDc | \\epsilon$.\n", "meta": {"hexsha": "154dbf43792d1cef5a565c02ad3b458c9a89f8bf", "size": 1739, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "grammars/ambiguity.tex", "max_stars_repo_name": "TiberioG/FLC-cheatsheet", "max_stars_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "grammars/ambiguity.tex", "max_issues_repo_name": "TiberioG/FLC-cheatsheet", "max_issues_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "grammars/ambiguity.tex", "max_forks_repo_name": "TiberioG/FLC-cheatsheet", "max_forks_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 86.95, "max_line_length": 294, "alphanum_fraction": 0.7395054629, "num_tokens": 511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696747, "lm_q2_score": 0.8056321913146127, "lm_q1q2_score": 0.6368766729304891}} {"text": "\n\\documentclass[a4paper,12pt]{article}\n\\usepackage{graphicx}\n%\\usepackage{stix}\n\\usepackage{amsmath}\n\\newcommand\\addtag{\\refstepcounter{equation}\\tag{\\theequation}}\n\\begin{document}\n\n\\title{Diagnosed Particle Disaggregation}\n\\author{Jacob A. Cram}\n\\maketitle\n\n\\section{Definitions and Units}\n\n\\begin{equation}\n m = C_m r^\\alpha %\\addtag\n \\label{eqn:m}\n\\end{equation}\n\nAs in DeVries et al. [2014] particle mass m is a function of radius r and scales with a fractal dimension $\\alpha$. $C_m$ is a constant.\n\n\\begin{equation}\n w = C_wr^\\gamma %\\addtag\n \\label{eqn:w}\n\\end{equation}\n\nSinking speed also scales with mass to another constant $\\gamma$. \nAccording to Guidi et al. [2008] $\\gamma = \\alpha - 1$, but we'll keep things in terms of $\\gamma$ going forward.\n\n\\begin{equation}\n F = nmw = n C_m C_w r^{\\alpha + \\gamma}\n\\end{equation}\n\nFlux F is a function of particle numbers, mass, and sinking speed.\n\n\n%\\includegraphics{ijFig2.png}\n\n\\begin{figure}[h]\n \\includegraphics{ijFig2.png}\n %\\includegraphics[natwidth=0.35\\textwidth]{ijFig2.png}\n \n \\centering\n \\caption{Some number of particles $\\Delta n_j$ of size ``j'' remineralize to size ``j-1'' as they sink from depth ``i-1'' to depth ``i''.}\n \\label{fig:boxes}\n\\end{figure}\n\nGoing forward we will determine the calculations for how many particles of size j in shallow depth i-1 remineralize into smaller particles of size j-1 in deeper depth i. We will call this term $\\Delta n_j$\n\n\\section{Conservation of particle number flux}\n\nIn the absence of disaggregation, the number of particles leaving a box of water is equal to the number of particles going into that box from above. In other words, particle \"number-flux\" is conserved. Thus the number of particles in the box is a function of the number of particles going into that box, and the difference in velocities between when the particle enters and when that particle leaves.\n\n\\begin{equation}\nn_{i-1, j-1} {{w_{j-1}}\\over{w_j}} + n_{i-1,j} = n_{i,j-1} {{w_{j-1}}\\over{w_j}} + n_{i,j}\n\\label{eqn:nfcons}\n\\end{equation}\n\nWhere $n_{i-1,j}$ is the number of particles of size j (the bigger size) at depth i-1 (the shallower depth). The subscripts correspond to locations in Figure \\ref{fig:boxes}.\n\nWe can re-arrange equation \\ref{eqn:nfcons}\n\n\\begin{equation}\nn_{i-1, j-1} w_{j-1} + n_{i-1,j} w_j = n_{i,j-1} w_{j-1} + n_{i,j} w_j\n\\label{eqn:renfcons}\n\\end{equation}\n\n\nSubstitue in equation \\ref{eqn:w} into equation \\ref{eqn:renfcons}.\n\n\\begin{equation}\nn_{i-1, j-1} r^\\gamma_{j-1} + n_{i-1,j} r^\\gamma_j = n_{i,j-1} r^\\gamma_{j-1} + n_{i,j} r^\\gamma_j\n\\label{eqn:nra}\n\\end{equation}\n\nRearrange equation \\ref{eqn:nra}\n\\begin{equation}\nr^\\gamma_{j-1} (n_{i-1, j-1} - n_{i, j-1} ) = r^\\gamma_{j} (n_{i, j} - n_{i-1, j} ) = \\Phi\n\\label{eqn:phi}\n\\end{equation}\n\nWhere $\\Phi$ is a placeholder standing for either side of equation \\ref{eqn:phi}, which I will subsequently substitute into things.\n\n\\medskip\n\nSolve for $\\Delta n_j$\n \\begin{equation}\n\\Delta n_j = n_{i, j} - n_{i-1, j} = {{r^\\gamma_{j-1}}\\over{r^\\gamma_{j}}} (n_{i-1,j-1}-n_{i,j-1})\n \\end{equation}\n\n\\section{Conservation of Mass Flux}\n\nTotal flux defined is the sum of flux in each (observed) particle size bin. Particles not in an observed bin don't count towards total flux.\n\n\\begin{equation}\n\\Delta F = \\sum_{j = 2}^n \\Delta f_j + \\Delta f_1\n\\label{eqn:df}\n\\end{equation}\n\nHere $\\Delta f_j$ is the flux attenuation from bin of size j and $\\Delta f_1$ is the loss that comes from particles in bin 1 becoming small enough that you can no longer see them with the UVP.\n\nThe flux attenuation in a bin is the product of the rate of flux attenuation with depth of each individual particle ${\\partial f} \\over {\\partial z}$, the depth interval over which the particles attenuate $\\Delta z$ and the number of particles in that bin at the top of the depth interval $n_{i-1,j}$\n\n\\begin{equation}\n\\Delta f_j = {{\\partial f} \\over {\\partial z}} \\Delta z n_{i-1,j}\n\\label{eqn:dfj}\n\\end{equation}\n\nFurthermore, the rate of flux attenuation with respect to depth is the product of the rate of mass attenuation with respect to time ${{\\partial m} \\over {\\partial t}}$ , the inverse of the sinking speed ${{\\partial t} \\over {\\partial z}}$ , and the deriviative of the flux to mass relationship ${{\\partial f} \\over {\\partial m}}$.\n\n\\begin{equation}\n{{\\partial f} \\over {\\partial z}} = \n{{\\partial m} \\over {\\partial z}} {{\\partial f} \\over {\\partial m}} = \n{{\\partial m} \\over {\\partial t}} \n{{\\partial t} \\over {\\partial z}} \n{{\\partial f} \\over {\\partial m}}\n\\label{eqn:dfdz}\n\\end{equation}\n\nIn PRiSM, fractional mass loss as a function of time is the same for all particles of all sizes.\n\n\\medskip\n\nNow we are going to come up with the values for each of these terms.\n\n\\medskip\n\nThe particle remineralization rate $C_r$ is the same for particles of all sizes.\n\n\\begin{equation}\n{{\\partial m} \\over {\\partial t}} = C_r * m = C_r C_m r^\\alpha\n\\label{eqn:dmdt}\n\\end{equation}\n\nSinking speed definition, substituting from equation \\ref{eqn:w}\n\\begin{equation}\n{{\\partial t} \\over {\\partial z}} = {1 \\over w} = {1 \\over C_w r^\\gamma}\n\\label{eqn:dtdz}\n\\end{equation}\n\nFlux for a given size class, substituting eqation \\ref{eqn:m}, and finally putting everything in terms of mass (rather than mass and radius, since the two are related)\n\\begin{equation}\nf = m w = m * C_w r^\\gamma = m * C_w ({m\\over C_m})^{\\gamma \\over \\alpha}\n\\label{eqn:f}\n\\end{equation}\n\nDerriving equation \\ref{eqn:f} with respect to mass, and substituting equation \\ref{eqn:m}\n\\begin{equation}\n{{\\partial f} \\over {\\partial m}} = \nCw(1+{\\gamma \\over \\alpha}) ({m \\over C_m}) ^ {\\gamma \\over \\alpha}=\nC_w(1+{\\gamma \\over \\alpha}) r ^ \\gamma\n\\label{eqn:dfdm}\n\\end{equation}\n\nFinally, we can construct our equation for flux attenuation by substituting equations \\ref{eqn:dmdt}, \\ref{eqn:dtdz} and \\ref{eqn:dfdm} into equation \\ref{eqn:dfdz}\n\n\\begin{equation}\n{{\\partial f} \\over {\\partial z}} = C_r C_m r^\\alpha(1+{\\gamma \\over \\alpha})\n\\end{equation}\n\n\nAnd now we can solve for equation \\ref{eqn:dfj}.\n\n\\begin{equation}\n\\Delta f_j = C_r C_m r^\\alpha (1+{\\gamma \\over \\alpha}) \\Delta z * n_{i-1,j}\n\\label{eqn:dfj}\n\\end{equation}\n\nWe also need to solve for $\\Delta f_1$ the flux ``attenuation'' that actually comes from particles leaving the smallest bin and escaping from what the UVP sees.\n\n\\begin{equation}\n\\Delta f_1 = \\Delta n_1 m_1 w_1 = \\Delta n_1 C_m C_w r_1^{\\alpha + \\gamma}\n\\label{eqn:df0}\n\\end{equation}\n\nHere, $\\Delta n_1$ is the number of particles leaving bin j = 1, but we haven't solved for that yet.\n\n\\section{Solving for $\\Delta n_j$}\n\nRecall that $\\Delta n_j$ is the number of particles that migrate between bin ``j'' and bin ``j-1'' as the particles sink from depth ``i-1'' to depth ``i''.\n\nThe flux at the shallower depth is equal to the flux at the deeper depth, plus the flux that attenuated between those two depths. Since $f = nmw$ and we know m and w\n\n\\begin{equation}\nn_{i-1,j-1} C_m C_w r_{j-1}^{\\alpha + \\gamma} + n_{i-1,j} C_m C_w r_{j}^{\\alpha + \\gamma} =\nn_{i,j-1} C_m C_w r_{j-1}^{\\alpha + \\gamma} + n_{i,j} C_m C_w r_{j}^{\\alpha + \\gamma} + \\Delta f_j\n\\label{eqn:n4f}\n\\end{equation}\n\nThis equation can be re-arranged, and we can substitute in equation \\ref{eqn:dfj} for $\\Delta f_j$. \n \n% \\begin{equation}\n% C_m C_w r_{j-1} ^{\\alpha + \\gamma} (n_{i-1,j-1}- n_{i,j-1}) = \n% C_m C_w r_{j}^{\\alpha + \\gamma} (n_{i,j} - n_{i-1,j}) + C_r C_m (1 + {\\gamma \\over \\alpha}) \\Delta z n_{i-1,j} r ^ \\alpha\n% \\end{equation}\n\nThe $C_m$ cancel out.\n\n\\begin{equation}\n C_w r_{j-1} ^{\\alpha + \\gamma} (n_{i-1,j-1}- n_{i,j-1}) = \n C_w r_{j}^{\\alpha + \\gamma} (n_{i,j} - n_{i-1,j}) + C_r (1 + {\\gamma \\over \\alpha}) \\Delta z n_{i-1,j} r ^ \\alpha\n\\end{equation}\n\nWe can then substitute in $\\Phi$ from equation \\ref{eqn:phi}.\n\\begin{equation}\nC_w r_{j-1} ^{\\alpha} \\Phi = \n C_w r_{j}^{\\alpha} \\Phi + C_r (1 + {\\gamma \\over \\alpha}) \\Delta z n_{i-1,j} r ^ \\alpha\n\\end{equation}\n\n\nRearrange\n% C_w \\Phi %(r_{j-1}^\\alpha - r_j^\\alpha)\n\n\\begin{equation}\nC_w\\Phi(r_{j-1}^\\alpha - r_j^\\alpha) = Cr(1 + {\\gamma \\over \\alpha}) \\Delta z r^{\\alpha} n_{i-1,j}\n\\end{equation}\n\nsolve for $\\Phi$\n\n\\begin{equation}\n\\Phi = {{{C_r \\over C_w} \\Delta z r^{\\alpha} n_{i-1,j}(1 + {\\gamma \\over \\alpha})} \\over {r_{j-1}^\\alpha - r_j^\\alpha}}\n\\end{equation}\n\n\\begin{equation}\n\\Delta n_j = {\\Phi \\over r_j^\\gamma} = {{{C_r \\over C_w} \\Delta z r^{\\alpha} n_{i-1,j}(1 + {\\gamma \\over \\alpha})} \\over {r_j^\\gamma(r_{j-1}^\\alpha - r_j^\\alpha})}\n\\label{eqn:dnj}\n\\end{equation}\n\n\\begin{equation}\n\\Delta n_{j-1} = {\\Phi \\over r_{j-1}^\\gamma} = {{\\Delta n_j r_j^\\gamma} \\over {r_{j-1}^\\gamma}}\n\\end{equation}\n\nAt this point, the only unsolved variable is $C_r$, which we can now calculate.\n\n\n\\section{Solving for $C_r$}\n\nWe can calculate $\\Delta F $, the attenuation of flux and can impose the size spectrum and all of the other constants. Here we find the $C_r$ that gives us the correct $\\Delta F $\n\nFirst, to solve equation \\ref{eqn:df} by substituting in equaitons \\ref{eqn:dfj} and \\ref{eqn:df0}\n\n\\begin{equation}\n\\Delta F = \\sum_{j = 2}^n \\Delta f_j + \\Delta f_1 = \n\\sum_{j = 2}^n \\left\\{C_r C_m r_j^\\alpha(1+{\\gamma \\over \\alpha}) \\Delta z n_{i-1,j} \\right\\} +\n\\Delta n_1 C_m C_w r_1^{\\alpha + \\gamma}\n\\end{equation}\n\nSubstitute equation \\ref{eqn:dnj} for $\\Delta n_j$ when j = 1 for $\\Delta n_1$\n\n\\begin{equation}\n\\Delta F = \\sum_{j = 2}^n \\Delta f_j + \\Delta f_1 = \n\\sum_{j = 2}^n \\left\\{C_r C_m r_j^\\alpha(1+{\\gamma \\over \\alpha}) \\Delta z n_{i-1,j} \\right\\} +\n {{{C_r \\over C_w} \\Delta z r_1^{\\alpha} n_{i-1,1}(1 + {\\gamma \\over \\alpha})} \\over {r_1^{\\gamma}(r_{0}^\\alpha - r_1^\\alpha})}\nC_m C_w r_1^{\\alpha + \\gamma}\n\\end{equation}\n\nIn the above, $r_0$ is the effective size of the particles smaller than the UVP can see. In principle this is arbitrary. Numbers closer to zero result in fewer particles in the smallest bin disapearing, larger ones to more of those particles disapearing. As $r_0$ approaches $r_1$ Cr approaches zero. They cannot be equal or the math breaks.\n\nPull what I can out of the sum operation, and cancel out $r^{\\gamma}$ and $C_w$ from the rightmost term\n\\begin{equation}\n\\Delta F =\nC_r C_m \\Delta z (1 + {\\gamma \\over \\alpha}) \\sum_{j = 2}^n \\left\\{r_j^\\alpha n_{i-1,j} \\right\\}\n+\n {{{C_r} \\Delta z r_1^{2 \\alpha} n_{i-1,1}(1 + {\\gamma \\over \\alpha})} \\over {(r_{0}^\\alpha - r_1^\\alpha})}\nC_m \n\\end{equation}\n\nNow we can solve for $C_r$\n\n\\begin{equation}\nC_r = {\n\\Delta F \\over \nC_m \\Delta z \n(1 + {\\gamma \\over \\alpha}) \\big[ \\sum_{j = 2}^n \\big\\{\nr_j^\\alpha n_{i-1,j}\n\\big\\} +\n{{r_1^{2\\alpha}n_{i-1,1}}\\over{r_0^\\alpha - r_1^\\alpha}}\n\\big]\n}\n\\end{equation}\n\nThus for a pair of profiles, we can estemate the flux attenuation, calculate Cr from that, and then plug Cr (and the profile) into the equation \\ref{eqn:dnj} for $\\Delta n_j$. We can thus compute $\\Delta n_j$ for each size class to see how many particles from that bin move to the next bin smaller.\n\n\n\n \\end{document}\n\n\n", "meta": {"hexsha": "c99b3f33ef5ae5e36f62ff70255dd0babca59829", "size": 10936, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "EularianPRiSM_Math/DeltaNjMathV2020.tex", "max_stars_repo_name": "cramjaco/POMZ-ETNP-UVP-2017", "max_stars_repo_head_hexsha": "a942b0a5ba082c025869c70fd6b30609b460567e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "EularianPRiSM_Math/DeltaNjMathV2020.tex", "max_issues_repo_name": "cramjaco/POMZ-ETNP-UVP-2017", "max_issues_repo_head_hexsha": "a942b0a5ba082c025869c70fd6b30609b460567e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "EularianPRiSM_Math/DeltaNjMathV2020.tex", "max_forks_repo_name": "cramjaco/POMZ-ETNP-UVP-2017", "max_forks_repo_head_hexsha": "a942b0a5ba082c025869c70fd6b30609b460567e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.8408304498, "max_line_length": 400, "alphanum_fraction": 0.6823335772, "num_tokens": 3834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891305219504, "lm_q2_score": 0.7745833945721304, "lm_q1q2_score": 0.6368540477000008}} {"text": "\\section{Temporal-Difference Learning}\n\n\\subsection{Exercise 6.1}\n\\subsubsection*{Q}\nIf $V$ changes during the episode, then (6.6) only holds approximately; what would the difference be between the two sides? Let $V_t$ denote the array of state values used at time $t$ in the TD error (6.5) and in the TD update (6.2). Redo the derivation above to determine the additional amount that must be added to the sum of TD errors in order to equal the Monte Carlo error.\n\\subsubsection*{A}\nWrite\n\\[\n \\delta_t \\doteq R_{t+1} + \\gamma V_t(S_{t+1}) - V_t(S_t),\n\\]\nthen\n\\begin{align}\n G_t - V_t(S_t) &= R_{t+1} + \\gamma G_{t+1} - V_t(S_t) \\\\\n &= \\delta_t + \\gamma [ G_{t+1} - V_{t+1}(S_{t+1})] + \\gamma [V_{t+1}(S_{t+1}) - V_t(S_{t+1})]\\\\\n &= \\delta_t + \\gamma [ G_{t+1} - V_{t+1}(S_{t+1})] + \\alpha \\gamma [R_{t+2} + \\gamma V_t(S_{t+2}) - V_t(S_{t+1})]\\\\\n &\\vdotswithin{=} \\notag\\\\\n &= \\sum_{k=t}^{T-1} \\gamma^{k-t} \\delta_k + \\alpha \\sum_{k=t}^{T-2} \\gamma^{k-t+1} [R_{k+2} + \\gamma V_k(S_{k+2}) - V_k(S_{k+1})].\n\\end{align}\n \n\\subsection{Exercise 6.2}\n\\subsubsection*{Q}\nThis is an exercise to help develop your intuition about why TD methods are often more efficient than Monte Carlo methods. Consider the driving home example and how it is addressed by TD and Monte Carlo methods. Can you imagine a scenario in which a TD update would be better on average than a Monte Carlo update? Give an example scenario—a description of past experience and a current state—in which you would expect the TD update to be better. Here’s a hint: Suppose you have lots of experience driving home from work. Then you move to a new building and a new parking lot (but you still enter the highway at the same place). Now you are starting to learn predictions for the new building. Can you see why TD updates are likely to be much better, at least initially, in this case? Might the same sort of thing happen in the original task?\n\\subsubsection*{A}\nTD updates incorporate prior information. Suppose we had a good value estimate for a trajectory $\\tau = S_1, S_2, \\dots, S_T$, then if we try to estimate the trajectory $\\tau = S_0, S_1, S_2, \\dots, S_T$ using MC then we need to see multiple episodes of this to get a good estimate of $V(S_0)$, not leveraging the info we already have on $\\tau$. TD would use info on $\\tau$ to back up the value of $S_0$ and hence converge much quicker. The key differences here are bootstrapping and online learning.\n\n\\subsection{Exercise 6.3}\n\\subsubsection*{Q}\nFrom the results shown in the left graph of the random walk example it appears that the first episode results in a change in only $V(A)$. What does this tell you about what happened on the first episode? Why was only the estimate for this one state changed? By exactly how much was it changed?\n\n\\subsubsection*{A}\nAll states apart from the terminal states were initialised to the same value (the terminal states must be initialised to 0) and the reward for non-terminal transitions is 0, so the TD(0) updates do nothing on the first pass to states that cannot lead directly to termination. \\\\\n\nIn the first run, the agent terminated on the left.\n\\begin{align}\n V_1(A) &= V_0(A) + \\alpha[ 0 + \\gamma \\times 0 + V_0(A)]\\\\\n &= (1-\\alpha) V_0(A) \\\\\n &= 0.9 \\times 0.5 \\\\\n &= \\frac{9}{20}.\n\\end{align} \nThe value of the estimate for $A$ reduced by $\\alpha V_0(A) = 0.05$.\n\n\\subsection{Exercise 6.4}\n\\subsubsection*{Q}\nThe specific results shown in the right graph of the random walk example are dependent on the value of the step-size parameter, $\\alpha$. Do you think the conclusions about which algorithm is better would be affected if a wider range of $\\alpha$ values were used? Is there a different, fixed value of $\\alpha$ at which either algorithm would have performed significantly better than shown? Why or why not?\n\\subsubsection*{A}\n\n\\begin{itemize}\n \\item General arguments given earlier about the benefits of TD are independent of $\\alpha$\n \\item Increases in $\\alpha$ make the curve more \n \\item Decreases in $\\alpha$ make the curve more smooth but make it converge slower.\n \\item We see enough of a range here to decide between the two methods\n\\end{itemize}\n\n\\subsection{*Exercise 6.5}\n\\subsubsection*{Q}\nIn the right graph of the random walk example, the RMS error of the TD method seems to go down and then up again, particularly at high $\\alpha$s. What could have caused this? Do you think this always occurs, or might it be a function of how the approximate value function was initialized?\n\\subsubsection*{A}\nThe state C happens to have een initialised to its true value. As training starts, updates occur on outer states (making them more accurate individually) which makes the error across all states reduce. This happens until the residual inaccuracies in the outer states propagate to C. The higher values of $\\alpha$ make this effect more pronounced, because the value estimate for C changes more readily in these cases.\n\n\\subsection{Exercise 6.6}\n\\subsubsection*{Q}\nIn Example 6.2 we stated that the true values for the random walk example are $\\frac16$, $\\frac26$, $\\frac36$, $\\frac46$, and $\\frac56$ for states A through E. Describe at least two different ways that these could have been computed. Which would you guess we actually used? Why?\n\\subsubsection*{A}\nCould have used DP, but probably did the following calculation.\\\\\n\nFirst note that\n\\[\n V(s) = \\E{}[\\mathds{1}\\{\\textsf{terminate on right from } S\\}] = \\P{}(\\textsf{terminate on right from } S).\n\\]\nAlso recognise that symmetry now implies $V(C) = 0.5$. Now\n\\begin{align*}\n V(E) &= \\frac12 \\times 1 + \\frac12 \\times V(D) \\\\\n &= \\frac12 + \\frac14[V(C) + V(E)],\n\\end{align*}\nso $V(E) = \\frac56$. We then get $V(D) = \\frac46$ and we can calculate the other states in the same way.\n\n\\subsection{*Exercise 6.7}\n\\subsubsection*{Q}\nDesign an off-policy version of the $TD(0)$ update that can be used with arbitrary target policy $\\pi$ and covering behavior policy $b$, using at each step $t$ the importance sampling ratio $\\rho_{t:t}$(5.3).\n\\subsubsection*{A}\nLet $G_t$ be returns from an episode generated by $b$. Then\n\\begin{align*}\n v_\\pi(s) &= \\E{}[\\rho_{t:T-1}G_t | S_t=s]\\\\\n &= \\E{}[\\rho_{t:T-1} R_{t+1} + \\gamma \\rho_{t:T-1}G_{t+1} | S_t=s] \\\\\n &= \\rho_{t:t} \\E{}[R_{t+1}|S_t=s] + \\gamma \\rho_{t:t}\\E{}[\\rho_{t+1:T-1}G_{t+1} |S_t=s] \\\\\n &= \\rho_{t:t} \\left( \\E{}[R_{t+1} | S_t=s] + \\E{}[\\rho_{t+1:T-1}G_t| S_t = s] \\right)\\\\\n &= \\rho_{t:t} \\left( r(s, A_t) + v_\\pi(S_{t+1})\\right).\n\\end{align*}\nSo the update for off-policy TD(0) (by sampling approximation) is \n\\begin{equation}\n V(S_t) \\leftarrow V(S_t) + \\alpha \\left[ \\rho_{t:t} R_{t+1} + \\rho_{t:t} \\gamma V(S_{t+1}) - V(S_t) \\right].\n\\end{equation}\n\n\\subsection{Exercise 6.8}\n\\subsubsection*{Q}\nShow that an action-value version of (6.6) holds for the action-value form of the TD error $\\delta_t = R_{t+1} + \\gamma Q(S_{t+1}, A_{t+1}) - Q(S_t, A_t)$, again assuming that the values don’t change from step to step.\n\\subsubsection*{A}\nWrite $\\delta_t \\doteq R_{t+1} + \\gamma Q(S_{t+1}, A_{t+1}) - Q(S_t, A_t)$. Then the Monte-Carlo error is\n\\begin{align*}\n G_t - Q(S_t, A_t) &= R_{t+1} + \\gamma G_{t+1} - Q(S_t, A_t) \\\\\n &= \\delta_t - \\gamma [Q(S_{t+1}, A_{t+1}) + G_{t+1}]\\\\\n &= \\sum_{k=t}^{T-1}\\gamma^{k-t}\\delta_k\n\\end{align*}\n\n\\subsection{Exercise 6.9 (programming): Windy Grid World with King's Moves}\n\\subsubsection*{Q}\nRe-solve the windy gridworld assuming eight possible actions, including the diagonal moves, rather than the usual four. How much better can you do with the extra actions? Can you do even better by including a ninth action that causes no movement at all other than that caused by the wind?\n\n\\subsubsection*{A}\n\\ProgrammingExercise\\\\\n\nOptimal trajectory is now 7 steps, rather than 15. Including the do-nothing action is not helpful in this example because the wind blows vertically and the goal position is not vertically separated from the start position. It could be useful in other wind environments though.\\\\\n\nBelow are the optimal trajectories for the book example (no diagonal moves) and for the exercise (king's moves). The numbers represent the wind strength in that position.\\\\\n\n\\includegraphics[width=\\textwidth]{\\ExerciseOutput/ex_6_9/optimal_no_diagonal_moves.png}\n\n\\includegraphics[width=\\textwidth]{\\ExerciseOutput/ex_6_9/optimal_with_diagonal_moves.png}\n\n\n\n\\subsection{Exercise 6.10 (programming): Stochastic Wind}\n\\subsubsection*{Q}\nRe-solve the windy gridworld task with King’s moves, assuming that the effect of the wind, if there is any, is stochastic, sometimes varying by 1 from the mean values given for each column. That is, a third of the time you move exactly according to these values, as in the previous exercise, but also a third of the time you move one cell above that, and another third of the time you move one cell below that. For example, if you are one cell to the right of the goal and you move left, then one-third of the time you move one cell above the goal, one-third of the time you move two cells above the goal, and one-third of the time you move to the goal.\n\n\\subsubsection*{A}\n\\ProgrammingExercise\\\\\n\nGreedy trajectory and learning curve shown below. Note that although the gradient of the learning curve becomes constant (so the algorithm converges), the greedy episode shown suffers from stochasticity in the wind.\n\n\\includegraphics[width=\\textwidth]{\\ExerciseOutput/ex_6_10/optimal.png}\n\n\\includegraphics[width=\\textwidth]{\\ExerciseOutput/ex_6_10/learning_curve.eps}\n\n\\subsection{Exercise 6.11}\n\\subsubsection*{Q}\nWhy is Q-learning considered an \\emph{off-policy} control method?\n\n\\subsubsection*{A}\nThe returns are sampled as if the agent followed the greedy policy with respect to $Q$.\n\n\\subsection{Exercise 6.12}\n\\subsubsection*{Q}\nSuppose action selection is greedy. Is Q-learning then exactly the same algorithm as Sarsa? Will they make exactly the same action selections and weight updates?\n\\subsubsection*{A}\nYes (?)\n\n\\subsection{Exercise 6.13}\n\\subsubsection*{Q}\nWhat are the update equations for Double Expected Sarsa with an $\\varepsilon$-greedy target policy?\n\\subsubsection*{A}\nExpected SARSA has the update \n\\[\n Q(S_t, A_t) \\leftarrow Q(S_t, A_t) + \\alpha \\left[ R_{t+1} + \\gamma \\E{}[Q(S_{t+1}) | S_{t+1}] - Q(S_t, A_t)\\right]\n\\]\n\nThe update for $S_t, A_t$ is \n\\[\n R_{t+1} + \\gamma \\sum_a \\pi(a|S_{t+1}) Q(S_{t+1}, a) - Q(S_t, A_t).\n\\]\nDouble expected SARSA would be keeping two $Q$ arrays and updating one of them each timestep, chosen with equal probability.\\\\\n\nFor an $\\varepsilon$-greedy policy we would increment $Q_1(S_t, A_t)$ by\n\\begin{equation}\n \\alpha \\left[ R_{t+1} + \\gamma \\left( \\frac{\\varepsilon}{|\\mathcal{A}(a)|} \\sum_a Q_2(S_{t+1}, a) + (1-\\varepsilon)\\max_a\\{Q_2(S_{t+1}, a)\\} \\right) - Q_1(S_t, A_t) \\right]\n\\end{equation}\nand the same with 1 and 2 reversed.\n\n\n\\subsection{Exercise 6.14}\n\\subsubsection*{Q}\nDescribe how the task of Jack’s Car Rental (Example 4.2) could be reformulated in terms of afterstates. Why, in terms of this specific task, would such a reformulation be likely to speed convergence?\n\n\\subsubsection*{A}\nOne might have coded this up initially with the states as the number of cars in each garage each evening. The agent then takes some action (moves some cars) and we transition stochastically to some state.\\\\\n\nAn alternative would be to introduce the number of cars in the morning (after the agent has moved cars) as an afterstate. This is because the agent is able to deterministically change the environment from evening to next morning (before rentals or returns).\\\\\n\nIn this case we would speed convergence by reducing the number of action-values to be calculated. For instance, we can now evaluate $(10, 0)$ moving one car and $(9, 1)$ moving no cars as the same afterstate $(9, 1)$.\n\n", "meta": {"hexsha": "0f63b67c4c9a2aaaaa05f0c5193b35d9a5a72cb7", "size": 11996, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises/chapters/chapter6/chapter6_content.tex", "max_stars_repo_name": "ElliotMunro200/reinforcement_learning_an_introduction", "max_stars_repo_head_hexsha": "c4fccb46a4bb00955549be3505144ec49f0132e5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 234, "max_stars_repo_stars_event_min_datetime": "2018-09-01T00:26:29.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T03:55:50.000Z", "max_issues_repo_path": "exercises/chapters/chapter6/chapter6_content.tex", "max_issues_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_issues_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2018-11-29T21:04:36.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-13T17:11:50.000Z", "max_forks_repo_path": "exercises/chapters/chapter6/chapter6_content.tex", "max_forks_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_forks_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 63, "max_forks_repo_forks_event_min_datetime": "2018-07-31T04:53:21.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T04:03:43.000Z", "avg_line_length": 66.2762430939, "max_line_length": 840, "alphanum_fraction": 0.7131543848, "num_tokens": 3417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649232, "lm_q2_score": 0.8221891283434876, "lm_q1q2_score": 0.636854037455907}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\\markright{tfrmh}\n\\section*{\\hspace*{-1.6cm} tfrmh}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nMargenau-Hill time-frequency distribution.\n\\end{minipage}\n\\vspace*{.2cm}\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[tfr,t,f] = tfrmh(x)\n[tfr,t,f] = tfrmh(x,t)\n[tfr,t,f] = tfrmh(x,t,N)\n[tfr,t,f] = tfrmh(x,t,N,trace)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.3cm}\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty tfrmh} computes the Margenau-Hill distribution of a\n discrete-time signal {\\ty x}, or the cross Margenau-Hill\n representation between two signals. This distribution has the\n following expression :\n\\begin{eqnarray*}\nMH_x(t,\\nu)&=&\\Re\\left\\{x(t)\\ X^*(\\nu)\\ e^{-j2\\pi \\nu t}\\right\\}\\\\\n&=&\\int_{-\\infty}^{+\\infty} \\frac{1}{2}\\ (x(t+\\tau)\\ x^*(t)+x(t)\\\nx^*(t-\\tau))\\ e^{-j2\\pi \\nu \\tau}\\ d\\tau.\n\\end{eqnarray*}\nIt corresponds to the real part of the Rihaczek distribution (see {\\ty\ntfrri}).\\\\\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{10cm} c}\nName & Description & Default\\\\\n\\hline\n {\\ty x} & signal if auto-MH, or {\\ty [x1,x2]} if cross-MH. {\\ty\n\t\t\t(Nx=length(x))}\\\\\n {\\ty t} & time instant(s) & {\\ty (1:Nx)}\\\\\n {\\ty N} & number of frequency bins & {\\ty Nx}\\\\\n {\\ty trace} & if nonzero, the progression of the algorithm is shown\n & {\\ty 0}\\\\\n \\hline {\\ty tfr} & time-frequency representation\\\\\n {\\ty f} & vector of normalized frequencies\\\\\n\n\\hline\n\\end{tabular*}\n\\vspace*{.1cm}\n\nWhen called without output arguments, {\\ty tfrmh} runs {\\ty tfrqview}.\n\\end{minipage}\n\\vspace*{.3cm}\n\n{\\bf \\large \\sf Example}\n\\begin{verbatim}\n sig=fmlin(128,0.1,0.4); tfrmh(sig,1:128,128,1);\n\\end{verbatim}\n\\vspace*{.3cm}\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nall the {\\ty tfr*} functions.\n\\end{minipage}\n\\vspace*{.3cm}\n\n{\\bf \\large \\sf Reference}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n[1] H. Margenhau, R. Hill ``Correlation between Measurements in Quantum\nTheory'', Prog. Theor. Phys. Vol. 26, pp. 722-738, 1961.\n\\end{minipage}", "meta": {"hexsha": "971e741c86ee273e25464c15de3a68e2b56f5ca3", "size": 2446, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/tfrmh.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/tfrmh.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/tfrmh.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 28.7764705882, "max_line_length": 75, "alphanum_fraction": 0.6161079313, "num_tokens": 917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189121808099, "lm_q2_score": 0.7745833841649232, "lm_q1q2_score": 0.6368540323937036}} {"text": "\\subsection{Partial Fractions}\r\n\r\nIf we have a function that is the quotient of two polynomials $f(x) = \\frac{P(x)}{Q(x)}$, it's often easier to break this quotient into a sum of parts where the denominator is a linear or quadratic factor and the numerator is always a smaller polynomial degree than the denominator.\r\n\r\n\\input{./backgroundReview/algebraPreCalc/partialFractions_example.tex}\r\n\r\n\r\nOne natural way to find these small denominators comes from the linear factors of the denominator where we keep complex roots in quadratic form.\r\nThis way, when making a common denominator, we get back the original big denominator.\r\nHowever, there are a few special cases we have to take care of.\r\n\r\n\\input{./backgroundReview/algebraPreCalc/linearFactors.tex}\r\n\\input{./backgroundReview/algebraPreCalc/repeatedLinearFactors.tex}\r\n\\input{./backgroundReview/algebraPreCalc/quadraticFactors.tex}\r\n\\input{./backgroundReview/algebraPreCalc/repeatedQuadraticFactors.tex}\r\n\\input{./backgroundReview/algebraPreCalc/improperFractions.tex}\r\n", "meta": {"hexsha": "d9c4a6556055b9fe37710343b5b2dadc2f2e2d4a", "size": 1027, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "calc/backgroundReview/algebraPreCalc/partialFractions.tex", "max_stars_repo_name": "rawsh/Math-Summaries", "max_stars_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "calc/backgroundReview/algebraPreCalc/partialFractions.tex", "max_issues_repo_name": "rawsh/Math-Summaries", "max_issues_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "calc/backgroundReview/algebraPreCalc/partialFractions.tex", "max_forks_repo_name": "rawsh/Math-Summaries", "max_forks_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.4117647059, "max_line_length": 283, "alphanum_fraction": 0.8033106134, "num_tokens": 225, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189121808099, "lm_q2_score": 0.7745833789613196, "lm_q1q2_score": 0.6368540281153573}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\\markright{tfrrgab}\n\\section*{\\hspace*{-1.6cm} tfrrgab}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nReassigned Gabor spectrogram time-frequency distribution.\n\\end{minipage}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[tfr,rtfr,hat] = tfrrgab(x) \n[tfr,rtfr,hat] = tfrrgab(x,t) \n[tfr,rtfr,hat] = tfrrgab(x,t,N) \n[tfr,rtfr,hat] = tfrrgab(x,t,N,Nh) \n[tfr,rtfr,hat] = tfrrgab(x,t,N,Nh,trace) \n[tfr,rtfr,hat] = tfrrgab(x,t,N,Nh,trace,k) \n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty tfrrgab} computes the Gabor spectrogram and its reassigned\n version. The analysis window $h$ used in this spectrogram is a\n gaussian window, which allows a 20 \\% faster algorithm than with\n the {\\ty tfrrsp} function (windows $\\ens{T}_h$ and $\\ens{D}_h$\n defined above are colinear in this case). The reassigned Gabor\n spectrogram is given by the following expressions :\n\\begin{eqnarray*}\n S_x^{(r)}(t',\\nu';h)=\\iint_{-\\infty}^{+\\infty} S_x(t,\\nu;h)\\\n\\delta(t'-\\hat{t}(x;t,\\nu))\\ \\delta(\\nu'-\\hat{\\nu}(x;t,\\nu))\\ dt\\ d\\nu,\n\\end{eqnarray*}\nwhere \n\\begin{eqnarray*}\n\\hat{t}(x;t,\\nu)=t-\\Re\\left\\{\\dfrac{F_x(t,\\nu;\\ens{T}_h)\\ F_x^*(t,\\nu;h)}\n{|F_x(t,\\nu;h)|^2}\\right\\} \\\\\n\\hat{\\nu}(x;t,\\nu)=\\nu+\\Im\\left\\{\\dfrac{F_x(t,\\nu;\\ens{D}_h)\\ F_x^*(t,\\nu;h)}\n{2\\pi\\ |F_x(t,\\nu;h)|^2}\\right\\} \n\\end{eqnarray*}\nwith $\\ens{T}_h(t)=t\\ h(t)$ and $\\ens{D}_h(t)=\\frac{dh}{dt}(t)$.\\\\\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty x} & analyzed signal ({\\ty Nx=length(x)})\\\\\n {\\ty t} & the time instant(s) & {\\ty (1:Nx)}\\\\\n {\\ty N} & number of frequency bins & {\\ty Nx}\\\\\n {\\ty Nh} & length of the gaussian window & {\\ty N/4}\\\\\n {\\ty trace} & if nonzero, the progression of the algorithm is shown\n & {\\ty 0}\\\\\n {\\ty k} & value at both extremities & {\\ty 0.001}\\\\\n\n\\hline \\end{tabular*} \\end{minipage}\n\n%\\newpage\n\n\\hspace*{1.5cm} \\begin{minipage}[t]{13.5cm} \n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty tfr, rtfr} & time-frequency representation and its reassigned\n version\\\\\n {\\ty hat} & complex matrix of the reassignment vectors\\\\\n\\hline\n\\end{tabular*}\n\\vspace*{.2cm}\n\nWhen called without output arguments, {\\ty tfrrgab} runs {\\ty tfrqview}.\n\\end{minipage}\n\\vspace*{1cm}\n\n{\\bf \\large \\sf Example}\n\\begin{verbatim}\n sig=fmlin(128,0.1,0.4); \n tfrrgab(sig,1:128,128,19,1);\n\\end{verbatim}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nall the {\\ty tfr*} functions.\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Reference}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n[1] F. Auger, P. Flandrin ``Improving the Readability of Time-Frequency and\nTime-Scale Representations by the Reassignment Method'' IEEE Transactions\non Signal Processing, Vol. 43, No. 5, pp. 1068-89, 1995.\n\\end{minipage}\n\n", "meta": {"hexsha": "39221a13bb7b0f08711a0ef4deb6e63b070192f7", "size": 3433, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/tfrrgab.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/tfrrgab.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/tfrrgab.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 31.495412844, "max_line_length": 77, "alphanum_fraction": 0.6163705214, "num_tokens": 1312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711832583695, "lm_q2_score": 0.7490872131147276, "lm_q1q2_score": 0.6367025448948395}} {"text": "\\section{Photons}\nSuppose you have a metal box at temperature $T$.\nThere is an electromagnetic field inside in equilibrium with the walls.\nThe field vanishes at the walls though.\n\nWhy should anybody study this?\nBecause all bodies which absorb all radiation shining on them (black bodies)\nemit radiation the same way when heated and a metal cavity is the best or\neasiest way to study.\n\nThe energy emitted by A equals the energy absorbed by a black body B equals the\nenergy emitted by B.\nThis is true frequency by frequency,\nso $u(\\nu)$ which is the energy density per unit is universal for every black\nbody.\n\nWe will find $u(\\nu)$ by applying equilibrium statistical mechanics to the\nelectromagnetic field inside the box.\n\nIn order to have a Hamiltonian formalism,\nwe'll use the variables\n$\\vec{A}(\\vec{r}, t)$ instead of $\\vec{E}$ and $\\vec{B}$.\n\nDifferent $\\vec{A}$'s differing by a gauge transformation,\ngive the same $\\vec{E}$ and $\\vec{B}$ and describe the same physics.\nWe'll impose the condition\n\\begin{align}\n \\vec{\\nabla}\\cdot{A} &= 0\\\\\n \\phi &= 0\n\\end{align}\nto select one $\\vec{A}$ for every physically distinct field $\\vec{E},\\vec{B}$.\nRecalling that we can recover the electric and magnetic fields with\n\\begin{align}\n \\vec{E} &= -\\frac{\\partial A}{c \\partial t} - \\vec{\\nabla}\\phi\n =\n -\\frac{1}{c}\\frac{\\partial \\vec{A}}{\\partial t}\\\\\n \\vec{B} &= \\vec{\\nabla}\\times\\vec{A}\n\\end{align}\nThe Lagrangian\n\\begin{align}\n L &=\n \\frac{1}{8\\pi} \\int d^3 r\\,\n \\left[ \n \\left( \\frac{\\partial \\vec{A}}{c \\partial t} \\right)^2\n - \\left( \\vec{\\nabla}\\times\\vec{A} \\right)^2\n \\right]\n\\end{align}\nleads to the equation of motion\n\\begin{align}\n \\frac{1}{c^2} \\frac{\\partial A}{\\partial t^2}\n - \\nabla^2 \\vec{A} &= 0\n\\end{align}\nwhich is equivalent to Maxwell's equations.\nThe Hamiltonian is\n\\begin{align}\n H &=\n \\int d^3r\\left[\n \\vec{\\Pi}\\cdot \\frac{\\partial \\vec{A}}{\\partial t} - H\n \\right]\\\\\n &=\n \\int d^3r\\,\\left[\n \\Pi^2 4\\pi c^2\n - \\frac{(4\\pi c)^2}{8\\pi} \\Pi^2\n + \\frac{\\vec{\\nabla}\\times\\vec{A}}{8\\pi}\n \\right]\n\\end{align}\nwhere\n\\begin{align}\n \\vec{\\Pi} &:=\n \\frac{\\partial H}{\\partial \\dot{\\vec{A}}(\\vec{r})}\n =\n \\frac{1}{4\\pi} \\frac{1}{c^2} \\frac{\\partial \\vec{A}}{\\partial t}\\\\\n\\end{align}\nso then\n\\begin{align}\n H\n &=\n \\frac{1}{8\\pi} \\int d^3r\\,\\left[ \n \\underbrace{\\left( 4\\pi c \\right)^2 \\Pi^2}_{E^2}\n +\n \\underbrace{\\vec{\\nabla}\\times\\vec{A}}_{B^2}\n \\right]\n\\end{align}\nApplying a change of variables to\n\\begin{align}\n \\vec{A}(\\vec{r}, t) &=\n \\frac{1}{\\sqrt{V}} \\sum_k\n \\vec{Q}(\\vec{k}, t) e^{i\\vec{k}\\cdot \\vec{r}}\\\\\n \\vec{\\Pi}(\\vec{r}, t) &=\n \\frac{1}{\\sqrt{V}} \\sum_k\n \\frac{\\vec{P}(\\vec{k}, t)}{\\sqrt{4\\pi} c^2} e^{i\\vec{k}\\cdot\\vec{r}}\n\\end{align}\nwhere $\\vec{k}=\\frac{2\\pi}{L}\\vec{n}$ with $\\vec{n}=0,\\pm 1,\\ldots$\nand\n\\begin{align}\n \\vec{Q}_{\\vec{k}}^* &= \\vec{Q}_{-\\vec{k}}\\\\\n \\vec{P}_{\\vec{k}}^* &= \\vec{P}_{-\\vec{k}}\n\\end{align}\nso then the Hamiltonian becomes decoupled as\n\\begin{align}\n H &=\n \\frac{1}{2}\\sum_{k}\\left[ \n \\vec{P}_{\\vec{k}}\\cdot\\vec{P}_{-\\vec{k}}\n +\n \\underbrace{c^2 k^2}_{=:\\omega_k^2}\n \\vec{Q}_{\\vec{k}}\\cdot\\vec{Q}_{-\\vec{k}}\n \\right]\n\\end{align}\nwhich is very much like the phonon Hamiltonian.\nClassically,\nthe energy density is given by the equipartition theorem\n\\begin{align}\n E &=\n \\sum_{k} \\frac{1}{2} k_B\n \\times \\underbrace{2}_{\\textrm{$P$ and $Q$ are quadratic}}\n \\times \\underbrace{2}_{\\textrm{2 independent polarizations}}\\\\\n &\\approx V \\int d^3 k 2k_B T\\\\\n &\\to \\infty\n\\end{align}\nThe divergence comes because contrary to the phonon case,\nthere is not cutoff at high momentum $k$.\nStill,\nthe contribution from every frequency range\n\\begin{align}\n \\frac{E}{V} &=\n \\frac{2k_B T}{2\\pi^2} \\int_{0}^{\\infty} dk\\, k^2\\\\\n &=\n \\int_{0}^{\\infty}\n \\underbrace{\\frac{k_B T}{\\pi^2} \\frac{(2\\pi)^3}{c^3} \\nu^2}_{u(\\nu)}\n \\, d\\nu\n\\end{align}\nand with $\\nu = \\frac{kc}{2\\pi}$,\nthe energy density between $\\nu$ and $\\nu + d\\nu $ is\n\\begin{align}\n u(\\nu) \\, d\\nu &=\n \\frac{8k_B T \\pi}{c^3} \\nu^2\\, d\\nu\n\\end{align}\nthis is known as the \\emph{Rayleigh-Jeans law},\nwhich agrees with experiments in the low frequency regime.\nBut not for high frequencies as it diverges!\n\nThe quantum theory is obtained by making\n$\\vec{A}$, $\\vec{\\Pi}$, $\\vec{Q}$ and $\\vec{P}$\ninto operators.\n\nAssuming canonical commutation relations for $\\vec{A}$ and $\\vec{\\Pi}$.\n\\begin{align}\n \\left[ \\hat{A}_i(\\vec{r}), \\hat{\\Pi}_j(\\vec{r}') \\right]\n =\n i\\hbar \\delta_{ij} \\delta\\left( \\vec{r} - \\vec{r}' \\right)\n\\end{align}\nwe find\n\\begin{align}\n \\left[ \\hat{A}_i(\\vec{r}), \\hat{\\Pi}_j(\\vec{r}') \\right] &=\n \\frac{1}{V} \\sum_{k,k'}\n e^{i\\left( \\vec{k}\\cdot\\vec{r} + \\vec{k}'\\cdot\\vec{r}' \\right)}\n \\underbrace{ \\left[ Q_i(\\vec{k}), P_j(\\vec{k}') \\right]}_{%\n i\\hbar \\delta_{ij} \\delta_{\\vec{k},-\\vec{k}'}\n }\\\\\n &=\n \\frac{i\\hbar}{V} \\sum_{\\vec{k}}\n e^{i\\vec{k}\\cdot\\left( \\vec{r} - \\vec{r}' \\right)} \\delta_{ij}\\\\\n &=\n i\\hbar \\delta_{ij} \\delta\\left( \\vec{r} - \\vec{r}' \\right)\n\\end{align}\nso then\n\\begin{align}\n \\left[ Q_i\\left( \\vec{r} \\right), P_j\\left( \\vec{r}' \\right) \\right]\n =\n i\\hbar \\delta_{ij} \\delta_{\\vec{k}, -\\vec{k}'}\n\\end{align}\nand all the other commutators between $Q_k$ and $P_k$ vanish.\nWe then define\n\\begin{align}\n \\hat{a}_{\\vec{k}}^{\\alpha} &=\n \\left( \n \\sqrt{\\frac{\\omega_k}{2\\hbar}} \\vec{Q}(\\vec{k})\n + \\frac{i}{\\sqrt{2\\omega_k\\hbar}} \\vec{P}(\\vec{k})\n \\right)\n \\cdot e_{\\vec{k}}^{\\alpha}\\\\\n {\\hat{a}_{\\vec{k}}^{\\alpha}}^\\dagger &=\n \\left( \n \\sqrt{\\frac{\\omega_k}{2\\hbar}} \\vec{Q}(\\vec{k})\n - \\frac{i}{\\sqrt{2\\omega_k\\hbar}} \\vec{P}(\\vec{k})\n \\right)\n \\cdot e_{\\vec{k}}^{\\alpha}\n\\end{align}\nwhere $e_{\\vec{k}}^1$ and $e_{\\vec{k}}^2$\nare two unit vectors orthogonal to $\\vec{k}$,\nwhich are the polarization directions.\nAnd the commutation relations\n\\begin{align}\n \\left[\n \\hat{a}_{\\vec{k}}^{\\alpha},\n {\\hat{a}_{\\vec{k}}^{\\beta}}^\\dagger,\n \\right]\n &=\n -i \\sqrt{\\frac{\\omega_k}{2\\hbar}}\n \\frac{1}{\\sqrt{2\\omega_{k'}\\hbar}}\n \\underbrace{\\left[ Q_i(\\vec{k}), P_j(\\vec{k}'), \\right]}_{%\n i\\hbar \\delta_{ij} \\delta_{k,k'}\n }\n e_i^{\\alpha}(k)\n e_j^{\\beta}(k)\\nonumber\\\\\n &\\qquad\n - \\sqrt{\\frac{\\omega_{k'}}{2\\hbar}}\n \\frac{1}{\\sqrt{2\\omega_k \\hbar}}\n \\underbrace{\\left[ P_j(\\vec{k}), Q_i(\\vec{k}') \\right]}_{%\n -i\\hbar \\delta_{ij} \\delta_{k,k'}\n }\n e_j^{\\alpha}(k)\n e_i^{\\alpha}(k')\\\\\n &=\n \\delta_{\\alpha\\beta} \\delta_{\\vec{k},\\vec{k}'}\n\\end{align}\nand also\n\\begin{align}\n \\left[ \\hat{a}_{\\vec{k}}^{\\alpha}, \\hat{a}_{\\vec{k}'}^{\\beta} \\right] =\n \\left[\n {\\hat{a}_{\\vec{k}}^{\\alpha}}^\\dagger,\n {\\hat{a}_{\\vec{k}'}^{\\beta}}^\\dagger\n \\right] = 0\n\\end{align}\nThus,\n$\\hat{a}^\\alpha(k)$ and ${\\hat{a}^{\\alpha}}^\\dagger(k)$\nare creation and annihilation operators for photons with momentum $k$ and\npolarization $e^{\\alpha}(k)$.\nIn terms of them,\nthe Hamiltonian is\n\\begin{align}\n \\hat{H} &=\n \\frac{1}{2} \\sum_{\\alpha=1,2} V \\int d^3k\\,\\left[\n -\\frac{\\hbar\\omega_k}{2}\n \\left( \n \\hat{a}_{\\vec{k}}^{\\alpha}\n - {\\hat{a}_{-\\vec{k}}^{\\alpha}}^\\dagger\n \\right)\n \\left( \n \\hat{a}_{-\\vec{k}}^{\\alpha}\n - {\\hat{a}_{\\vec{k}}^{\\alpha}}^\\dagger\n \\right)\n +\n \\omega_k^2\n \\left( \n \\hat{a}_{\\vec{k}}^{\\alpha}\n - {\\hat{a}_{-\\vec{k}}^{\\alpha}}^\\dagger\n \\right)\n \\left( \n \\hat{a}_{-\\vec{k}}^{\\alpha}\n - {\\hat{a}_{\\vec{k}}^{\\alpha}}^\\dagger\n \\right)\n \\right]\\\\\n &\\approx\n \\frac{1}{2} \\sum_{\\alpha=1,2} \\int d^3k\\left[ \n \\frac{\\hbar\\omega_k}{2} 2\n \\hat{a}_{k}^{\\alpha}\n {\\hat{a}_{k}^{\\alpha}}^\\dagger\n +\n {\\hat{a}_{k}^{\\alpha}}^\\dagger\n \\hat{a}_{k}^{\\alpha}\n 2\n \\frac{\\hbar\\omega_k}{2}\n \\right]\\\\\n &=\n \\sum_{\\alpha=1,2} \\int d^3k\\left[ \n {\\hat{a}_{k}^{\\alpha}}^\\dagger\n \\hat{a}_{k}^{\\alpha}\n +\n \\frac{1}{2}\n \\right]\n \\hbar\\omega_k\n\\end{align}\nso that's one harmonic oscillator for every $\\vec{k}$ and $\\alpha$.\n\nHow we can calculate the partition function\n\\begin{align}\n Z &= \\Tr e^{-\\beta \\hat{H}}\\\\\n &=\n \\prod_{\\vec{k},\\alpha} \\sum_{n_k=0}^{\\infty}\n e^{-\\beta\\hbar\\omega_k\\left( n_k + \\frac{1}{2} \\right)}\\\\\n &= \\prod_{\\vec{k}, \\alpha}\n \\frac{e^{-\\beta\\hbar\\omega_k/2}}{1 - e^{-\\beta\\omega_k\\hbar}}\\\\\n &= \\prod_{\\vec{k},\\alpha} \\frac{2}{\\sinh(\\beta\\hbar\\omega_k/2)}\n\\end{align}\nand the free energy\n\\begin{align}\n F &=\n -k_B T \\ln Z\\\\\n &=\n k_B T \\sum_{\\vec{k}, \\alpha} \\ln\\left( \n \\frac{1 - e^{-\\beta \\omega_k \\hbar}}{e^{-\\beta\\hbar\\omega_k/2}}\n \\right)\\\\\n &=\n k_B T \\sum_{\\vec{k},\\alpha}\\left[ \n \\ln\\left( \n 1 - e^{-\\beta\\hbar\\omega_k}\n \\right)\n +\n k_B T \\beta \\hbar\\omega_k\n \\right]\\\\\n &=\n k_B T \\sum_{\\vec{k},\\alpha}\\left[ \n \\ln\\left( \n 1 - e^{-\\beta\\hbar\\omega_k}\n \\right)\n +\n \\underbrace{\\hbar\\omega_k}_{\\textrm{$T$-independent zero-point energy}}\n \\right]\\\\\n &\\approx\n 2 k_B T V \\int d^3k \\ln\\left( \n 1 - e^{-\\beta \\hbar\\omega_k}\n \\right)\n + \\textrm{const}\n\\end{align}\nwhich is the free energy of an ideal gas of bosons with dispersion relation\n$\\omega_k=kc$ and two polarizations plus a constant.\n\nThe energy is then\n\\begin{align}\n \\frac{E}{V} &=\n 2 \\int d^3 k \\frac{\\hbar ck}{e^{\\beta \\hbar c k} - 1}\\\\\n &=\n \\frac{2}{\\left( \\beta\\hbar c \\right)^3 }\n \\frac{1}{2\\pi^2}\n \\frac{1}{\\beta}\n \\underbrace{\\int_{0}^{\\infty}\n dx\\,\n \\frac{x^3}{e^{x} - 1}}_{%\n \\pi^4/15\n }\\\\\n &=\n \\frac{\\pi^2 \\left( k_B T \\right)^4}{15 \\hbar^3 c^3}\n\\end{align}\nThis is the \\emph{Steffan-Boltzmann law}.\n\nAlso, to get the spectrum,\n\\begin{align}\n \\frac{E}{V} &=\n 2 \\frac{1}{2\\pi^2} \\int_{0}^{\\infty} dk\\,\n \\frac{k^2 \\hbar ck}{e^{\\beta\\hbar ck} - 1}\\\\\n &=\n \\frac{\\hbar c}{\\pi^2} \\left( \\frac{2\\pi}{c} \\right)^4\n \\int_{0}^{\\infty} d\\nu\\,\n \\frac{\\nu^3}{e^{\\beta h\\nu} - 1}\\\\\n &=\n \\int_{0}^{\\infty} d\\nu\\,\n \\underbrace{%\n \\frac{16\\pi^2\\hbar}{c^3}\n \\frac{\\nu^3}{e^{\\beta h \\nu} - 1}}_{%\n u(\\nu)\n }\n\\end{align}\nwhere\n\\begin{align}\n u(\\nu) = \n \\frac{16\\pi^2\\hbar}{c^3}\n \\frac{\\nu^3}{e^{\\beta h \\nu} - 1}\n\\end{align}\nis the energy per volume per frequency.\nThis is \\emph{Planck's law}.\n", "meta": {"hexsha": "a4da31424c15d2d2ad55a33cf745892cbc4e721c", "size": 10324, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "phys612/lecture22.tex", "max_stars_repo_name": "ehua7365/umdphysnotes", "max_stars_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-12-11T12:53:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-11T12:53:46.000Z", "max_issues_repo_path": "phys612/lecture22.tex", "max_issues_repo_name": "ehua7365/umdphysnotes", "max_issues_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "phys612/lecture22.tex", "max_forks_repo_name": "ehua7365/umdphysnotes", "max_forks_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.5983379501, "max_line_length": 79, "alphanum_fraction": 0.5700309957, "num_tokens": 4122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711604559846, "lm_q2_score": 0.7490872131147275, "lm_q1q2_score": 0.6367025278138644}} {"text": "\\subsection{The Space of Germs}\r\nLet $D\\subset\\mathbb C$ a domain.\r\n\\begin{definition}\r\n Let $(f,U)$ and $(g,V)$ be function elements on $D$.\r\n For any $z\\in U\\cap V$, write $(f,U)\\equiv_z(g,V)$ if $f,g$ agree on a neighbourhood of $z$.\r\n\\end{definition}\r\nEasily $\\equiv_z$ is an equivalence relation.\r\n\\begin{definition}\r\n Let $(f,U)$ be a function element and $z\\in U$.\r\n The equivalence class of $(f,U)$ under $\\equiv_z$ is called the germ of $f$ at $z$ and is denoted by $[f]_z$.\r\n\\end{definition}\r\nSo two germs $[f]_z$ and $[g]_w$ are equal iff $z=w$ and $f=g$ on a neighbourhood of $z=w$.\r\nWe want to study all possible germs on a domain $D$.\r\n\\begin{definition}\r\n The space of germ over $D$ is\r\n $$\\mathcal G=\\{[f]_z:z\\in D,(f,D)\\text{ a function element with $z\\in U$}\\}$$\r\n\\end{definition}\r\nNow we defined it as a set, it is natural to endow a topology on it.\r\nFor any function element $(f,U)$ on $D$, let $[f]_U=\\{[f]_z:z\\in U\\}$.\r\n\\begin{lemma}\r\n The collection $\\{[f]_U\\}$ for $U$ open in $D$ is a basis of a topology on $\\mathcal G$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Let $(f,U)$ and $(g,V)$ be function elements on $D$.\r\n For any $[h]_z\\in [f]_U\\cap [g]_V$, then $h$ agrees with $f,g$ on a neighbourhood $W$ of $z$, therefore $[h]_W\\subset [f]_U\\cap[g]_V$.\r\n\\end{proof}\r\nThis is the topology we want.\r\n\\begin{lemma}\r\n $\\mathcal G$ is Hausdorff.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Consider elements of $[f]_z,[g]_z\\in\\mathcal G$ with $[f]_z\\neq [g]_z$.\\\\\r\n If $z\\neq w$ then we can choose function elements $(f,U)\\in [f]_z,(g,V)\\in[g]_w$ such that $U\\cap V=\\varnothing$, therefore $[f]_U$ and $[g]_V$ are disjoint.\\\\\r\n If $z=w$, then we can choose an open neighbourhood $U$ such that $(f,U)\\in[f]_z$ and $(g,U)\\in [g]_z$.\r\n Unless $[f]_U\\cap [g]_U=\\varnothing$, there is a germ $[h]_z\\subset [f]_U\\cap[g]_U$.\r\n By the identity principle $f|_U=h|_U=g|_U$, which means $[f]_z=[h]_z=[g]_z$ due to the connectedness of $U$, contradiction.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $\\mathcal G$ be the space of germs over a domain $D$.\r\n The forgetful map $\\pi:\\mathcal G\\to D$ is defined by $\\pi([f]_z)=z$.\r\n\\end{definition}\r\n\\begin{lemma}\r\n For each component $G\\subset\\mathcal G$, the restriction $\\pi:G\\to D$ is a covering map.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Take an open $U\\subset D$.\r\n Then the pre-image of $U$ has to be\r\n $$\\pi^{-1}(U)=\\bigcup_{(f,V)\\text{ function element on }U}[f]_V$$\r\n which is open.\r\n So $\\pi$ is continuous.\\\\\r\n For each open set in the form $[f]_U$, we have\r\n $$(\\pi|_{[f]_U})^{-1}(z)=[f]_z$$\r\n which is a continuous inverse of $\\pi|_{[f]_U}$.\r\n This shows that $\\pi$ is a local homeomorphism, hence a covering map.\r\n\\end{proof}\r\nHence, by Lemma \\ref{covering_conformal}, $\\pi$ induces a well-defined conformal structure on $\\mathcal G$ (well, on each of its connected components) such that $\\pi$ is analytic.\r\nExplicitly, the atlas we have in mind consists of charts $(\\pi|_{[f]_U},[f]_U)$ across all the function elements $(f,U)$ on $D$.\r\n\\begin{definition}\r\n Let $\\mathcal G$ be the space of germs on a domain $D$.\r\n The evaluation map $\\mathcal E:\\mathcal G\\to\\mathbb C$ is defined by $\\mathcal E([f]_z)=f(z)$.\r\n\\end{definition}\r\nIn the chart $(\\pi|_{[f]_U},[f]_U)$, we have\r\n$$\\mathcal E\\circ(\\pi|_{[f]_U})^{-1}(z)=\\mathcal E([f]_z)=f(z)$$\r\nTherefore $\\mathcal E$ is analytic.", "meta": {"hexsha": "662d66ec7ca36471e730aa03e9c52ff401588aa4", "size": 3395, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "8/germs.tex", "max_stars_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_stars_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "8/germs.tex", "max_issues_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_issues_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "8/germs.tex", "max_forks_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_forks_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.046875, "max_line_length": 180, "alphanum_fraction": 0.6321060383, "num_tokens": 1234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.749087201911703, "lm_q2_score": 0.8499711718571775, "lm_q1q2_score": 0.6367025268321044}} {"text": "%auto-ignore\n%! TEX root = ../Text.tex\n\\providecommand{\\MainFolder}{..}\n\\documentclass[\\MainFolder/Text.tex]{subfiles}\n\n\\begin{document}\n\\section{Orientation, Poincar\\'e duality, Hodge decomposition}\n\\label{SubSec:CycStr}\n\nIn this section, a $\\DGA$ $V$ is a triple $(V,\\Dd,\\wedge)$, where \n\\begin{itemize}\n\\item $V = \\bigoplus_{n\\in \\Z} V^n$ is a $\\Z$-graded vector space,\n\\item $\\Dd: V\\rightarrow V$ a differential of degree $1$, and\n\\item $\\wedge: V\\otimes V \\rightarrow V$ an associative product of degree $0$ such that $\\Dd$ and $\\wedge$ satisfy the Leibnitz identity $\\Dd(v_1 \\wedge v_2) = \\Dd v_1 \\wedge v_2 + (-1)^{v_1} v_1 \\wedge\\Dd v_2$ for all homogenous $v_1$, $v_2\\in V$.\n%\\[ \\Dd(v_1 \\wedge v_2) = \\Dd v_1 \\wedge v_2 + (-1)^{v_1} v_1 \\wedge \\Dd v_2. \\]\n\\end{itemize}\nIn other words, we consider general, possibly non-unital and non-commutative, $\\Z$-graded $\\DGA$'s. We denote by $\\deg(v)$ the degree of a homogenous element $v\\in V$ and write $(-1)^v$ in the exponent.\n%We will consider two closely related notions on a $\\DGA$ --- an orientation and a cyclic structure.\n \n%We will not need degree shifts, and hence we will use $\\Abs{v}$ and $\\Deg(v)$ (and~$v$ in the exponent) to denote the degree of $v$ interchangeably.\n\n\\begin{Definition}[Orientations and cyclic structures]\\label{Def:CycStr} \nLet $(V,\\Dd)$ be a $\\Z$-graded cochain complex. We define the following:\n\\begin{itemize}\n\\item An \\emph{orientation in degree $n$} is a linear function $\\Or : V^n \\rightarrow \\R$ such that\n\\[\\Or \\neq 0\\quad\\text{and}\\quad\\Or \\circ \\Dd = 0. \\]\nIn other words, it is a surjective chain map $\\Or : (V,\\Dd) \\rightarrow (\\R[-n],0)$.\n\\item A~\\emph{cyclic structure of degree $n$} is a homogenous bilinear form $\\langle \\cdot, \\cdot \\rangle : V \\otimes V \\rightarrow \\R$ of degree $-n$ as a map such that for all homogenous $v_1$, $v_2\\in V$, we have\n\\begin{equation}\\label{Eq:CycStr}\n\\langle \\Dd v_1, v_2\\rangle = (-1)^{1+v_1 v_2} \\langle \\Dd v_2, v_1 \\rangle.\n\\end{equation}\n\\end{itemize}\nA cyclic structure on a $\\DGA$ $(V,\\Dd,\\wedge)$ is additionally required to satisfy \n\\begin{equation}\\label{Eq:CycStrII}\n\\langle v_1 \\wedge v_2, v_3 \\rangle = (-1)^{v_3(v_1 + v_2)}\\langle v_3\\wedge v_1, v_2 \\rangle\n\\end{equation}\nfor all homogenous $v_1$, $v_2$, $v_3\\in V$.\n%Morphisms of such objects are morphisms of $\\DGA$'s preserving the cyclic structure, resp.~the orientation. \n\\end{Definition}\n\nAnalogously, one can define a cyclic structure on an $\\AInfty$-algebra.\n\n\\begin{Remark}[Cyclic $\\DGA$]\\label{Rem:Eq}\nA cyclic $\\DGA$ $(V,\\Pair,\\mu_1,\\mu_2)$ of degree $n$ from Definition~\\ref{Def:CyclicAinfty} is the same as a $\\DGA$ equipped with a cyclic structure~$\\langle\\cdot,\\cdot\\rangle$ of degree $n$ which is symmetric, i.e., \n\\[ \\langle v_1, v_2 \\rangle = (-1)^{v_1 v_2} \\langle v_2, v_1\\rangle \\]\nfor all homogenous $v_1$, $v_2\\in V$, and non-degenerate (see Definition~\\ref{Def:PoincDual}). The correspondence is via the degree shift\n\\begin{align*}\n\\mu_1(\\SuspU v) &= \\SuspU \\Dd(v),\\\\\n\\mu_2(\\SuspU v_1, \\SuspU v_2) &= (-1)^{v_1}\\SuspU(v_1\\wedge v_2),\\\\\n\\Pair(\\SuspU v_1, \\SuspU v_2) & = (-1)^{v_1} \\langle v_1,v_2\\rangle,\n\\end{align*}\nwhere $\\SuspU$ is a formal symbol of degree $-1$. From this reason, we sometimes call a $\\DGA$ equipped with a non-degenerate symmetric cyclic structure a cyclic $\\DGA$, although there are degree shifts involved.\n\\end{Remark}\n\nThe homology $\\H(V)\\coloneqq \\H(V,\\Dd)$ of a $\\DGA$ $(V,\\Dd,\\wedge)$ is also a $\\DGA$ with the induced product $\\wedge$ and with zero differential. Given an orientation $\\Or$ or a cyclic structure $\\langle\\cdot,\\cdot\\rangle$ on~$V$, we define the maps $\\Or^\\H: \\H(V) \\rightarrow \\R$ and $\\langle\\cdot,\\cdot\\rangle^\\H: \\H(V)\\otimes\\H(V)\\rightarrow\\R$ for all closed $h_1$, $h_2\\in V$ by\n\\begin{equation}\\label{Eq:IndOnHom}\n\\begin{aligned}\n\\Or^\\H([h_1]) &\\coloneqq \\Or(h_1)\\text{ and}\\\\\n\\langle[h_1],[h_2]\\rangle^\\H &\\coloneqq \\langle h_1, h_2\\rangle,\n\\end{aligned}\n\\end{equation}\nrespectively, where $[\\cdot]$ denotes the cohomology class. It is easy to see that $\\langle\\cdot,\\cdot\\rangle^\\H$ is a cyclic structure on $\\H(V)$ and that $\\Or^\\H$ is an orientation on $\\H(V)$ provided that $\\Restr{\\Or}{\\Ker \\Dd} \\neq 0$. \n\n\\begin{Proposition}[Orientation on homology]\\label{Prop:OrOnHomG}\nWe have the following:\n\\begin{ClaimList}\n\\item Let $(V,\\Dd)$ be a $\\Z$-graded cochain complex, and let $n\\in \\Z$. Given an orientation $\\tilde{\\Or}: \\H^n(V) \\rightarrow \\R$, there is an orientation $\\Or^V: V^n \\rightarrow \\R$ such that~$\\tilde{\\Or} = \\Or^\\H$. If $\\Dd V^n = 0$, then we have the correspondence\n\\[\\text{Orientations on }V\\text{ in degree }n\\ \\overset{1:1}{\\simeq}\\ \\text{Orientations on }\\H(V)\\text{ in degree }n. \\] \n\\item If $(V_1,\\Dd_1,\\Or_1)$ and $(V_2,\\Dd_2,\\Or_2)$ are $\\Z$-graded cochain complexes oriented in degree $n$ with $\\Dd V_1^n = 0 = \\Dd V_2^n$, then a chain map $f: V_1 \\rightarrow V_2$ preserves the orientation if and only if the induced map $f_*: \\H(V_1) \\rightarrow \\H(V_2)$ preserves the induced orientation.\n\\end{ClaimList}\n\\end{Proposition}\n\\begin{proof}\n\\begin{ProofList}\n\\item We define $\\Or^V(v) \\coloneqq \\tilde{\\Or}([v])$ for all closed $v\\in V$ and extend it by $0$ to a complement of $\\Ker \\Dd$ in~$V$. It is obvious that $\\Or^\\H = \\tilde{\\Or}$. If $\\Dd V^n = 0$, any complement is trivial and $\\Or^V$ is uniquely determined by $\\Or^\\H$.\n\n\\item Let $v\\in V_1^n$. Because $\\Dd v = 0$, we have\n\\[ \\Or_2(f(v)) = \\Or^{\\H}_2([f(v)]) = \\Or^{\\H}_2(f_*[v]) = \\Or^{\\H}_1([v]) = \\Or_1(v). \\]\nThis finishes the proof.\\qedhere\n\\end{ProofList}\n\\end{proof}\n\nSuppose that $1$ is a unit for $(V,\\Dd,\\wedge)$, i.e., $1\\in V^0$, $\\Dd 1 = 0$ and $1\\wedge v = v \\wedge 1 = v$ for all $v\\in V$, and let $\\langle \\cdot,\\cdot\\rangle$ by a cyclic structure on $V$. For all homogenous $v_1$, $v_2\\in V$, we have \n\\begin{align*}\n\\langle v_1, v_2 \\rangle &= \\langle v_1\\wedge 1,v_2\\rangle \\\\\n& = (-1)^{v_1 v_2}\\langle 1\\wedge v_2, v_1 \\rangle \\\\\n& = (-1)^{v_1 v_2}\\langle v_2, v_1 \\rangle, \n\\end{align*}\nand hence $\\langle \\cdot,\\cdot\\rangle$ is automatically symmetric.\n\nRecall that a $\\DGA$ is called commutative if $v_1 \\wedge v_2 = (-1)^{v_1 v_2} v_2\\wedge v_1$ for all homogenous $v_1$, $v_2\\in V$. Commutativity of a $\\DGA$ and symmetry of a general cyclic structure on it seem to be unrelated.\n\n%The induced cyclic structure on the $\\DGA$ $(\\H(V), 0, \\wedge)$ is denoted by the same symbol and defined by \n%\\begin{equation}\\label{Eq:IndHom}\n%\\langle [v_1], [v_2] \\rangle = \\langle v_1, v_2 \\rangle\\quad\\text{for all closed }v_1, v_2\\in V. \n%\\end{equation}\n%\n%The (induced) orientation in degree $n$ on $\\H(V)$ is denoted by $\\int: \\H^n(V) \\rightarrow \\R$; by definition, we have\n%\\begin{equation}\\label{Eq:IndOr}\n%\\int [v] = \\Or(v)\\quad\\text{for all closed }in V.\n%\\end{equation}\n\n%\n%It turns out that, e.g., for unital $\\DGA$'s with the degree bounded from above by~$n$, cyclic structures $\\langle\\cdot,\\cdot\\rangle$ of degree $n$ on $V$ are in one-to-one correspondence with orientations $\\int: \\H^n(V) \\rightarrow \\R$ on the homology. The precise relation is stated in the following lemma.\n\\begin{Proposition}[Correspondence of orientations and cyclic structures on $\\DGA$'s]\\label{Prop:OrAndCyc}\nLet $(V,\\Dd,\\wedge)$ be a $\\DGA$. Then the following holds:\n\\begin{ClaimList}\n\\item If $\\wedge$ is commutative, then any orientation $\\Or$ in degree $n$ induces a cyclic structure~$\\langle\\cdot,\\cdot\\rangle$ of degree $n$ which is given for all homogenous $v_1$, $v_2\\in V$ by\n\\begin{equation}\\label{Eq:PairForm}\n\\langle v_1, v_2 \\rangle \\coloneqq \\begin{cases}\n \\Or(v_1 \\wedge v_2) & \\text{if }\\deg(v_1) + \\deg(v_2) = n, \\\\\n 0 & \\text{otherwise}. \\end{cases}\n\\end{equation}\n\\item If $1$ is a unit, then any non-zero cyclic structure $\\langle\\cdot,\\cdot\\rangle$ of degree $n$ induces an orientation~$\\Or$ in degree $n$ by defining \n\\begin{equation}\\label{Eq:OrForm}\n\\Or(v) \\coloneqq \\begin{cases}\n \\langle v, 1 \\rangle & \\text{for }v\\in V^n,\\\\\n 0 & \\text{otherwise.}\n \\end{cases}\n\\end{equation}\n\\item For a unital commutative $\\DGA$ $(V,\\Dd,\\wedge,1)$, formulas \\eqref{Eq:PairForm} and \\eqref{Eq:OrForm} define the correspondence\n\\[ \\text{Orientations in degree }n\\ \\overset{1:1}{\\simeq}\\ \\text{Non-zero cyclic structures of degree }n.\\]\n%If moreover $(V,\\Dd,\\wedge,1)$ is unital, then there is a canonical correspondence\n%\\[ \\text{cyc.~str.~on }V\\text{ of degree }n\\ \\overset{1:1}{\\simeq}\\ \\text{ cyc.~str.~on }\\H(V)\\text{ of degree }n. \\]\n%%In particular, a DGA-morphism $f: V \\rightarrow V'$ preserves the corresponding structure on the chain level if and only if $f_*: \\H(V) \\rightarrow \\H(V')$ preserves it on homology.\n\\end{ClaimList}\n\\end{Proposition}\n\\begin{proof}\n\\begin{ProofList}\n\\item Using the Leibnitz identity, properties of an orientation and commutativity, we check that\n\\begin{align*}\n\\langle \\Dd v_1, v_2 \\rangle &= \\Or(\\Dd v_1 \\wedge v_2) \\\\\n&= \\Or\\bigl(\\Dd(v_1\\wedge v_2) - (-1)^{v_1} v_1 \\wedge \\Dd v_2\\bigr) \\\\\n&= (-1)^{1+v_1}\\Or(v_1\\wedge\\Dd v_2)\\\\\n&= (-1)^{1+v_1 v_2} \\Or(\\Dd v_2 \\wedge v_1) \\\\\n&= (-1)^{1+v_1 v_2} \\langle \\Dd v_2, v_1\\rangle\n\\end{align*}\nand \n\\begin{align*}\n\\langle v_1\\wedge v_2, v_3\\rangle &= \\Or(v_1\\wedge v_2\\wedge v_3)\\\\\n&= (-1)^{v_3(v_1+v_2)} \\Or(v_3 \\wedge v_1 \\wedge v_2)\\\\\n&= (-1)^{v_3(v_1+v_2)} \\langle v_3\\wedge v_1, v_2\\rangle\n\\end{align*}\nfor all homogenous $v_1$, $v_2$, $v_3\\in V$.\n\\item For all $v\\in V$, we have\n\\begin{align*}\n\\Or(\\Dd v) = \\langle \\Dd v, 1 \\rangle = - \\langle \\Dd 1, v \\rangle = 0.\n\\end{align*}\nFrom $\\langle\\cdot,\\cdot\\rangle \\neq 0$ it follows that there are homogenous $v_1$, $v_2\\in V$ with $\\deg(v_1) + \\deg(v_2) = n$ such that $\\langle v_1, v_2 \\rangle \\neq 0$. Then $v_1\\wedge v_2\\in V^n$ and\n\\begin{align*}\n\\Or(v_1 \\wedge v_2) &= \\langle v_1\\wedge v_2,1\\rangle\\\\\n&= \\langle 1 \\wedge v_1, v_2\\rangle\\\\\n&=\\langle v_1,v_2\\rangle \\neq 0.\n\\end{align*}\nTherefore, $\\Or$ is an orientation on $V$.\n\\item This is a combination of (a) and (c) plus the uniqueness, which is easy to check.\\qedhere\n\\end{ProofList}\n\\end{proof}\n\n\\begin{Remark}[Volume forms]\\label{Rem:VolForms}\nIf $\\H^n(V) \\simeq \\R$, then orientations $\\Or: \\H^n(V)\\rightarrow\\R$ and elements $0\\neq [\\Vol] \\in \\H^n(V)$ called \\emph{volume forms} are in one-to-one correspondence via\n\\[ \\Or([\\Vol]) = 1. \\]\nA consequence is the following:\nSuppose that $(V_1,\\Dd_1,\\Or_1)$ and $(V_2,\\Dd_2,\\Or_2)$ are cochain complexes oriented in degree $n$ which satisfy \n\\begin{equation}\\label{Eq:SDDFG}\n\\H^n(V_1)\\simeq\\H^n(V_2)\\simeq\\R\\quad\\text{and}\\quad\\Dd_1 V_1^n \\simeq \\Dd_2 V_2^n = 0,\n\\end{equation}\nso that Proposition~\\ref{Prop:OrOnHomG} applies. Then a chain map $f: V_1 \\rightarrow V_2$ preserves orientation if and only if the induced map $f_*: \\H^n(V_1) \\rightarrow \\H^n(V_2)$ maps $[\\Vol_1]$ to $[\\Vol_2]$. In the category of unital commutative $\\DGA$'s satisfying \\eqref{Eq:SDDFG}, so that also Proposition~\\ref{Prop:OrAndCyc} holds, if the orientations come from cyclic structures, then $f$ preserves cyclic structure if and only if~$f_*$ preserves volume form.\n%If $V$ is an oriented augmented unital $\\DGA$ if the pairing is non-deg, one defines $\\Vol$ uniquely by requiring that $\\Vol \\perp \\bar{V}$ and $\\Or(\\Vol) = 1$.\n\\todo[noline,caption={DONE Augmented case}]{In the augmented unital case, if the pairing is non-deg, one defines $\\Vol$ uniquely by requiring that $\\Vol \\perp \\bar{V}$ and $\\Or(\\Vol) = 1$. Let us not write it.}\n\\end{Remark}\n\n\n%Observe that if the cyclic structure $\\langle\\cdot,\\cdot\\rangle$ comes from an orientation, which is always the case when the $\\DGA$ is unital by Proposition~\\ref{Prop:OrAndCyc}, then it is \\emph{graded symmetric} (shortly symmetric), i.e., it holds\n%\\[ \\langle v_1,v_2\\rangle = (-1)^{v_1 v_2}\\langle v_2,v_1\\rangle\\quad\\text{for all homogenous }v_1, v_2\\in V. \\]\n\n\\begin{Definition}[Non-degeneracy and Poincar\\'e duality]\\label{Def:PoincDual}\n%An orientation $\\Or: V \\rightarrow \\R$ on a graded vector space~$V$ is called \\emph{non-degenerate} if for every $v\\in V$, the following implication holds:\n%\\[ \\Or(v\\wedge w)= 0\\quad\\text{for all }w\\in V\\quad\\Implies\\quad v=0. \\]\nGiven a graded vector space~$V$, a homogenous bilinear form ($\\eqqcolon$\\,pairing) $\\langle\\cdot,\\cdot\\rangle: V\\otimes V \\rightarrow \\R$ of degree $-n$ as a map which is graded symmetric is called \\emph{non-degenerate} if for every $v\\in V$, the following implication holds:\n\\[ \\langle v,w \\rangle= 0\\quad\\text{for all }w\\in V\\quad\\Implies\\quad v=0. \\]\nWe say that $\\langle\\cdot,\\cdot\\rangle$ satisfies \\emph{Poincar\\'e duality} if the map $\\flat: V \\rightarrow V^{\\GD}$ (graded dual) defined by \n\\[\\flat(v)(w) \\coloneqq \\langle v,w\\rangle\\quad \\text{for all }v,w\\in V\\]\nis a graded isomorphism (it has degree $-n$ as a map) of graded vector spaces.\n\\end{Definition}\n\n\\begin{Remark}[On non-degeneracy and Poincar\\'e duality]\\phantomsection\\label{Rem:NonDegPD}\n\\begin{RemarkList}\n\\item Clearly, Poincar\\'e duality implies non-degeneracy. If the degree $k$ component $V^k$ of $V$ is finite-dimensional for every~$k\\in \\Z$ --- we say that $V$ is of \\emph{finite type} --- then the opposite is true as well. If $n=0$, then Poincar\\'e duality implies that $V$ is of finite type.\n%In this case, we obtain the well-known Poincar\\'e duality $V^k \\simeq V^{n-k}$.\n\\item If $V$ is non-negatively graded, then non-degeneracy of $\\langle\\cdot,\\cdot\\rangle: V\\otimes V \\rightarrow \\R$ implies $V = V^0 \\oplus \\dotsb \\oplus V^n$. Therefore, non-negatively graded vector spaces of finite type which admit a non-degenerate homogenous bilinear form are finite-dimensional.\n\\item The de Rham complex $(\\DR(M),\\Dd,\\wedge)$ of an oriented closed $n$-manifold~$M$ with the orientation $\\int: \\DR^n(M) \\rightarrow \\R$ is an oriented $\\DGA$ whose cyclic structure is non-degenerate but does not satisfy Poincar\\'e duality. On the other hand, the induced structure on homology $\\H(\\DR(M))$ satisfies Poincar\\'e duality.\n\\qedhere\n\\end{RemarkList}\n\\end{Remark}\n%\n%We now relate similar definitions from \\cite{Cieliebak2015}, \\cite{VanLe2019} and \\cite{Lambrechts2007} to our definitions:\n%\\begin{itemize}\n%\\item A ``cyclic $\\DGA$'' as defined in \\cite{Cieliebak2015} is a $\\DGA$ $(V,\\Dd,\\wedge)$ with a symmetric non-degenerate cyclic structure $\\langle \\cdot,\\cdot \\rangle : V \\otimes V \\rightarrow \\R$. Different signs in their definition occur because they consider the operations on the degree shift $V[1]$ (c.f., Definition~\\ref{Def:})\n%\\item A ``Poincar\\'e-DGCA of degree $n$'' as defined in \\cite{VanLe2019} is a unital commutative $\\DGA$ (shortly $\\uCDGA$) $(V,\\Dd,\\wedge,1)$ with $V=V^0\\oplus \\dotsb\\oplus V^n$ and with an orientation $\\int: \\H^n(V) \\rightarrow \\R$ such that $(\\H(V),\\wedge,\\int)$ is a finite-dimensional Poincar\\'e algebra.\n%\\item An ``oriented Poincar\\'e-CDGA'' as defined in \\cite{Lambrechts2007} is a $\\uCDGA$ with non-negatively graded $V$ of finite type such that $\\H(V)$ is a Poincar\\'e algebra.\n%\\end{itemize}\n%For unital cyclic $\\DGA$'s, cyclicity implies \\emph{graded symmetry} of $\\langle\\cdot,\\cdot\\rangle$. Indeed, for all homogenous $v_1$, $v_2\\in V$, we have\n%\\[ \\langle v_1, v_2 \\rangle = \\langle 1 \\wedge v_1, v_2 \\rangle = (-1)^{v_1 v_2} \\langle v_2 \\wedge 1, v_1 \\rangle = (-1)^{v_1 v_2} \\langle v_2, v_1\\rangle. \\]\n%Graded symmetry is also automatic for $\\langle\\cdot,\\cdot\\rangle$ coming from an orientation.\n%%For simplicity, we will assume from now on that $\\langle\\cdot,\\cdot\\rangle$ is symmetric.\n\nAn analog of the following definition is used in \\cite{Van2019} and also in \\cite{Lambrechts2007} (under the name ``set of orphans''). \n\n\\begin{Definition}[Degenerate subspace and non-degenerate quotient]\\label{Def:NonDegQ}\nGiven a symmetric pairing $\\langle\\cdot,\\cdot\\rangle: V\\otimes V\\rightarrow\\R$ on a graded vector space~$V$, we define the \\emph{degenerate subspace}~$V^\\perp\\subset V$ by \n\\begin{align*}\nV^\\perp\\coloneqq \\{ v\\in V \\mid \\langle w,v \\rangle = 0\\text{ for all }w\\in V\\}.\n\\end{align*}\nIf $(V,\\Dd,\\wedge)$ is a $\\DGA$ and $\\langle\\cdot,\\cdot\\rangle$ a cyclic structure, then \\eqref{Eq:CycStr} implies that $V^\\perp$ is a differential graded ideal in $V$, and thus we obtain the short exact sequence of $\\DGA$'s\n\\begin{equation}\\label{Eq:ImportantSES}\n\\begin{tikzcd}\n0 \\arrow{r} & V^\\perp \\arrow[hook]{r}{\\iota} & V \\arrow[two heads]{r}{\\pi^\\VansQuotient} & \\VansQuotient(V) \\coloneqq V / V^\\perp \\arrow{r} & 0,\n\\end{tikzcd}\n\\end{equation}\nwhere $\\iota$ is the inclusion and $\\pi^\\VansQuotient$ the canonical projection. We call the $\\DGA$ $\\VansQuotient(V)$ together with the induced non-degenerate cyclic structure $\\langle\\cdot,\\cdot\\rangle^\\VansQuotient$ such that $\\langle\\pi^\\VansQuotient(\\cdot),\\pi^\\VansQuotient(\\cdot)\\rangle^\\VansQuotient = \\langle\\cdot,\\cdot\\rangle$ the \\emph{non-degenerate quotient.}\n\\end{Definition}\n\nIt was observed in~\\cite{Van2019} that the question whether $(V^\\perp,\\Dd)$ is acyclic, and hence~$\\pi^\\VansQuotient$ is a quasi-isomorphism, turns out to be related to the existence of Hodge decomposition.\n\n\\begin{Definition}[Hodge decomposition]\\label{Def:HodgeDecomp}\nA cochain complex $(V,\\Dd)$ with a symmetric cyclic structure $\\langle\\cdot,\\cdot\\rangle: V\\otimes V \\rightarrow \\R$ is \\emph{of Hodge type} if there exist subspaces $\\Harm \\subset \\Ker \\Dd$ and $C\\subset V$ such that \n\\begin{equation}\\label{Eq:HodgeDecomp}\nV = \\Ker \\Dd \\oplus C, \\quad \\Ker \\Dd = \\Im \\Dd \\oplus \\Harm\\quad\\text{and}\\quad C \\perp \\Harm \\oplus C,\n\\end{equation}\nwhere $\\perp$ denotes the relation of being perpendicular with respect to $\\langle\\cdot,\\cdot\\rangle$. Such decomposition is called a \\emph{Hodge decomposition.} We call $\\Harm$ the \\emph{harmonic subspace} and $C$ the \\emph{coexact part}.\n\nGiven a Hodge decomposition, we define the \\emph{standard Hodge homotopy} $\\HtpStd: V \\rightarrow V$~by \n\\[\\HtpStd \\coloneqq \\begin{cases}\n -(\\Restr{\\Dd}{C})^{-1} & \\text{on }\\Im \\Dd, \\\\\n 0 & \\text{on } \\Harm \\oplus C.\n \\end{cases}\\]\nThen we have $\\Dd \\HtpStd = - \\pi_{\\Im \\Dd}$, $\\HtpStd \\Dd = -\\pi_{C}$, and hence \n\\[ [\\Dd,\\HtpStd] = \\Dd \\HtpStd + \\HtpStd \\Dd = \\pi_\\Harm - \\Id.\\]\n We call $(\\Harm,\\HtpStd)$ the \\emph{Hodge pair} associated to the Hodge decomposition \\eqref{Eq:HodgeDecomp}.\n\\end{Definition}\n\n\\begin{Proposition}[Non-deg., fin.~type implies Hodge type]\\label{Prop:NDegFin}\nAny cochain complex of finite type with a non-degenerate symmetric cyclic structure is of Hodge type.\n\\end{Proposition}\n\\begin{proof}\nThis is \\cite[Lemma~11.1]{Cieliebak2015}, and the proof uses formal Hodge theory.\n\\end{proof}\n\n\\begin{Remark}[Harmonic subspaces]\\label{Rem:RemarkHarm}\nIn the situation of Proposition~\\ref{Prop:NDegFin}, it was shown in \\cite[Remark~2.6]{Van2019} that for any complement $\\Harm$ of $\\im\\Dd$ in $\\ker\\Dd$ (in other words, the image of a section $\\H(V)\\rightarrow \\Ker\\Dd$) there is a coexact part $C$ such that $V=\\Harm \\oplus \\Im \\Dd \\oplus C$ is a Hodge decomposition. From this reason, we call any complement of $\\im \\Dd$ in $\\ker \\Dd$ a \\emph{harmonic subspace.}\\footnote{Given a Hodge decomposition $V=\\Harm\\oplus\\im\\Dd\\oplus C$ and a harmonic subspace $\\Harm'$, then it holds $\\Harm' = \\Graph(\\alpha: \\Harm \\rightarrow \\Dd V)$ because $\\Harm \\oplus \\Dd V = \\Harm' \\oplus \\Dd V$, and one can take $C'=\\Graph(- \\alpha^\\dagger - \\frac{1}{2}\\alpha\\alpha^\\dagger: C \\rightarrow \\Harm\\oplus\\Dd V)$.}\n\\end{Remark}\n\nThe following lemma will be used in the proof of Proposition~\\ref{Prop:HodgeAcyc}.\n\n\\begin{Lemma}[Complement of acyclic subcomplex over $\\R$]\\label{Lem:Pom}\nLet $f: V_1 \\rightarrow V_2$ be an injective chain map of cochain complexes $(V_1,\\Dd_1)$ and $(V_2,\\Dd_2)$ over $\\R$ such that $(V_1,\\Dd_1)$ is acyclic. Then there is a chain map $g: V_2 \\rightarrow V_1$ such that $g\\circ f=\\Id$.\\footnote{This lemma can be used to prove that over $\\R$, every surjective quasi-isomorphism is a deformation retraction and every injective quasi-isomorphism is a section of a deformation retraction.}\n\\end{Lemma}\n\n\\begin{proof}\nFor every $i\\in \\Z$, consider the diagram\n\\[\\begin{tikzcd}\n\\Ker \\Dd_1^i \\oplus C^i_1 \\arrow{r}{f^i} \\arrow{d}{\\Dd_1^i} & \\Ker \\Dd_2^i \\oplus C^i_2 \\arrow{d}{\\Dd_2^i} \\\\\n\\Ker \\Dd^{i+1}_1 \\oplus C^{i+1}_1 \\arrow{r}{f^{i+1}} & \\Ker \\Dd_2^{i+1} \\oplus C^{i+1}_2,\n\\end{tikzcd}\\]\nwhere $C^i_1$ is a complement of $\\Ker \\Dd_1^i$ in $V_1^i$ and $C^i_2$ is a complement of $\\Ker \\Dd_2^i$ in~$V_2^i$. With respect to this decomposition, we write\n\\begin{align*}\nf^i & = \\begin{pmatrix}\nf^{i}_{11} & f^i_{12} \\\\\nf^{i}_{21} & f^i_{22}\n\\end{pmatrix}, & g^i &= \\begin{pmatrix}\ng^{i}_{11} & g^i_{12} \\\\\ng^{i}_{21} & g^i_{22}\n\\end{pmatrix}, \\\\\n\\Dd^i_1 &= \\begin{pmatrix}\n0 & d^i_1 \\\\\n0 & 0\n\\end{pmatrix},& \\Dd^i_2 &= \\begin{pmatrix}\n0 & d^i_2 \\\\\n0 & 0\n\\end{pmatrix}.\n\\end{align*}\n\nThe assumption $\\H(V_1)=0$ implies that $d^i_1$ is an isomorphism. The fact that~$f$ is a chain map translates to\n\\begin{equation}\\label{Eq:EqEqEq}\nd_2^i f^i_{21} = 0,\\quad f^{i+1}_{21} d_1^i = 0,\\quad f^{i+1}_{11} d^i_1 = d^i_2 f^i_{22}.\n\\end{equation}\nFrom the second relation and surjectivity of $d_1^i$, we get that $f_{21}^{i+1} = 0$. Now,~$f^{i+1}_{11}$ has to be injective because it is the only possibly non-zero part of $f$ on $\\Ker d_1^{i+1}$. From the third relation of \\eqref{Eq:EqEqEq} and the fact that $d_1^i$ is injective, we get that $f_{22}^i$ is injective as well.\nRelations \\eqref{Eq:EqEqEq} hold also for $g$ with $d_1$ and $d_2$ switched. In particular, we have $g_{21}^{i}=0$. The relation $g \\circ f = \\Id$ translates using $f^i_{21} = g^i_{21} = 0$ to\n\\begin{equation}\\label{Eq:EqEq}\ng^i_{11} f^i_{11} = \\Id, \\quad g^i_{11} f^i_{12} + g^i_{12}f^i_{22}=0, \\quad g^i_{22} f^i_{22} = \\Id.\n\\end{equation}\nBecause $d_1^i$ is an isomorphism, the last equation is equivalent to \n\\[ \\Id = d_1^i g^i_{22}f^i_{22} (d_1^i)^{-1} = g^{i+1}_{11} d_2^i f^i_{22} (d_1^i)^{-1} = g^{i+1}_{11} f^{i+1}_{11} d_1^i (d_1^i)^{-1} = g^{i+1}_{11} f^{i+1}_{11}. \\]\n\nWe see that $g$ can be constructed as follows. For all $i\\in \\Z$, let $g_i^{11}$ be an arbitrary left inverse of $f_i^{11}$. Set $g_{22}^i \\coloneqq (d_1^i)^{-1}g_{11}^{i+1} d_2^i$ and $g_i^{21}\\coloneqq0$. Finally, $g_{12}^i$ has to be chosen such that the second equation of \\eqref{Eq:EqEq} is satisfied. This is possible since we can first define~$g_{12}^i$ on $\\Im f^i_{22}$ because~$f^i_{22}$ injective and then extend it by $0$ to a complement.\n\\end{proof}\n\nClaim (b) of the following proposition corresponds to \\cite[Lemma~2.8]{Van2019}. Claim (c) was suggested by Prof.~Hông Vân Lê via e-mail correspondence.\n\n\\begin{Proposition}[Hodge decomposition and acyclicity of $V^\\perp$]\\label{Prop:HodgeAcyc}\nLet $(V,\\Dd)$ be a cochain complex with a symmetric cyclic structure $\\langle\\cdot,\\cdot\\rangle: V\\otimes V\\rightarrow\\R$. If $V$ is of Hodge type, then the following implications hold:\n\\begin{ClaimList}\n\\item If $\\langle \\cdot,\\cdot \\rangle$ is non-degenerate, then $\\langle \\cdot,\\cdot \\rangle^\\H$ is non-degenerate.\n\\item If $\\langle \\cdot,\\cdot\\rangle^\\H$ is non-degenerate, then $(V^\\perp,\\Dd)$ is acyclic.\n\\end{ClaimList}\nMoreover, the following reverse implication holds:\n\\begin{ClaimList}[resume]\n\\item If $V$ is of finite type and $(V^\\perp,\\Dd)$ is acyclic, then $V$ is of Hodge type.\n\\end{ClaimList}\n\\end{Proposition}\n\\begin{proof}\n\\begin{ProofList}\n\\item Let $V = \\Im \\Dd \\oplus \\Harm \\oplus C$ be a Hodge decomposition. Then $\\Im \\Dd \\oplus C \\subset \\Harm^\\perp$, and hence\n\\[ \\langle \\Dd \\eta + b + c, b' \\rangle = \\langle b,b'\\rangle\\quad\\text{for all }\\eta\\in V, c\\in C\\text{ and }b, b'\\in \\Harm. \\]\nThe claim follows easily. Notice that having a Hodge decomposition, it holds\n\\[ \\Restr{\\langle \\cdot,\\cdot\\rangle}{\\Harm\\otimes \\Harm}\\text{ non-degenerate}\\quad\\Equiv\\quad \\Im\\Dd \\oplus C=\\Harm^\\perp. \\] \n\\item Consider a Hodge decomposition as above, and let $v\\in V^\\perp \\cap \\Ker \\Dd$ be a non-zero vector. Suppose that $v\\not\\in \\Im\\Dd$. Then $[v] \\neq 0$ in $\\H(V)$, and hence there is a $b\\in \\Harm$ such that $\\langle v,b \\rangle \\neq 0$ by non-degeneracy of $\\langle\\cdot,\\cdot\\rangle^\\H$. This is a contradiction with $v\\in V^\\perp$. Therefore, it holds $V^\\perp \\cap \\Ker \\Dd = V^\\perp \\cap \\Im \\Dd$. In particular, there is an $\\eta \\in C$ such that $v = \\Dd \\eta$. Now, for any $\\eta'\\in C$, $b\\in \\Harm$ and $c\\in C$, we have using $C\\perp \\Harm\\oplus C$ and $v\\in V^\\perp$ the following:\n\\begin{align*}\n\\langle \\Dd \\eta' + b + c, \\eta \\rangle &= \\langle \\Dd \\eta',\\eta \\rangle \\\\\n&=(-1)^{1+\\eta\\eta'}\\langle\\Dd\\eta,\\eta'\\rangle\\\\\n&= (-1)^{1+\\eta\\eta'}\\langle v,\\eta'\\rangle\\\\\n&= 0.\n\\end{align*}\nTherefore, it holds $\\eta\\in V^\\perp$, and we have shown that $V^\\perp \\cap \\Im\\Dd = \\Dd V^\\perp$. The claim follows.\n\\item Because $V^\\perp \\subset V$ is an acyclic subcomplex and we work over $\\R$, there is a complementary subcomplex $Z\\subset V$; i.e., $\\Dd Z\\subset Z$ and $V = V^\\perp\\oplus Z$. This follows from Lemma~\\ref{Lem:Pom} by setting $V^\\perp = \\Im f$ and $Z=\\Ker g$. \\Add[caption={Deformation retract over $\\R$ add},noline]{Add here deformation retract somewhere.}Now, the restriction of $\\langle\\cdot,\\cdot\\rangle$ to $Z$ is non-degenerate, and Proposition~\\ref{Prop:NDegFin} provides its Hodge decomposition $Z=\\Dd Z \\oplus \\Harm \\oplus D$. Let $E\\subset V^\\perp$ be a graded vector space which is complementary to $\\Dd V^\\perp$ in $V^\\perp$; i.e., $V^\\perp = \\Dd V^\\perp \\oplus E$. It is easy to check that $V=\\Im\\Dd\\oplus\\Harm\\oplus C$ with $C\\coloneqq D \\oplus E$ is a Hodge decomposition.\n\\qedhere\n\\end{ProofList}\n\\end{proof}\n\\todo[noline,caption={PD implies Hodge without fin type}]{Does the following hold without assuming finite type? I.e. does Poincar\\'e duality imply Hodge type?}\nThe following notions were taken from \\cite{Van2019}.\n\n\\begin{Definition}[Hodge subalgebra and small subalgebra]\\label{Def:SmallSubalg}\nConsider a $\\DGA$ $(V,\\Dd,\\wedge)$ with a symmetric cyclic structure $\\langle\\cdot,\\cdot\\rangle$. Suppose that it admits a Hodge decomposition with the Hodge pair $(\\Harm,\\HtpStd)$. A \\emph{Hodge subalgebra} is a differential graded subalgebra $W\\subset V$ which satisfies\n\\[ \\Harm\\subset W \\quad\\text{and}\\quad \\HtpStd W \\subset W. \\]\nWe denote the smallest Hodge subalgebra of $V$ by $\\VansSmall(V)$ and call it the \\emph{small subalgebra}. We stress that the definition of $\\VansSmall(V)$ depends on $(\\Harm,\\HtpStd)$!\n\\end{Definition}\n%In the situation of Definition~\\ref{Def:SmallSubalg}, it holds\n%\\[ \\VansSmall(V) = \\bigcap_{\\substack{W\\subset V\\\\\\text{Hodge subalgebra}}} W. \\]\n\nThe following is a version of \\cite[Proposition~3.3]{Van2019} which generalizes to the non-simply-connected case (see (iii) of Remark~\\ref{Rem:OnHodgeSubalg} below for the comparison).\n\n\\begin{Proposition}[Description of small subalgebra]\\label{Prop:SmallDescription}\nConsider the situation of Definition~\\ref{Def:SmallSubalg}. The small subalgebra $\\VansSmall(V)$ is generated as a graded vector space by Kontsevich-Soibelman--like evaluations of rooted binary trees with $k\\ge 1$ leaves labeled with homogenous elements of~$\\Harm$, interior vertices labeled with $\\wedge$ and interior edges labeled either with~$\\StdHtp$ or with~$\\Id$.\n\\end{Proposition}\n\\begin{proof}\nWe abbreviate $\\VansSmall \\coloneqq \\VansSmall(V)$, denote the set of labeled trees by $\\Trees$ and denote the vector space generated by evaluations of elements of $\\Trees$ by $\\langle \\Trees\\rangle$.\n\nClearly, we have $\\langle \\Trees\\rangle \\subset \\VansSmall$. For ``$=$'', it suffices to check that for any $T$, $T_1$, $T_2\\in\\Trees$, it holds $\\Dd T$, $\\HtpStd T$, $T_1 \\wedge T_2 \\in \\langle\\Trees\\rangle$, i.e., that $\\langle \\Trees\\rangle$ is a Hodge subalgebra.\n\nAs for $\\Dd T$, we imagine $\\Dd$ propagating from the root to the leaves. When it encounters $L\\wedge R$, where $L$ stands for the left and $R$ for the right sub-branch, it duplicates the tree and continues propagating in $L$ and $R$, respectively (we take the sum of the two copies in the end). This is justified by the Leibnitz identity $\\Dd(L\\wedge R) = \\Dd L\\wedge R + (-1)^L L \\wedge \\Dd R$. When it encouners $\\HtpStd$, it triples the tree and either goes past $\\HtpStd$ and continues propagating, or exchanges $\\HtpStd$ for $\\Id$ and stops, or exchanges $\\HtpStd$ for $\\pi_\\Harm$ and stops. This is justified by $\\Dd \\HtpStd = -\\HtpStd \\Dd - \\Id + \\pi_\\Harm$. If it encounters $\\Id$, nothing happens and it keeps propagating. If it reaches a leaf with $h\\in \\Harm$, then the corresponding tree evaluates to $0$. We see that we alway obtain an element of $\\langle\\Trees\\rangle$.\n\nAs for $\\HtpStd T$, we have either $\\HtpStd T = 0$ if the interior edge adjacent to the root ($\\eqqcolon$ the root edge) is labeled with $\\HtpStd$, or $\\HtpStd T$ is a new tree in $\\Trees$ which arises from $T$ by replacing $\\Id$ by $\\HtpStd$ on the root edge.\n\nAs for $T_1 \\wedge T_2$, using $\\Id = \\pi_\\Harm - \\Dd \\HtpStd - \\HtpStd \\Dd$ and the Leibnitz identity, we get\n\\begin{align*}\nT_1 \\wedge T_2 &= \\pi_\\Harm(T_1 \\wedge T_2) - (\\Dd \\HtpStd)(T_1 \\wedge T_2) - (\\HtpStd \\Dd)(T_1 \\wedge T_2) \\\\\n&= \\pi_\\Harm(T_1 \\wedge T_2) - \\Dd\\bigl(\\HtpStd(T_1 \\wedge T_2)\\bigr) - \\HtpStd(\\Dd T_1 \\wedge T_2)\n- (-1)^{T_1} \\HtpStd(T_1 \\wedge \\Dd T_2).\n\\end{align*}\nWe see that $T_1\\wedge T_2 \\in \\langle \\Trees\\rangle$.\n\\end{proof}\n\n\\begin{Remark}[On Hodge subalgebra and small subalgebra]\\phantomsection\\label{Rem:OnHodgeSubalg}\n\\begin{RemarkList}\n\\item Any Hodge subalgebra~$W$ inherits the Hodge decomposition\n\\begin{equation}\\label{Eq:InducedHodgeDecomp}\nW = \\Harm \\oplus \\Dd W \\oplus \\HtpStd W\n\\end{equation}\nwith the Hodge pair $(\\Harm,\\Restr{\\HtpStd}{W})$.\n\\item We will be in the situation of Definition~\\ref{Def:SmallSubalg} and use the notation of the proof of Proposition~\\ref{Prop:SmallDescription}. In addition, we suppose that $V$ is non-negatively graded. Having established $\\VansSmall = \\langle \\Trees \\rangle$, consider the Hodge decomposition \\eqref{Eq:InducedHodgeDecomp} for $W = \\VansSmall$. Let $\\widebar{\\VansSmall} = \\bigoplus_{k\\ge 1} \\VansSmall^k$ denote the reduced part of $\\VansSmall$, and let $\\langle\\widebar{\\VansSmall} \\wedge \\widebar{\\VansSmall}\\rangle$ denote the graded vector space generated by products $v_1 \\wedge v_2$ for $v_1$, $v_2\\in \\widebar{\\VansSmall}$. We have\n\\begin{align*}\n\\HtpStd\\langle\\Trees\\rangle &= \\bigl\\langle\\{ T\\in \\Trees \\text{ with }\\HtpStd\\text{ on the root edge}\\}\\bigr\\rangle \\\\\n&= \\HtpStd\\langle\\widebar{\\VansSmall} \\wedge \\widebar{\\VansSmall}\\rangle\n\\end{align*}\nand\n\\begin{align*}\n \\Dd\\langle\\Trees\\rangle &= \\Dd (\\pi_\\Harm - \\Dd \\HtpStd - \\HtpStd\\Dd)\\langle\\Trees\\rangle \\\\\n &= \\Dd \\HtpStd \\Dd \\langle\\Trees\\rangle \\\\\n &\\subset \\Dd \\HtpStd \\langle\\Trees\\rangle \\\\\n &\\subset \\Dd \\HtpStd\\langle\\widebar{\\VansSmall}\\wedge\\widebar{\\VansSmall}\\rangle.\n\\end{align*}\nIt holds even $\\Dd\\langle\\Trees\\rangle = \\Dd\\HtpStd\\langle\\widebar{\\VansSmall}\\wedge\\widebar{\\VansSmall}\\rangle$ due to \\eqref{Eq:InducedHodgeDecomp}. We can now write \\eqref{Eq:InducedHodgeDecomp} as \n\\begin{equation}\\label{Eq:VansFormula}\n\\VansSmall^k = \\Harm^k \\oplus \\Dd \\HtpStd \\langle\\widebar{\\VansSmall}\\wedge\\widebar{\\VansSmall}\\rangle^k \\oplus \\HtpStd \\langle\\widebar{\\VansSmall}\\wedge\\widebar{\\VansSmall}\\rangle^{k+1}.\n\\end{equation}\nIf $\\H^1 = 0$, this agrees with the formula from \\cite[Proposition~3.3]{Van2019}. In this case, it holds $\\VansSmall^1 = 0$, and hence $\\langle\\widebar{\\VansSmall}\\wedge\\widebar{\\VansSmall}\\rangle^{k+1}$ depends only on $\\VansSmall^i$ for $i\\lceil\\frac{n}{2}\\rceil$ and $(P_{l-1})$ holds.\nThen there is an $m\\in \\N_0$ and a Sullivan $\\DGA$ \n \\begin{equation}\\label{Eq:SullAlg}\n \\Lambda \\coloneqq \\Lambda(w_1,\\dotsc,w_m,z_1,\\dotsc,z_m)\n \\end{equation}\n specified by $\\deg w_i = l-1$, $\\deg z_i = l$, $\\Dd z_i = 0$ and $\\Dd w_i = z_i$ for all $i=1$, $\\dotsc$, $m$ such that the tensor product $\\DGA$ \n\\[\n \\hat{V}\\coloneqq V\\otimes \\Lambda\n \\]\n admits an orientation $\\hat{\\Or}: \\hat{V}\\to\\R$ which extends $\\Or: V \\rightarrow \\R$ on the canonical inclusion $V \\hookrightarrow\\hat{V}$, and there is a decomposition\n\\begin{equation}\\label{Eq:DevompOfVHat}\n\\hat{V} = \\hat{\\Harm}\\oplus \\Dd \\hat{V}\\oplus \\hat{C}\n\\end{equation}\nof type \\eqref{Eq:DecompOfV} for which $(P_{l})$ holds.\n\\end{Lemma}\n\n\\begin{proof}\nThe proof consists of a construction of $\\hat{\\Or}: \\hat{V}\\rightarrow\\R$, a construction of a harmonic subspace $\\hat{\\Harm}$ in $\\hat{V}$, a construction of a complement $\\hat{C}$ of $\\ker \\Dd$ in $\\hat{V}$, a degreewise description of $\\hat{C}$ and $\\hat{C}^\\perp$, a degreewise construction of a complement $\\hat{E}$ of $\\hat{C}^\\perp$ in $\\hat{C}$ and a proof of the property $(P_l)$ for the constructed decomposition.\n\\begin{description}[leftmargin=0pt,font=\\normalfont\\itshape]\n\\item[Construction of $\\hat{\\Or}$:]\nConsider the decomposition \\eqref{Eq:DecompOfV}.\nBecause $\\Dd : C \\rightarrow \\Dd V$ is an isomorphism, we can write\n\\begin{equation}\\label{Eq:VDecomp}\nV = \\Harm \\oplus \\underbrace{\\Dd E \\oplus \\Dd C^\\perp}_{\\displaystyle \\Dd V}\\oplus \\underbrace{E\\oplus C^\\perp}_{\\displaystyle C}.\n\\end{equation}\nThe restriction of $\\langle\\cdot,\\cdot\\rangle$ to $E$ is non-degenerate, and $V$ is of finite type by assumption; hence, $E$ is finite-dimensional.\nSet\n\\[\nm \\coloneqq \\dim E^l.\n\\]\nLet $\\xi_1$, $\\dotsc$, $\\xi_m$ be a basis of $E^{l}$ and $\\xi^1$, $\\dotsc$, $\\xi^m$ its dual basis in $E^{n-l}$.\nThe Sullivan algebra~$\\Lambda$ can be written as a direct sum \n\\begin{equation}\\label{Eq:LambdaDecomp}\n\\Lambda = \\bigoplus_{k=0}^\\infty \\Lambda_k\\quad\\text{with}\\quad\n\\Lambda_k = \\bigoplus_{\\substack{r, m \\ge 0 \\\\ r + m = k}}\\Lambda_r(w)\\otimes \\Lambda_m(z),\n\\end{equation}\nwhere $\\Lambda_r(w)$ and $\\Lambda_m(z)$ are the graded vector spaces generated by monomials $w_I = w_{i_1}\\dotsc w_{i_r}$ and $z_J = z_{j_1}\\dotsc z_{j_m}$ for all multiindices $I=\\{i_1, \\dotsc, i_r\\}$ and $J=\\{j_1,\\dotsc,j_m\\}$, respectively.\nThe direct sum decompositions~\\eqref{Eq:VDecomp} and~\\eqref{Eq:LambdaDecomp} induce a direct sum decomposition of $\\hat{V} = V \\otimes \\Lambda$ via the distributivity of $\\otimes$ and $\\oplus$.\nWe denote \n\\[\n\\hat{V}_k \\coloneqq V \\otimes \\Lambda_k\\quad\\text{for }k\\ge 0.\n\\]\nLet $\\hat{\\Or}: \\hat{V} \\to \\R$ be the linear map satisfying\n\\begin{align}\n\t\\hat{\\Or}(v) &\\coloneqq \\Or(v) && \\text{for all }v\\in V,\\\\\n\t\\hat{\\Or}(\\xi^i \\wedge z_j) &\\coloneqq \\Or(\\xi^i\\wedge \\xi_j) && \\text{and} \\\\\n\t\\hat{\\Or}(\\Dd \\xi^i \\wedge w_j) &\\coloneqq (-1)^{\\deg \\xi^i + 1}\\Or(\\xi^i \\wedge \\xi_j) && \\text{for all }i, j = 1, \\dotsc, m,\n\\end{align}\nand which is zero on $(\\Harm\\oplus\\Dd V\\oplus C^\\perp \\oplus \\bigoplus_{i\\ge 0, i \\neq n-l}E^i)\\otimes \\Lambda_1(z)$, on $(\\Harm\\oplus C \\oplus \\Dd C^\\perp \\oplus \\bigoplus_{i\\ge 0, i\\neq n-l} \\Dd E^i)\\otimes\\Lambda_1(w)$ and on $\\hat{V}_k$ for $k \\ge 2$.\n\nIn order to show that $\\hat{\\Or}$ is an orientation, we must check that $\\hat{\\Or}\\neq 0$ and $\\hat{\\Or}\\circ \\Dd = 0$.\nThe first condition is clear from $\\Restr{\\hat{\\Or}}{V} = \\Or \\neq 0$. \nAs for the second condition, $\\hat{V}$ is generated by elements $v \\wedge w_I \\wedge z_J$ for $v\\in V$ and multiindices $I$, $J$.\nIt holds $\\Dd \\hat{V}_k \\subset \\hat{V}_k$ for all~$k\\ge 0$, and hence $\\Dd \\hat{V} = \\bigoplus_{k=0}^\\infty \\Dd \\hat{V}_k$.\nFrom the definition of $\\hat{\\Or}$, we have immediately $\\Dd \\hat{V}_0 = \\Dd V \\subset \\ker \\Or \\subset \\ker \\hat{\\Or}$ and $\\bigoplus_{k=2}^\\infty \\Dd \\hat{V}_k \\subset \\ker\\hat{\\Or}$.\nAs for $\\Dd\\hat{V}_1$, we write $\\hat{V}_{1} = \\Span \\{v\\wedge w_j, v\\wedge z_j \\mid v\\in V, j=1,\\dotsc, m\\}$ as a graded vector space and compute\n\\begin{equation}\\label{Eq:DVI}\n\\begin{aligned}\n\t\\Dd \\hat{V}_1 &= \\Span\\{\\Dd(v\\wedge w_j), \\Dd(v\\wedge z_j) \\mid v\\in V, j=1, \\dotsc, m \\}\\\\\n\t&=\\Span\\{\\Dd v \\wedge w_j + (-1)^{\\Deg v} v \\wedge z_j \\mid v\\in V, j=1, \\dotsc, m\\}.\n\\end{aligned}\n\\end{equation}\nWrite $v\\in V^{n-l}$ as $v = h + \\Dd c + c^\\perp + \\sum_{i=1}^m \\alpha_i \\xi^i$ for $h\\in \\Harm^{n-l}$, $c\\in C^{n-l-1}$, $c^\\perp\\in C^{\\perp n-l}$ and $\\alpha_i\\in \\R$, and compute for every $j=1$, $\\dotsc$, $m$ the following:\n\\begin{align*}\n\\hat{\\Or}(\\Dd v\\wedge w_j) & = \\hat{\\Or}\\Bigl(\\Dd c^\\perp \\wedge w_j + \\sum_{i=1}^m \\alpha_i \\Dd \\xi^i \\wedge w_j\\Bigr)\\\\\n&= \\sum_{i=1}^m \\alpha_i \\hat{\\Or}(\\Dd \\xi^i \\wedge w_j)\\\\\n&= \\sum_{i=1}^m (-1)^{\\deg \\xi^i + 1}\\alpha_i \\hat{\\Or}(\\xi^i \\wedge \\xi_j) \\\\\n&= (-1)^{n-l+1}\\sum_{i=1}^m \\alpha_i \\hat{\\Or}(\\xi^i \\wedge z_j)\\\\\n&= (-1)^{n-l+1}\\hat{\\Or}\\Bigl((h + \\Dd c + c^\\perp)\\wedge z_j +\\sum_{i=1}^m \\alpha_i \\xi^i \\wedge z_j \\Bigr) \\\\\n&= (-1)^{\\deg v + 1}\\hat{\\Or}(v\\wedge z_j).\n\\end{align*}\nConsequently, $\\Dd \\hat{V}_{1}\\subset \\ker \\hat{\\Or}$.\nThis shows $\\hat{\\Or}\\circ\\Dd = 0$.\n\nThe inclusion $V\\hookrightarrow\\hat{V}$ is clearly orientation preserving. \n\n\\item[Construction of $\\hat{\\Harm}$:]\nIt holds $\\bar{\\H}(\\Lambda) = 0$ for the reduced homology, and hence $\\H(\\hat{V}) \\simeq \\H(V)\\otimes\\H(\\Lambda) = \\H(V)$ by K\\\"unneth's formula. Because $\\Harm\\subset \\ker \\Dd$, $\\Harm\\cap\\im\\Dd = 0$ and $\\dim(\\Harm) = \\dim \\H(\\hat{V})$, $\\Harm$ is a complement of $\\Dd\\hat{V}$ in $\\ker \\Dd$. Therefore, \n\\[\n\\hat{\\Harm} \\coloneqq \\Harm\n\\]\nis a harmonic subspace of $\\hat{V}$.\nAlso note that $\\H(\\hat{V})=\\bigoplus_{k=0}^\\infty \\H(\\hat{V}_k)$ and $\\dim \\H(\\hat{V}) = \\dim \\H(\\hat{V}_0)$, and hence $\\H(\\hat{V}_k) = 0$ for all $k\\ge 1$.\n\n\\item[Construction of $\\hat{C}$:]\nWe construct $\\hat{C}$ as a direct sum $\\hat{C} = \\bigoplus_{k=0}^\\infty \\hat{C}_k$. \nSet \n\\[\n\\hat{C}_0 \\coloneqq C.\n\\]\nFor $k=1$, define\n\\begin{equation}\\label{Eq:CIDef}\n\\begin{aligned}\n \\tilde{C}_1 &\\coloneqq \\Span\\{v \\wedge w_i \\mid v\\in V, i=1, \\dotsc,m\\} \\subset \\hat{V}_1, \\\\\n \\hat{C}_1 &\\coloneqq \\{\\tilde{c} - \\pi(\\tilde{c}) \\mid \\tilde{c}\\in \\tilde{C}_1\\}\\subset\\hat{V}_0\\oplus\\hat{V}_1,\n\\end{aligned}\n\\end{equation}\nwhere $\\pi: \\hat{V}\\rightarrow\\hat{\\Harm}$ is the orthogonal projection.\nFor all $k\\ge 2$, let $\\hat{C}_k \\subset \\hat{V}_k$ be an arbitrary complement of $\\ker\\Dd$ in $\\hat{V}_k$ as a graded vector space.\n\nWe show first that $\\hat{C}_i$ for $i\\ge 0$ are disjoint.\nClearly, $\\hat{C}_j \\cap \\sum_{i=0, i\\neq j}^\\infty \\hat{C}_i = 0$ for all $j\\ge 2$.\nBecause $(\\Harm + C)\\cap\\tilde{C}_1 = 0$ and $\\Harm \\cap C = 0$, it holds $\\hat{C}_1 \\cap \\hat{C}_0 = 0$, and $(\\hat{C}_0 + \\hat{C}_1) \\cap \\sum_{i\\ge 2} \\hat{C}_i = 0$ implies that $\\hat{C}_j \\cap \\sum_{i=0, i\\neq j}^\\infty \\hat{C}_i = 0$ holds also for $j=0$, $1$.\n\nWe show that $\\hat{C}=\\bigoplus_{k\\ge 0}\\hat{C}_k$ is a complement of $\\ker \\Dd$ in~$\\hat{V}$.\nFor $k\\ge 2$, $\\hat{C}_k$ are complements of $\\ker \\Dd$ in $\\hat{V}_k$ by construction.\nFor $k=0$, $\\hat{C}_0 \\oplus \\ker \\Dd \\cap \\hat{V}_0 = \\hat{V}_0$ follows from \\eqref{Eq:DecompOfV}. \nFor $k=1$, we compare \\eqref{Eq:DVI} and \\eqref{Eq:CIDef} to see that $\\tilde{C}_1 \\oplus \\Dd \\hat{V}_1 = \\hat{V}_1$.\nBecause $\\Harm\\subset \\ker \\Dd$, it is easy to see that $\\hat{C}_0\\oplus\\hat{C}_1$ is a complement of $\\ker \\Dd$ in $\\hat{V}_0\\oplus\\hat{V}_1$.\n\nFinally, $\\hat{C}_0 \\perp \\hat{\\Harm}$ holds by \\eqref{Eq:DecompOfV}, $\\hat{C}_1\\perp\\Harm$ holds by the construction of $\\hat{C}_1$ from $\\tilde{C}_1$ and $\\hat{C}_k \\perp \\Harm$ follows from the definition of $\\hat{\\Or}$.\n\n\\item[Degreewise description of $\\hat{C}$ and $\\hat{C}^\\perp$:]\nWe are interested in $\\hat{C}^i$ for $0\\le i\\le l$.\nFor all $k\\ge 2$, the graded vector space $\\hat{C}_k$ is concentrated in degrees $i\\ge 2(l-1)$.\nBut $2(l-1)>l$ due to $n\\ge 5$.\nTherefore, $\\hat{C}_k^i = 0$ for $k\\ge 2$ and $0\\le i \\le l$.\nWe denote $\\bar{V} \\coloneqq \\bigoplus_{i=1}^\\infty V^i$ and write $\\tilde{C}_1 = \\Lambda_1(w) \\oplus (\\bar{V}\\otimes \\Lambda_1(w))$.\nBecause $V^1=0$, the graded vector space $\\bar{V}\\otimes\\Lambda_1(w)$ is concentrated in degrees $i\\ge 2 + (l-1) = l + 1$. We obtain \n\\begin{equation}\\label{Eq:CDegreewise}\n\\hat{C}^i = (\\hat{C}_0 \\oplus \\hat{C}_1)^i =\n \\begin{cases}\n \tC^i & \\text{for }0 \\le i \\le l-1, \\\\\n\tC^i + \\Span\\{ w_j \\mid j = 1, \\dotsc, m\\} & \\text{for }i = l - 1, \\\\\n\tC^i & \\text{for }i = l.\n \\end{cases}\n\\end{equation}\nHere we used that $\\hat{C}_1^{l-1} = \\tilde{C}_1^{l-1} = \\Span\\{ w_i \\mid i = 1, \\dotsc, m\\}$, which is true because $\\Lambda_1(w)\\perp\\Harm$ from the definition of $\\hat{\\Or}$.\n\nWe are now interested in $\\hat{C}^{\\perp i}$ for $n-l\\le i \\le l$.\nNote that $n-l \\le i \\le l$ is equivalent to $n-l \\le n-i \\le l$.\nBy definition, $\\hat{\\Or}$ vanishes on $C\\wedge\\Lambda_1(w)=C\\otimes\\Lambda_1(w)$ and $\\Lambda_1(w)\\wedge\\Lambda_1(w) \\subset \\Lambda_2(w)$.\nLooking at \\eqref{Eq:CDegreewise}, we see the following: \n\\begin{equation}\\label{Eq:CPerpDegreewise}\n\t\\hat{C}^{\\perp i} =\n\t\t\\begin{cases}\n\t\t\tC^{\\perp i} & \\text{for }n-l \\le i \\le l - 2, \\\\\n\t\t\tC^{\\perp i} + \\Span\\{w_i\\mid i=1,\\dotsc,m\\} & \\text{for }i = l-1, \\\\\n\t\t\tC^{\\perp i} & \\text{for }i = l.\n\t\t\\end{cases}\t\t\n\\end{equation}\nBecause $\\hat{V}$ is non-negatively graded, it holds $\\hat{C}^{\\perp i} = \\hat{C}^i$ for $i>n$.\nNotice that $\\hat{C}^{\\perp i}$ might be smaller than $C^i$ for $0\\le i \\le n-l-1$.\nThe reason for this is a possible existence of $v_1$, $v_2\\in V$ such that $v_1 \\wedge v_2$ has a non-trivial $\\Dd E$-component and $\\langle v_1, v_2\\wedge w_i \\rangle = \\hat{\\Or}((v_1\\wedge v_2)\\wedge w_i) \\neq 0$.\n\n\\item[Construction of $\\hat{E}$:]\nBecause $\\hat{V}_i \\wedge \\hat{V}_j \\subset \\hat{V}_{i+j}$ and $\\hat{V}_k \\subset \\ker \\hat{\\Or}$ for $k\\ge 2$, we have $\\hat{C}_k \\subset \\hat{C}^\\perp$ for all $k\\ge 2$.\nIt follows that \n\\[\n\\hat{C}^\\perp = (\\hat{C}_0\\oplus\\hat{C}_1)\\cap\\hat{C}^\\perp \\oplus \\bigoplus_{k\\ge 2}\\hat{C}_k,\n\\]\nand hence it is enough to construct $\\hat{E} \\subset \\hat{C}_0\\oplus\\hat{C}_1$ such that $\\hat{E}\\oplus (\\hat{C}_0\\oplus\\hat{C}_1)\\cap\\hat{C}^\\perp = \\hat{C}_0\\oplus\\hat{C}_1$.\nBecause $E\\cap \\hat{C}^\\perp \\subset E \\cap C^\\perp = 0$, we can get $\\hat{E}$ by extending $E$.\nFor $0\\le i \\le n$, we define\n%For $n-l \\le i \\le l$, we set $\\hat{E}^i\\coloneqq E^i$.\n%For $i=0$, $\\dotsc$, $n-l-1$, we obtain $\\hat{E}^i$ from $E^i$ by adding elements from $C^i$.\n%For $i=l+1$, $\\dotsc$, $n$, we obtain $\\hat{E}^i$ from $E^i$ by adding elements from $\\hat{C}_0^i \\oplus \\hat{C}_1^i$.\n\\begin{equation}\\label{Eq:EDegreewise}\n\\hat{E}^i \\coloneqq\n\t\\begin{cases}\n\t\tE^i\\oplus \\Span\\{c_1^i,\\dotsc, c_{k_i}^i\\}\\text{ for }c_1^i, \\dotsc, c_{k_i}^i\\in C^i & \\text{for }0\\le i \\le n-l-1,\\\\\n\t\tE^i & \\text{for }n-l\\le i \\le l,\\\\\n\t\tE^i\\oplus\\Span\\{\\hat{v}_1,\\dotsc,\\hat{v}_k\\}\\text{ for }\\hat{v}_1^i,\\dotsc,\\hat{v}_{k_i}^i\\in(\\hat{C}_0\\oplus\\hat{C}_1)^i & \\text{for }l+1\\le i \\le n,\n\t\\end{cases}\n\\end{equation}\nwhere the existence of $c^i_j$ and $\\hat{v}^i_j$ and the fact that $\\hat{C} = \\hat{E}\\oplus\\hat{C}^\\perp$ are justified by \\eqref{Eq:CDegreewise}, \\eqref{Eq:CPerpDegreewise} and \\eqref{Eq:ConditionTempII}.\n\n\\item[Property $(P_l)$:] We define $\\hat{\\rho}: \\hat{E}^{\\lceil \\frac{n}{2}\\rceil}\\oplus\\dotsb\\oplus\\hat{E}^l \\rightarrow \\Dd\\hat{V}$ by $\\hat{\\rho} \\coloneqq \\rho$ on $\\hat{E}^{\\lceil \\frac{n}{2}\\rceil}\\oplus\\dotsb\\oplus\\hat{E}^{l-1}=E^{\\lceil \\frac{n}{2}\\rceil}\\oplus\\dotsb\\oplus E^{l-1}$ and by\n\\[\n\\hat{\\rho}(\\xi_i) \\coloneqq z_i\\quad\\text{for all }i=1, \\dotsc, m.\n\\]\nConditions \\eqref{Eq:ConditionTemp} and \\eqref{Eq:ConditionTempII} are checked easily using $(P_{l-1})$, \\eqref{Eq:CDegreewise}, \\eqref{Eq:CPerpDegreewise}, \\eqref{Eq:EDegreewise} and the definition of $\\hat{\\Or}$.\\qedhere\n%For $\\lceil\\frac{n}{2}\\rceil \\le i \\le l - 1$, \n%\\eqref{Eq:ConditionTemp} \\eqref{Eq:ConditionTempI}\n%\n%Given $\\lceil \\frac{n}{2} \\rceil \\le i \\le l - 1$ and $e\\in \\hat{E}^i = E^i$, there is $z\\in(\\im\\Dd)^i$ such that $\\langle e,f\\rangle = \\langle z,f\\rangle$ and $\\langle z,c^\\perp\\rangle=0$ for all $f\\in E^{n-i}$ and $c^\\perp\\in (C^\\perp)^{n-i}$. \n%Due to $j\\le l$ and $l\\le $\n%$n - j \\ge n - l$\n%\n%\n% we can write an $\\hat{e}\\in\\hat{E}^i$ as $\\hat{e} = e + c^\\perp + v \\wedge w + w$ for $e\\in E$, $c^\\perp\\in C^\\perp$, $v\\in V^{\\ge 2}$ and $w\\in \\Lambda_1(w)$.\n% It holds $\\deg(v\\wedge w) = l+1$, and hence $v=0$ for $k+1\\le i \\le l$; this summand is thus always zero.\n% It holds $\\deg w = l - 1$, and hence $w=0$ for $i\\neq l-1$.\n% In this case, $c^\\perp + w = \\hat{e}-e = \\hat{E}^{l-1}$ can pair non-trivially only with an element in degree $n - (l-1) \\le n - k = k+1 \\le l$.\n% However, since $\\hat{V}^l = V^l \\oplus \\Lambda_1(z)$, $\\hat{V}^{l-1} = V^{l-1}\\oplus\\Lambda_1(w)$ and $\\hat{V}^i = V^i$ for $i] (0,0,0) -- (1,0,0) node[anchor=north east]{$\\bsx_1$};\n% \\draw[thick,->] (0,0,0) -- (0,1,0) node[anchor=north west]{$\\bsx_2$};\n% \\draw[thick,->] (0,0,0) -- (0,0,1) node[anchor=south]{$\\bsx_3$};\n% \\tdplotsetcoord{P}{\\rvec}{\\thetavec}{\\phivec}\n% \\draw[-stealth,very thick,color=red] (O) -- (P);\n% \\draw[dashed, color=red] (O) -- (Pxy);\n% \\draw[dashed, color=red] (P) -- (Pxy);\n% \\tdplotdrawarc{(O)}{0.2}{0}{\\phivec}{anchor=north}{$\\theta_1$}\n% \\tdplotsetthetaplanecoords{\\phivec}\n% \\tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.5}{0}%\n% {\\thetavec}{anchor=south west}{$\\theta_2$}\n% \\draw[dashed,tdplot_rotated_coords] (\\rvec,0,0) arc (0:90:\\rvec);\n% \\draw[dashed] (\\rvec,0,0) arc (0:90:\\rvec);\n% \\tdplotsetrotatedcoords{\\phivec}{\\thetavec}{0}\n% \\tdplotsetrotatedcoordsorigin{(P)}\n% \\draw[dashed,blue,tdplot_rotated_coords,-] (-.4,0,0)\n% -- (.4,0,0) node[anchor=north west]{};\n% \\draw[dashed,blue,tdplot_rotated_coords,-] (0,-.4,0)\n% -- (0,.4,0) node[anchor=west]{};\n% \\draw[blue,tdplot_rotated_coords,-] (-.4,.4,0) -- (.4,.4,0) -- (.4,-.4,0) -- (-.4,-.4,0) -- (-.4,.4,0) node[anchor=north]{};\n% \\tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.2}{0}%\n% {30}{anchor=north west,color=black}{$\\theta_3$}\n% \\tdplotsetrotatedcoords{\\phivec}{\\thetavec}{30}\n% \\draw[thick,tdplot_rotated_coords,->] (0,0,0)\n% -- (.3,0,0) node[anchor=north west]{$\\bsy_1$};\n% \\draw[thick,tdplot_rotated_coords,->] (0,0,0)\n% -- (0,.3,0) node[anchor=west]{$\\bsy_2$};\n% \\node[blue] at (0.4,0.45,1.2) {$\\Omega_{\\mathrm{2D}}$};\n% \\node[red] at (1.5,0.75,1.2) {$\\bvth_{\\bth}$};\n% \\tdplotsetrotatedthetaplanecoords{45}\n% \\end{tikzpicture}\n% \\caption*{\\mdeff{Kept as reference for \\figref{imaging-geometry} as this tikz figure is beautiful. Will be commented out in the end.}}\n% \\end{figure}\n\n\\begin{figure}\n \\begin{minipage}[t]{0.48\\linewidth}\n \\centering\n \\includegraphics[height=0.7\\linewidth]{geomProj3D}\n \\caption{%\n % * (i) what we mean by a projection and an orientation (the two most important objects of our paper), and\n % * (ii) how a projection (=integration through z in the new coordinate system) is made from a 3D volume.\n % $\\p_i = \\mathbf{P} \\Rot(\\bth) \\x$ \\todo{($\\mathbf{P}$ is a projection/integration)}.\n Geometry of the imaging model defined in \\eqnref{imaging-model}.\n The 3D density $\\x$ in the coordinate system $(x_1, x_2, x_3)$ is imaged along the \\textit{orientation} $\\bth$ to produce the 2D \\textit{projection} $\\p$ in the coordinate system $(y_1, y_2)$ of the microscope's detector plane.\n The orientation $\\bth = (\\theta_3, \\theta_2, \\theta_1)$ is decomposed as the direction $(\\theta_2, \\theta_1) \\in [0,\\pi] \\times [0,2\\pi[$ (parameterizing the sphere $\\mathbb{S}^2$) and the in-plane rotation $\\theta_3 \\in [0,2\\pi[$ (parameterizing the circle $\\mathbb{S}^1$).\n In our work, we represent the orientation $\\bth$ as a unit quaternion $q$.\n %The 3D rotation $\\Rot(\\bth) = \\Rot(q) \\in \\SO(3)$ maps the object coordinate system to the projection coordinate system.\n %\\mdeff{$\\Rot$, $\\bth$, and $q$ all represent orientations. That's the problem with notation that separates the representation not the semantic (though we might need that separation later).}\n }\\label{fig:imaging-geometry}\n \\end{minipage}\n \\hfill\n \\begin{minipage}[t]{0.48\\linewidth}\n \\centering\n \\includegraphics[height=0.7\\linewidth]{intuition-method}\n \\caption{%\n Single-particle cryo-EM produces $P$ projections (with $P$ in the order of $10^5$) from unknown orientations: $\\{(\\p_i, q_i)\\}_{i=1}^P$.\n Observing that distances between orientations constrain the latter, we aim to \\textit{recover the orientations} $\\{q_i\\}$ from $\\{d_q(q_i, q_j)\\}$, where $d_q(q_i, q_j)$ is the distance (angle) between orientations $q_i$ and $q_j$.\n Observing that the similarity between projections depends on their relative orientation, we aim to \\textit{estimate the distance} $d_q(q_i, q_j)$ from the projections $(\\p_i, \\p_j)$.\n % \\todo{The projections don't look to be on the surface of the sphere: They should be on tangent planes.}\n % \\todo{The gray line to the center is the projection direction to be labeled $(\\theta_1, \\theta_2)$ with a color consistent with \\figref{imaging-geometry}.}\n % \\todo{The two green arcs seems too much like a single one.}\n }\\label{fig:intuition-method}\n \\end{minipage}\n\\end{figure}\n\n% RELATED WORKS\n\nA popular approach is to alternatively refine the 3D structure and estimated orientations~\\cite{penczek1994ribosome,Baker1996,Dempster1977,sigworth1998maximum,scheres2012bayesian,zehni2020joint}.\nYet, the outcome of these iterative-refinement procedures is often predicated on the quality of the initial reconstruction, or, equivalently, on the initial estimation of the orientations~\\cite{sorzano2006optimization,henderson2012outcome}.\n\nSeveral methods have been designed to produce a first rough \\textit{ab initio} structure for the refinement procedure~\\cite{singer2020computational}. \\textit{Moment-matching} techniques~\\cite{kam1980reconstruction, salzman1990method,goncharov1988integral,sharon2019method} reconstruct an initial structure such that the first few moments of the distribution of its theoretical measurements match the ones of its experimental projections; however, they typically remain sensitive to error in data and can require relatively high computational complexity. Based on the central-slice theorem, \\textit{common-lines} methods~\\cite{van1987angular, penczek1994ribosome,mallick2006structure,singer2010detecting,wang2013orientation,greenberg2017common,pragier2019common} aim at uniquely determining the orientations of each projection by identifying the common-lines between triplets of projections---a real challenge given the massive amount of noise. Alternatively, the marginalized maximum likelihood (ML) formulation of the reconstruction problem~\\cite{sigworth1998maximum}---classically used for the iterative-refinement procedures themselves---can be minimized using stochastic gradient descent~\\cite{punjani2017cryosparc}.\nThis permits to avoid the need for an initial volume estimate, at the possible cost of greater convergence instability.\n\nMore recently, the recovery of geometrical information from unknown view tomography of 2D point sources has been proposed~\\cite{zehni2019distance}, but the extension to 3D cryo-EM tomography is not straightforward.\nFinally,~\\cite{miolane2019estimation} proposed to recover the in-plane rotations by learning to embed projections in an appropriate latent space, but only after directions had been estimated through three rounds of 2D classification in RELION\\@.\n\n%Several methods have been designed to produce a first rough \\textit{ab initio} structure for the refinement procedure~\\cite{singer2020computational}.\n%An early approach~\\cite{kam1980reconstruction} proposed to reconstruct an initial structure such that the first few moments of the distribution of its theoretical measurements match the ones of its experimental projections.\n%Since then, \\textit{moment-matching} techniques have been refined and extended~\\cite{salzman1990method,goncharov1988integral,sharon2019method}, \\eg, to accommodate for non-uniform orientation configurations.\n%However, they typically remain sensitive to error in data and can require relatively high computational complexity.\n\n%Another popular approach relies on the central-slice theorem, which relates the Fourier transform of a projection to a plane (orthogonal to the projection direction) in the Fourier transform of the 3D object~\\cite{Natterer2001mathematics}.\n%Hence, every two projections \\textit{de facto} share a common 1D intersection in the 3D Fourier domain, and three projections theoretically suffice to define a coordinate system from which their orientations can be deduced~\\cite{van1987angular}.\n%Exploiting this principle, \\textit{common-lines} methods aim at uniquely determining the orientations of each projection by identifying the common-lines between triplets of projections~\\cite{penczek1994ribosome,mallick2006structure,singer2010detecting,wang2013orientation,greenberg2017common,pragier2019common}---a real challenge given the massive amount of noise.\n%If needs be for downside: sensitivity to high-noise levels, small particles, etc.\n\nDespite the aforementioned advances, providing a robust initial volume remains a challenge due to the high-dimensionality and ill-posedness of the underlying optimization problem.\n%\\mdeff{How is our method different from previous work? What's its singularity? Why do we expect it to solve the aforementioned issues? This isn't a survey: The related work is here to build a stage for our work to stand on.}\nOn the other hand, the remarkable ability of convolutional neural networks to capture relevant representations of images has had a profound influence in imaging~\\cite{lecun2015deep}.\n% and our ability to model the imaging / forward model / simulate training data.\n%%%\nIn this work, we present a learning-based approach to recover the unknown orientations directly from the acquired set of projections---without the need for an intermediate reconstruction procedure or an initial volume estimate.\n%, hence bypassing some of the aforementioned limitations.\n% \\mdeff{That's method:} The novelty consists in first estimating the relative distances between pairs of projections (by using a Siamese neural network previously trained on a synthetic dataset), and then recovering the absolute orientation of each projections from these relative distances.\n% By doing so, orientations are recovered without the need for an intermediate reconstruction procedure or an initial volume estimate, which has useful implications for SPA\\@.\n%\\mdeff{``bypassing some limitations'' and ``has useful implications'' are too vague IMO\\@. Could we be more specific? Not on the how (that's the method) but on the why. Think about an ML audience.}\n%\\mdeff{Idea: Our method is direct (as opposed to iterative refinement schemes) and does not require any prior model of the reconstructed protein.}\n%Our two-step method learns to estimate the unknown orientation $\\bth_i$ (represented by a unit quaternion $q_i$) associated to each projection $\\p_i$ in a single-particle cryo-EM dataset without relying on any intermediate reconstruction procedure.\n\n%\\mdeff{Content. (p1) Why is the general problem of protein reconstruction important and difficult. (p2) Short background on single-particle cryo-EM and how reconstruction is done. (p3) Previous work on orientation estimation (or initial structure estimation). If there's none (because people have researched other routes), state it and write why it's an interesting route to explore. How does it compare to initial rough structure estimation? Or whatever the other routes are. (p4) Our contribution on top of previous work.}\n", "meta": {"hexsha": "0025c4e1f25ebe0afdeecace561828c0696cc091", "size": 14540, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/1_introduction.tex", "max_stars_repo_name": "mdeff/paper-cryoem-orientation-recovery", "max_stars_repo_head_hexsha": "ebae0e4e28c706c233d88f5c34dd427b4274cb47", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-23T18:20:37.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-23T18:20:37.000Z", "max_issues_repo_path": "sections/1_introduction.tex", "max_issues_repo_name": "mdeff/paper-cryoem-orientation-recovery", "max_issues_repo_head_hexsha": "ebae0e4e28c706c233d88f5c34dd427b4274cb47", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/1_introduction.tex", "max_forks_repo_name": "mdeff/paper-cryoem-orientation-recovery", "max_forks_repo_head_hexsha": "ebae0e4e28c706c233d88f5c34dd427b4274cb47", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 107.7037037037, "max_line_length": 1220, "alphanum_fraction": 0.7435350757, "num_tokens": 4061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.8080672204860316, "lm_q1q2_score": 0.6367013711973895}} {"text": "\\section{The {\\tt simplesine} Model}\n\\label{sec:simplesine-model}\n\nIn the following chapters,\nwe use a simple sine wave model, {\\tt simplesine}, to illustrate \\TrickHLA\\ in action.\nUnderstanding the model and how it is used in a Trick \\sdefine file\nis important for those chapters.\nThis section introduces \\simplesine with that in mind.\n\n% -----------------------------------------------------------------------\n\\subsection{Description}\n\nThe system modeled by \\simplesine is an undamped harmonic oscillator,\nthe dynamics of which are governed by the differential equation\n\\begin{equation}\n\\ddot{x} + w^2 x = 0,\n\\label{eq:EOM}\n\\end{equation}\nwhich has an analytic solution of the form\n\\begin{subequations}\n\\label{eq:analytic-equations}\n\\begin{align}\nx(t) &= A \\sin {(\\omega t + \\phi)}, \\\\\n\\dot{x}(t) &= A \\omega \\cos {(\\omega t + \\phi)}.\n\\end{align}\n\\end{subequations}\n\nThe relevant model data are\nthe dynamic {\\em state}, $(x, \\dot{x})$ and\nthe constant system {\\em parameters}, $(A, \\phi, \\omega)$,\nwhere the parameters are specified as inputs\nand the state is calculated dynamically as simulation outputs.\\footnote{\nIn this model, the initial state cannot be specified as inputs explicitly\nbut rather through the parameters $A$ and $\\phi$.\n}\n\nThe \\simplesine model has functions that may be used to calculate the state\nanalytically based on equations~\\ref{eq:analytic-equations}.\nIt can also propagate the state approximately\nbased on numerical integration of the differential equations.\nThe integration involves the calculation of the derivative of\nthe 2-vector $\\boldsymbol{z}$ defined as\n\n\\begin{equation}\n\\label{eq:differential-equations}\n \\boldsymbol{z}(t)\n \\equiv \\left\\{\n \\begin{array}{c}\n x(t)\\\\ \\dot{x}(t)\n \\end{array}\n \\right\\} \\\\\n = \\left\\{\n \\begin{array}{r}\n A \\sin{(\\omega t + \\phi)} \\\\ A \\omega \\cos{(\\omega t + \\phi)}\n \\end{array}\n \\right\\}\n\\end{equation}\n\nSo that\n\n\\begin{equation}\n\\label{eq:derivative-equations}\n \\dot{\\boldsymbol{z}}(t)\n = \\left\\{\n \\begin{array}{c}\n \\dot{x}(t)\\\\ \\ddot{x}(t)\n \\end{array}\n \\right\\} \\\\\n = \\left\\{\n \\begin{array}{r}\n \\dot{x}(t)\\\\ - \\omega^2 x(t)\n \\end{array}\n \\right\\} \\\\\n = \\left\\{\n \\begin{array}{r}\n A w \\cos{(\\omega t + \\phi)} \\\\\n - A w^2 \\omega \\sin{(\\omega t + \\phi)}\n \\end{array}\n \\right\\}\n\\end{equation}\n\n% -------------------------------------\n\\subsection{Model}\n\nThe source code for the \\simplesine model is organized into three directories:\n{\\tt data}, {\\tt include} and {\\tt src}.\nThe files in these directories are discussed below.\n\n% ----------\n\\subsubsection{Include Files}\n\nThe \\simplesine C/C++ include files are in directory {\\tt simplesine/include}.\nA list of the files is shown below.\n\n{\n\\begin{center}\n\\scriptsize\n\\begin{tabular}{|l|l|}\n\\hline\n{\\em filename} \\\\\n\\hline\n\\hline\n{\\tt simplesine.h} \\\\\n\\hline\n{\\tt simplesine\\_InteractionHandler.h} \\\\\n\\hline\n{\\tt simplesine\\_LagCompensator.h} \\\\\n\\hline\n{\\tt simplesine\\_Packing.h} \\\\\n\\hline\n{\\tt simplesine\\_proto.h} \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n}\n\n\\paragraph{\\tt simplesine.h}\nThis file declares the fundamental \\simplesine data structures.\nThere is a single data structure that in turn holds state- and parameter-related\ndata structures.\nIn the simulations that follow, the {\\tt simplesine\\_T} structure will be\nfrequently declared as a {\\tt sim\\_object} in the \\sdefine files\nwhen the simulations need to model a sine wave.\n\nThe {\\tt simplesine\\_T} structure consists of a {\\em state} substructure,\nwhich holds $x$ and $\\dot{x}$\nand a parameters substructure which holds the constant {\\em parameters,}\n$A$, $\\phi$ and $\\omega$.\nIn this model, only the parameters\nmay be set from the input processor.\nThe state may only be calculated by a Trick job.\nThe purpose of this is to ensure that the state and parameters are never\ninitialized to inconsistent values.\\footnote{\n Of course, this requires that developers remember to explicitly\n call {\\tt simplesine\\_calc()} as an initialization job for every\n \\simplesine sim variable.\n}\n\nThe complete file is shown in Appendix~\\ref{sec:simplesine-h}.\n\n\\paragraph{\\tt simplesine\\_proto.h}\nThis file declares the C functions exported by the \\simplesine model.\nThese functions may be used in an \\sdefine file as Trick jobs.\nFunctions of particular interest are\n\\begin{itemize}\n \\item{\\tt simplesine\\_calc()}, which calculates the state, $(x, \\dot{x})$\n according to equations~\\ref{eq:analytic-equations},\\footnote{\n Since only the \\simplesine parameters have default values,\n this job may be used as an initialization job to initialize\n the state from the parameters.\n }\n \\item{\\tt simplesine\\_deriv()} and {\\tt simplesine\\_integ()}, which\n are used to numerically integrate equations~\\ref{eq:differential-equations}\n using the standard Trick integration scheme,\n \\item{\\tt simplesine\\_copyXXX()}, several routines which copy \\simplesine\n data from one data structure to another, and\n \\item{\\tt simplesine\\_calcError()}, which calculates the error between\none \\simplesine state and the true values based on equations~\\ref{eq:analytic-equations}.\n\\end{itemize}\n\n\nThe complete file is shown in Appendix~\\ref{sec:simplesine-proto-h}.\n\n\\paragraph{\\tt simplesine\\_InteractionHandler.hh}\nThis file declares the \\simplesine C++ class which acts as a\n\\TrickHLA\\ {\\em interaction handler}.\nIt is a subclass of {\\tt TrickHLAInteractionHandler},\nwhich is the \\TrickHLA\\ class which defines how interactions\nare sent and received.\n\nThe complete file is shown in Appendix~\\ref{sec:simplesine-InteractionHandler-hh}.\n\n\\paragraph{\\tt simplesine\\_LagCompensator.hh}\nThis file declares the \\simplesine C++ class which acts as a\n\\TrickHLA\\ {\\em lag compensator}.\nIt is a subclass of {\\tt TrickHLALagCompensator},\nwhich is the \\TrickHLA\\ class which defines how federates may compensate\nfor HLA-time lags created as a result of sending data to remote federates\nand transfering ownership between federates.\n\nThe complete file is shown in Appendix~\\ref{sec:simplesine-LagCompensator-hh}.\n\n\\paragraph{\\tt simplesine\\_Packing.hh}\nThis file declares the \\simplesine C++ class which may be optionally\nused by developers to {\\em pack} outbound data prior to sending via HLA\nand {\\em upack} is upon receipt from HLA.\nIt is a subclass of {\\tt TrickHLAPacking},\nwhich has {\\tt pack()} and {\\tt unpack()} virtual methods that implement\nthe application-specific packing and unpacking logic.\n\nThe complete file is shown in Appendix~\\ref{sec:simplesine-Packing-hh}.\n\n\n% ----------\n\\subsubsection{Source Files}\n\nThe \\simplesine C/C++ source files are in the directory {\\tt simplesine/src}.\n\nThe implementation of the \\simplesine functions and classes is in\n{\\tt .c} and {\\tt .cpp} files located in the {\\tt simplesine/src} directory.\nThe C functions are mainly implemented one function per file\\footnote{\n The copy functions\n ({\\tt simplesine\\_copyParams()},\n {\\tt simplesine\\_copyParams()}, and\n {\\tt simplesine\\_copyParams()})\n are located in a single file, {\\tt simplesine\\_copy.c}.\n},\nand the C++ classes are implemented one class per file.\nThe file names are shown in the table below:\n\n{\n\\begin{center}\n\\scriptsize\n\\begin{tabular}{|l|l|}\n\\hline\n{\\em filename} & {\\em implements what?} \\\\\n\\hline\n\\hline\n{\\tt simplesine\\_calc.c} & {\\tt simplesine\\_copy()} \\\\\n\\hline\n{\\tt simplesine\\_calcError.c} & {\\tt simplesine\\_calcError()} \\\\\n\\hline\n{\\tt simplesine\\_compensate.c} & {\\tt simplesine\\_compensate()} \\\\\n\\hline\n{\\tt simplesine\\_copy.c} & the {\\tt simplesine\\_copyXXX()} functions \\\\\n\\hline\n{\\tt simplesine\\_deriv.c} & {\\tt simplesine\\_deriv()} \\\\\n\\hline\n{\\tt simplesine\\_integ.c} & {\\tt simplesine\\_integ()} \\\\\n\\hline\n{\\tt simplesine\\_propagate.c} & {\\tt simplesine\\_propagate()} \\\\\n\\hline\n\\hline\n{\\tt simplesine\\_InteractionHandler.cpp} & the \\TrickHLA\\ interaction handler class \\\\\n\\hline\n{\\tt simplesine\\_LagCompensator.cpp} & the \\TrickHLA\\ lag compensator class \\\\\n\\hline\n{\\tt simplesine\\_Packing.cpp} & the \\TrickHLA\\ packing/unpacking class \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n}\n\n\n% ----------\n\\subsubsection{Data Files}\n\nThis {\\tt simplesine/data} directory\nconsists Trick input files for default \\simplesine data.\n\nDepending on how you build your \\sdefine file, these files may be used\nas ``fallback'' initializations for your data.\nIn the simulations that follow, the actual initial values\nwill often override the defaults in these files.\\footnote{\n Sine the \\simplesine state data are output-only,\n they cannot be set directly from the Trick input processor.\n Consequently, there is no default data file for\n {\\tt simplesine\\_state\\_T}.\n}\n\nThe files are shown below.\n\n{\n\\begin{center}\n\\scriptsize\n\\begin{tabular}{|l|l|}\n\\hline\n{\\em filename} & {\\em description} \\\\\n\\hline\n\\hline\n{\\tt integ.d}\n &\n Default numerical integration parameters.\n \\\\\n\\hline\n{\\tt simplesine.d}\n &\n Default \\simplesine parameters with uninitialized state.\n \\\\\n\\hline\n\\hline\n{\\tt simplesine\\_params.d}\n &\n Default \\simplesine parameters.\n \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n}\n", "meta": {"hexsha": "ea897b4602af19b541330f12038aa6df08c2ef00", "size": 9194, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/TrickHLA/LaTeX/TrickHLAUser-prelim-simplesine.tex", "max_stars_repo_name": "jiajlin/TrickHLA", "max_stars_repo_head_hexsha": "ae704b97049579e997593ae6d8dd016010b8fa1e", "max_stars_repo_licenses": ["NASA-1.3"], "max_stars_count": 18, "max_stars_repo_stars_event_min_datetime": "2020-03-04T14:23:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T10:47:21.000Z", "max_issues_repo_path": "docs/TrickHLA/LaTeX/TrickHLAUser-prelim-simplesine.tex", "max_issues_repo_name": "jiajlin/TrickHLA", "max_issues_repo_head_hexsha": "ae704b97049579e997593ae6d8dd016010b8fa1e", "max_issues_repo_licenses": ["NASA-1.3"], "max_issues_count": 57, "max_issues_repo_issues_event_min_datetime": "2020-06-04T16:03:44.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-17T20:54:35.000Z", "max_forks_repo_path": "docs/TrickHLA/LaTeX/TrickHLAUser-prelim-simplesine.tex", "max_forks_repo_name": "jiajlin/TrickHLA", "max_forks_repo_head_hexsha": "ae704b97049579e997593ae6d8dd016010b8fa1e", "max_forks_repo_licenses": ["NASA-1.3"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2020-08-25T05:51:05.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-01T18:37:38.000Z", "avg_line_length": 31.7034482759, "max_line_length": 89, "alphanum_fraction": 0.7054600827, "num_tokens": 2468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7879311981328134, "lm_q1q2_score": 0.6367013677464728}} {"text": "\\section{Predicate Logic \\& Correctness}\n\n\\begin{itemize}\n\t\n\t\\item Propositional operators\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item not ($ \\lnot $), and ($ \\land $), or ($ \\lor $)\n\t\t\n\t\t\\item implication ($ \\implies $) -- $ P \\implies Q \\equiv \\lnot P \\lor Q $\n\t\t\n\t\t\\item equivalence ($ \\iff $) -- $ P \\iff Q \\equiv (P \\implies Q) \\land (Q \\implies P)$\n\t\t\n\t\\end{itemize}\n\t\n\t\\item Operator precedence (tightest-binding first): $ \\lnot $, $ \\land $, $ \\lor $, $ \\implies $\n\t\n\t\\item Quantifiers\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item for all ($ \\forall $)\n\t\t\n\t\t\\item there exists ($ \\exists $)\n\t\t\n\t\t\\item Example: $ \\forall x \\in \\mathbb{N} \\cdot x \\ge 0 $\n\t\t\n\t\t\\item The type of the bound variable may be implicit, e.g. $ \\forall x \\cdot x \\ge 0 $\n\t\t\n\t\\end{itemize}\n\n\t\\item Entailment\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item If $ P \\implies Q $ is a tautology (always true) then $ P $ is stronger than $ Q $\n\t\t\n\t\t\\item Equivalently: $ P $ entails $ Q $ ($ P \\entails Q $)\n\t\t\n\t\\end{itemize}\n\n\t\\item Substitution\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item $ P[x \\backslash a] $ -- Substitute all occurrences of $ x $ by $ a $ in $ P $\n\t\t\n\t\t\\item $ P[x, y \\backslash a, b] $ -- Substitute $ a $ and $ b $ for $ x $ and $ y $ \\textbf{simultaneously}\n\t\t\n\t\\end{itemize}\n\n\t\\item Hoare triples -- $ \\hoare{P}{S}{Q} $\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item $ P $ is the precondition, $ S $ is the program and $ Q $ is the postcondition\n\t\t\n\t\t\\item If $ P $ is true before $ S $ executes, then $ S $ will terminate and $ Q $ will be true when it does\n\t\t\n\t\t\\item The program must terminate if started in $ \\text{States}_{_P} $ (total correctness)\n\t\t\n\t\\end{itemize}\n\t\n\t\\item Weakest preconditions\n\t\n\t\\begin{itemize}\n\t\t\n\t\t\\item For a program $ S $ and postcondition $ Q $, $ wp(S, Q) $ is the unique weakest possible precondition such that the triple $ \\hoare{P}{S}{Q} $ will be true\n\t\t\n\t\t\\item $ \\forall P \\cdot (\\hoare{P}{S}{Q}) \\implies (P \\entails wp(S, Q)) $\n\t\t\n\t\\end{itemize}\n\t\t\n\\end{itemize}", "meta": {"hexsha": "0beb71b11ae3effc3d185aff1736be9855aebc32", "size": 1929, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "CSSE3100/predicate.tex", "max_stars_repo_name": "mcoot/CourseNotes", "max_stars_repo_head_hexsha": "c643f46e32cdf4c567bf73d4a23784c834278803", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "CSSE3100/predicate.tex", "max_issues_repo_name": "mcoot/CourseNotes", "max_issues_repo_head_hexsha": "c643f46e32cdf4c567bf73d4a23784c834278803", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "CSSE3100/predicate.tex", "max_forks_repo_name": "mcoot/CourseNotes", "max_forks_repo_head_hexsha": "c643f46e32cdf4c567bf73d4a23784c834278803", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.72, "max_line_length": 163, "alphanum_fraction": 0.6018662519, "num_tokens": 672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527631, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.636701365734451}} {"text": "\\subsection{Simplifications}\n\n\\begin{frame}\n \\frametitle{Simplifications}\n\n \\scriptsize\n\n We use {\\bf simplification} rules to fight the two problems in previous slide,\n before bit-blasting everything\n \\vfill\n \\pause\n \\begin{center}\n \\scalebox{.4}{\\input{simplification.pdf_t}}\n \\end{center}\n \\vfill\n \\pause\n Simplifications exploit properties of \\bitvectors to try to reduce\n the complexity of the formula. We see here some examples, but many\n more rules do exist. Also, it is very important the way they are\n combined together \n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Trivial Simplifications}\n\n The following are trivial consequences of the semantic of \\bitvectors and\n \\formulae in general\n \\vfill\n \\pause\n \\bitvectors trivial simplifications\n \\begin{itemize}\n \\item $t = t\\ \\Rightarrow\\ \\top$\\ \\ \\ \\ for a generic term\n \\item $c = d\\ \\Rightarrow\\ \\bot$\\ \\ \\ \\ for two different constants $c$ and $d$\n \\item $t \\band 0\\ldots0\\ \\Rightarrow\\ 0\\ldots0$\\ \\ \\ \\ for a generic term\n \\item \\ldots\n \\end{itemize}\n \\vfill\n \\pause\n \\begin{itemize}\n \\item $\\varphi \\wedge \\varphi\\ \\Rightarrow\\ \\varphi$\\ \\ \\ \\ for a generic formula\n \\item $\\varphi \\wedge \\top\\ \\Rightarrow\\ \\varphi$\\ \\ \\ \\ for a generic formula\n \\item $\\varphi \\vee \\top\\ \\Rightarrow\\ \\top$\\ \\ \\ \\ for a generic formula\n \\item \\ldots\n \\end{itemize}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Ground Term evaluation}\n \n If a term is {\\bf ground}, i.e., it contains no variables, then it\n can be always simplified to a single constant\n \\vfill\n \\pause\n Examples:\n \\begin{itemize}\n \\item $0000 :: 1000\\ \\Rightarrow 00001000$\n \\item $0010[1:0]\\ \\Rightarrow 10$\n \\item $0100 + 0101\\ \\Rightarrow 1001$\n \\end{itemize}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Variable Elimination Rule}\n\n Suppose that the input formula $\\varphi$ is of the kind\n \\vfill\n $$\n \\varphi'\\ \\wedge\\ (\\w{x}{n} = \\w{t}{n})\n $$\n \\vfill\n where $\\w{x}{n}$ is a variable, and $\\w{t}{n}$ is a term {\\bf not containing} $\\w{x}{n}$\n \\pause\n \\vfill\n then we can rewrite $\\varphi$ as\n $$\n \\varphi'[\\w{t}{n}/\\w{x}{n}]\n $$\n \\vfill\n i.e., we replace every occurrence of $\\w{x}{n}$ by $\\w{t}{n}$. \\pause We save $n$\n Boolean variables in the reduction to SAT\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Concatenation Elimination Rule}\n\n Suppose that we have the equality\n \\vfill\n $$\n \\w{t}{n} :: \\w{s}{m} = \\w{r}{n} :: \\w{u}{m}\n $$\n \\vfill\n then, because the concatenations {\\bf match}, we can rewrite it as\n \\pause\n \\vfill\n $$\n (\\w{t}{n} = \\w{r}{n}) \\wedge (\\w{s}{m} = \\w{u}{m}) \n $$\n \\vfill\n this rewriting may give more opportunity for applications of previous rules\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Exercizes}\n\n \\begin{enumerate}\n\n \\item Complete the missing cases in procedures Bit-Blast-Term and Bit-Blast\n \\vfill\n \\item Bit-Blast the formula $\\neg(\\w{x}{3} = 000) \\wedge (\\w{x}{3} \\band \\w{y}{3}) = (\\w{x}{3} + \\w{y}{3})$ \n \\vfill\n \\item Simplify the formula $(\\w{x}{4} :: \\w{y}{4}) = (\\w{z}{4} :: \\w{x}{4}) \\wedge \\neg( \\w{y}{4} = \\w{z}{4})$ \n\n \\end{enumerate}\n\n\\end{frame}\n", "meta": {"hexsha": "ebe4234956d99da3f56c9327c6ff04c9f0dd718b", "size": 3117, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture2/simplification.tex", "max_stars_repo_name": "formalmethods/smtlectures", "max_stars_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-11-07T19:34:12.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-24T08:05:50.000Z", "max_issues_repo_path": "lecture2/simplification.tex", "max_issues_repo_name": "formalmethods/smtlectures", "max_issues_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture2/simplification.tex", "max_forks_repo_name": "formalmethods/smtlectures", "max_forks_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-05-06T00:40:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-06T00:40:41.000Z", "avg_line_length": 25.7603305785, "max_line_length": 115, "alphanum_fraction": 0.6374719281, "num_tokens": 1096, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7879311931529758, "lm_q1q2_score": 0.6367013637224292}} {"text": "\n\\subsection{Cubic polynomials}\n\nCubic polynomials are of the form \\(ax^3+bx^2+cx+d=0\\).\n\n\\subsection{Solving specific cases}\n\nWe start by solving when \\(b=0\\), that is:\n\n\\(aX^3+bx+c=0\\)\n\n\\subsection{Solving the general case}\n\n", "meta": {"hexsha": "fb26c82d5bb2a9d448bb735fd859ae7413bb48ec", "size": 227, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/algebra/polynomialsMulti/03-01-cubic.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/algebra/polynomialsMulti/03-01-cubic.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/algebra/polynomialsMulti/03-01-cubic.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 16.2142857143, "max_line_length": 55, "alphanum_fraction": 0.704845815, "num_tokens": 75, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357598021707, "lm_q2_score": 0.7341195269001831, "lm_q1q2_score": 0.6365078817915104}} {"text": "\\section{Average Value of a Function}{}{}\\label{sec:Average Value}\nThe average of some finite set of values is a familiar concept. If,\nfor example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6,\n3, 2, 7, 8, then the average score is the sum of these numbers divided\nby the size of the class:\n$$\n \\hbox{average score} = {10+ 9+ 10+ 8+ 7+ 5+ 7+ 6+\n 3+ 2+ 7+ 8\\over 12}={82\\over 12}\\approx 6.83.\n$$\nSuppose that between $t=0$ and $t=1$ the speed of an object is\n$\\sin(\\pi t)$. What is the average speed of the object over that time?\nThe question sounds as if it must make sense, yet we can't merely add\nup some number of speeds and divide, since the speed is changing\ncontinuously over the time interval.\n\nTo make sense of ``average'' in this context, we fall back on the idea\nof approximation. Consider the speed of the object at tenth of a\nsecond intervals: $\\sin 0$, $\\sin(0.1\\pi)$, $\\sin(0.2\\pi)$,\n$\\sin(0.3\\pi)$,\\dots, $\\sin(0.9\\pi)$. The average speed ``should'' be\nfairly close to the average of these ten speeds:\n$$\n {1\\over 10}\\sum_{i=0}^9 \\sin(\\pi i/10)\\approx {1\\over 10}6.3=0.63.\n$$\nOf course, if we compute more speeds at more times, the average of\nthese speeds should be closer to the ``real'' average. If we take the\naverage of $n$ speeds at evenly spaced times, we get:\n$${1\\over n}\\sum_{i=0}^{n-1} \\sin(\\pi i/n).$$\nHere the individual times are $\\ds t_i=i/n$, so rewriting slightly we have\n$${1\\over n}\\sum_{i=0}^{n-1} \\sin(\\pi t_i).$$\nThis is almost the sort of sum that we know turns into an integral;\nwhat's apparently missing is $\\Delta t$---but in fact, $\\Delta t=1/n$,\nthe length of each subinterval. So rewriting again:\n$$\n \\sum_{i=0}^{n-1} \\sin(\\pi t_i){1\\over n}=\n \\sum_{i=0}^{n-1} \\sin(\\pi t_i)\\Delta t.\n$$\nNow this has exactly the right form, so that in the limit we get\n$$\n \\hbox{average speed} = \\int_0^1 \\sin(\\pi t)\\,dt=\n \\left.-{\\cos(\\pi t)\\over\\pi}\\right|_0^1=\n -{\\cos(\\pi)\\over \\pi}+{\\cos(0)\\over\\pi}={2\\over\\pi}\\approx\n 0.6366\\approx 0.64.\n$$\n\nIt's not entirely obvious from this one simple example how to compute\nsuch an average in general. Let's look at a somewhat more complicated\ncase. Suppose that the velocity of an object is $\\ds 16\nt^2+5$ feet per second. \nWhat is the average velocity between $t=1$ and $t=3$? Again we\nset up an approximation to the average:\n$${1\\over n}\\sum_{i=0}^{n-1} 16t_i^2+5,$$\nwhere the values $\\ds t_i$ are evenly spaced \ntimes between 1 and 3. Once again we are ``missing'' $\\Delta t$, and\nthis time $1/n$ is not the correct value. What is $\\Delta t$ in\ngeneral? It is the length of a subinterval; in this case we take the\ninterval $[1,3]$ and divide it into $n$ subintervals, so each\nhas length $(3-1)/n=2/n=\\Delta t$. Now with the usual ``multiply and\ndivide by the same thing'' trick we can rewrite the sum:\n$$\n {1\\over n}\\sum_{i=0}^{n-1} 16t_i^2+5=\n {1\\over 3-1}\\sum_{i=0}^{n-1} (16t_i^2+5){3-1\\over n}=\n {1\\over 2}\\sum_{i=0}^{n-1} (16t_i^2+5){2\\over n}=\n {1\\over 2}\\sum_{i=0}^{n-1} (16t_i^2+5)\\Delta t.\n$$\nIn the limit this becomes\n$${1\\over 2}\\int_1^3 16t^2+5\\,dt={1\\over 2}{446\\over 3}={223\\over 3}.$$\nDoes this seem reasonable? Let's picture it: In\nFigure~\\xrefn{fig:average speed} we see the velocity function together\nwith the horizontal line $y=223/3\\approx 74.3$. Certainly \nthe height of the\nhorizontal line looks at least plausible for the average height of the\ncurve.\n\n\\figure[H]\n%\\texonly\n\\centerline{\\vbox{\\beginpicture\n\\normalgraphs\n%\\ninepoint\n\\setcoordinatesystem units <3truecm,0.4truemm>\n\\setplotarea x from 0 to 3.1, y from 0 to 150\n\\axis bottom ticks numbered from 0 to 3 by 1 /\n\\axis left ticks numbered from 0 to 150 by 25 /\n\\putrule from 1 74.3 to 3 74.3\n\\setquadratic\n\\plot 0.000 5.000 0.150 5.360 0.300 6.440 0.450 8.240 0.600 10.760 \n0.750 14.000 0.900 17.960 1.050 22.640 1.200 28.040 1.350 34.160 \n1.500 41.000 1.650 48.560 1.800 56.840 1.950 65.840 2.100 75.560 \n2.250 86.000 2.400 97.160 2.550 109.040 2.700 121.640 2.850 134.960 \n3.000 149.000 /\n\\setdashes\n\\putrule from 1 0 to 1 74.3\n\\putrule from 3 0 to 3 74.3\n\\endpicture}}\n%\\endtexonly\n%\\htmlfigure{Integration_applications-average_value.html}\n\\caption{\\label{fig:average speed}\nAverage velocity.\n}\n\\endfigure\n\nHere's another way to interpret ``average'' that may make our\ncomputation appear even more reasonable. The object of our example\ngoes a certain distance between $t=1$ and $t=3$. If instead the object\nwere to travel at the average speed over the same time, it should go\nthe same distance. At an average speed of $223/3$ feet per second for\ntwo seconds the object would go $446/3$ feet. How far does it actually\ngo? We know how to compute this:\n$$\\int_1^3 v(t)\\,dt = \\int_1^3 16t^2+5\\,dt={446\\over 3}.$$\nSo now we see that another interpretation of the calculation is:\n$${1\\over 2}\\int_1^3 16t^2+5\\,dt={1\\over 2}{446\\over 3}={223\\over 3}$$\nwhich is the total distance traveled divided by the time in transit, namely,\nthe usual interpretation of average speed.\n\nIn the case of speed, or more properly velocity, we can always\ninterpret ``average'' as total (net) distance divided by time. However, in\nthe case of a different sort of quantity this interpretation does not\nobviously apply, while the approximation approach always does. We might\ninterpret the same problem geometrically: What is the average height\nof $16x^2+5$ on the interval $[1,3]$? We approximate this in exactly\nthe same way, by adding up many sample heights and dividing by the\nnumber of samples. In the limit we get the same result:\n$$\n \\lim_{n\\to\\infty}{1\\over n}\\sum_{i=0}^{n-1} 16x_i^2+5=\n {1\\over 2}\\int_1^3 16x^2+5\\,dx={1\\over 2}{446\\over 3}={223\\over 3}.\n$$\nWe can interpret this result in a slightly different way. The area\nunder $y=16x^2+5$ above $[1,3]$ is\n$$\\int_1^3 16t^2+5\\,dt={446\\over 3}.$$ \nThe area under $y=223/3$ over the same interval $[1,3]$ is simply the\narea of a rectangle that is 2 by $223/3$ with area $446/3$. So the\naverage height of a function is the height of the horizontal line that\nproduces the same area over the given interval.\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\Opensolutionfile{solutions}[ex]\n\\section*{Exercises for \\ref{sec:Average Value}}\n\n\\begin{enumialphparenastyle}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the average height of $\\cos x$ over the intervals\n$[0,\\pi/2]$, $[-\\pi/2,\\pi/2]$, and $[0,2\\pi]$.\n\\begin{sol}\n $2/\\pi$; $2/\\pi$; $0$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the average height of $\\ds x^2$ over the interval\n$[-2,2]$.\n\\begin{sol}\n $4/3$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the average height of $\\ds 1/x^2$ over the interval\n$[1,A]$.\n\\begin{sol}\n $1/A$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Find the average height of $\\ds \\sqrt{1-x^2}$ over the interval\n$[-1,1]$.\n\\begin{sol}\n $\\pi/4$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n An object moves with velocity $\\ds v(t)=-t^2+1$ feet per second\nbetween $t=0$ and $t=2$. Find the average velocity and the average\nspeed of the object between $t=0$ and $t=2$.\n\\begin{sol}\n $-1/3$, $1$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n The observation deck on the 102nd floor of the Empire State Building\nis 1,224 feet above the ground. If a steel ball is dropped from the\nobservation deck its velocity at time $t$ is approximately $v(t)=-32t$\nfeet per second. Find the average speed between the time it is dropped\nand the time it hits the ground, and find its speed when it hits the\nground.\n\\begin{sol}\n $\\ds -4\\sqrt{1224}$ ft/s; $\\ds -8\\sqrt{1224}$ ft/s\n\\end{sol}\n\\end{ex}\n\n\\end{enumialphparenastyle}", "meta": {"hexsha": "165c46f14b0260cc79139efdd18eea664f6ad5c2", "size": 7487, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "8-applications-of-integration/8-4-average-value.org.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "8-applications-of-integration/8-4-average-value.org.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "8-applications-of-integration/8-4-average-value.org.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.435, "max_line_length": 76, "alphanum_fraction": 0.6890610391, "num_tokens": 2649, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.709019146082187, "lm_q2_score": 0.8976952934758465, "lm_q1q2_score": 0.6364831504222429}} {"text": "\n\\subsection{Modelling homogeneous preferences with multinomial logit}\n\n\\subsection{Recap}\n\nOur model is:\n\n\\(U_{ij}=\\Theta z_{ij} + \\epsilon_{ij}\\)\n\n\\(U_{ij}=\\alpha_i x_j -\\beta_i p_{ij} + \\theta_j d_i +\\epsilon_{ij}\\)\n\n\\subsection{Homogeneous model}\n\nWe model all customers as having the same preferences.\n\n\\(U_{ij}=\\alpha x_j -\\beta p_{ij} + \\theta_j d_i +\\epsilon_{ij}\\)\n\n\\subsection{The multinomial logit assumption}\n\nIf errors are IID and extreme we get:\n\n\\(P_{ij}=\\dfrac{e^{\\Theta z_j}}{\\sum_k e^{\\Theta z_k }}\\)\n\n\\subsection{The outside option}\n\nA user has the option of not buying anything.\n\n\\(U_0=0\\)\n\nThis gives us the following shares:\n\n\\(P_{ij}=\\dfrac{e^{\\Theta z_j}}{e^0+\\sum_{k=1} e^{\\Theta z_k }}\\)\n\n\\(P_{ij}=\\dfrac{e^{\\Theta z_j}}{1+\\sum_{k=1} e^{\\Theta z_k }}\\)\n\n\\subsection{Own-price elasticity of demand}\n\n\\(P_{ij}=\\dfrac{e^{\\Theta z_j}}{1+\\sum_{k=1}} e^{\\Theta z_k }\\)\n\n\\(P_{ij}=\\dfrac{e^{\\alpha x_j -\\beta p_j + \\theta_j d_i}}{1+\\sum_{k=1} e^{ \\alpha x_k -\\beta p_k +\\theta_k d_i}}\\)\n\n\\(\\dfrac{\\delta P_{ij} }{\\delta p_j}\\dfrac{p_j}{P_{ij}}=-\\beta p_j(1-P_{ij})\\)\n\n\\(\\dfrac{\\delta P_{ij} }{\\delta p_k}\\dfrac{p_j}{P_{ij}}=\\beta p_kP_{ij}\\)\n\nThis means that the lower the price, the lower the own price elasticity of demand.\n\nThis means that mark ups are higher for cheaper goods, which doesn't always match reality.\n\nThis can be adjusted by changing the form. For example we could use \\(\\ln p\\) or \\(p^2\\).\n\nHowever, we are still getting the shape by assumption.\n\n\\subsection{Cross-price elasticity of demand}\n\n\\subsection{Getting aggregate market shares}\n\n\\(s_j = \\dfrac{1}{n}\\sum_i P_{ij}\\)\n\n", "meta": {"hexsha": "417eeede2a296b1e01657b01c70ba4cd9296ce82", "size": 1614, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/economics/consumerDiscrete/03-01-discrete.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/economics/consumerDiscrete/03-01-discrete.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/economics/consumerDiscrete/03-01-discrete.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.9, "max_line_length": 114, "alphanum_fraction": 0.6790582404, "num_tokens": 548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8976952893703476, "lm_q2_score": 0.7090191460821871, "lm_q1q2_score": 0.6364831475113657}} {"text": "\\documentclass[letter]{article}\n\\renewcommand{\\baselinestretch}{1.25}\n\n\\usepackage[margin=1in]{geometry}\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{pythonhighlight}\n\\usepackage{hyperref}\n\n\\allowdisplaybreaks\n\n%opening\n\\title{MECH 6327 - Homework 2}\n\\author{Jonas Wagner}\n\\date{2021, March 3}\n\n\\begin{document}\n\n\\maketitle\n\n\\newpage\n\\textbf{(Probably not necessary... but its long)}\n\\tableofcontents\n\n\\newpage\n\\section{Problem Set 1: Convex Sets}\n\n\\subsection{Problem 2.5}\n\\textbf{Problem:}\\\\\nWhat is the distance between two parallel hyperplanes: $\\{x \\in \\real^n | a^T x = b_1\\}$ and $\\{x \\in \\real^n | a^T x = b_2\\}$?\\\\\n\n\\noindent\n\\textbf{Solution:}\\\\\nUnder the assumption that $a\\in \\real^n$ and $b_1,b_2 \\in \\real$, the quantity $a^T x_0$ represents the component of $x_0$ in the normal direction. Similarly, the quantities $b_1$ and $b_2$ represent the euclidean distance of the hyperplane from the origin (in the normal direction). Since the hyperplanes are parrellel, the distance between them is the difference between their offsets:\n\\begin{equation}\n\t\\text{Distance between hyperplanes: } b_1 - b_2\n\\end{equation}\n\n\n\\subsection{Problem 2.7}\n\\textbf{Problem:}\\\\\n\\textit{Voronoi description of halfspace.} Let $a$ and $b$ be distinct points in $\\real^n$. Show that the set of all points that are closer to $a$ than $b$ via the euclidean norm is a halfspace. Describe it explicitly as an inequality and draw a picture.\n\n\\noindent\n\\textbf{Solution:}\\\\\nThe set of all points closer to $a$ then $b$ can be defined as:\n\\begin{equation}\n\t\\{x \\in \\real^n \\ | \\ \\norm{x-a}_2 \\leq \\norm{x-b}_2\\}\n\\end{equation}\n\nThe boundary defining this halfspace will be a plane defined by the normal vector $c$ representing the distance between $a$ and $b$, and the offset coefficient $d$ describing intersection of the plane through the half-way point between $a$ and $b$. The quantities $c$ and $d$ can therefore be defined by:\n\\begin{equation}\n\t\\begin{aligned}\n\t\tc &= b - a\\\\\n\t\td &= \\frac{c^T a + c^T b}{2}\\\\\n\t\t &= \\frac{1}{2} c^T (a+b)\n\t\\end{aligned}\n\\end{equation}\n\nThe halfspace, that is equivalent to $x$, can be described by the following:\n\\begin{equation}\n\t\\{x\\in \\real^n \\ | \\ c^T x \\leq d\\}\n\\end{equation}\n\nThis can be visualized in two dimensions for $a = \\mqty[2\\\\4]$ and $b = \\mqty[-3\\\\7]$. The boundary (the red line) is calculated in the standard form using $$x_2 = \\frac{-1}{c_2} \\qty(c_1 * x_1 - d)$$ and then plotted. The half-space itself is the region below the boundary.\n\n\\begin{figure}[h]\n\t\\centering\n\t\\includegraphics[width = 0.7\\linewidth]{fig/pblm2_7}\n\t\\caption{Visualization of the boundary for the halfspace.}\n\t\\label{fig:pblm2_7}\n\\end{figure}\n\n\n%\\newpage\n\\subsection{Problem 2.12}\n\\textbf{Problem:}\\\\\nWhich of the following sets are convex?\\\\\n\n\\noindent\n\\textbf{Solution:}\n\\subsubsection{(a) - Slab}\nA slab defined as $$\\{x \\in \\real^n \\ | \\ \\alpha \\leq a^T x \\leq \\beta\\}$$ \\textbf{is convex} as it consists of the intersection of two halfspaces which themselves are complex.\n\n\\subsubsection{(b) - Rectangle/Hyperrectangle}\nA rectangle set defined as $$\\{x \\in \\real^n \\ | \\ \\alpha_i \\leq x_i \\leq \\beta_i, \\ i = 1,\\dots,n\\}$$ \\textbf{is convex} as it is composed of the intersections of half spaces which are themselves convex. This is similar to the polyhedrons/polytopes that by definition are also convex.\n\n\\subsubsection{(c) - Wedge}\nA wedge set given as $$\\{x \\in \\real^n \\ | \\ a_1^T x \\leq b_1, a_2^T x \\leq b_2\\}$$ \\textbf{is convex} as it is just an intersection of two halfspaces (a polyhedron).\n\n\\subsubsection{(d) - Closer to a point then a set}\nA set of points closer to a given point than a given set is defined as $$\\{x \\| \\norm{x-x_0}_2 \\leq \\norm{x - y}_2 \\ \\forall y \\in S\\}$$ where $S\\subset\\real^n$ \\textbf{is not convex} in general. This is because there is not enough information about $y$ for a conclusion to be made whether it is convex or not. A counter example would be if $y$ is a point in orbit around a convex shape $S$ that would end up generating a concave $x$.\n\n\\subsubsection{(e) - Closer to a set then another set}\nA set of points closer to a given set than another given set is defined as $$\\{x \\ | \\ \\textbf{dist}(x,S) \\leq \\textbf{dist}(x,T)\\}$$ where $S, T \\subset\\real^n$, and $$\\textbf{dist}(x,S) = \\inf\\{\\norm{x-z}_2 \\ | \\ z \\in S\\}$$ \\textbf{is not convex} in general. This is because there is not enough information about $S$ and $T$ for a conclusion to be made whether it is convex or not. A counter example includes if $S$ or $T$ themselves are a concave shave that causes the set $x$ to also be concave and therefore not convex.\n\n\n\\subsubsection{(f) - Set of the sum being within a convex set}\nThe set defined as $$\\{x \\ | \\ x + S_2 \\subseteq S_1\\}$$ with $S_1$ being convex \\textbf{is not complex} in general. This is because we do not know enough information about $S_2$ to conclude that $x$ is conncex purely due to the relationship with the convex set $S_1$.\\\\\n\n\\subsubsection{(g) - Set with weighted distances to two points}\nThe set of all points that is closer to $a$ then $b$ by at least a factor of $\\theta$, defined as $$\\{x \\in \\real^n \\ | \\ \\norm{x-a}_2 \\leq \\theta \\norm{x-b}_2\\}$$ with $a \\neq b$ and $0 \\leq \\theta \\leq 1$ \\textbf{is convex.} This is know because, as proven in a previous problem, a hyperplane is formed for a similarity stated problem which itself is convex. When the distance to $a$ must be less then a portion of the distance to $b$ it will cause the psudo-hyperplane to curve inwards and untimely remain convex.\n\n\n\\newpage\n\\subsection{Problem 2.28}\n\\textbf{Problem:}\\\\\nDefine the positive semi-definite cone ($S_+^n$) for $n = 1, 2, 3$ in terms of ordinary inequalities with the matrix coefficients themselves.\n\n\\noindent\n\\textbf{Solution:}\\\\\nThe positive semi-definite cone is defined for size $n$ as the set of all symmetric matrices that are positive semi-definite:\n\\begin{equation}\n\tS_+^n \\equiv \\{x \\in S^n \\ | \\ x \\succeq 0\\}\n\\end{equation}\nOne method to ensure that a matrix is positive semi-definite is to ensure that its leading principle minors are all non-negative (strictly positive for positive definite).\\\\\n\nFor $n=1$ the required inequalities are simple, \n\\begin{equation}\n\tX = \\mqty[x_1] \\in S_+^1 \\iff x_1 \\geq 0\n\\end{equation}\n\nFor $n=2$ the inequalities can be found by ensuring the leading principle minors are all non negative:\n\\begin{equation}\n\t\\begin{aligned}\n\t\tm_1 &= \\det[x_1]\\\\\n\t\t&= x_1\\\\\n\t\tm_2 &= \\det \\mqty[x_1 & x_2 \\\\ x_2 & x_3] \\\\\n\t\t&= x_1 x_3 - x_2^2\n\t\\end{aligned}\n\\end{equation}\nThese definitions of the minors can be then be used to construct inequalities such that all the minors are positive:\n\\begin{equation}\n\tX = \\mqty[x_1 & x_2 \\\\ x_2 & x_3] \\in S_+^2 \\ \\iff \\ \\mqty{x_1 \\geq 0\\\\ x_1 x_3 \\geq x_2^2}\n\\end{equation}\n\nFor $n=3$ the inequalities can be found by ensuring the leading principle minors are all non negative:\n\\begin{equation}\n\t\\begin{aligned}\n\t\tm_1 &= \\det[x_1]\\\\\n\t\t&= x_1\\\\\n\t\tm_2 &= \\det \\mqty[x_1 & x_2 \\\\ x_2 & x_4] \\\\\n\t\t&= x_1 x_4 - x_2^2\\\\\n\t\tm_3 &= \\det \\mqty[x_1 & x_2 & x_3\\\\ x_2 & x_4 & x_5\\\\ x_3 & x_5 & x_6]\\\\\n\t\t&= x_1 (x_1 x_4 - x_2^2) - x_2 (x_2 x_6 - x_3 x_5) + x_3 (x_2 x_5 - x_3 x_4)\\\\\n\t\t&= x_1^2 x_4 - x_1 x_2^2 - x_2^2 x_6 + x_2 x_3 x_5 + x_2 x_3 x_5 - x_3^2 x_4\n\t\\end{aligned}\n\\end{equation}\nThese definitions of the minors can be then be used to construct inequalities such that all the minors are positive:\n\\begin{equation}\n\tX = \\mqty[x_1 & x_2 & x_3\\\\ x_2 & x_4 & x_5\\\\ x_3 & x_5 & x_6] \\in S_+^2 \\ \\iff \\ \n\t\\mqty{x_1 \\geq 0\\\\\n\t\t\tx_1 x_4 \\geq x_2^2\\\\\n\t\t x_1 x_2^2 + x_2^2 x_6 + x_3^2 x_4 \\geq x_1^2 x_4 + 2 x_2 x_3 x_5}\n\\end{equation}\n\n\\subsection{Problem 2.33}\nThe monotone non-negative cone is defined as all the nonnegative vectors with components sorted in non-increasing order:\n\\begin{equation}\n\tK_{m+} = \\{x \\in \\real^n \\ | \\ x_1 \\geq x_2 \\geq \\cdots \\geq x_n \\geq n\\}\n\\end{equation}\n\n\\subsubsection{Part a}\n\\textbf{Problem:}\\\\\nShow that $K_{m+}$ is a proper cone.\n\n\\noindent\n\\textbf{Solution:}\\\\\nA set, $C \\subseteq \\real^n$, is considered a cone if\n\\begin{equation}\n\t\\theta x \\in C \\ \\forall x \\in C \\text{ and } \\theta \\geq 0\n\\end{equation}\n\nIt can be easily seen that the set $K_{m+}$ satisfies this condition as scaling each element of the matrix $x \\in K_{m+}$ will equally be scaled by the same amount and the conditions of non-increasing order will still apply. This guarantees that $K_{m+}$ is in fact a cone.\\\\\nTo ensure convexity, the definition of convexity and of a cone can be incorporated into the following test:\\\\\nThe set $C \\subseteq \\real^n$ is a convex cone iff\n\\begin{equation}\n\t\\theta_1 x_1 + \\theta_2 x_2 \\in C \\ \\forall x_1,x_2 \\in C, \\theta_1, \\theta_2 \\geq 0\n\\end{equation}\nIt is also clear that $K_{m+}$ will satisfy as if each element element of one of the matrices is scaled it maintains the nonincreasing order. The same is true for summing two $K_{m+}$ matrices as the nonincreasing order will be maintained.\\\\\nIt is also clear that the cone is closed because the complimentary set $K_{m+}'$ is clearly open.\\\\\nSimilarily, $K_{m+}$ is solid because its definition includes all all of the subspace above the cone's boundary.\\\\\nThe cone is also known to be pointed because the definition defines that all of the vectors contained within the set are nonnegative. It is therefore only ever possible for elements to be in the positive sector (quadrant/octant/etc.). This means the cone cannot contain a line.\\\\\nWe can finish stating that $K_{m+}$ is a proper cone because it is a cone, convex, closed, solid, and pointed.\n\n\\newpage\n\\subsubsection{Part b}\n\\textbf{Problem:}\\\\\nFind the dual cone, $K_{m+}^*$.\n\n\\noindent\n\\textbf{Solution:}\\\\\nA dual cone for $K$ is defined as:\n\\begin{equation}\n\tK^* = \\{y \\in \\real^n \\ | \\ x^ Ty \\geq 0 \\ \\forall x \\in K\\}\n\\end{equation}\nThe left side of the inequality that defines the dual cone can also be written in summation form as: $$x^T y = \\sum_{i=1}^{n} x_i y_i \\geq 0$$\nIt is known that the following is equivalent:\n\\begin{equation}\n\t\\begin{aligned}\n\t\t\\sum_{i=1}^{n} x_i y_i &= (x_1 - x_2) y_1 + (x_2 - x_3) (y_1 + y_2) + \\cdots \\\\\n\t\t\t\t\t\t&= + (x_{n-1} - x_n) (y_1 + \\cdots + y_{n-1}) + x_n (y_1 + \\cdots + y_n)\n\t\\end{aligned}\n\\end{equation}\nBy definition, $x_1 \\geq x_2 \\geq \\cdots \\geq x_n \\geq n$, thus it can be said that each $(x_i - x_j)$ term in the expanded form is positive. In order for the inequality to always hold true, each $(y_1 + \\dots + y_i)$, must be positive as well. This can be achieved by defining that the elements of y are never decreasing.\\\\\nThus, the dual cone of $K_{k+}$ can be defined as:\n\\begin{equation}\n\tK_{m+}^{*} = K_{m-} = \\{x \\in \\real_{+}^n \\ | \\ x_1 \\leq x_2 \\leq \\dots \\leq x_n\\} \n\\end{equation}\n\n\n\\newpage\n\\section{Problem Set 2: Convex Functions}\n\n\\subsection{Problem 3.6}\n(Note to self... remember to remove this time: Proofs don't really seem required... pretty simple question)\\\\\n\\textbf{Problem:}\\\\\nWhat is the epigraph for the following convex functions?\\\\\n\n\\noindent\n\\textbf{Solution:}\\\\\nThe epigraph is defined as \n\\begin{equation}\n\t\\textbf{epi} f = \\{(x,t) \\ | \\ x \\in \\textbf{dom} f, f(x) \\leq t\\}\n\\end{equation}\nwhere $\\textbf{epi} f \\subset R^{n+1}$. This is equivalent to saying the space 'above' the function.\n\n\\subsubsection{(a) - Epigraph of a halfspace}\nThe epigraph of a halfspace in $n$ dimensions is a halfspace in $n+1$ dimensions.\n\n\\subsubsection{(b) - Epigraph of a convex cone}\nThe epigraph of a convex cone is another convex cone of in a higher dimension. (It might be possible to generalize it to a triangular slab type thing)\n\n\\subsubsection{(c) - Epigraph of a polyhedron}\nThe epigraph of a polyhedron is another polyhedron.\n\n\\newpage\n\\subsection{Problem 3.16}\n\\textbf{Problem:}\\\\\nDetermine if the following functions are convex, concave, quasiconvex, or quasiconcave.\n\n\\noindent\n\\textbf{Solution:}\\\\\n\\subsubsection{(a) - $e^x - 1$}\nLet $$f(x) = e^x - 1$$ on $\\real$.\\\\\n$f(x)$ \\textbf{is convex} and can be proven in multiple ways. For instance, visualizing the epigraph of the function, it is clear that a convex set is produced. Additionally, $f(x)$ can be constructed by putting the convex function $e^x$ through the convex affine function $x - 1$.\n\n\\subsubsection{(b) - $x_1 x_2$}\nLet $$f(x_1,x_2) = x_1 x_2$$ over the domain $\\real_{++}^2$.\\\\\nThis function \\textbf{is convex} and can be demonstrated using the definition:\n\\begin{align}\n\tf(\\theta x + (1-\\theta)y) &= \\theta f(x) + (1-\\theta) f(y)\\\\\n\t (\\theta x_1 + (1-\\theta) y_1) (\\theta x_2 + (1-\\theta) y_2)&= \\theta x_1 x_2 + (1- \\theta) y_1 y_2\\nonumber\\\\\n\t\\theta^2 x_1 x_2 + \\theta(1-\\theta)y_1 x_2 + \\theta (1-\\theta) x_1 y_2 + (1-\\theta)^2 y_1 y_2 &= \\theta x_1 x_2 + (1- \\theta) y_1 y_2\\nonumber\\\\\n\t\\theta^2 x_1 x_2 + (\\theta-\\theta^2) (x_1 y_2 + x_2 y_1) + (1- 2 \\theta + \\theta^2) y_1 y_2 &= \\theta x_1 x_2 + (1- \\theta) y_1 y_2\n\\end{align}\nIf we analyze each set of terms it not explicitly clear by the result, but within the domain $\\real_{++}^2$ the equality holds true.\n\n\\subsubsection{(c) - $1/(x_1 x_2)$}\nLet $$f(x_1,x_2) = 1/(x_1 x_2)$$ be defined over the domain $R_{++}^2$.\\\\\n\nThis function \\textbf{is convex}.\n\n\\subsubsection{(d) - $x_1/x_2$}\nLet $$f(x_1,x_2) = x_1/x_2$$ be defined over the domain $R_{++}^2$.\\\\\nThis function \\textbf{is convex}.\n\n\\subsection{(e) - $x_1^2 / x_2$}\nLet $$f(x_1,x_2) = x_1^2 / x_2$$ be defined over the domain $R \\cross R_{+}$.\nThis function \\textbf{is convex}.\n\n\n\\subsection{(f) - $x_1^\\alpha x_2^{1-\\alpha}$}\nLet $$f(x_1,x_2) = x_1^\\alpha x_2^{1-\\alpha}$$ with $0 \\leq \\alpha 1$ be defined over the domain $\\real_{++}^2$.\\\\\nThis function \\textbf{is not convex}.\n\n\n\\newpage\n\\subsection{Problem 3.18a}\n\\textbf{Problem:}\\\\\nUsing the proof of concavity of the log-determinant function to show that $$f(X) = \\trace(X^{-1})$$ is convex over the domain $S_{++}^n$.\\\\\n\n\\noindent\n\\textbf{Solution:}\\\\\nA proof of the concavity for the $$f(X) = \\log \\det X$$ is given as follows:\\\\\nGiven an arbritrary line, $X = Z + t V$, with $Z, V \\in S^n$, the function $$g(t) = f(Z + t V)$$ can be defined within $\\{x \\ | \\ Z + t V \\succ 0\\}$. It can then be assumed (without loss of generalizty) that $t = 0$ is in the interval (so that $Z \\succ 0$ is defined). This then allows the following:\n\\begin{align}\n\tg(t) &= \\log \\det(Z+tV)\\\\\n\t\t &= \\log \\det(Z^{1/2} (I + t Z^{-1/2}V Z^{-1/2})Z^{1/2})\\\\\n\t\t &= \\sum_{i=1}^n \\log(1 + t \\lambda_i) + \\log \\det Z\n\\end{align}\nwith $\\lambda_1, \\dots, \\lambda_n$ being the eigenvalues of $Z^{-1/2} V Z^{-1/2}$.\\\\\nThis allows for the first and second derivatives to be computed as:\n\\begin{align}\n\tg'(t) &= \\sum_{i=1}^n \\cfrac{\\lambda_i}{1 + t \\lambda_i}\\\\\n\tg''(t) &= -\\sum_{i=1}^n \\cfrac{\\lambda_i^2}{(1 + t \\lambda_i)^2}\n\\end{align}\nSince $g''(t) \\leq 0$, it can be concluded that $f$ is concave.\\\\\n\nThis proof can then be applied to the trace(X{-1}) is also concave.\\\\\nFirst, Let\n\\begin{equation}\n\tf(X) = \\trace(X^{-1})\n\\end{equation}\nbe defined on the domain $\\mathbf{S}_{++}^n$.\\\\\nAn arbritrary line can then be defined as\n\\begin{equation}\n\tg(t) = f(Z + tV)\n\\end{equation}\nover the domain $\\{x \\ | \\ Z + t V \\succ 0\\}$.\nIt can then be assumed (without loss of generality) that $t = 0$ is in the interval (so that $Z \\succ 0$ is defined). This then allows the following:\n\\begin{align}\n\tg(t) &= \\trace\\qty(\\qty(Z+tV)^{-1})\\\\\n\t&= \\trace\\qty(\\qty(Z^{1/2} (I + t Z^{-1/2}V Z^{-1/2})Z^{1/2})^{-1})\\\\\n\t\\intertext{from the fact that $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$, this can be manipulated to be}\n\t&= \\trace\\qty(Z^{-1/2} \\qty(I + t Z^{-1/2}V Z^{-1/2})^{-1}Z^{-1/2})\\\\\n\t\\intertext{since $\\trace(ABC) = \\trace(CAB)$,}\n\t&= \\trace\\qty(Z^{-1/2} Z^{-1/2} \\qty(I + t Z^{-1/2} V Z^{-1/2})^{-1})\\\\\n\t\\intertext{by taking the eigenfactor decomposition of $Z^{-1/2} V Z^{-1/2} = Q \\Lambda Q^{-1}$, this can be rewritten as:}\n\t&= \\trace\\qty(Z^{-1} \\qty(I + t Q \\Lambda Q^{-1})^{-1})\\\\\n\t&= \\trace\\qty(Z^{-1} \\qty(Q \\qty( I + t \\Lambda )Q^{-1})^{-1})\\\\\n\t&= \\trace\\qty(Z^{-1} Q \\qty( I + t \\Lambda)^{-1}Q^{-1})\\\\\n\t\\intertext{since $\\trace(ABC) = \\trace(CAB)$,}\n\t&= \\trace\\qty(Q Z^{-1} Q^{-1}\\qty( I + t \\Lambda)^{-1})\\\\\n\t\\intertext{By the definition of the trace, we can rewite this as}\n\t&= \\sum_{i=1}^n \\qty(Q^T Z^{-1} Q)_{ii} \\qty(1 + t \\lambda_i)^{-1}\n\\end{align}\nThis allows for the first and second derivatives to be computed as:\n\\begin{align}\n\tg'(t) &= -\\sum_{i=1}^n \\qty(Q^T Z^{-1} Q)_{ii} \\cfrac{\\lambda_i}{\\qty(1 + t \\lambda_i)^{-2}}\\\\\n\tg''(t) &= 2\\sum_{i=1}^n \\qty(Q^T Z^{-1} Q)_{ii} \\cfrac{\\lambda_i^2}{\\qty(1 + t \\lambda_i)^{-3}}\n\\end{align}\nSince $g''(t) \\leq 0$, it can be concluded that $f$ is concave.\\\\\n\n\n\\newpage\n\\subsection{Problem 3.22}\n\\textbf{Problem:}\\\\\nUse various composition rules to show that the following functions are convex.\n\n\\noindent\n\\textbf{Solution:}\\\\\n\\subsubsection{(a) - double log functions}\nLet $$f(x) = -\\log(-\\log(\\sum_{i=1}^m e^{a_i^T x + b_i}))$$ be defined over the domain $\\{x \\ | \\ \\sum_{i=1}^m e^{a_i^T x + b_i} < 1\\}$.\\\\\nIt is known that $$\\log(\\sum_{i=1}^{n} e^{y_i})$$ is convex. Since all compositions of convex functions with affine functions are also convex, it can be said that $$\\sum_{i=1}^m e^{a_i^T x + b_i}$$ is also convex.\\\\\nAdditionally, it is known that the $\\log()$ function is concave, but when the sign changes it becomes convex, thus the composition of $-g(-g(h(x)))$ is convex for the concave function $g(x) = \\log(x)$.\\\\\nTherefore, the function $f(x)$ is convex.\n\n\\subsubsection{(b) - square root of some product sum}\nLet $$f(x,u,v) = - \\sqrt{uv - x^T x}$$ be defined over the domain $\\{(x,u,v) \\ | \\ uv > x^T x, u, v > 0\\}$.\\\\\nIt is known that $g(x) = x^x / u$ is convex for $u>0$ and that $h(x)-\\sqrt{x_1 x_2}$ is convex on $\\real_{++}^2$.\\\\\nThe function $f(x,u,v)$ can be manipulated as follows:\n\\begin{align*}\n\tf(x,u,v) \t&= - \\sqrt{uv - x^T x}\\\\\n\t\t\t\t&= - \\sqrt{u \\qty(v -\\cfrac{x^T x}{u})}\n\\end{align*}\nFrom what we know about the underling functions, it can be said that the convex function $g(x)$ is summed with $v$ (a convex combination) and then used as an input to the convex function $h(x)$, resulting in an overall convex function.\n\n\\newpage\n\\subsubsection{(c) - log of some product sum}\nLet $$f(x,u,v) = - \\log(uv - x^T x)$$ be defined over the domain $\\{(x,u,v) \\ | \\ uv > x^T x, u, v > 0\\}$.\\\\\nIt is known that $g(x) = x^x / u$ is convex for $u>0$ as well as that the $h(x) = \\log(x)$ function is concave.\\\\\nPerforming the same manipulation as in the previous problem, $f(x,u,v)$ can be written as: $$f(x,u,v) = -\\log(u \\qty(v - \\cfrac{x^Tx}{u}))$$\nFrom this it can be derived that the convex function $g(x)$ is summed with $v$ (a convex combination) and then used as an input to the concave function $h(x)$ but is then negated to result in an overall convex function.\n\n\n\\subsubsection{(d) - complicated root of a powered sum and norm}\nLet $$ f(x,t) = -\\qty(t^p - \\norm{x}_p^p)^{1/p}$$ be defined with $p>1$ over the domain $\\{(x,t) \\ | \\ t \\geq \\norm{x}_p\\}$.\\\\\nIt is known that $g(x,u) = \\norm{x}_p^p / u^{p-1}$ is convex for $u >0$ and that $h(x,y) = -x^{1/p} y^{1 - 1/p}$ is convex over $\\real_{++}^2$.\\\\\n$f(x,t)$ can be manipulated as follows:\n\\begin{align}\n\tf(x,t)\t&= -\\qty(t^p - \\norm{x}_p^p)^{1/p}\\\\\n\t\t\t&= -\\qty(t^{p-1}\\qty(t - \\cfrac{\\norm{x}_p^p}{t^{p-1}}))^{1/p}\\\\\n\t\t\t&= -t^{1 - 1/p}\\qty(t - \\cfrac{\\norm{x}_p^p}{t^{p-1}})^{1/p}\\\\\n\\end{align}\nFrom this it is clear that the known convex function $g(x,t)$ is put through a weighted sum (which over the domain is never negative) and then is composed with the known convex function $h(x,t)$. This means that the original function $f(x,t)$ is a convex function.\n\n\\newpage\n\\subsubsection{(e) - complicated log of a powered sum and norm}\nLet $$ f(x,t) = - \\log(t^p - \\norm{x}_p^p)$$ with $p >1$ be defined over the domain $\\{(x,t) \\ | \\ t \\geq \\norm{x}_p\\}$.\\\\\nIt is known that $g(x,u) = \\norm{x}_p^p / u^{p-1}$ is convex for $u > 0$ and that (with the negative sign) the function $h(x) = - log(x)$ is convex.\n$f(x,t)$ can be manipulated as follows:\n\\begin{align}\n\tf(x,t)\t&= -\\log(t^p - \\norm{x}_p^p)\\\\\n\t\t\t&= -\\log\\qty(t^{p-1}\\qty(t - \\cfrac{\\norm{x}_p^p}{t^{p-1}}))\\\\\n\t\t\t&= -\\log\\qty(t^{p-1}) - \\log\\qty(t - \\cfrac{\\norm{x}_p^p}{t^{p-1}})\n\\end{align}\nFrom this it is clear that the known convex function $g(x,t)$ is put through a weighted sum (which over the domain is never negative) and then is composed with the known convex function $h(x,t)$. This means that the original function $f(x,t)$ is a convex function.\n\n\\end{document}\n", "meta": {"hexsha": "0837d6c8728681e395c057e6bea0d8ef5053ed6f", "size": 20415, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homework/HW2/MECH6327-HW2.tex", "max_stars_repo_name": "jonaswagner2826/MECH6327", "max_stars_repo_head_hexsha": "2b55aaf6f9e1bcf5cc684f5c853cadec26acf9d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homework/HW2/MECH6327-HW2.tex", "max_issues_repo_name": "jonaswagner2826/MECH6327", "max_issues_repo_head_hexsha": "2b55aaf6f9e1bcf5cc684f5c853cadec26acf9d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homework/HW2/MECH6327-HW2.tex", "max_forks_repo_name": "jonaswagner2826/MECH6327", "max_forks_repo_head_hexsha": "2b55aaf6f9e1bcf5cc684f5c853cadec26acf9d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.0790816327, "max_line_length": 525, "alphanum_fraction": 0.6686749939, "num_tokens": 7271, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.8104789109591832, "lm_q1q2_score": 0.6364760434048118}} {"text": "%!TEX root = TTK4150-Summary.tex\n\\section{Feedback linearization}\nConsider a class of nonlinear systems of the form\n\n\\begin{equation}\\label{eq:nonlinear-system}\n\t\\begin{split}\n\t\t\\dot{x} &= f(x) + G(x)u \\\\\n\t\ty &= h(x) \\\\\n\t\tu &= \\alpha (x) + \\beta (x) v \\\\\n\t\tz &= T(x)\n\t\\end{split}\n\\end{equation}\nwhere $u$ is a state feedback conboller, and $T$ is a change of variables.\n\nTo be able to cancel nonlinearities with feedback the input and non-linearities must appear together as a sum $\\lambda (x) + u$ or as a product $\\lambda (x) u$, where the matrix $\\lambda(x)$ is non-singular in the domain of interest, and $u = \\beta(x) v, \\beta(x) = \\lambda^{-1}$.\n\n\\paragraph{Definition 13.1}\nA nonlinear system as \\eqref{eq:nonlinear-system} where $f:D \\to R^n$ and $G : D \\to R^{n \\times p}$ are sufficiently smooth on a domain $D \\subseteq R^n$, is said to be feedback linearizable (or input-state linearizable) if there exists a diffeomorphism $T:D \\to R^n$ such that $D_z = T(D)$ contains the origin and the change of variables $z = T(x)$ transforms \\eqref{eq:nonlinear-system} into the form\n\\begin{equation}\n\t\\dot{z} = Az + B\\lambda (x) [u - \\alpha(x)]\n\\end{equation}\nwith $(A,B)$ controllable and $\\lambda(x)$ nonsingular $\\forall x \\in D$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Input-output linearization}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nConsider \\eqref{eq:nonlinear-system} which satisfies Def. 13.1. The derivative $\\dot{y}$ is given by\n\\begin{equation}\n\t\\dot{y} = \\pd{h}{x} \\del{f(x) + g(x)} \\triangleq L_f h(x) + L_g h(x) u\n\\end{equation}\nwhere $L_f h(x) \\triangleq \\pd{h}{x} f(x) $ is the \\emph{Lie Derivative} of $h$ w.r.t. $f$.\n\n\\paragraph{Relative degree}\nThe relative degree is the number of times $y$ must be differentiated until $u \\in D_0 \\subseteq D$ appears. A system must have a well defined relative degree to be input-output linearizable. (It must also be minimum phase.)\n\n\\paragraph{Diffeomorphism}\nWikipedia: \\emph{In mathematics, a diffeomorphism is an isomorphism of smooth manifolds. It is an invertible function that maps one differentiable manifold to another such that both the function and its inverse are smooth.}\n\n\\paragraph{Theorem 13.1}\nConsider \\eqref{eq:nonlinear-system} with relative degree $\\rho \\leq n$ in D. If $\\rho = n$, then for every $x_0 \\in D$, a neighborhood N of $x_0$ exists such that the map\n\\begin{equation}\n\tT(x) =\n\t\\left[\n\t\\begin{array}{ccc}\n\t\th(x) \\\\\n\t\tL_f h(x) \\\\\n\t\t\\vdots \\\\\n\t\tL_f^{n-1} h(x)\n\t\\end{array}\n\t\\right]\n\\end{equation}\nrestricted to N, is a diffeomorphism on N. If $\\rho < n$, then, for every $x_0 \\in D$, a neighborhood N of $x_0$ and smooth function $\\phi_1 (x), \\dots , \\phi_{n-\\rho} (x)$ exist such that \n\\begin{equation}\n\t\\pd{\\phi_i}{x} g(x) = 0, \\mbox{ for } 1 \\leq i \\leq n-\\rho, \\forall x \\in D_0\n\\end{equation}\nis satisfied $\\forall x \\in N$ and the map\n\\begin{equation}\n\tz = T(x) =\n\t\\left[\n\t\\begin{array}{ccc}\n\t\t\\phi_1(x) \\\\\n\t\t\\vdots \\\\\n\t\t\\phi_{n-\\rho}(x) \\\\\n\t\t--- \\\\\n\t\th(x) \\\\\n\t\t\\vdots \\\\\n\t\tL_f^{\\rho-1} h(x)\n\t\\end{array}\n\t\\right]\n\t\\triangleq\n\t\\left[\n\t\\begin{array}{ccc}\n\t\t\\phi(x) \\\\\n\t\t--- \\\\\n\t\t\\psi(x)\n\t\\end{array}\n\t\\right]\n\t\\triangleq\n\t\\left[\n\t\\begin{array}{ccc}\n\t\t\\eta \\\\\n\t\t--- \\\\\n\t\t\\xi\n\t\\end{array}\n\t\\right]\n\\end{equation}\nrestricted to N, is a diffeomorphism on N.\n\n\\paragraph{Method}\n\\begin{enumerate}\n\t\\item Set system on following form $\\dot{x} = f(x) + g(x)u$\n\t\\item Find the relative degree $\\rho$, ($\\rho = n \\Rightarrow$ no internal dynamics)\n\t\\item Write the system in normal form (external and internal dynamics)\n\t\\item Choose $u$ to cancel the nonlinearities\n\t\\item Analyze the zero-dynamics\n\t\\item Choose $v$ to solve the control problem\n\\end{enumerate}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Full-state linearization}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{State feedback control}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "240b4f53551e480eb008d3dd2abbc228fed3af68", "size": 3965, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-linearization.tex", "max_stars_repo_name": "jakoblover/ntnu-course-summaries", "max_stars_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-05-30T09:19:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-11T02:42:40.000Z", "max_issues_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-linearization.tex", "max_issues_repo_name": "jakoblover/ntnu-course-summaries", "max_issues_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TTK4150 Nonlinear control systems/sec-feedback-linearization.tex", "max_forks_repo_name": "jakoblover/ntnu-course-summaries", "max_forks_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.4056603774, "max_line_length": 404, "alphanum_fraction": 0.6257250946, "num_tokens": 1269, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.8104789040926008, "lm_q1q2_score": 0.6364760380124257}} {"text": "% !TEX root = ./Basilisk-avsLibrary20170812.tex\n\n\\section{User Guide}\n\n\\subsection{{\\tt linearAlgebra} Library}\nThe linear algebra library provides numerous C-based functions to perform basic matrix math operations. For a complete list of functions supported, consult the {\\tt linearAlgebra.h} file. \n\n\\subsubsection{Vector Operations}\nThe vector related functions all begin with a letter '{\\tt v}' and are broken down in the following categories depending on if the linear algebra operation is performed on a matrix of a general size, or of a specific size of 2, 3, 4 or 6.\n\\begin{itemize}\n\t\\item if the function is {\\tt vXXXX()} the code works on a $n$x1 matrix of arbitrary length $n$. \n\t\\item if the function is {\\tt v2XXXX()} the code works on a matrix of length 2 defined as {\\tt double vec[2]}.\n\t\\item if the function is {\\tt v3XXXX()} the code works on a matrix of length 2 defined as {\\tt double vec[3]}.\n\t\\item if the function is {\\tt v4XXXX()} the code works on a matrix of length 2 defined as {\\tt double vec[4]}.\n\\end{itemize}\n\nThe most extensive set of linear algebra support functions are for operation on 3 and 4 dimensional matrices common with orbital mechanics (position and velocity vector representations) and attitude dynamics (working with 3- and 4-parameter attitude descriptions). The following list details what vector operation each function performs:\n\\begin{itemize}\n\t\\item {\\tt Copy}: Copies vector $\\bm v_{1}$ into vector $\\bm v_{2}$\n\t\\item {\\tt SetZero}: Returns a zero vector\n\t\\item {\\tt Set}: Sets the vector representation to specific values\n\t\\item {\\tt Add}: Sums $\\bm v_{1}$ and $\\bm v_{2}$\n\t\\item {\\tt Subtract}: Returns the difference $\\bm v_{1}$ - $\\bm v_{2}$\n\t\\item {\\tt Scale}: Returns $\\bm v_{1}$ multiplied by a scalar $\\alpha$\n\t\\item {\\tt Dot}: Returns the dot product $\\bm v_{1} \\cdot \\bm v_{2}$\n\t\\item {\\tt OuterProduct}: Returns the outer product $\\bm v_{1} \\bm v_{2}^{T}$\n\t\\item {\\tt Normalize}: Returns a normalized vector $\\bm v_{1}/ v_{1}$\n\t\\item {\\tt MaxAbs}: Returns the largest element of a vector\n\t\\item {\\tt IsEqual}: Checks if two vector represnetations are identical\t\n\t\\item {\\tt IsZero}: Checks if a vector is full of zero elements\n\t\\item {\\tt Print}: Prints the vector representation to a file\n\t\\item {\\tt Sort}: Sorts the vector elements by size\n\t\\item {\\tt Print}: Prints the vector representation to a file\n\t\\item {\\tt PrintScreen}: Prints the vector representation to the screen\n\\end{itemize}\n\n\n\n\n\n\\subsubsection{Matrix Operations}\nThe matrix related functions are all labeled with a letter '{\\tt M}' and are broken down into the following matrix dimension related categories:\n\\begin{itemize}\n\t\\item {\\tt mXXXXy()}: the code works on a $n\\times m$ matrix of arbitrary dimension $n$ and $m$.\n\t\\item {\\tt m22XXXX()}: the code works on a $2\\times 2$ matrix defined through {\\tt mat[2][2]}. \n\t\\item {\\tt m33XXXX()}: the code works on a $3\\times 3$ matrix defined through {\\tt mat[3]3]}. \n\t\\item {\\tt m44XXXX()}: the code works on a $4\\times 4$ matrix defined through {\\tt mat[4][4]}. \n\\end{itemize}\n\nThe following list provides an overview of the supported matrix functions. Note that not al dimensions have all functions provided, but the 2, 3, and 4 dimensional matrix support is pretty complete. \n\\begin{itemize}\n\t\\item {\\tt SetIdentity}: Returns an identity matrix\n\t\\item {\\tt SetZero}: Returns a zero matrix\n\t\\item {\\tt Set}: Creates a matrix with specific values\n\t\\item {\\tt Copy}: Copies the matrix $[M_{1}]$ into $[M_{2}]$\n\t\\item {\\tt m33MultM33}: Performs the matrix to matrix multiplication $[M_{1}][M_{2}]$\n\t\\item {\\tt m33MultM33t}: Performs the matrix to matrix multiplication $[M_{1}][M_{2}]^{T}$\n\t\\item {\\tt m33tMultM33}: Performs the matrix to matrix multiplication $[M_{1}]^{T}[M_{2}]$\n\t\\item {\\tt m33MultV3}: Computes $[M_{1}]\\bm v_{1}$ \n\t\\item {\\tt m33tMultV3}: Computes $[M_{1}]^{T}\\bm v_{1}$ \n\t\\item {\\tt v3tMultM33}: Computes $\\bm v_{1}^{T}[M_{1}]$ \n\t\\item {\\tt v3tMultM33t}: Computes $\\bm v_{1}^{T}[M_{1}]^{T}$\n\t\\item {\\tt Tilde}: Returns the skew-symmetric matrix $[\\tilde{\\bm v}_{1}]$\n\t\\item {\\tt Transpose}: Returns $[M_{1}]^{T}$\n\t\\item {\\tt Add}: Returns the sum $[M_{1}] + [M_{2}]$\n\t\\item {\\tt Subtract}: Returns the difference $[M_{1}] - [M_{2}]$\n\t\\item {\\tt Scale}: Returns the scaled matrix $\\alpha [M_{1}]$\n\t\\item {\\tt Trace}: Compute the matrix trace $\\sum M_{ii}$\n\t\\item {\\tt Determinant}: Returns the square matrix determinant\n\t\\item {\\tt IsEqual}: Checks if all matrix elements are equal\n\t\\item {\\tt IsZero}: Checks if all matrix elements are zero\n\t\\item {\\tt Print}: Prints the matrix representation to a file\n\t\\item {\\tt PrintScreen}: Prints the matrix representation to the screen\n\t\\item {\\tt Inverse}: Returns the matrix inverse $[M_{1}]^{-1}$\n\t\\item {\\tt SingularValues}: Computes the singular values of $[M_{1}]$\n\t\\item {\\tt EigenValues}: Computes the Eigenvalues of $[M_{1}]$\n\t\\item {\\tt ConditionNumber}: Computes the condition number of $[M_{1}]$\n\\end{itemize}\n\n\n\n\n\\subsection{{\\tt rigidBodyKinematics} Library}\n\nThe following discussion is a brief overview of the {\\tt rigidBodyKinematics} library function notation. Please see Appendix E in Reference~\\citenum{schaub} for a complete description:\n\\begin{itemize}\n\t\\item {\\tt XXX2YYY}: Converts the attitude coordinates {\\tt XXX} to {\\tt YYY}\n\t\\item {\\tt addXXX}: Add the two attitude descriptions $\\bm x_{1} = \\bm x_{\\cal B/F}$ and $\\bm x_{2}= \\bm x_{\\cal F/N}$ to return the sequential rotation of first $\\bm x_{1}$ and then $\\bm x_{2}$. Returns is the description $\\bm x_{3} = \\bm x_{\\cal B/N}$\n\t\\item {\\tt subXXX}: Subtract the two attitude descriptions $\\bm x_{1} = \\bm x_{\\cal F/N}$ and $\\bm x_{2}= \\bm x_{\\cal B/N}$ to return the relative orientation $\\bm x_{3} = \\bm x_{\\cal F/B}$\n\t\\item {\\tt BmatXXX}: Returns the matrix $[B]$ of the differential kinematic equations of the attitude parameters {\\tt XXX} in Reference~\\citenum{schaub}. Note the scalar multiplier is not included here. For example, with MRPs\\cite{survey} $[B]$ is defined as $$\\dot{\\bm \\sigma}_{\\cal B/N} = \\frac{1}{4}[B(\\bm\\sigma_{\\cal B/N})] \\leftexp{B}{\\bm\\omega}_{\\cal B/N}$$\n\t\\item {\\tt BinvXXX}: Returns the matrix inverse $[B]^{-1}$ of the attitude parameters {\\tt XXX}\n\t\\item {\\tt dX}: Returns the time derivatives of the attitude parameters {\\tt XXX} as a function of these parameters and the body angular velocity vector $\\bm \\omega$\n\t\\item {\\tt Mi}: Returns the single-axis DCM about the $\\hat{\\bm b}_{i}$ axis\n\t\\item {\\tt wrapToPi}: Makes sure that an angle lies within $\\pm\\pi$\n\\end{itemize}\n\n\n\n\n\\subsection{{\\tt orbitalMotion} Library}\n\n\\subsubsection{Orbit Anomaly Angle Conversions}\nTo convert between the various anomaly angles, the following functions are defined:\n\\begin{itemize}\n\t\\item {\\tt double f2E(double f, double e)}\n\t\\item {\\tt double E2f(double Ecc, double e)}\n\t\\item {\\tt double E2M(double Ecc, double e)}\n\t\\item {\\tt double M2E(double M, double e)}, with a change conversion tolerance of $10^{-13}$ and a maximum iteration limit of 200\n\t\\item {\\tt double f2H(double f, double e)}\n\t\\item {\\tt double H2f(double H, double e)}\n\t\\item {\\tt double H2N(double H, double e)}\n\t\\item {\\tt double N2H(double N, double e)}, with a change conversion tolerance of $10^{-13}$ and a maximum iteration limit of 200\n\\end{itemize}\n\n\n\n\\subsubsection{Orbit Element Conversion}\nTo convert from classical orbit elements to inertial Cartesian coordinates, the function\n\\begin{verbatim}\n\telem2rv(double mu, classicElements *elements, double *rVec, double *vVec)\n\\end{verbatim}\nis used where {\\tt mu} is the gravitational constant of the 2-body problem, the classical elements are defined through \n$$\n\t(a, e, i, \\Omega, \\omega, f)\n$$\nwhere the anomaly angle is typically given by $f$, unless the orbit is a rectilinear motion in which case the anomaly angle input is $E$. The function returns the inertial position and velocity vectors in the arrays {\\tt rVec} and {\\tt vVec}. \n\nTo convert from inertial Cartesian coordinates to classical orbit elements, the function\n\\begin{verbatim}\n\trv2elem(double mu, double *rVec, double *vVec, classicElements *elements)\n\\end{verbatim}\nBeyond the classical elements listed above, this routine also stores the radius of perapses $r_{p}$ and apoapses $r_{a}$, as well as $\\alpha = \\dfrac{1}{a}$. \n\n\n\\subsubsection{Space Environment Functions}\n\\paragraph{Atmospheric Density}\nThe Earth's atmospheric density $\\rho$ is computed using\\\\\n\\indent\t{\\tt double atmosphericDensity(double alt)} \\\\\nThe density is returned as a scalar value. \n\n\\paragraph{Mean Debye Length}\nThe mean Debye Length for the near-Earth environment is approximated, very crudely, through a polynomial fit. The function returns the scalar $\\lambda_{d}$ and is called through \\\\\n\\indent {\\tt double debyeLength(double alt)}\n\n\\paragraph{Atmospheric Drag Acceleration}\nTo compute an estimate of the Earth's atmospheric drag acceleration, use the function: \\\\\n\\indent {\\tt void atmosphericDrag(double Cd, double A, double m, double *rvec, double *vvec, double *advec)}\n\nThe inputs are the ballistic drag coefficient {\\tt Cd}, the velocity-projected cross-sectional area {\\tt A}, as well as the spacecraft mass {\\tt m}. Given the inertial position vectors {\\tt rvec} and {\\tt vvec}, the function returns the drag acceleration {\\tt advec}. \n\n\\paragraph{Gravitational Zonal Harmonics}\nThis function returns the inertial acceleration due to a planets zonal Harmonic. The function call is:\\\\\n\\indent {\\tt void jPerturb(double *rvec, int num, double *ajtot, ...)}\n\nIf not option argument is provided, then the zonal harmonics of Earth are simulated. About other celestial objects only the $J_{2}$ harmonic is implemented. Here the object is specified through the {\\tt CelestialObject\\_t} enumeration. For example, to get the $J_{2}$ zonal harmonic about Venus the argument {\\tt CELESTIAL\\_VENUS} is provided. \n\n\\paragraph{Solar Radiation Pressure Acceleration}\nTo compute the inertial disturbance acceleration due to the solar radiation pressure use the function\\\\\n\\indent {\\tt void solarRad(double A, double m, double *sunvec, double *arvec)}\n\nHere {\\tt A} is the projected surface area, {\\tt m} is the spacecraft mass, {\\tt sunvec} is the sun position vector to the Sun in units of AU. \n\n\n\n\n\n", "meta": {"hexsha": "73cefe623e32729e0c5543a88cf2c391c5fc14dc", "size": 10256, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/simulation/utilitiesSelfCheck/_Documentation/secUserGuide.tex", "max_stars_repo_name": "ian-cooke/basilisk_mag", "max_stars_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_stars_repo_licenses": ["0BSD"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/simulation/utilitiesSelfCheck/_Documentation/secUserGuide.tex", "max_issues_repo_name": "ian-cooke/basilisk_mag", "max_issues_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_issues_repo_licenses": ["0BSD"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-03-13T20:52:22.000Z", "max_issues_repo_issues_event_max_datetime": "2019-03-13T20:52:22.000Z", "max_forks_repo_path": "src/simulation/utilitiesSelfCheck/_Documentation/secUserGuide.tex", "max_forks_repo_name": "ian-cooke/basilisk_mag", "max_forks_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_forks_repo_licenses": ["0BSD"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.4131736527, "max_line_length": 366, "alphanum_fraction": 0.7233814353, "num_tokens": 2856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972751232808, "lm_q2_score": 0.7310585786300049, "lm_q1q2_score": 0.636457606510781}} {"text": "\\documentclass[epsfig,10pt,fullpage]{article}\n\n\\newcommand{\\LabNum}{7}\n\\newcommand{\\CommonDocsPath}{../../../common/docs}\n\\input{\\CommonDocsPath/preamble.tex}\n\n\\begin{document}\n\n\\centerline{\\huge Digital Logic}\n~\\\\\n\\centerline{\\huge Laboratory Exercise \\LabNum}\n~\\\\\n\\centerline{\\large Finite State Machines}\n~\\\\\n\n\\noindent\nThis is an exercise in using finite state machines.\n~\\\\\n\n\\section*{Part I}\n\\addcontentsline{toc}{1}{Part I}\nWe wish to implement a finite state machine (FSM) that recognizes two specific sequences \nof applied input symbols, namely four consecutive 1s or four consecutive 0s. There is an \ninput $w$ and an output $z$. Whenever $w = 1$\nor $w = 0$ for four consecutive clock pulses the value of $z$ has to be 1; \notherwise, $z = 0$. Overlapping sequences\nare allowed, so that if $w = 1$ for five consecutive clock pulses the output $z$\nwill be equal to 1 after the fourth and\nfifth pulses. Figure~\\ref{fig:sequence_diagram} illustrates the required relationship between $w$ and $z$.\n\n\\begin{figure}[H]\n\\centerline{\n\\includegraphics{figures/figure1.pdf}}\n\\caption{Required timing for the output $z$.}\n\\label{fig:sequence_diagram}\n\\end{figure}\n\nA state diagram for this FSM is shown in Figure~\\ref{fig:state_diagram}. For this part you are to manually\nderive an FSM circuit that implements this state diagram, including the logic expressions\nthat feed each of the state flip-flops. To\nimplement the FSM use nine state flip-flops called $y_8, \\ldots, y_0$ and the \none-hot state assignment given in Table~\\ref{tab:state_assignment}.\n\n\\begin{figure}[H]\n\\centerline{\n\\includegraphics{figures/figure2.pdf}}\n\\caption{A state diagram for the FSM.}\n\\label{fig:state_diagram}\n\\end{figure}\n\n\\begin{table}[H]\n\\begin{center}\n\\begin{tabular}{c|c}\n~ & State Code \\\\ \nName & $y_8 y_7 y_6 y_5 y_4 y_3 y_2 y_1 y_0$ \\\\ \\hline\n\\rule[-0.075in]{0in}{0.25in}{\\bf A} & $000000001$ \\\\ \n{\\bf B} & $000000010$ \\\\ \n{\\bf C} & $000000100$ \\\\ \n{\\bf D} & $000001000$ \\\\ \n{\\bf E} & $000010000$ \\\\ \n{\\bf F} & $000100000$ \\\\ \n{\\bf G} & $001000000$ \\\\ \n{\\bf H} & $010000000$ \\\\ \n{\\bf I} & $100000000$ \\\\ \n\\end{tabular}\n\\end{center}\n\\caption{One-hot codes for the FSM.}\n\\label{tab:state_assignment}\n\\end{table}\n~\\\\\n~\\\\\nDesign and implement your circuit on your DE-series board as follows:\n\n\\begin{enumerate}\n\\item Create a new Quartus\\textsuperscript{\\textregistered} project for the FSM circuit. \n\n\\item Write a Verilog file that instantiates the nine flip-flops in the circuit and which\nspecifies the logic expressions that drive the flip-flop input ports. Use only\nsimple {\\bf assign} statements in your Verilog code to specify the logic feeding the\nflip-flops. Note that the one-hot code enables you to derive these expressions by\ninspection.\n\nUse the toggle switch {\\it SW}$_0$ as an active-low synchronous reset input\nfor the FSM, use {\\it SW}$_1$ as the {\\it w} input, and the pushbutton {\\it KEY}$_0$ as the clock input which \nis applied manually. Use the red light {\\it LEDR}$_9$ as the output $z$, \nand assign the state flip-flop outputs to the red lights {\\it LEDR}$_8$ to {\\it LEDR}$_0$.\n\n\\item Include the Verilog file in your project, and assign the pins on the FPGA to \nconnect to the switches and the LEDs.\n\n\\item Simulate the behavior of your circuit.\n\n\\item Once you are confident that the circuit works properly as a result of your\nsimulation, download the circuit into the FPGA chip. Test the functionality of your \ndesign by applying the input sequences and observing the output LEDs. Make sure that the\nFSM properly transitions between states as displayed on the red LEDs, and that it produces\nthe correct output values on {\\it LEDR}$_9$.\n\n\\item Finally, consider a modification of the one-hot code given in Table~\\ref{tab:state_assignment}. It is often \ndesirable to set all flip-flop outputs to the value 0 in the reset state.\n\nTable~\\ref{tab:modified_state_assignment} shows a modified one-hot state assignment in which the reset state, {\\it A},\nuses all 0s. This is accomplished by inverting the state variable $y_0$. \nCreate a modified version of your Verilog code that implements this state\nassignment. ({\\it Hint}: you should need to make very few changes to the logic expressions\nin your circuit to implement the modified state assignment.) \n\n\\item Compile your new circuit and test it.\n\n\\begin{table}[H]\n\\begin{center}\n\\begin{tabular}{c|c}\n~ & State Code \\\\ \nName & $y_8 y_7 y_6 y_5 y_4 y_3 y_2 y_1 y_0$ \\\\ \\hline\n\\rule[-0.075in]{0in}{0.25in}{\\bf A} & $000000000$ \\\\ \n{\\bf B} & $000000011$ \\\\ \n{\\bf C} & $000000101$ \\\\ \n{\\bf D} & $000001001$ \\\\ \n{\\bf E} & $000010001$ \\\\ \n{\\bf F} & $000100001$ \\\\ \n{\\bf G} & $001000001$ \\\\ \n{\\bf H} & $010000001$ \\\\ \n{\\bf I} & $100000001$ \\\\\n\\end{tabular}\n\\end{center}\n\\caption{Modified one-hot codes for the FSM.}\n\\label{tab:modified_state_assignment}\n\\end{table}\n\n\\end{enumerate}\n\n\\section*{Part II}\n\\addcontentsline{toc}{2}{Part II}\nFor this part you are to write another style of Verilog code for the FSM in Figure 2. In\nthis version of the code you should not manually derive the logic expressions needed for\neach state flip-flop. Instead, describe the state table for the FSM by using a\nVerilog {\\bf case} statement in an {\\bf always} block, and use another {\\bf always} block to\ninstantiate the state flip-flops. You can use a third {\\bf always} block or simple assignment\nstatements to specify the output $z$. To implement the FSM, use four state flip-flops\n$y_3, \\ldots, y_0$ and binary codes, as shown in Table~\\ref{tab:bincodes}.\n\n\\begin{table}[H]\n\\begin{center}\n\\begin{tabular}{c|c}\n~ & State Code \\\\ \nName & $y_3 y_2 y_1 y_0$ \\\\ \\hline\n\\rule[-0.075in]{0in}{0.25in}{\\bf A} & $0000$ \\\\ \n{\\bf B} & $0001$ \\\\ \n{\\bf C} & $0010$ \\\\ \n{\\bf D} & $0011$ \\\\ \n{\\bf E} & $0100$ \\\\ \n{\\bf F} & $0101$ \\\\ \n{\\bf G} & $0110$ \\\\ \n{\\bf H} & $0111$ \\\\ \n{\\bf I} & $1000$ \\\\ \n\\end{tabular}\n\\end{center}\n\\caption{Binary codes for the FSM.}\n\\label{tab:bincodes}\n\\end{table}\n\nA suggested skeleton of the Verilog code is given in Figure~\\ref{fig:skeleton_code}.\n\n\\begin{figure}[H]\n\\begin{center}\n\\begin{minipage}[t]{12.5 cm}\n\\begin{tabbing}\nZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\kill\n{\\bf module} ~part2 (~$\\ldots$~);\\\\\n\\>$\\ldots$ define input and output ports\\\\\n~\\\\\n\\>$\\ldots$ define signals\\\\\n\\>{\\bf reg} [3:0] y\\_Q, Y\\_D; \\>\\>\\>\\>\\>\\>\\>\\>// y\\_Q represents current state, Y\\_D represents next state\\\\\n\\>{\\bf parameter} A = 4'b0000, B = 4'b0001, C = 4'b0010, D = 4'b0011, E = 4'b0100,\\\\\n\\>\\>F = 4'b0101, G = 4'b0110, H = 4'b0111, I = 4'b1000;\\\\\n~\\\\\n\\>{\\bf always} @(w, y\\_Q)\\\\\n\\>{\\bf begin}: state\\_table\\\\\n\\>\\>{\\bf case} (y\\_Q)\\\\\n\\>\\>\\>A:\t\\>{\\bf if} (!w) Y\\_D = B;\\\\\n\\>\\>\\>\\>{\\bf else} Y\\_D = F;\\\\\n\\>\\>\\>$\\ldots$ remainder of state table \\\\\n\\>\\>\\>{\\bf default}: Y\\_D = 4'bxxxx;\\\\\n\\>\\>{\\bf endcase}\\\\\n\\>{\\bf end} // state\\_table\\\\\n\\\\\n\\>{\\bf always} @({\\bf posedge} Clock)\\\\\n\\>{\\bf begin}: state\\_FFs\\\\\n\\>\\>$\\ldots$ \\\\\n\\>{\\bf end} // state\\_FFS\\\\\n\\\\\n\\>$\\ldots$ assignments for output z and the LEDs\\\\\n{\\bf endmodule}\n~\\rule{5.0in}{0in}\\\\\n\\end{tabbing}\n\\end{minipage}\n\\end{center}\n\\caption{Skeleton Verilog code for the FSM.}\n\\label{fig:skeleton_code}\n\\end{figure}\n~\\\\\nImplement your circuit as follows.\n\n\\begin{enumerate}\n\\item Create a new project for the FSM.\n\n\\item Include in the project your Verilog file that uses the style of code in Figure 3.\nUse the same switches, pushbuttons, and lights that were used in Part I.\n\n\\item Before compiling your code it is necessary to explicitly tell the Synthesis tool in\nQuartus that you wish to have the finite state machine implemented using the state\nassignment specified in your Verilog code. If you do not explicitly give this\nsetting to Quartus, the Synthesis tool will automatically use a state assignment of\nits own choosing, and it will ignore the state codes specified in your Verilog code. To\nmake this setting, choose {\\sf Assignments > Settings} in Quartus, and click on the\n{\\sf Compiler Settings} item on the left side of the window, then click on \nthe {\\sf Advanced Settings (Synthesis)} button. As indicated in Figure 4, change the \nparameter {\\sf State Machine Processing} to the setting {\\sf User-Encoded}.\n\n\\item Compile your project. To examine the circuit produced by Quartus open \nthe RTL Viewer tool. Double-click\non the box shown in the circuit that represents the finite state machine, and determine\nwhether the state diagram that it shows properly corresponds to the one in Figure 2.\nTo see the state codes used for your FSM, open the Compilation Report, select the {\\sf Analysis\nand Synthesis} section of the report, and click on {\\sf State Machines}.\n\n\\item Download the circuit into the FPGA chip and test its functionality.\n\n\\item In step 3 you instructed the Quartus Synthesis tool to use the state\nassignment given in your Verilog code. To see the result of removing this setting, open\nagain the Quartus settings window by choosing {\\sf Assignments > Settings}, and \nclick on the {\\sf Compiler Settings} item on the left side of the window, then click on \nthe {\\sf Advanced Settings (Synthesis)} button. Change the setting for \n{\\sf State Machine Processing} from {\\sf User-Encoded} to {\\sf One-Hot}. Recompile the\ncircuit and then open the report file, select the {\\sf Analysis\nand Synthesis} section of the report, and click on {\\sf State Machines}.\nCompare the state codes shown to those given in Table~\\ref{tab:modified_state_assignment}, and\ndiscuss any differences that you observe.\n\\end{enumerate}\n\n\\begin{figure}[H]\n\\centerline{\n\\includegraphics[scale = 0.75]{figures/figure4.png}}\n\\caption{Specifying the state assignment method in Quartus.}\n\\label{fig:fig7_4}\n\\end{figure}\n\n\\section*{Part III}\n\\addcontentsline{toc}{3}{Part III}\nThe sequence detector can be implemented in a straightforward manner using shift registers,\ninstead of using the more formal approach described above. Create Verilog code that\ninstantiates two 4-bit shift registers; one is for recognizing a sequence of four 0s, and\nthe other for four 1s. Include the appropriate logic expressions in your design\nto produce the output $z$. Make a Quartus project for your design and implement\nthe circuit on your DE-series board. Use the switches and LEDs on the board in a similar way as\nyou did for Parts I and II and observe the behavior of your shift registers and the\noutput $z$. Answer the following question: could you use just one 4-bit shift\nregister, rather than two? Explain your answer.\n\n~\\newpage\n\\section*{Part IV}\n\\addcontentsline{toc}{4}{Part IV}\nIn this part of the exercise you are to implement a Morse-code encoder using an FSM. The Morse code uses\npatterns of short and long pulses to represent a message. Each letter is represented as a sequence of dots\n(a short pulse), and dashes (a long pulse). For example, the first eight letters of the alphabet have the\nfollowing representation:\n~\\\\\n\\begin{center}\n\\begin{minipage}[t]{12.5 cm}\n\\begin{tabbing}\nZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\=ZZ\\kill\n\\>A\\>\\>{\\bf $\\bullet$ ---}\\\\\n\\>B\\>\\>{\\bf --- $\\bullet$ $\\bullet$ $\\bullet$}\\\\\n\\>C\\>\\>{\\bf --- $\\bullet$ --- $\\bullet$}\\\\\n\\>D\\>\\>{\\bf --- $\\bullet$ $\\bullet$}\\\\\n\\>E\\>\\>{\\bf $\\bullet$}\\\\\n\\>F\\>\\>{\\bf $\\bullet$ $\\bullet$ --- $\\bullet$}\\\\\n\\>G\\>\\>{\\bf --- --- $\\bullet$}\\\\\n\\>H\\>\\>{\\bf $\\bullet$ $\\bullet$ $\\bullet$ $\\bullet$}\\\\\n\\end{tabbing}\n\\end{minipage}\n\\end{center}\n\n~\\\\\nDesign and implement a Morse-code encoder circuit using an FSM. Your circuit should take as input one of the first eight letters of the alphabet and display the Morse code for it on a red LED. Use switches {\\it SW}$_{2-0}$ and pushbuttons {\\it KEY}$_{1-0}$ as inputs. When a user presses {\\it KEY}$_1$, the circuit should display the Morse code for a letter specified by {\\it SW}$_{2-0}$ (000 for A, 001 for B, etc.), using 0.5-second pulses to represent dots, and 1.5-second pulses to represent dashes. Pushbutton {\\it KEY}$_0$ should function as an asynchronous reset. \n\nA high-level schematic diagram of a possible circuit for the Morse-code encoder \nis shown in Figure~\\ref{fig:morse_code_cct}.\n\n\\begin{figure} [H]\n\\begin{center}\n\t\\includegraphics[scale = 0.9]{figures/fig_morse_code_circuit_schematic.pdf}\n\\end{center}\n\\caption{High-level schematic diagram of the circuit for Part IV.}\n\\label{fig:morse_code_cct}\n\\end{figure}\n\n\n\\input{\\CommonDocsPath/copyright.tex}\n\\end{document}\n", "meta": {"hexsha": "b40a339f48866e9b9e6bddbcc4c1eb2176a5a98d", "size": 12284, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "verilog/lab7/doc/verilog_lab7.tex", "max_stars_repo_name": "fpgacademy/Lab_Exercises_Digital_Logic", "max_stars_repo_head_hexsha": "f4119b617a5af228a032f8f0ff27a299b496ad78", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-03-09T23:21:40.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T23:21:40.000Z", "max_issues_repo_path": "verilog/lab7/doc/verilog_lab7.tex", "max_issues_repo_name": "fpgacademy/Lab_Exercises_Digital_Logic", "max_issues_repo_head_hexsha": "f4119b617a5af228a032f8f0ff27a299b496ad78", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "verilog/lab7/doc/verilog_lab7.tex", "max_forks_repo_name": "fpgacademy/Lab_Exercises_Digital_Logic", "max_forks_repo_head_hexsha": "f4119b617a5af228a032f8f0ff27a299b496ad78", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-15T16:44:27.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-15T16:44:27.000Z", "avg_line_length": 40.8106312292, "max_line_length": 571, "alphanum_fraction": 0.7163790296, "num_tokens": 3685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585786300049, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.6364575966927811}} {"text": "\\documentclass[letterpaper]{article}\n\n\\usepackage{fullpage}\n\\usepackage{nopageno}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage[utf8]{inputenc}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\n\\title{Notes}\n\\date{November 7, 2014}\n\\maketitle\n\\section*{5.3.J}\nLet $m(x,y)=\\max\\{x,y\\}$. prove that $m$ is continuous and then show that $h(x)=\\max\\{f(x),g(x)\\}$ is continuous if $f(x)$ and $g(x)$ are continuous.\n\ncase 1, $m(x,y)=x,m(x_1,y_1)=x_1$ then $|m(x,y)-m(x_1,y_1)|=|x-x_1|$. take $r=\\varepsilon$ and then $||(x,y)-(x_1,y_1)||y\\\\\n y_1>x_1\\\\\n -y_1<-x_1\\\\\n (x-y_1)<(x-x_1)\\\\\n\\end{align*}\nSo if $x-x_1<\\varepsilon$ then $(x-y_1)<\\varepsilon$. \n\\begin{align*}\n y-y1-\\varepsilon\\\\\n \\text{then }x-y1>-\\varepsilon\\\\\n\\end{align*}\nso $r=\\varepsilon$ will give us $|x-y1|<\\varepsilon$\n\nnow let $\\varepsilon\\circ f(x)=\\varepsilon_1(f(x)=(f(x),0)$ and $\\varepsilon_2\\circ g(x)=\\varepsilon_2(g(x)=(0,g(x))$. and $h(x)=m\\circ(\\varepsilon\\circ f(x)+\\varepsilon\\circ g(x))$\n\nbecause sum and composition of continuous functions are continuous, then h is continuous. induction gives us continuity for $\\max\\{f_1(x),\\dots,f_m(x)\\}$\n\n\\section*{5.4H,I\\& 5.5F}\nlet $f:\\mathbb{R}\\to\\mathbb{R}$ be periodic and continuous, ie $\\exists d>0$ such that $\\forall x\\in \\mathbb{R}, f(x+d)=f(x)$.\n\n\\subsection*{show it attains max min}\nlook at $y\\in[x,x+d]$ then $\\exists a,b\\in [x,x+d]$ such that $f(a)\\le f(y)\\le f(b)$. periodicity gives us same properties on all of $\\mathbb{R}$\n\\subsection*{show that it is bounded and uniformally continuous}\n\\subsection*{find a function $f:\\mathbb{R}^2\\to \\mathbb{R}$ such that $f(x+1,y)=f(x,y)\\forall(x,y)$ but $f$ is not bounded or does not attain it's max}\n\\end{document}\n", "meta": {"hexsha": "da5217af6f5069a896488431c9e202467e6b8dd4", "size": 1993, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "real analysis/analysis-notes-2014-11-07.tex", "max_stars_repo_name": "ylixir/school", "max_stars_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "real analysis/analysis-notes-2014-11-07.tex", "max_issues_repo_name": "ylixir/school", "max_issues_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "real analysis/analysis-notes-2014-11-07.tex", "max_forks_repo_name": "ylixir/school", "max_forks_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.86, "max_line_length": 219, "alphanum_fraction": 0.6588058204, "num_tokens": 819, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581741774411, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.6364561866951493}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Local B-Spline}\n\\label{local_b_spline}\n\nThis method creates the smooth, parametric representation of the surface over the entire modelling area by sequentially creating the n-th number of the control lattices with the increasing resolution, in each cell of which the spline surface is calculated, from which a value is assigned to each node in the grid. This is mine sentence and I want it to stay, anyway. So, let's merge it! An undeniable advantage of the method is the qualitative approximation of highly scattered, irregularly spaced set of the points in a fully automatic mode that does not require any complicated settings from the user and a significant increase in the productivity due to the use of B-Splines in the hierarchy of the control grids.\\\\\n\nThe method algorithm has the following steps:`\n\\begin{itemize}\n\t\\item \n\tThe $ \\Omega = {(x,y) | 0 \\le x < m, 0 \\le y < n} $ area is created, covering all $ P $ input data which equals $ {(x_{c},y_{c},z_{c})} $ in 3D space, where $ (x_{c},y_{c}) $ is the point of input data in 2D space of $ \\Omega $ area; \n\t\\item \n\tThe control lattices $ \\Phi_{n} $ are created sequentially with increasing resolution (i.e. the number of lattice cells is constantly increasing), covering the $ \\Omega $ area. \n\t\n\tThe lattice creation algorithm:\n\t\\begin{itemize}\n\t\t\\item \n\t\tThe $ \\Phi_{n} (m=1, n=1) $ lattice is created (see \\pic{phi_lattice}) containing $ (m+3)\\times(n+3) $ nodes (at the beginning $ (1+3)\\times(1+3) = 16 nodes $, then at each iteration the value of m and n is multiplied by 1.25 and the resulting number of nodes is rounded). Each lattice cell will be described in one spline surface, by calculating the bicubic spline function using the formula:\n\t\t\n\t\t$$\n\t\tf(x,y) = \\sum_{k=0}^3 \\sum_{l=0}^3 B_{k}(s)B_{l}(t)\\phi_{(i+k)(j+l)},\n\t\t$$\n\t\t\n\t\twhere $ i = \\lfloor x \\rfloor -1 $, $ \\lfloor y \\rfloor -1 $, $ s= x - \\lfloor x \\rfloor $, $ t = y - \\lfloor y \\rfloor $, $ B_{k} $ and $ B_{l} $ - uniform cubic B-spline basis functions defined as:\n\t\t\n\t\t$$\n\t\tB_{0}(t) = (1-t)^3 \\div 6,\n\t\t$$\n\t\t\n\t\t$$\n\t\tB_{1}(t) = (3t^3 - 6t^2 + 4) \\div 6,\n\t\t$$\n\t\t\n\t\t$$\n\t\tB_{2}(t) = (-3t^3 +3t^2 +3t + 1) \\div 6,\n\t\t$$\n\t\t\n\t\t$$\n\t\tB_{3}(t) = t^3 \\div 6.\n\t\t$$\n\t\t\n\t\t\\begin{figure}[h]\n\t\t\t\\center{\n\t\t\t\t\\includegraphics[width=0.6\\textwidth]{pics_geology/phi_lattice.png}\n\t\t\t}\n\t\t\t\\caption{Schematic representation of the conditional lattice $ \\Phi $ covering the $ \\Omega $ area with the set of the scattered points as well as the proximity data set $ \\phi $ with the control point $ \\phi_{ij} $. In this case, the resulting B-spline function value will be assigned to the node where the control point is located $ \\phi_{ij} $.}\n\t\t\t\\label{ris:phi_lattice}\n\t\t\\end{figure} \n\t\t\n\t\t\\clearpage\n\t\t\n\t\t\\item\n\t\tEach lattice cell of $ \\Phi_{n} $ will be presented as a spline patch, on the basis of which a value is assigned to each lattice node, thus enabling a parametric surface model to be created. For the approximation of the scattered source point data $ P $, a bicubic B-Spline function is defined inside the cell of the $ \\Phi $ lattice. Each cell of the $ \\Phi_{n} $ lattice will be presented as a spline surface, from which a value is assigned to each node in the grid;\n\t\t\n\t\tThe spline surface creation algorithm:\n\t\t\\begin{itemize}\n\t\t\t\\item \n\t\t\tThe proximity data set $ \\phi $ is created, consisting of 16 nodes, which allows you to limit the number of points affecting the value of a node to only those points that are inside this area (see \\pic{phi_lattice});\n\t\t\t\\item\n\t\t\tThe impact of each source point $ (x_{c},y_{c}) $ which has been placed in the proximity data set $ \\phi $ is sequentially calculated using the formula:\n\t\t\t\n\t\t\t$$\n\t\t\t\\phi_{ij} = \\frac{\\sum_{c} \\omega_{c}^2 \\phi_{c}}{\\sum_{c}, \\omega_{c}^2}\n\t\t\t$$\n\t\t\t\n\t\t\twhere $ \\omega_{c} = \\omega_{kl} = B_{k}(s)B_{l}(t) $, $ k = (i+1) - \\lfloor x_{c} \\rfloor $, $ l = (j+1) - \\lfloor y_{c} \\rfloor $, $ s = x_{c} - \\lfloor x_{c} \\rfloor $, $ t = y_{c} - \\lfloor y_{c} \\rfloor $, and $ \\phi_{c} $ in turn is calculated using the formula:\n\t\t\t\n\t\t\t$$\n\t\t\t\\phi_{c} = \\frac{\\omega_{c} z_{c}}{\\sum_{a=0}^3 \\sum_{b=0}^3 \\omega_{ab}^2}\n\t\t\t$$\n\t\t\t\n\t\t\t\\item\n\t\t\tThe value that affects the resulting function of the bicubic spline $ f $ is assigned to the coefficient $ \\phi_{ij} $.\n\t\t\\end{itemize}\n\t\t\\begin{info}\n\t\t\tCells with no source points inside it may appear while the lattice resolution increases. In this case, such cells will be represented by spline surfaces with zero values.\n\t\t\\end{info}\n\t\t\\item \n\t\tAfter estimating the obtained values in each $ \\Phi_{n} $ node, we subtract them from the source data (i.e. we calculate the discrepancies between the source data and the surface model $ \\Phi_{n} $);\n\t\t\\item\t\n\t\tThe algorithm continues to create the new lattices multiplying the number of cells by 1.25 at each subsequent iteration. The iterative process is repeated until the lattice $ \\Phi_{n} $ becomes as detailed as requested by the user at the \\textbf{Grid Properties};\n\t\\end{itemize}\n\t\\item \n\tThe algorithm then splits all lattices of the previous iterations $ \\Phi_{n-1} $ to achieve the resolution of the last $ \\Phi_{n} $ lattice, sums up the obtained values of all lattices, which results in the desired surface $ \\Phi_{ij} $.\n\\end{itemize}\n\nThe parametric surface model built on each iteration represents the difference between the source data and the sum of the surfaces obtained on all previous iterations (see \\pic{local_b_spline}).\n\n\\begin{figure}[h]\n\t\\center{\n\t\t\\includegraphics[width=0.5\\textwidth]{pics_geology/local_b_spline.png}\n\t}\n\t\\caption{Interpolation of points set using the Local B-spline method on the grid in 10 m steps, both X and Y.}\n\t\\label{ris:local_b_spline}\n\\end{figure}\n\n\\clearpage", "meta": {"hexsha": "2ebaa7c95e5ff5103ec1c7c9daf7bb8cefd6db00", "size": 5737, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "local_b_spline_eng.tex", "max_stars_repo_name": "aretmii/aretmii_test", "max_stars_repo_head_hexsha": "50dfaecf69fad82c262eab2e1402661654c77f87", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "local_b_spline_eng.tex", "max_issues_repo_name": "aretmii/aretmii_test", "max_issues_repo_head_hexsha": "50dfaecf69fad82c262eab2e1402661654c77f87", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "local_b_spline_eng.tex", "max_forks_repo_name": "aretmii/aretmii_test", "max_forks_repo_head_hexsha": "50dfaecf69fad82c262eab2e1402661654c77f87", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-10-30T22:53:28.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-30T22:53:28.000Z", "avg_line_length": 59.7604166667, "max_line_length": 718, "alphanum_fraction": 0.692347917, "num_tokens": 1736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916099737807, "lm_q2_score": 0.7371581626286833, "lm_q1q2_score": 0.6364561728372929}} {"text": "\n\\section{Numerical methods}\nRecall that we compute the buckling eigenmodes via computing the zeros\nof the determinant of linear system corresponding to a Nystr\\\"{o}m \ndiscretization of the integral operator $A(k)$ in~\\cref{eq:inteqfinal}.\nAs in section~\\cref{sec:spec}, let $\\tilde{A}_{k}$ denote the discretized\nlinear system.\nThus, the three main ingredients for the fast and accurate computation\nof buckling eigenvalues (and its corresponding eigenfunctions) are\nhigh order Nystr\\\"{o}m discretizations of the integral operator \n$A(k)$; for a given parameter $k$, the rapid evaluation of the determinant \n$\\tilde{A}_{k}$; and finding the zeros of $\\text{det}(\\tilde{A}_{k})$ (which\nis non-linear in $k$) while minimizing the function evaluations.\n\nFor discretizing the integral operator using the\nNystr\\\"{o}m method, we divide the boundary $\\Gamma$ into\npanels and represent the unknown density and the\nboundary data by their values at scaled Gauss-Legendre\nnodes on each panel. \nLet $n_{p}$ denote the number of Gauss-Legendre panels.\nWe discretize each panel using 16 scaled Gauss-Legendre nodes.\nThen $n_{d} = 16 n_{p}$ is the number of discretization points \non the boundary.\nLet $\\bx_{j}$ denote the discretization nodes, $w_{j}$ denote the\nappropriately scaled Gauss-Legendre quadrature weights for smooth functions,\nand $\\bmu_{j}$ denote the unknown density at $\\bx_{j}$.\nWhen forming the linear system, we use scaled unknowns, \n$\\bmu_{j} \\sqrt{w_{j}}$, so that the spectral properties of the \ndiscrete system with respect to the $l_2$ norm\nare approximations of the spectral properties of the continuous\nsystem as on operator on $L_2$ (for more on this point of view, \nsee \\cite{bremer2012}).\nThe integral kernels $K_{ij}$ in~\\cref{eq:inteqfinal} \nare either smooth or have a weak (logarithmic) singularity. \nFor the smooth kernels, we use standard \nGauss-Legendre quadrature rule.\nFor kernels with a logarithmic singularity, we use order $20$ \ngeneralized Gaussian quadrature rule~\\cite{bremer2010,bremer2010u}.\nNote that the resulting discretized linear system $\\tilde{A}_{k}$ \nis a dense matrix of size $M=2n_{d}+N+1$, \nwhere $N$ is the number of connected components.\n\nFor small problems ($M \\leq 10^3$), we directly compute\n$\\text{det}(\\tilde{A}_{k})$ using standard linear algebra routines\nfrom LAPACK. \nHowever, since the matrix $\\tilde{A}_{k}$ is dense,\nthe computational complexity of computing $\\text{det}(\\tilde{A}_{k})$ \nusing LAPACK routines is $O(M^3)$.\nMoreover, for evaluating the buckling eigenmodes, $\\text{det}(\\tilde{A}_{k})$\nmust be evaluated for many different values of $k$.\nThus, for larger problems, we use the HIFIE method for evaluating\n$\\text{det}(\\tilde{A}_{k})$. \nThis method exploits the rank structure of $\\tilde{A}_{k}$ (large blocks\nof $\\tilde{A}_{k}$ corresponding to interactions between well-separated\nclusters are low-rank) and evaluates the determinant in O($M \\cdot \\log{M})$ \nCPU-time.\n\n\n", "meta": {"hexsha": "86e6a2502aea4e37769fc9f9033f7f973efc2514", "size": 2940, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/draft-01/05numerics.tex", "max_stars_repo_name": "askhamwhat/biharm-evals", "max_stars_repo_head_hexsha": "d836302f544670b3d899bd91ea4cb49e9afb6a75", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/draft-01/05numerics.tex", "max_issues_repo_name": "askhamwhat/biharm-evals", "max_issues_repo_head_hexsha": "d836302f544670b3d899bd91ea4cb49e9afb6a75", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/draft-01/05numerics.tex", "max_forks_repo_name": "askhamwhat/biharm-evals", "max_forks_repo_head_hexsha": "d836302f544670b3d899bd91ea4cb49e9afb6a75", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.8305084746, "max_line_length": 77, "alphanum_fraction": 0.7595238095, "num_tokens": 808, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916099737806, "lm_q2_score": 0.7371581626286833, "lm_q1q2_score": 0.6364561728372928}} {"text": "\\section{Supplementary Information}\n\\subsection{Absorption Harmonic Oscillator Model}\nWe will find useful here the generic form of the Franck Condon Overlap factor for a displaced harmonic oscillator with the same curvature in the ground and excited states:\n\\begin{align}\n\tO_{m}^{n} &= \\left(-1\\right)^{n} \\sqrt{\\frac{e^{-S}S^{m+n}}{m!n!}} \\sum_{j=0}^{\\min \\left( m,n\\right)} \\frac{m!n!(-1)^j S^{-j}}{j!(m-j)!(n-j)!}\n\t\\label{eqn:overlap}\n\\end{align}\n\\subsubsection{Linear Correction}\nTo calculate the electric field ratio for a linear non-Condon correction for a system starting in the 0th state:\n\\begin{align*}\n\t\\mu_{0,\\lambda} =& \\mu_0 \\bra{0} \\left( 1 + c_1\\left(\\hat{a} + \\hat{a}^{\\dagger}\\right) \\right) \\ket{\\lambda} \\\\\n\t=& \\mu_0\\left( \\delta_{0,\\lambda} + c_1 \\delta_{1, \\lambda}\\right)\n\\end{align*}\n\n\\begin{align}\n\tH_{(1)}(0,k) &=\\frac{\\sum_{\\lambda,\\nu} \\left[ \\delta_{0,\\lambda} + c^*\\delta_{\\lambda, 1} \\right]\\left[ \\delta_{0,\\nu} + c \\delta_{\\nu, 1} \\right] O_{\\lambda}^{k+1} O_{\\nu}^{k+1} }{\\sum_{l,n} \\left[ \\delta_{0,l} + c^*\\delta_{l, 1} \\right]\\left[ \\delta_{0,n} + c \\delta_{n, 1} \\right] O_{l}^{k} O_{n}^{k} }\\\\\n\t&= \\frac{ \\left[ O_{0}^{k+1} + c^*O_{1}^{k+1} \\right]\\left[ O_{0}^{k+1} + c O_{1}^{k+1} \\right] }{\\left[ O_{0}^{k} + c^*O_{1}^{k} \\right]\\left[ O_{0}^{k} + c O_{1}^{k}\\right] } \\\\\n\t&= \\frac{\\left| O_{0}^{k+1} + cO_{1}^{k+1} \\right|^2 }{\\left|O_{0}^{k} + cO_{1}^{k} \\right|^2} \\\\\n\t&= \\left(\\frac{O_{0}^{k+1}}{O_{0}^{k}} \\right)^2 \\frac{ \\left|1 + c\\frac{O_{1}^{k+1}}{O_{0}^{k+1}} \\right|^2}{\\left|1 + c\\frac{O_{1}^{k} }{O_{0}^{k}}\\right|^2}\n\\end{align}\n\nFor this, several quantities can be computed\n\n\\begin{align}\n\tO_{0}^{k+1} &= \\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{k + 1}}{(k+1)!}} \\\\\n\tO_{1}^{k+1} &= \\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{1+k+1}}{(k+1)!}} \\sum_{j=0}^{\\min \\left( 1,k+1 \\right)} \\frac{(k+1)!(-1)^j S^{-j}}{j!(1-j)!(k+1-j)!}\\\\\n\t&= \\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{1+k+1}}{(k+1)!}} \\left[ \\frac{(k+1)! }{(k+1)!} - \\frac{(k+1)! S^{-1}}{k!}\\right]\\\\\n\t&= \\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{1+k+1}}{(k+1)!}} \\left[ 1 - \\frac{(k+1) }{S}\\right]\n\\end{align}\nthen the desired ratio is\n\\begin{align}\n\t\\frac{O_{1}^{k+1}}{ O_{0}^{k+1}} &= \\frac{\\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{1+k+1}}{(k+1)!}} \\left[ 1 - \\frac{(k+1) }{S}\\right]}{\\left(-1\\right)^{k+1} \\sqrt{\\frac{e^{-S}S^{k + 1}}{(k+1)!}}}\\\\\n\t&= \\frac{ \\sqrt{\\frac{e^{-S}S^{1+k+1}}{(k+1)!}} }{ \\sqrt{\\frac{e^{-S}S^{k + 1}}{(k+1)!}}} \\left[ 1 - \\frac{(k+1) }{S}\\right]\\\\\n\t&= \\sqrt{S} \\left[ 1 - \\frac{(k+1) }{S}\\right]\n\\end{align}\nas for the other ratio:\n\\begin{align}\n\tO_{0}^{k} &= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{k }}{k!}} \\\\\n\tO_{1}^{k} &= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{1+k}}{k!}} \\sum_{j=0}^{\\min \\left( 1,k \\right)} \\frac{k!(-1)^j S^{-j}}{j!(1-j)!(k-j)!}\n\\end{align}\nfirst we start with the case where $k\\geq1$\n\\begin{align}\n\tO_{1}^{k} &= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{1+k}}{k!}} \\left[ \\frac{k! }{k!} - \\frac{k! S^{-1}}{(k-1)!} \\right] \\\\\n\t&= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{1+k}}{k!}} \\left[ 1 - \\frac{k}{S} \\right]\n\\end{align}\nif $k=0$ then the summation above will be unity and the same equation will thus work for both. Now when we bring it back into the ratio:\n\\begin{align}\n\t\\frac{O_{1}^{k}}{O_{0}^{k}} &= \\frac{\\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{1+k}}{k!}} \\left[ 1 - \\frac{k}{S} \\right]}{\\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{k }}{k!}}}\\\\\n\t&= \\sqrt{S} \\left[ 1 - \\frac{k }{S}\\right]\n\\end{align}\nWhich after substitution leads to\n\\begin{align}\n\tH_{(1)}(0,k)&= \\frac{S }{k+1 } \\frac{\\left| 1 + c\\sqrt{S} \\left( 1 - \\frac{k+1}{S} \\right) \\right|^2}{\\left| 1 + c \\sqrt{S} \\left( 1 - \\frac{k}{S} \\right)\\right|^2 }\n\\end{align}\n\n\n\\subsubsection{Quadratic Correction}\nbut for a quadratic correction:\n\\begin{align*}\n\t\\mu^{(2)}_{0,\\lambda} =& \\mu_0 \\bra{0} \\left( 1 + c_2\\left(\\hat{a} + \\hat{a}^{\\dagger}\\right)^2\\right) \\ket{\\lambda}\\\\\n\t=& \\mu_0 \\bra{0} \\left( 1 + c_2\\left(\\hat{N} + 1 + \\hat{a}\\hat{a} + \\hat{a}^{\\dagger}\\hat{a}^{\\dagger}\\right)\\right) \\ket{\\lambda}\\\\\n\t=& \\mu_0 \\left( \\left(1 + c_2\\right) \\delta_{0,\\lambda} + c_2\\sqrt{2}\\delta_{2,\\lambda} \\right) \\\\\n\t=& \\frac{\\mu_0}{\\left(1 + c_2\\right)} \\left( \\delta_{0,\\lambda} + \\frac{c_2\\sqrt{2}}{\\left(1 + c_2\\right)} \\delta_{2,\\lambda} \\right) \\\\\n\t=& \\mu_0' \\left( \\delta_{0,\\lambda} + c_2' \\delta_{2,\\lambda} \\right)\n\\end{align*}\nwhich gives us for the peak ratios instead\n\\begin{align}\n\tH_{(2)}(0,k) &= \\frac{\\left| O_{0}^{k+1} + c_2' O_{2}^{k+1} \\right|^2 }{\\left|O_{0}^{k} + c_2' O_{2}^{k} \\right|^2} \\\\\n\t&= \\left(\\frac{O_{0}^{k+1}}{O_{0}^{k}} \\right)^2 \\frac{ \\left|1 + c_2'\\frac{O_{2}^{k+1}}{O_{0}^{k+1}} \\right|^2}{\\left|1 + c_2'\\frac{O_{2}^{k} }{O_{0}^{k}}\\right|^2}\n\\end{align}\n\nInstead of getting algebraic expressions for $O_{n\\geq2}^{k}$ as we did for $O_{1}^{k}$, we instead implement Equation \\ref{eqn:overlap} directly in our solver.\n\n\\subsubsection{3 Correction}\nThe cubic correction looks like so:\n\\begin{align*}\n\t\\mu^{(3)}_{0,\\lambda} =& \\mu_0 \\bra{0} \\left( 1 + c_3\\left(\\hat{a} + \\hat{a}^{\\dagger}\\right)^3\\right) \\ket{\\lambda}\\\\\n\t=& \\mu_0 \\bra{0} \\left( 1 + c_3 \\left(\\hat{a} + \\hat{a}^{\\dagger}\\right) \\left(\\hat{N} + 1 + \\hat{a}\\hat{a} + \\hat{a}^{\\dagger}\\hat{a}^{\\dagger}\\right)\\right)\\ket{\\lambda}\\\\\n\t=& \\mu_0 \\left( \\bra{0} + c_3 \\bra{1} \\left(\\hat{N} + 1 + \\hat{a}\\hat{a} + \\hat{a}^{\\dagger}\\hat{a}^{\\dagger}\\right)\\right)\\ket{\\lambda}\\\\\n\t=& \\mu_0 \\left( \\bra{0} + c_3 \\left(2\\bra{1} +\\sqrt{6}\\bra{3} \\right)\\right)\\ket{\\lambda}\\\\\n\t=& \\mu_0 \\left( \\delta_{0,\\lambda} + c_3 \\left(2\\delta_{1,\\lambda} +\\sqrt{6}\\delta_{3,\\lambda} \\right)\\right)\n\\end{align*}\nwhich indicates it makes more sense to treat (1,3) and not just (3):\n\\begin{align*}\n\t\\mu^{(1,3)}_{0,\\lambda} =& \\mu_0 \\left( \\delta_{0,\\lambda} + \\delta_{1,\\lambda} \\left(2c_3 + c_1\\right) +c_3 \\sqrt{6}\\delta_{3,\\lambda} \\right)\\\\\n\t=& \\mu_0 \\left( \\delta_{0,\\lambda} + c_1' \\delta_{1,\\lambda} +c_3 \\sqrt{6}\\delta_{3,\\lambda} \\right)\\\\\n\tc_1' =& \\left(2c_3 + c_1\\right)\\\\\n\tc_3' =& c_3 \\sqrt{6} \\\\\n\tc_1 =& \\left(c_1' - 2\\frac{c_3'}{\\sqrt{6}} \\right)\n\\end{align*}\n\\subsubsection{4 Correction}\nLooking at a quadratic correction:\n\\begin{align*}\n\t\\mu^{(4)}_{0,\\lambda} =& \\mu_0 \\bra{0} \\left( 1 + c_4\\left(\\hat{a} + \\hat{a}^{\\dagger}\\right)^4\\right) \\ket{\\lambda}\\\\\n\t=& \\mu_0 \\bra{0} \\left( 1 + c_4 \\left(\\hat{N} + 1 + \\hat{a}\\hat{a} + \\hat{a}^{\\dagger}\\hat{a}^{\\dagger}\\right)^2 \\right) \\ket{\\lambda} \\\\\n\t=& \\mu_0 \\left( \\delta_{0, \\lambda} + c_4 \\bra{0} \\left(1 + \\hat{a}\\hat{a}\\right)\\left(\\hat{N} + 1 + \\hat{a}\\hat{a} + \\hat{a}^{\\dagger}\\hat{a}^{\\dagger}\\right) \\ket{\\lambda}\\right) \\\\\n\t=& \\mu_0 \\left( \\delta_{0, \\lambda} + c_4\\left[ 3 \\delta_{0, \\lambda} + 4\\sqrt{2}\\delta_{2, \\lambda} + 2\\sqrt{6}\\delta_{4, \\lambda} \\right]\\right) \\\\\n\t=& \\mu_0 \\left( \\delta_{0, \\lambda} \\left( 1 + 3c_4 \\right) + 4\\sqrt{2}c_4\\delta_{2, \\lambda} + 2\\sqrt{6}c_4\\delta_{4, \\lambda}\\right)\\\\\n\t=& \\mu_0' \\left( \\delta_{0, \\lambda} + \\frac{4\\sqrt{2}c_4}{1 + 3c_4 }\\delta_{2, \\lambda} + \\frac{2\\sqrt{6}c_4}{1 + 3c_4 }\\delta_{4, \\lambda}\\right)\n\\end{align*}\nmuch like for (3) we had to consider (1,3) to be more accurate, it makes no sense to consider (4) on its own; we'll need to introduce (2) as well to get (2,4):\n\\begin{align*}\n\t\\mu^{(2,4)}_{0,\\lambda} =& \\mu_0 \\left( \\delta_{0, \\lambda} \\left( 1 + c_2 + 3c_4 \\right) + \\left( \\sqrt{2} c_2 + 4\\sqrt{2}c_4 \\right)\\delta_{2, \\lambda} + 2\\sqrt{6}c_4\\delta_{4, \\lambda}\\right)\\\\\n\t=& \\mu_0' \\left( \\delta_{0, \\lambda} + \\frac{\\sqrt{2} c_2 + 4\\sqrt{2}c_4 }{1 + c_2 + 3c_4 }\\delta_{2, \\lambda} + \\frac{2\\sqrt{6}c_4}{1 + c_2 + 3c_4 }\\delta_{4, \\lambda}\\right) \\\\\n\tc_2' &= \\frac{\\sqrt{2} c_2 + 4\\sqrt{2}c_4 }{1 + c_2 + 3c_4 } \\\\\n\tc_4' &= \\frac{2\\sqrt{6}c_4}{1 + c_2 + 3c_4 }\n\\end{align*}\nwhich when inverted turns into:\n\\begin{align*}\n\tc_2 &= -\\frac{2 \\left(\\sqrt{6} c_2'-2 \\sqrt{2} c_4'\\right)}{2 \\sqrt{6} c_2'-\\sqrt{2} c_4'-4 \\sqrt{3}}\\\\\n\tc_4 &= \\frac{c_4' \\left(3 \\sqrt{2} c_4'-4 \\sqrt{3}\\right)}{\\left(2 \\sqrt{6}-3 c_4'\\right) \\left(2 \\sqrt{6} c_2'-\\sqrt{2} c_4'-4 \\sqrt{3}\\right)}\n\\end{align*}\n\n\n\\subsubsection{Generic Correction}\nIt becomes clear that we should be able to define an arbitrary-order correction (at least for $\\eta=0$) rather simply by adding together generic $x^n$ operator\n\n\\begin{align*}\n\t\\mu_{0,\\lambda} = \\sum_n c_n \\mu^{(n)}_{0,\\lambda} =& \\sum_n c_n \\bra{0} \\left(\\hat{a} + \\hat{a}^{\\dagger}\\right)^n \\ket{\\lambda}\n\\end{align*}\nthis would then feed in to our expression for $H$:\n\\begin{align}\n\tH_{(\\{i\\})}(0,k)&= \\frac{S}{k+1} \\frac{ \\left|1 + \\sum_{l = \\{i\\}} c'_l \\frac{O_{l}^{k+1}}{O_{0}^{k+1}} \\right|^2}{\\left|1 + \\sum_{l = \\{i\\}} c'_l \\frac{O_{l}^{k} }{O_{0}^{k}}\\right|^2}\n\\end{align}\nwhere the transformation from $c'$ back into $c$ would have to be calculated for any individual $H_{(\\{i\\})}$ but is non-trivial. We saw above that the squared operator brought amplitude back to zero and in general, all even operators will bring back amplitude to all even operators in order below and the same with odd operators.\n\n% GRAVEYARD\n% \\begin{align}\n% \tO_{2}^{k} &= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{2+k}}{2!k!}} \\sum_{j=0}^{\\min \\left( 2,k\\right)} \\frac{2!k!(-1)^j S^{-j}}{j!(2-j)!(k-j)!}\n% \\end{align}\n% if $k>2$:\n% \\begin{align}\n% \tO_{2}^{k} &= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{2+k}}{2!k!}} \\sum_{j=0}^{2} \\frac{2!k!(-1)^j S^{-j}}{j!(2-j)!(k-j)!} \\\\\n% \t&= \\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{2+k}}{2!k!}} \\left(1 - \\frac{2 k}{S} + \\frac{k(k-1)}{S^2 } \\right)\n% \\end{align}\n% then one can see like we did above that this holds for $k=0,1$ and we can continue on to the ratios:\n% \\begin{align}\n% \t\\frac{O_{2}^{k}}{O_{0}^{k}} &= \\frac{\\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{2+k}}{2!k!}} \\left(1 - \\frac{2 k}{S} + \\frac{k(k-1)}{S^2 } \\right)}{\\left(-1\\right)^{k} \\sqrt{\\frac{e^{-S}S^{k }}{k!}}}\\\\\n% \t&= \\sqrt{\\frac{S^{2}}{2}} \\left(1 - \\frac{2 k}{S} + \\frac{k(k-1)}{S^2 } \\right)\\\\\n% \t\\frac{O_{2}^{k+1}}{O_{0}^{k+1}} &= \\sqrt{\\frac{S^{2}}{2}} \\left(1 - \\frac{2 (k+1)}{S} + \\frac{k(k+1)}{S^2 } \\right)\n% \\end{align}\n", "meta": {"hexsha": "69d9eb14c3b03549a9587c1084050d04ff3a1b21", "size": 10097, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "DetectingNonCondonPaper/supplement.tex", "max_stars_repo_name": "jgoodknight/dissertation", "max_stars_repo_head_hexsha": "012ad400e1246d2a7e63cc640be4f7b4bf56db00", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-21T06:20:42.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-21T06:20:42.000Z", "max_issues_repo_path": "DetectingNonCondonPaper/supplement.tex", "max_issues_repo_name": "jgoodknight/dissertation", "max_issues_repo_head_hexsha": "012ad400e1246d2a7e63cc640be4f7b4bf56db00", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "DetectingNonCondonPaper/supplement.tex", "max_forks_repo_name": "jgoodknight/dissertation", "max_forks_repo_head_hexsha": "012ad400e1246d2a7e63cc640be4f7b4bf56db00", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.1056338028, "max_line_length": 332, "alphanum_fraction": 0.5677924136, "num_tokens": 4738, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916029436189, "lm_q2_score": 0.7371581510799253, "lm_q1q2_score": 0.6364561576838511}} {"text": "\\chapter{Complex Problem Set}\n\\begin{abox}\n\tPractise Set-1\n\\end{abox}\n\n\n\\begin{enumerate}[label=\\color{ocre}\\textbf{\\arabic*.}]\n\t\\item The value of the integral $\\int_{C} d z z^{2} e^{z}$, where $C$ is an open contour in the complex $z$-plane as shown in the figure below, is:\n\t{\\exyear{NET/JRF(JUNE-2011)}}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=5cm,width=9cm]{diagram-20211005-crop}\n\t\\end{figure}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $\\frac{5}{e}+e$\n\t\t\\task[\\textbf{B.}] $e-\\frac{5}{e}$\n\t\t\\task[\\textbf{C.}] $\\frac{5}{e}-e$\n\t\t\\task[\\textbf{D.}] $-\\frac{5}{e}-e$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext { If we complete the contour, then by Cauchy integral theorem }\n\t\t\\int_{-1}^{1} d z z^{2} e^{z}+\\int_{C} d z z^{2} e^{z}&=0 \\Rightarrow \\int_{C} d z z^{2} e^{z}=-\\int_{-1}^{1} d z z^{2} e^{z}\\\\&=-\\left[z^{2} e^{z}-2 z e^{2}+2 e^{2}\\right]_{-1}^{1}=\\frac{5}{e}-e\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item Which of the following is an analytic function of the complex variable $z=x+i y$ in the domain $|z|<2 ?$\n\t{\\exyear{NET/JRF(JUNE-2011)}}\n\t\\begin{tasks}(2)\n\t\t\\task[\\textbf{A.}] $(3+x-i y)^{7}$\n\t\t\\task[\\textbf{B.}] $(1+x+i y)^{4}(7-x-i y)^{3}$\n\t\t\\task[\\textbf{C.}] $(1-x-i y)^{4}(7-x+i y)^{3}$\n\t\t\\task[\\textbf{D.}] $(x+i y-1)^{1 / 2}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\tPut $z=x+i y .$ If $\\bar{z}=x-i y$ appears in any of the expressions then that expression is non-analytic. For option (D) we have a branch point singularity as the power is $\\frac{1}{2}$ which is fractional. Hence only option (B) is analytic.\\\\\\\\\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item The first few terms in the Laurent series for $\\frac{1}{(z-1)(z-2)}$ in the region $1 \\leq|z| \\leq 2$ and around $z=1$ is\n\t{\\exyear{NET/JRF(JUNE-2012)}}\n\t\\begin{tasks}(1)\n\t\t\\task[\\textbf{A.}] $\\frac{1}{2}\\left[1+z+z^{2}+\\ldots\\right]\\left[1+\\frac{z}{2}+\\frac{z^{2}}{4}+\\frac{z^{3}}{8}+\\ldots .\\right]$\n\t\t\\task[\\textbf{B.}] $\\frac{1}{1-z}-z-(1-z)^{2}+(1-z)^{3}+\\ldots .$\n\t\t\\task[\\textbf{C.}] $\\frac{1}{\\mathrm{z}^{2}}\\left[1+\\frac{1}{\\mathrm{z}}+\\frac{1}{\\mathrm{z}^{2}}+\\ldots .\\right]\\left[1+\\frac{2}{\\mathrm{z}}+\\frac{4}{\\mathrm{z}^{2}}+\\ldots . .\\right]$\n\t\t\\task[\\textbf{D.}] $2(z-1)+5(z-1)^{2}+7(z-1)^{3}+\\ldots$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\frac{1}{(z-1)(z-2)}&=\\frac{1}{z-2}-\\frac{1}{z-1}=\\frac{1}{1-z}+\\frac{1}{(z-1)-1}\\\\&=\\frac{1}{1-z}-(1+(1-z))^{-1}\\\\\n\t\t&=\\frac{1}{1-z}-\\left[1+(1-z)+\\frac{(-1)(-2)}{2 !}(1-z)^{2}+\\frac{(-1)(-2)(-3)}{3 !}(1-z)^{3} \\ldots\\right]\\\\\n\t\t&=\\frac{1}{1-z}-\\left[z+(1-z)^{2}-(1-z)^{3}+\\ldots . .\\right]\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item Let $u(x, y)=x+\\frac{1}{2}\\left(x^{2}-y^{2}\\right)$ be the real part of analytic function $f(z)$ of the complex variable $z=x+i y$. The imaginary part of $f(z)$ is\n\t{\\exyear{NET/JRF(JUNE-2012)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $y+x y$\n\t\t\\task[\\textbf{B.}] $x y$\n\t\t\\task[\\textbf{C.}] $y$\n\t\t\\task[\\textbf{D.}] $y^{2}-x^{2}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tu(x, y)&=x+\\frac{1}{2}\\left(x^{2}-y^{2}\\right), v(x, y)=?\\\\\n\t\t\\text{Check }\\frac{\\partial u}{\\partial x}&=\\frac{\\partial v}{\\partial y}\\text{ and } \\frac{\\partial u}{\\partial y}=-\\frac{\\partial v}{\\partial x}\\\\\n\t\t\\Rightarrow \\frac{\\partial u}{\\partial x}&=\\frac{\\partial v}{\\partial y}, \\quad \\frac{\\partial v}{\\partial y}=1+x, \\\\ v&=y+x y+f(x)\\\\\n\t\t\\frac{\\partial u}{\\partial y}&=-\\frac{\\partial v}{\\partial x} \\Rightarrow \\frac{\\partial v}{\\partial x}=+y, \\\\ v&=y x+f(y)\\\\\n\t\ty+x y+f(x)&=y x+f(y)\\\\\n\t\t\\text{If }f(x)&=0\\quad \\quad\n\t\tf(y)=y\\\\\n\t\tv&=x y+y\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item The value of the integral $\\int_{C} \\frac{z^{3} d z}{\\left(z^{2}-5 z+6\\right)}$, where $C$ is a closed contour defined by the equation $2|z|-5=0$, traversed in the anti-clockwise direction, is\n\t{\\exyear{NET/JRF(DEC-2012)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $-16 \\pi i$\n\t\t\\task[\\textbf{B.}] $16 \\pi \\mathrm{i}$\n\t\t\\task[\\textbf{C.}] $8 \\pi i$\n\t\t\\task[\\textbf{D.}] $2 \\pi i$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tz^{2}-5 z+6&=0 \\Rightarrow z^{2}-2 z-3 z+6\\\\&=0 \\Rightarrow z(z-2)-3(z-2)=0 \\Rightarrow z=3,2\\\\\n\t\t2|z|&=5 \\Rightarrow|z|=2.5,\\text{ only 2 will be inside.}\\\\\n\t\t\\text{Residue }&=\\left.(z-2) \\frac{z^{3}}{(z-3)(z-2)}\\right|_{z=2}=\\frac{8}{2-3}\\\\&=-8 \\Rightarrow \\int \\frac{z^{3} d z}{z^{2}-5 z+6}=2 \\pi i(-8)=-16 \\pi i\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item With $z=x+i y$, which of the following functions $f(x, y)$ is NOT a (complex) analytic function of $z$ ?\n\t{\\exyear{NET/JRF(JUNE-2013)}}\n\t\\begin{tasks}(1)\n\t\t\\task[\\textbf{A.}] $f(x, y)=(x+i y-8)^{3}\\left(4+x^{2}-y^{2}+2 i x y\\right)^{7}$\n\t\t\\task[\\textbf{B.}] $f(x, y)=(x+i y)^{7}(1-x-i y)^{3}$\n\t\t\\task[\\textbf{C.}] $f(x, y)=\\left(x^{2}-y^{2}+2 i x y-3\\right)^{5}$\n\t\t\\task[\\textbf{D.}] $f(x, y)=(1-x+i y)^{4}(2+x+i y)^{6}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(x, y)&=(1-x+i y)^{4}(2+x+i y)^{6}\\\\&=\\{1-(x-i y)\\}^{4}(2+x+i y)^{6}\\\\\n\t\t\\text{Due to present of }\\bar{z}&=(x-i y)\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (D)}\n\t\\end{answer}\n\t\\item Which of the following functions cannot be the real part of a complex analytic function of $z=x+i y ?$\n\t{\\exyear{NET/JRF(DEC-2013)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $x^{2} y$\n\t\t\\task[\\textbf{B.}] $x^{2}-y^{2}$\n\t\t\\task[\\textbf{C.}] $x^{3}-3 x y^{2}$\n\t\t\\task[\\textbf{D.}] $3 x^{2} y-y-y^{3}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{ Let $x^{2} y$ be real part of a complex function. Use Milne Thomson's method to write analytic complex function. The real part of that function should be (1) but that is not the case. So this cannot be real part of an analytic function. Also,}\n\t\tz^{2}&=(x+i y)^{2}=x^{2}-y^{2}+2 i x y,\\text{ Real part option (2)}\\\\\n\t\tz^{3}&=(x+i y)^{3}=x^{3}-i y^{3}+3 i x y(x+i y)\\\\\n\t\t&=x^{3}-i y^{3}+3 i x^{2} y-3 x y^{2},\\text{ Real part option (3)}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item Given that the integral $\\int_{0}^{\\infty} \\frac{d x}{y^{2}+x^{2}}=\\frac{\\pi}{2 y}$, the value of $\\int_{0}^{\\infty} \\frac{d x}{\\left(y^{2}+x^{2}\\right)^{2}}$ is\n\t{\\exyear{NET/JRF(DEC-2013)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $\\frac{\\pi}{y^{3}}$\n\t\t\\task[\\textbf{B.}] $\\frac{\\pi}{4 y^{3}}$\n\t\t\\task[\\textbf{C.}] $\\frac{\\pi}{8 y^{3}}$\n\t\t\\task[\\textbf{D.}] $\\frac{\\pi}{2 y^{3}}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\int_{0}^{\\infty} \\frac{d x}{\\left(y^{2}+x^{2}\\right)^{2}}&=\\frac{1}{2} \\int_{-\\infty}^{\\infty} \\frac{d x}{\\left(y^{2}+x^{2}\\right)^{2}},\\text{ pole is of }2^{\\text {nd }}\\text{ order at }x=i y,\\text{ residue }=1 /\\left(4 i y^{3}\\right)\\\\\n\t\t\\text{Integral }&=\\left(\\frac{1}{2}\\right)(2 \\pi i) \\frac{1}{4 i y^{3}}=\\frac{\\pi}{\\left(4 y^{3}\\right)}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item If $C$ is the contour defined by $|z|=\\frac{1}{2}$, the value of the integral\n\t$$\n\t\\oint_{C} \\frac{d z}{\\sin ^{2} z}\n\t$$\n\tis\n\t{\\exyear{NET/JRF(JUNE-2014)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $\\infty$\n\t\t\\task[\\textbf{B.}] $2 \\pi i$\n\t\t\\task[\\textbf{C.}] 0\n\t\t\\task[\\textbf{D.}] $\\pi i$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\frac{1}{\\sin ^{2} z} \\quad\\left(|z|=\\frac{1}{2}\\right)\\\\\n\t\t\\sin z&=z-\\frac{z^{3}}{\\lfloor 3}+\\frac{z^{5}}{\\lfloor 5} \\ldots . \\Rightarrow \\frac{1}{\\sin ^{2} z}=\\frac{1}{\\left(z-\\frac{z^{3}}{\\frac{3}{3}}+\\frac{z^{5}}{5} \\cdots\\right)^{2}}\\\\\n\t\t\\Rightarrow \\frac{1}{\\sin ^{2} z}&=\\frac{1}{z^{2}}\\left[1-\\frac{z^{2}}{\\lfloor 3}+\\frac{z^{4}}{\\lfloor 5} \\ldots .\\right]^{-2} \\Rightarrow \\oint_{C} \\frac{d z}{\\sin ^{2} z}=0\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item The principal value of the integral $\\int_{-\\infty}^{\\infty} \\frac{\\sin (2 x)}{x^{3}} d x$ is\n\t{\\exyear{NET/JRF(DEC-2014)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $-2 \\pi$\n\t\t\\task[\\textbf{B.}] $-\\pi$\n\t\t\\task[\\textbf{C.}] $\\pi$\n\t\t\\task[\\textbf{D.}] $2 \\pi$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{Let }f(z)&=\\frac{e^{i 2 z}}{z^{3}}\\\\\n\t\t\\lim _{2 \\rightarrow 0}(z-0)^{3} f(z)&=\\lim _{z \\rightarrow 0}(z-0)^{3} \\frac{e^{i 2 z}}{z^{3}}\\\\&=1(\\text{ finite and }\\neq 0) \\Rightarrow z=0 \\text{is pole of order 3} .\\\\\n\t\t\\text{Residue }R&=\\frac{1}{2 !} \\lim _{z \\rightarrow 0} \\frac{d^{2}}{d z^{2}}\\left[(z-0)^{3} \\frac{e^{i 2 z}}{z^{3}}\\right]=-2\\\\\n\t\t\\Rightarrow \\int_{-\\infty}^{\\infty} f(x) d x&=\\pi i \\Sigma R=\\pi i(-2)=-2 \\pi i \\Rightarrow \\operatorname{Im} .\\text{ Part }\\\\&=-2 \\pi \\Rightarrow \\int_{-\\infty}^{\\infty} f(x) d x=-2 \\pi\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item The Laurent series expansion of the function $f(z)=e^{2}+e^{1 / 2}$ about $z=0$ is given by\n\t{\\exyear{NET/JRF(DEC-2014)}}\n\t\\begin{tasks}(2)\n\t\t\\task[\\textbf{A.}] $\\sum_{n=-\\infty}^{\\infty} \\frac{z^{n}}{n !}$ for all $|z|<\\infty$\n\t\t\\task[\\textbf{B.}] $\\sum_{n=0}^{\\infty}\\left(z^{n}+\\frac{1}{z^{n}}\\right) \\frac{1}{n !}$ only if $0<|z|<1$\n\t\t\\task[\\textbf{C.}] $\\sum_{n=0}^{\\infty}\\left(z^{n}+\\frac{1}{z^{n}}\\right) \\frac{1}{n !}$ for all $0<|z|<\\infty$\n\t\t\\task[\\textbf{D.}] $\\sum_{n=-\\infty}^{\\infty} \\frac{z^{n}}{n !}$ only if $|z|<1$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\te^{z}&=\\left(1+z+\\frac{z^{2}}{2 !}+\\ldots\\right)=\\sum_{n=0}^{\\infty} \\frac{z^{n}}{n !}\\text{ and }e^{1 / z}\\\\&=1+\\frac{1}{z}+\\frac{1}{2 !} \\frac{1}{z^{2}}+\\ldots .=\\sum_{n=0}^{\\infty} \\frac{1}{z^{n} n !}\\\\\n\t\t\\Rightarrow f(z)&=\\left(e^{z}+e^{1 / 2}\\right)=\\sum_{n=0}^{\\infty}\\left(z^{n}+\\frac{1}{z^{n}}\\right) \\frac{1}{n !},\\text{ for all }0<|z|<\\infty\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item Consider the function $f(z)=\\frac{1}{z} \\ln (1-z)$ of a complex variable $z=r e^{i \\theta}(r \\geq 0, \\quad-\\infty<\\theta<\\infty)$. The singularities of $f(z)$ are as follows:\n\t{\\exyear{NET/JRF(DEC-2014)}}\n\t\\begin{tasks}(1)\n\t\t\\task[\\textbf{A.}] Branch points at $z=1$ and $z=\\infty$; and a pole at $z=0$ only for $0 \\leq \\theta<2 \\pi$\n\t\t\\task[\\textbf{B.}] Branch points at $z=1$ and $z=\\infty$; and a pole at $z=0$ for all $\\theta$ other than $0 \\leq \\theta<2 \\pi$\n\t\t\\task[\\textbf{C.}] Branch points at $z=1$ and $z=\\infty$; and a pole at $z=0$ for all $\\theta$\n\t\t\\task[\\textbf{D.}] Branch points at $z=0, z=1$ and $z=\\infty$.\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{For }f(z)&=\\frac{1}{z} \\ln (1-z)=\\frac{1}{z}\\left(-z-\\frac{z^{2}}{2}-\\frac{z^{3}}{3}-\\ldots . .\\right)\\\\&=-1-\\frac{z}{2}-\\frac{z^{2}}{3}-\\ldots .\n\t\t\\intertext{There is no principal part and when $z \\rightarrow 0, f(z)=-1 .$ So there is removable singularity at $z=0$. Also $z=1$ and $z=\\infty$ is Branch point.}\n\t\t\\end{align*}\n\t\tNone of the above is correct\n\t\\end{answer}\n\t\\item The value of integral $\\int_{-\\infty}^{\\infty} \\frac{d x}{1+x^{4}}$\n\t{\\exyear{NET/JRF(JUNE-2015)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $\\frac{\\pi}{\\sqrt{2}}$\n\t\t\\task[\\textbf{B.}] $\\frac{\\pi}{2}$\n\t\t\\task[\\textbf{C.}] $\\sqrt{2} \\pi$\n\t\t\\task[\\textbf{D.}] $2 \\pi$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\int_{-\\infty}^{\\infty} \\frac{d z}{1+z^{4}} \\quad \\because|z|=R\\\\\n\t\t\\text{Now, pole }z&=e^{(2 n+1) \\frac{\\pi}{4}}\\\\\n\t\tn&=0, \\quad \\Rightarrow z_{0}=e^{\\frac{i \\pi}{4}}=\\frac{1}{\\sqrt{2}}+i \\frac{1}{\\sqrt{2}}, n\\\\&=2 \\Rightarrow z_{2}=\\frac{-1}{\\sqrt{2}}-i \\frac{1}{\\sqrt{2}}\\\\\n\t\tn&=1 \\Rightarrow z_{1}=e^{\\frac{i 3 \\pi}{4}}=\\frac{-1}{\\sqrt{2}}+i \\frac{1}{\\sqrt{2}}, n\\\\&=3 \\Rightarrow z_{3}=+\\frac{1}{\\sqrt{2}}-i \\frac{1}{\\sqrt{2}}\n\t\t\\intertext{only $z_{0}$ and $z_{1}$ lies in contour}\n\t\t\\text{i.e., residue at }\\left(z=e^{\\frac{i \\pi}{4}}\\right)&=\\frac{1}{4}\\left(-\\frac{1}{\\sqrt{2}}-i \\frac{1}{\\sqrt{2}}\\right)\\\\\n\t\t\\text{residue at }\\left(z=e^{\\frac{i 3 \\pi}{4}}\\right)&=\\frac{1}{4}\\left(\\frac{1}{\\sqrt{2}}-i \\frac{1}{\\sqrt{2}}\\right)\\\\\n\t\t\\text{now }\\int_{-\\infty}^{\\infty} \\frac{d x}{x^{4}+1}&=2 \\pi i \\Sigma \\operatorname{Re} S=\\frac{\\pi}{\\sqrt{2}}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item The function $\\frac{Z}{\\sin \\pi z^{2}}$ of a complex variable $z$ has\n\t{\\exyear{NET/JRF(DEC-2015)}}\n\t\\begin{tasks}(1)\n\t\t\\task[\\textbf{A.}] A simple pole at 0 and poles of order 2 at $\\pm \\sqrt{n}$ for $n=1,2,3 \\ldots$\n\t\t\\task[\\textbf{B.}] A simple pole at 0 and poles of order 2 at $\\pm \\sqrt{n}$ and $\\pm i \\sqrt{n}$ for $n=1,2,3 \\ldots$\n\t\t\\task[\\textbf{C.}] Poles of order 2 at $\\pm \\sqrt{n}, n=0,1,2,3 \\ldots$\n\t\t\\task[\\textbf{D.}] Poles of order 2 at $\\pm n, n=0,1,2,3 \\ldots$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\frac{z}{\\sin \\pi z^{2}}=\\frac{z}{\\pi z^{2} \\frac{\\sin \\pi z^{2}}{\\pi z^{2}}}\\\\\n\t\t\\text{\tat }z&=0,\\text{ it is a simple pole since,} \\lim _{z \\rightarrow 0} \\frac{\\sin \\pi z^{2}}{\\pi z^{2}}=1\\\\\n\t\t\\text{Also, }\\sin \\pi z^{2}&=\\sin n \\pi \\Rightarrow \\pi \\mathrm{z}^{2}\\\\&=\\pm n \\pi, z=\\pm \\sqrt{n}, \\pm i \\sqrt{n}\\\\\n\t\t\\lim _{z \\rightarrow \\sqrt{n}}&(z-\\sqrt{n})^{2} \\cdot \\frac{z}{\\sin \\pi z^{2}}, \\text{exists. So its pole of order 2}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item The value of the contour integral $\\frac{1}{2 \\pi i} \\oint_{C} \\frac{e^{4 z}-1}{\\cosh (z)-2 \\sinh (z)} d z$ around the unit circle $C$ traversed in the anti-clockwise direction, is\n\t{\\exyear{NET/JRF(JUNE-2016)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 0\n\t\t\\task[\\textbf{B.}] 2\n\t\t\\task[\\textbf{C.}] $\\frac{-8}{\\sqrt{3}}$\n\t\t\\task[\\textbf{D.}] $-\\tanh \\left(\\frac{1}{2}\\right)$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\frac{e^{4 z}-1}{\\cosh z-2 \\sinh z}=\\frac{e^{4 z}-1}{\\frac{e^{2}+e^{-z}}{2}-\\left(e^{z}-e^{-z}\\right)}\\\\&=\\frac{e^{42}-1}{-\\frac{e^{z}}{2}+\\frac{3}{2} e^{-z}}\\\\\n\t\t\\Rightarrow f(z)&=\\frac{2 e^{2}\\left(e^{4 z}-1\\right)}{\\left(3-e^{2 z}\\right)}=\\frac{2\\left(e^{5 z}-e^{z}\\right)}{\\left(3-e^{2 z}\\right)}\\\\\n\t\t\\text{For pole at }z&=z_{0}, 3-e^{2 \\xi_{0}}=0 \\Rightarrow e^{2 z_{0}}\\\\&=3 \\Rightarrow z_{0}=\\frac{\\ln 3}{2}\n\t\t\\intertext{It has simple pole at $z_{0}$}\n\t\t\\operatorname{Re}\\left(z_{0}\\right)&=\\lim _{z \\rightarrow z_{0}}\\left(z-z_{0}\\right) f(z)=\\lim _{2 \\rightarrow z_{0}}\\left(z-z_{0}\\right) \\frac{2\\left(e^{5 z}-e^{2}\\right)}{3-e^{22}}\\\\\n\t\t&=\\lim _{z \\rightarrow z_{0}} \\frac{\\left(z-z_{0}\\right) \\times 2\\left(5 e^{5 z}-e^{z}\\right)+2\\left(e^{5 z}-e^{z}\\right) \\times 1}{-2 e^{2 z}}\\\\&=-\\left(\\frac{e^{5 z_{0}}-e^{z_{0}}}{e^{2 z_{0}}}\\right)\\\\\n\t\t&=-\\left(\\frac{(\\sqrt{3})^{5}-\\sqrt{3}}{3}\\right)=-\\left(\\frac{9 \\sqrt{3}-\\sqrt{3}}{3}\\right)=-\\frac{8}{\\sqrt{3}}\\\\\n\t\t\\frac{1}{2 \\pi i} \\oint f(z) d z&=\\frac{1}{2 \\pi i} \\times 2 \\pi i \\sum\\text{ Residue } =-\\frac{8}{\\sqrt{3}}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item Let $u(x, y)=e^{a x} \\cos (b y)$ be the real part of a function $f(z)=u(x, y)+i v(x, y)$ of the complex variable $z=x+i y$, where $a, b$ are real constants and $a \\neq 0 .$ The function $f(z)$ is complex analytic everywhere in the complex plane if and only if\n\t{\\exyear{NET/JRF(JUNE-2017)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $b=0$\n\t\t\\task[\\textbf{B.}] $b=\\pm a$\n\t\t\\task[\\textbf{C.}] $b=\\pm 2 \\pi a$\n\t\t\\task[\\textbf{D.}] $b=a \\pm 2 \\pi$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{The function $f(z)$ will be analytic everywhere in the complex plane if and only if it satisfies the Cauchy Riemann equation in that region.}\n\t\t\\Rightarrow \\frac{\\partial u}{\\partial x}&=\\frac{\\partial v}{\\partial y}\\text{ and } \\frac{\\partial u}{\\partial y}=-\\frac{\\partial v}{\\partial x}\\\\\n\t\t\\text{Hence }a e^{a x} \\cos (b y)&=\\frac{\\partial v}{\\partial y}\\hspace{2cm}\\text{(i)}\\\\\n\t\t\\text{and }b e^{a x} \\sin (b y)&=\\frac{\\partial v}{\\partial x}\\hspace{2cm}\\text{(ii)}\n\t\t\\intertext{From equation (i)}\n\t\tv(x, y)&=\\frac{a e^{a x} \\sin (b y)}{b}+c(y)\\hspace{2cm}\\text{(iii)}\n\t\t\\intertext{Differentiating partially with $x$ gives}\n\t\t\\frac{\\partial v}{\\partial x}&=\\frac{a^{2} e^{a x} \\sin (b y)}{b}\\hspace{2cm}\\text{(iv)}\n\t\t\\intertext{From equation (iii) and (iv)}\n\t\tb e^{a x} \\sin (b y)&=\\frac{a^{2} e^{a x} \\sin (b y)}{b}\\\\\n\t\t\\Rightarrow b^{2}&=a^{2} \\Rightarrow b=\\pm a\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item The integral $\\oint_{\\Gamma} \\frac{z e^{i \\pi z / 2}}{z^{2}-1} d z$ along the closed contour $\\Gamma$ shown in the figure is\n\t{\\exyear{NET/JRF(JUNE-2017)}}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=5cm]{diagram-20211005(19)-crop}\n\t\\end{figure}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 0\n\t\t\\task[\\textbf{B.}] $2 \\pi$\n\t\t\\task[\\textbf{C.}] $-2 \\pi$\n\t\t\\task[\\textbf{D.}] $4 \\pi i$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\frac{z e^{i z \\pi / 2}}{(z+1)(z-1)}\\\\\n\t\t\\text{For }z&=+1\\text{ anti-clockwise}\\\\\n\t\tI&=2 \\pi i \\lim _{z \\rightarrow 1} \\frac{z e^{i \\pi z / 2}}{(z+1)}=\\frac{2 \\pi i}{2} e^{i \\pi / 2}=\\pi i e^{i \\pi / 2}\\\\\n\t\t\\text{For }z&=-1\\\\\n\t\tI&=-2 \\pi i \\lim _{z \\rightarrow-1} \\frac{z e^{i \\pi z / 2}}{(z-1)}=-2 \\pi i \\times \\frac{(-1) e^{-i \\pi / 2}}{(-2)}=-\\pi i e^{-i \\pi / 2}\\\\\n\t\t\\text{Integral }&=\\pi i \\frac{\\left(e^{i \\pi / 2}-e^{-i \\pi / 2}\\right)}{2 i} \\times 2 i=2 \\pi i^{2} \\sin \\frac{\\pi}{2}=-2 \\pi\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item What is the value of $a$ for which $f(x, y)=2 x+3\\left(x^{2}-y^{2}\\right)+2 i(3 x y+a y)$ is an analytic function of complex variable $z=x+i y$\n\t{\\exyear{NET/JRF(JUNE-2018)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 1\n\t\t\\task[\\textbf{B.}] 0\n\t\t\\task[\\textbf{C.}] 3\n\t\t\\task[\\textbf{D.}] 2\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(x, y)&=2 x+3\\left(x^{2}-y^{2}\\right)+2 i(3 x y+\\alpha y)\\\\\n\t\tu&=2 x+3\\left(x^{2}-y^{2}\\right), v=2(3 x y+\\alpha y)\\\\\n\t\t\\text{C-R conditions: }u_{x}&=v_{y}, u_{y}=-v_{x}\\\\\n\t\t2+3(2 x)&=2(3 x+\\alpha) \\Rightarrow \\alpha=1 \\Rightarrow-6 y=-6 y\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item The value of the integral $\\oint_{C} \\frac{d z}{z} \\frac{\\tanh 2 z}{\\sin \\pi z}$, where $C$ is a circle of radius $\\frac{\\pi}{2}$, traversed counter-clockwise, with centre at $z=0$, is\n\t{\\exyear{NET/JRF(DEC-2018)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 4\n\t\t\\task[\\textbf{B.}] $4 i$\n\t\t\\task[\\textbf{C.}] $2 i$\n\t\t\\task[\\textbf{D.}] 0\n\t\\end{tasks}\n\t\\begin{answer}$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4.5cm,width=5.5cm]{diagram-20211005(2)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t&\\oint_{C} \\frac{d z}{z} \\frac{\\tanh 2 z}{\\sin \\pi z} d z\\\\\n\t\tz&=0,1,-1, \\frac{\\pi i}{4}, \\frac{-\\pi i}{4}\\\\\n\t\tf(z)&=\\frac{2 z-\\frac{1}{3}(2 z)^{3}+\\frac{2}{15}(2 z)^{5} \\ldots .}{z\\left(\\pi z-\\frac{\\pi^{3} z^{3}}{3 !}+\\ldots\\right)}\\\\\n\t\t\\frac{2}{\\pi z}&\\left(1-\\frac{1}{2} z^{2}+\\ldots\\right)\\left(1-\\frac{\\pi^{2} z^{2}}{2 !}+\\ldots\\right)\\\\\n\t\tb_{1}&=\\frac{2}{\\pi}\\\\\n\t\t\\text{As Re } z&=1, \\frac{\\tanh ^{2}}{-\\pi}\\text{ and }\\operatorname{Re} z=-1, \\frac{\\tanh ^{2}}{-\\pi}\\\\\n\t\t\\operatorname{Re} z&=\\frac{i \\pi}{4}=-\\frac{1}{\\pi}\\left(2 \\operatorname{cosec} h \\frac{\\pi^{2}}{4}\\right)\\\\\n\t\t\\operatorname{Re} z&=\\frac{-i \\pi}{4}=-\\frac{1}{\\pi}\\left(2 \\operatorname{cosec} \\mathrm{h} \\frac{\\pi^{2}}{4}\\right)\n\t\t\\intertext{$I=2 \\pi i \\Sigma R=4 i$ only when 0 lies inside, otherwise wrong question.}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item The integral $I=\\int_{C} e^{z} d z$ is evaluated from the point $(-1,0)$ to $(1,0)$ along the contour $C$, which is an arc of the parabola $y=x^{2}-1$, as shown in the figure. The value of $I$ is\n\t{\\exyear{NET/JRF(DEC-2018)}}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4.5cm,width=5cm]{diagram-20211005(3)-crop}\n\t\\end{figure}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 0\n\t\t\\task[\\textbf{B.}] $2 \\sinh 1$\n\t\t\\task[\\textbf{C.}] $e^{2 i} \\sinh 1$\n\t\t\\task[\\textbf{D.}] $e+e^{-1}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\int_{C} f(z) d z&=2 \\pi i \\Sigma R\\\\\n\t\t\\int_{C} f(z) d z+\\int_{1}^{-1} e^{x} d x&=0\\\\\n\t\t\\int_{C} f(z) d z&=-\\int_{1}^{-1} e^{x} d x=\\int_{1}^{-1} e^{x} d x\\\\&=\\frac{\\left(e^{1}-e^{-1}\\right)}{2} \\cdot 2=2 \\sinh 1\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\t\\end{answer}\n\t\\item The contour $C$ of the following integral\n\t$$\n\t\\oint_{C} d z \\frac{\\sqrt{(z-1)(z-3)}}{\\left(z^{2}-25\\right)^{3}}\n\t$$\n\tin the complex $z$ plane is shown in the figure below.\\\\\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3.5cm,width=6cm]{diagram-20211005(8)-crop}\n\t\\end{figure}\n\tThis integral is equivalent to an integral along the contours\n\t{\\exyear{NET/JRF(DEC-2018)}}\n\t\\begin{tasks}(2)\n\t\t\\task[\\textbf{A.}] \\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=6.5cm]{diagram-20211005(4)-crop}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{B.}] \\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=6.5cm]{diagram-20211005(5)-crop}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{C.}] \\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=6.5cm]{diagram-20211005(6)-crop}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{D.}] \\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=6.5cm]{diagram-20211005(7)-crop}\n\t\t\\end{figure}\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{$z=1,3$ are branch points $\\infty$ is not a branch point 1 branch cut 3}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item Let $C$ be the circle of radius $\\frac{\\pi}{4}$ centered at $z=\\frac{1}{4}$ in the complex $z$-plane that is traversed counter-clockwise. The value of the contour integral $\\oint_{C} \\frac{z^{2}}{\\sin ^{2} 4 z} d z$ is\n\t{\\exyear{NET/JRF(DEC-2019)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 0\n\t\t\\task[\\textbf{B.}] $\\frac{i \\pi^{2}}{4}$\n\t\t\\task[\\textbf{C.}] $\\frac{i \\pi^{2}}{16}$\n\t\t\\task[\\textbf{D.}] $\\frac{i \\pi}{4}$\n\t\\end{tasks}\n\t\\begin{answer}$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=3cm]{diagram-20211026(16)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tf(z)&=\\left(\\frac{\\pi}{\\sin 4 z}\\right)^{2}\\\\\n\t\tz_{0}&=0, \\frac{\\pi}{4}\\text{ are poles}\\\\\n\t\t4 z&=n \\pi, z=0, \\frac{\\pi}{4}\n\t\t\\intertext{Others are outside the contour.}\n\t\t\\text{Residue at }z&=0\\text{ is }\\left[\\frac{\\pi}{4 z-\\frac{4^{3} z^{3}}{3 !}+\\ldots}\\right]^{2}\\\\\n\t\t&=\\left[\\frac{1}{4-\\frac{4^{3} z^{2}}{3 !}+\\ldots .}\\right]^{2}\\qquad \\text{ No terms for } \\frac{1}{z}, b_{1}=0\\\\\n\t\t&=\\left[4-\\frac{4^{3} z^{2}}{3 !}+\\ldots .\\right]^{-2}\\\\\n\t\t\\text{Residue for }z&=\\frac{\\pi}{4}\\\\\n\t\tz-\\frac{\\pi}{4}&=t\n\t\t\\intertext{$\\sin (4 t+\\pi)=-\\sin 4 t \\quad$ (But square so no effect)}\n\t\t&\\left[\\frac{t+\\frac{\\pi}{4}}{\\sin 4\\left(t+\\frac{\\pi}{4}\\right)}\\right]^{2}\\\\\n\t\t\\left(\\frac{t+\\frac{\\pi}{4}}{\\sin 4 t}\\right)^{2}&=\\frac{t^{2}+\\frac{\\pi^{2}}{4}+2 t \\cdot \\frac{\\pi}{4}}{\\sin ^{2} 4 t}\\\\\n\t\t\\frac{\\pi}{2} \\frac{t}{16 t^{2}[1-\\ldots .]^{2}}&=\\frac{\\pi}{32 t}[1-\\ldots .]^{-2} \\text{(from first term)}\\\\\n\t\tb_{1}&=\\frac{\\pi}{32}\\\\\n\t\t\\oint_{C} \\frac{z^{2}}{\\sin ^{2} 4 z} d z&=2 \\pi i\\left[0+\\frac{\\pi}{32}\\right]=\\frac{i \\pi^{2}}{16}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item A function of a complex variable $z$ is defined by the integral $f(z)=\\oint_{\\Gamma} \\frac{w^{2}-2}{w-z} d w$, where $\\Gamma$ is a circular contour of radius 3 , centred at origin, running counter-clockwise in the $w$ - plane. The value of the function at $z=(2-i)$ is\n\t{\\exyear{NET/JRF(JUNE-2020)}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] 0\n\t\t\\task[\\textbf{B.}] $1-4 i$\n\t\t\\task[\\textbf{C.}] $8 \\pi+2 \\pi \\mathrm{i}$\n\t\t\\task[\\textbf{D.}] $-\\frac{2}{\\pi}-\\frac{i}{2 \\pi}$\n\t\\end{tasks}\n\t\\begin{answer}$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4cm,width=4.6cm]{diagram-20211027-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tf(z)&=\\oint_{\\Gamma} \\frac{w^{2}-2}{w-z} d w\\\\\n\t\t\\omega&=z\\text{ is a simple pole.}\\\\\n\t\t\\text{Residue }\\lim _{\\omega \\rightarrow z}(\\omega-z) \\frac{\\left(\\omega^{2}-2\\right)}{(\\omega-z)}&=(2-i)^{2}-2 \\\\&=4-1-4 i-2=(1-4 i)\\\\\n\t\tf(z)&=\\oint_{\\Gamma} \\frac{w^{2}-2}{w-z} d w=2 \\pi i(1-4 i)=2 \\pi i+8 \\pi\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\\end{enumerate}\n\\colorlet{ocre1}{ocre!70!}\n\\colorlet{ocrel}{ocre!30!}\n\\setlength\\arrayrulewidth{1pt}\n\\begin{table}[H]\n\t\\centering\n\t\\arrayrulecolor{ocre}\n\t\\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}\n\t\t\\hline\n\t\t\\multicolumn{4}{|c|}{\\textbf{Answer key}}\\\\\\hline\\hline\n\t\t\\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\\\\hline\n\t\t1&\\textbf{C} &2&\\textbf{B}\\\\\\hline \n\t\t3&\\textbf{B} &4&\\textbf{A} \\\\\\hline\n\t\t5&\\textbf{A} &6&\\textbf{D} \\\\\\hline\n\t\t7&\\textbf{A}&8&\\textbf{-}\\\\\\hline\n\t\t9&\\textbf{C}&10&\\textbf{A}\\\\\\hline\n\t\t11&\\textbf{C} &12&\\textbf{-}\\\\\\hline\n\t\t13&\\textbf{A}&14&\\textbf{B}\\\\\\hline\n\t\t15&\\textbf{C}&16&\\textbf{B}\\\\\\hline\n\t\t17&\\textbf{C} &18&\\textbf{A}\\\\\\hline\n\t\t19&\\textbf{B}&20&\\textbf{B}\\\\\\hline\n\t\t21&\\textbf{C}&22&\\textbf{C}\\\\\\hline\n\t\t23&\\textbf{C}& &\\\\\\hline\n\t\t\n\t\\end{tabular}\n\\end{table}\n\\newpage\n\\begin{abox}\n\tPractise Set-2\n\\end{abox}\n\\begin{enumerate}[label=\\color{ocre}\\textbf{\\arabic*.}]\n\t\\item The value of the integral $\\oint_{C} \\frac{e^{z} \\sin (z)}{z^{2}} d z$, where the contour $C$ is the unit circle: $|z-2|=1$, is\n\t{\\exyear{GATE 2010}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $2 \\pi i$\n\t\t\\task[\\textbf{B.}] $4 \\pi i$\n\t\t\\task[\\textbf{C.}] $\\pi i$\n\t\t\\task[\\textbf{D.}] 0\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{$|z-2|=1 \\Rightarrow 10$\\\\\n\t$\\mathrm{R}$ : $\\ln z$ and 1 are linearly independent solutions for $m=0$\\\\\n\tS: $z^{m}$ and $\\ln z$ are linearly independent solutions for all values of $m$\\\\\n\tThe correct option for the combination of valid statements is\n\t{\\exyear{GATE 2015}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] P, R and S only\n\t\t\\task[\\textbf{B.}] P and R only\n\t\t\\task[\\textbf{C.}] $\\mathrm{Q}$ and $\\mathrm{R}$ only\n\t\t\\task[\\textbf{D.}] $\\mathrm{R}$ and $\\mathrm{S}$ only\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\ty^{\\prime \\prime}+\\frac{1}{z} y^{\\prime}-\\frac{m^{2}}{z^{2}} y&=0 \\Rightarrow z^{2} y^{\\prime \\prime}+z y^{\\prime}-m^{2} y\\\\&=0, m=0,1,2,3, \\ldots, \\quad z=e^{x}, D=\\frac{d}{d x}\\\\\n\t\t\\text{\tIf }m&=0 ; \\quad z^{2} y^{\\prime \\prime}+z y^{\\prime}=0,[D(D-1)+D] y\\\\&=0 \\Rightarrow\\left[D^{2}-D+D\\right] y=0\\\\\n\t\tD^{2} y&=0 \\Rightarrow y=c_{1}+c_{2} x \\Rightarrow y\\\\&=c_{1}+c_{2} \\ln z \\quad \\text{( $R$ is correct)}\\\\\n\t\t\\text{And if }m &\\neq 0, m>0,\\text{ then }m \\neq 0,\\text{ then }\\left(D^{2}-m^{2}\\right) y\\\\&=0 \\Rightarrow D=\\pm m\\\\\n\t\ty&=c_{1} e^{m x}+c_{2} e^{-m x}=c_{1} e^{m \\log z}+c_{2} e^{-m \\log z}\\\\&=c_{1} z^{m}+c_{2} z^{-m}\\\\\n\t\t\\text{or if }m &\\neq 0, m>0,\\text{ then}\\\\\n\t\ty&=c_{1} \\cosh (m \\log (z))+i c_{2} \\sinh (m \\log (x)), \\quad m>0\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)} \n\t\\end{answer}\n\t\\item Which of the following is an analytic function of $z$ everywhere in the complex plane?\n\t{\\exyear{GATE 2016}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $z^{2}$\n\t\t\\task[\\textbf{B.}] $\\left(z^{*}\\right)^{2}$\n\t\t\\task[\\textbf{C.}] $|z|^{2}$\n\t\t\\task[\\textbf{D.}] $\\sqrt{z}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tz^{2}&=(x+i y)^{2}=x^{2}-y^{2}+i(2 x y) \\Rightarrow u\\\\&=x^{2}-y^{2}\\text{ and }v=2 x y\\\\\n\t\t\\text{Cauchy Riemann equations }\\frac{\\partial u}{\\partial x}&=\\frac{\\partial v}{\\partial y}=2 x, \\quad \\frac{\\partial v}{\\partial x}=-\\frac{\\partial u}{\\partial y}=2 y\\text{ satisfies.}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\t\\item The contour integral $\\oint \\frac{d z}{1+z^{2}}$ evaluated along a contour going from $-\\infty$ to $+\\infty$ along the\n\treal axis and closed in the lower half-plane circle is equal to............... (up to two decimal places).\n\t{\\exyear{GATE 2017}}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\oint_{c} \\frac{1}{1+z^{2}} d z&=\\int_{-\\infty}^{+\\infty} \\frac{1}{1+x^{2}} d x+\\oint_{c} \\frac{1}{1+z^{2}} d z\\\\\n\t\t\\text{Poles, }1+z^{2}&=0 \\Rightarrow z=\\pm i, \\quad z=-i\\text{ is inside }C\\\\\n\t\t\\therefore \\operatorname{Res}(z=-i)&=\\lim _{z \\rightarrow-i}(z+i) \\frac{1}{(z-i)(z+i)}\\\\&=\\frac{1}{-i-i}=\\frac{1}{-2 i}\\\\\n\t\t\\int_{-\\infty}^{+\\infty} \\frac{1}{1+x^{2}} d x&=-\\frac{1}{2 i} \\times-2 \\pi i=\\pi\n\t\t\\intertext{(Since, here we use lower half plane i.e., we traversed in clockwise direction, hence we have to take $-2 \\pi i$ )}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item The imaginary part of an analytic complex function is $v(x, y)=2 x y+3 y .$ The real part of the function is zero at the origin. The value of the real part of the function at $1+i$ is ................. (up to two decimal places)\n\t{\\exyear{GATE 2017}}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{Solution: The imaginary part of the given analytic function is $v(x, y)=2 x y+3 y .$ From the\n\t\t\tCauchy - Riemann condition}\n\t\t\\frac{\\partial v}{\\partial y}&=\\frac{\\partial u}{\\partial x}=2 x+3\n\t\t\\intertext{Integrating partially gives}\n\t\tu(x, y)&=x^{2}+3 x+g(y)\\\\\n\t\t\\intertext{From the second Cauchy - Riemann condition}\n\t\t\\frac{\\partial u}{\\partial y}&=-\\frac{\\partial v}{\\partial x},\\text{ we obtain} \\frac{\\partial u}{\\partial y}\\\\&=-2 y, \\mu(x, y)=-y^{2}+g(x)\\\\\n\t\t\\frac{d g(y)}{d y}&=-2 y \\Rightarrow g(y)=-y^{2}+c\\\\\n\t\t\\text{\tHence, }u(x, y)&=x^{2}+3 x-y^{2}+c\n\t\t\\intertext{Since, the real part of the analytic function is zero at the origin.}\n\t\t\\text{Hence, }0&=0+0-0+c \\Rightarrow c=0\\\\\n\t\t\\text{Thus, }u(x, y)&=x^{2}+3 x-y^{2}\\\\\n\t\t\\therefore f(z)&=\\left(x^{2}+3 x-y^{2}\\right)+i(2 x y+3 y)\n\t\t\\intertext{Thus, the value of real part when}\n\t\tz&=1+i, i.e. x=1\\text{ and }y=1\\text{ is }u(x, y)\\\\&=(1)^{2}+3(1)-1=3\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item The absolute value of the integral\n\t$$\n\t\\int \\frac{5 z^{3}+3 z^{2}}{z^{2}-4} d z\n\t$$\n\tover the circle $|z-1.5|=1$ in complex plane, is $\\ldots$ (up to two decimal places).\n\t{\\exyear{GATE 2018}}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\frac{5 z^{3}+3 z^{2}}{(z-2)(z+2)}\\\\\n\t\t\\text{Pole, }z&=2,-2\\\\\n\t\tz&=-2 \\text{ is outside the center}\\\\\n\t\t|-2-1.5|>1&\\text{So, will not be considered}\\\\\n\t\t\\text{\tNow, }\\operatorname{Re} s(2)&=\\lim _{z \\rightarrow 2}(z-2) \\frac{\\left(5 z^{3}+3 z^{2}\\right)}{(z-2)(z+2)}\\\\&=\\frac{52^{3}+32^{2}}{4}=\\frac{40+12}{4}=13\\\\\n\t\tI&=2 \\pi i \\times residue =2 \\pi i \\times 13\\\\&=26 \\times 3.14 \\Rightarrow I=81.64\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item The pole of the function $f(z)=\\cot z$ at $z=0$ is\n\t{\\exyear{GATE 2019}}\n\t\\begin{tasks}(2)\n\t\t\\task[\\textbf{A.}] A removable pole\n\t\t\\task[\\textbf{B.}] An essential singularity\n\t\t\\task[\\textbf{C.}] A simple pole\n\t\t\\task[\\textbf{D.}] A second order pole\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tf(z)&=\\cot z\\text{ at }z=0\\\\\n\t\tf(z)&=\\frac{1}{\\tan z} \\quad z=0\\text{ is a simple pole }\\\\f(z)&=\\frac{1}{z}\\left[1-\\frac{1}{3} z^{2}+\\ldots .\\right]\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\t\\end{answer}\n\t\\item The value of the integral $\\int_{-\\infty}^{\\infty} \\frac{\\cos (k x)}{x^{2}+a^{2}} d x$, where $k>0$ and $a>0$, is\n\t{\\exyear{GATE 2019}}\n\t\\begin{tasks}(4)\n\t\t\\task[\\textbf{A.}] $\\frac{\\pi}{a} e^{-k a}$\n\t\t\\task[\\textbf{B.}] $\\frac{2 \\pi}{a} e^{-k a}$\n\t\t\\task[\\textbf{C.}] $\\frac{\\pi}{2 a} e^{-k a}$\n\t\t\\task[\\textbf{D.}] $\\frac{3 \\pi}{2 a} e^{-k a}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t&\\int_{-\\infty}^{\\infty} \\frac{\\cos k x}{x^{2}+a^{2}} d x\\\\\n\t\tf(z)&=\\frac{e^{i k x}}{z^{2}+a^{2}}=\\frac{e^{i k z}}{(z+i a)(z-i a)}\\\\\n\t\tI&=\\operatorname{Re} .2 \\pi i \\times \\frac{e^{i k(i a)}}{2 i a}=\\frac{\\pi e^{-k a}}{a}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\t\\end{answer}\n\\item The value of the integral $\\int_{0}^{\\infty} \\frac{\\ln x}{\\left(x^{2}+1\\right)^{2}} d x$ is\n{\\exyear{ JEST 2012}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]0\n\t\\task[\\textbf{b.}]$\\frac{-\\pi}{4}$\n\t\\task[\\textbf{c.}] $\\frac{-\\pi}{2}$\n\t\\task[\\textbf{d.}] $\\frac{\\pi}{2}$\n\\end{tasks}\n\\begin{answer}$\\left. \\right. $\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=4.6cm]{JEST 01 2012}\n\t\\end{figure}\n\t\\begin{align*}\n\t\\int_{0}^{\\infty} \\frac{\\ln x}{\\left(x^{2}+1\\right)^{2}} d x&=\\int_{0}^{\\infty} \\frac{\\ln z}{\\left(z^{2}+1\\right)^{2}} d z\\\\\n\t\\text{Let us }&\\text{consider new function }f(z)=\\left(\\frac{\\ln z}{z^{2}+1}\\right)^{2},\\text{ then }I=\\int_{0}^{\\infty}\\left(\\frac{\\ln z}{z^{2}+1}\\right)^{2} d z\\\\\n\\text{\tPole at }z&=\\pm i\\text{ is simple pole of second order.}\\\\\n\\text{\tResidue at }z&=i\\text{ is}\\\\\n&=\\frac{d}{d z}(z-i)^{2} \\frac{(\\ln z)^{2}}{(z-i)^{2}(z+i)^{2}}=\\frac{d}{d z} \\frac{(\\ln z)^{2}}{(z+i)^{2}}\\\\\n&=\\frac{(z+i)^{2} 2(\\ln z) \\cdot \\frac{1}{z}-(\\ln z)^{2} \\cdot 2(z+i)}{(z+i)^{4}}=\\frac{(z+i) 2 \\ln (z) \\frac{1}{z}-(\\ln z)^{2} \\cdot 2}{(z+i)^{3}}\\\\\n&=\\frac{(2 i) 2 \\times \\frac{1}{i} \\ln i-(\\ln i)^{2} \\cdot 2}{(2 i)^{3}}=\\frac{4 \\frac{i \\pi}{2}-\\left(\\frac{i \\pi}{2}\\right)^{2} \\times 2}{-8 i}=\\frac{2 \\pi i+\\frac{\\pi^{2}}{2}}{-8 i}\\\\\n\\left.\\Rightarrow \\operatorname{Res}\\right|_{z=i}&=\\frac{-\\pi}{4}+\\frac{\\pi^{2}}{16} i\\\\\n\\text{Similarly, at }z&=-i ;\\left.\\operatorname{Res}\\right|_{z=-i}=\\frac{-\\pi}{4}-\\frac{\\pi^{2}}{16} i\\\\\nI&=\\int_{0}^{\\infty}\\left(\\frac{\\ln z}{z^{2}+1}\\right)^{2} d z=2 \\pi i\\left(\\frac{-\\pi}{4}+\\frac{\\pi^{2}}{16} i-\\frac{\\pi}{4}-\\frac{\\pi^{2}}{16} i\\right)=-\\pi^{2} i\\\\\n-\\pi^{2} i&=\\left(\\int_{R} \\int_{A} \\int_{B}\\right) f(z) d z=\\left(\\iint_{A B}\\right) f(z) d z ;\\qquad \\left[\\because \\int_{A B}\\right.\\text{ vanish }]\\\\\n\\text{Along path }A ;& \\quad z=-x+i \\varepsilon\\text{ and along path }B ; \\quad z=-x-i \\varepsilon\\\\\n\\text{Thus }-\\pi^{2} i&=\\left(\\iint_{A B}\\right) f(z) d z=-\\int_{-\\infty}^{0}\\left[\\frac{\\ln (-x+i \\varepsilon)}{(-x+i \\varepsilon)^{2}+1}\\right] d x-\\int_{0}^{\\infty}\\left[\\frac{\\ln (-x-i \\varepsilon)}{(-x-i \\varepsilon)^{2}+1}\\right] d x\\\\\n\\Rightarrow-\\pi^{2} i&=\\int_{0}^{\\infty}\\left[\\frac{\\ln (-x+i \\varepsilon)}{(-x+i \\varepsilon)^{2}+1}\\right]^{2} d x-\\int_{0}^{\\infty}\\left[\\frac{\\ln (-x-i \\varepsilon)}{(-x-i \\varepsilon)^{2}+1}\\right]^{2} d x\\\\\n\\Rightarrow-\\pi^{2} i&=\\int_{0}^{\\infty}\\left[\\frac{\\ln (x)+i \\pi}{1+x^{2}}\\right]^{2} d x-\\int_{0}^{\\infty}\\left[\\frac{\\ln (x)-i \\pi}{1+x^{2}}\\right]^{2} d x ; \\quad \\varepsilon \\rightarrow 0\\\\\n\\Rightarrow-\\pi^{2} i&=\\int_{0}^{\\infty} \\frac{(\\ln (x)+i \\pi)^{2}-(\\ln (x)-i \\pi)^{2}}{\\left(1+x^{2}\\right)^{2}} d x\\\\&=4 \\pi i \\int_{0}^{\\infty} \\frac{\\ln x}{\\left(x^{2}+1\\right)^{2}} \\Rightarrow \\int_{0}^{\\infty} \\frac{\\ln x}{\\left(x^{2}+1\\right)^{2}}=\\frac{-i \\pi^{2}}{4 \\pi i}=\\frac{-\\pi}{4}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (B)}\n\\end{answer}\n\\item Compute $\\lim _{z \\rightarrow 0} \\frac{\\operatorname{Re}\\left(z^{2}\\right)+\\operatorname{Im}\\left(z^{2}\\right)}{z^{2}}$\n{\\exyear{ JEST 2013}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}] The limit does not exist\n\t\\task[\\textbf{b.}]1\n\t\\task[\\textbf{c.}]$-i$\n\t\\task[\\textbf{d.}] $-1$\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\t\\lim _{z \\rightarrow 0} \\frac{\\operatorname{Re}\\left(z^{2}\\right)+\\operatorname{Im}\\left(z^{2}\\right)}{z^{2}}&=\\lim _{z \\rightarrow 0} \\frac{x^{2}-y^{2}+2 x y}{x^{2}-y^{2}+2 i x y}=\\lim _{y=0 \\atop x \\rightarrow 0} \\frac{x^{2}-y^{2}+2 x y}{x^{2}-y^{2}+2 i x y}=1\\\\\n\t\\lim _{x=0 \\atop y \\rightarrow 0} \\frac{x^{2}-y^{2}+2 x y}{x^{2}-y^{2}+2 i x y}&=1\\text{ and }\\lim _{y=x \\atop x \\rightarrow 0} \\frac{x^{2}-y^{2}+2 x y}{x^{2}-y^{2}+2 i x y}=-i\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\\end{answer}\n\\item The value of limit\n$$\n\\lim _{z \\rightarrow i} \\frac{z^{10}+1}{z^{6}+1}\n$$\nis equal to\n{\\exyear{ JEST 2014}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]1\n\t\\task[\\textbf{b.}]0\n\t\\task[\\textbf{c.}] $\\frac{-10}{3}$\n\t\\task[\\textbf{d.}] $\\frac{5}{3}$\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\t\\lim _{z \\rightarrow i} \\frac{z^{10}+1}{z^{6}+1}=\\lim _{z \\rightarrow i} \\frac{10 z^{9}}{6 z^{5}}=\\lim _{z \\rightarrow i} \\frac{10 z^{4}}{6}=\\frac{10}{6}=\\frac{5}{3}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (D)}\n\\end{answer}\n\\item The value of integral\n$$\nI=\\oint \\frac{\\sin z}{2 z-\\pi} d z\n$$\nwith $c$ a circle $|z|=2$, is\n{\\exyear{ JEST 2014}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}] 0\n\t\\task[\\textbf{b.}]$2 \\pi i$\n\t\\task[\\textbf{c.}] $\\pi i$\n\t\\task[\\textbf{d.}] $-\\pi i$\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\tI&=\\oint_{C} \\frac{\\sin z}{2 z-\\pi},\\text{ for pole }2 z-\\pi=0 \\Rightarrow z=\\frac{\\pi}{2}\\\\\n\\text{\tResidue at }z&=\\frac{\\pi}{2}\n\t\\because|z|=2\\text{, so pole will lie within the contour}\\\\\n\tI&=\\oint_{C} \\frac{e^{i z}}{2\\left(z-\\frac{\\pi}{2}\\right)}=\\sum R \\times 2 \\pi i\\\\\n\t\\left.\\operatorname{Res}\\right|_{z=\\frac{\\pi}{2}}&=\\frac{\\left(z-\\frac{\\pi}{2}\\right) e^{i z}}{2\\left(z-\\frac{\\pi}{2}\\right)}=\\frac{e^{i \\pi / 2}}{2}=\\frac{i}{2}\\text{ (taking imaginary part); Residue }=\\frac{1}{2}\\\\\n\t\\text{Now, }I&=\\frac{1}{2} \\times 2 \\pi i=\\pi i\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\\end{answer}\n\\item Given an analytic function $f(z)=\\phi(x, y)+i \\psi(x, y)$, where $\\phi(x, y)=x^{2}+4 x-y^{2}+2 y$\nIf $C$ is a constant, which of the following relations is true?\n{\\exyear{ JEST 2015}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\psi(x, y)=x^{2} y+4 y+C$\n\t\\task[\\textbf{b.}]$\\psi(x, y)=2 x y-2 x+C$\n\t\\task[\\textbf{c.}]$\\psi(x, y)=2 x y+4 y-2 x+C$\n\t\\task[\\textbf{d.}] $\\psi(x, y)=x^{2} y-2 x+C$\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align}\n\tu&=\\phi(x, y)=x^{2}+4 x-y^{2}+2 y, v=\\psi\\notag\\\\\n\t\\text{From C.R. equation, }\\frac{\\partial u}{\\partial x}&=\\frac{\\partial v}{\\partial y}, \\Rightarrow \\frac{\\partial \\phi}{\\partial x}=\\frac{\\partial \\psi}{\\partial y}, \\frac{\\partial u}{\\partial y}=-\\frac{\\partial v}{\\partial x} \\Rightarrow \\frac{\\partial \\phi}{\\partial y}=-\\frac{\\partial \\psi}{\\partial x}\\notag\\\\\n\t\\text{Now, }\\frac{\\partial \\phi}{\\partial x}&=2 x+4=\\frac{\\partial \\psi}{\\partial y}\\notag\\\\\n\t\\Rightarrow \\psi&=2 x y+4 y+f(x)\\label{CN 01}\\\\\n\t\\text{and }\\frac{\\partial \\phi}{\\partial y}&=-2 y+2 \\Rightarrow \\frac{\\partial \\psi}{\\partial x}=+2 y-2\\notag\\\\\n\t\\psi&=2 x y+2 x+f(y)\\label{CN 02}\\\\\n\t\\text{From (\\ref{CN 01}) and (\\ref{CN 02}), }&2 x y+4 y+f(x)=2 x y-2 x+f(y)\\notag\\\\\n\tf(x)&=-2 x, \\quad f(y)=4 y \\notag\\\\\n\t\\psi&=2 x y+4 y-2 x+c\\notag\n\t\\end{align}\n\t\tSo the correct answer is \\textbf{Option (C)}\n\\end{answer}\n\\item Which one is the image of the complex domain $\\{z \\mid x y \\geq 1, x+y>0\\}$ under the mapping $f(z)=z^{2}$, if $z=x+i y ?$\n{\\exyear{ JEST 2017}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}] $\\{z \\mid x y \\geq 1, x+y>0\\}$\n\t\\task[\\textbf{b.}]$\\{z \\mid x \\geq 2, x+y>0\\}$\n\t\\task[\\textbf{c.}]$\\{z \\mid y \\geq 2 \\forall x\\}$\n\t\\task[\\textbf{d.}] $\\{z \\mid y \\geq 1 \\forall x\\}$\n\\end{tasks}\n\\item The integral $I=\\int_{1}^{\\infty} \\frac{\\sqrt{x-1}}{(1+x)^{2}} d x$ is\n{\\exyear{ JEST 2017}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\frac{\\pi}{\\sqrt{2}}$\n\t\\task[\\textbf{b.}]$\\frac{\\pi}{2 \\sqrt{2}}$\n\t\\task[\\textbf{c.}]$\\frac{\\sqrt{\\pi}}{2}$\n\t\\task[\\textbf{d.}]$\\sqrt{\\frac{\\pi}{2}}$\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\tI&=\\int_{1}^{\\infty} \\frac{\\sqrt{x-1}}{(1+x)^{2}} d x\\\\\n\\text{\tPut,} x&=\\left(1+z^{2}\\right), d x=2 z d z\\\\\n\t\\text{Hence, }I&=\\int_{0}^{\\infty} \\frac{2 z^{2} d z}{\\left(2+z^{2}\\right)^{2}}\\\\\n\t\\text{Here poles, }\\left(2+z^{2}\\right)&=0 \\Rightarrow(z+i \\sqrt{2})(z-i \\sqrt{2})=0\\\\\n\\text{\tOnly }&(z=i \\sqrt{2})\\text{ poles is allowed}\\\\\n\\text{Then }R(i \\sqrt{2})&=\\lim _{z \\rightarrow i \\sqrt{2}} \\frac{1}{\\sqrt{2-1}} \\frac{d}{d z}\\left[\\frac{2 z^{2}(z-i \\sqrt{2})^{2}}{(z-i \\sqrt{2})^{2}(z+i \\sqrt{2})^{2}}\\right]\\\\\n&=\\lim _{z \\rightarrow i \\sqrt{2}}\\left[\\frac{(z+i \\sqrt{2})^{2} \\cdot 4 z-2 z^{2} \\cdot 2(z+i \\sqrt{2})}{(z+i \\sqrt{2})^{4}}\\right]\\\\\n&=\\frac{(2 i \\sqrt{2})^{2} \\times 4(i \\sqrt{2})-2(i \\sqrt{2})^{2} \\cdot 2(2 i \\sqrt{2})}{(2 i \\sqrt{2})^{4}}\\\\&=-\\frac{32 \\sqrt{2} i+16 \\sqrt{2} i}{64}=-\\frac{16 \\sqrt{2} i}{64}=-\\frac{i}{2 \\sqrt{2}}\\\\\n\\text{Hence, }\\int_{-\\infty}^{\\infty} \\frac{2 z^{2}}{\\left(2+z^{2}\\right)^{2}} d z&=2 \\pi i\\left(-\\frac{i}{2 \\sqrt{2}}\\right)=\\frac{\\pi}{\\sqrt{2}}\\\\\n\\Rightarrow \\int_{0}^{\\infty} \\frac{2 z^{2}}{\\left(2+z^{2}\\right)^{2}} d z&=\\frac{\\pi}{2 \\sqrt{2}} \\Rightarrow \\int_{i}^{\\infty} \\frac{\\sqrt{x-1}}{(1+x)^{2}} d x=\\frac{\\pi}{2 \\sqrt{2}}\n\t\\end{align*}\n\tSo the correct answer is \\textbf{Option (B)}\n\\end{answer}\n\\item The integral\n$$\n\\int_{-\\infty}^{\\infty} \\frac{\\cos x}{x^{2}+1} d x \\text { is }\n$$\n{\\exyear{ JEST 2018}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\frac{\\pi}{e}$\n\t\\task[\\textbf{b.}] $\\pi e^{-2}$\n\t\\task[\\textbf{c.}]$\\pi$\n\t\\task[\\textbf{d.}] zero\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\tf(z)&=\\frac{e^{i z}}{z^{2}+1}=\\frac{e^{i z}}{(z+i)(z-i)}\\\\\n\t\\int_{-\\infty}^{\\infty} \\frac{\\cos x}{x^{2}+1} d x&=\\operatorname{Re} 2 \\pi i \\times \\frac{e^{i \\cdot i}}{z i}=\\frac{\\pi}{e}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (A)}\n\\end{answer}\n\\item Consider the function $f(x, y)=|x|-i|y| .$ In which domain of the complex plane is this function analytic?\n{\\exyear{ JEST 2019}}\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]First and second quadrants\n\t\\task[\\textbf{b.}]Second and third quadrants\n\t\\task[\\textbf{c.}]Second and fourth quadrants\n\t\\task[\\textbf{d.}] Nowhere\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\t f(x, y)&=|x|-i|y|\\\\\n\tf(x, y)&=x-i y=\\bar{z}\\\\\n\tf(x, y)&=-x-i y=-z\\\\\n\tf(x, y)&=-x+i y=-\\bar{z}\\\\\n\tf(x, y)&=x+i y=z\n\t\\intertext{We know $\\bar{z}$ is not analytic and $z$ and $-z$ are analytic. }\n\t\\end{align*}\n\tSo the correct answer is \\textbf{Option (C)}\n\\end{answer}\n\\end{enumerate}\n \\colorlet{ocre1}{ocre!70!}\n\\colorlet{ocrel}{ocre!30!}\n\\setlength\\arrayrulewidth{1pt}\n\\begin{table}[H]\n\t\\centering\n\t\\arrayrulecolor{ocre}\n\t\\begin{tabular}{|p{1.5cm}|p{1.7cm}||p{1.5cm}|p{1.5cm}|}\n\t\t\\hline\n\t\t\\multicolumn{4}{|c|}{\\textbf{Answer key}}\\\\\\hline\\hline\n\t\t\\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\\\\hline\n\t\t1&\\textbf{D} &2&\\textbf{C}\\\\\\hline \n\t\t3&\\textbf{3(NAT)} &4&\\textbf{C} \\\\\\hline\n\t\t5&\\textbf{A} &6&\\textbf{C} \\\\\\hline\n\t\t7&\\textbf{C}&8&\\textbf{A}\\\\\\hline\n\t\t9&\\textbf{$\\pi$(NAT)}&10&\\textbf{3(NAT)}\\\\\\hline\n\t\t11&\\textbf{81.64(NAT)} &12&\\textbf{C}\\\\\\hline\n\t\t13&\\textbf{A}& 14&\\textbf{B}\\\\\\hline\n\t\t15&\\textbf{A}&16 &\\textbf{D}\\\\\\hline\n\t\t17&\\textbf{C}&18&\\textbf{C}\\\\\\hline\n\t\t19&\\textbf{-} &20&\\textbf{B}\\\\\\hline\n\t\t21&\\textbf{A}&22&\\textbf{C}\\\\\\hline\n\t\\end{tabular}\n\\end{table}", "meta": {"hexsha": "de7b256ec826e15e935d0ebb9801bf328921ba90", "size": 46216, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "CSIR- Mathematical Physics/chapter/Complex Problem Set Solutions.tex", "max_stars_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_stars_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "CSIR- Mathematical Physics/chapter/Complex Problem Set Solutions.tex", "max_issues_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_issues_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "CSIR- Mathematical Physics/chapter/Complex Problem Set Solutions.tex", "max_forks_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_forks_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.8986784141, "max_line_length": 316, "alphanum_fraction": 0.5804916046, "num_tokens": 21072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.8244619306896956, "lm_q1q2_score": 0.6363555804186913}} {"text": "\\section{Data Description and Preprocessing}\n\\subsection{Fundamental Data Description}\n\nThe data used within this project was obtained from a study carried out as part of a research program for the \\textit{UK Forensic Science Service} \\cite{ForensicScienceInduction}. The data set contained a total of $214$ glass fragments that were obtained in a pre-split of $149$ training and $65$ test samples. For each glass fragment, a total of nine features were recorded, including a measure of the refractive index (RI), which describes how fast light travels through a material. It is a standard measure in glass analysis as it varies significantly for different types of glass. The remaining eight measured features described the chemical composition of the glass fragment in percent. Table \\ref{features} lists all measured features.\n\\newline\n\n\\begin{table}[!ht]\n \\footnotesize\n \\centering\n\\begin{tabular}{ c c c c c c c c c }\n \\toprule\n \\textbf{RI} & \\textbf{Na} & \\textbf{Mg} & \\textbf{Al} & \\textbf{Si} & \\textbf{K} & \\textbf{Ca} & \\textbf{Ba} & \\textbf{Fe} \\\\ \n \\midrule \n Refractive Index & Sodium & Magnesium & Aluminium & Silicon & Potassium & Calcium & Barium & Iron \\\\ \n \\bottomrule\n\\end{tabular}\n\\captionsetup{justification=centering,margin=2cm}\n\\caption{Observed Features for Glass Fragments}\n\\label{features}\n\\end{table}\n\nEach glass fragment in the data set belonged to one of six classes. In the data the classes were integer-encoded and mapped to the respective glass types depicted in Table \\ref{classes}.\n\\newline\n\n\\begin{table}[!ht]\n \\footnotesize\n \\centering\n\\begin{tabular}{ c c }\n \\toprule\n Integer Code & Glass Type \\\\\n \\midrule \n 1 & Window from building (float processed) \\\\\n 2 & Window from building (non-float processed) \\\\ \n 3 & Window from vehicle \\\\\n 5 & Container \\\\\n 6 & Tableware \\\\\n 7 & Headlamp \\\\\n \\bottomrule\n\\end{tabular}\n\\captionsetup{justification=centering,margin=2cm}\n\\caption{Glass Types and their Encoding}\n\\label{classes}\n\\end{table}\n\n\\subsection{Data Cleaning}\nAll data integrity checks carried out, such as checking for missing values, checking for the ranges of each feature, and adding up the percentages in the chemical composition in all glass fragments, did not report any major inconsistencies. Thus, no further data cleaning was necessary.\n\n% To get reproducible results, the training data was further divided into a training and validation split, that were used in the training of all models. Furthermore, all splits were normalised using a standard normalisation, meaning that each feature was transformed such that it had a zero mean and unit variance. This was crucial for ensuring that each feature contributed to the PCA analysis independently of its scale and the gradient descent training algorithm in the neural network to work properly.", "meta": {"hexsha": "30aa9c6b4b6657c6606b20887936b95b74d98005", "size": 2798, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/02_preproc.tex", "max_stars_repo_name": "jonas-mika/ml-project", "max_stars_repo_head_hexsha": "c052c33010033cd9fd596eb5ac4d270d1bf98ee3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/02_preproc.tex", "max_issues_repo_name": "jonas-mika/ml-project", "max_issues_repo_head_hexsha": "c052c33010033cd9fd596eb5ac4d270d1bf98ee3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/02_preproc.tex", "max_forks_repo_name": "jonas-mika/ml-project", "max_forks_repo_head_hexsha": "c052c33010033cd9fd596eb5ac4d270d1bf98ee3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-29T17:23:15.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-29T17:23:15.000Z", "avg_line_length": 58.2916666667, "max_line_length": 741, "alphanum_fraction": 0.7719799857, "num_tokens": 671, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355187, "lm_q2_score": 0.8244619177503206, "lm_q1q2_score": 0.6363555790854374}} {"text": "\\section{Stiefel-Whitney classes, immersions, cobordisms}\nThere is a result analogous to Theorem \\ref{chern-classes} for all vector\nbundles (not necessarily oriented):\n\\begin{theorem}\n There exist a unique family of characteristic classes $w_i:\\Vect_n(X) \\to\n H^n(X;\\FF_2)$ such that for $0\\leq i$ and $i>n$, we have $w_i=0$, and:\n \\begin{enumerate}\n\t\\item $w_0 = 1$;\n\t\\item $w_1(\\lambda) = e(\\lambda)$; and\n\t\\item the Whitney sum formula holds:\n\t \\begin{equation*}\n\t\tw_k(\\xi\\oplus\\eta) = \\sum_{i+j=k} w_i(\\xi)\\cup w_j(\\eta)\n\t \\end{equation*}\n \\end{enumerate}\n Moreover:\n $$\n H^\\ast(BO(n);\\FF_2) = \\FF_2[w_1,\\cdots,w_n],\n $$\n where $w_n = e_2$.\n\\end{theorem}\n\\begin{remark}\n We can express the Whitney sum formula simply by defining the \\emph{total\n Steifel-Whitney class}\n $$1 + w_1 + w_2 + \\cdots=:w.$$\n Then the Whitney sum formula is just\n $$\n w(\\xi\\oplus\\eta) = w(\\xi)\\cdot w(\\eta).\n $$\n Likewise, the Whitney sum formula can be stated by defining the total Chern\n class.\n\\end{remark}\n\\begin{remark}\\label{sw-stability}\n Again, the Steifel-Whitney classes are stable:\n $$w(\\xi\\oplus k\\epsilon) = w(\\xi).$$\n\\end{remark}\nAgain, Grothendieck's definition works since the splitting principle holds.\nThere is an injection $H^\\ast(BO(n))\\hookrightarrow H^\\ast(B(\\Z/2\\Z)^n)$. To\ncompute $H^\\ast(BO(n))$, our argument for computing $H^\\ast(BU(n))$ does not\nimmediately go through, although there is a fiber sequence\n$$S^{n-1} \\to EO(n)\\times_{O(n)} O(n)/O(n-1)\\to BO(n);$$\nthe problem is that $n-1$ can be even or odd. We still have a Gysin sequence,\nthough:\n$$\n\\cdots\\to H^{q-n}(BO(n))\\xar{e\\cdot} H^q(BO(n)) \\xar{\\pi^\\ast} H^q(BO(n-1)) \\to H^{q-n+1}(BO(n))\\to\\cdots\n$$\nIn order to apply our argument for computing $H^\\ast(BU(n))$ to this case, we\nonly need to know that $e$ is a nonzero divisor. The splitting principle gave a\nmonomorphism $H^\\ast(BO(n)) \\hookrightarrow H^\\ast((\\RP^\\infty)^n)$. The fact\nthat $e$ is a nonzero divisor follows from the observation that under this map,\n$$e_2=w_n\\mapsto e_2(\\lambda_1\\oplus\\cdots\\oplus\\lambda_n) = t_1\\cdots t_n,$$\nusing the same argument as in \\S \\ref{euler-multiplicativity}; however,\n$t_1\\cdots t_n$ {is} a nonzero divisor, since $H^\\ast((\\RP^\\infty)^n)$ is an\nintegral domain.\n\\subsection{Immersions of manifolds}\nThe theory developed above has some interesting applications to differential\ngeometry.\n\\begin{definition}\n Let $M^n$ be a smooth closed manifold. An \\emph{immersion} is a smooth map\n from $M^n$ to $\\RR^{n+k}$, denoted $f:M^n\\looparrowright \\RR^{n+k}$, such\n that $(\\tau_{M^n})_x \\hookrightarrow (\\tau_{\\RR^{n+k}})_{f(x)}$ for $x\\in\n M$.\n\\end{definition}\nInformally: crossings are allows, but not cusps.\n%It's a big problem in topology to try to eliminate the crossings.\n\\begin{example}\n There is an immersion $\\RP^2 \\looparrowright \\RR^3$, known as \\emph{Boy's\n surface}.\n\\end{example}\n\\begin{question}\n When can a manifold admit an immersion into an Euclidean space?\n\\end{question}\nAssume we had an immersion $i:M^n\\looparrowright \\RR^{n+k}$.\nThen we have an embedding $f:\\tau_M \\to i^\\ast \\tau_{\\RR^{n+k}}$ into a trivial\nbundle over $M$, so $\\tau_M$ has a $k$-dimensional complement, called $\\xi$\nsuch that\n$$\\tau_M\\oplus\\xi = (n+k)\\epsilon.$$\nApply the total Steifel-Whitney class, we have\n$$w(\\tau)w(\\xi) = 1,$$\nsince there's no higher Steifel-Whitney class of a trivial bundle. In\nparticular,\n$$\nw(\\xi) = w(\\tau)^{-1}.\n$$\n\\begin{example}\n Let $M = \\RP^n\\looparrowright \\RR^{n+k}$. Then, we know that\n $$\\tau_{\\RP^n}\\oplus\\epsilon\\simeq (n+1)\\lambda^\\ast \\simeq (n+1)\\lambda,$$\n where $\\lambda\\downarrow \\RP^n$ is the canonical line bundle. By Remark\n \\ref{sw-stability}, we have\n $$w(\\tau_{\\RP^n}) = w(\\tau_{\\RP^n}\\oplus\\eta) = w((n+1)\\lambda) =\n w(\\lambda)^{n+1}.$$\n It remains to compute $w(\\lambda)$. Only the first Steifel-Whitney class is\n nonzero. Writing $H^\\ast(\\RP^n) = \\FF_2[x]/x^{n+1}$, we therefore have\n $w(\\lambda) = x$. In particular,\n $$w(\\tau_{\\RP^n}) = (1+x)^{n+1} = \\sum^n_{i=0}\\binom{n+1}{i}x^i.$$\n It follows that\n $$w_i(\\tau_{\\RP^n}) = \\binom{n+1}{i}x^i.$$\n The total Steifel-Whitney class of the complement of the tangent bundle is:\n $$w(\\xi) = (1+x)^{-n-1}.$$\n The most interesting case is when $n$ is a power of $2$, i.e., $n=2^s$ for\n some integer $s$. In this case, since taking powers of $2$ is linear in\n characteristic $2$, we have\n $$w(\\xi) = (1+x)^{-1-2^s} = (1+x)^{-1}(1+x)^{-2^s} =\n (1+x)^{-1}(1+x^{2^s})^{-1}.\n %= (1+x+x^2+\\cdots)(1+x^{2^s}+\\cdots)\n $$\n As all terms of degree greater than $2^s$ are zero, we conclude that\n So\n $$w(\\xi) = 1+x+x^2+\\cdots+x^{2^s-1}+2x^s = 1+x+x^2+\\cdots+x^{2^s-1}.$$\n As $x^{2^s-1}\\neq 0$, this means that $k = \\dim\\xi \\geq 2^s-1$. We\n conclude:\n \\begin{theorem}\n\tThere is no immersion $\\RP^{2^s}\\looparrowright \\RR^{2\\cdot 2^{s}-2}$.\n \\end{theorem}\n The following result applied to $\\RP^{2^s}$ shows that the above result is\n sharp:\n \\begin{theorem}[Whitney]\n\tAny smooth compact closed manifold $M^n \\looparrowright \\RR^{2n-1}$.\n \\end{theorem}\n However, Whitney's result is \\emph{not} sharp for a general smooth compact\n closed manifold. Rather, we have:\n \\begin{theorem}[Brown--Peterson, Cohen]\n\tA closed compact smooth $n$-manifold $M^n \\looparrowright\n\t\\RR^{2n-\\alpha(n)}$, where $\\alpha(n)$ is the number of $1$s in the\n\tdyadic expansion of $n$.\n \\end{theorem}\n This result is sharp, since if $n=\\sum 2^{d_i}$ for the dyadic expansion,\n then $M = \\prod_i \\RP^{2^{d_i}} \\not \\looparrowright \\RR^{2n-\\alpha(n)-1}$.\n\\end{example}\n%There was a period of time in the late '60s, etc. when a lot of effort was put\n%in to stronger and stronger immersion results.\n\\subsection{Cobordism, characteristic numbers}\nIf we have a smooth closed compact $n$-manifold, then it embeds in $\\RR^{n+k}$\nfor some $k\\gg 0$. The normal bundle then satisfies\n$$\\tau_M\\oplus \\nu_M = (n+k)\\epsilon.$$\nA piece of differential topology tells us that if $k$ is large, then\n$\\nu_M\\oplus N\\epsilon$ is independent of the bundle for some $N$.\n\nThis example, combined with Remark \\ref{stability}, shows that $w(\\nu_M)$ is\nindependent of $k$. We are therefore motivated to think of Stiefel-Whitney\nclasses as coming from $H^\\ast(BO;\\FF_2) = \\FF_2[w_1,w_2,\\cdots]$, where $BO =\n\\varinjlim BO(n)$. Similarly, Chern classes should be thought of as coming from\n$H^\\ast(BU;\\Z) = \\Z[c_1,c_2,\\cdots]$. This exa\n\\begin{definition}\n The characteristic number of a smooth closed compact $n$-manifold\n $M$ is defined to be $\\langle w(\\nu_M),[M]\\rangle$.\n\\end{definition}\nNote that the fundamental class $[M]$ exists, since our coefficients are in\n$\\FF_2$, where everything is orientable.\n\nThis definition is very useful when thinking about cobordisms. \n\\begin{definition}\\label{cobordism}\n Two (smooth closed compact) $n$-manifolds $M, N$ are \\emph{(co)bordant} if\n there is an $(n+1)$-dimensional manifold $W^{n+1}$ with boundary such that\n $$\\partial W\\simeq M\\sqcup N.$$\n\\end{definition}\nFor instance, when $n=0$, the manifold $\\ast\\sqcup \\ast$ is \\emph{not}\ncobordant to $\\ast$, but it is cobordant to the empty set. However,\n$\\ast\\sqcup\\ast\\sqcup\\ast$ is cobordant to $\\ast$. Any manifold is cobordant\nto itself, since $\\partial(M\\times I) = M\\sqcup M$. In fact, cobordism forms\nan equivalence relation on manifolds.\n\n\\begin{example}\n A classic example of a cobordism is the ``pair of pants''; this is the\n following cobordism between $S^1$ and $S^1\\sqcup S^1$:\n \\todo{add image}\n\\end{example}\n\nLet us define\n$$\n\\Omega^O_n = \\{\\text{cobordism classes of $n$-manifolds}\\}.\n$$\nThis forms a group: the addition is given by disjoint union. Note that every\nelement is its own inverse. Moreover, $\\bigoplus_n \\Omega^O_n = \\Omega^O_\\ast$\nforms a graded ring, where the product is given by the Cartesian product of\nmanifolds. Our discussion following Definition \\ref{cobordism} shows that\n$\\Omega^O_0 = \\FF_2$.\n\\begin{exercise}\\label{nullbordant}\n Every $1$-manifold is nullbordant, i.e., cobordant to the point.\n\\end{exercise}\nThom made the following observation. Suppose an $n$-manifold $M$ is embedded\ninto Euclidean space, and that $M$ is nullbordant via some $(n+1)$-manifold\n$W$, so that $\\nu_W|_{M} = \\nu_M$. In particular,\n$$\\langle w(\\nu_M),[M]\\rangle = \\langle w(\\nu_W)|_{M},[M]\\rangle.$$\nOn the other hand, the boundary map $H_{n+1}(W,M) \\xar{\\partial} H_n(M)$ sends\nthe relative fundamental class $[W,M]$ to $[M]$. Thus\n$$\\langle w(\\nu_M),[M]\\rangle = \\langle w(\\nu_M),\\partial[W,M]\\rangle = \\langle\n\\delta w(\\nu_M),[W,M]\\rangle.$$\nHowever, we have an exact sequence\n$$H^n(W)\\xar{i^\\ast} H^n(M) \\xar{\\delta} H^{n+1}(W,M).$$\nSince $w(\\nu_M)$ is in the image of $i^\\ast$, it follows that $\\delta w(\\nu_M)\n= 0$. In particular, the characteristic number of a nullbordant manifold is\nzero. Thus, we find that ``Stiefel-Whitney numbers tell all'':\n\\begin{prop}\n Characteristic numbers are cobordism invariants. In other words,\n characteristic numbers give a map\n $$\\Omega^O_n \\to \\Hom(H^n(BO),\\FF_2)\\simeq H_n(BO).$$\n\\end{prop}\nMore is true:\n\\begin{theorem}[Thom, 1954]\\label{thom-sw}\n The map of graded rings $\\Omega^O_\\ast\\to H_\\ast(BO)$ defined by the\n characteristic number is an inclusion. Concretely, if $w(M^n) = w(N^n)$ for\n all $w\\in H^n(BO)$, then $M^n$ and $N^n$ are cobordant.\n\\end{theorem}\nThe way that Thom proved this was by expressing $\\Omega^O_\\ast$ is the graded\nhomotopy ring of some space, which he showed is the product of mod $2$\nEilenberg-MacLane spaces. Along the way, he also showed that:\n$$\n\\Omega^O_\\ast = \\FF_2[x_i:i\\neq 2^s-1] = \\FF_2[x_2,x_4,x_5,x_6,x_8,\\cdots]\n$$\nThis recovers the result of Exercise \\ref{nullbordant} (and so much more!).\n", "meta": {"hexsha": "1106befc2c52cab4349aa60aa1b8e6d1ea9abff9", "size": 9816, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "906/lec-73-steifel-whitney-classes.tex", "max_stars_repo_name": "ichung/algtop-notes", "max_stars_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-04-26T15:00:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T22:47:06.000Z", "max_issues_repo_path": "906/lec-73-steifel-whitney-classes.tex", "max_issues_repo_name": "ichung/algtop-notes", "max_issues_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-03-13T17:54:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-13T17:59:46.000Z", "max_forks_repo_path": "906/lec-73-steifel-whitney-classes.tex", "max_forks_repo_name": "ichung/algtop-notes", "max_forks_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-21T18:15:11.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-13T17:38:04.000Z", "avg_line_length": 46.5213270142, "max_line_length": 105, "alphanum_fraction": 0.6726772616, "num_tokens": 3390, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.8244619285331332, "lm_q1q2_score": 0.6363555787541626}} {"text": "\\section{Universal Hashing}\r\n\r\n\\def\\E{\\mathbb{E}}\r\n\\toclesssubsection{Introduction}\r\n\r\n\\begin{frame}{Universal Hashing}{Thought Experiment}\r\n \\textbf{Thought Experiment:}\\\\\r\n \\begin{itemize}\r\n \\item<1->\r\n A {\\color{MainA}hash function} is defined for a given\r\n {\\color{MainA}key set}\r\n \\item<2->\r\n Find a {\\color{MainA}set of keys} resulting in a degenerated\r\n {\\color{MainA}hash table}\r\n \\begin{itemize}\r\n \\item<3->\\textit{The {\\color{MainA}hash function} stays fixed}\\\\\r\n \\item<4->\\textit{For table size of 100: try $100 \\times (99 + 1)$ different numbers}\\\\\r\n \\item<5->\\textit{Worst case: all 100 {\\color{MainA}key sets} map to\r\n one bucket}\r\n \\end{itemize}\r\n \\item<5->\r\n \\textbf{Now:} find a solution to avoid that problem\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Idea}\r\n \\textbf{Solution:} {\\color{MainA}universal hashing}\r\n \\begin{itemize}\r\n \\item<1->\r\n Out of a set of hash functions we randomly choose one\r\n \\item<2->\r\n The {\\color{MainA}expected result} of the hash function is an equal\r\n distribution over the buckets\r\n \\item<3->\r\n This hash function stays fixed for the lifetime of table\\\\\r\n Optional: copy table with new hash when degenerated\r\n \\end{itemize}\r\n \\onslide<1->{\r\n \\vspace{-0.5em}\r\n \\begin{columns}\r\n \\begin{column}[T]{0.33\\linewidth}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{height=0.8\\linewidth}\r\n \\input{Images/UniversalHashing/Hash_1_Crossing.tikz}\r\n \\end{adjustbox}\r\n \\caption{Hash func. 1}\r\n \\label{fig:universal_hashing:hash_function_crossing}\r\n \\end{figure}\r\n \\end{column}\r\n \\begin{column}[T]{0.33\\linewidth}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{height=0.8\\linewidth}\r\n \\input{Images/UniversalHashing/Hash_2_Normal.tikz}\r\n \\end{adjustbox}\r\n \\caption{Hash func. 2}\r\n \\label{fig:universal_hashing:hash_function_normal}\r\n \\end{figure}\r\n \\end{column}\r\n \\begin{column}[T]{0.33\\linewidth}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{height=0.8\\linewidth}\r\n \\input{Images/UniversalHashing/Hash_3_Colliding.tikz}\r\n \\end{adjustbox}\r\n \\caption{Hash func. coll.}\r\n \\label{fig:universal_hashing:hash_function_colliding}\r\n \\end{figure}\r\n \\end{column}\r\n \\end{columns}}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Definition}\r\n \\textbf{Definition:}\r\n \\begin{itemize}\r\n \\item<1->\r\n We call {\\color{MainA}$\\mathbb{U}$} the set (universe) of\r\n possible keys\r\n \\item<2->\r\n The size {\\color{MainA}$m$} of the hash table {\\color{MainA}$T$}\r\n \\item<3->\r\n Set of hash functions\r\n {\\color{MainA}$\\mathbb{H} = \\{h_1, h_2,\\dots, h_n\\}$} with\r\n {\\color{MainA}$h_i: \\mathbb{U} \\to \\{0,\\dots, m - 1\\}$}\r\n \\item<5-> Idea: runtime should be\r\n $\\color{MainA} O(1+\\frac{\\vert \\mathbb{S}\\vert}{m})$,\r\n \\rlap{where $\\color{MainA}\\frac{\\vert \\mathbb{S} \\vert}{m}$ is the\r\n table load}\r\n \\end{itemize}%\\vspace*{-2em}\r\n \\begin{figure}[!b]%\r\n \\begin{adjustbox}{height=0.32\\linewidth}%\r\n \\input{Images/UniversalHashing/HashFunction.tikz}%\r\n \\end{adjustbox}\r\n \\vspace*{-1.0em}%\r\n \\onslide<3->\\caption{Hash function $h_1$}%\r\n \\label{fig:universal_hashing:hash_function_definition}\r\n \\end{figure}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Definition}\r\n \\begin{columns}\r\n \\begin{column}{0.5\\linewidth}\r\n \\begin{itemize}\r\n \\item<1->\r\n We choose two random keys\r\n {\\color{MainA}$x, y \\in \\mathbb{U} \\mid x \\neq y$}\r\n \\item<2->\r\n An average of 3 out of 15 functions produce collisions\r\n \\end{itemize}\r\n \\end{column}\r\n \\begin{column}{0.5\\linewidth}\r\n \\begin{figure}[!t]%\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\input{Images/UniversalHashing/HashUniverse.tikz}%\r\n \\end{adjustbox}\r\n \\caption{Set of hash functions $\\mathbb{H}$}%\r\n \\label{fig:universal_hashing:hash_universe}\r\n \\end{figure}\r\n \\end{column}\r\n \\end{columns}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Definition}\r\n \\textbf{Definition:}\r\n {\\color{MainA}$\\mathbb{H}$} is {\\color{MainA}$c$-universal} if\r\n {\\color{MainA}$\\forall x, y \\in \\mathbb{U} \\mid x \\neq y:$\r\n \\begin{displaymath}\r\n \\overbrace{\r\n \\frac{\r\n \\vert \\{h \\in \\mathbb{H}\\!: h(x) = h(y)\\} \\vert\r\n }{\r\n \\underbrace{\\vert \\mathbb{H} \\vert}_\\text{{\\color{black}Number of hash functions}}\r\n }\r\n }^\\text{\r\n {\\color{black}Number of hash functions that create collisions}\r\n }\r\n \\leq c \\cdot \\frac{1}{m}, \\hspace*{1.5em} c \\in \\mathbb{R}\r\n \\end{displaymath}}\r\n \\vspace{-1em}\r\n \\begin{itemize}\r\n \\item<2->\r\n In other words, given an arbitrary but fixed pair\r\n {\\color{MainA}$x, y$}.\\\\\r\n If {\\color{MainA}$h \\in \\mathbb{H}$} is chosen randomly then\\\\\r\n \\onslide<3->{\r\n \\begin{center}\r\n \\color{MainA}\r\n $Prob ( h(x) = h(y) ) \\leq c \\cdot \\frac{1}{m}$\r\n \\end{center}\r\n }\r\n \\end{itemize}\r\n \\onslide <4->\r\n \\begin{block}{Note: If the hash function assigns each key $x$ and $y$\r\n randomly to buckets then:}\r\n {\\onslide<5->$Prob( Collision ) = \\frac{1}{m} \\Leftrightarrow c = 1$}\r\n \\end{block}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Definition}\r\n \\begin{columns}\r\n \\begin{column}{0.5\\linewidth}\r\n \\begin{itemize}\r\n \\item\r\n {\\color{MainA}$\\mathbb{U}$}:\r\n key universe\r\n \\item\r\n {\\color{MainA}$\\mathbb{S}$}:\r\n used Keys\r\n \\item\r\n {\\color{MainA}$\\mathbb{S}_i \\subseteq \\mathbb{S}$}:\r\n keys mapping to Bucket {\\color{MainA}$i$} (\\enquote{synonyms})\r\n \\item\r\n Ideal would be\r\n {\\color{MainA}$\\vert \\mathbb{S}_i \\vert =\\dfrac{\\vert \\mathbb{S} \\vert}{m}$}\r\n \\end{itemize}\r\n \\end{column}\r\n \\begin{column}{0.5\\linewidth}\r\n \\begin{figure}[!h]%\r\n \\begin{adjustbox}{width=\\linewidth}%\r\n \\input{Images/UniversalHashing/HashFunctionToBucket.tikz}%\r\n \\end{adjustbox}\r\n \\caption{Hash function {\\color{MainA}$h \\in \\mathbb{H}$}}%\r\n \\label{fig:universal_hashing:hash_function_to_bucket}\r\n \\end{figure}\r\n \\end{column}\r\n \\end{columns}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Universal Hashing}{Definition}\r\n \\begin{itemize}\r\n \\item<1->\r\n Let {\\color{MainA}$\\mathbb{H}$} be a {\\color{MainA}$c$}-universal class of hash functions\r\n \\item<2->\r\n Let {\\color{MainA}$\\mathbb{S}$} be a set of keys and {\\color{MainA}$h \\in \\mathbb{H}$} selected randomly\r\n \\item<3->\r\n Let {\\color{MainA}$\\mathbb{S}_i$} be the key {\\color{MainA}$x$} for which {\\color{MainA}$h(x) = i$}\r\n \\item<4->\r\n The expected average number of elements to search through per \r\n bucket is {\\color{MainA}\\[\\E{}\\left[\\vert \\mathbb{S}_i \\vert\\right]\r\n \\leq 1 + c \\cdot \\frac{\\vert \\mathbb{S} \\vert}{m}\\]}\r\n \\item<5->\r\n Particulary: if {\\color{MainA}($m = \\Omega(\\vert \\mathbb{S} \\vert)$)} then\r\n {\\color{MainA}$\\E{}\\left[\\vert \\mathbb{S}_i \\vert\\right] = \\mathcal{O}(n)$}\r\n \\end{itemize}\r\n\\end{frame}\r\n", "meta": {"hexsha": "cd2d83b4af154dcd9e301606002d8f86db304ea0", "size": 7697, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture-5/Chapter/eng/030_UniversalHashing.tex", "max_stars_repo_name": "TobiOnline/AlgoDat", "max_stars_repo_head_hexsha": "565a9f03a9ed7ef354cb4f143959df77df89b726", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2016-12-16T17:42:34.000Z", "max_stars_repo_stars_event_max_datetime": "2020-08-26T11:07:16.000Z", "max_issues_repo_path": "Lecture-5/Chapter/eng/030_UniversalHashing.tex", "max_issues_repo_name": "TobiOnline/AlgoDat", "max_issues_repo_head_hexsha": "565a9f03a9ed7ef354cb4f143959df77df89b726", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 23, "max_issues_repo_issues_event_min_datetime": "2016-10-08T09:27:41.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-20T15:40:10.000Z", "max_forks_repo_path": "Lecture-5/Chapter/eng/030_UniversalHashing.tex", "max_forks_repo_name": "TobiOnline/AlgoDat", "max_forks_repo_head_hexsha": "565a9f03a9ed7ef354cb4f143959df77df89b726", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2016-10-07T11:55:23.000Z", "max_forks_repo_forks_event_max_datetime": "2021-04-05T08:36:38.000Z", "avg_line_length": 35.6342592593, "max_line_length": 111, "alphanum_fraction": 0.5530726257, "num_tokens": 2387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619134371954, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.6363555714294206}} {"text": "\\section{Differential Geometry of Space Curves}\r\n\\subsection{Parameterized Curve by Arc Length}\r\n\\begin{definition}\r\n A parameterized curve $C$ is the image of a continuous map $[a,b]\\to \\mathbb R^3$ sending $t\\mapsto \\underline{x}(t)$.\r\n We say $C$ is a differentiable parameterized curve if each component $x_i(t)$ is differentiable.\r\n We say $C$ is regular if $\\underline{x}^\\prime(t)\\neq\\underline{0}$ for any $t$.\\\\\r\n A regular and differentiable curve is called smooth.\r\n\\end{definition}\r\nSince it is an applied course (sadly), we will assert that our curve is as differentiable as we like.\\\\\r\nTo find the length of this curve, we partition the interval $[a,b]$ by $a=t_00$, so we can parameterize it by $(r\\cos t,r\\sin t,0), t\\in [0,2\\pi]$, and we unsurprisingly find that its length is $2\\pi r$.\\\\\r\n 2. Take $C$ be the same circle as in 1, we have\r\n $$\\int_C x^2y\\,\\mathrm ds=\\int_a^b (r\\cos t)^2(r\\sin t) r\\,\\mathrm dt=0$$\r\n\\end{example}\r\n\\begin{proposition}\r\n The way we define curve integrals is independent of parameterization.\r\n\\end{proposition}\r\n\\begin{example}\r\n If we parameterize the circle as $(r\\cos(2t),r\\sin(2t),0),t\\in [0,\\pi]$, we still get the same thing.\r\n\\end{example}\r\n\\begin{proof}\r\n Let $\\underline{s_1}(t),\\underline{s_2}(\\tau),t\\in[a,b],\\tau\\in[\\alpha,\\beta]$ be two different parameterizations of $C$, then there exists a function $\\tau\\to t(\\tau)$ such that $\\underline{s_1}(t(\\tau))=\\underline{s_2}(\\tau)$.\r\n Assume that $\\mathrm dt/\\mathrm d\\tau$ is nonzero and $t(\\tau)$ is a differentiable, invertible, and have differentiable inverse, then we have\r\n $$\\underline{s_2}^\\prime(\\tau)=\\frac{\\mathrm d\\underline{s_1}\\circ t}{\\mathrm d\\tau}=\\underline{s_1}^\\prime(t)t^\\prime(\\tau)$$\r\n If $\\mathrm dt/\\mathrm d\\tau>0$, we have\r\n $$\\int_\\alpha^\\beta|\\underline{s_2}^\\prime(\\tau)|\\,\\mathrm d\\tau=\\int_\\alpha^\\beta|\\underline{s_1}^\\prime(t(\\tau))|t^\\prime(\\tau)\\,\\mathrm d\\tau=\\int_a^b|\\underline{s_1}^\\prime(t)|\\,\\mathrm dt$$\r\n Similar for other cases.\r\n\\end{proof}\r\nWe now know the arc length is\r\n$$s(t)=\\int_{t_0}^t|\\underline{\\dot{x}}(u)|\\,\\mathrm du$$\r\nOn a regular curve, $\\mathrm ds/\\mathrm dt=|\\underline{\\dot{r}}(t)|>0$, this tells us that we can indeed parameterize each regular curve wrt arc length.\r\nThis is done by observing $\\mathrm dt/\\mathrm ds=1/|\\underline{\\dot{r}}(t)|$ which means we can write $\\underline{r}(s)=\\underline{r}(t(s))$, where we have\r\n$$\\frac{\\mathrm d\\underline{r}}{\\mathrm ds}=\\frac{\\underline{\\dot{r}}(t)}{|\\underline{\\dot{r}}(t)|}$$\r\nwhich is a unit vector.\r\nTherefore,\r\n\\begin{lemma}\r\n Any smooth curve $C$ has a parameterization $\\underline{r}(s)$ such that\r\n $$\\left|\\frac{\\mathrm dr}{\\mathrm ds}\\right|\\equiv 1$$\r\n\\end{lemma}\r\n\\begin{proof}\r\n Followed from above.\r\n\\end{proof}\r\n\\subsection{Curvature and Torsion}\r\nThroughout this section, we are only interested in smooth curves $C$ parameterized by arc length $\\underline{r}(s)$.\r\n\\begin{definition}\r\n The tangent vector is defined as $\\underline{t}(s)=\\underline{r}^\\prime(s)$.\r\n\\end{definition}\r\nNote that $\\underline{t}$ is always unit as it is an arc length parameterization.\r\n\\begin{definition}\r\n The curvature of $\\underline{r}(s)$ is defined as $\\kappa(s)=|\\underline{t}^\\prime(s)|=|\\underline{r}^{\\prime\\prime}(s)|$.\r\n\\end{definition}\r\nNote that if we differentiate $\\underline{t}\\cdot\\underline{t}=1$, then we have $\\underline{t}\\cdot\\underline{t}^\\prime=0$.\r\nThis shows that the unit vector in the direction $t^\\prime$ has a geometric interpretation as the normal to a curve, so we define\r\n\\begin{definition}\r\n The principal normal $\\underline{n}$ is the (unit) vector such that $\\underline{t}^\\prime=\\kappa\\underline{n}$.\r\n\\end{definition}\r\nNaturally, when we already have a pair of orthonormal vectors in $\\mathbb R^3$, adding a third one seems to be the next step to do.\r\n\\begin{definition}\r\n In $\\mathbb R^3$, the binormal $\\underline{b}$ is defined as $\\underline{b}=\\underline{t}\\times\\underline{n}$.\r\n\\end{definition}\r\nThen the vectors $\\underline{t},\\underline{n},\\underline{b}$ form an orthonormal basis for $\\mathbb R^3$.\r\nAgain we have $\\underline{b}\\cdot\\underline{b}^\\prime=0$ as $\\underline{b}$ is unit.\r\nBut we also have $\\underline{t}\\cdot\\underline{b}=0$, we get $\\underline{t}\\cdot\\underline{b}^\\prime=0$.\r\nHence $\\underline{n},\\underline{b}^\\prime$ are parallel.\r\n\\begin{definition}\r\n The torsion $\\tau$ is defined as such that $\\underline{b}^\\prime=-\\tau\\underline{n}$.\r\n\\end{definition}\r\nSo we have got there\r\n$$\\begin{cases}\r\n \\underline{t}^\\prime=\\kappa\\underline{n}=\\kappa(\\underline{b}\\times\\underline{t})\\\\\r\n \\underline{b}^\\prime=-\\tau\\underline{n}=\\tau(\\underline{t}\\times\\underline{b})\r\n\\end{cases}$$\r\nIntuitively and truthfully\r\n\\begin{proposition}\r\n The curvature and torsion uniquely defines a curve up to rigid motion.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Picard-Lindel\\\"of Theorem.\r\n\\end{proof}\r\nThe Taylor expansion of $\\underline{r}(t)$ around $0$ shows\r\n$$\\underline{r}(s)=\\underline{r}(0)+s\\underline{t}(0)+\\frac{1}{2}s^2\\kappa\\underline{n}(0)+o(s^2)$$\r\nNow we turn to consider the circle of best fit around $\\underline{r_0}$.\r\nParameterize the circle (with radiu $r$) by and expand to see that the second order term is somewhat like $s^2\\underline{n}/(2r)$, so it is natural to define\r\n\\begin{definition}\r\n The radius of curvature is defined as $r=1/\\kappa$.\r\n\\end{definition}", "meta": {"hexsha": "d24533db21a7188f3fde99ba9eb26c7d46069813", "size": 6651, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "1/curves.tex", "max_stars_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_stars_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "1/curves.tex", "max_issues_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_issues_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "1/curves.tex", "max_forks_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_forks_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.1588785047, "max_line_length": 233, "alphanum_fraction": 0.685460833, "num_tokens": 2109, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8397339696776499, "lm_q1q2_score": 0.6363456847764882}} {"text": "\\problemname{Limbo: Part 2}\n\n\\noindent Dominick Cobb and Arthur are ``extractors\" who perform corporate espionage. Using experimental military technology that gives them access to shared dream worlds, they infiltrate their targets' subconscious to extract valuable information. Cobb and his entourage are contacted by a mysterious Japanese syndicate and given a seemingly impossible task: instead of extracting information, do the opposite -- plant a new idea in the target's mind.\\\\\n\nCobb just figured out how to calculate the slowness of time in each dream level, but he is still missing a vital piece of the plan -- the architecture. For this challenging mission he has hired the brilliant young architect Ariadne to design a dream space, complete with all the aesthetic and tactile details. Cobb's target will be brought into that dream space, where they will fill it with details from their own subconscious and memories.\\\\\n\nAs extractors go into deeper dream levels, the perception of time slows down. This means subjects will be able to explore exponentially more space. Since dreams can go infinitely deep (onwards to Limbo), the architecture of a dream should be infinite in area. The master layout is denoted by a two-dimensional grid that starts at row $0$, column $0$, and extends infinitely downwards and to the right. Ariadne cannot simply draw a map of this space (since it is infinite), so she must come up with a program that \"generates\" the dream space to extend as large as the team needs to go deeper. She has an infinite number of building blocks, numbered $0, 1, 2, 3, 4, \\dots$, which must be used in that order.\\\\\n\nIn dream level $0$ (reality), the dream space consists of the single block $0$. With each deeper dream level, the area of the space doubles. To prevent the dream space from implosion due to growing too narrowly in a single direction, Ariadne has designed the space to double in \\textit{alternating directions}, with new blocks filling in the space sequentially based on the direction of doubling. For example, the first few levels of the dream map are depicted as follows:\\\\\n\n\\begin{center}\n\\begin{minipage}{8cm}\n\\begin{verbatim}\n 0 0 1 0 1 0 1 4 6\n 2 3 2 3 5 7\n\n(Reality) (Level 1) (Level 2) (Level 3)\n\n\n 0 1 4 6 0 1 4 6 16 20 24 28\n 2 3 5 7 2 3 5 7 17 21 25 29\n 8 9 10 11 8 9 10 11 18 22 26 30\n 12 13 14 15 12 13 14 15 19 23 27 31\n\n (Level 4) (Level 5)\n\n\\end{verbatim}\n\\end{minipage}\n\\end{center}\n\nAriadne's program must be able to generate any part of the map at will. Given only the row and column coordinates for a particular location in the dream space, can you determine the number of the building block that will be used?\n\n\\section*{Input}\nThe first line of input consists of a single integer $T$ ($1 \\leq T \\leq 100$), the number of test cases.\\\\\n$T$ lines follow, each of which is a test case consisting of two space-separated integers, $R$ and $C$ ($0 \\leq R, C \\leq 10^{9}$), specifying a particular row and column in the dream-space.\\\\\n\n\\section*{Output}\nFor each test case, print, on a separate line, the number of the building block at coordinates $(R, C)$.\\\\\nNote: the answer can be large, and may not necessarily fit in a 32-bit integer.\\\\\n", "meta": {"hexsha": "a48ae9ffecb4c721a21e2e1dc2c4a6e00081968a", "size": 3329, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "limbo2/problem_statement/problem.en.tex", "max_stars_repo_name": "csecutsc/utscode2", "max_stars_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2018-09-30T15:06:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-09T06:50:25.000Z", "max_issues_repo_path": "limbo2/problem_statement/problem.en.tex", "max_issues_repo_name": "csecutsc/utscode2", "max_issues_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "limbo2/problem_statement/problem.en.tex", "max_forks_repo_name": "csecutsc/utscode2", "max_forks_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-06-17T04:10:51.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-17T04:10:51.000Z", "avg_line_length": 83.225, "max_line_length": 707, "alphanum_fraction": 0.7308501051, "num_tokens": 884, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145997, "lm_q2_score": 0.839733963661418, "lm_q1q2_score": 0.6363456802174213}} {"text": "%% This file is part of the CroMagnon project\n%% Copyright 2015 David W. Hogg\n\n\\documentclass[12pt]{article}\n\\usepackage{url}\n\\input{vc}\n\n\\newcommand{\\like}{L}\n\\newcommand{\\dd}{\\mathrm{d}}\n\\newcommand{\\inverse}{^{-1}}\n\\newcommand{\\transpose}{^{\\mathsf{T}}}\n\n\\begin{document}\n\n\\section*{Notes on log-sum-exp%\n\\footnote{\\raggedright This document is available (in source-code form) at \\giturl.\nThis precise version has git hash \\texttt{\\githash~(\\gitdate)}.}}\n\\noindent\nDavid W. Hogg (NYU) (SCDA) (MPIA)\n\n\\bigskip\n\nGenerically in probabilistic inference problems we encounter quantities like\n\\begin{eqnarray}\n Q\n &=& \\ln\\sum_{k=1}^K\\exp q_k\n \\quad,\n\\end{eqnarray}\nwhere the $q_k$ are simple functions (for example, they might have simple\nderivatives with respect to parameters $\\theta$).\nLet's take the full derivative of $Q$:\n\\begin{eqnarray}\n \\frac{\\dd Q}{\\dd\\theta}\n &=& \\left[\\sum_{k=1}^K\\exp q_k\\right]\\inverse\n \\,\\frac{\\dd}{\\dd\\theta}\\left[\\sum_{k=1}^K\\exp q_k\\right] \\\\\n &=& \\left[\\sum_{k=1}^K\\exp q_k\\right]\\inverse\n \\,\\left[\\sum_{k=1}^K\\frac{\\dd q_k}{\\dd\\theta}\\exp q_k\\right]\n \\quad;\n\\end{eqnarray}\nthat is, the derivative of $Q$ is just the $\\exp q_k$-weighted average\nof derivatives of the $q_k$.\nThat's good!\n\nIn the particular case of cryo-EM, the objective function is related\nto a marginalized likelihood $\\like$ which looks like\n\\begin{eqnarray}\n Q\n &=&\n \\sum_{n=1}^N Q_n \\\\\n Q_n\n &=& -2\\,\\ln\\like_n \\\\\n &=& -2\\,\\ln\\sum_{k=1}^K P_k\\,\\exp(\\ln\\like_{nk}) \\\\\n &=& -2\\,\\ln\\sum_{k=1}^K P_k\\,\\exp(-\\frac{1}{2}\\,\\chi^2_{nk}) \\\\\n \\chi^2_{nk}\n &=& [y_n - \\mu_{nk}]\\transpose\\cdot C_n\\inverse\\cdot [y_n - \\mu_{nk}]\n \\quad,\n\\end{eqnarray}\nwhere the sum is over $k$ samples or grid locations in angles and\noffets, the $P_k$ are weights for those samples (typically roughly\n$1/K$), $y_n$ is the $n$th data point (tiny image), the $\\mu_{nk}$ are\nthe predictions for data point $n$ at angles and offsets $k$, and\n$C_n$ is a covariance matrix for the (Gaussian) noise model.\nThe predictions $\\mu_{nk}$ depend on the three-dimensional shape\nparameters or representation $\\theta$.\n\nTaking derivatives with respect to parameters,\n\\begin{eqnarray}\n \\frac{\\dd Q}{\\dd\\theta}\n &=&\n \\sum_{n=1}^N \\frac{\\dd Q_n}{\\dd\\theta} \\\\\n \\frac{\\dd Q_n}{\\dd\\theta}\n &=& -2\\,\\left[\\sum_{k=1}^K P_k\\,\\exp(-\\frac{1}{2}\\,\\chi^2_{nk})\\right]\\inverse\n \\,\\left[\\sum_{k=1}^K -\\frac{1}{2}\\,P_k\\,\\exp(-\\frac{1}{2}\\,\\chi^2_{nk})\\,\\frac{\\dd\\chi^2_{nk}}{\\dd\\theta}\\right] \\\\\n \\frac{\\dd\\chi^2_{nk}}{\\dd\\theta}\n &=& -2\\,[y_n - \\mu_{nk}]\\transpose\\cdot C_n\\inverse\\cdot\\frac{\\dd\\mu_{nk}}{\\dd\\theta}\n \\quad.\n\\end{eqnarray}\n\nThere is a literature on using the expectation-maximization (E-M)\nalgorithm to optimize marginalized likelihoods that have the logsumexp\nform.\nA good pedagogical (but not original) reference is Bishop (2006; ch\n9).\nThe E-M algorithm is clever because it optimizes a marginalized\nlikelihood of the form of the $Q$ defined above, without ever\nexplicitly constructing the marginalized likelihood or its\nderivatives.\nAnd the proof is beautiful!\nRumors fly, however, that E-M is not the best optimizer for many\nreal problems (Salakhutdinov 2003a, 2003b), and that if we are willing\nto compute the derivatives, we might get to the local optimum faster,\neven given the extra cost per iteration.\n\n\\begin{trivlist}\\raggedright\n\\item\nBishop,~C., 2006,\n\\textit{Pattern Recognition and Machine Learning},\nSpringer-Verlag, New York\n{\\footnotesize ISBN:978-0-387-31073-2}.\n\\item\nSalakhutdinov,~R., Roweis,~S., \\& Ghahramani,~Z., 2003a,\n``Optimization with EM and Expectation-Conjugate-Gradient'',\n\\textit{Proc.~20~ICML},\n{\\footnotesize \\url{http://www.cs.nyu.edu/~roweis/papers/emecgicml03.pdf}}.\n\\item\nSalakhutdinov,~R., Roweis,~S., \\& Ghahramani,~Z., 2003b,\n``On the Convergence of Bound Optimization Algorithms'',\n\\textit{Proc.~19~UAI (UAI2003)},\n{\\footnotesize \\url{http://arxiv.org/abs/1212.2490}}.\n\\end{trivlist}\n\n\\end{document}\n", "meta": {"hexsha": "dda5f7b3bb508975527eaded5f921c8c0df9f85f", "size": 3948, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "papers/logsumexp.tex", "max_stars_repo_name": "davidwhogg/CryoEM", "max_stars_repo_head_hexsha": "4ced7aa2216efeb9cdb657ab4ef47bf7ba0709e4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-06-18T15:48:00.000Z", "max_stars_repo_stars_event_max_datetime": "2019-06-18T15:48:00.000Z", "max_issues_repo_path": "papers/logsumexp.tex", "max_issues_repo_name": "davidwhogg/CroMagnon", "max_issues_repo_head_hexsha": "4ced7aa2216efeb9cdb657ab4ef47bf7ba0709e4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "papers/logsumexp.tex", "max_forks_repo_name": "davidwhogg/CroMagnon", "max_forks_repo_head_hexsha": "4ced7aa2216efeb9cdb657ab4ef47bf7ba0709e4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.9380530973, "max_line_length": 123, "alphanum_fraction": 0.696301925, "num_tokens": 1372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339596505965, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6363456725847159}} {"text": "\\section{Algorithms}\n\n\\subsection{Cristofeledes algorithm}\nObjective: solve TSP with $\\frac{3}{2}$-approximation\n\\begin{enumerate}\n \\item Construct an MST\n \\item Extract the odd degree vertices\n \\item Find a matching, using odd degree vertex from (2)\n \\item Combine the edges from the tree with those in the matching\n \\item Form a eularian circuit\n \\item Short the the traversal path to form a hamiltonian cycle\n\\end{enumerate}\n\n\\subsection{Double-tree algorithm}\n2-approximation\n\\begin{enumerate}\n \\item Construct MST\n \\item Give each node a redundant entry (graph can now be traversed like\n an~\\nameref{eularian} graph)\n \\item Traverse the nodes from start to finish (DFS), but only retain the first\n time occurence of a city\n\\end{enumerate}\n\n\\subsection{Dijkstra}\nAssign all nodes distance = $\\infty$\nStart at given node $s$. Assign all neighboring nodes their distance from $s$\nto $n$. Proceed to the lowest cost. Repeat. Whenever a better match is found,\nuse the new path instead. The result is a path from $s$ to $t$, or an MST.\\\n\n\\subsection{K-center}\n2-approximation. A form of clustering.\n\\begin{enumerate}\n \\item Pick arbitrary $i \\in V$\n \\item $S \\rightarrow \\left\\{i\\right\\}$\n \\item pick other node furthest away from k, add to $S$\n \\item repeat, furthest away from all previous picked nodes.\n\\end{enumerate}\n\n\\subsection{Kruskal's algorithm}\nAlways choose cheapest edge that does not include two pre-discovered nodes.\nReturns MST.\\\n\n\\subsection{Nearest addition}\n2-approximation\n\\begin{enumerate}\n \\item Find two closest cities\n \\item Traverse from i to j and back\n \\item Repeat, consider from each node\n\\end{enumerate}\n\n\\subsection{Prim's algorithm}\nStart at a given node $s$. Choose the cheapest edge. \nNow repeat, just that you consider $s$ and the node you added. Etc.\nNever choose an edge that leads to a pre-discovered vertex.\n\n\\subsection{List scheduling algorithm}\\label{listschedule}\n Whenever a machine is idle, assign it a job.\n This algorithm runs $\\leq 2\\times OPT$.\n\n \\begin{proof}\n Note that $OPT = \\frac{\\sum{p_{i}}}{m}$\n We know that the last job, $l$, starts at time $t$. Since we've always\n assigned jobs whenever we could, that means that all machines have \n so far been busy. Hence we know:\n $$\n t \\leq \\frac{\\sum{p_{j} - p_{l}}}{m} \\leq OPT - \\frac{p_{l}}{m}\n $$\n Since $OPT$ is lower bounded as the average work done by each machine.\n\n We can now bound the maximal finishing time: \n $$\n C_{\\max} \\leq t + p_{l} \\leq OPT + p_{l}\\times (1 - \\frac{1}{m}) \n \\leq (2 - \\frac{1}{m})\\times OPT\n $$\n \\end{proof}\n\n \\subsection{Longest processing time rule}\\label{srpt}\n As~\\nameref{listschedule}, just that you first sort the jobs in order of \n length; put the longest jobs first. This algorithm has approximation ratio\n $\\frac{4}{3} \\times OPT$\n\n \\subsection{Shortest remaining time, SRFT}\\label{sprt}\n In this scheduling algorithm, the process with the smallest amount of time\n remaining until completion is selected to execute.\n This algorithm is applied to~\\nameref{pre-emptive} schedules\n\n\\subsection{Knapsack DP}\n Fill out two arrays of $n \\times B$, where $B$ is the capacity of \n the knapsack. In row $i$, consider item $b_{i}$ against the capacity given\n in column $j$. If $j \\geq w_{i}$, compare the value of $A_{i,j}$ to \n $A_{i-1,j}$ and use whichever is greater. For all cases where $b_{i} \\geq\n j$, consider which is better: to use $b_{i} + A_{i-1,j-w_{i}}$,\n or retain the value above. The last equation is: to use current item + \n whatever we can fit in on the remaining weight, given by the line above.\n Make sure to simeoltaneously maintain a ``keep-array'' that gives 1 or zero,\n indicating whether or not you've used item $b_{i}$.\n\n In backtracking, start in the lower right of the keep array.\n Whenever you get a 1, include item $i$ and decrement $i$ and $j$ by one.\n Whenever you get a zero, decrement $i$ alone.\n\n The above mentioned runs in $O(n\\times W)$, where $W$ is the capasity of \n the knapsack. The algorithm is \\textit{exact}, but pseudopolynomial, since\n $W = \\sum{w_{i}}$, which in binary becomes $\\log_{2} W$, which hence \n becomes $O(\\log_{2}W^{n})$. An~\\nameref{FPTAS} exists, see book.\n", "meta": {"hexsha": "cf27fe45644ab0462f1f4c088562bf408f9621a7", "size": 4404, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/methods/algorithms.tex", "max_stars_repo_name": "andsild/NotusVitae", "max_stars_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/methods/algorithms.tex", "max_issues_repo_name": "andsild/NotusVitae", "max_issues_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/methods/algorithms.tex", "max_forks_repo_name": "andsild/NotusVitae", "max_forks_repo_head_hexsha": "8afc580cce2ece4f129c006af3879bb738bb2269", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.5471698113, "max_line_length": 82, "alphanum_fraction": 0.687102634, "num_tokens": 1223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339596505965, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6363456725847159}} {"text": "\n\\section{Model Description}\n\n\\subsection{Problem Statement}\n\n\\begin{figure}[!htb]\n\t\\centering\n\t\\includegraphics[width=0.49\\textwidth]{./Figures/Disegno1.eps}\n\t\\caption{Spacecraft with depleting mass and definition of frames and variables}\n\t\\label{fig:Spacecraft}\n\\end{figure}\n\nThe following derivation is a shortened version from Reference\\cite{Panicucci:2017fu}. To help define the problem, Figure~\\ref{fig:Spacecraft} is displayed. This problem involves a spacecraft consisting of a hub which is a rigid body and has a center of mass location labeled as point $B_c$. The hub has $M$ number of tanks and $N$ number of thrusters attached to it. The figure only shows one tank and one thruster but the analytical development is general. The $i_{\\text{th}}$ tank has a center of mass location labeled as $F_{c_i}$ and the $j_{\\text{th}}$ thruster is located at $B_j$. The body fixed reference frame $\\mathcal{B}$: $\\{\\bm{\\hat{b}}_1\\,, \\bm{\\hat{b}}_2\\,,\\bm{\\hat{b}}_3 \\}$ with origin $B$ can be oriented in any direction and point $B$ can be located anywhere fixed to the hub. This means that point $B$ and the center of mass location of the spacecraft, $C$, are not necessarily coincident. As a result, the vector $\\bm c$ defines the vector pointing from the body frame origin to the center of mass fo the spacecraft. The inertial reference frame $\\mathcal{N}$: $\\{\\bm{\\hat{n}}_1\\,, \\bm{\\hat{n}}_2\\,,\\bm{\\hat{n}}_3 \\}$ is centered at $N$ and is fixed in inertial space.\n\t\n\tThroughout this paper, vector calculus is used and the notation to define certain quantities needs to be introduced. A position vector, $ \\bm{r}_{C/N}$, is the vector pointing from $N$ to $C$. $\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}$ is the angular velocity of the $\\mathcal{B}$ frame with respect to the $\\mathcal{N}$ frame.\n\t$\\dot{\\bm{r}}$ denotes an inertial time derivate of vector $\\bm{r}$ and $\\bm{r}'$ defines a time derivate of $\\bm{r}$ with respect to the body frame. Using these definitions, the following describes the Reynolds transport theorem used in this formulation.\n\n\\subsection{Equations of Motion}\n\\subsubsection{Reynolds Transport Theorem and Continuity Equation}\nIn this section the main tool used for the development of the governing equations is presented and explained. The Reynolds transport theorem provides a basic tool to pass from a Lagrangian formulation, based on the analysis of particles moving in space, to an Eulerian one, which considers a fixed space volume where physical quantities are exchanged through the boundaries.\n\\begin{figure}[!htb]\n\t\\centering\n\t\\includegraphics[width=0.49\\textwidth]{./Figures/DefSyst.eps}\n\t\\caption{Division of the total system in spacecraft and exhausted gas. The control surface $\\mathcal{A}_{\\text{sc}}$ represents the exchanging surface between the two subsystems.}\n\t\\label{fig:DefSyst}\n\\end{figure}\nIn the present document, the Lagrangian system is labeled $\\textit{Body}$, the moving volume of the Eulerian approach is labeled $\\mathcal{V}_{\\text{sc}}$ and its surface $\\mathcal{A}_{\\text{sc}}$ are represented in Figure \\ref{fig:DefSyst}.\n\nBy using this notation, the Reynolds transport theorem affirms:\n\n\\begin{equation}\n\t\\frac{^\\mathcal{D}\\text{d}}{\\text{d}t}\\int_{\\text{Body}} \\rho\\,\\bm{f}\\text{d}\\mathcal{V} = \\frac{^\\mathcal{D} \\text{d}}{\\text{d}\\,t}\\int_{\\mathcal{V}_\\text{sc}} \\rho\\,\\bm{f}\\text{d}\\mathcal{V} + \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{f} \\left(\\bm{v}_{\\text{rel}}\\cdot \\hat{\\bm{n}}\\right)\\text{d}A\n\\end{equation}\n\\noindent\nwhere $\\bm{f}$ is a general vectorial quantity transported out from the control volume, $\\rho$ is the density of the infinitesimal mass $\\text{d}m$, $\\hat{\\bm{n}}$ the surface normal considered positive if exiting from the control volume, $\\mathcal{D}$ is a generic reference frame and $\\bm{v}_{\\text{rel}}$ is the relative velocity of the particles flowing out from the surface with respect to the control surface itself. This last quantity can be easily defined as $\\bm{v}_{\\text{rel}}(\\bm{x},t) =\\displaystyle\\frac{^\\mathcal{D}\\text{d}}{\\text{d}t}\\bm{r}_{M/B}(\\bm{x},t) - \\bm{v}_{\\text{surf}}(\\bm{x},t)$ where $\\displaystyle\\frac{^\\mathcal{D}\\text{d}}{\\text{d}t}\\bm{r}_{M/B}(\\bm{x},t)$ is the particles' velocity with respect to the $\\mathcal{D}$ frame and $\\bm{v}_{\\text{surf}}(\\bm{x},t)$ is the control surface velocity with respect to the $\\mathcal{D}$ reference frame.\\newline\nMoreover, if the control volume is fixed in the $\\mathcal{D}$ frame and a no deformable control volume is considered, the following relation is proved:\n\n\\begin{equation}\n\t\\frac{^\\mathcal{D}\\text{d}}{\\text{d}\\,t}\\int_{\\text{Body}} \\rho\\,\\bm{f}\\text{d}\\mathcal{V} = \\int_{\\mathcal{V}_\\text{sc}} \\frac{ ^\\mathcal{D}\\partial}{\\partial\\,t}\\left(\\rho\\,\\bm{f}\\right)\\text{d}\\mathcal{V}+ \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{f} \\left(\\frac{^\\mathcal{D}\\text{d}}{\\text{d}t}\\bm{r}_{M/B}\\right)\\cdot \\hat{\\bm{n}}\\,\\text{d}A\n\\end{equation}\n\nAn additional key equation that is used throughout the paper is the continuity equation. First, the continuity equation is gathered:\n\n\\begin{equation}\n\t\\frac{^\\mathcal{B}\\text{d}}{\\text{d}t}\\int_{\\text{Body}} \\rho\\, \\text{d}m = \\frac{^\\mathcal{B}\\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_\\text{sc}} \\rho\\, \\text{d}\\mathcal{V} + \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B} \\cdot \\hat{\\bm{n}}\\,\\text{d}\\mathcal{A}=0\n\\end{equation}\n\\noindent\nThus, by defining $\\dot{m}_\\text{sc} = \\frac{^\\mathcal{B}\\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_\\text{sc}} \\rho\\, \\text{d}\\mathcal{V}$:\n\n\\begin{equation}\\label{eq:cambVar}\n\t\\dot{m}_\\text{sc} = - \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B} \\cdot \\hat{\\bm{n}}\\,\\text{d}\\mathcal{A} \\qquad \\Rightarrow \\qquad \\text{d}\\dot{m} = \\rho\\,\\bm{r}'_{M/B} \\cdot \\hat{\\bm{n}}\\,\\text{d}\\mathcal{A}\n\\end{equation}\nThis definition will be used in the derivation of the EOMs. The translational EOM is developed in the following section.\n\n\\subsubsection{Translational Equation of Motion}\nThe derivation of the translational EOM must begins considering Newton's law for a closed system:\n\n\\begin{equation}\\label{eq:eq3}\n\t\\frac{^{\\mathcal{N}}\\text{d}}{\\text{d}\\,t}\\int_{\\text{Body}}\\dot{\\bm{r}}_{M/N}\\,\t\\text{d}m= \\bm{F}_{\\text{ext}}\n\\end{equation}\nwhere $\\dot{\\bm{r}}_{M/N}$ is the velocity of the particle at the $M$ point expressed with respect to the inertial reference frame and $\\bm{F}_{\\text{ext}}$ is the sum of the external forces experienced by the body.\n\nAs the total mass of the system is constant, the differentiation operator is brought inside the integration: \n\\begin{equation}\\label{eq:eq4}\n\t\\frac{^{\\mathcal{N}}\\text{d}}{\\text{d}\\,t}\\int_{\\text{Body}}\\dot{\\bm{r}}_{M/N}\\,\t\\text{d}m= \\int_{\\text{Body}}\\ddot{\\bm{r}}_{M/N}\\,\\text{d}m\n\\end{equation}\nThe acceleration of the origin of the $\\mathcal{B}$ frame is expressed as:\n\n\\begin{equation}\\label{eq:RcRbacc}\n\t\\ddot{\\bm{r}}_{M/N} = \\ddot{\\bm{r}}_{B/N} + \\ddot{\\bm{r}}_{M/B}\n\\end{equation}\nBy using the kinematic transport theorem, the expression of $\\ddot{\\bm{r}}_{M/B}$ is found:\n\n\\begin{equation}\\label{eq:rMB_ddot}\n\t\\dot{\\bm{r}}_{M/B} = \\bm{r}'_{M/B} + \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\times\\bm{r}_{M/B}\n\\end{equation}\n\\begin{equation}\\label{eq:rMB_dot}\n\t\\ddot{\\bm{r}}_{M/B} = \\bm{r}''_{M/B}+2\\,\\bm{\\omega}_{\\cal{B}/\\cal{N}}\\times\n\t\\bm{r}'_{M/B} + \\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\times \\bm{r}_{M/B} + \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\times\\left(\\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{B/M} \\right)\n\\end{equation}\n\nA Lagrangian formulation of the linear momentum equation is deduced by using Equations \\eqref{eq:eq3}, \\eqref{eq:eq4}, \\eqref{eq:RcRbacc} and \\eqref{eq:rMB_dot}: \n\n\\begin{multline}\\label{eq:eq11}\n\t\\int_{\\text{Body}}\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}m +\\\\+\n\t2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\times\\int_{\\text{Body}}\\bm{r}'_{M/B}\\text{d}m + \\int_{\\text{Body}}\\bm{r}''_{M/B}\\text{d}m = \\bm{F}_{\\text{ext}}\n\\end{multline}\n\nThe system mass is constant, therefore the derivative operator can be applied after the integration. This yields:\n\n\\begin{equation}\t\n\t\\int_{\\text{Body}}\\bm{r}'_{M/B}\\text{d}m = \\frac{^{\\mathcal{B}}\\text{d}}{\\text{d}t}\\int_{\\text{Body}}\\bm{r}_{M/B}\\text{d}m\n\\end{equation}\n\\begin{equation}\n\t\\int_{\\text{Body}}\\bm{r}''_{M/B}\\text{d}m = \\frac{^{\\mathcal{B}}\\text{d}^2}{\\text{d}t^2}\\int_{\\text{Body}}\\bm{r}_{M/B}\\text{d}m\n\\end{equation}\n\nBy using the Reynolds transport theorem, the two previous equations can be expressed in a space fixed volume, shown in Figure \\ref{fig:DefSyst}. Performing this conversion results in the following equations:\n\n\\begin{equation}\n\t\\frac{^{\\mathcal{B}}\\text{d}}{\\text{d}t}\\int_{\\text{Body}}\\bm{r}_{M/B} \\,\\text{d}m = \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B}\\,\\text{d}\\mathcal{V} + \\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}}\\,\\bm{r}_{M/B}\\text{d}A\n\\end{equation}\n\\begin{multline}\n\t\\frac{^{\\mathcal{B}}\\text{d}^2}{\\text{d}t^2}\\int_{\\text{Body}}\\bm{r}_{M/B}\\, \\text{d}m = \\frac{^{\\mathcal{B}} \\text{d}^2}{\\text{d}t^2}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\,\\text{d}\\mathcal{V} +\\\\+ \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t} \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}_{M/B}\\,\\text{d}A + \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}'_{M/B}\\,\\text{d}A\n\\end{multline}\n\\noindent\nwhere $\\bm{v}_{\\text{rel}} = \\bm{r}' _{M/B}$ because point $B$ is fixed with respect to the spacecraft. Equation \\eqref{eq:eq11} is re-organized by using the previous relations in order to convert it to an Eulerian approach, i.e. based on a volume-based derivation:\n\n\\begin{multline}\n\t\\int_{\\text{Body}}\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}m +\\\\+\n\t2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\times\\left(\\,\\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B}\\,\\text{d}\\mathcal{V} + \\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}}\\,\\bm{r}_{M/B}\\text{d}A\\right) \\;+ \t \\frac{^{\\mathcal{B}} \\text{d}^2}{\\text{d}t^2}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\,\\text{d}\\mathcal{V} +\\\\+ \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t} \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}_{M/B}\\,\\text{d}A + \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}'_{M/B}\\,\\text{d}A \n\t= \\bm{F}_{\\text{ext}}\n\\end{multline} \n\nAs explained in previous work, if all of the mass is contained in the control volume at the initial time, then a particular relation results because no mass is outside the control volume at $t=0$ and the dynamic quantities will be transported out during the integration. This relationship is quantified in the following equation:\n\n\\begin{multline}\n\t\\bm{F}_{\\text{ext}} - \\int_{\\text{Body}}\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}m \n\t= \\int_{\\mathcal{V}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{vol}}+\\\\ +\\int_{\\mathcal{A}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{surf}} - \\int_{\\mathcal{V}_{\\text{sc}}}\\rho\\,\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}\\mathcal{V}\n\\end{multline} \n\\noindent\nwhere the forces are divided into volumetric forces and the forces applied on the spacecraft surface. Rearranging this result, replacing the definition of $\\bm{F}_{\\text{ext}}$, and isolating the forces to the right hand side of the equation yields:\n\\begin{multline}\\label{eq:ComplTranslEOM}\n\t\\int_{\\mathcal{V}_{\\text{sc}}}\\rho\\,\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}\\mathcal{V} +\\\\+\n\t2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\times\\left(\\,\\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B}\\,\\text{d}\\mathcal{V}\\; + \\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}}\\,\\bm{r}_{M/B}\\text{d}A\\right) \\;+ \t \\frac{^{\\mathcal{B}} \\text{d}^2}{\\text{d}t^2}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\,\\text{d}\\mathcal{V} +\\\\+ \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t} \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}_{M/B}\\,\\text{d}A + \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}'_{M/B}\\,\\text{d}A \n\t= \\int_{\\mathcal{V}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{vol}} +\\int_{\\mathcal{A}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{surf}}\n\\end{multline}\n\nOne goal for this paper is to develop the EOMs of a spacecraft with depleting mass without the necessity of continuing to track the depleted mass once it has left the spacecraft. One aspect of achieving this goal, is to define the center of mass of the spacecraft with respect to point $B$, including the remaining fuel while disregarding the spent fuel. This variable, $\\bm{c} = \\bm{r}_{C/B}$, is defined as:\n\\begin{equation}\\label{eq:def_rCB}\n\t\\bm{c}=\\frac{m_{\\text{hub}}\\,\\bm{r}_{Bc/B} + \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\,\\bm{r}_{Fc_i/B}}{m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}}\n\\end{equation}\nwhere $m_{\\text{hub}}$ is the mass of the hub, $m_{\\text{fuel}_i}$ is the i-th tank's fuel mass and $\\bm{r}_{Fc_i/B}$ is the position of the center of mass of the i-th tank's fuel.\nIn order to infer the influence of the mass variation in the EOMs equation of motion the first and second time derivatives with respect to the body frame of $\\bm c$ is defined:\n\n\\begin{multline}\\label{eq:eq17}\n\t\\bm{c}' = \\frac{\\sum_{i=1}^{M}\\left(\\dot{m}_{\\text{fuel}_i}\\bm{r}_{Fc_i/B} + m_{\\text{fuel}_i}\\bm{r}'_{Fc_i/B}\\right)}{m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}} + \\vspace{5pt}\\\\\n\t- \\frac{\\left(\\sum_{i=1}^{M}\\dot{m}_{\\text{fuel}_i}\\right)\\left(m_{\\text{hub}}\\,\\bm{r}_{Bc/B} + \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\,\\bm{r}_{Fc_i/B}\\right)}{\\left(m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\right)^2}\n\\end{multline}\n\n\\begin{multline}\\label{eq:eq18}\n\t\\bm{c}'' = \\frac{\\sum_{i=1}^{M}\\left(\\ddot{m}_{\\text{fuel}_i}\\bm{r}_{Fc_i/B} +2\\,\\dot{m}_{\\text{fuel}_i}\\bm{r}'_{Fc_i/B}+ m_{\\text{fuel}_i}\\bm{r}''_{Fc_i/B}\\right)}{m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}} + \\vspace{5pt}\\\\\n\t- \\frac{\\left(\\sum_{i=1}^{M}\\ddot{m}_{\\text{fuel}_i}\\right)\\left(m_{\\text{hub}}\\,\\bm{r}_{Bc/B} + \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\,\\bm{r}_{Fc_i/B}\\right) }{\\left(m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\right)^2} +\\\\-\n\t\\frac{2\\,\\left(\\sum_{i=1}^{M}\\dot{m}_{\\text{fuel}_i}\\right)\\sum_{i=1}^{M}\\left(\\dot{m}_{\\text{fuel}_i}\\bm{r}_{Fc_i/B} + m_{\\text{fuel}_i}\\bm{r}'_{Fc_i/B}\\right) }{\\left(m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\right)^2} +\\\\\n\t+ \\frac{2\\,\\left(\\sum_{i=1}^{M}\\dot{m}_{\\text{fuel}_i}\\right)^2\\left(m_{\\text{hub}}\\bm{r}_{Bc/B} +\\sum_{i=1}^{M} m_{\\text{fuel}_i}\\bm{r}_{Fc_i/B}\\right)}{\\left(m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}\\right)^3}\n\\end{multline}\nUsing these definitions of $\\bm c$, the translational EOM can be simplified. Additionally, some assumptions need to be defined to further simplify the translational EOM. The hub is assumed to be rigid, therefore deformations are not considered. The mass flow within the tanks and the thrusters is assumed to be a second order effect and ignored for this paper. The particles are assumed to be accelerated instantaneously from the spacecraft velocity, $\\dot{\\bm{r}}_{B/N}$, to the exhausted velocity $\\bm{v}_{\\text{exh}}$ at the nozzle. And the exhausted velocity $\\bm{v}_{\\text{exh}}$ is considered constant and parallel to the nozzle's normal $\\hat{\\bm{n}}$.\n\nThe first integral in Eqn. \\eqref{eq:ComplTranslEOM} is computed using the fact that $\\bm{r}_{M/B} = \\bm{c} + \\bm{r}_{M/C}$ and the result is shown in the following equation:\n\n\\begin{multline}\\label{eq:eq19}\n\t\\int_{\\mathcal{V}_{\\text{sc}}}\\rho\\,\\left(\\ddot{\\bm{r}}_{B/N} + \\dot{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{r}_{M/B} + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{r}_{M/B}\\right)\\right)\\text{d}\\mathcal{V} = \\\\=\n\tm_{\\text{sc}} \\,\\bm{\\ddot{r}}_{B/N} + m_{\\text{sc}}\\,\\bm{\\dot{\\omega}}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{c} + m_{\\text{sc}}\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{c}\\right) \n\\end{multline}\n\\noindent \nwhere $m_{\\text{sc}}=m_{\\text{hub}}+ \\sum_{i=1}^{M}m_{\\text{fuel}_i}$ is the instantaneous mass of the spacecraft. The second and fourth integrals are computed and yield:\n\n\\begin{equation}\\label{eq:eq20}\n\t\\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B}\\,\\text{d}\\mathcal{V} = \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t}\\left(m_{\\text{sc}}\\,\\bm{c}\\right) = m_{\\text{sc}}\\,\\bm{c}' + \\dot{m}_{\\text{fuel}}\\bm{c}\n\\end{equation}\n\\begin{equation}\\label{eq:eq21}\n\t\\frac{^{\\mathcal{B}} \\text{d}^2}{\\text{d}t^2}\\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\,\\text{d}\\mathcal{V} = \\frac{^{\\mathcal{B}} \\text{d}^2}{\\text{d}t^2} \\left(m_{\\text{sc}}\\,\\bm{c}\\right) = m_{\\text{sc}}\\,\\bm{c}'' + 2\\,\\dot{m}_{\\text{fuel}}\\,\\bm{c}' + \\ddot{m}_{\\text{fuel}}\\bm{c}\n\\end{equation}\nwhere $\\dot{m}_{\\text{fuel}} = \\sum_{i=1}^{M}\\dot{m}_{\\text{fuel}_i}$ and $\\ddot{m}_{\\text{fuel}} = \\sum_{i=1}^{M}\\ddot{m}_{\\text{fuel}_i}$.\n\nIn order to find the term calculated on the reference surface seen in the third, fifth and sixth integrals, it is convenient to separate the integrals on the surface of each nozzle. Moreover, as the fuel's properties are flowing out of a surface plane, it is convenient to consider that $\\bm{r}_{M/B} = \\bm{r}_{M/Fc_j}+ \\bm{r}_{Fc_j/B} $ where $Fc_i$ is the area's geometric center. Finally, an appropriate variable transformation is given in Eqn. \\eqref{eq:cambVar}. Performing these calculations on the third integral results in:\n\\begin{equation}\\label{eq:eq22}\n\t\\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}}\\,\\bm{r}_{M/B}\\,\\text{d}A =-\\sum_{j=1}^{N} \\int_{\\dot{m}_{\\text{noz}_j}}\\left(\\bm{r}_{M/N_j}+ \\bm{r}_{N_j/B}\\right)\\text{d}\\dot{m} =-\\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j}\\bm{r}_{N_j/B}\n\\end{equation}\n\\noindent\nwhere the first part of the integral is null because of barycenter definition and $\\dot{m}_{\\text{noz}_j}$ is the mass flow of the $j-$th nozzle. The fifth integral in Eq.~\\eqref{eq:ComplTranslEOM} yields:\n\\begin{equation}\\label{eq:eq23}\n\t\\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t} \\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}_{M/B}\\,\\text{d}A = \\frac{^{\\mathcal{B}} \\text{d}}{\\text{d}t} \\left(-\\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j}\\bm{r}_{N_j/B}\\right) = -\\sum_{j=1}^{N}\\ddot{m}_{\\text{noz}_j}\\bm{r}_{N_j/B}\n\\end{equation}\n\\noindent\nUsing the assumption introduced earlier in this section, $\\bm{r}'_{M/B} = \\bm{v}_{\\text{exh}}$, the sixth is integral is found and can be in the following equation:\n\n\\begin{equation}\\label{eq:eq24}\n\t\\int_{\\mathcal{A}_\\text{sc}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{\\hat{n}} \\,\\bm{r}'_{M/B} \\,\\text{d}A =\\sum_{j=1}^{N}\\int_{\\mathcal{A}_{\\text{noz}_j}}\\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{n}\\,\\bm{r}'_{M/B}\\text{d}A = -\\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j}\\,\\bm{v}_{\\text{exh}_j}\n\\end{equation}\n\\noindent\nwhere $\\bm{v}_{\\text{exh}_j}$ is the exhausted velocity of a particle exiting from the $j-$th nozzle.\n\nThe two integrals on the right-hand-side of Eq.~\\eqref{eq:ComplTranslEOM} depends on the force model chosen. Therefore, to not lose generality, the resulting surface integral due to the pressure jump between the nozzle and the environment is the only term that is analytically computed seen in the following equation:\n\n\\begin{equation}\\label{eq:eq25}\n\t\\int_{\\mathcal{V}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{vol}} +\\int_{\\mathcal{A}_{\\text{sc}}} \\text{d}\\bm{F}_{\\text{surf}} = \\bm{F}_{\\text{ext, vol}} + \\bm{F}_{\\text{ext, surf}} + \\sum_{j=1}^{N}\\frac{\\bm{v}_{\\text{exh}_j}}{v_{\\text{exh}_j}}\\,A_{\\text{noz}_j}\\,(p_{\\text{exh}_j} - p_{\\text{atm}})\n\\end{equation}\n\\noindent\nwhere $\\bm{F}_{\\text{ext, vol}}$ are the external forces acting on the control volume, $\\bm{F}_{\\text{ext, surf}}$ are the external forces accelerating the control surface, $p_{\\text{exh}_j}$ is the particles' exhausted pressure at the $j-$th nozzle and $p_{\\text{atm}}$ is the atmospheric pressure.\n\nFinally, Equation \\eqref{eq:ComplTranslEOM} is rewritten considering the nozzles' geometry and fluid properties by using Equations~\\eqref{eq:eq20}-~\\eqref{eq:eq25}:\n\\begin{multline}\n\tm_{\\text{sc}} \\bm{\\ddot{r}}_{B/N} + m_{\\text{sc}}\\,\\bm{\\dot{\\omega}}_{\\mathcal{B}/\\mathcal{N}} \\times\\bm{c} + m_{\\text{sc}}\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times\\left(\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\times\\bm{c}\\right) + m_{\\text{sc}}\\,\\bm{c}'' + 2\\,\\dot{m}_{\\text{fuel}}\\,\\bm{c}' +\\\\+ \\ddot{m}_{\\text{fuel}}\\bm{c} + 2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times \\bigg(m_{\\text{sc}}\\,\\bm{c}'+ \\dot{m}_{\\text{fuel}}\\bm{c} - \\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j}\\bm{r}_{N_j/B} \\bigg) - \\sum_{j=1}^{N}\\ddot{m}_{\\text{noz}_j}\\bm{r}_{N_j/B}\t+\\\\-\n\t\\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j}\\,\\bm{v}_{\\text{exh}_j} = \\bm{F}_{\\text{ext, vol}} + \\bm{F}_{\\text{ext, surf}} + \\sum_{j=1}^{N}\\frac{\\bm{v}_{\\text{exh}_j}}{v_{\\text{exh}_j}}\\,A_{\\text{noz}_j}\\,(p_{\\text{exh}_j} - p_{\\text{atm}})\n\\end{multline}\nThe previous equation is modified by defining the following quantity:\n\n\\begin{equation}\\label{eq:IspRel}\n\t\\bm{F}_{\\text{thr}_j}= \\bm{v}_{\\text{exh}_j}\\,\\left(\\frac{A_{\\text{noz}_j}}{v_{\\text{exh}_j}}\\,(p_{\\text{exh}_j} - p_{\\text{atm}}) + \\dot{m}_{\\text{noz}_j}\\right) = I_{\\text{sp}_j}\\,g_0\\,\\dot{m}_{\\text{noz}_j}\\frac{\\bm{v}_{\\text{exh}_j}}{v_{\\text{exh}_j}}\n\\end{equation}\n\nFor further simplicity, the cross product is substituted with the associated skew symmetric matrix, and the translational equation is written in a more compact form:\n\n\\begin{multline}\\label{eq:eq29}\n\t\\ddot{\\bm{r}}_{B/N} + \\left[\\tilde{\\bm{c}}\\right]^T\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} = \\frac{\\bm{F}_{\\text{thr}}}{m_{\\text{sc}}} - 2\\,\\frac{\\dot{m}_{\\text{fuel}}}{m_{\\text{sc}}}\\,\\left(c' + \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\\times\\bm{c}\\right) - \\bm{c}''+2\\,\\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]^T\n\t\\bm{c}'\\\\- \\ddot{m}_{\\text{fuel}}\\,\\bm{c} + \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]^T \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\bm{c} +\\frac{2}{m_{\\text{sc}}}\\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j} \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\bm{r}_{N_j/B}\n\t\\\\+ \\frac{\t1}{m_{\\text{sc}}} \\sum_{j=1}^{N}\\ddot{m}_{\\text{noz}_j}\\bm{r}_{Fc_j/B} +\\frac{\\bm{F}_{\\text{ext, vol}}}{m_{\\text{sc}}} + \\frac{\\bm{F}_{\\text{ext, surf}}}{m_{\\text{sc}}} \n\\end{multline}\nMultiplying my \n\n\\begin{multline}\\label{eq:eq30}\nm_{\\text{sc}} \\ddot{\\bm{r}}_{B/N} - m_{\\text{sc}} \\left[\\tilde{\\bm{c}}\\right]\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} = \\bm{F}_{\\text{thr}} - 2\\,\\dot{m}_{\\text{fuel}} \\left(c' + \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\\times\\bm{c}\\right) - m_{\\text{sc}} \\bm{c}''-2 m_{\\text{sc}} \\,\\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\n\\bm{c}'\\\\- m_{\\text{sc}} \\ddot{m}_{\\text{fuel}}\\,\\bm{c} - m_{\\text{sc}} \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\bm{c} +2 \\sum_{j=1}^{N}\\dot{m}_{\\text{noz}_j} \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\bm{r}_{N_j/B}\n\\\\+ \\sum_{j=1}^{N}\\ddot{m}_{\\text{noz}_j}\\bm{r}_{Fc_j/B} +\\bm{F}_{\\text{ext, vol}} + \\bm{F}_{\\text{ext, surf}}\n\\end{multline}\nThis EOM is the translational equation for an open system subjected to external forces $\\bm{F}_{\\text{ext, vol}} \\text{ and } \\bm{F}_{\\text{ext, surf}}$ and thrust $\\bm{F}_{\\text{thr}}=\\sum_{j=1}^{N}\\bm{F}_{\\text{thr}_j}$ due to mass depletion of the spacecraft, represented in Figure \\ref{fig:Spacecraft}.\nFrom this equation, it can be deduced that the variation of the mass inside the spacecraft directly impacts the position of the satellite with respect to the origin as the body fixed point $B$ changes its state of motion according to the variation of the tanks' linear inertia. In the next section, the rotational EOM for the spacecraft is developed.\n\n\\subsubsection{Rotational Equation of Motion}\nThe goal of this section is to develop the EOM associated with attitude dynamics of a spacecraft with depleting mass due to thrusters pulling mass from fuel tanks. Beginning from Newton's equation:\n\n\\begin{equation}\n\t\\ddot{\\bm{r}}_{M/N}\\,\\text{\\text{d}}m=\\text{d}\\bm{F} \\quad \\Rightarrow \\quad\n\t\\bm{r}_{M/N} \\times \\ddot{\\bm{r}}_{M/N}\\,\\text{d}m =\\bm{r}_{M/N} \\times \\text{d}\\bm{F}\n\\end{equation}\nand performing an integration over the system:\n\n\\begin{equation}\n\t\\int_{\\text{Body}} \\bm{r}_{M/N} \\times \\ddot{\\bm{r}}_{M/N}\\,\\text{d}m = \\int_{\\text{Body}}\\bm{r}_{M/N} \\times \\text{d}\\bm{F}\n\\end{equation}\nThe term on the left-hand side of the previous equation is manipulated in order to define the momentum about point $B$. This manipulation can be seen in the following equation:\n\n\\begin{multline}\\label{eq:eq32}\n\t\\int_{\\text{Body}} \\rho\\,\\bm{r}_{M/N} \\times \\ddot{\\bm{r}}_{M/N}\\,\\text{d}\\mathcal{V} = \\int_{\\text{Body}} \\rho\\,\\bm{r}_{M/B} \\times \\ddot{\\bm{r}}_{M/B}\\,\\text{d}\\mathcal{V} +\\\\ + \\int_{\\text{Body}} \\rho\\,\\bm{r}_{B/N} \\times \\ddot{\\bm{r}}_{M/N}\\,\\text{d}\\mathcal{V} + \\int_{\\text{Body}} \\rho\\,\\bm{r}_{M/B} \\times \\ddot{\\bm{r}}_{B/N}\\,\\text{d}\\mathcal{V} = \\int_{\\text{Body}}\\bm{r}_{M/N} \\times \\text{d}\\bm{F}\n\\end{multline}\nKnowing that $\\ddot{\\bm{r}}_{M/N}\\,\\text{\\text{d}}m=\\text{d}\\bm{F}$, the torque caused by the forces acting on the body is easily defined:\n\n\\begin{equation}\n\t\\int_{\\text{Body}}\\bm{r}_{M/N} \\times \\text{d}\\bm{F}-\\int_{\\text{Body}} \\rho\\,\\bm{r}_{B/N} \\times \\ddot{\\bm{r}}_{M/N}\\,\\text{d}\\mathcal{V} \n\t= \\int_{\\text{Body}} \\left(\\bm{r}_{M/N}-\\bm{r}_{B/N}\\right) \\times \\text{d}\\bm{F} = \\bm{L}_{B}\n\\end{equation}\n\\noindent\nwhere $\\bm{L}_{B}$ is the torque with respect to the body-fixed point $B$.\n\nAs the mass of the system is constant, the derivative of the angular momentum about point $B$ is inferred from Equation~\\eqref{eq:eq32} due to a property of the cross product and the previously explained Reynold's transport theorem:\n\n\\begin{multline}\n\t\\int_{\\text{Body}} \\rho\\,\\bm{r}_{M/B} \\times \\ddot{\\bm{r}}_{M/B}\\,\\text{d}\\mathcal{V} = \\frac{^\\mathcal{N}\\text{d}}{\\text{d}t} \\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\times \\dot{\\bm{r}}_{M/B}\\,\\text{d}\\mathcal{V} + \\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B} \\cdot\\bm{\\hat{n}} \\left(\\bm{r}_{M/B} \\times \\dot{\\bm{r}}_{M/B}\\right)\\text{d}A\n\\end{multline} \n\nMoreover, similar to the translational equation, if all the mass of the system is assumed to be contained inside the control volume at the initial time, the following relationship results:\n\n\\begin{multline}\n\t\\int_{\\text{Body}} \\rho\\,\\bm{r}_{M/B} \\times \\ddot{\\bm{r}}_{B/N}\\,\\text{d}\\mathcal{V} - \\bm{L}_{B}= \\int_{\\mathcal{V}_{\\text{sc}}} \\rho\\,\\bm{r}_{M/B} \\times \\ddot{\\bm{r}}_{B/N}\\,\\text{d}\\mathcal{V} - \\int_{\\mathcal{V}_{\\text{sc}}}\\bm{r}_{M/B} \\times \\text{d}\\bm{F}_{\\text{vol}} +\\\\- \\int_{\\mathcal{A}_{\\text{sc}}} \\bm{r}_{M/B} \\times\\text{d}\\bm{F}_{\\text{surf}} = {m}_{\\text{sc}} \\bm{c} \\times \\ddot{\\bm{r}}_{B/N}-\\bm{L}_{B,\\,\\text{vol}}-\\bm{L}_{B,\\,\\text{surf}}\n\\end{multline}\n\\noindent\nwhere $\\bm{L}_{B,\\,\\text{vol}}\\text{ and }\\bm{L}_{B,\\,\\text{surf}}$ are the torques caused by the volume and surface forces about point $B$. The general rotational equation for a control volume in a rotating reference frame is reorganized:\n\n\\begin{multline}\\label{eq:eq36}\n\t\\dot{\\bm{H}}_{\\text{sc, }B} + \\int_{\\mathcal{A}_{\\text{sc}}} \\rho\\,\\bm{r}'_{M/B} \\cdot\\bm{\\hat{n}} \\left(\\bm{r}_{M/B} \\times \\dot{\\bm{r}}_{M/B}\\right)\\text{d}A +{m}_{\\text{sc}} \\bm{c} \\times \\ddot{\\bm{r}}_{B/N}=\\bm{L}_{B,\\,\\text{vol}}+\\bm{L}_{B,\\,\\text{surf}}\n\\end{multline}\n\nTo perform the inertial derivative of $\\bm{H}_{\\text{sc, }B}$, first the definition of $\\bm{H}_{\\text{sc, }B}$ is defined:\n\\begin{multline}\n\t\\bm{H}_{\\text{sc, }B}= \\left[I_{\\text{hub},\\,B_c}\\right] \\bm{\\omega}_{\\cal{B}/\\cal{N}} + \\bm{r}_{Bc/B} \\times m_\\text{hub} \\,\\dot{\\bm{r}}_{Bc/B} +\\\\ +\\sum_{i=1}^{M}\\left(\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right]\\, \\bm{\\omega}_{\\cal{B}/\\cal{N}} + \\bm{r}_{Fc_i/B} \\times m_{\\text{fuel}_i}\\, \\dot{\\bm{r}}_{Fc_i/B}\\right)\n\\end{multline}\nwhere $\\left[I_{\\text{hub},\\,B_c}\\right]$ is the hub's inertia about its center of mass, $B_c$, and $\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right]$ is the $i-$th tank's inertia about its center of mass, $Fc_i$. Furthermore, an analytical expression of mass depletion in the rotational motion is deduced:\n\\begin{multline}\\label{eq:HscB_dot}\n\t\\dot{\\bm{H}}_{\\text{sc, }B} = \\left[I_{\\text{hub},\\,Bc}\\right]\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\left(\\left[I_{\\text{hub},\\,Bc}\\right]\\bm{\\omega}_{\\cal{B}/\\cal{N}}\\right)+ \\bm{r}_{Bc/B} \\times m_\\text{hub}\\,\\ddot{\\bm{r}}_{Bc/B} +\\\\\n\t+ \\sum_{i=1}^{M}\\left(\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right] \\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\left(\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right]\\, \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\right) \n\t+ \\bm{r}_{Fc_i/B} \\times m_{\\text{fuel}_i}\\, \\ddot{\\bm{r}}_{Fc_i/B}+\\right.\\\\ \n\t\\left.+ \\bm{r}_{Fc_i/B} \\times \\dot{m}_{\\text{fuel}_i} \\,\\dot{\\bm{r}}_{Fc_i/B} + \\left[I_{\\text{fuel}_i,\\, Fc_i}\\right]' \\, \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\right)\n\\end{multline}\n\nIt should be noted here that any relative motion of particles inside the fuel tanks of the spacecraft has been neglected and, as a consequence, the effects both of the Coriolis' acceleration and of the whirling motion of the fuel on the spacecraft dynamics have not been considered.\n\nAdditionally, the inertial time derivate of the vectors $\\bm{r}_{Bc/B}$ and $\\bm{r}_{Fc_i/B}$ is computed using the transport theorem between the two reference frames given in Equations \\eqref{eq:rMB_ddot} and \\eqref{eq:rMB_dot} and, considering that the point $B_c$ is fixed in the $\\mathcal{B}$ frame, Equation \\eqref{eq:HscB_dot} is rewritten:\n\n\\begin{multline}\\label{eq:HscB_complic}\n\t\\dot{\\bm{H}}_{\\text{sc, }B} = \\left[I_{\\text{hub},\\,Bc}\\right]\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\left(\\left[I_{\\text{hub},\\,Bc}\\right]\\bm{\\omega}_{\\cal{B}/\\cal{N}}\\right) \n\t+ \\bm{r}_{Bc/B} \\times m_\\text{hub}\\,\\left(\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{Bc/B} +\\right.\\\\\n\t+\\left. \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\left(\\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{Bc/B}\\right) \\right) \n\t+\\sum_{i=1}^{M} \\bigl(\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right] \\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\left(\\left[I_{\\text{fuel}_i,\\,Fc_i}\\right] \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\right) +\\\\ \n\t+ \\bm{r}_{Fc_i/B} \\times m_{\\text{fuel}_i}\\, \\bigl( \\bm{r}''_{Fc_i/B} + 2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times \\bm{r}'_{Fc_i/B}\n\t+ \\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{Fc_i/B} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times\n\t\\left(\\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{Fc_i/B}\\right) \\bigr)+\\\\ \n\t+\n\t\\bm{r}_{Fc_i/B} \\times \\dot{m}_{\\text{fuel}_i} \\,\\left(\\bm{r}'_{Fc_i/B}+\\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{Fc_i/B}\\right) \n\t+ \\left[I_{\\text{fuel,}_i,\\,Fc_i}\\right]' \\, \\bm{\\omega}_{\\cal{B}/\\cal{N}}\\bigr)\n\\end{multline}\n\nIn order to simplify compact Equation \\eqref{eq:HscB_complic} the following inertia matrices are defined using the skew symmetric matrix to replace the cross product:\n\\begin{equation}\\label{eq:eq44}\n\t\\left[I_{\\text{hub},\\,B}\\right]= \\left[I_{\\text{hub},\\,Bc}\\right] + m_\\text{hub}\\, \\left[\\tilde{\\bm{r}}_{Bc/B}\\right]\\left[\\tilde{\\bm{r}}_{Bc/B}\\right]^T \n\\end{equation}\n\\begin{equation}\\label{eq:eq45}\n\t\\left[I_{\\text{fuel}_i,\\,B}\\right]= \\left[I_{\\text{fuel}_i,\\,Fc_i}\\right] + m_{\\text{fuel}_i}\\, \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]^T\n\\end{equation}\n\\begin{equation}\\label{eq:eq46}\n\t\\left[I_{\\text{sc, }B}\\right]= \\left[I_{\\text{hub},\\,B}\\right] + \\sum_{i=1}^{M}\\left[I_{\\text{fuel}_i,\\,B}\\right]\n\\end{equation}\nMoreover, using the Jacobi identity for the cross product $\n\\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )+\\mathbf {b} \\times (\\mathbf {c} \\times \\mathbf {a} )+\\mathbf {c} \\times (\\mathbf {a} \\times \\mathbf {b} )=\\mathbf {0}$, the body relative time derivative of the fuel inertia in the $\\mathcal{B}$ reference frame is introduced:\n\n\\begin{multline}\\label{eq:eq47}\n\t\\bm{r}_{Fc_i/B} \\times \\left(2\\,\\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times \\bm{r}'_{Fc_i/B}\\right) =-\\bm{r}_{Fc_i/B} \\times \\left( \\bm{r}'_{Fc_i/B} \\times \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\right) \n\t+\\\\-\\bm{r}'_{Fc_i/B} \\times \\left( \\bm{r}_{Fc_i/B} \\times \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}}\\right) + \\bm{\\omega}_{\\mathcal{B}/\\mathcal{N}} \\times \\left(\\bm{r}_{Fc_i/B}\\times \\bm{r}'_{Fc_i/B}\\right)\n\\end{multline}\n\\begin{multline}\\label{eq:eq48}\n\t\\left[I_{\\text{fuel}_i,\\,B}\\right]'= \\left[I_{\\text{fuel}_i,\\,Fc_i}\\right] + \\dot{m}_{\\text{fuel}_i}\\, \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]^T +\\\\+ m_{\\text{fuel}_i}\\, \\left(\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]\\left[\\tilde{\\bm{r}}'_{Fc_i/B}\\right]^T+ \\left[\\tilde{\\bm{r}}'_{Fc_i/B}\\right]\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]^T\\right)\n\\end{multline}\nBy substituting Equations \\eqref{eq:eq44}-\\eqref{eq:eq48} Eq.\\eqref{eq:HscB_complic} is simplified to:\n\\begin{multline}\\label{eq:eq49}\n\t\\dot{\\bm{H}}_{\\text{sc, }B} = \\left[I_{\\text{sc, }B}\\right]\\,\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\\left[I_{\\text{sc, }B}\\right]\\,\\bm{\\omega}_{\\cal{B}/\\cal{N}} + \\sum_{i=1}^{M}\\left(m_{\\text{fuel}_i}\\,\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right] \\bm{r}''_{Fc_i/B} +\\right. \\\\\n\t\\left.+ \\dot{m}_{\\text{fuel}_i} \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]\\bm{r}'_{Fc_i/B}+ \\left[I_{\\text{fuel}_i,\\,B}\\right]' \\, \\bm{\\omega}_{\\cal{B}/\\cal{N}}\n\t+ \\left[\\tilde{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\right] \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right] \\bm{r}'_{Fc_i/B}\\right)\n\\end{multline}\n\nConsidering that at the nozzles's exit, $\\dot{\\bm{r}}_{M/B} = \\bm{v}_{exh_j} + \\bm{\\omega}_{\\cal{B}/\\cal{N}} \\times \\bm{r}_{M/B}$ and $\\text{d}\\dot{m} =- \\rho\\,\\bm{r}_{M/B}'\\cdot\\hat{\\bm{n}} \\,\\text{d}A$, the surface integral is expressed in terms of the nozzles' surface:\n\\begin{multline}\\label{eq:eq50}\n\t\\int_{A_{\\text{exh}}} \\rho\\,\\bm{r}'_{M/B}\\cdot\\bm{n} \\left(\\bm{r}_{M/B} \\times \\dot{\\bm{r}}_{M/B}\\right)\\text{d}A =- \\sum_{j=1}^{N}\\int_{\\dot{m}_{\\text{noz}_j}}\\bm{r}_{M/B} \\times \\bm{v}_{\\text{exh}_j}\\,\\text{d}\\dot{m}+\\\\\n\t+ \\sum_{j=1}^{N}\\int_{\\dot{m}_{\\text{noz}_j}}\\bm{r}_{M/B} \\times\\left( \\bm{r}_{M/B}\\times \\bm{\\omega} _{\\mathcal{B}/\\mathcal{N}}\\right)\\,\\text{d}\\dot{m}\n\\end{multline}\nEquation~\\eqref{eq:eq36} is updated with Equations \\eqref{eq:eq49} and \\eqref{eq:eq50}:\n\\begin{multline}\n\t\\dot{\\bm{H}}_{\\text{sc, }B} = \\left[I_{\\text{sc, }B}\\right]\\,\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\\left[I_{\\text{sc, }B}\\right]\\,\\bm{\\omega}_{\\cal{B}/\\cal{N}} + \\sum_{i=1}^{M}\\left(m_{\\text{fuel}_i}\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right] \\bm{r}''_{Fc_i/B} + \\left[I_{\\text{fuel}_i,\\,B}\\right]' \\bm{\\omega}_{\\cal{B}/\\cal{N}}\n\t+\\right. \\\\\n\t\\left.+ \\dot{m}_{\\text{fuel}_i} \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]\\bm{r}'_{Fc_i/B}+ {m}_{\\text{fuel}_i}\\,\\left[\\tilde{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\right] \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right] \\bm{r}'_{Fc_i/B}\\right)\n\t+\\sum_{j=1}^{N}\\int_{\\dot{m}_{\\text{noz}_j}}\\left[\\tilde{\\bm{r}}_{M/B}\\right] ^T \\bm{v}_{\\text{exh}_j}\\,\\text{d}\\dot{m}+\\\\\n\t+\\sum_{j=1}^{N}\\int_{\\dot{m}_{\\text{noz}_j}}\\left[\\tilde{\\bm{r}}_{M/B}\\right] \\left[\\tilde{\\bm{r}}_{M/B}\\right] \\bm{\\omega} _{\\mathcal{B}/\\mathcal{N}} \\,\\text{d}\\dot{m} +\\left[\\tilde{\\bm{c}}\\right] m_{\\text{sc}}\\,\\ddot{\\bm{r}}_{B/N}=\\bm{L}_{B,\\,\\text{vol}}+\\bm{L}_{B,\\,\\text{surf}} \n\\end{multline}\n\nThe torque of each thruster nozzle is computed by the exhausting flow pressure distribution and by the lever arm distance from point $B$ and the application point of the force:\n\\begin{equation}\n\t\\bm{L}_{B_{\\text{thr}_j}} = \\bm{L}_{B_{\\text{sc, noz}_j}} + \\int_{\\dot{m}_{\\text{noz}_j}}\\bm{r}_{M/B} \\times \\bm{v}_{\\text{noz}_j}\\,\\text{d}\\dot{m}\n\\end{equation}\nFurthermore, a term taking into account the angular momentum variation caused by mass depletion is defined:\n\n\\begin{equation}\\label{eq:eq53}\n\t\\left[K\\right]= \\sum_{i=1}^{M}\\left[I_{\\text{fuel}_i,\\,B}\\right] '\t+\\sum_{j=1}^{N} \\int_{\\dot{m}_{\\text{noz}_j}}\\left[\\tilde{\\bm{r}}_{M/B}\\right] \\left[\\tilde{\\bm{r}}_{M/B}\\right]\\,\\text{d}\\dot{m} \n\\end{equation}\n\nThe second integral in Equation \\eqref{eq:eq53} is computed evaluating the momentum exchanged due to the fuel exiting the nozzle area (assuming a circular nozzle), coincident the interface surface between the spacecraft and the exhausted fuel:\n\n\\begin{multline}\n\t\\int_{\\dot{m}_{\\text{noz}_j}}\\left[\\tilde{\\bm{r}}_{M/B}\\right] \\left[\\tilde{\\bm{r}}_{M/B}\\right]\\,\\text{d}\\dot{m} \n\t=\\int_{\\dot{m}_{\\text{noz}_j}}\\left(\\left[\\tilde{\\bm{r}}_{N_j/B}\\right] + \\left[\\tilde{\\bm{r}}_{M/N_j}\\right]\\right)\\left(\\left[\\tilde{\\bm{r}}_{N_j/B}\\right] + \\left[\\tilde{\\bm{r}}_{M/N_j}\\right]\\right)\\,\\text{d}\\dot{m} =\\\\\n\t=-\\dot{m}_{\\text{noz}_j}\\left(\\left[\\tilde{\\bm{r}}_{N_j/B}\\right]\\left[\\tilde{\\bm{r}}_{N_j/B}\\right]^T + \\frac{A_{\\text{noz}_j}}{4\\,\\pi}\\left[BM_j\\right]\\left[\\begin{matrix}\n\t\t2&0&0\\\\\n\t\t0&1&0\\\\\n\t\t0&0&1\n\t\\end{matrix}\\right]\\left[BM_j\\right]^T\\right)\n\\end{multline}\n\\noindent\nwhere $A_{\\text{noz}_j}$ is the exiting area of the j-th nozzle and $\\left[BM_j\\right]$ is the direction cosine matrix (DCM) from from the j-th nozzle frame $\\mathcal{M}_j$, defined to have its origin at the $N_j$ point and its first axis in the exhausting velocity direction $\\bm{v}_{\\text{exh}_j}$, to the $\\mathcal{B}$ frame.\nFinally the rotational EOM is written as:\n\\begin{multline}\\label{eq:eq55}\n\t\\left[I_{\\text{sc, }B}\\right]\\,\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} + m_{\\text{sc}} \\left[\\tilde{\\bm{c}}\\right]\\ddot{\\bm{r}}_{B/N}= \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]^T\\left[I_{\\text{sc, }B}\\right]\\,\\bm{\\omega}_{\\cal{B}/\\cal{N}} - \\left[K\\right]\\,\\bm{\\omega} _{\\mathcal{B}/\\mathcal{N}}+\\\\+ \\sum_{i=1}^{M}\\left(m_{\\text{fuel}_i}\\,\\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]^T \\bm{r}''_{Fc_i/B} + \n\t{m}_{\\text{fuel}_i}\\,\n\t\\left[\\tilde{\\bm{\\omega}}_{\\mathcal{B}/\\mathcal{N}}\\right]^T \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right] \\bm{r}'_{Fc_i/B}+\\right.\\\\+\\left.\\dot{m}_{\\text{fuel}_i} \\left[\\tilde{\\bm{r}}_{Fc_i/B}\\right]^T\\bm{r}'_{Fc_i/B}\\right)\n\t+ \\bm{L}_{B,\\,\\text{vol}}+\\bm{L}_{B,\\,\\text{surf}} + \\sum_{j=1}^{N}\\bm{L}_{B_{\\text{thr}_j}} \n\\end{multline}\nThis concludes the derivation of the EOMs needed to describe the translational and rotational motion of a spacecraft with depleting mass due to thrusters. The following section describes the models used for the fuel tanks.\n\n\\subsection{Tank models}\nDifferent tank models could be developed to perfectly suit the needs of the spacecraft's fuel chain configuration. In the present paper five reservoir models will be considered as examples and their properties, such as inertia variation and barycenter motion, will be gathered.\\newline\nThe models and their main hypothesis are presented below:\n\n\\begin{itemize}\n\t\\item The constant tank's volume model where a spherical reservoir maintains a fixed geometry, i.e. a constant radius, and a fixed barycenter.\n\t\\item The constant fuel's density model where a spherical tank keeps its geometrical shape but gradually change its volume, so its radius, to maintain constant the density of the fuel and it has a fixed center of mass. \n\t\\item The emptying tank model where the fuel leaks out from an outlet in the spherical reservoir and the quantity of fuel decrease perpendicularly to the output direction modifying the barycenter position and the body's inertia accordingly to the mass distribution inside the tank.\n\t\\item The uniform burn cylinder model where a cylindrical tank does not change its geometrical shape and volume but the gas gradually decrease its density. As a consequence, the fuel barycenter remains fixed and the inertia varies accordingly to the mass variation.\n\t\\item The centrifugal burn cylinder model where a cylindrical tank is considered and the fuel burns radially from the center until the walls without breaking the tank's symmetry. The inertia tensor derivative is computed from these hypothesis and the barycenter remains in its initial position because the symmetry is conserved.\n\\end{itemize}\n\n\\subsubsection{The constant tank's volume model}\nThis model takes into account the variation of the fuel inside considering no variation of the volume off the tank. By looking at Figure \\ref{fig:CostTankVol}:\n\n\\begin{figure}[h!]\n\t\\centering\n\t\\includegraphics[width = 0.33\\textwidth]{./Figures/ConstVolTank.eps}\n\t\\caption{Geometrical properties of the constant volume sphere.}\n\t\\label{fig:CostTankVol}\n\\end{figure}\n\n\\begin{equation}\n\tV_{\\text{tank}}= const \\quad \\Rightarrow \\quad R_{\\text{tank}} = const\n\\end{equation}\n\\begin{equation}\n\t\\left[I_{\\text{fuel, }Tc}\\right] = \\frac{2}{5}\\,m_{\\text{fuel}} \\,R_{\\text{tank}}^2\\, \\left[I_{3\\times3}\\right]\n\\end{equation} \n\\begin{equation}\n\t\\left[I_{\\text{fuel, }Tc}\\right]' = \\frac{2}{5}\\,\\dot{m}_{\\text{fuel}} \\,R_{\\text{tank}}^2\\, \\left[I_{3\\times3}\\right]\n\\end{equation}\n\nMoreover the position of the center of mass of the tank does not change, so:\n\n\\begin{equation}\n\t\\bm{r}'_{Tc/B} = 0 \\qquad\\qquad \\bm{r}''_{Tc/B}=0\n\\end{equation}\n\\subsubsection{The constant fuel's density model}\nThe second model considers a shape-changing tank adapting itself to keep the fuel's density constant. Thus,according to Figure \\ref{fig:ConstDensTank}:\n\\begin{figure}[hb!]\n\t\\centering\n\t\\includegraphics[width = 0.33\\textwidth]{./Figures/ConstDensTank.eps}\n\t\\caption{Geometrical properties of the constant density sphere.}\n\t\\label{fig:ConstDensTank}\n\\end{figure}\n\\begin{equation}\n\t\\begin{cases}\n\t\t\\dot{V}_{\\text{tank}} = \\frac{\\dot{m}_{\\text{fuel}}}{\\rho_{\\text{fuel}}} \\vspace{3pt}\\\\\n\t\t\\dot{V}_{\\text{tank}} = 4\\,\\pi\\,R_{\\text{tank}}^2\\,\\dot{R}_{\\text{tank}}\n\t\\end{cases}\n\t\\quad \\Rightarrow \\quad \\dot{R}_{\\text{tank}} = \\frac{\\dot{m}_{\\text{fuel}}}{4\\pi\\,R_{\\text{tank}}^2\\,\\rho_{\\text{fuel}}}\n\\end{equation}\n\nAs a consequence:\n\n\\begin{equation}\n\t\\left[I_{\\text{fuel, }Tc}\\right]= \\frac{2}{5}\\,m_{\\text{fuel}} \\,R_{\\text{tank}}^2\\, \\left[I_{3\\times3}\\right]\n\\end{equation}\n\\begin{multline}\n\t\\left[I_{\\text{fuel, }Tc}\\right]' = \\frac{2}{5}\\left(R^2_{\\text{tank}}+\\frac{m_{\\text{fuel}}}{2\\pi\\,R_{\\text{tank}}\\,\\rho_{\\text{fuel}}}\\right)\\dot{m}_{\\text{fuel}}\\,\\left[I_{3\\times3} \\right] =\\\\\n\t= \\frac{2}{5}\\left(R^2_{\\text{tank}} + \\frac{2}{3}\\,R^2_{\\text{tank}}\\right)\\dot{m}_{\\text{fuel}}\\left[I_{3\\times3}\\right] = \\frac{2}{3}\\dot{m}_{\\text{fuel}}R^2_{\\text{tank}}\\left[I_{3\\times3}\\right]\n\\end{multline}\n\nAs in the previous model:\n\n\\begin{equation}\n\t\\bm{r}'_{Tc/B} = 0 \\qquad\\qquad \\bm{r}''_{Tc/B}=0\n\\end{equation}\n\n\\subsubsection{The emptying tank model}\nIn this case the mass variation stats from the opposite point to the outlet, that will be called from now on the pole, perpendicularly to the vector connecting the pole and the outlet.\\newline\nThe following notation will be used: $\\theta \\in (0\\,, \\pi)$ will be the latitude angle counted from the pole till the outlet, $\\phi \\in (0\\,,2\\,\\pi)$ will note the longitude angle, the radius will be $r \\in (0,\\,R_{\\text{tank}})$. Moreover the $\\theta^*$ will denote the angle between the pole and the circumference of the fuel's free surface. The volume $\\mathcal{V}$ and the center of mass of the tank can be computed using notations in figure \\ref{fig:EmpTank}:\n\n\\begin{figure}[hb!]\n\t\\centering\n\t\\includegraphics[width = 0.8\\textwidth]{./Figures/EmpTank.eps}\n\t\\caption{Geometrical properties of the emptying tank model.}\n\t\\label{fig:EmpTank}\n\\end{figure}\n\n\n\\begin{multline}\n\t\\mathcal{V}(\\theta^*)=\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}}r^2\\,\\sin{\\theta\\,\\text{d}\\theta}\\,\\text{d}\\phi\\,\\text{d}r\\;+\\\\\n\t+\\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}}r^2\\,\\sin{\\theta\\,\\text{d}\\theta}\\,\\text{d}\\phi\\,\\text{d}r= \\frac{2\\,\\pi\\,R_{\\text{tank}}^3}{3}\\left[1+\\frac{3}{2}\\cos{\\theta^*}-\\frac{1}{2}\\cos^3{\\theta^*}\\right]\n\\end{multline}\n\\vspace{3pt}\n\\begin{multline}\n\t\\bm{r}_{Tc/B}\\cdot\\bm{\\hat{k}}_3 = \\bm{r}_{Tc'/B}\\cdot\\bm{\\hat{k}}_3 + \\frac{1}{\\mathcal{V}(\\theta^*)}\\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}}r^3\\,\\sin{\\theta}\\cos{\\theta}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\;+\\right.\\\\\\left.+\\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}}r^3\\,\\sin{\\theta}\\cos{\\theta}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right)\n\t= \\frac{\\pi\\,R_{\\text{tank}}^4}\n\t{4\\,\\mathcal{V}(\\theta^*)}\\left[2\\,\\cos^2{\\theta^*} - \\cos^4{\\theta^*}-1\\right]\n\\end{multline}\n\\noindent\nwhere $\\bm{\\hat{k}}_3$ is the outlet-to-pole axis of the reference frame of the sphere and $\\bm{r}_{Tc'/B}$ the constant vector from $B$ to the center of the sphere.\\newline\nConsidering that $m_{\\text{fuel}}=\\rho_{\\text{fuel}}\\mathcal{V}\\,$, the derivatives in the $\\mathcal{B}$ reference frame can be performed:\n\n\\begin{multline}\n\t\\bm{r}'_{Tc/B}\\cdot\\bm{\\hat{k}}_3 =- \\frac{\\pi\\,R_{\\text{tank}}^4\\,\\rho_{\\text{fuel}}}{4\\,m^2_{\\text{fuel}}}\\left[4\\,m_{\\text{fuel}}\\dot{\\theta^*}\\,\\sin^3{\\theta^*}\\cos{\\theta^*} +\\right.\\\\\n\t\\left.+ \\dot{m}_{\\text{fuel}}\\left(2\\,\\cos^2{\\theta^*} - \\cos^4{\\theta^*}-1\\right)\\right]\n\\end{multline}\n\\begin{multline}\n\t\\bm{r}''_{Tc/B}\\cdot\\bm{\\hat{k}}_3 =- \\frac{\\pi\\,R_{\\text{tank}}^4\\,\\rho_{\\text{fuel}}}{2\\,m^3_{\\text{fuel}}}\\left[4\\,m_{\\text{fuel}}\\sin^3{\\theta^*}\\cos{\\theta^*} \\left(\\ddot{\\theta^*}\\,m_{\\text{fuel}} - 2\\,\\dot{\\theta^*}\\,\\dot{m}_{\\text{fuel}}\\right)+\\right.\\\\\n\t\\left.-4\\,m^2_{\\text{fuel}}\\,\\dot{\\theta^*}^2\\sin^2{\\theta^*}\\left(3\\,\\cos^2{\\theta^*} -\\sin^2{\\theta^*}\\right)+\\right.\\\\\n\t+\\left. \\left(2\\,\\cos^2{\\theta^*} - \\cos^4{\\theta^*}-1\\right)\\left(m_{\\text{fuel}}\\ddot{m}_{\\text{fuel}} - 2\\,\\dot{m}^2_{\\text{fuel}}\\right)\\right]\n\\end{multline}\n\nThe relation among $\\dot{m}_{\\text{fuel}}\\,$, $\\ddot{m}_{\\text{fuel}}\\, $, $\\dot{\\theta^*} $ and $\\ddot{\\theta^*} $ is deduced from the derivation of the relation between the volume $\\mathcal{V}$ and $m_{\\text{fuel}}$:\n\n\\begin{equation*}\n\tm_{\\text{fuel}}=\\rho_{\\text{fuel}}\\mathcal{V}(\\theta^*) \\quad \\Rightarrow \\quad \\dot{m}_{\\text{fuel}}=\\rho_{\\text{fuel}}\\dot{\\mathcal{V}}(\\theta^*)\n\\end{equation*}\n\\begin{equation}\n\t\\dot{m}_{\\text{fuel}} = -\\pi\\,\\rho_{\\text{fuel}} R_{\\text{tank}}^3 \\sin^3{\\theta^*}\\;\\dot{\\theta}^*\n\\end{equation}\n\n\\begin{equation}\n\t\\ddot{m}_{\\text{fuel}} = -\\pi\\,\\rho_{\\text{fuel}} R_{\\text{tank}}^3 \\sin^2{\\theta^*}\\left( \\ddot{\\theta^*}\\sin{\\theta^*} + 3\\, \\dot{\\theta^*}^2\\cos{\\theta^*}\\right)\n\\end{equation}\n\nFinally $\\theta^*$ can be found:\n\\begin{equation}\n\tm_{\\text{fuel}}=\\rho_{\\text{fuel}}\\mathcal{V}(\\theta^*) \\; \\Rightarrow \\; m_{\\text{fuel}} = \\frac{2}{3}\\pi\\, \\rho_{\\text{fuel}}\\,R_{\\text{tank}}^3\\left[1+\\frac{3}{2}\\cos{\\theta^*}- \\frac{1}{2}\\cos^3{\\theta^*}\\right]\n\\end{equation}\n\nAs far as the inertia concerns:\n\n\\begin{multline}\n\tI_{3\\,3} = \\rho_{\\text{fuel}} \\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\sin^3{\\theta}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r \\vspace{3pt}\\right.+\\\\\n\t\\left.+ \\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}} r^4\\,\\sin^3{\\theta}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right) =\\\\\n\t= \\frac{2}{5}\\pi\\,\\rho_{\\text{fuel}}R^5_{\\text{tank}}\\left[\\frac{2}{5} + \\frac{1}{4}\\cos{\\theta^*}\\sin^4{\\theta^*} -\\frac{1}{12}\\left(\\cos{3\\theta^*} - 9\\cos{\\theta^*}\\right)\\right]\n\\end{multline}\n\n\\begin{multline}\n\tI_{2\\,2} =\\rho_{\\text{fuel}} \\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\left(\\sin{\\theta}-\\sin^3{\\theta}\\sin^2{\\phi}\\right)\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r + \\vspace{3pt}\\right.\\\\\n\t\\left.+ \\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}} r^4\\,\\left(\\sin{\\theta}-\\sin^3{\\theta}\\sin^2{\\phi}\\right)\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right) =\n\t\\frac{2}{5}\\pi\\,\\rho_{\\text{fuel}}R^5_{\\text{tank}}\\left[\\frac{2}{3} +\\right.\\\\\n\t\\left.- \\frac{1}{4}\\cos^5{\\theta^*} +\\frac{1}{24}\\left(\\cos{3\\theta^*} -9\\cos{\\theta^*}\\right) +\\frac{5}{4}\\cos{\\theta^*} +\\frac{1}{8} \\cos{\\theta^*}\\sin^4{\\theta^*}\\right]\n\\end{multline}\n\n\\begin{multline}\n\tI_{1\\,1} =\\rho_{\\text{fuel}} \\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\left(\\sin{\\theta}-\\sin^3{\\theta}\\cos^2{\\phi}\\right)\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r+\\right.\\\\\n\t\\left.+ \\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}} r^4\\,\\left(\\sin{\\theta}-\\sin^3{\\theta}\\cos^2{\\phi}\\right)\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right) = I_{2\\,2}\n\\end{multline}\n\\vspace{3pt}\n\\begin{multline}\n\tI_{1\\,2} = \\rho_{\\text{fuel}} \\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\sin^3{\\theta}\\cos{\\phi}\\sin{\\phi}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r \\right.+\\\\\n\t\\left.+\\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}} r^4\\,\\sin^3{\\theta}\\cos{\\phi}\\sin{\\phi}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right)= 0\n\\end{multline}\n\\vspace{3pt}\n\\begin{multline}\n\tI_{1\\,3} = \\rho_{\\text{fuel}} \\left(\\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\sin^2{\\theta}\\cos{\\theta}\\cos{\\phi}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\right.+\\\\\n\t+\\int_{0}^{2\\,\\pi}\\int_{\\theta^*}^{\\pi}\\int_{0}^{R_{\\text{tank}}} r^4\\,\\sin^2{\\theta}\\cos{\\theta}\\cos{\\phi}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r\\left.\\right) = 0\n\\end{multline}\n\\vspace{3pt}\n\\begin{multline}\n\tI_{2\\,3} = \\rho_{\\text{fuel}} \\int_{0}^{2\\,\\pi}\\int_{0}^{\\theta^*}\\int_{0}^{R_{\\text{tank}}\\frac{\\cos{\\theta^*}}{\\cos{\\theta}}} r^4\\,\\sin^2{\\theta}\\cos{\\theta}\\sin{\\phi}\\,\\text{d}\\theta\\,\\text{d}\\phi\\,\\text{d}r = 0 \\vspace{5pt}\n\\end{multline}\n\\noindent\nbecause $\\int_{0}^{2\\,\\pi}\\cos{\\phi}\\sin{\\phi} =\\int_{0}^{2\\,\\pi}\\sin{\\phi} = \\int_{0}^{2\\,\\pi}\\cos{\\phi}=0 $.\\newline\nFrom those calculations the derivatives, in the tank reference frame, can be computed:\n\n\\begin{multline}\n\tI'_{3\\,3}\t= \\frac{2}{5}\\pi\\,\\rho_{\\text{fuel}}R^5_{\\text{tank}} \\, \\dot{\\theta^*}\\left[ \\cos^2{\\theta^*}\\sin^3{\\theta^*} -\\frac{1}{4}\\sin^5{\\theta^*} +\\frac{1}{4}\\sin{3\\theta^*} -\\frac{3}{4}\\sin{\\theta^*} \\right]\n\\end{multline}\n\\begin{multline}\n\tI'_{2\\,2}\t= I'_{1\\,1}= \\frac{2}{5}\\pi\\,\\rho_{\\text{fuel}}R^5_{\\text{tank}} \\, \\dot{\\theta^*}\\left[ \\frac{5}{4}\\sin{\\theta^*}\\cos{\\theta^*} -\\frac{5}{4}\\sin{\\theta^*} -\\frac{1}{8}\\sin{3\\theta^*} +\\right.\\\\\n\t\\left. + \\frac{3}{8} \\sin{\\theta^*} + \\frac{1}{2}\\cos^2{\\theta^*}\\sin^3{\\theta^*} -\\frac{1}{8} \\sin^5{\\theta^*} \\right]\n\\end{multline}\n\n\\subsubsection{Uniform burn cylinder}\nThis model consider a cylindrical tank whose geometry remains constant while fuel density changes.\n\\begin{figure}[htb!]\n\t\\centering\n\t\\includegraphics[width = 0.7\\textwidth]{./Figures/UnifCyl.eps}\n\t\\caption{Geometrical properties of the uniform burn cylinder}\n\t\\label{fig:UnfBurnCyl}\n\\end{figure}\nFrom these considerations and by looking at Figure \\ref{fig:UnfBurnCyl}, the inertia tensor and its derivative could be evaluated:\n\n\\begin{equation}\n\tI_{1\\,1}=I_{2\\,2} = m_{\\text{fuel}} \\left[\\frac{R^2}{4} + \\frac{h^2}{3}\\right]\\qquad I_{3\\,3} = m_{\\text{fuel}}\\,\\frac{R^2}{2}\n\\end{equation}\n\\begin{equation}\n\tI'_{1\\,1}=I_{2\\,2} = \\dot{m}_{\\text{fuel}} \\left[\\frac{R^2}{4} + \\frac{h^2}{3}\\right]\\qquad I'_{3\\,3} = \\dot{m}_{\\text{fuel}}\\,\\frac{R^2}{2}\n\\end{equation}\n\\noindent\nwhere $R$ is the cylinder radius and $h$ its half-height.\\newline\nMoreover, as the position of the center of mass of the tank does not change:\n\n\\begin{equation}\n\t\\bm{r}'_{Tc/B} = 0 \\qquad\\qquad \\bm{r}''_{Tc/B}=0\n\\end{equation}\n\n\\subsubsection{Centrifugal burn cylinder}\nThe present model consider a cylinder filled with propellant burning radially from the center to the edge. The geometry properties and their nomenclature can be seen in Figure \\ref{fig:CentrCyl}.\n\n\\begin{figure}[htb!]\n\t\\centering\n\t\\includegraphics[width = 0.7\\textwidth]{./Figures/CentrCyl.eps}\n\t\\caption{Geometrical properties of the centrifugal burn cylinder}\n\t\\label{fig:CentrCyl}\n\\end{figure}\nBy denoting $r$ the distance of the fuel surface from the axis of the cylinder, this quantity can be easily computed from the amount of mass in the tank:\n\n\\begin{equation}\n\tr =\\sqrt{R^2 - \\frac{m_\\text{fuel}}{2\\,\\pi\\,\\rho\\,h}}\n\\end{equation}\n\\noindent\nwhere $R$ is the cylinder radius, $h$ its half-height and $\\rho$ the fuel density.\\newline\nAs in the previous models, the time derivative of $r$ can be gathered from volume-mass relation:\n\n\\begin{equation}\n\t\\dot{m}_\\text{fuel} = - 4\\pi\\rho\\,h\\,r\\,\\dot{r}\n\\end{equation}\n\nAs a consequence:\n\\begin{equation}\n\tI_{1\\,1} = I_{2\\,2} = m_\\text{fuel}\\left[\\frac{R^2+ r^2}{4} +\\frac{h^2}{3}\\right]\n\\end{equation}\n\\begin{equation}\n\tI_{3\\,3} = m_\\text{fuel}\\left[\\frac{R^2+ r^2}{2} \\right]\n\\end{equation}\n\nMoreover, their time derivatives in the tank's reference frame can be computed:\n\\begin{equation}\n\tI'_{1\\,1} = I'_{2\\,2} = \\dot{m}_\\text{fuel}\\left[\\frac{r^2}{2}+\\frac{h^2}{3}\\right]\n\\end{equation}\n\\begin{equation}\n\tI'_{3\\,3} = \\dot{m}_\\text{fuel}\\,r^2\n\\end{equation}\n\nFinally, the tank's center of mass does not move as the mass variation is symmetric. Thus:\n\\begin{equation}\n\t\\bm{r}'_{Tc/B} = 0 \\qquad\\qquad \\bm{r}''_{Tc/B}=0\n\\end{equation}\n\n\\subsection{Update-Only Equations}\n\nEqs.\\eqref{eq:eq30} and \\eqref{eq:eq55} are the translational and rotational EOMs for the fully coupled dynamics of mass depletion. However, it is common to use an \"update-only\" method for mass depletion which does not consider mass depletion as a dynamical effect. In contrast to the fully coupled model it simply updates the mass properties of the spacecraft. The equations for the \"update-only\" can be simplified to:\n\n\\begin{equation}\\label{eq:eq31}\nm_{\\text{sc}} \\ddot{\\bm{r}}_{B/N} - m_{\\text{sc}} \\left[\\tilde{\\bm{c}}\\right]\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} = \\bm{F}_{\\text{thr}} - m_{\\text{sc}} \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right] \\bm{c} \n\\end{equation}\n\n\\begin{equation}\\label{eq:eq55}\nm_{\\text{sc}} \\left[\\tilde{\\bm{c}}\\right]\\ddot{\\bm{r}}_{B/N} + \\left[I_{\\text{sc, }B}\\right]\\,\\dot{\\bm{\\omega}}_{\\cal{B}/\\cal{N}} = - \\left[\\tilde{\\bm{\\omega}}_{\\cal{B}/\\cal{N}}\\right]\\left[I_{\\text{sc, }B}\\right]\\,\\bm{\\omega}_{\\cal{B}/\\cal{N}}\n+ \\bm{L}_{B} + \\sum_{j=1}^{N}\\bm{L}_{B_{\\text{thr}_j}} \n\\end{equation}\n\n\n\n", "meta": {"hexsha": "dbf0f088b987e36cbf1b11377c89e44a6fffc355", "size": 55036, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/simulation/dynamics/FuelTank/_Documentation/secModelDescription.tex", "max_stars_repo_name": "ian-cooke/basilisk_mag", "max_stars_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_stars_repo_licenses": ["0BSD"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/simulation/dynamics/FuelTank/_Documentation/secModelDescription.tex", "max_issues_repo_name": "ian-cooke/basilisk_mag", "max_issues_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_issues_repo_licenses": ["0BSD"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-03-13T20:52:22.000Z", "max_issues_repo_issues_event_max_datetime": "2019-03-13T20:52:22.000Z", "max_forks_repo_path": "src/simulation/dynamics/FuelTank/_Documentation/secModelDescription.tex", "max_forks_repo_name": "ian-cooke/basilisk_mag", "max_forks_repo_head_hexsha": "a8b1e37c31c1287549d6fd4d71fcaa35b6fc3f14", "max_forks_repo_licenses": ["0BSD"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 89.3441558442, "max_line_length": 1190, "alphanum_fraction": 0.6542081547, "num_tokens": 21920, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772384450968, "lm_q2_score": 0.727975460709318, "lm_q1q2_score": 0.6363067803525977}} {"text": "\\chapter{About prototype set models}\r\n\\label{ch_about}\r\n%\r\nThis report describes a supervised learning method we call the prototype set or `proset' algorithm.\r\nIt is an attempt to apply feature selection via the elastic net penalty to a nonlinear distribution model inspired by kernel regression.\r\nThe algorithm is implemented as Python package \\texttt{proset} using an interface compatible with machine learning library \\texttt{sklearn} \\cite{Pedregosa_11}.\\par\r\n%\r\nThe elastic net penalty was proposed by Zou and Hastie \\cite{Zou_05} as a combination of $L_1$- and $L_2$-regularization for linear regression.\r\nRegularization with either penalty has been studied by several authors in different contexts, usually with the purpose of preventing overfitting or finding a robust solution to an ill-conditioned problem.\r\nIn case of linear regression, the name `ridge regression' for applying an $L_2$-penalty appears to be due to Hoerl and Kennard \\cite{Hoerl_70}.\r\nThe LASSO algorithm that uses an $L_1$-penalty was proposed by Tibshirani \\cite{Tibshirani_96}.\\par\r\n%\r\nThe main advantage of an $L_1$-penalty over an $L_2$-penalty is that its derivative does not vanish near the origin.\r\nThus, a sufficiently large penalty weight selects features by forcing some model coefficients exactly to zero.\r\nThe main disadvantage is that the penalty term is not differentiable at zero, which means that model fitting usually requires specialized solvers.\r\nFor example, Andrew and Gao \\cite{Andrew_07} developed the OWL-QN algorithm to minimize an otherwise smooth objective function with an additional $L_1$-term.\r\nThis algorithm solves the optimization problem restricted to a particular orthant via backtracking line search.\r\nThe absolute value of each coefficient is represented as plus or minus identity depending on the orthant, so the restricted problem is everywhere differentiable.\r\nIf an optimum is found where any coefficients are exactly zero, the generalized Kuhn-Tucker conditions for the subgradient of the original term determine whether the solution is a local minimum or can be improved by moving to another orthant.\r\nIn case of proset, all model coefficients nonnegative, so optimization is naturally restricted to the first orthant.\r\nThus, we can rely on a standard algorithm for constrained continuous optimization like L-BFGS-B \\cite{Byrd_95}.\\par\r\n%\r\nZou and Hastie \\cite{Zou_05} show that a combined $L_1$- and $L_2$-penalty overcome some issues observed for the $L_1$-penalty applied to linear regression, but which we expect to affect other models as well.\r\nOne of these issues is that, in case of several highly correlated features, the LASSO tends to arbitrarily select only one of them.\r\nWith the elastic net penalty, highly correlated features are typically either included or excluded as a group.\r\nFurthermore, empirical results indicate that $L_2$-regularization improves the quality of the selected models.\r\nFor these reasons and because an $L_2$-penalty does not complicate the optimization problem, we decided to use an elastic net penalty for fitting proset models.\\par\r\n%\r\nTo derive a penalized version of negative log-likelihood, we follow the approach used by Friedman, Hastie, and Tibshirani \\cite{Friedman_10} for applying elastic net regularization to generalized linear models (GLMs).\r\nThis method is implemented as R package \\texttt{glmnet} \\cite{Friedman_10} for a variety of link functions.\r\nIn each case, a functional like this is minimized to determine model coefficients (with optional observation weights):\\par\r\n%\r\n\\begin{equation}\r\nf_{\\alpha,\\lambda}\\left(\\beta\\left|\\{(x_i,y_i)\\}_i\\right.\\right):=\\frac{1}{N}\\sum_{i=1}^Nl(y_i,\\beta_0+\\beta^Tx_i)+\\lambda\\left(\\frac{\\alpha}{2}\\sum_{j=1}^d\\beta_j^2+(1-\\alpha)\\sum_{j=1}^d|\\beta_j|\\right)\\label{eq_regularization}\r\n\\end{equation}\r\n%\r\nHere, $(x_i,y_i)$ are $N\\gg1$ observations with feature vectors $x_i\\in\\R^d$ and associated targets $y_i\\in\\R$, or some suitable subset thereof, e.g., the integers from $0$ to $K-1$ in case of classification with $K>1$ classes.\r\nThe link function $l$ is the negative log-likelihood for target $y_i$ conditional on a linear combination of $x_i$ with coefficients $\\beta\\in\\R^d$ and intercept $\\beta_0\\in\\R$.\r\nPenalty weights $\\lambda\\geq0$ and $\\alpha\\in[0,1]$ determine the overall magnitude of the elastic net penalty and the relative importance of the $L_1$- and $L_2$-terms.\r\nNote that the we have reversed the role of $\\alpha$ compared to the formulation used in \\cite{Friedman_10}, i.e., $\\alpha=1$ is pure $L_2$-regularization.\\par\r\n%\r\nFeature selection via $L_1$-penalty is a powerful technique.\r\nIt allows the user to evaluate a large number of potential input variables without having to worry about overfitting.\r\nUnfortunately, linear regression and GLMs are limited in the achievable model quality.\r\nAs the fitting procedure only deals with feature weights in the linear combination, modeling a nonlinear relationship requires that the effect is anticipated and explicitly coded as features.\r\nGeneralized additive models (GAMs) mitigate this somewhat by automatically applying nonlinear transform to the features.\r\nHowever, these transforms are restricted to a particular function class, e.g., splines with a predefined order and number of knots, and modeling higher order interactions can lead to a prohibitive number of terms.\r\nThe main reason for developing proset was to try overcoming these issues by using a different expression for the conditional distribution.\\par\r\n%\r\nThe distribution model for proset estimators is inspired by kernel-based methods from nonparametric statistics, in particular the Nadaraya-Watson estimator \\cite{Nadaraya_64}\\cite{Watson_64}.\r\nThis regression method uses a locally weighted average of the target values for training data to compute the conditional expectation of the target at an arbitrary point.\r\nLet $K:\\R\\rightarrow\\R$ be a kernel function, meaning that it is nonnegative, symmetric around the origin, and integrates to one.\r\nThen, for any bandwidth $h>0$, $K_h(x):=\\frac{1}{h}K\\left(\\frac{x}{h}\\right)$ is also a kernel function.\r\nGiven $N$ training samples $(x_n, y_n)\\in\\R^2$, the Nadaraya-Watson estimator with bandwidth $h$ at $x\\in\\R$ is defined as\r\n%\r\n\\begin{equation}\r\n\\hat{m}_h(x):=\\frac{\\sum_{n=1}^Ny_nK_h(x-x_n)}{\\sum_{n=1}^NK_h(x-x_n)}\\label{eq_nadaraya_watson}\r\n\\end{equation}\r\n%\r\nAs $K_h$ goes to zero for $x\\rightarrow\\pm\\infty$, samples where $x_n$ is close to $x$ have the most impact on $\\hat{m}(x)$.\r\nThe bandwidth controls the degree of smoothing, i.e., the impact of distant points increases with $h$.\r\nThe approach can be extended to multiple features using a multivariate kernel function with positive semi-definite bandwidth matrix $H$.\r\nUnlike GLMs or even GAMs, the Nadaraya-Watson estimator can approximate an arbitrary, smooth relationship between the features and target, as shown, e.g., by Devroye \\cite{Devroye_78}.\\par\r\n%\r\nLocally weighted averaging can also be used to estimate the conditional distribution itself instead of its expectation.\r\nThis requires replacing $y$ in (\\ref{eq_nadaraya_watson}) with a kernel function centered on $y$.\r\nA general framework that covers both continuous and discrete targets is proposed by Hall, Racine, and Li in \\cite{Hall_04}.\r\nThe prototype set algorithm uses a similar approach.\r\nA full description is given later, but two key features of proset models are:\r\n%\r\n\\begin{itemize}\r\n\\item The bandwidth matrix is limited to a diagonal structure.\r\nTo select features, we apply an elastic-net penalty to the inverse bandwidth term for each feature.\r\nThus, the penalty counteracts overfitting by enforcing smoothness and removing irrelevant features from the model altogether.\r\n%\r\n\\item Instead of including all samples with unit weight, the proset estimator includes only a subset of the training data with individual weights.\r\nThese are fitted subject to a separate elastic-net penalty term, which ensures that points are only included if they are suitably `representative' of their neighborhood.\r\n\\end{itemize}\r\n%\r\nWe refer to the set of training points selected for a proset model as `prototypes', which is how the method gets its name.\r\nNote that the term `prototype' has two slightly different meanings in English, both of which can be said to apply here:\r\n%\r\n\\begin{itemize}\r\n\\item A `prototype' can be a typical example in the sense of `archetype'.\r\nThe selected samples are considered typical for their neighborhood in the feature space.\r\n%\r\n\\item A `prototype' can be an incomplete version in the sense of `demonstrator'.\r\nDue to feature selection, each sample included in the model is reduced to its essential features.\r\n\\end{itemize}\r\n%\r\n\\section{Properties of proset}\r\n\\label{sec_properties}\r\n%\r\nThe first version of the \\texttt{proset} package implements a classifier -- a regressor is planned for a future release.\r\nChapter \\ref{ch_classifier} describes the model in detail, shows how to select good hyperparameters, and provides evaluation results for several public data sets and artificial test cases.\r\nFor comparison, we also trained k-nearest-neighbor (kNN)\\footnote{\r\nThe kNN method is commonly attributed to Fix and Hodges \\cite{Fix_51}, although they study nonparametric density estimators for discriminatory analysis, not supervised learning.\r\n}\r\nand extreme gradient boosting (XGBoost) \\cite{Chen_16} classifiers on the same data.\r\nThe former algorithm is conceptually close to proset as it relies on local averaging.\r\nThere is even an extension for $k=1$, Hart's `condensed nearest neighbor' algorithm \\cite{Hart_68}, concerned with extracting a subset of representative points from the training data.\r\nXGBoost has become a kind of industry standard for machine learning with highly informative features, i.e., those problems not requiring deep learning.\r\nEmpirical results shows that it is often very good at optimizing a model in terms of a chosen metric.\r\nBoth algorithms are capable of representing a nonlinear relationship between the features and target.\\par\r\n%\r\nIn the benchmark study, we optimize the models on each data set for minimal log-loss.\r\nFor the best performing algorithm, we compute the sum of cross-validation mean and standard deviation for log-loss using the optimal hyperparameters.\r\nIf another algorithm reaches a score less than or equal to this threshold, we consider it `equivalent' to the best model.\r\nTable \\ref{tab_classifier_comparison} summarizes the relative performance of the classifiers on eleven test cases.\r\nA model is rated as either `best' (lowest log-loss), `equivalent' (log-loss less than or equal to the threshold), `worse' (log-loss above the threshold), or `worst' (classifier returns constant estimator).\\par\r\n%\r\n\\begin{table}\r\n\\caption{Comparison of classification algorithms}\r\n\\label{tab_classifier_comparison}\r\n%\r\n\\begin{center}\r\n\\begin{tabular}{|lcccc|}\r\n\\hline\r\n&\\multicolumn{4}{c|}{\\textbf{Log-loss score}}\\\\\r\n\\textbf{Classifier}&\\textbf{Best}&\\textbf{Equivalent}&\\textbf{Worse}&\\textbf{Worst$^*$}\\\\\r\nproset&4&2&5&--\\\\\r\nkNN&1&3&6&1\\\\\r\nXGBoost&7&--&2&2\\\\\r\n\\multicolumn{5}{|l|}{$^*$ Algorithm returns constant estimator.}\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nFor the test cases -- which are of course not representative of anything in particular -- the performance of proset is in-between kNN and XGBoost.\r\nThere are also several instances where two or all three of the models are equivalent.\r\nThe exception are three of the four cases where XGBoost is `worse' or `worst'.\r\nThese are artificial data sets where the target is a deterministic function of five or six features.\r\nThe fourth data set is very small and performance on testing data highly variable.\\par\r\n%\r\nOur study contains artificial test cases relying on the interaction of three to six features.\r\nXGBoost provides very good results up to four features, is worse than the other two models for five features, and fails on six features.\r\nThis may be a limitation of the binary decision trees used as base learners.\r\nBoth kNN and proset models have some predictive value for the higher order interaction terms, although results are far from perfect given that the relationship between features and target is deterministic.\r\nIf we add six more irrelevant features to the sixth order interaction, kNN also fails.\r\nThe proset classifier is able to select only the relevant inputs and produce a nontrivial estimate as before.\\par\r\n%\r\nOne particular strength of proset models we found while studying the example cases is that they tend to be highly explainable.\r\nThis is due to feature selection and that fact that the distance-based approach allows for geometric interpretation.\r\nAnother important reason is that the estimate for a particular sample tends to depend meaningfully only on a few prototypes, although the total number can go into the hundreds or thousands.\r\nChapter \\ref{ch_explainability} `Explainability' shows examples for the following:\r\n%\r\n\\begin{itemize}\r\n\\item Feature selection makes it easier for humans to review the model structure.\r\nIf the number of relevant features is small, users can assess whether the choice is sensible and study low-dimensional representations like scatter plots or cuts through the decision space.\r\n%\r\n\\item Prototype selection simplifies reviewing the model structure even if the number of features is large.\r\nWe can perform weighted PCA on the feature matrix for the prototypes and use this to create low-dimensional maps of the data.\r\nAlso, a check whether the training data has labeling errors or artifacts can start with the smaller set of prototypes.\r\n%\r\n\\item The estimate for a particular sample can be explained by reviewing the prototypes with the highest impact.\r\nThis is an explanation in terms of similar training instances instead of more abstract properties, which can help nontechnical users to understand and trust the model.\r\n%\r\n\\item Proset rates new samples based on their absolute distance to the prototypes.\r\nThat means the algorithm can detect whether a new sample is far away from the training data and the estimate should not be relied on.\r\n\\end{itemize}\r\n%\r\nNote that the last three properties do not require the selected number of features to be small.\r\nFor example, proset could be used as the final stage in transfer learning and add its capabilities to a model of arbitrary complexity like a deep neural network.\r\n%\r\n\\section{The curse of dimensionality}\r\n\\label{sec_curse}\r\n%\r\nThe Nadaraya-Watson estimator is known to have poor convergence properties in feature spaces of high dimension.\r\nFor $n$ samples in a $d$-dimensional space, the asymptotic rate of convergence for the error is $n^{-\\frac{4}{4+d}}$ \\cite{Haerdle_04}.\r\nThis is due to the trade-off between the point-wise bias and variance.\r\nConsider the simple case where $H$ is a diagonal matrix with constant bandwidth $h_n$ in each dimension.\r\nTo reduce the bias, $h_n$ has to converge to zero as $n$ goes to infinity.\r\nTo reduce the variance, the number of training points with nonnegligible impact on the estimator at any one point has to go to infinity with $n$.\r\nAs the volume containing such points is proportional to $h_n^d$, the number of points is proportional to $nh_n^d$.\r\nThus, the bandwidth can only converge to zero very slowly with $n$ if $d$ is large, with a corresponding slow reduction in bias.\\par\r\n%\r\nWhat are the implications for proset?\r\nFeature selection is obviously beneficial as the convergence rate of the estimator should depend only on the number of features included in the model.\r\nThe impact of prototype selection is harder to quantify.\r\nConsider a target distribution that can be replicated exactly by proset, i.e., it is generated by local averaging using a finite number of kernel functions for weighting.\r\nIn this situation, the bandwidth does no longer have to vanish asymptotically with $n$ to yield a small bias.\r\nWe have established no theoretical results, but it would be interesting to study the convergence behavior for distributions that can be reproduced or at least closely approximated by proset.\\par\r\n%\r\nThe case with the largest dimension in the benchmark study has a 64 features.\r\nProset selects around 30 of these to build a model with reasonable performance, although XGBoost achieves a better log-loss.\r\nThis number of features is not particularly large for a classification problem.\r\nHowever, it appears to be quite large compared to problems that are typically considered feasible for local kernel methods.\r\nFor comparison, we found published examples with 8 features (local linear regression: Ormoneit and Hastie \\cite{Ormoneit_99}),\r\n7 to 17 features depending on the encoding of categorical variables (nonparametric conditional distribution: Bontemps, Racine, and Simoni \\cite{Bontemps_09}),\r\nor 24 variables, of which any one can serve as target (the largest sample data set for R package \\texttt{np}: Hayfield and Racine \\cite{Hayfield_08}).\r\n%\r\n\\section{Release history}\r\n\\label{sec_release_history}\r\n%\r\nThese versions of the \\texttt{proset} Python package have been released:\r\n%\r\n\\begin{description}\r\n\\item[0.1.0:] implementation of proset classifier using algorithm L-BFGS-B \\cite{Byrd_95} for parameter estimation;\r\nhelper functions for model fitting and plotting;\r\nbenchmark code for hyperparameter selection, comparison to other classifiers, and demonstration of explanatory features;\r\nfirst version of technical report.\r\n\\end{description}\r\n%\r\n\\endinput\r\n", "meta": {"hexsha": "76dda7050d9366cf66320aa4484e67179347ca7c", "size": 17460, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/about.tex", "max_stars_repo_name": "NRuf77/proset", "max_stars_repo_head_hexsha": "101d491e05c2423faddca31029232982f46d8831", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/about.tex", 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YES\n2. YES", "lm_q1_score": 0.8740772450055545, "lm_q2_score": 0.7279754371026367, "lm_q1q2_score": 0.636306764494387}} {"text": "% Chapter X\n\n\\chapter{Creating models} % Chapter title\n\n\\label{Creating models} % For referencing the chapter elsewhere, use \n\nThe previous section described the basic building blocks of Figaro models. Out of these building blocks, a wide variety of models can be created. This section describes how to build a range of models.\n\n\\section{Basic models}\n\nOne of the first things you can do with an element is to assign it to a Scala value:\n\n\\begin{flushleft}\n\\texttt{val burglary = Flip(0.01)}\n\\end{flushleft}\n\n\\marginpar{\\texttt{val} in Figaro represents an immutable value. When a thing is assigned to a val, data inside the thing can change but the reference stored in the val is constant}\n\nA \\texttt{val} represents a field (in this case burglary) that takes on an immutable value (in this case the element \\texttt{Flip(0.01)}). A field is not a variable; its value cannot be changed (Note that the scala assignment of the field \\texttt{burglary} cannot change, but the value of the Figaro element that is assigned to it, \\texttt{Flip(0.01)}, \\emph{can} change). You can use the value of a field by referring to its name:\n\n\\begin{flushleft}\n\\texttt{val alarm = If(burglary, Flip(0.9), Flip(0.1))}\n\\end{flushleft}\n\nRecall that an element defines a process that probabilistically produces a value. If an element is referred to multiple times, it must produce the same value everywhere it appears. Consider:\n\n\\begin{flushleft}\n\\texttt{val x = Flip(0.5)\n\\newline val y = x === x}\n\\end{flushleft}\n\nAlthough we don't know the value, \\texttt{x} must produce the same value on both sides of the equality test. Therefore, \\texttt{y} produces the value \\texttt{true} with probability 1. In contrast, in\n\n\\begin{flushleft}\n\\texttt{val y = Flip(0.5) === Flip(0.5)}\n\\end{flushleft}\n\nthe left and right hand sides are distinct elements (each call produces a new \\texttt{Flip}), so they need not produce the same value. Therefore, \\texttt{y} will produce \\texttt{true} with probability 0.5.\n\nWith the tools we have defined so far, we can easily create a Bayesian network. In the following code, \\texttt{CPD} is a library element (based on \\texttt{Chain}) that makes it easy to define conditional probability distributions:\n\n\\begin{flushleft}\n\\marginpar{This example is found in Burglary.scala}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline import com.cra.figaro.library.compound.CPD\n\\newline\n\\newline val burglary = Flip(0.01)\n\\newline\n\\newline val earthquake = Flip(0.0001)\n\\marginpar{Scala statements can be written on multiple lines}\n\\newline\n\\newline val alarm = CPD(burglary, earthquake,\n\\newline \\tab (false, false) -> Flip(0.001),\n\\newline \\tab (false, true) -> Flip(0.1), \n\\newline \\tab (true, false) -> Flip(0.9), \n\\newline \\tab (true, true) -> Flip(0.99))\n\\newline \n\\newline val johnCalls = CPD(alarm,\n\\newline \\tab false -> Flip(0.01),\n\\newline \\tab true -> Flip(0.7))}\n\\end{flushleft}\n\nWith \\texttt{CPD}, every single combination of values of the parents needs to be listed. \\texttt{RichCPD} provides a more flexible format that allows for specification of structures such as context specific independence. Each clause in a \\texttt{RichCPD} consists of a tuple of cases, one for each parent. A case can be \\texttt{OneOf} a set of values, \\texttt{NoneOf} a set of values (meaning that it matches all values except for the ones listed), or *, meaning that it accepts all values. For example:\n\n\\begin{flushleft}\n\\texttt{import com.cra.figaro.language.\\_}\n\\newline \\texttt{import com.cra.figaro.library.compound.RichCPD}\n\\newline \\texttt{val x1 = Select(0.1 -> 1, 0.2 -> 2, 0.3 -> 3, 0.4 -> 4)}\n\\newline \\texttt{val x2 = Flip(0.6) \n\\newline val x3 = Constant (5) \n\\newline val x4 = Flip(0.8)\n\\newline val y = RichCPD(x1, x2, x3, x4,\n\\newline \\tab (OneOf(1, 2), *, OneOf(5), *) -> Flip(0.1), \n\\newline \\tab (NoneOf(4), OneOf(false), *, *) -> Flip(0.7), \n\\newline \\tab (*, *, NoneOf(6, 7), OneOf(true)) -> Flip(0.9), \n\\newline \\tab (*, *, *, OneOf(false)) -> Constant (true))}\n\\end{flushleft}\n\nA particular combination of values of the parents is matched against each row in turn, and the first match is chosen. For example, the combination (1, false, 5, true) matches the first three rows, so the first result (\\texttt{Flip(0.1)}) is chosen. All possible values of the parent still need to be accounted for in the argument list using a combination of \\texttt{OneOf}, \\texttt{NoneOf} and *.\n\n\\section{Conditions and constraints}\n\nSo far, we have described models that generate the values of elements. It is also possible to influence the values of elements by imposing conditions or constraints on them.\n\nA \\emph{condition} represents something the value of the element must satisfy. Only values that satisfy the condition are possible. Every element has a condition, which is a function from a value of the element to a Boolean. If the element is of type \\texttt{Element[T]}, the condition is of type \\texttt{T => Boolean}. Conditions can have multiple purposes. One is to assert evidence, by specifying something that is known about an element. Alternatively, a condition can specify a structural property of a model, for example, that only one of two teams playing a game can be the winner.\n\nThe default condition of an element returns true for all values. The condition can be changed using \\texttt{setCondition}:\n\n\\begin{flushleft}\n\\texttt{val x1 = Select(0.1 -> 1, 0.2 -> 2, 0.3 -> 3, 0.4 -> 4)\nx1.setCondition((i: Int) => i == 1 || i == 4)}\n\\end{flushleft}\n\nwhich says that \\texttt{x1} must have value 1 or 4. We can add a condition on top of existing conditions using the \\texttt{addCondition} method. For example, the following code says that not only must \\texttt{x1} equal 1 or 4, it must also be odd:\n\n\\begin{flushleft}\n\\texttt{x1.addCondition((i: Int) => i \\% 2 == 1)}\n\\end{flushleft}\n\nThe \\texttt{observe} method provides an easy way to specify a condition that only allows a single value. For example, to specify that \\texttt{x1} must have the value 2, we can use\n\n\\begin{flushleft}\n\\texttt{x1.observe(2)}\n\\end{flushleft}\n\nNote that using \\texttt{observe} will remove all previous conditions on an element.\n\nA \\emph{constraint} provides a way to specify a potential or weighting over an element. It is a function from a value of the element to a Double, so if the element has type \\texttt{Element[T]}, the constraint is of type \\texttt{T => Double}.\n\nConstraints serve multiple purposes in Figaro. One is to specify soft evidence on an element. For example, if in the above Bayesian network we think we heard John call but we're not sure, we might introduce the constraint\n\n\\begin{flushleft}\n\\texttt{johnCalls.setConstraint((b: Boolean) => \n\\newline \\tab if (b) 1.0; else 0.1)}\n\\end{flushleft}\n\nThis line will have the effect of making John calling 10 times more likely than not, all else being equal. Another purpose of constraints is to define some probabilistic relationships conveniently that are more difficult to express without them. Consider the following example, in which we are modeling the process of firms bidding for a contract and one of them being selected as the winner.\n\n\\begin{flushleft}\n\\marginpar{This example is found in Firms.scala}\n\\texttt{import com.cra.figaro.language.\\_}\n\\newline \\texttt{import com.cra.figaro.library.atomic.\\_}\n\\newline \\texttt{import com.cra.figaro.library.compound.If}\n\\newline\n\\newline \\texttt{class Firm \\{\n\\newline \\tab val efficient = Flip(0.3)\n\\newline \\tab val bid = If(efficient, continuous.Uniform(5, 15),\n\\newline continuous.Uniform(10, 20))\n\\newline \\} \n\\newline\n\\newline \\texttt{val firms = Array.fill(20)(new Firm)\n\\newline val winner = discrete.Uniform(firms:\\_*)\n\\newline val winningBid = Chain(winner, (f: Firm) => f.bid)\n\\newline winningBid.setConstraint((d: Double) => 20 $-$ d)}\n}\n\\end{flushleft}\n\nThis example shows some new Scala features. First, we have a class definition (the \\texttt{Firm} class). A class creates a type that can be instantiated to create instances. The \\texttt{Firm} class has two fields, \\texttt{efficient} and \\texttt{bid}. Note that bid makes use of \\texttt{continuous.Uniform}. This is the continuous uniform element defined in the \\texttt{library.atomic.continuous} package, but we did not import the members of this package, only the members of the \\texttt{library.atomic} package. The reason we did things this way is that later in the example, we use the discrete uniform, and we want to be explicit about which uniform element we mean at each point.\n\nOnce we have defined the \\texttt{Firm} class, we create an array named \\texttt{firms} consisting of 20 instances of \\texttt{Firm.Array.fill(20)(new Firm)} creates an array filled with the result of 20 different invocations of \\texttt{new Firm}, each of which creates a separate instance of \\texttt{Firm} (and separate Figaro elements in each class). We then define the winner to be one of the firms, chosen uniformly. Note the notation \\texttt{firms:\\_*.} The element \\texttt{discrete.Uniform} takes as arguments an explicit sequence of values of variable length, for example, \\texttt{discrete.Uniform(1, 2, 5)} or \\texttt{discrete.Uniform(\"x\")}. Since \\texttt{firms} is a single field representing an array, we must convert it into a sequence of arguments, which is accomplished using the \\texttt{:\\_*} notation. The field \\texttt{winner} represents an \\texttt{Element[Firm]}; it is intended to mean the winning bidder, although so far we have done nothing to relate the winner to its bid.\n \nThe next line is interesting. It allows us to identify the bid of the winning bidder as an element with a name, even though we don't know who the winner is. We can do this because even though we don't know who the winner is, we can refer to the \\texttt{winner} field, and because the value of \\texttt{winner}, whatever it is, is a \\texttt{Firm} that has a \\texttt{bid} field, which is an element that can be referred to. It is important to realize that this \\texttt{Chain} does not create a new element but rather refers to the element \\texttt{f.bid} that was created previously.\n\nFinally, we introduce the constraint, which says that a winning bid of d has weight \\texttt{20 $-$ d}. This means that a winning bid of 5 is 15 times more likely than a winning bid of 19. The effect is to make the winning bid more likely to be low. Note that in this model, the winning bid is not necessarily the lowest bid. For various reasons, the lowest bidder might not win the contract, perhaps because they offer a poor quality service or they don't have the right connections. Using a constraint, the model is specified very simply using a discrete uniform selection and a simple constraint.\n\nConstraints are also useful for expressing undirected models such as relational Markov networks or Markov logic networks. To illustrate, we will use a version of the friends and smokers example. This example involves a number of people and their smoking habits. People have some propensity to smoke, and people are likely to have the same smoking habit as their friends.\n\n\\begin{flushleft}\n\\marginpar{This example is found in Smokers.scala}\n\\texttt{import com.cra.figaro.language.Flip\n\\newline import com.cra.figaro.library.compound.\\textasciicircum \\textasciicircum\n\\newline \n\\newline class Person \\{\n\\newline \\tab val smokes = Flip(0.6) \n\\newline \\} \n\\newline \n\\newline val alice, bob, clara = new Person\n\\newline val friends = List((alice, bob), (bob, clara))\n\\newline clara.smokes.observe(true) \n\\newline\n\\marginpar{Single line function definitions in Scala do not need bracketing}\n\\newline def smokingInfluence(pair: (Boolean, Boolean)) = \n\\newline \\tab if (pair.\\_1 == pair.\\_2) 3.0; else 1.0\n\\newline\n\\newline for \\{ (p1, p2) <- friends \\} \\{\n\\newline \\tab \\textasciicircum \\textasciicircum(p1.smokes, p2.smokes).setConstraint(smokingInfluence)\n\\newline \\} } \n\\end{flushleft}\n\nFirst, we create a \\texttt{Person} class with a \\texttt{smokes} field. We create three different people and a network of friends, represented by a list of pairs of people. We also observe that one of the people smokes.\n\nNow we create the constraint function \\texttt{smokingInfluence}. This function takes a pair of Booleans, and returns 3.0 if they are the same, 1.0 if different. The intended meaning of this function is to compare the smoking habit of two friends, and say that having the same smoking habit is three times as likely as a different smoking habit, all else being equal.\n\nFinally, we apply the constraint to all the pairs of friends. The code uses a Scala feature called a \"for comprehension\". The notation for \\texttt{\\{ (p1, p2) <- friends \\}} { \"do something\" } iterates through all pairs of people in the \\texttt{friends} list and executes \"do something\" for each pair. In this case, \"do something\" is \"add the constraint on their smoking habits to the pair of friends\". The notation \\texttt{\\textasciicircum \\textasciicircum(p1.smokes, p2.smokes)} takes each pair of friends and creates the pair element consisting of their smoking habits. We then assign the \\texttt{smokingInfluence} constraint to this pair.\n\n\\section{Classes, instances, and relationships}\n\nThe object-oriented nature of Scala makes Figaro ideal for representing probabilistic models involving objects and relationships such as probabilistic relational models (PRMs). \nIn the following example, we will see how to define general classes of object, and create instances of a class by using a subclass of the class specially designed for the instance.\n\nIn this example, we are given two possible sources and a sample that came from one of the sources, and want to determine which source the sample came from based on the strength of the match with each\nsource.\n\n\\begin{flushleft}\n\\marginpar{Abstract classes in Scala are similar as in Java; they cannot be instantiated}\n\\texttt{class Source(val name: String)\n\\newline\n\\newline abstract class Sample \\{\n\\newline \\tab val fromSource : Element[Source] \n\\newline \\}\n\\newline\n\\newline class Pair(val source: Source, val sample: Sample) \\{\n\\newline \\tab val isTheRightSource = Apply(sample.fromSource, (s: Source) => s == source)\n\\newline \\tab val distance = If(isTheRightSource, \n\\newline \\tab Normal(0.0, 1.0), \n\\newline \\tab Uniform(0.0, 10.0))\n\\newline \\}\n\\newline\n\\marginpar{Defining class contents at instantiation time will override undefined values}\n\\newline val source1 = new Source(\"Source 1\") \n\\newline val source2 = new Source(\"Source 2\") \n\\newline val sample1 = new Sample \\{ \n\\newline val fromSource = Select(0.5 -> source1, 0.5 -> source2)\n\\newline \\}\n\\newline val pair1 = new Pair(source1, sample1)\n\\newline val pair2 = new Pair(source2, sample1)\n\\newline \n\\newline pair1.distance.setCondition((d:Double) => (d > 0.15 \\&\\& d < 0.25))\n\\newline pair2.distance.setCondition((d:Double) => (d > 1.45 \\&\\& d < 1.55)) }\n\\end{flushleft}\n\nWe begin by creating classes representing sources and samples, where each sample comes from a source. Note that \\texttt{Sample} is an abstract class, because in this class we do not say anything about what source the sample came from (the from \\texttt{Source} field has not been assigned an \\texttt{Element[Source]} yet). We then create the \\texttt{Pair} class representing a pair of a source and a sample. \\texttt{Pair} has two fields: \\texttt{isTheRightSource}, which produces true if the sample is from the source in the pair, and \\texttt{distance}, which measures the closeness of the match between the sample and the source (lower distance means better match). The distance will tend to be smaller if the sample is from the right source but will not always be so.\n\nNow it's time to create some instances. Note that the \\texttt{Source} class takes an argument which is the name of the source. When we create instances \\texttt{source1} and \\texttt{source2} of this class, we supply the name argument. Next, we create an instance of \\texttt{Sample}. Since \\texttt{Sample} is abstract, we need to supply a definition of \\texttt{fromSource}. We can do that right in line here, specifying that \\texttt{sample1} could come either from \\texttt{source1} or \\texttt{source2}, each with probability 0.5. Finally, we create pairs pairing both of the sources to \\texttt{sample1} and create conditions about the distances. The conditions are ranges rather than exact observations because exact observations on continuous elements can be problematic for many types of inference algorithms.\n\nUsing similar techniques, we can create a PRM. The following example shows the classical actors and movies PRM. There are three classes: actors, movies, and appearances relating actors to movies. Whether an actor receives an award for an appearance depends on the fame of the actor and the quality of the movie. The Figaro code for this example is as follows:\n\n\\begin{flushleft}\n\\marginpar{This example can be found in SimpleMovie.scala}\n\\texttt{import com.cra.figaro.library.compound.CPD\n\\newline import com.cra.figaro.language.\\_\n\\newline\n\\newline class Actor \\{\n\\newline \\tab val famous = Flip(0.1)\n\\newline \\}\n\\newline\n\\newline class Movie \\{\n\\newline \\tab val quality = Select(0.3 -> 'low, 0.5 -> 'medium, 0.2 -> 'high)\n\\newline \\}\n\\newline\n\\marginpar{The ' in front of a string creates a Scala symbol, which are treated like \\texttt{String} constants}\n\\newline class Appearance(actor: Actor, movie: Movie) \\{\n\\newline \\tab def probAward(quality: Symbol, famous: Boolean) = \n\\newline \\tab \\tab (quality, famous) match \\{ \n\\newline \\tab \\tab case ('low, false) => 0.001\n\\newline \\tab \\tab case ('low, true) => 0.01 \n\\newline \\tab \\tab case ('medium, false) => 0.01\n\\newline \\tab \\tab case ('medium, true) => 0.05\n\\newline \\tab \\tab case ('high, false) => 0.05\n\\newline \\tab \\tab case ('high, true) => 0.2\n\\newline \\tab \\tab \\}\n\\newline \\tab val award = SwitchingFlip(Apply(movie.quality, actor.famous, (q: Symbol, f: Boolean) => probAward(q, f)))\n\\newline \\}\n\\newline\n\\newline val actor1 = new Actor \n\\newline val actor2 = new Actor \n\\newline val actor3 = new Actor \n\\newline val movie1 = new Movie \n\\newline val movie2 = new Movie\n\\newline val appearance1 = new Appearance(actor1, movie1) \n\\newline val appearance2 = new Appearance(actor2, movie2) \n\\newline val appearance3 = new Appearance(actor3, movie2) \n\\newline actor3.famous.observe(true) \n\\newline movie2.quality.observe('high)\n\\newline \n\\marginpar{The \\texttt{\\_.award} notation is Scala shorthand to retrieve the award value of each element of the map}\n\\newline // Ensure that exactly one appearance gets an award. \n\\newline def uniqueAwardCondition(awards: List[Boolean]) =\n\\newline \\tab awards.count((b: Boolean) => b) == 1\n\\newline val allAwards: Element[List[Boolean]] = \n\\newline \\tab Inject(appearances.map(\\_.award):\\_*)\n\\newline allAwards.setCondition(uniqueAwardCondition)}\n\\end{flushleft}\n\nThe code is self-explanatory except for the last few lines, which enforce the condition that an award is given to exactly one appearance. The function \\texttt{uniqueAwardCondition} takes a list of award Booleans and returns true if exactly one Boolean in the list is true. The count method counts the number of elements in the list that satisfy the predicate contained in its argument. In this case the predicate is \\texttt{(b: Boolean) => b} which is true precisely when the element of the list is true. So \\texttt{awards.count((b: Boolean) => b)} counts the number of elements in the list that are true.\n\nWe then define the \\texttt{allAwards} element to be the element over lists of Booleans consisting of the award field of all the appearances. Here we have a new notation: \\texttt{appearances.map(\\_.award)}. We have already seen the \\texttt{map} method, which applies a function to every element of a list and returns a new list consisting of the results. In this case, the argument to map is the function \\texttt{\\_.award}. This is shorthand for a function of one argument in which the argument appears once in the body and in which the type of the argument is known. Here, the type of the argument is clearly an appearance. We could have used \\texttt{appearance => appearance.award}. The notation \\texttt{\\_.award} is short for this. Finally, we impose the \\texttt{uniqueAwardCondition} on \\texttt{allAwards}, ensuring that exactly one appearance is awarded.\n\n\\section{Mutable fields}\n\nUp to this point, all our Figaro programs have been purely functional. All elements have been defined by a \\texttt{val}, and they have been immutable. In principle, all programs can be written in a purely functional style. However, this can make it quite inconvenient to represent situations in which different entities refer to each other. Scala supports both functional and non-functional styles of programming, allowing us to gain the benefits of both.\n\nFor example, let's expand the actors and movies example so that actors have a skill, and the quality of a movie depends on the skill of the actors in it. In turn, the fame of an actor depends on the quality of the movies in which he or she has appeared. We have created a mutual dependence of actors on movies which is hard to represent in a purely functional style. We can capture it in Figaro using the following code:\n\n\\begin{flushleft}\n\\marginpar{This example can be found in MutableMovie.scala}\n\\texttt{import com.cra.figaro.library.compound.CPD\n\\newline import com.cra.figaro.language.\\_\n\\newline \n\\newline class Actor \\{\n\\newline \\tab var movies: List[Movie] = List()\n\\newline \\tab lazy val skillful = Flip(0.1)\n\\newline \\tab lazy val famous =\n\\newline \\tab Flip(Apply(Inject(movies.map(\\_.quality):\\_*), probFamous \\_))\n\\newline \\tab private def probFamous(qualities: Seq[Symbol]) =\n\\newline \\tab if (qualities.count(\\_ == 'high) >= 2) 0.8; else 0.1\n\\newline \\}\n\\newline \n\\newline class Movie \\{\n\\newline \\tab var actors: List[Actor] = List()\n\\newline \\tab lazy val actorsAllGood = Apply(Inject(actors.map(\\_.skillful):\\_*), (s: Seq[Boolean]) => !(s.contains(false)))\n\\newline lazy val probLow = \n\\newline Apply(actorsAllGood, (b: Boolean) => if (b) 0.2; else 0.5)\n\\newline lazy val probHigh = \n\\newline Apply(actorsAllGood, (b: Boolean) => if (b) 0.5; else 0.2)\n\\newline lazy val quality = \n\\newline Select(probLow -> 'low, Constant(0.3) -> 'medium, probHigh -> 'high)\n\\newline \\}\n\\newline \n\\newline class Appearance(actor: Actor, movie: Movie) \\{\n\\newline \\tab actor.movies ::= movie\n\\marginpar{\\texttt{::=} is Scala shorthand for list concatenation}\n\\newline \\tab movie.actors ::= actor\n\\newline \n\\newline \\tab def probAward(quality: Symbol, famous: Boolean) =\n\\newline \\tab \\tab (quality, famous) match \\{\n\\newline \\tab \\tab case ('low, false) => 0.001\n\\newline \\tab \\tab case ('low, true) => 0.01\n\\newline \\tab \\tab case ('medium, false) => 0.01\n\\newline \\tab \\tab case ('medium, true) => 0.05\n\\newline \\tab \\tab case ('high, false) => 0.05\n\\newline \\tab \\tab case ('high, true) => 0.2\n\\newline \\tab \\tab \\}\n\\newline \\tab lazy val award = SwitchingFlip(Apply(movie.quality, actor.famous, (q: Symbol, f: Boolean) => probAward(q, f)))\n\\newline \\}\n\\newline \n\\newline val actor1 = new Actor \n\\newline val actor2 = new Actor \n\\newline val actor3 = new Actor \n\\newline val movie1 = new Movie \n\\newline val movie2 = new Movie\n\\newline val appearance1 = new Appearance(actor1, movie1)\n\\newline val appearance2 = new Appearance(actor2, movie2) \n\\newline val appearance3 = new Appearance(actor3, movie2) \n\\newline actor3.famous.observe(true) \n\\newline movie2.quality.observe('high)\n\\newline \n\\newline // Ensure that exactly one appearance gets an award. \n\\newline def uniqueAwardCondition(awards: List[Boolean]) =\n\\newline \\tab awards.count((b: Boolean) => b) == 1 \n\\newline val allAwards: Element[List[Boolean]] =\n\\newline \\tab Inject(appearances.map(\\_.award):\\_*)\n\\newline allAwards.setCondition(uniqueAwardCondition)\n}\n\\end{flushleft}\n\nFirst, note that the \\texttt{Actor} class has a \\texttt{movies} field, whose purpose is to indicate the list of movies the actor has appeared in. Likewise, the \\texttt{Movie} class has an \\texttt{actors} field to represent the actors who appear in it. If these fields were immutable, we would need to create all the movies an actor appears in before we create the actor, and we would need to create all the actors appearing in a movie before the movie, which is impossible. Therefore, we use mutable variables, which are indicated in Scala by the \\texttt{var} keyword.\n\nThe initial value of both \\texttt{movies} and \\texttt{actors} is an empty list. We add elements to them later. In fact, whenever we create an appearance, we make sure to add the movie to the actor's list of movies and vice versa. This is achieved by the first two lines of the \\texttt{Appearance} class. The notation \\texttt{actor.movies ::= movie} is short for \\texttt{actor.movies = movie :: actor.movies}, which prepends movie to the current \\texttt{actor.movies} list, and replaces the current list with the new list. The \\texttt{::=} notation is a variant of the familiar \\texttt{+=} notation common in many languages.\n\nThe \\texttt{Actor} class has \\texttt{skillful} and \\texttt{famous} fields. Rather than an ordinary \\texttt{val}, each of these fields is defined to be \\texttt{lazy val}, which means that their contents are not determined until they are required by some other computation. This is necessary for us because their contents can depend on the list of movies the actor appears in. For example, whether the actor is famous depends on whether at least two movies have high quality, as defined by \\texttt{probFamous}. (Comment on the notation: the underscore after \\texttt{probFamous} is required here to tell Scala that what is desired is the \\texttt{probFamous} function itself, not its application to arguments.) If \\texttt{famous} was an ordinary \\texttt{val}, its value (an \\texttt{Element[Boolean]}) would be computed at the point it is defined, so it would use an empty list of movies. Because we want to use the correct list of movies in defining it, we postpone evaluating it until the movies list has been filled. For \\texttt{actor3}, this will happen when we make the observation \\texttt{actor3.famous.observe(true)}, which we make sure to delay until after all the appearances have been created. For other actors, the \\texttt{famous} field will be evaluated even later, during inference. Care should be taken with declaring elements as lazy. Side effects and unintended consequences can occur if a lazy element declared outside a Chain is first required (i.e., created) during the execution of a Chain.\n\nDo not hesitate to use mutation if it will help you organize your program in a logical way. In one application, we have found it convenient to use a hash table that maps concepts to their associated elements. This allowed us to create the element associated with a concept as the concept was introduced. If we later had to refer to the same concept again, we could easily access its element.\n\n\\section{Universes}\n\nA central concept in Figaro is a \\emph{universe}. A universe is simply a collection of elements. Reasoning algorithms operate on a universe (or, as we shall see for dependent universe reasoning, on multiple connected universes). Most of the time while using Figaro, you will not need to create a new universe and can rely on the default universe, which is just called \\texttt{universe}. It can be accessed using\n\n\\begin{flushleft}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline import com.cra.figaro.language.Universe.\\_}\n\\end{flushleft}\n\nIf you do need a different universe, you can call \\texttt{Universe.create\\-New()}. This creates a new universe and sets the default universe to it. If you are going to need the old default universe, you will need a way to refer to it. You could use\n\n\\begin{flushleft}\n\\texttt{val u1 = Universe.universe\n\\newline val u2 = Universe.createNew()}\n\\end{flushleft}\n\n\\texttt{u1} will now refer to the old default universe while \\texttt{u2} refers to the new one. Every element belongs to exactly one universe. Ordinarily, when an element is created, it is assigned to the current default universe. As we will see below when we talk about element collections, it is possible to assign a particular element to a different universe from the current default.\n\nElements can be activated or deactivated. Elements that are inactive are not operated on by reasoning algorithms. Elements are active when created. To deactivate an element \\texttt{e} use \\texttt{e.deactivate()}; to reactivate it, use \\texttt{e.activate()}. When a compound element is created that uses a parent element, the parent must already be active.\n\nYou can get a list of all active elements in universe \\texttt{u} using \\texttt{u.active\\-Elements}. There are many more methods of a universe that are useful for writing reasoning algorithms. See the documentation in Universe.scala for more details.\n\n\\section{Names, element collections, and references}\n\nSuppose we want to create a PRM in which we are uncertain about the value of an attribute whose value is itself an instance of another class (which is called reference uncertainty). For example, suppose we have the following classes and instances:\n\n\\begin{flushleft}\n\\marginpar{This example can be found in CarAndEngine.scala}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline \n\\newline abstract class Engine \\{ val power : Element[Symbol] \\}\n\\newline class V8 extends Engine \\{\n\\newline \\tab val power = Select(0.8 -> 'low, 0.2 -> 'high) \n\\newline \\}\n\\newline class V6 extends Engine \\{\n\\newline \\tab val power = Select(0.8 -> 'low, 0.2 -> 'high)\n\\newline \\}\n\\newline \n\\newline object MySuperEngine extends V8 \\{\n\\newline \\tab val power = Constant('high)\n\\newline \\}\n\\newline class Car \\{\n\\newline \\tab val engine = Uniform[Engine](new V8, new V6, MySuperEngine)\n\\newline \\tab val speed = CPD(?,\n\\newline \\tab 'high -> Constant(90.0),\n\\newline \\tab 'medium -> Constant(80.0),\n\\newline \\tab 'low -> Constant(70.0))\n\\newline \\}\n}\n\\end{flushleft}\n\nWe want the speed of the car to depend on the power of its engine, but we have uncertainty over what the engine actually is. What should we put in place of the question mark? The obvious choice is \\texttt{engine.power}, but this does not work because engine is an \\texttt{Element\\-[Engine]}, not an instance of \\texttt{Engine}.\n\nTo get around this problem, Figaro provides \\emph{names} and \\emph{element collections}. Every element has a name and belongs to an element collection. By default, the name is the empty string and the element collection is the default universe at the time the element is created, which works because universes are element collections. So, most of the time, as in the tutorial to this point, you don't have to worry about the name and element collection of an element. To assign a name and element collection to an element explicitly, you provide an extra pair of arguments when creating it.\n\nWe can give the engine a name and make it belong to the car as an element collection as follows:\n\n\\begin{flushleft}\n\\texttt{class Car extends ElementCollection \\{\n\\newline \\tab val engine = Uniform[Engine](new V8, new V6, MySuperEngine)(\"engine\", this)\n\\newline \\}\n}\n\\end{flushleft}\n\nIn the first line we make the \\texttt{Car} class inherit from \\texttt{ElementCollect\\-ion}, so that every instance of \\texttt{Car} is an element collection. In the fourth line, we assign engine the name \"engine\" and add it to the instance of Car being created, which is referred to by \\texttt{this} within the \\texttt{Car} class. We similarly make the abstract \\texttt{Engine} class inherit element collections and assign \\texttt{power} the name \"power\" within \\texttt{V*, V6, and MySuperEngine} within each subclass of \\texttt{Engine}.\n\nAn element collection, like a universe, is simply a set of elements. The difference is that a universe is also a set of elements on which a reasoning algorithm operates. An element collection provides the ability to refer to an element by name. For example, if \\texttt{car} is an instance of \\texttt{Car}, we can use \\texttt{car.get[Engine](\"engine\")} to get at the element named \"engine\". The \\texttt{get} method takes a type parameter, which is the value type of the element being referred to. The notation \\texttt{[Engine]} specifies this type parameter, and serves to make sure that the expression \\texttt{car.get[Engine](\"engine\")} has type \\texttt{Element[Engine]}.\n\nThe key ability of element collections that allows them to solve our puzzle is their ability to get at elements embedded in the value of an element. It uses \\emph{references} to do this. A reference is a series of names separated by dots. For example, \"engine.power\" is a reference. When we call \\texttt{car.get[Symbol](\"engine.power\")}, it refers to the element named \"power\" within the \\emph{value} of the element named \"engine\" within the car. The value of this expression is a \\texttt{ReferenceElement} that captures the uncertainty about which power element is actually being referred to. In a particular state of the world, i.e., an assignment of values to all elements, it is possible to determine the value of engine and therefore which \\texttt{power} element is being referred to. So a \\texttt{ReferenceElement} is a deterministic element that defines a way to get its value in any possible world.\n\nSo, finally, the answer to our puzzle is that in place of the question mark, we put \\texttt{get[Symbol](\"engine.power\")}. This applies the get method to the instance of \\texttt{Car} being created. Here is the full example:\n\n\\begin{flushleft}\n\\marginpar{This example can be found in MultiValuedReferenceUncertainty.scala}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline\n\\newline abstract class Engine extends ElementCollection \\{\n\\newline \\tab val power : Element[Symbol]\n\\newline \\}\n\\newline class V8 extends Engine \\{\n\\newline \\tab val power = Select(0.8 -> 'low, 0.2 -> 'high)(\"power\", this)\n\\newline \\}\n\\newline class V6 extends Engine \\{\n\\newline \\tab val power = Select(0.8 -> 'low, 0.2 -> 'high)(\"power\", this)\n\\newline \\}\n\\newline object MySuperEngine extends V8 \\{\n\\newline \\tab val power = Constant('high)(\"power\", this)\n\\newline \\}\n\\newline class Car extends ElementCollection \\{\n\\newline \\tab val engine = Uniform[Engine](new V8, new V6, MySuperEngine)(\"engine\", this)\n\\newline \\tab val speed = CPD(\n\\newline \\tab get[Symbol](\"engine.power\"),\n\\newline \\tab 'high -> Constant(90.0),\n\\newline \\tab 'medium -> Constant(80.0),\n\\newline \\tab 'low -> Constant(70.0))\n\\newline )\n\\newline \\}\n}\n\\end{flushleft}\n\n\\section{Multi-valued references and aggregates}\n\nThe previous subsection described how to refer to elements using references that identify a single element. A feature of PRMs is the ability to define multi-valued relationships, where an entity is related to multiple entities via an attribute. In Figaro, we use multi-valued references and aggregates to capture these kinds of situations. For example:\n\n\\begin{flushleft}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline \n\\newline class Component extends ElementCollection \\{\n\\newline \\tab val f = Select(0.2 -> 2, 0.3 -> 3, 0.5 -> 5)(\"f\", this)\n\\newline \\}\n\\newline \n\\newline val specialComponent1 = new Component \n\\newline val specialComponent2 = new Component\n\\newline \n\\newline def makeComponent() =\n\\newline \\tab Select(0.1 -> specialComponent1,\n\\newline \\tab 0.2 -> specialComponent2,\n\\newline \\tab 0.7 -> new Component)\n\\newline \n\\newline class Container extends ElementCollection \\{\n\\newline \\tab val components = MakeList(Select(0.5 -> 1, 0.5 -> 2), makeComponent)(\"components\", this)\n\\newline \n\\marginpar{The body of the sum function is shorthand notation for Scala's fold function. Fold iterates through a sequence and applies a function to the previous\nresult and each new entry in turn. The (\\_ + \\_ ) notation will add the previous value to each value in \\texttt{xs}.}\n\\newline val sum = getAggregate((xs: MultiSet[Int]) => (0 /: xs)(\\_ + \\_))(\"components.f\") \n\\newline \\}\n}\n\\end{flushleft}\n\nFirst, we create a \\texttt{Component} class with an element named \"f\". We then define two specific instances of \\texttt{Component}. Next, we define a \\texttt{makeComponent} function that either produces one of the specific instances or a new instance of \\texttt{Component} that is distinct from all other instances. We then define a \\texttt{Container} class that contains components. Now, the contained components are a list that has either one or two elements, each produced by \\texttt{makeComponent}. We then create a \\texttt{sum} element that aggregates the values of all elements referred to by \"components.f\"; that is, the values of the elements named \"f\" in all the values of the element named \"components\".\n\nMulti-valued references have \"set semantics\". If the same element appears more than once as the target of the reference, it only contributes one value to the aggregate. So, if the components list has two components, both of which are \\texttt{specialComponent1}, whose value is 2, the value of the aggregate will be 2, not 4. On the other hand, if two different target elements both have the same value, both values contribute to the aggregate. For example, if the components are \\texttt{specialComponent1} and \\texttt{specialComponent2}, and both have value 2, the value of the aggregate is 4.\n\nA comment on the code: The code defining \\texttt{sum} might look mysterious. This code takes a list of integers and returns their sum. This is a standard Scala idiom that unfortunately is a bit obscure if you're not familiar with it. It is used to \"fold\" a function through a list. We begin with 0 and then repeatedly add the current result to the next element of the list until the list is exhausted. The notation \\texttt{(\\_ + \\_)} is shorthand for the function that takes two arguments and adds them. The notation \\texttt{(0 /: xs}) means that this function should be folded through \\texttt{xs}, starting from 0.\n\n\\section{Open Universe Models}\n\nWe close this section by showing how Figaro can be used to represent \"open universe\" situations. An open universe situation is one in which we don't know what objects are there, how many there are, which objects are the same as which other objects, and so on. In our example situation, there are an unknown number of sources that is geometrically distributed. Each source is uniformly distributed between 0 and 1. There is some number of observed samples, each drawn from a single unknown source. This is the classic data association problem in which we want to determine which sample comes from which source, and in particular which two samples actually come from the same source. The Figaro code for the example is as follows:\n\n\\begin{flushleft}\n\\marginpar{This example can be found in OpenUniverse.scala}\n\\texttt{import com.cra.figaro.language.\\_\n\\newline import com.cra.figaro.library.atomic.continuous.Uniform \n\\newline import com.cra.figaro.library.atomic.continuous.Normal \n\\newline import com.cra.figaro.library.atomic.discrete.Geometric \n\\newline import com.cra.figaro.library.compound.\\{MakeList, IntSelector\\}\n\\newline \n\\newline def source(): Element[Double] = Uniform(0.0, 1.0)\n\\newline \n\\newline val numSources = Geometric(0.9)\n\\newline \n\\newline val sources = MakeList(numSources, source \\_)\n\\newline \n\\newline class Sample \\{\n\\newline \\tab val sourceNum = IntSelector(numSources)\n\\newline \\tab val source =\n\\newline \\tab Apply(sources, sourceNum, (s: Seq[Double], i: Int) => s(i))\n\\newline \\tab val position =\n\\newline \\tab NonCachingChain(source, (x: Double) => Normal(x, 1.0))\n\\newline \\}\n\\newline \n\\newline val sample1 = new Sample \n\\newline val sample2 = new Sample\n\\newline \n\\newline val equal = sample1.source === sample2.source \n\\newline \n\\newline sample1.position.addCondition((y: Double) => y >= 0.7 \\&\\& y < 0.8)\n\\newline sample2.position.addCondition((y: Double) => y >= 0.7 \\&\\& y < 0.8) \n}\n\\end{flushleft}\n\nMost of this should be self-explanatory at this point. There are a couple of interesting new element classes being used. \\texttt{MakeList} takes an element over integers and a function that generates elements over a certain type (in this case doubles). It returns an element over lists of the appropriate type (in this case lists of doubles) whose length is distributed according to the first argument and in which each element is generated according to the second argument. In our example, \\texttt{sources} is a list of sources whose length is geometrically distributed and in which each source is generated according to the source model. A notable aspect of the \\texttt{MakeList} class, which is important for reasoning algorithms, is that the elements generating the values in the list are stored as an infinite lazy stream. Depending on the value of the first argument, the value of the \\texttt{MakeList} is a finite prefix of the values of elements in the stream. As a result of this design, we don't create a completely fresh list each time the length of the list changes. \\texttt{MakeList} could also have been used in the previous section's example to define the \\texttt{components} element of the \\texttt{Container} class.\n\nThe second new element class is \\texttt{IntSelector} which takes an element over integers and returns an element that produces uniformly a number between 0 and the value of its argument (exclusive). This element can be used to generate a random index into a list produced by \\texttt{MakeList}. \\texttt{IntSelector} also has an interesting implementation that has benefits for reasoning algorithms (especially Metropolis- Hastings). The \\texttt{Randomness} is an infinite stream of uniformly distributed doubles between 0 and 1. Given a particular value of the integer argument, the selected index is the one with the highest randomness value in the finite portion of the stream defined by the argument.\n\n\n\n\n", "meta": {"hexsha": "bab17c6fe391ab9332d664533c0d7db7cede3951", "size": 41704, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "FigaroLaTeX/Tutorial/Sections/4CreatingModels.tex", "max_stars_repo_name": "wkretschmer/figaro", "max_stars_repo_head_hexsha": "ab45d86d7f2b23c77d242b15396f0f704d40570c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "FigaroLaTeX/Tutorial/Sections/4CreatingModels.tex", "max_issues_repo_name": "wkretschmer/figaro", "max_issues_repo_head_hexsha": "ab45d86d7f2b23c77d242b15396f0f704d40570c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "FigaroLaTeX/Tutorial/Sections/4CreatingModels.tex", "max_forks_repo_name": "wkretschmer/figaro", "max_forks_repo_head_hexsha": "ab45d86d7f2b23c77d242b15396f0f704d40570c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 79.7399617591, "max_line_length": 1507, "alphanum_fraction": 0.7623968924, "num_tokens": 10566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637505099168, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6363010681017458}} {"text": "%!TEX root = ../notes.tex\n\\section{March 10, 2022}\n\\subsection{Jacobi Symbol \\emph{continued}}\n\\recall\n\\begin{definition*}[Jacobi Symbol]\n Let $b\\in\\ZZ_+$ be odd, and let $a\\in\\ZZ$. We write $b = p-1p_2\\cdots p_m$, where $p_i$ are primes (not necessarily distinct). Then we have\n \\[\\lege{a}{b} = \\lege{a}{p_1}\\lege{a}{p_2}\\cdots \\lege{a}{p_m}\\]\n is called the \\ul{Jacobi symbol}.\n\\end{definition*}\n\nThis generalizes the Legendre symbol. We have basic properties that\n\\begin{align*}\n \\lege{a_1a_2}{b} & = \\lege{a_1}{b}\\lege{a_2}{b} \\\\\n \\lege{a}{b_1b_2} & = \\lege{a}{b_1}\\lege{a}{b_2}\n\\end{align*}\nWe noted that $\\lege{a}{b} = -1$ implies that $a$ is \\emph{not} a quadratic residue mod $b$ but $\\lege{a}{b} = 1$ does not imply $a$ is a quadratic residue mod $b$.\n\nWe also stated analogues of the reciprocity laws for the Legendre symbol.\n\n\\begin{lemma}\\label{lemma:multiplicativity-for-jacobi}\n Let $r, s\\in\\ZZ_+$ be odd. Then\n \\begin{enumerate}[(a)]\n \\item $\\displaystyle\\frac{rs-1}{2}\\equiv \\frac{r-1}{2} + \\frac{s-1}{2}\\mod 2$.\n \\item $\\displaystyle\\frac{r^2s^2 - 1}{8} \\equiv \\frac{r^2-1}{8} + \\frac{s^2 - 1}{8}\\mod 2$.\n \\end{enumerate}\n\\end{lemma}\n\\begin{proof}\n ~\\begin{enumerate}[(a)]\n \\item $(r-1)(s-1)\\equiv 0\\mod 4$. Hence\n \\begin{align*}\n rs - 1 & \\equiv (r-1)(s-1) + r + s - 2\\mod 4 \\\\\n & \\equiv r + s - 2\\mod{4} \\\\\n & \\equiv (r-1) + (s-1)\\mod{4} \\\\\n & \\equiv \\frac{r-1}{2} + \\frac{s-1}{2} \\mod{2}\n \\end{align*}\n which gives (a).\n \\item We follow the same procedure, more or less. $(r^2-1)(s^2-1)\\equiv 0\\mod{16}$, so\n \\begin{align*}\n r^2s^2-1 & \\equiv (r^2-1)(s^2-1)+r^2+s^2-2\\mod{16} \\\\\n & \\equiv (r^2-1)+(s^2-1)\\mod{16} \\\\\n & \\equiv \\frac{r^2-1}{8} + \\frac{s^2-1}{8}\\mod{2}\n \\end{align*}\n which gives (b).\n \\end{enumerate}\n\\end{proof}\n\n\\begin{corollary}\n let $r_1, r_2, \\dots, r_m\\in\\ZZ_+$ be odd. Then\n \\begin{enumerate}[(a)]\n \\item $\\displaystyle\n \\sum_{i=1}^m \\frac{r_i-1}{2} \\equiv \\frac{r_1r_2\\cdots r_m - 1}{2}\\mod 2\n $.\n \\item $\\displaystyle\n \\sum_{i=1}^m \\frac{r_i^2 - 1}{2}\\equiv \\frac{r_1^2r_2^2\\cdots r_m^2 - 1}{8}\\mod 2\n $.\n \\end{enumerate}\n\\end{corollary}\n\\begin{proof}\n By induction on $m$ from \\cref{lemma:multiplicativity-for-jacobi}.\n\\end{proof}\n\nWe restate the reciprocity laws but for Jacobi symbols, \\cref{prop:5.2.2}:\n\\begin{proposition*}[5.2.2 of Text]\n We have the following properties about the Jacobi symbol:\n \\begin{enumerate}[(a)]\n \\item \\[\\lege{-1}{b} = (-1)^{\\frac{b-1}{2}}\\]\n \\item \\[\\lege{2}{b} = (-1)^{\\frac{b^2 - 1}{8}}\\]\n \\item If $a, b\\in\\ZZ_+$, then\n \\[\\lege{a}{b}\\lege{b}{a} = (-1)^{\\frac{a-1}{2}\\frac{b-1}{2}}\\]\n \\end{enumerate}\n\\end{proposition*}\n\\begin{proof}[Proof of \\cref{prop:5.2.2}]\n ~\\begin{enumerate}\n \\item[(a) $+$ (b)] are immediate from the corollary (factor $b$, sum exponents and take the parity of the exponent) and the supplemental laws of quadratic reciprocity.\n \\item[(c)] Let\n \\begin{align*}\n a & =q_1q_2\\cdots q_l \\\\\n b & = p_1p_2\\cdots p_m.\n \\end{align*}\n Then\n \\begin{align*}\n \\lege{a}{b}\\lege{b}{a} & = \\prod_{i}\\prod_{j} \\lege{q_i}{p_j}\\lege{p_i}{q_j} \\\\\n & = (-1)^{\\sum_i \\sum_j \\left(\\frac{q_i-1}{2}\\right)\\left(\\frac{p_j-1}{2}\\right)}\n \\intertext{Applying the corollary, }\n & = (-1)^{\\left(\\sum_i \\frac{q_i - 1}{2}\\right)\\left(\\sum_j \\frac{p_j - 1}{2}\\right)} \\\\\n & = (-1)^{\\left(\\frac{(\\sum_i q_i) - 1}{2}\\right)\\left(\\frac{(\\sum_j p_j) - 1}{2}\\right)} \\\\\n & = (-1)^{\\left(\\frac{a-1}{2}\\right)\\left(\\frac{b-1}{2}\\right)}\n \\end{align*}\n which is as desired!\n \\end{enumerate}\n\\end{proof}\n\\begin{example}\n We try to compute with the Jacobi symbol. Recall \\cref{example:legendre-symbol}\n \\[\\lege{219}{383}\\]\n where we repeatedly factored and flipped. With a Jacobi symbol, we don't need to start with factoring; we can forego factorization of top argument and simply repeatedly flip:\n \\begin{align*}\n \\lege{219}{383} & = - \\lege{383}{219} = -\\lege{164}{219} = - \\lege{4}{219}\\lege{41}{219} = - \\lege{41}{219} \\\\\n & = - \\lege{219}{41} = - \\lege{14}{41} = - \\lege{2}{41}\\lege{7}{41} = -\\lege{7}{41} \\\\\n & = -\\lege{41}{7} = -\\lege{-1}{7} = \\boxed{1}.\n \\end{align*}\n\\end{example}\nWhat we did here is to exploit the fact that \\emph{all} Legendre symbols agree with Jacobi symbols, we treat it as a Jacobi symbol and do `Jacobi-\\emph{like}' manipulations on it.\n\\begin{center}\n \\includegraphics[width=0.4\\textwidth]{images/jacobi_legendre_reveal.jpeg}\n\\end{center}\n\n\\emph{This marks the dividing line between the first half and latter half of the course! Everything up to this point is fair game on the midterm. We also now switch to Stewart and Tall.}\n\n\\subsection{Number Fields}\n\\begin{definition}[Algebraic Numbers]\n A complex number $x$ is called \\ul{algebraic} if it is algebraic over $\\QQ$, i.e., if it satisfies a nonzero polynomial equation over $\\QQ$.\n\n We denote the set of algebraic numbers over $\\QQ$ as $\\overline{\\QQ}$.\n\\end{definition}\n\\begin{proposition}\n The set $\\overline{\\QQ}$ of algebraic numbers is a subfield of $\\CC$. That is, addition and multiplication is closed, and we have inverses for nonzero elements.\n\\end{proposition}\n\\begin{proof}\n The key point is that if $L / K$ is a field extension, then $\\alpha\\in L$ is algebraic over $K$ iff $K(\\alpha)/K$ is finite.\n\n So suppose $\\alpha, \\beta\\in\\overline{\\QQ}$. Then $\\QQ(\\alpha)/\\QQ$ and $\\QQ(\\beta)/\\QQ$ are finite. Thus $\\QQ(\\alpha, \\beta)/\\QQ$ is finite (and all pieces of the associated diamond are finite extensions as well).\n\n \\[\n \\xymatrix{\n \\QQ \\ar[d] \\ar[r] &\\QQ(\\alpha)\\ar[d]\\\\\n \\QQ(\\beta) \\ar[r] &\\QQ(\\alpha, \\beta)}\n \\]\n\n\n Since $\\alpha + \\beta$, $\\alpha - \\beta$, $\\alpha\\beta$ and for $\\beta \\neq 0$, $\\alpha/\\beta \\in \\QQ(\\alpha, \\beta)$. This means that all of these elements are algebraic over $\\QQ$.\n\\end{proof}\n\n\\begin{definition}\n A \\ul{number field} is a subfield $K$ of $\\CC$ such that $[K : \\QQ] < \\infty$.\n\\end{definition}\n\nThus every element of a number field is algebraic, so $K\\subseteq \\overline{Q}$.\n\nBy the definition of a finite extension, every number field has the form\n\\[K = \\QQ(\\alpha_1, \\alpha_2, \\dots, \\alpha_N)\\text{ for some }\\alpha_1, \\dots, \\alpha_n\\in\\overline{\\QQ}\\]\n\nHowever, something stronger than this holds.\n\n\\begin{theorem}[Primitive Element Theorem]\n If $K$ is a number field, then $K = \\QQ(\\theta)$ for some $\\theta\\in\\overline{\\QQ}$.\n\\end{theorem}\n\\begin{proof}[Proof sketch]\n It is enough to show that if\n \\[K = K_1(\\alpha, \\beta), \\]\n then $K = K_1(\\theta)$ for some $\\theta\\in\\overline{\\QQ}$.\n\n Suppose the minimum polynomials (over $\\QQ$) of $\\alpha$ and $\\beta$ respectively are (factored over roots in $\\CC$):\n \\begin{align*}\n (t-\\alpha_1)(t-\\alpha_2)\\cdots (t-\\alpha_n)\\qquad \\alpha_1 & = \\alpha \\\\\n (t-\\beta_1)(t-\\beta_2)\\cdots (t-\\beta_n)\\qquad \\beta_1 & = \\beta \\\\\n \\end{align*}\n These polynomials are separable. Hence for each $i$ and each $k\\neq 1$, there exists at most one $x\\in K_1$ such that\n \\[\\alpha_i + x\\beta_k = \\alpha_1 + x\\beta_1.\\]\n (This only holds for more $x$ when you have $\\beta_k$ and $\\beta_1$ colliding). There are only finitely many of these equations, so we can choose a nonzero $c\\in K_1$ such that\n \\[\\alpha_i + c\\beta_k \\neq \\alpha_1 + c\\beta_1\\]\n for any $1\\leq i\\leq n$ and $2\\leq k\\leq m$.\n\n Define $\\theta = \\alpha + c\\beta$. We claim that\n \\[K_1(\\alpha, \\beta) = K_1(\\theta)\\]\n for which the proof is on page 39 of Stewart Tall.\n\\end{proof}", "meta": {"hexsha": "334d8f7d64a8b0221fad204938ee50fc0f467021", "size": 8315, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/2022-03-10.tex", "max_stars_repo_name": "jchen/math1560-notes", "max_stars_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_stars_repo_licenses": ["BSL-1.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2022-02-02T15:41:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-03T20:28:48.000Z", "max_issues_repo_path": "lectures/2022-03-10.tex", "max_issues_repo_name": "jchen/math1560-notes", "max_issues_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_issues_repo_licenses": ["BSL-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lectures/2022-03-10.tex", "max_forks_repo_name": "jchen/math1560-notes", "max_forks_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_forks_repo_licenses": ["BSL-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.0903614458, "max_line_length": 218, "alphanum_fraction": 0.5704149128, "num_tokens": 2827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.6363010641098987}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage{latexsym}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{graphicx}\n\\usepackage{float}\n\\usepackage{wrapfig}\n\\usepackage{pseudocode}\n\\usepackage{url}\n\\usepackage[backref, colorlinks=true, citecolor=red, urlcolor=blue, pdfauthor={Jyh-Ming Lien}]{hyperref}\n\\graphicspath{{images/}}\n\n\\newcommand{\\handout}[5]{\n \\noindent\n \\begin{center}\n \\framebox{\n \\vbox{\n \\hbox to 5.78in { {\\bf } \\hfill #2 }\n \\vspace{4mm}\n \\hbox to 5.78in { {\\Large \\hfill #5 \\hfill} }\n \\vspace{2mm}\n \\hbox to 5.78in { {\\em #3 \\hfill #4} }\n }\n }\n \\end{center}\n \\vspace*{4mm}\n}\n\n\\newcommand{\\lecture}[4]{\\handout{#1}{#2}{#3}{#4}{#1}}\n\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{corollary}[theorem]{Corollary}\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{observation}[theorem]{Observation}\n\\newtheorem{proposition}[theorem]{Proposition}\n\\newtheorem{definition}[theorem]{Definition}\n\\newtheorem{claim}[theorem]{Claim}\n\\newtheorem{fact}[theorem]{Fact}\n\\newtheorem{assumption}[theorem]{Assumption}\n\n% 1-inch margins, from fullpage.sty by H.Partl, Version 2, Dec. 15, 1988.\n\\topmargin 0pt\n\\advance \\topmargin by -\\headheight\n\\advance \\topmargin by -\\headsep\n\\textheight 8.9in\n\\oddsidemargin 0pt\n\\evensidemargin \\oddsidemargin\n\\marginparwidth 0.5in\n\\textwidth 6.5in\n\n\\parindent 0in\n\\parskip 1.5ex\n%\\renewcommand{\\baselinestretch}{1.25}\n\n\\begin{document}\n\n\\lecture{Voronoi Stippling Report}{Fall 2017}{Anthony Sermania}{Computational Geometry}\n\n\\section{Summary of the two methods}\n\n\\subsection{hedcuter method}\nThe hedcuter method is as follows:\n\nHedcut:\n\\begin{enumerate}\n\\item Create a set of n random initial points (keep on random probability, favoring darker areas)\n\\item Compute the Weighted Centroidal Voronoi Tessellation (CVT)\n\\item Generate Disks\n\\end{enumerate}\n\nComputing The CVT:\n\\begin{enumerate}\n\t\\item While (we have fewer iterations than the limit AND the max-displacement is greater than some threshold:\n\t\\begin{enumerate}\n\t\\item Computer the Voronoi\n\t\\item Shift Voronoi cell sites towards cell centroids (and record displacement\n\t\\end{enumerate}\n\\end{enumerate}\n\n\nCompute the Voronoi\n\\begin{enumerate}\n\\item For each input site\n\t\\begin{enumerate}\n\t\\item redirect site to corresponding position in resized image (fake enhanced resolution)\n\t\\item determine distance based on color of site\n\t\\item add site to open list/heap\n\t\\end{enumerate}\n\t\n\\item while heap is not empty\n\\begin{enumerate}\n\t\\item cell <- max heap element (pop)\n\t\\item if distance has been updated or cell has been visited\n\t\t\\begin{enumerate}\n\t\t\\item skip\n\t\t\\end{enumerate}\n\t\t\n\t\\item for each neighbor of cell\n\t\\begin{enumerate}\n\t\\item if cell.distance+neighbor.distance < neighbor.distance\n\t\\begin{enumerate}\n\t\\item set neighbor.distance <- cell.distance+neighbor.distance\n\t\t\t\\item make neighbor.root <- cell.root\n\t\t\t\\item push neighbor into heap\n\t\\end{enumerate}\n\t\t\t\n\t\\end{enumerate}\n\t\t\n\\end{enumerate}\n\t\n\\item For each cell in image\n\t\\begin{enumerate}\n\t\\item determine root of cell\n\t\\item add cell to cell.root's coverage\n\t\\end{enumerate}\n\t\n\\item Remove all cells with coverage.empty\n\\end{enumerate}\n\n\nMove voronoi cell sites to their new locations (and record displacement)\n\\begin{enumerate}\n\\item For each cell\n\\begin{enumerate}\n\\item calculate new position as average of positions in coverage\n\\item Generate high res img\n\\item newpos <- (0, 0), total <- 0\n\\item for each cell c in cell.coverage\n\\begin{enumerate}\n\\item newpos += c.distance*c.pos\n\\item total += c.distance\n\\end{enumerate}\n\n\\item newpos /= total to normalize\n\\item newpos /= subpixels to redirect to original image\n\\item store manhattan distance from new to old position\n\\end{enumerate}\n\n\\item return the max displacement (manhattan distance) or average of displacements (depending on termination mode)\n\\end{enumerate}\n\n\nGenerate Disks\n\\begin{enumerate}\n\\item Collect the darkness/color of the area in a voronoi cell\n\\item Create a disk at cell site (center)\n\\item Color/darken cell according to the average color/darkness of the pixels i the cell\n\\end{enumerate}\n\n\n\\subsection{voronoi method}\nThis one was a little bit tougher to work through mentally. At the highest level it uses Lloyd's Algorithm as so:\n\nWhile generating points (xi) not converged to centroid\n\\begin{enumerate}\n\\item compute voronoi diagram of xi\n\\item compute centroid Ci\n\\item move generating point xi to centroid Ci\n\\end{enumerate}\n\n\nThe Voronoi diagram computation is done differently here, utilizing an entire Voronoi Generation class that appears to implement a half-edge mechanism for doing it's work after some initial set up. It is extremely well commented; rather than just essentially replicating the comments, here is an overview of the half-edge voronoi generation:\n\n\\begin{enumerate}\n\\item While there are events:\n\\begin{enumerate}\n\\item If the lowest site is lower than the lowest vector intersection process the site\n\\begin{enumerate}\n\\item Get left and right half edges, plus the edge of the left halfedge (or the bottom site if there's no edge)\n\\item create a bisector for the left edge (or bottom site) and if it intersects the left edge shuffle the vertices so the left edge and bisector share a vertex.\n\\item Create a new half edge right of the first (only at the moment) bisector\n\\item if this new bisector intersects with the right halfedge, add the bisector halfedge to the ordered linked list of verts\n\\item get next site\n\\end{enumerate}\n\n\\item else process the lowest vector intersection\n\\begin{enumerate}\n\\item Get the halfedge for the intersection vector, as well as the surrounding halfedges\n\\item get the sites of the of the exterior halfedges\n\\item consider a circle passing through the triple of sites obtained\n\\item get the vertex that caused the event\n\\item set the endpoints of the left and right half edges to be this vertex\n\\item mark the lowest half edge for deletion\n\\item clear events for right halfedge\n\\item mark right halfedge for deletion\n\\item if the site to left of the event is higher than site to the right, swap them\n\\item create an edge between the two sites\n\\item create a half edge for this edge\n\\item insert the new bisector to the right of the left halfedge\n\\item cleanup (reposition half-edges not intersecting the bisector that should)\n\\end{enumerate}\n\n\\end{enumerate}\n\n\\end{enumerate}\n\n\nThat still got pretty verbose, but it's not quite done. Once this routine has run, the generator that ran it is used to collect the edges before completing (this process is straightforward and is not detailed here).\n\nNow it moves to step 2, redistributing the stipples (bringing them closer to the centroids of the Voronoi cells). This is done in parallel where applicable.\n\n\\begin{enumerate}\n\\item Do some setup (make pairs of points and edge lists)\n\\item For each pair:\n\\begin{enumerate}\n\\item Calculate Centroid\n\\item update Radius and vertex X/Y positions to be that of the calculated centroid\n\\item determine displacement of point\n\\end{enumerate}\n\\item average displacement of all points\n\\end{enumerate}\n\n\nCalculating a Centroid:\nAccording to the paper, \\textit{Weighted Voronoi Stippling} by Adrian Secord, there are some integrals being done here. The implementation provided however seems to work around that, operating primarily on distance constraints:\n\n\\begin{enumerate}\n\\item Compute the Clip Lines\n\\item For all points in this space if a point is inside it's polygon, it contributes to the area density\n\\item If a cell is not completely white, shift the centroid according to the weighted-spot:area density ratio (that's a weird way to put it, but it's what's happening)\n\\item Iterate over the available edges and determine the nearest and farthest, using one of those distances to determine the size of the stipple (nearest if NoOverlap, Farthest otherwise) and scale by the area:max-area ratio\n\\item Return the pair of the original point and the radius.\n\\end{enumerate}\n\n\n\n\\section{Comparison of the two methods}\nBoth methods saw fair success. I will however claim the voronoi code to be superior overall.\n\nLet's start with hedcut though. For one thing, hedcut was not consistent across multiple runs with the same parameters. Take for instance the following images:\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.4]{fairyeyes-10000-r4-1.png}\n\\includegraphics[scale=0.4]{fairyeyes-10000-r4-2.png}\n\\end{figure}\n\nWe can see here that the two images, run with the same parameters (radius 4, 10,000 stipples) produced different stipplings. By contrast, here are two images from the voronoi code with the equivalent parameters (4,000 stipples, color, overlap)\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.4]{fairyeyes-4k-color-2.png}\n\\includegraphics[scale=0.4]{fairyeyes-4k-color-3.png}\n\\end{figure}\n\nAnd two more from voronoi, this time with no overlap.\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.4]{fairyeyes-noverlap-4k-color.png}\n\\includegraphics[scale=0.4]{fairyeyes-noverlap-4k-color-2.png}\n\\end{figure}\n\nIt is worth noting I discovered later that hedcut might by consistent with sufficiently many iterations (see Improvements).\n\nIn terms of speed, hedcut may be faster in the simplest circumstances, coming out at around the same computatin time for 10,000 stipples as voronoi did for 4,000 and 8,000, however voronoi never seemed to vary much with number of dots- just by image. Hedcut on the other hand does take longer the more dots there are. The two images below are both 100,000 stipples. The image on the left is the hedcut image, created in 16.22 seconds; the one on the right is from voronoi, created in 4.43 seconds (not substantially slower than the 8,000 or 4,000 stipple runs for voronoi).\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.4]{fairyeyes-hedcut-100000.png}\n\\includegraphics[scale=0.4]{fairyeyes-voronoi-100000.png}\n\\end{figure}\n\nBeyond that time difference, the level of image fidelity with this many stipples is vastly superior in the voronoi image. I suspect the difference in speed comes from the use of so many loops in the hedcut code. The difference in quality is undoubtedly due to the stronger implementation of varied stipple sizes (based purely on the voronoi cell size) in the voronoi code. The hedcut code uses a base radius which is scaled, resulting in much weirder stipple sizes. It does however improve it's image fidelity with successive iterations (see below), perhaps even gaining on voronoi for image quality.\n\n\\section{Improvement of hedcuter method}\nWhile I technically made two improvements to hedcut, I can only claim one of them was significant. The insignificant one was some compiler optimizations; this ended up not amounting to any consistently measurable improvement, and moreover wasn't a significant revamp to the code, so I would understand considering it poorly. Given that I was working on Windows in the end, and this required some working knowledge of Visual Studio (2013), maybe it's worth something? I ultimately did not try a third attempt due to time- much of which was wasted either trying to get the project to run (copy-paste .dll saved the day) or interviewing CS 110 students one at a time.\n\nI did however make one very significant change that yielded marvelous improvements. There was a point in move-sites where the code was creating a higher resolution version of our source image to do centroid-relocation math in before redirecting the result back to the source image resolution. Being located in move-sites, it was being run for every voronoi cell, every iteration of the algorithm. This struck me as highly inefficient. So I relocated it to just before we began running the algorithm in full- we never change the source image, so the higher res version should never change either. I was initially confused because I was mistakenly attributing the benefits of greater iterations to this change, but I did discover my folly.\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.5]{sunbros-plain-itr1-100000.png}\n\\includegraphics[scale=0.5]{sunbros-plain-itr3-100000.png}\n\\caption{Original Code. Left is 1 iteration; Right is 3 iterations. A fair improvement can be seen by increasing the number of iterations.}\n\\end{figure}\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.5]{sunbros-lessrescale-itr1-100000.png}\n\\includegraphics[scale=0.5]{sunbros-lessrescale-itr3-100000.png}\n\\caption{The revised version. As before, left is 1 iteration and right is 3. The images are quite similar to their original counterparts. It is possible the hedcut algorithm gains consistency with increasing iterations- a costly but evidently worthwhile gain.}\n\\end{figure}\n\nSo, No significant changes in the images from this. However, consider for a moment the following tables, indicating run times for various runs using the sunbros.png image for 1 or 3 iterations, before and after the change.\n\n\\begin{figure}[H]\n\\includegraphics[scale=0.8]{hedcut-table.png}\n\\end{figure}\n\\begin{figure}[H]\n\\includegraphics[scale=0.8]{hedcut-results.png}\n\\end{figure}\n\nA 99 percent improvement! That image rescaling must have been taking a hefty toll to run for so many iterations, seemingly unnecessarily!\n\nAfter seeing the improvements made here, I decided to accept the negligible impact of the compiler optimizations; this now performs even better than the voronoi implementation in terms of run time.\n\nI did make several other efforts that ultimately were not completed. I made a few attempts to determine the stipple size solely using the voronoi cell area, but concluded that I would have to backtrack a litte further than I was comfortable doing and risk breaking other key parts of the code. I also swapped the random number generator for the uniform gaussian to the included MT19937 generator during the sample point creation. This had no noticeable difference however (pictures omitted because I'm already lost in ths LaTeX thing).\n\n\\bibliographystyle{plain}\n\\bibliography{report}\n\n\\end{document}\n\n\n", "meta": {"hexsha": "cd1336fe60dc4b2bb58373b535f84a0edf801f3f", "size": 13886, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/report.tex", "max_stars_repo_name": "TheOnePercent44/cs633-hedcut", "max_stars_repo_head_hexsha": "0d5fc6fb0af82dff315f311bd44b1c6b37dd278a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/report.tex", "max_issues_repo_name": "TheOnePercent44/cs633-hedcut", "max_issues_repo_head_hexsha": "0d5fc6fb0af82dff315f311bd44b1c6b37dd278a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/report.tex", "max_forks_repo_name": "TheOnePercent44/cs633-hedcut", "max_forks_repo_head_hexsha": "0d5fc6fb0af82dff315f311bd44b1c6b37dd278a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.9801324503, "max_line_length": 737, "alphanum_fraction": 0.7835229728, "num_tokens": 3502, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637361282707, "lm_q2_score": 0.740174367770488, "lm_q1q2_score": 0.6363010623839583}} {"text": "\\documentclass[11pt,a4paper]{report}\n\\usepackage{amsmath,amsfonts,amssymb,amsthm,epsfig,epstopdf,titling,url,array}\n\\usepackage{enumitem}\n\\usepackage{changepage}\n\\usepackage{graphicx}\n\\usepackage{caption}\n\\theoremstyle{plain}\n\\newtheorem{thm}{Theorem}[section]\n\\newtheorem{prop}[thm]{Proposition}\n\\newtheorem*{cor}{Corollary}\n\\theoremstyle{definition}\n\\newtheorem{defn}{Definition}[section]\n\\newtheorem{conj}{Conjecture}[section]\n\\newtheorem{exmp}{Example}[section]\n\\newtheorem{exercise}{Exercise}[section]\n\\theoremstyle{remark}\n\\newtheorem*{rem}{Remark}\n\\newtheorem*{lem}{Lemma}\n\\newtheorem*{note}{Note}\n\\def\\changemargin#1#2{\\list{}{\\rightmargin#2\\leftmargin#1}\\item[]}\n\\let\\endchangemargin=\\endlist\n\\begin{document}\n\n\n\\section*{Problem}\na) Let $x_0 , ... , x_{n-1}$ and $M$ be positive real numbers. Find real numbers $a_0, ..., a_n$ so that $\\sum_{i = 0}^{n-1} x_i a_i$ is maximized subject to the constraint that $\\sum_{i = 0}^{n-1} a_i ^2 = M$\n\\\\\nb) If you used the definition of the cosine between vectors as part of your solution in a), prove independently that your solutiion is correct (the definition depends on the result being established here).\n\n\\section*{Solution (contributed by Juan Carlos Ramirez)}\nConsider the equivalent problem (controlling over $a$):\nminimize $-x^Ta$\nsubject to $a^Ta=M$.\nLet is relax the constraint to consider the problem\nminimize $-a^Tx$\nsubject to $a^Ta \\leq M$.\nIf we find that the minimum of this relaxed version satisfies $a^Ta=M$ then it is also a minimum of the original problem (since it contains the feasible set for the original problem). The feasible set is a closed ball, so it is compact and since $f_0(a)=-x^Tc is$ linear, it is also continuous and it achieves a minimum in the closed ball (the minimum exists). Let us also note $f_0$ is convex and infinitely differentiable.\nLet $f_1(a) = a^Ta-M$. Then $f_1$ is also convex, inifinitely differentiable, and there is a point $a_s=0$ such that $f_1(a_s)<0$. So the problem satisfies Slater's condition, and since $f_1$ is convex, any local minimum of the problem will satisfy KKT conditions (In particular any global minima), i.e. $(a^*,\\lambda^*)$:\n$f_1(a^*)<=0$ (Primal Feasibility constraint)\n$\\lambda^*>=0$ (Dual Feasibility constraint)\n$\\lambda^* f_1(a^*)=0$ (Complimentary slackness CS)\n$\\nabla f_0(a^*)= -\\lambda^* f_1(a^*)$ (First order condition FOC)\nUsing the FOC, we see that $\\lambda^*$ cannot be 0, since then $-x=0$, but x is a prechosen vector with all positive entries. Therefore, $\\lambda^*>0$ and by CS condition $f_1(x^*)=0$ (this gives us that $*(a^*)^Ta^*=M$ and therefore both minimization problems are have the same minimum).\nUsing FOC again we get\n$$ -x=-\\lambda^*(2a*)$$\n\nor\n$$a^*=x/(2\\lambda^*)$$\n\nand therefore\n$$(a^*)^Ta^*=x^Tx/(2\\lambda^*)^2$$\n\nSo\n$$(\\lambda^*)=\\sqrt{x^Tx}/(2\\sqrt{M}$$\n\nand\n$$a^*=\\sqrt{M}x/\\sqrt{x^Tx}$$\n\n\n\n\\end{document}\n\n", "meta": {"hexsha": "f2d3c656ab47ed14175fd24338c7d3d9c69dae6d", "size": 2878, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "maxDot/maxDot.tex", "max_stars_repo_name": "psteitz/problems", "max_stars_repo_head_hexsha": "c231561593ef7de6264c21d2c78d736866c1b341", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "maxDot/maxDot.tex", "max_issues_repo_name": "psteitz/problems", "max_issues_repo_head_hexsha": "c231561593ef7de6264c21d2c78d736866c1b341", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-01-03T21:08:11.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-03T21:08:11.000Z", "max_forks_repo_path": "maxDot/maxDot.tex", "max_forks_repo_name": "psteitz/problems", "max_forks_repo_head_hexsha": "c231561593ef7de6264c21d2c78d736866c1b341", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.6825396825, "max_line_length": 425, "alphanum_fraction": 0.7213342599, "num_tokens": 927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837743174789, "lm_q2_score": 0.8333245973817158, "lm_q1q2_score": 0.6362298088405859}} {"text": "% Preamble.\n\\documentclass[12pt]{article}\n\\usepackage[margin=1.25in]{geometry}\n\\usepackage[fleqn]{amsmath}\n\\usepackage{textcomp}\n\\usepackage{gensymb}\n\\usepackage{amsfonts}\n\\usepackage{enumitem}\n%\\usepackage{tikz} % Include for figures.\n%\\usepackage{subfiles} % Include for subfiles.\n\n%% Title macros.\n\\newcommand{\\HOMEWORKNUM}{17}\n\\newcommand{\\NAME}{D. Choi}\n\\newcommand{\\DATE}{2020-06-18}\n\n\\title{\\vspace{-2\\baselineskip}MATH 225 - Homework \\#\\HOMEWORKNUM}\n\\author{\\NAME}\n\\date{\\DATE}\n\n%% Formatting options.\n%\\pagenumbering{gobble} % Include for single-page document.\n\n\n% Document.\n\\begin{document}\n\\maketitle\n\n\\section*{1.}\n\\textit{Find the matrix that rotates space $30 \\degree$ counterclockwise about\nthe vector $\\vec{n} = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix}$.}\n\\\\[\\baselineskip]\nThe plane normal to $\\vec{n}$ can be represented by the equation\n\\begin{equation*}\n\t0\n\t=\n\t\\vec{n} \\cdot \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\n\t=\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix}\n\t\\cdot\n\t\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\n\t=\n\tx - z\n\t.\n\\end{equation*}\nLet $\\vec{v}_1 = \\vec{n}$. \\\\\nA vector that is orthogonal to $\\vec{v}_1$ can be found on the plane\n$0 = x - z$:\n\\begin{equation*}\n\t\\vec{v}_2 = \\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\n\t.\n\\end{equation*}\nThe vector that is orthogonal to both $\\vec{v}_1$ and $\\vec{v}_2$ can be found\nby computing the cross product of $\\vec{v}_1$ and $\\vec{v}_2$:\n\\begin{equation*}\n\t\\vec{v}_3 = \\vec{v}_1 \\times \\vec{v}_2 =\n\t\\begin{pmatrix} 0 \\\\ -2 \\\\ 0 \\end{pmatrix}\n\t.\n\\end{equation*}\nA convenient basis for describing the rotation can be constructed from\nnormalizing the orthogonal triple $\\vec{v}_1$, $\\vec{v}_2$, and $\\vec{v}_3$.\nLet such a basis be $\\mathcal{B}$, where\n\\begin{equation*}\n\t\\mathcal{B}\n\t=\n\t\\left\\{ \\hat{v}_1, \\hat{v}_2, \\hat{v}_3 \\right\\}\n\t=\n\t\\left\\{\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{\\sqrt{2}} \\\\\n\t\t\t0 \\\\\n\t\t\t-\\frac{1}{\\sqrt{2}}\n\t\t\\end{pmatrix}\n\t\t,\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{\\sqrt{2}} \\\\\n\t\t\t0 \\\\\n\t\t\t\\frac{1}{\\sqrt{2}}\n\t\t\\end{pmatrix}\n\t\t,\n\t\t\\begin{pmatrix}\n\t\t\t0 \\\\\n\t\t\t-1 \\\\\n\t\t\t0\n\t\t\\end{pmatrix}\n\t\\right\\}\n\t,\n\\end{equation*}\nin contrast to the standard basis $\\mathcal{S}$, where\n\\begin{equation*}\n\t\\mathcal{S}\n\t=\n\t\\left\\{\n\t\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix},\n\t\t\\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix},\n\t\t\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\n\t\\right\\}\n\t.\n\\end{equation*}\n\\newpage \\noindent\nIn basis $\\mathcal{B}$ the rotation of interest is a $30 \\degree$\ncounterclockwise rotation about $\\hat{v}_1$. \\\\\nLet ${\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}$ be the matrix for such a rotation in\nbasis $\\mathcal{B}$ so that\n\\begin{equation*}\n\t{\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}\n\t=\n\t\\begin{pmatrix}\n\t\t1 & 0 & 0 \\\\\n\t\t0 & \\cos(30 \\degree) & -\\sin(30 \\degree) \\\\\n\t\t0 & \\sin(30 \\degree) & \\cos(30 \\degree)\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\nLet ${\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}}$ be the matrix with columns\n$\\hat{v}_1$, $\\hat{v}_2$, and $\\hat{v}_3$ so that\n\\begin{equation*}\n\t{\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}}\n\t=\n\t\\begin{pmatrix}\n\t\t\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\\n\t\t0 & 0 & -1 \\\\\n\t\t-\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\n${\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}}$ can be thought of as the matrix that\nconverts an input ${\\{\\vec{x}\\}}_{\\mathcal{B}}$ (a vector described in basis\n$\\mathcal{B}$) to ${\\{\\vec{x}\\}}_{\\mathcal{S}}$ (the same vector but\ndescribed in basis $\\mathcal{S}$). \\\\\nSimilarly, let ${\\{I\\}}_{\\mathcal{S}}^{\\mathcal{B}}$ be the matrix that\nconverts from basis $\\mathcal{S}$ to $\\mathcal{B}$, so that\n\\begin{equation*}\n\t{\\{I\\}}_{\\mathcal{S}}^{\\mathcal{B}}\n\t=\n\t{\\left( {\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}} \\right)}^{-1}\n\t=\n\t{\\left( {\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}} \\right)}^{\\text{T}}\n\t=\n\t\\begin{pmatrix}\n\t\t\\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} \\\\\n\t\t\\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\\n\t\t0 & -1 & 0\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\nLet ${\\{R\\}}_{\\mathcal{S}}^{\\mathcal{S}}$ be the rotation of interest in\nstandard basis. \\\\\n${\\{R\\}}_{\\mathcal{S}}^{\\mathcal{S}}$ can be computed as the composition of\nfirstly the transformation from basis $\\mathcal{S}$ to $\\mathcal{B}$, then the\nrotation of interest under basis $\\mathcal{B}$, and finally the transformation\nfrom basis $\\mathcal{B}$ back to $\\mathcal{S}$:\n\\begin{align*}\n\t{\\{R\\}}_{\\mathcal{S}}^{\\mathcal{S}}\n\t&=\n\t{{\\{I\\}}_{\\mathcal{B}}^{\\mathcal{S}}}\n\t\\cdot\n\t{{\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}}\n\t\\cdot\n\t{{\\{I\\}}_{\\mathcal{S}}^{\\mathcal{B}}}\n\t\\\\\n\t&=\n\t\\boxed{\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\\n\t\t\t0 & 0 & -1 \\\\\n\t\t\t-\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & 0 \\\\\n\t\t\t0 & \\cos(30 \\degree) & -\\sin(30 \\degree) \\\\\n\t\t\t0 & \\sin(30 \\degree) & \\cos(30 \\degree)\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} \\\\\n\t\t\t\\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\\n\t\t\t0 & -1 & 0\n\t\t\\end{pmatrix}\n\t}\n\t.\n\\end{align*}\n\n\\section*{2.}\n\\textit{Let $\\vec{a} = \\begin{pmatrix} 5 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and\n$\\vec{b} = \\begin{pmatrix} 0 \\\\ 3 \\\\ 4 \\end{pmatrix}$. \\\\\nFind the vector $\\vec{v}$ that is $\\vec{a}$ rotated $40 \\degree$ toward\n$\\vec{b}$.} \\\\[\\baselineskip]\nNote that $\\vec{a} \\cdot \\vec{b} = 0$, meaning that the two vectors are\northogonal. \\\\\nA vector $\\vec{c}$ in the axis of rotation can be found by computing\n\\begin{equation*}\n\t\\vec{c}\n\t=\n\t\\vec{a} \\times \\vec{b}\n\t=\n\t\\begin{pmatrix} 0 \\\\ -20 \\\\ 15 \\end{pmatrix}\n\t.\n\\end{equation*}\nLet the basis $\\mathcal{B}$ be constructed from the triple $\\vec{a}$,\n$\\vec{b}$, and $\\vec{c}$:\n\\begin{equation*}\n\t\\mathcal{B}\n\t=\n\t\\left\\{ \\vec{a}, \\vec{b}, \\vec{c} \\right\\}\n\t=\n\t\\left\\{\n\t\t\\begin{pmatrix} 5 \\\\ 0 \\\\ 0 \\end{pmatrix},\n\t\t\\begin{pmatrix} 0 \\\\ 3 \\\\ 4 \\end{pmatrix},\n\t\t\\begin{pmatrix} 0 \\\\ -20 \\\\ 15 \\end{pmatrix}\n\t\\right\\}\n\t.\n\\end{equation*}\nLet ${\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}$ be the matrix for the rotation of\ninterest about $\\vec{c}$ in basis $\\mathcal{B}$:\n\\begin{equation*}\n\t{\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}\n\t=\n\t\\begin{pmatrix}\n\t\t\\cos(40 \\degree) & -\\sin(40 \\degree) & 0 \\\\\n\t\t\\sin(40 \\degree) & \\cos(40 \\degree) & 0 \\\\\n\t\t0 & 0 & 1\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\nUsing ${\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}$, \n\\begin{gather*}\n\t{\\{\\vec{v}\\}}_{\\mathcal{B}}\n\t=\n\t{\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}\n\t{\\{\\vec{a}\\}}_{\\mathcal{B}}\n\t=\n\t{\\{R\\}}_{\\mathcal{B}}^{\\mathcal{B}}\n\t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\t\\cos(40 \\degree) \\\\\n\t\t\\sin(40 \\degree) \\\\\n\t\t0\n\t\\end{pmatrix}\n\t,\n\t\\\\\n\t\\vec{v}\n\t=\n\t\\cos(40 \\degree)\n\t\\begin{pmatrix} 5 \\\\ 0 \\\\ 0 \\end{pmatrix}\n\t+\n\t\\sin(40 \\degree)\n\t\\begin{pmatrix} 0 \\\\ 3 \\\\ 4 \\end{pmatrix}\n\t+\n\t0\n\t\\begin{pmatrix} 0 \\\\ -20 \\\\ 15 \\end{pmatrix}\n\t=\n\t\\boxed{\n\t\t\\begin{pmatrix}\n\t\t\t5 \\cos(40 \\degree) \\\\\n\t\t\t3 \\sin(40 \\degree) \\\\\n\t\t\t4 \\sin(40 \\degree)\n\t\t\\end{pmatrix}\n\t}\n\t.\n\\end{gather*}\n\n\\end{document}", "meta": {"hexsha": "d7c47583a6c0f1470e3f5fc02f1fd15ae21de9de", "size": 6810, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "usc-20202-math-225-39425/hw17/main.tex", "max_stars_repo_name": "Floozutter/coursework", "max_stars_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.8333245973817158, "lm_q1q2_score": 0.6362297998713579}} {"text": "\\chapter{Permutation representations}\r\n\\section {Basic Results}\r\n{\\bf Lemma 1:} \r\nLet $H \\le G$ and $Hg_1 , ..., Hg_n$ be the cosets; the map\r\n$\\pi(g): \\langle Hg_1, ..., Hg_n \\rangle \\mapsto \\langle Hg_1 g, ..., Hg_n g \\rangle $ \r\nis a map from $G$ to $\\Sigma_n$\r\nwhose kernel is the largest normal subgroup of $G$ in $H$; in fact,\r\n$ker(\\pi) = \\bigcap_{i=1}^n H^{x_i}$.\r\n\\\\\r\n\\\\\r\n{\\bf Corollary:} If $G$ is simple and $G > H$ with $|G:H| = k$ then\r\n$|G| \\mid k!$.\r\n\\section {Imprimitivity}\r\n{\\bf Definition 1:}\r\nA \\emph{system of imprimitivity for permutation \r\ngroup $G$:} is a set, ${\\cal B} = \\{ \\Delta_i\\}$ $|\\Delta_i| > 1$, with\r\nthe property that for $\\Delta \\in {\\cal B}, g \\in G$ either \r\n$\\Delta \\cap g \\Delta = \\emptyset$ or\r\n$\\Delta = g \\Delta$.\r\n\\\\\r\n\\\\\r\n{\\bf Theorem: 1}\r\nIf $H G_a$\r\nstabilises the block containing $a$. So if $G_a$ is not primitive then $G_a$ is not maximal.\r\nIf $G > M > G_a$ and $M$ is maximal, $Ma$ is a set of imprimitivity so $G$ is not primitive.\r\n\\end{quote}\r\n{\\bf Definition 4:}\r\n$G$ acts \\emph{regularly} on $\\Omega$ if $\\forall \\alpha, \\beta \\in \\Omega,\r\n\\exists ! g: \\alpha^g= \\beta$. \r\n\\\\\r\n\\\\\r\n{\\bf Theorem 4:} Let $G$ be $n-$fold transitive on $\\Omega$, $n \\ge 2$ and $N$ a regular\r\nnormal subgroup of $G$ then $n \\le 4$ and \r\n(1) If $n=2$, $N$ is an elementary abelian $p-$group; \r\n(2) If $n=3$, $N$ is an elementary abelian $2-$group or $N=C_3$ and $G=S_3$;\r\n(3) If $n=4$, $N= C_2 \\times C_2$ and $G=S_4$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nLet $n \\geq 2$ and $\\alpha \\in \\Omega$. $G_{\\alpha}$ is $(n-1)$-fold transitive on $\\Omega - \\{ \\alpha \\}$ and\r\nhence on $N^{\\#}$ so $\\exists g: x^g=y, x,y \\in N^{\\#}$. Thus every element of $N^{\\#}$ has the same order, $p$.\r\n$N= {\\mathbb Z}(N)$ and $N$ is abelian (and hence elementary abelian).\r\nIf $n \\geq 3$, $3 \\leq |N|=|\\Omega|$. If $n=3$, $G=S_3$. If $p \\geq 3$, $x_1 \\neq x_1^{-1}=x_2$. Let $x_3$ be another \r\nelement. Then $x_1^g= x_2, x_2^g=x_3$ for some $g$ and this is contradictory so $p=2$.\r\nIf $n \\geq 4$, $N$ is an elementary abelian $2$-group. Let $U= \\langle x_1 \\rangle \\times \\langle x_2 \\rangle$.\r\nIf $N \\neq U$, put $x_3=x_1 x_2$ and choose $x_4 \\notin U$. Then $\\exists g \\in G:$\r\n$x_1^g= x_1$,\r\n$x_2^g= x_2$, and\r\n$x_3^g= x_4$ which is again contradictory.\r\n\\end{quote}\r\n{\\bf Definition 5:}\r\nDefine ${\\cal G}(G, \\Omega )$ as the graph of $G$ acting on $\\Omega$ as follows:\r\n$G$ acts on $\\Omega \\times \\Omega$. Diagonal orbital is\r\n$\\Delta_1 = \\{ (\\alpha , \\alpha )\\}$. If $\\Delta = \\{ (\\alpha , \\beta ) \\}$,\r\n$\\Delta^*= \\{ (\\beta, \\alpha) \\}$. Self paired if $\\Delta^* = \\Delta$.\r\n$\\Delta(\\alpha)= \\{ \\beta : (\\alpha , \\beta ) \\in \\Delta \\}$ --- corresponds to \r\norbits of $G_{\\alpha}$. The \\emph{rank of the permutation group} is number of orbitals.\r\n$\\Delta^p= \\{ (y,x): (x, y) \\in \\Delta \\}$ is called the \\emph{paired orbit}.\r\n\\\\\r\n\\\\\r\n{\\bf Theorem 5:}\r\nOn a self-paired orbit $\\Delta$, the graph ${\\cal G}=(G, X,\\Delta)$ is \r\nsymmetric and $G$ is transitive on edges.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nSince $\\Delta$ is self-paired, the graph is symmetric. \r\n\\end{quote}\r\n{\\bf Theorem 6:}\r\n$G$ is primitive iff ${\\cal G}$ is connected.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\n$G$ is maximal so $G= \\langle G_x , G_y \\rangle , x \\ne y$.\r\n\\end{quote}\r\n{\\bf Definition 6:}\r\nA transitive permutation group is \\emph{regular} if $|X|= |G^X|$ or, equivalently \r\n$|G_x|=1, \\forall x \\in X$ and $G^X$, transitive.\r\n\\\\\r\n\\\\\r\n{\\bf Theorem 7:}\r\nLet $X$ be a faithful primitive $G-set$ with $G_x$ simple. Then either $G$ is simple or\r\nevery non-trivial normal subgroup $H$ of $G$ is a regular normal subgroup.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nSuppose $1 \\ne H \\lhd G$. Since $G$ is primitive, $H$ is transitive. $H_x= H \\cap G_x \\lhd G_x$\r\nso, since $G_x$ is simple, $H_x = 1$ of $H_x= G_x$. If $H_x=1$, $H$ is a regular normal subgroup\r\nof $G$. So $H_x = G_x$ and since $H$ is transitive, $|H:H_x|= |G:G_x|$, so $G$ is simple.\r\n\\end{quote}\r\n{\\bf Definition 7:} $A$ acts \\emph{semi-regularly} on $G$ if $C_G(a)=1, \\forall a \\in A^{\\#}$. \r\n\\\\\r\n\\\\\r\n{\\bf Theorem 8:}\r\nSuppose $A$ acts semi-regularly\r\non $G$. Then (1) $|G| = 1 \\jmod{|A|}$, (2) $A$ is semi-regular on each $A-$invariant\r\nsubgroups factor group of $G$, (3) $\\forall p \\in \\pi(G)$, $\\exists! A-$invariant\r\nSylow $p-$subgroup of $G$, (4) $\\forall a \\in A, g \\mapsto [g,a]$ is a permutation of\r\n$G$, (5) if $2 | |A|, \\exists t: |t|=2, t \\in A: g^t = g^{-1}, g \\in G$ and\r\n$G^{(1)}=1$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\n(1) each orbit of $A$ on $G^{\\#}$ has length $|A|$. (2) is clear (second part comes from coprime action).\r\n(3) By Schur, the set of $A$-invariant Sylow $p$-groups is non-empty. Since $C_G(A)= 1$ is transitive on this\r\nset, the subgroup is unique. (4) $[g,a]= [h,a]$ iff $h g^{-1} \\in C_G(a)$. The commutator map is already an\r\ninjection and since $G$ is finite, it is a bijection. (5) Let $t \\in Inv(A)$ and $g \\in G$. By previous result,\r\n$g= [h,t]$ for some $h \\in G$ and $g^t = (h^{-1} h^t)^t= g^{-1}$. Therefore $x^t =x^{-1}$ and $G$ is abelian.\r\nFinally, if $s \\in Inv(A)$, $s$ inverts $G$ so $st \\in C_A(G)=1$ and $t$ is unique.\r\n\\end{quote}\r\n{\\bf Theorem 9:}\r\nLet $\\Delta$ be an orbit of $G$ and let $\\delta \\in \\Delta$. For each $\\gamma \\in \\Delta$\r\nlet $v(\\gamma ) \\in G$ be such that $\\delta \\mapsto \\gamma$. Finally, suppose $S$\r\ngenerates $G$. \r\nThen $G_{\\delta}= \\langle v(\\gamma)sv(\\gamma^{s})^{-1} | \\gamma \\in \\Delta, s \\in S \\rangle$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nSee, Stellmacher.\r\n\\end{quote}\r\n\\section {Fixed point free automorphisms} \r\n{\\bf Fixed Point Free:} An automorphism, $\\phi$ acting on $G$ is fixed point free if $C_G(\\phi) =1$.\r\n\\\\\r\n\\\\\r\n{\\bf Lemma 1:} Let $\\phi$ be a fixed point free automorphism acting on $G$ with $|\\phi|=n$. Then\r\n(1) $y \\in G$ then $y= x^{-1}(x \\phi)$ for some $x$, and \r\n(2) $\\forall x \\in G, x(x \\phi) (x \\phi^2) \\ldots (x \\phi^{n-1}) = 1$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\n$x^{-1}(x\\phi) = y^{-1}(y\\phi)$ $\\rightarrow$ $y x^{-1} = (y x^{-1})\\phi$, so $x=y$. Thus\r\n$|{x^{-1}(x\\phi): x \\in G}| = |G|-1$ and (1) holds. Since $\\exists y: x= y^{-1}(y\\phi)$, so\r\n$x(x \\phi) (x \\phi^2) \\ldots (x \\phi^{n-1}) =\r\ny^{-1}(y \\phi) (y \\phi^2) \\ldots (y \\phi^{n-1})^{n-1}) = y^{-1}y = 1$.\r\n\\end{quote}\r\n{\\bf Lemma 2:} Let $\\phi$ be a fixed point free automorphism acting on $G$ then $\\phi$ leaves a unique\r\n$S_p$subgroup of $G$ invariant.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nLet $Q \\in S_p(G), (Q)\\phi = y^{-1}Qy$. $y^{-1}= (z\\phi)z^{-1}$ and $y = z(z^{-1}]phi)$.\r\n$Q\\phi = (z\\phi) (z^{-1}Q z) (z^{-1}\\phi)$ and $(z^{-1}Qz)\\phi= z^{-1}Qz$ and thus\r\n$\\phi$ leaves $P=z^{-1}Qz$ fixed. If both $P, Q$ are $\\phi$-invariant and \r\n$Q=x^{-1}Px$ then\r\n$Q=(x^{-1}\\phi)P(x(phi)$ and $(x\\phi)x^{-1} \\in N(P) = N$ but $N$ is $\\phi$-invariant and\r\n$\\phi$ is fixed point free on $N$ so $\\exists z: y=(z\\phi)z^{-1}$.\r\n$(z\\phi)z^{-1}=\r\n(x\\phi)x^{-1}$ $\\rightarrow$ $x=z$.\r\n\\end{quote}\r\n{\\bf Lemma 3:} Let $\\phi$ be a fixed point free automorphism acting on $G$, $H \\lhd G$ and $H= H\\phi$ then\r\n$\\phi$ is fixed point free on $G/H$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nLet ${\\overline G} = G/H$ and ${\\overline x} = {\\overline x}\\phi$. ${\\overline x}^{-1}({\\overline x}\\phi)$\r\nso $y= x^{-1}(x\\phi) \\in H$. Since $\\phi$ induces a fixed point free automorphism on $H$, $y= z^{-1}(z\\phi)$.\r\nSo $x = z$, $x \\in H$ and ${\\overline x} = 1$. Thus $\\phi$ induces a fixed point free automorphism on\r\n${\\overline G}$.\r\n\\end{quote}\r\n{\\bf Lemma 4:} Let $\\phi$ be a fixed point free automorphism acting on $G$ of order $2$, then $G$ is abelian.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nSince $x(x\\phi) = 1$, $x^{-1} = (x\\phi)$. If \r\n$x, y \\in G$, $y^{-1}x^{-1} = (xy)^{-1}= ((x\\phi)(y\\phi))^{-1} = x^{-1} y^{-1}$, so $G$ is abelian.\r\n\\end{quote}\r\n{\\bf Lemma 5:} Let $\\phi$ be a fixed point free automorphism acting on $G$ of order $3$, then $G$ is\r\nnilpotent and $[x, x\\phi]=1, \\forall x \\in G$.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\n$x(x\\phi)(x\\phi^2)=1$, so $x(x\\phi)= (x\\phi^2)^{-1}$ and $x$ commutes with $x\\phi$ for any $x \\in G$.\r\nLet $P$ be the unique $\\phi$-invariant $S_p$ subgroup of $G, \\forall p \\in \\pi(G)$. We show\r\n$P \\lhd G$.\r\nSuppose not and\r\n$Q \\ne P \\in S_p(G)$ and pick $x \\in Q \\setminus P$, put $H=\\langle x, x\\phi \\rangle$.\r\n$H$ is a $p$-group since $[x, x\\phi]=1$ and $H'=1$ and $H \\phi = H$. So $H \\subseteq P$ but then\r\n$x \\in P$.\r\n\\end{quote}\r\n{\\bf Thompson:}\r\nLet $G$ be a transitive permutation group on $X$ and $1 \\ne g \\in G$ fixes\r\nno more than one element then $N= \\{g: X_g= \\emptyset \\}$ is a normal subgroup of $G$.\r\nThompson showed any finite group having a fixed point free automorphism of prime order is nilpotent.\r\n\\begin{quote}\r\n\\emph{Proof:} \r\nLet $G$ be a counterexample of minimal order and $\\phi$ be a fixed point free automorphism of prime order $r$.\r\n$G$ has a proper normal subgroup, $H \\ne 1$ with $H\\phi=H$. By induction, $H$ is nilpotent and $\\phi$\r\nis a fixed point free automorphism on $G/H$ so $G/H$ is nilpotent and $G$ is solvable. If $G$ has no non-trivial\r\nnormal subgroup, which is $H$-invariant. Let $P \\in S_p(G), p \\ne 2$ with $P\\phi=P$. Put\r\n$N= N_G({\\mathbb Z}(J(P)))$, $ {\\mathbb Z}(J(P))) \\; char \\; P$. If $N < G$, $N$ is nilpotent:\r\n$N$ has a normal $p$-complement, $K$. By the Thompson $p$-complement theorem, $G=KP$, $K\\phi = K$ so\r\n$K = 1$ and $P=G$ so $G$ is nilpotent. We may assume $G$ is solvable.\r\nSuppose $H_1, H_2 \\lhd G$ $H_i\\phi = H_i$, $H_1 \\cap H_2 = 1$. ${\\overline G_i} = G/H_i$ is nilpotent\r\nso ${\\overline L} = {\\overline G_1} \\times {\\overline G_2}$ is too. $x\\phi = (H_1x, H_2x)$.\r\n$\\psi: G \\rightarrow {\\overline G_1} \\times {\\overline G_2}$. $G\\psi$ is nilpotent.\r\nLet $N$ a minimal normal subgroup of $G$, $N$ is elementary abelian and ${\\overline G} = G/N$ is nilpotent.\r\n${\\overline G}$ is not a $p$-group. Let ${\\overline Q} \\in S_q({\\overline G})$ with \r\n${\\overline Q} = {\\overline Q} \\phi$, $q \\ne p$. Let ${\\overline M}$ be a minimal $\\phi$-invariant\r\nsubgroup of $\\Omega_1({\\mathbb Z}({\\overline Q}))$. ${\\overline M} \\ne 1$ and ${\\overline M} \\lhd {\\overline G}$\r\nsince ${\\overline G}$ is nilpotent. Let $H$ be the inverse image of ${\\overline M}$ then\r\n$H=NM$ and $M$ is a non-trivial elementary abelian $q$-group. $H \\lhd G$, $H\\phi = H$ and $M\\phi=M$.\r\n$\\phi$ acts irreducibly on $M$. $H \\subset G$ then $H$ is nilpotent.\r\n$M \\; char \\; H \\lhd G$ and $M$ and $N$ are two minimal normal $\\phi$-invariant subgroup and\r\n$M \\cap N \\ne 1$. $C_M(N)\\phi = C_M(N)$ since $\\phi$ acts irreducibly on $M$. Either\r\n$C_M(N) = 1$ or $C_M(N) = N$.\r\nIf $C_M(N) = N$, $G$ is nilpotent so $C_M(N) = 1$. Let $G^*$ be the semidirect product of\r\n$M$ by $\\langle \\phi \\rangle$. $G^*$ acts irreducibly on $N$ as a vector space since\r\n$C_M(N) = 1$ and $\\phi$ is fixed point free. But $G^*$ is a $p'$-group $C_{M^*}(M) = M$ and\r\n$G^*/M$ has order $q$ so $C_N(\\phi) \\ne 1$.\r\n\\end{quote}\r\n{\\bf O-Nan-Scott:} Let $G$ be a finite primitive permutation group of degree $n$ and\r\n$H=soc(G)$. Then either (1) $H$ is a regular elementary abelian $p$ group for\r\nsome $p$ and $G$ is isomorphic to a subgroup of\r\n$AGL_m(P)$ ; or, (2) $H$ is isomorphic to $T^m$ where $T$ is a non-abelian simple group\r\nwith a bunch of conditions.\r\n\\section {Mathieu groups are simple}\r\nWe will use one result from a future section here.\r\n\\\\\r\n\\\\\r\n{\\bf Theorem 1:} \r\nLet $N$ be a finite group $G \\subseteq Aut(N)$. Then $G$ acts as a permutation group on $N^{\\#}$. Further, \\\\\r\n(i) If $G$ is transitive, then $N$ is an elementary abelian $p$-group.\r\n\\\\\r\n(ii) If $G$ is $2$-transitive, then $N$ is either an elementary abelian $2$-group or $|N|=3$.\r\n\\\\\r\n(iii) If $G$ is $3$-transitive, then $|N|=4$.\r\n\\\\\r\n(iv) $G$ cannot be $4$-transitive.\r\n\\\\\r\n\\begin{quote}\r\n\\emph{Proof:} These are all clear.\r\n\\end{quote}\r\n{\\bf Theorem 2:} \r\nLet $G$ be an transitive permutation group and $N \\lhd G$. Then $N$ is ${\\frac 1 2}$ transitive. If\r\n$N \\ne 1$ and $G$ is primitive then $N$ is transitive.\r\n\\begin{quote}\r\n\\emph{Proof:}\r\nLet $B$ be an orbit of $N$ of minimal length. $B^{Ng}= B^{gN} = B^g$, so $B^g$ is a union of orbits of $N$.\r\nBy minimality of $|B|$, $B^g$ is a single $N$ orbit. These form a system of imprimitivity.\r\n\\end{quote}\r\n{\\bf Theorem 3:} \r\nLet $G$ be primitive with no regular normal subgroups. If $G_a$ is simple then $G$ is simple.\r\n\\begin{quote}\r\n\\emph{Proof:}\r\nLet $N \\lhd G$. $N_a \\leq (N \\cap G_a) \\lhd G_a$ so either $N_a =1$ or $N=G_a$. If $N_a=1$, $N$\r\nis a regular normal subgroup.\r\n\\end{quote}\r\n{\\bf Theorem 4:} \r\nLet $G$ be $m$-transitive on $A$, $|A|=n$ with a regular normal subgroup, $N$.\\\\\r\n(i) if $m=2$, then $n = |N|= p^k$\r\n\\\\\r\n(ii) if $m=3$, then $n = 3$ or $n = 2^k$ or $n = 3$.\r\n\\\\\r\n(iii) if $m=4$, then $n = 4$.\r\n\\\\\r\n(iv) We cannot have $m>4$.\r\n\\begin{quote}\r\n\\emph{Proof:}\r\n\\end{quote}\r\n{\\bf Theorem 5:} Let $G$ be transitive and $H < G$ also be transitive. The $G= H G_a$\r\n\\begin{quote}\r\n\\emph{Proof:}\r\nClear.\r\n\\end{quote}\r\n{\\bf Theorem:} The Mathieu groups are simple.\r\n\\begin{quote}\r\n\\emph{Proof:}\r\n\\\\\r\nStep 1: $M_{11}$ is simple.\\\\\r\n\\\\\r\nPut $G = M_{11}$ so $|G| = 11 \\cdot 10 \\cdot 9 \\cdot 8$. Let $P=\\langle x\\rangle$ be a subgroup of order\r\n$11$. $P$ acts transitively on $\\Omega$. If $P \\subseteq A$, $A$, abelian, then $A$ is transitive and regualr so\r\n$A=P$. Thus $C_G(P)=P$. Since $|Aut(P)|=10$, $|N_G(P)/P| \\mid 10$.\r\nIf $2 \\mid N_G(P)$ let $y \\in N_G(P), |y|=2$. $y$ has a fixed point since $11$ is odd and $x^y = x^{-1}$. Thus\r\n$y$ is a product of $5$ transpositions thus $y \\notin A_{11}$ which is a contradiction. So $|N_G(P)/P|$ is $1$ or $5$.\r\nLet $1 \\ne H \\lhd G, H \\ne G$. Since $G$ is primitive, $H$ is transitive and $11 \\mid |H|, P \\subseteq H$.\r\n$G= H G_a$ and $G_a \\subseteq N_G(P)$ so $G= H N_G(P)$ and $N_G(P) \\nsubseteq H$ and $N_H(P)=P$. So, in $H$,\r\n$P$ is in the center of its normalizer and $H$ has a normal $11$ complement, $K$. $K \\lhd G$ and $11 \\nmid |K|$ so\r\n$K = 1$. Thus $H=K$ and $|G|=55$, which is a contradiction.\r\n\\\\\r\nStep 2: $PSL_3(4)$ is simple. This was already shown.\r\n\\\\\r\n\\\\\r\nStep 3:\r\nBy the above result, $M_{12}, M_{22}, M_{23}, M_{24}$ cannot ahve a regular normal subgroup.\r\nThese groups are all $3$-transitive. $M_{11} = (M_{12})_a$ so $M_{12}$ is simple.\r\n$PSL_3(4) = (M_{22})_a$, $M_{22} = (M_{23})_a$, and $M_{23} = (M_{24})_a$ proving their simplicity.\r\n\\end{quote}\r\n", "meta": {"hexsha": "8174683e23a1c894ec46be1ea53fe51c77559685", "size": 17353, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "groups/gtPermutations.tex", "max_stars_repo_name": "jlmucb/class_notes", "max_stars_repo_head_hexsha": "b8571df2dca933f6594a16eb02b581d38ca4ccfd", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "groups/gtPermutations.tex", "max_issues_repo_name": "jlmucb/class_notes", "max_issues_repo_head_hexsha": "b8571df2dca933f6594a16eb02b581d38ca4ccfd", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "groups/gtPermutations.tex", "max_forks_repo_name": "jlmucb/class_notes", "max_forks_repo_head_hexsha": "b8571df2dca933f6594a16eb02b581d38ca4ccfd", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.0382352941, "max_line_length": 122, "alphanum_fraction": 0.5947098484, "num_tokens": 6871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673223709251, "lm_q2_score": 0.7826624789529375, "lm_q1q2_score": 0.6362007535866648}} {"text": "\\documentclass[a4paper,10pt]{article}\n\\usepackage{mystyle}\n\n\\begin{document}\n\n\\section{Matrices and Vector Spaces}\n\n\\begin{defn}[Echelon Form]\n\tA matrix is in Echelon form if:\n\t\\begin{enumerate}\n\t\t\\item all non-zero rows come before zero rows\n\t\t\\item below each pivot is a column of zeros\n\t\t\\item each pivot is to the right of the previous pivot\n\t\\end{enumerate}\n\n\tIf each pivot is $1$ then the matrix is in row echelon form and if\n\tthere is a column of zeros above each pivot, the matrix is in reduced\n\trow echelon form (RRE).\n\\end{defn}\n\n\\begin{defn}[Rank]\n\tLet $A$ be an $n \\times m$ matrix. Its rank is equal to the number of\n\tnon-zero rows in its echelon form. This is denoted by $\\rho(A)$.\n\\end{defn}\n\n\\begin{defn}[Subspace]\n\tLet $U \\subset \\RR^n$. $U$ is a subspace if:\n\t\\begin{enumerate}\n\t\t\\item $\\bm{0} \\in U$\n\t\t\\item $U$ is closed under addition and scalar multiplication\n\t\\end{enumerate}\n\\end{defn}\n\n\\begin{defn}[Vector Space]\n\t$V$ is a vector space of a field $\\FF$ if:\n\t\\begin{enumerate}\n\t\t\\item addition is commutative and associative\n\t\t\\item neutral and inverse additive elements exist\n\t\t\\item scalar multiplication is commutative and associative\n\t\t\\item $1$ is the neutral scalar\n\t\t\\item $(a+b)\\bm{d} = a\\bm{d} + b\\bm{d} \\quad \\forall a,b \\in \\FF, \\bm{d} \\in V$\n\t\t\\item $(\\bm{a} + \\bm{b})d = \\bm{a}d + \\bm{b}d \\quad \\forall \\bm{a},\\bm{b} \\in V, d \\in \\FF$\n\t\\end{enumerate}\n\\end{defn}\n\n\\begin{defn}[Null Space]\n\tIf $A \\in M_{m \\times n}(\\RR)$, the null space is defined as:\n\t\\[\n\t\tN(A) = \\{ \\bm{x} \\in \\RR^n \\mid A\\bm{x} = \\bm{0} \\}\n\t\\]\n\\end{defn}\n\n\\begin{prop}\n\tIf $A \\in M_{m \\times n}(\\RR)$, the null space $N(A)$ is a\n\tsubspace of $\\RR^n$.\n\\end{prop}\n\n\\begin{proof}\n\tIt is clear that $\\bm{0} \\in N(A)$. It remains to be shown\n\tthat addition and scalar multiplication are closed.\n\n\tLet $x,y \\in \\RR^n$. We know that $\\RR^n$ is a vector space,\n\thence $A(x+y) = Ax + Ay = \\bm{0}$.\n\n\tLet $x \\in \\RR^n, \\, a \\in \\RR$. $A(ax) = a(Ax) = a\\bm{0} = \\bm{0}$.\n\\end{proof}\n\n\\begin{defn}[Image Space]\n\t$Im A = \\{ \\bm{y} \\in \\RR^m \\mid \\bm{y} = A\\bm{x}, \\bm{x} \\in \\RR^n \\}$\n\\end{defn}\n\n\\begin{prop}\n\t$Im A$ is a subspace of $\\RR^m$.\n\\end{prop}\n\n\\begin{proof}\n\tWe know that $\\bm{0} \\in \\RR^n$ hence $A\\bm{0} = 0 \\in Im A$.\n\n\tLet $y_1 = Ax_1, y_2 = Ax_2 \\in \\RR^m$.\n\t\\[\n\t\ty_1 + y_2 = Ax_1 + Ax_2 = A(x_1 + x_2)\n\t\\]\n\t\\[\n\t\tx_1 + x_2 \\in \\RR^n \\Rightarrow y_1 + y_2 \\in Im A\n\t\\]\n\n\tLet $k \\in \\RR, y \\in Im A$.\n\t\\[\n\t\tky = kAx = Akx\n\t\\]\n\t\\[\n\t\tkx \\in \\RR^n \\Rightarrow ky \\in Im A\n\t\\]\n\\end{proof}\n\n\\begin{ex}\n\tFind $N(A)$ and $Im(A)$ where\n\t\\[\n\t\tA =\n\t\t\\begin{pmatrix}\n\t\t\t1 & 3 & -2 \\\\\n\t\t\t-2 & 4 & 4\n\t\t\\end{pmatrix}\n\t\\]\n\n\tFirst we find the RRE of $A$:\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\t1 & 3 & -2 \\\\\n\t\t\t-2 & 4 & 4\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 3 & -2 \\\\\n\t\t\t0 & 10 & 0\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 3 & -2 \\\\\n\t\t\t0 & 1 & 0\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & -2 \\\\\n\t\t\t0 & 1 & 0\n\t\t\\end{pmatrix}\n\t\\]\n\n\t$N(A)$:\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & -2 \\\\\n\t\t\t0 & 1 & 0\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix}\n\t\t\tx_1 \\\\\n\t\t\tx_2 \\\\\n\t\t\tx_3\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t0 \\\\\n\t\t\t0 \\\\\n\t\t\t0\n\t\t\\end{pmatrix}\n\t\\]\n\t\\[\n\t\t\\Rightarrow x_2 = 0, \\quad x_1 = 2x_3\n\t\\]\n\t\\[\n\t\t\\Rightarrow N(A) = k\n\t\t\\begin{pmatrix}\n\t\t\t2 \\\\\n\t\t\t0 \\\\\n\t\t\t1\n\t\t\\end{pmatrix}\n\t\t, k \\in \\RR\n\t\\]\n\n\t$Im(A)$:\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\ty_1 \\\\\n\t\t\ty_2 \\\\\n\t\t\ty_3\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 3 & -2 \\\\\n\t\t\t-2 & 4 & 4\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix}\n\t\t\tx_1 \\\\\n\t\t\tx_2 \\\\\n\t\t\tx_3\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\tx_1 + 3x_2 - 2x_3 \\\\\n\t\t\t-2x_1 + 4x_2 + 4x_3\n\t\t\\end{pmatrix}\n\t\\]\n\t\\[\n\t\t= x_1\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t-2\n\t\t\\end{pmatrix}\n\t\t+ x_2\n\t\t\\begin{pmatrix}\n\t\t\t3 \\\\\n\t\t\t4\n\t\t\\end{pmatrix}\n\t\t+ x_3\n\t\t\\begin{pmatrix}\n\t\t\t-2 \\\\\n\t\t\t4\n\t\t\\end{pmatrix}\n\t\\]\n\n\tNote that the vectors in this expression are simply the columns of $A$, in fact in general:\n\t\\[ Im(A) = \\{x_1 a_{i1} + \\cdots + x_1 a_{in}\\} \\]\n\tHence the image space is sometimes referred to as the column space.\n\\end{ex}\n\nAnother example of a vector space is the space of polynomials over a field $\\FF$:\n\\[\n\t\\FF[x] =\n\t\\{a_0 + a_1 x + \\cdots + a_n x^2 \\mid a_i \\in \\FF, n \\in \\NN_0\\}\n\\]\ne.g. $\\RR[x]$ is a vs over $\\RR$.\nAddition and scalar multiplication work as expected.\n\nWe can also define a vs on a set of funtions:\n\\[\n\tV = \\{f:[0,1] \\to \\RR\\}\n\\]\nAddition:\n\\[\n\t(g+f)(x) = g(x) + f(x)\n\\]\nScalar multiplication:\n\\[\n\t(\\alpha f)(x) = \\alpha (f(x)), \\alpha \\in \\RR\n\\]\n\n% TODO: prove these examples ^^.\n\n\\begin{defn}[Transpose]\n\tIf $A = a(i,j) \\in M_{m \\times n}(\\RR)$, then its transpose is\n\t\\[ A^T = a(j,i) \\]\n\ti.e its columns are swapped with its rows.\n\\end{defn}\n\\begin{defn}[Symmetric]\n\tA matrix $A \\in M_{n}(\\RR)$ is called symmetric iff\n\t\\[ A^T = A. \\]\n\\end{defn}\n\nProperties of $A^T$:\n\\[\n\t{(A^T)}^T = A\n\\]\n\\[\n\t{(A+B)}^T = A^T + B^T\n\\]\n\\[\n\t{(AC)}^T = C^T A^T\n\\]\n\n\\section{Spanning sets, linear independence, bases}\n\n\\begin{defn}[Linear Combination]\n\tLet $V$ be a vector space over $\\RR$. A linear combination of vectors in $V$ is given as:\n\t\\[\n\t\t\\alpha_1 v_1 + \\alpha_2 v_2 + \\cdots + \\alpha_r v_r, \\quad\n\t\t\\alpha_i \\in \\RR, \\, v_i \\in V\n\t\\]\n\\end{defn}\n\n\\begin{prop}\n\t$U = \\{\n\t\t\\alpha_1 v_1 + \\alpha_2 v_2 + \\cdots + \\alpha_r v_r \\mid\n\t\t\\alpha_i \\in \\RR, \\, v_i \\in V\n\t\\}$ is a subspace.\n\\end{prop}\n\n\\begin{proof}\n\t\\[\n\t\t\\alpha_i = 0 \\, \\forall i \\Rightarrow 0 + \\cdots + 0 = 0\n\t\\]\n\t\\[\n\t\t\\Rightarrow \\bm{0} \\in U\n\t\\]\n\t\\[\n\t\ta = \\sum_{i=0}^r \\alpha_i v_i \\quad\n\t\tb = \\sum_{i=0}^r \\beta_i v_i \\\\\n\t\\]\n\t\\begin{align*}\n\t\t\\Rightarrow a + b &= \\sum (\\alpha_i v_i + \\beta_i v_i) \\\\\n\t\t&= \\sum (\\alpha_i + \\beta_i) v_i \\in U\n\t\\end{align*}\n\n\t\\[\n\t\ta = \\sum \\alpha_i v_i \\quad \\lambda \\in \\RR\n\t\\]\n\t\\[\n\t\t\\lambda a = \\sum \\lambda \\alpha_i v_i \\quad\n\t\t\\lambda \\alpha_i \\in \\RR \\, \\forall i\n\t\t\\Rightarrow \\lambda a \\in U\n\t\\]\n\\end{proof}\n\n\\begin{defn}[Spanning Set]\n\tLet $V$ be a vector space and $W$ be a subspace. A set of vectors\n\t$\\{v_1, \\ldots, v_r\\} \\subset V$ is a spanning set for $W$ if\n\t\\[\n\t\tW =\n\t\t\\{\\alpha_1 v_1 + \\cdots + \\alpha_r v_r \\mid \\alpha_i \\in \\RR\\}\n\t\t= sp\\{v_1, \\ldots, v_r\\}\n\t\\]\n\tWe say that $\\{v_1, \\ldots, v_r\\}$ spans $W$.\n\\end{defn}\n\n\\begin{ex}\n\tIf $A = \\begin{pmatrix}\n\t\t1 & 3 & -2 \\\\\n\t\t-2 & 4 & 4\n\t\\end{pmatrix}$ we know that\n\t\\[\n\t\tIm(A) = x_1\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t-2\n\t\t\\end{pmatrix}\n\t\t+ x_2\n\t\t\\begin{pmatrix}\n\t\t\t3 \\\\\n\t\t\t4\n\t\t\\end{pmatrix}\n\t\t+ x_3\n\t\t\\begin{pmatrix}\n\t\t\t-2 \\\\\n\t\t\t4\n\t\t\\end{pmatrix}\n\t\\]\n\n\tWe can also say that\n\t\\[\n\t\tIm(A) = sp\\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t-2\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t3 \\\\\n\t\t\t\t4\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t-2 \\\\\n\t\t\t\t4\n\t\t\t\\end{pmatrix}\n\t\t\\right\\}\n\t\\]\n\tsince every element in $Im(A)$ is a linear combination of these vectors.\n\\end{ex}\n\n\\begin{ex}\n\t\\[\n\t\t\\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t0 \\\\\n\t\t\t\t0\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t0 \\\\\n\t\t\t\t1 \\\\\n\t\t\t\t0\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t0 \\\\\n\t\t\t\t0 \\\\\n\t\t\t\t1\n\t\t\t\\end{pmatrix}\n\t\t\\right\\}\n\t\\]\n\tspans $\\RR^3$.\n\\end{ex}\n\n\\begin{ex}\n\tLet\n\t\\[\n\t\tW = \\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\ta + b + 2c \\\\\n\t\t\t\tb + c \\\\\n\t\t\t\ta + c\n\t\t\t\\end{pmatrix}\n\t\t\t\\mid a,b,c \\in \\RR\n\t\t\\right\\}\n\t\t\\subset \\RR^3\n\t\\]\n\tthen\n\t\\[\n\t\tw = a\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t0 \\\\\n\t\t\t1\n\t\t\\end{pmatrix}\n\t\t+ b\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t1 \\\\\n\t\t\t0\n\t\t\\end{pmatrix}\n\t\t+ c\n\t\t\\begin{pmatrix}\n\t\t\t2 \\\\\n\t\t\t1 \\\\\n\t\t\t1\n\t\t\\end{pmatrix}\n\t\t\\forall w \\in W\n\t\\]\n\t\\[\n\t\t\\Rightarrow W = sp\\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t0 \\\\\n\t\t\t\t1\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t1 \\\\\n\t\t\t\t0\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t2 \\\\\n\t\t\t\t1 \\\\\n\t\t\t\t1\n\t\t\t\\end{pmatrix}\n\t\t\\right\\}\n\t\\]\n\n\tAlso, since\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\t2 \\\\\n\t\t\t1 \\\\\n\t\t\t1\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t0 \\\\\n\t\t\t1\n\t\t\\end{pmatrix}\n\t\t+\n\t\t\\begin{pmatrix}\n\t\t\t1 \\\\\n\t\t\t1 \\\\\n\t\t\t0\n\t\t\\end{pmatrix},\n\t\\]\n\t\\[\n\t\tW = sp\\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t0 \\\\\n\t\t\t\t1\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t1 \\\\\n\t\t\t\t0\n\t\t\t\\end{pmatrix}\n\t\t\\right\\}\n\t\\]\n\\end{ex}\n\n\\begin{defn}[Linear Independence]\n\tLet $V$ be a vector space. A set of vectors\n\t$\\{v_1, \\ldots, v_r \\mid v_i \\in V \\}$\n\tis linearly independent if\n\t\\[\n\t\t\\alpha_1 v_1 + \\cdots + \\alpha_r v_r = 0\n\t\t\\implies \\alpha_1 = \\cdots = \\alpha_r = 0.\n\t\\]\n\\end{defn}\n\n\\begin{thm}\n\tLet $v_1, \\ldots, v_r \\in \\RR^n$ and $A$ be the matrix whose\n\tcolumns correspond to the vectors $v_i$, then the set\n\t$\\{v_1, \\ldots, v_r\\}$ is linearly independent iff\n\t$A\\bm{x} = 0$ has the unique solution $x = 0$ iff $\\rho(A) = r$.\n\\end{thm}\n\n\\begin{proof}\n\tBy definition of linear independence,\n\t\\[\n\t\tx_1 v_1 + \\cdots + x_r v_r = 0\n\t\t\\iff\n\t\tx_i = 0 \\, \\forall i\n\t\\]\n\tThis is clearly equivalent to saying that $\\bm x$ is the\n\tonly solution to $A\\bm x = 0$.\n\n\tA matrix equation like this can have a unique solution iff\n\t$\\rho(A) = r$, otherwise there would be free variables resulting\n\tin a range of solutions.\n\\end{proof}\n\nWe now know that any matrix whose rank is equal to the number of its\nrows forms a linearly independent spanning set for a subspace of\n$\\RR^n$.\n\n\\begin{cor}\n\tAny set of $m > n$ vectors in $\\RR^n$ is linearly dependent.\n\\end{cor}\n\n\\begin{cor}\n\tThe columns of the indentity matrix $I_n$ form a linearly\n\tindependent spanning set of $\\RR^n$.\n\\end{cor}\n\n\\begin{defn}[Basis]\n\tA subset $B$ of a vector space $V$ is called a basis of $V$ if $B$ is linearly independent and $B$ spans $V$.\n\\end{defn}\n\n\\begin{rem}\n\tWhen considering $\\RR^n$ or its subsets, a basis will always be\n\tfinite; however a basis can be infinite, for example\n\t$\\{1, x, x^2, \\ldots\\}$ is a basis of $\\RR[x]$.\n\\end{rem}\n\n\\begin{ex}\n\t\\[\n\t\t\\left\\{\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t2 \\\\\n\t\t\t\t3\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t1 \\\\\n\t\t\t\t-1 \\\\\n\t\t\t\t0\n\t\t\t\\end{pmatrix}\n\t\t\t,\n\t\t\t\\begin{pmatrix}\n\t\t\t\t5 \\\\\n\t\t\t\t4 \\\\\n\t\t\t\t4\n\t\t\t\\end{pmatrix}\n\t\t\\right\\}\n\t\\]\n\tis a basis of $\\RR^3$.\n\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\t1 & 1 & 5 \\\\\n\t\t\t2 & -1 & 4 \\\\\n\t\t\t3 & 0 & 4\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 1 & 5 \\\\\n\t\t\t0 & -3 & -6 \\\\\n\t\t\t0 & -3 & -11\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 1 & 5 \\\\\n\t\t\t0 & -3 & -6 \\\\\n\t\t\t0 & 0 & -5\n\t\t\\end{pmatrix}\n\t\\]\n\n\tWe can see that the rank is $3$ hence this set is linearly independent.\n\t\\[\n\t\t\\begin{pmatrix}\n\t\t\t1 & 1 & 5 \\\\\n\t\t\t0 & -3 & -6 \\\\\n\t\t\t0 & 0 & -5\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 1 & 5 \\\\\n\t\t\t0 & 1 & -2 \\\\\n\t\t\t0 & 0 & 1\n\t\t\\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix}\n\t\t\t1 & 0 & 0 \\\\\n\t\t\t0 & 0 & 0 \\\\\n\t\t\t0 & 0 & 1\n\t\t\\end{pmatrix}\n\t\\]\n\n\tThis shows that this set spans $\\RR^3$ and is therefore a basis.\n\\end{ex}\n\n\\begin{thm}\n\tLet $V$ be a vector space over $\\RR$. If $U = \\{v_1, \\ldots,\n\tv_r\\}$ spans $V$ and no proper subset of $U$ spans $V$, then $U$\n\tis a basis for $V$.\n\\end{thm}\n\n\\begin{proof}\n\tWe know that $U$ spans $V$ so we need only prove that $U$ is linearly independent. We will do so using the contrapositive method.\n\n\tIf $U$ is not linearly independent then $\\exists v_i \\in U$ such that\n\t\\[\n\t\tv_i = \\alpha_1 v_1 + \\cdots + \\alpha_{i-1} v_{i-1} +\n\t\t\\alpha_{i+1} v_{i+1} + \\cdots + \\alpha_r v_r\n\t\\]\n\t\\[\n\t\t\\implies\n\t\tW = {v_1, \\ldots, v_{i-1}, v_{i+1}, \\ldots, v_r}\n\t\t\\subset U\n\t\\]\n\tspans $V$.\n\n\tTherefore, if there is no such $W$, $U$ must be linearly\n\tindependent and hence a basis.\n\\end{proof}\n\n\\begin{thm}\n\tLet $V$ be a vector space over $\\RR$. If $U = \\{v_1, \\ldots, v_r\\}$\n\tis a linearly independent subset of $V$ and no proper, linearly\n\tindependent subset $W$ of $V$, where $W$ contains $U$, then $U$\n\tis a basis of $V$.\n\\end{thm}\n\n\\begin{proof}\n\tWe know $U$ is linearly independent so we need only show that it\n\tspans $V$.\n\n\tLet $v \\in V \\setminus U$. According to our assumption that no\n\tlinearly independent subset exists between $U$ and $V$, this\n\tmeans that $\\{v_1, \\ldots, v_r, v\\}$ is not linearly\n\tindependent; hence any element in this set can be expressed as a\n\tlinear combination of the rest.\n\n\t\\[\n\t\t\\implies\n\t\tv = \\sum_{i=1}^r \\alpha_i v_i \\quad \\forall v \\in V\n\t\\]\n\n\tTherefore, $U$ spans $V$ and $U$ is a basis for $V$.\n\\end{proof}\n\n%TODO: look into Mac Lane generalisation of this lemma.\n\\begin{lemma}[Exchange (Steinitz)]\n\tLet $V$ be a vector space. Let $S = \\{a_1, \\ldots, \\_s\\}$ be a\n\tspanning set for a subspace $U \\subset V$ and let $B = \\{b_1,\n\t\\ldots, b_r\\}$ be a linearly independent subset of $U$. The $r\n\t\\leq s$ and there exists a set $S_r$ containing $B$ and some\n\telements from $S$ such that $U$ is spanned by $S_r$.\n\\end{lemma}\n\n%TODO: be consistent with sp and span, declare them as operators:\n%\\DeclareMathOperator\\span{span}\n\\begin{proof}\n\t\\[\n\t\tb_1 \\in U = sp(S)\n\t\t\\implies b_1 = \\sum_{i=1}^s \\alpha_i a_i,\n\t\\]\n\twhere not all $a_i = 0$ (else $b_1 = 0$).\n\t\\[\n\t\t\\implies a_1 = \\alpha_1^{-1}\n\t\t( b_1 - \\alpha_2 a_2 - \\cdots - \\alpha_s a_s )\n\t\\]\n\n\tAssume we have replaced $(a_1, \\ldots, a_k)$ with\n\t$(b_1, \\ldots, b_k)$ in this manner, so that\n\t$U$ is spanned by $\\{ b_1, \\ldots, b_k, a_{k+1}, \\ldots, a_s \\}$.\n\n\tExchange $b_{k+1}$:\n\t\\[ b_{k+1} \\in U \\]\n\t\\[\n\t\t\\implies b_{k+1} =\n\t\t\\beta_1 b_1 + \\cdots + \\beta_k b_k\n\t\t+ \\alpha_{k+1} a_{k+1} + \\cdots + \\alpha_s a_s\n\t\\]\n\tagain, not all coefficients are zero.\n\n\tAt least one $\\alpha_i \\neq 0$ else $b_{k+1}$ would be a linear\n\tcombination of $\\{ b_1, \\ldots, b_k \\}$.\n\n\t\\[\n\t\t\\implies \\alpha_{k+1} = \\alpha_{k+1}^{-1}\n\t\t(\n\t\t\tb_{k+1} - \\beta_1 b_1 - \\cdots - \\beta_k b_k\n\t\t\t- \\alpha_{k+2} a_{k+2} - \\cdots - \\alpha_s a_s\n\t\t)\n\t\\]\n\n\tSo for all $1 \\leq k \\leq r$ we can replace $a_k$ with $b_k$, giving:\n\t\\[\n\t\tU = span \\{ b_1, \\ldots, b_{k+1}, a_{k+2}, \\ldots, a_s \\}\n\t\\]\n\\end{proof}\n\nThis result is significant because when we have a basis, $B$, for a\nvector space, $V$, it can be used as the linearly independent set or the\nspanning set. This tells us that any spanning set for $V$ has a greater\nor equal cardinality compared with $B$ and that any linearly independent\nsubset of $V$ has a cardinality smaller than or equal to that of $B$.\nThe next result follows from this.\n\n\\begin{cor}\n\tAny two bases for a vector space have the same number elements.\n\\end{cor}\n\n\\begin{defn}[Dimension]\n\tThe number of elements in a basis for a vector space is the dimension of the vector space.\n\\end{defn}\n\n\\begin{rem}\n\tLet $U$ be a vector space with dimension $n$ and $S \\subset U$\n\tsuch that $|S| = n$. If $S$ spans $U$ or if $S$ is linearly\n\tindependent, then $S$ must be a basis of $U$.\n\\end{rem}\n\n\\section{Rank and Nullity}\n\n\\begin{defn}[Nullity]\n\tThe nullity of a matrix $A$ is the dimension of its null space.\n\\end{defn}\n\n\\begin{defn}[Rowspace]\n\tLet $A \\in M_{m \\times n}$ an $a_{ij}$ be the $i^{th}$ row of $A$.\n\tThe rowspace of $A$ is:\n\t\\[\n\t\tsp \\{ a_{1j}, \\ldots, a{nj} \\}\n\t\\]\n\\end{defn}\n\nNote that the dimension of the rowspace is the number of non-zero rows\nin echelon form, i.e.\\ the (row) rank. We can refer to the column rank as\nthe dimension of the column/image space.\n\n\\begin{thm}[Row rank equals column rank]\n\tLet $A \\in M_{m \\times n}$, then $\\rho(A) = \\rho(A^T)$. In other\n\twords, the row rank and column rank are equal.\n\\end{thm}\n\n\\begin{proof}\n\tLet the column rank of $A$ be $r = \\rho(A^T)$. Let $C$ be the\n\tmatrix whose columns form a basis for the column space of $A$.\n\n\tEvery column of $A$ can be expressed as a linear combination of\n\tthe columns in $C$ hence there is a matrix $R$ such that\n\t$A = CR$.\n\n\tThis means that every column of $A$ can be given as a linear\n\tcombination of the rows of $R$, which implies that it is a\n\tspanning set for the row space of $A$.\n\n\tThere are $r$ columns in $C$, hence $r$ rows in $R$ so, by the\n\tExchange Lemma, we know that $\\rho(A) \\leq r = \\rho(A^T)$.\n\n\tWe can apply this same process to the matrix $A^T$, where the\n\trows and columns (and hence their ranks) are swapped. From this\n\twe see that $\\rho(A^T) \\leq \\rho({(A^T)}^T) = \\rho(A)$.\n\n\t\\[\n\t\t\\implies \\rho(A) = \\rho(A^T)\n\t\\]\n\\end{proof}\n\n\\begin{thm}\n\tLet $A \\in M_n(\\RR)$ with $Ax = b$, then the following statement\n\tare equivalent:\n\t\\begin{enumerate}\n\t\t\\item\n\t\t\t$Ax = b$ has a unique solution\n\t\t\\item\n\t\t\t$\\rho(A) = n$\n\t\t\\item\n\t\t\t$A$ is equivalent to $I_n$\n\t\t\\item\n\t\t\t$A$ is invertible\n\t\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\n\tWe have already seen $1 \\iff 2$ and $2 \\iff 3$ is obvious\n\tconsidering the RRE form of $A$.\n\n\t$3 \\implies 4$ because by reducing $A$ to $I_n$, we can use the\n\tproduct of the row operations to construct the inverse matrix.\n\n\t$4 \\implies 3$:\n\t\\[ \\exists A^{-1} \\implies A^{-1}A = I_n \\]\n\t$A^{-1}$ can be split into row operations that, when applied to\n\t$A$, yields $I_n$.\n\\end{proof}\n\n\\section{Determinants}\n\n\\begin{defn}\n\tThe $(i,j)$-minor of $A \\in M_n$ is the $(n-1)\\times(n-1)$\n\tmatrix obtained by removing the $i^{th}$ row and $j^{th}$\n\tcolumn. It is often denoted $M_{i,j}$.\n\\end{defn}\n\n\\begin{defn}[Determinant]\n\tThe determinant is defined recursively:\n\t\\[ n=1 \\implies \\det [a] = a \\]\n\t\\[ n=k \\implies \\det A = \\sum_{j=1}^k a_{1,j} C_{1,j} \\]\n\twhere\n\t\\[ C_{i,j} = {(-1)}^{i+j} \\det(M_{i,j}) \\]\n\tis the $(i,j)$-cofactor.\n\\end{defn}\n\n\\subsection{Properties of the Determinant}\n\n\\begin{enumerate}\n\t\\item\n\t\tIf any row is all zeros, $\\det A = 0$\n\t\\item\n\t\tIf $B$ is obtained by adding one row of $A$ to a\n\t\tdifferent one, $\\det A = \\det B$\n\t\\item\n\t\tIf $B$ is obtained by swapping one row of $A$ with\n\t\tanother, $\\det A = -\\det B$\n\t\\item\n\t\tIf $B$ is obtained by multiplying one row of $A$ by\n\t\t$\\alpha$, $\\alpha \\det A = \\det B$\n\t\\item\n\t\tIf $A$ is upper triangular, then $\\det A$ is the product\n\t\tof the diagonal elements\n\t\\item\n\t\t$\\det(A^T) = \\det(A)$\n\t\\item\n\t\t$\\det(AB) = \\det(A) \\det(B)$\n\\end{enumerate}\n\nWe will not prove most of these here, but property 5 can be deduced\nsimply enough using the others and induction.\n\n\\begin{proof}\nThe $n=1$ case is obvious, so we may assume the property holds for\n$n=k$.\n\n\\[\n\tA =\n\t\\begin{bmatrix}\n\t\ta_{1,1} & \\cdots & a_{1,k} & a_{1,k+1} \\\\\n\t\t\\vdots & \\ddots & \\vdots & \\vdots \\\\\n\t\ta_{k,1} & \\cdots & a{k,k} & a_{k,k+1} \\\\\n\t\ta_{k+1,1} & \\cdots & a_{k+1,k} & a_{k+1,K+1}\n\t\\end{bmatrix}\n\\]\n\nThe other row properties allow us to traverse any row in order to\ncalculate the determinant, so we can choose the bottom row. This is all\nzeros except for the element $a_{k+1,k+1}$. We can see that\n\\[\n\t\\det A = a_{k+1,K+1} \\det A_k\n\\]\nwhich is the product of the diagonals.\n\\end{proof}\n\nSome futher properties can be deduced from those we have already described:\n\\begin{itemize}\n\t\\item\n\t\tIf any two rows are identical then $\\det A = 0$ (using 1,2,4)\n\t\\item\n\t\tAll row properties also apply to columns (due to 6)\n\t\\item\n\t\tThe determinant is commutative (due to 7)\n\t\\item\n\t\t$\\det(\\alpha A) = \\alpha^n \\det A$ (due to 4)\n\t\\item\n\t\t$\\det I_n = 1$ (using 5)\n\\end{itemize}\n\n\\section{Eigenvalues and Eigenvectors}\n\n\\begin{defn}[Eigenvalues and Eigenvectors]\n\tLet $A \\in M_n(\\RR)$ and $\\lambda \\in \\RR$. If $\\exists x \\neq\n\t0$ such that $Ax = \\lambda x$, then $\\lambda$ is called an\n\tEigenvalue of $A$ and $x$ is called an Eigenvector of $A$.\n\\end{defn}\n\n\\begin{lemma}\n\t$\\exists A^{-1} \\iff |A| \\neq 0$\n\\end{lemma}\n\n%TODO: deduce the method for calculating the inverse (springer)\n\\begin{proof}\n\tAssume $\\exists A^{-1}$\n\t\\[ \\implies |A^{-1}A| = |I| = 1 \\]\n\t\\[ \\implies |A^{-1}||A| = 1 \\]\n\t\\[ \\implies |A| \\neq 0 \\]\n\n\tAssume $|A| \\neq 0$. This means we can use the standard method\n\tto calculate the inverse.\n\\end{proof}\n\n\\begin{lemma}\n\t$\\lambda$ is an Eigenvalue $\\iff |A - \\lambda I_n| = 0$.\n\\end{lemma}\n\n\\begin{proof}\n\t\\[ Ax = \\lambda x \\iff Ax = \\lambda I_n x \\iff (A - \\lambda I_n)x = 0 \\]\n\n\tBy definition, $x \\neq 0$ hence $(A - \\lambda I_n)$ cannot have an inverse.\n\n\t\\[ \\iff |A - \\lambda I_n| = 0 \\]\n\\end{proof}\n\n\\begin{defn}[Characteristic Equation]\n\t$|A - \\lambda I_n| = 0$ is called the characteristic equation of\n\t$A$. Solving it for $\\lambda$ yields eigenvalues for $A$, which\n\tcan be used to find eigenvectors from $(A - \\lambda I_n)x = 0$.\n\\end{defn}\n\n\\section{Linear Transformations}\n\n\\begin{defn}[Linear Transformation]\n\tLet $V$ and $W$ be vector spaces over $\\RR$. A function $T:V \\to\n\tW$ is a linear transformation if $\\forall v_1, v_2 \\in V, \\alpha\n\t\\in \\RR$ we have\n\t\\begin{enumerate}\n\t\t\\item\n\t\t\t$T(v_1 + v_2) = T(v_1) + T(v_2)$\n\t\t\\item\n\t\t\t$T(\\alpha v_1) = \\alpha T(v_1)$\n\t\\end{enumerate}\n\\end{defn}\n\n\\begin{defn}[Kernel]\n\t\\[ \\ker T = \\{ v \\in V \\mid T(v) = 0 \\} \\]\n\\end{defn}\n\n\\begin{defn}\n\t\\[ \\im T = \\{ w \\in W \\mid w = T(v), v \\in V \\} \\]\n\\end{defn}\n\n\\begin{prop}\n\t$\\ker T$ is a subspace of $V$ and $\\im T$ is a subspace of $W$.\n\\end{prop}\n\n\\begin{proof}\n\tLet $v_1, v_2 \\in \\ker T$, then\n\t\\[ T(0) = T(v_1-v_1) = T(v_1) - T(v_1) = 0 \\implies 0 \\in \\ker T \\]\n\t\\[ T(v_1 + v_2) = T(v_1) + T(v_2) = 0 \\]\n\t\\[ T(\\alpha v_1) = \\alpha T(v_1) = \\alpha \\cdot 0 = 0 \\]\n\n\t$T(v_1), T(v_2) \\in \\im T$\n\t\\[ T(0) = T(v_1 - v_1) = T(v_1) - T(v_1) = 0 \\]\n\t\\[ v_1 + v_2 \\in V \\implies T(v_1) + T(v_2) = T(v_1 + v_2) \\in \\im T \\]\n\t\\[ \\alpha v_1 \\in V \\implies \\alpha T(v_1) = T(\\alpha v_1) \\in \\im T \\]\n\\end{proof}\n\n\\begin{prop}\n\tLet $T$ be a linear map. $T$ is injective $\\iff \\ker T = 0$.\n\\end{prop}\n\n\\begin{proof}\n\t\\begin{align*}\n\t\t& &(T(x) = T(y) \\implies x = y) \\\\\n\t\t&\\iff &(T(x) - T(y) = 0 \\implies x - y = 0) \\\\\n\t\t&\\iff &(T(x-y) = 0 \\implies x-y = 0) \\\\\n\t\t&\\iff &(T(v) = 0 \\implies v = 0) \\\\\n\t\t&\\iff &\\ker T = 0\n\t\\end{align*}\n\\end{proof}\n\n\\begin{rem}\n\tIt is clear that $T: V \\to W$ is surjective $\\iff \\im T = W$.\n\\end{rem}\n\n\\begin{defn}[Isomorphism]\n\tA linear transformation $T:V \\to W$ is an isomorphism if $T$ is\n\tbijective i.e.\\ $\\ker T = 0$ and $\\im T = W$.\n\\end{defn}\n\n\\begin{defn}\n\tLet $T:V \\to W$ be a linear transformation. Let $v_1, \\ldots,\n\tv_n$ be a basis for $V$, then the matrix of the transformation\n\tis the matrix $M_T \\in M_{m \\times n}$ whose columns are\n\t$T(v_i)$.\n\\end{defn}\n\n\\begin{prop}\n\tLet $T:\\RR^n \\to \\RR^m$ be a linear transformation, then $T(x) =\n\tM_T \\cdot x$.\n\\end{prop}\n\n\\begin{proof}\n\tLet $\\{e_I\\}$ be the stadard basis for $\\RR^n$ (columns of $I_n$) and $T(e_i) = f_i$.\n\t\\begin{align*}\n\t\tT(x) &= T\n\t\t\\left(\n\t\t\t\\begin{pmatrix}\n\t\t\t\tx_1 \\\\\n\t\t\t\t\\vdots \\\\\n\t\t\t\tx_n\n\t\t\t\\end{pmatrix}\n\t\t\\right) \\\\\n\t\t&= T(x_1 e_1 + \\cdots + x_n e_n) \\\\\n\t\t&= x_1 T(e_1) + \\cdots + x_n T(e_n) \\\\\n\t\t&= x_1 f_1 + \\cdots + x_n f_n \\\\\n\t\t&=\n\t\t\\begin{pmatrix}\n\t\t\t\\uparrow & & \\uparrow \\\\\n\t\t\tf_1 & \\cdots & f_n \\\\\n\t\t\t\\downarrow & & \\downarrow\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix}\n\t\t\tx_1 \\\\\n\t\t\t\\vdots \\\\\n\t\t\tx_n\n\t\t\\end{pmatrix} \\\\\n\t\t&= M_T \\cdot x\n\t\\end{align*}\n\\end{proof}\n\n\\begin{cor}\n\tLet $T:V \\to W$ be a linear transformation and $M_T$ be its\n\tmatrix with respect to the bases $B_V$ and $B_W$, then\n\t\\begin{align*}\n\t\t\\ker T & = N(M_T) \\\\\n\t\t\\im T & = \\im M_T\n\t\\end{align*}\n\\end{cor}\n\n\\begin{defn}[Change of basis]\n\tThe matrix of $id:V \\to W$ with respect to the bases $B_1$ and\n\t$B_2$ is called the change of basis matrix, from $B_1$ to $B_2$.\n\\end{defn}\n\n\\begin{defn}[Eigenvalues and Eigenvectors]\n\tLet $T:V \\to W$ be a linear transformation. If $\\exists \\lambda\n\t\\in \\RR$ and $0 \\neq v \\in V$ such that $T(v) = \\lambda v$, then\n\t$\\lambda$ is and eigenvalue and $v$ is an eigenvector.\n\\end{defn}\n\n\\end{document}\n", "meta": {"hexsha": "c316888834a2585407667cab1ec12bbc0324e5d4", "size": 22834, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/linear_algebra.tex", "max_stars_repo_name": "judgedreads/maths", "max_stars_repo_head_hexsha": "51ff47883510cd0d8281a024dcdcd7fa634d23dd", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "notes/linear_algebra.tex", "max_issues_repo_name": "judgedreads/maths", "max_issues_repo_head_hexsha": "51ff47883510cd0d8281a024dcdcd7fa634d23dd", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/linear_algebra.tex", "max_forks_repo_name": "judgedreads/maths", "max_forks_repo_head_hexsha": "51ff47883510cd0d8281a024dcdcd7fa634d23dd", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 22.1045498548, "max_line_length": 130, "alphanum_fraction": 0.5904353158, "num_tokens": 9475, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.8128673133042217, "lm_q1q2_score": 0.6362007464904963}} {"text": "\\chapter{Statistics in Photometry}\n\nSo far we've seen how the photometry should be done. But we haven't seriously discussed how to analize the uncertainties, i.e., the error-bars of each measurement such as magnitude. \n\n\\section{Pixel-wise Uncertainty}\nAn object frame consist of at least bias, dark, flat, cosmic-ray, target's signal, and sky, as well as readout noise. For the $ j $-th pixel, they are denoted as:\n\\begin{itemize}\n\\item $ \\tilde{N}_j $: the raw pixel value of the object frame in ADU.\n\\item $ B_j $: the bias in ADU.\n\\item $ D_j $: the dark in ADU.\n\\item $ F_j $: the flat in ADU (\\textit{not necessarily} normalized; see \\textbf{Note} at the end).\n\\item $ C_j $: the cosmic-ray in ADU.\n\\item $ I_j $: the object signal count in ADU.\n\\item $ S_j $: the sky signal in ADU.\n\\end{itemize}\nIn some cases, such as infrared detectors, the gain and readout noise may differ for each pixel, so I will denote gain and readout noise of the $ j $-th pixel as $ g_j \\,[\\mathrm{e/ADU}]$ and $ R_j \\,[\\mathrm{e}] $. Then we define the bias-subtracted pixel value\n\\begin{equation}\n N_j = \\tilde{N}_j - B_j = D_j + (I_j + S_j + C_j) F_j\n\\end{equation}\nNote that the dark is not affected by the flat value. Since this pixel value should follow a Poisson distribution, which is approximated as a Gaussian distribution,\n\\begin{equation}\n N_j \n \\simdot \\mathcal{N} \\left (\n N_j ,~ \n \\frac{N_j}{g_j} + \\left (\\frac{R_j}{g_j} \\right )^2 \n \\right ) ~.\n\\end{equation}\n\n\\subsection{Dark Estimation}\nTo estimate dark $ D_j $, you should have taken nearly tens of dark frames, and combined it. From the median value of the frames, you may have obtained the estimation of the dark $ \\hat{D}_j $ (for brevity, I will just use $ D_j $ without the caret $ \\hat{\\cdot} $). During the combination, you can obtain the uncertainty of the dark at the $ j $-th pixel, $ \\Delta D_j $, as\n\\begin{equation}\n \\Delta D_j = \\mathrm{sstd} (D_j^{i=1..N_\\mathrm{Dark}}) \n := \\sqrt{ \\frac{\\sum_{i=1}^{N_\\mathrm{Dark}} (D_j^{i} - \\bar{D})^2}\n {N_\\mathrm{Dark} - 1}} \n \\equiv \\mathrm{std}(D_j^{i=1..N_\\mathrm{Dark}}) \n \\times \\sqrt{\\frac{N_\\mathrm{Dark}}{N_\\mathrm{Dark} - 1}} ~,\n\\end{equation}\nwhere $ D_j^{i} $ is the dark current (ADU) of the $ j $-th pixel at the $ i $-th dark frame, $ \\mathrm{sstd} $ and $ \\mathrm{std} $ are the \\textit{sample} standard deviation (\\texttt{np.std(x, ddof=1)}) and standard deviation (\\texttt{np.std(x)}) functions, respectively, and $ N_\\mathrm{Dark} $ is the number of dark framses used. Here it is assumed $ D_j $ roughly follows a Gaussian distribution so that $ \\mathrm{sstd} $ becomes an unbiased estimator\\footnote{An unbiased estimator of a random variable $ X $, $ \\hat{X} $ is defined such that the expectation value of $ \\hat{X} $ is the same as the true value $ X_\\mathrm{true} $.} of the true standard deviation. Note that you should \\emph{not} divide it by the number of dark frames (such as Thm \\ref{thm: clt}), because \\textbf{what you will need is not the uncertainty of the \\emph{mean} value, but the uncertainty of one pixel value that will have been added to the object frame}. \n%, and $ N_\\mathrm{Dark} $ is the number of dark frames. This is based on the CLT (Thm \\ref{thm: clt}). Note the strong point of the CLT helps us here: CLT holds regardless of the distribution of $ D_j^{i} $! Even if we don't know anything about its distribution, we can use CLT without caring about it.\n\nTherefore, to the first order,\n\\begin{equation}\n\\begin{aligned}\n \\tilde{O}_j = N_j - D_j \n &\\simdot \\mathcal{N} \n \\left ( \\tilde{O}_j,~ \n \\frac{N_j}{g_j} + \\left (\\frac{R_j}{g_j} \\right )^2 + (\\Delta D_j)^2 \\right ) \\\\\n \\mathrm{or}\n &\\simdot \\mathcal{N} \n \\left ( \\tilde{O}_j,~ \n \\frac{\\tilde{O}_j}{g_j} \n + \\frac{D_j}{g_j} \n + \\left (\\frac{R_j}{g_j} \\right )^2 \n + (\\Delta D_j)^2 \\right ) ~.\n\\end{aligned}\n\\end{equation}\nNote that we have been using $ N_j/g_j = (\\tilde{O}_j + D_j)/g_j $ term in $ \\Delta N_j $. This makes sence only if the Poissonity of dark is assumed.\n\n\\subsection{Flat Estimation}\nUnlike the dark current, which is a probabilistic electron generation, the flat, i.e., the pixel sensitivity, is a characterestic value of the detector and optics, so it should not change at each exposure. Therefore, following the CLT (Thm \\ref{thm: clt}): \n\\begin{equation}\n \\Delta F_j \n = \\frac{\\mathrm{sstd} (F_j^{i=1..N_\\mathrm{Flat}})} {\\sqrt{N_\\mathrm{Flat}}}\n := \\sqrt{ \\frac{1}{N_\\mathrm{Flat}}\n \\frac{\\sum_{i=1}^{N_\\mathrm{Flat}} (F_j^{i} - \\bar{F})^2}{N_\\mathrm{Flat} - 1}} \n \\equiv \\frac{\\mathrm{std}(F_j^{i=1..N})}{\\sqrt{N_\\mathrm{Flat} - 1}} ~,\n\\end{equation}\nwhere $ F_j^{i} $ is the flat frame pixel value (ADU) of the $ j $-th pixel at the $ i $-th flat frame, $ \\mathrm{sstd} $ is the sample standard deviation function and $ N_\\mathrm{Flat} $ is the number of the used flat frames. \n\nGenerally speaking, $ F_j $ must be about $ 10^4 $ ADU or more, so that the signal-to-noise ratio from Poisson statistics is larger than about 100. Moreover, you should have obtained, say, $ N_\\mathrm{flat} = 9 $, this will increase by a factor of 3. Therefore, the uncertainty in the flat $ \\Delta F_j $, is around 0.1 \\% order. In some observatories, people take hundreds of flats at one time to get flat as good as roughly 0.01 \\% order, assuming it won't change over certain period of time.\n\nAlthough there are mathematically known ratio distribution, i.e., the pdf of $ Z := X/Y $ where $ X, Y \\sim \\mathcal{N}(\\mu_{X, Y}, \\sigma_{X, Y}^2) $ and the covariance is zero, what we are interested in is only a rough estimation of the uncertainties\\footnote{If you are interested in, see FiellerEC (1932) Biometrika, 24, 428; HinkleyDV (1969) Biometrika, 56, 635; D\\'{i}az-Franc\\'{e}s+RubioFJ (2013) Statistical Papers, 54, 309, as well as \\href{https://en.wikipedia.org/wiki/Ratio_distribution}{Wikipedia}. For the ratio of $ N_j/F_j $, i.e., when the denominator has so high signal-to-noise ratio, the distribution is roughly a Gaussian with $ \\mathcal{N}(\\mu_X, \\sigma_X^2) $ as we assume in this section.}. From the propagation of error, \n\\begin{equation}\n\\begin{aligned}\n O_j = \\frac{N_j - D_j}{F_j} = I_j + S_j + C_j\n &\\simdot \\mathcal{N} \n \\left ( O_j,~ \n O_j^2 \\left [ \n \\left (\\frac{\\Delta \\tilde{O}_j}{\\tilde{O}_j} \\right )^2 \n + \\left (\\frac{\\Delta F_j}{F_j} \\right )^2 \\right ] \\right ) ~.\n\\end{aligned}\n\\end{equation}\nSubstituting previously obtained distribution of $ \\tilde{O}_j $,\n\\begin{equation}\n O_j \n \\simdot \\mathcal{N} \n \\left ( O_j,~ \n \\frac{O_j}{g_jF_j} \n + \\frac{D_j}{g_j F_j^2} \n + \\left (\\frac{R_j}{g_j F_j} \\right )^2 \n + \\left (\\frac{\\Delta D_j}{F_j} \\right )^2\n + O_j^2 \\left (\\frac{\\Delta F_j}{F_j} \\right )^2\n \\right ) ~.\n\\end{equation}\n\n%Note that this variance is also obtained by the following logic: $ F_j $ has too high signal-to-noise ratio, $ \\Delta F_j / F_j \\ll 1 $, we can just assume $ F_j $ is a perfectly known constant without any uncertainty. Then $ O_j = \\tilde{O}_j / F_j \\simdot \\mathcal{N}(O_j, (\\Delta \\tilde{O}_j)^2 / F_j^2) $, which is what we have above.\n\n\n\\subsection{Final Pixel-wise Uncertainty}\nMany times the $ C_j $ is removed by the so-called \\emph{cosmic-ray rejection} algorithms. \n\\begin{thm}[Pixel-wise Error]\nThe final, cosmic-ray removed pixel value will follow\n\\begin{equation} \\label{eq: pix error}\n\\begin{aligned}\n O_j^\\mathrm{cr} = \\frac{N_j - D_j}{F_j} - C_j = I_j + S_j \n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{O_j}{g_jF_j} \n + \\frac{D_j}{g_j F_j^2} \n + \\left (\\frac{R_j}{g_j F_j} \\right )^2 \n + \\left (\\frac{\\Delta D_j}{F_j} \\right )^2 \n + O_j^2 \\left (\\frac{\\Delta F_j}{F_j} \\right )^2\n \\right ) \\\\\n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{N_j}{g_jF_j^2} \n + \\left (\\frac{R_j}{g_j F_j} \\right )^2 \n + \\left (\\frac{\\Delta D_j}{F_j} \\right )^2 \n + O_j^2 \\left (\\frac{\\Delta F_j}{F_j} \\right )^2\n \\right ) \n\\end{aligned}\n\\end{equation}\nHere, $ O_j $ is the bias, dark, and flat corrected pixel value before the cosmic-ray removal, while $ N_j $ is only bias-subtracted pixel value. Usually the uncertainty from the estimation for $ C_j $ is ignored as it is too difficult to estimate. \n\\end{thm}\n\nIn practical application, the following approximations may be used:\n\\begin{enumerate}\n\\item Calculating the $ \\Delta D_j $ term all the time is annoying, and moreover, it is likely for most pixels $ \\Delta D_j \\ll \\Delta O_j $, except for hot/bad pixels\\footnote{Bad pixels must be masked properly prior to any sane calculation, as they are known to be ``wrong'' data. This is usually provided as a pixel mask file (e.g., \\texttt{.bpm} file) by the observatory.}. Therefore, people just ignore it and set it to 0. Theoretically, however, $ \\Delta D_j > (R_j/g_j) $, because $\\Delta D_j = \\mathrm{sstd} (D_j^{i}) $ and $ D_j^{i} \\simdot \\mathcal{N} \\left (D_j^{i},~ D_j^{i}/g_j + (R_j/g_j)^2 \\right ) $. Therefore, a more reasonable approximation, or the lower limit of the uncertainty, would be $ R_j/g_j $. \n\\item Frequently, $ \\Delta F_j/F_j $ is much smaller than $ \\Delta \\tilde{O}_j/\\tilde{O}_j $ (or we can increase $ N_\\mathrm{Flat} $ to force that this holds), so the flat-error term is negligible. StetsonPB, for instance, asked the user to give a constant $ \\Delta F_j/F_j \\equiv \\sigma_F $ for all pixels, such as 0, 0.01, or 0.0075, in \\texttt{DAOPHOT}.\n\\item Many times we use the normalized flat (for more discussion, see below) such that its mean or median is 1, so $ F_j \\sim 1 $ for all pixel. Then all $ F_j $ in \\cref{eq: pix error} can be just ignored.\n\\item The $ D_j $ term is negligible for many times. It is not negligible only for hot pixels where $ D_j \\gg 1 $, but likely the observer did not put the target of interest at where hot pixels present. Moreover, state-of-the-art CCDs, such as Subaru FOCAS for instance, has too small dark current $ D_j < 0.1 \\,\\mathrm{e/s} $. Therefore, most important parts in the object frames will have negligible $ D_j $ compared to $ O_j^\\mathrm{cr} $. Therefore, this term is also ignored many times.\n\\end{enumerate}\n\nSummarizing, \\cref{eq: pix error} is approximated as\n\\begin{equation} \n\\begin{aligned}\n O_j^\\mathrm{cr} \n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{O_j}{g_jF_j} \n + \\frac{D_j}{g_j F_j^2} \n + \\left (\\frac{R_j}{g_j F_j} \\right )^2 \n + \\left (\\frac{\\Delta D_j}{F_j} \\right )^2 \n + O_j^2 \\left (\\frac{\\Delta F_j}{F_j} \\right )^2\n \\right ) \\\\\n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{O_j}{g_jF_j} \n + \\frac{D_j}{g_j F_j^2} \n + 2 \\left (\\frac{R_j}{g_j F_j} \\right )^2\n + \\sigma_F O_j^2 \n \\right ) \\\\\n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{O_j}{g_j} \n + \\frac{D_j}{g_j} \n + 2 \\left (\\frac{R_j}{g_j} \\right )^2\n + \\sigma_F O_j^2\n \\right ) \\\\\n &\\simdot \\mathcal{N} \n \\left ( O_j^\\mathrm{cr},~ \n \\frac{O_j}{g_j} \n + 2 \\left (\\frac{R_j}{g_j} \\right )^2\n + \\sigma_F O_j^2\n \\right ) \\\\\n\\end{aligned}\n\\end{equation}\nMost frequently people even drop the factor 2 and $ \\sigma_F $, and write $ O_j^\\mathrm{cr} \\simdot \\mathcal{N} \\left ( O_j^\\mathrm{cr},~ O_j/g_j + (R_j/g_j)^2 \\right ) $. The first dropping ($ 2(R_j/g_j)^2 \\rightarrow (R_j/g_j)^2 $) may be justified because one $ (R_j/g_j)^2 $ came from the uncertainty in dark, and we usually ignore anything from dark\\footnote{Some high-spec observatories even skip taking dark frames (e.g., Subaru).}. \n\n\\subsubsection{Note}\n\\textbf{The flat frames are frequently assumed to be normalized}, i.e., the mean or median of $ F_j $ values of the frame is around 1. In the generalized error estimation given in this book, this normalization is not important. However, astronomers (including all tasks of IRAF) have conventionally used a very simplified version of error estimation, that is, the Poisson component of the pixel error is approximated as $ \\sqrt{O_j/g_j} $: $ O_j \\simdot \\mathcal{N} (O_j, O_j/g_j) $. Note the factor $ F_j $ is missing! The true signal-to-noise ratio is $ O_j / \\Delta O_j \\approx \\sqrt{O_j / (g_j F_j)} $, while in this approximation, it is $ \\sqrt{O_j / g_j} $. If the flat was normalized, $ F_j \\sim 1 $, so these two are similar. If it were not normalized, however, the signal-to-noise ratio in this classical approximation will be underestimated by a factor of $ \\sqrt{F_j} \\sim \\sqrt{10^4 \\,(\\mathrm{ADU})} = 100 $!! You will be calculating wrong uncertainty in that case.\n\n\nFrom STSDAS package, \\texttt{stsdas/pkg/hst\\_calib/wfpc/noise/fitnoise.x}: noise = R if pixel value not positive, otherwise, $ \\sqrt{R^2 + N/g + (N \\times \\texttt{scalen}/100)^2} $ where \\texttt{scalen} is maybe an uncertainty of pixel (flat uncertainty?) in percentage.\n\n%#################################################################################\n%#\t\t\t\t\t\t\t\t\t\t#\n%# NOISEFUNC --\tCalculate the square-root of the variance expected at the \t#\n%#\t\tgiven DN level, based upon the noise model parameters. \t#\n%#\t\t\t\t\t\t\t\t\t\t#\n%#\tInitial code:\t7/91 by RAShaw\t\t\t\t\t\t#\n%\n%real procedure noisefn(dn) \n%\n%include\t\"nmcom.h\"\n%\n%# Calling argument\n%real\tdn\t\t# Value of pixel in Data Numbers\n%\n%# Local variables\n%real\ttmp\n%real\tnoise\t\t# Function value\n%\n%begin\n%\ttmp = scalen / 100.\n%\tif (dn <= 0.) \n%\t noise = readn \n%\telse\n%\t noise = sqrt (readn * readn + dn / gain + dn * dn * tmp * tmp) \n%\treturn (noise)\n%end\n\n\n\\subsection{Signal-to-Noise Ratio (SNR)}\n\\textbf{SNR} $ := \\mathrm{signal}/\\mathrm{noise} $, is an estimation of the quality of the data. Simply speaking, the signal is the sky subtracted aperture summation. On the other hand, noise is the square root of the aperture summation on the \\textit{variance map} (the map of $ (\\Delta O_j)^2 $).\n\n\\begin{ex}[Maximum pixel-wise SNR]\nConsider a classic CCD, which loses linearity at around 60,000 ADU (bias subtracted value). Say gain is $ g_j = 1.0 $, there is neither readout noise ($ R_j = 0.0 $) nor dark ($ D_j = 0.0 $), and there is no flat pattern ($ F_j = 1.0 $, $ \\sigma_F = 0 $). Then only the Poisson noise term remains and thus the maximum SNR of the pixel is \n\\begin{equation}\n \\mathrm{SNR}_\\mathrm{max}^\\mathrm{pixel} \n \\approx \\frac{60000}{\\sqrt{60000}} = \\sqrt{60000} = 245 ~.\n\\end{equation}\nEven if the CCD has high bias, lose linearity earlier, high gain, etc, you should always get $ \\mathrm{SNR}_\\mathrm{max}^\\mathrm{pixel} \\sim 250 $.\n\\end{ex}\n\n\\begin{ex}[Maximum photometric SNR]\nConsider the same CCD and now you have a simple circular Gaussian star (i.e., perfect Gaussian PSF) in the image, almost saturated but not saturated. The peak value before the saturation will be roughly 60,000 ADU after the sky subtraction. That is, this star has a radial profile $ I(r) \\approx 60000 e^{-r^2/2\\sigma^2} $. Depending on the seeing size and pixel scale, $ \\sigma $ will change. \n\nFor the case if the $ \\mathrm{FWHM} \\ll 1 \\,\\mathrm{pix} $, $ \\mathrm{apsum} \\approx 60,000 $, so the maximum SNR is $ \\mathrm{SNR}_\\mathrm{max}^\\mathrm{star} \\approx \\mathrm{SNR}_\\mathrm{max}^\\mathrm{pixel} \\sim 250 $.\n \nIf the FWHM is much larger than 1 pixel (say, 5 pixel), 60,000 is roughly the peak value of the Gaussian function. The integration of such Gaussian function is $ \\sqrt{2 \\pi} \\sigma \\times 60,000 \\sim 60,000 \\times \\mathrm{FWHM} $, where FWHM is in pixel unit (note that $ \\sqrt{2\\pi}\\sigma = 2.51 \\sigma \\approx \\mathrm{FWHM} $). Therefore, \n\\begin{equation}\n \\mathrm{SNR}_\\mathrm{max}^\\mathrm{star} \n \\approx \\frac{60000 \\mathrm{FWHM}}{\\sqrt{60000 \\mathrm{FWHM}}}\n \\sim 250 \\sqrt{\\mathrm{FWHM}}\n\\end{equation}\n\nOtherwise, $ \\int_{-0.5}^{0.5}\\int_{-0.5}^{0.5} I(x, y) dx dy = 60,000 $, so it is a bit complicated to calculate (you may use the fact that the integration of Gaussian function is analytically expressed using erf). But you may simply use the above rule-of-thumb to check the maximum SNR.\n\nIn reality, uncertainty in sky, dark, readout noise will increase the noise and thus decrease the SNR. Gain larger than unity will have an effect of increasing SNR by factor of $ \\sqrt{g} $. The PSF usually have longer tail than Gaussian, and this will increase the SNR.\n\\end{ex}\n\nThe magnitude system used in astronomy uses the logarithmic scale. From Pogson's formula, $ I_1/I_2 = 10^{-0.4 \\Delta m} $ for two objects having fluxes $ I_1 $ and $ I_2 $, and magnitude difference $ \\Delta m $. When $ \\Delta m $ is small, $ \\Delta m $ is the fractional difference of the fluxes. For example, $ \\Delta m = 0.01 $ means $ I_1/I_2 \\approx 0.99 $ or the flux difference is 0.01 (1 \\%). See \\cref{fig:deltamag} for the validity of this approximation. \n\n\\begin{figure} [ht!]\n \\centering\n \\includegraphics[width=0.5\\linewidth]{figs/delta_mag}\n \\caption{The delta magnitude plot.}\n \\label{fig:deltamag}\n\\end{figure}\n\nDifferently put, if $ \\Delta m =0.01 $ was the magnitude \\textit{uncertainty} of the object, the uncertainty in flux is 1 \\%, or $ \\mathrm{SNR} = 100 $. Therefore, SNR can be directly converted to the magnitude uncertainty and vice versa. The $ \\mathrm{SNR}_\\mathrm{max}^\\mathrm{pixel} = 250 $ translates to $ \\Delta m \\sim 1/250 = 0.004 $.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "bb891482d55a2aadd3589a75f93fdb0ecc81e38d", "size": 17423, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Books/chaps/03_stats_phot.tex", "max_stars_repo_name": "ysBach/SNU_AOclass", "max_stars_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-03-23T06:14:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-14T01:49:51.000Z", "max_issues_repo_path": "Books/chaps/03_stats_phot.tex", "max_issues_repo_name": "ysBach/SNU_AOclass", "max_issues_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2020-05-04T17:21:49.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-24T11:41:55.000Z", "max_forks_repo_path": "Books/chaps/03_stats_phot.tex", "max_forks_repo_name": "ysBach/SNU_AOclass", "max_forks_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2020-05-10T14:19:34.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-14T09:18:08.000Z", "avg_line_length": 66.754789272, "max_line_length": 978, "alphanum_fraction": 0.6701486541, "num_tokens": 5595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424489603726, "lm_q2_score": 0.7520125793176223, "lm_q1q2_score": 0.6361593629969559}} {"text": "\\chapter{Hypothesis Testing and p-values}\n\n\\begin{ex}\n Suppose that the true value of $\\theta$ is $\\theta_\\star\\neq \\theta_0$. Then\n \\begin{align*}\n W\n =\\frac{\\thetahat-\\theta_0}{\\sehat}\n =\\frac{\\thetahat-\\theta_\\star+\\theta_\\star-\\theta_0}{\\sehat}\n =\\frac{\\thetahat-\\theta_\\star}{\\sehat}-\\frac{\\theta_0-\\theta_\\star}{\\sehat},\n \\end{align*}\n and therefore\n \\begin{align*}\n \\P{|W|>z_{\\alpha/2}}\n & =\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}-\\frac{\\theta_0-\\theta_\\star}{\\sehat}>z_{\\alpha/2}}\n +\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}-\\frac{\\theta_0-\\theta_\\star}{\\sehat}<-z_{\\alpha/2}} \\\\\n & =\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}>\\frac{\\theta_0-\\theta_\\star}{\\sehat} + z_{\\alpha/2}}\n +\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}<\\frac{\\theta_0-\\theta_\\star}{\\sehat}-z_{\\alpha/2}} \\\\\n & =1-\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}<\\frac{\\theta_0-\\theta_\\star}{\\sehat} + z_{\\alpha/2}}\n +\\P{\\frac{\\thetahat-\\theta_\\star}{\\sehat}<\\frac{\\theta_0-\\theta_\\star}{\\sehat}-z_{\\alpha/2}},\n \\end{align*}\n which, since $(\\thetahat-\\theta_\\star)/\\sehat \\approx N(0, 1)$, implies that\n \\[\n \\P{|W|>z_{\\alpha/2}}\\approx\n 1-\\Phi\\left(\\frac{\\theta_0-\\theta_\\star}{\\sehat} + z_{\\alpha/2}\\right)\n +\\Phi\\left(\\frac{\\theta_0-\\theta_\\star}{\\sehat}-z_{\\alpha/2}\\right).\n \\]\n\\end{ex}\n\n\\begin{ex}\n By Theorem 10.12,\n \\[\n p=\\mathbb{P}_{\\theta_0}{(T(X^n)\\geq T(x^n))}=1-F(T(x^n)),\n \\]\n where $T(X^n)\\sim F$ under $H_0:\\theta=\\theta_0$. Therefore,\n \\[\n \\P{p \\thetahat+\\sehat\\, z_{\\alpha/2}\\text{, or }\n \\theta_0< \\thetahat-\\sehat\\, z_{\\alpha/2}.\n \\]\n This is equivalent to\n \\[\n |\\theta_0-\\thetahat|> \\sehat\\, z_{\\alpha/2},\n \\]\n which is equivalent to\n \\[\n \\frac{|\\thetahat-\\theta_0|}{\\sehat} > z_{\\alpha/2},\n \\]\n but this is precisely the size $\\alpha$ Wald test.\n\\end{ex}\n\n\\begin{ex}\n Note that\n \\[\n p\\text{-value}\n =\\inf\\left\\{\\alpha \\mid T(X^n)\\geq c_\\alpha\\right\\},\n \\]\n where $\\alpha=\\sup_{\\theta\\in\\Theta_0}\\mathbb{P}_{\\theta}(T(X^n)\\geq\n c_\\alpha)$ is a decreasing function of $c_\\alpha$, and that therefore,\n having observed $x^n$, the smallest value of $\\alpha$ will be obtained for\n the largest $c_\\alpha$ such that we still reject the null, $c_\\alpha=T(x^n)$.\n The size of the test for which we reject the null is then\n \\[\n \\sup_{\\theta\\in\\Theta_0}\\mathbb{P}_{\\theta}(T(X^n)\\geq T(x^n))\n \\]\n and therefore\n \\[\n p\\text{-value}\n =\\sup_{\\theta\\in\\Theta_0}\\mathbb{P}_{\\theta}(T(X^n)\\geq T(x^n)),\n \\]\n or, in the case where $\\Theta_0=\\{\\theta_0\\}$, the supremum is over only a\n single element and therefore\n \\[\n p\\text{-value}\n =\\mathbb{P}_{\\theta_0}(T(X^n)\\geq T(x^n)).\n \\]\n\\end{ex}\n\n% 5\n\\begin{ex}\n Let $X_1,\\ldots,X_n\\sim\\text{Uniform}(0,\\theta)$ and\n $Y=\\max\\{X_1,\\ldots,X_n\\}$.\n \\begin{enumerate}[(a)]\n \\item We have\n \\begin{align*}\n \\beta(\\theta_0)\n & =\\mathbb{P}_{\\theta_0}(Y>c) \\\\\n & =1-\\mathbb{P}_{\\theta_0}(Y\\leq c) \\\\\n & =1-\\mathbb{P}_{\\theta_0}(X_1\\leq c)\\cdots\\mathbb{P}_{\\theta'}(X_n\\leq c) \\\\\n & =1-(c/\\theta_0)^n\n \\end{align*}\n for $c\\in[0,\\theta_0]$.\n \\item Solving\n \\[\n 0.05=1-(c/\\theta_0)^n\n \\]\n for $c$ we get\n \\[\n c=0.95^{1/n}\\theta_0,\n \\]\n or, substituting $0.5$ for $\\theta_0$,\n \\[\n c=0.5\\cdot 0.95^{1/n}.\n \\]\n \\item We have\n \\[\n p=1-(0.48/0.5)^{n},\n \\]\n which implies that $p\\approx 0.558$, which means that the test does\n not provide any evidence against $H_0$.\n \\item Note that $Y=0.52$ is outside the range $[0,\\theta_0]$, and therefore\n $\\mathbb{P}_{\\theta_0}(Y>0.52)=0$. The $p$-value is therefore $0$, and\n we can thus reject $H_0$ at any level.\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}\n Since we will need to construct a confidence interval, we will use the Wald\n test for a Binomial distribution. Let $H_0: p=1/2$ and $H_1:p\\neq 1/2$. By\n Exercise 10.15, $\\phat=X/n=922/1919$ with\n $\\sehat{\\phat}=\\sqrt{\\phat(1-\\phat)/n}$. The test statistic is\n \\[\n \\frac{|\\phat-p_0|}{\\sehat(\\phat)}\n \\]\n and thus by Theorem 10.13 the $p$-value is given by\n \\[\n \\P{|Z|>\\frac{|\\phat-p_0|}{\\sqrt{\\phat(1-\\phat)/n}}}\n =2\\Phi\\left(\n -\\frac{|\\phat-p_0|}{\\sqrt{\\phat(1-\\phat)/n}}\n \\right).\n \\]\n \\inputminted{python}{../code/10-06.py}\n \\inputminted{text}{../output/10-06.txt}\n\n The test indicates weak evidence against $H_0$.\n\\end{ex}\n\n\\begin{ex}~\n \\begin{enumerate}[(a)]\n \\item Let $X_1$ be the distribution of the proportion of three-letter words\n in the Twain essays, and let $X_2$ be the distribution in the\n Snodgrass essays. Recall from Example 10.8 that the plug-in estimator\n for a difference of means $d=\\mu_1-\\mu_2$ is given by the difference\n of the plug-in estimators of the means, the sample averages. Thus,\n \\[\n \\widehat{d}=\\Xbar_1-\\Xbar_2,\n \\]\n with\n \\[\n \\sehat(\\widehat{d})\n =\\sqrt{(\\sehat(\\muhat_1))^2+(\\sehat(\\muhat_2))^2},\n \\]\n where\n \\[\n \\sehat(\\muhat_j)=\\frac{1}{\\sqrt{n_j}}\n \\sqrt{\\frac{1}{n_j}\\sum_{i=1}^{n_j}(X_{j,i}-\\Xbar_j)^2}.\n \\]\n\n Note that $H_0: d=0$, and therefore the size $\\alpha$ Wald test for\n this hypothesis is obtained by checking whether\n \\[\n \\frac{\\left|\\Xbar_1-\\Xbar_2\\right|}{\n \\sqrt{(\\sehat(\\Xbar_1))^2+(\\sehat(\\Xbar_2))^2}}\n > z_{\\alpha/2}.\n \\]\n\n \\inputminted{python}{../code/10-07.py}\n \\inputminted{text}{../output/10-07.txt}\n\n We may conclude that there is very strong evidence to reject the null\n hypothesis, i.e.\\ that the proportion of three-letter words is\n similarly distributed in both sets of essays.\n \\item We obtain the same result from the permutation test: we have very\n strong evidence that the samples comes from sets with different means.\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots,X_n\\sim N(\\theta, 1)$. We are testing $H_0:\\theta=0$ versus\n $H_1:\\theta=1$, using the rejection region $R=\\{x^n \\mid T(x^n)>c\\}$ where\n $T(x^n)=\\Xbar$.\n \\begin{enumerate}[(a)]\n \\item Note that\n \\begin{align*}\n \\alpha\n & = \\mathbb{P}_{\\theta=0}\\left(\\Xbar > c \\right) \\\\\n & = \\mathbb{P}_{\\theta=0}\\left(\\sqrt{n}\\Xbar > \\sqrt{n}c \\right) \\\\\n & = \\mathbb{P}_{\\theta=0}\\left(Z > \\sqrt{n}c \\right) \\\\\n & = 1-\\Phi(\\sqrt{n}c),\n \\end{align*}\n and that therefore $c=z_{\\alpha}/\\sqrt{n}$.\n \\item We have\n \\begin{align*}\n \\beta(1)\n & = \\mathbb{P}_{\\theta=1}\\left(\\Xbar > c \\right) \\\\\n & = \\mathbb{P}_{\\theta=1}\\left(\\Xbar-1 > c-1 \\right) \\\\\n & = \\mathbb{P}_{\\theta=1}\\left(\\sqrt{n}(\\Xbar-1) > \\sqrt{n}(c-1) \\right) \\\\\n & = \\mathbb{P}_{\\theta=1}\\left(z > \\sqrt{n}(c-1) \\right) \\\\\n & = 1-\\Phi(\\sqrt{n}({c-1})) \\\\\n & = 1-\\Phi(z_\\alpha-\\sqrt{n}).\n \\end{align*}\n \\item Note that\n \\begin{align*}\n \\lim_{n\\to\\infty}\\beta_n(1)\n & =\\lim_{n\\to\\infty}[1-\\Phi(z_\\alpha-\\sqrt{n})] \\\\\n & =1-\\lim_{u\\to-\\infty}\\Phi(u) \\\\\n & =1.\n \\end{align*}\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}\n Note that\n \\begin{align*}\n \\beta(\\theta_1)\n & =\\mathbb{P}_{\\theta_1}\\left(\\left|\\frac{\\thetahat-\\theta_0}{\\sehat}\\right| > z_{\\alpha/2} \\right) \\\\\n & =\\mathbb{P}_{\\theta_1}\\left(\\frac{\\thetahat-\\theta_0}{\\sehat} > z_{\\alpha/2} \\right)\n +\\mathbb{P}_{\\theta_1}\\left(\\frac{\\thetahat-\\theta_0}{\\sehat} < -z_{\\alpha/2} \\right) \\\\\n & =1\n -\\mathbb{P}_{\\theta_1}\\left(\\frac{\\thetahat-\\theta_0}{\\sehat} < z_{\\alpha/2} \\right)\n +\\mathbb{P}_{\\theta_1}\\left(\\frac{\\thetahat-\\theta_0}{\\sehat} < -z_{\\alpha/2} \\right) \\\\\n & =1\n -\\mathbb{P}_{\\theta_1}\n \\left(\\frac{\\thetahat-\\theta_1}{\\sehat}\n +\\frac{\\theta_1-\\theta_0}{\\sehat}\n < z_{\\alpha/2} \\right)\n +\\mathbb{P}_{\\theta_1}\\left(\n \\frac{\\thetahat-\\theta_1}{\\sehat}\n +\\frac{\\theta_1-\\theta_0}{\\sehat}\n < -z_{\\alpha/2} \\right) \\\\\n & =1\n -\\mathbb{P}_{\\theta_1}\n \\left(\\frac{\\thetahat-\\theta_1}{\\sehat}\n < z_{\\alpha/2}\n -\\frac{\\theta_1-\\theta_0}{\\sehat}\n \\right)\n +\\mathbb{P}_{\\theta_1}\\left(\n \\frac{\\thetahat-\\theta_1}{\\sehat}\n < -z_{\\alpha/2}\n -\\frac{\\theta_1-\\theta_0}{\\sehat}\n \\right) \\\\\n & =1-\\lim_{n\\to\\infty}\\Phi\\left(z_{\\alpha/2}-\\sqrt{nI(\\thetahat)}(\\theta_1-\\theta_0)\\right)\n +\\lim_{n\\to\\infty}\\Phi\\left(-z_{\\alpha/2}-\\sqrt{nI(\\thetahat)}(\\theta_1-\\theta_0)\\right),\n \\end{align*}\n or, since $\\theta_1>\\theta_0$,\n \\[\n \\beta(\\theta_1)\n = 1 - \\lim_{u\\to-\\infty}\\Phi(u)+\\lim_{u\\to-\\infty}\\Phi(u)\n = 1.\n \\]\n\\end{ex}\n\n% 10\n\\begin{ex}\n We will use the Wald test to check whether there is a difference between the\n proportion of deaths before the Chinese Harvest Moon Festival between the two\n groups.\n \\inputminted{python}{../code/10-10.py}\n \\inputminted{text}{../output/10-10.txt}\n\n We find that there is little to no evidence against $H_0$.\n\\end{ex}\n\n\\begin{ex}~\n \\begin{enumerate}[(a)]\n \\item\n \\inputminted{python}{../code/10-11.py}\n \\inputminted{text}{../output/10-11.txt}\n Since only the $p$-value for Chlorpromazine is less than $0.05$, the\n only null hypothesis that we may reject is that the placebo is\n similarly effective as Chlorpromazine.\n \\item Our finding remains statistically significant at the $0.05$ level\n under both the Bonferroni or the Benjamini-Hochberg multiple\n testing method corrections.\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}~\n \\begin{enumerate}[(a)]\n \\item Let $X_1,\\ldots,X_n\\sim\\text{Poisson}(\\lambda)$. We begin by computing\n the maximum likelihood estimator for $\\lambda$. Recall that then\n \\[\n f(x;\\lambda)=e^{-\\lambda}\\frac{\\lambda^x}{x!},\n \\]\n and that therefore\n \\[\n \\ell_n(\\lambda)=\\sum_{i=1}^n-\\lambda+X_i\\log(\\lambda)+\\log(X_i!).\n \\]\n Hence,\n \\[\n \\frac{\\d\\ell_n(\\lambda)}{\\d\\lambda}\n =\\sum_{i=1}^n\\left[-1+\\frac{X_i}{\\lambda}\\right]\n =\\frac{n(\\Xbar-\\lambda)}{\\lambda}\n \\text{ implies }\n \\lambda=\\Xbar,\n \\]\n and it is clear by the second derivative test that this is a maximum.\n Hence, $\\widehat{\\lambda}=\\Xbar$, and\n $\\sehat(\\widehat{\\lambda})=\\sqrt{\\Xbar/n}$.\n\n Let $H_0:\\lambda=\\lambda_0$. Then, the size $\\alpha$ Wald test is\n given by rejecting $H_0$ when\n \\[\n \\frac{\\sqrt{n}|\\Xbar-\\lambda_0|}{\\sqrt{\\Xbar}}>z_{\\alpha/2}.\n \\]\n \\item\n \\inputminted{python}{../code/10-12.py}\n \\inputminted{text}{../output/10-12.txt}\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots, X_n\\sim N(\\mu, \\sigma^2)$. Recall from Example 9.11 that then\n \\[\n \\ell(\\mu, \\sigma)\n =-n\\log\\sigma -\\frac{nS^2}{2\\sigma^2}-\\frac{n(\\Xbar-\\mu)^2}{2\\sigma^2}.\n \\]\n Therefore,\n \\[\n \\frac{\\pd \\ell(\\mu, \\sigma)}{\\pd \\mu}\n =\\frac{n(\\Xbar-\\mu)}{\\sigma^2}\n \\text{ implies }\n \\muhat = \\overline{X},\n \\]\n and\n \\[\n \\frac{\\pd \\ell(\\Xbar, \\sigma)}{\\pd \\sigma}=-\\frac{n}{\\sigma}+\\frac{nS^2}{\\sigma^3}\n \\text{ implies }\n \\sigmahat = S.\n \\]\n We have $\\Theta_0=\\{(\\mu, \\sigma) \\mid \\mu=\\mu_0\\}$ and so for the likelihood\n ratio test we have\n \\begin{align*}\n \\lambda\n & = 2\\log\\left(\\frac{\\sup_{\\theta\\in\\Theta}\\L(\\theta)}{\\sup_{\\theta_0\\in\\Theta_0}\\L(\\theta_0)} \\right) \\\\\n & =2\\ell(\\Xbar, S)-2\\ell(\\mu_0, S) \\\\\n & =\\frac{n(\\Xbar-\\mu_0)^2}{S^2},\n \\end{align*}\n where, under $H_0$, we expect $\\lambda\\rightsquigarrow \\chi^2$, and\n therefore have $p$-value\n \\[\n \\P{\\chi^2>\\frac{n(\\Xbar-\\mu_0)^2}{S^2}}.\n \\]\n\n Recall that $\\se(\\muhat)=\\sigma/\\sqrt{n}$, and that therefore the $p$-value\n for the Wald test is given by\n \\[\n \\P{|Z|>\\frac{\\sqrt{n}|\\Xbar -\\mu_0|}{S}},\n \\]\n or, by squaring both sides,\n \\[\n \\P{Z^2>\\frac{n(\\Xbar -\\mu_0)^2}{S^2}},\n \\]\n which is equivalent to the likelihood-ratio test since $Z^2$ has a chi-squared\n distribution.\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots, X_n\\sim N(\\mu, \\sigma^2)$. Recall that\n \\[\n \\ell(\\mu, \\sigma)=-n\\log\\sigma -\\frac{nS^2}{2\\sigma^2}-\\frac{n(\\Xbar-\\mu)^2}{2\\sigma^2},\n \\]\n and by an identical argument to the previous problem,\n \\[\n \\muhat=\\Xbar\\text{ and }\\sigmahat=S.\n \\]\n We then have $\\Theta_0=\\{(\\mu, \\sigma) \\mid \\sigma=\\sigma_0\\}$, and therefore\n \\begin{align*}\n \\lambda\n & =2\\log\\left(\\frac{\\sup_{\\theta\\in\\Theta}\\L(\\theta)}{\\sup_{\\theta_0\\in\\Theta_0}\\L(\\theta_0)} \\right) \\\\\n & =2\\ell(\\Xbar, S)-2\\ell(\\Xbar, \\sigma_0) \\\\\n & = n\\left[2\\log\\frac{\\sigma_0}{S}+\\frac{S^2}{\\sigma_0^2}-1\\right] \\\\\n & = 2n\\log\\frac{\\sigma_0}{S}+\\frac{n(S^2-\\sigma_0^2)}{\\sigma_0^2},\n \\end{align*}\n where, under $H_0$, we expect $\\lambda\\rightsquigarrow \\chi^2$, and\n therefore have $p$-value\n \\[\n \\P{\\chi^2>\n 2n\\log\\frac{\\sigma_0}{S}+\\frac{n(S^2-\\sigma_0^2)}{\\sigma_0^2}\n }.\n \\]\n\n Recall that $\\se(\\sigmahat)=\\sigma/\\sqrt{2n}$, and that therefore the\n $p$-value for the Wald test is given by\n \\[\n \\P{|Z|>\\frac{\\sqrt{2n}|S -\\sigma_0|}{S}},\n \\]\n or,\n \\[\n \\P{Z^2>\\frac{2n(S -\\sigma_0)^2}{S^2}}.\n \\]\n\\end{ex}\n\n% 15\n\\begin{ex}\n Let $X \\sim \\text{Binomial}(n, p)$. Note that then\n \\[\n \\ell(n, p)=\\log\\binom{n}{X}+X\\log{p}+(n-X)\\log(1-p),\n \\]\n and therefore\n \\[\n \\frac{\\pd\\ell(n, p)}{\\pd{p}}\n =\\frac{X}{p}-\\frac{n-X}{1-p}\n \\text{ implies }\n \\phat =X/n.\n \\]\n\n We have $\\Theta_0=\\{p_0\\}$. Therefore,\n \\[\n \\lambda\n =2\\log\\left(\\frac{\\sup_{\\theta\\in\\Theta}\\L(\\theta)}{\\sup_{\\theta_0\\in\\Theta_0}\\L(\\theta_0)} \\right)\n =2\\ell(n, \\phat)-2\\ell(n, p_0)\n =2X\\log\\left(\\frac{\\phat}{p_0}\\right)+2(n-X)\\log\\left(\\frac{1-\\phat}{1-p_0}\\right),\n \\]\n where, under $H_0$, we expect $\\lambda\\rightsquigarrow \\chi^2$.\n\n Note that by Exercise 9.7, $\\sehat(\\phat)=\\sqrt{\\phat(1-\\phat)/n}$, and\n therefore the $p$-value of the Wald test is given by\n \\[\n \\P{|Z|>\\frac{\\sqrt{n}|\\phat-p_0|}{\\sqrt{\\phat(1-\\phat)}}}.\n \\]\n\\end{ex}\n\n\\begin{ex}\n Let $\\ell(\\theta)$ be a log-likelihood. Note that then the second degree\n Taylor polynomial expansion of $\\ell$ around the MLE $\\thetahat$ is given by\n \\[\n \\ell(\\theta)\n \\approx\n \\ell(\\thetahat)+\\ell'(\\thetahat)(\\theta-\\thetahat)\n +\\frac{1}{2}\\ell''(\\thetahat)(\\theta-\\thetahat)^2,\n \\]\n and therefore,\n \\[\n \\lambda\n =2\\ell(\\thetahat)-2\\ell(\\theta_0)\n \\approx -\\ell''(\\thetahat)(\\thetahat-\\theta_0)^2\n = I(\\thetahat)(\\thetahat-\\theta_0)^2\n = \\frac{(\\thetahat-\\theta_0)^2}{\\sehat^2(\\thetahat)},\n \\]\n however, this is precisely $W^2$.\n\n To complete the proof and show that $\\frac{W^2}{\\lambda}\\xrightarrow{P} 1$, we\n need to be able to show that the error of our quadratic approximation to the\n log-likelihood decreases with sample size, but this is not true in general,\n and it does not look like we have any of the relevant results in the book.\n\\end{ex}", "meta": {"hexsha": "d1e969c997c2ac064889b28cc445bac830311df3", "size": 16381, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/tex/ch10.tex", "max_stars_repo_name": "dtrifuno/all-of-stats-solutions", "max_stars_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/tex/ch10.tex", "max_issues_repo_name": "dtrifuno/all-of-stats-solutions", "max_issues_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/tex/ch10.tex", "max_forks_repo_name": "dtrifuno/all-of-stats-solutions", "max_forks_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.2279569892, "max_line_length": 110, "alphanum_fraction": 0.5436176058, "num_tokens": 6038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424217727027, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6361593378498861}} {"text": "\n\\subsection{Choice set}\n\nEconomic agents face options from some set. This could be consumption choices, numbers of hours to work, or how much capital to invest in at a factory.\n\n\\subsubsection{Utility functions}\n\nCalculating choices: Unrestricted choices\n\nThe choice of an agent is the selection which corresponds to the highest value of the utility function. Consider:\n\n\\(f=2(x-1)^2-10\\)\n\nWe can easily calculate that even if the agent can choose any real number \\(x\\), they will chose \\(1\\).\n\nThis approach can be used if there are not meaningful constraints, or those constraints are implicit in the utility function. For example a firm can be modelled as profit maximising, where profit is a function of revenue and costs, with a single maximising value.\n\nOther agents, such as a consumers, may instead have utility over consumption and leisure, and a separate constraint over this. This could be solved using simultaneous equations, but such an approach is not always desirable.\n\n", "meta": {"hexsha": "6bdb9feba4e0316a6b073c0caaa2bd65c7f9d948", "size": 986, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/ai/singleAgent/01-01-axiomsChoiceSet.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/ai/singleAgent/01-01-axiomsChoiceSet.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/ai/singleAgent/01-01-axiomsChoiceSet.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.3, "max_line_length": 263, "alphanum_fraction": 0.7900608519, "num_tokens": 206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8558511469672594, "lm_q2_score": 0.7431680143008301, "lm_q1q2_score": 0.6360411974287461}} {"text": "\\documentclass[a4paper,11pt,final]{article}\n\\usepackage{fancyvrb, color, graphicx, hyperref, amsmath, url}\n\\usepackage{palatino}\n\\usepackage[a4paper,text={16.5cm,25.2cm},centering]{geometry}\n\n\\hypersetup\n{ pdfauthor = {Matti Pastell},\n pdftitle={FIR filter design with Python and SciPy},\n colorlinks=TRUE,\n linkcolor=black,\n citecolor=blue,\n urlcolor=blue\n}\n\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{1.2ex}\n\n\n\n\\title{FIR filter design with Python and SciPy}\n\\author{Matti Pastell \\\\ \\url{http://mpastell.com}}\n\\date{15th April 2013}\n\n\\begin{document}\n\\maketitle\n\n\n\\section{Introduction}\n\nThis an example of a document that can be published using\n\\href{http://mpastell.com/pweave}{Pweave}. Text is written\nusing \\LaTeX{} and code between \\texttt{<<>>} and \\texttt{@} is executed\nand results are included in the resulting document.\n\nYou can define various options for code chunks to control code\nexecution and formatting (see\n\\href{http://mpastell.com/pweave/usage.html\\#code-chunk-options}{Pweave\ndocs}).\n\n\\section{FIR Filter Design}\n\nWe'll implement lowpass, highpass and ' bandpass FIR filters. If you\nwant to read more about DSP I highly recommend\n\\href{http://www.dspguide.com/}{The Scientist and Engineer's Guide to\nDigital Signal Processing} which is freely available online.\n\n\\subsection{Functions for frequency, phase, impulse and step response}\n\nLet's first define functions to plot filter properties.\n\n\n\n\\begin{verbatim}\nfrom pylab import *\nimport scipy.signal as signal\nimport warnings\nwarnings.simplefilter(\"ignore\")\n\n#Plot frequency and phase response\ndef mfreqz(b,a=1):\n w,h = signal.freqz(b,a)\n h_dB = 20 * log10 (abs(h))\n subplot(211)\n plot(w/max(w),h_dB)\n ylim(-150, 5)\n ylabel('Magnitude (db)')\n xlabel(r'Normalized Frequency (x$\\pi$rad/sample)')\n title(r'Frequency response')\n subplot(212)\n h_Phase = unwrap(arctan2(imag(h),real(h)))\n plot(w/max(w),h_Phase)\n ylabel('Phase (radians)')\n xlabel(r'Normalized Frequency (x$\\pi$rad/sample)')\n title(r'Phase response')\n subplots_adjust(hspace=0.5)\n\n#Plot step and impulse response\ndef impz(b,a=1):\n l = len(b)\n impulse = repeat(0.,l); impulse[0] =1.\n x = arange(0,l)\n response = signal.lfilter(b,a,impulse)\n subplot(211)\n stem(x, response)\n ylabel('Amplitude')\n xlabel(r'n (samples)')\n title(r'Impulse response')\n subplot(212)\n step = cumsum(response)\n stem(x, step)\n ylabel('Amplitude')\n xlabel(r'n (samples)')\n title(r'Step response')\n subplots_adjust(hspace=0.5)\n\\end{verbatim}\n\n\n\\subsection{Lowpass FIR filter}\n\nDesigning a lowpass FIR filter is very simple to do with SciPy, all you\nneed to do is to define the window length, cut off frequency and the\nwindow.\n\nThe Hamming window is defined as:\n$w(n) = \\alpha - \\beta\\cos\\frac{2\\pi n}{N-1}$, where $\\alpha=0.54$ and\n$\\beta=0.46$\n\nThe next code chunk is executed in term mode, see the source document\nfor syntax. Notice also that Pweave can now catch multiple\nfigures/code chunk.\n\n\n\\begin{verbatim}\nn = 61\na = signal.firwin(n, cutoff = 0.3, window = \"hamming\")\n#Frequency and phase response\nmfreqz(a)\n#Impulse and step response\np = figure(2)\nimpz(a)\n\\end{verbatim}\n\\includegraphics[width= \\linewidth]{figures/FIR_design_verb_figure2_1.pdf}\n\\includegraphics[width= \\linewidth]{figures/FIR_design_verb_figure2_2.pdf}\n\n\n\\subsection{Highpass FIR Filter}\n\nLet's define a highpass FIR filter:\n\n\n\\begin{verbatim}\nn = 101\na = signal.firwin(n, cutoff = 0.3, window = \"hanning\", pass_zero=False)\nmfreqz(a)\n\\end{verbatim}\n\\includegraphics[width= \\linewidth]{figures/FIR_design_verb_figure3_1.pdf}\n\n\n\\subsection{Bandpass FIR filter}\n\nNotice that the plot has a caption defined in code chunk options.\n\n\n\n\\begin{verbatim}\nn = 1001\na = signal.firwin(n, cutoff = [0.2, 0.5], window = 'blackmanharris', pass_zero = False)\nmfreqz(a)\n\\end{verbatim}\n\\begin{figure}[htpb]\n\\center\n\\includegraphics[width= \\linewidth]{figures/FIR_design_verb_figure4_1.pdf}\n\\caption{Bandpass FIR filter.}\n\\label{fig:None}\n\\end{figure}\n\n\n\\end{document}\n", "meta": {"hexsha": "fb08e872b1323721836fcdbbd5f132bce4e1d0a3", "size": 4036, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tests/weave/tex/FIR_design_verb_REF.tex", "max_stars_repo_name": "piccolbo/Pweave", "max_stars_repo_head_hexsha": "62a740e46c16b3aecce53aa951792b33ac93e1a2", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-12-14T17:48:36.000Z", "max_stars_repo_stars_event_max_datetime": "2018-12-14T17:48:36.000Z", "max_issues_repo_path": "tests/weave/tex/FIR_design_verb_REF.tex", "max_issues_repo_name": "piccolbo/Pweave", "max_issues_repo_head_hexsha": "62a740e46c16b3aecce53aa951792b33ac93e1a2", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2018-12-13T05:00:33.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-15T19:07:01.000Z", "max_forks_repo_path": "tests/weave/tex/FIR_design_verb_REF.tex", "max_forks_repo_name": "piccolbo/Pweave", "max_forks_repo_head_hexsha": "62a740e46c16b3aecce53aa951792b33ac93e1a2", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.3836477987, "max_line_length": 87, "alphanum_fraction": 0.7252229931, "num_tokens": 1186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431679972357831, "lm_q2_score": 0.8558511488056151, "lm_q1q2_score": 0.6360411841898131}} {"text": "%%%%%PLEASE CONSIDER CORRECTIONS AT PLACES INDICATED%%%%%%%%\n\\documentclass{discussion}\n%%%%%Packages Used, add more if necessary%%%%\n\\providecommand{\\tightlist}{%\n\\setlength{\\itemsep}{2pt}\\setlength{\\parskip}{0pt}} \n\n\n\\newenvironment{answer}\n {\\vspace{-1em}\\begin{proof}[Answer]}\n {\\end{proof}}\n\n\\usepackage[ruled]{algorithm2e}\n\\begin{document}\n\n%%%%%CHANGE HERE%%%%%%%%%\n%%%%%\\lecture{the ordinal number of the tutorial}{lecture title}{Instructor's name}%%%%%%%%\n\\lecture{6}{Ensemble Methods}{Chansoo Lee}{}\n\n%%%%%CHANGE HERE%%%%%%%\n%%%%%\\section{title of the section} similarly with the rest \\section{} or \\subsection{} or \\subsubsection{} etc\n\n \n\\renewcommand{\\R}{\\mathbb{R}}\n\\renewcommand{\\vec}[1]{\\mathbf{#1}}\n\\newcommand{\\Dtrain}{\\mathcal{D}_{\\mathrm{train}}}\n\\newcommand{\\tm}{\\mathbf{\\theta}_{\\text{model}}}\n\\newcommand{\\xtest}{\\mathbf{\\x}_{\\text{test}}}\n\\newcommand{\\test}{\\mathrm{test}}\n\\newcommand{\\pred}{\\mathrm{pred}}\n\n\\newcommand{\\D}{\\mathcal{D}}\n\\newcommand{\\F}{F}\n\\newcommand{\\WL}{\\mathrm{WL}}\n\\section{Important Concepts}\n\\paragraph{True vs Empirical Risk} Simply put, \\emph{risk} is the \\emph{expected loss}. If we take the expectation over true data distribution, we get true risk. If we take the expectation over \\emph{empirical probability density estimate} (which is uniform over a finite data, HW3 problem 2), we get empirical risk.\n\nNow, let us define these formally. Suppose the data is i.i.d. samples from a distribution $Q$. Then, the \\emph{true risk} (or simply \\emph{risk}) of hypothesis $h$ is the expectation of the loss function $\\ell$ over $Q$:\n\\[R(h) = \\mathbb{E}_{(\\mathbf{x},y) \\sim Q}[\\ell(h(\\mathbf{x}), y)].\\]\nWe have seen a very similar expression in homework 3.\n\nThe true risk cannot be computed, because $Q$ is unknown. The learning algorithm, however, can approximate the true risk using training examples $S = \\{(\\vec{x}_i, y_i)\\}_{i=1}^{n}$:\n\\[R_S(h) = \\frac{1}{n} \\sum_{i=1}^{n} \\ell(h(\\mathbf{x}_i), y_i).\\]\nAs $n$ goes to infinity, the empirical risk converges to the true risk.\n\n\nFor example, consider linear regression. The weight vector $\\mathbf{w}$ is the model parameter, and the prediction function $h(\\mathbf{x}) = \\mathbf{w}^\\top \\mathbf{x}$ is the associated hypothesis. Define our loss function as $\\ell(h(\\mathbf{x}), y) = (h(\\mathbf{x}) - y)^2$. The unregularized linear regression corresponds to \\emph{empirical risk minimization} where your \\emph{hypothesis class} is restricted to linear functions.\n\n\\paragraph{Bagging vs Boosting} Bagging focuses on aggregating outputs of multiple hypotheses independently trained on different (sub)samples of train data. \n\nBoosting is more holistic approach, which sequentially chooses the weights on hypotheses and the weights on train examples. \n\n\\begin{exercise}\nCan each hypothesis in the ensemble be trained in a parallel manner for bagging? How about for boosting?\n\\end{exercise}\n\\begin{answer}\nBagging can be, but boosting cannot be due to its sequential nature.\n\\end{answer}\n\n\n\\begin{exercise}\nWith access to distributed system, is bagging always computationally superior to boosting?\n\\end{exercise}\n\\begin{answer}\n Bagging reduces the variance only; its performance is essentially the average performance of the hypotheses in the ensemble. Each hypothesis in a boosted ensemble, on the other hand, needs to only satisfy weak learning hypothesis, which usually takes only a trivial amount of computing power.\n\\end{answer}\n\n\\section{AdaBoost}\nLet $S = \\{(\\vec{x}_i, y_i)\\}_{i=1}^{n}$ be the training set of examples. Let $\\WL$ be a $\\gamma$-weak learning algorithm that takes a distribution over input space (as a tuple of weights and samples) and returns a hypothesis (classifier) $f_t$ with error at most $\\frac{1}{2} - \\gamma$. Let $T$ be the number of iterations.\n\n\\begin{algorithm}\n\\caption{AdaBoost}\n\\KwIn{$S, \\WL, T$ as defined in the text}\n initialize $\\mathbf{w}^1 = (1,\\ldots,1)$ \\\\\n \\For{$t = 1,\\ldots, T$}{\n normalize $\\vec{w}^{t}$ \\\\\n invoke weak learner $f_t = \\WL(\\vec{w}^t, S)$ \\\\\n compute $r_t = \\sum_{i=1}^{n} w_i^t \\mathbf{1}[y_i \\neq f_t(\\vec{x}_i)]$ \\\\\n let $\\displaystyle\\alpha_t = \\frac{1}{2} \\log(\\frac{1}{r_t} - 1)$ \\\\\n update $\\displaystyle w_i^{t+1} = w_i^{t} \\exp(-\\alpha_t y_i f_t(\\vec{x}_i))$ \\\\\n }\n \\KwOut{Return $\\F_T = \\sum_{t=1}^{T} \\alpha_t f_t$}\n\\end{algorithm}\n\n\\begin{theorem}\n The training error (empirical risk) of AdaBoost satisfies:\n \\[R_S(\\F_T) := \\frac{1}{n} \\sum_{i=1}^{n} \\mathbf{1}[\\F_T(\\vec{x}_i) \\neq y_i] \\leq \\exp(-2 \\gamma^2 T).\\]\n\\end{theorem}\n\\begin{proof} For each $t$, let\n\\[\\textstyle \\F_t = \\sum_{s=1}^{t} \\alpha_s f_s\\]\nand let $z_t$ be the normalizing constant for $\\vec{w}^t$, which is (by inductive argument)\n\\[z_t = \\frac{1}{n}\\sum_{i=1}^{n} \\exp(-y_i \\F_t (\\vec{x}_i)).\\]\n\\emph{(Optional paragraph.)} The proof is left for exercise. Roughly speaking, we update $w_i$ in a multiplicative manner, adding $-\\alpha_t y_i f_t(\\mathbf{x}_i)$ to the exponent every round. So at round $t$, we have $\\exp(-\\sum_{s=1}^{t} \\alpha_s y_i f_s(\\mathbf{x}_i)) = \\exp(-y_i \\F_s(\\mathbf{x}_i))$. The base case for $t=1$ gives the $(1/n)$.\n\n\nNote that the exponential function acts as a \\emph{smooth} upper bound on the loss of a hypothesis $h$, as $\\mathbf{1}[h(\\vec{x}) \\neq y] \\leq \\exp(-y h(\\vec{x})).$ So,\n$R_S(\\F_T) \\leq z_T$ and it suffices to show that $z_T \\leq \\exp(-2\\gamma^2 T)$. In other words, our normalization constant $z_t$ coincides with an upper bound on the empirical risk of $\\F_t$.\n\n\nRewrite \\[z_T = \\frac{z_T}{z_{T-1}} \\cdots \\frac{z_2}{z_1}.\\]\nIt suffices to show that for every $t$, \\[\\frac{z_{t+1}}{z_t} \\leq \\exp(-2\\gamma^2)\\]\nThis means that the \\emph{upper bound} on the empirical risk decreases in a multiplicative manner (fraction of mistakes fixed as opposed to a fixed number of mistakes fixed) every round. The actual empirical risk, however, might oscillate. Higher $\\gamma$ means faster decay in the upper bound, as it should.\n\n\\emph{The rest of the proof is optional:} Now,\n\\begin{align*}\n \\frac{z_{t+1}}{z_t} &= \\frac{\\sum_{i=1}^{n} \\exp(-y_i \\F_{t+1}(\\vec{x}_i))}{z_t} \\\\\n &= \\frac{\\sum_{i=1}^{n} \\exp(-y_i \\F_{t}(\\vec{x}_i)) \\exp(-y_i \\alpha_{t+1} f_{t+1}(\\vec{x}_i))}{z_t} \\\\\n &=\\sum_{i=1}^{n} w_i^{t+1} \\exp(-y_i \\alpha_{t+1} f_{t+1}(\\vec{x}_i)) \\\\\n &= \\exp(-\\alpha_{t+1})(1-r_{t+1}) + \\exp(\\alpha_{t+1})r_{t+1} \\\\\n &= \\frac{1}{\\sqrt{1/r_{t+1} - 1}}(1-r_{t+1}) + \\sqrt{1/r_{t+1} - 1} \\ r_{t+1} \\\\\n &= 2\\sqrt{r_{t+1} (1-r_{t+1})} \\leq 2 \\sqrt{\\left(\\frac{1}{2} - \\gamma\\right) \\left(\\frac{1}{2} + \\gamma\\right)} = \\sqrt{1 - 4\\gamma^2}.\n\\end{align*}\nTo complete the proof, use the inequality $1 - a \\leq e^{-a}$.\n\\end{proof}\n\nGenerally, minimizing a all-or-nothing (0-or-1 or 0-or-infinity) loss function is \\emph{computationally} hard in the sense of NP-hardness. Hence, we use a \\emph{surrogate loss} function that is continuous and convex. The \\emph{hinge loss} for soft-SVM is one such example which converts a 0-or-$\\infty$ loss function to a linear-and-flat function. Similarly, AdaBoost can be considered an algorithm that minimizes the exponential loss function. The key difference is that the exponential loss directly upper bounds the classification error, whereas the hinge loss does not.\n\n\\begin{exercise}\nExplain the semantics of the variables used in AdaBoost: $\\mathbf{w}^t$, $f_t$, $r_t$, $\\alpha_t$ and $F_t$.\n\\end{exercise}\n\n\\begin{exercise} Answer the following questions:\n\\begin{itemize}\n \\item Convince yourself that $r_t$ is the risk of $f_t$ when the true distribution is $\\mathbf{w}^t$.\n \\[r_t = \\mathbb{E}_{i \\sim \\mathbf{w}^t}\\left[\\mathbf{1}[f_t(\\vec{x}_i) \\neq y_i]\\right]. \\]\n \\item Is $\\alpha_t$ an increasing or decreasing function of $r_t$? Why does it make sense, considering how we construct the final output $\\F_T$?\n \\begin{answer}\n Decreasing. $f_t$ that had higher risk should contribute less to the final output $F_T$.\n \\end{answer}\n \\item The update rule for $w_i$ shows that the amount of change in $w_i^t$ is increasing or decreasing in $\\alpha_t$? Why does it make sense?\n \\begin{answer}\n Increasing. The weight $w_i$ should reflect how difficult it is to classify the $i$-th example correctly, so we can focus on the difficult examples. If $\\alpha_t$ is large, then $f_t$ performed well overall and thus it is a ``trustworthy'' hypothesis. Its mistakes should affect our opinion on how difficult each example is, more so than a non-trustworthy hypothesis (low $\\alpha_t$) would.\n \\end{answer}\n\\end{itemize} \n\\end{exercise}\n\n\\begin{exercise}\nDoes AdaBoost guarantee a zero test error?\n\\end{exercise}\n\n\\begin{answer}\n No. AdaBoost guarantees that if $T$ is sufficently large, then the \\emph{empirical risk} ultimately becomes 0. But it says nothing about the test error aka true risk.\n\\end{answer}\n\n\\section{Generalized Boosting (Optional)}\n\n\\subsection{Coordinate Descent}\nCoordinate descent is a variation of gradient descent, where you change a single coordinate per iteration. For example, suppose we want to minimize \n\\[J(a,b) = (a - 2)^2 (b + 1)^2 \\]\n% with gradient\n% \\[\\nabla J(a,b) = (2(a-2)(b+1), (a-2)^2).\\]\n\nFix the learning rate $\\eta = .1$. Initialize (arbitrarily) $a=b=0$. In the first round, we change $a$. So, we look at the partial derivative \n\\[\\frac{\\partial J}{\\partial a} = 2(a-2)(b+1)^2\\]\nwhich we evaluate at $a=b=0$ and get $-4$. Now we perform the coordinate descent update and get $a=.4, b=0$. We repeat the process for $b$:\n\\[\\frac{\\partial J}{\\partial b} = 2(a-2)^2(b+1)\\]\nwhich we evaluate at the current point and get $5.12$. So we perform the update and get $a=.4,b= -.512$, we repeat this process, and eventually converge to $a=2, b=-1$.\n\n\n\\subsection{Boosting as Coordinate Descent}\nWe observe that boosting is a slight variation of coordinate descent on a function of $T$ variables, where we do a single sweep through each variable. We use the notation $\\hat{\\ell}$ (in lecture notes, its $\\phi$) to denote the surrogate loss function. Assume that $\\partial{\\ell}/\\partial{\\alpha_t}$ is always negative. We first define the objective function\n\\[J(\\alpha_1, \\ldots, \\alpha_T) = \\frac{1}{n} \\sum_{i=1}^{n}\\hat{\\ell}\\left(y_i \\sum_{t=1}^{T} \\alpha_{t}f_t (\\vec{x}_i)\\right) = \\frac{1}{n} \\sum_{i=1}^{n}\\hat{\\ell}\\left(y_i \\sum_{t=1}^{T} \\F_t (\\vec{x}_i)\\right).\\]\n\nWe initialize $\\alpha_t = 0$ for all $t$, and we do coordinate descent. At $t$-th iteration, we have\n\\[J(\\alpha_1, \\ldots, \\alpha_{T}) = \\frac{1}{n} \\sum_{i=1}^{n}\\hat{\\ell}\\left(y_i \\sum_{s=1}^{t-1} \\alpha_{s}f_s + y_i \\alpha_t f_t (\\vec{x}_i)\\right),\\]\nwhere $\\alpha_{t+1}, \\ldots, \\alpha_T$ terms are ommitted because they are still 0.\n\nHence, we update $\\alpha_t$ using the partial derivative:\n\\[\\frac{\\partial J}{\\partial \\alpha_t} = \\frac{1}{n} \\sum_{i=1}^{n} \\hat{\\ell}'_t(y_i \\F_{t-1}(\\vec{x}_i) + \\alpha_t y_i f_t(\\vec{x}_i)) y_i f_t(\\vec{x}_i)\\]\nwhere $\\hat{\\ell}'_t = \\frac{\\partial \\hat{\\ell}}{\\partial \\alpha_t}$ as a shorthand notation.\n\n\nBut we also have to choose $f_t$ which is not a number but a hypothesis. A smart way of choosing $f_t$ is such that the derivative is the largest, so we get the steepest coordinate descent update; that is, we choose $f_t$ to maximize the (directional) derivative:\n\\[\\arg\\max_{f_t} \\frac{\\partial J_t}{\\partial \\alpha_t}\\Big|_{\\alpha_t = 0} = \\arg\\max_{f_t} \\frac{1}{n} \\sum_{i=1}^{n} \\hat{\\ell}(y_i \\F_{t-1}) y_i f_t = \\arg\\min_{f_t} \\frac{1}{n} \\sum_{i=1}^{n} \\frac{\\hat{\\ell}'_t(y_i \\F_{t-1})}{\\sum_{i=1}^{n}\\hat{\\ell}_t'(y_i \\F_{t-1})} y_i f_t.\\]\n\nThe last equality is because $\\ell'$ is always negative. We cannot compute the exact minimizer (if that is the case, we don't even need boosting!). Instead, we have an algorithm $\\WL$ which will ``try'' to minimize the above and return a weak learning hypothesis $f_t$.\n\nOnce we fix $f_t$, it turns out that we can be very smart with how we update $\\alpha_t$, instead of doing the standard coordinate descent update. In particular, we update $\\alpha_t$ to be the minimizer:\n\\[\\alpha_t = \\arg\\min_{\\alpha} J(\\alpha_{1},\\ldots,\\alpha_{t-1},\\alpha).\\]\n\n% \\paragraph{AdaBoost as Coordinate Descent}\n% For AdaBoost, $\\hat{\\ell}$ is the exponential loss $\\hat{\\ell}(u) = \\exp(-u)$. So,\n% \\begin{align*}\n% \\sum w_i \\exp(-y_i f(\\vec{x}_i) \\alpha_t) \n% &\\leq \\sum_{i} \\frac{1 -y_i f(\\vec{x}_i)}{2} w_i \\exp(\\alpha^t) + \\sum_{i} \\frac{1 + y_i f(\\vec{x}_i)}{2} w_i \\exp(- \\alpha^t) \\\\\n% &= \\frac{1}{2} \\exp(\\alpha^{t}) - \\frac{1 - r_t}{2} \\exp(-\\alpha_t)\n% \\end{align*}\n\n\\end{document}\n", "meta": {"hexsha": "d28ded47f69fe22ca1cb7319269e5036d441e5a2", "size": 12456, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "discussion06_ensemble/discussion06_ensemble.tex", "max_stars_repo_name": "xipengwang/umich-eecs445-f16", "max_stars_repo_head_hexsha": "298407af9fd417c1b6daa6127b17cb2c34c2c772", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 97, "max_stars_repo_stars_event_min_datetime": "2016-09-11T23:15:35.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-22T08:03:24.000Z", "max_issues_repo_path": "discussion06_ensemble/discussion06_ensemble.tex", "max_issues_repo_name": "eecs445-f16/umich-eecs445-f16", "max_issues_repo_head_hexsha": "298407af9fd417c1b6daa6127b17cb2c34c2c772", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "discussion06_ensemble/discussion06_ensemble.tex", "max_forks_repo_name": "eecs445-f16/umich-eecs445-f16", "max_forks_repo_head_hexsha": "298407af9fd417c1b6daa6127b17cb2c34c2c772", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 77, "max_forks_repo_forks_event_min_datetime": "2016-09-12T20:50:46.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T14:41:23.000Z", "avg_line_length": 65.9047619048, "max_line_length": 573, "alphanum_fraction": 0.6897880539, "num_tokens": 4081, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.815232489352, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6358755579203013}} {"text": "\\documentclass[10pt]{article}\n\\author{Alex Peyrard}\n\\title{equivalences}\n\n\\begin{document}\n\\maketitle\n\n\\section{Equivalences}\n\\[(1) \\Leftrightarrow \\forall\\varepsilon>0, \\lim_{n \\to \\infty} Pr(|X_{n}-Y|<\\varepsilon)=1\\]\n\\[(4) \\Leftrightarrow \\forall\\varepsilon>0, \\lim_{n \\to \\infty} Pr(|X_{n}-Y|\\leq\\varepsilon)=1\\]\n\\section{Proof that $(1) \\Rightarrow (4)$}\n\\[\\lim_{n \\to \\infty} Pr(|X_{n}-Y|<\\varepsilon) \\leq \\lim_{n \\to \\infty} Pr(|X_{n}-Y|\\leq\\varepsilon)\\]\n\\[\\lim_{n \\to \\infty} Pr(|X_{n}-Y|<\\varepsilon) = 1\\]\nWe thus have\n\\[1 <= \\lim_{n \\to \\infty} Pr(|X_{n}-Y|\\leq\\varepsilon) <= 1\\]\nSo \\[ \\lim_{n \\to \\infty} Pr(|X_{n}-Y|\\leq\\varepsilon) = 1 \\]\nQED\n\\section{Proof that $(4) \\Rightarrow (1)$}\n\\end{document}", "meta": {"hexsha": "ed58d661b906371601d67e356c9b48d7732be755", "size": 717, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ICT/equivalences.tex", "max_stars_repo_name": "apeyrard/sjtu-work", "max_stars_repo_head_hexsha": "ca98fec3c83b81ed9091bdc968cb5ad8a74d1d6a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-03-26T10:04:05.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-26T10:04:05.000Z", "max_issues_repo_path": "ICT/equivalences.tex", "max_issues_repo_name": "apeyrard/sjtu-work", "max_issues_repo_head_hexsha": "ca98fec3c83b81ed9091bdc968cb5ad8a74d1d6a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ICT/equivalences.tex", "max_forks_repo_name": "apeyrard/sjtu-work", "max_forks_repo_head_hexsha": "ca98fec3c83b81ed9091bdc968cb5ad8a74d1d6a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-03-26T10:04:06.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-26T10:04:06.000Z", "avg_line_length": 37.7368421053, "max_line_length": 103, "alphanum_fraction": 0.6443514644, "num_tokens": 316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324938410783, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.635875557252491}} {"text": "\\subsection{Bonus: Construct the Reals from the Rationals}\r\nThe definition of convergence and Cauchy sequence makes sense in $\\mathbb Q$ as well.\r\nThe matters, there are sequences in $\\mathbb Q$ that ``want to converge'' (being Cauchy) but don't, like $1,1.4,1.41,1.414$.\r\nIn fact, every $x\\in\\mathbb R$ is a limit of a sequence in $\\mathbb Q$ (by e.g. decimal expansion).\r\nSo we can define\r\n$$\\mathbb R=\\{(x_n)\\in\\mathbb Q: (x_n)\\text{ is Cauchy}\\}/\\sim$$\r\nwhere $\\sim$ is an equivalence relation defined by $(x_n)\\sim (y_n)\\iff x_n-y_n\\to 0$.\r\n\\begin{proposition}\r\n This is indeed the real number with the intuitive definition of operations and order.\r\n\\end{proposition}\r\nIn particular $(x_n)\\ge0=(0,0,\\ldots)$ iff $\\exists N\\in\\mathbb n,\\forall n>N,x_n\\ge 0$.\r\n\\begin{proof}\r\n Other properties are trivial, so we shall check the least-upper-bound property.\r\n Note that the least-upper-bound property, the monotone sequence theorem and the statement that every Cauchy sequence converges are equivalent.\r\n So we shall prove that every Cauchy sequence converges.\r\n Suppose $(x_{m,n})=((x_m)_n)$ is a Cauchy sequence in $\\mathbb R$ defined the way we made it.\r\n Then the sequence $(y_k)=x_{k,k}$ goes to the limit we want.\r\n\\end{proof}", "meta": {"hexsha": "ae3c821c47342974d253b5ba4287d1fb9d137971", "size": 1252, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/rq.tex", "max_stars_repo_name": "david-bai-notes/IA-Analysis-I", "max_stars_repo_head_hexsha": "4209ac010e35cfcd72799530eeed7d96d6706a3c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2/rq.tex", "max_issues_repo_name": "david-bai-notes/IA-Analysis-I", "max_issues_repo_head_hexsha": "4209ac010e35cfcd72799530eeed7d96d6706a3c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2/rq.tex", "max_forks_repo_name": "david-bai-notes/IA-Analysis-I", "max_forks_repo_head_hexsha": "4209ac010e35cfcd72799530eeed7d96d6706a3c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.5555555556, "max_line_length": 147, "alphanum_fraction": 0.71485623, "num_tokens": 367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.815232489352, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.6358755537510418}} {"text": "\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{titletoc}\n\\usepackage{titlesec}\n\\usepackage{geometry} \n\\usepackage{fontspec, xunicode, xltxtra}\n\\usepackage{float}\n\\usepackage{cite}\n\\usepackage{amsmath}\n\\usepackage{listings}\n\\usepackage{titletoc}\n\n\\geometry{left=3cm,right=3cm,top=3cm,bottom=3cm}\n\\DeclareMathOperator*{\\argmin}{argmin}\n\\DeclareMathOperator*{\\argmax}{argmax}\n\\DeclareMathOperator*{\\logit}{logit}\n\\DeclareMathOperator*{\\var}{var}\n\\DeclareMathOperator*{\\cov}{cov}\n\\DeclareMathOperator*{\\expec}{E}\n\\DeclareMathOperator*{\\deriv}{d}\n\\DeclareMathOperator*{\\const}{constant}\n\n\\begin{document}\n\\title{\\textsf{Homework 5 for Bayesian Data Analysis}}\n\\author{Fan JIN\\quad (2015011506)}\n\\maketitle\n\n\\section*{Question 6.1a}\n{\n Under the model assumption of identical effect for each group, we fix the hyperparameter $\\tau=0$ in our normal hierarchical model. Plug it in and we obtain $\\theta_j \\sim N(7.7, 4.1^2)$. The posterior predictive distribution is simulated as follows: \\\\\n 1) generate $\\theta \\sim N(7.7, 4.1^2)$, and \\\\\n 2) generate $y_j^\\mathrm{rep} \\sim N(\\theta, \\sigma_j^2)$ for each $j$, followed by \\\\\n 3) calculate the order stastistics out of $y_j^\\mathrm{rep}$ for each $j$.\n\n The observed order statistics are approximately the order statistics calculated from the replicated simulations above. \n \\begin{figure}[H]\n \\centering\n \\includegraphics[width = 1.0\\linewidth]{data.vs.png}\n \\caption{Observed order statistics v.s. Simulations order statistics}\n \\end{figure}\n \\begin{lstlisting}\n > mod = lm((data.simulated - data.observed) ~ data.observed)\n > summary(mod)\n \n Call:\n lm(formula = (data.simulated - data.observed) ~ data.observed)\n \n Residuals:\n Min 1Q Median 3Q Max \n -4.4555 -1.1794 0.3315 2.2663 2.6837 \n \n Coefficients:\n Estimate Std. Error t value Pr(>|t|)\n (Intercept) -1.30976 1.33122 -0.984 0.363\n data.observed 0.07826 0.10150 0.771 0.470\n \n Residual standard error: 2.805 on 6 degrees of freedom\n Multiple R-squared: 0.09014,\tAdjusted R-squared: -0.06151 \n F-statistic: 0.5944 on 1 and 6 DF, p-value: 0.47\n \\end{lstlisting}\n The p-value is 0.47, so the null hypothesis of unit slope cannot be rejected. We therefore conclude that the model fits the aspect of data here. \n}\n\n\\section*{Question 6.1b}\n{\n The model of identical effect for each group assumes that $\\theta_j = \\theta$ for all $j$, and that school A has better effect than school C simply \\emph{by chance}. But the model assumption can be skeptical given the observation, as we wonder whether there is other factors that cause school A to perform better than schoold C. In this case, the likelihood of the assumption of identical effects is far from being convincing, especially compared to the hierarchical models. \n}\n\n\\section*{Question 6.6a}\n{\n We have 7 ones and 13 zeros out of the $n=20$ observations. \n \\begin{itemize}\n \\item \\textbf{Assume stops at 20th observation:}\\quad The likelihood is proportional to $\\theta^{7} (1-\\theta)^{13}$, according to the binomial model with number of observations fixed to 20.\n \\item \\textbf{Assume stops at 13th zero:}\\quad The likelihood is also proportional to $\\theta^{7} (1-\\theta)^{13}$, \\emph{up to a different scale}, according to the negative binomial model with number of failures fixed to 13 with the last trial fixed to zero. \n \\end{itemize}\n There is no difference with the likelihood except for a constant factor. Therefore, the posterior distribution does not change, either.\n}\n\n\\section*{Question 6.6b}\n{\n Generate $\\theta$ from $p(\\theta) \\sim \\theta^{7} (1-\\theta)^{13}$ for 10000 times, and generate $y^\\mathrm{rep}$ under the new protocol (stops at the 13th zero) each time. \n \\begin{figure}[H]\n \\centering\n \\includegraphics[width = 1.0\\linewidth]{switches.histogram.png}\n \\caption{Histogram of number of switches}\n \\end{figure}\n\n First, the statistic T has a distribution with a heavier tail. This is because the number of observations can exceed the limit of $n=20$ under the new protocol. \n\n Second, we find that T is more likely to be an even number than an odd number when $T < 10$. A possible explanation is: The last few observations must be zeros when $T$ is small, since the last one must be zero and there is few switches. Note that there must be $13$ zeros in total. Therefore, the distribution of $T$ is highly related to the specific value of $T$, i.e. $p(T=k)$ is more sensitive to $k$ when $k$ is relatively small. \n}\n\n\\section*{Source Code in R}\n{\n \\begin{lstlisting}[language=R]\n # Question 6.1\n data.observed = c(28, 18, 12, 8, 7, 1, -1, -3)\n data.simulated = c(26, 19, 14, 10, 6, 2, -3, -9)\n \n png(\"data.vs.png\", width=800, height=600)\n plot(data.observed, data.simulated, main=\"Simulated v.s. simulated order statistics\")\n dev.off()\n \n mod = lm((data.simulated - data.observed) ~ data.observed)\n summary(mod)\n \n \n # Question 6.6\n T0 = 3 # number of switches\n N = 10000\n T = 1:N\n theta = rbeta(N, shape1=7+1, shape2=13+1)\n for (j in 1:N) {\n count.switches = 0\n y.last = rbinom(1, size=1, prob=theta[j])\n count.zeros = 1 - y.last\n while (1) {\n y = rbinom(1, size=1, prob=theta[j])\n if (y != y.last) {\n count.switches = count.switches + 1\n }\n count.zeros = count.zeros + 1 - y.last\n if (count.zeros > 12.5) {\n break;\n }\n y.last = y\n }\n T[j] = count.switches\n }\n png(\"switches.histogram.png\", width=800, height=600)\n hist(T, breaks=seq(-0.5, max(T)+0.5))\n dev.off()\n \\end{lstlisting}\n}\n\n\\clearpage\n\\end{document}\n", "meta": {"hexsha": "a073d8fad13512fc157e9f9a9e2a7485b8fe675f", "size": 5807, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW5/Homework5.tex", "max_stars_repo_name": "goldsail/BayesianHomework", "max_stars_repo_head_hexsha": "d5506faccbf4d0b7b696c7c2bcb42d020bb0d357", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-07-07T18:55:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-07T18:55:43.000Z", "max_issues_repo_path": "HW5/Homework5.tex", "max_issues_repo_name": "kingium/BayesianHomework", "max_issues_repo_head_hexsha": "d5506faccbf4d0b7b696c7c2bcb42d020bb0d357", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW5/Homework5.tex", "max_forks_repo_name": "kingium/BayesianHomework", "max_forks_repo_head_hexsha": "d5506faccbf4d0b7b696c7c2bcb42d020bb0d357", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.0797101449, "max_line_length": 479, "alphanum_fraction": 0.6729808851, "num_tokens": 1732, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799928900257126, "lm_q2_score": 0.8152324938410783, "lm_q1q2_score": 0.6358755489139716}} {"text": "\\subsubsection{Direction Matters}\r\n\\noindent\r\n\\begin{lemma}\r\n\t\\begin{equation*}\r\n\t\t\\int\\limits_{-C}{\\vec{F} \\cdot \\mathrm{d}\\vec{r}} = -\\int\\limits_{C}{\\vec{F}\\mathrm{d}\\vec{r}}\r\n\t\\end{equation*}\r\n\\end{lemma}\r\nThat is, the direction in which one takes a line integral matters. You can think of this as the wind helping you sail in one direction (positive line integral) but fighting against you in the opposite direction (negative line integral).", "meta": {"hexsha": "57fad3ae7fe1939b32fd82715ac4e6d35bddccab", "size": 446, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "multiCalc/lineSurfaceIntegrals/directionMatters.tex", "max_stars_repo_name": "aneziac/Math-Summaries", "max_stars_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2020-03-26T06:20:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T17:38:45.000Z", "max_issues_repo_path": "multiCalc/lineSurfaceIntegrals/directionMatters.tex", "max_issues_repo_name": "aneziac/Math-Summaries", "max_issues_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 26, "max_issues_repo_issues_event_min_datetime": "2020-03-28T17:44:18.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T04:47:03.000Z", "max_forks_repo_path": "multiCalc/lineSurfaceIntegrals/directionMatters.tex", "max_forks_repo_name": "aneziac/Math-Summaries", "max_forks_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2020-04-10T05:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-17T15:21:12.000Z", "avg_line_length": 55.75, "max_line_length": 236, "alphanum_fraction": 0.7286995516, "num_tokens": 132, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.7690802476562641, "lm_q1q2_score": 0.635807711324228}} {"text": "% Created 2020-11-11 Wed 19:46\n% Intended LaTeX compiler: pdflatex\n\\documentclass[11pt]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{graphicx}\n\\usepackage{grffile}\n\\usepackage{longtable}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{amssymb}\n\\usepackage{capt-of}\n\\usepackage{hyperref}\n\\author{Muaz Alhaider and Zakariya Ahmed}\n\\date{\\today}\n\\title{Program 2}\n\\hypersetup{\n pdfauthor={Muaz Alhaider and Zakariya Ahmed},\n pdftitle={Program 2},\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 27.1 (Org mode 9.3)}, \n pdflang={English}}\n\\begin{document}\n\n\\maketitle\n\n\\section*{Modules}\n\\label{sec:orgeb98485}\n\n\\subsection*{Transitional Probability}\n\\label{sec:org7d7cbf4}\n \\begin{math}\n Transitional\\ Probability(Pos_i, Dir) = \\\\\n (Pos_{from\\ left}:Drift(Left), Pos_{from\\ straight}:Drift(Straight), Pos_{from\\ right}:Drift(Right)): (\\text{For smoothing})\n\\end{math}\n\n\\newline\n\\begin{math}\n\\\\ \\\\\nTransitional\\ Probability(Pos_i, Dir) =\\\\ (Pos_{from\\ left}:Drift(Left), Pos_{from\\ straight}:Drift(Straight), Pos_{from\\ right}:Drift(Right), Pos_{from\\ behind}:Drift(Straight)): (\\text{For prediction})\n\\\\ \\\\\n\\end{math}\n\n\n\\emph{Left}, \\emph{Right}, and \\emph{Straight} are all positions defined in relation to what \\emph{Dir} is: if \\emph{Dir} is EAST, then \\emph{Left} is SOUTH, \\emph{Right} is NORTH, and Straight is EAST for example.\n\nHere, transitional probability has two forms: one which is all paths that converge to a point, and another where they diverge from a point. Prediction (as the name implies) wants all the possible paths to a point, which is why we include \\(Pos_{from\\ behind}\\). Smoothing however, does not use that, which is why it's not included.\n\n\n\\subsection*{Prediction}\n\\label{sec:org59a7289}\n$$Prediction(Grid, Direction) = \\left\\{pos_i \\in Grid \\mid \\sum_{(Pos_j, DriftProb)}^{Transition\\ Probability(pos_i, direction)} DriftProb \\cdot P(Pos_j) \\right \\}$$\n\n\\(Prediction(Grid, Direction)\\) (as the name implies) attempts to predict where the agent will be given previous information. It does this by transforming the grid by the expression \\(\\sum^{Transition\\ Probability(pos_i, direction)}_{(Pos_j, DriftProb)} DriftProb \\cdot P(Pos_j)\\).This gets the probability of an agent drifting (or if direction is straight, accurately going to) a point, and what is the probability the agent would be at the point \\(Pos_j\\).\n\n\n\\subsection*{Evidence Conditional Probability}\n\\label{sec:org7e2fa92}\n \\begin{math}\n Evidence\\ Conditional\\ Probability(Pos_i, Evidence)= \\\\\n \\prod^{\\text{Directions} }_{dir=W} Sense(evidence[pos_i dir], actual[pos_i+dir])\n\\end{math}\n\nThis is the expression we use to get the evidence conditional probability: it's the product of each the evidecne's value at a direction times what's actually in the value of the direction. So if \\emph{Left} has opened, but evidence says it's closed, it's 0.2. Taking the product of all direction's sensed value and actual value, it will result in the Evidence Conditional Probability at \\(Pos_i\\) given \\(Evidence\\)\n\n\\subsection*{Filtering}\n\\label{sec:org2604945}\n\n\n \\begin{math}\nFiltering(Grid, Evidence) = \\{pos_i \\in Grid \\mid\\\\\n\\frac{P(pos_i) \\cdot Evidence\\ Conditional\\ Probability(pos_i, Evidence) }{\\sum^{all\\ positions}_{pos} P(pos_i) \\cdot Evidence\\ Conditional\\ Probability(pos_i, evidence) } \\}\n\\end{math}\n\n\n\nFiltering is a transformation upon the grid: each value gets transformed by the expression \\(\\frac{P(pos_i) \\cdot Evidence\\ Conditional\\ Probability(pos_i, Evidence) }{\\sum^{all\\ positions}_{pos} P(pos_i) \\cdot Evidence\\ Conditional\\ Probability(pos_i, evidence) }\\), which for purposes of making it easier to talk about, will be expressed as \\(Filter\\ Step(pos_i, Evidence)\\). \\(Filter\\ Step\\) is conditional probability of each point times what probabilty of the point previously, and then dividing it by the sum of all points on the grid. \n\n\\section*{Results}\n\\label{sec:org4a73001}\nThe code outputs the following:\n\\begin{verbatim}\njulia SUBMIT.jl\n\\end{verbatim}\n\n\\section*{Screenshots}\n\\label{sec:org3466924}\n\\begin{center}\n\\includegraphics[width=.9\\linewidth]{data/a9/3abd3f-f652-4b14-a3ef-d46d087ebe0c/screenshot-20201111-135323.png}\n\\end{center}\n\\begin{center}\n\\includegraphics[width=.9\\linewidth]{data/a9/3abd3f-f652-4b14-a3ef-d46d087ebe0c/screenshot-20201111-135329.png}\n\\end{center}\n\n\\section*{Who did what}\n\\label{sec:org8a29d99}\n\\subsection*{Zakariya}\n\\label{sec:org52942fd}\n\\begin{itemize}\n\\item Transitional probability/prediction algorithm, code and report.\n\\end{itemize}\n\\subsection*{Muaz}\n\\label{sec:org5f24d77}\n\\begin{itemize}\n\\item Conditional Evidence probability, filtering and smoothing algorithm and code.\n\\end{itemize}\n\\end{document}\n", "meta": {"hexsha": "99a3c0725c0b71b7c2b296903a1a6f6f63f92346", "size": 4786, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "submit.tex", "max_stars_repo_name": "MuazAlhaidar/Fall-2020---Robot-Localization-with-HMM-Algorithm---Console---Group-Project", "max_stars_repo_head_hexsha": "ca6daf03f9b4f5a5d6d0a047478fcf7215c5587e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "submit.tex", "max_issues_repo_name": "MuazAlhaidar/Fall-2020---Robot-Localization-with-HMM-Algorithm---Console---Group-Project", "max_issues_repo_head_hexsha": "ca6daf03f9b4f5a5d6d0a047478fcf7215c5587e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "submit.tex", "max_forks_repo_name": "MuazAlhaidar/Fall-2020---Robot-Localization-with-HMM-Algorithm---Console---Group-Project", "max_forks_repo_head_hexsha": "ca6daf03f9b4f5a5d6d0a047478fcf7215c5587e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-10-31T15:20:42.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-31T15:20:42.000Z", "avg_line_length": 42.7321428571, "max_line_length": 544, "alphanum_fraction": 0.7540743836, "num_tokens": 1402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.7690802476562641, "lm_q1q2_score": 0.635807711324228}} {"text": "\\subsection{Solving}\n\n\\begin{frame}\n \\frametitle{Solving}\n\n Suppose that no negative cycle exists in $G(V,E)$ for $\\babst{\\mu}$, \n how do we find a model $\\mu$ for the integer variables ?\n \\vfill\n \\pause\n State-of-the-art methods employ SSSP (Single Source Shortest Paths) algorithms,\n such as the {\\bf Bellman-Ford} algorithm (or its variations), as follows:\n \\begin{itemize}\n \\item we add an artificial vertex $I$ to $V$, and add $\\{ (I,x_1;0), \\ldots, (I,x_n;0) \\}$ to $E$\n (notice that this never creates negative cycles)\n \\vfill\n\n \\item Compute the shortest paths from source $I$ (run Bellman-Ford). Let $\\pi(x)$ be the shortest path\n from $I$ to $x$, for all $x \\in V$\n \\vfill\n\n \\item The model $\\mu$ can be computed as $\\mu(x) = -\\pi(x)$ for all $x \\in V$ (see later why)\n \\end{itemize}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Bellman-Ford} \n\n $\\pi(x)$: current distance of $x$ from $I$\\\\\n $TBV$: queue of vertexes To Be Visited\\\\\n\n \\begin{tabbing}\n as \\= a \\= a \\= a \\= as \\= asdfasdfasdfasdfasdfasdfasdfasdf \\= \\kill\n 1 \\> $\\pi(x) = \\infty$ for all $x \\in V, x \\not= I$ \\\\\n 2 \\> $\\pi(I) = 0$ \\\\\n 3 \\> $TBV.pushBack( I )$ \\\\\n 4 \\> while ( $TBV.size(\\ ) > 0$ ) \\\\\n 5 \\> \\> $s = TBV.popFront(\\ )$ \\\\\n 6 \\> \\> foreach $(s,t;w) \\in E$ \\> \\> \\> \\> // for each outgoing edge \\\\\n 7 \\> \\> \\> if ( $\\pi(t) - \\pi(s) > w$ ) \\> \\> \\> // is too far ? \\\\\n 8 \\> \\> \\> \\> $\\pi(t) = \\pi(s) + w$ \\> \\> // relax (decrease $\\pi(t)$) \\\\\n 9 \\> \\> \\> \\> if ( $TBV.has( t ) == false$ ) \\\\ \n 10 \\> \\> \\> \\> \\> $TBV.pushBack( t )$ \\> // enqueue t if not there \\\\\n \\end{tabbing}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Bellman-Ford, Example} \n\n \\scriptsize\n \n $\\pi(x)$: current distance of $x$ from $I$\\\\\n $TBV$: queue of vertexes To Be Visited\\\\\n\n \\begin{columns}\n\n \\begin{column}{.45\\textwidth}\n \\begin{tabbing}\n as \\= a \\= a \\= a \\= as \\= asdfasdfasdfasdfasdfasdfasdfasdf \\= \\kill\n \\> \\coloneat{$\\pi(x) = \\infty$ for all $x \\in V, x \\not= I$}{3|handout:0} \\\\\n \\> \\coloneat{$\\pi(I) = 0$}{3|handout:0} \\\\\n \\> \\coloneat{$TBV.pushBack( I )$}{3|handout:0} \\\\\n \\> while ( $TBV.size(\\ ) > 0$ ) \\\\\n \\> \\> \\coloneat{$s = TBV.popFront(\\ )$}{4,7,10|handout:0} \\\\\n \\> \\> \\coloneat{foreach $(s,t;w) \\in E$}{5,8,11|handout:0} \\\\\n \\> \\> \\> \\coloneat{if ( $\\pi(t) - \\pi(s) > w$ )}{5,8,11|handout:0} \\\\\n \\> \\> \\> \\> \\coloneat{$\\pi(t) = \\pi(s) + w$}{5,8,11|handout:0} \\\\\n \\> \\> \\> \\> \\coloneat{if ( $TBV.has( t ) == false$ )}{5,8,11|handout:0} \\\\ \n \\> \\> \\> \\> \\> \\coloneat{$TBV.pushBack( t )$}{5,8,11|handout:0} \\\\\n \\end{tabbing}\n\n $TBV = $ [ \\only<3|handout:0>{$I$}\\only<5,6|handout:0>{$x$}\\only<6|handout:0>{$,\\,$}\\only<6-9|handout:0>{$y,$}\\only<6-12|handout:0>{$z,w,t$} ] \\\\\n Current vertex: \\only<4,5|handout:0>{$I$}\\only<7,8|handout:0>{$x$}\\only<10,11|handout:0>{$y$} \\\\\n Current edge: \\only<5|handout:0>{$(I,x;0)$}\\only<11|handout:0>{$(y,z;-1)$} \n \\end{column}\n\n \\begin{column}{.55\\textwidth}\n \\begin{center}\n \\begin{overlayarea}{.55\\textwidth}{3cm}\n \\only<1|handout:0>{\\input{example_bf}}\n \\only<2|handout:0>{\\input{example_bf_1}}\n \\only<3|handout:0>{\\input{example_bf_2}}\n \\only<4|handout:0>{\\input{example_bf_3}}\n \\only<5|handout:0>{\\input{example_bf_4}}\n \\only<6|handout:0>{\\input{example_bf_5}}\n \\only<7,8|handout:0>{\\input{example_bf_6}}\n \\only<9|handout:0>{\\input{example_bf_5}}\n \\only<10|handout:0>{\\input{example_bf_7}}\n \\only<11|handout:0>{\\input{example_bf_8}}\n \\only<12|handout:0>{\\input{example_bf_9}}\n \\only<13|handout:0>{\\input{example_bf_10}}\n \\only<14>{\\input{example_bf_11}}\n \\end{overlayarea}\n \\end{center}\n \\end{column}\n\n \\end{columns}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Bellman-Ford, consideration}\n\n \\scriptsize\n\n \\begin{exampleblock}{Invariant}\n At the end of BF, $\\pi(x)$ holds the shortest distance from $I$ to $x$, for all $x \\in V$ \\\\\n \\end{exampleblock}\n \\vfill\n \\pause\n \\begin{lemma}[Shortest Path]\n At the end of BF, $\\pi(y) - \\pi(x) \\leq c$ holds for all $(x,y;c) \\in E$\n \\end{lemma}\n \\vfill\n \\pause\n \\begin{proof}\n Suppose, for the sake of contradiction, that for an edge $(x,y;c)$, we have $\\pi(y) - \\pi(x) > c$\n \\begin{center}\n \\input{arc_proof}\n \\end{center}\n \\vspace{-30pt}\n $\\pi(x)$ is the shortest dist. from $I$ to $x$ (by Invariant). But since $\\pi(y) > \\pi(x) + c$, \n the shortest path from $I$ to $y$ is $\\pi(x) + c$. So $\\pi(y)$ is not the shortest dist. Contradiction.\n \\end{proof}\n\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Finding a model $\\mu$}\n\n Because of the previous lemma we have that\n \\vfill\n \\begin{center}\n $\\pi(y) - \\pi(x) \\leq c\\quad\\quad$ holds for all $(x,y;c) \\in E$\n \\end{center}\n \\vfill\n So, if we take $\\quad\\quad\\mu(x) = -\\pi(x)\\quad\\quad$ we have that\n \\vfill\n \\begin{center}\n $\\mu(x) - \\mu(y) \\leq c\\quad\\quad$ holds for all constraints in $\\babst{\\mu}$\n \\end{center}\n \\vfill\n and therefore $\\mu$ is a model\n\n\\end{frame}\n", "meta": {"hexsha": "6a87c636820eeaf706acb4de02aaa10816cbfdec", "size": 5057, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture5/solving.tex", "max_stars_repo_name": "formalmethods/smtlectures", "max_stars_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-11-07T19:34:12.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-24T08:05:50.000Z", "max_issues_repo_path": "lecture5/solving.tex", "max_issues_repo_name": "formalmethods/smtlectures", "max_issues_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture5/solving.tex", "max_forks_repo_name": "formalmethods/smtlectures", "max_forks_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-05-06T00:40:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-06T00:40:41.000Z", "avg_line_length": 34.4013605442, "max_line_length": 149, "alphanum_fraction": 0.5653549535, "num_tokens": 2042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.8267118026095991, "lm_q1q2_score": 0.6358077091401595}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% CS624: Analysis of Algorithms\n% Copyright 2015 Pejman Ghorbanzade \n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section*{Question 4}\n\nProve that all problems that are NP-complete are polynomially equivalent, in the sense that if $A$ and $B$ are NP-complete then $A \\leq_P B$ and $B \\leq_P A$.\n\n\\subsection*{Solution}\n\nBy definition, $A$ is NP-complete if and only if $A$ is in NP and for \\textbf{any} problem $B$ in NP, $B\\leq_P A$.\nProof is immediately attained by using assumptions that $A$ and $B$ are NP-complete, which means they are in NP.\nSince $B$ is in NP, using the assumption that $A$ is NP-hard, we can obtain $B \\leq_P A$.\nSimilarly, since $A$ is in NP, using the assumption that $B$ is NP-hard, we can write $A \\leq_P B$.\nTherefore, both NP-complete problems are equivalent.\n", "meta": {"hexsha": "13cbf21288997ed6712b359c9ff3fc9fee1d21d2", "size": 1033, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "umb-cs624-2015s/src/tex/hw06/hw06q04.tex", "max_stars_repo_name": "ghorbanzade/beacon", "max_stars_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-11-13T20:00:10.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-01T11:16:51.000Z", "max_issues_repo_path": "umb-cs624-2015s/src/tex/hw06/hw06q04.tex", "max_issues_repo_name": "ghorbanzade/beacon", "max_issues_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "umb-cs624-2015s/src/tex/hw06/hw06q04.tex", "max_forks_repo_name": "ghorbanzade/beacon", "max_forks_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-09-20T05:58:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T17:18:05.000Z", "avg_line_length": 54.3684210526, "max_line_length": 158, "alphanum_fraction": 0.6398838335, "num_tokens": 276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7690802264851918, "lm_q2_score": 0.8267117962054049, "lm_q1q2_score": 0.6358076954636326}} {"text": "\\epigraph{``Since Newton, mankind has come to realise that the law of physics are always expressed in the language of differential equations\"}{Steven Strogatz}\n\n\\section{Finite Difference Method}\\label{fdm_section}\n\\subsection{Introduction}\nThe use of numerical models for the simulation of dynamics within the atmosphere typically involves the solution of a set of partial differential equations. These equations generally describe three processes: advection, adjustment and diffusion. \n\n\\begin{definition}\nAdvection is the transport of a substance by bulk motion.\n\\end{definition}\n\n\\begin{definition}\nAdjustment is how the mass and wind fields adjust to one another.\n\\end{definition}\n\n\\begin{definition}\nDiffusion is the movement of a liquid, or gas from an area of high concentration to an area of low concentration.\n\\end{definition}\n\nMost meteorological problems involving partial differential equations generally fall into three distinct categories: initial value problems, boundary value problems, and eigenvalue problems. The meteorological problems associated with this project are solely initial value problems\\cite{numerical_methods}. An initial value problem is a situation where you want to predict the future state of a given system, given the necessary initial conditions. Unfortunately, the equations describing the evolution of the atmosphere do not have exact analytical solutions.\n\n\\begin{definition}\nAn analytical function is the most precise way of representing a physical field as it gives us the value of this field at any point in space, and at any instant in time.\n\\end{definition}\n\nWhen an analytical solution does not exist, an approximate numerical solution is found using a specified computational technique\\cite{numerical_methods}. For the purposes of this project, the finite difference method will be utilised. It must be noted, however, that there are several other methods to acquire an approximate numerical solution, with the finite element method and the spectral method being just a few.\n\nThe idea of the finite difference method is to approximate the derivatives in the partial differential equations with differences between adjacent points in space and in time. The advantages of this being that the problem becomes an algebraic one, and that a continuous problem becomes a discrete one.\n\n\\subsection{Derivation of the Finite Difference Method}\n\nIn order to derive the finite difference method, it is necessary to look at the Taylor series expansion of the function.\n\n\\begin{definition}\nThe Taylor series of a function is the limit of that function's Taylor polynomials as the degree increases.\n\\end{definition}\n\nThe Taylor series expansion of $f(x)$ can be represented as the following:\n\n\\begin{equation}\n f(x + \\Delta x) = f(x) + \\frac{d f}{d x}(x) \\frac{\\Delta x}{1!} + \\frac{d^2 f}{d x^2}(x) \\frac{\\Delta x^2}{2!} + ... + \\frac{d^n f}{d x^n}(x) \\frac{\\Delta x^n}{n!} + ...\n\\end{equation}\n\\begin{equation}\n f(x - \\Delta x) = f(x) + \\frac{d f}{d x}(x) \\frac{-\\Delta x}{1!} + \\frac{d^2 f}{d x^2}(x) \\frac{(-\\Delta x)^2}{2!} + ... + \\frac{d^n f}{d x^n}(x) \\frac{(-\\Delta x)^n}{n!} + ...\n\\end{equation}\n\nThe higher order terms, which will be represented as $\\mathcal{O}(\\Delta x)$ from this point on, become less important as $\\Delta x$ approaches zero.\n\n\\begin{equation}\n \\Rightarrow f(x + \\Delta x) = f(x) + \\frac{d f}{d x}(x) \\frac{\\Delta x}{1!} + \\frac{d^2 f}{d x^2}(x) \\frac{\\Delta x^2}{2!} + [\\mathcal{O}(\\Delta x^3)] \n \\label{fds}\n\\end{equation}\n\\begin{equation}\n \\Rightarrow f(x - \\Delta x) = f(x) - \\frac{d f}{d x}(x) \\frac{\\Delta x}{1!} + \\frac{d^2 f}{d x^2}(x) \\frac{\\Delta x^2}{2!} + [\\mathcal{O}(\\Delta x^3)]\n \\label{bds}\n\\end{equation}\n\nThese higher order terms are neglected, and the following approximation for the derivative of $f(x)$ is found:\n\n\\begin{equation}\n \\Rightarrow \\frac{d f}{d x}(x) = \\frac{f(x + \\Delta x) - f(x)}{\\Delta x} + \\mathcal{O}(\\Delta x)\n\\end{equation}\n\\begin{equation}\n \\Rightarrow \\frac{d f}{d x}(x) = \\frac{f(x - \\Delta x) - f(x)}{\\Delta x} + \\mathcal{O}(\\Delta x)\n\\end{equation}\n\nThese are called the forward and backward difference schemes respectively, and by keeping only the leading order terms, an error of order $\\mathcal{O}(\\Delta x)$ is occurred. It is possible to obtain a better approximation by subtracting \\ref{bds} from \\ref{fds}, which yields equation \\ref{eq_fds_bds_subtract}.\n\n\\begin{equation}\n \\frac{d f}{d x}(x) = \\frac{f(x + \\Delta x) - f(x - \\Delta x)}{\\Delta x} + \\mathcal{O}(\\Delta x^2)\n \\label{eq_fds_bds_subtract}\n\\end{equation}\n\nThis particular approximation is called the central difference scheme, and has an error of order $\\mathcal{O}(\\Delta x^2)$. Therefore, this scheme is more accurate than the previously mentioned forward and backward difference schemes. It is possible to take more and more terms from the Taylor series expansion, however, there is an inherent trade off between accuracy and computational efficiency. \n\nIn relation to the atmosphere, what this method does is divide the atmosphere into several discrete horizontal layers, and each layer is divided up into grid cells. Following which, each equation is evaluated at the centre of the cell. Similarly, the time interval under consideration is sliced into a number of discrete time steps. The size of the grid step $\\Delta x$ and time step $\\Delta t$ determines the accuracy of the scheme, with accuracy increasing as $\\Delta x$ and $\\Delta t$ approach zero. On a synoptic scale, $\\Delta x$ is generally equal to 500 km. For higher resolutions, the grid-size is smaller, which corresponds to a greater computational burden. As such, there is a trade off between accuracy and computational performance. For Eulerian schemes, the typical time step is 2 minutes. As such, since the software will use an Eulerian scheme, the time step will be 2 minutes\\cite{leapfrog_slides_one}.\n\n\\subsection{FTCS Scheme}\\label{ftcs_section}\n\nGiven the information mentioned in the previous section, the most obvious scheme to approximate a differential equation, which will be used to predict the future state of the atmosphere, would be to combine the central difference scheme for space and the forward difference scheme for time (FTCS). This scheme would allow us access to the increased accuracy of the central difference scheme, while maintaining two time variable unknowns. If only it was that simple! Let's take the example of the 1-D linear advection equation for temperature. This equation is represented as the following:\n\n\\begin{equation}\n \\frac{\\partial T}{\\partial t} + u \\frac{\\partial T}{\\partial x} = 0\n \\label{1d_temp_eq}\n\\end{equation}\n\nUsing the FTCS scheme mentioned above, this equation can be approximated as:\n\n\\begin{equation}\n \\frac{T^{n+1}_{i} - T^{n}_{i}}{\\Delta t} + u \\frac{T^{n}_{i+1} - T^{n}_{i-1}}{2 \\Delta x} = 0\n\\end{equation}\n\nIt can be shown, by using Fourier Series, that:\n\n\\begin{equation}\n |\\lambda_j|^2 = 1 + \\alpha^2(\\sin{j \\Delta x}^2)\n\\end{equation}\n\nTherefore, $|\\lambda_j|^2 \\geq 1$, and so the scheme is said to be absolutely unstable. What it means for a scheme to be unstable is that if there is a slight change in the initial value, the result of the computation will change dramatically. The stability of a scheme is important in meteorological problems because if slight deviations from the mathematical model caused by unavoidable errors in measurement do not have a correspondingly slight effect on the approximate numerical solution, the mathematical equations describing the problem will not accurately predict the future outcome\\cite{ftcs_leapfrog}. For a more detailed technical explanation of the stability of this scheme and the leapfrog scheme (discussed in section \\ref{leapfrog}), please see the following article: \\url{https://www.ecmwf.int/sites/default/files/elibrary/2002/16948-numerical-methods.pdf}.\n\n\\subsection{Leapfrog Scheme}\\label{leapfrog}\n\nThis scheme is probably the most common scheme used for meteorological problems. The \"leapfrog\" refers to the centred\ntime difference which is used in conjunction with centred space differences. \n\nTaking the 1-D linear advection equation for temperature seen in equation \\ref{1d_temp_eq}, applying this scheme results in:\n\n\\begin{equation}\n \\frac{T^{n+1}_{i} - T^{n-1}_{i}}{2 \\Delta t} + u \\frac{T^{n}_{i+1} - T^{n}_{i-1}}{2 \\Delta x} = 0\n\\end{equation}\n\nIt can be shown that this scheme is stable using a similar technique mentioned in section \\ref{ftcs_section}. This equation can then be rearranged for the forecast value $T^{n+1}_{i}$\\cite{ftcs_leapfrog}:\n\n\\begin{equation}\n T^{n+1}_{i} = T^{n-1}_{i} - u \\frac{2 \\Delta t}{2 \\Delta x}(T^{n}_{i+1} - T^{n}_{i-1})\n\\end{equation}\n\nFor the physical equation, a single initial condition $T^{0}$ is sufficient to determine the solution. One problem with the leapfrog scheme is that two values of $T$ are required to start the computation. In addition to the physical initial condition $T^{0}$, a computational initial condition $T^{1}$ is required. This cannot be obtained using the leapfrog scheme, so a non-centred step is used to provide the value at $t = \\Delta t$. From which point on, the leapfrog scheme is used, however, the errors of the first step will persist. This method, however, still retains an error of order $\\mathcal{O}(\\Delta t^2)$. If you also use half of the time step for the forward time step, followed by leapfrog time steps; this will reduce the error introduced in the first step\\cite{leapfrog_slides_two}. This will be the method utilised within the software.\n\n\\subsection{Nonlinear Instability}\nA major problem which occurs while dealing with nonlinear partial differential equations is nonlinear instability. This is a problem where there is a nonlinear interaction between atmospheric waves\\cite{nonlinear_instability}. \n\n\\begin{definition}\nAn atmospheric wave is a periodic disturbance in the fields of atmospheric variables (like geopotential height, temperature, or wind velocity) which may either propagate (travelling wave) or not (standing wave).\n\\end{definition}\n\nIf one of the waves involved in this nonlinear interaction have a wavelength less than $4 \\Delta x$ something called aliasing causes a channelling of energy towards the small wavelengths. The continuous feedback of energy leads to a catastrophic rise in the kinetic energy of wavelengths between $2 \\Delta x$ and $4 \\Delta x$. Within the software, a smoothing operator, which reduces the amplitude of the short waves while having little effect on the meteorologically important waves, is utilised\\cite{nonlinear_instability}.\n\nAnother problem to mention before moving on is that for nonlinear equations, the leapfrog scheme has a tendency to increase the amplitude of the computational mode with time This can separate the space dependence between the even and odd time steps. This problem can be rectified by applying a Robert-Asselin Time Filter. After $T^{n+1}$ is obtained a slight time smoothing is applied to $T^{n}$, where $\\gamma$ is on the order of 0.1\\cite{leapfrog_slides_two}:\n\n\\begin{equation}\n T^{n} = T^{n} + \\gamma(T^{n+1} - 2 T^{n} + T^{n-1})\n\\end{equation}\n\n\\section{Ensemble Prediction System}\n\\subsection{Introduction}\nEnsemble Prediction Systems (EPS) are numerical weather prediction systems that allow for the estimation of uncertainty in a weather forecast, as well as, providing a better prediction for the future state of the atmosphere. Instead of running a atmospheric dynamical simulation once (this would be regarded as deterministic), the simulation is run many different time with slightly different initial conditions. Due to the high computational resources required to run these simulations, they are often run at half the resolution of an equivalent deterministic simulation. The ensemble prediction system has a control simulation that doesn't have any perturbations to the initial conditions. Each simulation that makes up the system is called an ensemble member\\cite{intro_efs}.\n\n\\subsection{Advantages of EPS}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.8\\linewidth]{Images/efs.jpg}\n \\caption{Visualisation of the Advantages of an Ensemble Prediction System}\n \\label{efs}\n\\end{figure}\n\nAs a consequence of Chaos Theory, a tiny difference in the initial conditions in a large system, such as the atmosphere, can result in drastically different forecasted events, so that even with a tiny error, it can become a large error in the forecasted future state of the atmosphere. Even with the most accurate observations, error cannot be avoided. Therefore, it is simply not possible to make a better forecast or simulation. This is why an ensemble prediction system is utilised. In an ensemble simulation, small perturbations are made to the initial conditions, after which, the simulation is re-run. If there is a large degree of overlap between the ensemble members, there will be a higher degree of confidence in the ensemble forecast, with the opposite also holding true\\cite{intro_efs}.\n\n\\subsection{Global EPS}\nThere are three distinct types of ensemble prediction systems: global, regional and convective-scale. Each system address different timescales, and different grid-sizes. A global ensemble prediction system is designed and used for medium-range forecasting between 3 and 15 days into the future. They use synoptic simulation models and are run at relatively low resolutions. Although they are primarily designed for use in the medium range, their global coverage means that they can also be used to provide short-range EPS forecasts in regions of the globe where no other EPS are currently available, and may be the only available option for certain countries. Considering the software is focused on synoptic scale simulations, this will be the system of interest\\cite{intro_efs}, The default number of ensemble members in the software is fifteen, which is typical for a global ensemble prediction system. The grid-size, used by the software, can be specified by the end-user in order to get a more detailed forecast (this will increase the amount of computational resources required to run the simulation), however, the default grid-size is of the scale of 1000 km. There is also a hard limit of a $5^{\\circ} \\times 5^{\\circ}$ cell, as a smaller cell size would result in inaccurate simulations due to the fact that the dynamical equations utilised by the software do not work on this scale. This will be discussed at greater depth in chapter \\ref{4}.\n\n\\section{Recurrent Neural Network}\n\\subsection{Introduction}\nWeather forecasting has traditionally been done by physical models of the atmosphere, which are unstable to perturbations, and thus are inaccurate for large periods of time\\cite{why_rnn}. Since machine learning techniques are more robust to perturbations, it would be logical to combine a neural network with a physical model. Weather forecasting is a sequential data problem, therefore, a recurrent neural network is the most suitable option for this task. \n\n\\begin{definition}\nA recurrent neural network is a class of artificial neural networks where connections between nodes form a directed graph along a temporal sequence.\n\\end{definition}\n\nBefore, we delve into the specific example of using a recurrent neural network to predict the future state of the atmosphere, it is necessary to review what a recurrent neural network is. Recurrent Neural Networks (RNNs) are neural networks that are used in situations where data is presented in a sequence. For example, let's say you want to predict the future position of a fast-moving ball. Without information on the previous position of the ball, it is only possible to make an inaccurate guess. If you had, however, a large number of snapshots of the previous position, you are then able to predict the future position of the ball with some certainty. RNNs excel at modelling sequential data such as these. This is due to sequential memory.\n\nIn order to intuitively understand sequential memory, the prime example would be the alphabet. While it is easy to say the alphabet from A-Z, it is much harder to go from Z-A. There is a logical reason why this is difficult. As a child, you learn the alphabet in a sequence. Sequential memory is a mechanism that makes it easier for your brain to recognise sequence patterns.\n\nIn a traditional neural network, there is a input layer, hidden layer, and a output layer. In a recurrent neural network, a loop is added that can be added to pass information forward as seen in the diagram below (provided by Towards Data Science)\\cite{intro_rnn}:\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.2\\linewidth]{Images/rnn.png}\n \\caption{Visualisation of a Recurrent Neural Network}\n\\end{figure}\n\nThe information that is forwarded is the hidden layer, which is a representation of previous inputs. How this works in practise is that you initialise your network layers and the hidden the initial hidden state. The shape and dimension of the hidden state will be dependent on the shape and dimension of your recurrent neural network. Then you loop through your inputs, pass the relevant parameter and hidden state into the RNN. The RNN returns the output and a modified hidden state. Last you pass the output to the output layer, and it returns a prediction. \n\nThere is, however, a major problem known as short-term memory. Short-term memory is caused by something known as the vanishing gradient problem, which is also prevalent in other neural network architectures. As the RNN processes more steps, it has troubles retaining information from previous steps. Short-Term memory and the vanishing gradient is due to the nature of back-propagation. This can be comprehended through understanding how a neural network is trained\\cite{intro_rnn}.\n\n\\begin{definition}\nBack-propagation is an algorithm used to train and optimise neural networks.\n\\end{definition}\n\nTo train a recurrent neural network, you use an application of back-propagation called back-propagation through time. Training a neural network has three major steps. First, the relevant data vector is normalised between 0 and 1, the vector is feed into the RNN, and it goes through an activation function. The activation function utilised in the software is the rectified linear activation function\\cite{lstm_rnn}. \n\n\\begin{definition}\nThe rectified linear activation function is a piece-wise linear function that will output the input directly if is positive, otherwise, it will output zero.\n\\end{definition}\n\nThe function is linear for values greater than zero, meaning it has a lot of the desirable properties of a linear activation function when training a neural network using back-propagation. Yet, it is a nonlinear function as negative values are always output as zero. As a result, the rectified function is linear for half of the input domain and nonlinear for the other half, it is referred to as a piece-wise linear function\\cite{relu}. This nonlinear element is extremely important if the system has a nonlinear component, for example in predicting the evolution of the future state of the atmosphere.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.65\\linewidth]{Images/relu.png}\n \\caption{Sketch of the Rectified Linear Activation Function}\n\\end{figure}\n\nSecond, it outputs the results. Third, it compares the prediction to the ground truth using a loss function.\n\n\\begin{definition}\nA loss function outputs an error value which is an estimate of how poorly the network is performing.\n\\end{definition}\n\nThe lost function that will be utilised in the software will be the function for mean squared error. The reason for choosing this particular function is that it heavily penalises large errors, as it squares the difference between the predicted and actual value. A large error in a weather forecast is highly undesirable, hence, the use of this function. The function is represented below:\n\n\\begin{equation}\n MSE = \\frac{1}{n}\\sum_{i=1}^n(Y_i-\\hat{Y_i})^2\n\\end{equation}\n\nIf a vector of $n$ predictions is generated from a sample of $n$ data points on all variables, and $Y$ is the vector of observed values of the variable being predicted, with $\\hat{Y_i}$ being the predicted values.\n\n\\begin{definition}\nMean squared error is the average squared difference between the estimated values and the actual value.\n\\end{definition}\n\nReturning to the training of the RNN, it uses that error value from the loss function. to do back propagation which calculates the gradients for each time step in the network. The gradient is the value used to adjust the networks internal weights, allowing the network to learn. The bigger the gradient, the bigger the adjustments and vice versa. Here is where the problem lies. When doing back propagation, the gradient of the current time step is calculated with respect to the effects of the gradients, in the time step before it. So if the adjustments to the time step before it is small, then adjustments to the current time step will be even smaller. The gradient values will exponentially shrink as it propagates through each time step. That causes gradients to exponentially shrink as it back propagates down. The earlier layers fail to do any learning as the internal weights are barely being adjusted due to extremely small gradients.\n\nBecause of vanishing gradients, the RNN doesn’t learn the long-range dependencies across time steps. So not being able to learn on earlier time steps causes the network to have a short-term memory. In order to combat this, a long short-term memory is used\\cite{intro_rnn}.\n\n\\subsection{LSTM}\nLSTM's were created as a solution to the short-term memory problem. They have internal mechanisms called gates that can regulate the flow of information. These gates can learn which data in a sequence is important to keep or throw away. By doing that, it can pass relevant information down the long chain of sequences to make predictions. For example, if you were interested in buying a particular, you might read a review in order to determine if the purchase of the product is a good decision. When you read a review, your brain subconsciously only remembers important keywords. You pick up words like ``amazing\", ``superb\", or ``awful\", you don't remember words such as \"the\", \"as\", or \"because\". This is what an LSTM does, it learns to keep only the relevant information to make predictions.\n\nAn LSTM has a similar control flow as a recurrent neural network. It processes data passing on information as it propagates forward. The differences are the operations within the LSTM’s cells. The core concept of LSTM’s are the cell state, and it’s various gates. The cell state is the method by which information is transferred down the sequence chain. The cell state, in theory, can carry relevant information throughout the processing of the sequence. So even information from the earlier time steps can make it’s way to later time steps, reducing the effects of short-term memory. As the cell state goes on its journey, information get’s added or removed to the cell state via gates\\cite{lstm_rnn}.\n\n\\begin{definition}\nA gate is an electric circuit with an output which depends on the combination of several inputs.\n\\end{definition}\n\nGates contain the sigmoid activation function. The sigmoid activation function squishes values between 0 and 1. That is helpful to update or forget data because any number getting multiplied by 0 is 0, causing values to disappears or be ``forgotten\". Any number multiplied by 1 is the same value therefore that value stay’s the same or is ``kept\".\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.65\\linewidth]{Images/sigmoid.png}\n \\caption{Sketch of the Sigmoid Activation Function}\n\\end{figure}\n\nThere are three types of gates utilised within a neural network: a forget gate, an input gate, and an output gate. A forget gate decides what information should be thrown away or kept. Information from the previous hidden state and information from the current input is passed through the sigmoid function. An input gate is where the previous hidden state and current input into a sigmoid function. The output gate decides what the next hidden state should be. The hidden state is also used for predictions. First, we pass the previous hidden state and the current input into a sigmoid function. Then we pass the newly modified cell state to the rectified linear activation function. We multiply the rectified linear activation function output with the sigmoid output to decide what information the hidden state should carry. The output is the hidden state. The new cell state and the new hidden is then carried over to the next time step\\cite{lstm_rnn}.\n\n\\subsection{Implementation}\\label{implement_rnn}\nThe data set for the initial conditions (which is discussed in section \\ref{noaa_initial_conditions}) consists of three features: geopotential height, air temperature, and relative humidity. For the purposes of this specific project, the RNN will solely be trained on air temperature and relative humidity. Unfortunately, due to the COVID-19, there was a time constraint on the developed of the RNN, which resulted in the inability to also train the RNN on geopotential height. This is due to the lack of computational resources at my disposable. The data set in question is updated every six hours by the National Oceanic and Atmospheric Administration. This means for a single day, there will be four observations. The goal for this project will be to, first predict the relevant atmospheric parameter in seven days time given the last thirty days of data and combine this RNN prediction with the physical model prediction in an attempt to make a more accurate prediction overall. In order to make such predictions, it is necessary to create a window of the last 120 ($30 \\times 4$) observations to train the model\\cite{time_series}.\n\nAt the start, a seed is set in order to ensure reproducibility. As mentioned previously, it is important to scale features before training a neural network. Normalisation is a common way of doing this scaling by subtracting the mean and dividing by the standard deviation of each feature. In order for the most optimal performance, the method ``MinMaxScaler\" from the library, scikit-learn, is utilised within the software\\cite{scikit-learn}. An LSTM requires a 1-dimensional sequence, however, the atmosphere is a 3-dimensional system. Hence, it is necessary to flatten the 3-dimensional vector that represents the state of the atmosphere. This is done in order to avoid the need of repeatably running the RNN. Batches are then created to split the data into manageable sequences. The diagram on the following page shows how the data is represented after flattening the data and batching it (provided by Tensorflow)\\cite{time_series}.\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.5\\linewidth]{Images/data_rnn.png}\n \\caption{Visualisation of how the data is represented after flattening and batching.}\n\\end{figure}\n\nFollowing this process, the data is feed into the RNN. The LSTM model is built using Keras in TensorFlow, which is an free and open-source software library for machine learning. It was developed by the Google Brain Team\\cite{tensorflow}. It is apparent that a multi-step model is needed as the model needs to learn to predict a range of future values. The source code for the LSTM model developed for the software is shown below:\n\n\\begin{minted}[mathescape,linenos,frame=lines]{python}\n# Prepossessed historical data, which has been flatten and batched.\nx_data, y_data = prepossessing_function(input_data)\n# Prepossessed initial conditions, which has been flatten and batched.\ninitial_conditions = prepossessing_function(input_initialconditions)\n\n# The network is shown data from the last 15 days.\npast_history = 15 * 4\n\n# The network predicts the next 7 days worth of steps.\nfuture_target = 7 * 4\n\n# Create, and train models.\n# Optimiser.\nopt = Adam(lr=1e-6, decay=1e-10, clipvalue=0.6)\n# Create model.\nmodel = Sequential()\nmodel.add(\n LSTM(\n 400, activation='relu', input_shape=(past_history, features)\n )\n)\nmodel.add(RepeatVector(future_target))\nmodel.add(LSTM(400, activation='relu', return_sequences=True))\nmodel.add(LSTM(400, activation='relu', return_sequences=True))\nmodel.add(LSTM(400, activation='relu', return_sequences=True))\nmodel.add(TimeDistributed(Dense(features)))\nmodel.compile(\n optimizer=opt, loss='mse', metrics=['mean_absolute_error']\n)\n\n# Train.\nmodel.fit(\n x_data, y_data, epochs=epochs, batch_size=10\n)\n\n# Predict.\nfuture_state = model.predict(initial_conditions)\n# Invert normalisation, and flattening.\nfuture_state = inverse_prepossessing(future_state)\n\\end{minted}\n\nThe model consists of four LSTM layers, which in combination are able to produce a more accurate and reliable prediction than a single LSTM layer. As is evident, the activation function for each LSTM is the rectified linear activation function, which is built into Keras. The number of epochs can be specified by the end user depending on the computational resources they have and what they need. More epochs will evidently lead to a more accurate neural network.\n\n\\begin{definition}\nAn epoch is one forward pass and one backward pass of all the training examples.\n\\end{definition}\n\n\\section{Initial Conditions}\\label{noaa_initial_conditions}\n\\subsection{Global Data Assimilation System}\nThe initial conditions utilised by the software are from the Global Data Assimilation System (GDAS), which is provided by the National Oceanic and Atmospheric Adminstration in the United States. The primary reason for utilising this data is that it is freely available to the general public. In an ideal world, data from the European Centre for Medium-Range Weather Forecasts would be utilised, however unfortunately, there data is not freely available to the general public. This fundamentally violates the software's open source principles (these principles are discussed in chapter \\ref{5}).\n\nThe GDAS is a model to place observations into a gridded model space for the purpose of initialising weather forecasts with observed data. This system is utilised by the National Center for Environmental Prediction for such a purpose. GDAS adds the following types of observations to a gridded, 3-D, model space: surface observations, balloon data, wind profiler data, aircraft reports, buoy observations, radar observations, and satellite observations\\cite{gdas}. The initial conditions provided by the GDAS to the software have a vertical pressure co-ordinate, or the vertical co-ordiante is pressure. This co-ordinate system is known as isobaric co-ordinates.\n\n\\subsection{Isobaric Co-Ordinates}\nIn coordinate systems applied to the earth, the vertical coordinate describes position in the vertical direction (that is, parallel to the force of effective gravity). In meteorology, pressure can be a more convenient vertical coordinate than altitude. One reason is that until recently, radiosondes, which are the primary means of gathering observations of weather variables above the earth’s surface, measure and reported pressure, temperature, and humidity, but not altitude, as they rise through the atmosphere\\cite{isobar_i}.\n\n\\begin{definition}\nA radiosonde is an instrument carried by balloon to various levels of the atmosphere and transmitting measurements by radio.\n\\end{definition}\n\nAnother reason is that on scales large enough for the hydrostatic approximation to be valid (this is discussed in section \\ref{hydro_balance}), the pressure-gradient force in the equations of motion becomes simpler and density no longer becomes an explicit variable in the tendency equations\\cite{isobar_i}. Thus,a given geopotential gradient implies the same geostrophic wind at any height, whereas a given horizontal pressure gradient implies different values of the geostrophic wind depending on the density\\cite{isobar_ii}.\n\nFrom the Global Data Assimilation System, three prognostic variables are chosen: geopotential height, air temperature, and relative humidity.\n\n\\begin{definition}\nGeopotential Height is the height above sea level of a pressure level. For example, if a station reports that the 500 hPa height at its location is 5600 m, it means that the level of the atmosphere over that station at which the atmospheric pressure is 500 hPa is 5600 meters above sea level.\n\\end{definition}\n\nGeophysical sciences such as meteorology often prefer to express the horizontal pressure gradient force as the gradient of geopotential along a constant-pressure surface, because then it has the properties of a conservative force. For example, the primitive equations which weather forecast models solve use hydrostatic pressure as a vertical coordinate, and express the slopes of those pressure surfaces in terms of geopotential height. As such, this will be a parameter of great interest. From the aforementioned three selected parameters, any other parameter that is needed in the software can be calculated, including the wind. This will be discussed in greater depth in the next chapter. \n", "meta": {"hexsha": "cdfdaae05097b70a39dab6e4e8ca18a42429f219", "size": 33147, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "scifest/online/project-book/Chapters/Dynamical_Core.tex", "max_stars_repo_name": "amsimp/papers", "max_stars_repo_head_hexsha": "a212b3f65140f0292d51055be324a7c1b084e121", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-05-15T10:06:17.000Z", "max_stars_repo_stars_event_max_datetime": "2020-05-15T10:06:17.000Z", "max_issues_repo_path": "scifest/online/project-book/Chapters/Dynamical_Core.tex", "max_issues_repo_name": "amsimp/papers", "max_issues_repo_head_hexsha": "a212b3f65140f0292d51055be324a7c1b084e121", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "scifest/online/project-book/Chapters/Dynamical_Core.tex", "max_forks_repo_name": "amsimp/papers", "max_forks_repo_head_hexsha": "a212b3f65140f0292d51055be324a7c1b084e121", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 104.5646687697, "max_line_length": 1451, "alphanum_fraction": 0.7887591637, "num_tokens": 7354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117769928211, "lm_q2_score": 0.7690802423634963, "lm_q1q2_score": 0.6358076938143955}} {"text": "\\section{Computational Problems and Complexity}\n\\label{cc-sect}\n\nWe base this section on the definitions in Papadimitriou \\cite{papad-cc},\nto which we refer for further details.\n\nA {\\em deterministic Turing machine} $\\mathcal{M}$ (we will imply the\n``deterministic'' from now on) is a representation of an\nalgorithm that takes an {\\em input}, runs a algorithm\nmanipulating the input on a {\\em tape}, and returns an\n{\\em output} when it comes to a halting state (if it ever does).\n\nThe tape and its manipulation can be seen as follows: initially the tape\ncontains the input, given by a {\\em first symbol} $\\rhd$ followed by a\nfinite string of symbols $x\\in (\\Sigma \\setminus \\{ \\sqcup \\})^\\ast$,\nwhere $\\Sigma$ is the {\\em alphabet} of symbols of $\\mathcal{M}$\nand $\\sqcup$ is a special\nsymbol representing the {\\em blank} string.\nThe program is then carried on: a {\\em cursor} starts at the first symbol\nand follows the {\\em algorithm} given by a\n{\\em transition function} $\\delta$. The transition function depends on the\ncurrent {\\em state} $p\\in K$ and the current symbol $\\sigma\\in \\Sigma$: the\nformer is the ``instruction'' to follow, the latter the symbol in the\nposition of the cursor. The transition function returns the next\nstate $q\\in K$, the symbol $\\tau\\in \\Sigma$ to be written at the position\nof the cursor, and a direction in which the cursor will move on the tape.\nThe possible directions are ``left'', ``right'', or ``stay'', represented\nrespectively\nby $\\leftarrow,\\rightarrow,-$. The first symbol is never\noverwritten, and from there the cursor can only go left.\n\nIn some cases, the machine reaches the state\n{\\sc Yes} (the machine {\\em accepts} the input), {\\sc No} (the machine\n{\\em rejects} the input) or the {\\em halting state} $h$\nafter a finite amount of time. The output\nis then given by $\\mathcal{M}(x)=\\text{\\sc Yes}$ or\n$\\mathcal{M}(x)=\\text{\\sc No}$ if the machine accepts or rejects the input.\nIf the halting state $h$ is reached, the output is $\\mathcal{M}(x)=y$,\nwhere $y$ is the finite (possibly empty) string of symbols that is left on\nthe tape after the symbol $\\rhd$ and before any string of $\\sqcup$'s at the\nend. It is also possible that the machine doesn't halt on input $x$; we\ndenote this case as $\\mathcal{M}(x)=\\nearrow$.\n\nA problem $P$ that requires an output that is either ``{\\sc Yes}'' or\n``{\\sc No}'' is a {\\em decision problem}; its {\\em complement} is the\nproblem $\\overline{P}$ that outputs ``{\\sc No}'' for each instance\nof $P$ that outputs ``{\\sc Yes}'', and vice versa.\nA problem where the output can be a string $\\mathcal{M}(x)=y$ on the\noutput tape as above is a {\\em function problem}; note that no guarantee\non the halting state is required. If the output is a\nstring $\\mathcal{M}(x)=y$ satisfying a relation $R(x,y)$\nor the input is rejected if it is not\npossible to find any such string, the problem is a {\\em search problem}.\nA search problem where the input is never rejected is a {\\em total\nfunction problem}. By Theorem \\ref{nash-thm}, the problem $n$-Nash of\nTable \\ref{n-nash} is a total function problem.\n\n\\begin{problem}\n{$n$-Nash}\n{A $n$-player game.}\n{A Nash equilibrium of the game.}\n\\label{n-nash}\n\\end{problem}\n\nFormally, a Turing machine is $\\mathcal{M}=(K,\\Sigma,\\delta,s)$,\nwhere $K$ and $\\Sigma$ are finite, $\\Sigma\\cap K=\\varnothing$ and $\\Sigma$\nalways contains the symbols $\\sqcup$ and $\\rhd$.\nThe {\\em transition function} is $\\delta$, defined as follows,\nwhere $\\{ \\leftarrow,\\rightarrow,- \\}\\nsubseteq(K\\cup\\Sigma)$:\n\\[\n\\delta:K\\times \\Sigma\\longrightarrow (K\\cup\\{ h,\\ \\text{{\\sc Yes}},\\ \\text{{\\sc No}} \\})%\n\\times \\Sigma \\times \\{ \\leftarrow,\\rightarrow,- \\}.\n\\]\n\nA slightly different, although equivalent, model allows for\n{\\em multiple strings}: $k\\in \\naturals$ cursors move\non $k$ strings $\\sigma_i\\in \\Sigma^\\ast$; the states are still\nrepresented as $p\\in K$. The input is given in the tape of the first string;\nassuming the machine halts, the output is given in the tape of the $k$-th\nstring.\n\nA {\\em language} is a set of {\\em strings of symbols}\n$L\\subseteq(\\Sigma\\setminus\\{ \\sqcup \\})^\\ast$; a {\\em class} is a\nset of languages. Let $\\mathcal{M}$ be a Turing machine, and let\n$x\\in (\\Sigma\\setminus\\{ \\sqcup \\})^\\ast$.\nWe say that $\\mathcal{M}$ {\\em decides} $L$ if\neither $\\mathcal{M}(x)=\\text{{\\sc Yes}}$ if $x\\in L$ or\n$\\mathcal{M}(x)=\\text{{\\sc No}}$\nif $x\\notin L$. A Turing\nmachine $\\mathcal{M}$ {\\em accepts} $L$ if\n$x\\in L$ is a necessary and sufficient condition for\n$\\mathcal{M}(x)=\\text{{\\sc Yes}}$, and if $x\\notin L$ then\n$\\mathcal{M}(x)=\\text{{\\sc No}}$ or $\\mathcal M$\ndoes not halt.\nFinally, $\\mathcal{M}$ {\\em computes} a function\n$f:(\\Sigma\\setminus\\{ \\sqcup \\})^\\ast\\to\\Sigma^\\ast$ if\n$\\mathcal{M}(x)=f(x)$ for every $x\\in (\\Sigma\\setminus\\{ \\sqcup \\})^\\ast$.\n\nLet $P_1$ be a problem and let $x$ be an instance of $P_1$ that is encoded\nin $|x|$ bits. $P_1$ {\\em reduces to the problem $P_2$ in polynomial time}\nif there exists a function $f: \\{0,1\\}^\\ast \\to \\{0,1\\}^\\ast$, a\nTuring machine $\\mathcal{M}$, and a polynomial $p$\nsuch that for all $x\\in\\{0,1\\}^\\ast$\n\\begin{enumerate}\n\\item $x\\in P_1\\quad\\iff\\quad f(x)\\in P_2$;\n\\item $\\mathcal{M}$ computes $f(x)$;\n\\item $\\mathcal{M}$ stops after $p(|x|)$ steps.\n\\end{enumerate}\n\nFor any class $\\mathrm{C}$ of decision problems, the class of all complements\nof the problems in $\\mathrm{C}$ is the {\\em complement class}\n$\\mathrm{co-C}$. A problem $P$ is {\\em hard} for a class $\\mathrm{C}$\nif every problem in $\\mathrm{C}$ is polynomial-time reducible to $P$; that is,\nif $P$ is at least as hard to solve as every problem in $\\mathrm{C}$. A\n{\\em complete} problem for the class $\\mathrm{C}$ is a $\\mathrm{C}-hard$ problem\nthat is also in $\\mathrm{C}$.\nIntuitively, if $P_1$ is polynomial-time reducible to $P_2$, it takes\npolynomial time to ``translate'' $P_1$ to $P_2$, and then to\n``translate back'' a solution of $P_2$ as a solution of $P_1$.\nThis is particularly useful if $P_2$ is hard: then\nthe problem $P_1$ is at least as ``difficult''. Furthermore, if the problem\n$P_2$ is also complete in a class $\\mathrm{C}$, it can be used as a test\nof belonging to the class $C$.\n\nThe complexity class $\\mathrm{\\mathbf{P}}$ ({\\em polynomial-time})\ncontains all the\n{\\em polynomially decidable problems}, that is, all problems $P$ such that\nthere exists a Turing machine $\\mathcal{M}$ that outputs either {\\sc Yes} or\n{\\sc No} for all inputs $x\\in\\{0,1\\}^\\ast$ of $P$ after $p(|x|)$ steps,\nwhere $p$ is a polynomial.\nA problem $P$ belongs to the class $\\mathrm{\\mathbf{NP}}$\n({\\em non-deterministic polynomial-time}) if there exists a\nTuring machine $\\mathcal{M}$ and polynomials $p_1,p_2$ such that\n\\begin{enumerate}\n\\item for all $x\\in P$ there exists a {\\em certificate} $y\\in \\{0,1\\}^\\ast$\nsuch that \\mbox{$|y|\\leq p_1(|x|)$};\n\\item $\\mathcal{M}$ accepts the combined input $xy$, stopping after at most\n$p_2(|x| + |y|)$ steps;\n\\item for all $x\\notin P$ there does not exist $y\\in \\{0,1\\}^\\ast$ such\nthat $\\mathcal{M}$ accepts the combined input $xy$.\n\\end{enumerate}\nAn equivalent definition gives $\\mathrm{\\mathbf{NP}}$ as the class of all\nlanguages $L$ for which the binary relation $R(x,y)$ such\nthat $L=\\{ x\\ |\\ R(x,y) \\text{ holds for some } y \\}$ satisfies\nthe following conditions:\n\\begin{enumerate}\n\\item {\\em (polynomially balanced)} if $(x,y)\\in R$ then $|y|\\leq |x|^k$ for\nsome $k\\in \\naturals$;\n\\item {\\em (polynomially decidable)} if there is a Turing machine that\ndecides $L$ in polynomial time.\n\\end{enumerate}\nInformally, a decision problem is in\n$\\mathrm{\\mathbf{P}}$ if the answer to its question can be found in a\nnumber of steps that is polynomial in the input of the problem, and\na decision problem is in $\\mathrm{\\mathbf{NP}}$ if it takes\npolynomial time to verify whether the ``certificate'' $y$ is,\nindeed, a correct answer to the question posed by the problem.\nNotice that there may be many different certificates for each instance\nof a problem. Consider for instance the $\\mathrm{\\mathbf{NP}}$ problem\n{\\sc Hamilton Path},\ndefined as ``given an input graph $x$, is it possible to find a Hamiltonian\npath $y$ of $x$?'' There are graphs with more than one possible\nHamiltonian path, and each one of these can be used as certificate.\n\nIt is quite simple to see that $\\mathbf{P}\\subseteq \\mathbf{NP}$, but it is\nan important open\nproblem whether the inclusion is strict. If this were\nnot the case, it could be\nargued that there isn't any substantial difference between finding a\nsolution and verifying its validity. This seems to contradict most of\nthe human intellectual experience, where the value\nof an ``original'' discovery is perceived as fundamental; the\n``conventional'' view is therefore that $\\mathbf{P}\\neq\\mathbf{NP}$.\nAnother open problem is whether $\\mathbf{NP} = \\mathbf{co-NP}$; again,\nthis is widely believed to be false. Notice that if\n$\\mathbf{NP} \\neq \\mathbf{co-NP}$ then\nalso $\\mathbf{P} \\subsetneq \\mathbf{NP}$, but not vice versa.\n\nThe classes $\\mathbf{FP}$ and $\\mathbf{FNP}$ are analogous to\n$\\mathbf{P}$ and $\\mathbf{NP}$, respectively, but for function problems\ninstead of decision problems.\nFormally, {\\bf FNP} ({\\em function non-deterministic polynomial-time})\nis defined\nin Megiddo and Papadimitriou \\cite{megiddo-papad} as the class\nof problems of the form ``given an $x\\in\\Sigma^\\ast$,\nfind $y\\in \\Sigma^\\ast$ such that $(x,y)\\in R$, where R is a\npolynomially balanced binary relation, or reject the input.''\n$\\mathbf{FP}$ ({\\em function polynomial-time}) is the\nclass of all $\\mathbf{FNP}$ problems that can be solved in polynomial time.\nAs in the case of decision problems, it is not known whether\n$\\mathbf{FP} = \\mathbf{FNP}$; the question is actually equivalent\nto whether $\\mathbf{P} = \\mathbf{NP}$.\nMegiddo and Papadimitriou \\cite{megiddo-papad} also define the class\n$\\mathbf{TFNP}$ ({\\em total function non-deterministic polynomial-time})\nas the class of all $\\mathbf{FNP}$ problems where for every $x\\in\\Sigma^\\ast$\na solution $y\\in\\Sigma^\\ast$ is guaranteed to exist.\nFrom another point of view, $\\mathbf{TFNP} = \\mathbf{F(NP\\cap co-NP)}$,\nand the existence of a $\\mathbf{TFNP}$-complete problem would imply\nthat $\\mathbf{NP} = \\mathbf{co-NP}$.\nThe lack of $\\mathbf{TFNP}$-complete problems has led to define\na number of new classes; we will see more of this in\nsection \\ref{ppad-sect}.\n\nThe class $\\mathrm{\\mathbf{\\# P}}$ is the class of all function\nproblems that output the\nnumber of possible certificates\nfor a decision problem in $\\mathrm{\\mathbf{NP}}$.\nFormally: the\n{\\em counting problem} associated with a binary\nrelation $Q(x,y)$ is ``given $x$, how many $y$ are there such\nthat $(x,y)\\in Q$?'' Then $\\mathrm{\\mathbf{\\# P}}$ is the class of\nall\ncounting problems associated with binary relations that are both\npolynomially\nbalanced and polynomially decidable.\nIt is interesting to notice that there are\n$\\mathrm{\\mathbf{\\# P}}$-complete\nproblems for which the corresponding search problem can\nbe solved\nin polynomial time. An example close to the topic\nof this thesis is given in Brightwell \\cite{brightwell}: although finding\nan Eulerian circuit in an\nundirected graph takes polynomial time, counting the number of possible\ncircuits in the same graph is complete in $\\mathrm{\\mathbf{\\# P}}$.\n\nFinally, the class {\\bf PSPACE} is the set of decision problems that can be\nsolved by a Turing machine using an amount of tape space that is polynomial\nin the size of the input; although it can be proven that\n$\\mathbf{NP}\\subseteq \\mathbf{PSPACE}$, the possibility of an identity\nis yet another open problem.\n", "meta": {"hexsha": "4acfc5a24c109c8d912a27ea97e837fb657cb96a", "size": 11542, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/chapter-1/complexity-classes.tex", "max_stars_repo_name": "mmcasetti/mphil-thesis", "max_stars_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_stars_repo_licenses": ["OLDAP-2.4"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "thesis/chapter-1/complexity-classes.tex", "max_issues_repo_name": "mmcasetti/mphil-thesis", "max_issues_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_issues_repo_licenses": ["OLDAP-2.4"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/chapter-1/complexity-classes.tex", "max_forks_repo_name": "mmcasetti/mphil-thesis", "max_forks_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_forks_repo_licenses": ["OLDAP-2.4"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.9653679654, "max_line_length": 89, "alphanum_fraction": 0.7170334431, "num_tokens": 3443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117855317474, "lm_q2_score": 0.7690802317779601, "lm_q1q2_score": 0.6358076916303276}} {"text": "%\n% STAT 100: Chance and Data Analysis - A Course Overview\n% Section: Probability\n%\n% Author: Jeffrey Leung\n%\n\n\\section{Probability}\n\t\\label{sec:probability}\n\\subsection{Introduction}\n\t\\label{subsec:probability:introduction}\n\\begin{easylist}\n\n\t& \\emph{Experiment:} Process which creates a bag of outcomes\n\t& \\emph{Sample space:} Set of all possible outcomes\n\t\t&& Denoted by $S$\n\t& \\emph{Event:} Possible outcome which satisfies a given condition(s)\n\t\t&& Denoted by $E$\n\n\t& \\emph{Probability:} Value between 0 and 1 (inclusive) representing the proportion of times an outcome/event is expected\n\t\t&& Denoted by $P(event)$\n\t\t&& Calculation:\n\t\t\\begin{math}\n\t\t\tP(A) = \\frac{|A|}{|S|}\n\t\t\\end{math}\n\t\twhere |X| is the number of outcomes for X\n\t\t&& 0 means the event is impossible; 1 means the event is certain\n\t\t&& Represented either by a fraction or a decimal value\n\t\t\n\t\t&& E.g. The probability of getting an odd number when rolling a 6-sided die:\n\t\t\n\t\t\\medskip\n\t\t\\Deactivate\n\t\t\\begin{center}\n\t\t\t\\begin{tabular}{ r @{ = } l }\n\t\t\t\t$S$ & Results of rolling a 6-sided die = 1, 2, 3, 4, 5, 6 \\\\\n\t\t\t\t$|S|$ & 6 \\\\\n\t\t\t\t$E$ & Getting an odd number = 1, 3, 5 \\\\\n\t\t\t\t$|E|$ & 3 \\\\\n\t\t\t\t$P(E)$ & $\\frac{|E|}{|S|}$ = $\\frac{3}{6}$ = 0.5\n\t\t\t\\end{tabular}\n\t\t\t\\Activate\n\t\t\\end{center}\n\t\t\n\t& %TODO\n\n\n\\end{easylist}\n\\clearpage", "meta": {"hexsha": "faeb58c5693b3836182f163a4fafac0173d0df49", "size": 1301, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "stat-100-chance-and-data-analysis/tex/probability.tex", "max_stars_repo_name": "AmirNaghibi/notes", "max_stars_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2019-08-11T08:45:10.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T02:37:39.000Z", "max_issues_repo_path": "stat-100-chance-and-data-analysis/tex/probability.tex", "max_issues_repo_name": "AmirNaghibi/notes", "max_issues_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "stat-100-chance-and-data-analysis/tex/probability.tex", "max_forks_repo_name": "AmirNaghibi/notes", "max_forks_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-18T09:17:46.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-27T21:44:56.000Z", "avg_line_length": 26.5510204082, "max_line_length": 122, "alphanum_fraction": 0.6456571868, "num_tokens": 447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8807970748488297, "lm_q2_score": 0.7217432122827968, "lm_q1q2_score": 0.6357093101706854}} {"text": "\\section{Reduced Order Modeling}\n\\label{sec:ROMs}\nTo provide a very simple idea of a ROM, assume that the final\nresponse space of a physical system is governed by the transfer\nfunction $H \\left ( \\overline{x}\\right)$ (see Section\n~\\ref{sec:mathBackground}), which, from a practical point of\nview, represents the outcome of the system based on the initial\nconditions $\\overline{x}$. Now, sample the domain of variability of the\ninitial conditions $\\overline{x}$ to create a\nset of $N$ realizations of the input and response space $ \\left ( \\left (\n\\overline{x}_{i}, H \\left ( \\overline{x}_{i}\\right) \\right), i=1,N \\right)$,\nnamed ``training'' set. Based on the data set generated, it is possible\nto construct a mathematical representation $G\\left ( \\overline{x}:\n\\overline{x}_{i}\\right)$ of the\nreal system $H \\left ( \\overline{x}\\right)$, which will approximate its\nresponse (see Figure~\\ref{fig:ROMexampleOfPhysicalSystem}):\n\\begin{equation}\n\\label{eq:regressor}\nG\\left ( \\overline{x} \\right ):\\overline{x}_{i} \\rightarrow G\\left ( \\overline{x}_{i} \\right ) \\cong H\\left ( \\overline{x}_{i} \\right )\n\\end{equation}\nThe ROMs reported above are generally named ``regressors'', among\nwhich all the most common data fitting algorithms are found (e.g.,\nleast square for construction of linear models).\nAn important class of ROMs for the work presented here after is the\none containing the so called ``classifiers''. A classifier is a ROM that is\ncapable of representing the system behavior from a binary point of\nview (e.g., event happened/not happened or failure/success). It is a\nmodel (set of equations) that identifies to which category an object\nbelongs in the feature (input) space. Referring to the example that\nbrought to Equation ~\\ref{eq:regressor}, a classifier can be formally represented as follows (see\nFigure~\\ref{fig:ROMClassifierExampleOfPhysicalSystem}):\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=1.0\\textwidth] {pics/ROMexampleOfPhysicalSystem.png}\n \\caption{Example of reduced order model representation of physical system (regression).}\n \\label{fig:ROMexampleOfPhysicalSystem}\n\\end{figure}\n\\begin{figure}[h!]\n \\centering\n \\includegraphics[width=1.0\\textwidth] {pics/ROMClassifierExampleOfPhysicalSystem.png}\n \\caption{Example of reduced order model representation of physical system (classifier).}\n \\label{fig:ROMClassifierExampleOfPhysicalSystem}\n\\end{figure}\n\n\\begin{equation}\n\\label{eq:classifier}\nG\\left ( \\overline{x} \\right ):\\overline{x}_{i} \\rightarrow G\\left ( \\overline{x}_{i} \\right ) \\cong\nC \\left ( H\\left ( \\overline{x}_{i} \\right ) \\right )\n\\end{equation}\n\nThe function $C\\left ( H\\left ( \\overline{x}_{i} \\right ) = \\overline{\\theta}\n\\right ) $ is the so called ``goal'' function that is able to recast the\nresponse of the system $H\\left ( \\overline{x}_{i} \\right )$ into a binary\nform (e.g., failure/success). As an example, referring to\nFigure~\\ref{fig:ROMClassifierExampleOfPhysicalSystem}, the\n``goal'' function would be:\n\\begin{equation}\n\\label{eq:goalFunctionClassifier}\nC\\left ( \\overline{\\theta} \\right ) = \\left\\{\\begin{matrix}\n1 & if \\: \\overline{\\theta}>1.0 \\\\\n0 & if \\: \\overline{\\theta} \\leq 1.0\n\\end{matrix}\\right.\n\\end{equation}\nHence, the ROM of type classifier $G\\left ( \\overline{x} \\right )$ will operate in the space transformed through the ``goal'' function $C\\left ( \\overline{\\theta} \\right )$.\n\\\\The classifiers and regressors can be categorized into two main classes:\n\\begin{itemize}\n \\item Model-based algorithms\n \\item Data-based algorithms\n\\end{itemize}\nIn the first class, the created ROM aims to approximate the response\nof the system as a function of the input parameters. These algorithms\nconstruct a functional representation of the system. Examples of such ROM type are Support Vector Machines (SVMs), Kriging-based regressors, discriminant-based models, and polynomial chaos.\n\nOn the other side, data-based algorithms do not build a response-\nfunction-based ROM but classify or predict the response of the\nsystem from the neighborhood graph constructed from the training\ndata, without any dependencies on a particular prediction model.\nThese algorithms directly build a neighborhood structure as the\nROM (e.g., a relaxed Gabriel graph) on the initial training data. Examples of such ROM type are nearest neighbors and decision trees.\n\n\\textcolor{red}{\\\\It is important to NOTE that RAVEN uses a Z-score normalization of the training data before\n constructing most of the ROMs:\n\\begin{equation}\n \\mathit{\\mathbf{X}} = \\frac{(\\mathit{\\mathbf{X}}-\\mu )}{\\sigma }\n\\end{equation}\n }\nIn order to identify which ROMs get trained with data normalized by the previous reported normalization approach, please refer to the RAVEN user manual \\cite{RAVENuserManual}.\n\nRAVEN has support of several different ROMs,\nsuch as:\n\\begin{enumerate}\n \\item \\textit{Nearest Neighbors approaches}\n \\item \\textit{Support Vector Machines}\n \\item \\textit{Inverse Weight regressors}\n \\item \\textit{Spline regressors }, etc.\n\\end{enumerate}\nIn this section only few of them are going to be explained. \n\n\\subsection{Gaussian Process Models}\n\\label{sec:GPM}\nGaussian Processes (GPs)~\\cite{Rasmussen_GPM} are algorithms that extend multivariate Gaussian distributions to infinite dimensionality. A Gaussian process generates a data set located throughout some domain such that any finite subset of the range follows a multivariate Gaussian distribution. Now, the n observations in an arbitrary data set, $y={y_1,\\ldots,y_n}$, can always be imagined as a single point sampled from some multivariate ($n$-variate) Gaussian distribution.\nWhat relates one observation to another in such cases is just the covariance function, $k(x,x')$. A popular choice is the squared exponential:\n\\begin{equation}\nk(x,x')=\\sigma_{f}^{2} exp \\left [ \\frac{-(x-x')^2}{2 l^2} \\right ]\n\\end{equation}\nwhere the maximum allowable covariance is defined as $\\sigma_{f}^{2}$; this should be high for functions that cover a broad range on the y axis. If $x \\simeq x'$, then $k(x,x')$ approach this maximum meaning $f(x)$ is very correlated to $f(x')$. On the other hand, if $x$ is very distant from $x'$, then $k(x,x' ) \\simeq 0$ (i.e., the two points cannot see each other.\nSo, for example, during interpolation at new $x$ values, distant observations will have negligible effect). How much effect this separation has will depend on the length parameter $l$.\nEach observation $y$ can be thought of as related to an underlying function $f(x)$ through a Gaussian noise model:\n\\begin{equation}\ny=f(x)+N(0,\\sigma_{n}^{2})\n\\end{equation}\nThe new kernel function can be written as:\n\\begin{equation}\nk(x,x')=\\sigma_{f}^{2} exp \\left [ \\frac{-(x-x')^2}{2 l^2} \\right ] + \\sigma_{n}^{2} \\delta(x,x')\n\\end{equation}\nSo given $n$ observations $y$, the objective is to predict the value $y_*$ at the new point $x_*$. This process is performed by following this sequence of steps:\n\\begin{enumerate}\n\\item Calculate three matrices:\n\\begin{equation}\nK=\\begin{bmatrix}\nk(x_1,x_1) & \\ldots & k(x_1,x_n)\\\\\n\\vdots & \\ddots &\\vdots \\\\\nk(x_n,x_1) & \\ldots & k(x_n,x_n)\n\\end{bmatrix}\n\\end{equation}\n\\begin{equation}\nK_*= \\begin{bmatrix}\nk(x_*,x_1) & \\ldots & k(x_*,x_n)\n\\end{bmatrix}\n\\end{equation}\n\\begin{equation}\nK_{**}=k(x_*,x_*)\n\\end{equation}\n\\item The basic assumption of GPM is that:\n\\begin{equation}\n\\begin{bmatrix}\ny\\\\\ny_*\n\\end{bmatrix}\n=\\mathcal{N}(0,\\begin{bmatrix}\nK & K_{*}^{T}\\\\\nK_* & K_{**}\n\\end{bmatrix})\n\\end{equation}\n\\item The estimate $\\bar{y_*} $ for $y_*$ is the mean of this distribution\n\\begin{equation}\n\\bar{y_*}=K_* K^{-1}y\n\\end{equation}\n\\item The uncertainty associated to the estimate $\\bar{y_*} $ can be expressed in terms of variance of $y_*$:\n\\begin{equation}\nvar(y_*)=K_{**}-k_* K^{-1} K_{*}^{T}\n\\end{equation}\n\\end{enumerate}\n\n\\subsection{Support Vector Machines}\n\\label{sec:SVM}\nThe Support Vector Machine (SVM)~\\cite{SVM_Burges} classifier is a methodology that aims to determine the optimal separation hyperplane between data sets having different labels.\nThe training data consist of $N$ data points $(x_i,y_i)$ $i=1,\\ldots,N$ where $x_i \\in \\mathbb{R}^M$ and $y_i \\in {-1,1}$.\nAssuming a linear property of the hyperplane then its definition is:\n\\begin{equation}\n\\left \\{ x: f(x)=x^T\\beta+\\beta_0=0 \\right \\}\n\\end{equation}\nwhere $\\beta$ is a unit vector.\n\nThe SVM parameters $\\beta$ and $\\beta_0$ are determined by solving this optimization problem:\n\\begin{equation}\n\\left\\{\\begin{matrix}\n\\underset{\\beta,\\beta_0}{min} \\left \\| \\beta \\right \\|\\\\\n\\text{subject to } y_i(x_{i}^{T}\\beta+\\beta_0)\\geq 1 , \\quad i=1,\\dots,N\n\\end{matrix}\\right.\n\\end{equation}\n\nOnce the SVM parameters $\\beta$ and $\\beta_0$ are determined then the classification of a new point $\\bar{x}$ is given by:\n\\begin{equation}\nG(\\bar{x})=sign(\\bar{x}^T\\beta+\\beta_0)\n\\end{equation}\n\n\n\\subsection{KNN Classifier and KNR Regressor}\n\\label{sec:KNN_KNR}\nThe K Nearest Neighbor algorithm~\\cite{altman_KNN} (KNN) is a non-parametric method used for both regression and classification. The only input parameter is the variable $K$ which indicates the number of neighbors to be considered in the classification/regression process. The special case where the class is predicted to be the class of the closest training sample (i.e. when $K = 1$) is called the nearest neighbor algorithm. In binary (two class) classification problems, it is helpful to choose k to be an odd number as this avoids tied votes. The output depends on whether KNN is used for classification or regression:\n\\begin{itemize}\n\\item In KNN classification, the output is a class membership. An object is classified by a majority vote of its neighbors, with the object being assigned to the class most common among its $K$ nearest neighbors ($K$ is a positive integer, typically small). If $K = 1$, then the object is simply assigned to the class of that single nearest neighbor.\n\\item In KNN regression, the output is the property value for the object. This value is the average of the values of its $K$ nearest neighbors.\n\\end{itemize}\nBoth for classification and regression, it can be useful to assign weight to the contributions of the neighbors, so that the nearer neighbors contribute more to the average than the more distant ones. For example, a common weighting scheme consists in giving each neighbor a weight of $1/d$, where $d$ is the distance to the neighbor.\n\n\\subsection{Multi-Dimensional Interpolation}\n\\label{sec:ND_interp}\nThis section covers the methods that have been implemented in the CROW statistical library:\n\\begin{itemize}\n\\item Shepard's Method (see Section~\\ref{sec:shepard})\n\\item Multi-Dimensional Spline method (see Section~\\ref{sec:ND_spline}).\n\\end{itemize}\n\nThese two methods are interpolation methods that can be used in any dimension.\nIn RAVEN they are employed in two major applications:\n\\begin{enumerate}\n\\item ROMs\n\\item Multi-dimensional distributions.\n\\end{enumerate}\nFor both applications, given a set of $N$ data points $ (x_i,u_i )$ $i=1,\\ldots,N$ where $x_i$ are the coordinate in the input space $D \\subset \\mathbb{R}^M$ and $u_i \\in \\mathbb{R}$ is the outcome, the methods predicts the outcome $\\tilde{u}$ for a new coordinate $\\tilde{x}\\in \\mathbb{R}^n$.\n\n\n\\subsubsection{Shepard's Method}\n\\label{sec:shepard}\nThe Shepard interpolator~\\cite{Shepard} is also know as Inverse Distance Weighting (IDW) interpolator.\nThe starting point is a set of $N$ data points $ (x_i,u_i )$ for $i=1,\\ldots,N$.\nThe Inverse-Weight interpolator can be represented as a function $f_{IDW}(x)$ that, given a new coordinate in the input space $x$, generates a prediction on $u$ such that\n\\begin{equation}\nu:x \\in \\mathbb{R}^M \\rightarrow f_{IDW}(x) \\in \\mathbb{R}\n\\end{equation}\nbased on the distance $d(x,x_i)$ in the euclidean space between $x$ and $x_i$.\n\nSuch prediction $u=f_{IDW}(x)$ is performed by summing all data points $x_i$ $i=1,\\ldots,N$ weighted by a weighting parameter $w_i (x)$ as follows:\n\\begin{equation}\nf_{IDW}(x) =\n\\left\\{\n\\begin{matrix}\n\\sum_{i=1}^{N} w(x_i) u_i & \\text{if } d(x,x_i) \\neq 0 \\\\\n u_i & \\text{if } d(x,x_i) = 0\n\\end{matrix}\\right.\n\\end{equation}\nwhere\n\\begin{equation}\nw(x_i) =\\frac{w_i}{\\sum_{i=1}^{N} w_i}\n\\end{equation}\nand\n\\begin{equation}\nw_i = \\left ( \\frac{1}{d(x,x_i)} \\right )^p\n\\end{equation}\nLarge values of $p$ assign greater weight $w_i$ to data points $x_i$ closest to $x$, with the result turning into a mosaic of tiles (i.e., Voronoi diagram) with nearly constant interpolated value.\n\n\\subsubsection{Multi-Dimensional Spline}\n\\label{sec:ND_spline}\nThe Multi-Dimensional Spline (MDS)~\\cite{MD_spline} is a method that requires the sampled points $x_i$ to be lying in multi-dimensional cartesian grid.\nA generic grid $\\Delta_m$ for each dimension $m$ will be indicated as follows:\n\\begin{equation}\n\\Delta_m = \\{x_{0_m},x_{1_m},\\ldots,x_{p_m}\\} \\text{ for } m=1,\\ldots,M\n\\end{equation}\nThis methods construct a $M$-dimensional cubic spline so that, given a coordinate in the input space $x=(x_1,x_2,\\ldots,x_M)$, generates a prediction on $u$ such that\n\\begin{equation}\nu:x \\in \\mathbb{R}^M \\rightarrow f_{MDS}(x) \\in \\mathbb{R}\n\\end{equation}\nwhere\n\\begin{equation}\nf_{MDS}(x)=\\sum_{i_1=1}^{p_1+3} \\sum_{i_2=1}^{p_2+3} \\ldots \\sum_{i_M=1}^{p_M+3} c_{i_1,i_2,\\ldots,i_p} \\prod_{m=1}^{M} u_{i_j} (x_m)\n\\end{equation}\nwhere\n\\begin{equation}\nu_{i_j} (x_m) = \\Phi\\left ( \\frac{x_m-x_{0_m}}{h_j}+2-i_j \\right )\n\\end{equation}\n\nThe cubic kernel $\\Phi(t)$ is defined as:\n\\begin{equation}\n\\Phi(t) = \\left\\{\\begin{matrix}\n(2-\\left | t \\right |)^3 & 1\\leq \\left | t \\right |\\leq 2 \\\\\n4-6\\left | t \\right |^2+3\\left | t \\right |^3 & \\left | t \\right |\\leq 1\\\\\n0 & \\text{elsewhere}\n\\end{matrix}\\right.\n\\end{equation}\n\nThe set of $\\prod_{m=1}^{M}(p_m+3)$ coefficients $c_{i_1,i_2,\\ldots,i_p}$ is determined when the interpolator is initialized.\n\n", "meta": {"hexsha": "f31b29c53a2b630fa4df6d8551a3d42356deed61", "size": 13827, "ext": "tex", "lang": "TeX", "max_stars_repo_path": 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["Apache-2.0"], "max_forks_count": 95, "max_forks_repo_forks_event_min_datetime": "2017-03-24T21:05:03.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-08T17:30:22.000Z", "avg_line_length": 52.1773584906, "max_line_length": 623, "alphanum_fraction": 0.7365299776, "num_tokens": 4113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8807970654616712, "lm_q2_score": 0.7217432003123989, "lm_q1q2_score": 0.6357092928520761}} {"text": "\\chapter{Linear Models: Analysis of variance}\n\\label{ch:ANOVA}\n\nAims of this chapter \\footnote{Here you work with the script file {\\tt anova.R}}:\n\\begin{compactitem}\n\t\\item Plotting boxplots and barplots using factors\n\t\\item Fitting factors in linear models using analysis of variance\n\t\\item Diagnostic plots for explanatory factors\n\t\\item Exploring differences between levels of a factor\n\\end{compactitem}\n\n\\section{What is ANOVA?}\n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=.4\\textwidth]{ANOVA_is_LM.pdf} \n\n\t\\caption{A dataset where an ANOVA would be appropriate. performing an \n\tANOVA test on this dataset is the same as fitting the linear model \n\t$y = \\beta_1 + \\beta_2 x_s + \\beta_3 x_a$, where $x_s$ and $x_a$ \n\tare two levels. There are three ``treatments'' here with the first \n\ttreatment, the control, captured by the baseline value $\\beta_1$ \n\t(the sample with the lowest value, on the far left)}\n\n\t\\label{fig:anova1} \n\n\\end{figure}\n\nA {\\it One-way analysis of variance} (one-way ANOVA) is a technique \nused to compare means of two or more samples representing numerical, \ncontinuous data. \n\nANOVA tests the null hypothesis that samples from two or more groups \nare drawn from populations with the {\\it same mean value}. To do this, \nANOVA uses the {\\it F}-statistic --- the ratio of the variance \ncalculated across the samples (groups) (the null hypothesis) to the \nvariance within the samples (groups). If the null hypothesis that the \ngroup means are drawn from populations with the same mean is indeed \ntrue, the between-group variance (numerator in the F-ratio) should be \nlower than the within-group variance (denominator). A higher ratio (and \n{\\it F} value) therefore implies that the samples were drawn from \npopulations with different mean values.\n\nThis is same as asking whether a linear model with a predictor (or \nexplanatory variable) with at least two categorical levels (or \nfactors), better accounts for the variance (Explained Sum of Squares, \nESS) than a null model of the form $y = \\beta_1$ (Figure \n\\ref{fig:anova1}). Thus, ANOVA is just a type of linear model. \n\nBy the end of this chapter, it will make more sense to you how/why \nlinear regression models that we covered in Chapter \\ref{ch:regress}, \nof the form $y = \\beta_1 + \\beta_2 x$ (where $x$ is a continuous \npredictor variable), require ANOVA to determine if the the model \nbetter fits than a null model of the form $y = \\beta_1$.\n\nTypically, one-way ANOVA is used to test for differences among at least \nthree groups, since the two-group (or levels or factors) case can be \ncovered by a $t$-test (see Chapter~\\ref{ch:t_F_tests}). When there \nare only two means to compare, the $t$-test and the F-test are \nequivalent; the relation between ANOVA and t is given by $F = t^2$. \n\nAn extension of one-way ANOVA is two-way analysis of variance that \nexamines the influence of two different categorical independent \nvariables on one dependent variable --- we will look at multiple \npredictor variables in Chapter \\ref{ch:MulExpl} onwards.\n\n\\section{Calculating the ANOVA test statistic}\n\nANOVA ``partitions'' variability in your data as follows: \n\n\\begin{description}\n\n\t\\item[Total sum of squares (TSS)] This is sum of the squared \n\tdifference between the observed dependent variable ($y$) and the mean \n\tof the response variable $y$ (denoted by $\\bar{y}$), i.e., \n\t$$\\text{TSS} = \\sum_{i=1}^{n}(y_i - \\bar{y})^2$$ TSS tells us how \n\tmuch variation there is in the dependent variable without having any \n\tother information (your null model). You might notice that TSS is the \n\tnumerator of the sample variance you learned about in Chapter \n\t\\ref{ch:ExpDesign}.\n \n\t\\item [Explained sum of squares (ESS)] Sum of the squared differences \n\tbetween the predicted $y$'s (denoted $\\hat{y}$'s) and $\\bar{y}$, or, \n\t$$\\text{ESS} = \\sum_{i=1}^{n} (\\hat{y}_i - \\bar{y})^2$$ ESS tells us \n\thow much of the variation in the dependent variable our alternative \n\t(linear) model was able to explain. That is, it's the reduction in \n\tuncertainty that occurs when the linear model is used to predict the \n\tresponses.\n\t\n\t\\item [Residual sum of squares (RSS)] Sum of the squared differences \n\tbetween the observed $y$'s (denoted by $y_i$) and the predicted \n\t$\\hat{y}$, or, $$\\text{RSS} = \\sum_{i=1}^{n} (\\hat{y}_i - y_i)^2$$ \n\tRSS tells us how much of the variation in the dependent variable our \n\tmodel could not explain. That is, it's the uncertainty that remains \n\teven after the linear model is used. The linear model is considered \n\tto be statistically significant if it can account for a large amount \n\tof variability in the response.\n\n\\end{description} \n\nAnd of course, TSS = ESS + RSS; the OLS method ``decomposes'' the total variation in the dependent variable into an explained component (ESS; explained by the predictor) and an unexplained or residual component (the RSS). \n\nThese sums of squares can then be used to calculate the statistical \nsignificance of the linear model (Regression, ANOVA, etc) through the \nF-Value (or F-Ratio), as follows: \n\n\\begin{center}\n\\def\\arraystretch{1.5}\n\\begin{tabular}{|>{\\centering}m{2cm}|>{\\centering}m{2.1cm}|>{\\centering}m{2cm}|>{\\centering}m{2cm}|p{1.5cm}|}\n\\hline\nType of Sum of Squares (SS) & Calculation & Degrees of Freedom (DF) & \nMean Sum of Squares (MSS) & F-Value \\\\ \\hline\n\nTSS & $\\sum_{i=1}^{n}(y_i - \\bar{y})^2$ & $n-1$ &$\\frac{TSS}{n-1}$& \n\\multirow{3}{*}{ \n$\\frac{\\left(\\frac{ESS}{n_c-1}\\right)}{\\left(\\frac{RSS}{n-n_c}\\right)}$} \\\\ \\cline{1-4}\n\nESS & $\\sum_{i=1}^{n} (\\hat{y}_i - \\bar{y})^2$ & $n_c-1$ & $\\frac{ESS}{n_c-1}$ & \\\\ \\cline{1-4}\n\nRSS & $\\sum_{i=1}^{n} (\\hat{y}_i - y_i)^2$ & $n-n_c$ & $\\frac{RSS}{n-n_c}$ & \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\subsection{Degrees of freedom}\nThus each sum of squares has a corresponding degrees of freedom (DF) \nassociated with it that gives the Mean Sum of Squares (MSS) --- the Sums of Squares divided by the corresponding degrees of freedom.\n\nThe TSS DF is one less than the number of observations $n-1$. This is because calculating TSS, needs $\\bar y$ , which imposes loss of one degree of freedom. Note that MSS is thus nothing but the sample variance.\n\nThe ESS DF is one less than the number of coefficients ($n_c$) \n(estimated parameters) in the model: $n_c-1$. Note that in the case \nwhere the linear model is an ANOVA, it the number of coefficients \nequals the number of ``treatments'' (the categories or levels in the \npredictor). So for example, in Fig. \\ref{fig:anova1}, there are three \ntreatments (predictors) and therefore three coefficients ($\\beta_1$, \n$\\beta_2$, $\\beta_3$), which means that the ESS degrees of freedom \nthere is $n_c-1 = 2$. \n\nThe RSS DF is the sample size $n$ minus the number of coefficients $n_c$, that is, $n - n_c$, because each estimated coefficient is an unknown parameter.\n\n\\subsection{The F-Value (or Ratio)}\n\nFinally, The F-Value or F-Ratio, the test statistic used to decide \nwhether the linear model fit is statistically significant, is the ratio \nof the Mean ESS to the Mean RSS. The null hypothesis is rejected if the \nF-ratio is large --- the model explains a significant amount of \nvariance. The p-value is then calculated from the F-distribution as you \nlearned before, in Chapter \\ref{ch:t_F_tests} (see Fig. \\ref{fig:fdist}). \n\nAlso note that the Root Mean Square Error (RMSE), also known as the \nstandard error of the estimate, is the square root of the Mean RSS. It \nis the standard deviation of the data about the Linear model, rather \nthan about the sample mean.\n\n\\subsection{The $R^{2}$}\n\nFinally, $R^{2}$, also called the Coefficient of Determination, is the \nproportion of total error (TSS) explained by the model (ESS), so the \nratio ESS/TSS. That is it is the proportion of the variability in the \nresponse that is explained by by the fitted model. Since TSS = ESS + \nRSS, $R^{2}$ can be rewritten as (TSS-RSS)/TSS = 1 - RSS/TSS. If a \nmodel has perfectly fits the data, $R^{2}=1$, and if it has no \npredictive capability $R^{2}=0$. In \nreality, $R^{2}$ will never be exactly 0 because even a null model will \nexplain some variance just by chance due to sampling error. Note that \n$R$, the square root of $R^2$, is the multiple correlation coefficient: \nthe correlation between the observed values ($y$), and the predicted \nvalues ($\\hat{y}$).\n\nAs additional predictors (end therefore linear model coefficients) are \nadded to a linear model, $R^2$ increases even when the new predictors \nadd no real predictive capability. The adjusted-$R^2$ tries to addresses this \nproblem of over-specification or over-fitting by including the degrees \nof freedom: Adjusted $R^2$ = 1 - (RSS/$n-n_c-2$)/(TSS/$n-1$) \n\\footnote{That is, it is 1 minus the ratio of the square of the \nstandard error of the estimate to the sample variance of the response}. \nThus additional predictors with little explanatory capability will increase \nthe ESS (and reduce the RSS), but they will also have lower RSS degrees of \nfreedom (because of the additional number of fitted coefficients, \n$n_c$'s)\\footnote{i.e., Standard error of the estimate won't \ndecrease}. Thus if the additional predictors have poor predictive \ncapability, these two reductions will cancel each other out. In other \nwords, the Adjusted $R^2$ penalizes the addition of new predictors to \nthe linear model, so you should always have a look at the Adjusted \n$R^2$ as a corrected measure of $R^2$. \n\n\\section{A new dataset}\n\nIn this Chapter, we will use a new dataset of genome size and life \nhistory in mammals to try out a one-way ANOVA. The dataset is a \ncomposite of data taken from an online database of genome sizes and a \npublished database of mammalian life history:\n\n\\begin{compactdesc}\n\t\\item[Genome size] Average genome sizes for available mammal species \n\tare taken from the online database \n\t\\href{www.genomesize.com}{www.genomesize.com}.\n\t\\item[Life history] Trait data for these species are taken from: \n\t\\href{http://esapubs.org/archive/ecol/e090/184/metadata.htm}{\n\tJones, K. E. {\\it et al.} (2009) PanTHERIA: a species-level database \n\tof life history, ecology, and geography of extant and recently \n\textinct mammals. Ecology 90, 2648--2648}.\n\\end{compactdesc}\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Download the file {\\tt MammalData.csv} from bitbucket and save \n\tto your {\\tt Data} directory.\n\t\\item Create a new blank script called {\\tt ANOVA\\_Prac.R} and add \n\tsome introductory comments.\n\t\\item Use {\\tt read.csv} to load the data in the data frame \n\t{\\tt mammals} and then {\\tt str} and {\\tt summary} to examine \n\tthe data.\n\\end{compactitem}\n\nThere are nine variables. The first two are the latin binomial and \ntaxonomic order of each species, followed by the species mean genome \nsize (`C value', picograms), adult body mass (g), diet breadth, habitat \nbreadth, litter size and then two factors showing whether the species \nare ground dwelling and their trophic level. For more information, see \nthe link above.\n\nYou will see from the output of {\\tt summary} that there is lots of \nmissing data for the life history traits.\n\n\\section{Exploring the data with boxplots}\n\nWe are interested in finding out whether the mean C value for species \nvaries predictably for different levels of life history traits (a \ntypical one-way ANOVA question). For example: \n\n\\begin{compactitem}\n\\item Do carnivores or herbivores have larger genome sizes?\n\\item Do ground dwelling mammals have larger or smaller genome sizes?\n\\end{compactitem}\n\nBefore we fit any models, we want to plot the data to see if the means \nwithin these groupings look different. We also want to check whether \nthe variance looks similar for each group: {\\it constant normal \nvariance}! A simple way is to look at box and whisker plots, showing \nthe median and range of the data:\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Use {\\tt plot(meanCvalue \\textasciitilde{} TrophicLevel, \n\tdata= mammals)} to generate a boxplot of the differences in genome \n\tsizes between trophic levels.\n\t\\item Looking at the plots, it is clear that there is more spread in \n\tthe data above the median than below. Create a new variable {\\tt \n\tlogCvalue} in the {\\tt mammals} data frame containing logged C \n\tvalues.\n\t\\item Create a boxplot of log C values within trophic groups.\n\t\\item Repeat the two plot commands to look at differences between \n\tground dwelling and other species.\n\\end{compactitem}\n\n\\section{Differences in means with barplots}\n\nBox and whisker plots show the median and spread in the data very \nclearly, but we want to test whether the means are different. This is \n$t$ test territory --- how different are the means given the standard \nerror --- so it is common to use barplots and standard error bars to \nshow these differences.\n\nWe're going to use some R code to construct a barplot by hand. We need \nto calculate the means and standard errors within trophic groups, but \nbefore we can do that, we need a new functions to calculate the \nstandard error of a mean:\n\n\\begin{lstlisting}\n\t\n# get standard error of the mean from a set of values (x)\n\nseMean <- function(x){\n\tx <- na.omit(x) # get rid of missing values\n\n\tse <- sqrt(var(x)/length(x)) # calculate the standard error\n\n\treturn(se) \t# tell the function to return the standard error\n}\n\n\\end{lstlisting}\n\t\nNow we can use the function {\\tt tapply}: it splits a variable up into \ngroups from a factor and calculates statistics on each group using a \nfunction.\n\\begin{lstlisting}\ntrophMeans <- tapply(mammals$logCvalue, mammals$TrophicLevel, FUN = \nmean, na.rm = TRUE)\n\nprint(trophMeans)\n\n Carnivore Herbivore Omnivore \n 1.085 1.197 1.236 \n\\end{lstlisting}\n\n\\begin{lstlisting}\ntrophSE <- tapply(mammals$logCvalue, mammals$TrophicLevel, FUN = seMean)\n\nprint(trophSE)\n\n Carnivore Herbivore Omnivore \n 0.03983 0.02206 0.01844 \n\\end{lstlisting}\n\nNow we have to put these values together on the plot:\n\n\\begin{lstlisting}\n# get the upper and lower limits of the error bars\nupperSE <- trophMeans + trophSE\nlowerSE <- trophMeans - trophSE\n\n# get a barplot\n# - this function can report where the middle of the bars are on the x-axis\n# - set the y axis limits to contain the error bars\n\nbarMids <- barplot(trophMeans, ylim=c(0, max(upperSE)), ylab = 'log C value (pg)')\n\n# Now use the function to add error bars\n# - draws arrows between the points (x0,y0) and (x1,y1)\n# - arrow heads at each end (code=3) and at 90 degree angles\n\narrows(barMids, upperSE, barMids, lowerSE, ang=90, code=3)\n\n\\end{lstlisting}\n\n\\begin{center}\n\t\\includegraphics[width=0.7\\textwidth]{TLbarplot.pdf}\n\\end{center} \n\nNow we need to draw all these pieces together into a script and get \nused to using them.\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Copy all the lines of code from this section into your script.\n\t\\item Run it and check you get the graph above.\n\t\\item Use the second two chunks as a model to plot a similar graph \n\tfor {\\tt GroundDwelling}. You should get something like the plot \n\tbelow.\n\\end{compactitem}\n\n\\begin{center}\n\t\\includegraphics[width=0.6\\textwidth]{GDbarplot.pdf}\t\n\\end{center}\n \n\\section{An alternative to barplots}\n\nThat is a lot of work to go through for a plot. Doing it the hard way \nuses some useful tricks, but one strength of R is that there is a huge \nlist of add-on packages that you can use to get new functions that \nother people have written.\n\nWe will use the {\\tt gplots} package to create plots of group means \nand confidence intervals. Rather than plotting the means $\\pm$ 1 \nstandard error, the option {\\tt p=0.95} uses the standard error and the \nnumber of data points to get 95\\% confidence intervals. The default \n{\\tt connect=TRUE} option adds a line connecting the means, which isn't \nuseful here. \n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Replicate the code below into your script and run it to get the \n\tplots below.\n\\end{compactitem}\n\n\\begin{lstlisting}\n#Load the gplots package\n> library(gplots)\n\n# Get plots of group means and standard errors\n> par(mfrow=c(1,2))\n> plotmeans(logCvalue ~ TrophicLevel, data=mammals, p=0.95, connect=FALSE)\n> plotmeans(logCvalue ~ GroundDwelling, data=mammals, p=0.95, connect=FALSE)\n\\end{lstlisting}\n\n\n\\begin{center}\n\t\\includegraphics[width = \\textwidth]{test.pdf}\t\n\\end{center}\n\n\\section{Analysis of variance}\n\nHopefully, those plots should convince you that there are differences \nin genome size between different trophic groups and between ground \ndwelling and other mammals. We'll now use a linear model to test \nwhether those differences are significant.\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Using your code from Chapter~\\ref{ch:regress} as a guide, \n\tcreate a linear model called {\\tt trophicLM} which models log C value \n\tas a function of trophic group.\n\t\\item Use {\\tt anova} and {\\tt summary} to look at the analysis of \n\tvariance table and then the coefficients of the model. \n\\end{compactitem}\n\nThe ANOVA table for the model should look like the one below: trophic \nlevel explains highly significant variation in genome size ($F= 7.22, \n\\textrm{df}=2 \\textrm{ and } 300, p =0.0009$). {\\it Note the style of \nreporting the result} - the statistic ($F$), degrees of freedom and $p$ \nvalue are all provided in support. It is common to contract this style \nto this: $F_{2,300}=7.22, p=0.0009$. \n\\begin{lstlisting}\n> anova(trophicLM)\n \n Analysis of Variance Table\n \n Response: logCvalue\n Df Sum Sq Mean Sq F value Pr(>F) \n TrophicLevel 2 0.83 0.413 7.22 0.00087 ***\n Residuals 300 17.18 0.057 \n ---\n Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 \n\n\\end{lstlisting}\n\nHowever, look at the sum of squares column. Of a total of $17.18+0.83 = \n18.01$ units of sums of squares, only 0.83 are explained by trophic \nlevel: $0.83/18.01 \\approx 0.046$ or 4.6\\%. This ratio is called \n$r^2$, a measure of explanatory power, and shows that, although the \nmodel is very significant, it isn't very explanatory. We care about \nexplanatory power or effect size, {\\it not} $p$ values.\n \nThe coefficients table for the model looks like this:\n\n\\begin{lstlisting}\n> summary(trophicLM)\n \n Call:\n lm(formula = logCvalue ~ TrophicLevel, data = mammals)\n \n Residuals:\n Min 1Q Median 3Q Max \n -0.5038 -0.1635 -0.0038 0.1511 0.9313 \n \n Coefficients:\n Estimate Std. Error t value Pr(>|t|) \n (Intercept) 1.0851 0.0335 32.38 < 2e-16 ***\n TrophicLevelHerbivore 0.1119 0.0396 2.83 0.00503 ** \n TrophicLevelOmnivore 0.1513 0.0399 3.80 0.00018 ***\n ---\n Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 \n \n Residual standard error: 0.239 on 300 degrees of freedom\n (76 observations deleted due to missingness)\n Multiple R-squared: 0.0459,\tAdjusted R-squared: 0.0396 \n F-statistic: 7.22 on 2 and 300 DF, p-value: 0.000866 \n \n\\end{lstlisting} \n\nIt shows the following:\n\n\\begin{compactitem}\n\n\t\\item The reference level (or intercept) is for carnivores. Their \n\tmean genome size is significantly different from zero - this is not \n\tan exciting finding!\n\n\t\\item The mean genome size for both herbivores and omnivores are both \n\tsignificantly different from carnivores. Both larger in fact: \n\therbivore mean genome size = $1.085 + 0.112 = 1.197$ and omnivore \n\tmean genome size = $1.085 + 0.151 = 1.236$. These are the same group \n\tmeans we found above.\n\n\t\\item The $r^2$ is shown and is the 4.6\\% we calculated above. The \n\t{\\it adjusted} $r^2$ reduces the raw $r^2$ to account for the number \n\tof variables included in the model. That 4.6\\% would be even less \n\timpressive if we needed 6 explanatory variables to get it\\ldots\n\n\t\\item The $F$ statistic, as in the ANOVA table above.\n\n\\end{compactitem}\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Repeat the analysis of variance above to look at the effects of \n\tground dwelling on genome size.\n\\end{compactitem}\n\n\\section{Model criticism}\n\nThe next question must be --- and actually, we should do this before \nwe go anywhere near the model summaries --- is the model appropriate to \nthe data.\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Using Chapter \\ref{ch:regress} to guide you, get the four model \n\tdiagnostic plots for the trophic level model on a single figure.\n\\end{compactitem}\n\nThe four plots are:\n\\begin{center}\n\t\\includegraphics[width=0.8\\textwidth]{modelDiag.pdf}\t\n\\end{center}\n\nNote that in regression, the predicted (or fitted) values from the \nmodel take a range along the relationship $y=a + bx$ (as we saw in the \nFigures \\ref{fig:DiagModDragon} \\& \\ref{fig:DiagModDamsel}). For ANOVA, \nthere are only a few predicted values --- one for each group mean. This \nmeans that the plots above look different but we are looking for the \nsame things: is there constant variance at each fitted value and are \nthe residuals normally distributed? The answer for this model looks to \nbe yes.\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Check the ground dwelling model in the same way.\n\\end{compactitem}\n \n\\section{Testing pairwise differences between levels}\n\nThe one thing that the trophic level model does not tell us is whether \nthere is a difference in genome size between omnivores and herbivores \n--- both are compared to carnivores, but not to each other. This is \nbecause of the multiple pairwise testing problem mentioned in Chapter \n\\ref{ch:t_F_tests} --- if you do \nlots of tests then you may find small $p$ values by chance and say \nsomething important is going on when it is just random chance. This is \ncalled a false positive or Type I error.\n \nWith a 95\\% confidence interval, there is a 5\\% chance of a false \npositive {\\it per test} but there are ways of getting a 5\\% chance \nacross a set (or family) of tests. For linear models, we can use \nTukey's Honest Significant Difference test. We have to convert the {\\tt \nlm} object into an {\\tt aov} object first.\n\n\\begin{lstlisting}\n\n> TukeyTroph <- TukeyHSD(aov(trophicLM))\n> print(TukeyTroph)\n\t\n Tukey multiple comparisons of means\n 95% family-wise confidence level\n \n Fit: aov(formula = trophicLM)\n \n $TrophicLevel\n diff lwr upr p adj\n Herbivore-Carnivore 0.11186 0.01863 0.2051 0.0139\n Omnivore-Carnivore 0.15128 0.05741 0.2452 0.0005\n Omnivore-Herbivore 0.03942 -0.03161 0.1104 0.3923\n \n\\end{lstlisting}\n\nThe table shows the following:\n\\begin{compactitem}\n\n\t\\item The differences between the three possible pairs and then the \n\tlower and upper bounds of the 95\\% confidence interval for the \n\tdifference and a $p$ value. \n\n\t\\item In each case, we want to know if the difference could be zero: \n\tdoes the 95\\% confidence interval for each pair include zero.\n\n\t\\item For the first two pairs, carnivores versus omnivores and \n\therbivores, the confidence intervals do not include zero, so they are \n\tsignificantly different. For the comparison between herbivores and \n\tomnivores, the interval does include zero (difference = 0.039, 95\\% \n\tCI's limits are -0.032 \\& 0.110), so these groups are not \n\tsignificantly different.\n\n\t\\item The $p$ values for the top two pairs are both larger (less \n\tsignificant) than in the summary table. The test has made it harder \n\tto find significant results.\n\n\\end{compactitem}\n\nYou can visualise these confidence intervals by plotting the Tukey \ntest. You have to tweak the graphics parameters to get a clean plot \nthough.\n\n\\begin{lstlisting}\n> par(las=1, mar=c(4,10,3,1))\n# las= 1 turns labels horizontal\n# mar makes the left margin wider (bottom, left, top, right)\n> plot(TukeyTroph)\n\\end{lstlisting} \n\nThe result should be:\n\n{\\centering \\includegraphics[width=0.8\\textwidth]{TukeyPLot.pdf} }\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Run the Tukey test in your script for both the trophic level \n\tand ground dwelling models.\n\\end{compactitem}\n\n\\section{Are the factors independent?}\n\nWe've looked at two models, using trophic level and ground dwelling. It \nis worth asking whether these are independent factors. What if, for \nexample, our herbivores are all big, ground dwellers? This is important \nto know because otherwise, a two-way ANOVA would not be appropriate. We \nwill look at interactions in Chapter \\ref{ch:MulExplInter}.\n\nOK, so we want to know whether the two factors are independent. This is \na job for the $\\chi^2$ test!\n\n\\subsection{The Chi-square test and count data}\n\nThe Chi-square test, also known as $\\chi^{2}$ test or chi-square test, \nis designed for scenarios where you want to statistically test how \nlikely it is that an observed distribution of values is due to chance. \nIt is also called a ``goodness of fit'' statistic, because it measures \nhow well the observed distribution of data fits with the distribution \nthat is expected if the variables of which measurements are made are \nindependent. In our mammals example below, the two variables are \ntrophic level and ground dwelling.\n\nNote that a $\\chi^{2}$ test is designed to analyze categorical data. \nThat is the data have been counted (count data) and divided into \ncategories. It is not meant for continuous data (such as body weight, \ngenome size, or height). For example, if you want to test whether \nattending class influences how students perform on an exam, using test \nscores (from 0-100) as data would not be appropriate for a Chi-square \ntest. However, arranging students into the categories ``Pass'' and \n``Fail'' and counting up how many fall in each categories would be \nappropriate. Additionally, the data in a Chi-square table (see below) \nshould not be in the form of percentages -- only count data are \nallowed! \n\n\\subsubsection{The Chi-square test with the mammals data}\n\nWe can easily build a table for a Chi-square test on the mammals \ndata as follows:\n \n\\begin{lstlisting}\n> factorTable <- table(mammals$GroundDwelling, mammals$TrophicLevel)\n> print(factorTable) \n\n Carnivore Herbivore Omnivore\n No 26 45 64\n Yes 22 62 40\n\\end{lstlisting}\n\nNow let's run the test:\n\n\\begin{lstlisting}\n> chisq.test(factorTable)\n \n \tPearson's Chi-squared test\n \n data: factorTable \n X-squared = 8.12, df = 2, p-value = 0.01725\n\\end{lstlisting}\n\nThe ``{\\tt X-squared}'' value is the $\\chi^{2}$ {\\it test statistic}, akin to the \nt-value of the t-test or W value in the Wilcox test. \n\nThe $\\chi^{2}$ statistic is calculated as the sum of the quantity \n$$ \\frac{(\\mathrm{Observed} - \\mathrm{Expected})^2}{\\mathrm{Expected}} $$\nacross all the cells/categories in the table (so the sum would be over \n6 categories in our current mammals example).\n \n``Observed'' is the observed proportion of data that fall in a \ncertain category. For example, there are 26 species observed in the \n{\\tt Carnivore}, {\\tt No} category, and 22 in the {\\tt Carnivore}, {\\tt \nYes} category. \n\n``Expected'' is what count would be expected if the values in each \ncategory we truly independent. Each cell has its own expected value, \nwhich is simply calculated as the count one would expect in each \ncategory if the value were generated in proportion to the total number \nseen in that category. So in our example, the expected value for the \n{\\tt Carnivore}, {\\tt No} category would be\n$$26+22 \\mathrm{~(Total~number~of~carnivore~species)} \n\\times \\frac{26+45+64 \\mathrm{~(Total~number~in~the~''No''~category)}}{ \n26+22+45+62+64+40 \\mathrm{~(Total~number~of~species)}}$$ \n$$= 48 \\times \\frac{135}{259} = 25.02$$\n\nThe sum of all six (one for each cell in the table above) such \ncalculations would be the $\\chi^{2}$ value that R gave you through the \n{\\tt chisq.test()} above --- try it!\n\nNow back to the R output from the {\\tt chisq.test()} above. Why df = 2? \nThis is calculated as $DF = (r - 1) * (c - 1)$ where $r$ and $c$ are \nthe number of rows and columns in the $\\chi^{2}$ table, respectively. \nThe same principle you learned before applies here; you lose one degree \nof freedom for each new level of information you need to estimate: \nthere is uncertainity about the information (number of categories) in \nboth rows and columns, so you need to lose one degree of freedom for \neach. \n\nFinally, note that the p-value is significant --- we can conclude that the \nfactors aren't independent. From the table, carnivores can be either \nground dwelling or not, but herbivores tend to be ground dwelling and \nomnivores tend not to be. Ah well... it's OK. We will look at a better \nway to analyze these data using ``interactions'' in Chapter \n\\ref{ch:MulExplInter}.\n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Include and run the $\\chi^2$ test in your script.\n\\end{compactitem}\n\n\\section{Saving data}\n\nThe last thing to do is to save a copy of the mammal data, including \nour new column of log data, for use in later chapters. \n\n\\begin{compactitem}[$\\quad\\star$]\n\t\\item Use this code in your script to create the saved data in you \n\t{\\tt Data} directory :\n\\end{compactitem}\n\n\\begin{lstlisting}\nsave(mammals, file='../Data/mammals.Rdata')\n\\end{lstlisting}\t\n", "meta": {"hexsha": "ad7fea5418b54f83b9aa6b9b6e783f328eca122b", "size": 28824, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "archived/SilCompBioStat/anova.tex", "max_stars_repo_name": "mathemage/TheMulQuaBio", "max_stars_repo_head_hexsha": "63a0ad6803e2aa1b808bc4517009c18a8c190b4c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-10-12T13:33:14.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-12T13:33:14.000Z", "max_issues_repo_path": "archived/SilCompBioStat/anova.tex", "max_issues_repo_name": "OScott19/TheMulQuaBio", "max_issues_repo_head_hexsha": "197d710f76163469dfc7fa9d2d95ba3a739eccc7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "archived/SilCompBioStat/anova.tex", "max_forks_repo_name": "OScott19/TheMulQuaBio", "max_forks_repo_head_hexsha": "197d710f76163469dfc7fa9d2d95ba3a739eccc7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.1184022825, "max_line_length": 222, "alphanum_fraction": 0.7345267832, "num_tokens": 8013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519528132451417, "lm_q2_score": 0.7461389986757757, "lm_q1q2_score": 0.6356752189937401}} {"text": "\\section{Graphs}\r\n\\begin{comment}\r\n\tUsing adjacency list:\r\n\t\\code{graphs/graph_template_adjlist.cpp}\r\n\tUsing adjacency matrix:\r\n\t\\code{graphs/graph_template_adjmat.cpp}\r\n\tUsing edge list:\r\n\t\\code{graphs/graph_template_edgelist.cpp}\r\n\\end{comment}\r\n\t\\subsection{Single-Source Shortest Paths}\r\n\t\t\\subsubsection{Dijkstra}\r\n\t\t\t\\code{graphs/shortest_paths/dijkstra.cpp}\r\n\t\t\\subsubsection{Bellman-Ford}\r\n\t\t\t\\code{graphs/shortest_paths/bellman_ford.cpp}\r\n \\subsubsection{Shortest Path Faster Algorithm}\r\n \\code{graphs/shortest_paths/spfa.cpp}\r\n\t\\subsection{All-Pairs Shortest Paths}\r\n\t\t\\subsubsection{Floyd-Washall}\r\n\t\t\t\\code{graphs/shortest_paths/floyd_warshall.cpp}\r\n\t\\subsection{Strongly Connected Components}\r\n\t\t\\subsubsection{Kosaraju}\r\n \\code{graphs/scc/kosaraju.cpp}\r\n \\subsubsectionRed{Tarjan's Offline Algorithm}\r\n \\code{graphs/scc/tarjan.cpp}\r\n \\subsectionRed{Minimum Mean Weight Cycle}\r\n Run this for each strongly connected component\r\n \\code{graphs/min_mean_cycle.cpp}\r\n\t\\subsection{Biconnected Components}\r\n\t \\subsubsection{Bridges and Articulation Points}\r\n \\code{graphs/bcc/bridges_artics.cpp}\r\n \\subsubsection{Block Cut Tree}\r\n \\code{graphs/bcc/block_cut_tree.cpp}\r\n\t\t\\subsubsection{Bridge Tree}\r\n \\code{graphs/bcc/bridge_tree.cpp}\r\n\t\\subsection{Minimum Spanning Tree}\r\n\t\t\\subsubsection{Kruskal}\r\n \\code{graphs/mst/kruskal.cpp}\r\n\t\t\\subsubsection{Prim}\r\n \\code{graphs/mst/prim.cpp}\r\n\t\\subsectionRed{Euler Path/Cycle}\r\n \\subsubsectionRed{Euler Path/Cycle in a Directed Graph}\r\n \\code{graphs/euler_path.cpp}\r\n \\subsubsectionRed{Euler Path/Cycle in an Undirected Graph}\r\n \\code{graphs/euler_path_undirected.cpp}\r\n\t\\subsectionRed{Bipartite Matching}\r\n\t\t\\subsubsectionRed{Alternating Paths Algorithm}\r\n \\code{graphs/bipartite_matching/bipartite_matching.cpp}\r\n\t\t\\subsubsectionRed{Hopcroft-Karp Algorithm}\r\n \\code{graphs/bipartite_matching/hopcroft_karp.cpp}\r\n \\subsubsectionRed{Minimum Vertex Cover in Bipartite Graphs}\r\n \\code{graphs/bipartite_matching/bipartite_mvc.cpp}\r\n\t\\subsection{Maximum Flow}\r\n\t\t\\subsubsectionBlack{Edmonds-Karp} $O(VE^2)$\r\n\\begin{comment}\r\n\t\t\t\\code{graphs/max_flow/edmonds_karp.cpp}\r\n\\end{comment}\r\n\t\t\\subsubsection{Dinic}\r\n\t\t\t$O(V^2E)$\r\n\t\t\t\\code{graphs/max_flow/dinic.cpp}\r\n\t\t\\subsubsection{Push-relabel}\r\n\t\t\t$\\omega(VE+V^2\\sqrt{E})$, $O(V^3)$\r\n\t\t\t\\code{graphs/max_flow/push_relabel.cpp}\r\n \\subsubsection{Gomory-Hu (All-pairs Maximum Flow)}\r\n $O(V^3E)$, possibly amortized $O(V^2E)$ with a big constant factor.\r\n \\code{graphs/max_flow/gomory_hu_tree.cpp}\r\n \\subsection{Minimum Cost Maximum Flow}\r\n \\code{graphs/max_flow/mcst.cpp}\r\n \\subsubsection{Hungarian Algorithm}\r\n \\code{graphs/max_flow/hungarian.cpp}\r\n \\subsectionRed{Minimum Arborescence}\r\n Given a weighted directed graph, finds a subset of edges of minimum\r\n total weight so that there is a unique path from the root $r$ to each\r\n vertex. Returns a vector of size $n$, where the $i$th element is the\r\n edge for the $i$th vertex. The answer for the root is undefined!\r\n \\code{graphs/arborescence.cpp}\r\n \\subsectionRed{Blossom algorithm}\r\n Finds a maximum matching in an arbitrary graph in $O(|V|^4)$ time. Be\r\n vary of loop edges.\r\n \\code{graphs/blossom.cpp}\r\n \\subsection{Maximum Density Subgraph}\r\n Given (weighted) undirected graph $G$. Binary search density. If $g$ is\r\n current density, construct flow network: $(S, u, m)$, $(u, T,\r\n m+2g-d_u)$, $(u,v,1)$, where $m$ is a large constant (larger than sum\r\n of edge weights). Run floating-point max-flow. If minimum cut has empty\r\n $S$-component, then maximum density is smaller than $g$, otherwise it's\r\n larger. Distance between valid densities is at least $1/(n(n-1))$. Edge\r\n case when density is $0$. This also works for weighted graphs by\r\n replacing $d_u$ by the weighted degree, and doing more iterations (if\r\n weights are not integers).\r\n \\subsection{Maximum-Weight Closure}\r\n Given a vertex-weighted directed graph $G$. Turn the graph into a flow\r\n network, adding weight $\\infty$ to each edge. Add vertices $S,T$. For\r\n each vertex $v$ of weight $w$, add edge $(S,v,w)$ if $w\\geq 0$, or edge\r\n $(v,T,-w)$ if $w<0$. Sum of positive weights minus minimum $S-T$ cut is\r\n the answer. Vertices reachable from $S$ are in the closure. The\r\n maximum-weight closure is the same as the complement of the\r\n minimum-weight closure on the graph with edges reversed.\r\n \\subsection{Maximum Weighted Ind. Set in a Bipartite Graph}\r\n This is the same as the minimum weighted vertex cover. Solve this by\r\n constructing a flow network with edges $(S,u,w(u))$ for $u\\in L$,\r\n $(v,T,w(v))$ for $v\\in R$ and $(u,v,\\infty)$ for $(u,v)\\in E$. The\r\n minimum $S,T$-cut is the answer. Vertices adjacent to a cut edge are\r\n in the vertex cover.\r\n \\subsection{Synchronizing word problem}\r\n A DFA has a synchronizing word (an input sequence that moves all states\r\n to the same state) iff.\\ each pair of states has a synchronizing word.\r\n That can be checked using reverse DFS over pairs of states. Finding the\r\n shortest synchronizing word is NP-complete.\r\n \\subsection{Max flow with lower bounds on edges}\r\n % TODO: Test this!\r\n Change edge $(u,v,l\\leq f\\leq c)$ to $(u,v,f\\leq c-l)$. Add edge\r\n $(t,s,\\infty)$. Create super-nodes $S$, $T$. Let $M(u) = \\sum_{v}\r\n l(v,u) - \\sum_{v} l(u,v)$. If $M(u)<0$, add edge $(u,T,-M(u))$, else\r\n add edge $(S,u,M(u))$. Max flow from $S$ to $T$. If all edges from $S$\r\n are saturated, then we have a feasible flow. Continue running max flow\r\n from $s$ to $t$ in original graph.\r\n % TODO: Was there something similar for vertex capacities that we should add?\r\n \\subsection{Tutte matrix for general matching}\r\n Create an $n\\times n$ matrix $A$. For each edge $(i,j)$, $i c\\right\\},\n%\\]\n%where the critical value $c$ is chosen appropriately. \n%\\end{solution}\n%\\end{example}\n\n%-----------------------------\n\\subsection{The size (or significance level) of a test}\n\nWe would like to choose a critical region $C$ that minimises the probability of making both Type I and Type II errors. These are conflicting objectives, as illustrated by the following extreme cases.\n\\bit\n\\it If we choose $C=\\emptyset$ we will never reject $H_0$ (because the random sample never falls into $C$), so we never make Type I errors when $\\theta\\in\\Theta_0$ but always make Type II errors when $\\theta\\in\\Theta_1$.\n\\it If we choose $C=D$ we will always reject $H_0$ (because the random sample always falls into $C$), so we never make Type II errors when $\\theta\\in\\Theta_1$ but always make Type I errors when $\\theta\\in\\Theta_0$.\n\\eit\n\n% remark: status quo\n\\begin{remark}[Conservatie testing]\nThe null hypothesis represents the \\textit{status quo}. From a conservative standpoint, rejecting the status quo incorrectly (Type I error) is worse than retaining the status quo incorrectly (Type II error). As a result, hypothesis tests usually proceed in two stages:\n\\ben\n\\it find a set of tests for which $\\prob(\\text{Type I error})$ is bounded above by some acceptable value, then\n\\it choose one of these tests so that $\\prob(\\text{Type II error})$ is as small as possible.\n\\een\n\\end{remark}\n\n%We compute these probabilities over all possible realisations of the sample $\\mathbf{X}=(X_1,X_2,\\ldots,X_n)$. If $F(x;\\theta)$ is the common CDF of the $X_i$, it can be shown that for every $\\theta\\in\\Theta$, there exits a unique probability measure \n\\begin{definition}\nLet $\\mathbf{X}=(X_1,X_2,\\ldots,X_n)$ be a random sample and let $F_{\\theta}(x)$ be the common CDF of the component variables $X_i$. It can be shown that for every $\\theta\\in\\Theta$ there exits a unique probability measure $\\prob_\\theta$ on subsets of $\\R^n$ for which the component variables $X_i$ are independent and identically distribted according to $F_{\\theta}(x)$. This is called the \\emph{probability distribution induced by $\\mathbf{X}$} on the sample space and is denoted by $\\prob_{\\theta}(A)=\\prob_{\\theta}(\\mathbf{X}\\in A)$, which is the probability that %$\\mathbf{X}$\nthe random sample falls into the set $A\\subseteq\\R^n$ when the parameter value is $\\theta$.%: we often write $\\prob_{\\theta}(\\mathbf{X}\\in A)$ instead of $\\prob_{\\theta}(A)$.\n\\end{definition}\n\n% defn: size\n\\begin{definition}\nThe \\emph{size} of a critical region (also called the \\emph{significance level} of the test) is the maximum probability of making a Type I error. This is usually denoted by $\\alpha$,\n\\[\n\\alpha = \\max_{\\strut\\theta\\in\\Theta_0}\\ \\prob_{\\theta}(\\mathbf{X}\\in C).\n\\]\nIf $H_0$ is a simple hypothesis, say $H_0:\\theta=\\theta_0$, this reduces to $\\alpha = \\prob_{\\theta_0}(\\mathbf{X}\\in C)$.\n\\end{definition}\n\n\\begin{remark}[$p$-values]\nLet $T:D\\to\\R$ be a test statistic, let $C = \\{\\mathbf{x} : T(\\mathbf{x}) \\leq c\\}$ be a critical region and suppose that $\\mathbf{x}_{\\text{obs}}$ is the observed sample realisation. Then the \\emph{empirical size} (or $p$-value) of the test is defined to be\n\\[\np = \\max_{\\theta\\in\\Theta_0}\\,\\prob_{\\theta}\\big[ T(\\mathbf{X})\\leq T(\\mathbf{x}_{\\text{obs}})\\big].\n\\]\nIf $H_0$ is a simple hypothesis, say $H_0:\\theta=\\theta_0$, this reduces to \n\\[\np = \\prob_{\\theta_0}\\big[ T(\\mathbf{X})\\leq T(\\mathbf{x}_{\\text{obs}})\\big].\n\\] \nFor a test of size $\\alpha$, we reject $H_0$ if the empirical size satisfies $p\\leq\\alpha$.\n\\end{remark}\n\n\n%-----------------------------\n\\subsection{The power of a test} \n\nAmong all critical regions of size $\\alpha$ we would like to choose one that minimises the probability $\\prob_{\\theta}(\\mathbf{X}\\notin C)$ of making Type II errors for every $\\theta\\in\\Theta_1$, or equivalently a critical region that maximises $\\prob_{\\theta}(\\mathbf{X}\\in C)$ for every $\\theta\\in\\Theta_1$. \n\n\\begin{definition}\nLet $C$ be a critical region for testing $H_0:\\theta\\in\\Theta_0$ against $H_1:\\theta\\in\\Theta_1$. \nThe \\emph{power function} of the associated test is \n\\[\n\\gamma(\\theta) = \\prob_{\\theta}(\\mathbf{X}\\in C) \\quad\\text{which is defined for all $\\theta\\in\\Theta_1$.}\n\\]\nThe value $\\gamma(\\theta)$ is called the \\emph{power of the test to detect the alternative hypothesis at $\\theta\\in\\Theta_1$}.\n\\end{definition}\n\n% remark: beta\n\\begin{remark}\nThe probability of making Type II errors is often denoted by $\\beta$:\n\\begin{align*}\n\\beta(\\theta) \n\t& = \\prob_{\\theta}(\\mathbf{X}\\notin C) \\\\\n\t& = 1-\\gamma(\\theta).% \\quad \\text{for $\\theta\\in\\Theta_1$.}\n\\end{align*}\nTo maximise the power $\\gamma(\\theta)$ is to minimise the probability $\\beta(\\theta)$ of making a Type II error.\n\\end{remark}\n\n%% more powerful\n%Let $C_1$ and $C_2$ be two critical regions of size $\\alpha$, and let $\\gamma_1$ and $\\gamma_2$ respectively denote the power functions of the associated tests. If $\\gamma_1(\\theta) > \\gamma_2(\\theta)$ for all $\\theta\\in\\Theta_1$, we say that $C_1$ is a \\emph{more powerful test} than $C_2$. Later we prove that the \\emph{simple likelihood ratio test} is at least as powerful as any other test of a simple null hypothesis against a simple alternative.\n%\n\\begin{example}\nLet $X_1,X_2,\\ldots,X_8$ be a random sample from the $\\text{Poisson}(\\theta)$ distribution, where $\\theta>0$ is unknown. We reject the simple null hypothesis $H_0:\\theta=0.5$ in favour of the alternative $H_1:\\theta>0.5$ whenever the observed sum satisfies $\\sum_{i=1}^8 X_i \\geq 8$.\n\\ben\n\\it Compute the size of the test.\n\\it Compute the power of the test at $\\theta = 0.75$, $\\theta = 1.0$ and $\\theta = 1.25$.\n\\een\nUse the fact that if $X\\sim\\text{Poisson}(\\theta_1)$ and $Y\\sim\\text{Poisson}(\\theta_2)$ then $X+Y\\sim\\text{Poisson}(\\theta_1+\\theta_2)$.\n\\end{example}\n\n\\begin{solution}\nThe critical region of the test is \n\\[\nC = \\big\\{\\mathbf{x} : T(\\mathbf{x}) \\geq 8\\big\\}\n\\text{\\quad where\\quad}\nT(\\mathbf{X}) = \\sum_{i=1}^8 X_i.\n\\]\nBy the hint, if $X_i\\sim\\text{Poisson}(\\theta)$, then $T(\\boldX)\\sim\\text{Poisson}(8\\theta)$. This is our test statistic.\n\\ben\n\\it % << (i)\nUnder $H_0:\\theta=0.5$, we have $T\\sim\\text{Poisson}(4)$ so\n\\begin{align*}\n\\alpha\t= \\prob_{0.5}\\big(\\boldX\\in C\\big) \n\t\t& = \\prob(T\\geq 8) \\text{ where } T\\sim\\text{Poisson}(4) \\\\\n\t\t& = 1 - \\prob(T\\leq 7) \\text{ where } T\\sim\\text{Poisson}(4) \\\\\n\t\t& = 1 - 0.9489 = 0.0511 \\quad\\text{(from tables).}\n\\end{align*}\n\\it % << (ii)\nUnder $H_1:\\theta=0.75$, we have $T\\sim\\text{Poisson}(6)$ so\n\\begin{align*}\n\\gamma(0.75) = \\prob_{0.75}\\big(\\boldX\\in C\\big) \n\t\t\t& = \\prob(T\\geq 8)\\text{ where }T\\sim\\text{Poisson}(6) \\\\\n\t\t\t& = 1 - \\prob\\big(T\\leq 7)\\text{ where }T\\sim\\text{Poisson}(6) \\\\\n\t\t\t& = 1 - 0.7440 = 0.2560 \\quad\\text{(from tables).}\n\\end{align*}\nUnder $H_1:\\theta=1$, we have $T\\sim\\text{Poisson}(8)$ so\n\\begin{align*}\n\\gamma(1.0)\t= \\prob_{1.0}\\big(\\boldX\\in C\\big) \n\t\t\t& = \\prob(T\\geq 8)\\text{ where }T\\sim\\text{Poisson}(8) \\\\\n\t\t\t& = 1 - \\prob\\big(T\\leq 7)\\text{ where } T\\sim\\text{Poisson}(8) \\\\\n\t\t\t& = 1 - 0.4530 = 0.4570 \\quad\\text{(from tables).}\n\\end{align*}\nUnder $H_1:\\theta=1.25$, we have $T\\sim\\text{Poisson}(10)$ so\n\\begin{align*}\n\\gamma(1.25) = \\prob_{1.25}\\big(\\boldX\\in C;\\theta\\big)\n\t\t\t& = \\prob(T\\geq 8)\\text{ where } T\\sim\\text{Poisson}(10) \\\\\n\t\t\t& = 1 - \\prob(T\\leq 7)\\text{ where } T\\sim\\text{Poisson}(10) \\\\\n\t\t\t& = 1 - 0.2202 = 0.7798 \\quad\\text{(from tables).}\n\\end{align*}\nNotice that the power of the test to detect the alternative hypothesis \\emph{increases} as $\\theta$ moves away from the value $\\theta=0.5$ specified by the null hypothesis.\n\\een\n\\end{solution}\n\n\n", "meta": {"hexsha": "c517c8f1db132c3daecde1dacfb4c4b991e989ed", "size": 14796, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/09A_nhst.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/09A_nhst.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/09A_nhst.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 52.6548042705, "max_line_length": 581, "alphanum_fraction": 0.6865369019, "num_tokens": 5018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8354835350552604, "lm_q1q2_score": 0.6355110962409118}} {"text": "\\newpage\n\\section{Group extensions}\nToday our goal is to understand how addition of 2-digit numbers is equivalent to another abstract construction in group theory called group extensions.\nUsing this interpretation and some facts from group cohomology, we will eventually be able to answer the question,\n\\begin{q*}\n What are all the normalized, symmetric, 2-cocycles $c:\\bbz/10 \\times \\bbz/10 \\rightarrow \\bbz/10$?\n\\end{q*}\n\\begin{mdframed}\n \\adjustbox{scale=1,center}{%\n \\begin{tikzcd}\n &\\mbox{Adding 2-digit numbers}\n \\ar[ddddr, leftrightarrow]\n &\n \\\\\\\\\\\\\\\\\n \\mbox{\\color{white}2-cocycle condition}\n & & \\text{Group extensions}\n \\end{tikzcd}\n }\n\\end{mdframed}\n\n\n\n\n\n\\subsection{Review: Group theory}\nAll our groups will be abelian unless otherwise specified.\n\nA map between abelian groups $\\varphi: G_1 \\longrightarrow G_2$ is a group homomorphism if it satisfies\n\\begin{align*}\n \\varphi(a + b) = \\varphi(a) + \\varphi(b)\n\\end{align*}\n\n\\begin{qbox}(Practice problem)\n \\label{q:groupHomsExamples}\n Find all the group homomorphisms\n \\begin{enumerate}\n \\item $\\bbz \\longrightarrow \\bbz$,\n \\item $\\bbz \\longrightarrow \\bbz/n$,\n \\item $\\bbz/n \\longrightarrow \\bbz$,\n \\item $\\bbz/n \\longrightarrow \\bbz/n^2$,\n \\item $\\bbz/n^2 \\longrightarrow \\bbz/n$,\n \\end{enumerate}\n where $n$ is a positive integer.\n\\end{qbox}\n\n\\begin{qbox}\n Let $M$ be an abelian group. Describe all the group homomorphisms $ \\bbz/n \\longrightarrow M$, where $n$ is a positive integer.\n (This set will show up again when we discuss group cohomology.)\n\\end{qbox}\n\n\\begin{definition}\n The \\emph{kernel} of a group homomorphism $\\varphi: G_1 \\longrightarrow G_2$ is the set of elements $g \\in G_1$ such that $\\varphi(g) = 0$.\n \\begin{align*}\n \\ker \\varphi = \\set{ g \\in G_1 : \\varphi(g) = 0}\n \\end{align*}\n\\end{definition}\n\n\\begin{definition}\n The \\emph{image} of a group homomorphism $\\varphi: G_1 \\longrightarrow G_2$ is the set of elements $\\varphi(g) \\in G_2$ where $g \\in G_1$.\n \\begin{align*}\n \\im \\varphi = \\set{ \\varphi(g) : g \\in G_1}\n \\end{align*}\n\\end{definition}\n\n\n\\begin{qbox}\n Find the kernel and image of the group homomorphisms you found in Q.\\ref{q:groupHomsExamples}.\n\\end{qbox}\n\n\\begin{qbox}\n Which of the group homomorphisms in Q.\\ref{q:groupHomsExamples} are\n \\begin{enumerate}\n \\item injective (=one-to-one)?\n \\item surjective (=onto)?\n \\item isomorphisms (=one-to-one and onto)?\n \\end{enumerate}\n\\end{qbox}\n\n\\subsection*{Optional practice problems}\n\n\\begin{qbox}\n Show that the image of a group homomorphism $\\varphi: G_1 \\longrightarrow G_2$ is a subgroup of $G_2$.\n\\end{qbox}\n\n\\begin{qbox}\n Show that the kernel of a group homomorphism $\\varphi: G_1 \\longrightarrow G_2$ is a subgroup of $G_1$.\n\\end{qbox}\n\n\\begin{qbox}\n Show that for a group homomorphism $\\varphi: G_1 \\longrightarrow G_2$ we have $G_1 / \\ker \\varphi \\cong \\im \\varphi$.\n\\end{qbox}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\newpage\n\\subsection{Group extensions}\n\n\\begin{qbox}\n \\label{q:inclusionMaps}\n Consider the two inclusion maps,\n \\begin{align*}\n i_u: \\bbz/10 &\\longrightarrow \\bbz/100 &&& i_t: \\bbz/10 & \\longrightarrow \\bbz/100 \\\\\n b &\\longmapsto \\tens{0}\\units{b} &&& a &\\longmapsto \\tens{a}\\units{0}\n \\end{align*}\n Which of these two maps is a group homomorphism?\n\\end{qbox}\n\n\\begin{qbox}\n \\label{q:projectionMaps}\n Consider the two projection maps,\n \\begin{align*}\n p_u: \\bbz/100 &\\longrightarrow \\bbz/10 &&& p_t: \\bbz/100 &\\longrightarrow \\bbz/10 \\\\\n \\tens{a}\\units{b} &\\longmapsto {b} &&& \\tens{a}\\units{b} &\\longmapsto a\n \\end{align*}\n Which of these two maps is a group homomorphism?\n\\end{qbox}\n\n\n\\begin{qbox}\n \\label{q:SES1}\n For the $i$ and $p$ in Questions \\ref{q:inclusionMaps} and \\ref{q:projectionMaps} that are group homomorphisms, check that\n \\begin{enumerate}\n \\item $i$ is injective,\n \\item $p$ is surjective,\n \\item $\\im i = \\ker p$.\n \\end{enumerate}\n\\end{qbox}\n\n\\begin{definition}\n An \\emph{extension} of a group $K$ by $H$ is a group $G$ along with a pair of maps\n \\begin{align*}\n i: H \\longrightarrow G && p: G \\longrightarrow K\n \\end{align*}\n such that\n \\begin{enumerate}\n \\item $i$ is injective,\n \\item $p$ is surjective,\n \\item $\\im i = \\ker p$.\n \\end{enumerate}\n\\end{definition}\n\nThis is often written as a \\emph{short exact sequence}\n\\begin{equation*}\n \\begin{tikzcd}\n 0 \\ar[r] & H \\ar[r] & G \\ar[r] & K \\ar[r] & 0\n \\end{tikzcd}\n\\end{equation*}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\newpage\nQ.\\ref{q:SES1} is saying that the following is a SES\n\\begin{equation*}\n \\begin{tikzcd}\n 0 \\ar[r] & \\bbz/10 \\ar[r,\"i\"] & \\bbz/100 \\ar[r,\"p\"] & \\bbz/10 \\ar[r] & 0 \\\\\n & \\mbox{tens} \\ar[u, equal] & \\mbox{2-digit numbers} \\ar[u,equal] & \\mbox{units} \\ar[u, equal] &\n \\end{tikzcd}\n\\end{equation*}\n\nIn fact, every group extension arises in this manner.\nConsider an extension of abelian groups,\n\\begin{equation*}\n \\begin{tikzcd}\n 0 \\ar[r] & H \\ar[r,\"i\"] & G \\ar[r, \"p\"] & K \\ar[r] & 0.\n \\end{tikzcd}\n\\end{equation*}\ni.e.\n\\begin{enumerate}\n \\item $i$ is injective,\n \\item $p$ is surjective,\n \\item $\\im i = \\ker p$.\n\\end{enumerate}\n\nFor an element $a \\in H$, denote by $\\tens{a}\\units{0} \\in G$ the element $i(a)$.\nFor an element $b \\in K$, let $\\tens{0}\\units{b}$ be \\emph{some} element in $G$ such that $p(\\tens{0}\\units{b}) = b$.\nDefine $\\tens{a} \\units{b}$ to be the element $\\tens{a}\\units{0} + \\tens{0}\\units{b}$ in $G$.\n\n\\begin{qbox}\n Show that $p(\\tens{a}\\units{0}) = 0$ for any $a \\in H$.\n\\end{qbox}\n\n% \\begin{qbox}\n% % \\begin{enumerate}\n% % \\item\n% Check that $\\tens{a_1}\\units{0} + \\tens{a_2}\\units{0} = \\tens{a_1 + a_2}\\units{0}$.\n% % \\item Is it true that $\\tens{0}\\units{b_1} + \\tens{0}\\units{b_2} = \\tens{0}\\units{b_1 + b_2}$?\n% % \\end{enumerate}\n% \\end{qbox}\n\n\\begin{qbox}\n \\label{q:uniquenessOfDigits}\n Let $a_1$, $a_2$ be elements in $H$ and let $b_1$, $b_2$ be elements in $K$, show that\\hint{Apply $p$ to both sides of $[a_1][b_1] = [a_2][b_2]$.}\n\\begin{align*}\n [a_1][b_1] = [a_2][b_2] \\implies a_1 = a_2 \\mbox{ and } b_1 = b_2.\n\\end{align*}\n\\end{qbox}\n\n\\begin{qbox}\n \\label{q:existenceOfDigits}\n Show that for every element $g \\in G$, there exist unique elements $a \\in H$ and $ b \\in K $ such that $g = \\units{a}\\tens{b}$.\n \\hint{Look at $p(g)$ and $g - \\tens{p(g)}\\units{0}$.}\n\\end{qbox}\n", "meta": {"hexsha": "a3e321fce9a21d4d8965257c2efbeec6914b0060", "size": 6341, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "02.tex", "max_stars_repo_name": "apurvnakade/mc2019-group-cohomology", "max_stars_repo_head_hexsha": "14a7f5f0e2ae64f3ceaa602b50fa80269e3800a5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "02.tex", "max_issues_repo_name": "apurvnakade/mc2019-group-cohomology", "max_issues_repo_head_hexsha": "14a7f5f0e2ae64f3ceaa602b50fa80269e3800a5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.841825655188238, "lm_q2_score": 0.7549149978955811, "lm_q1q2_score": 0.635506812714875}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath,amssymb}\n\\newcommand\\set[1]{\\left\\{#1\\right\\}}\n\n\\begin{document}\n\\section{Probability method}\n\nTake a die. If the outcome is $5$ or $6$, roll again.\nIf the outcome is between $1$ and $4$, the number denotes the student's group.\n\nIf one group has reached cardinality $4$, it is excluded just like $5$ and $6$.\n\n\\textbf{Rationale:} As the result of $5$ or $6$ does not influence\nthe probability of outcomes $1$ to $4$, this defines a uniform discrete probability distribution.\n\n\\section{Probability space}\n\n\\[ \\Omega = \\operatorname{permutations}(\\set{1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4}) \\]\nwhere the groups are implicitly defined by indices 1--4, 5--8, 9--12 and 13--16.\n\\[ \\mathcal A = \\mathcal P(\\Omega) \\]\n\\[ \\mathbb P(A) = \\frac14^{|A|} \\]\n\n\\section{In a group of friends}\n\nLet Peter and Alice be Moritz' friends.\nLet $\\mathbb P[M=1]$ be the probability that Moritz is assigned to group 1.\nLet $\\mathbb P[P=N]$ and $\\mathbb P[A=N]$ be correspondingly.\n\n\\[ \\mathbb P[M=1] = \\frac14 \\]\n\\[ \\mathbb P[M=N] = \\frac14 \\]\n\n\\[ \\mathbb P[M=1, P=1] = \\frac14 \\cdot \\frac14 \\]\n\\[ \\mathbb P[M=N, P=N] = \\sum_{N=1}^4 \\frac14 \\cdot \\frac14 = \\frac14 \\]\n\n\\[ \\mathbb P[M=1, P=1, A=1] = \\frac14 \\cdot \\frac14 \\cdot \\frac14 = \\frac14^3 \\]\n\\[ \\mathbb P[M=N, P=N, A=N] = \\sum_{N=1}^4 \\frac14^3 = 4 \\frac14^3 \\]\n\n\\section{Blackboard solution}\n\nApproach: All children draw balls from one urn.\n\n\\[ \\Omega = \\set{\\text{set partitions of $\\{1, \\dots, 16\\}$ in $4$-tary subsets}} \\]\n$(k_1, \\dots, k_{16})$\n\nWe can arrange 16 elements ($16!$).\nIn each group, you can arrange the students as well ($(4!)^5$)\n\n\\[ \\frac{16!}{(4!)^5} = 2627625 \\]\n\nMoritz' group consists of members:\n\\[ \\set{M, F_1, F_2, R} \\]\n\nWe can $13$ choices for the remaining child.\n\\[ \\frac{13 {12 \\choose 4} {8 \\choose 4} {4 \\choose 4} \\cdot 4}{{16 \\choose 4} {12 \\choose 4} {8 \\choose 4} {4 \\choose 4}} = \\frac{1}{35} \\]\n\nYou can also draw probability trees.\nYou get one branch $\\left(\\frac14 \\cdot \\frac3{15} \\cdot \\frac2{14}\\right) \\cdot 4 = \\frac1{35}$\n\n\\end{document}\n", "meta": {"hexsha": "d02d1993345244af4c503912efc1fe13d178c659", "size": 2090, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "probability_theory_practicals/ex6/solution.tex", "max_stars_repo_name": "prokls/math-lecture-notes", "max_stars_repo_head_hexsha": "d1a94e128d13ce4399a9cc55323b2f8e0d9494fd", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2015-11-25T01:49:55.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-26T14:47:36.000Z", "max_issues_repo_path": "probability_theory_practicals/ex6/solution.tex", "max_issues_repo_name": "prokls/math-lecture-notes", "max_issues_repo_head_hexsha": "d1a94e128d13ce4399a9cc55323b2f8e0d9494fd", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2019-05-22T07:56:03.000Z", "max_issues_repo_issues_event_max_datetime": "2019-07-02T09:32:40.000Z", "max_forks_repo_path": "probability_theory_practicals/ex6/solution.tex", "max_forks_repo_name": "prokls/math-lecture-notes", "max_forks_repo_head_hexsha": "d1a94e128d13ce4399a9cc55323b2f8e0d9494fd", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2019-03-24T14:42:30.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-25T11:00:11.000Z", "avg_line_length": 34.262295082, "max_line_length": 140, "alphanum_fraction": 0.6488038278, "num_tokens": 773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.841825659156573, "lm_q2_score": 0.7549149923816048, "lm_q1q2_score": 0.6355068110688238}} {"text": "\\newpage\n\\subsection{Sample generators}\n\\label{sec:samplers}\nWhen rendering an image, Mitsuba has to solve a high-dimensional integration problem that involves the\ngeometry, materials, lights, and sensors that make up the scene. Because of the mathematical complexity\nof these integrals, it is generally impossible to solve them analytically --- instead, they\nare solved \\emph{numerically} by evaluating the function to be integrated at a large number of\ndifferent positions referred to as \\emph{samples}. Sample generators are an essential ingredient to this\nprocess: they produce points in a (hypothetical) infinite dimensional hypercube $[0,1]^{\\infty}$ that\nconstitute the canonical representation of these samples.\n\nTo do its work, a rendering algorithm, or \\emph{integrator}, will send many queries to the sample generator.\nGenerally, it will request subsequent 1D or 2D components of this infinite-dimensional ``point'' and map\nthem into a more convenient space (for instance, positions on surfaces). This allows it to construct\nlight paths to eventually evaluate the flow of light through the scene.\n\nSince the whole process starts with a large number of points in the abstract space $[0,1]^{\\infty}$,\nit is natural to consider different ways of positioning them. Desirable properties of a sampler are\nthat it ``randomly'' covers the whole space evenly with samples, but without placing samples too close\nto each other. This leads to such notions as \\emph{stratified sampling} and \\emph{low-discrepancy}\nnumber sequences. The samplers in this section make different guarantees on the quality of generated\nsamples based on these criteria. To obtain intuition about their behavior, the provided point plots\nillustrate the resulting sample placement.\n", "meta": {"hexsha": "97762b709593069620fb41ae38ba37f4d931031f", "size": 1757, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "mitsuba-af602c6fd98a/doc/section_samplers.tex", "max_stars_repo_name": "NTForked-ML/pbrs", "max_stars_repo_head_hexsha": "0b405d92c12d257e2581366542762c9f0c3facce", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 139, "max_stars_repo_stars_event_min_datetime": "2017-04-21T00:22:34.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-16T20:33:10.000Z", "max_issues_repo_path": "mitsuba-af602c6fd98a/doc/section_samplers.tex", "max_issues_repo_name": "NTForked-ML/pbrs", "max_issues_repo_head_hexsha": "0b405d92c12d257e2581366542762c9f0c3facce", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2017-08-15T18:22:59.000Z", "max_issues_repo_issues_event_max_datetime": "2019-07-01T05:44:41.000Z", "max_forks_repo_path": "mitsuba-af602c6fd98a/doc/section_samplers.tex", "max_forks_repo_name": "NTForked-ML/pbrs", "max_forks_repo_head_hexsha": "0b405d92c12d257e2581366542762c9f0c3facce", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 30, "max_forks_repo_forks_event_min_datetime": "2017-07-21T03:56:45.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-11T06:55:34.000Z", "avg_line_length": 73.2083333333, "max_line_length": 108, "alphanum_fraction": 0.8064883324, "num_tokens": 380, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256393148982, "lm_q2_score": 0.7549149868676283, "lm_q1q2_score": 0.6355067914482391}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\\markright{tfrpwv}\n\\section*{\\hspace*{-1.6cm} tfrpwv}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nPseudo Wigner-Ville time-frequency distribution.\n\\end{minipage}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[tfr,t,f] = tfrpwv(x)\n[tfr,t,f] = tfrpwv(x,t)\n[tfr,t,f] = tfrpwv(x,t,N)\n[tfr,t,f] = tfrpwv(x,t,N,h)\n[tfr,t,f] = tfrpwv(x,t,N,h,trace)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.5cm}\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty tfrpwv} computes the pseudo Wigner-Ville distribution of a\n discrete-time signal {\\ty x}, or the cross pseudo Wigner-Ville\n distribution between two signals. The pseudo Wigner-Ville\n distribution writes\n\\[PW_x(t,\\nu)=\\int_{-\\infty}^{+\\infty} h(\\tau)\\ x(t+\\tau/2)\\ x^*(t-\\tau/2)\\\ne^{-j2\\pi \\nu \\tau}\\ d\\tau.\\]\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty x} & signal if auto-PWV, or {\\ty [x1,x2]} if cross-PWV\n\t\t\t({\\ty Nx=length(x)}) \\\\ \n {\\ty t} & time instant(s) & {\\ty (1:Nx)}\\\\\n {\\ty N} & number of frequency bins & {\\ty Nx}\\\\\n {\\ty h} & frequency smoothing window, in the time-domain,\n {\\ty h(0)} being forced to {\\ty 1} & {\\ty window(odd(N/4))}\\\\ \n {\\ty trace} & if nonzero, the progression of the algorithm is shown\n & {\\ty 0}\\\\\n \\hline {\\ty tfr} & time-frequency representation \\\\\n {\\ty f} & vector of normalized frequencies\\\\\n\n\\hline\n\\end{tabular*}\n\\vspace*{.2cm}\n\nWhen called without output arguments, {\\ty tfrpwv} runs {\\ty tfrqview}.\n\\end{minipage}\n\\vspace*{1cm}\n\n{\\bf \\large \\sf Example}\n\\begin{verbatim}\n sig=fmlin(128,0.1,0.4); \n tfrpwv(sig);\n\\end{verbatim}\n\n\\newpage\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nall the {\\ty tfr*} functions.\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Reference}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n[1] T. Claasen, W. Mecklenbrauker ``The Wigner Distribution - A Tool for\nTime-Frequency Signal Analysis'' {\\it 3 parts} Philips\nJ. Res., Vol. 35, No. 3, 4/5, 6, pp. 217-250, 276-300, 372-389, 1980.\n\\end{minipage}\n", "meta": {"hexsha": "3670bf9af8e3ff325b64653aa901579491e208bd", "size": 2535, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/tfrpwv.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/tfrpwv.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/tfrpwv.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 28.8068181818, "max_line_length": 80, "alphanum_fraction": 0.608678501, "num_tokens": 939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256393148982, "lm_q2_score": 0.7549149758396752, "lm_q1q2_score": 0.6355067821646254}} {"text": "% Copyright 2018 Melvin Eloy Irizarry-Gelpí\n\\chapter{Dual Complex Numbers}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nAn exo-1 binion has the form\n\\begin{equation}\n a_{0} + a_{1} A + a_{2} W + a_{3} AW\n\\end{equation}\nThese follow from a parabolic Cayley-Dickson construct on the binions.\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Arithmetic Operations}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Multiplication}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Conjugate Operations}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Asterisk Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Quadrance}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Cloak Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Dagger Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Hodge Star}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Differential Operators}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...", "meta": {"hexsha": "d43762714164d063d030f11dad3e250cb4be1dc7", "size": 2090, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter/I.tex", "max_stars_repo_name": "meirizarrygelpi/plexifications", "max_stars_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter/I.tex", "max_issues_repo_name": "meirizarrygelpi/plexifications", "max_issues_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapter/I.tex", "max_forks_repo_name": "meirizarrygelpi/plexifications", "max_forks_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.5, "max_line_length": 80, "alphanum_fraction": 0.1947368421, "num_tokens": 223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9173026505426831, "lm_q2_score": 0.6926419958239132, "lm_q1q2_score": 0.6353623386464496}} {"text": "\\section{Leader Value Calculation}\n\\label{sec:leader-value-calc}\n\nThis section details how we determine whether a node is entitled to lead (under\nthe Praos protocol) given the output of its verifiable random function\ncalculation.\n\n\\begin{figure}\n \\emph{Values associated with the leader value calculations}\n \\begin{equation*}\n \\begin{array}{r@{~\\in~}lr}\n \\var{certNat} & \\{n | n \\in \\N, n \\in [0,2^{512})\\} & \\text{Certified natural value from VRF} \\\\\n \\var{f} & [0,1] & \\text{Active slot coefficient} \\\\\n \\sigma & [0,1] & \\text{Stake proportion}\n \\end{array}\n \\end{equation*}\n\\end{figure}\n\n\\subsection{Computing the leader value}\n\nThe verifiable random function gives us a 64-byte random output. We interpret\nthis as a natural number $\\var{certNat}$ in the range $[0,2^{512})$.\n\n\\subsection{Node eligibility}\n\nAs per \\cite{ouroboros_praos}, a node is eligible to lead when its leader value\n$p < 1 - (1 - f)^\\sigma$. We have\n\n\\begin{align*}\n p & < 1 - (1 -f)^\\sigma \\\\\n \\iff \\left(\\frac{1}{1-p}\\right) & < \\exp{(-\\sigma \\cdot \\ln{(1-f)})}\n\\end{align*}\n\nThe latter inequality can be efficiently computed through use of its Taylor\nexpansion and error estimation to stop computing terms once we are certain that\nthe result will be either above or below the target value.\n\nWe carry out all computations using fixed precision arithmetic (specifically, we\nuse 34 decimal bits of precision, since this is enough to represent the fraction\nof a single lovelace.)\n\nAs such, we define the following:\n\n\\begin{align*}\n p & = \\frac{\\var{certNat}}{2^{512}} \\\\\n q & = 1 - p \\\\\n c & = \\ln{(1 - f)}\n\\end{align*}\n\nand define the function \\textit{checkLeaderVal} as follows:\n\n\\begin{equation*}\n \\fun{checkLeaderVal}~\\var{certNat}~\\sigma~\\var{f} =\n \\left\\{\n \\begin{array}{l@{~}r}\n \\mathsf{True}, & f = 1 \\\\\n \\frac{1}{q} < \\exp{(-\\sigma \\cdot c)}, & \\text{ otherwise}\n \\end{array}\n \\right.\n\\end{equation*}\n", "meta": {"hexsha": "b70a264d00384f9ff54601c738fc78ea81f2f3b0", "size": 1934, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "shelley/chain-and-ledger/formal-spec/leader-value.tex", "max_stars_repo_name": "ilap/cardano-ledger-specs", "max_stars_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 108, "max_stars_repo_stars_event_min_datetime": "2019-03-24T02:26:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-30T05:27:16.000Z", "max_issues_repo_path": "shelley/chain-and-ledger/formal-spec/leader-value.tex", "max_issues_repo_name": "ilap/cardano-ledger-specs", "max_issues_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1266, "max_issues_repo_issues_event_min_datetime": "2019-03-18T20:23:28.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-04T12:50:51.000Z", "max_forks_repo_path": "shelley/chain-and-ledger/formal-spec/leader-value.tex", "max_forks_repo_name": "ilap/cardano-ledger-specs", "max_forks_repo_head_hexsha": "6474f68b24d05175fc3fd44a9bdfa95bda703a25", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 86, "max_forks_repo_forks_event_min_datetime": "2019-03-29T06:53:05.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-04T17:17:15.000Z", "avg_line_length": 31.7049180328, "max_line_length": 100, "alphanum_fraction": 0.6747673216, "num_tokens": 604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9086179043564153, "lm_q2_score": 0.6992544147913993, "lm_q1q2_score": 0.6353550809797328}} {"text": "\\subsection{Complex Roots}\r\nConsider the $k^{th}$ power map $p_k:z\\mapsto z^k$.\r\nWe can ``invert'' it by $\\sqrt[k]{z}=\\exp(k^{-1}\\log z)$.\r\nThis multi-valued function can be analysed analogously to how we analyse $\\log$.\r\nTake $I(n)=((n-1)\\pi/2,(n+1)\\pi/2)$ as usual and $U_{I(n)}$ as before equipped with the same $f_{I(n)}:U_{I(n)}\\to\\mathbb C$ we did for $\\log$.\r\nThen consider $g_{I(n)}(z)=\\exp(k^{-1}f_{I(n)}(z))$ and the function elements $G_{I(n)}=(g_{I(n)},U_{I(n)})$.\r\nNow $G_{I(n)}$ only depends on $n\\bmod k$, so WLOG we can think of $n\\in\\mathbb Z/k\\mathbb Z$.\r\nEverything else is similar to what we did before, and we can get $G_{I(n)}\\sim G_{I(m)}$ iff $n-m\\equiv 0,\\pm 1\\pmod{k}$.\r\nFurthermore, a similar gluing constructio defines a path-connected Hausdorff space $R_k$ and maps\r\n\\[\r\n \\begin{tikzcd}\r\n R_k\\arrow{r}{g}\\arrow[swap]{dr}{\\pi}&\\mathbb C_\\star\\arrow{d}{p_k}\\\\\r\n &\\mathbb C_\\star\r\n \\end{tikzcd}\r\n\\]\r\nSo $g$ gives what we want from the $k^{th}$ root.", "meta": {"hexsha": "4b5cc568cce8d1efc884caad4c0739e70197f59b", "size": 995, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/roots.tex", "max_stars_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_stars_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2/roots.tex", "max_issues_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_issues_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2/roots.tex", "max_forks_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_forks_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.1875, "max_line_length": 144, "alphanum_fraction": 0.6311557789, "num_tokens": 373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388083214156, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6352105437481881}} {"text": "\\section{Permutations}\n\n\\prob{https://artofproblemsolving.com/community/c5h1434000p8108658}{USAMO 2017\n P5}{M}{\n Let $m_1, m_2, \\ldots, m_n$ be a collection of $n$ positive integers, not\n necessarily distinct. For any sequence of integers $A = (a_1, \\ldots,\n a_n)$ and any permutation $w = w_1, \\ldots, w_n$ of $m_1, \\ldots, m_n$,\n define an $A$-inversion of $w$ to be a pair of entries $w_i, w_j$ with $i\n < j$ for which one of the following conditions holds:\n \\[\\begin{aligned}\n a_i \\ge w_i > w_j,\\quad\n w_j > a_i \\ge w_i,\\quad\n w_i > w_j > a_i\n \\end{aligned}\\]\n Show that, for any two sequences of integers $A = (a_1, \\ldots, a_n)$ and $B =\n (b_1, \\ldots, b_n)$, and for any positive integer $k$, the number of\n permutations of $m_1, \\ldots, m_n$ having exactly $k$ $A$-inversions is equal\n to the number of permutations of $m_1, \\ldots, m_n$ having exactly $k$ $B$-inversions.\n\n \\index[cat]{Permutations!USAMO 2017 P5}\n \\index[strat]{Bijection!USAMO 2017 P5}\n}\n\n\\begin{solution}\n Notice that if we take $B$ as a sequence with all elements greater than\n all $w_i$, then we have the $B$-inversions to be normal inversions wrt\n $M$. So we need to show that there exists a bijection between $A$-inversion and\n normal inversion. \\\\\n\n So we can either show that there for a permutation $w$ with $k$ normal\n inversions, there is a permutation $p$ with $k$ $A$-inversions. But we\n soon figure out it is pretty hard.\\\\\n\n If we try the other way, show that for every $w$ with $k$ $A$-inversions,\n there is a $p$ with the same $k$ normal inversions, and if we can show\n injectivity, we will be done. It turns out that this is much easier.\n\\end{solution}\n\n\n\\prob{https://artofproblemsolving.com/community/c6h271833p1472110}\n{ISL 2008 C2}{E}{\n Let $n \\in \\mathbb N$ and $A_n$ set of all permutations $(a_1, \\ldots,\n a_n)$ of the set $\\{1, 2, \\ldots , n\\}$ for which\n \\[k\\ |\\ 2(a_1 + \\cdots+ a_k), \\text{ for all } 1 \\leq k \\leq n.\\]\n Find the number of elements of the set $A_n$.\n\n \\index[cat]{Permutations!ISL 2008 C2}\n \\index[strat]{Induction!ISL 2008 C2}\n}\n\n\\begin{solution}\n First we try some smaller cases: $ |A_1| = 1$, $ |A_2| = 2$, $ |A_3| = 6$,\n $ |A_4| = 12$, which has a clear pattern. So we proceed with induction.\\\\\n\n With induction, we focus on $a_n$ only, it can have values eiher $n, 1$ or\n $\\frac{n+1}{2}$. But the later case is impossible, and so we only have two\n options for $a_n$, which gives us our desired inductive relation.\n\\end{solution}\n", "meta": {"hexsha": "2e687ce0596885b5f23d9dc04ce70122c979c86e", "size": 2569, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "combi/sec13_permutation.tex", "max_stars_repo_name": "M-Ahsan-Al-Mahir/BCS_Question_Bank", "max_stars_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 48, "max_stars_repo_stars_event_min_datetime": "2020-10-14T17:15:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-13T19:47:04.000Z", "max_issues_repo_path": "combi/sec13_permutation.tex", "max_issues_repo_name": "AnglyPascal/BCS_Question_Bank", "max_issues_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "combi/sec13_permutation.tex", "max_forks_repo_name": "AnglyPascal/BCS_Question_Bank", "max_forks_repo_head_hexsha": "83ff9b542999386ea182863e4f25f0b488d3984f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-10-15T08:59:33.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-27T15:19:26.000Z", "avg_line_length": 43.5423728814, "max_line_length": 90, "alphanum_fraction": 0.6605683145, "num_tokens": 890, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936484231889, "lm_q2_score": 0.8289388083214156, "lm_q1q2_score": 0.6352105437481881}} {"text": "\\documentclass[11pt, oneside]{article} \t% use \"amsart\" instead of \"article\" for AMSLaTeX format\n\\usepackage{fullpage}\n\\usepackage{graphicx}\n\\usepackage{amssymb}\n\n\\title{Brief Article}\n\\author{The Author}\n%\\date{}\t\t\t\t\t\t\t% Activate to display a given date or no date\n\n\\begin{document}\n%\\maketitle\n\n\\section{Mapping cylinder to 4 sides of cube} \n\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{cube_cylinder} \n\\end{center} \n \nGiven the cylinder $x^2 + y^2 = 1$ inscribed inside a $2 \\times 2 \\times 2$ cube\ncentered at the origin, we can project the cylindrical panorama onto the corresponding\nfour faces of the cube as follows.\nFor every pixel in the output image we cast a ray from the cube's center at $(0,0,0)$\nthrough the corresponding point on one of the cube's faces and determine where that\nray intersects the cylinder. We then map this point on the cylinder to the corresponding\npoint in the input image and (bilinearly) sample the input image at the point. \nThe resulting pixel color is assigned to the current output pixel.\n\nWe parameterize the ray emanating from \nthe cube's center $(0,0,0)$ through a point $(u,v,w)$ on a face of the cube as\n\\begin{equation}\n \\mathbf{r}(t) = (0,0,0) + t\\cdot(u,v,w). \\label{eq:r}\n\\end{equation}\nThis ray intersects the cylinder where\n\\begin{equation}\n (ut)^2 + (vt)^2 = 1.\n\\end{equation}\nSolving we get $t = 1/\\sqrt{u^2 + v^2},$ which we plug back into Equation~\\ref{eq:r} and\nget the cylindrical point\n\\begin{equation}\n \\mathbf{p} = (x,y,z) = \\frac{(u,v,w)}{\\sqrt{u^2 + v^2}}. \\label{eq:p-cyl}\n\\end{equation}\nIn cylindrical coordinates we have\n\\begin{eqnarray}\n\\mathbf{p} = \\left(\\theta, z\\right), \\ \\ \\ \n \\theta = \\mbox{atan2}\\left(y,x\\right) + \\pi/4. \\label{eq:theta}\n\\end{eqnarray}\nWe added $\\pi/4$ to $\\theta$ so that the left edge of the image corresponds to\nthe left edge of the cube's $x = 1$ face (instead of the middle of the face);\nThen we adjust for the appropriate image ``wrap-around:''\n\\begin{equation}\n\\theta' = \\left\\{\\begin{array}{ll}\n \\theta & \\theta \\geq 0 \\\\\n \\theta + 2\\pi & \\theta < 0\n \\end{array}\\right.\n\\end{equation}\nNow the left edge of the input image corresponds to the left edge of the output image.\nThe resulting pixel coordinate in the input image is\n\\begin{equation}\n(r,c) = \\left(\\frac{H}{2}\\left(z + 1\\right),\\ \\frac{W}{2\\pi}\\cdot \\theta'\\right).\n\\end{equation}\n\n\n\n\n\\begin{verbatim}\nfor row = 0 .. outImage.H-1 {\n for col = 0 .. outImage.W-1 {\n s = 4.0*col/outImage.W\n face = floor(s) // face = 0,1,2,3\n f = s - face // f = frac(s)\n if face == 0 { // (u,v,w) = point on cube\n u = 1\n v = 2*f - 1\n } else if face == 1 {\n u = 1 - 2*f\n v = 1\n } else if face == 2 {\n u = -1\n v = 1 - 2*f\n } else {\n u = 2*f - 1\n v = -1\n }\n w = 2.0*row/outImage.H - 1\n (x,y,z) = (u,v,w)/sqrt(u*u + v*v) // project onto cylinder\n theta = atan2(y,x) + pi/4; // cyl. coords, -3pi/4 <= theta <= 5pi/4\n if theta < 0 // map to [0,2*pi)\n theta += 2*pi\n r = inImage.H*(z + 1)/2.0 // map to input pixel\n c = imImage.W*theta/(2*pi)\n outImage(row,col) = inImage.sample(r,c)\n }\n}\n\\end{verbatim}\n\n\\section{Mapping sphere to 6 sides of cube} \n\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{cube_sphere} \n\\end{center} \n\nGiven the sphere $x^2 + y^2 + z^2= 1$ inscribed inside a $2 \\times 2 \\times 2$ cube\ncentered at the origin, we can project the spherical panorama onto the corresponding\nfour faces of the cube as follows.\nFor every pixel in the output image we cast a ray from the cube's center at $(0,0,0)$\nthrough the corresponding point on one of the cube's faces and determine where that\nray intersects the sphere. We then map this point on the sphere to the corresponding\npoint in the input image and (bilinearly) sample the input image at the point. \nThe resulting pixel color is assigned to the current output pixel.\n\nWe parameterize the ray emanating from \nthe cube's center $(0,0,0)$ through a point $(u,v,w)$ on a face of the cube \nand find where it intersects the unit sphere. We use the mapping defined at\n\\begin{verbatim}\nhttp://mathproofs.blogspot.com/2005/07/mapping-cube-to-sphere.html\n\\end{verbatim}\nyielding\n\\begin{eqnarray}\nx &=& u\\sqrt{1 - \\frac{v^2}{2} - \\frac{w^2}{2} + \\frac{v^2 w^2}{3}}, \\\\\ny &=& v\\sqrt{1 - \\frac{w^2}{2} - \\frac{u^2}{2} + \\frac{w^2 u^2}{3}}, \\\\\nz &=& w\\sqrt{1 - \\frac{u^2}{2} - \\frac{v^2}{2} + \\frac{u^2 v^2}{3}}.\n\\end{eqnarray}\nWe convert $(x,y,z)$ to the spherical coordinate $(\\theta, \\phi)$ where\nthe azimuthal angle is\n\\begin{equation}\n\\theta = \\mbox{atan2}\\left(y,x\\right)\n\\end{equation}\nand the elevation angle is\n\\begin{equation}\n\\phi = \\mbox{atan2}\\left(\\sqrt{x^2+y^2},\\ z\\right).\n\\end{equation}\n\n\\pagebreak\n\nWe divvy up the output image into $3 \\times 4 = 12$ squares where only 6 of the\nsquares actually map to the actual faces of the cube:\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{cube-coords} \n\\end{center} \nWe pick $(u,v,w)$ coordinates for the unused corners so that the image has a toroidal topology.\n\n\\begin{verbatim}\nfor row = 0 .. outImage.H-1 {\n for col = 0 .. outImage.W-1 {\n a = 4.0*col/outImage.W\n b = 3.0*row/outImage.H\n (i,j) = (floor(a),floor(b)) // map to one of 12 squares\n (s,t) = (a - i, b - j) // uvw interpolation values\n (u,v,w) = bilerp(corners[j][i]), s,t) // get cube coords w/in square\n (x,y,z) = cubeToSphereCoords(u,v,w) \n (theta,phi) = cartesianToSpherical(x,y,z)\n r = theta*inImage.W/(2*pi)\n c = (phi + pi/2)*inImage.H/pi\n outImage(row,col) = inImage.sample(r,c)\n }\n}\n\\end{verbatim}\n\n\n\\end{document} ", "meta": {"hexsha": "6308739fcdecf5e7cd6ecb06b0f4706cfd1194f9", "size": 5814, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pano_to_cube.tex", "max_stars_repo_name": "wcochran/pano_to_cube", "max_stars_repo_head_hexsha": "5d5e6852f6b2917f27edb578428ccf28de601e60", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2017-04-21T18:59:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T08:59:28.000Z", "max_issues_repo_path": "pano_to_cube.tex", "max_issues_repo_name": "wcochran/pano_to_cube", "max_issues_repo_head_hexsha": "5d5e6852f6b2917f27edb578428ccf28de601e60", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pano_to_cube.tex", "max_forks_repo_name": "wcochran/pano_to_cube", "max_forks_repo_head_hexsha": "5d5e6852f6b2917f27edb578428ccf28de601e60", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-02-20T06:33:08.000Z", "max_forks_repo_forks_event_max_datetime": "2019-03-14T06:45:31.000Z", "avg_line_length": 36.3375, "max_line_length": 97, "alphanum_fraction": 0.6357069143, "num_tokens": 1905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388083214156, "lm_q2_score": 0.7662936377487305, "lm_q1q2_score": 0.6352105348997152}} {"text": "\\documentclass[epsfig,10pt,fullpage]{article}\n\n\\newcommand{\\LabNum}{2}\n\\newcommand{\\CommonDocsPath}{../../common/docs}\n\\input{\\CommonDocsPath/preamble.tex}\n\n\\begin{document}\n\n\\centerline{\\huge OpenCL}\n~\\\\\n\\centerline{\\huge Laboratory Exercise \\LabNum}\n~\\\\\n\\centerline{\\large Accelerating Computer Vision Techniques}\n~\\\\\n\nThis exercise introduces you to accelerating computer vision techniques using the Intel FPGA SDK for OpenCL. \n\n\\section*{The Canny Edge-detection Technique}\n\n\\noindent\nIn this exercise, we will implement a variation of the \\textit{Canny edge detector}, which is \na widely-used edge-detection scheme in computer vision applications. Figures~\\ref{fig:tracks} and \\ref{fig:tracks_edges} \nshow a sample image that is provided as the input to a Canny edge detector, as well as the \nresulting edge-detected output image. \n\n~\\\\\n\\noindent\nThe Canny edge-detection algorithm involves five stages which are applied to the input image in \nsuccession. The details of these stages are given below. As well, we will see how the sample \ninput image from Figure~\\ref{fig:tracks} is transformed as it passes through \neach stage.\n\n~\\\\\n~\\\\\n\\begin{figure}[h]\n\\centering\n\\begin{minipage}[b]{0.475\\textwidth}\n\t\\includegraphics[width=\\textwidth]{figures/tracks.png}\n\t\\caption{Original image.}\n\t\\label{fig:tracks}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[b]{0.475\\textwidth}\n\t\\includegraphics[width=\\textwidth]{figures/tracks_edges.png}\n\t\\caption{Edge-detected image.}\n\t\\label{fig:tracks_edges}\n\\end{minipage}\n\\end{figure}\n\n\\pagebreak\n\n\\noindent\n\\section*{Stage 1: Grayscale Conversion}\n\nFigure~\\ref{fig:sample_stage1} shows the state of our sample image at the end of the grayscale \nconversion stage. This stage converts the input 24-bit bitmap color image (8 bits each for red, \ngreen, and blue) into an 8-bit grayscale image. The grayscale value at each pixel is calculated \nas the average of the three 8-bit color values of the original image.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_stage1_grayscale.png}\n \\end{center}\n \\caption{The sample input image after the grayscale conversion stage.}\n\t\\label{fig:sample_stage1}\n\\end{figure}\n\n\\noindent\n\\section*{Stage 2: Gaussian Smoothing}\n\nFigure~\\ref{fig:sample_stage2} shows the state of our sample image at the end of the gaussian \nsmoothing stage. In this stage, a gaussian filter is used to smooth out the image, by \nmodifying noisy pixels (pixels that are unlike their neighbouring pixels) to be more like \ntheir neighbours. Shown below is the 5 x 5 gaussian filter operation that is applied to the \nimage. Note that the * denotes convolution, A is the original image, and B is the resulting \nfiltered image. The effect of this operation is that each pixel gets assigned the weighted \naverage value of the 5 x 5 grid of pixels surrounding each pixel.\n\n\\[\nB = 1/159\n\\begin{bmatrix}\n 2 & 4 & 5 & 4 & 2 \\\\\n 4 & 9 & 12 & 9 & 4 \\\\\n 5 & 12 & 15 & 12 & 5 \\\\\n 4 & 9 & 12 & 9 & 4 \\\\\n 2 & 4 & 5 & 4 & 2\n\\end{bmatrix} * A\n\\]\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_stage2_gaussian.png}\n \\end{center}\n \\caption{The sample input image after the gaussian smoothing stage.}\n\t\\label{fig:sample_stage2}\n\\end{figure}\n\n\\noindent\n\\section*{Stage 3: Sobel Operator} \n\nFigure~\\ref{fig:sample_stage3} shows the state of our sample image at the end of the sobel operator stage. This stage overwrites each pixel with the overall intensity gradient at that pixel. To calculate the overall intensity gradient, the gradient is first calculated in the horizontal ($C_x$) and vertical ($C_y$) directions across the pixel, using the matrices below:\n\n\\[\nC_x = \n\\begin{bmatrix}\n -1 & 0 & 1 \\\\\n -2 & 0 & 2 \\\\\n -1 & 0 & 1 \n\\end{bmatrix} * B\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ C_y = \n\\begin{bmatrix}\n -1 & -2 & -1 \\\\\n 0 & 0 & 0 \\\\\n 1 & 2 & 1\n\\end{bmatrix} * B\n\\]\n\n~\\\\\n\\noindent\nThe magnitudes of the two gradients are then added to calculate the overall gradient intensity value for each pixel, resulting in the image C:\n\n\\[\nC = 0.5|C_x| + 0.5|C_y|\n\\]\n\n~\\\\\n\\noindent\nIn image C, which is the final output of this stage, the edges of the original image are highlighted as brighter pixels. Non-edges, which are areas with low intensity gradients, appear as darker pixels.\n\n~\\\\\n~\\\\\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_stage3_sobel.png}\n \\end{center}\n \\caption{The sample input image after the sobel operator stage.}\n\t\\label{fig:sample_stage3}\n\\end{figure}\n\n~\\\\\n\\noindent\nTo illustrate the effect of the sobel operator, let us examine different image boundaries that may exist in an image as shown in Figure~\\ref{fig:image_boundaries}. For each 3 x 3 image shown, we can use the sobel operator to calculate the intensity gradient at the center pixel. Let us examine the vertical boundary example, where there is a boundary between darker pixels on the left side of the image, and brighter pixels on the right side. In this example, $|C_x|$ of the center pixel can be calculated as |\\{-1*2 + 0*98 + 1*181\\} + \\{-2*1 + 0*94 + 2*178\\} + \\{-1*6 + 0*91 + 1*184\\}| = 711. In the vertical direction, $|C_y|$ for this pixel is calculated to be 7. The intensity gradient for this pixel is high in the horizontal direction, and low in the vertical direction, which is to be expected along a vertical boundary. The total intensity gradient for this pixel, is $0.5*711 + 0.5*7 = 359$ which saturates the 8-bit grayscale channel to value 255. The sobel operator would therefore detect this boundary as a strong edge, and the center pixel would become bright with a value of 255.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.80]{figures/fig_image_boundaries.pdf}\n \\end{center}\n \\caption{Examples of boundaries in an image.}\n\t\\label{fig:image_boundaries}\n\\end{figure}\n\n~\\\\\n\\noindent\nIn the horizontal boundary case, we see that the vertical gradient is high ($|C_y|$ = 726) and the horizontal gradient is low ($|C_x|$ = 2). For the diagonal boundary case, the gradient is high in both directions with $|C_x|$ = 516 and $|C_y|$ = 552. Finally, for the non-boundary case, the gradients are low in both directions with $|C_x|$ = 4 and $|C_y|$ = 36. As the gradients are low, the center pixel would become dark with a value of $0.5*4 + 0.5*36 = 20$.\n\n\\noindent\n\\section*{Stage 4: Non-Maximum Suppression} \n\n\\noindent\nFigure~\\ref{fig:sample_stage4} shows the state of our sample image at the end of the non-maximum suppression stage. This stage aims to thin the thick and/or blurry edges that may have resulted from the sobel operator stage. Thick edges are problematic as many applications of edge detection benefit from the edges being as thin as possible. For example, to accurately calculate the surface area of an object, thin edges are desired as to not overlap with the surface. The non-maximum suppression stage thins the edges by removing the weaker (non-maximum) pixels of each edge, and keeping only the maxima. Figure~\\ref{fig:nonmaximum_suppression} shows the effect of non-maximum suppression on a sample image containing a blurry vertical line. Notice that the vertical line, which is originally three-pixels wide, becomes one-pixel wide. \n\n~\\\\\n~\\\\\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_stage4_nonmax_suppression.png}\n \\end{center}\n \\caption{The sample input image after the non-maximum suppression stage.}\n\t\\label{fig:sample_stage4}\n\\end{figure}\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.8]{figures/fig_nonmaximum_suppression.pdf}\n \\end{center}\n \\caption{The effect of nonmaximum suppression on a blurred vertical line.}\n\t\\label{fig:nonmaximum_suppression}\n\\end{figure}\n\n\\noindent\n\\section*{Stage 5: Hysteresis}\n\nFigure~\\ref{fig:sample_stage5} shows the end result of the hysteresis stage. The goal of the hysteresis stage is to remove pixels that do not belong to an edge and weak edges altogether. This stage uses two user-defined thresholds: the high threshold and the low threshold. The hysteresis algorithm examines each pixel to determine whether: \n\n\\begin{enumerate}\n\\item the pixel exceeds the high threshold, or\n\\item the pixel exceeds the low threshold value and there exists at least one adjacent pixel (horizontally, vertically, or diagonally) that exceeds the high threshold.\n\\end{enumerate}\n\nIf at least one of the two criteria are met, the pixel is preserved. Otherwise, the pixel is removed by turning it black.\n \n~\\\\\n~\\\\\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_stage5_hysteresis.png}\n \\end{center}\n \\caption{The sample input image after the hysteresis stage.}\n\t\\label{fig:sample_stage5}\n\\end{figure}\n\n\\newpage\n\\section*{Part I}\n\nWrite a C++-language program that implements the five stages of the canny edge detector as described in the preceding sections. Start with the skeleton code provided in \\textit{/design\\_files/part1/}, which contains functionality for loading and storing 24-bit color BMP image files. Once a 24-bit color image is loaded into memory, your code should transform the pixels according to the five stages, then store the resulting edge-detected image. The skeleton also contains code that measures the runtime of your program, which we will use to compare with the OpenCL version. Test your program on the sample BMP \nimages provided in \\textit{/design\\_files/}\n\n\\section*{Part II}\n\nBefore we proceed with implementing a Canny edge detector using OpenCL, let us devise an efficient memory architecture for\nstoring the pixel values as they undergo transformations at each Canny stage. Recall that you have two types of memory at your disposal: global memory outside \nthe FPGA, and local memories inside the FPGA. The host program provides input data to the accelerator by placing it in global memory.\nSince global memory accesses are slow, your accelerator should cache the required pixels into local memory before using them.\nThe question then, is how to configure the FPGA's local memory resources to best cache the pixels. \n\nTo determine the best memory configuration, let us consider the pixel usage pattern for box operations.\nNote that the Gaussian smoothing, Sobel operator, non-max suppression, and hysteresis stages of the detector are all \ntypes of box operations, as they work on a box (or frame) of pixels to determine each output pixel. \nFigure~\\ref{fig:box_operation} depicts a 3x3 box operation (such as the 3x3 sobel operation) on three adjacent pixels of a 10-pixel-wide image. \nNotice that there is significant overlap (6 pixels) in the frames of successive operations.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_box_operation}\n \\end{center}\n \\caption{A 3x3 box operation on three adjacent pixels of a 10-pixel-wide image.}\n\t\\label{fig:box_operation}\n\\end{figure}\n\nTo take advantage of the overlap in successive frames of box operations, we will use the FPGA's local memory resources to construct \nthe shift register design shown in Figure~\\ref{fig:shift_register}. The figure shows the shift register design for a 3x3 box operation that\nworks on 10-pixel wide images, but the idea can be extended for arbitrary box sizes and image widths.\nOnce a sufficient number of pixels have been loaded, the shift register provides a new 3x3 frame for the box operation at every cycle by simply\nshifting in a new pixel and shifting out the oldest pixel which is no longer be used.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[scale = 0.85]{figures/fig_shift_register}\n \\end{center}\n \\caption{The shift register for 3x3 box operations on 10-pixel wide images providing 3 frames over 3 cycles.}\n\t\\label{fig:shift_register}\n\\end{figure}\n\nHow many shift registers will you require for your canny edge detector? How large should each of these shift registers be, given that\nyour circuit works on images that are 720 pixels wide? How many pixels must be loaded into each shift register before the corresponding box operation\ncan start? Using such a shift register design requires you to zero pad the input image before \nshifting in its pixels, to properly operate on the boundaries of the image. \nWhy is zero padding necessary, and what is the necessary padding size for a given box size?\n\n\n\\section*{Part III}\n\nCreate an OpenCL application that implements the five stages of the canny edge detector.\nUse shift registers to hold the pixels as they undergo the Canny stages, and apply the box operations on the pixels\nwhile they are in flight through the shift registers. The shift registers should be the minimum length necessary to hold\nrequired pixels.\nWhen operating in its steady state, your accelerator should have a throughput of one pixel per clock cycle.\nThe accelerator is considered to be in its steady state when its shift registers are full. \nFor simplicity, do not zero pad the boundaries of the input image and accept some error along the boundary edge pixels.\nYour application must be able to operate on images that are 720 pixels wide with variable height.\nStart with the skeleton code provided in \\textit{/design\\_files/part3/}.\n\n~\\\\\nThe shift-register design methodology requires the use of Intel FPGA SDK for OpenCL's ability to infer shift registers from \nyour OpenCL code. This feature is described in Section \\textit{Inferring a Shift Register} of the document \\textit{Intel FPGA SDK for OpenCL Programming Guide}.\n\n\n\\input{\\CommonDocsPath/copyright.tex}\n\\end{document}\n", "meta": {"hexsha": "660472ab40961c66fa00d3d8c433ca14482ca390", "size": 13554, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lab2/doc/opencl_lab2.tex", "max_stars_repo_name": "fpgacademy/Lab_Exercises_OpenCL", "max_stars_repo_head_hexsha": "923005e0f727cfd922d3904ad21a47faa86c7a51", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lab2/doc/opencl_lab2.tex", "max_issues_repo_name": "fpgacademy/Lab_Exercises_OpenCL", "max_issues_repo_head_hexsha": "923005e0f727cfd922d3904ad21a47faa86c7a51", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lab2/doc/opencl_lab2.tex", "max_forks_repo_name": "fpgacademy/Lab_Exercises_OpenCL", "max_forks_repo_head_hexsha": "923005e0f727cfd922d3904ad21a47faa86c7a51", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.2, "max_line_length": 1093, "alphanum_fraction": 0.7599232699, "num_tokens": 3495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936324115012, "lm_q2_score": 0.8289388083214156, "lm_q1q2_score": 0.6352105304754787}} {"text": "\\section{Exercises in Simulation}\\label{S:xsSimulation}\n\\begin{ExerciseList}\n\\Exercise Suppose the continuous RV $X$ has PDF:\n\\[\nf_X(x) = \\left(\\pi(1+x^2) \\right)^{-1}\n\\]\nDevise an algorithm to transform samples from $\\uniform(0,1)$ RV to those from $X$. Present your answer as pseudo-code.\n\\Answer\nThis is nothing but the inversion sampler for the standard Cauchy RV $X$.\n\\end{ExerciseList}\n\n", "meta": {"hexsha": "368c7b9fa59257f79293e922a58befd0e831ca2b", "size": 393, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "matlab/csebook/ExsInSimulation.tex", "max_stars_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_stars_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-02-19T07:54:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-14T13:55:18.000Z", "max_issues_repo_path": "matlab/csebook/ExsInSimulation.tex", "max_issues_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_issues_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "matlab/csebook/ExsInSimulation.tex", "max_forks_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_forks_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2018-07-18T07:47:52.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-19T11:28:24.000Z", "avg_line_length": 32.75, "max_line_length": 119, "alphanum_fraction": 0.7353689567, "num_tokens": 114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8289387998695209, "lm_q2_score": 0.7662936377487304, "lm_q1q2_score": 0.635210528423082}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\n\\title{Debugging your Neural Nets and checking your Gradients}\n\\author{Rishit Dagli}\n\\date{April 2020}\n\n\\begin{document}\n\nNote this only contains major formulas and notes, the complete article could be found here - https://medium.com/@rishit.dagli/debugging-your-neural-nets-and-checking-your-gradients-f4d7f55da167\n\n\\maketitle\n\\section{Introduction}\nThis document contains a brief and all the major formulas discussed in my blog \"Debugging your Neural Nets and checking your Gradients\"\n\n\\maketitle\n\\section{Double sided derivatives}\n\nA formula derived to compute double sided derivative\n$$\\frac{f(\\theta + \\epsilon)-f(\\theta - \\epsilon)}{2 \\cdot \\epsilon} \\approx g(\\theta)$$\n\nWorking thee double sided derivative with example numbers\n$$\\frac{(1.01)^3 - (0.99)^3}{2 \\cdot 0.01} = 3.0001$$\n\nActual derivative for $f(\\theta ^ 3)$\n$$g(\\theta) = 3 \\cdot \\theta ^2 = 3$$\n\nOne sided derivative\n$$g(\\theta) = 3.0301$$\n\nComparing methods\n\\vspace{10}\n\n1. One sided derivative\n\nApproximation error = $0.0301$\n\\vspace{5}\n\n2. Double sided derivative\n\nApproximation error = $0.0001$\n\n\\maketitle\n\\section{Going a bit deeper}\n\n$$f^{'}=\\lim_{\\epsilon \\to 0} \\frac{f(\\theta + \\epsilon)-f(\\theta - \\epsilon)}{2 \\cdot \\epsilon} \\approx g(\\theta)$$ \n\niff $\\lim_{\\epsilon \\to 0}$\n\\vspace{5}\n\nif $\\epsilon \\neq 0$ then $error = O(\\epsilon ^2)$\n\\vspace{5}\n\nAgain $\\epsilon$ is a very small number of course less than 1, so $\\epsilon >> \\epsilon ^2$\n\nThe error function in case of double sided approach would be\n$$error = O(\\epsilon ^2)$$\n\nAnd for single sided approach it would be\n$$error = O(\\epsilon)$$\n\n\\maketitle\n\\section{Gradient Checking}\n\nConcatenate all $w^{[i]}$ and $b^{[i]}$ into one single huge vector called $\\theta$\n\nSo, now our cost function\n$$J(w^{[1]}, b^{[1]}, w^{[3]} ... w^{[n]}, b^{[n]}) = J(\\theta) \\hspace{20} \\cdot\\cdot\\cdot\\cdot\\cdot(*)$$\n\nNow you can also write \n$$J(\\theta) = J(\\theta_1, \\theta_2, \\theta_3 ... \\theta_n)$$\n\nExample for single iteration of $\\theta_5$\n$$d \\theta^{[5]} _{approx.} = \\frac{J(\\theta_1, \\theta_2 ... \\theta_5 + \\epsilon ... \\theta_n) - J(\\theta_1, \\theta_2 ... \\theta_5 - \\epsilon ... \\theta_n)}{2 \\cdot \\epsilon}$$\n\nConsidering the double sided discussion earlier\n$$d \\theta^{[5]} _{approx.} = \\frac{J(\\theta_1, \\theta_2 ... \\theta_5 + \\epsilon ... \\theta_n) - J(\\theta_1, \\theta_2 ... \\theta_5 - \\epsilon ... \\theta_n)}{2 \\cdot \\epsilon} \\approx d \\theta^{[i]} = \\frac{\\partial J}{\\partial \\theta_5}$$\n\nHaving this clear to us we can now repeat the same process not just for θ₅ but for all $i$ such that $i \\in (1, n)$ . So a pseudo Python code for this would look something like this-\n\n\\vspace{10}\nfor i in range(1, n+1):\n$$d \\theta^{[i]} _{approx.} = \\frac{J(\\theta_1, \\theta_2 ... \\theta_i + \\epsilon ... \\theta_n) - J(\\theta_1, \\theta_2 ... \\theta_i - \\epsilon ... \\theta_n)}{2 \\cdot \\epsilon}$$\n\n\\maketitle\n\\section{Approaching an outcome}\n\nComputing similarity\n$$\\frac{||d\\theta_{approx.} - d\\theta||_2}{||d\\theta_{approx.}||_2 + ||d\\theta||_2}$$\n\n\\maketitle\n\\section{Implementing in practice}\n\n\\begin{table}[]\n\\begin{tabular}{ll}\nDifference & Comments \\\\\n \\\\\n \\\\\n\\textless{}= 10\\textasciicircum{}\\{-7\\} & Keep up the good work, you are doing just wonderful! \\\\\n \\\\\n10\\textasciicircum{}\\{-5\\} & Okay, Double check components of my vectors \\\\\n \\\\\n & and check that none of the components are too large \\\\\n \\\\\n=\\textgreater 10\\textasciicircum{}\\{-3\\} & Something to worry about, there might be a bug \n\\end{tabular}\n\\end{table}\n\\end{document}\n", "meta": {"hexsha": "c69104c3446244af95edae26d0fbc1f076b2bbff", "size": 3796, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main.tex", "max_stars_repo_name": "Rishit-dagli/Debugging-Neural-Nets", "max_stars_repo_head_hexsha": "fcc9666c8f3b285b6b0bedf58df1f731246fe837", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-05-13T10:13:03.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-24T03:49:37.000Z", "max_issues_repo_path": "main.tex", "max_issues_repo_name": "Rishit-dagli/Debugging-Neural-Nets", "max_issues_repo_head_hexsha": "fcc9666c8f3b285b6b0bedf58df1f731246fe837", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "main.tex", "max_forks_repo_name": "Rishit-dagli/Debugging-Neural-Nets", "max_forks_repo_head_hexsha": "fcc9666c8f3b285b6b0bedf58df1f731246fe837", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-10-15T14:12:31.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-15T14:12:31.000Z", "avg_line_length": 34.8256880734, "max_line_length": 238, "alphanum_fraction": 0.6319810327, "num_tokens": 1170, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147275, "lm_q2_score": 0.847967764140929, "lm_q1q2_score": 0.6352018092514551}} {"text": "Essential programming concepts and data structures are described in this chapter. The rest of the thesis relies on these terms.\n\n\\section{Related Data Structures}\n\n\\subsection{Binary Search Tree}\n\nA binary search tree is a data structure that implements \\findop, \\insertop, and \\removeop operations on a set of keys. Key~$K$ can be of any type, that has a total order. Throughout the thesis, trees with distinct keys are discussed, although of course, it is not the necessary condition.\n\nA binary tree (not to be confused with a binary search tree) consists of nodes. A node~$N$ consists of a key (defined as~\\func{key}{N}) and two references to other nodes~-- left child (defined as~\\func{left}{N}) and right child (defined as~\\func{right}{N}). A reference may contain a link to an existing node or a special value $nil$, that means that the reference is empty. A node that contains a link to a child is called a \\emph{parent} of this child node. Nodes may contain other attributes as well, but those are not substantial for this explanation. A node that has no children (\\(\\func{left}{N} = nil \\land \\func{right}{N} = nil\\)) is called a \\emph{leaf}.\n\nFrom the perspective of the graph theory, the binary tree is a simple oriented acyclic graph, where vertices are represented as nodes and edges are represented as links between a node and its left and right children. Every vertex in such graph has at most one incoming edge, i.e. every node can have at most one parent. Moreover, only one node has no parent~-- this node is called the \\emph{root} of a binary tree. The length of the longest path from any leaf to the root is known as the \\emph{height} of a binary tree. The \\emph{subtree} with a root in $N$ defined as $N$ and a set of nodes that can be reached from $N$ by child links.\n\nA binary \\emph{search} tree (\\emph{BST}) is a binary tree that satisfies the following condition: for each $N$, subtrees with the root in \\func{left}{N} and with the root in \\func{right}{N} contain only nodes with keys, that are less or equal than \\func{key}{N} and larger or equal than \\func{key}{N}, respectively. Using this property, it is possible to implement a fast lookup of a key $K$ in a binary search tree (see \\Cref{alg:bst_find}).\n\n%\\begin{float}{tbp}\n\\begin{algorithm}[t]\n\\caption{Lookup in $BST$}\\label{alg:bst_find}\n\\begin{algorithmic}[1]\n \\Procedure{Find}{$root,K$}\\Comment{The node with key = $K$ or $nil$}\n \\State $node\\gets root$\n \\While{$node\\not= nil$}\n \\If{$K = \\func{key}{node}$}\n \\State \\textbf{return} $node$\n \\ElsIf{$K < \\func{key}{node}$}\n \\State $node \\gets \\func{left}{node}$\n \\Else \\Comment{$K > \\func{key}{node}$}\n \\State $node \\gets \\func{right}{node}$\n \\EndIf\n \\EndWhile\n \\State \\textbf{return} $nil$\\Comment{The node was not found}\n \\EndProcedure\n\\end{algorithmic}\n\\end{algorithm}\n%\\end{float}\n\nThe complexity of this search algorithm is~$\\mathcal{O}(\\func{height}{T})$ (assuming that key comparison takes~$\\mathcal{O}(1)$). Furthermore, two remaining operations of a binary search tree, \\insertop and \\removeop, are implemented in the same fashion, and both of those operations have complexity~$\\mathcal{O}(\\func{height}{T})$ as well. Their implementation is described in detail in~\\cite[p.~327]{sedgewick}.\n\nThe tree height can vary between $\\mathcal{O}(\\func{size}{T})$,\n and $\\mathcal{O}(\\log(\\func{size}{T}))$ in case of a complete binary tree~\\cite{complete_bt}, where~$\\func{size}{T}$ is the count of nodes in~$T$. BST will maintain optimal operation time only if its structure is close to a complete binary tree and the height is bounded by $c \\times \\log(\\func{size}{T})$, where $c$ is a constant factor greater or equal~$1$.\n\n To keep the height logarithmic, even in a worst-case scenario, the \\emph{tree rebalancing} has been invented. The idea is that a tree keeps track of its structure and if it is not optimal, then the rebalancing is applied to restore optimal structure. The rebalancing can be achieved with the \\emph{tree rotation}\\cite[p.~435]{sedgewick}~-- the operation, that swaps a node with its parent in a way, that preserves the $BST$ property.\n\n\n\\subsection{AVL Tree}\n\nAVL tree was invented in 1962 by Georgy Adelson-Velsky and Evgenii Landis\\cite{avl_tree}. It is a classic example of a self-balancing $BST$. In fact, the height of the AVL tree is never greater than $1.4405\\times \\log(\\func{size}{T}) - 0.3277$ \\cite[p.~460]{knuth3}.\n\nSelf-balancing is achieved with the following approach: every node holds the difference between the heights of its left and right subtrees; this difference is called \\emph{balance factor}.\n\\emph{AVL property} requires the balance factor of every node to be in range of values $-1$, $0$ and~$1$.\n\nAfter every operation, that modifies the tree structure~-- \\insertop and \\removeop, balance factors are updated.\nIf at any step the balance factor happens to be $-2$ or $2$, a rotation or a double rotation is applied.\nThe rotation adjusts the heights of the left and right subtrees and, consequently, restores the AVL property. The exact AVL tree implementation is described in \\cite[p.~458]{knuth3}.\n\n\\subsection{Splay Tree}\n\\label{sssec:splay}\nAnother approach on tree balancing is presented in the Sleator and Tarjan work\\cite{splay_tree}~-- ``The efficiency of splay trees comes not from an explicit structural constraint, as with balanced trees, but from applying a simple restructuring heuristic, called splaying, whenever the tree is accessed.'' Term \\emph{splaying} stands for the process of using rotations (similar to ones in the AVL tree) to bring the last accessed node to the root.\n\nSleator and Tarjan proved that by using this approach, all three basic operations (\\findop, \\insertop and \\removeop) have a logarithmic time bound. Another benefit of splaying is that the most frequently accessed items tend to gather near the root, therefore improving access speed, especially on skewed input sequences~-- the sequences, in which only a small number of items are accessed often while other items occur rarer. This property is exploited in the Splay eviction policy (\\Cref{sssec:spolicy}).\n\nEven though splay trees show several interesting theoretical properties, in practice they are outperformed by more conventional BSTs, like AVL or Red-Black tree\\cite{splay_overview} (the performance evaluation of \\numdbname (\\Cref{sec:secanalysis}) reaffirms this statement). This is due to the fact that in the splay tree the structure of the tree is altered on every operation, including find operation, while AVL, for instance, modifies the tree only during insertions and removals.\n\nThe typical use scenario for those data structures is a scenario, where a vast majority of operations is the search operation, while updates are not so often. AVL and Red-Black trees happen to be faster because they execute fewer instructions per find operation. Moreover, they do not make any writes to memory during the lookup, and, as a consequence, there is lower load on the memory bus and the system cache.\n\nFurther researches on splay trees were focused in the main on how to reduce the number of rotations during splaying. An extensive overview of those optimizations is provided in \\cite{splay_overview}. One of the described techniques, the partial splaying is a modification of a conventional splay tree, where every node contains a counter that denotes a total count of accesses to this node. As usual, splaying is performed on every access, but the node is splayed only until its access count is less than the access count of its parent.\n\nW. Klostermeyer showed that this modification does not gain any noticeable advantage over a standard splay tree \\cite{partial_splaying}. However, partial splaying and other derived modifications can have some interesting properties specifically in application to a numerical database. It will be discussed in \\Cref{ch:alt}.\n\n\\subsection{Hash Table}\n\nA hash table is another popular data structure that implements dictionary abstract data type. It uses the entirely different approach on item storage and lookup. A hash table allocates a contiguous array $A$, which size is bounded by the expected number of items to be stored, often multiplied by the \\mbox{\\emph{load~factor}~$\\alpha$}. The items are stored in $A$.\n\nFirstly, let’s look at the simplified case: the key $K$ that is used in a hash table is of an integer type. Having $A$, $K$ and the value $V$, associated with $K$, it is possible to use a remainder of the division of $K$ by the size of $A$ as an index in $A$. Then, $V$ will be stored in $A$ at this index. This approach would give the best performance possible, as the $V$ can be retrieved immediately and \\emph{the~lookup time does not depend on~the~total count of~items} in the hash table. However, since the $modulo$ operation has been used, there can be several keys that point at the same index in $A$. This circumstance is called \\emph{collision}.\n\nTo deal with a collision, it is necessary to store $K$ itself together with $V$, so that in case of a collision it would be possible to tell if the stored $V$ is actually associated with the $K$ or another $K'$, that collides with $K$. Secondly, one must pick a strategy on how to deal with the case when two different keys, $K_1$ and $K_2$, that point at the same index are inserted. There are two main approaches:\n\\begin{description}\n\\item[Separate chaining (open hashing)]-- each element in $A$ is a linked list (or another data structure), that stores all pairs $\\langle K, V\\rangle$ that collides.\n\\pagebreak\n\\item[Linear probing (closed hashing)]-- if during insertion of $K$ in $A$ at the index $i$ a collision occurs ($i$ is already occupied), a special function $F$ is used to determine the second index at which $K$ can be inserted. If it is also occupied, 3rd and all consequent positions, generated by $F$, are used to try to insert the element.\n\\end{description}\n\nThe approach described above can be generalized on keys of any type $T$. It is achieved with the help of a \\emph{hash function}. This function takes an argument of type $T$ and maps it to an integer, called \\emph{hash}. This function must satisfy two properties:\n\n\\begin{block-description}\n\\blockitem[Determinism]-- it should \\emph{always} map the same input to the same hash.\n\\blockitem[Uniformity]-- if used with a uniformly generated random sequence of objects on input, the hash function should produce a uniformly distributed sequence of hashes.\n\\end{block-description}\n\nIn \\cite[p.~464]{sedgewick} Sedgewick and Wayne provide the detailed explanation of collision avoidance strategies as well as general information about hash tables. More information about hash function properties and hash function construction is presented in \\cite{knott}.\n\n\\section{Thread Synchronization}\n\n\n\n\\subsection{Coarse-grained Locking}\n\\label{sec:cgl}\nA trivial way to parallelize a sequential data structure is to eliminate a concurrent access at all. It can be achieved with a single mutual exclusion lock~-- \\emph{mutex}.\nWhile a thread holds a mutex, no other threads can lock the same mutex.\n\nThe sequential data structure is wrapped into the helper type, that locks the mutex in the beginning of every operation and releases it in the end, so that only one thread can access the data structure at a time, no matter how many threads are involved.\nThis approach is called the \\emph{coarse-grained locking}, in contrast with the \\emph{fine-grained locking}, where many locks are used and each lock protects only a part of the data structure, so other threads can freely access other parts.\n\nPros of this approach is a very trivial implementation and the absence of any special requirements on the underlying data structure. However, coarse-grained locking is only suitable when a data structure has a support role in the program and is used occasionally. If the data structure is the key element of the application, then a single lock becomes the bottleneck in the software, drastically decreasing program scalability. In this case one should use more sophisticated parallelization approaches.\n\n\\subsection{Binning}\n\\label{sec:pre_bin}\nThe evolution of coarse-grained locking is the \\emph{binning}. The main drawback of the previous approach is that a single lock becomes the main point of contention between threads. One way to cope with this is to increase the number of locks. In contrast with the fine-grained locking, the binning does not involve any modifications of the underlying container.\n\nFirstly, the numbers of bins~-- independent data structure instances~-- is chosen. Then a mapping between the item domain and a bin number is introduced. The mapping should yield a uniform distribution of mapped values. Every item is stored only in its assigned bin. Every bin has its own mutex, therefore the access to every bin is serialized. But since items are mapped uniformly, it is expected to produce much less contention than in case of a single lock.\n\n\\subsection{Fine-grained Locking}\n\\label{sec:fgl}\nThe fine-grained locking usually offers better scalability, than the previously discussed approaches. Instead of a single lock, many mutexes are used simultaneously. Every mutex protects its part of data. The contention between threads is lower as it is unlikely that several threads will access the same portion of data at the same time.\n\nHowever, this is true only if every portion has the same probability of being accessed (like in a concurrent hash table). In some data structures, typically binary trees, there are some nodes that are accessed (and are locked before access) oftener than others, e.g. the root in a binary tree.\n\n\\emph{Substantial modifications} got to be made to the data structure to integrate the fine-grained locking. Sometimes the overhead added by this approach is so big, that it brings to naught any potential speed-up. Fine-grained locking is not a silver bullet, but usually it offers a reasonable trade-off between implementation complexity and application scalability.\n", "meta": {"hexsha": "f734d730b5cb1acf75079c310528f29e66d45819", "size": 14023, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Text/tex/1_Preliminaries.tex", "max_stars_repo_name": "metopa/bachelors_thesis", "max_stars_repo_head_hexsha": "7937368e8d34eb68b1e90a6097737d48ca72c174", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Text/tex/1_Preliminaries.tex", "max_issues_repo_name": "metopa/bachelors_thesis", "max_issues_repo_head_hexsha": "7937368e8d34eb68b1e90a6097737d48ca72c174", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Text/tex/1_Preliminaries.tex", "max_forks_repo_name": "metopa/bachelors_thesis", "max_forks_repo_head_hexsha": "7937368e8d34eb68b1e90a6097737d48ca72c174", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 118.8389830508, "max_line_length": 663, "alphanum_fraction": 0.7739428082, "num_tokens": 3317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677622198947, "lm_q2_score": 0.7490872131147275, "lm_q1q2_score": 0.6352018078124329}} {"text": "\\section{Classifying spaces of groups}\\label{classifying-g-bundles}\n%I learned this attitude towards classifying spaces from Segal.\n%An article is on the website.\n%\nThe constructions of the previous sections can be summarized in a single\ndiagram:\n\\begin{equation*}\n \\xymatrix{\n\t\\Cat\\ar[r]^{\\text{nerve}} & s\\Set\\ar[d]^{|-|}\\\\\n\t\\mathbf{Gp}\\ar@{^(->}[u]\\ar[r]_{B} & \\Top\n }\n\\end{equation*}\nThe bottom functor is defined as the composite along the outer edge of the\ndiagram. The space $BG$ for a group $G$ is called the \\emph{classifying space\nof $G$}. At this point, it is far from clear what $BG$ is classifying. The goal\nof the next few sections is to demystify this definition.\n\n\\begin{lemma}\\label{conjugation-homotopic}\n Let $G$ be a group, and $g\\in G$. Let $c_g:G\\to G$ via $x\\mapsto\n gxg^{-1}$. Then the map $Bc_g:BG\\to BG$ is homotopic to the identity.\n\\end{lemma}\n\\begin{proof}\n The homomorphism $c_g$ is a functor from $G$ to itself. It suffices to\n prove that there is a natural transformation $\\theta$ from the identity to\n $c_g$. This is rather easy to define: it sends the only object to the only\n object: we define $\\theta_\\ast:\\ast\\to \\ast$ to be the map given by\n $\\ast\\xrightarrow{g}\\ast$ specified by $g\\in \\Hom_G(\\ast,\\ast) = G$. In\n order for $\\theta$ to be a natural transformation, we need the following\n diagram to commute, which it obviously does:\n \\begin{equation*}\n\t\\xymatrix{\n\t \\ast\\ar[d]_1\\ar[r]^g & \\ast\\ar[d]^{gxg^{-1}}\\\\\n\t \\ast\\ar[r]_{g} & \\ast.\n\t }\n \\end{equation*}\n\\end{proof}\nGroups are famous for acting on objects. Viewing groups as categories allows\nfor an abstract definition a group action on a set: it is a functor $G\\to\n\\Set$. More generally, if $\\cc$ is a category, an action of $\\cc$ is a functor\n$\\cc\\xrightarrow{X}\\Set$. We write $X_c = X(c)$ for an object $c$ of $\\cc$.\n\\begin{definition}\n The ``translation'' category $X\\cc$ has objects given by\n $$\\mathrm{ob}(X\\cc) = \\coprod_{c\\in \\cc}X_c,$$\n and morphisms defined via $\\Hom_{X\\cc}(x\\in X_c, y\\in X_d) = \\{f:c\\to\n d:f_\\ast(x) = y\\}$.\n\\end{definition}\nThere is a projection $X\\cc\\to \\cc$. (For those in the know: this is a special\ncase of the Grothendieck construction.)\n\\begin{example}\n The group $G$ acts on itself by left translation. We will write\n $\\widetilde{G}$ for this $G$-set. The translation category $\\widetilde{G}G$\n has objects as $G$, and maps $x\\to y$ are elements $yx^{-1}$. This category\n is ``unicursal'', in the sense that there is exactly one map from one\n object to another object. Every object is therefore initial and terminal,\n so the classifying space of this category is trivial by the discussion at\n the end of \\S \\ref{classifying-space-properties}. We will denote by $EG$\n the classifying space $B(\\widetilde{G}G)$. The map $\\widetilde{G}G\\to G$\n begets a canonical map $EG\\to BG$.\n\\end{example}\nThe $G$ also acts on itself by right translation. Because of associativity, the\nright and left actions commute with each other. It follows that the right\naction is equivariant with respect to the left action, so we get a right action\nof $G$ on $EG$.\n\\begin{claim}\n This action of $G$ on $EG$ is a principal action, and the orbit projection\n is $EG\\to BG$.\n\\end{claim}\nTo prove this, let us contemplate the set $N(\\widetilde{G}G)_n$. An element is\na chain of composable morphisms. In this case, it is actually just a sequence\nof $n+1$ elements in $G$, i.e., $N(\\widetilde{G}G)_n = G^{n+1}$. The right\naction of $G$ is simply the diagonal action. We claim that this is a free\naction. More precisely:\n\\begin{lemma}[Shearing]\n If $G$ is a group and $X$ is a $G$-set, and if $X\\times^\\Delta G$ has the\n diagonal $G$-action and $X\\times G$ has $G$ acting on the second factor by\n right translation, then $X\\times^\\Delta G\\simeq X\\times G$ as $G$-sets.\n\\end{lemma}\n\\begin{proof}\n Define a bijection $X\\times^\\Delta G\\mapsto X\\times G$ via $(x,g)\\mapsto\n (xg^{-1},g)$. This map is equivariant since $(x,g)\\cdot h = (xh, gh)$,\n while $(xg^{-1}, g) \\cdot h = (xg^{-1}, gh)$. The element $(xh, gh)$ is\n sent to $(xh(gh)^{-1}, gh)$, as desired.\n The inverse map $X\\times G\\to X\\times^\\Delta G$ is given by $(x,g)\\mapsto\n (xg, g)$.\n\\end{proof}\nWe know that $G$ acts freely on $N(\\widetilde{G}G)_n$, soo a nonidentity group\nelement is always going to send a simplex to another simplex. It follows that\n$G$ acts freely on $EG$.\n\nTo prove the claim, we need to understand the orbit space. The shearing lemma\nshows that quotienting out by the action of $G$ simply cancels out one copy of\n$G$ from the product $N(\\widetilde{G}G) = G^n$. In symbols:\n$$N(\\widetilde{G} G)/G\\simeq G^n\\simeq (NG)_n.$$\nOf course, it remains to check the compatibility with the face and degeneracy\nmaps. We will not do this here; but one can verify that everything works out:\nthe realization is just $BG$!\n\nWe need to be careful: the arguments above establish that $EG/G \\simeq BG$ when\n$G$ is a finite group. The case when $G$ is a topological group is more\ncomplicated. To describe this generalization, we need a preliminary categorical\ndefinition.\n\nLet $\\cc$ be a category, with objects $\\cc_0$ and morphisms $\\cc_1$. Then we\nhave maps $\\cc_1\\times_{\\cc_0}\\cc_1\\xrightarrow{\\text{compose}}\\cc_1$ and two\nmaps (source and target) $\\cc_1\\to \\cc_0$, and the identity $\\cc_0\\to\\cc_1$.\nOne can specify the same data in any category $\\cd$ with pullbacks. Our\ninterest will be in the case $\\cd = \\Top$; in this case, we call $\\cc$ a\n``category in $\\Top$''.\n\n%The story described above works just as well if we were to consider functors\n%$X:\\cc\\to \\Top$ instead of $X:\\cc\\to \\Set$. Let $G$ be a topological group. Define $\\widetilde{G}$ to be a category in $\\Top$.\n%\n%An example is a topological group -- that's a category in $\\Top$.\nLet $G$ be a topological group acting on a space $X$. We can again define $XG$,\nalthough it is now a category in $\\Top$. Explicitly, $(XG)_0 = X$ and $(XG)_1 =\nG\\times X$ as spaces. The nerve of a topological category begets a simplicial\nspace. In general, we will have\n$$(N\\cc)_n = \\cc_1\\times_{\\cc_0}\\cc_1\\times\\cdots\\times_{\\cc_0}\\cc_1.$$\nThe geometric realization functor works in exactly the same way, so the\nrealization of a simplicial space gets a topological space.\n%\n%You need to put some mild topological conditions on the group for the theorem we proved above to be true.\n%Again, we have $\\widetilde{G}$ with $G$ acting on it.\n%Again, we let $EG = B\\widetilde{G}G$.\nThe above discussion passes through with some mild topological conditions on\n$G$ (namely, if $G$ is an absolute neighborhood retract of a Lie group); we\nconclude:\n\\begin{theorem}\n Let $G$ be an absolute neighborhood retract of a Lie group. Then $EG$ is\n contractible, and $G$ acts from the right principally. Moreover, the map\n $EG\\to BG$ is the orbit projection.\n\\end{theorem}\nA generalization of this result is:\n\\begin{exercise}\n Let $X$ be a $G$-set. Show that\n $$EG\\times_G X \\simeq B(XG).$$\n\\end{exercise}\n", "meta": {"hexsha": "e140429ab0866fbf6d5d942db54ab5e38a9442ea", "size": 7037, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "906/lec-58-bundles-classifying-spaces.tex", "max_stars_repo_name": "ichung/algtop-notes", "max_stars_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-04-26T15:00:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T22:47:06.000Z", "max_issues_repo_path": "906/lec-58-bundles-classifying-spaces.tex", "max_issues_repo_name": "ichung/algtop-notes", "max_issues_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-03-13T17:54:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-13T17:59:46.000Z", "max_forks_repo_path": "906/lec-58-bundles-classifying-spaces.tex", "max_forks_repo_name": "ichung/algtop-notes", "max_forks_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-21T18:15:11.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-13T17:38:04.000Z", "avg_line_length": 50.9927536232, "max_line_length": 127, "alphanum_fraction": 0.6983089385, "num_tokens": 2217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147276, "lm_q2_score": 0.8479677602988601, "lm_q1q2_score": 0.6352018063734105}} {"text": "% Preamble\n\\documentclass[11pt]{PyRollDocs}\n\\usepackage{textcomp}\n\n\\addbibresource{refs.bib}\n\n% Document\n\\begin{document}\n\n \\title{The Wusatowski Spreading PyRoll Plugin}\n \\author{Max Weiner}\n \\date{\\today}\n\n \\maketitle\n\n This plugin provides a spreading modelling approach with Wusatowski's formula for flat rolling, adapted on groove rolling by an equivalent rectangle approach.\n\n\n \\section{Model approach}\\label{sec:model-approach}\n\n \\subsection{Wusatowski's spread equation}\\label{subsec:wusatowski's-spread-equation}\n\n \\textcite{Wusatowski1969} proposed \\autoref{eq:wusatowski} for estimation of spreading in flat rolling,\n where $\\gamma = \\frac{h_1}{h_0}$ is the compression. $h$ and $b$ are height and width of the workpiece with the indices\n 0 and 1 denoting the incoming respectively the outgoing profile. $a$, $c$, $d$ and $f$ are correction\n coefficients for temperature, velocity, material and friction, respectively.\n\n \\begin{equation}\n \\beta = \\frac{b_1}{b_0} = a \\times c \\times d \\times f \\times \\gamma^{-w}\n \\label{eq:wusatowski}\n \\end{equation}\n\n \\noindent The velocity coefficient $c$ can be assumed as below in dependence on the velocity $v$.\n\n \\begin{equation}\n c = \\left(-0.002958 + 0.00341 \\gamma \\right) v + 1.07168 - 0.10431 \\gamma\n \\label{eq:velocity-coefficient}\n \\end{equation}\n\n $w$ is the spread exponent, many different expressions were given by various authors for its value.\n The original expression by \\textcite{Wusatowski1969} is given in \\autoref{eq:exponent}, where $R$ is the roll radius.\n\n \\begin{equation}\n w = 10^{ \\num{-1.269} \\left( \\frac{h_0}{2 R} \\right)^{\\num{0.56}} \\frac{b_0}{h_0} }\n \\label{eq:exponent}\n \\end{equation}\n\n \\subsection{Equivalent rectangle approach}\\label{subsec:equivalent-rectangle-approach}\n\n Wusatowskis spreading model~\\cite{Wusatowski1969} was originally built for flat rolling.\n A common approach for groove rolling is to calculate some equivalent rectangular profile to be able to use flat rolling models~\\cite{Hensel1978, Spittel1984}.\n \\autoref{fig:equivalent_rectangle} shows 3 variants of calculating an equivalent rectangle of a profile.\n\n \\begin{figure}\n \\centering\n \\includegraphics[width=\\linewidth]{equivalent_rectangle}\n \\caption{Three methods of defining an equivalent rectangle of an oval groove}\n \\label{fig:equivalent_rectangle}\n \\end{figure}\n\n The first variant is to keep the width constant and calculate the height $h'$ so that the cross section $A$ is\n equal:\n\n \\[\n h' = \\frac{A}{b}\n \\]\n\n The second variant is to keep the height constant and calculate the width $b'$ so that the cross section $A$ is\n equal:\n\n \\[\n b' = \\frac{A}{h}\n \\]\n\n Both represent the geometry of the profile poorly.\n A better way is to keep the aspect ratio equal as propose by \\textcite{Spittel1984}:\n\n \\[\n h'' = \\sqrt{\\frac{A h}{b}}\n \\]\n\n \\[\n b'' = \\sqrt{\\frac{A b}{h}}\n \\]\n\n This variant is used in the current implementation.\n So $h$ and $b$ in Wusatowski's model are replaced with $h''$ and $b''$.\n In the end, $b_1$ can be obtained from $b_1''$ by:\n\n \\[\n b_1 = \\frac{b_1'' h_1}{h_1''}\n \\]\n\n\n \\section{Usage instructions}\\label{sec:usage-instructions}\n\n The plugin can be loaded under the name \\texttt{pyroll\\_wusatowski\\_spreading}.\n\n An implementation of the \\lstinline{spread} hook on \\lstinline{RollPass} is provided,\n calculating the spread using the equivalent rectangle approach and Wusatowski's model.\n\n Several additional hooks on \\lstinline{RollPass} are defined, which are used in spread calculation, as listed in \\autoref{tab:hookspecs}.\n Base implementations of them are provided, so it should work out of the box.\n For \\lstinline{wusatowski_exponent} and \\lstinline{wusatowski_velocity_coefficient} the equations~\\ref{eq:exponent} and~\\ref{eq:velocity-coefficient} are implemented.\n The others default to \\num{1}.\n Provide your own hook implementations or set attributes on the \\lstinline{RollPass} instances to alter the spreading behavior.\n\n \\begin{table}\n \\centering\n \\caption{Hooks specified by this plugin. Symbols as in \\autoref{eq:wusatowski}.}\n \\label{tab:hookspecs}\n \\begin{tabular}{ll}\n \\toprule\n Hook name & Meaning \\\\\n \\midrule\n \\texttt{wusatowski\\_temperature\\_coefficient} & temperature correction coefficient $a$ \\\\\n \\texttt{wusatowski\\_velocity\\_coefficient} & velocity correction coefficient $c$ \\\\\n \\texttt{wusatowski\\_material\\_coefficient} & material correction coefficient $d$ \\\\\n \\texttt{wusatowski\\_friction\\_coefficient} & friction correction coefficient $f$ \\\\\n \\texttt{wusatowski\\_exponent} & spread exponent $w$ \\\\\n \\bottomrule\n \\end{tabular}\n \\end{table}\n\n \\printbibliography\n\n\\end{document}", "meta": {"hexsha": "8f96685ffb3fb5bcd0d1aa86af0cfbb2b0a3a1bb", "size": 5130, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/docs.tex", "max_stars_repo_name": "pyroll-project/pyroll-wusatowski-spreading", "max_stars_repo_head_hexsha": "902ab86d666b0e9533b36da3cdc9c6ebdef8af53", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "docs/docs.tex", "max_issues_repo_name": "pyroll-project/pyroll-wusatowski-spreading", "max_issues_repo_head_hexsha": "902ab86d666b0e9533b36da3cdc9c6ebdef8af53", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/docs.tex", "max_forks_repo_name": "pyroll-project/pyroll-wusatowski-spreading", "max_forks_repo_head_hexsha": "902ab86d666b0e9533b36da3cdc9c6ebdef8af53", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.3937007874, "max_line_length": 170, "alphanum_fraction": 0.6723196881, "num_tokens": 1366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677468516188, "lm_q2_score": 0.749087201911703, "lm_q1q2_score": 0.6352017868004504}} {"text": "\\section{Grammar as Network of FSA}\n\nGiven a EBNF grammar where every nonterminal has a unique rule. Any transition labelled with a nonterminal $B$ is a call to the automaton $M_B$.\n\n\\begin{description}\n \\item[Machine] The FSA of a nonterminal\n \\item[Automaton] the PDA that accepts $L(G)$\n \\item[Net] the set of all machines\n\\end{description}\n\nThe machine $M_A$ must be \\emph{normalized}, initial state $0_A$ must have no entering arcs.\n\n\\subsection{Followers of nonterminal}\nGiven a final state of a machine, the \\emph{set of followers} if the set of all the terminal characters that can follow exiting the machine (including $\\dashv$). Given a state $q_A$:\n\\[\n Ini(q_A) = Ini(L(q_A)) = \\{ a \\in \\Sigma | a\\Sigma^* \\cap L(q_A) \\ne \\emptyset \\}\n\\]\n\n\\textbf{Note}: $Ini$ cannot contain $\\epsilon$, it is empty iff $L(q_A)=\\{\\epsilon\\}$.\nTerminal $a$ is in $Ini(q_A)$ iff:\n\\begin{align*}\n & \\exists \\text{ arc } q_A \\xrightarrow{a} r_A \\\\\n & \\exists \\text{ arc } q_A \\xrightarrow{B} r_A \\land a \\in Ini(0_B) \\\\\n & \\exists \\text{ arc } q_A \\xrightarrow{B} r_A \\land B \\text{ is nullable } \\land a \\in Ini(r_A)\n\\end{align*}\n", "meta": {"hexsha": "f3e84599338c10222bac7a19b65202b243bbd084", "size": 1142, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "parsing/network-of-fsa.tex", "max_stars_repo_name": "TiberioG/FLC-cheatsheet", "max_stars_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-01-13T14:36:20.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-18T16:22:18.000Z", "max_issues_repo_path": "parsing/network-of-fsa.tex", "max_issues_repo_name": "TiberioG/FLC-cheatsheet", "max_issues_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "parsing/network-of-fsa.tex", "max_forks_repo_name": "TiberioG/FLC-cheatsheet", "max_forks_repo_head_hexsha": "d86e8ba9c80fece75ffccf47c273334b677f4934", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-05-21T11:05:53.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-17T14:59:50.000Z", "avg_line_length": 43.9230769231, "max_line_length": 182, "alphanum_fraction": 0.6830122592, "num_tokens": 364, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240895276223, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6351778942336018}} {"text": "\\chapter{Units and Conversions}\n\nAt this point, you are working with a lot of units: grams for weight,\njoules for energy, newtons for force, meters for distance, seconds for\ntime, etc. For each type of measurement, there are several different\nunits; for example, distance can be measured in feet, miles,\nand light-years.\n\n\\begin{mdframed}[style=important, frametitle={Some Equalencies}]\n\n\\begin{tabular}{r | l}\n \\hline\n \\multicolumn{2}{c}{\\textbf{Distance}}\\\\\n 1 mile & 1.6093 kilometers \\\\\n 1 foot & 0.3048 meters \\\\\n 1 inch & 2.54 centimeters \\\\\n 1 light-year & $9.461 \\times 10^{12}$ kilometers\\\\\n \\hline\n \\multicolumn{2}{c}{\\textbf{Volume}}\\\\\n 1 milliliter & 1 cubic centimeter \\\\\n 1 quart & 0.9461 liters \\\\\n 1 gallon & 3.7854 liters \\\\\n 1 fluid ounce & 29.6 milliliters \\\\\n \\hline\n \\multicolumn{2}{c}{\\textbf{Mass}}\\\\\n 1 pound & 0.4535924 kilograms\\\\\n 1 ounce & 0.4535924 grams\\\\\n 1 metric ton & 1000 kilograms \\\\\n \\hline\n \\multicolumn{2}{c}{\\textbf{Force}}\\\\\n 1 newton & 1 kilogram meter per sec$^2$\\\\\n \\hline\n \\multicolumn{2}{c}{\\textbf{Pressure}}\\\\\n 1 pascal & 1 newton per square meter \\\\\n 1 bar & 0.98692 atmosphere \\\\\n 1 pound per square inch & 6897 pascals \\\\\n \\hline\n \\multicolumn{2}{c}{\\textbf{Energy}}\\\\\n 1 joule & 1 newton meter \\\\\n 1 calorie & 4.184 joules \\\\\n 1 kilowatt-hour & $3.6 \\times 10^{6}$ joules \\\\\n\\end{tabular}\\index{units table}\n\n(You don't need to memorize these! Just remember that this page is here.)\n\n\\end{mdframed}\n\nIn the metric system, prefixes are often used to express a multiple. Here are the common prefixes:\\index{metric system!prefixes}\n\n\\begin{mdframed}[style=important, frametitle={Common Prefixes for Metric Units}]\n\n\\begin{tabular}{r | l}\ngiga & $\\times 10^{9}$\\\\\nmega & $\\times 10^{6}$\\\\\nkilo & $\\times 10^{3}$\\\\\nmilli & $\\div 10^{3}$\\\\\nmicro & $\\div 10^{6}$\\\\\nnano & $\\div 10^{9}$\\\\\n\\end{tabular}\n\n(These are worth memorizing.)\n\n\\end{mdframed}\n\n\\section{Conversion Factors}\n\nHere is a really handy trick to remembering how to do conversions\nbetween units.\\index{conversion factors}\n\nOften, you will a table like the one above, and someone will ask you\n``How many miles are in 0.23 light-years?'' You know that 1 mile = 1.6093\nkilometers and that 1 light-year is $9.461 \\times 10^{12}$ kilometers.\nHow do you do the conversion?\n\nThe trick is to treat the two parts of the equality as a fraction that equals 1. That is, you think:\n\n$$\\frac{1 \\text{ miles}}{1.6093 \\text{ km}} = \\frac{1.6093 \\text{ km}}{1 \\text{ miles}} = 1$$\n\nand\n\n$$\\frac{1 \\text{ light-years}}{9.461 \\times 10^{12} \\text{ km}} = \\frac{9.461 \\times 10^{12} \\text{ km}}{1 \\text{ light-years}} = 1$$\n\nWe call these fractions \\textit{conversion factors}.\n\nNow, your problem is\n\n$$0.23 \\text{ light-years} \\times \\textit{ Some conversion factors} = ? \\text{ miles}$$\n\nNote that when you multiply fractions together, things in the numerators can cancel with things in the denominator:\n\n$$\\left( \\frac{31\\pi}{47} \\right) \\left( \\frac{11}{37\\pi}\\right) = \\left(\\frac{31\\cancel{\\pi}}{47}\\right) \\left( \\frac{11}{37\\cancel{\\pi}}\\right) = \\left(\\frac{31}{47} \\right) \\left( \\frac{11}{37} \\right)$$\n\nWhen working with conversion factors, you will do the same with the units:\n\n\\begin{multline*}\n 0.23 \\text{ light-years} \\left( \\frac{9.461 \\times 10^{12} \\text{ km}}{1 \\text{ light-years}} \\right) \\left( \\frac{1 \\text{ miles}}{1.6093 \\text{ km}} \\right) = \\\\\n 0.23 \\text{ \\cancel{light-years}} \\left( \\times \\frac{9.461 \\times 10^{12} \\text{ \\cancel{km}}}{1 \\text{ \\cancel{light-years}}} \\right) \\left( \\frac{1 \\text{ miles}}{1.6093 \\text{ \\cancel{km}}}\\right) = \\frac{(0.23)(9.461 \\times 10^{12})}{1.6093} \\text{ miles}$$\n\\end{multline*}\n\n\\begin{Exercise}[title={Simple Conversion Factors}, label=simple_conversion_factors]\n\n How many calories are in 4.5 killowatt-hours?\n \n\\end{Exercise}\n\\begin{Answer}[ref=simple_conversion_factors]\n\n $$4.5 \\text{ \\cancel{kWh}} \\left( \\frac{3.6 \\times 10^{6} \\text{ \\cancel{joules}}}{1 \\text{ \\cancel{kWh}}} \\right) \\left( \\frac{1 \\text{ calories}}{4.184 \\text{ \\cancel{joules}}}\\right) = \\frac{(4.5)(3.6 \\times 10^6)}{4.184} = 1.08 \\times 10^6 \\text {calories}$$\n \n\\end{Answer}\n\n\\section{Conversion Factors and Ratios}\n\nConversion factors also work on ratios. For example, if you are told\nthat a bug is moving 0.5 feet every 120 milliseconds. What is that in\nmeters per second?\n\nThe problem, then is\n\n$$\\frac{0.5 \\text{ feet}}{120 \\text{ milliseconds}} = \\frac{\\text{? m}}{second}$$\n\nSo you will need conversion factors to replace the ``feet'' with ``meters'' and to replace ``milliseconds'' with ``seconds'':\n\n\\begin{multline*}\n\\left(\\frac{0.5 \\text{ \\cancel{feet}}}{120 \\text{ \\cancel{milliseconds}}}\\right) \\left( \\frac{0.3048 \\text{ meters}}{1 \\text{ \\cancel{feet}}} \\right) \\left( \\frac{ 1000 \\text{ \\cancel{milliseconds}}} {1 \\text{ second}}\\right) = \\frac{(0.5)(0.3048)(1000)}{120}\\text{ m/second}\n\\end{multline*}\n\n\\begin{Exercise}[title={Conversion Factors}, label=conversion_factors]\n\nThe hole in the bottom of the boat lets in 0.1 gallons every 2 minutes. How many milliliters per second is that?\n \n\\end{Exercise}\n\\begin{Answer}[ref=onversion_factors]\n\n \\begin{multline*}\n \\frac{0.1 \\text{ \\cancel{gallons}}}{2 \\text{ \\cancel{minutes}}}\n \\left( \\frac{3.7854 \\text{ \\cancel{liters}}}{1 \\text{ \\cancel{gallons}}} \\right)\n \\left( \\frac{1000 \\text{ milliliters}}{1\\text{ \\cancel{liters}}}\\right)\n \\left( \\frac{1 \\text{ \\cancel{minutes}}}{60 \\text{ seconds}} \\right) = \\\\\n \\frac{(0.1)(3.7854)(1000)}{(2)(60)} \\text{ ml/second} = 3.1545 \\text{ ml/second}\n \\end{multline*}\n \n\\end{Answer}\n\n\\section{When Conversion Factors Don't Work}\n\nConversion factors only work when the units being converted are\nproportional to each other. Gallons and liters, for example are\nproportional to each other: If you have $n$ gallons, you have $n\n\\times 3.7854$ liters.\n\nDegrees celsius and degrees farenheit are \\textit{not} proportional to\neach other. If your food is $n$ degrees celsius, it is $n \\times\n\\frac{9}{5} + 32$ degrees farenheit. You can't use conversion factors\nto convert celsius to farenheit.\n", "meta": {"hexsha": "19974b3b705688e8ce15bcc2e3b6331e1acfd9da", "size": 6097, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Modules/MatterEnergy/units_conversions-en_US.tex", "max_stars_repo_name": "rajivjhoomuck/sequence", "max_stars_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Modules/MatterEnergy/units_conversions-en_US.tex", "max_issues_repo_name": "rajivjhoomuck/sequence", "max_issues_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Modules/MatterEnergy/units_conversions-en_US.tex", "max_forks_repo_name": "rajivjhoomuck/sequence", "max_forks_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.335483871, "max_line_length": 275, "alphanum_fraction": 0.6808266361, "num_tokens": 2100, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195152660688, "lm_q2_score": 0.8652240947405564, "lm_q1q2_score": 0.6351778930274604}} {"text": "\\documentclass[twocolumn]{article}\n\\usepackage{amsmath}\n\n\\begin{document}\n\n\\section*{Proof of extremum}\n\\davidsays{I think this section should probably move to supplementary information, if we're going to send to Nature. It suddenly gets way more mathematical here, and we start introducing Fourier space, etc.}\n\nTo show that a uniform potential gives the highest deliverable capacity, we consider a potential of interaction between guest and substrate $\\V(\\vec r)$ that varies with space. In this proof we will actually make use of the Fourier transform of this potential:\n\\begin{align}\n \\Vk \\equiv \\iiint \\V(\\vec r) e^{-i\\vec k\\cdot \\vec r} d\\vec r\n\\end{align}\nThe Fourier transform of a uniform potential is a Dirac delta function $\\tilde{\\V}(\\vec k)\\propto\\delta(\\vec k)$. Therefore, to show that a uniform potential extremizes the deliverable capacity, we must show that the functional derivative of the deliverable capacity with respect to $\\Vk$ is zero for \\emph{nonzero} values of $\\vec k$, i.e.\n\\begin{align}\n \\frac{\\delta D}{\\delta \\Vk} &= 0, \\text{ if } \\vec k\\ne 0.\n\\end{align}\nWe note that this functional derivative may be non-zero for $\\vec k=0$ because we separately maximize with respect to the particular uniform potential $\\V$.\nThis means that\n\\begin{align}\n \\frac{\\delta N_H}{\\delta \\V(\\vec k)} &= -\\frac{\\delta N_L}{\\delta \\Vk}\n\\end{align}\nwhere $N_H$ and $N_L$ are the number of particles at the low and high pressure.\n\nBecause the chemical potential $\\mu$ varies monotonically with $N$ at fixed temperature, we can consider how the chemical potential varies as we change $\\Vk$ with the number of molecules held fixed. We demonstrate this using the cyclic chain rule, which shows us that\n\\begin{align}\n \\left(\\frac{\\delta N}{\\delta \\Vk}\\right)_{\\mu} &=\n -\\left(\\frac{\\delta \\mu}{\\delta \\V(\\vec k)}\\right)_{N}\n \\left(\\frac{\\partial N}{\\partial \\mu}\\right)_{\\Vk}.\n\\end{align}\nSince changing the chemical potential changes the number of molecules in the general case, if we can show that $\\left(\\frac{\\delta \\mu}{\\delta \\V(\\vec k)}\\right)_{N}=0$ then we will have shown that $\\left(\\frac{\\delta N}{\\delta \\V(\\vec k)}\\right)_{\\mu}=0$. Thus we consider\n\\begin{align}\n \\left(\\frac{\\delta \\mu}{\\delta \\tilde\\V(\\vec k)}\\right)_N\n &= \\left(\\frac{\\delta \\left(\\frac{\\partial F}{\\partial N}\\right)_{\\volume}}{\\delta \\Vk}\\right)_N\n \\\\\n &= \\left(\\frac{\\partial \\left(\\frac{\\delta F}{\\delta \\Vk}\\right)_{\\volume}}{\\partial N}\\right)_{\\volume}\n \\label{eq:dmudpot}\n\\end{align}\nwhere we have made use of the derivative relationship between $\\mu$ and the Helmholtz free energy $F$, and have then reorderd the functional and partial derivatives.\nLet us consider the interior derivative first. The derivative of the Helmholtz free energy with respect to the external potential $\\Vk$ just gives the number density:\n\\begin{align}\n \\frac{\\delta F}{\\delta \\Vk} &= \\rho(\\vec k)\n\\end{align}\nThe number density is itself homogeneous for any system that is stable in a fluid state at this density (i.e. does not spontaneously crystallize), and thus has a Fourier transform that is proportional to a Dirac $\\delta$-function. Thus the functional derivative $\\frac{\\delta F}{\\delta \\V(\\vec r)}$ is actually a uniform function.\nWe can insert this expression into Eq.~\\ref{eq:dmudpot} to find that\n\\begin{align}\n \\frac{\\delta \\mu}{\\delta \\Vk} &\\propto \\delta(\\vec k) \\\\\n \\frac{\\delta N}{\\delta \\Vk} &\\propto \\delta(\\vec k)\n\\end{align}\nThus the functional derivative of both the chemical potential and $N$ with regard to $\\V(\\vec r)$ are itself homogeneous. Since we already maximize $D$ with respect to the homogeneous component of the potential (i.e. $\\vec k=0$), the derivative of $D$ with respect to any change of potential is zero.\n\nThis demonstrates that a homogeneous potential leads to an extremum value of the deliverable capacity. This proof is insufficient, however, to show that it must be a true maximum.\n\n\\end{document}", "meta": {"hexsha": "a02de73967d4b666bf85d4e0a2bfece06090dc56", "size": 3971, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "gas-adsorption/supplemental-proof.tex", "max_stars_repo_name": "SimonEnsemble/thesis-pommerenck-1", "max_stars_repo_head_hexsha": "c546b981b0fa7cebbe80e32d45dee5e8714ea89c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-24T00:42:10.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-24T00:42:10.000Z", "max_issues_repo_path": "gas-adsorption/supplemental-proof.tex", "max_issues_repo_name": "SimonEnsemble/thesis-pommerenck-1", "max_issues_repo_head_hexsha": "c546b981b0fa7cebbe80e32d45dee5e8714ea89c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "gas-adsorption/supplemental-proof.tex", "max_forks_repo_name": "SimonEnsemble/thesis-pommerenck-1", "max_forks_repo_head_hexsha": "c546b981b0fa7cebbe80e32d45dee5e8714ea89c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-23T18:42:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-23T18:42:16.000Z", "avg_line_length": 74.9245283019, "max_line_length": 340, "alphanum_fraction": 0.7280282045, "num_tokens": 1083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.89181104831338, "lm_q2_score": 0.7122321842389469, "lm_q1q2_score": 0.6351765308686635}} {"text": "In this section, the governing equations of general dynamic problems in elastic-plastic solids are first written with a tensorial formalism based on the fourth-order stiffness tensor.\nThis framework leads on the one hand to the same characteristic structure as in \\cite{Raniecki,mandel_book}, so that well known results can be used.\nOn the other hand, it allows easy specialization to problems involving less than three space dimensions, such as the one-dimensional ones already treated in the literature.\n% This framework is more generic than the works mentioned in the introduction, which are formulated in matrix form based on the elastoplastic compliances, since it can be easily specialized to problems already treated in the literature.\n%\nSecond, the spectral analysis of the hyperbolic system written in an arbitrary direction is performed, which is motivated by the solution of Riemann problems in one space dimension in numerical schemes such as the Finite Volume Method (FVM) \\cite{Leveque,Toro}.\nThat approach furthermore avoids the seeking of bi-characteristics as proposed by \\textsc{Clifton} in order to build elastoplastic finite difference schemes \\cite{Clifton_thesis}.\nIt must be emphasized that the present work aims at highlighting a sufficient amount of information so as to enable numerical schemes to mimic the analytical behavior, and does not require the complete bi-characteristic structure.\n\n\\subsection{Governing equations}\nWe consider the isothermal deformation of a solid body of mass density $\\rho$ in the linearized geometrical framework.\nThe balance equation of linear momentum with neglected body forces, and the geometrical balance equations \\cite{Plohr,Gil_HE} are: \n\\begin{equation}\n \\label{eq:balance_equations}\n %\\left\\lbrace\n \\begin{aligned}\n & \\rho \\dot{\\vect{v}} - \\nablav \\cdot \\tens{\\sigma} = \\vect{0} \\\\\n & \\dot{\\tens{\\eps}} - \\nablav \\cdot \\(\\frac{\\vect{v}\\otimes \\tens{I} + \\tens{I} \\boxtimes \\vect{v}}{2}\\) = \\vect{0} \n \\end{aligned}\n%\\right.\n\\end{equation}\nwhere the operator $ \\boxtimes $ refers to the transpose on second and third indices of the classical tensor product, namely: $\\tens{I} \\boxtimes \\vect{v} = \\delta_{ik} v_j\\vect{e}_i \\otimes \\vect{e}_j \\otimes \\vect{e}_k$.\n%\nFurthermore, $\\tens{I}$, $\\vect{v}$, $\\tens{\\sigma}$ and $\\tens{\\eps}$ denote respectively the second-order identity tensor, the velocity vector, the Cauchy stress tensor and the linearized strain tensor.\nThe latter additively decomposes into an elastic strain $\\tens{\\eps}^e$ and a plastic strain $\\tens{\\eps}^p$ in the small strain case.\nAssuming Cartesian coordinates, system \\eqref{eq:balance_equations} can be written as:\n\\begin{equation}\n \\label{eq:conservative_form}\n \\drond{\\Ucb}{t} + \\drond{\\Fcb\\cdot \\vect{e}_i}{x_i}=\\tens{0}\n\\end{equation}\nwith the vector of conserved quantities $\\Ucb$ and the flux vectors $\\Fcb_i=\\Fcb\\cdot \\vect{e}_i$:\n\\begin{equation}\n \\label{eq:vectors}\n \\Ucb =\\matrice{\\rho\\vect{v} \\\\ \\tens{\\eps}} \\quad ; \\quad \\Fcb_i = \\matrice{-\\tens{\\sigma}\\cdot\\vect{e}_i\\\\-\\frac{\\vect{v}\\otimes\\vect{e}_i +\\vect{e}_i \\otimes\\vect{v} }{2} }\n\\end{equation}\nAlternatively, the introduction of an auxiliary vector of conserved quantities $\\Qcb$ allows rewriting equation \\eqref{eq:conservative_form} as a quasi-linear form by means of the chain rule \\cite{Trangenstein91}:\n\\begin{equation}\n \\label{eq:quasi-linear_form}\n \\drond{\\Qcb}{t} + \\Absf^i \\drond{\\Qcb}{x_i}=\\tens{0}\n\\end{equation}\nIn particular, by setting $\\Qcb=\\matrice{\\vect{v} \\\\ \\tens{\\sigma}}$, one writes:\n\\begin{equation}\n \\label{eq:matrix-quasi}\n \\Absf^i = \\(\\drond{\\Ucb}{\\Qcb}\\)^{-1}\\drond{\\Fcb_i}{\\Qcb} = -\\matrice{\\tens{0}^2 & \\frac{1}{\\rho}\\(\\frac{\\tens{I}\\otimes\\vect{e}_i + \\vect{e}_i\\otimes \\tens{I}}{2}\\)\\\\ \\Hbb\\cdot \\vect{e}_i & \\tens{0}^4},\n\\end{equation}\n$\\tens{0}^q$ being a $q$th-order zero tensor and $\\Hbb=\\drond{\\tens{\\sigma}}{\\tens{\\eps}}$ the fourth-order tangent modulus tensor.\nIt then appears that the characteristic structure of the hyperbolic problem, which is driven by the matrices $\\Absf^i$, depends on the nature of the deformation (\\textit{i.e. elastic or elastic-plastic}) through the tangent modulus.\n\n% On the other hand, following the Generalized Standard Material framework \\cite{GSM} a Helmholtz free-energy $\\psi$ depending on $\\tens{\\eps}$ and the set of additional internal variables $\\Vcb$ is assumed to govern the evolution of the microstructure.\nFollowing the Generalized Standard Material framework \\cite{GSM}, the elastic-plastic constitutive response is described by a Helmholtz free-energy $\\psi$, convex with respect to $\\tens{\\eps}$ and the set of additional internal variables $\\Vcb$, and a yield surface $f$ in the case of associative plasticity. \nAssuming elastic isotropy, reversible evolutions are governed by the elastic law derived from Helmholtz's free energy:\n\\begin{equation}\n \\label{eq:elastic_law}\n \\tens{\\sigma} = \\Cbb: \\tens{\\eps}^e = 2\\mu \\tens{\\eps}^e + \\lambda \\tr \\tens{\\eps}^e \\: \\tens{I} \n\\end{equation}\nin which $(\\lambda,\\mu)$ are Lam{\\'e}'s parameters.\nThe inverse constitutive law can also be written based on Young's modulus and Poisson's ratio $(E,\\nu)$:\n\\begin{equation} \n \\label{eq:elastic_inverse}\n \\tens{\\eps}^e = \\Cbb^{-1}:\\tens{\\sigma} = \\frac{1+\\nu}{E} \\tens{\\sigma} - \\frac{\\nu}{E} \\tr \\tens{\\sigma} \\: \\tens{I}\n\\end{equation}\n\n\nNext, restricting ourselves to isotropic hardening, the set $\\Vcb$ consists of the cumulated plastic strain $p$.\n%\nIn addition, the von-Mises yield surface is considered:\n\\begin{equation}\n \\label{eq:von-Mises_yield}\n f\\(\\tens{\\sigma},R \\)= \\sqrt{\\frac{3}{2}}\\norm{\\tens{s}} - \\(R(p)+\\sigma^y\\) \\leq 0 \\\\\n\\end{equation}\nwhere $\\tens{s}$ denotes the deviatoric part of the Cauchy stress tensor, $\\sigma^y$ is the yield stress in tension, and $-R(p)$ is the thermodynamical force conjugate to the cumulated plastic strain through some hardening law.\n% The evolution of the cumulated plastic strain follows from the consistency condition:\n% \\begin{equation}\n% %\\label{eq:consistance}\n% \\dot{f}=0 \\quad \\Leftrightarrow \\quad \\drond{f}{\\tens{\\sigma}}:\\tens{\\dot{\\sigma}} - \\drond{f}{R}\\:\\dot{R} =0\n% \\end{equation}\n% so that one writes:\n% \\begin{equation}\n% \\label{eq:consistance}\n% \\dot{R}=\\sqrt{\\frac{3}{2}}\\:\\frac{\\tens{s}:\\tens{\\dot{\\sigma}}}{\\norm{\\tens{s}}}\n% \\end{equation}\n% Then, the plastic flow rule:\n% \\begin{equation}\n% \\label{eq:plastic_flow_rule}\n% \\tens{\\dot{\\eps}}^p = \\dot{p}\\drond{f}{\\tens{\\sigma}}\n% \\end{equation}\nThe evolution of the plastic strain tensor and the cumulated plastic strain are governed by the following flow rule and hardening law respectively:\n\\begin{align}\n \\label{eq:plastic_flow_rule}\n & \\tens{\\dot{\\eps}}^p = \\lambda \\drond{f}{\\tens{\\sigma}} \\\\\n \\label{eq:hardening_rule}\n & \\dot{p} = -\\lambda \\drond{f}{R} %= \\lambda\n\\end{align}\nwhere the plastic multiplier $\\lambda$ and the yield surface obey the Kuhn-Tucker complementarity conditions:\n\\begin{equation}\n \\label{eq:Kuhn-Tucker}\n \\lambda \\geq 0 \\quad ;\\quad f \\leq 0 \\quad ; \\quad \\lambda f =0\n\\end{equation}\nalong with the consistency condition:\n\\begin{equation}\n \\label{eq:consistency}\n \\lambda \\dot{f}=0 %\\quad \\Leftrightarrow \\quad \\drond{f}{\\tens{\\sigma}}:\\tens{\\dot{\\sigma}} - \\drond{f}{R}\\:\\dot{R} =0\n\\end{equation}\n% From the above conditions, one deduces:\n% \\begin{equation}\n% %\\label{eq:consistance}\n% \\dot{f}=0 \\quad \\Leftrightarrow \\quad \\drond{f}{\\tens{\\sigma}}:\\tens{\\dot{\\sigma}} - \\drond{f}{R}\\:\\dot{R} =0\n% \\end{equation}\n% in such a way that:\n% \\begin{equation}\n% \\label{eq:consistance}\n% \\dot{R}=\\sqrt{\\frac{3}{2}}\\:\\frac{\\tens{s}:\\tens{\\dot{\\sigma}}}{\\norm{\\tens{s}}}\n% \\end{equation}\nTherefore, combining equations \\eqref{eq:plastic_flow_rule} and \\eqref{eq:hardening_rule}, the flow rule can be rewritten as:\n\\begin{equation}\n \\label{eq:plastic_strain_rate}\n \\tens{\\dot{\\eps}}^p = \\dot{p}\\:\\sqrt{\\frac{3}{2}}\\frac{\\tens{s}}{\\norm{\\tens{s}}}\n\\end{equation}\n\nGiven this description of internal processes, the thermodynamical framework then leads to the elastic-plastic constitutive equations during irreversible deformations by combining the elastic law \\eqref{eq:elastic_law}, the additive decomposition of the strain tensor and the plastic flow rule \\eqref{eq:plastic_strain_rate} \\cite{Simo}:\n\\begin{align}\n \\label{eq:elastoplastic_tangent}\n & \\tens{\\dot{\\sigma}} = \\Cbb^{ep}:\\tens{\\dot{\\eps}}=\\(\\Cbb - \\beta\\:\\tens{s}\\otimes\\tens{s} \\):\\tens{\\dot{\\eps}} \\\\\n \\label{eq:plastic_flow}\n % & \\beta = \\frac{6\\mu^2}{3\\mu +C}\\times\\frac{1}{\\tens{s}:\\tens{s}}\n & \\beta = \\frac{6\\mu^2}{3\\mu +R'}\\times\\frac{1}{\\tens{s}:\\tens{s}}\n\\end{align}\nwhere the elastoplastic tangent modulus $\\Cbb^{ep}$ can be decomposed into the elasticity tensor $\\Cbb$ and another part depending on the direction of the plastic flow.\nTherefore, plastic evolutions involve $\\Hbb \\equiv \\Cbb^{ep}$ while elastic ones involve $\\Hbb\\equiv\\Cbb$.\n%$\n% \\begin{remark}\n% In what follows, only linear isotropic hardening is considered by setting $R(p)=C p$ so that $R'=C$, with $C$ the hardening modulus.\n% \\end{remark}\n\n% Given this description of the internal processes, the thermodynamical framework then leads to the elastic-plastic constitutive equations for the components of the Cauchy stress tensor during irreversible deformations \\cite{Simo}:\n% \\begin{align}\n% \\label{eq:elastoplastic_tangent}\n% & \\tens{\\dot{\\sigma}} = \\Cbb^{ep}:\\tens{\\dot{\\eps}}=\\(\\Cbb - \\beta\\:\\tens{s}\\otimes\\tens{s} \\):\\tens{\\dot{\\eps}} \\\\\n% \\label{eq:plastic_flow}\n% % & \\beta = \\frac{6\\mu^2}{3\\mu +C}\\times\\frac{1}{\\tens{s}:\\tens{s}}\n% & \\beta = \\frac{6\\mu^2}{3\\mu +R'}\\times\\frac{1}{\\tens{s}:\\tens{s}}\n% \\end{align}\n% where the elastoplastic tangent modulus $\\Cbb^{ep}$ can be decomposed into the elasticity tensor $\\Cbb$ and another part depending on the direction of the plastic flow.\n% In addition, the evolution of the cumulated plastic strain follows from the consistency condition:\n% \\begin{equation}\n% %\\label{eq:consistance}\n% \\dot{f}=0 \\quad \\Leftrightarrow \\quad \\drond{f}{\\tens{\\sigma}}:\\tens{\\dot{\\sigma}} - \\drond{f}{R}\\:\\dot{R} =0\n% \\end{equation}\n% so that one writes:\n% \\begin{equation}\n% \\label{eq:consistance}\n% C\\dot{p}=\\sqrt{\\frac{3}{2}}\\:\\frac{\\tens{s}:\\tens{\\dot{\\sigma}}}{\\norm{\\tens{s}}}\n% \\end{equation}\n% %Then, assuming the existence of a dissipation pseudo-potential $\\phi(\\tens{\\sigma},-R(p))$, the plastic flow rule:\n% Then, the plastic flow rule:\n% \\begin{equation}\n% \\label{eq:plastic_flow_rule}\n% \\tens{\\dot{\\eps}}^p = \\dot{p}\\drond{f}{\\tens{\\sigma}}\n% \\end{equation}\n% finally leads to the evolution of the plastic part of the strain tensor:\n% \\begin{equation}\n% \\label{eq:plastic_strain_rate}\n% \\tens{\\dot{\\eps}}^p = \\dot{p}\\:\\sqrt{\\frac{3}{2}}\\frac{\\tens{s}}{\\norm{\\tens{s}}}\n% \\end{equation}\n\n% On the other hand, reversible evolutions are governed by the elastic law:\n% \\begin{equation}\n% \\label{eq:elastic_law}\n% \\tens{\\sigma} = \\Cbb: \\tens{\\eps}^e = 2\\mu \\tens{\\eps}^e + \\lambda \\tr \\tens{\\eps}^e \\: \\tens{I} \n% \\end{equation}\n% in which $(\\lambda,\\mu)$ are Lam{\\'e}'s parameters.\n% The inverse constitutive law also can be written based on Young's modulus and Poisson's ratio $(E,\\nu)$:\n% \\begin{equation} \n% \\label{eq:elastic_inverse}\n% \\tens{\\eps}^e = \\Cbb^{-1}:\\tens{\\sigma} = \\frac{1+\\nu}{E} \\tens{\\sigma} - \\frac{\\nu}{E} \\tr \\tens{\\sigma} \\: \\tens{I}\n% \\end{equation}\n% Therefore, plastic evolutions involve $\\Hbb \\equiv \\Cbb^{ep}$ while elastic ones involve $\\Hbb\\equiv\\Cbb$.\n\n\\subsection{Spectral analysis}\n\\label{sec:spectral-analysis}\n\nConsidering an arbitrary direction of space $\\vect{n}$, the quasi-linear form \\eqref{eq:quasi-linear_form} reads: \n% The quasi-linear form of the sets of equations \\eqref{eq:balance_equations} and \\eqref{eq:plasticity_equations} in a Cartesian coordinate system and an arbitrary direction $\\vect{n}$ is:\n\\begin{equation}\n \\label{eq:quasilinear_normal}\n \\drond{\\Qcb}{t} + \\Jbsf \\drond{\\Qcb}{x_n} = \\vect{0} \n\\end{equation}\nwhere $x_n=\\vect{x}\\cdot\\vect{n}$ and $\\Jbsf=n_i\\Absf^i$ is the Jacobian matrix.\nSimple waves are solutions for which the vector $\\Qcb$ is constant along each curve of the one-parameter family $\\eta^K(x_n,t)=\\text{const}$.\nFor such \\textit{self-similar} solutions, system \\eqref{eq:quasilinear_normal} reads:\n\\begin{equation}\n \\label{eq:simple_system}\n \\(\\Jbsf - c_K \\Ibsf \\vphantom{\\Qcb'(\\eta^K)}\\)\\Qcb'(\\eta^K) = \\vect{0} \n\\end{equation}\nwhere $\\Ibsf$ is the $9\\times 9$ identity matrix and $c_K = -\\drond{\\eta^K}{t} / \\drond{\\eta^K}{x_n}$ appears to be the $K$th eigenvalue of the Jacobian matrix.\nThe problem therefore admits nontrivial solutions if $\\Jbsf$ has real eigenvalues and distinct left eigenvectors $\\Lcb^K= \\[ \\vect{v}^K \\: , \\: \\tens{\\sigma}^K \\]$ satisfying:\n\\begin{equation}\n \\label{eq:eigen_system}\n \\vect{\\Lc}^K \\(\\Jbsf - c_K \\Ibsf\\) = \\vect{0} \\qquad K=1,\\cdots,9\n\\end{equation}\n%The characteristic structure of the problem is given by the eigenvalues $c_K$ and the associated left eigenvectors $\\Lcb^K= \\[ \\vect{v}^K \\: , \\: \\tens{\\sigma}^K \\]$ of the Jacobian matrix satisfying:\n% \\begin{equation}\n% \\label{eq:eigen_system}\n% \\vect{\\Lc}^K \\(\\Jbsf - c_K \\Ibsf\\) = \\vect{0} \\qquad K=1,\\cdots,9\n% \\end{equation}\n%with $\\Ibsf$, the $9\\times 9$ identity matrix.\nThus, for non-zero eigenvalues one gets:\n\\begin{align}\n \\label{eq:eigen_left_stress}\n & -\\tens{\\sigma}^K:\\(\\Hbb\\cdot \\vect{n}\\) - c_K \\vect{v}^K =\\vect{0} \\\\\n \\label{eq:eigen_left_velo}\n & -\\frac{1}{\\rho}\\vect{v}^K\\otimes\\vect{n} - c_K \\tens{\\sigma}^K = \\tens{0}\n\\end{align}\nSubstitution of $\\tens{\\sigma}^K$ obtained from \\eqref{eq:eigen_left_velo} in \\eqref{eq:eigen_left_stress} leads to:\n\\begin{equation}\n \\label{eq:acoustic_eigen}\n (\\vect{v}^K\\otimes\\vect{n}):\\(\\Hbb\\cdot \\vect{n}\\) - \\rho c_K^2 \\vect{v}^K = \\tens{0}\n\\end{equation}\nwhich is the left eigensystem of the acoustic tensor $A_{ij}=n_k H_{ik j l} n_l$.\nDue to the symmetry of $\\tens{A}$, system \\eqref{eq:acoustic_eigen} is equivalent to the right eigensystem:\n\\begin{equation}\n \\label{eq:acoustic_eigen_system_lambda}\n \\( n_k H_{ik j l} n_l - \\rho c_K^2 \\delta_{ij} \\) v_j^K =0\n\\end{equation}\nor alternatively, with the eigenvalues $\\omega_p$ and associated left eigenvectors of the acoustic tensor $\\vect{l}^q\\: \\: (q=1,2,3)$:\n\\begin{equation}\n \\label{eq:acoustic_eigen_system}\n \\vect{l}^q \\cdot \\( \\tens{A} - \\omega_q \\tens{I} \\) = \\vect{0}\n\\end{equation}\nThe condition for system \\eqref{eq:quasilinear_normal} to be hyperbolic (real eigenvalues and independent eigenvectors) is thus ensured by the positive definiteness of the acoustic tensor, also known as the \\textit{strong ellipticity} condition \\cite{Foundation_of_elasticity}:\n\\begin{equation}\n \\label{eq:strong_ellipticity}\n (\\vect{m}\\otimes \\vect{n}): \\Hbb: (\\vect{n}\\otimes \\vect{m}) > 0 \\quad \\forall \\vect{n},\\vect{m} \\in \\Rbb^3 \\: ; \\: \\vect{n},\\vect{m} \\ne \\vect{0}\n\\end{equation}\nIf the condition holds, the acoustic tensor admits $3$ couples of eigenvalue--eigenvector $\\{\\omega_q,\\vect{l}^q\\}$ leading to $6$ couples $\\{c_K,\\Lcb^K\\}$ for the Jacobian matrix, the $3$ other eigenvalues being null \\cite{Kluth}.\nThe couples $\\{c_K,\\Lcb^K\\}$ are referred to as the \\textit{left characteristic fields}.\nNotice that since the elastic stiffness tensor $\\Cbb$ or the elastoplastic tangent modulus $\\Cbb^{ep}$ may be involved in equation \\eqref{eq:quasi-linear_form}, three left characteristic fields are obviously associated with elastic and elastic-plastic evolutions respectively.\nThe left eigenvectors related to the non-zero eigenvalues of the Jacobian matrix are obtained by using equation \\eqref{eq:eigen_left_velo} so that the following $6$ eigenfields of the quasi-linear form \\eqref{eq:quasilinear_normal} can be defined:\n\\begin{equation}\n \\label{eq:left_eigenfields}\n \\left\\lbrace \\pm \\sqrt{\\frac{\\omega_q}{\\rho_0}} ; \\: \\[ \\pm \\rho_0\\sqrt{\\frac{\\omega_q}{\\rho_0}} \\vect{l}^q , -\\vect{l}^q\\otimes \\vect{n} \\] \\right\\rbrace ,\\: q=1,2,3\n\\end{equation}\nAt last, three independent left eigenvectors associated with the null eigenvalue of multiplicity $3$ can be found by solving equation \\eqref{eq:eigen_left_stress} for the null eigenvalue:\n\\begin{equation}\n \\label{eq:left_null_eigenvectors}\n \\tens{\\sigma}^K:\\(\\Hbb\\cdot \\vect{n}\\) =\\vect{0},\\quad K=1,...,3\n\\end{equation}\n\nSince the above formulation is based on the elastoplastic stiffnesses rather than compliances, it differs from those of \\textsc{Bleich} \\cite{Bleich}, \\textsc{Clifton} \\cite{Clifton}, and hence from these of \\textsc{Ting} and \\textsc{Nan} \\cite{Ting68} and \\textsc{Ting} \\cite{Ting69}.\n%\nThe use of the tangent modulus $\\Hbb$ enables the specialization of equations \\eqref{eq:left_eigenfields} and \\eqref{eq:left_null_eigenvectors} to the plane strain and the plane stress cases, as we shall see in section \\ref{sec:2dproblem}.\n\n\\subsection{Known properties of the plastic waves}\n\\label{sec:plastic_speeds}\n\nLet us now recall some of the important results of \\textsc{Mandel} \\cite{Mandel62,Mandel69_thermoWaves} for which proofs can also be found in \\cite{mandel_book}.\n% eq (6.13)\nDenoting the elastic acoustic tensor as $\\tens{A}^e = \\vect{n} \\cdot \\Cbb \\cdot \\vect{n}$, the eigenvalue problem \\eqref{eq:acoustic_eigen_system} can be rewritten as:\n\\begin{equation}\n \\label{eq:mod_eigen_acoustic}\n \\vect{l}^q \\cdot \\( \\tens{B} - \\beta \\vect{a}\\otimes\\vect{a} \\) = \\vect{0}\n\\end{equation}\nin which $\\tens{B}=\\tens{A}^e- \\omega_q \\tens{I}$ and $\\vect{a}=\\tens{s}\\cdot\\vect{n}$.\nThen, assuming that the coordinate axes coincide with the eigenbasis of $\\tens{A}^e$, the determinant of system \\eqref{eq:mod_eigen_acoustic} reads:\n\\begin{equation}\n \\label{eq:polynomial_principal}\n F(\\omega_q)=(A^e_1-\\omega_q)(A^e_2-\\omega_q)(A^e_3-\\omega_q) - \\beta\\[ (A^e_2-\\omega_q)(A^e_3-\\omega_q){a}^2_1 + (A^e_1-\\omega_q)(A^e_3-\\omega_q){a}^2_2 + (A^e_1-\\omega_q)(A^e_2-\\omega_q){a}^2_3 \\] =0\n\\end{equation}\nWith $\\beta>0$ and the eigenvalues of $\\tens{A^e}$ satisfying $A^e_1 \\geq A^e_2 \\geq A^e_3$, it comes out that:\n\\begin{equation}\n \\label{eq:polynomial_property}\n F\\(A^e_1\\) \\leq 0\\quad ; \\quad F\\(A^e_2\\) \\geq 0 \\quad ;\\quad F(A^e_3)\\leq 0 \\quad ; \\quad F(\\infty) < 0\n\\end{equation}\nfrom which the following conclusion is drawn.\n\\begin{property}[First Mandel's inequality]\n \\label{pr:mandel_inequality}\n For a unit normal vector $\\vect{n}$, the speeds of plastic waves $c_I^p\\geq c_{II}^p \\geq c_{III}^p$ are bounded by the elastic speeds $c_I^e\\geq c_{II}^e \\geq c_{III}^e$ according to:\n \\begin{equation}\n \\label{eq:Mandels_inequality}\n c_I^e\\geq c_I^p\\geq c_{II}^e\\geq c_{II}^p \\geq c_{III}^e \\geq c_{III}^p \n \\end{equation}\n\\end{property}\n\nConsider now equation \\eqref{eq:simple_system}, from which one gets for plastic evolutions:\n\\begin{align}\n \\label{eq:right_eigen1}\n & \\tens{\\sigma}' = -\\frac{1}{c_K} \\Hbb : \\( \\vect{n}\\otimes \\vect{v}' \\) \\\\\n \\label{eq:right_eigen2}\n & \\( \\tens{A}^{ep} - \\rho c_K^2 \\tens{I} \\) \\vect{v}' = \\vect{0}\n\\end{align}\nIf one of the plastic speed is equal to an elastic speed, the expansion of equation \\eqref{eq:right_eigen2} leads to:\n\\begin{equation}\n \\label{eq:zero_dot_prod}\n \\beta \\vect{a}\\cdot \\vect{v}' =0\n\\end{equation}\nThen, the projection of the variation $\\tens{\\sigma}'$ from equation \\eqref{eq:right_eigen1} onto the normal to the yield surface is:\n\\begin{equation}\n \\label{eq:nutral_wave}\n % \\drond{f}{\\tens{\\sigma}}:\\tens{\\sigma}' = -\\frac{1}{c_K}\\sqrt{\\frac{3}{2}} \\frac{\\tens{s}}{\\norm{\\tens{s}}} : \\Hbb : \\( \\vect{n}\\otimes \\vect{v}' \\) = \\frac{1}{c_K}\\sqrt{\\frac{3}{2}} \\frac{\\mu - \\beta \\norm{\\tens{s}}}{\\norm{\\tens{s}}} \\vect{a}\\cdot\\vect{v}' = 0\n \\drond{f}{\\tens{\\sigma}}:\\tens{\\sigma}'= \\frac{1}{c_K}\\sqrt{\\frac{3}{2}} \\frac{\\mu - \\beta \\norm{\\tens{s}}}{\\norm{\\tens{s}}} \\vect{a}\\cdot\\vect{v}' = 0\n\\end{equation}\nHence, if condition \\eqref{eq:zero_dot_prod} is satisfied, the stress increment is tangent to the yield surface.\n% Neutral waves\n\\begin{property}\n \\label{pr:neutral_wave}\n A plastic wave traveling with the speed of an elastic wave results in a stress path that is tangent to the yield surface in the stress space.\n Such a wave is called a \\textit{neutral wave}.\n\\end{property}\n\n%In what follows, the above equations are specified to plane strain and plane stress cases.\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"manuscript\"\n%%% End:\n", "meta": {"hexsha": "22c9da2f094ed3b114dfe7f72ffd77393402dba0", "size": 20380, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "papJmPs/characteristic_analysis.tex", "max_stars_repo_name": "adRenaud/research", "max_stars_repo_head_hexsha": "2f0062a1800d7a17577bbfc2393b084253d567f4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-06-18T14:52:03.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-18T14:52:03.000Z", "max_issues_repo_path": "papJmPs/characteristic_analysis.tex", "max_issues_repo_name": "adRenaud/research", "max_issues_repo_head_hexsha": "2f0062a1800d7a17577bbfc2393b084253d567f4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-01-07T13:11:11.000Z", "max_issues_repo_issues_event_max_datetime": "2019-01-07T13:11:11.000Z", "max_forks_repo_path": "papJmPs/characteristic_analysis.tex", "max_forks_repo_name": "adRenaud/research", "max_forks_repo_head_hexsha": "2f0062a1800d7a17577bbfc2393b084253d567f4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.4936708861, "max_line_length": 336, "alphanum_fraction": 0.7075564279, "num_tokens": 6844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8918110368115781, "lm_q2_score": 0.7122321842389469, "lm_q1q2_score": 0.6351765226767101}} {"text": "\\chapter{Tree}\n\n\\section{Binary Tree}\n\\subsection{Introductions}\n\\runinhead{Get parent ref.} To get a parent reference (implicitly), \\textit{return the Node} of the current recursion function to its parent to maintain the path. Sample code:\n\\begin{java}\nNode deleteMin(Node x) {\n if (x.left == null) return x.right;\n x.left = deleteMin(x.left);\n // x.count = 1+size(x.left)+size(x.right);\n return x;\n}\n\\end{java}\n\\runinhead{Construct path from root to target.} To search a node in binary tree (not necessarily BST), use dfs:\n\\begin{python}\ndef dfs(self, root, t, path, found):\n # post-call check\n if not root: return \n if found[0]: return \n\n path.append(root)\n if root == t:\n found[0] = True\n\n self.dfs(root.left, t, path, found)\n self.dfs(root.right, t, path, found)\n if not found[0]:\n path.pop() # 1 pop() corresponds to 1 append()\n\\end{python}\nThe `found` is a wrapper for boolean to keep it referenced by all calling stack. \n\n\\runinhead{Lowest common ancestor.} In BST, the searching is straightforward. In normal binary tree, construct the path from root to $node_1$ and $node_2$ respectively, and \\textbf{diff} the two paths.\n\n\\runinhead{Find all paths.} Find all paths from root to leafs. For every currently visiting node, add itself to path; search left, search right and pop itself. Record current result when reaching the leaf.\n\\begin{python}\ndef dfs_path(self, cur, path, ret):\n if not cur: return\n\n path.append(cur)\n if not cur.left and not cur.right:\n ret.append(\"->\".join(map(lambda x: str(x.val), path)))\n\n self.dfs_path(cur.left, path, ret)\n self.dfs_path(cur.right, path, ret)\n path.pop()\n\\end{python}\n\\subsection{Morris Traversal} \nTraversal with O(1) space. \\footnote{\\href{http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html}{ref}}\n\nTime complexity $O(3n).$ - find \\pyinline{pre} twice, \\pyinline{cur} traverse once. \\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[height=0.9in]{morris_time}}\n\\caption{Morris traversal time complexity}\n\\label{fig:morrisTime}\n\\end{figure}\n\n\\subsubsection{Inoder}\nAssign the current node's in-order predecessor's right child to itself (threading). Two ptr \\pyinline{cur}, \\pyinline{pre}. \n\nProcess:\n\\begin{enumerate}\n\\item If no left, \\textit{consume} \\pyinline{cur}, go right \n\\item If left, find in-order predecessor \\pyinline{pre}\n\\begin{enumerate}\n\\item If no thread (i.e. no \\pyinline{pre} right child), assign it to \\pyinline{cur}; go left\n\\item If thread, \\textit{consume} \\pyinline{cur}, go right. ($\\equiv$ no left). \n\\end{enumerate}\n\\end{enumerate}\n\n\\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[height=1.6in]{morris_inorder}}\n\\caption{Morris inorder traversal}\n\\label{fig:morrisInorder}\n\\end{figure}\n\\newpage\nCode:\n\\begin{python}\ndef morris_inorder(self, root):\n cur = root\n while cur:\n if not cur.left:\n self.consume(cur)\n cur = cur.right\n else:\n pre = cur.left\n while pre.right and pre.right != cur:\n pre = pre.right\n\n if not pre.right:\n pre.right = cur\n cur = cur.left\n else:\n pre.right = None\n self.consume(cur)\n cur = cur.right\n\\end{python}\n\\subsubsection{Preoder}\nSimilar to inorder. \n\nProcess:\n\\begin{enumerate}\n\\item If no left, \\textit{consume} \\pyinline{cur}, go right \n\\item If left, find in-order predecessor \\pyinline{pre}\n\\begin{enumerate}\n\\item If no thread (i.e. no \\pyinline{pre} right child), assign it to \\pyinline{cur}; \\textit{consume} \\pyinline{cur}, go left\n\\item If thread, go right. ($\\equiv$ no left, but no \\textit{consume}, since consume before). \n\\end{enumerate}\n\\end{enumerate}\n\n\\subsubsection{Postorder}\nMore tedious.\n\\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[height=1.6in]{morris_postorder}}\n\\caption{Morris inorder traversal}\n\\label{fig:morrisInorder}\n\\end{figure}\n\nTODO\n\n\\section{Binary Search Tree (BST)}\n\\runinhead{Array and BST.}Given either the \\textbf{preorder} or \\textbf{postorder} (but not inorder) traversal of a BST containing N distinct keys, it is possible to reconstruct the shape of the BST. \n\\subsection{Rank}\n\\runinhead{Calculates rank.}\n\\begin{enumerate}\n\\item When inserting: \n \\begin{enumerate}\n \\item insert to an existing node: \\pyinline{node.cnt_this += 1}\n \\item insert to left subtree: \\pyinline{node.cnt_left += 1}\n \\item insert to right subtree: do nothing. \n\\end{enumerate}\n\\item When querying rank:\n \\begin{enumerate}\n \\item query equals current node: \\pyinline{return node.cnt_left}\n \\item query goes to \\textbf{left} node: \\pyinline{return rank(node.left, val)};\n \\item query goes to \\textbf{right} node: \\pyinline{return node.cnt_left} \\pyinline{+ node.cnt_this + rank(node.right, val)}\n \\end{enumerate}\nNotice that the \\pyinline{rank} calculates a val's rank in a subtree.\n\\end{enumerate}\n\n\\runinhead{Count of smaller number before itself.} Given an array $A$. For each element $A_i$ in the array, count the number of element before this element $A_i$ is smaller than it and return count number array. Average $O(n \\log n)$\n\\\\\nClues:\n\\begin{enumerate}\n\\item Put $A[:i+1]$ into a BST; so as to count the rank of $A[i]$ in the BST\n\\end{enumerate}\nCodes:\n\\begin{python}\nclass Node(object):\n def __init__(self, val):\n \"\"\"Records the left subtree size\"\"\"\n self.val = val\n self.cnt_left = 0\n self.cnt_this = 0\n self.left, self.right = None, None\n\n\nclass BST(object):\n def __init__(self):\n self.root = None\n\n def insert(self, root, val):\n \"\"\"\n :return: subtree's root after insertion\n \"\"\"\n if not root:\n root = Node(val)\n\n if root.val == val:\n root.cnt_this += 1\n elif val < root.val:\n root.cnt_left += 1\n root.left = self.insert(root.left, val)\n else:\n root.right = self.insert(root.right, val)\n\n return root\n\n def rank(self, root, val):\n \"\"\"\n Rank in the root's subtree\n :return: number of items smaller than val\n \"\"\"\n if not root:\n return 0\n if root.val < val:\n return (root.cnt_this+root.cnt_left+\n self.rank(root.right, val))\n elif root.val == val:\n return root.cnt_left\n else:\n return self.rank(root.left, val)\n\n\nclass Solution(object):\n def countOfSmallerNumberII(self, A):\n tree = BST()\n ret = []\n for a in A:\n tree.root = tree.insert(tree.root, a)\n ret.append(tree.rank(tree.root, a))\n\n return ret\n\\end{python}\nNotice: if worst case $O(n \\log n)$ is required, need to use Red-Back Tree - Section \\ref{rbtree}. However, there is a more elegant way using Segment Tree - Section \\ref{segmentTreeInversionCount}.\n\n\n\\subsection{Range search}\n\\runinhead{1-d range count}\n\\begin{java}\nint size(Key lo, Key hi) {\n if (contains(hi)) return rank(hi)-rank(lo)+1;\n else return rank(hi)-rank(lo);\n}\n\\end{java}\n\n\\runinhead{Closest value} Find the value in BST that is closet to the \\pyinline{target}.\n\\\\\nClues:\n\\begin{enumerate}\n\\item Find the value just $\\leq$ the target.\n\\item Find the value just $\\geq$ the target.\n\\end{enumerate}\n\\\n\\\\\nCode for finding either the lower value or higher value:\n\\begin{python}\ndef find(self, root, target, ret, lower=True):\n \"\"\"ret: result container\"\"\"\n if not root: return\n\n if root.val == target:\n ret[0] = root.val\n return\n\n if root.val < target:\n if lower:\n ret[0] = max(ret[0], root.val)\n\n self.find(root.right, target, ret, lower)\n else:\n if not lower:\n ret[0] = min(ret[0], root.val)\n\n self.find(root.left, target, ret, lower)\n\\end{python}\n\n\\runinhead{Closet values} Find $k$ values in BST that are closet to the \\pyinline{target}.\n\\\\\\\\\nClues:\n\\begin{enumerate}\n\\item Find the predecessors $\\triangleq \\{node | node.value \\leq target\\}$. Store in the stack. \n\\item Find the successors $\\triangleq \\{node | node.value \\geq target\\}$. Store in the stack.\n\\item Merge the predecessors and successors as in merge in MergeSort to get he $k$ values. \n\\end{enumerate}\n\\\n\\\\\nCode for finding the predecessors:\n\\begin{python}\ndef predecessors(self, root, target, stk):\n if not root: return\n\n self.predecessors(root.left, target, stk)\n if root.val <= target:\n stk.append(root.val)\n self.predecessors(root.right, target, stk)\n\\end{python}\n\n\n\\section{Binary Index Tree (BIT)}\\label{BIT}\n\\subsection{Introduction}\nCompared to Segment Tree \\ref{section:segmentTree}, BIT is shorter and more elegant. BIT can do most of things that Segment Tree can do and it is easier to code. BIT updates and queries $$i\\rightarrow prefixSum$$ in $O(\\log n)$ time; however, BIT CANNOT query $$prefixSum \\rightarrow i$$\n\\subsection{Implementation}\nGiven an array $A$ of length $n $ starting from $1$. prefix sum $s[i]\\triangleq A_1+...+A_i$. BIT uses binary to maintain the array of prefix sum for querying and updating. For $i$-th node in the BIT, \n$$\nN[i]=A_{j+1}+...+A_i\n$$\n, where $j=i-lowbit(i)$, i.e. set $i$'s lowest bit 1 to 0. $lowbit(i)$ can be defined as \\pyinline{return i & -i}, using 2's complement. Notice that the summation ends with $A_i$ since easier to \\pyinline{set}.\n\nFor the range, we use $(j, i]$ here instead of $[j, i)$ since more elegant for \\pyinline{get(i)} and \\pyinline{set(i)}\n\\\\\\\\\nClues:\n\\begin{enumerate}\n\\item Binary \n\\item Low bit\n\\item BIT uses array index starting from \\textbf{1}, because 0 doesn't have $lowbit$.\n\\end{enumerate}\n\\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[height=1.1in]{BITget}}\n\\caption{Binary Indexed Tree \\textit{get} Operation}\n\\label{fig:LABEL}\n\\end{figure}\n\n\\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[height=1.05in]{BIT}}\n\\caption{Binary Indexed Tree \\textit{set} Operation}\n\\label{fig:LABEL}\n\\end{figure}\n\nTime complexity, longest update is along the leftmost branch, which takes $O(\\log_2 n)$ (e.g. 1, 10, 100, 1000, 10000); longest query is along a branch starting with node with all 1's (e.g. 1111, 1110, 1100, 1000), which also takes $O(\\log_2 n)$.\n\\newpage\nCode:\n\\begin{python}\nclass BIT(object):\n def __init__(self, n):\n \"\"\"BIT uses index starting from 1\"\"\"\n self.N = [0 for _ in xrange(n+1)]\n\n def lowbit(self, i):\n return i & -i\n\n def get(self, i):\n ret = 0\n while i > 0:\n ret += self.N[i]\n i -= self.lowbit(i)\n\n return ret\n \n def set(self, i, val):\n while i < len(self.N):\n self.N[i] += val\n i += self.lowbit(i)\n\\end{python}\n\n\n\\section{Segment Tree}\\label{section:segmentTree}\n\\subsection{Introduction}\nSegment Tree is specially built for \\textit{range queries}. \n\nThe structure of Segment Tree is a binary tree which each node has two attributes start and end denote an segment/interval. \n\nNotice that by practice, the interval is normally $[start, end)$ but sometimes it can be $[start, end]$, which depends on the question definition. \n\nStructure: \n\\begin{lstlisting}[columns=flexible]\n# a Count Segment Tree\n [0, 4, count=3]\n / \\\n [0,2,count=1] [2,4,count=2]\n / \\ / \\\n [0,1,count=1] [1,2,count=0] [2,3,count=1], [3,4,count=1]\n\\end{lstlisting}\nVariants:\n\\begin{enumerate}\n\\item Sum Segment Tree.\n\\item Min/Max Segment Tree.\n\\item Count Segment Tree. \n\\end{enumerate}\n\nFor a Maximum Segment Tree, which each node has an extra value max to store the maximum value in this node's interval.\n\n\\subsection{Operations}\nSegment Tree does a decent job for range queries.\n\\\\\nComponents in Segment Tree operations:\n\\begin{enumerate}\n\\item Build\n\\item Query \n\\item Modify\n\\item Search \n\\end{enumerate}\nNotice:\n\\begin{enumerate}\n\\item Only build need to change the start and end recursively.\n\\item Pre-check is preferred in recursive calls.\n\\end{enumerate}\nCode: Notice the code has abstracted out segment tree functions of sum, min/max or count, by abstracting the subtree combine function to \\pyinline{lambda}.\n\\begin{python}\nDEFAULT = 0\nf = lambda x, y: x+y\n\n\nclass Node(object):\n def __init__(self, start, end, m):\n self.start, self.end, self.m = start, end, m\n self.left, self.right = None, None\n\n\nclass SegmentTree(object):\n def __init__(self, A):\n self.A = A\n self.root = self.build_tree(0, len(self.A))\n\n def build_tree(self, s, e):\n \"\"\"\n segment: [s, e)\n Either check s+1==e or have root.right \n only if have root.left\n \"\"\"\n if s >= e: return None\n if s+1 == e: return Node(s, e, self.A[s])\n\n left = self.build_tree(s, (s+e)/2)\n right = self.build_tree((s+e)/2, e)\n\n val = DEFAULT\n if left: val = f(val, left.m)\n if right: val = f(val, right.m)\n root = Node(s, e, val)\n root.left = left\n root.right = right\n\n return root\n\n def query(self, root, s, e):\n \"\"\"\n :type root: Node\n \"\"\"\n if not root:\n return DEFAULT\n\n if s <= root.start and e >= root.end:\n return root.m\n\n if s >= root.end or e <= root.start:\n return DEFAULT\n\n l = self.query(root.left, s, e)\n r = self.query(root.right, s, e)\n return f(l, r)\n\n def modify(self, root, idx, val):\n \"\"\"\n :type root: Node\n \"\"\"\n if not root or idx >= root.end or idx < root.start:\n return\n\n if idx == root.start and idx == root.end-1:\n root.m = val\n self.A[idx] = val\n return\n\n self.modify(root.left, idx, val)\n self.modify(root.right, idx, val)\n\n val = DEFAULT\n if root.left: val = f(val, root.left.m)\n if root.right: val = f(val, root.right.m)\n \n root.m = val\n\\end{python}\nConcrete example - Count Segment Tree \\ref{inversionReconstruct}. \n\n\\newpage\n\\section{Trie}\n\\subsection{Basic}\nTrie is aka radix tree, prefix tree. \n\\begin{figure}[hbtp]\n\\centering\n\\subfloat{\\includegraphics[scale=.30]{trie.jpg}}\n\\caption{Trie}\n\\label{fig:trie} \n\\end{figure}\n\\runinhead{Notice:}\n\\begin{enumerate}\n\\item Children are stored in HashMap rather than ArrayList. \n\\item self.word to stores the word and indicates whether a word ends at the current\nnode. \n\\end{enumerate}\nCodes:\n\\begin{python}\nclass TrieNode(object):\n def __init__(self, char):\n self.char = char\n self.word = None\n self.children = {} # map from char to TrieNode\n\n\nclass Trie(object):\n def __init__(self):\n self.root = TrieNode(None)\n\n def add(self, word):\n word = word.lower()\n cur = self.root\n for c in word:\n if c not in cur.children:\n cur.children[c] = TrieNode(c)\n cur = cur.children[c]\n cur.word = word\n\\end{python}\n\n\\subsection{Advanced}\nImplicit storage of word in TrieNode: \n\\begin{enumerate}\n\\item Implicitly stores the current word. \n\\item Implicitly stores the current char. \n\\item When insert new word, do not override the existing TrieNode. A flag to indicate\nwhether there is a word ending here.\n\\end{enumerate}\n\\newpage\nCode:\n\\begin{python}\nclass TrieNode:\n def __init__(self):\n \"\"\"Implicit storage\"\"\"\n self.ended = False\n self.children = {}\n\n\nclass Trie:\n def __init__(self):\n self.root = TrieNode()\n\n def insert(self, word):\n cur = self.root\n for w in word:\n if w not in cur.children: # not override\n cur.children[w] = TrieNode()\n cur = cur.children[w]\n\n cur.ended = True\n\n def search(self, word):\n cur = self.root\n for w in word:\n if w in cur.children:\n cur = cur.children[w]\n else:\n return False\n\n if not cur.ended: # not ended here\n return False\n\n return True\n\n def startsWith(self, prefix):\n cur = self.root\n for w in prefix:\n if w in cur.children:\n cur = cur.children[w]\n else:\n return False\n\n return True\n\\end{python}\n\\subsection{Applications}\n\\begin{enumerate}\n\\item Word search in matrix.\n\\item Word look up in dictionary.\n\\end{enumerate}\n \n \n", "meta": {"hexsha": "56491e1834f358e30d6c0b92daf72074c034aa5f", "size": 16223, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapterTree.tex", "max_stars_repo_name": "li77leprince/Algo-Quicksheet", "max_stars_repo_head_hexsha": "1736bc0ee1d73b0b06dcf3823a65ea0f8c286108", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapterTree.tex", "max_issues_repo_name": "li77leprince/Algo-Quicksheet", "max_issues_repo_head_hexsha": "1736bc0ee1d73b0b06dcf3823a65ea0f8c286108", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapterTree.tex", "max_forks_repo_name": "li77leprince/Algo-Quicksheet", "max_forks_repo_head_hexsha": "1736bc0ee1d73b0b06dcf3823a65ea0f8c286108", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.5500910747, "max_line_length": 287, "alphanum_fraction": 0.647475806, "num_tokens": 4461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7122321720225278, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.6351765158779495}} {"text": "\\section{The Whitney sum formula}\n%Let me just repeat what I was rushing through at the end on Wednesday.\n%I wrote slightly incorrect things.\n%Let's look at $\\cc^n$.\n%Of course, $U(n)$ acts on this, and preserves the inner product.\n%Recall that $Fr(\\cc^n)$ is the space of orthonormal bases of $\\cc^n$.\n%Thus there's an action of $U(n)$ on this, and this is simply transitive, i.e., it's free and transitive.\n%Some would say that this is a ``torsor for $U(n)$''.\n%There's a basepoint in here, namely the standard basis, but you may forget about that.\n%\n%Now I can make various constructions.\n%For instance, think about $\\Fl(\\cc^n)$, which is the space of ordered orthonormal sets of $n$ lines in $\\cc^n$.\n%There's lots of isotropy now.\n%The isotropy group of the $U(n)$ action on the lines through the standard basis is now $T_n=U(1)^n$, i.e., diagonal matrices with roots of unity on the diagonal.\n%This is called the \\emph{(complete) flag manifold}.\n%\n%Another thing you could do as well -- you can forget the ordering.\n%This doesn't really have a name, I guess, but what I'm doing is considering $\\Fl(\\cc^n)/\\Sigma_n$, which is the space of unordered orthonormal lines in $\\cc^n$.\n%This is a homogeneous space, too, and it's $U(n)$ quotiented by permutation matrices where $1$ is any complex number of norm $1$.\n%For instance, if $N_{U(n)}(T^n) = \\Sigma_n\\cdot T^n$ is the normalizer, then $\\Fl(\\cc^n)/\\Sigma_n = U(n)/N(T^n)$.\n%I can consider the cofiber sequence $T^n\\hookrightarrow \\Sigma_n\\cdot T^n \\to W_n$, and the $W_n$ is called the Weyl group.\n%Something nice here is that\n%\\begin{equation*}\n% \\xymatrix{\n%\tT^n\\ar@{^(->}[r] & \\Sigma_n\\cdot T^n\\ar[r] & W_n\\\\\n%\t& \\Sigma_n\\ar@{^(->}[u]\\ar[ur]^\\simeq & \n% }\n%\\end{equation*}\n%To get projective space I consider the space of splittings of $\\cc^n$ into a line and its orthogonal complement -- but that's just giving a line because the metric determines the orthogonal complement.\n%\n%Consider a complex $n$-plane bundle $\\xi^n\\downarrow X$, with a metric.\n%Then $\\Fr(\\xi^n)\\to X$ is a principal $U(n)$-bundle over $X$.\n%I can form $E(\\xi) = \\Fr(\\xi)\\times_{U(n)}\\cc^n$, and $\\PP(\\xi) = \\Fr(\\xi)\\times_{U(n)} U(n)/(U(1)\\times U(n-1))$.\n%Note that $U(n)/(U(1)\\times U(n-1)) = \\CP^{n-1}$.\n%I can also consider $\\Fl(\\xi) = \\Fr(\\xi)\\times_{U(n} U(n)/T^n$.\n%\n%We found that $H^\\ast(X) \\hookrightarrow H^\\ast(\\Fl(\\xi))$.\n%This is the splitting principle.\n%\n%Let's look at the universal case $\\xi_n\\downarrow BU(n)$.\n%So, $\\Fl(\\xi_n) = EU(n)\\times_{U(n)} U(n)/T^n$, so this is $BT^n$.\n%It maps down to $BU(n)$, and the map is the one induced by the inclusion $T^n\\hookrightarrow U(n)$.\n%So, for any ring, we find that $H^\\ast(BU(n))\\hookrightarrow H^\\ast(BT^n)$.\n%We know that \n%$$\n%H^\\ast(BT^n) = \\Z[t_1,\\cdots,t_n]\n%$$\n%where $t_i = e(\\lambda_i)$ and $|t_i| = 2$, where $\\lambda_i = pr_i^\\ast\\lambda$ and $\\lambda\\downarrow \\CP^\\infty$ is the universal line bundle.\nAs we saw in the previous section, there is an injection\n$H^\\ast(BU(n))\\hookrightarrow H^\\ast(BT^n)$. What is the image of this map?\n\nThe symmetric group sits inside of $U(n)$, so it acts by conjugation on $U(n)$.\nThis action stabilizes this subgroup $T^n$. By naturality, $\\Sigma_n$ acts on\nthe classifying space $BT^n$. Since $\\Sigma_n$ acts by conjugation on $U(n)$,\nit acts on $BU(n)$ in a way that is homotopic to the identity (Lemma\n\\ref{conjugation-homotopic}). However, each element $\\sigma\\in \\Sigma_n$ simply\npermutes the factors in $BT^n = (\\CP^\\infty)^n$; we conclude that\n$H^\\ast(BU(n);R)$ actually sits inside the invariants\n$H^\\ast(BT^n;R)^{\\Sigma_n}$.\n\nRecall the following theorem from algebra:\n\\begin{theorem}\n Let $\\Sigma_n$ act on the polynomial algebra $R[t_1,\\cdots,t_n]$ by\n permuting the generators. Then\n $$R[t_1,\\cdots,t_n]^{\\Sigma_n} = R[\\sigma_1^{(n)},\\cdots,\\sigma_n^{(n)}],$$\n where the $\\sigma_i$ are the {elementary symmetric polynomials}, defined\n via\n $$\\prod^n_{i=1}(x-t_i) = \\sum^n_{j=0} \\sigma_i^{(n)}x^{n-i}.$$\n\\end{theorem}\nFor instance,\n$$\\sigma_1^{(n)} = -\\sum t_i,\\ \\sigma_n^{(n)} = (-1)^n\\prod t_i.$$\nIf we impose a grading on $R[t_1,\\cdots,t_n]$ such that $|t_i| = 2$, then\n$|\\sigma_i^{(n)}| = 2i$. It follows from our discussion in \\S\n\\ref{homology-bun} that the ring $H^\\ast(BT^n)^{\\Sigma_n}$ has the same size as\n$H^\\ast(BU(n))$.\n%they the same Poincar\\'{e} series. You could imagine that maybe it embeds as a\n%lattice inside the integers?\n\nConsider an injection of finitely generated abelian groups $M\\hookrightarrow\nN$, with quotient $Q$. Suppose that, after tensoring with any field, the map\n$M\\to N$ an isomorphism. If $Q\\otimes k = 0$, then $Q = 0$.\nIndeed, if $Q\\otimes \\QQ = 0$ then $Q$ is torsion. Similarly, if $Q\\otimes\n\\FF_p = 0$, then $Q$ has no $p$-component. In particular, $M\\simeq N$. Applying\nthis to the map $H^\\ast(BU(n)\\to H^\\ast(BT^n)^{\\Sigma_n}$, we find that\n$$\nH^\\ast(BU(n);R) \\xrightarrow{\\simeq} H^\\ast(BT^n;R)^{\\Sigma_n} =\nR[\\sigma_1^{(n)},\\cdots,\\sigma_n^{(n)}].\n$$\nWhat happens as $n$ varies? There is a map $R[t_1,\\cdots,t_n] \\to\nR[t_1,\\cdots,t_{n-1}]$ given by sending $t_n\\mapsto 0$ and $t_i\\mapsto t_i$ for\n$i\\neq n$. Of course, we cannot say that this map is equivariant with respect\nto the action of $\\Sigma_n$. However, it \\emph{is} equivariant with respect to\nthe action of $\\Sigma_{n-1}$ on $R[t_1,\\cdots,t_n]$ via the inclusion of\n$\\Sigma_{n-1}\\hookrightarrow \\Sigma_n$ as the stabilizer of\n$n\\in\\{1,\\cdots,n\\}$. Therefore, the $\\Sigma_n$-invariants sit inside the\n$\\Sigma_{n-1}$-invariants, giving a map\n$$\nR[t_1,\\cdots,t_n]^{\\Sigma_n} \\to R[t_1,\\cdots,t_n]^{\\Sigma_{n-1}} \\to\nR[t_1,\\cdots,t_{n-1}]^{\\Sigma_{n-1}}.\n$$\nWe also find that for $i>h$, one can approximate $R+h\\rightarrow R$ on the right-hand side, thus\n\\[\nh(\\theta=\\pi)-h(\\theta=0)=\\frac{\\omega^2R^2}{2g}.\n\\]\nThis come out a bit less than 11 km, or a difference of near 22 km for the diameter of the Earth in the equatorial plane compared to a diameter between the poles. In reality, the difference is approximately 41 km. The discrepancy comes from the assumption that the true gravitational force can be treated as if it came from a point at Earth's center. This would be true if the distribution of mass was radially symmetric. However, Earth's center is molten and the rotation distorts the mass distribution. Remarkably this effect nearly doubles the elliptic distortion of Earth's shape. Due to this distortion, the top of Mount Everest is not the furthest point from the center of the Earth. That belongs to the top of a volcano, Chimborazo, in Equador, which is one degree in latitude below the Equator. Chimborazo is about 8500 ft lower than Everest when measured relative to sea level, but is 7700 feet further from the center of the Earth.\n\n\\exampleend\n\n\\subsection{Coriolis Force}\n\nConsider some vector $\\vec{A}$ according to an observer in a frame rotating about the $z$ axis with angular velocity $\\vec{\\omega}=\\omega\\hat{z}$. To an observer in the laboratory frame (the primed frame) the vector will change even if the vector appears fixed to the rotating observer. For a rotation of $\\Delta\\theta=\\omega\\Delta t$, the change of the vector due to the rotation is\n\\begin{eqnarray}\n\\Delta\\vec{A}'=\\vec{\\omega}\\times\\vec{A}\\Delta t.\n\\end{eqnarray}\nIf one includes the fact that $A$, the vector measured in the rotating frame, might be changing as a function of time,\n\\begin{eqnarray}\n\\label{eq:dAdtrot}\n\\Delta\\vec{A}'&=&\\Delta\\vec{A}+\\vec{\\omega}\\times\\vec{A}\\Delta t,\\\\\n\\nonumber\n\\frac{d}{dt}\\vec{A}'&=&\\frac{d}{dt}\\vec{A}+\\vec{\\omega}\\times\\vec{A}.\n\\end{eqnarray}\nIf the vector happens to be the position $\\vec{r}$,\n\\begin{eqnarray}\n\\dot{\\vec{r}}'&=&\\vec{v}'=\\vec{v}+\\vec{\\omega}\\times\\vec{r}.\n\\end{eqnarray}\nHere, the first term on the r.h.s. corresponds to the vector $\\vec{r}$ not being fixed, but changing with time, $\\dot{\\vec{r}}=\\vec{v}$. One can now use $\\vec{v}'=\\vec{v}+\\vec{\\omega}\\times\\vec{r}$ in place of $\\vec{A}$ in Eq. (\\ref{eq:dAdtrot}), and see\n\\begin{eqnarray}\n\\dot{\\vec{v}}'&=&\\frac{d}{dt}(\\vec{v}+\\vec{\\omega}\\times\\vec{r})+\\vec{\\omega}\\times(\\vec{v}+\\vec{\\omega}\\times\\vec{r})\\\\\n\\nonumber\n&=&\\dot{\\vec{v}}+\\vec{\\omega}\\times\\dot\\vec{r}+\\vec{\\omega}\\times\\left(\\vec{v}+\\vec{\\omega}\\times\\vec{r}\\right)\\\\\n\\nonumber\n&=&\\dot{\\vec{v}}+2\\vec{\\omega}\\times\\vec{v}+\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r}).\n\\end{eqnarray}\nBecause $\\dot{\\vec{v}}'$ is $\\vec{F}/m$,\n\\begin{eqnarray}\n\\label{eq:FmaRotatingFrame}\n\\vec{F}&=&m\\left\\{\\vec{a}+2\\vec{\\omega}\\times\\vec{v}+\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r})\\right\\},\\\\\n\\nonumber\nm\\vec{a}&=&\\vec{F}-2m\\vec{\\omega}\\times\\vec{v}-m\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r}).\n\\end{eqnarray}\nThe extra terms on the right behave like additional forces. Like gravitational forces, they are proportional to the mass, so the mass cancels for many problems. \n\nThe last term, $-m\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r})$, represents the centrifugal force. Using the vector identity, \n$\\vec{A}\\times(\\vec{B}\\times\\vec{C})=\\vec{B}(\\vec{A}\\cdot{\\vec{C}})-\\vec{C}(\\vec{A}\\cdot\\vec{B})$,\n\\begin{equation}\n-\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r})=\\omega^2\\vec{r}+(\\omega\\cdot\\vec{r})\\vec{\\omega}.\n\\end{equation}\nIf $\\vec{\\omega}$ is in the $z$ direction,\n\\begin{equation}\n-\\vec{\\omega}\\times(\\vec{\\omega}\\times\\vec{r})=\\omega^2(x\\hat{x}+y\\hat{y}).\n\\end{equation}\nThe centrifugal force points outward in the $x-y$ plane, and its magnitude is $m\\omega^2r_{\\rm\\perp}$, where $r_{\\rm\\perp}=\\sqrt{x^2+y^2}$.\n\nThe second term is Eq. (\\ref{eq:FmaRotatingFrame}) represents the Coriolis force. It does not enter problems like the shape of the Earth above because in that case the water was not moving relative to the rotating frame. Once an object is moving in a rotating frame, the particle is no longer being described in a single accelerating frame because at each point the acceleration is $-\\omega^2\\vec{r}$. \n\n\\example\nA ball is dropped from a height $h=500$m above Minneapolis. Due to the Coriolis force, it is deflected by an amount $\\delta x$ and $\\delta y$. Find the deflection. Ignore the centrifugal terms.\n\n{\\bf Solution}: The equations of motion are:\n\\begin{eqnarray*}\n\\frac{dv_x}{dt}&=&-2(\\omega_yv_z-\\omega_zv_y),\\\\\n\\frac{dv_y}{dt}&=&-2(\\omega_zv_x-\\omega_xv_z),\\\\\n\\frac{dv_z}{dt}&=&-g-2(\\omega_xv_y-\\omega_yv_x),\\\\\n\\omega_z&=&\\omega\\cos\\theta,~~~\\omega_y=\\omega\\sin\\theta,~~~\\omega_x=0.\n\\end{eqnarray*}\nHere the coordinate system is $\\hat{x}$ points east, $\\hat{y}$ points north and $\\hat{z}$ points upward.\n\nOne can now ignore all the Coriolis terms on the right-hand sides except for those with $v_z$. The other terms will all be doubly small. One can also throw out terms with $\\omega_x$. This gives\n\\begin{eqnarray*}\n\\frac{dv_x}{dt}&\\approx& -2\\omega v_z\\sin\\theta,\\\\\n\\frac{dv_y}{dt}&\\approx& 0,\\\\\n\\frac{dv_z}{dt}&\\approx& -g.\n\\end{eqnarray*}\nThere will be no significant deflection in the $y$ direction, $\\delta y=0$, but in the $x$ direction one can substitute $v_z=-gt$ above,\n\\begin{eqnarray*}\nv_x&\\approx&\\int_0^t dt'~2\\omega gt'\\sin\\theta=\\omega gt^2\\sin\\theta,\\\\\n\\delta x&\\approx& \\int_0^t dt'~v_x(t')=\\frac{g\\omega\\sin\\theta t^3}{3}.\n\\end{eqnarray*}\nOne can find the deflections by using $h=\\frac{1}{2}gt^2$, to find the time, and using the all-knowing internet to see that the latitude of Minneapolis is $44.6^\\circ$ or $\\theta=45.4^\\circ$.\n\\begin{eqnarray*}\nt&=&\\sqrt{2h/g}=10.1~{\\rm s},\\\\\n\\omega&=&\\frac{2\\pi}{3600\\cdot 24~{\\rm s}}=7.27\\times 10^{-5}~{\\rm s}^{-1},\\\\\n\\delta x&=&17.4~{\\rm cm}~~{\\rm(east)}.\n\\end{eqnarray*}\n\n\\exampleend\n\n\n\\subsection{The Foucault Pendulum}\nThe Foucault Pendulum is simply a regular pendulum moving in both horizontal directions, and with the Coriolis force included. In this case,\n\\begin{eqnarray*}\nm\\ddot{\\vec{r}}&=&\\vec{T}+m\\vec{g}-2m\\vec{\\Omega}\\times\\vec{v},\n\\end{eqnarray*}\nas the centrifugal force term is absorbed into the definition of $\\vec{g}$. The magnitude of the tension, $\\vec{T}$, is considered constant because we consider only small oscillations. Then $T\\approx mg$, and the components, using $\\hat{x},\\hat{y}$ to correspond to east and north respectively, are\n\\begin{eqnarray*}\nT_x=-mgx/L,~~~T_y=-mgy/L. \n\\end{eqnarray*}\nIf $\\Omega$ is the rotation of the earth, and if $\\theta$ is the polar angle, $\\pi-$lattitude, \n\\begin{eqnarray*}\n\\ddot{x}&=&-gx/L+2\\dot{y}\\Omega_z,\\\\\n\\ddot{y}&=&-gy/L-2\\dot{x}\\Omega_z.\n\\end{eqnarray*}\nHere we have used the fact that the oscillations are sufficiently small so we can ignore $v_z$. Using $\\omega_0\\equiv\\sqrt{k/m}$,\n\\begin{eqnarray*}\n\\ddot{x}-2\\Omega_z\\dot{y}+\\omega_0^2x&=&0\\\\\n\\ddot{y}+2\\Omega_z\\dot{x}+\\omega_0^2y&=&0,\n\\end{eqnarray*}\nwhere $\\Omega_z=|\\vec{\\Omega}|\\cos\\theta$, with $\\theta$ being the polar angle (zero at the north pole). The terms linear in time derivatives are what make life difficult. This will be solved with a trick. We will incorporate both differential equations into a single complex equation where the first/second are the real/imaginary parts.\n\\begin{eqnarray*}\n\\eta\\equiv x+iy,\\\\\n\\ddot{\\eta}+2i\\Omega_z\\dot{\\eta}+\\omega_0^2\\eta&=&0. \n\\end{eqnarray*}\n\nNow, we guess at a form for the solutions, $\\eta(t)=e^{-i\\alpha t}$, which turns the differential equation into\n\\begin{eqnarray*}\n-\\alpha^2+2\\Omega_z\\alpha+\\omega_0^2&=&0,\\\\\n\\alpha&=&\\Omega_z\\pm \\sqrt{\\Omega_z^2+\\omega_0^2},\\\\\n&\\approx&\\Omega_z\\pm \\omega_0.\n\\end{eqnarray*}\nThe solution with two arbitrary constants is then\n\\begin{eqnarray*}\n\\eta&=&e^{-i\\Omega_zt}\\left[C_1e^{i\\omega_0t}+C_2e^{-i\\omega_0t}\\right].\n\\end{eqnarray*}\nHere, $C_1$ and $C_2$ are complex, so they actually represent four arbitrary numbers. These four numbers should be fixed by the four initial conditions, i.e. $x(t=0), \\dot{x}(t=0), y(t=0)$ and $\\dot{y}(t=0)$. With some lengthy algebra, one can rewrite the expression as\n\\begin{eqnarray*}\n\\label{eq:precmess}\n\\eta&=&e^{-i\\Omega_zt}\\left[A\\cos(\\omega_0t+\\phi_A)+iB\\cos(\\omega_0t+\\phi_B)\\right].\n\\end{eqnarray*}\nHere, the four coefficients are represented by the two real arbitrary real amplitudes, $A$ and $B$, and two arbitrary phases, $\\phi_A$ and $\\phi_B$. For an initial condition where $y=0$ at $t=0$, one can see that $B=0$. This then gives\n\\begin{eqnarray*}\n\\eta(t)&=&Ae^{-i\\Omega_zt}\\cos(\\omega_0t+\\gamma)\\\\\n\\nonumber\n&=&A\\cos\\Omega_zt\\cos(\\omega_0t+\\gamma)+iA\\sin\\Omega_zt\\cos(\\omega_0t+\\gamma).\n\\end{eqnarray*}\nTranslating into $x$ and $y$,\n\\begin{eqnarray}\nx&=&A\\cos\\Omega_zt\\cos(\\omega_0t+\\gamma),\\\\\n\\nonumber\ny&=&A\\sin\\Omega_zt\\cos(\\omega_0t+\\gamma).\n\\end{eqnarray}\nAssuming the pendulum's frequency is much higher than Earth's rotational frequency, $\\omega_0>>\\Omega_z$, one can see that the plane of the pendulum simply precesses with angular velocity $\\Omega_z$. This means that in this limit the pendulum oscillates only in the $x$-direction with frequency many times before the phase $\\Omega_zt$ becomes noticeable. Eventually, when $\\Omega_zt=\\pi/2$, the motion is along the $y$-direction. If you were at the north pole, the motion would switch from the $x$-direction to the $y$ direction every 6 hours. Away from the north pole, $\\Omega_z\\ne|\\vec{\\Omega}|$ and the precession frequency is less. At the equator it does not precess at all. If one were to repeat for the solutions where $A=0$ and $B\\ne 0$ in Eq. (\\ref{eq:precmess}), one would look at motions that started in the $y$-direction, then precessed toward the $-x$ direction. Linear combinations of the two sets of solutions give pendulum motions that resemble ellipses rather than simple back-and-forth motion.\n\n\\subsection{Exercises}\n\n\\begin{enumerate}\n\n\\item Consider a pail of water spinning about a vertical axis at the center of the pail with frequency $\\omega$. Find the height of the water (within a constant) as a function the radius $r_\\perp$ from the axis of rotation. Use the concept of a centrifugal potential in the rotating frame.\n\n\\item A high-speed cannon shoots a projectile with an initial velocity\n of 1000 m/s in the east direction. The cannon is situated in\n Minneapolis (latitude of 45 degrees) The projectile velocity is\n nearly horizontal and it hits the ground after a distance $x=3000$\n m. Find the alteration of the point of impact in the north-south\n ($y$) direction due to the Coriolis force. Assume the effect is\n small so that you can approximate the eastward ($x$) component of\n the velocity as being constant. Be sure to indicate whether the\n deflection is north or south.\n\n\\item Someone wishes to use a Foucault pendulum as a crude clock. If the person lives in Minneapolis (latitude of $45^\\circ$), how much time will pass between having the pendulum swinging in the east-west direction until it swings in the north-south direction.\n\n\\end{enumerate}\n%\\end{document}\n", "meta": {"hexsha": "4008fcbe11fce634a7d563df259fb449a9574fa4", "size": 13604, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/AdminBackground/lectures/chapter5.tex", "max_stars_repo_name": "Shield94/Physics321", "max_stars_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2020-01-09T17:41:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T00:48:58.000Z", "max_issues_repo_path": "doc/AdminBackground/lectures/chapter5.tex", "max_issues_repo_name": "Shield94/Physics321", "max_issues_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2020-01-08T03:47:53.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-15T15:02:57.000Z", "max_forks_repo_path": "doc/AdminBackground/lectures/chapter5.tex", "max_forks_repo_name": "Shield94/Physics321", "max_forks_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 33, "max_forks_repo_forks_event_min_datetime": "2020-01-10T20:40:55.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-11T20:28:41.000Z", "avg_line_length": 66.3609756098, "max_line_length": 1011, "alphanum_fraction": 0.7280211702, "num_tokens": 4284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.7981867801399694, "lm_q1q2_score": 0.6350837769816946}} {"text": "\\documentclass{article}\n\n\\usepackage{mathrsfs,amssymb,amsmath}\n\n\\begin{document}\n\n\\section{Why does Theorem VIII.1.14 not contradict Example VII.5.1}\n\nTake $\\mathbb{T}$ as the unit circle in the complex plane, and take $\\mathscr{A} = C(\\mathbb{T})$ and $\\mathscr{B}\\subset \\mathscr{A}$ the closure in $\\mathscr{A}$ of polynomials in $z$. $\\sigma_{\\mathscr{A}}(z)=\\mathbb{T}$ and $\\sigma_{\\mathscr{B}}(z)= cl \\mathbb{T} =$ unit disk.\n\nThis appears to contradict VIII.1.14 since $\\mathbb{B}$ is a subset of $\\mathscr{A}$ with a common identity (namely the function that maps $\\mathbb{T}$ to 1) and common norm (namely the sup norm), yet the spectra of a shared point (in this case z) differ.\n\nHowever, the premise of VIII.1.14 requires both $\\mathscr{A}$ and $\\mathscr{B}$ to be C* algebras, and while $\\mathscr{A}$ is closed under taking the adjoint, $\\mathscr{B}$ is not.\n\nTake z*, which as a function equals $z^{-1}/||z||$. The problem thus reduces to on what subset of the complex plane the function $1/z$ is holomorphic.\n\nHere my complex analysis is too weak, but certainly the whole unit disk would be no good since it contains a pole. I believe the circle is also out because of some fact about winding numbers that I can't remember.\n\n\\end{document}\n", "meta": {"hexsha": "eb5f3d9e7e546c127b6c4078c30a7381f598c526", "size": 1256, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "analysis/8_C_star_Algebras/hw.tex", "max_stars_repo_name": "lukemassa/math-exercises", "max_stars_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "analysis/8_C_star_Algebras/hw.tex", "max_issues_repo_name": "lukemassa/math-exercises", "max_issues_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "analysis/8_C_star_Algebras/hw.tex", "max_forks_repo_name": "lukemassa/math-exercises", "max_forks_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.8, "max_line_length": 281, "alphanum_fraction": 0.722133758, "num_tokens": 367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867873410141, "lm_q2_score": 0.7956580927949807, "lm_q1q2_score": 0.6350837769099041}} {"text": "\\section{Recursion Equations}\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Equation}\r\n \\textbf{Recursion equation:}\r\n \\begin{itemize}\r\n \\item\r\n Runtime description for recursive functions:\r\n \\end{itemize}\r\n \\vspace{2em}\r\n \\begin{displaymath}\r\n {\\color{MainA}T(n)} =\r\n \\begin{cases}\r\n \\hfill\r\n \\smash{\r\n \\overbrace{f_0(n)}^\\text{%\r\n \\footnotesize%\r\n trivial case for $n_0$\r\n }\r\n }\r\n \\vphantom{f_0(n)} & n = n_0\\\\[0.5em]\r\n \\smash{\r\n \\underbrace{\r\n a \\cdot {\\color{MainA}T\\left(\\frac{n}{b}\\right)}\r\n }_\\text{\r\n \\footnotesize%\r\n \\begin{tabular}{c}\r\n \\text{solving of {\\color{MainA}$a$}}\\\\\r\n \\text{subproblems}\\\\\r\n \\text{with reduced}\\\\\r\n \\text{input size {\\color{MainA}$\\frac{n}{b}$}}\r\n \\end{tabular}\r\n }\r\n } \\hspace{0.5em} + \\hspace{1.5em}\r\n \\smash{\r\n \\underbrace{\r\n f(n)\r\n \\vphantom{\\left(\\frac{n}{b}\\right)}\r\n }_\\text{\r\n \\makebox[0pt][c]{\r\n \\footnotesize%\r\n \\begin{tabular}{c}\r\n slicing and\\\\\r\n splicing of\\\\\r\n subsolutions\r\n \\end{tabular}\r\n }\r\n }\r\n }\r\n \\vphantom{\\left(\\frac{n}{b}\\right)}\r\n &n > n_0\r\n \\end{cases}\r\n \\end{displaymath}\r\n \\vspace{4em}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Equation}\r\n \\textbf{Recursion equation:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Runtime descripion for recursive functions:\r\n \\begin{displaymath}\r\n {\\color{MainA}T(n)} = \\begin{cases}\r\n f_0(n) & n = n_0\\\\\r\n a \\cdot {\\color{MainA}T\\left(\\frac{n}{b}\\right)} + f(n) & n > n_0\r\n \\end{cases}\r\n \\end{displaymath}\r\n \\item<3->\r\n {\\color{MainA}$n_0$} is usually small, {\\color{MainA}$f_0(n_0) \\; \\in \\; \\Theta(1)$}\r\n \\item<4->\r\n Usually, {\\color{MainA}$a > 1$} and {\\color{MainA}$b > 1$}\r\n \\item<5->\r\n Dependent on the strategy of solving {\\color{MainA}$T(n)$}\r\n {\\color{MainA}$f_0$} is ignored\r\n \\item<6->\r\n {\\color{MainA}$T(n)$} is only defined for integers of\r\n {\\color{MainA}$\\frac{n}{b}$}, which is often ignored in benefit of\r\n a simpler solution\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\subsection{Substitution Method}\r\n\r\n\\begin{frame}{Recursion Equations}{Substitution Method}\r\n \\textbf{Substitution Method:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Guess the solution and prove it with induction\r\n \\item<3->\r\n Example: \\color{MainA}\r\n \\begin{displaymath}\r\n T(n) = \\begin{cases}\r\n \\hfill 1 & n = 1\\\\\r\n 2 \\cdot T\\left(\\frac{n}{2}\\right) + n & n > 1\r\n \\end{cases}\r\n \\end{displaymath} \\color{black}\r\n \\item<4->\r\n Assumption: {\\color{MainA}$T(n) = n + n \\cdot \\log_2 n$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Substitution Method}\r\n \\textbf{Induction:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Induction basis (for {\\color{MainA}$n = 1$}):\r\n ${\\color{MainA}T(1)} = 1 + 1 \\cdot \\log_2 1 = 1$\r\n \\item<3->\r\n Induction step (from {\\color{MainA}$\\frac{n}{2}$}\r\n to {\\color{MainA}$n$}):\r\n \\begin{align*}\r\n \\onslide<4->{{\\color{MainA}T(n)} & =\r\n 2 \\cdot {\\color{MainA}T\\left(\\dfrac{n}{2}\\right)} + n\\\\}\r\n \\onslide<5->{{} & \\stackrel{IA}{=} 2 \\cdot \\left(\r\n \\dfrac{n}{2} + \\dfrac{n}{2} \\cdot \\log_2 \\dfrac{n}{2}\r\n \\right) + n \\\\}\r\n \\onslide<6->{{} & = 2 \\cdot \\left(\r\n \\dfrac{n}{2} + \\dfrac{n}{2} \\cdot \\left(\\log_2 n - 1\\right)\r\n \\right) + n}\\\\\r\n \\onslide<7->{{} & = n + n \\, \\log_2 n - n + n\\\\}\r\n \\onslide<8->{{} & = n + n \\, \\log_2 n}\r\n \\end{align*}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Substitution Method}\r\n \\textbf{Substitution Method:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Alternative assumption\r\n \\item<3->\r\n Example:\r\n \\begin{displaymath}\r\n \\color{MainA}\r\n T(n) = \\begin{cases}\r\n \\hfill 1 & n = 1\\\\\r\n 2 \\cdot T\\left(\\frac{n}{2}\\right) + n & n > 0\r\n \\end{cases}\r\n \\end{displaymath}\r\n \\item<4->\r\n Assumption:\r\n {\\color{MainA}$T(n) \\, \\in \\, O(n \\, \\log n)$}\r\n \\item<5->\r\n Solution: Find {\\color{MainA}$c > 0$} with\r\n {\\color{MainA}$T(n) \\leq c \\cdot n \\, \\log_2 n$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Substitution Method}\r\n \\textbf{Induction:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Solution: Find {\\color{MainA}$c > 0$} with\r\n {\\color{MainA}$T(n) \\leq c \\cdot n \\, \\log_2 n$}\r\n \\item<3->\r\n Induction step (from {\\color{MainA}$\\frac{n}{2}$} to\r\n {\\color{MainA}$n$}):\r\n \\begin{align*}\r\n \\onslide<4->{{\\color{MainA}T(n)} & =\r\n 2 \\cdot {\\color{MainA}T\\left(\\dfrac{n}{2}\\right)} + n\\\\}\r\n \\onslide<5->{{} & \\leq 2 \\cdot \\left(\r\n c \\cdot \\dfrac{n}{2} \\, \\log_2 \\dfrac{n}{2}\\right) + n\\\\}\r\n \\onslide<6->{{} & = c \\cdot n \\, \\log_2 n - c \\cdot n \\, \\log_2 2 + n\\\\}\r\n \\onslide<7->{{} & = c \\cdot n \\, \\log_2 n - c \\cdot n + n\\\\}\r\n \\onslide<8->{{} & \\leq c \\cdot n \\, \\log_2 n,\r\n \\hspace{0.5em} {\\color{MainA}c \\geq 1}}\r\n \\end{align*}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\subsection{Recursion Tree Method}\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Tree Method}\r\n \\textbf{Recursion tree method:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Can be used to make assumptions about the runtime\r\n \\item<3->\r\n Example:\\\\\r\n \\begin{displaymath}\r\n {\\color{MainA}T(n)}\r\n = 3 \\cdot {\\color{MainA}T\\left(\\dfrac{n}{4}\\right)} + \\Theta(n^2)\r\n \\leq 3 \\cdot {\\color{MainA}T\\left(\\dfrac{n}{4}\\right)} + c \\cdot \r\n n^2\r\n \\end{displaymath}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Tree Method}\r\n \\begin{figure}\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\input{Images/RecursionEquations/Example_RecursionTree.tikz}\r\n \\end{adjustbox}\r\n \\caption{recursion tree of example}\r\n \\label{fig:recursion_equations:example_recursion_tree}\r\n \\end{figure}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Tree Method}\r\n \\begin{figure}\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\input{Images/RecursionEquations/Example_RecursionTree_Levels.tikz}\r\n \\end{adjustbox}\r\n \\caption{levels of the recursion tree}\r\n \\label{fig:recursion_equations:example_recursion_tree_layer}\r\n \\end{figure}\r\n\\end{frame}\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Tree Method Costs}\r\n \\textbf{Costs of connecting the partial solutions:}\\\\\r\n \\hspace{1.5em}(excludes the last layer)\r\n \\begin{itemize}\r\n \\item<2->\r\n Size of partial problems on {\\color{MainA}level $i$}:\r\n \\begin{math}\r\n \\color{MainA}{\r\n \\mathrm{s}_i(n) = \\left(\\frac{1}{4}\\right)^i \\cdot n\r\n }\r\n \\end{math}\r\n \\item<3->\r\n Costs of partial problems on {\\color{MainA}level $i$}:\r\n \\begin{displaymath}\r\n \\color{MainA}{\r\n T_{i_p}(n) =\r\n c \\cdot \\left(\\left(\\tfrac{1}{4}\\right)^i \\cdot n\\right)^2\r\n }\r\n \\end{displaymath}\r\n \\item<4->\r\n Number of partial problems on {\\color{MainA}level $i$}:\r\n {\\color{MainA}$n_{i} = 3^{i}$}\r\n \\item<5->\r\n Costs on {\\color{MainA}level $i$}:\r\n \\color{MainA}\r\n \\begin{displaymath}\r\n T_i(n) = 3^{i} \\cdot c \\cdot \\left(\r\n \\left(\\tfrac{1}{4}\\right)^{i} \\cdot n\r\n \\right)^2\r\n = \\left(\\tfrac{3}{16}\\right)^{i} \\cdot c \\cdot n^2\r\n \\end{displaymath}\r\n \\color{black}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Recursion Tree Method Costs}\r\n \\textbf{Costs of solving partial solutions:} (only the last layer)\r\n \\begin{itemize}\r\n \\item<2->\r\n Size of partial problems on the {\\color{MainA}last level}:\r\n {\\color{MainA}$\\mathrm{s}_{i+1}(n) = 1$}\r\n \\item<3->\r\n Costs of partial problem on the {\\color{MainA}last level}:\r\n {\\color{MainA}$T_{i+1_p}(n) = d$}\r\n \\item<4->\r\n With this the depth of the tree is:\r\n \\begin{displaymath}\r\n \\left(\\tfrac{1}{4}\\right)^i \\cdot n = 1\r\n \\hspace{1.5em}\\Rightarrow n = 4^i\r\n \\hspace{1.5em}\\Rightarrow\\color{MainA} i = \\log_4 n\r\n \\end{displaymath}\r\n \\vspace{-1em}\r\n \\item<5->\r\n Number of partial problems on the {\\color{MainA}last level}:\r\n \\begin{displaymath}\r\n \\color{MainA}{n_{i+1} = 3^{\\log_4 n} \\onslide<6-> \\quad=\r\n n^{\\log_4 3}} \\quad \\leftarrow \\text{next slide}\r\n \\end{displaymath}\r\n \\item<7->\r\n Costs on the {\\color{MainA}last level}:\r\n {\\color{MainA}$T_{i+1}(n) = d \\cdot n^{\\log_4 3}$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Fun with logarithm}{Logarithm}\r\n \\begin{itemize}\r\n \\item\r\n Transforming $\\color{MainA}3^{\\log_4 n}$ using general log rules\r\n {\\color{MainA}\r\n \\begin{align*}\r\n \\log_4 n & = \\log_4 \\left(3^{\\log_3 n} \\right) &&\r\n \\qquad \\text{\\color{black}using } n= 3^{\\log_3 n}\\\\\r\n \\onslide<2->{& = {\\log_3 n \\cdot \\log_4 3} &&\r\n \\qquad \\text{\\color{black}using } \\log a^b = b \\cdot \\log a}\r\n \\end{align*}}\r\n \\item<3->\r\n This proves the general log rule\r\n {\\color{MainA}$\\log_b c = \\log_a c \\cdot \\log_b a$}\r\n \\item<4->\r\n Now the whole expression:\r\n\\color{MainA}\r\n \\begin{align*}\r\n 3^{\\log_4 n} & = 3^{\\log_3 n \\cdot \\log_4 3} &&\r\n \\qquad \\text{\\color{black}using reformulation above}\\\\\r\n \\onslide<5->{& = {\\left(3^{\\log_3 n}\\right)}^{\\log_4 3} &&\r\n \\qquad \\text{\\color{black}using }x^{a \\cdot b} = (x^a)^b}\\\\\r\n \\onslide<6->{& = n^{\\log_4 3}}\r\n \\end{align*}\r\n \\item<7->\r\n {\\color{gray}This term will recur in the master theorem}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Total costs}\r\n \\textbf{Total costs:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Costs of {\\color{MainA}level i}:\r\n $T_i(n) = \\left(\\frac{3}{16}\\right)^i \\cdot c \\cdot n^2$\r\n \\item<3->\r\n Costs of {\\color{MainA}last level}:\r\n $T_{i+1}(n) = d \\cdot n^{\\log_4 3}$\r\n \\end{itemize}\r\n \\vspace{0.5em}\r\n \\onslide<4->\r\n \\begin{displaymath}\r\n \\mathclap{T(n) =}\\underbrace{\r\n \\sum\\limits_{i = 0}^{(\\log_4 n) - 1} \\left(\\tfrac{3}{16}\\right)^i\r\n \\mathrlap{\\cdot c \\cdot n^2}\r\n }_{\r\n \\text{\r\n \\footnotesize\r\n \\begin{tabular}{c}\r\n geometric series,\\\\\r\n constant\\\\\r\n \\color{gray}\r\n $\\left(\\begin{array}{c}\r\n \\text{even with}\\\\\r\n \\text{infinite elements}\r\n \\end{array}\\right)$\r\n \\end{tabular}\r\n }\r\n } \\hspace{1.25em} + \\hspace{0.5em} \\underbrace{\r\n d \\cdot n^{\\log_4 3}\r\n \\vphantom{\\sum\\limits_{i = 0}^{(\\log_4 n) - 1}}\r\n }_{\r\n \\text{\r\n \\makebox[0pt][c]{\r\n \\footnotesize\r\n \\begin{tabular}{c}\r\n {\\color{MainA}$\\log_4 3 < 1$},\\\\\r\n grows a lot\\\\\r\n slower than {\\color{MainA}$n^2$}\r\n \\end{tabular}\r\n }\r\n }\r\n } ~ \\in \\mathcal{O}(n^2)\r\n \\end{displaymath}\r\n \\begin{itemize}\r\n \\item<5->\r\n Here: The costs of connecting the partial problems dominate\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Geometric Series}\r\n \\begin{itemize}\r\n \\item\r\n \\textbf{Geometric progression:}\\\\\r\n Quotient of two neighboring sequence parts is constant\r\n \\begin{displaymath}\r\n 2^0, 2^1, 2^2, \\ldots, 2^k\r\n \\end{displaymath}\r\n \\item<2->\r\n \\textbf{Geometric series:}\\\\\r\n The series (cumulative sum) of a geometric sequence\\\\\r\n \\item<3->\r\n For $\\mid q \\mid < 1$:\r\n \\begin{displaymath}\r\n \\sum\\limits^{\\infty}_{k=0} a_0 \\cdot q^k = \\dfrac{a_0}{1 - q}\r\n \\hspace{1.5em}\\Rightarrow\\text{constant}\r\n \\end{displaymath}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Proof of $O(n^2)$}\r\n \\textbf{Proof of $\\mathcal{O}(n^2)$:}\r\n \\begin{itemize}\r\n \\item<2->\r\n We know:\r\n \\begin{align*}\r\n T(n) &= 3T\\left(\\frac{n}{4}\\right) + \\Theta(n^2)\\\\\r\n {} &\\leq 3T \\left(\\frac{n}{4}\\right) + c \\cdot n^2\r\n \\end{align*}\r\n \\item<3->\r\n Assumption:\r\n {\\color{MainA}$T(n) \\, \\in \\, \\mathcal{O}(n^2)$},\r\n so there exists a {\\color{MainA}$k > 0$} with\r\n \\begin{displaymath}\r\n \\color{MainA}\r\n T(n) \\leq k \\cdot n^2\r\n \\end{displaymath}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Proof of $O(n^2)$}\r\n \\textbf{Proof of $\\mathcal{O}(n^2)$:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Presumption:\r\n {\\color{MainA}$T(n) \\, \\in \\, \\mathcal{O}(n^2)$},\r\n so there exists a {\\color{MainA}$k > 0$} with\r\n \\begin{displaymath}\r\n \\color{MainA}{T(n) < k \\cdot n^2}\r\n \\end{displaymath}\r\n \\item<3->\r\n Substitution method:\r\n \\begin{align*}\r\n \\onslide<4->{T(n) & \\leq 3 \\cdot T \\left( \\frac{n}{4}\\right) + c \\cdot n^2\\\\}\r\n \\onslide<5->{{} & \\leq 3 \\, k \\cdot \\left( \\frac{n}{4}\\right)^2 + c \\cdot n^2\\\\}\r\n \\onslide<6->{{} & = \\frac{3}{16} \\, k \\cdot n^2 + c \\cdot n^2\\\\}\r\n \\onslide<7->{{} & \\leq k \\cdot n^2\r\n \\hspace{6em}\\text{for } k \\geq \\frac{16}{13} \\, c}\r\n \\end{align*}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\subsection{Master theorem}\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem}\r\n \\textbf{Master theorem:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Solution approach for a recursion equation of the form:\r\n \\begin{displaymath}\r\n \\color{MainA}{\r\n T(n) = a \\cdot T\\left(\\frac{n}{b}\\right) + f(n),%\r\n \\hspace{1.5em}a \\geq 1, b > 1\r\n }\r\n \\end{displaymath}\r\n \\item<3->\r\n {\\color{MainA}$T(n)$} is the runtime of an algorithm $\\ldots$\r\n \\begin{itemize}\r\n \\item<4->\r\n $\\ldots$ which divides a {\\color{MainA}problem of size $n$}\r\n in {\\color{MainA}$a$ partial problems}\r\n \\item<5->\r\n $\\ldots$ which solves each partial problem recursively\\newline\r\n \\hphantom{$\\ldots$} with a\r\n {\\color{MainA}runtime of $T\\left(\\tfrac{n}{b}\\right)$}\r\n \\item<6->\r\n $\\ldots$ which takes {\\color{MainA}$f(n)$} steps to\r\n merge all partial solutions\r\n \\end{itemize} \r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\subsubsection{Master theorem (Simple Form)}\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\textbf{Master theorem:}\r\n \\begin{itemize}\r\n \\item<2->\r\n In the examples we have seen that $\\ldots$\r\n \\begin{itemize}\r\n \\item<3->\r\n Either the runtime of {\\color{MainA}connecting the solutions}\r\n dominates\r\n \\item<4->\r\n Or the runtime of {\\color{MainA}solving the problems} dominates\r\n \\item<5->\r\n Or both have {\\color{MainA}equal influence on runtime}\r\n \\end{itemize}\r\n \\item<6->\r\n \\textbf{Simple form:}\r\n Special case with runtime of connecting the solutions\r\n {\\color{MainA}$f(n) \\in O(n)$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\textbf{Simple form:}\r\n \\onslide<2->\r\n \\begin{displaymath}\r\n \\color{MainA}{\r\n T(n) = a \\cdot T\\left(\\frac{n}{b}\\right) +\r\n \\underbrace{c \\cdot n}_{\r\n \\text{\r\n \\clap{\r\n \\footnotesize\\color{black}\r\n \\begin{tabular}{c}\r\n Is any $f(n)$\\\\\r\n in general form\r\n \\end{tabular}\r\n }\r\n }\r\n },\r\n \\hspace{1.5em}a \\geq 1, b > 1, c > 0\r\n }\r\n \\end{displaymath}\r\n \\begin{itemize}\r\n \\item<3->\r\n This yields a runtime of:\r\n \\end{itemize}\r\n \\onslide<4->\r\n \\vspace{2em}\r\n \\begin{displaymath}\r\n T(n) = \\begin{cases}\r\n \\Theta({\\color{MainA}\r\n \\smash{\\overbrace{n^{\\log_b a}}^{\r\n \\text{\\clap{Number of leaves}}\r\n }}\r\n }) & \\text{if } a > b\\\\\r\n \\Theta(n \\, \\log n) & \\text{if } a = b\\\\\r\n \\Theta(n) & \\text{if } a < b\r\n \\end{cases}\r\n \\end{displaymath}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\def\\AlgoREDivide{3}% 3\r\n \\def\\AlgoRESize{0.5}% 1/2\r\n \\def\\AlgoREScale{4.444}% 15 / (3/2)^3\r\n \\input{Images/RecursionEquations/ProblemSizeTree.tikz}\r\n \\end{adjustbox}\r\n \\caption{simple recursion equation with {\\color{MainA}$a = 3, b = 2$}}\r\n \\label{fig:recursion_equations:master_theorem_tree_3_2}\r\n \\end{figure}\r\n \\onslide<2->\r\n \\textbf{Case 1:} {\\color{MainA}$a > b$}\r\n \\begin{itemize}\r\n \\item<3->\r\n Three partial problems with $\\frac{1}{2}$ the size\r\n \\item<4->\r\n Solving the partial problems dominates (last layer, leaves)\r\n \\item<5->\r\n Runtime of {\\color{MainA}$\\Theta(n^{\\log_b a})$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\def\\AlgoREDivide{2}% 2\r\n \\def\\AlgoRESize{0.5}% 1/2\r\n \\def\\AlgoREScale{15}% 15\r\n \\input{Images/RecursionEquations/ProblemSizeTree.tikz}\r\n \\end{adjustbox}\r\n \\caption{simple recursion equation with {\\color{MainA}$a = 2, b = 2$}}\r\n \\label{fig:recursion_equations:master_theorem_tree_2_2}\r\n \\end{figure}\r\n \\onslide<2->\r\n \\textbf{Case 2:} {\\color{MainA}$a = b$}\r\n \\begin{itemize}\r\n \\item<3->\r\n Two partial problems with $\\frac{1}{2}$ the size\r\n \\item<4->\r\n Each layer has equal costs, $\\log n$ layers\r\n \\item<5->\r\n Runtime of {\\color{MainA}$\\Theta(n \\, \\log n)$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n%\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n% \\begin{figure}[!h]\r\n% \\begin{adjustbox}{width=\\linewidth}\r\n% \\def\\AlgoREDivide{2}% 2\r\n% \\def\\AlgoRESize{0.25}% 1/4\r\n% \\def\\AlgoREScale{15}% 15\r\n% \\input{Images/RecursionEquations/ProblemSizeTree.tikz}\r\n% \\end{adjustbox}\r\n% \\caption{simple recursion equation with\r\n% {\\color{MainA}$a = 2, b = 4$}}\r\n% \\label{fig:recursion_equations:master_theorem_tree_2_4}\r\n% \\end{figure}\r\n% \\textbf{Case 3:} {\\color{MainA}$a < b$}\r\n% \\begin{itemize}\r\n% \\item\r\n% Two partial problems with $\\frac{1}{4}$ the size\r\n% \\item\r\n% Connecting all partial solutions dominates (first layer, root)\r\n% \\item\r\n% Runtime of {\\color{MainA}$\\Theta(n)$}\r\n% \\end{itemize}\r\n%\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\begin{figure}[!h]\r\n \\begin{adjustbox}{width=\\linewidth}\r\n \\def\\AlgoREDivide{2}% 2\r\n \\def\\AlgoRESize{0.33}% 1/3\r\n \\def\\AlgoREScale{15}% 15\r\n \\input{Images/RecursionEquations/ProblemSizeTree.tikz}\r\n \\end{adjustbox}\r\n \\caption{simple recursion equation with {\\color{MainA}$a = 2, b = 3$}}\r\n \\label{fig:recursion_equations:master_theorem_tree_2_3}\r\n \\end{figure}\r\n \\onslide<2->\r\n \\textbf{Case 3:} {\\color{MainA}$a < b$}\r\n \\begin{itemize}\r\n \\item<3->\r\n Two partial problems with $\\frac{1}{3}$ the size\r\n \\item<4->\r\n Connecting all partial solutions dominates (first layer, root)\r\n \\item<5->\r\n Runtime of {\\color{MainA}$\\Theta(n)$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (Simple Form)}\r\n \\textbf{For a recursion equation like}\r\n \\begin{displaymath}\r\n \\color{MainA}{%\r\n T(n) = a \\cdot T\\left(\\frac{n}{b}\\right) + c \\cdot n,%\r\n \\hspace{1.5em}a \\geq 1, b > 1, c > 0%\r\n }%\r\n \\end{displaymath}\r\n \\vspace{-1.0em}\r\n \\begin{itemize}\r\n \\item<2->\r\n $\\ldots$ yields a runtime of:\r\n \\end{itemize}\r\n \\onslide<3->\r\n \\begin{displaymath}\r\n T(n) = \\begin{cases}\r\n \\Theta(n^{\\log_b a}) & \\text{if } a > b\\\\\r\n \\Theta(n \\, \\log_b n) & \\text{if } a = b\\\\\r\n \\Theta(n) & \\text{if } a < b\r\n \\end{cases}\r\n \\end{displaymath}\r\n \\begin{itemize}\r\n \\item<4->\r\n Proof with \\textit{geometric series}:\r\n Number of operations per layer grows / shrinks by constant factor\r\n $\\tfrac{a}{b}$\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\subsubsection{Master theorem (General Form)}\r\n\r\n\\begin{frame}{\\LectureToC}\r\n \\tableofcontents[currentsection,\r\n currentsubsection,\r\n subsubsectionstyle=show/shaded/shaded/shaded]\r\n\\end{frame}\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form)}\r\n \\textbf{Master theorem (general form):}\r\n \\onslide<2->\r\n \\begin{displaymath}\r\n \\color{MainA}{%\r\n T(n) = a \\cdot T\\left(\\frac{n}{b}\\right) + f(n),%\r\n \\hspace{1.5em}a \\geq 1, b > 1%\r\n }%\r\n \\end{displaymath}\r\n \\vspace{-1.0em}\r\n \\begin{itemize}\r\n \\item<3->\r\n \\textbf{Case 1:}\r\n {\\color{MainB}$T(n) \\in \\Theta(n^{\\log_b a})$}\r\n \\hfill if {\\color{MainA}\r\n $f(n) \\in \\mathcal{O}(n^{\\log_b a-\\varepsilon}), ~ \\varepsilon > 0$\r\n }\\\\\r\n {\\color{gray}Solving the partial problems dominates\\\\\r\n (last layer, leaves)}\r\n \\vspace{1.5em}\r\n \\item<4->\r\n \\textbf{Case 2:}\r\n {\\color{MainB}$T(n) \\in \\Theta(n^{\\log_b a} \\log n)$}\r\n \\hspace{1.5em} if {\\color{MainA}\r\n $f(n) \\in \\Theta(n^{\\log_b a})$\r\n }\\\\\r\n {\\color{gray}Each layer has equal costs, $\\log_b n$ layers}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form)}\r\n \\textbf{Master theorem (general form):}\r\n \\begin{itemize}\r\n \\item\r\n \\textbf{Case 3:}\r\n {\\color{MainB}$T(n) \\in \\Theta(f(n))$}\r\n \\hfill if {\\color{MainA}\r\n $f(n) \\in \\Omega(n^{\\log_b a+\\varepsilon}), ~ \\varepsilon > 0$\r\n }\\\\\r\n {\\color{gray}Connecting all partial solutions in first layer (root) dominates}\\\\[1.0em]\r\n {\\color{MainA}Regularity condition:}\r\n \\begin{align*}\r\n a \\cdot f\\left(\\frac{n}{b}\\right) \\leq c \\cdot f(n), \\hspace{1.5em} &\r\n 0 \\leq c \\leq 1,\\\\\r\n {} & n > n_0\r\n \\end{align*}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form) - Case 1}\r\n \\textbf{Case 1 - Example:}\r\n \\onslide<2->{{\\color{MainB}$T(n) \\in \\Theta(n^{\\log_b a})$}}\r\n \\hfill if {\\color{MainA}\r\n $f(n) \\in O(n^{\\log_b a-\\varepsilon}), ~ \\varepsilon > 0$\r\n }\\\\\r\n {\\color{gray}Solving the partial problems dominates (last layer, leaves)}\r\n \\vspace{1.0em}\r\n \\begin{itemize}\r\n \\item<3->\r\n $T(n) = 8 \\cdot T(\\tfrac{n}{2}) + 1000 \\cdot n^2$\r\n \\vspace{-0.5em}\r\n \\begin{displaymath}\r\n a = 8, ~ b = 2, ~ f(n) = 1000 \\cdot n^2, ~\r\n \\smash{\\underbrace{\r\n \\log_b a = \\log_2 8 = 3\r\n }_{\\text{$n^3$ leaves}}}\r\n \\end{displaymath}\r\n {\\color{MainA}\\begin{math}\r\n f(n) \\in \\mathcal{O}(n^{3-\\varepsilon})$\r\n $\\Rightarrow$\r\n $T(n) \\in \\Theta(n^3)\r\n \\end{math}}\r\n \\vspace{1.5em}\r\n \\item<4->\r\n $T(n) = 9 \\cdot T(\\tfrac{n}{3}) + 17 \\cdot n$\r\n \\vspace{-0.5em}\r\n \\begin{displaymath}\r\n a = 9, ~ b = 3, ~ f(n) = 17 \\cdot n, ~\r\n \\smash{\\underbrace{\r\n \\log_b a = \\log_3 9 = 2\r\n }_{\\text{$n^2$ leaves}}}\r\n \\end{displaymath}\r\n {\\color{MainA}\\begin{math}\r\n f(n) \\in \\mathcal{O}(n^{2-\\varepsilon})$\r\n $\\Rightarrow$\r\n $T(n) \\in \\Theta(n^2)\r\n \\end{math}}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form) - Case 2}\r\n \\textbf{Case 2:}\r\n \\onslide<2->{{\\color{MainB}$T(n) \\in \\Theta(n^{\\log_b a} \\log n)$}}\r\n \\hspace{1.5em} if {\\color{MainA}\r\n $f(n) \\in \\Theta(n^{\\log_b a})$\r\n }\\\\\r\n {\\color{gray}Each layer has equal costs, $\\log n$ layers}\r\n \\vspace{1.0em}\r\n \\begin{itemize}\r\n \\item<3->\r\n $T(n) = 2 \\cdot T(\\tfrac{n}{2}) + 10 \\cdot n$\r\n \\vspace{-0.5em}\r\n \\begin{displaymath}\r\n a = 2, ~ b = 2, ~ f(n) = 10 \\cdot n, ~\r\n \\smash{\\underbrace{\r\n \\log_b a = \\log_2 2 = 1\r\n }_{\\text{$n^1$ leaves}}}\r\n \\end{displaymath}\r\n {\\color{MainA}\\begin{math}\r\n f(n) \\in \\Theta(n^{\\log_2 2})$\r\n $\\Rightarrow$\r\n $T(n) \\in \\Theta(n \\, \\log n)\r\n \\end{math}}\r\n \\vspace{1.0em}\r\n \\item<4->\r\n $T(n) = T(\\tfrac{2\\,n}{3}) + 1$\r\n \\vspace{-0.5em}\r\n \\begin{displaymath}\r\n a = 1, ~ b = \\tfrac{3}{2}, ~ f(n) = 1, ~\r\n \\smash{\\underbrace{\r\n \\log_b a = \\log_{\\nicefrac{3}{2}} 1 = 0\r\n }_{\\text{$n^0$ leaves $=$ 1 leaf}}}\r\n \\end{displaymath}\r\n \\hfill\\\\[1.0em]\r\n {\\color{MainA}\\begin{math}\r\n f(n) \\in \\Theta(n^{\\log_{\\nicefrac{3}{2}} 1})$\r\n $\\Rightarrow$\r\n $T(n) \\in \\Theta(n^0 \\log n) = \\Theta(\\log n)\r\n \\end{math}}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form) - Case 3}\r\n \\textbf{Case 3:}\r\n \\onslide<2->{{\\color{MainB}$T(n) \\in \\Theta(f(n))$}}\r\n \\hfill if {\\color{MainA}\r\n $f(n) \\in \\Omega(n^{\\log_b a+\\varepsilon}), ~ \\varepsilon > 0$\r\n }\\\\\r\n {\\color{gray}Connecting all partial solutions in first layer (root) dominates}\r\n \\vspace{1.0em}\r\n \\begin{itemize}\r\n \\item<3->\r\n $T(n) = 2 \\cdot T(\\tfrac{n}{2}) + n^2$\r\n \\vspace{-0.5em}\r\n \\begin{displaymath}\r\n a = 2, ~ b = 2, ~ f(n) = n^2, ~\r\n \\smash{\\underbrace{\r\n \\log_b a = \\log_2 2 = 1\r\n }_{\\text{$n^1$ leaves}}}\r\n \\end{displaymath}\r\n {\\color{MainA}$f(n) \\in \\Omega(n^{1+\\varepsilon})$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form) - Case 3}\r\n \\textbf{Case 3:}\r\n {\\color{MainB}$T(n) \\in \\Theta(f(n))$}\r\n \\hfill if {\\color{MainA}\r\n $f(n) \\in \\Omega(n^{\\log_b a+\\varepsilon}), ~ \\varepsilon > 0$\r\n }\\\\\r\n {\\color{gray}Connecting all partial solutions in first layer (root) dominates}\r\n \\vspace{1.0em}\r\n \\begin{itemize}\r\n \\item\r\n $T(n) = 2 \\cdot T(\\tfrac{n}{2}) + n^2$\r\n \\item<3->\r\n {\\color{MainA}$f(n) \\in \\Omega(n^{1+\\varepsilon})$}\r\n \\item<4->\r\n Check if {\\color{MainA}regularity condition} also holds:\\\\\r\n \\begin{align*}\r\n a \\cdot f\\left(\\frac{n}{b}\\right) &\\leq c \\cdot f(n)\\\\\r\n 2 \\cdot \\left(\\frac{n}{2}\\right)^2 &\\leq c \\cdot n^2\r\n \\hspace{1.5em} \\Rightarrow \\frac{1}{2} \\cdot n^2 \\leq c \\cdot n^2\r\n \\hspace{1.5em} \\Rightarrow c \\geq \\frac{1}{2}\r\n \\end{align*}\r\n {\\color{MainA}$\\Rightarrow T(n) \\in \\Theta(n^2)$}\r\n \\end{itemize}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem (General Form)}\r\n \\textbf{Master theorem:}\r\n \\begin{itemize}\r\n \\item<2->\r\n Not always applicable:\r\n {\\color{MainA}$T(n) = 2 \\cdot T(\\frac{n}{2}) + n \\, \\log n$}\r\n \\end{itemize}\r\n \\onslide<3->\r\n \\vspace{1.0em}\r\n \\begin{displaymath}\r\n a = 2, ~ b = 2, ~ f(n) = n \\, \\log n, ~\r\n \\underbrace{\\log_b a = \\log_2 2 = 1}_{\\text{$n^1$ leaves}}\r\n \\end{displaymath}\r\n \\begin{itemize}\r\n \\item<4->\r\n \\textbf{Case 1:}\r\n $f(n) \\notin O(n^{1-\\varepsilon})$\r\n \\item<5->\r\n \\textbf{Case 2:}\r\n $f(n) \\notin \\Theta(n^1)$\r\n \\item<6->\r\n \\textbf{Case 3:}\r\n $f(n) \\notin \\Omega(n^{1+\\varepsilon})$\r\n \\end{itemize}\r\n \\onslide<7->\r\n \\begin{center}\r\n {\\color{MainA}$n \\, \\log n$} is \\textit{asymptotically} larger than\r\n {\\color{MainA}$n$},\\\\ but not \\textit{polynominal} larger\r\n \\end{center}\r\n\\end{frame}\r\n\r\n%-------------------------------------------------------------------------------\r\n\r\n\\begin{frame}{Recursion Equations}{Master theorem - Summary}\r\n \\textbf{Master theorem:}\r\n \\onslide<2->\r\n \\begin{displaymath}\r\n \\color{MainA}{\r\n T(n) = a \\cdot T\\left(\\frac{n}{b}\\right) + f(n)\r\n }\r\n \\end{displaymath}\r\n \\vspace{-1.0em}\r\n \\begin{itemize}\r\n \\item<3->\r\n Three cases depending on the dominance of the terms\r\n \\item<4->\r\n \\textbf{Case 1:}\r\n Solving the partial problems is \\textit{polynominal} bigger than merging\r\n all solutions\\\\\r\n {\\color{MainA}$T(n) \\in \\Theta(n^{\\log_b a})$}, \\hspace{3.5em}\r\n {\\color{gray}$T(n) \\in \\Theta(\\text{number of leaves})$}\r\n \\vspace{0.5em}\r\n \\item<5->\r\n \\textbf{Case 2:}\r\n Each layer has equal costs\\\\\r\n {\\color{MainA}$T(n) \\in \\Theta(n^{\\log_b a} \\log n)$}, \\hspace{1.5em}\r\n {\\color{gray}$\\log n$ layers}\r\n \\vspace{0.5em}\r\n \\item<6->\r\n \\textbf{Case 3:}\r\n Connecting all partial solutions is \\textit{polynominal} bigger than\r\n solving all partial problems\\\\\r\n {\\color{MainA}$T(n) \\in \\Theta(f(n))$}\r\n \\end{itemize}\r\n\\end{frame}\r\n", "meta": {"hexsha": "75557b573ffd3b756aeae6cb8fd8a8e4dc0adf53", "size": 30874, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture-9/Chapter/eng/020_RecusionEquations.tex", "max_stars_repo_name": "TobiOnline/AlgoDat", "max_stars_repo_head_hexsha": "565a9f03a9ed7ef354cb4f143959df77df89b726", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2016-12-16T17:42:34.000Z", "max_stars_repo_stars_event_max_datetime": "2020-08-26T11:07:16.000Z", "max_issues_repo_path": "Lecture-9/Chapter/eng/020_RecusionEquations.tex", "max_issues_repo_name": "TobiOnline/AlgoDat", "max_issues_repo_head_hexsha": "565a9f03a9ed7ef354cb4f143959df77df89b726", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 23, "max_issues_repo_issues_event_min_datetime": "2016-10-08T09:27:41.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-20T15:40:10.000Z", "max_forks_repo_path": 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YES\n2. YES", "lm_q1_score": 0.7981867681382279, "lm_q2_score": 0.7956581000631542, "lm_q1q2_score": 0.6350837674324118}} {"text": "\\section{State Error Objective}\n\n\t\\begin{frame}\n\t\t\\frametitle{Robot Models Used}\n\t\t\\begin{columns}[T]\n\t\t\t\\begin{column}{0.45\\textwidth}\n\t\t\t\t\\onslide<2->{\n\t\t\t\t\\begin{block}{Cartesian Co-ordinate Model}\n\t\t\t\t\t\\[\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\dot{x} \\\\\n\t\t\t\t\t\t\\dot{y} \\\\\n\t\t\t\t\t\t\\dot{\\theta}\n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t=\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\cos(\\theta) & 0 \\\\\n\t\t\t\t\t\t\\sin(\\theta) & 0 \\\\\n\t\t\t\t\t\t0 & 1\n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\upsilon \\\\\n\t\t\t\t\t\t\\omega \n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t\\]\n\t\t\t\t\\end{block}\n\t\t\t\t}\n\t\t\t\t\\onslide<3->{\n\t\t\t\t\\begin{block}{Kino-dynamic Model}\n\t\t\t\t\t\\[\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\dot{x} \\\\\n\t\t\t\t\t\t\\dot{y} \\\\\n\t\t\t\t\t\t\\dot{\\theta} \\\\\n\t\t\t\t\t\t\\dot{\\upsilon} \\\\\n\t\t\t\t\t\t\\dot{\\omega}\n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t=\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\cos(\\theta) & 0 & 0 & 0 \\\\\n\t\t\t\t\t\t\\sin(\\theta) & 0 & 0 & 0 \\\\\n\t\t\t\t\t\t0 & 1 & 0 & 0 \\\\\n\t\t\t\t\t\t0 & 0 & 1 & 0 \\\\\n\t\t\t\t\t\t0 & 0 & 0 & 1\n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\t\\upsilon \\\\\n\t\t\t\t\t\t\\omega \\\\\n\t\t\t\t\t\ta \\\\\n\t\t\t\t\t\t\\alpha\n\t\t\t\t\t\\end{bmatrix}\n\t\t\t\t\t\\]\n\t\t\t\t\\end{block}\n\t\t\t\t}\n\t\t\t\\end{column}\n\t\t\t\\begin{column}{0.49\\textwidth}\n\t\t\t\t\\centering\n\t\t\t\t\\onslide<1->{\\includegraphics[scale=0.35]{pictures/robot_cart.pdf}}\n\t\t\t\\end{column}\n\t\t\\end{columns}\n\t\\end{frame}\n\n\t\\begin{frame}\n\t\t\\frametitle{Robot Models Used}\n\t\t\\begin{columns}[T]\n\t\t\t\\begin{column}{0.45\\textwidth}\n\t\t\t\t\\onslide<2->{\n\t\t\t\t\\begin{block}{Polar Co-ordinate Model}\n\t\t\t\t\t\\begin{align*}\n\t\t\t\t\t\t\\rho& = \\sqrt{(x-x_g)^2+(y-y_g)^2} \\\\[0.4cm]\n\t\t\t\t\t\t\\phi& = \\arctan2\\big((y-y_g),(x-x_g)\\big) \\\\[0.4cm]\n\t\t\t\t\t\t\\alpha&= \\theta - \\phi\n\t\t\t\t\t\\end{align*}\n\t\t\t\t\\end{block}\n\t\t\t\t}\n\t\t\t\\end{column}\n\t\t\t\\begin{column}{0.49\\textwidth}\n\t\t\t\t\\centering\n\t\t\t\t\\onslide<1->{\\includegraphics[scale=0.35]{pictures/robot_polar.pdf}}\n\t\t\t\\end{column}\n\t\t\\end{columns}\n\t\\end{frame}\n\n\t\\begin{frame}\n\t\t\\frametitle{NMPC General Formulation}\n\t\t\\begin{block}{NMPC Optimization Problem}\n\t\t\t\\begin{align*}\n\t\t\t\t&\\underset{\\mathbf{X}^*,\\mathbf{U}^*}{\\text{min }}\n\t\t\t\t\\tikzmark{a9}J_{N_p}(\\mathbf{x}_k) = \\sum_{j=0}^{N_p}\\tikzmark{a11}\\ell (\\mathbf{x}_{k + j},\\mathbf{u}_{k + j}), \\\\\n\t\t\t\t&\\quad \\text{subject to:} \\notag \\\\\n\t\t\t\t&\\quad\\quad \\tikzmark{a12}\\mathbf{x}_{k+j+1} = \\mathcal{F}(\\mathbf{x}_{k+j},\\mathbf{u}_{k+j}), & \\forall\\text{ } 0\\leq j\\leq N_p-1 \\\\ \n\t\t\t\t&\\quad\\quad \\tikzmark{a13}\\mathbf{x}_{k+j} \\in \\mathbb{X}, & \\forall \\text{ }0\\leq j\\leq N_p \\\\\n\t\t\t\t&\\quad\\quad \\tikzmark{a14}\\mathbf{u}_{k+j} \\in \\mathbb{U}, & \\forall \\text{ }0\\leq j\\leq N_p-1 \\\\\n\t\t\t\t&\\quad\\quad \\tikzmark{a15}\\mathbf{d}_{\\imath,k+j} \\geq \\xi_o, & \\forall\\text{ } 0\\leq j\\leq N_p-1, \\text{ }\\forall\\text{ } \\imath \\in I_s\n\t\t\t\\end{align*}\n\t\t\\end{block}\n\t\n\t\t\\begin{tikzpicture}[overlay, remember picture]\n\t\t\t\\coordinate (A9) at ($({pic cs:a9})+(1ex, 1ex)$);\n\t\t\t\\coordinate (A11) at ($({pic cs:a11})+(1ex, 1ex)$);\n\t\t\t\\coordinate (A12) at ($({pic cs:a12})+(0ex, 0.5ex)$);\n\t\t\t\\coordinate (A13) at ($({pic cs:a13})+(0ex, 0.5ex)$);\n\t\t\t\\coordinate (A14) at ($({pic cs:a14})+(0ex, 0.5ex)$);\n\t\t\t\\coordinate (A15) at ($({pic cs:a15})+(1ex, -0.5ex)$);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A9), baseline] (n9) {};\n\t\t\t\\node[cyan, above of= n9, node distance = 3em] (t9_1) {\\small cost};\n\t\t\t\\node[cyan, above of= n9, node distance = 2em] (t9) {\\small function};\n\t\t\t\\draw [<-, cyan] (n9.north) to [below] (t9.south);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A11), baseline] (n11) {};\n\t\t\t\\node[cyan, above of= n11, node distance = 3em] (t11_1) {\\small regulatory};\n\t\t\t\\node[cyan, above of= n11, node distance = 2em] (t11) {\\small term};\n\t\t\t\\draw [<-, cyan] (n11.north) to [below] (t11.south);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A12), baseline] (n12) {};\n\t\t\t\\node[cyan, left of= n12, node distance = 4em] at (2.8,3) (t12) {\\small model};\n\t\t\t\\node[cyan, below of= t12, node distance = 1em] (t12_1) {\\small constraints};\n\t\t\t\\draw [<-, cyan] (n12.west) to [right] (t12.east);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A13), baseline] (n13) {};\n\t\t\t\\node[cyan, left of= n13, node distance = 4em] (t13) {\\small state};\n\t\t\t\\node[cyan, below of= t13, node distance = 1em] (t13_1) {\\small boundries};\n\t\t\t\\draw [<-, cyan] (n13.west) to [right] (t13.east);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A14), baseline] (n14) {};\n\t\t\t\\node[cyan, below of= t13, node distance = 3em] (t14) {\\small input};\n\t\t\t\\node[cyan, below of= t14, node distance = 1em] (t14_1) {\\small boundries};\n\t\t\t\\draw [<-, cyan] (n14.west) to [right] (t14.east);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A15), baseline] (n15) {};\n\t\t\t\\node[cyan, below of= n15, node distance = 3em] (t15_1) {\\small static};\n\t\t\t\\node[cyan, below of= n15, node distance = 2em] (t15) {\\small obstacles};\n\t\t\t\\draw [<-, cyan] (n15.south) to [above] (t15.north);\n\t\t\\end{tikzpicture}\n\t\\end{frame}\n\t\n\t\\begin{frame}\n\t\t\\frametitle{NMPC General Formulation}\n\t\t\\begin{block}{Regulatory Terms (e.g., Cartesian coordinate)}\n\t\t\t\\[\n\t\t\t\t\\begin{aligned}\n\t\t\t\t\t\\ell (\\mathbf{x}_{k + j},\\mathbf{u}_{k + j}) = \\enspace\n\t\t\t\t\t\\tikzmark{a1}\n\t\t\t\t\t(\\mathbf{x}_{k + j} - \\mathbf{x}_{ref})^T \\enspace \\mathbf{Q_1} \\enspace (\\mathbf{x}_{k + j} - \\mathbf{x}_{ref})\n\t\t\t\t\t\\tikzmark{a2}\n\t\t\t\t\t\\enspace + \\enspace\n\t\t\t\t\t\\tikzmark{a3}\n\t\t\t\t\t\\Delta\\mathbf{u}_{k + j}^T \\enspace \\mathbf{R} \\enspace \\Delta\\mathbf{u}_{k + j}\n\t\t\t\t\t\\tikzmark{a4}\n\t\t\t\t\t\\enspace \\tikzmark{a7}-\\tikzmark{a8} \\enspace\n\t\t\t\t\t\\tikzmark{a5} \n\t\t\t\t\t\\sum_{i=0}^{I} \\mathbf{d}_{i,k + j}^T \\enspace \\mathbf{Q_2} \\enspace \\mathbf{d}_{i,k + j}\n\t\t\t\t\t\\tikzmark{a6}\n\t\t\t\t\\end{aligned}\n\t\t\t\\]\n\t\t\\end{block}\n\t\t\\vfill\n\t\t\\vfill\n\t\t\\vfill\n\t\t\\begin{itemize}\n\t\t\t\\item $\\mathbf{d}_{i,k + j}$: distance between the robot and $i^{th}$ detected moving obstacle.\n\t\t\\end{itemize}\n\t\t\n\t\t\\begin{tikzpicture}[overlay, remember picture]\n\t\t\t\\coordinate (A1) at ($({pic cs:a1})+(+0.1ex, 2ex)$);\n\t\t\t\\coordinate (A2) at ($({pic cs:a2})+(-0.1ex,-0.5ex)$);\n\t\t\t\\coordinate (A3) at ($({pic cs:a3})+(+0.1ex, 2ex)$);\n\t\t\t\\coordinate (A4) at ($({pic cs:a4})+(-0.1ex,-0.5ex)$);\n\t\t\t\\coordinate (A5) at ($({pic cs:a5})+(+0.1ex, 3.5ex)$);\n\t\t\t\\coordinate (A6) at ($({pic cs:a6})+(-0.1ex,-2.5ex)$);\n\t\t\t\\coordinate (A7) at ($({pic cs:a7})+(+1ex, 0.5ex)$);\n\t\t\t\\coordinate (A8) at ($({pic cs:a8})+(-1ex, 0.5ex)$);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A1)(A2), draw=tugreen, thick, rounded corners] (n1) {};\n\t\t\t\\node[overlay, below of= n1, node distance = 6em] (t1) {State Error};\n\t\t\t\\node[overlay, below of= n1, node distance = 7em] (t1_1) {Objective};\n\t\t\t\\draw [thick, -latex, tugreen] (n1.south) to [above] (t1.north);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A3)(A4), draw=tugreen, thick, rounded corners] (n2) {};\n\t\t\t\\node[overlay, below of= n2, node distance = 6em] (t2) {Input Penealty};\n\t\t\t\\node[overlay, below of= n2, node distance = 7em] (t2_1) {Objective};\n\t\t\t\\draw [thick, -latex, tugreen] (n2.south) to [above] (t2.north);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A5)(A6), draw=tugreen, thick, rounded corners] (n3) {};\n\t\t\t\\node[overlay, below of= n3, node distance = 5.7em] (t3) {Dyn Obstacles};\n\t\t\t\\node[overlay, below of= n3, node distance = 6.7em] (t3_1) {Objective};\n\t\t\t\\draw [thick, -latex, tugreen] (n3.south) to [above] (t3.north);\n\t\t\t\\pause\n\t\t\t\\node [fit=(A7)(A8), draw=red, circle] (n4) {};\n\t\t\t\\node[overlay, red, above of= n4, node distance = 8em] (t4) {Maximization};\n\t\t\t\\draw [thick, -latex, red] (n4.north) to [below] (t4.south);\n\t\t\\end{tikzpicture}\n\t\t\n%\t\t\\tikz[baseline=(n1.base)]{\n%\t\t\t\\node[draw=tugreen, ,thick, rounded corners, inner sep=5pt] \n%\t\t\t(n1){$xfoz$};\n%\t\t\t\\node[overlay, below of= n1, node distance = 6em] (t1) {State Error};\n%\t\t\t\\draw [thick, -latex, tugreen] (n1.south) to [above] (t1.north);\n%\t\t}\n\t\\end{frame}\n\t\n%\t\\begin{frame}\n%\t\t\\frametitle{Cartesian Coordinate Model Block Diagram}\n%\t\t\\begin{figure}[hbtp]\n%\t\t\t\\centering\n%\t\t\t\\includegraphics[scale=0.7]{pictures/block_diagram_cart.pdf}\n%\t\t\t\\caption{NMPC Control Loop (Cartesian coordinate Model)}\n%\t\t\\end{figure}\n%\t\\end{frame}\n\n\t\\begin{frame}\n\t\t\\frametitle{NMPC General Formulation}\n\t\t\\begin{block}{Regulatory Terms (e.g., Polar coordinate)}\n\t\t\t\\parbox[c][4\\baselineskip][t]{\\textwidth}{\n\t\t\t\\begin{align*}\n\t\t\t\t\\ell_p (\\vartheta_{k + j},\\mathbf{u}_{k + j}) &= \n\t\t\t\t\\pmb{[} \\ell_{p,1}(\\cdot,\\cdot) \\enspace \\ell_{p,2}(\\cdot,\\cdot) \\enspace \\ell_{p,3}(\\cdot,\\cdot) \\pmb{]}\n\t\t\t\t\\enspace\n\t\t\t\t\\mathbf{Q_1} \\enspace\n\t\t\t\t\\begin{bmatrix}\n\t\t\t\t\t\\ell_{p,1}(\\cdot,\\cdot) \\\\\n\t\t\t\t\t\\ell_{p,2}(\\cdot,\\cdot) \\\\\n\t\t\t\t\t\\ell_{p,3}(\\cdot,\\cdot)\n\t\t\t\t\\end{bmatrix}\n\t\t\t\t+ \\cdots\n\t\t\t\\end{align*}\n\t\t\t}\n\t\t\\end{block}\n\t\tError state objective is the minimization of: \\\\[0.5cm]\n\t\t\\begin{itemize}\n\t\t\t\\item Euclidean distance: $\\ell_{p,1}(\\cdot,\\cdot) = \\sqrt{(x_{k+j} - x_{ref})^2+(y_{k+j} - y_{ref})^2}$ \\\\ [0.2cm]\n\t\t\t\\item Heading towards goal: $\\ell_{p,2}(\\cdot,\\cdot) = \\Big(\\theta_{k+j} - \\arctan2\\big((y_{k+j} - y_{ref}),(x_{k+j} - x_{ref})\\big)\\Big)^2$ \\\\ [0.2cm]\n\t\t\t\\item Reference orientation: $\\ell_{p,3}(\\cdot,\\cdot) = (\\theta_{k+j} - \\theta_{ref})^2$\n\t\t\\end{itemize}\n\t\\end{frame}\n\n%\t\\begin{frame}\n%\t\t\\frametitle{Polar Coordinate Model Block Diagram}\n%\t\t\\begin{figure}[hbtp]\n%\t\t\t\\centering\n%\t\t\t\\includegraphics[scale=0.64]{pictures/block_diagram_polar.pdf}\n%\t\t\t\\caption{NMPC Control Loop (Polar coordinate Model)}\n%\t\t\\end{figure}\n%\t\\end{frame}\n\n%\t\\begin{frame}\n%\t\t\\frametitle{Kino-dynamic Model Block Diagram}\n%\t\t\\begin{figure}[hbtp]\n%\t\t\t\\centering\n%\t\t\t\\includegraphics[scale=0.64]{pictures/block_diagram_kino.pdf}\n%\t\t\t\\caption{NMPC Control Loop (Kino-dynamic Model)}\n%\t\t\\end{figure}\n%\t\\end{frame}\n\n\t\\begin{frame}\n\t\t\\frametitle{Potential Collision Zone}\n\t\t\\centering\n\t\t\\includegraphics[scale=0.6]{pictures/robots_pot_coll.pdf}\n\t\\end{frame}\n\n", "meta": {"hexsha": "770e8a8bce9c050fbe64228db15138ba9438669b", "size": 9337, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Kapitel_4.tex", "max_stars_repo_name": "smababou/latex_presentation", "max_stars_repo_head_hexsha": "c38b668cff733fc6feacbc135b5a785e6c7b32ef", "max_stars_repo_licenses": ["Beerware"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Kapitel_4.tex", "max_issues_repo_name": "smababou/latex_presentation", "max_issues_repo_head_hexsha": "c38b668cff733fc6feacbc135b5a785e6c7b32ef", "max_issues_repo_licenses": ["Beerware"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Kapitel_4.tex", "max_forks_repo_name": "smababou/latex_presentation", "max_forks_repo_head_hexsha": "c38b668cff733fc6feacbc135b5a785e6c7b32ef", "max_forks_repo_licenses": ["Beerware"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.6374045802, "max_line_length": 154, "alphanum_fraction": 0.5866980829, "num_tokens": 3964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7745833893685269, "lm_q1q2_score": 0.635075759052575}} {"text": "\\section*{Web Appendix F}\n\nRecall that a density $f(x;\\theta)$ is identifiable if $f(x;\\theta_{1})=f(x;\\theta_{2})$\nfor all $x$ implies that $\\theta_{1}=\\theta_{2}.$ Call a density\n$f(x;\\theta)$ \\textit{strongly identifiable} if $f(x;\\theta_{1})/f(x;\\theta_{2})$\nbeing constant for all $x$ in an open interval $I$ implies that\n$\\theta_{1}=\\theta_{2}$.\n\\paragraph{Proposition A.}\\label{prop:identifiable}\n\nLet $f(x;\\theta)$ be a family of densities on $\\mathbb{R}$, $-\\infty=a_{1} d = 2011 - n$$\n$$\\frac{n} {d} < \\frac{1}{3}$$\n$$\\frac{n}{2011 - n} < \\frac{1}{3}$$\n$$3n < 2011 - n$$\n$$4n < 2011$$\n$$n < 502.75$$\nDue to the constraint that $n$ and $d$ are positive integers, the maximum value for $n$ is $502$. Therefore $d = 2011 - n = 2011 - 502 = 1509$.\n\n\\section{Question 2}\n\nLet X be the random variable modeling the outcome of the bet, where $x_1$ is the amount by which one's money, $m$, is increased in the event that the bet is favorable, and $x_2$ is the amount by which $m$ is increased if the bet is not favorable. We know that $x_1$ and $x_2$ are complements. Therefore, the expectation of such a variable $X$ is:\n\n$$E[X; m] = \\sum_i x_iP(x_i) = 3mP(x_1) - mP(x_2) = 3mP(x_1) - m(1 - P(x_1)) = 4mP(x_1) - m$$\n$$= m(4P(x_1) - 1)$$\nIn order for the bet to be favorable, we would need the expectation of $X$ to be positive, i.e.,:\n\n$$E[X; m] = m(4P(x_1) - 1) > 0$$\n$$P(x_1) > 0.25$$\nLet $n$ be the number of face cards in the first 5 cards, and let $c$ be the event where the sixth card is a face card, i.e., one's money is increased. Then:\n\n$$P(c | n = i) = \\frac{12 - i}{47} $$\nTherefore:\n$$P(c | n = 0) = 0.255$$\n$$P(c | n = 1) = 0.234$$\nThus we can see that the only way the bet is favorable (i.e., $P(x_1) = P(c | n = i) > 0.25$) is when $i = 0$. We now need to figure out the probability of such an occurrence:\n\n$$P(n = 0) = \\frac{{12 \\choose 0}{40 \\choose 5}}{{52 \\choose 5}} = 0.25318$$\n\n\\section{Question 3}\n\nWe have the following parameters for the Poisson distribution:\n\n$$\\lambda = 100 / 5 = 20$$\n$$\\alpha = 0.05$$\n$$n = 60$$\n$$k = n\\lambda = (60)(20) = 1200$$\n\n$$CI = [\\frac{1}{2}\\chi^2(\\frac{\\alpha}{2}; 2k), \\frac{1}{2}\\chi^2(1 - \\frac{\\alpha}{2}; 2k + 2)]\n= [\\frac{1}{2}\\chi^2(0.025; 2400), \\frac{1}{2}\\chi^2(0.975; 2402)]\n= [1133.1, 1269.9]$$\n\n\\section{Question 4}\n\nThis is a one-sided Z-test.\n\n$$Z^* = \\frac{\\hat{p} - p_0}{\\sqrt{\\frac{p_0(1-p_0)}{n}}} = \\frac{(\\frac{26}{200} - 0.1)}{\\sqrt{\\frac{0.1(0.9)}{200}}} = 1.414$$\n$$p = P(Z \\geq Z^*) = 0.0787 > 0.05$$\nTherefore we fail to reject the null hypothesis.\n\n\\end{document}", "meta": {"hexsha": "9aa9142aaae1d7b39075215a0f0a3d0d526614c0", "size": 2375, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "scratch_work.tex", "max_stars_repo_name": "damoose717/stats_problems", "max_stars_repo_head_hexsha": "abbb4faf19a46c210a79d07f9d363aadac35cef3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "scratch_work.tex", "max_issues_repo_name": "damoose717/stats_problems", "max_issues_repo_head_hexsha": "abbb4faf19a46c210a79d07f9d363aadac35cef3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "scratch_work.tex", "max_forks_repo_name": "damoose717/stats_problems", "max_forks_repo_head_hexsha": "abbb4faf19a46c210a79d07f9d363aadac35cef3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.9264705882, "max_line_length": 346, "alphanum_fraction": 0.6168421053, "num_tokens": 965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8577680995361899, "lm_q1q2_score": 0.6348979558516188}} {"text": "\\documentclass{article}\n\n\\begin{document}\n\n\\def\\subI{${}_I$}\n\\def\\subF{${}_F$}\n\\def\\subD{${}_D$}\n\\def\\ifso{\\mathbin{?}}\n\n\\section{Rounding}\n\nConverting a rational number $x \\in {\\cal Q}$ to a nearby\ninteger $n \\in {\\cal Z}$ can be done in surprisingly many\nways. We shall take the floor operation\n\\[ \\lfloor x \\rfloor = \\max \\{n \\in {\\cal Z} | n \\le x\\} \\]\nas the basic operation. We also need $\\bot$ (undefined),\neven$(n)$ testing whether $n$ is a multiple of 2,\n$p \\ifso t : f$ (if $p$ then $t$ else $f$), and $\\sigma(x)$ =\n$x\\ge 0 \\ifso 1 : -1$.\n\n\\begin{tabular}{lll}\nexact & $x=\\lfloor x\\rfloor \\ifso \\lfloor x \\rfloor : \\bot$ &\n exception if not integral\\\\\ndown & $\\lfloor x \\rfloor$ & floor\\\\\nup & $\\lceil x \\rceil = -\\lfloor -x \\rfloor$ & ceiling\\\\\nin & $\\sigma(x)\\times\\lfloor|x|\\rfloor$ & truncate\\\\\nout & $\\sigma(x)\\times\\lfloor|x|+\\frac12\\rfloor$ & away from zero\\\\\neven & $\\mbox{even}(\\lfloor x\\rfloor) \\ifso \\lfloor x \\rfloor\n : \\lceil x \\rceil$ & round to even\\\\\nodd & $\\mbox{even}(\\lfloor x\\rfloor) \\ifso \\lceil x \\rceil\n : \\lfloor x \\rfloor$ & round to odd (not used in practice)\n\\end{tabular}\n\nNone of these is rounding. For that we need\n\\[\\mbox{nearest} f x = d > \\frac12 \\ifso \\lceil x \\rceil :\n d < \\frac12 \\ifso \\lfloor x \\rfloor : f(x) \\]\n\\noindent where\n$ d = x - \\lfloor x \\rfloor $\n\nThis higher-order function can be combined with the seven conversions\nabove to make seven ``round to nearest'' modes such as nearest out\n(banker's rounding) or nearest even (IEEE rounding).\n\nThe claim of this paper is that executable specifications are important\nto find and correct mistakes. Did you notice the mistake(s) above?\nNeither did two pre-publication readers. The problem is that rounding\n$x$ to an even (odd) integer is not well defined when $x$ is an odd (even)\ninteger. This issue was discovered by testing. The solution for even\nis simple: if $x$ is exactly equal to an odd integer, round it to the\nnearest multiple of 4. There is no sensible way to round 0 to an odd\ninteger, so we strike odd rounding from the list.\n\n\\section{Existing standards}\n\n\\subsection{ISO 10967 Language-independent arithmetic}\n\nRoughly speaking, the three parts of LIA cover everything\nin Fortran 95 and C99 and a little bit more.\n\nWe have to discuss ISO 10967 for two reasons: it is the\nstandard that is meant to serve as a reference for arithmetic\nin programming languages, and it does define fixed point\nnumbers (but no operations on them). Any definition of\nfixed point arithmetic should be practically compatible with it.\n\nFurther, the 1998 edition provided axioms (though no evidence\nthat the system described was consistent with them) which could\nserve as models for checking a fixed-point specification.\n\nThe Language-independent arithmetic standard has three parts.\n\n\\subsubsection{Part: 1 Integer and floating point arithmetic}\nAn integer data type $I$ has parameters {\\it bounded}\\subI{}\ntrue if the set is finite, {\\it minint}\\subI{} being the\nsmallest value or $-\\infty$, and {\\it maxint}\\subI{} being the\nlargest value or $\\infty$. If {\\it bounded}\\subI{} is true,\nthen {\\it minint}\\subI{} can be 0, -{\\it maxint}\\subI{}, or\n-({\\it maxint}\\subI{}+1). Curiously, while both signed and\nunsigned bounded integers are allowed, unbounded integers\nmust be signed; ``unbounded natural numbers are not covered\nby this document.'' The usual 6 comparison operations are\ndefined.\n\nThe arithmetic operations are {\\it neg}\\subI{} (unary negation),\n{\\it add}\\subI{}, {\\it sub}\\subI{}, {\\it mul}\\subI{},\nflooring division {\\it quot}\\subI{} and the corresponding\nremainder {\\it mod}\\subI{}, {\\it abs}\\subI{},\nand {\\it signum}\\subI{} taking values 1 and -1. The 1998 edition\nincluded {\\it sign}\\subI{} with values 1, 0, and -1, but that was\ndropped. The 1998 edition included\ntruncating division {\\it div}${}^t_I$ and the corresponding\nremainder {\\it rem}${}^t_I$ and two additional remainder functions\n{\\it mod}${}^a_I$ and {\\it mod}${}^p_I$, all of which were deleted\nin the 2012 edition.\n\nLIA-1 integer arithmetic must handle overflow by causing\nan overflow indication. This includes conversion to a \nsmaller integer type. Wrapping arithmetic is covered in\npart 2.\n\nA floating point data type $F$ has parameters {\\it r}\\subF{}\n(radix), {\\it p}\\subF{} (precision), {\\it emax}\\subF{} (largest\nexponent), {\\it emin}\\subF{} (smallest exponent),\n{\\it denorm}\\subF{} (whether IEEE-style subnormals are supported),\nand {\\it ice\\_60559}\\subF{} (whether $F$ conforms to the IEEE\nstandard). The usual 6 comparison operations are defined,\ntaking care to dodge around infinities and NaNs.\n\nAdditional tests {\\it isnegzero}\\subF{}, {\\it istiny}\\subF{},\n{\\it isnan}\\subF{}, and {\\it issignan}\\subF{} express tests that\ncannot be done using ordinary comparison against zero.\n\nThree sets of basic operations (rounding to nearest even, down,\nor up) are defined if {\\it denorm}\\subF{} is true, just the\nrounded to nearest set if {\\it denorm}\\subF{} is false.\nThe basic operations are {\\it add}, {\\it sub}, {\\it mul},\n{\\it div}, and {\\it sqrt}.\n\nSome arithmetic operations are not sensitive to rounding\nmode: {\\it abs}\\subF{}, two-valued {\\it signum}\\subF{},\nthe IEEE remainder {\\it residue}\\subF{}, {\\it intpart}\\subF{},\n{\\it fractpart}\\subF{}, {\\it succ}\\subF{} (next higher),\n{\\it pred}\\subF{} (next lower), and {\\it ulp}\\subF{}.\n\nThere are three conversions from floating point to integer:\n{\\it ceiling}${}_{F\\rightarrow I}$,\n{\\it floor}${}_{F\\rightarrow I}$, and\n{\\it rounding}${}_{F\\rightarrow I}$.\n\nSection 5.3.5 defines floating point to fixed point conversion\n(for the sake of format conversions like C's {\\tt \\%.$n$f}) and in\npassing defines unbounded fixed point types, with parameters\n{\\it r}\\subD{} (radix), {\\it d}\\subD{} (``density'' = scale),\nand {\\it dmax}\\subD{} (a positive element of $\\cal R$ or\nundefined). Two sets are defined:\n\\begin{eqnarray*}\nD^* &=& \\{n/(r_D^{d_D}) | n \\in {\\cal Z}\\}\\\\\nD &=& D^* \\mbox{if unbounded}\\\\\nD &=& D^* \\cap [-dmax_D,dmax_D] \\mbox{if bounded}\n\\end{eqnarray*}\nThree rounding functions {\\it nearest}\\subD{} (our nearest even),\n{\\it up}\\subD{} (ceiling), and\n{\\it down}\\subD{} (floor), all ${\\cal R} \\rightarrow D^*$, are\ndefined. Conversion from floating point to fixed point and\nfrom fixed point to floating point uses one of these rounding\nmodes just like the basic arithmetic operations.\n\nHowever, fixed point exists in LIA {\\it only} for I/O conversion;\nthere are no comparison or arithmetic operations on $D$.\n\nFloating point arithmetic with truncation is conspiciously\nabsent from the 2012 edition of LIA-1. What it describes is\naccurately rounded arithmetic and the building blocks for\ninterval arithmetic.\n\n\\subsubsection{Part: 2 Elementary numerical functions}\n\nThis part was issued in 2001. There are some slight ``continuity\nerrors'' due to the revision of LIA-1: {\\it iec\\_559}\\subF{} is\nnow {\\it iec\\_60559}\\subF{}, for example. This is another example\nof why standards need to be machine-checked.\n\nThe additional integer operations include maximum, minimum,\ngreatest common divisor, and least common multiple\nof a pair or sequence, if $x\\ge y$ then $x-y$ else 0 (diminish),\ninteger power (but broken for $0^0$ so of very little use),\nfloored shifting by a power of 2 or 10, floored square root,\ndivisibility, evenness, oddness, floored ({\\it quot}\\subI{},\n{\\it mod}\\subI{}), rounded ({\\it ratio}\\subI{}, {\\it residue}\\subI{}),\nand ceilinged ({\\it group}\\subI{}, {\\it pad}) quotient and remainder.\n(The {\\it pad} operation is actually $\\lceil x/y \\rceil\\times y - x$.)\n\nThere are also wrapping versions of {\\it add}, {\\it sub}, and\n{\\it mul}, plus ``overflow'' versions giving you the high bits\nlost by wrapping so that multiprecision arithmetic can be written.\n\nThe additional floating point operations include maximum and\nminimum of a pair or sequence, diminish, conversion to integral\nin floating form ({\\it floor}\\subF{}, {\\it rounding}\\subF{}, and\n{\\it ceiling}\\subF{}) together with {\\it *\\_rest}\\subF{} operations\nproviding the lost fraction parts exactly, square root (already\nin part 1) and reciprocal square root. There is an operation\nfor multiplying two floats giving a result in a wider format with\na single rounding, just right for dot product.\n\nThe elementary transcendental operations include raising a\nfloating point number to an integer or floating point power,\nplus the usual exponential, logarithmic, trigonometric, and\nhyperbolic functions in several forms.\n\n\\subsubsection{Part 3: Complex integer and floating point\narithmetic and complex elementary numerical functions}\n\nThis part was issued in 2006. It defines imaginary and\ncomplex integer and imaginary and complex floating point\ndata types, basic operations on them, and elementary\ntranscendental operations. Maximum and minimum are extended\nto imaginary but not to complex types.\n\nPart 3 ends with recommended changes to some of the operations\ndefined in Part 2 ``so that the returned angle is for the\nproper quadrant''. Again this suggests that a standard on which\ngreat care was taken might have benefited from testing.\n\n\\section{ISO/IEC/IEEE 60559:2011}\n\nThis is the standard for floating point arithmetic.\nIt is the ISO/IEC edition of IEEE 754:2008, which in\nturn was an update of classic IEEE 754 arithmetic to\ninclude decimal floating point. It specifies\n32-, 64-, and 128-bit binary formats that can hold about 7,\n16, and 34 decimal digits respectively\nand 64- and 128-bit decimal formats that can hold\n16 and 34 decimal digits respectively (table 3.2).\n\nIt is common belief that IEEE 754 arithmetic is all but\nuniversal. However, few systems support 128-bit binary floats\n(the popular ARM, x86, and x86-64 architectures do not; SPARC\ndoes), and even fewer support decimal floating-point natively\n(recent IBM z/Series and POWER machines do).\n\nIEEE 754 originally defined four rounding modes, up, down, in,\nand nearest even in our terminology. The current standard\nadds nearest out (banker's rounding).\n\n\\subsection{COBOL}\n\nCOBOL is one of the oldest programming languages still in use.\nIt was designed to support accounting applications on small\nmachines. For example, the Siemens 3003, regarded as ``a large-scale\n\\dots computer'', had 8000 words of\nstorage; the COBOL 61 compiler for that machine needed 10 passes.\nBrinch Hansen and House [ref] report that ``The major problem of\nimplementation turned out to be the numerous definition problems\ncreated by the vagueness of the official Cobol report.''\n\nOne of the major features of COBOL from the beginning was its\nuse of decimal fixed point arithmetic rather than floating point.\nAccounting applications required answers that were exact or\ncomputed according to known rounding rules. Binary integers\nand floats were eventually adopted in 2002, but had been common\nextensions for many years. \n\nA number is characterised by (a) whether it is signed or not,\n(b) its precision, and (c) its scale. A declaration like\\\\\n77 SOME-NUMBER PICTURE S9$(b)$V$9(a)$\\\\\ndeclares a signed variable (S) with a precision of $b+a$\ndecimal digits and a scale of $a$. The letter V designates\na virtual (not actually stored) decimal point. In terms of\nLIA-1, this corresponds to $r_D = 10$, $d_D = a$,\n{\\it dmax}${}_D = (10^{a+b}-1)/10^b$.\n\nIn LIA-1, arithmetic is only defined between values of the\nsame type. In COBOL, it is extremely common for the\noperands and result of an operation to be of different types.\nA typical instruction is\n\\begin{tabbing}\nwww\\=\\kill\nMULTIPLY X BY Y GIVING Z ROUNDED\\\\\n\\> ON SIZE ERROR {\\it statements}\\\\\n\\> NOT SIZE ERROR {\\it statements}\\\\\nEND MULTIPLY \n\\end{tabbing}\n\\noindent where the product $X\\times Y$ is computed\naccording to exact rational arithmetic, rounded to the\nscale of $Z$, and if the result is too large to fit,\nthe ON SIZE ERROR statements are executed, otherwise\nthe result is stored in Z and the NOT SIZE ERROR\nstatements are executed.\n\n\\subsubsection{Four semantics for arithmetic}\n\nThe previous paragraph is a lie. The current COBOL standard does not\noffer a semantics for arithmetic. It offers {\\it four} semantics for\narithmetic. A COBOL program may contain an OPTIONS paragraph near\nthe beginning:\n\\begin{tabbing}\nwww\\=wwwwwwwwwwwwwwww\\=\\kill\nOPTIONS.\\\\\n\\> [ARITHMETIC IS \\{NATIVE $|$ STANDARD $|$\\\\\n\\> \\> STANDARD-BINARY $|$ STANDARD-DECIMAL \\}]\\\\\n\\> [DEFAULT ROUNDED MODE IS one of eight]\\\\\n\\> \\dots\\\\\n\\> [INTERMEDIATE ROUNDING IS one of four]\\\\\n\\end{tabbing}\n\nIn our terminology, the rounding modes are out, nearest out,\nnearest even, nearest in, exact, up, down, and in.\n\nThere is a subtle difference between these statements:\n\\begin{tabbing}\nMULTIPLY X BY Y GIVING Z.\\\\\nCOMPUTE Z = X * Y.\n\\end{tabbing}\nThe latter involves an arithmetic expression which has an\nintermediate result that is then assigned to the destination.\nThe former does not. It appears that the MULTIPLY statement\ninvolves a single rounding whereas the COMPUTE statement may\ninvolve a double rounding.\n\n{\\it Native} arithmetic is whatever the implementor says it is.\n\n{\\it Standard} arithmetic is what used to be standard arithmetic in\nCOBOL-85. It does not cover binary or floating-point operations,\nwhich are treated as native. Standard arithmetic is declared to\nbe obsolete. ``A standard intermediate data item is \\dots the\nunique value zero or an abstract, signed, normalized decimal\nfloating-point temporary data item \\dots with a precision of\n32 decimal digits'' and a scale of -999 to 999 (roughly speaking).\nIntermediate values are rounded in certain circumstances.\n\nIn expressions,\naddition, subtraction, multiplication, and\ndivision are are ``the exact [result] truncated\nto 32 significant digits, normalized, and stored in a standard\nintermediate data item''.\n\n{\\it Standard-binary} arithmetic requires arithmetic operations\nto be done according to IEEE binary floating-point\narithmetic (128-bit format).\n\n{\\it Standard-decimal} arithmetic requires arithmetic operations\nto be done according to IEEE decimal floating-point arithmetic\n(128-bit format).\n\nThe result is that the meaning of an arithmetic statement or\nexpressions can be changed by a line thousands of lines away.\n\nThe standard tries very hard to avoid using\nmathematical notation. Let us take Format 5 of the\nDIVIDE statement as an example:\n\\begin{quote}\n\\begin{tabbing}\nwww\\=\\kill\nDIVIDE \\{identifier-2$|$literal-2\\} BY \\{identifier-1$|$literal-1\\}\\\\\n\\> GIVINE identifier-3 [rounded-prhase]\\\\\n\\> REMAINDER identifier-4\\\\\n\\> [ON SIZE ERROR imperative-statement-1]\\\\\n\\> [NOT SIZE ERROR imperative-statement-2]\\\\\n\\> [END-DIVIDE]\n\\end{tabbing}\n\n1) When native arithmetic is in effect, the quotient is the result\nof dividing the dividend by the divisor. When standard arithmetic,\nstandard-decimal arithmetic, or standard-binary arithmetic is in\neffect, the quotient is the result of the arithmetic expression\n$$(\\mbox{dividend} / \\mbox{divisor})$$\nwhere the values of dividend and divisor are ...\n\n6b) The quotient is then formed as specified in general rule 1\nand stored in identifier-3 \\dots\n\n6c) If the size error condition is not raised, a subsidiary\nquotient is developed that is signed and derived from the\nquotient by truncation of digits at the least significant end\nand that has the same number of digits and the same decimal point\nlocation as the data item references by identifier-3. The\nremainder is calculated as indicated in general rules 7 and is\nstored in the data item referenced by identifier-4 unless\nstoring the value would cause a size error condition \\dots\n\n7) When native arithmetic is in effect, the remainder is the\nresult of multiplying the subsidiary quotient and the divisor\nand subtracting the product from the dividend. When standard\nstandard arithmetic, standard-decimal arithmetic, or\nstandard-binary arithmetic is in effect, the remainder is the\nresult of the arithmetic expression\n$$(\\mbox{divided} - (\\mbox{subsidiary-quotient} * \\mbox{divisor}))$$\nwhere \\dots\n\\end{quote}\n\nThe reason the standard {\\it does} use some expressions here is\nso that the rules for intermediate rounding apply to them. With\n4 arithmetics, 8 intermediate roundings, and 4 default roundings,\na simple DIVIDE statement could mean 128 different things.\n\n\\subsection{PL/I}\n\nPL/I was introduced in the mid-1960s as a blend of Algol,\nFortran, and COBOL. The ECMA-50 (1976), ANSI X3.53-1976, BS 7148:1990,\nNF Z 65-500, and ISO 6160:1979 standards for PL/I are all basically the\nsame. They have expired. BS ISO/IEC 6522:1992 (Information technology,\nProgramming languages, PL/I general purpose subset) is still current.\n\nPL/I fixed point types are characterised by a {\\it base} (2 or 10),\na {\\it precision} (total number of digits), and a {\\it scale} (number\nof digits after the radix point). There are actually two sets of\nrules for determining the precision and scale of a result (ANSI and\nIBM). For FIXED DECIMAL they coincide:\n\\begin{itemize}\n\\item\n$(p_1,s_1) \\pm (p_2,s_2) \\Rightarrow (p,s)$ where\\\\\n$s = s_1 \\vee s_2$\\\\\n$p = 1+s+(p_1-s_1 \\vee p_2-s_2)$\n\\item\n$(p_1,s_1) \\times (p_2,s_2) \\Rightarrow (p,s)$ where\\\\\n$s = s_1 + s_2$\\\\\n$p = 1+p_1+p_2$\n\\item\n$(p_1,s_1) / (p_2,s_2) \\Rightarrow (p,s)$ where\\\\\n$s = N-p_1+s_1-s+2$\\\\\n$p = N$ and $N$ is the implementation-defined maximum precision.\n\\item\nMOD$((p_1,s_1), (p_2,s_2)) \\Rightarrow (p,s)$ where\\\\\n$s - s_1 \\vee s_2$\\\\\n$p = N \\wedge (p_2 - s_2 + s)$.\n\\item\n$(p_1,s_1) ** k \\Rightarrow (p,s)$ where\\\\\n$s = s_1\\times k$\\\\\n$p = (p_1 + 1) \\times k - 1$, if $k$ is a literal integer.\n\\item\nROUND$((p_1,s_1), s) \\Rightarrow (p,s)$ where\\\\\n$s$ is given as the second argument\\\\\n$p = 1 \\vee (p_1-s_1+1-s \\wedge N)$\n\\end{itemize}\nThe rule for division can lead to some nasty surprises.\nFor example, 25+01/3 yields 25.3333333333333 but\n25+1/3 results in an overflow. So there are special functions\nADD(x, y, p[, s]), SUBTRACT(x, y, p[, s]), MULTIPLY(x, y, p[, s]),\nDIVIDE(x, y, p[, s]), \ncomputing a result to a specified precision and scale (default s = 0)\nin a single rounding. There is no version of MOD with precision and\nscale parameters. In assignment\nstatements, ``padding or truncation can occur on the left or the right.\nIf nonzero \\dots digits on the left are lost, the SIZE condition is\nraised.'' This means that a statement like\\\\\nX = X * Y;\\\\\nuses the ``in'' rounding mode. Unlike COBOL, PL/I does not offer\ncontrol over rounding, not even using the special functions, so there is\nno way to change this.\n\n\\subsection{SQL}\n\nThe ISO/IEC 9075-1:2003 standard, SQL 2003, part 1,\nsays in section 4.4.3.1 that ``There are two classes of numeric\ntype: {\\it exact numeric}, which includes integer types and types\nwith specified precision and scale; and {\\it approximate numeric},\nwhich is essentially floating point, and for which a precision may\noptionally be specified.'' The result of an arithmetic operation\n``is of a numeric type that depends only on the numeric type of the\noperands. If the result cannot be represented exactly in the result\ntype, then whether it is rounded or truncated is implementation-defined.\nAn exception condition is raised if the result is outside the range of\nnumeric values of the result type, or if the arithmetic operation is not\ndefined for the operands.''\n\nNumbers and arithmetic are defined in\nsection 4.4 of part 2.\n\\begin{quote}\n`An exact numeric type has a precision\n$P$ and a scale $S$. $P$ is a positive integer that determines\nthe number of significant digits in a particular radix $R$,\nwhere $R$ is either 2 or 10. $S$ is a non-negative integer.\nEvery value of an exact numeric type of scale $S$ is of the\nform $n\\times 10^{-S}$,\\footnote{It is possible that 10 should\nhave been $R$, but only INTEGER types with $S=0$ may have $R=2$.}\nwhere $n$ is an integer such that $-R^P \\le n < R^P$.\n[Note 13 --- Not every value in that range is necessarily a\nvalue of the type in question.]''\n\\end{quote}\nThe intent of note 13 may be to allow both sign-and-magnitude\nrepresentation for NUMERIC types and $R$'s-complement representation\nfor INTEGER types; it's not obvious why any other value would be\nexcluded.\n\nConversion may be done by rounding or truncation; it is not the\nprogrammer who chooses.\n\\begin{quote}\n``If an assignment of some number would result in a loss of its most\nsignificant digit, an exception condition is raised. If least\nsignificant digits are lost, implementation-defined rounding or\ntruncating occurs, with no exception condition being raised.''\n\n``An approximation obtained by truncation of a numeric value $N$ for an\n$\\langle$exact numeric type$\\rangle$ $T$ is a value $V$ in $T$\nsuch that $N$ is not closer to\nzero than is $V$ and there is no value in $T$ between $V$ and $N$.\n\nAn approximation obtained by rounding of a numeric value $N$ for an\n$\\langle$exact numeric type$\\rangle$ $T$ is a value $V$ in $T$\nsuch that the absolute value of the difference between $N$\nand the numeric value of $V$ is not greater than half\nthe absolute value of the difference between two successive numeric\nvalues in $T$. If there is more than one such value $V$, then it is\nimplementation-defined which one is taken.''\n\\end{quote}\n\nAside from the non-portable nature of rounding,\narithmetic is supposed to be unsurprising:\n\\begin{quote}\n``Operations on numbers are performed according to the normal\nrules of arithmetic, within implementation-defined limits \\dots''\n\n``Numbers are compared with respect to their algebraic value.'' (8.2)\n\\end{quote}\n\nParaphrasing sections 4.4.3 and 6.26,\n\\begin{itemize}\n\\item The POWER function never returns an exact answer (unlike PL/I).\n\\item The floor and ceiling functions return exact integers $S=0$.\n\\item Unary plus, unary minus, and the absolute value function return a result\nof the same type as their argument.\n\\item The MOD function only accepts integer arguments.\n\\item WIDTH\\_BUCKET$(x, l, u, n)$ appears to mean\n\\begin{tabbing}\nwww\\=www\\=\\kill\nif $l < u$ then\\\\\n\\> if $x else if $x \\ge u$ then $n+1$\\\\\n\\> else $\\lfloor n\\times((x-l)/(u-l)) \\rfloor$\\\\\nelse if $l > u$ then\\\\\n\\> if $x > l$ then 0\\\\\n\\> else if $x < u$ then $n+1$\\\\\n\\> else $\\lfloor n\\times((x-u)/(l-u)) \\rfloor$\\\\\n\\end{tabbing}\nand the result has the same type as $n$, which must be integral.\nA system that can implement this function ought to be able to\nimplement MOD on non-integral exact numbers.\n\\item If $x$ has radix $R$, precision $P_x$, and scale $S_x$,\nand $y$ has radix $R$, precision $P_y$, and scale $S_y$,\nthen $x+y$ and $x-y$ have scale max$(S_x,S_y)$ and\nimplementation-defined precision;\n\\item $x*y$ has scale $S_x+S_y$ and implementation-defined precision;\n\\item and $x/y$ has implementation-defined precision and\nimplementation-defined scale.\n\\item ``If the operator is not division and the mathematical result of\nthe operation is not exactly representable with the precision and scale\nof the result data type, then an exception condition is raised.''\n\\item ``If the operator is division and the approximate mathematical\nresult of the operation represented with the precision and scale of the\nresult data type loses one or more leading significant digits after\nrounding or truncating if necessary, then an exception condition is\nraised \\dots The choice of whether to round or truncate is\nimplementation-defined.''\n\\end{itemize}\n\nCOBOL and PL/I have fixed-point decimal types in order to support\naccounting.\nSQL was given fixed-point NUMERIC types in order to inter-operate\nwith COBOL and PL/I, and is very much used for accounting data.\nThe problem is that SQL arithmetic is {\\it not} guaranteed\nto be consistent with COBOL or PL/I or accounting rules:\n\\begin{itemize}\n\\item there is no programmer control over rounding;\n\\item there are no precisions that have to be supported,\nnot even older COBOL's 18 digits;\n\\item there is no fixed-point remainder; and\n\\item if you want to calculate the price of one\napple at \\$2.72 per dozen, $2.72/12$ could give\nyou the answer 0 if an implementation defined the\nscale for division to always be $S=0$.\n\\end{itemize}\n\n\\subsection{Smalltalk}\n\nSmalltalk-80 is the programming language that brought Object-Oriented\nprogramming and Integrated Development ENvironments to the world.\nIt got an ANSI standard in 1989. The standard includes unbounded\nintegers, three sizes of floating point, rational numbers, and\na ScaledDecimal class.\n\nFor the semantics of operations on ScaledDecimal numbers,\nthe Smalltalk standard defers to ISO 10967 (LIA). Unfortunately, LIA has,\nby intent and quite explicitly, nothing to say about fixed point\narithmetic.\n\nThe result is that implementations use one of four interpretations\nof ScaledDecimal:\n\\begin{itemize}\n\\item An arbitrary precision integer scaled by a power of 10.\nThis agrees well with the standard, which says that\n``Scaled decimal objects provide a precise representation of decimal\nfractions with an explicitly specified number of fractional digits.''.\n\\item An integer scaled by a power of 10, represented as a 16-byte\npacked decimal value. This also agrees well with the standard.\n\\item An arbitrary precision rational number that is not in general\na whole multiple of a power of 10, rounded to a specified number of\ndigits when and only when printing.\n\\item The same, but with truncation instead of rounding.\n\\end{itemize}\nFor example,\n\\begin{tabbing}\n(2/3) asScaledDecimal: 2 $\\Longrightarrow$ 0.66s2\\\\\n``and'' 0.66s2 * 3 $\\Longrightarrow$ 1.98s2\\\\\n``but'' ((2/3) asScaledDecimal: 2) * 3 $\\Longrightarrow$ 2.00s2\n\\end{tabbing}\nin several Smalltalks (several others yield 0.67s2), while\n\\begin{tabbing}\n(2/3) asScaledDecimal: 2 $\\Longrightarrow$ 0.67s2\\\\\n0.67s2 * 3 $\\Longrightarrow$ 2.01s2\\\\\n((2/3) asScaledDecimal: 2) * 3 $\\Longrightarrow$ 2.01s2\n\\end{tabbing}\nin Smalltalks using a scaled-integer interpretation.\n\n% Standard: #+ #- #* #/ result scale is at least receiver scale.\n% Majority rule: result scale is max(receiver scale, operand scale).\n% astc: #+ #- => max(r,o); #* => r+o #/ ->\n% astc: x \\\\ y = x - ((x // y) * y) => #- rule\n% x rem: y = x - ((x quo: y) * y) => #- rule.\n\n\\subsection{C$\\sharp$}\n\nSummary: decimal arithmetic is standard in C$\\sharp$.\nSuch numbers are value types, as in COBOL and PL/I.\nC$\\sharp$ does not support sufficient precision or sufficient scale\nto allow unproblematic conversion of COBOL 2002 code or\ndata, though it is adequate for COBOL 85. There is no\nprogrammer control over rounding. Scale is part of the\nvalue, not part of the type, and the rule for division\nmakes it hard to track scale at compile time. Above all,\ndecimal arithmetic in C$\\sharp$ is exact {\\it except when\nit isn't}. The compiler will not tell you when unexpected\nrounding {\\it might} happen, nor does the runtime system\ntell you when it {\\it has} happened.\n\nISO/IEC 23270:2006 defines C$\\sharp$ as it was in 2005.\nISO/IEC 23231:2012 defines the Common Language Infrastructure\nThe CLI standard omits the ``Extended numerics library'' (section\nIV.5.7), including System.Decimal, but the language standard\nhas much to say.\n\nSection 8.2.1 says\n\\begin{quote}\nThe {\\tt decimal} type is appropriate for calculations in which\nrounding errors caused by floating-point representations are\nunacceptable.\n\\end{quote}\n\nExcept for rounding errors introduced silently by C$\\sharp$ itself.\n\nSection 11.1.7 defines the {\\tt decimal} type.\n\\begin{quote}\nThe {\\tt decimal} type is a 128-bit data type suitable for\nfinancial and monetary calcuations. The {\\tt decimal} type\ncan represent values including those in the range $1\\times 10^{-28}$\nthrough $1\\times 10^{28}$ with at least 28 significant digits.\nThe finite set of values of of type {\\tt decimal} are of the\nform $(-1)^s\\times c\\times 10^{-e}$, where the sign $s$ is 0 or 1,\nthe coefficient $c$ is given by $0\\le c X, Y, Z\n 000: 1\n 001: 1\n 010: 0\n 011: 0\n 100: 0\n 101: 1\n 110: 0\n 111: 1 \n\\end{verbatim}\n\nNotes: \n\n\\begin{itemize}\n\t\\item Any line that starts with a hash mark (``\\#'') is a comment and will be ignored by \\ac{KARMA}. \n\t\\item The various inputs are named before they are used. In the example, there are three inputs: $ X $, $ Y $, and $ Z $. \n\t\\item Each row in the truth table is shown, along with the output expected. So, in the example above, an input of $ 000 $ should yield an output of $ 1 $. \n\t\\item An output of ``-'' is permitted and means ``don't care.'' \n\\end{itemize}\n\n\\paragraph{Integer}\n\\label{ASM:para:karma_integer}\n\nIn \\ac{KARMA}, an integer can be used to define the outputs of the truth table, so it is ``shorthand'' for an entire truth table input. Here is the example of the ``integer'' type input.\n\n\\begin{verbatim}\n #Sample Integer Input\n inputs -> A, B, C, D\n onset -> E81A base 16 \n\\end{verbatim}\n\nNotes: \n\n\\begin{itemize}\n\t\\item Any line that starts with a hash mark (``\\#'') is a comment and will be ignored by \\ac{KARMA}. \n\t\\item Input variables are defined first. In this example, there are four inputs: $ A $, $ B $, $ C $, and $ D $. \n\t\\item The ``onset'' line indicates what combinations of inputs should yield a \\emph{True} on a truth table. In the example, the number $ E81A $ is a hexadecimal number that is written like this in binary: \n\\end{itemize}\n\n\\begin{verbatim}\n 1110 1000 0001 1010\n E 8 1 A \n\\end{verbatim}\n\nThe least significant bit of the binary number, $ 0 $ in this example, corresponds to the output of the first row in the truth table; thus, it is false. Each bit to the left of the least significant bit corresponds to the next row, counting from $ 0000 $ to $ 1111 $. Here is the truth table generated by the hexadecimal integer $ E81A $: \n\n\\begin{table}[H]\n\t\\sffamily\n\t\\newcommand{\\head}[1]{\\textcolor{white}{\\textbf{#1}}}\t\t\n\t\\begin{center}\n\t\t\\rowcolors{2}{gray!10}{white} % Color every other line a light gray\n\t\t\\begin{tabular}{ccccc} \n\t\t\t\\rowcolor{black!75}\n\t\t\t\\multicolumn{4}{c}{\\head{Inputs}} & \\head{Output} \\\\\n\t\t\tA & B & C & D & Y \\\\\n\t\t\t\\hline\n\t\t\t0 & 0 & 0 & 0 & 0 \\\\\n\t\t\t0 & 0 & 0 & 1 & 1 \\\\\n\t\t\t0 & 0 & 1 & 0 & 0 \\\\\n\t\t\t0 & 0 & 1 & 1 & 1 \\\\\n\t\t\t0 & 1 & 0 & 0 & 1 \\\\\n\t\t\t0 & 1 & 0 & 1 & 0 \\\\\n\t\t\t0 & 1 & 1 & 0 & 0 \\\\\n\t\t\t0 & 1 & 1 & 1 & 0 \\\\\n\t\t\t1 & 0 & 0 & 0 & 0 \\\\\n\t\t\t1 & 0 & 0 & 1 & 0 \\\\\n\t\t\t1 & 0 & 1 & 0 & 0 \\\\\n\t\t\t1 & 0 & 1 & 1 & 1 \\\\\n\t\t\t1 & 1 & 0 & 0 & 0 \\\\\n\t\t\t1 & 1 & 0 & 1 & 1 \\\\\n\t\t\t1 & 1 & 1 & 0 & 1 \\\\\n\t\t\t1 & 1 & 1 & 1 & 1 \\\\\n\t\t\\end{tabular}\n\t\\end{center}\n\t\\caption{Truth Table for KARMA}\n \\label{03:tab:truth_table_for_karma}\n\\end{table}\n\nThe ``Output'' column contains the binary integer $ 1110\\;1000\\;0001\\;1010 $ (or $ E81A_{16} $) from bottom to top. \n\n\\paragraph{Terms}\n\\label{ASM:para:karma_terms}\n\nData input can be defined by using the minterms for the Boolean expression. Following is an example minterm input.\n\n\\begin{verbatim}\n #Sample Minterms\n inputs -> A, B, C, D\n onset -> 0, 1, 2, 3, 5, 10 \n\\end{verbatim}\n\n\\begin{itemize}\n\t\\item Any line that starts with a hash mark (``\\#'') is a comment and will be ignored by \\ac{KARMA}. \n\t\\item The inputs, $ A $, $ B $, $ C $, and $ D $, are defined first. \n\t\\item The ``onset'' line indicates the minterm numbers that yield a \\emph{True} output. \n\t\\item This is similar to a \\ac{SOP} expression, and the digits in that expression could be directly entered on the onset line. For example, the onset line above would have been generated from the Sigma expression in Equation \\ref{ASM:eq:KARMA Input}.\n\\end{itemize}\n\n\\begin{align}\n\t\\label{ASM:eq:KARMA Input}\n\t\\int(A,B,C,D) &= \\sum(0,1,2,3,5,10)\n\\end{align}\n\n\\subsubsection{Truth Table and K-Map Input}\n\\label{ASM:subsubsec:karma_truth_table_and_kmap_input}\n\nWhile \\ac{KARMA} will accept a number of different input methods, as described above, one of the easiest to use is the Truth Table, and the related Karnaugh Map, and these are displayed by default when the Karnaugh Map function is selected. The value of any of the cells in the \\emph{Out} column in the Truth Table, or cells in the Karnaugh Map, can be cycled through $ 0 $, $ 1 $, and ``don't care'' (indicated by a dash) on each click of the mouse in the cell. The Truth Table and Karnaugh Map are synchronized as cells are clicked. The number of input variables can be adjusted by changing the \\emph{Var} setting at the top of the screen. Also, the placement of those variables on the Karnaugh Map can be adjusted as desired. \n\n\\subsubsection{Solution}\n\\label{ASM:subsubsec:karma_truth_table_and_kmap_input}\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\maxwidth{.95\\linewidth}]{gfx/07_04}\n\t\\caption{Karma Solution}\n\t\\label{fig:07_04}\n\\end{figure}\n\nTo simplify the Karnaugh Map, click the \\emph{Minimize} button. A number of windows will pop up (illustrated in Figure \\ref{ASM:fig:karma_solution}), each showing the circuit simplification in a slightly different way. Note: the following Boolean expression was entered to generate the illustrated simplification: $ A'C + A'B + AB'C' + B'C'D' $. \n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\maxwidth{.95\\linewidth}]{gfx/07_05}\n\t\\caption{Karma Minimized Solution}\n\t\\label{fig:07_05}\n\\end{figure}\n\n\\marginpar{KARMA includes parenthesis for clarity, but the groups are obvious when the expression is written in normal Boolean form.}\n\nIn this solution, a \\emph{NOT} term is identified by a leading exclamation point; thus, the minimized expression is: $ A'D' + A'B + AB'C' + A'C $. \n\n\\paragraph{BDDeiro}\n\\label{ASM:para:karma_bddeiro}\n\nThe BDDeiro window is a visualization of a \\ac{BDD}, which graphically represents the solution to a logic network. \n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\maxwidth{.95\\linewidth}]{gfx/07_06}\n\t\\caption{Karma BDDeiro Map}\n\t\\label{fig:07_06}\n\\end{figure}\n\nIn a \\ac{BDD}, each circle represents an input and the squares at the bottom of the diagram represent the two possible outputs: \\emph{False} and \\emph{True}. The lines are paths from the inputs to either a \\emph{False} or \\emph{True} output; thus, a truth table can be viewed graphically. A \\ac{BDD} is useful because it provides a compact, visual representation of a Boolean expression, and for any given Boolean expression there is one, and only one, \\ac{BDD} representing it. One disadvantage to using a \\ac{BDD} is its size, there are potentially two nodes for each input (except the first), and that can lead to a very large diagram.\n\nFigure \\ref{ASM:fig:karma_bddeiro_map} is actually a ``Reduced Order'' \\ac{BDD} and a number of nodes and paths have been consolidated to make the diagram as simple as possible. The top node represents the start of the decision diagram: input $ a $. If that input is \\emph{False}, then follow the dotted line down to node $ d $. (Note: the lines are color-coded to aid in their use; \\emph{False} lines are blue and \\emph{True} lines are red.) If node $ d $ is false, then that leads directly to output \\emph{True}. Thus $ A'D' $ gives a \\emph{True} output, and that is one of the minimized solutions. To follow one other path, if $ a $ is \\emph{True} (follow the solid line down and right), $ b $ is \\emph{False}, $ c $ is \\emph{False}, the output is \\emph{True}. Thus, $ AB'C' $ is \\emph{True}. In a similar way, all four \\emph{True} outputs, and three false outputs, can be traced from the top to bottom of the diagram. \n\n\\paragraph{Quine-McCluskey}\n\\label{ASM:para:quine-mccluskey}\n\n\\ac{KARMA} includes the complete Quine-McCluskey solution data. Several tables display the various Implicants and show how they are derived. \n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\maxwidth{.95\\linewidth}]{gfx/07_07}\n\t\\caption{Karma Quine-McCluskey Solution}\n\t\\label{fig:07_07}\n\\end{figure}\n\n\\ac{KARMA} also displays the Covering Table for a Quine-McCluskey solution.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\maxwidth{.95\\linewidth}]{gfx/07_08}\n\t\\caption{Karma Quine-McCluskey Covering Table}\n\t\\label{fig:07_08}\n\\end{figure}\n\nEach of the minterms (down the left column) can be turned on or off by clicking on it. The smaller blue dots in the table indicate prime implicants and the larger red dots (if any) indicate essential prime implicants. Because this table is interactive, various different solutions can be attempted by clicking some of the colored dots to achieve the best possible simplification. \n\n\\subsubsection{Practice Problems}\n\\label{ASM:subsubsec:karma_practice_problems}\n\nThe following problems are presented as practice for using \\ac{KARMA} to simplify a Boolean expression. Note: designers can select different Prime Implicants so the simplified expression could vary from what is presented below. \n\n\\begin{table}[H]\n\t\\sffamily\n\t\\begin{center}\n\t\t\\begin{tabular}{c c p{6cm} }\n\t\t\t\\multirow{2}{*}{\\textbf{1}} \n\t\t\t& Expression & $ \\int(A,B,C,D) = \\sum(5,6,7,9,10,11,13,14) $ \\\\\n\t\t\t& \\cellcolor{gray!10} Simplified \n\t\t\t& \\cellcolor{gray!10} $ BC'D+A'BC+ACD'+AB'D $ \\\\\n\t\t\t\\hline\n\t\t\t\\multirow{2}{*}{\\textbf{2}} \n\t\t\t& Exression & $ A'BC'D+A'BCD'+A'BCD+AB'C'D+AB'CD'+AB'CD+ABC'D+ABCD' $ \\\\\n\t\t\t& \\cellcolor{gray!10} Simplified \n\t\t\t& \\cellcolor{gray!10} $ BC'D+A'BC+ACD'+AB'D $ \\\\\n\t\t\t\\hline\n\t\t\t\\multirow{2}{*}{\\textbf{3}} \n\t\t\t& Exression & A 4-variable Karnaugh Map where cells 5, 6, 7, 9, and 10 are True and 13, 14 are ``Don't Care'' \\\\\n\t\t\t& \\cellcolor{gray!10} Simplified \n\t\t\t& \\cellcolor{gray!10} $ BC'D+AC'D+A'BC+ACD' $ \\\\\n\t\t\t\\hline\n\t\t\t\\multirow{2}{*}{\\textbf{3}} \n\t\t\t& Exression & $ \\int(A,B,C,D,E) = \\sum(0, 3, 4, 12, 13, 14, 15, 24, 25, 28, 29, 30) $ \\\\\n\t\t\t& \\cellcolor{gray!10} Simplified \n\t\t\t& \\cellcolor{gray!10} $ ABD'+A'B'C'DE+BCD'+A'BC+A'B'D'E' $ \\\\\n\t\t\\end{tabular}\n\t\\end{center}\n\t\\caption{KARMA Practice Problems}\n \\label{ASM:tab:karma_practice_problems}\n\\end{table}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "a0b2c255cb9623105f49853c2b13e86ae78677b1", "size": 44500, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chapters/07_Adv_Simp_Methods.tex", "max_stars_repo_name": "grself/CIS221_Text", "max_stars_repo_head_hexsha": "e5d78f880e0bae66b33823ee22c68ed16696af3c", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2019-03-10T15:35:42.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-06T17:09:02.000Z", "max_issues_repo_path": "Chapters/07_Adv_Simp_Methods.tex", "max_issues_repo_name": "grself/CIS221_Text", "max_issues_repo_head_hexsha": "e5d78f880e0bae66b33823ee22c68ed16696af3c", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-05-09T19:18:59.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-09T19:18:59.000Z", "max_forks_repo_path": "Chapters/07_Adv_Simp_Methods.tex", "max_forks_repo_name": "grself/CIS221_Text", "max_forks_repo_head_hexsha": "e5d78f880e0bae66b33823ee22c68ed16696af3c", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-01-20T17:30:54.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-20T06:06:00.000Z", "avg_line_length": 48.7938596491, "max_line_length": 1340, "alphanum_fraction": 0.6260898876, "num_tokens": 15824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7248702761768249, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.634831966733071}} {"text": "\\section*{Introduction}\n\nIn this part, we present a development for the \\coq{} proof assistant, after the work of Kirby and Paris. This formalization contains the following main parts:\n\n\\begin{itemize}\n\\item Representation in \\coq{} of hydras and hydra battles\n\\item A proof that every battle is finite and won by Hercules. This proof is based on a \\emph{variant} which maps any hydra to an ordinal strictly less than $\\epsilon_0$ and is strictly decreasing along any battle.\n\n\\item Using a combinatorial toolkit designed by J.~Ketonen and R.~Solovay~\\cite{KS81}, we prove that, for any ordinal $\\mu<\\epsilon_0$, there exists no such variant mapping any hydra to an ordinal stricly less than $\\mu$. Thus, the complexity of $\\epsilon_0$ is really needed in the previous proof.\n\n\\item We prove a relation between the length of a ``classic'' class of battles \\footnote{This class is also called \\emph{standard} in this document (text and proofs). The \\emph{replication factor} of the hydra is exactly $i$ at the $i$-th round of the battle (see Sect~\\vref{sect:replication-def}).}\nand the Wainer-Hardy hierarchy of ``rapidly growing functions'' $H_\\alpha$~\\cite{Wainer1970}. The considered class of battles, which we call \\emph{standard} is the most considered one in the scientific litterature(including popularization).\n\\end{itemize}\n\n\nSimply put, this document tries to combines the scientific interest of two articles~\\cite{KP82, KS81} and a book~\\cite{schutte} with the playful activity of truly proving theorems.\nWe hope that such a work, besides exploring a nice piece of discrete maths, \nwill show how \\coq{} and its standard library are well fitted to help us to understand some non-trivial mathematical developments, and also to experiment the constructive parts of the proof through functional programming.\n\n We also hope to provide a little clarification on infinity (both potential and actual) through the notions of function, computation, limit,\n types and proofs.\n\n\n\n%\\section{Remarks}\n\n\\subsection*{Difference from Kirby and Paris's Work}\nIn~\\cite{KP82}, Kirby and Paris showed that there is no proof of termination of all hydra battles in Peano Arithmetic (PA).\nSince we are used to writing proofs in higher order logic, the restriction to PA was quite unnatural for us. So we chosed to prove another statement without any reference to PA, by considering a class of proofs indexed by ordinal numbers upto $\\epsilon_0$.\n\n\\subsection*{State of the Development}\nThe \\coq{} scripts herein are in constant development since our contribution~\\cite{CantorContrib} on notations for the ordinals $\\epsilon_0$ and $\\Gamma_0$.\nWe added new material : axiomatic definition of countable ordinals after Schütte~\\cite{schutte}, combinatorial aspects of $\\epsilon_0$, after Ketonen and Solovay~\\cite{KS81} and Kirby and Paris~\\cite{KP82}, recent \\coq{} technology: type classes, equations, etc.\n\nWe are now working in order to make clumsy proofs more readable, simplify definitions, and ``factorize'' proofs as much as possible. \nMany possible improvements are suggested as ``todo''s or ``projects'' in this text.\n\n\n\\section*{Future Work (projects)}\n\\index{Projects}\n\nThis document and the proof scripts are far from being complete.\n\nFirst, there must be a lot of typos to correct, references and index items to add. Many proofs are too complex and should be simplified, etc.\n\nThe following extensions are planned, but help is needed:\n\n\\begin{itemize}\n\\item Semi automatic tactics for proving inequalities $\\alpha < \\beta$, even when $\\alpha$ and $\\beta$ are not closed terms.\n\\item Extension to $\\Gamma_0$ (in Veblen normal form)\n\\item More lemmas about hierarchies of rapidly growing functions, and their relationship \n with primitive recursive functions and provability in Peano arithmetic \n(following~\\cite{KS81, KP82}).\n\\item From \\coq's point of view, this development could be used as an illustration of the evolution of the software, every time new libraries and sets of tactics could help to simplify the proofs.\n\\end{itemize}\n\n\\subsection*{Main References}\n\nIn our development, we adapt the definitions and prove many theorems which\nwe found in the following articles. \n\\begin{itemize}\n\\item ``Accessible independence results for Peano arithmetic'' by Laurie Kirby and Jeff Paris~\\cite{KP82}\n\\item ''Rapidly growing Ramsey Functions'' by Jussi Ketonen and Robert Solovay~\\cite{KS81}\n\\item ``The Termite and the Tower'', by Will Sladek~\\cite{Sladek07thetermite}\n\\item Chapter V of ``Proof Theory'' by Kurt Schütte~\\cite{schutte}\n\\end{itemize}\n\n\n\n\n\n\n\\chapter{Hydras and Hydra Games}\n\n\\label{sec:orgheadline91}\n\n\n\n\n\nThis chapter is dedicated to the representation of hydras and rules of the hydra game in \\coq's specification language:~\\gallina. \n\nTechnically, a \\emph{hydra} is just a finite ordered tree, each node of which \nhas any number of sons. Note that, contrary to the computer science tradition, we will show the hydras \nwith the heads up and the foot (i.e., the root of the tree) down.\nFig.~\\ref{fig:Hy} represents such a hydra, which will be referred to as \\texttt{Hy} in our examples (please look at the \nmodule~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}). \n\n\\begin{figure}[h]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.6]\n\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {$\\bullet$};\n\\node (N3) at (2,6) {$\\bullet$};\n\\node (H0) at (0,2) {$\\Smiley[2][vertfluo]$};\n\\node (H1) at (0,8) {$\\Smiley[2][vertfluo]$};\n\\node (H2) at (4,8) {$\\Smiley[2][vertfluo]$};\n\\node (H4) at (4,2) {$\\Smiley[2][vertfluo]$};\n\\node (H5) at (4,4) {$\\Smiley[2][vertfluo]$};\n\\draw (foot) -- (N1)[very thick] ;\n\\draw (N1) -- (N2);\n\\draw (N2) -- (N3);\n\\draw (N3) to [bend left= 10] (H1) ;\n\\draw (N3) to [bend right= 16] (H2);\n\\draw (foot) to [bend left= 10] (H0) ;\n\\draw (foot) to [bend right = 10] (H4) ;\n\\draw (N1) to [bend right= 16] (H5);\n\\end{tikzpicture}\n\\caption{The hydra Hy \\label{fig:Hy}}\n\\end{figure}\n\n\n\nWe use a specific vocabulary for talking about hydras. Table~\\ref{tab:hyd2tree} shows the correspondance between our terminology and the usual vocabulary for trees in computer science.\n\n\n\\begin{figure}[h]\n \\centering\n \\begin{tabular}{ll}\nHydras & Finite rooted trees\\\\\n\\hline\nfoot & root\\\\\nhead & leaf\\\\\nnode & node\\\\\nsegment & (directed) edge \\\\\nsub-hydra & subtree\\\\\ndaughter & immediate subtree\\\\\n\\end{tabular}\n \\caption{Translation from hydras to trees}\n \\label{tab:hyd2tree}\n\\end{figure}\n\n\nThe hydra \\texttt{Hy} has a \\emph{foot} (below), five \\emph{heads}, and eight \\emph{segments}. \nWe leave it to the reader to define various parameters such as the height, the size, the highest arity (number of sons of a node) of a hydra. In our example, these parameters have the respective values : $4$, $9$ and $3$.\n\n\n\n\n\\subsection{The Rules of the Game}\n\n\\label{sec:orgheadline44}\n\\label{sect:replication-def}\n\nA \\emph{hydra battle} is a fight between Hercules and the Hydra. \nMore formally, a battle is a sequence of \\emph{rounds}.\nAt each round:\n\\begin{itemize}\n\\item If the hydra is composed of just one head, the battle is finished\nand Hercules is the winner.\n\\item Otherwise, Hercules chops off \\emph{one} head of the hydra,\n\n\\begin{itemize}\n\\item If the head is at distance 1 from the foot, the head is just lost by the hydra, with no more reaction.\n\\item Otherwise, let us denote by \\(r\\) the node that was at distance \\(2\\) from \nthe removed head in the direction of the foot, and consider the sub-hydra \\(h'\\) of \\(h\\), whose root is \\(r\\) \\footnote{$h'$ will be called ``the wounded part of the hydra'' in the subsequent text. In Figures~\\vref{fig:Hy2} and ~\\vref{fig:Hy4}, this sub-hydra is displayed in red.}. Let $n$ be some natural number.\nThen $h'$ is replaced by $n+1$ of copies of \\(h'\\) which share the same root $r$.\n The \\emph{replication factor} $n$ may be different (and generally is) at each round of the fight.\nIt may be chosen by the hydra, according to its strategy, or imposed by some \nparticular rule. In many presentations of hydra battles, this number is increased by $1$ at each round. In the following presentation, we will also consider battles where the hydra is free to choose its ~replication factor at each round of the battle\\footnote{Let us recall that, if the chopped-off head was at distance 1 from the foot, the replication factor is meaningless.}.\n\\end{itemize}\n\\end{itemize}\n\n\n\nNote that the description given in~\\cite{KP82} of the replication process in hydra battles is also semi-formal. \n\n\\label{original-rules}\n\n\\begin{quote}\n ``From the node that used to be attached to the head which was just chopped off, traverse one \nsegment towards the root until the next node is reached. From this node sprout $n$ replicas of \nthat part of the hydra (after decapitation) which is ``above'' the segment just traversed, i.e., those \nnodes and segments from which, in order to reach the root, this segment would have to be \ntraversed. If the head just chopped off had the root of its nodes, no new head is grown. ''\n\\end{quote}\n\nMoreover, we note that this description is in \\emph{imperative} terms. In order to build a formal study of the properties of hydra battles, we prefer to use a mathematical vocabulary, i.e., graphs, relations, functions, etc.\nThus, the replication process will be represented as a binary relation on a data type \\texttt{Hydra},\nlinking the state of the hydra \\emph{before} and \\emph{after} the transformation.\nA battle will thus be represented as a sequence of terms of type \\texttt{Hydra}, respecting the rules of the game.\n\n\n\n\n\n\\subsection{Example}\nLet us start a battle between Hercules and the hydra \\texttt{Hy} of Fig.~\\ref{fig:Hy}.\n\nAt the first round, Hercules choses to chop off the rightmost head of \\texttt{Hy}.\nSince this head is near the floor, the hydra loses this head. Let us call \n \\texttt{Hy'} the resulting state of the hydra, represented in Fig.~\\vref{fig:Hy-prime}.\n\nNext, assume Hercules choses to chop off one of the two highest heads of \\texttt{Hy'}, for instance the rightmost one. Fig.~\\vref{fig:Hy2} represents the rotten neck in dashed lines, and the part that will be replicated in red. Assume also that the hydra decides to add 4 copies of the red part\\footnote{In other words, the replication factor at this round is equal to $4$.}. We obtain a new state \\texttt{Hy''} depicted in Fig.~\\ref{fig:Hy3}.\n\n\n\n\\begin{figure}[h]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.6]\n\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {$\\bullet$};\n\\node (N3) at (2,6) {$\\bullet$};\n\\node (H0) at (0,2) {$\\Smiley[2][vertfluo]$};\n\\node (H1) at (0,8) {$\\Smiley[2][vertfluo]$};\n\\node (H2) at (4,8) {$\\Smiley[2][vertfluo]$};\n\\node (H5) at (4,4) {$\\Smiley[2][vertfluo]$};\n%\\node (H4) at (6,0) {$\\Xey[2][lightgray]$};\n\\draw (foot) -- (N1)[very thick] ;\n\\draw (N1) -- (N2);\n\\draw (N2) -- (N3) ;\n\\draw (N3) to [bend left= 10] (H1) ;\n\\draw (N3) to [bend right= 16] (H2);\n\\draw (foot) to [bend left= 10] (H0) ;\n\\draw (N1) to [bend right= 16] (H5);\n\\end{tikzpicture}\n\n\\caption{Hy': the state of Hy after one round \\label{fig:Hy-prime}}\n\\end{figure}\n\n\n\\begin{figure}[hp]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {{\\color{lightred}$\\bullet$}};\n\\node (N3) at (2,6) {{\\color{lightred}$\\bullet$}};\n\\node (H0) at (0,2) {$\\Smiley[2][vertfluo]$};\n\\node (H1) at (0,8) {$\\Sey[2][lightred]$};\n%\\node (H2) at (5,0) {$\\Xey[2][lightgray]$};\n\\node (H5) at (4,4) {$\\Smiley[2][vertfluo]$};\n\\node (ex) at (5,8) {};\n\\draw (foot) -- (N1)[very thick] ;\n\\draw (N1) -- (N2);\n\\draw (N2) -- (N3)[draw=lightred];\n\\draw (N3) to [bend left= 10](H1) [draw=lightred];\n\\draw [dashed] (N3) to [bend left= 10](ex);\n\\draw (foot) to [bend left= 10] (H0) ;\n\\draw (N1) to [bend right= 16] (H5);\n\\end{tikzpicture}\n\\caption{A second beheading}\n\\label{fig:Hy2}\n\\end{figure}\n\n\\begin{figure}[hp]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.6]\n\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {$\\bullet$};\n\\node (N3) at (2,6) {{\\color{lightred}$\\bullet$}};\n\\node (H1) at (0,8) {$\\Smiley[2][vertfluo]$};\n\\node (H11) at (2,8) {$\\Smiley[2][vertfluo]$};\n\\node (H12) at (4,8) {$\\Smiley[2][vertfluo]$};\n\\node (H13) at (6,8) {$\\Smiley[2][vertfluo]$};\n\\node (H14) at (8,8) {$\\Smiley[2][vertfluo]$};\n\n\\node (N3) at (1,6) {$\\bullet$};\n\\node (N31) at (2,6) {$\\bullet$};\n\\node (N32) at (3,6) {$\\bullet$};\n\\node (N33) at (4,6) {$\\bullet$};\n\\node (N34) at (5,6) {$\\bullet$};\n\n\\node (H0) at (0,2) {$\\Smiley[2][vertfluo]$};\n\\node (H5) at (4,4) {$\\Smiley[2][vertfluo]$};\n\\draw (foot) -- (N1)[very thick] ;\n\\draw (N1) -- (N2);\n\\draw (N2) -- (N3);\n\\draw (N2) -- (N31);\n\\draw (N2) -- (N32);\n\\draw (N2) -- (N33);\n\\draw (N2) -- (N34);\n\\draw (N3) to [bend left= 10](H1) ;\n\\draw (N31) to [bend left= 10](H11) ;\n\\draw (N32) to [bend left= 10](H12) ;\n\\draw (N33) to [bend left= 10](H13) ;\n\\draw (N34) to [bend left= 10](H14) ;\n\\draw (foot) to [bend left= 10] (H0) ;\n\\draw (N1) to [bend left= 10] (H5) ;\n\\end{tikzpicture}\n\\caption{Hy'', the state of Hy after two rounds \\label{fig:Hy3}}\n\\end{figure}\n\nFigs.~\\ref{fig:Hy4} and~\\vref{fig:Hy5} represent a possible third round of the battle, with a replication factor equal to $2$. Let us call \\texttt{Hy'''} the state of the hydra after that third round.\n\n\\begin{figure}[hp]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.6]\n\n\\node (foot) at (2,0) {{\\color{lightred}$\\bullet$}};\n\\node (N1) at (2,2) {{\\color{lightred}$\\bullet$}};\n\\node (N2) at (2,4) {{\\color{lightred}$\\bullet$}};\n\\node (N3) at (2,6) {{\\color{lightred}$\\bullet$}};\n\\node (exN4) at (4,4) {};\n\\node (H1) at (0,8) {$\\Sey[2][lightred]$};\n\\node (H11) at (2,8) {$\\Sey[2][lightred]$};\n\\node (H12) at (4,8) {$\\Sey[2][lightred]$};\n\\node (H13) at (6,8) {$\\Sey[2][lightred]$};\n\\node (H14) at (8,8) {$\\Sey[2][lightred]$};\n\n\\node (N3) at (1,6) {{\\color{lightred}$\\bullet$}};\n\\node (N31) at (2,6) {{\\color{lightred}$\\bullet$}};\n\\node (N32) at (3,6) {{\\color{lightred}$\\bullet$}};\n\\node (N33) at (4,6) {{\\color{lightred}$\\bullet$}};\n\\node (N34) at (5,6) {{\\color{lightred}$\\bullet$}};\n\n\\node (H0) at (0,2) {$\\Smiley[2][vertfluo]$};\n%\\node (H5) at (4,0) {$\\Xey[2][lightgray]$};\n\\draw (foot) -- (N1)[very thick,draw=lightred] ;\n\\draw (N1) -- (N2)[draw=lightred];\n\\draw (N2) -- (N3)[draw=lightred];\n\\draw (N2) -- (N31)[draw=lightred];\n\\draw (N2) -- (N32)[draw=lightred];\n\\draw (N2) -- (N33)[draw=lightred];\n\\draw (N2) -- (N34)[draw=lightred];\n\\draw (N3) to [bend left= 10](H1) [draw=lightred];\n\\draw (N31) to [bend left= 10](H11) [draw=lightred];\n\\draw (N32) to [bend left= 10](H12) [draw=lightred];\n\\draw (N33) to [bend left= 10](H13) [draw=lightred];\n\\draw (N34) to [bend left= 10](H14) [draw=lightred];\n\\draw (foot) to [bend left= 10] (H0) ;\n\\draw [dashed] (N1) to [bend left= 10](exN4);\n\\end{tikzpicture}\n\\caption{A third beheading (wounded part in red) \\label{fig:Hy4}}\n\\end{figure}\n\n\\begin{figure}[hp]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.4]\n\n\\node (foot) at (10,0) {$\\bullet$};\n\n\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {$\\bullet$};\n\\node (N3) at (2,6) {{\\color{lightred}$\\bullet$}};\n\\node (H1) at (0,8) {$\\Smiley[1][vertfluo]$};\n\\node (H11) at (2,8) {$\\Smiley[1][vertfluo]$};\n\\node (H12) at (4,8) {$\\Smiley[1][vertfluo]$};\n\\node (H13) at (6,8) {$\\Smiley[1][vertfluo]$};\n\\node (H14) at (8,8) {$\\Smiley[1][vertfluo]$};\n\n\\node (N3) at (1,6) {$\\bullet$};\n\\node (N31) at (2,6) {$\\bullet$};\n\\node (N32) at (3,6) {$\\bullet$};\n\\node (N33) at (4,6) {$\\bullet$};\n\\node (N34) at (5,6) {$\\bullet$};\n\n\\node (H0) at (-3,3) {$\\Smiley[1][vertfluo]$};\n\n\\draw (foot) to [bend left=10] (N1)[very thick] ;\n\\draw (N1) -- (N2);\n\\draw (N2) -- (N3);\n\\draw (N2) -- (N31);\n\\draw (N2) -- (N32);\n\\draw (N2) -- (N33);\n\\draw (N2) -- (N34);\n\\draw (N3) to [bend left= 10](H1) ;\n\\draw (N31) to [bend left= 10](H11) ;\n\\draw (N32) to [bend left= 10](H12) ;\n\\draw (N33) to [bend left= 10](H13) ;\n\\draw (N34) to [bend left= 10](H14) ;\n\\draw (foot) to [bend left = 15] (H0) ;\n\n\n% second copy \n\\node (N01) at (12,2) {$\\bullet$};\n\\node (N02) at (12,4) {$\\bullet$};\n\\node (N03) at (12,6) {{\\color{lightred}$\\bullet$}};\n\\node (H001) at (10,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0011) at (12,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0012) at (14,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0013) at (16,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0014) at (18,8) {$\\Smiley[1][vertfluo]$};\n\n\\node (N03) at (11,6) {$\\bullet$};\n\\node (N031) at (12,6) {$\\bullet$};\n\\node (N032) at (13,6) {$\\bullet$};\n\\node (N033) at (14,6) {$\\bullet$};\n\\node (N034) at (15,6) {$\\bullet$};\n\n\\draw (foot) -- (N01)[very thick] ;\n\\draw (N01) -- (N02);\n\\draw (N02) -- (N03);\n\\draw (N02) -- (N031);\n\\draw (N02) -- (N032);\n\\draw (N02) -- (N033);\n\\draw (N02) -- (N034);\n\\draw (N03) to [bend left= 10](H001) ;\n\\draw (N031) to [bend left= 10](H0011) ;\n\\draw (N032) to [bend left= 10](H0012) ;\n\\draw (N033) to [bend left= 10](H0013) ;\n\\draw (N034) to [bend left= 10](H0014) ;\n\n% third copy \n\\node (N001) at (22,2) {$\\bullet$};\n\\node (N002) at (22,4) {$\\bullet$};\n\\node (N003) at (22,6) {{\\color{lightred}$\\bullet$}};\n\\node (H001) at (20,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0011) at (22,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0012) at (24,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0013) at (26,8) {$\\Smiley[1][vertfluo]$};\n\\node (H0014) at (28,8) {$\\Smiley[1][vertfluo]$};\n\n\\node (N003) at (21,6) {$\\bullet$};\n\\node (N0031) at (22,6) {$\\bullet$};\n\\node (N0032) at (23,6) {$\\bullet$};\n\\node (N0033) at (24,6) {$\\bullet$};\n\\node (N0034) at (25,6) {$\\bullet$};\n\n\\draw (foot) -- (N001)[very thick] ;\n\\draw (N001) -- (N002);\n\\draw (N002) -- (N003);\n\\draw (N002) -- (N0031);\n\\draw (N002) -- (N0032);\n\\draw (N002) -- (N0033);\n\\draw (N002) -- (N0034);\n\\draw (N003) to [bend left= 10](H001) ;\n\\draw (N0031) to [bend left= 10](H0011) ;\n\\draw (N0032) to [bend left= 10](H0012) ;\n\\draw (N0033) to [bend left= 10](H0013) ;\n\\draw (N0034) to [bend left= 10](H0014) ;\n\\end{tikzpicture}\n\\caption{The configuration Hy''' of Hy \\label{fig:Hy5}}\n\\end{figure}\n\\FloatBarrier\n\nWe leave it to the reader to guess the following rounds of the battle \\dots\n % Please keep in mind that, in this \n% the hydra is free to chose any number of replications at each time, whereas\n% Hercules chops only one head per round.\n\n% Let us precise that, in this game, Hercules wins if the hydra is eventually reduced \n% to a single head. \n% We know from~\\cite{KP82} that, whichever the initial configuration of the\n% hydra, and the strategies of both players, Hercules eventually wins. The \n% aforementionned paper shows also that there do not exist any \\emph{simple} proof of this result.\n\n\n\\section{Hydras and their Representation in \\emph{Coq}}\n\\label{sec:orgheadline48}\n\n\nIn order to describe trees where each node can have an arbitrary (but finite) number of sons, it usual to define a type where each node carries a \\emph{forest}, \\emph{i.e} a list of trees\n(see for instance Chapter 14, pages 400-406 of \\cite{BC04}).\n\nFor this purpose, we define two mutual \\emph{ad-hoc} inductive types, where \\texttt{Hydra} is the main type, and \\texttt{Hydrae} is a helper for describing finite sequences of hydra.\n\\label{types:Hydra}\n\\label{types:Hydrae}\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#Hydra}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nInductive Hydra : Set :=\n| node : Hydrae -> Hydra\nwith Hydrae : Set :=\n| hnil : Hydrae\n| hcons : Hydra -> Hydrae -> Hydrae.\n\\end{Coqsrc}\n\n%\\index{To do}\n\\index{Projects}\n\n\\begin{project}\nLook for an existing library on trees with nodes of arbitrary arity, in order to replace this ad-hoc type with something more generic.\n\\end{project}\n\n\n\\index{Projects}\n\n\\begin{project}\n\n Another very similar representation could use the \\texttt{list} type family instead of the specific \ntype \\texttt{Hydrae}:\n\n\n\\begin{Coqalt}\nModule Alt.\n\nInductive Hydra: Set :=\n hnode (daughters : list Hydra).\n\nEnd Alt.\n\\end{Coqalt}\n\nUsing this representation, re-define all the constructions of this chapter.\nYou will probably have to use patterns described for instance in~\\cite{BC04} or the archives of the Coq-club~\\cite{Coq}.\n\n \n\\end{project}\n\n\n\\index{Projects}\n\n\\begin{project}\nThe type \\texttt{Hydra} above describes hydras as \\emph{plane trees}, i.e., as drawn on a sheet of paper or computer screen. Thus, hydras are \\emph{oriented},\nand it is appropriate to consider a \\emph{leftmost} or \\emph{rightmost} head of\nthe beast. It could be interesting to consider another representation, in which\nevery non-leaf node has a \\emph{multi-set} -- not an ordered list -- of daughters.\n\\end{project}\n\n\\subsubsection{Abbreviations}\n\nWe provide several notations for hydra patterns which occur often in our developments. \n\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#head}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\n(** heads *)\nNotation head := (node hnil).\n \n(** nodes with 1, 2 or 3 daughters *)\nNotation hyd1 h := (node (hcons h hnil)).\nNotation hyd2 h h' := (node (hcons h (hcons h' hnil))).\nNotation hyd3 h h' h'' := \n (node (hcons h (hcons h' (hcons h'' hnil)))).\n\\end{Coqsrc}\n\n\nFor instance, the hydra \\texttt{Hy} of Figure~\\vref{fig:Hy} is defined in \\emph{Gallina} as follows:\n\n\\vspace{4mm}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html\\#Hy}{Hydra.Hydra\\_Examples}}\n\n\\begin{Coqsrc}\nExample Hy := hyd3 head\n (hyd2\n (hyd1 \n (hyd2 head head))\n head) \n head.\n\\end{Coqsrc}\n\n\n\nHydras quite frequently contain multiple copies of the same pattern. The following functions\nwill help us to describe and reason about replications in hydra battles.\n\n\\vspace{4pt}\n\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#hcons_mult}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nFixpoint hcons_mult (h:Hydra)(n:nat)(s:Hydrae):Hydrae :=\n match n with \n | O => s\n | S p => hcons h (hcons_mult h p s)\n end.\n\n(** hydra with n copies of the same daughter *)\n\nDefinition hyd_mult h n :=\n node (hcons_mult h n hnil).\n\\end{Coqsrc}\n\n\\vspace{4mm}\n\n\n\nFor instance, the hydra $Hy''$ of Fig~\\vref{fig:Hy3} can be defined in \\coq{} as follows:\n\n\\vspace{4pt}\n\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}}\n\n\\begin{Coqsrc}\nExample Hy'' := \n hyd2 head\n (hyd2 (hyd_mult (hyd1 head) 5)\n head).\n\\end{Coqsrc}\n\n\n\n\n\\subsubsection{Recursive Functions on type \\texttt{Hydra}}\n\\label{sec:orgheadline41}\n\\label{sec:hsize-def}\n\n\n\n\nWhen defining a recursive function over the type \\texttt{Hydra}, one has to consider the three constructors \n\\texttt{node}, \\texttt{hnil} and \\texttt{hcons} of the mutually inductive types \\texttt{Hydra} and \\texttt{Hydrae}. \nLet us define for instance the function which computes the number of nodes of any hydra:\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html}{Hydra.Hydra\\_Definitions}}\n\n\n\\begin{Coqsrc}\nFixpoint hsize (h:Hydra) : nat :=\n match h with node l => S (lhsize l)\n end\nwith lhsize l : nat :=\n match l with hnil => 0\n | hcons h hs => hsize h + lhsize hs \n end.\n\n Compute hsize Hy.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = 9\n : nat \n\\end{Coqanswer}\n\n\nLikewise, the \\emph{height} (maximum distance between the foot and a head) \nis defined by mutual recursion:\n\n\\begin{Coqsrc}\nFixpoint height (h:Hydra) : nat :=\n match h with node l => lheight l\n end\nwith lheight l : nat :=\n match l with \n | hnil => 0\n | hcons h hs => Max.max (S (height h)) (lheight hs)\n end.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nCompute height Hy.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = 4\n : nat \n\\end{Coqanswer}\n\n\\index{Exercises}\n\n\\begin{exercise}\nDefine a function \\texttt{max\\_degree: Hydra $\\arrow$ nat} which returns the highest degree of a node in any hydra. For instance, the evaluation of the term \\texttt{(max\\_degree Hy)} should return $3$.\n\\end{exercise}\n\n\\subsection{Induction Principles for Hydras}\n\\label{sec:orgheadline42}\n\n\nIn this section, we show how induction principles are used to prove properties on the type \n\\texttt{Hydra}. Let us consider for instance the following statement:\n\\begin{quote}\n `` The height of any hydra is strictly less than its size. ''\n\\end{quote}\n\n\n\n\\subsubsection{A failed Attempt}\n\nOne may try to use the default tactic of proof by induction, which corresponds to an application of the automatically generated induction principle for type \\texttt{Hydra}:\n\n\\begin{Coqanswer}\nHydra_ind :\nforall P : Hydra -> Prop,\n(forall h : Hydrae, P (node h)) -> forall h : Hydra, P h\n\\end{Coqanswer}\n\nLer us start a simple proof by induction.\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}}\n\n\\begin{Coqbad}\nModule Bad.\n\nLemma height_lt_size (h:Hydra) :\n height h <= hsize h.\nProof.\n induction h as [s].\n\\end{Coqbad}\n\n\\begin{Coqanswer}\n1 subgoal, subgoal 1 (ID 11)\n \n s : Hydrae\n ============================\n height (node s) <= hsize (node s)\n\\end{Coqanswer}\n\nWe might be tempted to do an induction on the sequence \\texttt{s}:\n\n% \\begin{Coqbad}\n% induction s as [ | h s'].\n% - cbn; auto with arith.\n% - \n% \\end{Coqbad}\n\n\n\\begin{Coqanswer}\n 1 focused subgoal\n(unfocused: 0), subgoal 1 (ID 19)\n \n h : Hydra\n s' : Hydrae\n IHs' : height (node s') <= hsize (node s')\n ============================\n height (node (hcons h s')) <= hsize (node (hcons h s'))\n\n\\end{Coqanswer}\n\nNote that the displayed subgoal does not contain any assumption on \\texttt{h}, thus there is no way to \ninfer any property about the height and size of the hydra \\texttt{(hcons h t)}.\n\n\\begin{Coqbad}\nAbort.\n\nEnd Bad.\n\\end{Coqbad}\n\n\\subsubsection{A Principle of Mutual Induction}\nIn order to get an appropriate induction scheme for the types \n\\texttt{Hydra} and \\texttt{Hydrae}, we can use \\coq{}'s command \\texttt{Scheme}.\n\n\n\\index{Coq!Techniques!Mutually inductive types}\n\\index{Coq!Commands!Scheme}\n\n\\begin{Coqsrc}\nScheme Hydra_rect2 := Induction for Hydra Sort Type\nwith Hydrae_rect2 := Induction for Hydrae Sort Type.\n\\end{Coqsrc} \n\n\n\\begin{Coqsrc}\nCheck Hydra_rect2.\n\\end{Coqsrc}\n\n\n\\begin{Coqanswer}\nHydra_rect2\n : forall (P : Hydra -> Type) (P0 : Hydrae -> Type),\n (forall h : Hydrae, P0 h -> P (node h)) ->\n P0 hnil ->\n (forall h : Hydra, P h -> \n forall h0 : Hydrae, P0 h0 -> P0 (hcons h h0)) ->\n forall h : Hydra, P h\n\\end{Coqanswer} \n\n\n\n\n\\subsubsection{A Correct Proof}\n\nLet us now use \\texttt{Hydra\\_rect2} for proving that the height of any hydra is strictly less than its size.\nUsing this scheme requires an auxiliary predicate, called \\texttt{P0} in \\texttt{Hydra\\_rect2}'s statement. \nLet us begin by defining an ad-hoc version of \\texttt{List.Forall}.\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}}\n\n\\begin{Coqsrc}\n(** All elements of s satisfy P *)\n\nFixpoint h_forall (P: Hydra -> Prop) (s: Hydrae) :=\n match s with\n hnil => True\n | hcons h s' => P h /\\ h_forall P s'\n end.\n \\end{Coqsrc}\n\n \\begin{Coqsrc}\nLemma height_lt_size (h:Hydra) :\n height h < hsize h.\nProof.\n induction h using Hydra_rect2 with \n (P0 := h_forall (fun h => height h < hsize h)).\n \\end{Coqsrc}\n\n\\begin{enumerate}\n\\item The first subgoal is as follows:\n\n\\begin{Coqanswer}\n\n h: Hydrae\n IHh : h_forall (fun h : Hydra => height h < hsize h) h\n ============================\n height (node s) < hsize (node s) \n\n\\end{Coqanswer}\n\nThis goal is easily solvable, using some arithmetic. We let the reader look at the source.\n\n\\item The second subgoal is trivial:\n\n\\begin{Coqanswer}\n\n ============================\n h_forall (fun h : Hydra => height h < hsize h) hnil\n\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\n reflexivity.\n\\end{Coqsrc}\n\n\\item Finally, the last subgoal is also easy to solve:\n\n\n\n\\begin{Coqanswer}\n h : Hydra\n h0 : Hydrae\n IHh : height h < hsize h\n IHh0 : h_forall (fun h : Hydra => height h < hsize h) h0\n ============================\n h_forall (fun h1 : Hydra => height h1 < hsize h1) \n (hcons h h0)\n\\end{Coqanswer} \n\n\n\n\\begin{Coqsrc}\n split;auto. \nQed. \n\\end{Coqsrc}\n\n\\end{enumerate}\n\n\\index{Exercises}\n\n\\begin{exercise}\nIt happens very often that, in the proof of a proposition of the form \n(\\texttt{$\\forall\\,$ h:Hydra, $P$ h}), the predicate \\texttt{P0}\nis (\\texttt{h\\_forall $P$}). Design a tactic for induction on hydras that frees the user from binding explicitly \\texttt{P0}, and solves trivial subgoals. Apply it for writing a shorter proof of \\texttt{height\\_lt\\_size}.\n\\end{exercise}\n \n\n\n\\section{Relational Description of Hydra Battles}\n\n\nIn this section, we represent the rules of hydra battles as a binary relation associated with\na \\emph{round}, i.e., an interaction composed of the two following actions:\n\\begin{enumerate}\n\\item Hercules chops off one head of the hydra\n\\item Then, the hydra replicates the wounded part (if the head is at distance $\\geq 2$ from the foot).\n\\end{enumerate}\nThe relation associated with each round of the battle is parameterized by the \\emph{expected} replication factor (irrelevant if the chopped head is at distance 1 from the foot,\nbut present for consistency's sake).\n\nIn our description, we will apply the following naming convention: if $h$ represents the configuration of the hydra before a round, then the configuration of $h$ after this round will be called $h'$.\n Thus, we are going to define a proposition (\\texttt{round\\_n $n\\;h\\;h'$}) whose intended meaning will be `` the hydra $h$ is transformed into $h'$ in a single round of a battle, with the expected replication factor $n$ ''.\n\n\nSince the replication of parts of the hydra depends on the distance of the chopped head from the foot, we decompose our description into two main cases, under the form of a bunch of [mutually] inductive predicates over the types \\texttt{Hydra} and \\texttt{Hydrae}.\n\nThe mutually exclusive cases we consider are the following:\n\\begin{itemize}\n\\item \\textbf{R1}: The chopped off head was at distance 1 from the foot.\n\\item \\textbf{R2}: The chopped off head was at a distance greater than or equal to $2$ from the foot.\n\\end{itemize}\n\n\n\n\\subsection{Chopping off a Head at Distance 1 from the Foot (Relation R1)}\n\nIf Hercules chops off a head near the floor, there is no replication at all. We use an auxiliary \npredicate \\texttt{S0}, associated with the removing of one head from a sequence of hydras.\n\n\n\\vspace{4pt}\\emph{From Module\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nInductive S0 : relation Hydrae :=\n| S0_first : forall s, S0 (hcons head s) s\n| S0_rest : forall h s s', S0 s s' ->\n S0 (hcons h s) (hcons h s').\n\nInductive R1 : Hydra -> Hydra -> Prop :=\n| R1_intro : forall s s', S0 s s' -> R1 (node s) (node s').\n\\end{Coqsrc}\n\n\\subsubsection{Example}\n\\label{sec:orgheadline45}\n\nLet us represent in \\coq{} the transformation of the hydra of Fig.~\\vref{fig:Hy} into\nthe configuration represented in Fig.~\\ref{fig:Hy-prime}.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}}\n\n\n\\begin{Coqsrc}\nExample Hy_1 : R1 Hy Hy'.\nProof. \n split; right; right; left.\nQed.\n\\end{Coqsrc}\n\n\n\\subsection{Chopping off a Head at Distance \\texorpdfstring{$\\geq 2$}{>= 2} from the Foot (relation R2) }\n\n\nLet us now consider beheadings where the chopped-off head is at distance greater than or equal to $2$ from the foot. All the following relations are parameterized by the replication factor $n$.\n\n Let $s$ be a sequence of hydras. \nThe proposition (\\texttt{S1 n s s'}) holds if $s'$ is obtained by replacing some element $h$ of $s$ by \n$n+1$ copies of $h'$, where the proposition (\\texttt{R1 h h'}) holds, in other words, $h'$ is just $h$, without the chopped-off head. \\texttt{S1} is an inductive relation with two constructors that allow us to choose the position in $s'$ of the wounded sub-hydra $h$.\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#S1}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nInductive S1 (n:nat) : Hydrae -> Hydrae -> Prop :=\n| S1_first : forall s h h' , R1 h h' -> \n S1 n (hcons h s) (hcons_mult h' (S n) s)\n| S1_next : forall h s s', S1 n s s' ->\n S1 n (hcons h s) (hcons h s').\n\\end{Coqsrc}\n\n\nThe rest of the definition is structured as two mutually inductive relations on hydras and sequences of hydras. The first constructor of \\texttt{R2} describes the case where the chopped head is exactly at height $2$. The others constructors allow us to consider beheadings at height strictly greater than $2$.\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#R2}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nInductive R2 (n:nat) : Hydra -> Hydra -> Prop :=\n| R2_intro : forall s s', S1 n s s' -> R2 n (node s) (node s')\n| R2_intro_2 : forall s s', S2 n s s' -> R2 n (node s) (node s')\n\nwith S2 (n:nat) : Hydrae -> Hydrae -> Prop :=\n| S2_first : forall h h' s ,\n R2 n h h' -> \n S2 n (hcons h s) (hcons h' s)\n| S2_next : forall h r r',\n S2 n r r' ->\n S2 n (hcons h r) (hcons h r'). \n\\end{Coqsrc}\n\n\n\\subsubsection{Example}\nLet us prove the transformation of \\texttt{Hy'} into \\texttt{Hy''} (see Fig.~\\vref{fig:Hy3}). We use an experimental set of tactics for specifying the place where the \ninteraction between Hercules and the hydra holds. \n\n\n\\vspace{4pt}\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Examples.html}{Hydra.Hydra\\_Examples}}. \n\n\\begin{Coqsrc}\nExample R2_example: R2 4 Hy' Hy''.\nProof.\n (** move to 2nd sub-hydra (0-based indices) *) R2_up 1. \n (** move to first sub-hydra *) R2_up 0.\n (** we're at distance 2 from the to-be-chopped-off head \n let's go to the first daughter, \n then chop-off the leftmost head *) r2_d2 0 0. \nQed.\n\\end{Coqsrc}\n\nThe reader is encouraged to look at all the successive subgoals of this example.\n\\emph{Please consider also exercise~\\vref{exo:interactive-battle}.}\n\n\n\\subsection{Relation Associated with a Round}\n\nWe combine the two cases above into a single relation.\nFirst, we define the relation \\texttt{(round\\_n n h h')} where \\texttt{n} is the expected number of replications (irrelevant in the case of an \\texttt{R1}-transformation).\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#round_n}{Hydra.Hydra\\_Definitions}}\n\n\\index{Predicates!round\\_n}\n\n\\begin{Coqsrc}\nDefinition round_n n h h' := R1 h h' \\/ R2 n h h'. \n\\end{Coqsrc}\n\nBy abstraction over \\texttt{n}, we define a \\emph{round} (small step) of a battle:\n\n\\index{Predicates!round}\n\\label{sect:infix-round}\n\\begin{Coqsrc}\nDefinition round h h' := exists n, round_n n h h'.\n\nInfix \"-1->\" := round (at level 60).\n\\end{Coqsrc}\n\n\\index{Projects}\n\n\\begin{project}\nGive a direct translation of Kirby and Paris's description of hydra battles (quoted on page~\\pageref{original-rules}) and prove that our relational description is consistent with theirs.\n\\end{project}\n\n\n\\subsection{Rounds and Battles}\n\n\nUsing library \\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Relations.Relation_Operators.html}{Relations.Relation\\_Operators}, we define \\texttt{round\\_plus}, the transitive closure of \\texttt{round}, and \\texttt{round\\_star}, the reflexive and transitive closure of \\texttt{round}.\n\n\\label{sect:infix-rounds} \n\n\\begin{Coqsrc}\nDefinition round_plus := clos_trans_1n Hydra round.\nInfix \"-+->\" := rounds (at level 60).\n\nDefinition round_star h h' := h = h' \\/ round_plus h h'.\nInfix \"-*->\" := round_star (at level 60).\n\\end{Coqsrc}\n\n\n\\index{Exercises}\n\n\\begin{exercise}\nProve the following lemma:\n\n\\begin{Coqsrc}\nLemma rounds_height : forall h h', \n h -+-> h' -> height h' <= height h. \n\\end{Coqsrc}\n \n\\end{exercise}\n\n\\begin{remark}\n\\label{remark:transitive-closure}\n\\coq's library \\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Relations.Relation_Operators.html}{Coq.Relations.Relation\\_Operators} \ncontains three logically equivalent definitions of the transitive closure of a binary relation. This equivalence is proved in \n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Relations.Operators_Properties.html}{Coq.Relations.Operators\\_Properties} . \n\nWhy three definitions for a single mathematical concept?\nEach definition generates an associated induction principle. \n According to the form of statement one would like to prove, there is a ``best choice'':\n\n\\begin{itemize}\n\\item For proving $\\forall y, x\\,R^+\\,y \\;\\arrow\\; P\\,y$, prefer \n\\texttt{clos\\_trans\\_n1}\n\\item For proving $\\forall x,\\,x\\,R^+\\,y \\;\\arrow\\; P\\,x$, prefer \\texttt{clos\\_trans\\_1n}\n\\item For proving $\\forall x\\,y, \\,x\\,R^+\\,y \\;\\arrow\\;P\\,x\\,y$, \nprefer \\texttt{clos\\_trans},\n\\end{itemize}\nBut there is no ``wrong choice'' at all: the equivalence lemmas in \\linebreak \n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Relations.Operators_Properties.html}{Coq.Relations.Operators\\_Properties} \n allow the user\nto convert any one of the three closures into another one before applying the corresponding elimination tactic.\nThe same remark also holds for reflexive and transitive closures. \n\\end{remark}\n\n\\index{Exercises}\n\n\\begin{exercise}\nDefine a restriction of \\coqsimple{round}, where Hercules always chops off\nthe leftmost among the lowest heads.\n\nProve that, if $h$ is not a simple head, then there exists a unique $h'$ such that \\texttt{h} is transformed into \\texttt{ h'} in one round, according to this restriction.\n\n\n\\end{exercise}\n\n\\index{Exercises}\n\n\\begin{exercise}[Interactive battles]\n\\label{exo:interactive-battle}\nGiven a hydra \\texttt{h}, the specification of a hydra battle for \\texttt{h} is the type \n\\Verb@{h':Hydra | h -*-> h'}@. In order to avoid long sequences of \\texttt{split}, \\texttt{left}, and \n\\texttt{right}, design a set of dedicated tactics for the interactive building of a battle.\nYour tactics will have the following functionalities:\n\\begin{itemize}\n\\item Chose to stop a battle, or continue\n\\item Chose an expected number of replications\n\\item Navigate in a hydra, looking for a head to chop off.\n\\end{itemize}\n\nUse your tactics for simulating a small part of a hydra battle, for instance the rounds which lead from\n\\texttt{Hy} to \\texttt{Hy'''} (Fig.~\\vref{fig:Hy5}).\n\n\\textbf{Hints:} \n\\begin{itemize}\n\n\\item Please keep in mind that the last configuration of your interactively built battle is known only at the end of the battle. Thus, you will have to create and solve subgoals with existential variables. For that purpose, the tactic \\texttt{eexists}, applied to the \ngoal \\Verb@{h':Hydra | h -*-> h'}@ generates the subgoal \\Verb|h -*-> ?h'|.\n\\item You may use Gérard Huet's \\emph{zipper} data structure~\\cite{zipper} for writing tactics associated with Hercule's interactive search of a head to chop off.\n\\end{itemize}\n\n\n\n\n\n\n\\end{exercise}\n\n\n\n\n\\subsection{Classes of Battles}\n\\label{sect:battle-classes}\n\nIn some presentations of hydra battles, e.g.~\\cite{KP82, bauer2008}, the transformation associated with the $i$-th round may depend on $i$. For instance, in these articles, the replication factor at the $i$-th round is equal to $i$. In other examples, one can allow the hydra to apply any replication factor at any time. In order to be the most general as possible, we define the type of predicates which relate the state of the hydra before and after the $i$-th round of a battle.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html}{Hydra.Hydra\\_Definitions}}\n\\label{types:Battle}\n\n\\begin{Coqsrc}\nDefinition dep_round_t := nat -> Hydra -> Hydra -> Prop.\n\nClass Battle := {battle_r : dep_round_t;\n battle_inclusion : forall i h h',\n battle_r i h h' -> round h h'}.\n\n\\end{Coqsrc}\n\nThe most general class of battles is \\texttt{free}, which allows the hydra to chose any replication factor at every step:\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#free}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nProgram Instance free : Battle :=\n (Build_Battle ( fun _ h h' => round h h') _).\n\\end{Coqsrc}\n\nWe chosed to call \\emph{standard} the kind of battles which appear most often in the litterature and correspond to an arithmetic progression of the replication factor : $0,1,2,3, \\dots$\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#standard}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nProgram Instance standard : Battle := (Build_Battle round_n _).\nNext Obligation.\n now exists i. \nDefined.\n\\end{Coqsrc}\n\n\n\\subsection{Big Steps}\n\nLet $B$ be any instance of class \\texttt{Battle}. It is easy to define inductively the relation between the $i$-th and the $j$-th steps of a battle of type $B$.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#fight}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nInductive battle (B:Battle) : nat -> Hydra -> nat -> Hydra -> Prop :=\n\n| battle_1 : forall i h h', battle_r B i h h' -> \n battle B i h (S i) h'\n| battle_n : forall i h j h' h'', battle_r B i h h'' ->\n battle B (S i) h'' j h' ->\n battle B i h j h'.\n\\end{Coqsrc}\n\n% \\begin{remark}\n% The class \\texttt{free} is strongly related with the transitive closure \\texttt{round\\_plus}, as expressed by the following lemmas.\n\n% \\vspace{4pt}\n% \\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Lemmas.html}{Hydra.Hydra\\_Lemmas}}\n\n% \\begin{Coqsrc}\n% Lemma battle_free_equiv1 : forall i j h h', \n% battle free i h j h' -> h -+-> h'.\n \n% Lemma battle_free_equiv2 : forall h h',\n% h -+-> h' ->\n% forall i, exists j, battle free i h j h'.\n% \\end{Coqsrc}\n\n% \\end{remark}\n\n\n\n\\section{A Long Battle}\n\\label{sect:big-battle}\n\n\nIn this section we consider a simple example of battle, starting with a small hydra,\nshown on figure~\\vref{fig:hinit}, with a simple strategy for both players:\n\n\\begin{itemize}\n\\item At each round, Hercules chops off the rightmost head of the hydra.\n\\item The battle is standard: at the round number $i$, the expected replication is $i$.\n\\end{itemize}\n\n\n\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[thick, scale=0.30]\n \\node (foot) at (6,0) {$\\bullet$};\n\\node (n1) at (3,3) {$\\bullet$};\n\\node (h1) at (1,6) {$\\Smiley[1][green]$};\n\\node (h2) at (3,6) {$\\Smiley[1][green]$};\n\\node (h3) at (6,6) {$\\Smiley[1][green]$};\n\\node (h4) at (6,6) {$\\Smiley[1][green]$};\n\\node (h5) at (6,3) {$\\Smiley[1][green]$};\n\\node (h6) at (9,3) {$\\Smiley[1][green]$};\n\\draw (foot) -- (n1);\n\\draw (n1) to [bend left=20] (h1);\n\\draw (n1) to (h2);\n\\draw (n1) to [bend right=20] (h3);\n\\draw (foot) -- (h5);\n\\draw (foot) to [bend right=20] (h6);\n\\end{tikzpicture}\n\n \\caption{The hydra hinit}\n \\label{fig:hinit}\n\\end{figure}\n\n\n\\begin{Coqsrc}\nDefinition hinit := hyd3 (hyd_mult head 3) head head. \n\\end{Coqsrc}\n\n\n\nThe lemma we would like to prove is ``The considered battle lasts exactly $N$ rounds'',\nwith $N$ being a natural number we gave to guess.\n\nBut the battle is so long that no \\emph{test} can give us an estimation of its length, and we do need the expressive power of logic to compute this length. However, in order to guess this length, we made some experiments, computing with \\gallina{}, \\coq{}'s functional programming language.\nThus, we can consider this development as a collaboration of proof with computation.\n\nIn the following lines, we try to show faithfully how we found the value of the number $N$.\n\nThe complete proof is in file \\url{../theories/html/hydras.Hydra.BigBattle.html}. \n\n\\subsection{The beginning of Hostilities}\nDuring the two first rounds, our hydra loses its two rightmost heads. Thus just before the third round, it looks like in figure~\\vref{fig:hinit-plus2}.\n\n\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[thick, scale=0.30]\n \\node (foot) at (3,0) {$\\bullet$};\n\\node (n1) at (3,3) {$\\bullet$};\n\\node (h1) at (1,6) {$\\Smiley[1][green]$};\n\\node (h2) at (3,6) {$\\Smiley[1][green]$};\n\\node (h3) at (6,6) {$\\Smiley[1][green]$};\n\\node (h4) at (6,6) {$\\Smiley[1][green]$};\n\\draw (foot) -- (n1);\n\\draw (n1) to [bend left=20] (h1);\n\\draw (n1) to (h2);\n\\draw (n1) to [bend right=20] (h3);\n\\end{tikzpicture}\n\n \\caption{The hydra (hyd1 h3)}\n \\label{fig:hinit-plus2}\n\\end{figure}\n\nThe following lemma is a formal description of these first rounds, in terms of the\n\\texttt{battle} predicate.\n\n\\begin{Coqsrc}\nLemma L_0_2 : battle standard 0 hinit 2 (hyd1 h3). \n\\end{Coqsrc}\n\n\n\\subsection{Looking for Regularities}\n\n\nA first study with pencil and paper suggested us that, after three rounds, the hydra always looks like in figure~\\vref{fig:hinit-plusn} (with a variable number of \nsubtrees of height 1 or 0).\nThus, we introduce handy notations.\n\n\\begin{Coqsrc}\nNotation h3 := (hyd_mult head 3).\nNotation h2 := (hyd_mult head 2).\nNotation h1 := (hyd1 head).\n\nDefinition hyd a b c := \n node (hcons_mult h2 a\n (hcons_mult h1 b\n (hcons_mult head c hnil))).\n\\end{Coqsrc}\n\n\nFor instance Fig~\\vref{fig:hinit-plusn} shows the hydra (\\texttt{hyd 3 4 2}). The hydra (\\texttt{hyd 0 0 0}) is the ``final'' hydra of any terminating battle, {i.e.},\na tree whith exactly one node and no edge.\n\n\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[thick, scale=0.30]\n \\node (foot) at (15,0) {$\\bullet$};\n\\node (a) at (3,4) {$\\bullet$};\n\\node (b) at (6,4) {$\\bullet$};\n\\node (c) at (9,4) {$\\bullet$};\n\\node (d) at (13,4) {$\\bullet$};\n\\node (e) at (16,4) {$\\bullet$};\n\\node (f) at (19,4) {$\\bullet$};\n\\node (g) at (22,4) {$\\bullet$};\n\\node (h) at (25,4) {$\\Smiley[1][green]$};\n\\node (i) at (28,4) {$\\Smiley[1][green]$};\n\\node (aa) at (2.5,8) {$\\Smiley[1][green]$};\n\\node (ab) at (3.5,8) {$\\Smiley[1][green]$};\n\\node (ba) at (5.5,8) {$\\Smiley[1][green]$};\n\\node (bb) at (6.5,8) {$\\Smiley[1][green]$};\n\\node (ca) at (8.5,8) {$\\Smiley[1][green]$};\n\\node (cb) at (9.5,8) {$\\Smiley[1][green]$};\n\\node (da) at (13,8) {$\\Smiley[1][green]$};\n\\node (ea) at (16,8) {$\\Smiley[1][green]$};\n\\node (fa) at (19,8) {$\\Smiley[1][green]$};\n\\node (ga) at (22,8) {$\\Smiley[1][green]$};\n\\draw (foot) -- (a);\n\\draw (foot) -- (b);\n\\draw (foot) -- (c);\n\\draw (foot) -- (d);\n\\draw (foot) -- (e);\n\\draw (foot) -- (f);\n\\draw (foot) -- (g);\n\\draw (foot) -- (h);\n\\draw (foot) -- (i);\n\\draw (a) -- (aa);\n\\draw (a) -- (ab);\n\\draw (b) -- (ba);\n\\draw (b) -- (bb);\n\\draw (c) -- (ca);\n\\draw (c) -- (cb);\n\\draw (d) -- (da);\n\\draw (e) -- (ea);\n\\draw (f) -- (fa);\n\\draw (g) -- (ga);\n% \\node (a) at (3,4) {$\\bullet$};\n% \\node (h1) at (1,6) \n% \\node (h2) at (3,6) {$\\Smiley[1][green]$};\n% \\node (h3) at (6,6) {$\\Smiley[1][green]$};\n% \\node (h4) at (6,6) {$\\Smiley[1][green]$};\n% \\draw (foot) -- (n1);\n% \\draw (n1) to [bend left=20] (h1);\n% \\draw (n1) to (h2);\n% \\draw (n1) to [bend right=20] (h3);\n\\end{tikzpicture}\n\n \\caption{The hydra (hyd 3 4 2)}\n \\label{fig:hinit-plusn}\n\\end{figure}\n\n\nWith these notations, we get a formal description of the first three rounds.\n\n\n\\begin{Coqsrc}\nLemma L_2_3 : battle standard 2 (hyd1 h3) 3 (hyd 3 0 0).\n\nLemma L_0_3 : battle standard 0 hinit 3 (hyd 3 0 0).\n\\end{Coqsrc}\n\n\n\\subsection{Computing \\dots}\nIn order to study \\emph{experimentally} the different configurations of the battle, we will use a simple datatype for representing the states as tuples composed of\nthe round number, and the respective number of daughters \\texttt{h2}, \\texttt{h1}, and heads\nof the current hydra.\n\n\n\n\n\\begin{Coqsrc}\n Record state : Type :=\n mks {round: nat ; n2 : nat ; n1 : nat ; nh : nat}.\n\\end{Coqsrc}\n\nThe following function returns the next configurarion of the game. \nNote that this function is defined only for making experiments and is not ``certified''. Formal proofs about our battle will only start with the lemma\n\\texttt{lemma:step-battle}, page~\\pageref{lemma:step-battle}.\n\n\n\\begin{Coqsrc}\nDefinition next (s : state) :=\n match s with\n | mks round a b (S c) => mks (S round) a b c\n | mks round a (S b) 0 => mks (S round) a b (S round)\n | mks round (S a) 0 0 => mks (S round) a (S round) 0\n | _ => s\n end.\n\\end{Coqsrc}\n\nWe can make bigger steps through iterations of \\texttt{next}.\nThe functional \\texttt{iterate}, similar to Standard Library's \\texttt{Nat.iter},\nis defined and studied in~\\href{../theories/html/hydras.Prelude.Iterates.html\\#iterate}{Prelude.Iterates}.\n\n\\label{Functions:iterate}\n\n\\begin{Coqsrc}\nFixpoint iterate {A:Type}(f : A -> A) (n: nat)(x:A) :=\n match n with\n | 0 => x\n | S p => f (iterate f p x)\n end.\n\\end{Coqsrc}\n\n\n\nThe following function computes the state of the battle at the $n$-th round.\n\n\n\\begin{Coqsrc}\nDefinition test n := iterate next (n-3) (mks 3 3 0 0).\n\nCompute test 3.\n (**\n = {| round := 3; n2 := 3; n1 := 0; nh := 0 |}\n : state\n *)\n\nCompute test 4.\n (*\n = {| round := 4; n2 := 2; n1 := 4; nh := 0 |}\n : state\n *)\n\nCompute test 5.\n (*\n = {| round := 5; n2 := 2; n1 := 3; nh := 5 |}\n : state\n *)\n\nCompute test 2000.\n(*\n = {| round := 2000; n2 := 1; n1 := 90; nh := 1102 |}\n : state\n*)\n\\end{Coqsrc}\n\n\nThe battle we are studying seems to be awfully long. Let us concentrate our\ntests on some particular events : the states where $\\texttt{nh}=0$.\nFrom the value of \\texttt{test 5}, it is obvious that at the 10-th round, the counter \\texttt{nh} is equal to zero.\n\n \\begin{Coqsrc}\n Compute test 10.\n(*\n\n = {| round := 10; n2 := 2; n1 := 3; nh := 0 |}\n : state\n*)\n \\end{Coqsrc}\n\nThus, $ (1 + 11)$ rounds later, the \\texttt{n1} field is equal to $2$, and \n\\texttt{nh} to $0$. \n\n\n\\begin{Coqsrc}\nCompute test 22.\n(*\n\n = {| round := 22; n2 := 2; n1 := 2; nh := 0 |}\n : state\n*)\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\n Compute test 46.\n(*\n\n = {| round := 46; n2 := 2; n1 := 1; nh := 0 |}\n : state\n\n*)\n\nCompute test 94.\n\n(*\n\n = {| round := 94; n2 := 2; n1 := 0; nh := 0 |}\n : state\n\n*) \n\\end{Coqsrc}\n\n\nNext round, we decrement \\texttt{n2} and set \\texttt{n1} to $95$.\n\n\n\\begin{Coqsrc}\n Compute test 95.\n\n(*\n\n = {| round := 95; n2 := 1; n1 := 95; nh := 0 |}\n : state\n\n*)\n\\end{Coqsrc}\n\nWe now have some intuition of the sequence.\nIt looks like the next ``\\texttt{nh}=0'' event will happen at the $192=2(95+1)$-th round, then at the $2(192+1)$-th round, etc.\n\n\n\\begin{Coqsrc}\nDefinition doubleS (n : nat) := 2 * (S n).\n\nCompute test (doubleS 95).\n\n(**\n = {| round := 192; n2 := 1; n1 := 94; nh := 0 |}\n : state\n *)\n\n\nCompute test (iterate doubleS 2 95).\n\n(*\n = {| round := 386; n2 := 1; n1 := 93; nh := 0 |}\n : state\n*)\n\\end{Coqsrc}\n\n\\subsection{Proving \\dots}\nWe are now able to reason about the sequence of transitions defined by our hydra battle. Instead of using the data-type \\texttt{state} we study the relationship\nbetween different configurations of the battle.\n\nLet us define a binary relation associated with every round of the battle.\nIn the following definition \\texttt{i} is associated with the round number (or date, if we consider a discrete time), and \\texttt{a}, \\texttt{b}, \\texttt{c} respectively associated with the number of \\texttt{h2}, \\texttt{h1} and heads connected to the hydra's foot.\n\n\\begin{Coqsrc}\nInductive one_step (i: nat) :\n nat -> nat -> nat -> nat -> nat -> nat -> Prop :=\n| step1: forall a b c, one_step i a b (S c) a b c\n| step2: forall a b, one_step i a (S b) 0 a b (S i)\n| step3: forall a, one_step i (S a) 0 0 a (S i) 0.\n\\end{Coqsrc}\n\nThe relation between \\texttt{one\\_step} and the rules of hydra battle is asserted by the following lemma. \n\n\\label{lemma:step-battle}\n\n\\begin{Coqsrc}\nLemma step_battle : forall i a b c a' b' c', \n one_step i a b c a' b' c' ->\n battle standard i (hyd a b c) (S i) (hyd a' b' c').\n\\end{Coqsrc}\n\nNext, we define ``big steps'' as the transitive closure of \\texttt{one\\_step},\nand reachability (from the initial configuration of figure~\\ref{fig:hinit} at time $0$).\n\n\n\n\\begin{Coqsrc}\n Inductive steps : nat -> nat -> nat -> nat ->\n nat -> nat -> nat -> nat -> Prop :=\n| steps1 : forall i a b c a' b' c',\n one_step i a b c a' b' c' -> steps i a b c (S i) a' b' c'\n| steps_S : forall i a b c j a' b' c' k a'' b'' c'',\n steps i a b c j a' b' c' ->\n steps j a' b' c' k a'' b'' c'' ->\n steps i a b c k a'' b'' c''.\n\nDefinition reachable (i a b c : nat) : Prop :=\n steps 3 3 0 0 i a b c.\n\\end{Coqsrc}\n\n\nThe following lemma establishes a relation between \\texttt{steps} and the predicate \\texttt{battle}.\n\n\\begin{Coqsrc}\n Lemma steps_battle : forall i a b c j a' b' c', \n steps i a b c j a' b' c' ->\n battle standard i (hyd a b c) j (hyd a' b' c').\n\\end{Coqsrc}\n\nThus, any result about \\texttt{steps} will be applicable to standard battles.\nUsing the predicate \\texttt{steps} our study of the length of the considered battle\ncan be decomposed into three parts:\n\n\\begin{enumerate}\n\\item Characterization of regularities of some events\n\\item Study of the beginning of the battle\n\\item Computing the exact length of the battle.\n\\end{enumerate}\n\nFirst, we prove that, if at round $i$ the hydra is equal to\n(\\texttt{hyd a (S b) 0}), then it will be equal to (\\texttt{hyd a b 0}) at the $2(i+1)$-th round. \n\n\\begin{Coqsrc}\nLemma LS : forall c a b i, steps i a b (S c) (i + S c) a b 0.\nProof.\n induction c.\n - intros; replace (i + 1) with (S i).\n + repeat constructor.\n + ring.\n - intros; eapply steps_S.\n + eleft; apply rule1.\n + replace (i + S (S c)) with (S i + S c) by ring; apply IHc.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma doubleS_law : forall a b i, steps i a (S b) 0 (doubleS i) a b 0.\nProof.\n intros; eapply steps_S.\n + eleft; apply step2.\n + unfold doubleS; replace (2 * S i) with (S i + S i) by ring; \n apply LS.\nQed.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nLemma reachable_S : forall i a b, reachable i a (S b) 0 ->\n reachable (doubleS i) a b 0.\nProof.\n intros; right with (1 := H); apply doubleS_law.\nQed.\n\\end{Coqsrc}\n\nFrom now on, the lemma \\texttt{reachable\\_S} allows us to watch larger steps of \nthe battle.\n\n\n\\begin{Coqsrc}\n Lemma L4 : reachable 4 2 4 0.\nProof.\n left; constructor.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma L10 : reachable 10 2 3 0.\nProof.\n change 10 with (doubleS 4).\n apply reachable_S, L4.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma L22 : reachable 22 2 2 0.\nProof.\n change 22 with (doubleS 10).\n apply reachable_S, L10.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma L46 : reachable 46 2 1 0.\nProof.\n change 46 with (doubleS 22); apply reachable_S, L22.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma L94 : reachable 94 2 0 0.\nProof.\n change 94 with (doubleS 46); apply reachable_S, L46.\nQed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma L95 : reachable 95 1 95 0.\nProof.\n eapply steps_S.\n - eexact L94.\n - repeat constructor.\nQed.\n\\end{Coqsrc}\n\n\\subsection{Giant Steps}\n\nWe are now able to make bigger steps in the simulation of the battle.\nFirst, we iterate the lemma \\texttt{reachable\\_S}.\n\n\\begin{Coqsrc}\nLemma Bigstep : forall b i a , reachable i a b 0 ->\n reachable (iterate doubleS b i) a 0 0.\n Proof.\n induction b.\n - trivial.\n - intros; simpl; apply reachable_S in H.\n rewrite <- iterate_comm; now apply IHb.\n Qed.\n\\end{Coqsrc}\n\nApplying lemmas \\texttt{BigStep} and \\texttt{L95} we make a first jump.\n\n\n\\begin{Coqsrc}\n Definition M := (iterate doubleS 95 95).\n\nLemma L2_95 : reachable M 1 0 0.\nProof.\n apply Bigstep, L95.\nQed.\n\\end{Coqsrc}\n\nFigure~\\ref{fig:HM} represents the hydra at the $M$-th round.\nAt the $(M+1)$-th round, it will look like in fig~\\ref{fig:HM-plus1}.\n\n\n\n\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (3,4) {$\\Smiley[2][green]$};\n\\node (N3) at (1,4) {$\\Smiley[2][green]$};\n\\draw (foot) -- (N1);\n\\draw (N1) to [bend right =15] (N2);\n\\draw (N1) to [bend left=15](N3);\n\\end{tikzpicture}\n\\caption{\\label{fig:HM}}\nThe state of the hydra after $M$ rounds.\n% The hydra \\texttt{h} of the proof that \\(\\omega^2\\) is too small for proving Hercules' victory\n\n\\end{figure}\n\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\\node (foot) at (10,0) {$\\bullet$};\n\\node (N1) at (0,5) {$\\bullet$};\n\\node (N12) at (0,8) {$\\Smiley[2][green]$};\n\\node (N2) at (2,5) {$\\bullet$};\n\\node (N22) at (2,8) {$\\Smiley[2][green]$};\n\\node (N3) at (4,5) {$\\bullet$};\n\\node (N32) at (4,8) {$\\Smiley[2][green]$};\n\\node (N4) at (6,5) {$\\bullet$};\n\\node (N42) at (6,8) {$\\Smiley[2][green]$};\n\n\\node (Ndots) at (12,8) {\\Huge $\\dots$};\n\\node (Ndots2) at (12,5) {\\Huge $\\dots$};\n\n\\node (N8) at (18,5) {$\\bullet$};\n\\node (N82) at (18,8) {$\\Smiley[2][green]$};\n\\node (N9) at (20,5) {$\\bullet$};\n\\node (N92) at (20,8) {$\\Smiley[2][green]$};\n\n\n\\draw (foot) -- (N1);\n\\draw (foot) -- (N2);\n\\draw (foot) -- (N3);\n\\draw (foot) -- (N4);\n\\draw (foot) -- (N8);\n\\draw (foot) -- (N9);\n\\draw (N1) to (N12);\n\\draw (N2) to (N22);\n\\draw (N3) to (N32);\n\\draw (N4) to (N42);\n\\draw (N8) to (N82);\n\\draw (N9) to (N92);\n\\end{tikzpicture}\n\\caption{\\label{fig:HM-plus1}}\nThe state of the hydra after $M+1$ rounds (with $M+1$ heads). \n\n\\end{figure}\n\n\\begin{Coqsrc}\nLemma L2_95_S : reachable (S M) 0 (S M) 0.\nProof.\n eright.\n - apply L2_95.\n - left; constructor 3.\nQed.\n\\end{Coqsrc}\n\n\nThen, applying once more the lemma \\texttt{BigStep}, we get the exact time when\nHercules wins!\n\n\n\\begin{Coqsrc}\nDefinition N := iterate doubleS (S M) (S M).\n\nTheorem SuperbigStep : reachable N 0 0 0 .\nProof.\n apply Bigstep, L2_95_S.\nQed.\n\\end{Coqsrc}\n\nWe are now able to prove formally that the considered battle is \ncomposed of $N$ steps.\n\n\\begin{Coqsrc}\nLemma Almost_done :\n battle standard 3 (hyd 3 0 0) N (hyd 0 0 0).\nProof. \n apply steps_battle, SuperbigStep.\nQed.\n\nTheorem Done :\n battle standard 0 hinit N head.\nProof.\n eapply battle_trans.\n - apply Almost_done.\n - apply L_0_3.\nQed.\n\\end{Coqsrc}\n\n\n\n\\subsection{A Minoration Lemma}\n\nNow, we would like to get an intuition of how big the number $N$ is.\nFor that purpose, we use a minoration of the function \\texttt{doubleS} by the\nfunction (\\texttt{fun n => 2 * n}).\n\n\\begin{Coqsrc}\nDefinition exp2 n := iterate (fun n => 2 * n) n 1.\n\\end{Coqsrc}\nUsing some facts (proven in \n\\href{../theories/html/hydras.Hydra.BigBattle.html}{hydras.Hydra.BigBattle}),we get several minorations.\n\n\\begin{Coqsrc}\nLemma minoration_0 : forall n, 2 * n <= doubleS n.\n\nLemma minoration_1 : forall n x, exp2 n * x <= iterate doubleS n x.\n\nLemma minoration_2 : exp2 95 * 95 <= M.\n\nLemma minoration_3 : exp2 (S M) * S M <= N.\n\nLemma minoration : exp2 (exp2 95 * 95) <= N.\n\\end{Coqsrc}\n\n\nThe number $N$ is greater than or equal to $2^{2^{95}\\times 95}.$ If we wrote $N$ in base $10$, $N$ would require at least $10^{30}$ digits!\n\n\n\\section{Generic Properties}\n\n\nThe example we just studied shows that the termination of any battle may take a very long time. If we want to study hydra battles in general, we have to consider \nany hydra and any strategy, both for Hercules and the hydra itself. So, we first give some definitions, generally borrowed from transition systems vocabulary (see~\\cite{tel_2000} for instance).\n\n\n\\subsection{Liveliness}\n\n\nLet $B$ be an instance of \\texttt{Battle}. We say that $B$ is \\emph{alive} if\nfor any configuration $(i,h)$, where $h$ is not a head, there exists a further step in class $B$.\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Definitions.html\\#Alive}{Hydra.Hydra\\_Definitions}}\n\n\\begin{Coqsrc}\nDefinition Alive (B : Battle) :=\n forall i h, \n h <> head -> {h' : Hydra | B i h h'}.\n\\end{Coqsrc}\n\nThe theorems \\texttt{Alive\\_free} and \\texttt{Alive\\_standard} of the module \n\\url{../theories/html/hydras.Hydra.Hydra_Theorems.html} show that the classes \\texttt{free} and \\texttt{standard} satisfy this property.\n\n\\begin{Coqsrc}\nTheorem Alive_free: Alive free.\n\nTheorem Alive_standard: Alive standard. \n\\end{Coqsrc}\n\nBoth theorems are proved with the help of the following strongly specified function:\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Lemmas.html\\#next_round_dec}{Hydra.Hydra\\_Lemmas}}\n\n\\begin{Coqsrc}\nDefinition next_round_dec n :\n forall h , (h = head) + {h' : Hydra & {R1 h h'} + {R2 n h h'}}.\n\\end{Coqsrc}\n\n\n\\subsection{Termination}\n\nThe termination of all battles is naturally expressed by the predicate \\texttt{well\\_founded} defined in the module \\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Init.Wf.html}{Coq.Init.Wf} \n of the Standard Library.\n\n\\index{Predicates!Termination}\n\n\\begin{Coqsrc}\nDefinition Termination := well_founded (transp _ round).\n\\end{Coqsrc}\n\n\nLet $B$ be an instance of class \\texttt{Battle}. A \\emph{variant} for $B$ consists\nin a well-founded relation $<$ on some type \\texttt{A}, and a function\n(also called a \\emph{measure}) \\texttt{m:Hydra->A} such that for any successive steps $(i,h)$ and $(1+i,h')$ of a battle in $B$, the inequality $m(h') A): Prop :=\n {variant_decr :forall i h h',\n h <> head ->\n battle_r B i h h' -> Lt (m h') (m h)}.\n\\end{Coqsrc} \n\n\\index{Exercises}\n\n\\begin{exercise}\n Prove that, if there is an instance of (\\texttt{Hvariant Lt wf\\_Lt $B$ $m$}), then there exists no infinite battle in $B$.\n\\end{exercise}\n\n\n\n\n\\subsection{A Small Proof of Impossibility}\n\\index{Maths!Proofs of impossibility}\n\n\\label{omega-case}\n\nWhen one wants to prove a termination theorem with the help of a variant, \none has to consider first a well-founded set $(A,<)$, then a strictly decreasing measure on this set. The following two lemmas show that if the order structure $(A,<)$ is too simple, it is useless to look for a convenient measure, which simply no exists. Such kind of result is useful, because it saves you time and effort.\n\n\nThe best known well-founded order is the natural order on the set $\\mathbb{N}$ of natural numbers (the type \\texttt{nat} of Standard library). It would be interesting to look for some measure $m:\\texttt{nat}\\arrow\\texttt{nat}$ and prove it is a variant.\n\nUnfortunately, we can prove that \n\\emph{no} instance of class (\\texttt{WfVariant round Peano.lt $m$}) can be built, where\n$m$ is \\emph{any} function of type \\texttt{Hydra $\\arrow$ nat}.\n\n\nLet us present the main steps of that proof, the script of which is in the module ~\\href{../theories/html/hydras.Hydra.Omega_Small.html}{Hydra/Omega\\_Small.v} \\footnote{ The name of this file means ``the ordinal $\\omega$ is too small for proving the termination of [free] hydra battles ''. In effect, the elements of $\\omega$, considered as a set, are just the natural numbers (see next chapter for more details)}.\n\n%\\subsubsection{Preliminaries}\n\n\nLet us assume there exists some variant $m$ from \\texttt{Hydra} into \\texttt{nat} for proving\n the termination of all hydra battles.\n\n\\begin{Coqsrc}\nSection Impossibility_Proof.\n Variable m : Hydra -> nat.\n Hypothesis Hvar : Hvariant lt_wf free m.\n\\end{Coqsrc}\n\nWe define an injection $\\iota$ from the type \\texttt{nat} into \\texttt{Hydra}.\nFor any natural number $i$, $\\iota(i)$ is the hydra composed of a foot and\n$i+1$ heads at height $1$. For instance, Fig.~\\ref{fig:flower} represents the hydra $\\iota(3)$.\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\\node (foot) at (4,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\Smiley[2][green]$};\n\\node (N2) at (4,2) {$\\Smiley[2][green]$};\n\\node (N3) at (6,2) {$\\Smiley[2][green]$};\n\\node (N4) at (8,2) {$\\Smiley[2][green]$};\n\\draw (foot) to [bend left =25] (N1);\n\\draw (foot) to [bend left =15] (N2);\n\\draw (foot) to [bend right =15] (N3);\n\\draw (foot) to [bend right =25] (N4);\n\\end{tikzpicture}\n\\caption{\\label{fig:flower}\nThe hydra $\\iota(3)$}\n\\end{figure}\n\n \\begin{Coqsrc}\n Let iota (i: nat) := hyd_mult head (S i). \n \\end{Coqsrc}\n\nLet us consider now some hydra \\texttt{big\\_h} out of the range of the injection $\\iota$ (see Fig.~\\vref{fig:h-omega-omega}).\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (2,4) {$\\Smiley[2][green]$};\n\\draw (foot) -- (N1);\n\\draw (N1) to (N2);\n\\end{tikzpicture}\n\\caption{\\label{fig:h-omega-omega}}\n The hydra \\texttt{big\\_h}.\n\\end{figure}\n\n \\begin{Coqsrc}\n Let big_h := hyd1 (hyd1 head).\n \\end{Coqsrc}\n\n Using the functions $m$ and $\\iota$, we define a second hydra \\texttt{small\\_h}, and show\n there is a one-round battle that transforms \\texttt{big\\_h} into \\texttt{small\\_h}. Please note that,\ndue to the hypothesis \\texttt{Hvar}, we are interested in the termination of \\emph{free} battles. \nThere is no problem to consider a round with (\\texttt{m big\\_h}) as the replication factor.\n\n \\begin{Coqsrc}\n Let small_h := iota (m big_h).\n \n Fact big_to_small : big_h -1-> small_h.\n Proof.\n exists (m big_h); right; repeat constructor. \n Qed.\n \\end{Coqsrc}\n \nBut, by hypothesis, $m$ is a variant. Hence, we infer the following inequality.\n\n\n \\begin{Coqsrc}\nLemma m_lt : m small_h < m big_h.\n \\end{Coqsrc}\n\nIn order to get a contradiction, it suffices to prove the inequality\n\\texttt{m big\\_h <= m small\\_h}, i.e., \\texttt{m big\\_h <= m (iota (m big\\_h))}.\n\nMore generally, we prove the following lemma: \n\n\\begin{Coqsrc}\nLemma m_ge : forall i:nat, i <= m (iota i).\n\\end{Coqsrc}\n\nIntuitively, it means that, from any hydra of the form (\\texttt{iota $i$}), the battle will \ntake (at least) $i$ rounds. Thus the associated measure cannot be less than $i$.\nTechnically, we prove this lemma by Peano induction on $i$.\n\n\\begin{itemize}\n\\item The base case $i=0$ is trivial\n\\item Otherwise, let $i$ be any natural number and assume the inequality\n $i \\leq m(\\iota(i))$.\n \\begin{enumerate}\n \\item But the hydra $\\iota(S(i))$ can be transformed in one round into\n $\\iota(i)$ (by losing its righmost head, for instance)\n \\item Since $m$ is a variant, we have $m(\\iota(i)) < m(\\iota(S(i)))$,\n hence $i< m(\\iota(S(i)))$, which implies $S(i)\\leq m(\\iota(S(i)))$.\n \\end{enumerate}\n\\end{itemize}\n\n Then our proof is almost finished.\n \n \\begin{Coqsrc}\nTheorem Contradiction : False.\nProof.\n apply (Nat.lt_irrefl (m big_h));\n apply Lt.le_lt_trans with (m small_h).\n - apply m_ge.\n - apply m_lt.\nQed. \n\nEnd Impossibility_Proof.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\nProve that there exists no variant $m$ from \\texttt{Hydra} into \\texttt{nat} for proving\n the termination of all \\emph{standard} battles.\n\\end{exercise}\n\n\n\n\n\n\n\\subsubsection{Conclusion}\n\nIn order to build a variant for proving the termination of all hydra battles, we need to consider order structures more complex than the usual order on type \\texttt{nat}. \nThe notion of \\emph{ordinal number} provides a catalogue of well-founded order types.\nFor a reasonably large bunch of ordinal numbers, \\emph{ordinal notations} are data-types which allow the \\coq{} user to define functions, to compute and prove some properties, for instance by reflection.\n\nThe next chapter is dedicated to a generic implementation of ordinal notations, and chapter~\\ref{chap:T1} to a proof of termination of all hydra battles with the help of an ordinal notation for the interval $[0,\\epsilon_0)$.\n\n\n%--------------------------------------------------------------\n\n\\chapter{Introduction to Ordinal Numbers and Ordinal Notations}\n\n\nThe proof of termination of all hydra battles presented in~\\cite{KP82} is based\non \\emph{ordinal numbers}.\nFrom a mathematical point of view, an ordinal is a representant of an equivalence class for isomorphims of totally ordered well-founded sets.\n\nFor the computer scientist, ordinals are tools for proving the totality of a given recursive function, or termination of a transition system. \\emph{Ordinal arithmetic} \nprovides a set of functions whose properties, like \\emph{monotony}, allow to define \\emph{variants}, \\emph{i.e.} strictly decreasing measures used in proofs of termination.\n\n\nLet us have a look at Figure~\\ref{fig:ordinal-sequence}. It presents a small sequence of ordinal numbers, which extends the sequence of natural numbers. \n\n\n\n\n\\begin{figure}[h]\n \\centering\n\\fbox{\\Large\n \\begin{minipage}{1.0\\linewidth}\n \\begin{align*}\n &\\textcolor{blue}{0},\\,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,\\ldots\\\\\n&\\textcolor{red}{\\omega},\\,\\omega+1,\\omega+2,\\omega+3,\\ldots\\\\\n&\\textcolor{red}{\\omega\\times 2},\\,\\omega\\times 2+1,\\ldots, \\textcolor{red}{\\omega\\times 3},\\,\\omega\\times 3+1,\\ldots, \\textcolor{red}{\\omega\\times 4},\\ldots,\\\\\n&\\textcolor{red}{\\omega^2},\\ldots, \\textcolor{red}{\\omega^2\\times 42},\\ldots,\\textcolor{red}{\\omega^3},\\ldots, \\textcolor{red}{\\omega^4},\\omega^4+1,\\ldots,\\\\\n&\\textcolor{red}{\\omega^\\omega},\\ldots, \\textcolor{red}{\\omega^\\omega+\\omega^7\\times 8},\\ldots,\\textcolor{red}{\\omega^\\omega\\times 2},\\omega^\\omega\\times 2+1, \\ldots,\\\\\n&\\textcolor{red}{\\omega^{\\omega^\\omega}},\\ldots, \\textcolor{red}{\\omega^{\\omega^\\omega}+\\omega^\\omega\\times 42+ \\omega^{55}+\\omega}, \\ldots, \\textcolor{red}{\\omega^{\\omega^{\\omega+1}}}, \\omega^{\\omega^{\\omega+1}}+1,\\dots\\\\\n& \\textcolor{red}{\\epsilon_0 (= \\omega^{\\epsilon_0)}}, \\epsilon_0+1, \\epsilon_0+2, \\epsilon_0+3, \\ldots,\\\\\n& \\textcolor{red}{\\epsilon_1}, \\ldots, \\textcolor{red}{\\epsilon_2}, \\ldots, \\textcolor{red}{\\epsilon_\\omega},\\ldots \\\\\n& \\textcolor{red}{\\Gamma_0}, \\Gamma_0+1, \\Gamma_0+2, \\Gamma_0+3,\\ldots, \\textcolor{red}{\\Gamma_0+\\omega}, \\ldots,\\\\\n&\\ldots\n \\end{align*} \n \\end{minipage}}\n \n \n \\caption{A short overview of the sequence of ordinal numbers}\n \\label{fig:ordinal-sequence}\n\\end{figure}\n\n\nLet us comment some features of this figure:\n\n\\begin{itemize}\n\\item The ordinals are listed in a strictly increasing order. \n\\item Dots : ``$\\ldots$'' stand for infinite sequences of ordinals, not shown for lack of space. For instance, the ordinal $42$ is not shown in the first line, but it exists, between $17$ and $\\omega$.\n\\item Each ordinal printed in black is the immediate successor of another ordinal. We call it a \\emph{successor} ordinal. For instance, $12$ is the successor of $11$, and $\\omega^4+1$ the successor of $\\omega^4$.\n\\item Ordinals (displayed in red) that follow immediately dots are called \\emph{limit ordinals}. With respect to the order induced by this sequence, any limit ordinal $\\alpha$ is the least upper bound of the set $\\mathbb{O}_\\alpha$ of all ordinals strictly less than $\\alpha$.\n\\item\nFor instance $\\omega$ is the least upper bound of the set of all finite ordinals (in the first line). It is also the first limit ordinal, and the first infinite ordinal, in the sense that \nthe set $\\mathbb{O}_\\omega$ is infinite.\n\\item The ordinal $\\epsilon_0$ is the first number which is equal to its own exponential of base $\\omega$. It plays an important role in proof theory, and is particularly studied in chapters~\\ref{chap:T1} to \\ref{chap:alpha-large}.\n\\item Any ordinal is either the ordinal \\textcolor{blue}{$0$},\na successor ordinal, or a \\textcolor{red}{limit ordinal}.\n\\end{itemize}\n\n\n\n\n\\section{The Mathematical Point of View}\n\n\\subsection{Well-ordered Sets}\nLet us start with some definitions.\nA \\emph{well-ordered set} is a set provided with a binary relation $<$ which has the following properties.\n\\begin{description}\n\\item[irreflexivity] : $\\forall x\\in A, x\\not< x$\n\\item[transitivity] : $\\forall x\\,y\\,z\\in A, xa_2> \\dots a_n> a_{n+1}\\dots$ in $A$. In contrast, \\coq's standard library contains\nan inductive definition of a predicate \\texttt{Acc} which allows us to write \nconstructive proofs of accessibility (See \\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Init.Wf.html}{Coq.Init.Wf}).}.\n\\end{description}\n\nThe best known examples of well-ordered sets are the set $\\mathbb{N}$ of natural numbers (with the usual order $<$), as well as any finite segment $[0,i)=\\{j\\in\\mathbb{N}\\,|\\,j Irreflexive R ;\n StrictOrder_Transitive :> Transitive R }.\n\\end{Coqsrc}\n\n\n\n\\subsection{A Class for Ordinal Notations}\n\nThe following class definition, parameterized with a type $A$, a binary relation \\texttt{lt} on $A$, specifies that \\texttt{lt} is a well-founded strict order. The reflexive closure of \\texttt{lt}, (called \\texttt{le}, for ``less or equal than'') is a total decidable order, implemented through a comparison function \\texttt{compare}. The correctness of this function is expressed through Stdlib's type \n\\texttt{Datatypes.CompareSpec}.\n\n\n\\begin{Coqsrc}\nInductive CompareSpec (Peq Plt Pgt : Prop) : comparison -> Prop :=\n CompEq : Peq -> CompareSpec Peq Plt Pgt Eq\n | CompLt : Plt -> CompareSpec Peq Plt Pgt Lt\n | CompGt : Pgt -> CompareSpec Peq Plt Pgt Gt\n\\end{Coqsrc}\n\n\\vspace{4pt}\n\\noindent\\emph{From Library~\\href{../theories/html/hydras.OrdinalNotations.Definitions.html}{OrdinalNotations.Definitions}}\n\n\\label{types:ON}\n\n\\begin{Coqsrc}\nClass ON {A:Type}(lt: relation A)\n (compare : A -> A -> comparison) :=\n {\n sto :> StrictOrder lt;\n wf : well_founded lt;\n compare_correct :\n forall alpha beta:A,\n CompareSpec (alpha=beta) (lt alpha beta) (lt beta alpha)\n (compare alpha beta);\n }.\n\\end{Coqsrc}\n\nThe following definitions allow us to make implicit several guessable arguments.\n\\label{sect:on-lt-notation}\n\\label{sect:on-le-notation}\n\\begin{Coqsrc}\nDefinition on_t {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare} := A.\n\nDefinition ON_compare {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare} := compare.\n\nDefinition ON_lt {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare} := lt.\nInfix \"o<\" := ON_lt : ON_scope.\n\nDefinition ON_le {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare} :=\n clos_refl _ ON_lt.\n\nInfix \"o<=\" := ON_le : ON_scope.\n\\end{Coqsrc}\n\n\n\\begin{remark}\nThe infix notations \\texttt{o<} and \\texttt{o<=} were defined in order to make apparent the distinction between the various notation scopes that may co-exist in a same statement. So the infix \\texttt{<} and \\texttt{<=} are reserved to the natural numbers. In the mathematical formulas, we still use $<$ and $\\leq$ for comparing ordinals.\n\\end{remark}\n\n\n\\subsection{Ordinal Notations and Measures for Proving Termination}\n\\label{sect:measure-ON}\n\nThe following lemma (together with the type class mechnism) allows us to use simply measures towards an ordinal notation. It is just an application of the libraries \\texttt{Coq.Wellfounded.Inverse\\_Image}\nand \\texttt{Coq.Wellfounded.Inclusion}. \n\n\\begin{Coqsrc}\nDefinition measure_lt {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare}\n {B : Type} (m : B -> A) : relation B :=\n fun x y => on_lt (m x) (m y).\n \nLemma wf_measure {A:Type}(lt: relation A)\n {compare : A -> A -> comparison}\n {on : ON lt compare}\n {B : Type}\n (m : B -> A): well_founded (measure_lt m). \n\\end{Coqsrc}\n\nA simple example of application is given in Sect.~\\vref{sect:merge-example}.\n\n\n\\section{Example: The Ordinal \\texorpdfstring{$\\omega$}{omega}}\n\n\n\n\nThe simplest example of ordinal notation is built over the type \\texttt{nat} of \\coq's standard library. We have only to apply already proven lemmas about Peano numbers.\n\n\\vspace{4pt}\n\\noindent\\emph{From Library~\\href{../theories/html/hydras.OrdinalNotations/ON_Omega.html}{OrdinalNotations.ON\\_Omega}}\n\n\n\\begin{Coqsrc}\nGlobal Instance Omega : ON Peano.lt Nat.compare.\nProof.\n split.\n - apply Nat.lt_strorder.\n - apply Wf_nat.lt_wf.\n - apply Nat.compare_spec.\nQed.\n\nCompute ON_compare 6 9.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = Lt\n : comparison\n\\end{Coqanswer}\n\n\\section{Sum of Ordinal Notations}\n\nLet \\texttt{NA} and \\texttt{NB} be two ordinal notations, on the respective types \\texttt{A} and \\texttt{B}.\n\n We consider a new strict order\non the disjoint sum of the associated types, by putting all elements of \\texttt{A} before the elements of \\texttt{B} (thanks to Standard Library's relation operator \\texttt{le\\_AsB}).\n\n\n\\noindent\n\\emph{From Library~\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Relations.Relation_Operators.html}{Relations.Relation\\_Operators}}.\n\n\\begin{Coqanswer}\nInductive\nle_AsB (A B : Type) (leA : A -> A -> Prop) (leB : B -> B -> Prop)\n : A + B -> A + B -> Prop :=\n| le_aa : forall x y : A, leA x y -> le_AsB A B leA leB (inl x) (inl y)\n| le_ab : forall (x : A) (y : B), le_AsB A B leA leB (inl x) (inr y)\n| le_bb : forall x y : B, leB x y -> le_AsB A B leA leB (inr x) (inr y)\n\\end{Coqanswer}\n\n\n\\vspace{4pt}\n\\noindent\\emph{From Library~\\href{../theories/html/hydras.OrdinalNotations/ON_plus.html}{OrdinalNotations.ON\\_plus}}\n\n\n\\begin{Coqsrc}\nSection Defs.\n\n Context `(ltA: relation A)\n (compareA : A -> A -> comparison)\n (NA: ON ltA compareA).\n Context `(ltB: relation B)\n (compareB : B -> B -> comparison)\n (NB: ON ltB compareB).\n\n\nDefinition t := (A + B)%type.\nArguments inl {A B} _.\nArguments inr {A B} _.\n\nDefinition lt : relation t := le_AsB _ _ ltA ltB.\n\\end{Coqsrc}\n\nBefore building an instance of \\texttt{ON}, we have to define a comparison function.\n\n\n\\begin{Coqsrc}\nDefinition compare (alpha beta: t) : comparison :=\n match alpha, beta with\n inl _, inr _ => Lt\n | inl a, inl a' => compareA a a'\n | inr b, inr b' => compareB b b'\n | inr _, inl _ => Gt\n end.\n\nLemma compare_correct alpha beta :\n CompareSpec (alpha = beta) (lt alpha beta) (lt beta alpha)\n (compare alpha beta).\n\\end{Coqsrc}\n\nThe Lemma \\texttt{Wellfounded.Disjoint\\_Union.wf\\_disjoint\\_sum} of Standard Library\nhelps us to prove that our order \\texttt{lt} is well-founded.\n\n\n\\begin{Coqsrc}\nGlobal Instance ON_plus : ON lt compare.\nProof.\n split.\n - apply lt_strorder.\n - apply lt_wf.\n - apply compare_correct.\nQed.\n\\end{Coqsrc}\n\n\n\n\n\n\n\\subsection{The Ordinal \\texorpdfstring{$\\omega+\\omega$}{omega + omega}}\n\nThe ordinal $\\omega+\\omega$ (also known as $\\omega\\times 2$) may be represented as the concatenation \nof two copies of $\\omega$ (Figure~\\ref{fig:omega-plus-omega}).\n\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[very thick, scale=0.5]\n\\begin{scope}[color=blue]\n\\node(A0) at (2,0)[label=below:$0$]{$\\bullet$};\n\\node(A1) at (3,0)[label=below:$1$]{$\\bullet$};\n\\node(A2) at (4,0)[label=below:$2$]{$\\bullet$};\n\\node (Adots) at (6,0) {$\\ldots$};\n\\node(An) at (8,0)[label=below:$n$]{$\\bullet$};\n\\node(A2) at (10,0)[label=below:$n+1$]{$\\bullet$};\n\\node (Adots1) at (12,0) {$\\ldots$};\n\\end{scope}\n\\begin{scope}[color=red]\n\\node(B0) at (14,0)[label=below:$0$,label=above:\\textcolor{red}{$\\omega$}]{$\\bullet$};\n\\node(B1) at (16,0)[label=below:$1$, label=above:$\\omega+1$]{$\\bullet$};\n\\node(B2) at (18,0)[label=below:$2$,label=above:$\\omega+2$]{$\\bullet$};\n\\node (Bdots) at (20,0) {$\\ldots$};\n\\node (Bn) at (22,0) [label=below:$p$, label=above:$\\omega+p$]{$\\bullet$};\n\\node (Bdots2) at (24,0) {$\\ldots$};\n\\end{scope}\n\\end{tikzpicture}\n \\caption{\\textcolor{blue}{$\\omega+{\\color{red}\\omega}$}}\n \\label{fig:omega-plus-omega}\n \\end{figure}\n\nWe can define this notation in \\coq{} as an instance of \\texttt{ON\\_plus}.\n\n\n\\vspace{4pt}\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Omega_plus_omega.html}{OrdinalNotations.ON\\_Omega\\_plus\\_omega}}\n\n\\begin{Coqsrc}\nDefinition Omega_plus_Omega := ON_plus Omega Omega.\n\nExisting Instance Omega_plus_Omega.\nDefinition t := @ON_plus.t nat nat.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nExample ex1 : inl 7 o< inr 0.\nProof. constructor. Qed.\n\\end{Coqsrc}\n\nWe can now define abbreviations. For instance, the finite ordinals are represented by terms built with the constructor \\texttt{inl}, and the first infinite ordinal $\\omega$ by the term \\texttt{(inr 0)}.\n\n\\begin{Coqsrc}\nDefinition fin (i:nat) : t := inl i.\nCoercion fin : nat >-> t.\n\nNotation \"'omega'\" := (inr 0:t).\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nExample ex1' : fin 7 o< omega.\nProof. constructor. Qed.\n\nLemma lt_omega alpha : \n alpha o< omega <-> exists n:nat, alpha = fin n.\n(* ... *)\n\\end{Coqsrc}\n\n\n% \\label{warning:coercions}\n% \\index{Coq!Techniques!Coercions} \n% \\begin{remark}\n% Beware of coercions and notation scopes!\n% Let us consider the following goal:\n\n% \\begin{Coqsrc}\n% Goal (6 o< 8).\n% auto with arith.\n% \\end{Coqsrc}\n\n\n% \\begin{Coqanswer}\n% 1 subgoal (ID 9)\n \n% ============================\n% 6 o< 8\n% \\end{Coqanswer}\n\n% Please keep in mind that the current notation scope interprets the infix \\texttt{``<''} as the predicate \\texttt{Omega\\_plus\\_omega.lt} and not \\texttt{Nat.lt}. More, the coercion mechanism converts the terms \\texttt{6:nat} [resp. \\texttt{8:nat} ]\n% into \\texttt{inl 6} [resp. \\texttt{inl 8}]. So, the initial goal is correctly interpreted by \\coq{}, but not as an inequality between two natural numbers.\n\n\n% \\begin{Coqsrc}\n% Set Printing All.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% 1 subgoal (ID 337)\n \n% ============================\n% @on_lt nat Peano.lt Nat.compare Omega (S (S (S (S (S (S O))))))\n% (S (S (S (S (S (S (S (S O))))))))\n% \\end{Coqanswer}\n\n\n% Anyway, the initial goal is provable, using \\texttt{le\\_AsB}'s first constructor.\n\n% \\begin{Coqsrc}\n% constructor; auto with arith.\n% Qed.\n% \\end{Coqsrc}\n\n% \\end{remark}\n%\n\n\n\n\\section{Limits and Successors}\n\nLet us look again at our implementation of $\\omega+\\omega$. We can distinguish between the three kinds of ordinals seen in Fig~\\ref{fig:ordinal-sequence}:\n\n\\begin{itemize}\n\\item The least ordinal, \\texttt{(inl 0)}, also written \\texttt{(fin 0)}.\n\\item The limit ordinal $\\omega$.\n\\item The successor ordinals, either of the form \\texttt{(inl (S $i$))} or \\texttt{(inr (S $i$))}\n\\end{itemize}\n\n\\subsection{Definitions}\nIt would be interesting to specify at the most generic level, what is a zero, a successor or a limit ordinal. Let $<$ be a strict order on a type $A$.\n\n\\begin{itemize}\n\\item A \\emph{least} element is a minorant (in the large sense) of the full set on $A$,\n\\item $y$ is a \\emph{successor} of $x$ if $x z < y -> False).\n\nDefinition Limit {sto : StrictOrder lt} (x:A) :=\n (exists w:A, w < x) /\\\n (forall y:A, y < x -> exists z:A, y < z /\\ z < x).\n\\end{Coqsrc}\n\n\\index{Exercises}\n\\begin{exercise}\nProve, that, in any ordinal notation system, that every ordinal has at most one predecessor, and at most one successor. \n\\end{exercise}\n\n\\index{Exercises}\n\\begin{exercise}\nProve, that, in any ordinal notation system, that if $\\beta$ is a successor of $\\alpha$,\nthen for any $\\gamma$, $\\gamma<\\beta$ implies \n$\\gamma\\leq\\alpha$.\n\\end{exercise}\n\n\n\n\n\\subsection{Limits and Successors in \\texorpdfstring{$\\omega+\\omega$}{omega+omega}}\n\nUsing the definitions above, we can prove the following lemma:\n\n\\vspace{4pt}\n\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Omega_plus_omega.html}{OrdinalNotations.ON\\_Omega\\_plus\\_omega}}\n\n\\begin{Coqsrc}\nLemma limit_iff (alpha : t) : Limit alpha <-> alpha = omega.\n\\end{Coqsrc}\n\nRegarding successors, let us introduce the following definition:\n\n\\begin{Coqsrc}\nDefinition succ (alpha : t) :=\n match alpha with\n inl n => inl (S n)\n | inr n => inr (S n)\n end.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nLemma Successor_correct alpha beta : Successor beta alpha <->\n beta = succ alpha.\n\\end{Coqsrc}\n\nWe can also check whether an ordinal is a successor by a simple pattern matching:\n\n\\begin{Coqsrc}\nDefinition succb (alpha: t) : bool := match alpha with\n | inr (S _) | inl (S _) => true\n | _ => false\n end.\n\nLemma succb_correct (alpha : t) :\n succb alpha <-> exists beta: t, alpha = succ beta.\n\\end{Coqsrc}\n\n\nFinally, the nature of any ordinal is decidable(inside this notation system) :\n\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.Generic.html}{OrdinalNotations.Generic}}\n\\begin{Coqsrc}\n Definition ZeroLimitSucc_dec {A:Type}{lt: relation A}\n {compare : A -> A -> comparison}\n {on : ON lt compare} :=\n forall alpha,\n {Least alpha} +\n {Limit alpha} +\n {beta: A | Successor alpha beta}.\n(* ... *)\n\\end{Coqsrc}\n\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Omega_plus_omega.html}{OrdinalNotations.ON\\_Omega\\_plus\\_omega}}\n\n\\begin{Coqsrc}\nDefinition Zero_limit_succ_dec : ZeroLimitSucc_dec.\n\\end{Coqsrc}\n\n\\section{Product of Ordinal Notations}\n\nLet \\texttt{NA} and \\texttt{NB} be two ordinal notations, on the respective ordered types \\texttt{A} and \\texttt{B}. The product of \\texttt{NA} and \\texttt{NB} is considered as the concatenation of $B$ copies of $A$, ordered by the lexicographic order on $B\\times A$.\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_mult.html}{OrdinalNotations.ON\\_mult}}\n\n\\begin{Coqsrc}\nSection Defs.\n\n Context `(ltA: relation A)\n (compareA : A -> A -> comparison)\n (NA: ON ltA compareA).\n Context `(ltB: relation B)\n (compareB : B -> B -> comparison)\n (NB: ON ltB compareB).\n\nDefinition t := (B * A)%type.\nDefinition lt : relation t := lexico ltB ltA.\nDefinition le := clos_refl _ lt.\n\nDefinition compare (alpha beta: t) : comparison :=\n match compareB (fst alpha) (fst beta) with\n | Eq => compareA (snd alpha) (snd beta)\n | c => c\n end.\n\nLemma compare_reflect alpha beta :\n match (compare alpha beta)\n with\n Lt => lt alpha beta\n | Eq => alpha = beta\n | Gt => lt beta alpha\n end.\n\nGlobal Instance ON_mult : ON lt compare.\n\nEnd Defs.\n\\end{Coqsrc}\n\n\\section{The Ordinal \\texorpdfstring{$\\omega^2$}{omega^2}}\n\nThe ordinal $\\omega^2$ (also called $\\phi_0(2)$, see Chap.~\\ref{chap:schutte}), is an instance of the multiplication presented in the last section.\n\n\\vspace{4pt}\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Omega2.html}{OrdinalNotations.ON\\_Omega2}}\n\n\\begin{Coqsrc}\nDefinition Omega2 := ON_mult Omega Omega.\nExisting Instance Omega2.\nDefinition t := ON_mult.t nat nat.\n\\end{Coqsrc}\n\n\n\n\\begin{Coqsrc}\nDefinition zero: t := (0,0).\n\nDefinition fin (n:nat) : t := (0, n).\n\nCoercion fin : nat >-> t.\n\nNotation \"'omega'\" := (1,0) : ON_scope.\n\nDefinition limitb (alpha : t) := \n match alpha with\n | (S _, 0) => true\n | _ => false\nend.\n\\end{Coqsrc}\n\n\n\n\n\\subsection{Arithmetic of \\texorpdfstring{$\\omega^2$}{omega^2}} \n\n\\subsubsection{Successor}\n\nThe successor of any ordinal is defined by a simple pattern-matching.\nThis function is proved to be correct w.r.t. the \\texttt{Successor} predicate.\n\n\\begin{Coqsrc}\nDefinition succ (alpha : t) := (fst alpha, S (snd alpha)).\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma lt_succ_le alpha beta : alpha o< beta <-> succ alpha o<= beta.\n\nLemma lt_succ alpha : alpha o< succ alpha.\n\nLemma succ_ok alpha beta : Successor beta alpha <-> beta = succ alpha.\n\\end{Coqsrc}\n\n\\subsubsection{Addition}\n\nWe can define on \\texttt{Omega2} an addition which extends the addition on \\texttt{nat}.\n\n\\begin{Coqsrc}\nDefinition plus (alpha beta : t) : t :=\n match alpha,beta with\n | (0, b), (0, b') => (0, b + b')\n | (0,0), y => y\n | x, (0,0) => x\n | (0, b), (S n', b') => (S n', b')\n | (S n, b), (S n', b') => (S n + S n', b')\n | (S n, b), (0, b') => (S n, b + b')\n end.\n\nInfix \"+\" := plus : o2_scope.\n\\end{Coqsrc}\n\nPlease note that this operation is not commutative:\n\n\\begin{Coqsrc}\nExample non_commutativity_of_plus : omega + 3 <> 3 + omega.\nProof.\n cbn.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n1 subgoal (ID 237)\n \n ============================\n(1, 3) <> omega.\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\n discriminate.\nQed.\n\\end{Coqsrc}\n\n\n\\subsubsection{Multiplication}\n\nThe restriction of ordinal multiplication to the segment $[0,\\omega^2)$ is not a total function.\nFor instance $\\omega\\times\\omega= \\omega^2$ is outside the set of represented values.\nNevertheless, we can define two operations mixing natural numbers and ordinals.\n\n\\begin{Coqsrc}\n(** multiplication of an ordinal by a natural number *)\n\nDefinition mult_fin_r (alpha : t) (p : nat): t :=\n match alpha, p with\n | (0,0), _ => zero\n | _, 0 => zero\n | (0, n), p => (0, n * p)\n | ( n, b), n' => ( n * n', b)\n end.\nInfix \"*\" := mult_fin_r : o2_scope.\n\n(** multiplication of a natural number by an ordinal *)\n\nDefinition mult_fin_l (n:nat)(alpha : t) : t :=\n match n, alpha with\n | 0, _ => zero\n | _, (0,0) => zero\n | n , (0,n') => (0, (n*n')%nat)\n | n, (n',p') => (n', (n * p')%nat)\n end.\n\nExample e1 : (omega * 7 + 15) * 3 = omega * 21 + 15.\nProof. reflexivity. Qed.\n\nExample e2 : mult_fin_l 3 (omega * 7 + 15) = omega * 7 + 45.\nProof. reflexivity. Qed.\n\\end{Coqsrc}\n\nMultiplication with a finite ordinal and addition are related through the following lemma:\n\n\\begin{Coqsrc}\nLemma unique_decomposition alpha : \n exists! i j: nat, alpha = omega * i + j.\n\\end{Coqsrc}\n\n\\subsection{A Proof of Termination using \\texorpdfstring{$\\omega^2$}{omega^2}} \n\\label{sect:merge-example}.\n\nUsing the lemma of Sect.~\\vref{sect:measure-ON}, we can define easily a total function which merges two lists.\n\n\\index{Coq!Commands!Function}\n\n\\begin{Coqsrc}\n(* adapted from Pascal Manoury et al. *)\n\n\nRequire Import Coq.Program.Wf List.\nRequire Import FunInd Recdef.\n\nSection Merge.\n\n Variable A: Type.\n\n Local Definition m (p : list A * list A) :=\n omega * length (fst p) + length (snd p).\n\n Function merge (ltb: A -> A -> bool)\n (xys: list A * list A)\n {wf (measure_lt m) xys} :\n list A :=\n match xys with\n (nil, ys) => ys\n | (xs, nil) => xs\n | (x :: xs, y :: ys) =>\n if ltb x y then x :: merge ltb (xs, (y :: ys))\n else y :: merge ltb ((x :: xs), ys)\n end.\n\n - intros; unfold m, measure_lt; cbn; destruct xs0; simpl; left; lia.\n - intros; unfold m, measure_lt; cbn; destruct ys0; simpl; right; lia.\n - auto.\n Defined.\n\nEnd Merge.\n\nGoal forall l, merge nat Nat.leb (nil, l) = l.\n intro; now rewrite merge_equation.\nQed.\n\\end{Coqsrc}\n \n\n\\subsection{Yet Another Proof of Impossibility}\n\\label{omega2-case}\n\nIn Sect.~\\vref{omega-case}, we proved that there exists no variant towards \\texttt{nat}\n(\\emph{i.e.} the ordinal $\\omega$) for proving the termination of all hydra battles.\nWe prove now that the ordinal $\\omega^2$ is also insufficient for this purpose. \n\nThe proof we are going to develop has exactly the same structure as in Section~\\ref{omega-case}.\n Nevertheless, the proof of technical lemmas is a little more complex, due to \n the structure of the lexicographic order on $\\mathbb{N}\\times\\mathbb{N}$. \nConsider for instance that there exists an infinite number of ordinals between\n$\\omega$ and $\\omega\\times 2$.\n\n\n\nThe detailed proof script is in the file \\url{../theories/html/hydras.Hydra.Omega2_Small.html}.\n\n\\subsubsection{Preliminaries}\nLet us assume there is a variant from \\texttt{Hydra} into $\\omega^2$ for proving the termination of all hydra battles.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Omega2_Small.html}{Hydra.Omega2\\_Small}}\n\n\n\\begin{Coqsrc}\nSection Impossibility_Proof.\n \n Variable m : Hydra -> ON_Omega2.t.\n \n Context (Hvar : Hvariant (Generic.wf (ON:=Omega2)) free m).\n\\end{Coqsrc}\n\n\nLet us follow the same pattern as in Sect.~\\ref{omega-case}.\nFirst, we define an injection $\\iota$ from type \\texttt{t} into \\texttt{Hydra}, by\n associating to each ordinal $\\omega\\times i+ j = (i,j)$ the hydra with $i$ branches of length $2$ and\n$j$ branches of length $1$.\n\n%% revenir ici\n\n\\vspace{4pt}\n\\emph{From Module ~\\href{../theories/html/hydras.Hydra.Omega2_Small.html\\#iota}{Hydra.Omega2\\_Small}}\n\n\\begin{Coqsrc}\n Let iota (p: ON_Omega2.t) := \n node (hcons_mult (hyd1 head) (fst p)\n (hcons_mult head (snd p) hnil)).\n \\end{Coqsrc}\n\nFor instance, Figure~\\vref{fig:essai2} shows the hydra associated to the ordinal \n$(3,5)$, a.k.a. $\\omega\\times 3 + 5$.\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.4]\n\\node (foot) at (6,0) {$\\bullet$};\n\\node (N1) at (1,3) {$\\bullet$};\n\\node (N2) at (3,3) {$\\bullet$};\n\\node (N3) at (5,3) {$\\bullet$};\n\\node (N4) at (8,3) {$\\Smiley[2][green]$};\n\\node (N5) at (11,3) {$\\Smiley[2][green]$};\n\\node (N6) at (14,3) {$\\Smiley[2][green]$};\n\\node (N7) at (17,3){$\\Smiley[2][green]$};\n\\node (N8) at (20,3){$\\Smiley[2][green]$};\n\\node (N9) at (0,5) {$\\Smiley[2][green]$};\n\\node (N10) at (2,5) {$\\Smiley[2][green]$};\n\\node (N11) at (4,5) {$\\Smiley[2][green]$};\n\\draw (foot) to [bend left=10] (N1);\n\\draw (foot) -- (N2);\n\\draw (foot) -- (N3);\n\\draw (foot) -- (N4);\n\\draw (foot) -- (N5);\n\\draw (foot) -- (N6);\n\\draw (foot) to [bend right=10] (N7);\n\\draw (foot) to [bend right=15] (N8);\n\\draw (N1) to [bend left=10] (N9);\n\\draw (N2) -- (N10);\n\\draw (N3) -- (N11);\n\\end{tikzpicture}\n\\caption{\\label{fig:essai2}\nThe hydra $\\iota(\\omega\\times 3+5)$}\n\\end{figure}\n\n\n\n\nLike in Sect.~\\ref{omega-case}, we build a hydra out of the range of \\texttt{iota} (represented in Fig.~\\vref{fig:h-omega2-small}).\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5]\n\\node (foot) at (2,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (3,4) {$\\Smiley[2][green]$};\n\\node (N3) at (1,4) {$\\Smiley[2][green]$};\n\\draw (foot) -- (N1);\n\\draw (N1) to [bend right =15] (N2);\n\\draw (N1) to [bend left=15](N3);\n\\end{tikzpicture}\n\\caption{\\label{fig:h-omega2-small}}\n The hydra \\texttt{big\\_h}.\n\\end{figure}\n\n\n\\begin{Coqsrc}\n Let big_h := hyd1 (hyd2 head head). \n\\end{Coqsrc}\n \n In a second step, we build a ``smaller'' hydra.\n \n\\begin{Coqsrc}\n Let small_h := iota (m big_h).\n\\end{Coqsrc}\n\nLike in Sect.~\\ref{omega-case}, we prove the double inequality \\texttt{m big\\_h o<= m small\\_h o< m big\\_h}, which is impossible.\n\n\\subsubsection{Proof of the Inequality \\texttt{m small\\_h o< m big\\_h}}\n\nIn order to prove the inequality \\texttt{m\\_lt: m small\\_h o< m big\\_h}, it suffices to\nbuild a battle transforming \\texttt{big\\_h} into \\texttt{small\\_h}.\n\nFirst we prove that \\texttt{small\\_h} is reachable from \\texttt{big\\_h} in one or two steps. Let us decompose \\texttt{m big\\_h} as $(i,j)$.\nIf $j=0$, then one round suffices to transform \\texttt{big\\_h} into $\\iota(i,j)$.\nIf $j>0$, then a first round transforms \\texttt{big\\_h} into $\\iota(i+1,0)$ and a second round into $\\iota(i,j)$. So, we have the following result.\n\n\\begin{Coqsrc}\n Lemma big_to_small: big_h -+-> small_h.\n\\end{Coqsrc}\n\nSince $m$ is a variant, we infer the following inequality:\n\n\\begin{Coqsrc}\n Corollary m_lt : m small_h o< m big_h.\n\\end{Coqsrc}\n\n\n\\subsubsection{Proof of the Inequality \\texttt{m big\\_h o<= m small\\_h} }\n\n\nThe proof of the inequality \\texttt{m big\\_h o<= m small\\_h} is quite more complex than in Sect~\\ref{omega-case}. If we consider any ordinal $\\alpha=(i,j)$, where $i>0$, there exists an infinite number of\nordinals stricly less than $\\alpha$, and there exists an infinite number of battles that start from\n$\\iota(\\alpha)$. Indeed, at any configuration $\\iota(k,0)$, where $k>0$, the hydra can freely choose any replication number. Intuitively, the measure of such a hydra must be large enough for taking into account\nall the possible battles issued from that hydra.\nLet us now give more technical details.\n\n\\begin{itemize}\n\\item The proof of the lemma \\texttt{m\\_ge : m big\\_h o<= m small\\_h} uses well-founded induction on \\texttt{big\\_h}.\n\n\\item For any pair $p$, we have to distinguish between three cases, according to the value of $p$'s components.\n \\begin{itemize}\n \\item $p=(0,0)$\n \\item $p=(i,0)$, where $i>0$\\,: $p$ corresponds to a limit ordinal\n \\item $p=(i,j)$, where $j>0$\\,: $p$ is the successor of $(i,j-1)$.\n \\end{itemize}\n\\end{itemize}\n\n\nLet us define the notion of elementary ``step'' of decreasing sequences in\n\\texttt{t}\n\n\n\\begin{Coqsrc}\n Inductive step : t -> t -> Prop :=\n | succ_step : forall i j, step (i, S j) (i, j)\n | limit_step : forall i j, step (S i, 0) (i, j).\n\\end{Coqsrc}\n\nThe following lemma establishes a correspondance between the relation\n\\texttt{step} and hydra battles.\n\n\\begin{Coqsrc}\n Lemma step_to_battle : forall p q, step p q -> iota p -+-> iota q.\n\\end{Coqsrc}\n\n\\index{Maths!Transfinite induction}\n\nThus, starting from any inequality $q < p$ on type \\texttt{t}, we can build \nby \\emph{transfinite induction} (\\emph{i.e.} well-founded) over \\texttt{p} a battle \nthat transforms the hydra $\\iota(p)$ into $\\iota(q)$.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Omega2_Small.html\\#m_ge}{Hydra.Omega2\\_Small}}\n\n\\begin{Coqsrc}\n Lemma m_ge : forall p : t, p o<= m (iota p).\n Proof.\n unfold small_h; pattern (m big_h) . \n apply well_founded_induction with (R := lt) (1:= lt_wf).\n intro p ; pattern p;\n apply well_founded_induction with \n (R := lt2) (1:= wf_lexico lt_wf lt_wf);\n intros (i,j) IHij. \n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n i, j : nat\n IHij : forall y : t, y o< (i, j) -> y o<= m (iota y)\n ============================\n (i, j) o<= m (iota (i, j)) \n\\end{Coqanswer}\n\n\nThen we have three cases to consider, according to the value of $p$.\n\\begin{itemize}\n\\item If $p=(0,0)$ then obviously, $\\iota(p)\\geq p = (0,0)$\n\\item If $p=(i+1,0)$ for some $i\\in\\mathbb{N}$, we\n remark that $p$ is strictly greater than any pair $ (i, j)$, where $j$ \nis any natural number.\n\nApplying the battle rules, for any $j$, we have $\\iota(i+1,j) {\\round} \\iota(i, j) $, thus $m(\\iota(p)) > m(\\iota(i,j)$ since $m$ is assumed to be a variant.\n\nApplying the induction hypothesis, we get the inequality\n $ m(\\iota(i,j)) \\geq (i,j)$ for any $j$. \n\nThus, $m(\\iota(p)) > (i,j)$ for any $j$.\nApplying the lemma \\texttt{limit\\_is\\_lub}, we get the inequality\n$m(\\iota(i+1,0))\\geq (i+1,0)$\n\n\\item If $p=(i,j+1)$ with $j\\in\\mathbb{N}$, we have $\\iota(p) {\\round} \\iota(i, j) $,\nhence $m(\\iota(p))> m(\\iota(i,j)) \\geq (i,j)$, thus $m(\\iota(p))\\geq (i,j+1)=p$\n\n\\end{itemize}\n\n\\begin{Coqsrc}\n (* ... *)\nQed.\n\\end{Coqsrc}\n\n\\subsubsection{End of the Proof}\nFrom \\texttt{m\\_ge}, we get \\texttt{m big\\_h o<= m small\\_h = m (iota (m big\\_h)) }. \nSince $<$ is a strict order (irreflexive and transitive), this inequality is incompatible with the strict inequality \\texttt{m small\\_h o< m big\\_h} (lemma \\texttt{m\\_lt}).\n\n\n\\vspace{4pt}\n\\noindent\nIn \\coq (Module~\\href{../theories/html/hydras.Hydra.Omega2_Small.html\\#Impossible}{Hydra.Omega2\\_Small}):\n\n\\begin{Coqsrc}\n Theorem Impossible: False.\n Proof.\n destruct (StrictOrder_Irreflexive (m big_h)).\n apply le2_lt2_trans with (m small_h).\n - unfold small_h; apply m_ge.\n - apply m_lt. \nQed. \nEnd Impossibility_Proof.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\nProve that there exists no variant $m$ from \\texttt{Hydra} into $\\omega^2$ for proving\n the termination of all \\emph{standard} battles.\n\\end{exercise}\n\n\n\n\\begin{remark}\nIn Chapter~\\ref{ks-chapter}, we prove a generalization of the impossibility lemmas of\nSect.~\\ref{omega-case} and this section, with the same proof structure, but with much more \ncomplex technical details.\n \\end{remark}\n\n% \\index{Exercises}\n% \\begin{exercise}\n\n% \\label{sec:orgheadline63}\n% Write \\emph{direct} proofs ({i.e.}, without applying the result and tools of Chap.~\\ref{ks-chapter}) that the following data structures are too simple for defining a variant for any hydra battle.\n\n% \\begin{itemize}\n% \\item $\\omega^n$ : the set of all $n$-uples of natural numbers, ordered by \n% lexicographic ordering\n% \\item $\\omega^\\omega$: the set of all decreasing sequences (with respect to $\\le$) of natural numbers, ordered by lexicographic ordering on lists.\n\n% For instance, the following inequality holds:\n% \\[\\langle 4,3,3,3,3,3,3,2,2,2 \\rangle\\,<\\,\\langle 4,4,2 \\rangle\\]\n% \\end{itemize}\n\n \n% \\end{exercise}\n\n\n\\section{A Notation for Finite Ordinals}\n\n\nLet $n$ be some natural number. The segment associated with $n$ is the interval \n$[0,n)\\,=\\,\\{0,1,\\dots,n-1\\}$. \nOne may represent the ordinal $n$ by a sigma type.\n\n\n\\vspace{4pt}\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Finite.html}{OrdinalNotations.ON\\_Finite}}\n\n\\label{def: Finite-ord-type}\n\\begin{Coqsrc}\nCoercion is_true: bool >-> Sortclass.\n\nDefinition t (n:nat) := {i:nat | Nat.ltb i n}.\n\\end{Coqsrc}\n\nThe order on type (\\texttt{t $n$}) is defined through the projection on \\texttt{nat}.\n\n\n\\begin{Coqsrc}\nDefinition lt {n:nat} : relation (t n) :=\n fun alpha beta => Nat.ltb ( proj1_sig alpha) (proj1_sig beta).\n\\end{Coqsrc}\n\nFor instance, let us build two elements of the segment $[0, 7)$, \\emph{i.e.} two\ninhabitants of type (\\texttt{t 7}), and prove a simple inequality (see Fig.~\\ref{fig:O7}).\n\n\\begin{figure}[h]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.6]\n\n\\node (N0) at (0,0) {$\\bullet$};\n\\node (i0) at (0,1) {$0$};\n\\node (N1) at (2,0) {$\\bullet$};\n\\node (i1) at (2,1) {$1$};\n\\node (N2) at (4,0) {$\\bullet$};\n\\node (i2) at (4,1) {$2$};\n\\node (N3) at (6,0) {$\\bullet$};\n\\node (i3) at (6,1) {$3$};\n\\node (N4) at (8,0) {$\\bullet$};\n\\node (i4) at (8,1) {$4$};\n\\node (N5) at (10,0) {$\\bullet$};\n\\node (i5) at (10,1) {$5$};\n\\node (N6) at (12,0) {$\\bullet$};\n\\node (i6) at (12,1) {$6$};\n\\node(alpha1) at (4,-1) {$\\alpha_1$};\n\\node(alpha2) at (10,-1) {$\\beta_1$};\n\\end{tikzpicture}\n\n\\caption{The segment $\\mathbb{O}_7$\\label{fig:O7}}\n\\end{figure}\n \n\\index{Coq!Commands!Program}\n\n\\begin{Coqsrc}\nProgram Example alpha1 : t 7 := 2.\n\nProgram Example beta1 : t 7 := 5.\n\nExample i1 : lt alpha1 beta1.\nProof. now compute. Qed.\n\\end{Coqsrc}\n\n\n\n\nNote that the type (\\texttt{t 0}) is empty, and that, for any natural number\n $n$, $n$ does not belong to (\\texttt{t $n$}).\n\n\\begin{Coqsrc}\nLemma t0_empty (alpha: t 0): False.\nProof.\n destruct alpha.\n destruct x; cbn in i; discriminate.\nQed.\n\n\nProgram Definition bad : t 10 := 10.\nNext Obligation.\n compute.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n1 subgoal (ID 162)\n \n ============================\n false = true\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nAbort.\n\\end{Coqsrc}\n\nNote also that attempting to compare a term of type (\\texttt{t $n$}) with a term of\ntype (\\texttt{t $p$}) leads to an error if $n$ and $p$ are not convertible.\n\n\\begin{Coqsrc}\n\nProgram Example gamma1 : t 8 := 7.\n\nFail Goal lt alpha1 gamma1.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n The command has indeed failed with message:\nThe term \"gamma1\" has type \"t 8\" while it is expected to have type \"t 7\".\n\\end{Coqanswer}\n\n\nIn order to build an instance of \\texttt{OrdinalNotation}, we define a comparison function, by delegation to standard library's \\texttt{Nat.compare}, and prove its correction.\n\n\\begin{Coqsrc}\nDefinition compare {n:nat} (alpha beta : t n) :=\n Nat.compare (proj1_sig alpha) (proj1_sig beta).\n\nLemma compare_correct {n} (alpha beta : t n) :\n CompareSpec (alpha = beta) (lt alpha beta) (lt beta alpha)\n (compare alpha beta).\n\\end{Coqsrc}\n\n\\begin{remark}\n The proof of \\texttt{compare\\_correct} uses a well-known pattern of \\coq{}.\nLet us consider the following subgoal.\n\n\\begin{Coqanswer}\n 1 subgoal (ID 110)\n \n n, x0 : nat\n i, i0 : x0 i1 <=? n) x0 i =\n exist (fun i1 : nat => i1 <=? n) x0 i0\n\\end{Coqanswer}\n\nApplying the tactic \\texttt{f\\_equal} generates a simpler subgoal.\n\n\\begin{Coqanswer}\n1 subgoal (ID 112)\n \n n, x0 : nat\n i, i0 : x0 y) -> \nforall (y : A) (p1 p2 : x = y), p1 = p2\n\\end{Coqanswer}\n\nThus unicity of proofs of \\texttt{Nat.ltb x0 (S n)} comes from the decidability of\nequality on type \\texttt{bool}.\nThis is why we used the boolean function \\texttt{Nat.ltb} instead of the inductive predicate \\texttt{Nat.lt} in the definition of type \\texttt{t $n$} (see page~\\pageref{def: Finite-ord-type}).\nFor more information about this pattern, please look at the numerous mailing lists and \nFAQs on \\coq{}).\n\n\n\n\\end{remark}\n\n\nApplying lemmas of the libraries \\texttt{Coq.Wellfounded.Inverse\\_Image}, \\linebreak\n \\texttt{Coq.Wellfounded.Inclusion}, and \\texttt{Coq.Arith.Wf\\_nat}, we prove that our\nrelation \\texttt{lt} is well founded.\n\n\\begin{Coqsrc}\nLemma lt_wf (n:nat) : well_founded (@lt n).\n\\end{Coqsrc}\n\nNow we can build our instance of \\texttt{OrdinalNotation}.\n\n\\begin{Coqsrc}\nGlobal Instance sto n : StrictOrder (@lt n).\n\nGlobal Instance FinOrd (n:nat) : OrdinalNotation (sto n) compare.\nProof.\n split.\n - apply compare_correct.\n - apply lt_wf.\nQed.\n\\end{Coqsrc}\n\n\\begin{remark}\nIt is important to keep in mind that the integer $n$ is not an ``element'' of \\texttt{FinOrd $n$}. In set-theoretic presentations of ordinals, the set associated with the ordinal $n$ is $\\{0,1,\\dots,n-1\\}$. \nIn our formalization, the interpretation of an ordinal as a set is realized by the following definition\n(in ~\\href{../theories/html/hydras.OrdinalNotations.Generic.html}{OrdinalNotations.Generic}).\n\n\\begin{Coqsrc}\nDefinition bigO `{nA : @OrdinalNotation A ltA stoA compareA}\n (a: A) : Ensemble A :=\n fun x: A => ltA x a.\n\\end{Coqsrc}\n\\end{remark}\n\n\n\\begin{remark}\n There is no interesting arihmetic on finite ordinals, since functions like successor, addition, etc., cannot be represented in \\coq{} as \\emph{total} functions.\n\\end{remark}\n\n\\begin{remark}\nFinite ordinals are also formalized in MathComp~\\cite{SSR}. See also Adam Chlipala's \\emph{CPDT}~\\cite{chlipalacpdt2011} for a thorough study of the use of dependent types. \n\\end{remark}\n\n\n\n%%%\n\n\\section{Comparing two Ordinal Notations}\n\nIt is sometimes useful to compare two ordinal notations with respect to expressive power\n(the segment of ordinals they represent). \n\nThe following class specifies a strict inclusion of segments. The notation \\texttt{OA} describes a segment $[0,\\alpha($, and \\texttt{OB} is a larger segment (which contains a notation for $\\alpha$, whilst $\\alpha$ is not represented in \\texttt{OA}). We require also that the comparison functions of the two notation systems are compatible.\n\nIf \\texttt{OB} is presumed to be correct, the we can consider that \\texttt{OA} ``inherits'' its correcteness from the bigger notation system \\texttt{OB}.\n\n\n\\label{types:SubON}\n\n\\begin{Coqsrc}\nClass SubON \n `(OA : @ON A ltA compareA)\n `(OB : @ON B ltB compareB)\n (alpha : B)\n (iota : A -> B):=\n {\n SubON_compare: forall x y : A, compareB (iota x) (iota y) =\n compareA x y;\n SubON_incl : forall x, ltB (iota x) alpha;\n SubON_onto : forall y, ltB y alpha -> exists x:A, iota x = y}.\n\\end{Coqsrc}\n\nFor instance, we prove that \\texttt{Omega} is a sub-notation of\n\\texttt{Omega\\_plus\\_Omega} (with $\\omega$ as the first ``new'' ordinal, and \\texttt{fin} as the injection).\n\n\\begin{Coqsrc}\nInstance Incl : SubON Omega Omega_plus_Omega omega fin.\n\\end{Coqsrc}\n\n\n\nWe can also show that, if $i,thin] (A0) -- node [auto] {$\\iota$} (B0);\n\\draw [->,thin] (A1) -- node [auto] {$\\iota$} (B1);\n\\draw [->,thin] (A2) -- node [auto] {$\\iota$} (B2);\n\\draw [->,thin] (Adots) -- node [auto] {$\\iota$} (Bdots);\n\\end{scope}\n\\end{tikzpicture}\n \\caption{\\textcolor{blue}{$A$} is a sub-segment of \\textcolor{red}{$B$}}\n \\label{fig:subsegment}\n \\end{figure}\n\n\n\n\n\\index{Coq!Commands!Program}\n\nWe are now able to build an instance of \\texttt{SubON}. \n\n\\vspace{4pt}\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.OrdinalNotations.ON_Finite.html}{OrdinalNotations.ON\\_Finite}}\n\n\\begin{Coqsrc}\nSection Inclusion_ij.\n\n Variables i j : nat.\n Hypothesis Hij : (i < j)%nat.\n\n Remark Ltb_ij : Nat.ltb i j.\n\n Program Definition iota_ij (alpha: t i) : t j := alpha.\n \n Let b : t j := exist _ i Ltb_ij.\n \n Global Instance F_incl_ij : SubON (FinOrd i) (FinOrd j) b iota_ij.\n (* ... *)\n\n End Inclusion_ij.\n\\end{Coqsrc}\n \n\n\n\\index{Exercises}\n\\begin{exercise}\nProve that \\texttt{Omega\\_plus\\_Omega} cannot be a sub-notation of \\texttt{Omega}.\n\\end{exercise}\n\n\\index{Projects}\n\\begin{project}\nAdapt the definition of \\texttt{Hvariant} (Sect.~\\ref{sect:hvariant-def}) in order to\nhave an ordinal notation as argument. Prove that if $O_A$ is a sub-notation of $O_B$, then any variant defined on $O_A$ can be automatically transformed into \na variant on $O_B$.\n\\end{project}\n\n\n\n\n\\section{Comparing an Ordinal Notation with Schütte's Model}\n\nFinally, it may be interesting to compare an ordinal notation with the more theoretical model from Schütte (well, at least with our formalization of that model). This would be a relative proof of correctenss of the considered ordinal notation.\n\nThe following class specifies that a notation \\texttt{OA} describes a segment $[0,\\alpha)$,\nwhere $\\alpha$ is a countable ordinal \\emph{à la} Schütte.\n\n\n\\label{types:ON-for}\n\n\\begin{Coqsrc}\nClass ON_correct `(alpha : Schutte_basics.Ord)\n `(OA : @ON A ltA compareA)\n (iota : A -> Schutte_basics.Ord) :=\n { ON_correct_inj : forall a, Schutte_basics.lt (iota a) alpha;\n ON_correct_onto : forall beta, Schutte_basics.lt beta alpha ->\n exists b, iota b = beta;\n On_compare_spec : forall a b:A,\n match compareA a b with\n Datatypes.Lt => Schutte_basics.lt (iota a) (iota b)\n | Datatypes.Eq => iota a = iota b\n | Datatypes.Gt => Schutte_basics.lt (iota b) (iota a)\n end}.\n\\end{Coqsrc}\n\n\n\nFor instance, the following theorem tells that \\texttt{Epsilon0}, our notation system for the segment $[0,\\epsilon0)$ is a correct implementation of the theoretically defined ordinal $\\epsilon_0$\n(see chapter~\\ref{chap:schutte} for more details).\n\n\n\\begin{Coqsrc}\nInstance Epsilon0_correct :\n ON_correct epsilon0 Epsilon0 (fun alpha => inject (cnf alpha)).\n\\end{Coqsrc}\n\n\\index{Projects}\n\n\\begin{project}\n When you have read Chapter~\\ref{chap:schutte}, prove that the sum of two ordinal notations \\texttt{ON\\_plus} implements the addition of ordinals.\n\\end{project}\n\n\n\n\n\n\\section{Isomorphism of Ordinal Notations}\n\n\nIn some cases we want to show that two notation systems describe the same segment (for instance $[0,3+\\omega)$ and $[0,\\omega)$\\;). For this purpose, one may prove that the two notation systems are order-isomorphic.\n\n\n\\label{types:ON-iso} \n\\begin{Coqsrc}\nClass ON_Iso \n `(OA : @ON A ltA compareA)\n `(OB : @ON B ltB compareB)\n (f : A -> B)\n (g : B -> A):=\n {\n iso_compare: forall x y : A, \n compareB (f x) (f y) = compareA x y;\n iso_inv1 : forall a, g (f a)= a;\n iso_inv2 : forall b, f (g b) = b}.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\nLet $i$ be some natural number. Prove that the notation systems \n\\texttt{Omega} and (\\texttt{ON\\_plus (OrdFin $i$) Omega}) are isomorphic.\n\n{\\it \\textbf{Note:} This property reflects the equality $i+\\omega=\\omega$, that we prove also in larger notation systems, as well as in Schütte's model.}\n\nThis exercise is partially solved for $i=3$ (in ~\\href{../theories/html/hydras.OrdinalNotations.Example_3PlusOmega.html}{OrdinalNotations.Example\\_3PlusOmega}).\n\n\\end{exercise}\n\n\\index{Projects}\n\\label{exo:ON-mult}\n\\begin{project}\n% Define in \\coq{} the product of two ordinal notations $N_A$ and $N_B$.\n% If $A$ [resp. $B$] is the underlying type of $N_A$ [resp. $N_B$], the\n% product \\texttt{ON\\_mult $N_A$ $N_B$} is implemented over the cartesian product $B\\times A$ (with the lexicographic ordering).\n\nThis exercise is about the non-commutativity of the multiplication of ordinals, reflected in ordinal notations.\n\nFor instance, the\nelements of the product (\\texttt{ON\\_mult Omega (FinOrd 3)}) are ordered as follows.\n\\[(0,0),(0,1),(0,2),(0,3),(0,4),\\dots,{\\color{red}(1,0),} (1,1),(1,2),\\dots, {\\color{red}(2,0)},(2,1),(2,2),\\dots\\]\n\nNote that the elements of (\\texttt{ON\\_mult (FinOrd 3) Omega}) are differently ordered (without limit ordinals):\n\\[(0,0),(1,0),(2,0),(0,1),(1,1),(2,1),(0,2),(1,2),(2,2),(0,3),\\dots\\]\n\n\nProve formally that \\texttt{ON\\_mult (FinOrd $i$) Omega} is isomorphic to\n\\texttt{Omega} whilst\n\\texttt{Omega} is a sub-notation of \\texttt{ON\\_mult Omega (FinOrd $i$)},\nfor any strictly positive $i$.\n\\end{project}\n\n\\index{Projects}\n\\begin{project}\nConsider two isomorphic ordinal notations \\texttt{OA} and \\texttt{OB}.\nProve that, if \\texttt{OA} [resp. \\texttt{OB}] is a correct implementation \nof $\\alpha$ [resp. $\\beta$], then $\\alpha=\\beta$.\n\\end{project}\n\n\n\\index{Projects}\n\\begin{project}\n\\label{project:succ-limit-dec}\nAdd to the class \\texttt{ON} the requirement that for any $\\alpha$ it is decidable whether $\\alpha$ is $0$, a successor or a limit ordinal.\n\n\n\\textbf{Hint:} Beware of the instances associated with sum and product of notations!\n You may consider additional fields \nto make the sum and product of notations ``compositional''.\n\n\\end{project}\n\n\\index{Projects}\n\\begin{project}\n\\label{project:on-setoid}\nReconsider the class \\texttt{ON}, with an equivalence instead of Leibniz equality.\n\\end{project}\n\n\n\n\n\n%%%% ICI ICI\n\n\\section{Other Ordinal Notations}\n\n\\index{Projects}\n\n\\begin{project}\nThe directory \\texttt{theories/OmegaOmega} contains an ad-hoc formalization of $\\omega^\\omega$, contributed by Pascal Manoury. Every ordinal $\\alpha$ is represented by a list $l$ whose elements are the coefficients of $\\omega$ in the Cantor normal form of $\\alpha$ (in reverse order). For instance, the ordinal \n$\\omega^{8}\\times 5 + \\omega^{6}\\times 8 + \\omega^2\\times 10 + \\omega + 7$ is represented by the list \\texttt{[5;0;8;0;0.0;10,1,7]}. \n\n\n Develop this representation and compare it with the other ordinal notations.\n\n\n\n\\end{project}\n\n\\index{Projects}\n\n\\begin{project}\nLet $N_A$ be a notation system for ordinals strictly less than $\\alpha$, \nwith the strict order $(A,<_A)$. Please build the notation system\n\\texttt{ON\\_Expl $N_A$}, on the type of multisets of elements of $A$\n(or, if preferred, the type of non-increasing finite sequences on $A$,\nprovided with the lexicographic ordering on lists).\n\nFor instance, let us take $N_A=\\texttt{Omega}$, and take $\\alpha=\\langle 4,4,2,1,0\\rangle$,\n $\\beta=\\langle 4,3,3,3,3,3,2\\rangle$, and $\\gamma=\\langle 5\\rangle$. Then $\\beta<\\alpha<\\gamma$. \n\nIn contrast the list $\\langle5,6,3,3\\rangle$ is not non-increasing (\\emph{i.e.} sorted w.r.t. $\\geq$), so it is not to be considered.\n\nNote that if the notation $N_A$ implements the ordinal \n$\\alpha$, the new notation $\\omega^{N_A}$ must implement the ordinal $\\phi_0(\\alpha)$, a.k.a. $\\omega^\\alpha$ (see chapter~\\ref{chap:schutte})\n\n\\end{project}\n\n\n\n\\begin{remark}\n The set of ordinal terms in Cantor normal form (see Chap.~\\ref{chap:T1}) and \nin Veblen normal form (see \n\\href{../theories/html/hydras.Gamma0.Gamma0.html}{Gamma0.Gamma0}) are shown to be ordinal notation systems, but there is a lot of work to be done in order to unify ad-hoc definitions and proofs which were written before the definition of the \\texttt{ON} type class.\n\\end{remark}\n\n\n\n\n\n\n\n\n\n\n%------------------------------------------------------------------------\n\n\\chapter[A proof of termination, using epsilon0]{A Proof of Termination, using Ordinals below \\texorpdfstring{$\\epsilon_0$}{Epsilon0}}\n\n\\label{cnf-math-def}\n\\label{chap:T1}\n\nIn this chapter, we adapt to \\coq{} the well-known~\\cite{KP82} proof that Hercules eventually wins every battle, whichever the strategy of each player.\nIn other words, we present a formal and self contained proof of termination of all [free] hydra battles.\nFirst, we take from Manolios and Vroon~\\cite{Manolios2005} a representation of the ordinal $\\epsilon_0$ as terms in Cantor normal form. Then, we define a variant for hydra battles as a measure that maps any hydra to some ordinal strictly less than $\\epsilon_0$.\n\n\n\n\\section{The Ordinal \\texorpdfstring{\\(\\epsilon_0\\)}{epsilon0}}\n\\label{sec:epsilon0-intro}\n\n\\subsection{Cantor Normal Form}\n\\index{Maths!Cantor normal form}\n\nThe ordinal \\(\\epsilon_0\\) is the least ordinal number that satisfies \nthe equation \\(\\alpha = \\omega^\\alpha\\), where \\(\\omega\\) is \nthe least infinite ordinal. Thus, we can consider \\(\\epsilon_0\\) as an\n\\emph{infinite} \\(\\omega\\)-tower.\nNevertheless, \nany ordinal strictly less that \\(\\epsilon_0\\) \ncan be finitely represented by a unique \\emph{Cantor normal form}, \nthat is, an expression which is either the ordinal \\(0\\) or \na sum \\(\\omega^{\\alpha_1} \\times n_1 + \\omega^{\\alpha_2} \\times n_2 + \n \\dots + \\omega^{\\alpha_p} \\times n_p\\) where all the \\(\\alpha_i\\) \nare ordinals in Cantor normal form, \\(\\alpha_1 > \\alpha_2 > \\alpha_p\\), \nand all the \\(n_i\\) are positive integers.\n\nAn example of Cantor normal form is displayed in Fig \\ref{fig:cnf-example}:\nNote that any ordinal of\nthe form \\(\\omega^0 \\times i + 0\\) is just written \\(i\\).\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[scale=2, every node/.style={transform shape}]\n\\node[color=blue]{$\\omega^{(\\omega^\\omega\\,+\\, \\omega^2 \\times 8 \\,+\\, \\omega)}+ \\omega^\\omega + \\omega^4+ 6$};\n\\end{tikzpicture}\n\\caption{\\label{fig:cnf-example}\nAn ordinal in Cantor normal form}\n\\end{figure}\n\n\n\n\nIn the rest of this section, we define an inductive type for representing in \\texttt{Coq}\nall the ordinals strictly less than \\(\\epsilon_0\\), then extend some arithmetic operations\nto this type, and finally prove that our representation fits well with \nthe expected mathematical properties: the order we define is a well order, \nand the decomposition into Cantor normal form is consistent \nwith the implementation of the arithmetic operations of exponentiation of base \\(\\omega\\) \nand addition.\n\n\\paragraph*{Remark}\n\\label{sec:orgheadline65}\nUnless explicitly mentionned, the term ``ordinal\" will be used instead of\n``ordinal strictly less than \\(\\epsilon_0\\)\" (except in Chapter~\\ref{chap:schutte} where it stands for ``countable ordinal'').\n\n\n\n\\subsection{A Data Type for Ordinals in Cantor Normal Form}\n\\label{sec:orgheadline72}\n\\label{sec:T1-inductive-def}\n\n\n\n% Our user contribution~\\cite{CantorContrib} represents \n% the set of ordinals strictly less than $\\epsilon_0$ in Cantor normal form as in~\\cite{Manolios2005}, and also the set\n% of ordinals strictly less than $\\Gamma_0$ in Veblen normal form.\n\n\n Let us define an inductive type whose \nconstructors are respectively associated\nwith the ways to build Cantor normal forms:\n\n\\begin{itemize}\n\\item the ordinal \\(0\\)\n\\item the construction \\((\\alpha,\\, n,\\,\\beta) \\mapsto \\omega^\\alpha \\times (n + 1)+ \\beta \\quad (n\\in\\mathbb{N})\\)\n\\end{itemize}\n\n\n\\vspace{4pt}\n\\noindent\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#T1}{Epsilon0.T1}}\n\n\\label{types:T1}\n%\\index{Constants!zero:T1}\n\n\n\\begin{Coqsrc}\nInductive T1 : Set :=\n| zero : T1\n| ocons : T1 -> nat -> T1 -> T1.\n\\end{Coqsrc}\n\n\n\n\\subsubsection*{Remark}\n\\label{sec:orgheadline66}\n\n\nThe name \\texttt{T1} we gave to this data-type is proper to this development and refers\nto a hierarchy of ordinal notations. For instance, in~\\cite{CantorContrib}, the following type is used to represent ordinals strictly less than \\(\\Gamma_0\\), in Veblen normal form (see also~\\cite{schutte}).\n\n\n\\emph{Please look also at the library\n\\href{../theories/html/hydras.Gamma0.T2.html}{Gamma0.T2.html}}.\n\n\n\\begin{Coqsrc}\nInductive T2 : Set :=\n zero : T2\n| gcons : T2 -> T2 -> nat -> T2 -> T2.\n\\end{Coqsrc}\n\n\\subsubsection{Example}\n\n\\label{alpha0-def}\nFor instance, the ordinal $\\omega^\\omega+\\omega^3\\times 5+2$ is represented by the following term:\n\n\\begin{Coqsrc}\nExample alpha_0 : T1 :=\n ocons (ocons (ocons zero 0 zero)\n 0\n zero)\n 0\n (ocons (ocons zero 2 zero)\n 4\n (ocons zero 1 zero)).\n\\end{Coqsrc}\n\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.5, level 1/.style={sibling distance=6cm},\nlevel 2/.style={sibling distance=35mm}, \nlevel 3/.style={sibling distance=17mm}]\n\\node {ocons}\n child { node {ocons}\n child { node {ocons} child {node {zero}} child {node{0}} child{node{zero}}}\n child {node {0}}\n child {node {zero}}}\n child {node {0}}\n child {node {ocons} \n child { node {ocons} child {node {zero}} child {node{2}} child{node{zero}}}\n child {node {4}}\n child {node {ocons} child {node {zero}} child {node{1}} child{node{zero}}}};\n\n\\end{tikzpicture}\n\n\\caption{The tree-like representation of the ordinal $\\omega^\\omega+\\omega^3\\times 5 +2$\\label{fig:cnf-tree}}\n\n\\end{figure}\n\n\n\n\\paragraph{Remark}\nFor simplicity's sake, we chosed to forbid expressions of the form $\\omega^\\alpha\\times 0 + \\beta$. Thus, the contruction (\\texttt{ocons $\\alpha$ $n$ $\\beta$}) is intented to represent the\nordinal $\\omega^\\alpha\\times(n+1)+\\beta$ and not $\\omega^\\alpha\\times n+\\beta$.\nIn a future version, we should replace the type \\texttt{nat} with \\texttt{positive} in \\texttt{T1}'s \ndefinition. But this replacement would take a lot of time \\dots{}\n\n\\subsection{Abbreviations}\n\nSome abbreviations may help to write more concisely complex ordinal terms.\n\n\\subsubsection{Finite Ordinals}\n\\label{sec:orgheadline67}\n\nFor representing finite ordinals, \\emph{i.e.} natural numbers, we first introduce a notation for terms of the form $n+1$, then define a coercion from type \\texttt{nat} into \\texttt{T1}.\n\\label{sect:notation-FS}\n\n\\begin{Coqsrc}\nNotation \"'FS' n\" :=\n (ocons zero n zero) (at level 10) : t1_scope.\n\\end{Coqsrc}\n\n\\label{sect:notation-F}\n\n\\begin{Coqsrc}\nDefinition fin (n:nat) : T1 := \n match n with 0 => zero | S p => FS p end. \n\nCoercion fin : nat >-> T1.\n\nExample ten : T1 := 10. \n\\end{Coqsrc}\n\n% \\index{Coq!Techniques!Coercions}\n% \\index{Functions!Coercions@Coercions (from nat to ordinal types)}\n% \\begin{remark}\n% Please refer to the remark~\\pageref{warning:coercions} about the use of coercions.\n% % The use of coercions like \\texttt{fin} allow us to be close to the mathematical tradition where natural numbers are ordinals too.\n% % Nevertheless, it may happen that a goal like \\texttt{3 < 5} could be \n% % interpreted as \\texttt{(lt (fin 3) (fin 5))}, depending on the current notation scope. \n% % When this misinterpretation happens, tactics like \\texttt{auto with arith}, \\texttt{lia} do not work!\n% % Thus, it is useful to write \\texttt{(3 < 5)\\%nat} an inequality between two natural numbers. \n% \\end{remark}\n\n\n\\subsubsection{The Ordinal \\(\\omega\\)}\n\\label{sec:orgheadline68}\n\n Since \\(\\omega\\)'s Cantor normal form is\ni.e. \\(\\omega^{\\omega^0}\\times 1+ 0\\), we can define the following abbreviation:\n\n\\label{sect:omega-notation2}\n\\begin{Coqsrc}\nNotation omega := (ocons (ocons zero 0 zero) 0 zero): t1_scope.\n\\end{Coqsrc}\n\nNote that \\texttt{omega} is not an identifier, thus any tactic like \\texttt{unfold omega} would fail.\n\n\n\\subsubsection{The Ordinal \\(\\omega^\\alpha\\), a.k.a. \\(\\phi_0(\\alpha)\\)}\n\\label{sec:orgheadline70}\n\nWe provide also a notation for ordinals of the form $\\omega^\\alpha$.\n\n\\label{sect:notation-phi0}\n\n\\index{Notations!phi0@phi0 (exponential of base omega)}\n\n\\begin{Coqsrc}\nNotation \"'phi0' alpha\" := (ocons alpha 0 zero) (at level 29) : t1_scope.\n\\end{Coqsrc}\n\n\\index{Maths!Additive principal ordinals}\n\n\\begin{remark}\n\\label{sec:orgheadline69}\nThe name \\(\\phi_0\\)\n comes from ordinal numbers theory. In~\\cite{schutte}, Schütte defines \n$\\phi_0$ as the ordering (\\emph{i.e.} enumerating) function of the set of \\emph{additive principal ordinals} \\emph{i.e.} strictly positive ordinals $\\alpha$ that verify $\\forall \\beta<\\alpha, \\beta+\\alpha=\\alpha$. For Schütte, $\\omega^\\alpha$ is just a notation for $\\phi_0(\\alpha)$. See also Chapter~\\ref{chap:schutte} of this document.\n\\end{remark}\n\n\n\n \n\\subsubsection{The Hierarchy of \\(\\omega\\)-towers:}\n\\label{sec:orgheadline71}\n\nThe ordinal $\\epsilon_0$, although not represented by a finite term in Cantor normal form, is approximed by the sequence of $\\omega$-towers (see also Sect~\\vref{sect:epsilon0-as-limit} ).\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html}{Epsilon0.T1}}\n\n\\begin{Coqsrc}\nFixpoint omega_tower (height:nat) : T1 := \n match height with \n | 0 => 1 \n | S h => phi0 (omega_tower h)\n end.\n\\end{Coqsrc}\n\nFor instance, Figure~\\ref{fig:tower7} represents the ordinal returned by the\n evaluation of the term \\texttt{omega\\_tower 7}.\n\n\\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[scale=2, every node/.style={transform shape}]\n\\node[color=blue]{$\\omega^{{{\\omega}^{{{\\omega}}^{{{\\omega}}^{{\\omega^{{\\omega}^{\\omega}}}}}}}}$};\n\\end{tikzpicture}\n\\caption{\\label{fig:tower7}\nThe $\\omega$-tower of height 7}\n\\end{figure}\n\n\n\n\\subsection{Comparison between Ordinal Terms}\n\\label{sec:orgheadline73}\n\n\n% Our formalisation of Cantor Normal Form will take two steps:\n% 1 Definition of a strict order \\texttt{o<} on the type \\texttt{T1}, \n% 2 Using \\texttt{o<} for characterizing terms in normal form.\n\nIn order to compare two terms of type \\texttt{T1}, we define a recursive function \\texttt{compare} that maps two ordinals $\\alpha$ and $\\beta$ to a value of type \\texttt{comparison}. This type is defined in \\coq's standard library \n\\texttt{Init.Datatypes} and\ncontains three constructors: \\texttt{Lt} (less than), \\texttt{Eq} (equal), and\n\\texttt{Gt} (greater than).\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#compare}{Epsilon0.T1}}\n\n\n\\begin{Coqsrc}\nFixpoint compare (alpha alpha':T1):comparison :=\n match alpha, alpha' with\n zero, zero => Eq\n | zero, ocons a' n' b' => Lt\n | _ , zero => Gt\n | (ocons a n b),(ocons a' n' b') =>\n (match compare a a' with \n | Lt => Lt\n | Gt => Gt\n | Eq => (match lt_eq_lt_dec n n'\n with\n inleft (left _) => Lt\n | inright _ => Gt\n | _ => compare b b'\n end)\n end)\n end.\n\\end{Coqsrc}\n \nIt is now easy to define the boolean predicate \\texttt{lt\\_b $\\alpha$ $\\beta$}: \n`` $\\alpha$ is strictly less than $\\beta$ ''. By coercion to sort \\texttt{Prop} we define also the predicate \\texttt{lt}.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html}{Epsilon0.T1}}\n\n\n\\begin{Coqsrc}\nDefinition lt_b alpha beta : bool :=\n match compare alpha beta with\n Lt => true\n | _ => false\n end.\n\nDefinition lt alpha beta : Prop := lt_b alpha beta.\n\\end{Coqsrc}\n\n\\label{Predicates:lt-T1}\nPlease note that this definition of \\texttt{lt} makes it easy to write proofs by reflection, as shown by the following exampgles.\n\n\\begin{Coqsrc}\nExample E1 : lt (ocons omega 56 zero) (tower 3).\nProof. reflexivity. Qed.\n\nExample E2 : ~ lt (tower 3) (tower 3).\nProof. discriminate. Qed.\n\\end{Coqsrc}\n\nThe following lemmas establish relations between \\texttt{compare}, \nthe predicate \\texttt{lt} and Leibniz equality \\texttt{eq}.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#compare_refl}{Epsilon0.T1}}\n\n\n\\begin{Coqsrc}\nLemma compare_refl : forall alpha, compare alpha alpha = Eq.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma compare_reflect : forall alpha beta,\n match compare alpha beta with\n | Lt => lt alpha beta\n | Eq => alpha = beta\n | Gt => lt beta alpha\n end.\n\\end{Coqsrc}\n\nWe prove also that the relation \\texttt{lt} is a strict total order.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#lt_irrefl}{Epsilon0.T1}}\n\n \n\\begin{Coqsrc}\nTheorem lt_irrefl (alpha: T1): ~ lt alpha alpha.\n\nTheorem lt_trans (alpha beta gamma : T1) :\n lt alpha beta -> lt beta gamma -> lt alpha gamma.\n\nDefinition lt_eq_lt_dec :\n forall alpha beta : T1, \n {lt alpha beta} + {alpha = beta} + {lt beta alpha}.\n\\end{Coqsrc}\n\n\nNote that the order \\texttt{lt} is not reflected \nin the structure (size and/or height) of the terms of \\texttt{T1}. \nFor instance the ordinal of Fig \\ref{fig:cnf-example} is strictly less\nthan the structurally simpler \\(\\omega^{\\omega^\\omega}\\times 2\\).\n\n\\subsubsection{A Predicate for Characterizing Normal Forms}\n\\label{sect:t1-nf}\n\n\\label{sec:orgheadline74}\n\\label{sec:orgheadline75}\nOur data-type \\texttt{T1} allows us to write expressions that\nare not properly in Cantor normal form as specified in Section \\ref{sec:epsilon0-intro}.\nFor instance, consider the following term of type \\texttt{T1}. \n\n\\begin{Coqbad}\nExample bad_term : T1 := ocons 1 1 (ocons omega 2 zero).\n\\end{Coqbad}\n\nThis term would have been written \\(\\omega^1\\times 2 + \\omega^\\omega \\times 3\\) in the usual mathematical notation. We note that the exponents of $\\omega$ are not in the right (strictly decreasing) order.\n\nWith the help of the order \\texttt{lt} on \\texttt{T1}, we are now able to characterize\nthe set of all well-formed ordinal terms:\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#nf_b}{Epsilon0.T1}}\n\n\\label{Predicates:nf-T1}\n\n\\begin{Coqsrc}\nFixpoint nf_b (alpha : T1) : bool :=\n match alpha with\n | zero => true\n | ocons a n zero => nf_b a\n | ocons a n ((ocons a' n' b') as b) =>\n (nf_b a && nf_b b && lt_b a' a)%bool\n end. \n\nDefinition nf alpha: Prop := nf_b alpha.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\n Compute nf_b alpha_0.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = true \n : bool\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\n Compute nf_b bad_term.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = false \n : bool\n\\end{Coqanswer}\n\n\n\n\\subsection{Making Normality Implicit}\n We would like to get rid of terms of type \\texttt{T1} which are not in Cantor normal form.\nA simple way to do this is to consider statements of the form \n\\texttt{forall alpha: T1, nf alpha -> $P$ alpha}, where $P$ is a predicate over type \\texttt{T1}, like in the following lemma \\footnote{Ordinal addition is formally defined a little later (page~\\ref{sect:infix-plus-T1})}.\n\n\\begin{Coqsrc}\nLemma plus_is_zero alpha beta :\n nf alpha -> nf beta ->\n alpha + beta = zero -> alpha = zero /\\ beta = zero.\n\\end{Coqsrc}\n\nBut this style leads to clumsy statements, and generates too many sub-goals in interactive proofs (although often solved with \\texttt{auto} or \\texttt{eauto}).\n\nOne may encapsulate conditions of the form \\texttt{(nf $\\alpha$)} in\nthe most used predicates. For instance, we introduce the restriction of \\texttt{lt} to terms in normal form, and provide a handy notation for this restriction.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Prelude.Restriction.html}{Ordinals.Prelude.Restriction}}\n\n\\begin{Coqsrc}\nDefinition restrict {A:Type}(E: Ensemble A)(R: relation A) :=\n fun a b => E a /\\ R a b /\\ E b.\n \\end{Coqsrc}\n\n \n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#LT}{Epsilon0.T1}}\n\n\\begin{Coqsrc}\nDefinition LT := restrict nf lt.\nInfix \"t1<\" := LT : t1_scope.\n\nDefinition LE := restrict nf le.\nInfix \"t1<=\" := LE : t1_scope.\n\\end{Coqsrc}\n\n\n\\label{Predicates:LT-T1}\n \n\nFor instance, in the following lemma, the condition that $\\alpha$ is in normal form is included in the condition $\\alpha< 1$.\n\n\\begin{Coqsrc}\nLemma LT_one : forall alpha, alpha t1< one -> alpha = zero.\n\\end{Coqsrc}\n\n \n\\subsubsection{A Sigma Type for \\texorpdfstring{$\\epsilon_0$}{epsilon0}}\n\nAs we noticed in Sect.~\\ref{sect:t1-nf}, the type \\texttt{T1} is not a correct ordinal notation, since it contains terms that are not in Cantor normal form. In certain contexts (for instance in Sections~\\ref{sect:L-equations}, \\ref{sect:hardy},\nand \\ref{sect:wainer}), we need to define total recursive functions on well-formed ordinal terms less than $\\epsilon_0$, using the \\texttt{Equations} plug-in~\\cite{sozeau:hal-01671777}.\n In order to define a type whose inhabitants represent just ordinals, we build a type gathering a term of type \\texttt{T1} and a proof that this term is in normal form.\n \n\n\\label{sect:E0-def}\n\\label{types:E0}\n\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.E0.html}{Epsilon0.E0}}\n\n\n\n\\begin{Coqsrc}\nClass E0 : Type := t1_2o {cnf : T1; cnf_ok : nf cnf}.\n\\end{Coqsrc}\n\nMany constructs : types, predicates, functions, notations, etc., on type \\texttt{T1} are adapted to \\texttt{E0}.\n\nFirst, we declare a notation scope for \\texttt{E0}.\n\n\\begin{Coqsrc}\nDeclare Scope E0_scope.\nDelimit Scope E0_scope with e0.\nOpen Scope E0_scope.\n\\end{Coqsrc}\n\nThen we redefine the predicates of comparison.\n\n\\label{Predicates:Lt-E0}\n\n\\begin{Coqsrc}\nDefinition Lt (alpha beta : E0) := T1.LT (@cnf alpha) (@cnf beta).\nDefinition Le (alpha beta : E0) := T1.LE (@cnf alpha) (@cnf beta).\n\nInfix \"o<\" := Lt : E0_scope.\nInfix \"o<=\" := Le : E0_scope.\n\\end{Coqsrc}\n \n\nEquality in \\texttt{E0} is just Leibniz equality. Note that, since \\texttt{nf} is\ndefined by a Boolean function, for any term $\\alpha:\\texttt{T1}$, there exists at most one proof of \\texttt{nf $\\alpha$}, thus two ordinals of type \\texttt{E0} are\nequal if and only iff their projection to \\texttt{T1} are equal (see also Sect.~\\vref{sect:eq-proof-unicity}).\n\n\\index{Coq!Techniques!Unicity of equality proofs}\n\n\n\\begin{Coqsrc}\nRequire Import Logic.Eqdep_dec.\n\nLemma nf_proof_unicity :\n forall (alpha:T1) (H H': nf alpha), H = H'.\n\nLemma E0_eq_iff alpha beta : alpha = beta <-> cnf alpha = cnf beta.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\\begin{exercise}\nIn earlier versions of this development, the predicate \\texttt{nf} was defined inductively, with various constructors describing all possible cases.\n\\begin{enumerate}\n\\item Please give such a definition, in a dedicated module.\n\\item Prove the logical equivalence between your definition and ours.\n\\item Define a variant of the type \\texttt{E0} (with your definition of \\texttt{nf}).\n\\item Can you still prove a lemma like \\texttt{E0\\_eq\\_iff} ? With the help of an axiom from some module of the standard library ?\n\\end{enumerate}\n\\end{exercise}\nFor upgrading constants and fonctions of \\texttt{T1}, we have to prove that \nthe term they build is in normal form.\nFor instance, let us represent the ordinals $0$ and $\\omega$ as instances of the class \\texttt{E0}.\n\n\\label{sect:omega-T1}\n%\\index{Constants!zero:T1}\n\n\\begin{Coqsrc}\nInstance Zero : E0.\nProof.\n now exists T1.zero.\nDefined.\n\nInstance _Omega : E0.\nProof. now exists omega%t1. \nDefined.\n\nNotation \"'omega'\" := _Omega : E0_scope.\n\\end{Coqsrc}\n\n\n\n\\subsection{Syntactic Definition of Limit and Successor Ordinals}\n\nPattern matching and structural recursion allow us to define the notions of successor and limit ordinal with the help of boolean functions on type \\texttt{T1}. \n\n \\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#succb}{Epsilon0.T1}}\n\n\\begin{Coqsrc}\n Fixpoint succb alpha :=\n match alpha with\n zero => false\n | ocons zero _ _ => true\n | ocons alpha n beta => succb beta\n end.\n\nFixpoint limitb alpha :=\n match alpha with\n zero => false\n | ocons zero _ _ => false\n | ocons alpha n zero => true\n | ocons alpha n beta => limitb beta\n end.\n\\end{Coqsrc}\n\n\n\n\\begin{Coqsrc}\n Compute limitb omega.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = true\n : bool\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nCompute succb 42.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = true\n : bool\n \\end{Coqanswer}\n\nThe correctness of these definitions with respect to the mathematical notions of\nlimit and successor ordinals is established through several lemmas. For instance,\nLemma \\texttt{canonS\\_limit}, page~\\pageref{lemma:canonS-limit}, shows that\nif $\\alpha$ is (syntactically) a limit ordinal, then it is the least upper bound of\na strictly increasing sequence of ordinals.\n\n\n The following function is very useful in constructions by cases (proofs and function definitions).\n \n\\begin{Coqsrc}\nDefinition zero_succ_limit (alpha: T1) :\n {succb alpha} + {limitb alpha} + {alpha=zero}.\n (* ... *)\nDefined.\n\\end{Coqsrc}\n\n\n\n\\subsection{Arithmetic on \\texorpdfstring{$\\epsilon_0$}{epsilon0}}\n\\subsubsection{Successor}\n\nThe successor of any ordinal $\\alpha< \\epsilon_0$ is defined by structural \nrecursion on its Cantor normal form.\n\n\\label{Functions:succ-T1}\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.T1.html\\#succ}{Epsilon0.T1}}\n\n\\begin{Coqsrc}\nFixpoint succ (alpha:T1) : T1 :=\n match alpha with \n | zero => 1\n | ocons zero n _ => ocons zero (S n) zero\n | ocons beta n gamma => ocons beta n (succ gamma)\n end.\n\\end{Coqsrc}\n\n\nThe following lemma establishes the connection between the function\n\\texttt{succ} and the Boolean predicate \\texttt{succb}.\n\n\n\\begin{Coqsrc}\n Lemma succb_iff alpha (Halpha : nf alpha) :\n succb alpha <-> exists beta : T1, nf beta /\\ alpha = succ beta.\n\\end{Coqsrc}\n\n\\index{Exercises}\n \\begin{exercise}\nProve in \\coq{} that for any ordinal $\\alpha<\\epsilon_0$, $\\alpha$ is a limit if \nand only if for all $\\beta<\\alpha$, the interval $[\\beta,\\alpha)$ is infinite.\n \\end{exercise}\n\n\n\\subsubsection{Addition and Multiplication}\n\nOrdinal addition and multiplication are also defined by structural recursion over the type \\texttt{T1}. Please note that they use the \\texttt{compare} function on some subterms of their arguments.\n\n\\label{sect:infix-plus-T1}\n\n\\begin{Coqsrc}\nFixpoint plus (alpha beta : T1) : T1 :=\n match alpha,beta with\n | zero, y => y\n | x, zero => x\n | ocons a n b, ocons a' n' b' =>\n (match compare a a' with\n | Lt => ocons a' n' b'\n | Gt => (ocons a n (plus b (ocons a' n' b')))\n | Eq => (ocons a (S(n+n')) b')\n end)\n end\nwhere \"alpha + beta\" := (plus alpha beta) : t1_scope.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nFixpoint mult (alpha beta : T1) :T1 :=\n match alpha,beta with\n | zero, y => zero\n | x, zero => zero\n | ocons zero n _, ocons zero n' _ => \n ocons zero (Peano.pred((S n) * (S n'))) zero\n | ocons a n b, ocons zero n' b' => \n ocons a (Peano.pred((S n) * (S n'))) b\n | ocons a n b, ocons a' n' b' =>\n ocons (a + a') n' ((ocons a n b) * b')\n end\nwhere \"alpha * beta\" := (mult alpha beta) : t1_scope.\n\\end{Coqsrc}\n\n\n\\subsubsection{Examples}\n\nThe following examples are instances of \\emph{proofs by computation}. Please note that addition and multiplication on \\texttt{T1}\nare not commutative. Moreover, both operations fail to be strictly monotonous in their first argument.\n\n\n\\begin{Coqsrc}\nExample e2 : 6 + omega = omega.\nProof. reflexivity. Qed.\n\nExample e'2 : omega t1< omega + 6.\nProof. now compute. Qed.\n\nExample e''2 : 6 * omega = omega.\nProof. reflexivity. Qed.\n\nExample e'''2 : omega t1< omega * 6.\nProof. now compute. Qed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma plus_not_monotonous_l : exists alpha beta gamma : T1,\n alpha t1< beta /\\ alpha + gamma = beta + gamma.\nProof.\n exists 3, 5, omega; now compute.\nQed.\n\nLemma mult_not_monotonous : exists alpha beta gamma : T1,\n alpha t1< beta /\\ alpha * gamma = beta * gamma.\nProof.\n exists 3, 5, omega; now compute.\nQed.\n\\end{Coqsrc}\n\n\n\\subsection{Pretty Printing Ordinals in Cantor Normal Form}\n\\label{sect:ppT1}\n\nLet us consider again the ordinal $\\alpha_0$ defined in section~\\vref{alpha0-def}\nIf we ask \\coq{} to print its normal form, we get a hardly readable term of type \\texttt{T1}.\n\n\\begin{Coqsrc}\nCompute alpha_0.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = ocons omega 0 (ocons (FS 2) 4 (FS 1))\n : T1\n\\end{Coqanswer}\n\nThe following data type defines an abstract syntax for more readable ordinals terms in Cantor normal form:\n\n\\label{types:ppT1}\n\\index{Functions!pp@ pp (pretty printing terms in Cantor normal form)}\n\n\\begin{Coqsrc}\nInductive ppT1 : Set :=\n P_fin : nat -> ppT1\n | P_add : ppT1 -> ppT1 -> ppT1\n | P_mult : ppT1 -> nat -> ppT1\n | P_exp : ppT1 -> ppT1 -> ppT1\n | P_omega : ppT1\n\\end{Coqsrc}\n\nThe function \\texttt{pp: T1 -> ppT1} converts any closed term of type \\texttt{T1} into a human-readable expression. For instance, let us convert the term \\texttt{alpha\\_0}.\n\n\\begin{Coqsrc}\nCompute pp alpha_0.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = (omega ^ omega + omega ^ 3 * 5 + 2)%pT1\n : ppT1\n\\end{Coqanswer}\n\n\\index{Projects}\n\\begin{project}\nDesign (in \\ocaml?) a set of tools for systematically pretty printing ordinal terms in Cantor normal form.\n\\end{project}\n\n\n\\subsubsection{Arithmetic on Type \\texttt{E0}}\n\n We define an addition in type \\texttt{E0}, using the function{T1.plus}\nand its properties.\n\n\\begin{Coqsrc}\nSearch (nf (_ + _)%t1).\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nplus_nf: forall a : T1, nf a -> forall b : T1, nf b -> nf (a + b)\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nInstance plus (alpha beta : E0) : E0.\nProof.\n refine (@mkord (T1.plus (@cnf alpha) (@cnf beta))_ );\n apply plus_nf; apply cnf_ok.\nDefined.\n\nInfix \"+\" := plus : E0_scope.\n\nCheck omega + omega.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nomega + omega\n : E0\n\\end{Coqanswer}\n\n\\index{Exercises}\n \\begin{exercise}\nLet $\\alpha$ be an ordinal. We say that $\\alpha$ is \\emph{infinite} iff the segment $[0,\\alpha)$ is an infinite set. \n\\begin{enumerate}\n\\item Adapt this definition to the type \\texttt{E0}.\n\\item Prove that \\emph{being infinite} is decidable on type \\texttt{E0}.\n\\item Prove that $\\alpha$ is infinite if and only if for all finite ordinal $i$,\n$i+\\alpha=\\alpha$.\n\\end{enumerate}\n \\end{exercise}\n\n\\section{Well-foundedness and Transfinite Induction}\n\n\\index{Maths!Transfinite induction}\n\n\\subsection{About Well-foundedness}\n\\label{sec:orgheadline82}\n In order to use \\texttt{T1} for proving termination results,\nwe need to prove that our order \\texttt{<} is well-founded. Then we will get \\emph{transfinite induction} for free.\n\n\nThe proof of well-foundedness of the strict order $<$ on Cantor normal forms is already \navailable in the Cantor contribution by Castéran and Contejean~\\cite{CantorContrib}. That proof relies on a library on recursive path orderings written by\nE. Contejean. We present here a direct proof of the same result, which does not require any knowledge on r.p.o.s.\n\n\\index{Exercises}\n\n\\begin{exercise}\nProve that the \\emph{total} order \\texttt{lt} on \\texttt{T1} is not well-founded. \n\\textbf{Hint:} You will have to build a counter-example with terms of type \\texttt{T1}\nwhich are not in Cantor normal form.\n\\end{exercise}\n\n% \\subsubsection{The total order \\texttt{lt} on \\texttt{T1} is \\emph{not} well-founded}\n% \\label{sec:orgheadline76}\n\n% Let us recall that the data type \\texttt{T1} contains too many inhabitants, including\n% terms which are not in Cantor normal form. Thus, the following result is not \n% very surprising.\n\n% \\begin{Coqsrc}\n% Section lt_not_well_founded.\n \n% (* let us build the sequence of terms :\n% omega + omega + .... + omega ^ 2 *)\n% Let f := (fix f (i:nat): T1 :=\n% match i with 0 => phi0 2\n% | S i => ocons 1 1 (f i)\n% end).\n\n \n% Lemma f_decreases : forall i, f (S i) < f i.\n% Proof.\n% induction i; compute; auto with T1.\n% Qed.\n\n% Theorem lt_not_wf : ~ well_founded lt.\n% Proof. \n% intro wf; case (not_decreasing _ lt);auto.\n% exists f; apply f_decreases.\n% Qed.\n\n% End lt_not_well_founded.\n% \\end{Coqsrc}\n\n% Thus, we have \n\n\\subsubsection{A First Attempt}\n\\label{sec:orgheadline77}\n\\index{Coq!Techniques!Well-founded induction}\n\nIt is natural to try to prove by structural induction over \\texttt{T1} \nthat every term in normal form is accessible through \\texttt{LT}.\n\nUnfortunately, it won't work. Let us consider some well-formed term\n $\\alpha=\\texttt{ocons $\\beta\\;n\\;\\gamma$}$, and assume that \\(\\beta\\) and \\(\\gamma\\) are accessible\n through \\texttt{LT}. For proving the accessibility of $\\alpha$, we have to consider\nany well formed term \\(\\delta\\) such that \\(\\delta<\\alpha\\). \nBut nothing guarantees that \\(\\delta\\) is strictly less than \\(\\beta\\) nor \\(\\gamma\\), and we cannot use the induction hypotheses on \\(\\beta\\) nor \\(\\gamma\\).\n\n\\begin{Coqbad}\nSection First_attempt.\n\n Lemma wf_LT : forall alpha, nf alpha -> Acc LT alpha. \n Proof.\n induction alpha as [| beta IHbeta n gamma IHgamma].\n - split.\n inversion 1.\n destruct H2 as [H3 _];not_neg H3.\n - split; intros delta Hdelta.\n\\end{Coqbad}\n\n\\begin{Coqanswer}\n1 subgoal (ID 560)\n \n beta : T1\n n : nat\n gamma : T1\n IHbeta : nf beta -> Acc LT beta\n IHgamma : nf gamma -> Acc LT gamma\n H : nf (ocons beta n gamma)\n delta : T1\n Hdelta : delta t1< ocons beta n gamma\n ============================\n Acc LT delta\n \\end{Coqanswer}\n\n\\begin{Coqbad}\n Abort.\n\\end{Coqbad}\n\nThe problem comes from the hypothesis \\texttt{Hdelta}. It does not prevent \\(\\delta\\) to be bigger that \\(\\beta\\) or\n\\(\\gamma\\);\nfor instance \\(\\delta\\) may be of the form\n\\texttt{ocons $\\beta'$ $p'$ $\\gamma'$},\nwhere \\(\\beta' \\leq \\beta\\) and \\(p' < n\\).\nThus, the induction hypotheses \\texttt{IHbeta} and \\texttt{IHgamma} are useless for finishing our proof.\n\n\\subsubsection{Using a Stronger Inductive Predicate.}\n\\label{sec:orgheadline78}\n Instead of trying to prove directly that any ordinal term \\(\\alpha\\) in normal form is accessible\nthrough \\texttt{LT}, we propose to show first that any well formed \nterm of the form \\(\\omega^\\alpha\\times(n+1)+\\beta\\) is accessible (which is a stronger result).\n\n\\begin{Coqsrc}\n Let Acc_strong (alpha:T1) :=\n forall n beta, \n nf (ocons alpha n beta) -> Acc LT (ocons alpha n beta).\n\\end{Coqsrc}\n\nThe following lemma is an application of the strict inequality \n\\showmath{\\alpha < \\omega ^\\alpha}. If \\showmath{\\omega^\\alpha} is accessible,\nthen \\showmath{\\alpha} is \\emph{a fortiori} accessible.\n\n\\begin{Coqsrc}\n Lemma Acc_strong_stronger : forall alpha, \n nf alpha -> Acc_strong alpha -> Acc LT alpha.\n Proof.\n intros alpha H H0; apply acc_imp with (phi0 alpha).\n - repeat split; trivial.\n + now apply lt_a_phi0_a.\n - apply H0; now apply single_nf.\nQed.\n\\end{Coqsrc}\n\nThus, it remains to prove that every ordinal strictly less than \\showmath{\\epsilon_0} \nis strongly accessible.\n\n% \\subsubsection{Structure of the proof of well-foundedness of \\texttt{LT}}\n\n\\label{sec:orgheadline81}\n\\label{proof-wf-epsilon0}\n\\paragraph{A helper}\n\\label{sec:orgheadline79}\n\nFirst, we prove that, for any \\texttt{LT}-accessible term \\showmath{\\alpha}, \\showmath{\\alpha} is \nstrongly accessible too (\\emph{i.e.} any well formed\nterm (\\texttt{ocons $\\alpha$ $n$ $\\beta$}) is accessible).\n\n\\begin{Coqsrc}\nLemma Acc_implies_Acc_strong : \n forall alpha, Acc LT alpha -> Acc_strong alpha.\n\\end{Coqsrc}\n\n\nThe proof is structured as an induction on \\showmath{\\alpha}'s accessibility. Let us consider\nan accessible term $\\alpha$.\n\n\n\n\\begin{Coqanswer}\n subgoal 1 \n\n alpha : T1\n Aalpha : forall y : T1, y t1< alpha -> Acc LT y\n IHalpha : forall y : T1,\n LT y alpha ->\n forall (n : nat) (beta : T1),\n nf (ocons y n beta) -> Acc LT (ocons y n beta)\n ============================\n forall (n : nat) (beta : T1),\n nf (ocons alpha n beta) -> Acc LT (ocons alpha n beta)\n\\end{Coqanswer}\n\nLet \\texttt{n:nat} and \\texttt{beta:T1} such that \\texttt{ocons alpha n beta} is in normal form. \nWe prove first that \\texttt{beta} is accessible, which allows us to by well-founded induction on \\texttt{beta}, \nand natural induction on \\texttt{n}, that (\\texttt{ocons alpha n beta}) is accessible.\nThe proof, quite long, can be consulted in \\url{../theories/html/hydras.Epsilon0.T1.html} \n\n\\paragraph{Accessibility of any well-formed ordinal term}\n\\label{sec:orgheadline80}\n\nOur goal is still to prove accessibility of any well formed ordinal term.\nThanks to our previous lemmas, we are almost done.\n\n\\begin{Coqsrc}\n(* A (last) structural induction *)\n\nTheorem nf_Acc : forall alpha, nf alpha -> Acc LT alpha.\nProof.\n induction alpha.\n- intro; apply Acc_zero.\n - intros; eapply Acc_implies_Acc_strong;auto.\n apply IHalpha1;eauto.\n apply nf_inv1 in H; auto. \nDefined.\n\nCorollary T1_wf : well_founded LT.\n\\end{Coqsrc}\n\n\\index{Maths!Transfinite induction}\n\n\\begin{Coqsrc}\n\nDefinition transfinite_recursor :\n forall (P:T1 -> Type),\n (forall x:T1, \n (forall y:T1, nf x -> nf y -> lt y x -> P y) -> P x) ->\n forall alpha:T1, P alpha.\nProof.\n intros; apply well_founded_induction_type with LT.\n - exact T1_wf;auto.\n - intros. apply X. intros; apply X0. repeat split;auto. \nDefined.\n\\end{Coqsrc}\n\nThe following tactic starts a proof by transfinite induction on any ordinal \\mathcolor{$\\alpha<\\epsilon_0$}.\n\n\\begin{Coqsrc}\nLtac transfinite_induction alpha :=\n pattern alpha; apply transfinite_recursor;[ | try assumption].\n\\end{Coqsrc}\n\n\n\\begin{remark}\n\\label{remark:a3pat}\nThe alternate proof of well-foundedness using \\'Evelyne Contejean's work on recursive path ordering~\\cite{DershowitzRPO, a3pat} is available in the library \\href{../theories/html/hydras.Epsilon0.Epsilon0rpo.html}{Epsilon0rpo}.\n \\end{remark}\n\n\n\\subsection{An Ordinal Notation for \\texorpdfstring{$\\epsilon_0$}{epsilon0}}\n\nWe build an instance of \\texttt{ON}, and prove its correction w.r.t. Schutte's model.\n\n\\label{instance-epsilon0}\n\\begin{Coqsrc}\nInstance Epsilon0 : ON Lt compare. \n(* ... *)\n\\end{Coqsrc}\n\n\n\\emph{From Module~\\href{../theories/html/hydras.Schutte.Schutte.Correctness_E0.html}{Schutte.Schutte.Correctness\\_E0}}\n\n\\begin{Coqsrc}\nInstance Epsilon0_correct :\n ON_correct epsilon0 Epsilon0 (fun alpha => inject (cnf alpha)).\n\\end{Coqsrc}\n\n\\index{Projects}\n\\begin{project}\n \\emph{This exercise is a continuation of Project~\\vref{exo:ON-mult}.}\nUse \\texttt{ON\\_mult} to define an ordinal notation \\texttt{Omega2} for $\\omega^2=\\omega\\times\\omega$.\n\nProve that \\texttt{Omega2} is a sub-notation of \\texttt{Epsilon0}.\n\nDefine on \\texttt{Omega2} an addition compatible with the addition on \\texttt{Epsilon0}.\n\n\\textbf{Hint}. You may use the following definition (in \n\\href{../theories/html/hydras.OrdinalNotations.Definitions.html}{OrdinalNotations.Definitions}).\n\n\\begin{Coqsrc}\nDefinition SubON_same_op `{OA : @ON A ltA compareA}\n `{OB : @ON B ltB compareB}\n {iota : A -> B} \n {alpha: B}\n {_ : SubON OA OB alpha iota}\n (f : A -> A -> A)\n (g : B -> B -> B)\n :=\n forall x y, iota (f x y) = g (iota x) (iota y).\n\\end{Coqsrc}\n\n\\end{project}\n\n\\index{Projects}\n\\begin{project}\nThe class \\texttt{ON} of ordinal notations has been defined long after this \nchapter, and is not used yet in the development of the type \\texttt{E0}. \nA better integration of both notions should simplify the development on ordinals in Cantor normal form. This integration is planned for the future versions.\n\n\\end{project}\n\n\n\\section{A Variant for Hydra Battles}\n\n\\subsection{Natural Sum (a.k.a. Hessenberg's Sum)}\n\\label{sec:orgheadline87}\n\\label{hydra-variant}\n\nNatural sum (Hessenberg's sum) is a commutative and monotonous version of\naddition. It is used as an auxiliary operation for defining variants\nfor hydra battles, where Hercules is allowed to chop off any head of the hydra.\n\nIn the litterature, the natural sum of ordinals \\(\\alpha\\) and \\(\\beta\\) \nis often denoted by \\(\\alpha \\# \\beta\\) or \\(\\alpha \\oplus \\beta\\).\nThus we called \\texttt{oplus} the associated \\emph{Coq} function.\n\n\\subsubsection{Definition of \\texttt{oplus}}\n\\label{sec:orgheadline84}\n%\\index{Functions!oplus @ oplus (Hessenberg commutative sum)}\n\nThe definition of \\texttt{oplus} is recursive in both of its \narguments, which makes a structural recursive definition a little \ncomplex.\nWe used the same pattern as for the \\texttt{merge} function on lists of library\n\\texttt{Coq.Sorting.Mergesort}.\n\n\\begin{enumerate}\n\\item Define a nested recursive function, using the \\texttt{Fix} \n construct\n\n\\item Build a principle of induction dedicated to \\texttt{oplus}\n\n\\item Establish equations associated to each case of the definition.\n\\end{enumerate}\n\n\\paragraph{Nested recursive definition}\n\\label{sec:orgheadline83}\n\nThe following definition is composed of \n\\begin{itemize}\n\\item A main function \\texttt{oplus}, structurally recursive in its \nfirst argument \\texttt{alpha}\n\\item An auxiliary function \\texttt{oplus\\_aux} within the scope of \\texttt{alpha},\nstructurally recursive in its argument \\texttt{beta}; \\texttt{oplus\\_aux beta} \n is supposed to compute \\texttt{oplus alpha beta}.\n\\end{itemize}\n \n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Hessenberg.html\\#oplus}{Epsilon0.Hessenberg}}\n\n\\label{sect:infix-oplus}\n\n\\begin{Coqsrc}\nFixpoint oplus (alpha beta : T1) : T1 :=\n let fix oplus_aux beta {struct beta} :=\n match alpha, beta with\n | zero, _ => beta\n | _, zero => alpha\n | ocons a1 n1 b1, ocons a2 n2 b2 =>\n match compare a1 a2 with\n | Gt => ocons a1 n1 (oplus b1 beta)\n | Lt => ocons a2 n2 (oplus_aux b2)\n | Eq => ocons a1 (S (n1 + n2)%nat) (oplus b1 b2)\n end\n end\n in oplus_aux beta.\n\nInfix \"o+\" := oplus (at level 50, left associativity).\n\\end{Coqsrc}\n\n\nThe reader will note that each recursive call of the functions\n\\texttt{oplus} and \\texttt{oplus\\_aux} satisfies \\emph{Coq}'s constraint\non recursive definitions. The function \\texttt{oplus} is recursively called on a sub-term of its first argument,\nand \\texttt{oplus\\_aux} on a sub-term of its unique argument.\nThus, \\texttt{oplus}'s definition is accepted by \\coq{} as a structurally recursive function.\n\n\\subsubsection{Rewriting Lemmas}\n\\label{sec:orgheadline86}\n\n\\emph{Coq}'s constraints on recursive definitions result in \nthe quite complex form of \\texttt{oplus}'s definition.\nProofs of properties of this function can be simpler if we\n \\emph{derive} rewriting lemmas that will help to simplify \nexpressions of the form (\\texttt{oplus $a$ $ b$}).\n\nA first set of lemmas correspond to the various cases of \\texttt{oplus}'s \ndefinition. They can be proved almost immediately, using \\texttt{cbn} \nand \\texttt{reflexivity} tactics.\n\n\n\n\\begin{Coqsrc}\nLemma oplus_alpha_0 (alpha : T1) : alpha o+ zero = alpha.\nProof.\n destruct alpha; reflexivity.\nQed.\n\nLemma oplus_0_beta (beta : T1): zero o+ beta = beta.\nProof.\n destruct beta; reflexivity.\nQed.\n\\end{Coqsrc}\n\n\n% \\subsubsection{A hand-made induction principle}\n% \\label{sec:orgheadline85}\n\n% \\index{Coq!Commands!Functional Scheme}\n\n% \\emph{Coq} contains a command \\texttt{Functional Scheme} that \n% generates induction principles which correspond to recursive functions.\n% Unfortunately, the current version ( \\texttt{8.11.0} ) doesn't work on \\texttt{oplus},\n% probably because of the inner \\texttt{Fix}.\n\n% \\begin{Coqsrc}\n% Functional Scheme oplus_ind := Induction for oplus Sort Prop.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% Error: Anomaly \"todo.\" Please report at http://coq.inria.fr/bugs/.\n% \\end{Coqanswer}\n\n\n% Fortunately, it's a good exercise for a semi-experienced user, to write\n% her/him-self induction principles similar to the ones returned by\n% \\texttt{Functional Scheme}.\n\n% \\begin{itemize}\n% \\item First, we choose to write a version for sort \\texttt{Type}, since versions\n% for sorts \\texttt{Prop} and \\texttt{Set} can be easily derived from\n% the former one. According to \\emph{Coq}'s naming politics, we will call our \n% principle \\texttt{oplus\\_rect}\n\n% \\item The conclusion of \\texttt{oplus\\_rect} will be (\\texttt{$P$ a b (oplus a b)}),\n% where $P$ is an arbitrary function of type \n% \\texttt{T1 -> T1 -> T1 -> Type}\n\n% \\item The premises of \\texttt{oplus\\_rect} will describe how to build an induction \n% on the graph of \\texttt{oplus}.\n% \\end{itemize}\n\n% We are now ready to state and prove \\texttt{oplus\\_rect}, and the reader\n% will note that the statement is longer than the proof script itself,\n% which is a standard proof by induction, simplification and case-analysis \n% that follows \\texttt{oplus}'s definition.\n\n% We associate also a tactic to the application of \\texttt{oplus\\_rect}.\n\n% \\begin{Coqsrc}\n% Lemma oplus_rect:\n% forall P: T1 -> T1 -> T1 -> Type, \n% (forall a:T1, P zero a a) ->\n% (forall a: T1, P a zero a) ->\n% (forall a1 n1 b1 a2 n2 b2 o,\n% compare a1 a2 = Gt ->\n% P b1 (ocons a2 n2 b2) o ->\n% P (ocons a1 n1 b1) (ocons a2 n2 b2)\n% (ocons a1 n1 o)) ->\n% (forall a1 n1 b1 a2 n2 b2 o,\n% compare a1 a2 = Lt ->\n% P (ocons a1 n1 b1) b2 o ->\n% P (ocons a1 n1 b1) (ocons a2 n2 b2) \n% (ocons a2 n2 o)) ->\n% (forall a1 n1 b1 a2 n2 b2 o,\n% compare a1 a2 = Eq ->\n% P b1 b2 o ->\n% P (ocons a1 n1 b1) (ocons a2 n2 b2)\n% (ocons a1 (S (n1 + n2)%nat) o)) ->\n% forall a b, P a b (oplus a b).\n% Proof with auto.\n% induction a.\n% - intro; simpl; destruct b;auto.\n% - induction b.\n% + apply X0.\n% + case_eq (compare a1 b1).\n% * intro Comp; unfold oplus; rewrite Comp.\n% cbn; apply X3 ...\n% * intro Comp; cbn; rewrite Comp; apply X2...\n% * intro Comp; cbn; rewrite Comp ...\n% Defined.\n\n\n% Ltac oplus_induction a b:= pattern (oplus a b); apply oplus_rect.\n% \\end{Coqsrc}\n\n% \\index{Exercises}\n\n% \\begin{exercise}\n% The induction principle \\texttt{oplus\\_rect} is still unused in our development. \n% Please build some nice examples of application.\n% \\end{exercise}\n\n\\index{Projects}\n\\begin{project}\nCompare \\texttt{oplus}'s definition (with inner fixpoint) with other possibilities\n(\\texttt{coq-equations}, \\texttt{Function}, etc.).\n\\end{project}\n\\subsection{More Theorems on Hessenberg's Sum}\n\nWe need to prove some properties of $\\oplus$, particularly about \nits relation with the order $<$ on \\texttt{T1}.\n\n\\subsubsection{Boundedness}\nIf $\\alpha$ and $\\beta$ are both strictly less than $\\omega^\\gamma$, then so is their natural sum\n$\\alpha \\oplus \\beta$. This result can be proved by structural induction on $\\gamma$.\n\n\n\\begin{Coqsrc}\nLemma lt_phi0_oplus : forall gamma alpha beta,\n lt_phi0 alpha gamma ->\n lt_phi0 beta gamma ->\n lt_phi0 (alpha o+ beta) gamma.\n\nProof with auto.\n induction gamma; destruct alpha, beta. \n(* ... *)\n\\end{Coqsrc}\n\n\\subsubsection{Commutativity, Associativity}\n\nWe prove the commutativity of $\\oplus$ in two steps. \n\nFirst, we prove by transfinite induction on $\\alpha$ that the restriction of $\\oplus$ to the\ninterval $[0..\\alpha)$ is commutative.\n\n\\index{Maths!Transfinite induction}\n\n\\begin{Coqsrc}\nLemma oplus_comm_0 : forall alpha, nf alpha ->\n forall a b, nf a -> nf b ->\n lt a alpha ->\n lt b alpha ->\n a o+ b = b o+ a.\n Proof with eauto with T1.\n intros alpha Halph; transfinite_induction alpha.\n(* rest of proof omitted *) \n\\end{Coqsrc}\n\nThen, we infer $\\oplus$'s commutativity for any pair of ordinals:\nLet $\\alpha$ and $\\beta$ be two ordinals strictly less than $\\epsilon_0$. Both ordinals $\\alpha$ and $\\beta$ are\nstrictly less than $\\textrm{max}(\\alpha,\\beta)+1$.\n Thus, we have just to apply the lemma \\coqsimple{oplus\\_comm\\_0}.\n\n\\begin{Coqsrc}\n Lemma oplus_comm : forall alpha beta, \n nf alpha -> nf beta ->\n alpha o+ beta = beta o+ alpha.\n Proof with eauto with T1.\n intros alpha beta Halpha Hbeta;\n apply oplus_comm_0 with (succ (max alpha beta)) ... \n (* ... *)\n\\end{Coqsrc}\n\nThe associativity of Hessenberg's sum is proved the same way.\n\n\n\\begin{Coqsrc}\n Lemma oplus_assoc_0 :\n forall alpha,\n nf alpha ->\n forall a b c, nf a -> nf b -> nf c ->\n lt a alpha ->\n lt b alpha -> lt c alpha ->\n a o+ (b o+ c) = (a o+ b) o+ c.\n Proof with eauto with T1.\n intros alpha Halpha.\n transfinite_induction alpha.\n (* ... *)\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\n Lemma oplus_assoc : forall alpha beta gamma,\n nf alpha -> nf beta -> nf gamma ->\n alpha o+ (beta o+ gamma) =\n alpha o+ beta o+ gamma.\n Proof with eauto with T1.\n intros;\n apply oplus_assoc_0 with (succ (max alpha (max beta gamma))) ...\n (* ... *) \n\\end{Coqsrc}\n\n\n\\subsubsection{Monotonicity}\n\nAt last, we prove that $\\oplus$ is strictly monotonous in both of its arguments.\n\n\\begin{Coqsrc}\nLemma oplus_strict_mono_LT_l (alpha beta gamma : T1) :\n nf gamma -> alpha t1< beta ->\n alpha o+ gamma t1< beta o+ gamma.\n\nLemma oplus_strict_mono_LT_r (alpha beta gamma : T1) :\n nf alpha -> beta t1< gamma ->\n alpha o+ beta t1< alpha o+ gamma.\n\\end{Coqsrc}\n\n\\index{Projects}\n\n\\begin{project}\nThe library \\texttt{Hessenberg} looks too long (proof scripts and compilation).\nPlease try to make it simpler and more efficient!\nThanks!\n\\end{project}\n\n\\subsection{A Measure for Hydra-battle Termination}\n\n\\label{sec:hydra-measure}\n\nLet us define a measure from type \\texttt{Hydra} into \\texttt{T1}.\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Hydra_Termination.html\\#m}{Hydra.Hydra\\_Termination}}\n\n\\begin{Coqsrc}\nFixpoint m (h:Hydra) : T1 :=\n match h with head => zero\n | node hs => ms hs\nend \nwith ms (s:Hydrae) : T1 :=\n match s with hnil => zero\n | hcons h s' => phi0 (m h) o+ ms s'\n end. \n\\end{Coqsrc}\n\nFirst, we prove that the measure $m(h)$ of any hydra $h$ is a well-formed ordinal term of type \\texttt{T1}.\n\n\\begin{Coqsrc}\nLemma m_nf : forall h, nf (m h).\nProof.\n intro h; elim h using Hydra_rect2 \n with (P0 := fun s => nf (ms s)).\n (* ... *)\n\nLemma ms_nf : forall s, nf (ms s).\nProof with auto with T1.\n(* ... *)\n\\end{Coqsrc}\n\nFor proving the termination of all hydra battles, we have to prove that\n\\texttt{m} is a variant. First, a few technical lemmas follow the decomposition of \\texttt{round} into several relations. Then the lemma \\texttt{round\\_decr} gathers all the cases.\n\n\\begin{Coqsrc}\nLemma S0_decr :\n forall s s', S0 s s' -> ms s' t1< ms s.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma R1_decr : forall h h',\n R1 h h' -> m h' t1< m h.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma S1_decr n:\n forall s s', S1 n s s' -> ms s' t1< ms s.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nLemma R2_decr n : forall h h', R2 n h h' -> m h' t1< m h.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nLemma round_decr : forall h h', h -1-> h' -> m h' t1< m h.\nProof.\n destruct 1 as [n [H | H]].\n - now apply R1_decr.\n - now apply R2_decr with n.\nQed.\n\\end{Coqsrc}\n\nFinally, we prove the termination of all (free) battles.\n\n\\label{thm:every-battle-terminates}\n\n\\begin{Coqsrc}\nGlobal Instance HVariant : Hvariant lt_wf free var.\nProof.\n split; intros; eapply round_decr; eauto.\nQed.\n\nTheorem every_battle_terminates: Termination.\nProof. \n red; apply Inclusion.wf_incl with \n (R2 := fun h h' => m h t1< m h').\n red; intros; now apply round_decr.\n apply Inverse_Image.wf_inverse_image, T1_wf.\nQed.\n\\end{Coqsrc}\n\n\n\\section*{Conclusion}\n\nLet us recall three results we have proved so far.\n\\begin{itemize}\n\\item There exists a strictly decreasing variant which maps \\texttt{Hydra} into \nthe segment $[0,\\epsilon_0)$ for proving the termination of any hydra battle\n\\item There exists \\emph{no} such variant from \\texttt{Hydra} into \n$[0,\\omega^2)$, \\emph{a fortiori} into $[0,\\omega)$.\n\\end{itemize}\n\nSo, a natural question is `` Does there exist any strictly decreasing variant mapping\ntype \\texttt{Hydra} into some interval $[0,\\alpha[$ (where $\\alpha <\\epsilon_0$) for proving the termination of all hydra battles''. The next chapter is dedicated to a formal proof that there exists no such $\\alpha$, even if we consider a restriction to the set of ``standard'' battles.\n\n\n\n\n\n\n%\\include{epsilon0}\n\n\n%\\include{impossibility-proofs}\n\n\n\n\n%-------------------------------------------------------------------\n\n\\chapter[The Ketonen-Solovay machinery]{Strolling inside \\texorpdfstring{$\\epsilon_0$}{Epsilon0}: The Ketonen-Solovay Machinery\\label{ks-chapter}}\n\\label{chap:ketonen}\n\\index{Maths!Ketonen-Solovay machinery}\n\n\\section{Introduction}\nThe reader may think that our proof of termination in the previous chapter requires a lot of mathematical tools and may be too complex. So, the question is ``is there any simpler proof'' ?\n\nIn their article~\\cite{KP82}, Kirby and Paris show that this result cannot be proved in Peano arithmetic. Their proof uses some knowledge about model theory and non-standard models of Peano arithmetic. In this chapter, we focus on a specific class of proofs of termination of hydra battles: construction of some variant mapping the type \\texttt{Hydra} into a given initial segment of ordinals. Our proof relies only on the Calculus of Inductive Constructions and is a natural complement of the results proven in the previous chapter.\n\n\\begin{itemize}\n\\item There is no variant mapping the type \\texttt{Hydra} into the interval $[0,\\omega^2)$ (section ~\\vref{omega2-case}), and a fortiori \n$[0,\\omega)$ (section ~\\vref{omega-case}).\n\n\\item There exists a variant which maps the type \\texttt{Hydra} into the\ninterval $[0,\\epsilon_0)$ (theorem \\texttt{every\\_battle\\_terminates}, in section~\\vref{thm:every-battle-terminates}).\n\\end{itemize}\n\n\nThus, a very natural question is the following one:\n\\begin{quote}\n `` Is there any variant from\n\\texttt{Hydra} into some interval $[0,\\mu)$, where $\\mu<\\epsilon_0$, for proving the termination of all hydra battles ?''\n\\end{quote}\n\nWe prove in \\coq{} the following result:\n\n\\begin{quote}\nThere is no variant for proving the termination of all hydra battles\nfrom \\texttt{Hydra} into the interval $[0..\\mu)$, where\n$\\mu< \\epsilon_0$.\nThe same impossibility holds even if we consider only standard battles (with the successive replication factors $0,1,2,\\dots,t,t+1,\\dots$).\n\\end{quote}\n\nOur proofs are constructive and require no axioms: they are closed terms of the CIC, and are mainly composed on function definitions and proofs of properties of these functions. \nThey share much theoretical material with Kirby and Paris', although they do not use any knowledge about Peano arithmetic nor model theory. The combinatorial arguments we use and implement\ncome from \n an article by J.~Ketonen and R.~Solovay~\\cite{KS81}, already cited in the work\n by L.~Kirby et J.~Paris.% on the termination of Goodstein sequences and hydra battles~\\cite{KP82}.\n Section $2$ of this article: ''A hierarchy of probably recursive functions'', contains a systematic study of \\emph{canonical sequences}, which are closely related to\nrounds of hydra battles. \nNevertheless, they have the same global structure as the simple proofs described in\nsections~\\vref{omega-case} and \\vref{omega2-case}. \nWe invite the reader to compare the three proofs step by step, lemma by lemma.\n\n\\section{Canonical Sequences}\n\\label{ketonen-solovay-sect}\n\\index{Maths!Canonical sequences}\n\nCanonical sequences are functions that associate an ordinal $\\canonseq{\\alpha}{i}$ to every ordinal $\\alpha<\\epsilon_0$ and positive integer $i$. They satisfy several nice properties:\n\n\\index{Maths!Transfinite induction}\n\\begin{itemize}\n\\item If $\\alpha\\not=0$, then $\\canonseq{\\alpha}{i}<\\alpha$. Thus canonical sequences can be used for proofs by transfinite induction or function definition by transfinite recursion\n\\item If $\\lambda$ is a limit ordinal, then $\\lambda$ is the least upper bound of the set \n$\\{\\canonseq{\\lambda}{i}\\;|\\,i\\in\\mathbb{N}_1\\}$\n\n\n\\item If $\\beta<\\alpha<\\epsilon_0$, then there is a ``path'' from $\\alpha$ to $\\beta$, \\emph{i.e.} a\nsequence $\\alpha_0=\\alpha, \\alpha_1, \\dots, \\alpha_n=\\beta$, where for every $k nat -> T1} such that\n(\\texttt{canonS $\\alpha$ $i$}) is equal to $\\canonseq{\\alpha}{i+1}$.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Canon.html\\#canonS}{Epsilon0.Canon}}\n\n\\label{Functions:canonS}\n\\label{Functions:canon}\n\\begin{Coqsrc}\nFixpoint canonS alpha (i:nat) :=\n match alpha with\n zero => zero\n | ocons zero 0 zero => zero\n | ocons zero (S k) zero => FS k\n | ocons gamma 0 zero =>\n match pred gamma with\n Some gamma' => ocons gamma' i zero\n | None => ocons (canonS gamma i) 0 zero\n end\n | ocons gamma (S n) zero =>\n match pred gamma with\n Some gamma' => ocons gamma n (ocons gamma' i zero)\n | None => ocons gamma n (ocons (canonS gamma i) 0 zero)\n end\n | ocons alpha n beta => ocons alpha n (canonS beta i)\n end.\n\\end{Coqsrc}\n\n\nThe following function computes $\\canonseq{\\alpha}{i}$, except for the case $i=0$, where it simply returns $0$\\;\\footnote{This restriction did not prevent us from proving all the main theorems of~\\cite{KS81, KP82}. Nevertheless, in a future version of this development, we may define $\\canonseq{\\alpha}{0}$ exactly as \nin~\\cite{KS81}. But we are afraid this would be done at the cost of making some proofs much more complex.}.\n\n\\begin{Coqsrc}\n Definition canon alpha i := \n match i with 0 => zero | S j => canonS alpha j end. \n\\end{Coqsrc}\n\nFor instance \\coq's computing facilities allow us to verify the equalities\\linebreak \n\\mathcolor{$\\canonseq{\\omega^\\omega}{3} = \\omega^3$} and\n\\mathcolor{$\\canonseq{\\omega^\\omega*3}{42} = \\omega^\\omega*2 + \\omega^{42}$}.\n\n\n\\begin{Coqsrc}\nCompute (canon (omega ^ omega) 3).\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = phi0 (FS 2) : T1\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nExample canon3 : canon (omega ^ omega) 3 = omega ^ 3.\nProof. reflexivity. Qed.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nCompute pp (canon (omega ^ omega * 3) 42). \n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = (omega ^ omega * 2 + omega ^ 42)%pT1\n : ppT1\n\\end{Coqanswer}\n\n\\index{Projects}\n\\begin{project}\nMany lemmas presented in this chapter were stated and proved before the introduction of \nthe type class \\texttt{ON} of ordinal notations, and in particular its instance \\texttt{Epsilon0}.\nThus definitions and lemmas refer to the type \\texttt{T1} of possibly not well-formed terms.\nThis should be fixed in a future version.\n\\end{project}\n\n\n\\subsection{Basic Properties of Canonical Sequences}\n\nWe did not try to prove that our definition really implements Ketonen and Solovay's \\cite{KS81}'s canonical sequences. The most important is that we were able to prove the \nabstract properties of canonical sequences that are really used in our proof. The complete proofs are in the module\n~\\href{../theories/html/hydras.Epsilon0.Canon.html}{Epsilon0.Canon}\n\n\nProving the equality $\\canonseq{\\alpha+1}{i}=\\alpha$ is not \nas simple as suggested by the equations of definition~\\ref{def:canonseq-math}\\,.\nNevertheless, we can prove it by plain structural induction on $\\alpha$.\n\n\\begin{Coqsrc}\nLemma canonS_succ i alpha :\n nf alpha -> canonS (succ alpha) i = alpha.\nProof.\n induction alpha.\n (* ... *)\n\\end{Coqsrc}\n\n\\subsubsection{Canonical Sequences and the Order $<$}\n\n\\index{Maths!Transfinite induction}\n\nWe prove by transfinite induction over $\\alpha$ that $\\canonseq{\\alpha}{i+1}$ is an ordinal strictly less than $\\alpha$ (assuming $\\alpha\\not=0$). This property allows us to use the function \\texttt{canonS} and its derivates in function definitions by transfinite recursion.\n\n\\label{lemma:canonS_LT}\n\\begin{Coqsrc}\nLemma canonS_LT i alpha :\n nf alpha -> alpha <> zero -> canonS alpha i t1< alpha.\n\\end{Coqsrc}\n\n\n\\subsubsection{Limit Ordinals are Really Limits}\nThe following theorem states that any limit ordinal $\\lambda<\\epsilon_0$ \nis the limit of the sequence \\showmath{\\canonseq{\\lambda}{i}\\;(1\\le i)}.\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Canon.html\\#canonS_limit_strong}{Epsilon0.Canon}}\n\n\n\\begin{Coqsrc}\nLemma canonS_limit_strong (lambda : T1) : \n nf lambda ->\n limitb lambda ->\n forall beta, beta t1< lambda ->\n {i:nat | beta t1< canonS lambda i}.\n\nProof.\n transfinite_induction_LT lambda.\n (* ... *)\nDefined.\n\\end{Coqsrc}\n\n\\label{lemma:canonS-limit}\n\n\nNote the use of \\coq's \\texttt{sig} type in the theorem's statement, which\nrelates the boolean function \\texttt{limitb} defined on the \\texttt{T1} data-type with a constructive view of the limit of a sequence: for any $\\beta<\\lambda$, we can compute an item of the canonical sequence of $\\lambda$ which is greater than $\\beta$.\nWe can also state directly that $\\lambda$ is a (strict) least upper bound of the elements of its canonical sequence.\n\n\n\\begin{Coqsrc}\nLemma canonS_limit_lub (lambda : T1) :\n nf lambda -> limitb lambda ->\n strict_lub (fun i => canonS lambda i) lambda.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\\label{exo:simply-typed-canonseq}\nInstead of using the \\texttt{sig} type, define a simply typed function that, given two ordinals $\\alpha$ and $\\beta$, returns a natural number $i$ such that, if $\\alpha$ is a limit ordinal and $\\beta<\\alpha$, then $\\beta< \\canonseq{\\alpha}{i+1}$. Of course, you will have to prove the correctness of your function.\n\\end{exercise}\n\n\n\n\n\n\n\\section{Accessibility inside \\texorpdfstring{$\\epsilon_0$}{epsilon0} : Paths}\n\\index{Maths!Accessibility inside epsilon0}\n\\label{sect:pathes-intro}\n\nLet us consider a kind of accessibility problem inside $\\epsilon_0$: given two ordinals $\\alpha$ and $\\beta$, where $\\beta<\\alpha<\\epsilon_0$, find a \\emph{path} consisting of a finite sequence $\\gamma_0=\\alpha,\\dots,\\gamma_l=\\beta$,\nwhere, for every $i alpha <> zero /\\ beta = canonS alpha i.\n\nDefinition transition i : relation T1 :=\n match i with 0 => fun _ _ => False | S j => transition_S j end.\n\nInductive path_to (beta: T1) : list nat -> T1 -> Prop :=\n path_to_1 : forall (i:nat) alpha , \n i <> 0 ->\n transition i alpha beta ->\n path_to beta (i::nil) alpha\n| path_to_cons : forall i alpha s gamma,\n i <> 0 ->\n transition i alpha gamma ->\n path_to beta s gamma ->\n path_to beta (i::s) alpha.\n\\end{Coqsrc}\n\n\\begin{remark}\nThe definition above is parameterized with the \\emph{destination} of the path and indexed by the origin, hence the name \\texttt{path\\_to}. The rationale behind this choice is a personal preference of the developer for the kind of eliminators generated by \\coq{} in this case. The symmetric option could have been also considered (see also Remark~\\vref{remark:transitive-closure}).\n\\end{remark}\n\n\n\n\\begin{remark}\nIn the present version of our library, we use a variant \\texttt{path\\_toS} of\n\\texttt{path\\_to}, where the proposition\n(\\texttt{path\\_toS $\\beta$ $s$ $\\alpha$}) is equivalent to\n(\\texttt{path\\_to $\\beta$ (shift $s$) $\\alpha$}). This variant is scheduled to be deprecated.\n\\end{remark}\n\n\\index{Exercises}\n\n\\begin{exercise}\nWrite a tactic for solving goals of the form (\\texttt{path\\_to $\\beta$ $s$ $\\alpha$})\nwhere $\\alpha$, $\\beta$ and $s$ are closed terms. \nYou should solve automatically the following goals:\n\n\\begin{Coqsrc}\n path_to omega (2::2::2::nil) (omega * 2).\n\n path_to omega (3::4::5::6::nil) (omega * 2).\n\n path_to zero (interval 3 14) (omega * 2).\n\n path_to zero (repeat 3 8) (omega * 2).\n\\end{Coqsrc}\n\n\\end{exercise}\n\n\n\n\\subsection{Existence of a Path}\n\n\\index{Maths!Transfinite induction}\n\nBy transfinite induction on $\\alpha$, we prove that for any $\\beta<\\alpha$, \none can build a path from $\\alpha$ to $\\beta$ (in other terms, $\\beta$ is accessible from $\\alpha$).\n\n\\begin{Coqsrc}\nLemma LT_path_to (alpha beta : T1) :\n beta t1< alpha -> {s : list nat | path_to beta s alpha}.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}[continuation of exercise~\\vref{exo:simply-typed-canonseq}]\nDefine a simply typed function for computing a path from $\\alpha$ to $\\beta$.\n\\end{exercise}\n\n\n\\noindent \nFrom the lemma \\texttt{canonS\\_LT}~\\vref{lemma:canonS_LT}, we can convert any path into an inequality on ordinals (by induction on paths).\n\n\n\\begin{Coqsrc}\nLemma path_to_LT beta s alpha :\n path_to beta s alpha -> nf alpha -> beta t1< alpha.\n\\end{Coqsrc}\n\n\\subsection{Paths and Hydra Battles}\n\\label{KS-o2h}\n\nIn order to apply our knowledge about ordinal numbers less than $\\epsilon_0$ to the study of hydra battles, we define an injection\nfrom the interval $[0,\\epsilon_0)$ into the type \\texttt{Hydra}.\n\n\\vspace{4pt}\n\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.O2H.html}{Hydra.O2H}}\n\n\n\\begin{Coqsrc}\nFixpoint iota (alpha : T1) : Hydra :=\n match alpha with\n | T1.zero => head\n | ocons gamma n beta => \n node (hcons_mult (iota gamma) (S n) (iotas beta))\n end \nwith iotas (alpha : T1) : Hydrae :=\n match alpha with\n | T1.zero => hnil\n | ocons alpha0 n beta => \n hcons_mult (iota alpha0) (S n) (iotas beta)\n end.\n\\end{Coqsrc} \n\n\n\n\nFor instance Fig.~\\ref{fig:iota-example} shows the image by $\\iota$ of the ordinal \\textcolor{black}{$\\omega^{\\omega+2}+\\omega^\\omega \\times 2 + \\omega + 1$}\n\n \\begin{figure}[htb]\n\\centering\n\\begin{tikzpicture}[very thick, scale=0.3]\n\\node (foot) at (10,0) {$\\bullet$};\n\\node (N1) at (2,2) {$\\bullet$};\n\\node (N2) at (10,2) {$\\bullet$};\n\\node (N22) at (7,2) {$\\bullet$};\n\\node (N3) at (14,2) {$\\bullet$};\n\\node (N4) at (18,2) {$\\Smiley[2][green]$};\n\\node (N5) at (0,4) {$\\bullet$};\n\\node (N6) at (2,5) {$\\Smiley[2][green]$};\n\\node (N7) at (4,6) {$\\Smiley[2][green]$};\n\\node (N88) at (7,4) {$\\bullet$};\n\\node (N8) at (10,4) {$\\bullet$};\n\\node (N9) at (14,6) {$\\Smiley[2][green]$};\n\\node (N10) at (0,8) {$\\Smiley[2][green]$};\n\\node (N11) at (10,7) {$\\Smiley[2][green]$};\n\\node (N111) at (7,7) {$\\Smiley[2][green]$};\n\\draw (foot) to [bend left=10] (N1);\n\\draw (foot) -- (N2);\n\\draw (foot) -- (N22);\n\\draw (foot) -- (N3);\n\\draw (foot) -- (N4);\n\\draw (N1) to (N5);\n\\draw (N1) to [bend left=10] (N6);\n\\draw (N1) to [bend right=20] (N7);\n\\draw (N2) to (N8);\n\\draw (N22) to (N88);\n\\draw (N8) to (N11);\n\\draw (N88) to (N111);\n\\draw (N3) to (N9);\n\\draw (N5) to (N10);\n\\end{tikzpicture}\n\\caption{The hydra $\\iota(\\omega^{\\omega+2}+\\omega^\\omega \\times 2 + \\omega + 1$) \\label{fig:iota-example}}\n\n\\end{figure}\n\n\nThe following lemma (proved in ~\\href{../theories/html/hydras.Hydra.O2H.html}{Hydra.O2H.v}) maps canonical sequences to rounds of hydra battles.\n\n\n\\label{lemma:canonS-iota}\n\n\\begin{Coqsrc}\nLemma canonS_iota i alpha :\n nf alpha -> alpha <> 0 ->\n iota alpha -1-> iota (canonS alpha i).\n\\end{Coqsrc}\n \n\nThe next step of our development extends this relationship to\nthe order $<$ on $[0,\\epsilon_0)$ on one side, and hydra battles on the other side.\n\n\n\\begin{Coqsrc}\nLemma path_to_battle alpha s beta :\n path_to beta s alpha -> nf alpha ->\n iota alpha -+-> iota beta.\n\\end{Coqsrc}\n\nAs a corollary, we are now able to transform any inequality $\\beta<\\alpha<\\epsilon_0$ into a (free) battle.\n\n\\begin{Coqsrc}\nLemma LT_to_battle alpha beta :\n beta t1< alpha -> iota alpha -+-> iota beta.\n\\end{Coqsrc}\n\n\\section{A Proof of Impossibility}\n\nWe now have the tools for proving that there exists no variant bounded by some $\\mu<\\epsilon_0$ for proving the termination of all battles. The proof we are going to show is a proof by contradiction. It can\n be considered as a generalization of the\nproofs described in sections~\\vref{omega-case} and \\vref{omega2-case}.\n\n\n\nIn the module \\href{../theories/html/hydras.Hydra.Epsilon0_Needed_Generic.html}{Hydra.Epsilon0\\_Needed\\_Generic}, we assume there exists some variant $m$ bounded by some ordinal $\\mu<\\epsilon_0$. This part of the development is parameterized by some class $B$ of battles, which will be instantiated later to \\texttt{free} or \\texttt{standard}.\n\n\n\n\n\\begin{Coqsrc}\nClass BoundedVariant (B:Battle) :=\n {\n mu:T1 ;\n m: Hydra -> T1;\n mu_nf: nf mu;\n Hvar: Hvariant T1_wf B m;\n m_bounded: forall h, m h t1< mu\n }.\n\\end{Coqsrc}\n\nLet us assume there exists such a variant:\n\n\\begin{Coqsrc}\nSection Bounded.\n Context (B: Battle) (Hy : BoundedVariant B).\n\n Hypothesis m_decrease : forall i h h',\n round_n i h h' -> m h' t1< m h.\n\\end{Coqsrc}\n\n\\label{remark:m-decrease}\n\\begin{remark}\n The hypothesis \\texttt{m\\_decrease} is not provable in general, but is satisfied by\nthe \\texttt{free} and \\texttt{standard} kinds of battles. This trick allows to \n``factorize'' our proofs of impossibility.\n\\end{remark}\n\n\\index{Maths!Transfinite induction}\n\nFirst, we prove that $m(\\iota(\\alpha))$ is always greater than or equal to $\\alpha$, by transfinite induction over $\\alpha$.\n\n\\begin{Coqsrc}\nLemma m_ge_0 alpha: nf alpha -> alpha t1<= m (iota alpha).\n\\end{Coqsrc}\n\n\n\\begin{itemize}\n\\item If $\\alpha=0$, the inequality trivially holds\n\\item If $\\alpha$ is the successor of some ordinal $\\beta$, the inequality $\\beta \\leq m(\\iota(\\beta))$ holds (by induction hypothesis). But the hydra $\\iota(\\alpha)$ is transformed in one round into \n$\\iota(\\beta)$, thus $m(\\iota(\\beta)) small_h.\n\\end{Coqsrc}\n\\item From the hypotheses on $m$, we infer:\n\n\\begin{Coqsrc}\nLemma m_lt : m small_h t1< m big_h.\n\\end{Coqsrc}\n\n\n\\item From lemmas \\texttt{m\\_ge} and \\texttt{m\\_lt}, and the irreflexivity of $<$, we get a contradiction. \n\n \\begin{Coqsrc}\nTheorem Impossibility_free : False.\n\nEnd Impossibility_Proof.\n\\end{Coqsrc}\n\n\n\\end{enumerate}\n\nWe have now proved there exists no bounded variant for the class of free battles.\n\n \n\\begin{Coqsrc}\nCheck Impossibility_free.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n Impossibility_free\n : BoundedVariant free -> False\n\\end{Coqanswer}\n%%% ICI\n \n\n\n\n\\section{The Case of Standard Battles}\n\\label{sec:standard-intro}\\label{std-case}\nOne may wonder if our theorem holds also in the framework of standard battles. Unfortunately, its proof relies on the lemma \\texttt{LT\\_to\\_round\\_plus} of\nModule~\\href{../theories/html/hydras.Hydra.O2H.html}{Hydra.O2H}.\n\n\\begin{Coqsrc}\nLemma LT_to_round_plus alpha beta :\n beta t1< alpha -> iota alpha -+-> iota beta.\n\\end{Coqsrc}\n\nThis lemma builds a battle out of any inequality $\\beta<\\alpha$. \nIt is a straightforward application of \\texttt{LT\\_path\\_to} of\nModule~\\href{../theories/html/hydras.Epsilon0.Paths.html}{Epsilon0.Paths}:\n\n\\begin{Coqsrc}\nLemma LT_path_to (alpha beta : T1) :\n beta t1< alpha -> {s : list nat | path_to beta s alpha}.\n\\end{Coqsrc}\n\nThe sequence $s$, used to build the sequence of replication factors of the battle depends on \n$\\beta$, so we cannot be sure that the generated battle is a genuine standard battle.\n\n\nThe solution of this issue comes once again from Ketonen and Solovay's article~\\cite{KS81}. Instead of considering plain paths, i.e. sequences \n$\\alpha_0=\\alpha,\\alpha_1,\\dots,\\alpha_k=\\beta$ where $\\alpha_{j+1}$ is equal\nto $\\canonseq{\\alpha_j}{i_j}$ where $i_j$ is \\emph{any} natural number, \nwe consider various constraints on these sequences.\nIn particular, a path is called \\emph{standard} if $i_{j+1} = i_j + 1$ for every $j T1 -> Prop :=\n std_1 : forall i alpha, \n beta = canon alpha i -> j = S i ->\n standard_pathR j beta i alpha\n| std_S : forall i alpha, \n standard_pathR j beta (S i) (canon alpha i) ->\n standard_pathR j beta i alpha.\n\nDefinition standard_path i alpha j beta := \n standard_pathR j beta i alpha.\n\\end{Coqsrc}\n\nIn the mathematical text and figures, we shall use the notation \n$\\alpha \\xrightarrow[i,j]{}\\beta$ for the proposition \n(\\texttt{standard\\_path $i$ $\\alpha$ $j$ $\\beta$}).\nIn~\\cite{KS81} the notation is\n$\\alpha \\xrightarrow[i]{*}\\beta$\nfor \nthe proposition $\\exists j, i0$. \n\n\nLike in~\\cite{KS81}, we shall use the notation $\\alpha \\xrightarrow[i]{} \\beta$ for denoting such a path, also called an $i$-path.\n\n\\begin{Coqsrc}\nDefinition const_pathS i :=\n clos_trans_1n T1 (fun alpha beta => beta = canonS alpha i).\n\nDefinition const_path i alpha beta :=\n match i with\n 0 => False\n | S j => const_pathS j alpha beta\nend.\n\\end{Coqsrc}\n\n% Paths with a given index can be effectively computed.\n% Given $i$, $\\alpha$ and $l$, the following function returns the ordinal $\\beta$ such that there exists a path \n% $\\alpha \\xrightarrow [i+1] {} \\beta$ of length $l$. \n\n% \\begin{Coqsrc}\n% Fixpoint const_funS (i:nat)(alpha : T1)(l:nat): T1 :=\n% match l\n% with\n% | 0 => alpha\n% | S m => const_funS i (canonS i alpha) m\n% end.\n% \\end{Coqsrc}\n\n% The following computations show applications of \\texttt{constS\\_fun} to the \n% ordinal $\\omega^\\omega$, with various values of $i$ and $l$.\n\n% \\begin{Coqsrc}\n% Compute (const_funS 2 (omega ^omega) 55).\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = zero\n% : T1 \n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute pp (const_funS 2 (omega ^omega) 15).\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = (omega ^ 2 * 2)%pT1\n% : ppT1 \n% \\end{Coqanswer}\n\n\n% \\begin{Coqsrc}\n% Compute pp (const_funS 4 (omega^omega) 100).\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = (omega ^ 4 * 4 + omega ^ 3 * 4 + omega ^ 2 + omega * 4 + 4)%pT1\n% : ppT1\n% \\end{Coqanswer}\n\n\n\n\nA most interesting property of $i$-paths is that we can ``upgrade'' their index, as stated by K.\\&S.'s Corollary 12.\n\n\\index{Maths!Transfinite induction}\n\n\\begin{Coqsrc}\nCorollary Cor12 (alpha : T1) : nf alpha ->\n forall beta i n, beta t1< alpha ->\n i < n ->\n const_pathS i alpha beta ->\n const_pathS n alpha beta.\nProof.\n transfinite_induction_lt alpha.\n (* (long) proof skipped *)\n\\end{Coqsrc}\n\nWe also use a version of \\texttt{Cor12} with large inequalities.\n\n\n\\begin{Coqsrc}\nCorollary Cor12_1 (alpha : T1) :\n nf alpha ->\n forall beta i n, beta t1< alpha ->\n i <= n ->\n const_pathS i alpha beta ->\n const_pathS n alpha beta.\n\\end{Coqsrc}\n\n\n\\subsubsection{Sketch of Proof of \\texttt{Cor12}}\n\\index{Maths!Transfinite induction}\n\nWe prove this lemma by transfinite induction on $\\alpha$.\nLet us consider a path $\\alpha \\xrightarrow [i]{} \\beta$ $(i>0)$. Its first step is\nthe pair $(\\alpha,\\canonseq{\\alpha}{i})$, We have $\\canonseq{\\alpha}{i}<\\alpha$ and\n$\\canonseq{\\alpha}{i} \\xrightarrow [i]{} \\beta$. \nLet $n$ be any natural number such that $n>i$.\nBy the induction hypothesis, there exists a path $\\canonseq{\\alpha}{n} \\xrightarrow[i]{} \\beta$.\n\\begin{itemize}\n\\item If $\\alpha$ is a successor ordinal $\\gamma+1$, then $\\canonseq{\\alpha}{n} =\n\\canonseq{\\alpha}{i}=\\gamma$. Thus we have a path \n$\\alpha \\xrightarrow [n]{} \\gamma \\xrightarrow [n]{} \\beta$\n\\item If $\\alpha$ is a limit ordinal, we apply the following theorem (numbered \\texttt{2.4} in Ketonen and Solovay's article). \n\n% \\begin{theorem}\n% Let $\\lambda$ be a limit ordinal, then for any pair of indices $0\n limitb lambda ->\n forall i j, (i < j)%nat ->\n const_pathS 0 (canonS lambda j)\n (canonS lambda i). \n \\end{Coqsrc}\n\n We build the following paths :\n\n \\begin{enumerate}\n \\item $\\alpha \\xrightarrow[n]{} \\canonseq{\\alpha}{n}$\n \\item $\\canonseq{\\alpha}{n} \\xrightarrow[1]{} \\canonseq{\\alpha}{i}$ (by \\texttt{Theorem\\_2\\_4}),\n\\item $\\canonseq{\\alpha}{n} \\xrightarrow[n]{} \\canonseq{\\alpha}{i}$ (applying the induction hypothesis to the preceding path);\n\\item $\\canonseq{\\alpha}{i} \\xrightarrow[n]{} \\beta$ (applying the induction hypothesis)\\item $\\alpha \\xrightarrow[n]{} \\beta$ (by composition of 1, 3, and 4).\n\n\n \\end{enumerate}\n\n\n\\end{itemize}\n\n\n\n\n\n\\begin{remark}\n \\texttt{Cor12} ``casts'' $i$-paths into $n$-paths for any $n>i$.\nBut the obtained $n$-path can be much longer than the original $i$-path.\nThe following exercise will give an idea of this increase. \n\\end{remark}\n\n\\index{Exercises}\n\\begin{exercise}\n Prove that the length of the $i+1$-path from\n $\\omega^\\omega$ to $\\omega^i$ is $1 + (i+1)^{(i+1)}$, for any $i$. Note that the $i$-path from\n $\\omega^\\omega$ to $\\omega^i$ is only one step long.\n \\end{exercise}\n\n\nWhy is \\texttt{Cor12} so useful? \nLet us consider two ordinals $\\beta<\\alpha<\\epsilon_0$. By induction on $\\alpha$,\nwe decompose any inequality $\\beta<\\alpha$ into $\\beta < \\canonseq{\\alpha}{i}< \\alpha$, where $i$ is some integer. Applying collorary \\texttt{Cor12'} we build a $n$-path from $\\beta$ to $\\alpha$,\nwhere $n$ is the maximum of the indices $i$ met in the induction.\n\n Lemma 1, Section 2.6 of~\\cite{KS81} is naturally expressed in terms of \\coq's\n\\verb@sig@ construct.\n\n\\label{lemma:L-2_6-1}\n\\index{Coq!Techniques!Sigma types}\n\n\\begin{Coqsrc}\nLemma Lemma2_6_1 (alpha : T1) : \n nf alpha -> forall beta, beta t1< alpha ->\n {n:nat | const_pathS n alpha beta}.\nProof.\n transfinite_induction alpha.\n (* ... *)\n\\end{Coqsrc}\n\n\n\nIntuitively, lemma \\texttt{L2\\_6\\_1} shows that if $\\beta<\\alpha<\\epsilon_0$, then there exists a battle from $\\iota(\\alpha)$ to $\\iota(\\beta)$ where the replication factor is constant, although large enough. \n\n\n\n\n\n\n\n\\subsection{Casting Paths with a Constant Index into a Standard Path}\n\n%%% traduire la v.f. (voir %%% A traduire %%%% )\n\nThe article~\\cite{KS81} contains \nthe following lemma, the proof of which is quite complex, which allows to simulate $i$-paths by $[i+1,j]$-paths, where $j$ is large enough.\n\n\n\\begin{Coqsrc}\n(* Lemma 1 page 300 of [KS] *)\n\nLemma constant_to_standard_path \n (alpha beta : T1) (i : nat):\n nf alpha -> const_pathS i alpha beta -> zero t1< alpha ->\n {l:nat | standard_path (S i) alpha j beta}.\n\\end{Coqsrc}\n\n \n\n\\subsubsection{Sketch of Proof of \\texttt{constant\\_to\\_standard\\_path}}\n\nOur proof follows the proof by Ketonen and Solovay, including its organization as a sequence of lemma. Since it is a non-trivial proof, we will comment its main steps below.\n\n\\subsubsection*{Préliminaries}\n\n\nPlease note that, given an ordinal $\\alpha:\\texttt{T1}$, and two natural numbers $i$ and $l$, there exists at most a standard path $\\alpha \\xrightarrow [i,i+l]{*} \\beta$.\nThe following function computes $\\beta$ from $\\alpha$, $i$ and $l$.\n\n\\begin{Coqsrc}\nFixpoint standard_gnaw (i:nat)(alpha : T1)(l:nat): T1 :=\n match l with\n | 0 => alpha\n | S m => standard_gnaw (S i) (canon alpha i) m\n end.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\n Compute standard_gnaw 2 omega 15.\n(* = zero\n : T1 *)\nCompute pp (standard_gnaw 2 (omega^omega) 10).\n(*\n= (omega + 7)%pT1\n : ppT1\n*)\nCompute pp (standard_gnaw 4 (omega^omega) 100).\n(*\n = (omega ^ 3 * 4 + omega ^ 2 * 5 + omega * 3 + 39)%pT1\n : ppT1 *)\n\\end{Coqsrc}\n\n\\index{Maths!Transfinite induction}\n\nBy transfinite induction over $\\alpha$, we prove that the ordinal $0$ is reachable from any ordinal $\\alpha<\\epsilon_0$ by some standard path.\n\n\n\\begin{Coqsrc}\nLemma standard_path_to_zero :\n forall alpha i, nf alpha ->\n {j: nat | standard_path (S i) alpha j zero}.\n\\end{Coqsrc}\n\n\\paragraph*{}\nNoq, let us consider two ordinals $\\beta<\\alpha<\\epsilon_0$. Let $p$ be some $(n+1)$-path from $\\alpha$ to $\\beta$.\n\n\\begin{Coqsrc}\n Section Constant_to_standard_Proof.\n\n Variables (alpha beta: T1) (n : nat).\n Hypotheses (Halpha: nf alpha) (Hpos : zero t1< beta)\n (p : const_pathS n alpha beta).\n\\end{Coqsrc}\n\nApplying \\texttt{standard\\_path\\_to\\_zero}, $0$ is reachable from $\\alpha$ by some standard path (see figure~\\vref{fig:belle-preuve-1}).\n\n\\begin{figure}[h]\n \\centering\n \n\\begin{tikzpicture}[very thick, scale=0.25]\n\\node (alpha) at (0,0) {$\\alpha$};\n \\node (beta) at (32, 0){$\\beta$};\n \n\n \\draw[->, very thick,blue] (alpha)-- node [below]{$n+1$} node [above] {$+$} (beta);\n\n \\node (alpha1) at (5,5) {};\n \\node (alpha2) at (13,5) {};\n \\node (alpha3) at (20,5) {};\n \\node (alphalast) at (35,5) {};\n \\node (zero) at (45,0) {$0$};\n \\draw [->, dashed,very thick,blue] (alpha)-- node [below, rotate=40]{$n+1$} (alpha1);\n \\draw [->, dashed,very thick,blue] (alpha1)-- node [below]{$n+2$} (alpha2);\n \\draw [->, dashed,very thick,blue] (alpha2)-- node [below]{$n+3$} (alpha3);\n \n \\node (dots) at (24,5) {$\\dots$};\n \\draw [->, dashed, very thick,blue] (alphalast)-- node [below, rotate=-26]{$n+p+1$} (zero);\n\n\\end{tikzpicture}\n\\caption{A nice proof (1)}\n \\label{fig:belle-preuve-1}\n\\end{figure}\n\n\n\\paragraph*{}\n\n\n\n\nSince comparison on \\texttt{T1} is decidable, one can compute the last step $\\gamma$ of the standard path from $(\\alpha,n+1)$ such that $\\beta\\leq \\gamma$.\nLet $l$ be the length of the path from $\\alpha$ to $\\gamma$. \nThis step of the proof is illustrated in figure~\\vref{fig:belle-preuve-2}.\n\n\n\n\\begin{figure}[h]\n \\centering\n \n\n\\begin{tikzpicture}[very thick, scale=0.25]\n\\node (alpha) at (0,0) {$\\alpha$};\n \\node (beta) at (32, 0){$\\beta$};\n \n\n\n\n \\node (alpha1) at (5,5) {};\n \\node (alpha2) at (13,5) {};\n \\node (dots) at (17,5) {$\\ldots$};\n \\node (alpha3) at (20,5) {};\n \\node (gamma) at (24,0) {$\\gamma$};\n \\node (delta) at (38,0) {$\\delta$};\n \\draw [->, dashed,very thick,blue] (alpha)-- node [below,rotate=35]{$n+1$} (alpha1);\n \\draw [->, dashed,very thick,blue] (alpha1)-- node [below]{$n+2$} (alpha2);\n \\draw [->, dashed,very thick,blue] (alpha3)-- node [below,rotate = -48]{\\tiny $n+l$} (gamma);\n \\draw [->, dashed, blue] (gamma) to [bend left=80] node [below]{$n+l+1$} (delta);\n \\draw[->, very thick,blue] (alpha) to [bend right=34] node [below]{$n+1$} node [above] {$+$} (beta);\n \\draw[thick] (alpha)-- (gamma);\n \\draw[thick] (gamma)-- node [above] {$\\geq$} (beta);\n \\draw[thick] (beta)-- node [above] {$>$} (delta);\n\\end{tikzpicture}\n\n\\caption{A nice proof (2)}\n \\label{fig:belle-preuve-2}\n\\end{figure}\n\n\\paragraph*{}\n\n\\begin{itemize}\n\\item If $\\beta=\\gamma$, its OK! We have got a standard path\nfrom \n$\\alpha$ to $\\beta$ with successive indices $n+1, n+2, \\dots, n+l+1$\n\n\\item Otherwise, $\\beta < \\gamma$. Let us consider $\\delta=\\canonseq{\\gamma}{n+l+1}$.\nBy applying several times lemma \\texttt{Cor12}, one converts every path of Fig~\\ref{fig:belle-preuve-2} into\n a $n+l+1$-path (see figure~\\ref{fig:belle-preuve-3}).\n\n\nBut $\\gamma$ is on the $n+l+1$-path from $\\alpha$ to $\\beta$.\nAs shown by figure~\\vref{fig:fin-belle-preuve}, the ordinal $\\delta$, reachable from\n$\\gamma$ in one single step, must be greater than or equal to $\\beta$, which contradicts our hypothesis $\\beta < \\gamma$.\n\n\n\\begin{figure}[h]\n \\centering\n \n\\begin{tikzpicture}[very thick, scale=0.25]\n\\node (alpha) at (0,0) {$\\alpha$};\n \\node (beta) at (32, 0){$\\beta$};\n \\node (alpha1) at (5,5) {};\n \\node (alpha2) at (13,5) {};\n \\node (dots) at (17,5) {$\\ldots$};\n \\node (alpha3) at (20,5) {};\n \\node (gamma) at (24,0) {$\\gamma$};\n \\node (delta) at (38,0) {$\\delta$};\n \\draw [->, dashed,very thick,blue] (alpha)-- node [below, rotate = 40] {\\tiny $n+l+1$} node [above, rotate = 40]{\\tiny $+$} (alpha1);\n \\draw [->, dashed,very thick,blue] (alpha1)-- node [below]{\\tiny $n+l+1$} node [above]{\\tiny $+$} (alpha2);\n \\draw [->, dashed,very thick,blue] (alpha3)-- node [below, rotate = -48]{\\tiny $n+l+1$} node [above, rotate = -36]{\\tiny $+$} (gamma);\n \\draw [->, dashed, blue] (gamma) to [bend left=80] node [below]{\\tiny $n+l+1$} node [above]{\\color{red} $1$} (delta);\n \\draw[->, very thick,blue] (alpha) to [bend right=34] node [below]{\\small $n+l+1$} node [above] {\\tiny $+$} (beta);\n \\draw[thick] (gamma)-- node [above]{\\color{red} $>$}(beta);\n \\draw[thick] (alpha)-- (gamma);\n \n \\draw[thick] (beta)-- node [above] {$>$} (delta);\n\n \n \n\\end{tikzpicture}\n\n\\caption{A nice proof (3)}\n \\label{fig:belle-preuve-3}\n\\end{figure}\n\n\n\\begin{figure}[h]\n \\centering\n\\begin{tikzpicture}[very thick, scale=0.25]\n\\node (alpha) at (0,0) {$\\alpha$};\n \\node (beta) at (32, 0){$\\beta$};\n \n \\node (alpha1) at (5,5) {};\n \\node (alpha2) at (13,5) {};\n \\node (dots) at (15,5) {$\\ldots$};\n \\node (alpha3) at (18,5) {};\n \\node (gamma) at (24,0) {$\\gamma$};\n \\node (delta) at (42,0) {$\\delta$};\n \\draw [->, dashed,very thick,blue] (alpha)-- node [below, rotate = 40] {\\tiny $n+l+1$} node [above, rotate = 40]{\\tiny $+$} (alpha1);\n \\draw [->, dashed,very thick,blue] (alpha1)-- node [below]{\\tiny $n+l+1$} node [above]{\\tiny $+$} (alpha2);\n \\draw [->, dashed,very thick,blue] (alpha3)-- node [below, rotate = -36]{\\tiny $n+l+1$} node [above, rotate = -36]{\\tiny $+$} (gamma);\n \\draw [->, dashed, blue] (gamma) to [bend left=80] node [below]{\\small $n+l+1$} node [above]{\\color{red} $1$} (delta);\n \\draw[->, very thick,blue] (alpha) to [bend right=34] node [below]{\\small $n+l+1$} node [above] {\\tiny $+$} (beta);\n \\draw[thick] (alpha)-- (gamma);\n \n \\draw[thick] (gamma)-- node [below]{\\tiny $n+l+1$} node [above]{\\color{red} $+$}(beta);\n \\draw[thick] (beta)-- node [above] {$>$} (delta);\n\n\\end{tikzpicture}\n\n\\caption{A nice proof (4)}\n \\label{fig:fin-belle-preuve}\n\\end{figure}\n\n\n\\end{itemize}\n The only possible case is thus $\\beta=\\gamma$, so we have got a standard path from $\\alpha$ to $\\beta$.\n\n\n\\begin{Coqsrc}\n Lemma constant_to_standard_0 : \n {l : nat | standard_fun (S n) alpha l = beta}.\n (* ... *)\n\nEnd Constant_to_standard_Proof.\n\\end{Coqsrc}\n\nHere is the full statement of the conversion from constant to standard paths.\n\n\\begin{Coqsrc}\nLemma constant_to_standard_path \n (alpha beta : T1) (i : nat):\n nf alpha -> const_pathS i alpha beta -> zero t1< alpha ->\n {j:nat | standard_path (S i) alpha j beta}.\n\\end{Coqsrc}\n\n\n\n\nApplying \\texttt{Lemma2\\_6\\_1} and \\texttt{constant\\_to\\_standard\\_path}, we get the following corollary.\n\n\\begin{Coqsrc}\nCorollary LT_to_standard_path (alpha beta : T1) :\n beta t1< alpha ->\n {n : nat & {j:nat | standard_path (S n) alpha j beta}}.\n\\end{Coqsrc}\n\n\n\\subsection{Back to Hydras}\n\\label{sec:standard-battles-cases}\nWe are now able to complete our proof that there exists no bounded variant for proving the termination of standard hydra battles. This proof can\nbe consulted in the module \n\\url{../theories/html/hydras.Hydra.Epsilon0_Needed_Std.html}.\nPlease note that it has the same global structure as in section\\ref{sec:free-battles-case} \n% ICI !\nApplying the lemmas \\texttt{Lemma2\\_6\\_1} of the module \n\\href{../theories/html/hydras.Epsilon0.Paths.html\\#Lemma2_6_1}%\n{\\texttt{Lemma2\\_6\\_1}} and \n\\href{../theories/html/hydras.Epsilon0.Paths.html\\#constant_to_standard_path}%\n{\\texttt{constant\\_to\\_standard\\_path}},\nwe can convert any inequality $\\beta<\\alpha<\\epsilon_0$ into a standard path from\n$\\alpha$ to $\\beta$, then into a fragment of a standard battle from \n$\\iota(\\alpha)$ to $\\iota(\\beta)$.\n\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Hydra.Epsilon0_Needed_Std.html\\#LT_to_standard_battle}{Hydra.Epsilon0\\_Needed\\_Std}}\n\n\\begin{Coqsrc}\nLemma LT_to_standard_battle :\n forall alpha beta,\n beta t1< alpha ->\n exists n i, battle standard n (iota alpha) i (iota beta).\n\\end{Coqsrc}\n\n\nNext, please consider the following context:\n\n\\begin{Coqsrc}\nSection Impossibility_Proof.\n \n Context (Var : BoundedVariant standard).\n \\end{Coqsrc}\n\nIn the same way as for free battles, we import a large inequality \nfrom \nthe module \\href{../theories/html/hydras.Hydra.Epsilon0_Needed_Generic.html}{Hydra.Epsilon0\\_Needed\\_Generic}.\n\n\n\\begin{Coqsrc}\n Lemma m_ge : m big_h t1<= m small_h.\n\\end{Coqsrc}\n\n\\paragraph*{} If remains to prove the following strict inequality, in order to have a contradiction.\n\n\\begin{Coqsrc}\nLemma m_lt : m small_h t1< m big_h.\n\\end{Coqsrc}\n\n\n\\paragraph*{Sketch of proof:} Let us recall that $\\texttt{big\\_h} = \\iota(\\mu)$\n and $\\texttt{small\\_h} = \\iota (m (\\texttt{big\\_h}))$.\n\nSince $m(\\texttt{big\\_h})< \\mu$, there exists a standard path from $\\mu$ to\n$m(\\texttt{big\\_h})$, hence a standard battle from $\\iota(\\mu)$ to\n$\\iota(m(\\texttt{big\\_h}))$, i.e. from \\texttt{big\\_h} to \\texttt{small\\_h}.\n\nSince $m$ is assumed to be a variant for standard battles, we get the inequality $m(\\texttt{small\\_h}) < m(\\texttt{big\\_h})$.\n\n\n\n\\subsection{Remarks}\n\nWe are grateful to \n J. Ketonen and R. Solovay for the high quality of their explanations and proof details.\nOur proof follows tightly the sequence of lemmas in their article, with a focus on \nconstructive aspects.\nRoughly steaking, our implementation \\emph{builds}, out of a hypothetic \n variant $m$, bounded by some ordinal $\\mu<\\epsilon_0$, a hydra \\texttt{big\\_h} which verifies the impossible inequality $m(\\texttt{big\\_h})< m(\\texttt{big\\_h})$.\n\n\n\nOn may ask whether the preceding results are not too restrictive, since they \nrefer to a particular data type \\texttt{T1}.\nIn fact, our representation of ordinals strictly less than \n $\\epsilon_0$ is faithful to their mathematical definition, at least \nKurt Schütte's~\\cite{schutte}, as proved in Chapter~\\vref{chap:schutte}.\n(please see also \\href{../theories/html/hydras.Schutte.Correctness_E0.html}{the module \\texttt{Ordinals.Schutte.Correctness\\_E0}}).\n\nThus, we can infer that our theorems can be applied to any well order.\n\n\\index{Projects}\n\\begin{project}\nStudy a possible modification of the definition of a variant (for standard battles).\n\n\\begin{itemize}\n\\item The variant is assumed to be strictly decreasing \\emph{on configurations \nreachable from some initial configuration where the replication factor is equal to $0$}\n\\item The variant may depend on the number of the current round.\n\\end{itemize}\n\nIn other words, its type should be \\texttt{nat -> Hydra -> T1}, and it must \nverify the inequality $m\\, (S\\,i)\\, h' < m\\,i\\, h$ whenever the configuration \n$(i,h)$ is reachable from some initial configuration $(0,h_0)$\nand \\texttt{h} is transformed into \\texttt{h'} in the considered round.\nCan we still prove the theorems of section~\\ref{std-case} with this new definition?\n\n\\end{project}\n\n\n\n%---------------------------------------------------------------------\n\\chapter{Large Sets and Rapidly Growing Functions}\\label{chap:alpha-large}\n\n\n%\\section{Introduction}\n\nIn this chapter, we try to feel how long a standard battle can be.\nTo be precise, for any ordinal $\\alpha<\\epsilon_0$ and any positive integer $k$,\nwe give a minoration of the number of steps of a standard battle which\nstarts with the hydra $\\iota(\\alpha)$ and the replication factor $k$.\n\nWe express this number in terms of the Hardy hierarchy of fast-growing \nfunctions~\\cite{BW85, Wainer1970, KS81, Promel2013}.\n From the \\coq{} user's point of view, such functions are very \nattractive: they are defined as functions in \\gallina{}, and we can apply them \\emph{in theory}, but they are so complex that you will never be able to look at the result of the computation.\n Thus, our knowledge on these functions must rely on \\emph{proofs}. In our development, we use often the rewriting rules generated by \\coq's \\texttt{Equations} plug-in.\n\n\n\\section{Definitions}\n\n%\\subsection{Definition}\n\n\\begin{definition}\nLet $0<\\alpha<\\epsilon_0$ be any ordinal, and $s$ a sequence of strictly positive natural numbers. \nWe say that $s$ is \\emph{minimally $\\alpha$-large} (in short:\n\\emph{$\\alpha$-mlarge}) if $s$ if $s$ is $\\alpha$-large \n and every strict prefix of $s$ leads to a non-zero ordinal (\\emph{cf} Sect.~\\vref{sect:path-to-def}).\n\n\\index{Maths!Large sequences}\n\\index{Maths!Minimal large sequences}\n\n\\end{definition}\n\n\\vspace{4pt}\n\n\\noindent\n\\emph{From Module~ \\href{../theories/html/hydras.Epsilon0.Large_Sets.html\\#mlarge}{Epsilon0.Large\\_Sets}}\n\n\n\\index{Predicates!mlarge @ mlarge (minimal large sequences)}\n\\begin{Coqsrc}\nDefinition mlarge alpha (s:list nat) := path_to zero s alpha.\n\\end{Coqsrc}\n\n\n\n\\begin{remark}\n Ketonen and Solovay~\\cite{KS81} consider large finite \\emph{sets} of natural numbers, but they are mainly used as sequences. Thus, we chosed to represent them explicitely as (sorted) lists. \n\nThey also consider large (but not minimally large) sets. They can be defined in\n\\coq{} as follows:\n\n\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Paths.html\\#gnaw}{Epsilon0.Paths}}\n\n\\begin{Coqsrc}\nFixpoint gnaw (alpha : T1) (s: list nat) :=\n match s with\n | nil => alpha\n | (0::s') => gnaw alpha s'\n | (S i :: s') => gnaw (canonS i alpha) s'\n end.\n\nDefinition large alpha (s:list nat) := gnaw alpha s = zero.\n\\end{Coqsrc}\n\\end{remark}\n\n\n\nLet us consider two integers $k$ and $l$, such that $0 nil -> ~ In 0 (x::s) ->\n mlarge alpha (x::s) <-> gnaw alpha (but_last x s) = one.\n \\end{Coqsrc}\n\n\nFor instance, we can verify that the interval $[6,510]$ is $\\omega^2$-mlarge.\n\n\\vspace{4pt}\n \\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Large_Sets_Examples.html}{Epsilon0.Large\\_Sets\\_Examples}}\n\\begin{Coqsrc}\nExample Ex1 : mlarge (omega * omega) (interval 6 510).\n\\end{Coqsrc}\n\n\n\\section{The Length of Minimal Large Sequences}\n\nNow, consider any natural number $k>0$. We would like to compute\na number $l$ such that the interval $[k,l]$ is $\\alpha$-mlarge. So, \nthe standard battle starting with $\\iota(\\alpha)$ and the replication factor $k$ will end after $(l-k+1)$ steps.\n\n\n\nFirst, we notice that this number $l$ exists, since the segment $[0,\\epsilon_0)$ is well-founded and $\\canonseq{\\alpha}{i}<\\alpha$ for any $i$ and $\\alpha>0$.\nMoreover, it is unique:\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Large_Sets.html}{Epsilon0.Large\\_Sets}}\n\\begin{Coqsrc}\nLemma mlarge_unicity alpha k l l' : \n mlarge alpha (interval (S k) l) ->\n mlarge alpha (interval (S k) l') ->\n l = l'.\n\\end{Coqsrc}\n\nThus, it seems obvious that there must exist a function, parameterized by $\\alpha$ which associates to any strictly positive integer $k$ the number $l$ such that\nthe interval $[k,l]$ is $\\alpha$-mlarge. It would be fine to write in \\gallina{} a definition like this:\n\n\\begin{Coqbad}\nFunction L (alpha: E0) (i:nat) : nat := ...\n\\end{Coqbad}\n\nBut we do not know how to fill the dots yet \\dots{} In the next section, we will \nuse \\coq{} to reason about the \\emph{specification} of \\texttt{L},\nprove properties of any function which satisfies this specification.\nIn Sect.~\\ref{sect:L-equations}, we use the \\texttt{coq-equations} plug-in\nto define a function \\texttt{L\\_}, and prove its correctness w.r.t. its specification.\n\n\n\n%\\subsection{Looking for a function definition}\n\nLet $0<\\alpha<\\epsilon_0$ be an ordinal term. We consider the functions which associate to each stritly positive integer $k$ the number $l$, where \nthe interval $[k,l)$ is $\\alpha$-mlarge.\n\n\\begin{remark}\nThe upper bound of the considered interval has been chosen to be $l-1$ and not $l$, in order to simplify some statements and proofs in composition lemmas associated to ordinals of the form $\\alpha\\times i$ and \n$\\omega^\\alpha\\times i + \\beta$.\nIn order to consider any ordinal below $\\epsilon_0$, we consider a special case for $\\alpha=0$.\n\\end{remark}\n\n\n\\subsection{Formal Specification}\n\nOur specification of the function \\texttt{L} is as follows:\n\n\\begin{Coqsrc}\nInductive L_spec : T1 -> (nat -> nat) -> Prop :=\n| L_spec0 : forall f, (forall k, f k = k) -> L_spec zero f\n| L_spec1 : forall alpha f,\n alpha <> zero ->\n (forall k, mlarge alpha (interval (S k) (Nat.pred (f (S k))))) ->\n L_spec alpha f.\n\\end{Coqsrc}\n\n\nNote that, for $\\alpha\\not=0$, the value of $f(0)$ is not specified.\nNevertheless, the restriction of $f$ to the set of strictly positive integers is unique (up to extensionnality).\n\n\\begin{Coqsrc}\nLemma L_spec_unicity alpha f g :\n L_spec alpha f -> L_spec alpha g -> forall k, f (S k) = g (S k).\n\\end{Coqsrc}\n\n\n\\subsection{Abstract Properties}\n\n\n\nLet us now prove properties of any function $f$ (if any) which satisfies \n\\texttt{L\\_spec}. We are looking for properties which could be used for writing \\emph{equations} and prove the correctness of the function generated by the \\texttt{coq-equations} plug-in. Moreover, they will give us some examples of\n$L_\\alpha$ for small values of $\\alpha$. \n\n\nOur exploration of the $L_\\alpha$s follows the usual scheme : transfinite induction, and proof-by-cases : zero, successors and limit ordinals.\n\n\\index{Maths!Transfinite induction}\n\n\\subsubsection{The Ordinal Zero}\n\\label{sect:L-spec-zero}\nThe base case is directly a consequence of the specification.\n\n\\begin{Coqsrc}\nLemma L_zero_inv f : L_spec zero f -> forall k, f (S k) = S k.\n\\end{Coqsrc}\n\n\\subsubsection{Successor Ordinals}\n\\label{sect:L-spec-succ}\nLet $\\beta$ be some ordinal, and assume the arithmetic function $f$ satisfies \nthe specification $(\\texttt{L\\_spec}\\;\\beta)$. Let $k$ be any natural number.\nAny path from $\\texttt{succ}\\,\\beta$ to $0$ starting at $k+1$ can be decomposed into a first step from $\\texttt{succ}\\,\\beta$ to $\\beta$, then a path from\n$\\beta$ at $k+2$ to $0$. \nBy hypothesis the interval $[k+2, f(k+2)-1]$ is $\\beta$-mlarge.\nBut the interval $[k+1, f(k+2)-1]$ is the concatenation of the singleton\n$\\{k+1\\}$ and the interval $[k+2, f(k+2)-1]$.\nSo, the function $\\lambda\\,k.\\,f(k+1)$ satisfies the specification $\\texttt{L\\_spec}\\,\\beta$.\n\n\nNote that our decomposition of intervals works only if the intervals we consider are not empty. In order to ensure this property, we assume that $f\\;k$ is always greater than $k$, which we note \\texttt{S <<= f}, or \\texttt{(fun\\_le S f)} (defined \nin~\\href{../theories/html/hydras.Prelude.Iterates.html\\#fun_le}{Prelude.Iterates}).\n\n\\begin{Coqsrc}\nDefinition fun_le f g := forall n:nat, f n <= g n.\n\\end{Coqsrc}\n\nIt looks also natural to show that the functions we consider are strictly monotonous. The section on successor ordinals has thus the following structure.\n\n\\begin{Coqsrc}\nSection succ.\n Variables (beta : T1) (f : nat -> nat).\n\n Hypotheses (Hbeta : nf beta)\n (f_mono : strict_mono f)\n (f_Sle : S <<= f)\n (f_ok : L_spec beta f).\n\n Definition L_succ := fun k => f (S k).\n\n Lemma L_succ_mono : strict_mono L_succ.\n\n Lemma L_succ_Sle : S <<= L_succ.\n \n Lemma L_succ_ok : L_spec (succ beta) L_succ.\n \nEnd succ.\n\n\\end{Coqsrc}\n\n\\subsubsection{Limit Ordinals}\n\\label{sect:L-spec-lim}\n\nLet $\\lambda<\\epsilon_0$ be any limit ordinal. In a similar way as for successors, we decompose any path from $\\lambda$ (at $k$) into a step to\n$\\canonseq{\\lambda}{k}$, then to $0$. In the following section, we assume that there exists à correct function for $\\canonseq{\\lambda}{k}$, \\emph{for any strictly positive $k$}.\n\n\\begin{Coqsrc}\nSection lim.\n Variables (lambda : T1)\n (Hnf : nf lambda)\n (Hlim : limitb lambda)\n (f : nat -> nat -> nat)\n (H : forall k, L_spec (canonS lambda k) (f (S k))).\n \n Let L_lim k := f k (S k).\n\n Lemma L_lim_ok : L_spec lambda L_lim.\n \nEnd lim.\n\\end{Coqsrc}\n\n\\subsection{First Results}\n\nApplying the previous lemmas on successors and limit ordinals, \nwe get several correct implementations of \\texttt{(L\\_spec $\\alpha$)} for small values of $\\alpha$.\n\n\\subsubsection{Finite Ordinals}\n\nBy iterating the functional \\texttt{L\\_succ}, we get a realization of\n\\texttt{(L\\_spec (fin $i$))} for any natural number $i$. \n\n\\begin{Coqsrc}\nDefinition L_fin i := fun k => (i + k)%nat.\n\nLemma L_fin_ok i : L_spec (fin i) (L_fin i).\n\\end{Coqsrc}\n\n\\subsubsection{The First Limit Ordinal \\texorpdfstring{$\\omega$}{omega}}\n\nThe lemmas \\texttt{L\\_fin\\_ok} and \\texttt{L\\_lim\\_ok} allow us to get \nby diagonalization a correct implementation for \n\\texttt{L\\_spec omega}.\n\n\\begin{Coqsrc}\nDefinition L_omega k := S (2 * k)%nat.\n\nLemma L_omega_ok : L_spec omega L_omega.\n\\end{Coqsrc}\n\n\\subsubsection{Towards \\texorpdfstring{$\\omega^2$}{omega*omega}}\n\nWe would like to get exact formulas for the ordinal $\\omega^2$, a.k.a.\n$\\phi_0(2)$. This ordinal is the limit of the sequence $\\omega\\times i\\;(i \\in \\mathbb{N}$. Thus, we have to study ordinals of this form, then use \nour lemma on limits.\n\nThe following lemma establishes a path from $\\omega\\times ( i+1)$ to\n$\\omega \\times i$.\n\n\\begin{Coqsrc}\nLemma path_to_omega_mult (i k:nat) :\n path_to (omega * i) (interval (S k) (2 * (S k))) (omega * (S i)).\n\\end{Coqsrc}\n\nLet us consider a path from $\\omega\\times(i+1)$ to $0$ starting at $k+1$.\nA first ``big step'' will lead to $\\omega\\times i$ at $2(k+1)$. If $i>0$, the\nnext jump leads to $\\omega\\times(i-1)$ at $2(2(k+1))+1$, etc.\n\n\nThe following lemma expresses the length of the mlarge sequences associated with the finite multiples of $\\omega$.\n\n\n\\begin{Coqsrc}\nLemma omega_mult_mlarge_0 i : forall k,\n mlarge (omega * (S i))\n (interval (S k)\n (Nat.pred (iterate (fun p => S (2 * p)%nat)\n (S i)\n (S k)))).\n\\end{Coqsrc}\n\nThus, we infer the following result:\n\n\\emph{From Module~ \\href{../theories/html/hydras.Epsilon0.Large_Sets.html\\#L_omega_mult}{Epsilon0.Large\\_Sets}}\n\n\\begin{Coqsrc}\nDefinition L_omega_mult i (x:nat) := iterate L_omega i x.\n\nLemma L_omega_mult_ok (i: nat) : L_spec (omega * i) (L_omega_mult i).\n\\end{Coqsrc}\n\nFor instance, let us consider the ordinal $\\omega\\times 8$, and a sequence \nstarting at $k=5$.\n\n\\begin{Coqsrc}\nCompute L_omega_mult 8 5.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n= 1535\n : nat\n\\end{Coqanswer}\n\nMore generally, we prove the equality $L_{\\omega\\times i}(k)=2^i\\times(k+1)-1$.\n\n\\begin{Coqsrc}\nLemma L_omega_mult_eqn (i : nat) :\n forall (k : nat), (0 < k)%nat ->\n L_omega_mult i k = (exp2 i * S k - 1)%nat.\n\\end{Coqsrc}\n\n\nBy diagonalization, we obtain a simple formula for $L_{\\omega^2}$.\n\n\\begin{Coqsrc}\nDefinition L_omega_square k := iterate (fun z => S (2 * z)%nat)\n k\n (S k).\n\nLemma L_omega_square_eqn k :\n (0 < k)%nat ->\n L_omega_square k = (exp2 k * (k + 2) - 1)%nat.\n\n\nLemma L_omega_square_ok: L_spec (omega * omega) \n L_omega_square.\n\nCompute L_omega_square 8.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = 2559\n : nat\n\\end{Coqanswer}\n\n\n%%%% ICI \n\n\n\\subsubsection{Going Further}\nLet us consider a last example, ``computing'' $L_{\\omega^3}$.\nSince the canonical sequence associated with this ordinal is composed of the\n$\\omega^2\\times i\\;(i\\in\\mathbb{N}_1)$, we have to study this sequence.\n\nTo this end, we prove a generic lemma, which expresses $L_{\\omega^\\alpha\\times i}$ as an iterate of $L_{\\omega^\\alpha}$. Note that in this lemma, we assume that the fonction associated with $\\alpha$ is stritly monotonous and \ngreater or equal than the successor function, and prove that $L_{\\omega^\\alpha\\times i}$satisfies the same properties.\n\n\\begin{Coqsrc}\nSection phi0_mult.\n Variables (alpha : T1) (f : nat -> nat).\n Hypotheses (Halpha : nf alpha)\n (f_mono : strict_mono f)\n (f_Sle : S <<= f)\n (f_ok : L_spec (phi0 alpha) f).\n\n Definition L_phi0_mult i := iterate f i.\n\nLemma L_phi0_mult_ok i: \n L_spec (ocons alpha i zero) (L_phi0_mult (S i)).\n\n Lemma L_phi0_mult_smono i: strict_mono (L_phi0_mult i).\n\n Lemma L_phi0_mult_Sle i: S <<= L_phi0_mult (S i).\n\nEnd phi0_mult.\n\\end{Coqsrc}\n\nLet us look \nat the ordinal $\\omega^2\\times i$, using \\texttt{L\\_phi0\\_mult}\n\n\\begin{Coqsrc}\n Definition L_omega_square_times i := iterate L_omega_square i.\n\n Lemma L_omega_square_times_ok i : \n L_spec (ocons 2 i zero) (L_omega_square_times (S i)).\n Proof.\n apply L_phi0_mult_ok.\n - auto with T1.\n - apply L_omega_square_Sle.\n - apply L_omega_square_ok.\n Qed.\n\\end{Coqsrc}\n\n\nWe are now ready to get an exact formula for $L_{\\omega^3}$. \n\\begin{Coqsrc}\nDefinition L_omega_cube := L_lim L_omega_square_times .\n\nLemma L_omega_cube_ok : L_spec (phi0 3) L_omega_cube.\n\\end{Coqsrc}\n\n\nThe function $L_{\\omega^3}$ is just obtained by diagonalization upon $L_{\\omega^2\\times i}$.\n\n\\begin{Coqsrc}\nLemma L_omega_cube_eqn i : \n L_omega_cube i = L_omega_square_times i (S i).\nProof. reflexivity. Qed.\n\\end{Coqsrc}\n\nThus, for instance, $L_{\\omega^3}(3)=L_{\\omega^2\\times 4}(3)$.\nThus, we obtain an exact expression of this number.\n\n\n\\begin{Coqsrc}\nLemma L_omega_cube_3_eq:\n let N := exp2 95 in\n let P := (N * 97 - 1)%nat in\n L_omega_cube 3 = (exp2 P * (P + 2) - 1)%nat.\n\\end{Coqsrc}\n\n\nThis number is quite big. Using \\texttt{Ocaml}'s \\texttt{float} arithmetic,\nwe can [under-]approximate it by $2^{3.8\\times10^{30}}\\times 3.8\\times{10^{30}}$.\n\n\\begin{Coqsrc}\n# let exp2 x = 2.0 ** x;;\n\nval exp2 : float -> float = \n# exp2 95.0 *. 97.0 -. 1.0;;\n- : float = 3.84256588194182037e+30\n# let n = exp2 95.0 ;;\n# let p = n *. 97.0 -. 1.0;;\nval p : float = 3.84256588194182037e+30\n\nEstimation :\n2 ** (3.84 e+30) * 3.84 e+30.\n\\end{Coqsrc}\n\n\n\\subsection{Using \\texttt{Equations}}\n\\label{sect:L-equations}\n\nNote that we did not define any function $L_\\alpha$ \\emph{for any $\\alpha<\\epsilon_0$} yet. We have got no more than a collection of proved realizations of $\\texttt{L\\_spec}\\;\\alpha$ for several values of $\\alpha$.\n\n\\index{Coq!Plug-ins!Equations}\n\nUsing the \\texttt{coq-equations} plug-in by \nM. Sozeau~\\cite{sozeau:hal-01671777}, we will now define a function \\texttt{L\\_} which maps any ordinal $\\alpha<\\epsilon_0$ to a proven realization of \n$\\texttt{L\\_spec}\\;\\alpha$. \n\\subsection{Definition}\n\nIn order to get a total function, we use our type \\texttt{E0} of well-formed ordinal terms,(see Sect~\\vref{sect:E0-def}). Our definition is structured along a well-founded recursion and\na case-study (zero, limit and successor ordinals). \n\n\n\\vspace{4pt}\n\\noindent\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.L_alpha.html\\#L_}{L\\_alpha}).}\n\n\\label{Functions:L-alpha}\n \n\\begin{Coqsrc}\nFrom Equations Require Import Equations.\nRequire Import ArithRing Lia.\n\nInstance Olt : WellFounded Lt := Lt_wf.\nHint Resolve Olt : E0.\n\n(** Using Coq-Equations for building a function which satisfies \n Large_sets.L_spec *)\n\nEquations L_ (alpha: E0) (i:nat) : nat by wf alpha Lt :=\n L_ alpha i with E0_eq_dec alpha Zero :=\n { | left _ => i ;\n | right nonzero\n with Utils.dec (Limitb alpha) :=\n { | left _ => L_ (Canon alpha i) (S i) ;\n | right notlimit => L_ (Pred alpha) (S i)}}.\n\nSolve All Obligations with auto with E0.\n\\end{Coqsrc}\n\nIt is worth looking at the answer from \\texttt{Equations} and check (with \\texttt{About} ) all the lemmas this plug-in gives you for free. We show here only a part of \\coq's anwer.\n\n\n\\begin{Coqanswer}\nL__obligations_obligation_1 is defined\nL__obligations_obligation_2 is defined\nL__obligations is defined\nL__clause_1 is defined\nL__functional is defined\nL_ is defined\n...\nL__equation_1 is defined\nL__graph_mut is defined\nL__graph_rect is recursively defined\nL__graph_correct is defined\nL__elim has type-checked, generating 1 obligation\nL__elim is defined\nFunctionalElimination_L_ is defined\nFunctionalInduction_L_ is defined\n\\end{Coqanswer}\n\nSometimes, these automatically generated statements may look cryptic. \n\n\\begin{Coqsrc}\nAbout L__equation_1.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nL__equation_1 :\nforall (alpha : E0) (i : nat),\nL_ alpha i = L__unfold_clause_1 alpha (E0_eq_dec alpha Zero) i\n\\end{Coqanswer}\n\nIn most cases, it may be useful to write human-readable paraphrases of these statements.\n\n\\begin{Coqsrc}\nLemma L_zero_eqn : forall i, L_ Zero i = i.\nProof. intro i; now rewrite L__equation_1. Qed.\n\nLemma L_lim_eqn alpha i : Limitb alpha -> L_ alpha i =\n L_ (Canon alpha i) (S i).\n\nLemma L_succ_eqn alpha i : L_ (Succ alpha) i = L_ alpha (S i).\n\nHint Rewrite L_zero_eqn L_succ_eqn : L_rw.\n\\end{Coqsrc}\n\nUsing these three lemmas as rewrite rules, we can prove more properties of the functions \\texttt{L\\_$\\alpha$}.\n\n\\begin{Coqsrc}\nLemma L_finite : forall i k :nat, L_ i k = (i+k)%nat. \n(* Proof by induction on i, using L_zero_eqn and L_succ_eqn *)\n\nLemma L_omega : forall k, L_ omega%e0 k = S (2 * k)%nat.\n(* Proof using L_finite and L_lim_eqn *)\n\\end{Coqsrc}\n\nBy well-founded induction on $\\alpha$, we prove the following lemmas:\n\n\\begin{Coqsrc}\nLemma L_ge_S alpha : alpha <> Zero -> S <<= L_ alpha.\n\nTheorem L_correct alpha : L_spec (cnf alpha) (L_ alpha).\n\\end{Coqsrc}\n\nPlease note that the proof of \\texttt{L\\_correct} applies the lemmas proven in Sections~\\ref{sect:L-spec-zero}, ~\\ref{sect:L-spec-succ} and ~\\ref{sect:L-spec-lim}.\nOur previous study of \\texttt{L\\_spec} allowed us to pave the way for the definition by \\texttt{Equations} and the correctness proof.\n\n\n\n\\subsubsection{Back to Hydra Battles}\n\nTheorem \\texttt{battle\\_length\\_std } of\nModule~\\href{../theories/html/hydras.Hydra.Hydra_Theorems.html\\#battle_length_std}{Hydra.Hydra\\_Theorems} relates the length of standard battles with the functions $L_\\alpha$.\n\n\\begin{Coqsrc}\nTheorem battle_length_std (alpha : E0) :\n alpha <> Zero ->\n forall k, (1 <= k)%nat ->\n battle_length standard k (iota (cnf alpha))\n (L_ alpha (S k) - k).\n\\end{Coqsrc}\n\n\n\\index{Projects}\n\\begin{project}\nInstead of considering standard paths and battles, consider ``constant'' paths and the corresponding battles. Please use \\texttt{Equations} in order to define the function that computes the length of the $k$-path which leads from $\\alpha$ to $0$.\nProve a few exact formulas and minoration lemmas.\n\\end{project}\n\n\\section{The Wainer-Hardy Hierarchy (Functions \\texorpdfstring{$H_\\alpha$}{\\texttt{H\\_alpha}})}\n\n\\label{sect:hardy}\nIn order to give an idea of the complexity of the functions $L_\\alpha$s, we compare them with a better known family of functions, the so called \\emph{Wainer-Hardy hierarchy} of fast growing functions,\npresented for instance in~\\cite{Promel2013}. \n\\index{Maths!Hardy Hierarchy}\n\nFor each ordinal $\\alpha$ below $\\epsilon_0$, $H_\\alpha$ is a \ntotal arithmetic function, defined by transfinite recursion on $\\alpha$, according to three cases:\n\n\\index{Maths!Transfinite induction}\n\n\\begin{itemize}\n\\item If $\\alpha=0$, then $H_\\alpha (k)= k$ for any natural number $k$.\n\\item If $\\alpha=\\textrm{succ}(\\beta)$, then \n$H_\\alpha(k)=H_\\beta(k+1)$ for any $k \\in \\mathbb{N}$\n\\item If $\\alpha$ is a limit ordinal, then \n$H_\\alpha(k) = H_{(\\canonseq{\\alpha}{k+1})}(k)$ for any $k\\in \\mathbb{N}$.\n\\end{itemize}\n\n\\subsection{Hardy Functions in \\texttt{Coq}}\n\n\nWe define a function \\texttt{H\\_} of type \\texttt{E0 -> nat -> nat} by transfinite induction over the type \\texttt{E0} of the well formed ordinals below $\\epsilon_0$.\n\n\\vspace{4pt}\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.H_alpha.html\\#H_}{Epsilon0.H\\_alpha}}\n\n\n\\index{Coq!Plug-ins!Equations}\n\\label{Functions:H-alpha}\n\n\\begin{Coqsrc}\nEquations H_ (alpha: E0) (i:nat) : nat by wf alpha lt :=\n H_ alpha i with E0_eq_dec alpha Zero :=\n { | left _ => i ;\n | right nonzero\n with Utils.dec (Limitb alpha) :=\n { | left _ => H_ (Canon alpha (S i)) i ;\n | right notlimit => H_ (Pred alpha) (S i)}}. \n\nSolve All Obligations with auto with E0.\n\\end{Coqsrc} \n \n\n\n\\begin{Coqsrc}\nLemma H_eq1 : forall i, H_ Zero i = i.\nProof. intro i; now rewrite H__equation_1. Qed.\n\nLemma H_eq2 alpha i : Is_Succ alpha ->\n H_ alpha i = H_ (Pred alpha) (S i).\n\nLemma H_eq3 alpha i : Limitb alpha ->\n H_ alpha i = H_ (Canon alpha (S i)) i.\n\nLemma H_eq4 alpha i : H_ (Succ alpha) i = H_ alpha (S i).\n\\end{Coqsrc}\n\n\n\\subsection{First Steps of the Hardy hierarchy}\nUsing rewrite rules from \\texttt{H\\_eq1} to \\texttt{H\\_eq4}, we can explore the functions $H_\\alpha$ for some small values of $\\alpha$.\n\n\\subsubsection{Finite Ordinals} \n\nBy induction on $i$, we prove a simple expression of \\texttt{H\\_ (Fin i)}, where \n\\texttt{Fin $i$} is the $i$-th finite ordinal.\n\n\\begin{Coqsrc}\nLemma H_Fin : forall i k: nat, H_ (Fin i) k = (i+k)%nat.\nProof with eauto with E0.\n induction i.\n - intros; simpl OF; simpl; autorewrite with H_rw E0_rw ... \n - intros ;simpl; autorewrite with H_rw E0_rw ... \n rewrite IHi; lia. \nQed.\n\\end{Coqsrc}\n\n\\subsubsection{Multiples of \\texorpdfstring{$\\omega$}{omega}}\n\nSince the canonical sequence of $\\omega$ is composed of finite ordinals, \nit is easy to get the formula associated with $H_\\omega$.\n\n\n\\begin{Coqsrc}\nLemma H_omega : forall k, H_ Omega k = S (2 * k)%nat.\nProof with auto with E0.\n intro k; rewrite H_eq3 ...\n - replace (Canon omega (S k)) with (Fin (S k)).\n + rewrite H_Fin; lia.\n + now autorewrite with E0_rw.\nQed.\n\\end{Coqsrc}\n\n\nBefore going further, we prove a useful rewriting lemma:\n\n\\begin{Coqsrc}\nLemma H_Plus_Fin alpha : forall i k : nat,\n H_ (alpha + i)%e0 k = H_ alpha (i + k)%nat.\n(* Proof by induction on i *)\n\\end{Coqsrc}\n\n\nThen, we get easily formulas for $H_{\\omega+i}$, and $H_{\\omega\\times i}$ for any natural number $i$.\n\n\\begin{Coqsrc}\nLemma H_omega_double k : H_ (omega * 2)%e0 k = (4 * k + 3)%nat.\nProof.\n rewrite H_lim_eqn; simpl Canon.\n - ochange (CanonS (omega * 2)%e0 k) (omega + (S k))%e0.\n + rewrite H_Plus_Fin, H_omega; lia.\n - now compute.\nQed.\n\nLemma H_omega_3 k : H_ (omega * 3)%e0 k = (8 * k + 7)%nat.\n\nLemma H_omega_4 k : H_ (omega * 4)%e0 k = (16 * k + 15)%nat.\n\nLemma H_omega_i i : forall k,\n H_ (omega * i)%e0 k = (exp2 i * k + Nat.pred (exp2 i))%nat.\n\\end{Coqsrc}\n\nCrossing a new limit, we prove the following equality: \n$$H_{\\omega^2} (k) = 2 ^ {k+1} \\times (k+1) - 1$$.\n\n\\begin{Coqsrc}\n\nLemma H_omega_sqr : forall k,\n H_ (Phi0 2)%e0 k = (exp2 (S k ) * S k - 1)%nat.\nProof.\n intro k; \n rewrite H_lim_eqn; auto with E0.\n - ochange (Canon (Phi0 2) (S k)) (omega * (S k))%e0.\n + rewrite H_omega_i; simpl (exp2 (S k)).\n * rewrite Nat.add_pred_r.\n -- lia. \n -- generalize (exp2_not_zero k); lia.\n + cbn; f_equal; lia.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection{New Limits}\n\nOur next step would be to prove an exact formula for $H_{\\omega^\\omega}(k)$.\nSince the canonical sequence of $\\omega^\\omega$ is composed of all the\n$\\omega^i$, we first need to express $H_{\\omega^i}$ for any natural number $i$.\n\nLet $i$ and $k$ be two natural numbers. \nThe ordinal $\\canonseq{\\omega^(i+1)}{k}$ is the product\n$\\omega^i \\times k$, so we need also to consider ordinals of this form.\n\n\\begin{enumerate}\n\\item First, we express $H_{\\omega^\\alpha \\times (i+2)}$ in terms of\n$H_{\\omega^\\alpha \\times (i+1)}$.\n\n\\begin{Coqsrc}\nLemma H_Omega_term_1 : alpha <> Zero -> forall k, \n H_ (Omega_term alpha (S i)) k =\n H_ (Omega_term alpha i) (H_ (Phi0 alpha) k).\n\\end{Coqsrc}\n\n\\item\nThen, we prove by induction on $i$ that $H_{\\omega^\\alpha \\times (i+1)}$ is just the\n$(i+1)$-th iterate of $H_{\\omega^\\alpha}$.\n\n\n\\begin{Coqsrc}\nLemma H_Omega_term (alpha : E0) :\nalpha <> Zero -> forall i k, \n H_ (Omega_term alpha i) k = iterate (H_ (Phi0 alpha)) (S i) k.\n\\end{Coqsrc}\n\n\\item In particular, we have got a formula for $H_{\\omega^{i+1}}$.\n\n\\begin{Coqsrc}\nDefinition H_succ_fun f k := iterate f (S k) k.\n\nLemma H_Phi0_succ alpha : alpha <> Zero -> forall k,\n H_ (Phi0 (Succ alpha)) k = H_succ_fun (H_ (Phi0 alpha)) k. \n\nLemma H_Phi0_Si : forall i k,\n H_ (Phi0 (S i)) k = iterate H_succ_fun i (H_ omega) k. \n\\end{Coqsrc}\n\n\\end{enumerate}\nWe get now a formula for $H_{\\omega^3}$:\n\n\\begin{Coqsrc}\nLemma H_omega_cube : forall k,\n H_ (Phi0 3)%e0 k = iterate (H_ (Phi0 2)) (S k) k.\nProof.\n intro k; rewrite <-FinS_eq, -> Fin_Succ, H_Phi0_succ; auto.\n compute; injection 1; discriminate.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection{A Numerical Example}\n\nIt seems hard to capture the complexity of this function by looking only at this\n``exact'' formula. \nLet us consider a simple example, the number $H_{\\omega^3}(3)$. \n\n\\begin{Coqsrc}\nSection H_omega_cube_3.\n \nLet f k := (exp2 (S k ) * (S k) - 1)%nat.\n\nRemark R0 k : H_ (Phi0 3)%e0 k = iterate f (S k) k.\n\\end{Coqsrc}\n\nThus, the number $H_{\\omega^3}(3)$ can be written as four nested applications of $f$.\n\n\\begin{Coqsrc}\nFact F0 : H_ (Phi0 3) 3 = f (f (f (f 3))).\n rewrite R0; reflexivity. \nQed.\n\\end{Coqsrc}\n\nIn order to make this statement more readable, we can introduce a local définition.\n\n\\begin{Coqsrc}\nLet N := (exp2 64 * 64 - 1)%nat.\n\\end{Coqsrc}\n\nThis number looks quite big; let us compute an approximation in \\texttt{Ocaml}:\n\n\n\\begin{Coqsrc}\n# (2.0 ** 64.0 *. 64.0 -. 1.0);; \n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n- : float = 1.1805916207174113e+21\n\\end{Coqanswer}\n\n\n\\begin{Coqsrc}\nFact F1 : H_ (Phi0 3) 3 = f (f N).\nProof.\n rewrite H_omega_cube_0; reflexivity. \nQed.\n\n\nLemma F1_simpl : H_ (Phi0 3) 3 =\n (exp2 (exp2 (S N) * S N) * (exp2 (S N) * S N) - 1)%nat.\n\n\\end{Coqsrc}\n\n\nIn a more classical writing, this number is displayed as follows:\n\n{\\Large\n$$\nH_{\\omega^3}(3) = 2 ^ {(2 ^ {N + 1} \\times (N+1) } ) \\times (2 ^ {N+1} \\times ( N +1) ) - 1\n$$\n}\n\n\nWe leave as an exercise to determine the best approximation as possible of\n the size of this number (for instance its number of digits). For instance, if\nwe do not take into account the multiplications in the formula above,\nwe obtain that, in base $2$, the number $H_{\\omega^3}(3)$ has at least\n$2^{10^{21}}$ digits. But it is still an under-approximation !\n\n\n\\begin{Coqsrc}\nEnd H_omega_cube_3.\n\\end{Coqsrc}\n\n\n\n\nNow, we have got at last an exact formula for $H_{\\omega^\\omega}$.\n\n\\begin{Coqsrc}\nLemma H_Phi0_omega : forall k, H_ (Phi0 omega) k =\n iterate H_succ_fun k (H_ omega) k.\nProof with auto with E0.\n intro k; rewrite H_lim_eqn, <- H_Phi0_Si ...\n - rewrite CanonS_Canon, CanonS_Phi0_lim; f_equal ...\nQed.\n\\end{Coqsrc}\n\nUsing extensionality of the functional \\texttt{iterate}, we can get a closed formula.\n\n\\begin{Coqsrc}\nLemma H_Phi0_omega_closed_formula k :\n H_ (Phi0 omega) k =\n iterate (fun (f: nat -> nat) (l : nat) => iterate f (S l) l)\n k\n (fun k : nat => S (2 * k)%nat)\n k.\n\\end{Coqsrc}\n\n\n\n\nNote that this short formula contains two occurences of the functional \\texttt{iterate}, the outer one is in fact a second-order iteration (on type \\texttt{nat -> nat)}\nand the inner one first-order (on type \\texttt{nat}). \n\n\n\\subsection{Abstract Properties of H-functions}\n~\\label{sect:H-alpha-prop} \n\nSince pure computation seems to be useless for dealing with expressions of the form $H_\\alpha(k)$, even for small values of $\\alpha$ and $k$, we need to prove theorems for comparing $H_\\alpha(k)$ and $H_\\beta(l)$, in terms of comparison\nbetween $\\alpha$ and $\\beta$ on the one hand, $k$ and $l$ on the other hand.\n\nBut beware of non-theorems! For instance, one could believe that $H$ is monotonous in its first argument. The following proof shows this is false.\n\n\\begin{Coqsrc}\nRemark H_non_mono1 :\n ~ (forall alpha beta k, (alpha o<= beta)%e0 ->\n (H_ alpha k <= H_ beta k)%nat).\nProof.\n intros H ;specialize (H 42 omega 3).\n assert (H0 :(42 o<= omega)%e0) by (repeat split; auto). \n apply H in H0; rewrite H_Fin, H_omega in H0; lia.\nQed.\n\\end{Coqsrc}\n\nOn the contrary, the fonctions of the Hardy hierarchy have the following five properties~\\cite{KS81}: for any $\\alpha < \\epsilon_0$,\n\\begin{itemize}\n\\item the function $H_\\alpha$ is strictly monotonous :\n For all $n,p \\in\\mathbb{N}, n < p \\Rightarrow H_\\alpha(n)< H_\\alpha(p)$.\n\\item If $\\alpha \\not= 0$, then for every $n$, $ng(p)$ for any all sufficiently large $p$.\n\\end{definition}\n\n\n\\begin{Coqsrc}\nSection Proof_of_Abstract_Properties.\n Record P (alpha:E0) : Prop :=\n mkP {\n PA : strict_mono (H_ alpha);\n PB : alpha <> Zero -> forall n, (n < H_ alpha n)%nat;\n PC : H_ alpha <<= H_ (Succ alpha);\n PD : dominates_from 1 (H_ (Succ alpha)) (H_ alpha);\n PE : forall beta n, Canon_plus n alpha beta -> \n (H_ beta n <= H_ alpha n)%nat}.\n\n\nTheorem P_alpha : forall alpha, P alpha.\n Proof.\n intro alpha; apply well_founded_induction with lt.\n (* rest of proof skipped *)\n\nSection Proof_of_Abstract_Properties.\n\\end{Coqsrc}\n\n\n\\subsection{Comparison between \\texttt{L\\_} and \\texttt{H\\_} }\n\nBy well-founded induction on $\\alpha$, we prove that our $L$ hierachy is ``almost'' the Hardy hierarchy (up to a small shift).\n\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.L_alpha.html\\#H_L_}{Epsilon0.L\\_alpha}}\n\n\\begin{Coqsrc}\n Theorem H_L_ alpha: forall i:nat, (H_ alpha i <= L_ alpha (S i))%nat.\n\\end{Coqsrc}\n \n\\subsubsection{Back to Hydras}\n\nThe following theorem relates the length of (standard) battles with the Hardy hierarchy.\n\n\\emph{From Module~\\href{../theories/html/hydras.Epsilon0.L_alpha.html}{Epsilon0.L\\_alpha}}\n\n\\begin{Coqsrc}\nTheorem battle_length_std_Hardy (alpha : E0) :\n alpha <> Zero ->\n forall k , 1 <= k -> exists l: nat, \n H_ alpha k - k <= l /\\\n battle_length standard k (iota (cnf alpha)) l. \n\\end{Coqsrc}\n\n\n\n\\section{The Wainer Hierarchy (Functions \\texorpdfstring{$F_\\alpha$}{F\\_alpha})}\n\\label{sect:wainer}\n\n\\index{Maths!Wainer Hierarchy}\n\nThe Wainer hierarchy~\\cite{BW85, Wainer1970, KS81}, is also a family of fast growing functions, indexed by ordinals below $\\epsilon_0$, by the following equations:\n\n\\label{F_equations}\n\\begin{itemize}\n\\item $F_0(i)=i+1$\n\\item $F_{\\beta+1}(i)= (F_\\beta)^{(i+1)}(i)$, where $f^{(i)}$ is the $i$-th iterate of $f$.\n\\item $F_\\alpha(i) = F_{\\canonseq{\\alpha}{i}} (i)$ if $\\alpha$ is a limit ordinal.\n\\end{itemize}\n\nA first attempt is to write a definition of $F_\\alpha$ by equations, in the same as for $H\\_alpha$. We use the functional \\texttt{iterate} defined in \nModule~\\href{../theories/html/hydras.Prelude.Iterates.html\\#iterate}{Prelude.Iterates}.\n\n\\begin{Coqsrc}\nFixpoint iterate {A:Type}(f : A -> A) (n: nat)(x:A) :=\n match n with\n | 0 => x\n | S p => f (iterate f p x)\n end.\n\\end{Coqsrc}\n\nThe following code comes from \n \\url{../theories/html/hydras.Epsilon0.F_alpha.html}.\n\n\n\\index{Coq!Plug-ins!Equations}\n\\begin{Coqsrc}\nFail Equations F_ (alpha: E0) (i:nat) : nat by wf alpha Lt :=\n F_ alpha i with E0_eq_dec alpha Zero :=\n { | left _ => i ;\n | right nonzero\n with Utils.dec (Limitb alpha) :=\n { | left _ => F_ (Canon alpha i) i ;\n | right notlimit => iterate (F_ (Pred alpha)) (S i) i}}.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nThe command has indeed failed with message:\nIn environment\nalpha : E0\nnotlimit : Limitb alpha = false\nnonzero : alpha <> Zero\ni : nat\nF_ : forall x : E0, nat -> x o< alpha -> nat\nThe term \"F_ (Pred alpha) ?x\" has type \"Pred alpha o< alpha -> nat\"\nwhile it is expected to have type \"Pred alpha o< alpha -> Pred alpha o< alpha\"\n(cannot unify \"nat\" and \"Pred alpha o< alpha\").\n\\end{Coqanswer}\n\n\nWe presume that this error comes from the recursive call of \\texttt{F\\_} inside\nan application of \\texttt{iterate}. The workaround we propose is to define first \nthe iteration of \\texttt{F\\_} as an helper $F^*$, then to define the function $F$ as a ``iterating $F^*$ once''.\n\n\\texttt{Equations} accepts the following definition, relying on lexicographic ordering on pairs $(\\alpha,n)$.\n\n\n\\label{sect:F-equations}\n\n\\index{Coq!Plug-ins!Equations}\n\\label{Functions:F-alpha}\n\\index{Maths! Fast growing functions}\n \n\\begin{Coqsrc}\nDefinition call_lt (c c' : E0 * nat) :=\n lexico Lt (Peano.lt) c c'.\n\nLemma call_lt_wf : well_founded call_lt.\n unfold call_lt; apply Inverse_Image.wf_inverse_image, wf_lexico.\n - apply E0.Lt_wf.\n - unfold Peano.lt; apply Nat.lt_wf_0. \nQed.\n\nInstance WF : WellFounded call_lt := call_lt_wf.\n\nEquations F_star (c: E0 * nat) (i:nat) : nat by wf c call_lt :=\n F_star (alpha, 0) i := i;\n F_star (alpha, 1) i\n with E0_eq_dec alpha Zero :=\n { | left _ => S i ;\n | right nonzero\n with Utils.dec (Limitb alpha) :=\n { | left _ => F_star (Canon alpha i,1) i ;\n | right notlimit =>\n F_star (Pred alpha, S i) i}};\n F_star (alpha,(S (S n))) i :=\n F_star (alpha, 1) (F_star (alpha, (S n)) i).\n\n(* Finally, F_ alpha is defined as its first iterate ! *)\n\nDefinition F_ alpha i := F_star (alpha, 1) i.\n\\end{Coqsrc}\n\nIt is quite easy to prove that our function \\texttt{F\\_} satisfies the equations on page~\\pageref{sect:F-equations}.\n\n\\begin{Coqsrc}\nLemma F_zero_eqn : forall i, F_ Zero i = S i.\n\nLemma F_lim_eqn : forall alpha i, Limitb alpha ->\n F_ alpha i = F_ (Canon alpha i) i.\n\nLemma F_succ_eqn : forall alpha i,\n F_ (Succ alpha) i = iterate (F_ alpha) (S i) i.\n\\end{Coqsrc}\n\nAs for the Hardy functions, we can use these equalities as rewrite rules for\n``computing'' some values of $F_\\alpha(i)$, for small values of $\\alpha$.\n\n\\begin{Coqsrc}\nLemma LF1 : forall n, F_ 1 n = S (2 * n).\n\nLemma LF2 : forall i, (exp2 i * i < F_ 2 i)%nat.\n\\end{Coqsrc}\n\n\nLike in Sect~\\ref{sect:H-alpha-prop}, we prove by induction the following properties (see~\\cite{KS81}). \n\n\\begin{Coqsrc}\nTheorem F_alpha_mono alpha : strict_mono (F_ alpha).\n \nTheorem F_alpha_ge_S alpha : forall n, (n < F_ alpha n)%nat.\n\nTheorem F_alpha_Succ_le alpha : F_ alpha <<= F_ (Succ alpha).\n\nTheorem F_alpha_dom alpha : dominates_from 1 (F_ (Succ alpha)) (F_ alpha).\n\nTheorem F_alpha_beta alpha : forall beta n, Canon_plus n alpha beta -> \n (F_ beta n <= F_ alpha n)%nat.\n\\end{Coqsrc}\n\nAs a corollary, we prove the following proposition, p. 284 of~\\cite{KS81}.\n\n\\begin{quote}\n If $\\beta<\\alpha$, $F_\\alpha$ dominates $F_\\beta$.\n\\end{quote}\n\n\\begin{Coqsrc}\nLemma Propp284 : forall alpha beta : E0, \n beta o< alpha -> dominates (F_ alpha) (F_ beta).\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\nLet us quote a theorem from ~\\cite{KS81} (page 297).\n\n\\begin{quote}\n\\begin{align*}\n H_{\\omega^\\alpha}(n+1) &\\geq F_{\\alpha}(n) \\quad (n\\geq 1, \\alpha<\\epsilon_0) \\\\\n F_{\\alpha}(n+1) &\\geq H_{\\omega^\\alpha}(n) \\quad (n\\geq 1, \\alpha<\\epsilon_0) \n\\end{align*}\nThus $H_{\\omega^\\alpha}$ and $F_{\\alpha}$ have essentially the same order of growth.\n\n\\end{quote}\n\n But, before trying to prove these facts, look at the definition of function $H$ in Ketonen and Solovay's paper ! Is it really the same as the definition we quote from Pr{\\H o}mel's chapter~\\cite{Promel2013},\nwhereas \\cite{KS81} define $H_\\alpha(n)$ as ``the least integer $k$ such that $[n,k]$ is $\\alpha$-large''. Thus, it may be useful to adapt the statement above.\n\n\n\n\\end{exercise}\n\n\\index{Exercises}\n\n\\begin{exercise}\nProve the following result~\\cite{KS81}(p. 298).\n\n\\begin{quote}\n For $n\\geq 2$ and $\\alpha \\geq 3$, $F_\\alpha(n+1)\\geq 2^{F_\\alpha(n)}$.\n\\end{quote}\n\\end{exercise}\n\n\n\n% \\subsection{Gnawing ordinals}\n\n% \\begin{definition}[After~\\cite{KS81}]\n% Let $S$ be a finite set of positive integers, and $\\alpha$ be an ordinal strictly less than $\\epsilon_0$. Let us denote by $s=s_1,s_2, \\dots, s_N$ the sequence of \n% the elements of $S$, enumerated in strictly increasing order.\n\n% We consider the sequence of ordinals $\\alpha_o=\\alpha, \\alpha_1=\\canonseq{\\alpha_0}{s_1},\\dots,\\alpha_{i+1}=\\canonseq{\\alpha_i}{s_{i+1}},\\dots, \n% \\alpha_{N}=\\canonseq{\\alpha_{N-1}}{s_N}$.\n% We denote by $\\gnaw{s}{\\alpha}$ the last ordinal of the sequence, \\emph{i.e.} $\\alpha_N$.\n% \\end{definition}\n\n\n% The following function computes $\\gnaw{s}{\\alpha}$ by recursion on $s$.\n% \\vspace{4pt}\n\n\n\n% For instance, let us consider the ordinal $\\omega^\\omega$, and try some sequence of integers. \n\n\n% \\begin{Coqsrc}\n% Compute pp (gnaw (omega ^ omega) (1::nil)). \n% \\end{Coqsrc}\n\n\n% \\begin{Coqanswer}\n% = omega%pT1 : ppT1\n% \\end{Coqanswer}\n\n\n% \\begin{Coqsrc}\n% Compute pp (gnaw (omega ^ omega) (1::2::nil)). \n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = P_fin 2\n% : ppT1\n% \\end{Coqanswer}\n\n% Likewise, we can verify that $\\gnaw{\\omega^\\omega}{\\langle 1,2,3 \\rangle}=1$\n% and $\\gnaw{\\omega^\\omega}{\\langle 1,2,3,4 \\rangle}=0$.\n\n% \\begin{Coqsrc}\n% Example omega_omega_1_4 : gnaw (omega ^omega) (interval 1 4) = 0.\n% Proof. trivial. Qed.\n\n% Example omega_omega_1_3 : gnaw (omega ^omega) (interval 1 3) = 1.\n% Proof. trivial. Qed.\n% \\end{Coqsrc}\n\n% \\begin{remark}\n% \\label{remark:gnaw-vs-battles}\n% Let us consider an hydra battle, where Hercules always chops off the rightmost head of the hydra. Let $\\alpha<\\epsilon_0$ be an ordinal number, and \n% $s=\\langle i_1 true\n% | _ => false\n% end.\n\n\n% Definition large (alpha : T1) (s : list nat) : Prop :=\n% largeb alpha s.\n% \\end{Coqsrc}\n\n\n\n\n% For instance, the sequence $\\langle 1,2,3,4 \\rangle$ is $\\omega^\\omega$-large but not\n% $\\omega^{\\omega+1}$-large, since $\\gnaw{\\omega^{\\omega+1}}{\\langle 1,2,3,4\\rangle}$ is equal to $\\omega \\times 2 + 4$.\n\n\n% %\\omega^\\omega+\\omega^2\\times 3 + \\omega\\times 4+ 5$.\n\n\n% \\begin{Coqsrc}\n% Compute largeb (omega^omega) (interval 1 4). \n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = true : bool\n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute largeb (omega^(omega+1)) (interval 1 4).\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = false : bool\n% \\end{Coqanswer}\n\n\n% \\begin{remark}\n% A sequence $s$ is $\\alpha$-large if (still considering Hercules ``righmost-head'', tactic), it leads to Hercules' victory.\n% \\end{remark}\n\n\n% \\subsection{A little game} \n\n% Let $\\alpha<\\epsilon_0$ be an ordinal, and $i$ a positive integer. We want to guess the least natural number $j$ such that the interval $[i,i+j]$ is $\\alpha$-large.\n% Equivalently, we should have $\\gnaw{\\alpha}{[i, i+j-1]}=1$.\n\n% The following functions takes three arguments: an ordinal $\\alpha$, and two positive natural numbers $i$ and $j$ (we assume, but not verify that $i$ and $j$ are strictly positive). It returns one of the three possible answers:\n\n% \\begin{itemize}\n% \\item \\texttt{Ok} if $j$ is the smallest integer such that the interval $[i,i+j]$ is $\\alpha$-large\n% \\item \\texttt{Too\\_far} if $[i,i+j]$ is $\\alpha$-large, but $j$ is not the smallest such positive integer\n% \\item \\texttt{(Remaining $\\beta$)} if $j$ is too small, and gnawing $\\alpha$ with\n% $[i,i+j]$ is still equal to $\\beta$, instead of $0$\n% \\end{itemize}\n\n\n% \\vspace{4pt}\n% \\noindent\n% \\emph{From Module~\\href{../theories/html/hydras.Epsilon0.Large_Sets_Demo.html}{Ordinals.Epsilon0.Large\\_Sets\\_Demo}} \n\n% \\begin{Coqsrc}\n% Inductive answer : Set := You_won | Too_far | Remaining (rest : ppT1).\n\n% Definition large_set_check alpha i j :=\n% let beta := gnaw alpha (interval i (Nat.pred j))\n% in match beta with\n% | one => Ok\n% | zero => Too_far\n% | _ => Remaining (pp (canonseq j beta))\n% end.\n% \\end{Coqsrc}\n\n% \\subsubsection{\\texorpdfstring{$\\omega$}{omega}-large intervals}\n\n\n% For instance, let us consider the ordinal $\\omega$ and start with $i=1$.\n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 1 2.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n% Let us give greater values of $i$, still with the ordinal $\\omega$.\n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 2 3.\n% \\end{Coqsrc}\n\n\n% \\begin{Coqanswer}\n% = Remaining 1\n% : answer\n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 3 3.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Remaining 3\n% : answer\n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 2 4.\n% \\end{Coqsrc}\n\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 3 6.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n% It looks like every request to compute (\\texttt{large\\_set\\_check omega $i$ $2\\times i$}) will succeed. Let us try an example. \n\n% \\begin{Coqsrc}\n% Compute large_set_check omega 49 98.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n% \\subsubsection{\\texorpdfstring{$\\omega^2$}{omega\\^{}2}-large intervals}\n\n% Still using \\texttt{Compute} and \\texttt{large\\_set\\_check}, we obtained the \n% following values of $j$, the least integer such that the interval $[i,j]$ is $\\omega^2$ large.\n\n% $$\n% \\begin{array}{|c|c|}\n% \\hline\n% i & j \\\\\n% \\hline \n% 1 & 4 \\\\\n% 2 & 14 \\\\\n% 3 & 38 \\\\\n% 4 & 94 \\\\\n% 5 & 222 \\\\\n% % 6 & 510 \n% \\hline\n% \\end{array}\n% $$\n\n% \\begin{exercise}\n% Please give the 6-th and 7-th line of the array above.\n% \\end{exercise}\n\n\n% \\subsubsection{The limits of (pure) computation}\n\n% Let us now play with bigger ordinals, for instance $\\alpha=\\omega^\\omega$ or\n% $\\alpha=\\omega^{\\omega + 1}$. We notice that, even for small values of $i$, it is hard\n% to guess values of $j$ such that $[i,i+j]$ is $\\alpha$-large.\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ omega) 1 4.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ omega) 2 38.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Ok\n% : answer\n% \\end{Coqanswer}\n\n\n\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ omega) 3 1000.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Remaining (omega ^ 2 * 2 + omega * 220 + 798)%pT1\n% : answer\n% \\end{Coqanswer}\n\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ omega) 3 1798.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Remaining (omega ^ 2 * 2 + omega * 220)%pT1\n% : answer\n% \\end{Coqanswer}\n\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ omega) 3 5000.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Remaining (omega ^ 2 * 2 + omega * 218 + 2198)%pT1\n% : answer\n% \\end{Coqanswer}\n\n\n\n\n\n\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ (omega + 1)) 3 5000.\n% \\end{Coqsrc}\n\n% \\begin{Coqanswer}\n% = Remaining\n% (omega ^ omega * 2 + omega ^ 3 * 4 + \n% omega ^ 2 * 4 + omega * 1148 + 4222)%pT1\n% : answer \n% \\end{Coqanswer}\n\n% \\begin{Coqsrc}\n% Compute large_set_check (omega ^ (omega + 1)) 3 10000.\n% \\end{Coqsrc}\n\n\n% \\begin{Coqanswer}\n% Warning: To avoid stack overflow, large numbers in nat are interpreted as\n% applications of Init.Nat.of_uint. [abstract-large-number,numbers]\n\n% = Remaining\n% (omega ^ omega * 2 + omega ^ 3 * 4 + \n% omega ^ 2 * 4 + omega * 1147 +\n% 8446)%pT1\n% : answer\n% \\end{Coqanswer}\n\n% Since computation is not enough, let us comme back to proofs, or, better, proofs \\emph{and} computations.\n\n% \\subsection{Proving largeness}\n\n\n\n% %%% ICI %%%\n\n% \\subsection{$n$-large and $\\omega$-large intervals }\n\n% A finite ordinal is just made out of \\texttt{zero} and \\texttt{succ} constructions.\n% Thus any sequence of strictly positive integers of length greater than or equal to $n$ will completely gnaw the ordinal $n$.\n\n% Concerning the first limit ordinal $\\omega$ the largeness of a sequence $s$ depends only on its first element and the length of the rest of the sequence (please keep in mind that the argument $i$ of $\\canonseq{\\alpha}{i}$ is meaningful only if $\\alpha$ is a limit ordinal).\n% The following proposition is labeled $4.2$ in~\\cite{KS81}.\n\n% \\begin{proposition}\n% For $n<\\omega$, a set $X$id $n$-large if and only if $|X|\\geq n$.\n% A finite set $X$ is $\\omega$-large if $|X|>\\min{X}$.\n% \\end{proposition}\n\n% Rewritten in terms of list of strictly positive integers, we get the following statements:\n\n% \\begin{Coqsrc}\n% Lemma large_n_iff : forall s (n:nat) , ~ In 0 s ->\n% large n s <-> (n <= List.length s)%nat.\n% \\end{Coqsrc}\n\n\n% \\begin{Coqsrc}\n% Lemma large_omega_iff : forall s n, ~In 0 (n::s) -> \n% large omega (n::s) <->\n% (n <= List.length s)%nat.\n% \\end{Coqsrc}\n\n\n\n% % \\begin{Coqsrc}\n% % (* sorted list of natural numbers greater than or equal to n *)\n\n% % Inductive sorted_ge (n: nat) : list nat -> Prop :=\n% % | sorted_ge_nil : sorted_ge n nil\n% % | sorted_ge_one : forall p, n<=p ->\n% % sorted_ge n (p::nil)\n% % | sorted_ge_cons: forall p q s, n<=p -> p\n% % sorted_ge p (q::s) ->\n% % sorted_ge n (p::q::s).\n% % \\end{Coqsrc}\n\n\n\n% \\subsection{Mimimal large sequences}\n\n\n\n% Let us consider \\emph{minimal} large sequences, \\emph{i.e.} large sequences \n% the strict prefix of which do not lead to $0$. In other words, only the last ordinal \n% defined by the sequence is null. For this purpose, we use the predicate \\texttt{path\\_to} introduced page~\\pageref{path-to-definition}.\n\n% \\begin{Coqsrc}\n% Definition mlarge alpha s := path_to zero s alpha.\n% \\end{Coqsrc}\n\n\n\n\n\n\n\n%----------------------------------------------------------------------\n\\chapter[Countable Ordinals (after Sch\\\"{u}tte)]{Kurt Schütte's Axiomatic Definition of Countable Ordinals}\n\n\\label{chap:schutte} \n%ON\n\nIn the present chapter, we compare our implementation of the segment $[0,\\epsilon_0)$ with a mathematical text in order to ``validate'' our constructions.\nOur reference here is the axiomatic definition of the set of countable ordinals,\nin chapter V of Kurt Schütte's book `` Proof Theory ''~\\cite{schutte}.\n\n\\begin{remark}\n\\emph{In all this chapter, the word ``ordinal'' will be considered as a synonymous of\n``countable ordinal''} \n\\end{remark}\n\n\n\nSchütte's definition of countable ordinals relies on the following three axioms:\n\nThere exists a strictly ordered set , such that\n\\begin{enumerate}\n\\item $(\\mathbb{O},<)$ is well-ordered\n\\item Every bounded subset of $\\mathbb{O}$ is countable\n\\item Every countable subset of $\\mathbb{O}$ is bounded.\n\\end{enumerate}\n\nStarting with these three axioms, Schütte re-defines the vocabulary about ordinal numbers: the null ordinal $0$, limits and successors, the addition of ordinals, the infinite ordinals $\\omega$, $\\epsilon_0$, $\\Gamma_0$, etc.\n\nThis chapter describes an adaptation to \\coq{} of Schütte's axiomatization. \n Unlike the rest of our libraries, our library\n\\href{../theories/html/hydras.Schutte.Schutte.html}{Ordinals.Schutte}\nis not constructive, and relies on several axioms.\n\n\\begin{itemize}\n\\item First, please keep in mind that the set of countable ordinals is not countable. Thus, we cannot hope to represent all countable ordinals as finite terms of an inductive type, which was possible with the set of ordinals strictly less than $\\epsilon_0$ (resp. $\\Gamma_0$)\n\\item We tried to be as close as possible to K. Schütte's text, which uses ``classical'' mathematics : excluded middle, Hilbert's $\\epsilon$ (choice) and Russel's $\\iota$ (definite description) operators. Both operators allow us to write definitions close to the natural mathematical language, such as ``$\\textrm{succ}$ is \\emph{the} least ordinal strictly greater than $\\alpha$''\n\\item Please note that only the library \\href{../theories/html/hydras.Schutte.Schutte.html}{Schutte/*.v} is ``contaminated'' by axioms, and that the rest of our libraries remain constructive.\n\\end{itemize}\n\n\\section{Declarations and Axioms}\n\nLet us declare a type \n\\texttt{Ord} for representing countable ordinals, and a binary relation\n \\texttt{lt}. Note that, in our development, \\texttt{Ord} is a type, while the \\emph{set} of countable ordinals (called $\\mathbb{O}$ by Schütte) \nis the full set over the type \\texttt{Ord}.\n\n\\label{types:Ord} \n\nWe use Florian Hatat's library on countable sets, written as he was a student of \\emph{\\'Ecole Normale Supérieure de Lyon}. A set $A$ is countable if there is an injective function from $A$ to $\\mathbb{N}$ (see \nLibrary \\href{../theories/html/hydras.Schutte.Countable.html}%\n{\\texttt{Schutte.Countable}}).\n\n\n\\vspace{6pt}\n\n\\emph{From Module\\href{../theories/html/hydras.Schutte.Schutte_basics.html}%\n{\\texttt{Schutte.Schutte\\_basics}}}\n\n\\begin{Coqsrc}\nParameter Ord : Type.\nParameter lt : relation Ord.\nInfix \"<\" := lt : schutte_scope.\n\nDefinition ordinal := Full_set Ord.\n\\end{Coqsrc}\n\nSchütte's first axiom tells that \\texttt{lt} is a well order on the set \n\\texttt{ordinal} (The class \\texttt{WO} is defined in\nModule~\\href{../theories/html/hydras.Schutte.Well_Orders.html}%\n{\\texttt{Well\\_Orders.v}}).\n\n\\label{types:WO}\n\n\\begin{Coqsrc}\nVariables (M:Type)\n (Lt : relation M).\n \nClass WO : Type:=\n {\n Lt_trans : Transitive Lt;\n Lt_irreflexive : forall a:M, ~ (Lt a a);\n well_order : forall (X:Ensemble M)(a:M),\n In X a ->\n exists a0:M, least_member X a0\n }.\n\\end{Coqsrc}\n\n\n\n\\begin{Coqsrc}\n Axiom AX1 : WO lt.\n\\end{Coqsrc}\n\nThe second and third axioms say that a subset $X$ of $\\mathbb{O}$ is\n(strictly) bounded if and only if it is countable. \n\n\n\n\\begin{Coqsrc}\nAxiom AX2 : forall X: Ensemble Ord, \n (exists a, (forall y, In X y -> y < a)) ->\n countable X.\n\nAxiom AX3 : forall X : Ensemble Ord,\n countable X -> \n exists a, forall y, In X y -> y < a.\n\\end{Coqsrc}\n\n\\texttt{AX2} and \\texttt{AX3} could have been replaced by a single axiom (using the \\texttt{iff} connector), but we decide to respect as most as possible the structure of Schütte's definitions.\n\n\\section{Additional Axioms}\n\nThe adaptation of Schütte's mathematical discourse to \\coq{} led us to\nimport a few axioms from the standard library. We encourage the reader to consult \\coq{}'s FAQ about the safe use of axioms\n \\url{https://github.com/coq/coq/wiki/The-Logic-of-Coq#axioms}.\n\n\\subsubsection{Classical Logic}\n\nIn order to work with classical logic, we import the module\n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Logic.Classical.html}{Coq.Logic.Classical} of \\coq{}'s standard library, specifially the following axiom:\n\n\\begin{Coqsrc}\n Axiom classic : forall P:Prop, P \\/ ~P.\n\\end{Coqsrc}\n\n\n\\subsubsection{Description Operators}\n\nIn order to respect Schütte's style, we imported also the library \n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Logic.Epsilon.html}{\\texttt{Coq.Logic.Epsilon}}. The rest of this section presents a few examples of\nhow Hilbert's choice operator and Church's definite description allow us\n to write understandable definitions (close to the mathematical natural language).\n\n\n\\subsubsection{The Definition of zero}\n\nAccording to the definition of a well order, every non-empty subset of \\texttt{Ord} has a least element. Furthermore, this least element is unique.\n\n\n\\begin{Coqsrc}\nRemark R : exists! z : Ord, least_member lt ordinal z.\nProof.\n destruct inh_Ord as [a]; apply (well_order (WO:=AX1)) with a .\n split.\nQed.\n\\end{Coqsrc}\n\nAssume we want to call this element \\texttt{zero}.\n\n\n\n\\begin{Coqsrc}\nDefinition zero : Ord.\nProof.\n Fail destruct R.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nThe command has indeed failed with message:\nCase analysis on sort Type is not allowed for inductive \ndefinition ex.\n\\end{Coqanswer}\n\n\nIndeed, the basic logic of \\coq{} does not allow us to eliminate a proof of a proposition \n$\\exists\\,x:A,\\,P(x)$ for building a term whose type lies in the sort \\texttt{Type}. \nThe reasons for this impossibility are explained in many documents~\\cite{BC04, chlipalacpdt2011, Coq}.\n\nLet us import the library \\texttt{Coq.Logic.Epsilon}, which contains the following axiom and lemmas.\n\n\n\\begin{Coqsrc}\nAxiom epsilon_statement:\n forall (A : Type) (P : A->Prop), inhabited A ->\n {x : A | (exists x, P x) -> P x}.\n\\end{Coqsrc}\n\nHilbert's $\\epsilon$ \\emph{operator} is derived from this axiom.\n\n\\begin{Coqsrc}\n Definition epsilon (A : Type) (i:inhabited A) (P : A->Prop) : A\n := proj1_sig (epsilon_statement P i).\n\nLemma constructive_indefinite_description :\n forall (A : Type) (P : A->Prop),\n (exists x, P x) -> { x : A | P x }.\n\\end{Coqsrc}\n\n\n\n\nIf we consider the \\emph{unique existential} quantifier $\\exists!$, we obtain\nChurch's \\emph{definite description operator}.\n\n\\begin{Coqsrc}\nDefinition iota (A : Type) (i:inhabited A) (P : A->Prop) : A\n := proj1_sig (iota_statement P i).\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\n Lemma constructive_definite_description :\n forall (A : Type) (P : A->Prop),\n (exists! x, P x) -> { x : A | P x }.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nDefinition iota_spec (A : Type) (i:inhabited A) (P : A->Prop) :\n (exists! x:A, P x) -> P (iota i P)\n := proj2_sig (iota_statement P i).\n\\end{Coqsrc}\n\n\n\nIndeed, the operators \\texttt{epsilon} and \\texttt{iota} allowed us to make our definitions \nquite close to Schütte's text. Our libraries \\href{../theories/html/hydras.Schutte.MoreEpsilonIota.html}%\n{\\texttt{Schutte.MoreEpsilonIota}}\nand\n\\href{../theories/html/hydras.Schutte.PartialFun.html}%\n{\\texttt{Schutte.PartialFun}} are extensions of \\texttt{Coq.logic.Epsilon} for making easier \nsuch definitions. See also an article in french~\\cite{PCiota}. \n\n\n\n\\begin{Coqsrc}\nClass InH (A: Type) : Prop :=\n InHWit : inhabited A.\n\nDefinition some {A:Type} {H : InH A} (P: A -> Prop) := \n epsilon (@InHWit A H) P.\n\nDefinition the {A:Type} {H : InH A} (P: A -> Prop) := \n iota (@InHWit A H) P.\n\\end{Coqsrc}\n\nIn order to use these tools, we had to tell \\coq{} that the type \\texttt{Ord} is not empty:\n\n\\begin{Coqsrc}\nAxiom inh_Ord : inhabited Ord.\n\\end{Coqsrc}\n\n\nWe are now able to define \\texttt{zero} as the least ordinal. For this purpose,\nwe define a function returning the least element of any [non-empty] subset.\n\n\n\\begin{Coqsrc}\nDefinition the_least {M: Type} {Lt}\n {inh : InH M} {WO: WO Lt} (X: Ensemble M) : M :=\n the (least_member Lt X ).\n\\end{Coqsrc}\n\n\n\\vspace{4pt}\n\nFrom Module \\href{../theories/html/hydras.Schutte.Schutte_basics.html}%\n{\\texttt{~Schutte.Schutte\\_basics}}\n\n\\label{Constants:zero:Ord}\n\n\\begin{Coqsrc}\nDefinition zero: Ord :=the_least ordinal.\n\\end{Coqsrc}\n\nWe want to prove now that zero is less than or equal to any ordinal number.\n\n\\begin{Coqsrc}\nLemma zero_le (alpha : Ord) : zero <= alpha.\nProof.\n unfold zero, the_least, the; apply iota_ind.\n\\end{Coqsrc}\n\nAccording to the use of the description operator \\texttt{iota}, we have to solve two trivial sub-goals.\n\\begin{enumerate}\n\\item Prove that there exists a unique least member of \\texttt{Ord}\n\\item Prove that being a least member of \\texttt{Ord} entails the announced inequality \n\\end{enumerate}\n\n\n\\begin{Coqanswer}\n2 subgoals (ID 155)\n \n alpha : Ord\n ============================\n exists ! x : Ord, least_member lt ordinal x\n\nsubgoal 2 (ID 156) is:\n forall a : Ord, unique (least_member lt ordinal) a -> \n a <= alpha\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\n - apply the_least_unicity, Inh_ord.\n - destruct 1 as [[_ H1] _]; apply H1; split. \nQed.\n\\end{Coqsrc}\n\n\\subsubsection{Remarks on \\texttt{epsilon} and \\texttt{iota}}\n\n What would happen in case of a misuse of \\texttt{epsilon} or \\texttt{iota} ?\nFor instance, one could give a unsatisfiable specification to \\texttt{epsilon} or \na specification for \\texttt{iota} that admits several realizations.\n\nLet us consider an example:\n\n\\begin{Coqbad}\nModule Bad.\n\n Definition bottom := the_least (Empty_set Ord).\n\\end{Coqbad}\n\n\\begin{Coqanswer}\n bottom is defined\n\\end{Coqanswer}\n\nSince we won't be able to prove the proposition\n\\linebreak \\Verb|{exists! a: Ord, least_member (Empty_set Ord) a|, the only properties we would be able to prove about \\texttt{bottom} would be \\emph{trivial} properties, \n\\emph{i.e.}, satisfied by \\emph{any} element of type \\texttt{Ord}, like for instance\n\\texttt{bottom = bottom}, or \\texttt{zero <= bottom}.\n\n\\begin{Coqbad}\nLemma le_zero_bottom : zero <= bottom. \nProof. apply zero_le. Qed.\n\nLemma bottom_eq : bottom = bottom.\nProof. trivial. Qed.\n\nLemma le_bottom_zero : bottom <= zero.\nProof.\n unfold bottom, the_least, the; apply iota_ind.\n\\end{Coqbad}\n\n\\begin{Coqanswer}\n2 subgoals (ID 413)\n \n ============================\n exists ! x : Ord, least_member lt (Empty_set Ord) x\n\nsubgoal 2 (ID 414) is:\n forall a : Ord, unique (least_member lt (Empty_set Ord)) a -> \n a <= zero\n\\end{Coqanswer}\n\n\\begin{Coqbad}\nAbort.\nEnd Bad.\n\\end{Coqbad}\n\n\nIn short, using \\texttt{epsilon} and \\texttt{iota} in our implementation of countable ordinals after Schütte has two main advantages.\n\n\n\\begin{itemize}\n\\item It allows us to give a \\emph{name} (using \\texttt{Definition}) two witnesses \nof existential quantifiers (let us recall that, in classical logic, one may consider non-constructive proofs of existential statements)\n\\item By separating definitions from proofs of [unique] existence, one may make definitions more concise and readable. Look for instance at \nthe definitions of \\texttt{zero}, \\texttt{succ}, \\texttt{plus}, etc. in the rest of this chapter.\n\\end{itemize}\n%%%% ICI ICI \n\n\\section{The Successor Function}\n\nThe definition of the function \\texttt{succ:Ord -> Ord} is very concise. The successor of any ordinal $\\alpha$ is the smallest ordinal strictly greater than $\\alpha$.\n\n\\label{Functions:succ-sch}\n\n\\begin{Coqsrc}\nDefinition succ (alpha : Ord) := the_least (fun beta => alpha < beta).\n\\end{Coqsrc}\n\nUsing \\texttt{succ}, we define the folloing predicates.\n\n\\begin{Coqsrc}\nDefinition is_succ (alpha:Ord) := exists beta, alpha = succ beta.\n\nDefinition is_limit (alpha:Ord) := alpha <> zero /\\ ~ is_succ alpha.\n\\end{Coqsrc}\n\n\n\n% \\begin{remark}\n% Please look at remark~\\vref{warning:coercions}.\n% \\end{remark}\n\nHow do we prove properties of the successor function?\nFirst, we make its specification explicit.\n\n\\begin{Coqsrc}\nDefinition succ_spec (alpha:Ord) :=\n least_member lt (fun z => alpha < z).\n\\end{Coqsrc}\n\nThen, we prove that our function \\texttt{succ} meets this specification. \n\n\n\\begin{Coqsrc}\nLemma succ_ok : forall alpha, succ_spec alpha (succ alpha).\nProof.\n intros; unfold succ, the_least, the; apply iota_spec.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n1 subgoal (ID 172)\n \n alpha : Ord\n ============================\n exists ! x : Ord, succ_spec alpha x\n\\end{Coqanswer}\n\nWe have now to prove that the set of all ordinals strictly greater than $\\alpha$ has a unique least element. But the singleton set $\\{\\alpha\\}$ is countable, hence bounded (by the axiom \\texttt{AX3}). Hence; the set $\\{\\beta\\in\\mathbb{O}|\\alpha < \\beta\\}$ is not empty\nand therefore has a unique least element.\n\nThe \\coq{} proof script is quite short.\n\n\\begin{Coqsrc}\n destruct (@AX3 (Singleton _ alpha)).\n - apply countable_singleton.\n - unfold succ_spec; apply the_least_unicity; exists x; intuition.\nQed. \n\\end{Coqsrc}\n\n\nWe can ``uncap'' the description operator for proving properties of the\n\\texttt{succ} function.\n\n\\begin{Coqsrc}\nLemma lt_succ (alpha : Ord) : alpha < succ alpha.\nProof.\n destruct (succ_ok alpha); tauto.\nQed.\n\nHint Resolve lt_succ : schutte.\n\nLemma lt_succ_le (alpha beta : Ord):\n alpha < beta -> succ alpha <= beta.\nProof with eauto with schutte.\n intros H; pattern (succ alpha); apply the_least_ok ... \n exists (succ alpha); red;apply lt_succ ...\nQed.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nLemma lt_succ_le_2 (alpha beta : Ord):\n alpha < succ beta -> alpha <= beta.\n\nLemma succ_mono (alpha beta : Ord):\n alpha < beta -> succ alpha < succ beta.\n\nLemma succ_monoR (alpha beta : Ord) :\n succ alpha < succ beta -> alpha < beta.\n\nLemma lt_succ_lt (alpha beta : Ord) :\n is_limit beta -> alpha < beta -> succ alpha < beta.\n\\end{Coqsrc}\n\n\\section{Finite Ordinals}\n\nUsing \\texttt{succ}, it is now easy to define recursively all the finite ordinals.\n\n\\label{sect:notation-F-sch}\n\n\\begin{Coqsrc}\nReserved Notation \"'F' n\" (at level 29) .\n\nFixpoint finite (i:nat) : Ord :=\n match i with \n | 0 => zero\n | S i => succ (F i)\n end\nwhere \"'F' i\" := (finite i) : schutte_scope.\n\nCoercion finite : nat >-> Ord.\n\\end{Coqsrc}\n\n\\section{The Definition of \\texttt{omega}}\nIn order to define $\\omega$, the first infinite ordinal, we use an operator which\n``returns'' the least upper bound (if it exists) of a subset $X\\subseteq \\mathbb{O}$.\nFor that purpose, we first use a predicate:\n(\\texttt{is\\_lub $D$ \\textit{lt} $X$ $a$}) if $a$ belongs to $D$ and is the least \nupper bound of $X$ (with respect to \\textit{lt}).\n\n\n\\begin{Coqsrc}\nDefinition is_lub (M:Type)\n (D : Ensemble M)\n (lt : relation M)\n (X:Ensemble M)\n (a:M) :=\n In _ D a /\\ upper_bound D lt X a /\\\n (forall y, In _ D y -> upper_bound D lt X y -> \n y = a \\/ lt a y).\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nDefinition sup_spec X lambda := is_lub ordinal lt X lambda.\n\nDefinition sup (X: Ensemble Ord) : Ord := the (sup_spec X).\n\nNotation \"'|_|' X\" := (sup X) (at level 29) : schutte_scope.\n\\end{Coqsrc}\n\n\n\nThen, we define the function \\texttt{omega\\_limit} which returns the least upper bound \nof the (denumerable) range of any sequence \\texttt{s: nat -> Ord}. \nBy \\texttt{AX3} this range is bounded, hence the set of its upper bounds is not empty and has a least element.\n\n\n\\begin{Coqsrc}\nDefinition omega_limit (s:nat->Ord) : Ord \n := |_| (seq_range s).\n\\end{Coqsrc}\n\nThen we define \\texttt{omega} as the limit of the sequence of finite ordinals.\n\n\n\\label{sect:notation-omega}\n\\begin{Coqsrc}\nDefinition _omega := omega_limit finite.\n\nNotation \"'omega'\" := (_omega) : schutte_scope.\n\\end{Coqsrc}\n\n\n\nAmong the numerous properties of the ordinal $\\omega$, les us quote the following ones\n(proved in Module \n\\href{../theories/html/hydras.Schutte.Schutte_basics.html\\#finite_lt_omega}{\\texttt{Schutte.Schutte\\_basics}})\n\n\\begin{Coqsrc}\nLemma finite_lt_omega : forall i: nat, i < omega.\n\nLemma lt_omega_finite alpha : Ord) : \n alpha < omega -> exists i:nat, alpha = i.\n\nLemma is_limit_omega : is_limit omega.\n\\end{Coqsrc}\n\n\n\\subsection{Ordering Functions and Ordinal Addition}\n\nAfter having defined the finite ordinals and the infinite ordinal $\\omega$, we define the sum $\\alpha+\\beta$ of two countable ordinals.\nSchütte's definition looks like the following one:\n\n\\begin{quote}\n``$\\alpha+\\beta$ is the $\\beta$-th ordinal greater than or equal to $\\alpha$''\n\\end{quote}\n\n\nThe purpose of this section is to give a meaning to the construction\n``the $\\alpha$-th element of $X$'' where $X$ is any non-empty subset of $\\mathbb{O}$.\nWe follow Schütte's approach, by defining the notion of \\emph{ordering functions},\na way to associate a unique ordinal to each element of $X$.\nComplete definitions and proofs can be found in Module\n \\href{../theories/html/hydras.Schutte.Ordering_Functions.html}%\n{\\texttt{Schutte.Ordering\\_Functions}} ).\n\n\\subsection{Definitions}\n\nA \\emph{segment} is a set $A$ of ordinals such that, whenever $\\alpha\\in A$ and\n$\\beta<\\alpha$, then $\\beta\\in A$; a segment is \\emph{proper} if it strictly included in $\\mathbb{O}$.\n\n\\begin{Coqsrc}\n Definition segment (A: Ensemble Ord) :=\n forall alpha beta, In A alpha -> beta < alpha -> In A beta.\n\nDefinition proper_segment (A: Ensemble Ord) :=\n segment A /\\ ~ Same_set A ordinal.\n\\end{Coqsrc}\n\n\nLet $A$ be a segment, and $B$ a subset of $\\mathbb{O}$ : an \\emph{ordering function for $A$ and $B$} is a strictly increasing bijection from $A$ to $B$.\nThe set $B$ is said to be an \\emph{ordering segment} of $A$.\nOur definition in \\coq{} is a direct translation of the mathematical text of~\\cite{schutte}.\n\n\\index{Maths!Ordering functions}\n\n\\begin{Coqsrc}\nDefinition ordering_function (f : Ord -> Ord)(A B : Ensemble Ord) :=\n segment A /\\\n (forall a, In A a -> In B (f a)) /\\\n (forall b, B b -> exists a, In A a /\\ f a = b) /\\\n forall a b, In A a -> In A b -> a < b -> f a < f b.\n\nDefinition ordering_segment (A B : Ensemble Ord) :=\n exists f : Ord -> Ord, ordering_function f A B.\n\\end{Coqsrc}\n\n\nWe are now able to associate with any subset $B$ of $\\mathbb{O}$ its ordering segment and ordering function.\n\n\\begin{Coqsrc}\nDefinition the_ordering_segment (B : Ensemble Ord) :=\n the (fun x => ordering_segment x B).\n\nDefinition ord (B : Ensemble Ord) := \n some (fun f => ordering_function f (the_ordering_segment B) B).\n\\end{Coqsrc}\n\nThus (\\texttt{ord $B \\;\\alpha$}) is the $\\alpha$-th element of $B$.\nPlease note that the last definition uses the epsilon-based operator \\texttt{some} and\nnot \\texttt{the}. This is due to the fact that we cannot prove the unicity (w.r.t. Leibniz' equality) of the ordering function of a given set. \nBy contrast, we admit the axiom \\texttt{Extensionality\\_Ensembles}, from the library \n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Sets.Ensembles.html}{Coq.Sets.Ensembles}, so we use the operator \\texttt{the} in the definition of\n\\texttt{the\\_ordering\\_segment}.\n\nOne of the main theorems of\n\\href{../theories/html/hydras.Schutte.Ordering_Functions.html\\#ordering_function_ex}%\n{\\texttt{Ordering\\_Functions}} \nassociates a unique segment and a unique (up to extensionality) ordering function to every subset $B$ of $\\mathbb{O}$.\n\n\\begin{Coqsrc}\nAbout ordering_function_ex.\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\nforall B : Ensemble Ord,\n exists ! S : Ensemble Ord, \n exists f : Ord -> Ord, ordering_function f S B\n\\end{Coqanswer}\n\n\n\\begin{Coqanswer}\nordering_function_unicity :\nforall (B S1 S2 : Ensemble Ord) (f1 f2 : Ord -> Ord),\nordering_function f1 S B ->\nordering_function f2 S2 B -> \nfun_equiv f1 f2 S1 S2\n\\end{Coqanswer}\n\nThus, our function \\texttt{ord} which enumerates the elements of $B$ is defined in a non-ambiguous way.\nLet us quote the following theorems (see Library\n\\href{../theories/html/hydras.Schutte.Ordering_Functions.html}%\n{\\texttt{Schutte.Ordering\\_Functions}} for more details).\n \n\n\\begin{Coqsrc}\nTheorem ordering_le : forall f A B,\n ordering_function f A B ->\n forall alpha, In A alpha -> alpha <= f alpha.\n\nTh_13_5_2 :\nforall (A B : Ensemble Ord) (f : Ord -> Ord),\nordering_function f A B -> Closed B -> continuous f A B\n\\end{Coqsrc}\n\n\n\\subsection{Ordinal Addition}\n\nWe are now ready to define and study addition on the type \\texttt{Ord}.\nThe following definitions and proofs can be consulted in Module\n\\href{../theories/html/hydras.Schutte.Addition.html}%\n{\\texttt{Schutte.Addition.v}}.\n\n\\begin{Coqsrc}\nDefinition plus alpha := ord (ge alpha).\nNotation \"alpha + beta \" := (plus alpha beta) : schutte_scope.\n\\end{Coqsrc}\n\nIn other words, $\\alpha + \\beta$ is the $\\beta$-th ordinal greater than or equal to $\\alpha$. \nThanks to generic properties of ordering functions, we can show the following \nproperties of addition on $\\mathbb{O}$. First, we prove a useful lemma:\n\n\\begin{Coqsrc}\nLemma plus_elim (alpha : Ord) :\n forall P : (Ord->Ord)->Prop,\n (forall f: Ord->Ord, \n ordering_function f ordinal (ge alpha)-> P f) ->\n P (plus alpha).\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nLemma alpha_plus_zero (alpha: Ord): alpha + zero = alpha.\nProof.\n pattern (plus alpha); apply plus_elim; eauto.\n \\end{Coqsrc}\n\n \\begin{Coqanswer}\n 1 subgoal (ID 24)\n \n alpha : Ord\n ============================\n forall f : Ord -> Ord,\n ordering_function f ordinal (ge alpha) -> \n f zero = alpha\n \\end{Coqanswer}\n\n \\begin{Coqsrc}\n (* rest of proof skipped *)\n \\end{Coqsrc}\n\nThe following lemmas are proved the same way.\n\n \\begin{Coqsrc}\n\nLemma zero_plus_alpha (alpha : Ord) : zero + alpha = alpha.\n\nLemma le_plus_l (alpha beta : Ord) : alpha <= alpha + beta.\n\nLemma le_plus_r (alpha beta : Ord) : beta <= alpha + beta.\n\nLemma plus_mono_r (alpha beta gamma : Ord) : \n beta < gamma -> alpha + beta < alpha + gamma.\n\nLemma plus_of_succ (alpha beta : Ord) :\n alpha + (succ beta) = succ (alpha + beta).\n\nTheorem plus_assoc (alpha beta gamma : Ord) :\n alpha + (beta + gamma) = (alpha + beta) + gamma.\n\nLemma one_plus_omega : 1 + omega = omega.\n\nLemma finite_plus_infinite (n : nat) (alpha : Ord) :\n omega <= alpha -> n + alpha = alpha.\n\\end{Coqsrc} \n\n\nIt isinteresting to compare the proof of these lemmas with the \ncomputational proofs of the corresponding statements in Module\n\\href{../theories/html/hydras.Epsilon0.T1.html}%\n{\\texttt{Epsilon0.T1}}. \nFor instance, the proof of the lemma \n\\texttt{one\\_plus\\_omega} uses the continuity of ordering functions (applied to \\texttt{(plus 1)}) and compares the limit of the $\\omega$-sequences $i_{(i \\in \\mathbb{N})}$ and\n$(1+i)i_{(i \\in \\mathbb{N})}$, whereas in the library \\texttt{Epsilon0/T1}, the equality \n$1+\\omega=\\omega$ is just proved with \\texttt{reflexivity}!\n\n\n\n\\subsubsection{Multiplication by a Natural Number}\n\nThe multiplication of an ordinal by a natural number is defined in terms of addition.\nThis operation is useful for the study of Cantor normal forms.\n\n\\begin{Coqsrc}\nFixpoint mult_Sn (alpha:Ord)(n:nat){struct n}: Ord :=\n match n with \n | 0 => alpha\n | S p => mult_Sn alpha p + alpha\n end.\n\nDefinition mult_fin_r alpha n :=\n match n with\n 0 => zero\n | S p => mult_Sn alpha p\n end.\n\nNotation \"alpha * n\" := (mult_fin_r alpha n) : schutte_scope.\n\\end{Coqsrc}\n\n\\section{The Exponential of Basis \\texorpdfstring{$\\omega$}{omega}}\n\nIn this section, we define the function which maps any $\\alpha\\in\\mathbb{O}$ to\nthe ordinal $\\omega^\\alpha$, also written \n$\\phi_0(\\alpha)$. \nIt is an opportunity to apply the definitions and results of the preceding section. \nIndeed, Schütte first defines a subset of $\\mathbb{O}$: the set of additive principal ordinals, and $\\phi_0$ is just defined as the ordering function of this set.\n\n\\subsection{Additive Principal Ordinals}\n\n\\index{Maths!Additive principal ordinals}\n\n\\begin{definition}\nA non-zero ordinal $\\alpha$ is said to be \\emph{additive principal} if, for all $\\beta<\\alpha$, $\\beta+\\alpha$ is equal to $\\alpha$.\nWe call \\texttt{AP} the set of additive principal ordinals.\n\n\\end{definition}\n\n\n\n\\noindent\\emph{From Module \\href{../theories/html/hydras.Schutte.AP.html}%\n{\\texttt{Schutte.AP}}}\n\n\\begin{Coqsrc}\nDefinition AP : Ensemble Ord :=\n fun alpha => \n zero < alpha /\\\n (forall beta, beta < alpha -> beta + alpha = alpha).\n\\end{Coqsrc}\n\n\\subsection{The Function \\texttt{phi0}}\n\nLet us call $\\phi_0$ the ordering function of \\texttt{AP}.\nIn the mathematical text, we shall use indifferently the notations $\\omega^\\alpha$ and$\\phi_0(\\alpha)$. \n\n\n\\begin{Coqsrc}\nDefinition phi0 := ord AP.\n\nNotation \"'omega^'\" := phi0 (only parsing) : schutte_scope.\n\\end{Coqsrc}\n\n\\subsection{Omega-towers and the Ordinal \\texorpdfstring{$\\epsilon_0$}{epsilon0}}\n\n\nUsing $\\phi_0$, we can define recursively the set of finite omega-towers.\n\n\n\\begin{Coqsrc}\nFixpoint omega_tower (i : nat) : Ord :=\n match i with\n 0 => 1\n | S j => phi0 (omega_tower j)\n end.\n\\end{Coqsrc}\n\n\\label{sect:epsilon0-as-limit}\nThen, the ordinal $\\epsilon_0$ is defined as the limit of the sequence of all finite towers (a kind of infinite tower).\n\n\\begin{Coqsrc}\nDefinition epsilon0 := omega_limit omega_tower.\n\\end{Coqsrc}\n\nThe rest of our library \\texttt{AP} is devoted to the proof of properties of additive principal ordinals, hence of the ordering function $\\phi0$ and the ordinal $\\epsilon_0$ (which we could not express within the type \\texttt{T1}).\n\n\\subsection{Properties of the Set \\texttt{AP}}\n\nThe set of additive principal ordinals is not empty: it contains at least the ordinals $1$ and $\\omega$. \n\n\\begin{Coqsrc}\nLemma AP_one : In AP 1.\n\nLemma AP_omega : In AP omega.\n\\end{Coqsrc}\n\nMoreover, $1$ is the least principal ordinal and $\\omega$ is the second element of\n\\texttt{AP}.\n\n\n\\begin{Coqsrc}\nLemma least_AP: least_member lt AP 1. \n\nLemma omega_second_AP :\n least_member lt \n (fun alpha => 1 < alpha /\\ In AP alpha)\n omega.\n\\end{Coqsrc}\n\nThe set \\texttt{AP} is \\emph{closed} under addition, and unbounded.\n\n\\begin{Coqsrc}\nLemma AP_plus_closed (alpha beta gamma : Ord): \n In AP alpha -> beta < alpha -> gamma < alpha ->\n beta + gamma < alpha.\n\nTheorem AP_unbounded : Unbounded AP.\n\\end{Coqsrc}\n\nFinally, \\texttt{AP} is (topologically) \\emph{closed} and ordered by the segment of all countable ordinals.\n\n\\index{Predicates!Closed}\n\n\\begin{Coqsrc} \nDefinition Closed (B : Ensemble Ord) : Prop := \n forall M, Included M B -> Inhabited _ M -> \n countable M -> In B (|_| M).\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nTheorem AP_closed : Closed AP.\n\nLemma AP_o_segment : the_ordering_segment AP = ordinal.\n\\end{Coqsrc}\n\n\\subsubsection{Properties of the Function \\texorpdfstring{$\\phi_0$}{phi0}}\n \nThe ordering function $\\phi_0$ of the set \\texttt{AP} is defined on the full set $\\mathbb{O}$ and is continuous (Schütte calls such a function \\emph{normal}).\n\n\\begin{Coqsrc}\nTheorem normal_phi0 : normal phi0 AP.\n\\end{Coqsrc}\n\nThe following properties come from the definition of $\\phi_0$ as the ordering function of \\texttt{AP}. It may be interesting to compare these proofs with the computational ones described in Chapter ~\\ref{chap:T1}.\n\n\\begin{Coqsrc}\nLemma AP_phi0 (alpha : Ord) : In AP (phi0 alpha).\n\nLemma phi0_zero : phi0 zero = 1.\n\nLemma phi0_mono (alpha beta : Ord) :\n alpha < beta -> phi0 alpha < phi0 beta.\n\nLemma phi0_inj (alpha beta : Ord) :\n phi0 alpha = phi0 beta -> alpha = beta.\n\nLemma phi0_sup : forall (U: Ensemble Ord),\n Inhabited _ U -> countable U -> phi0 (|_| U) = |_| (image U phi0).\n\nLemma is_limit_phi0 (alpha : Ord) :\n zero < alpha -> is_limit (phi0 alpha).\n\nLemma omega_eq : omega = phi0 1. \n\nLemma phi0_le (alpha : Ord) : alpha <= phi0 alpha.\n\\end{Coqsrc}\n\nPlease note that the lemma \\texttt{omega\\_eq} above, is consistent with the interpretation of the ordering function $\\phi_0$ as the exponential of basis $\\omega$. Indeed we could have written this lemma with our alternative notation:\n\n\\begin{Coqsrc}\n Lemma omega_eq : omega = omega^ 1.\n\\end{Coqsrc}\n\n\\section{More about \\texorpdfstring{$\\epsilon_0$}{\\texttt{epsilon0}}}\n\nLet us recall that the limit ordinal $\\epsilon_0$ cannot be written within the type \\texttt{T1}. Since we are now considering the set of all countable ordinals, we can now prove some properties of this ordinal.\n\n\nWe prove the inequality $\\alpha<\\omega^\\alpha$ whenever $\\alpha < \\epsilon_0$.\n\\emph{Note that this condition was implicit in Module \\href{../theories/html/hydras.Epsilon0/T1.html\\#lt_phi0}%\n{\\texttt{Epsilon0.T1}}.}\n\n\\begin{Coqsrc}\nLemma lt_phi0 (alpha : Ord):\n alpha < epsilon0 -> alpha < phi0 alpha.\n\\end{Coqsrc}\n\nThe proof is as follows:\n\\begin{enumerate}\n\\item Since $\\alpha<\\epsilon_0$, consider the least $i$ such that $\\alpha$ is strictly less than the omega-tower of height $i$.\n\\item\n \\begin{itemize}\n \\item If $i=0$, then the result is trivial (because $\\alpha=0$)\n \\item Otherwise let $i=j+1$; \n $\\alpha$ is greater than or equal to the omega-tower of height $j$.\n By monotonicity, $\\phi_0(\\alpha)$ is greater than or equal to \n the omega-tower of height $j+1$, thus strictly greater than $\\alpha$\n \\end{itemize}\n \\end{enumerate}\n\nMoreover, $\\epsilon_0$ is the least ordinal $\\alpha$ that verifies the equality \n$\\alpha = \\omega^\\alpha$, in other words the least fixpoint of the function $\\phi_0$.\n\n\\begin{Coqsrc}\nTheorem epsilon0_lfp : least_fixpoint lt phi0 epsilon0.\n\\end{Coqsrc}\n\n\n\\section{Critical Ordinals}\n\n\\index{Maths!Critical ordinals}\n\nFor any (countable) ordinal $\\alpha$, the set $\\textit{Cr}(\\alpha)$ is inductively defined \nas follows by Schütte (p.81 of~\\cite{schutte}).\n\n\\begin{quote}\n \\begin{itemize}\n \\item $\\textit{Cr}(0)$ is the set \\textit{AP} of additive principal ordinals.\n \\item If $0<\\alpha$, then $\\textit{Cr}(\\alpha)$ is the intersection of all the sets of fixpoints of the $\\textit{Cr}(\\beta)$ for $\\beta<\\alpha$.\n \\end{itemize}\n\\end{quote}\n\nThis definition is translated in \\coq{} in \nModule \\href{../theories/html/hydras.Schutte.Critical.html}%\n{\\texttt{Schutte.Critical}}, as the least fixpoint of a functional. \n\n\n\\begin{Coqsrc}\nDefinition Cr_fun : forall alpha : Ord,\n (forall beta : Ord, beta < alpha -> Ensemble Ord) ->\n Ensemble Ord \n:= \n fun (alpha :Ord)\n (Cr : forall beta, \n beta < alpha -> Ensemble Ord) \n (x : Ord) => (\n (alpha = zero /\\ AP x) \\/\n (zero < alpha /\\\n forall beta (H:beta < alpha),\n the_ordering_segment (Cr beta H) x /\\ ord (Cr beta H) x = x)).\n\nDefinition Cr (alpha : Ord) : Ensemble Ord := \n (Fix all_ord_acc (fun (_:Ord) => Ensemble Ord) Cr_fun) alpha.\n\\end{Coqsrc}\n\n\\label{sect:phi-schutte}\n\n\\begin{Coqsrc}\nDefinition phi (alpha : Ord) : Ord -> Ord \n := ord (Cr alpha).\n\nDefinition A (alpha : Ord) : Ensemble Ord :=\n the_ordering_segment (Cr alpha).\n\\end{Coqsrc}\n\nFor instance, we prove that $\\textit{Cr}(0)$ is the set of additive principals and that $\\epsilon_0$\nbelongs to $\\textit{Cr}(1)$.\n\n\\begin{Coqsrc}\nLemma Cr_zero_AP : Cr 0 = AP\n\nLemma epsilon0_Cr1 : In (Cr 1) epsilon0.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\n Prove that $\\epsilon_0$ is the least element of $\\textit{Cr}(1)$.\n\\end{exercise}\n\n\n\\subsection{A flavor of Infinity}\n\n\n\nThe family of the $\\textit{Cr}(\\alpha)$s is made of infinitely many unbounded (hence infinite) sets.\nLet us quote Lemma 5, p. 82 of~\\cite{schutte}:\n\\begin{quote}\n For all $\\alpha$, the set $\\textit{Cr}(\\alpha)$ is closed (for the least upper bound of non-empty countable sets) and unbounded.\n\\end{quote}\n\nWe prove this result by transfinite induction on $\\alpha$ of both properties.\n\nThe proof is still quite long, by transfinite induction over $\\alpha$.\n\n\n\\index{Coq!Techniques!Transfinite induction}\n\n\\begin{Coqsrc}\nSection Proof_of_Lemma5.\n Let P (alpha:Ord) := Unbounded (Cr alpha) /\\ Closed (Cr alpha).\n \n Lemma Lemma5 : forall alpha, P alpha.\n(* ... *)\n End Proof_of_Lemma5.\n\nCorollary Unbounded_Cr alpha : Unbounded (Cr alpha).\nProof.\n now destruct (Lemma5 alpha).\nQed.\n\nCorollary Closed_Cr alpha : Closed (Cr alpha).\nProof.\n now destruct (Lemma5 alpha).\nQed.\n\\end{Coqsrc}\n\n\\section{Cantor Normal Form}\n\nThe notion of Cantor normal form is defined for all countable ordinals.\nNevertheless, note that, contrary to the implementation based on type \\texttt{T1},\nthe Cantor normal form of an ordinal $\\alpha$ may contain $\\alpha$ as a \nsub-term\\footnote{This would prevent us from trying to represent Cantor normal forms as finite trees (like in Sect.~\\ref{sec:T1-inductive-def})}.\n\n\nWe represent Cantor normal forms as lists of ordinals.\nA list $l$ is a Cantor normal form of a given ordinal $\\alpha$ if it satisfies two conditions:\n\n\n\n\\begin{itemize}\n\\item The list $l$ is sorted (in decreasing order) w.r.t. the order $\\leq$\n\\item The sum of all the $\\omega^{\\beta_i}$ where the $\\beta_i$ are the terms of $l$ (in this order) is equal to $\\alpha$.\n\\end{itemize}\n\n\n\n\\vspace{4pt}\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Schutte.CNF.html\\#cnf_t}%\n{\\texttt{Schutte.CNF}}}\n\n\n\\begin{Coqsrc}\n Definition cnf_t := list Ord.\n\nFixpoint eval (l : cnf_t) : Ord :=\n match l with nil => zero\n | beta :: l' => phi0 beta + eval l'\n end.\n\nDefinition sorted (l: cnf_t) :=\n LocallySorted (fun alpha beta => beta <= alpha) l.\n\nDefinition is_cnf_of (alpha : Ord)(l : cnf_t) : Prop :=\n sorted l /\\ alpha = eval l.\n\\end{Coqsrc}\n\n\\index{Maths!Transfinite induction}\n\nBy transfinite induction on $\\alpha$, we prove that every countable ordinal $\\alpha$ \n has at least a Cantor normal form.\n\n\\begin{Coqsrc}\nTheorem cnf_exists (alpha : Ord) :\n exists l: cnf_t, is_cnf_of alpha l.\n\\end{Coqsrc}\n\nBy structural induction on lists, we prove that this normal form is unique.\n\n\\begin{Coqsrc}\n Lemma cnf_unicity : forall l alpha, \n is_cnf_of alpha l -> \n forall l', is_cnf_of alpha l' -> l=l'.\nProof.\n induction l.\n (* ... *)\n\nTheorem cnf_exists_unique (alpha:Ord) :\n exists! l: cnf_t, is_cnf_of alpha l.\n\\end{Coqsrc}\n\n\nFinally, the following two lemmas relate $\\epsilon_0$ with Cantor normal forms.\n\nIf $\\alpha<\\epsilon_0$, then the Cantor normal form of $\\alpha$ is made of ordinals strictly less than $\\alpha$.\n\n\\begin{Coqsrc}\nLemma cnf_lt_epsilon0 : \n forall l alpha, \n is_cnf_of alpha l -> alpha < epsilon0 ->\n Forall (fun beta => beta < alpha) l.\n\\end{Coqsrc}\n\n\\index{Exercises}\n\n\\begin{exercise}\nPlease consider the following statement :\n\n\\begin{Coqsrc}\nLemma cnf_lt_epsilon0_iff : \n forall l alpha, \n is_cnf_of alpha l -> \n (alpha < epsilon0 <-> Forall (fun beta => beta < alpha) l).\n\\end{Coqsrc}\n\nIs it true ?\n\\end{exercise}\n\nFinally, the Cantor normal form of $\\epsilon_0$ is just $\\omega^{\\epsilon_0}$.\n\n\\begin{Coqsrc}\nLemma cnf_of_epsilon0 : is_cnf_of epsilon0 (epsilon0 :: nil).\nProof.\n split.\n - constructor. \n - simpl; now rewrite alpha_plus_zero, epsilon0_fxp.\nQed.\n\\end{Coqsrc}\n\n\\index{Projects}\n\n\\begin{project}\nImplement pages 82 to 85 of~\\cite{schutte} (critical, strongly critical, maximal critical ordinals, Feferman's ordinal $\\Gamma_0$).\n\\end{project}\n\n\\begin{remark}\nThe sub-directory \\href{../theories/html/hydras.Gamma0.html}%\n{\\texttt{theories/Gamma0}} contains an (incomplete, still undocumented) implementation of the set of ordinals below $\\Gamma_0$, represented in Veblen normal form. \n\\end{remark}\n\n\\section{An Embedding of \\texttt{T1} into \\texttt{Ord}}\n\n\nOur library \n\\href{../theories/html/hydras.Schutte.correctness_E0.html}%\n{\\texttt{Schutte.Correctness\\_E0}} establishes the link between two very different modelizations of ordinal numbers. In other words, it ``validates'' a data structure in terms of\na classical mathematical discourse considered as a model. \nFirst, we define a function from \\texttt{T1} into \\texttt{Ord} by structural recursion.\n\n\\begin{Coqsrc}\nFixpoint inject (t:T1) : Ord :=\n match t with \n | T1.zero => zero\n | T1.ocons a n b => AP.phi0 (inject a) * S n + inject b\n end. \n\\end{Coqsrc}\n\nThis function enjoys good commutation properties with respect to the main operations which\nallow us to build Cantor normal form.\n\n\\begin{Coqsrc}\nTheorem inject_of_zero : inject T1.zero = zero.\n\nTheorem inject_of_finite (n : nat):\n inject (T1.fin n) = n.\n\nTheorem inject_of_phi0 (alpha : T1):\n inject (phi0 alpha) = AP.phi0 (inject alpha).\n\nTheorem inject_plus (alpha beta : T1): nf alpha -> nf beta ->\n inject (alpha + beta)%t1 = inject alpha + inject beta.\n\nTheorem inject_mult_fin_r (alpha : T1) :\n nf alpha -> forall n:nat , inject (alpha * n)%t1 = inject alpha * n.\n\nTheorem inject_mono (beta gamma : T1) :\n T1.lt beta gamma -> \n T1.nf beta -> T1.nf gamma -> \n inject beta < inject gamma.\n\nTheorem inject_injective (beta gamma : T1) : nf beta -> nf gamma ->\n inject beta = inject gamma -> beta = gamma.\n\\end{Coqsrc}\n\nFinally, we prove that \\texttt{inject} is a bijection from the set of all terms of \\texttt{T1} in normal form to the set \n\\texttt{members epsilon0} of the elements of \\texttt{Ord} strictly less than $\\epsilon_0$.\n\n\\begin{Coqsrc}\nTheorem inject_lt_epsilon0 (alpha : T1):\n inject alpha < epsilon0.\n\nTheorem embedding : \n fun_bijection (nf: Ensemble T1) (members epsilon0) inject.\n \\end{Coqsrc}\n\n\\subsection{Remarks}\nLet us recall that the library \\href{../theories/html/hydras.Schutte.Schutte.html}%\n{\\texttt{Schutte}} depends on five \\emph{axioms} and lies explicitly in the \nframework of classical logic with a weak version of the axiom of choice\n(please look at the documentation of\n\\href{https://coq.inria.fr/distrib/current/stdlib/Coq.Logic.ChoiceFacts.html}{\\texttt{Coq.Logic.ChoiceFacts}}).\nNevertheless, the other modules:\n\\href{../theories/html/hydras.Epsilon0.Epsilon0.html}%\n{\\texttt{Epsilon0}},\n\\href{../theories/html/hydras.Hydra.Hydra.html}%\n{\\texttt{Hydra}}, et \n\\href{../theories/html/hydras.Gamma0.Gamma0.html}%\n{\\texttt{Gamma0}}\ndo not import any axioms and are really constructive.\n\n\\index{Projects}\n\\begin{project}\nThere is no construction of ordinal multiplication in~\\cite{schutte}. \nIt would be interesting to derive this operation from Schütte's axioms,\nand prove its consistence with multiplication in ordinal notations for \n$\\epsilon_0$ and $\\Gamma_0$.\n\\end{project}\n\n\\section{Related Work}\n\nIn~\\cite{grimm:hal-00911710}, José Grimm establishes the consistency between our ordinal notations (\\texttt{T1} and \\texttt{T2} (Veblen normal form) and his implementation\nof ordinal numbers after Bourbaki's set theory.\n\n\n\\chapter{The Ordinal \\texorpdfstring{$\\Gamma_0$}{Gamma0} (first draft)}\n\n\n\\emph{This chapter and the files it presents are still very incomplete, considering the impressive properties of $\\Gamma_0$~\\cite{Gallier91}. We hope to add new material soon, and accept contributions!}\n\n\n\\section{Introduction}\nWe present a notation system for the ordinal $\\Gamma_0$, following Chapter V, Section 14 of~\\cite{schutte}: ``A notation system for the ordinals $<\\Gamma_0$''.\nWe try to be as close as possible to Schütte's text and usual practices of \\coq{} developments.\n\nThe ordinal $\\Gamma_0$ is defined in Section 13 of ~\\cite{schutte} as the least \\emph{strongly critical ordinal}. It is widely known as the \\emph{Feferman-Schütte ordinal}.\n\n\nSection V, 13 of~\\cite{schutte} defines \\emph{strongly critical} and\n\\emph{maximal $\\alpha$-critical} ordinals: \n\n\\begin{itemize}\n\\item $\\alpha$ is strongly critical if\n$\\alpha$ is $\\alpha$-critical,\n\\item $\\gamma$ is maximal $\\alpha$-critical if $\\gamma$ is $\\alpha$-critical, and, for all $\\xi>\\alpha$, $\\gamma$ is not $\\xi$-critical.\n\n\\end{itemize}\n\n\n\n\n\n\\vspace{4pt}\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Schutte.Critical.html\\#strongly_critical}%\n{\\texttt{Schutte.Critical}}}\n\n\\begin{Coqsrc}\nDefinition strongly_critical alpha := In (Cr alpha) alpha.\n\nDefinition maximal_critical alpha : Ensemble Ord :=\n fun gamma =>\n In (Cr alpha) gamma /\\\n forall xi, alpha < xi -> ~ In (Cr xi) gamma.\n\nDefinition Gamma0 := the_least strongly_critical.\n\\end{Coqsrc}\n\n\\index{Projects}\n\\begin{project}\nProve that a (countable) ordinal $\\alpha$ is strongly critical iff \n$\\phi_\\alpha(0)=\\alpha$ (Theorem 13.13 of~\\cite{schutte} ). \n\\end{project}\n\n\n\\index{Projects}\n\\begin{project}\nProve that the set of strongly critical ordinals is unbounded and closed (Theorem 13.14 of~\\cite{schutte} ). Thus this set is not empty, hence has a least element. Otherwise, the definition of $\\Gamma_0$ above would be useless.\n\\end{project}\n\n\n\n\nIn the present version of this development, we only study $\\Gamma_0$ as a notation system, much more powerful than the ordinal notation for $\\epsilon_0$.\n\n%\\index{Projects}\n%\n%\\begin{project}\n% Schûtte's section 13 of~\\cite{schutte} contain several definitions and lemmas which give another view on $\\Gamma_0$. We leave it as a project to implement them in \\coq{}. \n%\\end{project}\n\n\n\n\n\\section{The Type \\texttt{T2} of Ordinal Terms}\n\nThe notation system for ordinals less than $\\gamma_0$ comes from the following theorem of~\\cite{schutte}, where $\\psi\\,\\alpha$ is the ordering function \nof the set of maximal $\\alpha$-critical ordinals.\n\n\n\\begin{quote}\n Any ordinal $\\not= 0$ which is not strongly critical can be expressed in terms of $+$ and $\\psi$.\n\\end{quote}\n\n\\index{Projects}\n\\begin{project}\nThis theorem is not formally proved in this development yet. It should be!\n\\end{project}\n\n\nLike in Chapter~\\ref{chap:T1}, we define an inductive type with two constructors, one for $0$, the other for the construction $\\psi(\\alpha,\\beta)\\times(n+1)+\\gamma$, adapting a Manolios-Vroon-like notation~\\cite{Manolios2005} to\n\\emph{Veblen normal forms}.\n\\label{types:T2}\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Gamma0.T2.html\\#T2}%\n{\\texttt{Gamma0.T2}}}\n\\begin{Coqsrc}\n(** [gcons alpha beta n gamma] is : [psi(alpha,beta)*(S n)+ gamma] *)\n\nInductive T2 : Set :=\n| zero : T2\n| gcons : T2 -> T2 -> nat -> T2 -> T2.\n\nNotation \"[ alpha , beta ]\" := (gcons alpha beta 0 zero)\n (at level 0): t2_scope.\n\\end{Coqsrc}\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=11cm]{epsilon0.jpg}\n \\caption{Veblen normal form}\n \\label{fig:gamma0}\n\\end{figure}\n\nLike in chapter~\\ref{chap:T1}, we get familiar with the type \\texttt{T2} by recognising simple constructs like finite ordinals, $\\omega$, etc., as inhabitants of \\texttt{T2}.\n\n\\begin{Coqsrc}\nNotation \"'one'\" := [zero,zero] : T2_scope.\n\n(** The (n+1)-th finite ordinal *)\nNotation \"'FS' n\" := (gcons zero zero n zero) (at level 10) : T2_scope.\n\n(** the [n]-th ordinal *)\nDefinition fin (n:nat) := match n with 0 => zero | S p => FS p end.\n\nNotation \"'omega'\" := [zero,one] : T2_scope.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nNotation \"'epsilon0'\" := ([one,zero]) : T2_scope.\n\nDefinition epsilon alpha := [one, alpha].\n\\end{Coqsrc}\n\n\\section{How Big is \\texorpdfstring{$\\Gamma_0$}{\\texttt{Gamma0}}?}\n\nLet us define a strict order on type \\texttt{T2}. The following definition is \nan adaptation of Schütte's, taking into account the multiplications by a natural number (inspired by~\\cite{Manolios2005}, and also present in \\texttt{T1}).\n\n\\label{sect:t2-lt-def}\n\n\\begin{Coqsrc}\nInductive lt : T2 -> T2 -> Prop :=\n| (* 1 *) \n lt_1 : forall alpha beta n gamma, zero t2< gcons alpha beta n gamma\n| (* 2 *)\n lt_2 : forall alpha1 alpha2 beta1 beta2 n1 n2 gamma1 gamma2, \n alpha1 t2< alpha2 ->\n beta1 t2< gcons alpha2 beta2 0 zero ->\n gcons alpha1 beta1 n1 gamma1 t2<\n gcons alpha2 beta2 n2 gamma2\n| (* 3 *)\n lt_3 : forall alpha1 beta1 beta2 n1 n2 gamma1 gamma2, \n beta1 t2< beta2 ->\n gcons alpha1 beta1 n1 gamma1 t2<\n gcons alpha1 beta2 n2 gamma2\n\n| (* 4 *)\n lt_4 : forall alpha1 alpha2 beta1 beta2 n1 n2 gamma1 gamma2, \n alpha2 t2< alpha1 ->\n [alpha1, beta1] t2< beta2 ->\n gcons alpha1 beta1 n1 gamma1 t2<\n gcons alpha2 beta2 n2 gamma2\n\n| (* 5 *)\nlt_5 : forall alpha1 alpha2 beta1 n1 n2 gamma1 gamma2, \n alpha2 t2< alpha1 ->\n gcons alpha1 beta1 n1 gamma1 t2<\n gcons alpha2 [alpha1, beta1] n2 gamma2\n\n| (* 6 *)\nlt_6 : forall alpha1 beta1 n1 n2 gamma1 gamma2, (n1 < n2)%nat ->\n gcons alpha1 beta1 n1 gamma1 t2< \n gcons alpha1 beta1 n2 gamma2\n\n| (* 7 *)\n lt_7 : forall alpha1 beta1 n1 gamma1 gamma2, gamma1 t2< gamma2 ->\n gcons alpha1 beta1 n1 gamma1 t2<\n gcons alpha1 beta1 n1 gamma2\nwhere \"o1 t2< o2\" := (lt o1 o2): T2_scope.\n\nHint Constructors lt : T2.\n\\end{Coqsrc}\n\nSeven constructors! In order to get accustomed with this definition, let us look at a small set of examples, covering all the constructors of \\texttt{lt}.\n\n\n\\subsection{Examples}\n\n\\subsubsection*{Proof of $0<\\epsilon_0$}\n\n\\begin{Coqsrc}\nExample Ex1: 0 t2< epsilon0.\nProof. constructor 1. Qed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\omega<\\epsilon_0$}\n\n\\begin{Coqsrc}\nExample Ex2: omega t2< epsilon0.\nProof. info_auto with T2. (* uses lt_1 and lt_2 *) Qed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\psi(\\omega,8)\\times 13+56 < \\psi(\\omega,8)\\times 13+57 $}\n\n\\begin{Coqsrc}\nExample Ex3: gcons omega 8 12 56 t2< gcons omega 8 12 57.\nProof.\n constructor 7; constructor 6; auto with arith.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\epsilon_0<\\psi(2,1)$}\n\n\\begin{Coqsrc}\nExample Ex4: epsilon0 t2< [2,1].\nProof.\n constructor 2; auto with T2.\n - constructor 6; auto with arith.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\psi(2,1)<\\psi(2,3)$}\n\n\\begin{Coqsrc}\nExample Ex5 : [2,1] t2< [2,3].\nProof.\n constructor 3; auto with T2.\n - constructor 6; auto with arith.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\psi(1,0)\\times 13+ \\omega < \\psi(0,\\psi(2,1))$}\n\\label{sect:ex6-first-proof}\n\n\\begin{Coqsrc}\nExample Ex6 : gcons 1 0 12 omega t2< [0,[2,1]].\nProof.\n constructor 4.\n - constructor 1.\n - constructor 2.\n + constructor 6; auto with arith.\n + constructor 1.\nQed.\n\\end{Coqsrc}\n\n\\subsubsection*{Proof of $\\psi(2,1)\\times 43 + \\epsilon_0 < \\psi(1,\\psi(2,1))$}\n\n\\begin{Coqsrc}\nExample Ex7 : gcons 2 1 42 epsilon0 t2< [1, [2,1]].\nProof.\n constructor 5.\n constructor 6; auto with arith.\nQed.\n\\end{Coqsrc}\n\n\\index{Projects}\n\\begin{project}\nWrite a tactic that solves automatically goals of the form (\\texttt{$\\alpha$ t2< $\\beta$}), where $\\alpha$ and $\\beta$ are closed terms of type \\texttt{T2}.\n\\end{project}\n\n\\section{Veblen Normal Forms}\n\\begin{definition}\n A term of the form $\\psi(\\alpha_1,\\beta_1)\\times n_1+ \\psi(\\alpha_2,\\beta_2)\\times n_2+\\dots+\\psi(\\alpha_k,\\beta_k)\\times n_k$ is said to be in\n \\emph{[Veblen] normal form} if for every $i Prop :=\n| zero_nf : nf zero\n| single_nf : forall a b n, nf a -> nf b -> nf (gcons a b n zero)\n| gcons_nf : forall a b n a' b' n' c', \n [a', b'] t2< [a, b] -> \n nf a -> nf b -> \n nf(gcons a' b' n' c')-> \n nf(gcons a b n (gcons a' b' n' c')).\n\\end{Coqsrc}\n\nLet us look at some positive examples (we have to prove some inversion lemmas before proving counter-examples).\n\n\n\\begin{Coqsrc}\nLemma nf_fin i : nf (fin i).\nProof.\n destruct i.\n - auto with T2.\n - constructor 2; auto with T2.\nQed.\n\nLemma nf_omega : nf omega.\nProof. compute; auto with T2. Qed.\n\nLemma nf_epsilon0 : nf epsilon0.\nProof. constructor 2; auto with T2. Qed.\n\nLemma nf_epsilon : forall alpha, nf alpha -> nf (epsilon alpha).\nProof. compute; auto with T2. Qed.\n\nExample Ex8: nf (gcons 2 1 42 epsilon0).\nProof.\n constructor 3; auto with T2.\n - apply Ex4.\n - apply nf_fin.\n - apply nf_fin.\nQed.\n\\end{Coqsrc}\n\n\n\\subsection{Length of a Term}\n\nThe notion of \\emph{term length} is introduced by Schütte as a helper for proving (at least) the \\emph{trichotomy} property and transitivity of the strict order \\texttt{lt} on \\texttt{T2}. These properties are proved by induction on length.\n\n\\begin{Coqsrc}\nFixpoint nbterms (t:T2) : nat :=\n match t with zero => 0\n | gcons a b n v => (S n) + nbterms v\n end.\n\nFixpoint t2_length (t:T2) : nat :=\n match t with \n zero => 0\n | gcons a b n v => \n nbterms (gcons a b n v) + \n 2 * (Max.max (t2_length a)\n (Max.max (t2_length b) \n (t2_length_aux v)))\n end\nwith t2_length_aux (t:T2) : nat :=\n match t with \n | zero => 0\n | gcons a b n v =>\n Max.max (t2_length a) \n (Max.max (t2_length b) (t2_length_aux v))\n end.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nCompute t2_length (gcons 2 1 42 epsilon0).\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = 48 : nat\n\\end{Coqanswer}\n\n\\subsection{Trichotomy}\n\n\\emph{Trichotomy} is another name for the well-known property of decidable total ordering (like Standard Library's \\texttt{Compare\\_dec.lt\\_eq\\_lt\\_dec}).\n\nWe first prove by induction on $l$ the following lemma:\n\n\\vspace{4pt}\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Gamma0.Gamma0\\#tricho_aux}%\n{\\texttt{Gamma0.Gamma0}}}\n\n\\begin{Coqsrc}\nLemma tricho_aux (l: nat) : forall t t' :T2,\n t2_length t + t2_length t' < l ->\n {t t2< t'} + {t = t'} + {t' t2< t}.\n\\end{Coqsrc}\n\nThen we get our version of \\texttt{lt\\_eq\\_lt\\_dec}, and derive a comparison function;\n\n\\begin{Coqsrc}\nDefinition lt_eq_lt_dec (t t': T2) : {t t2< t'}+{t = t'}+{t' t2< t}.\nProof.\n eapply tricho_aux.\n eapply lt_n_Sn.\nDefined.\n\nDefinition compare (t1 t2 : T2) : comparison := \n match lt_eq_lt_dec t1 t2 with\n | inleft (left _) => Lt\n | inleft (right _) => Eq\n | inright _ => Gt\n end.\n\\end{Coqsrc}\n\nWith the help of \\texttt{compare}, we get a boolean version of \\texttt{nf}\n(being in Veblen normal form).\n\n\\begin{Coqsrc}\nFixpoint nfb (alpha : T2) : bool :=\n match alpha with\n zero => true\n | gcons a b n zero => andb (nfb a) (nfb b)\n | gcons a b n ((gcons a' b' n' c') as c) =>\n match compare [a', b'] [a, b] with\n Lt => andb (nfb a) (andb (nfb b) (nfb c))\n | _ => false\n end\nend.\n\\end{Coqsrc}\n\n\n\\begin{Coqsrc}\nCompute compare (gcons 2 1 42 epsilon0) [2,2].\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = Lt\n : comparison\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nCompute nfb (gcons 2 1 42 epsilon0).\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = true\n : bool\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nCompute nfb (gcons 2 1 42 (gcons 2 2 4 epsilon0)).\n\\end{Coqsrc}\n\n\\begin{Coqanswer}\n = false\n : bool\n\\end{Coqanswer}\n\n\\begin{remark}\nThe connexion between the predicate \\texttt{nf} and the relation \\texttt{lt} on one part, and the functions \\texttt{nfb} and \\texttt{compare} on the other, is expressed by the following lemmas:\n\n\\begin{Coqsrc}\nLemma nfb_equiv gamma : nfb gamma = true <-> nf gamma.\n\nLemma compare_correct alpha beta :\n CompareSpec (alpha = beta) (lt alpha beta) (lt beta alpha)\n (compare alpha beta).\n\\end{Coqsrc}\n\nThe function \\texttt{compare} helps to make easier proofs of inequalities of\nclosed terms of type \\texttt{T2}.\n\nFirst, we prove a lemma:\n\n\\begin{Coqsrc}\nLemma compare_Lt : forall alpha beta, compare alpha beta = Lt -> \n alpha t2< beta.\nProof.\n intros alpha beta; destruct (compare_correct alpha beta);\n trivial; discriminate. \nQed.\n\\end{Coqsrc}\n\nThen, we give another version of the proof of Sect.~\\vref{sect:ex6-first-proof}.\n\n\\begin{Coqsrc}\nExample Ex6 : gcons 1 0 12 omega t2< [0,[2,1]].\nProof. now apply compare_Lt. Qed.\n\\end{Coqsrc}\n\n\\end{remark}\n\n\n\\section{Main Functions on \\texttt{T2}}\n\n\\subsection{Successor}\nThe successor function is defined by structural recursion.\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Gamma0.T2.html\\#succ}%\n{\\texttt{Gamma0.T2}}}\n\\begin{Coqsrc}\nFixpoint succ (a:T2) : T2 :=\n match a with zero => one\n | gcons zero zero n c => fin (S (S n))\n | gcons a b n c => gcons a b n (succ c)\n end.\n\\end{Coqsrc}\n\n\\subsection{Addition}\n\nLike for Cantor normal forms (see Sect.~\\ref{sect:infix-plus-T1}), the definition of addition in \\texttt{T2} requires comparison between ordinal terms.\n\n\n\\begin{Coqsrc}\nFixpoint plus (t1 t2 : T2) {struct t1}:T2 :=\n match t1,t2 with\n | zero, y => y\n | x, zero => x\n | gcons a b n c, gcons a' b' n' c' =>\n (match compare (gcons a b 0 zero)\n (gcons a' b' 0 zero) with\n | Lt => gcons a' b' n' c'\n | Gt => gcons a b n (c + gcons a' b' n' c')\n | Eq => gcons a b (S(n+n')) c'\n end)\n end\nwhere \"alpha + beta\" := (plus alpha beta): T2_scope.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\nExample Ex7 : 3 + epsilon0 = epsilon0.\nProof. trivial. Qed.\n\\end{Coqsrc}\n\n\\subsection{The Veblen Function \\texorpdfstring{$\\phi$}{\\texttt{phi}}}\n\nThe enumeration function of critical ordinals, presented in Sect.~\\vref{sect:phi-schutte}, is recursively defined in type \\texttt{T2}.\n\n\\begin{Coqsrc}\nDefinition phi (alpha beta : T2) : T2 :=\n match beta with zero => [alpha, beta] \n | [b1, b2] => \n (match compare alpha b1\n with Datatypes.Lt => [b1, b2 ]\n | _ => [alpha,[b1, b2]]\n end)\n | gcons b1 b2 0 (gcons zero zero n zero) => \n (match compare alpha b1\n with Datatypes.Lt => \n [alpha, (gcons b1 b2 0 (fin n))]\n | _ => [alpha, (gcons b1 b2 0 (fin (S n)))]\n end)\n | any_beta => [alpha, any_beta]\n end.\n\\end{Coqsrc}\n\nDespite its complexity, the function \\texttt{phi} is well adapted to proofs by simplification or computation.\n\\begin{Coqsrc}\nExample Ex8: phi 1 (succ epsilon0) = [1, [1,0] + 1].\nProof. reflexivity. Qed.\n\\end{Coqsrc}\n\n\\begin{Coqsrc}\n(** All epsilons are fixpoints of phi 0 *)\n\nTheorem epsilon_fxp : forall beta, phi zero (epsilon beta) =\n epsilon beta.\nProof. reflexivity. Qed.\n\nTheorem epsilon0_fxp : epsilon0 = phi zero epsilon0.\nProof. apply epsilon_fxp. Qed.\n\\end{Coqsrc}\n\n\nThe relation between the constructor $\\psi$ and the function $\\phi$ is\nstudied in~\\cite{schutte}, and partially implemented in this development.\n\\emph{Please contribute!}\n \nFor instance, the following theorem states that, if $\\gamma$ is the sum of a limit ordinal $\\beta$ and a finite ordinal $n$, and $\\beta$ is a fixpoint of\n$\\phi(\\alpha)$, then $\\psi(\\alpha,\\gamma)=\\phi_\\alpha(\\gamma+1)$.\n\n\\begin{Coqanswer}\nphi_psi :\nforall (alpha : T2) [beta gamma : T2] [n : nat],\nnf gamma ->\nlimit_plus_fin beta n gamma ->\nphi alpha beta = beta -> [alpha, gamma] = phi alpha (succ gamma)\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nExample Ex9 : [zero, epsilon 2 + 4] = phi 0 (epsilon 2 + 5).\nProof. trivial. Qed.\n\\end{Coqsrc}\n\nOn the other hand, $\\phi$ can be expressed in terms of $\\psi$.\n\n\\begin{Coqanswer}\nphi_of_psi:\n forall a b1 b2 : T2,\n phi a [b1, b2] = (if lt_ge_dec a b1 then [b1, b2] else [a, [b1, b2]])\n\\end{Coqanswer}\n\n\\begin{Coqsrc}\nExample Ex10 : phi omega [epsilon0, 5] = [epsilon0, 5].\nProof. reflexivity. Qed.\n\\end{Coqsrc}\n\n\\index{Projects}\n\\begin{project}\nPlease study a way to pretty print ordinal terms in Veblen normal form (see Section~\\vref{sect:ppT1}).\n\\end{project}\n\n\\section{An Ordinal Notation for \\texorpdfstring{$\\Gamma_0$}{\\texttt{Gamma0}}}\n\nIn order to consider type \\texttt{T2} as an ordinal notation, we have to build an instance of class \\texttt{ON} (See Definition page~\\pageref{types:ON}).\n\nFirst, we define a type that contains only terms in Veblen normal form, and redefine \\texttt{lt} and \\texttt{compare} by delegation (see for comparison the construction of type \\texttt{E0} in Sect.~\\vref{sect:E0-def}).\n\n\\begin{Coqsrc}\nModule G0.\n\nClass G0 := mkg0 {vnf : T2; vnf_ok : nfb vnf}.\n\nDefinition lt (alpha beta : G0) := T2.lt (@vnf alpha) (@vnf beta).\n\nDefinition compare alpha beta := Gamma0.compare (@vnf alpha) (@vnf beta).\n\\end{Coqsrc}\n\nThen, we prove that \\texttt{lt} is a well-founded strict order and that the\nfunction \\texttt{compare} is correct.\n\n\\begin{Coqsrc}\nInstance lt_sto : StrictOrder lt.\n\nLemma lt_wf : well_founded lt.\n\nLemma compare_correct alpha beta :\n CompareSpec (alpha = beta) (lt alpha beta) (lt beta alpha)\n (compare alpha beta).\n\nInstance Gamma0: ON lt compare.\nProof.\n split.\n - apply lt_sto.\n - apply lt_wf. \n - apply compare_correct.\nQed.\n\\end{Coqsrc}\n\n\n\\begin{remark}\nThe proof of \\texttt{lt\\_wf} has been written by \\'Evelyne Contejean, using her library on the recursive path ordering (see also remark~\\vref{remark:a3pat}).\n\\end{remark}\n\n\\index{Projects}\n\\begin{project}\nProve that \\texttt{Epsilon0} (page~\\pageref{instance-epsilon0})\nis a sub-notation system of \\texttt{Gamma0}.\n\nProve that the implemantations of \\texttt{succ}, \\texttt{+}, $\\phi_0$, etc.\nare compatible in both notation systems.\n\nNote that a function \\texttt{T1\\_inj} from \\texttt{T1} to \\texttt{T2} has already been defined. It may help to complete the task.\n\n\n\n\\noindent\\emph{From \\href{../theories/html/hydras.Gamma0.T2.html\\#T1_to_T2}%\n{\\texttt{Gamma0.T2}}}\n\\begin{Coqsrc}\n(* injection from T1 *)\n\nFixpoint T1_to_T2 (alpha :T1) : T2 :=\n match alpha with\n | T1.zero => zero\n | T1.ocons a n b => gcons zero (T1_to_T2 a) n (T1_to_T2 b)\n end.\n\\end{Coqsrc}\n\n\\end{project}\n\n\\begin{project}\nProve that the notation system \\texttt{Gamma0} is a correct implementation \nof the segment $[0,\\Gamma_0)$ of the set of countable ordinals.\n\\end{project}\n\n\n\n\n\n\n", "meta": {"hexsha": "e9cfdda34f78b67cadd4b958814bb84d71dd18e1", "size": 323336, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/part-hydras.tex", "max_stars_repo_name": "ybertot/hydra-battles", "max_stars_repo_head_hexsha": "597de4e718176bcf5eb8d3c97b060ddfe61b5367", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/part-hydras.tex", "max_issues_repo_name": "ybertot/hydra-battles", "max_issues_repo_head_hexsha": "597de4e718176bcf5eb8d3c97b060ddfe61b5367", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/part-hydras.tex", "max_forks_repo_name": "ybertot/hydra-battles", "max_forks_repo_head_hexsha": "597de4e718176bcf5eb8d3c97b060ddfe61b5367", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.5680902498, "max_line_length": 534, "alphanum_fraction": 0.6811180939, "num_tokens": 103186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.78793120560257, "lm_q2_score": 0.8056321843145404, "lm_q1q2_score": 0.6347827382591877}} {"text": "%!TEX root = paper.tex\n\nWe next moved on to develop a continuous erosion model that operates on smooth surfaces, in contrast to our previous discrete model that operated on piecewise linear surfaces. Physically this new model represents a surface being worn down gradually, such as a bar of soap in water or an ice cube melting. \n\n\\subsection*{Continuous Surface}\n\nWe will start by introducing a set of parametric equations to represent a surface. A parametric representation was chosen over an implicit one, so that we can use spectral methods and the FFT when integrating Equation \\ref{eq:erosion-diff-eq}, which we will discuss in detail in Section \\ref{sec:numerical-modeling}. \n\nLet us introduce $s \\in [0, 2\\pi)$ as our parameter with $s$ increasing counterclockwise along the surface. The coordinates of any point on the surface can be represented at time $t \\ge 0$ by two functions $x(t, s), y(t, s)$. Coordinates can be written more compactly as\n\n\\[ \n \\bvec{r}(t, s) \\coloneqq \\paren{x(t, s), y(t, s)}\n\\]\n\n\\textbf{Note:} The reader should keep in mind that derivative operators still apply to each coordinate individually, e.g.\n\n\\[\n \\frac{\\partial \\bvec{r}(t, s)}{\\partial t} \\coloneqq \\paren{\\frac{\\partial x(t, s)}{\\partial t}, \\frac{\\partial y(t, s)}{\\partial t}}\n\\]\n\nTo simplify our equations, we introduce the following vector calculus notation\n\n\\begin{align*}\n \\dot{\\bvec{r}}(s)& \\coloneqq \\frac{\\partial \\bvec{r}(s)}{ds}\\\\\n \\ddot{\\bvec{r}}(s)& \\coloneqq \\frac{\\partial^2 \\bvec{r}(s)}{ds^2}\n\\end{align*}\n\nThe equations for the initial shape that we will use in our simulations are shown below and a plot of the shape is shown in Figure \\ref{fig:blob-shape} ($T_i$ is the i\\textsuperscript{th} Chebyshev polynomial)\n\n\\begin{align*}\n x(0, s) = \\cos(s) \\paren{2 T_0\\paren{\\frac{s-\\pi}{\\pi}} + T_2\\paren{\\frac{s-\\pi}{\\pi}}}\\\\\n y(0, s) = \\sin(s) \\paren{2 T_0\\paren{\\frac{s-\\pi}{\\pi}} + T_4(\\paren{\\frac{s-\\pi}{\\pi}})}\n\\end{align*}\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{blob_shape.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{\\label{fig:blob-shape} Example shape.}\n\\end{figure}\n\nThe method used to represent the shape numerically will be explained in Section \\ref{sec:numerical-modeling}, but first we will take a moment to explain the various erosion processes that we will model in this section.\n\n\\subsection*{Continuous Erosion Processes}\n\nAll of the processes that we will look at in this section, will move points on the surface in the direction of the surface normal, which is defined as\n\n\\[\n \\hat{n}(\\bvec{r}(s)) = (-\\dot{y}(s), \\dot{x}(s))\n\\]\n\nWe can now model our erosion process with the following differential equation, where $g(\\bvec{r}(s))$ is a function specific to the erosion process being modeled.\n\n\\begin{equation}\n \\label{eq:erosion-diff-eq}\n \\frac{\\partial \\bvec{r}(t, s)}{\\partial t} = g(\\bvec{r}(s)) \\; \\hat{n}(\\bvec{r}(s))\n\\end{equation}\n\n\\subsubsection*{Smoothing Process}\n\nOne of the processes that we will look at is a soap bar being worn down by a person holding it in their hand, which we will call the \\textit{Smoothing Process}. In this process, bumps on the object will get worn down quickest, because they are the first parts to come into contact with the person's hand. Technically the furthest protruding bumps will be worn down quickest, but to simplify the problem, we assume that every point on the surface wears down at a rate proportional to the curvature at the point. We will use a signed curvature $\\kappa$, so that a bump will have positive curvature, and an indentation will have negative curvature. \n\n\\[\n \\kappa(\\bvec{r}(s)) = \\frac{\\dot{\\bvec{r}}(s) \\times \\ddot{\\bvec{r}}(s)}{\\norm{\\dot{\\bvec{r}}(s)}^3}\n\\]\n\nTo make bumps erode faster than indentations, we will use the following damping function\n\n\\begin{equation}\n g(\\bvec{r}(s)) = \\tan^{-1}(\\beta \\, (\\kappa(\\bvec{r}(s)) - \\alpha)) + \\frac{\\pi}{2}\n\\end{equation}\n\nFor our simulations we chose $\\alpha = 1, \\; \\beta = 5$ so that $g(\\bvec{r}(s)) \\approx 1$ at the corners on our initial surface. This restriction is not based on the physics of the problem, and was chosen purely for the aesthetic look of the resulting simulation.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{g_1.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{\\label{fig:g-1} Plot of $g(\\kappa)$ for \\textit{Smoothing Process}.}\n\\end{figure}\n\n\\subsubsection*{Bottom Recession Process}\n\nOur second process is an object sitting in a bath of acid, which we will call our \\textit{Bottom Recession Process}. In this process, the lowest points (smallest values of $y(s)$) on the surface will erode quickest.\n\n\\begin{gather}\n f(s) = \\frac{1}{\\alpha \\, (y(s) - y_{min}) + \\beta} - \\gamma \\notag\\\\\n g(s) = \\begin{cases}\n f(s) \\qquad &\\text{for} \\quad f(s) > 0\\\\\n 0 \\qquad &\\text{otherwise}\n \\end{cases}\n\\end{gather}\n\nFor our simulations we chose $\\alpha = 10, \\; \\beta = \\gamma = 0.1$ so that $g(\\bvec{r}(s)) \\approx 1$ for $y(s) \\approx y_{min}$.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{g_2.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{\\label{fig:g-2} Plot of $g(y)$ for \\textit{Bottom Recession Process}.}\n\\end{figure}\n\n\n\\subsection*{Numerical Modeling}\\label{sec:numerical-modeling}\n\nWe chose to model the parametric surface functions $x, y$ with discrete Fourier series. This is because spectral methods using the DFT will maintain the smoothness of the surface much better than a local approximation of the surface using finite differences which only acts locally at points. The DFT and surface derivatives can be calculated quickly using the Fast Fourier Transform (refer to Appendix \\ref{sec:dft} for more details).\n\nTo integrate Equation \\ref{eq:erosion-diff-eq}, we sample the surface at parameter values $S_n = 2 \\pi \\, \\frac{n}{N} \\; , \\; n = 0, \\dotsc, N-1$, which gives us a matrix of initial evaluation points\n\n\\begin{align*} \n X_0 =& \\bvec{r}\\paren{0, S}\\\\\n =& \\begin{bmatrix}\n x(0, S_1)& y(0, S_1)\\\\\n \\vdots& \\vdots\\\\\n x(0, S_N)& y(0, S_N)\\\\ \n \\end{bmatrix}\n\\end{align*}\n\n\\textbf{Note:} When refering to vectors and matrices in our numerical models, we will use uppercase letters to distinguish them from continuous functions.\n\nForward time integration of Equation \\ref{eq:erosion-diff-eq} can then be implemented with standard ODE time stepping integration methods. In this paper, we used SciPy's {\\tt odeint} method for integrating Equation \\ref{eq:erosion-diff-eq}.\n\n\\subsection*{Point Collision Problems}\n\nUnfortunately, the simplicity gained from modeling the surface with a Fourier series is soon lost when we start time stepping. As can be seen in Figure \\ref{fig:remove-lowest-blob-bad}, the surface forms loops as time is increased (a physical impossibility).\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{remove_lowest_blob_bad.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{Time integration of \\ref{eq:erosion-diff-eq} showing loop artifact.\\label{fig:remove-lowest-blob-bad}}\n\\end{figure}\n\nTo investigate why these loop artifacts occur, we will show a zoomed in portion of the same figure, with the evaluation points shown. \n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{remove_lowest_blob_bad_zoom.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{Expanded view of loop artifact.\\label{fig:remove-lowest-blog-bad-zoom}}\n\\end{figure}\n\nLooking closely, it can be seen that areas of high curvature form when evaluation points get close to each other, and this eventually leads to surface looping. Our guess is that this occurs because we are using discrete time stepping, so there are errors in the approximation of the shape's change with time. This is fine when the evaluation points are far from each other, but when they get close to each other, the errors become significant and affect the direction of the surface unit normal vectors. The folding causes unit normal vectors of nearby points to point at each other and which subsequently causes the points to cross over each other.\n\n\\subsection*{Redistributing Points}\n\nA seemingly obvious way to fix this problem would be to resample to get a set of evenly spaced evaluation points on the surface. This wouldn't be too difficult of a task, since we can integrate and sample at any point on the surface via the FFT. Unfortunately, if we pick points on the surface at arbitrary values of $s$, we lose spectral accuracy, because we are now sampling at unevenly spaced values of our parameter $s$, which we will call $\\tilde{S}$. The Discrete Fourier transform assumes that our points are sampled at evenly spaced values of $s$. Although we are working with the discrete Fourier transform, we need to return to the continuous equations to understand the problem. Redistribution in the continuous sense is a parameter transformation\n\n\\[ \n \\tilde{s} = h(s)\n\\]\n\nwith the condition of even spacing being\n\n\\[\n \\norm{\\frac{\\partial \\bvec{r}(s)}{\\partial \\tilde{s}}} = constant\n\\]\n\nOur shape function now becomes\n\n\\[ \\tilde{x}(s) = (x \\circ h)(s) \\]\n\nThere is absolutely no guarentee that the Fourier series of this composite function will decay at a reasonable rate, so using the DFT with our chosen resolution N could very poorly capture the surface shape. The method is not withough hope, since the technique of parameter transformation is utilized extensively in spectral methods such as the transformation $h(s) = \\cos(s)$ used for Chebyshev spectral methods to cluster evaluation points near the endpoints of the domain. We need to find a way to generate a parameter transformation $h(s)$ where $(x \\circ h)(s)$ has a rapidly decaying Fourier series.\n\n\\subsection*{Continuous Redistribution}\n\nIf we fix a time $t = t_1$, then we can define a differential equation that will allow us to redistribute the points and ensure that our redistributed points will still have a rapidly decaying Fourier series.\n\nPhysically $\\dot{\\bvec{r}}(t_1, s)$ is the surface tangent vector, and $\\ddot{\\bvec{r}}(t_1, s)$ is the rate of change of this surface tangent vector with respect to $s$. More importantly though, $\\norm{\\dot{\\bvec{r}}(t_1, s)}$ is inversely proportional to the density of evaluation points on the surface at $s$. We would like to shift the evaluation points away from denser areas, thus we need move the evaluation points in the direction of larger $\\norm{\\dot{\\bvec{r}}(t_1, s)}$, which is the projection of $\\ddot{\\bvec{r}}(t_1, s)$ onto the surface tangent vector.\n\n\\begin{equation}\n a_t(t, s) = \\ddot{\\bvec{r}}(t, s) \\cdot \\dot{\\bvec{r}}(t, s)\n\\end{equation}\n\n\nWe can now set up a differential equation for this redistribution process.\n\n\\begin{equation}\n \\label{eq:redist-diff-eq}\n \\frac{\\partial \\bvec{r}(t_1, \\tilde{s}(t'))}{\\partial t'} = a_t(t_1, \\tilde{s}(t')) \\frac{\\dot{\\bvec{r}}(t_1, \\tilde{s}(t'))}{\\norm{\\dot{\\bvec{r}}(t_1, \\tilde{s}(t'))}} \\; , \\quad \\tilde{s}(0) = s\n\\end{equation}\n\nTo simplify the numerical computation, we can combine Equations \\ref{eq:erosion-diff-eq} and \\ref{eq:redist-diff-eq} to get a single ODE\n\n\\begin{equation}\n \\frac{\\partial (\\bvec{r}(t, s))}{\\partial t} = g(t, s) \\; \\bhat{n}(t, s) + a_t(t, s) \\; \\frac{\\dot{\\bvec{r}}(t_1, s)}{\\norm{\\dot{\\bvec{r}}(t_1, s)}}\n\\end{equation}\n\nAs can be seen in Figure \\ref{fig:remove-points-blob-good}, this modification dramatically improves our algorithm.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{remove_points_blob_good.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{\\label{fig:remove-points-blob-good}\\textit{Bottom recession process} simulation with point redistribution.}\n\\end{figure}\n\nIf we plot the evaluation points (Figure \\ref{fig:remove-points-blob-good-points}), it can be seen that the points nicely redistribute themselves as time increases to stay evenly spaced on the surface.\n\n\\begin{figure}[H]\n \\begin{center}\n \\includegraphics[keepaspectratio, width=4in]{remove_points_blob_good_points.pdf}\n \\end{center}\n \\vspace{-.2in} % corrects bad spacing\n \\caption{\\label{fig:remove-points-blob-good-points}\\textit{Smoothing process} simulation with evaluation points shown.}\n\\end{figure}\n\n\\subsection*{Higher dimensions}\n\nA next logical step would be to generalize the algorithm to a 2-dimensional surface that would be parameterized by $s_1, s_2 \\in [0, 2\\pi)$. Much of the algorithm would be similar, except that we would need to use the 2-dimensional discrete Fourier transform (refer to Appendix \\ref{sec:dft}).\n\nThe surface normal vector equation for a surface in 3-dimensional space is the following\n\n\\[\n \\bvec{n}(t, s_1, s_2) = \\frac{\\partial \\bvec{r}(t, s_1, s_2)}{\\partial s_1} \\times \\frac{\\partial \\bvec{r}(t, s_1, s_2)}{\\partial s_2} \n\\]\n\nCurvature $\\kappa$ could be generalized to mean curvature $H$\n\n\\begin{align*}\n H(t, s_1, s_2)& = \\frac{\\nabla \\cdot \\bvec{n}(t, s_1, s_2)}{2}\\\\\n &= \\frac{1}{2} \\paren{\\frac{\\partial \\bvec{n}_1(t, s_1, s_2)}{\\partial s_1} + \\frac{\\partial \\bvec{n}_2(t, s_1, s_2)}{\\partial s_2}}\n\\end{align*}\n\nAvoiding singularities when mapping the 2-dimensional surface to an N x N array would likely be difficult, but the mapping only needs to be used to sample the initial points on the object, and the redistribution algorithm can be oblivious to the mapping.\n\n", "meta": {"hexsha": "5ebbfc7f87547d7d5a81b56b2d6b39a5233ffbef", "size": 13513, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/continuous_surface.tex", "max_stars_repo_name": "wangjohn/shape_of_stones", "max_stars_repo_head_hexsha": "c4d37ba76b93c12d31cf6efeac2f7ffdc8b64099", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-06-24T02:43:32.000Z", "max_stars_repo_stars_event_max_datetime": "2019-06-24T02:43:32.000Z", "max_issues_repo_path": "paper/continuous_surface.tex", "max_issues_repo_name": "wangjohn/shape_of_stones", "max_issues_repo_head_hexsha": "c4d37ba76b93c12d31cf6efeac2f7ffdc8b64099", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/continuous_surface.tex", "max_forks_repo_name": "wangjohn/shape_of_stones", "max_forks_repo_head_hexsha": "c4d37ba76b93c12d31cf6efeac2f7ffdc8b64099", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.9957081545, "max_line_length": 758, "alphanum_fraction": 0.7298897358, "num_tokens": 3853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812553, "lm_q2_score": 0.7879311956428946, "lm_q1q2_score": 0.6347827339124029}} {"text": "%!TEX root = ../../report.tex\n\n\\subsubsection{Cellular Automaton} % (fold)\n\\label{ssub:cellular_automaton}\n\nIt's a model of a system of cells within a grid with a determined shape, each of this cells can be on one of a finite set of states. It evolves during a finite amount of time steps with a set of simple rules according with the state of the neighbouring cells.\nThe neighbourhood of the cell can be defined in many different ways, the most common is the use of the adjacent cells. \n\nIn the case where each cell have two possible states and the next generation state depends only on the previous state of the cell and the two immediate neighbors is called an \\emph{elementary cellular automaton}. In this case we have $2^3 = 8$ possible patterns for a neighborhood and $2^8 = 256$ sets of possible different rules. This rules are reffered by their \\emph{Wolfram code}, defined by Wolfram. \n\nA common initial state for yhis elementary cellular automata is a random line. But to able to compare the results between rules and get clean results other option is to start with a line with zeros except the middle cell with one. Applying this second option and the following set of rules (the rule 30):\n\n\\begin{figure}[htbp]\n\t\\centering\n\t\\includegraphics[width=0.85\\textwidth]{img/Theory/Cellular_A/Rules.png}\n\t\\caption{Example Production Rules\\cite{Shiffman2012}}\n\t\\label{fig:label}\n\\end{figure}\n\nwe get the pattern in the Figure~\\ref{fig:resultCA} that represents the evolution of this Cellular automaton over generations:\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=0.75\\textwidth]{img/Theory/Cellular_A/Result.png}\n \\caption{Sierpiński Triangle, rule 90}\n \\label{fig:resultCA}\n\\end{figure}\n\n\nIn Figure~\\ref{fig:resultCA} each line represents an iteration of the system with the application of the rules. With this set of rules a Sierpiński triangle is reproduced.\n\nCellular automata are used mainly to model phenomena that occurs in the physical world, most of them can only express the basic idea of a phenomenon but some are accurate enough to be able to make predictions.\n\nIn this context, cellular automata are used to model natural shapes and textures, the Figure~\\ref{fig:CAshell} shows a natural texture on a Textile Cone Snail that looks like the patterns formed with the cellular automaton in the Figure~\\ref{fig:CArule30}.\n\n\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[b]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{img/Theory/Cellular_A/shell.jpeg}\n \\caption{a)}\n\t\t\t\t\\label{fig:CAshell}\n \\end{subfigure}%\n %~ %add desired spacing between images, e. g. ~, \\quad, \\qquad, \\hfill etc.\n %(or a blank line to force the subfigure onto a new line)\n\n \\begin{subfigure}[b]{0.9\\textwidth}\n \\includegraphics[width=\\textwidth]{img/Theory/Cellular_A/Rule30.png}\n\t\t\t\t\\caption{b)}\n\t\t\t\t\\label{fig:CArule30}\n \\end{subfigure}\n \\caption{Example of the representation of natural patterns with cellular automata. a) Natural Shell, image from \\cite{Shiffman2012}. b) Pattern formed with the rule 30.}\n\t\t\\label{fig:CArule30shell}\n\\end{figure}\n\n\n% subsubsection cellular_automaton (end)", "meta": {"hexsha": "eb74da67b437d5af79a98c0c7e4938651479a1d0", "size": 3211, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/overview/2-cellular_automaton.tex", "max_stars_repo_name": "arturalkaim/ProceduralGeneration", "max_stars_repo_head_hexsha": "736fcb8a15291ede1db069ad968527508bc081c4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sections/overview/2-cellular_automaton.tex", "max_issues_repo_name": "arturalkaim/ProceduralGeneration", "max_issues_repo_head_hexsha": "736fcb8a15291ede1db069ad968527508bc081c4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/overview/2-cellular_automaton.tex", "max_forks_repo_name": "arturalkaim/ProceduralGeneration", "max_forks_repo_head_hexsha": "736fcb8a15291ede1db069ad968527508bc081c4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.3620689655, "max_line_length": 405, "alphanum_fraction": 0.7461849891, "num_tokens": 825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7879311981328135, "lm_q1q2_score": 0.6347827285642612}} {"text": "\n\\subsection{Hard and soft classifiers}\n\nA hard classifier can return a sparce vector with \\(1\\) in the relevant classification.\n\nA soft classifier returns probabilities for each entry in the vector.\n\nThe vector represents \\(P(Y=k|X=x)\\)\n\n\\subsection{Transforming soft classifiers into hard classifiers}\n\nWe can use a cutoff.\n\nIf there are more than two classes we can choose the one with the highest score.\n\n", "meta": {"hexsha": "437608ffbc1e2af712a8a39e10fafbbf220c5995", "size": 410, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/parametric/02-04-hardSoft.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/parametric/02-04-hardSoft.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/parametric/02-04-hardSoft.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.625, "max_line_length": 87, "alphanum_fraction": 0.7756097561, "num_tokens": 86, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8056321889812552, "lm_q2_score": 0.787931185683219, "lm_q1q2_score": 0.6347827258885675}} {"text": "\\chapter{Classical Problems}\n\\section{Review of Basic Concept}\n\\begin{enumerate}\n\t\\item At the moment $t=0$ a stationary particle of mass $m$ experiences a time-dependent force $F=k t\\left(t^{\\prime}-t\\right)$, where $k$ is a constant vector, $t^{\\prime}$ is the time during which the given force acts. Find :\\\\\n\t(a) the momentum of the particle when the action of the force discontinued;\\\\\n\t(b) the distance covered by particle while the force acted.\n\t\\begin{answer}\n\t(a) Since $F=k t\\left(t^{\\prime}-t\\right)$, the direction of the force is fixed, as $k$ is a constant vector and the particle starts from rest, the motion will be in the direction of the force.\n\t\\begin{align*}\n\tm \\frac{d v}{d t}&=k t\\left(t^{\\prime}-t\\right),\\text{ from Newton's II law}\\\\\n\t\\text{Integrating, }\\int_{0}^{v} d v&=\\frac{k}{m} \\int_{0}^{t^{\\prime}} t\\left(t^{\\prime}-t\\right) d t=\\frac{k}{m}\\left[t^{\\prime} \\frac{t^{2}}{2}-\\frac{t^{3}}{3}\\right]_{0}^{t^{\\prime}} \\quad v=\\frac{k}{m} \\cdot \\frac{t^{\\prime 3}}{6}\\\\\n\\text{\tTherefore, momentum of the particle }p&=m v=k t \\frac{t^{\\prime 3}}{6}.\\\\\n\\text{(b) }d s=v d t \\Rightarrow s&=\\int_{0}^{s} d s=\\int_{0}^{t^{\\prime}} v d t=\\frac{k t^{\\prime 4}}{12 m}\n\t\\end{align*}\n\t\t\\end{answer}\n\t\\item A block of mass $m_{1}=4 \\mathrm{~kg}$ on a smooth inclined plane of $30^{\\circ}$ is connected by a cord over a small, frictionless pulley to a second block of mass $m_{2}=5 \\mathrm{~kg}$ hanging vertically. Calculate the acceleration with which the block moves and also the tension in the cord. Take $g=10 \\mathrm{~m} / \\mathrm{sec}^{2}$.\n\t\\begin{answer}\n\t\t\tThe different forces acting on the masses are shown in figure.\n\t\t\t\\begin{figure}[H]\n\t\t\t\t\\centering\n\t\t\t\t\\includegraphics[height=3.5cm,width=6cm]{diagram-20220216-crop}\n\t\t\t\\end{figure}\n\t\t\\begin{align}\n\\text{We have }T-m_{1} g \\sin \\theta&=m_{1} a \\label{1} \\\\\n\\text{ and }m_{2} g-T&=m_{2} a \\label{2}\\\\\n\\intertext{Solving equations (\\ref{1}) and (\\ref{2}), we get}\na&=\\frac{\\left(m_{2}-m_{1} \\sin \\theta\\right) g}{\\left(m_{1}+m_{2}\\right)}\\label{3}\\\\\n\\text { and } \\quad T & =m_{2} g\\left[1-\\frac{\\left(m_{2}-m_{1} \\sin \\theta\\right)}{\\left(m_{1}+m_{2}\\right)}\\right] \\notag\\\\\n \\text { or } \\quad T & =\\frac{m_{1} m_{2}(1+\\sin \\theta) g}{\\left(m_{1}+m_{2}\\right)}\\\\\n \\text{Here }m_{1}&=4 \\mathrm{~kg}, m_{2}=5 \\mathrm{~kg}, \\theta=30^{\\circ} \\text{and }g=10 \\mathrm{~m} / \\mathrm{s}\\notag\\\\\n \\intertext{Substituting these values in equation (\\ref{3}), we get}\n T&=\\frac{5 \\times 4\\left(1+\\frac{1}{2}\\right) 10}{9}=\\frac{300}{9}=33.33 \\mathrm{~N}\\notag\n\\end{align}\n\t\\end{answer}\n\t\\item A particle of mass $m$ is connected to two springs of unstretched length ' $l$ ' and spring constant $k$ as show in figure. Calculate acceleration of particle if it is slightly displaced along X-direction.\n\t\\begin{answer}\n\t\t\tLet the particle is displaced by a distance $x$ along $+x$ direction as shown in figure.\n\t\t\t\\begin{figure}[H]\n\t\t\t\t\\centering\n\t\t\t\t\\includegraphics[height=4.5cm,width=2cm]{diagram-20220216(1)-crop}\n\t\t\t\\end{figure}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=2cm]{diagram-20220216(2)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\\text{\tTherefore, stretched length of spring }&=\\sqrt{l^{2}+x^{2}}\\\\\n\t\t\\text{elongation in springs }&=\\sqrt{l^{2}+x^{2}}-l\\\\\n\t\\text{\tRestoring force on the particle due to each spring} F&=k\\left(\\sqrt{l^{2}+x^{2}}-l\\right)\n\t\\intertext{ Because of restoring force the particle moves back towards its initial position.}\n\\text{\tEquation of motion of particle }F_{x}&=m \\frac{d^{2} x}{d t^{2}}\\\\\n\t\\text{or }-2 F \\cos \\theta&=m \\frac{d^{2} x}{d t^{2}}\\text{ or }-2 k\\left(\\sqrt{l^{2}+x^{2}}-l\\right) \\cdot \\frac{x}{\\sqrt{l^{2}+x^{2}}}=m \\frac{d^{2} x}{d t^{2}}\\\\\n\t\\text{Therefore, }\\frac{d^{2} x}{d t^{2}}&=-\\frac{2 k x}{m}\\left(1-\\frac{l}{\\sqrt{l^{2}+x^{2}}}\\right)\\text{ or }\\frac{d^{2} x}{d t^{2}}\\\\&=-\\frac{2 k x}{m}\\left[1-\\left(1+\\frac{x^{2}}{l^{2}}\\right)^{-1 / 2}\\right]\\\\\n\\text{\tSince, }&x$ 0 and $b>0$. Find the equilibrium points and find frequency of oscillation about the stable equilibrium.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{At equilibrium point, }\\left.\\frac{d V}{d x}\\right|_{x=x_{0}}=0\\\\\n\t\t\\therefore-a+2 b x_{0}=0 \\quad\\text{ or }\\quad x_{0}&=\\frac{a}{2 b}\\\\\n\t\\text{\tThus, }x_{0}&=\\frac{a}{2 b}\\text{ is an equilibrium point.}\n\\intertext{\tTo know whether it is stable or To know whether it is stable or unstable equilibrium point let us calculate second derivative of potential at equilibrium point.}\n\t\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=x_{0}}&=2 b>0\\\\\n\t\\text{Therefore }x_{0}=\\frac{a}{2 b}&\\text{ is stable equilibrium point, force constant }k=\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=x_{0}}=2 b\n\t\\intertext{Therefore, frequency of oscillation is}\n\t\\omega&=\\sqrt{\\frac{k}{m}}=\\sqrt{\\frac{2 b}{m}}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Potential corresponding to force between the atoms of a diatomic molecule is $V(r)=\\frac{a}{r^{12}}-\\frac{b}{r^{6}}$ where $a$ and $b$ are positive constants and $r$ is separation between the atoms. Calculate bond length for stable configuration and also calculate frequency of oscillation of atoms if mass of each atom be $m$.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\text{\tFor stable configuration }\\left.\\frac{d V}{d r}\\right|_{r=r_{0}}&=0\\\\\n\t\t-\\frac{12 a}{r_{0}^{13}}+\\frac{6 b}{r_{0}^{7}}=0 \\Rightarrow r_{0}&=\\left(\\frac{2 a}{b}\\right)^{1 / 6}\\\\\n\t\t\\text{Therefore bond length for stable configuration is }&\\left(\\frac{2 a}{b}\\right)^{1 / 6}\\\\\n\t\t\\text{Force constant }k=\\left.\\frac{d^{2} V}{d r^{2}}\\right|_{r=r_{0}}&=\\frac{12 \\times 13 a}{r_{0}^{14}}-\\frac{6 \\times 7 b}{r_{0}^{8}}=\\frac{1}{r_{0}^{8}}\\left(\\frac{12 \\times 13 a}{r_{0}^{6}}-6 \\times 7 b\\right)\\\\\n\t\t&=\\left(\\frac{b}{2 a}\\right)^{8 / 6}\\left(\\frac{12 \\times 13 a}{2 a / b}-42 b\\right)\\\\&=\\left(\\frac{b}{2 a}\\right)^{4 / 3} \\cdot 36 b=\\frac{18}{2^{1 / 3}} \\cdot \\frac{b^{7 / 3}}{a^{4 / 3}}\\\\\n\t\t\\text{Reduced mass of system }\\mu&=\\frac{m \\cdot m}{m+m}=m / 2\\\\\n\t\t\\text{Frequency of oscillation} \\omega=\\sqrt{\\frac{k}{\\mu}}&=\\sqrt{\\frac{18}{2^{1 / 3}} \\cdot \\frac{b^{7 / 3}}{m / 2 a^{4 / 3}}}=6\\left(\\frac{b^{7}}{2 m^{3} a^{4}}\\right)^{1 / 6}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\n\t\\item A particle of mass ' $m$ ' is moving under potential $V(x)=a x^{3}-b x^{2}$. Initially the particle is at at stable point. What minimum speed be given to the particle so that it reaches unstable point. Plot potential versus $x$.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{For equilibrium point }\\left.\\frac{d V}{d x}\\right|_{x=x_{0}}&=0\\\\\n\t\t3 a x_{0}^{2}-2 b x_{0}&=0 \\Rightarrow x_{0}=0, \\frac{2 b}{3 a}\\\\\n\t\t\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=x_{0}}&=6 a x_{0}-2 b,\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=0}=-2 b<0, \\therefore x_{0}=0\\text{ is unstable point }\\\\\n\t\t\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=\\frac{2 b}{3 a}}&=2 b>0, \\therefore x_{0}=\\frac{2 b}{3 a}\\text{ is stable point}\n\t\t\\intertext{To calculate speed let us apply conservation of energy.}\n\t\t\\text{Total energy initial }&=\\text{ Total energy final}\n\t\t\\intertext{(Kinetic Energy + Potential Energy) at $x=\\frac{2 b}{a}=($ Kinetic Energy $+$ Potential Energy) at $x=0$ for minimum speed $(u)$ at stable point the particle will just reach unstable point and stops there.}\n\t\t\t\\therefore \\frac{1}{2} m u^{2}+V\\left(x=\\frac{2 b}{3 a}\\right)&=\\frac{1}{2} m .0^{2}+V(x=0)\\\\\n\t\t\\frac{1}{2} m u^{2}+a\\left(\\frac{2 b}{3 a}\\right)^{3}-b\\left(\\frac{2 b}{3 a}\\right)^{2}&=0+0\\\\\n\t\tu^{2}=-\\frac{2}{m}\\left(\\frac{2 b}{3 a}\\right)^{2}\\left(\\frac{2 b}{3}-b\\right)&=\\frac{2}{m} \\cdot \\frac{4 b^{2}}{9 a^{2}} \\cdot \\frac{b}{3}=\\frac{8 b^{3}}{27 m a^{2}}\\\\\n\t\t\\therefore u&=\\sqrt{\\frac{8 b^{3}}{27 m a^{2}}}\n\t\t\\end{align*}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=5cm]{stability001}\n\t\t\\end{figure}\n\t\\end{answer}\n\t\\item A particle is moving under potential $V(r)=\\frac{a}{r^{2}}-\\frac{b}{r} .$ Calculate the minimum value of potential energy.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\text{\tFor potential to be minimum }\\left.\\frac{d V}{d r}\\right|_{r=r_{0}}=0\\\\\n\t -\\frac{2 a}{r_{0}^{3}}+\\frac{b}{r_{0}^{2}}=0 \\quad \\therefore r_{0}&=\\frac{2 a}{b}\\\\\n\t \\left.\\frac{d^{2} V}{d r^{2}}\\right|_{r=r_{0}}=\\frac{6 a}{r_{0}^{4}}-\\frac{2 b}{r_{0}^{3}}&=\\frac{1}{r_{0}^{3}}\\left(\\frac{6 a}{r_{0}}-2 b\\right)=\\frac{1}{r_{0}^{3}}\\left(\\frac{6 a}{2 a / b}-2 b\\right)=\\frac{b}{r_{0}^{3}}>0\\\\\n\t \\text{Therefore at }r&=r_{0}\\text{ potential is minimum}\\\\\n\t \\therefore \\quad V_{\\min }&=V\\left(r_{0}\\right)=\\frac{1}{r_{0}}\\left(\\frac{a}{r_{0}}-b\\right)=-\\frac{b^{2}}{4 a}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A cube is placed on the top of a fixed hemisphere as shown in figure. What should be relation between length of side of cube and radius of hemisphere so that cube has stable equilibrium.\n\t\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=6cm]{diagram-20220219(2)-crop}\n\t\\end{figure}\n\t\\begin{answer}\n\t\t$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=5.5cm,width=9cm]{diagram-20220219(3)-crop}\n\t\t\\end{figure}\n\t\tTo discuss equilibrium of cube we first write its potential energy as function of angle from its equilibrium position. As shown in the figure below, initially point A was in contact with spherical surface but now in displaced position point $\\mathrm{B}$ is in contact. Therefore,\n\t\t\\begin{align*}\n\t\tA B&=R \\theta\n\t\t\\intertext{Height of centre of cube from centre level of hemisphere is}\n\t\th&=O C+O^{\\prime} D \\\\\n\t\t&=O O^{\\prime} \\sin \\theta+O^{\\prime} S \\cos 0 \\\\\n\t\t&=A B \\sin \\theta+\\left(O^{\\prime} B+B S\\right) \\cos \\theta \\\\\n\t\t&=R \\theta \\sin \\theta+\\left(\\frac{L}{2}+R\\right) \\cos \\theta\n\t\t\\intertext{Potential energy of the cube}\n\t\tV(\\theta)&=M g h=M g\\left[R \\theta \\sin \\theta+\\left(\\frac{L}{2}+R\\right) \\cos \\theta\\right]\n\t\t\\intertext{$\\theta=0$ is equilibrium position of the cube. For this position to be stable equilibrium position, $\\left.\\frac{d^{2} V}{d \\theta^{2}}\\right|_{\\theta=0}>0$}\n\t&\\therefore \\frac{d^{2}}{d \\theta^{2}} M g\\left[R \\theta \\sin \\theta+\\left(\\frac{L}{2}+R\\right) \\cos \\theta\\right]_{\\theta=0}>0\\\\\n&\\text{\tor }\\frac{d}{d \\theta}\\left[R \\theta \\cos \\theta+R \\sin \\theta-\\left(\\frac{L}{2}+R\\right) \\sin \\theta\\right]_{\\theta=0}>0\\\\\n\t&\\text{or }\\left[R \\cos \\theta-R \\theta \\sin \\theta+R \\cos \\theta-\\left(\\frac{L}{2}+R\\right) \\cos \\theta\\right]_{\\theta=0}>0\\\\\n\t&\\text{or }\\left[2 R-\\left(\\frac{L}{2}+R\\right)\\right]>0 \\quad \\therefore R-\\frac{L}{2}>0\\text{ or }\\quad 2 R>L\n\t\\intertext{Thus, cube can be in stable equilibrium position if its side length is less than diameter of hemisphere}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item For the mass pulley system shown in figure, what should be relation between $m$ and $M$ so that system remains in stable equilibrium position. Pulley are smooth and strings are tight and inextensible\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=5cm]{diagram-20220218(4)-crop}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tThe pulleys are fixed. Therefore we can write potential energy of the system by specifying position of blocks with respect to pulleys.\\\\\n\t\tLet ' $l$ ' be length of string and ' $d$ ' be the half distance between two pulleys. Therefore, $l$ and $d$ are constants If $x$ be distance of $m$ below the pulley as shown in figure then potential energy of system is\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3cm,width=6cm]{diagram-20220219(4)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align}\n\t\tV(x)=-2 m g x-M g \\sqrt{(l-x)^{2}-d^{2}} \\quad &\\therefore \\frac{d V}{d x}=-2 m g+\\frac{M g(l-x)}{\\sqrt{(l-x)^{2}-d^{2}}}\\notag\\\\\n\t\t\\text{for equilibrium }\\left.\\frac{d V}{d x}\\right|_{x=x_{0}}=0 \\quad &\\therefore-2 m g+\\frac{M g\\left(l-x_{0}\\right)}{\\sqrt{\\left(l-x_{0}\\right)^{2}-d^{2}}}=0\\notag\\\\\n\t\\text{\tor }\\frac{2 m}{M}&=\\frac{l-x_{0}}{\\sqrt{\\left(l-x_{0}\\right)^{2}-d^{2}}}\\label{23}\\\\\n\t\\left.\\frac{d^{2} V}{d x^{2}}\\right|_{x=x_{0}}&=\\frac{M g d^{2}}{\\left[\\left(l-x_{0}\\right)^{2}-d^{2}\\right]^{3 / 2}}>0\\text{ for all values of $d>0$}\\notag\\\\\n\t\\text{since, }\\frac{l-x_{0}}{\\sqrt{\\left(l-x_{0}\\right)^{2}-d^{2}}}>1 \\quad &\\therefore\\text{ from (\\ref{23}) }\\frac{2 m}{M}>1\\text{ or }2 m>M\\notag\n\t\t\\end{align}\n\t\\end{answer}\n\\item The potential energy between two atoms are given $v(r)=\\frac{a}{r^{12}}-\\frac{b}{r^{6}}$ where $a, b$ positive constants.\n(i) Find the equilibrium distance of two atoms.\\\\\n(ii) Plot the potential\\\\\n(iii) Calculate the frequency of small oscillation.\\\\\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{(i) For equilibrium potential energy should be minimum.}\n\t\t\\frac{\\mathrm{dU}}{\\mathrm{dr}}=-\\frac{12 \\mathrm{a}}{\\mathrm{r}^{13}}+\\frac{6 \\mathrm{~b}}{\\mathrm{r}^{7}}&=0 \\Rightarrow \\mathrm{r}^{6}=\\frac{2 \\mathrm{a}}{\\mathrm{b}} \\\\\n\t\t\\mathrm{r}=\\left(\\frac{20}{\\mathrm{~b}}\\right)^{1 / 6}&=\\mathrm{r}_{0} \\\\\n\t\t\\mathrm{U}\\left(\\mathrm{r}_{0}\\right)=\\frac{\\mathrm{ab}^{2}}{4 \\mathrm{a}^{2}}-\\frac{\\mathrm{b} \\cdot \\mathrm{b}}{2 \\mathrm{a}}&=\\frac{\\mathrm{b}^{2}}{4 \\mathrm{a}}-\\frac{\\mathrm{b}^{2}}{2 \\mathrm{a}}=-\\frac{\\mathrm{b}^{2}}{4 \\mathrm{a}}\\\\\n\t\t\\frac{\\mathrm{dU}}{\\mathrm{dr}}=-\\frac{12 \\mathrm{a}}{\\mathrm{r}^{13}}+\\frac{6 \\mathrm{~b}}{\\mathrm{r}^{7}}&=0 \\Rightarrow \\mathrm{r}^{6}=\\frac{2 \\mathrm{a}}{\\mathrm{b}}\\\\ \\mathrm{r}=\\left(\\frac{20}{\\mathrm{~b}}\\right)^{1 / 6}&=\\mathrm{r}_{0} \\quad[\\text{ Position of stability equilibrium }]\\\\ \\mathrm{U}\\left(\\mathrm{r}_{0}\\right)=\\frac{\\mathrm{ab}^{2}}{4 \\mathrm{a}^{2}}-\\frac{\\mathrm{b} \\cdot \\mathrm{b}}{2 \\mathrm{a}}&=\\frac{\\mathrm{b}^{2}}{4 \\mathrm{a}}-\\frac{\\mathrm{b}^{2}}{2 \\mathrm{a}}=-\\frac{\\mathrm{b}^{2}}{4 \\mathrm{a}}\n\t\t\t\\end{align*}\n\t\t\t\\begin{figure}[H]\n\t\t\t\t\\centering\n\t\t\t\t\\includegraphics[height=5cm,width=9cm]{diagram-20220219(5)-crop}\n\t\t\t\\end{figure}\n\t\t\t\t\\begin{align*}\n\t\t\\mathrm{U}\\left(\\mathrm{r}_{0}\\right)=\\frac{-\\mathrm{b}^{2}}{4 \\mathrm{a}} ; \\mathrm{U}(\\mathrm{r})&=\\mathrm{U}\\left(\\mathrm{r}_{0}\\right)+\\left.\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right) \\frac{\\mathrm{dU}}{\\mathrm{dr}}\\right|_{\\mathrm{r}_{0}}+\\left.\\frac{\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right)^{2}}{2} \\frac{\\mathrm{d}^{2} \\mathrm{U}}{\\mathrm{dr}^{2}}\\right|_{\\mathrm{c}_{0}}\\\\\n\t\t\\frac{\\mathrm{d}^{2} \\mathrm{U}}{\\mathrm{dr}^{2}}=\\frac{12 \\times 13 \\mathrm{a}}{\\mathrm{r}^{14}}-\\frac{42 \\mathrm{~b}}{\\mathrm{r}^{8}} ;\\left.\\frac{\\mathrm{d}^{2} \\mathrm{U}}{\\mathrm{dr}^{2}}\\right|_{\\mathrm{k}_{0}}&=\\frac{156 \\mathrm{a}}{\\mathrm{r}^{14}}-\\frac{42 \\mathrm{~b}}{\\mathrm{r}^{8}}=\\frac{156 \\mathrm{a}}{(2 \\mathrm{a})^{1 / 6}}-\\frac{42 \\mathrm{~b}}{\\left(\\frac{2 \\mathrm{a}}{\\mathrm{h}}\\right)^{3 / 6}}\\\\\n\t\t&=\\frac{156 \\mathrm{a}}{\\left(\\frac{2 \\mathrm{a}}{\\mathrm{b}}\\right)^{1 / 3}}-\\frac{42 \\mathrm{~b}}{\\left(\\frac{2 \\mathrm{a}}{\\mathrm{b}}\\right)^{1 / 3}}=\\text { constant }=c\\\\\n\t\\intertext{\t(Force constant equivalent to spring constant)}\n\t\tU(r)&=U\\left(r_{0}\\right)+\\frac{c}{2}\\left(r-r_{0}\\right)^{2}\\\\\n\t\t\\text{So, the force }&=-\\frac{\\mathrm{dU}}{\\mathrm{dr}}=-\\frac{\\mathrm{c}}{2} \\cdot 2\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right)^{2}=-\\mathrm{c}\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right)\\\\\n\t\t\\text{Force }&=-\\nabla U ; U=U(r)\\text{ only}\\\\\n\t\t\\overrightarrow{\\mathrm{F}}&=\\mathrm{m} \\overline{\\mathrm{a}} \\text{(Here introduced to harmonic oscillator maynecessary).}\\\\\n\t\t \\text{Equation of motion, }&\\mathrm{m} \\frac{\\mathrm{d}^{2} \\mathrm{r}}{\\mathrm{dt}^{2}}=-\\mathrm{c}\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right) ; \\frac{\\mathrm{d}^{2} \\mathrm{r}}{\\mathrm{dt}^{2}}+\\frac{\\mathrm{c}}{\\mathrm{m}}\\left(\\mathrm{r}-\\mathrm{r}_{0}\\right)=0\\\\\n\t\t \\text{ Take }\\mathrm{c}&=m \\omega^{2} ; \\omega=\\sqrt{\\frac{\\mathrm{c}}{\\mathrm{m}}}\n\t\t \\intertext{A particle of mass ' $\\mathrm{m}$ ' moving in a potential} \\mathrm{V}(\\mathrm{x})&=\\frac{1}{2} \\mathrm{~m} \\omega_{0}^{2} \\mathrm{x}^{2}+\\frac{\\mathrm{a}}{2 \\mathrm{~m} \\mathrm{x}^{2}} \\quad\\left(\\omega_{0} \\&\\right. \\text{are constant })\n\t\t \\intertext{ Find the angular trequency of small oscillation.}\n\t\t\\frac{\\mathrm{dv}}{\\mathrm{dx}}&=m \\omega_{0}^{2} \\mathrm{x}-\\frac{\\mathrm{a}}{\\mathrm{m} \\mathrm{x}^{3}}=0\\\\\n\t\tm \\omega_{0}^{2} x_{0}-\\frac{a}{m x^{3}}&=0 ; x_{0}^{4}=\\frac{a}{m^{2} \\omega_{0}^{2}} ; \\chi_{0}=\\left(\\frac{a}{m^{2} \\omega_{0}^{2}}\\right)^{1 / 4}\\\\\n\t\t\\text{Equilibrium distance, }\\frac{d^{2} z}{d x^{2}}&=m \\omega_{0}^{2}+\\frac{3 a}{m x^{4}}\\\\\n\t\t\\left.\\frac{\\mathrm{d}^{2} \\mathrm{r}}{\\mathrm{dx}^{2}}\\right|_{\\mathrm{x}=\\mathrm{x}_{0}}&=m \\omega_{0}^{2}+\\frac{3 \\mathrm{a}}{\\mathrm{ma}} \\mathrm{m}^{2} \\omega^{2}=4 \\mathrm{~m} \\omega_{0}^{2} \\quad\\left(\\mathrm{x}_{0}^{4}=\\frac{\\mathrm{a}}{\\mathrm{m}^{2} \\omega_{0}^{2}}\\right)\\\\\n\t\t\\mathrm{U}(\\mathrm{x})&=\\mathrm{U}\\left(\\mathrm{x}_{0}\\right)+\\left.\\left(\\mathrm{x}-\\mathrm{x}_{0}\\right) \\frac{\\mathrm{dU}}{\\mathrm{dx}}\\right|_{\\mathrm{x}=\\mathrm{x}_{0}}+\\left.\\frac{\\left(\\mathrm{x}-\\mathrm{x}_{0}\\right)^{2}}{2} \\frac{\\mathrm{d}^{2} \\mathrm{U}}{\\mathrm{dx}^{2}}\\right|_{\\mathrm{x}=\\mathrm{x}_{0}}\\\\\n\t\t&=\\mathrm{U}\\left(\\mathrm{x}_{0}\\right)+\\left(\\mathrm{x}-\\mathrm{x}_{0}\\right)^{2} 2 \\mathrm{~m} \\omega_{0}^{2}\\\\\n\t\t\\text{Force }&=-\\frac{\\mathrm{dU}}{\\mathrm{dx}}=-4 \\mathrm{~m} \\omega_{0}^{2}\\left(\\mathrm{x}-\\mathrm{x}_{0}\\right)\\\\\n\t\\text{\tEquation of motion, }\\mathrm{m} \\frac{\\mathrm{d}^{2} \\mathrm{x}}{\\mathrm{dt}^{2}}&=\\mathrm{F}=-4 \\omega_{0}^{2}\\left(\\mathrm{x}-\\mathrm{x}_{0}\\right)\\\\\n\t\t\\frac{d^{2} x}{d t^{2}}+4 \\omega_{0}^{2}\\left(x-x_{0}\\right)&=0\\\\\n\t\t\\text{Hence, the frequency of small oscillation, }\\omega&=\\sqrt{4 \\omega_{0}^{2}}=2 \\omega_{0}.\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A particle of mass ' $m$ ' is constrained to move in one dimension under a potential $u(x)=\\frac{\\alpha}{x^{2}}-\\frac{\\beta}{x}$, where $\\alpha$ and $\\beta$ are positive constants. Show that period of small oscillations about the equilibrium point is\n\t$T=4 \\pi \\sqrt{\\frac{2 \\alpha^{3} m}{\\beta^{4}}}$\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\text{\tAt equilibrium point }&\\left(x=x_{0}\\right)\\\\\n\t\t\\left.\\frac{d u}{d x}\\right|_{x=x_{0}}=0 \\\\\n\t\t-\\frac{2 \\alpha}{x_{0}^{3}}+\\frac{\\beta}{x_{0}^{2}}&=0\n\t\t\\hspace{1cm}\\therefore x_{0}=\\frac{2 \\alpha}{\\beta}=\n\t\t \\intertext{time period of oscillation is given by}\n\t\tT&=2 \\pi \\sqrt{\\frac{m}{k}}\n\t\\intertext{\twhere $k$ is force constant and is given by}\n\t\tk&=\\left.\\frac{d^{2} u}{d x^{2}}\\right|_{x=x_{0}} =\\frac{6 \\alpha}{x_{0}^{4}}-\\frac{2 \\beta}{x_{0}^{3}}=6 \\alpha \\cdot \\frac{\\beta^{4}}{16 \\alpha^{4}}-\\frac{2 \\beta^{4}}{8 \\alpha^{3}} \\\\\n\t\t&=\\frac{3}{8} \\cdot \\frac{\\beta^{4}}{\\alpha^{3}}-\\frac{2}{8} \\cdot \\frac{\\beta^{4}}{\\alpha^{3}} \\Rightarrow k=\\frac{\\beta^{4}}{8 \\alpha^{3}}\\\\\n\t\t\\therefore T&=2 \\pi \\sqrt{\\frac{8 \\alpha^{3}}{\\beta^{4}} \\cdot m}=4 \\pi \\sqrt{\\frac{2 \\alpha^{3} m}{\\beta^{4}}}\n\t\t\\end{align*}\n\t\\end{answer}\n\\end{enumerate}\n\\section{Central Force Motion}\n\\begin{enumerate}\n\t\\item Equation of the orbit of a particle moving under central force is $r \\theta=\\beta$, where $\\beta$ is a constant. Find the force acting on the particle.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tr \\theta=\\beta,\\text{ therefore, }u&=\\frac{1}{r}=\\frac{\\theta}{\\beta} \\quad \\therefore \\frac{\\partial^{2} u}{\\partial \\theta^{2}}=0\\\\\n\t\t\\text{\tDifferential equation of orbit, }\\frac{\\partial^{2} u}{\\partial \\theta^{2}}+u&=\\frac{-m f(r)}{L^{2} u^{2}}\\\\\n\t\t\\therefore 0+u=\\frac{-m f(r)}{L^{2} u^{2}} \\quad \\therefore \\quad f(r)&=\\frac{-L^{2} u^{3}}{m}=\\frac{-L^{2}}{m r^{3}}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Equation of orbit of a particle moving under central force is $r^{n}=a \\cos n \\theta .$ Find the force on the particle.\n\t\\begin{answer}\n\t\t\\begin{align}\n\t\t\\text{\tGiven, }r^{n}=a \\cos n \\theta,\\text{ therefore, }u^{n}=\\frac{1}{a \\cos n \\theta}&=\\frac{1}{a} \\sec n \\theta \\label{25}\\\\\n\t\t\\text{ Taking $\\ln$ both sides we get, }n \\ln u&=\\ln \\left(\\frac{1}{a}\\right)+\\ln \\sec n \\theta\\notag\\\\\n\t\t\\text{Differentiating w.r.t. $\\theta$ we get, }\\frac{n}{u} \\frac{\\partial u}{\\partial \\theta}&=n \\tan n \\theta \\quad \\therefore \\frac{\\partial u}{\\partial \\theta}=u \\tan n \\theta\\notag\\\\\n\t\t\\text{Differentiating again w.r.t. $\\theta$ we get }\\frac{\\partial^{2} u}{\\partial \\theta^{2}}&=\\frac{\\partial u}{\\partial \\theta} \\tan n \\theta+u n \\sec ^{2} n \\theta\\notag\\\\&=u \\tan ^{2} n \\theta+u n \\sec ^{2} n \\theta\\notag\\\\\n\t\t\\text{ Differential equation of orbit is }\\frac{\\partial^{2} u}{\\partial \\theta^{2}}+u&=\\frac{-m f(r)}{L^{2} u^{2}}\\notag\\\\\n\t\tu \\tan ^{2} n \\theta+u n \\sec ^{2} n \\theta+u&=\\frac{-m f(r)}{L^{2} u^{2}}, u \\sec ^{2} n \\theta+u n \\sec ^{2} n \\theta=\\frac{-m f(r)}{L^{2} u^{2}}\\notag\\\\\n\t\t\\therefore f(r)&=-\\frac{L^{2} u^{3}(1+n) \\sec ^{2} n \\theta}{m}\\notag\\\\\n\t\t\\text{from (\\ref{25}) }\\sec ^{2} n \\theta&=a^{2} u^{2 n}\\notag\\\\\n\t\t\\therefore f(r)&=\\frac{-L^{2}(n+1) a^{2} u^{2 n+3}}{m}\\notag\\\\\n\t\t\\therefore \\quad f(r)&=\\frac{-L^{2}(n+1) a^{2}}{m} \\cdot \\frac{1}{r^{2 n+3}}\\notag\\\\\n\t\t\\text{Or }\n\t\tf(r) \\propto \\frac{1}{r^{2 n+3}}&\n\t\t\\text{and the force is attractive in nature.}\\notag\n\t\t\\end{align}\n\t\\end{answer}\n\t\\item A particle of mass $m$ is moving under a central force. The angular momentum of the partix $\\mathrm{L}$ and its equation of orbit is $r=A e^{k 0}$. Calculate potential energy of the particle.\n\t\\begin{answer}\n\t\t\\begin{align}\n\t\t\\text{Potential energy is given as }V(r)&=-\\int f(r) d r\\label{26}\n\t\t\\intertext{\tTherefore, let us first find $f(r)$.}\\notag\n\t\t\\text{Given }r=A e^{k \\theta},\\text{ thereforc, }u&=\\frac{1}{A} e^{-k \\theta} \\quad \\therefore \\frac{\\partial^{2} u}{\\partial \\theta^{2}}=k^{2} u\\notag\\\\\n\t\t\\text{Differential equation of orbit is }\\frac{\\partial^{2} u}{\\partial \\theta^{2}}+u&=\\frac{-m f(r)}{L^{2} u^{2}} \\quad \\therefore\\left(k^{2}+1\\right) u=\\frac{-m f(r)}{L^{2} u^{2}}\\notag\\\\\n\t\t\\therefore f(r)&=\\frac{-L^{2}\\left(k^{2}+1\\right) u^{3}}{m} \\text { or } f(r)=\\frac{-L^{2}\\left(k^{2}+1\\right)}{m r^{3}}\\notag\\\\\n\t\t\\text{Therefore, from (\\ref*{26}) }V(r)&=-\\int f(r) d r=\\frac{-L^{2}\\left(k^{2}+1\\right)}{2 m r^{2}}\\notag\n\t\t\\end{align}\n\t\\end{answer}\n\t\\item For a particle moving under gravitational force pericentre distance in parabolic orbit is $r_{p}$ while the radius of the circular orbit with same angular momentum is $r_{c}$. What is relation between $r_{p}$ and $r_c$ ?\n\t\\begin{answer}\n\t\tPericentre distance is the minimum distance $(\\cos \\theta=\\max =1)$ and for parabolic orbit $e=1 .$\n\t\t\\begin{align*}\n\t\t\\text{Therefore, }r_{\\min }&=\\frac{l}{1+e \\cos \\theta}=\\frac{l}{2}=r_{p}\\\\\n\t\t\\text{for circular orbit $e=0$, therefore }r_{c}&=\\frac{l}{1+e \\cos \\theta}=l \\quad \\therefore r_{p}=\\frac{r_{c}}{2}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Ratio of maximum to minimum speed of a planet revolving around the sun in an elliptical orbit is $2: 1$, What is eccentricity of the orbit?\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{\tGiven }\\frac{v_{\\max }}{v_{\\min }}=\\frac{2}{1} \\therefore \\frac{\\sqrt{\\frac{G M}{a}\\left(\\frac{1+e}{1-e}\\right)}}{\\sqrt{\\frac{G M}{a}\\left(\\frac{1-e}{1+e}\\right)}}=\\frac{2}{1} \\quad\\text{ or }\\quad \\frac{1+e}{1-e}=\\frac{2}{1} \\quad \\therefore e=\\frac{1}{3}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A planet is revolving around the sun in a circular orbit. Due to some reason the speed of the planet suddenly becomes double. What is new orbit of the planet.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{\tOrbital speed of the planet is }&\\sqrt{\\frac{G M}{r}}\\\\\n\t\t\\text{New speed of the planet }&=2 \\sqrt{\\frac{G M}{r}}\\\\\n\t\t\\text{Therefore, new energy of the planet }&=\\frac{1}{2} m v^{2}-\\frac{G M m}{r}=\\frac{1}{2} m \\cdot \\frac{4 G M}{r}-\\frac{G M m}{r}=\\frac{2 G M m}{r}>0\n\t\t\\intertext{\tTotal energy of the planet becomes positive on doubling its speed therefore new orbit of the planet will be hyperbolic.}\n\t\t\\end{align*}\n\t\\end{answer}\n\\end{enumerate}\n\\begin{enumerate}\n\t\\item A planet of mass $m$ and angular momentum $\\mathrm{L}$ moves in a circular orbit in a potential, $V(r)=-k / r$ where $k$ is a positive constant find the radius of stable circular orbit. If the planet is slightly perturbed, find angular frequency of radial oscillation.\n\t\\begin{answer}$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=1.5cm,width=6cm]{diagram-20220221(2)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tV_{e f f}=\\frac{L^{2}}{2 m r^{2}}+V(r)&=\\frac{L^{2}}{2 m r^{2}}-\\frac{k}{r}\\\\\n\t\t\\text{For stable point (orbit) }\\left.\\frac{\\partial V_{e f f}}{\\partial r}\\right|_{r=r_{0} .}&=0\\\\\n\t\t\\therefore-\\frac{L^{2}}{m r_{0}^{3}}+\\frac{k}{r_{0}^{2}}&=0 \\text { or } r_{0}=\\frac{L^{2}}{m k}\\\\\n\t\t\\text{\tTherefore radius of stable circular orbit is }L^{2} / \\mathrm{mk}\n\t\t\\end{align*}\n\t\tAngular frequency of oscillation about stable point:\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=0.7cm,width=6.4cm]{diagram-20220221-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t\\omega&=\\sqrt{\\frac{\\left.\\frac{\\partial^{2} V_{e f f}}{\\partial r^{2}}\\right|_{r=r_{0}}}{m}}=\\sqrt{\\frac{\\left(\\frac{3 L^{2}}{m r_{0}^{4}}-\\frac{2 k}{r_{0}^{3}}\\right)}{m}}\\\\\n\t\t&=\\sqrt{\\frac{\\frac{1}{r_{0}^{3}}\\left(\\frac{3 L^{2}}{m \\cdot \\frac{L^{2}}{m k}}-2 k\\right)}{m}}=\\sqrt{\\frac{k}{m r_{0}^{3}}}=\\sqrt{\\frac{k^{4} m^{2}}{L^{6}}}=\\frac{m k^{2}}{L^{3}}\n\t\t\\end{align*}\n\t\tactual radial oscillation is as shown in figure.\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4cm,width=4cm]{diagram-20220221(1)-crop}\n\t\t\\end{figure}\n\t\\end{answer}\n\t\\item A small asteroid is approaching a massive star with a speed $v$ from very large distance, at an impact parameter ' $b$ ' as shown in figure. If the mass of the star is $M$ and its radius is $R$, then what is the minimum value of $b$ such that the asteroid will miss the star?\n\t\\begin{answer}\n\t\tAngular momentum of asteroid about centre of star is $L=m v b$\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3.5cm,width=5cm]{diagram-20220221(3)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t\\text{Effective potential energy }V_{e f f}&=\\frac{-G M m}{r}+\\frac{L^{2}}{2 m r^{2}}=\\frac{-G M m}{r}+\\frac{m v^{2} b^{2}}{2 r^{2}}\n\t\t\\end{align*}\n\t\tFor minimum value of $b$ we will have to assume that asteroid just misses the star as shown in figure. Equivalent 1-d problem is also shown in figure\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.5cm,width=12cm]{diagram-20220221(7)-crop}\n\t\t\\end{figure}\n\t\t\\text{Total energy is conserved in central force motion }\n\t\t\\begin{align*}\n\t\t\\text{Therefore, }E_{1}&=E_{2}\\\\\n\t\t\\text{or }&\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{e f f}\\right)_{\\mathrm{at} r=\\infty}=\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{e f f}\\right)_{\\mathrm{at} r=R}\\\\\n\t\t\\frac{1}{2} m v^{2}-\\frac{G M m}{\\infty}+\\frac{m v^{2} b^{2}}{\\infty}&=\\left(\\frac{1}{2} m o^{2}-\\frac{G M m}{R}+\\frac{m v^{2} b^{2}}{2 R^{2}}\\right)\\\\\n\t\t\\therefore v^{2}+\\frac{2 G M}{R}&=\\frac{v^{2} b^{2}}{R^{2}} \\quad \\text { or } b=R \\sqrt{1+\\frac{2 G M}{R v^{2}}}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A particle is thrown in radially outward direction from earth's surface with initial speed $\\sqrt{\\frac{3 G M}{4 R}}$ where $M$ is mass of earth $R$ is radius of earth. Find the height upto which particle goes.\n\t\\begin{answer}\n\t\tParticle has been thrown radially outward \n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4cm,width=6.5cm]{diagram-20220221(6)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t\\text{therefore }L&=0, V_{e f f}=V(r)+\\frac{L^{2}}{2 m r^{2}}\\\\\n\t\t\\therefore V_{e f f}&=\\frac{-G M m}{r}+0\\\\\n\t\t\\intertext{from conservation of energy we get }E_{1}&=E_{2}\\\\\n\t\t\\text { or }\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{e f f}\\right)_{r=R}&=\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{d f}\\right)_{r=r} \\\\\n\t\t\\frac{1}{2} m \\cdot \\frac{3 G M}{4 R}-\\frac{G M m}{R}&=0-\\frac{G M m}{r} \\\\\n\t\t\\frac{3}{8 R}-\\frac{1}{R}&=-\\frac{1}{r} \\quad \\therefore r=\\frac{8 R}{5}\n\t\t\\end{align*}\n\t\tTherefore, maximum height from earth's surface $=r-R=\\frac{3 R}{5}$\n\t\\end{answer}\n\t\\item A particle is thrown from the earth's surface with speed $\\sqrt{\\frac{G M}{R}}$ where $M=$ mass of earth\n\t$R=$ radius of earth. If direction of initial velocity makes an angle $\\alpha$ with the outward radial direction, find the maximum distance of the particle from centre of earth.\n\t\\begin{answer}\n\t\tAngular momentum of the particle is\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=5cm,width=8cm]{diagram-20220221(5)-crop}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tL &=m v \\sin \\alpha R=m \\sqrt{\\frac{G M}{R}} \\sin \\alpha R \\\\\n\t\t&=m \\sqrt{G M R} \\sin \\alpha \\\\\n\t\t\\therefore V_{e f f} &=V(r)+\\frac{L^{2}}{2 m r^{2}}=\\frac{-G M m}{r}+\\frac{G M m R}{2 r^{2}} \\sin ^{2} \\alpha\n\t\t\\end{align*}\n\t\tFrom conservation of energy we get $\\mathrm{E}_{1}=\\mathrm{E}_{2}$\n\t\t\\begin{align*}\n\t\t\\text{\tor }\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{e f f}\\right)_{r=R}&=\\left(\\frac{1}{2} m \\dot{r}^{2}+V_{e f f}\\right)_{r=r}\\text{ (see the figure)}\\\\\n\t\t\\frac{1}{2} m \\frac{G M}{R} \\cos ^{2} \\alpha-\\frac{G M m}{R}+\\frac{G M m}{2 R} \\sin ^{2} \\alpha&=\\frac{-G M m}{r}+\\frac{G M m R}{2 r^{2}} \\sin ^{2} \\alpha\\\\\n\t\t\\text{Divide both sides by }\\frac{G M m}{2 R}\\text{ to get }&\\cos ^{2} \\alpha-2+\\sin ^{2} \\alpha=-2 \\frac{R}{r}+\\frac{R^{2}}{r^{2}} \\sin ^{2} \\alpha\\\\\n\t\t\\text{or }-1=-2\\left(\\frac{R}{r}\\right)+\\left(\\frac{R}{r}\\right)^{2} \\sin ^{2} \\alpha \\quad &\\therefore\\left(\\frac{R}{r}\\right)^{2} \\sin ^{2} \\alpha-2\\left(\\frac{R}{r}\\right)+1=0\\\\\n\t\t\\therefore \\frac{R}{r}=\\frac{+2 \\pm \\sqrt{4-4 \\sin ^{2} \\alpha}}{2 \\sin ^{2} \\alpha} \\quad \\frac{R}{r}&=\\frac{+1 \\pm \\cos \\alpha}{\\sin ^{2} \\alpha} \\\\\n\t\t\\therefore r=\\frac{R \\sin ^{2} \\alpha}{1 \\pm \\cos \\alpha}=\\frac{R\\left(1-\\cos ^{2} \\alpha\\right)}{(1 \\pm \\cos \\alpha)} \\quad &\\therefore r=R(1-\\cos \\alpha) \\quad \\text { or } \\quad r=R(1+\\cos \\alpha)\n\t\t\\end{align*}\n\t\tSince $r$ must be greater than $r$. Therefore, $r=R(1+\\cos \\alpha)$\n\t\\end{answer}\n\t\\item A particle moves under a central potential $V(r)=\\frac{-k}{r^{m}} .$ What should be value of $m$ for its orbit to be stable.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{Corresponding is }f(r)&=-\\frac{\\partial V}{\\partial r}=\\frac{-m k}{r^{m+1}}=-m k r^{-(m+1)}\\\\\n\t\t\\text{\tWe know that for }f&=-k r^{n}\\text{ condition for stability is $n>-3$.} \n\t\t\\intertext{Therefore for orbit to be stable under given potential we must have.}\n\t\t-(m+1)>-3&\\text{ or }m+1<3 \\quad \\therefore m<2\n\t\t\\end{align*}\n\t\\end{answer}\n\\end{enumerate}\n\\section{Lagrangian}\n\\begin{enumerate}\n\t\\item The Lagrangian of a particle of charge e and mass $m$ in applied electric and magnetic fields is given by $L=\\frac{1}{2} \\mathrm{~m} \\vec{v}^{2}+\\mathrm{e} \\overrightarrow{\\mathrm{A}} \\cdot \\vec{v}-\\mathrm{e} \\phi$, where $\\overrightarrow{\\mathrm{A}}$ and $\\phi$ are the vector and scalar potentials corresponding to the magnetic and electric fields, respectively. Which of the following statements is correct?\n\t \\begin{tasks}(1)\n\t\t\\task[\\textbf{a.}] The carionically conjugate momentum of the particle is given by $\\overrightarrow{\\mathrm{p}}=\\mathrm{m} \\overrightarrow{\\mathrm{v}}$\n\t\t\\task[\\textbf{b.}]The Hamiltonian of the particle is given by $\\mathrm{H}=\\frac{\\overrightarrow{\\mathrm{p}}^{2}}{2 \\mathrm{~m}}+\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\overrightarrow{\\mathrm{A}} \\cdot \\overrightarrow{\\mathrm{p}}+\\mathrm{e} \\phi$\n\t\t\\task[\\textbf{c.}]L remains unchanged under a gauge transformation of the potentials.\n\t\t\\task[\\textbf{d.}] Under a gauge transformation of the potentials, $\\mathrm{L}$ changes by the total time derivative of a function of $\\overrightarrow{\\mathrm{r}}$ and $\\mathrm{t}$.\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\mathrm{L}&=\\frac{1}{2} \\mathrm{~m} \\overrightarrow{\\mathrm{v}}-\\mathrm{e} \\phi+\\mathrm{e} \\overrightarrow{\\mathrm{A}} \\cdot \\overrightarrow{\\mathrm{v}}\n\t\\intertext{\tWe know that $\\vec{E}$ and $\\vec{B}$ fields are in variant under gauge transformation.}\n\t\t\\overrightarrow{\\mathrm{A}}(\\overrightarrow{\\mathrm{x}}, \\mathrm{t}) \\rightarrow \\overrightarrow{\\mathrm{A}}^{\\prime}&=\\overrightarrow{\\mathrm{A}}+\\vec{\\nabla} \\lambda(\\overrightarrow{\\mathrm{x}}, \\mathrm{t}) ; \\quad \\varphi(\\overrightarrow{\\mathrm{x}}, \\mathrm{t}) \\rightarrow \\varphi^{\\prime}=\\varphi-\\frac{\\partial \\lambda}{\\partial \\mathrm{t}}(\\overrightarrow{\\mathrm{x}}, \\mathrm{t})\n\t\\intertext{\twhere $\\lambda(\\vec{x}, t)$ is an arbitrary scalar function.}\n\t\t\\therefore \\quad \\mathrm{L} \\rightarrow \\mathrm{L}^{\\prime}&=\\mathrm{L}+\\mathrm{e}\\left(\\frac{\\partial \\lambda}{\\partial \\mathrm{t}}(\\overrightarrow{\\mathrm{x}}, \\mathrm{t})+\\overrightarrow{\\mathrm{v}} \\cdot \\vec{\\nabla} \\lambda(\\overrightarrow{\\mathrm{x}}, \\mathrm{t})\\right)\n\t\t\\end{align*}\n\t\tThe expression in bracket is just the total time derivative of $\\lambda(\\vec{x}, t)$.\\\\\n\t\tIf we add a total time derivative of a function of $\\overrightarrow{\\mathrm{x}}$ and $\\mathrm{t}$ to the Lagrangian, the equations of motion do not change.\\\\\n\t\tCorrect answer is option \\textbf{(d)}\n\t\\end{answer}\n\t\\item The Hamiltonian o fa system with $n$ degrees of freedom is given by \\\\$H\\left(q_{i}, \\ldots \\ldots \\ldots ., q_{n} ; p_{i}, \\ldots \\ldots \\ldots \\ldots, p_{n} ; t\\right)$\\\\\n\twith an explicit dependence on the time $t$. Which of the following is correct?\n\t \\begin{tasks}(1)\n\t\t\\task[\\textbf{a.}]Different phase trajectories cannot intersect each other\n\t\t\\task[\\textbf{b.}]H always represents the total energy of the system and is a constant of the motion.\n\t\t\\task[\\textbf{c.}]The equations $\\dot{\\mathrm{q}}_{\\mathrm{i}}=\\partial \\mathrm{H} / \\partial \\mathrm{p}_{\\mathrm{i}}, \\dot{\\mathrm{p}}_{\\mathrm{i}}=-\\partial \\mathrm{H} / \\partial \\mathrm{q}_{\\mathrm{i}}$ are not valid since H has explicit time dependence.\n\t\t\\task[\\textbf{d.}] Any initial volume element in phase space remains unchanged in magnitude under time evolution.\n\t\\end{tasks}\n\\begin{answer}\n\tAccording to Liouville's theorem, the phase volume occupied by a collection of system evolve according to - Hamilton's equation of motion, and will be preserved in time.\\\\\n\tCorrect answer is option \\textbf{(d)}\n\\end{answer}\n\t\\item A double pendulum consists of two point masses $m$ attached by massless strings of length $l$ as shown in the figure:\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=4.5cm]{CM-02}\n\t\\end{figure}\n\tThe kinetic energy of the pendulum is :\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\frac{1}{2} \\mathrm{~m} \\ell^{2}\\left[\\dot{\\theta}_{1}^{2}+\\dot{\\theta}_{2}^{2}\\right]$\n\t\t\\task[\\textbf{b.}]$\\frac{1}{2} m \\ell^{2}\\left[2 \\dot{\\theta}_{1}^{2}+\\dot{\\theta}_{2}^{2}+2 \\dot{\\theta}_{1} \\dot{\\theta}_{2} \\cos \\left(\\dot{\\theta}_{1}-\\dot{\\theta}_{2}\\right)\\right]$\n\t\t\\task[\\textbf{c.}]$\\frac{1}{2} m \\ell^{2}\\left[\\dot{\\theta}_{1}^{2}+2 \\dot{\\theta}_{2}^{2}+2 \\dot{\\theta}_{1} \\dot{\\theta}_{2} \\cos \\left(\\dot{\\theta}_{1}-\\dot{\\theta}_{2}\\right)\\right]$\n\t\t\\task[\\textbf{d.}] $\\frac{1}{2} \\mathrm{~m} \\ell^{2}\\left[2 \\dot{\\theta}_{1}^{2}+\\dot{\\theta}_{2}^{2}+2 \\dot{\\theta}_{1} \\dot{\\theta}_{2} \\cos \\left(\\dot{\\theta}_{1}+\\dot{\\theta}_{2}\\right)\\right]$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\mathrm{x}_{1}&=\\ell \\sin \\theta_{1}, \\mathrm{x}_{2}=\\ell \\sin \\theta_{1}+\\ell \\sin \\theta_{2} \\\\ \\mathrm{y}_{1}&=-\\ell \\cos \\theta_{1}, \\mathrm{y}_{2}=-\\ell \\cos \\theta_{1}-\\ell \\cos \\theta_{2} \\\\ \\dot{\\mathrm{x}}_{1}&=\\ell \\cos \\theta_{1} \\dot{\\theta}_{1}, \\dot{\\mathrm{y}}_{1}=\\ell \\sin \\theta_{1} \\dot{\\theta}_{1}\n\t\t\\end{align*}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4.5cm,width=6.5cm]{CM-03}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t\\dot{x}_{2}&=\\ell \\cos \\theta_{1} \\dot{\\theta}_{1}+\\ell \\cos \\theta_{2} \\dot{\\theta}_{2} ; \\quad \\dot{y}_{2}=\\ell \\sin \\theta_{1} \\dot{\\theta}_{1}+\\ell \\sin \\theta_{2} \\dot{\\theta}_{2},\\\\\n\t\tT&=\\frac{m}{2}\\left(\\dot{x}_{1}^{2}+\\dot{y}_{1}^{2}\\right)+\\frac{m}{2}\\left(\\dot{x}_{2}^{2}+\\dot{y}_{2}^{2}\\right)\\\\\n\t\t&=\\frac{\\mathrm{m}}{2} \\dot{\\theta}_{1}^{2} \\ell^{2}+\\frac{\\mathrm{m}}{2}\\left[\\dot{\\theta}_{1}^{2} \\ell^{2}+\\dot{\\theta}_{2}^{2} \\ell^{2}+2 \\ell^{2} \\dot{\\theta}_{1} \\dot{\\theta}_{2} \\cos \\left(\\theta_{1}-\\theta_{2}\\right)\\right]\\\\\n\t\t&=\\frac{\\mathrm{m} \\ell^{2}}{2}\\left[2 \\dot{\\theta}_{1}^{2}+\\dot{\\theta}_{2}^{2}+2 \\dot{\\theta}_{1} \\dot{\\theta}_{2} \\cos \\left(\\theta_{1}-\\theta_{2}\\right)\\right]\n\t\t\\end{align*}\n\t\tCorrect answer is \\textbf{(b)}\n\t\\end{answer}\n\t\\item A particle of mass ' $\\mathrm{m}$ ' moves inside a bowl. If the surface of the bowl is given by the equation $\\mathrm{z}=\\frac{1}{2} \\mathrm{a}\\left(\\mathrm{x}^{2}+\\mathrm{y}^{2}\\right)$, where $a$ is a constant, the Lagrangian of the particle is:\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}] $\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\phi}^{2}-g a r^{2}\\right)$\n\t\t\\task[\\textbf{b.}] $\\frac{1}{2} m\\left[\\left(1+a^{2} r^{2}\\right) \\dot{r}^{2}+r^{2} \\dot{\\phi}^{2}\\right]$\n\t\t\\task[\\textbf{c.}]$\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\theta}^{2}+r^{2} \\sin ^{2} \\dot{\\phi}^{2}-g a r^{2}\\right)$\n\t\t\\task[\\textbf{d.}] $\\frac{1}{2} m\\left[\\left(1+a^{2} r^{2}\\right) \\dot{r}^{2}+r^{2} \\dot{\\phi}^{2}-g a r^{2}\\right]$\n\t\\end{tasks}\n\t\\begin{answer}$\\left. \\right. $\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=1.5cm,width=2cm]{CM-04}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tz&=\\frac{1}{2} a\\left(x^{2}+y^{2}\\right)=\\frac{1}{2} a r^{2} \\quad \\Rightarrow \\dot{z}=a r \\dot{r}\\\\\n\t\\text{\tK.E. }&=T=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)\\\\\n\t&=\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\varphi}^{2}+\\dot{z}^{2}\\right)\\\\\n\t&=\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\varphi}^{2}+a^{2} r^{2} \\dot{r}^{2}\\right) \\qquad\\text{ P.E.} =V=m g z=\\frac{1}{2} m g a r^{2}\\\\\n\tL&=T-V=\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\varphi}^{2}+a^{2} r^{2} \\dot{r}^{2}-g a r^{2}\\right)=\\frac{1}{2} m\\left[\\left(1+a^{2} r^{2}\\right) r^{2}+r^{2} \\dot{\\varphi}^{2}-g a r^{2}\\right]\n\t\t\\end{align*}\n\t\tCorrect answer is option\\textbf{(d)}\n\t\\end{answer}\n\t\\item A mass point glides without friction on a cycloid, which is given by $x=a(\\vartheta-\\sin \\vartheta)$ and $y=a(1+\\cos \\vartheta)$ (with $0 \\leq \\vartheta \\leq 2 \\pi$ ). Determine\\\\\n\t(a) the Lagrangian, and\\\\\n\t(b) the equation of motion\\\\\n\t(c) Solve the equation of motion\n\t\\begin{answer}\n\t\tThe cycloid is represented by\n\t\t\\begin{align*}\n\t\tx=a(\\vartheta-\\sin \\vartheta), \\quad y=a(1+\\cos \\vartheta),\n\t\t\\end{align*}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=3.2cm,width=9cm]{CM-05}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\\text{\twhere }&0 \\leq \\vartheta \\leq 2 \\pi.\\text{ The kinetic energy is}\\\\\n\t\tT=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)&=\\frac{1}{2} m a^{2}\\left\\{[(1-\\cos \\vartheta) \\dot{\\vartheta}]^{2}+[-(\\sin \\vartheta) \\dot{\\vartheta}]^{2}\\right\\},\\\\\n\t\\text{\tand the potential energy is }V&=m g y=m g a(1+\\cos \\vartheta).\\\\\n\t\t\\text{The Lagrangian is given by }L&=T-V=m a^{2}(1-\\cos \\vartheta) \\dot{\\vartheta}^{2}-m g a(1+\\cos \\vartheta).\\\\\n\t\t\\text{The equation of motion then reads }&\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{\\vartheta}}\\right)-\\frac{\\partial L}{\\partial \\vartheta}=0,\\\\\n\t\\text{\ti.e., }\\quad &\\frac{d}{d t}\\left[2 m a^{2}(1-\\cos \\vartheta) \\dot{\\vartheta}\\right]-\\left[m a^{2}(\\sin \\vartheta) \\dot{\\vartheta}^{2}+m g a \\sin \\vartheta\\right]=0\\\\\n\t\\text{ or }\\quad &\\frac{d}{d t}[(1-\\cos \\vartheta) \\dot{\\vartheta}]-\\frac{1}{2}(\\sin \\vartheta) \\dot{\\vartheta}^{2}-\\frac{8}{2 a} \\sin \\vartheta=0,\\\\\n\t\\text{ i.e.,} \\quad&(1-\\cos \\vartheta) \\ddot{\\vartheta}+\\frac{1}{2}(\\sin \\vartheta) \\dot{\\vartheta}^{2}-\\frac{g}{2 a} \\sin \\vartheta=0.\\\\\n\\text{\tBy setting }u&=\\cos \\left(\\frac{\\vartheta}{2}\\right), \\text{one has }\\frac{d u}{d t}=-\\frac{1}{2} \\sin \\left(\\frac{\\vartheta}{2}\\right) \\dot{\\vartheta}\\text{ and } \\frac{d^{2} u}{d t^{2}}\\\\&=-\\frac{1}{2} \\sin \\left(\\frac{\\vartheta}{2}\\right) \\ddot{\\vartheta}-\\frac{1}{4} \\cos \\left(\\frac{\\vartheta}{2}\\right) \\dot{\\vartheta}^{2} .\\\\ \\text{Since }&\\cot \\left(\\frac{\\vartheta}{2}\\right)=\\sin \\frac{\\vartheta}{(1-\\cos \\vartheta)},\\text{ we can write as}\\\\\n\\ddot{\\vartheta}+\\frac{1}{2} \\cot \\left(\\frac{\\vartheta}{2}\\right) \\dot{\\vartheta}^{2}-\\frac{g}{2 a} \\cot \\left(\\frac{\\vartheta}{2}\\right)&=0,\\text{ and therefore,} \\frac{d^{2} u}{d t^{2}}+\\frac{g}{4 a} u=0. \n\\intertext{The solution of this differential equation is}\nu&=\\cos \\left(\\frac{\\vartheta}{2}\\right)=C_{1} \\cos \\sqrt{\\frac{g}{4 a}} t+C_{2} \\sin \\sqrt{\\frac{g}{4 a}} t \n\\intertext{ The motion is just like the vibration of an ordinary pendulum of length $l=4 a$. The arrangement is therefore called a \"cycloid pendulum\".}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item \tThree particles of equal mass ' $m$ ' are connected by two identical massless springs of stiffiness constant ' $k$ ' as shown in the figure.\\\\\n\tIf $x_{1}, x_{2}$ and $x_{3}$ denote the displacements of the masses from their respective equilibrium positions, the potential energy of the system is:\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\frac{1}{2} k\\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\right)$\n\t\t\\task[\\textbf{b.}]$\\frac{1}{2} k\\left[x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{2}\\left(x_{1}+x_{3}\\right)\\right]$\n\t\t\\task[\\textbf{c.}]$\\frac{1}{2} k\\left[x_{1}^{2}+2 x_{2}^{2}+x_{3}^{2}+2 x_{2}\\left(x_{1}+x_{3}\\right)\\right]$\n\t\t\\task[\\textbf{d.}] $\\frac{1}{2} k\\left[x_{1}^{2}+2 x_{2}^{2}+x_{3}^{2}-2 x_{2}\\left(x_{1}+x_{3}\\right)\\right]$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{Potential energy }&=\\frac{1}{2} k\\left(x_{2}-x_{1}\\right)^{2}\\hspace{2cm}\n\t\t\\text{first spring}\\\\\n\t\\text{\tPotential energy }&=\\frac{1}{2} k\\left(x_{3}-x_{2}\\right)^{2}\\hspace{2cm}\n\t\\text{\tsecond spring}\n\t\\intertext{Potential energy of the system,}\n\t&=\\frac{1}{2} k\\left(x_{2}-x_{1}\\right)^{2}+\\frac{1}{2} k\\left(x_{3}-x_{2}\\right)^{2}=\\frac{1}{2} k\\left[x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}+x_{2}^{2}+x_{3}^{2}-2 x_{2} x_{3}\\right]\\\\\n\t&=\\frac{1}{2} k\\left[x_{1}^{2}+2 x_{2}^{2}+x_{3}^{2}-2 x_{2}\\left(x_{1}+x_{3}\\right)\\right]\n\t\t\\end{align*}\n\t\tCorrect answer is option \\textbf{(d)}\n\t\\end{answer}\n\t\\item The Lagrangian of a particle of mass $m$ moving in one dimensions is given by\n\t$$\n\t\\mathrm{L}=\\frac{1}{2} \\mathrm{~m} \\dot{\\mathrm{x}}^{2}-\\mathrm{bx}\n\t$$\n\twhere $b$ is a positive constant. The coordinate of the particle $x(t)$ at time $t$ is given by: (in the following $c_{1}$ and $c_{2}$ are constants)\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$-\\frac{b}{2 m} t^{2}+c_{1} t+c_{2}$\n\t\t\\task[\\textbf{b.}]$c_{1} t+c_{2}$\n\t\t\\task[\\textbf{c.}]$c_{1} \\cos \\left(\\frac{b t}{m}\\right)+c_{2} \\sin \\left(\\frac{b t}{m}\\right)$\n\t\t\\task[\\textbf{d.}] $c_{1} \\cosh \\left(\\frac{b t}{m}\\right)+c_{2} \\sin h\\left(\\frac{b t}{m}\\right)$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tL&=\\frac{1}{2} m \\dot{x}^{2}-b x=\\mathrm{T}-\\mathrm{V}\\\\\n\t\t\\therefore \\mathrm{V}&=b x\n\t\t\\intertext{ This is same as uniform gravitational potential. Therefore, solution must be of the form. This is same as uniform} \\mathrm{S}&=\\mathrm{S}_{0}+\\mathrm{ut}+\\frac{1}{2} \\mathrm{at}^{2}\n\t\t\\end{align*}\n\t\tCorrect answer is option \\textbf{(a)}\n\t\\end{answer}\n\t\\item A particle moves in a potential $V=x^{2}+y^{2}+\\frac{z^{2}}{2}$. Which component (s) of the angular momentum is $/$ are constant (s) of motion?\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]None\n\t\t\\task[\\textbf{b.}]$L_{x}, L_{y}$ and $L_{z}$\n\t\t\\task[\\textbf{c.}]Only $L_{x}$ and $L_{y}$\n\t\t\\task[\\textbf{d.}] Only $\\mathrm{L}_{z}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{Given }V(x, y, z)&=x^{2}+y^{2}+\\frac{z^{2}}{2}\n\t\\intertext{\tIn polar co-ordinate (spherical polar)}\n\tV&=r^{2} \\sin ^{2} \\theta+\\frac{r^{2} \\cos ^{2} \\theta}{2} \\quad\\left[\\begin{array}{l}x=r \\cos \\phi \\sin \\theta \\\\ y=r \\sin \\phi \\sin \\theta \\\\ z=r \\cos \\theta\\end{array}\\right.\n\\intertext{\tTherefore, Lagrangian of system is}\n\t\\mathrm{L}&=\\mathrm{T}-\\mathrm{V} \\\\\n\tL&=\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\theta}^{2}+r^{2} \\sin ^{2} \\theta \\dot{\\phi}^{2}\\right)-r^{2} \\sin ^{2} \\theta-\\frac{r^{2} \\cos ^{2} \\theta}{2}\n\t\t\\end{align*}\n\t\t$\\phi$ is cyclic, therefore $p_{\\phi}$ is constant of motion.\\\\\n\t\t$p_{\\phi}$ is equal to $L_{z}$. Therefore, $L_{z}$ is constant of motion.\\\\\n\t\tCorrect answer is option \\textbf{(d)}\n\t\\end{answer}\n\t\\item\tA particle of mass $m$ is projected with an initial velocity $u$ at an angle $\\alpha$ with the horizontal. Use Lagran equations to describe the motion of projectile. The resistance of air is negligible.\n\t\\begin{answer}\n\t\t\tLet $x, y$ be the coordinate of the particle at any instant. The system is holonomic because its positic confined to a plane and conservative because the only external force acting on the projectile is conse tive.\n\t\t\\begin{align*}\n\t\tT&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)\\text{ and }V=m g y\\\\\n\t\t\\therefore L&=T-V=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)-m g v\\\\\n\t\t\\frac{\\partial L}{\\partial \\dot{x}}&=m \\dot{x}\\text{ and } \\frac{\\partial L}{\\partial x}=0\\\\\n\t\t\\text{By Lagrange's equation, }\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}}\\right)&=\\frac{\\partial L}{\\partial x}\\\\\n\t\t\\therefore \\quad \\frac{d}{d t}(m \\dot{x})&=0 \\text { or } \\ddot{x} =0 \\\\\n\t\t\\frac{\\partial L}{\\partial \\dot{y}}&=m y \\text { and } \\frac{\\partial L}{\\partial y} =-m g\\\\\n\t\t\\text{By Lagrange's equation }\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{y}}\\right)&=\\frac{\\partial L}{\\partial y}\\\\\n\t\t \\therefore \\quad \\frac{d}{d t}(m y)&=-m g\\text{ or }\\ddot{y}=-g\n\t\t \\intertext{Thus, $\\ddot{x}=0, \\ddot{y}=-g$ are the two equations describing the motion of a projectile. Proceeding with the first equation,}\n\t\t \\frac{d}{d t}\\left(\\frac{d x}{d t}\\right)&=0 \\text { or } d\\left(\\frac{d x}{d t}\\right)=0\\\\\n\t\t \\text{Integrating, }\\quad \\frac{d x}{d t}&=c_{1}(\\text{ a constant} )\\\\\n\t\t\\text{ At }t&=0, \\quad \\frac{d x}{d t}=u \\cos \\alpha ; \\quad \\therefore u \\cos \\alpha=c_{1}\\\\\n\t\t\\therefore \\quad \\frac{d x}{d t}&=u \\cos \\alpha, \\quad \\text { or } d x=u \\cos \\alpha d t\\\\\n\t\\text{\tIntegrating, }x&=(u \\cos \\alpha) t+x_{0}(\\text{ a constant })\\\\\n\\text{\tAt $t=0, x=0$, }&\\text{therefore, }x_{0}=0 ; x=(u \\cos \\alpha) t\n \\intertext{Proceeding with the second equation,}\n\t\\frac{d}{d t}\\left(\\frac{d y}{d t}\\right)&=-g, \\text { or } d\\left(\\frac{d y}{d t}\\right)=-g d t\\\\\n\\text{\tIntegrating, }\\frac{d y}{d t}&=-g t+c_{2}(\\mathrm{a}\\text{ constant })\\\\\n\t\\text{At }t&=0, \\frac{d y}{d t}=u \\sin \\alpha ; \\therefore u \\sin \\alpha=c_{2}\\\\\n\t\\therefore \\quad \\frac{d y}{d t}&=-g t+u \\sin \\alpha, o r d y=-g t d t+u \\sin \\alpha d t\\\\\n\\text{\tIntegrating, }y&=-\\frac{1}{2} g t^{2}+(u \\sin \\alpha) t+y_{0}\\\\\n\\text{\tAt }t&=0, y=0\\text{ and therefore, $y_{0}=0$}\\\\\n\t\\therefore \\quad y&=(u \\sin \\alpha) t-\\frac{1}{2} g t^{2}\n\t\t\\end{align*}\n\t\tThus, $x=(u \\cos \\alpha) t, y=(u \\sin \\alpha) t-\\frac{1}{2} g t^{2}$ are the two solved equation describing the motion of a projectile.\n\t\\end{answer}\n\t\\item A particle of mass $m$ is moving in a plane under the influence of a force directed towards a fixed point and varying inversely as the square of the distance from that point. Set up the Lagrangian and equations of motion of the particle.\n\t\\begin{answer}\n\t\tThe problem is best solved in polar coordinates with respect to the fixed point as origin and any line through it as the reference line. Let $(r, \\theta)$ be the polar coordinates of the particle at any time. Since there are two coordinates there will be two Lagrange's equations describing the motion of the particle. Motion is holonomic because the position of the particle is confined to a plane and conservative because the force is derivable from energy or energy is derivable from the force.\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.5cm,width=5cm]{CM-06}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\tT&=\\frac{1}{2} m \\dot{r}^{2} \\text{(linear kinetic energy)} +\\frac{1}{2} I \\dot{\\theta}^{2} \\text{(rotational $k$. energy)}\\\\\n\t\t\\text{Or, }\\quad T&=\\frac{1}{2} m \\dot{r}^{2}+\\frac{1}{2}\\left(m r^{2}\\right) \\dot{\\theta}^{2} \\quad\\left(\\because I=m r^{2}\\right)\n\t\t\\intertext{Since force varies inversely as the square of the distance and it is directed towards the origin, $F=-\\left(k / r^{2}\\right)$ where $k$ is a constant. But}\n\t\tF&=-\\left(\\frac{d V}{d r}\\right)\\\\\n\t\t\\therefore\\quad&=\\frac{k}{r^{2}}=-\\frac{d V}{d r}\\text{, or } d V=\\frac{k d r}{r^{2}}\\\\\n\t\t\\text{Integrating, }V&=-\\frac{k}{r}+V_{0}\\text{ (a constant)}\\\\\n\t\t\\text{At }r&=\\infty, V=0\\text{ and therefore, }V_{0}=0 ; \\therefore V=-\\left(\\frac{k}{r}\\right)\\\\\n\t\t\\therefore \\quad L&=T-V=\\frac{1}{2} m\\left(\\dot{r}^{2}+r^{2} \\dot{\\theta}^{2}\\right)+\\frac{k}{r}\\\\\n\t\t\\frac{\\partial L}{\\partial \\dot{r}}&=m \\dot{r}\\text{ and }\\frac{\\partial L}{\\partial r}=m r \\dot{\\theta}^{2}-\\frac{k}{r^{2}}\\\\\n\t\t\\frac{\\partial L}{\\partial \\dot{\\theta}}&=m r^{2} \\dot{\\theta}\\text{ and} \\frac{\\partial L}{\\partial \\theta}=0\\\\\n\t\t\\text{By Lagrange's equations of motion, }&\\left[\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial q_{k}}\\right)=\\frac{\\partial L}{\\partial q_{k}}\\right]\\text{ we have }\\\\\n\t\t\\frac{d}{d t}(m \\dot{r})&=m r \\dot{\\theta}^{2}-\\frac{k}{r^{2}}\\text{ and } \\frac{d}{d t}\\left(m r^{2} \\dot{\\theta}\\right)=0\\\\\n\t\t\\text{Or, }\\quad \\ddot{r}&=r \\dot{\\theta}^{2}-\\frac{k}{m r^{2}}\\text{ and} r^{2} \\dot{\\theta}=\\text{ constant}\\\\\n\t\\text{\tOr, }\\quad \\ddot{r}-r \\dot{\\theta}^{2}&=-\\frac{k}{m r^{2}}\\text{ or }\\ddot{r}-r \\dot{\\theta}^{2}=-\\frac{\\omega^{2}}{r^{2}}\\\\\n\\text{\twhere, }\\omega^{2}&=\\frac{k}{m}\\text{ and }r^{2} \\dot{\\theta}=h \\text{(another constant).}\n\\intertext{\tThus equations describing the motion of the particles are} \\ddot{r}-r \\dot{\\theta}^{2}&=-\\frac{\\omega^{2}}{r^{2}}\\text{ and }r^{2} \\dot{\\theta}=h\\text{ where }\\omega^{2}\\text{ and $h$ are constants.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Obtain the Lagrangian of a linear simple harmonic oscillator and the equation of motion in one dimension.\n\t\\begin{answer}\n\t\tIn a simple harmonic oscillator the particle is acted on by a force which is directed towards a fixed point (position of equilibrium) and whose magnitude varies linearly with the distance from the position ofequilibrium. That is, $F=-k x$ where $\\mathrm{x}$ is the distance of the particle from the equilibrium position.\n\t\t\\begin{align*}\n\t\t\\text{But}\n\t\tF&=-\\left(\\frac{d V}{d x}\\right) ; \\quad \\because d V=-F d x=h x d x\\\\\n\t\t\\text{Integrating, }\\quad V&=\\frac{1}{2} k x^{2}+V_{0}(a\\text{ constant })\\\\\n\t\\text{\tAt }x=0, V&=0\\text{ and therefore, }V_{0}=0\\\\\n\t\t\\therefore \\quad V&=\\frac{1}{2} k x^{2} ; \\quad T=\\frac{1}{2} m \\dot{x}^{2}\\\\\n\t\t\\therefore \\quad L&=T-V=\\frac{1}{2} m x^{2}-\\frac{1}{2} k x^{2}\\\\\n\t\\text{\tNow, }\\quad \\frac{\\partial L}{\\partial \\dot{x}}&=m \\dot{x}\\text{ and } \\frac{\\partial L}{\\partial \\dot{x}}=-k x\\\\\n\t\t\\text{By Lagrange's equation, }&\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}}\\right)=\\frac{\\partial L}{\\partial x}\\\\\n\t\t\\therefore\\quad\\frac{d}{d t}(m x)&=-k x, \\text { or } m \\ddot{x}=-k x\n\t\t\\intertext{This is the equation of a simple harmonic oscillator in one dimension.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Figure below shows a solid cylinder with centre $G$ and radius $a$ rolling on the rough inside surface of a fixed cylinder with centre $O$ and radius $b>a$. Find the Lagrange equation of motion and deduce the period of small oscillations about the equilibrium position.\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4cm,width=7cm]{Lagrangian 01}\n\t\\end{figure}\n\t\\begin{answer}\n\t\tIf the cylinder were not obliged to roll, the system would have two degrees of freedom with generalised coordinates $\\theta$ (the angle between $O G$ and the downward vertical) and $\\phi$ (the rotation angle of the cylinder measured from some reference position).\\\\\n\t\tThe roling condition imposes the kinematical constraint.\n\t\t\\begin{align*}\n\t\t(b-a) \\dot{\\theta}-a \\dot{\\phi}&=o\n\t\\intertext{\tThis constraint is integrable and is equivalent to the geometrical constraint.}\n\t(b-a) \\theta-a \\phi&=o\n\\intertext{\ton taking $\\phi=0$ when $\\theta=0$. Thus the rolling cylinder is a standard conservative system with one degree of freedom.}\n\\intertext{Take $\\theta$ as the generalised coordinate. Then the kinetic energy is given by}\nT &=\\frac{1}{2} m((b-a) \\dot{\\theta})^{2}+\\frac{1}{2}\\left(\\frac{1}{2} m a^{2}\\right) \\dot{\\phi}^{2} \\\\\n&=\\frac{1}{2} m((b-a) \\dot{\\theta})^{2}+\\frac{1}{2}\\left(\\frac{1}{2} m a^{2}\\right)\\left(\\frac{b-a}{a}\\right)^{2} \\dot{\\theta}=\\frac{3}{4} m(b-a)^{2} \\dot{\\theta}^{2}\\\\\n\\text{and the potential energy by }V&=-m g(b-a) \\cos \\theta\n\\intertext{There is only one Lagrange equation, namely}\n\\frac{d}{d t}\\left[\\frac{3}{2} m(b-a)^{2} \\dot{\\theta}\\right]-0*=-m g(b-a) \\sin \\theta\n\\intertext{which simplifies to give}\n\\ddot{\\theta}+\\frac{2 g}{3(b-a)} \\sin \\theta&=0\n\\intertext{Interestingly, this equation is identical to the exact equation for the oscillations of a simple pendulum of length $3(b-a) / 2$ as obtaind.}\n\\intertext{The linearised equation governing small oscillations of the cylinder about $\\theta=0$ is is}\n\\ddot{\\theta}+\\frac{2 g}{3(b-a)} \\theta&=0\n\\intertext{so that the period $\\tau$ of small oscillation is given by}\n\\tau&=2 \\pi\\left(\\frac{3(b-a)}{2 g}\\right)^{1 / 2}\n\t\t\\end{align*}\n\\end{answer}\n\\item A pendulum of mass $m$ is attached to a block of mass $M$. The block slides on a horizontal frictionless surface. Find the Lagrangian and equation of motion of the pendulum. For small amplitude oscillations, derive an expression for periodic time.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=4cm,width=5cm]{Lagrangian 02}\n\\end{figure}\n\\begin{answer}\n\t\tLet at any time $t$ the coordinates of $M$ and $m$ be $\\left(x_{1}, 0\\right)$ and $\\left(x_{2}, y_{2}\\right)$ respectively.\n\t\\begin{align}\n\t\\text{Here, }\\quad x_{2}&=x_{1}+\\ell \\sin \\theta\\text{ and }ay_{2}=-\\ell \\cos \\theta.\\text{ The suitable generalized coordinates are $x_{1}$ and $\\theta$.}\\notag\\\\\n\t T &=\\frac{1}{2} M \\dot{x}_{1}^{2}+\\frac{1}{2} m\\left(\\dot{x}_{2}^{2}+\\dot{y}_{2}^{2}\\right) \\notag\\\\ &=\\frac{1}{2} M \\dot{x}_{1}^{2}+\\frac{1}{2} m\\left[\\left(\\dot{x}_{1}+\\ell \\dot{\\theta} \\cos \\theta\\right)^{2}+(\\ell \\dot{\\theta} \\sin \\theta)^{2}\\right] \\notag\\\\ &=\\frac{1}{2} M \\dot{x}_{1}^{2}+\\frac{1}{2} m\\left(\\dot{x}_{1}^{2}+\\ell^{2} \\dot{\\theta}^{2}+2 \\ell \\dot{x}_{1} \\dot{\\theta} \\cos \\theta\\right)\\notag\\\\\n\t (\\text{because }\\dot{x}_{2}&=\\dot{x}_{1}+\\ell \\cos \\theta \\dot{\\theta} \\text{and }\\dot{y}_{2}=\\ell \\sin \\theta \\dot{\\theta} )\\notag\\\\\n\t V&=-m g \\cos \\theta\\notag\\\\\n\t\\text{ Hence, }\\quad L&=T-V=\\frac{1}{2}(M+m) \\dot{x}_{1}^{2}+\\frac{1}{2} m \\ell^{2} \\dot{\\theta}^{2}+m \\ell \\dot{x}_{1} \\dot{\\theta} \\cos \\theta+m g \\ell \\cos \\theta\\notag\n\t\\intertext{We see that $x_{1}$ is cyclic coordinate, so}\\notag\n\t\\text{Here, }\\frac{\\partial L}{\\partial x_{1}}&=0\\text{ and }\\frac{\\partial L}{\\partial \\dot{x}_{1}}=(M+m) \\dot{x}_{1}+m \\ell \\dot{\\theta} \\cos \\theta\\text{ is conserved.}\\notag\\\\\n\t\\frac{\\partial L}{\\partial \\theta}&=m \\ell\\left(\\dot{x}_{1} \\dot{\\theta}+g\\right)(-\\sin \\theta) \\text { and } \\frac{\\partial L}{\\partial \\dot{\\theta}}=m \\ell^{2} \\dot{\\theta}+m \\ell \\dot{x}_{1} \\cos \\theta\\notag\n\\intertext{\tEquation of motion in $\\theta$ is}\\notag\n\tm \\ell^{2} \\ddot{\\theta}&+m \\ell \\ddot{x}_{1} \\cos \\theta+m \\ell(-\\sin \\theta) \\dot{\\theta} \\dot{x}_{1}-m \\ell(-\\sin \\theta) \\dot{x}_{1} \\dot{\\theta}+m g \\ell \\sin \\theta=0\\notag\\\\\n\\text{\tOr, }\\quad m \\ell^{2} \\ddot{\\theta}&+m \\ell \\cos \\theta \\ddot{x}_{1}+m g \\ell \\sin \\theta=0\\notag\\\\\n\t\\text{If }\\theta\\text{ is small, }&\\sin \\theta \\approx \\theta\\text{ and also }\\cos \\theta \\approx 1\\text{, then}\\notag\\\\\n\tm \\ell^{2} \\ddot{\\theta}&+m \\ell \\ddot{x}_{1}+m g \\ell \\theta=0\\notag\\\\\n\\text{\tOr,}\\quad \n\t\\ddot{\\theta}+\\frac{\\ddot{x}_{1}}{\\ell}&+\\frac{g}{\\ell} \\theta=0 \\label{CMP-25}\n\\intertext{\tEquation of motion in $x_{1}$ is}\\notag\n(M+m) \\ddot{x}_{1}&+m \\ell\\left(\\ddot{\\theta} \\cos \\theta-\\dot{\\theta}^{2} \\sin \\theta\\right)=0\\notag\\\\\n\\text{For small }\\theta,&\\left(\\cos \\theta \\cong 1, \\sin \\theta \\cong \\theta\\right.\\text{ and }\\dot{\\theta}^{2} \\theta\\text{ is negligible })\\notag\\\\\n(M+m) \\ddot{x}_{1}&+m \\ell \\ddot{\\theta}=0\\label{CMP-26}\n\\intertext{Fromequations (\\ref{CMP-25}) and (\\ref{CMP-26}), we have}\\notag\n\\ddot{\\theta}-\\frac{m \\ddot{\\theta}}{M+m}&+\\frac{g}{\\ell} \\theta=0\\notag\\\\\n\\text{Hence,}\\quad\n\\ddot{\\theta}&=-\\left[\\frac{M+m}{M}\\right] \\frac{g}{\\ell} \\theta\\notag\n\\intertext{This is the equation of simple harmonic motion whose period is given by}\\notag\nT&=2 \\pi \\sqrt{\\frac{\\ell}{g}} \\sqrt{\\frac{M}{M+m}}\\notag\n\t\\end{align}\n\\end{answer}\n\\item Auniform disc of radius ' $\\dot{a}$ ' and mass $m$, rotates about a fixed axis. A mass less rope is fixed to a points on the out side 'circumference' and leads to mass less spring which is intwin fastened to a fixed point. At a radius $a / 2$ another cord is fastenped to a spring which connects to a $m a, m$. Set up the Lagrange's eq. of the DISC and the mass.\n\\begin{answer}$\\left. \\right. $\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=6cm,width=5cm]{Lagrangian 03}\n\t\\end{figure}\n\tLet $b_{1}$ and $b_{2}$ be the original length of the spring 1 and 2 . Which the whole system is in eq. and the DISC is stationary.\\\\\n\tLet $\\theta$ be the angle the DISC has twined away from the eq. position. Then extension of the spring $-1$ is\n\t\\begin{align*}\n\tx_{1}&=a \\theta\n\t\\intertext{If the spring $-2$ is stretched by a distance $x_{2}$ the mass $m$ is low ered by a distance}\n\tx_{3}&=\\frac{a}{2} \\theta+x_{2}\\\\\n\tK.E.\\hspace{2cm}\n\t\\mathrm{T}&=\\mathrm{K} \\cdot \\mathrm{E} \\cdot\\text{ of }\\mathrm{DISC}+\\mathrm{K} \\cdot \\mathrm{E} \\cdot\\text{ of mass}\\\\\n\t&=\\frac{1}{2}(\\dot{\\theta})^{2}+\\frac{1}{2} m \\dot{x}_{3}^{2}=\\frac{1}{2} m a^{2} \\cdot \\dot{\\theta}^{2}+\\frac{1}{2} m\\left(\\frac{a}{2} \\dot{\\theta}+\\dot{x}_{2}\\right)^{2}\\\\\n\tP.E.\n\t\\hspace{2cm}V&=\\frac{1}{2} k x_{1}{ }^{2}+\\frac{1}{2} k x_{2}{ }^{2}-m g x_{3} .\n\\intertext{\tHence the Lagrangian for the system is}\n\\mathrm{L}&=\\mathrm{T}-\\mathrm{V} \\\\\n\\mathrm{L}&=\\frac{1}{2} m a^{2} \\dot{\\theta}^{2}+\\frac{1}{2} m\\left(\\frac{a^{2}}{4} \\dot{\\theta}^{2}+a \\dot{\\theta} \\dot{x}_{2}+\\dot{x}_{2}^{2}\\right)-\\frac{1}{2} k x_{1}^{2}-\\frac{1}{2} k x_{2}^{2}+m g\\left(\\frac{a}{2} \\theta+x_{2}\\right)\\\\\n\\text{Lagrangian eq. of motion, }&\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}_{2}}\\right)-\\left(\\frac{\\partial L}{\\partial x_{2}}\\right)=0 \\Rightarrow \\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{\\theta}}\\right)-\\left(\\frac{\\partial L}{\\partial \\theta}\\right)\\\\\n\\text { Or } \\quad \\frac{d}{d t}\\left[\\frac{1}{2} m a \\theta+m \\dot{x}_{2}\\right]&-\\left[-k x_{2}+m g\\right]=0 \\Rightarrow m \\ddot{x}_{2}+k x_{2}+\\frac{1}{2} m(a \\ddot{\\theta}-2 g)=0 \\\\ \n\\text { Or } \\quad \\ddot{x}+\\left(\\frac{k}{m}\\right) x_{2}&+\\frac{1}{2}(a \\ddot{\\theta}-2 g)=0 \\quad \\Rightarrow \\frac{d}{d t}\\left[\\frac{1}{2} m a^{2} \\dot{\\theta}+\\frac{1}{2} m a^{2}+\\dot{\\theta} \\max _{2}\\right]-\\left[-k a^{2} \\theta+\\frac{m g a}{2}\\right]\\\\\n\\text{Or }\\quad \\frac{3}{4} m a^{2} \\ddot{\\theta}&+R a^{2} \\theta+\\frac{1}{2} m a\\left(\\ddot{x}_{2}-g\\right)=0\\\\\n\\text{Or }\\quad \\ddot{\\theta}&+\\left(\\frac{4}{3} \\frac{k}{m}\\right) \\theta+\\frac{2}{3 a}\\left(\\ddot{x}_{2}-g\\right)=0\n\\intertext{Which are the required equation of motion.}\n\t\\end{align*}\n\\end{answer}\n\\item Set up the lagrangian for the following system, where the pully is massless.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=6cm,width=4cm]{Lagrangian 04}\n\\end{figure}\n\\begin{answer}\n\t\\begin{align*}\n\t\\mathrm{T}&=\\frac{1}{2} m_{1} \\dot{x}_{1}^{2}+\\frac{1}{2} m_{2} \\dot{x}_{2}^{2}\\\\\n\t\\mathrm{~V}&=-m_{1} g x_{1}-m_{2} g x_{2}+\\frac{1}{2} k x_{1}^{2}\\\\\n\t\\mathrm{~L}&=\\frac{1}{2} m_{1} \\dot{x}_{1}^{2}+\\frac{1}{2} m_{2} \\dot{x}_{2}^{2}+m_{1} g x_{1}+m_{2} g x_{2}-\\frac{1}{2} k x_{1}^{2}\\\\\n\t\\text{Thus, equation of motion is }&=\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}_{1}}\\right)-\\left(\\frac{\\partial L}{\\partial x_{1}}\\right)=0\\\\\n\t\\text{And }\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}_{2}}\\right)-\\left(\\frac{\\partial L}{\\partial x_{2}}\\right)=0\\\\\n\t\\Rightarrow \\quad m_{1} \\ddot{x}_{1}-m_{1} g+k x_{1}=0\\\\\n\t\\text{Say, }m_{2} \\ddot{x}_{2}-m_{2} g=0\n\t\\end{align*}\n\\end{answer}\n\\item Set up the lagrangian for the following system, the disk also have mass ' $m$ '.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=6cm,width=2.7cm]{Lagrangian 11}\n\\end{figure}\n\\begin{answer}$\\left. \\right. $\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=6cm,width=4cm]{Lagrangian 04}\n\t\\end{figure}\n\t\tLet $\\theta$ be the angle one disc has turned away from one equilibrium position. Then extension of the spring $x_{1}=a \\theta$\\\\\n\tMass $m$ is lowered by a distance $x_{2}=\\mathrm{a} \\theta$\n\t\\begin{align*}\n\\therefore \\quad \\mathrm{T} &=\\frac{1}{2} m(a \\dot{\\theta})^{2}+\\frac{1}{2} \\mathrm{I} \\dot{\\theta}^{2} \\\\ \\mathrm{~V} &=-\\mathrm{mga} \\theta+\\frac{1}{2} k \\theta^{2} a^{2} \\\\ \\mathrm{~L} &=\\frac{1}{2} m a^{2} \\dot{\\theta}^{2}+\\frac{1}{4} m a^{2} \\dot{\\theta}^{2}+m g a \\theta-\\frac{1}{2} k a^{2} \\theta^{2} \\\\\n\\text{Eq. of motion is}&\n\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\theta}\\right)-\\frac{\\partial L}{\\partial \\theta}=0\\\\\n\\Rightarrow \\quad &m a^{2} \\ddot{\\theta}+\\frac{1}{2} m a^{2} \\ddot{\\theta}-m g a+k a^{2} \\theta=0\\\\\n\\Rightarrow \\quad &\\frac{3}{2} m a \\ddot{\\theta}-m g+k a \\theta=0\\\\\n\\Rightarrow \\quad &\\ddot{\\theta}-\\frac{2}{3} g / a+\\frac{2}{3}\\left(\\frac{k}{m}\\right) \\theta=0\n\t\\end{align*}\n\\end{answer}\n\\item Two blocks connected by a spring of spring constant $k$ are free to slide frictionlessly along a horizontal surface, as shown in figure. The unstretched length of the spring is $a$.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=1cm,width=6cm]{Lagrangian 06}\n\\end{figure}\nTwo masses connected by a spring sliding horizontally along a frictionless surface.\\\\\n(a) Identify a set of generalized coordinates and write the Lagrangian.\n\\begin{answer}\n\tAs generalized coordinates I choose $X$ and $u$, where $X$ is the position of the right edge of the block of mass $M$, and $X+u+a$ is the position of the left edge of the block of mass $m$, where $a$ is the unstretched length of the spring. Thus, the extension of the spring is $u$. The Lagrangian is then\n\t\\begin{align*}\n\tL&=\\frac{1}{2} M \\dot{X}^{2}+\\frac{1}{2} m(\\dot{X}+\\dot{u})^{2}-\\frac{1}{2} k u^{2}=\\frac{1}{2}(M+m) \\dot{X}^{2}+\\frac{1}{2} m \\dot{u}^{2}+m \\dot{X} \\dot{u}-\\frac{1}{2} k u^{2}\\\\\n\t&\\text{(b) Find the equation of motion.}\\\\\n\t&\\text{The canonical momenta are}\\\\\n\tp_{X} &\\equiv \\frac{\\partial L}{\\partial \\dot{X}}=(M+m) \\dot{X}+m \\dot{u}, p_{u} \\equiv \\frac{\\partial L}{\\partial \\dot{u}}=m(\\dot{X}+\\dot{u})\\\\\n\t&\\text{The corresponding equations of motion are then}\\\\\n\t&\\dot{p}_{X}=F_{X}=\\frac{\\partial L}{\\partial X} \\quad \\Rightarrow \\quad(M+m) \\ddot{X}+m \\ddot{u}=0 \\\\\n\t&\\dot{p}_{u}=F_{u}=\\frac{\\partial L}{\\partial u} \\quad \\Rightarrow \\quad m(\\ddot{X}+\\ddot{u})=-k u .\n\t\\end{align*}\n\t(c) Find all conserved quantities.\\\\\n\tThere are two conserved quantities. One is $p_{X}$ itself, as is evident from the fact that $L$ is cyclic in $X$. This is the conserved 'charge' $\\Lambda$ associated with the continuous symmetry $X \\rightarrow X+\\zeta$. i.e. $\\Lambda=p_{X}$. The other conserved quantity is the Hamiltonian $H$, since $L$ is cyclic in $t$. Furthermore, because the kinetic energy is homogeneous of degree two in the generalized velocities, we have that $H=E$, with\n\t\\begin{align*}\n\tE&=T+U=\\frac{1}{2}(M+m) \\dot{X}^{2}+\\frac{1}{2} m \\dot{u}^{2}+m \\dot{X} \\dot{u}+\\frac{1}{2} k u^{2} .\\\\\n\t\\intertext{It is possible to eliminate $\\dot{X}$, using the conservation of $\\Lambda$ :}\n\t\\dot{X}&=\\frac{\\Lambda-m \\dot{u}}{M+m} .\n\t\\intertext{This allows us to write}\n\tE&=\\frac{\\Lambda^{2}}{2(M+m)}+\\frac{M m \\dot{u}^{2}}{2(M+m)}+\\frac{1}{2} k u^{2}\n\t\\end{align*}\n\t(d) Find a complete solution to the equations of motion. As there are two degrees of freedom, your solution should involve 4 constants integration. You need not match initial conditions and you need not choose the quantities in part (c) to be among the constants.\\\\\n\tUsing conservation of $\\Lambda$, we may write $\\ddot{X}$ in terms of $\\ddot{x}$, in which case\n\t\\begin{align*}\n\t\\frac{M m}{M+m} \\ddot{u}&=-k u \\Rightarrow u(t)=A \\cos (\\Omega t)+B \\sin (\\Omega t)\\\\\n\\text{}\\Omega&=\\sqrt{\\frac{(M+m) k}{M m}} .\n\\intertext{\tFor the $X$ motion, we integrate equation above, obtaining}\nX(t)&=X_{0}+\\frac{\\Lambda t}{M+m}-\\frac{m}{M+m}(A \\cos (\\Omega t)-A+B \\sin (\\Omega t))\n\\intertext{There are thus four constant: $X_{0}, \\Lambda, A$ and $B$. Note that conservation of energy says}\nE&=\\frac{\\Lambda^{2}}{2(M+m)}+\\frac{1}{2} k\\left(A^{2}+B^{2}\\right)\n\t\\end{align*}\n\t\\textbf{Alternate solution :} We could choose $X$ as the position of the left block and $x$ as the position of the right block. In this case,\n\t\\begin{align*}\n\tL&=\\frac{1}{2} M \\dot{X}^{2}+\\frac{1}{2} m \\dot{x}^{2}-\\frac{1}{2} k(x-X-b)^{2} .\n\\intertext{\tHere, $b$ includes the unstretched length $a$ of the spring, but may also include the size of the blocks if say, $X$ and $x$ are measured relative to the blocks midpoint. The canonical momenta are}\np_{X}&=\\frac{\\partial L}{\\partial \\dot{X}}=M \\dot{X}, \\quad p_{x}=\\frac{\\partial L}{\\partial \\dot{x}}=m \\dot{x}\n\\intertext{The equation of motion are then}\n\\dot{p}_{X}&=F_{X}=\\frac{\\partial L}{\\partial X} \\quad \\Rightarrow \\quad M \\ddot{X}=k(x-X-b) \\\\\n\\dot{p}_{x}&=F_{x}=\\frac{\\partial L}{\\partial x} \\quad \\Rightarrow \\quad m \\ddot{x}=-k(x-X-b) .\n\\intertext{The one parameter family which leaves $L$ invariant is $X \\rightarrow X+\\zeta$ and $x \\rightarrow x+\\zeta$, i.e. simultaneous and identical displacement of both of the generalized coordinates. Then}\n\\Lambda&=M \\dot{X}+m \\dot{x}\n\\intertext{which is simply the $x$-components of the total momentum. Again, the energy is conserved.}\nE&=\\frac{1}{2} M \\dot{X}^{2}+\\frac{1}{2} m \\dot{x}^{2}+\\frac{1}{2} k(x-X-b)^{2} .\n\\intertext{We can combine the equations of motion to yield}\nM m \\frac{d^{2}}{d t^{2}}(x-X-b)&=-k(M+m)(x-X-b),\n\\intertext{which yields}\nx(t)-X(t)&=b+A \\cos (\\Omega t)+B \\sin (\\Omega t)\n\\intertext{From the conservation of $\\Lambda$, we have}\nM X(t)+m x(t)&=\\Lambda t+C\n\\intertext{where $C$ is another constant. Thus, we have the motion of the system in terms of four constant: $A, B, \\Lambda$ and $C$ }\nX(t)&=-\\frac{m}{M+m}(b+A \\cos (\\Omega t)+B \\sin (\\Omega t))+\\frac{\\Lambda t+C}{M+m}\\\\\nx(t)&=-\\frac{M}{M+m}(b+A \\cos (\\Omega t)+B \\sin (\\Omega t))+\\frac{\\Lambda t+C}{M+m}\n\t\\end{align*}\n\\end{answer}\n\\item A particle of charge $e$ moves in three dimensions in the presence of a uniform magnetic field $B=B_{0} \\hat{z}$ and a uniform electric field $E=E_{0} \\hat{x}$. The potential energy is\n$$\nU(r, \\dot{r})=-e E_{0} x-\\frac{e}{c} B_{0} x \\dot{y},\n$$\nwhere we have chosen the gauge $A=B_{0} x \\hat{y}$.\\\\\n(a) Find the canonical momenta $p_{x}, p_{y}$ and $p_{z}$\\\\\n(b) Identify all conserved quantities\\\\\n(c) Find a complete, general solution for the motion of the system $\\{x(t), y(t), x(t)\\}$.\n\\begin{answer}\n\t\\begin{align*}\n\t\\text{( The Lagrangian is )}L&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)+\\frac{e}{c} B_{0} x \\dot{y}+e E_{0} x. \n\t\\intertext{The canonical momenta are}\n\tp_{x}=\\frac{\\partial L}{\\partial \\dot{x}}&=m \\dot{x} ; \\quad p_{y}=\\frac{\\partial L}{\\partial \\dot{y}}=m \\dot{y}+\\frac{e}{c} B_{0} x ; \\quad p_{x}=\\frac{\\partial L}{\\partial \\dot{z}}=m \\dot{z}\n\\intertext{\t(b) There are three conserved quantities. First is the momentum $p_{y}$, since $F_{y}=\\frac{\\partial L}{\\partial y}=0 .$ Second is the momentum $p_{z}$, since $F_{y}=\\frac{\\partial L}{\\partial z}=0$. The third conserved quantity is the Hamiltonian, since $\\frac{\\partial L}{\\partial t}=0$. We have }\nH&=p_{x} \\dot{x}+p_{y} \\dot{y}+p_{z} \\dot{z}-L\\\\\n\\Rightarrow \\quad H&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)-e E_{0} x\\\\\n\\text{(c) The equations of motion are }\\ddot{x}-\\omega_{c} \\dot{y}&=\\frac{e}{m} E_{0} ; \\ddot{y}+\\omega_{c} \\dot{x}=0 ; \\ddot{z}=0.\n\\intertext{The second equation can be integrated once to yield $\\dot{y}=\\omega_{c}\\left(x_{0}-x\\right)$, where $x_{0}$ is a constant. Substituting this into the first equation gives}\n\\ddot{x}+\\omega_{c}^{2} x&=\\omega_{c}^{2} x_{0}+\\frac{e}{m} E_{0} .\n\\intertext{This is the equation of a constantly forced harmonic oscillator. We can therefore write the generalsolution as}\nx(t)&=x_{0}+\\frac{e E_{0}}{m \\omega_{c}^{2}}+A \\cos \\left(\\omega_{c} t+\\delta\\right) \\\\\ny(t)&=y_{0}-\\frac{e E_{0}}{m \\omega_{c}} t-A \\sin \\left(\\omega_{c} t+\\delta\\right) \\\\\nz(t)&=z_{0}+\\dot{z}_{0} t\n\\intertext{Note that there are six constants, $\\left\\{A, \\delta, x_{0}, y_{0}, z_{0}, \\dot{z}_{0}\\right\\}$, are required for the general solution of three coupled second order ODEs.}\n\t\\end{align*}\n\\end{answer}\n\\item A point mass $m$ slides frictionlessly, under the influence of gravity, along a massive ring of radius $a$ and mass $M$. The ring is affixed by horizontal springs to two fixed vertical surfaces, as depicted in figure. Allmotion is within the plane of the figure.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=2cm,width=6cm]{Lagrangian 07}\n\\end{figure}\nA point mass $m$ slides frictionlessly along a massive ring of radius $a$ and mass $M$, which is affixed by horizontal springs to two fixed vertical surfaces.\n\\begin{answer}\n\t(a) Choose as generalized coordinates the horizontal displacement $X$ of the center of the ring with respect to equilibrium, and the angle $\\theta$ a radius to the mass $m$ makes with respect to the vertical (see figure). You may assume that at $X=0$ the spring are both unstretched. Find the Lagrangian $L(X, \\theta, \\dot{X}, \\dot{\\theta}, t)$.\\\\\n\tThe coordinates of the mass point are $x=X+a \\sin \\theta, y=-a \\cos \\theta$.\\\\\n\tThe kinetic energy is\n\t\\begin{align*}\nT &=\\frac{1}{2} M \\dot{X}^{2}+\\frac{1}{2} m(\\dot{X}+a \\cos \\theta \\dot{\\theta})^{2}+\\frac{1}{2} m a^{2} \\sin ^{2} \\theta \\dot{\\theta}^{2} \\\\ &=\\frac{1}{2}(M+m) \\dot{X}^{2}+\\frac{1}{2} m a^{2} \\dot{\\theta}^{2}+m a \\cos \\theta \\dot{X} \\dot{\\theta} .\\\\\n\\text{The potential energy is }U&=k X^{2}-m g a \\cos \\theta.\\\\\n\\text{Thus, the Lagrangian is}&=\\frac{1}{2}(M+m) \\dot{X}^{2}+\\frac{1}{2} m a^{2} \\dot{\\theta}^{2}+m a \\cos \\theta \\dot{X}-k X^{2}+m g a \\cos \\theta.\n\\intertext{(b) Find the generalized momenta $p_{X}$ and $p_{\\theta}$, and the generalized forces $F_{X}$ and $F_{\\theta}$}\n\\text{We have }p_{X}&=\\frac{\\partial L}{\\partial \\dot{X}}=(M+m) \\dot{X}+m a \\cos \\theta \\dot{\\theta}, p_{\\theta}=\\frac{\\partial L}{\\partial \\dot{\\theta}}=m a^{2} \\dot{\\theta}+m a \\cos \\theta \\dot{X}.\\\\\n\\text{For the forces, }F_{X}&=\\frac{\\partial L}{\\partial X}=-2 k X, \\quad F_{\\theta}=\\frac{\\partial L}{\\partial \\theta}=-m a \\sin \\theta \\dot{X} \\dot{\\theta}-m g a \\sin \\theta.\n\\intertext{(c) Derive the equations o motion.}\n\\text{The equations of motion arc }&\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{q}_{\\sigma}}\\right)=\\frac{\\partial L}{\\partial q_{\\sigma}}, \\text{for each generalized coordinate }q_{\\sigma} .\\text{ For $X$ we have }\\\\\n(M+m) \\ddot{X}+m a \\cos \\theta \\ddot{\\theta}-m a \\sin \\theta \\dot{\\theta}^{2}&=-2 k X .\\\\\n\\text{For }\\theta, m a^{2} \\ddot{\\theta}+m a \\cos \\theta \\ddot{X}&=-m g a \\sin \\theta.\n\\intertext{(d) Find expression for all conserved $_{1}$ uantities.}\n\\intertext{Horizontal and vertical translationa $^{1}$ symmetries are broken by the springs and by gravity, respectively. The remaining symmetry is that of time translation. From $\\frac{d H}{d t}=-\\frac{\\partial L}{\\partial t}$, we have that $H=\\sum_{\\sigma} p_{\\sigma} \\dot{q}_{\\sigma}-L$ is conserved. For this problem, the kinetic energy is a homogeneous function of degree 2 in the generalized velocities, and the potential is velocity-independent. Thus,}\nH=T+U&=\\frac{1}{2}(M+m) \\dot{X}^{2}+\\frac{1}{2} m a^{2} \\dot{\\theta}^{2}+m a \\cos \\theta \\dot{X} \\dot{\\theta}+k X^{2}-m g a \\cos \\theta\n\t\\end{align*}\n\\end{answer}\n\\item A point particle of mass $m$ moves in three dimensions in a helical potential\n$$\nU(\\rho, \\phi, z)=U_{0} \\rho \\cos \\left(\\phi-\\frac{2 \\pi z}{b}\\right)\n$$\nWe call $b$ the pitch of the helix.\n\\begin{answer}\n\t\\begin{align*}\n\t\\intertext{(a) Write down the Lagrangian, choosing $(\\rho, \\phi, z)$ as generalized coordinates.}\n\\text{\tThe Lagrangian is }L&=\\frac{1}{2} m\\left(\\dot{\\rho}^{2}+\\rho^{2} \\dot{\\phi}^{2}+\\dot{z}^{2}\\right)-U_{0} \\rho \\cos \\left(\\phi-\\frac{2 \\pi z}{b}\\right)\n\\intertext{\t(b) Find the equation of motion.}\n\t\\text{Clearly }p_{\\rho}&=m \\dot{\\rho} ; p_{\\phi}=m \\rho^{2} \\dot{\\phi} ; p_{z}=m \\dot{z},\\text{ and}\\\\\n\tF_{\\rho}=m \\rho \\dot{\\phi}^{2}-U_{0} \\cos \\left(\\phi-\\frac{2 \\pi z}{b}\\right), F_{\\phi}&=U_{0} \\rho \\sin \\left(\\phi-\\frac{2 \\pi z}{b}\\right), F_{z}=-\\frac{2 \\pi U_{0}}{b} \\rho \\sin \\left(\\phi-\\frac{2 \\pi z}{b}\\right) \\text {. }\n\\intertext{\tThus, the equation of motion are}\n\tm \\ddot{\\rho}&=m \\rho \\dot{\\phi}^{2}-U_{0} \\cos \\left(\\phi-\\frac{2 \\pi z}{b}\\right) \\\\\n\tm \\rho^{2} \\ddot{\\phi}+2 m \\rho \\dot{\\rho} \\dot{\\phi}&=U_{0} \\rho \\sin \\left(\\phi-\\frac{2 \\pi z}{b}\\right)\\\\\n\tm \\ddot{z}&=-\\frac{2 \\pi U_{0}}{b} \\rho \\sin \\left(\\phi-\\frac{2 \\pi z}{b}\\right)\n\t\\intertext{(c) Show that there exists a continuous one-parameter family of coordinate transformations which leaves $L$ invariant. Find the associated conserved quantity, $\\Lambda$. Is anything else conserved?}\n\t\\intertext{Due to the helical symmetry, we have that $\\phi \\rightarrow \\phi+\\zeta, z \\rightarrow z+\\frac{b}{2 \\pi} \\zeta$ is such a continuous one-parameter}\n\t\\intertext{family of coordinate transformations. Since it leaves the combination $\\phi-\\frac{2 \\pi z}{\\dot{b}}$ unchanged, we have that $\\frac{d L}{d \\zeta}=0$, and}\n\\Lambda &=\\left.p_{\\rho} \\frac{\\partial \\rho}{\\partial \\zeta}\\right|_{\\zeta=0}+\\left.p_{\\phi} \\frac{\\partial \\phi}{\\partial \\zeta}\\right|_{\\zeta=0}+\\left.p_{z} \\frac{\\partial z}{\\partial \\zeta}\\right|_{\\zeta=0} \\\\ &=p_{\\phi}+\\frac{b}{2 \\pi} p_{z} \\\\ &=m \\rho^{2} \\dot{\\phi}+\\frac{m b}{2 \\pi} \\dot{z}\n\\intertext{is the conserved Noether 'charge'. The other conserved quantity is the Hamiltonian,}\nH&=\\frac{1}{2} m\\left(\\dot{\\rho}^{2}+\\rho^{2} \\dot{\\phi}^{2}+\\dot{z}^{2}\\right)+U_{0} \\rho \\cos \\left(\\phi-\\frac{2 \\pi z}{b}\\right)\n\\intertext{Note that $H=T+U$ because $T$ is homogeneous of degree 2 and $U$ is homogeneous of degree 0 in the generalized velocities.} \n\t\\end{align*}\n\\end{answer}\n\\textbf{Statement for Linked Answer Q.12 and Q.13 :}\\\\\nA particle of mass $m$ is constrained to move in a vertical plane along a trajectory given by $x=\\mathrm{A} \\cos \\theta$, $y=\\mathrm{A} \\sin \\theta$, where $\\mathrm{A}$ is a constant.\n\\item The Lagrangian of the particle is\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\frac{1}{2} m \\mathrm{~A}^{2} \\dot{\\theta}^{2}-m g \\mathrm{~A} \\cos \\theta$\n\t\\task[\\textbf{b.}]$\\frac{1}{2} m \\mathrm{~A}^{2} \\dot{\\theta}^{2}-m g \\mathrm{~A} \\sin \\theta$\n\t\\task[\\textbf{c.}] $\\frac{1}{2} m \\mathrm{~A}^{2} \\dot{\\theta}^{2}$\n\t\\task[\\textbf{d.}] $\\frac{1}{2} m \\mathrm{~A}^{2} \\dot{\\theta}^{2}+m g \\mathrm{~A} \\cos \\theta$\n\\end{tasks}\n\\begin{answer}$\\left. \\right. $\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=4.2cm,width=4.5cm]{Lagrangian 08}\n\t\\end{figure}\n\t\\begin{align*}\n\tL&=T-V\\\\\n\tL&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)-m g y^{x}\\\\\n\tx&=A \\cos \\theta, \\quad y=A \\sin \\theta\\\\\n\t\\dot{x}&=-A \\sin \\theta \\dot{\\theta}, \\quad \\dot{y}=A \\cos \\theta \\dot{\\theta}\\\\\n\t \\therefore \\quad L &=\\frac{1}{2} m\\left(A^{2} \\sin ^{2} \\theta \\dot{\\theta}^{2}+A^{2} \\cos ^{2} \\theta \\dot{\\theta}^{2}\\right)-m g A \\cos \\theta \\\\ L &=\\frac{1}{2} m A^{2} \\dot{\\theta}^{2}-m g A \\cos \\theta \n\t\\end{align*}\n\t Correct answer is option \\textbf{(a)}\n\\end{answer}\n\\item The equation of motion of the particle is\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\ddot{\\theta}-\\frac{g}{\\mathrm{~A}} \\cos \\theta=0$\n\t\\task[\\textbf{b.}]$\\ddot{\\theta}+\\frac{g}{\\mathrm{~A}} \\sin \\theta=0$\n\t\\task[\\textbf{c.}] $\\ddot{\\theta}=0$\n\t\\task[\\textbf{d.}] $\\ddot{\\theta}-\\frac{g}{\\mathrm{~A}} \\sin \\theta=0$\n\\end{tasks}\n\\begin{answer}\n\tEquation of motion\n\t\\begin{align*}\n\t\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{\\theta}}\\right)-\\frac{\\partial L}{\\partial \\theta}&=0\\\\\n\t\\frac{d}{d t}\\left(m A^{2} \\dot{\\theta}\\right)-m g A \\sin \\theta&=0\\\\\n\tm A^{2} \\ddot{\\theta}-m g A \\sin \\theta&=0\\\\\n\t\\ddot{\\theta}-\\frac{g}{A} \\sin \\theta&=0\n\t\\end{align*}\n\tCorrect answer is option \\textbf{(d)}\n\\end{answer}\n\\item A particle of mass $M$ is attached to two identical springs of unstretched length $L_{0}$ and spring constant $k$. The entire system is placed on a horizontal frictionless table as shown in the figure. The mass is slightly puled along the surface of the table and perpendicular to the lengths of the springs and then let go. Using the Lagrangian equation (s) of motion, show whether the mass will execute simple harmonic motion. If so, find the time period.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[height=2.5cm,width=7cm]{Lagrangian 09}\n\\end{figure}\n\\begin{answer}$\\left. \\right. $\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=2.5cm,width=6cm]{Lagrangian 10}\n\t\\end{figure}\n\t\\begin{align*}\n\\text{\tKinetic energy }T&=\\frac{1}{2} m \\dot{x}^{2}\\\\\n\\text{\tpotential energy }V&=\\frac{1}{2} k\\left(\\sqrt{\\ell_{0}^{2}+x^{2}}-\\ell_{0}\\right)^{2}\n\\intertext{Number gravitational potential energy term because motion is taking place in horizontal plane.}\n\\text{Lagrangian }L&=T-V=\\frac{1}{2} m \\dot{x}^{2}-\\frac{1}{2} k\\left(\\sqrt{\\ell_{0}^{2}+x^{2}}-\\ell_{0}\\right)^{2}\n\\intertext{Lagrange's equation}\n&\\frac{d}{d t}\\left(\\frac{\\partial L}{\\partial \\dot{x}}\\right)-\\frac{\\partial L}{\\partial x}=0 \\\\\n&m \\ddot{x}+2 k\\left(\\sqrt{\\ell_{0}^{2}+x^{2}}-\\ell_{0}\\right) \\frac{x}{\\sqrt{\\ell_{0}^{2}+x^{2}}}=0 \\\\\n&m \\ddot{x}+2 k x\\left(1-\\frac{\\ell_{0}}{\\sqrt{\\ell_{0}^{2}+x^{2}}}\\right)=0\\\\\n\\text{for }&x<<\\ell_{0} \\quad m \\ddot{x}+2 k x\\left(1-\\left(1+\\frac{x^{2}}{\\ell_{0}^{2}}\\right)^{-1 / 2}\\right)=0\\\\\nm \\ddot{x}+2 k x\\left(1-1+\\frac{x^{2}}{2 \\ell_{0}^{2}}\\right)&=0 \\Rightarrow m \\ddot{x}+\\frac{k x^{3}}{\\ell_{0}^{2}}=0 \\ddot{x} \\alpha-x^{3}\n\\intertext{Motion is not simple harmonic.}\n\t\\end{align*}\n\\end{answer}\n\\end{enumerate}\n\\section{Hamiltonian Mechanics}\n\\begin{enumerate}\n\t\\item Show that $\\frac{d H}{d t}=\\frac{\\partial H}{\\partial t}$\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{Since }H&=H\\left(q_{j}, p_{j}, t\\right)\\\\\n\t\t\\therefore \\quad \\frac{d H}{d t} & =\\frac{\\partial H}{\\partial t}+\\sum_{j} \\frac{\\partial H}{\\partial q_{j}}+\\sum_{j} \\frac{\\partial H}{\\partial p_{j}} \\dot{p}_{j} \\Rightarrow \\frac{d H}{d t}=\\frac{\\partial H}{\\partial t}+\\sum_{j}\\left(-\\dot{p}_{j} \\dot{q}_{j}\\right)+\\sum_{k}\\left(\\dot{q}_{j} \\dot{p}_{j}\\right) \\\\\n\t\t\\therefore \\quad \\frac{d H}{d t} & =\\frac{\\partial H}{\\partial t}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Obtain the Hamiltonian function of a compound pendulum and hence obtain the equation describing its motion.\n\t\\begin{answer}\n\t\tThe pendulum is a harmonic conservative system. It is holonomic because the constraint is on the position of the pendulum in the sense that it must vibrate in its own plane and conservative because the gravitational force acting on it is conservative. With reference to a frame fixed at the point of suspension with downward vertical as the reference line, we have\n\t\t\\begin{align*}\n\t\tT&=\\frac{1}{2} I \\dot{\\theta}^{2}\n\t\\intertext{\twhere $I$ is the moment of inertia of the pendulum about the axis passing through the point suspension and perpendicular to its plane.}\n\t\tV&=-m g \\ell \\cos \\theta\n\t\\intertext{\twhere $\\ell$ is the distance of the centre of gravity from the centre of suspension.}\n\\therefore & H=T+V=\\frac{1}{2} I \\dot{\\theta}^{2}-m g \\ell \\cos \\theta \\\\ \\text { Now, } & p_{\\theta}=\\frac{\\partial T}{\\partial \\dot{\\theta}}=I \\dot{\\theta} \\\\ \\therefore & H=\\frac{1}{2} I \\frac{p_{\\theta}^{2}}{I^{2}}-m g \\ell \\cos \\theta \\\\ \\text { Or, } & H=\\frac{p_{\\theta}^{2}}{I^{2}}-m g \\ell \\cos \\theta \\\\ & \\frac{\\partial H}{\\partial p_{\\theta}}=\\frac{p_{\\theta}}{I} \\text { and } \\frac{\\partial H}{\\partial p_{\\theta}}=-m g \\ell(-\\sin \\theta)=m g \\ell \\sin \\theta\\\\\n\\text{By Hamilton's general equations }&\\left(\\dot{p}_{k}=-\\frac{\\partial H}{\\partial q_{k}}\\right.\\text{ and }\n\\left.\\dot{q}_{k}=\\frac{\\partial H}{\\partial p_{k}}\\right)\\\\\n\\text{We have here }\\dot{p}_{\\theta}&=-\\frac{\\partial H}{\\partial \\theta}=-m g \\ell \\sin \\theta\\text{ and }\\dot{\\theta}=\\frac{\\partial H}{\\partial p_{\\theta}}=\\frac{p_{\\theta}}{I}\n\\intertext{These are the two equations describing the motion of a compound pendulum. However, these two equaions can be combined into a signle equation. Differentiating the second equation and eliminating $\\dot{p}_{\\theta}$,}\n\\intertext{we have}\n\\ddot{\\theta}&=\\frac{\\dot{p}_{\\theta}}{I} \\sin \\theta ; \\text { or } \\ddot{\\theta}+\\frac{m g I}{I} \\sin \\theta=0\n\\intertext{This is the required single equation that describes the motion of a compound pendulum.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A particle moves in the $x y$ plane under the influence of a central force depending on its distance from the origin.\\\\\n\t(a) Set up the Hamiltonian for the system\\\\\n\t(b) Obtain Hamilton's equations of motion\n\t\\begin{answer}\n\t\\begin{align}\n\t\\text{(a) }T \\text{(kinetic energy of of the particle) }&=\\frac{1}{2 m \\dot{r}^{2}}+\\frac{1}{2} I \\dot{\\theta}^{2}=\\frac{1}{2} m \\dot{r}^{2}+\\frac{1}{2} m r^{2} \\dot{\\theta}^{2}\\quad\n\t\\left(\\because I=m r^{2}\\right)\\notag\\\\\n\\therefore & H=T+V=\\frac{1}{2} m \\dot{r}^{2}+\\frac{1}{2} m r^{2} \\dot{\\theta}^{2}+V(r)\\notag \\\\ \\text { Now, } & p_{r}=\\frac{\\partial T}{\\partial \\dot{r}}=m \\dot{r} \\text { and } p_{\\theta}=\\frac{\\partial T}{\\partial \\dot{\\theta}}=I \\dot{\\theta}=m r^{2} \\dot{\\theta} \\notag\\\\ \\therefore & H=\\frac{1}{2} m \\cdot \\frac{p_{r}^{2}}{m^{2}}+\\frac{1}{2} m r^{2} \\frac{p_{\\theta}^{2}}{m^{2} r^{4}}+V(r) \\notag\\\\ \\text { Or, } & H=\\frac{p_{r}^{2}}{2 m}+\\frac{p_{\\theta}^{2}}{2 m r^{2}}+V(r)\\notag\n\\intertext{(b) We have from Hamilton's equation,}\n&\\left(\\dot{p}_{k}=-\\frac{\\partial H}{\\partial q_{k}} \\text { and } \\dot{q}_{k}=\\frac{\\partial H}{\\partial p_{k}}\\right) \\notag\\\\\n&\\dot{p}_{r}=-\\frac{\\partial H}{\\partial r}=-\\left(-\\frac{p_{\\theta}^{2}}{m r^{3}}+\\frac{\\partial V}{\\partial r}\\right)=\\frac{p_{\\theta}^{2}}{m r^{3}}-\\frac{\\partial V}{\\partial r} \\label{HM-29}\\\\\n&\\dot{r}=\\frac{\\partial H}{\\partial p_{r}}=\\frac{p_{r}}{m} \\label{HM-30}\\\\\n&\\dot{p}_{\\theta}=-\\frac{\\partial H}{\\partial \\theta}=0 \\label{HM-31}\\\\\n&\\dot{\\theta}=\\frac{\\partial H}{\\partial p_{\\theta}}=\\frac{p_{\\theta}}{m r^{2}}\\label{HM-32}\n\\intertext{Equations (\\ref{HM-29}), (\\ref{HM-30}), (\\ref{HM-31}) and (\\ref{HM-32}) are the required Hamilton's equations. However they can be combined to reduce the number of equations describing the motion. Equations (\\ref{HM-29}) and (\\ref{HM-30}) combine into (after elimination of $\\dot{p}_{r}$ )}\\notag\n\\ddot{r}&=\\frac{p_{\\theta}^{2}}{m^{2} r^{3}}-\\frac{1}{m} \\frac{\\partial V}{\\partial r}=\\frac{m^{2} r^{4} \\dot{\\theta}^{2}}{m^{2} r^{2}}-\\frac{1}{m} \\frac{\\partial V}{\\partial r}\\notag\n\\intertext{and (\\ref{HM-31}) and (\\ref{HM-32}) combine into (after elimination of $\\dot{p}_{\\theta}$ ) $\\ddot{\\theta}=0$ Thus the equations describing motion of the particle are}\n\\ddot{r}-r \\dot{\\theta}^{2}&=-\\frac{1}{m} \\frac{\\partial V}{\\partial r} \\text { and } \\ddot{\\theta}=0\\notag\\\\\n\\text{Or, }\\quad \\ddot{r}-r \\dot{\\theta}^{2}&=-\\frac{1}{m} \\frac{\\partial V}{\\partial r}\\text{ and }\\dot{\\theta}=\\frac{p_{\\theta}}{m r^{2}}=\\frac{a \\text { constant }}{m r^{2}}\\notag\\\\\n\\quad\\left(\\because p_{\\theta}=a\\right.&\\text{ constant from }\\left.(i i i)\\right)\\text{\n\tOr, }\\quad m r^{2} \\dot{\\theta}=a\\text{ constant}\\notag\n\t\\end{align}\n\t\\end{answer}\n\t\\item The Hamiltonian of a simple pendulum consisting of a mass ' $\\mathrm{m}$ ' attached to a massless string of length $l$ is $\\mathrm{H}=\\frac{\\mathrm{p}_{\\theta}^{2}}{2 \\mathrm{~m} \\ell^{2}}+\\mathrm{mg} \\ell(1-\\cos \\theta)$. If L denotes the Lagrangian, the value of $\\frac{\\mathrm{dL}}{\\mathrm{dt}}$ is :\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$-\\frac{2 g}{\\ell} p_{\\theta} \\sin \\theta$\n\t\t\\task[\\textbf{b.}]$-\\frac{\\mathrm{g}}{\\ell} \\mathrm{p}_{\\theta} \\sin 2 \\theta$\n\t\t\\task[\\textbf{c.}] $\\frac{\\mathrm{g}}{\\ell} \\mathrm{p}_{\\theta} \\cos \\theta$\n\t\t\\task[\\textbf{d.}] $\\ell \\mathrm{p}_{\\theta}^{2} \\cos \\theta$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\mathrm{H}&=\\frac{\\mathrm{p}_{\\theta}^{2}}{2 \\mathrm{~m} \\ell^{2}}+\\mathrm{mg} \\ell(1-\\cos \\theta) \\Rightarrow \\mathrm{L}=\\sum_{\\mathrm{i}} \\mathrm{p}_{\\mathrm{i}} \\dot{\\mathrm{q}}_{\\mathrm{i}}-\\mathrm{H}=\\mathrm{p}_{\\theta} \\dot{\\theta}-\\mathrm{H}\\\\\n\t\t\\dot{\\theta}&=\\frac{\\partial \\mathrm{H}}{\\partial \\mathrm{p}_{\\theta}}=\\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}}\\text{ and }\\dot{\\mathrm{p}}_{\\theta}=-\\frac{\\partial \\mathrm{H}}{\\partial \\theta}=-\\mathrm{mg} \\ell \\sin \\theta\\\\\n\t\t\\text{Therefore, }\\mathrm{L}&=\\mathrm{p}_{\\theta} \\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}}-\\left[\\frac{\\mathrm{p}_{\\theta}^{2}}{2 \\mathrm{~m} \\ell^{2}}+\\mathrm{mg} \\ell(1-\\cos \\theta)\\right] \\Rightarrow \\mathrm{L}=\\frac{\\mathrm{p}_{\\theta}^{2}}{2 \\mathrm{~m} \\ell^{2}}-\\mathrm{mg} \\ell(1-\\cos \\theta)\\\\\n\t\t\\text{Since, }L&=L\\left(\\theta, p_{\\theta}\\right)\\text{Since, }\n\t\t -\\frac{L}{1}=\\frac{\\partial L}{\\partial p_{\\theta}} \\dot{p}_{\\theta}+\\frac{\\partial L}{\\partial \\theta} \\dot{\\theta}\\\\\n\t\t \\text{Now, }\\frac{\\partial \\mathrm{L}}{\\partial \\mathrm{p}_{\\theta}}&=\\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}}, \\quad \\dot{\\mathrm{p}}_{\\theta}=-1.1 \\mathrm{~g} \\ell \\sin \\theta, \\quad \\frac{\\partial \\mathrm{L}}{\\partial \\theta}=-\\mathrm{mg} \\ell \\sin \\theta, \\quad \\dot{\\theta}=\\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}} \\\\\n\t\t \\text{Then} \\frac{\\mathrm{dL}}{\\mathrm{dt}}&=\\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}}(-\\mathrm{mg} \\ell \\sin \\theta)-\\mathrm{mg} \\ell \\quad \\mathrm{n} \\theta \\frac{\\mathrm{p}_{\\theta}}{\\mathrm{m} \\ell^{2}}=-\\frac{2 \\mathrm{~g}}{\\ell} \\mathrm{p}_{\\theta} \\sin \\theta\n\t\t\\end{align*}\n\t\tCorrect option is \\textbf{(a)}\n\t\\end{answer}\n\t\\item The particel of mass $m$ is constrained to move on the surface of a cylinder of radius $a$ under an attractive central force $F$, given by\n\t$$\n\tF=-k r\n\t$$\n\twhere $k$ is the force constant. This force is proportional to the distance of the particle from the origin\n\\begin{answer}\n\t\tThe motion of the particle can be described in terms of the cartesian coordinates $\\mathrm{x}, \\mathrm{y}, \\mathrm{z}$ or cylindrical $\\rho, \\theta, z$ (See figure). The equation of constraint is\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=7cm,width=4.5cm]{Lagrangian 12}\n\t\t\\end{figure}\n\t\\begin{align}\n\t\\rho^{2}&=x^{2}+y^{2}=a^{2}\\label{HM-33-}\n\\intertext{\tTo find velocity in cylinderical coordinates, we can proceed as follows}\\notag\n\\text{The change in position along }\\rho&=d \\rho\\notag\\\\\n\\text{The change in position along }\\theta&=\\rho d \\theta\\notag\\\\\n\\text{The change in position along }z&=d z\n\\text{So, }v_{\\rho}=\\dot{\\rho}, v_{\\theta}=\\rho \\dot{\\theta}\\text{and} v_{z}=\\dot{z}\\notag\\\\\n\\text{So, we have }v^{2}&=v_{\\rho}^{2}+v_{\\theta}^{2}+v_{z}^{2}.\\notag\\\\\n\\text{The kinetic energy, }T&=\\frac{1}{2} m v^{2}=\\frac{1}{2}\\left(\\dot{\\rho}^{2}+\\rho^{2} \\dot{\\theta}^{2}+\\dot{z}^{2}\\right)\\notag\\\\\n\\text{Here, }\\rho&=a\\notag\\\\\n\\text{Therefore, }\\dot{\\rho}&=0,\\text{ and hence,}\\notag\\\\\nT &=\\frac{1}{2} m\\left(a^{2} \\dot{\\theta}^{2}+\\dot{z}^{2}\\right) \\notag\\\\ \\text{and}\\quad V &=\\frac{1}{2} k r^{2}=\\frac{1}{2} k\\left(x^{2}+y^{2}+z^{2}\\right)=\\frac{1}{2} k\\left(a^{2}+z^{2}\\right)\\notag\\\\\n\\therefore \\quad L=T-V&=\\frac{1}{2} m\\left(a^{2} \\dot{\\theta}^{2}+\\dot{z}^{2}\\right)-\\frac{1}{2} k\\left(a^{2}+z^{2}\\right)\\label{HM-34-}\\\\\n\\text{Hence,}\np_{\\theta}&=\\frac{\\partial L}{\\partial \\dot{\\theta}}=m a^{2} \\dot{\\theta}\\text{ or} \\dot{\\theta}=\\frac{p_{\\theta}}{m a^{2}}\\notag\\\\\n\\text{and }\\quad p_{z}&=\\frac{\\partial L}{\\partial \\dot{z}}=m \\dot{z}\\text{ or} \\dot{z}=\\frac{p_{z}}{m}\\notag\\\\\n\\text{Therefore, }\\quad H&=T+V=\\frac{1}{2} m\\left(a^{2} \\dot{\\theta}+\\dot{z}^{2}\\right)+\\frac{1}{2} k\\left(a^{2}+z^{2}\\right)\\notag\\\\\n\\text { Or, } \\quad H&=\\frac{p_{\\theta}^{2}}{2 m a^{2}}+\\frac{p_{z}^{2}}{m}+\\frac{1}{2} k\\left(a^{2}+z^{2}\\right)\\label{HM-35-}\n\\intertext{Hence, the Hamilton's equation are}\\notag\\\\\n\\dot{z}&=\\frac{\\partial H}{\\partial p_{z}}=\\frac{p_{z}}{m}\\text{ or }p_{z}=m \\dot{z}\\label{HM-36-}\\\\\n\\dot{\\theta}&=\\frac{\\partial H}{\\partial p_{\\theta}}=\\frac{p_{\\theta}}{m a^{2}}\\text{ or }p_{\\theta}=m a^{2} \\dot{\\theta}\\label{HM-37-}\\\\\n-\\dot{p}_{z}&=\\frac{\\partial H}{\\partial z}=k z \\quad\\text{ or }\\quad \\dot{\\mathrm{p}}_{z}=-k z\\label{HM-38-}\\\\\n-\\dot{p}_{\\theta}&=\\frac{\\partial H}{\\partial \\theta}=0 \\quad\\text{ or }\np_{\\theta}= \\text{constant}\\label{HM-39-}\n\\intertext{From equation (\\ref{HM-36-}) and (\\ref{HM-38-}), we get}\nm \\ddot{z}+k z&=0\\notag\n\\intertext{which shows that the motion of the particle in z direction is simple harmonic with period $T$, given by}\\notag\nT&=2 \\pi \\sqrt{\\frac{m}{k}}\\notag\n\\intertext{From equation (\\ref{HM-37-}) and (\\ref{HM-39-}), we get}\\notag\np_{\\theta}&=m a^{2} \\dot{\\theta}=\\text { constant }\\notag\n\\intertext{Thus the angular momentum about Z-axis is a constant of motion.}\\notag\n\t\\end{align}\n\\end{answer}\n\t\\item Find equations of motion of particle moving near the surface of earth.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{Let us consider z-axis along upward vertical direction, the kinetic energy is}\n\t\tT&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)\n\t\\intertext{\tFurther the applied force on the body is its weight acting in negative z-direction, i.e.,}\n\t\tF&=F_{z}-m g=-\\frac{\\partial V}{\\partial Z}\n\t\t\\intertext{This gives $V=m g Z$, on setting additive constant to zero Now} \n\t\\intertext{\tLagrangian is}\n\t\tL&=T-V=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)-m g Z\\\\\n\t\t\\text{So that }\\frac{\\partial L}{\\partial \\dot{x}}&=\\frac{\\partial T}{\\partial \\dot{x}}=p_{x}=m \\dot{x}\\text{ giving }\\dot{x}=\\frac{p_{x}}{m}\\\\\n\t\\text{\tAlso, }p_{y}&=m \\dot{y}, p_{z}=m \\dot{z}\\\\\n\t\t\\therefore \\quad \\dot{y}&=\\frac{p_{y}}{m}, \\dot{z}=\\frac{p_{z}}{m}\\\\\n\t\\intertext{\tHamiltonian for such a system is conserved, i.e.,}\n\t\tH&=T+V=\\frac{1}{2}\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)+m g Z=\\frac{1}{2 m}\\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\\right)+m g Z \\\\\n\t\t\\intertext { giving equations of motion }\n\t\t\\dot{p}_{x}&=-\\frac{\\partial H}{\\partial x}=0, \\quad \\dot{p}_{y}=-\\frac{\\partial H}{\\partial y}=0\\\\\n\t\t\\dot{p}_{z}&=-\\frac{\\partial H}{\\partial Z}=-m g\\\\\n\t\t\\therefore \\quad \\dot{x}&=\\frac{\\partial H}{\\partial p_{x}}=\\frac{p_{x}}{m}, \\quad \\dot{y}=\\frac{\\partial H}{\\partial p_{y}}=\\frac{p_{y}}{m}, \\dot{z}=\\frac{\\partial H}{\\partial Z}=\\frac{p_{z}}{m}\n\t\\intertext{ so we finally get,}\n\t\t\\ddot{x}&=\\frac{\\dot{p}_{y}}{m}=0, \\quad \\ddot{y}=\\frac{\\dot{p}_{y}}{m}=0, \\quad \\ddot{z}=\\frac{\\dot{p}_{z}}{m}=-g\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A mechanical system is described by the Hamiltonian $H(q, p)=\\frac{p^{2}}{2 m}+\\frac{1}{2} m \\omega^{2} q^{2}$. As a result of the canonical transformation generated by $F(q, Q)=-\\frac{Q}{q}$, the Hamiltonian in the new coordinate $Q$ and momentum $P$ becomes\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\frac{1}{2 m} Q^{2} P^{2}+\\frac{m \\omega^{2}}{2} Q^{2}$\n\t\t\\task[\\textbf{b.}]$\\frac{1}{2 m} Q^{2} P^{2}+\\frac{m \\omega^{2}}{2} P^{2}$\n\t\t\\task[\\textbf{c.}]$\\frac{1}{2 m} P^{2}+\\frac{m \\omega^{2}}{2} Q^{2}$\n\t\t\\task[\\textbf{d.}] $\\frac{1}{2 m} Q^{2} P^{4}+\\frac{m \\omega^{2}}{2} P^{-2}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tF(q, Q)&=-\\frac{Q}{q}\n\t\\intertext{\tDifferential relation for $F(q, Q)$}\n\t\tp&=\\frac{\\partial F}{\\partial q}, \\quad P=-\\frac{\\partial F}{\\partial Q}\\\\\n\t\t\\text{So, }\\quad p&=\\frac{Q}{q^{2}}, P=\\frac{1}{q} \\quad \\Rightarrow p=Q P^{2}\n\t\\intertext{\tNew Hamiltonian,}\n\t\tH^{\\prime}&=H+\\frac{\\partial F}{\\partial t}=H+0\\\\\n\t\tH^{\\prime}(Q, P)&=H(q, p)=\\frac{p^{2}}{2 m}+\\frac{1}{2} m \\omega^{2} q^{2}=\\frac{1}{2 m} Q^{2} P^{4}+\\frac{1}{2} m \\omega^{2} P^{-2}\n\t\t\\end{align*}\n\t\tCorrect answer is option \\textbf{(d)}\n\t\\end{answer}\n\\end{enumerate}\n\\section{Small Oscillations}\n\\begin{enumerate}\n\t\\item A particle of mass $m$ moves in one dimension under the influence of a potential energy\n\t$$\n\tV(x)=-a\\left(\\frac{x}{\\ell}\\right)^{2}+b\\left(\\frac{x}{\\ell}\\right)^{4}\n\t$$\n\twhere $a$ and $b$ are positive constants and $\\ell$ is a characteristic length. The frequency of small oscillations about a point of stable equilibrium is:\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}] $\\frac{1}{2 \\pi \\ell} \\sqrt{\\frac{b}{m}}$\n\t\t\\task[\\textbf{b.}]$\\frac{2 b}{\\pi \\ell} \\sqrt{\\frac{1}{m a}}$\n\t\t\\task[\\textbf{c.}]$\\frac{1}{\\pi \\ell} \\sqrt{\\frac{a^{2}}{m b}}$\n\t\t\\task[\\textbf{d.}] $\\frac{1}{\\pi \\ell} \\sqrt{\\frac{a}{m}}$\n\t\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\text{\tWe have,}\n\t\tV(x)&=-a\\left(\\frac{x}{\\ell}\\right)^{2}+b\\left(\\frac{x}{\\ell}\\right)^{4}\n\t\t\\intertext{Here first we need to find the stable equilibrium. At equilibrium}\n\t\t\\frac{\\partial V(x)}{\\partial x}&=0 \\quad \\Rightarrow \\frac{-2 a x}{\\ell^{2}}+\\frac{4 b x^{3}}{\\ell^{4}}=0 \\quad \\Rightarrow x\\left(\\frac{2 b x^{2}}{\\ell^{2}}-a\\right)=0\\\\\n\t\t\\Rightarrow \\quad x&=0, x=\\pm \\ell \\sqrt{\\frac{a}{2 b}}\\\\\n\t\t\\text{Now, }\\quad \\frac{\\partial^{2} V}{\\partial x^{2}}&=\\frac{-2 a}{\\ell^{2}}+\\frac{12 b x^{2}}{\\ell^{4}}\\\\\n\t\t\\text{We see that for }x&=0, \\frac{\\partial^{2} V}{\\partial x^{2}}<0\n\t\t\\text{hence, here there is unstable equilibrium}\\\\\n\t\t\\text{For}\\qquad x&=\\pm \\ell \\sqrt{\\frac{a}{2 b}},\\left(\\frac{\\partial^{2} V}{\\partial x^{2}}\\right)_{x=\\pm \\ell \\sqrt{\\frac{a}{2 b}}}=-\\frac{2 a}{\\ell^{2}}+\\frac{12 b}{\\ell^{4}} \\ell^{2} \\times \\frac{a}{2 b}=\\frac{4 a}{\\ell^{2}}>0\n\t\t\\intertext{hence, $x=\\pm \\ell \\sqrt{\\frac{a}{2 b}}$ corresponds to stable equilibrium.. Now to find the frequeny, we see that the problemis one dimensional, so matrices $\\mathbf{V}$ and $\\mathbf{T}$ both have one element each, i.e. $V_{11}$ and $T_{11}$ respectively.}\n\t\tV_{11}&=\\left(\\frac{\\partial^{2} V}{\\partial x^{2}}\\right)_{x=\\pm f \\sqrt{\\frac{a}{2 b}}}=\\frac{4 a}{\\ell^{2}}\\\\\n\t\t\\text{Now, the kinetic energy }&=\\frac{1}{2} m \\dot{x}^{2},\\text{ so }T_{11}=m\\\\\n\t\\text{\tNow, we have, }V_{11}-\\omega^{2} T_{11}&=0 \\qquad\\Rightarrow \\omega^{2}=\\frac{V_{11}}{T_{11}}=\\frac{4 a}{m \\ell^{2}} \\qquad\\Rightarrow \\omega=\\frac{2}{\\ell} \\sqrt{\\frac{a}{m}}\\\\\n\t\\text{So, the frequency }v&=\\frac{\\omega}{2 \\pi}=\\frac{1}{\\pi \\ell} \\sqrt{\\frac{a}{m}}\n\t\\intertext{Note: As we have seen in this problem, first we should find the equilibrium configuration of the system then expand potential energy about this configuration in Taylor series.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item A particle of mass $m$ is moving in a potential of the form $V(x, y, z)=\\frac{1}{2} m \\omega^{2}\\left(3 x^{2}+3 y^{2}+2 z^{2}+2 x\\right)$. The oscillation frequencies of the three normal modes of the particles are given by\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{The potential is, }V(x, y, z)&=\\frac{1}{2} m \\omega^{2}\\left(3 x^{2}+3 y^{2}+2 z^{2}+2 x y\\right)\\\\\n\t\\text{\tNow, }\\quad \\frac{\\partial V}{\\partial x}&=0 \\quad \\Rightarrow 6 x+2 y=0 \\quad \\Rightarrow y=-3 x\\\\\n\t\\frac{\\partial V}{\\partial y}&=0 \\quad \\Rightarrow 6 y+2 x=0 \\quad \\Rightarrow y=\\frac{-x}{3}\\\\\n\t\\frac{\\partial V}{\\partial z}&=0 \\quad \\Rightarrow z=0\\\\\n\t\\text{Now, }\\quad \\frac{\\partial^{2} V}{\\partial x^{2}}&=3 m \\omega^{2}>0 \\text{for all $x$}\\\\\n\t\\frac{\\partial^{2} V}{\\partial y^{2}}&=3 m \\omega^{2}>0\\text{ for all $y$}\\\\\n\t\t\\text{and }\\frac{\\partial^{2} V}{\\partial z^{2}}&=4>0\\text{ for all $z$}\n\t\t\\intertext{All these conditions tell us that equilibrium point is $(0,0,0)$, because $y=-3 x=-\\frac{x}{3}$ satisfies only $x=y=0$ So, the given potential is in the form of expansion about $(0,0,0)$ and the matrix $V_{i j}$ can be written just by inspection.}\n\t\tV &=\\frac{1}{2} m \\omega^{2}\\left(3 x^{2}+3 y^{2}+2 z^{2}+2 x y\\right) \\\\ &=\\frac{1}{2} m \\omega^{2}\\left[\\begin{array}{lll}x & y & z\\end{array}\\right]\\left[\\begin{array}{lll}3 & 1 & 0 \\\\ 1 & 3 & 0 \\\\ 0 & 0 & 2\\end{array}\\right]\\left[\\begin{array}{l}x \\\\ y \\\\ z\\end{array}\\right]\\\\\n\\text{\t\tbecause}\\quad\n\t\t&V=\\frac{1}{2}\\left(3 m \\omega^{2} x^{2}+3 m \\omega^{2} y^{2}+2 m \\omega^{2} z^{2}+m \\omega^{2} x y+m \\omega^{2} y x\\right) \\\\\n\t\t\\text{or,}\\quad&V=\\frac{1}{2}\\left(V_{11} x^{2}+V_{22} y^{2}+V_{33} z^{2}+V_{12} x y+V_{21} y x\\right)\\\\\n\t\\text{\tKinetic energy }T&=\\frac{1}{2} m\\left(\\dot{x}^{2}+\\dot{y}^{2}+\\dot{z}^{2}\\right)=\\frac{1}{2}\\left[\\begin{array}{lll}\\dot{x} & \\dot{y} & \\dot{z}\\end{array}\\right]\\left[\\begin{array}{lll}m & 0 & 0 \\\\ 0 & m & 0 \\\\ 0 & 0 & m\\end{array}\\right]\\left[\\begin{array}{l}\\dot{x} \\\\ \\dot{y} \\\\ \\dot{z}\\end{array}\\right]\\\\\n\t\\text{Now,}\n\t\\left|\\mathbf{V}-\\Omega^{2} \\mathbf{T}\\right|&=0\\quad \\text{where $\\Omega$ is normal mode freuency.}\\\\\n\t\\Rightarrow \\quad&\\left|\\begin{array}{ccc}3 m \\omega^{2}-\\Omega^{2} m & m \\omega^{2} & 0 \\\\ m \\omega^{2} & 3 m \\omega^{2}-\\Omega^{2} m & 0 \\\\ 0 & 0 & 2 m \\omega^{2}-\\Omega^{2} m\\end{array}\\right|=0\\\\\n\t\\Rightarrow \\quad\\left(2 \\omega^{2}-\\Omega^{2}\\right)&\\left[\\left(3 \\omega^{2}-\\Omega^{2}\\right)^{2}-\\omega^{4}\\right]=0 \\quad \\Rightarrow \\Omega_{1}^{2}=2 \\omega^{2}\\\\\n\\text{\tand }\\quad \\Omega^{2}&=3 \\omega^{2} \\pm \\omega^{2} \\quad \\Rightarrow \\Omega_{2}^{2}=2 \\omega^{2}\\text{ and }\\Omega_{3}^{2}=4 \\omega^{2}\n\\intertext{So, the frequencies are $\\omega \\sqrt{2}, \\omega \\sqrt{2}$ and $2 \\omega$}\n\\intertext{\tNote that frequencies are always positive, hence we shouldn't write $\\Omega_{1}=\\pm \\omega \\sqrt{2}$ etc.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item \\textbf{Coupled mass point on a circle:}\\\\\n\tFour mass points of mass $m$ move on a circle of radius $R$. Each mass point is coupled to its two neighboring points by a spring with spring constant $k$ (see figure below). Find the Lagrangian of the system and derive the equations of motion of the system. Calculate the eigenfrequencies of the system, and discuss the related eigenvibrations.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\intertext{\tThe kinetic energy of the system is given by}\n\t\tT&=\\frac{1}{2} m \\sum_{v=1}^{4} \\dot{s}_{v}^{2}\n\t\\intertext{\tFor small displacement from the equilibrium position, the potential reads}\n\t\tV&=\\frac{1}{2} k \\sum_{v=1}^{4}\\left(s_{v+1}-s_{v}\\right)^{2}, \\quad s_{4+1}=s_{1}\n\t\\intertext{\tWe set $s_{v}=R \\varphi_{v}$, and take the angles $\\varphi_{v}$ as generalized coordinates. Then the Lagrangian is}\n\t\tL&=T-V=\\frac{1}{2} m R^{2} \\sum_{v=1}^{4} \\dot{\\varphi}_{v}^{2}-\\frac{1}{2} k R^{2} \\sum_{v=1}^{4}\\left(\\varphi_{v+1}-\\varphi_{v}\\right)^{2}\n\t\\intertext{\tFrom the Lagrange equations}\n\t\t\\frac{d}{d t} \\frac{\\partial L}{\\partial \\dot{\\varphi}_{v}}&=\\frac{\\partial L}{\\partial \\varphi_{v}}\n\t\t\\end{align*}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=4.8cm,width=5cm]{small oscillations-01}\n\t\t\\end{figure}\n\t\t\\begin{align*}\n\t\t\\intertext{we find the equations of motion:}\n\t\t\\frac{d}{d t} \\frac{\\partial L}{\\partial \\dot{\\varphi}_{v}}=m R \\ddot{\\varphi}_{v}=-\\frac{1}{2} k R^{2}\\left[2\\left(\\varphi_{v}-\\varphi_{v+1}\\right)+2\\left(\\varphi_{v}-\\varphi_{v+1}\\right)\\right]&=\\frac{\\partial L}{\\partial \\varphi_{v}} .\n\t\\intertext{\tFor the case of four mass points, we then obtain}\n\t\\ddot{\\varphi}_{1}=\\frac{k}{m}\\left(\\varphi_{2}-2 \\varphi_{1}+\\varphi_{4}\\right), \\quad \\ddot{\\varphi}_{2}&=\\frac{k}{m}\\left(\\varphi_{3}-2 \\varphi_{2}+\\varphi_{1}\\right)\\\\\n\t\\ddot{\\varphi}_{3}=\\frac{k}{m}\\left(\\varphi_{4}-2 \\varphi_{3}+\\varphi_{2}\\right), \\ddot{\\varphi}_{4}&=\\frac{k}{m}\\left(\\varphi_{1}-2 \\varphi_{4}+\\varphi_{3}\\right)\n\t\\intertext{With the ansatz $\\varphi_{v}=A_{v} \\cos \\omega t, \\ddot{\\varphi}_{v}=-A_{v} \\omega^{2} \\cos \\omega t$, we are led to the following linear system of equations:}\n\t\\left(\\begin{array}{cccc}2 \\frac{k}{m}-\\omega^{2} & -\\frac{k}{m} & 0 & -\\frac{k}{m} \\\\ -\\frac{k}{m} & 2 \\frac{k}{m}-\\omega^{2} & -\\frac{k}{m} & 0 \\\\ 0 & -\\frac{k}{m} & 2 \\frac{k}{m}-\\omega^{2} & -\\frac{k}{m} \\\\ -\\frac{k}{m} & 0 & -\\frac{k}{m} & 2 \\frac{k}{m}-\\omega^{2}\\end{array}\\right)\\left(\\begin{array}{l}A_{1} \\\\ A_{2} \\\\ A_{3}\\end{array}\\right)&=0\n\t\\intertext{For the nontrivial solutions, the determinant of the coefficient matrix must vanish. This condition leads to the determining equation for the eigenfrequencies:}\n\t\\left(2 \\frac{k}{m}-\\omega^{2}\\right)^{2}\\left(4 \\frac{k}{m}-\\omega^{2}\\right)\\left(-\\omega^{2}\\right)=0\n\t\\intertext{The frequencies are}\n\t\\omega_{1}^{2}=0, \\quad \\omega_{2}^{2}&=4 \\frac{k}{m}, \\quad \\omega_{3}^{2}=\\omega_{4}^{2}=2 \\frac{k}{m}\n\\intertext{\tTo calculate the related eigenvibrations, we insert these frequencies into the system of equations:}\n\t\\end{align*}\n\\begin{align*}\n\t&\\text{(1) }\\omega_{1}^{2}=0: A_{1}=A_{2}=A_{3}=A_{4} : \\text{The system does not vibrate but performs a uniform rotation}\\\\\n\t&\\text{(2) }\\omega_{2}^{2}=4 \\frac{k}{m}: A_{1}=A_{3}=-A_{2}=-A_{4}:\\text{ Two neighboring mass points perform an out-of-phase vibration}\\\\\n\t&\\text{(3) }\\omega_{3}^{2}=\\omega_{4}^{2}=2 \\frac{k}{m}: A_{1}=A_{2}=-A_{3}=-A_{4}\\text{ or }A_{1}=A_{4}=-A_{2}=-A_{3} :\\text{ Two neighboring mass points }\\\\&\\text{vibrate in phase}\n\t\\end{align*}\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=8cm,width=11cm]{small oscillations-02}\n\t\\end{figure}\n\t\\end{answer}\n\t\\item Two equal masses coupled by two equal springs\n\tTwo equal masses move without friction on a plate. They are connected to each other and to the wall by two springs, as is indicated by Figure $7.3$. The two spring constants are equal, and the motion shall berestricted to a straight line (one-dimensional motion).\n\tFind\\\\\n\t(a) the equations of motion,\\\\\n\t(b) the normal frequencies, and\\\\\n\t(c) the anplitude ratios of the normal vibrations and the general solution.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\intertext{(a) Let $x_{1}$ and $x_{2}$ be the displacements from the rest positions. The cquations of motion then read}\n\t\tm \\ddot{x}_{1}&=-k x_{2}+k\\left(x_{2}-x_{1}\\right) \\\\\n\t\tm \\dot{x}_{2}&=-k\\left(x_{2}-x_{1}\\right)\n\t\t\\intertext{(b) For determining the normal frequencies, we use the ansatz}\n\t\tx_{1}&=A_{1} \\cos \\omega t, \\quad x_{2}=A_{2} \\cos \\omega t\n\t\t\\end{align*}\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2cm,width=6.5cm]{small oscillations-03}\n\t\t\\end{figure}\n\t\t\\begin{align}\n\t\\intertext{\tand thereby get from $\\color{red}{(7.9)}$ and $(\\color{red}{7.10})$ the equations}\\notag\\\\\n\t\t&\\left(2 k-m \\omega^{2}\\right) A_{1}-k A_{2}=0 \\label{so-05}\\\\\n\t\t&-k A_{1}+\\left(k-m \\omega^{2}\\right) A_{2}=0\\notag\n\t\t\\intertext{From therequirement for nontrivial solutions of the system of equations, it follows that the determinant of coefficient vanishes:}\\notag\n\t\tD&=\\left|\\begin{array}{cc}\n\t\t2 k-m \\omega^{2} & -k \\notag\\\\\n\t\t-k & k-m \\omega^{2}\n\t\t\\end{array}\\right|=0\n\t\\intertext{\tFrom this follows the determining equation for the eigenfrequencies,}\\notag\n\t\t\\omega^{4}-3 \\frac{k}{m} \\omega^{2}+\\frac{k^{2}}{m^{2}}&=0\\notag\n\t\t\\intertext{with the positive solutions}\\notag\n\t\t\\omega_{1}=\\frac{\\sqrt{5}+1}{2} \\sqrt{\\frac{k}{m}} \\text { and } \\omega_{2}&=\\frac{\\sqrt{5}-1}{2} \\sqrt{\\frac{k}{m}}, \\omega_{1}>\\omega_{2} .\\notag\n\t\t\\intertext{By inserting the eigenfrequencies in (\\ref*{so-05}) one sees that the higher frequency $\\omega_{1}$ corresponds to the oppositephase mode, and the lower frequency $\\omega_{2}$ to the equal-phase normal vibration:}\\notag\n\t\t\\text{with }\\omega_{1}^{2}&=\\frac{1}{2}(3+\\sqrt{5}) \\frac{k}{m}, \\quad\\text{ it follows from (\\ref{so-05}) that }A_{2}=-\\frac{\\sqrt{5}-1}{2} A_{1}\\notag\\\\ \\text{with }\\omega_{2}^{2}&=\\frac{1}{2}(3-\\sqrt{5}) \\frac{k}{m},\\text{ it follows from (\\ref{so-05}) that }A_{2}=\\frac{\\sqrt{5}+1}{2} A_{1} .\\notag\n\t\t\\intertext{Since the two mass points are fixed in different ways, we find amplitudes of different magnitudes. The general solution is obtained as a superposition of the normal vibrations, using the calculated amplitudes ratios:}\\notag\n\t\tx_{1}(t)&=C_{1} \\cos \\left(\\omega_{1} t+\\varphi_{1}\\right)+C_{2} \\cos \\left(\\omega_{2} t+\\varphi_{2}\\right)\\notag\\\\\n\t\tx_{2}(t)&=-\\frac{\\sqrt{5}-1}{2} C_{1} \\cos \\left(\\omega_{1} t+\\varphi_{1}\\right)+\\frac{\\sqrt{5}+1}{2} C_{2} \\cos \\left(\\omega_{2} t+\\varphi_{2}\\right)\\notag\n\t\t\\end{align}\n\t\\end{answer}\n\t\\item The Lagrangian of a system is given by $L=\\frac{1}{2} m \\dot{q}_{1}^{2}+2 m \\dot{q}_{2}^{2}-k\\left(\\frac{5}{4} q_{1}^{2}+2 q_{2}^{2}-2 q_{1} q_{2}\\right)$ where $m$ and $k$ are positive constants. The frequencies of its normal modes are\n \\begin{tasks}(2)\n\t\\task[\\textbf{a.}]$\\sqrt{\\frac{k}{2 m}}, \\sqrt{\\frac{3 k}{m}}$\n\t\\task[\\textbf{b.}]$\\sqrt{\\frac{k}{2 m}}(13 \\pm \\sqrt{73})$\n\t\\task[\\textbf{c.}]$\\sqrt{\\frac{5 k}{2 m}}, \\sqrt{\\frac{k}{m}}$\n\t\\task[\\textbf{d.}] $\\sqrt{\\frac{k}{2 m}}, \\sqrt{\\frac{6 k}{m}}$\n\\end{tasks}\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t L &=\\frac{1}{2} m \\dot{q}_{1}^{2}+2 m \\dot{q}_{2}^{2}-k\\left(\\frac{5}{4} q_{1}^{2}+2 q_{2}^{2}-2 q_{1} q_{2}\\right) \\\\ &=\\frac{1}{2} m \\dot{q}_{1}^{2}+\\frac{1}{2} 4 m \\dot{q}_{2}^{2}-\\frac{1}{2} k\\left(\\frac{5}{2} q_{1}^{2}+4 q_{2}^{2}-4 q_{1} q_{2}\\right) \\\\ \\hat{T} &=\\left(\\begin{array}{cc}m & 0 \\\\ 0 & 4 m\\end{array}\\right), \\hat{V}=\\left(\\begin{array}{cc}\\frac{5}{2} k & -2 k \\\\ -2 k & 4 k\\end{array}\\right) \n\t\t \\intertext{For frequencies of normal modes:}\n\t\t \\operatorname{det}\\left|\\omega^{2} \\hat{T}-\\hat{V}\\right|&=0\\\\\n\t\t \\left|\\begin{array}{cc}\\left(m \\omega^{2}-\\frac{5}{2} k\\right) & 2 k \\\\ 2 k & \\left(4 m \\omega^{2}-4 k\\right)\\end{array}\\right|&=0 \\Rightarrow 4\\left(m \\omega^{2}-k\\right) \\frac{\\left(2 m \\omega^{2}-5 k\\right)}{2}-4 k^{2}=0\\\\\n\t\t \\Rightarrow 2 m^{2} \\omega^{4}+5 k^{2}-7 k m \\omega^{2}-2 k^{2}&=0\\\\\n\t\t \\Rightarrow 2\\left(m \\omega^{2}\\right)^{2}-7 k\\left(m \\omega^{2}\\right)+3 k^{2}&=0\\\\\n\t\t \\Rightarrow m \\omega^{2}&=\\frac{7 k \\pm \\sqrt{49 k^{2}-24 k^{2}}}{4}=\\frac{7 k \\pm \\sqrt{49 k^{2}-24 k^{2}}}{4}\\\\&=\\frac{7 k \\pm 5 k}{4}=3 k, \\frac{k}{2}\\\\\n\t\t \\therefore \\omega&=\\sqrt{\\frac{3 k}{m}}, \\sqrt{\\frac{k}{2 m}}\n\t\t\\end{align*}\n\t\t Correct answer is option \\textbf{(a)}\n\t\\end{answer}\n\\end{enumerate}\n\\section{Canonical Transformations}\n\\begin{enumerate}\n\t\\item Show that for a transformation to be canonical function \n\t$\\Sigma\\left(p_{i} d q_{i}-P_{i} d Q_{i}\\right)$ must be a perfect differential if the generating function does not contain time. \n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\\intertext{\tWe have by the property of generating function in terms of $q, Q$ that is, $G=G\\left(q_{i}, Q_{i}, t\\right)$}\n\tp_{i}&=\\frac{\\partial G}{\\partial q_{i}}, P=-\\frac{\\partial G}{\\partial Q_{i}}\\text{ and }\\bar{H}-H=\\frac{\\partial G}{\\partial t}\\\\\n\\text{\tNow,}\\qquad\n\td G&=\\Sigma\\left(\\frac{\\partial G}{\\partial q_{i}} d q_{i}+\\frac{\\partial G}{\\partial Q_{i}} d Q_{i}\\right)+\\frac{\\partial G}{\\partial t} d t=\\Sigma\\left(p_{i} d q_{i}-P_{i} d Q_{i}\\right)+0 \\qquad\\left(\\because \\frac{\\partial G}{\\partial t}=0\\right)\\\\\n\t\\text{Hence, }&\\Sigma\\left(p_{i} d q_{i}-P_{i} d Q_{i}\\right)\\text{ is a perfect differential.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Show that $P=\\frac{1}{2}\\left(p^{2}+q^{2}\\right), Q=\\tan ^{-1} \\frac{q}{p}$ is canonical.\n\t\\begin{answer}\n\t\t Since the transformation equations do not contain time, the generating function also does not contain time. Hence, the condition for transformation to be canonical is that $(p d q-P d Q)$ be a perfect differential. Here we consider only one term in the summation because the system for which transformation equations are given is of one degree of freedom.\n\t\t\\begin{align*}\n\t\t\\text{Now, }\\quad p d q-P d Q&=p d q-\\frac{p^{2}+q^{2}}{2} d\\left(\\tan ^{-1} \\frac{q}{p}\\right)\\\\\n\t\t&=p d q-\\frac{p^{2}+q^{2}}{2} \\frac{p d-q d p}{p^{2}+q^{2}} \\quad\\left(\\because d\\left(\\tan ^{-1} x\\right)=\\frac{d x}{1+x^{2}}\\right)\\\\\n\t\t&=p d q-\\frac{1}{2} p d q+\\frac{1}{2} q d p=\\frac{1}{2} p d q+\\frac{1}{2} q d p=d\\left(\\frac{1}{2} p q\\right)\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Find the values of $\\alpha$ and $\\beta$ so that $Q=q^{\\alpha} \\cos p \\beta$ and $P=q^{\\alpha} \\sin p \\beta$ may be canonical. Also find the generating function, $G=G(p, Q)$ is perfect differential.\n\t\\begin{answer}\n\t\tHere the transformation equations do not contain time. Hence the condition for canonical transformation is that $p d q-P d Q$.\n\t\t\\begin{align*}\n\t\\text{\tHere,}\\quad\n\t\tp d q-P d Q&=p d q-q^{\\alpha} \\sin p d\\left(q^{\\alpha} \\cos p \\beta\\right)\\\\\n\t\t&=p d q-q^{\\alpha} \\sin p \\beta\\left(\\alpha q^{\\alpha-1} \\cos p \\beta-\\beta q^{\\alpha} \\sin p \\beta d p\\right)\\\\\n\t\t&=\\left(p-\\alpha q^{2 \\alpha-1} \\sin p \\beta \\cos p \\beta\\right) d q+\\beta q^{2 \\alpha} \\sin ^{2} p \\beta d p\n\t\t\\intertext{Now, $M(x, y) d x+N(x, y) d y$ is an exact differential when $\\frac{\\partial M}{\\partial y}=\\frac{\\partial N}{\\partial x}$. By this principle $p d q-P d Q$ is a perfect differential if}\n\t\t\\frac{\\partial}{\\partial p}\\left(p-\\alpha q^{2 \\alpha-1} \\sin p \\beta \\cos p \\beta\\right)&=\\frac{\\partial}{\\partial q}\\left(\\beta q^{2 \\alpha} \\sin ^{2} p \\beta\\right)\\\\\n\t\t\\text{or}\\quad1-\\alpha \\beta q^{2 \\alpha-1}\\left(1-2 \\sin ^{2} p \\beta\\right)&=2 \\alpha \\beta q^{2 \\alpha-1} \\sin ^{2} p \\beta\\\\\n\t\t\\text{or}\\quad1-\\alpha \\beta q^{2 \\alpha-1}+2 \\alpha \\beta q^{2 \\alpha-1} \\sin ^{2} p&=2 \\alpha \\beta q^{2 \\alpha-1} \\sin ^{2} p \\beta\\\\\n\t\t\\text{or}\\quad\\alpha \\beta q^{2 \\alpha-1}&=1\n\t\t\\intertext{Since, RHS is a constant, this equation is satisfied only when the power of $q$ is zero. Hence, $2 \\alpha-1=0$ or $\\alpha=\\frac{1}{2}$ and $\\beta=2$. Thus, $\\alpha=\\frac{1}{2}$ and $\\beta=2$ are the required values of $\\alpha$ and $\\beta$}\n\t\t\\end{align*}\n\t\t\\textbf{Generating function :} When the generating function is assumed to be $G=G(p, Q)$, we have\n\t\t\\begin{align*}\n\t\tq&=-\\frac{\\partial G}{\\partial p}\\text{ and }P=-\\frac{\\partial G}{\\partial Q}\\\\\n\t\t\\text{Here,}\\quad\n\t\tQ&=q^{1 / 2} \\cos 2 p \\text { or } q=\\left(\\frac{Q}{\\cos 2 p}\\right)^{2}\\\\\n\t\t\\text{ And }\\quad P&=q^{1 / 2} \\sin 2 p=\\left(\\frac{Q}{\\cos 2 p}\\right) \\sin 2 p=Q \\tan 2 p\\\\\n\t\t\\therefore \\quad \\frac{\\partial G}{\\partial p}&=-q=-\\left(\\frac{Q Q}{\\cos 2 p}\\right)^{2}\\text{ or }G=-Q^{2} \\int \\frac{\\partial p}{\\cos ^{2} 2 p}\\\\\n\t\t\\text{or}\\quad G&=-Q^{2} \\int \\sec ^{2} 2 p \\partial p=-\\frac{1}{2} Q^{2} \\tan 2 p\\\\\n\t\\text{\tThus, }G&=-\\frac{1}{2} Q^{2} \\tan 2 p\\text{ is the required generating function.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Show that the generating function for the transformation $p=\\frac{1}{Q}$ and $q=P Q^{2}$ is $G=\\frac{q}{Q}$\n\t\\begin{answer}\n\t\t Since we have to show that $G=\\frac{q}{Q}$ we assume that the generating function needed is a function of $q, Q$ that is, $G=G(q, Q)$. When $q$ and $Q$ are independent variables, the other two variables $p$ and $P$ are given by $p=\\frac{\\partial G}{\\partial q}$ and $P=-\\frac{\\partial G}{\\partial Q}$.\n\t\t\\begin{align*}\n\t\t\\text{Here, }\\quad p&=\\frac{1}{Q}\\text{ and hence here }\\frac{1}{Q}=\\frac{\\partial G}{\\partial q}\\text{ or }G=\\int \\frac{\\partial q}{Q}.\\\\\n\t\t\\text{Or }G&=\\frac{q}{Q}\n\t\\intertext{\tFrom the second relation also we can show that}\n\t\tG&=\\frac{q}{Q}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Show that if there exists a generating function $\\mathrm{G}$ such that $\\frac{d G}{d t}=L-\\bar{L}$ where $\\mathrm{L}$ and $\\bar{L}$ are the Lagrangians in old and new set of coordinates, then the transformation is canonical.\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\text{By Hamilton's principle, }\\frac{d G}{d t}&=L-\\delta \\int_{t_{1}}^{t_{2}} L d t=0\\text{ and }\\delta \\int_{t_{1}}^{t_{2}} \\bar{L} d t=0\\\\\n\t\t\\text{Subtracting we have, }\\delta \\int_{t_{1}}^{t_{2}}(L-\\bar{L}) d t&=0\\\\\n\t\t\\text{By virtue of the given relation i.e. }\\frac{d G}{d t}&=L-\\bar{L}\\\\\n\t\t\\text{We have, }\\delta \\int_{t_{1}}^{t_{2}} \\frac{d G}{d t} d t=0\\text{, or} \\delta\\left[G\\left(t_{2}\\right)-G\\left(t_{1}\\right)\\right]&=0\\text{ which is the condition for canonical transformation.}\n\t\t\\intertext{Hence under the given condition the transformation is canonical.}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\\item Show that $G=-\\Sigma Q_{i} p_{i}$ generating on identity transformation.\n\t\\begin{answer}\n\t\t Let us define an identity transformation. When the coordinates of old set are equal to the coordinates of the new set, it is called an identity transformation.\\\\\n\t\t \\text { It is given here that the generating function is a function of } p, Q \\text { i.e. } G=G(p, Q)\n\t\t\\begin{align*}\n\t\t\\text{In this case we have, }q_{i}&=-\\frac{\\partial G}{\\partial p_{i}}\\text{ and }P_{i}=-\\frac{\\partial G}{\\partial Q_{i}}\n\t\t\\intertext{Here it is given that $G=-\\Sigma p_{i} Q_{i}$. As we have to take derivative of $G$ with respect to $p_{i}$, we change the suffix in the summation to $j$ i.e. we take $G=-\\Sigma p_{j} Q_{j}$.}\n\t\t\\therefore \\quad \\frac{\\partial G}{\\partial p_{i}}&=-\\Sigma\\left(p_{j} \\frac{\\partial Q_{j}}{\\partial p_{i}}+Q_{j} \\frac{\\partial p_{j}}{\\partial p_{i}}\\right)\\\\\n\t\\text{\tNow, }\\quad \\frac{\\partial Q_{j}}{\\partial p_{i}}&=0\\text{ because, $p, Q$ are selected as independent variables}\\\\\n\t\\text{here and }\\quad \\frac{\\partial p_{j}}{\\partial p_{i}}&=0\\text{ if $i \\neq j=1$ if $i=j$}\\\\\n\t\\text{Here, }\\quad \\frac{\\partial G}{\\partial p_{i}}&=-Q_{i} .\\text{ But }\\frac{\\partial G}{\\partial p_{i}}=-q_{i} (\\text{in general}).\\\\\n\t\\therefore \\quad q_{i}&=Q_{i}\\\\\n\t\\frac{\\partial G}{\\partial Q_{i}}&=-\\Sigma\\left(Q_{j} \\frac{\\partial p_{j}}{\\partial Q_{i}}+p_{j} \\frac{\\partial Q_{j}}{\\partial Q_{i}}\\right)=-p_{i}\\\\\n\t(\\text{Because }\\frac{\\partial p_{j}}{\\partial Q_{i}}&=0\\text{ and }\\frac{\\partial Q_{j}}{\\partial Q_{i}}=1\\text{ where }\\mathrm{i}=\\mathrm{j}\\text{, otherwise zero}).\\\\\n\t\\therefore \\quad p_{i}&=P_{i}\n\t\t\\end{align*}\n\t\\end{answer}\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\\end{enumerate}", "meta": {"hexsha": "68d1aa55dd571b761691f14251351183511714b4", "size": 133289, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Classical Mechanics -CSIR/chapter/Classical Problems.tex", "max_stars_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_stars_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Classical Mechanics -CSIR/chapter/Classical Problems.tex", "max_issues_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_issues_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Classical Mechanics -CSIR/chapter/Classical Problems.tex", "max_forks_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_forks_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 80.9775212637, "max_line_length": 534, "alphanum_fraction": 0.6338257471, "num_tokens": 52542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321703143954, "lm_q2_score": 0.7879312006227324, "lm_q1q2_score": 0.6347827232161192}} {"text": "\\section{Limits and Continuity}\r\nIf we want to differentiate, we need to make sense of a limit, so here goes.\r\n\\subsection{Limit of a function}\r\n\\begin{definition}\r\n Suppose $a\\in\\mathbb C$ and $f:\\mathbb C\\setminus\\{a\\}\\to\\mathbb C$.\r\n We say $\\lim_{z\\to a}f(z)=c\\in\\mathbb C$ if\r\n $$\\forall\\epsilon>0,\\exists\\delta>0,|z-a|<\\delta\\implies |f(z)-c|<\\epsilon$$\r\n Sometimes we write $f(z)\\to c$ as $z\\to a$.\r\n\\end{definition}\r\nWe say $\\lim_{z\\to a}f(z)$ is undefined if there is no $c$ with $\\lim_{z\\to a}f(z)=c$.\r\n\\begin{proposition}\r\n If $\\lim_{z\\to a}f(z)=c$ and $\\lim_{z\\to a}f(z)=d$, then $c=d$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\n\\begin{example}\r\n 1. If $f\\equiv c$, then $\\lim_{z\\to a}f(z)=c$ for any $a$, just take for example $\\delta=1$.\\\\\r\n 2. If $f(z)=z$, then $\\lim_{z\\to a}f(z)=a$, as we can take $\\delta=\\epsilon$.\\\\\r\n 3. If $f(z)=1_{\\mathbb C\\setminus\\{0\\}}$, then $\\lim_{z\\to 0}f(z)=1$, taking for example $\\delta=1$.\\\\\r\n 4. If\r\n $$f(z)=\\begin{cases}\r\n 1\\text{, if $\\operatorname{Re}(z)>0$}\\\\\r\n 0\\text{, if $\\operatorname{Re}(z)=0$}\\\\\r\n -1\\text{, if $\\operatorname{Re}(z)<0$}\r\n \\end{cases}$$\r\n then $\\lim_{z\\to 0}f(z)$ does not exist.\\\\\r\n 5. If $f(z)=z^2$, then $\\lim_{z\\to a}f(z)=a^2=f(a)$ by choosing $\\delta=\\min\\{\\epsilon/(2|a|+1),1\\}$.\r\n\\end{example}\r\n\\begin{proposition}\r\n $\\lim_{z\\to a}f(z)=c$ if and only if $\\forall (z_n)\\in\\mathbb C$ with $z_n\\to a$, we have $f(z_n)\\to c$\r\n\\end{proposition}\r\n\\begin{proof}\r\n If the limit of the function is $c$, given $\\epsilon>0$, we can find $\\delta$ such that $|z-a|<\\delta\\implies |f(z)-c|<\\epsilon$.\r\n Given a sequence $z_n\\to a$, we can find $N\\in\\mathbb N$ such that $\\forall n>N,|z_n-a|<\\delta$, so using this $N$, $\\forall n>N,|z_n-a|<\\delta\\implies |f(z_n)-c|<\\epsilon$, so $f(z_n)\\to c$.\\\\\r\n Conversely, if $c$ is not the limit of that function, then $\\exists\\epsilon>0,\\forall\\delta>0,\\exists z\\in\\mathbb C,|z-a|<\\delta,|f(z)-c|>\\epsilon$.\r\n So we choose such $z_n$ for $\\delta=1/n$, then $z_n\\to a$ but $|f(z_n)-c|>\\epsilon$ for any $n$, hence $f(z_n)$ does not tend to $c$.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Let $f,g:\\mathbb C\\setminus\\{a\\}\\to\\mathbb C$, if $\\lim_{z\\to a}f(z)=c$ and $\\lim_{z\\to a}f(z)=d$, then\\\\\r\n 1. $\\lim_{z\\to a}(f(z)+g(z))=c+d$.\\\\\r\n 2. $\\lim_{z\\to a}f(z)g(z)=cd$.\\\\\r\n 3. If $c\\neq 0$, then $\\lim_{z\\to a}1/f(z)=1/c$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Using the preceding proposition and laws of limits for sequences.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Suppose $f,g:\\mathbb C\\setminus\\{a\\}\\to\\mathbb R$ such that\r\n $$\\lim_{z\\to a}f(z)=c,\\lim_{z\\to a}g(z)=d$$\r\n then,\\\\\r\n 1. If $\\forall z,f(z)\\ge g(z)$, then $c\\ge d$.\\\\\r\n 2. (Squeeze Rule) Let $h:\\mathbb C\\setminus\\{a\\}$ be such that $f\\ge h\\ge g$.\r\n Then if $c=d$, we have $\\lim_{z\\to a}h(z)=c$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\nWhat if $f:A\\to\\mathbb C$ where $A$ is a subset of $\\mathbb C$?\r\nDoes it still makes sense?\r\nDoes it change the limit?\r\nThe answer to the second question if obviously no.\r\nBut we can make sense of the limit.\r\n\\begin{definition}\r\n Suppose $A$ is a subset of $\\mathbb C$.\r\n We say $a\\in\\mathbb C$ is a limit point of $A$, if $\\forall\\delta>0,\\exists z\\in A,0<|z-a|<\\delta$.\r\n\\end{definition}\r\n\\begin{example}\r\n Any $a\\in [0,\\infty)$ is a limit point of $(0,\\infty)$, and $\\mathbb Z$ has no limit point.\r\n\\end{example}\r\n\\begin{definition}\r\n If $f:A\\to\\mathbb C$ and $a\\in\\mathbb C$ is a limit point of $A$, we say $\\lim_{z\\to a}f(z)=c$ if $\\forall\\epsilon>0,\\exists\\delta>0,z\\in A,0<|z-a|<\\delta\\implies |f(z)-c|<\\epsilon$.\r\n\\end{definition}\r\nEquivalently, $f(z_n)\\to c$ for any sequence $(z_n)\\in A$ such that $z_n\\to a$.\r\nAlso, it is also easy to show that the limit is unique.\r\n\\begin{definition}\r\n For $f:\\mathbb R\\to\\mathbb R$, we define\r\n $$\\lim_{x\\to a^+}f(x)=\\lim_{x\\to a}f|_{(a,\\infty)}(x),\\lim_{x\\to a^-}f(x)=\\lim_{x\\to a}f|_{(-\\infty,a)}(x)$$\r\n\\end{definition}\r\n\\subsection{Continuity}\r\n\\begin{definition}\r\n Suppose $A\\subset\\mathbb C$, and $f:A\\to\\mathbb C$.\r\n We say $f$ is continuous at $a\\in A$ if\r\n $$\\forall\\epsilon>0,\\exists\\delta>0,z\\in A,|z-a|<\\delta\\implies |f(z)-f(a)|<\\epsilon$$\r\n\\end{definition}\r\n\\begin{remark}\r\n $|f(a)-f(a)|=0<\\epsilon$, so this condition is equivalent to say $|f(z)-f(z)|<\\epsilon$ whenever $0<|z-a|<\\delta$.\r\n\\end{remark}\r\n\\begin{proposition}\r\n Suppose $f:A\\to\\mathbb C$ and $a\\in A$, then the following conditions are equivalent:\\\\\r\n 1. $f$ is continuous at $a$.\\\\\r\n 2. Either $a$ is not a limit point of $A$ or $\\lim_{z\\to a}f(z)=f(a)$.\\\\\r\n 3. $f(z_n)\\to f(a)$ whenever $z_n\\to a$ is a sequence in $A$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Verify from definition.\r\n\\end{proof}\r\n\\begin{example}\r\n 1. Constant functions are continuous at every point.\r\n\\end{example}\r\n\\begin{definition}\r\n We say $f:A\\to\\mathbb C$ is continuous if $f$ is continuous at every $a\\in A$.\r\n\\end{definition}\r\n\\begin{definition}\r\n 2. The identity function is continous.\\\\\r\n 3. (non-example) $f:\\mathbb R\\to\\mathbb R$ by $f=1_{\\mathbb Q}$. is not continuous at any point.\\\\\r\n 4. Consider $f:\\mathbb R\\to\\mathbb R$ by\r\n $$f(x)=\\begin{cases}\r\n 1/q\\text{, if $x$ is rational and $q=\\min\\{q\\in\\mathbb N:\\exists p\\in\\mathbb Z,x=p/q\\}$}\\\\\r\n 0\\text{, otherwise}\r\n \\end{cases}$$\r\n Then $f$ is not continuous at any $x\\in\\mathbb Q\\setminus\\{0\\}$ but it is continuous at $a\\notin\\mathbb Q$.\\\\\r\n 5. $f:\\mathbb R_{\\ge 0}\\to\\mathbb R_{\\ge 0}$ by $x\\mapsto \\sqrt{x}$ is continuous.\r\n\\end{definition}\r\n\\begin{proposition}\r\n Suppose $f,g:A\\to\\mathbb C$ for some $A\\subset\\mathbb C$ and they are both continuous at $a\\in A$.\r\n Then $f+g,fg$ are continuous at $a$.\r\n And if $f(a)\\neq 0$, then $1/f$ is continuous at $a$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\n\\begin{example}\r\n Everywhere in $\\mathbb C$ we have any constant function is continous, the identity is continuous, hence any linear function is continuous, therefore any polynomial is continuous.\r\n Also any rational function is continuous at anywhere it is defined.\r\n\\end{example}\r\nThe function $\\exp$ is continuous, but we will not prove it now.\r\n\\begin{proposition}\r\n Let $A,B,C\\subset\\mathbb C$ and $f:A\\to B,g:B\\to C$ such that $f$ is continuous at $a$ and $g$ continuous at $f(a)$, then $g\\circ f$ is continuous at $a$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Straightforward.\r\n\\end{proof}\r\n\\begin{example}\r\n $g:(0,\\infty)\\to (0,\\infty),g(x)=\\sqrt{x}$ and $f:(0,\\infty)\\to(0,\\infty),f(x)=x^2+1$, so $x\\mapsto (g\\circ f)(x)=\\sqrt{x^2+1}$ is continuous.\r\n\\end{example}\r\n\\begin{lemma}\r\n Suppose $(x_n)$ is a sequence in $[a,b]$ and $x_n\\to x\\in\\mathbb R$, then $x\\in [a,b]$.\r\n\\end{lemma}\r\nNote that this is false if we replace $[a,b]$ by $(a,b)$.\r\n\\begin{proof}\r\n Obvious.\r\n\\end{proof}\r\n\\begin{theorem}[Intermediate Value Theorem]\r\n If $f:[a,b]\\to\\mathbb R$ is continuous, and $f(a)\\le0,f(b)\\ge0$, then there is some $c\\in [a,b]$ such that $f(c)=0$.\r\n\\end{theorem}\r\nThis is false for $\\mathbb Q$.\r\n\\begin{remark}\r\n Note that it does not matter if it is $f(a)\\le0,f(b)\\ge0$ or $f(b)\\le0,f(a)\\ge0$.\r\n Also it suffices to show the case where the inequalities are both strict, so we will assume that in the proof.\r\n\\end{remark}\r\n\\begin{proof}\r\n For $n\\in\\mathbb N$, we define $Q_n=\\{a+k(b-a)/n:0\\le k\\le n\\}$ and $P_n\\subset Q_n$ such that $\\forall x\\in P_n,f(x)\\le 0$, so $P_n$ is never empty.\r\n Define $x_n=\\max P_n$, then $a\\le x_n\\le b-(b-a)/n$ and $f(x_n)\\le 0$.\r\n We let $y_n=x_n+(b-a)/n=\\min Q_n\\setminus P_n\\subset [a,b]$, then by maximality of $x_n$ we have $f(y_n)>0$.\r\n By Bolzano-Weierstrass Theorem, there is a subsequence $x_{n_k}$ that converges to some $c\\in [a,b]$ by the preceding lemma, so by continuity of $f$, $f(c)=\\lim_{n\\to\\infty}f(x_n)\\le 0$.\r\n Therefore $y_{n_k}\\to c$ as well since $(b-a)/n_k\\to 0$, but then $f(c)=\\lim_{n\\to\\infty}f(y_n)\\ge 0$.\r\n So $f(c)=0$.\r\n\\end{proof}\r\n\\begin{corollary}\r\n An odd degree polynomial has a root in $\\mathbb R$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\n\\begin{corollary}\r\n If $f:[a,b]\\to\\mathbb R$ is continuous, and $f(a)=x,f(b)=y$, then for all $z$ in between $x,y$ there is some $c\\in [a,b]$ such that $f(c)=z$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n $g:x\\mapsto x-z$ is continuous.\r\n Then apply the preceding theorem.\r\n\\end{proof}\r\n\\begin{theorem}\r\n Continuous function on bounded closed interval is bounded and attains its bounds.\r\n\\end{theorem}\r\nAgain it is not true in general for open intervals.\r\n\\begin{proof}\r\n Suffice to show this for upper bounds.\r\n Let $f:[a,b]\\to\\mathbb R$ be continuous.\r\n Let $A=f([a,b])$.\r\n Suppose $A$ is not bounded above, then $\\forall n\\in\\mathbb N,\\exists x_n\\in[a,b],f(x_n)>n$.\r\n Then choose a convergent subsequence $(x_{n_k})$ of $(x_n)$, then $x_{n_k}\\to x\\in [a,b]$, but by continuity $f(x)=\\lim_{k\\to\\infty}f(x_k)\\ge\\lim_{k\\to\\infty}n_k\\to\\infty$, contradiction.\\\\\r\n Now that $A$ is bounded above, we choose $M=\\sup A$, then $\\forall n\\in\\mathbb N$, we can choose $x_n\\in [a,b]$ such that $M-1/n 0$. \n%and $\\widetilde{\\bcM}^{n}_{i}$ is realizable. \nIn Eq.~\\eqref{eq:widetildeM}, if $\\beta \\in [0,1]$, $\\widetilde{\\bcM}^{n}_{i}$ is expressed as a convex combination of $\\bcM_{i}^{n}$ and the expression in the square brackets on the right-hand side of Eq.~\\eqref{eq:widetildeM}. \nIt follows that $\\widetilde{\\bcM}^{n}_{i}$ is realizable if the expression inside the square brackets is realizable. \nIt can be shown (Lemma~2 in~\\cite{chu_etal_2018}) that the expression in square brackets is realizable for a distribution satisfying $f\\in[0,1]$. \nFor the two-moment model considered here, realizability depends on the algebraic closure. \nSpecifically, if $\\bcM_{i}^{n}$ is realizable and the Eddington factor satisfies the bounds in Eq.~\\eqref{eq:eddingtonFactorBounds}, then $\\bcM^{n+1}_{i}$ is realizable provided $\\beta \\in [0,1]$. \nThus, realizability of $\\bcM^{n+1}_{i}$ requires both a closure based on Fermi-Dirac statistics and a Courant-Friedrichs-Lewy (CFL) condition $\\dt\\le\\dx$. ", "meta": {"hexsha": "c80ebf0074ae5770f6aef8e74f5279443ea7ac3f", "size": 3943, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documents/M1/Astronum_2018/sections/SpatialDiscretization.tex", "max_stars_repo_name": "srichers/thornado", "max_stars_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-12-08T16:16:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-24T19:31:21.000Z", "max_issues_repo_path": "Documents/M1/Astronum_2018/sections/SpatialDiscretization.tex", "max_issues_repo_name": "srichers/thornado", "max_issues_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2019-07-10T20:13:15.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-11T13:21:00.000Z", "max_forks_repo_path": "Documents/M1/Astronum_2018/sections/SpatialDiscretization.tex", "max_forks_repo_name": "srichers/thornado", "max_forks_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2018-11-14T01:13:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-24T02:08:20.000Z", "avg_line_length": 77.3137254902, "max_line_length": 247, "alphanum_fraction": 0.6796855186, "num_tokens": 1446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438950986284991, "lm_q2_score": 0.7520125793176222, "lm_q1q2_score": 0.6346197297931168}} {"text": "\\chapter{Cumulative distribution function of the Durbin-Watson statistic}\n\nLet us consider the linear regression model \\[ y_i = \\beta_0 +\n\\beta_1x_{i1} + \\ldots + \\beta_kx_{ik} + \\xi_i, \\] $i = 1, \\ldots, n$,\n$k \\geq 1$, $n > k + 1$, where $\\xi_i$ are independent identically\ndistributed normal variables with zero mean, or $y = X\\beta + \\xi$. The\nleast squares estimate of $\\beta$ is $b = (X^TX)^{-1}X^Ty$ and the\nvector of residuals from the regression is $e = y - Xb$. The\nDurbin-Watson $d$ statistic is defined by \\[ d = \\frac{\\sum_{i = 2}^n\n (e_i - e_{i - 1})^2}{\\sum_{i = 1}^n e_i^2} \\] It is shown in\n\\cite[p.~426]{durbin-watson-1950} that $d_L \\leq d \\leq d_U$, where \\[\nd_L = \\frac{\\sum_{i = 1}^{n - k - 1} \\lambda_i \\zeta_i^2} {\\sum_{i =\n 1}^{n - k - 1} \\zeta_i^2}, \\] \\[ d_U = \\frac{\\sum_{i = 1}^{n - k -\n 1} \\lambda_{i + k} \\zeta_i^2} {\\sum_{i = 1}^{n - k - 1}\n \\zeta_i^2}, \\] \\[ \\lambda_i = 2\\left(1 - \\cos \\frac{\\pi i}{n}\\right),\n\\quad i = 1, \\ldots, n - 1, \\] and $\\zeta_1, \\ldots, \\zeta_{n - 1}$ are\nindependent standard normal variables. The cumulative distribution\nfunctions of $d_L$ and $d_U$ are\n\\begin{align} \\label{eq:dwcdfdL}\n \\Pr(d_L < x) &= \\Pr\\left(\\sum_{i = 1}^{n - k - 1} (x - \\lambda_i)\n \\zeta_i^2 > 0 \\right), \\\\\n \\label{eq:dwcdfdU}\n \\Pr(d_U < x) &= \\Pr\\left(\\sum_{i = 1}^{n - k - 1} (x - \\lambda_{i + k})\n \\zeta_i^2 > 0 \\right).\n\\end{align}\n\nIn both cases, the probabilities~(\\ref{eq:dwcdfdL})\nand~(\\ref{eq:dwcdfdU}) are of the form \\[ \\Pr\\left(\\sum_{i = 1}^m w_i\n\\zeta_i^2 > 0 \\right), \\] where $w_i$, $i = 1, \\ldots, m$, $m > 0$, are\nthose $x - \\lambda_i$ or $x - \\lambda_{i + k}$ which are not equal to 0.\nThey can be calculated by the formula (3.2) of\n\\cite[p.~422]{imhof-1961}, which states that if $w_i \\neq 0$, $i = 1,\n\\ldots, m$, then \\[ \\Pr\\left(\\sum_{i = 1}^m w_i \\zeta_i^2 > 0 \\right) =\n\\frac{1}{2} + \\frac{1}{\\pi} \\int_0^{+\\infty} f(u) du, \\] where\n\\begin{align*}\n f(u) &= \\frac{\\sin \\theta(u)}{u \\rho(u)}, \\\\\n \\theta(u) &= \\frac{1}{2} \\sum_{i = 1}^m \\arctan(w_iu), \\\\\n \\rho(u) &= \\prod_{i = 1}^m (1 + w_i^2u^2)^{1 / 4}.\n\\end{align*}\nIt is shown in (3.3) of \\cite[p.~423]{imhof-1961} that\n\\begin{align}\n \\nonumber\n \\lim_{u \\rightarrow 0} f(u) &= \\frac{1}{2} \\sum_{i = 1}^m w_i, \\\\\n \\nonumber\n \\lim_{u \\rightarrow \\infty} \\theta(u) &= \\frac{\\pi}{4}\n \\sum_{i = 1}^m \\sgn w_i, \\\\\n \\label{eq:dwcdftail}\n |\\frac{1}{\\pi} \\int_u^{+\\infty} f(u) du| & \\leq\n \\frac{1}{\\pi} \\frac{2}{m} u^{-m / 2} \\prod_{i = 1}^m |w_i|^{-1 / 2}.\n\\end{align}\n\nTo achieve the result $\\mathit{prob}$ and the accuracy $\\mathit{eps}$\nsuch that\n\\begin{equation} \\label{eq:dwcdfprob}\n \\mathit{prob} - \\mathit{eps} \\leq \\Pr \\left( \\sum_{i = 1}^m w_i\n \\zeta_i^2 > 0 \\right) < \\mathit{prob} + \\mathit{eps}\n\\end{equation}\nwe put\n\\begin{equation} \\label{eq:dwcdfupperlimit}\n u \\geq (\\frac{1}{\\pi} \\frac{2}{m} (\\prod_{i = 1}^m |w_i|^{-1 / 2}) /\n (\\mathit{eps} / 2))^{2 / m}\n\\end{equation}\nso that by~(\\ref{eq:dwcdftail}) we have \\[ |\\frac{1}{\\pi}\n\\int_u^{+\\infty} f(u) du| \\leq \\mathit{eps} / 2 \\] and then calculate\n\\begin{equation} \\label{eq:dwcdfintegral}\n \\frac{1}{\\pi} \\int_0^u f(u) du\n\\end{equation}\nby the Simpson method until two consecutive approximations differ no\nmore than $\\mathit{eps} / 2$. To have four significant digits after the\ncomma, we put $\\mathit{eps} = 0.000049$.\n\nThe function should not be used for low values of $n - k - 1$, say 1, 2,\n3, as integration~(\\ref{eq:dwcdfintegral}) fails. For $n - k - 1 = 1$ we\nhave\n\\begin{align*}\n \\Pr(d_L < x) = 0 & \\quad \\mathrm{for} \\quad x \\leq \\lambda_1, \\\\\n \\Pr(d_L < x) = 1 & \\quad \\mathrm{for} \\quad x > \\lambda_1, \\\\\n \\Pr(d_U < x) = 0 & \\quad \\mathrm{for} \\quad x \\leq \\lambda_{1 + k}, \\\\\n \\Pr(d_U < x) = 1 & \\quad \\mathrm{for} \\quad x > \\lambda_{1 + k}.\n\\end{align*}\nFor possible solution to this problem, see~\\cite{farebrother-1984}. See\nalso~\\cite{savin-white-1977}.\n", "meta": {"hexsha": "c7b5b6e5bbd0b8abc7b91a46b7716b2ac491c7ac", "size": 3872, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/drbnwtsn.tex", "max_stars_repo_name": "shgalus/shg", "max_stars_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2015-05-21T04:14:50.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-31T17:15:15.000Z", "max_issues_repo_path": "doc/drbnwtsn.tex", "max_issues_repo_name": "shgalus/shg", "max_issues_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2015-05-21T05:31:04.000Z", "max_issues_repo_issues_event_max_datetime": "2015-05-21T05:31:04.000Z", "max_forks_repo_path": "doc/drbnwtsn.tex", "max_forks_repo_name": "shgalus/shg", "max_forks_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-05-21T04:14:44.000Z", "max_forks_repo_forks_event_max_datetime": "2020-08-18T12:35:22.000Z", "avg_line_length": 46.6506024096, "max_line_length": 73, "alphanum_fraction": 0.5973657025, "num_tokens": 1694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355188, "lm_q2_score": 0.8221891261650248, "lm_q1q2_score": 0.6346013396545273}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\n\\section{Sheet 6}\n\n\\subsection{Sphere pole Riemann coordinates}\n\nRecall from section \\ref{sec:spherical-surface-curvature} the spherical surface metric in the coordinates \\((\\theta, \\varphi )\\): \n%\n\\begin{subequations}\n\\begin{align}\n g_{\\mu \\nu } = \\left[\\begin{array}{cc}\n R^2 & 0 \\\\ \n 0 & R^2 \\sin^2 \\theta \n \\end{array}\\right]\n\\,,\n\\end{align}\n\\end{subequations}\n%\nand the Christoffel symbols, which are all zero except for \\(\\Gamma^{0}_{11}= - \\sin \\theta \\cos \\theta \\) and \\(\\Gamma^{1}_{01} = \\Gamma^{1}_{10} = 1/ \\tan \\theta \\).\n\nWe can plug these into the geodesic equation \\(u^{\\mu } \\nabla_{\\mu } u^{\\nu }\\), which comes out to be \n%\n\\begin{subequations}\n\\begin{align}\n \\ddot{\\theta } &= + \\dot{\\varphi }^2 \\sin \\theta \\cos \\theta \\\\\n \\ddot{\\varphi } &= - 2 \\frac{\\dot{\\varphi } \\dot{\\theta } }{\\tan(\\theta )} \n\\,,\n\\end{align}\n\\end{subequations}\n%\nfor a trajectory \\((\\theta (s), \\varphi (s))\\) with velocity \\((\\dot{\\theta }, \\dot{\\varphi })\\): dots denote derivatives with respect to \\(s\\).\n\nNow, we want to check whether parallels and meridians are geodesics. First of all, we want to choose a parameter such that the velocity is of constant norm \\(1\\): the equation to satisfy is \n%\n\\begin{subequations}\n\\begin{align}\n R^2 \\left[\\begin{array}{cc}\n \\dot{\\theta} & \\dot{\\varphi}\n \\end{array}\\right]\n \\left[\\begin{array}{cc}\n 1 & 0 \\\\ \n 0 & \\sin^2 \\theta \n \\end{array}\\right]\n \\left[\\begin{array}{c}\n \\dot{\\theta} \\\\ \n \\dot{\\varphi}\n \\end{array}\\right]\n \\equiv 1\n\\,,\n\\end{align}\n\\end{subequations}\n%\nor \\(\\dot{\\theta}^2 + \\dot{\\varphi}^2 \\sin^2 \\theta = R^{-2}\\).\nMeridians have constant \\(\\varphi \\): for them, then, \\(\\dot{\\theta}^2 = R^{-2} \\), so an appropriate parametrization is \\((\\theta(s), \\varphi(s)) = (s/R, \\varphi_0 )\\).\nParallels have constant \\(\\theta \\): by an analogous line of reasoning, we parametrize them as \\((\\theta (s), \\varphi (s) )= (\\theta_0, s / R \\sin\\theta )\\). \n\nOne can readily check that for meridians both of the geodesic equations are identities, while for parallels the second one is an identity but the first reads \\(0= \\cos \\theta / \\sin \\theta \\): it can only be satisfied if \\(\\theta = \\pi /2\\).\nThis makes sense: geodesics on a sphere are great circles, and the only parallel which is a great circle is the equator. \n\n\\subsubsection{Riemann coordinates}\n\nThe coordinates we wish to use are \n%\n\\begin{subequations}\n\\begin{align}\n x &= R \\theta \\cos \\varphi \\\\\n y &= R \\theta \\sin \\varphi \n\\,.\n\\end{align}\n\\end{subequations}\n%\n% whose tangent vectors at the north pole (i.e. \\(\\theta = 0\\)) are:\n% %\n% \\begin{subequations}\n% \\begin{align}\n% \\dv[]{}{x} &= R \\cos \\varphi \\dv[]{}{\\theta } \n% - R \\theta \\sin \\varphi \\dv[]{}{\\varphi } \n% = R \\cos \\varphi \\dv[]{}{\\theta } \\\\ \n% \\dv[]{}{y} &= R \\sin \\varphi \\dv[]{}{\\theta } \n% + R \\theta \\cos \\varphi \\dv[]{}{\\varphi }\n% \\,,\n% \\end{align}\n% \\end{subequations}\n% %\n\nThey are in the form \\(x^{\\alpha } = \\theta n^{\\alpha }\\), for vectors \\(n^{\\alpha } = R (\\cos \\varphi, \\sin \\varphi )\\).\nAn orthonormal basis for these vectors can be found by selecting \\(\\varphi = 0, \\pi /2\\). \n\nAs we have shown before, the coordinates \\(x^{\\alpha }\\) describe geodesics if we consider them for fixed \\(\\varphi \\) and with parameter \\(\\theta \\), since they are meridians.\n\n\\subsubsection{Metric computation}\n\nThe metric transforms as \n%\n\\begin{align}\n g_{\\mu \\nu }^{\\prime } \n = \\pdv{x^{\\alpha }}{x^{\\prime \\mu }}\n \\pdv{x^{\\beta }}{x^{\\prime \\nu }}\n g_{\\alpha \\beta }\n\\,,\n\\end{align}\n%\nso we need the inverse Jacobian, which is expressed in terms of the new coordinates \\(x^{\\prime \\mu } = (x, y)\\) and the old ones \\(x^{\\mu } = (\\theta , \\varphi )\\):\n%\n\\begin{subequations}\n\\begin{align}\n \\pdv{x^{\\alpha }}{x^{\\prime \\mu }}\n = \\left[\\begin{array}{cc}\n \\frac{1}{R} \\frac{x}{\\sqrt{x^2 + y^2}} & \n \\frac{1}{R} \\frac{y}{\\sqrt{x^2 + y^2}} \\\\ \n \\frac{-y / x^2}{1 + (y/x)^2} & \n \\frac{1/x}{1 + (y/x)^2}\n \\end{array}\\right]\n\\,,\n\\end{align}\n\\end{subequations}\n%\nso we get, using the expansion \\(\\sin^2\\theta \\sim \\theta^2 - \\theta^{4} /3 + O (\\theta^{6})\\) and the identification \\(R^2\\theta^2 = x^2+y^2\\):\n%\n\\begin{subequations}\n\\begin{align}\n g_{00}^{\\prime } &= R^2 \\qty(\\pdv{x^{0}}{x^{\\prime 0}})^2\n + R^2 \\sin^2 \\theta \\qty(\\pdv{x^{1}}{x^{\\prime 0}})^2 \\\\\n &= \\frac{x^2}{x^2+y^2} + R^2 \\sin^2 \\theta \\frac{y^2}{x^{4} \\qty(1 + (y/x)^2)^2} \\\\\n &=\\frac{1}{x^2+y^2} \\qty(x^2 + \\frac{R^2 \\sin^2\\theta y^2}{x^2+y^2}) \\\\\n &= 1 - \\frac{y^2}{3R^2} + O((x^2+y^2)y^2)\n\\,,\n\\end{align}\n\\end{subequations}\n%\nwhile for the off-diagonal elements \\(g_{01}^{\\prime } = g_{10}^{\\prime }\\): \n%\n\\begin{subequations}\n\\begin{align}\n g_{01}^{\\prime } &= R^2 \\pdv{x^{0}}{x^{\\prime 0}} \\pdv{x^{0}}{x^{\\prime 1}} \n + R^2 \\sin^2 \\theta \\pdv{x^{1}}{x^{\\prime 0}} \\pdv{x^{1}}{x^{\\prime 1}} \\\\\n &= \\frac{xy}{x^2+y^2} + R^2 \\sin^2\\theta \\qty(- \\frac{xy}{(x^2+y^2)^2}) \\\\\n &= \\frac{xy}{x^2+y^2} \\qty(1 - \\frac{R^2 \\sin^2\\theta }{x^2+y^2}) \\\\\n &= \\frac{xy}{x^2+y^2} \\qty(\\frac{(x^2+y^2)^2}{3R^2 (x^2+y^2)}) + O(x^2+y^2) \\\\\n &= \\frac{xy}{3R^2} + O(x^2+y^2)\n\\,,\n\\end{align}\n\\end{subequations}\n%\nand lastly for the element \\(g_{11}^{\\prime }\\): \n%\n\\begin{subequations}\n\\begin{align}\n g_{11}^{\\prime } &= R^2 \\qty(\\pdv{x^{0}}{x^{\\prime 1}})^2\n + R^2 \\sin^2 \\theta \\qty(\\pdv{x^{1}}{x^{\\prime 1}})^2 \\\\\n &= \\frac{y^2}{x^2+y^2} + R^2 \\sin^2 \\theta \\frac{1}{x^{2} \\qty(1 + (y/x)^2)^2} \\\\\n &= \\frac{1}{x^2+y^2} \\qty(y^2 + R^2 \\sin^2\\theta \\frac{x^2}{x^2+y^2}) \\\\\n &= 1 - \\frac{x^2}{3R^2} + O ((x^2+y^2) x^2)\n \\,.\n\\end{align}\n\\end{subequations}\n\nAt the north pole \\(x=y=0\\), so there \\(g^{\\prime }_{\\mu \\nu }= \\delta_{\\mu \\nu }\\), and all the first derivatives calculated there vanish since there are no first order terms.\n\n\\subsubsection{Scalar curvature calculation}\n\nThe expression we have for the scalar curvature in a LIF is given in equation \\eqref{eq:ricci-scalar-LIF}.\n\nWe can evaluate it for \\(g_{\\mu \\nu }^{\\prime }\\). Do note that the non-differentiated metric can be identified with the identity, and derivatives with upper and lower indices are the same. So, we get:\n%\n\\begin{align}\n R _{\\text{Ric}} = \\tensor{g}{_{\\alpha \\nu , }^{\\alpha \\nu }} - \\delta^{\\mu \\nu } \\square g_{\\mu \\nu }\n\\,,\n\\end{align}\n%\nwhere the only nonvanishing terms are: \n%\n\\begin{align}\n \\tensor{g}{_{\\alpha \\nu,}^{\\alpha \\nu }} = 2 \\pdv[2]{}{x}{y} \\qty(\\frac{xy}{3R^2}) = \\frac{2}{3R^2}\n\\,\n\\end{align}\n%\nand what was denoted as the Dalambertian before is just the Laplacian: \\(\\square = \\delta^{\\mu \\nu }\\partial_{\\mu } \\partial_{\\nu }= \\partial^2_{xx} + \\partial^2_{yy}\\):\n%\n\\begin{align}\n \\delta^{\\mu \\nu } \\square g_{\\mu \\nu }\n = \\partial_{xx}^2 g_{11} + \\partial_{yy}^2 g_{00}\n = - 2 \\frac{2}{3R^2} \n\\,,\n\\end{align}\n%\nso in the end we get \n%\n\\begin{align}\n R _{\\text{Ric}} = \\frac{2}{3R^2} + \\frac{4}{3R^2} = \\frac{2}{R^2}\n\\,,\n\\end{align}\n%\nwhich means that the curvature decreases as the radius increases, as we might expect.\n\n\\subsection{Schwarzschild metric curvature}\n\n\\subsubsection{Christoffel symbols}\n\nThe computation is tedious and not particularly enlightening: we start from the metric\n\\footnote{There is a typo in the homework assignment: the coefficients are written as functions of time.}\n%\n\\begin{align}\n \\dd{s^2} = - A(r) \\dd{t^2}\n + B(r) \\dd{r^2}\n + r^2 \\qty(\\dd{\\theta^2} + \\sin^2\\theta \\dd{\\varphi^2})\n\\,,\n\\end{align}\n%\nand compute the Christoffel symbols with the usual formula: \n%\n\\begin{align}\n \\Gamma^{\\mu }_{\\nu \\rho } =\n \\frac{1}{2} g^{\\mu \\alpha }\\qty(g_{\\alpha \\nu , \\rho }\n + g_{\\alpha \\rho, \\nu } - g_{\\nu \\rho , \\alpha })\n\\,,\n\\end{align}\n%\nwhere fortunately, since the metric is diagonal, we only need to compute one term in the sum (that is, \\(\\mu \\equiv \\alpha \\) always), and the two indices in the metric must be equal in order for the term to not vanish. \n\nThe metric only depends on \\(\\theta \\) and \\(r\\), so any derivatives with respect to \\(t\\) and \\(\\varphi \\) are to be discarded. \n\nWith this out of the way, we start computing the \\(40\\) independent symbols and find that the nonzero ones are (denoting differentiation with respect to \\(r\\) with a prime): \n%\n\\begin{subequations} \\label{eq:schwarzschild-christoffel}\n\\begin{align}\n \\Gamma^{t}_{rt} &= \\frac{A^{\\prime }}{2A} &\n \\Gamma^{r}_{rr} &= \\frac{B^{\\prime }}{2B} \\\\\n \\Gamma^{r}_{tt} &= \\frac{A^{\\prime }}{2B} &\n \\Gamma^{\\theta }_{r \\theta } &= \\frac{1}{r} \\\\\n \\Gamma^{r}_{\\theta \\theta } &= -\\frac{r}{B} &\n \\Gamma^{\\varphi }_{r \\varphi } &= \\frac{1}{r} \\\\\n \\Gamma^{r}_{\\varphi \\varphi } &= - \\frac{r}{B} \\sin^2 \\theta &\n \\Gamma^{\\theta }_{\\varphi \\varphi } &= - \\sin \\theta \\cos \\theta \\\\\n \\Gamma^{\\varphi }_{\\theta \\varphi } &= \\frac{\\cos \\theta }{\\sin \\theta }\n\\,.\n\\end{align}\n\\end{subequations}\n\n\\subsubsection{Ricci component}\n\nWe want to compute \\(R^{\\mu }_{t \\mu t }\\), and in order to do so we must find the three components \\(R^{i}_{tit}\\) with varying \\(i\\) and sum them (here \\(i = r, \\theta,\\varphi \\)), since \\(R^{t}_{ttt}\\) vanishes by antisymmetry. \n\nIn general we have: \n%\n\\begin{subequations}\n\\begin{align}\n R^{i}_{tit} &= 2 \\qty( \\Gamma^{i}_{[t| t, |i]} + \\Gamma^{\\alpha }_{t[t} \\Gamma^{i}_{i] \\alpha }) \\\\\n &= \\Gamma^{i}_{tt, i} - \\cancelto{}{\\Gamma^{i}_{it,t}}\n + \\Gamma^{\\alpha }_{tt} \\Gamma^{i}_{i\\alpha }\n - \\Gamma^{\\alpha }_{ti} \\Gamma^{i}_{t\\alpha }\n\\,.\n\\end{align}\n\\end{subequations}\n\nNote that the index \\(i\\) is consider not to be summed here, we are writing a formula for the components of the Riemann tensor; although the expression holds when summing over \\(i\\) as well. \n\nSo, we can compute this for the specific values of \\(i\\): for \\(i=r\\) we have \n%\n\\begin{subequations}\n\\begin{align}\n R^{r}_{trt} &= \\Gamma^{r}_{tt, r} + \\Gamma^{\\alpha}_{tt} \n \\Gamma^{r}_{r \\alpha } - \\Gamma^{\\alpha }_{tr} \\Gamma^{r}_{t \\alpha } \\\\\n &= \\qty(\\frac{A^{\\prime }}{2B})^{\\prime } + \\frac{A^{\\prime }}{2B} \\frac{B^{\\prime }}{2B} - \\frac{A^{\\prime }}{2A} \\frac{A^{\\prime }}{2B} \\\\\n &= \\frac{A^{\\prime \\prime}}{2B} - \\frac{A^{\\prime }}{2B^2} B^{\\prime } + \\frac{A^{\\prime }}{4B} \\qty(\\frac{B^{\\prime }}{B} - \\frac{A^{\\prime}}{A}) \\\\ \n &= \\frac{A^{\\prime \\prime}}{2B} - \\frac{A^{\\prime }}{4B} \\qty(\\frac{A^{\\prime}}{A} + \\frac{B^{\\prime }}{B})\n \\,,\n\\end{align}\n\\end{subequations}\n%\nfor \\(i= \\theta \\) instead \n%\n\\begin{subequations}\n\\begin{align}\n R^{\\theta }_{t \\theta t} &= \\cancelto{}{\\Gamma^{\\theta}_{tt, \\theta}} + \\Gamma^{\\alpha}_{tt} \n \\Gamma^{\\theta}_{\\theta \\alpha } - \\cancelto{}{\\Gamma^{\\alpha }_{t\\theta} \\Gamma^{i}_{t \\alpha }} \\\\\n &= \\frac{A^{\\prime }}{2B} \\frac{1}{r}\n \\,,\n\\end{align}\n\\end{subequations}\n%\nand for \\(i = \\varphi \\): \n%\n\\begin{subequations}\n \\begin{align}\n R^{\\varphi }_{t \\varphi t} &= \\cancelto{}{\\Gamma^{\\varphi}_{tt, \\varphi}} + \\Gamma^{\\alpha}_{tt} \n \\Gamma^{\\varphi}_{\\varphi \\alpha } - \\cancelto{}{\\Gamma^{\\alpha }_{t\\varphi} \\Gamma^{i}_{t \\alpha }} \\\\\n &= \\frac{A^{\\prime }}{2B} \\frac{1}{r}\n \\,,\n \\end{align}\n\\end{subequations}\n%\nso our final solution is \n%\n\\begin{subequations}\n\\begin{align}\n R_{00} = \\sum_i R^{i}_{0i0} &= \\frac{A^{\\prime \\prime}}{2B} \n - \\frac{A^{\\prime }}{4B} \\frac{(AB)^{\\prime }}{AB} + \\frac{A^{\\prime }}{Br} \\\\\n &= \\frac{A''}{2B} - \\frac{A^{\\prime 2} }{4AB} - \\frac{A' B' }{4B^2} + \\frac{A'}{Br}\n\\,.\n\\end{align}\n\\end{subequations}\n%\n\n\\subsection{Schwarzschild geometry orbits}\n\nThe derivation up to the equation for the perturbed orbit equation is documented in the lecture notes, I might copy it here later, but for now one can find it there.\n\nDuring the lecture we got up to the first order equation for the perturbation \\(w\\) for the orbit \\(u\\), written in the form \\(u(\\varphi ) = u_{c} \\qty(1+w(\\varphi ))\\): \n%\n\\begin{align}\n \\dv[2]{w}{\\varphi } = (6GMu_c-1) w\n\\,,\n\\end{align}\n%\nwhich is in the form \\(\\ddot{w} + \\omega^2 w = 0\\), for \\(\\omega^2 = 1- 6GMu_c\\). Now, we know that the first order equation must be complemented by the zeroth order one: \n%\n\\begin{align}\n u_c = \\frac{GM}{l^2} + 3GM u_c^{2}\n\\,,\n\\end{align}\n%\nwhich can be solved for \\(u_c\\) to yield: \n%\n\\begin{align}\n u_c = \\frac{1 \\pm \\sqrt{1 - 3 \\times 4 \\frac{G^2 M^2}{l^2}}}{6GM}\n\\,,\n\\end{align}\n%\ntherefore the square angular velocity of the perturbation's evolution is: \n%\n\\begin{align}\n \\omega^2 = 1 - \\cancelto{}{6GM} \\qty(\\frac{1 \\pm \\sqrt{1 - 12 \\frac{G^2 M^2}{l^2}}}{ \\cancelto{}{6GM}})\n = \\pm \\sqrt{1 - 12 \\frac{G^2 M^2}{l^2}}\n\\,.\n\\end{align}\n\nThe solution with the minus sign has no meaning for us, since the solution we want to consider must be stable, with positive \\(\\omega^2\\). So, the angular velocity is \n%\n\\begin{align}\n \\omega = \\qty(1 - 12 \\frac{G^2M^2}{l^2})^{1/4}\n\\,,\n\\end{align}\n%\nand we know that angular velocity and period are related by \\(T = 2 \\pi / \\omega \\): therefore we get \n%\n\\begin{align}\n T = 2 \\pi \\qty(1 - 12 \\frac{G^2M^2}{l^2})^{-1/4}\n\\,,\n\\end{align}\n%\nwhich we can Taylor expand: at \\(x=0\\) we have \n%\n\\begin{align}\n (1-12x)^{-1/4} = 1 - \\frac{1}{4} (1-12 \\times 0)^{-5/4} (-12x) + O(x^2) = 1 + 3x + O(x^2 )\n\\,.\n\\end{align}\n%\n\nTherefore: \n%\n\\begin{align}\n T = 2 \\pi \\qty(1 + 3 \\qty(\\frac{GM}{l})^2) + O\\qty(\\qty(\\frac{GM}{l})^{4})\n\\,,\n\\end{align}\n%\nwhich is approximately \\(2 \\pi \\) as we should expect: the Newtonian approximation is \\(l \\gg GM\\), and Newtonian orbits have a period of exactly \\(2 \\pi \\). Then we can read off the first-order correction directly from the first term in the expansion: it is \n%\n\\begin{align}\n \\delta \\varphi = 6 \\pi \\qty(\\frac{GM}{l})^2\n\\,.\n\\end{align}\n%\n\nThe \\(M\\) here is the mass of the central object, while \\(l\\) is the angular momentum of the orbit. \n\n\n\n\\end{document}", "meta": {"hexsha": "9539bce43b56fe69c5bfd9bbe7ad637d6865538d", "size": 13628, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ap_first_semester/gr_exercises/sheet6.tex", "max_stars_repo_name": "jacopok/notes", "max_stars_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-10-10T13:10:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T14:52:50.000Z", "max_issues_repo_path": "ap_first_semester/gr_exercises/sheet6.tex", "max_issues_repo_name": "jacopok/notes", "max_issues_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ap_first_semester/gr_exercises/sheet6.tex", "max_forks_repo_name": "jacopok/notes", "max_forks_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-03T16:20:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-06T16:11:07.000Z", "avg_line_length": 35.8631578947, "max_line_length": 259, "alphanum_fraction": 0.603977106, "num_tokens": 5244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189134878876, "lm_q2_score": 0.7718434925908525, "lm_q1q2_score": 0.6346013334351631}} {"text": "%%\n%% Hyperbolic Trig Identities\n%%\n%% Author: William Gao\n%% 09/04/2019\n%%\n\n\\documentclass[12pt]{article}\n\\setlength{\\parindent}{0pt} % disable indent globally\n\n\\RequirePackage[\n letterpaper,\n left=0.8in,\n right=0.8in,\n top=0.70in,\n bottom=0.55in,\n nohead\n]{geometry}\n\n% Packages\n\\usepackage{amssymb}\n\\usepackage[fleqn]{amsmath}\n\\usepackage{hyperref}\n\\hypersetup{\n colorlinks=true,\n urlcolor=blue\n}\n\n\\newcommand{\\hzline}{\\noindent\\makebox[\\linewidth]{\\rule{0.8\\paperwidth}{0.4pt}}}\n\n\\title{Hyperbolic Trig Identities}\n%\\author{math2.org}\n\\author{Version 1.1}\n\\date{September 4, 2019}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Introduction}\nA compiled list of the most commonly used hyperbolic trig identities. Enjoy. \\\\\n\nAll credits goes to \\href{http://math2.org/math/trig/hyperbolics.htm}{math2.org}. If you have comments, corrections, or clarifications, please submit an issue or pull request on \\href{https://github.com/chemclub/calculus}{GitHub}.\n\n\\large\n\\section{Hyperbolic Definitions}\n\n\\begin{align*}\n& \\sinh x = \\frac{e^x-e^{-x}}{2} \\\\\n& \\operatorname{csch}x = \\frac{1}{\\sinh x} = \\frac{2}{e^x-e^{-x}}\n\\end{align*}\n\n\\begin{align*}\n& \\cosh x = \\frac{e^x+e^{-x}}{2} \\\\\n& \\operatorname{sech}x = \\frac{1}{\\cosh x} = \\frac{2}{e^x+e^{-x}}\n\\end{align*}\n\n\\begin{align*}\n& \\tanh x = \\frac{\\sinh x}{\\cosh x} = \\frac{e^x-e^{-x}}{e^x+e^{-x}} \\\\\n& \\operatorname{coth}x = \\frac{1}{\\tanh x} = \\frac{e^x+e^{-x}}{e^x-e^{-x}}\n\\end{align*}\n\n\n\\begin{align*}\n& \\cosh^2x-\\sinh^2x = 1 \\\\\n& \\tanh^2x+\\operatorname{sech}^2x = 1 \\\\\n& \\coth^2x-\\operatorname{csch}^2x = 1\n\\end{align*}\n\n\n\\section{Inverse Hyperbolic Definitions}\n\n\\begin{align*}\n& \\operatorname{arcsinh}z=\\ln\\left(z+\\sqrt{z^2+1}\\right) \\\\\n& \\operatorname{arccosh}z=\\ln\\left(z\\pm\\sqrt{z^2-1}\\right) \\\\\n& \\operatorname{arctanh}z=\\frac{1}{2}\\ln\\left(\\frac{1+z}{1-z}\\right)\n\\end{align*}\n\n\\begin{align*}\n& \\operatorname{arccsch}z=\\ln\\left(\\frac{1+\\sqrt{1+z^2}}{z}\\right) \\\\\n& \\operatorname{arcsech}z=\\ln\\left(\\frac{1+\\sqrt{1-z^2}}{z}\\right) \\\\\n& \\operatorname{arccoth}z=\\frac{1}{2}\\ln\\left(\\frac{z+1}{z-1}\\right)\n\\end{align*}\n\n\n\\section{Relations to Trigonometric Functions}\n\n\\begin{align*}\n& \\sinh z=-i\\sin\\left(iz\\right) \\\\\n& \\operatorname{csch}z=i\\csc\\left(iz\\right)\n\\end{align*}\n\n\\begin{align*}\n& \\cosh z=\\cos\\left(iz\\right) \\\\\n& \\operatorname{sech}z=\\sec\\left(iz\\right)\n\\end{align*}\n\n\\begin{align*}\n& \\tanh z=-i\\tan\\left(iz\\right) \\\\\n& \\coth z=i\\cot\\left(iz\\right)\n\\end{align*}\n\n\\end{document}\n\n", "meta": {"hexsha": "e556eef08bd6077ad8dccfb775703ef410ac664d", "size": 2481, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "references/hyperbolic.tex", "max_stars_repo_name": "w-gao/calculus", "max_stars_repo_head_hexsha": "6048f127e4ff42ad7b03885a60c6025ae84e5a87", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "references/hyperbolic.tex", "max_issues_repo_name": "w-gao/calculus", "max_issues_repo_head_hexsha": "6048f127e4ff42ad7b03885a60c6025ae84e5a87", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "references/hyperbolic.tex", "max_forks_repo_name": "w-gao/calculus", "max_forks_repo_head_hexsha": "6048f127e4ff42ad7b03885a60c6025ae84e5a87", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.6285714286, "max_line_length": 230, "alphanum_fraction": 0.6545747682, "num_tokens": 985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680199891789, "lm_q2_score": 0.8539127585282744, "lm_q1q2_score": 0.6346006539989556}} {"text": "\\documentclass{article}\n\n\\usepackage{geometry}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\n\\newcommand{\\n}{^{\\mathrm{min}}}\n\\newcommand{\\x}{^{\\mathrm{max}}}\n\n\\begin{document}\n\n\\section{Timers}\n\nThese host-ticks timers are a bit weird. In fact, originally I had\nthem \\emph{wrong}.\n\nLet $i\\n$ be the minimum interval, $i\\x$ the maximum. $t_j$ is a\nticker value, $r_j$ is a remainder, $d_j$ is a duration of time\npassage, $\\delta_j$ is the amount by which the ticker advances.\n\nConsider\n\\[\n\\mathrm{Ticker}(t_0,r_0)\\xrightarrow{d_1}\n\\mathrm{Ticker}(t_1,r_1)\\xrightarrow{d_2}\n\\mathrm{Ticker}(t_2,r_2)\n\\]\nwhere $t_1 = t_0+\\delta_1$, $t_2=t_1+\\delta_2 =\nt_0+\\delta_1+\\delta_2$.\n\nThe original rule was:\n\\[\n\\exists i_j.\\;\ni\\n\\leq i_j \\wedge i_j\\leq i\\x \\wedge\n\\delta_j = \\left\\lfloor \\frac{r_{j-1}+d_j}{i_j} \\right\\rfloor \\wedge\nr_j = r_{j-1}+d_j - \\delta_j i_j\n\\]\n\nA crucial property is the following (roughly \\textbf{S1} and\n\\textbf{S2} in Lynch and Vaandrager 1996), which I shall call\nadditivity:\n\\[\n\\mathrm{Ticker}(t_0,r_0)\\xrightarrow{d_1}\n\\mathrm{Ticker}(t_1,r_1)\\xrightarrow{d_2}\n\\mathrm{Ticker}(t_2,r_2)\n\\Leftrightarrow\n\\mathrm{Ticker}(t_0,r_0)\\xrightarrow{d_1+d_2}\n\\mathrm{Ticker}(t_2,r_2)\n\\]\n\nThis does not hold for the original rule! If $d_1 > d_2$, and we\nchoose $i_1=i\\n$ and $i_2=i\\x$, it can happen that even though $r_2 <\ni_2$, it is the case that $r_2 > i\\x_{12}$, where $i\\x_{12}$ is the\ngreatest interval consistent with $t_2$, i.e., $i\\x_{12} =\n\\frac{r_0+d_1+d_2-r_2}{\\delta_1+\\delta_2}$. This means that no\n$i_{12}$ satisfying the condition of the rule exists.\n\nFor example: let $r_0 = 14$, $d_1=364$, $d_2=141$, $i_1=56$,\n$i_2=112$. Then $\\delta_1=6$, $r_1=42$, $\\delta_2=1$, $r_2=71$. But\nto achieve $r_{12}=71/56$ we must have\n$\\delta_{12}i_{12}=r_0+d_{12}-r_{12}=448$; and since $\\delta_{12}=7$,\nwe must have $i_{12}=64$. But $\\delta_{12}=\\left\\lfloor \\frac{519}{64}\n\\right\\rfloor = 8 \\neq 7$, which is a CONTRADICTION.\n\nA new possible rule is as follows:\n\\[\n\\begin{array}{r@{}c@{}l}\n\\exists \\delta\\n_j \\delta\\x_j.\\;\\\\\n\\delta\\n_j &=& \\left\\lfloor \\frac{r_{j-1}+d_j}{i\\x} \\right\\rfloor \\wedge\\\\[1ex]\n\\delta\\x_j &=& \\left\\lfloor \\frac{r_{j-1}+d_j}{i\\n} \\right\\rfloor \\wedge\\\\\n\\delta\\n_j &{}\\leq \\delta_j\\leq{}&\\delta\\x_j \\wedge\\\\\nr_{j-1}+d_j-\\delta_j i\\x &{}\\leq r_j \\leq {}&r_{j-1}+d_j-\\delta_j i\\n \\wedge\\\\\n0 &{}\\leq r_j <{}& i\\x\n\\end{array}\n\\]\nwhere the $\\delta\\n_j, \\delta\\x_j$ are now naturals as well as\n$\\delta_j$ (note the strict inequality on $r_j < i\\x$ is still\nnecessary for the model).\n\nDoes this have the desired properties?\n\nArgh, I'm an idiot! The reason I was having trouble proving the\nproperties was that the above rule is overconstrained. It is\nsufficient to require:\n\\[\n\\begin{array}{r@{}c@{}l}\nr_{j-1}+d_j-\\delta_j i\\x &{}\\leq r_j \\leq {}&r_{j-1}+d_j-\\delta_j i\\n \\wedge\\\\\n0 &{}\\leq r_j <{}& i\\x\n\\end{array}\n\\]\nThe following results follow:\n\\begin{itemize}\n\\item $\\delta_j$ lies in the range $[\\left\\lfloor\n\\frac{r_{j-1}+d_j}{i\\x} \\right\\rfloor,\\left\\lfloor\n\\frac{r_{j-1}+d_j}{i\\n} \\right\\rfloor]$.\n\\item $r_j$ can then be determined\nfrom the inequations above---it is usually free in $[0,i\\x]$, but when\n$\\delta_j$ is chosen at or near the boundaries, it is restricted by\nthe other inequation.\n\\item The additivity property holds (this is trivial to verify: add\nthe first inequations for $j=1$ and $j=2$, and subtract $r_1$\nthrough).\n\\end{itemize}\n\n\n\\end{document}\n", "meta": {"hexsha": "69fefd68ecfd5730c9543e9add68d03f3596ff97", "size": 3411, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "unmaintained/Net/TCP/Notes1/timers.tex", "max_stars_repo_name": "rems-project/netsem", "max_stars_repo_head_hexsha": "0b02450eee334fc42f74ad3da09d3ce1a0698a48", "max_stars_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_stars_count": 34, "max_stars_repo_stars_event_min_datetime": "2018-02-26T20:37:44.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-24T01:49:55.000Z", "max_issues_repo_path": "unmaintained/Net/TCP/Notes1/timers.tex", "max_issues_repo_name": "PeterSewell/netsem", "max_issues_repo_head_hexsha": "0b02450eee334fc42f74ad3da09d3ce1a0698a48", "max_issues_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "unmaintained/Net/TCP/Notes1/timers.tex", "max_forks_repo_name": "PeterSewell/netsem", "max_forks_repo_head_hexsha": "0b02450eee334fc42f74ad3da09d3ce1a0698a48", "max_forks_repo_licenses": ["BSD-2-Clause-FreeBSD"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-04-24T15:26:34.000Z", "max_forks_repo_forks_event_max_datetime": "2017-12-12T23:39:13.000Z", "avg_line_length": 32.179245283, "max_line_length": 79, "alphanum_fraction": 0.683963647, "num_tokens": 1347, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127529517044, "lm_q2_score": 0.7431680199891789, "lm_q1q2_score": 0.634600649854627}} {"text": "\\section{Learning spectral clustering}\n\\label{ch:jordan04}\n\n\\textit{Learning spectral clustering} by Francis R. Bach and Michael I. Jordan. \\\\\nCited by 313. \\textit{Advances in Neural Information Processing Systems 16 (2004)}.\n\\newline\n\n\\textbf{Main point} is that the author derived \\begin{inparaenum}[\\itshape a\\upshape)]\n\\item a cost function for spectral clustering that minimize the error between a given partition and the minimum normalized cut\n\\item with respect to two algorithms one for spectral clustering and one for learning the similarity matrix.\n\\end{inparaenum}.\n\n\\subsection{Objective way of similarity measure}\nA general framework is provided for learning the similarity matrix for spectral clustering. New cost function $J(W,e)$ that characterizes how close the eigen structure of a similarity matrix $W$ is to a partition $e$. \n%\n%\\subsection{Spectral clustering}\n%Given a dataset $I$ of $P$ points in space $X$ and a $P \\times P$ similarity matrix $W$ that measures the similarity between the $P$ points. \n\n\\subsection{Learning the similarity matrix}\n\n\\begin{figure}[ht]\n\\begin{mdframed}\n\\begin{description}\n\\item[Assumption] \\hfill \\\\\ngiven $N$ datasets $D_n, n \\in \\{ 1, \\cdots, N \\}$ of points in $R^F$ where $D_n$ is composed of $P_n$ points $x_{np}, p \\in \\{ 1, \\cdots, P_n \\}$.\n\\item[Step 1] \\hfill \\\\\nfor each $n$ we know the partition $e_n$, so that target matrix $ \\Pi (e_n, \\alpha)$ can be computed from each dataset.\n\\item[Step 2] \\hfill \\\\\nfor each $n$, we have a similarity matrix $W_n (\\alpha)$\n\\item[Step 3] \\hfill \\\\\nThe cost function $H(\\alpha) = \\frac{1}{N} \\sum_n F( W_n(\\alpha), \\Pi_0 (e_n, \\alpha)) + C || \\alpha ||_1$.\n\\item[Step 4] \\hfill \\\\\nThe learning algorithm is the minimization of $H(\\alpha)$ with respect to $\\alpha \\in R_{+}^F$ using the method of conjugate gradient with line search.\n\\end{description}\n\\end{mdframed}\n\\caption{Learning similarity matrix}\n\\end{figure}\n\n", "meta": {"hexsha": "29e7738bee75f13e90f17a729b424cf7f7c3f298", "size": 1921, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/references/reference_research/jordan04.tex", "max_stars_repo_name": "wsgan001/AnomalyDetection", "max_stars_repo_head_hexsha": "397673dc6ce978361a3fc6f2fd34879f69bc962a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/references/reference_research/jordan04.tex", "max_issues_repo_name": "wsgan001/AnomalyDetection", "max_issues_repo_head_hexsha": "397673dc6ce978361a3fc6f2fd34879f69bc962a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/references/reference_research/jordan04.tex", "max_forks_repo_name": "wsgan001/AnomalyDetection", "max_forks_repo_head_hexsha": "397673dc6ce978361a3fc6f2fd34879f69bc962a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-03-16T21:50:52.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-16T21:50:52.000Z", "avg_line_length": 49.2564102564, "max_line_length": 218, "alphanum_fraction": 0.7319104633, "num_tokens": 550, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127455162773, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.6346006443288555}} {"text": "% !TEX root = as_grf_sopt.tex\n\n\\subsection{DISCUSSIONS}\n\n\nOur approach and two other popular criteria, information gain from \\cite{srinivas2012information} and V-optimality of \\cite{mingji}, can also be connected by functions of the eigenvalues of the covariance matrix at each iteration.\n% Particularly, we are interested in primitives of monotone functions as described in \\cite{spectral_submod}.\n\n% From the rank-one update rule, \\eqref{eq:var_update}, the posterior covariance matrices at different iterations can be ordered using convex cone of postive semi-definite matrix (Loewner order) as \n% $\\bC_0\\succ\\bC_1\\succ\\dotsc\\succ\\bC_T\\succ\\mathbf{0}$. With this ordering, we can define monotone spectral operators, $h(\\bC_t)$, that maps Hermitian matrices to real values, such that \n% \\begin{equation*}\n% h(\\bC_t)>h(\\bC_{t'}) \\Leftrightarrow \\bC_t\\succ\\bC_{t'}\n% \\quad(\\Leftrightarrow t0, \\forall s>0.\n%\\end{align}\n%Correspondingly, assuming an observation of $v$, the exploration heuristic is then,\n\\begin{equation}\n\t\\Delta h_t(v) = \n\th\\bigl(\\blambda_{t}\\bigr)\n\t-\n\th\\bigl(\\blambda_{t+1\\mid v}\\bigr) \n\t% \\\\\n\t% &{\\rm s.t. } \\;\n\t% h'(s)>0, \\forall s>0\n\\end{equation}\nwhere $h(\\blambda):\\mathbb{R}^n\\to\\mathbb{R}$ is a multivariate function defined on the eigenvalues.\nFurther, by the one-step update rule of \\eqref{eq:var_update}, $\\bC_t$ has Loewner order as \n$\\bC_0\\succ\\bC_1\\succ\\dotsc\\succ\\bC_T\\succ\\mathbf{0}$. \n% for any monotone increasing function $h:\\mathbb{R}^n\\to\\mathbb{R}$. \nIt is thus often desirable to require $h(\\cdot)$ to be monotone with respect to this ordering, i.e. $\\bC_t\\succ\\bC_{t'} \\Rightarrow h(\\blambda_t)\\geq h(\\blambda_{t'})$.\n% Notice that $\\bC_t$ is positive semi-definite and $\\lambda_{t,(j)}$ is monotone decreasing as $t$ increases, \n% by applying Weyl's inequality to rank-one updates at each iteration.\n\n% where the difference inside $h^{-1}(\\cdot)$ is nonnegative, because we can prove using induction and definition of eigen-vectors, for example with $j=1$\n% and \n% $\\bq_{t+1,(1)}$ being the eigen-vector corresponding to $\\lambda_{t+1,(1)}$ in the posterior model,\n% $\n\t% \\lambda_{t,(j)}^2 - \\lambda_{t+1,(1)}^2 \n\t% \\geq \n\t% \\nicefrac{\\langle \\bq_{t+1,(1)}, \\bc_{t+1}(v)\\rangle^2}{(\\sigma_n^2 - \\sigma_{t+1}^2(v))}\\geq0$.\n\n%For example, some reasonable choices of $s_t(v)$ are,\n%\\begin{numcases}{\\hspace{-2em}}\n%\t\\sqrt{ 1+\\nicefrac{\\sigma_t^2(v)}{\\sigma_n^2}}\n%\t & $h_{ig}(s) = -log(s)$\n%\t \\label{eq:s-ig}\n%\t\\\\\t \n%\t \\sqrt{\\tr(\\bC_{t}) - \\tr(\\bC_{t+1})},\n%\t & $h_2(s) = s^2$\n%\t \\label{eq:s-variance}\n%\t\\\\\n%\t \\lambda_{\\max}(\\bC_{t}) - \\lambda_{\\max}(\\bC_{t+1}),\n%\t \\hspace{-1em}\n%\t & $h_\\infty(s) = s^p, p\\to\\infty$\n%\t \\label{eq:s-spectral}\n%\\end{numcases}\n\n\\textbf{Case 1}. $\n\th(\\blambda) = \\sum_j \\log(\\lambda_{(j)}).$ \n\t% the differential entropy of $\\mathcal{N}(\\bmu_t,\\bC_t)$. \n\tThen,\n$\t\\Delta h_t(v) \n\t= 2\\mathcal{I}_t(\\bff; y(v)) = \\log(1+\\frac{\\sigma_t^2(v)}{\\sigma_n^2}),\n\t% = \\sqrt{ 1+\\nicefrac{\\sigma_t^2(v)}{\\sigma_n^2}}.\n$ twice the information gain from $\\bff\\sim\\mathcal{N}(\\bmu_t, \\bC_t)$ to $\\mathcal{N}(\\bmu_{t+1\\mid v}, \\bC_{t+1\\mid v})$. \nThis metric is important to \\textbf{GP-UCB} \\citep{srinivas2012information}, which set $s_t(v)=\\sigma_t(v)$ and used the inequality, $\\log(1+\\frac{\\sigma_m^2}{\\sigma_n^2}) \\frac{\\sigma_t^2(v)}{\\sigma_m^2}\\leq\\log(1+\\frac{\\sigma_t^2(v)}{\\sigma_n^2}),$ where $\\sigma_m=\\max_{v,t}\\sigma_t(v)$, in its proofs.\n\n% This heuristic adds biases to maximize the differential information gain of the joint distribution of node values, turns out to pay too much attention to the graph periphery, which actually prevents information gathering in the true problem against intuition. Precisely, differential entropy is sensitive to tails of the distribution, which happens to be the place of the biggest model mismatch of our \\textsc{grf} models. \n\n\n\n\\textbf{Case 2}. $\n\th(\\blambda) = \\sum_j \\lambda_{(j)} $ \n\tgives\n\t$\n\t\\Delta h_t(v) = \\tr(\\bC_{t}) - \\tr(\\bC_{t+1\\mid v}) = \\nicefrac{ \\| \\bc_t(v) \\|_2^2 }{(\\sigma_t^2(v) + \\sigma_n^2)}.\n$ For $\\sigma_n= 0$, $\\Delta h_t(v)$ is used as the greedy \\textbf{V-optimal} criterion for design of experiments by \\cite{mingji}.\n%, which though alleviates the situation by adding independence assumptions on the nodes and measuring the sum of the marginal variances, cannot completely address the selection bias at graph peripheries, because the self-variance term usually dominates the sum of squares of $\\|\\bc_t(v)\\|_2^2$.\n\n\n\\textbf{Case 3}. $\n\th(\\blambda) = \\lambda_{(1)}$ connects to the greedy design for \\textbf{E-optimality} \\citep{pukelsheim1993optimal}. To some extent, it is also related to greedy \\textbf{$\\Sigma$-Optimality}. First,\n\tapproximate $\\Delta h_t(v)$ by $\\partial\\lambda_{(j)} = \\bq_{(j)}^\\top \\partial(\\bC) \\bq_{(j)}$ around $\\bC=\\bC_t$, as\n\\vspace{-.5em}\n\\begin{equation*}\n\t\\Delta h_t(v) \n\t% &\n\t\\approx \n\t\\bq_{t,(1)}^\\top (\\bC_t-\\bC_{t+1|v}) \\bq_{t,(1)}\n\t= \n\t \\bigg( \n\t\t\\frac{|\\bc_t(v)^\\top \\bq_{t,(1)}|}\n\t{\\sqrt{\\sigma_t(v)^2 + \\sigma_n^2}}\n\t \\bigg)^2\n\t% \\\\\n\t% &=\n\t% \\bigg( \n\t% \t\\frac{\\lambda_{(1)} \\cdot |q_{t(1)}(v)|}\n\t% {\\sqrt{\\sigma_t(v)^2 + \\sigma_n^2}}\n\t% \\bigg)^2\n\t\\end{equation*}\nThe above resembles \\eqref{eq:sopt} if $\\bq_{t,(1)}\\propto \\bone$, which holds true for $t=0$ and $\\omega_0=0$ and approximately so for small $t$s.\n\n\n\nIn all these cases, exploration is measured by how much the objective, $h(\\blambda_T)$, is eventually decreased after $T$ iterations. \nEach definition of $h(\\blambda_t)$ aggregates the eigenvalues of the posterior covariance matrices in a different way, which \naffects the relative importance of large and small eigenvalues. In \\textbf{Case 1}, since $\\frac{\\partial \\log(\\lambda)}{\\partial \\lambda}=\\frac{1}{\\lambda}$, the same change introduced to a smaller $\\lambda$ will have a relatively larger impact on the objective.\n % than a larger $\\lambda$. \nSuch an effect is not evident in the other two cases. \nParticularly in \\textbf{Case 3}, changes to small eigenvalues are ignored unless they become the largest eigenvalue. \n\n\nEstablishing biases to penalize larger eigenvalues more has the benefit of improving global robustness because the posterior marginal variance of every node is upper-bounded by $\\lambda_{t,(1)}$. \nCompared with \\textbf{Cases 2} and \\textbf{3}, \\textbf{Case 1} is more sensitive to changes in small eigenvalues, which may be another explanation\nof \\textbf{GP-UCB}'s strong tendency to select peripheral nodes, as seen in Figure 4 of \\cite{MIG} or Figure 1(d) of \\cite{gotovos2013active}. \n\nAlthough Algorithm~\\ref{alg:main} is not built around the concept of functions on eigenvalues, it still establishes strong biases to penalize large eigenvalues in its initial explorations, per analysis in \\textbf{Case 3}. \nNote that \\textbf{$\\Sigma$-optimality} achieves a more complex goal than \\textbf{E-optimality}; exact execution of \\textbf{E-optimality} may over-simplify the model and select nodes between clusters for separation rather than inside them.\n\n\n% First, assuming that the principal eigen-vector of $\\bC_t$ is $\\bq_t$, then\n% $\n% \t\\lambda_{\\max}^2(\\bC_{t+1} \\mid v) \\approx \\lambda_{\\max}^2(\\bC_t) - \n% \t\\frac{\\langle \\bq_t, \\bc_t(v)\\rangle^2}\n% \t{ (\\sigma_t^2(v) + \\sigma_n^2) }\n% $ and, compounding the square-root operator,\n% $\n% \ts_t(v) \\approx \\frac{1}{2\\lambda_{\\max}(\\bC_t)}\n% \t\\frac{\\bc_t(v)^\\top \\bq_t}\n% \t{\\sqrt{\\sigma_t(v)^2 + \\sigma_n^2}}.\n% $\n\n\n% This approximation works for the very first selections. Note that $\\bC_0^{-1}=\\bD-\\bA+\\omega_0\\bI$ has its smallest eigen-vector (with respect to $\\omega_0$) very close to $\\frac{1}{n}\\mathbf{1}$, that same vector carries to be $\\bq_0$ for the largest eigen-value of $\\bC_0$. At this point, $s_t(v)$ is our Sigma-optimality up to a selection-independent constant.\n\n\n% In fact, this approximation can be valid for larger $t$'s. Further break the graph down to different (relatively isolated) connected components, where each individual component is relatively unexplored, and therefore contains a principal eigen-vector, relative to the component, which will approximate\n% $\\bq_{t,(c)}\\approx\\mathbf{1}_\\cC$, where $c$ is the rank of this eigen-vector and $\\cC$ the subset of nodes of this connected component. The more under-explored the component is, the more likely that $\\bq_{t,(c)}$ becomes the principal eigen-vector, $\\bq_t$ and also $\\bq_{t,(c)}$ gets close to $\\mathbf{1}_\\cC$.\n\n% In the meantime, every column on the current covariance matrix $\\bc_t(v)$ will also reflect independence between these (relatively isolated) components. Thus, the inner product can be roughly approximated as, \n% $ %\\begin{equation}\n% \t\\bq_t^\\top \\bc_t(v) \\approx \\mathbf{1}_\\cC^\\top \\bc_t(v) + \\mathbf{1}_{\\bar{\\cC}}^\\top \\mathbf{0} = \\mathbf{1}^\\top \\bc_t(v),\n% $ %\\end{equation}\n% where $\\bar{\\cC}$ is the complement of $\\cC$. \n% Again, Sigma-optimality approximates the difference of the spectral norm between prior and one-step look-ahead covariance matrices.\n\n", "meta": {"hexsha": "e7441fae432eae9a4b1e6b686b5b6dde7dd5e85e", "size": 9837, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "texts/sec_discussions.tex", "max_stars_repo_name": "AutonlabCMU/active-search-gp-sopt", "max_stars_repo_head_hexsha": "45d75dc0fe33d3d68784c30ba7f6ecd7b1718c31", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-10-23T03:53:13.000Z", "max_stars_repo_stars_event_max_datetime": "2018-10-23T03:53:13.000Z", "max_issues_repo_path": "texts/sec_discussions.tex", "max_issues_repo_name": "AutonlabCMU/active-search-gp-sopt", "max_issues_repo_head_hexsha": "45d75dc0fe33d3d68784c30ba7f6ecd7b1718c31", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "texts/sec_discussions.tex", "max_forks_repo_name": "AutonlabCMU/active-search-gp-sopt", "max_forks_repo_head_hexsha": "45d75dc0fe33d3d68784c30ba7f6ecd7b1718c31", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-12-22T23:55:18.000Z", "max_forks_repo_forks_event_max_datetime": "2019-11-14T15:33:27.000Z", "avg_line_length": 61.48125, "max_line_length": 434, "alphanum_fraction": 0.7007217648, "num_tokens": 3243, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127529517043, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6346006401399197}} {"text": "\\chapter{Experimental}\n\n\\section{Further development of the maximum proof}\nStarting point is:\n\\[\nv_{n+2}=\\frac{2}{\\left(2-\\prod_{i=1}^{n}\\beta_i\\right)\\cdot2^{\\alpha_{n+1}}}\n\\]\n\n\\par\\medskip\nIn order to show that $v_{n+2}<1$ we consider the worst case that maximizes $v_{n+2}$ by using the inserting the Engel expansion into the product:\n\\[\n\\prod_{i=1}^{n}\\beta_i=\\prod_{i=1}^{n}\\left(1+\\frac{1}{3v_i}\\right)=\\prod_{i=1}^{n}\\left(1+\\frac{1}{3v_i}\\right)=\\prod_{i=1}^{n}\\left(1+\\frac{1}{3\\frac{3^{i-1}(v_1+1)-2^{i-1}}{2^{i-1}}}\\right)\n=\\prod_{i=1}^{n}\\frac{3^i(v_1+1)-2^i}{3^i(v_1+1)-3*2^{i-1}}\n=\\frac{1}{v_1}-\\frac{1}{v_1}\\left(\\frac{2}{3}\\right)^n+1\n\\]\n\n\\par\\medskip\nThis leads to:\n\\[\nv_{n+2}=\\frac{2}{\\left(2-\\left(\\frac{1}{v_1}-\\frac{1}{v_1}\\left(\\frac{2}{3}\\right)^n+1\\right)\\right)\\cdot2^{\\alpha_{n+1}}}\n\\]\n\n\\par\\medskip\nMultiplying leads to\n\\[\n2^{\\alpha_{n+1}}v_1v_{n+2}-2^{\\alpha_{n+1}}v_{n+2}+2^{\\alpha_{n+1}}v_{n+2}\\left(\\frac{2}{3}\\right)^n-2v_1=0\n\\]\n\n\\par\\medskip\nNow we remove the variable $v_1$, because we know the relatonship between $v_1$ and $v_{n+2}$ in an Engel expansion (see \\ref{eq:asc_continued_fraction}):\n\\[\nv_1=\\frac{(v_{n+2}+1)2^{n+1}}{3^{n+1}}-1\n\\]\n\n\\par\\medskip\nNow we substitute the term above into our equation:\n\\[\n2^{\\alpha_{n+1}}\\left(\\frac{(v_{n+2}+1)2^{n+1}}{3^{n+1}}-1\\right)v_{n+2}-2^{\\alpha_{n+1}}v_{n+2}+2^{\\alpha_{n+1}}v_{n+2}\\left(\\frac{2}{3}\\right)^n-2\\left(\\frac{(v_{n+2}+1)2^{n+1}}{3^{n+1}}-1\\right)=0\n\\]\n\n\\par\\medskip\nWe know that in an Engel expansion $\\alpha_{n+1}=1$ (we make onla one division by two between vertices). Via substitution $v_{n+2}=y$ we obtain a simple equation, which has for $y\\ge1$ only complex solutions having an imaginary part:\n\\[\n\\left(\\frac{(y+1)2^{n+1}}{3^{n+1}}-1\\right)y-y+y\\left(\\frac{2}{3}\\right)^n-\\left(\\frac{(y+1)2^{n+1}}{3^{n+1}}-1\\right)=0\n\\]\n\nLet us now substitute $\\left(\\frac{2}{3}\\right)^n$ with $z$:\n\\[\n\\left((y+1)\\frac{2}{3}z-1\\right)y-y+yz-\\left((y+1)\\frac{2}{3}z-1\\right)=0\n\\]\n\nHence $y=v_{n+2}$ must be smaller than 1.", "meta": {"hexsha": "655f537846518f9d008ed0e068d38cbca5b92563", "size": 2008, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "01 Graph Theory/TeX/v6.0/chapter/xx_experimental.tex", "max_stars_repo_name": "Sultanow/collatz", "max_stars_repo_head_hexsha": "d8a5137af508be19da371fff787c114f1b5185c3", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-04-01T15:12:10.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-01T15:54:55.000Z", "max_issues_repo_path": "01 Graph Theory/TeX/v4.1/chapter/xx_experimental.tex", "max_issues_repo_name": "Sultanow/collatz", "max_issues_repo_head_hexsha": "d8a5137af508be19da371fff787c114f1b5185c3", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "01 Graph Theory/TeX/v4.1/chapter/xx_experimental.tex", "max_forks_repo_name": "Sultanow/collatz", "max_forks_repo_head_hexsha": "d8a5137af508be19da371fff787c114f1b5185c3", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-05-06T20:44:07.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-06T20:44:07.000Z", "avg_line_length": 38.6153846154, "max_line_length": 233, "alphanum_fraction": 0.6309760956, "num_tokens": 920, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127492339907, "lm_q2_score": 0.7431680086124812, "lm_q1q2_score": 0.634600637377034}} {"text": "\\chapter{Semisimple algebras}\nIn what follows, \\textbf{assume the field $k$ is algebraically closed}.\n\nFix an algebra $A$ and suppose\nyou want to study its representations.\nWe have a ``direct sum'' operation already.\nSo, much like we pay special attention to prime numbers,\nwe're motivated to study irreducible representations\nand then build all the representations of $A$ from there.\n\nUnfortunately, we have seen (\\Cref{exer:irred_not_indecomp})\nthat there exists a representation which is not irreducible,\nand yet cannot be broken down as a direct sum (indecomposable).\nThis is \\emph{weird and bad}, so we want to give a name\nto representations which are more well-behaved.\nWe say that a representation is \\vocab{completely reducible}\nif it doesn't exhibit this bad behavior.\n\nEven better, we say a finite-dimensional algebra $A$\nis \\vocab{semisimple} if all its finite-dimensional\nrepresentations are completely reducible.\nSo when we study finite-dimensional representations of\nsemisimple algebras $A$,\nwe just have to figure out what the irreps are,\nand then piecing them together will give all\nthe representations of $A$.\n\nIn fact, semisimple algebras $A$ have even nicer properties.\nThe culminating point of the chapter is when we prove that\n$A$ is semisimple if and only if $A \\cong \\bigoplus_i \\Mat(V_i)$,\nwhere the $V_i$ are the irreps of $A$\n(yes, there are only finitely many!).\n\n\\section{Schur's lemma continued}\n\\prototype{For $V$ irreducible,\n\t$\\Homrep(V^{\\oplus 2}, V^{\\oplus 2}) \\cong k^{\\oplus 4}$.}\n\\begin{definition}\n\tFor an algebra $A$ and representations $V$ and $W$,\n\twe let $\\Homrep(V,W)$ be the set of intertwining operators between them.\n\t(It is also a $k$-algebra.)\n\\end{definition}\n\nBy Schur's lemma (since $k$ is algebraically closed,\nwhich again, we are taking as a standing assumption),\nwe already know that if $V$ and $W$ are irreps, then\n\\[\n\t\\Homrep(V,W) \\cong\n\t\\begin{cases}\n\t\tk & \\text{if $V \\cong W$} \\\\\n\t\t0 & \\text{if $V \\not\\cong W$}.\n\t\\end{cases}\n\\]\nCan we say anything more?\nFor example, it also tells us that\n\\[ \\Homrep(V, V^{\\oplus 2}) = k^{\\oplus 2}. \\]\nThe possible maps are $v \\mapsto (c_1v_1, c_2v_2)$ for some choice of $c_1, c_2 \\in k$.\n\nMore generally, suppose $V$ is an irrep and consider\n$\\Homrep(V^{\\oplus m}, V^{\\oplus n})$.\nIntertwining operators are determined completely\n$T : V^{\\oplus m} \\to V^{\\oplus n}$ by the $mn$ choices of compositions\n\\begin{diagram}\n\tV & \\rInj & V^{\\oplus m} & \\rTo^T & V^{\\oplus n} & \\rSurj & V\n\\end{diagram}\nwhere the first arrow is inclusion to the $i$th component of $V^{\\oplus m}$\n(for $1 \\le i \\le m$) and the second arrow is inclusion to the $j$th\ncomponent of $V^{\\oplus n}$ (for $1 \\le j \\le n$).\nHowever, by Schur's lemma on each of these compositions,\nwe know they must be constant.\n\nThus, $\\Homrep(V^{\\oplus n}, V^{\\oplus m})$ consist of $n \\times m$ ``matrices''\nof constants, and the map is provided by\n\\[\n\t\\begin{bmatrix}\n\t\tc_{11} & c_{12} & \\dots & c_{1(n-1)} & c_{1n} \\\\\n\t\tc_{21} & c_{22} & \\dots & c_{2(n-1)} & c_{1n} \\\\\n\t\t\\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\\n\t\tc_{m1} & c_{m2} & \\dots & c_{m(n-1)} & c_{mn}\n\t\\end{bmatrix}\n\t\\begin{bmatrix} v_1 \\\\ v_2 \\\\ \\vdots \\\\ v_n \\end{bmatrix}\n\t\\in V^{\\oplus n}\n\\]\nwhere the $c_{ij} \\in k$ but $v_i \\in V$; note the type mismatch!\nThis is \\emph{not} just a linear map $V^{\\oplus n_i} \\to V^{\\oplus m_i}$;\nrather, the outputs are $m$ \\emph{linear combinations} of the inputs.\n\nMore generally, we have:\n\\begin{theorem}\n\t[Schur's lemma for completely reducible representations]\n\t\\label{thm:compred_schur}\n\tLet $V$ and $W$ be completely reducible representations,\n\tand set $V = \\bigoplus V_i^{\\oplus n_i}$, $W = \\bigoplus V_i^{\\oplus m_i}$\n\tfor integers $n_i, m_i \\ge 0$, where each $V_i$ is an irrep.\n\tThen\n\t\\[ \\Homrep(V, W)\n\t\t\\cong \\bigoplus_i \\Mat_{n_i \\times m_i}(k) \\]\n\tmeaning that an intertwining operator $T : V \\to W$\n\tamounts to, for each $i$, an $n_i \\times m_i$ matrix of constants\n\twhich gives a map $V_i^{\\oplus n_i} \\to V_i^{\\oplus m_i}$.\n\\end{theorem}\n\n\\begin{corollary}\n\t[Subrepresentations of completely reducible representations]\n\t\\label{cor:subrep_schur}\n\tLet $V = \\bigoplus V_i^{\\oplus n_i}$ be completely reducible.\n\tThen any subrepresentation $W$ of $V$ is isomorphic\n\tto $\\bigoplus V_i^{\\oplus m_i}$ where $m_i \\le n_i$ for each $i$,\n\tand the inclusion $W \\injto V$ is given\n\tby the direct sum of inclusion $V_i^{\\oplus m_i} \\injto V_i^{\\oplus n_i}$,\n\twhich are $n_i \\times m_i$ matrices.\n\\end{corollary}\n\\begin{proof}\n\tApply Schur's lemma to the inclusion $W \\injto V$.\n\\end{proof}\n\n\n\n\\section{Density theorem}\nWe are going to take advantage of the previous result to prove that\nfinite-dimensional algebras have finitely many irreps.\n\n\\begin{theorem}\n\t[Jacobson density theorem]\n\tLet $(V_1, \\rho_1)$, \\dots, $(V_r, \\rho_r)$ be pairwise nonisomorphic\n\tfinite-dimensional representations of $A$.\n\tThen there is a surjective map of vector spaces\n\t\\[ \\bigoplus_{i=1}^r \\rho_i : A \\surjto \\bigoplus_{i=1}^r \\Mat(V_i). \\]\n\\end{theorem}\nThe right way to think about this theorem is that\n\\begin{moral}\n\tDensity is the ``Chinese remainder theorem''\n\tfor irreps of $A$.\n\\end{moral}\nRecall that in number theory, the Chinese remainder theorem tells us\nthat given lots of ``unrelated'' congruences, we can find a single $N$\nwhich simultaneously satisfies them all.\nSimilarly, given lots of different nonisomorphic representations of $A$,\nthis means that we can select a single $a \\in A$ which induces any tuple\n$(\\rho_1(a), \\dots, \\rho_r(a))$ of actions we want --- a surprising result,\nsince even the $r=1$ case is not obvious at all!\n\n\\begin{diagram}\n\t&&& \\rho_1(a) &= M_1 \\in \\Mat(V_1) \\\\\n\t&& \\ruTo(3,2) & \\rho_2(a) &= M_2 \\in \\Mat(V_2) \\\\\n\t\\boxed{a \\in A} && \\ruTo(3,1) && \\vdots \\\\\n\t& \\rdTo(3,1) && \\rho_r(a) &= M_r \\in \\Mat(V_r) \\\\\n\\end{diagram}\n\nThis also gives us the non-obvious corollary\n\\begin{corollary}\n\t[Finiteness of number of representations]\n\tAny finite-dimensional algebra $A$ has at most $\\dim A$ irreps.\n\t\\label{cor:finiteness}\n\\end{corollary}\n\\begin{proof}\n\tIf $V_i$ are such irreps then \n\t$A \\surjto \\bigoplus_i V_i^{\\oplus \\dim V_i}$,\n\thence we have the inequality $\\sum (\\dim V_i)^2 \\le \\dim A$.\n\\end{proof}\n\n\\begin{proof}[Proof of density theorem]\n\tLet $V = V_1 \\oplus \\dots \\oplus V_r$, so $A$\n\tacts on $V = (V, \\rho)$ by $\\rho = \\bigoplus_i \\rho_i$.\n\tThus by \\Cref{prob:reg_mat}, we can instead consider $\\rho$\n\tas an \\emph{intertwining operator}\n\t\\[ \\rho : \\Reg(A) \\to \\bigoplus_{i=1}^r \\Mat(V_i)\n\t\t\\cong \\bigoplus_{i=1}^r V_i^{\\oplus d_i}. \\]\n\tWe will use this instead as it will be easier to work with.\n\n\tFirst, we handle the case $r = 1$.\n\tFix a basis $e_1$, \\dots, $e_n$ of $V = V_1$.\n\tAssuming for contradiction that the map is not surjective.\n\tThen there is a map of representations (by $\\rho$ and the isomorphism)\n\t$\\Reg(A) \\to V^{\\oplus n}$ given by $a \\mapsto (a \\cdot e_1, \\dots, a \\cdot e_n)$.\n\tBy hypothesis is not surjective:\n\tits image is a \\emph{proper} subrepresentation of $V^{\\oplus n}$.\n\tAssume its image is isomorphic to $V^{\\oplus m}$ for $m < n$,\n\tso by \\Cref{thm:compred_schur} there is a matrix of constants $X$ with\n\t\\begin{diagram}\n\t\t\\Reg(A) & \\rTo & V^{\\oplus n} & \\lInj^{X \\cdot -} & V^{\\oplus r} \\\\\n\t\ta & \\rMapsto & (a \\cdot e_1, \\dots, a \\cdot e_n) && \\\\\n\t\t1_A & \\rMapsto & (e_1, \\dots, e_n) & \\lMapsto & (v_1, \\dots, v_m)\n\t\\end{diagram}\n\twhere the two arrows in the top row have the same image;\n\thence the pre-image $(v_1, \\dots, v_m)$ of $(e_1, \\dots, e_n)$ can be found.\n\tBut since $r < n$ we can find constants $c_1, \\dots, c_n$ not all zero\n\tsuch that $X$ applied to the column vector $(c_1, \\dots, c_n)$ is zero:\n\t\\[\n\t\t\\sum_{i=1}^n c_ie_i\n\t\t=\n\t\t\\begin{bmatrix} c_1 & \\dots & c_n \\end{bmatrix}\n\t\t\\begin{bmatrix} e_1 \\\\ \\vdots \\\\ e_n \\end{bmatrix}\n\t\t=\n\t\t\\begin{bmatrix} c_1 & \\dots & c_n \\end{bmatrix}\n\t\tX\n\t\t\\begin{bmatrix} v_1 \\\\ \\vdots \\\\ v_m \\end{bmatrix}\n\t\t= 0\n\t\\]\n\tcontradicting the fact that $e_i$ are linearly independent.\n\tHence we conclude the theorem for $r=1$.\n\n\tAs for $r \\ge 2$, the image $\\rho\\im(A)$ is necessarily of the form\n\t$\\bigoplus_i V_i^{\\oplus r_i}$ (by \\Cref{cor:subrep_schur})\n\tand by the above $r_i = \\dim V_i$ for each $i$.\n\\end{proof}\n\n\\section{Semisimple algebras}\n\n\\begin{definition}\n\tA finite-dimensional algebra $A$ is a \\vocab{semisimple}\n\tif every finite-dimensional representation of $A$ is completely reducible.\n\\end{definition}\n\n\\begin{theorem}\n\t[Semisimple algebras]\n\tLet $A$ be a finite-dimensional algebra.\n\tThen the following are equivalent:\n\t\\begin{enumerate}[(i)]\n\t\t\\ii $A \\cong \\bigoplus_i \\Mat_{d_i}(k)$ for some $d_i$.\n\t\t\\ii $A$ is semisimple.\n\t\t\\ii $\\Reg(A)$ is completely reducible.\n\t\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\n\t(i) $\\implies$ (ii) follows\n\tfrom \\Cref{thm:rep_1mat} and \\Cref{prop:rep_direct_sum}.\n\t(ii) $\\implies$ (iii) is tautological.\n\n\tTo see (iii) $\\implies$ (i), we use the following clever trick.\n\tConsider\n\t\\[ \\Homrep(\\Reg(A), \\Reg(A)). \\]\n\tOn one hand, by \\Cref{prob:regA_intertwine},\n\tit is isomorphic to $A\\op$ ($A$ with opposite multiplication),\n\tbecause the only intertwining operators $\\Reg(A) \\to \\Reg(A)$\n\tare those of the form $- \\cdot a$.\n\tOn the other hand, suppose that we have set\n\t$ \\Reg(A) = \\bigoplus_i V_i^{\\oplus n_i} $.\n\tBy \\Cref{thm:compred_schur}, we have\n\t\\[ A\\op \\cong \\Homrep(\\Reg(A), \\Reg(A))\n\t\t= \\bigoplus_i \\Mat_{n_i \\times n_i}(k). \\]\n\tBut $\\Mat_n(k)\\op \\cong \\Mat_n(k)$ (just by transposing),\n\tso we recover the desired conclusion.\n\\end{proof}\n\nIn fact, if we combine the above result with\nthe density theorem (and \\Cref{cor:finiteness}), we obtain:\n\\begin{theorem}\n\t[Sum of squares formula]\n\tFor a finite-dimensional algebra $A$ we have\n\t\\[ \\sum_{i} \\dim(V_i)^2 \\le \\dim A \\]\n\twhere the $V_i$ are the irreps of $A$;\n\tequality holds exactly when $A$ is semisimple,\n\tin which case \n\t\\[ \\Reg(A) \\cong \\bigoplus_i \\Mat(V_i)\n\t\t\\cong \\bigoplus_I V_i^{\\oplus \\dim V_i}. \\]\n\\end{theorem}\n\\begin{proof}\n\tThe inequality was already mentioned in \\Cref{cor:finiteness}.\n\tIt is equality if and only if the map $\\rho : A \\to \\bigoplus_i \\Mat(V_i)$\n\tis an isomorphism; this means all $V_i$ are present.\n\\end{proof}\n\n\\begin{remark}\n\t[Digression]\n\tFor any finite-dimensional $A$, the kernel of the map\n\t$\\rho : A \\to \\bigoplus_i \\Mat(V_i)$ is denoted $\\opname{Rad}(A)$\n\tand is the so-called \\vocab{Jacobson radical} of $A$;\n\tit's the set of all $a \\in A$ which act by zero in all irreps of $A$.\n\tThe usual definition of ``semisimple'' given in books is that\n\tthis Jacobson radical is trivial.\n\\end{remark}\n\n\\section{Maschke's theorem}\nWe now prove that the representation theory of groups is as nice as possible.\n\\begin{theorem}\n\t[Maschke's theorem]\n\tLet $G$ be a finite group, and $k$ an algebraically closed\n\tfield whose characteristic does not divide $|G|$.\n\tThen $k[G]$ is semisimple.\n\\end{theorem}\nThis tells us that when studying representations of groups,\nall representations are completely reducible.\n\\begin{proof}\n\tConsider any finite-dimensional representation $(V, \\rho)$ of $k[G]$.\n\tGiven a proper subrepresentation $W \\subseteq V$,\n\tour goal is to construct a supplementary $G$-invariant subspace $W'$\n\twhich satisfies \\[ V = W \\oplus W'. \\]\n\tThis will show that indecomposable $\\iff$ irreducible,\n\twhich is enough to show $k[G]$ is semisimple.\n\n\tLet $\\pi : V \\to W$ be any projection of $V$ onto $W$,\n\tmeaning $\\pi(v) = v \\iff v \\in W$.\n\tWe consider the \\emph{averaging} map $P : V \\to V$ by\n\t\\[ \n\t\tP(v) = \\frac{1}{\\left\\lvert G \\right\\rvert}\n\t\t\\sum_{g \\in G} \\rho(g\\inv) \\circ \\pi \\circ \\rho(g).\n\t\\]\n\tWe'll use the following properties of the map:\n\t\\begin{exercise}\n\t\tShow that the map $P$ satisfies:\n\t\t\\begin{itemize}\n\t\t\t\\ii For any $w \\in W$, $P(w) = w$.\n\t\t\t\\ii For any $v \\in V$, $P(w) \\in W$.\n\t\t\t\\ii The map $P : V \\to V$ is an intertwining operator.\n\t\t\\end{itemize}\n\t\\end{exercise}\n\tThus $P$ is idempotent (it is the identity on its image $W$),\n\tso by \\Cref{prob:idempotent} we have $V = \\ker P \\oplus \\img P$,\n\tbut both $\\ker P$ and $\\img P$ are subrepresentations as desired.\n\\end{proof}\n\\begin{remark}\n\tIn the case where $k = \\CC$, there is a shorter proof.\n\tSuppose $B : V \\times V \\to \\CC$ is an arbitrary bilinear form.\n\tThen we can ``average'' it to obtain a new bilinear form\n\t\\[ \\left< v,w \\right> \\defeq \\frac{1}{|G|} \\sum_{g \\in G} B(g \\cdot v, g \\cdot w). \\]\n\tThe averaged form $\\left< -,- \\right>$ is $G$-invariant,\n\tin the sense that $\\left< v,w \\right> = \\left< g \\cdot v, g \\cdot w\\right>$.\n\tThen, one sees that if $W \\subseteq V$ is a subrepresentation,\n\tso is its orthogonal complement $W^\\perp$.\n\tThis implies the result.\n\\end{remark}\n\n\\section{Example: the representations of $\\CC[S_3]$}\nWe compute all irreps of $\\CC[S_3]$.\nI'll take for granted right now there are exactly three such representations\n(which will be immediate by the first theorem in the next chapter:\nwe'll in fact see that the number of representations of $G$\nis exactly equal to the number of conjugacy classes of $G$).\n\nGiven that, if the three representations of have dimension $d_1$, $d_2$, $d_3$ ,\nthen we ought to have\n\\[ d_1^2 + d_2^2 + d_3^2 = |G| = 6. \\]\nFrom this, combined with some deep arithmetic,\nwe deduce that we should have $d_1 = d_2 = 1$ and $d_3 = 2$\nor some permutation.\n\nIn fact, we can describe these representations explicitly.\nFirst, we define:\n\\begin{definition}\n\tLet $G$ be a group.\n\tThe complex \\vocab{trivial group representation} of a group $G$\n\tis the one-dimensional representation $\\Ctriv = (\\CC, \\rho)$\n\twhere $g \\cdot v = v$ for all $g \\in G$ and $v \\in \\CC$\n\t(i.e.\\ $\\rho(g) = \\id$ for all $g \\in G$).\n\\end{definition}\n\\begin{remark}\n\t[Warning] The trivial representation of an \\emph{algebra} $A$\n\tdoesn't make sense for us:\n\twe might want to set $a \\cdot v = v$ but this isn't linear in $A$.\n\t(You \\emph{could} try to force it to work by\n\tdeleting the condition $1_A \\cdot v = v$ from our definition;\n\tthen one can just set $a \\cdot v = 0$.\n\tBut even then $\\Ctriv$ would not be the trivial representation of $k[G]$.)\n\\end{remark}\n\nThen the representations are:\n\\begin{itemize}\n\t\\ii The one-dimensional $\\Ctriv$;\n\teach $\\sigma \\in S_3$ acts by the identity.\n\n\t\\ii There is a nontrivial one-dimensional representation\n\t$\\Csign$ where the map $S_3 \\to \\CC^\\times$ is given\n\tby sending $\\sigma$ to the sign of $\\sigma$.\n\tThus in $\\Csign$ every $\\sigma \\in S_3$ acts as $\\pm 1$.\n\tOf course, $\\Ctriv$ and $\\Csign$ are not isomorphic\n\t(as one-dimensional representations are never isomorphic\n\tunless the constants they act on coincide for all $a$,\n\tas we saw in \\Cref{prob:one_dim}).\n\n\t\\ii Finally, we have already seen the two-dimensional representation,\n\tbut now we give it a name.\n\tDefine $\\refl_0$ to be the representation whose vector space is\n\t$\\{ (x,y,z) \\mid x+y+z = 0 \\}$,\n\tand whose action of $S_3$ on it is permutation of coordinates.\n\t\\begin{exercise}\n\t\tShow that $\\refl_0$ is irreducible, for example by showing directly\n\t\tthat no subspace is invariant under the action of $S_3$.\n\t\\end{exercise}\n\tThus $V$ is also not isomorphic to the previous two representations.\n\\end{itemize}\nThis implies that these are all the irreps of $S_3$.\nNote that, if we take the representation $V$ of $S_3$ on $k^{\\oplus 3}$,\nwe just get that $V = \\refl_0 \\oplus \\CC_{\\text{triv}}$.\n\n\\section\\problemhead\n\n\\begin{problem}\n\tFind all the irreps of $\\CC[\\Zc n]$.\n\t\\begin{hint}\n\t\tThey are all one-dimensional, $n$ of them.\n\t\tWhat are the homomorphisms $\\Zc n \\to \\CC^\\times$?\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}\n\t[Maschke requires $|G|$ finite]\n\tConsider the representation of the group $\\RR$\n\ton $\\CC^{\\oplus 2}$ under addition by a homomorphism\n\t\\[ \n\t\t\\RR \\to \\Mat_2(\\CC)\n\t\t\\quad\\text{by}\\quad\n\t\tt \\mapsto \n\t\t\\begin{bmatrix} 1 & t \\\\ 0 & 1 \\end{bmatrix}.\n\t\\]\n\tShow that this representation is not irreducible,\n\tbut it is indecomposable.\n\t\\begin{hint}\n\t\tThe span of $(1,0)$ is a subrepresentation.\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}\n\tProve that all irreducible representations\n\tof a finite group are finite-dimensional.\n\t\\begin{hint}\n\t\tThis is actually easy.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tPick any $v \\in V$, then the subspace\n\t\tspanned by elements $g \\cdot v$ for $v \\in V$\n\t\tis $G$-invariant;\n\t\tthis is a finite-dimensional subspace,\n\t\tso it must equal all of $V$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t\\gim\n\tDetermine all the complex irreps of $D_{10}$.\n\t\\begin{hint}\n\t\tThere are only two one-dimensional ones\n\t\t(corresponding to the only two\n\t\thomomorphisms $D_{10} \\to \\CC^\\times$).\n\t\tSo the remaining ones are two-dimensional.\n\t\\end{hint}\n\\end{problem}\n", "meta": {"hexsha": "c8443330946789d4ede02085dfab70873c43f3e3", "size": 16509, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/rep-theory/semisimple.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/rep-theory/semisimple.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/rep-theory/semisimple.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.7780320366, "max_line_length": 87, "alphanum_fraction": 0.6898055606, "num_tokens": 5613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124812, "lm_q2_score": 0.853912747375134, "lm_q1q2_score": 0.6346006359955911}} {"text": "This module tries to recognize syntactic symmetries in the input formula and adds symmetry breaking constraints.\nThe core functionality is provided by \\texttt{carl} through \\texttt{carl::formula::breakSymmetries()} which internally encodes the formula as a graph and uses \\texttt{bliss} to find automorphisms on this graph.\n\n\\paragraph{Efficiency}\nFinding automorphisms is as difficult as determining whether two graphs are isomorphic, and it is not known whether this problem can be solved in polynomial or exponential time.\nIn practice, current solvers like \\texttt{bliss} perform very good on large graphs and we therefore assume this module to be sufficiently fast.\n", "meta": {"hexsha": "c755145e216f7afc906e05367ff0b33df8fd8af9", "size": 670, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/smtrat-modules/SymmetryModule/SymmetryModule.tex", "max_stars_repo_name": "minemebarsha/smtrat", "max_stars_repo_head_hexsha": "eaada50cdf9bbfe4dd4f6a54776387484c37b0f2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/smtrat-modules/SymmetryModule/SymmetryModule.tex", "max_issues_repo_name": "minemebarsha/smtrat", "max_issues_repo_head_hexsha": "eaada50cdf9bbfe4dd4f6a54776387484c37b0f2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/smtrat-modules/SymmetryModule/SymmetryModule.tex", "max_forks_repo_name": "minemebarsha/smtrat", "max_forks_repo_head_hexsha": "eaada50cdf9bbfe4dd4f6a54776387484c37b0f2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 95.7142857143, "max_line_length": 210, "alphanum_fraction": 0.8194029851, "num_tokens": 145, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8539127492339909, "lm_q2_score": 0.7431680029241321, "lm_q1q2_score": 0.6346006325196802}} {"text": "\n\\subsection{Endomorphisms on real functions}\n\nWe start with our vector \\(f(x)\\).\n\n\\(h(x)=f(x)g(x)\\)\n\nThe equivalent of the identity matrix is where \\(g(x)=1\\).\n\nThese are similar to endomorphisms where all off diagonal elements are \\(0\\).\n\n\\subsubsection{Differentiation}\n\n\\(h(x)=\\dfrac{\\delta }{\\delta x}f(x)\\)\n\n\\subsubsection{Integration}\n\n\\(h(x)=\\int_{-\\infty }^x f(z) dz\\)\n\n", "meta": {"hexsha": "1704f0dc45aa51c4b98806cb5f2425e8809bd6f2", "size": 379, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/geometry/functionalAnalysis/02-01-endomorphisms.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/geometry/functionalAnalysis/02-01-endomorphisms.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/geometry/functionalAnalysis/02-01-endomorphisms.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 18.95, "max_line_length": 77, "alphanum_fraction": 0.6781002639, "num_tokens": 116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8991213799730774, "lm_q2_score": 0.7057850278370112, "lm_q1q2_score": 0.6345864081931504}} {"text": "\\section{Neural Networks}\n\n% ##############################################################################\n\\subsection{Artificial Neural Networks}\n\\label{ssec:ArtificialNeuralNetworks}\n\nNeural networks are computing systems that are inspired by, but not identical to, biological neural networks that constitute animal brains. Such systems essentially \\emph{learn} to perform tasks by considering multiple samples, generally without being programmed with task-specific rules. They form a basis for deep machine learning~\\cite{goodfellow2016dl}.\n\nThe goal of a neural network is to approximate some unknown function $f$. For instance, when considering a classifier, the transformation $y = \\func{f}{\\vx}$ maps the given input $\\vx$ to a category $y$. Such a network, therefore, defines a mapping and learns the value of the parameters that result in the best function approximation.\n\nThese models can be described with a directed acyclic graph denoting the sequential composition of several functions. More concretely, we might have three functions $\\suprbrackets{f}{1}$, $\\suprbrackets{f}{2}$ and $\\suprbrackets{f}{3}$, forming a chain, $\\func{f}{\\vx} = \\func{\\suprbrackets{f}{3}}{\\func{\\suprbrackets{f}{2}}{\\func{\\suprbrackets{f}{1}}{\\vx}}}$. These chain structures are the most commonly used structures in neural networks. Deep machine learning consists of multiple such layers of neurons that are trained using the \\emph{backpropagation} algorithm~\\cite{rumelhart1986backprop}.\n\n% ##############################################################################\n\\subsection{Convolutional Neural Networks}\n\\label{ssec:ConvolutionalNeuralNetworks}\n\nThis type of neural network has gained popularity in the computer vision community thanks to a never-seen-before performance on image classification task~\\cite{krizhevsky2012classification}. This approach processes images or other high dimensional, grid-like input and then learns the importance (weights and biases) of various aspects of the input data.\n\nThe successful ability of these networks to capture spatial properties via learned convolutional filters is the fundamental principle. Let $f$ and $g$ be functions. Then, the operation of convolution denoted by $\\star$ produces a third function, as a result of the following computation (demonstrating the commutativity property, too)~\\cite{goodfellow2016dl}:\n\\begin{equation}\n \\label{eq:ConvolutionContinuous}\n \\func{\\rbrackets{f \\star g}}{t} =\n \\int_{-\\infty}^{\\infty}\n \\func{f}{\\tau}\n \\func{g}{t - \\tau}\n d\\tau =\n \\int_{-\\infty}^{\\infty}\n \\func{f}{t -\\tau}\n \\func{g}{\\tau}\n d\\tau.\n\\end{equation}\nIn this setting, $f$ is the input, $g$ is the kernel, and the output of this operation is a feature map. During training, the aim is to learn the weights of the kernel matrix that produces a feature map based on which the model can solve the given task.\n\nLet $I$ be a two-dimensional input image and $K$ be a two-dimensional kernel. Then, for a given position $\\rbrackets{i, j}$ in the input image $I$, the discrete convolution can be written as\n\\begin{equation}\n \\label{eq:ConvolutionDiscrete}\n \\func{\\rbrackets{f \\star g}}{i, j} =\n \\sum_{m}\n \\sum_{n}\n \\func{I}{m, n}\n \\func{K}{i - m, j - n}.\n\\end{equation}\nMany machine learning libraries implement the cross-correlation operation, not the convolution operation by its strict definition. This operation is the same except for the fact that the kernel is not flipped. Thus, the result of the cross-correlation is given by\n\n\\begin{equation}\n \\label{eq:CrossCorrelation}\n \\func{\\rbrackets{f \\star g}}{i, j} =\n \\sum_{m}\n \\sum_{n}\n \\func{I}{i + m, j + n}\n \\func{K}{m, n}.\n\\end{equation}\n\nAn indispensable outcome of \\glspl{cnn} is the ability to capture hierarchical relations. Layers placed near the input of the model capture low-level features such as edges, colors, gradient orientations, and so on. On the other hand, layers placed further, deeper in the model, highlight semantic, abstract features that are specific to the task at hand.\n\nAnother prominent use case of \\glspl{cnn} is \\emph{transfer learning}, where a pre-trained model is adopted for a new task, utilizing the already learned features. These pre-trained models may come in various flavors, but typical ones are pre-trained for an image classification task using the \\datasetname{ImageNet} dataset~\\cite{deng2009imagenet}. The reasoning is that visual features such as edges and contours are vital to general object recognition tasks, hence it is not needed to learn coarse, rudimentary, low-level features from scratch all the time.\n", "meta": {"hexsha": "a31cf7276e54b3ebd9a3bab90c58400e32af94ff", "size": 4622, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/chapters/theoretical_foundations/sections/neural_networks.tex", "max_stars_repo_name": "mondrasovic/phd_thesis", "max_stars_repo_head_hexsha": "68a3a6d1687ea43dc6cdfafcd5e6d9ce35f424e8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/chapters/theoretical_foundations/sections/neural_networks.tex", "max_issues_repo_name": "mondrasovic/phd_thesis", "max_issues_repo_head_hexsha": "68a3a6d1687ea43dc6cdfafcd5e6d9ce35f424e8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/chapters/theoretical_foundations/sections/neural_networks.tex", "max_forks_repo_name": "mondrasovic/phd_thesis", "max_forks_repo_head_hexsha": "68a3a6d1687ea43dc6cdfafcd5e6d9ce35f424e8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 81.0877192982, "max_line_length": 597, "alphanum_fraction": 0.7345305063, "num_tokens": 1137, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8991213745668094, "lm_q2_score": 0.7057850278370112, "lm_q1q2_score": 0.6345864043774874}} {"text": "\\section{Algebraic Manipulations}\n\\subsection{Definitions}\nSophie Germain: \\[x^4 + 4y^4 = (x^2-2xy+y^2)(x^2+2xy+y^2)\\] \\\\\n$(x+y)^3 = x^3+3xy(x+y)+y^3 = x^3+3x^2y+3xy^2+y^3$. \\\\\n$(x-y)^3 = x^3-3xy(x-y)+y^3 = x^3-3x^2y+3xy^2-y^3$. \\\\\n$x^3 + y^3 = (x+y)(x^2-xy+y^2)$. \\\\\n$x^3 - y^3 = (x-y)(x^2+xy+y^2)$. \\\\\n$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2−xy−yz−zx)$.\n\n\\subsection{Practice Problems}\n\\begin{enumerate}\n \\item Given that $x$ and $y$ are distinct nonzero real numbers such that $x+2/x=y+2/y$, what is $xy$?\n \\\\ \\textbf{Solution}: \n \\begin{align*}\n x - y &= 2/y - 2/x \\\\\n x - y &= \\frac{2x - 2y}{xy} \\\\\n &= \\frac{2(x-y)}{xy} \\\\\n &= \\frac{2}{xy} = 1 \\\\\n xy &= \\boxed{2}.\n \\end{align*}\n \\item Suppose that real number $x$ satisfies $\\sqrt{49-x^2} - \\sqrt{25 - x^2} = 3$. What is the value of $\\sqrt{49-x^2} + \\sqrt{25 - x^2}$\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n &(\\sqrt{49-x^2} - \\sqrt{25 - x^2})(\\sqrt{49-x^2} + \\sqrt{25 - x^2}) \\\\\n &= (49-x^2) - (25 - x^2) \\\\\n &= 24 = 3(\\sqrt{49-x^2} + \\sqrt{25 - x^2})\n (\\sqrt{49-x^2} + \\sqrt{25 - x^2}) = \\boxed{8}.\n \\end{align*}\n \\item What is the minimum value of the expression $x^2 + 8x + 13$ for any real $x$?\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n x^2 + 8x + 13 = (x + 4)^2 - 3 \\\\\n \\min((x + 4)^2 - 3) = \\boxed{-3}.\n \\end{align*}\n \\item Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$. What is $x+y$?\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n (x^2-10x) + (y^2+6y) = -34 \\\\\n (x-5)^2+(y+3)^2 = 0 \\\\\n x = 5, y = -3 \\\\\n x+y = \\boxed{2}.\n \\end{align*}\n \\item There is a positive integer $n$ such that $(n+1)!+(n+2)! = n!\\cdot 440$. What is the sum of the digits of $n$?\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n n!(n+1)^2(n+2) &= n!(440) \\\\\n (n+1)(n+3) &= 440 \\\\\n ((n+2)+1)((n+2)-1) &= 440 \\\\\n (n+2)^2-1^2 &= 440 \\\\\n (n+2)^2 &= 441 \\\\\n n + 2 &= 21 \\\\\n n &= \\boxed{19}.\n \\end{align*}\n \\item For all integers $n \\geq 9$, the value of $\\frac{(n+2)! - (n+1)!}{n!}$ is always which of the following?\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n \\frac{n!((n+2)(n+1)-(n+1))}{n!} = (n+1)^2.\n \\end{align*}\n The expression must always be a perfect square.\n \\item Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum\n $f \\left(\\frac{1}{2019} \\right)-f \\left(\\frac{2}{2019} \\right)+f \\left(\\frac{3}{2019} \\right)-f \\left(\\frac{4}{2019} \\right)+\\cdots + f \\left(\\frac{2017}{2019} \\right) - f \\left(\\frac{2018}{2019} \\right)?$\n \\\\ $\\textbf{(A) }0\\qquad\\textbf{(B) }\\frac{1}{2019^{4}}\\qquad\\textbf{(C) }\\frac{2018^{2}}{2019^{4}}\\qquad\\textbf{(D) }\\frac{2020^{2}}{2019^{4}}\\qquad\\textbf{(E) }1$\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n \\left( f \\left(\\frac{1}{2019} \\right) - f \\left(\\frac{1}{2019} \\right) \\right) + \\left( f \\left(\\frac{2}{2019} \\right) - f \\left(\\frac{2}{2019} \\right) \\right) + \\cdots \\\\ + \\left( f \\left(\\frac{1009}{2019} \\right) - f \\left(\\frac{1009}{2019} \\right) \\right)\n \\end{align*}\n The answer is $\\boxed{\\textbf{(A) }0}$.\n \\item Given that $x+\\frac{1}{x}=3$, find \n \\begin{enumerate}\n \\item $x^2+\\frac{1}{x^2} = (x + \\frac{1}{x})^2 = x^2 + 2 + \\frac{1}{x^2} = 3^2 - 2 = \\boxed{7}$.\n \\item $x^3 + \\frac{1}{x^3} = (x+\\frac{1}{x})(x^2-1+\\frac{1}{x^2}) = 3(7-1)=\\boxed{18}$.\n \\item $x^4 + \\frac{1}{x^4} = 7^2 + 2 = 49 + 2 = \\boxed{51}$.\n \\end{enumerate}\n \\item Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \\cdot y = -2$. What is the value of\\[x + \\frac{x^3}{y^2} + \\frac{y^3}{x^2} + y?\\]\n $\\textbf{(A)}\\ 360\\qquad\\textbf{(B)}\\ 400\\qquad\\textbf{(C)}\\ 420\\qquad\\textbf{(D)}\\ 440\\qquad\\textbf{(E)}\\ 480$\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n &\\frac{x^3}{x^2}+\\frac{x^3}{y^2}+\\frac{y^3}{x^2}+\\frac{y^3}{y^2} \\\\\n &= \\frac{x^3+y^3}{x^3}+\\frac{x^3+y^3}{y^2} \\\\\n &= \\frac{88}{x^2} = \\frac{88(x^2+y^2)}{4} = \\boxed{44}.\n \\end{align*}\n \\item Let $r$, $s$, and $t$ be the three roots of the equation\\[8x^3 + 1001x + 2008 = 0.\\]Find $(r + s)^3 + (s + t)^3 + (t + r)^3$.\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n (r+s)^3 + (s+t)^3 + (t+r)^3 \\\\ = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3) \\\\\n r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0 \\\\\n r^3+s^3+t^3 = 3rst = -251, -251 \\cdot -3 = \\boxed{753}.\n \\end{align*}\n \\item Find $3x^2y^2$ is $x$ and $y$ are integers such that $y^2+3x^2y^2=30x^2+517$.\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n a = x^2, b = y^2 \\\\\n b+3ab = 30a+517 \\\\\n ab + \\frac{b}{3} - 10a = \\frac{517}{3} \\\\\n (a+\\frac{1}{3})(b-10) = \\frac{507}{3} \\\\\n (3x^2 + 1)(y^2 - 10) = 507 \\\\\n (x, y) = (\\pm 2, \\pm 7)\n \\end{align*}\n $3x^2y^2 = 3 \\cdot 4 \\cdot 49 = \\boxed{588}$.\n \\item What is the remainder when $s^{202} + 202$ is divided by $2^{101} + 2^{51} + 1$.\n \\item Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\\frac {a}{b} + \\frac {b}{a} - ab$?\n\n \\textbf{Solution}:\n \\begin{align*}\n \\frac{a^2 + b^2}{ab} - ab &= \\frac{(a-b)^2+2ab}{ab} - ab \\\\\n &= \\frac{(ab)^2 + 2ab}{ab} - ab \\\\\n &= ab + 2 - ab = \\boxed{2} \\\\\n \\end{align*}\n \\item If $x + \\frac{1}{x} = 4$, then what is the value of $x^8 + \\frac{1}{x^8}$?\n\n \\textbf{Solution}:\n \\begin{align*}\n x^2 + \\frac{1}{x^2} = 4^2 - 2 = 14 \\\\\n x^4 + \\frac{1}{x^4} = 14^2 - 2 = 194 \\\\\n x^8 = \\frac{1}{x^8} = 194^2 - 2 = \\boxed{37624}.\n \\end{align*}\n \\item Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\\dfrac{a^3-b^3}{(a-b)^3} = \\dfrac{73}{3}.$ What is $a-b?$\n \\\\ \\textbf{Solution}:\n \\begin{align*}\n \\frac{a^3-b^3}{(a-b)^3} &= \\frac{(a-b)(a^2+ab+b^2)}{(a-b)^3} \\\\\n &= \\frac{a^2+ab+b^2}{(a-b)^2} \\\\\n &= \\frac{(a-b)^2 - 3ab}{(a-b)^2} \\\\\n &= 1+\\frac{3ab}{(a-b)^2} \\\\\n \\frac{3ab}{(a-b)^2} &= \\frac{70}{3} \\\\\n 9ab &= 70(a-b)^2 \\\\\n 3 \\sqrt{\\frac{ab}{70}} &= a - b \\\\\n ab | 70&, 3 | a - b.\n \\end{align*}\n Thus $a = 10, b = 7$ so $a - b = \\boxed{3}$. \n \\item If $(a - 1/a)^2 = 4$, what is the absolute value of $a^3 - 1/a^3$? \n\n \\textbf{Solution}:\n \\begin{align*}\n a - \\frac{1}{a} &= 2 \\\\\n (a - \\frac{1}{a})^3 &= a^3 - \\frac{1}{a^3} - 3(x - \\frac{1}{x}) \\\\\n 2^3 &= a^3 - \\frac{1}{a^3} - 6 \\\\\n a^3 - \\frac{1}{a^3} &= \\boxed{14}\n \\end{align*}\n\\end{enumerate}\n", "meta": {"hexsha": "74c56b15d0ea01fd37bdd841c890e0a7020fa46b", "size": 6215, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/algebra/algebraic-manipulations/algebraic-manipulations.tex", "max_stars_repo_name": "coderinblack08/math-binder", "max_stars_repo_head_hexsha": "5126211d519de4835e5350babdb6bb2c752a0b62", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-12-07T01:58:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-21T17:58:32.000Z", "max_issues_repo_path": "tex/algebra/algebraic-manipulations/algebraic-manipulations.tex", "max_issues_repo_name": "coderinblack08/math-binder", "max_issues_repo_head_hexsha": "5126211d519de4835e5350babdb6bb2c752a0b62", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/algebra/algebraic-manipulations/algebraic-manipulations.tex", "max_forks_repo_name": "coderinblack08/math-binder", "max_forks_repo_head_hexsha": "5126211d519de4835e5350babdb6bb2c752a0b62", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.3928571429, "max_line_length": 263, "alphanum_fraction": 0.4962188254, "num_tokens": 2976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8991213718636754, "lm_q2_score": 0.7057850278370112, "lm_q1q2_score": 0.6345864024696558}} {"text": "\\documentclass[main.tex]{subfiles}\r\n\\begin{document}\r\n\\chapter{Set Theory}\r\n\r\n\\epigraph{Sets are wild fam.}{Justin Goodman}\r\n\r\n\\minitoc\r\n\r\n\\section{Introduction}\r\n\r\nSets were introduced by Georg Cantor in the late 1800s. Cantor is the grandfather of set theory and continuity. We will see continuity topics later, including Cantor's famous diagonal argument. We can define all of discrete mathematics using sets. What is a set though?\r\n\r\n\\begin{defn}[Set\\index{Set}]\r\n\tAn unordered collection of unique objects. We denote sets using curly braces \\(\\{\\}\\) with objects appearing in them -- e.g.\\ \\(\\{\\circ, 3, \\pi, \\blacktriangle\\}\\)\r\n\\end{defn}\r\n\r\nEach part of this definition is important -- we do not have a set unless it satisfies the entire definition. We now examine the definition.\r\n\r\n\\begin{itemize}\r\n\t\\item A set is a \\textbf{collection} of stuff. Think of a set like a box. You can put things in your box, and you can take them out. Your box is special -- it can expand/contract to fit anything you like.\r\n\t\\item A set is \\textbf{unordered}. This simply means that any different ordering we give to a set does not change the equality property of the set -- \\(\\{1,2\\} = \\{2,1\\}\\)\r\n\t\\item A set contains \\textbf{objects}. A set can contain anything you want.\r\n\t\\item A set contains \\textbf{unique} objects. For any two distinct objects in a set, the objects cannot be equal. Sometimes we see books describe the sets \\(\\{1,1,2,3\\}\\) and \\(\\{1,2,3\\}\\) as equal, however we argue that the first set is not even a set!\\footnote{We would classify the set \\(\\{1,1,2,3\\}\\) as a \\textit{multi-set}} We recommend you ask your instructor about this distinction, and follow what they prefer.\r\n\\end{itemize}\r\n\r\nThis chapter includes an overview of set theory -- sets, operations, binary relations, and theorems.\r\n\r\n\\section{Building Sets}\r\n\r\nBefore we dive into set theory concepts, we first introduce a few ways to denote sets.\r\n\r\nTo start, you can simply denote a set by just writing each element inside some curly braces. For example, \\[\\{1,2,3\\}\\] describes the set containing 1, 2, and 3. This method is not useful when describing big sets, though. If your set has \\(2^{64}\\) elements, you would never be able to write them all out!\r\n\r\nTo solve this, we can use \\textit{ellipses} -- \\(\\cdots\\) -- three dots in a row. Ellipses inside a set simply mean that you take the implicit pattern described in the set already, and continue it (possibly indefinitely). For example, \\[\\{1,2,3,\\cdots,10\\} = \\{1,2,3,4,5,6,7,8,9,10\\}\\]\r\nHere, the pattern is described in the \\(1,2,3\\) part -- increase by 1. Now, \r\n\r\nSo far, the presented notations lack power. To combat this, we introduce set-builder notation. As the name implies, this is a way of building sets.\r\n\r\n\\begin{defn}[Set-builder Notation\\index{Set-builder Notation}]\r\n\tA way of defining sets. Syntax: \\(\\{\\,\\mbox{element} \\mid \\mbox{condition(s)}\\,\\}\\), read as, ``element \\textit{such that} condition(s) is (are) satisfied.'' For example, the set of even integers (which we will learn about soon) can be written as: \\(\\{e \\mid (\\exists k \\in \\Z)[e = 2k]\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{rem}\r\n\tYou may also see the \\(:\\) character instead of the \\(\\mid\\) character in set-builder notation. These are equivalent. \\(\\{e \\mid (\\exists k \\in \\Z)[e = 2k]\\} = \\{e : (\\exists k \\in \\Z)[e = 2k]\\}\\).\r\n\\end{rem}\r\n\r\nIt is pivotal that you know how to read and create sets this way. The syntax is very flexible because the conditions can be almost anything you like. The conditions should, however, relate to the set itself (otherwise the set would be trivially pointless).\r\n\r\n\\exsol{\r\n\tBuild a set that contains all square roots of even integers (for now, use the definition of even integers given previously).\r\n}{\r\n\t\\[S = \\{x \\mid (\\exists k \\in \\Z)[x^2 = 2k]\\}\\]\r\n}\r\n\r\n\\exsol{\r\n\tBuild a set that contains everything except for the object \\(\\star\\).\r\n}{\r\n\t\\[S = \\{y : y \\neq \\star\\}\\]\r\n}\r\n\r\nWith set-builder notation, we can describe the following set short-cut: \\[[n] = \\{i \\in \\Z \\mid 0 \\leq i \\leq n\\}\\].\r\nWe do not know what \\(\\Z\\) is just yet, but we will get there soon enough.\r\n\r\n\\begin{rem}\r\n\tSome authors denote \\([3] = \\{1,2,3\\}\\) since they do not include 0 as part of the natural numbers. More on this later. For now, just follow whatever your professor is doing.\r\n\\end{rem}\r\n\r\n\\begin{rem}\r\n\tSome authors also reserve \\([x]\\) to mean the floor function -- the greatest integer smaller than \\(x\\). As with anything in math, context is key. We will not use brackets to indicate the floor function, but you should be aware that different notation conventions exist.\r\n\\end{rem}\r\n\r\nFinally, we can describe continuous-interval sets. You may be familiar with the real number line -- this is a continuous line because there are no ``breaks'' between any two numbers you pull from it. Given any two numbers on the real number line, we can always take the midpoint to get another real number! How do we denote these intervals? We use interval notation:\r\n\\begin{itemize}\r\n\t\\item \\((x,y) = \\{r \\in \\R : x < r < y\\}\\)\r\n\t\\item \\((x,y] = \\{r \\in \\R : x < r \\leq y\\}\\)\r\n\t\\item \\([x,y) = \\{r \\in \\R : x \\leq r < y\\}\\)\r\n\t\\item \\([x,y] = \\{r \\in \\R : x \\leq r \\leq y\\}\\)\r\n\\end{itemize}\r\n\r\nA parenthesis means we \\textit{exclude} the associated number from the continuous interval, and a bracket means we \\textit{include} the associated number.\r\n\r\n\\section{Definitions}\r\n\r\nWe include a handful of definitions and notations we use in our study of set theory.\r\n\r\nBinary relations:\r\n\r\n\\begin{defn}[Member/Element]\r\n\tAn object that is part of a set. Symbol: \\(\\in\\). Example: \\(5 \\in \\{1,2,3,4,5\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Subset]\r\n\tA set of elements that are all members of another set. A subset CAN be equal to its parent set. Symbol: \\(\\subseteq\\). Example: for sets \\(S\\) and \\(T\\), \\(S \\subseteq T \\Leftrightarrow (\\forall s \\in S)[s \\in T]\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Proper Subset]\r\n\tA set of elements that are all members of another set, but the subset is NOT equal to the parent set. Symbol: \\(\\subset\\). Example: for sets \\(S\\) and \\(T\\), \\(S \\subset T \\Leftrightarrow (\\forall s \\in S)[(s \\in T) \\land (S \\neq T)]\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Superset]\r\n\tA flipped version of subset. Symbol: \\(\\supseteq\\). Example: for sets \\(S\\) and \\(T\\), \\(S \\supseteq T \\Leftrightarrow (\\forall t \\in T)[t \\in S]\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Proper Superset]\r\n\tA flipped version of proper subset. Symbol: \\(\\supset\\). Example: for sets \\(S\\) and \\(T\\), \\(S \\supset T \\Leftrightarrow (\\forall t \\in T)[(t \\in S) \\land (S \\neq T)]\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Equality]\r\n\tTwo sets are equal if and only if both sets are subsets of each other. Notation: for sets \\(S\\) and \\(T\\), \\(S = T \\Leftrightarrow (S \\subseteq T) \\land (T \\subseteq S)\\). To prove this, you can use set-builder notation \\& theorems (given later in this chapter) to show equivalence, or you can prove \\((e \\in S \\Rightarrow e \\in T) \\land (e \\in T \\Rightarrow e \\in S)\\)\r\n\\end{defn}\r\n\r\nThings:\r\n\r\n\\begin{defn}[Cardinality]\r\n\tThe number of elements in a set. Notation: for a set \\(S\\), cardinality is denoted by \\(|S|\\). For example, $|\\{2,3,4,5,6\\}| = 5$\r\n\\end{defn}\r\n\r\n\\begin{defn}[Empty/Null Set]\r\n\tThe set containing zero elements. Notation: $\\emptyset$ or $\\{\\}$ -- these symbols are \\textit{interchangable}. \\textbf{Note}: \\(|\\emptyset| = 0\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Universal Set]\r\n\tThe set of all possible sets. Notation: \\(U\\) is the Universal Set\r\n\\end{defn}\r\n\r\n\\begin{defn}[Power Set]\r\n\tThe set of all possible subsets of a set. Notation: of a set \\(S\\), the power set is denoted \\(\\mathcal{P}(S)\\). For example, \\(\\mathcal{P}(\\{1,2,3\\}) = \\{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\}, \\{1,3\\}, \\{2,3\\}, \\{1,2,3\\}\\}\\). \\textbf{Note}: for any set \\(S\\), \\(|\\mathcal{P}(S)| = 2^{|S|}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Disjoint]\r\n\tTwo sets are disjoint if and only if both sets have no members in common. For two sets \\(S\\) and \\(T\\), this is equivalent to \\(S \\cap T = \\emptyset\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Partition]\r\n\t(of a set) A set of sets \\(T\\) where the union of every element in \\(T\\) equals the original set, and all elements in \\(T\\) are disjoint. For example, \\(\\{\\{1,2\\}, \\{3\\}, \\{4,5,6\\}\\}\\) is a partition of \\(\\{1,2,3,4,5,6\\}\\)\r\n\\end{defn}\r\n\r\nOperations:\r\n\r\n\\begin{defn}[Union]\r\n\t(of two sets) A set that includes all elements from both sets (discounting duplicates, since a set must contain unique elements). Notation: for sets \\(S\\) and \\(T\\), \\(S \\cup T = \\{e \\mid (e \\in S) \\lor (e \\in T)\\}\\). For example, \\(\\{1,2,3\\} \\cup \\{3,4,5\\} = \\{1,2,3,4,5\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Intersection]\r\n\t(of two sets) A set that includes only elements from both sets. Notation: for sets \\(S\\) and \\(T\\), \\(S \\cap T = \\{e \\mid (e \\in S) \\land (e \\in T)\\}\\). For example, \\(\\{1,2,3\\} \\cap \\{3,4,5\\} = \\{3\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Subtraction]\r\n\t(of two sets) A set representing the elements of the second set taken out of the first set. \\textbf{Note}: subtraction order \\textit{matters} (i.e.\\ subtraction is not commutative). Notation: for sets \\(S\\) and \\(T\\), \\(S - T = \\{e \\mid (e \\in S) \\land (e \\not\\in T)\\}\\). You may also see set subtraction represented as \\(S \\setminus T\\). For example, \\(\\{1,2,3\\} \\setminus \\{3,4,5\\} = \\{1,2\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Compliment]\r\n\t(of a set) A set containing every element from the universal set that is NOT contained in the original set. \\textbf{Note}: you can take the compliment with respect to different universes so long as you specify which one -- by default the universal set is assumed. Notation: for a set \\(S\\), the complement is noted as \\(S^{\\mathsf{c}}\\), or \\(S^{'}\\), or \\(\\overline{S}\\); with the universe \\(U\\), \\(S^{'} = \\{e \\mid e \\in (U - S)\\}\\)\r\n\\end{defn}\r\n\r\n\\begin{defn}[Cross Product]\r\n\t(of two sets) The set of all ordered pairings of two sets. Notation: for sets \\(S\\) and \\(T\\), \\(S \\times T = \\{(s,t) \\mid s \\in S \\land t \\in T\\}\\).\r\n\t\r\n\tNote: you can take the cross product of multiple sets. For sets \\(A_1 \\cdots A_n\\), the cross product \\(A_1 \\times A_2 \\times \\cdots \\times A_n = \\{(a_1, a_2, \\cdots, a_n) \\mid a_i \\in A_i\\}\\). When \\(A_1 = \\cdots = A_n = A\\) then we let \\(A^n = A_1 \\times A_2 \\times \\cdots \\times A_n\\)\r\n\\end{defn}\r\n\r\n\\section{Theorems}\r\n\r\nWe will not go into depth on the axioms of Zermelo–Fraenkel set theory. We do include a handful of nice theorems that will aid in proving statements about sets. Sometimes these are referred to as axioms, however we argue that the following statements can be derived from ZF set theory axioms and should hence be called theorems. It does not really matter though.\r\n\r\nYou are not required to memorize these theorems -- they will be given to you as a table.\r\n\\marginpar{See appendix \\ref{appendix:tables}.}\r\n\r\n\\begin{thm}[Commutativity]\r\n\tFor any sets \\(A\\) and \\(B\\) the union and intersection operations are commutative: \\[A \\cup B = B \\cup A\\] \\[A \\cap B = B \\cap A\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Associativity]\r\n\tFor any sets \\(A\\), \\(B\\), and \\(C\\) the union and intersection operations are associative: \\[(A \\cup B) \\cup C = A \\cup (B \\cup C)\\] \\[(A \\cap B) \\cap C = A \\cap (B \\cap C)\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Distributivity]\r\n\tFor any sets \\(A\\), \\(B\\), and \\(C\\) the union and intersection operations are distributive: \\[A \\cap (B \\cup C) = (A \\cap B) \\cup (A \\cap C)\\] \\[A \\cup (B \\cap C) = (A \\cup B) \\cap (A \\cup C)\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Identity]\r\n\tFor any set \\(A\\) and universe \\(U\\) the following hold: \\[A \\cup \\emptyset = A\\] \\[A \\cap U = A\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Inverse]\r\n\tFor any set \\(A\\) and universe \\(U\\) the following hold: \\[A \\cup A^{\\mathsf{c}} = U\\] \\[A \\cap A^{\\mathsf{c}} = \\emptyset\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Double Compliment]\r\n\tFor any set \\(A\\) the following holds: \\[(A^{\\mathsf{c}})^{\\mathsf{c}} = A\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Idempotence]\r\n\tFor any set \\(A\\) the following hold: \\[A \\cup A = A\\] \\[A \\cap A = A\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Universal Bound (Domination)]\r\n\tFor any set \\(A\\) and universe \\(U\\) the following hold: \\[A \\cup U = U\\] \\[A \\cap \\emptyset = \\emptyset\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[De Morgan's]\r\n\tFor any sets \\(A\\) and \\(B\\) the following hold: \\[(A \\cup B)^{\\mathsf{c}} = A^{\\mathsf{c}} \\cap B^{\\mathsf{c}}\\] \\[(A \\cap B)^{\\mathsf{c}} = A^{\\mathsf{c}} \\cup B^{\\mathsf{c}}\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Absorption]\r\n\tFor any sets \\(A\\) and \\(B\\) the following hold: \\[A \\cup (A \\cap B) = A\\] \\[A \\cap (A \\cup B) = A\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Absolute Compliment]\r\n\tFor a given universe \\(U\\) the following hold: \\[\\emptyset^{\\mathsf{c}} = U\\] \\[U^{\\mathsf{c}} = \\emptyset\\]\r\n\\end{thm}\r\n\r\n\\begin{thm}[Set Subtraction Equality]\r\n\t\\label{set-sub-eq}\r\n\tFor any sets \\(A\\) and \\(B\\) the set subtraction operation satisfies the following: \\[A - B = A \\cap B^{\\mathsf{c}}\\]\r\n\tThis establishes a relationship between the relative and absolute compliment\r\n\\end{thm}\r\n\r\nThe aforementioned theorems are helpful for simplifying complicated sets.\r\n\r\n\\exsol{\r\n\tSimplify the following expression: \\[((A \\cup B) \\cap C) \\cup ((A^{\\mathsf{c}} \\cap B^{\\mathsf{c}}) \\cup D^{\\mathsf{c}})^{\\mathsf{c}}\\]\r\n}{\r\n\tLots of compliments is a good indication for using De Morgan's.\r\n\t\\begin{align*}\r\n\t& ((A \\cup B) \\cap C) \\cup ((A^{\\mathsf{c}} \\cap B^{\\mathsf{c}}) \\cup D^{\\mathsf{c}})^{\\mathsf{c}} \\\\\r\n\t&= ((A \\cup B) \\cap C) \\cup ((A \\cup B)^{\\mathsf{c}} \\cup D^{\\mathsf{c}})^{\\mathsf{c}} & \\text{De Morgan's} \\\\\r\n\t&= ((A \\cup B) \\cap C) \\cup ((A \\cup B) \\cap D) & \\text{De Morgan's} \\\\\r\n\t&= (A \\cup B) \\cap (C \\cup D) & \\text{Distributivity}\r\n\t\\end{align*}\r\n}\r\n\r\n\\exsol{\r\n\t\\label{ex-prove-equivalence}\r\nShow the following equivalence: \\[A \\cup (B \\cup (A \\cap C)) = A \\cup B\\]\r\n}{\r\n\tSometimes just trying random things works out in your favor.\r\n\t\\begin{align*}\r\n\tA \\cup (B \\cup (A \\cap C)) &= A \\cup ((A \\cap C) \\cup B) & \\text{Commutativity} \\\\\r\n\t&= (A \\cup (A \\cap C)) \\cup B & \\text{Associativity} \\\\\r\n\t&= A \\cup B & \\text{Absorption}\r\n\t\\end{align*}\r\n}\r\n\r\n\\begin{rem}\r\n\tThere is a stark similarity between set and Boolean simplification.\r\n\\end{rem}\r\n\r\n\\section{Important Sets}\r\n\r\nWe introduce some notation for a handful of important sets.\r\n\r\n\\begin{defn}[The Natural Numbers -- \\(\\N\\)]\r\n\tThe standard discrete numbers with which you count. Some math classes start the naturals at 1, however in computer science we start the naturals at 0. Just remember simply that arrays utilize 0-indexing, so we do the same. The set looks like so: \\(\\{0,1,2,3,4,5,\\cdots\\}\\)\r\n\\end{defn}\r\n\r\nIn mathematics, we define the natural numbers \\textit{inductively} -- we will discuss induction in a few chapters. We introduce the inductive definition here. You do not need to know this, however we think it is interesting.\r\n\r\n\\begin{prop}[Inductive Definition of \\(\\N\\)]\r\n\t\\label{ind-defn-N}\r\n\tDefine a set \\(S \\subseteq \\R\\) to be \\textit{inductive} if and only if the following conditions hold:\r\n\t\\begin{itemize}\r\n\t\t\\item \\(0 \\in S\\)\r\n\t\t\\item if \\(x \\in S\\) then \\(x + 1 \\in S\\)\r\n\t\\end{itemize}\r\n\t\r\n\t\\noindent Define \\(\\N\\) as the intersection of all inductive sets.\r\n\\end{prop}\r\n\r\nUnderstandably you may be confused on the notation of \\(\\R\\) -- we will come back to this in a moment. For now, we continue to the integers.\r\n\r\n\\begin{defn}[The Integers -- \\(\\Z\\)]\r\n\tThe standard \\textit{signed} discrete numbers with which you count. The integers include all of the natural numbers as well as all of their negatives\\footnote{0 is neither positive nor negative, so we cannot take its negation, however we let \\(0 \\in \\Z\\)}. The set looks like so: \\(\\{\\cdots,-3,-2,-1,0,1,2,3,\\cdots\\}\\)\r\n\\end{defn}\r\n\r\nThe next set, which you may be familiar with, is the rationals.\r\n\r\n\\begin{defn}[The Rationals -- \\(\\Q\\)]\r\n\tThe set of numbers that can be written as a ratio (\\(\\Q\\)uotient) of integers. \\(\\Q = \\{x = \\frac{a}{b} \\mid a \\in \\Z \\land b \\in \\Z^{\\neq 0}\\}\\)\r\n\\end{defn}\r\n\r\nNaturally we can define somewhat of an `opposite' to the rationals.\r\n\r\n\\begin{defn}[The Irrationals -- \\(\\R \\setminus \\Q\\)]\r\n\tThe set of real numbers that do not satisfy the rational property. A provable example is that \\(\\sqrt{2} \\in \\R \\setminus \\Q\\)\r\n\\end{defn}\r\n\r\nThe real numbers are not the main focus in discrete mathematics, however we still provide a definition.\r\n\r\n\\begin{defn}[The Reals -- \\(\\R\\)]\r\n\tThe continuous interval \\((-\\infty, \\infty)\\). Any number that does not have the form \\(a+bi\\), where \\(i = \\sqrt{-1}\\)\r\n\\end{defn}\r\n\r\nIn an analytical mathematics course, the reals and irrationals are more strongly defined. The above definitions are enough for this course.\r\n\r\n\\section{Sets to Logic}\r\n\r\nWe can use set-builder notation along with our familiar logic rules to prove things about sets. This proof technique can be used to prove the theorems we showed earlier.\r\n\r\n\\exproof{\r\n\tProve theorem \\ref{set-sub-eq}: \\[A - B = A \\cap B^{\\mathsf{c}}\\]\r\n}{\r\n\t\\begin{align*}\r\n\tA - B &= \\{x \\mid x \\in A \\land x \\not\\in B\\} & \\text{defn of subtraction} \\\\\r\n\t&= \\{x \\mid x \\in A \\land x \\in B^{\\mathsf{c}}\\} & \\text{defn of compliment} \\\\\r\n\t&= A \\cap B^{\\mathsf{c}} & \\text{defn of intersection}\r\n\t\\end{align*}\r\n}\r\n\r\n\\exproof{\r\n\tProve example \\ref{ex-prove-equivalence} \\[A \\cup (B \\cup (A \\cap C)) = A \\cup B\\]\r\n}{\r\n\t\\begin{align*}\r\n\t& A \\cup (B \\cup (A \\cap C)) \\\\\r\n\t&= \\{x \\mid x \\in A \\lor (x \\in B \\lor (x \\in A \\land x \\in C))\\} & \\text{defn of } \\cup/\\cap \\\\\r\n\t&= \\{x \\mid x \\in A \\lor ((x \\in A \\land x \\in C) \\lor x \\in B)\\} & \\text{Commutativity (logic)} \\\\\r\n\t&= \\{x \\mid (x \\in A \\lor (x \\in A \\land x \\in C)) \\lor x \\in B\\} & \\text{Associativity (logic)} \\\\\r\n\t&= \\{x \\mid x \\in A \\lor x \\in B\\} & \\text{Absorption (logic)} \\\\\r\n\t&= A \\cup B & \\text{defn of } \\cup\r\n\t\\end{align*}\r\n}\r\n\r\n\\section{Summary}\r\n\r\n\\begin{itemize}\r\n\t\\item Sets are unordered collections of unique objects\r\n\t\\item Many operations and theorems exist in set theory\r\n\t\\item \\(\\N, \\Z, \\Q, \\R \\setminus \\Q, \\R\\) are all important sets\r\n\\end{itemize}\r\n\r\n\\section{Practice}\r\n\r\n\\begin{enumerate}\r\n\t\\item Provide an example of a valid set, and an invalid set.\r\n\t\\item Build a set of natural numbers that are all multiples of 3.\r\n\t\\item Explain why the cardinality of a finite set is a natural number.\r\n\t\\item For two sets \\(A\\) and \\(B\\), is it necessarily the case that \\(|A \\cup B| = |A| + |B|\\)?\r\n\t\\item Explain intuitively why \\(|\\mathcal{P}(A)| = 2^{|A|}\\).\r\n\t\\item Is \\(\\{\\Z^-, 0, \\Z^+\\}\\) a valid partition of \\(\\Z\\)? Explain. If it is not, change the partition to make it valid.\r\n\t\\item Is \\(\\{\\{1,3,5,7,9,\\cdots\\} , \\{0,2,4,6,8,\\cdots\\}\\}\\) a valid partition of \\(\\N\\)? Explain. If it is not, change the partition to make it valid.\r\n\t\\item Provide a set \\(T\\) which is a valid partition of \\(\\N\\) such that \\(|T| = 5\\).\r\n\t\\item For each of the following, answer true or false:\r\n\t\\begin{enumerate}\r\n\t\t\\item \\(\\Z \\supset \\N\\)\r\n\t\t\\item \\(\\N \\in \\Z\\)\r\n\t\t\\item \\(0 \\in \\emptyset\\)\r\n\t\t\\item \\(\\emptyset \\in \\emptyset\\)\r\n\t\t\\item \\(\\emptyset \\subseteq \\emptyset\\)\r\n\t\t\\item \\(\\emptyset \\subset \\emptyset\\)\r\n\t\t\\item \\(\\mathcal{P}(\\emptyset) = \\emptyset\\)\r\n\t\\end{enumerate}\r\n\t\\item Simplify the expression \\(A^{\\mathsf{c}} \\cup (B \\cup A)^{\\mathsf{c}}\\).\r\n\t\\item Prove the two Absorption theorem statements are equivalent.\r\n\\end{enumerate}\r\n\r\n%\\section{Solutions}\r\n%\r\n%\\begin{enumerate}\r\n%\t\\item Valid: \\(\\{1,2,3\\}\\). Invalid: \\(1,2,3\\) (this is invalid because it is missing the \\(\\{\\}\\))\r\n%\t\\item Let \\(S = \\{x \\mid (\\exists k \\in \\N)[x = 3k]\\}\\)\r\n%\t\\item Consider that \\(\\N\\) contains all of the \\textit{counting} numbers. By taking the cardinality, you are simply \\textit{counting} the number of elements.\r\n%\t\\item The only case when \\(|A \\cup B| = |A| + |B|\\) is if \\(A \\cap B = \\emptyset\\). For example, \\(|\\{1,2,3\\} \\cup \\{3,4,5\\}| = |\\{1,2,3,4,5\\}| = 5 \\neq 6 = |\\{1,2,3\\}| + |\\{3,4,5\\}|\\)\r\n%\t\\item Consider each subset as a binary string, where each position in the string represents an element. 1 means the element is in the subset, and 0 means the element is not in the subset. The amount of strings total is \\(2^{\\text{length of each string}}\\). Since the length of each string is just the number of elements in \\(A\\), we recover that \\(|\\mathcal{P}(A)| = 2^{|A|}\\). The intuition here is to think of each element as being in or out of a subset.\r\n%\t\\item No -- each element in a partition must be a set, and 0 is not a set.\r\n%\t\\item Yes -- the first set contains all odd naturals, and the second set contains all even naturals. Since there is no overlap between odds and evens, the partition is valid. We will learn about this soon.\r\n%\t\\item One possibility: \\(T = \\{\\{0,1,2\\}, \\{4,5\\}, \\{3,6,9\\}, \\{7,8\\}, \\N^{\\geq 10}\\}\\)\r\n%\t\\item \r\n%\t\\begin{enumerate}\r\n%\t\t\\item True\r\n%\t\t\\item False\r\n%\t\t\\item False\r\n%\t\t\\item False\r\n%\t\t\\item True\r\n%\t\t\\item False\r\n%\t\t\\item False\r\n%\t\\end{enumerate}\r\n%\t\\item \r\n%\t\\begin{align*}\r\n%\tA^{\\mathsf{c}} \\cup (B \\cup A)^{\\mathsf{c}} &= A^{\\mathsf{c}} \\cup (B^{\\mathsf{c}} \\cap A^{\\mathsf{c}}) & \\text{De Morgan's} \\\\\r\n%\t&= A^{\\mathsf{c}} \\cup (A^{\\mathsf{c}} \\cap B^{\\mathsf{c}}) & \\text{Commutativity} \\\\\r\n%\t&= A^{\\mathsf{c}} & \\text{Absorption} \\\\\r\n%\t\\end{align*}\r\n%\t\\item \r\n%\t\\begin{align*}\r\n%\tA \\cup (A \\cap B) &= (A \\cup A) \\cap (A \\cup B) & \\text{Distributivity} \\\\\r\n%\t&= A \\cap (A \\cup B) & \\text{Idempotence}\r\n%\t\\end{align*}\r\n%\\end{enumerate}\r\n\\end{document}\r\n", "meta": {"hexsha": "21ba81a8f759838318d24122dcdaa32a8aecbd9a", "size": 21257, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ch-settheory.tex", "max_stars_repo_name": "jugoodma/250-textbook", "max_stars_repo_head_hexsha": "ebfcd8e9d15079fe8924bf562a194ed057aed302", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-04-22T03:33:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-27T14:39:11.000Z", "max_issues_repo_path": "ch-settheory.tex", "max_issues_repo_name": "jugoodma/250-textbook", "max_issues_repo_head_hexsha": "ebfcd8e9d15079fe8924bf562a194ed057aed302", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ch-settheory.tex", "max_forks_repo_name": "jugoodma/250-textbook", "max_forks_repo_head_hexsha": "ebfcd8e9d15079fe8924bf562a194ed057aed302", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-06-19T22:24:49.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-19T22:24:49.000Z", "avg_line_length": 54.5051282051, "max_line_length": 459, "alphanum_fraction": 0.648727478, "num_tokens": 7042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672227971211, "lm_q2_score": 0.7853085808877581, "lm_q1q2_score": 0.634582123996719}} {"text": "%!TEX root = ../notes.tex\n\\section{May 2, 2022}\n\\subsection{LLL Reduction}\n\\ul{Problem}: Given a lattice, find a good basis for the given lattice. Let's say we have\n\\begin{align*}\n \\mathcal{B} & = (\\bvec{v}_1, \\bvec{v}_2, \\dots, \\bvec{v}_n) \\\\\n \\mathcal{B}' & = (\\bvec{v}_1', \\bvec{v}_2', \\dots, \\bvec{v}_n')\n\\end{align*}\nwhere $\\mathcal{B}'$ is the corresponding vector space obtained from Gram-Schmidt. Note that this is \\emph{not} a basis of the lattice.\n\nWe have\n\\[\\bvec{v}_i' = \\bvec{v}_i - \\sum_{j < i}\\mu_{ij}\\bvec{v}_j'\\]\nwhere $\\mu_{ij} = \\frac{\\bvec{v}_i\\cdot \\bvec{v}_j'}{\\bvec{v}_i'\\cdot\\bvec{v}_i'}$.\n\nWhat do we know about $\\det\\{\\bvec{v}_i\\}$ and $\\det\\{\\bvec{v}_i'\\}$? We note that $\\det\\{\\bvec{v}_i\\} = \\det\\{\\bvec{v}_i'\\}$. This is as our row/column operations (except scaling) don't change determinant.\n\n\\ul{Ideas}:\n\nWe want some $\\bvec{v}_1, \\bvec{v}_2, \\dots, \\bvec{v}_n$ are approximately orthogonal.\n\nWe want $\\bvec{v}_1 \\leq \\bvec{v}_2 \\leq \\dots \\leq \\bvec{v}_n$ (approximately sorted).\n\n\\begin{definition}[LLL Reduction]\n $\\mathcal{B}$ is LLL reduced if\n \\begin{enumerate}\n \\item \\emph{We want a measure of orthogonality, so we bound $\\mu_{ij}$ from Gram-Schmidt to a certain value. }\n\n We have that\n \\[\\boxed{|\\mu_{ij}| = \\left\\lvert\\frac{\\bvec{v}_i\\cdot \\bvec{v}_j'}{\\bvec{v}_i'\\cdot\\bvec{v}_i'}\\right\\rvert\\leq \\frac{1}{2}}\\]\n \\item \\emph{We want our measure in a way that failure of exact orthogonality doesn't contribute.}\n\n We take the projection of $\\bvec{v}_{i-1}$ onto\n \\[\\langle\\bvec{v}_1, \\dots, \\bvec{v}_{i-2}\\rangle^\\perp\\]\n and compare that with the length of the projection of $\\bvec{v}_i$ onto\n \\[\\langle\\bvec{v}_1, \\dots, \\bvec{v}_{i-1}\\rangle^\\perp\\]\n We want\n {\\small\n \\[\\frac{3}{4}\\left|\\left|\\ \\text{Projection of $\\bvec{v}_{i-1}$ onto } \\langle\\bvec{v}_1, \\dots, \\bvec{v}_{i-2}\\rangle^\\perp\\right|\\right|\\leq \\left|\\left|\\ \\text{Projection of $\\bvec{v}_{i}$ onto } \\langle\\bvec{v}_1, \\dots, \\bvec{v}_{i-1}\\rangle^\\perp\\right|\\right|\\]\n }\n \\emph{Noting that $\\frac{3}{4}$ was arbitrarly chosen as a `fudge factor'.}\n\n So we have\n \\[\\frac{3}{4}||\\bvec{v}_{i-1}||^2\\leq ||\\bvec{v}_i' + \\mu_{i-1, i}\\cdot\\bvec{v}_{i-1}'||^2 = ||\\bvec{v}_i||^2 + \\mu_{i-1, i}^2\\cdot||\\bvec{v}_{i-1}'||\\]\n so then\n \\[\\boxed{||\\bvec{v}_i^2||\\geq\\left( \\frac{3}{4} - \\mu_{i-1, i}^2 \\right)\\cdot ||\\bvec{v}_{i-1}'||^2}\\]\n \\end{enumerate}\n\\end{definition}\n\nOur goal is that we want an algorithm to find an LLL basis. Additionally, once we do find an LLL basis, we want to know that it will be sufficiently orthogonal.\n\nCode in \\textsf{lll.ipynb}.", "meta": {"hexsha": "97aba3bb29c2110606b8fba55c19a88bd23b6500", "size": 2811, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/2022-05-02.tex", "max_stars_repo_name": "jchen/math1580-notes", "max_stars_repo_head_hexsha": "9784be9e0faa57bbb3c421d8a104daadebf99a2f", "max_stars_repo_licenses": ["BSL-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-02-14T15:03:38.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-14T15:03:38.000Z", "max_issues_repo_path": "lectures/2022-05-02.tex", "max_issues_repo_name": "jchen/math1580-notes", "max_issues_repo_head_hexsha": "9784be9e0faa57bbb3c421d8a104daadebf99a2f", "max_issues_repo_licenses": ["BSL-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lectures/2022-05-02.tex", "max_forks_repo_name": "jchen/math1580-notes", "max_forks_repo_head_hexsha": "9784be9e0faa57bbb3c421d8a104daadebf99a2f", "max_forks_repo_licenses": ["BSL-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.1176470588, "max_line_length": 290, "alphanum_fraction": 0.5919601565, "num_tokens": 1054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7853085733507947, "lm_q1q2_score": 0.6345821106466725}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage{fullpage}\n\n\\usepackage{latexsym} % for some symbols\n\\usepackage{amsmath} % for maths\n\\usepackage{amssymb} % for Real number symbol\n\\usepackage{graphicx}\n\\usepackage{algorithm}\n\\usepackage{algorithmic}\n\\usepackage{subfigure}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Notation\n\n\\newcommand{\\normal}{\\mathcal{N}}\n\\newcommand{\\thetaall}{\\tilde{\\Theta}}\n\n\\newcommand{\\vect}[2]{\\begin{bmatrix} #1 \\\\ #2 \\end{bmatrix}}\n\\newcommand{\\mat}[4]{\\begin{bmatrix} #1 & #2\\\\ #3& #4 \\end{bmatrix}}\n\n\\newcommand{\\dotprod}[2]{\\langle #1 , #2 \\rangle}\n\\newcommand{\\trace}{\\mathrm{tr}}\n\\newcommand{\\deter}{\\mathrm{det}}\n\n% end notation\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\title{Distance between two 2D Gaussians}\n\\author{Cheng Soon Ong}\n\\date{24 March 2017}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Motivation}\n\nIn astronomy cross identification is the task of finding the same object in the sky in two images.\nThe location of each pixel is known, but the challenge lies in the fact that the two images are\nmeasured in different wavelengths.\nWhen doing radio cross identification we need to compare two distributions, one from the optical\nimage and a second from the radio image.\nWe assume here that each empirical distribution is well approximated by a single two dimensional\nGaussian.\nThe first step to cross identification is then to compute the distance between the two Gaussians.\n\n\\section{General abstract setup}\n\nOne way to measure the difference between two distributions is by using the Kullback-Leibler (KL)\ndivergence. This is asymmetric, and one often looks at the average of the KL divergence in\nboth directions. This section selects the relevant results\nfrom~\\cite{nielsen09clumnd,nielsen11staefd}.\n\nLet $\\normal(\\mu, S)$ denote the $d$ dimensional normal distribution with mean $\\mu$ and covariance\nmatrix $S$. We rewrite this Gaussian in its canonical decomposition in terms of its natural\nparameters. The sufficient statistics are stacked onto a tuple containing a $d$ dimensional vector\nand a $d\\times d$ matrix\n\\[\n \\tilde{x} = (x, -\\frac{1}{2}xx^\\top)\n\\]\nassociated with the natural parameters\n\\begin{equation}\n \\label{eq:nat-param}\n \\thetaall = (\\theta, \\Theta) = \\left(S^{-1}\\mu, \\frac{1}{2} S^{-1}\\right)\n\\end{equation}\nThe KL divergence between two Gaussian distributions\n\\[\nN_p = \\normal(\\mu_p, S_p)\\qquad\\mbox{and}\\qquad N_q = \\normal(\\mu_q, S_q)\n\\]\nis given by the Bregman divergence with generator $F$\n\\begin{align}\n KL(N_p||N_q) &= D_F(\\thetaall_q || \\thetaall_p)\\\\\n &= F(\\thetaall_q) - F(\\thetaall_p) - \\dotprod{(\\thetaall_q - \\thetaall_p)}{\\nabla F(\\thetaall_p)}\n \\label{eq:bregman-div}\n\\end{align}\nThe function $F$ turns out to be the log normaliser specifying the exponential family\n(of the Gaussian)\n\\begin{equation}\n \\label{eq:log-normaliser}\n F(\\thetaall) = \\frac{1}{4}\\trace(\\Theta^{-1} \\theta\\theta^\\top)\n -\\frac{1}{2} \\log \\deter \\Theta + \\frac{d}{2}\\log 2\\pi.\n\\end{equation}\nThe gradient of $F$ is given by\n\\begin{equation}\n \\label{eq:grad-log-normaliser}\n \\nabla F(\\thetaall) = \\left(\n \\frac{1}{2} \\Theta^{-1}\\theta\n ,\n -\\frac{1}{2} \\Theta^{-1} - \\frac{1}{4} (\\Theta^{-1}\\theta)(\\Theta^{-1}\\theta)^\\top\n \\right).\n\\end{equation}\nThe inner product in \\eqref{eq:bregman-div} is the sum of the inner products of the\nvectors and matrices,\n\\[\n \\dotprod{\\thetaall_p}{\\thetaall_q} = \\dotprod{\\theta_p}{\\theta_q} + \\dotprod{\\Theta_p}{\\Theta_q}\n\\]\nwhere the matrix inner product is $\\dotprod{\\Theta_p}{\\Theta_q} = \\trace(\\Theta_p\\Theta_q^\\top)$.\n\n\\section{Two dimensional Gaussian}\n\nWe explicitly show each element of the vector and matrix for a 2 dimensional Gaussian\nmean $\\mu$ and variance $S$,\n\\[\n \\normal\\left( \\vect{\\mu_1}{\\mu_2}, \\mat{s_{11}}{s_{12}}{s_{21}}{s_{22}}\\right).\n\\]\nThe determinant and inverse of $S$ is given by\n\\begin{equation}\n \\label{eq:det-S}\n a := \\deter S = |s_{11} s_{22} - s_{21} s_{22}|\n\\end{equation}\nand\n\\begin{equation}\n \\label{eq:inv-S}\n S^{-1} = \\frac{1}{a}\\mat{s_{22}}{-s_{12}}{-s_{21}}{s_{11}}\n\\end{equation}\nrespectively.\nWe can then explicitly compute the parameters for the KL divergence in the previous section.\nStarting from the right most term in \\eqref{eq:bregman-div},\nsince $d=2$ the constant is $\\log 2\\pi$.\nRecall the relationship between determinants and inverses:\n\\[\n \\deter S^{-1} = \\frac{1}{\\deter S}\n\\]\nBecause we are only considering a two dimensional problem, constants are squared in the determinant,\ni.e. $\\deter(c S) = c^2 \\deter S$.\nBy the definition of $\\Theta$, we have\n\\begin{align*}\n \\frac{1}{2} \\log \\deter \\Theta &= \\frac{1}{2} \\log \\deter (\\frac{1}{2} S^{-1}) \\\\\n &= \\frac{1}{2} \\log \\frac{1}{4} \\frac{1}{\\deter S}\\\\\n &= \\frac{1}{2} \\log \\frac{1}{4a}\n\\end{align*}\nwhere the last line substitutes \\eqref{eq:det-S}.\nBy the definition of $\\theta$,\n\\begin{equation}\n \\label{eq:theta}\n \\theta = S^{-1}m = \\frac{1}{a}\\vect{s_{22}\\mu_1 - s_{12}\\mu_2}{s_{11}\\mu_2 - s_{21}\\mu_1}.\n\\end{equation}\n\nWe also require\n\\begin{align}\n \\Theta^{-1}\\theta &= 2SS^{-1}m = 2m\\nonumber\\\\\n &=\\vect{2\\mu_1}{2\\mu_2}\\label{eq:2mean},\n\\end{align}\nwhich allows us to compute terms in the gradient. By multiplying\n\\eqref{eq:2mean} and \\eqref{eq:theta} we have\n\\begin{align*}\n \\Theta^{-1}\\theta\\theta^\\top &= \\frac{1}{a}\\vect{2\\mu_1}{2\\mu_2}\n \\vect{s_{22}\\mu_1 - s_{12}\\mu_2}{s_{11}\\mu_2 - s_{21}\\mu_1}^\\top\\\\\n &=\\frac{2}{a}\\mat{\\mu_1(s_{22}\\mu_1 - s_{12}\\mu_2)}{\\mu_1(s_{11}\\mu_2 - s_{21}\\mu_1)}{\\mu_2(s_{22}\\mu_1 - s_{12}\\mu_2)}{\\mu_2(s_{11}\\mu_2 - s_{21}\\mu_1)},\n\\end{align*}\nallowing us to calculate the first term of the log normaliser\n\\[\n \\frac{1}{4}\\trace(\\Theta^{-1} \\theta\\theta^\\top)= \\frac{1}{2a}\n \\left(\\mu_1(s_{22}\\mu_1 - s_{12}\\mu_2) + \\mu_2(s_{11}\\mu_2 - s_{21}\\mu_1)\\right).\n\\]\nSubstituting into \\eqref{eq:log-normaliser}, we have\n\\begin{equation}\n \\label{eq:gauss-log-normaliser}\n F(\\thetaall) = \\frac{1}{2a}\n \\left(\\mu_1(s_{22}\\mu_1 - s_{12}\\mu_2) + \\mu_2(s_{11}\\mu_2 - s_{21}\\mu_1)\\right)\n - \\frac{1}{2} \\log \\frac{1}{4a} + \\log 2\\pi.\n\\end{equation}\nSubstituting into \\eqref{eq:grad-log-normaliser}, we have\n\\begin{equation}\n \\label{eq:gauss-grad-log-normaliser-vec}\n \\frac{1}{2}\\Theta^{-1}\\theta = \\mu\n\\end{equation}\nand\n\\begin{equation}\n \\label{eq:gauss-grad-log-normaliser-mat}\n -\\frac{1}{2} \\Theta^{-1} - \\frac{1}{4} (\\Theta^{-1}\\theta)(\\Theta^{-1}\\theta)^\\top\n =\n - S - \\mu\\mu^\\top.\n\\end{equation}\nIt is likely that \\eqref{eq:gauss-log-normaliser} and \\eqref{eq:gauss-grad-log-normaliser-mat}\nare true in general for all Gaussians (not just 2D Gaussians).\n\n\\bibliographystyle{alpha}\n\\bibliography{gaussian}\n\\end{document}\n", "meta": {"hexsha": "285152a3210792de53203ca73f1ee441b1ee203f", "size": 6598, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/notes/two-gaussians.tex", "max_stars_repo_name": "chengsoonong/crowdastro", "max_stars_repo_head_hexsha": "ce14432c36de0574b73d813304365b74446a61f8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 13, "max_stars_repo_stars_event_min_datetime": "2015-11-07T15:24:44.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-20T05:29:58.000Z", "max_issues_repo_path": "tex/notes/two-gaussians.tex", "max_issues_repo_name": "chengsoonong/crowdastro", "max_issues_repo_head_hexsha": "ce14432c36de0574b73d813304365b74446a61f8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 234, "max_issues_repo_issues_event_min_datetime": "2016-02-21T23:53:16.000Z", "max_issues_repo_issues_event_max_datetime": "2019-06-27T00:26:08.000Z", "max_forks_repo_path": "tex/notes/two-gaussians.tex", "max_forks_repo_name": "chengsoonong/crowdastro", "max_forks_repo_head_hexsha": "ce14432c36de0574b73d813304365b74446a61f8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2015-11-07T00:20:09.000Z", "max_forks_repo_forks_event_max_datetime": "2018-10-03T13:37:15.000Z", "avg_line_length": 36.2527472527, "max_line_length": 156, "alphanum_fraction": 0.6841467111, "num_tokens": 2295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6345642203170545}} {"text": "\\chapter{Geometric Parameters}\n\n\\section{Wing Basic Geometric Parameters}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=120mm]{eps/wing_geometric_parameters.eps}\n \\caption{Wing basic geometric parameters}\n\\end{figure}\n\nAspect ratio is given by the following formula: \\cite{Raymer1992}\n\\begin{equation}\n A = \\frac{b^2}{S}\n\\end{equation}\n\nTaper ratio is given by the following formula. \\cite{Raymer1992}\n\\begin{equation}\n \\lambda = \\frac{c_t}{c_r}\n\\end{equation}\n\n\\section{Mean Aerodynamic Chord}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=120mm]{eps/wing_mean_aerodynamic_chord.eps}\n \\caption{Mean aerodynamic chord}\n\\end{figure}\n\nFor tapper wing mean aerodynamic chord can be calculated using following formula: \\cite{Corke2003, Galinski2016}\n\\begin{equation}\n \\hat c = \\frac{2}{3} c_r \\frac{1+\\lambda+\\lambda^2}{1+\\lambda}\n\\end{equation}\n\nFor more complex shapes mean aerodynamic chord is given as follows: \\cite{Paturski02}\n\\begin{equation}\n \\hat c = \n \\left(\n \\int_{-\\frac{b}{2}}^{\\frac{b}{2}} \\left( c \\left( y \\right) \\right)^2 dy\n \\right)\n \\div\n \\left(\n \\int_{-\\frac{b}{2}}^{\\frac{b}{2}} \\left( c \\left( y \\right) \\right) dy\n \\right)\n\\end{equation}\n\n\\section{Wing Aerodynamic Center}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=120mm]{eps/wing_aerodynamic_center.eps}\n \\caption{Wing aerodynamic center}\n\\end{figure}\n\nPosition of wing aerodynamic center ${\\vec r}_{AC}$ is at 25\\% of the mean aerodynamic chord and its lateral coordinate is given by the following formula. \\cite{Raymer1992, Corke2003, Galinski2016, Torenbeek1982}\n\\begin{equation}\n y_{AC} =\n \\frac{ b \\left( 1 + 2 \\lambda \\right) }{ 6 \\left( 1 + \\lambda \\right) }\n\\end{equation}\n", "meta": {"hexsha": "74b26eea01e37254f2df0fa8a655b75335fbd264", "size": 1700, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/data_1.tex", "max_stars_repo_name": "marek-cel/mscsim-docs", "max_stars_repo_head_hexsha": "9984f33c84787c4420f11f2834bb35e040e1f36f", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2019-12-01T02:27:28.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-09T07:02:20.000Z", "max_issues_repo_path": "tex/data_1.tex", "max_issues_repo_name": "marek-cel/mscsim-docs", "max_issues_repo_head_hexsha": "9984f33c84787c4420f11f2834bb35e040e1f36f", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/data_1.tex", "max_forks_repo_name": "marek-cel/mscsim-docs", "max_forks_repo_head_hexsha": "9984f33c84787c4420f11f2834bb35e040e1f36f", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-12-01T10:56:23.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-01T19:41:05.000Z", "avg_line_length": 28.813559322, "max_line_length": 212, "alphanum_fraction": 0.7205882353, "num_tokens": 563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6345642158441807}} {"text": "\\chapter{Workbook Assignment 1: Basic Probabilities and Visualisations}\n\n\\section{Bernoulli Distribution}\n\nAssumed a binary vote with outcome for or against is described by a Bernoulli distribution, with $P(vote = \"for\") = 0.69$. \nInterpreting this as a one-trial Bernoulli experiment with the two outcomes for and against - represented by success and no success respectively - we can visualize it as displayed in figure~\\eqref{fig:1a}.\n\n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=16cm]{pics/1a.pdf}\n\\caption{Bernoulli distribution of one trial}\n\\label{fig:1a}\n\\end{figure}\n\\FloatBarrier\n\n\nAs this is a one-trial Bernoulli experiment, we can only observe the two outcomes $success = for$ and $no~success = against$. With provided information that $P(vote = \"for\") = 0.69$, we can calculate that $P(vote=”against”) = 1 - P(vote = \"for\") = 0.31$. This is visualized by the two bars in figure~\\eqref{fig:1a}. Following Hogg et al. \\cite[Chapter~3.1]{hogg}, the expected value of X, described by a Bernoulli Distribution, is: \n\\begin{equation} \\mu = E(X) = (0)(1-p) + (1)(p) = p\n\\label{eq:expvaluebernoulli}\n\\end{equation}\nApplied to our problem, $ \\mu = p = P(vote = \"for\") = 0.69 $.\nThis means if we would repeat the experiment one hundred times, we expect to observe 69 successes. \nThis result was achieved using Python 3.9 and the following code:\n\\begin{minted}{python}\nfrom scipy.stats import bernoulli\nimport numpy as np\nimport pandas as pd\nimport plotly.express as px\nimport plotly.graph_objects as go\nfrom plotly.subplots import make_subplots\n\n#Dataset creation. Done manually as only one trial\np_for = 0.69\np_against = 0.31 #p_against = 1-p_for\ndf = pd.DataFrame([[1,p_for],[0,p_against]])\n\n#Visualisation\ntrace1 = go.Bar(x = df[0],y = df[1],width = 0.1,text = df[1], \n textposition = 'auto')\nfig = make_subplots(x_title = 'Number of Successes',y_title = 'P(X = x)')\nfig.add_trace(trace1)\nfig.show()\nfig.write_image(\"figures/1a.pdf\")\n\\end{minted}\n\n\\section{Poisson Distribution}\n\nAssumed we try to model the number of meteorites falling on an ocean each year - why would a Poisson Distribution be a natural candidate?\n\n\nThe Poisson distribution is a discrete probability distribution representing the number of events occurring in a defined period. It assumes the events occur with a known constant mean rate and that the events are independent of the time since the last event \\cite[Chapter~4.6]{illowsky2018introductory}.\n\nThe event of a meteor impacting the world in a given year fulfills all requirements for such distribution. The event itself is of discrete nature as one cannot have a fractional count of impacts, and we assume meteorite impacts occur with a known constant mean rate and the impacts themselves are independent of the time since the last event. We do not know, whether the variance of the distribution is equal to the mean, but even if it was, we could relax this assumption by using the Poisson-based Gamma-Poisson distribution. For this exercise, we assume $\\sigma^2 = \\mu$. Additionally assuming $\\mu = 64$ allows an exemplary visualisation as in figure \\eqref{fig:1b}.\n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=16cm]{pics/1b.pdf}\n\\caption{Poisson distribution of annual meteorite impacts with $\\mu = 64$}\n\\label{fig:1b}\n\\end{figure}\n\\FloatBarrier\n\nTo ensure readability of the graphic, probabilities $<0.5$ percent are excluded. Additionally, the variance ($s^2 = 64.36$) is visualised as its root, the sample standard deviation $s$ ($s = 8.02$). Looking at mean ($\\bar{x} = 64.01$) and median ($m = 64$), one can barely see a difference, which is to be expected as the distribution displayed is not skewed. \n\nThe calculations for this exercise were done in python 3.9 using Jupyter notebooks with the following code: \n\n\\begin{minted}{python}\nfrom scipy.stats import poisson\nimport numpy as np\nimport pandas as pd\nimport plotly.graph_objects as go\nfrom plotly.subplots import make_subplots\n\n#creating a sample with fixed random state\nsize = 100000\nraw = pd.Series(data = poisson.rvs(64, size=size, random_state = 42))\n\n# calculating median, mean, variance, stdev\nmedian = raw.median()\nvar = raw.var() # sample assumed\nmean = raw.mean()\nstdev = np.sqrt(var)\nstdev_range = pd.DataFrame([[raw.mean() - stdev, 0.025],\n [raw.mean() + stdev, 0.025]])\n\n# Filter to only display probabilities >=0.5%\ndf = pd.DataFrame(raw.value_counts())\ndf[0] = df[0].div(size)\ndf = df[df[0] >= 0.005]\n\n# Display measures\nprint(var)\nprint(stdev)\nprint(mean)\nprint(median)\n\n#creating graphic\ntrace1 = go.Bar(x = df.index, y = df[0],showlegend = False)\ntrace2 = go.Bar(x = [median], y = [0.055],width = [0.1], name = 'Median', \ntext = [median], textposition = 'outside')\ntrace3 = go.Line(x = stdev_range[0], y = stdev_range[1], \nname = 'Standard Deviation', text = ['-1 s', '+1 s'],\nmode =\"lines+markers+text\",textposition = \"bottom center\")\ntrace4 = go.Bar(x = [mean], y = [0.025],width = [0.15], name = 'Mean', \ntext = [mean], textposition = 'outside')\n\nfig = make_subplots(x_title = 'Annual Meteorite Impacts',\ny_title = 'P(X = x)')\nfig.add_trace(trace1)\nfig.add_trace(trace2)\nfig.add_trace(trace4)\nfig.add_trace(trace3)\nfig['layout'].update(barmode = 'overlay')\nfig.show()\n\nfig.write_image(\"figures/1b.pdf\")\n\\end{minted}\n\n\\section{Exponential Distributions}\nAssumed the probability that you need to wait more than the random variable of y hours to hear an owl from your room is described by:\n\\begin{equation} f(Y > y) = 0.2\\overline{8}\\mathrm{e}^{-0.5y} + 0.7\\overline{1}\\mathrm{e}^{-0.25y}.\n\\label{eq:initcdf}\n\\end{equation}\nTo facilitate calculations, we reformulate the periodic numbers to fractions:\n\\begin{equation} f(Y > y) = \\dfrac{13}{45}\\mathrm{e}^{-0.5y} + \\dfrac{32}{45}\\mathrm{e}^{-0.25y}.\n\\label{eq:initcdffractioned}\n\\end{equation}\nSuspecting that such events are usually exponentially distributed, we reformulate the probability in equation \\eqref{eq:truecdf} to the conventional expression \\cite{deisenroth}. \n\\begin{equation} F_Y(y) = f(Y \\leq y) =1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y} + \\dfrac{32}{45}\\mathrm{e}^{-0.25y} \\right)\n\\label{eq:truecdf}\n\\end{equation}\nAs $Y$ and $y$ are defined as time and time cannot be negative in this case, the domain of the function we are interested in is $[0,+\\infty]$.\n\\begin{equation} f'_Y(y) = f''(Y \\leq y) = \n\\dfrac{-8\\mathrm{e}^{-\\frac{y}{4}}-13\\mathrm{e}^{-\\frac{y}{2}}}{180}\n\\label{eq:secderivative}\n\\end{equation}\n\nAs \\eqref{eq:secderivative}, the second derivative of formula \\eqref{eq:truecdf}, additionally has no real-valued solution for $y=0$, we can describe $F_Y(y)$ of \\eqref{eq:truecdf} as being upwards continuous monotonic increasing. Furthermore, $\\lim_{y\\to\\infty} F_Y(y)= 1$ and $F_Y(0)= 0$ for the support of $y$. This is sufficient evidence to assume that the provided distribution of the random variable $y$ is a cumulative distribution function (CDF) of the family of exponential distributions. \n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=16cm]{pics/1c1.pdf}\n\\caption{Visualisation of provided and reformulated CDF of task assignment}\n\\label{fig:1c1}\n\\end{figure}\n\\FloatBarrier\n\n\nFor any calculation regarding the probability to hear an owl within a time interval, we use the CDF. If we for example wanted to know $P(2 \\leq Y \\leq 4)$, the probability of needing to wait two to four hours to hear an owl, we would calculate it as described in equation \\eqref{eq:prob24}.\n\n\\begin{equation} P(2 \\leq Y \\leq 4) = P(Y \\leq 4) - P(Y \\leq 2)\n\\label{eq:prob24}\n\\end{equation}\nApplied to our distribution, we calculate\n\\begin{equation} P (2 \\leq Y \\leq 4) =1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-2} + \\dfrac{32}{45}\\mathrm{e}^{-1}\\right) -\\left[ 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-1} + \\dfrac{32}{45}\\mathrm{e}^{-0.5}\\right)\\right] \t\\approx 0.23688696\n\\label{eq:probcalc24}\n\\end{equation}\nConsequently, we can say that in 23.69 percent of cases, we have to wait in between two and four hours to hear an owl.\n\\\\\n\nNext to the CDF, we are also often interested in the corresponding probability density function (PDF). Generally, the PDF $f_Y(y)$ of a CDF $F_Y(y)$ is given by its first derivative \\cite[Chapter~4.3]{montgomery2010applied}:\n\\begin{equation}\nf_Y(y) = \\frac{df(Y \\leq y)}{dy}\n\\label{eq:firstderivative}\n\\end{equation}\nFor our CDF, the PDF is described by:\n\\begin{equation}\nf_Y(y) = \\dfrac{16\\mathrm{e}^{-\\frac{y}{4}}+13\\mathrm{e}^{-\\frac{y}{2}}}{90}\n\\label{eq:pdf}\n\\end{equation}\n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=17cm]{pics/1c2.pdf}\n\\caption{Visualisation of $f_Y(y)$ (PDF), the first two moments, and applicable quartiles}\n\\label{fig:1c2}\n\\end{figure}\n\\FloatBarrier\n\nThe graphical representation of the PDF in figure \\eqref{fig:1c2} also includes the mean value. This is calculated using equation \\eqref{eq:ExpValueHogg}, following the definition of an expectation for a random variable in Hogg et al. \\cite[Definition~1.8.1]{hogg}.\n\\begin{equation}\nE(X) = \\int^{+\\infty}_{-\\infty} x f_X(x) ~ dx\n\\label{eq:ExpValueHogg}\n\\end{equation}\nConditions for this to be true are that equation \\eqref{eq:ExpValueHoggcondition} is fulfilled and that $X$ is a continuous random variable with $f_X(x)$ as its PDF.\n\\begin{equation}\n\\int^{+\\infty}_{-\\infty} |x| f_X(x) ~ dx < \\infty\n\\label{eq:ExpValueHoggcondition}\n\\end{equation}\nBy the task assignment, we know that $Y$ is a continuous random variable. By equation \\eqref{eq:pdf}, we calculated $f_Y(y)$ as the PDF of $y$. Lastly, as $y \\in [0,\\infty)$, the left hand side of inequality \\eqref{eq:ExpValueHoggcondition} reduces to the right hand side of equation \\eqref{eq:ExpValueHogg} and the lower limit of the integral changes to 0. Consequently, if the mean or expected value $E(Y) < \\infty$, it is valid. The calculation for our random variable $y$ in \\eqref{eq:ExpValueCalculation} returns $E(Y) < \\infty$, we can therefore accept $3.4\\overline{2}$ hours as our expected value $E(Y)$ for the waiting time to hear an owl. \n%\n\\begin{equation}\nE(Y) = \\int^{+\\infty}_{0} y f_Y(y) ~ dy = \\int^{+\\infty}_{0} y\\left(\\dfrac{16\\mathrm{e}^{-\\frac{y}{4}}+13\\mathrm{e}^{-\\frac{y}{2}}}{90} \\right) ~dy = \\frac{154}{45} = 3.4\\overline{2}\n\\label{eq:ExpValueCalculation}\n\\end{equation}\nFollowing Hogg et al., the first moment or mean $\\mu = E(X)$, if $X$ is a random variable with existing expectation \\cite[Definition~1.9.1]{hogg}. This is the case for $Y$. Consequently, we can also assume that the mean value $\\mu = E(X) = 3.4\\overline{2}$. \n\\\\\nTo calculate the second moment, the variance $\\sigma^2$, we use definition 1.9.2 by Hogg et al. \\cite{hogg} as equation \\eqref{eq:HoggVariance}.\n\\begin{equation}\n\\sigma^2 = E(X^2) - \\mu^2\n\\label{eq:HoggVariance}\n\\end{equation}\n%\nThe variance of $y$ is calculated by \\eqref{eq:YVariance}, using the same assumptions as for equation \\eqref{eq:ExpValueCalculation}.\n\\begin{equation}\n\\sigma^2 = \\int^{+\\infty}_{0} y^2 f(y) ~ dy - \\mu^2\n= \\int^{+\\infty}_{0} \\dfrac{y^2\\left(16\\mathrm{e}^{-\\frac{y}{4}}+13\\mathrm{e}^{-\\frac{y}{2}}\\right)}{90} - \\left(\\frac{154}{45}\\right)^2\n= \\frac{27044}{2025} \\approx 13.3551\n\\label{eq:YVariance}\n\\end{equation}\n\nThe variance is also depicted in figure \\eqref{fig:1c2} as its root, the standard deviation $\\sigma \\approx 3.6545$. \n\nBoth figures \\eqref{fig:1c1} and \\eqref{fig:1c2} furthermore display the first three quartiles of the PDF. Hogg et al. \\cite[Chapter~1.7.1]{hogg} define the second quartile, the median, as the point dividing area under the PDF into two areas of equal size, which again are divided into two areas of equal size by the first and third quartile. As the CDF is the integral of the PDF and consequently by nature describes the area under the curve of the PDF, we can simply use equation \\eqref{eq:truecdf} for the calculation of quartiles. We know that each quartile has to cover a quarter of the area under the curve of PDF \\eqref{eq:pdf}. This means we can substitute 0.25, 0.5 and 0.75 for $(F_Y \\leq y)$ in equation \\eqref{eq:truecdf} and solve for $y$ to get the quartiles. This is exemplarily done in equation \\eqref{eq:firstquartile} for the first quartile.\n\n\\begin{equation}\n\\begin{split}\n0.25 &\n=1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y_1} + \\dfrac{32}{45}\\mathrm{e}^{-0.25y_1} \\right)\\\\\ny_1&=\\ln \\left(\\frac{413529488+7788544\\sqrt{2779}}{332150625}\\right)\\\\\ny_1 &\\approx 0.9087 \n\\end{split}\n\\label{eq:firstquartile}\n\\end{equation}\n\nThe second and third quartile can be calculated in the same way, resulting in $y_2 \\approx 2.2430$ for the second quartile and $y_3 \\approx 4.6588$ for the third quartile. The calculations were done using symbolab \\cite{symbolab} with the following commands:\n\\begin{lstlisting}\nsolve 0.25 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}+\\dfrac{32}{45}\n\\mathrm{e}^{-0.25y} \\right)\n\nsolve 0.5 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}+\\dfrac{32}{45}\n\\mathrm{e}^{-0.25y} \\right)\n\nsolve 0.75 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}+\\dfrac{32}{45}\n\\mathrm{e}^{-0.25y} \\right)\n\\end{lstlisting}\n\nAs $\\lim_{y\\to\\infty} F_Y(y)= 1$, we would only get $\\infty$ as a result if we tried to calculate the fourth quartile $y_4$. The first three quartiles are visualised in figure \\eqref{fig:1c1} as points whereas in figure \\eqref{fig:1c2}, they are displayed as vertical lines to indicate that we want to highlight that the individual quartile $y_i$ has $i\\cdot25$ percent of PDF \\eqref{eq:pdf}'s area under the curve to its left. \n\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=17cm]{pics/1c4.pdf}\n\\caption{Histogram of $f_Y(y)$ by minute}\n\\label{fig:1c4}\n\\end{figure}\n\\FloatBarrier\n\nLastly, figure \\eqref{fig:1c4} displays the PDF as a histogram per minute for one observation. If we for example had 1'000 observations, we would have approximately five observations where we hear an owl in the first minute. As the PDF is per hour while the histogram should be per minute, we need to divide every value $f_Y(y_i)$ for our 600 data points by 60. The standard y-axis comprises 10 hours for all graphics, we have 600 columns which might be too many. However, the graphics are created with python and Plotly, which allows for magnification, if desired. It can also be magnified in this document as it is a vector graphic.\nCalculations and visualisations were done in Jupyter notebooks using the following python code:\n\\begin{minted}{python}\nimport numpy as np\nimport pandas as pd\nimport plotly.graph_objs as go\nfrom plotly.subplots import make_subplots\nimport plotly\nimport plotly.express as px\n\n# Define initial function as written in task assignment (P(Y>y))\ndef func(y): # cdf\n return ((13 / 45) * np.exp(-0.5 * y) + (32 / 45) * np.exp(-0.25 * y))\n\n\n# Reformulate function so that you have a standard exponential function (P(Y<=y)\ndef realfunc(y): # cdf\n return 1 - func(y)\n\n\n# Define pdf of realfunc(x)\ndef pdf_realfunc(y):\n # First derivative of realfunc(x) with respect to x\n return (16 * np.exp(-y / 4) + 13 * np.exp(-y / 2)) / (90)\n\n\n# create datasets\nfunc_data = []\nrealfunc_data = []\npdf_realfunc_data = []\n\nfor i in np.arange(0, 10, 1 / 60):\n func_data.append([i, func(i)])\n realfunc_data.append([i, realfunc(i)])\n pdf_realfunc_data.append([i, pdf_realfunc(i)])\n\n# create dataframes\nfunc_df = pd.DataFrame(func_data)\nrealfunc_df = pd.DataFrame(realfunc_data)\npdf_realfunc_df = pd.DataFrame(pdf_realfunc_data)\n\n# Calculate mean value\ny_mean = pdf_realfunc(154 / 45)\n\n# Calculate quartiles\n# symbolab.com: \"solve 0.25 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}\n# + \\dfrac{32}{45}\\mathrm{e}^{-0.25y} \\right)\"\nq1 = np.log(413529488 / 332150625 + 7788544 * np.sqrt(2779) / 332150625)\n\n# symbolab.com: \"solve 0.5 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}\n# + \\dfrac{32}{45}\\mathrm{e}^{-0.25y} \\right)\"\nq2 = np.log(19342148 / 4100625 + 411904 * np.sqrt(2194) / 4100625)\n\n# symbolab.com: \"solve 0.75 = 1-\\left( \\dfrac{13}{45}\\mathrm{e}^{-0.5y}\n# + \\dfrac{32}{45}\\mathrm{e}^{-0.25y} \\right)\"\nq3 = np.log(216370448 / 4100625 + 5392384 * np.sqrt(1609) / 4100625)\n\n# Visualisation of initial function and reformulated function\ncolor = px.colors.qualitative.Plotly\ntrace1 = go.Line(\n x=func_df[0], \n y=func_df[1], \n name='P(Y > y)'\n)\ntrace2 = go.Line(\n x=realfunc_df[0], \n y=realfunc_df[1], \n name='P(Y <= y) (CDF)'\n)\n# Additionally, the PDF of P(Y <= y) can be displayed.\n#trace3 = go.Line(x=pdf_realfunc_df[0],\n# y=pdf_realfunc_df[1],\n# name='PDF of P(Y <= y)'\n#)\ntrace9 = go.Scatter(\n x=[q1],\n y=[0.25],\n showlegend=True,\n name='Quartiles 1-3',\n marker_color=color[9],\n hovertext=['1st Quartile'],\n)\ntrace10 = go.Scatter(\n x=[q2],\n y=[0.5],\n showlegend=False,\n marker_color=color[9],\n hovertext=['2nd Quartile (Median)'],\n)\ntrace11 = go.Scatter(\n x=[q3],\n y=[0.75],\n showlegend=False,\n marker_color=color[9],\n hovertext=['3rd Quartile'],\n)\n# Mean (can also be displayed additionally)\n#trace6 = go.Scatter(x=[154 / 45], y=[y_mean], name='Mean')\n\n# Creating graphic\nfig = make_subplots(x_title=\"y (Hours)\", y_title=\"Probability\")\nfig.add_trace(trace1)\nfig.add_trace(trace2)\n#fig.add_trace(trace3)\n#fig.add_trace(trace6)\nfig.add_trace(trace9)\nfig.add_trace(trace10)\nfig.add_trace(trace11)\nfig.show()\nfig.write_image(\"figures/1c1.pdf\")\n\n# Visualisation of PDF\ncolor = px.colors.qualitative.Plotly\ntrace3 = go.Line(x=pdf_realfunc_df[0],\n y=pdf_realfunc_df[1],\n name='f_Y(y) (PDF)',\n marker_color=color[0])\ntrace6 = go.Scatter(x=[154 / 45],\n y=[y_mean],\n showlegend=True,\n name='Mean',\n marker_color=color[2])\ntrace7 = go.Bar(x=[154 / 45],\n y=[y_mean],\n showlegend=True,\n name='Mean',\n marker_color=color[2],\n width=0.05)\ntrace8 = go.Line(x=[154 / 45, 154 / 45 + 3.66],\n y=[y_mean, y_mean],\n name='Standard Deviation',\n text=['Mean', '+1 std. dev.'],\n mode=\"lines+markers+text\",\n textposition=\"top center\",\n fillcolor=color[3])\ntrace9 = go.Bar(\n x=[q1],\n y=[pdf_realfunc(q1)],\n showlegend=True,\n name='Quartiles 1-3',\n marker_color=color[9],\n width=0.05,\n text=['1st Quartile'],\n textposition='outside',\n)\ntrace10 = go.Bar(\n x=[q2],\n y=[pdf_realfunc(q2)],\n showlegend=False,\n marker_color=color[9],\n width=0.05,\n text=['2nd Quartile (Median)'],\n textposition='outside',\n)\ntrace11 = go.Bar(\n x=[q3],\n y=[pdf_realfunc(q3)],\n showlegend=False,\n marker_color=color[9],\n width=0.05,\n text=['3rd Quartile'],\n textposition='outside',\n)\n\nfig = make_subplots(x_title=\"y (Hours)\", y_title=\"P(y)\")\nfig.add_trace(trace3)\nfig.add_trace(trace8)\nfig.add_trace(trace6)\n#fig.add_trace(trace7) #Additionally, this line can be uncommented\n#to get a vertical line for the mean for better visibility. This\n#might however be confusing as the mean is no indicator for area\n#under the curve, contrary to quartiles, which are displayed as\n#columns. Thus, this column was left in as comment, in case some-\n#one wants better visuability\nfig.add_trace(trace9)\nfig.add_trace(trace10)\nfig.add_trace(trace11)\n\nfig.update_layout(bargap=0.0, barmode='overlay')\nfig.update_traces(marker_line_width=0)\n\nfig.show()\nfig.write_image(\"figures/1c2.pdf\")\n\n# Zooming in on the mean, if wanted\n#fig.update_xaxes(range=[3.37, 3.47])\n#fig.update_yaxes(range=[0.099, 0.105])\n#fig.show()\n#fig.write_image(\"figures/1c3.pdf\")\n\n#Creating the Histogram per minute\n\ntrace4 = go.Bar(y=pdf_realfunc_df[1] / 60,\n name=\"Histogram of f_Y(y)\",\n marker_color='#3399FF')\n\nfig = make_subplots(x_title=\"y\", y_title=\"P(y)\")\n\nfig.add_trace(trace4)\n\nfig.update_layout(bargap=0.0, barmode='overlay')\nfig.update_traces(marker_line_width=0)\n\nfig.show()\nfig.write_image(\"figures/1c4.pdf\")\n\\end{minted}\n\n\\chapter{Workbook Assignment 2: Basic Probabilities and Visualizations }\t\n\n\\section{Visualisation, Expectations and (Co-)Variances}\nAssumed one has recorded mappings of X and Y values, what is an appropriate way of visualisation? How do you find the sample covariance as well as the expectation values and variances per variable; assuming the data comes in the following format?\\\\\n\\\\\n(1.754, 720.15), (-7.385, -260.86), (1.396, -340.56), (-3.304, 954.75), \n(-12.159, -370.12),\\\\ (-6.767, -259.89), (1.233, 261.89), (-7.056, 400.72), \n(-1.222, -94.33), (-0.722, -492.02),\\\\ (-5.065, -789.42), (-9.333, 301.86), \n(7.27, -17.04), (-1.989, 257.91), (3.582, -229.89), \\\\(-1.2, -349.18), \n(4.118, 226.05), (-1.834, -534.54), (-1.882, 610.91), (8.677, 70.11) \\\\\n\nAs the variables in the data are both continuous and the question addresses the calculation of correlation, the optimal visualisation is a scatter diagram as for example figure~\\eqref{fig:WorkbookAssignment2a}. Other possibilities are excluded as there is not much more information provided on the data set. A line graph for example would be appropriate if we had information that the observations show some development over some time domain. \n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=16cm]{pics/WorkbookAssignment2a.pdf}\n\\caption{Visualisation of provided X and Y values as scatter plot}\n\\label{fig:WorkbookAssignment2a}\n\\end{figure}\n\\FloatBarrier\nLooking at the scatter plot, one can observe that the data is quite evenly dispersed; no pattern can be determined at first glance. This rather evenly distributed data also suggests that the expected value for X and Y is rather close to 0. Calculating the expected value for each variable can be understood at the calculation of the expected value of a discrete random variable as the number of observations is discrete. Equation~\\eqref{eq:ExpValDiscrete} is therefore the formula to calculate the expected value \\cite{mathcenter}. \n\\begin{equation} \nE(X) = \\sum \\left[x * P(x)\\right]\n\\label{eq:ExpValDiscrete}\n\\end{equation}\nThe results are consequently $E(X) = -1.5944$ and $E(Y) = 3.3250$.\nFollowing Bruce and Bruce \\cite{bruce2017practical}, the variance for a sample is calculated by \n\\begin{equation} \ns^2 = \\frac{\\sum{(x-\\bar{x})^2}}{n-1}.\n\\label{eq:ExpValDiscrete}\n\\end{equation}\nThis computes to $s^2_x = -1.5944$ and $s^2_y = 3.3250$, where $\\bar{x} = E(X)$ and $\\bar{y} = E(Y)$.\nThe sample covariance is defined in Equation~\\eqref{eq:COV} (adapted from Bruce and Bruce \\cite{bruce2017practical}).\n\\begin{equation} s_{xy} = \\frac{\\sum(x_{i}-\\bar{x})(y_{i}-\\bar{y})}{n-1}\n\\label{eq:COV}\n\\end{equation}\nApplying the formula to the provided data, $s_{xy} = 284.23$ can be observed. This indicates a positive relationship.\nThe calculations were done using Python 3.9 and the following code:\n\\begin{minted}{python}\nimport numpy as np\nimport pandas as pd\nimport seaborn as sns\nimport matplotlib.pyplot as plt\ndata = (1.754, 720.15), (-7.385, -260.86), (1.396, -340.56), (-3.304, 954.75),\n(-12.159, -370.12), (-6.767, -259.89), (1.233, 261.89), (-7.056, 400.72), \n(-1.222, -94.33), (-0.722, -492.02), (-5.065, -789.42), (-9.333, 301.86),\n(7.27, -17.04), (-1.989, 257.91), (3.582, -229.89), (-1.2, -349.18), \n(4.118, 226.05), (-1.834, -534.54), (-1.882, 610.91), (8.677, 70.11)\ndata = pd.DataFrame(data)\ndata.columns = [\"X\", \"Y\"]\nprint(\"The variance of the variables is: \\n{}\".format(data.var()))\nprint(\"The expectation value for the variables is: \\n{}\".format(data.mean()))\nprint(\"The sample covariance is {}\".format(data.cov().iloc[0,1]))\n\ndata = data.set_index(\"X\")\n\nsns.scatterplot(data = data, x = \"X\", y = \"Y\", color = \"black\").set(\n title = \"Visualisation of X and Y Values\")\nplt.savefig('./figures/2a.pdf')\n\\end{minted}\n\n\n\\section{Transformations}\nAssumed we have a ball being thrown at a random angle $\\theta$ [0,360) (in degrees) and a random radius $r$ $\\in$ [0,1] (in meters) both independent and uniform. What is the density of the variables $X$ and $Y$, which are the Cartesian coordinates of the point at angle $\\theta$ and radius $r$? And, furthermore, what is their expectation and variance?\n\nFollowing Illowsky and Dean \\cite[Chapter~5.2]{illowsky2018introductory}, the pdf of an uniformly distributed variable is $f(x) = \\frac{1}{b-a}$ for $a \\leq x \\leq b$. \n\nAs $\\theta,r \\sim \\mathcal{U}$, we can formulate the marginal distributions as described in \\eqref{eq:theta_untransformed} (using radiants) and \\eqref{eq:r_untransformed}.\n\n\\begin{equation}\nf_\\Theta(\\theta) = \n\\begin{cases} \n \\frac{1}{2\\pi} & for~\\theta \\in [0,2\\pi)\\\\\n 0&elsewhere\n \\end{cases}\n\\label{eq:theta_untransformed}\n\\end{equation}\n\n\\begin{equation}\nf_R(r) = \n\\begin{cases} \n 1 & for~r\\in[0,1]\\\\\n 0&elsewhere.\n \\end{cases}\n\\label{eq:r_untransformed}\n\\end{equation}\n\nJoining these distributions returns the following joint pdf:\n\\begin{equation}\n\\begin{split}\nf_{R,\\Theta}(r,\\theta) = \n\\begin{cases} \n \\frac{1}{2\\pi} & \\theta \\in [0,2\\pi),~ r\\in[0,1]\\\\\n 0&elsewhere.\n \\end{cases}\n\\end{split}\n\\label{eq:joint_pdf_untransformed}\n\\end{equation}\n\nFollowing Dawkins \\cite{dawkins}, we can transform to Cartesian $x$ and $y$ using $x = rcos(\\theta)$ and $y=rsin(\\theta)$. For the transformation, we furthermore need their inverses, which calculate to \n\n\\begin{align*}\nx^2=r^2cos^2(\\theta)& \\qquad y^2=r^2sin^2(\\theta)\\\\\nx^2+y^2&=r^2cos^2(\\theta)sin^2(\\theta)\\\\\nx^2+y^2&=r^2\\\\\nr &= \\sqrt{x^2+y^2} = u_1(x,y)\\\\\n\\label{eq:r-1}\n\\end{align*}\nand\n\\begin{align*}\n r = \\frac{x}{cos(\\theta)}& \\qquad r = \\frac{y}{sin(\\theta)}\\\\\n \\frac{x}{cos(\\theta)} &= \\frac{y}{sin(\\theta)}\\\\\n \\frac{x}{cos(\\theta)} &= \\frac{y}{sin(\\theta)}\\\\\n \\frac{y}{x} &= \\frac{cos(\\theta)}{sin(\\theta)}\\\\\n \\frac{y}{x} &= tan(\\theta)\\\\\n \\theta &= tan^{-1}\\left(\\frac{y}{x}\\right) = u_2(x,y).\n\\label{eq:theta-1}\n\\end{align*}\n\nUsing the transformation theorem proved in Hogg et al. \\cite[Chapter~2.7]{hogg}, we get\n \n\\begin{equation}\n\\begin{split}\nf_{X,Y}(x,y) = f_{R,\\Theta}\\left(\\sqrt{x^2+y^2},tan^{-1}\\left(\\frac{y}{x}\\right)\\right) \\cdot |J|,\n%f_{X,Y}(x,y) = f_{r,\\theta}\\left(u_1,u_2\\right)\\right) \\cdot |J|\n\\end{split}\n\\label{eq:transtheorem}\n\\end{equation}\nwhere \n\\begin{equation}\n\\begin{split}\n|J| &=\n\\left|\n\\begin{matrix}\n\\frac{\\partial u_1}{\\partial x} & \n\\frac{\\partial u_1}{\\partial y} \\\\[2ex] % <-- 1ex more space between rows of matrix\n\\frac{\\partial u_2}{\\partial x} & \n\\frac{\\partial u_2}{\\partial y} \\\\[2ex]\n\\end{matrix}\n\\right|\n= \n\\left|\n\\begin{matrix}\n\\frac{x}{\\sqrt{x^2+y^2}}& \n\\frac{y}{\\sqrt{x^2+y^2}} \\\\[2ex] % <-- 1ex more space between rows of matrix\n-\\frac{y}{y^2+x^2} & \n\\frac{x}{y^2+x^2} \\\\[2ex]\n\\end{matrix}\n\\right|\\\\[2ex]\n&= \\frac{x}{\\sqrt{x^2+y^2}} \\cdot \\frac{x}{y^2+x^2} -\\frac{y}{\\sqrt{x^2+y^2}} \\cdot \\left(-\\frac{y}{y^2+x^2}\\right)\\\\[2ex]\n&= \\frac{x^2+y^2}{(x^2+y^2)^{3/2}}\\\\[2ex]\n&=\\frac{1}{\\sqrt{x^2+y^2}},\n\\end{split}\n\\label{eq:jacobiandet}\n\\end{equation}\n\nwith the derivatives being calculated by Symbolab \\cite{symbolab} with the following commands:\n\\begin{lstlisting}\n\\frac{\\partial }{\\partial \\:x}\\left(\\sqrt{x^2+y^2}\\right)\n\\frac{\\partial }{\\partial \\:y}\\left(\\sqrt{x^2+y^2}\\right)\n\\frac{\\partial }{\\partial \\:x}\\left(tan^{-1}\\left(\\frac{y}{x}\\right)\\right)\n\\frac{\\partial }{\\partial \\:y}\\left(tan^{-1}\\left(\\frac{y}{x}\\right)\\right)\n\\end{lstlisting}\n\nSubstituting \\eqref{eq:jacobiandet} into \\eqref{eq:transtheorem}, we get\n\\begin{equation}\n\\begin{split}\nf_{X,Y}(x,y) &= f_{r,\\theta}\\left(\\sqrt{x^2+y^2},tan^{-1}\\left(\\frac{y}{x}\\right)\\right) \\cdot \\frac{1}{\\sqrt{x^2+y^2}}\\\\[2ex]\nf_{X,Y}(x,y) &= \\frac{1}{2\\pi} \\cdot \\frac{1}{\\sqrt{x^2+y^2}} = \\frac{1}{2\\pi\\sqrt{x^2+y^2}},\n%\\begin{cases} \n% \\frac{1}{2\\pi} & \\theta \\in [0,2\\pi),~ r\\in[0,1]\\\\\n% 0&elsewhere\n% \\end{cases}\n\\end{split}\n\\label{eq:transtheoremcomplete}\n\\end{equation}\n\nwhich represents the joint PDF of $X$ and $Y$.\\\\\nTo calculate their individual densities, the marginal distributions are needed. As proofed by Larsen and Marx \\cite[Theorem~3.7.2]{larsen2005introduction}, the marginal distribution for $X$ can be obtained by \n\n\\begin{equation}\n\\begin{split}\nf_{X}(x) = \\int_{0}^{\\sqrt{1-x^2}}\\frac{1}{2\\pi\\sqrt{x^2+y^2}} dy,\n\\end{split}\n\\label{eq:marginalX}\n\\end{equation}\n\nwhich calculates to \n\n\\begin{equation}\n\\begin{split}\nf_{X}(x) = \\frac{1}{2\\pi }\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right) \\quad x \\in (0,1)\n%https://www.symbolab.com/solver/step-by-step/%5Cint_%7B0%7D%5E%7B%5Csqrt%7B1-x%5E2%7D%7D%5Cfrac%7B1%7D%7B2%5Cpi%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D%20dy\n\\end{split}\n\\label{eq:marginalXcalculated}\n\\end{equation}\n\nusing Symbolab \\cite{symbolab} with the following code:\n\\begin{lstlisting}\n\\int _0^{\\sqrt{1-x^2}}\\frac{1}{2\\pi \\sqrt{x^2+y^2}}\\:dy\n\\end{lstlisting}\n\nPDF \\eqref{eq:transtheoremcomplete} is symmetric in $x$ and $y$ as $f_{X,Y}(x,y) = f_{Y,X}(y,x)$. Consequently, $f_Y(y)$ has similar form:\n\\begin{equation}\n\\begin{split}\nf_{Y}(y) = \\frac{1}{2\\pi }\\ln\\left(\\frac{\\sqrt{1-y^2}+1}{y}\\right) \\quad y \\in (0,1).\n\\end{split}\n\\label{eq:marginalYcalculated}\n\\end{equation}\nTo calculate the expected value, equation \\eqref{eq:ExpValueHogg} is used again:\n\\begin{equation}\n\\begin{split}\nE(X) &= \\int_0^1 \\frac{x}{2\\pi }\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right) dx\\\\\nE(X) &= \\frac{1}{2\\pi}\\int_0^1 x\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right) dx\\\\\nE(X) &= \\frac{1}{4\\pi } \\cdot \\left[x^2\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right)-\\sqrt{1-x^2}\\right]_0^1\\\\\nE(X)&= \\frac{1}{4\\pi} \\approx 0.07958.\n\\end{split}\n\\label{eq:expvalF_X}\n\\end{equation}\n$E(X) = E(Y)$, as again the same value calculates for $Y$ due to the symmetry of the distribution.\nWolframAlpha \\cite{wolfalph} was used for this calculation with the following input:\n\\begin{lstlisting}\n\\int_0^1 \\frac{x}{2\\pi }\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right)\n\\end{lstlisting}\n\nTo calculate the second moment, the variance - equation \\eqref{eq:HoggVariance} - is recalled:\n\\[\nVar(x) = E(X^2)-(E(X))^2,\n\\]\nwhere\n\\begin{equation}\n\\begin{split}\nE(X^2) = \\frac{1}{2\\pi}\\int_0^1 x^2\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right) dx = \\frac{1}{24}\n\\end{split}\n\\label{eq:E(X^2)}\n\\end{equation}\nThis calculation was done in WolframAlpha \\cite{wolfalph} with the following input:\n\\begin{lstlisting}\n\\int_0^1 \\frac{x^2}{2\\pi }\\ln \\left(\\frac{\\sqrt{1-x^2}+1}{x}\\right)\n\\end{lstlisting}\nConsequently, \n\\begin{equation}\n\\begin{split}\nVar(X) = Var(Y) = \\frac{1}{24}- \\left(\\frac{1}{4\\pi}\\right)^2 = \\frac{2\\pi ^2-3}{48\\pi ^2} \\approx 0.1581.\n\\end{split}\n\\label{eq:VarXcalculated}\n\\end{equation}\n\n\n\n\\chapter{Workbook Assignment 3: A Simple Parameter Estimation}\t\n\nSupposed we have a network router with a total bandwith until it breaks down described by the random variable $S$, which is exponentially distributed:\n\\begin{equation}\n\\begin{split}\n%s(t) = f_S(t) = \\frac{1}{\\theta} e^{-\\frac{t}{\\theta}}\ns(t) = \\frac{1}{\\theta} e^{-\\frac{t}{\\theta}}.\n\\end{split}\n\\label{eq:exphardwarefailure}\n\\end{equation}\n\n\\section{Bandwith Total to Failure}\nAssumed we have another backup device, the bandwidth total to failure of both devices is described by $T$, which is the sum of $S_1$ and $S_2$, which each are described by \\eqref{eq:exphardwarefailure}. T is consequently\n\\begin{equation}\n\\begin{split}\nT = S_1 + S_2.\n\\end{split}\n\\label{eq:T}\n\\end{equation}\nBoth variables are random variables by task assignment and it can furthermore be assumed that they are independent as the performance of one device has no influence on the performance of another device. Furthermore, bandwidth is a continuous variable that cannot take negative values. With these assumptions about the variables, the sum of both variables is consequently calculated by convolution, following Larsen and Marx \\cite[Theorem~3.8.3]{larsen2005introduction}.\n\\begin{equation}\n\\begin{split}\nt(z) = \\int_{-\\infty}^\\infty s(t)s(z-t)~dt.\n\\end{split}\n\\label{eq:tz1}\n\\end{equation}\nAs only positive values can be obtained for the bandwidth, \\eqref{eq:tz1} simplifies to \n\\begin{equation}\n\\begin{split}\nt(z) = \\int_{0}^zs(t)s(z-t)~dt.\n\\end{split}\n\\label{eq:tz2}\n\\end{equation}\nInserting the density of $S$ yields\n\\begin{equation}\n\\begin{split}\nt(z) \n&= \\int_{0}^z \\left(\\frac{1}{\\theta} e^{-\\frac{t}{\\theta}}\\right)\n\\left(\\frac{1}{\\theta} e^{-\\frac{z-t}{\\theta}}\\right)~dt\\\\\n&= \\frac{1}{\\theta^2} e^{-\\frac{z}{\\theta}}\\int_{0}^z~dt\\\\\n&= \\frac{z}{\\theta^2} e^{-\\frac{z}{\\theta}}\\\\\n&= f_T(z;~\\theta).\n\\end{split}\n\\label{eq:tz3}\n\\end{equation}\n\n\\section{Likelihood Function}\nGiven an experiment with the dual-router-system yielding a sample $T_1$, $T_2$, ..., $T_n$, calculate the likelihood function for $\\theta$. Propose a transformation of this likelihood function whose maximum is the same and can be computed easily.\n\nFollowing Definition 5.2.1 in Larsen and Marx \\cite{larsen2005introduction}, the likelihood function is described by\n\\begin{equation}\n\\begin{split}\nL(\\theta) = \\prod_{i=1}^n f_Y(y_i;~\\theta),\n\\end{split}\n\\label{eq:LikelyhoodFunctionDefinition}\n\\end{equation}\nwhich, in the case of this tasks, is \n\\begin{equation}\n\\begin{split}\nL(\\theta) \n&= \\prod_{i=1}^n f_T(z_i;~\\theta)\\\\\n&= \\prod_{i=1}^n \\frac{z_i}{\\theta^2} e^{-\\frac{z_i}{\\theta}}\\\\\n\\end{split}\n\\label{eq:LikelyhoodFunction2}\n\\end{equation}\nand simplifies to \n\\begin{equation}\n\\begin{split}\nL(\\theta) \n&= \\theta^{-2n}\\left(\\prod_{i=1}^n z_i\\right)e^{-\\frac{1}{\\theta}\\sum_{i=1}^n z_i}.\n\\end{split}\n\\label{eq:LikelyhoodFunction3}\n\\end{equation}\nTo calculate the maximum likelihood, equation \\eqref{eq:LikelyhoodFunction3} must be derived with respect to $\\theta$ and equated to $0$. In its current form, such calculation quickly becomes computationally expensive and eventually inaccurate. To surpass such problem, \\eqref{eq:LikelyhoodFunction3} can be transformed using the natural logarithm as the value maximising $L(\\theta)$ also maximises $\\ln L(\\theta)$:\n\\begin{equation}\n\\begin{split}\n\\ln~L(\\theta) \n&= -2n\\ln\\theta + \\ln\\left(\\prod_{i=1}^n z_i\\right)-\\frac{1}{\\theta}\\sum_{i=1}^n z_i.\n\\end{split}\n\\label{eq:LikelyhoodFunction4}\n\\end{equation}\nThe derivation then yields\n\\begin{equation}\n\\begin{split}\n\\frac{d\\ln~L(\\theta)}{d\\theta}\n&= \\frac{-2n}{\\theta}+\\frac{1}{\\theta^2}\\sum_{i=1}^n z_i,\n\\end{split}\n\\label{eq:LikelyhoodFunction5}\n\\end{equation}\nwhich results in the maximum likelihood function when equated to $0$ and solved for $\\theta$:\n\\begin{equation}\n\\begin{split}\n\\frac{-2n}{\\theta}+\\frac{1}{\\theta^2}\\sum_{i=1}^n z_i = 0.\n\\end{split}\n\\label{eq:LikelyhoodFunction6}\n\\end{equation}\nThe maximum likelihood estimator $\\theta_e$ is consequently \n\\begin{equation}\n\\begin{split}\n\\theta_e =\\frac{1}{2n}\\sum_{i=1}^n z_i.\n\\end{split}\n\\label{eq:MLE}\n\\end{equation}\nIn an experiment with $n=5$ observations measuring the total bandwidth to failure, the following results were observed:\n\\begin{center}\n16, 86, 96, 17, 339.\n\\end{center}\nTo estimate the model-parameter with the maximum likelihood, the values are inserted into \\eqref{eq:MLE}:\n\\begin{equation}\n\\begin{split}\n\\theta_e =\\frac{1}{2\\cdot5}(16+86+96+17+339) = 55.4.\n\\end{split}\n\\label{eq:MLECalculation}\n\\end{equation}\nTo calculate the expectation of the bandwidth, we reuse equation \\eqref{eq:ExpValueHogg} which, applied to this task, is\n\\begin{equation}\n\\begin{split}\nE(T) = \\int_0^\\infty z\\frac{z}{\\theta^2} e^{-\\frac{z}{\\theta}}~dz,\\\\\n\\end{split}\n\\label{eq:expvalue3}\n\\end{equation}\nand calculates to \n\\begin{equation}\n\\begin{split}\nE(T) \n&= \\int_0^\\infty z\\frac{z}{55.4^2} e^{-\\frac{z}{55.4}}~dz\\\\\n&= 110.8.\n\\end{split}\n\\label{eq:expvalue4}\n\\end{equation}\nEquation \\eqref{eq:expvalue4} was calculated using symbolab \\cite{symbolab} with the following command:\n\\begin{lstlisting}\n\\int _0^{\\infty }\\:z\\frac{z}{55.4^2}\\:e^{-\\frac{z}{55.4}}\n\\end{lstlisting}\n\n\\chapter{Workbook Assignment 4: Hypothesis Test}\t\n\nAssumed we are producing high-quality hammers in our factory and want to model the production quality. From the last 1000 produced hammers, we observed an average weight of $\\mu_1=971~g$ and a standard deviation of $\\sigma_1 = 42~g$. As $n = 1000$, we assume that we can treat the point estimates for the mean and the standard deviation as population parameters.\n\nProduction inaccuracies are usually normally dispersed, and as we neither have access to the raw data of individual observations nor have information on any skewness or non-normal behaviour, a normal distribution $N(\\mu, \\sigma^2)$ is assumed.\nThe normal distribution is defined as follows \\cite[Definition~3.4.1]{hogg}: \n\n\\begin{equation} \\frac{1}{{\\sigma \\sqrt {2\\pi } }}e^{{{ - \\left( {x - \\mu } \\right)^2 } \\mathord{\\left/ {\\vphantom {{ - \\left( {x - \\mu } \\right)^2 } {2\\sigma ^2 }}} \\right. \\kern-\\nulldelimiterspace}{2\\sigma ^2 }}}.\n\\label{eq:normaldist}\n\\end{equation}\n\n\nIt depends on two parameters, the population standard deviation $\\sigma$ and the population mean $\\mu$, which both are provided. In general, any normal distribution has a bell-shaped form, indicating that it is symmetric and not skewed to any side. Apart from the mean and the standard deviation, we are not provided any further data. The assumption of such model of course does not hold anymore if the data of the 1000 produced hammers introduces any kind of bias or skewness. \n\n\nOther distributions are excluded as they are not applicable to this case or are less fitting. A Binomial Distribution for example is not applicable as it is binary, whereas the independent variable central to this exercise - hammer weight - is continuous. \\\\Describing the hammer weight as a repeated Bernoulli trial would also be an option to model the distribution, however such endeavor would firstly approximate a normal distribution as well and would secondly assume that the weight measurements are integer variables, whereas they are of continuous nature. \\\\\nAnother popular distribution, the Poisson distribution, is also not applicable as it models the average number of events per time interval \\cite{bruce2017practical}. \\\\\nUsing an exponential distribution to model the weight dispersion of hammer production would imply that we expect for example continuously increasing counts of hammers for increasing weights, which is not a senseful assumption. \\\\\nConsequently, we assume that the normal distribution is the best approximation.\n\nUsing the normal distribution as approximation implies accepting several assumptions. We assume that the model has the x-axis as an horizontal asymptote, that its mean is equal to its median, that its mode is the same as its mean and that it is symmetric about the mean \\cite[Chapter~3.4]{hogg}.\n\nUsing \\eqref{eq:normaldist} with the provided parameters of $\\mu_1 = 971$ and $\\sigma_1 = 42$, the following model is proposed: \n\n\\begin{equation} \\frac{1}{{42 \\sqrt {2\\pi } }}e^{{{ - \\left( {x - 971 } \\right)^2 } \\mathord{\\left/ {\\vphantom {{ - \\left( {x - 971 } \\right)^2 } {2(42) ^2 }}} \\right. \\kern-\\nulldelimiterspace}{2(42) ^2 }}} ~~~=~~~ \\frac{1}{{42 \\sqrt {2\\pi } }}e^{{{ - \\left( {x - 971 } \\right)^2 } \\mathord{\\left/ {\\vphantom {{ - \\left( {x - 971 } \\right)^2 } {7056 }}} \\right. \\kern-\\nulldelimiterspace}{3528 }}}.\n\\label{eq:normaldistadapted}\n\\end{equation}\n\n\n\nAssumed a new production system for hammer production is tested. A random sample output of hammers of this new production output returns the following weights in grams: \\\\\n\n\\begin{center}\n 1009, 1069, 966, 952, 1084, 897, 909, 995, 975, 983.\n\\end{center}\n\nHow do we know if this new system's weight accuracy is better than the old system's accuracy, i.e. how do we find out whether the weight variance of the new system is lower than the variance of the old system?\n\nTo answer this question, we have to perform a statistical test comparing the variances of both samples, which requires hypotheses. Overall, we expect that the new system reduces weight variance and is more accurate. The hypotheses are consequently \n\n\\begin{center}\n$H_0~~:~~ \\sigma^2_2~=~\\sigma^2_1$\\\\ \n$H_1~~:~~ \\sigma^2_2~<~\\sigma^2_1$.\n\\end{center}\n\nwhere $\\sigma^2_1$ is defined as the population variance of the old production system and $\\sigma^2_2$ is defined as the population variance of the new production system.\n\nFollowing Larsen and Marx \\cite[Chapter 7.5]{larsen2005introduction}, the chi square distribution is needed for inferences on variances. To test our sample variance $s_2^2 = \\frac{32594.9}{9} \\approx 3621.66 $ against the variance $\\sigma^2_1 = 42^2 = 1764$, we can now use the Chi Square Test or the F-Test. As we earlier assumed that $\\sigma_1$ is the population variance instead of a sample variance, we do not need to rely on an F-Test and can work with the Chi Square Test. The hypotheses can then be formulated as \n\n\\begin{center}\n$H_0~~:~~ \\sigma^2_2~=~1764$\\\\ \n$H_1~~:~~ \\sigma^2_2~<~1764$.\n\\end{center}\n\nFollowing Larsen and Marx \\cite[Chapter 7.5]{larsen2005introduction}, the Chi Square test statistic is given by \n\n\\begin{equation}\n\\chi^2 = \\frac{(n-1)s_2^2}{\\sigma_1^2},\n\\label{eq:chisquarestatistic}\n\\end{equation}\n\nwith the decision rule of rejecting $H_0$ if \n\\\\\n\\begin{equation}\n\\chi^2 \\leq \\chi^2_{1-\\alpha,n-1},\n\\label{eq:decisionrule}\n\\end{equation}\n\\\\\nwhere our error probability $\\alpha$ is our level of significance. An usual assumption is a 95 percent confidence interval, which results in $\\alpha = 0.05$. As the losses, which an error in this test might incur, can be expected to be quite minor, there is no need to increase the confidence interval and to work with a smaller $\\alpha$.\n\n\nInserting the sample figures into \\eqref{eq:chisquarestatistic}, we observe \n\\begin{equation}\n\\chi^2 = \\frac{(10-1) \\cdot \\frac{32594.9}{9}}{1764} = \\frac{32594.9}{1764} \\approx 18.48.\n\\label{eq:chisqstatcalc}\n\\end{equation}\n\nUsing the lower-tail critical values of chi-square distribution of the Engineering Statistics Handbook \\cite{nist}, the critical value is approximated by \n\\\\\n\\begin{equation}\n\\chi^2_{0.95,9}\\approx 3.33,\n\\label{eq:criticalvalue}\n\\end{equation}\n\\\\\nwith the corresponding p-value of\n\\\\\n\\begin{equation}\np = P(\\chi^2 < 18.48) = 1-0.03 = 0.97,\n\\label{eq:pvalue}\n\\end{equation}\n\\\\\nusing the p-Value Calculator for a Chi-Square Test \\cite{freepvalue}. \n\nConsequently, as \n\\\\\n\\begin{equation}\n\\chi^2 \\nleq \\chi^2_{0.95,9}\n\\label{eq:criticalvalue}\n\\end{equation}\n\\\\\nor, put differently, the p-value is much larger than our $\\alpha$, we can not reject $H_0$ and can not say that the new model makes more constant weights. Looking at the large initial difference in variances $s_2^2 \\approx 3621.66 $ and $\\sigma^2_1 = 1764$, this result is not too surprising. \n\n\n\n\\chapter{Workbook Assignment 5: Sufficient Statistics}\t\nLet $X_1,~X_2,~...,~X_n$ be the random sample of a positive random variable $X$ having the density function \n\\begin{equation}\n\\begin{split}\nf_X(x;~\\theta) = \\frac{1}{2!~\\theta^3}x^2e^{-\\frac{x}{\\theta}}\n\\end{split}\n\\label{eq:suffpdf}\n\\end{equation}\nwith one parameter $\\theta \\in \\mathbb{R}+$. Find an estimator for $\\theta$ that is sufficient. \n\\\\\nLarsen and Marx define sufficient statistics as follows: \"Let $X_1 = k_1, . . . , X_n = k_n$ be a random sample of size $n$ from $p_X(k;~\\theta)$. The statistic $\\hat{\\theta} = h(X_1, . . . , X_n)$ is sufficient for $\\theta$ if the likelihood function, $L(\\theta)$, factors into the product of the pdf for $\\hat{\\theta}$ and a constant that does not involve $\\theta$ — that is, if\n\\begin{equation}\n\\begin{split}\nL(\\theta) \n= \\prod_{i=1}^n p_X(k_i;\\theta)\n=p_{\\hat{\\theta}}(\\theta_e;\\theta)b(k_1,...,k_n)\n\\end{split}\n\\label{eq:SufficientLikelihood}\n\\end{equation}\nA similar statement holds if the data consist of a random sample $Y_1~=~y_1,...,Y_n~=~ y_n$ drawn from a coninuous pdf $f_Y(y;\\theta)$.\" \\cite{larsen2005introduction}.\nThis can be constructed for the given density function \\eqref{eq:suffpdf}. Firstly, to calculate the likelihood function, its definition \\eqref{eq:LikelyhoodFunctionDefinition} is used.\n\\begin{equation}\n\\begin{split}\nL(\\theta) \n&= \\prod_{i=1}^n f_X(x_i;~\\theta)\\\\\n&= \\prod_{i=1}^n \\frac{1}{2!~\\theta^3}x_i^2e^{-\\frac{x_i}{\\theta}}.\\\\\n\\end{split}\n\\label{eq:SufficientLikelihood2}\n\\end{equation}\nThis can be simplified to \n\\begin{equation}\n\\begin{split}\nL(\\theta) \n&= \\frac{1}{2^n\\theta^{3n}}\\prod_{i=1}^n x_i^2e^{-\\frac{x_i}{\\theta}}\\\\\n&= \\frac{1}{2^n\\theta^{3n}} e^{-\\frac{1}{\\theta}\\sum_{i=1}^nx_i} \\left(\\prod_{i=1}^n x_i\\right)^2.\\\\\n\\end{split}\n\\label{eq:SufficientLikelihood3}\n\\end{equation}\nNow\n\\begin{equation}\n\\begin{split}\np_{\\hat{\\theta}}(\\theta_e;\\theta) = \\frac{1}{2^n\\theta^{3n}} e^{-\\frac{1}{\\theta}\\sum_{i=1}^nx_i}\n\\end{split}\n\\label{eq:SufficientLikelihood4}\n\\end{equation}\nand\n\\begin{equation}\n\\begin{split}\nb(k_1,...,k_n) = \\left(\\prod_{i=1}^n x_i\\right)^2.\n\\end{split}\n\\label{eq:SufficientLikelihood5}\n\\end{equation}\nAs \\eqref{eq:SufficientLikelihood5} is not a function of $\\theta$, $\\sum_{i=1}^nx_i$ is a sufficient estimator for $\\theta$.\n\n\n\n\n\n\n\\chapter{Workbook Assignment 6: Bayesian Estimates}\t\nLet $X_1, X_2, ... , X_{10}$ be a random sample from a gamma distribution with $\\alpha =3$ and $\\beta =1/\\theta$. Suppose we believe that $\\theta$ follows a gamma-distribution with $\\alpha =3$ and $\\beta = 2$.\n\n\\section{Posterior Distribution}\nTo calculate the posterior distribution, the following Bayesian model can be formulated from the task assignment:\n\\begin{equation}\n\\begin{split}\nX_i|\\theta &\\sim \\Gamma\\left(3,\\frac{1}{\\theta}\\right)\\\\\n\\Theta &\\sim \\Gamma(3,2).\n\\end{split}\n\\label{eq:Bayesmodel}\n\\end{equation}\nThe Gamma distribution which both $X$ and $\\Theta$ follow is given by Hogg et al. \\cite[Equation~3.3.2]{hogg}:\n\\begin{equation}\n\\begin{split}\nf(x) = \\frac{x^{\\alpha-1}e^{-\\frac{x}{\\beta}}}{\\Gamma(\\alpha)\\beta^\\alpha}\\quad x>0.\n\\end{split}\n\\label{eq:GammaDist}\n\\end{equation}\nGenerally, the posterior pdf $k(\\theta|X)$ is given by Hogg et al. as: \\cite[Equation~11.1.5]{hogg} \n\\begin{equation}\n\\begin{split}\nk(\\theta|x) = \\frac{L(x|\\theta)h(\\theta)}{g_1(x)}\n\\end{split}\n\\label{eq:Posterior}\n\\end{equation}\nwhere \n\\begin{equation}\n\\begin{split}\ng_1(x) = \\int_{-\\infty}^\\infty L(x|\\theta)h(\\theta) ~d\\theta.\n\\end{split}\n\\label{eq:g_1}\n\\end{equation}\nFollowing Hogg et al. \\cite[Chapter 11.1.1]{hogg} the simplified notation of \n\\begin{equation}\n\\begin{split}\nk(\\theta|x) \\propto L(x|\\theta)h(\\theta)\n\\end{split}\n\\label{eq:Posterior2}\n\\end{equation}\nis sufficient to calculate the posterior distribution as $g_1(x)$ is a normalising factor in this context. The calculation of $g_1(x)$ can consequently be dropped. $L(x|\\theta)$ is obtained by inserting the upper model of \\eqref{eq:Bayesmodel} into \\eqref{eq:GammaDist}:\n\\begin{equation}\n\\begin{split}\nL(x|\\theta) \n&= \\prod_{i=1}^n \\frac{x_i^{3-1}e^{-\\frac{x_i}{\\frac{1}{\\theta}}}}{\\Gamma(3)\\left(\\frac{1}{\\theta}\\right)^3} \\\\\n&= \\prod_{i=1}^n \\frac{x_i^{2}\\theta^3e^{-x_i\\theta}}{\\Gamma(3)}\\\\\n&= \\frac{x^{2n}\\theta^{3n}e^{-\\theta\\sum_{i=1}^nx_i}}{\\Gamma(3)^n}.\n\\end{split}\n\\label{eq:bayeslikelyhood}\n\\end{equation}\nThe prior pdf is given by the lower part of \\eqref{eq:Bayesmodel}:\n\\begin{equation}\n\\begin{split}\nh(\\theta) = \\frac{\\theta^{3-1}e^{-\\frac{\\theta}{2}}}{\\Gamma(3)2^3}\n=\\frac{\\theta^{2}e^{-\\frac{\\theta}{2}}}{8\\Gamma(3)} .\n\\end{split}\n\\label{eq:BayesPrior}\n\\end{equation}\nConsequently, \n\\begin{equation}\n\\begin{split}\nk(\\theta|x) \n\\propto \\frac{x^{2n}\\theta^{3n}e^{-\\theta\\sum_{i=1}^nx_i}}{\\Gamma(3)^n} \\frac{\\theta^{2}e^{-\\frac{\\theta}{2}}}{8\\Gamma(3)},\n\\end{split}\n\\label{eq:calculatedposterior}\n\\end{equation}\nassuming $0<\\theta<\\infty$. Dropping the constants results in \n\\begin{equation}\n\\begin{split}\nk(\\theta|x) \n&\\propto \\theta^{3n}\\theta^{2}e^{-\\theta\\sum_{i=1}^nx_i} e^{-\\frac{\\theta}{2}}\\\\\n&\\propto \\theta^{3n+2}e^{-\\theta(\\frac{1}{2}+\\sum_{i=1}^nx_i)}\n\\end{split}\n\\label{eq:calculatedposterior2}\n\\end{equation}\nLooking at the form of \\eqref{eq:calculatedposterior2}, it is apparent that it is a $\\Gamma$ function. $\\alpha$ and $\\beta$ can now be calculated using \\eqref{eq:calculatedposterior2}, \\eqref{eq:GammaDist} and $n=10$ from the assignment.\n\\begin{equation}\n\\begin{split}\n\\alpha-1 &= 3n+2\\\\\n\\alpha &= 33\n\\end{split}\n\\label{eq:postalpha}\n\\end{equation}\n\\begin{equation}\n\\begin{split}\n-\\frac{\\theta}{\\beta} &= -\\theta\\left(\\frac{1}{2}+\\sum_{i=1}^nx_i\\right)\\\\\n\\beta &= \\frac{1}{\\frac{1}{2}+\\sum_{i=1}^nx_i}\n\\end{split}\n\\label{eq:postbeta}\n\\end{equation}\nThe posterior distribution is consequently \n\\begin{equation}\n\\begin{split}\n\\Theta|X &\\sim \\Gamma\\left(33,\\frac{1}{\\frac{1}{2}+\\sum_{i=1}^nx_i}\\right).\n\\end{split}\n\\label{eq:Posteriormodel}\n\\end{equation}\n\n\\section{Bayes Point Estimate by Square-Error Loss Function}\nAssumed, $\\bar{x} = 16.5$ is observed, what is the Bayes point estimate associated with the square-error loss function?\nFollowing Theorem 5.8.1 \\cite{larsen2005introduction}, the Bayes estimate for $\\theta$ is the mean of the posterior if a square-error loss function is used. As the distribution is a Gamma-distribution, the mean or expected value is calculated by \n\\begin{equation}\n\\begin{split}\nE(X|\\Theta) = \\alpha\\beta,\n\\end{split}\n\\label{eq:expvalgamma}\n\\end{equation}\nas proven in Chapter 3.3 \\cite{hogg}. Using $\\alpha = 33$ and \n\\begin{equation}\n\\begin{split}\n\\beta &= \\frac{1}{\\frac{1}{2}+\\sum_{i=1}^nx_i}\\\\\n&=\\frac{1}{\\frac{1}{2}+n\\bar{x}}\\\\\n&=\\frac{1}{\\frac{1}{2}+10\\cdot16.5}\\\\\n&=\\frac{1}{165.5},\n\\end{split}\n\\label{eq:postbeta}\n\\end{equation}\nthe mean computes to \n\\begin{equation}\n\\begin{split}\nE(X|\\Theta) = \\frac{33}{165.5} \\approx 0.1994.\n\\end{split}\n\\label{eq:expvalgammacalc}\n\\end{equation}\n\n\\section{Point Estimate by Mode}\nWhat is the Bayes point estimate using the mode of the posterior distribution?\\\\\nAs the mode of a distribution is simply its highest point, it is the maximum of the posterior distribution. Consequently, the maximum is found by deriving the posterior distribution with respect to $\\theta$, equating it to 0 and solving it. The posterior distribution is\n\\begin{equation}\n\\begin{split}\nf(\\theta) = \\frac{\\theta^{32}e^{-\\theta(0.5+\\sum_{i=1}^n x_i)}}\n{\\Gamma(33)\\beta^\\alpha},\n\\end{split}\n\\label{eq:postdistinserted}\n\\end{equation}\nwhere the denominator is constant and can be excluded. The derivative computes to\n\\begin{equation}\n\\begin{split}\nf_\\theta'(\\theta) = 32\\theta^{31}e^{-\\theta\\left(0.5+\\sum_{i=1}^n x_i\\right)}+e^{-\\theta\\left(0.5+\\sum_{i=1}^n x_i\\right)}\\left(-0.5-\\sum_{i=1}^n x_i\\right)\\theta^{32}\n\\end{split}\n\\label{eq:postdistderived}\n\\end{equation}\nand results in the following roots if equated to 0:\n\\begin{equation}\n\\begin{split}\n\\theta_1 = 0 \\qquad \\theta_2 = \\frac{32}{0.5+\\sum_{i=1}^n x_i}.\n\\end{split}\n\\label{eq:roots}\n\\end{equation}\n$\\theta_1 = 0$ is no solution for a gamma distribution. Consequently, $\\theta_2$ is the only possible solution. Inserting $\\bar{x} = 16.5$ from previous section results in \n\\begin{equation}\n\\begin{split}\n\\theta_2 = \\frac{32}{0.5+10\\cdot16.5} = \\frac{32}{165.5} \\approx 0.1934.\n\\end{split}\n\\label{eq:rootsN}\n\\end{equation}\n", "meta": {"hexsha": "2d202e25aac2aa603323e0a667720fa85104edb6", "size": 50616, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "020-Body.tex", "max_stars_repo_name": "Paul1911/Advanced-Statistics", "max_stars_repo_head_hexsha": "54163a8575235197a82594a722f3eaff2602b0dc", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "020-Body.tex", "max_issues_repo_name": "Paul1911/Advanced-Statistics", "max_issues_repo_head_hexsha": "54163a8575235197a82594a722f3eaff2602b0dc", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "020-Body.tex", "max_forks_repo_name": "Paul1911/Advanced-Statistics", "max_forks_repo_head_hexsha": "54163a8575235197a82594a722f3eaff2602b0dc", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.6721311475, "max_line_length": 859, "alphanum_fraction": 0.7021495179, "num_tokens": 16836, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.831143031127974, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.634564209457508}} {"text": "% CREATED BY DAVID FRISK, 2016\n\\section{Theory}\nIn the following the section the underlying theory of the models used in the project are introduced and explained, along with relevant background and examples. \n\n\\subsection{CPG Model}\n\n\\subsubsection{Mathematical Model of Physical Behaviour}\n%CPG is a mathematical model of human/animal behaviour. Limit cycle, standing wave. Robust, can handle disturbance. (Model different different joints as CPG. Model different DoF as CPG.)\n\nBiological behaviour in mammals, such as walking or swimming, are usually repetative and oscillatory creating energy efficient movements. This can be seen in Figure \\ref{fig:walkingCycle}, which shows the walking cycle of a human, measured as the acceleration in Z-direction whilst walking. Due to the rhytmic patterns in the behaviour, it is possible to construct a mathematical model to mimic this type of behaviour. \n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=.75\\textwidth]{include/figure/reference_right_hip_z.png}\n \\caption{Walking cycle of a human. Where \\textit{Z} is the horizontal direction of the reference system.}\n \\label{fig:walkingCycle}\n\\end{figure}\n\nA Central Pattern Generator (CPG) is a mathematical model of a neural oscillator, a neural network that produces rhytmic patterned outputs without sensory feedback \\cite{CPGHauert}. The output of a CPG can be used as a model of biological rhytmic behaviour, for instance the activation of a certain muscle.\n\nThe activity of one muscle in an animal affect the activation of other muscles in its body automatically in order to distribute load and to achieve balance. Similarily CPGs can be modelled as a network, to account for more than a single muscle as well as the interaction between muscles.\n\n\n\n\\subsubsection{Model of Neural Oscillator} \\label{modelOfNeuralOscillator}\n%The different existing models. We chose the Matsuoka, show matsuoka sketch, describe parameters.\n\nThere exists many different models of CPGs \\cite{CPGmodels}, the model to be examined throughout this project is the half-centre model. The half-centre model proposes to account for the alternate activation of the extensor and flexor muscles in the limbs, originally inspired by a cat walking. A mathimatical model was proposed by K. Matsuoka \\cite{matsuoka}, using two mutually inhibatory neurons in each oscillator to account for the extensor-flexor behaviour. The model ensures that when one neuron is active, the other is surpressed, creating the oscillatory behaviour.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.75\\textwidth]{include/figure/matsuoka.png}\n \\caption{The Matsuoka model of a half-centre CPG \\cite{CPGwolff}, with parameters explained in Table \\ref{tab:parameters}.}\n \\label{fig:matsuoka}\n\\end{figure}\n\nThe setup of the Matsuoka model is shown in Figure \\ref{fig:matsuoka}, with the parameters as described in Table \\ref{tab:parameters}. The behaviour of the model is described by the following equations \\cite{CPGequations}:\n\n\\begin{align}\n \\tau_u \\dot{u}_i &= -u_i - \\beta v_i + \\sum^{n}_{j=1} w_{ij}y_j + u_0, \\\\\n \\tau_v \\dot{v_i} &= -v_i + y_i, \\\\\n y_i &= \\text{max}(0, u_i), \\\\\n o &= y_2 - y_1\n\\end{align}\n\nwhere the model is simulated by integrating over time using Euler integration and timestep $\\Delta t$ and $\\dot{u}_i$, $\\dot{v}_i$ from above in:\n\n\\begin{align}\n u_i(t + \\Delta t) &= u_i(t) + \\Delta t \\dot{u}_i, \\\\\n v_i(t + \\Delta t) &= v_i(t) + \\Delta t \\dot{v}_i.\n\\end{align}\n\n\\begin{table}[htbp]\n \\centering\n \\begin{tabular}{|c|p{6.5cm}|c|c|}\n \\hline\n\\textbf{Parameter} & \\textbf{Description} & \\textbf{Initial value} & \\textbf{Comment} \\\\ \\hline\n$z_i$ & input of neuron i & & oscillating \\\\ \\hline\n$u_0$ & external tonic input & 1.0 & \\\\ \\hline\n$u_1$ & inner state of neuron 1 & 0.0 & fixed \\\\ \\hline\n$u_2$ & inner state of neuron 2 & 1.0 & fixed \\\\ \\hline\n$v_1$ & auxiliary variable measuring the degree of self-inhibition of neuron 1 & 1.0 & fixed \\\\ \\hline\n$v_2$ & auxiliary variable measuring the degree of self-inhibition of neuron 2 & 0.0 & fixed \\\\ \\hline\n$\\beta$ & modulation parameter & 2.5 & \\\\ \\hline\n$w_{ij}$ & weights connecting neuron j to neuron i & -2.0 & \\\\ \\hline\n$y_i$ & the output of neuron i & 0 & \\\\ \\hline\n$\\tau_u$ & time constant & 0.025 & \\\\ \\hline\n$\\tau_v$ & time constant & 0.3 & \\\\ \\hline\n$o$ & the output of the oscillator & & \\\\ \\hline\n \\end{tabular}\n \\caption{The parameters used in the Matsuoka model of a half-centre CPG.}\n \\label{tab:parameters}\n\\end{table}\n\n\n\\subsubsection{Model of CPG Network} \\label{CPGnetwork}\n%Combining different CPG neurons on a network to model a human where each Neuron is a DoF.\n\nCoupling multiple CPGs into a network makes it possible to model more complex structures, such as an animal or a human. This can be done by connecting the output of one neuron to the input of another. This project investigatex the walking behaviour of a human implemented on a humanoid robot, thus human behaviour can be modelled by looking at the movement in different joints of the body, corresponding to the joints of the humanoid robot. Since the output of a CPG is one-dimensional over time, and human movement is three dimensional, one can imagine modelling the movement in each dimension with a separate CPG. The output signals can then be superimposed to obtain the total movement.\n\n\n\\subsubsection{Choice of Input to Oscillator} \\label{choiseOfInputToOscillator}\n%The different models (summing output and weights). Asyncronous vs syncronous.\n%2.1.3 in old report\nEach of the neurons in a CPG network have inputs ($z_i$) that are affected by the output of the other neurons ($o$). One can imagine that all neurons are fully connected, but by setting the weight between two neurons to zero, that weight is \\textit{removed}. The input to the neurons can be calculated in different ways, one way inspired by Liu et.al. \\cite{liu} is to define the inputs to neuron $i$ as\n\\begin{align}\nz_1 &= \\sum_{j \\text{ neighbour of } i} a_{ij}u_j, \\\\\nz_2 &= 0.\n\\end{align}\nAnother implementation inspired by Shan et.al. \\cite{shan} defines the inputs as\n\\begin{align}\nz_1 &= \\sum_{j \\text{ neighbour of } i} a_{ij}[y_j]^+, \\\\\nz_2 &= \\sum_{j \\text{ neighbour of } i} a_{ij}[y_j]^-,\n\\end{align}\nwhere \n\\begin{align}\n[y_j]^+ &:= \\text{max}(y_j,0), \\\\\n[y_j]^- &:= \\text{min}(y_j,0).\n\\end{align}\n\nNeurons can also be affected by an external tonic input ($u_0$). The amplitude of the oscillation in the CPG is proportional to the tonic input. If the input is oscillatory, the oscillator will lock to the frequency of the input. When the input is removed, the oscillator will smoothly return to the original frequency \\cite{CPGwolff}.\n\n\n\\subsection{Optimisation Using Genetic Algorithm}\nA CPG alone might not achieve satisfactory results, as the parameters in the CPG or network might need to be tweaked and optimised to achieve the desired behaviour. A commonly used technique for optimising behaviour is to use reinforcement learning to improve the model based on an external measure of success. This project uses a Genetic algorithm (GA) as a reinforcement learning technique to achieve desired walking behaviour. A genetic algorithm is an evoutionary algorithm that takes inspration from natural selection implemented as an optimisation problem \\cite{GAHauert}. \n\n\n\\subsubsection{Genome}\n%Difference between using only weights or the entire CPG network as genome.\n\n\nThe \\textit{genome} is a set of parameters that defines a proposed solution to the problem that the GA is trying to solve. One can imagine various different genomes when optimising the behaviour of a CPG, for instance the internal parameters in the CPG, shown in Table \\ref{tab:parameters}. The connection weights between neurons in a CPG network, as described in Section \\ref{CPGnetwork}, could also be used to account relation between CPGs in a network.\n\n\\subsubsection{Operators}\n%Init population, simulate behaviour, evaluation, selection, mutation, permutation\nThe basic implementation of a GA is in summary; initialise the population, simulate the behaviour, evaluate success, selecion process, mutation and permutations, repeat until converge. A more detailed description can be found in \\cite{wahde}. The evaluation process, also known as the \\textit{fitness function}, produces a fitness score which is the measurement the success of the individual. Since this project investigates walking behaviour in a humanoid robot, a fitness function may include e.g.; distance walked, time walked, stability, falling, walking style, etc.", "meta": {"hexsha": "4268542c73a519bb2ff715e346373c11cf761972", "size": 8591, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/include/Theory.tex", "max_stars_repo_name": "horken7/bioloid-cpg", "max_stars_repo_head_hexsha": "d0e16e6699e5f6c19a9b0a6c11fc29798fc2b1d9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-11-14T09:30:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-14T11:45:35.000Z", "max_issues_repo_path": "Report/include/Theory.tex", "max_issues_repo_name": "horken7/bioloid-cpg", "max_issues_repo_head_hexsha": "d0e16e6699e5f6c19a9b0a6c11fc29798fc2b1d9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/include/Theory.tex", "max_forks_repo_name": "horken7/bioloid-cpg", "max_forks_repo_head_hexsha": "d0e16e6699e5f6c19a9b0a6c11fc29798fc2b1d9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-08-06T05:47:58.000Z", "max_forks_repo_forks_event_max_datetime": "2020-08-06T05:47:58.000Z", "avg_line_length": 74.0603448276, "max_line_length": 689, "alphanum_fraction": 0.7602141776, "num_tokens": 2217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8791467706759584, "lm_q2_score": 0.7217432062975979, "lm_q1q2_score": 0.6345182090738453}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n\n\n \n \n\n\\subsection*{dynamics\\_cdp.m} \n\n\\begin{par}\n\\textbf{Summary:} Implements ths ODE for simulating the cart-double pendulum dynamics.\n\\end{par} \\vspace{1em}\n\n\\begin{verbatim} function dz = dynamics_cdp(t,z,f)\\end{verbatim}\n \\begin{par}\n\\textbf{Input arguments:}\n\\end{par} \\vspace{1em}\n\n\\begin{lstlisting}\n%\t\tt current time step (called from ODE solver)\n% z state [6 x 1]\n% f (optional): force f(t)\n%\n% *Output arguments:*\n%\n% dz if 3 input arguments: state derivative wrt time\n% if only 2 input arguments: total mechanical energy\n%\n% Note: It is assumed that the state variables are of the following order:\n% x: [m] position of cart\n% dx: [m/s] velocity of cart\n% dtheta1: [rad/s] angular velocity of inner pendulum\n% dtheta2: [rad/s] angular velocity of outer pendulum\n% theta1: [rad] angle of inner pendulum\n% theta2: [rad] angle of outer pendulum\n%\n%\n% A detailed derivation of the dynamics can be found in:\n%\n% M.P. Deisenroth:\n% Efficient Reinforcement Learning Using Gaussian Processes, Appendix C,\n% KIT Scientific Publishing, 2010.\n%\n% Copyright (C) 2008-2013 by\n% Marc Deisenroth, Andrew McHutchon, Joe Hall, and Carl Edward Rasmussen.\n%\n% Last modified: 2013-03-05\n\nfunction dz = dynamics_cdp(t,z,f)\n\\end{lstlisting}\n\n\n\\subsection*{Code} \n\n\n\\begin{lstlisting}\n% set up the system\nm1 = 0.5; % [kg] mass of cart\nm2 = 0.5; % [kg] mass of 1st pendulum\nm3 = 0.5; % [kg] mass of 2nd pendulum\nl2 = 0.6; % [m] length of 1st pendulum\nl3 = 0.6; % [m] length of 2nd pendulum\nb = 0.1; % [Ns/m] coefficient of friction between cart and ground\ng = 9.82; % [m/s^2] acceleration of gravity\n\nif nargin == 3\n\n A = [2*(m1+m2+m3) -(m2+2*m3)*l2*cos(z(5)) -m3*l3*cos(z(6))\n -(3*m2+6*m3)*cos(z(5)) (2*m2+6*m3)*l2 3*m3*l3*cos(z(5)-z(6))\n -3*cos(z(6)) 3*l2*cos(z(5)-z(6)) 2*l3];\n b = [2*f(t)-2*b*z(2)-(m2+2*m3)*l2*z(3)^2*sin(z(5))-m3*l3*z(4)^2*sin(z(6))\n (3*m2+6*m3)*g*sin(z(5))-3*m3*l3*z(4)^2*sin(z(5)-z(6))\n 3*l2*z(3)^2*sin(z(5)-z(6))+3*g*sin(z(6))];\n x = A\\b;\n\n dz = zeros(6,1);\n dz(1) = z(2);\n dz(2) = x(1);\n dz(3) = x(2);\n dz(4) = x(3);\n dz(5) = z(3);\n dz(6) = z(4);\n\nelse\n\n dz = (m1+m2+m3)*z(2)^2/2+(m2/6+m3/2)*l2^2*z(3)^2+m3*l3^2*z(4)^2/6 ...\n -(m2/2+m3)*l2*z(2)*z(3)*cos(z(5))-m3*l3*z(2)*z(4)*cos(z(6))/2 ...\n +m3*l2*l3*z(3)*z(4) *cos(z(5)-z(6))/2+(m2/2+m3)*l2*g*cos(z(5)) ...\n +m3*l3*g*cos(z(6))/2;\n\n% I2 = m2*l2^2/12; % moment of inertia around pendulum midpoint (1st link)\n% I3 = m3*l3^2/12; % moment of inertia around pendulum midpoint (2nd link)\n%\n%\n% dz = m1*z(2)^2/2 + m2/2*(z(2)^2-l2*z(2)*z(3)*cos(z(5))) ...\n% + m3/2*(z(2)^2 - 2*l2*z(2)*z(3)*cos(z(5)) - l3*z(2)*z(4)*cos(z(6))) ...\n% + m2*l2^2*z(3)^2/8 + I2*z(3)^2/2 ...\n% + m3/2*(l2^2*z(3)^2 + l3^2*z(4)^2/4 + l2*l3*z(3)*z(4)*cos(z(5)-z(6))) ...\n% + I3*z(4)^2/2 ...\n% + m2*g*l2*cos(z(5))/2 + m3*g*(l2*cos(z(5))+l3*cos(z(6))/2);\n\nend\n\\end{lstlisting}\n", "meta": {"hexsha": "c59d9d09721186a33c04ef8268bd87ba5eb5f8d4", "size": 3215, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/tex/dynamics_cdp.tex", "max_stars_repo_name": "SJTUGuofei/pilco-matlab", "max_stars_repo_head_hexsha": "a0b48b7831911837d060617903c76c22e4180d0b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 53, "max_stars_repo_stars_event_min_datetime": "2016-12-17T15:15:48.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-09T16:59:27.000Z", "max_issues_repo_path": "doc/tex/dynamics_cdp.tex", "max_issues_repo_name": "sahandrez/quad_pilco", "max_issues_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-04-24T11:02:23.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-24T11:09:45.000Z", "max_forks_repo_path": "doc/tex/dynamics_cdp.tex", "max_forks_repo_name": "sahandrez/quad_pilco", "max_forks_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 36, "max_forks_repo_forks_event_min_datetime": "2017-04-19T06:55:25.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-19T10:19:12.000Z", "avg_line_length": 30.3301886792, "max_line_length": 86, "alphanum_fraction": 0.5583203733, "num_tokens": 1328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9073122313857378, "lm_q2_score": 0.6992544273261175, "lm_q1q2_score": 0.6344420947636159}} {"text": "% !TEX root = ../main.tex\n\\newpage\n\\section*{Nomenclature}\n\\addcontentsline{toc}{subsection}{Nomenclature}\n\\vspace{-.5cm}\n\\begin{alignat*}{2}\n& \\ic &&\\text{Imaginary unit.}\\\\\n&e \\text{ (or} \\exp\\text{)} &&\\text{Euler's number.}\\\\ \\\\\n& n&&\\text{Network node.}\\\\\n& N &&\\text{Network degree. The number of nodes in the network.}\\\\\n& A_{ij} &&\\text{Adjacency matrix. Models which node $i$ is connected to node $j$ and vice-versa.}\\\\\n& \\kmean &&\\text{Average node degree in the network.}\\\\\n& \\degree(n) = \\k &&\\text{Degree of node $n$, as a vector of the in- and out-degree $\\left(\\kin, \\kout \\right)$.}\\\\\n& \\kinb, \\koutb && \\text{Node degree vector of all in- and out degrees of the network.} \\\\\n& M_{\\k} &&\\text{Number of unique node degrees in the network. Cardinality of $\\K$.}\\\\\n& P(k), P(\\k) &&\\text{Univariate and bivariate network degree distribution.}\\\\\n&\\kmin, \\kmax &&\\text{Lowest and highest degree in a network.}\\\\ \n&\\gamma &&\\text{Degree exponent of a scale-free network.}\\\\ \n&p &&\\text{Probability threshold of forming a link in random networks.}\\\\ \\\\\n&\\theta_i(t) &&\\text{Phase variable function of the theta model (of neuron $i$).}\\\\\n& \\Pulse_s (\\theta)&&\\text{Pulse-shaped synaptic coupling function with sharpness } s \\text{.}\\\\\n&\\kappa &&\\text{Macroscopic coupling strength.}\\\\\n&\\eta_i, I_i(t) &&\\text{Excitability threshold and input current (of neuron $i$).}\\\\\n&g(\\eta \\rvert \\k) \\quad &&\\text{Excitability threshold distribution with mean $\\eta_0(\\k)$ and width $\\sigma(\\k)$.}\\\\ \\\\\n& Z(t) &&\\text{Kuramoto order parameter function.}\\\\\n& z(\\k, t) &&\\text{Synchronisation dynamics function for nodes of degree $\\k$.}\\\\\n& \\bar{Z}(t) &&\\text{Mean field order parameter function for arbitrary networks.}\\\\ \\\\\n& S_i^{\\rm in}(t), S_i^{\\rm out}(t)\\quad &&\\text{Spike trains received and emitted by neuron $i$ as a sum of delta functions in time.}\\\\ \n& K_{ij} &&\\text{Coupling matrix. Strength of the synaptic connections between neurons $i$ and $j$.}\\\\\n& \\Delta t_{ij} &&\\text{Time difference between spikes of neurons $i$ and $j$.}\\\\\n& W(t) &&\\text{Learning window. Correlation between change in synaptic strength and spike times.}\\\\ \n& \\phi(t) &&\\text{IP learning function. Correlation between excitability strength and spike times.}\\\\ \\\\\n& \\T &&\\text{Set of angles in [-$\\pi$, $\\pi$[.}\\\\\n& \\K &&\\text{Set of $M_{\\k}$ unique degrees in a network, support of $P$.}\\\\ \n& \\R, \\c &&\\text{Set of real and complex numbers.}\\\\ \n& \\C && \\text{Set in the complex unit circle so that $\\C = \\{ z \\in \\c : \\: \\rvert z \\rvert \\leq 1 \\} $.} \\\\ \\\\\n& \\permute(v), \\permuteinv(v) &&\\text{Random permutation and inverse permutation of the elements of a vector $v$.}\\\\ \n\\end{alignat*}\n", "meta": {"hexsha": "562688d496113a4a2048335da2886e12d5c4563d", "size": 2679, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Writing/Frontmatter/Nomenclature.tex", "max_stars_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_stars_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Writing/Frontmatter/Nomenclature.tex", "max_issues_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_issues_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Writing/Frontmatter/Nomenclature.tex", "max_forks_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_forks_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 68.6923076923, "max_line_length": 137, "alphanum_fraction": 0.6621873834, "num_tokens": 854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8947894632969137, "lm_q2_score": 0.7090191276365462, "lm_q1q2_score": 0.6344228446851511}} {"text": "% declare document class and geometry\n\\documentclass[12pt]{article} % use larger type; default would be 10pt\n\\usepackage[margin=1in]{geometry} % handle page geometry\n\n% import packages and commands\n\\input{../header2.tex}\n\n\n\\title{Math 217 -- Geometry and Physics -- Lec06}\n\\author{UCLA, Fall 2014}\n\\date{\\formatdate{15}{10}{2014}} % Activate to display a given date or no date (if empty),\n % otherwise the current date is printed \n\n\\begin{document}\n\\maketitle\n\n\n\\section{More stuff}\n\nLet $E^r \\rightarrow M$ be a complex or real vector bundle on a compact closed manifold. \n\n\\begin{definition}\nDefine \n\\begin{eqn}\n\\Gamma(M,E) = \\Gamma(E) = \\set{ s : M \\rightarrow E \\text{ such that } \\pi \\circ s = \\opname{id}_M}.\n\\end{eqn}\n\\end{definition}\n\nFor example, we have\n\\begin{align}\n\\Gamma(T^*M) &= \\set{ \\sum_i f_i dx_i \\text{ on $U_\\alpha$}} \\\\\n\\Gamma(TM) &= \\set{ \\sum_i f_i \\pd{}{x_i} }.\n\\end{align}\n\n\\begin{definition}\nWe define $S^p E$ the symmetric products, $\\Lambda^p E$ the exterior products, and $\\Omega^p(M) = \\Gamma(\\Lambda^p T^*M)$.\n\\end{definition}\n\n\\begin{definition}\nWe define a connection on $E$ as a map\n\\begin{eqn}\n\\nabla : \\Gamma(E) \\rightarrow \\Gamma(T^*M \\otimes E)\n\\end{eqn}\nsuch that\n\\begin{enumerate}\n\\item $\\nabla(s_1 + s_2) = \\nabla s_1 + \\nabla s_2$, and\n\\item $\\nabla(fs) = (\\dif{f}) s + f (\\nabla s)$ for $f \\in C^\\infty (M)$. \n\\end{enumerate}\n\\end{definition}\n\n\\begin{definition}\nWe define the curvature of $\\nabla$ as\n\\begin{eqn}\n\\Omega = \\nabla^2 : \\Gamma(E) \\rightarrow \\Gamma(\\Lambda^2 T^* M \\otimes E).\n\\end{eqn}\n\\end{definition}\n\n\\begin{claim} \n$\\Omega$ is a $C^\\infty(M)$-linear operator (tensor). \n\\end{claim}\n\n\\begin{proof}\nWe have\n\\begin{align}\n\\Omega(fs) &= \\nabla^2 (fs) = \\nabla(\\nabla(fs)) \\\\\n\t&= \\nabla( \\dif{f} \\, s) + \\nabla(f \\nabla s) \\\\\n\t&= [ \\underbrace{\\dif (\\dif{f})}_{=0} s + \\cancel{(-1) \\dif{f} \\nabla s}] + [\\cancel{\\dif{f} \\nabla s} + f \\nabla^2 s] \\\\\n\t&= f \\Omega(s).\n\\end{align}\n\\end{proof}\n\n\\begin{definition}\nGiven a curvature $\\Omega^\\nabla$, we can expand\n\\begin{eqn}\n\\det(t I_r + \\frac{i}{2\\pi} \\Omega^\\nabla) = \\underbrace{c_0^\\nabla(E)}_{1} t^r + \\underbrace{c_1^\\nabla(E)}_\\text{2-form} t^{r-1} + \\dots + \\underbrace{c_r^\\nabla(E)}_\\text{$2r$-form}.\n\\end{eqn}\nWe call $\\set{c_k^\\nabla(E)}$ the Chern forms of $(E,\\nabla)$. \n\\end{definition}\n\n\\begin{theorem}[Chern-Weil]\nLet $M$ be a compact closed manifold, $E$ a complex vector bundle. Then\n\\begin{enumerate}\n\\item $\\dif c_k^\\nabla(E) = 0$, which implies that\n\\item $[c_k^\\nabla(E)] \\in H^{2k}(M)$. Furthermore, $[c_k^\\nabla(E)]$ is independent of $\\nabla$, so we define $c_k(E) = [c_k^\\nabla(E)]$ as the \\textbf{Chern class} of $E$. \n\\end{enumerate}\n\\end{theorem}\n\nWe will see that Chern-Simons follows from Chern-Weil. \n\n\\begin{remark}\nGiven a function with series expansion\n\\begin{align}\nf(x) &= a_0 + a_1 x + \\dots + a_r x^r + \\dots \\in \\C[[x]], \\\\\n\\shortintertext{we have}\nf(\\Omega) &= a_0 I + a_1 \\Omega + \\dots + a_r \\Omega^r \\in \\C[\\Omega],\n\\end{align}\nthus,\n\\begin{eqn}\n\\tr f(\\Omega) = \\sum_{k=0}^n a_k \\tr \\Omega^k.\n\\end{eqn}\n\\end{remark}\n\n\\begin{lemma}\nWe find that\n\\begin{enumerate}\n\\item $\\dif \\tr f(\\Omega) = 0$, and\n\\item if $\\nabla'$ is another connection, then \n\\begin{eqn}\n\\tr f(\\Omega') - \\tr f(\\Omega) = \\dif{\\eta}\n\\end{eqn}\nfor some $\\eta \\in \\Omega^*(M)$.\n\\end{enumerate}\n\\end{lemma}\n\nThis lemma implies Chern-Weil. We also find that\n\\begin{align}\n\\det (I + \\frac{i}{2\\pi} \\Omega) &= \\exp \\tr \\log (I + \\frac{i}{2\\pi} \\Omega) \\\\\n\t&= \\exp \\log \\det (I + \\frac{i}{2\\pi} \\Omega),\n\\end{align}\nfrom the linear algebra identity $\\tr \\log A = \\log \\det A$. \n\n\\begin{remark}\nNow, given\n\\begin{eqn}\n\\omega, \\eta \\in \\Omega^*(M), \\qquad A, B \\in \\Gamma(\\opname{Hom}(E,E)) = \\opname{End}(E),\n\\end{eqn}\nwe have\n\\begin{eqn}\n[\\omega A, \\eta B] = (\\omega A)(\\eta B) - (-1)^{\\deg \\omega \\deg \\eta} (\\eta B) (\\omega A).\n\\end{eqn}\nThen we find\n\\begin{eqn}\n\\tr [\\omega A, \\eta B] = 0.\n\\end{eqn}\n\\end{remark}\n(Show this as an exercise.)\n\n\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "262ba0b582bec2fbf347f99ea8b11347e98874fa", "size": 3984, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "geometry/lec06.tex", "max_stars_repo_name": "paulinearriaga/phys-ucla", "max_stars_repo_head_hexsha": "48084dbbac2f8a4748c1fdaaf63a4cebaae16809", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "geometry/lec06.tex", "max_issues_repo_name": "paulinearriaga/phys-ucla", "max_issues_repo_head_hexsha": "48084dbbac2f8a4748c1fdaaf63a4cebaae16809", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "geometry/lec06.tex", "max_forks_repo_name": "paulinearriaga/phys-ucla", "max_forks_repo_head_hexsha": "48084dbbac2f8a4748c1fdaaf63a4cebaae16809", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.6666666667, "max_line_length": 185, "alphanum_fraction": 0.6388052209, "num_tokens": 1538, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.8104788995148791, "lm_q1q2_score": 0.6343314246333641}} {"text": "\\chapter{Ordinals}\n\\label{ch:ordinal}\n\\section{Counting for preschoolers}\nIn preschool, we were told to count as follows.\nWe defined a set of symbols $1$, $2$, $3$, $4$, \\dots.\nThen the teacher would hold up three apples and say:\n\\begin{quote}\n\t``One . . .\\ two . . .\\ three! There are three apples.''\n\\end{quote}\n\n\\begin{center}\n\t\\includegraphics[height=5cm]{media/three-apples.jpg}\n\t\\\\ \\scriptsize Image from \\cite{img:apples}\n\\end{center}\n\nThe implicit definition is that the \\emph{last} number said is the final answer.\nThis raises some obvious problems if we try to count infinite sets,\nbut even in the finite world,\nthis method of counting fails for the simplest set of all:\nhow many apples are in the following picture?\n\n\\begin{center}\n\t\\includegraphics[height=7cm]{media/velociraptor.jpg}\n\t\\\\ \\scriptsize Image from \\cite{img:velociraptor}\n\\end{center}\n\nAnswer: $0$. There is nothing to say, and our method of counting has failed\nfor the simplest set of all: the empty set.\n\n\\section{Counting for set theorists}\n\\prototype{$\\omega+1 = \\{0,1,2,\\dots,\\omega\\}$ might work.}\nRather than using the \\emph{last} number listed, I propose instead\nstarting with a list of symbols $0$, $1$, $2$, \\dots\\ and making\nthe final answer the \\emph{first} number which was \\emph{not} said.\nThus to count three apples, we would say \n\\begin{quote}\n\t``Zero . . .\\ one . . .\\ two! There are three apples.''\n\\end{quote}\nWe will call these numbers \\emph{ordinal numbers} (rigorous definition later).\nIn particular, we'll \\emph{define} each ordinal to be the set of things we say:\n\\begin{align*}\n\t0 &= \\varnothing \\\\\n\t1 &= \\{0\\} \\\\\n\t2 &= \\{0,1\\} \\\\\n\t3 &= \\{0,1,2\\} \\\\\n\t&\\vdotswithin=\n\\end{align*}\nIn this way we can write out the natural numbers.\nYou can have some fun with this, by saying things like\n\\[\n\t4 \\defeq\n\t\\left\\{ \n\t\t\\left\\{ \\right\\},\n\t\t\\left\\{ \\left\\{ \\right\\} \\right\\},\n\t\t\\left\\{ \\left\\{ \\right\\}, \\left\\{ \\left\\{ \\right\\} \\right\\} \\right\\},\n\t\t\\left\\{ \n\t\t\t\\left\\{ \\right\\},\n\t\t\t\\left\\{ \\left\\{ \\right\\} \\right\\},\n\t\t\t\\left\\{ \\left\\{ \\right\\}, \\left\\{ \\left\\{ \\right\\} \\right\\} \\right\\}\n\t\t\\right\\}\n\t\\right\\}.\n\\]\nIn this way, we soon write down all the natural numbers.\nThe next ordinal, $\\omega$,\\footnote{As mentioned in the last chapter,\n\tit's not immediate that $\\omega$ is a set;\n\tits existence is generally postulated by the $\\Infinity$ axiom.\n} is defined as\n\\begin{align*}\n\t\\omega &= \\left\\{ 0, 1, 2, \\dots \\right\\} \\\\\n\t\\intertext{Then comes}\n\t\\omega+1 &= \\left\\{ 0, 1, 2, \\dots, \\omega \\right\\} \\\\\n\t\\omega+2 &= \\left\\{ 0, 1, 2, \\dots, \\omega, \\omega+1 \\right\\} \\\\\n\t\\omega+3 &= \\left\\{ 0, 1, 2, \\dots, \\omega, \\omega+1, \\omega+2 \\right\\} \\\\\n\t&\\vdotswithin= \\\\\n\t\\intertext{And in this way we define $\\omega+n$, and eventually reach}\n\t\\omega \\cdot 2 = \\omega+\\omega &= \\left\\{ 0, 1, 2 \\dots, \\omega, \\omega+1, \\omega+2, \\dots \\right\\} \\\\\n\t\\omega \\cdot 2 + 1 &= \\left\\{ 0, 1, 2 \\dots, \\omega, \\omega+1, \\omega+2, \\dots, \\omega \\cdot 2 \\right\\}.\n\\end{align*}\nIn this way we obtain\n\\begin{align*}\n\t0,\\; & 1,\\; 2,\\; 3,\\; \\dots,\\; \\omega \\\\\n\t& \\omega+1,\\; \\omega+2,\\; \\dots,\\; \\omega+\\omega \\\\\n\t& \\omega \\cdot 2 +1,\\; \\omega \\cdot 2 +2,\\; \\dots,\\; \\omega \\cdot 3,\\; \\\\\n\t& \\vdots \\\\\n\t& \\omega^2 + 1,\\; \\omega^2+2,\\; \\dots \\\\\n\t& \\vdots \\\\\n\t& \\omega^3,\\; \\dots,\\; \\omega^4,\\; \\dots,\\; \\omega^\\omega \\\\\n\t& \\vdots \\\\\n\t& \\omega^{\\omega^{\\omega^{\\dots}}} \\\\\n\\end{align*}\n\nThe first several ordinals can be illustrated in a nice spiral.\n\\begin{center}\n\t\\includegraphics[scale=0.60]{media/500px-Omega-exp-omega-labeled.png}\n\\end{center}\n\n\n\\begin{remark}\n\t(Digression)\n\tThe number $\\omega^{\\omega^{\\omega^{\\dots}}}$ has a name, $\\eps_0$;\n\tit has the property that $\\omega^{\\eps_0} = \\eps_0$.\n\tThe reason for using ``$\\eps$'' (which is usually used to denote small quantities)\n\tis that, despite how huge it may appear, it is actually a countable set.\n\tMore on that later.\n\\end{remark}\n\n\\section{Definition of an ordinal}\nOur informal description of ordinals gives us a chain\n\\[ 0 \\in 1 \\in 2 \\in \\dots \\in \\omega \\in \\omega+1 \\in \\dots. \\]\nTo give the actual definition of an ordinal, I need to define two auxiliary terms first.\n\\begin{definition}\n\tA set $x$ is \\vocab{transitive} if whenever $z \\in y \\in x$, we have $z \\in x$ also.\n\\end{definition}\n\\begin{example}\n\t[$7$ is transitive]\n\tThe set $7$ is transitive: for example, $2 \\in 5 \\in 7 \\implies 2 \\in 7$.\n\\end{example}\n\\begin{ques}\n\tShow that this is equivalent to: whenever $y \\in x$, $y \\subseteq x$.\n\\end{ques}\nMoreover, recall the definition of ``well-ordering'': a strict linear order\nwith no infinite descending chains.\n\\begin{example}\n\t[$\\in$ is a well-ordering on $\\omega \\cdot 3$]\n\tIn $\\omega \\cdot 3$, we have an ordering\n\t\\[ 0 \\in 1 \\in 2 \\in \\dots \\in \\omega \\in \\omega+1 \\in \\dots\n\t\t\\in \\omega \\cdot 2 \\in \\omega \\cdot 2 + 1 \\in \\dots. \\]\n\twhich has no infinite descending chains.\n\tIndeed, a typical descending chain might look like\n\t\\[ \\omega \\cdot 2 + 6 \\ni \\omega \\cdot 2 \\ni\n\t\t\\omega + 2015 \\ni \\omega+3 \\ni \\omega \\ni 1000 \\ni 256 \\ni 42 \\ni 7 \\ni 0. \\]\n\tEven though there are infinitely many elements, there is no way\n\tto make an infinite descending chain.\n\\end{example}\n\\begin{exercise}\n\t(Important)\n\tConvince yourself there are no infinite\n\tdescending chains of ordinals at all,\n\twithout using the $\\Foundation$ axiom.\n\\end{exercise}\n\n\\begin{definition}\n\tAn \\vocab{ordinal} is a transitive set which is well-ordered by $\\in$.\n\tThe class of all ordinals is denoted $\\On$.\n\\end{definition}\n\n\\begin{ques}\n\tSatisfy yourself that this definition works.\n\\end{ques}\n\nWe typically use Greek letters $\\alpha$, $\\beta$, etc.\\ for ordinal numbers.\n\\begin{definition}\n\tWe write\n\t\\begin{itemize}\n\t\t\\ii $\\alpha < \\beta$ to mean $\\alpha \\in \\beta$,\n\t\tand $\\alpha > \\beta$ to mean $\\alpha \\ni \\beta$.\n\t\t\\ii $\\alpha \\le \\beta$ to mean $\\alpha \\in \\beta$ or $\\alpha = \\beta$,\n\t\tand $\\alpha \\ge \\beta$ to mean $\\alpha \\ni \\beta$ or $\\alpha = \\beta$,\n\t\\end{itemize}\n\\end{definition}\n\n\\begin{theorem}[Ordinals are strictly ordered]\n\tGiven any two ordinal numbers $\\alpha$ and $\\beta$,\n\teither $\\alpha < \\beta$, $\\alpha = \\beta$ or $\\alpha > \\beta$.\n\\end{theorem}\n\\begin{proof}\n\tSurprisingly annoying, thus omitted.\n\\end{proof}\n\\begin{theorem}[Ordinals represent all order types]\n\tSuppose $<$ is a well-ordering on a set $X$.\n\tThen there exists a unique ordinal $\\alpha$\n\tsuch that there is a bijection $\\alpha \\to X$\n\twhich is order preserving.\n\\end{theorem}\nThus ordinals represent the possible \\emph{equivalence classes} of order types.\nAny time you have a well-ordered set, it is isomorphic to a unique ordinal.\n\nWe now formalize the ``$+1$'' operation we were doing:\n\\begin{definition}\n\tGiven an ordinal $\\alpha$, we let $\\alpha+1 = \\alpha \\cup \\{\\alpha\\}$.\n\tAn ordinal of the form $\\alpha+1$ is called a \\vocab{successor ordinal}.\n\\end{definition}\n\\begin{definition}\n\tIf $\\lambda$ is an ordinal which is neither zero nor a successor ordinal,\n\tthen we say $\\lambda$ is a \\vocab{limit ordinal}.\n\\end{definition}\n\\begin{example}\n\t[Sucessor and limit ordinals]\n\t$7$, $\\omega+3$, $\\omega\\cdot2+2015$ are successor ordinals,\n\tbut $\\omega$ and $\\omega \\cdot 2$ are limit ordinals.\n\\end{example}\n\n\\section{Ordinals are ``tall''}\nFirst, we note that:\n\\begin{theorem}\n\t[There is no set of all ordinals]\n\t$\\On$ is a proper class.\n\\end{theorem}\n\\begin{proof}\n\tAssume for contradiction not.\n\tThen $\\On$ is well-ordered by $\\in$ and transitive, so $\\On$ is an ordinal,\n\ti.e.\\ $\\On \\in \\On$, which violates $\\Foundation$.\n\\end{proof}\n\\begin{exercise}\n\t[Unimportant] Give a proof without $\\Foundation$ by considering $\\On+1$.\n\\end{exercise}\n\nFrom this we deduce:\n\\begin{theorem}\n\t[Sets of ordinals are bounded]\n\tLet $A \\subseteq \\On$.\n\tThen there is some ordinal $\\alpha$ such that $A \\subseteq \\alpha$\n\t(i.e.\\ $A$ must be bounded).\n\\end{theorem}\n\\begin{proof}\n\tOtherwise, look at $\\bigcup A$.\n\tIt is a set.\n\tBut if $A$ is unbounded it must equal $\\On$,\n\twhich is a contradiction.\n\\end{proof}\nIn light of this, every set of ordinals has a \\vocab{supremum},\nwhich is the least upper bound. We denote this by $\\sup A$.\n\n\\begin{ques}\n\tShow that\n\t\\begin{enumerate}[(a)]\n\t\t\\ii $\\sup (\\alpha+1) = \\alpha$ for any ordinal $\\alpha$.\n\t\t\\ii $\\sup \\lambda = \\lambda$ for any limit ordinal $\\lambda$.\n\t\\end{enumerate}\n\\end{ques}\n\nThe pictorial ``tall'' will be explained in a few sections.\n\n\\section{Transfinite induction and recursion}\nThe fact that $\\in$ has no infinite descending chains means that induction and recursion still work verbatim.\n\\begin{theorem}[Transfinite induction]\n\tGiven a statement $P(-)$, suppose that\n\t\\begin{itemize}\n\t\t\\ii $P(0)$ is true, and\n\t\t\\ii If $P(\\alpha)$ is true for all $\\alpha < \\beta$, then $P(\\beta)$ is true.\n\t\\end{itemize}\n\tThen $P(\\alpha)$ is true for every ordinal $\\alpha$.\n\\end{theorem}\n\\begin{theorem}\n\t[Transfinite recursion]\n\tTo define a sequence $x_\\alpha$ for every ordinal $\\alpha$,\n\tit suffices to\n\t\\begin{itemize}\n\t\t\\ii define $x_0$, then\n\t\t\\ii for any $\\beta$, define $x_\\beta$ for any $\\alpha < \\beta$.\n\t\\end{itemize}\n\\end{theorem}\n\nThe difference between this and normal induction lies in the \\emph{limit ordinals}.\nIn real life, we might only do things like ``define $x_{n+1} = \\dots$''.\nBut this is not enough to define $x_\\alpha$ for all $\\alpha$,\nbecause we can't hit $\\omega$ this way.\nSimilarly, the simple $+1$ doesn't let us hit the ordinal $2\\omega$,\neven if we already have $\\omega+n$ for all $n$.\nIn other words, simply incrementing by $1$ cannot get us past limit stages,\nbut using transfinite induction to jump upwards lets us sidestep this issue.\n\nSo a transfinite induction or recursion is very often broken up into three cases.\nIn the induction phrasing, it looks like\n\\begin{itemize}\n\t\\ii (Zero Case) First, resolve $P(0)$.\n\t\\ii (Successor Case) Show that from $P(\\alpha)$ we can get $P(\\alpha+1)$.\n\t\\ii (Limit Case) For $\\lambda$ a limit ordinal,\n\tshow that $P(\\lambda)$ holds given $P(\\alpha)$ for all $\\alpha < \\lambda$.\n\\end{itemize}\nSimilarly, transfinite recursion often is split into cases too.\n\\begin{itemize}\n\t\\ii (Zero Case) First, define $x_0$.\n\t\\ii (Successor Case) Define $x_{\\alpha+1}$ from $x_\\alpha$.\n\t\\ii (Limit Case) Define $x_\\lambda$ from $x_\\alpha$ for all $\\alpha < \\lambda$,\n\twhere $\\lambda$ is a limit ordinal.\n\\end{itemize}\nIn both situations, finite induction only does the first two cases,\nbut if we're able to do the third case we can climb above the barrier $\\omega$.\n\n\\section{Ordinal arithmetic}\n\\prototype{$1+\\omega=\\omega \\neq \\omega+1$.}\nTo give an example of transfinite recursion, let's define addition of ordinals.\nRecall that we defined $\\alpha+1 = \\alpha \\cup \\{\\alpha\\}$.\nBy transfinite recursion, let\n\\begin{align*}\n\t\\alpha + 0 &= \\alpha \\\\\n\t\\alpha + (\\beta + 1) &= (\\alpha + \\beta) + 1 \\\\\n\t\\alpha + \\lambda &= \\bigcup_{\\beta < \\lambda} (\\alpha + \\beta).\n\\end{align*}\nHere $\\lambda \\neq 0$.\n\nWe can also do this explicitly:\nThe picture is to just line up $\\alpha$ after $\\beta$.\nThat is, we can consider the set\n\\[\n\tX = \n\t\\left( \\left\\{ 0 \\right\\} \\times \\alpha \\right)\n\t\\cup\n\t\\left( \\left\\{ 1 \\right\\} \\times \\beta \\right)\n\\]\n(i.e.\\ we tag each element of $\\alpha$ with a $0$, and\neach element of $\\beta$ with a $1$).\nWe then impose a well-ordering on $X$ by a lexicographic ordering $\\llex$\n(sort by first component, then by second).\nThis well-ordering is isomorphic to a unique ordinal, \n\\begin{example}\n\t[$2+3=5$]\n\tUnder the explicit construction for $\\alpha = 2$ and $\\beta = 3$, we get the set\n\t\\[\n\t\tX = \\left\\{ (0,0) < (0,1) < (1,0) < (1,1) < (1,2) \\right\\}\n\t\\]\n\twhich is isomorphic to $5$.\n\\end{example}\n\n\\begin{example}[Ordinal arithmetic is not commutative]\n\tNote that $1 + \\omega = \\omega$!\n\tIndeed, under the transfinite definition, we have\n\t\\[ 1 + \\omega = \\cup_n (1+n) = 2 \\cup 3 \\cup 4 \\cup \\dots = \\omega. \\]\n\tWith the explicit construction, we have\n\t\\[ X = \\left\\{ (0,0) < (1,0) < (1,1) < (1,2) < \\dots \\right\\} \\]\n\twhich is isomorphic to $\\omega$.\n\\end{example}\n\\begin{exercise}\n\tShow that $n+\\omega = \\omega$ for any $n \\in \\omega$.\n\\end{exercise}\n\n\\begin{remark}\n\tOrdinal addition is not commutative.\n\tHowever, from the explicit construction\n\twe can see that it is at least associative.\n\\end{remark}\n\nSimilarly, we can define multiplication in two ways.\nBy transfinite induction:\n\\begin{align*}\n\t\\alpha \\cdot 0 &= 0 \\\\\n\t\\alpha \\cdot (\\beta + 1) &= (\\alpha \\cdot \\beta) + \\alpha \\\\\n\t\\alpha \\cdot \\lambda &= \\bigcup_{\\beta < \\lambda} \\alpha \\cdot \\beta.\n\\end{align*}\nWe can also do an explicit construction: $\\alpha \\cdot \\beta$\nis the order type of\n\\[ \\llex \\text{ applied to } \\beta \\times \\alpha. \\]\n\\begin{example}[Ordinal multiplication is not commutative]\n\tWe have $\\omega \\cdot 2 = \\omega + \\omega$,\n\tbut $2 \\cdot \\omega = \\omega$.\n\\end{example}\n\\begin{exercise}\n\tProve this.\n\\end{exercise}\n\\begin{exercise}\n\tVerify that ordinal multiplication\n\t(like addition) is associative but not commutative.\n\t(Look at $\\gamma \\times \\beta \\times \\alpha$.)\n\\end{exercise}\n\nExponentiation can also be so defined, though the explicit construction is less natural.\n\\begin{align*}\n\t\\alpha^0 &= 1 \\\\\n\t\\alpha^{\\beta+1} &= \\alpha^{\\beta} \\cdot \\alpha \\\\\n\t\\alpha^{\\lambda} &= \\bigcup_{\\beta < \\lambda} \\alpha^\\beta.\n\\end{align*}\n\\begin{exercise}\n\tVerify that $2^\\omega = \\omega$.\n\\end{exercise}\n\n\n\\section{The hierarchy of sets}\nWe now define the \\vocab{von Neumann Hierarchy} by transfinite recursion.\n\\begin{definition}\n\tBy transfinite recursion, we set\n\t\\begin{align*}\n\t\tV_0 &= \\varnothing \\\\\n\t\tV_{\\alpha + 1} &= \\PP(V_\\alpha) \\\\\n\t\tV_\\lambda &= \\bigcup_{\\alpha<\\lambda} V_\\alpha\n\t\\end{align*}\n\\end{definition}\nBy transfinite induction, we see $V_\\alpha$ is transitive\nand that $V_\\alpha \\subseteq V_\\beta$ for all $\\alpha < \\beta$.\n\n\\begin{example}[$V_\\alpha$ for $\\alpha \\le 3$]\n\tThe first few levels of the hierarchy are:\n\t\\begin{align*}\n\t\tV_0 &= \\varnothing \\\\\n\t\tV_1 &= \\left\\{ 0 \\right\\} \\\\\n\t\tV_2 &= \\left\\{ 0, 1 \\right\\} \\\\\n\t\tV_3 &= \\left\\{ 0, 1, 2, \\left\\{ 1 \\right\\} \\right\\}.\n\t\\end{align*}\n\tNotice that for each $n$, $V_n$ consists of only finite sets,\n\tand each $n$ appears in $V_{n+1}$ for the first time.\n\tObserve that\n\t\\[ V_\\omega = \\bigcup_{n \\in \\omega} V_n \\]\n\tconsists only of finite sets; thus $\\omega$ appears for the first time\n\tin $V_{\\omega+1}$.\n\\end{example}\n\\begin{ques}\n\tHow many sets are in $V_5$?\n\\end{ques}\n\n\\begin{definition}\n\tThe \\vocab{rank} of a set $y$, denoted $\\rank(y)$,\n\tis the smallest ordinal $\\alpha$ such that $y \\in V_{\\alpha+1}$.\n\\end{definition}\n\\begin{example}\n\t$\\rank(2) = 2$, and actually $\\rank(\\alpha)=\\alpha$\n\tfor any ordinal $\\alpha$ (problem later).\n\tThis is the reason for the extra ``$+1$''.\n\\end{example}\n\\begin{ques}\n\tShow that $\\rank(y)$ is the smallest ordinal $\\alpha$\n\tsuch that $y \\subseteq V_\\alpha$.\n\\end{ques}\n\nIt's not yet clear that the rank of a set actually exists, so we prove:\n\\begin{theorem}[The von Neumann hierachy is complete]\n\tThe class $V$ is equal to $\\bigcup_{\\alpha \\in \\On} V_\\alpha$.\n\tIn other words, every set appears in some $V_\\alpha$.\n\\end{theorem}\n\\begin{proof}\n\tAssume for contradiction this is false.\n\tThe key is that because $\\in$ satisfies $\\Foundation$,\n\twe can take a $\\in$-minimal counterexample $x$.\n\tThus $\\rank(y)$ is defined for every $y \\in x$,\n\tand we can consider (by $\\Replacement$) the set\n\t\\[ \\left\\{ \\rank(y) \\mid y \\in x \\right\\}. \\]\n\tSince it is a set of ordinals, it is bounded.\n\tSo there is some large ordinal $\\alpha$ such that $y \\in V_\\alpha$\n\tfor all $y \\in x$, i.e.\\ $x \\subseteq V_\\alpha$,\n\tso $x \\in V_{\\alpha+1}$.\n\\end{proof}\n\nThis leads us to a picture of the universe $V$:\n\\begin{center}\n\t\\begin{asy}\n\t\tsize(11cm);\n\t\tpair A = (12,30);\n\t\tpair B = -conj(A);\n\t\tpair M = midpoint(A--B);\n\t\tpair O = origin;\n\t\tMP(\"V\", A, dir(10));\n\t\tdraw(A--O--B);\n\n\t\tfill(A--O--B--cycle, opacity(0.3)+palecyan);\n\n\t\tMP(\"V_0 = \\varnothing\", origin, dir(-20));\n\t\tMP(\"V_1 = \\{\\varnothing\\}\", 0.05*A, dir(0));\n\t\tMP(\"V_2 = \\{\\varnothing, \\{\\varnothing\\} \\}\", 0.10*A, dir(0));\n\n\t\tdraw(MP(\"V_n\", 0.3*A, dir(0))--0.3*B);\n\t\tdraw(MP(\"V_{n+1} = \\mathcal P(V_n)\", 0.35*A, dir(0))--0.35*B);\n\t\tDrawing(\"n\", 0.35*M, dir(45));\n\n\t\tdraw(MP(\"V_\\omega = \\bigcup V_n\", 0.5*A, dir(0))--0.5*B);\n\t\tdraw(MP(\"V_{\\omega+1} = \\mathcal P(V_{\\omega})\", 0.55*A, dir(0))--0.55*B);\n\t\tDrawing(\"\\omega\", 0.55*M, dir(45));\n\t\tdraw(MP(\"V_{\\omega+2} = \\mathcal P(V_{\\omega+1})\", 0.6*A, dir(0))--0.6*B);\n\t\tDrawing(\"\\omega+1\", 0.6*M, dir(45));\n\n\t\tdraw(MP(\"V_{\\omega+\\omega}\", 0.8*A, dir(0))--0.8*B);\n\n\t\tdraw(origin--M);\n\t\tMP(\"\\mathrm{On}\", M, dir(90));\n\n\t\\end{asy}\n\\end{center}\n\nWe can imagine the universe $V$ as a triangle,\nbuilt in several stages or layers,\n$V_0 \\subsetneq V_1 \\subsetneq V_2 \\subsetneq \\dots$.\nThis universe doesn't have a top: but each of the $V_i$ do.\nHowever, the universe has a very clear bottom.\nEach stage is substantially wider than the previous one.\n\nIn the center of this universe are the ordinals:\nfor every successor $V_\\alpha$, exactly one new ordinal appears, namely $\\alpha$.\nThus we can picture the class of ordinals as a thin line\nthat stretches the entire height of the universe.\nA set has rank $\\alpha$ if it appears at the same stage that $\\alpha$ does.\n\n\nAll of number theory, the study of the integers, lives inside $V_\\omega$.\nReal analysis, the study of real numbers, lives inside $V_{\\omega+1}$, since a real number\ncan be encoded as a subset of $\\NN$ (by binary expansion).\nFunctional analysis lives one step past that, $V_{\\omega+2}$.\nFor all intents and purposes, most mathematics does not go beyond $V_{\\omega+\\omega}$.\nThis pales in comparison to the true magnitude of the whole universe.\n\n\\section\\problemhead\n\\begin{problem}\n\tProve that $\\rank(\\alpha) = \\alpha$ for any $\\alpha$\n\tby transfinite induction.\n\\end{problem}\n\n\\begin{problem}\n\t[Online Math Open]\n\tCount the number of transitive sets in $V_5$.\n\\end{problem}\n\n\\begin{problem}\n\t[Goodstein]\n\tLet $a_2$ be any positive integer.\n\tWe define the infinite sequence $a_2$, $a_3$, \\dots recursively as follows.\n\tIf $a_{n} = 0$, then $a_{n+1} = 0$.\n\tOtherwise, we write $a_n$ in base $n$,\n\tthen write all exponents in base $n$, and so on until all\n\tnumbers in the expression are at most $n$.\n\tThen we replace all instances of $n$ by $n+1$\n\t(including the exponents!), subtract $1$,\n\tand set the result to $a_{n+1}$.\n\tFor example, if $a_2 = 11$ we have\n\t\t\\begin{align*}\n\t\t\ta_2 &= 2^{3} + 2 + 1 = 2^{2+1} + 2 + 1 \\\\\n\t\t\ta_3 &= 3^{3+1}+3+1-1 = 3^{3+1} + 3\\\\\n\t\t\ta_4 &= 4^{4+1} + 4 - 1 = 4^{4+1} + 3 \\\\\n\t\t\ta_5 &= 5^{5+1} + 3 - 1 = 5^{5+1} + 2\n\t\t\\end{align*}\n\tand so on. 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YES", "lm_q1_score": 0.8104789040926008, "lm_q2_score": 0.7826624738835051, "lm_q1q2_score": 0.634331424107507}} {"text": "\n\\documentclass[a4paper]{article}\n%\\documentclass[11pt]{article}\n%\\documentclass[12pt]{article}\n%LATEX HEADER\n\n\\ifx\\pdfoutput\\undefined\n% we are running LaTeX, not pdflatex\n\\usepackage{graphicx}\n\\usepackage{overpic}\n\\else\n% we are running pdflatex, so convert .eps files to .pdf\n%\\usepackage[pdftex]{graphicx}\n\\usepackage{graphicx}\n\\usepackage{overpic}\n\\usepackage{epstopdf}\n\\fi \n\n\\usepackage{mathrsfs}\n\\usepackage{color} \n\\usepackage{verbatim}\n\\usepackage{epsfig}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{float}\n\\usepackage{setspace}\n\\usepackage{moreverb}\n\\usepackage{alltt}\n\\usepackage{ulem}\n\\usepackage{amsthm}\n\\usepackage{cancel} \n\\usepackage{verbatim}\n\n\n\n\n\\setlength{\\textwidth}{17cm}\n% for US letter\n%\\setlength{\\textheight}{22.5cm}\n% for A4\n\n\\setlength{\\textheight}{24cm}\n\\setlength{\\parskip}{0.3cm plus 0.1 cm minus 0.1 cm}\n\\setlength{\\oddsidemargin}{-0.5cm}\n\\setlength{\\topmargin}{0.5cm}\n\\setlength{\\headsep}{0cm}\n\\setlength{\\headheight}{0cm}\n\\setlength{\\parindent}{0.5cm}\n% remove extra space at top and bottom of lists\n%\\setlength{\\partopsep}{-2ex}\n\n\n% section,subsection and subsubsection \n% numbering enable\n\\setcounter{secnumdepth}{3}\n\\setcounter{tocdepth}{3}\n\n\\newcommand{\\bit}{\\begin{itemize}}\n\\newcommand{\\eit}{\\end{itemize}}\n\\newcommand{\\bnum}{\\begin{enumerate}}\n\\newcommand{\\enum}{\\end{enumerate}}\n\\newcommand{\\bt}{\\begin{tabular}}\n\\newcommand{\\et}{\\end{tabular}}\n\\newcommand{\\be}{\\begin{equation}}\n\\newcommand{\\ee}{\\end{equation}}\n\\newcommand{\\pdv}[2]{\\frac{\\partial #1}{\\partial #2}}\n\\newcommand{\\dv}[2]{\\frac{d #1}{d #2}}\n\\newcommand{\\ds}{\\displaystyle}\n\\newcommand{\\vect}[1]{\\mbox{\\boldmath $#1$}}\n\\newcommand{\\subs}[1]{$_{#1}$}\n\\newcommand{\\sups}[1]{$^{#1}$}\n%\\pagestyle{empty}% no page numbering\n\n%in order to obtain doublspacing put following 2 lines in the\n%preamble and the last line just above the place where the\n%double-spacing should commence.\n\\def\\doublespaced{\\baselineskip=1.5\\normalbaselineskip}\n\\let\\doublespace=\\doublespaced\n\n\\begin{document}\n\n\n% for double spacing\n\\title{\\Huge A Wrapper Class For LAPACK and BLAS }\n% \\Large\n\\date{\\today}\n\\author{Minh Tran\\\\Email: minh.t.tran@adelaide.edu.au}\n\\maketitle\n% %\\tableofcontents\n% %\\clearpage\n\n \n\n\n\n\\section{Introduction} \\label{Introduction}\nThe Matrix class is a wrapper class for LAPACK and BLAS. The class implements matrix solvers and operators typically used in mathematical modelling of physical systems. It is the hope of the author that the class will be useful for anyone working with C++ projects that require matrix operations.\n\n\n\n\\section{Examples} \\label{Examples}\nThe use of class will be demonstrated through examples.\n\n\n \n\n\n\\subsection{Example 1}\n \n\n \n$\\boldsymbol{A}_{1}$ = $\\begin{bmatrix}\n5 & 7 & 6 & 5 \\\\\n7 & 10 & 8 & 7 \\\\\n6 & 8 & 10 & 9 \\\\\n5 & 7 & 9 & 10 \n \\end{bmatrix}$ \\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{b}_{1}$ = $\\begin{bmatrix}\n23 \\\\ \n32 \\\\\n33 \\\\\n31 \n\\end{bmatrix}$\n\n\\vspace{0.5cm}\n\nConsider the system,\n\\begin{equation*}\n\\begin{aligned}\n\\left(5*\\boldsymbol{A}_{1}+\\boldsymbol{A}_{1}*15\\right) * \\boldsymbol{x}_{1} = \\boldsymbol{b}_{1}\n\\end{aligned}\n\\end{equation*}\n\n\n\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{x}_{1} &= \\left(5*\\boldsymbol{A}_{1}+\\boldsymbol{A}_{1}*15\\right)^{-1} * \\boldsymbol{b}_{1} \\\\\n\\boldsymbol{x}_{1} &= \n\\begin{bmatrix} \n\t\t0.050 \\\\\n\t\t0.050 \\\\\n\t\t0.050 \\\\\n\t\t0.050\n\\end{bmatrix}\n\\end{aligned}\n\\end{equation*}\n\n\n\nBelow are codes to compute $\\boldsymbol{x}_{1}$. The result is stored in the variable \\texttt{example1}.\n\\begin{verbatim} \ndouble A1[] = { 5 , 7 , 6 , 5 ,\n 7 , 10 , 8 , 7 ,\n 6 , 8 , 10 , 9 ,\n 5 , 7 , 9 , 10 };\ndouble b1[] = { 23 , 32 , 33 , 31 };\nMatrix P1(A1,4); \nMatrix example1 = (5*P1+P1*15) | b1; \n\\end{verbatim}\n\n\n\n\\subsection{Example 2}\n\n\n$\\boldsymbol{A}_{2}$ = $\\begin{bmatrix}\n0 & 1 & 2 \\\\\n3 & 4 & 5 \\\\ \n6 & 7 & 0 \\\\\n\\end{bmatrix}$ \\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{b}_{2}$ = $\\begin{bmatrix}\n1 \\\\ \n2 \\\\\n3 \\\\ \n\\end{bmatrix}$\n\n\\vspace{0.5cm}\n\nConsider the system,\n\\begin{equation*}\n\\begin{aligned}\n\\left( \\boldsymbol{A}_{2}*\\boldsymbol{A}_{2}*\\boldsymbol{A}_{2} \\right) * \\boldsymbol{x}_{2} = \\boldsymbol{b}_{2}\n\\end{aligned}\n\\end{equation*}\n\n\n\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{x}_{2} &= \\left(\\boldsymbol{A}_{2}*\\boldsymbol{A}_{2}*\\boldsymbol{A}_{2} \\right)^{-1} * \\boldsymbol{b}_{2} \\\\\n\\boldsymbol{x}_{2} &\\approx \n\\begin{bmatrix} \n-3.088 \\\\\n2.694 \\\\\n-0.569 \\\\\n\\end{bmatrix}\n\\end{aligned}\n\\end{equation*}\n\n\n\nBelow are codes to compute $\\boldsymbol{x}_{2}$. The result is stored in the variable \\texttt{example2}.\n\n\\begin{verbatim} \ndouble A2[] = { 0 , 1 , 2 ,\n 3 , 4 , 5 , \n 6 , 7 , 0 };\ndouble B2[] = { 1 , 2 , 3 }; \nMatrix P2(A2, 3);\nMatrix example2 = (P2*P2*P2) | B2; \n\\end{verbatim}\n\n\n \n\n\n\n\n\\subsection{Example 3 \\& 4} \n\n$\\boldsymbol{A}_{3}$ = $\\begin{bmatrix}\n 0 & 1 & 0 & 0\\\\\n-1 & 0 & 1 & 0\\\\ \n 0 & -1 & 0 & 1\\\\\n 0 & 0 & -1 & 0\\\\\n\\end{bmatrix}$ \\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{d}$ = $\\begin{bmatrix}\n1 \\\\ \n2 \\\\\n3 \\\\ \n4 \\\\ \n\\end{bmatrix}$\n\n\\vspace{0.5cm}\n\nConsider the system,\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{A}_{3} * \\boldsymbol{x}_{3} = \\boldsymbol{d} \n\\end{aligned}\n\\end{equation*}\n\n\n\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{x}_{3} &= \\boldsymbol{A}_{3} ^{-1} * \\boldsymbol{d} \\\\\n\\boldsymbol{x}_{3} &\\approx \n\\begin{bmatrix} \n-6 \\\\\n1 \\\\\n-4 \\\\\n4\\\\\n\\end{bmatrix}\n\\end{aligned}\n\\end{equation*}\n\n\n\nBelow are codes to compute $\\boldsymbol{x}_{3}$. The result is stored in the variables \\texttt{example3}. The system for \\texttt{example4} is a similar to \\texttt{example3}. The main difference is $\\boldsymbol{d}$ is an array and $\\boldsymbol{Q}_{3}$ is a matrix object.\n \n\n\\begin{verbatim} \ndouble a[] = { -1, -1, -1 };\ndouble b[] = { 0, 0, 0, 0};\ndouble c[] = { 1, 1, 1 };\ndouble d[] = { 1, 2, 3, 4 };\nMatrix Q3(d, 4,1); \nMatrix P3(a,b,c,4); \nMatrix example3 = P3 | d;\nMatrix example4 = P3 | Q3;\n\\end{verbatim}\n\n\n\n\n\n\n\n\n\\subsection{Example 5 \\& 6}\n\nThis example demonstrates the use of pseudo inverse to compute a solution of the system below.\n\n$\\boldsymbol{A}_{5}$ = $\\begin{bmatrix}\n -74 & 80 & 18 & -11 & -4 \\\\\n 14 & -69 & 21 & 28 & 0 \\\\ \n 66 & -72 & -5 & 7 & 1 \\\\\n -12 & 66 & -30 & -23 & 3 \\\\\n 3 & 8 & -7 & -4 & 1 \\\\\n 4 & -12 & 4 & 4 & 0\\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{d}$ = $\\begin{bmatrix}\n51 \\\\\n-61 \\\\\n-56\\\\\n 69\\\\\n 10\\\\\n -12\\\\\n\\end{bmatrix}$\n\n\\vspace{0.5cm}\n\nConsider the system,\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{A}_{5} * \\boldsymbol{x}_{5} = \\boldsymbol{d} \n\\end{aligned}\n\\end{equation*}\n\n\n\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{x}_{5} &= \\boldsymbol{A}_{5} ^{-1} * \\boldsymbol{d} \\\\\n\\boldsymbol{x}_{5} &=\n\\begin{bmatrix} \n1 \\\\\n2 \\\\\n-1 \\\\\n3 \\\\\n-4 \\\\\n\\end{bmatrix}\n\\end{aligned}\n\\end{equation*}\n\n\n\nBelow are codes to compute $\\boldsymbol{x}_{5}$. The result is stored in the variables \\texttt{example5}. The system for \\texttt{example6} is a similar to \\texttt{example5}. The main difference is $\\boldsymbol{b}_{5}$ is an array and $\\boldsymbol{Q}_{5}$ is a matrix object.\n\n\n\\begin{verbatim} \ndouble A5[30] = { -74 , 80 , 18 , -11 , -4 ,\n 14 , -69 , 21 , 28 , 0 , \n 66 , -72 , -5 , 7 , 1 ,\n -12 , 66 , -30 , -23 , 3 ,\n 3 , 8 , -7 , -4 , 1 ,\n 4 , -12 , 4 , 4 , 0};\ndouble b5[6] = { 51 , -61 , -56 , 69 , 10 , -12 };\nMatrix P5(A5,6,5); \nMatrix example5 = P5 | b5;\nMatrix Q5(b5, 6, 1);\nMatrix example6 = P5 | Q5;\n\\end{verbatim}\n\n\n\n\n\n\n\n\n\\subsection{Example 7}\n\nConsider solving the system below,\n\n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x},\\boldsymbol{y}}{\\min} \\hspace{0.2cm} || \\boldsymbol{y} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{d}_{7} = \\boldsymbol{A}_{7}\\boldsymbol{x} + \\boldsymbol{B}_{7}\\boldsymbol{y}\n\\end{aligned}\n\\end{equation*}\n\n\n\n$\\boldsymbol{A}_{7}$ = $\\begin{bmatrix}\n1 & 2 & 1 & 4 \\\\\n-1 & 1 & 1 & 1 \\\\ \n-1 & -2 & -1 & 1 \\\\ \n-1 & 2 & -1 & -1 \\\\ \n1 & 1 & 1 & 2 \\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{B}_{7}$ = $\\begin{bmatrix}\n1 & 2 & 2 \\\\\n-1 & 1 & -2 \\\\ \n3 & 1 & 6 \\\\\n2 & -2 & 4 \\\\\n1 & -1 & 2\\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{d}_{7}$ = $\\begin{bmatrix}\n7.99 \\\\\n0.98 \\\\\n-2.98 \\\\\n3.04 \\\\\n4.02 \\\\\n\\end{bmatrix}$ \n\n\nSolution:\n\n\n$\\boldsymbol{x}$ = $\\begin{bmatrix}\n1.002951 \\\\\n2.001436 \\\\\n-0.987798 \\\\\n0.990908 \\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{y}$ = $\\begin{bmatrix}\n0.003436 \\\\\n-0.004417 \\\\\n0.006871 \\\\\n\\end{bmatrix}$ \n\n\nBelow are codes to compute $\\boldsymbol{x}$ and $\\boldsymbol{y}$. The variable \\texttt{info} is used to flag whether the algorithm is able to obtain a solution to the system.\n\n\n\n\\begin{verbatim} \ndouble *x7 = new double[4];\ndouble *y7 = new double[3];\ndouble A7[] = { 1 , 2 , 1 , 4 ,\n -1 , 1 , 1 , 1 , \n -1 , -2 , -1 , 1 , \n -1 , 2 , -1 , -1 , \n 1 , 1 , 1 , 2} ;\ndouble B7[] = { 1 , 2 , 2 ,\n -1 , 1 , -2 , \n 3 , 1 , 6 ,\n 2 , -2 , 4 ,\n 1 , -1 , 2 };\ndouble d7[] = { 7.99 , 0.98 , -2.98 , 3.04 , 4.02 };\nMatrix P7(A7, B7, d7, 5, 4, 3); \nint info7 = P7.solve(x7,y7); \n\\end{verbatim}\n\n\n\n\n\\subsection{Example 8}\n\n\nConsider solving the system below,\n\n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x}_{8}}{\\min} \\hspace{0.2cm} ||\\boldsymbol{d}_{8} - \\boldsymbol{A}_{8}\\boldsymbol{x}_{8} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{H}_{8}\\boldsymbol{x}_{8} = \\boldsymbol{f}_{8}\n\\end{aligned}\n\\end{equation*}\n\n\n\n$\\boldsymbol{A}_{8}$ = $\\begin{bmatrix}\n1 & 1 & 1 \\\\ \n1 & 3 & 1 \\\\ \n1 & -1 & 1 \\\\\n1 & 1 & 1 \\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{d}_{8}$ = $\\begin{bmatrix}\n1 \\\\ \n2 \\\\ \n3 \\\\ \n4\\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{H}_{8}$ = $\\begin{bmatrix}\n1 & 1 & 1 \\\\\n1 & 1 & -1 \\\\\n\\end{bmatrix}$ \n\\hspace{0.5cm} , \\hspace{0.5cm} \n$\\boldsymbol{f}_{8}$ = $\\begin{bmatrix}\n7 \\\\\n4 \\\\\n\\end{bmatrix}$ \n\nSolution:\n\n\n$\\boldsymbol{x}_{8}$ = $\\begin{bmatrix}\n5.75 \\\\\n-0.25 \\\\\n1.5 \\\\\n\\end{bmatrix}$ \n\n\n\nBelow are codes to compute a solution to the system. The variable \\texttt{info} is used to flag whether the algorithm is able to obtain a solution to the system.\n\n\n\n\n\\begin{verbatim} \ndouble A8[] = { 1 , 1 , 1 , \n 1 , 3 , 1 , \n 1 , -1 , 1 , \n 1 , 1 , 1};\ndouble H8[] = { 1 , 1 , 1 ,\n 1 , 1 , -1 };\ndouble D8[] = { 1 , 2 , 3 , 4 };\ndouble F8[] = { 7 ,\n4 };\ndouble *x8 = new double[3];\nMatrix P8(A8, D8, H8, F8, 4, 3, 2);\nint info8 = P8.solve(x8);\nprintf(\"\\n X8 = \\n \");\nMatrix::printMatrix(x8,3,1);\n\\end{verbatim}\n\n\n\\section{Constructor}\nBelow is a description of the syntax used to create a single matrix or a matrix system.\n\n\n\\subsection{Create a Single Matrix Object}\n\n\n\n\\subsubsection{\\texttt{Matrix( double Ain[], int nd )}}\n \nThis is used to create a square matrix with dimension of $n_{d}*n_{d}$.\nThe variable $A_{in}$ is an array storing entries of the matrix. The entries are read in row by row. \n\n \n\\subsubsection{\\texttt{Matrix(double Ain[], int nd , int md)}}\n\n\nThis is used to create a general matrix with dimension of $n_{d}*m_{d}$. \nThe variable $A_{in}$ is an array storing entries of the matrix. The entries are read in row by row. \n\n \n\\subsubsection{\\texttt{Matrix(double ain[], double bin[], double cin[], int nd)}}\n\nCreate a tridiagonal matrix with dimension of $n_{d}*n_{d}$.\nThe array $b_{in}$ is the main diagonal with $n_{d}$ number of entries. The array $a_{in}$ is the subdiagonal with $n_{d}-1$ number of entries. The array $c_{in}$ is the superdiagonal with $n_{d}-1$ number of entries. \n\n\n\\subsubsection{\\texttt{Matrix(double Ain[], int nd, int md, bool isTranspose)}}\n\n\nThis is used to create a general matrix with dimension of $n_{d}*m_{d}$. The variable $A_{in}$ is an array storing entries of the matrix. The entries are read in row by row if \\texttt{isTranspose} is \\texttt{false}. On the other hand, if \\texttt{isTranspose} is \\texttt{true} entries are read in column by column. \n\n\n\\subsubsection{\\texttt{Matrix(double Ain[], int nd, int md, bool isTranspose, int flag)}}\n\nThis is used to create a general matrix with dimension of $n_{d}*m_{d}$. The variable $A_{in}$ is an array storing entries of the matrix. The entries are read in row by row if \\texttt{isTranspose} is \\texttt{false}. On the other hand, if \\texttt{isTranspose} is \\texttt{true} entries are read in column by column. The parameter input \\texttt{flag} is used to set the default \\texttt{info} about the matrix. That is, if \\texttt{flag} = 0, then the matrix is non-singular. It should be noted that \\texttt{info} is an instance field of the Matrix class.\n\n \n\\subsection{Create a Matrix System}\n\n\n\n\\subsubsection{\\texttt{Matrix(double Ain[], double din[], double Hin[], double fin[], int nd , int md, int pd)}}\n\nThis is used to construct the following matrix system.\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x}}{\\min} \\hspace{0.2cm} ||\\boldsymbol{d} - \\boldsymbol{A}\\boldsymbol{x} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{H}\\boldsymbol{x} = \\boldsymbol{f}\n\\end{aligned}\n\\end{equation*}\n\nDimension of the matrix: \n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{A} &\\hspace{0.7cm}&& \\text{dimension } n_{d}*m_{d} \\\\\n\\boldsymbol{H} &\\hspace{0.7cm}&& \\text{dimension } p_{d}*m_{d}\\\\\n\\boldsymbol{d} &\\hspace{0.7cm}&& \\text{dimension } n_{d}*1\\\\\n\\boldsymbol{f} &\\hspace{0.7cm}&& \\text{dimension } p_{d}*1\\\\\n\\boldsymbol{x} &\\hspace{0.7cm}&& \\text{dimension } m_{d}*1\\\\\n\\end{aligned}\n\\end{equation*}\n\n\nThe method ``\\texttt{int Matrix::solve(double xinout[])}\" is used to solve the system.\n\n\n\\subsubsection{\\texttt{Matrix(double Ain[], double Bin[], double din[], int nd, int md, int pd)}}\n\n\nThis is used to construct the following matrix system.\n\n\n\n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x},\\boldsymbol{y}}{\\min} \\hspace{0.2cm} || \\boldsymbol{y} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{d} = \\boldsymbol{A} \\boldsymbol{x} + \\boldsymbol{B} \\boldsymbol{y}\n\\end{aligned}\n\\end{equation*}\n\n\nDimension of the matrix: \n\\begin{equation*}\n\\begin{aligned}\n\\boldsymbol{A} &\\hspace{0.7cm}&& \\text{dimension } n_{d}*m_{d} \\\\\n\\boldsymbol{B} &\\hspace{0.7cm}&& \\text{dimension } n_{d}*p_{d} \\\\\n\\boldsymbol{d} &\\hspace{0.7cm}&& \\text{dimension } n_{d}*1 \\\\\n\\boldsymbol{x} &\\hspace{0.7cm}&& \\text{dimension } m_{d}*1 \\\\\n\\boldsymbol{y} &\\hspace{0.7cm}&& \\text{dimension } p_{d}*1 \\\\\n\\end{aligned}\n\\end{equation*}\n \nThe method ``\\texttt{int Matrix::solve(double xinout[], double yinout[])}'' is used to solve the system.\n\n\n\\section{Methods} \n\n\\subsubsection{\\texttt{int Matrix::solve(double xinout[])}}\n\nThis method is used to solve the system \n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x}}{\\min} \\hspace{0.2cm} ||\\boldsymbol{d} - \\boldsymbol{A}\\boldsymbol{x} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{H}\\boldsymbol{x} = \\boldsymbol{f}\n\\end{aligned}\n\\end{equation*}\n\nSolution of the system is stored in the array $x_{inout}$.\n\n \n\\subsubsection{\\texttt{int Matrix::solve(double xinout[], double yinout[])}}\n \nThis method is used to solve the system \n \n\n\\begin{equation*}\n\\begin{aligned}\n\\underset{\\boldsymbol{x},\\boldsymbol{y}}{\\min} \\hspace{0.2cm} || \\boldsymbol{y} ||_{2} \\hspace{0.5cm} \\text{ subject to } \\hspace{0.5cm} \\boldsymbol{d} = \\boldsymbol{A} \\boldsymbol{x} + \\boldsymbol{B} \\boldsymbol{y}\n\\end{aligned}\n\\end{equation*}\n\n \n Solution of the system is stored in the array $x_{inout}$ and $y_{inout}$ .\n\n\n\\subsubsection{\\texttt{ void Matrix::printMatrix()}}\n\nThis method is used to print to console all the matrix entries stored in the Matrix object. \n\n\n\\subsubsection{\\texttt{static void Matrix::printMatrix(double input[], int nd, int md)}}\n\nThis method is used to print to console all the matrix entries stored in an array \\texttt{input}. The dimension of the matrix is $n_{d}*m_{d}$.\n\n\n\\section{Operators} \n\t\\begin{tabular}{|c|c|l|} \n\t\t\\hline\n\t\tOperator & Example & Description \\\\ \n\t\t\\hline\n\t\t+& $\\boldsymbol{A} + \\boldsymbol{B} $ & Adding two matrix object. \\\\\n\t\t-& $\\boldsymbol{A} - \\boldsymbol{B}$ & Substract two matrix object. \\\\\n\t\t*& $c*\\boldsymbol{A}$ & multiply matrix $\\boldsymbol{A}$ with a scalar $c$ \\\\\n\t\t*& $\\boldsymbol{A}*c$ & multiply matrix $\\boldsymbol{A}$ with a scalar $c$ \\\\\n\t\t*& $\\boldsymbol{A}*\\boldsymbol{B}$ & multiply matrix two matrix together \\\\\n\t\t$|$& $\\boldsymbol{A} \\hspace{0.1cm} | \\hspace{0.1cm} \\boldsymbol{b}$ & $\\boldsymbol{b}$ is an array, compute $\\boldsymbol{A}^{-1} \\boldsymbol{b}$. If $\\boldsymbol{A}$ is not a square, matrix pseudo inverse is used. \\\\\n\t\t$|$& $\\boldsymbol{A} \\hspace{0.1cm} | \\hspace{0.1cm} \\boldsymbol{B}$ & $\\boldsymbol{B}$ is matrix, compute $\\boldsymbol{A}^{-1} \\boldsymbol{B}$. If $\\boldsymbol{A}$ is not a square matrix, pseudo inverse is used. \\\\\n\t\t\\hline\n\t\\end{tabular} \n\n\n\n\n\\end{document}\n\n \n\n\n\n%\\font size\n% 1. \\tiny\n% 2. \\scriptsize\n% 3. \\footnotesize\n% 4. \\small\n% 5. \\normalsize\n% 6. \\large\n% 7. \\Large\n% 8. \\LARGE\n% 9. \\huge\n% 10. \\Huge ", "meta": {"hexsha": "ea080c430d6a3243f8cfd41cf033952459ccee21", "size": 17320, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documentation/BLAS and LAPACK.tex", "max_stars_repo_name": "tran0208/BLAS-And-LAPACK-Wrapper", "max_stars_repo_head_hexsha": "c99ed6c3b92b20c620072cdac16e6020c8d3bc7c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-06-26T07:51:41.000Z", "max_stars_repo_stars_event_max_datetime": "2020-06-26T07:51:41.000Z", "max_issues_repo_path": "Documentation/BLAS and LAPACK.tex", "max_issues_repo_name": "tran0208/BLAS-And-LAPACK-Wrapper", "max_issues_repo_head_hexsha": "c99ed6c3b92b20c620072cdac16e6020c8d3bc7c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Documentation/BLAS and LAPACK.tex", "max_forks_repo_name": "tran0208/BLAS-And-LAPACK-Wrapper", "max_forks_repo_head_hexsha": "c99ed6c3b92b20c620072cdac16e6020c8d3bc7c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.3958944282, "max_line_length": 553, "alphanum_fraction": 0.6258083141, "num_tokens": 6420, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624688140726, "lm_q2_score": 0.8104789040926008, "lm_q1q2_score": 0.6343314199988389}} {"text": "\\documentclass{ximera}\n\\input{../preamble}\n\\title{Exercises: Substitution and Tables}\n%%%%%\\author{Philip T. Gressman}\n\n\\begin{document}\n\\begin{abstract}\nVarious exercises relating to substitution and the use of integral tables.\n\\end{abstract}\n\\maketitle\n\nCompute the indefinite integrals below. Since there are many possible answers (which differ by constant values), use the given instructions if needed to choose which possible answer to use.\n\n\\begin{exercise}%[APEX0601SUBS06]\n\\[ \\int (12 x+14) \\left(3 x^2+7 x-1\\right)^5 dx = \\answer{\\frac{1}{3}(3 x^2+7 x-1)^6}+C \\]\n(Add a constant to your answer if needed so that it equals $1/3$ at $x = 0$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS11]\n\\[ \\int \\frac{e^{\\sqrt{x}}}{\\sqrt{x}} dx = \\answer{2 e^{\\sqrt{x}}}+C \\]\n(Add a constant to your answer if needed so that it equals $2$ at $x = 0$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS09]\n\\[ \\int \\frac{x}{\\sqrt{x+3}} dx = \\answer{\\frac{2}{3} (x-6) \\sqrt{x+3}}+C\\]\n(Add a constant to your answer if needed so that it equals $0$ at $x=6$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS33]\n\\[ \\int \\frac{\\ln |x|}{x} dx = \\answer{\\frac{1}{2}\\ln^2 |x|}+C\\]\nRemember absolute value in your logarithm.\n(Add a constant to your answer if needed so that it equals $0$ at $x = 1$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS57]\n\\[ \\int \\sin x \\sqrt{\\cos x} dx = \\answer{-\\frac{2}{3} \\cos ^{\\frac{3}{2}}(x)}+C\\]\n(Add a constant to your answer if needed so that it equals $-2/3$ at $x = 0$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS63]\n\\[ \\int \\frac{9 (2 x+3)}{3 x^2+9 x+7} dx = \\answer{3 \\ln \\left|3 x^2+9 x+7\\right|}+C \\]\nUse absolute values as needed in logarithms. (Add a constant to your answer if needed so that it equals $3 \\ln 7$ at $x = 0$.)\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS65]\n\\[ \\int \\frac{x}{x^4+81} dx = \\answer{\\frac{1}{18} \\arctan \\left(\\frac{x^2}{9}\\right)}+C \\]\n(Add a constant to your answer if needed so that it equals $0$ at $x = 0$.)\n\\begin{hint}\nMake a substitution $x^2 = 9 u$.\n\\end{hint}\n%\n%\n\\end{exercise}\n\n\\begin{exercise}%[APEX0601SUBS83]\nEvaluate the definite integral \\(\\displaystyle \\int_{-2}^{-1} (x+1)e^{x^2+2x+1}\\ dx. \\)\n\\begin{center}\n\\begin{prompt}\nValue = \\(\\answer{(1-e)/2}\\)\n\\end{prompt}\n\\end{center}\n%\n%\n\\end{exercise}\n\n\\section*{Sample Exam Questions}\n\n\\begin{question}%[2016.13]\nEvaluate the integral $\\displaystyle \\int_1^3 \\left( x - \\sqrt{4 x^2 - 8 x + 13} \\right) dx$ using the fact that $\\displaystyle \\int_0^4 \\sqrt{x^2 + 9} ~ dx = \\frac{20 + 9 \\ln 3}{2}$.\n(Hints will not be displayed until you have chosen a response.)\n\\begin{multiplechoice}\n\\choice{\\(\\displaystyle -\\frac{4 + 9 \\ln 3}{6}\\)}\n\\choice[correct]{\\(\\displaystyle -\\frac{4 + 9 \\ln 3}{4}\\)}\n\\choice{\\(\\displaystyle -\\frac{4 + 9 \\ln 3}{2}\\)}\n\\choice{\\(\\displaystyle -4 - 9 \\ln 3\\)}\n\\choice{\\(\\displaystyle -8 - 18 \\ln 3\\)}\n\\choice{\\(\\displaystyle -12 - 27 \\ln 3\\)}\n\\end{multiplechoice}\n\\begin{feedback}\nFirst complete the square:\n\\[ 4 x^2 - 8 x + 13 = 4 x^2 - 8 x + 4 + (13 - 4) = (2x - 2)^2 + 9. \\] \\begin{hint} \nNext substitute $u = 2x-2$ (i.e., $x = \\frac{u}{2} + 1$). Then $du = 2 dx$ and $x=1 \\leftrightarrow u = 0$, $x=3 \\leftrightarrow u = 4$, so\n\\[ \\int_1^3 \\left( x - \\sqrt{4x^2 - 8x + 13} \\right) dx = \\int_0^4 \\left( \\frac{u}{2} + 1 - \\sqrt{u^2 + 9} \\right) \\frac{du}{2}. \\] \\begin{hint}\nNow use linearity of the integral to finish:\n\\[ \\int_0^4 \\frac{u}{4} ~ du + \\int_0^4 \\frac{1}{2} ~ du - \\int_0^4 \\frac{1}{2} \\sqrt{u^2+9} ~ du = \\left. \\frac{u^2}{8} \\right|_0^4 + \\left. \\frac{u}{2} \\right|_0^4 - \\frac{20 + 9 \\ln 3}{4} = 4 - \\frac{20+9 \\ln 3}{4} = - \\frac{4 + 9 \\ln 3}{4}.\\] \\end{hint}\\end{hint}\n\\end{feedback}\n\\end{question}\n\n\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "56bbbcd58398934cedfa6e756c9af890b85e15d8", "size": 3724, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "techniques/08substitutionpractice.tex", "max_stars_repo_name": "ptgressman/math104", "max_stars_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "techniques/08substitutionpractice.tex", "max_issues_repo_name": "ptgressman/math104", "max_issues_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "techniques/08substitutionpractice.tex", "max_forks_repo_name": "ptgressman/math104", "max_forks_repo_head_hexsha": "3b797f5622f6c7b93239a9a2059bd9e7e1f1c7c0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.8037383178, "max_line_length": 267, "alphanum_fraction": 0.6299677766, "num_tokens": 1484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.746138993030751, "lm_q2_score": 0.8499711718571775, "lm_q1q2_score": 0.6341966342746819}} {"text": "\\section{Model}\n\\label{sec:model}\n\n\\subsection{Model components}\n\nThe core of this term project is to demonstrate the working of the human capital accumulation mechanism in light of demographic transitions. For this, I use a standard overlapping generations model with household heterogeneity by age. Household's maximize discounted lifetime utility from consumption by choosing between consumption, saving and effort invested in human capital accumulation in every model period subject to a budget constraint, borrowing constraint, time constraint and the human capital formation technology.\n\nHuman capital is produced with a technology commonly used in the literature and first introduced by \\cite{Ben-Porath1967}. Human capital next period is a function of the current level of human capital, $h_t$, and the effort exerted to produce additional human capital, $e_t$,\n$$ h_{t+1} = (1-\\delta_h) h_t + \\zeta (e_t h_t)^\\psi$$\nwhere $\\delta_h$ is the depreciation rate on human capital, $\\zeta$ is a scaling factor capturing average learning ability and $\\psi$ is the curvature parameter of the production technology.\n\nThe model features neither idiosyncratic income risk, nor aggregate risk of any kind, i.e. the only risk households are facing is mortality risk. Households are assumed to have perfect foresight regarding survival and idiosyncratic survival risk washes out in the aggregate. Accidental bequests, i.e. asset holdings of deceased households are taxed 100\\% and spent on non-productive government expenditures. The latter is a deviation from the framework in \\citet{LudwigSchelkleVogel2012}, where accidental bequests are redistributed to households via transfers. While altering the budget constraint of households, the assumption taken for this term paper simplifies computation. In the original framework, income from transfers affects the consumption/ savings/ investment decision and, therefore, the level of accidental bequests. In effect, transfers act as an additional variable (next to aggregate capital, aggregate labor and pension benefits), that needs to be first guessed as input and then verified as results of aggregation. Thus, (wasteful) spending of tax revenue from accidental bequests removes one equlibrium condition.\n\nIn contrast to the original paper by \\cite{LudwigSchelkleVogel2012}, pension benefits are in this paper not linked to income histories. In other words, every retired agent qualifies for the same level of pension benefits. While the level of pension benefits remains an aggregate variable to be solved for by guess and verify, this assumption simplifies computation, as it is no longer necessary to keep track of a households earnings history to calculate pension benefit claims, effectively eliminating a state variable from the household problem. In the specific model used here, the difference is not too material, as in the absence of idiosyncratic risk other than mortality risk, all households of a given age are identical in their earnings history. However, this simplifying assumption becomes truly relevent when introducing idiosyncratic income risk to the model. Pension benefits are thus given by\n$$ b_j = \\begin{cases}\n\t\t\t\t0 \\;\\text{if}\\; j 0 \\), the value \\( S(x^*, A, \\alpha) \\) is called a \\term{slice} of \\( A \\), where\n \\begin{balign*}\n & S: X^* \\times \\pow(X) \\times \\BbbR_{>0} \\mapsto \\pow(A) \\\\\n & S(x^*, A, \\alpha) \\coloneqq \\{ x \\in A \\colon \\inprod {x^*} x > \\sigma_A^*(x^*) - \\alpha \\}.\n \\end{balign*}\n\n We define a weak* slice of \\( A^* \\subseteq X^* \\) as \\( S^*(x, A^*, \\alpha) \\), where\n \\begin{balign*}\n & S^*: X \\times \\pow(X) \\times \\BbbR_{>0} \\mapsto \\pow(A) \\\\\n & S^*(x, A^*, \\alpha) \\coloneqq \\{ x^* \\in A^* \\colon \\inprod {x^*} x > \\sigma_{A^*}(x) - \\alpha \\}.\n \\end{balign*}\n\n If we need to make the underlying space explicit, we will use \\( S_X(x^*, A, \\alpha) \\) and \\( S_X^*(x, A^*, \\alpha) \\).\n\\end{definition}\n\n\\begin{proposition}\n If \\( \\{ a_k \\}_{k=1}^\\infty \\) and \\( \\{ b_k \\}_{k=1}^\\infty \\) are sequences a in a \\hyperref[def:banach_space]{Banach} \\hyperref[def:algebra_over_ring]{algebra} \\( X \\), that converge to \\( a \\) and \\( b \\), correspondingly, then \\( a_k b_k \\to a b \\).\n\\end{proposition}\n\\begin{proof}\n Let \\( \\delta > 0 \\) and let \\( k_0 \\) be an index such that for \\( k \\geq k_0 \\) we have both \\( \\norm{a - a_k} < \\delta \\) and \\( \\norm{b - b_k} < \\delta \\). Then\n \\begin{balign*}\n ab - a_k b_k\n & =\n (ab - a b_k) + (a b_k - a_k b) + (a_k b - a_k b_k)\n = \\\\ &=\n a (b - b_k) + (a b_k - ab + ab - a_k b) + (-a_k)(b_k - b)\n = \\\\ &=\n a (b - b_k) + a \\underbrace{(b_k - b)} + (a - a_k) b + (-a_k)\\underbrace{(b_k - b)}\n = \\\\ &=\n a \\underbrace{(b - b_k)}_{\\in B(0, \\delta)} + \\underbrace{(a - a_k)}_{\\in B(0, \\delta)} \\underbrace{(b_k - b)}_{\\in B(0, \\delta)} + \\underbrace{(a - a_k)}_{\\in B(0, \\delta)} b.\n \\end{balign*}\n\n Therefore, \\( \\norm{ab - a_k b_k} < \\delta^2 + \\norm{a + b} \\delta \\). If we require \\( \\delta \\) to be strictly less than \\( 1 \\), we obtain \\( \\delta^2 < \\delta \\) and \\( \\norm{ab - a_k b_k} < (1 + \\norm{a + b}) \\delta \\).\n\n Given an arbitrary \\( \\varepsilon > 0 \\), we can choose \\( \\delta = \\tfrac {\\min \\{\\varepsilon, 1 \\}} {1 + \\norm{a + b}} \\) in order to have \\( \\norm{ab - a_k b_k} < \\varepsilon \\) for some large enough \\( k \\).\n\n Therefore, \\( a_k b_k \\to a b \\).\n\\end{proof}\n", "meta": {"hexsha": "c8ce8cf35677c7124101b547cf0a99b8497a4b12", "size": 5853, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/banach_spaces.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/banach_spaces.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/banach_spaces.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.0256410256, "max_line_length": 257, "alphanum_fraction": 0.5803861268, "num_tokens": 2226, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619436290698, "lm_q2_score": 0.76908023177796, "lm_q1q2_score": 0.6340773826983523}} {"text": "\\subsection{SMT-LIB}\n\n\\begin{frame}[fragile]\n \\frametitle{SMT-LIB (v2) \\url{http://www.smtlib.org}}\n\n The SMT-LIB initiative \n \\begin{itemize}\n \\item defines a standard input language for \\smtsolvers\n \\item defines theories and logics in which \\formulae can be written\n \\item collects benchmarks \n \\end{itemize}\n\n \\vfill\n \\pause\n The SMT-LIB language allows to write \\formulae in a lisp-like format. E.g.:\n \\begin{verbatim}\n (< (+ x y) 0)\n (= (f x y) (g z))\n \\end{verbatim} \n stand for $x + y < 0$ and $f(x,y) = g(z)$ respectively\n\n \\vfill\n \\pause\n An SMT-LIB file looks more similar to a {\\bf set of commands} for\n an \\smtsolver, rather then a logic formula\n\n\\end{frame}\n\n\\begin{frame}[fragile]\n \\frametitle{SMT-LIB Theories}\n\n \\scriptsize\n\n An SMT-LIB theory consists of some {\\bf sorts}, (e.g., \\SInt) \n and of some functions (e.g., $-,+$). \n Predicates are also considered functions, with codomain in \\SBoo\n (e.g., $<,\\leq$). \\pause For instance\n\n \\vfill\n \\begin{tabular}{cc}\n \\begin{minipage}{.4\\textwidth}\n \\tiny\n \\begin{verbatim}\n (theory Ints\n\n :sorts ((Int 0))\n \n :funs ((NUMERAL Int)\n (- Int Int)\n (- Int Int Int :left-assoc)\n (+ Int Int Int :left-assoc) \n (* Int Int Int :left-assoc)\n (div Int Int Int :left-assoc)\n (mod Int Int Int)\n (abs Int Int)\n (<= Int Int Bool :chainable)\n (< Int Int Bool :chainable)\n (>= Int Int Bool :chainable)\n (> Int Int Bool :chainable)\n )\n\n [...]\n )\n \\end{verbatim}\n \\end{minipage}\n &\n \\begin{minipage}{.4\\textwidth}\n \\tiny\n \\begin{verbatim}\n (theory Core\n\n :sorts ((Bool 0))\n\n :funs ((true Bool) \n (false Bool)\n (not Bool Bool)\n (=> Bool Bool Bool :right-assoc)\n (and Bool Bool Bool :left-assoc)\n (or Bool Bool Bool :left-assoc)\n (xor Bool Bool Bool :left-assoc)\n (par (A) (= A A Bool :chainable))\n (par (A) (distinct A A Bool :pairwise))\n (par (A) (ite Bool A A A))\n )\n\n [...]\n )\n \\end{verbatim}\n \\end{minipage} \\\\\n \\multicolumn{2}{c}{\\tiny These definitions can be found at \\url{www.smtlib.org}}\n \\end{tabular}\n \\vfill\n \\pause\n\n The sorts and the function symbols declared in a theory are always {\\bf interpreted}.\n This means that a to specify a model for a formula $\\varphi$, we just need to specify \n the assignment of the variables to the concrete values in the sorts.\n\n\\end{frame}\n\n\\begin{frame}[fragile]\n \\frametitle{SMT-LIB Logics}\n\n \\scriptsize\n The difference between ``logic'' and ``theory'' might look very subtle. \n An SMT-LIB logic includes a theory definition, plus it describes some\n restrictions on how \\formulae can be built. \\pause\n\n \\vfill\n \\hspace{-25pt}\n \\begin{tabular}{ccc}\n \\begin{minipage}{.4\\textwidth}\n \\tiny\n \\begin{verbatim}\n (logic QF_LIA\n \n :theories (Ints)\n \n :language \n \"Closed quantifier-free formulas built \n over an arbitrary expansion of the\n Ints signature with free constant symbols, \n but whose terms of sort Int are all linear, \n that is, have no occurrences of the function \n symbols *, /, div, mod, and abs, except as \n specified the :extensions attribute.\n \"\n \n :extensions\n \"Terms with _concrete_ coefficients are also \n allowed, that is, terms of the form c, (* c x), \n or (* x c) where x is a free constant and c \n is a term of the form n or (- n) for some numeral n.\n \"\n )\n \\end{verbatim}\n \\end{minipage}\n & ~~~~ & \\pause\n \\begin{minipage}{.4\\textwidth}\n \\tiny\n \\begin{verbatim}\n (logic QF_IDL\n\n :theories (Ints)\n\n :language\n \"Closed quantifier-free formulas with \n atoms of the form:\n - q\n - (op (- x y) n),\n - (op (- x y) (- n)), or\n - (op x y)\n where\n - q is a variable or free constant symbol of sort Bool,\n - op is <, <=, >, >=, =, or distinct,\n - x, y are free constant symbols of sort Int, \n - n is a numeral. \n \"\n )\n \\end{verbatim}\n \\end{minipage}\n \\end{tabular}\n\n \\vfill\n \\pause\n\n In the following we will not be so strict, and we will not make any distinction \n between ``theories'' and ``logics'', calling both ``theories''. \\pause \n For instance when we will say that we reason modulo the theory \\Lia we mean \n that we are working with {\\tt QF\\_LIA} \\formulae\n\n\\end{frame}\n\n\\begin{frame}[fragile]\n \\frametitle{Writing an SMT-LIB file} \n \\scriptsize\n\n The logic can be specified with the command\n \\begin{verbatim}\n (set-logic QF_LIA)\n \\end{verbatim}\n \\vfill\n \\pause\n Variables are declared with\n \\begin{verbatim}\n (declare-fun x ( ) Int)\n \\end{verbatim}\n \\vfill\n \\pause\n A formula is specified with \n \\begin{verbatim}\n (assert (<= (+ x y) 0))\n \\end{verbatim}\n \\vfill\n \\pause\n Asks the tool to compute satisfiability of assertions\n \\begin{verbatim}\n (check-sat)\n \\end{verbatim}\n \\vfill\n \\pause\n Asks the tool to return a model (in case of sat result)\n \\begin{verbatim}\n (set-option :produce-models true)\n ...\n (get-value (x y))\n \\end{verbatim}\n \\vfill\n \\pause\n Disable annoying printouts\n \\begin{verbatim}\n (set-option :print-success false)\n \\end{verbatim}\n\n\\end{frame}\n\n\\begin{frame}[fragile]\n \\frametitle{Example}\n\n \\begin{verbatim}\n (set-logic QF_LIA)\n (declare-fun x ( ) Int)\n (declare-fun y ( ) Int)\n (declare-fun a ( ) Bool)\n (assert (<= (+ x y) 0))\n (assert (= x 0))\n (assert (or (not a) (= x 1) (>= y 0)))\n (assert (not (= (+ y 1) 0)))\n (check-sat)\n (exit)\n \\end{verbatim} \n\n which stands for the \\Lia formula\n $$\n (x + y \\leq 0) \\wedge (x = 0) \\wedge ((\\neg a \\vee (x = 1) \\vee (y \\geq 0)) \\wedge \\neg(y + 1 = 0)\n $$\n\n\\end{frame}\n", "meta": {"hexsha": "cd33ba6066b9035af7f42c9c733940504f7918fa", "size": 5925, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture1/smtlib.tex", "max_stars_repo_name": "formalmethods/smtlectures", "max_stars_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-11-07T19:34:12.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-24T08:05:50.000Z", "max_issues_repo_path": "lecture1/smtlib.tex", "max_issues_repo_name": "formalmethods/smtlectures", "max_issues_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture1/smtlib.tex", "max_forks_repo_name": "formalmethods/smtlectures", "max_forks_repo_head_hexsha": "d4ec5f7eb377d26427ecc34c72906c85eafe8631", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-05-06T00:40:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-06T00:40:41.000Z", "avg_line_length": 24.7907949791, "max_line_length": 100, "alphanum_fraction": 0.5891983122, "num_tokens": 1848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802264851919, "lm_q2_score": 0.8244619220634456, "lm_q1q2_score": 0.6340773617489713}} {"text": "\\section{Multi-armed Bandits}\n\n\\subsection{Exercise 1}\n$1 - \\epsilon = \\frac{1}{2}$.\n\n\\subsection{Exercise 2}\nExploration definitely happened at $t = 2$, since we switched from action 1 to action 2 despite\n$Q_2(1) = 1 > Q_2(2) = 0$. Similarly, exploration also definitely happened at $t = 5$, since at that\npoint action 2 would have been the greedy choice. Depending on how ties are broken during greedy selection,\nit is possible that exploration also happened at $t = 3$, since at that point actions 1 and 2 would have\nbeen tied.\n\n\\subsection{Exercise 3}\nIntuitively, it seems clear that the $\\epsilon = 0.01$ strategy will perform the best both in terms of\ncumulative reward and probability of selecting the best action over the long run, as it will eventually\nfind the optimal action and choose it 99\\% of the time.\n\nTo put this more quantitatively, we can consider the performance of each of the 3 methods after $N$ trials,\nsuch that $N$ is large enough for the two $\\epsilon > 0$ approaches to have found the optimal action with a\nhigh degree of certainty (we can choose $N$ for any degree of certainty we like, courtesy of the law of\nlarge numbers). After $N$ trials, the probability that the $\\epsilon = 0.1$ strategy selects the best action is\n0.9, whereas for $\\epsilon = 0.01$ the probability is 0.99. For greedy, the probability is either 1 or 0 after\n$N$ trials, since there is no guarantee that it found the optimal action. Since it is unlikely (\\textbf{TODO:}\nmake this more precise) that greedy converged to the optimal action, the expected difference in cumulative\nreward after $N$ trials will quickly grow to favor the $\\epsilon = 0.01$ strategy.\n", "meta": {"hexsha": "039a1f4b27f8b556674d9b0712051200484bc008", "size": 1665, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Reinforcement_Learning_Sutton_Barto/chapter_2.tex", "max_stars_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_stars_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-09-19T07:33:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-19T07:33:25.000Z", "max_issues_repo_path": "Reinforcement_Learning_Sutton_Barto/chapter_2.tex", "max_issues_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_issues_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Reinforcement_Learning_Sutton_Barto/chapter_2.tex", "max_forks_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_forks_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.0384615385, "max_line_length": 111, "alphanum_fraction": 0.7597597598, "num_tokens": 431, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8128673110375458, "lm_q1q2_score": 0.6340307314579325}} {"text": "\\chapter{Background}\\label{ch:background}\nIn this chapter we provide the background needed for different concepts that are discussed in this thesis. In section \\ref{sec:cnn} we introduce convolutional neural networks and how they are trained. Moreover we define the necessary terminology and provide the mathematical background required to navigate this thesis. Section \\ref{sec:synth} provides the terminology that we use throughout this thesis to describe small parts, synthetic images and real images.\n\n\n\\section{Convolutional Neural Networks}\\label{sec:cnn}\nA \\textbf{neural network} is a massively parallel distributed processor made up of single processing units, which has a natural propensity for storing experiential knowledge and making it available for use \\cite{haykin1994neural}. Convolutional neural networks are a special type of neural network. They are typically used to solve computer vision problems like image classification and object recognition.\n\nRegular neural networks consist of an \\textbf{input layer}, \\textbf{hidden layers} and an \\textbf{output layer}. Every layer is made up of a set of \\textbf{neurons}, where each neuron is fully connected to all neurons in the layer before. A neuron which has an input $x$ of size $n$ calculates the weighted sum $z$ as follows: \\[ z = \\sum_{i=1}^n w_ix_i + b_i \\] where $w$ is called \\textbf{weight} and $b$ is called \\textbf{bias}. The weights and biases of all the neurons in a network are called the network \\textbf{parameters}. Each neuron calculates its output $y$ by applying a differentiable non-linear function $\\varphi(\\cdot)$, called the \\textbf{activation function}, to the weighted sum of its input signals $z$. \\[y = \\varphi(z)\\] A neural networks relays the output of its neurons through a series of hidden layers. Finally, the last fully-connected layer, called the output layer, computes the predictions of the neural network. Figure \\ref{fig:NN} shows an example structure of a regular neural network.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.75\\textwidth]{NN}\n\\caption[A neural network with an input layer, two hidden layers and an output layer. Each neuron is fully connected to all the neurons in the preceding layer.]{A neural network with an input layer, two hidden layers and an output layer. Each neuron is fully connected to all the neurons in the preceding layer.\\footnotemark}\n\\label{fig:NN}\n\\end{figure}\n\\footnotetext{Image: http://cs231n.github.io/convolutional-networks/}\n\nConvolutional neural networks (CNNs) are structured differently. CNNs have 3 types of layers: \\textbf{convolution layers}, \\textbf{pooling layers} and \\textbf{fully connected} layers. Convolution layers slide a weighted matrix (called \\textbf{filter} or \\textbf{kernel}) over the previous layer, and calculate the sum of products. The size of the step that the filter takes while sliding is called a \\textbf{stride}. Figure \\ref{fig:CONV} shows how a convolution layer applies a 3x3 filter to the previous layer. A non-linear activation function is applied to the output of the convolution operation to create a \\textbf{feature map}. It is common to add a pooling layer after the convolution layer for subsampling, which reduces the convolution layer dimensionality. A frequently used type of pooling is \\textbf{max pooling} \\cite{weng1992cresceptron}. Max pooling divides the input layer into sections and computes the maximum activation of each section. Figure \\ref{fig:MP} shows how a max pooling layer applies a 2x2 filter to the preceding layer. Fully connected layers are similar to layers in a regular neural network. Each neuron in a fully connected layer is connected to all neurons in the previous layer.\n\nFigure \\ref{fig:lenet5} shows the LeNet-5 convolutional neural network, which was designed by LeCun et al. \\cite{lecun1998gradient} for handwritten digit recognition. LeNet-5 is an example of a CNN \\textbf{architecture}. A CNN architecture is a set up of different types of layers which are combined to predict the class of the input image.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\textwidth]{CONV}\n\\caption[A 3x3 filter is applied to the source pixel. Each weight in the filter is multiplied by the source's neighboring pixel in the corresponding position. The sum of products is used to calculate the destination pixel's value in the output feature map.]{A 3x3 filter is applied to the source pixel. Each weight in the filter is multiplied by the source's neighboring pixel in the corresponding position. The sum of products is used to calculate the destination pixel's value in the output feature map.\\footnotemark}\n\\label{fig:CONV}\n\\end{figure}\n\\footnotetext{Image: https://towardsdatascience.com/applied-deep-learning-part-4-convolutional-neural-networks-584bc134c1e2}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.7\\textwidth]{MP}\n\\caption[A max-pooling filter of size 2x2 and stride 2 is applied to the source feature map. The maximum value of each filter is used in the destination feature map.]{A max-pooling filter of size 2x2 and stride 2 is applied to the source feature map. The maximum value of each filter is used in the destination feature map.\\footnotemark}\n\\label{fig:MP}\n\\end{figure}\n\\footnotetext{Image: http://cs231n.github.io/convolutional-networks/}\n\n\\begin{figure}[H]\n\\centering\n\\makebox[\\textwidth][c]{\\includegraphics[width=1.3\\textwidth]{lenet5}}\n\\caption[Architecture of the LeNet-5 convolutional neural network.]{Architecture of the LeNet-5 convolutional neural network.\\cite{lecun1989backpropagation}}\n\\label{fig:lenet5}\n\\end{figure}\n\nThe LeNet-5 network uses 3x3 filters with a stride of 1 for convolution layers and 2x2 filters with a stride of 2 for subsampling layers. LeNet-5 uses \\textbf{average pooling}, which calculates the average of the values in the filter.\n\nThe network takes a 32x32 image as input. The first layer is a convolution layer (C1), which applies 6 filters, followed by a subsampling layer (S2). Next comes another convolution layer (C3) which applies 16 filter, followed by a subsampling layer (S4). A final convolution layer (C5) that applies 120 filters follows S4. Afterwards comes a fully connected (F6) layer with 84 neurons. Finally, an output layer of size 10 is connected to F6. Each neuron in the output layer corresponds to the probability of the input image to be one of the 10 digits (0-9).\n\n\\subsection{Training a Convolutional Neural Network}\nWe train a CNN using supervised learning. Supervised learning entails that our network learns from examples. The network is presented with a \\textbf{dataset} of images. The dataset is divided into a \\textbf{training set}, a \\textbf{validation set} and a \\textbf{testing set}. Each set contains images and their corresponding \\textbf{labels} (also referred to as \\textbf{classes}). We train a CNN to predict the labels of images as seen in the training set. The validation set is used for intermediate assessment of the CNN performance during the training phase, while the testing set is used to assess the performance of the CNN after the training phase is done.\n\nCNNs process the training set in \\textbf{mini-batches}. While the use of large mini-batches increases the available computational parallelism, small batch training has been shown to provide improved generalization performance \\cite{masters2018revisiting}. It is also common for CNNs to process the whole training set multiple times. Each single iteration through the training set is called an \\textbf{epoch}. The batch size and number of epochs are examples of CNN \\textbf{hyperparameters}. Hyperparameters are network parameters that are not learnable, but rather optimized manually.\n\nTraining a convolutional neural network is the process of learning the right parameters (filter weights) to compute the correct labels from the input images in the training set. To do so, each CNN has a \\textbf{loss function}. A loss function calculates the error of the predictions made by the CNN. The loss of a CNN which predicts an output \\textit{y} on a training set that has the correct labels $\\hat{y}$ is calculated as follows: \\[L(y, \\hat{y}) = \\dfrac{1}{m}\\sum_{i=1}^m \\mathscr{L}(y_i, \\hat{y}_i)\\] where \\textit{m} is the number of examples in the training set, and $\\mathscr{L}(\\cdot)$ is a distance function that calculates the difference between pairwise instances in $y$ and $\\hat{y}$.\n\nAn \\textbf{optimization function} (or \\textbf{optimizer}) uses \\textbf{back propagation} \\cite{lecun1989backpropagation} to minimize the loss of a CNN during the training phase. The optimizer propagates the output of the loss function backwards by calculating the partial derivative of the loss function with respect to weights and using it to update the values of the weights.\n\nThe \\textbf{stochastic gradient descent (SGD)} optimizer updates the weights of a CNN that uses the loss function $L$ thusly: \\[w_{new}=w_{old}-\\alpha\\dfrac{\\partial L}{\\partial w_{old}}\\] where $\\alpha$ is called the \\textbf{learning rate}. The learning rate controls how fast the weights are updated during training. The learning rate is a tricky hyperparameter to tune. If it is too large, the optimizer might overshoot the optimum parameter values. If it is too small, the optimizer might never reach the optimal value.\n\n\\textbf{Transfer learning} \\cite{pan2010survey} is sometimes employed to speed up the training of convolutional neural networks. Transfer learning involves pre-training a CNN on images from a different domain, and then retraining the network on an image set that is specific to the desired classification task. The advantage of transfer learning is the ability to learn features from a different dataset. This is useful when we have a classification task in one domain of interest, but we only have sufficient training data in another domain. A common technique is to freeze the weights of earlier layers while training a CNN, meaning that the weights are set to be untrainable. The number of \\textbf{frozen layers} is a network hyperparameter.\n\n\n\\section{Small Part Classification}\\label{sec:synth}\nWe describe a system for the automatic classification of \\textbf{small parts} in aircraft engines. Small parts refer to the fasteners used to build the engine like screws, bolts and nuts.\n\nThere is a set of classification challenges that are specific to small parts. A number of small parts can only be separated by subtle distinctions. This difference can be a slight change in size, which means that small parts, unlike scale invariant categories, are sensitive to differences in size, length and width. Figure \\ref{fig:similar_small_parts} shows 2 screws that are identical except for a 6mm difference in length.\n\n\\begin{figure}[h]\n\\centering\n \\includegraphics[width=0.5\\textwidth]{similar_small_parts}\n\\caption{Screw (a) is identical to screw (b) except that screw (a) is 6mm shorter than screw (b). An overhead light shines on the 2 screws. The reflection changes the natural color of their surface.}\n\\label{fig:similar_small_parts}\n\\end{figure}\n\nAircraft fasteners are metallic and have a shiny surface. The light reflection off the fastener surface disturbs the natural surface color as shown in figure \\ref{fig:similar_small_parts}, and may hide important features. Furthermore, the effect of the light reflection depends on the direction of the light, which may cause variance between images of a single small part.\n\nUnlike cases where the background can provide information, the background is non-informative to the classification of small parts. Additionally, exposure to a regular lighting condition, like an overhead light in a room, causes the fasteners to cast a shadow on their background. A fastener shadow is not a discriminative feature. Both the background and the fastener shadow are a source of noise for the classifier.\n\n\\subsection{Real and Synthetic Images}\nTo automatically classify small parts, we build a convolutional neural network that is trained on both \\textbf{real images} and \\textbf{synthetic images}. We define real images as the natural images of the small parts, taken using a camera. Synthetic images, on the other hand, are artificially created images. They are 2-dimensional renditions of \\textbf{3D models} of small parts. To generate synthetic images, we must create a \\textbf{synthetic scene}: an artificial environment created in 3D modeling software. Figure \\ref{fig:synthetic_scene} shows an example of a synthetic scene. A 3D model is placed in a synthetic scene in 3D modeling software. The software renders a 2D image of the environment to create a synthetic scene.\n\nThe real and synthetic images should capture the small parts and their corresponding 3D models from different angles and in different positions. To do so, we apply \\textbf{transformations} to the small parts and 3D models. There are two types of transformations: \\textbf{translation} to change the position of the target object, and \\textbf{rotation} to change the angle. Each transformation has a range. This prevents the target objects from being translated or rotated away from the camera view.\n\n\\begin{figure}[h]\n\\centering\n \\includegraphics[width=0.75\\textwidth]{synthetic_scene}\n\\caption{A synthetic scene where a screw lies on a horizontal plane. (a) shows the isometric view of the scene while (b) shows the side view.}\n\\label{fig:synthetic_scene}\n\\end{figure}\n", "meta": {"hexsha": "5fc14eb0812f35cb12912d408a6216983239f679", "size": 13382, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/2-Background.tex", "max_stars_repo_name": "AmrARaouf/master-thesis", "max_stars_repo_head_hexsha": "9c9bea771d07d17498fd5dd5605cfd47c3707da5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/2-Background.tex", "max_issues_repo_name": "AmrARaouf/master-thesis", "max_issues_repo_head_hexsha": "9c9bea771d07d17498fd5dd5605cfd47c3707da5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/2-Background.tex", "max_forks_repo_name": "AmrARaouf/master-thesis", "max_forks_repo_head_hexsha": "9c9bea771d07d17498fd5dd5605cfd47c3707da5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 148.6888888889, "max_line_length": 1214, "alphanum_fraction": 0.7978628008, "num_tokens": 3108, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339797047029, "lm_q2_score": 0.7549149868676284, "lm_q1q2_score": 0.633927766261077}} {"text": "\n\\subsection{Breaking monoalphabetic substitution ciphers with frequency analysis}\n\n(need to identify algorithm and needs to identify key)\n\nfinding substitution cyphers\n\nSearch space is larger, \\(26!=4*10^26\\). need alternative to brute force.\n\n\nLetter popularity. Compare against popularity for corpus. \nMonogram (ie letters); ngrams(ie n letter in a row frequency); common words.\n\n\nSingle letter words are I or A. More generally. corpus smaller for fewer letters\n\nCan test substitution cypher by matching each word against a corpus\n\n\n", "meta": {"hexsha": "2fd2b976939a27c57a6e8e7b9d67d9c0e1819afc", "size": 536, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/encryptionClassical/04-02-frequency.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/encryptionClassical/04-02-frequency.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/encryptionClassical/04-02-frequency.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.8, "max_line_length": 81, "alphanum_fraction": 0.7910447761, "num_tokens": 113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8333246118695629, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.6338689144321233}} {"text": "\\chapter{Inequalities}\n\n\\begin{ex}\n Let $X\\sim\\text{Exponential}(\\beta)$, and recall that $\\mu_X=\\beta$ and\n $\\sigma^2_X=\\beta^2$. Then, for any $k>1$,\n \\begin{align*}\n \\P{|X-\\mu_X|\\geq k\\sigma_X}\n & =\\P{|X-\\beta|\\geq k\\beta} \\\\\n & =\\P{X-\\beta\\geq k\\beta}+\\P{X-\\beta\\leq -k\\beta} \\\\\n & =\\P{X\\geq (k+1)\\beta}+\\P{X\\leq -(k+1)\\beta} \\\\\n & =\\P{X\\geq (k+1)\\beta} \\\\\n & =\\int_{(k+1)\\beta}^\\infty \\frac{1}{\\beta}e^{-x/\\beta}\\,\\d{x} \\\\\n & =\\int_{k+1}^\\infty e^{-u}\\,\\d{u} \\\\\n & =e^{-(k+1)}.\n \\end{align*}\n\n The comparable bound obtained from Chebyshev's inequality is\n \\begin{align*}\n \\P{|X-\\mu_X|\\geq k\\sigma_X}<\\frac{\\beta^2}{k^2\\beta^2}=\\frac{1}{k^2}.\n \\end{align*}\n This is a weaker bound, since it decays quadratically in $k$, instead of\n exponentially.\n\\end{ex}\n\n\\begin{ex}\n Let $X\\sim\\text{Poisson}(\\lambda)$. Recall that $\\mu_X=\\lambda$ and\n $\\sigma_X^2=\\lambda$. Note that\n \\begin{align*}\n \\P{|X-\\mu_X|\\geq \\lambda}\n & =\\P{X-\\lambda\\geq \\lambda}+\\P{X-\\lambda\\leq -\\lambda} \\\\\n & =\\P{X\\geq \\lambda + \\lambda}+\\P{X\\leq -\\lambda + \\lambda} \\\\\n & =\\P{X\\geq 2\\lambda}+\\P{X\\leq 0} \\\\\n & =\\P{X\\geq 2\\lambda}.\n \\end{align*}\n However, by Chebyshev's inequality for $t=\\lambda$,\n \\[\n \\P{X\\geq 2\\lambda}\n =\\P{|X-\\mu_X|\\geq \\lambda}\n \\leq \\frac{1}{\\lambda}.\n \\]\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots,X_n\\sim\\text{Bernoulli}(p)$. Then\n \\[\n \\E{\\overline{X}_n}=\\frac{np}{n}=p,\\quad\n \\var{\\overline{X}_n}=\\frac{np(p-1)}{n^2}=\\frac{p(p-1)}{n}.\n \\]\n Thus, by Chebyshev's inequality,\n \\[\n \\P{|\\overline{X}_n-p|>\\epsilon}\n \\leq \\frac{\\frac{p(p-1)}{n}}{\\epsilon^2}\n =\\frac{p(p-1)}{n\\epsilon^2}.\n \\]\n\n By Theorem 4.5, we also have\n \\[\n \\P{|\\overline{X}_n-p|>\\epsilon}\n \\leq 2e^{-2n\\epsilon^2}.\n \\]\n\n Note that the bound from Chebyshev's inequality decays at the rate of $1/n$\n as $n\\to\\infty$, while the bound from Hoeffding's inequality does so at an\n exponential rate. Hence, for $n$ sufficiently large, Hoeffding's inequality\n will provide a tighter bound.\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots X_n\\sim\\text{Bernoulli}(p)$. For any $\\alpha>0$ define\n \\[\n \\epsilon_n=\\sqrt{\\frac{1}{2n}\\log\\left(\\frac{2}{\\alpha}\\right)},\n \\]\n and let $C_n=(\\widehat{p}_n-\\epsilon_n, \\widehat{p}_n+\\epsilon_n)$.\n \\begin{enumerate}[(a)]\n \\item Note that\n \\begin{align*}\n \\P{C_n\\text{ contains } p}\n & =\\P{|\\widehat{p}_n-p|\\leq \\epsilon_n} \\\\\n & =1-\\P{|\\widehat{p}_n-p|> \\epsilon_n} \\\\\n & \\geq 1-2e^{-2n\\epsilon_n^2} \\\\\n & =1-2e^{-\\log(2/\\alpha)} \\\\\n & =1 - \\alpha.\n \\end{align*}\n \\item\n \\inputminted{python}{../code/04-04b.py}\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.9]{../images/04-04b}\n \\caption{Graph of simulated coverage versus $n$ for $p=0.4$ and\n $\\alpha=0.05$, with 50,000 simulations per $n$.}\n \\end{figure}\n \\item\n \\inputminted{python}{../code/04-04c.py}\n \\inputminted{text}{../output/04-04c.txt}\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.9]{../images/04-04c}\n \\caption{Graph of interval width versus $n$ for $\\alpha=0.05$.}\n \\end{figure}\n \\end{enumerate}\n\\end{ex}\n\n\\begin{ex}\n Let $Z\\sim N(0, 1)$. Then\n \\[\n \\P{|Z|>t}\n =2\\P{Z>t}\n =\\sqrt{\\frac{2}{\\pi}}\\frac{e^{-t^2/2}}{t},\n \\]\n since\n \\begin{align*}\n t\\P{Z>t}\n & =t\\int_t^\\infty \\frac{1}{\\sqrt{2\\pi}}e^{-x^2/2}\\,\\d{x} \\\\\n & \\leq \\int_t^\\infty \\frac{x}{\\sqrt{2\\pi}}e^{-x^2/2}\\,\\d{x} \\\\\n & =-\\frac{1}{\\sqrt{2\\pi}}e^{-x^2/2}\\bigg\\rvert_{x=t}^\\infty \\\\\n & =\\frac{1}{\\sqrt{2\\pi}}e^{-t^2/2}.\n \\end{align*}\n\\end{ex}\n\n\\begin{ex}\n Let $k\\geq 3$ be an integer. Note that\n \\begin{align*}\n \\E{|Z|}\n & =\\frac{2}{\\sqrt{2\\pi}}\\int_{0}^\\infty\\!xe^{-\\frac{1}{2}x^2}\\,\\d{x}\n =-\\frac{2}{\\sqrt{2\\pi}}\\int_{0}^\\infty\\!e^{-u}\\,\\d{u}\n =\\sqrt{\\frac{2}{\\pi}}, \\\\\n \\E{|Z|^2}\n & =\\E{Z^2}=\\var{Z}+\\E{Z}^2=1,\n \\end{align*}\n \\begin{align*}\n \\E{|Z|^k}\n & =\\frac{2}{\\sqrt{2\\pi}}\\int_{0}^\\infty\\!x^ke^{-\\frac{1}{2}x^2}\\,\\d{x} \\\\\n & =\\frac{2}{\\sqrt{2\\pi}}\\frac{1}{2}\\int_0^\\infty\\!u^{(k-1)/2}e^{-u/2}\\,\\d{u} \\\\\n & =\\frac{2}{\\sqrt{2\\pi}}\\frac{1}{2}\\left[-2e^{-u/2}u^{(k-1)/2}\\bigg\\rvert_{u=0}^\\infty+\n (k-1)\\int_0^\\infty e^{-u/2}u^{(k-3)/2}\\right] \\\\\n & =(k-1)\\frac{2}{\\sqrt{2\\pi}}\\frac{1}{2}\\int_0^\\infty e^{-u/2}u^{(k-3)/2}\\,\\d{u} \\\\\n & =(k-1)\\E{|Z|^{k-2}},\n \\end{align*}\n and that therefore\n \\[\n \\E{|Z|^k}=\\begin{cases}\n \\sqrt{\\frac{2}{\\pi}}(k-1)!! & \\text{if $k$ is odd}, \\\\\n (k-1)!! & \\text{if $k$ is even}.\n \\end{cases}\n \\]\n\n Hence,\n \\[\n \\P{|Z|>t}\n =\\P{|Z|^k>t^k}\n \\leq \\frac{\\E{|Z|^k}}{t^k}\n =\\begin{cases}\n \\sqrt{\\frac{2}{\\pi}}\\frac{(k-1)!!}{t^k} & \\text{if $k$ is odd}, \\\\\n \\frac{(k-1)!!}{t^k} & \\text{if $k$ is even},\n \\end{cases}\n \\]\n or, in particular, $\\P{|Z|>t}t}\n \\leq \\sqrt{\\frac{2}{\\pi}}\\frac{e^{-t^2/2}}{t}.\n \\]\n\n Finally,\n \\begin{align*}\n \\P{|Z|>t}\n =\\frac{2}{\\sqrt{2\\pi}}\\int_{t}^\\infty\\!e^{-\\frac{1}{2}x^2}\\,\\d{x}\n =\\erfc\\left(\\frac{t}{\\sqrt{2}}\\right).\n \\end{align*}\n\n \\inputminted{python}{../code/04-06.py}\n\n \\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{../images/04-06}\n \\caption{Graph of various lower bounds for $\\P{|Z|>t}$.}\n \\end{figure}\n\\end{ex}\n\n\\begin{ex}\n Let $X_1,\\ldots,X_n\\sim N(0,1)$. Then $X_1+X_2+\\cdots+X_n\\sim N(0, n)$ and\n therefore\n \\[\n \\sqrt{n}\\cdot \\overline{X}_n=\\frac{\\sqrt{n}}{n}\\left(X_1+X_2+\\cdots+X_n\\right)\n \\]\n is a standard normal random variable. Thus, by Mill's inequality,\n \\[\n \\P{|\\overline{X}_n|>t}\n =\\P{|\\sqrt{n}\\cdot \\overline{X}_n|>t\\sqrt{n}}\\leq \\sqrt{\\frac{2}{\\pi}}\\frac{\\exp(-t^2n/2)}{t\\sqrt{n}},\n \\]\n while, since $\\overline{X}_n\\sim N(0,1/n)$, by the Chebyshev bound,\n \\[\n \\P{|\\overline{X}_n|>t}\\leq \\frac{1}{nt^2}.\n \\]\n\\end{ex}", "meta": {"hexsha": "f24f044a3f73c3f2d302a544077fc7d5bbba2bbb", "size": 6564, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/tex/ch04.tex", "max_stars_repo_name": "dtrifuno/all-of-stats-solutions", "max_stars_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/tex/ch04.tex", "max_issues_repo_name": "dtrifuno/all-of-stats-solutions", "max_issues_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/tex/ch04.tex", "max_forks_repo_name": "dtrifuno/all-of-stats-solutions", "max_forks_repo_head_hexsha": "0572cdae22b128e71c1c6c7ead2bf3b259875bc9", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.6567164179, "max_line_length": 106, "alphanum_fraction": 0.491773309, "num_tokens": 2786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506635289836, "lm_q2_score": 0.8333245911726382, "lm_q1q2_score": 0.6338689032104863}} {"text": "\\input{../newcommands}\n\nAlthough full PIC codes are powerful tools, which capture a wide range\nof physical phenomena, they also require large computational ressources.\nThis is partly due to the use of a 3D Cartesian grid, which\nleads to a very large number of grid cells. (Typical 3D simulations of\nlaser-wakefield acceleration require $\\sim 10^6$--$ 10^8$ grid\ncells.) For this reason, these algorithms need to be highly parallelized, and\nhigh-resolution simulations can only be run on costly large-scale\ncomputer facilities. However, when the driver is\ncylindrically-symmetric, it is possible to take advantage of the\nsymmetry of the problem to reduce the computational cost of the algorithm \\cite{godfrey1985iprop,LifschitzJCP2009,DavidsonJCP2015,Lehe2016}.\n\n\\subsection{Azimuthal decomposition}\nLet us consider the fields $\\vec{E}$, $\\vec{B}$, $\\vec{J}$ and $\\rho$\n in cylindral coordinates $(r,\\theta,z)$, expressed as a Fourier series in $\\theta$:\n%\n\\begin{equation}\nF(r,\\theta,z) = \\mathrm{Re}\\left[ \\sum_{\\ell=0}^\\infty\n \\tilde{F}_{\\ell}(r,z) e^{-i\\ell\\theta} \\right]\n\\label{eq:chap2:azimuthal}\n\\end{equation}\n%\n\\begin{equation}\n\\mathrm{with} \\qquad \\tilde{F}_{\\ell} = C_\\ell \\int_0^{2\\pi} d\\theta\n\\,F(r,\\theta,z)e^{i\\ell\\theta} \\qquad\n\\label{eq:chap2:Fourier-coeffs}\n\\end{equation}\n\\begin{equation}\n\\mathrm{and} \\;\n\\left \\{ \\begin{array}{l l}\nC_{0} = 1/2\\pi &\\\\\nC_\\ell = 1/\\pi &\\mathrm{for}\\,\\ell > 0\n\\end{array} \\right.\n\\end{equation}\n%\nwhere $F$ represents any of the quantities $E_r$,\n$E_\\theta$, $E_z$, $B_r$, $B_\\theta$, $B_z$, $J_r$, $J_\\theta$, $J_z$\nare $\\rho$, and where the\n$\\tilde{F}_\\ell$ are the associated Fourier components ($\\ell$ is the\nindex of the corresponding azimuthal mode). In the general case, this\nazimuthal decomposition does not simplify the problem, since an\ninfinity of modes have to be considered in (\\ref{eq:chap2:azimuthal}). However, in the case of a\ncylindrically-symmetric laser pulse, only the very first modes have\nnon-zero components. For instance, the wakefield is represented\nexclusively by the mode $\\ell = 0$. (This is because the quantities $E_r$,\n$E_\\theta$, $E_z$, $B_r$, $B_\\theta$, $B_z$, $J_r$, $J_\\theta$, $J_z$\nand $\\rho$ associated with the\nwakefield are independent of $\\theta$.) On the other hand, the field\nof the laser pulse \\emph{does}\ndepend on $\\theta$, in cylindrical coordinates. For example, for a\ncylindrically-symmetric pulse propagating along $z$ and polarized along $\\vec{e}_\\alpha = \\cos(\\alpha)\\vec{e}_x + \\sin(\\alpha)\\vec{e}_y$:\n%\n\\begin{align}\n\\vec{E} &= E_0(r,z)\\vec{e}_\\alpha \\\\\n& = E_0(r,z) [\\; \\cos(\\alpha)(\\cos(\\theta)\\vec{e}_r - \\sin(\\theta)\\vec{e}_\\theta) \\; \\nonumber \\\\\n& + \\; \\sin(\\alpha)(\\sin(\\theta)\\vec{e}_r + \\cos(\\theta)\\vec{e}_\\theta) \\; ]\\\\\n& = \\mathrm{Re}[ \\; E_0(r,z) e^{i\\alpha} e^{-i\\theta} \\; ]\\vec{e}_r \\; \\nonumber \\\\\n& + \\; \\mathrm{Re}[ \\; -i E_0(r,z) e^{i\\alpha} e^{-i\\theta} \\; ]\\vec{e}_\\theta.\n\\end{align}\n%\nHere the amplitude $E_0$ does not depend on $\\theta$ because the pulse was assumed\nto be cylindrically symmetric. In this case, the above relation shows\nthat the fields $E_r$ and $E_\\theta$ of the laser are represented\nexclusively by the mode $\\ell = 1$. A similar calculation shows that\nthe same holds for $B_r$ and $B_\\theta$. On the whole, only the modes\n$\\ell = 0$ and $\\ell = 1$ are a priori necessary to model\nlaser-wakefield acceleration.\nUnder those conditions, the infinite sum in\n(\\ref{eq:chap2:azimuthal}) is truncated at a chosen $\\ell_{max}$. In\nprinciple, $\\ell_{max} = 1$ is sufficient for laser-wakefield\nacceleration. However, $\\ell_{max}$ is kept as a free parameter in the algorithm, in order to verify that\nhigher modes are negligible, as well as to allow for less-symmetric configurations.\nBecause codes based on this algorithm are able to take into account the modes with $\\ell > 0$, they are said to be\n``quasi-cylindrical'' (or ``quasi-3D'' by some authors \\cite{DavidsonJCP2015}), in contrast to cylindrical codes, which\nassume that all fields are independent of $\\theta$, and thus only\nconsider the mode $\\ell = 0$.\n\n\\subsection{Discretized Maxwell equations} When the Fourier expressions\nof the fields are injected into the Maxwell equations (written in\ncylindrical coordinates), the different azimuthal modes\ndecouple. In this case, the Maxwell-Amp\\`ere and Maxwell-Faraday equations\n-- which are needed to update the fields in the PIC cycle -- can be written separately\nfor each azimuthal mode $\\ell$:\n\\begin{subequations}\n\\begin{align}\n\\frac{\\partial \\tilde{B}_{r,\\ell} }{\\partial t} &=\n\\frac{i\\ell}{r}\\tilde{E}_{z,\\ell} + \\frac{\\partial\n \\tilde{E}_{\\theta,\\ell}}{\\partial z} \\\\[3mm]\n\\frac{\\partial \\tilde{B}_{\\theta,\\ell} }{\\partial t} &=\n - \\frac{\\partial \\tilde{E}_{r,\\ell}}{\\partial z} + \\frac{\\partial\n \\tilde{E}_{z,\\ell}}{\\partial r} \\\\[3mm]\n\\frac{\\partial \\tilde{B}_{z,\\ell} }{\\partial t} &=\n- \\frac{1}{r} \\frac{\\partial (r\\tilde{E}_{\\theta,\\ell})}{\\partial r} - \\frac{i\\ell}{r}\\tilde{E}_{r,\\ell} \\\\[3mm]\n\\frac{1}{c^2} \\frac{\\partial \\tilde{E}_{r,\\ell} }{\\partial t} &=\n-\\frac{i\\ell}{r}\\tilde{B}_{z,\\ell} - \\frac{\\partial\n \\tilde{B}_{\\theta,\\ell}}{\\partial z} - \\mu_0 \\tilde{J}_{r,\\ell} \\\\[3mm]\n\\frac{1}{c^2}\\frac{\\partial \\tilde{E}_{\\theta,\\ell} }{\\partial t} &=\n \\frac{\\partial \\tilde{B}_{r,\\ell}}{\\partial z} - \\frac{\\partial\n \\tilde{B}_{z,\\ell}}{\\partial r} - \\mu_0 \\tilde{J}_{\\theta,\\ell} \\\\[3mm]\n\\frac{1}{c^2}\\frac{\\partial \\tilde{E}_{z,\\ell} }{\\partial t} &=\n \\frac{1}{r} \\frac{\\partial (r\\tilde{B}_{\\theta,\\ell})}{\\partial r} +\n \\frac{i\\ell}{r}\\tilde{B}_{r,\\ell} - \\mu_0 \\tilde{J}_{z,\\ell}\n\\end{align}\n\\end{subequations}\n%\\begin{figure}\n%\\input{./Chap2/Circ_lattice.tex}\n%\\caption{Representation of the lattice in \\CCirc. The\n% table shows at which\n% position each component of the fields is defined ($j$,$k$ and\n% $n$ are integers ; $\\Delta r$ and $\\Delta z$ are the\n% spatial steps of the grid). The above sketch represents one grid\n% cell, and the positions of the fields within it.}\n%\\label{fig:chap2:Circ_lattice}\n%\\end{figure}\nIn order to discretize these equations, each azimuthal mode is\nrepresented on a two-dimensional grid,\n%(The two dimensions correspond\n%to $r$ and $z$.) \\Cref{fig:chap2:Circ_lattice} summarizes the\n%positions of the different fields within one grid cell, as well as the\n%corresponding notations for these fields. Using these notations,\non which the discretized\nMaxwell-Amp\\`ere and Maxwell-Faraday equations are given by\n%\\begin{strip}\n%\\begin{align*}\n%\n%\\frac{ \\tBr{n+\\hf}{j,\\ell,k+\\hf}- \\tBr{n-\\hf}{j,\\ell,k+\\hf}\n%}{\\Delta t} =& \\frac{i\\,\\ell}{j\\Delta r}\\tEz{n}{j,\\ell,k+\\hf} + (D_z \\tilde{E}_{\\theta}^n)_{j,\\ell,k+\\hf} \\\\\n%\n%\\frac{ \\tBt{n+\\hf}{j+\\hf,\\ell,k+\\hf}- \\tBt{n-\\hf}{j+\\hf,\\ell,k+\\hf} }{\\Delta t} =& -(D_z \\tilde{E}_r^n)_{j+\\hf,\\ell,k+\\hf} + (D_r \\tilde{E}_z^{n})_{j+\\hf,\\ell,k+\\hf} \\\\\n%\n%\\frac{ \\tBz{n+\\hf}{j+\\hf,\\ell,k}- \\tBz{n-\\hf}{j+\\hf,\\ell,k} }{\\Delta t} =&\n% -\\frac{(j+1)\\tEt{n}{j+1,\\ell,k} - j\\tEt{n}{j,\\ell,k}}{(j+\\hf)\\Delta r} -\n%\\frac{i\\,\\ell}{(j+\\hf) \\Delta r}\\tEr{n}{j+\\hf,\\ell,k} \\\\\n%\n%\\frac{ \\tEr{n+1}{j+\\hf,\\ell,k}- \\tEr{n}{j+\\hf,\\ell,k}}{c^2 \\Delta t}\n%=& -\\frac{i\\,\\ell}{(j+\\hf)\\Delta r}\\tBz{n+\\hf}{j+\\hf,\\ell,k} - (D_z \\tilde{B}_{\\theta}^{n+\\hf})_{j+\\hf,\\ell,k} - \\mu_0\\tJr{n+\\hf}{j+\\hf,\\ell,k} \\\\\n%\n%\\frac{ \\tEt{n+1}{j,\\ell,k}- \\tEt{n}{j,\\ell,k}}{c^2 \\Delta t}\n%=& (D_z \\tilde{B}_r^{n+\\hf})_{j,\\ell,k} - (D_r \\tilde{B}_z^{n+\\hf})_{j,\\ell,k} - \\mu_0\\tJt{n+\\hf}{j,\\ell,k} \\\\\n%\n%\\frac{ \\tEz{n+1}{j,\\ell,k+\\hf}- \\tEz{n}{j,\\ell,k+\\hf}}{c^2 \\Delta t}\n%=& \\frac{\\left(j+\\hf\\right)\\tBt{n+\\hf}{j+\\hf,\\ell,k+\\hf} - \\left(j-\\hf\\right)\\tBt{n+\\hf}{j-\\hf,\\ell,k+\\hf}}{j\\Delta r} \\\\\n%& \\qquad \\qquad + \\frac{i\\,\\ell}{j\\Delta r}\\tBr{n+\\hf}{j,\\ell,k+\\hf} - \\mu_0\\tJz{n+\\hf}{j,\\ell,k+\\hf}\n%\\end{align*}\n%\\end{strip}\n\n\\begin{subequations}\n\\begin{align}\n%\nD_{t}\\tilde{B}_r|_{j,\\ell,k+\\hf}^{n} \\nonumber\n=& \\frac{i\\,\\ell}{j\\Delta r}\\tEz{n}{j,\\ell,k+\\hf} \\\\\n& + D_z \\tilde{E}_{\\theta}|^n_{j,\\ell,k+\\hf} \\\\\n%\nD_{t}\\tilde{B}_\\theta|_{j+\\hf,\\ell,k+\\hf}^{n} \\nonumber\n=& -D_z \\tilde{E}_r|^n_{j+\\hf,\\ell,k+\\hf} \\\\\n& + D_r \\tilde{E}_z|^{n}_{j+\\hf,\\ell,k+\\hf} \\\\\n%\nD_{t}\\tilde{B}_z|_{j+\\hf,\\ell,k}^{n} =& \\nonumber\n -\\frac{(j+1)\\tEt{n}{j+1,\\ell,k} }{(j+\\hf)\\Delta r} \\\\ \\nonumber\n & +\\frac{ j\\tEt{n}{j,\\ell,k}}{(j+\\hf)\\Delta r} \\\\\n & - \\frac{i\\,\\ell}{(j+\\hf) \\Delta r}\\tEr{n}{j+\\hf,\\ell,k}\n\\end{align}\n\\end{subequations}\nfor the magnetic field components, and\n\\begin{subequations}\n\\begin{align}\n%\n\\frac{1}{c^2}D_{t}\\tilde{E}_r|_{j+\\hf,\\ell,k}^{n+\\hf} \\nonumber\n=& -\\frac{i\\,\\ell}{(j+\\hf)\\Delta r}\\tBz{n+\\hf}{j+\\hf,\\ell,k} \\\\\n& - D_z \\tilde{B}_{\\theta}|^{n+\\hf}_{j+\\hf,\\ell,k} \\nonumber\\\\\n& - \\mu_0\\tJr{n+\\hf}{j+\\hf,\\ell,k} \\\\\n%\n\\frac{1}{c^2}D_{t}\\tilde{E}_\\theta|_{j,\\ell,k}^{n+\\hf} \\nonumber\n=& D_z \\tilde{B}_r|^{n+\\hf}_{j,\\ell,k} - D_r \\tilde{B}_z|^{n+\\hf}_{j,\\ell,k} \\\\\n& - \\mu_0\\tJt{n+\\hf}{j,\\ell,k} \\\\\n%\n\\frac{1}{c^2}D_{t}\\tilde{E}_z|_{j,\\ell,k+\\hf}^{n+\\hf} \\nonumber\n=& \\frac{\\left(j+\\hf\\right)\\tBt{n+\\hf}{j+\\hf,\\ell,k+\\hf} }{j\\Delta r} \\\\\n=& -\\frac{\\left(j-\\hf\\right)\\tBt{n+\\hf}{j-\\hf,\\ell,k+\\hf}}{j\\Delta r} \\nonumber\\\\\n& + \\frac{i\\,\\ell}{j\\Delta r}\\tBr{n+\\hf}{j,\\ell,k+\\hf} \\nonumber\\\\\n& - \\mu_0\\tJz{n+\\hf}{j,\\ell,k+\\hf}\n\\end{align}\n\\end{subequations}\nfor the electric field components.\n\nThe numerical operator $D_r$ and $D_z$ are defined by\n\\begin{align*}\n(D_r F)_{j',\\ell,k'} = \\frac{F_{j'+\\hf,\\ell,k'}-F_{j'-\\hf,\\ell,k'} }{\\Delta r} \\\\\n(D_z F)_{j',\\ell,k'} = \\frac{F_{j',\\ell,k'+\\hf}-F_{j',\\ell,k'-\\hf} }{\\Delta z} \\\\\n\\end{align*}\nwhere $j'$ and $k'$ can be integers or half-integers. Notice\nthat these discretized Maxwell equations are not valid on-axis (i.e. for $j=0$), due to\nsingularities in some of the terms. Therefore, on the axis, these equations are replaced by specific boundary conditions, which are based on the symmetry properties of the fields (see \\cite{LifschitzJCP2009} for details).\n\nCompared to a 3D Cartesian calculation with $n_x\\times n_y \\times n_z$\ngrid cells, a quasi-cylindrical calculation with two modes ($l=0$ and $l=1$)\nwill require only $3 \\,n_r \\times n_z$ grid cells. Assuming $n_x=n_y=n_r=100$\nas a typical transverse resolution, a quasi-cylindrical calculation is typically\nover an order of magnitude less computationally demanding than its 3D Cartesian\nequivalent.\n", "meta": {"hexsha": "3bbbf9f8bd96380c8b8ca41255da48a1fbba521e", "size": 10226, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Doxygen/pages/latex_theory/Cylindrical/Cylindrical.tex", "max_stars_repo_name": "LukeMcCulloch/PICSAR", "max_stars_repo_head_hexsha": "4c29ad7f923d158137296c6776f01011049075f0", "max_stars_repo_licenses": ["BSD-3-Clause-LBNL"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2020-06-22T17:38:17.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-11T17:20:30.000Z", "max_issues_repo_path": "Doxygen/pages/latex_theory/Cylindrical/Cylindrical.tex", "max_issues_repo_name": "LukeMcCulloch/PICSAR", "max_issues_repo_head_hexsha": "4c29ad7f923d158137296c6776f01011049075f0", "max_issues_repo_licenses": ["BSD-3-Clause-LBNL"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2020-11-03T10:55:37.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-07T17:00:36.000Z", "max_forks_repo_path": "Doxygen/pages/latex_theory/Cylindrical/Cylindrical.tex", "max_forks_repo_name": "LukeMcCulloch/PICSAR", "max_forks_repo_head_hexsha": "4c29ad7f923d158137296c6776f01011049075f0", "max_forks_repo_licenses": ["BSD-3-Clause-LBNL"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2020-06-23T13:54:59.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-28T21:51:38.000Z", "avg_line_length": 50.6237623762, "max_line_length": 221, "alphanum_fraction": 0.6582241346, "num_tokens": 3831, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245953120233, "lm_q2_score": 0.7606506418255928, "lm_q1q2_score": 0.6338688882731429}} {"text": "\\chapter{Flavors of rings}\nWe continue our exploration of rings by considering\nsome nice-ness properties that rings or ideals can satisfy,\nwhich will be valuable later on.\nAs before, number theory is interlaced as motivation.\nI guess I can tell you at the outset what the completed table\nis going to look like, so you know what to expect.\n\n\\begin{center}\n\t\\begin{tabular}[h]{lll}\n\t\tRing noun & Ideal adjective & Relation \\\\ \\hline\n\t\tPID & principal &\n\t\t\t$R$ is a PID $\\iff$ $R$ is an integral domain, \\\\\n\t\t\t&& \\qquad and every $I$ is principal \\\\\n\t\tNoetherian ring & finitely generated &\n\t\t\t$R$ is Noetherian $\\iff$ every $I$ is fin.\\ gen. \\\\\n\t\tfield & maximal & $R/I$ is a field $\\iff$ $I$ is maximal \\\\\n\t\tintegral domain & prime & $R/I$ is an integral domain\n\t\t\t$\\iff$ $I$ is prime \\\\\n\t\\end{tabular}\n\\end{center}\n\n\\section{Fields}\n\\prototype{$\\QQ$ is a field, but $\\ZZ$ is not.}\n\nWe already saw this definition last chapter:\na field $K$ is a nontrivial ring for which every nonzero element is a unit.\n\nIn particular, there are only two ideals in a field:\nthe ideal $(0)$, which is maximal, and the entire field $K$.\n\n\\section{Integral domains}\n\\prototype{$\\ZZ$ is an integral domain.}\n\nIn practice, we are often not so lucky that we have a full-fledged field.\nNow it would be nice if we could still conclude the zero product property:\nif $ab = 0$ then either $a = 0$ or $b = 0$.\nIf our ring is a field, this is true: if $b \\neq 0$,\nthen we can multiply by $b\\inv$ to get $a = 0$.\nBut many other rings we consider like $\\ZZ$ and $\\ZZ[x]$ also have this property,\ndespite not having division.\n\nNot all rings though: in $\\Zc{15}$,\n\\[ 3 \\cdot 5 \\equiv 0 \\pmod{15}. \\]\nIf $a, b \\neq 0$ but $ab=0$ then we say $a$ and $b$ are \\vocab{zero divisors}\nof the ring $R$.\nSo we give a name to such rings.\n\\begin{definition}\n\tA nontrivial ring with no zero divisors\n\tis called an \\vocab{integral domain}.\\footnote{Some\n\t\tauthors abbreviate this to ``domain'', notably Artin.}\n\\end{definition}\n\\begin{ques}\n\tShow that a field is an integral domain.\n\\end{ques}\n\\begin{exercise}\n\t[Cancellation in integral domains]\n\tSuppose $ac = bc$ in an integral domain, and $c \\neq 0$.\n\tShow that that $a = b$.\n\t(There is no $c\\inv$ to multiply by,\n\tso you have to use the definition.)\n\\end{exercise}\n\n\\begin{example}\n\t[Examples of integral domains]\n\tEvery field is an integral domain,\n\tso all the previous examples apply.\n\tIn addition:\n\t\\begin{enumerate}[(a)]\n\t\t\\ii $\\ZZ$ is an integral domain, but it is not a field.\n\t\t\\ii $\\RR[x]$ is not a field,\n\t\tsince there is no polynomial $P(x)$ with $xP(x) = 1$.\n\t\tHowever, $\\RR[x]$ is an integral domain,\n\t\tbecause if $P(x) Q(x) = 0$ then one of $P$ or $Q$ is zero.\n\t\t\\ii $\\ZZ[x]$ is also an example of an integral domain.\n\t\tIn fact, $R[x]$ is an integral domain for any integral domain $R$ (why?).\n\t\t\\ii $\\Zc n$ is a field (hence integral domain)\n\t\texactly when $n$ is prime.\n\t\tWhen $n$ is not prime, it is a ring but not an integral domain.\n\t\\end{enumerate}\n\tThe trivial ring $0$ is \\emph{not} considered an integral domain.\n\\end{example}\n\nAt this point, we go ahead and say:\n\\begin{definition}\n\tAn integral domain where all ideals are principal\n\tis called a \\vocab{principal ideal domain (PID)}.\n\\end{definition}\nThe ring $\\ZZ/6\\ZZ$ is an example of a ring\nwhich is a principal ideal ring, but not an integral domain.\nAs we alluded to earlier, we will never really use ``principal ideal ring''\nin any real way: we typically will want to strengthen it to PID.\n\n\\section{Prime ideals}\n\\prototype{$(5)$ is a prime ideal of $\\ZZ$.}\n\nWe know that every integer can be factored (up to sign)\nas a unique product of primes; for example $15 = 3 \\cdot 5$\nand $-10 = -2 \\cdot 5$.\nYou might remember the proof involves the so-called B\\'ezout's lemma,\nwhich essentially says that $(a,b) = (\\gcd(a,b))$;\nin other words we've carefully used the fact that $\\ZZ$ is a PID.\n\nIt turns out that for general rings, the situation is not as nice\nas factoring elements because most rings are not PID's.\nThe classic example of something going wrong is\n\\[ 6 = 2 \\cdot 3 = \\left( 1-\\sqrt{-5} \\right)\\left( 1+\\sqrt{-5} \\right) \\]\nin $\\ZZ[\\sqrt{-5}]$.\nNonetheless, we can sidestep the issue\nand talk about factoring \\emph{ideals}:\nsomehow the example $10 = 2 \\cdot 5$ should be $(10) = (2) \\cdot (5)$,\nwhich says ``every multiple of $10$ is the product of a\nmultiple of $2$ and a multiple of $5$''.\nI'd have to tell you then how to multiply two ideals, which I do\nin the chapter on unique factorization.\n\nLet's at least figure out what primes are.\nIn $\\ZZ$, we have that $p \\neq 1$ is prime if whenever $p \\mid xy$,\neither $p \\mid x$ or $p \\mid y$.\nWe port over this definition to our world of ideals.\n\\begin{definition}\n\t\\label{def:prime_ideal}\n\tA \\emph{proper} ideal $I \\subsetneq R$ is a \\vocab{prime ideal}\n\tif whenever $xy \\in I$, either $x \\in I$ or $y \\in I$.\n\\end{definition}\nThe condition that $I$ is proper is analogous to the\nfact that we don't consider $1$ to be a prime number.\n\n\\begin{example}[Examples and non-examples of prime ideals]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii The ideal $(7)$ of $\\ZZ$ is prime.\n\t\t\\ii The ideal $(8)$ of $\\ZZ$ is not prime,\n\t\tsince $2 \\cdot 4 = 8$.\n\n\t\t\\ii The ideal $(x)$ of $\\ZZ[x]$ is prime.\n\t\t\\ii The ideal $(x^2)$ of $\\ZZ[x]$ is not prime,\n\t\tsince $x \\cdot x = x^2$.\n\n\t\t\\ii The ideal $(3,x)$ of $\\ZZ[x]$ is prime.\n\t\tThis is actually easiest to see\n\t\tusing \\Cref{thm:prime_ideal_quotient} below.\n\n\t\t\\ii The ideal $(5) = 5\\ZZ + 5i\\ZZ$ of $\\ZZ[i]$\n\t\tis not prime, since the elements\n\t\t$3+i$ and $3-i$ have product $10 \\in (5)$,\n\t\tyet neither is itself in $(5)$.\n\t\\end{enumerate}\n\\end{example}\n\\begin{remark}\n\t\\label{rem:unit_sign_issue}\n\tIdeals have the nice property that they get rid of ``sign issues''.\n\tFor example, in $\\ZZ$, do we consider $-3$ to be a prime?\n\tWhen phrased with ideals, this annoyance goes away: $(-3) = (3)$.\n\tMore generally, for a ring $R$, talking about ideals\n\tlets us ignore multiplication by a unit.\n\t(Note that $-1$ is a unit in $\\ZZ$.)\n\\end{remark}\n\n\\begin{exercise}\n\tWhat do you call a ring $R$ for which the zero ideal $(0)$ is prime?\n\\end{exercise}\n\nWe also have:\n\\begin{theorem}[Prime ideal $\\iff$ quotient is integral domain]\n\t\\label{thm:prime_ideal_quotient}\n\tAn ideal $I$ is prime if and only if $R/I$ is an integral domain.\n\\end{theorem}\n\\begin{exercise}\n\t[Mandatory]\n\tConvince yourself the theorem is true;\n\tit is just definition chasing.\n\t(A possible start is to consider $R = \\ZZ$ and $I = (15)$.)\n\\end{exercise}\n\nI now must regrettably inform you that unique factorization is still\nnot true even with the notion of a ``prime'' ideal\n(though again I haven't told you how to multiply two ideals yet).\nBut it will become true with some additional assumptions\nthat will arise in algebraic number theory\n(relevant buzzword: Dedekind domain).\n\n\\section{Maximal ideals}\n\\prototype{The ideal $(x,5)$ is maximal in $\\ZZ[x]$, by quotient-ing.}\n\nHere's another flavor of an ideal.\n\\begin{definition}\n\tA proper ideal $I$ of a ring $R$ is \\vocab{maximal} if\n\tit is not contained in any other proper ideal.\n\\end{definition}\n\n\\begin{example}\n\t[Examples of maximal ideals]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii The ideal $I = (7)$ of $\\ZZ$ is maximal, because\n\t\tif an ideal $J$ contains $7$\n\t\tand an element $n$ not in $I$\n\t\tit must contain $\\gcd(7,n) = 1$, and hence $J = \\ZZ$.\n\t\t\\ii The ideal $(x)$ is \\emph{not} maximal in $\\ZZ[x]$,\n\t\tbecause it's contained in $(x,5)$ (among others).\n\t\t\\ii On the other hand, $(x,5)$ is indeed maximal in $\\ZZ[x]$.\n\t\tThis is actually easiest to verify using\n\t\t\\Cref{thm:max_ideal_quotient} below.\n\t\t\\ii Also, $(x)$ is maximal in $\\CC[x]$,\n\t\tagain appealing to \\Cref{thm:max_ideal_quotient} below.\n\t\\end{enumerate}\n\\end{example}\n\n\\begin{exercise}\n\tWhat do you call a ring $R$ for which the zero ideal $(0)$ is maximal?\n\\end{exercise}\n\nThere's an analogous theorem to the one for prime ideals.\n\\begin{theorem}\n\t[$I$ maximal $\\iff$ $R/I$ field]\n\t\\label{thm:max_ideal_quotient}\n\tAn ideal $I$ is maximal if and only if $R/I$ is a field.\n\\end{theorem}\n\\begin{proof}\n\tA ring is a field if and only if $(0)$ is the only maximal ideal.\n\tSo this follows by \\Cref{prob:inclusion_preserving}.\n\\end{proof}\n\n\\begin{corollary}\n\t[Maximal ideals are prime]\n\tIf $I$ is a maximal ideal of a ring $R$, then $I$ is prime.\n\\end{corollary}\n\\begin{proof}\n\tIf $I$ is maximal, then $R/I$ is a field,\n\thence an integral domain, so $I$ is prime.\n\\end{proof}\n\nIn practice, because modding out by generated ideals is pretty convenient,\nthis is a very efficient way to check whether an ideal is maximal.\n\\begin{example}\n\t[Modding out in $\\ZZ{[x]}$]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii This instantly implies that $(x,5)$ is a maximal ideal\n\t\tin $\\ZZ[x]$, because if we mod out by $x$ and $5$ in $\\ZZ[x]$,\n\t\twe just get $\\FF_5$, which is a field.\n\t\t\\ii On the other hand, modding out by just $x$ gives $\\ZZ$,\n\t\twhich is an integral domain but not a field; that's why $(x)$ is\n\t\tprime but not maximal.\n\t\\end{enumerate}\n\\end{example}\n\nAs we saw, any maximal ideal is prime.\nBut now note that $\\ZZ$ has the special property that\nall of its nonzero prime ideals are also maximal.\nIt's with this condition and a few other minor conditions\nthat you get a so-called \\emph{Dedekind domain}\nwhere prime factorization of ideals \\emph{does} work.\nMore on that later.\n\n\\section{Field of fractions}\n\\prototype{$\\Frac(\\ZZ) = \\QQ$.}\nAs long as we are here, we take the time to introduce a useful\nconstruction that turns any integral domain into a field.\n\n\\begin{definition}\n\tGiven an integral domain $R$,\n\twe define its \\vocab{field of fractions} or \\vocab{fraction field}\n\t$\\Frac(R)$ as follows:\n\tit consists of elements $a / b$, where $a,b \\in R$ and $b \\neq 0$.\n\tWe set $a / b \\sim c / d$ if and only if $bc = ad$.\n\tAddition and multiplication is defined by\n\t\\begin{align*}\n\t\t\\frac ab + \\frac cd &= \\frac{ad+bc}{bd} \\\\\n\t\t\\frac ab \\cdot \\frac cd &= \\frac{ac}{bd}.\n\t\\end{align*}\n\\end{definition}\nIn fact everything you know about $\\QQ$ basically carries over by analogy.\nYou can prove if you want that this indeed a field, but \nconsidering how comfortable we are that $\\QQ$ is well-defined,\nI wouldn't worry about it\\dots\n\n\\begin{definition}\n\tLet $k$ be a field.\n\tWe define $k(x) = \\Frac(k[x])$\n\t(read ``$k$ of $x$''),\n\tand call it the \\vocab{field of rational functions}.\n\\end{definition}\n\n\\begin{example}\n\t[Examples of fraction fields]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii By \\emph{definition}, $\\Frac(\\ZZ) = \\QQ$.\n\t\t\\ii The field $\\RR(x)$ consists of rational functions in $x$:\n\t\t\\[ \\RR(x) = \\left\\{ \\frac{f(x)}{g(x)} \\mid f,g \\in \\RR[x] \\right\\}. \\]\n\t\tFor example, $\\frac{2x}{x^2-3}$ might be a typical element.\n\t\\end{enumerate}\n\\end{example}\n\n\\begin{example}\n\t[Gaussian rationals]\n\t\\label{ex:gaussian_rationals}\n\tJust like we defined $\\ZZ[i]$ by abusing notation,\n\twe can also write $\\QQ(i) = \\Frac(\\ZZ[i])$.\n\tOfficially, it should consist of\n\t\\[ \\QQ(i) = \\left\\{ \\frac{f(i)}{g(i)} \\mid g(i) \\ne 0 \\right\\} \\]\n\tfor polynomials $f$ and $g$ with rational coefficients.\n\tBut since $i^2=-1$ this just leads to\n\t\\[ \\QQ(i) = \\left\\{ \\frac{a+bi}{c+di} \\mid a,b,c,d \\in \\QQ,\n\t\t(c,d) \\ne (0,0) \\right\\}. \\]\n\tAnd since $\\frac{1}{c+di} = \\frac{c-di}{c^2+d^2}$ we end up with\n\t\\[ \\QQ(i) = \\left\\{ a+bi \\mid a,b \\in \\QQ \\right\\}. \\]\n\\end{example}\n\n\n\\section{Unique factorization domains (UFD's)}\n\\prototype{$\\ZZ$ and polynomial rings in general.}\n\nHere is one stray definition that will be important\nfor those with a number-theoretic inclination.\nOver the positive integers, we have a fundamental theorem of arithmetic,\nstating that every integer is uniquely the product of prime numbers.\n\nWe can even make an analogous statement in $\\ZZ$ or $\\ZZ[i]$,\nif we allow representations like $6 = (-2)(-3)$ and so on.\nThe trick is that we only consider everything \\emph{up to units};\nso $6 = (-2)(-3) = 2 \\cdot 3$ are considered the same.\n\nThe general definition goes as follows.\n\\begin{definition}\n\tA nonzero non-unit of an integral domain $R$ is\n\t\\vocab{irreducible} if it cannot be written as the product of two non-units.\n\n\tAn integral domain $R$ is a \\vocab{unique factorization domain}\n\tif every nonzero non-unit of $R$ can be written\n\tas the product of irreducible elements,\n\twhich is unique up to multiplication by units.\n\\end{definition}\n\\begin{ques}\n\tVerify that $\\ZZ$ is a UFD.\n\\end{ques}\n\n\\begin{example}\n\t[Examples of UFD's]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Fields are a ``degenerate'' example of UFD's:\n\t\tevery nonzero element is a unit,\n\t\tso there is nothing to check.\n\n\t\t\\ii $\\ZZ$ is a UFD.\n\t\tThe irreducible elements are $p$ and $-p$,\n\t\tfor example $5$ or $-17$.\n\n\t\t\\ii $\\QQ[x]$ is a UFD:\n\t\tpolynomials with rational coefficients\n\t\tcan be uniquely factored, up to scaling by constants\n\t\t(as the units of $\\QQ[x]$ are just the rational numbers).\n\n\t\t\\ii $\\ZZ[x]$ is a UFD.\n\n\t\t\\ii The Gaussian integers $\\ZZ[i]$ turns out to be a UFD too\n\t\t(and this will be proved in the chapters on algebraic number theory).\n\n\t\t\\ii $\\ZZ[\\sqrt{-5}]$ is the classic non-example of a UFD:\n\t\tone may write\n\t\t\\[ 6 = 2 \\cdot 3 = \\left( 1-\\sqrt{-5} \\right)\n\t\t\t\\left( 1+\\sqrt{-5} \\right) \\]\n\t\tbut each of $2$, $3$, $1 \\pm \\sqrt{-5}$ is irreducible.\n\t\t(It turns out the right way to fix this is\n\t\tby considering prime \\emph{ideals} instead,\n\t\tand this is one big motivation for \\Cref{part:algnt1}.)\n\n\t\t\\ii Theorem we won't prove: every PID is a UFD.\n\t\t\\ii Theorem we won't prove: if $R$ is a UFD,\n\t\tso is $R[x]$ (and hence by induction so is\n\t\t$R[x,y]$, $R[x,y,z]$, \\dots).\n\t\\end{enumerate}\n\\end{example}\n\n\n\n\\section{\\problemhead}\nNot olympiad problems, but again the spirit is very close\nto what you might see in an olympiad.\n\n\\begin{problem}\n\tConsider the ring\n\t\\[ \\QQ[\\sqrt2] = \\left\\{ a + b\\sqrt2 \\mid a,b \\in \\QQ \\right\\}. \\]\n\tIs it a field?\n\t\\begin{hint}\n\t\tYes.\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}\n\t[Homomorphisms from fields are injective]\n\t\\label{prob:field_hom}\n\tLet $K$ be a field and $R$ a ring.\n\tProve that any homomorphism $\\psi \\colon K \\to R$\n\tis injective.\\footnote{Note that $\\psi$\n\t\tcannot be the zero map for us,\n\t\tsince we require $\\psi(1_K) = 1_R$.\n\t\tYou sometimes find different statements in the literature.}\n\t\\begin{hint}\n\t\tThe kernel is an ideal of $K$!\n\t\\end{hint}\n\\end{problem}\n\n\\begin{sproblem}\n\t[Pre-image of prime ideals]\n\t\\label{prob:prime_preimage}\n\tSuppose $\\phi \\colon R \\to S$ is a ring homomorphism,\n\tand $I \\subseteq S$ is a prime ideal.\n\tProve that $\\phi\\pre(I)$ is prime as well.\n\t\\begin{hint}\n\t\tThis is just a definition chase.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tConsider $ab \\in \\phi\\pre(I)$,\n\t\tmeaning $\\phi(ab) = \\phi(a) \\phi(b) \\in I$.\n\t\tSince $I$ is prime, either $\\phi(a) \\in I$ or $\\phi(b) \\in I$.\n\t\tIn the former case we get $a \\in \\phi\\pre(I)$ as needed;\n\t\tthe latter case we get $b \\in \\phi\\pre(I)$.\n\t\\end{sol}\n\\end{sproblem}\n\n\\begin{sproblem}\n\t\\gim\n\tLet $R$ be an integral domain with finitely many elements.\n\tProve that $R$ is a field.\n\t\\label{prob:finite_domain_field}\n\t\\begin{hint}\n\t\tFermat's little theorem type argument;\n\t\tcancellation holds in integral domains.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tLet $x \\in R$ with $x \\neq 0$.\n\t\tLook at the powers $x$, $x^2$, \\dots.\n\t\tBy pigeonhole, eventually two of them coincide.\n\t\tSo assume $x^m = x^n$ where $m < n$, or equivalently\n\t\t\\[ 0 = x \\cdot x \\cdot \\dots \\cdot x\n\t\t\t\\cdot \\left( x^{n-m} - 1 \\right). \\]\n\t\tSince $x \\ne 0$, we get $x^{n-m} - 1 = 0$,\n\t\tor $x^{n-m} = 1$.\n\t\tSo $x^{n-m-1}$ is an inverse for $x$.\n\n\t\tThis means every nonzero element has an inverse,\n\t\tergo $R$ is a field.\n\t\\end{sol}\n\\end{sproblem}\n\n\\begin{sproblem}\n\t[Krull's theorem]\n\t\\label{prob:krull_max_ideal}\n\tLet $R$ be a ring and $J$ a proper ideal.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Prove that if $R$ is Noetherian,\n\t\tthen $J$ is contained in a maximal ideal $I$.\n\t\t\\ii Use Zorn's lemma (\\Cref{ch:zorn})\n\t\tto prove the result even if $R$ isn't Noetherian.\n\t\\end{enumerate}\n\t\\begin{hint}\n\t\tJust keep on adding in elements to get an ascending chain.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tFor part (b), look at the poset of \\emph{proper} ideals.\n\t\tApply Zorn's lemma (again using a union trick to verify the condition;\n\t\tbe sure to verify that the union is proper!).\n\t\tIn part (a) we are given no ascending infinite chains,\n\t\tso no need to use Zorn's lemma.\n\t\\end{sol}\n\\end{sproblem}\n\n\\begin{problem}\n\t[{$\\Spec k[x]$}]\n\tDescribe the prime ideals of $\\CC[x]$ and $\\RR[x]$.\n\t\\begin{hint}\n\t\tUse the fact that both are PID's.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tThe ideal $(0)$ is of course prime in both.\n\t\tAlso, both rings are PID's.\n\n\t\tFor $\\CC[x]$ we get a prime ideal $(x-z)$ for each $z \\in \\CC$.\n\n\t\tFor $\\RR[x]$ a prime ideal $(x-a)$ for each $a \\in \\RR$\n\t\tand a prime ideal $(x^2 - ax + b)$ for each quadratic\n\t\twith two conjugate real roots.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\tProve that any nonzero prime ideal of $\\ZZ[\\sqrt 2]$ is also a maximal ideal.\n\t%(This is starred since the main idea of the solution will get used extensively in\n\t%the algebraic NT chapters, even though the problem itself won't.)\n\t\\label{prob:dedekind_sample}\n\t\\begin{hint}\n\t\tShow that the quotient $\\ZZ[\\sqrt2]/I$ has finitely many elements\n\t\tfor any nonzero prime ideal $I$.\n\t\tTherefore, the quotient is an integral domain, it is also a field,\n\t\tand thus $I$ was a maximal ideal.\n\t\\end{hint}\n\\end{problem}\n\n", "meta": {"hexsha": 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YES\n2. YES", "lm_q1_score": 0.7310585786300048, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.6338539314382297}} {"text": "\\documentclass[a4paper,11pt]{article}\r\n\\usepackage{a4wide}\r\n\\usepackage{amsmath}\r\n\\usepackage{amssymb}\r\n\\begin{document}\r\n\\begin{center}\r\n {\\LARGE\\bf Supplemental note for Week 2 Part 3}\r\n \\end{center}\r\n\\begin{flushright}\r\n {\\large\\bf ver. 20170418-01}\\\\\r\n \\ \\\\\r\n{\\large\\bf Ryoichi Yamamoto}\\\\\r\n\\end{flushright}\r\n\\section{Binomial distribution $\\rightarrow$ Normal distribution}\r\n\r\nStart from Eq.(C6) shown below.\r\n\\begin{eqnarray}\r\n P(n)&=&\\frac{M!}{n!(M-n)!}p^n(1-p)^{M-n}\\\\\r\n \\ln P(n)&=&\\ln M! -\\ln n! - \\ln (M-n)!+n\\ln p +(M-n)\\ln(1-p)\\\\\r\n &=&M(\\ln M -1) -n(\\ln n -1) - (M-n)(\\ln (M-n)-1)\\\\& & + n\\ln p +(M-n)\\ln (1-p)\r\n\\end{eqnarray}\r\nHere we used Stirling's approximation valid for large $n$.\r\n\\begin{equation}\r\n\\ln n!\\simeq n(\\ln n -1)\r\n\\end{equation}\r\nTake the 1st derivative of the above equation in terms of $n$.\r\n\\begin{eqnarray}\r\n \\frac{d \\ln P}{dn}&=&-(\\ln n +1) + (\\ln(M-n)+1) + \\ln p -ln(1-p)\\\\\r\n &=&-\\ln n + \\ln(M-n)+\\ln p -\\ln(1-p)\\\\\r\n &=&\\ln \\left[\\frac{M-n}{n} \\right] - \\ln \\left[\\frac{1-p}{p} \\right] \r\n\\end{eqnarray}\r\nWhen $n$ and $M$ are both large, the peak in $P(n)$ and also in $\\ln P(n)$ is very sharp around the mean value $n=\\langle N\\rangle$, and thus\r\n\\begin{eqnarray}\r\n \\left.\\frac{d \\ln P}{dn}\\right|_{n=\\langle N\\rangle} &=&\\ln \\left[\\frac{M-\\langle N\\rangle}{\\langle N\\rangle} \\right] - \\ln \\left[\\frac{1-p}{p} \\right] =0\\\\\r\n \\therefore \\frac{M-\\langle N\\rangle}{\\langle N\\rangle} &=& \\frac{1-p}{p} \\\\\r\n \\langle N\\rangle&=&Mp.\r\n\\end{eqnarray}\r\n\\\\\r\nConsider a Taylor expansion of $\\ln P(n)$ around the mean $\\langle N\\rangle$ by defining $n=\\langle N\\rangle+\\delta n$.\r\n\\begin{equation}\r\n \\ln P(n) = \\ln (\\langle N\\rangle) + \\left.\\frac{d \\ln P}{dn}\\right|_{n=\\langle N\\rangle}\\delta n\r\n + \\frac{1}{2}\\left.\\frac{d^2 \\ln P}{dn^2}\\right|_{n=\\langle N\\rangle}\\delta n^2 + \\cdots\r\n \\end{equation}\r\nFrom Eq.(7),\r\n\\begin{eqnarray}\r\n \\frac{d^2 \\ln P}{dn^2}&=&-\\frac{1}{n}-\\frac{1}{M-n}\\\\\r\n \\left.\\frac{d^2 \\ln P}{dn^2}\\right|_{n=\\langle N\\rangle}&=&-\\frac{1}{Mp}-\\frac{1}{M-Mp}=-\\frac{1}{Mp(1-p)}=-\\sigma^{-2}.\r\n\\end{eqnarray}\r\nTerminate Eq.(12) at the 2nd order in terms of $\\delta n$, and use Eqs.(9) and (14).\r\n\\begin{equation}\r\nP(n)=const.\\times\\exp\\left(-\\frac{\\delta n^2}{2\\sigma^2}\\right)\r\n\\end{equation}\r\nDetermine $const.$ so that that $\\int_{-\\infty}^{\\infty}P(n)dn=1$, we finally obtain the following normal distribution function with $\\langle N\\rangle=Mp$ and $\\sigma^2=Mp(1-p)$.\r\n\\begin{equation}\r\nP(n)=\\frac{1}{\\sqrt{2\\pi\\sigma^2}}\\exp\\left(-\\frac{(n-\\langle N\\rangle)^2}{2\\sigma^2}\\right)\r\n\\end{equation}\r\n\r\n\\section{Binomial distribution $\\rightarrow$ Poisson distribution}\r\n\r\nWe now consider the limit of $M\\rightarrow\\infty$ while $\\langle N\\rangle=Mp=a$ remains constant.\r\nNotice that the following approximations hold.\r\n\\begin{eqnarray}\r\n%p=\\frac{a}{M-1}&\\simeq&\\frac{a}{M}\\\\\r\nM-n&\\simeq& M\\\\\r\n \\frac{M!}{(M-n)!}&=&M(M-1)\\cdots(M-n+1)\\simeq M^n\r\n\\end{eqnarray}\r\nAgain start from Eq.(C6), we can derive Poisson distribution as shown below\r\nwith $\\langle N\\rangle=Mp=a$ and $\\sigma^2=Mp(1-p)\\simeq a$.\r\n\\begin{eqnarray}\r\n P(n)&=&\\frac{M!}{n!(M-n)!}p^n(1-p)^{M-n}\\\\\r\n &\\simeq&\\frac{1}{n!}M^n\\left(\\frac{a}{M}\\right)^n\\left(1-\\frac{a}{M}\\right)^M\\\\\r\n &\\simeq&\\frac{a^n e^{-a}}{n!}\r\n\\end{eqnarray}\r\nHere we used\r\n\\begin{equation}\r\n\\lim_{M\\rightarrow \\infty}\\left(1-\\frac{a}{M}\\right)^M= e^{-a}.\r\n\\end{equation}\r\n\\end{document}\r\n\r\n", "meta": {"hexsha": "ed28e83ca3637651a6b4cbf94cbc464125b96a57", "size": 3396, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "edx-stochastic-data-analysis/downloaded_files/02/Supplemental_note_2-3.tex", "max_stars_repo_name": "mirandagil/extra-courses", "max_stars_repo_head_hexsha": "51858f5089b10b070de43ea3809697760aa261ec", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "edx-stochastic-data-analysis/downloaded_files/02/Supplemental_note_2-3.tex", "max_issues_repo_name": "mirandagil/extra-courses", "max_issues_repo_head_hexsha": "51858f5089b10b070de43ea3809697760aa261ec", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "edx-stochastic-data-analysis/downloaded_files/02/Supplemental_note_2-3.tex", "max_forks_repo_name": "mirandagil/extra-courses", "max_forks_repo_head_hexsha": "51858f5089b10b070de43ea3809697760aa261ec", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.45, "max_line_length": 179, "alphanum_fraction": 0.6192579505, "num_tokens": 1355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.734119526900183, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.6338326376630498}} {"text": "\\lab{IVP Methods}{IVP Methods}\n\\label{lab:IVP}\n\n\\objective{Implement several basic numerical methods for initial value problems, and use them to study harmonic oscillators.}\n\n\\section*{Methods for initial value problems}\nConsider the initial value problem \n\\begin{align*}\ny' &= f(x,y),\\,\\, a \\leq x \\leq b, \\\\\ny(a) &= y_0,\n\\end{align*}\nwhere $f$ is a continuous function. A solution is a continuously differentiable function $y(x)$ that satisfies the equation $y' = f(x,y)$ on the interval $[a,b]$ and for which $y(a) = y_0$. \n\nThere are many initial value problems (IVPs) where it is impossible to find a closed form (analytic) expression for the solution. \nFor other IVPs there is a closed form expression for the solution, but it may be difficult to interpret. \nIn either case, we can usually make good use of various methods of numerical approximation to study the solution. (It is still important to know mathematically that a solution exists, even when we cannot find an explicit solution!)\n\nConsider the initial value problem \n\\begin{align*}\ny'(x) &= \\sin y(x), \\\\\ny(0) &= y_0.\n\\end{align*}\nThis IVP does have solution, which is given implicitly by \n\\[x = \\ln \\left|\\frac{\\cos y_0 + \\cot y_0}{\\csc y + \\cot y} \\right|.\\]\nIn this case, to understand the general solution it helps to use an IVP solver to plot the integral curves for some initial values. It is easy to show that this differential equation has constant solutions $y_n(x) = n \\pi, n \\in \\mathbb{N}$. Knowing this, and after plotting several integral curves (see Figure \\ref{ivp:int_curves} ), it is obvious how solutions of this IVP will behave.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{Fig2.pdf}\n\\caption{Several integral curves for the differential equation $y' =\\sin y$, using the python solver \\li{dopri5}. }\n\\label{ivp:int_curves}\n\\end{figure}\n\n\n\nNumerical methods for solving intial value problems require us to approximate the solution on a set of grid points $a = x_0< x_1<\\hdots< x_n = b$ in our interval. For simplicity we will assume that each of the $n$ subintervals $[x_{i-1},x_i]$ has equal length $h = (b-a)/n$. $h$ is called the \\textit{step size}. We then look for values $y_0,y_1, \\hdots, y_n$ that approximate our solution ( so $y_i \\approx y(x_i)$). \n\nFor a fixed value of $i$, $ 1 \\leq i \\leq n$, Taylor's theorem says that \n\\begin{align*}\ny(x_{i+1}) &= y(x_{i}) + h y'(x_i) + \\frac{h^2}{2} y''(\\xi_i)\\text{ for some }\\xi_i \\in [x_i,x_{i+1}].\n\\end{align*}\nFor small values of $h$ the quantity $\\frac{h^2}{2} y''(\\xi_i)$ will be negligible and so we will have\n\\begin{align*}\ny(x_{i+1}) &\\approx y(x_{i}) + h y'(x_i) ,\\\\\n&\\approx y(x_{i}) + h f(x_i,y(x_i)).\n\\end{align*}\nThis approximation leads to a first order ($\\mathcal{O}(h^1)$) method called Euler's method: Let $y_0 = y(a)$, and for $i = 0, 1, \\hdots, n-1$, let $y_{i+1} = y_i +hf(x_i,y_i)$. \n% \\begin{enumerate}\n% \\item Let $y_0 = y(a)$. \n% \\item For $i = 0, 1, \\hdots, n-1$, let $y_{i+1} = y_i +hf(x_i,y_i)$. \n% \\end{enumerate}\n\nTaylor's theorem also says that \n\\begin{align*}\ny(x_{i}) &= y(x_{i+1}) - h y'(x_{i+1}) + \\frac{h^2}{2} y''(\\xi_i) \\text{ for some } \\xi_i \\in [x_i,x_{i+1}], \\\\\n\\end{align*}\nso that for small $h$\n\\begin{align*}\ny(x_{i+1}) &\\approx y(x_{i}) + h f(x_{i+1},y(x_{i+1})).\n\\end{align*}\nThis approximation leads to the backwards Euler method, another first order method: Let $y_0 = y(a)$ and for $i = 0, 1, \\hdots, n-1$, solve $y_{i} = y_{i+1}-hf(x_{i+1},y_{i+1})$ for $y_{i+1}$.\n\nNote that for both the Euler and backwards Euler methods, only $y_i, f, $ and other points in the interval $[x_i, x_{i+1}]$ are needed to find $y_{i+1}$. Because of this these are called \\textit{one-step methods}. \n\nEuler's method is an explicit method. The backwards Euler method is an implicit method since an equation must be solved at each step to find $y_{i+1}$. Explicit and implicit methods usually have their own advantages and disadvantages. While implicit methods require an equation to be solved at each time step, they also often have bettern stability properties than the corresponding explicit methods.\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\textwidth]{Fig1.pdf}\n\\caption{The solution of $y' -y= -2x+4, y(0) = 0$, is $y(x) = -2+2x + 2e^x.$ This is a plot of the solution, alongside approximations with Euler's method for several stepsizes.}\n\\label{ivp:euler}\n\\end{figure}\n\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\textwidth]{Fig3.pdf}\n\\caption{The solution of $y' -y= -2x+4,$ $ y(0) = 0$, is $y(x) = -2+2x + 2e^x.$ This is a loglog plot of the relative error in numerically approximating $y(2)$, using stepsizes $h = 0.2,$ $0.1,$ $0.05,$ $0.025,$ and $0.0125$. The plot shows first, second, and fourth order convergence for the Euler method, Midpoint method, and the RK4 method, respectively.}\n\\label{ivp:relative_error}\n\\end{figure}\n\n\\begin{problem} The solution of the IVP\n\\begin{align*}\ny' + y &= 2-2x,\\,\\, 0 \\leq x \\leq 2, \\\\\ny(0) &= 0,\n\\end{align*}\nis given by $y(x) = 4-2x -4e^{-x}$. \nUse Euler's method to numerically approximate the solution with step sizes $h = 0.4, 0.2$, and $0.1.$ Plot your results.\n\\end{problem}\n\nSo how do we come up with numerical methods with higher order accuracy? Using Taylor's theorem (as we did for Euler's method) to create higher-order one-step methods would lead to numerically approximating derivatives of $f(t,y)$ - not necessarily desirable. \n\nLet us look for a second order method of the form\n\n$y_{i+1} = y_i + a f(x_i+b, y_i+c).$\n\nBy expanding $a f(x+b, y+c)$ with Taylor's theorem and matching constants in the equation\n\\begin{align*}\nf(x,y) + \\frac{h}{2}f'(x,y) &= f(x,y) + \\frac{h}{2}\\frac{\\partial f}{\\partial x}(x,y) + \\frac{h}{2}\\frac{\\partial f}{\\partial y}(x,y) \\cdot f(x,y),\n\\end{align*}\nwe find that $a = h, b = h/2,$ and $c = h/2$. This method is called the Midpoint method. IVP solvers with this general form are called \\textit{Runge-Kutta methods}. Another second order Runge-Kutta method is the modified Euler method: \n\n$y_{i+1} = y_i + \\frac{h}{2}[ f(x_i, y_i) + f(x_{i+1}, y_i+ hf(x_i, y_i))]$\n\nThere are many Runge-Kutta methods with varying orders of accuracy. In practice, the methods of order four, such as the following, are most commonly used: \n\\begin{enumerate}\n\\item Apply initial conditions.\n\\item $K_1 = f(x_i,y_i),$\n\\item $K_2 = f(x_i + \\frac{h}{2}, y_i + \\frac{h}{2} K_1),$\n\\item $K_3 = f(x_i + \\frac{h}{2} , y_i + \\frac{h}{2} K_2),$\n\\item $K_4 = f(x_{i+1} , y_i + h K_3),$\n\\item $y_{i+1} = y_i + \\frac{h}{6}(K_1 + 2K_2 + 2K_3 + K_4)$\n\\end{enumerate}\n\n\\begin{comment}\n\\begin{problem}\nSuppose a differential equation is given by\n\\[ y' = f(t).\\]\nIf we were to solve this by just integrating, which integration approximation method would correspond to Euler's method, backward Euler's method, modified Euler's method, the Midpoint method, and the fourth order Runge-Kutta method (RK4)? \n\\end{problem}\n\\end{comment}\n\nNotice that these methods are doing a type of quadrature where we are sampling the function at different points and then performing computation using the samples and some inherent weights.\n\n\\begin{problem} Consider the IVP given by \n\\begin{align*}\ny' + y &= 2-2x,\\,\\, 0 \\leq x \\leq 2, \\\\\ny(0) &= 0.\n\\end{align*}\nUse Euler's method, the Midpoint method, and RK4 to approximate the value of the solution at $x = 2$, with a stepsize of $h = 0.2,$ $ 0.1,$ $0.05 $, $0.025,$ and $0.0125.$ Create a log-log plot of the relative error of each approximation using the \\li{loglog} function in \\li{matplotlib}.(see Figure \\ref{ivp:relative_error}).\n\\end{problem}\n\n% \\begin{problem}\n% Plot the solutions of \n% \\[ y' + y = 2-2x\\,\\, 0 \\leq x \\leq 2, \\] \n% with initial conditions $y(0) = 0, 2, 4, 6, $ and $8$. Use RK4 to compute the solutions. \n% \\end{problem}\n\n\\section*{Harmonic Oscillators and Resonance} Harmonic oscillators show up often in classical mechanics. \nA few examples include the pendulum (with small \n displacement), spring-mass systems, and the flow of electric current through various types of circuits. \nA harmonic oscillator can be described by an initial value problem of the form \n\\begin{align*}\n\tmy'' + \\gamma y' + ky &= f(t) ,\\\\\n\ty(0) &= y_0,\\\\\n\ty'(0) &= y'_0.\n\\end{align*}\n\nWe will describe the construction of this mathematical model in the context of a spring-mass system.\n\nSuppose an object with mass $m$ is placed at the end of a horizontal spring. \nThe natural position of the object is called the \\textit{equilibrium position} for the system.\nIf the object is displaced from its equilibrium position and given an initial velocity, \nit will act like a harmonic oscillator.\nThe principal property of a harmonic oscillator $y(t)$ is that once $y$ leaves its equilibrium value $y = 0$, it experiences a restoring force $F_r = -ky.$ \nThis force pushes $y$ back towards its equilibrium. \nHooke's law says that this holds true for a \nspring-mass system if the displacement $y$ is small.\n\nOften there is an additional damping force $F_d$, often due to some type of friction. This force is usually proportional to the $y'$, is always in the opposite direction of $y'$ (the \\emph{velocity}. You can think of it as drag), and represents energy leaving the system. \nThus we have $F_d = -\\gamma y', $ where $ \\gamma \\geq 0$ is constant. \nWe may also need to consider an additional external force $f(t)$, or a driving force, that is interacting with our spring-mass system.\n\nBy using Newton's law we obtain\n\\begin{align*}\nma &= F = F_r + F_d + f(t),\\\\\nmy'' &= -ky -\\gamma y' + f(t).\n\\end{align*}\n\n\\section*{Simple harmonic oscillators}\nA simple harmonic oscillator is a harmonic oscillator that is not damped ($\\gamma =0$), and is free ($f(t) \\equiv 0$) rather than forced ($f(t) \\not = 0$). A simple harmonic oscillator can described by the IVP\n\\begin{align*}\nmy'' + ky &= 0,\\\\\ny(0) &= y_0,\\\\\ny'(0) &= y_0'.\n\\end{align*} \nThe solution of this IVP is $y = c_1\\cos (\\omega_0 t) + c_2 \\sin (\\omega_0 t)$ where $\\omega_0 = \\sqrt{k/m}$ is the natural frequency of the oscillator and $c_1$ and $c_2$ are determined by applying the initial conditions. This in turn can be written in the form \n\\[y = A\\sin (\\omega_0 t + \\delta) .\\]\n\nTo solve this IVP using the fourth order Runge Kutta method (RK4), we need to write this system in the form \n\\[z'(t) = f(t,z(t)) \\]\nWe can do this by letting $z_1 = y, z_2 = y'$. Then we have \\[ z'= \n \\left[\\begin{array}{c}z_1 \\\\z_2\\end{array}\\right]' = \\left[\\begin{array}{c}z_2 \\\\\\frac{-k}{m}z_1\\end{array}\\right]= f(z).\\]\n\n\n\\begin{problem} Use the RK4 method to solve for the simple harmonic oscillator \ndescribed by \n\\begin{align*}\nmy'' + ky &= 0,\\,\\, 0 \\leq t \\leq 20, \\\\\ny(0) &= 2, \\\\\ny'(0) &= -1,\n\\end{align*} \nfor $m = 1$ and $k =1$. Note that in your implementation of RK4, the constants $K_1, K_2, K_3,$ and $K_4$ become vectors with $n$ entries, where $n$ is the number of equations in the first-order system. \n\nPlot your solution $y(t)$. Compare this with the solution of the IVP if $m = 3$ and $k =1$. Consider: Why does the difference in solutions make sense physically?\n\\end{problem}\n\n\n\\section*{Damped free harmonic oscillators} We now consider damped free harmonic oscillators. These systems are described by the differential equation\n\\[my''(t) +\\gamma y'(t) + ky(t) = 0.\\]\nFor fixed values of $m$ and $k$, it is interesting to study the effect of the damping coefficient $\\gamma$. \n\nThe roots of the characteristic equation are \\[r_1,r_2 = \\frac{-\\gamma \\pm \\sqrt{\\gamma^2 -4km}}{2m} .\\]\nNote that the real parts of $r_1$ and $r_2$ are always negative, and so any solution $y(t)$ will decay over time due to a dissipation of the system energy. There are several cases to consider for the general solution of this equation: \n\\begin{enumerate}\n\\item If $\\gamma^2 > 4km$, then the general solution is $y(t) = c_1 e^{r_1t} + c_2e^{r_2t}$. Here the system is said to be $\\textit{overdamped}$. Notice from the general solution that there is no oscillation in this case.\n\\item If $\\gamma^2 = 4km$, then the general solution is $y(t) = c_1 e^{\\gamma t/2m} + c_2 te^{\\gamma t/2m}$. Here the system is said to be $\\textit{critically damped}$. \n\\item If $\\gamma^2 < 4km$, then the general solution is \n\\begin{align*}\ny(t) &= e^{-\\gamma t/2m} \\left[c_1\\cos(\\mu t) + c_2 \\sin (\\mu t)\\right],\\\\\n&= R e^{-\\gamma t/2m} \\sin (\\mu t + \\delta),\n\\end{align*}\nwhere $R$ and $\\delta$ are fixed, and $\\mu = \\sqrt{4km-\\gamma^2}/2m.$ This system does oscillate.\n\\end{enumerate}\n\n\\begin{problem}\nUse RK4 method to solve for the damped free harmonic oscillator \n\\begin{align*}\ny'' +\\gamma y'+ y &= 0, \\,\\, 0 \\leq x \\leq 20,\\\\\ny(0) &= 1, \\\\\ny'(0) &= -1.\n\\end{align*} \nFor $\\gamma = 1/2,$ and $\\gamma = 1$, simultaneously plot the solutions $y(t)$ and find $y(20)$ accurate to four significant digits. (Check that the relative error is less than $5 \\cdot 10^{-5}$.) How many subintervals do you need?\n\\end{problem}\n\n\\section*{Forced harmonic oscillators without damping}\nLet's look at the systems described by the differential equation\n\\begin{align}\nmy''(t) + ky(t) &= F(t). \\label{Forced_harm_osc}\n\\end{align}\nIn many instances the external force $F(t)$ is periodic, so let us assume that $F(t) = F_0 \\cos(\\omega t)$. If $\\omega_0 = \\sqrt{k/m} \\not = \\omega,$ then the general solution of \\ref{Forced_harm_osc} is given by \n\\[y(t) = c_1 \\cos (\\omega_0 t) + c_2\\sin (\\omega_0 t) + \\frac{F_0}{m(\\omega_0^2 - \\omega^2)} \\cos (\\omega t).\\]\nIf $\\omega_0 = \\omega$, then the general solution is \n\\[y(t) = c_1 \\cos (\\omega_0 t) + c_2\\sin (\\omega_0 t) + \\frac{F_0}{2m\\omega_0} t \\sin (\\omega_0 t).\\]\nThis last solution contains a term that grows arbitrarily large as $t \\to \\infty$. \nIf we included damping then the solution would be bounded, but will still be large for small $\\gamma$ and $\\omega$ close to $\\omega_0$. \nConsider a physical spring-mass system. \nEquation \\ref{Forced_harm_osc} holds only for small oscillations (this is where Hooke's law is applicable). \nFor larger oscillations, this equation will not hold. \nHowever, the fact that the equation predicts large oscillations suggests the spring-mass system could fall apart as a result of the external force. Mechanical resonance has \nbeen known to cause failure of bridges, buildings, and airplanes.\n\n\\begin{problem}\nUse the RK4 method to solve for the undamped forced harmonic oscillator\n\\begin{align*}\n2y'' + \\gamma y' + 2y &= 2 \\cos (\\omega x), \\,\\, 0 \\leq x \\leq 40,\\\\\ny(0) &= 2, \\\\\ny'(0) &= -1.\n\\end{align*} \nFor the following values of $\\gamma$ and $\\omega,$ plot the solution $y(t)$ and find $y(40)$ correct to four decimal places: \n\\begin{align*}\n\t(\\gamma, \\omega) &= (0.5, 1.5),\\\\\n\t &= (0.1, 1.1), \\\\\n\t&= (0.0, 1.0).\n\\end{align*}\n\\end{problem}\n\n", "meta": {"hexsha": "59ab80cac6ae83dcf632523c45a219d59648ae33", "size": 14653, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Labs/IVP/IVP.tex", "max_stars_repo_name": "rachelwebb/numerical_computing", "max_stars_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Labs/IVP/IVP.tex", "max_issues_repo_name": "rachelwebb/numerical_computing", "max_issues_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Labs/IVP/IVP.tex", "max_forks_repo_name": "rachelwebb/numerical_computing", "max_forks_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-12-08T01:19:23.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-08T01:19:23.000Z", "avg_line_length": 57.0155642023, "max_line_length": 420, "alphanum_fraction": 0.691940217, "num_tokens": 4722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.727975460709318, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.6337734561163697}} {"text": "\\chapter{Centrality measures}\n\\section{Some famous centrality measures}\nThe centrality of a node is a common problem of various fields such as biology, computer science, sociology, psychology. There are some old and popular measures to compute the centrality:\n\\begin{itemize}\n\t\\item Closeness\n\t\\item Betweenness\n\\end{itemize}\nNext we examine some algorithms for computing them\n\\section{Closeness}\n\\subsection{Definition}\nIn words, the closeness says how a node is distant from the other nodes.\\\\\nThe formula for calculate is:\n\\begin{align*}\nc(v) = \\dfrac{n-1}{\\sum_{u \\in V} d(v,u)}\n\\end{align*}\nA variant of this definition for non strongly connected components is\n\\begin{align*}\nc(v) = \\dfrac{r(v) -1}{f(v)} \\dfrac{r(v)-1}{n-1} = \\dfrac{(r(v) - 1)^2}{(n-1)f(v)}\n\\end{align*}\n where $ r(v) $ is the set of reachable nodes from v. If the graph is strongly connected we have the previous definition, in fact\n \\begin{align*}\n \t\\dfrac{(n - 1)^2}{(n-1)f(v)} = \\dfrac{n-1}{f(v)}\n \\end{align*}\n \\subsection{A simple approach}\n The initial idea for computing the closeness is doing a BFS from all nodes to obtain all the distances but this have time complexity $ O(nm) $ which is too much for real graph.\n \\section*{The pruned breadth first search}\n A similar problem is find the top k nodes with lower closeness but for doing this we need before the pruned breadth first search.\\\\\n \\subsection{The upper bound on closeness}\n The pruned breadth first search use a lower bound on the farness of the nodes from v, that is equivalent to an upper bound of the closeness.\\\\\n Let's define the following terms:\n \\begin{itemize}\n \t\\item $ \\Gamma_d(v) = $ set of nodes at level \\textit{d} of the BFS tree starting from \\textit{v}\n \t\\item $ \\gamma_d(v) = |\\Gamma_d(v)| $\n \t\\item $ f_d(v) = \\sum_{i=1}^{d} i \\cdot \\gamma_i(v) $\n \t\\item $ n_d(v) = \\sum_{i=1}^{d} \\gamma_i(v) $\n\\end{itemize}\nNow let's define a lower bound on closeness:\n\\begin{center}\n\t$ f(v) \\geq f_d(v) + (d+1) \\cdot \\gamma_{d+1}(v) + (d+2)(r(v) - n_{d+1}(v))$\n\\end{center}\nSince $ n_{d+1}(v) = \\gamma_{d+1}(v) + n_d(v) $ we have\n\\begin{align*}\n\t f(v) &\\geq f_d(v) + (d+1) \\cdot \\gamma_{d+1}(v) + (d+2)\\left[r(v) - (\\gamma_{d+1}(v) + n_d(v))\\right] \\\\\n\t &= f_d(v) + (d+1) \\cdot \\gamma_{d+1}(v) - (d+2) \\cdot \\gamma_{d+1}(v) + (d+2)(r(v) - n_d(v)) \\\\\n\t &= f_d(v) - \\gamma_{d+1}(v) + (d+2)(r(v) - n_d(v)) \n\\end{align*}\nWe must consider that at the level \\textit{d} we don't know how much is $ \\gamma_{d+1}(v) $ so we use an upper bound that is the sum of degree of nodes on level \\textit{d}:\n\\begin{align*}\n\t\\gamma_{d+1}(v) \\leq \\sum_{u \\in \\Gamma_d(v)} degree(u) := \\tilde{\\gamma}_{d+1}(v)\n\\end{align*}\nSo we use the latter approximation on lower bound of farness $ f(v) $ \n\\begin{align*}\n\t f(v) &\\geq \\tilde{f}_d(v, r(v)) := f_d(v) - \\tilde{\\gamma}_{d+1}(v) + (d+2)(r(v) - n_d(v))\n\\end{align*}\nthat is equivalent to an upper bound on closeness:\n\\begin{align*}\n\tc(v) \\leq \\dfrac{\\left[r(v) - 1 \\right]^2}{(n-1) \\cdot \\tilde{f}_d(v, r(v)) } := \\tilde{c}_d(v) \n\\end{align*}\n\\subsubsection{How to prune the breadth first search}\nThe pruned breadth first search receive in input the following arguments:\n\\begin{itemize}\n\t\\item \\textit{v}: the starting node of pruned BFS\n\t\\item $ x_k $: the k-th biggest closeness value found until now (0 if not are processed at least k vertices)\n\\end{itemize}\nthen at each level \\textit{d} of \\textit{BFS} tree we check if $ x_k > \\tilde{c}_d(v) \\geq c_d(v) $, and, if it is true, it means that it is not in the \\textit{k} lowest values of closeness in the graph, so we can stop the \\textit{BFS}.\n\\subsection{Compute the upper bound on closeness}\nOnce we have defined the algorithm in the latter section, then it remains one thing: calculate $ r(v) $ for each node.\\\\\nWe have four different cases:\n\\begin{itemize}\n\t\\item Graph indirected and strongly connected: $ r(v) \\rightarrow n \\,\\,\\forall v \\in V $\n\t\\item Graph indirected but not strongly connected: A linear time scan graph is needed then set $ r(v) = |SCC| : v \\in SCC \\,\\, \\forall v \\in V $ that mean set $ r(v) $ to the size of strongly connected component where it belongs\n\t\\item Graph directed and strongly connected : $ r(v) \\rightarrow n \\,\\,\\forall v \\in V $\n\t\\item Graph directed but not strongly connected : see below\n\\end{itemize}\n\\subsubsection{Graph directed but not strongly connected}\nSuppose that $ r(v) $ is included in the range $ [\\alpha(v) , \\omega(v)] $ a lower bound for $ \\dfrac{1}{c(v)} $ is obtained simply bi reverting the upper bound in the latter section:\n\\begin{align*}\n \\dfrac{1}{c(v)} \\geq \\dfrac{(n-1) \\cdot \\tilde{f}_d(v, r(v))}{(r(v) -1)^2}\n\\end{align*}\nThen if we define $ a = d+2 $ and $ b = \\tilde{\\gamma}_{d+1}(v) + a(n_d(v) -1) - f_d(v) $ we have that\n\\begin{align*}\n\tf(v) \\geq \\tilde{f}_d(v, r(v)) &= f_d(v) - \\tilde{\\gamma}_{d+1}(v) + (d+2)(r(v) - n_d(v)) \\\\\n\t&= f_d(v) - \\tilde{\\gamma}_{d+1}(v) + a \\cdot (r(v) - n_d(v)) \\\\\n\t&= f_d(v) - \\tilde{\\gamma}_{d+1}(v) + a \\cdot (r(v) - n_d(v) + 1 - 1) \\\\\n\t&= f_d(v) - \\tilde{\\gamma}_{d+1}(v) + a \\cdot (r(v) - 1)) - a \\cdot n_d(v) + a\\\\\n\t&= f_d(v) - \\tilde{\\gamma}_{d+1}(v) + a \\cdot (r(v) - 1)) + a (- n_d(v) + 1)\\\\\n\t&= a \\cdot (r(v) - 1)) - \\tilde{\\gamma}_{d+1}(v) - a ( n_d(v) -1) + f_d(v)\\\\\n\t&= a \\cdot (r(v) - 1)) - b\n\\end{align*}\n\nso we can write the lower bound as \n\\begin{align*}\n\\dfrac{1}{c(v)} \\geq \\dfrac{(n-1) \\cdot \\tilde{f}_d(v, r(v))}{(r(v) -1)^2} =(n-1)\\dfrac{a \\cdot (r(v) - 1)) - b}{(r(v) -1)^2}\n\\end{align*}\nNow consider the functions in the form $ g(x) = \\dfrac{ax - b}{x^2} $ and the derivative $ g'(x) = \\dfrac{-ax + 2b}{x^3} $: we can easily see that we have a zero in the derivative with $ x = \\dfrac{2b}{a} $ and it is a local maximum because x is positive in range $ 0 < x < \\frac{2b}{a} $ and negative in $ x > \\frac{2b}{a} $. Then this it implies that for $ x > 0 $ there aren't local minimum so for any interval $ [x_1, x_2] $ the local minimum in the interval is $ x_1 $ or $ x_2 $ that in formula is $ \\forall x \\in [x_1, x_2] \\,\\, g(x) \\geq \\min(g(x_1), g(x_2))$\n\\begin{figure}[H]\n\t\\includegraphics[width=0.7\\linewidth]{img/local_maximum}\n\t\\caption{Plot of $ g(x) = \\dfrac{ax -b}{x^2} $ with $ a = b = 1 $}\n\t\\label{fig:localmaximum}\n\\end{figure}\nReturning to the lower bound of $ \\frac{1}{c(v)} $ we can use the same reasoning with $ (n-1)\\dfrac{a \\cdot (r(v) - 1)) - b}{(r(v) -1)^2} $ so we can say that \n\\begin{align*}\n\\dfrac{1}{c(v)} \\geq (n-1) \\cdot \\dfrac{\\tilde{f}_d(v, r(v))}{(r(v) -1)^2} \\geq (n-1) \\cdot \\min\\left( \\dfrac{\\tilde{f}_d(v, \\alpha(v))}{(\\alpha(v) -1)^2}, \\dfrac{\\tilde{f}_d(v, \\omega(v))}{(\\omega(v) -1)^2} \\right)\n\\end{align*}\n\n\\subsection{Compute $ \\alpha(v) \\text{ and } \\omega(v) $}", "meta": {"hexsha": "50ae9718ed649d9616e0619ef82e7031b6951046", "size": 6712, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/centrality.tex", "max_stars_repo_name": "Michedev/AAGM_resume", "max_stars_repo_head_hexsha": "31bcd6b58a39b19aa03a5aa13f8ad3e4d8f21ff3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/centrality.tex", "max_issues_repo_name": "Michedev/AAGM_resume", "max_issues_repo_head_hexsha": "31bcd6b58a39b19aa03a5aa13f8ad3e4d8f21ff3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/centrality.tex", "max_forks_repo_name": "Michedev/AAGM_resume", "max_forks_repo_head_hexsha": "31bcd6b58a39b19aa03a5aa13f8ad3e4d8f21ff3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.320754717, "max_line_length": 569, "alphanum_fraction": 0.6373659118, "num_tokens": 2509, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754489059775, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.6337734287313743}} {"text": "\\section{Representations of Counterpoint and Harmony}\n\\label{sec:cpharmony}\n\nHarmony refers to simultaneously sounding tones (chords) and harmonic progression\na series of chords. It is in essence the of dual of counterpoint. A\nfundamental representation used in Music Tools is a grid of\n\\texttt{Point}s, where a \\texttt{Point} represents either a pitch, a\nhold of a previous pitch, or a rest.\n\n\\begin{alltt}\ndata Point : Set where\n tone : Pitch \\(\\rightarrow\\) Point\n hold : Pitch \\(\\rightarrow\\) Point\n rest : Point\n\\end{alltt}\n\nMusic is thus considered to be quantized to some minimal duration,\nwith the horizontal dimension representing time and the vertical\ndimension representing pitch. A slice in the horizontal direction is a\n\\texttt{Melody}, represented as a vector of points of some length $n$. Similarly a slice\nin the vertical direction is a \\texttt{Chord}, again of some fixed length.\n\n\\begin{alltt}\ndata Melody (n : \\(\\mathbb{N}\\)) : Set where\n melody : Vec Point n \\(\\rightarrow\\) Melody n\n\ndata Chord (n : \\(\\mathbb{N}\\)) : Set where\n chord : Vec Point n \\(\\rightarrow\\) Chord n\n\\end{alltt}\n\nOne can think of the grid as several melodies in parallel of the same\nduration. This is essentially counterpoint, and thus given that\nname. Note the crucial use of dependent types here to ensure all\nmelodies have the same duration $d$. The counterpoint is specified to\nhave $v$ voices.\n\n\\begin{alltt}\ndata Counterpoint (v : \\(\\mathbb{N}\\)) (d : \\(\\mathbb{N}\\)): Set where\n cp : Vec (Melody d) v \\(\\rightarrow\\) Counterpoint v d\n\\end{alltt}\n\nDually one can think of the grid as a series of chords, in another\nword a harmonic progression, although this is named \\texttt{Harmony}\nin the library for conciseness. Again dependent types enforce every\nchord has the same number of voices.\n\n\\begin{alltt}\ndata Harmony (v : \\(\\mathbb{N}\\)) (d : \\(\\mathbb{N}\\)): Set where\n harmony : Vec (Chord v) d \\(\\rightarrow\\) Harmony v d\n\\end{alltt}\n\nConverting between \\texttt{Counterpoint} and \\texttt{Harmony} is\nsimply matrix transposition. These are fundamental representations\nthat can be used directly for analysis, for example, and there are\nfunctions to convert them to MIDI for sound generation.\n\nThere are many other ways to represent music, which have advantages in\ncertain contexts. For example in Section~\\ref{sec:cp}\n\\texttt{PitchInterval} and \\texttt{PitchInterval2} were used instead\nof \\texttt{Counterpoint} since they more explictly represent the\nstructure of the music in species counterpoint. It is straightforward\nto convert between the representations, and worth using the most\nappropriate type for a given situation.\n", "meta": {"hexsha": "54cef406d886bee88b665c29dffc512bb9699bb3", "size": 2635, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/icfp20/cpharmony.tex", "max_stars_repo_name": "halfaya/MusicTools", "max_stars_repo_head_hexsha": "04896c61b603d46011b7d718fcb47dd756e66021", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 28, "max_stars_repo_stars_event_min_datetime": "2017-04-21T09:08:52.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-04T18:04:07.000Z", "max_issues_repo_path": "doc/icfp20/cpharmony.tex", "max_issues_repo_name": "halfaya/MusicTools", "max_issues_repo_head_hexsha": "04896c61b603d46011b7d718fcb47dd756e66021", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-11-13T01:26:20.000Z", "max_issues_repo_issues_event_max_datetime": "2020-11-17T00:58:55.000Z", "max_forks_repo_path": "doc/icfp20/cpharmony.tex", "max_forks_repo_name": "halfaya/MusicTools", "max_forks_repo_head_hexsha": "04896c61b603d46011b7d718fcb47dd756e66021", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-01-12T17:02:36.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-10T04:04:40.000Z", "avg_line_length": 41.171875, "max_line_length": 88, "alphanum_fraction": 0.7616698292, "num_tokens": 685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637577007394, "lm_q2_score": 0.7371581568543043, "lm_q1q2_score": 0.6337081511411223}} {"text": "\\chapter{Graphs}\n\n\\pgfplotsset{width=1\\columnwidth,compat=1.9}\n\n\\begin{tikzpicture}\n\\begin{axis}\n\\addplot[color=red]{exp(x)};\n\\end{axis}\n\\end{tikzpicture}\n\n\\begin{tikzpicture}\n\\begin{axis}\n\\addplot3[surf,]\n{exp(-x^2-y^2)*x};\n\\end{axis}\n\\end{tikzpicture}\n\n\\hskip 10pt\n\n\\begin{tikzpicture}\n\\begin{axis}[\naxis lines = left,\nxlabel = $x$,\nylabel = {$f(x)$},\n]\n%Below the red parabola is defined\n\\addplot [\ndomain=-10:10, \nsamples=100, \ncolor=red,\n]\n{x^2 - 2*x - 1};\n\\addlegendentry{$x^2 - 2x - 1$}\n%Here the blue parabloa is defined\n\\addplot [\ndomain=-10:10, \nsamples=100, \ncolor=blue,\n]\n{x^2 + 2*x + 1};\n\\addlegendentry{$x^2 + 2x + 1$}\n\n\\end{axis}\n\\end{tikzpicture}\n\n", "meta": {"hexsha": "dcbbdd68fd202424d492be404366dceba8cddf75", "size": 664, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "graph.tex", "max_stars_repo_name": "christianhauschel/latex-summary", "max_stars_repo_head_hexsha": "54a14ecadc17209ddd6eafe2ffa9211538355570", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2019-05-01T08:11:48.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-11T19:21:00.000Z", "max_issues_repo_path": "graph.tex", "max_issues_repo_name": "ch101112/latex-summary", "max_issues_repo_head_hexsha": "2661049d27bb8f6fc6dbc64594dad1635003bbf1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "graph.tex", "max_forks_repo_name": "ch101112/latex-summary", "max_forks_repo_head_hexsha": "2661049d27bb8f6fc6dbc64594dad1635003bbf1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 14.4347826087, "max_line_length": 44, "alphanum_fraction": 0.6596385542, "num_tokens": 278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637469145054, "lm_q2_score": 0.7371581568543044, "lm_q1q2_score": 0.633708143189962}} {"text": "\\chapter{Feedforward Neural Network}\n\nStarting from the late 2000's, neural networks have been experiencing an exciting resurgence in the big data era. Given recent improvements in processing speeds, large-scale deep neural networks (DNNs), a.k.a. deep learning, are able to deliver highly accurate results when tackling several problem domains, including computer vision, natural language processing, speech recognition, and other AI challenges. For instance, the systems using DNNs have reported breakthroughs in object recognition accuracy on the ImageNet dataset, and have even achieved human-level performance for face recognition. Motivated by these encouraging results, there has been a growing interest from both academia and industry (Google, Facebook, Amazon, etc.), both of whom have been devoted significant resources to investigate, improve and promote exploration into DNNs. DNNs are revolutionizing a number of traditional and emerging real-world applications, such as self-driving systems, automatic machine translations, drug discovery and toxicology.\n\nIn this chapter we will introduce the basics of deep feedforward neural networks, define related terminologies, and analyze how to train and use neural network model.\n\\section{Feedforward Networks}\nA \\textit{network model} is a model expressed through compositions of function, where the composition relations forms an directed acyclic graph. Figure \\ref{fig1} gives an example of a network model. In this example, $f_1$ and $f_3$ are composited by $f_2$ and $f_4$ separately, which in turn is composited $f_5$. Thus the expression of the entire model is\n\\begin{align*}\n\\label{enetwork}\nf(\\bm{x}) = f_5(f_2(f_1(\\bm{x})), f_4(f_3(\\bm{x})))\n\\end{align*}\nIn a network model, the input vector is denoted as $\\bm{x}\\in\\mathbb{R}^m$. If the information flows through the function that evaluates $\\bm{x}$, through each of the subsequent function, and finally goes to output vector $\\bm{y}\\in\\mathbb{R}^n$. Such network is called a \\textit{feedforward network}, as there is no feedback connections in which outputs of the model are fed back into itself.\n\n\\begin{wrapfigure}{r}{0.3\\textwidth}\n\\centering\n\\includegraphics[width=0.3\\textwidth]{Fig1}\n\\caption{Feedforward Network}\n\\label{fig1}\n\\end{wrapfigure}\n\n\n\nFunctions may form one or more layers according to which values they take as inputs (cf. Figure \\ref{fig2}). The functions that directly take input $\\bm{x}$ form the \\textit{input layer}, functions that take values from input-layer functions form the \\textit{first hidden layer}, and so on, until the last layer (call \\textit{output layer}) which returns the output of the entire model. The total number of layers is called \\textit{the depth} of the network. The \\textit{width} of a layer is defined to be the number of functions on the layer. The functions are sometimes referred to as units, neurons, or activation functions, depending on the context. Nowadays neural networks with more than 1000 layers were used for tasks in computer vision, which justifies the name \"deep learning\". The training data do not provide information on what values the intermediate layers should produce, thus these layers are referred to as \\textit{hidden layers}.\n\nThe information usually flow into a function from multiple sources. This is typically achieved by a weighted summation of the input values, possibly added with a bias value. For a single unit with activation function $\\sigma$, the relation between the input vector $\\bm x$ and the output vector $\\bm y$ is given in the form of\n\\begin{equation}\n \\bm y = \\bm \\sigma(\\bm{W^Tx}+\\bm\\theta).\n\\end{equation}\nHere $\\bm W$ is a matrix of parameters representing the weighted summation of the input vector, $\\bm\\theta$ is a vector representing the bias value, and $\\bm\\sigma$ is the function that applies the activation function $\\sigma$ to each of its input coordinates. For the ease of notation, we use new notations\n$\\bm x' = (\\bm x^T, 1)^T$ and $\\bm W' = (\\bm W, \\bm\\theta^T)^T$ and simplify the notation above as\n\\begin{equation}\n \\bm y = \\bm \\sigma(\\bm{(W')^Tx'}).\n\\end{equation}\nFrom now on we will drop the primes and stick with the simplified notations. As an example, the overall expression of the model illustrated in Figure \\ref{fig2} can be expressed as\n\n\\begin{align*}\n\\bm{y} = \\bm{W_3^T\\cdot\\sigma(W_2^T\\cdot\\sigma\n(W_1^Tx))}\n\\end{align*}\n\nThe weight matrix related to the input of $j$-th layer is referred to as the weight matrix \\textit{of that layer}. Every layer except for the input layer has a corresponding weight matrix.\n\n\\begin{wrapfigure}{r}{0.5\\textwidth}\n\\centering\n\\includegraphics[width=0.5\\textwidth]{Fig2}\n\\caption{Feedforward Network}\n\\label{fig2}\n\\end{wrapfigure}\n%{\\bf complicated network}\n%Not every network model of interest is a layered feedforward network, as the topology of an acyclic graph can be quite complicated. Note that since there is no loop in the graph, one can still find a suitable order to let the information flow from input to output. Also there are network models that do contain directed cycles. An important class is called \\textit{recurrent neural network}. We will study recurrent neural network at a later stage of the seminar.\n\nThe goal of a feedforward network is to approximate a function $g(\\bm{x})$. More precisely, the goal is to find an appropriate set of weights so that the function representing the network is close to the target function $g$ (under certain measure). The challenge is to find a network model and a training method that is capable to well approximate the target function with limited memory space and computational power. The purpose the the next two sections is to introduce the basic setting of linear network models and neural network models, analyze their training procedures, and discuss the challenges of applying these models. \n\n\\section{Linear Feedforward Neural Network}\nOne way to understand feedforward networks is to begin with linear models and consider how to overcome their limitations. Linear models, such as logistic regression and linear regression, are appealing because they may be fit efficiently and reliably, either in closed form or with convex optimization. On the other hand, linear models also have the obvious defect that the model capacity is limited to linear functions, so the model cannot understand the interaction between any two input variables.\n\n{\\bf Linear Regression Model}. Linear Regression is the simplest form of regression. We model our system with a linear combination of features to produce one output. That is,\n\\begin{equation}\n f(\\bm x, \\bm w) = \\bm w^T\\bm x\n\\end{equation}\nHere $\\bm x$ is the input vector, and $w_0$, $\\bm w$ are parameters. Our task is then to find the values for the parameters that provide the best fit to our training data. One way to measure the fitness of the parameters is to calculate the least squares error (loss function) over our dataset $S = \\{(\\bm x^1, y^1), ..., (\\bm x^N, y^N)\\}$\n\\begin{equation}\n L(S, \\bm w) = \\frac{1}{2}\\sum_{i=1}^N(f(\\bm x^i, \\bm w)-y^i)^2\n\\end{equation}\n\nIn order to find the best fit, we must minimize $L(S, \\bm w)$. The approach using gradient descent takes the following form:\n\\begin{equation}\n \\bm w \\leftarrow \\bm w-\\eta\\nabla_{\\bm w}L(S, \\bm w)\n\\end{equation}\nThe step size $\\eta$ may be application-dependent. The gradient of the loss function explicitly is expressed as\n\\begin{equation}\n \\nabla_{\\bm w}L(S, \\bm w) = \\sum_i(\\bm w^T\\bm{x^i} - y^i)\\bm x^i = (\\bm{w^TX-y})\\cdot\\bm X\n\\end{equation}\nHere $\\bm X = (\\bm x^1,...,\\bm x^N)$ and $\\bm y=(y^1,...,y^N)$.\nNote that there exists a closed form solution for linear regression, thus this gradient descent method is rarely used in practice.\n\n{\\bf Logistic regression model}. Another example on linear feedforward network is logistic regression model. We model our system as\n\\begin{equation}\n% \\tanh(\\bm w^T \\bm x + b), \\textit{ where } \\tanh(a) = \\frac{1-\\exp(a)}{1+\\exp(a)}\n\\frac{1}{1+\\exp(\\bm w^T \\bm x)}\n\\end{equation}\nWith training set $S = \\{(\\bm x^1, y^1), ..., (\\bm x^N, y^N)\\}$, the loss function is defined by the negative log-likelihood\n\\begin{equation}\n L(\\bm w) = \\sum_{i=1}^N\\log(1+\\exp(-y^i(\\bm w^T\\bm x^i)))\n\\end{equation}\nThe loss function is minimized by the gradient descent\n\\begin{equation}\n \\bm w = \\bm w-\\eta\\nabla_{\\bm w}L(\\bm w)\n\\end{equation}\nwith $\\nabla_{\\bm w}L(\\bm w) = -\\sum_{i=1}^N y^i\\frac{\\exp(-y^i(\\bm w^T\\bm x^i+b))}{1+\\exp(-y^i(\\bm w^T\\bm x^i+b))}\\bm x^i$.\n\nThe convexity of the loss function guarantees that the gradient descent algorithm will find the minimum value.\n\n{\\bf Summary}.The key elements of a network model can be summarized as follows: \n\\begin{enumerate}\n \\item A function representing a network structure.\n \\item A loss function.\n \\item An optimization algorithm that minimizes the value of the loss function.\n\\end{enumerate}\n\n{\\bf Limitations of Linear Network Model}. One important application of network models is the classification problem. A linear network is only capable of separating the dataset with hyperplanes. We are going to show a classical result on the limited capability of linear network models on classifying $p$ points in general position.\n\n\\begin{wrapfigure}{r}{0.4\\textwidth}\n\\centering\n\\includegraphics[width=0.4\\textwidth]{nonlinearseparable}\n\\caption{A non-linear dichotomy}\n\\label{fig4}\n\\end{wrapfigure}\n\nWhen $n$ points are partitioned into two classes, we call such classification a \\textit{dichotomy}. Moreover, we use the term \\textit{linearly separable dichotomies} to represent the dichotomies that are realizable by a linear classifier (the boundary between the classes is a hyperplane). Figure \\ref{fig4} shows an example of a non-linear dichotomy, as no straight line can separate the two blue points from the two red ones. In general, suppose that we have $p$ points in general position in $\\mathbb{R}^N$. There are $2^p$ ways to partition these $p$ points into two classes, i.e., the $p$ points can form $2^p$ different dichotomies. The following theorem shows that the proportion of linear separable dichotomies will converge to 0 as $p\\rightarrow\\infty$. \n\n% Cover's function counting theorem (1966)\n% Suppose we have $p$ points in $\\mathbb{R}^N$. Consider all possible partitions of these points into two classes. We have $2^p$ such partitions. How many of there partitions yield linearly separable classes, i.e., where the two classes can be perfectly separated by an $(N-1)$-dimensional hyperplane? The only thing we assume about the points is that they are in general position, which means that any subset of $N$ or fewer points is linearly independent. \n\n\n% A dichotomy is a partition of a whole (or a set) into two parts (subsets). In other words, this couple of parts must be jointly exhaustive: everything must belong to one part or the other, and mutually exclusive: nothing can belong simultaneously to both parts.\n% Such a partition is also frequently called a bipartition.\n\n\n% The two parts thus formed are complements. In logic, the partitions are opposites if there exists a proposition such that it holds over one and not the other.\n\n\n% Treating continuous variables or multicategorical variables as binary variables is called dichotomization. The discretization error inherent in dichotomization is temporarily ignored for modeling purposes.\n\n\\begin{theorem}[Cover, 1965]\nLet $C(p, N)$ be the number of dichotomies of $p$ points in $N$-dimensional space that can be separated by a hyperplane. Then\n\\begin{align*}\n\\lim_{p\\rightarrow\\infty}\\frac{C(p,N)}{2^p}=0.\n\\end{align*}\n\\end{theorem}\n\n\\begin{proof}\nFirstly, we will find an explicit expression for $C(p, N)$ through induction. Suppose there are $p$ points located on the plane in general position, and we are trying to add one more point. There are two possibilities:\n\n1) There exists a separating hyperplane that passes through the new point;\n\n2) There does not exist a separating hyperplane that passes through the new point.\n\nIn case 1), No matter which class the new point belongs to, one can always shift the hyperplane infinitesimally so that this point locates on the same side of the hyperplane with other members of the class. Thus in this case each separable partitions in $p$ points gives rise to two separable partitions in $(p+1)$ points. \n\nNow consider case 2). For any separating hyperplane, the new point must be on the same side with the same class of the previous $p$ points, as otherwise one can always construct a separating hyperplane that passes through the new point. Thus in this case each separable partitions in $p$ points will give rise to exactly on separable partitions in $(p+1)$ points. \n\nMoreover, the number of separable partitions in case 1) is equal to $C(p-1, N)$, as one degree of freedom is eliminated by requiring passing through the new point. Therefore we obtain the following recursive expression\n\n\\begin{align*}\nC(p+1, N) = 2C(p,N-1) + [C(p,N)-C(p, N-1)] = C(p,N) + C(p, N-1).\n\\end{align*}\n\nMoreover, we have $C(1, N)=0$ for all $N<1$ and $C(1, N)=2$ for all $N\\ge 1$. By induction we can prove\n\n\\begin{align*}\nC(p+1,N) = 2\\sum_{i=0}^{N-1}\\big(^p_i\\big).\n\\end{align*}\nHere by convention we have $\\big(^p_i\\big)=0$ for $p 0, then, when < s < 2,\n\n/ - 7\" dx = \\ u\\ f- cosec (J7rs)/r (s), y - and, when < s < 1,\n\n/ * cosfto; \\ J (a. .)/p (5), \\addexamplecitation{Euler.}\n\n\\ / .\n\n13. If < /i < 1, prove that\n\njj(l+.t'-cos -rf.r=r(,0 cos (f -1) - +,\n\n\\addexamplecitation{Peterhouse, 1895.}\n\n14. By taking as contour of integration a parabola with its vertex at\nthe origin, derive from the formula\n\n1 r** \"\")\n\n2 sin an \\ the result\n\n1 f\n\nr(a)= - -. / e- ''x\"--' l+:c''-) [Ziim x + a?(.YCcot(-Jc)\n\n2 sui utt J )\n\n+ sin x + a - 2) arc cot ( - x) ] dx, the arc cot denoting an obtuse\nangle.\n\n(Bourguet, Acta Math. i. p. 367.)\n\n15. Shew that, if the real part of a-n is positive and 2 l/a is\nconvergent, then\n\nn = l\n\nd' is convergent when m > 2, where \\//-( ) (2) = -r log r (z). (Math.\nTrip. 1907.)\n\n16. Prove that\n\n\"jo 1'\n\nd\\ ogr z)\\ /\"e- g-e-g'\n\n< 2\n\n= r (l+a)-i-(l+a)- ---), Jo a\n\n/ 1 .2 - 1 \\ ] = / dx-y. \\addexamplecitation{Legendre.}\n\nJo x-l\n\n%\n% 261\n%\n\n17. Prove that, when R z)>0,\n\n\\ ogr z)=r i- -xiz-l)] f- . \\addexamplecitation{Bmet.}\n\n18. Prove that, for all values of z except negative real values,\n\nlog r (2) = (3 - 1) log 2 - 2 + Uog (27r)\n\nJ 1 1 2 J; \\ 3\\ = J ]\n\n19. Prove that, when (2) > 0,\n\n-rlogr(2) = log2- / r j l-a.-+log.r .\n\ndz ° ° Jo (l-.r)log.r\n\n20. Prove that, when R (2) > 0,\n\nc/2' °\n\n21. If\n\nshew that\n\n/2+1 logr(Oo? = ?',\n\nand deduce from \\hardsubsectionref{12}{3}{3} that, for all values of 2 except negative real\nvalue.s,\n\nu = z log z - z + i log (2n-).\n\n\\addexamplecitation{Raabe, Jovnial fur Math, xxv.}\n\n22. Prove that, for all wilues of 2 except negative real values,\n\nsin 2nTrx\n\n00 f dx' logr(2) = (2-i)log2-2 + Uog(27r)+ 2 / --\n\nn=lj *+'\n\n23. Prove that\n\n(Bourguet*.\n\nBip,p)B p+h,p+h)= y \\addexamplecitation{Binet.}\n\n24. Prove that, when -(\\,\n\nB p, q) + Bip + \\, q) + B p + 2, q) + ...=B p, q- ) .\n\n26. Prove that, when p-a>0,\n\nB p-ci,q) aq a(a + l) g(g + l)\n\n (i, ?) ' it;4-?'*' 1. 2. (/> + (/) (io + ? + l)\n\n27. Prove that\n\nB p, q)B p q, r) = B q, r)B q+r, p). \\addexamplecitation{Euler.}\n\n28. Shew that\n\nn,,, .,, d.r r(a)r(b) 1\n\nJo \" '' (.r+?>) + '' r(a + 6) (l+jo) jo\"'\n\nif a > 0, 6 > 0, p > 0. \\addexamplecitation{Trinity, 1908.}\n\n* This result is attributed to Bourguet by Stieltjes, Journal de Math,\n(i), v. p. 432.\n\n%\n% 262\n%\n\n29. Shew that, if m > 0, n>0, then\n\nn ( i+. )2 >-i (1 - -y n-i, \\ r(m)r n)\n\nand deduce that, when a is real and not an integer multiple of ijr,\n\n'i /cos 6 + sin \\ cos 2a\n\n/'\n\n\\ i,r veos C - SUl\n\nd6=.\n\nand\n\n30. Shew that, if a > 0, > 0,\n\n/:\n\n2 sin (tt cos a) '\n\n\\addexamplecitation{St John's, 1904.}\n\n\\addexamplecitation{Kummer.}\n\n31 . Shew that, if a > 0, + 6 > 0,\n\nfr(a)r(S) r(a + b)r 8)\n\nr..r--(i- ) \\,, lrw\n\nJo I-''*'' 5-0 I (a\n\nl= / (a+6)- /'(a).\n\n' + 8) T a + b + 8) Deduce that, if in addition a + c>0, a- b + c>0,\n\n/l a-l (l\\ b)(l\\ . .c) r (t ).! ( ± \\ +\\ ' )\n\njo \"(i-.r)(-log.r) * ~ *r(a + 6)r(a+c)\" 32. Shew that, if a, b, c be\nsuch that the integral converges,\n\nfUl-x\")(l-afi)(l-x'),, T b + c+l)T c + a + l)T a+b + l) ' a.v=los\n\nJO (i-A-)(-log.r) '\" '' °r(a + ' ) r b + l)r c + l)r a + h + c+l)' 33.\nBy the substitution cos = 1-2 tan <, shew that\n\n(3-cos )5 4v/7r\n\n\\addexamplecitation{St John's, 1896.}\n\nf sinP.v\n\n34. Evaluate in terms of Gamma-functions the integral / ' dx, when /)\nis a\n\nJ '*-\"\n\nfraction greater than unity whose numerator and denominator are both\nodd integers.\n\n[Shew that the integral is h I sin'' x\\~+ 2 ( - )\" ( 1 ) [ d.v.]\n\n'- Jo l n=i \\ x+mr x-mrj)\n\n35. Shew that\n\n\\addexamplecitation{Clare, 1898.}\n\n/:\n\n93?-\n\n2\" + 2 r=o2/'!(n-r)\n\n-a-mi-\n\n36. Prove that\n\nlog Bip,g) log ( +i) 4- I ' \\: '\"\\ \\ l '\" dv. \\addexamplecitation{Euler.}\n\ns v/', -/; °\\ pq J j(, (1-V)l0g 7\n\n37. Prove that, if p>0, p + s>0, then\n\nB(p,p)\n\n '\n\ns s~l), s(s-l)(s-2)(s-3)\n\n (P>P + )=- f7 ' l+ ( ) + 2.4.(2 +\n\nfd i,-- - <--\n\n38. The curve r\"*=2'\"~i a'\"cos ??i is composed of ?n equal closed\nloops. Shew that the length of the arc of half of one of the loops is\n\ni~ a I (i cos x) ' dx, Jo\n\nand hence that the total perimeter of the curve is\n\na iV\n\nlm)\\\n\n%\n% 263\n%\n\n39. Draw the straight line joiniug the points ±i, and the semicircle\nof \\ z\\ = \\ which lies on the right of this line. Let C be the contour\nformed by indenting this figure at\n\n- ?', 0, i. By considering / 2P- -i z + z- )p \"-- dz, shew that, if p\n+ q>l, q 'l.\n\n\\addexamplecitation{Cauchy.}\n\n40. If s is positive (not necessarily integral), and - in .r hn, shew\nthat\n\nMild draw graphs of the series and of the function cos*.i'.\n\n41. Obtain the expansion\n\ncos .f-2, r (*+i)[r(| +ia+i)r(i5- a+i) \" r(h+?,a+i)raj- a+iy-]'\n\nand find the values of x for which it is applicable. \\addexamplecitation{Cauchy.}\n\n42. Prove that, if /> > A,\n\n22p-i r 2/>\"- f 12 12.32 1 \"|2\n\n'\"'''\"=-A \"\"Wi'U + l i' 2(2yT3-) + 2.4.(2p+3)(2f+5)+ . '\n\n\\addexamplecitation{Binet.}\n\n43. Shew that, if .c < 0, ./ + -- > 0, then\n\nr(-.f) [-.f,(-..-)(l-.r) (-.r)(l- )(2-f ) 1 r(5) 1 5 - s(l+2) \" z\nl+z) 2 + z) j\n\nand deduce that, when x + r> 0,\n\n Ina m +* ) \\ f \\ 4 -1), X '( -l)(- -2) \\ f/-' r>) z z(z +\nl) - z z + l) z + 2)\n\n44. Using the result of example 43, prove that\n\nlogr(2 + a) = logr(0 + rtlogs\n\n2z\n\nfa\n\ndt\n\na [\\ \\ { t)(2-t) ...(n-t)dt- f\" t l-t) 2-t) ...(n-t)'\n\n\\ 5 .' .'o\n\nnti n + l)z z+l) z + 2) ... z + n)\n\ninvestigating the region of convergence of the .series.\n\n(Binet, Journal de V Ecole polytechnique, xvi. (1839), p. 256.)\n\n45. Prove that, if /> > 0, > 0, then\n\n-. P-\n\nB ip, q) = V- i (2 )* * \"' '\n\n%\n% 264\n%\n\nwhere\n\nII (p, g) = 2p - - T arc tan - - -. - -, -,\n\nand p2 = 2 + q +pq-\n\n46. If 6 =2*-'7r(l-i'r), F=2 - 7r(i-iA'),\n\nand if the function F (,r) be defined by the equation\n\nshew (1) that F o:) satisfies the equation\n\nF x+ ) =xF x) +\n\nr(l-A-)'\n\n(2) that, for all positive integral values of x,\n\nF x) = rix\\\n\n(3) that F(x) is analytic for all finite values of x,\n\n1 (7 \\ 2\n\n(4) that \"- \"* '\n\n47. Expand\n\nF(x)= ~- r -r- log- 7\n\nas a series of ascending powei's of a.\n\n(Various evaluations of the coefficients in this expansion have been\ngiven by Bourguet, Bull des Set. Math. v. (1881), p. 43; Bourguet,\nActa Math. ll. (1883), p. 261; Schlomilch,' Zeitschrift fiir Math,\nund Phys. xxv. (1880), pp. 35, 351.)\n\n48. Prove that the G-function, defined by the equation\n\nG(z + ) = (2nf'e- ' + - n |(i+i)%-'+-'- /(2\")|,\n\nis an integral function which satisfies the relations\n\n0 z + ) = V z)G z), (?(1) = 1,\n\n(n !)'V6-' (n + 1) = 11 . 22 . 33 ... w\". \\addexamplecitation{Alexeiewsky.}\n\n(The most important properties of the G-function are discussed in\nBarnes' memoir, Quarterly Journal, xxxi.)\n\n49. Shew that and deduce that\n\nlog Q J = / s cot Tvzdz-z log (27r).\n\n50. Shew that\n\nlogrri + l)c <=i3log(2rr)-|3(2+i;+3logr(s + l)-logG'(s + l).\n", "meta": {"hexsha": "eb7b2162c08742999eb737e0ad930ddc25360fe1", "size": 44622, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/tex/wandw-ch12.tex", "max_stars_repo_name": "CdLbB/Whittaker-and-Watson", "max_stars_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/tex/wandw-ch12.tex", "max_issues_repo_name": "CdLbB/Whittaker-and-Watson", "max_issues_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/tex/wandw-ch12.tex", "max_forks_repo_name": "CdLbB/Whittaker-and-Watson", "max_forks_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.1679316888, "max_line_length": 100, "alphanum_fraction": 0.6142261665, "num_tokens": 16406, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872075132152, "lm_q2_score": 0.8397339656668287, "lm_q1q2_score": 0.6290339713953629}} {"text": "\\chapter{Some Properties of Trees}\n\\label{sct_defining_clades}\n\n\\begin{figure}[t]\n\\centering\n\\includegraphics{oriented_graph.png}\n % oriented_graph.png: 485x393 pixel, 300dpi, 4.11x3.33 cm, bb=0 0 116 94\n \\caption{An oriented graph (not a tree!), with edge orientation shown as arrows. Two paths are highlighted (cyan and blue). The cyan path starts at node A and ends at node E. Nodes B, C, D, and E are reachable from A. Node C is the direct successor of B in the cyan path; in the blue path the direct successor of B is F. The segment highlighted in red is \\emph{not} a path.}\n \\label{fig_oriented_graph}\n\\end{figure}\n\n\n\nIn the following, $T$ will be a rooted tree. We'll recall that a tree (rooted or not) is a connected, acyclic graph. By choosing one node to be the root (which we'll note $R$), and orienting all edges away from the root, we get an oriented graph. A \\textit{path} along such a graph is a sequence of nodes such that there is an edge between two successive nodes, and that these nodes are ordered in the sequence as they are in the edge. The first node in the path is called the \\textit{start} of the path, and the last node is its \\textit{end}. We speak of a path \\textit{from} the start \\textit{to} the end, equivalently we say that the end is \\textit{reachable} from the start. The nodes in a path are ordered: if $a$ and $b$ are in a path, then either $a$ is reachable from $b$, or $b$ is reachable from $a$, or $a = b$; furthermore if $a$ is reachable from $b$, then $b$ is \\textit{not} reachable from $a$. If an edge connects $a$ to $b$, then $b$ is the \\textit{direct successor} of $a$, and $a$ is the \\textit{direct predecessor} of $b$\nSee figure {\\ref{fig_oriented_graph}.\n\nAlthough paths are not sets, we will (somewhat abusively, perhaps) use set notation with paths, unless there is a risk of confusion. Thus is $P$ is a path and $n$ is a node, then $n \\in P$ means that $n$ is ``in'' $P$ or ``belongs to'' $P$ (strictly speaking, $n$ is reachable from the start of $P$, and the end of $P$ is reachable from $n$).\n\nNow a few definitions - these are mostly renamings of graph-theoretical concepts, to make them more intuitive for phylogenetic trees. \n\n\\begin{dfn}\n\\label{def_ancestor}\nLet $a$ and $b$ be two different nodes of $T$. We say that $a$ is an \\textit{ancestor} of $b$, or (equivalently) that $b$ is a \\textit{descendant} of $a$, if and only if there is a path from $a$ to $b$. \n\\end{dfn}\n\n\\begin{dfn}\n\\label{def_parent}\nLet $P$ be a path in $T$, and $a, b$ two nodes in $P$ such that $b$ is the direct successor of $a$. We will call $a$ the \\textit{parent} of $b$. We can say \\textit{the} parent (rather than \\textit{a} parent) of $b$ because i) $a$ lies on the path from $R$ to $b$ (which is unique), and ii) only one node on any path can be another node's direct predecessor. We also see that any node's parent is also its ancestor (which shouldn't be too surprising\\ldots), since $b$ is reachable from $a$\n\\end{dfn}\n\n\\begin{dfn}\n\\label{def_child}\nLet $P$ be a path in $T$, and $a, b$ two nodes in $P$ such that $b$ is the direct successor of $a$. We say that $b$ is a \\textit{child} of $a$.\n\\end{dfn}\n\nNote that a node has exactly one parent, except the root which has none. A node can have zero or more children, but we rarely ever encounter nodes with a single child. A node with no children is called a \\textit{leaf}.\n\n\\begin{dfn}\n\\label{def_lineage}\nA path that starts from the root we call a \\emph{lineage}. By the definition of a rooted tree, there is always a lineage to $n$ for any node $n$, and this lineage is unique. So we can unambiguously talk of \\textit{the} lineage of $n$ to mean the path from $R$ to $n$. We say that a set $L$ of nodes form a lineage if there exists $n \\in L$ such that the path from $R$ to $n$ contains all and only elements of $L$.\n\\end{dfn}\n\n\\begin{figure}[b]\n \\centering\n \\includegraphics{prop_tree.png}\n \\caption{A rooted tree}\n \\label{fig_app_tree_prop}\n\\end{figure}\n\n\\begin{prop}\n\\label{prop_lineage_ancestors_equivalence}\nLet $n$ be a node of $T$, and $L$ its lineage. Then all ancestors of $n$ belong to $L$, and all nodes of $L$, except $n$ itself, are ancestors of $n$.\n\\end{prop}\n\n\\begin{proof}\n\n$\\Rightarrow$ Let $a$ be an ancestor of $n$. By definition \\ref{def_ancestor}, there is a path from $a$ to $n$. By tree properties, there is a path from $R$ to $a$. Hence $a$ belongs to the path from $R$ to $n$, which is the lineage of $n$.\n\\end{proof}\n\n\\noindent{}$\\Leftarrow$ Let $l \\neq n$ be a node in $L$. Since $L$ is the path from $R$ to $n$, $n$ is reachable from $a$, and hence $a$ is an ancestor of $n$ by definition \\ref{def_ancestor}.\n\n\\begin{dfn}\nIf $L$ and $M$ are lineages of $T$, then the \\textit{intersection}\nof $L$ and $M$, noted $L \\cap M$, is the set of nodes that belong to both $L$ and $M$.\n\\end{dfn} \n\n\\begin{lemma}\n\\label{lem_ancestors_in_lineage}\nLet $n$ be a node of $T$, $L$ its lineage, and $l$ a node of $L$. Then $l$'s ancestors also belong to $L$.\n\\end{lemma}\n\\begin{proof}\nIf $l = n$ then the ancestors of $l$ are those of $n$, which belong to $L$ by proposition \\ref{prop_lineage_ancestors_equivalence}. If $l \\neq n$ then let $a$ be an ancestor of $l$. By definition \\ref{def_ancestor}, $l$ is reachable from $a$. By proposition \\ref{prop_lineage_ancestors_equivalence}, $l$ is an ancestor of $n$ and again $n$ is reachable from $l$. Therefore, $n$ is reachable from $a$, so $a$ is an ancestor of $n$, and so belongs in $L$.\n\\end{proof}\n\n\\begin{lemma}\nLet $L$ and $M$ be lineages of $T$, and $q \\in L \\cap M, q \\neq R$. Then $q$'s parent also belongs to $M \\cap L$.\n\\end{lemma}\n\\begin{proof}\nCall $p$ the parent of $q$. $p$ is an ancestor of $q$, and since $q$ belongs to $L$ (by hypothesis), $p$ belongs to $L$ by lemma \\ref{lem_ancestors_in_lineage}. By the same reasoning, $p$ belongs to $M$, therefore $p \\in L \\cap M$.\n\n\\end{proof}\n\n\n\\begin{prop}\n\\label{prop_lineage_intersection}\nLet $L$ and $M$ be lineages of $T$. Then $L \\cap M$ also forms a lineage.\n\\end{prop}\n\\begin{proof}\nLet $N = L \\cap M$. Because $N$ is a subset of a lineage (in fact it is a subset of at least two lineages, $L$ and $M$), its elements are ordered, so we can pick the last one and call it $n$. Now all other elements of $N$ are ancestors of $n$, so by lemma \\ref{lem_ancestors_in_lineage} they belong in $L$ and in $M$. So far we have proven \n\\end{proof}\n\n\\begin{prop}\nIn a rooted tree, you can always define a clade by specifying a single node: the clade consists of the specified node and all its descendants.\n\\end{prop}\n\n\\begin{prop}\nIn a rooted tree $T$ with unique labels and a node $n$ in $T$, it is always possible to choose two nodes $n_1$ and $n_2$ such that lca($n_1$,$n_2$) = $n$.\n\\end{prop}", "meta": {"hexsha": "efe192da512790308ea5c93310c8263dd5b936f9", "size": 6711, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/tree_properties.tex", "max_stars_repo_name": "Cactusolo/newick_utils", "max_stars_repo_head_hexsha": "da121155a977197cab9fbb15953ca1b40b11eb87", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 62, "max_stars_repo_stars_event_min_datetime": "2015-01-08T22:22:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-07T09:12:51.000Z", "max_issues_repo_path": "doc/tree_properties.tex", "max_issues_repo_name": "Cactusolo/newick_utils", "max_issues_repo_head_hexsha": "da121155a977197cab9fbb15953ca1b40b11eb87", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 24, "max_issues_repo_issues_event_min_datetime": "2015-01-22T19:34:50.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-27T10:53:41.000Z", "max_forks_repo_path": "doc/tree_properties.tex", "max_forks_repo_name": "Cactusolo/newick_utils", "max_forks_repo_head_hexsha": "da121155a977197cab9fbb15953ca1b40b11eb87", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 26, "max_forks_repo_forks_event_min_datetime": "2015-05-07T09:23:34.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T02:43:50.000Z", "avg_line_length": 68.4795918367, "max_line_length": 1041, "alphanum_fraction": 0.7042169572, "num_tokens": 2124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.7956581049086031, "lm_q1q2_score": 0.6289918552255076}} {"text": "\\mychapter{8}{Lesson 8} %181019\n\n\\section{Domain extension}\n\nUp until now, encryption has been dealt with messages of fixed size around a polynomial function to $\\lambda$. How to deal with messages with arbitrary size? Setting a maximum bound to message length seems impractical, both for waste reasons when messages are too short, and for practicality when messages eventually get too long. The solution takes the form of a ``block-cipher'', where a message of a given size is split into equally-sized blocks, and then encrypted using a fixed-size encryption scheme. Various instances of this technique, called \\emph{modes}, have been devised.\n\n\\subsection{Electronic Codebook mode}\n\nThe operation of \\textsc{ecb}-mode is straightforward: Given a message split into blocks $(m_1, \\dots, m_t)$, apply the scheme's encryption routine to each block, as shown in figure \\ref{fig:ecb}:\n\\[\n c_i = F_k(r) \\oplus m_i \\quad \\forall i \\in \\{0, \\dots, t\\}\n\\]\nDecryption is trivially implemented by \\textsc{xor}-ing the ciphered blocks with $F_k(r)$.\n\n\n\\begin{figure}[ht]\n \\centering\n \\begin{tikzpicture}[node distance = 12mm, >=latex']\n\n \\path[->]\n node (m1) {$m_1$}\n node (x1) [right of = m1] {$\\oplus$}\n node (c1) [below of = x1] {$c_1$}\n node (f1) [box, above of = x1] {$f_k$}\n node (r1) [above of = f1] {$r$}\n\n node (m2) [right of = x1] {$m_2$}\n node (x2) [right of = m2] {$\\oplus$}\n node (c2) [below of = x2] {$c_2$}\n node (f2) [box, above of = x2] {$f_k$}\n node (r2) [above of = f2] {$r$}\n\n node (xd) [right of = x2, node distance = 15mm] {$\\dots$}\n node (cd) [right of = c2, node distance = 15mm] {$\\dots$}\n node (fd) [right of = f2, node distance = 15mm] {$\\dots$}\n node (rd) [right of = r2, node distance = 15mm] {$\\dots$}\n\n node (mt) [right of = xd] {$m_t$}\n node (xt) [right of = mt] {$\\oplus$}\n node (ct) [below of = xt] {$c_t$}\n node (ft) [box, above of = xt] {$f_k$}\n node (rt) [above of = ft] {$r$}\n\n (r1) edge (f1)\n (r2) edge (f2)\n (rt) edge (ft)\n\n (f1) edge (x1)\n (f2) edge (x2)\n (ft) edge (xt)\n\n (m1) edge (x1)\n (m2) edge (x2)\n (mt) edge (xt)\n \n (x1) edge (c1)\n (x2) edge (c2)\n (xt) edge (ct)\n ;\n\n \\end{tikzpicture}\n \\caption{\\textsc{ecb}-mode block-cipher in action, using a \\prf{} as the encryption routine}\n \\label{fig:ecb}\n\\end{figure}\n\nThis approach has the advantage of being completely parallelizable, as each block can clearly be encrypted separately; however there is a dangerous flaw in being not \\cpa-secure, even when using a \\prf-based encryption scheme. To understand why, observe that random nonces for ciphertext randomization are chosen per-message; this means the encryption of message blocks become deterministic in the message scope, enabling an adversary to attack the scheme within a single plaintext. It is sufficient to choose an all-0 or all-1 message to realize that all its blocks would encrypt to the same ciphered block.\n\n\\subsection{Cipher block chaining mode (\\textsc{cbc})}\n\n% AP190105: The explanation anticipates PRPs, which may be a problem. Should be reviewed at later times...\nThis mode serializes block encryption by using the preceding ciphered block in the formula:\n\\[\n c_i = P_k(r) \\oplus m_i \\quad \\forall i \\in \\{0, \\dots, t\\}\n\\]\nThis time, a \\emph{pseudorandom permutation}(\\prp) is used instead of a \\prf; they will be discussed later on. The diagram in figure \\ref{fig:cbc} shows a general view of \\textsc{cbc}-mode's operation. The decryption process is analogous but in a reversed fashion, by computing the preimage of a ciphered block and \\textsc{xor}-ing it with the preceding ciphered block:\n\\[\n m_i = P_k^{-1}(c_i) \\oplus c_{i-1}\n\\]\n\n\\begin{figure}[ht]\n \\centering\n \\begin{tikzpicture}[node distance = 12mm, > = latex']\n\n \\draw[->]\n node (r) {$r$}\n node (c0) [below of = r, node distance = 24mm] {$c_0$}\n\n node (c1) [right of = c0, node distance = 24mm] {$c_1$}\n node (p1) [box, above of = c1] {$P_k$}\n node (x1) [above of = p1] {$\\oplus$}\n node (m1) [above of = x1] {$m_1$}\n \n node (c2) [right of = c1, node distance = 24mm] {$c_2$}\n node (p2) [box, above of = c2] {$P_k$}\n node (x2) [above of = p2] {$\\oplus$}\n node (m2) [above of = x2] {$m_2$}\n\n node (cd) [right of = c2, node distance = 24mm] {$\\dots$}\n node (pd) [above of = cd] {$\\dots$}\n node (xd) [above of = pd] {$\\dots$}\n node (md) [above of = xd] {$\\dots$}\n\n node (ct) [right of = cd, node distance = 24mm] {$c_t$}\n node (pt) [box, above of = ct] {$P_k$}\n node (xt) [above of = pt] {$\\oplus$}\n node (mt) [above of = xt] {$m_t$}\n \n (r) edge (c0)\n\n (m1) edge (x1)\n (m2) edge (x2)\n (mt) edge (xt)\n\n (x1) edge (p1)\n (x2) edge (p2)\n (xt) edge (pt)\n\n (p1) edge (c1)\n (p2) edge (c2)\n (pt) edge (ct)\n\n (c0) -| +(12mm, 24mm) edge (x1)\n (c1) -| +(12mm, 24mm) edge (x2)\n (c2) -| +(12mm, 24mm) edge (xd)\n (cd) -| +(12mm, 24mm) -- (xt)\n ;\n\n \\end{tikzpicture}\n \\caption{\\textsc{cbc}-mode block-cipher in action, using a \\prp{} as the encryption routine}\n \\label{fig:cbc}\n\\end{figure}\n\n\\subsection{Counter mode}\n\nAlso denoted as \\textsc{ctr} in short, this mode closely resembles \\textsc{ecb}-mode but uses a ``rolling'' nonce instead of a static one, as shown in figure \\ref{fig:ctr}. At each successive block, the nonce is incremented by 1 and then used in a single block encryption. Since the nonce is in $\\binary^n$, the increment is done modulo $2^n$ so that the value will wrap around to 0 if it ever overflows. Decryption is analogous.\n\n\\begin{figure}[ht]\n \\centering\n\n \\begin{tikzpicture}[node distance = 12mm, > = latex']\n\n \\draw[->]\n node (c0) {$c_0$}\n\n node (c1) [right of = c0, node distance = 24mm] {$c_1$}\n node (x1) [above of = c1] {$\\oplus$}\n node (m1) [left of = x1] {$m_1$}\n node (f1) [box, above of = x1] {$f_k$}\n node (r1) [above of = f1] {$r$}\n\n node (c2) [right of = c1, node distance = 24mm] {$c_2$}\n node (x2) [above of = c2] {$\\oplus$}\n node (m2) [left of = x2] {$m_2$}\n node (f2) [box, above of = x2] {$f_k$}\n node (r2) [above of = f2] {$r + 1$}\n\n node (cd) [right of = c2, node distance = 19mm] {$\\dots$}\n node (xd) [above of = cd] {$\\dots$}\n node (fd) [above of = xd] {$\\dots$}\n node (rd) [above of = fd] {$\\dots$}\n\n node (ct) [right of = cd, node distance = 24mm] {$c_t$}\n node (xt) [above of = ct] {$\\oplus$}\n node (mt) [left of = xt] {$m_t$}\n node (ft) [box, above of = xt] {$f_k$}\n node (rt) [above of = ft] {$r + t - 1$}\n \n (r1) edge (r2)\n (r2) edge (rd)\n (rd) edge (rt)\n\n (r1) edge (f1)\n (r2) edge (f2)\n (rt) edge (ft)\n\n (f1) edge (x1)\n (f2) edge (x2)\n (ft) edge (xt)\n \n (m1) edge (x1)\n (m2) edge (x2)\n (mt) edge (xt)\n\n (x1) edge (c1)\n (x2) edge (c2)\n (xt) edge (ct)\n\n (r1) -| (c0)\n ;\n\n \\end{tikzpicture}\n \\caption{Counter-mode block-cipher in action, using a \\prf{} as the encryption routine}\n \\label{fig:ctr}\n\\end{figure}\n\nThis apparently innocuous change to \\textsc{ebc} is enough to ensure \\cpa-security, at the cost of perfect parallelization.\n\n\\begin{theorem}\n Assume $f_k$ is a \\prf, then the counter-mode block cipher is \\cpa-secure for variable length messages\\footnotemark.\n\\end{theorem}\n\\footnotetext{\\emph{Variable length messages} exactly means every message $m = (m_1, \\dots, m_t)$ is made of $t$ blocks, and $t$ can change from any message to a different one.}\n\n\\begin{proof}\n Figure \\ref{cryptogame:ctrcpa} models a \\cpa{} attack to a counter-mode block-cipher. The proof will proceed by hybrid argument starting from this game, therefore the statement to verify will be $\\cryptog{cpa}[\\textsc{ctr}](\\lambda, 0) \\compindist \\cryptog{cpa}[\\textsc{ctr}](\\lambda, 1)$.\n\n \\begin{cryptogame}\n {ctrcpa}\n {A chosen-plaintext attack to counter-mode block-cipher}\n {cpa}\n \n \\cseqbeginloop\n\n \\send{}{$m = (m_1, \\dots, m_t)$}{}\n \\receive{\\shortstack[l]{\n $r = c_0 \\pickUAR \\binary^n$ \\\\\n $c_i = f_k(r + i - 1) \\oplus m_i \\quad \\forall i$\n }}\n {$c = (c_0, c_1, \\dots, c_t)$}{}\n\n \\cseqendloop\n\n \\cseqdelay\n \n \\send{$|\\mu_0| = |\\mu_1| \\in \\mathcal{M}$}{$\\mu_0, \\mu_1$}{}\n \\receive{\\shortstack[l]{\n $\\rho = \\gamma_0 \\pickUAR \\binary^n$ \\\\\n $b \\pickUAR \\binary$ \\\\\n $(\\gamma_b)_i = f_k(\\rho + i - 1) \\oplus (\\mu_b)_i \\quad \\forall i$\n }}\n {$\\gamma_b = (\\gamma_0, \\gamma_1, \\dots, \\gamma_t)$}{}\n\n \\cseqdelay\n \n \\cseqbeginloop\n\n \\send{}{$m = (m_1, \\dots, m_t)$}{}\n \\receive{\\shortstack[l]{\n $r = c_0 \\pickUAR \\binary^n$ \\\\\n $c_i = f_k(r + i - 1) \\oplus m_i \\quad \\forall i$\n }}\n {$c = (c_0, c_1, \\dots, c_t)$}{}\n\n \\cseqendloop\n\n \\cseqdelay\n \n \\send{}{$b'$}{\\textsc{Output 1 iff} $b' = b$}\n\n \\end{cryptogame}\n\n Define the two hybrid games from the original \\cpa{} game as follows:\n\n \\begin{itemize}\n \\item $\\hybridg{1}[\\textsc{ctr}](\\lambda, b)$: A random function $R$ is chosen \\uar{} from $\\mathfrak{R}(n, n)$ at the beginning of the game, and is used in place of $F_k$ in all block encryptions;\n \\item $\\hybridg{2}[\\textsc{ctr}](\\lambda, b)$: The challenger will pick random values from $\\binary^n$ as ciphered blocks, disregarding any encryption routine.\n \\end{itemize}\n\n \\begin{lemma}\n $\\cryptog{cpa}[\\textsc{ctr}](\\lambda, b) \\compindist \\hybridg{1}[\\textsc{ctr}](\\lambda, b) \\quad \\forall b \\in \\binary$\n \\end{lemma}\n\n \\begin{proof} The proof is left as exercise.\n \n \\emph{Hint}: Since the original game and the first hybrid are very similar, we can use a distinguisher which plays the \\cpa-game; since this is a lemma, our goal in the reduction is to break the precondition contained in the theorem statement.\n \\end{proof}\n\n \\begin{lemma}\n $\\hybridg{1}[\\textsc{ctr}](\\lambda, b) \\compindist \\hybridg{2}[\\textsc{ctr}](\\lambda, b) \\quad \\forall b \\in \\binary$\n \\end{lemma}\n\n \\begin{proof}\n \n Since $m_i$ doesn't affect the distribution of the result at all, for any $i$, if $R(r^{*})$ behaves like a true random extractor, then the two hybrids are indistinguishable in the general case ($R(r + i) \\oplus m_i \\approx R(r + i)$). However, there is a sneaky issue: if in both games it happens that a given nonce $r_i$ is used in both one query encryption and the challenge message encryption at any step, the subsequent encrypted blocks will be completely random in the second hybrid, whereas in the first hybrid the function's images, albeit random, become predictable, enabling a \\cpa.\n\n Nevertheless, it can be proved that these ``collisions'' happen with negligible probability within $\\hybridg{1}[\\textsc{ctr}]$. Let:\n \\begin{itemize}\n \\item $q$ = number of encryption queries in a game run\n \\item $t_i$ = number of blocks for the $i$-th query\n \\item $\\tau$ = number of blocks for the challenge ciphertext\n \\item \\textsc{Overlap} event: $\\exists i, j, \\iota : r_i + j = \\rho + \\iota$\n \\end{itemize}\n\n The \\textsc{Overlap} event exactly models our problematic scenario. Now it suffices to show that it occurs negligibly. For simplicity, assume the involved messages are of the same length, that is $t_i = \\tau =: t$. Denote with $\\textsc{Overlap}_i$ to be the event that the $i$-th query overlaps the challenge sequence as specified above.\n\n Fix some $\\rho$. One can see that $\\textsc{Overlap}_i$ happens if:\n \\[\n \\rho - t + 1 \\leq r_i \\leq \\rho + t - 1\n \\]\n which means that $r_i$ should be chosen \\emph{at least} in a way that:\n \\begin{itemize}\n \\item the sequence $\\rho, \\dots, \\rho + t - 1$ comes before the sequence $r_i, \\dots, r_i + t - 1$, and they overlap just for the last element $\\rho + t - 1 = r_i$ or\n \\item the sequence $r_i, \\dots, r_i + t - 1$ comes before the sequence the sequence $\\rho, \\dots, \\rho + t - 1$, and they overlap just for the last element $r_i + t - 1 = \\rho$.\n \\end{itemize}\n\n Then:\n\n \\begin{align*}\n \\Pr[\\textsc{Overlap}_i] &= \\frac{(\\rho + t - 1) - (\\rho - t + 1) + 1}{2^n} \\\\\n &= \\frac{2t-1}{2^n} \\\\\n \\Pr[\\textsc{Overlap}] &\\leq \\sum_{i = 1}^{t} \\Pr[\\textsc{Overlap}_i] \\\\\n &\\leq 2 \\frac{t^2}{2^n} \\in \\negl{\\lambda} \n \\end{align*}\n \n which proves that our collision scenario happens with negligible probability, thus the two hybrids are indistinguishable.\n\n \\end{proof}\n \n Having proven the indistinguishability between the hybrids, the conclusion is reached:\n \\[\n \\cryptog{cpa}[\\textsc{ctr}](\\lambda, 0) \\compindist \\cryptog{cpa}[\\textsc{ctr}](\\lambda, 1)\n \\]\n \n\\end{proof}\n", "meta": {"hexsha": "ac91ad7f39e228e45a8478bee1ef698736a7da05", "size": 13812, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lessons/lesson_8.tex", "max_stars_repo_name": "Project2100/Cryptography-2018_19", "max_stars_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-10-15T09:22:45.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-15T09:22:45.000Z", "max_issues_repo_path": "lessons/lesson_8.tex", "max_issues_repo_name": "Project2100/cryptography_1819", "max_issues_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-07-18T15:45:10.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-27T20:36:12.000Z", "max_forks_repo_path": "lessons/lesson_8.tex", "max_forks_repo_name": "Project2100/cryptography_1819", "max_forks_repo_head_hexsha": "da5dcf51b0396bd26d7fd0445feceef365950757", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-07-17T14:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-03T15:23:22.000Z", "avg_line_length": 43.9872611465, "max_line_length": 608, "alphanum_fraction": 0.5728352158, "num_tokens": 4305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7905303186696747, "lm_q2_score": 0.7956580952177051, "lm_q1q2_score": 0.6289918475645588}} {"text": "% Preamble.\n\\documentclass[12pt]{article}\n\\usepackage[margin=1.25in]{geometry}\n\\usepackage[fleqn]{amsmath}\n\\usepackage{textcomp}\n\\usepackage{gensymb}\n\\usepackage{amsfonts}\n\\usepackage{enumitem}\n%\\usepackage{tikz} % Include for figures.\n%\\usepackage{subfiles} % Include for subfiles.\n\n%% Title macros.\n\\newcommand{\\HOMEWORKNUM}{21}\n\\newcommand{\\NAME}{D. Choi}\n\\newcommand{\\DATE}{2020-06-24}\n\n\\title{\\vspace{-2\\baselineskip}MATH 225 - Homework \\#\\HOMEWORKNUM}\n\\author{\\NAME}\n\\date{\\DATE}\n\n%% Formatting options.\n%\\pagenumbering{gobble} % Include for single-page document.\n\n% Macros.\n%% Contextualized by input/output bases.\n\\newcommand{\\based}[3]{{\\{#1\\}}_{#2}^{#3}}\n\n\n% Document.\n\\begin{document}\n\\maketitle\n\n\\section*{1.}\n\\begin{equation*}\n\tA = \\begin{pmatrix} 5 & 1 \\\\ 1 & 5 \\end{pmatrix}\n\\end{equation*}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item \\textit{Let}\n\t\\begin{equation*}\n\t\t\\vec{a} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}, \\quad\n\t\t\\vec{b} = \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix}, \\quad\n\t\t\\vec{c} = \\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix},\\quad\n\t\t\\vec{d} = \\begin{pmatrix} 5 \\\\ -5 \\end{pmatrix},\\quad\n\t\t\\vec{e} = \\begin{pmatrix} 10 \\\\ 0 \\end{pmatrix}.\n\t\\end{equation*}\n\t\\textit{Which are eigenvectors of $A$? Find their corresponding\n\teigenvalues.} \\\\[\\baselineskip]\n\tFor a square matrix $A$, a vector $\\vec{v}$ is an eigenvector of $A$ if\n\tthere is a scalar $\\lambda$ such that $A\\vec{v} = \\lambda \\vec{v}$.\n\t\\\\[\\baselineskip]\n\tFor $\\vec{a} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$,\n\t\\begin{equation*}\n\t\tA \\vec{a}\n\t\t=\n\t\t\\begin{pmatrix} 5 & 1 \\\\ 1 & 5 \\end{pmatrix}\n\t\t\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} 6 \\\\ 6 \\end{pmatrix}\n\t\t=\n\t\t6\n\t\t\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t6 \\vec{a}\n\t\t.\n\t\\end{equation*}\n\tThus, $\\vec{a}$ is an eigenvector of $A$ with a corresponding eigenvalue\n\tof \\boxed{6}.\n\t\\\\[\\baselineskip]\n\tFor $\\vec{b} = \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix}$,\n\t\\begin{equation*}\n\t\t\\vec{b} = 2 \\vec{a}\n\t\t.\n\t\\end{equation*}\n\tBecause $\\vec{b}$ is a scalar multiple of $\\vec{a}$, an eigenvector of\n\t$A$, $\\vec{b}$ is also an eigenvector of $A$. Furthermore, $\\vec{b}$\n\tshares $\\vec{a}$'s corresponding eigenvalue of \\boxed{6}.\n\t\\\\[\\baselineskip]\n\tFor $\\vec{c} = \\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$,\n\t\\begin{equation*}\n\t\tA \\vec{c}\n\t\t=\n\t\t\\begin{pmatrix} 5 & 1 \\\\ 1 & 5 \\end{pmatrix}\n\t\t\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} -4 \\\\ 4 \\end{pmatrix}\n\t\t=\n\t\t4\n\t\t\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t4 \\vec{c}\n\t\t.\n\t\\end{equation*}\n\tThus, $\\vec{c}$ is an eigenvector of $A$ with a corresponding eigenvalue\n\tof \\boxed{4}.\n\t\\\\[\\baselineskip]\n\tFor $\\vec{d} = \\begin{pmatrix} 5 \\\\ -5 \\end{pmatrix}$,\n\t\\begin{equation*}\n\t\t\\vec{d} = -5 \\vec{c}\n\t\t.\n\t\\end{equation*}\n\tBecause $\\vec{d}$ is a scalar multiple of $\\vec{c}$, an eigenvector of\n\t$A$, $\\vec{d}$ is also an eigenvector of $A$. Furthermore, $\\vec{d}$\n\tshares $\\vec{c}$'s corresponding eigenvalue of \\boxed{4}.\n\t\\\\[\\baselineskip]\n\tFor $\\vec{e} = \\begin{pmatrix} 10 \\\\ 0 \\end{pmatrix}$,\n\t\\begin{equation*}\n\t\tA \\vec{e}\n\t\t=\n\t\t\\begin{pmatrix} 5 & 1 \\\\ 1 & 5 \\end{pmatrix}\n\t\t\\begin{pmatrix} 10 \\\\ 0 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} 50 \\\\ 10 \\end{pmatrix}\n\t\t\\neq\n\t\t\\lambda\n\t\t\\begin{pmatrix} 10 \\\\ 0 \\end{pmatrix}\n\t\t\\text{ for any $\\lambda$.}\n\t\\end{equation*}\n\tThus, $\\vec{e}$ is not an eigenvector of $A$.\n\t\n\t\\item \\textit{Let $\\mathcal{S}$ be the standard basis, and let}\n\t\\begin{equation*}\n\t\t\\based{T}{\\mathcal{S}}{\\mathcal{S}} = A.\n\t\\end{equation*}\n\t\\textit{Find a basis $\\mathcal{B}$ made of the eigenvectors of $A$, and\n\tfind $\\based{T}{\\mathcal{B}}{\\mathcal{B}}$.} \\\\[\\baselineskip]\n\tLet\n\t\\begin{equation*}\n\t\t\\mathcal{B}\n\t\t=\n\t\t\\left\\{\n\t\t\t\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix},\n\t\t\t\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}\n\t\t\\right\\}\n\t\t.\n\t\\end{equation*}\n\tUsing $\\mathcal{B}$,\n\t\\begin{align*}\n\t\t\\based{T}{\\mathcal{B}}{\\mathcal{B}}\n\t\t&=\n\t\t\\based{T}{\\mathcal{S}}{\\mathcal{B}}\n\t\t\\based{T}{\\mathcal{S}}{\\mathcal{S}}\n\t\t\\based{I}{\\mathcal{B}}{\\mathcal{S}}\n\t\t\\\\\n\t\t&=\n\t\t\\begin{pmatrix}\n\t\t\t\\frac{1}{2} & \\frac{1}{2} \\\\\n\t\t\t-\\frac{1}{2} & \\frac{1}{2}\n\t\t\\end{pmatrix}\n\t\t\\begin{pmatrix} 5 & 1 \\\\ 1 & 5 \\end{pmatrix}\n\t\t\\begin{pmatrix} 1 & -1 \\\\ 1 & 1 \\end{pmatrix}\n\t\t\\\\\n\t\t&=\n\t\t\\boxed{\n\t\t\t\\begin{pmatrix} 6 & 0 \\\\ 0 & 4 \\end{pmatrix}\n\t\t}\n\t\t.\n\t\\end{align*}\n\tNote the diagonal elements of $\\based{T}{\\mathcal{B}}{\\mathcal{B}}$ are\n\tthe eigenvalues of $A$.\n\\end{enumerate}\n\\newpage\n\n\\section*{2.}\n\\textit{Suppose that $A$ is a $3 \\times 4$ matrix.}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item \\textit{Is it possible that $\\text{dim}(\\text{nul}(A)) = 0$?}\n\t\\\\[\\baselineskip]\n\t\\boxed{\\text{No.}} By the rank-nullity theorem,\n\t\\begin{equation*}\n\t\t\\text{rank}(A) + \\text{nullity}(A) = \\text{dim}(\\text{domain}(A))\n\t\t.\n\t\\end{equation*}\n\tThe dimension of the domain of $A$ is $4$, because $A$ has $4$ columns. \\\\\n\tThe rank of $A$ is at maximum 3, because $A$ is a $3 \\times 4$ matrix. \\\\\n\tAs a result of this, the nullity of $A$ must be at least $1$.\n\t\n\t\\item \\textit{If $\\text{dim}(\\text{nul}(A)) = 1$, what can you say about\n\t$\\text{col}(A)$?} \\\\[\\baselineskip]\n\t\\boxed{\\text{dim}(\\text{col}(A)) = \\text{rank}(A) = 4 - 1 = 3.}\n\n\t\\item \\textit{If $\\text{dim}(\\text{nul}(A)) = 2$, what can you say about\n\t$\\text{col}(A)$?} \\\\[\\baselineskip]\n\t\\boxed{\\text{dim}(\\text{col}(A)) = \\text{rank}(A) = 4 - 2 = 2.}\n\t\n\t\\item \\textit{If $\\text{dim}(\\text{nul}(A)) = 3$, what can you say about\n\t$\\text{col}(A)$?} \\\\[\\baselineskip]\n\t\\boxed{\\text{dim}(\\text{col}(A)) = \\text{rank}(A) = 4 - 3 = 1.}\n\t\n\t\\item \\textit{If $\\text{dim}(\\text{nul}(A)) = 4$, what can you say about\n\t$\\text{col}(A)$?} \\\\[\\baselineskip]\n\t\\boxed{\\text{dim}(\\text{col}(A)) = \\text{rank}(A) = 4 - 4 = 0.}\n\\end{enumerate}\n\\newpage\n\n\\section*{3.}\n\\textit{Let}\n\\begin{equation*}\n\tA = \\begin{pmatrix} 8 & -10 \\\\ 5 & -7 \\end{pmatrix},\n\t\\qquad\n\t\\vec{v}_1 = \\begin{pmatrix} 2 \\\\ 1 \\end{pmatrix}, \\quad\n\t\\vec{v}_2 = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}.\n\\end{equation*}\n\\begin{enumerate}[label=(\\alph*)]\n\t\\item \\textit{Show that the vectors $\\vec{v}_1$ and $\\vec{v}_2$ are\n\teigenvectors of $A$ and find their associated eigenvalues.}\n\t\\begin{gather*}\n\t\tA \\vec{v}_1\n\t\t=\n\t\t\\begin{pmatrix} 8 & -10 \\\\ 5 & -7 \\end{pmatrix}\n\t\t\\begin{pmatrix} 2 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} 6 \\\\ 3 \\end{pmatrix}\n\t\t=\n\t\t3 \\begin{pmatrix} 2 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t3 \\vec{v}_1\n\t\t, \\\\\n\t\tA \\vec{v}_2\n\t\t=\n\t\t\\begin{pmatrix} 8 & -10 \\\\ 5 & -7 \\end{pmatrix}\n\t\t\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} -2 \\\\ -2 \\end{pmatrix}\n\t\t=\n\t\t-2 \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\n\t\t=\n\t\t-2 \\vec{v}_2\n\t\t.\n\t\\end{gather*}\n\tThus, $\\boxed{\\lambda_1 = 3}, \\boxed{\\lambda_2 = -2}$.\n\t\n\t\\item \\textit{Express the vector}\n\t\\begin{equation*}\n\t\t\\vec{x} = \\begin{pmatrix} -4 \\\\ -1 \\end{pmatrix}\n\t\\end{equation*}\n\t\\textit{as a linear combination of $\\vec{v}_1$ and $\\vec{v}_2$.}\n\t\\\\[\\baselineskip]\n\tSolving\n\t\\begin{equation*}\n\t\t\\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \\end{pmatrix}\n\t\t\\begin{pmatrix} a \\\\ b \\end{pmatrix}\n\t\t=\n\t\t\\begin{pmatrix} -4 \\\\ -1 \\end{pmatrix}\n\t\t,\n\t\\end{equation*}\n\t$a = -3$, and $b = 2$. \\\\\n\tThus,\n\t\\begin{equation*}\n\t\t\\boxed{\n\t\t\t\\vec{x}\n\t\t\t=\n\t\t\t-3 \\vec{v}_1 + 2 \\vec{v}_2\n\t\t}\n\t\t.\n\t\\end{equation*}\n\\end{enumerate}\n\\newpage\n\n\\section*{Extra.}\n\\textit{Consider the Fibonacci sequence and the following:}\n\\begin{equation*}\n\t\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}\n\t\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}\n\t=\n\t\\begin{pmatrix} 3 \\\\ 5 \\end{pmatrix}\n\t, \\quad\n\t\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}\n\t\\begin{pmatrix} 3 \\\\ 5 \\end{pmatrix}\n\t=\n\t\\begin{pmatrix} 5 \\\\ 8 \\end{pmatrix}\n\t.\n\\end{equation*}\n\\textit{Find the 1000th Fibonacci number.} \\\\[\\baselineskip]\nLet $0$ be the $0$th Fibonacci number.\n\\begin{equation*}\n\t\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}^n\n\t\\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\tn\\text{th Fibonacci number} \\\\\n\t\t(n+1)\\text{th Fibonacci number}\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\nThus, let\n\\begin{equation*}\n\t\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}^n\n\t\\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}\n\t=\n\t\\begin{pmatrix}\n\t\ta \\\\\n\t\t\\_\n\t\\end{pmatrix}\n\t.\n\\end{equation*}\n\\boxed{a} is the 1000th Fibonacci number.\n\n\n\\end{document}", "meta": {"hexsha": "10c4540ccf22455bf8e4aee865fad80aae49f493", "size": 8073, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "usc-20202-math-225-39425/hw21/main.tex", "max_stars_repo_name": "Floozutter/coursework", "max_stars_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "usc-20202-math-225-39425/hw21/main.tex", "max_issues_repo_name": "Floozutter/coursework", "max_issues_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "usc-20202-math-225-39425/hw21/main.tex", "max_forks_repo_name": "Floozutter/coursework", "max_forks_repo_head_hexsha": "244548f415553f058098cae84ccdd4ce3f58c245", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.0, "max_line_length": 75, "alphanum_fraction": 0.6070853462, "num_tokens": 3512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956580903722561, "lm_q2_score": 0.7905303211371898, "lm_q1q2_score": 0.6289918456973828}} {"text": "\\subsection{Heuristic approach}\nA heuristic method has been developed basing on imitating the visual selection of L-shapes (see figure)\n\\begin{itemize}\n\\item Overimpose a $3\\times 3$ grid on the scatterplot.\n\\item Score points on each subgrid in such a way that\n\\begin{itemize}\n\t\\item Points in permitted regions increase score\n\t\\item Points in non-desired regions decrease score\n\t\\item Points in non-allowed regions set score to $\\-infty$.\n\\end{itemize}\n\\item Use cross-validation to tune scoring parameters.\n\\end{itemize}\n\n\\subsection{The Naive method}\n\nTraditionally selection has been based on searching for genes with negative correlation betweem expression and correlation. This ``Naïve approach\" is used to compare with other approaches.\n", "meta": {"hexsha": "1a7c22a74f4a9476e340cf13c1286aac6a2c9e34", "size": 745, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Treballs_nostres/2019-07-UseR-A Shiny_App_for_SGRM/2019-07-UseR-Poster/sectionsUseR/methods3.tex", "max_stars_repo_name": "bertamiro/Selecting_GRM", "max_stars_repo_head_hexsha": "f7d91df489cb5bd6b6fd6447be9c7a1002705158", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Treballs_nostres/2019-07-UseR-A Shiny_App_for_SGRM/2019-07-UseR-Poster/sectionsUseR/methods3.tex", "max_issues_repo_name": "bertamiro/Selecting_GRM", "max_issues_repo_head_hexsha": "f7d91df489cb5bd6b6fd6447be9c7a1002705158", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Treballs_nostres/2019-07-UseR-A Shiny_App_for_SGRM/2019-07-UseR-Poster/sectionsUseR/methods3.tex", "max_forks_repo_name": "bertamiro/Selecting_GRM", "max_forks_repo_head_hexsha": "f7d91df489cb5bd6b6fd6447be9c7a1002705158", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.8235294118, "max_line_length": 188, "alphanum_fraction": 0.7973154362, "num_tokens": 179, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.879146761176671, "lm_q2_score": 0.7154239957834733, "lm_q1q2_score": 0.6289626887611128}} {"text": "\\documentclass[]{article}\n\n%opening\n\\title{MTH 343 Numerical Analysis: Lecture 12}\n\\author{Sheikh Abdul Raheem Ali}\n\n\\begin{document}\n\n\\maketitle\n\n\\section*{Interpolation}\n\n\t\\begin{enumerate}\n\t\t\\item Solving Systems of Equations\n\t\t\\item Lagrange Polynomials\n\t\\end{enumerate}\n\n\\begin{tabular}{c c}\n\t$ x_i $ & $ f_i $ \\\\\n\t$ x_0 $ & $ f_0 $ \\\\\n\t$ x_1 $ & $ f_1 $ \\\\\n\t$ x_2 $ & $ f_2 $ \n\\end{tabular}\n\n\\[ P_2(x) = \\frac{(x-x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)}f_0 + \\frac{(x-x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)}f_1 + \\frac{(x-x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)}f_2 \\]\n\n\n\\begin{tabular}{c c}\n\t$ x_i $ & $ f_i $ \\\\\n\t$ 0 $ & $ 1 $ \\\\\n\t$ -1 $ & $ 2 $ \\\\\n\t$ 2 $ & $ 3 $ \n\\end{tabular}\n\n\\[ P_2(x) = \\frac{(x+1)(x - 2)}{(0 + 1)(0 - 2)} + \\frac{(x-0)(x - 2)}{(-1 - 0)(-1 - 2)}2 + \\frac{(x-0)(x + 1)}{(2 - 0)(2 + 1)}3 \\]\n\n$ P_n(x) = \\sum_{0}^{n}L_{n,i}f(x_i) $\n\n\\section*{Lagrange Polynomial Error Function}\n\n\\[ E(x) = |f(x) - P_n(x)| = \\frac{|(x-x_0)(x-x_1)\\dots(x-x_n)f^{n+1}(\\xi)|}{(n+1)!} \\]\n\nWhere $ \\xi $ is the element in the smallest interval containing $ x_0,x_1,\\dots,x_n $\n\n\\end{document}\n", "meta": {"hexsha": "944b9cbd0515f71c94d11adb9bae4ddf88441862", "size": 1096, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Numerical/Lecture_12/Lecture12.tex", "max_stars_repo_name": "sheikheddy/aus-files", "max_stars_repo_head_hexsha": "0c38d15d560ccbb8231c8ef210916ea94a0f004b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Numerical/Lecture_12/Lecture12.tex", "max_issues_repo_name": "sheikheddy/aus-files", "max_issues_repo_head_hexsha": "0c38d15d560ccbb8231c8ef210916ea94a0f004b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Numerical/Lecture_12/Lecture12.tex", "max_forks_repo_name": "sheikheddy/aus-files", "max_forks_repo_head_hexsha": "0c38d15d560ccbb8231c8ef210916ea94a0f004b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.8260869565, "max_line_length": 171, "alphanum_fraction": 0.5346715328, "num_tokens": 524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867825403177, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.6289162659133477}} {"text": "%!TEX root = main.tex\n\\section{Simulation setup}\n\\subsection{Variance reduction}\nBecause the simulation can be time consuming to run, we want to lower the variance of the estimated parameters as much as possible relative to the computational resources used.\n\n\\subsubsection{Control Variates}\nControl variates is a variance reduction technique which can be used to reduce the variance of the estimated expected value $\\hat{\\mu} = \\mathbb{E}[Y]$. It works by drawing/calculating a variable $X_i$ such that $Y_i \\approx f(X_i)$. One can then estimate $\\hat{\\mu} = \\mathbb{E}[Z]$ where $Z$ is given by:\n\\begin{align}\nZ_i = Y_i - c (X_i - \\mathbb{E}[X])\n\\end{align}\n\nChoosing $c = \\frac{Cov(X, Y)}{Var(X)}$ gives the largest variance reduction of:\n\\begin{equation}\n\\frac{Var[Z]}{Var[Y]} = \\frac{Var[Y]}{Var[Y]} - \\frac{Cov(X, Y)^2}{Var[X] Var[Y]} = 1 - \\rho_{x, y}^2 \\label{eq:variance-reduction}\n\\end{equation}\n\nIn our case there is no analytical expression for the covariance or variance of the random variables. Instead $c$ is estimated from the dataset itself. This introduces some overfitting, thus one should take care to reduce the degrees of freedom in the variance, covariance estimates:\n\\begin{equation}\n\\hat{c} = \\frac{Cov_{N-1}(X, Y)}{Var_{N-1}(X)}, \\quad \\hat{\\sigma}^2 = Var_{N-2}(Z)\n\\end{equation}\n\nThe subscript denotes the degrees of freedom, and $\\hat{\\sigma}^2$ has $N - 2$ degrees of freedom, because $\\hat{c}$ is estimated from the dataset.\n\n\\subsubsection{Estimating infection rate}\n\nFrom \\eqref{eq:variance-reduction} it seen that to get a high variance reduction, $X_i$ just need to be heavily correlated with $Y_i$. In our case we would like to estimate the infection peak time and how many people where infected at the peak. From the theoretical differential equation model those values should be correlated with the infection rate $\\beta$.\n\nAssuming the stochastic model have a behaviour somewhat similar to the differential equation model, there exists a fairly simple method for estimating $\\beta$ and $\\gamma$ \\cite{wiki-sir}.\n\\begin{equation}\n\\gamma = \\frac{R'_{max}}{I_{max}} \\text{ and } \\beta = -\\frac{\\ln(p)}{1 - p} \\gamma \\quad \\text{where: } p = \\frac{N - R_\\infty}{N} \n\\end{equation}\n\n$R'_{max}$ is the maximum removed difference, and $I_{max}$ is the maximum infected. Due to the construction of the differential equation, these are found at the same time $t$. $R_\\infty$ is how many people where removed in the end, thus $p$ is the survival rate.\n\n\\subsection{Effect of Global Events}\nWe want to estimate what effect a global event can have on a virus outbreak. We define an event $e$ as simply being a transfer of people from one or more regions $r_{i}$ to region $r_t$ at a time $t_0$, and the reverse transfer of people from $r_t$ to $r_{i}$ at time $t_1$.\n\n\\subsubsection{Motivation: Zika virus}\nRecently, a new virus has been observed in South and Latin America. The Zika virus affects pregnant women's fetuses and can cause serious birth defects. The Zika virus is primarily spread by mosquitoes, but can also spread from a man to his sex partners (and from mother to fetus). Some political discussions have been made as to whether or not the Olympic Games should be canceled to prevent the spreading of the virus.\n\nIn this report we will run a \"hypothetical\" simulation, showing how the a virus could spread assuming inter-human infections. As such this simulation is primarily an attempt of showing how such a model would work and not a realistic Zika virus predictor.\n\n\\subsubsection{Event definition: Olympic Games in Rio 2016}\nIn our simulation we define the Olympic Games to take place during $t_0=$ `2016-07-03' and $t_1=$ `2016-07-21'. According to The Gaurdian \\cite{theguardian-olympics} Rio expects to recieve 380,000 tourist. We choose to transfer people from all over the world to Rio, such that the transfer is proportional to the population size of each region:\n\n\\begin{align}\n\t\\text{Transfer}_0(i, j) &= 380000 \\cdot \\frac{\\text{population}_i}{N} \\\\\n\t\\text{Transfer}_1(j, i) &= \\text{transfer}_0(i, j)\n\\end{align}\n\nwhere $\\text{Transfer}_0(i, j)$ denotes the number of people transferred from region $i$ to region $j$ at the start of the Olympics and $ \\text{Transfer}_1(j, i) $ the people transferred from $j$ to $i$ at the end of the Olympics.\n", "meta": {"hexsha": "e183cf4895e7065f8c07bde2206f68189b825f1e", "size": 4303, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/simulation.tex", "max_stars_repo_name": "FrederikWR/course-02443-stochastic-virus-outbreak", "max_stars_repo_head_hexsha": "4f1d7f1fa4aa197b31ed86c4daf420d5a637974e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/simulation.tex", "max_issues_repo_name": "FrederikWR/course-02443-stochastic-virus-outbreak", "max_issues_repo_head_hexsha": "4f1d7f1fa4aa197b31ed86c4daf420d5a637974e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/simulation.tex", "max_forks_repo_name": "FrederikWR/course-02443-stochastic-virus-outbreak", "max_forks_repo_head_hexsha": "4f1d7f1fa4aa197b31ed86c4daf420d5a637974e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 82.75, "max_line_length": 420, "alphanum_fraction": 0.747153149, "num_tokens": 1156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867825403176, "lm_q2_score": 0.7879311956428946, "lm_q1q2_score": 0.6289162659133476}} {"text": "%Note for this test case:\n%Simple water flow example using ANUGA: Water flowing down a channel.\n%It was called \"steep_slope\" in an old validation test.\n\n\n\\section{Shallow flow down a mild slope}\nThis case simulates very shallow flow running down a mild slope topography. It represents an idealisation of the rainfall-runoff problem, which will often involve very shallow flows down such a topography. This case has an analytical solution, and in particular, we consider the steady-uniform solution with the values of bed slope and friction slope are the same. \n\nSuppose that we are given a one dimensional domain. The steady state conditions with a contant water depth everywhere make the shallow water equations to the single identity\n\\begin{equation}\nz_x = - S_f, \n\\end{equation}\nwhere $z_x$ is the bed slope, and $S_f$ is the symbol for the force of bottom friction. We take Manning's friction\n\\begin{equation}\nS_f = n^2 \\frac{q|q|}{h^{10/3}}\n\\end{equation}\nwhere $n$ is the Manning's coefficient and $q$ is the discharge $uh$. \nIf $q$, $n$, and $z_x$ are given, then the analytical solution for $u$ and $h$ is\n\\begin{equation}\nu(x)= \\left[- n^{-2} q^{4/3} z_x\\right]^{3/10},\n\\end{equation}\n\\begin{equation}\nh(x)= \\frac{q}{u}\\,.\n\\end{equation}\n\n\\subsection{Results}\nFor our test, we consider a square dimensional domain with length and width 100. We take $q=0.2$, $n=0.03$, and $z_x=-0.1$.\nThe topography is\n\\begin{equation}\nz(x, y)= -0.1 x\\,.\n\\end{equation}\nThe initial condition is $u=v=0$ and\n\\begin{equation}\nw(x,y,0)= -0.1 x + 0.01\\,.\n\\end{equation}\n\n\nSome simulation results are as follows.\nFigures~\\ref{fig:depthdownchan} shows the steady state depth in the downstream direction. \nThere should be a good agreement with the analytical solution, at least away from the boundaries. \n\nFigures~\\ref{fig:depthacrosschan} shows the steady state depth across the slope around the line x = 50m. \nThere should be a good agreement with the analytical solution, at least away from the boundaries. \n\nFigures~\\ref{fig:xvelscrosschan} and~\\ref{fig:yvelscroschan} show the steady state $x$- and $y$-velocities, along a slice in the cross slope direction (near $x=50$). The $x$-velocities should agree well with the analytical solution, and the $y$-velocities should be zero. \n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{depth_x.png}\n\\caption{Depth in the downstream direction}\n\\label{fig:depthdownchan}\n\\end{center}\n\\end{figure}\n \n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{depth_y.png}\n\\caption{Depth across the slope around $x$ = 50m}\n\\label{fig:depthacrosschan}\n\\end{center}\n\\end{figure}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{x_velocity.png}\n\\caption{$x$-velocity along the cross-section $x=50$ (i.e. a cross-section with constant bed elevation)}\n\\label{fig:xvelscrosschan}\n\\end{center}\n\\end{figure}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{y_velocity.png}\n\\caption{$y$-velocity along the cross-section $x=50$ (i.e. a cross-section with constant bed elevation)}\n\\label{fig:yvelscroschan}\n\\end{center}\n\\end{figure}\n\n\n\\endinput\n", "meta": {"hexsha": "6621bb067a966f6f4b7f1b43bbc19c807f89e738", "size": 3135, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "validation_tests/analytical_exact/rundown_mild_slope/results.tex", "max_stars_repo_name": "samcom12/anuga_core", "max_stars_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_stars_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_stars_count": 136, "max_stars_repo_stars_event_min_datetime": "2015-05-07T05:47:43.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-16T03:07:40.000Z", "max_issues_repo_path": "validation_tests/analytical_exact/rundown_mild_slope/results.tex", "max_issues_repo_name": "samcom12/anuga_core", "max_issues_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_issues_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_issues_count": 184, "max_issues_repo_issues_event_min_datetime": "2015-05-03T09:27:54.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-20T04:22:48.000Z", "max_forks_repo_path": "validation_tests/analytical_exact/rundown_mild_slope/results.tex", "max_forks_repo_name": "samcom12/anuga_core", "max_forks_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_forks_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_forks_count": 70, "max_forks_repo_forks_event_min_datetime": "2015-03-18T07:35:22.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-01T07:07:29.000Z", "avg_line_length": 38.7037037037, "max_line_length": 367, "alphanum_fraction": 0.7499202552, "num_tokens": 926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428947, "lm_q2_score": 0.7981867801399695, "lm_q1q2_score": 0.6289162640220385}} {"text": "% !TEX root = ../thesis.tex\n\n\\chapter{Preliminaries}\n\\label{chap:preliminaries}\nThere are several topological spaces that play a key role\nwhen describing properties of the Hopf map $h : S^3 \\to S^2\\!$.\nThese include of course the domain $S^3$ and codomain $S^2\\!$,\nwhich are traditionally defined as subspaces of $\\R^4$ and $\\R^3$ respectively.\nAs we will see later,\nit is useful to consider $S^3$ and $S^2$ as quotient spaces of $\\C^2 \\setminus \\set{0}$\nor subspaces of the quaternion algebra $\\H$ instead.\nBoth $\\C^2$ and $\\H$ are isomorphic to $\\R^4$ as a real vector space,\nbut their additional structure sheds light on various properties of the Hopf map.\nWe will therefore briefly examine these spaces before defining the Hopf map.\nIn later chapters we will explore the differentiable structure of these spaces,\nbut for now we focus on their topological properties.\nFurthermore, group actions are used extensively throughout this chapter,\nso we quickly recall some of the terms involved.\n\n\\section{Definitions}\n\n\\definition\nLet $X$ be a set with additional structure,\nsuch as a vector space or a topological space.\nThe \\emph{automorphism group} of $X$,\ndenoted $\\Aut(X)$ is the group of bijections $X \\to X$\nthat preserve its structure.\nMore formally we can say that $X$ is an object in a concrete category $\\mathcal{A}$,\na category equipped with a faithful functor $\\mathcal{A} \\to \\mathbf{Sets}$,\nthe forgetful functor.\nThe automorphism group is the group of invertible morphisms $X \\to X$\nin this category.\n\n\\example\nFor a group $G$, $\\Aut(G)$ is the group of group isomorphisms $G \\to G$.\nFor a module $M$ over a ring $R$,\nor for a vector space $V$ over a field $F$,\nthe automorphism group consists of $R$-linear bijections $M \\to M$\nand $F$-linear bijections $V \\to V$ respectively.\nFor a topological space $X$,\n$\\Aut(X)$ is the group of homeomorphisms $X \\to X$.\nFor sets, the automorphism group is simply the group of bijections\nfrom the set to itself.\n\n\\definition\nLet $G$ be a group and $X$ a set with additional structure.\nA \\emph{group action} of $G$ on $X$ is a group homomorphism $\\phi : G \\to \\Aut(X)$.\nIf $X$ is an object in the category $\\mathcal{A}$ \\,(e.g. the category of vector spaces,\ntopological spaces, etc.),\nwe say $G$ acts on $X$ in this category.\nGiven an element $x \\in X$ and $g \\in G$,\nwe will often write $g \\cdot x$ for the element $\\phi(g)(x)$.\n\nIn the definition above,\nthe group $G$ acts \\emph{from the left} on $X$.\nFor $g, h \\in G$ and $x \\in X$,\nwe have $(gh) \\cdot x = g \\cdot (h \\cdot x)$.\nSometimes we encounter a natural \\emph{antihomomorphism} $G \\to \\Aut(X)$.\nIn this case,\nwe say that $G$ acts \\emph{from the right} on $X$ in the category $\\mathcal{A}$,\nand to make the notation more natural we write $x \\cdot g$ instead of $g \\cdot x$,\nso that $x \\cdot (gh) = (x \\cdot g) \\cdot h$.\n\n\\definition\nLet $G$ and $X$ be as before, and $x \\in X$.\nThe \\emph{orbit} of $x$ is the set\n\\[ Gx = \\set{ g \\cdot x \\mid g \\in G } \\]\nHaving the same orbit defines an equivalence relation on $X$,\nand the quotient with respect to this relation is called the \\emph{orbit space},\nwritten $X/G$.\n\n\\definition\nLet $G$ and $X$ be as before, and $x \\in X$.\nThe \\emph{stabiliser} of $x$ is the subgroup\n\\[ G_x = \\set{ g \\in G \\mid g \\cdot x = x } \\]\n\n\\proposition[prop:open-quotient-map]\nLet $X$ be a topological space, $G$ a group that acts on $X$.\nWhen $X/G$ is endowed with the quotient topology,\nthe quotient map $q : X \\surj X/G$ is an open map.\n\n\\proof\nLet $U \\subseteq X$ be open,\ndefine $V = q(U)$.\n$V$ is open if and only if $q^{-1}(V)$ is open\nby definition of the quotient topology.\nWe have\n\\[ q^{-1}(V)\n = \\bigcup_{g \\in G} g \\cdot U \\]\nBecause the automorphisms of the group action are homeomorphisms,\nthey are open maps,\nso $g \\cdot U$ is open for all $g \\in G$.\nHence, $q^{-1}(V)$ is the union of open sets,\nso it is open.\n\\qed\n\n\\definition\nA \\emph{topological group} is a group $G$ that is also a Hausdorff space,\nsuch that the map $G \\times G \\to G$, $(g, h) \\mapsto gh^{-1}$ is continuous\nwhen $G \\times G$ is endowed with the product topology.\nThis is equivalent to the statement that multiplication and inversion are continuous;\nsee for example \\parencite[ch.~\\textsc{iii}, \\S~1.1]{bourbaki1971} or \\parencite[p.~276]{szekeres2004}.\n\n\\definition\nLet $G$ be a topological group and $X$ a topological space,\nsuch that $G$ acts on $X$ in the category of sets.\nThe action is said to be \\emph{continuous} if\n\\[ G \\times X \\longto X,\n\\quad (g, x) \\longmapsto g \\cdot x \\]\nis a continuous map.\nIt follows immediately from this definition that\n$x \\mapsto g \\cdot x$ is a homeomorphism for all $g \\in G$ when $G$ acts continuously on $X$,\ntherefore $G$ acts on $X$ in the category of topological spaces;\nthe group homomorphism $G \\to \\Aut(X)$ is actually a group homomorphism $G \\to \\Homeo(X)$.\n\n\\theorem{Universal property of the quotient topology}[thm:universal-property-quotient-topology]\nLet $X$ and $Y$ be topological spaces and $\\sim$ an equivalence relation on $X$.\nDenote by $q : X \\surj X\\modsim$ the quotient map.\nLet $f : X \\to Y$ be a continuous map such that\nfor all $x, y \\in X$ it holds that $x \\sim y$ implies $f(x) = f(y)$.\n(Such $f$ is said to be \\emph{compatible} with the equivalence relation.)\nThen there exists a unique continuous map $g : X\\modsim \\to Y$ that makes the following diagram commute:\n\\vspace{-\\parskip}\n\\begin{center}\n\\begin{tikzcd}[column sep = small] &\nX \\ar[dl, two heads, swap, \"q\"]\n \\ar[dr, \"f\"] & \\\\\nX\\modsim \\ar[rr, dashed, swap, \"\\exists_! g\"] & & Y\n\\end{tikzcd}\n\\end{center}\n\n\\proof\nSee for example \\parencite[ch.~\\textsc{i}, \\S~3.4]{bourbaki1971}.\n\n\\proposition\nLet $G_1$ and $G_2$ be groups, and $X$ a set.\nSuppose that $G_1 \\times G_2$ acts on $X$ in the category of sets.\nThis implies that the subgroups $G_1$ and $G_2$ act on $X$ individually as well.\nThen the following holds:\n\\begin{enumerate}\n\\item There is a natural action of $G_1$ on $X / G_2$ in the category of sets.\n\\item If $X$ is a topological space and $G_1 \\times G_2$ acts in the category of topological spaces,\n $G_1$ acts on $X / G_2$ in this category.\n\\item If $G_1$ and $G_2$ are topological groups such that $G_1 \\times G_2$ acts continuously on $X$,\n then $G_1$ acts continuously on $X / G_2$.\n\\end{enumerate}\n\n\\proof\nLet $x \\in X$ such that $[x] \\in X / G_2$, and $g \\in G_1$.\nDefine $g \\cdot [x] = [g \\cdot x]$.\nWe have to show that this action is well-defined.\nSuppose that $[x] = [y]$ for some $y \\in X$.\nThen there exists an $h \\in G_2$ such that $x = h \\cdot y$.\nBecause $h$ and $g$ commute in $G_1 \\times G_2$,\nwe have\n\\[ g \\cdot x = g \\cdot (h \\cdot y) = (g, h) \\cdot y = h \\cdot (g \\cdot y) \\]\nThus, we find $[g \\cdot x] = [g \\cdot y]$.\nThat this defines a homomorphism $G_1 \\to \\Aut(X / G_2)$ follows\nfrom the fact that $G_1 \\to \\Aut(X)$ is a homomorphism.\nThis proves statement \\textbbf{i}.\n\nSuppose that $X$ is a topological space\nand $G_1 \\times G_2$ acts on $X$ in the category of topological spaces.\nLet $g \\in G_1$, then $g$ induces a homeomorphism $\\phi : X \\to X$,\nand a bijection $\\psi : X / G_2 \\to X / G_2$.\nDenote by $q : X \\surj X / G_2$ the quotient map,\nthen $q \\circ \\phi$ is a continuous map $X \\to X / G_2$\nthat satisfies $\\psi \\circ q = q \\circ \\phi$ due to statement \\textbbf{i}.\nThis means that $q \\circ \\phi$ is compatible with the quotient map,\nso by the universal property of the quotient topology (theorem \\ref{thm:universal-property-quotient-topology}),\nthere exists a unique continuous map $\\psi'$ such that $q \\circ \\phi = \\psi' \\circ q$.\nUniqueness implies that $\\psi' = \\psi$,\ntherefore $\\psi$ is continuous.\nThe same argument applies to $\\psi^{-1}$,\nhence $\\psi$ is a homeomorphism.\nThis shows that $G_1$ acts on $X / G_2$ in the category of topological spaces,\nwhich proves statement \\textbbf{ii}.\n\nTo prove statement \\textbbf{iii},\nwe will use the following diagram:\n\\begin{center}\n\\begin{tikzcd}[row sep = large]\n&\nX \\ar[dr, two heads, \"q\"] & \\\\\nG_1 \\times X \\ar[ur, \"a\"]\n \\ar[r, \"f\"]\n \\ar[dr, two heads, swap, \"r\"] &\nG_1 \\times {(X / G_2)} \\ar[r, \"T\"] &\nX / G_2 \\\\ &\n{(G_1 \\times X)} / G_2 \\ar[u, dashed, \"\\exists_! \\phi\"]\n \\ar[ur, dashed, swap, \"\\exists_! \\psi\"] &\n\\end{tikzcd}\n\\end{center}\nThe map $a : G_1 \\times X \\to X$,\n$(g, x) \\mapsto g \\cdot x$\nis continuous because it is the restriction of $G_1 \\times G_2 \\times X \\to X$\nthat is continuous by assumption.\nLet $q: X \\surj X / G_2$ denote the quotient map.\nDefine $f : G_1 \\times X \\to G_1 \\times {(X / G_2)}$ as $f = (\\id, q)$.\nThis map is continuous because both of its coordinates are.\n(See proposition 1 of \\parencite[ch.~\\textsc{i}, \\S~4.1]{bourbaki1971}.)\nFurthermore $f$ is open,\nfor $\\id$ and $q$ are open. (See proposition~\\ref{prop:open-quotient-map}.)\nLet $G_2$ act on $G_1 \\times X$ by $h \\cdot (g, x) = (g, h \\cdot x)$\nwhere $g \\in G_1, h \\in G_2, x \\in X$,\nand let $r$ denote the quotient map.\n$f$ is compatible with $r$,\nso by the universal property of the quotient topology (theorem \\ref{thm:universal-property-quotient-topology}),\nthere exists a unique continuous map\n$\\phi$ that makes the bottom left triangle of the diagram commute.\nThe map is given by $[g, x] \\mapsto (g, [x])$\nand its inverse is given by $(g, [x]) \\mapsto [g, x]$.\nAn open set in $(G_1 \\times X) / G_2$\nis the image under $r$ of an open set in $G_1 \\times X$,\nso from commutativity it follows that $\\phi$ is an open map.\nHence, $\\phi$ is a homeomorphism.\n\nOn the top of the diagram, we have the map $q \\circ a : G_1 \\times X \\to X / G_2$,\ngiven by $(g, x) \\mapsto [g \\cdot x]$.\nAs the composition of continuous maps it is continuous,\nand it is compatible with $r$.\nThus, by the universal property of the quotient topology,\nthere exists a unique continuous map $\\psi$\nsuch that $\\psi \\circ r = q \\circ a$.\nComposing with $\\phi^{-1}$, we find that the map\n\\[ T : G_1 \\times {(X / G_2)} \\longto X / G_2,\n\\quad (g, [x]) \\longmapsto [g \\cdot x] \\]\nis continuous,\nwhich proves claim \\textbbf{iii}.\nFurthermore, the above diagram commutes.\n\\qed\n\n\\theorem[thm:quotient-map-factors]\nLet $G_1$ and $G_2$ be groups and $X$ a topological space\nsuch that $G_1 \\times G_2$ acts on $X$.\nThen $X / (G_1 \\times G_2)$ is canonically homeomorphic to $(X / G_1) / G_2$.\nIn particular, the quotient map $X \\surj X / (G_1 \\times G_2)$ factors over $X / G_1$.\n\n\\proof\nLet\n$q_1 : X \\surj X / G_1$,\n$q_2 : (X / G_1) \\surj (X / G_1) / G_2$,\nand $q_{12} : X \\surj X / (G_1 \\times G_2)$ denote the quotient maps.\nThen we have the following commutative diagram:\n\\begin{center}\n\\begin{tikzcd}[row sep = huge, column sep = large]\n X \\ar[r, two heads, \"q_1\"]\n \\ar[d, two heads, \"q_{12}\"] &\n X / G_1 \\ar[d, two heads, \"q_2\"]\n \\ar[dl, dashed, bend right = 7, swap, \"\\phi_1\"] \\\\\n X / (G_1 \\times G_2) \\ar[r, dashed, bend left = 13, \"\\phi_{12}\"] &\n(X / G_1) / G_2 \\ar[l, dashed, bend left = 13, \"\\phi_2\"]\n\\end{tikzcd}\n\\end{center}\nThe map $q_2 \\circ q_1$ is continuous and compatible with $q_{12}$,\nso by the universal property of the quotient topology (theorem \\ref{thm:universal-property-quotient-topology})\nthere exists a unique continuous map $\\phi_{12}$ that makes the diagram commute.\nBecause $q_{12}$ is compatible with $q_1$,\nthere exists a unique continuous map $\\phi_1$ such that $q_{12} = \\phi_1 \\circ q_1$.\nIt follows that $\\phi_1$ is compatible with $q_2$,\nso there exists a unique continuous map $\\phi_2$ that makes the diagram commute.\nNow we see that $\\phi_{12}$ and $\\phi_2$ are continuous inverses of one another,\nhence $X / (G_1 \\times G_2)$ and $(X / G_1) / G_2$ are homeomorphic.\n\\qed\n\n\\subsection*{Physical interpretation}\nGroups are prevalent in mathematics.\nIn physics, groups are often encountered in the context of symmetries.\nIn that case one may think of a group as a set of transformations of a system,\ntransformations under which a certain property is invariant.\nFor instance, angular momentum is invariant under rotation of space,\nand four-momentum is invariant under Lorentz transformations.\nA \\emph{group action} generalises this idea.\nElements of the group induce a transformation of a system.\nBy applying all possible transformations to a point,\nwe obtain the \\emph{orbit} of a point.\nFor instance, when we let the Lorentz group act on Minkowski space,\nthe orbit of a timelike vector is all of the light cone (past and future).\nOften, a group encodes transformations that we are \\emph{not} interested in.\nThe \\emph{orbit space} is what remains if we consider points that differ\nby such a transformation to be equal.\nFor example, the orbit space of the Lorentz group action on Minkowski space\nconsists of four elements:\nthe origin, the class of null (or light-like) vectors,\nthe class of timelike vectors,\nand the class of spacelike vectors.\nThe \\emph{stabiliser} of a point is the subgroup of transformations under which the point is invariant.\n\nTopology is the branch of mathematics that studies abstract properties of space.\nIt gives us the tools to study properties that do not depend\non exact distances, but rather on overall shape.\nFor instance, one would like to think of a garden hose\nas a one dimensional system where water can move back and forth,\nregardless of how the hose is bent or twisted.\nTopology allows us to ignore the bending and twisting.\nVirtually all spaces that occur in physics are topological spaces:\n$\\R^3\\!$, Minkowski space, Hilbert spaces, etc.\nOften these spaces have additional structure\nsuch as a metric or inner product,\nbut many properties can be derived from the topology alone.\nAn important example of such a property\nis \\emph{continuity} of a map between topological spaces,\na notion that is prevalent throughout physics.\nMany of the groups encountered in this thesis happen to have a natural topology as well.\nIn this case, an action on another topological space can be \\emph{continuous}.\nThe definition given in this section codifies our intuition:\nif two group elements that are near act on a point,\nthe resulting points should be near as well.\n\n\\section{Projective space}\n\\label{sec:projective-space}\nIt is possible to identify $\\R^4$ and $\\C^2$ as four-dimensional real vector spaces,\nby identifying the standard basis $(e_1, e_2, e_3, e_4)$ with the basis $((1, 0), (i, 0), (0, 1), (0, i))$.\nThe space $\\C^2 \\setminus \\set{0}$ will be prevalent in the rest of this section,\nso we introduce a shorthand notation. Furthermore, we embed $S^1$ in $\\C$.\n\n\\definition\n$\\CZ = \\C^2 \\setminus \\set{0}$.\n\n\\definition\nThe \\emph{unit circle} is defined by\n\\[ S^1 = \\set{ \\, z \\in \\C \\mid 1 = |\\,z\\,| \\, } \\]\nThis is a group under multiplication.\n\n\\definition[def:s3-real]\nThe \\emph{three-sphere} is defined by\n\\[ S^3 = \\set{ \\, x \\in \\R^4 \\mid 1 = \\nsq{x} } \\]\nHere $\\|{}\\cdot{}\\|$ denotes the regular Euclidean norm.\nBy identifying $\\R^4$ with $\\C^2$ as above,\nwe can consider $S^3$ to be a subset of $\\CZ$.\n\nConsider the multiplicative group $\\Rpos$ of positive real numbers.\nIt acts continuously on $\\C^2$ (in the category of real vector spaces) by scalar multiplication,\nand this action can be restricted to $\\CZ$ (in the category of sets).\nThis allows us to give an alternative definition of $S^3$ as a quotient:\n\n\\definition[def:s3-complex]\n$\\SC$ is the orbit space of $\\CZ$ with respect to the $\\Rpos$ action.\nDenote by $r : \\CZ \\surj \\SC$ the quotient map.\n$\\C^2$ is endowed with its regular topology induced by the Euclidean metric,\nand $\\SC$ is endowed with the quotient topology.\n\nIntuitively, this definition is not that different from definition~\\ref{def:s3-real}.\nEvery point $p$ at the three-sphere defines a ray from the origin through $p$.\nThis ray, except for the origin, is the orbit of $p$ under the $\\Rpos$ action.\nIn other words, every orbit can be represented by a point at unit distance from the origin.\nThe quotient map $r$ corresponds to projection onto the sphere.\n\n\\proposition[prop:s3-equivalence]\n$S^3$ and $\\SC$ as defined in definition~\\ref{def:s3-real} and \\ref{def:s3-complex} are homeomorphic.\n\n\\proof\n\\newcommand*{\\RZ}{{\\R^4_{\\,\\circ}}}\nWrite $\\R^4 \\setminus \\set{0} = \\RZ$.\nLet $i : S^3 \\to \\RZ$ be the inclusion,\nand let $\\phi : \\R^4 \\to \\C^2$ be the vector space isomorphism\ninduced by the identification of the bases given earlier in this section.\nThe inclusion $i$ is continuous, and the restriction $\\phi|_\\RZ = \\phi_\\circ$ is a homeomorphism.\nTherefore, the composition $\\psi = r \\circ \\phi_\\circ \\circ i : S^3 \\to \\SC$ is continuous.\nConsider the map\n\\vspace{-0.3\\parskip}\n\\[ \\CZ \\longto S^3,\n\\quad x \\longmapsto \\phi_\\circ^{-1} \\left( \\frac{x}{\\|x\\|} \\right) \\vspace{0.3\\parskip} \\]\nwhich is continuous and compatible with $r$.\nBy the universal property of the quotient topology (theorem~\\ref{thm:universal-property-quotient-topology}),\nthis map induces a unique continuous map $\\psi^{-1} : \\SC \\to S^3$ that is the inverse of $\\psi$.\nThus, $\\psi$ is a homeomorphism.\n\\qed\n\nConsider the multiplicative group $\\C^*$ (the complex plane minus the origin).\nIt acts continuously on $\\C^2$ (in the category of complex vector spaces) by scalar multiplication,\nand this action can be restricted to $\\CZ$.\nThis allows us to define the projective space:\n\n\\definition[def:complex-projective-line]\nThe \\emph{complex projective line} $\\PC$ is the orbit space of $\\CZ$ with respect to the $\\C^*$ action.\nDenote by $q : \\CZ \\surj \\PC$ the quotient map.\n$\\PC$ is endowed with the quotient topology.\n\nElements of $\\PC$ are indicated by \\emph{homogeneous coordinates}:\nif $(z_1, z_2) \\in \\C^2$ is nonzero,\nthen we write $(z_1 : z_2)$ for $q(z_1, z_2)$.\nWe can embed $\\C$ in $\\PC$ via $z \\mapsto (z : 1)$.\nThe only point that is not reached in this manner is $(1 : 0)$.\n\n\\theorem[thm:s2-homeom-p1c]\nThere exists a homeomorphism between $S^2$ and $\\PC$.\n\n\\proof\nWe will postpone the proof until section \\ref{sec:hopf-quaternionic},\nand prove this with the aid of quaternions in theorem~\\ref{thm:hopf-map-equivalence}.\nFor an alternative proof, see \\parencite[ch.~\\textsc{viii}, \\S~4.3]{bourbaki1974}.\n\nThe general linear group $\\GLC$\nof invertible complex $2 \\times 2$ matrices\nacts on $\\C^2$ by matrix multiplication.\nThis induces a group action of $\\GLC$ on $\\CZ$.\nFurthermore, the groups $\\Rpos$, $S^1$, and $\\C^*$ are isomorphic to subgroups of $\\GLC$:\ngiven an element $z \\in \\C^*\\!$,\nwe can identify it with the matrix\n\\[ \\begin{pmatrix} z & 0 \\\\ 0 & z \\end{pmatrix} \\]\nin the centre of $\\GLC$.\n$\\C^*$ is isomorphic to the direct product $\\Rpos \\times S^1$:\nthis is the decomposition of a complex number into its modulus and argument.\nIt follows that $S^1$ and $\\Rpos$ are central in $\\GLC$, because their elements correspond to scalar matrices.\nConsequently, $S^1$ and $\\Rpos$ are normal in $\\GLC$.\n\n\\subsection*{Informal summary}\nThe \\emph{projective space} $\\PC$ is a construction with several interpretations.\nFor starters, $\\PC$ can be thought of as $\\C$ with one extra point,\na point “at infinity”.\nThis allows us to talk about $z_1/z_2$ even when $z_2$ is zero.\nInstead of $z_1/z_2$, we write $(z_1 : z_2)$, called \\emph{homogeneous coordinates}.\nSecondly, theorem \\ref{thm:s2-homeom-p1c} tells us that $\\PC$ can be thought of as the unit sphere $S^2\\!$.\n(In fact, $\\PC$ is sometimes called the \\emph{Riemann sphere}.)\nA \\emph{homeomorphism} between two spaces\nis a function, both one-to-one and onto,\nthat preserves all topological properties.\nFrom a topological point of view, $\\PC$ and $S^2$ are the same space.\nThis means that when we formulate the Hopf map later on\n— a function from $S^3$ to $S^2$ —\nwe can express it as a function to $\\PC$.\nThis expression is significantly simpler than the one involving Cartesian coordinates on $S^2\\!$.\n\n\\section{Quaternions}\n\\label{sec:quaternions}\n\n\\definition\nThe \\emph{quaternion algebra} $\\H$ is\nthe real noncommutative algebra with basis $(1, i, j, k)$.\nMultiplication is given by the identities\n\\[ i^2 = j^2 = k^2 = -1,\n\\quad i\\!j = k, \\ jk = i, \\ ki = j,\n\\quad j i = -k, \\ kj = -i, \\ ik = -j \\]\nand $1$ commutes with all elements.\nIn particular, $\\H$ is a ring and a four-dimensional real vector space.\nAnalogously to complex numbers,\nthis algebra has an involution $\\overline{\\raisebox{0pt}[0.5em]{${}\\cdot{}$}}$\ncalled \\emph{conjugation}\nthat flips the sign of the $i$, $j$, and $k$ components.\n\n\\definition\nThe \\emph{trace} is the map $\\Tr : \\H \\to \\R,\\ q \\mapsto q + \\overline{q}$.\nBecause the imaginary parts cancel, the trace of a quaternion is real.\nFurthermore, the trace is $\\R$-linear.\n\nThe reals commute with all quaternions,\nso $\\Tr(q)$ commutes with $q$ for all $q \\in \\H$.\nBecause $\\overline{q} = \\Tr(q) - q$,\nit follows that $q$ and $\\overline{q}$ commute.\n\n\\definition\nThe standard inner product on $\\H$ is given by\n\\[ \\inp{{}\\cdot{}}{{}\\cdot{}} : \\H \\times \\H \\longto \\R,\n\\quad (p, q) \\longmapsto \\tfrac{1}{2} \\Tr(p\\overline{q}) = \\tfrac{1}{2}(p\\overline{q} + q \\overline{p}) \\]\nSymmetry is clear from the definition,\nand bilinearity follows from the linearity of the trace.\nFor positive definiteness,\nremark that for $q = a + bi + cj + dk$,\nwe have $q \\overline{q} = a^2 + b^2 + c^2 + d^2\\!$.\nTherefore $\\inp{q}{q} \\geq 0$,\nand $\\inp{q}{q} = 0 \\implies a = b = c = d = 0 \\implies q = 0$.\n\n\\definition\nThe \\emph{norm} of $q \\in \\H$ is given by $\\nsq{q} = q \\overline{q}$.\nBecause $q$ and $\\overline{q}$ commute,\n$q\\overline{q} = \\smallfrac{1}{2}(q \\overline{q} + \\overline{q} q)$,\nso the norm is induced by the inner product.\nThis norm coincides with the Euclidean norm on $\\H$\nas real vector space with orthonormal basis $(1, i, j, k)$.\nTherefore,\n$\\H$ with the topology induced by the norm is homeomorphic to $\\R^4\\!$.\n\nBecause conjugation reverses the order of multiplication,\nthe norm is multiplicative:\nfor $p, q \\in \\H$, we have\n\\[ \\nsq{pq}\n = (pq)\\overline{(pq)}\n = p \\, q \\overline{q} \\, \\overline{p}\n = p \\nsq{q} \\, \\overline{p}\n = \\nsq{q} p \\overline{p}\n = \\nsq{q} \\nsq{p} \\]\n\nBecause $q \\overline{q} = \\nsq{q}\\!$,\nwe have $q^{-1} = \\overline{q} \\, \\|q\\|^{-2}$ for $\\|q\\| \\neq 0$.\nTherefore, $\\H$ is a division algebra:\nevery nonzero element has an inverse.\n\n\\proposition\n$\\H^* = \\H \\setminus \\set{0}$ is a topological group.\n\n\\proof\nMultiplication is continuous,\nbecause for $p, q \\in \\H^*\\,$,\nthe components of the product $pq$\ncan be written as a polynomial in the components of $p$ and $q$.\nInversion is continuous,\nbecause the components of $q^{-1}$ are rational functions of the components of $q$,\nwhich do not vanish because $\\nsq{q} \\neq 0$.\nSee also \\parencite[ch.~\\textsc{viii}, \\S~1.4]{bourbaki1974}.\n\\qed\n\nWith this machinery,\nwe can give a quaternionic definition of $S^3$ and $S^2\\!$.\nWhereas the definitions in section \\ref{sec:projective-space} emphasise\nhow $S^3$ and $S^2$ are quotients with respect to a group action,\nthe quaternionic definitions emphasise the group structure on the three-sphere itself,\nand the action of $S^3$ on $S^2\\!$.\n\nLet us revisit the three-sphere as defined in definition~\\ref{def:s3-real}.\nBy identifying $\\H$ with $\\R^4$ as a normed real vector space\nvia the basis given earlier in this section,\nwe can consider $S^3$ to be a subset of $\\H$,\nthe set of quaternions with unit norm:\n\\[ S^3 = \\set{ q \\in \\H \\mid 1 = \\nsq{q} } \\]\nThis set is closed under multiplication due to the multiplicativity of the norm,\nand it contains $1$.\nTherefore, this is a subgroup of $\\H^*\\!$.\nWe can embed $S^2$ in $S^3\\!$,\nbut in $\\R^4$ there is no preferred way of doing so.\nFor quaternions, there is one natural choice:\n\n\\definition[def:s2-quaternion]\nThe \\emph{two-sphere} $S^2 = \\set{ q \\in S^3 \\mid \\Tr(q) = 0 }$,\nthe set of pure imaginary quaternions with unit norm.\nThis definition coincides with the conventional definition of $S^2$\nwhen $\\R^3$ is identified with the subspace of $\\H$ spanned by $i$, $j$, and $k$.\n$S^2$ may alternatively be written as $\\set{q \\in S^3 \\mid \\inp{1}{q} = 0} = 1^\\perp \\cap S^3$.\n\nThe group $\\H^*$ acts\non $\\H$ in the category of $\\R$-algebras via the following homomorphism:\n\\[ \\phi : \\H^* \\longto \\Aut(\\H),\n \\quad\n p \\longmapsto (q \\mapsto p q p^{-1}) \\]\nBecause quaternion multiplication is continuous,\nthis is a continuous action.\nBy restriction to the subgroup $S^3\\!$,\nwe get a continuous action of $S^3$ on $\\H$.\n\n\\proposition[prop:invariant-inner-product]\nThe inner product on $\\H$ is invariant under the action of $\\H^*\\!$.\n\n\\proof\nLet $p \\in \\H^*\\!$, $q_1, q_2 \\in \\H$,\nthen we have\n\\begin{align*}\n 2 \\, \\inp{p \\cdot q_1}{p \\cdot q_2}\n&= p q_1 p^{-1} \\overline{p q_2 p^{-1}} + p q_2 p^{-1} \\overline{p q_1 p^{-1}} \\\\\n&= p q_1 \\nsq{p^{-1}} \\overline{q_2} \\, \\overline{p} + p q_2 \\nsq{p^{-1}} \\overline{q_1} \\, \\overline{p} \\\\\n&= \\nsq{p^{-1}} p (q_1 \\overline{q_2} + q_2 \\overline{q_1}) \\overline{p} \\\\\n&= \\nsq{p^{-1}} \\, \\nsq{p} \\, 2 \\, \\inp{q_1}{q_2} \\\\\n&= 2 \\, \\inp{q_1}{q_2}\n \\tag*{\\qed}\n\\end{align*}\n\n\\corollary[cor:s3-action]\nIdentify $\\R^3$ with the subspace of $\\H$ spanned by $i$, $j$, and $k$.\nThen $\\R^3 = 1^\\perp$ and $S^2 = 1^\\perp \\cap S^3$ are invariant under the action of $\\H^*\\!$,\nwhich means $\\H^*$ and its subgroup $S^3$ act continuously on $\\R^3$ and $S^2\\!$.\n\n$\\C$ is a commutative subring of~$\\H$.\nAs real vector spaces with bases $(1, i)$ and $(1, i, j, k)$,\n$\\C$ can be identified with the subspace of $\\H$ spanned by $1$ and $i$.\nThe stabiliser of $i \\in \\H$ consists of the nonzero elements that commute with $i$.\nThese elements are linear combinations of $1$ and $i$,\nso we have $\\H^*_i = \\C^*$ and $S^3_i = S^1\\!$.\n\n\\proposition[prop:s3-isom-su2c]\n$S^3$ is isomorphic to $\\SUC$, the group of unitary $2 \\times 2$ matrices with determinant $1$.\n\n\\proof\nDefine the unitary matrices\n\\[ I = \\begin{pmatrix}1 & 0 \\\\ 0 & 1\\end{pmatrix}\n\\qquad \\sigma_1 = \\begin{pmatrix}0 & 1 \\\\ 1 & 0\\end{pmatrix}\n\\qquad \\sigma_2 = \\begin{pmatrix}0 & -i \\\\ i & 0\\end{pmatrix}\n\\qquad \\sigma_3 = \\begin{pmatrix}1 & 0 \\\\ 0 & -1\\end{pmatrix} \\]\nThese matrices are sometimes called the \\emph{Pauli spin matrices}.\nLet $\\phi: \\H \\to \\Mat(2 \\times 2, \\C)$ be the $\\R$-linear extension of\n\\[ 1 \\longmapsto I,\n\\quad i \\longmapsto i \\sigma_1,\n\\quad j \\longmapsto i \\sigma_2,\n\\quad k \\longmapsto i \\sigma_3 \\]\nLet $\\psi$ be the restriction of $\\phi$ to $S^3\\!$.\nAll matrices in the image of $\\psi$ are unitary,\nand a little computation shows that for $q \\in S^3\\!$,\n$\\det \\psi(q) = 1$.\nThe matrices $I, i\\sigma_1, i\\sigma_2, i\\sigma_3$\nsatisfy the same multiplication rules as $1, i, j, k$.\nThat is, $i\\sigma_1 i\\sigma_2 = i\\sigma_3$, etc.\nTherefore, $\\psi$ is a group homomorphism $S^3 \\to \\SUC$.\nThis homomorphism is surjective (see \\parencite[p.~173]{szekeres2004}),\nand $1$ is the only element in its kernel.\nTherefore, $\\psi$ is an isomorphism.\n\\qed\n\n\\theorem[thm:s3-surj-so3r]\nThe map\n\\[ \\phi: S^3 \\longto \\SOR,\n\\quad q \\longmapsto (x \\mapsto q \\cdot x) \\]\nis a surjective group homomorphism with kernel $\\set{\\pm 1}$.\nHere $x \\in \\R^3 \\cong \\Span(i, j, k)$.\n\n\\proof\nThe map $x \\mapsto q \\cdot x$ is linear,\nand orthogonality follows from\nthe fact that the inner product is invariant under the action,\nas shown in proposition~\\ref{prop:invariant-inner-product}.\nTo show that $x \\mapsto q \\cdot x$ is not a reflection,\nnote that $\\det : \\textup{O}_3(\\R) \\to \\R$ is a continuous map\n(see for example \\parencite[p.~281]{hatcher2002}).\nWe can express $\\phi$ as a polynomial on all coordinates\nwhen elements of $\\SOR$ are written as matrices,\nso $\\phi$ is continuous.\nBy composition we get a continuous map $S^3 \\to \\set{\\pm 1}$.\nBecause $S^3$ is connected, this map must be constant.\n% (See for instance proposition~3.4.11 of \\parencite[p.~92]{runde2005}.)\nThe determinant of $\\id$ is $1$,\nso all $q \\in S^3$ induce an orthogonal map with positive determinant.\n\nTo show that the kernel of $\\phi$ is $\\set{\\pm 1}$,\nsuppose that $q \\in S^3$ is such that $q \\cdot x = q x q^{-1} = x$ for all $x \\in \\R^3\\!$.\nThen $q$ commutes with all $x \\in \\R^3\\!$,\nso $q$ must be real.\nBecause $\\nsq{q} = 1$, it follows that $q = 1$ or $q = -1$.\n\nTo prove surjectivity,\nsuppose that $\\rho \\in \\SOR$ is an anticlockwise rotation\nof $\\alpha$ radians about an axis spanned by $u \\in \\R^3\\!$,\nwhere $\\nsq{u} = 1$.\nThen the quaternion $q = \\cos(\\smallfrac{1}{2}\\alpha) + u \\sin(\\smallfrac{1}{2}\\alpha)$\nwill map to $\\rho$.\nTo see this,\nnote that all points on the axis of rotation are fixed points,\nfor $u$ commutes with $q$.\nFurthermore, suppose that $v \\in \\R^3$ is such that $\\inp{u}{v} = 0$.\nSet $q_0 = \\cos(\\smallfrac{1}{2}\\alpha)$ and $\\vec{q} = u \\sin(\\smallfrac{1}{2}\\alpha)$.\nBy using identities from \\parencite[p.~157]{szekeres2004},\nwe find\n\\newcommand*{\\vv}{v} % Could switch to v with an arrow, but to me it is just noise.\n\\newcommand*{\\vq}{\\vec{q}}\n\\begin{align*}\nq \\cdot v &= (q_0 + \\vq) \\vv (q_0 - \\vq)\n% \\\\ &= (q_0 + \\vq) (\\inp{v}{\\vq} + q_0 \\vec{v} - \\vec{v} \\times \\vq)\n \\\\ &= (q_0 + \\vq) (q_0 \\vv - \\vv \\times \\vq)\n \\\\ &= -\\inp{\\vq}{q_0 \\vv - \\vv \\times \\vq} + q_0 (q_0 \\vv - \\vv \\times \\vq) + \\vq \\times {(q_0 \\vv - \\vv \\times \\vq)}\n \\\\ &= q_0^2 \\vv - q_0 \\vv \\times \\vq + q_0 \\vq \\times \\vv - \\vq \\times {(\\vv \\times \\vq)}\n \\\\ &= q_0^2 \\vv - 2 q_0 \\vv \\times \\vq - \\vv \\inp{\\vq}{\\vq} + \\vq \\inp{\\vq}{\\vv}\n \\\\ &= (\\cos^2(\\smallfrac{1}{2}\\alpha) - \\sin^2(\\smallfrac{1}{2}\\alpha)) v\n - 2 \\cos(\\smallfrac{1}{2}\\alpha)\\sin(\\smallfrac{1}{2}\\alpha) \\, {v \\times u}\n \\\\ &= \\cos(\\alpha) v + \\sin(\\alpha) \\, {u \\times v}\n\\end{align*}\nThis demonstrates that $q$ rotates $v$ anticlockwise by $\\alpha$ radians about $u$.\nWe saw already that $x \\mapsto q \\cdot x$ is an orthogonal map with determinant $1$.\nTherefore, $q$ maps to $\\rho$.\n\\qed\n\n\\corollary[cor:transitive-s3-action]\n$S^3$ acts transitively on $S^2\\!$,\nfor every point on $S^2$ can be mapped\ninto any other point on $S^2$ by a rotation of the sphere.\n\nThe proof of theorem~\\ref{thm:s3-surj-so3r} gives us a way to explicitly get\na $q \\in S^3$ such that $q \\cdot i = p$ for any $p \\in S^2$:\nwe rotate $i$ onto $p$ with a rotation of $\\R^3\\!$.\nIf $p = -i$, $q = j$ will suffice,\nso suppose $p \\neq -i$.\nThen an axis that we can rotate about is the one spanned by $i + p$,\nwhich bisects the angle between $i$ and $p$,\nso we need to rotate by $\\pi$ radians.\nWe find\n\\begin{equationref}\n\\label{eqn:transitive-s3-action}\nq = \\frac{i + p}{\\|i + p\\|}\n\\end{equationref}\nTo verify that this works,\nnote that for $p \\in S^2$ we have\n$p \\overline{p} = 1$ and $\\overline{p} = -p$,\nso $p^2 = -1$.\nIt then follows that\n\\[ p^2 = -1\n% \\implies pi + p^2 = -1 + pi\n\\enskip \\implies \\enskip p(i + p) = (i + p)i\n\\enskip \\implies \\enskip p(i + p)\\overline{(i + p)} = (i + p)i\\overline{(i + p)} \\]\nMultiplying by $\\|i + p\\|^{-2}$ on both sides then yields $p = q i q^{-1}$.\n\n\\subsection*{Physical interpretation}\nJust like complex numbers are an extension of the real numbers,\nquaternions are an extension of the complex numbers.\nThese extensions come at a cost:\nwhen going from $\\R$ to $\\C$, you have to give up the ordering.\nWhen going from $\\C$ to $\\H$, you have to give up commutativity.\nApart from their rich structure that is interesting in its own right,\nquaternions have many useful applications.\nBy considering $S^3$ as a subset of $\\H$,\nit inherits a group structure.\nTheorem~\\ref{thm:s3-surj-so3r} tells us that this group is in a sense\ntwice $\\SOR$: every rotation of $\\R^3$ is represented by two antipodal quaternions.\nWhen traversing a great circle through $1$ in $S^3\\!$,\nthe points $1$ and $-1$ both correspond to the identity in $\\SOR$.\nThis path in $S^3$ corresponds to a $4\\pi$ rotation of $\\R^3\\!$,\nand after a $2\\pi$ rotation we will have moved from $1 \\in S^3$ to $-1 \\in S^3\\!$.\nThis property is reminiscent of \\emph{spinors},\nand indeed proposition~\\ref{prop:s3-isom-su2c} links the unit quaternions to the Pauli spin matrices.\n", "meta": {"hexsha": "08a9bdc9e35fc3524e2723e40ab5cdf889bf1905", "size": 31817, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/preliminaries.tex", "max_stars_repo_name": "ruud-v-a/bscthesis", "max_stars_repo_head_hexsha": "7dfc0003ed842b556d530f2ffb6313856ce17721", "max_stars_repo_licenses": ["CNRI-Python", "Naumen", "Condor-1.1", "MS-PL"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/preliminaries.tex", "max_issues_repo_name": 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YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7879311956428946, "lm_q1q2_score": 0.6289162621307292}} {"text": "\n\\chapter{Filtering}\n\\label{sec:filtering}\n\\section{Curvelet Coefficient Thresholding}\nWe now apply our digital transforms for\nremoving noise from image data. The methodology is standard and is\noutlined mainly for the sake of clarity and self-containedness.\n\nSuppose that one is given noisy data of the form\n\\[\nx_{i,j} = f(i,j) + \\sigma z_{i,j},\n\\]\nwhere $f$ is the image to be recovered and $z$ is white noise, i.e.\n$z_{i,j} \\stackrel{i.i.d.}{\\sim} N(0,1)$. Unlike FFTs or FWTs, our\ndiscrete ridgelet (resp. curvelet) transform is not norm-preserving\nand, therefore, the variance of the noisy ridgelet (resp. curvelet)\ncoefficients will depend on the ridgelet (resp. curvelet) index\n$\\lambda$. For instance, letting $F$ denote the discrete curvelet\ntransform matrix, we have $F z \\stackrel{i.i.d.}{\\sim} N(0,F F^T)$\nwhere $T$ denotes transpose.\nBecause the computation of $F F^T$ is prohibitively expensive, we\ncalculated an approximate value $\\tilde{\\sigma}^2_\\lambda$ of the\nindividual variances using Monte Carlo simulations where the diagonal\nelements of $F F^T$ are simply estimated by evaluating the curvelet\ntransforms of a few standard white noise images.\n\nLet $y_\\lambda$ be the noisy curvelet coefficients\n($y = F x$). We use the following hard-thresholding rule\nfor estimating the unknown curvelet coefficients:\n\\begin{eqnarray}\n\\hat y_\\lambda = y_\\lambda & \\mbox{ if } &\n|y_\\lambda|/\\sigma \\geq k \\tilde{\\sigma}_\\lambda\\\\\n\\hat y_\\lambda = 0 & \\mbox{ if } & |y_\\lambda|/\\sigma\n< k \\tilde{\\sigma}_\\lambda.\n\\end{eqnarray}\nIn our experiments, we actually chose a scale-dependent value for $k$;\nwe have $k = 4$ for the first scale $(j = 1)$ while $k = 3$\nfor the others $(j > 1)$.\n\n\\subsection*{Poisson Observations}\nAssume now that we have Poisson data $x_{i,j}$ with unknown mean\n$f(i,j)$. The Anscombe transformation \\cite{rest:anscombe48}\n\\begin{equation}\n\\tilde{x} = 2\\sqrt{x + \\frac{3}{8}}\n\\end{equation}\nstabilizes the variance and we have $\\tilde{x} = 2 \\sqrt{f} +\n\\epsilon$ where $\\epsilon$ is a vector with independent and\napproximately standard normal components. In practice, this is a good\napproximation whenever the number of counts is large enough, greater\nthan 30 per pixel, say.\n \nFor a small number of counts, a possibility is to compute the Radon\ntransform of the image, and then to apply the Anscombe transformation\nto the Radon data. The rationale is that, roughly speaking, the\nRadon transform corresponds to a summation of pixel values over lines\nand the sum of independent Poisson random variables is a Poisson\nrandom variable with intensity equal to the sum of the individual\nintensities. Hence, the intensity of the sum may be quite large (hence\nvalidating the Gaussian approximation) even though the individual\nintensities may be small. This might be viewed as an interesting\nfeature since, unlike wavelet transforms, the ridgelet and curvelet\ntransforms tend to average data over elongated and rather large\nneighborhoods.\n\n\\section{Filtering Experiments}\n\n\\subsubsection{Recovery of Linear Features}\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{\n\\psfig{figure=fig_line.ps,bbllx=1cm,bblly=12cm,bburx=15cm,bbury=26cm,width=7.5cm,height=7.5cm,clip=}\n\\psfig{figure=fig_line_g0p5.ps,bbllx=1cm,bblly=12cm,bburx=15cm,bbury=26cm,width=7.5cm,height=7.5cm,clip=}\n}\n\\hbox{\n\\psfig{figure=fig_f24_line_g0p5.ps,bbllx=1cm,bblly=12cm,bburx=15cm,bbury=26cm,width=7.5cm,height=7.5cm,clip=}\n\\psfig{figure=fig_cur_line_g0p5.ps,bbllx=1cm,bblly=12cm,bburx=15cm,bbury=26cm,width=7.5cm,height=7.5cm,clip=}\n}}}\n\\caption{The top panels display a geometric image and that same image\n contaminated with Gaussian white noise. The bottom left and right\n panels display the restored images using the undecimated wavelet\n transform and the curvelet transform, respectively.}\n\\label{fig_cur_line}\n\\end{figure}\nThe next experiment (Figure~\\ref{fig_cur_line}) consists of an\nartificial image containing a few bars, lines and a square. The\nintensity is constant along each individual bar; from left to right,\nthe intensities of the ten vertical bars (these are in fact thin\nrectangles which are 4 pixels wide and 170 pixels long) are equal to\n${32 \\over 2^i}$, $i = 0, \\ldots 9$. The intensity along all the\nother lines is equal to 1, and the noise standard deviation is $1/2$.\nDisplayed images have been log-transformed in order to better see\nthe results at low signal to noise ratio.\n\nThe curvelet reconstruction of the nonvertical lines is obviously\nsharper than that obtained using wavelets. The curvelet transform\nalso seems to go one step further as far as the reconstruction of the\nvertical lines is concerned. Roughly speaking, for those templates,\nthe wavelet transform stops detecting signal at a SNR equal to 1\n(we defined here the SNR as the intensity level of the pixels\non the line, divided by the noise standard deviation of the noise)\nwhile the cut-off value equals 0.5 for the curvelet approach. It is\nimportant to note that the horizontal and vertical lines correspond to\nprivileged directions for the wavelet transform, because the underlying\nbasis functions are direct products of functions varying solely in\nthe horizontal\nand vertical directions. Wavelet methods will\ngive even poorer results on lines\nof the same intensity but tilting substantially\naway from the Cartesian axes.\nCf.\\ the reconstructions of the faint diagonal\nlines in the image.\n\n\\subsubsection{Recovery of Curves}\nIn this experiment (Figure~\\ref{fig_cur_picasso}), we have added \nGaussian noise to ``War and Peace,'' a drawing from Picasso which\ncontains many curved features. Figure \\ref{fig_cur_picasso} bottom left\nand right show respectively the restored images by the undecimated\nwavelet transform and the curvelet transform. Curves are more sharply\nrecovered with the curvelet transform.\n\nThe authors are working on new methods (some of which will be based on\nthe curvelet transform) to extract and recover curves from noisy data\nwith greater accuracy and, therefore, this example is merely to be\ntaken for illustrative purposes.\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{\n\\psfig{figure=fig_picasso.ps,bbllx=2cm,bblly=13cm,bburx=18cm,bbury=25cm,width=8cm,height=6cm,clip=}\n\\psfig{figure=fig_picasso_g50.ps,bbllx=2cm,bblly=13cm,bburx=18cm,bbury=25cm,width=8cm,height=6cm,clip=}\n}\n\\hbox{\n\\psfig{figure=fig_picasso_wt.ps,bbllx=2cm,bblly=13cm,bburx=18cm,bbury=25cm,width=8cm,height=6cm,clip=}\n\\psfig{figure=fig_picasso_cur.ps,bbllx=2cm,bblly=13cm,bburx=18cm,bbury=25cm,width=8cm,height=6cm,clip=}\n}}}\n\\caption{The top panels display a Picasso picture (War and Peace) \n and that same image\n contaminated with Gaussian white noise. The bottom left and right\n panels display the restored images using the undecimated wavelet\n transform and the curvelet transform respectively.}\n\\label{fig_cur_picasso}\n\\end{figure}\n\n\n\\subsubsection{Denoising of a Color Image}\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{\n\\psfig{figure=fig_cmp_lena_rgb.ps,bbllx=2cm,bblly=13cm,bburx=20cm,bbury=25cm,width=6cm,height=4.cm,clip=}\n\\psfig{figure=fig_cmp_pepper_rgb.ps,bbllx=2cm,bblly=13cm,bburx=20cm,bbury=25cm,width=6cm,height=4.cm,clip=}\n}\n\\hbox{\n\\psfig{figure=fig_cmp_mandrill_rgb.ps,bbllx=2cm,bblly=13cm,bburx=20cm,bbury=25cm,width=6cm,height=4.cm,clip=}\n\\psfig{figure=fig_cmp_barbara_rgb.ps,bbllx=2cm,bblly=13cm,bburx=20cm,bbury=25cm,width=6cm,height=4.cm,clip=}\n}\n}}\n\\caption{PSNR versus noise standard deviation using different filtering\n methods. YUV and curvelet, YUV and undecimated wavelet, and YUV and\n decimated wavelet transforms are represented respectively with a\n continuous, dashed, and dotted line. The upper left panel\n corresponds to {\\tt Lena} (RGB), the upper right to {\\tt pepper}\n (RGB), the bottom left to {\\tt Baboon} (RGB), and the bottom right\n to {\\tt Barbara} (RGB).}\n\\label{fig_exp_rgb_curv}\n\\end{figure}\n\nIn a wavelet based denoising scenario, color RGB images are generally\nmapped into the YUV space, and each YUV band is then filtered\nindependently from the others. The goal here is to see whether the\ncurvelet transform would give improved results. We used four of the\nclassical color images, namely {\\tt Lena}, {\\tt Peppers}, {\\tt\n Baboon}, and {\\tt Barbara} (all images except perhaps {\\tt Barbara}\nare available from the USC-SIPI Image Database \\cite{SIPI}).\nWe performed\na series of experiments and\nsummarized our findings in Figure~\\ref{fig_exp_rgb_curv} which again\ndisplays the PSNR (peak signal-to-noise ratio) \nversus the noise standard deviation for the four\nimages.\n\nIn all cases, the curvelet transform outperforms the wavelet\ntransforms in terms of PSNR -- at least for moderate and large values\nof the noise level. In addition, the curvelet transform outputs\nimages that are visually more pleasant.\n \n\\subsubsection{Saturn Rings}\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{\n\\psfig{figure=SATURN/fig_sat_sub_g20.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n\\psfig{figure=SATURN/fig_sat_sub_fil_owt.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n}\n\\hbox{\n\\psfig{figure=SATURN/fig_sat_sub_fil_atrou.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n\\psfig{figure=SATURN/fig_sat_sub_fil_cur.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n}}\n}\n\\caption{Top left, part of Saturn image with Gaussian noise. Top right,\nfiltered image using the undecimated bi-orthogonal wavelet transform.\nBottom left and right, filtered image by the \\`a trous wavelet \ntransform algorithm and the curvelet transform.}\n\\label{fig_sub_saturn_cur_filter}\n\\end{figure}\n\nGaussian white noise with a standard deviation\nfixed to 20 was added to the {\\tt Saturn} image.\nWe employed several methods to filter the noisy image:\n\\begin{enumerate}\n\\item Thresholding of the Curvelet transform.\n\\item Bi-orthogonal undecimated wavelet de-noising methods using the\n Dau\\-che\\-chies-An\\-to\\-ni\\-ni 7/9 fil\\-ters (FWT-7/9) and hard thresholding.\n\\item A trous wavelet transform algorithm and hard thresholding.\n\\end{enumerate}\nOur experiments are reported in Figure~\\ref{fig_sub_saturn_cur_filter}. The\ncurvelet reconstruction does not contain the quantity of disturbing artifacts\nalong edges that one sees in wavelet reconstructions. An examination\nof the details of the restored images % (Figure~\\ref{fig_sub_saturn_cur_filter})\nis instructive. One notices that the decimated wavelet transform\nexhibits distortions of the boundaries and suffers substantial loss of\nimportant detail. The \\`a trous wavelet transform gives better\nboundaries, but completely omits to reconstruct certain ridges. \nIn addition, it exhibits numerous small-scale embedded\nblemishes; setting higher thresholds to avoid these blemishes would\ncause even more of the intrinsic structure to be missed.\n\n% Further results are visible at the following URL: {\\tt\n% http://www-stat.stanford.edu/$\\sim$jstarck}.\n\n\\subsubsection{Supernova with Poisson noise}\n\n\\begin{figure}[htb]\n\\centerline{ \n\\hbox{\n\\psfig{figure=fig_kepler1604.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n\\psfig{figure=fig_kepler1604_fil_rid.ps,bbllx=1.8cm,bblly=12.7cm,bburx=14.5cm,bbury=25.4cm,width=7cm,height=7cm,clip=}\n}}\n\\caption{Left, XMM/Newton image of the Kepler SN1604 supernova.\nRight, ridgelet filtered image.}\n\\label{fig_sn1604}\n\\end{figure}\nFigure~\\ref{fig_sn1604} shows an example of an X-ray image filtering by the\nridgelet transform using such an approach. Figure~\\ref{fig_sn1604} left\nand right \nshow respectively the XMM/Newton image of the Kepler SN1604 supernova and\nthe ridgelet filtered image (using a five sigma hard thresholding). \n\n\n% \\clearpage\n% \\newpage\n", "meta": {"hexsha": "173585ec09c88d2bfc17f3b2363578e3ec91ada0", "size": 11788, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr4/ch_curfilter.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr4/ch_curfilter.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr4/ch_curfilter.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.8677042802, "max_line_length": 124, "alphanum_fraction": 0.7792670512, "num_tokens": 3455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.798186768138228, "lm_q2_score": 0.7879311906630568, "lm_q1q2_score": 0.6289162505906513}} {"text": "\\BoSSSopen{tutorial9-SIP/sip}\n\\graphicspath{{tutorial9-SIP/sip.texbatch/}}\n\n\\BoSSScmd{\n/// \\section*{What's new}\n/// \\begin{itemize}\n/// \\item Symmetric Interior Penalty method (SIP)\n/// \\item investigating matrix properties\n/// \\end{itemize}\n/// \\section*{Prerequisites}\n/// \\begin{itemize}\n/// \\item basics SIP method\n/// \\item spatial operator, chapter \\ref{SpatialOperator}\n/// \\item implementing numerical fluxes and convergence study, chapter \\ref{NumFlux}\n/// \\end{itemize}\n///\n///%_______________________________________________________\n/// \\section{Problem statement}\n/// We consider the 2D Poisson problem:\n/// \\begin{equation*}\n/// \\Delta u = f(x,y)\n/// \\end{equation*}\n/// with $f(x,y)\\neq 0$ is an arbitrary function of $x$ and/or $y$ or constant.\n/// Within this exercise, we are going to investigate the Symmetric Interior Penalty discretization method (SIP) for the Laplace operator:\n/// \\begin{equation*}\n/// a_{\\text{sip}}(u,v)\n/// = \\int_{\\domain} \\underbrace{\\nabla u \\cdot \\nabla v}_{\\text{Volume\\ term}}\\dV\n/// - \\oint_{\\Gamma \\setminus \\Gamma_{N }} \\underbrace{\n/// \\mean {\\nabla u} \\cdot n_{\\Gamma}\\jump{v}\n/// }_{\\text{consistency\\ term}} + \\underbrace{\n/// \\mean{\\nabla v} \\cdot \\vec{n}_{\\Gamma} \\jump{u}\n/// }_{\\text{symmetry\\ term}} \\dA\n/// + \\oint_{\\Gamma \\setminus \\Gamma_{N}} \\underbrace{\n/// \\eta \\jump{u}\\jump{v}\n/// }_{\\text{penalty\\ term}} \\dA\n/// \\end{equation*}\n/// The use of these fluxes including a penalty term stabilizes the DG-discretization for the Laplace operator.\n///\n///%____________________________________________\n/// \\section{Solution within the \\BoSSS{} framework}\n/// First, we initialize the new worksheet:\n }\n\\BoSSSexeSilent\n\\BoSSScmd{\nrestart;\n }\n\\BoSSSexeSilent\n\\BoSSScmd{\n/// We need the following packages:\nusing ilPSP.LinSolvers;\\newline \nusing ilPSP.Connectors.Matlab;\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// BoSSScmdSilent BoSSSexeSilent\nusing NUnit.Framework;\n }\n\\BoSSSexeSilent\n\\BoSSScmd{\n/// \\subsection{Implementation of the SIP fluxes}\n/// We are going to implement the SIP-form\n/// \\begin{equation*}\n/// a_{\\text{sip}}(u,v)\n/// = \\int_{\\domain} \\underbrace{\\nabla u \\cdot \\nabla v}_{\\text{volume\\ term}}\\dV\n/// - \\oint_{\\Gamma \\setminus \\Gamma_{N }} \\underbrace{\n/// \\mean {\\nabla u} \\cdot n_{\\Gamma}\\jump{v}\n/// }_{\\text{consistency\\ term}} + \\underbrace{\n/// \\mean{\\nabla v} \\cdot \\vec{n}_{\\Gamma} \\jump{u}\n/// }_{\\text{symmetry\\ term}} \\dA\n/// + \\oint_{\\Gamma \\setminus \\Gamma_{N}} \\underbrace{\n/// \\eta \\jump{u}\\jump{v}\n/// }_{\\text{penalty\\ term}} \\dA\n/// \\end{equation*}\n/// First, we need a class in which the integrands are defined.\n/// This also includes some technical aspects like the \\code{TermActivationFlags}.\npublic class SipLaplace :\\newline \n\\btab \\btab BoSSS.Foundation.IEdgeForm, // edge integrals\\newline \n\\btab \\btab BoSSS.Foundation.IVolumeForm // volume integrals\\newline \n\\{\\newline \n /// We do not use parameters (e.g. variable viscosity, ...)\n /// at this point: so this can be null\n\\btab public IList ParameterOrdering \\{ \\newline \n\\btab \\btab get \\{ return new string[0]; \\} \\newline \n\\btab \\} \\newline \n /// but we have one argument variable, $u$ (our trial function)\n\\btab public IList ArgumentOrdering \\{ \\newline \n\\btab \\btab get \\{ return new string[] \\{ \"u\" \\}; \\} \\newline \n\\btab \\}\\newline \n /// The \\code{TermActivationFlags} tell \\BoSSS\\ which kind of terms, \n /// i.e. products of $u$, $v$, $\\nabla u$, and $\\nabla v$\n /// the \\code{VolumeForm(...)} actually contains.\n /// This additional information helps to improve the performance.\n\\btab public TermActivationFlags VolTerms \\{\\newline \n\\btab \\btab get \\{\\newline \n\\btab \\btab \\btab return TermActivationFlags.GradUxGradV;\\newline \n\\btab \\btab \\}\\newline \n\\btab \\}\\newline \n /// activation flags for the 'InnerEdgeForm(...)'\n\\btab public TermActivationFlags InnerEdgeTerms \\{\\newline \n\\btab \\btab get \\{\\newline \n\\btab \\btab \\btab return (TermActivationFlags.AllOn);\\newline \n\\btab \\btab \\btab // if we do not care about performance, we can activate all terms.\\newline \n\\btab \\btab \\}\\newline \n\\btab \\}\\newline \n\\btab public TermActivationFlags BoundaryEdgeTerms \\{\\newline \n\\btab get \\{\\newline \n\\btab \\btab return TermActivationFlags.AllOn;\\newline \n\\btab \\btab \\}\\newline \n\\btab \\}\\newline \n /// For the computation of the penalty factor $\\eta$,\n /// we require\n /// some length scale for each cell and \n /// the polynomial degree of the DG approximation.\n\\btab public MultidimensionalArray cj;\\newline \n\\btab public int PolynomialDegree;\\newline \n /// The safety factor for the penalty factor should be in the order of 1.\n /// A very large penalty factor increases the condition number of the \n /// system, but without affecting stability.\n /// A very small penalty factor yields to an unstable discretization.\n\\btab public double PenaltySafety = 2.2; \\newline \n /// The actual computation of the penalty factor, which should be \n /// used in the \\code{InnerEdgeForm} and \\code{BoundaryEdgeForm} functions.\n /// Hint: for the parameters \\code{jCellIn}, \\code{jCellOut} and \\code{g},\n /// take a look at\n /// \\code{CommonParams} and \\code{CommonParamsBnd}.\n\\btab double PenaltyFactor(int jCellIn, int jCellOut) \\{\\newline \n\\btab \\btab double cj\\_in = cj[jCellIn];\\newline \n\\btab \\btab double penalty\\_base = PenaltySafety*PolynomialDegree*PolynomialDegree;\\newline \n\\btab \\btab double eta = penalty\\_base * cj\\_in;\\newline \n\\btab \\btab if(jCellOut >= 0) \\{\\newline \n\\btab \\btab \\btab double cj\\_out = cj[jCellOut];\\newline \n\\btab \\btab \\btab eta = Math.Max(eta, penalty\\_base * cj\\_out);\\newline \n\\btab \\btab \\}\\newline \n\\btab \\btab return eta;\\newline \n\\btab \\}\\newline \n /// The following functions cover the actual math.\n /// For any discretization of the Laplace operator, we have to specify:\n /// \\begin{itemize}\n /// \\item a volume integrand,\n /// \\item an edge integrand for inner edges, i.e. on $\\Gamma_i$,\n /// \\item an edge integrand for boundary edges, i.e. on $\\partial \\Omega$.\n /// \\end{itemize}\n /// The integrand for the volume integral:\n\\btab public double VolumeForm(ref CommonParamsVol cpv, \\newline \n\\btab \\btab double[] U, double[,] GradU, // the trial-function u \\newline \n\\btab \\btab // (i.e. the function we search for) and its gradient\\newline \n\\btab \\btab double V, double[] GradV // the test function; \\newline \n\\btab \\btab ) \\{\\newline \n \\newline \n\\btab \\btab double acc = 0;\\newline \n\\btab \\btab for(int d = 0; d < cpv.D; d++)\\newline \n\\btab \\btab \\btab acc += GradU[0, d] * GradV[d];\\newline \n\\btab \\btab return acc;\\newline \n\\btab \\}\\newline \n /// The integrand for the integral on the inner edges,\n /// \\[\n /// -( \\mean{ \\nabla u } \\jump{ v }) \\cdot \\vec{n}_{\\Gamma} \n /// -( \\mean{ \\nabla v } \\jump{ u }) \\cdot \\vec{n}_{\\Gamma} \n /// + \\eta \\jump{ u } \\jump{v} :\n /// \\]\n\\btab public double InnerEdgeForm(ref CommonParams inp, \\newline \n\\btab \\btab double[] U\\_IN, double[] U\\_OT, double[,] GradU\\_IN, double[,] GradU\\_OT, \\newline \n\\btab \\btab double V\\_IN, double V\\_OT, double[] GradV\\_IN, double[] GradV\\_OT) \\{\\newline \n \\newline \n\\btab \\btab double eta = PenaltyFactor(inp.jCellIn, inp.jCellOut);\\newline \n \\newline \n\\btab \\btab double Acc = 0.0;\\newline \n\\btab \\btab for(int d = 0; d < inp.D; d++) \\{ // loop over vector components \\newline \n\\btab \\btab \\btab // consistency term: -(\\{\\{ \\textbackslash /u \\}\\} [[ v ]])*Normale\\newline \n\\btab \\btab \\btab // index d: spatial direction\\newline \n\\btab \\btab \\btab Acc -= 0.5 * (GradU\\_IN[0, d] + GradU\\_OT[0, d])*(V\\_IN - V\\_OT)\\newline \n\\btab \\btab \\btab \\btab \\btab * inp.Normale[d];\\newline \n \\newline \n\\btab \\btab \\btab // symmetry term: -(\\{\\{ \\textbackslash /v \\}\\} [[ u ]])*Normale\\newline \n\\btab \\btab \\btab Acc -= 0.5 * (GradV\\_IN[d] + GradV\\_OT[d])*(U\\_IN[0] - U\\_OT[0])\\newline \n\\btab \\btab \\btab \\btab \\btab * inp.Normale[d];;\\newline \n\\btab \\btab \\}\\newline \n \\newline \n\\btab \\btab // penalty term: eta*[[u]]*[[v]]\\newline \n\\btab \\btab Acc += eta*(U\\_IN[0] - U\\_OT[0])*(V\\_IN - V\\_OT);\\newline \n\\btab \\btab return Acc;\\newline \n\\btab \\} \\newline \n /// The integrand on boundary edges, i.e. on $\\partial \\Omega$, is\n /// \\[ \n /// -( \\mean{ \\nabla u } \\jump{ v }) \\cdot \\vec{n}_{\\Gamma} \n /// -( \\mean{ \\nabla v } \\jump{ u }) \\cdot \\vec{n}_{\\Gamma} \n /// + \\eta \\jump{ u } \\jump{v} .\n /// \\]\n /// For the boundary we have to consider the special definition for \n /// the mean-value operator $\\mean{-}$ and the jump operator \n /// $\\jump{-}$ on the boundary.\n\\btab public double BoundaryEdgeForm(ref CommonParamsBnd inp, \\newline \n\\btab \\btab double[] U\\_IN, double[,] GradU\\_IN, double V\\_IN, double[] GradV\\_IN) \\{\\newline \n \\newline \n\\btab \\btab double eta = PenaltyFactor(inp.jCellIn, -1);\\newline \n\\btab \\btab double Acc = 0.0;\\newline \n\\btab \\btab for(int d = 0; d < inp.D; d++) \\{ // loop over vector components \\newline \n\\btab \\btab \\btab // consistency term: -(\\{\\{ \\textbackslash /u \\}\\} [[ v ]])*Normale\\newline \n\\btab \\btab \\btab // index d: spatial direction\\newline \n\\btab \\btab \\btab Acc -= (GradU\\_IN[0, d])*(V\\_IN) * inp.Normale[d];\\newline \n \\newline \n\\btab \\btab \\btab // symmetry term: -(\\{\\{ \\textbackslash /v \\}\\} [[ u ]])*Normale\\newline \n\\btab \\btab \\btab Acc -= (GradV\\_IN[d])*(U\\_IN[0]) * inp.Normale[d];;\\newline \n\\btab \\btab \\}\\newline \n \\newline \n\\btab \\btab // penalty term: eta*[[u]]*[[v]]\\newline \n\\btab \\btab Acc += eta*(U\\_IN[0])*(V\\_IN);\\newline \n \\newline \n \\newline \n\\btab \\btab return Acc;\\newline \n\\btab \\}\\newline \n\\}\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\subsection{Evaluation of the poisson operator in 1D}\n/// We consider the following problem:\n/// \\begin{equation*}\n/// \\Delta u = 2,\\quad -1 2);\\newline \n/// We have to ensure to set the \\code{PolynomialDegree} in the \\emph{SipLaplace}-object.\nvar i\\_SipLaplace = new SipLaplace();\\newline \ni\\_SipLaplace.PolynomialDegree = DGBasisOn1D.Degree;\\newline \ni\\_SipLaplace.cj = gdata1D.Cells.cj;\\newline \nvar Operator\\_SipLaplace = i\\_SipLaplace.Operator();\\newline \n/// We now want to calculate the residual after inserting the exact solution as well as a wrong solution. \n/// The implementation of the exact solution:\nvar u\\_ex = new SinglePhaseField(DGBasisOn1D, \"$u\\_\\{ex\\}$\");\\newline \nu\\_ex.ProjectField((double x) => 1.0 - x*x);\\newline \n/// The implementation of a spurious, i.e. a wrong solution; we take the exact solution and add random values in each cell:\nvar u\\_wrong = new SinglePhaseField(DGBasisOn1D, \"$u\\_\\{wrong\\}$\");\\newline \nu\\_wrong.ProjectField((double x) => 1.0 - x*x);\\newline \nRandom R = new Random();\\newline \nfor(int j = 0; j < gdata1D.Cells.NoOfLocalUpdatedCells; j++)\\{\\newline \n\\btab double ujMean = u\\_wrong.GetMeanValue(j);\\newline \n\\btab ujMean += R.NextDouble();\\newline \n\\btab u\\_wrong.SetMeanValue(j, ujMean);\\newline \n\\btab \\}\\newline \n/// Evaluating the Laplace operator using the different solutions:\nvar Residual = new SinglePhaseField(DGBasisOn1D,\"Resi1\");\\newline \nvar ResidualNorm = new List();\\newline \nforeach(var u in new DGField[] \\{u\\_ex, u\\_wrong\\}) \\{\\newline \n\\btab Residual.Clear();\\newline \n\\btab Operator\\_SipLaplace.Evaluate(u, Residual); // evaluate\\newline \n\\btab Residual.Acc(-1.0, RHS); \\newline \n\\btab double ResiNorm = Residual.L2Norm();\\newline \n\\btab ResidualNorm.Add(ResiNorm);\\newline \n\\btab Console.WriteLine(\"Residual for \" + u.Identification + \" = \" + ResiNorm); \\newline \n\\}\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// tests BoSSScmdSilent\nAssert.LessOrEqual(ResidualNorm[0], 1e-10);\\newline \nAssert.GreaterOrEqual(ResidualNorm[1], 1e-1);\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\subsection{The matrix of the Poisson Operator}\n/// If we do not know the exact solution, we have to solve a linear system.\n/// Therefore, we not only need to evaluate the operator,\n/// but we need its matrix.\n/// The \\emph{Mapping} controls which degree-of-freedom of the DG approximation\n/// is mapped to which row, resp. column of the matrix.\nvar Mapping = new UnsetteledCoordinateMapping(DGBasisOn1D);\\newline \nvar Matrix\\_SipLaplace = Operator\\_SipLaplace.ComputeMatrix(Mapping,null,Mapping);\n }\n\\BoSSSexe\n\\BoSSScmd{\nMatrix\\_SipLaplace.NoOfCols;\n }\n\\BoSSSexe\n\\BoSSScmd{\nMatrix\\_SipLaplace.NoOfRows;\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We see that the matrix has 27 rows and columns.\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\paragraph{Matrix rank and determinant of the matrix \n/// \\code{Matrix\\_SipLaplace}:}\n/// Use the functions \\emph{rank} and \\emph{det} to analyze the matrix (warning: this can get costly \n/// for larger matrices!).\\\\\n///Interpret the results:\n/// \\begin{itemize}\n/// \\item What does it mean, when a matrix has full rank?\n/// \\item How many solutions can a linear system have?\n/// \\end{itemize}\ndouble rank = Matrix\\_SipLaplace.rank(); \\newline \nConsole.WriteLine(\"Matrix rank = \" + rank);\\newline \n \\newline \ndouble det = Matrix\\_SipLaplace.det(); \\newline \nConsole.WriteLine(\"Determinante = \" + det);\\newline \n/// So the matrix of the SIP discretization has a unique solution.\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// tests BoSSScmdSilent\nAssert.AreEqual(rank, Matrix\\_SipLaplace.NoOfCols);\\newline \nAssert.Greater(det, 1.0);\n }\n\\BoSSSexe\n\\BoSSScmd{\n % \n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\section{Advanced topics}\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// %========================================================\n/// %========================================================\n/// \\subsection{The penalty parameter of the SIP and stability in 2D}\n/// %========================================================\n/// %========================================================\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We define a two-dimensional grid:\nvar grd2D = Grid2D.Cartesian2DGrid(GenericBlas.Linspace(-1,1,21), \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab GenericBlas.Linspace(-1,1,16));\\newline \nvar gdata2D = new GridData(grd2D);\\newline \nvar DGBasisOn2D = new Basis(gdata2D, 5);\\newline \nvar Mapping2D = new UnsetteledCoordinateMapping(DGBasisOn2D);\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We are going to choose the \\code{PenaltySafety} for the \\code{SipLaplace}\n/// from the following list\ndouble[] SFs = new double[] \\newline \n\\btab \\{0.001, 0.002, 0.01, 0.02, 0.1, 0.2, 1, 2, 10, 20, 100\\};\\newline \n/// and compute the condition number as well as the determinate.\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We consider the example \n/// \\[\n/// -\\Delta u = \\pi^2 0.5 \\cos(x/2) \\cos(y/2) \n/// \\text{ with } \n/// (x,y) \\in (-1,1)^2\n/// \\]\n/// and $u = 0$ on the boundary.\n/// The exact solution is $u_{Ex}(x,y) = \\cos(x/2) \\cos(y/2)$.\nFunc exSol = \\newline \n\\btab \\btab (X => Math.Cos(X[0]*Math.PI*0.5)*Math.Cos(X[1]*Math.PI*0.5));\\newline \nFunc exRhs = \\newline \n\\btab \\btab (X => (Math.PI*Math.PI*0.5*Math.Cos(X[0]*Math.PI*0.5)\\newline \n\\btab \\btab \\btab \\btab \\btab *Math.Cos(X[1]*Math.PI*0.5))); // == - /\\textbackslash exSol\\newline \nSinglePhaseField RHS = new SinglePhaseField(DGBasisOn2D, \"RHS\");\\newline \nRHS.ProjectField(exRhs);\\newline \ndouble[] RHSvec = RHS.CoordinateVector.ToArray();\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We check our discretization once more in 2D; the residual should be low,\n/// but not exactly (resp. up to $10^{-12}$) since the solution is not \n/// polynomial and cannot be fulfilled exactly.\nSinglePhaseField u = new SinglePhaseField(DGBasisOn2D,\"u\");\\newline \nu.ProjectField(exSol);\\newline \ni\\_SipLaplace.PolynomialDegree = DGBasisOn2D.Degree;\\newline \ni\\_SipLaplace.cj = gdata2D.Cells.cj;\\newline \nvar Matrix\\_SIP\\_sf = Operator\\_SipLaplace.ComputeMatrix(Mapping2D,\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab null,\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab Mapping2D);\\newline \nSinglePhaseField Residual = new SinglePhaseField(DGBasisOn2D,\"Residual\");\\newline \nResidual.Acc(1.0, RHS);\\newline \nMatrix\\_SIP\\_sf.SpMV(-1.0, u.CoordinateVector, 1.0, Residual.CoordinateVector);\\newline \nConsole.WriteLine(\"Residual L2 norm: \" + Residual.L2Norm());\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// We also check that the matrix is symmetric:\nvar checkMatrix = Matrix\\_SIP\\_sf - Matrix\\_SIP\\_sf.Transpose();\\newline \ncheckMatrix.InfNorm();\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// tests BoSSScmdSilent\nAssert.LessOrEqual(checkMatrix.InfNorm(), 1e-8);\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\paragraph{Matrix properties for different penalty factors.}\n/// Now, we assemble the matrix of the SIP for different \n/// \\code{PenaltySafety}-factors. We also try to solve the linear system\n/// using an iterative method. As Matlab is called multiple times during this \n/// command, it can take some minutes until it is done.\nint cnt = 0;\\newline \nvar Results = new List>();\\newline \nforeach(double sf in SFs) \\{\\newline \n\\btab cnt++;\\newline \n\\btab i\\_SipLaplace.PenaltySafety = sf;\\newline \n\\btab i\\_SipLaplace.PolynomialDegree = DGBasisOn2D.Degree;\\newline \n\\btab var Matrix\\_SIP\\_sf = Operator\\_SipLaplace.ComputeMatrix(\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab Mapping2D,null,Mapping2D);\\newline \n\\btab double condNo1 = Matrix\\_SIP\\_sf.condest(); \\newline \n\\btab bool definite = Matrix\\_SIP\\_sf.IsDefinite();\\newline \n \\newline \n /// We solve the system \n /// \\[\n /// \\text{\\tt Matrix\\_SIP\\_sf} \\cdot u = \\text{\\tt RHS}\n /// \\]\n /// using a an iterative solver, the so-called \n /// conjugate gradient (CG) method.\n /// CG requires a positive definite matrix. \n /// The function \\code{Solve\\_CG} returns the number of iterations.\n\\btab SinglePhaseField u = new SinglePhaseField(DGBasisOn2D,\"u\");\\newline \n\\btab u.InitRandom();\\newline \n\\btab int NoOfIter = Matrix\\_SIP\\_sf.Solve\\_CG(u.CoordinateVector, RHSvec);\\newline \n \\newline \n\\btab SinglePhaseField Error = new SinglePhaseField(DGBasisOn2D,\"Error\");\\newline \n\\btab Error.ProjectField(exSol);\\newline \n\\btab Error.Acc(-1.0, u);\\newline \n \\newline \n\\btab double L2err = u.L2Error(exSol);\\newline \n \\newline \n\\btab Console.WriteLine(sf + \"\\textbackslash t\" + condNo1.ToString(\"0.#E-00\") \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab + \"\\textbackslash t\" + NoOfIter \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab + \"\\textbackslash t\" + L2err.ToString(\"0.#E-00\") \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab + \"\\textbackslash t\" + definite);\\newline \n\\btab Results.Add(new Tuple(sf, condNo1, NoOfIter,\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab L2err, definite));\\newline \n\\}\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// tests BoSSScmdSilent\nforeach(var r in Results) \\{\\newline \n\\btab if(r.Item1 >= 1 && r.Item1 <= 20) \\{\\newline \n\\btab \\btab Assert.LessOrEqual(r.Item2, 1e7); // cond No.\\newline \n\\btab \\btab Assert.LessOrEqual(r.Item3, 6000); // iter\\newline \n\\btab \\btab Assert.LessOrEqual(r.Item4, 1e-4); // L2 err\\newline \n\\btab \\btab Assert.IsTrue(r.Item5); // definite \\newline \n\\btab \\}\\newline \n\\btab if(r.Item1 <= 0.1) \\{\\newline \n\\btab \\btab Assert.IsFalse(r.Item5); // indefinite \\newline \n\\btab \\}\\newline \n\\}\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\paragraph{Plotting:}\n/// Plot the number of conjugate gradient iterations versus the \n/// \\code{PenaltySafety}.\n/// A logarithmic scale is used for the horizontal axis.\nvar format = new PlotFormat(lineColor: LineColors.Blue, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab pointSize: 2, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab dashType: DashTypes.DotDashed, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab Style: Styles.LinesPoints, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab pointType:PointTypes.OpenCircle);\\newline \nGnuplot gp = new Gnuplot(baseLineFormat:format);\\newline \ngp.PlotLogXY(Results.Select(r => r.Item1), \\newline \n\\btab \\btab \\btab Results.Select(r => ((double)(r.Item3))));\\newline \ngp.PlotNow();\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\paragraph{Convergence study, indefinite vs. definite.}\n/// We are going to solve the SIP-system for different grid resolutions,\n/// comparing an insufficient penalty to a penalty which is large enough.\ndouble[] Resolution = new double[] \\{ 8, 16, 32, 64, 128, 256 \\};\\newline \nList L2Error\\_indef = new List();\\newline \nList L2Error\\_posdef = new List();\\newline \nforeach(int Res in Resolution) \\{\\newline \n\\btab var grd2D = Grid2D.Cartesian2DGrid(GenericBlas.Linspace(-1,1,(int)Res + 1), \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab GenericBlas.Linspace(-1,1,(int)Res + 1));\\newline \n\\btab var gdata2D = new GridData(grd2D);\\newline \n\\btab var DGBasisOn2D = new Basis(gdata2D, 2);\\newline \n\\btab var Mapping2D = new UnsetteledCoordinateMapping(DGBasisOn2D);\\newline \n \\newline \n\\btab SinglePhaseField RHS = new SinglePhaseField(DGBasisOn2D, \"RHS\");\\newline \n\\btab RHS.ProjectField(exRhs);\\newline \n\\btab SinglePhaseField uEx = new SinglePhaseField(\\newline \n\\btab \\btab new Basis(gdata2D, DGBasisOn2D.Degree*2),\\newline \n\\btab \\btab \"Error\");\\newline \n\\btab uEx.ProjectField(exSol);\\newline \n \\newline \n\\btab i\\_SipLaplace.PolynomialDegree = DGBasisOn2D.Degree;\\newline \n\\btab i\\_SipLaplace.cj = gdata2D.Cells.cj;\\newline \n \\newline \n\\btab i\\_SipLaplace.PenaltySafety = 0.01;\\newline \n\\btab var Matrix\\_SIP\\_indef = Operator\\_SipLaplace.ComputeMatrix(\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab Mapping2D,null,Mapping2D);\\newline \n \\newline \n\\btab SinglePhaseField u\\_indef = new SinglePhaseField(DGBasisOn2D,\"u\");\\newline \n\\btab Matrix\\_SIP\\_indef.Solve\\_Direct(u\\_indef.CoordinateVector, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab RHS.CoordinateVector);\\newline \n\\btab var Error\\_indef = uEx.CloneAs();\\newline \n\\btab Error\\_indef.AccLaidBack(-1.0, u\\_indef);\\newline \n\\btab L2Error\\_indef.Add(Error\\_indef.L2Norm());\\newline \n \\newline \n /// In order to have a positive definite system, we are\n /// using $\\text{\\tt PenaltySafety} = 2$!\n\\btab i\\_SipLaplace.PenaltySafety = 2.0;\\newline \n\\btab var Matrix\\_SIP\\_posdef = Operator\\_SipLaplace.ComputeMatrix(\\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab Mapping2D,null,Mapping2D);\\newline \n \\newline \n\\btab SinglePhaseField u\\_posdef = new SinglePhaseField(DGBasisOn2D,\"u\");\\newline \n\\btab Matrix\\_SIP\\_posdef.Solve\\_Direct(u\\_posdef.CoordinateVector, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab \\btab RHS.CoordinateVector);\\newline \n\\btab var Error\\_posdef = uEx.CloneAs();\\newline \n\\btab Error\\_posdef.AccLaidBack(-1.0, u\\_posdef);\\newline \n\\btab L2Error\\_posdef.Add(Error\\_posdef.L2Norm());\\newline \n \\newline \n\\btab Console.WriteLine(L2Error\\_indef.Last().ToString(\"0.#E-00\") \\newline \n\\btab \\btab \\btab \\btab \\btab + \"\\textbackslash t\" + L2Error\\_posdef.Last().ToString(\"0.#E-00\"));\\newline \n\\}\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\paragraph{Convergence plot:}\n/// The convergence plot unveils that there is something wrong if the\n/// penalty factor is set too low.\n/// While the solution of the indefinite system may look right at the first\n/// glance, we see that we do not obtain grid convergence for \n/// \\code{Error\\_indef}.\n/// The error of the positive definite system, \\code{Error\\_posdef}, where the \n/// penalty is chosen sufficiently large converges with the expected \n/// rate.\nvar format = new PlotFormat(lineColor: LineColors.Blue, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab pointSize: 2, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab dashType: DashTypes.DotDashed, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab Style: Styles.LinesPoints, \\newline \n\\btab \\btab \\btab \\btab \\btab \\btab \\btab pointType:PointTypes.OpenCircle);\\newline \nGnuplot gp = new Gnuplot(baseLineFormat:format);\\newline \ngp.PlotLogXLogY(Resolution, L2Error\\_indef);\\newline \ngp.PlotLogXLogY(Resolution, L2Error\\_posdef);\\newline \ngp.PlotNow();\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// Finally, we are going to compute the convergence rate of the SIP\n/// discretization: we take the logarithm of the resolution and the error:\n }\n\\BoSSSexe\n\\BoSSScmd{\nvar log\\_h = Resolution.Select(h => Math.Log10(h)).ToArray();\\newline \nvar log\\_Err = L2Error\\_posdef.Select(h => Math.Log10(h)).ToArray();\n }\n\\BoSSSexe\n\\BoSSScmd{\nvar LeastSquaresSys = MultidimensionalArray.Create(log\\_h.Length,2);\\newline \nLeastSquaresSys.SetColumn(0, log\\_h.Length.ForLoop(i => 1.0));\\newline \nLeastSquaresSys.SetColumn(1, log\\_h);\\newline \ndouble[] dk = new double[2]; // intercept and slope of best-fit line\\newline \nLeastSquaresSys.LeastSquareSolve(dk, log\\_Err);\n }\n\\BoSSSexe\n\\BoSSScmd{\ndk;\n }\n\\BoSSSexe\n\\BoSSScmdSilent{\n/// tests BoSSScmdSilent\nAssert.LessOrEqual(dk[1], -2.9);\n }\n\\BoSSSexe\n\\BoSSScmd{\n/// \\section{Further reading}\n/// \\begin{itemize}\n/// \\item\n/// \\bibentry{DiPietroErn2011}\n/// \\item\n/// \\bibentry{Arnold_1982} \n/// \\end{itemize}\n }\n\\BoSSSexe\n", "meta": {"hexsha": "52b1d23140c85120b54e3a362f28a040a4115701", "size": 25686, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/handbook/tutorial9-SIP/sip.tex", "max_stars_repo_name": "leyel/BoSSS", "max_stars_repo_head_hexsha": "39f58a1a64a55e44f51384022aada20a5b425230", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-12-20T10:55:58.000Z", "max_stars_repo_stars_event_max_datetime": "2018-12-20T10:55:58.000Z", "max_issues_repo_path": "doc/handbook/tutorial9-SIP/sip.tex", "max_issues_repo_name": "leyel/BoSSS", "max_issues_repo_head_hexsha": "39f58a1a64a55e44f51384022aada20a5b425230", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/handbook/tutorial9-SIP/sip.tex", "max_forks_repo_name": "leyel/BoSSS", "max_forks_repo_head_hexsha": "39f58a1a64a55e44f51384022aada20a5b425230", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.3626943005, "max_line_length": 138, "alphanum_fraction": 0.6803316982, "num_tokens": 8188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.628838870221482}} {"text": "\\section{Asymmetric-Key Encryption (Public-Key Encryption)}\\label{s:public-key-encryption}\nThe major problem that traditional symmetric-key algorithms face is that they require the parties to exchange keys, or more specifically to agree to a secret common private key, prior to communicating.\nThis needs a secure (physical or digital) communication channel accessible even for a short period of time, in order for the secret key to be exchanged. Such a channel could be a piece of paper delivered by hand or by a trusted courier, or a short in person communication.\n\nAn asymmetric or public-key cryptosystem is one where different keys are employed for the operations in the cryptosystem (\\textit{e.g.}, encryption and decryption), and where one of the keys can be made public without compromising the secrecy of the other key \\cite{Kaliski2011}. That way there is no need for a secure channel or prior agreement.\n\nUsually three algorithms are needed to define a public key cryptosystem. These are a \\textit{Key Generation} algorithm which produces the public and private keys (based on a security parameter $\\eta$), an \\textit{Encryption} and a \\textit{Decryption} algorithm.\n\nIn the classic model such an encryption scheme involves a \\textit{public key} which is used for encryption and a \\textit{private key} which is used for decryption.\nMore formally, given a plaintext $P$, a public key $PubKey$, and a private key $PrivKey$ we can produce a ciphertext $C$ such that:\n\\begin{equation}\n \\label{eq:public-key-encryption}\n \\begin{aligned}\n C = Enc(P, PubKey)\\\\\n P = Dec(C, PrivKey)\n \\end{aligned}\n\\end{equation}\n\nWhere $Enc$ and $Dec$ are the \\textit{Encryption} and \\textit{Decryption} algorithms respectively.\n\nIf Alice wants to send a message to Bob, she needs Bob's public key which could be publicly available (\\textit{e.g.} on Bob's website).\nAlice encrypts her message with Bob's public key and sends the resulted ciphertext to Bob. Bob then needs his private key which is only in his possession in order to decrypt the ciphertext Alice sent and retrieve the original message.\nIf Bob wishes to reply, he needs Alice's public key etc.\nGenerally, the public (encryption) key could be shared to any party that wants to communicate with the owner of that public key, while the private (decryption) key should be kept secret and used by the recipient of the ciphertext to recover the original plaintext.\n\nThe ability for two users to establish a shared secret over an insecure communication channel, despite having no prior communication or information exchange was first proposed in 1976 by Whitfield Diffie and Martin Hellman.\nThat method is named Diffie-Hellman (DH) key exchange \\cite{diffie1976new} after its authors. The security of the algorithm is based on the difficulty of solving the Discrete Log Problem (DLP). The DLP is stated as follows: Given $g, p$ and $g^k \\pmod{p}$, find $k$.\n\nSome of the most widely used public key cryptosystems are the RSA (Rivest -- Shamir -- Adleman) \\cite{rivest1978method}, the El Gamal \\cite{elgamal1985public} and the Paillier \\cite{paillier1999public} Cryptosystems.\n\n\\subsection{RSA}\\label{s:pk-rsa}\n\nOne if the first and most widely adopted public key cryptosystem is RSA . The algorithm's difficulty reduces to the difficulty of factoring the product of two large prime numbers. Its three basic algorithms are described below.\n\n\\begin{itemize}\n \\item \\textit{Key Generation}\n \\begin{itemize}\n \\item Randomly select two large primes $p$ and $q$.\n \\item Calculate modulus $n = p \\cdot q$.\n \\item Calculate $\\phi(n) = (p-1) \\cdot (q-1)$.\n \\item Randomly select $e : gcd(e,\\phi(n)) = 1$.\n \\item Calculate reverse $d = e ^ {-1} \\pmod{\\phi(n)}$.\\\\\n $d \\cdot e = 1 \\pmod{\\phi(n)}$.\n \\item The public key is $(n, e)$, while the private key is $d$. $p, q$ and $\\phi(n)$ must also be kept private.\n \\end{itemize}\n \\item \\textit{Encryption}\n \\begin{itemize}\n \\item Given a plaintext message $m$, we get a ciphertext $c = m ^ e \\pmod{n}$.\n \\end{itemize}\n \\item \\textit{Decryption}\n \\begin{itemize}\n \\item Given a ciphertext $c$, we get back the plaintext $m$ as follows: $c^d \\pmod{n} = m^{ed} \\pmod{n} = m$.\n \\end{itemize}\n\\end{itemize}\n\n\n\\subsection{ElGamal}\\label{s:pk-elgamal}\nThe ElGamal encryption system is a public-key cryptosystem based on the Diffie-Hellman key exchange. Its three basic algorithms are described below.\n\n\\begin{itemize}\n \\item \\textit{Key Generation}\n \\begin{itemize}\n \\item Select two large primes $p$ and $ q : q \\mid (p - 1)$.\n \\item Select a generator $g$ of group $\\mathbb{G}$ which is a large enough order-$q$ subgroup of the multiplicative group $\\mathbb{Z}_{p}^{*}$ of integers between $1$ and $p - 1$.\n \\item Randomly select $ x \\in_{R} \\mathbb{Z}_{q}$.\n \\item Calculate $y = g^{x} \\pmod{p}$.\n \\item The public key is $y$, while the private key is $x$.\n \\end{itemize}\n \\item \\textit{Encryption}\n \\begin{itemize}\n \\item Given a plaintext message $m$, we get a ciphertext $c$ as follows.\n \\item Randomly select $r \\in_{R} \\mathbb{Z}_{q}$.\n \\item Calculate $G = g^{r} \\pmod{p}$.\n \\item Calculate $M = m \\cdot y^{r} \\pmod{p}$.\n \\item Return $c = (G,M)$\n \\end{itemize}\n \\item \\textit{Decryption}\n \\begin{itemize}\n \\item Given a ciphertext $c = (G,M)$, we get back the plaintext $m$ as follows: \\\\$m = M/G^{x}\\pmod{p}$.\n \\end{itemize}\n\\end{itemize}\n\n\\subsection{Paillier}\\label{s:pk-paillier}\nThe Pallier cryptosystem is a public-key cryptosystem that relies its security upon the decisional composite residuosity assumption \\cite{paillier1999public}.\n\n\\begin{itemize}\n \\item \\textit{Key Generation}\n \\begin{itemize}\n \\item Randomly and independently select two large primes $p$ and \\\\$ q :gcd(p-1, q-1) = 1$ If both primes have the same length this property holds.\n \\item Calculate the RSA modulus $n = p \\cdot q$.\n \\item Calculate $\\lambda = lcm(p-1, q-1)$.\n \\item Select generator $g \\in \\mathbb{Z}_{n^{2}}^{*}$ such that the order of $g$ is a non zero multiple of $n$.\n \\item Calculate $\\mu = (L(g^{\\lambda} \\pmod{n^{2}}))^{-1} \\pmod{n}$, where $L(x) = \\frac{x-1}{n}$.\n \\item The public key is $ (n,g) $, while the private key is $ (\\lambda, \\mu) $.\n \\end{itemize}\n \\item \\textit{Encryption}\n \\begin{itemize}\n \\item Given a plaintext message $m$, we get a ciphertext $c$ as follows.\n \\item Encode $m$ into $\\mathbb{Z}_{n}$.\n \\item Randomly select $ r \\in_{R} \\mathbb{Z}_{n}^{*}$.\n \\item Return $c = g^{m} \\cdot r^{n} \\pmod{n^{2}}$.\n \\end{itemize}\n \\item \\textit{Decryption}\n \\begin{itemize}\n \\item Given a ciphertext $c \\in \\mathbb{Z}_{n^{2}}$, we get back the plaintext $m$ as follows: \\\\$m = L(c^{\\lambda} \\pmod{n^{2}}) \\cdot \\mu \\pmod{n}$.\n \\end{itemize}\n\\end{itemize}\n", "meta": {"hexsha": "71cb587e42b2739d0cb46ef6ea72291cac39f69d", "size": 6779, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pk-encryption.tex", "max_stars_repo_name": "jimouris/master-thesis", "max_stars_repo_head_hexsha": "e424cdd458cb7ff964bebcaaecfb7cad5b3ea525", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2018-08-29T07:51:22.000Z", "max_stars_repo_stars_event_max_datetime": "2020-10-09T12:09:24.000Z", "max_issues_repo_path": "pk-encryption.tex", "max_issues_repo_name": "jimouris/master-thesis", "max_issues_repo_head_hexsha": "e424cdd458cb7ff964bebcaaecfb7cad5b3ea525", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pk-encryption.tex", "max_forks_repo_name": "jimouris/master-thesis", "max_forks_repo_head_hexsha": "e424cdd458cb7ff964bebcaaecfb7cad5b3ea525", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2018-08-28T14:33:15.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-28T17:09:24.000Z", "avg_line_length": 62.7685185185, "max_line_length": 346, "alphanum_fraction": 0.7110193244, "num_tokens": 1966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.628838870221482}} {"text": "% Part: first-order-logic\n% Chapter: introduction\n% Section: formulas\n\n\\documentclass[../../../include/open-logic-section]{subfiles}\n\n\\begin{document}\n\n\\olfileid{fol}{int}{fml}\n\n\\section{\\usetoken{P}{formula}}\n\nHere is the approach we will use to rigorously specify !!{sentence}s\nof first-order logic and to deal with the issues arising from the use\nof !!{variable}s. We first define a \\emph{different} set of\nexpressions: !!{formula}s. Once we've done that, we can consider the\nrole !!{variable}s play in them---and on the basis of some other\nideas, namely those of ``free'' and ``bound'' !!{variable}s, we can\ndefine what !!a{sentence} is (namely, !!a{formula} without free\n!!{variable}s). We do this not just because it makes the definition of\n``!!{sentence}'' more manageable, but also because it will be crucial\nto the way we define the semantic notion of satisfaction.\n\nLet's define ``!!{formula}'' for a simple first-order language, one\ncontaining only a single !!{predicate}~$\\Obj P$ and a single\n!!{constant}~$\\Obj a$, and only the logical symbols $\\lnot$, $\\land$,\nand~$\\lexists$. Our full definitions will be much more general:\nwe'll allow infinitely many !!{predicate}s and !!{constant}s. In fact,\nwe will also consider !!{function}s which can be combined with\n!!{constant}s and !!{variable}s to form ``terms.'' For now, $\\Obj a$\nand the variables will be our only terms. We do need infinitely many\n!!{variable}s. We'll officially use the symbols $\\Obj v_0$, $\\Obj\nv_1$, \\dots, as variables.\n\n\\begin{defn}\nThe set of \\emph{!!{formula}s}~$\\Frm$ is defined as follows:\n\\begin{enumerate}\n\\item\\ollabel{fmls-atom} $\\Atom{\\Obj P}{\\Obj a}$ and $\\Atom{\\Obj\n P}{\\Obj v_i}$ are !!{formula}s ($i \\in \\Nat$).\n\n\\tagitem{prvNot}{If $!A$ is !!a{formula}, then $\\lnot !A$ is\n !!{formula}.}{}\n\n\\tagitem{prvAnd}{If $!A$ and $!B$ are !!{formula}s, then $(!A \\land\n !B)$ is !!a{formula}.}{}\n\n\\tagitem{prvEx}{If $!A$ is !!a{formula} and $x$ is !!a{variable},\n then $\\lexists[x][!A]$ is !!a{formula}.}{}\n\n\\tagitem{limitClause}{\\ollabel{fmls-limit}Nothing else is !!a{formula}.}{}\n\\end{enumerate}\n\\end{defn}\n\n\\olref{fmls-atom} tell us that $\\Atom{\\Obj P}{\\Obj a}$ and $\\Atom{\\Obj\nP}{\\Obj v_i}$ are !!{formula}s, for any $i \\in \\Nat$. These are\nthe so-called \\emph{atomic} !!{formula}s. They give us something to\nstart from. The other clauses give us ways of forming new\n!!{formula}s from ones we have already formed. So for instance, we get\nthat $\\lnot \\Atom{\\Obj P}{\\Obj v_2}$ is !!a{formula}, since\n$\\Atom{\\Obj P}{\\Obj v_2}$ is already !!a{formula} by\n\\olref{fmls-atom}, and then we get that $\\lexists[\\Obj v_2][\\lnot\n\\Atom{\\Obj P}{\\Obj v_2}]$ is another !!{formula}, and so on.\n\\olref{fmls-limit} tells us that \\emph{only} strings we can form in\nthis way count as !!{formula}s. In particular, $\\lexists[\\Obj\nv_0][\\Atom{\\Obj P}{\\Obj a}]$ and $\\lexists[\\Obj v_0][\\lexists[\\Obj\nv_0][\\Atom{\\Obj P}{\\Obj a}]]$ \\emph{do} count as !!{formula}s, and\n$(\\lnot \\Atom{\\Obj P}{\\Obj a})$ does not.\n\nThis way of defining !!{formula}s is called an \\emph{inductive\ndefinition}, and it allows us to prove things about !!{formula}s using\na version of proof by induction called \\emph{structural induction}.\nThese are discussed in a general way in \\olref[mth][ind][idf]{sec} and\n\\olref[mth][ind][sti]{sec}, which you should review before delving\ninto the proofs later on. Basically, the idea is that if you want to\ngive a proof that something is true for all !!{formula}s you show\nfirst that it is true for the atomic !!{formula}s, and then that\n\\emph{if} it's true for any !!{formula}~$!A$ (and~$!B$), it's\n\\emph{also} true for $\\lnot !A$, $(!A \\land !B)$, and\n$\\lexists[x][!A]$. For instance, this proves that it's true for\n$\\lexists[\\Obj v_2][\\lnot \\Atom{\\Obj P}{\\Obj v_2}]$: from the first\npart you know that it's true for the atomic !!{formula}~$\\Atom{\\Obj\nP}{\\Obj v_2}$. Then you get that it's true for $\\lnot \\Atom{\\Obj\nP}{\\Obj v_2}$ by the second part, and then again that it's true for\n$\\lexists[\\Obj v_2][\\lnot \\Atom{\\Obj P}{\\Obj v_2}]$ itself. Since all\n!!{formula}s are inductively generated from atomic !!{formula}s, this\nworks for any of them.\n\n\\end{document}\n", "meta": {"hexsha": "f903d77fee15946be593bb422b674259f9d2d5cc", "size": 4153, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/first-order-logic/introduction/formulas.tex", "max_stars_repo_name": "jzc/OpenLogic", "max_stars_repo_head_hexsha": "5948483c1d08c25664dc12ac8350e9ae34986b31", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 754, "max_stars_repo_stars_event_min_datetime": "2015-01-13T20:57:28.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T23:18:26.000Z", "max_issues_repo_path": "content/first-order-logic/introduction/formulas.tex", "max_issues_repo_name": "jzc/OpenLogic", "max_issues_repo_head_hexsha": "5948483c1d08c25664dc12ac8350e9ae34986b31", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 229, "max_issues_repo_issues_event_min_datetime": "2015-01-12T23:00:15.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-16T19:14:08.000Z", "max_forks_repo_path": "content/first-order-logic/introduction/formulas.tex", "max_forks_repo_name": "jzc/OpenLogic", "max_forks_repo_head_hexsha": "5948483c1d08c25664dc12ac8350e9ae34986b31", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 241, "max_forks_repo_forks_event_min_datetime": "2015-02-28T22:05:36.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-26T18:47:05.000Z", "avg_line_length": 46.6629213483, "max_line_length": 74, "alphanum_fraction": 0.6874548519, "num_tokens": 1341, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506581031359, "lm_q2_score": 0.8267117876664789, "lm_q1q2_score": 0.6288388653501271}} {"text": "\\documentclass[../PHYS306Notes.tex]{subfiles}\n\n\\begin{document}\n\\subsection{Worksheet - Review of Damped \\& Driven Harmonic Oscillators}\n\\begin{p}\nAn undamped harmonic oscillator has general solution $x(t) = B_1\\cos(\\omega t) + B_2\\sin(\\omega t)$. Show this can be recast as $x(t) = A\\cos(\\omega t - \\delta)$, where $A = \\sqrt{A^2 + B^2}$.\n\\end{p}\n\\begin{s}\nConsider a right-angle triangle with side lengths $B_1$, $B_2$ and hypotenuse $A + \\sqrt{B_1^2 + B_2^2}$. Then, let $\\cos\\delta = \\frac{B_1}{A}$ and $\\sin\\delta = \\frac{B_2}{A}$. Looking at the equation for $x(t)$ above, we can multiply it by one in a clever way:\n\\[x(t) = A\\left[\\frac{B_1}{A}\\cos\\omega t + \\frac{B_2}{A}\\sin\\omega t\\right] = A\\left[\\cos\\delta\\cos\\omega t + \\sin\\delta \\sin\\omega t\\right] = A\\cos(\\omega t - \\delta)\\]\nWhere the last equality follows by a trigonometric identity.\n\\end{s}\n\n\\begin{p}\nShow that the kinetic and potential energy of a simple undamped oscillator have the same amplitude but are out of phase, such that the total energy is conserved. \n\\end{p}\n\\begin{s}\nUsing our equation for $x(t)$ above (where $x(t)$ is the displacement from equilibrium), the total energy is given by:\n\\[E = U + T = \\frac{1}{2}kx^2 + \\frac{1}{2}m\\dot{x}^2 = \\frac{1}{2}kA^2\\cos^2(\\omega t - \\delta) + \\frac{1}{2}mA^2\\omega^2\\sin^2(\\omega t - \\delta)\\]\nWe have that $\\omega^2 = \\frac{k}{m}$ so:\n\\[\\frac{k}{2}A^2\\cos^2(\\omega t - \\delta) + \\frac{k}{2}A^2\\sin^2(\\omega t - \\delta) = \\frac{kA^2}{2}\\]\nAnd hence the total energy is a constant; we can therefore see that $\\od{E}{t} = 0$ and that the total energy is conserved.\n\\end{s}\n\n\\begin{p}\nDamped oscillatons are described by the ODE $\\ddot{x} + 2\\beta\\dot{x} + \\omega_0^2x = 0$. Show that for overdamped motion, the decay \"constant\" (decay parameter) for a damped oscillator decreases with increasing friction. Sketch the decay parameters vs. $\\beta$ for the whole range of $\\beta$.\n\\end{p}\n\\begin{s}\nWe guess a solution of the form $x(t) = \\exp(rt)$. We can therefore generate a characteristic equation for $r$, by substituting in the solution into the ODE and then cancelling out the exponential terms (as these can never be zero). Hence, we have:\n\\[r^2 + 2\\beta r + \\omega_0^2 = 0\\]\nThis is a quadratic equation with solutions:\n\\[r = -\\beta \\pm \\sqrt{\\beta^2 - \\omega_0^2}\\]\nThere are now different possibilities for the damping depending on the ratio of $\\beta$ and $\\omega_0$. For $\\beta > \\omega_0$, the oscillator is overdamped. For $\\beta < \\omega_0$, the oscillator is underdamped. For $\\beta = \\omega_0$, the oscillator is critically damped. Let us now consider the overdamped case. Then, we have that the discriminant in the equation above $\\beta^2 - \\omega_0^2$ is positive, and the general solution is the sum of decaying exponentials:\n\\[x(t) = C_1\\exp(r_1 t) + C_2\\exp(r_2 t) \\]\nwhere $r_1 = -\\beta + \\sqrt{\\beta^2 - \\omega^2}, r_2 = -\\beta - \\sqrt{\\beta^2 - \\omega^2}$. The dominant term will be the $\\exp(r_1 t)$ term as this decays more slowly. Therefore, the decay parameter is:\n\\[-\\beta + \\sqrt{\\beta^2 -\\omega_0^2}\\]\nWhich decreases with increasing $\\beta$. We will also have a decay parameter for the underdamped case. In this case, the decay parameter is just $\\beta$, as the discriminant is negative, and hence the solution $x(t)$ is composed of an oscillating part (the imaginary exponential part from the square root) and a real exponentially decaying part ($\\exp(-\\beta t)$). Hence the decay parameter increases linearly in this regime. Overall, we obtain a plot that looks like the follows:\n\nWhere the decay parameter is maximized when $\\beta = \\omega_0$, where we have critical damping.\n\\begin{center}\n \\includegraphics[scale=0.5]{Lecture-9/w9-img1.png}\n\\end{center}\n\\end{s}\n\n\\begin{p}\nShow that for critical damping, the solution $x(t) = te^{rt}$ solves the ODE $\\ddot{x} + 2\\beta\\dot{x} + \\omega_0^2x = 0$. Find $r$. Write the general solution.\n\\end{p}\n\\begin{s}\nFor critical damping, we have that $\\beta = \\omega_0$ and we can calculate $r$ to be:\n\\[r = -\\beta \\pm \\sqrt{\\beta^2 - \\beta^2} = -\\beta\\]\nHence one solution is given by:\n\\[x_1(t) = C_1\\exp(-\\beta t)\\]\nAnd we check the Ansatz provided in the question to verify it as a second solution. The first time derivative is given by:\n\\[\\dot{x} = \\exp(rt) + rt\\exp(rt)\\]\nAnd the second time derivative as:\n\\[\\ddot{x} = r\\exp(rt) + r\\exp(rt) + r^2t\\exp(rt) = 2r\\exp(rt) + r^2t\\exp(rt)\\]\nSo substituting this into the ODE, with $r = -\\beta$:\n\\[-2\\beta\\exp(-\\beta t) + \\beta^2t\\exp(-\\beta t) + 2\\beta\\left(\\exp(-\\beta t) - \\beta t\\exp(-\\beta t)\\right) + \\beta^2t\\exp(-\\beta t) = 0\\]\nSo we can see that this is indeed a solution! Hence the general solution is given by:\n\\[x(t) = C_1\\exp(-\\beta t) + C_2 t\\exp(-\\beta t)\\]\n\\end{s}\n\n\\begin{p}\nFind the condition on $C$ for $z(t) = Ce^{i\\omega t}$ to be a solution to $\\ddot{z} + 2\\beta\\dot{z} + \\omega_0^2z = f_0e^{i\\omega t}$. Express the coefficient $C$ as $Ae^{-i\\delta}$ and find $A$ and $\\delta$.\n\\end{p}\n\\begin{s}\nLet us plug in $z(t)$ into the ODE:\n\\[(-\\omega^2 + 2i\\beta\\omega + \\omega_0^2)C\\exp(i\\omega t) = f_0\\exp(i\\omega t)\\]\nWe may cancel out the exponentials on both sides:\n\\[(-\\omega^2 + 2i\\beta\\omega + \\omega_0^2) =f_0\\]\nHence we find:\n\\[C = \\frac{f_0}{-\\omega^2 + 2i\\beta\\omega + \\omega_0^2} = A\\exp(i\\delta)\\]\nWe have that $A^2 = CC^*$ and so:\n\\[A^2 = \\frac{f_0}{-\\omega^2 + 2i\\beta\\omega + \\omega_0^2}\\frac{f_0}{-\\omega^2 - 2i\\beta\\omega + \\omega_0^2} = \\frac{f_0^2}{(\\omega_0^2 - \\omega^2)^2 + 4\\beta^2\\omega^2}\\]\nTo get the phase, we see that:\n\\[f_0\\exp(i\\delta) = A(\\omega_0^2 - \\omega^2 + 2i\\beta\\omega)\\]\nAnd solving (using some knowledge of complex numbers and extracting their phase) we get:\n\\[\\delta = \\arctan(\\frac{\\Im}{\\Re}) = \\arctan(\\frac{2\\beta \\omega}{\\omega_0^2 - \\omega^2})\\]\n\\end{s}\n\n\\begin{p}\nWrite the general solution to $\\ddot{x} + 2\\beta\\dot{x} + \\omega_0^2x = f_0\\cos(\\omega t)$ for the underdamped case. What are the two undetermined constants?\n\\end{p}\n\\begin{s}\nWriting the general solution, we have:\n\\[x(t) = A\\cos(\\omega t -\\delta) + C_1\\exp(r_1 t) + C_2\\exp(r_2 t)\\]\nWhere the first term is the particular solution (comes from the driving) and the exponentials come from the homogenous solution. The former is the long-term oscillatory behavior, the latter is the transient solution (the exponentials decay away with time).\n\\end{s}\n\n\\begin{p}\nShow that the Q-factor given by the ratio of the width to the mean of the resonance curve is equal to $\\pi$ times the number of oscillations in one decay time.\n\\end{p}\n\\begin{s}\nWhen $\\omega \\sim \\omega_0$ (when the driving frequency approaches the natural frequency), we notice a resonance phenomenon, where the amplitude of the driven oscillator gets very large (technically it is slightly off from $\\omega = \\omega_0$, but to good approximation it is at the natural frequency of the system). This is pictured below:\n\\begin{center}\n \\includegraphics[scale=0.5]{Lecture-9/w9-img2.png}\n\\end{center}\nWe can also consider the width of the resonance peak. It is defined as:\n\\[Q = \\frac{\\omega_0}{2\\beta}\\]\nStronger damping implies a smaller quality factor and a narrower peak, and weak damping implies a larger quality factor and a broadened peak. To prove the claim provided in the question, we consider that the decay time is given by $\\tau = \\frac{1}{\\beta}$ and the period is $T = \\frac{2\\pi}{\\omega_0}$, so plugging this in we can immediately see that:\n\\[Q = \\frac{\\omega_0}{2\\beta} = \\frac{\\frac{1}{\\beta}\\pi}{\\frac{2\\pi}{\\omega_0}} = \\pi\\frac{\\tau}{T}\\]\nwhich is the desired result.\n\\end{s}\n\n\\begin{p}\nFind the phase shift at resonance when the driving frequency $\\omega$ is varied. Sketch the phase shift $\\delta$ vs. $\\omega$.\n\\end{p}\n\\begin{s}\nThe phase shift is given by $\\delta = \\frac{\\pi}{2}$ (perfectly out of phase). We can see this as the phase shift is given above by:\n\\[\\delta = \\arctan(\\frac{2\\beta\\omega}{\\omega_0^2 - \\omega^2})\\]\nAnd at resonance we have $\\omega \\sim \\omega_0$, so therefore the argument fo the arctan goes to infinity, and hence the value of $\\delta$ goes to $\\frac{\\pi}{2}$.\n\\end{s}\n\\end{document}", "meta": {"hexsha": "6333f6ae6153e5a0ca96a7d0c1f3fc9dd551fd5f", "size": 8081, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture-9/Worksheet-9.tex", "max_stars_repo_name": "RioWeil/PHYS306-notes", "max_stars_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Lecture-9/Worksheet-9.tex", "max_issues_repo_name": "RioWeil/PHYS306-notes", "max_issues_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture-9/Worksheet-9.tex", "max_forks_repo_name": "RioWeil/PHYS306-notes", "max_forks_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.5132743363, "max_line_length": 480, "alphanum_fraction": 0.6924885534, "num_tokens": 2676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506418255928, "lm_q2_score": 0.8267118026095991, "lm_q1q2_score": 0.6288388632597843}} {"text": "\\hypertarget{haskell}{%\n\\section{Haskell - Recursive Functions}\\label{haskell}}\n\nIn Haskell, functions can also be defined in terms of themselves. Such\nfunctions are called recursive.\n\n\\begin{itemize}\n\\tightlist\n\\item\n Some functions, such as factorial, are simpler to define in terms of\n other functions.\n\\item\n As we shall see, however, many functions can naturally be defined in\n terms of themselves.\n\\item\n Properties of functions defined using recursion can be proved using\n the simple but powerful mathematical technique of induction.\n\\end{itemize}\n\n\\hypertarget{recursion-on-lists}{%\n\\subsection{Recursion on Lists}\\label{recursion-on-lists}}\n\n\\begin{lstlisting}[language=Haskell]\nproduct :: Num a -> [a] -> a\nproduct [] = 1\nproduct (n:ns) = n * product ns\n\nIn execution:\n\nproduct [2,3,4]\n2 * product [3,4]\n2 * (3 * product [4])\n2 * (3 * (4 * product []))\n2 * (3 * (4 * 1))\n24\n\n--or\n-- The maximum function takes a list of things that can be ordered (e.g. instances of the Ord typeclass) and returns the biggest of them.\n\nmaximum' :: (Ord a) => [a] -> a \nmaximum' [] = error \"maximum of empty list\" \nmaximum' [x] = x \nmaximum' (x:xs) \n | x > maxTail = x \n | otherwise = maxTail \n where maxTail = maximum' xs\n\\end{lstlisting}\n\n\\hypertarget{multiple-arguments}{%\n\\subsection{Multiple Arguments}\\label{multiple-arguments}}\n\nFunctions with more than one argument can also be defined using\nrecursion.\n\n\\begin{lstlisting}[language=Haskell]\nzip :: [a] -> [b] -> [(a,b)]\nzip [] _ = []\nzip _ [] = []\nzip (x:xs) (y:ys) = (x,y) : zip xs ys\n\\end{lstlisting}\n\n\\subsection{Some Examples}\n\n\\begin{lstlisting}[language=Haskell]\ntake' :: (Num i, Ord i) => i -> [a] -> [a] \ntake' n _ \n | n <= 0 = [] \ntake' _ [] = [] \ntake' n (x:xs) = x : take' (n-1) xs \n\nreverse' :: [a] -> [a] \nreverse' [] = [] \nreverse' (x:xs) = reverse' xs ++ [x]\n\\end{lstlisting}\n\n\\clearpage\n\\subsection{Quick Sort}\n\nWe have a list of items that can be sorted. Their type is an instance of the \\textbf{Ord} typeclass. The edge condition is the empty list. Now a sorted list is a list that has all the values smaller than (or equal to) the head of the list in front (and those values are sorted), then comes the head of the list in the middle and then come all the values that are bigger than the head (they're also sorted).\n\n\\begin{lstlisting}[language=Haskell]\nquicksort :: (Ord a) => [a] -> [a] \nquicksort [] = [] \nquicksort (x:xs) = \n let smallerSorted = quicksort [a | a <- xs, a <= x] \n biggerSorted = quicksort [a | a <- xs, a > x] \n in smallerSorted ++ [x] ++ biggerSorted \n\\end{lstlisting}\n\nSo if we have, say [5,1,9,4,6,7,3] and we want to sort it, this algorithm will first take the head, which is 5 and then put it in the middle of two lists that are smaller and bigger than it. So at one point, you'll have [1,4,3] ++ [5] ++ [9,6,7]. We know that once the list is sorted completely, the number 5 will stay in the fourth place since there are 3 numbers lower than it and 3 numbers higher than it. Now, if we sort [1,4,3] and [9,6,7], we have a sorted list! We sort the two lists using the same function. Eventually, we'll break it up so much that we reach empty lists and an empty list is already sorted in a way, by virtue of being empty.\n\n\\clearpage", "meta": {"hexsha": "edb08d3fd864c3ae3b450430ede5ed779ea80877", "size": 3299, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TSM_AdvPrPa/Summary/07_Haskell06.tex", "max_stars_repo_name": "nortismo/mse-documentations", "max_stars_repo_head_hexsha": "cc67637785237d630f077a863edcd5f49aa52b59", "max_stars_repo_licenses": ["Beerware"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "TSM_AdvPrPa/Summary/07_Haskell06.tex", "max_issues_repo_name": "nortismo/mse-documentations", "max_issues_repo_head_hexsha": "cc67637785237d630f077a863edcd5f49aa52b59", "max_issues_repo_licenses": ["Beerware"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TSM_AdvPrPa/Summary/07_Haskell06.tex", "max_forks_repo_name": "nortismo/mse-documentations", "max_forks_repo_head_hexsha": "cc67637785237d630f077a863edcd5f49aa52b59", "max_forks_repo_licenses": ["Beerware"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-09-15T07:10:24.000Z", "max_forks_repo_forks_event_max_datetime": "2020-09-15T07:10:24.000Z", "avg_line_length": 35.8586956522, "max_line_length": 651, "alphanum_fraction": 0.6732343134, "num_tokens": 980, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8267117876664789, "lm_q1q2_score": 0.628838860864515}} {"text": "\\chapter{Dimension Reduction}\n\\emph{“你们都不是49年的国军，你们是50年的国军。”}\n\\newpage\n\n\n\\section{Principal Component Analysis}\n \\emph{aka. Discrete Karhunen Loeve Transform; Hotelling Transform}\n\n Let $y \\in \\mathbb{R}^K$ be a feature vector computed from $x \\in \\mathbb{R}^D$ where $K \\ll D$. PCA determines an orthogonal matrix $W \\in \\mathbb{R}^{D \\times K}$ s.t.\n \\[ y = W^Tx \\]\n\n Assume in this section that $x$ has a distribution with mean $\\mu=0$ and covariance matrix $\\Sigma$.\n\n \\subsection{Objective Funtion}\n PCA minimizes the reconstruction error between original $x$ and reconstructed $\\hat{x}$.\n\n Notice that $W$ is orthogonal, so $x$ can be reconstructed by\n \\[ \\hat{x} = Wy \\]\n\n Let the reconstruction error be\n \\[ \\epsilon = x - \\hat{x} = x - WW^Tx \\]\n\n We want to minimize the reconstruction error\n \\[ \\|\\epsilon\\|^2 = \\epsilon^T\\epsilon = x^Tx - x^TWW^Tx \\]\n\n The minimization is performed over all possible $x$\n \\[ \\int p(x)\\epsilon(x)^T\\epsilon(x)\\mathrm{d}x \\]\n\n \\[ \\mathbb{E}[\\epsilon^T\\epsilon] = \\mathbb{E}[x^Tx] - \\mathbb{E}[x^TWW^Tx]\\]\n\n \\subsection{Simple Case}\n If $K=1$, $w$ is a vector and $y$ is a scalar.\n \\[\\mathbb{E}[\\epsilon^T\\epsilon] = \\mathbb{E}[x^Tx]-w^T\\mathbb{E}[xx^T]w = \\mathbb{E}[x^Tx] - w^T \\Sigma w\\]\n where $\\Sigma = XX^T$ (recall that $X$ is assumed to have zero mean).\n\n We maximize the normalized second term\n \\[ \\max_w \\quad J \\triangleq \\frac{w^TXX^Tw}{w^Tw} \\]\n\n Or it can also be formulated as a equality constrained maximization and solved by Lagrangian.\n\n The solution is\n \\[ XX^Tw=Jw \\]\n\n So the problem becomes finding the eigenvalues of $XX^T$, to maximize $J$, we find the largest eigenvalue of $XX^T$.\n\n Other dimensions turn out to be the remaining eigenvectors.\n\n \\subsection{Geometric Interpretation}\n \\begin{itemize}\n \\item PCA is a shift and rotation of axis.\n \\item $w_1$ is the direction of greatest elongation\n \\item $w_2$ is the second largest elongation that is orthogonal to $w_1$\n \\end{itemize}\n\n\n\\section{Multi-Dimensional Scaling}\n MDS aims to keep the distances of samples in the mapped space that same as the distances in the original space.\n\n In this section we also assume the data samples have $\\mu=0$.\n\n \\[ \\min_{\\hat{x}} \\left(\\sum_{i 1\\).\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= 2e^{-j \\omega} \\frac{1}{1 - (-2) e^{-j \\omega}} \\\\\n \\overset{\\mathcal{F}^{-1}}{\\implies} y[n] &= 2 (-(-2)^{n - 1} u[-(n - 1) - 1]) \\\\\n &= (-2)^n u[-n]\n\\end{align}\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n H(z) = \\frac{1 - z^{-1}}{1 - 0.25z^{-2}} = \\frac{1 - z^{-1}}{(1 - 0.5z^{-1}) (1 + 0.5z^{-1})}\n\\end{equation}\n\n\\subsection{}\n\nUsing the \\(z\\)-transform multiplication property,\n\\begin{equation}\n Y(z) = H(z) X(z) = \\frac{\\cancel{1 - z^{-1}}}{(1 - 0.5z^{-1}) (1 + 0.5z^{-1}) \\cancel{(1 - z^{-1})}} = \\frac{1}{(1 - 0.5z^{-1}) (1 + 0.5z^{-1})}\n\\end{equation}\nThen, using partial fraction decomposition,\n\\begin{align}\n Y(z) &= \\frac{1}{(1 - 0.5z^{-1}) (1 + 0.5z^{-1})} = \\frac{A}{1 - 0.5z^{-1}} + \\frac{B}{1 + 0.5z^{-1}} \\\\\n &= 1 = A(1 + 0.5z^{-1}) + B(1 - 0.5z^{-1}) \\\\\n\\end{align}\nLetting \\(z = -0.5\\), we get \\(B = \\frac{1}{2}\\).\nLetting \\(z = 0.5\\), we get \\(A = \\frac{1}{2}\\).\nThen,\n\\begin{align}\n Y(z) &= \\frac{1}{2} \\frac{1}{1 - 0.5z^{-1}} + \\frac{1}{2} \\frac{1}{1 + 0.5z^{-1}} \\\\\n \\overset{\\mathcal{Z}^{-1}}{\\implies} y[n] &= \\frac{1}{2} \\left(\\frac{1}{2}\\right)^n u[n] + \\frac{1}{2} \\left(-\\frac{1}{2}\\right)^n u[n]\n\\end{align}\n\n\\subsection{}\n\n\\begin{equation}\n y[n] = \\delta[n] - \\delta[n - 1] \\overset{\\mathcal{Z}}{\\implies} Y(z) = 1 - z^{-1} = H(z) X(z)\n\\end{equation}\nmeaning that \\(X(z) = 1 - 0.25z^{-2} = (1 - 0.5z^{-1}) (1 + 0.5z^{-1})\\).\nBy definition of the \\(z\\)-transform,\n\\begin{equation}\n x[n] = \\delta[n] - \\frac{1}{4} \\delta[n - 2]\n\\end{equation}\n\n\\subsection{}\n\nTreating the input as the real part of \\(z = e^{j 0.5 \\pi}\\),\n\\begin{equation}\n H(e^{j 0.5 \\pi}) = \\frac{1 - e^{-j 0.5 \\pi}}{1 - 0.25e^{-j\\pi}} = 0.8 \\sqrt{2} e^{j \\frac{\\pi}{4}}\n\\end{equation}\nMeaning that the final output is\n\\begin{equation}\n y[n] = 0.8 \\sqrt{2} \\cos\\left(0.5 \\pi n + \\frac{\\pi}{4}\\right)\n\\end{equation}\n\n\\newpage\n\\section{}\n\n\\begin{align}\n x[n] &= -\\frac{1}{3} \\left(\\frac{1}{2}\\right)^n u[n] - \\frac{4}{3} 2^n u[-n - 1] \\\\\n Y(z) &= \\frac{1 - z^{-2}}{\\left(1 - \\frac{1}{2} z^{-1}\\right) (1 - 2z^{-1})}\n\\end{align}\n\n\\subsection{}\n\n\\begin{equation}\n X(z) = -\\frac{1}{3} \\frac{1}{1 - \\frac{1}{2} z^{-1}} + \\frac{4}{3} \\frac{1}{1 - 2z^{-1}}\n\\end{equation}\n\n\\subsection{}\n\n\\begin{equation}\n R_y: \\frac{1}{2} < |z| < 2\n\\end{equation}\n\n\\subsection{}\n\nSimplifiying \\(X(z)\\),\n\\begin{align}\n X(z) &= \\frac{1}{3} \\frac{-(1 - 2z^{-1}) + 4 \\left(1 - \\frac{1}{2} z^{-1}\\right)}{\\left(1 - \\frac{1}{2} z^{-1}\\right) (1 - 2z^{-1})} \\\\\n &= \\frac{1}{3} \\frac{-1 + \\cancel{2z^{-1}} + 4 - \\cancel{2z^{-1}}}{\\left(1 - \\frac{1}{2} z^{-1}\\right) (1 - 2z^{-1})} \\\\\n &= \\frac{1}{\\left(1 - \\frac{1}{2} z^{-1}\\right) (1 - 2z^{-1})} \\\\\n \\implies Y(z) &= (1 - z^{-2}) X(z) = X(z) - z^{-2} X(z) \\\\\n \\overset{\\mathcal{Z}^{-1}}{\\implies} y[n] &= x[n] - x[n - 2]\n\\end{align}\n\n\\subsection{}\n\n\\begin{equation}\n H(z) = \\frac{Y(z)}{X(z)} = 1 - z^{-2} \\overset{\\mathcal{Z}^{-1}}{\\implies} h[n] = \\delta[n] - \\delta[n - 2]\n\\end{equation}\n\n\\newpage\n\\section{}\n\n\\begin{align}\n X(z) &= \\frac{z^{-2}}{1 - 2.3z^{-1} + 1.6z^{-2} - 0.3z^{-3}} = \\frac{2.04}{1 + 0.3z^{-1}} - \\frac{3.47}{1 - z^{-1}} + \\frac{1.43}{(1 - z^{-1})^2} \\\\\n \\overset{\\mathcal{Z}^{-1}}{\\implies} x[n] &= (-0.3)^n u[n] - 3.47 u[n] + 1.43 (n + 1) u[n + 1]\n\\end{align}\n\n\\newpage\n\\section{}\n\n\\begin{enumerate}\n \\item System A, causal, stable, ROC: \\(|z| > 0.9\\)\n \\item System B, non-causal, stable, ROC: \\(|z| < 1.111\\)\n \\item System A, non-causal, unstable, ROC: \\(|z| < 0.9\\)\n\\end{enumerate}\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n Z(e^{j \\omega}) = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi X(e^{j \\theta}) Y(e^{j (\\omega - \\theta)}) \\, d\\theta\n\\end{equation}\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= \\pi \\sum_{k = -\\infty}^\\infty \\delta(\\omega - \\pi + 2\\pi k) + \\delta(\\omega + \\pi + 2\\pi k) \\\\\n X(e^{j \\omega}) \\ast Y(e^{j \\omega}) &= \\frac{1}{2} \\int_{-\\pi}^\\pi X(e^{j \\theta}) \\sum_{k \\in \\Z} \\delta((\\omega - \\theta) - \\pi + 2\\pi k) + \\delta((\\omega - \\theta) + \\pi + 2\\pi k) \\, d\\theta \\\\\n &= \\frac{1}{2} \\sum_{k \\in \\Z} \\int_{-\\pi}^\\pi X(e^{j \\omega}) (\\delta(\\omega - \\theta - \\pi + 2\\pi k) + \\delta(\\omega + \\theta + \\pi + 2\\pi k)) \\, d\\theta \\\\\n &= \\frac{1}{2} \\sum_{k \\in \\Z} (X(e^{j (\\omega - \\pi + 2\\pi k)}) + X(e^{j (\\omega + \\pi + 2\\pi k)}))\n\\end{align}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel={\\(\\frac{\\omega}{\\pi}\\)}, ylabel={\\(Z(e^{j \\omega})\\)},\n title={DTFT of Product},\n axis lines=middle,\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=index, y=8a\n ]{q8.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= \\pi \\sum_{k = -\\infty}^\\infty \\delta\\left(\\omega - \\frac{\\pi}{2} + 2\\pi k\\right) + \\delta\\left(\\omega + \\frac{\\pi}{2} + 2\\pi k\\right) \\\\\n X(e^{j \\omega}) \\ast Y(e^{j \\omega}) &= \\frac{1}{2} \\sum_{k \\in \\Z} \\left(X\\left(e^{j \\left(\\omega - \\frac{\\pi}{2} + 2\\pi k\\right)}\\right) + X\\left(e^{j \\left(\\omega + \\frac{\\pi}{2} + 2\\pi k\\right)}\\right)\\right)\n\\end{align}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel={\\(\\frac{\\omega}{\\pi}\\)}, ylabel={\\(Z(e^{j \\omega})\\)},\n title={DTFT of Product},\n axis lines=middle,\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=index, y=8b\n ]{q8.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= \\pi \\sum_{k = -\\infty}^\\infty \\delta\\left(\\omega - \\frac{\\pi}{4} + 2\\pi k\\right) + \\delta\\left(\\omega + \\frac{\\pi}{4} + 2\\pi k\\right) + \\delta\\left(\\omega - \\frac{3\\pi}{4} + 2\\pi k\\right) + \\delta\\left(\\omega + \\frac{3\\pi}{4} + 2\\pi k\\right) \\\\\n X(e^{j \\omega}) \\ast Y(e^{j \\omega}) &= \\frac{1}{2} \\sum_{k \\in \\Z} \\left(X\\left(e^{j \\left(\\omega - \\frac{\\pi}{2} + 2\\pi k\\right)}\\right) + X\\left(e^{j \\left(\\omega + \\frac{\\pi}{2} + 2\\pi k\\right)}\\right)\\right)\n\\end{align}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel={\\(\\frac{\\omega}{\\pi}\\)}, ylabel={\\(Z(e^{j \\omega})\\)},\n title={DTFT of Product},\n axis lines=middle,\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=index, y=8c\n ]{q8c.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= \\pi \\sum_{k = -\\infty}^\\infty \\cancelto{1}{e^{j \\frac{\\pi}{2}}} \\delta(\\omega - \\pi + 2\\pi k) + \\cancelto{-1}{e^{-j \\frac{\\pi}{2}}} \\delta(\\omega + \\pi + 2\\pi k) \\\\\n X(e^{j \\omega}) \\ast Y(e^{j \\omega}) &= \\frac{1}{2} \\sum_{k \\in \\Z} (X(e^{j (\\omega - \\pi + 2\\pi k)}) - X(e^{j (\\omega + \\pi + 2\\pi k)}))\n\\end{align}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel={\\(\\frac{\\omega}{\\pi}\\)}, ylabel={\\(Z(e^{j \\omega})\\)},\n title={DTFT of Product},\n axis lines=middle,\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=index, y=8d\n ]{q8.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n H(e^{j \\omega}) =\n \\begin{cases}\n j & \\omega \\in (-\\pi, 0) \\\\\n 0 & \\omega = 0 \\\\\n -j & \\omega \\in (0, \\pi)\n \\end{cases}\n\\end{equation}\n\n\\subsection{}\n\nFrom Table 2.1 of O\\&S, the symmetry properties of the DTFT imply that the Hilbert filter impulse response is purely \\emph{real and odd}.\n\n\\subsection{}\n\n\\begin{align}\n X(e^{j \\omega}) &= \\pi (\\delta(\\omega - \\omega_0) + \\delta(\\omega + \\omega_0)) \\\\\n Y(e^{j \\omega}) &= -j \\pi (\\delta(\\omega - \\omega_0) - \\delta(\\omega + \\omega_0)) \\\\\n \\overset{\\mathcal{F}^{-1}}{\\implies} y[n] &= \\sin(\\omega_0 n)\n\\end{align}\n\n\\subsection{}\n\n\\begin{align}\n Y(e^{j \\omega}) &= H(e^{j \\omega})^2 X(e^{j \\omega}) =\n \\begin{cases}\n -X(e^{j \\omega}) & \\omega \\in (-\\pi, 0) \\\\\n 0 & \\omega = 0 \\\\\n -X(e^{j \\omega}) & \\omega \\in (0, \\pi)\n \\end{cases} \\\\\n &= -X(e^{j \\omega}) \\overset{\\mathcal{F}^{-1}}{\\implies} y[n] = -x[n]\n\\end{align}\n\n\\subsection{}\n\n\\begin{center}\n \\includegraphics[width=0.6\\textwidth]{q9d.png}\n\\end{center}\nwhere \\(\\mathcal{H}\\) is the Hilbert transform.\n\n\\end{document}\n", "meta": {"hexsha": "8bc71dc75904fccbfdb04e71e611a47b37c02621", "size": 9368, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw02/hw02.tex", "max_stars_repo_name": "bdngo/ee-123", "max_stars_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw02/hw02.tex", "max_issues_repo_name": "bdngo/ee-123", "max_issues_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw02/hw02.tex", "max_forks_repo_name": "bdngo/ee-123", "max_forks_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.0821917808, "max_line_length": 267, "alphanum_fraction": 0.5042698548, "num_tokens": 3954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.8006920044739461, "lm_q1q2_score": 0.628790301761609}} {"text": "%\\section{Stabilization and Weak Formulation}\n\\label{app:stab}\n\nThis appendix details the weak formulation of the Navier Stokes\nequations instantiated in the software GRINS, and provides a\nderivation of the $\\tau$ stabilization terms shown in\nEquation~\\ref{eq:tau}. \n\nIn brief, our process is the following: \n\n\\begin{itemize}\n \\item Cast Navier Stokes + Boussinesq equations into weak form\n \\item Prepare as an operator $L\\,{\\bf c}={\\bf f}$\n \\item Calculate Fr\\'echet derivative\n \\item Separate into differential (P) and constant (Z) components,\n $L'[{\\bf c}] = P + Z$\n \\item Choose stabilization operator such that $S = -P^*$\n \\item Then stabilization has form, $a_h({\\bf c},{\\bf \\phi}) = a({\\bf c},{\\bf \\phi}) + \\langle\n L{\\bf c},S{\\bf \\phi} \\rangle_\\tau$\n\\end{itemize}\n\nThis is essentially the least-squares stabilization proposed by Hughes\nand extended to natural convection by Becker and Braack. \n\n%\n% start the real work\n%\n\n\\section{Weak Formulation of the Equations of Interest}\n\nWe begin with the incompressible Navier-Stokes equations with Boussinesq\nbuoyancy,\n\\begin{align}\n \\nabla \\cdot {\\bf u} &= \\, 0 \\label{eq_cont}\\\\\n \\frac{\\partial {\\bf u}}{\\partial t} + {\\bf u} \\cdot \\nabla {\\bf u} &= -\\frac{1}{\\rho}\n \\nabla p + \\nu \\nabla^2 {\\bf u} + g \\frac{T'}{T_0} \\label{eq_mom}\\\\\n \\rho c_p \\frac{\\partial T}{\\partial t} + {\\bf u} \\cdot \\nabla T &= \\nabla\n \\cdot (k \\nabla T) \\label{eq_energy}\n\\end{align}\ne.g. the continuity, momentum and energy equations, respectively. Our \nstate vector is ${\\bf c} = \\left[p,{\\bf u},T \\right]$. To cast these into\nweak form we multiply by appropriate test \nfunctions ${\\bf \\phi} = \\left[q,{\\bf v},w \\right] \\in H^1_0(\\Omega)$ and\nintegrate over the domain, $\\Omega \\in \\mathbb{R}^n$. Our system of\nequations now appears as, \n\\begin{align}\n \\bigintsss_\\Omega q \\nabla \\cdot {\\bf u} \\, dx &= 0 \\\\\n \\bigintsss_\\Omega \\dot {\\bf u} \\cdot {\\bf v} \\, dx +\n \\bigintsss_\\Omega ({\\bf u} \\cdot \\nabla) \\, {\\bf u} \\cdot {\\bf v} \\, dx &=\n \\bigintsss_\\Omega \\frac{p}{\\rho} \\nabla \\cdot {\\bf v} \\, dx -\n \\bigintsss_\\Omega \\nu \\, \\nabla {\\bf u} \\cdot \\nabla {\\bf v}\n \\,dx + \\bigintsss_\\Omega g \\frac{T'}{T_0} \\cdot {\\bf v} \\, dx \\\\ \n \\rho c_p \\bigintsss_\\Omega \\dot T \\cdot w \\, dx + \\bigintsss_\\Omega ({\\bf u}\n \\cdot \\nabla) T \\cdot w \\, dx &= -\\bigintsss_\\Omega (k \\nabla T) \\cdot\n \\nabla w \\, dx\n\\end{align}\n%\nwhere an ``over-dot'' denotes time differentiation, e.g. $\\dot {\\bf u} =\n\\frac{\\partial {\\bf u}}{\\partial t}$. Note that both the pressure term as well\nas the viscous term were integrated by parts to reduce the required\norder of the solution on those state variables. \n\nThe inner product is abbreviated by the shorthand notation $({\\bf u},{\\bf v}) =\n\\bigintsss_\\Omega {\\bf u}\\cdot {\\bf v}\\, dx $, giving our equations the form, \n\\begin{align}\n (\\nabla \\cdot {\\bf u}, q) &= 0 \\\\\n (\\dot {\\bf u},{\\bf v}) + ({\\bf u} \\cdot \\nabla {\\bf u}, {\\bf v}) -\n (p,\\nabla \\cdot {\\bf v}) + (\\nu \\, \\nabla\n {\\bf u}, \\nabla {\\bf v}) &= (g \\frac{T'}{T_0},{\\bf v}) \\\\\n \\rho c_p (\\dot T,w) + ({\\bf u} \\cdot \\nabla T,w) + (k \\nabla T,\\nabla w) &= 0.\n\\end{align}\n%\nThis defines our weak form operator, $ a({\\bf c},{\\bf \\phi})$. Our full equations will\nalso include a stabilization term such that, \n\\begin{equation}\n a_h({\\bf c},{\\bf \\phi}) = a({\\bf c},{\\bf \\phi}) + \\langle L\\,{\\bf c},S{\\bf \\phi} \\rangle_\\tau. \n\\end{equation}\n%\nThe subsequent section will define the operators L and S, so that we\nmight then fully define the stabilization term $\\langle L\\,{\\bf c},S{\\bf \\phi}\n\\rangle_\\tau$. \n\n%\n% section\n%\n\\section{The Stabilization Operators, L and S}\n\n\nTo form the stabilization terms, \n\n\\begin{equation}\n \\langle L\\,{\\bf c},S{\\bf \\phi} \\rangle_\\tau\n\\end{equation}\n\nwe must define the operators L and S. The operator L is simply the PDEs\nin Equations \\ref{eq_cont} - \\ref{eq_energy} written in operator form. S is\ndefined as the negative adjoint of the differential terms in L, e.g.\n\\begin{align}\n L'[{\\bf c}] = P + Z \\\\\n S = -P^*. \n\\end{align}\nWhere P are the differential terms, and Z the constant terms. \n\nOur objective is now to construct the adjoint operator of L. This is\naccomplished using the Fr\\'echet derivative, which defines the\nfunctional derivative on L. In general this is accomplished by taking\nthe first variation of a function $\\Pi({\\bf u})$ around a base state, ${\\bf u}$,\n\\begin{equation}\n \\delta\\, \\Pi({\\bf u}) = \\lim_{\\epsilon \\to 0} \\frac{\\Pi({\\bf\n u}+\\epsilon \\, \\hat {\\bf u}) -\n \\Pi({\\bf u})}{\\epsilon} =\n \\frac{\\partial \\Pi({\\bf u} +\\epsilon \\, \\hat {\\bf u})}{\\partial \\epsilon}\n \\bigg|_{\\epsilon = 0}\n\\end{equation}\n$\\forall \\hat {\\bf u}, \\epsilon > 0$ with ${\\bf u} + \\epsilon \\, \\hat {\\bf u} \\in\nH^1_0(\\Omega)$. This is recognizable as the G\\^{a}teaux\nderivative of the functional. \n\nWe now consider the first variation of state for the momentum equation\nterm by term. The convective term is, \n\\begin{align}\n \\frac{\\partial}{\\partial \\epsilon} ({\\bf u} + \\epsilon \\, \\hat {\\bf u}) &\\cdot \\nabla\n ({\\bf u} + \\epsilon \\hat {\\bf u}) \\\\\n = \\lim_{\\epsilon \\to 0} \\, \\hat {\\bf u} &\\cdot \\nabla ({\\bf u} + \\epsilon \\,\\hat {\\bf u}) \\\\\n = \\hat {\\bf u} & \\cdot \\nabla {\\bf u} \\\\\n = - {\\bf u} &\\cdot \\nabla \\hat {\\bf u}\n\\end{align}\nand the viscous term is, \n\\begin{align}\n \\frac{\\partial}{\\partial \\epsilon} \\nabla^2 ({\\bf u} + \\epsilon \\, \\hat {\\bf u}) \\\\\n = \\nabla^2 \\hat {\\bf u}\n\\end{align}\nwhile the buoyancy term is, \n\\begin{align}\n \\delta \\left(-g \\frac{T'}{T_0}\\right) &= \\delta \\left( -g\n \\frac{T-T_0}{T_0} \\right) \\\\\n &= -g \\frac{\\partial}{\\partial \\epsilon} \\left( \\frac{T-T_0+\\epsilon\n \\hat T}{T_0} \\right) \\\\\n &= -g \\left( \\frac{\\hat T}{T_0} \\right) \n\\end{align}\n%\n%\n%Lagrangian is therefore, $\\mathcal{L}$\nand thus the full adjoint equation for momentum appears as,\n\\begin{align}\n - {\\bf u} &\\cdot \\nabla \\hat {\\bf u} - \\nabla^2 \\hat {\\bf u} = -\\frac{1}{\\rho} \\nabla p. \n\\label{eq_adjmom}\n\\end{align}\n%\nThe continuity equation is straightforward, \n\\begin{align}\n \\frac{\\partial}{\\partial \\epsilon} \\nabla \\cdot ({\\bf u} + \\epsilon \\,\n \\hat {\\bf u}) &= 0, \\\\\n\\nabla \\cdot \\hat {\\bf u} &= 0.\n\\label{eq_adjcont}\n\\end{align}\n%\nFinally, consider the convective term of the energy equation, \n\\begin{align}\n \\frac{\\partial}{\\partial \\epsilon} {\\bf u} \\cdot \\nabla (T + \\epsilon\n \\, \\hat T)\n = {\\bf u} \\cdot \\nabla \\hat T\n\\end{align}\nand the thermal diffusion term, \n\\begin{align}\n \\frac{\\partial}{\\partial \\epsilon} \\cdot (-k \\nabla (T + \\epsilon \\hat\n T)) = \\nabla \\cdot (-k \\nabla \\hat T).\n\\end{align}\nThe full adjoint energy equation is therefore, \n\\begin{align}\n {\\bf u} \\cdot \\nabla \\hat T + \\nabla \\cdot (k \\nabla \\hat T) = 0.\n\\label{eq_adjen}\n\\end{align}\n\nWe are now in a position to define the matrices L and S. L comes\ndirectly from the PDEs in Equations \\ref{eq_cont} - \\ref{eq_energy} and is\ndefined as thus, \n\n\\begin{equation}\n\\renewcommand\\arraystretch{2}\n L = \n \\begin{pmatrix}\n 0 & \\nabla \\cdot () & 0 \\\\\n \\nabla \\,() & {\\bf u} \\cdot \\nabla() - \\nu \\nabla^2() & -g \\frac{()}{T_0} \\\\\n 0 & 0 & {\\bf u} \\cdot \\nabla() - \\nabla \\cdot (k \\nabla() \\,)\n \\end{pmatrix}.\n\\end{equation}\n%\nWhile the S matrix is constructed from Equations~\\ref{eq_adjmom},\n\\ref{eq_adjcont}, and \\ref{eq_adjen}, and must be, \n\\begin{equation}\n\\renewcommand\\arraystretch{2}\n S = -P^* = \n \\begin{pmatrix}\n 0 & \\nabla \\cdot () & 0 \\\\\n \\nabla \\,() & {\\bf u} \\cdot \\nabla() + \\nu \\nabla^2() & -g \\frac{()}{T_0} \\\\\n 0 & 0 & {\\bf u} \\cdot \\nabla() + \\nabla \\cdot k \\nabla()\n \\end{pmatrix}.\n\\end{equation}\n\n%\n% tau!\n%\n\\section{Tau Stabilization Terms}\n\nFinally, we may now form the $\\tau$ stabilization terms, \n\\begin{equation}\n \\langle L\\, {\\bf c},S{\\bf \\phi} \\rangle_\\tau. \n\\end{equation}\nWhere the operator $ \\langle \\cdot,\\cdot \\rangle_\\tau. $ is shorthand \nand denotes\n\\begin{equation}\n \\langle {\\bf u},{\\bf v} \\rangle_\\tau = \\sum_K \\tau_K ({\\bf u},{\\bf v})_K.\n\\end{equation}\nWhere $K$ denotes the FEM cells. Now, through what Becker and Braack\ncontemptibly referred to as ``elementary calculus'', we arrive at our\nstabilization terms, \n\\begin{align*}\n \\langle L\\,{\\bf c},S{\\bf \\phi} \\rangle_\\tau = \\sum_K \\{ \\quad &\\tau_p (\\nabla \\cdot {\\bf u},\n \\nabla \\cdot {\\bf v}) \\\\\n +\\quad &\\tau_u \\,(\\nabla p + {\\bf u} \\cdot \\nabla {\\bf u} - \\nu\n \\nabla^2 {\\bf u} - g \\frac{T'}{T_0},\n \\nabla q) \\\\\n +\\quad &\\tau_u \\, (\\nabla p + {\\bf u} \\cdot \\nabla {\\bf u} - \\nu \\nabla^2 {\\bf u}- g \\frac{T'}{T_0},\n \\nabla {\\bf u} \\cdot \\nabla {\\bf v} + \\nu \\nabla^2 {\\bf v}) \\\\\n +\\quad &\\tau_T \\, ({\\bf u} \\cdot T - \\nabla \\cdot (k \\nabla T), \\nabla {\\bf u} \\cdot \\nabla\n w + \\nabla \\cdot (k \\nabla w)) \\quad \\}.\n\\end{align*}\n\n", "meta": {"hexsha": "4404f10d1893501a0f7dbbb33b43f3026cb6ec18", "size": 8590, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "disputatio/dissertation/appendix-derivation.tex", "max_stars_repo_name": "nicholasmalaya/paleologos", "max_stars_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-04T17:49:42.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-04T17:49:42.000Z", "max_issues_repo_path": "disputatio/dissertation/appendix-derivation.tex", "max_issues_repo_name": "nicholasmalaya/paleologos", "max_issues_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "disputatio/dissertation/appendix-derivation.tex", "max_forks_repo_name": "nicholasmalaya/paleologos", "max_forks_repo_head_hexsha": "11959056caa80d3c910759b714a0f8e42f986f0f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-01-04T16:08:18.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-16T19:34:24.000Z", "avg_line_length": 37.347826087, "max_line_length": 101, "alphanum_fraction": 0.6332945285, "num_tokens": 3118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.800691997339971, "lm_q2_score": 0.7853085758631159, "lm_q1q2_score": 0.6287902921360464}} {"text": "\\documentclass{amsproc}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage[cmtip,all]{xy}\n\n\\copyrightinfo{2016}{Todd D. Vance}\n\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{lemma}[theorem]{Lemma}\n\n\\theoremstyle{definition}\n\\newtheorem{definition}[theorem]{Definition}\n\\newtheorem{example}[theorem]{Example}\n\\newtheorem{xca}[theorem]{Exercise}\n\n\\theoremstyle{remark}\n\\newtheorem{remark}[theorem]{Remark}\n\n\\numberwithin{equation}{section}\n\n\\begin{document}\n\\title{A Geometric Algebra Primer}\n\\author[T. D. Vance]{Todd D. Vance}\n%\\address{15 Lilac Dr \\\\ Moorefield, WV 26836}\n%\\email{tdvance@gmail.com}\n\\date{\\today}\n\n\\begin{abstract}\n A brief introduction to Geometric Algebra with an emphasis on algorithms.\n\\end{abstract}\n\n\\maketitle{}\n\n\\newcommand{\\term}[1]{\\emph{#1}}\n\n\\newcommand{\\ga}{\\ensuremath{\\mathop{\\mathrm{GA}}}}\n\n\\section{definitions}\n\n\\begin{definition}[Geometric Algebra]\n The \\term{geometric algebra} $\\ga(n)$ of \\term{ambient rank} $n$ is\n the quotient algebra $\\mathbf{R}\\{[1], [2], \\dots, [n]\\}/I$, where\n $\\mathbf{R}\\{[1], [2], \\dots, [n]\\}$ is the free (unital,\n associative, noncommutative) algebra over the real field\n $\\mathbf{R}$ generated by noncommuting elements $[1], [2], \\dots,\n [n]$, and $I$ is the (two-sided) ideal of $\\mathbf{R}\\{[1], [2],\n \\dots, [n]\\}$ generated by the set of elements of the form $[i][i] -\n 1$ and of the form $[i][j] - [j][i]$ when $i\\ne{j}$.\n\\end{definition}\n\nThe multiplication operator in $\\ga(n)$ is called the \\term{geometric\n product} and is written by juxtaposition: $ab$ is the product of $a$\nand $b$, in that order. An (equivalence class of the) element of\n$\\ga(n)$ of the form $[i_1][i_2]\\cdots[i_m]$ is abbreviated\n$[i_1i_2\\cdots{i}_m]$ and is called a \\term{basis blade}. As a\nspecial case, the empty basis blade $[]$ is defined to be the\nmultiplicative identity $1$.\n\n\\begin{example}\nA commonly-used geometric algebra in graphics applications is $\\ga[3]$\nof ambient rank $3$. The algebra is generated by the elements $[1]$,\n$[2]$, and $[3]$. Clearly $[1][1] = [] = 1$, $[1][2]=-[2][1]$, and so\non. From this one finds that $\\ga(3)$ is an eight-dimensional vector\nspace with basis $\\{[], [1], [2], [3], [12], [13], [23], [123]\\}$. A\ntypical element of $\\ga(3)$ is then equal to $s[] + a[1] + b[2] + c[3]\n+ A[23] + B[13] + C[12] + S[123]$ where $s, a, b, c, A, B, C,$ and $S$\nare real numbers.\n\nWe consider elements of $\\ga(3)$ of the form $a[1]+b[2]+c[3]$ to be\nvectors (or rank one elements) and this three-dimensional vector space\nis the geometric entity $\\ga(3)$ models. Elements of the form $s[]$,\nalso just written $s$, are scalars (or rank zero elements), so in a\ngeometric algebra, one can add a scalar to a vector. Elements of the\nform $A[23] + B[13] + C[12]$ are bivectors (rank two elements), which\nmodel oriented plane regions with an area magnitude.\n\nJust as a vector has direction and magnitude, and one considers two\nvectors having the same direction and magnitude to be equal even if\nthey have different locations in three-space, a bivector has a\ncontaining plane, one of two orientations on that plane, and an area,\nso two planar regions having the same area, contained in parallel or\nequal planes, and given the same orientation (which may be thought of\nas clockwise versus counterclockwise) are considered to be the same\nbivector.\n\nElements of the form $S[123]$ are considered volume elements (rank\nthree elements), also called pseudoscalars because they have one\ndegree of freedom in three-dimensional space. A volume element only\nhas magnitude and orientation (positive or negative, which may be\nthought of as being a right-handed or left-handed coordinate system).\n\nAddition of two elements expressed in the linear combination form\nmentioned above is done by combining like terms. If $x = 4 + 3[1] +\n4[2] + 5[12]$ and $y = 3 + 2[1] + 3[2] + 4[12]$, then\n$x+y = 7 +5[1] +7[2] + 9[12]$.\n\nMultiplication works like polynomial multiplication, except without\ncommutativity, and using the rules of converting products of blades\ninto the standard ones $[]=1$, $[1]$, $[2]$, $[12]$, $[3]$, $[13]$,\n$[23]$, and $[123]$. So, using the same $x$ and $y$ as above, to find\n$xy$, one multiplies every term of $x$ by every term of $y$, then\nreduces and combines like terms:\n\\begin{alignat*}{4}\n xy &=&&12 + 8[1] + 12[2] +16[12]\\\\\n && +~ & 9[1] + 6[1][1] + 9[1][2] + 12[1][12]\\\\\n && +~ & 12[2] + 8[2][1] + 12[2][2] + 16[2][12]\\\\\n && +~ & 15[12] + 10[12][1] + 15[12][2] + 20[12][12]\\\\\n &=&&12 + 8[1] + 12[2] + 16[12]\\\\\n && +~ &9[1] + 6 + 9[12] + 12[2]\\\\\n && +~ & 12[2] - 8[12] + 12 - 16[1]\\\\\n && +~ & 15[12] - 10[2] + 15[1] - 20\\\\\n &=&& 10 + 16[1] + 26[2] + 32[12]\n\\end{alignat*}\n\\end{example}\n\n\\subsection{Standard Basis Blades}\n\nGeneralizing from the example, we see that swapping adjacent unequal\nindices of a basis blade changes its sign and deleting pairs of\nadjacent, equal indices does not change the element. Thus, all basis\nblades in $\\ga(n)$ are either plus or minus a \\term{standard} basis\nblade $[i_1i_2\\cdots{i}_m]$ ($m \\le n$) in which the indices are in\nincreasing order and there are no duplicate indices. In this case,\nthe \\term{rank} of the basis blade is $m$. Any element of $\\ga(n)$\ncan be written uniquely (up to ordering of terms) as a linear\ncombination of standard basis blades (including $[]$, the rank-zero\nempty basis blade equal to $1$ in the algebra), and the rank of that\nelement is the maximum rank among the basis blades in the linear\ncombination. The rank of the $0$ element is defined to be $-\\infty$.\nThus, $\\ga(n)$ is an algebra filtered by rank.\n\nA nonzero element of $\\ga(n)$ that is a linear combination of standard\nbasis blades of the same rank $r$ is said to be \\term{purely} of rank\n$r$. Rank 0 elements (that is, pure rank 0 elements along with the 0\nelement) are called \\term{scalars}. Pure rank 1 elements (along with\n0) are called \\term{vectors}. Pure rank two elements (along with 0)\nare called \\term{bivectors}, and so on. All elements are called\n\\term{multivectors}. When the ambient rank is $n$, rank $n$ elements\n(along with 0) are called \\term{pseudoscalars}, rank $n-1$ elements\n(along with 0) are called \\term{pseudovectors}, and so on.\n\nThe algebra $\\ga(n)$ is generated, as a vector space, by the standard\nbasis blades $[]$, $[1]$, $[2]$, $\\dots$, $[n]$, $[12]$, $\\dots$\n$[123\\dots{n}]$, of which there are $2^n$, so as a real vector\nspace, $\\ga(n)$ has dimension $2^n$.\n\nIn addition to the geometric product, other products are defined in\n$\\ga(n)$. The \\term{inner product} is the symmetric component of the\ngeometric product: $a\\cdot{b} = \\frac{ab+ba}{2}$. The \\term{outer\n product} is the antisymmetric component of the geometric product:\n$a\\wedge{b}=\\frac{ab-ba}{2}$. As the names imply,\n$a\\cdot{b}=b\\cdot{a}$, and $a\\wedge{b} = -b\\wedge{a}$. Then, $ab =\na\\cdot{b} + a\\wedge{b}$.\n\n\\subsection{Additional Operations}\n\nNote that for vectors, $a\\cdot{b}$ is exactly the standard dot\nproduct, using $\\{[1], [2], \\dots, [n]\\}$ as the standard ordered\nbasis. For vectors in $\\ga(3)$, the wedge product has the same\nmagnitude as the cross product, but is a bivector (pseudovector)\nrather than a vector. Also, the wedge product is defined in any\nnumber of dimensions, not just three, and for non-vector elements as\nwell.\n\nThe \\emph{standard pseudoscalar} $I$ of $\\ga(n)$ is defined to be the\nrank-$n$ standard basis blade $[123\\dots{n}]$. Then, $I^2 = II$ must\neither be $-1$ or $1$, so $I^4=1$ regardless of $n$. In particular,\nthe subalgebra of $\\ga(2)$ generated by $[]$ and $I$ is isomorphic to\nthe complex number field. In fact $I^2=-1$ in $\\ga(2)$ and the\nisomorphism maps $1$ to $[]$ and $\\imath$ to $I$. In addition,\n$\\ga(3)$ contains a subalgebra isomorphic to the quanternion algebra.\nThis subalgebra is generated by $[]$, $i=[23]$, $j=[13]$, and\n$k=[12]$, with $[]$ maping to 1 and $i$, $j$, and $k$ mapping to the\nsame-named elements of the quaternion ring.\n\nThe \\term{norm} $|a|$ of an element $a\\in\\ga(n)$ is defined to be the\nsquare root of the sum of the squares of the coefficients of the\nstandard basis blades making up the linear combination. Note for\nvectors this reduces to the usual norm for vectors.\n\nThe \\term{reversion} $\\tilde{a}$ of an element $a\\in \\ga(n)$ is formed\nfrom $a$ by taking the linear combination of standard basis blades\ncomposing $a$, and then reversing the order of multiplication of the\nalgebra generators in each term: $[i_{j_1}][i_{j_2}]\\cdots[i_{j_m}]$\n$\\rightarrow$ $[i_{j_m}][i_{j_{m-1}}]\\cdots[i_{j_1}]$. Equivalently,\neach term of $a$ of pure rank $r$, where $r$ is congruent to $2 $ or\n$3$ modulo $4$, is negated.\n\nThe \\term{grade projection} $\\mathrm{proj}_k(a)$ of an element $a\\in \\ga(n)$ is\nformed from $a$ by deleting all terms of pure rank not equal to $k$.\n\nThe \\term{left inverse} of $a\\in \\ga(n)$ is an element $b$ satisfying\n$ba=1=[]$ and the \\term{right inverse} $c$ likewise satisfies $ac=1$.\nWe use $a^{-1}$ to represent an element that is both the left and right\ninverse of $a$. If $a{\\tilde{a}}$ is a nonzero scalar, then\n$\\frac{a}{a{\\tilde{a}}}$ is well-defined by dividing each coefficient of\n$a$ by the scalar denominator. In such a case, $a$ does have a\ntwo-sided inverse and $a^{-1} = \\frac{a}{a{\\tilde{a}}}$.\n\nThe \\term{dual} of an element $a$ of $\\ga(n)$ is $aI^{-1}$. Note that\n$I^{-1}$ exists and is equal to $I^3$, which is either $I$ or $-I$.\nNote that the dual of a scalar is a psuedoscalar, the dual of a vector\nis a pseudovector, and so on. The double dual of $a$ is either $a$ or\n$-a$.\n\nNote that in $\\ga(3)$, $a\\wedge{b}I^{-1}$ is a vector, and is equal to\nthe vector space cross product $a\\times{b}$. Thus, we define a\n\\term{generalized cross product} on any $\\ga(n)$ to be\n$a\\times{b}=a\\wedge{b}I^{-1}$, the dual of the wedge product. In\ndimensions other than 3, the cross product of vectors is no longer a\nvector. However, in $n$ dimensions, one can take the dual of the\nwedge product of $n-1$ vectors to give a vector which geometrically is\nsimilar to the cross product in three dimensions.\n\nThe \\term{exponential} of $a\\in{\\ga}(n)$ is computed as follows. If\n$a=0$, $e^0=1$ by definition. Otherwise, let $\\theta=|a|$ and\n$b=a/\\theta$. Then use DeMoivre's formula with $b$ taking the role of\nthe imaginary unit: $e^a = \\cos\\theta + b\\sin\\theta$. The inverse of\nthis operation, a geometric algebraic logarithm, is difficult and\nmultivalued, even more so than with complex numbers.\n\nThe main use for the exponential is in computing a \\term{rotor}. If\n$v$ and $w$ are linearly-independent unit vectors (that is,\n$|v|=|w|=1$) in the $n$-dimensional vector space whose basis is the\nrank-one standard blades, then $vw$ is a bivector that represents an\noriented plane in the $n$-dimensional space. A rotation about the\norigin in the direction from $v$ to $w$ along an angle $theta$ is then\nmodeled by the \\term{rotor} $r$ in $\\ga(n)$ that is $r =\ne^{\\frac12vw\\theta}$. Then if $a$ is any element (vector or\notherwise) of $\\ga(n)$, its rotation is then $ra{\\tilde{r}}$. If $a$\nis a vector, so is $ra{\\tilde{r}}$, and this is an ordinary vector\nrotation. If $a$ is a bivector representing an oriented plane,\n$ra{\\tilde{r}}$ is the rotation of the plane. And so on.\n\nThe \\term{meet} of two elements $a$ and $b$ in $\\ga(n)$ is defined by\n$aI^{-1}b$, or the product of the dual of $a$ with $b$.\n\n\\section{Algorithms}\n\\newcommand{\\n}{\\ensuremath{\\mathop{\\mathrm{N}}}} An apparent way to\nrepresent a standard basis blade $a=[i_1i_2\\dots{i}_m]$ is via the\ninteger $\\n(a)$ defined to be $\\sum_{k=1}^{m}2^{i_k}$. Then, the\nproduct $ab$ of $a=[i_1i_2\\dots{i}_m]$ and $b=[j_1j_2\\dots{j}_r]$\ncan be computed as follows:\n\n\\begin{enumerate}\n\\item If $m=0$, let $c$ be the multivector $b$ and return $c$\n\\item If $r=0$, let $c$ be the multivector $a$ and return $c$\n\\item Let $c$ be the blade for which $\\n(c)$ is the bitwise exclusive\n or of $\\n(a)$ and $\\n(b)$.\n\\item Let $s$ be 1.\n\\item Let $p$ be the maximum of $\\n(a)$ and $\\n(b)$.\n\\item Let $d$ be $m$\n\\item Let $e$ be $1$\n\\item while $e\\le{p}$, do the following:\n \\begin{enumerate}\n \\item Compute the bitwise and of $e$ and $\\n(a)$.\n \\item If the result of the bitwise and is nonzero, then replace $d$ with $d-1$.\n \\item If $d$ is odd, Compute the bitwise and of $e$ and $\\n(b)$.\n \\item If $d$ is odd and the result of the bitwise and is nonzero\n then replace $s$ with $-s$.\n \\item Replace $e$ with $2e$.\n \\end{enumerate}\n\\item Let $c$ be the multivector that is the real coefficient $s$\n multiplied by the blade given by $c$, and return $c$.\n\\end{enumerate}\n\nA standard basis blade is simply a nonnegative integer in this\nrepresentation. A choice needs to be made to represent a multivector,\na real linear combination of these blades.\n\nThe \\term{dense} representation of a multivector in $\\ga(n)$ is an\narray $A$ of dimension $2^n$, thus having elements $A_0$, $A_1$, and\nso on up to $A_{2^n-1}$. Then, a multivector $\\sum_{i=1}^{l}c_ib_i$,\nwhere the $c_i$ are real numbrers and the $b_i$ are standard basis\nblades in $\\ga(n)$, is represented by the array $A$ in which for\n$1\\le{i}\\le{l}$, $A_{\\n(b_i)} = c_i$, and all other elements of $A$\nare zero.\n\nThe dense representation requires little computation, but uses\n$O(2^n)$ memory per element of $\\ga(n)$, so it is useful for low\ndimensionality, such as in representing the commonly-used $\\ga(3)$.\n\nFor larger dimensions, a \\term{sparse} representation is needed. Let\n$D$ be a \\term{dictionary}, algorithmically represented by a hash\ntable in most cases, but mathematically works like the array: for any\nnonnegative integer $i$, $D_i$ is a real number. For most $i$, this\nreal number is zero and thus $i$ has no entry in the hash table. Only\nwhen $D_i$ is nonzero is there an entry in the hash table with key $i$\nand value $D_i$. \n \nThen, a multivector $\\sum_{i=1}^{l}c_ib_i$, where the $c_i$ are real\nnumbrers and the $b_i$ are standard basis blades in $\\ga(n)$, is\nrepresented by the dictionary $D$ in which for $1\\le{i}\\le{l}$,\n$D_{\\n(b_i)} = c_i$, and no other entries are present in $D$ (so that\nthe dictionary returns the default value of 0 for those).\n\nThe sparse representation requires a small amount of computation for\nhashing and, sometimes, rebuilding the table, but only requires $O(l)$\nmemory for an element of $\\ga(n)$ having $l$ nonzero terms. Thus, the\nsparse representation is most useful for large $n$ when many elements\nare such that most standard basis blades have a zero coefficient.\n\nIn both cases, the addition procedure is evident and involves adding\nthe values from one array or dictionary to the same-indexed values of\nthe other.\n\nThe geometric product, however, requires an all-to-all mapping of\narray or dictionary elements. We illustrate the algorithm for the\narray case, with the dictionary case being nearly identical, just\niterating over keys in the hash table instead of over all integers in\nthe range from $0$ to $2^{n-1}$.\n\nSuppose $A$ and $B$ are arrays representing multivectors in $\\ga(n)$.\nWe compute a new array $C$ with $2^n$ entries representing the\ngeometric product of the multivectors $A$ and $B$.\n\n\\begin{enumerate}\n\\item Initialize $C$ to $C_0=C_1=\\cdots=c_{2^{n-1}}=0$.\n\\item For each $i$ from 0 to $2^{n-1}$ such that the coefficient $A_i$\n is nonzero\n \\begin{enumerate}\n \\item For each $j$ from 0 to $2^{n-1}$ such that the coefficient\n $B_j$ is nonzero\n \\begin{enumerate}\n \\item Let $k$ be the integer that is the result of multiplying\n together the blades represented by $i$ and $j$, and let $s$ be\n the coefficient, $1$ or $-1$, of the result.\n \\item Let $C_k$ be $C_k + A_iB_js$, an ordinary sum and product\n of real numbers.\n \\end{enumerate}\n \\end{enumerate}\n \\item Return the multivector $C$.\n\\end{enumerate}\n\n\\bibliographystyle{amsplain}\n\n\\end{document}\n", "meta": {"hexsha": "8b2d251b5293ebab67302c4ebdc604591e0dc1b9", "size": 15849, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/ga.tex", "max_stars_repo_name": "tdvance/GeometricAlgebra", "max_stars_repo_head_hexsha": "3e75c09674a817c1c1cc45a746a2089d1466ac08", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-06T08:09:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-11T02:27:49.000Z", "max_issues_repo_path": "doc/ga.tex", "max_issues_repo_name": "tdvance/GeometricAlgebra", "max_issues_repo_head_hexsha": "3e75c09674a817c1c1cc45a746a2089d1466ac08", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/ga.tex", "max_forks_repo_name": "tdvance/GeometricAlgebra", "max_forks_repo_head_hexsha": "3e75c09674a817c1c1cc45a746a2089d1466ac08", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.8905325444, "max_line_length": 81, "alphanum_fraction": 0.6915262793, "num_tokens": 5086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920020959544, "lm_q2_score": 0.7853085708384736, "lm_q1q2_score": 0.62879029184777}} {"text": "% !TEX root = lectures.tex\n\\section{Oscillations}\n\\bigskip\n\n\\subsection{Harmonic Oscillator}\n\nThe harmonic oscillator is omnipresent in physics. Although students think of this as being related to springs, it, or an equivalent mathematical representation, appears in just about any problem where a mode is sitting near its potential energy minimum. At that point, $\\partial_x V(x)=0$, and the first non-zero term (aside from a constant) in the potential energy is that of a harmonic oscillator. In a solid, sound modes (phonons) are built on a picture of coupled harmonic oscillators, and in relativistic field theory the fundamental interactions are also built on coupled oscillators positioned infinitesimally close to one another in space. The phenomena of a resonance of an oscillator driven at a fixed frequency plays out repeatedly in atomic, nuclear and high-energy physics, when quantum mechanically the evolution of a state oscillates according to $e^{-iEt}$ and exciting discrete quantum states has very similar mathematics as exciting discrete states of an oscillator.\n\nThe potential energy for a single particle as a function of its position $x$ can be written as a Taylor expansion about some point $x_0$\n\\begin{equation}\nV(x)=V(x_0)+(x-x_0)\\left.\\partial_xV(x)\\right|_{x_0}+\\frac{1}{2}(x-x_0)^2\\left.\\partial_x^2V(x)\\right|_{x_0}\n+\\frac{1}{3!}\\left.\\partial_x^3V(x)\\right|_{x_0}+\\cdots\n\\end{equation}\nIf the position $x_0$ is at the minimum of the resonance, the first two non-zero terms of the potential are\n\\begin{eqnarray}\nV(x)&\\approx& V(x_0)+\\frac{1}{2}(x-x_0)^2\\left.\\partial_x^2V(x)\\right|_{x_0},\\\\\n\\nonumber\n&=&V(x_0)+\\frac{1}{2}k(x-x_0)^2,~~~~k\\equiv \\left.\\partial_x^2V(x)\\right|_{x_0},\\\\\n\\nonumber\nF&=&-\\partial_xV(x)=-k(x-x_0).\n\\end{eqnarray}\n\nPut into Newton's 2nd law (assuming $x_0=0$),\n\\begin{eqnarray}\nm\\ddot{x}&=&-kx,\\\\\nx&=&A\\cos(\\omega_0 t-\\phi),~~~\\omega_0=\\sqrt{k/m}.\n\\end{eqnarray}\nHere $A$ and $\\phi$ are arbitrary. Equivalently, one could have written this as $A\\cos(\\omega_0 t)+B\\sin(\\omega_0 t)$, or as the real part of $Ae^{i\\omega_0 t}$. In this last case $A$ could be an arbitrary complex constant. Thus, there are 2 arbitrary constants (either $A$ and $B$ or $A$ and $\\phi$, or the real and imaginary part of one complex constant. This is the expectation for a second order differential equation, and also agrees with the physical expectation that if you know a particle's initial velocity and position you should be able to define its future motion, and that those two arbitrary conditions should translate to two arbitrary constants.\n\nA key feature of harmonic motion is that the system repeats itself after a time $T=1/f$, where $f$ is the frequency, and $\\omega=2\\pi f$ is the angular frequency. The period of the motion is independent of the amplitude. However, this independence is only exact when one can neglect higher terms of the potential, $x^3, x^4\\cdots$. Once can neglect these terms for sufficiently small amplitudes, and for larger amplitudes the motion is no longer purely sinusoidal, and even though the motion repeats itself, the time for repeating the motion is no longer independent of the amplitude.\n\nOne can also calculate the velocity and the kinetic energy as a function of time,\n\\begin{eqnarray}\n\\dot{x}&=&-\\omega_0A\\sin(\\omega_0 t-\\phi),\\\\\n\\nonumber\nT&=&\\frac{1}{2}m\\dot{x}^2=\\frac{m\\omega_0^2A^2}{2}\\sin^2(\\omega_0t-\\phi),\\\\\n\\nonumber\n&=&\\frac{k}{2}A^2\\sin^2(\\omega_0t-\\phi).\n\\end{eqnarray}\nThe total energy is then\n\\begin{equation}\nE=T+U=\\frac{1}{2}m\\dot{x}^2+\\frac{1}{2}kx^2=\\frac{1}{2}kA^2.\n\\end{equation}\nThe total energy then goes as the square of the amplitude.\n\n\\example\nA pendulum is an example of a harmonic oscillator. By expanding the kinetic and potential energies for small angles find the frequency for a pendulum of length $L$ with all the mass $m$ centered at the end by writing the eq.s of motion in the form of a harmonic oscillator.\n\n{\\bf Solution:} The potential energy and kinetic energies are (for $x$ being the displacement)\n\\begin{eqnarray*}\nU&=&mgL(1-\\cos\\theta)\\approx mgL\\frac{x^2}{2L^2},\\\\\nT&=&\\frac{1}{2}mL^2\\dot{\\theta}^2\\approx \\frac{m}{2}\\dot{x}^2.\n\\end{eqnarray*}\nFor small $x$ Newton's 2nd law becomes\n\\[\nm\\ddot{x}=-\\frac{mg}{L}x,\n\\]\nand the spring constant would appear to be $k=mg/L$, which makes the frequency equal to $\\omega_0=\\sqrt{g/L}$. Note that the frequency is independent of the mass.\n\n\\exampleend\n\n\\subsection{Damped Oscillators}\n\nIn this chapter we consider only the case where the damping force is proportional to the velocity. This is counter to dragging friction, where the force is proportional in strength to the normal force and independent of velocity, and is also inconsistent with wind resistance, where the magnitude of the drag force is proportional the square of the velocity. Rolling resistance does seem to be mainly proportional to the velocity. However, the main motivation for considering damping forces proportional to the velocity is that the math is more friendly. This is because the differential equation is linear, i.e. each term is of order $x$, $\\dot{x}$, $\\ddot{x}\\cdots$, or even terms with no mention of $x$, and there are no terms such as $x^2$ or $x\\ddot{x}$. The equations of motion for a spring with damping force $-b\\dot{x}$ are\n\\begin{equation}\nm\\ddot{x}+b\\dot{x}+kx=0.\n\\end{equation}\nJust to make the solution a bit less messy, we rewrite this equation as\n\\begin{equation}\n\\label{eq:dampeddiffyq}\n\\ddot{x}+2\\beta\\dot{x}+\\omega_0^2x=0,~~~~\\beta\\equiv b/2m,~\\omega_0\\equiv\\sqrt{k/m}.\n\\end{equation}\nBoth $\\beta$ and $\\omega$ have dimensions of inverse time. To find solutions (see appendix C in the text) you must make an educated guess at the form of the solution. To do this, first realize that the solution will need an arbitrary normalization $A$ because the equation is linear. Secondly, realize that if the form is\n\\begin{equation}\nx=Ae^{rt}\n\\end{equation}\nthat each derivative simply brings out an extra power of $r$. This means that the $Ae^{rt}$ factors out and one can simply solve for an equation for $r$. Plugging this form into Eq. (\\ref{eq:dampeddiffyq}),\n\\begin{equation}\nr^2+2\\beta r+\\omega_0^2=0.\n\\end{equation}\nBecause this is a quadratic equation there will be two solutions,\n\\begin{equation}\nr=-\\beta\\pm\\sqrt{\\beta^2-\\omega_0^2}.\n\\end{equation}\nWe refer to the two solutions as $r_1$ and $r_2$ corresponding to the $+$ and $-$ roots. As expected, there should be two arbitrary constants involved in the solution,\n\\begin{equation}\nx=A_1e^{r_1t}+A_2e^{r_2t},\n\\end{equation}\nwhere the coefficients $A_1$ and $A_2$ are determined by initial conditions.\n\nThe roots listed above, $\\sqrt{\\omega_0^2-\\beta_0^2}$, will be imaginary if the damping is small and $\\beta<\\omega_0$. In that case, $r$ is complex and the factor $e{rt}$ will have some oscillatory behavior. If the roots are real, there will only be exponentially decaying solutions. There are three cases:\n\\begin{enumerate}\\itemsep 0pt\n\\item Underdamped: $\\beta<\\omega_0$\n\\begin{eqnarray}\nx&=&A_1e^{-\\beta t}e^{i\\omega't}+A_2e^{-\\beta t}e^{-i\\omega't},~~\\omega'\\equiv\\sqrt{\\omega_0^2-\\beta^2}\\\\\n\\nonumber\n&=&(A_1+A_2)e^{-\\beta t}\\cos\\omega't+i(A_1-A_2)e^{-\\beta t}\\sin\\omega't.\n\\end{eqnarray}\nHere we have made use of the identity $e^{i\\omega't}=\\cos\\omega't+i\\sin\\omega't$. Because the constants are arbitrary, and because the real and imaginary parts are both solutions individually, we can simply consider the real part of the solution alone:\n\\begin{eqnarray}\n\\label{eq:homogsolution}\nx&=&B_1e^{-\\beta t}\\cos\\omega't+B_2e^{-\\beta t}\\sin\\omega't,\\\\\n\\nonumber \n\\omega'&\\equiv&\\sqrt{\\omega_0^2-\\beta^2}.\n\\end{eqnarray}\n\n\\item Critical dampling: $\\beta=\\omega_0$\\\\\nIn this case the two terms involving $r_1$ and $r_2$ are identical because $\\omega'=0$. Because we need to arbitrary constants, there needs to be another solution. This is found by simply guessing, or by taking the limit of $\\omega'\\rightarrow 0$ from the underdamped solution. The solution is then\n\\begin{equation}\n\\label{eq:criticallydamped}\nx=Ae^{-\\beta t}+Bte^{-\\beta t}.\n\\end{equation}\nThe critically damped solution is interesting because the solution approaches zero quickly, but does not oscillate. For a problem with zero initial velocity, the solution never crosses zero. This is a good choice for designing shock absorbers or swinging doors.\n\\item Overdamped: $\\beta>\\omega_0$\n\\begin{eqnarray}\nx&=&A_1e^{-(\\beta+\\sqrt{\\beta^2-\\omega_0^2})t}+A_2e^{-(\\beta-\\sqrt{\\beta^2-\\omega_0^2})t}\n\\end{eqnarray}\nThis solution will also never pass the origin more than once, and then only if the initial velocity is strong and initially toward zero.\n\\end{enumerate}\n\n\\example\nGiven $b$, $m$ and $\\omega_0$, find $x(t)$ for a particle whose initial position is $x=0$ and has initial velocity $v_0$ (assuming an underdamped solution).\n\n{\\bf Solution:} The solution is of the form,\n\\begin{eqnarray*}\nx&=&e^{-\\beta t}\\left[A_1\\cos(\\omega' t)+A_2\\sin\\omega't\\right],\\\\\n\\dot{x}&=&-\\beta x+\\omega'e^{-\\beta t}\\left[-A_1\\sin\\omega't+A_2\\cos\\omega't\\right].\\\\\n\\omega'&\\equiv&\\sqrt{\\omega_0^2-\\beta^2},~~~\\beta\\equiv b/2m.\n\\end{eqnarray*}\nFrom the initial conditions, $A_1=0$ because $x(0)=0$ and $\\omega'A_2=v_0$. So \n\\[\nx=\\frac{v_0}{\\omega'}e^{-\\beta t}\\sin\\omega't.\n\\]\n\\exampleend\n\n\\subsection{Sinusoidally Driven Oscillators}\n\nHere, we consider the force\n\\begin{equation}\nF=-kx-b\\dot{x}+F_0\\cos\\omega t,\n\\end{equation}\nwhich leads to the differential equation\n\\begin{equation}\n\\label{eq:drivenosc}\n\\ddot{x}+2\\beta\\dot{x}+\\omega_0^2x=(F_0/m)\\cos\\omega t.\n\\end{equation}\nConsider a single solution with no arbitrary constants, which we will call a {\\it particular solution}, $x_p(t)$. It should be emphasized that this is {\\bf A} particular solution, because there exists an infinite number of such solutions because the general solution should have two arbitrary constants. Now consider solutions to the same equation without the driving term, which include two arbitrary constants. These are called either {\\it homogenous solutions} or {\\it complementary solutions}, and were given in the previous section, e.g. Eq. (\\ref{eq:homogsolution}) for the underdamped case. The homogenous solution already incorporates the two arbitrary constants, so any sum of a homogenous solution and a particular solution will represent the {\\it general solution} of the equation. The general solution incorporates the two arbitrary constants $A$ and $B$ to accommodate the two initial conditions. One could have picked a different particular solution, i.e. the original particular solution plus any homogenous solution with the arbitrary constants $A_p$ and $B_p$ chosen at will. When one adds in the homogenous solution, which has adjustable constants with arbitrary constants $A'$ and $B'$, to the new particular solution, one can get the same general solution by simply adjusting the new constants such that $A'+A_p=A$ and $B'+B_p=B$. Thus, the choice of $A_p$ and $B_p$ are irrelevant, and when choosing the particular solution it is best to make the simplest choice possible.\n\nTo find a particular solution, one first guesses at the form,\n\\begin{equation}\n\\label{eq:partform}\nx_p(t)=D\\cos(\\omega t-\\delta),\n\\end{equation}\nand rewrite the differential equation as\n\\begin{equation}\nD\\left\\{-\\omega^2\\cos(\\omega t-\\delta)-2\\beta\\omega\\sin(\\omega t-\\delta)+\\omega_0^2\\cos(\\omega t-\\delta)\\right\\}=\\frac{F_0}{m}\\cos(\\omega t).\n\\end{equation}\nOne can now use angle addition formulas to get\n\\begin{eqnarray}\nD\\left\\{(-\\omega^2\\cos\\delta+2\\beta\\omega\\sin\\delta+\\omega_0^2\\cos\\delta)\\cos(\\omega t)\\right.\\hspace*{36pt}&&\\\\\n\\nonumber\n\\left.+(-\\omega^2\\sin\\delta-2\\beta\\omega\\cos\\delta+\\omega_0^2\\sin\\delta)\\sin(\\omega t)\\right\\}\n&=&\\frac{F_0}{m}\\cos(\\omega t).\n\\end{eqnarray}\nBoth the $\\cos$ and $\\sin$ terms need to equate if the expression is to hold at all times. Thus, this becomes two equations\n\\begin{eqnarray}\nD\\left\\{-\\omega^2\\cos\\delta+2\\beta\\omega\\sin\\delta+\\omega_0^2\\cos\\delta\\right\\}&=&\\frac{F_0}{m}\\\\\n\\nonumber\n-\\omega^2\\sin\\delta-2\\beta\\omega\\cos\\delta+\\omega_0^2\\sin\\delta&=&0.\n\\end{eqnarray}\nAfter dividing by $\\cos\\delta$, the lower expression leads to\n\\begin{equation}\n\\tan\\delta=\\frac{2\\beta\\omega}{\\omega_0^2-\\omega^2}.\n\\end{equation}\nUsing the identities $\\tan^2+1=\\csc^2$ and $\\sin^2+\\cos\\^2=1$, one can also express $\\sin\\delta$ and $\\cos\\delta$,\n\\begin{eqnarray}\n\\sin\\delta&=&\\frac{2\\beta\\omega}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2}},\\\\\n\\nonumber\n\\cos\\delta&=&\\frac{(\\omega_0^2-\\omega^2)}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2}}\n\\end{eqnarray}\nInserting the expressions for $\\cos\\delta$ and $\\sin\\delta$ into the expression for $D$,\n\\begin{equation}\n\\label{eq:Ddrive}\nD=\\frac{F_0/m}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2}}.\n\\end{equation}\n\nFor a given initial condition, e.g. initial displacement and velocity, one must add the homogenous solution then solve for the two arbitrary constants. However, because the homogenous solutions decay with time as $e^{-\\beta t}$, the particular solution is all that remains at large times, and is therefore the steady state solution. Because the arbitrary constants are all in the homogenous solution, all memory of the initial conditions are lost at large times, $t>>1/\\beta$.\n\nThe amplitude of the motion, $D$, is linearly proportional to the driving force ($F_0/m$), but also depends on the driving frequency $\\omega$. For small $\\beta$ the maximum will occur at $\\omega=\\omega_0$. This is referred to as a resonance. In the limit $\\beta\\rightarrow 0$ the amplitude at resonance approaches infinity. \n\n\\subsection{Alternative Derivation for Driven Oscillators}\n\nHere, we derive the same expressions as in Eq.s (\\ref{eq:partform}-\\ref{eq:Ddrive}) but express the driving forces as\n\\begin{eqnarray}\nF(t)&=&F_0e^{i\\omega t},\n\\end{eqnarray}\nrather than as $F_0\\cos\\omega t$. The real part of $F$ is the same as before. For the differential equation,\n\\begin{eqnarray}\n\\label{eq:compdrive}\n\\ddot{x}+2\\beta\\dot{x}+\\omega_0^2x&=&\\frac{F_0}{m}e^{i\\omega t},\n\\end{eqnarray}\none can treat $x(t)$ as an imaginary function. Because the operations $d^2/dt^2$ and $d/dt$ are real and thus do not mix the real and imaginary parts of $x(t)$, Eq. (\\ref{eq:compdrive}) is effectively 2 equations. Because $e^{\\omega t}=\\cos\\omega t+i\\sin\\omega t$, the real part of the solution for $x(t)$ gives the solution for a driving force $F_0\\cos\\omega t$, and the imaginary part of $x$ corresponds to the case where the driving force is $F_0\\sin\\omega t$. It is rather easy to solve for the complex $x$ in this case, and by taking the real part of the solution, one finds the answer for the $\\cos\\omega t$ driving force.\n\nWe assume a simple form for the particular solution\n\\begin{equation}\nx_p=De^{i\\omega t},\n\\end{equation}\nwhere $D$ is a complex constant.\n\nFrom Eq. (\\ref{eq:compdrive}) one inserts the form for $x_p$ above to get\n\\begin{eqnarray}\nD\\left\\{-\\omega^2+2i\\beta\\omega+\\omega_0^2\\right\\}e^{i\\omega t}=(F_0/m)e^{i\\omega t},\\\\\n\\nonumber\nD=\\frac{F_0/m}{(\\omega_0^2-\\omega^2)+2i\\beta\\omega}.\n\\end{eqnarray}\nThe norm and phase for $D=|D|e^{-i\\delta}$ can be read by inspection,\n\\begin{equation}\n|D|=\\frac{F_0/m}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}},~~~~\\tan\\delta=\\frac{2\\beta\\omega}{\\omega_0^2-\\omega^2}.\n\\end{equation}\nThis is the same expression for $\\delta$ as before. One then finds $x_p(t)$,\n\\begin{eqnarray}\n\\label{eq:fastdriven1}\nx_p(t)&=&\\Re\\frac{(F_0/m)e^{i\\omega t-i\\delta}}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}\\\\\n\\nonumber\n&=&\\frac{(F_0/m)\\cos(\\omega t-\\delta)}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}.\n\\end{eqnarray}\nThis is the same answer as before.\nIf one wished to solve for the case where $F(t)= F_0\\sin\\omega t$, the imaginary part of the solution would work\n\\begin{eqnarray}\n\\label{eq:fastdriven2}\nx_p(t)&=&\\Im\\frac{(F_0/m)e^{i\\omega t-i\\delta}}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}\\\\\n\\nonumber\n&=&\\frac{(F_0/m)\\sin(\\omega t-\\delta)}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}.\n\\end{eqnarray}\n\n\\example\nConsider the damped and driven harmonic oscillator worked out above. Given $F_0, m,\\beta$ and $\\omega_0$, solve for the complete solution $x(t)$ for the case where $F=F_0\\sin\\omega t$ with initial conditions $x(t=0)=0$ and $v(t=0)=0$. Assume the underdamped case.\n\n{\\bf Solution:}\nThe general solution including the arbitrary constants includes both the homogenous and particular solutions,\n\\begin{eqnarray*}\nx(t)&=&\\frac{F_0}{m}\\frac{\\sin(\\omega t-\\delta)}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}\n+A\\cos\\omega't e^{-\\beta t}+B\\sin\\omega't e^{-\\beta t}.\n\\end{eqnarray*}\nThe quantities $\\delta$ and $\\omega'$ are given earlier in the section, $\\omega'=\\sqrt{\\omega_0^2-\\beta^2}, \\delta=\\tan^{-1}(2\\beta\\omega/(\\omega_0^2-\\omega^2)$. Here, solving the problem means finding the arbitrary constants $A$ and $B$. Satisfying the initial conditions for the initial position and velocity:\n\\begin{eqnarray*}\nx(t=0)=0&=&-\\eta\\sin\\delta+A,\\\\\nv(t=0)=0&=&\\omega\\eta\\cos\\delta-\\beta A+\\omega'B,\\\\\n\\eta&\\equiv&\\frac{F_0}{m}\\frac{1}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}.\n\\end{eqnarray*}\nThe problem is now reduced to 2 equations and 2 unknowns, $A$ and $B$. The solution is\n\\begin{eqnarray}\nA&=& \\eta\\sin\\delta ,~~~B=\\frac{-\\omega\\eta\\cos\\delta+\\beta\\eta\\sin\\delta}{\\omega'}.\n\\end{eqnarray}\n\n\\exampleend\n\n\\subsection{Resonance Widths; the $Q$ factor}\n\nFrom the previous two sections, the particular solution for a driving force, $F=F_0\\cos\\omega t$, is\n\\begin{eqnarray}\nx_p(t)&=&\\frac{F_0/m}{\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2}}\\cos(\\omega_t-\\delta),\\\\\n\\nonumber\n\\delta&=&\\tan^{-1}\\left(\\frac{2\\beta\\omega}{\\omega_0^2-\\omega^2}\\right).\n\\end{eqnarray}\nIf one fixes the driving frequency $\\omega$ and adjusts the fundamental frequency $\\omega_0=\\sqrt{k/m}$, the maximum amplitude occurs when $\\omega_0=\\omega$ because that is when the term from the denominator $(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2$ is at a minimum. This is akin to dialing into a radio station. However, if one fixes $\\omega_0$ and adjusts the driving frequency one minimize with respect to $\\omega$, e.g. set \n\\begin{equation}\n\\frac{d}{d\\omega}\\left[(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2\\right]=0,\n\\end{equation}\nand one finds that the maximum amplitude occurs when $\\omega=\\sqrt{\\omega_0^2-2\\beta^2}$. If $\\beta$ is small relative to $\\omega_0$, one can simply state that the maximum amplitude is\n\\begin{equation}\nx_{\\rm max}\\approx\\frac{F_0}{2m\\beta \\omega_0}.\n\\end{equation}\n\\begin{figure}\n\\centerline{\\includegraphics[width=0.32\\textwidth]{figs/qfactor}}\n\\caption{\\label{fig:qfactor}\nThe maximum amplitude squared of the steady-state motion of a sinusoidally driven harmonic oscillator is shown as a function of the driving frequency $\\omega$. This peaks near the fundamental frequency $\\omega_0$ and the full-width-half-maximum, $FWHM$, is given by the damping, $2\\beta$.}\n\\end{figure}\nFigure \\ref{fig:qfactor} displays the maximum amplitude squared as a function of $\\omega$, and one can see the peak at $\\omega\\approx\\omega_0$. The squared amplitude is usually more the quantity of interest than the amplitude because the power being absorbed by the oscillator is proportional to the square of the amplitude, not the amplitude. The width of the peak is quantified by the full-width-half-maximum, sometimes called $FWHM$ and can be found by finding the frequency difference, $\\omega-\\omega_0$, such that the maximum response falls by a factor of two,\n\\begin{eqnarray}\n\\frac{4\\omega^2\\beta^2}{(\\omega_0^2-\\omega^2)^2+4\\omega^2\\beta^2}=\\frac{1}{2}.\n\\end{eqnarray}\nFor small damping this occurs when $\\omega=\\omega_0\\pm \\beta$, so the $FWHM\\approx 2\\beta$. For the purposes of tuning to a specific frequency, one wants the width to be as small as possible. The ratio of $\\omega_0$ to $FWHM$ is known as the {\\it quality} factor, or $Q$ factor,\n\\begin{equation}\nQ\\equiv \\frac{\\omega_0}{2\\beta}.\n\\end{equation}\n\n\\subsection{Principal of Superposition and Periodic Forces (Fourier Transforms)}\n\nIf one has several driving forces, $F(t)=\\sum_n F_n(t)$, one can find the particular solution to each $F_n$, $x_{pn}(t)$, and the particular solution for the entire driving force is \n\\begin{equation}\nx_p(t)=\\sum_nx_{pn}(t).\n\\end{equation}\nThis is known as the principal of superposition. It only applies when the homogenous equation is linear. If there were an anharmonic term such as $x^3$ in the homogenous equation, then when one summed various solutions, $x=(\\sum_n x_n)^2$, one would get cross terms. Superposition is especially useful when $F(t)$ can be written as a sum of sinusoidal terms, because the solutions for each sinusoidal term is analytic, and are given in the previous two subsections. \n\nDriving forces are often periodic, even when they are not sinusoidal. Periodicity implies that for some time $\\tau$\n\\begin{eqnarray}\nF(t+\\tau)=F(t). \n\\end{eqnarray}\nOne example of a non-sinusoidal periodic force is a square wave. Many components in electric circuits are non-linear, e.g. diodes, which makes many wave forms non-sinusoidal even when the circuits are being driven by purely sinusoidal sources.\n\nFor the sinusoidal example studied in the previous subsections the period is $\\tau=2\\pi/\\omega$. However, higher harmonics can also satisfy the periodicity requirement. In general, any force that satisfies the periodicity requirement can be expressed as a sum over harmonics,\n\\begin{equation}\nF(t)=\\frac{f_0}{2}+\\sum_{n>0} f_n\\cos(2n\\pi t/\\tau)+g_n\\sin(2n\\pi t/\\tau).\n\\end{equation}\nFrom the previous subsection, one can write down the answer for $x_{pn}(t)$, by substituting $f_n/m$ or $g_n/m$ for $F_0/m$ into Eq.s (\\ref{eq:fastdriven1}) or (\\ref{eq:fastdriven2}) respectively. By writing each factor $2n\\pi t/\\tau$ as $n\\omega t$, with $\\omega\\equiv 2\\pi/\\tau$,\n\\begin{equation}\n\\label{eq:fourierdef1}\nF(t)=\\frac{f_0}{2}+\\sum_{n>0}f_n\\cos(n\\omega t)+g_n\\sin(n\\omega t).\n\\end{equation}\nThe solutions for $x(t)$ then come from replacing $\\omega$ with $n\\omega$ for each term in the particular solution in Eq.s (\\ref{eq:partform}-\\ref{eq:Ddrive}),\n\\begin{eqnarray}\nx_p(t)&=&\\frac{f_0}{2k}+\\sum_{n>0} \\alpha_n\\cos(n\\omega t-\\delta_n)+\\beta_n\\sin(n\\omega t-\\delta_n),\\\\\n\\nonumber\n\\alpha_n&=&\\frac{f_n/m}{\\sqrt{((n\\omega)^2-\\omega_0^2)+4\\beta^2n^2\\omega^2}},\\\\\n\\nonumber\n\\beta_n&=&\\frac{g_n/m}{\\sqrt{((n\\omega)^2-\\omega_0^2)+4\\beta^2n^2\\omega^2}},\\\\\n\\nonumber\n\\delta_n&=&\\tan^{-1}\\left(\\frac{2\\beta n\\omega}{\\omega_0^2-n^2\\omega^2}\\right).\n\\end{eqnarray}\nBecause the forces have been applied for a long time, any non-zero damping eliminates the homogenous parts of the solution, so one need only consider the particular solution for each $n$.\n\nThe problem will considered solved if one can find expressions for the coefficients $f_n$ and $g_n$, even though the solutions are expressed as an infinite sum. The coefficients can be extracted from the function $F(t)$ by\n\\begin{eqnarray}\n\\label{eq:fourierdef2}\nf_n&=&\\frac{2}{\\tau}\\int_{-\\tau/2}^{\\tau/2} dt~F(t)\\cos(2n\\pi t/\\tau),\\\\\n\\nonumber\ng_n&=&\\frac{2}{\\tau}\\int_{-\\tau/2}^{\\tau/2} dt~F(t)\\sin(2n\\pi t/\\tau).\n\\end{eqnarray}\n\nTo check the consistency of these expressions and to verify Eq. (\\ref{eq:fourierdef2}), one can insert the expansion of $F(t)$ in Eq. (\\ref{eq:fourierdef1}) into the expression for the coefficients in Eq. (\\ref{eq:fourierdef2}) and see whether\n\\begin{eqnarray}\nf_n&=?&\\frac{2}{\\tau}\\int_{-\\tau/2}^{\\tau/2} dt~\\left\\{\n\\frac{f_0}{2}+\\sum_{m>0}f_m\\cos(m\\omega t)+g_m\\sin(m\\omega t)\n\\right\\}\\cos(n\\omega t).\n\\end{eqnarray}\nImmediately, one can throw away all the terms with $g_m$ because they convolute an even and an odd function. The term with $f_0/2$ disappears because $\\cos(n\\omega t)$ is equally positive and negative over the interval and will integrate to zero. For all the terms $f_m\\cos(m\\omega t)$ appearing in the sum, one can use angle addition formulas to see that $\\cos(m\\omega t)\\cos(n\\omega t)=(1/2)(\\cos[(m+n)\\omega t]+\\cos[(m-n)\\omega t]$. This will integrate to zero unless $m=n$. In that case the $m=n$ term gives\n\\begin{equation}\n\\int_{-\\tau/2}^{\\tau/2}dt~\\cos^2(m\\omega t)=\\frac{\\tau}{2},\n\\end{equation}\nand\n\\begin{eqnarray}\nf_n&=?&\\frac{2}{\\tau}\\int_{-\\tau/2}^{\\tau/2} dt~f_n/2\\\\\n\\nonumber\n&=&f_n~\\checkmark.\n\\end{eqnarray}\nThe same method can be used to check for the consistency of $g_n$.\n\n\\example\nConsider the driving force:\n\\begin{equation}\nF(t)=At/\\tau,~~-\\tau/20.\n\\end{equation}\nHere, $\\omega'=\\sqrt{\\omega_0^2-\\beta^2}$. For an impulse $I_i$ that occurs at time $t_i$ the trajectory would be\n\\begin{equation}\nx(t)=\\frac{I_i}{m\\omega'}e^{-\\beta (t-t_i)}\\sin[\\omega'(t-t_i)] \\Theta(t-t_i),\n\\end{equation}\nwhere $\\Theta(t-t_i)$ is a step function, i.e. $\\Theta(x)$ is zero for $x<0$ and unity for $x>0$. If there were several impulses linear superposition tells us that we can sum over each contribution,\n\\begin{equation}\nx(t)=\\sum_i\\frac{I_i}{m\\omega'}e^{-\\beta(t-t_i)}\\sin[\\omega'(t-t_i)]\\Theta(t-t_i)\n\\end{equation} \n\nNow one can consider a series of impulses at times separated by $\\Delta t$, where each impulse is given by $F_i\\Delta t$. The sum above now becomes an integral,\n\\begin{eqnarray}\\label{eq:Greeny}\nx(t)&=&\\int_{-\\infty}^\\infty dt'~F(t')\\frac{e^{-\\beta(t-t')}\\sin[\\omega'(t-t')]}{m\\omega'}\\Theta(t-t')\\\\\n\\nonumber\n&=&\\int_{-\\infty}^\\infty dt'~F(t')G(t-t'),\\\\\n\\nonumber\nG(\\Delta t)&=&\\frac{e^{-\\beta\\Delta t}\\sin[\\omega' \\Delta t]}{m\\omega'}\\Theta(\\Delta t)\n\\end{eqnarray}\nThe quantity $e^{-\\beta(t-t')}\\sin[\\omega'(t-t')]/m\\omega'\\Theta(t-t')$ is called a Green's function, $G(t-t')$. It describes the response at $t$ due to a force applied at a time $t'$, and is a function of $t-t'$. The step function ensures that the response does not occur before the force is applied. One should remember that the form for $G$ would change if the oscillator were either critically- or over-damped.\n\nWhen performing the integral in Eq. (\\ref{eq:Greeny}) one can use angle addition formulas to factor out the part with the $t'$ dependence in the integrand,\n\\begin{eqnarray}\n\\label{eq:Greeny2}\nx(t)&=&\\frac{1}{m\\omega'}e^{-\\beta t}\\left[I_c(t)\\sin(\\omega't)-I_s(t)\\cos(\\omega't)\\right],\\\\\n\\nonumber\nI_c(t)&\\equiv&\\int_{-\\infty}^t dt'~F(t')e^{\\beta t'}\\cos(\\omega't'),\\\\\n\\nonumber\nI_s(t)&\\equiv&\\int_{-\\infty}^t dt'~F(t')e^{\\beta t'}\\sin(\\omega't').\n\\end{eqnarray}\nIf the time $t$ is beyond any time at which the force acts, $F(t'>t)=0$, the coefficients $I_c$ and $I_s$ become independent of $t$. \n\n\\example\nConsider an undamped oscillator ($\\beta\\rightarrow 0$), with characteristic frequency $\\omega_0$ and mass $m$, that is at rest until it feels a force described by a Gaussian form,\n\\begin{eqnarray*}\nF(t)&=&F_0 \\exp\\left\\{\\frac{-t^2}{2\\tau^2}\\right\\}.\n\\end{eqnarray*}\nFor large times ($t>>\\tau$), where the force has died off, find $x(t)$.\\\\\n{\\bf Solution:} Solve for the coefficients $I_c$ and $I_s$ in Eq. (\\ref{eq:Greeny2}). Because the Gaussian is an even function, $I_s=0$, and one need only solve for $I_c$,\n\\begin{eqnarray*}\nI_c&=&F_0\\int_{-\\infty}^\\infty dt'~e^{-t^{\\prime 2}/(2\\tau^2)}\\cos(\\omega_0 t')\\\\\n&=&\\Re F_0 \\int_{-\\infty}^\\infty dt'~e^{-t^{\\prime 2}/(2\\tau^2)}e^{i\\omega_0 t'}\\\\\n&=&\\Re F_0 \\int_{-\\infty}^\\infty dt'~e^{-(t'-i\\omega_0\\tau^2)^2/(2\\tau^2)}e^{-\\omega_0^2\\tau^2/2}\\\\\n&=&F_0\\tau \\sqrt{2\\pi} e^{-\\omega_0^2\\tau^2/2}.\n\\end{eqnarray*}\nThe third step involved completing the square, and the final step used the fact that the integral\n\\begin{eqnarray*}\n\\int_{-\\infty}^\\infty dx~e^{-x^2/2}&=&\\sqrt{2\\pi}.\n\\end{eqnarray*}\nTo see that this integral is true, consider the square of the integral, which you can change to polar coordinates,\n\\begin{eqnarray*}\nI&=&\\int_{-\\infty}^\\infty dx~e^{-x^2/2}\\\\\nI^2&=&\\int_{-\\infty}^\\infty dxdy~e^{-(x^2+y^2)/2}\\\\\n&=&2\\pi\\int_0^\\infty rdr~e^{-r^2/2}\\\\\n&=&2\\pi.\n\\end{eqnarray*}\nFinally, the expression for $x$ from Eq. (\\ref{eq:Greeny2}) is\n\\begin{eqnarray*}\nx(t>>\\tau)&=&\\frac{F_0\\tau}{m\\omega_0} \\sqrt{2\\pi} e^{-\\omega_0^2\\tau^2/2}\\sin(\\omega_0t).\n\\end{eqnarray*}\n\n\n\\exampleend\n\n\\subsection{Solving Equations of Motion Numerically}\n%If one has a known force $F(t)$, and initial conditions $x(t=0)=x_0$ and $v(t=0)=v_0$, one can solve for the equations of motion numerically. First, one must choose a small time step $\\Delta t$, smaller than any characteristic time scale of the problem. One can then define a mesh $x_n$, where $n=0$ so some maximum number of steps $N$. We will defines these values as\n\\begin{eqnarray}\nx_n=x(n\\Delta t).\n\\end{eqnarray}\nAssume that one knows $x_n$ for all $i<\\le j$. One can express the velocities and accelerations in Newton's equations of motion for a mass $m$ as:\n\n\\begin{eqnarray}\nv_n&=&\\frac{x_{n+1}-x_{n-1}}{2\\Delta t},\\\\\n\\nonumber\na_n&=&\\frac{v_{n+1/2}-v_{n-1/2}}{\\Delta t}\\\\\n\\nonumber\n&=&\\frac{x_{n+1}-2x_n+x_{n-1}}{(\\Delta t)^2}.\n\\end{eqnarray}\nThe equations of motion at $t_n=n\\Delta t$ become\n\\begin{eqnarray}\nF(x_n)&=&ma_n.\n\\end{eqnarray}\nIf one knows $x_{n-1}$ and $x_n$ the only unknown in the equation is $x_{n+1}$. One can solve for it, then move onto the next $n$. Note that to proceed one needs to know $x$ at two points. This can be tricky depending on how the initial conditions are specified. \n\n\\example\nWrite a numerical program to solve the evolution of a particle of mass $m$ feeling a spring force, $-kx$ and a drag force, $-bv$. Assume there is also some external force, $F(t)=F_0\\sin\\omega t$. Let the initial conditions be $x(t=0)=0$ and $v(t=0)=v_0$.\n\n{\\bf Solution:} The differential equation,\n\\begin{eqnarray*}\nm\\ddot{x}+b\\dot{x}+kx&=&0,\\\\\n\\end{eqnarray*}\nbecomes\n\\begin{eqnarray*}\n\\frac{m}{(\\Delta t)^2}\\left(x_{n+1}-2x_n+x_{n-1}\\right)+\\frac{b}{2\\Delta t}\\left(x_{n+1}-x_{n-1}\\right)+kx_n&=&F(n\\Delta t).\n\\end{eqnarray*}\nSolving for $x_{n+1}$,\n\\begin{eqnarray*}\n\\left\\{\\frac{m}{(\\Delta t)^2}+\\frac{b}{2\\Delta t}\\right\\}x_{n+1}&=&\\left[\\frac{2m}{(\\Delta t)^2}-k\\right]x_n+F(n\\Delta t)\n+\\left[-\\frac{m}{(\\Delta t)^2}+\\frac{b}{2\\Delta t}\\right]x_{n-1},\\\\\nx_{n+1}&=&\\frac{\\left[\\frac{2m}{(\\Delta t)^2}-k\\right]x_n+F(n\\Delta t)\n+\\left[-\\frac{m}{(\\Delta t)^2}+\\frac{b}{2\\Delta t}\\right]x_{n-1}}{\\frac{m}{(\\Delta t)^2}+\\frac{b}{2\\Delta t}}.\n\\end{eqnarray*}\nIn order to iterate forward to find $x_{n+1}$, one must know $x_{n-1}$ and $x_n$. However, to get started, one only knows one value of $x$, along with the velocity being zero. To get $x$ at two time steps, one must can use the above two pieces of information,\n\\begin{eqnarray*}\nv_0&=&\\frac{x_1-x_{-1}}{2\\Delta t},\\\\\n-bv_0-kx_0&=&m\\frac{x_1-2x_0+x_{-1}}{\\Delta t^2},\n\\end{eqnarray*}\nthen combine the equations to eliminate $x_{-1}$ to solve for $x_1$. The fact that $x_0=0$ makes the algebra simpler,\n\\begin{eqnarray*}\nx_1&=&v_0\\Delta t-\\frac{bv_0\\Delta t^2}{2m}.\n\\end{eqnarray*}\n\n\n\nThe program might read:\n{\\tt\\begin{verbatim}\ndouble F(double t){\n double F0=????,omega=????;\n return F0*sin(omega*t);\n}\nvoid main(){\n const int Ntsteps=???;\n double x[Ntsteps+1],dt=???,k=???,m=???,b=???,v0=???;\n x[0] = 0.0;\n x[1] = v0*dt-0.5*b*v0/m;\n for(int it=1;it\\tau\\end{array}\n\\right.\n\\]\nFind $x(t)$ for $t>\\tau$ using Eq. (\\ref{eq:Greeny2}).\n\n\\item Consider a particle of mass $m$ in a harmonic oscillator with angular frequency $\\omega_0$ and no damping. It experiences an external force,\n\\[\nF(t)=f_0\\Theta(t)e^{-\\gamma t}.\n\\]\nA ``Theta'' function is a step function, and is zero for negative arguments and unity for positive arguments.\n\\begin{enumerate}\n\\item Find a particular solution, $x_p(t)$, assuming it is proportional to $e^{-\\gamma t}$. \n\\item For a particle initially at rest at the origin at $t=0$, find $x(t)$ by adding in the homogenous solutions and matching the BC, determine the arbitrary constants.\n\\item Use Eq. (\\ref{eq:Greeny}) or Eq. (\\ref{eq:Greeny2}) to find $x(t)$. Check that you get the same result as (b).\n\\end{enumerate}\n\n\\end{enumerate} \n\n", "meta": {"hexsha": "54ca41583b4cb257aad3c4bbfb6fe82d4dbab8e2", "size": 37434, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/AdminBackground/lectures/test.tex", "max_stars_repo_name": "Shield94/Physics321", "max_stars_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2020-01-09T17:41:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T00:48:58.000Z", "max_issues_repo_path": "doc/AdminBackground/lectures/test.tex", "max_issues_repo_name": "Shield94/Physics321", "max_issues_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2020-01-08T03:47:53.000Z", "max_issues_repo_issues_event_max_datetime": "2020-12-15T15:02:57.000Z", "max_forks_repo_path": "doc/AdminBackground/lectures/test.tex", "max_forks_repo_name": "Shield94/Physics321", "max_forks_repo_head_hexsha": "9875a3bf840b0fa164b865a3cb13073aff9094ca", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 33, "max_forks_repo_forks_event_min_datetime": "2020-01-10T20:40:55.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-11T20:28:41.000Z", "avg_line_length": 60.4749596123, "max_line_length": 1493, "alphanum_fraction": 0.715899984, "num_tokens": 12330, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744939732856, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.6287803856479551}} {"text": "\\documentclass[a4paper]{article}\n\n\\input{temp}\n\n\\begin{document}\n\n\\title{Metric and Topological spaces}\n\\date{Easter 2016}\n\n\\maketitle\n\n\\newpage\n\n\\tableofcontents\n\n\\newpage\n\n\\section{Introduction}\nSuppose $\\left(x_n\\right)$ is a sequence of real numbers. $\\left(x_n\\right)$ converges to $x\\in \\R$ if for every $\\varepsilon>0$, there exists some $N$ such that $|x_n-x|<\\varepsilon$ whenever $n>N$. We know that a function $f:\\R \\to \\R$ is \\emph{continuous} if $\\left(f\\left(x_n\\right)\\right)\\to f\\left(x\\right)$ whenever $\\left(x_n\\right) \\to x$. But what does it mean for a sequence $\\left(x_n\\right)$ to converge if $x_n\\in X \\neq \\R$?\n\n\\begin{eg}\nLet\n\\begin{equation*}\n\\begin{aligned}\nX=\\left\\{A\\in M_{n\\times n} \\left(\\R\\right) | A^T =A\\right\\}\n\\end{aligned}\n\\end{equation*}\nand let $f\\left(A\\right)$ be the largest eigenvalue of $A$. Is $f$ continuous?\n\\end{eg}\n\n\\begin{eg}\nLet\n\\begin{equation*}\n\\begin{aligned}\n&X=\\left\\{f:\\left[0,1\\right] \\to \\R | f \\text{ is continuous} \\right\\} \\\\\n&F\\left(f\\right) = f\\left(\\frac{1}{2}\\right)\n\\end{aligned}\n\\end{equation*}\n\\end{eg}\nIs $F$ continuous?\n\n\\begin{thm} (Intermediate value theorem)\\\\\nIf $f:\\left[0,1\\right]\\to \\R$ is continuous and $f\\left(0\\right)\\left<0, f\\left(1\\right)\\right>0$, then $\\exists x \\in \\left[0,1\\right]$ with $f\\left(x\\right)=0$.\n\\end{thm}\n\n\\begin{thm} (Maximum value theorem)\\\\\nIf $f:\\left[0,1\\right] \\to \\R$ is continuous, then $\\exists x \\in \\left[0,1\\right]$ such that $f\\left(x\\right) \\geq f\\left(y\\right)$ for all $y\\in\\left[0,1\\right]$.\n\\end{thm}\n\nThese are both theorems about $\\left[0,1\\right]$. However, they might not be true if we change to another domain. For example, neither of these theorems hold for $\\left[0,1\\right] \\cap \\Q$.\n\n\\begin{eg}\nLet\n\\begin{equation*}\n\\begin{aligned}\nf\\left(x\\right) = x^2-\\frac{1}{2}\n\\end{aligned}\n\\end{equation*}\nThen $f\\left(0\\right)\\left<0, f\\left(1\\right)\\right>0$, but there is no $x\\in\\left[0,1\\right] \\cap \\Q$ such that $f\\left(x\\right) = 0$.\n\\end{eg}\n\n\\newpage\n\\section{Metric spaces}\nSuppose $\\left(\\mathbf{v}_n\\right) = \\left(\\left(x_n,y_n\\right)\\right)$ is a sequence in $\\R^2$. What should it mean for $\\left(\\mathbf{v}_n\\right)$ to converge to $\\mathbf{v} = \\left(x,y\\right)$?\\\\\nA natural thought would be that the sequence ocnverges to $\\left(x,y\\right)$ if $\\left(x_n\\right) \\to x$ and $\\left(y_n\\right) \\to y$. That is then equivalent to\n\\begin{equation*}\n\\begin{aligned}\n&\\left(x_n-x\\right)^2 + \\left(y_n-y\\right)^2 \\to 0 \\iff\n&|\\mathbf{v}_n-\\mathbf{v}| \\to 0.\n\\end{aligned}\n\\end{equation*}\n\n\\begin{defi} (Metric space)\\\\\nA \\emph{metric space} is a pair $\\left(X,d\\right)$ where $X$ is a set and $d:X\\times X \\to \\R$, called the \\emph{metric}, or \\emph{distance function}, that satisfies the following properties:\n\\begin{equation*}\n\\begin{aligned}\n&\\bullet d\\left(x,y\\right) \\geq 0 \\forall x,y\\in X;\\\\\n&\\bullet d\\left(x,y\\right) = 0 \\iff x=y;\\\\\n&\\bullet d\\left(x,y\\right) = d\\left(y,x\\right);\\\\\n&\\bullet d\\left(x,z\\right) \\leq d\\left(x,y\\right) + d\\left(y,z\\right).\n\\end{aligned}\n\\end{equation*}\n\\end{defi}\n\n\\begin{eg}(Euclidean metric)\\\\\nLet $X=\\R^n$, $d\\left(\\mathbf{x},\\mathbf{y}\\right) = |\\mathbf{x}-\\mathbf{y}|$, where\n\\begin{equation*}\n\\begin{aligned}\n|\\mathbf{v}| = \\left(\\sum_{i=1}^n v_i^2\\right)^\\frac{1}{2}\n\\end{aligned}\n\\end{equation*}\nis the \\emph{Euclidean metric} on $\\R^n$. Note that this satisfies the triangle inequality.\n\\end{eg}\n\n\\begin{eg} (Metric subspace)\\\\\nSuppose $\\left(X,d_X\\right)$ is a metric space and $Y\\subset X$. Then $\\left(Y,d_Y\\right)$ is a metric space, where $d_Y\\left(y_1,y_2\\right) = d_X\\left(y_1,y_2\\right)$, i.e. $d_Y=d_X |_{Y\\times Y}$.\\\\\nWe say $\\left(Y,d_Y\\right)$ is a \\emph{subspace} of $\\left(X,d_X\\right)$, or just say $Y$ is a subspace of $X$.\n\\end{eg}\n\n\\begin{eg}\nConsider\n\\begin{equation*}\n\\begin{aligned}\n\\left\\{A \\in M_{n\\times n} \\left(\\R\\right) | A^T = A \\right\\} \\subset \\R^{n^2}.\n\\end{aligned}\n\\end{equation*} \n\\end{eg}\n\n\\begin{defi} (Convergence)\\\\\nIf $\\left(X,d\\right)$ is a metric space and $\\left(x_n\\right)$ is a sequence in $X$, we say $\\left(x_n\\right)$ converges to $x$ if for every $\\varepsilon >0$ there exists $N$ such that $d\\left(x_n,x\\right) < \\varepsilon$ whenever $n>N$.\n\\end{defi}\n\n\\begin{prop}\nSuppose $\\left(x_n\\right)$ is a sequence in a metric space $\\left(X,d\\right)$ that $\\left(x_n\\right) \\to x$ and $\\left(x_n\\right) \\to y$. Then $x=y$.\n\\begin{proof}\nGiven $\\varepsilon>0$, pick $N_1,N_2$ with\n\\begin{equation*}\n\\begin{aligned}\n&n>N_1 \\implies d\\left(x_n,x\\right) < \\frac{\\varepsilon}{2},\\\\\n&n>N_2 \\implies d\\left(x_n,y\\right) < \\frac{\\varepsilon}{2}\n\\end{aligned}\n\\end{equation*}\nThen if $n>\\max\\left\\{N_1,N_2\\right\\}$,\n\\begin{equation*}\n\\begin{aligned}\n0\\leq d\\left(x,y\\right)\\leq d\\left(x,x_n\\right) + d\\left(x_n,y\\right) = d\\left(x,x_n\\right) + d\\left(y,x_n\\right) < \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon.\n\\end{aligned}\n\\end{equation*}\nSince $\\varepsilon$ is arbitrary, $d\\left(x,y\\right)=0$. So $x=y$.\n\\end{proof}\n\\end{prop}\n\n\\begin{defi} (Continuity)\\\\\nIf $\\left(X,d_X\\right)$ and $\\left(Y,d_Y\\right)$ are metric spaces, then $f:X\\to Y$ is \\emph{continuous} if $\\left(f\\left(x_n\\right)\\right) \\to f\\left(x\\right)$ whenever $\\left(x_n\\right) \\to x$.\n\\end{defi}\n\n\\begin{prop}\nIf $\\left(X,d_X\\right), \\left(Y,d_Y\\right)$ and $\\left(Z,d_Z\\right)$ are metric spaces, and $f:X\\to Y$ and $g:Y\\to Z$ are both continuous, then $g \\circ f:X\\to Z$ is also continuous.\n\\begin{proof}\nSuppose $\\left(x_n\\right) \\to x$ in $X$. Since $f$ is continuous, \n\\begin{equation*}\n\\begin{aligned}\n\\left(f\\left(x_n\\right)\\right) \\to f\\left(x\\right)\n\\end{aligned}\n\\end{equation*}\nin $Y$. Since $g$ is also continuous,\n\\begin{equation*}\n\\begin{aligned}\n\\left(g\\left(f\\left(x_n\\right)\\right)\\right) \\to g\\left(f\\left(x\\right)\\right)\n\\end{aligned}\n\\end{equation*}\nin $Z$, i.e.\n\\begin{equation*}\n\\begin{aligned}\n\\left(g\\circ f\\left(x_n\\right)\\right) \\to g\\circ f\\left(x\\right).\n\\end{aligned}\n\\end{equation*}\nSo $g\\circ f$ is continuous.\n\\end{proof}\n\\end{prop}\n\n\\begin{eg} (Discrete metric)\\\\\nLet $X$ be any set, and let\n\\begin{equation*}\n\\begin{aligned}\nd\\left(x,y\\right) = \\left\\{\n\\begin{array}{ll}\n1 & x\\neq y\\\\\n0 & x=y\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nThis is a metric space.\n\\end{eg}\n\n\\begin{eg}\nLet $X=\\R^2$, $\\mathbf{x}=\\left(x_1,x_2\\right)$, $\\mathbf{x}'=\\left(x_1',x_2'\\right)$, and\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{x}'\\right) = |x_1-x_1'| + |x_2-x_2'|.\n\\end{aligned}\n\\end{equation*}\nTriangle inequality follows by:\n\\begin{equation*}\n\\begin{aligned}\nd\\left(x,z\\right) &= |x_1-z_1| + |x_2-x_2| \\\\\n&\\leq |x_1-y_1| + |y_1-z_1| + |x_2 - y_2| + |y_2 - z_2| \\\\\n&= d\\left(x,y\\right) + d\\left(y,z\\right).\n\\end{aligned}\n\\end{equation*}\n\\end{eg}\n\n\\begin{eg} (British Railway metric)\\\\\nLet $X=\\R^2$, and\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{y}\\right) = \\left\\{\n\\begin{array}{ll}\n|\\mathbf{x}-\\mathbf{y}| & \\mathbf{x}=k\\mathbf{y}\\\\\n|\\mathbf{x}| + |\\mathbf{y}| & \\text{ otherwise}\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\n(Unless you, your destination and London are collinear, you always have to take a train to London first and transfer there.)\n\\end{eg}\n\n\\begin{defi}\nIf $V$ is a real vector space, a \\emph{norm} on $V$ is a map $||\\cdot|| : V \\to \\R$ satisfying\n\\begin{equation*}\n\\begin{aligned}\n&\\bullet ||\\mathbf{v}|| \\geq 0 \\forall \\mathbf{v}\\in V;\\\\\n&\\bullet ||\\mathbf{v}|| = 0 \\iff \\mathbf{v}=\\mathbf{0} \\in V;\\\\\n&\\bullet ||\\lambda \\mathbf{v}|| = |\\lambda| ||\\mathbf{v}||, \\lambda \\in \\R, \\mathbf{v} \\in V;\\\\\n&\\bullet ||\\mathbf{v}+\\mathbf{w}|| \\leq ||\\mathbf{v}|| + ||\\mathbf{w}||.\n\\end{aligned}\n\\end{equation*}\n\\end{defi}\n\n\\begin{eg}\nLet $V = \\R^n$,\n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}|| = \\sum_{i=1}^n |v_i|\n\\end{aligned}\n\\end{equation*}\n(this is called the Manhattan norm).\n\\end{eg}\n\n\\begin{lemma}\nIf $||\\cdot||$ is a norm on $V$, then \n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{y}\\right) = ||\\mathbf{x}-\\mathbf{y}||\n\\end{aligned}\n\\end{equation*}\ndefines a metric on $V$.\n\\begin{proof}\nFirstly\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{y}\\right) = ||\\mathbf{x}-\\mathbf{y}|| \\geq 0\n\\end{aligned}\n\\end{equation*}\nby our definition of norm. The second holds by\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{y}\\right) = 0 &\\iff ||\\mathbf{x}-\\mathbf{y}|| = 0\\\\\n&\\iff \\mathbf{x}-\\mathbf{y}=0\\\\\n&\\iff \\mathbf{x} = \\mathbf{y}.\n\\end{aligned}\n\\end{equation*}\nFor the third condition,\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{y}-\\mathbf{x}\\right) &= ||\\mathbf{y}-\\mathbf{x}|| \\\\\n&= ||\\left(-1\\right)\\left(\\mathbf{x}-\\mathbf{y}\\right)||\\\\\n&= |-1| ||\\mathbf{x}-\\mathbf{y}||\\\\\n&= d\\left(\\mathbf{x},\\mathbf{y}\\right)\n\\end{aligned}\n\\end{equation*}\nLastly\n\\begin{equation*}\n\\begin{aligned}\nd\\left(\\mathbf{x},\\mathbf{y}\\right) + d\\left(\\mathbf{y},\\mathbf{z}\\right) &= ||\\mathbf{x}-\\mathbf{y}|| + ||\\mathbf{y}-\\mathbf{z}||\\\\\n&\\geq \\mathbf{x}-\\mathbf{z}\\\\\n&= d\\left(\\mathbf{x},\\mathbf{z}\\right)\n\\end{aligned}\n\\end{equation*}\n\\end{proof}\n\\end{lemma}\n\nThe Manhattan metric on $\\R^2$ is induced by the Manhattan norm (exercise).\n\n\\begin{defi}\nLet $V$ be a real vector space. An \\emph{inner product} on $V$ is a map $\\left<\\cdot, \\cdot\\right>: V\\times V \\to \\R$, satisfying\n\\begin{equation*}\n\\begin{aligned}\n&\\bullet \\left<\\mathbf{v},\\mathbf{v}\\right> \\geq 0 \\forall \\mathbf{v}\\in V;\\\\\n&\\bullet \\left<\\mathbf{v},\\mathbf{v}\\right> = 0 \\iff \\mathbf{v}=\\mathbf{0};\\\\\n&\\bullet \\left<\\mathbf{v},\\mathbf{w}\\right> = \\left<\\mathbf{w},\\mathbf{v}\\right>;\\\\\n&\\bullet \\left<\\mathbf{v},\\mathbf{w_1}+\\lambda \\mathbf{w_2}\\right> = \\left<\\mathbf{v},\\mathbf{w_1}\\right> + \\lambda\\left<\\mathbf{v},\\mathbf{w_2}\\right>.\n\\end{aligned}\n\\end{equation*}\n\\end{defi}\n\nThe standard inner product on $\\R^n$ is given by\n\\begin{equation*}\n\\begin{aligned}\n\\left<\\mathbf{v},\\mathbf{w}\\right> = \\mathbf{v}\\cdot\\mathbf{w}=\\sum_{i=1}^n v_i w_i\n\\end{aligned}\n\\end{equation*}\n\n\\begin{thm} (Cauchy-Schwartz inequality)\\\\\nIf $\\left<\\cdot,\\cdot\\right>$ is an inner product on $V$, then\n\\begin{equation*}\n\\begin{aligned}\n\\left<\\mathbf{v},\\mathbf{w}\\right>^2 \\leq \\left<\\mathbf{v},\\mathbf{v}\\right>\\left<\\mathbf{w},\\mathbf{w}\\right>.\n\\end{aligned}\n\\end{equation*}\n\\begin{proof}\nConsider $\\left<\\mathbf{v}+\\lambda\\mathbf{w},\\mathbf{v}+\\lambda\\mathbf{w}\\right>\\geq 0$ by definition. So\n\\begin{equation*}\n\\begin{aligned}\n\\left<\\mathbf{v},\\mathbf{v}\\right> + 2\\lambda\\left<\\mathbf{v},\\mathbf{w}\\right>+\\lambda^2\\left<\\mathbf{w},\\mathbf{w}\\right> \\geq 0\n\\end{aligned}\n\\end{equation*}\nNow if $ax^2 + bx + c\\geq 0$ for all $x\\in \\R$, then $ax^2+bx+c=0$ has at most one real root. So $b^2-4ac \\leq 0$.\\\\\nNow we've just seen $a\\lambda^2+b\\lambda+c\\geq 0$ for all $\\lambda\\in\\R$, where $a=\\left<\\mathbf{w},\\mathbf{w}\\right>$, $b=2\\left<\\mathbf{v},\\mathbf{w}\\right>$, $c=\\left<\\mathbf{v},\\mathbf{v}\\right>$. So\n\\begin{equation*}\n\\begin{aligned}\n4\\left<\\mathbf{v},\\mathbf{w}\\right>^2 \\leq 4\\left<\\mathbf{w},\\mathbf{w}\\right>\\left<\\mathbf{v},\\mathbf{v}\\right>\n\\end{aligned}\n\\end{equation*}\nwhich is the desired result.\n\\end{proof}\n\\end{thm}\n\n\\begin{coro}\nIf $\\left<,\\right>$ is an inner product on $V$, then $||\\mathbf{v}|| = \\left(\\left<\\mathbf{v},\\mathbf{v}\\right>\\right)^{\\frac{1}{2}}$ is a norm on $V$.\nFor the proof we just need to verify all the conditions again. Cauchy-Schwartz inequality is required for verifying the fourth condition.\n\\end{coro}\n\nAnother exercise: The norm on $\\R^n$ induced by the standard inner product is Euclidean length:\n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}|| = |\\mathbf{v}| = \\left(\\sum_{i=1}^n v_i^2\\right)^\\frac{1}{2}\n\\end{aligned}\n\\end{equation*}\nThe metric induced by the norm is the Euclidean norm.\\\\\nAnd another exercise: is the Manhattan norm induced by an inner product on $\\R^n$?\n\n\\begin{eg}\nLet $V=\\R^n$.\\\\\n1) \n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}||_1=\\sum_{i=1}^n |v_i|\n\\end{aligned}\n\\end{equation*}\nThe Manhattan norm;\\\\\n2)\n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}||_2 = \\left(\\sum_{i=1}^n v_i^2\\right)^\\frac{1}{2}\n\\end{aligned}\n\\end{equation*}\nThe Euclidean norm;\\\\\n3)\n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}||_\\infty = \\max_{1\\leq i \\leq n |v_i|}\n\\end{aligned}\n\\end{equation*}\n(In example sheet:\n\\begin{equation*}\n\\begin{aligned}\n||\\mathbf{v}||_p = \\left(\\sum_{i=1}^n |v_i|^p \\right)^\\frac{1}{p}\n\\end{aligned}\n\\end{equation*}\nis a norm for all $1\\leq p < \\infty$.)\n\\end{eg}\n\n\\begin{eg}\nLet\n\\begin{equation*}\n\\begin{aligned}\nV=C\\left[0,1\\right] := \\left\\{ f:\\left[0,1\\right] \\to \\R | f \\text{ is continuous} \\right\\}\n\\end{aligned}\n\\end{equation*}\nwhere\n\\begin{equation*}\n\\begin{aligned}\n&\\left(\\lambda f\\right)\\left(x\\right) = \\lambda f\\left(x\\right),\\\\\n&\\left(f+g\\right) \\left(x\\right) = f\\left(x\\right) + g\\left(x\\right)\n\\end{aligned}\n\\end{equation*}\n\\end{eg}\n1)\n\\begin{equation*}\n\\begin{aligned}\n||f||_1 = \\int_0^1 |f\\left(x\\right)| dx\n\\end{aligned}\n\\end{equation*}\n2)\n\\begin{equation*}\n\\begin{aligned}\n||f||_2 = \\left(\\int_0^1 f\\left(x\\right)^2\\right)^\\frac{1}{2} dx\n\\end{aligned}\n\\end{equation*}\n3)\n\\begin{equation*}\n\\begin{aligned}\n||f||_\\infty = \\max_{x \\in \\left[0,1\\right]} |f\\left(x\\right)|\n\\end{aligned}\n\\end{equation*}\nThis is usually called the 'sup norm'.\\\\\nHow do we know that the first two norms satisfy the second condition, i.e. the norm is zero if and only if the function is zero? This is actually a standard result in analysis:\n\n\\begin{lemma}\nSuppose $f\\left[0,1\\right]\\to\\R$ is continuous. If $f \\not\\cong 0$, then\n\\begin{equation*}\n\\begin{aligned}\n\\int_0^1 |f\\left(x\\right)|>0\n\\end{aligned}\n\\end{equation*}\n\\begin{proof}\nPick $x_0\\in\\left[0,1\\right]$ with $f\\left(x_0\\right) \\neq 0.$ Since $f$ is continuous, there exists $\\delta > 0$ such that\n\\begin{equation*}\n\\begin{aligned}\n|f\\left(x\\right) - f\\left(x_0\\right)| < \\frac{|f\\left(x_0\\right)|}{2}\n\\end{aligned}\n\\end{equation*}\nwhenever $|x-x_0| < \\delta$. Then for $|x-x_0|<\\delta$, we have\n\\begin{equation*}\n\\begin{aligned}\n|f\\left(x\\right)| > |f\\left(x_0\\right)| - \\frac{|f\\left(x_0\\right)|}{2} \\geq \\frac{|f\\left(x_0\\right)|}{2}\n\\end{aligned}\n\\end{equation*}\nSo if we define\n\\begin{equation*}\n\\begin{aligned}\ng\\left(x\\right) = \\left\\{\n\\begin{array}{ll}\n\\frac{|f\\left(x_0\\right)|}{2} & |x-x_0| < \\delta\\\\\n0 & \\text{ otherwise}\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nThen $|f\\left(x\\right)| \\geq g\\left(x\\right)$ for all $x\\in\\left[0,1\\right]$.\\\\\nSo\n\\begin{equation*}\n\\begin{aligned}\n\\int_0^1 |f\\left(x\\right)| dx \\geq \\int_0^1 |g\\left(x\\right)| dx = 2\\delta \\frac{|f\\left(x_0\\right)|}{2} = \\delta |f\\left(x_0\\right)|>0.\n\\end{aligned}\n\\end{equation*}\n\\end{proof}\n\\end{lemma}\n\n\\begin{eg}\n\\begin{equation*}\n\\begin{aligned}\nf_n\\left(x\\right) = \\left\\{ \n\\begin{array}{ll}\n1-nx & x\\in\\left[0,\\frac{1}{n}\\right]\\\\\n0 & x\\in\\left[\\frac{1}{n},1\\right]\n\\end{array}\n\\right.\n\\end{aligned}\n\\end{equation*}\nThen\n\\begin{equation*}\n\\begin{aligned}\n||f_n||_1 = \\int_0^1 |f\\left(x\\right)| dx = \\frac{1}{2}\\left(\\frac{1}{n}\\right)\\cdot 1 = \\frac{1}{2n} \\to 0\n\\end{aligned}\n\\end{equation*}\nas $n\\to \\infty$.\\\\\nSo $\\left(f_n\\right) \\to 0$ with respect to the metric induced by $|| \\cdot ||_1$.\\\\\nOn the other hand,\n\\begin{equation*}\n\\begin{aligned}\n||f_n||_\\infty = 1\n\\end{aligned}\n\\end{equation*}\nregardless of $n$, since $f\\left(0\\right)=1$.\\\\\nThis means that $\\left(f_n\\right)$ does not converge to 0 as $n\\to \\infty$ with respect to the metric induced by $||\\cdot||_\\infty$, i.e. the sup norm.\n\\end{eg}\n\nExercise: is it possible to find a sequence that converges with respect to $||\\cdot||_1$ on $\\R^2$, but not with respect to $||\\cdot||_\\infty$?\n\n\\begin{eg} (The \\emph{p-adic} norm)\\\\\nLet $p \\in \\N$ be a prime number. If $a \\in \\Z$, let $|a|_p = p^{-k}$, where $p^k | a$, but $p^{k+1} \\nmid a$.\\\\\nSo\n\\begin{equation*}\n\\begin{aligned}\n&|0|_p = 0,\\\\\n&|ab|_p = |a|_p |b|_p,\\\\\n&|a+b|_p \\leq \\max\\left\\{|a|_p, |b|_p\\right\\} \\leq |a|_p + |b|_p\n\\end{aligned}\n\\end{equation*}\n\nIf $r=\\frac{a}{b} \\in \\Q$, define\n\\begin{equation*}\n\\begin{aligned}\n|r|_p = \\frac{|a|_p}{|b|_p}\n\\end{aligned}\n\\end{equation*}\nThen the \\emph{p-adic metric} on $\\Q$ is defined by\n\\begin{equation*}\n\\begin{aligned}\nd_p \\left(r,s\\right) = |r-s|_p\n\\end{aligned}\n\\end{equation*}\n\\end{eg}\n\nExercise: 1,2,4,8,16, ... $\\to 0$ with respect to $d_2$.\n\n\\newpage\n\\section{Open and Closed Subsets}\nLet $\\left(X,d\\right)$ be a metric space.\n\n\\begin{defi}\nIf $x\\in X$ and $r\\in \\R$ with $r>0$, let\n\\begin{equation*}\n\\begin{aligned}\nB_r\\left(x\\right) = \\left\\{ y\\in X | d\\left(x,y\\right) < r \\right\\}\n\\end{aligned}\n\\end{equation*}\nis the \\emph{open ball} of radius $r$ with centre $x$,\n\\begin{equation*}\n\\begin{aligned}\n\\bar{B}_r \\left(x\\right) = \\left\\{y \\in X | d\\left(x,y\\right) \\leq r \\right\\}\n\\end{aligned}\n\\end{equation*}\nis the \\emph{closed ball} of radius $r$ with centre $x$.\n\\end{defi}\n\n\\begin{eg}\nLet $X=\\R$ with the usual metric. Then\n\\begin{equation*}\n\\begin{aligned}\n&B_r\\left(a\\right) = \\left(a-r,a+r\\right),\\\\\n&\\bar{B}_r\\left(a\\right) = \\left[a-r,a+r\\right].\n\\end{aligned}\n\\end{equation*}\nThese are just the open and closed interval respectively.\n\\end{eg}\n\n\\begin{eg}\nLet $X=\\R^3$, with the Euclidean metric. Then\n\\begin{equation*}\n\\begin{aligned}\n\\bar{B}_1\\left(0\\right) = \\left\\{ \\mathbf{v}\\in \\R^3 | |\\mathbf{v}| \\leq 1\\right\\}\n\\end{aligned}\n\\end{equation*}\nwhich is a sphere with centre 0 and radius 1.\n\\end{eg}\n\n\\begin{eg}\nLet $X=\\R^2$, with the metric induced by\\\\ 1) the Manhattan norm, $||\\cdot ||_1$;\\\\\n2) the Euclidean norm, $||\\cdot ||_2$;\\\\\n3) the sup norm, $||\\cdot ||_\\infty$.\\\\\nNow consider $\\bar{B}_1\\left(0\\right)$. For the Manhattan norm, this is a square with with centre 0 and edges parallel to the axes(so with edge length 2); for the Euclidean norm this is a disc with centre 0 and radius 1; for the sup norm this is a square with centre 0 but edges making an angle of $\\frac{\\pi}{4}$ to the axes (so with edge length $\\sqrt{2}$).\n\\end{eg}\n\n\\begin{eg}\nLet $X=\\R^2$, $d$ is induced by an inner product $\\left<,\\right>$: if $\\mathbf{v}=x\\mathbf{e}_1+y\\mathbf{e}_2$, then\n\\begin{equation*}\n\\begin{aligned}\n\\left<\\mathbf{v},\\mathbf{v}\\right> = ax^2 + 2bxy + cy^2\n\\end{aligned}\n\\end{equation*}\nThen\n\\begin{equation*}\n\\begin{aligned}\nB_1\\left(0\\right) = \\left\\{\\left(x,y\\right) | ax^2+2bxy+cy^2 \\leq 1\\right\\}\n\\end{aligned}\n\\end{equation*}\nwhere\n\\begin{equation*}\n\\begin{aligned}\na=\\left<\\mathbf{e}_1,\\mathbf{e}_1\\right>,\\\\\nb=\\left<\\mathbf{e}_1,\\mathbf{e}_2\\right>,\\\\\nc=\\left<\\mathbf{e}_2,\\mathbf{e}_2\\right>.\n\\end{aligned}\n\\end{equation*}\nThis is an ellipse.\\\\\nSo we know that $||\\cdot ||_\\infty$, $||\\cdot ||_1$ are not induced from an inner product on $\\R^2$.\n\\end{eg}\n\n\\begin{notation}\nIf $A\\subset X$ is a subset, then $X\\backslash A$, or $X-A$, is defined as\n\\begin{equation*}\n\\begin{aligned}\nX-A = \\left\\{x\\in X | x\\not\\in A\\right\\}\n\\end{aligned}\n\\end{equation*}\nwhich is called the complement of $A$ (in $X$).\n\\end{notation}\n\n\\begin{defi} (Open and closed subset)\\\\\nIf $A\\subset X$, we say $A$ is an \\emph{open subset} if for every $x\\in A$, $\\exists r>0$ such that $B_r\\left(x\\right) \\subset A$.\\\\\nIf $C\\subset X$, we say $C$ is a \\emph{closed subset} if $X-C$ is an open subset of $X$.\n\\end{defi}\n\n\\begin{prop}\nIf $X$ is a metric space, then $B_r\\left(x\\right)$ is an open subset of $X$, and $\\bar{B}_r\\left(x\\right)$ is a closed subset of $X$.\n\\begin{proof}\nSuppose $b\\in B_r\\left(a\\right)$. Then $d\\left(a,b\\right) < r$. So $r_0 = r-d\\left(a,b\\right)>0$. Now if $c\\in B_{r_0}\\left(b\\right)$, then\n\\begin{equation*}\n\\begin{aligned}\nd\\left(a,c\\right) \\leq d\\left(a,b\\right)+d\\left(b,c\\right)r\\right\\}\n\\end{aligned}\n\\end{equation*}\nis open. The proof is analogous to the previous proof.\n\\end{proof}\n\\end{prop}\n\n\\begin{rem}\nNote that $A\\subset X$ may be \\emph{neither} open nor closed. In fact if we take a random subset from a random metric space $X$, this is most likely the case. For example, consider $X=\\R$ with the usual metric. then an obvious example is $\\left[0,1\\right)$,which is neither open nor closed; $\\Q \\subset \\R$ is a more subtle example.\n\\end{rem}\n\n\\begin{rem}\n$A\\subset X$ can also be \\emph{both} open and closed. Consider\n\\begin{equation*}\n\\begin{aligned}\nX=\\left[-1,1\\right]-\\left\\{0\\right\\}\n\\end{aligned}\n\\end{equation*}\nas a subspace of $\\R$. Then\n\\begin{equation*}\n\\begin{aligned}\nA=\\left[-1,0\\right)=B_1\\left(-1\\right) \\subset X\n\\end{aligned}\n\\end{equation*}\nis open. On the other hand, this is also a closed ball, since it is also\n\\begin{equation*}\n\\begin{aligned}\n\\bar{B}_{\\frac{1}{2}} \\left(-\\frac{1}{2}\\right) \\subset X\n\\end{aligned}\n\\end{equation*}\nSo $A$ is both closed and open in $X$.\n\\end{rem}\n\n\\begin{rem}\nNote that being open or closed is \\emph{not} a property of a space, but a property of a subset of a space.\\\\\nWe've seen that $\\left[-1,0\\right)$ is not a closed subset of $\\R$, but it \\emph{is} a closed subset of $\\left[-1,1\\right] - \\left\\{0\\right\\}$.\n\\end{rem}\n\n\\begin{defi}\nIf $\\left(X,d\\right)$ is a metric space and $x\\in X$, then $U$ is an \\emph{open neighbourhood} of $x$ if $U\\subset X$ is open and $x\\in U$.\n\\end{defi}\n\n\\begin{prop}\nSuppose sequence $\\left(x_n\\right) \\to x$ and $U$ is an open neighbourhood of $x$. Then there exists $N$ such that $x_n\\in U$ for all $n>N$.\n\\begin{proof}\nSince $U\\subset X$ is open, $x\\in U$, $\\exists \\varepsilon > 0$ with\n\\begin{equation*}\n\\begin{aligned}\nB_\\varepsilon \\left(x\\right) \\subset U.\n\\end{aligned}\n\\end{equation*}\nSince $\\left(x_n\\right) \\to x$, $\\exists N$ such that $d\\left(x_n,x\\right) < \\varepsilon$ whenever $n>N$. This implies that\n\\begin{equation*}\n\\begin{aligned}\nx_n \\in B_\\varepsilon \\left(x\\right)\n\\end{aligned}\n\\end{equation*}\nwhenever $n>N$. So $x_n\\in U$ whenever $n>N$.\n\\end{proof}\n\\end{prop}\n\n\\begin{defi} (Limit point)\\\\\nIf $\\left(X,d\\right)$ is a metric space and $A \\subset X$, we say $x\\in X$ is a \\emph{limit point} of $A$ if there is a sequence $\\left(x_n\\right) \\to x$ with $x_n \\in A$ for all $n$.\n\\end{defi}\n\n\\begin{eg}\nIf $a\\in A$, then $a$ is a limit point of $A$. We can just take a constant sequence with value $a$, which clearly converges to $a$.\n\\end{eg}\n\n\\begin{eg}\nLet $X = \\R$ with the usual metric, $A=\\left(0,1\\right)$. Then\n\\begin{equation*}\n\\begin{aligned}\n\\left(\\frac{1}{n}\\right) \\to 0\n\\end{aligned}\n\\end{equation*}\nand $\\frac{1}{n} \\in A$ for all $n\\geq 1$. So $0$ is a limit point of $A$.\n\\end{eg}\n\n\\begin{prop}\nSuppose $C\\subset X$ is a closed subset and $x$ is a limit point of $C$. Then $x \\in C$.\n\\begin{proof}\nSuppose $\\left(x_n\\right) \\to x$, where $x_n \\in C$, but $x\\not\\in C$.\\\\\nSince $C$ is closed, $U=X-C$ is open. So\n\\begin{equation*}\n\\begin{aligned}\nx \\not\\in C \\implies X \\in U\n\\end{aligned}\n\\end{equation*}\nSo $U$ is an open neighbourhood of $x$.\\\\\nNow $\\left(x_n\\right)$ converges to $x$ and $U$ is an open neighbourhood. So by the previous proposition, there exists an $N$ such that $x_n \\in U$ for all $n>N$. But $x_n \\in C$ for all $n$. Contradiction.\n\\end{proof}\n\\end{prop}\n\n\\begin{thm}\nSuppose $f:\\left(X,d_X\\right) \\to \\left(Y,d_Y\\right)$. Then the following conditions are equivalent:\\\\\n1) $f$ is continuous;\\\\\n2) $f^{-1}\\left(U\\right)$ is open in $X$ whenever $U$ is open in $Y$;\\\\\n3) $f^{-1}\\left(C\\right)$ is closed in $X$ whenever $U$ is closed in $Y$.\n\\end{thm}\n\n\n\\end{document}", "meta": {"hexsha": "4aa29db44ad27b93e72ddcc6a0758c680989a4d7", "size": 23390, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/Metric and Topological spaces.tex", "max_stars_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_stars_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-25T17:34:25.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-25T17:34:25.000Z", "max_issues_repo_path": "Notes/Metric and Topological spaces.tex", "max_issues_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_issues_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/Metric and Topological spaces.tex", "max_forks_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_forks_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.2716927454, "max_line_length": 439, "alphanum_fraction": 0.6580589996, "num_tokens": 9387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744850834648, "lm_q2_score": 0.7690802423634963, "lm_q1q2_score": 0.6287803831382017}} {"text": "\\hypertarget{classnumpp_1_1differentiation_1_1symbolic_1_1differentiate}{}\\section{numpp\\+:\\+:differentiation\\+:\\+:symbolic\\+:\\+:differentiate$<$ Function, Order $>$ Class Template Reference}\n\\label{classnumpp_1_1differentiation_1_1symbolic_1_1differentiate}\\index{numpp\\+::differentiation\\+::symbolic\\+::differentiate$<$ Function, Order $>$@{numpp\\+::differentiation\\+::symbolic\\+::differentiate$<$ Function, Order $>$}}\n\n\nAllows differentiation of N-\\/th order. \n\n\n\n\n{\\ttfamily \\#include $<$functions.\\+hpp$>$}\n\n\\subsection*{Public Types}\n\\begin{DoxyCompactItemize}\n\\item \n\\mbox{\\Hypertarget{classnumpp_1_1differentiation_1_1symbolic_1_1differentiate_ac0c973a49c1b2e262edd5345c4f569a7}\\label{classnumpp_1_1differentiation_1_1symbolic_1_1differentiate_ac0c973a49c1b2e262edd5345c4f569a7}} \n{\\footnotesize template$<$std\\+::size\\+\\_\\+t Variable, std\\+::size\\+\\_\\+t... Variables$>$ }\\\\using {\\bfseries with\\+\\_\\+respect\\+\\_\\+to} = typename \\hyperlink{classnumpp_1_1differentiation_1_1symbolic_1_1differentiate}{differentiate}$<$ typename Function\\+::template derivative$<$ Variable $>$, Order-\\/1 $>$\\+::template with\\+\\_\\+respect\\+\\_\\+to$<$ Variables... $>$\n\\end{DoxyCompactItemize}\n\n\n\\subsection{Detailed Description}\n\\subsubsection*{template$<$typename Function, std\\+::size\\+\\_\\+t Order$>$\\newline\nclass numpp\\+::differentiation\\+::symbolic\\+::differentiate$<$ Function, Order $>$}\n\nAllows differentiation of N-\\/th order. \n\n\n\\begin{DoxyTemplParams}{Template Parameters}\n{\\em Function} & Function which should be differentiated, see /ref numpp\\+\\_\\+differentiation\\+\\_\\+symbolic \\char`\\\"{}tutorial\\char`\\\"{} for more info \\\\\n\\hline\n\\end{DoxyTemplParams}\n\n\\begin{DoxyParams}{Parameters}\n{\\em Order} & order of the differentiation, e.\\+g. 3 for 3rd order\\\\\n\\hline\n\\end{DoxyParams}\n\n\\begin{DoxyCode}\n\\textcolor{preprocessor}{#include\"numpp/differentiation/symbolic.hpp\"}\n\\end{DoxyCode}\n \n\nThe documentation for this class was generated from the following file\\+:\\begin{DoxyCompactItemize}\n\\item \ndifferentiation/symbolic/functions.\\+hpp\\end{DoxyCompactItemize}\n", "meta": {"hexsha": "3ac5af7bfb6faab067e17d2de069669648257ea5", "size": 2061, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1differentiate.tex", "max_stars_repo_name": "szymonmaszke/numpp", "max_stars_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2018-06-06T01:51:17.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-02T15:17:00.000Z", "max_issues_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1differentiate.tex", "max_issues_repo_name": "vyzyv/numpp", "max_issues_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-11-28T12:15:46.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-16T00:03:38.000Z", "max_forks_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1differentiate.tex", "max_forks_repo_name": "szymonmaszke/numpp", "max_forks_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-08-06T13:58:27.000Z", "max_forks_repo_forks_event_max_datetime": "2018-04-06T06:45:22.000Z", "avg_line_length": 45.8, "max_line_length": 366, "alphanum_fraction": 0.7622513343, "num_tokens": 675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.7690802476562641, "lm_q1q2_score": 0.6287803806284479}} {"text": "\\chapter{Mathematical Model of Vehicle Dynamics}\n\nThis chapter will detail the mathematical model that was developed to describe the motion of the uDrone. \n\n\\section{Reference Frames and State Space}\n\nConventions for creating this model will follow those laid out by Thor Fossen in the 2011 Handbook of Marine Craft Hydrodynamics and Control. The fixed world frame is defined using the North-East-Down (NED) coordinate system, represented as $\\{n\\} = (x_n, y_n, z_n)$. in this system, $x_n$ points North, $y_n$ points East, and $z_n$ points down. \n\nThe body-fixed frame $\\{b\\} = (x_b, y_b, z_b)$ is fixed its origin point, $o_b$, to the uDrone. For purposes of this model, $o_b$ is set at the point on the plane of the motors at the back of the uDrone that is along its center-line. $x_b$ runs along the center-line of the vehicle, pointing from the aft (back) of the uDrone to the fore (front). $z_b$ runs from top to bottom and, following the right-hand rule, $y_b$ runs towards the starboard (right). Further, following convention, roll is defined as rotation about $x_b$, pitch as rotation about $y_b$, and yaw as rotation about $z_b$, with counter-clockwise (CCW) being the positive direction. \n\nThis choice of body frame origin reduces the complexity of modeling the forces produced by the motors. The drawback is some added complexity caused by the moments of the center of buoyancy and center of gravity relative to the chosen origin, but this is more straight forward than the complexity of calculating motor forces about a different point.\n\nThe position of the vehicle, or the body-fixed frame $\\{b\\}$, with respect to the world frame $\\{n\\}$ is expressed as $\\bm{p} = [N, E, D]^T$. The attitude of the vehicle is expressed as $\\bm{\\Theta} = [\\phi, \\theta, \\psi]^T$ using Euler angles. Together these create the 6-dimensional position/orientation vector $\\bm{\\eta} = [\\bm{p}, \\bm{\\Theta]}^T$. The linear and angular velocities are expressed with respect to the vehicles fixed frame $\\{b\\}$. The linear velocity is $\\bm{v} = [u, v, w]^T$ and angular velocity $\\bm{\\omega} = [p, q, r]^T$. when combine, these form the combine 6-dimensional velocity vector $\\bm{\\nu}= [\\bm{v}, \\bm{\\omega]}^T$. All together, $\\boldsymbol{\\eta}$ and $\\boldsymbol{\\nu}$ represent the 12-dimensional state space describing the state of the uDrone \\parencite{thor_kin}. \n\n% \\begin{align*}\n% \\boldsymbol{p}^{n}_{b}=\\left[\\begin{array}{c}\n% N \\\\ E \\\\ D\n% \\end{array}\\right]\n% \\bm{\\Theta}^{n}_{b}=\\left[\\begin{array}{c}\n% \\phi \\\\ \\theta \\\\ \\psi\n% \\end{array}\\right]\n% \\bm{\\eta}=\\left[\\begin{array}{c}\n% \\bm{p} \\\\ \\bm{\\Theta}\n% \\end{array}\\right]\n% \\end{align*}\n% \\begin{align*}\n% \\boldsymbol{v}^{b}_{b}=\\left[\\begin{array}{c}\n% u \\\\ v \\\\ w\n% \\end{array}\\right]\n% \\bm{\\omega}^{b}_{b}=\\left[\\begin{array}{c}\n% p \\\\ q \\\\ r\n% \\end{array}\\right]\n% \\bm{\\nu}=\\left[\\begin{array}{c}\n% \\bm{v} \\\\ \\bm{\\omega}\n% \\end{array}\\right]\n% \\end{align*}\n\n\n\n\\section{Equations of Motions}\n\nThe general equation of motion for the uDrone can be derived as a function of the combine linear and angular velocity vector $\\boldsymbol{\\nu}$ \\parencite{thor_mod}. This is shown in equation \\ref{eqm}.\n\n\\begin{gather}\n\\underbrace{\\boldsymbol{M}_{R B} \\dot{\\boldsymbol{\\nu}}+\\boldsymbol{C}_{R B}(\\boldsymbol{\\nu}) \\boldsymbol{\\nu}}_{\\text {rigid-body forces}}+\\underbrace{\\boldsymbol{M}_{A} \\dot{\\boldsymbol{\\nu}}+\\boldsymbol{C}_{A}\\left(\\boldsymbol{\\nu}\\right) \\boldsymbol{\\nu}+\\boldsymbol{D}\\left(\\boldsymbol{\\nu}\\right) \\boldsymbol{\\nu}}_{\\text {hydrodynamic forces}}+\\underbrace{\\boldsymbol{g}(\\boldsymbol{\\eta})}_{\\text{hydrostatic forces}}=\\boldsymbol{\\tau}\n\\label{eqm}\n\\end{gather}\nWith the terms defined as:\n\\begin{align*}\n \\boldsymbol{\\eta}&=\\text{combine position and orientation}\\\\\n \\boldsymbol{\\nu}&=\\text{combine linear and angular velocity}\\\\\n \\dot{\\boldsymbol{\\nu}}&=\\text{combine linear and angular acceleration}\\\\\n \\boldsymbol{M}_{R B}&=\\text{rigid-body system inertia matrix}\\\\\n \\boldsymbol{C}_{R B}(\\boldsymbol{\\nu})&=\\text{rigid-body Coriolis matrix}\\\\\n \\boldsymbol{M}_{A}&=\\text{added mass system inertia matrix}\\\\\n \\boldsymbol{C}_{A}(\\boldsymbol{\\nu})&=\\text{added mass Coriolis matrix}\\\\\n \\boldsymbol{D}(\\boldsymbol{\\nu})&=\\text{damping matrix}\\\\\n \\boldsymbol{g}(\\boldsymbol{\\eta})&=\\text{gravitational and buoyant forces}\\\\\n \\boldsymbol{\\tau}&=\\text{control inputs}\n\\end{align*}\n% \\begin{align*}\n% \\boldsymbol{\\eta}&=\\text{combine position and orientation}\\\\\n% \\boldsymbol{\\nu}&=\\text{combine linear and angular velocity}\\\\\n% \\dot{\\boldsymbol{\\nu}}&=\\text{combine linear and angular acceleration}\\\\\n% \\boldsymbol{\\tau}&=\\text{control inputs}\\\\\n% \\boldsymbol{M}&=\\text{system inertia matrix}\\\\\n% \\boldsymbol{C}(\\boldsymbol{\\nu})&=\\text{Coriolis matrix}\\\\\n% \\boldsymbol{D}(\\boldsymbol{\\nu})&=\\text{damping matrix}\\\\\n% \\boldsymbol{g}(\\boldsymbol{\\eta})&=\\text{gravitational and buoyant forces}\\\\\n% \\boldsymbol{\\tau}&=\\text{control inputs}\n% \\end{align*}\n% \\begin{align*}\n% \\boldsymbol{M}&=\\boldsymbol{M}_{R B}+\\boldsymbol{M}_{A}\\\\\n% \\boldsymbol{C}&=\\boldsymbol{C}_{R B}+\\boldsymbol{C}_{A}\n% \\end{align*}\n\n\\section{Rigid-Body Forces}\n\nThe exact values of the system inertia ($\\boldsymbol{M}$), Coriolis ($\\boldsymbol{C}$), and Drag ($\\boldsymbol{D}$) matrices can be determined either through intensive hydrodynamic modeling or experimentation. Due to the complexity of hydrodynamic modeling, experimentation is typically used to determine these values in underwater vehicles similar to the uDrone \\parencite{hipp_pen}. It is possible, however, to estimate the rigid body matrix values using information about the vehicle generated from the 3D model. This section details these estimations and explains how they were calculated. \n\nThe rigid body system inertia matrix about the center of origin can be calculated using equation \\ref{mrb} which is found in \\cite{thor_rb}.\n\\begin{gather}\n \\boldsymbol{M}_{R B}^{C O}=\\left[\\begin{array}{cc}\n m \\boldsymbol{I}_{3 \\times 3} & -m \\boldsymbol{S}\\left(\\boldsymbol{r}_{g}^{b}\\right) \\\\\n m \\boldsymbol{S}\\left(\\boldsymbol{r}_{g}^{b}\\right) & \\boldsymbol{I}_{g}-m \\boldsymbol{S}^{2}\\left(\\boldsymbol{r}_{g}^{b}\\right)\n \\end{array}\\right]\n \\label{mrb}\n\\end{gather}\n% \\begin{align*}\n% \\boldsymbol{M}_{R B}^{C O}&=\\text{rigid body system inertia matrix about the body fixed-frame origin}\\\\\n% \\boldsymbol{I}_{3 \\times 3}&=\\text{3 by 3 identity matrix}\\\\\n% m&=\\text{mass}\\\\\n% \\boldsymbol{S}(\\cdot)&=\\text{cross product operator}\\\\\n% \\boldsymbol{r}_{g}^{b}&=\\text{ vector from the body fixed-frame origin to the center of gravity}\\\\\n% \\boldsymbol{I}_{g}&=\\text{inertia matrix}\n% \\end{align*}\n\nIn this equation, $m$ is the mass, $\\boldsymbol{I}_{3 \\times 3}$ is the three-by-three identity matrix, $\\boldsymbol{I}_{g}$ is the inertia matrix and $\\boldsymbol{r}_{g}^{b}$ is the vector from the body fixed-frame origin to the center of gravity. $\\boldsymbol{S}$ is the cross product operator matrix. Before the final vehicle was constructed, m, $\\boldsymbol{I}_{g}$, and $\\boldsymbol{r}_{g}^{b}$ were taken from the 3D model. Those estimates are shown in equation \\ref{m_matrix}.\n\n\\begin{equation}\n\\boldsymbol{M}_{R B}^{C O}=\\left[\\begin{array}{cccccc}\n9.9000 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 9.9000 & 0 & 0 & 0 & 1.4850 \\\\\n0 & 0 & 9.9000 & 0 & -1.4850 & 0 \\\\\n0 & 0 & 0 & 0.3420 & 0.0010 & -0.0001 \\\\\n0 & 0 & -1.4850 & 0.0010 & 0.5577 & -0.0015 \\\\\n0 & 1.4850 & 0 & -0.0001 & -0.0015 & 0.3167\n\\end{array}\\right]\n\\label{m_matrix}\n\\end{equation}\n\n\nWith a completed vehicle these numbers can be verified and varied if needed. For example, with no ballast, the uDrone is positively buoyant. Therefore, to maintain its depth without using power to stay underwater ballast will need to be added. By placing the ballast closer to the center of the vehicle, the moments of inertia will decrease. Conversely, by placing it further from the center the moments can be increased. Additionally, the ballast can be used to change the center of gravity relative to the center of buoyancy. This will create a moment in the vehicle that can help it stay oriented or level in specific ways. This relationship is discussed further in the section \\ref{hydrostatics}. \n\nThe Coriolis matrix is a function of the angular velocity of the vehicle and can be derived as shown in equation \\ref{crb} from \\cite{thor_rb}.\n\\begin{gather}\n \\boldsymbol{C}_{RB}^{CO}(\\boldsymbol{\\nu})=\\left[\\begin{array}{cc}\n m \\boldsymbol{S}\\left(\\boldsymbol{\\omega}^{b}\\right) & -m \\boldsymbol{S}\\left(\\boldsymbol{\\omega}^{b}\\right) \\boldsymbol{S}\\left(\\boldsymbol{r}_{g}^{b}\\right) \\\\\n m \\boldsymbol{S}\\left(\\boldsymbol{r}_{g}^{b}\\right) \\boldsymbol{S}\\left(\\boldsymbol{\\omega}^{b}\\right) & -\\boldsymbol{S}\\left(\\left(\\boldsymbol{I}_{g}-m \\boldsymbol{S}^{2}\\left(\\boldsymbol{r}_{g}^{b}\\right)\\right) \\boldsymbol{\\omega}^{b}\\right)\n \\end{array}\\right]\n \\label{crb}\n\\end{gather}\n% \\begin{align*}\n% \\boldsymbol{C}_{R B}^{C O}&=\\text{rigid body Coriolis matrix about the body fixed-frame origin}\\\\\n% m&=\\text{mass}\\\\\n% \\boldsymbol{S}(\\cdot)&=\\text{cross product operator}\\\\\n% \\boldsymbol{\\omega}^b&=\\text{body-fixed frame angular velocity}\\\\\n% \\boldsymbol{r}_{g}^{b}&=\\text{ vector from the body fixed-frame origin to the center of gravity}\\\\\n% \\boldsymbol{I}_{g}&=\\text{inertia matrix}\n% \\end{align*}\n\nWhere $\\boldsymbol{\\omega}^{b}$ is the body-fixed frame angular velocity vector. Since this matrix is a function of the vehicle state, it will change with motion. The actual calculations of this matrix are handled by a MatLab program written as a companion to Thor Forson’s book \\parencite{mss}. The values based on the system inertia matrix in equation \\ref{m_matrix} along with a velocity state vector $\\bm{\\nu}= [1,0,0,0,0,0]^T$, which represents only forward motion, is shown in equation \\ref{c_matrix}.\n\n\\begin{equation}\n\\boldsymbol{C}_{RB}^{CO}(\\boldsymbol{\\nu})=\\left[\\begin{array}{cccccc}\n0 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 0 & 0 & 9.9000 \\\\\n0 & 0 & 0 & 0 & -9.9000 & 0 \\\\\n0 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 9.9000 & 0 & 0 & 0 \\\\\n0 & -9.9000 & 0 & 0 & 0 & 0\n\\end{array}\\right]\n\\label{c_matrix}\n\\end{equation}\n\n\\section{Hydrostatic Forces} \\label{hydrostatics}\n\nThe hydrostatic term $\\boldsymbol{g}(\\boldsymbol{\\eta})$ accounts for the forces on the uDrone caused by gravity and buoyancy. The force of gravity, or weight ($W$) is equal to the mass of the uDrone times the acceleration due to gravity. The buoyant force ($B$) is the weight of the water displaced by the uDrone. This is calculated by multiplying the volume of the uDrone, the density of water, and the acceleration due to gravity. If the location of the center of gravity and center of buoyancy (or center of volume) of the uDrone are not in the same spot then there will be a moment on the whole vehicle. Similarly, if the two forces are not equal then there will also be a net linear force on it. This value is a function of the orientation of the vehicle and can be seen in equation \\ref{bigG} \\parencite{thor_rb}. \n\n\\begin{equation}\\label{bigG}\n\\boldsymbol{g}(\\boldsymbol{\\eta})=\\left[\\begin{array}{llll} \n& (W-B) \\sin (\\theta) \\\\\n- & (W-B) \\cos (\\theta) \\sin (\\phi) \\\\\n- & (W-B) \\cos (\\theta) \\cos (\\phi) \\\\\n- & \\left(y_{g} W-y_{b} B\\right) \\cos (\\theta) \\cos (\\phi) & + & \\left(z_{g} W-z_{b} B\\right) \\cos (\\theta) \\sin (\\phi) \\\\\n& \\left(z_{g} W-z_{b} B\\right) \\sin (\\theta) & + & \\left(x_{g} W-x_{b} B\\right) \\cos (\\theta) \\cos (\\phi) \\\\\n- & \\left(x_{g} W-x_{b} B\\right) \\cos (\\theta) \\sin (\\phi) & - & \\left(y_{g} W-y_{b} B\\right) \\sin (\\theta)\n\\end{array}\\right]\n\\end{equation}\n\n% \\begin{equation}\n% \\boldsymbol{g}(\\boldsymbol{\\eta})=\\left[\\begin{array}{llll} \n% 0\\\\0\\\\0\\\\0\\\\0\\\\0\n% \\end{array}\\right]\n% \\end{equation}\n\nBallasting can be used to adjust the weight and center of mass on the uDrone. To simplify this equation the ballast is set so that the weight and buoyancy are equal and the center of mass and buoyancy are the same. This means the $\\boldsymbol{g}$ vector will be zeros for all $\\boldsymbol{\\eta}$. These forces can be seen interacting on the uDrone in figure \\ref{cord_frame}. Maintaining overlapped centers of buoyancy and gravity also makes the vehicle more nimble as there are less moments to overcome when rotating. This also has the benefit of making the uDrone more maneuverable. Equal weight and buoyancy reduces energy used as the uDrone does not need to overcome a force making it sink or float as it cruises through the water column. \n\nThere might be reasons to adjust the ballasting of the uDrone to impart other properties on it. For example, if the vehicle is slightly positively buoyant than it will float in the event of a power failure, allowing for it to be recovered more easily. Also, if the center of gravity is placed below the center of buoyancy, but still in line with it in the z body direction, then the vehicle will be self righting in roll. This could make control easier as roll, which should be at zero most of the time, can be ignored.\n\n\\section{Hydrodynamic Forces} \\label{hydrodynamics}\n\nThe added mass system inertia matrix and added mass Coriolis matrix can only be determined through experimentation or complex fluid dynamic simulation and are not calculated in the scope of this thesis. The damping matrix, however, can be estimated using drag calculations. \n\nThe vehicle is modeled as a flat-faced cylinder moving through water in a direction along its axis. This estimation neglects two factors: 1) drag from the motors, and 2) rotational damping. Both of these forces are small in comparison to the drag of the flat-faced cylinder, so for the purposes of an approximate model, they can be ignored. \n\nThe force of drag in a single direction can be calculated using the drag equation:\n\n\\begin{gather}\nF_{D}=C_{D} A \\frac{\\rho V^{2}}{2}\n\\end{gather}\n\\begin{align*}\n F_D&=\\text{drag Force}\\\\\n C_D&=\\text{coefficient of drag}\\\\\n A&=\\text{cross-sectional area normal to velocity}\\\\\n \\rho&=\\text{fluid density}\\\\\n V&=\\text{velocity}\\\\\n\\end{align*}\n\nBased on the ratio of the diameter to the length of the main body of the uDrone, the Coefficient of drag in the direction of motion is estimated at 0.8. The density of salt water is 1024kg/m$^3$ and the surface area is calculated from the diameter of the eight inch cylinder, which is 0.037m$^2$. Putting this all together, damping force can be calculated as a function of velocity. This is:\n\n\\begin{gather}\n \\boldsymbol{D}(\\boldsymbol{\\nu})\\boldsymbol{\\nu}=\\left[\\begin{array}{c}\n -15.0u ^2 \\\\ \\sim 0 \\\\ \\sim 0 \\\\ D_{roll} \\\\ D_{pitch} \\\\ D_{yaw}\n \\end{array}\\right]\n\\end{gather}\n\n$u$ is the velocity in the x body direction. The last three values which refer to the rotational damping. For the purposes of calculations in the next section these values were assumed to be zero. The second and third terms, however, will always be zero, or nearly zero, as the vehicle cannot independently move in the Y or Z direction.\n\n\\section{Control Inputs}\\label{control_inputs}\n\nThe control inputs are the forces put on the uDrone from the thrusters. The four motor of the uDrone are arranged on a plane, all facing the same direction. The center point between all the thrusters along that plane is the Center of Origin $C_O$ for the vehicle. Thrusters one and three spin in a counterclockwise direction while thrusters two and four spin clockwise for forward thrust. This ensures that the vehicle has no roll moment when moving forward. Figure \\ref{cord_frame} shows the uDrone with the $C_O$ and rotation direction of the thrusters. \n\n\\begin{figure}[h]\n\\includegraphics[width=\\maxwidth{\\textwidth}]{img/cord_frame.png}\n\\caption{uDrone Coordinate Frame and Motor Directions}\n\\label{cord_frame}\n\\end{figure}\n\nUsing this convention it is possible to determine the actuation matrix, tau, as a function of thruster inputs.\n\n\\begin{gather}\n \\boldsymbol{\\tau}=\\left[\\begin{array}{c}\n F_1+F_2+F_3+F_4 \\\\ 0 \\\\ 0 \\\\ M_1-M_2+M_3-M_4 \\\\ (F_2+F_3-F_1-F_4) \\frac{L}{\\sqrt{2}} \\\\ (F_1+F_2-F_3-F_4) \\frac{L}{\\sqrt{2}}\n \\end{array}\\right]\n \\label{tau}\n\\end{gather}\n\nIn equation \\ref{tau} $F_x$ indicates the linear force produced by the $x^{th}$ motor and $M_x$ indicates the angular momentum produced by the $x^{th}$. $L$ is 0.115m, the length from $C_O$ to the center of the motor, which is the same for each motor. Unfortunately, there is no data from Blue Robotics about the moment of the thrusters, so the relationship between control input to the roll moment will need to be determined experimentally. Data is available, however, correlating motor power and thrust force. \n\n\\begin{figure}[h]\n\\includegraphics[width=\\maxwidth{\\textwidth}]{img/force_amps.png}\n\\caption{Force Output vs. Current Draw for T200 Thruster at 14 Volts}\n\\label{force_amp}\n\\legend{\\emph{Source}: \\iftoggle{usebiblatex}{\\textcite{t200}}{\\citet{t200}}}% See: https://upload.wikimedia.org/wikipedia/commons/8/89/Antidorcas_marsupialis%2C_male_%28Etosha%2C_2012%29.jpg\n% \\legend{\\emph{Note}: Here is a note that is especially long to show what happens when it extends to more than one line.}\n\\end{figure}\n\n\nUsing the figure \\ref{force_amp}, along with data about the uDrone battery, it is possible to calculate the maximum velocity, running time at this velocity, and running time at the operational velocity of 1m/s. While the uDrone can carry two 14V, 18Ah batteries, the intent is to use one for thrusters and the other to control all electronics. Therefore, all longevity calculations are done assuming 18Ah are available for thruster actuation. \n\nIn order to calculate maximum velocity, the drag equation is set equal to the maximum thruster force. The T200 thrusters produce a maximum thrust of 41.68N each. This is 166.7N of total thrust. Setting this equal to $15u^2$ and solving yields a maximum speed of 3.33m/s. Maintaining this velocity takes 20.29A, for a total of 81.16A. The uDrone could only maintain this speed for approximately 13.3 minute with an 18Ah battery.\n\nTypically, however, the uDrone will be cruising at a speed of 1m/s. At this speed, the drag force is approximately 15N. To maintain this thrust each motor must produce about 3.75N of force. To maintain this force each motor will use approximately 0.5A, for a total of 2A. In order to stay conservative, a 50\\% safety factor will be added to this value to account for uncalculated drag forces and actuation that affect orientations, not just forward motion. With this addition, the vehicle will use 3A for actuation during normal operation. With a single 18Ah battery used for thrusters, this will yield a dive time of approximately 6 hours. \n", "meta": {"hexsha": "d677b8a40c5a21476fc7e7aa3a47ab022f745cfd", "size": 18832, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "model.tex", "max_stars_repo_name": "auman66/asu-thesis", "max_stars_repo_head_hexsha": "402c17282b3e4b590539238aca405b1c071305af", "max_stars_repo_licenses": ["Info-ZIP", "Linux-OpenIB"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "model.tex", "max_issues_repo_name": "auman66/asu-thesis", "max_issues_repo_head_hexsha": "402c17282b3e4b590539238aca405b1c071305af", "max_issues_repo_licenses": ["Info-ZIP", "Linux-OpenIB"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "model.tex", "max_forks_repo_name": "auman66/asu-thesis", "max_forks_repo_head_hexsha": "402c17282b3e4b590539238aca405b1c071305af", "max_forks_repo_licenses": ["Info-ZIP", "Linux-OpenIB"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 80.1361702128, "max_line_length": 821, "alphanum_fraction": 0.7163870008, "num_tokens": 5493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424528443251, "lm_q2_score": 0.7431680143008301, "lm_q1q2_score": 0.6286773728930908}} {"text": "\n\\section*{Motivation}\n\n\\subsection*{Why are we interested in the Ising model?}\n%\\subsection{The Ising model: a model found in statistical mechanics}\n\n\\begin{frame}\n\t \tThe Ising model is a mathematical model of ferromagnetism in statistical mechanics.\n\n Probability of a configuration $\\sigma$ ({\\color{TurkishRose}\\textit{Boltzmann distribution}}):\n \\begin{align*}\n P_\\beta(\\sigma) & = e^{-\\beta H(\\sigma)} / Z(G), \\\\ \n Z(G) & = \\sum_{\\sigma \\colon V \\to \\{+,-\\}} e^{- \\beta H(\\sigma)}.\n \\end{align*}\n \n \\begin{itemize}\n \t\\item $Z(G)$ is \\textit{the partition function of the Ising model}.\n \t\\item $Z(G)$ encodes information about the physical system.\n \\end{itemize}\n\\end{frame}\n\n\n\\begin{frame}\n\t\t \\begin{align*}\n\t\t P_\\beta(\\sigma) & = e^{-\\beta H(\\sigma)} / Z(G), \\\\ \n\t\t Z(G) & = \\sum_{\\sigma \\colon V \\to \\{+,-\\}} e^{- \\beta H(\\sigma)}.\n\t\t \\end{align*}\n\t\t \n\t\\begin{enumerate}\t\t\t\n\t\t\\item {\\color{TurkishRose} \\textbf{Problem:}} What is the complexity of evaluating $Z(G)$? \\\\\n\t\t%{\\color{TurkishRose}\\textbf{Answer:}}\n\t\t$\\# \\mathfs{P}$-hard in some cases.\n\t\t\n\t\t\\item For some of those cases, $Z(G)$ admits a \\textit{fully polynomial-time randomised approximation scheme} \\\\\n\t\t(Jerrum and Sinclair, 1993, G\\\"odel prize in 1996).\n\t\t\n\t\t\\item {\\color{TurkishRose} \\textbf{Problem:}} What is the complexity of approximating $Z(G)$?\n\t\t\n\t\t\\item {\\color{TurkishRose} \\textbf{Application:}} Strongly simulating \\textit{instantaneous quantum polynomial-time circuits} is equivalent to approximating $|Z(G)|$ for some parameters.\n\t\\end{enumerate}\n\\end{frame}\n\n", "meta": {"hexsha": "67c4e71d93f473ec00b4573e3a4fadf4feebb656", "size": 1603, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Slides/Sections/motivation.tex", "max_stars_repo_name": "andreshp/LatexTemplates", "max_stars_repo_head_hexsha": "64fdc06ddf76702b9392e871b1fdd0aee6000b30", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 23, "max_stars_repo_stars_event_min_datetime": "2015-10-14T17:46:47.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-22T20:02:09.000Z", "max_issues_repo_path": "Slides/Sections/motivation.tex", "max_issues_repo_name": "andreshp/PlantillasLatex", "max_issues_repo_head_hexsha": "64fdc06ddf76702b9392e871b1fdd0aee6000b30", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2016-04-11T09:14:42.000Z", "max_issues_repo_issues_event_max_datetime": "2016-04-11T09:21:16.000Z", "max_forks_repo_path": "Slides/Sections/motivation.tex", "max_forks_repo_name": "andreshp/LatexTemplates", "max_forks_repo_head_hexsha": "64fdc06ddf76702b9392e871b1fdd0aee6000b30", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2015-10-14T17:54:14.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-15T06:39:01.000Z", "avg_line_length": 37.2790697674, "max_line_length": 189, "alphanum_fraction": 0.6500311915, "num_tokens": 501, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424411924673, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6286773594217868}} {"text": "\\section{Error Calculation}\n\\label{sec:Error_Calculation}\nThis section contains the error calculation of the measured values. The error calculation is done for the fitted values width $w$, diameter $d$ and grid constant $g$ (see sections \\ref{subsec:Slit} to \\ref{subsec:Cross-Grid}).\n\n\\subsection{Uncertainties}\n\\label{subsec:Uncertainties}\nAll conducted measurements in this experiment have uncertainties. Although it was possible to read the position on the y-axis to half a millimeter, the uncertainty is still one millimeter. This is due to the difficulty to find the exact center of the minima or maxima. The systematic uncertainties are shown in table \\ref{tab:equipment} ($s_x = 1$ mm and $s_y = 1$ mm).\n\n\\subsection{Uncertainty of the Angle $\\varphi$}\n\\label{subsec:Uncertainty_Angle}\nThe uncertainty $s_\\varphi$ of the angle $\\varphi$ is calculated with the following equation \\ref{eq:uncertainty_angle}:\n\n\\begin{equation}\ns_{\\varphi}=\\sqrt{\\left(\\frac{\\partial \\varphi}{\\partial x}\\Biggr|_{\\varphi}\\cdot s_{x}\\right)^2 + \\left(\\frac{\\partial \\varphi}{\\partial y}\\Biggr|_{\\varphi}\\cdot s_{y}\\right)^2}\n\\label{eq:uncertainty_angle}\n\\end{equation}\n\nwith:\n\n\\[\n\\frac{\\partial \\varphi}{\\partial x}\\Biggr|_{\\varphi}=-\\frac{y}{x^2+y^2} \\qquad , \\qquad \\frac{\\partial \\varphi}{\\partial y}\\Biggr|_{\\varphi}=\\frac{x}{x^2+y^2}\n\\]\n\nwhere:\n\\begin{conditions}\n\ts_{\\varphi} & uncertainty of $\\varphi$ \\\\\n\ts_x & uncertainty of x \\\\\n\ts_y & uncertainty of y \\\\\n\tx & distance between the aperture and the projection surface (see figure \\ref{fig:experimental_arrangement}) \\\\\n\ty & position of the maxima or the minima\n\\end{conditions}\n\n\\subsection{Uncertainty of the Width $w$}\n\\label{subsec:Uncertainty_Width}\nTo derive the total uncertainty of the width $w$ the following equations are used. The statistical uncertainty $s_{w,\\ \\text{stat}}$ is obtained from the fits (from QtiPlot).\n\n\\begin{equation}\ns_{w,\\ \\text{tot}}=\\sqrt{s_{w,\\ \\text{syst}}^2+s_{w,\\ \\text{stat}}^2}\n\\label{eq:total_uncertainty_w}\n\\end{equation}\n\nwith:\n\\begin{equation}\ns_{w,\\ \\text{syst}}=\\sqrt{\\left(\\frac{\\partial w}{\\partial \\varphi}\\Biggr|_{w}\\cdot s_{\\varphi}\\right)^2}\n\\label{eq:syst_uncertainty_w}\n\\end{equation}\n\nwhere:\n\\begin{conditions}\n\ts_{w,\\ \\text{tot}} & total uncertainty of $w$ \\\\\n\ts_{w,\\ \\text{syst}} & systematic uncertainty of $w$ \\\\\n\ts_{w,\\ \\text{stat}} & statistical uncertainty of $w$ \\\\\n\ts_{\\varphi} & uncertainty of $\\varphi$\n\\end{conditions}\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\\subsubsection{Calculated Uncertainties of the Width $w$ (Slit)}\n\\label{subsubsec:Uncertainty_Width_Slit}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{3}}\nThe systematic uncertainty was calculated by using equation \\ref{eq:slit_maxima} and equation \\ref{eq:syst_uncertainty_w}. The statistical uncertainty was obtained from table \\ref{tab:Slit}. The total uncertainty was calculated by using equation \\ref{eq:total_uncertainty_w}. All calculations were done in MATLAB (see appendix \\ref{sec:MATLAB_Error_Calculation}).\n\n\\begin{table}[H]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{1.3}\n\t\\begin{tabular}{r||c|c|c}\n\t\t& \\textbf{Systematic} ($\\mu$m) & \\textbf{Statistical} ($\\mu$m) & \\textbf{Total} ($\\mu$m) \\\\\n\t\t\\hline\\hline\n\t\t\\textbf{Slit 40 $\\mu$m} & -0.92 & 0.10 & 0.93 \\\\\n\t\t\\textbf{Slit 100 $\\mu$m} & -6.77 & 0.14 & 6.77 \\\\\n\t\\end{tabular}\n\t\\caption{Calculated Uncertainties for the Width $w$ of the Slit}\n\t\\label{tab:Calculated_Uncertainties_Width_Slit}\n\\end{table}\n\nTo calculate the uncertainty of the 40 $\\mu$m slit and the 100 $\\mu$m slit the 1st order ($m = 1$) was used. It is generally best to use a low order number (greater uncertainty).\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\\subsubsection{Calculated Uncertainties of the Width $w$ (Anti-Slit)}\n\\label{subsubsec:Uncertainty_Width_Anti-Slit}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{3}}\nThe systematic uncertainty was calculated by using equation \\ref{eq:slit_maxima} and equation \\ref{eq:syst_uncertainty_w}. The statistical uncertainty was obtained from table \\ref{tab:Anti-Slit}. The total uncertainty was calculated by using equation \\ref{eq:total_uncertainty_w}. All calculations were done in MATLAB (see appendix \\ref{sec:MATLAB_Error_Calculation}).\n\n\\begin{table}[H]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{1.3}\n\t\\begin{tabular}{r||c|c|c}\n\t\t& \\textbf{Systematic} ($\\mu$m) & \\textbf{Statistical} ($\\mu$m) & \\textbf{Total} ($\\mu$m) \\\\\n\t\t\\hline\\hline\n\t\t\\textbf{Anti-Slit 230 $\\mu$m} & -23.33 & 0.34 & 23.34 \\\\\n\t\t\\textbf{Anti-Slit 124 $\\mu$m} & -6.80 & 0.08 & 6.80 \\\\\n\t\\end{tabular}\n\t\\caption{Calculated Uncertainties for the Width $w$ of the Anti-Slit}\n\t\\label{tab:Calculated_Uncertainties_Width_Anti-Slit}\n\\end{table}\n\nTo calculate the uncertainty of the 230 $\\mu$m anti-slit the 6th order was used. To calculate the uncertainty of the 124 $\\mu$m anti-slit the 1st order was used.It is generally best to use a low order number (greater uncertainty).\n\n\\newpage\n\n\\subsection{Uncertainty of the Diameter $d$}\n\\label{subsec:Uncertainty_Diameter}\nTo derive the total uncertainty of the diameter $d$ the following equations are used. The statistical uncertainty $s_{d,\\ \\text{stat}}$ is obtained from the fits (from QtiPlot).\n\n\\begin{equation}\ns_{d,\\ \\text{tot}}=\\sqrt{s_{d,\\ \\text{syst}}^2+s_{d,\\ \\text{stat}}^2}\n\\label{eq:total_uncertainty_d}\n\\end{equation}\n\nwith:\n\\begin{equation}\ns_{d,\\ \\text{syst}}=\\sqrt{\\left(\\frac{\\partial d}{\\partial \\varphi}\\Biggr|_{d}\\cdot s_{\\varphi}\\right)^2}\n\\label{eq:syst_uncertainty_d}\n\\end{equation}\n\nwhere:\n\\begin{conditions}\n\ts_{d,\\ \\text{tot}} & total uncertainty of $d$ \\\\\n\ts_{d,\\ \\text{syst}} & systematic uncertainty of $d$ \\\\\n\ts_{d,\\ \\text{stat}} & statistical uncertainty of $d$ \\\\\n\ts_{\\varphi} & uncertainty of $\\varphi$\n\\end{conditions}\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\\subsubsection{Calculated Uncertainties of the Diameter $d$ (Circular Aperture)}\n\\label{subsubsec:Uncertainty_Diameter_Circular-Aperture}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{3}}\n\nThe systematic uncertainty was calculated by using equation \\ref{eq:circular_aperture} and equation \\ref{eq:syst_uncertainty_d}. The statistical uncertainty was obtained from table \\ref{tab:Circular_Apertures}. The total uncertainty was calculated by using equation \\ref{eq:total_uncertainty_d}. All calculations were done in MATLAB (see appendix \\ref{sec:MATLAB_Error_Calculation}).\n\n\\begin{table}[H]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{1.3}\n\t\\begin{tabular}{r||c|c|c}\n\t\t& \\textbf{Systematic} ($\\mu$m) & \\textbf{Statistical} ($\\mu$m) & \\textbf{Total} ($\\mu$m) \\\\\n\t\t\\hline\\hline\n\t\t\\textbf{Circular Aperture 150 $\\mu$m} & -10.21 & 0.83 & 10.24 \\\\\n\t\t\\textbf{Circular Aperture 100 $\\mu$m} & -4.57 & 0.55 & 4.60 \\\\\n\t\\end{tabular}\n\t\\caption{Calculated Uncertainties for the Diameter $d$}\n\t\\label{tab:Calculated_Uncertainties_Diameter}\n\\end{table}\n\nTo calculate the uncertainty of the 150 $\\mu$m circular aperture and the 100 $\\mu$m circular aperture the 2nd Bessel coefficient ($c = 2.233$) was used (see table \\ref{eq:coeffs}). It is generally best to use a low \"order\" number (greater uncertainty).\n\n\\newpage\n\n\\subsection{Uncertainty of the Grid Constant $g$}\n\\label{subsec:Uncertainty_Grid_Constant}\nTo derive the total uncertainty of the grid constant $g$ the following equations are used. The statistical uncertainty $s_{g,\\ \\text{stat}}$ is obtained from the fits (from QtiPlot).\n\n\\begin{equation}\ns_{g,\\ \\text{tot}}=\\sqrt{s_{g,\\ \\text{syst}}^2+s_{g,\\ \\text{stat}}^2}\n\\label{eq:total_uncertainty_g}\n\\end{equation}\n\nwith:\n\\begin{equation}\ns_{g,\\ \\text{syst}}=\\sqrt{\\left(\\frac{\\partial g}{\\partial \\varphi}\\Biggr|_{g}\\cdot s_{\\varphi}\\right)^2}\n\\label{eq:syst_uncertainty_g}\n\\end{equation}\n\nwhere:\n\\begin{conditions}\n\ts_{g,\\ \\text{tot}} & total uncertainty of $g$ \\\\\n\ts_{g,\\ \\text{syst}} & systematic uncertainty of $g$ \\\\\n\ts_{g,\\ \\text{stat}} & statistical uncertainty of $g$ \\\\\n\ts_{\\varphi} & uncertainty of $\\varphi$\n\\end{conditions}\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\\subsubsection{Calculated Uncertainties of the Grid Constant $g$ (Cross-Grid)}\n\\label{subsubsec:Uncertainty_Grid_Constant_Cross-Grid}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{3}}\n\nThe systematic uncertainty was calculated by using equation \\ref{eq:cross-grid} and equation \\ref{eq:syst_uncertainty_g}. The statistical uncertainty was obtained from table \\ref{tab:Cross-Grid}. The total uncertainty was calculated by using equation \\ref{eq:total_uncertainty_g}. All calculations were done in MATLAB (see appendix \\ref{sec:MATLAB_Error_Calculation}).\n\n\\begin{table}[H]\n\t\\centering\n\t\\renewcommand{\\arraystretch}{1.3}\n\t\\begin{tabular}{r||c|c|c}\n\t\t & \\textbf{Systematic} ($\\mu$m) & \\textbf{Statistical} ($\\mu$m) & \\textbf{Total} ($\\mu$m) \\\\\n\t\t\\hline\\hline\n\t\t\\textbf{Cross-Grid 28 $\\mu$m} & -0.36 & 0.44 & 0.57 \\\\\n\t\t\\textbf{Cross-Grid 50 $\\mu$m} & -0.30 & 0.06 & 0.30 \\\\\n\t\\end{tabular}\n\t\\caption{Calculated Uncertainties for the Grid Constant $g$}\n\t\\label{tab:Calculated_Uncertainties_Grid_Constant}\n\\end{table}\n\nTo calculate the uncertainty of the 28 $\\mu$m cross-grid and the 50 $\\mu$m cross-grid the 1st order ($m = 1$) was used. It is generally best to use a low order number (greater uncertainty).\n", "meta": {"hexsha": "913c27ea5c8af260c98302f7fb9d099c174ef83a", "size": 9210, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "glaL3_O_9_Interference_and_Diffraction/sections/error_calculation.tex", "max_stars_repo_name": "MuellerDominik/Physics-Laboratory-Notebooks", "max_stars_repo_head_hexsha": "02836870e6d97a29b1857c956fbd58eb5933eede", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "glaL3_O_9_Interference_and_Diffraction/sections/error_calculation.tex", "max_issues_repo_name": "MuellerDominik/Physics-Laboratory-Notebooks", "max_issues_repo_head_hexsha": "02836870e6d97a29b1857c956fbd58eb5933eede", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "glaL3_O_9_Interference_and_Diffraction/sections/error_calculation.tex", "max_forks_repo_name": "MuellerDominik/Physics-Laboratory-Notebooks", "max_forks_repo_head_hexsha": "02836870e6d97a29b1857c956fbd58eb5933eede", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.219895288, "max_line_length": 383, "alphanum_fraction": 0.7327904452, "num_tokens": 2919, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8459424295406088, "lm_q1q2_score": 0.6286773507624984}} {"text": "% Copyright 2018-2021 Melvin Eloy Irizarry-Gelpí\n\\setcounter{chapter}{2}\n\\chapter{Circuits in Series and Parallel}\n%\nIn this experiment you will learn about circuits connected in series or in parallel.\n%\n\\section{Preliminary}\n%\nA \\textbf{resistor} is any component in an electric circuit with \\textbf{electrical resistance}. The SI unit for resistance is the ohm:\n\\begin{equation}\n\t1 \\ \\text{ohm} = 1 \\ \\text{V/A}\n\\end{equation}\nOhm's law is a statement that relates the voltage $V$ across a resistor to the amount of current $I$ leaving (or equivalently, entering) a resistor:\n\\begin{equation}\n\tV = R I\n\\end{equation}\nHere, $R$ is the amount of electrical resistance. Solving for $R$ gives\n\\begin{equation} \\label{eq.03.ROhmLaw}\n\tR = \\frac{V}{I}\n\\end{equation}\nFor \\textbf{ohmic resistors}, the ratio of voltage and current is fixed (i.e. constant) and \\textbf{always} gives the same value. This is \\textbf{not true} for \\textbf{non-ohmic resistors}.\n\nAn electric circuit can contain many components that are connected with wires. The electric properties of the circuit will depend on \\textbf{how} these components are connected. You are going to look at two ways of connecting two components in a circuit to a battery: \\textbf{series} and \\textbf{parallel}.\n%\n\\subsection{Circuits in Series}\n%\nA collection of two resistors in \\textbf{series} can be replaced by an \\textbf{equivalent resistor} with a resistance given by the sum of the individual resistances:\n\\begin{equation} \\label{eq.03.RSeries}\n\tR_{\\text{eq}} = R_{1} + R_{2}\n\\end{equation}\nWhen two resistors are connected in \\textbf{series} to a battery, the \\textbf{current} $I_{0}$ leaving the battery is the \\textbf{same} as the \\textbf{current} $I_{1}$ leaving resistor 1, and also the same as the \\textbf{current} $I_{2}$ leaving resistor 2:\n\\begin{equation} \\label{eq.03.ISeries}\n\tI_{0} = I_{1} = I_{2}\n\\end{equation}\nWhen two resistors are connected in \\textbf{series} to a battery, the \\textbf{voltage} $V_{0}$ across the battery is \\textbf{split} between the \\textbf{voltage} $V_{1}$ across resistor 1, and the \\textbf{voltage} $V_{2}$ across resistor 2:\n\\begin{equation} \\label{eq.03.VSeries}\n\tV_{0} = \\left| V_{1} + V_{2} \\right|\n\\end{equation}\nThe absolute value is needed because the voltage across each resistor is \\textbf{negative} since it correspond to an \\textbf{electric potential drop}.\n%\n\\subsection{Circuits in Parallel}\n%\nA collection of two resistors in \\textbf{parallel} can be replaced by an \\textbf{equivalent resistor} with a resistance that satisfies the relation\n\\begin{equation}\n\t\\frac{1}{R_{\\text{eq}}} = \\frac{1}{R_{1}} + \\frac{1}{R_{2}}\n\\end{equation}\nSolving for the equivalent resistance gives\n\\begin{equation} \\label{eq.03.RParallel}\n\tR_{\\text{eq}} = \\frac{R_{1} R_{2}}{R_{1} + R_{2}}\n\\end{equation}\nWhen two resistors are connected in \\textbf{parallel} to a battery, the \\textbf{current} $I_{0}$ leaving the battery is split between the \\textbf{current} $I_{1}$ leaving resistor 1, and the \\textbf{current} $I_{2}$ leaving resistor 2:\n\\begin{equation} \\label{eq.03.IParallel}\n\tI_{0} = I_{1} + I_{2}\n\\end{equation}\nWhen two resistors are connected in \\textbf{parallel} to a battery, the \\textbf{voltage} $V_{0}$ across the battery has the same magnitude as the \\textbf{voltage} $V_{1}$ across resistor 1, and also the same magnitude as the \\textbf{voltage} $V_{2}$ across resistor 2:\n\\begin{equation} \\label{eq.03.VParallel}\n\tV_{0} = \\left| V_{1} \\right| = \\left| V_{2} \\right|\n\\end{equation}\nAgain, remember that the voltage across a resistor (measured in the direction along the current) is negative because it correspond to a drop in electric potential.\n%\n\\subsection{Power}\n%\nThe power $P$ input or output of a component in an electric circuit can be found by multiplying the voltage $V$ across that component, and the amount of current $I$ leaving that component:\n\\begin{equation}\n\tP = VI\n\\end{equation}\nThe SI unit for power is the \\textbf{watt} (W). One watt is equivalent to one joule of energy consumed or generated per second:\n\\begin{equation}\n\t1 \\ \\text{W} = 1 \\ \\text{J/s}\n\\end{equation}\nSince power equals voltage multiplied by current, you also have\n\\begin{equation}\n\t1 \\ \\text{W} = 1 \\ \\text{V {\\textperiodcentered} A}\n\\end{equation}\n%\n\\section{Experiment}\n%\nThere are four experiments. The first two experiments involve using \\textbf{two ohmic} resistors. The other two experiments use \\textbf{two non-ohmic} resistors (light bulbs).\n%\n\\subsection{Part 1: Resistors in Series}\n%\nIn this part, you connect two \\textbf{resistors} in \\textbf{series} to a battery.\n%\n\\subsection{Part 2: Resistors in Parallel}\n%\nIn this part, you connect two \\textbf{resistors} in \\textbf{parallel} to a battery.\n%\n\\subsection{Part 3: Light Bulbs in Series}\n%\nIn this part, you connect two \\textbf{light bulbs} in \\textbf{series} to a battery.\n%\n\\subsection{Part 4: Light Bulbs in Parallel}\n%\nIn this part, you connect two \\textbf{light bulbs} in \\textbf{parallel} to a battery.\n%\n\\section{Analysis}\n%\nEach run is a measurement of either \\textbf{voltage} (electric potential difference between two points) or \\textbf{current}. If the voltage and current sensors were zeroed correctly, then the measurement over time should correspond to the measurement of the desired quantity, and no offset or baseline needs to be taken into account. One way to determine a single value is to take the \\textbf{time-average}.\n\nNote that each \\textbf{voltage measurement has three decimal figures}, and each \\textbf{current measurement has four decimal figures}. You should round your time-averaged values appropriately.\n\nIn each part there are three components: one battery and two resistors (or two light bulbs). There are \\textbf{six quantities} that describe the system:\n\\begin{itemize}\n\t\\item Time-averaged voltage across the battery: $V_{0}$\n\t\\item Time-averaged voltage across resistor/bulb 1: $V_{1}$\n\t\\item Time-averaged voltage across resistor/bulb 2: $V_{2}$\n\t\\item Time-averaged current leaving the battery: $I_{0}$\n\t\\item Time-averaged current leaving resistor/bulb 1: $I_{1}$\n\t\\item Time-averaged current leaving resistor/bulb 2: $I_{2}$\n\\end{itemize}\nThese six quantities satisfy certain mathematical relations. Note that $V_{1}$ and $V_{2}$ are measured to be negative.\n%\n\\subsection{Part 1: Two Resistors in Series}\n%\nFor the two \\textbf{resistors in series}, you find the time average for all six quantities above and fill a table similar to Table \\ref{table.03.resistors.series}.\n\nYou should check that relations Equation (\\ref{eq.03.ISeries}) and Equation (\\ref{eq.03.VSeries}) hold true. The percent difference between the labeled resistance and the prediction Equation (\\ref{eq.03.ROhmLaw}) from Ohm's law should be very small. The resistance of the battery should be very close to the \\textbf{series equivalent} resistance in Equation (\\ref{eq.03.RSeries}).\n%\n\\subsection{Part 2: Two Resistors in Parallel}\n%\nFor the two \\textbf{resistors in parallel}, you find the time average for all six quantities above and fill a table similar to Table \\ref{table.03.resistors.parallel}.\n\nYou should check that relations Equation (\\ref{eq.03.IParallel}) and Equation (\\ref{eq.03.VParallel}) hold true. The percent difference between the labeled resistance and the prediction Equation (\\ref{eq.03.ROhmLaw}) from Ohm's law should be very small. The resistance of the battery should be very close to the \\textbf{parallel equivalent} resistance in Equation (\\ref{eq.03.RParallel}).\n%\n\\subsection{Part 3: Two Light Bulbs in Series}\n%\nFor two \\textbf{light bulbs in series}, you can find the time average for all six quantities mentioned above, and fill a table similar to Table \\ref{table.03.bulbs.series}. Note that the fourth column consist of the ratio of voltage and current for each component. This is a quantity with units of resistance (ohms) but it does not correspond to the actual resistance of the light bulbs. As you will find, the light bulbs are not ohmic, and thus the ratio of the voltage across the light bulb and the current leaving the light bulb is not constant. However, you should find that the relations Equation (\\ref{eq.03.ISeries}) and Equation (\\ref{eq.03.VSeries}) again hold.\n%\n\\subsection{Part 4: Two Light Bulbs in Parallel}\n%\nFor two \\textbf{light bulbs in parallel}, you can find the time average for all six quantities mentioned above, and table similar to Table \\ref{table.03.bulbs.parallel}. Note that the fourth column consist of the ratio of voltage and current for each component. You can verify that the light bulbs are not ohmic by showing that the values in the fourth column in Tables \\ref{table.03.bulbs.series} and \\ref{table.03.bulbs.parallel} are very different. However, you should find that the relations Equation (\\ref{eq.03.IParallel}) and Equation (\\ref{eq.03.VParallel}) again hold.\n%\n\\section{My Data}\n%\nFor parts 1 and 2, I used \\textbf{three} resistors, instead of two. The resistances are given by\n\\begin{align}\n\tR_{1} = 10 \\text{ ohm} && R_{2} = 51 \\text{ ohm} && R_{3} = 68 \\text{ ohm}\n\\end{align}\nThe equivalent resistance in \\textbf{series} with three resistors is given by\n\\begin{equation}\n\tR_{\\text{eq}} = R_{1} + R_{2} + R_{3}\n\\end{equation}\nIn parallel, the equivalent resistant satisfies the following relation:\n\\begin{equation}\n\t\\frac{1}{R_{\\text{eq}}} = \\frac{1}{R_{1}} + \\frac{1}{R_{2}} + \\frac{1}{R_{3}}\n\\end{equation}\nSolving for the equivalent resistance in \\textbf{parallel} gives\n\\begin{equation}\n\tR_{\\text{eq}} = \\frac{R_{1} R_{2} R_{3}}{R_{1} R_{2} + R_{1} R_{3} + R_{2} R_{3}}\n\\end{equation}\nAs you can see from my spreadsheet, the results are in good agreement with expectations.\n%\n\\subsection{Part 1: Three Resistors in Series}\n%\nTable \\ref{table.03.3resistors.series} has the results for three resistors in series. You can check that the values for the current in the third column are all very close. Indeed,\n\\begin{equation}\n\tI_{0} \\approx I_{1} \\approx I_{2} \\approx I_{3}\n\\end{equation}\nThis is the three-resistor analog of Equation (\\ref{eq.03.ISeries}). Also, you can check that\n\\begin{equation}\n\t| V_{1} + V_{2} + V_{3} | = 3.075 \\text{ V}\n\\end{equation}\nwhich is very close to the value of $V_{0}$ (the voltage across the battery). This is the three-resistor analog of Equation (\\ref{eq.03.VSeries}).\n\nThe last column in Table \\ref{table.03.3resistors.series} contains the power generated/consumed by the corresponding component. As you can see, the rate of energy generation for the battery is very close to the sum of the rate of consumption for each resistor. Note that, in \\textbf{series}, the resistor with the \\textbf{largest} resistance consumes the most energy.\n%\n\\subsection{Part 2: Three Resistors in Parallel}\n%\nTable \\ref{table.03.3resistors.parallel} has the results for three resistors in parallel. You can check that, ignoring the signs, the values for the voltage in the second column are all very close. Indeed,\n\\begin{equation}\n\tV_{0} \\approx |V_{1}| \\approx |V_{2}| \\approx |V_{3}|\n\\end{equation}\nThis is the three-resistor analog of (\\ref{eq.03.VParallel}). Also, you can check that\n\\begin{equation}\n\tI_{1} + I_{2} + I_{3} = 0.3532 \\text{ A}\n\\end{equation}\nwhich is very close to the value of $I_{0}$ (the current leaving the battery). This is the three-resistor analog of Equation (\\ref{eq.03.IParallel}).\n\nThe last column in Table \\ref{table.03.3resistors.parallel} contains the power generated/consumed by the corresponding component. As you can see, the rate of energy generation for the battery is very close to the sum of the rate of consumption for each resistor. Note that, in \\textbf{parallel}, the resistor with the \\textbf{smallest} resistance consumes the most energy.\n\nOne last thing about resistors: proof that they \\textbf{are ohmic} (at least for the amount of voltage and current that we used) is that the values in the fourth column in Tables \\ref{table.03.3resistors.series} and \\ref{table.03.3resistors.parallel} are consistent (except for the first value, because it corresponds to the equivalent resistance). That is, it does not matter if they are connected in series or parallel, the amount of current going through the resistor is proportional to the amount of voltage across the resistor.\n%\n\\subsection{Part 3: Two Light Bulbs in Series}\n%\nThe six runs in the text file \\texttt{bulbs-serial.txt} are for two light bulbs connected in series.\n\\begin{itemize}\n\t\\item Run 1: measurement of voltage across battery ($V_{0}$).\n\t\\item Run 2: measurement of voltage across long light bulb ($V_{L}$).\n\t\\item Run 3: measurement of voltage across round light bulb ($V_{R}$).\n\t\\item Run 4: measurement of current leaving battery ($I_{0}$).\n\t\\item Run 5: measurement of current leaving long light bulb ($I_{L}$).\n\t\\item Run 6: measurement of current leaving round light bulb ($I_{R}$).\n\\end{itemize}\nThe \\textbf{long} light bulb is connected \\textbf{first}, and the \\textbf{round} light bulb is connected \\textbf{second}.\n%\n\\subsection{Part 4: Two Light Bulbs in Parallel}\n%\nThe six runs in the text file \\texttt{bulbs-parallel.txt} are for two light bulbs connected in parallel.\n\\begin{itemize}\n\t\\item Run 1: measurement of voltage across battery ($V_{0}$).\n\t\\item Run 2: measurement of voltage across long light bulb ($V_{L}$).\n\t\\item Run 3: measurement of voltage across round light bulb ($V_{R}$).\n\t\\item Run 4: measurement of current leaving battery ($I_{0}$).\n\t\\item Run 5: measurement of current leaving long light bulb ($I_{L}$).\n\t\\item Run 6: measurement of current leaving round light bulb ($I_{R}$).\n\\end{itemize}\n%\n\\section{Your Data}\n%\nIf your data for resistors in series and/or parallel is not good, let me know via email and I will make data available to you.\n%\n% \\newpage\n% \\section{Your Lab Report}\n% %\n% In your lab report you should include:\n% \\begin{enumerate}\n% \t\\item One table like Table \\ref{table.03.resistors.series} for two resistors in series.\n% \t\\item One table like Table \\ref{table.03.resistors.parallel} for two resistors in parallel.\n% \t\\item One table like Table \\ref{table.03.bulbs.series} for two light bulbs in series.\n% \t\\item One table like Table \\ref{table.03.bulbs.parallel} for two light bulbs in parallel.\n% \\end{enumerate}\n% You should also:\n% \\begin{enumerate}\n% \t\\item Verify that equations (\\ref{eq.03.ISeries}) and (\\ref{eq.03.VSeries}) hold for two resistors, and also for two light bulbs, connected in series.\n% \t\\item Verify that the voltage across the battery divided by the current leaving the battery is close to (\\ref{eq.03.RSeries}) for two resistors in series.\n% \t\\item Verify that equations (\\ref{eq.03.IParallel}) and (\\ref{eq.03.VParallel}) hold for two resistors, and also for two light bulbs, connected in parallel.\n% \t\\item Verify that the voltage across the battery divided by the current leaving the battery is close to (\\ref{eq.03.RParallel}) for two resistors in parallel.\n% \t\\item Confirm that the long light bulb is non-ohmic by comparing the ratio of the voltage and the current $V_{1} / I_{1}$ in series and in parallel.\n% \t\\item Confirm that the round light bulb is non-ohmic by comparing the ratio of the voltage and the current $V_{2} / I_{2}$ in series and in parallel.\n% \t\\item Find which resistor consumes more power when two resistors are connected in series.\n% \t\\item Find which resistor consumes more power when two resistors are connected in parallel.\n% \t\\item Find which light bulb consumes more power when two light bulbs are connected in series.\n% \t\\item Find which light bulb consumes more power when two light bulbs are connected in parallel.\n% \\end{enumerate}\n%\n\\newpage\n\\section{Tables}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r}\n \\textbf{Quantity} & \\textbf{Battery} & \\textbf{Resistor 1} & \\textbf{Resistor 2} \\\\\n \\hline\n Voltage (V) & $V_{0}$ & $V_{1}$ & $V_{2}$ \\\\\n Current (A) & $I_{0}$ & $I_{1}$ & $I_{2}$ \\\\\n \\hline\n Observed Resistance (ohm) & $V_{0} / I_{0}$ & $V_{1} / I_{1}$ & $V_{2} / I_{2}$ \\\\\n Expected Resistance (ohm) & $R_{1} + R_{2}$ & $R_{1}$ & $R_{2}$ \\\\\n Percent Difference (\\%) & & & \\\\\n \\hline\n Power (W) & $V_{0} I_{0}$ & $V_{1} I_{1}$ & $V_{2} I_{2}$ \\\\\n \\hline\n \\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and resistance for two resistors in series.}\n\t\\label{table.03.resistors.series}\n\\end{table}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r}\n \\textbf{Quantity} & \\textbf{Battery} & \\textbf{Resistor 1} & \\textbf{Resistor 2} \\\\\n \\hline\n Voltage (V) & $V_{0}$ & $V_{1}$ & $V_{2}$ \\\\\n Current (A) & $I_{0}$ & $I_{1}$ & $I_{2}$ \\\\\n \\hline\n Observed Resistance (ohm) & $V_{0} / I_{0}$ & $V_{1} / I_{1}$ & $V_{2} / I_{2}$ \\\\\n Expected Resistance (ohm) & $R_{1} R_{2} / (R_{1} + R_{2})$ & $R_{1}$ & $R_{2}$ \\\\\n Percent Difference (\\%) & & & \\\\\n \\hline\n Power (W) & $V_{0} I_{0}$ & $V_{1} I_{1}$ & $V_{2} I_{2}$ \\\\\n \\hline\n \\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and resistance for two resistors in parallel.}\n\t\\label{table.03.resistors.parallel}\n\\end{table}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r}\n \\textbf{Quantity} & \\textbf{Battery} & \\textbf{Long bulb} & \\textbf{Round bulb} \\\\\n \\hline\n Voltage (V) & $V_{0}$ & $V_{L}$ & $V_{R}$ \\\\\n Current (A) & $I_{0}$ & $I_{L}$ & $I_{R}$ \\\\\n \\hline\n Observed Resistance (ohm) & $V_{0} / I_{0}$ & $V_{L} / I_{L}$ & $V_{R} / I_{R}$ \\\\\n Power (W) & $V_{0} I_{0}$ & $V_{L} I_{L}$ & $V_{R} I_{R}$ \\\\\n \\hline\n \\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and power for two light bulbs in series.}\n\t\\label{table.03.bulbs.series}\n\\end{table}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r}\n \\textbf{Quantity} & \\textbf{Battery} & \\textbf{Long bulb} & \\textbf{Round bulb} \\\\\n \\hline\n Voltage (V) & $V_{0}$ & $V_{L}$ & $V_{R}$ \\\\\n Current (A) & $I_{0}$ & $I_{L}$ & $I_{R}$ \\\\\n \\hline\n Observed Resistance (ohm) & $V_{0} / I_{0}$ & $V_{L} / I_{L}$ & $V_{R} / I_{R}$ \\\\\n Power (W) & $V_{0} I_{0}$ & $V_{L} I_{L}$ & $V_{R} I_{R}$ \\\\\n \\hline\n \\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and power for two light bulbs in parallel.}\n\t\\label{table.03.bulbs.parallel}\n\\end{table}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r|r}\n\t\t\t\\textbf{Quantity} & \\textbf{Battery} & \\textbf{Resistor 1} & \\textbf{Resistor 2} & \\textbf{Resistor 3} \\\\\n\t\t\t\\hline\n\t\t\tVoltage (V) & 4.430 & \\textminus 0.338 & \\textminus 1.757 & \\textminus 2.345 \\\\\n\t\t\tCurrent (A) & 0.0346 & 0.0346 & 0.0347 & 0.0347 \\\\\n\t\t\t\\hline\n\t\t\tObserved Resistance (ohm) & 128.05 & 9.76 & 50.70 & 67.68 \\\\\n\t\t\tExpected Resistance (ohm) & 129 & 10 & 51 & 68 \\\\\n\t\t\tPercent Difference (\\%) & \\textminus 0.73 & \\textminus 2.36 & \\textminus 0.58 & \\textminus 0.46 \\\\\n\t\t\t\\hline\n\t\t\tPower (W) & 0.15 & \\textminus 0.01 & \\textminus 0.06 & \\textminus 0.08 \\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and resistance for three resistors in series.}\n\t\\label{table.03.3resistors.series}\n\\end{table}\n%\n\\begin{table}[ht]\n\t\\begin{center}\n\t\t\\begin{tabular}{l|r|r|r|r}\n\t\t\t\\textbf{Quantity} & \\textbf{Battery} & \\textbf{Resistor 1} & \\textbf{Resistor 2} & \\textbf{Resistor 3} \\\\\n\t\t\t\\hline\n\t\t\tVoltage (V) & 3.731 & \\textminus 3.707 & \\textminus 3.736 & \\textminus 3.734 \\\\\n\t\t\tCurrent (A) & 0.4904 & 0.3722 & 0.0739 & 0.0554 \\\\\n\t\t\t\\hline\n\t\t\tObserved Resistance (ohm) & 7.61 & 9.96 & 50.55 & 67.37 \\\\\n\t\t\tExpected Resistance (ohm) & 7.45 & 10 & 51 & 68 \\\\\n\t\t\tPercent Difference (\\%) & 2.20 & \\textminus 0.39 & \\textminus 0.88 & \\textminus 0.92 \\\\\n\t\t\t\\hline\n\t\t\tPower (W) & 1.83 & \\textminus 1.38 & \\textminus 0.28 & \\textminus 0.21 \\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\\end{center}\n\t\\caption{Time-averages and resistance for three resistors in parallel.}\n\t\\label{table.03.3resistors.parallel}\n\\end{table}\n%\n\\FloatBarrier\n\\newpage\n\\section{Figures}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/03-serial-parallel/resistor-V.pdf}\n\t\\caption{Voltage across ohmic resistor}\n\t\\label{figure.03.resistor.v}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/03-serial-parallel/resistor-I.pdf}\n\t\\caption{Current through ohmic resistor}\n\t\\label{figure.03.resistor.i}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/03-serial-parallel/bulb-V.pdf}\n\t\\caption{Voltage across non-ohmic light bulb}\n\t\\label{figure.03.bulb.v}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/03-serial-parallel/bulb-I.pdf}\n\t\\caption{Current through non-ohmic light bulb}\n\t\\label{figure.03.bulb.i}\n\\end{figure}", "meta": {"hexsha": "498f7d13065ee293c8a78343062c88b603e294bf", "size": 20846, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter/03-serial-parallel.tex", "max_stars_repo_name": "meirizarrygelpi/phys-208L", "max_stars_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter/03-serial-parallel.tex", "max_issues_repo_name": "meirizarrygelpi/phys-208L", "max_issues_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapter/03-serial-parallel.tex", "max_forks_repo_name": "meirizarrygelpi/phys-208L", "max_forks_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 54.4281984334, "max_line_length": 670, "alphanum_fraction": 0.7037321309, "num_tokens": 6412, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624890918021, "lm_q2_score": 0.8031738034238806, "lm_q1q2_score": 0.6286140081610642}} {"text": "\\subsection{Theoretical Results}\n\n\\hogwild's original article \\cite{2011NRRW} manages to prove that the\nasynchronous noise generated by \\hogwild, under certain sparsity assumptions of\nthe $f_i$'s, converged at the same rate as seen in the stochastic gradient\nmethod. However, a newer article from 2015 by De Sa et al. \\cite{2015SZOR}\npresented a new framework in the context of martingale theory, which drops said\nsparsity assumptions and generalizes to certain non-convex formualations. Given\nthat I just learned about martingale theory essentially two weeks ago in Basic\nProbability, and that much of my synthetic tests are on dense datasets, I feel\ncompelled to present this argument instead. \n\nThe following is a cleaned up summary of the arguments presented in the article,\nexcept I focus on just demonstrating how martingale theory can be used to\ngeneralize convergence arguments for a sequential stochastic gradient method to\nthe asynchronous case, and cut out the generalization required for the\nnon-convex situation.\n\n\\subsubsection{Initial Machinery}\n\nFirst, recall the definition of a martingale:\n\\begin{definition} \\label{def:martingale}\n \\cite{2004JP} A sequence of random variables $(X_n)_{n \\geq 0}$ is called\n a martingale, or an $(\\mathcal{F}_n)$-martingale, if\n \\begin{enumerate}[(i)]\n \\item $\\E{|X_n|} < \\infty, \\forall n$.\n \\item $X_n$ is $\\mathcal{F}_n$ measurable, $\\forall n$.\n \\item $\\E{X_n \\mid \\mathcal{F}_m} = X_m$ a.s., $\\forall m \\leq n$.\n \\end{enumerate}\n furthermore a supermartingale (submartingale) is one satisfying (i), (ii)\n exactly, and (iii) with $\\leq\\ (\\geq)$ instead.\n\\end{definition}\nUsing this, the idea is to model our convergence with a non-negative\nsupermartingale $W_t(x_t, \\dots, x_0)$ which is a function of the previous\nstochastic gradient iterates. These $W_t$, when used in the theory later, will\nbe associated with specific stochastic algorithms, as an example below we'll see\nit applied to serial stochastic gradient. However, given such a supermartingale,\nand given a bounded stopping time $B$ (in literature known as a horizon), if our\nstochastic gradient iterates are written as $x_{t+1} = x_t - \\eta \\nabla\nf_t(x_t)$, where $\\eta$ is the learning rate, and $f_t$ denotes the random\nfunction chosen at time step $t$, then we see that condition (iii) in the above\ndefinition implies that:\n\\[\n W_{t+1}(x_t - \\eta \\nabla f_t(x_t), x_t, \\dots, x_0)\n \\leq\n W_t(x_t, \\dots, x_0), \\forall t \\leq B\n\\]\nwhich certainly makes sense if our $-\\nabla f_t(x_t)$ is a sufficient search\ndirection. Furthermore, letting our success region be denoted as $S\n= B_\\e(x^*)$, where $x^*$ is the minimizer of our optimization problem, if we\nimpose that if $x_t \\not\\in S, \\forall t \\leq T$ then:\n\\[\n W_T(x_T, \\dots, x_0) \\geq T\n\\]\nthen we call $W_t$ a {\\it rate supermartingale}. To simplify notation, let $F_t$\nbe the event where $\\nexists t \\leq T$ such that $x_t \\in S$. \n\nA good example of the power of this machinery is proving a convergence bound on\nthe serial version of stochastic gradient: using (iii) of definition\n\\ref{def:martingale}, one can see that considering $F_T$:\n\\[\n \\E{W_0(x_0)}\n \\underbrace{\\geq}\n \\E{W_T}\n =\n \\Prob{F_t}\n \\underbrace{\\Prob{F_T}\\E{W_T | F_T}}_{W_T \\geq T} + \n \\underbrace{\\Prob{F_T^c}\\E{W_T | F_T^c}}_{W_T \\geq 0}\n \\geq\n \\Prob{F_T}T\n\\]\nwhere the first inequality is by Doob's optional sampling theorem. Thus for\na simple serial stochastic gradient method: $\\Prob{F_t} = \\E{W_0} / T$.\n\nNow that we can characterize serial stochastic gradient in this model, we need\na method to analyze asynchronous noise. Recall that since we've guaranteed in\nthe description of \\hogwild\\ that writes to the iteration variable $x_t$ are\ndone atomically, the only race condition possible is when updates to entries of\n$x_t$ are interleaved with either the other thread's updates or its reads on\n$x_t$.\n\nIndeed, when going to update the $i$th component of $x_{t+1}$, the variable used\nto compute the gradient may have long since changed, making our iteration look\nmore like $x_{t+1} = x_t - \\nabla f_t(v_t)$ where $v_t$'s entries were the\nentries of some previous iterate $x$. Let $\\tau_{i,t}$ denote the lag for the\nupdate of the $i$th component of $x_{t+1}$, i.e. $(v_t)_i = ( x_{t-\\tau_{i,t}}\n)_i$. Then we recognize that this lag for each component $i$ (supposing the\ncomputer hasn't crashed) must be bounded; let $\\tau'$ be the the maximum over\n$i$ of such bounds and let $\\tau = \\E{\\tau'}$. This is known in literature as\nthe {\\it worst-case expected delay}.\n\nFinally, we need one last definition, mainly one of convinience, to proceed\nonward. This describes the main conditions upon a rate supermartingale necessary\nto prove that the asynchronous noise error is irrelevant to the convergence\nrate.\n\\begin{definition}\n An algorithm with associated rate supermartingale $W$ is $(H, R, \\xi)$-bounded\n if the following conditions hold.\n \\begin{enumerate}[(1)]\n \\item $W$ must be Lipschitz continuous in the current iterate with parameter\n $H$, i.e.\n \\[\n \\norm{W_t(u, x_{t-1}, \\dots, x_0) - W_t(v, x_{t-1}, \\dots, x_0)}\n \\leq\n H\\norm{u-v}, \\forall t, u, v, x_t, \\dots, x_0.\n \\]\n \\item $\\nabla f$ must be Lipschitz continuous in expectation with parameter\n $R$, i.e.\n \\[\n \\E{||\\nabla f(x) - \\nabla f(y)||} \\leq R\\norm{u-v}\n \\]\n \\item The expected magnitude of the update must be bounded by $\\xi$, i.e.\n \\[\n \\E{||\\nabla f(x)||} \\leq \\xi\n \\]\n \\end{enumerate}\n Note that these look very familiar to the conditions in the stochastic\n gradient method convergence theorem (theorem \\ref{thm:sgd}).\n\\end{definition}\n\n\\subsubsection{Convergence of \\hogwild}\n\nFinally, now that all of the machinery has been defined, we can get to the main\nresult:\n\\begin{theorem} \\label{thm:convergence}\n Suppose we have an asynchronous stochastic algorithm with associated rate\n supermartingale $W$ which is $(H,R,\\xi)$ bounded with horizon $B$.\n Furthermore, assume that $HR\\xi \\tau < 1$; then $\\forall T \\leq B$:\n \\[\n \\Prob{F_T} \\leq \\frac{\\E{W(0, x_0)}}{(1-HR\\xi\\tau)T}\n \\]\n\\end{theorem}\nBefore we prove it, we briefly discuss how to use this theorem in practice.\nSuppose we have a loss function $f$ which we want to use \\hogwild\\ on to\nminimize. Then first we need to obtain a rate supermartingale proving the\nproblem%\n\\footnote{\n In the article, De Sa et al. describes this as being no more difficult than\n proving serial convergence, but the proof (in the convex case) doesn't seem to\n have any similarity between it and serial convergence, so I can't validate\n this claim.\n}, determine $H, R, \\xi$ such that $W$ is $(H,R,\\xi)$-bounded, and then apply\nthis theorem to get a proper rate of convergence. This is a very powerful\ntheorem, as we'll be able to state later, given strongly convex $f$ and with the\nother required bounds, we can blanket prove that \\hogwild\\ works on them.\nFurthermore, for nicer non-convex problems, such as the low-rank least-squared\nmatrix completion problem presented in the paper, it's easy to derive a proper\nrate supermartingale as well.\n\nNow, with respect to the proof, we only outline it because it's mainly just\na repeated application of the above bounds we have in order to lower bound away\nnoise terms, and then it follows the path outlined in the serial stochastic\ngradient method proof.\n\\begin{proof}\n \\begin{enumerate}[(i)]\n \\item With $W$ defined exactly as in the serial case, it's not a rate\n supermartingale. Instead, from it we construct rate supermartingale $V_t$,\n where $\\forall t, x$ where $x_u$ not converged $\\forall u < t$:\n \\[\n V_t(x_t, \\dots, x_0) = W_t(x_t, \\dots, x_0) \n - \\underbrace{HR\\xi \\tau t}_{(1)}\n + \\underbrace{\n HR \\sum_{k=1}^\\infty \\norm{x_{t-k+1}-x_{t-k}} \\sum_{m=k}^\\infty\n \\Prob{\\tau' \\geq m}\n }_{(2)}\n \\]\n where (1) allows for longer iteration counts (as HOGWILD needs to allow\n for given noise corruption), and (2) measures distance between recent\n iterates. Otherwise if $x_u$ is converged, then we let $V_t(x_t, \\dots,\n x_0) = V_u(x_u, \\dots, x_0)$. Basically, we've defined $V$ a stopped\n process.\n \\item Show $V_t$ is a rate supermartingale for \\hogwild\n \\item Using a similar process to the serial stochastic gradient proof, show\n that $\\E{V_T} \\leq \\E{V_0} = \\E{W_0}$, then using the law of total\n expectation on this with $F_T$, and recalling that $\\E{W_T | F_T} \\geq T$,\n we recieve the desired result.\n \\end{enumerate}\n\\end{proof}\nGiven this, the general theorem for the convex case is:\n\\begin{theorem} \\label{thm:convexconv}\n Consider trying to minimize $f$, which is:\n \\begin{itemize}\n \\item Strongly convex with parameter $c$.\n \\item $\\nabla f_k$ Continuously differentiable in $||\\cdot||_1$ with\n Lipschitz constant $L$.\n \\item Upper bounded second moment of the gradient by $M^2$\n \\item Success criteria $\\norm{x-x^*}^2 \\leq \\e$, for some $\\e > 0$.\n \\end{itemize}\n Then we can construct rate supermartingale $W_t$ such that it's $(H,R,\\xi)$\n bounded with $H = 2\\sqrt{\\e}(2\\eta c\\e - \\eta^2 M^2)^{-1}, R = \\eta L,\n \\xi = \\eta M$. Then choosing step size $\\eta$ (for some $\\nu \\in (0,1)$):\n \\[\n \\eta = \\frac{c\\e \\nu}{M^2 + 2LM\\tau\\sqrt{\\e}}\n \\]\n then we recieve:\n \\[\n \\Prob{F_T} \\leq \\frac{M^2 + 2LM\\tau\\sqrt{\\e}}{c^2\\e\\nu T} \\log\n \\left(\n e \\norm{x_0 -x^*}^2 \\e^{-1}\n \\right)\n \\]\n\\end{theorem}\n", "meta": {"hexsha": "39483b6eb998cc37bbd7d823ca700e986536932e", "size": 9503, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/TeXsrc/src/convergence/theory.tex", "max_stars_repo_name": "abhijit-c/HOGWILD", "max_stars_repo_head_hexsha": "1ae85888fb5c33f0cf01043064d30106e7a3de39", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Report/TeXsrc/src/convergence/theory.tex", "max_issues_repo_name": "abhijit-c/HOGWILD", "max_issues_repo_head_hexsha": "1ae85888fb5c33f0cf01043064d30106e7a3de39", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/TeXsrc/src/convergence/theory.tex", "max_forks_repo_name": "abhijit-c/HOGWILD", "max_forks_repo_head_hexsha": "1ae85888fb5c33f0cf01043064d30106e7a3de39", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.8128078818, "max_line_length": 80, "alphanum_fraction": 0.7013574661, "num_tokens": 2884, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569016, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6286139985583695}} {"text": "\\chapter{Calculation Objects} \\label{Ch:CalculationObjects}\r\n\r\nGMAT has the ability to calculate numerous quantities that are\r\ndependent upon the states of objects, coordinate systems, and the\r\nmission sequence. These calculation objects can range from the\r\nspacecraft state, to the local atmospheric density, to the\r\npositions of celestial bodies with respect to spacecraft, or other\r\ncelestial bodies. In chapter, we present how GMAT performs these\r\ncalculations by showing the mathematical algorithms.\r\n\r\nThe chapter begins by discussing different orbit state\r\nrepresentations. Each of the orbit state representations\r\navailable in GMAT are defined. Next we present the algorithms\r\nused to convert between different state representations. These\r\ninclude the Keplerian elements, modified Keplerian elements,\r\nCartesian state, spherical state, and the equinoctial elements. In\r\nthe second section we present how GMAT calculates all calculation\r\nparameters. Examples include the orbit period, percent shadow, and\r\nenergy. The algorithms to calculate all parameters are included\r\nand described in detail. We conclude this chapter with a\r\npresentation of the algorithms used to calculate libration point\r\nand barycenter locations.\r\n\r\n\\section{Spacecraft State Representations} \\index{Orbit state\r\nrepresentations}\r\n\r\nThere are several state representations that can be used in GMAT\r\nto define the state of a spacecraft object. These include the\r\nKeplerian elements, Cartesian state, equinoctial elements,\r\nspherical elements, and the modified Keplerian elements. In the\r\nfollowing subsections, we discuss the definitions of these states\r\ntypes, and show how GMAT converts between the different state\r\nrepresentations.\r\n\r\n\\subsection{Definitions}\r\n\r\nThe Keplerian elements are one of the most commonly used state\r\nrepresentations. They provide a way to define the spacecraft\r\nstate in way that provides an intuitive understanding of the\r\nmotion of spacecraft in orbit. The Keplerian elements are denoted\r\n$a$, $e$, $i$, $\\omega$, $\\Omega$, and $\\nu$. They are defined in\r\ndetail in Table \\ref{Table:KeplerianElements} and illustrated in\r\nFig. \\ref{fig:KeplerianElements}. Sections \\ref{Sec:Cart2Kep} and\r\n\\ref{Sec:Kep2Cart} show the algorithms that GMAT uses to convert\r\nbetween the Keplerian elements and the cartesian state.\r\n\\index{Keplerian elements!definition}\r\n%\r\n\r\n\\begin{figure*}[htb]\r\n\\index{Keplerian elements} \\centerline{\r\n\\begin{picture}(100,545)\r\n\\special{psfile=Images/OrbitElements.eps hoffset= -195 voffset= 75\r\nhscale=80\r\nvscale=80}\\makebox(-78,600){$\\hat{\\mathbf{x}}_I$,$\\Upsilon$}\r\n\\makebox(40,610){$\\hat{\\mathbf{x}}_{ep}$} \\makebox(10,685){$\\Omega$}\r\n\\makebox(215,795){$\\omega$} \\makebox(-120,840){$y_I$}\r\n\\makebox(-135,968){$\\mathbf{x}_p$}\\makebox(-255,980){$\\nu$}\r\n\\makebox(-485,1083){$z_I$} \\makebox(-540,1028){$i$}\r\n \\makebox(-610,1060){$\\mathbf{h}$, $\\hat{\\mathbf{z}}_{ep}$}\r\n \\makebox(-340,610){$\\mathbf{N}$}\r\n\\end{picture}}\\vskip -3.95 in \\caption{ The Keplerian Elements } \\label{fig:KeplerianElements}\r\n\\end{figure*}\r\n\r\nThe cartesian state is another common state representation and is\r\noften used in the numerical integration of the equations of\r\nmotion. The cartesian state with respect to a given coordinate\r\nsystem is described in detail in Table\r\n\\ref{Table:CartesianStates}. \\index{Cartesian state!definition}\r\n\r\n\r\nThe equinoctial elements are a set of non-singular elements that\r\ncan be used to describe the state of a spacecraft. Because they\r\nare nonsingular, they are useful for expressing the equations of\r\nmotion in Variation of Parameters (VOP) form. The elements can be\r\nunintuitive to use however. The equinoctial elements are\r\ndescribed in detail in Table \\ref{Table:EquinoctialElements}.\r\n%\r\n\r\n\\begin{figure*}[htb]\r\n\\index{Spherical elements} \\centerline{\r\n\\begin{picture}(100,420)\r\n\\special{psfile=Images/SphericalElements.eps hoffset= -100 voffset=\r\n75\r\nhscale=55 vscale=55} \\makebox(-280,360){$\\mathbf{x}$}%\r\n\\makebox(140,590){$\\mathbf{y}$}%\r\n\\makebox(-435,820){$\\mathbf{z}$}%\r\n\\makebox(-410,530){$\\lambda$}%\r\n\\makebox(-273,636){$\\delta$}%\r\n\\makebox(-330,695){$\\mathbf{r}$}%\r\n\\makebox(-250,770){$\\mathbf{v}$}%\r\n\\makebox(90,710){$\\mathbf{v}$}%\r\n\\makebox(10,675){$\\hat{\\mathbf{r}}$}%\r\n\\makebox(-350,650){North}%\r\n\\makebox(-275,635){$\\psi$}%\r\n\\makebox(-180,597){$\\alpha_f$}%\r\n\\end{picture}}\\vskip -2.5 in \\caption{ The Spherical Elements } \\label{fig:SphericalElements}\r\n\\end{figure*}\r\n%\r\n\r\nThe modified Keplerian elements are similar to the Keplerian\r\nelements except $a$ and $e$ are replaced with the radius of\r\nperiapsis $r_p$, and the radius of apoapsis $r_a$. $r_p$ and $r_a$\r\nare often more convenient and intuitive for describing the\r\ndimensions of a Keplerian orbit than $a$ and $e$. The modified\r\nKeplerian elements are defined in detail in Table\r\n\\ref{Table:ModKeplerianElements}. Note that both the Keplerian and\r\nmodified Keplerian elements are undefined for parabolic orbits\r\nbecause the semimajor axis is infinite. Currently, GMAT does not\r\nsupport parabolic orbits for this reason. Let's begin by looking\r\nat how GMAT converts from the Cartesian state to the Keplerian\r\nelements.\r\n\r\n\\index{Modified Keplerian elements!definition}\r\n%\r\n\r\n\\subsection{Cartesian State to Keplerian Elements}\\label{Sec:Cart2Kep} \\index{Cartesian state!to\r\nKeplerian} \\index{Keplerian elements!from cartesian}\r\n\r\nThe conversion from the Cartesian state to the Keplerian elements\r\nhas four special cases: elliptic inclined, circular inclined,\r\nelliptic equatorial, and circular equatorial. Certain orbital\r\nelements are undefined for some of the cases. For example, the\r\nright ascension of the ascending node, $\\Omega$, is undefined for\r\nequatorial orbits. However, computer systems don't handle undefined\r\nvalues gracefully. In this section, we'll see how the orbital\r\nelements are defined for each of the special cases, and how GMAT\r\ncalculates the orbital elements for each case\\cite{Vallado:02}.\r\n\r\n%\r\n\\begin{table*} \\caption{The Cartesian State}\r\n\\centering \\index{Cartesian state!definition}\r\n\\begin{tabular}{p{.5 in} p{2.0 in}}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Symbol & Description \\\\\r\n \\hline\r\n $x$ & $x$-component of position\\\\\r\n%\r\n $y$ & $y$-component of position \\\\\r\n %\r\n $z$ & $z$-component of position \\\\\r\n %\r\n $\\dot{x}$ & $x$-component of velocity\\\\\r\n %\r\n $\\dot{y}$ & $y$-component of velocity\\\\\r\n %\r\n $\\dot{z}$ & $z$-component of velocity \\\\\r\n \\hline\\hline \\label{Table:CartesianStates}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\n\\begin{table*} \\caption{The Keplerian Elements (also see Fig. \\ref{fig:KeplerianElements})}\r\n\\centering \\index{Keplerian elements!definition} (See Table\r\n\\ref{Table:ElementSpecialCases} for definitions of elements for near\r\ncircular and near equatorial orbits.)\r\n\\begin{tabular}{p{.5 in} p{1.5 in} p{3.5 in}}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Symbol & Name & Description \\\\\r\n \\hline\r\n $a$ & semimajor axis & The semimajor contains information on the type and size of an orbit.\r\n If $a>0$ the orbit is elliptic. If $a < 0$ the orbit is hyperbolic. $a = \\infty$ for parabolic orbits. \\\\\r\n %\r\n $e$ & eccentricity & The eccentricity contains information on the shape of an orbit. If $e = 0$, then the orbit is circular. If $0 < e < 1$ the orbit is\r\n elliptical. If $e = 1$ the orbit is parabolic. If $e > 1$ then the orbit is hyperbolic. \\\\\r\n %\r\n $i$ & inclination & The inclination is the angle between the $\\hat{\\mathbf{z}}_I$ axis and the orbit normal direction $\\mathbf{h}$. If $i \\leq 90^{\\circ}$ then the orbit is prograde.\r\n If $i > 90^{\\circ}$ then the orbit is retrograde. \\\\\r\n %\r\n $\\omega$ & argument of periapsis & The argument of periapsis is the angle between a vector pointing at periapsis and a vector pointing in the direction of the line of nodes. The argument of periapsis is undefined for circular orbits.\\\\\r\n %\r\n $\\Omega$ & right ascension of the ascending node& $\\Omega$ is defined as the angle between $\\hat{\\mathbf{x}}_I$ and $\\mathbf{N}$ measured\r\n counterclockwise.\r\n $\\mathbf{N}$ is defined as the vector pointing from the center of the central body to the spacecraft, when the spacecraft crosses the bodies equatorial plane\r\n from the southern to the northern hemisphere. $\\Omega$ is undefined for equatorial orbits. \\\\\r\n %\r\n $\\nu$ & true anomaly & The true anomaly is defined as the angle between a vector pointing at periapsis and a vector pointing at the spacecraft.\r\n The true anomaly is undefined for circular orbits. \\\\\r\n \\hline\\hline \\label{Table:KeplerianElements}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\n%\r\n\\begin{table*} \\caption{Keplerian Elements for Special Cases}\r\n \\centering \\index{Keplerian elements!special cases}\r\n \\label{Table:ElementSpecialCases}\r\n\\begin{tabular}{p{1.25 in} p{1.35 in} p{3.2 in} }\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Orbit Type & Numerical Threshold & Description \\\\\r\n \\hline \\\\\r\n %--------------\r\n Elliptic Inclined & $e \\geq 10^{-11}$, $i \\geq 10^{-11}$ & $\\Omega$ is the angle between the $x$-axis and the line of nodes. $\\omega$ is the angle between the line of nodes and the eccentricity vector, $\\nu$ is\r\n the angle between the eccentricity vector and the spacecraft position vector.\\\\\r\n %--------------\r\n Elliptic Equatorial & $e \\geq 10^{-11}$, $i < 10^{-11}$ & $\\Omega = 0$, $\\omega$ is the angle between the $x$-axis and the eccentricity vector, $\\nu$ is\r\n the angle between the eccentricity vector and the spacecraft position vector.\\\\\r\n %--------------\r\n Circular Inclined & $e < 10^{-11}$, $i \\geq 10^{-11}$ & $\\Omega$ is the angle between the $x$-axis and the line of nodes, $\\omega = 0$, $\\nu$ is\r\n the angle between the line of nodes and the spacecraft position vector.\\\\\r\n %--------------\r\n Circular Equatorial & $e < 10^{-11}$, $i < 10^{-11}$ & $\\Omega = 0$, $\\omega = 0$, $\\nu$ is the angle between the $x$-axis and the spacecraft position vector. \\\\\r\n \\hline\\hline \\label{Table:KeplerianElementsSpecialCases}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$\\\\\r\n\r\n\\noindent Find: $a$, $e$, $i$, $\\omega$, $\\Omega$, and $\\nu$\r\n\r\nWe begin by calculating the specific angular momentum and its\r\nmagnitude.\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\|\\mathbf{h} \\|\r\n\\end{equation}\r\n%\r\nA vector pointing in the direction of the line of nodes is\r\n%\r\n\\begin{equation}\r\n \\mathbf{n} = [ \\hspace{0.05 in} 0 \\hspace{0.1 in} 0 \\hspace{0.1 in} 1 \\hspace{0.05\r\n in}]^T \\times \\mathbf{h}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n n = \\| n\\|\r\n\\end{equation}\r\n%\r\nThe orbit eccentricity and energy are calculated using\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{e} = \\displaystyle\\frac{(v^2 - \\displaystyle\\frac{\\mu}{r} )\\mathbf{r} - (\\mathbf{r}\\cdot\\mathbf{v} )\\mathbf{v}}{\\mu}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e = \\| \\mathbf{e} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\xi = \\frac{v^2}{2} - \\frac{\\mu}{r}\r\n\\end{equation}\r\n%\r\nFor parabolic orbits, the semimajor axis is infinite\r\nand the energy is zero. Here we check to see if the orbit is near\r\nparabolic. If $|1 - e| < 10^{-7}$ an error message is reported\r\nand conversion is aborted.\r\n\r\nThe semimajor axis is computed using\r\n%\r\n\\begin{equation}\r\n a = -\\frac{\\mu}{2\\xi}\r\n\\end{equation}\r\n%\r\nHere we check to see if the conic section is nearly singular.\r\nIf $|a(1-e)| < .001 (km)$ then an error message is thrown and conversion is aborted.\r\n\r\nIf the above tests pass, then we continue and calculate the inclination.\r\n%\r\n\\begin{equation}\r\n i = \\cos^{-1}\\left(\\frac{h_z}{h}\\right)\r\n\\end{equation}\r\n\r\nThere are four special cases for $\\Omega$, $\\omega$, and $\\nu$ and\r\neach case is treated differently.\r\n\r\n\\noindent\\textit{Special Case 1: Non-circular, Inclined Orbit }\r\n\r\n\\noindent if $(e \\geq 10^{-11})$ and $(i \\geq 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = \\cos^{-1}\\left(\\frac{n_x}{n}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\Omega$: if $n_y < 0$, then $\\Omega = 2\\pi -\r\n\\Omega$\r\n%\r\n\\begin{equation}\r\n \\omega = \\cos^{-1}\\left( \\frac{\\mathbf{n}\\cdot\\mathbf{e}}{ne}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\omega $: if $e_z < 0$, then $\\omega = 2\\pi -\r\n\\omega$\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{\\mathbf{e}\\cdot\\mathbf{r}}{er}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu $: if $\\mathbf{r} \\cdot \\mathbf{v} < 0$, then\r\n$\\nu = 2\\pi - \\nu$\r\n\r\n\\noindent\\textit{Special Case 2: Non-circular, Equatorial Orbit\r\n}\r\n\r\n\\noindent if $(e \\geq 10^{-11})$ and $(i < 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = 0\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\omega = \\cos^{-1}{\\frac{e_x}{e}}\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\omega $: if $e_y < 0$, then $\\omega = 2\\pi -\r\n\\omega$\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{\\mathbf{e}\\cdot\\mathbf{r}}{er}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu $: if $\\mathbf{r} \\cdot \\mathbf{v} < 0$, then\r\n$\\nu = 2\\pi - \\nu$\r\n%\r\n\r\n\\noindent\\textit{Special Case 3: Circular, Inclined Orbit }\r\n\r\n\\noindent if $(e < 10^{-11})$ and $(i \\geq 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = \\cos^{-1}\\left(\\frac{n_x}{n}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\Omega$: if $n_y < 0$, then $\\Omega = 2\\pi -\r\n\\Omega$\r\n\\begin{equation}\r\n \\omega = 0\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{\\mathbf{n}\\cdot\\mathbf{r}}{nr}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu$: if $r_z < 0$, then $\\nu = 2\\pi - \\nu$\r\n\r\n\\noindent\\textit{Special Case 4: Circular, Equatorial Orbit }\r\n\r\n\\noindent if $(e < 10^{-11})$ and $(i < 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = 0\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\omega = 0\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{r_x}{r}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu$: if $r_y < 0$, then $\\nu = 2\\pi - \\nu$\r\n%\r\n\r\nIn the next section, we look at how to perform the inverse\r\ntransformation and convert from Keplerian elements to the\r\nCartesian state vector.\r\n\r\n\\subsection{Keplerian Elements to Cartesian \\\\ State}\r\n\r\n\r\n\\label{Sec:Kep2Cart} \\index{Keplerian elements!to\r\ncartesian} \\index{Cartesian state!from Keplerian}\r\n\r\nThe transformation from the Keplerian elements to the Cartesian\r\nstate is one of the most common state transformations in\r\nastrodynamics. We previously defined both state types and refer you\r\nto Tables \\ref{Table:CartesianStates} and\r\n\\ref{Table:KeplerianElements} for their definitions. Below we show\r\nthe algorithm that GMAT uses to convert from the Keplerian elements\r\nto the Cartesian state\\cite{Vallado:02}.\r\n\r\n\\noindent Give: $a$, $e$, $i$, $\\Omega$, $\\omega$, $\\nu$, and\r\n$\\mu$\r\n\r\n\\noindent Find: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\nFirst check to ensure the keplerian elements are not singular. If $|a(1-e)| < .001 (km)$ then the following error message is reported and conversion is aborted: ``Warning: A nearly singular conic section was encountered while converting from the Keplerian elements to the Cartesian state. The radius of periapsis must be greater than 1 meter.\"\r\n\r\nNext check that the magnitude of the position vector is\r\nnot infinite. If ($1+e\\cos{\\nu}$ $<$ 1e-30) then the following error message is reported and conversion is aborted:\r\n``Warning: A near infinite radius was encountered while converting from the Keplerian elements to the Cartesian state. \"\r\n\r\nCheck that the orbit is not parabolic, in which case, $p$ is undefined. If ($|1-e|< 1e-7$ then the following error message is reported and conversion is aborted:\r\n``Warning: A nearly parabolic orbit was encountered while converting from the Keplerian elements to the Cartesian state. The Keplerian elements are undefined for a parabolic orbit.\".\r\n\r\nFinally, define $\\nu_M$ as the true anomaly placed between $-\\pi$ and $\\pi$. If $e > 1$ and $\\|\\nu_M\\| \\geq \\pi - acos(1/e)$ the following error message is reported and conversion is aborted:\\\\\r\n``Error: The TA value is not physically possible for a hyperbolic orbit with the input values of SMA and ECC.\\\\\r\n The allowed values are: $-limitTA < TA < limitTA$ (degrees).\r\n or equivalently\\\\\r\n The allowed values are $TA < limitTA$ or $TA > 360 - limitTA$.\r\n\" where the text ``limitTA\"' is replaced by the numeric value computed below in degrees:\r\n%\r\n\\begin{equation}\r\n limitTA = \\pi - acos(1/e)\r\n\\end{equation}\r\n%\r\n\r\nIf the previous tests pass, we continue by\r\ncalculating the semilatus rectum, and the radius.\r\n%\r\n\\begin{equation}\r\n p = a(1-e^2);\r\n\\end{equation}\r\n%\r\nIf $1+e\\cos{\\nu} < 1e-10$, then the following warning is displayed but computation proceeds. ``Warning: The orbital radius is large in the conversion from Keplerian to Cartesian state and the state may be near a singularity causing numerical errors in the conversion.\"\r\n%\r\n\\begin{equation}\r\n r = \\frac{p}{1+e\\cos{\\nu}}\r\n\\end{equation}\r\n%\r\nThe position components of the cartesian state vector are\r\ncalculated using the following three equations.\r\n%\r\n\\begin{equation}\r\n x = r\\left(\\cos{(\\omega + \\nu)}\\cos{\\Omega} - \\cos{i}\\sin{(\\omega + \\nu)}\\sin{\\Omega}\\right)\r\n\\end{equation}\r\n\\begin{equation}\r\n y = r\\left(\\cos{(\\omega + \\nu)}\\sin{\\Omega} + \\cos{i}\\sin{(\\omega + \\nu)}\\cos{\\Omega}\\right)\r\n\\end{equation}\r\n\\begin{equation}\r\n z = r\\left(\\sin{(\\omega + \\nu)}\\sin{i}\\right)\r\n\\end{equation}\r\n%\r\n\r\nBefore calculating the velocity components we check to ensure the\r\norbit is not parabolic. This avoids another possible division by\r\nzero.\r\n\r\n\\noindent if ($\\| p\\| < 1e-30$), then error and return: ``Warning:\r\nGMAT does not support parabolic orbits in conversion from\r\nkeplerian to cartesian elements\".\r\n\r\n\\noindent If the orbit is not parabolic, we continue and calculate\r\nthe velocity components using\r\n%\r\n\\begin{equation}\\begin{split}\r\n \\dot{x} = \\sqrt{\\frac{\\mu}{p}} & \\left( \\cos{\\nu} + e \\right)\r\n \\left( -\\sin{\\omega}\\cos{\\Omega} - \\cos{i}\\sin{\\Omega}\\cos{\\omega}\r\n \\right)- \\\\ &\\sqrt{\\frac{\\mu}{p}} \\sin{\\nu}\\left(\\cos{\\omega}\\cos{\\Omega} - \\cos{i}\\sin{\\Omega}\\sin{\\omega} \\right)\r\n \\end{split}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\\begin{split}\r\n \\dot{y} = \\sqrt{\\frac{\\mu}{p}}& \\left( \\cos{\\nu} + e \\right)\r\n \\left( -\\sin{\\omega}\\sin{\\Omega} + \\cos{i}\\cos{\\Omega}\\cos{\\omega}\r\n \\right)- \\\\\\ & \\sqrt{\\frac{\\mu}{p}}\\sin{\\nu}\\left(\\cos{\\omega}\\sin{\\Omega} + \\cos{i}\\cos{\\Omega}\\sin{\\omega} \\right)\r\n \\end{split}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\dot{z} = \\sqrt{\\frac{\\mu}{p}} \\left[ \\left( \\cos{\\nu} + e \\right)\\sin{i}\\cos{w} - \\sin{\\nu}\\sin{i}\\sin{\\omega} \\right]\r\n\\end{equation}\r\n\r\nNow let's look at how to calculate the cartesian state given the\r\nequinoctial elements.\r\n\r\n\\subsection{Equinoctial Elements to Cartesian State} \\index{Equinoctial elements!to\r\ncartesian}\\index{Cartesian state!from Equinotial}\r\n\\label{Sec:EquinoctialtoCartesian}\r\n\r\nThe equinoctial elements used in GMAT are defined in Table\r\n\\ref{Table:EquinoctialElements}. The algorithm to convert from\r\nequinoctial elements to the cartesian state was taken from the\r\nGTDS Mathematical Theory \\cite{GTDS}.\r\n\r\n\\begin{table*} \\caption{The Equinoctial Elements}\r\n\\centering \\index{Equinoctial elements!definition}\r\n\\begin{tabular}{p{.5 in} p{5.0 in}}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Symbol & Description \\\\\r\n \\hline\r\n $a$ & The semimajor contains information on the type and size of an orbit. If $a>0$ the orbit is elliptic. If $a < 0$ the orbit is hyperbolic. \\\\\r\n%\r\n $h$ & The projection of the eccentricity vector onto the $\\hat{\\mathbf{y}}_{ep}$ axis. \\\\\r\n %\r\n $k$ & The projection of the eccentricity vector onto the $\\hat{\\mathbf{x}}_{ep}$ axis. \\\\\r\n %\r\n $p$ & The projection of $\\mathbf{N}$ onto the $\\hat{\\mathbf{y}}_{ep}$ axis.\\\\\r\n %\r\n $q$ & The projection of $\\mathbf{N}$ onto the $\\hat{\\mathbf{x}}_{ep}$ axis. \\\\\r\n %\r\n $\\lambda$ & The mean longitude. \\\\\r\n \\hline\\hline \\label{Table:EquinoctialElements}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\n\\noindent Given: $a$, $h$, $k$, $p$, $q$, $\\lambda$, and $\\mu$\r\n\r\n\\noindent Find: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\nWe begin by using the mean longitude, $\\lambda$, to find the true\r\nlongitude $F$. The equation relating the two is transcendental:\r\n\\begin{equation}\r\n \\lambda = F + h\\cos{F} - k \\sin{F}\r\n\\end{equation}\r\n%\r\nWe use the Newton-Raphson method to solve for $F$, using $\\lambda$\r\nas the initial guess. We iterate on the following equation until\r\n$|F(i+1) - F(i)| < 10^{-10}$.\r\n%\r\n\\begin{equation}\r\n F(i+1) = F(i) - \\frac{f(F)}{f'(F)}\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{eqnarray}\r\n f(F) &=& F + h \\cos(F) - k\\sin(F) - \\lambda \\\\\r\n %\r\n f'(F) &=& 1 - h \\sin(F) - k \\cos(F)\r\n\\end{eqnarray}\r\n\r\nOnce the true longitude is calculated, we continue with\r\n%\r\n\\begin{equation}\r\n \\beta = \\frac{1}{1 + \\sqrt{1 - h^2 - k^2}} \\label{Eq:Beta}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n n = \\sqrt{\\frac{\\mu}{a^3}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n r = a(1 - k\\cos{F} - h\\sin{f})\r\n\\end{equation}\r\n\r\nThe cartesian components expressed in the equinoctial coordinate\r\nsystem can be calculated using.\r\n%\r\n\\begin{equation}\r\n X_1 = a\\left[ ( 1 - h^2 \\beta) \\cos{F} + h k \\beta \\sin{F} - k \\right]\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n Y_1 = a\\left[ ( 1 - k^2 \\beta) \\sin{F} + h k \\beta \\cos{F} - h \\right]\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\dot{X}_1 = \\frac{n a^2}{r}\\left[h k \\beta \\cos{F} - (1 - h^2 \\beta)\\sin{F} \\right]\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\dot{Y}_1 = \\frac{n a^2}{r}\\left[ (1 - k^2 \\beta)\\cos{F} - h k \\beta \\sin{F} \\right]\r\n\\end{equation}\r\n%\r\n\r\nThe transformation from the equinoctial system to the inertial\r\nCartesian system is given by\r\n%\r\n\\begin{equation}\r\n \\mathbf{r} = X_1 \\hat{\\mathbf{f}} + Y_1 \\hat{\\mathbf{g}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{v} = \\dot{X}_1 \\hat{\\mathbf{f}} + \\dot{Y}_1 \\hat{\\mathbf{g}}\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{equation}\r\n \\left[ \\hat{\\mathbf{f}} \\hspace{.1 in} \\hat{\\mathbf{g}} \\hspace{.1 in} \\hat{\\mathbf{w}}\r\n \\right]= \\frac{1}{1+p^2+q^2}\\hspace{.1 in} \\mathbf{Q}\r\n\\end{equation}\r\n%\r\nand\r\n%\r\n\\begin{equation} \\mathbf{Q} =\r\n \\begin{pmatrix}\r\n 1 - p^2 + q^2 &\r\n 2 p q j &\r\n 2 p \\\\\r\n%\r\n 2 p q&\r\n \\left( 1 + p^2 - q^2 \\right)j &\r\n -2 q \\\\\r\n %\r\n - 2 p j&\r\n 2 q &\r\n \\left(1 - p^2 - q^2 \\right)j \\\\\r\n %\r\n \\end{pmatrix}\r\n\\end{equation}\r\n%\r\nand\r\n%\r\n\\noindent j = 1 for direct orbits ( $0\\leq i \\leq 90^{\\circ}$ )\\\\\r\n j = -1 for retrograde orbits ( $90 < i \\leq 180^{\\circ}$ )\\\\\r\n\r\nNow let's look at how to calculate the cartesian state given the\r\nequinoctial elements.\r\n\r\n\\subsection{Cartesian State to Equinoctial Elements} \\index{Equinoctial elements!from cartesian}\r\n\\index{Cartesian state!to Equinotial}\r\n\\label{Sec:CartesiantoEquinoctial}\r\n\r\nThe equinoctial elements used in GMAT are defined in Table\r\n\\ref{Table:EquinoctialElements}. The algorithm to convert from\r\nthe cartesian state to the equinoctial elements was taken from the\r\nGTDS Mathematical Theory \\cite{GTDS}.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$\r\n\r\n\\noindent Fine: $a$, $h$, $k$, $p$, $q$, $\\lambda$, and $\\mu$\r\n\r\nThe orbit eccentricity and energy are calculated using\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{e} = \\displaystyle\\frac{(v^2 - \\displaystyle\\frac{\\mu}{r} )\\mathbf{r} - (\\mathbf{r}\\cdot\\mathbf{v} )\\mathbf{v}}{\\mu}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e = \\| \\mathbf{e} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\xi = \\frac{v^2}{2} - \\frac{\\mu}{r}\r\n\\end{equation}\r\n%\r\nFor parabolic orbits, the semimajor axis is infinite\r\nand the energy is zero. Here we check to see if the orbit is near\r\nparabolic. If $|1 - e| < 10^{-7}$ an error message is reported\r\nand conversion is aborted.\r\n\r\nThe semimajor axis is computed using\r\n%\r\n\\begin{equation}\r\n a = -\\frac{\\mu}{2\\xi}\r\n\\end{equation}\r\n%\r\nHere we check to see if the conic section is nearly singular.\r\nIf $|a(1-e)| < .001 (km)$ then an error message is thrown and conversion is aborted.\r\n\r\nThe angular momentum unit vector is\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{h}} = \\frac{\\mathbf{r} \\times \\mathbf{v} }{ \\| \\mathbf{r} \\times \\mathbf{v} \\|}\r\n\\end{equation}\r\n%\r\nThe unit vectors that define the equinoctial coordinate system can\r\nbe calculated using\r\n%\r\n\\begin{equation}\r\n f_x = 1 - \\frac{\\hat{h}_x^2}{ 1 + \\hat{h}_z^j}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n f_y = -\\frac{\\hat{h}_x \\hat{h}_y}{ 1 + \\hat{h}_z^j}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n f_z = -\\hat{h}_x^j\r\n\\end{equation}\r\n%\r\nwhere $j = 1$ if $ \\hat{h}_z > 0 $ and $j = -1$ otherwise. \\\\\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{g}} = \\hat{\\mathbf{h}} \\times \\hat{\\mathbf{f}}\r\n\\end{equation}\r\n%\r\n\r\nWe now have the necessary information to calculate the elements\r\n$h$, $k$, $p$, and $q$ using the following relationships.\r\n%\r\n\\begin{equation}\r\n h = \\mathbf{e} \\cdot \\hat{\\mathbf{g}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n k = \\mathbf{e} \\cdot \\hat{\\mathbf{f}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n p = \\frac{\\hat{h}_x}{1+\\hat{h}_z^j}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n q = - \\frac{\\hat{h}_y}{1+\\hat{h}_z^j}\r\n\\end{equation}\r\n%\r\n\r\nThe final element to calculate is the mean longitude, $\\lambda$.\r\nWe begin by computing the eccentric longitude, $F$, using\r\n%\r\n\\begin{equation}\r\n X_1 = \\mathbf{r} \\cdot \\hat{\\mathbf{f}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n Y_1 = \\mathbf{r} \\cdot \\hat{\\mathbf{g}}\r\n\\end{equation}\r\n%\r\nand\r\n%\r\n\\begin{equation}\r\n \\cos{F} = k + \\frac{ \\left( 1 - k^2 \\beta \\right) X_1 - h k \\beta Y_1}{a\\sqrt{1 - h^2 - k^2}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\sin{F} = h + \\frac{ \\left( 1 - h^2 \\beta \\right) Y_1 - h k \\beta X_1}{a\\sqrt{1 - h^2 - k^2}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n F = \\tan^{-1}_2\\left( \\frac{\\sin{F}}{\\cos{F}}\\right)\r\n\\end{equation}\r\n%\r\nwhere $\\beta$ is given by Eq.~\\ref{Eq:Beta}. The mean longitude\r\nis computed using the generalized Kepler equation\r\n%\r\n\\begin{equation}\r\n \\lambda = F + h \\cos{F} - k \\sin{F}\r\n\\end{equation}\r\n\r\nNow let's look at transformations involving the spherical\r\nelements.\r\n\r\n\\subsection{Cartesian State to SphericalAZFPA State}\r\n\\label{Sec:CarttoSphericalAZFPA} \\index{Cartesian state!to\r\nSphericalAZFPA} \\index{SphericalAZFPA! from Cartesian state }\r\n\r\nThe spherical state, with azimuth, $\\alpha_f$, and flight path\r\nangle, $\\psi$, is described in Table \\ref{Table:SphericalElements}\r\nand Fig. \\ref{fig:SphericalElements}. The algorithm below shows\r\nhow GMAT converts from the cartesian state to the spherical state\r\nwith azimuth and flight path angle.\r\n\r\n\\noindent Given: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\n\\noindent Find: $r$, $\\lambda$, $\\delta$, $v$, $\\psi$, and\r\n$\\alpha_f$\r\n\r\n\\begin{table*} \\caption{The Spherical Elements}\r\n\\centering \\index{Spherical elements!definition}\r\n\\begin{tabular}{p{.5 in} p{1.5 in} p{3.5 in}}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Symbol & Name & Description \\\\\r\n \\hline\r\n $r$ & $r$ & Magnitude of the position vector, $\\|\\mathbf{r}\\|$ \\\\\r\n %\r\n $\\lambda$ & Right Ascension & The angle between the projection of $\\mathbf{r}$ into the $xy$-plane and the $x$-axis measured counterclockwise. \\\\\r\n%\r\n $\\delta$ & Declination & The angle between $\\mathbf{r}$ and the $xy$-plane. \\\\\r\n\r\n %\r\n $v$ & $v$ & Magnitude of the velocity vector, $\\|\\mathbf{v}\\|$. \\\\\r\n %\r\n $\\psi$ & Vertical flight path angle & The angle measured from a plane normal to $\\mathbf{r}$ to the velocity vector $\\mathbf{v}$, measured in the plane formed by $\\mathbf{r}$ and $\\mathbf{v}$ \\\\\r\n %\r\n $\\alpha_f$ & Flight path azimuth & The angle measured from vector perpendicular $\\mathbf{r}$ and pointing north, to the projection of $\\mathbf{v}$ into a plane normal to $\\mathbf{r}$. \\\\\r\n %\r\n $\\lambda_v$ & Right ascension of velocity & The angle between the projection of $\\mathbf{v}$ into the $xy$-plane and the $x$-axis measured counterclockwise. \\\\\r\n %\r\n $\\delta_v$ & Declination of velocity & The angle between the velocity vector and the $xy$-plane. \\\\\r\n \\hline\\hline \\label{Table:SphericalElements}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\nWe begin by calculating the right ascension $\\lambda$, and the\r\ndeclination $\\delta$.\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\| \\label{Eq:Sphericalr}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\lambda= \\tan^{-1}_2(y,x) \\label{Eq:Sphericallambda}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\delta = \\sin^{-1}(\\frac{z}{r}) \\label{Eq:Sphericaldelta}\r\n\\end{equation}\r\n%\r\nThe magnitude of the velocity vector is simply\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\| \\label{Eq:Sphericalv}\r\n\\end{equation}\r\n%\r\nWe calculate the vertical flight path angle, $psi$, using\r\n%\r\n\\begin{equation}\r\n \\psi = \\cos^{-1}\\left( \\frac{\\mathbf{r} \\cdot \\mathbf{v} }{r v} \\right)\r\n\\end{equation}\r\n%\r\nTo calculate the azimuth angle, $\\alpha_z$, we begin by\r\ncalculating the rotation matrix from the frame in which the\r\ncartesian state is expressed in, $\\mathcal{F}_{i}$, to a local\r\nframe, $\\mathcal{F}_{\\ell}$, where $\\hat{\\mathbf{z}}$ is a unit\r\nvector that points north. The basis vectors of\r\n$\\mathcal{F}_{\\ell}$ expressed in $\\mathcal{F}_{i}$ can be\r\ncalculated using\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{x}} =\r\n \\begin{pmatrix}\r\n \\cos(\\delta)\\cos(\\lambda)\\\\\r\n \\cos(\\delta)\\sin(\\lambda)\\\\\r\n \\sin(\\delta)\r\n \\end{pmatrix} \\label{Eq:SphericalAZFPAx}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{y}} =\r\n \\begin{pmatrix}\r\n \\cos(\\lambda + \\pi/2)\\\\\r\n \\sin(\\lambda + \\pi/2)\\\\\r\n 0\r\n \\end{pmatrix} \\label{Eq:SphericalAZFPAy}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{z}} =\r\n \\begin{pmatrix}\r\n -\\sin(\\delta)\\cos(\\lambda)\\\\\r\n -\\sin(\\delta)\\sin(\\lambda)\\\\\r\n \\cos(\\delta)\r\n \\end{pmatrix} \\label{Eq:SphericalAZFPAz}\r\n\\end{equation}\r\n%\r\nWe can write the tranformation matrix that goes from\r\n$\\mathcal{F}_{i}$ to $\\mathcal{F}_{\\ell}$, $\\mathbf{R}_{\\ell i}$,\r\nas\r\n%\r\n\\begin{equation}\r\n \\mathbf{R}_{\\ell i} = [\\hspace{.05 in} \\hat{\\mathbf{x}} \\hspace{.1 in}\r\n \\hat{\\mathbf{y}} \\hspace{.1in} \\hat{\\mathbf{z}} \\hspace{.05 in}]^T\r\n\\end{equation}\r\n%\r\nThe velocity in the local frame, $\\mathbf{v}'$, can be written as\r\n%\r\n\\begin{equation}\r\n \\mathbf{v}' = \\mathbf{R}_{\\ell i}\\mathbf{v}\r\n\\end{equation}\r\n%\r\nFinally, we calculate the azimuth angle using\r\n\\begin{equation}\r\n \\alpha_f = \\tan^{-1}_2( v'_y, v'_z )\r\n\\end{equation}\r\n\r\nNow that we have looked at how to convert from the Cartesian state\r\nto the spherical state, let's look at the inverse transformation\r\nthat converts from the spherical state (with $\\psi$ and\r\n$\\alpha_f$) to the cartesian state.\r\n\r\n\\subsection{SphericalAZFPA State to Cartesian State} \\index{Cartesian\r\nstate!from SphericalAZFPA} \\index{SphericalAZFPA! to Cartesian\r\nstate } \\label{Sec:SphericalAZFPAtoCartesian}\r\n\r\nIn this section we present the algorithm used to convert from the\r\nspherical state (with $\\psi$ and $\\alpha_f$) to the cartesian\r\nstate.\r\n\r\n\\noindent Given: $r$, $\\lambda$, $\\delta$, $v$, $\\psi$, and\r\n$\\alpha_f$\r\n\r\n\\noindent Find: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\nThe components of the position vector are calculated using\r\n%\r\n\\begin{eqnarray}\r\n x &=& r\\cos{\\delta} \\cos{ \\lambda } \\label{Eq:Sphericalx}\\\\\r\n %\r\n y &=& r\\cos{\\delta}\\sin{\\lambda} \\label{Eq:Sphericaly}\\\\\r\n %\r\n z &=& r\\sin{\\delta} \\label{Eq:Sphericalz}\r\n\\end{eqnarray}\r\n%\r\n\r\nWe can write the velocity vector in terms of $v$, $\\psi$, and\r\n$\\alpha_f$ as,\r\n%\r\n\\begin{equation}\r\n \\mathbf{v} = v\\left[\\cos( \\psi )\\hat{\\mathbf{x}} + \\sin( \\psi )\\sin(\\alpha_f)\\hat{\\mathbf{y}} +\r\n \\sin( \\psi )\\cos(\\alpha_f)\\hat{\\mathbf{z}}\\right]\r\n \\label{Eq:SphericalAZFPAv}\r\n\\end{equation}\r\n%\r\nwhere, $\\hat{\\mathbf{x}}$, $\\hat{\\mathbf{y}}$, and\r\n$\\hat{\\mathbf{x}}$ are given in Eqs.~(\\ref{Eq:SphericalAZFPAx}),\r\n(\\ref{Eq:SphericalAZFPAy}), and (\\ref{Eq:SphericalAZFPAz})\r\nrespectively. Breaking down Eq.~(\\ref{Eq:SphericalAZFPAv}) into\r\ncomponents gives us\r\n%\r\n\\begin{equation}\r\n\\begin{split}\r\n v_x = v[ \\cos{\\psi}\\cos{\\delta}\\cos{\\lambda}- &\\sin{\\psi}(\\sin{\\alpha_f}\\sin{\\lambda}\r\n \\\\\r\n + &\\cos{\\alpha_f}\\sin{\\delta}\\cos{\\lambda})]\r\n\\end{split}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n\\begin{split}\r\n v_y = v[\r\n \\cos{\\psi}\\cos{\\delta}\\sin{\\lambda}+&\\sin{\\psi}(\\sin{\\alpha_f}\\cos{\\lambda}\r\n \\\\\r\n -&\\cos{\\alpha_f}\\sin{\\delta}\\sin{\\lambda})]\r\n\\end{split}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v_z = v[\\cos{\\psi}\\sin{\\delta}+\\sin{\\psi}\\cos{\\alpha_f}cos{\\delta}]\r\n\\end{equation}\r\n%\r\n\r\n\\subsection{Cartesian State to SphericalRADEC State}\\index{Cartesian state! to\r\nSphericalRADEC} \\index{SphericalRADEC! from Cartesian State }\r\n\r\nThe conversion form the Cartesian state to the spherical state\r\nwith right ascension of velocity, $\\lambda_v$, and declination of\r\nvelocity, $\\delta_v$, is very similar to the transformation shown\r\nin Sec. \\ref{Sec:CarttoSphericalAZFPA}. The algorithm to\r\ncalculate $\\lambda_v$ and $\\delta_v$ is shown below.\r\n\r\n\\noindent Given: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\n\\noindent Find: $r$, $\\lambda$, $\\delta$, $v$, $\\lambda_v$, and\r\n$\\delta_v$\r\n\r\nTo calculate $r$, $\\lambda$, $\\delta$, and $v$ we use\r\nEqs.~(\\ref{Eq:Sphericalr}), (\\ref{Eq:Sphericallambda}),\r\n(\\ref{Eq:Sphericaldelta}), and (\\ref{Eq:Sphericalv}) respectively.\r\nThe right ascension of velocity, $\\lambda_v$, and declination of\r\nvelocity, $\\delta_v$, are calculated using\r\n%\r\n\\begin{equation}\r\n \\lambda_v= \\tan^{-1}_2(v_y,v_x) \\label{Eq:Sphericallambdav}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\delta_v = \\sin^{-1}(\\frac{v_z}{v}) \\label{Eq:Sphericaldeltav}\r\n\\end{equation}\r\n%\r\n\r\nIn the next section, we show the the transformation from the\r\nspherical state with right ascension of velocity, $\\lambda_v$,\r\nand declination of velocity, $\\delta_v$, to the cartesian state.\r\n\r\n\\subsection{SphericalRADEC State to Cartesian State} \\index{Cartesian state!from SphericalRADEC }\r\n\r\nThis transformation is similar to the conversion presented in Sec\r\n\\ref{Sec:SphericalAZFPAtoCartesian}. The primary difference is\r\nhow the velocity is represented.\r\n\r\n\\noindent Given: $r$, $\\lambda$, $\\delta$, $v$, $\\lambda_v$, and\r\n$\\delta_v$\r\n\r\n\\noindent Find: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\nThe position components are calculated using Eqs.\r\n(\\ref{Eq:Sphericalx}), (\\ref{Eq:Sphericaly}), and\r\n(\\ref{Eq:Sphericalz}). The velocity components are calculated\r\nusing\r\n%\r\n\\begin{eqnarray}\r\n v_x &=& v\\cos{\\lambda_v}\\cos{\\delta_v}\\\\\r\n v_y &=& v_x\\tan{\\lambda_v}\\\\\r\n v_z &=& v\\sin{\\delta_v}\r\n\\end{eqnarray}\r\n\r\nIn the last few subsections, we have looked at transformations\r\ninvolving the spherical elements. Now let's look at\r\ntransformations involving the modified Keplerian elements.\r\n\r\n\\subsection{Keplerian or Cartesian, to Modified Keplerian Elements} \\index{Modified Keplerian elements!from Keplerian} \\index{Keplerian elements!to modified Keplerian}\r\n\r\nThe modified Keplerian elements, described in Table\r\n\\ref{Table:ModKeplerianElements}, are similar to the classical\r\nKeplerian elements. The modified Keplerian elements use the\r\nradius of apoapsis, $r_a$, and the radius of periapsis, $r_p$, to\r\ndescribe the size and shape of an orbit. The remaining elements,\r\n$i$, $\\Omega$, $\\omega$, and $\\nu$, are the same for both the\r\nKeplerian and modified Keplerian elements. The modified Keplerian\r\nelements, like the Keplerian elements, are undefined for parabolic\r\norbits. Let's look at how GMAT calculates the modified\r\n\r\n\r\n\\begin{table*} \\caption{The Modified Keplerian Elements}\r\n\\centering\\index{Modified Keplerian elements!definition}\r\n\\begin{tabular}{p{.5 in} p{1.5 in} p{3.5 in}}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Symbol & Name & Description \\\\\r\n \\hline\r\n $r_p$ & radius of periapsis & The radius of periapsis is the radius at the spacecrafts closest approach to the central body. The radius of periapsis must be greater than zero, parabolic orbits are not currently supported. \\\\\r\n%\r\n $r_a$ & radius of apoapsis & For an elliptic orbit $r_a$ is the radius at the spacecrafts farthest distance from the central body and $r_a>r_p$. For hyperbolic orbits, $r_a 90^{\\circ}$ then the orbit is retrograde. \\\\\r\n %\r\n $\\omega$ & argument of periapsis & The argument of periapsis is the angle between a vector pointing at periapsis,\r\n $\\mathbf{x}_p$, and a vector pointing at the spacecraft. The argument of periapsis is undefined for circular orbits.\\\\\r\n %\r\n $\\Omega$ & right ascension of the ascending node& $\\Omega$ is defined as the angle between $\\hat{\\mathbf{x}}_I$ and $\\mathbf{N}$ measured\r\n counterclockwise.\r\n $\\mathbf{N}$ is defined as the vector pointing from the center of the central body to the spacecraft, when the spacecraft crosses the bodies equatorial plane\r\n from the southern to the northern hemisphere. $\\Omega$ is undefined for equatorial orbits. \\\\\r\n %\r\n $\\nu$ & true anomaly & The true anomaly is defined as the angle between a vector pointing at periapsis, $\\mathbf{x}_p$, and a vector pointing at the spacecraft.\r\n The true anomaly is undefined for circular orbits. \\\\\r\n \\hline\\hline \\label{Table:ModKeplerianElements}\r\n\\end{tabular}\r\n\\end{table*}\r\n\r\n\r\n\\noindent Given: $a$, $e$, $i$, $\\omega$, $\\Omega$, and $\\nu$, or\r\n$\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$\r\n\r\n\\noindent Find: $r_p$ and $r_a$\r\n\r\nIf we are given the Cartesian state, we first calculate the\r\norbital elements using the algorithm in Sec. \\ref{Sec:Kep2Cart}.\r\nKnowing the Keplerian elements, we calculate $r_a$ and $r_p$ using\r\n%\r\n\\begin{equation}\r\n r_a = a(1+e)\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n r_p = a(1-e)\r\n\\end{equation}\r\n\r\nNow let's look at the inverse transformation.\r\n\r\n\\subsection{Modified Keplerian Elements to Keplerian Elements} \\index{Modified\r\nKeplerian elements!to Keplerian} \\index{Keplerian elements!from\r\nmodified Keplerian}\r\n\r\nThe conversion from modified Keplerian elements to the Keplerian\r\nelements is discussed below. To perform the conversion, we use\r\nrelationships that allow us to write the semimajor axis, $a$, and\r\nthe eccentricity, $e$, in terms of $r_a$ and $r_p$.\r\n\r\n\\noindent Given: $r_p$, $r_a$, $i$, $\\omega$, $\\Omega$, and $\\nu$\r\n\r\n\\noindent Find: $a$ and $e$\r\n\r\nWe begin by calculating the eccentricity using\r\n%\r\n\\begin{equation}\r\n e = \\frac{1 - \\displaystyle\\frac{r_p}{r_a}}{1 + \\displaystyle\\frac{r_p}{r_a}}\r\n\\end{equation}\r\n%\r\nThe semimajor axis is calculated using\r\n%\r\n\\begin{equation}\r\n a = \\displaystyle\\frac{r_p}{1 - e}\r\n\\end{equation}\r\n\r\nThis concludes our discussion of state transformations. In the\r\nlast few subsections we presented the algorithms used to convert\r\nbetween different orbit state representations used in GMAT. These\r\ninclude the Cartesian state, the Keplerian elements, the modified\r\nKeplerian elements, and two spherical state parameterizations. In\r\nthe next section, we present the algorithms used to calculate\r\n properties such as orbit period, beta angle, and mean motion to\r\n name a few.\r\n\r\n\\section{Simple Parameters}\r\n\r\nSimple parameters, which we will abbreviate as simply ``parameters\",\r\nare properties of spacecraft or other objects that are only\r\ndependent upon one of the following: CoordinateSystem, CentralBody,\r\nor None. An example of a simple parameter is the magnitude of a\r\nspacecrafts velocity vector. The spacecrafts velocity vector is\r\ndependent upon the coordinate system in which it is expressed. Once\r\nwe have specified a coordinate system, it is trivial to calculate\r\nthe velocity vector, and therefore its magnitude, in that coordinate\r\nsystem.\r\n\r\nIn GMAT, the syntax to specify a simple parameter is\r\n\r\n\\st{ObjectName.Dependency.ParameterName}\r\n\r\n\\noindent So, to calculate the magnitude of the velocity, of a\r\nspacecraft named \\st{Sat}, in the Earth Fixed frame, we would use\r\n\r\n\\st{Sat.EarthFixed.VMAG}\r\n\r\n\\noindent GMAT has the ability to calculate many parameters in\r\naddition to \\st{VMAG}. In the following subsections, we present the\r\nalgorithms used to calculate all parameters in GMAT. We begin each\r\nsubsection with a description of the parameter, and then give the\r\ntype of dependency.\r\n\r\n\\subsection{A1Gregorian} \\index{A1Gregorian}\r\n\r\n\\noindent \\textit{Description}: \\st{A1Gregorian} is the epoch of an\r\nobject, in the A1 time system, given in the Gregorian date format.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\nThe A1 date, in modified Julian date format is the current\r\nindependent variable for time in GMAT. Therefore, it is not\r\nnecessary to convert the date to another system for this parameter.\r\nThe only calculation required for this parameter is to use the\r\nalgorithm in Sec. \\ref{Sec:GregorianData} to convert from Modified\r\nJulian date format to Gregorian date format.\r\n\r\n\\subsection{A1ModJulian} \\index{A1ModJulian}\r\n\r\n\\noindent \\textit{Description}: \\st{A1ModJulian} is the epoch of an\r\nobject, in the A1 time system, given in the modified Julian date\r\nformat.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\nThe A1 date, in modified Julian date format is the current\r\nindependent variable for time in GMAT. There are no calculations\r\nrequired for this parameter.\r\n\r\n\\subsection{Altitude} \\index{Altitude}\r\n\r\n\\noindent \\textit{Description}: \\st{Altitude} is the distance\r\nbetween a spacecraft and a plane tangent to the surface of the body\r\nat the sub-satellite point. GMAT assumes the body is an ellipsoid.\r\nThe equatorial radius, and properties of the ellipsoid depend upon\r\nthe particular body chosen by the user.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$ in $\\mathcal{F}_1$\r\n\r\n\\noindent Find: $A$\r\n\r\n\\noindent Definitions:\r\n\\begin{itemize}\r\n %\r\n \\item $\\mathcal{F}_1$ is the coordinate system in which GMAT originally knows $\\mathbf{r}$\r\n %\r\n \\item $\\mathcal{F}_F$ is body fixed system of the central body selected by the user.\r\n %\r\n \\item $f$ is the bodies flattening coefficient\r\n %\r\n \\item $R$ is the bodies mean equatorial radius\r\n %\r\n % \\item $\\phi_{gc}$ is the geocentric latitude of the spacecraft\r\n %in the body fixed frame.\r\n %\r\n \\item $\\phi_{gd}$ is the geodedic latitude of the spacecraft\r\n in the body fixed frame.\r\n %\r\n \\item $h$ is the \\st{Altitude} parameter\r\n %\r\n\\end{itemize}\r\n\r\nFirst we calculate $\\phi_{gd}$ using the algorithm in Sec.\r\n\\ref{Sec:Latitude}. However, to calculate $h$, GMAT does not\r\nconvert to degrees, or use the modulo function.\r\n\r\nThen, with $\\mathbf{r}$ expressed in $\\mathcal{F}_F$, we perform\r\n%\r\n\\begin{equation}\r\n r_{xy} = \\sqrt{ x^2 + y^2 }\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e^2 = 2f-f^2\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\frac{ r_{xy}}{ \\cos(\\phi_{gd})} - \\frac{R} { \\sqrt{ 1 - e^2 \\sin^{2}{ \\phi_{gd} } } };\r\n\\end{equation}\r\n\r\n\\subsection{AOP} \\index{AOP}\r\n\r\n\\noindent \\textit{Description}: \\st{AOP} is the argument of\r\nperiapsis of a spacecraft. The argument of periapsis is the angle\r\nbetween the eccentricity vector and a vector in the direction of the\r\nright ascension of the ascending node. See below for treatment of\r\ncircular and equatorial orbits. This algorithm is adopted from\r\nVallado\\cite{vallado2}.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$ and $\\mathbf{v}$\r\n\r\n\\noindent Find: $\\omega$\r\n\\[\r\nr = \\| \\mathbf{r} \\|\r\n\\]\r\n%\r\n\\[\r\nv = \\| \\mathbf{v} \\|\r\n\\]\r\n%\r\n\\[\r\n \\mathbf{e} = \\displaystyle\\frac{\\left( v^2 - \\displaystyle\\frac{\\mu}{r} \\right)\\mathbf{r} - (\\mathbf{r} \\cdot \\mathbf{v})\\mathbf{v}}{\\mu}\r\n\\]\r\n%\r\n\\[\r\n e = \\| \\mathbf{e} \\|\r\n\\]\r\n%\r\n\\textit{Special Case: Circular Orbit}\\\\\r\nif $ e < 10^{-11}$ then, $\\omega = 0.0$ and return.\r\n\r\n\r\n\\noindent Otherwise continue,\r\n%\r\n\\[ \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v} \\]\r\n%\r\n\\[ h = \\| \\mathbf{h} \\| \\]\r\n%\r\n\\[ i = \\cos^{-1}{\\left(\\frac{h_z}{h} \\right)}\\]\r\n\r\n\\noindent \\textit{Special Case: Elliptic, Equatorial Orbit}\\\\\r\nif $ i < 10^{-11}$ then,\r\n%\r\n\\begin{equation}\r\n \\omega = \\cos^{-1}\\left(\\frac{e_x}{e} \\right)\r\n\\end{equation}\r\n%\r\nwhere $e_x$ is the first component of the eccentricity vector.\\\\\r\nFix quadrant for $\\omega $: if $e_y < 0$, then $\\omega = 2\\pi -\r\n\\omega$\r\n\r\n\\noindent Otherwise continue\r\n\r\n\\noindent \\textit{Special Case: Elliptic, Inclined Orbit}\\\\\r\n%\r\n\\[ \\mathbf{n} = [\\hspace{.05 in}0 \\hspace{.1 in} 0 \\hspace{.1 in} 1 \\hspace{.05 in}]^T \\times \\mathbf{h} \\]\r\n%\r\n\\[ \\omega = \\cos^{-1}\\frac{\\mathbf{n}\\cdot \\mathbf{e}}{\\| \\mathbf{n}\\|\\|\r\n\\mathbf{e}\\|}\\]\r\n%\r\nFix quadrant for $\\omega $: if $e_z < 0$, then $\\omega = 2\\pi -\r\n\\omega$. \\\\Finally, $\\omega$ is converted to degrees.\r\n%\r\n\\subsection{Apoapsis} \\index{Apoapsis}\r\n\r\n\\noindent \\textit{Description}: \\st{Apoapsis} is the parameter used\r\nin stopping conditions to allow the stopping condition algorithm to\r\nlocate the time when a spacecraft is at apoapsis. Apoapsis is\r\ndefined as a point, along an orbital path, when the component of\r\nvelocity, in the spacecraft position vector direction, changes from\r\npositive to negative. The \\st{Apoapsis} parameter is defined as the\r\ndot product of the position and velocity vectors.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ in $\\mathcal{F}_1$\r\n\r\n\\noindent Find: $A$\r\n\r\n\\noindent Definitions:\r\n%\r\n\\begin{itemize}\r\n %\r\n \\item $\\mathcal{F}_1$ is the coordinate system in which GMAT originally knows $\\mathbf{r}$ and $\\mathbf{v}$\r\n %\r\n \\item $\\mathcal{F}_2$ is a system with the MJ2000Eq axes, centered at the central body selected by the user.\r\n %\r\n \\item $A$ is the \\st{Apoapsis} parameter\r\n %\r\n\\end{itemize}\r\n%\r\n\r\nif ($\\mathcal{F}_1 \\neq \\mathcal{F}_2)$ convert $\\mathbf{r}$ and\r\n$\\mathbf{v}$ to $\\mathcal{F}_2$. Then,\r\n%\r\n\\begin{equation}\r\n A = \\mathbf{r} \\cdot \\mathbf{v}\r\n\\end{equation}\r\n\r\n\\subsection{AZI} \\index{AZI}\r\n\r\n\\noindent \\textit{Description}: \\st{AZI} is the azimuth angle of a\r\nspacecraft, as shown in Fig. \\ref{fig:KeplerianElements} using the\r\nsymbol $\\alpha_f$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ and $\\mathcal{F}$\r\n\r\n\\noindent Find: $\\alpha_f$\r\n\r\n\r\nAZI is calculated using the algorithm shown in Sec.\r\n\\ref{Sec:CarttoSphericalAZFPA}. There is little benefit using a\r\nroutine that calculates only $\\alpha_f$ and not $\\psi$.\r\n\r\n\\subsection{BdotT and BdotR} \\index{BdotT} \\index{BdotR}\r\n\r\n\\noindent \\textit{Description}: The ``B\" vector, $\\mathbf{B}$, is\r\nonly defined for hyperbolic orbits and is the vector from the center\r\nof mass of the central body, to the incoming hyperoblic asysmptote,\r\nsuch that the length of $\\mathbf{B}$ is a minimum. Another way to\r\nsay this is that $\\mathbf{B}$ is perpendicular to the incoming\r\nasymptote. Let's define $\\mathbf{S}$ as a unit vector in the\r\ndirection of the incoming asymptote. Then, $\\mathbf{T}$ is a unit\r\nvector perpendicular to $\\mathbf{S}$, that lies in the $xy$-plane of\r\nthe coordinate system, $\\mathcal{F}_B$, chosen by the user.\r\n$\\mathbf{R}$ is a unit vector perpendicular to both $\\mathbf{S}$ and\r\n$\\mathbf{T}$. Finally, \\st{BdotT} is the dot product of $\\mathbf{B}$\r\nand $\\mathbf{T}$, and \\st{BdotR} is the dot product of $\\mathbf{B}$\r\nand $\\mathbf{R}$. The method below was adopted from work by\r\nKizner\\cite{Kizner:59}.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and definition of\r\n$\\mathcal{F}_B$\r\n\r\n\\noindent Find: $B_R$ and $B_T$\r\n\r\n\\noindent Definitions:\r\n\\begin{itemize}\r\n %\r\n \\item $\\mathcal{F}_1$ is the coordinate system in which GMAT originally knows\r\n $\\mathbf{r}$ and $\\mathbf{v}$\r\n %\r\n \\item $\\mathcal{F}_B$ is the coordinate system in which to perform B-plane\r\n calculations. GMAT will place $\\mathbf{T}$ in the $xy$-plane\r\n of $\\mathcal{F}_B$. $\\mathcal{F}_B$ must have a gravitational\r\n body at its origin.\r\n %\r\n \\item $\\mu$ is the gravitational parameter of the central body\r\n at the origin of $\\mathcal{F}_B$\r\n %\r\n \\item $B_R$ is the dot product of $\\mathbf{B}$ and\r\n $\\mathbf{R}$\r\n %\r\n \\item $B_T$ is the dot product of $\\mathbf{B}$ and\r\n $\\mathbf{T}$\r\n\\end{itemize}\r\n\r\n\r\n\r\nif the selected coordinate system does not have a celestial body as\r\nits origin, then exit and throw an error message.\\\\\r\n%\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,200)(55,340)\r\n \\includegraphics[scale=1]{Images/BPlaneGeometry1.eps}\r\n \\makebox(-520,845){$\\theta_B$}\r\n \\makebox(-615,820){$\\mathbf{R}$}\r\n \\makebox(-500,810){$\\mathbf{B}$}\r\n \\makebox(-560,890){$\\mathbf{T}$}\r\n \\makebox(-760,920){$xy$-plane of $\\mathcal{F}_B$ }\r\n \\makebox(-720,990){Central Body}\r\n \\makebox(-570,1030){$z$-axis of $\\mathcal{F}_B$}\r\n \\makebox(-580,760){$\\mathbf{B}\\cdot\\mathbf{T}$}\r\n \\makebox(-485,830){$\\mathbf{B}\\cdot\\mathbf{R}$}\r\n \\end{picture}\r\n \\caption{Geometry of the B-Plane as Seen From a Viewpoint Perpendicular to the B-Plane}\r\n \\label{fig:BPlaneGeometry1}\r\n\\end{figure}\r\n%\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,240)(125,340)\r\n \\includegraphics[scale=1]{Images/BPlaneGeometry2.eps}\r\n \\makebox(-420,850){$\\mathbf{B}$}\r\n \\makebox(-580,990){Central Body}\r\n \\makebox(-570,850){Incoming Trajectory}\r\n \\makebox(-570,760){Incoming}\r\n \\makebox(-570,740){Asymptote}\r\n \\end{picture}\r\n \\caption{The B-Vector as Seen From a Viewpoint Perpendicular to Orbit Plane}\r\n \\label{fig:BPlaneGeometry2}\r\n\\end{figure}\r\n\r\n\\noindent if $\\mathcal{F}_1 \\neq \\mathcal{F}_B$ convert $\\mathbf{r}$\r\nand $\\mathbf{v}$ from $\\mathcal{F}_1$ to $\\mathcal{F}_B$\r\n\r\n\\[\r\nr = \\| \\mathbf{r} \\|\r\n\\]\r\n%\r\n\\[\r\nv = \\| \\mathbf{v} \\|\r\n\\]\r\n%\r\nCalculate eccentricity related information\r\n%\r\n\\[\r\n \\mathbf{e} = \\displaystyle\\frac{\\left( v^2 - \\displaystyle\\frac{\\mu}{r} \\right)\\mathbf{r} - (\\mathbf{r} \\cdot \\mathbf{v})\\mathbf{v}}{\\mu}\r\n\\]\r\n%\r\n\\[\r\n e = \\| \\mathbf{e} \\|\r\n\\]\r\n%\r\n\\[\r\n\\hat{\\mathbf{e}} = \\frac{\\mathbf{e}}{e}\r\n\\]\r\n%\r\nIf $e\\leq 1$, then the method fails and returns.\r\n\r\nNow let's calculate the angular momentum and orbit normal vectors.\r\n%\r\n\\[\r\n\\mathbf{h} = \\mathbf{r}\\times \\mathbf{v}\r\n\\]\r\n%\r\n\\[\r\nh = \\| \\mathbf{r}\\times \\mathbf{v} \\|\r\n\\]\r\n%\r\n\\[\r\n\\hat{\\mathbf{h}} = \\frac{\\mathbf{h}}{h}\r\n\\]\r\n\r\nA unit vector normal to both the eccentricity vector and the orbit\r\nnormal vector is defined as:\r\n%\r\n\\[\r\n \\hat{\\mathbf{n}} = \\hat{\\mathbf{h}} \\times \\hat{\\mathbf{e}}\r\n\\]\r\nThe following relations are only true for hyperbolic orbits: The\r\nsemiminor axis, $b$, can be calculated using\r\n%\r\n\\[\r\n b = \\frac{h^2}{\\mu \\sqrt{e^2 - 1}}\r\n\\]\r\n%\r\nThe incoming asymptote is defined using\r\n%\r\n\\[\r\n \\mathbf{S} = \\frac{\\hat{\\mathbf{e}} }{e} + \\sqrt{1 - \\left(\\frac{1}{e}\\right)^2\r\n }\\hat{\\mathbf{n}}\r\n\\]\r\n%\r\nThe B-vector, $\\mathbf{B}$, is calculated using\r\n%\r\n\\[\r\n \\mathbf{B} = b \\left(\\sqrt{1 - \\left(\\frac{1}{e}\\right)^2 }\\hat{\\mathbf{e}} - \\frac{1}{e} \\hat{\\mathbf{n}} \\right)\r\n\\]\r\n%\r\nThe remaining vectors, $\\mathbf{T}$ and $\\mathbf{R}$ are found using\r\n%\r\n\\[\r\n \\mathbf{T} = \\frac{[\\hspace{.03 in} S_y \\hspace{.1 in} -S_x \\hspace{.1 in} 0 \\hspace{.03 in}]^T }{\\sqrt{S_x^2 + S_y^2}}\r\n\\]\r\n%\r\n\\[\r\n \\mathbf{R} = \\mathbf{S} \\times \\mathbf{T}\r\n\\]\r\n%\r\nFinally, the desired quantities are found using\r\n%\r\n\\[\r\n B_T = \\mathbf{B} \\cdot \\mathbf{T}\r\n\\]\r\n%\r\n\\[\r\n B_R = \\mathbf{B} \\cdot \\mathbf{R}\r\n\\]\r\n\r\n\r\n\r\n\\noindent if $\\mathcal{F}_1 \\neq \\mathcal{F}_2$ , convert\r\n$\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}_2$\r\n%\r\n\\begin{equation}\r\n A = \\mathbf{r}\\cdot\\mathbf{v}\r\n\\end{equation}\r\n\r\n\\subsection{BetaAngle} \\index{BetaAngle}\r\n\r\nDefinition: The Beta angle, $\\beta$, is defined as the angle\r\nbetween the orbit normal vector, and the vector from the celestial\r\nbody to the sun.\r\n\r\n\\[\r\n \\hat{\\mathbf{h}} = \\frac{ \\mathbf{r}_\\oplus \\times \\mathbf{v}_\\oplus }{\\|\\mathbf{r}_\\oplus \\times \\mathbf{v}_\\oplus \\|}\r\n\\]\r\n%\r\n\\[\r\n\\hat{\\mathbf{r}}_{s \\oplus} = \\frac{\\mathbf{r}_{s \\oplus}}{\\|_{s\r\n\\oplus } \\|}\r\n\\]\r\n%\r\n\\begin{equation}\r\n \\beta = \\sin^{-1}\\left(\\hat{\\mathbf{h}}\\cdot \\hat{\\mathbf{r}}_{s\\oplus } \\right)\r\n\\end{equation}\r\n\r\n\\begin{itemize}\r\n \\item $\\mathbf{r}_\\oplus$: Position vector of spacecraft\r\n with respect to celestial body, in the EarthMJ2000Eq system.\r\n\r\n \\item $\\mathbf{v}_\\oplus$: Velocity vector of spacecraft\r\n with respect to celestial body, in the EarthMJ2000Eq system.\r\n\r\n \\item $\\mathbf{r}_{s\\oplus}$: Position vector from celestial\r\n body, to the sun.\r\n\\end{itemize}\r\n\r\n\\subsection{BVectorAngle and BVectorMag} \\index{BVectorAngle}\r\n\\index{BVectorMag}\r\n\r\nTo avoid code reduplication, the magnitude and angle of the B\r\nvector, $\\|\\mathbf{B}\\|$ and $\\theta_B$ respectively, are\r\ncalculated from the outputs of the B-Plane coordinates algorithm.\r\nThe equations for $\\|\\mathbf{B}\\|$ and $\\theta_B$ are\r\n%\r\n\\begin{equation}\r\n \\|\\mathbf{B}\\| = \\sqrt{ B_T^2 + B_R^2}\r\n\\end{equation}\r\n%\r\n%\r\n\\begin{equation}\r\n \\theta_B = \\tan^{-1}{\\frac{B_R}{B_T}}\r\n\\end{equation}\r\n%\r\nwhich is implemented using $atan2( B_R , B_T )$\r\n\r\n\\subsection{C3Energy}\\index{Spacecraft properties!C3Energy}\r\n\\index{C3Energy}\r\n\r\nGiven: $a$, and $\\mu$\r\n\r\n\\noindent Find: $C_3$\r\n\r\n\\begin{equation}\r\n C_3 = -\\frac{\\mu}{a}\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ is calculated from the satellite\r\ncartesian state as shown in Section \\ref{Sec:Cart2Kep}, and $\\mu$ is\r\nassociated with the specified central body.\r\n\r\n\\subsection{DEC} \\index{DEC}\r\n\r\n\\noindent \\textit{Description}: \\st{DEC} is the declination of a\r\nspacecraft, as shown in Fig. \\ref{fig:SphericalElements} using the\r\nsymbol $\\delta$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ and $\\mathcal{F}$\r\n\r\n\\noindent Find: $\\delta$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\delta = \\sin^{-1}(\\frac{z}{r})\r\n\\end{equation}\r\n\r\n\\subsection{DECV} \\index{DECV}\r\n\r\n\\noindent \\textit{Description}: \\st{DECV} is the declination of\r\nvelocity of a spacecraft.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ and $\\mathcal{F}$\r\n\r\n\\noindent Find: $\\delta_v$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\delta_v = \\sin^{-1}(\\frac{v_z}{v})\r\n\\end{equation}\r\n\r\n\\subsection{ECC} \\index{ECC}\r\n\r\n\\noindent \\textit{Description}: \\st{ECC} is the eccentricity of an\r\norbit and must be greater than or equal to zero. The eccentricity\r\ncontains information on the shape of an orbit. If ECC is zero then\r\nthe orbit is circular. If ECC is greater than zero, but less than\r\none, the orbit is elliptic. If ECC equals one, the orbit is\r\nparabolic. Finally, if ECC is greater than one, the orbit is\r\nhyperbolic. The algorithm used in GMAT to calculate SMA is adopted\r\nfrom Vallado\\cite{vallado2}.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$ (Central\r\nBody)\r\n\r\n\\noindent Find: $e$\r\n\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{e} = \\displaystyle\\frac{(v^2 - \\displaystyle\\frac{\\mu}{r} )\\mathbf{r}\r\n - (\\mathbf{r}\\cdot\\mathbf{v} )\\mathbf{v}}{\\mu} \\label{Eq:EccentricityVector}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e = \\| \\mathbf{e} \\|\r\n\\end{equation}\r\n\r\n\\subsection{FPA} \\index{FPA}\r\n\r\n\\noindent \\textit{Description}: \\st{FPA} is the orbit vertical\r\nFlight Path Angle as as shown in Fig. \\ref{fig:SphericalElements}\r\nusing the symbol $\\psi$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and coordinate system\r\n$\\mathcal{F}$.\r\n\r\n\\noindent Find: $\\psi$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\psi = \\cos^{-1}\\left( \\frac{\\mathbf{r} \\cdot \\mathbf{v} }{r v} \\right)\r\n\\end{equation}\r\n%\r\n\r\n\\subsection{EA} \\label{sec:EccentricAnomaly}\r\n\\index{EA}\r\n\r\nGiven: $\\nu$, $e$\r\n\r\n\\noindent Find: $E$\\\\\r\n%\r\n\r\n\\noindent If e $>$ ( 1 - $1e^{-11}$ ) then $E = 0$, return.\\\\\r\n%\r\n\r\n\r\n\\noindent Otherwise,\r\n%\r\n\\begin{eqnarray}\r\n \\sin(E) & = & \\frac{\\sqrt{1 - e^2} \\sin(\\nu)}{1+e \\cos{\\nu}} \\\\ %Vallado pg. 213, Eq. 4-9\r\n %\r\n \\cos(E) & = & \\frac{ e + \\cos{\\nu} }{1+e \\cos{\\nu}} \\\\ %Vallado pg. 213, Eq. 4-9\r\n %\r\n E & = & \\mbox{atan2}(\\sin{E},\\cos{E})\r\n\\end{eqnarray}\r\n%\r\n\r\n\\subsection{Energy}\\index{Spacecraft properties!Energy}\r\n\\index{Energy}\r\n\r\n\\noindent \\textit{Description}: \\st{Energy} is the orbit energy.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and central body.\r\n\r\n\\noindent Find: $\\xi$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to a coordinate\r\nsystem with the origin equal to the central body defined by the\r\nuser, and the MJ2000Eq axis system. Then,\r\n%\r\n\\begin{equation}\r\n r = \\mathbf{r}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\xi = \\frac{v^2}{2} - \\frac{\\mu}{r}\r\n\\end{equation}\r\n\r\n\r\n\\subsection{HMAG} \\index{HMAG}\r\n\r\n\\noindent \\textit{Description}: \\st{HMAG} is the magnitude of the\r\norbit angular momentum.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and central body.\r\n\r\n\\noindent Find: $h$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to a coordinate\r\nsystem with the origin equal to the central body defined by the\r\nuser, and the MJ2000Eq axis system. Then,\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\| \\mathbf{h} \\|\r\n\\end{equation}\r\n\r\n\\subsection{HX,HY, and HZ} \\index{HX} \\index{HY} \\index{HZ}\r\n\r\n\\noindent \\textit{Description}: \\st{HX,HY,} and \\st{HZ} are the\r\ncomponents of the orbit angular momentum vector.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and coordinate system\r\n$\\mathcal{F}$.\r\n\r\n\\noindent Find: $h_x$, $h_y$, and $h_z$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v} = [\\hspace{.05 in} h_x\r\n \\hspace{.1 in} h_y \\hspace{.1 in} h_z \\hspace{.05 in} ]^T\r\n\\end{equation}\r\n\r\n\r\n\\subsection{HA} \\label{sec:HyperbolicAnomaly}\r\n\\index{HA}\r\n\r\n\\noindent \\textit{Description}: \\st{HA} is the orbit Hyperbolic\r\nAnomaly and is only defined for hyperbolic orbits. For\r\nnon-hyperbolic orbits, \\st{HA} returns a value of zero.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n Given: $\\nu$, $e$\r\n\r\n\\noindent Find: $H$\\\\\r\n%\r\n\r\n\\noindent If e $<$ ( 1 + $1e^{-11}$ ) then $H = 0$, return.\\\\\r\n%\r\n\r\n\r\n\\noindent Otherwise,\r\n%\r\n\\begin{eqnarray}\r\n \\sinh(H) & = & \\frac{ \\sin(\\nu) \\sqrt{e^2 - 1}}{1+e \\cos{\\nu}} \\\\ %Vallado pg. 213, Eq. 4-9\r\n %\r\n H & = & \\mbox{asinh}(\\sinh(H))\r\n\\end{eqnarray}\r\n%\r\n\r\n\\subsection{INC} \\index{INC}\r\n\r\n\\noindent \\textit{Description}: \\st{INC} is the inclination of an\r\norbit in the chosen coordinate system.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and coordinate system\r\n$mathcal{F}$.\r\n\r\n\\noindent Find: $e$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\mathbf{h}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n i = \\cos^{-1}( \\frac{h_z}{h} )\r\n\\end{equation}\r\n\r\n\r\n\\subsection{Latitude} \\label{Sec:Latitude} \\index{Latitude}\r\n\r\n\\noindent \\textit{Description}: \\st{Latitude} is the geodetic\r\nlatitude of a spacecraft. The geodedic latitude is defined as the\r\nthe angle $\\phi_{gc}$, as shown in Fig. ( ), where the\r\nsub-satellite point is defined by the interscection of a line drawn\r\nfrom the spacecraft and perpendicular to a plane tangent to the\r\nsurface of the body. GMAT assumes the body is an ellipsoid. The\r\nequatorial radius, and properties of the ellipsoid depend upon the\r\nparticular body chosen by the user. The algorithm in GMAT is taken\r\nfrom Vallado\\cite{vallado2}.\r\n\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,200)(75,320)\r\n \\includegraphics[scale=1]{Images/GeodeticDiagram.eps}\r\n \\makebox(-575,880){$h$}\r\n \\makebox(-640,680){$\\phi_{gd}$}\r\n \\makebox(-730,670){$\\phi_{gc}$}\r\n \\end{picture}\r\n \\caption{Geocentric and Geodetic Latitude}\r\n \\label{fig:GeodeticDiagram}\r\n\\end{figure}\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$ in $\\mathcal{F}_1$\r\n\r\n\\noindent Find: $\\phi_{gc}$\r\n\r\n\\noindent Definitions:\r\n\\begin{itemize}\r\n %\r\n \\item $\\mathcal{F}_1$ is the coordinate system in which GMAT originally knows $\\mathbf{r}$\r\n %\r\n \\item $\\mathcal{F}_F$ is body fixed system of the central body selected by the user.\r\n %\r\n \\item $f$ is the bodies flattening coefficient\r\n %\r\n \\item $R$ is the bodies mean equatorial radius\r\n %\r\n \\item $\\phi_{gd}$ is the geodedic latitude of the spacecraft\r\n in the body fixed frame.\r\n %\r\n\\end{itemize}\r\n%\r\nif $\\mathcal{F}_1 \\neq \\mathcal{F}_F$ convert $\\mathbf{r}$ from\r\n$\\mathcal{F}_1$ to $\\mathcal{F}_F$. Then,\r\n%\r\n\\begin{equation}\r\n r_{xy} = \\sqrt{ x^2 + y^2 }\r\n\\end{equation}\r\n%\r\nCalculate the geocentric latitude to use as an initial guess to find\r\nthe geodetic latitude\r\n%\r\n\\begin{equation}\r\n \\phi_{gd} \\approx \\mbox{atan2}(z, r_{xy} );\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e^2 = 2f-f^2\r\n\\end{equation}\r\n%\r\n\r\n\\noindent Set $\\delta = 1.0$ to initialize the loop, then,\r\n\r\n\r\n\\noindent While ( $\\delta > 10^{-7}$ )\r\n%\r\n\\begin{eqnarray}\r\n \\phi' & = & \\phi_{gd}\\\\\r\n %c & = & \\frac{R} { \\sqrt{1 - e^2\\sin{\\phi}} }\\\\\r\n \\phi_{gd} & = & \\mbox{atan2}\\left(z + \\frac{Re^2\\sin^{2}{\\phi_{gd}}} { \\sqrt{1 - e^2\\sin{\\phi_{gd}}} }, r_{xy}\r\n \\right)\\\\\r\n \\delta & = & | \\phi_{gd} - \\phi' |\r\n\\end{eqnarray}\r\n%\r\nEndWhile\\\\\r\n%\r\n\r\nAfter convergence, $\\phi_{gd}$ is converted to degrees, and\r\nconverted to fall between $-90^\\circ$ and $+90^\\circ$ degrees.\r\n\r\n\\subsection{Longitude} \\label{Sec:CalcObjectLongitude}\r\n\\index{Longitude}\r\n\r\n\\noindent \\textit{Description}: \\st{Longitude} is the longitude of\r\nan object, in the body fixed frame of the central body chosen by the\r\nuser.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, central body.\r\n\r\n\\noindent Find: $\\phi$\r\n\r\nBegin by converting $\\mathbf{r}$ to the body fixed system of the\r\ncentral body defined by the user. Then,\r\n%\r\n\\begin{equation}\r\n \\phi = \\tan_2^{-1}(y,x);\r\n\\end{equation}\r\n%\r\nThe calculation is completed by converting to degrees and setting\r\nthe value to such that $-180 \\leq \\phi < 180$.\r\n\r\n\\subsection{LST} \\index{LST}\r\n\r\n\\noindent \\textit{Description}: \\st{LST} is the local sidereal time\r\nof an object, with respect to the selected central body. The local\r\nsidereal time is the sum of the longitude in the bodies fixed frame,\r\nand the mean sidereal time. This is illustrated in Fig.\r\n\\ref{fig:SiderealTimeDiagram}, where $\\mathcal{F}_I$ is the body's\r\nequatorial inertial system (as described in Sec. \\ref{Sec:Equator}),\r\n$\\mathcal{F}_F$ is the body's fixed system (as described in Sec.\r\n\\ref{Sec:Fixed}). $\\lambda$ is the longitude of the object, in this\r\ncase a spacecraft, and $\\theta_{MST}$ is the mean sidereal time of\r\nthe prime meridian.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $t_i$ (epoch of spacecraft in\r\ninternal time system), and central body\r\n\r\n\\noindent Find: $\\theta_{LST}$\r\n\r\n\\noindent Definitions:\r\n\\begin{itemize}\r\n %\r\n \\item $\\mathcal{F}_I$ equatorial inertial system (as described in Sec. \\ref{Sec:Equator}) of selected central\r\n body.\r\n %\r\n \\item $\\mathcal{F}_F$ is the central body's fixed system (as described in Sec. \\ref{Sec:Fixed})\r\n %\r\n \\item $\\lambda$ is the longitude of the object in $\\mathcal{F}_F$\r\n %\r\n \\item $\\theta_{MST}$ is the mean sidereal time of the central body's prime meridian.\r\n %\r\n \\item $t_i$ (epoch of spacecraft in internal time system)\r\n\\end{itemize}\r\n\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,230)(55,460)\r\n \\includegraphics[scale=1]{Images/SiderealTimeDiagram.eps}\r\n \\makebox(-350,1066){$\\hat{\\mathbf{x}}_I$}\r\n \\makebox(-380,1277){$\\hat{\\mathbf{x}}_F$}\r\n \\makebox(-617,1330){$\\hat{\\mathbf{y}}_I$}\r\n \\makebox(-800,1240){$\\hat{\\mathbf{y}}_F$}\r\n \\makebox(-575,1059){$\\theta_{MST}$}\r\n \\makebox(-600,1120){$\\lambda$}\r\n \\makebox(-580,1220){$\\theta_{LST}$}\r\n \\makebox(-425,1125){$\\theta_{MHA}$}\r\n \\end{picture}\r\n \\caption{Local Sidereal Time Geometry}\r\n \\label{fig:SiderealTimeDiagram}\r\n\\end{figure}\r\n\r\nWe begin by calculating $\\lambda$ using the algorithm described in\r\nSec. \\ref{Sec:CalcObjectLongitude}. The mean sidereal time\r\n$\\theta_{MST}$ is calculated differently for Earth than for other\r\ncentral bodies. If the central body is Earth, then we use the\r\nfollowing equations to calculate $\\theta_{MST}$.\r\n\r\nFirst, convert $t_i$, which is the spacecraft epoch in the interal\r\ntime system (A1 Modified Julian Date), to $T_{UT1}$, which is the\r\nnumber elapsed Julian centuries from the J2000 epoch.\r\n%\r\n\\begin{equation}\r\n T_{UT1} = \\frac{t_{ut1} - 21544.5}{36525}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\\begin{split}\r\n \\theta_{MST} = & 67310.54841^s + \\\\ &( 876600^h \\left( \\frac{3600 s}{1h}\\right) + 8640184.812866)T_{UT1} + \\\\& 0.093104T_{UT1}^2 - 6.2\\times10^{-6}T_{UT1}^3\r\n \\end{split}\r\n\\end{equation}\r\n\r\n\\subsection{MA} \\index{MA}\r\n\r\nGiven: $\\nu$, $e$\r\n\r\n\\noindent Find: $M$\r\n\r\n\\noindent If e $<$ ( 1 - $1e^{-11}$ ) then calculate $E$ using\r\nalgorithm in Sec. \\ref{sec:EccentricAnomaly}. Then $M$ is\r\ncalculated using\r\n%\r\n\\begin{equation}\r\n M = E - e \\sin{E}\r\n\\end{equation}\r\n%\r\nNote: $E$ must be expressed in radians in the above equation, and\r\nresults in $M$ in radians.\r\n\r\n\\noindent If e $>$ ( 1 + $1e^{-11}$ ) then calculate $H$ using\r\nalgorithm in Sec. \\ref{sec:HyperbolicAnomaly}. Then $M$ is\r\ncalculated using\r\n%\r\n\\begin{equation}\r\n M = e \\sinh{H} - H\r\n\\end{equation}\r\n%\r\nNote: $H$ must be expressed in radians in the above equation, and\r\nresults in $M$ in radians. GMAT outputs \\st{MA} in degrees.\r\n\r\n\\noindent If neither of the above conditions are satisfied, $M = 0$,\r\nand output ``Warning: Orbit is near parabolic in mean anomaly\r\ncalculation. Setting MA = 0\".\r\n\r\n\\subsection{MHA}\r\n\r\n\\subsection{MM}\\index{Spacecraft properties!mean motion}\r\n\\index{MM}\r\n\r\nGiven: $a$, $e$, and $\\mu$\r\n\r\n\\noindent Find: $n$\r\n\r\n\\noindent The orbit is considered either circular or elliptic ( both\r\norbit types use the same equation to calculate $n$) if $ e < 1 -\r\n1e^{-11}$. In this case the mean motion, $n$, is calculated using\r\n%\r\n\\begin{equation}\r\n n = \\sqrt{\\frac{\\mu}{a^3}}\r\n\\end{equation}\r\n%\r\nThe orbit is considered hyperbolic if $ e > 1 + 1e^{-11}$. In this\r\ncase the mean motion, $n$, is calculated using\r\n%\r\n\\begin{equation}\r\n n = \\sqrt{-\\frac{\\mu}{a^3}}\r\n\\end{equation}\r\n%\r\nIf neither of the above two conditions are met, the mean motion is\r\ncalculated using\r\n%\r\n%\r\n\\begin{equation}\r\n n = 2\\sqrt{\\mu}\r\n\\end{equation}\r\n%\r\n\r\n\\noindent \\textit{Comment}: $a$ and $e$ are calculated from the\r\nsatellite cartesian state as shown in Section \\ref{Sec:Cart2Kep},\r\nand $\\mu$ is associated with the specified central body.\r\n\r\n\r\n\\subsection{OrbitPeriod}\\index{Spacecraft properties!OrbitPeriod}\r\n\\index{OrbitPeriod}\r\n\r\nGiven: $a$, and $\\mu$\r\n\r\n\\noindent Find: $T$\r\n\r\n\\noindent If a $ < 0$, then $T = 0$, return.\\\\\r\n%\r\n\r\n\r\n\\noindent Otherwise,\r\n\\begin{equation}\r\n T = 2\\pi\\sqrt{\\frac{a^3}{\\mu}}\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ is calculated from the satellite\r\ncartesian state as shown in Section \\ref{Sec:Cart2Kep}, and $\\mu$ is\r\nassociated with the specified central body.\r\n\r\n\\subsection{PercentShadow } \\label{sec:PercentShadow}\r\n\\index{PercentShadow}\r\n\r\nThe \\st{PercentShadow} parameter calculates the percentage of the\r\napparent solar disk that is in view from the perspective of a\r\nspacecraft. The algorithm used in GMAT was adapted from\r\nMontenbruck\\cite{Montenbruck:Gill:05} pgs. 80-83.\r\n\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,220)(65,275)\r\n \\includegraphics[scale=1]{Images/ShadowGeometry.eps}\r\n \\makebox(-520,850){$\\mathbf{r}_B$}\r\n \\makebox(-460,750){$\\mathbf{r}_\\odot$}\r\n \\makebox(-690,800){$\\mathbf{r}$}\r\n \\makebox(-620,940){$\\mathbf{s}$}\r\n \\makebox(-450,890){$\\mathbf{s}_\\odot$}\r\n \\end{picture}\r\n \\caption{Shadow Geometry}\r\n \\label{fig:ShadowGeometry}\r\n\\end{figure}\r\n%\r\n\\begin{itemize}\r\n \\item $R_\\odot$ = Radius of the Sun\r\n \\item $R_B$ = Radius of occulting body\r\n \\item $R'_\\odot$ = Apparent radius of the Sun\r\n \\item $R'_B$ = Apparent radius of occulting body\r\n \\item $\\mathbf{r}_\\odot$ = Vector from central body to Sun\r\n \\item $\\mathbf{r}_B$ = Vector from central body to occulting body\r\n \\item $\\mathbf{r}$ = Vector from central body to s/c\r\n\\end{itemize}\r\n%\r\n\r\nWe begin by calculating the vector from the occulting body to the\r\nspacecraft, $\\mathbf{s}$, using\r\n%\r\n\\begin{equation}\r\n \\mathbf{s} = \\mathbf{r} - \\mathbf{r}_B\r\n\\end{equation}\r\n%\r\nand the vector from the occulting body to the sun,\r\n$\\mathbf{s}_\\odot$, using\r\n%\r\n\\begin{equation}\r\n \\mathbf{s}_\\odot = \\mathbf{r}_\\odot - \\mathbf{r}_B\r\n\\end{equation}\r\n%\r\n(Note that when the occulting body is the same as the central body,\r\n$\\mathbf{s} = \\mathbf{r}$, and $\\mathbf{s}_\\odot =\r\n\\mathbf{r}_\\odot$)\r\n\r\nNext we calculate the apparent radius of the Sun and occulting body\r\nusing\r\n%\r\n\\begin{eqnarray}\r\n R'_\\odot & = & \\sin^{-1}\\frac{R_\\odot}{\\|\\mathbf{r}_\\odot - \\mathbf{r} \\|}\\\\\r\n R'_B & = & \\sin^{-1}\\frac{R_B}{\\|\\mathbf{r} - \\mathbf{r}_B\\|}\r\n\\end{eqnarray}\r\n%\r\nWe can calculate the apparent separation of the two bodies, $D'$,\r\nusing\r\n%\r\n\\begin{equation}\r\n D' = \\cos^{-1}\\left(\\frac{-\\mathbf{s}^T\\left( \\mathbf{r}_\\odot - \\mathbf{r} \\right)}{s \\|\\mathbf{r}_\\odot - \\mathbf{r}\r\n \\|}\\right)\r\n\\end{equation}\r\n%\r\n\r\nIf $ D' \\geq R'_\\odot + R'_B $, then the spacecraft is not in the\r\nbody's shadow and\r\n%\r\n\\begin{equation}\r\n p = 0;\r\n\\end{equation}\r\n%\r\n\r\nIf $D' \\leq R'_B - R'_\\odot$, then the spacecraft is in full shadow\r\nand\r\n%\r\n\\begin{equation}\r\n p = 100;\r\n\\end{equation}\r\n%\r\n\r\nIf neither of the above conditions are met, the spacecraft is in\r\npartial shadow.\r\n%\r\n\\begin{figure}[htb]\r\n \\begin{picture}(100,135)(35,355)\r\n \\includegraphics[scale=1]{Images/ShadowIllustration.ps}\r\n \\makebox(-650,895){$R'_B$}\r\n \\makebox(-550,870){$R'_\\odot$}\r\n \\makebox(-670,820){$D'$}\r\n \\makebox(-560,750){Sun}\r\n \\makebox(-720,765){Occulting Body}\r\n \\end{picture}\r\n \\caption{Occultation Geometry in Calculation of PercentShadow}\r\n \\label{fig:ShadowIllustration}\r\n\\end{figure}\r\n%\r\n\r\nIf $|R'_s - R'_B| < D' < R'_s + R'_B$, then we can calculate the\r\npercentage of shadow by calculating the area of overlap, $A$, of\r\nthe two apparent disks as shown in Fig.\r\n\\ref{fig:ShadowIllustration}.\r\n%\r\n\\begin{equation}\r\n A = R^{'\\mbox{} 2}_\\odot\\cos^{-1}\\left(\\frac{c_1}{R'_\\odot}\\right) +\r\n R^{'\\mbox{} 2}_B\\cos^{-1}\\left(\\frac{D' - c_1}{R'_B}\\right) -\r\n D'c_2\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{equation}\r\n c_1 = \\frac{ D^{'\\mbox{} 2} + R^{'\\mbox{} 2}_\\odot - R^{'\\mbox{} 2}_B }{2D'}\r\n\\end{equation}\r\n%\r\nand\r\n%\r\n\\begin{equation}\r\n c_2 = \\sqrt{ R^{'\\mbox{} 2}_\\odot - c_1^2 }\r\n\\end{equation}\r\n%\r\nThe percent shadow can be calculated using\r\n%\r\n\\begin{equation}\r\n p = 100 \\frac{A}{\\pi R^{'\\mbox{} 2}_\\odot}\r\n\\end{equation}\r\n\r\nIf the condition $|R'_\\odot - R'_B| < D' < R'_\\odot + R'_B$ is not\r\nsatisfied, then the eclipse is annular and we use\r\n%\r\n\\begin{equation}\r\n p = 100 \\frac{R^{'\\mbox{} 2}_B}{R^{'\\mbox{} 2}_\\odot}\r\n\\end{equation}\r\n\r\n\r\n\\subsection{RA}\\index{RA}\r\n\r\n\\noindent \\textit{Description}: \\st{RA} is the right ascension of a\r\nspacecraft, as shown in Fig. \\ref{fig:SphericalElements} using the\r\nsymbol $\\lambda$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ and $\\mathcal{F}$\r\n\r\n\\noindent Find: $\\lambda$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\lambda= \\tan^{-1}_2(y,x)\r\n\\end{equation}\r\n\r\n\\subsection{RAV} \\index{RAV}\r\n\r\n\\noindent \\textit{Description}: \\st{RAV} is the right ascension of\r\nvelocity of a spacecraft, as shown in Fig.\r\n\\ref{fig:SphericalElements} using the symbol $\\lambda_v$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$ and $\\mathcal{F}$\r\n\r\n\\noindent Find: $\\lambda_v$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n \\lambda_v= \\tan^{-1}_2(v_y,v_x)\r\n\\end{equation}\r\n\r\n\r\n\\subsection{RAAN} \\index{RAAN}\r\n\r\n\\noindent \\textit{Description}: \\st{RAAN} is the right ascension of\r\nthe ascending node as shown in Fig. \\ref{fig:KeplerianElements}\r\nusing the symbol $\\Omega$.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and coordinate system\r\n$\\mathcal{F}$.\r\n\r\n\\noindent Find: $e$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%First calculate the specific angular momentum and its magnitude.\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\|\\mathbf{h} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{n} = [ \\hspace{0.05 in} 0 \\hspace{0.1 in} 0 \\hspace{0.1 in} 1 \\hspace{0.05\r\n in}]^T \\times \\mathbf{h}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n n = \\| n\\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n i = \\cos^{-1}\\left(\\frac{h_z}{h}\\right)\r\n\\end{equation}\r\n%\r\n\r\n\r\n\\noindent if $(i \\geq 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = \\cos^{-1}\\left(\\frac{n_x}{n}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\Omega$: if $n_y < 0$, then $\\Omega = 2\\pi -\r\n\\Omega$\r\n\r\n\r\n\\noindent if $(i < 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\Omega = 0\r\n\\end{equation}\r\n\r\n\r\n\\subsection{RadApo}\\index{Spacecraft properties!RadApo}\r\n\\index{RadApo}\r\n\r\nGiven: $a$, and $e$\r\n\r\n\\noindent Find: $r_a$\r\n\r\n\\noindent if $ 1 - e < 10^{-12}$ then $r_a = 0$. Note, this\r\nmeans that for parabolica, and hyperbolic orbits, GMAT outputs a\r\nvalue of zero for \\st{RadApo}.\r\n%\r\n\\noindent Otherwise,\r\n%\r\n\\begin{equation}\r\n r_a = a(1+e)\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ and $e$ are calculated from the\r\nsatellite cartesian state as shown in Section \\ref{Sec:Cart2Kep}.\r\n\r\n\\subsection{RadPer}\\index{Spacecraft properties!RadPer}\r\n\\index{RadPer}\r\n\r\nGiven: $a$, and $e$\r\n\r\n\\noindent Find: $r_p$\r\n\r\n\\begin{equation}\r\n r_p = a(1-e)\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ and $e$ are calculated from the\r\nsatellite cartesian state as shown in Section \\ref{Sec:Cart2Kep}.\r\n\r\n\\subsection{RLA and DLA} \\index{RLA} \\index{DLA}\r\n\r\n\\noindent \\textit{Description}: \\st{RLA} ($\\lambda_s$) is the right ascenstion of the outgoing\r\nasypmptote of a hyperbolic trajectory. \\st{DLA} ($\\delta_s$)is the declination of the outgoing\r\nasypmptote of a hyperbolic trajectory.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{r}$ and $\\mathbf{v}$ and desired coordinate system.\r\n\r\n\\noindent Find: $\\lambda_s$ and $\\delta_s$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to the desired coordinate system (an excpeption\r\nis thrown if the origin of the requested coordinate system is not a celestial body).\r\nThe eccentricity vector $\\mathbf{e}$ is computed using Eq.~(\\ref{Eq:EccentricityVector}).\r\nIf $\\|\\mathbf{e}\\| < 1 + tol$ then RLA = DLA = NaN and return. Otherwise:\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\| \\mathbf{h} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n C_3 = v^2 - \\frac{2\\mu}{r}\r\n\\end{equation}\r\n%\r\nwhere $\\mu$ is gravitational parameter of the central body at the origin of the given coordinate system. The outgoing asmympote unit vector, $\\hat{\\mathbf{s}} $, is computed using\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{s}} = \\frac{1}{ 1 + C_3 \\displaystyle\\left(\\frac{h}{\\mu}\\right)^2}\\left( \\frac{\\sqrt{C_3}}{\\mu} \\left(\\mathbf{h} \\times \\mathbf{e} \\right) - \\mathbf{e} \\right)\r\n\\end{equation}\r\n%\r\nRLA and DLA are computed from\r\n%\r\n\\begin{equation}\r\n \\lambda_s = \\tan^{-1}(s_y,s_x)\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\delta_s = \\sin^{-1}(s_z)\r\n\\end{equation}\r\n\r\n\\subsection{RMAG} \\index{RMAG}\r\n\r\n\\noindent \\textit{Description}: \\st{RMAG} is the magnitude of the\r\nspacecraft's position vector.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$ and central body.\r\n\r\n\\noindent Find: $r$\r\n\r\nBegin by converting $\\mathbf{r}$ to a coordinate system with the\r\norigin equal to the central body defined by the user, and the\r\nMJ2000Eq axis system. Then,\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n\r\n\\subsection{SemilatusRectum} \\index{SemilatusRectum}\r\n\r\n\\noindent \\textit{Description}: \\st{SemilatusRectum} is the orbit\r\nsemilatus rectum, which is the magnitude of the position vector,\r\nwhen at true anomaly of $90^{\\circ}$.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$ (central\r\nbody).\r\n\r\n\\noindent Find: $p$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to a coordinate\r\nsystem with the origin equal to the central body defined by the\r\nuser, and the MJ2000Eq axis system. Then,\r\n%\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\| \\mathbf{h} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n p = \\frac{h^2}{\\mu}\r\n\\end{equation}\r\n\r\n\r\n\\subsection{SMA} \\index{SMA}\r\n\r\n\\noindent \\textit{Description}: \\st{SMA} is the semimajor axis of\r\nan orbit. The SMA contains information on the size and type of an\r\norbit. If the SMA is positive, the orbit is elliptic. If the SMA\r\nis negative the orbit is hyperbolic. The SMA is undefined for\r\nparabolic orbits. The algorithm used in GMAT to calculate SMA is\r\nadopted from Vallado\\cite{vallado2}.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and $\\mu$ (Central\r\nBody)\r\n\r\n\\noindent Find: $a$\r\n\r\n\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\xi = \\frac{v^2}{2} - \\frac{\\mu}{r}\r\n\\end{equation}\r\n%\r\nif $|1 - e| > 10^{-30}$, then\r\n\\begin{equation}\r\n a = -\\frac{\\mu}{2\\xi}\r\n\\end{equation}\r\n%\r\notherwise, report error and return. Error: ``Warning: Orbit is\r\nnear parabolic and SMA is undefined\".\r\n\r\n\\subsection{TA} \\index{TA}\r\n\r\n\\noindent \\textit{Description}: \\st{TA} is the orbit true anomaly as\r\nshown in Fig. \\ref{fig:KeplerianElements} using the symbol $\\nu$.\r\n\r\n\\noindent \\textit{Dependency}: Central Body.\r\n\r\n\\noindent Given: $\\mathbf{r}$, $\\mathbf{v}$, and coordinate system\r\n$\\mathcal{F}$.\r\n\r\n\\noindent Find: $\\nu$\r\n\r\nBegin by converting $\\mathbf{r}$ and $\\mathbf{v}$ to $\\mathcal{F}$\r\nif necessary. Then,\r\n%First calculate the specific angular momentum and its magnitude.\r\n\\begin{equation}\r\n \\mathbf{h} = \\mathbf{r} \\times \\mathbf{v}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n h = \\|\\mathbf{h} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{n} = [ \\hspace{0.05 in} 0 \\hspace{0.1 in} 0 \\hspace{0.1 in} 1 \\hspace{0.05\r\n in}]^T \\times \\mathbf{h}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n n = \\| n\\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n r = \\| \\mathbf{r} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{e} = \\displaystyle\\frac{(v^2 - \\displaystyle\\frac{\\mu}{r} )\\mathbf{r} - (\\mathbf{r}\\cdot\\mathbf{v} )\\mathbf{v}}{\\mu}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n e = \\| \\mathbf{e} \\|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n i = \\cos^{-1}\\left(\\frac{h_z}{h}\\right)\r\n\\end{equation}\r\n%\r\nThere are three special cases, and they are treated differently.\r\n\r\n\\noindent\\textit{Special Case 1: Elliptic Orbit }\r\n\r\n\\noindent if $(e \\geq 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{\\mathbf{e}\\cdot\\mathbf{r}}{er}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu $: if $\\mathbf{r} \\cdot \\mathbf{v} < 0$, then\r\n$\\nu = 2\\pi - \\nu$\r\n\r\n\\noindent\\textit{Special Case 2: Circular, Inclined Orbit }\r\n\r\n\\noindent if $(e < 10^{-11})$ and $(i \\geq 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{\\mathbf{n}\\cdot\\mathbf{r}}{nr}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu$: if $r_z < 0$, then $\\nu = 2\\pi - \\nu$\r\n\r\n\\noindent\\textit{Special Case 3: Circular, Equatorial Orbit }\r\n\r\n\\noindent if $(e < 10^{-11})$ and $(i < 10^{-11})$, then\r\n%\r\n\\begin{equation}\r\n \\nu = \\cos^{-1}\\left( \\frac{r_x}{r}\\right)\r\n\\end{equation}\r\n%\r\nFix quadrant for $\\nu$: if $r_y < 0$, then $\\nu = 2\\pi - \\nu$\r\n%\r\n\r\n\\subsection{TAIModJulian}\\label{Sec:TAIModJulian}\\index{TAIModJulian}\r\n\r\n\\noindent \\textit{Description}: \\st{TAIModJulian} is the epoch in\r\nthe TAI time system, expressed in the modified Julian date format.\r\nSee Sec. \\ref{Sec:AtomicTime} and \\ref{Sec:JDFormat} for more\r\ndetails.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\n\\noindent Given: $A1$ (epoch in the internal, A1 time system).\r\n\r\n\\noindent Find: $TAI$\r\n\r\nTo convert from A1 to TAI we use the following equation\r\n%\r\n\\begin{equation}\r\n TAI = A1 - 0.0343817 \\mbox{sec}\r\n\\end{equation}\r\n\r\n\\subsection{TTModJulian}\\label{Sec:TTModJulian}\\index{TTModJulian}\r\n\r\n\\noindent \\textit{Description}: \\st{TTModJulian} is the epoch in\r\nthe TT time system, expressed in the modified Julian date format.\r\nSee Sec. \\ref{Sec:DynamicTime} and \\ref{Sec:JDFormat} for more\r\ndetails.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\n\\noindent Given: $A1$ (epoch in the internal, A1 time system).\r\n\r\n\\noindent Find: $TT$\r\n\r\nTo convert from A1 to TT we use the following equation\r\n%\r\n\\begin{equation}\r\n TT = A1 - 0.0343817 \\mbox{sec} + 32.184 \\mbox{sec}\r\n \\label{Eq:TTModJulian}\r\n\\end{equation}\r\n\r\n\\subsection{TTGregorian}\\label{Sec:TTGregorian}\\index{TTGregorian}\r\n\r\n\\noindent \\textit{Description}: \\st{TTGregorian} is the epoch in\r\nthe TT time system, expressed in the Gregorian date format. See\r\nSec. \\ref{Sec:DynamicTime} and \\ref{Sec:GregorianDateFormat} for\r\nmore details.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\n\\noindent Given: $A1$ (epoch in the internal, A1 time system).\r\n\r\n\\noindent Find: $TT$\r\n\r\nTo convert from A1 to TT we use Eq.~(\\ref{Eq:TTModJulian}). Then,\r\nknowing the epoch in the TT time system in the modified Julian\r\ndate format, we use the algorithm in Sec. \\ref{Sec:JDFormat} to\r\nobtain the Gregorian date.\r\n\r\n\r\n\\subsection{Umbra and Penumbra } \\index{Umbra} \\index{Penumbra}\r\n\r\nThe Umbra and Penumbra parameters are used to determine if a\r\nspacecraft is in the shadow of a celestial body. The algorithm used\r\nin GMAT is adapted from Montenbruck\\cite{Montenbruck:Gill:05} pgs.\r\n80-81. For both functions, if the value is less than 1, then the\r\nbody is in shadow, if the function is greater than 1, then the body\r\nis not in shadow.\r\n%\r\n\\begin{figure*}[htb]\r\n\\index{Shadow!Umbra}\\index{Shadow!Penumbra} \\centerline{\r\n\\begin{picture}(200,420)\r\n\\special{psfile=Images/UmbraPenumbraGeom.eps hoffset= -235 voffset=\r\n-180 hscale=105 vscale=105}\\makebox(-40,535){$\\alpha_p$}\r\n\\makebox(580,550){$\\alpha_u$} \\makebox(-640,770){Penumbra (Annular\r\nEclipse)} \\makebox(-950,800){Umbra (Total Eclipse)}\r\n\\makebox(-1140,780){ Penumbra } \\makebox(-970,590){ $s$\r\n}\\makebox(-870,610){ $d$ } \\makebox(-995,655){ $\\ell$ }\r\n\\end{picture}}\\vskip -3.25 in \\caption{ Geometry of Umbra and Penumbra Regions} \\label{fig:UmbraPenumbraGeom}\r\n\\end{figure*}\r\n\r\nFor definitions of see Sec. \\ref{sec:PercentShadow}.\r\n%\r\n\\begin{equation}\r\n \\ell = \\frac{-\\mathbf{s}^T\\mathbf{s}_\\odot}{s_\\odot}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n d = \\sqrt{ s^2 - l^2 }\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\sin{\\alpha_p} = \\frac{R_\\odot + R_B}{s_\\odot}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\sin{\\alpha_u} = \\frac{R_\\odot - R_B}{s_\\odot}\r\n\\end{equation}\r\n%\r\nThe radii of the umbra and penumbra cones, $r_p$ and $r_u$, at\r\ndistance $\\ell$, are respectively\r\n%\r\n\\begin{equation}\r\n r_p = \\tan{\\alpha_p}\\left( \\ell +\r\n \\frac{R_B}{\\sin{\\alpha_p}}\\right)\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n r_u = \\tan{\\alpha_u}\\left( \\ell -\r\n \\frac{R_B}{\\sin{\\alpha_u}}\\right)\r\n\\end{equation}\r\n%\r\nFinally, if $\\ell \\geq 0$\r\n%\r\n\\begin{equation}\r\n d_p = d - r_p\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n d_u = d - |r_u|\r\n\\end{equation}\r\n%\r\nIf $\\ell > 0$ $d_u < 0$ and $r_u < 0$, then the object is in the\r\ntotal umbral\r\neclipse region.\\\\\r\n%\r\nIf $\\ell > 0$ $d_u < 0$ and $r_u \\geq 0$, then the object is in the\r\nannular umbral eclipse region.\\\\\r\n%\r\nIf $\\ell < 0$, then the object is on the day side of the occulting\r\nbody and is not in shadow and\r\n%\r\n\\begin{equation}\r\n d_p = |d - r_p|\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n d_u = |d - |r_u||\r\n\\end{equation}\r\n%\r\n\r\n\\subsection{UTCModJulian} \\label{Sec:UTCModJulian}\\index{UTCModJulian}\r\n\r\n\\noindent \\textit{Description}: \\st{UTCModJulian} is the epoch in\r\nthe UTC time system, expressed in the modified Julian date format.\r\nSee Sec. \\ref{Sec:UniversalTime} and \\ref{Sec:JDFormat} for more\r\ndetails.\r\n\r\n\\noindent \\textit{Dependency}: None.\r\n\r\n\\noindent Given: $A1$ (epoch in the internal, A1 time system).\r\n\r\n\\noindent Find: $UTC$\r\n\r\nTo convert from A1 to UTC we use the following equation\r\n%\r\n\\begin{equation}\r\n UTC = A1 - 0.0343817 \\mbox{sec} - \\Delta AT\r\n\\end{equation}\r\n%\r\nThe default is to read $\\Delta AT$ from the file named\r\n\\textit{tai-utc.dat}. $\\Delta AT$ is the accumulated leap seconds\r\nsince Jan. 1961.\r\n\r\n\r\n\\subsection{VelApoapsis}\\index{Spacecraft properties!VelApoapsis}\r\n\\index{VelApoapsis}\r\n\r\nGiven: $a$, $e$, and $\\mu$\r\n\r\n\\noindent Find: $v_a$\r\n\r\n\\noindent If e $>$ ( 1 - $1e^{-12}$ ) then $v_a = 0$.\r\n%\r\n\r\n\\noindent Otherwise,\r\n%\r\n\\begin{equation}\r\n v_a = \\sqrt{ \\frac{\\mu}{a} \\left(\\frac{1-e}{1+e}\\right)}\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ and $e$ are calculated from the\r\nsatellite cartesian state as shown in Section \\ref{Sec:Cart2Kep},\r\nand $\\mu$ is associated with the specified central body.\r\n\r\n\\subsection{VelPeriapsis} \\index{Spacecraft properties!VelPeriapsis}\r\n\\index{VelPeriapsis}\r\n\r\nGiven: $a$, $e$, and $\\mu$\r\n\r\n\\noindent Find: $v_p$\r\n\r\n\\begin{equation}\r\n v_p = \\sqrt{ \\frac{\\mu}{a} \\left(\\frac{1+e}{1-e}\\right)}\r\n\\end{equation}\r\n\r\n\\noindent \\textit{Comment}: $a$ and $e$ are calculated from the\r\nsatellite cartesian state as shown in Section \\ref{Sec:Cart2Kep},\r\nand $\\mu$ is associated with the specified central body.\r\n\r\n\\subsection{VMAG} \\index{VMAG}\r\n\r\n\\noindent \\textit{Description}: \\st{VMAG} is the magnitude of the\r\nspacecraft's velocity vector, when the velocity is expressed in the\r\nchosen coordinate system.\r\n\r\n\\noindent \\textit{Dependency}: Coordinate System.\r\n\r\n\\noindent Given: $\\mathbf{v}$ and coordinate system $\\mathcal{F}$.\r\n\r\n\\noindent Find: $v$\r\n\r\nBegin by converting $\\mathbf{v}$ to coordinate system $\\mathcal{F}$\r\nif necessary. Then,\r\n%\r\n\\begin{equation}\r\n v = \\| \\mathbf{v} \\| = \\sqrt{v_x^2 + v_y^2 + v_z^2 }\r\n\\end{equation}\r\n%\r\n\r\n\\section{ Other Calculations }\r\n\r\n\\subsection{MA to TA}\r\n\r\n\\noindent \\textit{Description}: This algorithm shows how to\r\ncalculate $\\nu$ given $M$ and $e$ and is taken from\r\nVallado\\cite{vallado2}.\r\n\r\n\\noindent Given: $M$ and $e$.\r\n\r\n\\noindent Find: $\\nu$\r\n\r\nThe algorithm is different for elliptic and hyperbolic orbits. Let's\r\nfirst look at what happens for elliptic orbits.\r\n\r\n\\noindent\\textit{Elliptic Orbit Case}\r\n\r\n\\noindent If $e <= 1$ then use the following algorithm:\r\n\r\n\\noindent Determine initial guess for the Eccentric anomaly\\\\\r\n\\noindent If ( - $\\pi < M < 0$ ) or $M > \\pi$\\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} E = M - e$ \\\\\r\n%\r\nElse\\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} E = M + e$ \\\\\r\n%\r\nEnd\r\n\r\n\\noindent Iterate to determine the eccentric anomaly:\\\\\r\n\r\n\r\n \\noindent Iterate On: $E_{n+1} = E_{n} +\r\n\\displaystyle\\frac{M - E_n + e\\sin{E_n}}{1 -\r\ne\\cos{E_n}}$ \\\\\r\nUntil: $| E_{n+1} - E_n | < 1e^{-8}$\r\n\r\n\\noindent Finally we convert the eccentric anomaly to the true\r\nanomaly using the algorithm given in sec. \\ref{Sec:EAtoTA}\r\n\r\n\\noindent \\textit{Hyperbolic Orbit Case}\r\n\r\n\\noindent If $e > 1$ then use the following algorithm:\r\n\r\nWe begin by choosing the initial guess for the hyperbolic anomaly.\r\nThe initial guess depends on the value of the mean anomaly and the\r\neccentricity:\r\n\r\n\\noindent If $e < 1.6$\r\n\r\n\\begin{minipage}{2.0 in}\r\n\\noindent If ( - $\\pi < M < 0$ ) or $M > \\pi$\\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} H = M - e$ \\\\\r\n%\r\nElse\\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} H = M + e$ \\\\\r\n%\r\nEnd\r\n%\r\n\\end{minipage}\\\\\r\n\r\n\\noindent Else\r\n\r\n\\begin{minipage}{2.25 in}\r\n\\noindent If ($ e < 3.6$ \\& $|M|>\\pi$ ) \\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} H = M - \\mbox{sign}(M)e$ \\\\\r\n%\r\nElse\\\\\r\n%\r\n$\\mbox{}\\hspace{ .25 in} H = \\frac{M}{e-1}$ \\\\\r\n%\r\nEnd\r\n%\r\n\\end{minipage}\r\n\r\n\\noindent End\r\n\r\n\\noindent Iterate to determine the Hyperbolic Anomaly:\r\n\r\n\\noindent Iterate On: $H_{n+1} = H_{n} + \\displaystyle\\frac{M + H_n\r\n-\r\ne\\sinh{H}_n }{e\\cosh{H_n} - 1}$ \\\\\r\nUntil: $| H_{n+1} - H_n | < 1e^{-8}$\r\n\r\n\\noindent Convert the hyperolic anomaly to the true anomaly using\r\nthe algorithm given in sec. \\ref{Sec:HAtoTA}\r\n\r\n\r\n\\subsection{EA to TA} \\label{Sec:EAtoTA}\r\n\r\n\\noindent \\textit{Description}: This algorithm shows how to\r\ncalculate $\\nu$ given $E$ and $e$ and is taken from\r\nVallado\\cite{vallado2}.\r\n\r\n\\noindent Given: $E$ and $e$.\r\n\r\n\\noindent Find: $\\nu$\r\n\r\n\\begin{equation}\r\n \\sin{\\nu} = \\frac{\\sqrt{1 - e^2}\\sin(E)}{1-e\\cos{E}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\cos{\\nu} = \\frac{\\cos{E} - e}{ 1 - e\\cos{E} }\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\nu = \\mbox{atan2}(\\sin{\\nu},\\cos{\\nu})\r\n\\end{equation}\r\n\r\n\\subsection{HA to TA} \\label{Sec:HAtoTA}\r\n\r\n\\noindent \\textit{Description}: This algorithm shows how to\r\ncalculate $\\nu$ given $H$ and $e$ and is taken from\r\nVallado\\cite{vallado2}.\r\n\r\n\\noindent Given: $H$ and $e$.\r\n\r\n\\noindent Find: $\\nu$\r\n\r\n\\begin{equation}\r\n \\sin{\\nu} = -\\frac{\\sqrt{e^2 - 1}\\sinh(H)}{1-e\\cosh{H}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\cos{\\nu} = \\frac{\\cosh{H} - e}{ 1 - e\\cosh{H} }\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\nu = \\mbox{atan2}(\\sin{\\nu},\\cos{\\nu})\r\n\\end{equation}\r\n\r\n\\section{ Libration Points } \\index{Libration Points}\r\n\r\nWe begin by assuming that the planets move in circular orbits about\r\nthe sun, and the mass of a spacecraft is negligible compared to the\r\nmass of the planets. For illustrative purposes, lets consider the\r\nEarth and its orbit about the Sun. In this case, the libration\r\npoints are locations in space where a spacecraft will stay fixed\r\nwith respect to the Earth and Sun. Figure\r\n\\ref{fig:LibrationPointFig1} shows a simple illustration. We see\r\nthe Sun, the Earth's position with respect to the Sun, and the\r\nLibration points $L_1$ and $L_2$ at two different epochs. Notice\r\nthat at $t_1$, the points $L_1$ and $L_2$ are on the Earth-Sun line.\r\nAt a later time, $t_2$, although the Earth has moved with respect to\r\nthe sun, $L_1$ and $L_2$ still lie on the Earth-Sun line.\r\n%\r\n\\begin{figure*}[htb]\r\n\\index{Libration Points!Definition} \\centerline{\r\n\\begin{picture}(100,600)\r\n\\special{psfile=Images/LibrationDescription2.eps hoffset= -295\r\nvoffset= 45 hscale=105 vscale=105}\\makebox(390,1020){$\\mbox{Earth },\r\nL_1, L_2 \\mbox{ at } t_1$} \\makebox(-480,1130){$\\mbox{Earth }, L_1,\r\nL_2 \\mbox{ at } t_2$}\r\n\\end{picture}}\\vskip -4.5 in \\caption{ Geometry of Libration Points } \\label{fig:LibrationPointFig1}\r\n\\end{figure*}\r\n\r\n\r\nThe preceding example gives a brief qualitative description of two\r\nof the Earth-Sun libration points. In general, there are five\r\nlibration points for a given three body system. To determine the\r\nlocations of the libration points, it is convenient to work in a\r\nrotating coordinate system rather than the inertial system shown in\r\nFig. \\ref{fig:LibrationPointFig1}. The system we use is constructed\r\nas follows:\r\n\r\n\\begin{itemize}\r\n\\item Define the primary as the heavier of the two bodies, the\r\nsecondary as the lighter.\r\n\r\n\\item Define the coordinate system x-axis as the axis pointing\r\nfrom the primary to the secondary.\r\n\r\n\\item Define the y-axis to be orthogonal to the x-axis in the\r\nplane of the secondary's motion about the primary, pointing in the\r\ndirection the secondary moves about the primary.\r\n\r\n\\item Define the z-axis orthogonal to the x and y axes to form a\r\nright-handed system.\r\n\r\n\\item Place the origin at center-of-mass of the system.\r\n\r\n\\end{itemize}\r\n\\begin{figure*}[htb]\r\n\\index{Libration Points!Definition} \\centerline{\r\n\\begin{picture}(100,555)\r\n\\special{psfile=Images/LibrationPointFig.eps hoffset= -235 voffset=\r\n-125 hscale=105 vscale=105}\\makebox(-60,820){$\\mu^*$}\r\n\\makebox(157,820){$1 - \\mu^*$} \\makebox(-50,920){$\\gamma_1$}\r\n\\makebox(50,920){$\\gamma_2$} \\makebox(-630,920){$\\gamma_3$}\r\n\\makebox(-695,850){$L3$}\\makebox(-220,850){$L1$}\r\n\\makebox(-20,850){$L2$} \\makebox(-340,640){$L4$}\r\n\\makebox(-350,1073){$L5$} \\makebox(-385,890){$x$}\r\n\\makebox(-485,997){$y$} \\makebox(-670,780){Primary}\r\n\\makebox(-100,780){Secondary}\r\n\\end{picture}}\\vskip -4.5 in \\caption{ Location of Libration Points } \\label{fig:LibrationPointFig}\r\n\\end{figure*}\r\n\r\nThis coordinates system is illustrated in Fig.\r\n\\ref{fig:LibrationPointFig}. The locations of the libration points\r\nin the rotating coordinate system can be found by calculating the\r\nvalues of $\\gamma$ that solve the following equations:\r\n%\r\n\\begin{eqnarray}\r\n\\begin{split}\r\n\\gamma _1^5 - \\left( {3 - \\mu^*} \\right)\\gamma _1^4 & + \\left( {3 -\r\n2\\mu^*} \\right)\\gamma _1^3 - \\mu^*\\gamma _1^2 \\\\ &+ 2\\mu^*\\gamma\r\n_1 - \\mu^* = 0 \\hspace{.1 in} \\mbox{(For L1)} \\label{Eq:L1quintic}\r\n\\end{split}\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\r\n\\begin{split}\r\n \\gamma _2^5 + \\left( {3 - \\mu^*}\r\n\\right)\\gamma _2^4 & + \\left( {3 - 2\\mu^*} \\right)\\gamma _2^3 -\r\n\\mu^*\\gamma _2^2 \\\\ &- 2\\mu^*\\gamma _2 - \\mu^* = 0 \\hspace{.1\r\nin} \\mbox{(For L2)} \\label{Eq:L2quintic}\r\n\\end{split}\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\r\n\\begin{split}\r\n \\gamma _3^5 + & \\left( {2 + \\mu^*} \\right)\\gamma _3^4 + \\left(\r\n{1 + 2\\mu^*} \\right)\\gamma _3^3 - \\left( {1 - \\mu^*} \\right)\\gamma\r\n_3^2 \\\\ & - 2\\left( {1 - \\mu^*} \\right)\\gamma _3 - \\left( {1 -\r\n\\mu^*} \\right) = 0 \\hspace{.1 in} \\mbox{(For L3)}\r\n\\label{Eq:L3quintic}\r\n\\end{split}\r\n\\end{eqnarray}\r\n%equations:\r\n%\\begin{eqnarray}\r\n%\\gamma _1^5 - \\left( {3 - \\mu^*} \\right)\\gamma _1^4 + \\left( {3 -\r\n%2\\mu^*} \\right)\\gamma _1^3 - \\mu^*\\gamma\r\n%_1^2 + 2\\mu^*\\gamma _1 - \\mu^* & = 0 \\hspace{.1 in} \\mbox{(For L1)} \\label{Eq:L1quintic} \\\\\r\n%%\r\n% \\gamma _2^5 + \\left( {3 - \\mu^*}\r\n%\\right)\\gamma _2^4 + \\left( {3 - 2\\mu^*} \\right)\\gamma _2^3 -\r\n%\\mu^*\\gamma _2^2 - 2\\mu^*\\gamma _2 - \\mu^* & = 0 \\hspace{.1 in}\r\n%\\mbox{(For L2)} \\label{Eq:L2quintic}\r\n%\\\\\r\n%%\r\n% \\gamma _3^5 + \\left( {2 + \\mu^*}\r\n%\\right)\\gamma _3^4 + \\left( {1 + 2\\mu^*} \\right)\\gamma _3^3 -\r\n%\\left( {1 - \\mu^*} \\right)\\gamma _3^2 - 2\\left( {1 - \\mu^*}\r\n%\\right)\\gamma _3 - \\left( {1 - \\mu^*} \\right) & = 0 \\hspace{.1\r\n%in} \\mbox{(For L3)} \\label{Eq:L3quintic}\r\n%\\end{eqnarray}\r\n%%\r\nwhere\r\n%\r\n\\begin{equation}\r\n\\mu^* = \\frac{{m_2 }} {{m_1 + m_2 }}\r\n\\end{equation}\r\n%\r\n\r\nEquations~(\\ref{Eq:L1quintic})-(\\ref{Eq:L3quintic}) do not have\r\nexact analytic solutions.%\r\n% Szebehely\\cite{Szebehely67} presents series solutions for\r\n%$\\gamma_1$ and $\\gamma_2$, and $\\gamma_3$. The expansions for\r\n%$\\gamma_1$ and $\\gamma_2$ are:\r\n%%\r\n%\\begin{equation}\r\n% \\gamma_1 = \\left(\\frac{\\mu}{3}\\right)^{1/3} -\r\n% \\frac{1}{3}\\left(\\frac{\\mu}{3}\\right)^{2/3} -\r\n% \\frac{1}{9}\\left(\\frac{\\mu}{3}\\right)^{1} -\r\n% \\frac{23}{81}\\left(\\frac{\\mu}{3}\\right)^{4/3} +\r\n% \\frac{151}{243}\\left(\\frac{\\mu}{3}\\right)^{5/3} -\r\n% \\frac{1}{9}\\left(\\frac{\\mu}{3}\\right)^{2}\r\n%\\end{equation}\r\n%%\r\n%\\begin{equation}\r\n% \\gamma_2 = \\left(\\frac{\\mu}{3}\\right)^{1/3} +\r\n% \\frac{1}{3}\\left(\\frac{\\mu}{3}\\right)^{2/3} -\r\n% \\frac{1}{9}\\left(\\frac{\\mu}{3}\\right)^{1} -\r\n% \\frac{31}{81}\\left(\\frac{\\mu}{3}\\right)^{4/3} -\r\n% \\frac{119}{243}\\left(\\frac{\\mu}{3}\\right)^{5/3} -\r\n% \\frac{1}{9}\\left(\\frac{\\mu}{3}\\right)^{2}\r\n%\\end{equation}\r\n%%\r\n%However,\r\nSzebehely\\cite{Szebehely67} notes that they are most easily solved\r\nusing an iterative method with the following as the initial guesses:\r\n%\r\n\\begin{equation}\r\n\\gamma _1 = \\gamma _2 = \\left( {\\frac{\\mu^*} {{3\\left( {1 - \\mu^*}\r\n\\right)}}} \\right)^{{\\raise0.7ex\\hbox{$1$} \\!\\mathord{\\left/\r\n {\\vphantom {1 3}}\\right.\\kern-\\nulldelimiterspace}\r\n\\!\\lower0.7ex\\hbox{$3$}}}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n\\gamma _3 = 1\r\n\\end{equation}\r\n\r\nGMAT uses the Newton-Raphson method to solve for the roots of the\r\nequations by iterating on\r\n%\r\n\\begin{equation}\r\n \\gamma(i+1) = \\gamma(i) - \\frac{F(\\gamma(i))}{F'(\\gamma(i))}\r\n\\end{equation}\r\n%\r\nuntil the the difference $|\\gamma(i+1) - \\gamma(i)| < 10^{-8}$. The\r\nderivative $F'(\\gamma)$ for each libration point is shown below.\r\n%\r\n\\begin{eqnarray}\\begin{split}\r\nF'(\\gamma) = 5\\gamma _1^4 - 4\\left( {3 - \\mu^*} \\right)&\\gamma _1^3\r\n+ 3\\left( {3 - 2\\mu^*} \\right)\\gamma _1^2\\\\& - 2\\mu^*\\gamma\r\n_1 + 2\\mu^* \\hspace{.1 in} \\mbox{(For L1)} \\label{Eq:L1Deriv} \\\\\r\n\\end{split}\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\\begin{split}\r\nF'(\\gamma) = 5\\gamma _2^4 + 4\\left( {3 - \\mu^*} \\right)&\\gamma_2^3\r\n+ 3\\left( {3 - 2\\mu^*} \\right)\\gamma _2^2\\\\& - 2\\mu^*\\gamma _2 -\r\n2\\mu^* \\hspace{.1 in} \\mbox{(For L2)}\r\n \\label{Eq:L2Deriv}\\end{split}\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\\begin{split}\r\nF'(\\gamma) = 5\\gamma _3^4 &+ 4\\left( {2 + \\mu^*} \\right)\\gamma _3^3\r\n+ 3\\left( {1 + 2\\mu^*} \\right)\\gamma _3^2\\\\& - 2\\left( {1 - \\mu^*}\r\n\\right)\\gamma _3 - 2\\left( {1 - \\mu^*} \\right) \\hspace{.1 in}\r\n\\mbox{(For L3)} \\label{Eq:L3Deriv}\\end{split}\r\n\\end{eqnarray}\r\n\r\n\\begin{table}[htb]\r\n\\centering \\caption{Location of Libration Points in RLP Frame, with\r\nthe Origin at the Primary Body} \\index{Libration Points!Location}\r\n\\begin{tabular}{ccc}\r\n \\hline\\hline\r\n % after \\\\: \\hline or \\cline{col1-col2} \\cline{col3-col4} ...\r\n Point & $x$-Position & $y$-Position \\\\\r\n \\hline\r\n L1 & $1 - \\gamma_1$ & 0 \\\\\r\n L2 & $1 + \\gamma_2$ & 0 \\\\\r\n L3 & $-\\gamma_3$ & 0 \\\\\r\n L4 & $1/2$ & $\\sqrt{3}/2$ \\\\\r\n L5 & $1/2$ & -$\\sqrt{3}/2$ \\\\\r\n \\hline\r\n \\end{tabular}\r\n\\end{table}\r\n\r\nWe now need to redimensionalize the results found in the rotating\r\nsystem, and perform the necessary transformations to obtain the\r\nresults in the MJ2000 system. Let's assume that $\\mathbf{r}_s$ ,\r\n$\\mathbf{v}_s$, and $\\mathbf{a}_s$ are the position, velocity, and\r\nacceleration vectors respectively of the secondary body, with\r\nrespect to the primary body, expressed in the FK5 system. Then, the\r\nposition of the $i^{th}$ libration point can be expressed in the\r\nrotating system with the origin centered on the primary body as\r\n%\r\n\\begin{equation}\r\n \\mathbf{r}^i = r_s\\left[x_i \\hspace{.2 in} y_i \\hspace{.2 in} 0 \\right]^T\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{equation}\r\n r_s = \\| \\mathbf{r}_s\\|\r\n\\end{equation}\r\n%\r\nThe velocity of the $i^{th}$ libration point can be expressed in the\r\nrotating system with the origin centered on the primary body as\r\n%\r\n\\begin{equation}\r\n \\mathbf{v}^i = \\frac{\\mathbf{v}_s\\cdot\\mathbf{r}_s}{r_s}\\left[x_i \\hspace{.2 in} y_i \\hspace{.2 in} 0 \\right]^T\r\n\\end{equation}\r\n%\r\nNow we have the redimensionalized position and velocity vectors of\r\nthe libration point in the rotating coordinate system defined by the\r\nmotion of the secondary body with respect to the primary body. To\r\ndetermine the position and velocity vectors in the FK5 system, with\r\nthe origin located at the primary body, we need to determine the\r\nrotation matrix and its derivative as follows:\r\n%\r\n\\begin{equation}\r\n \\mathbf{R}^{Ii} = \\begin{pmatrix}\r\n \\hat{x}_1 & \\hat{y}_1 & \\hat{z}_1 \\\\\r\n \\hat{x}_2 & \\hat{y}_2 & \\hat{z}_2 \\\\\r\n \\hat{x}_3 & \\hat{y}_3 & \\hat{z}_3\r\n \\label{Eq:RObjectReferenced}\r\n \\end{pmatrix}\r\n\\end{equation}\r\nand\r\n\\begin{equation}\r\n \\dot{\\mathbf{R}}^{Ii} = \\begin{pmatrix}\r\n \\dot{\\hat{x}}_1 & \\dot{\\hat{y}}_1 & \\dot{\\hat{z}}_1 \\\\\r\n \\dot{\\hat{x}}_2 & \\dot{\\hat{y}}_2 & \\dot{\\hat{z}}_2 \\\\\r\n \\dot{\\hat{x}}_3 & \\dot{\\hat{y}}_3 & \\dot{\\hat{z}}_3\r\n \\label{Eq:RdotObjectReferenced}\r\n \\end{pmatrix}\r\n\\end{equation}\r\n%\r\nwhere\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{x}} = \\frac{\\mathbf{r}_s}{r_s}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{z}} = \\frac{\\mathbf{r}_s\\times\\mathbf{v}_s}{\\| \\mathbf{r}_s\\times\\mathbf{v}_s \\|}\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\hat{\\mathbf{y}} = \\hat{\\mathbf{z}} \\times \\hat{\\mathbf{x}}\r\n\\end{equation}\r\n%\r\nand\r\n\\begin{equation}\r\n \\dot{\\hat{\\mathbf{x}}} = \\dot{\\hat{\\mathbf{r}}}_s = \\frac{\\mathbf{v}_s}{r_s} -\r\n \\frac{\\hat{\\mathbf{r}}_s}{r_s}\r\n \\left(\\hat{\\mathbf{r}}_s \\cdot\r\n \\mathbf{v}_s \\right)\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\dot{\\hat{\\mathbf{z}}} =\\frac{\\mathbf{r}_s \\times \\mathbf{a}_s}{\\| \\mathbf{r}_s\\times\\mathbf{v}_s \\|} - \\frac{\\hat{\\mathbf{z}}}{\\| \\mathbf{r}_s\\times\\mathbf{v}_s \\|}\r\n \\left( \\mathbf{r}_s\\times\\mathbf{a}_s \\cdot \\hat{\\mathbf{z}} \\right)\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\dot{\\hat{\\mathbf{y}}} = \\dot{\\hat{\\mathbf{z}}} \\times\r\n \\hat{\\mathbf{x}} +\\hat{\\mathbf{z}} \\times\r\n \\dot{\\hat{\\mathbf{x}}}\r\n\\end{equation}\r\n%\r\nGMAT currently assumes that the terms $\\mathbf{r}_s \\times\r\n\\mathbf{a}_s$ are zero.\r\n\r\n\r\n We finally arrive at the position of\r\nthe Libration Point in the FK5 system with the origin at the primary\r\nby performing the calculations:\r\n%\r\n\\begin{equation}\r\n \\mathbf{r} = \\mathbf{R}^{Ii}\\mathbf{r}^i\r\n\\end{equation}\r\n%\r\n\\begin{equation}\r\n \\mathbf{v} = \\dot{\\mathbf{R}}^{Ii}\\mathbf{r}^i + \\mathbf{R}^{Ii}\\mathbf{v}^i\r\n\\end{equation}\r\n%\r\n%Note that in the GMAT script language, we can define this\r\n%coordinate system using\r\n%%\r\n%\\begin{quote}\r\n%\\begin{verbatim}\r\n% Create CoordinateSystem CS;\r\n% CS.Origin = PrimaryBodyName;\r\n% CS.Primary = PrimaryBodyName;\r\n% CS.Secondary = SecondaryBodyName;\r\n% CS.XAxis = R;\r\n% CS.ZAxis = N;\r\n% \\end{verbatim}\r\n%\\end{quote}\r\n\\section{ Barycenter } \\index{Barycenter}\r\n\r\nThe barycenter of a system of point masses, $\\mathbf{r}_b$, is also\r\ncalled the center of mass. If we have a system of $n$ bodies, and\r\nwe know the position of the $i^{th}$ body with respect to a common\r\nreference system, then we can calculate the barycenter of the system\r\nusing\r\n%\r\n\\begin{equation}\r\n \\mathbf{r}_b = \\displaystyle\\frac{\\displaystyle\\sum_{i=1}^n m_i \\mathbf{r}_i}{\\displaystyle\\sum_{i=1}^n m_i}\r\n\\end{equation}\r\n%\r\nSimilarly, we can calculate the velocity of the barycenter using the\r\nfollowing equation\r\n%\r\n\\begin{equation}\r\n \\mathbf{v}_b = \\displaystyle\\frac{\\displaystyle\\sum_{i=1}^n m_i \\mathbf{v}_i}{\\displaystyle\\sum_{i=1}^n m_i}\r\n\\end{equation}\r\n\r\n\\section{Ground Station Model}\r\n\r\nThis section contains algorithms for converting the station location\r\nparameters to body fixed and inertial cartesian coordinates. Ground\r\nstations are defined in the body fixed frame of the\r\n\\st{CentralBody}. The user can define the ground system location\r\nusing several methods. GMAT internally uses the cartesian location\r\nof the ground station with respect to the body fixed coordinate\r\nsystem. If the user provides the ground station location in\r\ncartesian elements, no transformation is required to obtain the body\r\nfixed representation. For spherical elements with respect to the\r\nGeocentric reference, GMAT uses the following transformation to\r\ncalculate the cartesian station location in the body fixed frame,\r\n%\r\n\\begin{eqnarray}\r\n \\mathbf{r}_F = \\left(\r\n \\begin{array}{cc}\r\n \\cos{\\phi}\\cos{\\lambda}\\\\\r\n \\cos{\\phi}\\sin{\\lambda}\\\\\r\n \\sin{\\phi}\r\n \\end{array}\r\n \\right)\r\n\\end{eqnarray}\r\n%\r\nIf the station location is defined in terms of the Geodetic\r\nreference, then the cartesian location is calculated as\r\nfollows\\cite{Vallado}:\r\n%\r\n\\begin{eqnarray}\r\n C &=& \\displaystyle\\frac{R_b}{\\sqrt{1 - e_b^2\\sin^2{\\phi}}}\\\\\r\n%\r\n S &=& C\\left(1 - e_b^2\\right)\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\r\n r_{xy} &=& \\left( C + h_{e\\ell\\ell}\\right)\\cos{\\phi}\\\\\r\n%\r\n r_{z} &=& \\left( S + h_{e\\ell\\ell}\\right)\\sin{\\phi}\r\n\\end{eqnarray}\r\n%\r\n\\begin{eqnarray}\r\n \\mathbf{r}_F = \\left(\r\n \\begin{array}{cc}\r\n r_{xy}\\cos{\\lambda}\\\\\r\n r_{xy}\\sin{\\lambda}\\\\\r\n r_{z}\r\n \\end{array}\r\n \\right)\r\n\\end{eqnarray}\r\n%\r\nTo calculate the body-centered MJ2000Eq representation, we calculate\r\nthe rotation matrix from fixed to inertial, $\\mathbf{R}_{IF}$, using\r\nthe algorithm in Sec \\ref{Sec:Fixed} of the GMAT Mathematical\r\nSpecification. Then,\r\n%\r\n\\begin{equation}\r\n \\mathbf{r}_I = \\mathbf{R}_{IF}\\mathbf{r}_F\r\n\\end{equation}\r\n", "meta": {"hexsha": "f339da3e099647bddc30cca8a3f5c666131b9113", "size": 106112, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/SystemDocs/MathematicalSpecification/CalculationObjects.tex", "max_stars_repo_name": "ddj116/gmat", "max_stars_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_stars_repo_licenses": ["NASA-1.3"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-05-16T16:58:21.000Z", "max_stars_repo_stars_event_max_datetime": "2020-05-16T16:58:21.000Z", "max_issues_repo_path": "doc/SystemDocs/MathematicalSpecification/CalculationObjects.tex", "max_issues_repo_name": "ddj116/gmat", "max_issues_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_issues_repo_licenses": ["NASA-1.3"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/SystemDocs/MathematicalSpecification/CalculationObjects.tex", "max_forks_repo_name": "ddj116/gmat", "max_forks_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_forks_repo_licenses": ["NASA-1.3"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.1941747573, "max_line_length": 345, "alphanum_fraction": 0.6529987183, "num_tokens": 35184, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.863391617003942, "lm_q2_score": 0.727975460709318, "lm_q1q2_score": 0.6285279101610076}} {"text": "\\section{Attitude} \\label{Ch:Attitude}\nThe attitude of a spacecraft can be defined qualitatively as how the spacecraft\nis oriented in inertial space, and how that orientation changes in time. GMAT\nhas the ability to model the orientation and rate of rotation of a spacecraft\nusing several different mathematical models. Currently, GMAT assumes that a\nspacecraft is a rigid body.\n\nThere are many ways to quantitatively describe the orientation and rate of\nrotation of a spacecraft, just like there are many ways we can quantitatively\ndescribe an orbit state. Let's define any set of numbers that can uniquely\ndefine the spacecraft attitude as an \\emph{attitude parameterization}. GMAT\nallows its users to employ several common attitude parameterizations including\nquaternions, Euler angles, the Direction Cosine Matrix (DCM) or Attitude Matrix\n\\textbf{A}, Euler angle rates, and the angular velocity vector. Given an initial\nattitude state, GMAT can propagate spacecraft attitude using one of several\nkinematic or dynamic attitude propagation models.\n\nOur notation will be based on a compromise that spans multiple disciplinary\nfields, from astrodynamics and aircraft flight simulation to computer graphics,\nand is informed by the AIAA and CCSDS standards. We want a consistent\ndefinition in the inner core algorithms of GMAT combined with the capability of\ntranslating internal variables into interface variables that are easy for users\nto understand and work with. For example, if a user wishes to visualize the\nattitude of a specific mission spacecraft using quaternions extracted from real\nworld telemetry, GMAT should be flexible enough for that user to define all four\nquaternion components, their locations with respect to each other, and their\nnumerical signs. The user should also be able to specify the two reference\nframes the attitude parameterization is relating, and the direction the attitude\nparameters take a transformation. In other words, following the CCSDS standard,\nthe user should be able to specify which is the ``from\" frame and which is the\n``to\" frame.\n\nWe should also point out that the parameters presented in this document can\ndescribe not only the ``transformation\" of a vector between two reference frames,\nbut also the ``rotation\" of a vector from one orientation to another in the same\nreference frame. The literature uses the definition of ``transformation\" and\n``rotation\" quite liberally, and a detailed search has found little agreement on\nterminology and notation. Almost all authors agree that the components of the\n3$\\times$3 matrix relating vectors to each other are ``Direction Cosines\". They\nare the dot products of the basis axes unit vectors of one reference frame with\nthe basis axes unit vectors of the other reference frame. They are the cosines\nof the angles between each of the basis axes.\n\nFor now it is suggested that the matrix we use to transform or rotate vectors be\nlabeled \\textbf{R} to remain consistent with the mathematical development\nearlier in this specification in Chapter~\\ref{Ch:CoordinateSystems}. The\n3$\\times$3 matrix \\textbf{R} will relate various reference frames to each other\nand can be used to rotate vectors between frames such as in the FK5 reduction\nprocess. Meanwhile, if we wish to explicitly model the attitude or orientation\nof a spacecraft, we can represent this with the 3$\\times$3 matrix \\textbf{A}.\nThe letter brings the word ``Attitude\" to mind, and is used by several well known\nattitude dynamics authors including Landis Markley who wrote the\n\"Parameterization of the Attitude\" section of ``Spacecraft Attitude\nDetermination and Control\" by James R. Wertz, a very popular text in the field.\nTo be consistent with Chapter~\\ref{Ch:CoordinateSystems}, we should specify the\ndirection of the transformation with additional notation.\n\nMeanwhile, if we intentionally keep our notation as clutter free as possible,\nthe definition of \\textbf{A} will be as follows.\n\n\\begin{equation}\n \\textbf{ b = Ar}\n \\label{Eq:FundamentalAMat}\n\\end{equation}\n\nIn this equation \\textbf{b} is a vector in the spacecraft body frame, and\n\\textbf{r} is a vector in the reference frame. The attitude matrix \\textbf{A}\ntransforms vectors from the reference frame to the body frame.\n\nThe human mind has some difficulty visualizing the orientation of a spacecraft\nwhen presented with the nine component attitude matrix. Likewise, it is\ndifficult to know if an aircraft or spacecraft is ``upside-down\" given the four\ncomponents of a quaternion vector. For human interface purposes, the attitude\ndescriptor of choice is the set of three Euler Angles, such as the classic\n``roll\", ``pitch\", and ``yaw\". The CCSDS specification avoids mentioning the DC\nmatrix altogether. Meanwhile, GMAT will use it as an intermediate form of\nattitude parameterization when transforming from internal attitude propagation\nequations using quaternions, and human interpretable Euler Angles. GMAT will\nalso use it for both Coordinate System Fixed and Spinning Spacecraft modes of\nattitude propagation.\n\nIn this chapter, we discuss the attitude parameterizations supported in GMAT,\nand how to convert between the different types. We discuss the internal state\nparameterization that GMAT uses. Next we investigate the types of attitude\nmodes in GMAT and discuss in detail how GMAT propagates the spacecraft attitude\nin all of the Kinematic and Dynamic attitude modes. We conclude the chapter\nwith a discussion of how GMAT converts between different attitude\nparameterizations.\n\n\\subsection{Attitude Propagation}\n\nGiven a set of initial conditions that define the attitude, GMAT can propagate\nthe attitude using several methods. Currently, GMAT supports both Kinematic\nattitude and Dynamic attitude propagation. In Kinematic mode, the attitude is\ndefined by describing the desired orientation with respect to other objects such\nas spacecraft or celestial bodies. With this information, GMAT can calculate\nthe required attitude to satisfy the desired geometrical configuration. In\nDynamic mode, the user must also supply spacecraft body inertial mass properties,\nand tell GMAT how to apply torques dynamically to the spacecraft. With these\nconditions specified, GMAT will numerically integrate quaternion attitude\nequations of motion combined with Euler's moment equations. The resulting\nattitude motion will display basic physics-based phenomena such as precession\nand nutation. Dynamic mode will allow a user to model the implementation of\nattitude control laws, as well as the effects of asymmetric propellant mass\ndepletion and dynamic spacecraft component articulation. It will also\neventually support the modeling of attitude perturbations such as Gravity\nGradient, Aerodynamic Drag and Solar Photon Pressure Torques, and spacecraft\nmagnetic interaction with planetary magnetic fields. This section presents the\ndifferent Kinematic and Dynamic attitude modes, and how GMAT calculates the\nattitude state in each mode. Let's begin by looking at the internal attitude\nstate representations and how the user can define initial conditions.\n\n\\vspace{- .1 in} \\subsubsection{Internal State Representation and Attitude\nInitial Conditions}\n\nAs mentioned in the introduction, certain attitude parameterizations are more\nuseful for attitude propagation, while other attitude parameterizations are\nmore intuitive for providing attitude initial conditions or output. GMAT uses\ndifferent internal parameterizations of the attitude depending on the attitude\nmode. The type of parameterization is chosen to make the attitude propagation\nalgorithms natural and convenient. For the Kinematic modes, GMAT uses the\nDirection Cosine Matrix (DCM) or \"Attitude\" matrix that represents the rotation\nfrom the inertial reference frame to the body frame. The notation for the\nAttitude matrix, including the direction of the rotation it parameterizes is\n\\textbf{A}$_{BI}$. The subscripts denote the direction of the rotation from\nInertial to Body frames. For Dynamic mode, six degree of freedom (6DOF)\nattitude propagation, GMAT uses the quaternion that represents the rotation from\nthe inertial reference frame to the body frame. The notation for the quaternion,\nincluding the direction of the rotation it parameterizes, is\n\\textbf{\\emph{q}}$_{BI}$. Once again, the subscripts denote the direction of\nthe rotation from Inertial to Body frames. GMAT uses the angular velocity of\nthe body with respect to the inertial frame, expressed in the body frame,\n$\\{\\boldsymbol\\omega_{IB}\\}_B$, as the rate portion of the state vector.\n\nFor convenience, the user can choose a coordinate system in which to define the\ninitial attitude state. Let's call this system $\\mathcal{F}_i$. The user can\ndefine the initial attitude with respect to $\\mathcal{F}_i$ using Euler angles,\nthe DCM or attitude matrix \\textbf{A}, or quaternions. The user can define the\nbody rate with respect to $\\mathcal{F}_i$ by defining the angular velocity in\n$\\mathcal{F}_i$, $\\{\\boldsymbol\\omega_{IB}\\}_i$, by defining the angular\nvelocity in the inertial frame as measured in body coordinates\n$\\{\\boldsymbol\\omega_{IB}\\}_B$ or by defining the Euler angle rates. Note that\nnot all attitude modes require these three pieces of information. The specific\ninputs for each attitude mode are discussed below, along with details about how\nattitude propagation is performed in each mode.\n\n\\subsubsection{Kinematic Attitude Propagation}\n\nThe Kinematic attitude mode allows a user to define a geometrical configuration\nbased on the relative position of a spacecraft with respect to other spacecraft\nor celestial bodies, and with respect to different coordinate systems. In\nKinematic mode, GMAT does not integrate the attitude equations of motion, but\nrather calculates the attitude based on the geometrical definition provided by\nthe user. There are two Kinematic modes to choose from. These are Coordinate\nSystem Fixed and Spinning Spacecraft. The different modes allow the user to\nconveniently define the spacecraft attitude depending on the type of attitude\nprofile needed for a specific mission. To begin, let's look at how GMAT\ncalculates the attitude state in the Coordinate System Fixed attitude mode\n(CSFixed).\n\n\\subsubsection{Coordinate System Fixed Mode}\n\nIn the CSFixed attitude mode, the user specifies a coordinate system $\\mathcal{F}_i$ in which to fix the attitude.\n$\\mathcal{F}_i$ can be any of the default coordinate systems or any user defined\ncoordinate system. GMAT calculates the attitude matrix of $\\mathcal{F}_i$ using the\nfollowing equation.\n%\n\\begin{equation}\n \\mathbf{A}_{BI} = \\mathbf{R}_{iI}\n \\label{Eq:CSFixedRotationMatrix}\n\\end{equation}\n%\n$\\mathbf{R}_{iI}$ is the DCM matrix relating $\\mathcal{F}_I$ to $\\mathcal{F}_i$\nand GMAT knows how to calculate this matrix for all allowable $\\mathcal{F}_i$.\nFor details on the calculation of this matrix for all coordinate systems in GMAT\nsee Ch.~\\ref{Ch:CoordinateSystems}. To calculate $\\{\\mathbf{\\boldsymbol\\omega}_{IB}\\}_B$, we start from Euler's\nequation:\n%\n\\begin{equation}\n \\dot{\\mathbf{A}}_{BI} = -\\{\\mathbf{\\boldsymbol\\omega^\\times}_{IB}\\}_B\\mathbf{A}_{BI}\n \\label{Eq:CSFixedKinematics}\n\\end{equation}\n%\n% Note from Dunn: We want the \"where\" statement below to stay above the\n% equation and not be orphaned on the preceding page. Adding \\\\* makes the\n% \"where\" line stay with the equation below. Then we have to add the [0.5cm]\n% AND skip the white line between this line and the \\begin{equation} command so\n% the vertical white space looks good. If we change to 10 pt when integrating\n% back into Steve's main document, the [0.5cm] command might need a tweak.\nwhere\\\\*[0.5cm]\n%\n\\begin{equation}\n \\{ \\mathbf{\\boldsymbol\\omega^\\times}_{IB}\\}_B\\hspace{0.082cm}\\equiv\n \\begin{bmatrix}\n 0 & -\\omega_{3} & \\omega_{2} \\\\\n \\omega_{3} & 0 & -\\omega_{1} \\\\\n -\\omega_{2} & \\omega_{1} & 0\n \\end{bmatrix}\n \\equiv \\hspace{0.082cm}\\boldsymbol \\Omega\n \\label{Eq:OmegaCrossMatrix}\n\\end{equation}\n%\nand $\\{\\mathbf{\\boldsymbol\\omega}_{IB}\\}_B$ is the skew symmetric angular\nvelocity ``cross product matrix\" of $\\mathcal{F_B}$ with respect to\n$\\mathcal{F}_I$, expressed in $\\mathcal{F}_B$. Solving Equation\n\\ref{Eq:CSFixedKinematics} for $\\{\\mathbf{\\boldsymbol\\omega}_{IB}^\\times\\}_B$\nwe obtain\n%\n\\begin{equation}\n \\{\\mathbf{\\boldsymbol\\omega^\\times}_{IB}\\}_B = -\\dot{\\mathbf{A}}_{BI}\\mathbf{A}_{BI}^{T}\n \\label{Eq:CSFixedKinematics2}\n\\end{equation}\n%\nwhere \n%\n\\begin{equation}\n \\dot{\\mathbf{A}}_{BI} = \\dot{\\mathbf{R}}_{iI}\n \\label{Eq:CSFixedKinematics2}\n\\end{equation}\n%Taking the derivative of Equation \\ref{Eq:CSFixedRotationMatrix} with respect to\n%time yields\n%%\n%\\begin{equation}\n% {\\mathbf{\\dot A}_{BI}} = \\mathbf{A}_{Bi}{\\mathbf{\\dot R}_{iI}}\n% \\label{Eq:CSFixedTimeDerivative}\n%\\end{equation}\n%%\n%because by definition, for the CSFixed mode,\n%$ \\mathbf{\\dot A}_{Bi} = \\mathbf{0}$.\n%Substituting Eq. \\ref{Eq:CSFixedTimeDerivative} into\n%Eq. \\ref{Eq:CSFixedKinematics2} we obtain\n%%\n%\\begin{equation}\n% \\{\\mathbf{\\boldsymbol\\omega^\\times}_{IB}\\}_B = -\\mathbf{A}_{Bi}{\\mathbf{\\dot R}_{iI}} \\mathbf{A}_{BI}^{T}\n% \\label{Eq:CSFixedKinematics3}\n%\\end{equation}\n%%\n%where $\\mathbf{A}_{Bi}$ is known from user input, and $\\mathbf{A}_{BI}$ is known\n%from Eq. \\ref{Eq:CSFixedRotationMatrix}. GMAT knows how to calculate\n%$\\mathbf{\\dot{R}}_{iI}$ for all allowable $\\mathcal{F}_i$ and details are contained\n%in Ch.~\\ref{Ch:CoordinateSystems}. Next we would like to get the angular\n%velocity vector $\\{\\mathbf{\\boldsymbol\\omega}_{IB}\\}_B$ in terms of the\n%components of the 3$\\times$3 cross product matrix in the middle of\n%Eq. \\ref{Eq:OmegaCrossMatrix}. This matrix, in symbol notation also shows up on\n%the left hand side of Eq. \\ref{Eq:CSFixedKinematics3}.\n%%\n%% Note - Dunn discovered, through trial and error, that you can't add white lines\n%% between matrix components in the equation below to enhance readability. LaTeX\n%% actually crashes. But you can add a % comment to create spacing.\n%\\begin{equation}\n%\t\\{\\boldsymbol \\omega_{IB}\\}_{B} =\n% %\n% \\begin{bmatrix}\n% \\omega_{1} \\\\\n% \\omega_{2} \\\\\n% \\omega_{3}\n% \\end{bmatrix} = \\hspace{0.082cm}\n% %\n% \\begin{bmatrix}\n% \\Omega_{32} \\\\\n% \\Omega_{13} \\\\\n% \\Omega_{21}\n% \\end{bmatrix}\n% %\n% \\label{Eq:OmegaComponents}\n%\\end{equation}\n%%\n%Each of the three components of the angular velocity vector, measured in the\n%spacecraft body frame, can be found in the cross product matrix defined in\n%Eq. \\ref{Eq:OmegaCrossMatrix}. The subscripts on the right hand side of\n%Eq. \\ref{Eq:OmegaComponents} tell us where to find these components after\n%multiplying the three 3$\\times$3 matrices on the right hand side of\n%Eq. \\ref{Eq:CSFixedKinematics3}.\n\n\n\\subsubsection{Spinning Spacecraft Mode}\n\nIn spinning spacecraft mode, GMAT propagates the attitude by assuming the spin\naxis direction is fixed in inertial space. The spacecraft attitude at some time,\n$t$, is kinematically propagated from the attitude initial conditions.\nThe user defines the initial attitude with respect to $\\mathcal{F}_i$ by providing $\\mathbf{A}_{BI}(t_o)$ or an equivalent\nparameterization that is then converted to the Attitude matrix. The user also\nprovides the angular velocity of the body axes with respect to the inertial axes\nexpressed in $\\mathcal{F}_i$, $\\{\\boldsymbol\\omega_{IB}\\}_i$. \n\nTo calculate $\\mathbf{A}_{BI}(t)$ where $t$ is an arbitrary epoch,\nwe begin by calculating $\\mathbf{A}_{B_{o}I}$, understanding that\n$\\mathbf{A}_{B_{o}I} = \\mathbf{A}_{BI}(t_o)$. We calculate\n$\\mathbf{A}_{B_{o}I}$ using\n%\n\\begin{equation}\n \\mathbf{A}_{B_{o}I} = \\mathbf{A}_{Bi}\\mathbf{R}_{iI}(t_o)\n \\label{Eq:ABoI}\n\\end{equation}\n%\nwhere $\\mathbf{A}_{Bi}$ comes from user provided data, and\n$\\mathbf{R}_{iI}(t_o)$ is calculated by GMAT and is dependent upon\n$\\mathcal{F}_i$. See Ch.~\\ref{Ch:CoordinateSystems} for details on how GMAT\ncalculates $\\mathbf{R}_{iI}$ for all allowable coordinate systems in GMAT.\n\nBefore calculating $\\mathbf{A}_{BI}(t)$ we must determine the spin axis in the\nbody frame, $\\{\\boldsymbol\\omega_{IB}\\}_B$. If the user has chosen to provide\nthe angular velocity vector in body frame coordinates, then we already have what\nwe need. Otherwise, the user provides $\\{\\boldsymbol\\omega_{IB}\\}_i$.\nIn spinning mode we assume the rotation rate magnitude is constant and also that\nthe spin axis direction is constant in inertial space. This assumption is only\nrealistic if the angular velocity is aligned with either the maximum or minimum\nprimary axis of moment of inertia in the spacecraft body reference frame. If\nangular velocity is aligned with the axis of minimum moment of inertia, then we\nmust also assume there is an active attitude control system holding it there.\nIf there is no control system, and the spacecraft dissipates rotational kinetic\nenergy (all real bodies will) then the direction of the angular velocity vector\nwill migrate slowly until it is aligned with the axis of maximum moment of\ninertia. Understanding these assumptions we can state\n$\\{ \\boldsymbol\\omega_{IB} \\}_B (t)$ $ = \\{\\boldsymbol\\omega_{IB} \\}_B (t_o) =\n\\{ \\boldsymbol\\omega_{IB} \\}_B$. We can find the spin axis in the body frame\n(if we need to) using $\\mathbf{A}_{Bi}$ as follows\n%\n\\begin{equation}\n \\{ \\boldsymbol\\omega_{IB}\\}_B = \\mathbf{A}_{Bi} \\{ \\boldsymbol\\omega_{IB}\\}_i\n \\label{Eq:AngularVelocityBody}\n\\end{equation}\n%\nOnce calculated, GMAT saves the parameters $\\mathbf{A}_{B_{o}I}$ and\n$\\{\\boldsymbol\\omega_{IB}\\}_B$ for use in calculating the attitude and rate at\nother epochs.\n\nGMAT calculates $\\mathbf{A}_{BI}(t)$ using the Euler axis/angle to Attitude\nmatrix conversion algorithm in Sec.\\ref{Sec:AttitudeParameterizations}. The\nEuler axis is simply the unitized angular velocity vector or,\n%\n\\begin{equation}\n \\mathbf{a} = \\frac{ \\{ \\boldsymbol\\omega_{IB} \\}_B }{\\omega_{IB} }\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\n \\omega_{IB} = \\| \\{\\boldsymbol{\\omega}_{IB} \\}_B \\|\n\\end{equation}\n%\n\\newpage\nThe Euler angle $\\phi$ is calculated using\n%\n\\begin{equation}\n \\phi(t) = \\omega_{IB}(t -t_o)\n\\end{equation}\n%\nwhere $t$ is the current epoch, and $t_o$ is the spacecraft's initial epoch.\nLet's define the Attitude matrix that results from the Euler axis/angle to\nAttitude matrix conversion algorithm using $\\mathbf{a}$ and $\\phi(t)$, as\n$\\mathbf{A}_{BB_{o}}(t)$. We can calculate $\\mathbf{A}_{BI}(t)$ using\n%\n\\begin{equation}\n \\mathbf{A}_{BI}(t) =\n \\mathbf{A}_{BB_{o}}(t)\\mathbf{A}_{B_{o}I}\n\\end{equation}\n%\nTo summarize, in spinning mode the user provides $\\mathbf{A}_{Bi}$ and\n$\\{ \\boldsymbol\\omega_{IB}\\}_i$ or $\\{ \\boldsymbol\\omega_{IB}\\}_B$. GMAT\nassumes that that the spin axis direction is constant in both inertial and body\nspace, and uses the Euler axis/angle method to propagate the attitude to find\n$\\mathbf{A}_{BI}$. Now let's look at dynamic attitude propagation in GMAT.\n\n\n\\subsubsection{Dynamic Attitude Propagation}\nIn Dynamic mode, GMAT propagates quaternion attitude equations of motion along\nwith Euler's moment equations. This allows the user to provide detailed mass\nmoment of inertia properties, and to model spacecraft attitude control and\nenvironmental perturbing torques. The resulting numerically integrated attitude\nwill exhibit basic dynamics properties such as precession and nutation.\nDetailed tracking of the dynamic inertia tensor will also allow the user to\nproperly model the effect on the attitude trajectory of propellant mass\ndepletion, articulation of various components, and implementation of attitude\ncontrol laws. For modes where control laws cause the spacecraft to slew to a\ntargeted pointing direction, GMAT will model line of sight angular acceleration,\ndeceleration, overshoot, settling, and drift behavior. There are several\nDynamic modes to choose from. The different modes allow the user to\nconveniently define the spacecraft attitude and controlled pointing depending on\nthe type of attitude profile needed for a specific mission. To begin, let's\nlook at how GMAT calculates the attitude state in the simplest possible dynamic\nmode, Rigid Torque Free attitude mode (RigidTorqueFreeQuat).\\\\*[0.1cm]\n%\n\\vspace{-0.2 in}\n\\begin{figure*}[htb]\n \\begin{center}\n \\includegraphics[width=0.8\\textwidth]{Images/QuaternionX42DefB&W}\n \\caption{ Euler Axis and Angle} \\label{fig:EulerAxisAndAngle}\n \\end{center}\n\\end{figure*}\n%\n\\subsubsection{Rigid Torque Free (Quaternion) Mode}\nThe fundamental equations of attitude motion run most efficiently using the four\ncomponent quaternion. There are equations of motion for Euler Angles, but these\nhave singularities that cause problems in implementation. There are also\nequations of motion for the nine components of the Attitude matrix. Numerically\nintegrating nine equations for the Attitude matrix is not as efficient as four\nequations for the quaternion. The quaternion is the most popular attitude\nparameterization in use today for propagating attitude using numerical\nintegration schemes. Let's begin by defining the quaternion.\n\nAs shown above in Figure \\ref{fig:EulerAxisAndAngle}, any spacecraft attitude\ncan be described as the rotation about a single axis via a fixed angle from an\ninertial frame orientation to the current body orientation. The single axis is\ncalled the Euler Axis, and the angle is the Euler Angle. In the figure, the\nEuler Axis is given the vector label $\\vec{e}$. This is not the same as\nthe three Euler Angles used as another of our attitude parameterizations. We\nwill use the CCSDS Attitude Data Message standard to define the GMAT quaternion.\nThe definition is based on the Euler Axis/Angle parameterization and is\n%\n\\begin{equation}\n\t\\boldsymbol q \\equiv\n %\n \\begin{bmatrix}\n q_1 \\\\\n q_2 \\\\\n q_3 \\\\\n q_c\n \\end{bmatrix}\n %\n \\equiv %\\hspace{0.082cm}\n %\n \\begin{bmatrix}\n \\mathbf{q} \\\\\n q_c \\\\\n \\end{bmatrix}\n %\n \\equiv %\\hspace{0.082cm}\n %\n \\begin{bmatrix}\n \\mathbf{a}_1 \\sin \\frac{\\phi}{2} \\\\\n \\mathbf{a}_2 \\sin \\frac{\\phi}{2} \\\\\n \\mathbf{a}_3 \\sin \\frac{\\phi}{2} \\\\\n \\cos \\frac{\\phi}{2} \\\\\n \\end{bmatrix}\n %\n \\label{Eq:QuaternionDefinition}\n\\end{equation}\n%\nwhere $\\mathbf{a}_1$, $\\mathbf{a}_2$, and $\\mathbf{a}_3$ are components of the\nunit vector aligned with the Euler Axis or $\\mathbf{\\hat{a}}$. This unit vector\nis shown near the spacecraft center of mass in Figure \\ref{fig:EulerAxisAndAngle}.\nIn Eq.~\\ref{Eq:QuaternionDefinition} the Euler Angle is denoted using the symbol\n $\\phi$. There are actually two possible values of $\\phi$, since the rotation\nshown in Figure \\ref{fig:EulerAxisAndAngle} can be the long way around or the\nshort way around. The two possible values will add to $360\\,^{\\circ}$. Our\ndefinition of quaternions assumes the smaller angle.\n\nThe subscript ``c\" on the fourth component of the quaternion comes from the\nCCSDS standard and most likely stands for ``cosine\". The CCSDS has cleverly\navoided specifying whether the cosine ``scalar\" component should be placed\nbefore or after the vector component. This standard has thus eliminated the\nconfusion of which component of the quaternion should be defined as $q_1$ and\nwhich as $q_4$. The CCSDS standard does place its initial definition for $q_c$\nafter the definitions for $q_1$, $q_2$, and $q_3$. This lets us stack the four\ncomponents in the vector shown in the second part of the definition in\nEq.~\\ref{Eq:QuaternionDefinition}. If $q_4$ were substituted for $q_c$, we would\nhave Landis Markley's definition from the ``Parameterization of the Attitude\"\nsection in ``Spacecraft Attitude Determination and Control\" by Wertz. The third\npart of the definition above shows the 3$\\times$1 vector component of the\nquaternion as a single symbol $\\mathbf{q}$, placed above $q_c$ to form a second\n4$\\times$1 matrix or vector. This also parallels Markley's definition and will\nconveniently fit into the GMAT equations relating quaternions to the Attitude\nmatrix shown later in this section. Note that for the final part of the\ndefinition in Eq. \\ref{Eq:QuaternionDefinition}, the CCSDS standard uses\n$\\mathbf{e}$ to denote the unit vector aligned with the Euler Axis instead of\n$\\mathbf{a}$. We have chosen $\\mathbf{a}$ to represent ``axis\", and because\nthat is how GMAT is using it later in the attitude parameterization conversion\nSection \\ref{Sec:AttitudeParameterizations}.\n\nIn torque free quaternion attitude propagation mode, the user provides four\npieces of information. They first choose a coordinate system, $\\mathcal{F}_i$,\nin which to define the initial conditions. Secondly, they define the initial\nattitude with respect to $\\mathcal{F}_i$ by providing $\\mathbf{A}_{Bi}$ or an\nequivalent parameterization that is then converted to the Attitude matrix. GMAT\nwill then use $\\mathbf{A}_{Bi}$ to calculate $\\mathbf{A}_{BI}$ using\nEq.~\\ref{Eq:CSFixedRotationMatrix}. From this Attitude matrix the\nparameterization will next be converted by GMAT to the quaternion\n$\\boldsymbol q_{Bi}$. Thirdly, the user provides the angular velocity of the body\naxes with respect to the inertial axes expressed in $\\mathcal{F}_i$, or\n$\\{ \\boldsymbol\\omega_{IB}\\}_i$. If it is more convenient for the user to\nprovide angular velocity expressed in the spacecraft body frame, GMAT will accept\nthis input as well. If the user provides angular velocity in $\\mathcal{F}_i$,\nGMAT will use Eq.~\\ref{Eq:AngularVelocityBody} to convert it to\n$\\{\\boldsymbol\\omega_{IB}\\}_B$. Fourthly, the user provides the mass moment of\ninertia properties. These can be in the form of the three principal moments of\ninertia $I_{xx}$, $I_{yy}$, and $I_{zz}$, or a full inertia tensor with the\noff-diagonal products of inertia $I_{xy}$, $I_{xz}$, and $I_{yz}$ included. In\na future release of GMAT the user will be able to use the GUI or scripting\nlanguage to assemble a complex spacecraft model from components. Locating and\norienting these components and describing how those that articulate can move,\nthe user will provide enough information for GMAT to calculate a dynamic inertia\ntensor. This future capability will also include dynamically adjusting the\ncomponents of the inertia tensor as propellant is consumed from tanks.\n\n\\newpage\nGMAT will assemble the data provided by the user into a state vector that can be\npropagated numerically. In this case, GMAT will need to create an initial state,\nand then provide its numerical integrators with the state vector derivative\nequations.\n\nWe will define an attitude state variable $\\mathbf{x}$ such that\n%\n\\begin{equation}\n\t\\mathbf{x} =\n %\n \\begin{bmatrix}\n \\boldsymbol{q}^T & \\boldsymbol\\omega^T\n \\end{bmatrix}^T\n %\n =\n %\n \\begin{bmatrix}\n q_1 & q_2 & q_3 & q_c & \\omega_1 & \\omega_2 & \\omega_3\n \\end{bmatrix}^T\n %\n \\label{Eq:QuatStateVec}\n\\end{equation}\n%\nthen taking the derivative we arrive at\n%\n\\begin{equation}\n\t\\mathbf{\\dot{x}} =\n %\n \\begin{bmatrix}\n \\boldsymbol{\\dot{q}}^T & \\boldsymbol{\\dot{\\omega}}^T\n \\end{bmatrix}^T\n %\n =\n %\n \\begin{bmatrix}\n \\dot{q_1} & \\dot{q_2} & \\dot{q_3} & \\dot{q_c} & \\dot{\\omega_1} & \\dot{\\omega_2} & \\dot{\\omega_3}\n \\end{bmatrix}^T\n %\n \\label{Eq:QuatStateVecDot}\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\n\t\\boldsymbol{\\dot{q}} =\n %\n \\frac{1}{2}\\boldsymbol{\\Omega(\\omega)q}\n %\n \\label{Eq:QDot}\n\\end{equation}\n%\nand where\n%\n\\begin{equation}\n\t\\boldsymbol{\\Omega(\\omega)} =\n %\n \\begin{bmatrix}\n -[\\omega\\times] & \\omega \\\\\n -\\omega^T & 0 \\\\\n \\end{bmatrix}\n %\n \\label{Eq:CapOmega1}\n\\end{equation}\n%\nIn Eq.~\\ref{Eq:CapOmega1}, $[\\omega\\times]$ is the skew symmetric angular\nvelocity cross product matrix originally introduced in\nEq.~\\ref{Eq:OmegaCrossMatrix}. As in Eq.~\\ref{Eq:OmegaCrossMatrix} the angular\nvelocity used here represents the body with respect to the inertial frame. If\nwe substitute Eq.~\\ref{Eq:OmegaCrossMatrix} into Eq.~\\ref{Eq:CapOmega1} we get\n%\n\\begin{equation}\n \\boldsymbol{\\Omega}(\\omega) =\n %\n \\begin{bmatrix}\n 0 & \\omega_3 & -\\omega_2 & \\omega_1 \\\\\n -\\omega_3 & 0 & \\omega_1 & \\omega_2 \\\\\n \\omega_2 & -\\omega_1 & 0 & \\omega_3 \\\\\n -\\omega_1 & -\\omega_2 & -\\omega_3 & 0 \\\\\n \\end{bmatrix}\n %\n \\label{Eq:CapOmega2}\n\\end{equation}\n%\nSubstituting Eq.~\\ref{Eq:CapOmega2} into Eq.~\\ref{Eq:QDot} and multiplying the\nterms together we get\n%\n\\begin{equation}\n% The \"arraystretch\" command prevents the fractions from touching vertically\n\\renewcommand{\\arraystretch}{1.4}\n \\boldsymbol{\\dot{q}} =\n %\n \\frac{1}{2}\\boldsymbol{\\Omega(\\omega)q} =\n %\n \\begin{bmatrix}\n \\frac{1}{2}( \\hspace{0.35cm}\\omega_3 q_2 -\\omega_2 q_3 +\\omega_1 q_c )\\\\\n \\frac{1}{2}( -\\omega_3 q_1 +\\omega_1 q_3 +\\omega_2 q_c )\\\\\n \\frac{1}{2}( \\hspace{0.35cm}\\omega_2 q_1 -\\omega_1 q_2 +\\omega_3 q_c )\\\\\n \\frac{1}{2}( -\\omega_1 q_1 -\\omega_2 q_2 -\\omega_3 q_3 )\\\\\n \\end{bmatrix}\n %\n \\label{Eq:QuatEOMs}\n\\end{equation}\n%\nNext let's evaluate the time derivative of the body angular velocity terms.\nThese are the last three components of the state vector derivative in\nEquation~\\ref{Eq:QuatStateVecDot}. Following the Markley/Wertz convention, if\nwe let $\\mathbf{L}$ represent angular momentum we can start with a simple\nequation for angular momentum measured in the Inertial reference frame.\n%\n\\begin{equation}\n \\mathbf{L} =\n %\n \\mathbf{I} \\cdot \\boldsymbol{\\omega}\n %\n \\label{Eq:AngMom}\n\\end{equation}\n%\nEuler's second law says that, in an inertial reference frame, the time\nderivative of $\\mathbf{L}$ is the applied torque, $\\mathbf{T}$. We write this\nas\n%\n\\begin{equation}\n\t\\mathbf{\\dot{L}} =\n %\n \\frac{d}{dt}(\\mathbf{I}\\cdot\\boldsymbol{\\omega})\n %\n =\n %\n \\mathbf{T}\n %\n \\label{Eq:Eulers2ndLaw}\n\\end{equation}\n%\nwhere $\\mathbf{I}$ is the body oriented Inertia Tensor. This is a 3$\\times$3\nmatrix with the principal moments of inertia on the diagonal and the products of\ninertia on the off-diagonal. It is written as follows.\n%\n\\begin{equation}\n \\mathbf{I} =\n %\n \\begin{bmatrix}\n I_{11} & I_{12} & I_{13} \\\\\n I_{21} & I_{22} & I_{23} \\\\\n I_{31} & I_{32} & I_{33} \\\\\n \\end{bmatrix}\n %\n \\label{Eq:FullInertiaTensor}\n\\end{equation}\n%\nwhere the moments and products of inertia are defined as follows.\n%\n\\begin{equation}\n I_{11} \\equiv \\sum_{i=1}^{n} m_i ( \\rho_{i2}^2 + \\rho_{i3}^2 )\n \\label{Eq:Moment11}\n\\end{equation}\n%\n\\begin{equation}\n I_{22} \\equiv \\sum_{i=1}^{n} m_i ( \\rho_{i3}^2 + \\rho_{i1}^2 )\n \\label{Eq:Moment22}\n\\end{equation}\n%\n\\begin{equation}\n I_{33} \\equiv \\sum_{i=1}^{n} m_i ( \\rho_{i1}^2 + \\rho_{i2}^2 )\n \\label{Eq:Moment33}\n\\end{equation}\n%\n\\begin{equation}\n I_{12} = I_{21} \\equiv -\\sum_{i=1}^{n} m_i \\rho_{i1} \\rho_{i2}\n \\label{Eq:Product12}\n\\end{equation}\n%\n\\begin{equation}\n I_{23} = I_{32} \\equiv -\\sum_{i=1}^{n} m_i \\rho_{i2} \\rho_{i3}\n \\label{Eq:Product23}\n\\end{equation}\n%\n\\begin{equation}\n I_{31} = I_{13} \\equiv -\\sum_{i=1}^{n} m_i \\rho_{i3} \\rho_{i1}\n \\label{Eq:Product31}\n\\end{equation}\n%\nThe signs for the products of inertia depend on how products of inertia are\nthemselves defined. They can be positive or negative depending on individual\nauthors. Meanwhile, each regular geometric shape, if constructed of a uniform\nsolid material, will have an analytic formula for each of its moments and\nproducts of inertia. A spacecraft model composed of multiple elements can sum\nmoments and products of inertia for all components into a single Inertia Tensor.\nFor the current RigidTorqueFreeQuat mode we will assume ours is diagonal.\n%\n\\begin{equation}\n \\mathbf{I} =\n %\n \\begin{bmatrix}\n I_{11} & 0 & 0 \\\\\n 0 & I_{22} & 0 \\\\\n 0 & 0 & I_{33} \\\\\n \\end{bmatrix}\n %\n \\label{Eq:DiagonalInertiaTensor}\n\\end{equation}\n%\nIn this initial development, we will assume our spacecraft body frame is aligned\nwith its principal axes of inertia, and that it is a completely rigid body. This\nwill mean our inertia tensor is both diagonal and has a time derivative of zero.\nLet's apply these facts to Eq.~\\ref{Eq:Eulers2ndLaw}.\n%\n\\begin{equation}\n \\frac{d}{dt}(\\mathbf{I}\\cdot\\boldsymbol{\\omega}) =\n %\n (\\mathbf{\\dot{I}}\\cdot\\boldsymbol{\\omega})+(\\mathbf{I}\\cdot\\boldsymbol{\\dot{\\omega}})\n %\n \\label{Eq:TimeDerivEulersEquation}\n\\end{equation}\n%\nSince we already stated that the time derivative of I is zero, the first term\non the RHS of Eq.~\\ref{Eq:TimeDerivEulersEquation} will drop out. We can now\nwrite Eq.~\\ref{Eq:Eulers2ndLaw} as follows.\n%\n\\begin{equation}\n\t\\mathbf{\\dot{L}} =\n %\n \\mathbf{T} =\n %\n (\\mathbf{I}\\cdot\\boldsymbol{\\dot{\\omega}})\n %\n \\label{Eq:TimeDerivEulersEquation2}\n\\end{equation}\n%\nIn Eq.~\\ref{Eq:TimeDerivEulersEquation2} the angular momentum term is measured\nin the inertial frame. Let's express it now in the spacecraft body frame.\nSuperscripts to the left will indicate the reference frame.\n%\n\\begin{equation}\n ^{I}\\mathbf{\\dot{L}} = \\hspace{0.082cm}\n %\n ^{B}\\mathbf{\\dot{L}} + \\hspace{0.082cm} ^{B}\\boldsymbol{\\omega} \\times \\hspace{0.03cm} ^{B}\\mathbf{L}\n %\n \\label{Eq:BasicAttDynamics1}\n\\end{equation}\n%\nCombining Equations~\\ref{Eq:Eulers2ndLaw} and ~\\ref{Eq:BasicAttDynamics1} gives\nus\n%\n\\begin{equation}\n ^{B}\\mathbf{T} = \\hspace{0.082cm}\n %\n ^{B}\\mathbf{\\dot{L}} + \\hspace{0.082cm} ^{B}\\boldsymbol{\\omega} \\times \\hspace{0.03cm} ^{B}\\mathbf{L}\n %\n \\label{Eq:BasicAttDynamics2}\n\\end{equation}\n%\nSubstituting~\\ref{Eq:TimeDerivEulersEquation2} for $\\mathbf{\\dot{L}}$ on the RHS\nof~\\ref{Eq:BasicAttDynamics2}, and~\\ref{Eq:AngMom} for $\\mathbf{L}$ on the RHS,\nwe get the following.\n%\n\\begin{equation}\n ^{B}\\mathbf{T} = \\hspace{0.082cm}\n %\n ^{B}(\\mathbf{I}\\cdot\\boldsymbol{\\dot{\\omega}}) +\n \\hspace{0.082cm} ^{B}\\boldsymbol{\\omega} \\times \\hspace{0.03cm} ^{B}\\mathbf{I} \\cdot \\boldsymbol{\\omega}\n %\n \\label{Eq:BasicAttDynamics3}\n\\end{equation}\n%\nAssuming our Inertia Tensor is diagonal, if we multiply out the terms on the RHS\nof Eq.~\\ref{Eq:BasicAttDynamics3} we will get three equations for torque along\nthe principal axes of the spacecraft body.\n%\n\\begin{equation}\n T_1 = I_{11}\\dot{\\omega_1}+(I_{22}-I_{33}) \\omega_2 \\omega_3\n \\label{Eq:TorqueAxis1}\n\\end{equation}\n%\n\\begin{equation}\n T_2 = I_{22}\\dot{\\omega_2}+(I_{33}-I_{11}) \\omega_3 \\omega_1\n \\label{Eq:TorqueAxis2}\n\\end{equation}\n%\n\\begin{equation}\n T_3 = I_{33}\\dot{\\omega_3}+(I_{11}-I_{22}) \\omega_1 \\omega_2\n \\label{Eq:TorqueAxis3}\n\\end{equation}\n%\nThese equations are equivalent to Equations 16-50a through 16-50c on Page 522 of\nWertz.\n\nSolving equations~\\ref{Eq:TorqueAxis1} through~\\ref{Eq:TorqueAxis3} for\nangular acceleration, we get\n%\n\\begin{equation}\n \\dot{\\omega_1} = \\frac{T_1 - (I_{22}-I_{33}) \\omega_2 \\omega_3}{I_{11}}\n \\label{Eq:OmegaDot1}\n\\end{equation}\n%\n\\begin{equation}\n \\dot{\\omega_2} = \\frac{T_2 - (I_{33}-I_{11}) \\omega_3 \\omega_1}{I_{22}}\n \\label{Eq:OmegaDot2}\n\\end{equation}\n%\n\\begin{equation}\n \\dot{\\omega_3} = \\frac{T_3 - (I_{11}-I_{22}) \\omega_1 \\omega_2}{I_{33}}\n \\label{Eq:OmegaDot3}\n\\end{equation}\n%\nThese are the three angular velocity time derivative terms we needed for the\nstate derivative vector in Equation~\\ref{Eq:QuatStateVecDot}. They are also\nknown as Euler's Moment Equations. As long as we are rotating torque free, we\nwill set the torque $\\mathbf{T}$ terms in Equations~\\ref{Eq:OmegaDot1}\nthrough~\\ref{Eq:OmegaDot3} to zero. We now have everything GMAT needs to\nintegrate our quaternion attitude equations of motion for a torque free rigid\ntumbling body.\n\nThese equations will model several types of rotational motion. They are\n``motionless hang\", ``flat spin\", ``spin with precession\", and ``spin with\nprecession and nutation\" which is the same as a full 3-axis tumble. Next, let's\nadd a level of complexity and look at Articulated Torque Free attitude mode.\n\n\\subsubsection{Articulated Torque Free (Quaternion) Mode}\nLet's assume our spacecraft is still rotating without the influence of control\nor environmental torques, but now it is moving articulated appendages. We can\nno longer assume our inertia tensor $\\mathbf{I}$ is constant. It is also\nunlikely, now that objects are moving, that $\\mathbf{I}$ will remain diagonal.\nLet's develop new equations to handle our dynamic inertia tensor. The\ndefinition of angular momentum is\n%\n\\begin{equation}\n \\mathbf{L} \\equiv\n \\mathbf{I} \\boldsymbol{\\omega}\n \\label{Eq:AngMomDef}\n\\end{equation}\n%\nwhich applies to an inertia tensor with off-diagonal products of inertia. The\ninitial conditions for both parameters on the RHS of Eq.~\\ref{Eq:AngMomDef} will\nbe provided by the User. Multiplying both sides by the inverse of $\\mathbf{I}$,\nand then solving for angular velocity we get\n%\n\\begin{equation}\n \\boldsymbol{\\omega} =\n %\n \\mathbf{I^{-1}L}\n \\label{Eq:Omega}\n\\end{equation}\n%\nIf GMAT calculates changes in $\\mathbf{I}$ based on articulation, and for later\nmodes includes propellant mass consumption, and it integrates $\\mathbf{\\dot{L}}$\nto get $\\mathbf{L}$, during each integration step we can use Eq.~\\ref{Eq:Omega}\nto calculate current angular velocity $\\mathbf{\\omega}$. Solving\nEq.~\\ref{Eq:BasicAttDynamics2} for $\\mathbf{\\dot{L}}$, we get\n%\n\\begin{equation}\n ^{B}\\mathbf{\\dot{L}} = \\hspace{0.082cm}\n %\n ^{B}\\mathbf{T}- \\hspace{0.082cm} ^{B}\\boldsymbol{\\omega} \\times \\hspace{0.03cm} ^{B}\\mathbf{L}\n %\n \\label{Eq:BasicAttDynamics4}\n\\end{equation}\n%\nIf the user inputs all the same initial conditions as they did in\nRigidTorqueFreeQuat mode, with the addition of a full inertia tensor with\nproducts, and GMAT tracks articulation of components and adjusts the inertia\ntensor dynamically, the state vector to calculate will now be\n%\n%\n\\begin{equation}\n\t\\mathbf{x} =\n %\n \\begin{bmatrix}\n \\boldsymbol{q}^T & \\mathbf{L}^T\n \\end{bmatrix}^T\n %\n =\n %\n \\begin{bmatrix}\n q_1 & q_2 & q_3 & q_c & L_1 & L_2 & L_3\n \\end{bmatrix}^T\n %\n \\label{Eq:QuatStateVec}\n\\end{equation}\n%\nwhere the initial condition of $\\mathbf{L}$ is calculated from\nEq.~\\ref{Eq:AngMomDef}. To get angular velocity, rather than integrating Euler's\nMoment Equations, we will use Eq.~\\ref{Eq:Omega} which means we need to invert a\n3$\\times$3 matrix. Since a full inertia tensor is still symmetric, we can use\nCramer's rule to invert it with no numerical pathologies. The time derivative\nof $\\mathbf{x}$, which we need to send to the GMAT numerical integrators will be\n%\n\\begin{equation}\n\t\\mathbf{\\dot{x}} =\n %\n \\begin{bmatrix}\n \\boldsymbol{\\dot{q}}^T & \\mathbf{\\dot{L}}^T\n \\end{bmatrix}^T\n %\n =\n %\n \\begin{bmatrix}\n \\dot{q_1} & \\dot{q_2} & \\dot{q_3} & \\dot{q_c} & \\dot{L_1} & \\dot{L_2} & \\dot{L_3}\n \\end{bmatrix}^T\n %\n \\label{Eq:QuatStateVecDot}\n\\end{equation}\n%\nWe get $\\mathbf{\\dot{q}}$ from Eq.~\\ref{Eq:QuatEOMs} and $\\mathbf{\\dot{L}}$ from\nEq.~\\ref{Eq:BasicAttDynamics4}. For this torque free mode we can choose to set\nthe torque vector in Eq.~\\ref{Eq:BasicAttDynamics4} to zero. We now have\neverything GMAT needs to integrate our quaternion attitude equations of motion\nfor a torque free articulated tumbling body.\n\n\\subsubsection{Attitude Hold Mode}\nUnder Construction\n\n\\subsubsection{Seek Coordinate System Fixed Mode}\nUnder Construction\n\n\\subsubsection{Target Pointing Slew Mode}\nUnder Construction\n\n\\subsubsection{Torque Perturbations Mode}\nTorque Perturbations are higher order terms that model environmental torques or\ntorques applied as the result of human operator control inputs. These can be\nattitude maneuvers selected by a ground controller from a mouse activated GUI\nmenu, or input through joysticks by an astronaut pilot. Full 6DOF maneuvering\ninvolved in manual or automated rendezvous, proximity operations, and docking\nwill involve translational forces and equations of motion and is covered in\nCh. TBD. Simple environmental torques include Gravity Gradient, Aerodynamic Drag\nand Solar Photon Pressure Torques, and spacecraft magnetic interaction with\nplanetary magnetic fields. These torques can be added to the basic Euler Moment\nEquations presented above in the Torque Free Mode section.\n\n\n\\subsection{Attitude Parameterizations and Conversions}\n\\label{Sec:AttitudeParameterizations}\nThis section details how GMAT converts between different attitude\nparameterizations. For each conversion type, any singularities that may occur\nare addressed. The orientation parameterizations in GMAT include the DCM or A\nMatrix, Euler Angles, quaternions, and Euler axis/angle. The body rate\nparameterizations include Euler angle rates and angular velocity. We begin with\nthe algorithm to transform from the quaternions to the Attitude matrix.\n\n\\subsubsection{Conversion: Quaternions to Attitude Matrix}\\label{sec:QuatToAMat}\n\\index{Attitude Parameterization!Quaternions to Attitude Matrix}\n\nGiven: $\\mathbf{q}$, $q_c$\n\n\\noindent Find: $\\mathbf{A}$\n\n\\noindent Name: \\emph{QuatsToAMat}\n\n\\begin{equation}\n \\mathbf{q} = \\left[ q_1 \\hspace{.1 in} q_2 \\hspace{.1 in} q_3 \\right]^T\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n \\mathbf{q}^{\\times} = \\begin{bmatrix}\n 0 & -q_3 & q_2 \\\\\n q_3 & 0 & -q_1 \\\\\n -q_2 & q_1 & 0 \\\\\n \\end{bmatrix}\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n c = \\frac{1}{q_1^2 + q_2^2 + q_3^2 + q_c^2}\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n \\mathbf{A} = c\\left[ (q_c^2 - \\mathbf{q}^T\\mathbf{q})\\mathbf{I}_3 +\n 2\\mathbf{q}\\mathbf{q}^T -2q_c\\mathbf{q}^{\\times}\\right]\n \\label{Eq:QuatsToAMat}\n\\end{equation}\n% After an equation, and before text, I can't get \\medskip to work. So I'm just\n% adding a blank line.\n\n%\nwhere $\\mathbf{I_3}$ is a 3$\\times$3 identity matrix. Multiplying out\nEq.~\\ref{Eq:QuatsToAMat} we get\n%\n% Inspection of the comments in the equation below will reveal that Dunn would\n% like to implement prettier matrix alignment but hasn't figured out how yet.\n\\begin{multline}\n %\\begin{align}\n \\mathbf{A} =\n c\\left(\n %& % Start Algigment at left bracket of top matrix\n \\begin{bmatrix}\n (q_c^2-q_1^2-q_2^2-q_3^2) & 0 & 0 \\\\\n 0 & (q_c^2-q_1^2-q_2^2-q_3^2) & 0 \\\\\n 0 & 0 & (q_c^2-q_1^2-q_2^2-q_3^2) \\\\\n \\end{bmatrix} + \\right. \\\\\n %\n \\left.\n %& % Align Left bracket of this matrix with bracket above\n \\begin{bmatrix}\n q_1^2 & q_1 q_2 & q_1 q_3 \\\\\n q_2 q_1 & q_2^2 & q_2 q_3 \\\\\n q_3 q_1 & q_3 q_2 & q_3^2 \\\\\n \\end{bmatrix} +\n %\n \\begin{bmatrix}\n 0 & -q_c q_3 & q_c q_2 \\\\\n q_c q_3 & 0 & -q_c q_1 \\\\\n -q_c q_2 & q_c q_1 & 0 \\\\\n \\end{bmatrix}\n \\right)\n %\n \\label{Eq:QuatsToAMat2}\n %\\end{align}\n\\end{multline}\n%\nAdding terms in Eq.~\\ref{Eq:QuatsToAMat2} we get\n%\n\\begin{equation}\n \\mathbf{A} = c\n %\n \\begin{bmatrix}\n q_1^2-q_2^2-q_3^2+q_c^2 & 2(q_1q_2+q_3q_c) & 2(q_1q_3+q_2q_c) \\\\\n 2(q_1q_2-q_3q_c) & -q_1^2+q_2^2-q_3^2+q_c^2 & 2(q_2q_3+q_1q_c) \\\\\n 2(q_1q_3+q_2q_c) & 2(q_2q_3-q_1q_c) & -q_1^2-q_2^2+q_3^2+q_c^2 \\\\\n \\end{bmatrix}\n %\n \\label{Eq:AMatQuats}\n\\end{equation}\n%\nIf we substitute the traditional symbol $q_4$ in for the CCSDS symbol $q_c$, we\nwill get the exact identity shown in Equation 12-13a on Page 414 of Wertz, except\nfor the constant term c. That is there to normalize the Attitude matrix if the\nquaternion magnitude is not exactly equal to 1.\n\nIt may seem a waste of space to multiply out all these terms when the math was\nso compact in the form shown above in Eq.~\\ref{Eq:QuatsToAMat}. However, when\ncoding this algorithm, several multiply by zero and add to zero operations will\nbe avoided if Eq.~\\ref{Eq:AMatQuats} is used instead. So it may be a waste of\nspace, but it is a savings in time when running GMAT.\n\n\n\n\\subsubsection{Conversion: Attitude Matrix to Quaternions} \\label{sec:AMatToQuat}\n\\index{Attitude Parameterization!Attitude Matrix to Quaternions}\n\nGiven: $\\mathbf{A}$\n\n\\noindent Find: $\\mathbf{q}$, $q_c$\n\n\\noindent Name: \\emph{AMatToQuats}\n\nDefine following vector\n%\n\\begin{equation}\n \\mathbf{v} = [ \\hspace{.02 in} A_{11} \\hspace{.1 in} A_{22}\\hspace{.1 in}\n A_{33} \\hspace{.1 in} \\mbox{trace}(\\mathbf{A}) \\hspace{.02 in}]\n\\end{equation}\n%\nwhere the trace of $\\mathbf{A}$ is\n%\n\\begin{equation}\n \\text{trace}(\\mathbf{A}) =\n %\n A_{11}+A_{22}+A_{33}\n\\end{equation}\n%\nDefine $i_m$ as the index of the maximum component of $\\mathbf{v}$. Then use\nthe following logic\n%\n\\noindent if $i_m = 1$\n%\n\\begin{equation}\n \\mathbf{q}'' = \\begin{pmatrix}\n 2v_{i_m} + 1 - \\mbox{trace}(\\mathbf{A})\\\\\n A_{12} + A_{21}\\\\\n A_{13} + A_{31}\\\\\n A_{23} - A_{32}\\\\\n \\end{pmatrix}\n\\end{equation}\n%\nif $i_m = 2$\n%\n\\begin{equation}\n \\mathbf{q}'' = \\begin{pmatrix}\n A_{21} + A_{12}\\\\\n 2v_{i_m} + 1 - \\mbox{trace}(\\mathbf{A})\\\\\n A_{23} + A_{32}\\\\\n A_{31} - A_{13}\\\\\n \\end{pmatrix}\n\\end{equation}\n%\nif $i_m = 3$\n%\n\\begin{equation}\n \\mathbf{q}'' = \\begin{pmatrix}\n A_{31} + A_{13}\\\\\n A_{32} + A_{23}\\\\\n 2v_{i_m} + 1 - \\mbox{trace}(\\mathbf{A})\\\\\n A_{12} - A_{21}\\\\\n \\end{pmatrix}\n\\end{equation}\n%\nif $i_m = 4$\n%\n\\begin{equation}\n \\mathbf{q}'' = \\begin{pmatrix}\n A_{23} - A_{32}\\\\\n A_{31} - A_{13}\\\\\n A_{12} - A_{21}\\\\\n 1 + \\mbox{trace}(\\mathbf{A})\\\\\n \\end{pmatrix}\n\\end{equation}\n%\nWe normalize $\\mathbf{q}''$ using\n%\n\\begin{equation}\n \\mathbf{q}' = \\frac{\\mathbf{q}''}{\\| \\mathbf{q}'' \\|}\n\\end{equation}\n%\nFinally,\n%\n\\begin{equation}\n \\mathbf{q} = [\\hspace{.05 in} q_{1}' \\hspace{.2 in} q_{2}' \\hspace{.2 in} q_3'\n \\hspace{.05 in}]^T\n\\end{equation}\n%\nand\n%\n\\begin{equation}\n q_c = q_c'\n\\end{equation}\n%Note: There is not a unique quaternion for a given AMat. GMAT\n%assumes that the ``+\" sign is used in Eq.~(\\ref{Eq:q_4}).\n\n\\subsubsection{Conversion: Attitude Matrix to Euler Axis/Angle}\n\n\\index{Attitude Parameterization!AMat to Axis/Angle}\n\nGiven: $\\mathbf{A}$\n\n\\noindent Find: $\\mathbf{a}$, $\\phi$\n\n\\noindent Name: \\emph{AMatToEulAxisAngle}\n\n\\begin{equation}\n \\mathbf{A} = \\begin{pmatrix}\n A_{11} & A_{12} & A_{13}\\\\\n A_{21} & A_{22} & A_{23}\\\\\n A_{31} & A_{32} & A_{33}\\\\\n \\end{pmatrix}\n\\end{equation}\n%\n\\begin{equation}\n \\phi = \\cos^{-1}\\left( \\frac{1}{2}\\left(\\mbox{trace}(\\mathbf{A}) -\n 1 \\right)\\right)\n\\end{equation}\n%\n\\begin{equation}\n \\mathbf{a} = \\frac{1}{2\\sin{\\phi}}\\begin{pmatrix}\n A_{23} - A_{32}\\\\\n A_{31} - A_{13}\\\\\n A_{12} - A_{21}\\\\\n \\end{pmatrix}\n\\end{equation}\n%\nIf $\\|\\sin{\\phi} \\| < 10^{-14}$, then we assume\n%\n\\begin{equation}\n \\mathbf{a} = \\left[\\hspace{.05 in} 1 \\hspace{.1 in} 0 \\hspace{.1 in}\n 0 \\hspace{.05 in} \\right]^T\n\\end{equation}\n%\nNote that if $\\|\\sin{\\phi} \\| < 10^{-14}$ then $\\cos{\\phi} \\approx\n1 $ and we arrive at an Attitude matrix of $\\mathbf{I}_3$.\n\n\\subsubsection{Conversion: Euler Axis/Angle to Attitude Matrix} \\index{Attitude\nParameterization!Axis/Angle to Attitude Matrix} \\label{Sec:EulerAxis/AngleToAMat}\n\nGiven: $\\mathbf{a}$, $\\phi$\n\n\\noindent Find: $\\mathbf{A}$\n\n\\noindent Name: \\emph{EulAxisAngleToAMat}\n%\n\\begin{equation}\n \\mathbf{a}^{\\times} = \\begin{pmatrix}\n 0 & -a_3 & a_2 \\\\\n a_3 & 0 & -a_1 \\\\\n -a_2 & a_1 & 0 \\\\\n \\end{pmatrix}\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n \\mathbf{A} = \\cos{\\phi}\\mathbf{I}_3 +\n (1 - \\cos{\\phi})\\mathbf{a}\\mathbf{a}^T -\n \\sin{\\phi}\\mathbf{a}^{\\times}\n \\label{Eq:AMat}\n\\end{equation}\n%\nMultiplying out the terms in Eq.~\\ref{Eq:AMat} we get\n%\n\\begin{equation}\n \\mathbf{A} =\n \\begin{pmatrix}\n \\cos{\\phi} & 0 & 0 \\\\\n 0 & \\cos{\\phi} & 0 \\\\\n 0 & 0 & \\cos{\\phi} \\\\\n \\end{pmatrix} +\n (1 - \\cos{\\phi})\n \\begin{pmatrix}\n a_1^2 & a_1 a_2 & a_1 a_3 \\\\\n a_2 a_1 & a_2^2 & a_2 a_3 \\\\\n a_3 a_1 & a_3 a_2 & a_3^2 \\\\\n \\end{pmatrix} -\n \\sin{\\phi}\n \\begin{pmatrix}\n 0 & -a_3 & a_2 \\\\\n a_3 & 0 & -a_1 \\\\\n -a_2 & a_1 & 0 \\\\\n \\end{pmatrix}\n \\label{Eq:AMat2}\n\\end{equation}\n%\nContinuing to multiply and sum terms, Eq.~\\ref{Eq:AMat2} becomes nine separate\nequations, one for each of the elements of the Attitude Matrix.\n%\n\\begin{equation}\n \\begin{aligned}\n \\mathbf{A}_{11} &= \\cos{\\phi} + a_1^2 - a_1^2\\cos{\\phi} \\\\\n \\mathbf{A}_{12} &= a_1 a_2 - a_1 a_2 \\cos{\\phi} + a_3 \\sin{\\phi} \\\\\n \\mathbf{A}_{13} &= a_1 a_3 - a_1 a_3 \\cos{\\phi} - a_2 \\sin{\\phi} \\\\\n %\n \\mathbf{A}_{21} &= a_2 a_1 - a_2 a_1 \\cos{\\phi} - a_3 \\sin{\\phi} \\\\\n \\mathbf{A}_{22} &= \\cos{\\phi} + a_2^2 - a_2^2\\cos{\\phi} \\\\\n \\mathbf{A}_{23} &= a_2 a_3 - a_2 a_3 \\cos{\\phi} + a_1 \\sin{\\phi} \\\\\n %\n \\mathbf{A}_{31} &= a_3 a_1 - a_3 a_1 \\cos{\\phi} + a_2 \\sin{\\phi} \\\\\n \\mathbf{A}_{32} &= a_3 a_2 - a_3 a_2 \\cos{\\phi} - a_1 \\sin{\\phi} \\\\\n \\mathbf{A}_{33} &= \\cos{\\phi} + a_3^2 - a_3^2\\cos{\\phi}\n \\end{aligned}\n \\label{Eq:NineAMatEquations}\n\\end{equation}\n%\nIt may seem a waste of space to multiply out all these terms when the math was so\ncompact in the form shown above in Eq.~\\ref{Eq:AMat}. However, when coding this\nalgorithm, several multiply by zero operations will be avoided if\nEquations~\\ref{Eq:NineAMatEquations} are used instead. So it may be a waste of\nspace, but it is a savings in time when running GMAT.\n%\n\\subsubsection{Conversion: Euler Angles to Attitude Matrix}\n\\label{sec:EulerAnglestoAMat}\n\nGiven: Sequence order ( i.e. 123, 121, ...$\\boldsymbol{321}$,...313),\n$\\theta_1$, $\\theta_2$, $\\theta_3$\n\n\\noindent Find: $\\mathbf{A}$\n\n\\noindent Name: \\emph{EulerAnglesToAMat}\n\nWe'll give an example for a 321 rotation, and then present results\nfor the remaining 11 Euler angle sequences. First, let's define\n$\\mathbf{A}_3(\\theta_1)$, $\\mathbf{A}_2(\\theta_2)$, and\n$\\mathbf{A}_1(\\theta_3)$.\n%\n\\begin{equation}\n \\mathbf{A}_3(\\theta_1) =\n \\begin{pmatrix}\n \\cos{\\theta_1} & \\sin{\\theta_1} & 0 \\\\\n -\\sin{\\theta_1} & \\cos{\\theta_1} & 0 \\\\\n 0 & 0 & 1\n \\end{pmatrix}\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n \\mathbf{A}_2(\\theta_2) =\n \\begin{pmatrix}\n \\cos{\\theta_2} & 0 & -\\sin{\\theta_2} \\\\\n 0 & 1 & 0 \\\\\n \\sin{\\theta_2} & 0 & \\cos{\\theta_2}\n \\end{pmatrix}\n\\end{equation}\n\\medskip\n%\n\\begin{equation}\n \\mathbf{A}_1(\\theta_3) =\n \\begin{pmatrix}\n 1 & 0 & 0 \\\\\n 0 & \\cos{\\theta_3} & \\sin{\\theta_3} \\\\\n 0 & -\\sin{\\theta_3} & \\cos{\\theta_3}\n \\end{pmatrix}\n\\end{equation}\n%\nNow we can write\n%\n\\begin{equation}\n \\mathbf{A}_{321} = \\mathbf{A}_1(\\theta_3)\\mathbf{A}_2(\\theta_2)\\mathbf{A}_3(\\theta_1) = \\\\\n %\n \\medskip\n \\begin{pmatrix}\n 1 & 0 & 0 \\\\\n 0 & c_3 & s_3 \\\\\n 0 & -s_3 & c_3\n \\end{pmatrix}\n %\n \\begin{pmatrix}\n c_2 & 0 & -s_2 \\\\\n 0 & 1 & 0 \\\\\n s_2 & 0 & c_2\n \\end{pmatrix}\n %\n \\begin{pmatrix}\n c_1 & s_1 & 0 \\\\\n -s_1 & c_1 & 0 \\\\\n 0 & 0 & 1\n \\end{pmatrix}\n\\end{equation}\n%\nwhere $c_1 =\\cos{\\theta_1}$, $s_1 = \\sin{\\theta_1}$ etc. We can\nrewrite $\\mathbf{A}_{321} $ as\n%\n\\begin{equation}\n \\mathbf{A}_{321} =\n \\begin{pmatrix}\n c_2 c_1 & c_2 s_1 & -s_2 \\\\\n -c_3s_1 + s_3s_2c_1 & c_3c_1 + s_3s_2s_1 & s_3c_2 \\\\\n s_3s_1 + c_3s_2c_1 & -s_3c_1 + c_3s_2s_1 & c_3c_2\n \\end{pmatrix}\n \\label{Eq:A321}\n\\end{equation}\n%\nEquation~\\ref{Eq:A321} is equivalent to the matrix at the bottom of\nthe left hand column of Table E-1 on Page 764 of Wertz. The approach for\nobtaining the Attitude matrix is similar for the remaining 11 Euler angle\nsequences. Rather than derive the Attitude matrices for the remaining 11\nsequences, we present them in Table \\ref{table:EulerAnglestoDCM}.\n\n\\subsubsection{Conversion: Attitude Matrix to Euler Angles}\n\\label{sec:AttitudeMatrixtoEulerAngles}\n\nGiven: Sequence order ( i.e. 123, 121, ...$\\boldsymbol{321}$,...313), $\\mathbf{A}$\n\n\\noindent Find: $\\theta_1$, $\\theta_2$, $\\theta_3$\n\n\\noindent Name: \\emph{AMatToEulerAngles}\n\nWe'll give an example for a 321 rotation, and then present results\nfor the remaining 11 Euler angle sequences. Examining,\nEq.~\\ref{Eq:A321}, we see that\n%\n\\begin{equation}\n \\frac{A_{21} }{ A_{11}} = \\frac{\\cos{\\theta_2}\\sin{\\theta_1}}\n {\\cos{\\theta_2}\\cos{\\theta_1}}\n\\end{equation}\n%\nFrom this we can see that\n%\n\\begin{equation}\n \\theta_1 = \\tan^{-1}{\\frac{ A_{21} } { A_{11} }}\n\\end{equation}\n%\nFurther inspection of Eq.~\\ref{Eq:A321} shows us that\n%\n\\begin{equation}\n \\theta_2 = \\sin^{-1}{A_{13}}\n\\end{equation}\n%\nAt first glance, we may choose to calculate $\\theta_3$ using\n$\\theta_3 = \\tan^{-1}{(A_{23}/A_{33})}$. However, in the case that\n$\\theta_2 = 90^\\circ$, this would result in the indeterminate case,\n$\\theta_3 =$ $\\tan^{-1}(A_{23}/A_{33})$ $= \\tan^{-1}(0/0)$. An\nimproved method, found in the ADEAS mathematical specifications\ndocument, is to determine $\\theta_3$ using\n%\n\\begin{equation}\n \\theta_3 = \\tan^{-1} \\left(\\frac{ A_{31} \\sin{\\theta_1} - A_{32} \\cos{\\theta_1} }\n { -A_{21} \\sin{\\theta_1} + A_{22} \\cos{\\theta_1}} \\right)\n \\label{Eq:Atotheta3}\n\\end{equation}\n%\nSubstituting values from Eq.~\\ref{Eq:A321} into\nEq.~\\ref{Eq:Atotheta3}, and using abbreviated notation, we see\nthat\n%\n\\begin{equation}\n \\theta_3 = \\tan^{-1} \\left( \\frac{ s_1( s_3s_1 + c_3s_2c_1) - c_1(-s_3c_1 + c_3s_2s_1 )}\n { s_1(c_3s_1 - s_3s_2c_1 ) + c_1( c_3c_1 + s_3s_2s_1 ) } \\right)\n\\end{equation}\n%\nNow, if $\\theta_2 = 90^\\circ$, and we substitute $c_2 = 0$ and $s_2 = 1$ into\nthe above equation, we see we get a determinate form.\nResults for all twelve Euler Sequences are shown in Table\n\\ref{table:AMatToEulerAngles}.\n\n\\noindent Note: For all $\\tan^{-1}$ we need to use a quadrant check ( equivalent\nto atan2 ) to make sure the the correct quadrant is chosen.\n\n\\subsubsection{Conversion: Angular Velocity to \\\\ Euler Angles\nRates}\n\nGiven: Sequence ( i.e. 123, 121, .... 313), $\\theta_2$,\n$\\theta_3$ $\\boldsymbol\\omega$\n\n\\noindent Find: $\\dot\\theta_1$, $\\dot\\theta_2$, $\\dot\\theta_3$\n\n\\noindent Name: \\emph{AngVelToEulerAngles}\n%\n\\begin{equation}\n \\begin{pmatrix}\n \\dot\\theta_1\\\\\n \\dot\\theta_2\\\\\n \\dot\\theta_3\n \\end{pmatrix}\n %\n = \\mathbf{S}^{-1}(\\theta_2,\\theta_3)\\boldsymbol\\omega\n\\end{equation}\n%\n$\\mathbf{S}^{-1}(\\theta_2,\\theta_3)$ is dependent upon the Euler\nsequence. Table \\ref{table:EulerAngleKinematics} contains the\ndifferent expressions for $\\mathbf{S}^{-1}(\\theta_2,\\theta_3)$ for\neach of the 12 unique Euler sequences.\n\nNote: Each of the forms of $\\mathbf{S}^{-1}$ have a possible\nsingularity due to the appearance of either $\\sin{\\theta_2}$ or\n$\\cos{\\theta_2}$ in the denominator. If GMAT encounters a\nsingularity, an error message is thrown, and the zero vector is\nreturned.\n\n\\subsubsection{Conversion: Euler Angles Rates to Angular Velocity}\n\n\\noindent Given: Sequence ( i.e. 123, 121, .... 313), $\\theta_2$,\n$\\theta_3$, $\\dot\\theta_1$, $\\dot\\theta_2$, $\\dot\\theta_3$\n\n\\noindent Find: $\\boldsymbol\\omega$\n\n\\noindent Name: \\emph{EulerAnglesToAngVel}\n%\n\\begin{equation}\n \\boldsymbol\\omega = \\mathbf{S}(\\theta_2,\\theta_3)\n %\n \\begin{pmatrix}\n \\dot\\theta_1\\\\\n \\dot\\theta_2\\\\\n \\dot\\theta_3\n \\end{pmatrix}\n\\end{equation}\n%\n$\\mathbf{S}(\\theta_2,\\theta_3)$ is dependent upon the Euler\nsequence. Table \\ref{table:EulerAngleKinematics} contains the\ndifferent expressions for $\\mathbf{S}^{-1}(\\theta_2,\\theta_3)$ for\neach of the 12 unique Euler sequences.\n\n\\newpage\n\\subsubsection{Conversion: Quaternions to Euler Angles}\n\n\\noindent Given: $\\mathbf{q}$, $q_4$, Euler Sequence\n\n\\noindent Find: $\\theta_1$, $\\theta_2$, and $\\theta_3$\n\n\\noindent Name: \\emph{QuatsToEulerAngles}\n\nThere is not a direct transformation to convert from the quaternions to the Euler\nAngles. GMAT first converts from the quaternion to the Attitude matrix using the\nalgorithm presented above called ``QuatsToAMat\". % in Sec.~\\ref{sec:QuatToAMat}.\nThe Attitude matrix is then used to\ncalculate the Euler Angles for the given Euler angle sequence using the algorithm\ncalled ``AMatToEulerAngles:. % in Sec.~\\ref{sec:AttitudeMatrixtoEulerAngles}.\n\n\n\\subsubsection{Conversion: Euler Angles to Quaternions}\n\n\\noindent Given: $\\theta_1$, $\\theta_2$, and $\\theta_3$, Euler\nSequence\n\n\\noindent Find: $\\mathbf{q}$, $q_4$\n\n\\noindent Name: \\emph{EulerAnglesToQuats}\n\nThere is not a direct transformation to convert from Euler Angles to quaternions.\nGMAT first converts from the Euler Angles to the Attitude matrix using the\nalgorithm above called ``EulerAnglesToAMat\". % in Sec.~\\ref{sec:EulerAnglestoAMat}.\nThe Attitude matrix is then used to\ncalculate the quaternions using the algorithm\ncalled ``AMatToQuats\". % in Sec.~\\ref{sec:AMatToQuat}.\n\n\n\\newpage\n\\begin{table}[h]\n \\centering\n \\vspace{0 pt}\n \\caption{Attitude Matrices for 12 Unique Euler Angle Rotation Sequences}\n \\begin{tabular}{clccccccc} \\hline \\hline \\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.25 in} c_3 c_2 & \\hspace{0.3 in} c_3 s_2 s_1 + s_3 c_1 & \\hspace{0.1 in} -c_3 s_2 c_1 + s_1 s_3 \\\\\n \\hspace{0.2 in} -s_3 c_2 & \\hspace{0.3 in} -s_3 s_2 s_1 + c_3 c_1 & \\hspace{0.1 in} s_3 s_2 c_1 + c_3 s_1 \\\\\n \\hspace{0.3 in} s_2 & \\hspace{0.3 in} -c_2 s_1 & \\hspace{0.1 in} c_2 c_1 \\\\\n \\end{pmatrix}\n \\vspace{.1 in}$ \\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} c_3 c_2 & \\hspace{0.4 in} c_3 s_2 c_1 + s_1 s_3 & \\hspace{0.2 in} c_3 s_2 s_1 - s_3 c_1 \\\\\n \\hspace{0.3 in} -s_2 & \\hspace{0.4 in} c_2 c_1 & \\hspace{0.2 in} c_2 s_1 \\\\\n \\hspace{0.3 in} s_3 c_2 & \\hspace{0.4 in} s_3s_2c_1 - c_3s_1 & \\hspace{0.2 in} s_3 s_2 s_1 + c_3 c_1 \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.0 in} c_2 c_1 & \\hspace{0.3 in} s_2 & \\hspace{0.3 in} -c_2 s_1 \\\\\n \\hspace{0.0 in} -c_3 s_2 c_1 + s_3 s_1 & \\hspace{0.3 in} c_3 c_2 & \\hspace{0.3 in} c_3 s_2 s_1 + s_3 c_1 \\\\\n \\hspace{0.0 in} s_3 s_2 c_1 + c_3 s_1 & \\hspace{0.3 in} -s_3 c_2 & \\hspace{0.3 in} -s_3 s_2 s_1 + c_3 c_1 \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.0 in} c_3 c_1 + s_3 s_2 s_1 & \\hspace{0.4 in} s_3 c_2 & \\hspace{0.3 in} -c_3 s_1 + s_3 s_2 c_1 \\\\\n \\hspace{0.0 in} -s_3 c_1 + c_3 s_2 s_1 & \\hspace{0.4 in} c_3 c_2 & \\hspace{0.3 in} s_3 s_1 + c_3 s_2 c_1 \\\\\n \\hspace{0.0 in} c_2 s_1 & \\hspace{0.4 in} -s_2 & \\hspace{0.3 in} c_2 c_1 \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.1 in} c_3 c_1 - s_3 s_2 s_1 & \\hspace{0.1 in} c_3 s_1 + s_3 s_2 c_1 & \\hspace{0.4 in} -s_3 c_2 \\hspace{0.2 in} \\\\\n \\hspace{0.1 in} -c_2 s_1 & \\hspace{0.1 in} c_2 c_1 & \\hspace{0.4 in} s_2 \\hspace{0.2 in} \\\\\n \\hspace{0.1 in} s_3 c_1 + c_3 s_2 s_1 & \\hspace{0.1 in} s_3 s_1 - c_3 s_2 c_1 & \\hspace{0.4 in} c_3 c_2 \\hspace{0.2 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.0 in} c_2 c_1 & \\hspace{0.1 in} c_2 s_1 & \\hspace{0.4 in} -s_2 \\hspace{0.2 in} \\\\\n \\hspace{0.0 in} -c_3 s_1 + s_3 s_2 c_1 & \\hspace{0.1 in} c_3 c_1 + s_3 s_2 s_1 & \\hspace{0.4 in} s_3 c_2 \\hspace{0.2 in} \\\\\n \\hspace{0.0 in} s_3 s_1 + c_3 s_2 c_1 & \\hspace{0.1 in} -s_3 c_1 +c_3 s_2 s_1 & \\hspace{0.4 in} c_3 c_2 \\hspace{0.2 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} c_2 & \\hspace{0.4 in} s_2 s_1 & -s_2 c_1 \\\\\n \\hspace{0.3 in} s_3 s_2 & \\hspace{0.4 in} c_3 c_1 - s_3 c_2 s_1 & c_3 s_1 + s_3 c_2 c_1 \\\\\n \\hspace{0.3 in} c_3 s_2 & \\hspace{0.4 in} -s_3 c_1 - c_3 c_2 s_1 & -s_3 s_1 + c_3 c_2 c_1 \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.2 in} c_2 & \\hspace{0.4 in} s_2 c_1 & s_2 s_1 \\\\\n \\hspace{0.2 in} -c_3 s_2 & \\hspace{0.4 in} c_3 c_2 c_1 - s_3 s_1 & c_3 c_2 s_1 + s_3 c_1 \\\\\n \\hspace{0.2 in} s_3 s_2 & \\hspace{0.4 in} -s_3 c_2 c_1 - c_3 s_1 & -s_3 c_2 s_1 + c_3 c_1 \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 c_1 - s_3 c_2 s_1 & \\hspace{0.4 in} s_3 s_2 & \\hspace{0.2 in} -c_3 s_1 - s_3 c_2 c_1 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} s_2 s_1 & \\hspace{0.4 in} c_2 & \\hspace{0.2 in} s_2 c_1 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} s_3 c_1 + c_3 c_2 s_1 & \\hspace{0.4 in} -c_3 s_2 & \\hspace{0.2 in} -s_3 s_1 + c_3 c_2 c_1 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %-------------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 c_2 c_1 - s_3 s_1 & \\hspace{0.45 in} c_3 s_2 & \\hspace{0.25 in} -c_3 c_2 s_1 - s_3 c_1 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} -s_2 c_1 & \\hspace{0.45 in} c_2 & \\hspace{0.25 in} s_2 s_1 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} s_3 c_2 c_1 + c_3 s_1 & \\hspace{0.45 in} s_3 s_2 & \\hspace{0.25 in} -s_3 c_2 s_1 + c_3 c_1 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %-------------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 c_1 - s_3 c_2 s_1 & \\hspace{0.05 in} c_3 s_1 + s_3 c_2 c_1 & \\hspace{0.35 in} s_3 s_2 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} -s_3 c_1 - c_3 c_2 s_1 & \\hspace{0.05 in} -s_3 s_1 + c_3 c_2 c_1 & \\hspace{0.35 in} c_3 s_2 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} s_2 s_1 & \\hspace{0.05 in} -s_2 c_1 & \\hspace{0.35 in} c_2 \\hspace{0.25 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %-------------------------------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{A} = \\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1) = $\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 c_2 c_1 -s_3 s_1 & \\hspace{0.05 in} c_3 c_2 s_1 + s_3 c_1 & \\hspace{0.3 in} -c_3 s_2 \\hspace{0.2 in} \\\\\n \\hspace{0.05 in} -s_3 c_2 c_1 - c_3 s_1 & \\hspace{0.05 in} -s_3 c_2 s_1 + c_3 c_1 & \\hspace{0.3 in} s_3 s_2 \\hspace{0.2 in} \\\\\n \\hspace{0.05 in} s_2 c_1 & \\hspace{0.05 in} s_2 s_1 & \\hspace{0.3 in} c_2 \\hspace{0.2 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n\n \\hline \\hline\n \\normalsize\n \\end{tabular}\n \\label{table:EulerAnglestoDCM}\n\\end{table}\n\n\n\\begin{table}[h]\n \\centering\n \\caption{Kinematics of Euler Angle Rotation Sequences}\n \\begin{tabular}{cllcccccc} \\hline \\hline \\\\\n Euler Sequence & \\hspace{0.2 in} $\\mathbf{S}(\\theta_2,\\theta_3)$ &\n $\\hspace{0.7 in} \\mathbf{S}^{-1}(\\theta_2,\\theta_3)$\\\\\n \\hline\n \\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n c_3 c_2 & s_3 & 0 \\\\\n -s_3 c_2 & c_3 & 0 \\\\\n s_2 & 0 & 1 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 / c_2 & \\hspace{0.05 in} -s_3 / c_2 & \\hspace{0.3 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} s_3 & \\hspace{0.05 in} c_3 & \\hspace{0.3 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} -s_2 c_3 / c_2 & \\hspace{0.05 in} s_3 s_2 / c_2 & \\hspace{0.3 in} 1 \\hspace{0.25 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n c_3 c_2 & -s_3 & 0 \\\\\n -s_2 & 0 & 1 \\\\\n s_3 c_2 & c_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.1 in} c_3 / c_2 & \\hspace{0.3 in} 0 & \\hspace{0.25 in} s_3 / c_2 \\hspace{0.1 in} \\\\\n \\hspace{0.1 in} -s_3 & \\hspace{0.3 in} 0 & \\hspace{0.25 in} c_3 \\hspace{0.1 in} \\\\\n \\hspace{0.1 in} s_2 c_3 / c_2 & \\hspace{0.3 in} 1 & \\hspace{0.25 in} s_3 s_2 / c_2 \\hspace{0.1 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n s_2 & 0 & 1 \\\\\n c_3 c_2 & s_3 & 0 \\\\\n -s_3 c_2 & c_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} 0 & \\hspace{0.2 in} c_3 / c_2 & \\hspace{0.05 in} -s_3 / c_2 \\hspace{0.1 in} \\\\\n \\hspace{0.3 in} 0 & \\hspace{0.2 in} s_3 & \\hspace{0.05 in} c_3 \\hspace{0.1 in} \\\\\n \\hspace{0.3 in} 1 & \\hspace{0.2 in} -s_2 c_3 / c_2 & \\hspace{0.05 in} s_3 s_2 / c_2 \\hspace{0.1 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n s_3 c 2 & c_3 & 0 \\\\\n c_3 c_2 & -s_3 & 0 \\\\\n -s_2 & 0 & 1 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.1 in} s_3 / c_2 & \\hspace{0.15 in} c_3 / c_2 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.1 in} c_3 & \\hspace{0.15 in} -s_3 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.1 in} s_3 s_2 / c_2 & \\hspace{0.15 in} s_2 c_3 / c_2 & \\hspace{0.25 in} 1 \\hspace{0.25 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n -s_3 c_2 & c_3 & 0 \\\\\n s_2 & 0 & 1 \\\\\n c_3 c_2 & s_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.1 in} -s_3 / c_2 & \\hspace{0.3 in} 0 & \\hspace{0.2 in} c_3 / c_2 \\hspace{0.05 in} \\\\\n \\hspace{0.1 in} c_3 & \\hspace{0.3 in} 0 & \\hspace{0.2 in} s_3 \\hspace{0.05 in} \\\\\n \\hspace{0.1 in} s_3 s_2 / c_2 & \\hspace{0.3 in} 1 & \\hspace{0.2 in} -s_2 c_3 / c_2 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n -s_2 & 0 & 1 \\\\\n s_3 c_2 & c_3 & 0 \\\\\n c_3 c_2 & -s_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} 0 & \\hspace{0.3 in} s_3 / c_2 & \\hspace{0.05 in} c_3 / c_2 \\hspace{0.1 in} \\\\\n \\hspace{0.3 in} 0 & \\hspace{0.3 in} c_3 & \\hspace{0.05 in} -s_3 \\hspace{0.1 in} \\\\\n \\hspace{0.3 in} 1 & \\hspace{0.3 in} s_3 s_2 / c_2 & \\hspace{0.05 in} s_2 c_3 / c_2 \\hspace{0.1 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n c_2 & 0 & 1 \\\\\n s_3 s_2 & c_3 & 0 \\\\\n c_3 s_2 & -s_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} 0 & \\hspace{0.2 in} s_3 / s_2 & \\hspace{0.0 in} c_3 / s_2 \\hspace{0.05 in} \\\\\n \\hspace{0.3 in} 0 & \\hspace{0.2 in} c_3 & \\hspace{0.0 in} -s_3 \\hspace{0.05 in} \\\\\n \\hspace{0.3 in} 1 & \\hspace{0.2 in} -s_3 c_2 / s_2 & \\hspace{0.0 in} -c_3 c_2 / s_2 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n c_2 & 0 & 1 \\\\\n -c_3 s_2 & s_3 & 0 \\\\\n s_3 s_2 & c_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.3 in} 0 & \\hspace{0.3 in} -c_3 / s_2 & \\hspace{0.0 in} s_3 / s_2 \\hspace{0.05 in} \\\\\n \\hspace{0.3 in} 0 & \\hspace{0.3 in} s_3 & \\hspace{0.0 in} c_3 \\hspace{0.05 in} \\\\\n \\hspace{0.3 in} 1 & \\hspace{0.3 in} c_3 c_2 / s_2 & \\hspace{0.0 in} -s_3 c_2 / s_2 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n s_3 s_2 & c_3 & 0 \\\\\n c_2 & 0 & 1 \\\\\n -c_3 s_2 & s_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} s_3 / s_2 & \\hspace{0.25 in} 0 & \\hspace{0.25 in} -c_3 / s_2 \\hspace{0.1 in} \\\\\n \\hspace{0.05 in} c_3 & \\hspace{0.25 in} 0 & \\hspace{0.25 in} s_3 \\hspace{0.1 in} \\\\\n \\hspace{0.05 in} -s_3 c_2 / s_2 & \\hspace{0.25 in} 1 & \\hspace{0.25 in} c_3 c_2 / s_2 \\hspace{0.1 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n c_3 s_2 & -s_3 & 0 \\\\\n c_2 & 0 & 1 \\\\\n s_3 s_2 & c_3 & 0 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} c_3 / s_2 & \\hspace{0.25 in} 0 & \\hspace{0.2 in} s_3 / s_2 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} -s_3 & \\hspace{0.25 in} 0 & \\hspace{0.2 in} c_3 \\hspace{0.05 in} \\\\\n \\hspace{0.05 in} -c_3 c_2 / s_2 & \\hspace{0.25 in} 1 & \\hspace{0.2 in} -s_3 c_2 / s_2 \\hspace{0.05 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n s_3 s_2 & c_3 & 0 \\\\\n c_3 s_2 & -s_3 & 0 \\\\\n c_2 & 0 & 1 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.05 in} s_3 / s_2 & \\hspace{0.0 in} c_3 / s_2 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} c_3 & \\hspace{0.0 in} -s_3 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.05 in} -s_3 c_2 / s_2 & \\hspace{0.0 in} -c_3 c_2 / s_2 & \\hspace{0.25 in} 1 \\hspace{0.25 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\begin{pmatrix}\n -c_3 s_2 & s_3 & 0 \\\\\n s_3 s_2 & c_3 & 0 \\\\\n c_2 & 0 & 1 \\\\\n \\end{pmatrix}$\n &\n \\footnotesize\n $\\begin{pmatrix}\n \\hspace{0.1 in} -c_3 / s_2 & \\hspace{0.05 in} s_3 / s_2 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.1 in} s_3 & \\hspace{0.05 in} c_3 & \\hspace{0.25 in} 0 \\hspace{0.25 in} \\\\\n \\hspace{0.1 in} c_3 c_2 / s_2 & \\hspace{0.05 in} -s_3 c_2 / s_2 & \\hspace{0.25 in} 1 \\hspace{0.25 in} \\\\\n \\end{pmatrix} \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\hline \\hline\n \\end{tabular}\n \\label{table:EulerAngleKinematics}\n\\end{table}\n\n\n\\begin{landscape}\n\\begin{table}[h]\n \\centering\n \\vspace{0 pt}\n \\caption{ Computation of Euler Angles from Attitude Matrix}\n \\begin{tabular}{llllllll} \\hline \\hline \\\\\n % Note the blanks between $$ symbols below help align the equations in the table\n Euler Sequence \\hspace{0.5 in} & $ $ & Euler Angle Computations \\hspace{-1.1 in} & $ $ \\\\\n \\hline \\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n \\hspace{.1 in}\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(-A_{32}/A_{33})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(A_{31})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{13}\\sin{\\theta_1}+\n A_{12}\\cos{\\theta_1}}{A_{23}\\sin{\\theta_1}+A_{22}\\cos{\\theta_1}}\\hspace{0.15 in}\\right ) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1)$ \\hspace{.1 in}\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{23}/A_{22})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(-A_{21})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{12}\\sin{\\theta_1}- A_{13}\\cos{\\theta_1}}{-A_{32}\\sin{\\theta_1}+\n A_{33}\\cos{\\theta_1}}\\hspace{0.05 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(-A_{13}/A_{11})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(A_{12})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{21}\\sin{\\theta_1}+ A_{23}\\cos{\\theta_1}}{A_{31}\\sin{\\theta_1}+\n A_{33}\\cos{\\theta_1}}\\hspace{0.15 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{31}/A_{33})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(-A_{32})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{23}\\sin{\\theta_1}- A_{21}\\cos{\\theta_1}}{-A_{13}\\sin{\\theta_1}+\n A_{11}\\cos{\\theta_1}}\\hspace{0.05 in}\\right) \\vspace{.1 in} $\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(-A_{21}/A_{22})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(A_{23})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{32}\\sin{\\theta_1}+ A_{31}\\cos{\\theta_1}}{A_{12}\\sin{\\theta_1}+\n A_{11}\\cos{\\theta_1}}\\hspace{0.15 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{12}/A_{11})$\n &\n \\footnotesize\n $\\theta_2 = \\sin^{-1}(-A_{13})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{A_{31}\\sin{\\theta_1}- A_{32}\\cos{\\theta_1}}{-A_{21}\\sin{\\theta_1}+\n A_{22}\\cos{\\theta_1}}\\hspace{0.05 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{12}/-A_{13})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{11})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{-A_{33}\\sin{\\theta_1}- A_{32}\\cos{\\theta_1}}{A_{23}\\sin{\\theta_1}+\n A_{22}\\cos{\\theta_1}}\\hspace{0.05 in}\\right)\\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_1(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_1(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{13}/A_{12})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{11})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{-A_{22}\\sin{\\theta_1} + A_{23}\\cos{\\theta_1}}{-A_{32}\\sin{\\theta_1}+\n A_{33}\\cos{\\theta_1}}\\hspace{0.05 in}\\right)$ \\vspace{.1 in}\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{21}/A_{23})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{22})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{-A_{33}\\sin{\\theta_1} + A_{31}\\cos{\\theta_1}}{-A_{13}\\sin{\\theta_1}+\n A_{11}\\cos{\\theta_1}}\\hspace{0.05 in}\\right)$ \\vspace{.1 in}\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_2(\\theta_3)\\mathbf{R}_3(\\theta_2)\\mathbf{R}_2(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{23}/-A_{21})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{22})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\displaystyle\\frac{-A_{11}\\sin{\\theta_1} - A_{13}\\cos{\\theta_1}}{A_{31}\\sin{\\theta_1}+\n A_{33}\\cos{\\theta_1}}\\hspace{0.10 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_1(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{31}/-A_{32})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{33})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{-A_{22}\\sin{\\theta_1} - A_{21}\\cos{\\theta_1}}{A_{12}\\sin{\\theta_1}+\n A_{11}\\cos{\\theta_1}}\\hspace{0.05 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\footnotesize\n $\\mathbf{R}_3(\\theta_3)\\mathbf{R}_2(\\theta_2)\\mathbf{R}_3(\\theta_1)$\n &\n \\footnotesize\n $\\theta_1 = \\tan^{-1}(A_{32}/A_{31})$\n &\n \\footnotesize\n $\\theta_2 = \\cos^{-1}(A_{33})$\n &\n \\footnotesize\n $\\theta_3 = \\tan^{-1}\\left(\\hspace{0.05 in}\\displaystyle\\frac{-A_{11}\\sin{\\theta_1} + A_{12}\\cos{\\theta_1}}{-A_{21}\\sin{\\theta_1}+\n A_{22}\\cos{\\theta_1}}\\hspace{0.05 in}\\right) \\vspace{.1 in}$\\\\\n %\n %---------------------------------------------------------------------------------------------------------------------\n %\n \\hline \\hline\n \\end{tabular}\n \\label{table:AMatToEulerAngles}\n\\end{table}\n\\end{landscape}\n\n%\\end{document}\n%\n%After End Notes - These don't need comments since they are ignored completely.\n%If I do two carriage returns in a row, I get a new paragraph. I get\n%automatic indenting at the beginning of each paragraph. Here are the\n%rules as I learn them.\n%\n%1) Install WinEdt and follow its instructions. I had to pay $\\$$70.00.\n%I think the activation code will be e-mailed to me.\n%\n%2) After WinEdt, install MiKTeX. WinEdt has all its settings pre-configured\n%to support MiKTeX. MiKTeX is free but they would like a donation.\n%\n%3) Don't try to use dollar signs. They trigger something else in LaTeX.\n%\n%4) I used File$>$New to start this document.\n%\n%5) To get the right carrot to show in the line above, I had to put a\n%dollar sign on either side of it. I'm going to try that with dollar sign now.\n%It turns out three dollar signs in a row does not work well. So then I\n%randomly tried an opening dollar sign, a backslash prefix, the dollar sign\n%that I want to print, and a closing dollar sign. I get the following - $\\$$\n%\n%6) Up in Item 1 above, I used to say \"70 bucks\" because I hadn't figured\n%out the dollar sign slash double dollar sign trick yet. I fixed it. But\n%you didn't know that till now! ;-)\n%\n%7) The emoticon above printed fine with no backslashes or dollar signs.\n%\n%8) To set up a document so it will print at all you need two command lines\n%before your text and one after your text. The first two lines at the top\n%go like this -\n%\n%\\textbackslash documentclass\\{article\\}\n%\n%\\textbackslash begin\\{document\\}\n%\n%The final line at the bottom goes like this -\n%\n%\\textbackslash end\\{document\\}\n%\n%9) To get a leading backslash to print, like shown above, you just type a\n%backslash and then the words \"textbackslash\"\n%\n%10) And now, let's try an image!\n%\n%\\begin{figure}[h!]\n% \\centering\n% \\includegraphics[width=0.82\\textwidth]{QuaternionX42DefB&W}\n% \\caption{Physical Definition of Quaternions}\n%\\end{figure}\n%\n%11) To prevent orphaning a line, say one that describes a following equation,\n%use the following command.\n%\\\\*[0.5cm]\n%\n", "meta": {"hexsha": "f24f7e034247f257a7d351ce520c1357c0c8511e", "size": 89973, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/SystemDocs/MathematicalSpecification/AttitudeD5.tex", "max_stars_repo_name": "Randl/GMAT", "max_stars_repo_head_hexsha": "d6a5b1fed68c33b0c4b1cfbd1e25a71cdfb8f8f5", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-01-01T13:14:57.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-09T07:05:07.000Z", "max_issues_repo_path": "doc/SystemDocs/MathematicalSpecification/AttitudeD5.tex", "max_issues_repo_name": "ddj116/gmat", "max_issues_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_issues_repo_licenses": ["NASA-1.3"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2018-03-15T08:58:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-20T20:11:26.000Z", "max_forks_repo_path": "doc/SystemDocs/MathematicalSpecification/AttitudeD5.tex", "max_forks_repo_name": "ddj116/gmat", "max_forks_repo_head_hexsha": "39673be967d856f14616462fb6473b27b21b149f", "max_forks_repo_licenses": ["NASA-1.3"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-13T10:26:49.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-09T07:06:55.000Z", "avg_line_length": 42.0631136045, "max_line_length": 148, "alphanum_fraction": 0.5901659387, "num_tokens": 29408, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.863391617003942, "lm_q2_score": 0.727975460709318, "lm_q1q2_score": 0.6285279101610076}} {"text": "\\section{Fourier transform and Fourier series}\nWe make use of the theory of tempered distributions (see\n\\cite{strichartz2003guide} for an introduction)\nand we begin by collecting some results of independent interest, which\nwill also be important later. \n\\subsection{Fourier transform}\nBefore studying the Fourier transform, we first consider Schwartz space which is defined below.\n\\begin{definition} \\label{def:schwarz}\nThe Schwartz space $\\mathcal{S}\\left(\\mathbb{R}^{n}\\right)$ is the topological vector space of functions $f: \\mathbb{R}^{n} \\rightarrow \\mathbb{C}$ such that $f \\in C^{\\infty}\\left(\\mathbb{R}^{n}\\right)$ and\n$$\nx^{\\alpha} \\partial^{\\beta} f(x) \\rightarrow 0 \\quad \\text { as }|x| \\rightarrow \\infty\n$$\nfor every pair of multi-indices $\\alpha, \\beta \\in \\mathbb{N}_{0}^{n} .$ For $\\alpha, \\beta \\in \\mathbb{N}_{0}^{n}$ and $f \\in \\mathcal{S}\\left(\\mathbb{R}^{n}\\right)$ let\n(5.10)\n$$\n\\|f\\|_{\\alpha, \\beta}=\\sup _{\\mathbb{R}^{n}}\\left|x^{\\alpha} \\partial^{\\beta} f\\right|\n$$\nA sequence of functions $\\left\\{f_{k}: k \\in \\mathbb{N}\\right\\}$ converges to a function $f$ in $\\mathcal{S}\\left(\\mathbb{R}^{n}\\right)$ if\n$$\n\\left\\|f_{n}-f\\right\\|_{\\alpha, \\beta} \\rightarrow 0 \\quad \\text { as } k \\rightarrow \\infty\n$$\nfor every $\\alpha, \\beta \\in \\mathbb{N}_{0}^{n}$.\n\\end{definition}\nThe Schwartz space consists of smooth functions whose derivatives and the function itself decay at infinity faster than any power. Schwartz functions are rapidly decreasing. When there is no ambiguity, we will write $\\mathcal{S}\\left(\\mathbb{R}^{n}\\right)$ as $\\mathcal{S}$.\nRoughly speaking, tempered distributions grow no faster than a polynomial at infinity.\n\n\\begin{definition}\nA tempered distribution $T$ on $\\mathbb{R}^{n}$ is a continuous linear functional $T: \\mathcal{S}\\left(\\mathbb{R}^{n}\\right) \\rightarrow \\mathbb{C} .$ The topological vector space of tempered distributions is denoted by $\\mathcal{S}^{\\prime}\\left(\\mathbb{R}^{n}\\right)$ or $\\mathcal{S}^{\\prime} .$ If $\\langle T, f\\rangle$ denotes the value of $T \\in \\mathcal{S}^{\\prime}$ acting on $f \\in \\mathcal{S}$\nthen a sequence $\\left\\{T_{k}\\right\\}$ converges to $T$ in $\\mathcal{S}^{\\prime}$, written $T_{k} \\rightarrow T$, if\n$$\n\\left\\langle T_{k}, f\\right\\rangle \\rightarrow\\langle T, f\\rangle\n$$\nfor every $f \\in \\mathcal{S}$.\n\\end{definition}\nSince $\\mathcal{D} \\subset \\mathcal{S}$ is densely and continuously imbedded, we have $\\mathcal{S}^{\\prime} \\subset \\mathcal{D}^{\\prime} .$ Moreover, a distribution $T \\in \\mathcal{D}^{\\prime}$ extends uniquely to a tempered distribution $T \\in \\mathcal{S}^{\\prime}$ if and only if it is continuous on $\\mathcal{D}$ with respect to the topology on $\\mathcal{S}$. Every function $f \\in L_{\\text {loc }}^{1}$ defines a regular distribution $T_{f} \\in \\mathcal{D}^{\\prime}$ by\n$$\n\\left\\langle T_{f}, \\phi\\right\\rangle=\\int f \\phi d x \\quad \\text { for all } \\phi \\in \\mathcal{D}.\n$$\nIf $|f| \\leq p$ is bounded by some polynomial $p,$ then $T_{f}$ extends to a tempered distribution $T_{f} \\in \\mathcal{S}^{\\prime}$, but this is not the case for functions $f$ that grow too rapidly at infinity.\n\nThe Schwartz space is a natural one to use for the Fourier transform. Differentiation and multiplication exchange roles under the Fourier transform and therefore so do the properties of smoothness and rapid decrease. As a result, the Fourier transform is an automorphism of the Schwartz space. By duality, the Fourier transform is also an automorphism of the space of tempered distributions.\n\n\\begin{definition}\\label{def:fourier1}\nThe Fourier transform of a function $f \\in \\mathcal{S}\\left(\\mathbb{R}^{n}\\right)$ is the function $\\hat{f}: \\mathbb{R}^{n} \\rightarrow \\mathbb{C}$ defined by \n$$\n\\hat{f}(\\omega)= \\int f(x) e^{-2 \\pi i\\omega \\cdot x} d x.\n$$\nThe inverse Fourier transform of $f$ is the function $\\check{f}: \\mathbb{R}^{n} \\rightarrow \\mathbb{C}$ defined by\n$$\n\\check{f}(x)=\\int f(\\omega) e^{2 \\pi i\\omega \\cdot x} d k.\n$$\n\\end{definition}\n\n\\begin{definition}\\label{def:fourier2}\nThe Fourier transform of a tempered distribution $f \\in \\mathcal{S}'$ is defined by \n$$\n\\langle \\hat{f}, \\phi\\rangle = \\langle f, \\hat \\phi\\rangle,\\quad \\forall \\phi\\in \\mathcal{S}.\n$$ \n\\end{definition}\n\nThe support of a continuous function $f$ is the closure of the set $\\{x\\in \\mathbb{R}: f(x)\\neq 0\\}$.\n\\begin{properties}\nThe Fourier transform has the following properties\n\\begin{enumerate}\n\\item If $f\\in \\mathcal{S}'$ and the support of $\\hat f$ is $\\{0\\}$, then $f$ is a polynomial.\n\\item If $f\\in \\mathcal{S}'$ and the support of $\\hat f$ is a single point $\\{a\\}$, then $f(x)=e^{2\\pi iax}P(x)$, where $P(x)$ is a polynomial.\n\\end{enumerate}\n\\end{properties}\n\n\n\n\n\n\n\n\\subsection{Poisson summation formula}\n\\input{6DL/PoissonSum}\n\\subsection{A special cut-off function}\n\\input{6DL/Cut-off-function}\n", "meta": {"hexsha": "90c93cc7d0883a251a32ddf282e41a234d76529c", "size": 4813, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/Fourier.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/Fourier.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/Fourier.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.1625, "max_line_length": 473, "alphanum_fraction": 0.7016413879, "num_tokens": 1542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916099737806, "lm_q2_score": 0.7279754371026367, "lm_q1q2_score": 0.6285278846614122}} {"text": "\\documentclass[a4paper]{article}\n\n\\addtolength{\\oddsidemargin}{-10pt}\n\\addtolength{\\textwidth}{50pt}\n\\addtolength{\\textheight}{70pt}\n\\addtolength{\\voffset}{-40pt}\n\n\\title{Mobile antenna model description}\n\\author{Jan Šimbera}\n\\date{\\today}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage[pdftex,unicode,hidelinks]{hyperref}\n\\usepackage[pdftex]{graphicx}\n\n\\begin{document}\n\\maketitle\n\\begin{abstract}\nThis document outlines a model that describes the spread of cell phone signal\nfor the purposes of mapping cell phone antenna coverage areas and detecting\nthe user's location somewhat more accurately.\n\\end{abstract}\n\n\n\\section{Model description} \\label{sec:antmodel}\n\n\\subsection{Assumptions}\nThe connection of the cell phone to the mobile network is facilitated through\na system of antennas. Usually, the operator shares very little information\nabout these antennas; nevertheless, some parameters, such as antenna location,\nare almost always available, sometimes through crowdsourced registers. We thus\nassume the antenna locations $\\vec{x_0} = (x_0,y_0)$ are fixed and known; we\nalso assume a planar coordinate system for simplicity.\n\n\\subsection{Signal spread model}\\label{sec:sigmodel}\nWe assume three orthogonal effects on the signal strength $S_i(\\vec{x})$\nof antenna $i$ for a given location $\\vec{x} = (x,y)$, that is:\n\\begin{itemize}\n\\item antenna power $P_i$,\n\\item distance from the antenna $d_i(\\vec{x}) = ||\\vec{x} - \\vec{x_i}||$,\n\\item azimuth of the vector from the antenna to the location\n $\\varphi_i(\\vec{x}) = \\arctan{\\frac{y - y_i}{x - x_i}}$.\n\\end{itemize}\n\nWe can employ a probabilistic approach to the problem: we first model the\nprobability of a user being at a given location $\\vec{x}$ if we know that he\nor she is connected to an antenna $i$ and then use the Bayesian rule to\ndetermine signal strength as the inverse conditional probability:\n\\begin{equation}\nS_i(\\vec{x}) = P(i|\\vec{x}) = P(\\vec{x}|i) \\frac{P(i)}{P(\\vec{x})}\n\\end{equation}\nWe can isolate the $P(i)$ as the antenna power $P_i$ and amalgamate the\nremaining two terms to make the model agnostic to an apriori distribution of\nusers (i.e. population/settlement density):\n\\begin{equation}\nS_i(\\vec{x}) = P_i \\frac{P(\\vec{x}|i)}{P(\\vec{x})} = P_i \\cdot L_i(\\vec{x})\n\\end{equation}\nThe resulting term $L_i(\\vec{x})$ should capture the remaining two effects of\ndistance and azimuth. As we assume the effects to be orthogonal, we can\nexpress it as\n\\begin{equation}\nL_i(\\vec{x}) = f_d(d_i(\\vec{x})) \\cdot f_a(\\varphi_i(\\vec{x}))\n\\end{equation}\n\n\\subsubsection{Distance decay}\nAs we operate on macroscopic scales, the distance decay of an EM signal is\ngoverned by an inverse square law:\n\\begin{equation}\nS_i(\\vec{x}) \\propto d_i(\\vec{x})^{-2}\n\\end{equation}\nThis, however, would imply the signal strength goes to infinity near the\nlocation of the antenna. Therefore, it is useful to employ a small correcting\nfactor. Using this, we arrive at a probabilistic description by a Cauchy\ndistribution, which is useful:\n\\begin{equation}\nd_i(\\vec{x}) \\sim \\mathrm{Cauchy}(0,\\gamma_i)\n\\end{equation}\nwhere $\\gamma_i$ is the distance decay scale factor. We can use the probability\ndensity function of the Cauchy distribution as our distance decay function:\n\\begin{equation}\nf_d(d_i(\\vec{x})) = \\frac{1}{\n \\pi \\gamma_i \\left[1 + \\left( \\frac{d_i(\\vec{x})}{\\gamma_i} \\right)^2 \\right]\n} = \\frac{\\gamma_i}{\\pi [d_i(\\vec{x})^2 + \\gamma_i^2]}\n\\end{equation}\n\n\\subsubsection{Angular distribution}\nTo model the angular dispersion of the signal, we can use the von Mises\ndistribution in a manner similar to the Cauchy distribution for distance decay.\nThe von Mises distribution is a circular analogue of the normal distribution:\n\\begin{equation}\nf(\\varphi) = \\frac{\\exp{[\\kappa \\cos{(\\varphi - \\alpha)}]}}{2\\pi I_0(\\kappa)}\n\\end{equation}\nwhere $I_0(\\kappa)$ is the modified Bessel function of order zero, $\\alpha$ is\nthe principal angle and $\\kappa$ is the directional concentration.\n\nUsing this, we define\n\\begin{equation}\nf_a(\\varphi_i(\\vec{x})) = \\frac\n {\\exp{[\\kappa_i \\cos{(\\varphi_i(\\vec{x}) - \\alpha_i)}]}}\n {2\\pi I_0(\\kappa_i)}\n\\end{equation}\nwhich leaves us with two parameters for the antenna: orientation (principal\nangle) $\\alpha$ and angular signal concentration (narrowness) $\\kappa$.\n\n\\subsubsection{Final signal strength equation}\nCombining the above results, we arrive at the final signal model result\n\\begin{align}\nS_i(\\vec{x}) &= P_i \\cdot f_d(d_i(\\vec{x})) \\cdot f_a(\\varphi_i(\\vec{x}))\\\\\n&= P_i \\frac\n {\\gamma_i \\exp{[\\kappa_i \\cos{(\\varphi_i(\\vec{x}) - \\alpha_i)}]}}\n {2\\pi^2 I_0(\\kappa_i) [d_i(\\vec{x})^2 + \\gamma_i^2]}\\\\\n&= \\frac{P_i \\gamma_i}{2\\pi^2 I_0(\\kappa_i)} \\cdot \\frac\n {\\exp{[\\kappa_i \\cos{(\\varphi_i(\\vec{x}) - \\alpha_i)}]}}\n {d_i(\\vec{x})^2 + \\gamma_i^2}\\\\\n&= K_i \\cdot \\frac\n {\\exp{[\\kappa_i \\cos{(\\varphi_i(\\vec{x}) - \\alpha_i)}]}}\n {d_i(\\vec{x})^2 + \\gamma_i^2}\n\\end{align}\n\nSince the signal strengths are relative to each other, we should specify a\ncondition on their absolute values; this can be formulated e.g. as\n\\begin{equation} \\label{eq:powersumcond}\n\\sum_{a=1}^{n_a} P_a \\gamma_a = 1\n\\end{equation}\n\n\\subsection{Antenna parameters}\nThe signal strength model uses four additional parameters (in addition to\nlocation) to describe the antennas:\n\\begin{itemize}\n\\item Antenna orientation azimuth $\\alpha \\in [0;2\\pi)$ (with zero pointing\n north).\n\\item Antenna power or overall signal strength $P \\in \\mathbb{R}^{+}$. This\n is somehow related to the power of the antenna transmitter.\n\\item Distance decay scale parameter $\\gamma \\in \\mathbb{R}^{+}$, defining\n the rate of signal strength diminishing with increasing distance from the\n antenna. This can be related e.g. to the vertical orientation of the\n antenna with respect to the surface.\n\\item Angular concentration of antenna signal $\\kappa \\in \\mathbb{R}^{+}_{0}$.\n The higher the concentration (narrowness), the more the signal power is\n concentrated along the direction of the antenna orientation. $\\kappa = 0$\n means the antenna is isotropic and radiates the signal uniformly in all\n directions.\n\\end{itemize}\nSometimes, some of the parameters may be known a priori, such as the antenna\nazimuths. The rest of the parameters need to be estimated from the antenna\nconnection data, which should be the next large task.\n\n\n\n\\section{Model parameter estimation}\nTo estimate the antenna parameters, several variants of data sources may be\nused. In this estimation scenario, we consider a set of $n_p$ places $p$ with\nlocations $\\vec{x}_p$. We denote the signal strength of a given antenna\n$a$ at that place as $S_{ap}$, the distance $d_{ap}$ and angle $\\varphi_{ap}$\nof a given place from it as in \\ref{sec:sigmodel}.\n\nIn addition to the place data, we enter data about a set of $n_u$ users and\ntheir simultaneous connections to antennas (that is, how often they are\nconnected to what antennas while not moving themselves).\nTherefore, we propose using night-time signalling (SS.7 logs) or CDR\ndata where antenna connection variability is most likely to come from antenna\nsignal strength variation, not from user movements. Given this, we may derive\nfor each user a set of connection time fractions $f_{ua}$ denoting the\nlikelihood of the connection of the user to the respective antennas, satisfying\n$\\sum_{a=1}^{n_a} f_{ua} = 1 \\,\\forall u$.\n\n\\subsection{Estimation with known locations} \\label{sec:locest}\nIf we have an additional data source that allows us to determine the actual\nnight-time positions of the users (such as independent mobility tracker app\nlogs), the model estimation becomes simpler because we can compute the user\nlocation probabilities with respect to the places $p$ directly as $p_{up}$.\n\nThe algorithm works as an EM procedure, trying to estimate the signal strengths\nfor all places and antennas with antenna parameters as proxy variables.\nThe E step estimates the antenna parameters from signal strengths using OLS,\nwith the initial values obtained from location averaging, and the M step\nestimates the signal strengths from the antenna parameters using the equations\nfrom \\ref{sec:antmodel}.\n\n\\subsubsection{Antenna dominance fractions}\nDominance fractions denote how much the given place is dominated by a signal\nfrom a given antenna, an therefore, the probability of a user at a given\nlocation being connected to the antenna:\n\\begin{equation}\n\\Psi_{ap} = \\frac\n {\\sum_{u=1}^{n_u} p_{up} f_{ua}}\n {\\sum_{u=1}^{n_u} p_{up}}\n\\end{equation}\nThis ensures that $\\sum_{a=1}^{n_a} \\Psi_{ap} = 1 \\,\\forall p$.\n\n\\subsubsection{Expectation step}\nIn this step, we compute the optimal antenna parameters using OLS on a\nlinearized signal strength function (with parameters from the previous round)\nset as equal to the E-computed signal strength:\n\\begin{align}\n\\frac{1}{P_a^{(i)}} &\\cdot P_a^{(i+1)} \\nonumber \\\\\n+ \\frac\n {d_{ap}^2 - {\\gamma_a^{(i)}}^2}\n {\\gamma_a^{(i)} (d_{ap}^2 + {\\gamma_a^{(i)}}^2)}\n &\\cdot \\gamma_a^{(i+1)} \\nonumber \\\\\n+ \\kappa_a^{(i)} \\sin{(\\varphi_{ap} - \\alpha_a^{(i)})}\n &\\cdot \\alpha_a^{(i+1)} \\nonumber \\\\\n+ \\left[\n \\cos{(\\varphi_{ap} - \\alpha_a^{(i)})}\n - \\frac{I_1(\\kappa_a^{(i)})}{I_0(\\kappa_a^{(i)})}\n \\right] &\\cdot \\kappa_a^{(i+1)} =\\\\\n\\frac{S_{ap}^{(i+1)}}{S_{ap}^{(i)}}\n+ 1\n+ \\frac\n {d_{ap}^2 - {\\gamma_a^{(i)}}^2}\n {d_{ap}^2 + {\\gamma_a^{(i)}}^2}\n+ \\kappa_a^{(i)} \\alpha_a^{(i)} \\sin{(\\varphi_{ap} - \\alpha_a^{(i)})}\n&+ \\kappa_a^{(i)} \\left[\n \\cos{(\\varphi_{ap} - \\alpha_a^{(i)})}\n - \\frac{I_1(\\kappa_a^{(i)})}{I_0(\\kappa_a^{(i)})}\n \\right] \\nonumber\n\\end{align}\n\n\\subsubsection{Maximization step}\nIn this step, we estimate the signal strengths for places from their antenna\ndominances and antenna parameters:\n\\begin{equation}\nS_{ap}^{(i+1)} = \\Psi_{ap} \\sum_{b=1}^{n_a} \\frac\n {P_b^{(i)} \\gamma_b^{(i)} \\exp{[\n \\kappa_b^{(i)} \\cos{(\\varphi_{bp} - \\alpha_b^{(i)})}\n ]}}\n {2\\pi^2 I_0(\\kappa_b^{(i)}) [d_{bp}^2 + {\\gamma_b^{(i)}}^2]}\\\\\n\\end{equation}\n\n\\subsubsection{Initial antenna parameter estimations}\nThe initial values can be obtained as follows:\n\n\\paragraph{Antenna principal angles} $\\alpha_a$ -- if not known -- can be\ndetermined as weighted circular means\n\\begin{equation}\n\\alpha_a = \\arctan \\frac\n {\\sum_{p=1}^{n_p} \\Psi_{ap} \\sin \\varphi_{ap}}\n {\\sum_{p=1}^{n_p} \\Psi_{ap} \\cos \\varphi_{ap}}\n\\end{equation}\n\n\\paragraph{Antenna angular concentrations} $\\kappa_a$ can be determined\nusing the von Mises-Fisher iterative estimation procedure from the parameter\n$\\bar{R}_a$:\n\\begin{equation}\n\\bar{R}_a = \\frac\n {\\left(\\sum_{p=1}^{n_p} \\Psi_{ap} \\sin \\varphi_{ap}\\right)^2\n + \\left(\\sum_{p=1}^{n_p} \\Psi_{ap} \\cos \\varphi_{ap}\\right)^2}\n {\\left(\\sum_{p=1}^{n_p} \\Psi_{ap}\\right)^2}\n\\end{equation}\nWith this parameter, we can produce the initial estimate for $\\kappa_a$ as\n\\begin{equation}\n\\kappa_a^{(0)} = \\frac\n {\\bar{R}_a (2 - \\bar{R}_a^2)}\n {1 - \\bar{R}_a^2}\n\\end{equation}\nand then repeat the following equation until convergence\n\\begin{equation}\n\\kappa_a^{(i+1)} = \\kappa_a^{(i)} - \\frac\n {A_p(\\kappa_a^{(i)}) - \\bar{R}_a}\n {1 - A_p(\\kappa_a^{(i)})^2 - \\frac{A_p(\\kappa_a^{(i)})}{\\kappa_a^{(i)}}}\n\\end{equation}\nwhere $A_p(x) = \\frac{I_1(x)}{I_0(x)}$ is the ratio of modified Bessel functions\nof the first and zeroth order respectively. Three iterations are usually\nsufficient.\n\n\\paragraph{Antenna distance decay parameter} $\\gamma_a$ may be estimated as\nthe weighted mean distance of a dominated place from the antenna:\n\\begin{equation}\n\\gamma_a = \\frac\n {\\sum_{p=1}^{n_p} \\Psi_{ap} d_{ap}}\n {\\sum_{p=1}^{n_p} \\Psi_{ap}}\n\\end{equation}\n\n\\paragraph{Strengths of the antennas} $P_a$, to honor the condition from\n\\eqref{eq:powersumcond}, can be estimated as the relative size of the antenna's\narea of dominance to the total area of study. We need to factor out the\ngeneral influence of the distance decay parameter, too:\n\\begin{equation}\nP_a = \\frac\n {\\sum_{p=1}^{n_p} \\Psi_{ap} A_p}\n {\\gamma_a \\sum_{p=1}^{n_p} A_p}\n\\end{equation}\nwhere $A_p$ is the area corresponding to place $p$ -- if the places are only\ngiven as points, the areas may be computed e. g. using Voronoi polygons.\n\n\n\\subsection{Estimation with unknown locations}\nIf we do not have data about real user locations $p_{up}$, we can supply them by\nemploying an EM estimation procedure over the proposed approach.\n\nTo do this, it is useful to know at least roughly the expected user densities\nover the places used $w_p$. These densities may be derived from an apriori\npopulation distribution measure such as a census grid or, alternatively, a land\nuse layer such as GHSL aggregated to a reasonable level.\n\n\\subsubsection{Expectation step}\nThis step of the procedure estimates the user locations\nfrom signal strengths in those places by computing place affinities as\n\\begin{equation}\na_{up}^{(i+1)} = \\left(\n 1 + \\frac\n {w_p - \\sum_{u=1}^{n_u} p_{up}^{(i)}}\n {w_p + \\sum_{u=1}^{n_u} p_{up}^{(i)}}\n\\right)^{\\frac{1}{k_c}}\n\\sqrt{\n \\frac{1}{n_a} \\sum_{a=1}^{n_a} \\left(\n f_{ua} - \\frac{S_{ap}^{(i)}}{\\sum_{b=1}^{n_a} S_{bp}^{(i)}}\n \\right)^2\n}\n\\end{equation}\nwhere $k_c$ is the crowding tolerance coefficient specifying the degree of\nadherence to the apriori place weights $w_p$. Its higher values mean a more\ndifferent spatial distribution of users can arise.\n\nThen the actual localization probabilities are produced by normalizing the\naffinities to sum to one:\n\\begin{equation}\np_{up}^{(i+1)} = \\frac\n {a_{up}^{(i+1)}}\n {\\sum_{q=1}^{n_p} a_{uq}^{(i+1)}}\n\\end{equation}\n\n\\subsubsection{Maximization step}\nThis step of the procedure uses the method from\n\\ref{sec:locest} to derive $S_{ap}^{(i)}$ from $p_{up}^{(i)}$.\n\\end{document}", "meta": {"hexsha": "0024beb9b62302864348bc0b7544980e6d9e6a06", "size": 13724, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/signal_model.tex", "max_stars_repo_name": "simberaj/mobilib", "max_stars_repo_head_hexsha": "ae350d095a34f53704bd4aaaf7f45e573bda779a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "docs/signal_model.tex", "max_issues_repo_name": "simberaj/mobilib", "max_issues_repo_head_hexsha": "ae350d095a34f53704bd4aaaf7f45e573bda779a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/signal_model.tex", "max_forks_repo_name": "simberaj/mobilib", "max_forks_repo_head_hexsha": "ae350d095a34f53704bd4aaaf7f45e573bda779a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.9694189602, "max_line_length": 81, "alphanum_fraction": 0.7113815214, "num_tokens": 4212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637577007393, "lm_q2_score": 0.7310585903489892, "lm_q1q2_score": 0.6284645748788176}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{examples}\n\\usepackage{fancyhdr}\n\\usepackage{footer}\n\n\\begin{document}\n\n% ============================================================================================\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n import cdblib\n checkpoint_file = 'tests/semantic/output/example-13.json'\n cdblib.create (checkpoint_file)\n checkpoint = []\n\\end{cadabra}\n\\egroup\n\n\\clearpage\n\n% =============================================================================\n\\section*{Example 13a The Weyl tensor vanishes in 3d -- direct proof}\n\n\\begin{cadabra}\n {x,y,z}::Coordinate.\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,u,v,w#}::Indices (values={x,y,z},position=independent).\n\n \\partial{#}::PartialDerivative.\n\n g_{a b}::Metric.\n g^{a b}::InverseMetric.\n\n {\\partial_{a b}{g_{c d}},\\partial_{a}{g_{b c}},g_{a b},g^{a b}}::SortOrder.\n\n GammaU := \\Gamma^{a}_{b c} -> 1/2 g^{a d} ( \\partial_{b}{g_{d c}}\n + \\partial_{c}{g_{b d}}\n - \\partial_{d}{g_{b c}}). # cdb(Gamma.000,GammaU)\n\n GammaD := \\Gamma_{a b c} -> 1/2 ( \\partial_{b}{g_{a c}}\n + \\partial_{c}{g_{b a}}\n - \\partial_{a}{g_{b c}}). # cdb(Gamma.010,GammaD)\n\n Rabcd := R_{a b c d} -> \\partial_{c}{\\Gamma_{a b d}}\n - \\partial_{d}{\\Gamma_{a b c}}\n + \\Gamma_{e a d} \\Gamma^{e}_{b c}\n - \\Gamma_{e a c} \\Gamma^{e}_{b d}. # cdb (Rabcd.000,Rabcd)\n\n Rab := R_{a b} -> g^{c d} R_{a c b d}. # cdb (Rab.000,Rab)\n\n Rscalar := R -> g^{a b} R_{a b}. # cdb (R.000,Rscalar)\n\n # Weyl in 3-dimensions\n\n Cabcd := R_{a b c d} - (R_{a c} g_{b d} - R_{a d} g_{b c})\n - (g_{a c} R_{b d} - g_{a d} R_{b c})\n + 1/2 R (g_{a c} g_{b d} - g_{a d} g_{b c}). # cdb (ex-13a.100,Cabcd)\n\n # Use 8 Cabcd to clear the fractions\n\n EightCabcd := 8 @(Cabcd). # cdb (ex-13a.110,EightCabcd)\n\n substitute (Cabcd,Rscalar)\n substitute (Cabcd,Rab)\n substitute (Cabcd,Rabcd)\n substitute (Cabcd,GammaU)\n substitute (Cabcd,GammaD)\n\n distribute (Cabcd)\n\n sort_product (Cabcd)\n rename_dummies (Cabcd)\n canonicalise (Cabcd) # cdb (ex-13a.101,Cabcd)\n\n EightCabcd := 8 @(Cabcd). # cdb (ex-13a.111,EightCabcd)\n\n gab := {g_{x x} = gxx, g_{x y} = gxy, g_{x z} = gxz,\n g_{y x} = gxy, g_{y y} = gyy, g_{y z} = gyz,\n g_{z x} = gxz, g_{z y} = gyz, g_{z z} = gzz}.\n\n complete (gab, $g^{a b}$)\n evaluate (Cabcd,gab) # cdb (ex-13a.102,Cabcd)\n evaluate (EightCabcd,gab) # cdb (ex-13a.112,EightCabcd)\n\n\\end{cadabra}\n\n\\clearpage\n\n% \\begin{dgroup*}\n% \\Dmath*{ C_{abcd} = \\Cdb*{ex-13a.100}\n% = \\Cdb*{ex-13a.101}\n% = \\Cdb*{ex-13a.102}}\n% \\end{dgroup*}\n\n\\begin{dgroup*}\n \\Dmath*{ 8C_{abcd} = \\Cdb*{ex-13a.110}\n = \\Cdb*{ex-13a.111}\n = \\Cdb*{ex-13a.112}}\n\\end{dgroup*}\n\n\\clearpage\n\n% =============================================================================\n\\section*{Example 13b The Weyl tensor vanishes in 3d -- orthonormal basis}\n\n\\begin{cadabra}\n {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w#}::Indices(position=independent).\n\n g_{a b}::Metric.\n g^{a b}::InverseMetric.\n\n R_{a b c d}::RiemannTensor.\n\n ex{#}::LaTeXForm{\"e_x\"}.\n ey{#}::LaTeXForm{\"e_y\"}.\n ez{#}::LaTeXForm{\"e_z\"}.\n\n {R_{a b c d}, g_{a b}, g^{a b}}::SortOrder.\n\n Rab := R_{a b} -> g^{c d} R_{a c b d}.\n\n Rscalar := R -> g^{a b} R_{a b}.\n\n gab := g^{a b} -> ex^{a} ex^{b} + ey^{a} ey^{b} + ez^{a} ez^{b}.\n\n ortho := {ex^{a} ex^{b} g_{a b} -> 1, ey^{a} ey^{b} g_{a b} -> 1, ez^{a} ez^{b} g_{a b} -> 1,\n ex^{a} ey^{b} g_{a b} -> 0, ex^{a} ez^{b} g_{a b} -> 0,\n ey^{a} ex^{b} g_{a b} -> 0, ey^{a} ez^{b} g_{a b} -> 0,\n ez^{a} ex^{b} g_{a b} -> 0, ez^{a} ey^{b} g_{a b} -> 0}.\n\n # Weyl in 3-dimensions\n\n Cabcd := R_{a b c d} - (R_{a c} g_{b d} - R_{a d} g_{b c})\n - (g_{a c} R_{b d} - g_{a d} R_{b c})\n + 1/2 R (g_{a c} g_{b d} - g_{a d} g_{b c}). # cdb (ex-13b.100,Cabcd)\n\n\n substitute (Cabcd, Rscalar) # cdb(ex-13b.101,Cabcd)\n substitute (Cabcd, Rab) # cdb(ex-13b.102,Cabcd)\n distribute (Cabcd) # cdb(ex-13b.103,Cabcd)\n\n Cabcd := C_{a b c d} -> @(Cabcd).\n\n Cxyxy := C_{a b c d} ex^{a} ey^{b} ex^{c} ey^{d}. # cdb(ex-13b.104,Cxyxy)\n\n substitute (Cxyxy,Cabcd) # cdb(ex-13b.105,Cxyxy)\n distribute (Cxyxy) # cdb(ex-13b.106,Cxyxy)\n\n substitute (Cxyxy, ortho, repeat=True) # cdb(ex-13b.107,Cxyxy)\n\n substitute (Cxyxy, gab) # cdb(ex-13b.108,Cxyxy)\n distribute (Cxyxy) # cdb(ex-13b.109,Cxyxy)\n\n sort_product (Cxyxy) # cdb(ex-13b.110,Cxyxy)\n rename_dummies (Cxyxy) # cdb(ex-13b.111,Cxyxy)\n canonicalise (Cxyxy) # cdb(ex-13b.112,Cxyxy)\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\Dmath*{\\cdb*{ex-13b.101}}\n \\Dmath*{\\cdb*{ex-13b.102}}\n \\Dmath*{\\cdb*{ex-13b.103}}\n\\end{dgroup*}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\Dmath*{\\cdb{ex-13b.104}\n = \\Cdb*[\\hfill\\hskip2.5cm]{ex-13b.105}\n = \\Cdb*{ex-13b.106}\n = \\Cdb*{ex-13b.107}\n = \\Cdb*[\\hfill\\hskip2.0cm]{ex-13b.108}\n = \\Cdb*{ex-13b.109}\n = \\Cdb*{ex-13b.110}\n = \\Cdb*{ex-13b.111}\n = \\Cdb*{ex-13b.112}}\n\\end{dgroup*}\n\n\\clearpage\n\n% =============================================================================\n\\section*{Example 13c The Weyl tensor vanishes in 3d -- orthonormal basis}\n\n\\begin{cadabra}\n Cxyxz := C_{a b c d} ex^{a} ey^{b} ex^{c} ez^{d}. # cdb(ex-13c.101,Cxyxz)\n\n substitute (Cxyxz,Cabcd) # cdb(ex-13c.102,Cxyxz)\n\n distribute (Cxyxz) # cdb(ex-13c.103,Cxyxz)\n\n substitute (Cxyxz, ortho, repeat=True) # cdb(ex-13c.104,Cxyxz)\n\n substitute (Cxyxz, gab) # cdb(ex-13c.105,Cxyxz)\n distribute (Cxyxz) # cdb(ex-13c.106,Cxyxz)\n\n sort_product (Cxyxz) # cdb(ex-13c.107,Cxyxz)\n rename_dummies (Cxyxz) # cdb(ex-13c.108,Cxyxz)\n canonicalise (Cxyxz) # cdb(ex-13c.109,Cxyxz)\n\\end{cadabra}\n\n\\clearpage\n\n\\begin{dgroup*}\n \\Dmath*{\\cdb{ex-13c.101}\n = \\Cdb*[\\hfill\\hskip2.5cm]{ex-13c.102}\n = \\Cdb*{ex-13c.103}\n = \\Cdb*{ex-13c.104}\n = \\Cdb*{ex-13c.105}\n = \\Cdb*{ex-13c.106}\n = \\Cdb*{ex-13c.107}\n = \\Cdb*{ex-13c.108}\n = \\Cdb*{ex-13c.109}}\n\\end{dgroup*}\n\n\\clearpage\n\n% ============================================================================================\n% export to json format\n\n\\bgroup\n\\CdbSetup{action=hide}\n\\begin{cadabra}\n for i in range( len(checkpoint) ):\n cdblib.put ('check{:03d}'.format(i),checkpoint[i],checkpoint_file)\n\\end{cadabra}\n\\egroup\n\n\\end{document}\n", "meta": {"hexsha": "1d35e70c9a5abc13ca514c0af1b66d31bc6942ac", "size": 7850, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "source/cadabra/example-13.tex", "max_stars_repo_name": "leo-brewin/cadabra-tutorial", "max_stars_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 20, "max_stars_repo_stars_event_min_datetime": "2019-12-20T07:49:47.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T22:55:47.000Z", "max_issues_repo_path": "source/cadabra/example-13.tex", "max_issues_repo_name": "leo-brewin/cadabra-tutorial", "max_issues_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "source/cadabra/example-13.tex", "max_forks_repo_name": "leo-brewin/cadabra-tutorial", "max_forks_repo_head_hexsha": "5b428ae158b5346315ab6c975dee9de933e5c3d7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-12-22T13:52:19.000Z", "max_forks_repo_forks_event_max_datetime": "2019-12-22T13:52:19.000Z", "avg_line_length": 34.1304347826, "max_line_length": 103, "alphanum_fraction": 0.4278980892, "num_tokens": 2672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637612961506, "lm_q2_score": 0.7310585727705126, "lm_q1q2_score": 0.6284645623956945}} {"text": "\\subsection{k-means}\n\\begin{figure}\n \\begin{minipage}{0.5\\textwidth}\n \\centering\n \\includegraphics[width=.9\\linewidth]{elbowds1.png}\n \\caption{Timing curve vs silhouette coeff. dataset 1}\\label{Fig:K-means vs silhouette coeff. dataset 1}\n \\end{minipage}\\hfill\n \\begin{minipage}{0.5\\textwidth}\n \\centering\n \\includegraphics[width=.9\\linewidth]{elbowds2.png}\n \\caption{Timing curve vs silhouette coeff. dataset 2}\\label{Fig:K-means vs silhouette coeff. dataset 2}\n \\end{minipage}\n\\end{figure}\nThe two clustering techniques I explored for these datasets are K-Means clustering and Gaussian Mixture Models (Expectation Maximization)\nIn order to choose an appropriate $k$ to do k-means clustering, I used the elbow method\\cite{developers_2020}.\nUnless otherwise specified the standard distance measure I used was Euclidean distance.\nComparing all cluster cohesion metrics (mean distortion, silhouette coefficient, and Calinski-Harabasz score) here were\nthe elbow results:\n\\begin{center}\n \\begin{tabular}{|c| c | c | c |}\n \\hline\n & Metric & Cluster Count & Value \\\\\n \\hline\n \\hline\n Dataset 1 & Avg Silhouette Coefficient & 4 & 0.558 \\\\\n \\hline\n Dataset 2 & Avg Silhouette Coefficient & 3 & 0.687 \\\\\n \\hline\n \\end{tabular}\n\\end{center}\nLooking at figure~\\ref{Fig:K-means vs silhouette coeff. dataset 1} and figure~\\ref{Fig:K-means vs silhouette coeff. dataset 2} you can\nsee similar numbers in terms of the best \"k\".\nI chose silhouette coeff over the other cluster cohesion metrics because it has a bias towards preferring well separated\nclusters (coeffs closer to 1), which I anticipate would improve generalization and overall accuracy during a machine\nlearning process where the clusters are inputs to the learner.\nDataset 2 appears to have better separation between its 3 clusters than dataset 1's 4 clusters.\nDoing further silhouette analysis on DS1 reveals that the clusters when plotted on it to the surface do not appear to\nhave a \"cluster\" round shape and instead are ovular, whereas data set two had similar characteristics to data set one\nexcept that cluster two is actually interspersed in between cluster zero and cluster one, however this was only for the\nfirst two features and others could have had better separation/shape.\nBoth clustering methods On both data sets appeared to have two very strongly defined clusters, and then up to two additional weakly defined clusters.\nThis more or less aligns with the fact that this is a binary classification problem and that samples could be placed into one of two clusters.\nThe clusters were more well defined and well separated on the second dataset which also makes sense given that all supervised learning methods performed better in terms of accuracy on this dataset relative to the first.\nOn closer inspection, the third and fourth clusters appear to be randomly dispersed throughout the dataset to capture outliers.\nThese clusters likely are noise capturers.\n\\begin{center}\n \\begin{tabular}{|c| c | c | c |}\n \\hline\n & Normalized Mutual Information & Homogeneity Score & Completeness Score \\\\\n \\hline\n \\hline\n Dataset 1 & 0.001 & 0.002 & 0.001 \\\\\n \\hline\n Dataset 2 & 0.189 & 0.236 & 0.158 \\\\\n \\hline\n \\end{tabular}\n\\end{center}\nValues closer to 1 are better than 0 in the above table for each score metric.\nIt seems that the performance of K-means is slightly better on dataset 2 than on 1.\nTo improve the performance of this we would likely need to include a dimensionality reduction technique, or modify the\ndataset so that clusters are more hyper-spherical and separated from each other in the feature space.\n\n\\subsection{Expectation Maximization}\\label{subsec:expectation-maximization}\n\\begin{figure}\n \\begin{minipage}{0.5\\textwidth}\n \\centering\n \\includegraphics[width=.9\\linewidth]{gmmcomponentsds1.png}\n \\caption{n\\_components vs AIC/BIC DS1}\\label{Fig:GMM DS1}\n \\end{minipage}\\hfill\n \\begin{minipage}{0.5\\textwidth}\n \\centering\n \\includegraphics[width=.9\\linewidth]{gmmcomponentsds2.png}\n \\caption{n\\_components vs AIC/BIC DS2}\\label{Fig:GMM DS2}\n \\end{minipage}\n\\end{figure}\nFor expectation maximization, I used the scikit-learn Gaussian mixture model function.\nBased on my exploration, I do not believe that the data are distributed in a hyper-spherical format, therefore a GMM would\nbe able to better group samples based on its ability to accommodate non-spherical hyper-surfaces.\nIn order to select a best value for the number of estimators in the model, I plotted the Bayesian Information Criterion\nvs the Akaike Information Criterion, where they begin to diverge is likely the best place to stop in terms of n\\_estimators.\nIn this case looking at figures~\\ref{Fig:GMM DS1} and~\\ref{Fig:GMM DS2} it appears to be the same as the optimal K found\npreviously (4 and 3 respectively.) After finding these and running best fits and comparing results I got:\n\\begin{center}\n \\begin{tabular}{|c| c | c | c |}\n \\hline\n & Normalized Mutual Information & Homogeneity Score & Completeness Score \\\\\n \\hline\n \\hline\n Dataset 1 & 0.002 & 0.002 & 0.002 \\\\\n \\hline\n Dataset 2 & 0.133 & 0.174 & 0.108 \\\\\n \\hline\n \\end{tabular}\n\\end{center}\nThese clusters more or less make sense much in the same way that K-means did.\nFor dataset 1, GMM outperformed in terms of mutual information, homogeneity, and completeness shared by the clustering techniques and the actual labels\nwhereas for dataset 2 K-means outperformed, which indicates that the data was slightly more spherical than I anticipated.\nHowever, neither of these clustering techniques did very well since all the values were close to 0 indicating that assigment\nto clusters was independent and that the clusters themselves were not very independent, nor complete.\n", "meta": {"hexsha": "a5e0e5ad009270f0002036e5b190d3fb7ad26204", "size": 6208, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assignment3/clustering-discussion.tex", "max_stars_repo_name": "zparnold/cs7641", "max_stars_repo_head_hexsha": "e37e7b9259237adffbeb36ccc8dd17f67892286a", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assignment3/clustering-discussion.tex", "max_issues_repo_name": "zparnold/cs7641", "max_issues_repo_head_hexsha": "e37e7b9259237adffbeb36ccc8dd17f67892286a", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assignment3/clustering-discussion.tex", "max_forks_repo_name": "zparnold/cs7641", "max_forks_repo_head_hexsha": "e37e7b9259237adffbeb36ccc8dd17f67892286a", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.08, "max_line_length": 219, "alphanum_fraction": 0.7160115979, "num_tokens": 1495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929053683038, "lm_q2_score": 0.80563219364797, "lm_q1q2_score": 0.6283873953817201}} {"text": "\\documentclass[journal]{IEEEtran}\n\\usepackage[utf8]{inputenc}\n\\usepackage{minted}\n\\usepackage{booktabs}\n\\usepackage{color}\n\\usepackage{mathtools}\n\\usepackage{mathrsfs}\n\\usepackage{multirow}\n\n\\usepackage[binary-units=true]{siunitx}\n\n\\newcommand{\\scala}[1]{\\mintinline{scala}{#1}}\n\n\\title{Advanced Algorithms (\\texttt{LINGI2266}) \\\\ Assignment 1: Knapsack problem}\n\\author{Gilles Peiffer}\n\\date{October 11, 2019}\n\n\\begin{document}\n\n\\maketitle\n\n\\begin{abstract}\n\tAn analysis of the solution to the knapsack problem which takes into account various dimensions along which to judge algorithms for this problem is given.\n\tVarious heuristics are given, some of which are based on a theoretical understanding of the mathematics behind the knapsack problem, while others are highly-specialized in order to profit from the input formulation as much as possible.\n\tSome implementation fragments in Scala are also presented.\n\\end{abstract}\n\n\\section{Introduction}\n\\label{sec:intro}\nThe goal for this assignment was to solve various instances of the \\emph{knapsack problem}, defined as follows:\n\\[\n\\begin{array}{rrcl@{\\quad}l}\n\\textnormal{maximize} & \\sum_{i \\in \\{1, \\ldots, n\\}} x_i v_i &&& \\\\\n\\textnormal{such that} & \\sum_{i \\in \\{1, \\ldots, n\\}} x_i w_i & \\le & C &\\\\\n& x_i & \\in & \\{0, 1\\} & \\forall i \\in \\{1, \\ldots, n\\}\n\\end{array}\n\\]\n\nThe two main types of algorithms used to solve this class of problems are\n\\begin{itemize}\n\t\\item \\emph{dynamic programming} and\n\t\\item \\emph{branch and bound}.\n\\end{itemize}\nEach of these has its own advantages and disadvantages, which usually allow one to make an informed choice as to which algorithm is best-suited for a particular instance of the knapsack problem.\n\n\\section{I/O}\nIn order to be easily tested on INGInious, the input and output for this particular instance have a very specific fixed format.\n\\subsection{Reading the Input}\nThe program is given a \\scala{String} as input which contains the path to the data file.\nThis file is opened and read according to the format specified in the problem description on INGInious, and uses Scala code written for this purpose by Pierre Schaus.\n\n\\subsection{Output Formatting}\nThe \\scala{toString} method definition is overridden in order to produce output with the correct format.\nThis requires some internal bookkeeping, explained in slightly more detail in Sections~\\ref{sec:dpoutput} (for dynamic programming) and \\ref{sec:bnboutput} (for branch and bound).\n\n\\section{Dynamic Programming}\nDynamic programming, or DP for short, is an algorithmic technique which uses memoization in order to improve computation time at the cost of memory space.\n\n\\subsection{Recurrence Equations}\nOne has to define an \\emph{objective function} \\(\\mathscr{O}(j, k)\\), which, for the knapsack problem, gives the optimal value that can be obtained using items 1 through \\(j\\) and with a capacity \\(k\\).\nThis function can be defined recursively as follows:\n\\[\n\\mathscr{O}(j, k) = \\left\\{\n\\begin{array}{ll}\n\\max\\big\\{ \\mathscr{O}(j-1, k), & \\multirow{2}{*}{\\(\\textnormal{if } w_j \\le k,\\)}\\\\\n\\qquad v_j + \\mathscr{O}(j-1, k-w_j) \\big\\}, & \\\\\\\\\n\\mathscr{O}(j-1, k), & \\textnormal{otherwise,}\\\\\\\\\\\\\n0, & \\textnormal{if } j = 0.\n\\end{array}\n\\right.\n\\]\nBy this definition, the final solution to the problem is \\(\\mathscr{O}(n, C)\\).\n\\subsection{Implementation}\nTurning these equations into Scala is fairly straightforward.\nThe following is what one could consider the ``core'' of the implementation:\n\\begin{minted}[fontsize=\\scriptsize]{scala}\nval cache = collection.mutable.Map.empty[(Int, Int), Int]\nval tkn = collection.mutable.Map.empty[(Int, Int), Boolean]\ndef O(j: Int, k: Int): Int = {\n if (j < 0) 0\n else {\n val (vj, wj) = items(j)\n if (wj > k) {\n tkn.update((j, k), false)\n O_(j-1,k)\n } else {\n if (O_(j-1,k) > vj + O_(j-1,k-wj)) {\n tkn.update((j, k), false)\n O_(j-1,k)\n } else {\n tkn.update((j, k), true)\n vj + O_(j-1,k-wj)\n }\n }\n }\n}\ndef O_(j: Int, k: Int): Int = {\n cache.getOrElseUpdate((j,k), O(j,k))\n}\nprintln(O(n-1, c))\n\\end{minted}\n\\subsubsection{Recurrence}\nThe recurrence equation given earlier is implemented in the form of a multitude of nested \\scala{if}/\\scala{else} branches, and a function call to \\scala{O_}, which is explained in Section~\\ref{sec:dpmemo}.\nOne should also note that instead of the solution being \\(\\mathscr{O}(n, C)\\), one must call \\scala{O(n-1, C)}; this is explained by the fact that in Scala, array indexing starts at 0, causing a potential off-by-one error that must be kept in mind.\n\\subsubsection{Memoization}\n\\label{sec:dpmemo}\nThe main goal of DP is to speed up computation time by storing values that were already computed--also called \\emph{memoization}--and reusing those values whenever applicable.\nIn the code fragment, \\scala{cache} is responsible for this: whenever a value of \\scala{O}, the objective function, is needed, the program calls \\scala{O_} which serves as an alias for \\scala{cache.getOrElseUpdate}.\nThe latter's name describes what it does quite aptly: informally, it tells the computer to go and check if the value that is needed is already stored in \\scala{cache}.\nIf it is, \\scala{O_} simply returns that value; on the other hand, if it was not cached yet, it is computed (using the recursive property of the objective function) and the result is stored in \\scala{cache} for future reference.\n\\subsubsection{Output Formatting}\n\\label{sec:dpoutput}\nThe \\scala{Map} called \\scala{tkn} stores whether the optimal solution for a given \\scala{j} and \\scala{k} uses the \\scala{j}th element.\nThis is not related to the actual knapsack problem, but is necessary to be able to construct the output according to the specifications.\n\n\\subsection{Advantages and Disadvantages}\n\\subsubsection{Advantages}\nDP has several advantages:\n\\begin{itemize}\n\t\\item The algorithm is very easy to understand and easy to debug.\n\t\\item \\emph{For small values of \\(C\\)}, the algorithm is very fast and can very quickly solve problems which would be nigh impossible with a brute force search.\n\\end{itemize}\n\\subsubsection{Disadvantages}\nOn the other hand, there are some major downsides to the DP algorithm as well:\n\\begin{itemize}\n\t\\item \\emph{For large values of \\(C\\)}, the algorithm's pseudo-polynomiality makes it unusable.\n\tA variation based on the value \\(V\\) is also possible, but the same restrictions apply when \\(V\\) is large.\n\tThis is for example what caused Instance B to take far too long using DP: for reference, while there are only 56 items to select, the maximum capacity and optimal value are both 104 723 596, meaning neither variation of the algorithm performs adequately.\n\\end{itemize}\nTaking all of this into account, DP makes for a good first try at a solution, and can serve as an easy way to obtain solutions to a real-life problem to see whether other, more complex algorithms are correctly implemented and thought out.\nFor this assignment, two instances were solved using DP (D and E).\n\n\\section{Branch and Bound}\nBranch and bound, or BnB for short, is an algorithm which uses upper-bounding and tree search in order to discard parts of a huge binary tree which have been shown not to contain an optimal solution.\nMore formally, consider the knapsack problem of Section~\\ref{sec:intro}.\nThis problem, in a way, searches for a tuple of binary values which maximizes the objective function.\nConsider the search tree formed by the possible value assignments to this tuple.\nAt the root, all variables are ``free'' to be either 1 or 0, but every time the tree branches in two, the left subtree has a variable set to 1, while the right subtree has that same variable set to 0.\nIf one has a procedure to efficiently find an upper bound for the subtree rooted at a certain node, the optimal solution can be obtained by first finding a feasible solution, and then trying to find a better one while discarding subtrees with an upper bound lower than or equal to the best known value of the feasible solution.\nThis way, one can avoid having to explore the entire search tree, whose size grows exponentially with the number of decision variables.\n\n\\subsection{Design Choices}\nThe explanation above is purposely vague, since many design choices still need to be taken based on the specific instance of the problem.\nThe most important is of the upper-bounding procedure.\nWhile technically not necessary to find a solution, this is required to help discard subtrees faster and thus make the algorithm solve problems much more efficiently.\nIf the upper-bounding procedure is not sufficiently tight (or takes a long time to compute), branch and bound is not much better than a brute force approach.\n\n\\subsubsection{Upper-bounding}\nThe upper-bounding procedure of choice, which gives great results in the majority of cases, is called \\emph{linear relaxation}.\nWhat LR does is, instead of forcing variables to be binary, it allows them to be anywhere in \\([0, 1]\\).\nBy then sorting free items (i.e., whose decision variable has not been set yet) by their utility ratio, that is, the ratio between their value and weight, and selecting as many of these items (and potentially, fractions of items) as possible until the capacity constraint is tight, we can guarantee that the optimal objective will be lower than or equal to this upper bound.\n\nTo implement this, we do the following (suppose that \\scala{selectable} has been \\textbf{sorted} already):\n\\begin{minted}[fontsize=\\scriptsize]{scala}\noverride val upperBound: Double = {\n // Linear relaxation.\n var i: Int = 0\n var ub: Double = obj\n var c: Int = capa\n if (c > 0) {\n while (i < selectable.length\n && c >= weight(selectable(i))) {\n c = c - weight(selectable(i))\n ub = ub + value(selectable(i))\n i += 1\n }\n if (i < selectable.length) {\n val d: Double = c.toDouble / weight(selectable(i))\n ub = ub + value(selectable(i)) * d\n }\n }\n ub\n}\n\\end{minted}\nIn this code fragment, one can distinguish two main steps:\n\\begin{enumerate}\n\t\\item first, use all the free variables (those in \\scala{selectable}) in order to find the \\emph{critical item} which would make the knapsack too full;\n\t\\item add whatever you can of the critical item, maxing out the capacity.\n\\end{enumerate}\n\nAdditionally, though not necessary, the flooring function can be applied to this final result, since one can prove that if the weights and values are integers, then so is the optimal solution.\n\n\\subsubsection{Traversal algorithm}\nThis section mentions the traversal algorithms which can be used to decide which nodes are to be explored first.\nConcretely, two options exist, each with their own advantages and disadvantages:\n\\begin{itemize}\n\t\\item \\emph{Best-first search}.\n\tThis traversal looks at the most promising nodes (highest upper bound) first.\n\tThis search strategy can be implemented with a \\scala{PQueue}; the code used for this is heavily influenced by the one on the course's BitBucket repository, written by Pierre Schaus.\n\tThe main advantage of this strategy is also its main drawback: it depends heavily on the quality of the upper-bounding procedure in order to be successful.\n\tAnother big drawback is that the user has no control over the number of open nodes which are kept in memory.\n\t\\item \\emph{Depth-first search}.\n\tDFS looks at the deepest, leftmost node first, regardless of the upper bound (as long as exploring the subtree makes sense, i.e. as long as the subtree potentially contains an optimal solution).\n\tThis can be implemented with a \\scala{Stack}.\n\tThis strategy, while not being as good in general at finding optimal solutions, does fare much better in the memory department, since memory consumption is proportional to the height of the tree, which is typically linear.\n\\end{itemize}\nIn the code, the only change that needs to be made is in the \\scala{OpenNodes} implementation.\n\nFor this assignment, a best-first search algorithm using linear relaxation for upper bounding was used to solve instances A, C and F, which did either not get solved within the memory and time constraints by the DP algorithm (C) or because BnB was implemented before DP (A and F, where both algorithms arrive at the correct solution).\n\n\\subsection{Output formatting}\n\\label{sec:bnboutput}\nWhen sorting the elements with the goal of being able to apply LR, the initial order gets lost.\nThis is quite problematic for this particular assignment, since the output format makes use of the initial order.\nTo get around this problem, one can add the original weight and value vectors to the \\scala{KnapsackNode}, using these to identify used items.\nThe assumption that two items cannot be identical was verified beforehand, meaning this is a correct way for the instances of the assignment to recover the initial ordering.\n\n\\subsection{The Peculiar Case of Instance B}\nInstance B has the property that all its items have a unit utility ratio.\nThis means that sorting them does nothing, other than potentially shuffle them (since quicksort is not a stable sort).\nAdd in the fact that the LR upper bound, while tight, gives little extra information in this case, and one quickly gets a problem which takes a very long time to get solved.\n\nHowever, seeing as sorting by utility ratio does not change the ordering, and that the upper bound, despite being tight, does not allow best-first search to solve the problem optimally fast enough, an algorithm using DFS based on a \\scala{Stack} was implemented, with elements being sorted based on their value (most valuable items being taken first).\n\nThis algorithm outperforms the other BnB one (and the DP solver as well) by quite some margin, finding the optimal solution in a matter of seconds (in fact, it happens to go down the ``optimal'' path almost immediately due to the order in which variables are examined).\n\n\\section{Conclusion}\nThe knapsack problem is one of the most important problems in combinatorial optimization, and DP and BnB are only some of the many possible ways to solve instances of it.\nThanks to this assignment, we learnt that in some cases, it is very hard to predict which algorithm is best-suited to a given instance.\n\nImplementing the various algorithms helped to better understand why and how they work, while playing around with them on various instances helped to identify their strengths and shortcomings.\n\\end{document}", "meta": {"hexsha": "3bb6130e9bef965f3351fcb0c9e01a195bf91ec8", "size": 14366, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/assignment1/report.tex", "max_stars_repo_name": "Peiffap/lingi2266-implementations", "max_stars_repo_head_hexsha": "f069d8cc7872c7d8b22cf69b4c77d4d7bb83fd16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/assignment1/report.tex", "max_issues_repo_name": "Peiffap/lingi2266-implementations", "max_issues_repo_head_hexsha": "f069d8cc7872c7d8b22cf69b4c77d4d7bb83fd16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/assignment1/report.tex", "max_forks_repo_name": "Peiffap/lingi2266-implementations", "max_forks_repo_head_hexsha": "f069d8cc7872c7d8b22cf69b4c77d4d7bb83fd16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.7117117117, "max_line_length": 374, "alphanum_fraction": 0.761937909, "num_tokens": 3484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812554, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.6283873917417158}} {"text": "\\subsection{Groups}\\label{subsec:groups}\n\n\\begin{definition}\\label{def:unital_magma_inverse_element}\n Let \\( \\mscrM \\) be a \\hyperref[def:unital_magma]{unital magma}. We say that \\( y \\) is the \\term{left inverse} (resp. \\term{right inverse}) of \\( x \\) if \\( yx = e \\) (resp. \\( xy = e \\)).\n\n If \\( y \\) is simultaneously a left and right inverse of \\( x \\), we call a \\term{two-sided inverse} or simply an \\term{inverse} of \\( x \\) and denote it by \\( x^{-1} \\). It is unique by \\fullref{def:unital_magma_inverse_element_unique}. This notation is consistent with \\fullref{def:unital_magma/exponentiation}\n\\end{definition}\n\n\\begin{proposition}\\label{def:unital_magma_inverse_element_unique}\n For every element \\( x \\) of any unital magma, the (two-sided) \\hyperref[def:unital_magma_inverse_element]{inverse} \\( x^{-1} \\) of \\( x \\) is unique.\n\\end{proposition}\n\\begin{proof}\n If \\( y \\) and \\( z \\) are both inverses of \\( x \\), then\n \\begin{equation*}\n y = ey = zxy = ze = z.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:group}\n A \\term{group} is a \\hyperref[def:unital_magma/monoid]{monoid} in which every element has an \\hyperref[def:unital_magma_inverse_element]{inverse}. Groups are the most well-studied and most well-behaved magmas. Many useful properties like \\hyperref[thm:def:group/properties/cancellative]{cancellation} rely on associativity, so we do not consider non-associative groups.\n\n \\begin{thmenum}\n \\thmitem{def:group/theory} We can construct the \\hyperref[def:first_order_theory]{theory of groups} by adding a unary \\hyperref[def:first_order_language/func]{functional symbol} \\( (\\anon)^{-1} \\) and the axiom\n \\begin{equation}\\label{eq:def:group/theory/inverse_axiom}\n \\qforall \\xi (\\xi \\cdot \\xi^{-1} = e \\wedge \\xi^{-1} \\cdot \\xi = e)\n \\end{equation}\n to the theory of \\hyperref[def:unital_magma/monoid]{monoids}.\n\n \\thmitem{def:group/function_parity} A \\hyperref[def:function]{function} \\( \\varphi: \\mscrG \\to \\mscrH \\) between two groups is called \\term{even} if, for every \\( x \\in \\mscrG \\), we have\n \\begin{equation}\\label{eq:def:group/function_parity/even}\n \\varphi(x^{-1}) = \\varphi(x)\n \\end{equation}\n and \\term{odd} if\n \\begin{equation}\\label{eq:def:group/function_parity/odd}\n \\varphi(x^{-1}) = \\varphi(x)^{-1}.\n \\end{equation}\n\n \\thmitem{def:group/homomorphism} A \\hyperref[def:first_order_homomorphism]{first-order homomorphism} between the groups \\( \\mscrG \\) and \\( \\mscrH \\) is an odd \\hyperref[def:unital_magma/homomorphism]{unital magma homomorphism}.\n\n As shown in \\fullref{thm:group_homomorphism_single_condition}, however, the conditions \\eqref{eq:def:pointed_set/homomorphism} and \\eqref{eq:def:group/function_parity/odd} are redundant.\n\n \\thmitem{def:group/submodel} The set \\( S \\subseteq \\mscrG \\) is a \\hyperref[def:first_order_substructure]{first-order substructure} of \\( \\mscrG \\) if it is a \\hyperref[def:unital_magma/submodel]{unital submagma} and if \\( S^{-1} = S \\), where\n \\begin{equation*}\n S^{-1} = \\set{ s^{-1} \\given s \\in S }.\n \\end{equation*}\n\n \\thmitem{def:group/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} group is the \\hyperref[def:pointed_set/trivial]{trivial pointed set} \\( \\set{ e } \\).\n\n \\thmitem{def:group/category} The \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} of groups \\( \\ucat{Grp} \\) is \\hyperref[def:concrete_category]{concrete} with respect to \\hyperref[def:unital_magma/monoid]{\\( \\ucat{Mon} \\)}.\n\n \\thmitem{def:group/exponentiation} We extend \\hyperref[def:unital_magma/exponentiation]{unital magma exponentiation} to all integers by setting\n \\begin{equation*}\n x^{-n} \\coloneqq (x^n)^{-1}.\n \\end{equation*}\n\n This operation behaves well as shown in \\fullref{thm:def:group/properties/negative_power}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:group_homomorphism_single_condition}\n A function \\( \\varphi: \\mscrG \\to \\mscrH \\) between the groups \\( \\mscrG \\) and \\( \\mscrH \\) is a \\hyperref[def:group/homomorphism]{homomorphism} if and only if it satisfies \\eqref{eq:def:magma/homomorphism}.\n\\end{proposition}\n\\begin{proof}\n \\SufficiencySubProof \\eqref{eq:def:magma/homomorphism} is required to hold by definition.\n\n \\NecessitySubProof Let the function \\( \\varphi \\) satisfy \\eqref{eq:def:magma/homomorphism}. Then it preserves identities (i.e. is a \\hyperref[def:pointed_set/homomorphism]{pointed set homomorphism}) since\n \\begin{equation*}\n e_{\\mscrH} \\varphi(e_{\\mscrG}) = \\varphi(e_{\\mscrG}) = \\varphi(e_{\\mscrG} e_{\\mscrG}) = \\varphi(e_{\\mscrG}) \\varphi(e_{\\mscrG})\n \\end{equation*}\n and by \\fullref{thm:def:group/properties/cancellative}, the operation is cancellative.\n\n Inverses are preserved (i.e. \\eqref{eq:def:group/function_parity/odd} holds) because\n \\begin{equation*}\n \\varphi(x^{-1})\n =\n \\varphi(x^{-1}) e_{\\mscrH}\n =\n \\varphi(x^{-1}) \\varphi(x) \\varphi(x)^{-1}\n =\n \\varphi(x^{-1} x) \\varphi(x)^{-1}\n =\n e_{\\mscrH} \\varphi(x)^{-1}\n =\n \\varphi(x)^{-1}.\n \\end{equation*}\n\n Therefore, \\( \\varphi \\) is indeed a group homomorphism.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:group/properties}\n Every \\hyperref[def:group]{group} \\( \\mscrG \\) has the following basic properties:\n \\begin{thmenum}\n \\thmitem{thm:def:group/properties/cancellative} The (binary) group operation is \\hyperref[def:magma/cancellative]{cancellative}.\n \\thmitem{thm:def:group/properties/identity_inverse} The identity \\( e \\) is its own inverse.\n \\thmitem{thm:def:group/properties/inverse_composition} \\( (xy)^{-1} = y^{-1} x^{-1} \\).\n \\thmitem{thm:def:group/properties/double_inverse} \\( x = (x^{-1})^{-1} \\)\n \\thmitem{thm:def:group/properties/negative_power} For any positive integer \\( n \\), \\( (x^n)^{-1} = (x^{-1})^n \\)\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:group/properties/cancellative} If \\( x = y \\), obviously \\( xz = yz \\) and \\( zx = zy \\). Now if \\( xz = yz \\), we have\n \\begin{equation*}\n x = x(zz^{-1}) = (xz)z^{-1} = (yz)z^{-1} = y(zz^{-1}) = y.\n \\end{equation*}\n\n The case \\( zx = zy \\) is analogous.\n\n \\SubProofOf{thm:def:group/properties/identity_inverse} \\( ee = e \\).\n \\SubProofOf{thm:def:group/properties/inverse_composition}\n \\begin{equation*}\n (xy) (y^{-1} x^{-1})\n =\n x (y y^{-1}) x^{-1}\n =\n e\n =\n y^{-1} (x^{-1} x) y\n =\n (y^{-1} x^{-1}) (xy).\n \\end{equation*}\n\n \\SubProofOf{thm:def:group/properties/double_inverse}\n \\begin{equation*}\n (x^{-1})^{-1}\n =\n x x^{-1} (x^{-1})^{-1}\n =\n x.\n \\end{equation*}\n\n \\SubProofOf{thm:def:group/properties/negative_power} Using \\fullref{thm:def:group/properties/double_inverse},\n \\begin{equation*}\n x^{-n}\n =\n (x^n)^{-1}\n =\n x^{-1} \\cdots x^{-1}\n =\n (x^{-1})^n.\n \\end{equation*}\n\\end{proof}\n\n\\begin{example}\\label{ex:power_set_is_not_a_group}\n The \\hyperref[def:magma/power_set]{power set magma} \\( \\pow(\\mscrG) \\) of a group \\( \\mscrG \\) is a monoid, but it is not a group unless \\( \\mscrG \\) is trivial.\n\\end{example}\n\n\\begin{definition}\\label{def:group_cosets}\n Let \\( \\mscrH \\subseteq \\mscrG \\) be a subgroup of \\( \\mscrG \\) and let \\( x \\in \\mscrG \\). The sets\n \\begin{equation*}\n x \\mscrH \\coloneqq \\set{ xh \\colon h \\in \\mscrH }\n \\quad\\quad\n \\mscrH x \\coloneqq \\set{ hx \\colon h \\in \\mscrH }\n \\end{equation*}\n are called the left and right \\term{cosets} of \\( \\mscrH \\) with respect to \\( x \\). The name is justified by \\fullref{thm:group_coset_partition}.\n\n The \\hyperref[def:cardinal]{cardinality} of the set of all left cosets is called the \\term{index} of \\( \\mscrH \\) and is denoted by \\( [\\mscrG : \\mscrH] \\). By \\fullref{thm:lagranges_theorem_for_groups}, the index can analogously be defined as the cardinality of all right cosets.\n\\end{definition}\n\n\\begin{lemma}\\label{thm:group_coset_partition}\n The \\hyperref[def:group_cosets]{left cosets} of a subgroup of \\( \\mscrG \\) \\hyperref[def:set_partition]{partition} \\( \\mscrG \\). The same holds for right cosets.\n\\end{lemma}\n\\begin{proof}\n To each element \\( x \\in \\mscrG \\) there corresponds a coset \\( x \\in x\\mscrH \\) (since, as a subgroup, \\( \\mscrH \\) contains the identity).\n\n Two cosets \\( x\\mscrH \\) and \\( y\\mscrH \\) are either disjoint or equal. Indeed, if they are not disjoint, then there exists \\( z \\in x\\mscrH \\cap y\\mscrH \\) and thus \\( z = xa = yb \\) for some \\( a, b \\in \\mscrH \\). Thus,\n \\begin{equation*}\n x = x a a^{-1} = y b a^{-1}\n \\end{equation*}\n and since \\( b a^{-1} \\in \\mscrH \\), we have that \\( x \\in y\\mscrH \\). Furthermore, for any \\( c \\in \\mscrH \\), we have\n \\begin{equation*}\n xc = y(b a^{-1} c) \\in y\\mscrH,\n \\end{equation*}\n hence \\( x\\mscrH \\subseteq y\\mscrH \\). After obtaining the converse inclusion, we conclude \\( x\\mscrH = y\\mscrH \\).\n\\end{proof}\n\n\\begin{lemma}\\label{thm:group_coset_bijection}\n Any two left cosets in a group are \\hyperref[def:equinumerosity]{equinumerous}. The same holds for right cosets.\n\\end{lemma}\n\\begin{proof}\n Let \\( \\mscrH \\) be a subgroup of \\( \\mscrG \\) and let \\( x, y \\in \\mscrG \\). Then \\( z \\mapsto y x^{-1} z \\) sends \\( x\\mscrH \\) into \\( y\\mscrH \\). By \\fullref{thm:group_multiplication_is_bijection}, this function is a bijection.\n\\end{proof}\n\n\\begin{definition}\\label{def:normal_subgroup}\n Let \\( \\mscrN \\) be a subgroup of \\( \\mscrG \\). We say that \\( \\mscrN \\) is a normal subgroup if any of the following equivalent conditions hold:\n \\begin{thmenum}\n \\thmitem{def:normal_subgroup/direct} For every element \\( x \\) of \\( \\mscrG \\), we have the set equality\n \\begin{equation}\\label{eq:def:normal_subgroup/direct}\n x \\mscrN x^{-1} = \\mscrN.\n \\end{equation}\n\n \\thmitem{def:normal_subgroup/cosets} The partitions induced by the left and rights cosets of \\( \\mscrN \\) coincide.\n \\thmitem{def:normal_subgroup/kernel} \\( \\mscrN \\) is the \\hyperref[def:pointed_set_kernel]{kernel} of some group homomorphism.\n \\end{thmenum}\n\n In particular, kernels are always normal subgroups.\n\\end{definition}\n\\begin{proof}\n This is the group-theoretic analog to \\fullref{thm:equivalence_partition}.\n\n \\ImplicationSubProof{def:normal_subgroup/direct}{def:normal_subgroup/cosets} For any \\( x \\in \\mscrG \\)\n \\begin{equation*}\n \\mscrN x = (x \\mscrN x^{-1})x = x \\mscrN(x^{-1}x) = x \\mscrN,\n \\end{equation*}\n thus every left coset is a right coset and vice versa.\n\n \\ImplicationSubProof{def:normal_subgroup/cosets}{def:normal_subgroup/kernel} We can take the \\hyperref[def:quotient_group]{canonical projection} \\( \\pi(x) \\coloneqq x \\mscrN \\) as the homomorphism. The proof of correctness in \\fullref{def:quotient_group} only uses \\fullref{def:normal_subgroup/cosets} and therefore does not cause circular references.\n\n \\ImplicationSubProof{def:normal_subgroup/kernel}{def:normal_subgroup/direct} Let \\( \\varphi: \\mscrG \\to \\mscrH \\) be a group homomorphism and fix any \\( x \\in \\mscrG \\). Denote \\( \\mscrN \\coloneqq \\ker(f) \\). Then \\( x \\mscrN = \\mscrN x \\) since\n \\begin{equation*}\n \\varphi(x \\mscrN)\n =\n \\varphi(x) \\varphi(\\mscrN)\n =\n \\varphi(x) \\varphi(e_{\\mscrG})\n =\n \\varphi(x)\n =\n \\varphi(\\mscrN) \\varphi(x)\n =\n \\varphi(\\mscrN x)\n =\n e_{\\mscrH}.\n \\end{equation*}\n\n Thus,\n \\begin{equation*}\n \\varphi^{-1}(e_{\\mscrH}) = \\mscrN = xx^{-1}\\mscrN = x \\mscrN x^{-1}.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:quotient_group}\n Let \\( \\mscrG \\) be a group and \\( \\mscrN \\) be a normal subgroup of \\( \\mscrG \\). Define the \\term{quotient group}\n \\begin{equation*}\n \\mscrG / \\mscrN \\coloneqq \\{ x \\mscrN \\colon x \\in \\mscrG \\}\n \\end{equation*}\n with the group operation\n \\begin{equation*}\n x \\mscrN \\star y \\mscrN \\coloneqq xy \\mscrN.\n \\end{equation*}\n\n Define the canonical projection homomorphism\n \\begin{align*}\n &\\pi: \\mscrG \\to \\mscrG / \\mscrN \\\\\n &\\pi(x) \\coloneqq x \\mscrN.\n \\end{align*}\n\n The kernel of \\( \\pi \\) is precisely \\( \\mscrN \\).\n\\end{definition}\n\\begin{proof}\n This definition is used in the proof of equivalence in \\fullref{def:normal_subgroup}. This is why it is important to only use \\fullref{def:normal_subgroup/cosets} as the definition for a normal subgroup.\n\n We first check that the group operations is well-defined, that is, does not depend on the choice of coset representatives. Fix \\( x_1, x_2 \\in \\mscrG \\) and \\( y_1, y_2 \\in \\mscrG \\), so that\n \\begin{equation*}\n x_1 \\mscrN = y_1 \\mscrN\n \\end{equation*}\n and\n \\begin{equation*}\n x_2 \\mscrN = y_2 \\mscrN.\n \\end{equation*}\n\n Since the left and right cosets coincide, we have\n \\begin{equation*}\n x_1 x_2 \\mscrN = x_1 \\mscrN x_2 = x_1 \\mscrN y_2 = y_1 \\mscrN y_2 = y_1 y_2 \\mscrN.\n \\end{equation*}\n\n Thus, the operation is well-defined.\n\n It follows from the definition that the identity is \\( e \\mscrN = \\mscrN \\) and the inverse of \\( x \\mscrN \\) is \\( x^{-1} \\mscrN \\). Therefore, \\( \\mscrG / \\mscrN \\) is indeed a group. The fact that \\( \\pi \\) is a homomorphism is also part of the definition of \\( \\star \\).\n\n It remains to prove that \\( \\mscrN = \\ker(\\pi) \\). Obviously \\( \\pi(\\mscrN) = \\mscrN \\), so \\( \\mscrN \\subseteq \\ker(\\pi) \\). To see that the converse holds, assume that there exists \\( x \\in \\ker(\\pi) \\setminus \\mscrN \\), i.e. \\( \\pi(x) = x\\mscrN = \\mscrN \\), but \\( x \\not\\in N \\). Then there exists \\( y \\in N \\) such that \\( xy \\in N \\). The subgroup \\( \\mscrN \\) is closed under multiplication and inverses, hence \\( x = xyy^{-1} \\in N \\). This contradicts our assumption that \\( x \\not\\in N \\). Therefore, \\( \\mscrN = \\ker \\pi \\).\n\\end{proof}\n\n\\begin{theorem}[Lagrange's theorem for groups]\\label{thm:lagranges_theorem_for_groups}\n Let \\( \\mscrH \\) be a subgroup of \\( \\mscrG \\). We have the following equality\n \\begin{equation}\\label{eq:thm:lagranges_theorem_for_groups/index}\n \\card(\\mscrG) = \\card(\\mscrH) \\cdot [\\mscrG : \\mscrH].\n \\end{equation}\n\n If \\( \\mscrH \\) is a \\hyperref[def:normal_subgroup]{normal subgroup}, then \\( [\\mscrG : \\mscrH] = \\card(\\mscrG / \\mscrH) \\) and\n \\begin{equation}\\label{eq:thm:lagranges_theorem_for_groups/card}\n \\card(\\mscrG) = \\card(\\mscrH) \\cdot \\card(\\mscrG / \\mscrH).\n \\end{equation}\n\n This demonstrates that there exists a bijective function between the \\hyperref[def:group_direct_product]{direct product} \\( \\mscrH \\times \\mscrG / \\mscrH \\) and \\( \\mscrH \\), however this may not be a group homomorphism --- see \\fullref{ex:lagranges_theorem_for_groups/direct_product_zn}.\n\\end{theorem}\n\\begin{proof}\n Follows from \\fullref{thm:group_coset_partition} and \\fullref{thm:group_coset_bijection}\n\\end{proof}\n\n\\begin{theorem}[Homomorphism theorem for groups]\\label{thm:homomorphism_theorem_for_groups}\n For any \\hyperref[def:group/homomorphism]{group homomorphism} \\( \\varphi: \\mscrG \\to \\mscrH \\), we have the isomorphism\n \\begin{equation*}\n \\mscrG / \\ker \\varphi \\cong \\img \\varphi.\n \\end{equation*}\n\\end{theorem}\n\\begin{proof}\n Denote \\( \\mscrN \\coloneqq \\ker \\varphi \\). Define the function\n \\begin{align*}\n &\\psi: \\img \\varphi \\to \\pow(\\mscrG) \\\\\n &\\psi(y) \\coloneqq \\varphi^{-1}(y) \\mscrN.\n \\end{align*}\n\n We will show that \\( \\psi \\) is the desired isomorphism.\n\n Fix any \\( y \\in \\img \\varphi \\) and \\( x_1, x_2 \\in \\varphi^{-1}(y) \\). We will first show that \\( x_1 \\mscrN = x_2 \\mscrN \\). Note that\n \\begin{equation*}\n \\varphi(x_1^{-1} x_2)\n =\n \\varphi(x_1)^{-1} \\varphi(x_2)\n =\n \\varphi(x_2)^{-1} \\varphi(x_2)\n =\n e_{\\mscrH},\n \\end{equation*}\n therefore \\( x_1^{-1} x_2 \\in N \\). Thus,\n \\begin{equation*}\n x_2 \\mscrN = x_1 x_1^{-1} x_2 \\mscrN = x_1 \\cdot \\mscrN \\cdot \\mscrN = x_1 \\mscrN.\n \\end{equation*}\n\n Hence, \\( \\varphi^{-1}(y) \\mscrN \\) is a coset in \\( \\mscrG / \\mscrN \\) formed by any of the elements of \\( \\varphi^{-1}(y) \\).\n\n Furthermore, if \\( x_1 \\in x_2 \\mscrN \\), then there exists \\( n \\in N \\) such that\n \\begin{equation*}\n x_1 = x_2 n.\n \\end{equation*}\n\n But \\( \\mscrN \\) is closed under taking inverses, hence\n \\begin{equation*}\n x_2 = x_1 n^{-1} \\in x_1 \\mscrN,\n \\end{equation*}\n that is,\n \\begin{equation*}\n x_1 \\mscrN = x_2 \\mscrN.\n \\end{equation*}\n\n This shows that \\( \\psi \\) is injective. It is obviously surjective because if \\( x \\mscrN \\) is a coset, then \\( \\varphi(x) \\in \\img \\varphi \\). Therefore, \\( \\varphi \\) is bijective.\n\n It remains to show that \\( \\psi \\) is a homomorphism. Indeed, if \\( y_1, y_2 \\in \\img \\varphi \\) and\n \\begin{equation*}\n x_k \\in \\varphi^{-1}(y_k), k = 1, 2,\n \\end{equation*}\n we have\n \\begin{balign*}\n \\psi(y_1) \\psi(y_2)\n &=\n \\varphi^{-1}(y_1) \\mscrN \\varphi^{-1}(y_2) \\mscrN\n = \\\\ &=\n x_1 \\mscrN x_2 \\mscrN\n \\reloset {\\eqref{eq:def:normal_subgroup/direct}} = \\\\ &=\n (x_1 x_2) \\mscrN\n = \\\\ &=\n \\varphi^{-1}(y_1 y_2) \\mscrN\n = \\\\ &=\n \\psi(y_1 y_2).\n \\end{balign*}\n\\end{proof}\n", "meta": {"hexsha": "80f819d7a08247e125ea8601496020d2c6e79444", "size": 16918, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/groups.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/groups.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/groups.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.3893557423, "max_line_length": 538, "alphanum_fraction": 0.6623714387, "num_tokens": 6040, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6283873907814009}} {"text": "\\newpage\r\n\\section{Roots Package}\r\n\\begin{Introduction}{Roots Package}\r\n\\index{roots}\\index{polynomial}\r\nThe root finding package is designed so that it can \r\nbe used to find some or all of the roots of univariate\r\npolynomials with real or complex coefficients, to the accuracy\r\nspecified by the user. \r\n\r\nNot all operators of \\name{roots package} are described here. For using\r\nthe operators \r\n\r\n\\nameindex{isolater} (intervals isolating real roots)\r\n\r\n\\nameindex{rlrootno} (number of real roots in an interval)\r\n\r\n\\nameindex{rootsat-prec} (roots at system precision)\r\n\r\n\\nameindex{rootval} (result in equation form)\r\n\r\n\\nameindex{firstroot} (computing only one root)\r\n\r\n\\nameindex{getroot} (selecting roots from a collection)\r\n\r\nplease consult the full documentation of the package.\r\n\r\n\\end{Introduction}\r\n\r\n\\begin{Operator}{MKPOLY}\r\n\\index{polynomial}\\index{roots}\\index{interpolation}\r\nGiven a roots list as returned by \\nameref{roots},\r\nthe operator \\name{mkpoly} constructs a \r\npolynomial which has these numbers as roots.\r\n\\begin{Syntax}\r\n\\name{mkpoly} \\meta{rl}\r\n\\end{Syntax}\r\nwhere \\meta{rl} is a \\nameref{list} with equations, which\r\nall have the same \\nameref{kernel} on their left-hand sides\r\nand numbers as right-hand sides.\r\n\r\n\\begin{Examples}\r\nmkpoly{x=1,x=-2,x=i,x=-i};&\r\nx**4 + x**3 - x**2 + x - 2\\\\\r\n\\end{Examples}\r\n\r\nNote that this polynomial is unique only up to a numeric\r\nfactor.\r\n\\end{Operator}\r\n\r\n\r\n\\begin{Operator}{NEARESTROOT}\r\n\\index{roots}\\index{solve}\r\nThe operator \\name{nearestroot} finds one root of a polynomial\r\nwith an iteration using a given starting point.\r\n\r\n\\begin{Syntax}\r\n\\name{nearestroot}\\(\\meta{p}\\,\\meta{pt}\\)\r\n\\end{Syntax}\r\n\r\nwhere \\meta{p} is a univariate polynomial\r\nand \\meta{pt} is a number. \r\n\r\n\\begin{Examples}\r\nnearestroot(x^2+2,2);&\\{x=1.41421*i\\}\\\\\r\n\\end{Examples}\r\nThe minimal accuracy of the result values is controlled by \r\n\\nameref{rootacc}.\r\n\\end{Operator}\r\n\r\n\\begin{Operator}{REALROOTS}\r\n\\index{roots}\\index{solve}\r\nThe operator \\name{realroots} finds that real roots of a polynomial \r\nto an accuracy that is sufficient to separate them and which is\r\na minimum of 6 decimal places. \r\n\r\n\\begin{Syntax}\r\n\\name{realroots}\\(\\meta{p}\\) or \\\\\r\n\\name{realroots}\\(\\meta{p}\\,\\meta{from},\\meta{to}\\)\r\n\\end{Syntax}\r\n\r\nwhere \\meta{p} is a univariate polynomial. \r\nThe optional parameters \\meta{from} and \\meta{to} classify\r\nan interval: if given, exactly the real roots in this\r\ninterval will be returned. \\meta{from} and \\meta{to}\r\ncan also take the values \\name{infinity} or \\name{-infinity}.\r\nIf omitted all real roots will be returned.\r\nResult is a \\nameref{list}\r\nof equations which represent the roots of the polynomial at the\r\ngiven accuracy.\r\n\r\n\\begin{Examples}\r\nrealroots(x^5-2);&\\{x=1.1487\\}\\\\\r\nrealroots(x^3-104*x^2+403*x-300,2,infinity);&\\{x=3.0,x=100.0\\}\\\\\r\nrealroots(x^3-104*x^2+403*x-300,-infinity,2);&\\{x=1\\}\\\\\r\n\\end{Examples}\r\nThe minimal accuracy of the result values is controlled by \r\n\\nameref{rootacc}.\r\n\\end{Operator}\r\n\r\n\r\n\\begin{Operator}{ROOTACC}\r\n\\index{roots}\\index{accuracy}\r\nThe operator \\name{rootacc} allows you to set the accuracy\r\nup to which the roots package computes its results.\r\n\\begin{Syntax}\r\n\\name{rootacc}\\(\\meta{n}\\)\r\n\\end{Syntax}\r\nHere \\meta{n} is an integer value. The internal accuracy of\r\nthe \\name{roots} package is adjusted to a value of\r\n\\name{max(6,n)}. The default value is \\name{6}.\r\n\\end{Operator}\r\n\r\n\\begin{Operator}{ROOTS}\r\n\\index{roots}\\index{solve}\\index{polynomial}\r\nThe operator \\name{roots} \r\nis the main top level function of the roots package.\r\nIt will find all roots, real and complex, of the polynomial p\r\nto an accuracy that is sufficient to separate them and which is\r\na minimum of 6 decimal places. \r\n\r\n\\begin{Syntax}\r\n\\name{roots}\\(\\meta{p}\\)\r\n\\end{Syntax}\r\n\r\nwhere \\meta{p} is a univariate polynomial. Result is a \\nameref{list}\r\nof equations which represent the roots of the polynomial at the\r\ngiven accuracy. In addition, \\name{roots} stores\r\nseparate lists of real roots and complex roots in the global\r\nvariables \\nameref{rootsreal} and \\nameref{rootscomplex}. \r\n\r\n\\begin{Examples}\r\nroots(x^5-2);&\\begin{multilineoutput}{3cm}\r\n\\{x=-0.929316 + 0.675188*i,\r\n x=-0.929316 - 0.675188*i,\r\n x=0.354967 + 1.09248*i,\r\n x=0.354967 - 1.09248*i, \r\n x=1.1487\\}\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\nThe minimal accuracy of the result values is controlled by \r\n\\nameref{rootacc}.\r\n\\end{Operator}\r\n\r\n\\begin{Operator}{ROOT\\_VAL}\r\n\\index{roots}\\index{solve}\\index{polynomial}\r\nThe operator \\name{root\\_val} computes the roots of a \r\nunivariate polynomial at system precision\r\n(or greater if required for root separation) and presents \r\nits result as a list of numbers.\r\n\\begin{Syntax}\r\n\\name{roots}\\(\\meta{p}\\)\r\n\\end{Syntax}\r\n\r\nwhere \\meta{p} is a univariate polynomial.\r\n\r\n\\begin{Examples}\r\nroot_val(x^5-2);&\\begin{multilineoutput}{3cm}\r\n\\{-0.929316490603 + 0.6751879524*i,\r\n -0.929316490603 - 0.6751879524*i,\r\n 0.354967313105 + 1.09247705578*i,\r\n 0.354967313105 - 1.09247705578*i,\r\n 1.148698355\\}\r\n\\end{multilineoutput}\\\\\r\n\\end{Examples}\r\n\\end{Operator} \r\n\r\n\\begin{Variable}{ROOTSCOMPLEX}\r\n\\index{roots}\\index{complex}\r\nWhen the operator \\nameref{roots} is called the complex\r\nroots are collected in the global variable \\name{rootscomplex}\r\nas \\nameref{list}.\r\n\\end{Variable}\r\n\r\n\r\n\\begin{Variable}{ROOTSREAL}\r\n\\index{roots}\\index{complex}\r\nWhen the operator \\nameref{roots} is called the real\r\nroots are collected in the global variable \\name{rootreal}\r\nas \\nameref{list}.\r\n\\end{Variable}\r\n", "meta": {"hexsha": "fedddc594ab46ffa1e44fc71f9259ef5e038d32b", "size": 5594, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "pk-roots.tex", "max_stars_repo_name": "nilqed/REDHELP", "max_stars_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "pk-roots.tex", "max_issues_repo_name": "nilqed/REDHELP", "max_issues_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "pk-roots.tex", "max_forks_repo_name": "nilqed/REDHELP", "max_forks_repo_head_hexsha": "0c98a32bf21fa060ccd67ce82f638d6a1bc47a52", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.9060773481, "max_line_length": 72, "alphanum_fraction": 0.7121916339, "num_tokens": 1643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.7799928900257127, "lm_q1q2_score": 0.6283873866612391}} {"text": "\\documentclass[../../main.tex]{subfiles}\n\\begin{document}\n\\subsubsection{Remove}\nThe inverse of the $append$ Operations.\n\\begin{zed}\n remove(coll, idx) = \\inv append(coll, idx)\n\\end{zed}\nThe operation $remove$ will return a Collection minus the Value removed from the specified Numeric Index\n\\begin{schema}{Remove[Collection, \\nat]}\n coll?, coll! : Collection \\\\\n idx? : \\nat \\\\\n remove~\\_ : Collection \\cross \\nat \\surj Collection\n \\where\n \\#~ idx? = 1 \\\\\n coll! = remove(coll?, idx?) @ \\\\\n \\t1 let ~ ~ coll' == front(\\{~ i : \\nat ~| ~ i \\in 0~..~idx?\\} \\extract coll?) \\\\\n \\t1 \\ \\ \\ ~~~ coll'' == tail(\\{~ j : \\nat ~| ~ j \\in idx?~..~\\#~coll?\\} \\extract coll?) \\\\\n \\t1 = coll' \\cat coll'' \\implies \\\\\n \\t2 (coll?_{idx?} \\not \\in coll') ~\\land \\\\\n \\t2 (coll?_{idx?} \\not \\in coll'') ~\\land \\\\\n \\t2 (~\\#~ coll! = \\#~ coll? - 1)\n\\end{schema}\nsuch that\n\\begin{argue}\n X = \\langle x_{0}, x_{1}, x_{2} \\rangle \\\\\n \\t1 x_{0} = 0 \\\\\n \\t1 x_{1} = foo \\\\\n \\t1 x_{2} = baz \\\\\n remove(X, 0) = \\langle foo, baz \\rangle & 0 was removed from $X$\\\\\n remove(X, 1) = \\langle 0, baz \\rangle & foo was removed from $X$ \\\\\n remove(X, 2) = \\langle 0, foo \\rangle & baz was removed from $X$ \\\\\n remove(X, 3) = \\langle 0, foo, baz \\rangle = X & nothing at 3, X unaltered\n\\end{argue}\n\\end{document}\n", "meta": {"hexsha": "6428aaf169f27feb27cd71a30aafa21f949e7fa6", "size": 1297, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/operations/collections/remove.tex", "max_stars_repo_name": "yetanalytics/dave", "max_stars_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2018-08-17T00:38:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T02:32:37.000Z", "max_issues_repo_path": "docs/operations/collections/remove.tex", "max_issues_repo_name": "adlnet/dave", "max_issues_repo_head_hexsha": "9339713fac747118e462e4fc7e1ecd54e5d916e3", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 95, "max_issues_repo_issues_event_min_datetime": "2018-08-31T18:57:51.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-15T16:40:01.000Z", "max_forks_repo_path": "docs/operations/collections/remove.tex", "max_forks_repo_name": "yetanalytics/dave", "max_forks_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2019-09-28T06:48:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-31T16:20:47.000Z", "avg_line_length": 37.0571428571, "max_line_length": 104, "alphanum_fraction": 0.5797995374, "num_tokens": 484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8902942319436397, "lm_q2_score": 0.7057850216484838, "lm_q1q2_score": 0.628356333765862}} {"text": "\\documentclass{article}\n\\usepackage{mathrsfs,amsmath}\n\\usepackage{xcolor}\n\\usepackage{titlesec}\n\\usepackage{listings}\n\\usepackage{syntax}\n\\usepackage{pythonhighlighting}\n\\usepackage{fancyvrb}\n\n\\usepackage[margin=1.4in]{geometry}\n\n\\title{Homework \\#1 | Fall 2021} \n\\author{Jared Dyreson\\\\ \n California State University, Fullerton}\n\n\\DeclareRobustCommand{\\bowtie}{%\n \\mathrel\\triangleright\\joinrel\\mathrel\\triangleleft}\n\n\n\\usepackage [english]{babel}\n\\usepackage [autostyle, english = american]{csquotes}\n\\MakeOuterQuote{\"}\n\n\\titlespacing*{\\section}\n{0pt}{5.5ex plus 1ex minus .2ex}{4.3ex plus .2ex}\n\\titlespacing*{\\subsection}\n{0pt}{5.5ex plus 1ex minus .2ex}{4.3ex plus .2ex}\n\n\\usepackage{hyperref}\n\\hypersetup{\n colorlinks,\n citecolor=black,\n filecolor=black,\n linkcolor=black,\n urlcolor=black\n}\n\n\\begin{document}\n\n\\maketitle\n\\tableofcontents\n\n\\newpage\n\n\n\\section{Questions}\n\n\\begin{enumerate}\n\n\\item Order the following functions by asymptotic growth rate: \\\\\n\n$$4n, 2^{\\log(n)}, 4n \\times \\log(n) + 2n, 2^{10}, 3n + 100\\times \\log(n) \\times, 2^{n}, n^{2} + 10n, n^{3}, n \\times \\log(n)$$ \\\\\nState the asymptotic growth rate for each of the functions in terms of Big-Oh and explicitly order them from least to greatest.\n\n\n\\begin{itemize}\n\\item $2 ^ {10} \\implies O(1)$. Running time is independent of the input size.\n\\item $2^{\\log(n)} \\implies O(n)$ ? I would assume the log and 2 would cancel each other out in an event like this: $2^{\\log_2(n)}$. Unsure about this one.\n\\item $3n + 100 \\times log(n) \\implies O(n)$\n\\item $4n \\implies O(n)$\n\\item $n \\times \\log(n) \\implies O(n \\times \\log(n))$\n\\item $4n \\times \\log(n) + 2n \\implies O(n \\times \\log(n))$\n\\item $n^{2} + 10n \\implies O(n^{2})$\n\\item $n^{3} \\implies O(n^{3})$\n\\item $2^{n} \\implies O(2^{n}) \\rightarrow$ exponential.\n\\end{itemize}\n\n\\item Write a problem definition and pseudocode for each of the following problems.\n\\begin{itemize}\n\\item Computing a square root\n\\begin{table}[!htbp]\n\\centering\n\\begin{tabular}{|l|}\n\\hline\n\\textit{computing a square root} \\\\ \\hline\n\\textbf{input}: an real integer that is greater than or equal to 0 \\\\ \\hline\n\\textbf{output}: a floating point integer that is the resultant of a square root operation \\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\begin{python}\n# Compute square root\ndef square_root_function(value):\n # Driver code should recognize that -1 denotes failure\n if value is less than 0, then return -1\n else return value ** 0.5 # value raised to the 1/2 power\n\\end{python}\n\n\\item Determining whether an integer is even or odd\n\n\\begin{table}[!htbp]\n\\centering\n\\begin{tabular}{|l|}\n\\hline\n\\textit{determine whether integer is even or odd} \\\\ \\hline\n\\textbf{input}: a real integer that is greater than or equal to 0 \\\\ \\hline\n\\textbf{output}: a boolean value denoting status (true for even, false for odd) \\\\ \\hline\n\\end{tabular}\n\\end{table}\n\n\\begin{python}\n# Check if integer is even or odd\ndef even_or_odd(value):\n # even values should not have a remainder, thus equal to 0\n return (value % 2 == 0)\n\n\\end{python}\n\n\\newpage\n\n\\item Determining whether every element in a sequence is identical\n\n\\begin{table}[!htbp]\n\\centering\n\\begin{tabular}{|l|}\n\\hline\n\\textit{determine whether all sequence elements are identical} \\\\ \\hline\n\\textbf{input}: a non-empty sequence that contains elements that are comparable with one another \\\\ \\hline\n\\textbf{output}: a boolean value denoting status of homogeneity \\\\ \\hline\n\\end{tabular}\n\n\n\\end{table}\n\n\\begin{python}\n# Check if sequence has all the same elements\ndef contains_same_elements(container: list[int]):\n if container is empty:\n return None\n top_most = container[0]\n x = 0\n for element in container:\n if(container[x] != top_most)\n return False\n top_most = container[x]\n x = x + 1\n return True\n\\end{python}\n\n\n\\end{itemize}\n\n\\item Consider the following algorithm:\n\\begin{Verbatim}\nAlgorithm Calc(a, n):\n Input: two integers, a and n\n Output: ?\n k <- 0\n b <- 1\n while k < n do\n k <- k + 1\n b <- b * a\n return b\n\\end{Verbatim}\n\n\\begin{python}\ndef calc(a: int, n: int) -> int:\n k, b = 0, 1\n while(k < n):\n k += 1\n b *= a\n # two atomics, this is what a for loop is when broken down into smaller components\n return b\n\\end{python}\n\n\\begin{itemize}\n\\item What does the algorithm calculate?: The value of $b$ continuously multiplied by a factor of $a$, $n$ times. $k$ is simply an index counter, and this entire expression can be written as a for loop:\n\\begin{python}\ndef calc(a: int, n: int) -> int:\n b = 1 # 1\n for _ in range(0, n): # one atomic\n b *= a\n # two atomics\n return b\n\\end{python}\n\\item Analyze its worst-case running time and express it using Big-Oh notation: Well we can tell that this function will at max run $n$ times. Coutning each instruction in this algorithm will allow us to determine that $T(n) = 3n + 3$. We can assume this is a linear operation, as seen in the for/while loop condition. Thus classifying it as $O(n)$.\n\\end{itemize}\n\n\\end{enumerate}\n\n\n\n\\end{document}\n\n", "meta": {"hexsha": "16d67d6fc02050af9ae7536ef34bf9e2dedd1ee7", "size": 5107, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "1/Homework_1.tex", "max_stars_repo_name": "JaredsAlgorithms/Assignments", "max_stars_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "1/Homework_1.tex", "max_issues_repo_name": "JaredsAlgorithms/Assignments", "max_issues_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "1/Homework_1.tex", "max_forks_repo_name": "JaredsAlgorithms/Assignments", "max_forks_repo_head_hexsha": "1efa28b481033384ff566cdda0e93ee269230b26", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.6054054054, "max_line_length": 349, "alphanum_fraction": 0.6880751909, "num_tokens": 1587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195152660687, "lm_q2_score": 0.855851148805615, "lm_q1q2_score": 0.6282970305010861}} {"text": "\\section{Unknown Values}\n\\label{unknowns}\n\n\\begin{quote}\nWhen you come to a fork in the road, take it \\footnote{attributed to Yogi Berra}\n\\end{quote}\nAssume that we have \\(X\\) instances of which the fraction which have class\n``Heads'' is \\(p_H\\) and Tails is \\(p_T\\).\nAssume that we are considering splitting using\na feature \\(f\\) such that \\(X' \\subseteq X\\) of the instances have unknown\nvalues for this feature. \nthe form \\(x \\leq y\\). \n\nIn terms of the training process, unknown values affect the process \nas follows. With each instance \\(x_i \\in X\\) is associated a weight \\(w_i\\),\ninitialized to 1. Let \\(X'\\) be the instances where the value of the splitting feature is unknown.\n\n\\be\n\\item Benefit benefit calculation of Section~\\ref{XXX}. \nFor all \\(x_i \\in X'\\), we treat it as if it were two instances, one with a class of Heads and a weight of \\(w_i \\times p_H\\) and the other with a class of Tails and a weight of \\(w_i \\times p_T\\)\n\\item Partitioning the subsets of the data passed to the left child and right child. All elements of \\(X'\\) are passed to the left child with a weight of \\(w_i \\times p_H\\) and to the right child with a weight of \\(w_i \\times p_T\\). See Figure~\\ref{fig_unknown_values}.\n\\ee\n\nIn terms of the testing process, the change in evaluation is as follows.\nHitherto, a test instance given to a decision tree ends up at precisely one leaf\nnode. The proportion of heads and tails at that leaf is used to make a\nprediction, the simplest being that \\(p_H = \\frac{n_H}{n_H+n_T}\\). Now, if\ndecision requires knowing the value of a feature that is unknown, we send the\ninstance down to both left and right childs and combine the resulting decision,\nweighted by the proportions of heads and tails at that node. In\nFigure~\\ref{fig_unknown_accuracy}, we show how the accuracy of classification\nfalls off as the proportion of unknown values in the data set increases. We\nchange the proportion of unknown values by randomly selecting instances and\nrandomly selecting features to artificially turn to null values. The results are\nreported for the data sets of Table~\\ref{tbl_data_sets}.\n\n\n\\begin{figure}\n\\centering\n\\fbox{\n\\begin{minipage}{35cm}\nTODO \n\\end{minipage}\n}\n\\label{fig_unknown_values}\n\\caption{Change in weight of datum when split involves unknown values}\n\\end{figure}\n\n\n\n", "meta": {"hexsha": "154babff46bfb4a668427cb2908f41e3ab640e29", "size": 2308, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ML/DT/doc/unknown.tex", "max_stars_repo_name": "subramon/qlu", "max_stars_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ML/DT/doc/unknown.tex", "max_issues_repo_name": "subramon/qlu", "max_issues_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2020-07-29T16:48:25.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-26T23:47:22.000Z", "max_forks_repo_path": "ML/DT/doc/unknown.tex", "max_forks_repo_name": "subramon/qlu", "max_forks_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2015-05-14T22:34:13.000Z", "max_forks_repo_forks_event_max_datetime": "2015-05-14T22:34:13.000Z", "avg_line_length": 45.2549019608, "max_line_length": 271, "alphanum_fraction": 0.756932409, "num_tokens": 593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511322604133, "lm_q2_score": 0.7341195210831258, "lm_q1q2_score": 0.6282970233334656}} {"text": "\\documentclass[]{article}\n\\usepackage{caption,subcaption,graphicx,float,url,amsmath,amssymb,tocloft,wasysym,amsthm,thmtools,textcomp,listings,amsfonts,cancel}\n\\usepackage[hidelinks]{hyperref}\n\\usepackage[toc,acronym,nonumberlist]{glossaries}\n\\usepackage[]{algorithm2e}\n\\setacronymstyle{long-short}\n\\usepackage{glossaries-extra}\n\\graphicspath{{figs/}} \n\\setlength{\\cftsubsecindent}{0em}\n\\setlength{\\cftsecnumwidth}{3em}\n\\setlength{\\cftsubsecnumwidth}{3em}\n\\newcommand\\numberthis{\\addtocounter{equation}{1}\\tag{\\theequation}}\n\\newtheorem{thm}{Theorem}\n\\newtheorem{cor}[thm]{Corollary}\n\\setcounter{tocdepth}{1}\n\\usepackage[toc,page]{appendix}\n%opening\n\\title{\n\tComputation in Complex Systems\\\\\n\tWeek 2\\\\\n\t Algorithms \\& Landscapes\n}\n\n%\\makeglossaries\n\n\n\\begin{document}\n\n\\maketitle\n\n\\tableofcontents\n%\\listofalgorithms\n\\section{Maximum independent set}\n\n\\subsection{What is the maximum value of the independent set?}\n\n\\begin{figure}[H]\n\t\\begin{center}\n\t\t\\caption{What is the maximum value of the independent set?}\n\t\t\\includegraphics[width=0.6\\textwidth]{mwisQ}\n\t\\end{center}\n\\end{figure}\n\nObservations:\n\\begin{enumerate}\n\t\\item A tree with $N$ nodes has $2^N-1$ non empty subsets, so a brute-force search is exponential. Let's not go there!\n\t\\item Either the maximal independent set includes the root of the tree, or it doesn't.\n\t\\item If the the maximal independent set doesn't include the root of the tree, each subtree must be maximal.\n\t\\item If the maximal independent set includes the root of the tree, each subtree must have a set that excludes the node.\n\\end{enumerate}\n\nSo we can solve this problem as follows:\n\\begin{enumerate}\n\t\\item Either the root node, $3$ is in $M$ or it isn't.\n\t\\item If $3$ is in the tree, we have to exclude {4,1,5}, which gives a score of $3+(1+2)+2+(1+1)=9$\n\t\\item If $3$ isn't in the tree, the maximum is the sum of the maxima for the 4 subtrees:\n\t\\begin{enumerate}\n\t\t\\item (4,(1,2)): since $1+2<4$, the maximum value is 4.\n\t\t\\item (1,(2)): since $1<2$, the maximum value is 2. \n\t\t\\item ((5),(1,1)): since $1+1<5$, the maximum value is 5.\n\t\\end{enumerate}\n\t\\item Putting this together, if $3$ isn't in the tree, the maximum is $4+2+5=10$\n\\end{enumerate}\n\n\\subsection{Algorithm for maximum independent set}\n\nAlgorithm \\ref{alg:find:maximum:independent:set} is a description is pseudo-code, and Appendix \\ref{sect:ref:implementation} provides a reference implementation in Python.\n\n\\begin{algorithm}[H]\n\t\\KwData{A binary tree $T$ with a value $V(n)$ assigned to each node $n$}\n\t\\KwResult{A set $S\\subseteq N(T) | I(S) \\land \\sum_{s\\in S} V(s) \\ge \\sum_{s\\in S^\\prime} V(s) \\forall S^\\prime \\subset N(T) \\text{ with } I(S^\\prime)$ }\n\tinitialization\\;\n\tStarting at the root of the tree, traverse it top down to build two data structures: a stack \\emph{unprocessed}, which has the \\emph{root} at the bottom, and the $leaves$ at the top; and a dictionary \\emph{score}, which is initially empty, but will map each node to the value of the maximum independent set starting at that node.\\;\n\t\\While {$length(unprocessed)>0$}{\n\t\t$node \\leftarrow pop(unprocessed)$\\;\n\t\t\\eIf {leaf(node)}{\n\t\t\t$score(node)\\leftarrow value(node)$ \\;}{\n\t\t\t$val_{out} \\leftarrow \\sum_{\\forall child(node)} score(child)$\\;\n\t\t\t$val_{in} \\leftarrow value(node) + \\sum_{\\forall grandchild(node)} value(grandchild)$\\;\n\t\t\t$score(node)\\leftarrow max(val_{out},val_{in})$\\;\n\t\t}\n\t}\n\t\\caption{Find the maximum independent set from a tree}\\label{alg:find:maximum:independent:set}\n\\end{algorithm}\n\n\\section{Reductions \\& Translations}\n\nTable \\ref{table:transform} illustrates the transformation of ASTRO to START. It exploits the common subsequence S-T-R to produce a minimum set of transformations.\n\\begin{table}[H]\n\t\\begin{center}\n\t\t\\caption{Transforming ASTRO to START}\\label{table:transform}\n\t\t\\begin{tabular}{|c|c|c|c|c|c|c|l|l|} \\hline\n\t\t\t\\textbf{\\cancel{A}}&S&T&&R&O&&delete&$\\rightarrow$\\\\\\hline\n\t\t\t&\\textbf{S}&T&&R&O&&match&$\\searrow$\\\\\\hline\n\t\t\t&S&\\textbf{T}&&R&O&&match&$\\searrow$\\\\\\hline\n\t\t\t&S&T&\\textbf{A}&R&O&&insert&$\\downarrow$\\\\\\hline\n\t\t\t&S&T&A&\\textbf{R}&O&&match&$\\searrow$\\\\\\hline\n\t\t\t&S&T&A&R&\\textbf{\\cancel{O}}&&delete&$\\rightarrow$\\\\\\hline\n\t\t\t&S&T&A&R&&\\textbf{T}&insert&$\\downarrow$\\\\\\hline\n\t\t\\end{tabular}\n\t\\end{center}\n\\end{table}\n\n\\subsection{Assign Weights}\n\nFigure \\ref{fig:Shortest:path} shows the path that gives rise to the transformations in Table \\ref{table:transform}. In order to make this the shortest path we need the following rules:\n\\begin{itemize}\n\t\\item If two symbols are equal, accept them and mover diagonally;\n\t\\item If two symbols don't match, insert or delete; disfavour mismatches (point mutations). \n\\end{itemize}\n\nThe simplest way to achieve this is with the weights shown in Figure \\ref{fig:weights}.\n\\begin{figure}[H]\n\t\\begin{center}\n\t\t\\caption{Weights}\\label{fig:weights}\n\t\t\\includegraphics[width=\\textwidth]{q2-weights}\n\t\\end{center}\n\\end{figure}\n\n\\subsection{Shortest path}\n\nCopying the weights from Figure \\ref{fig:weights} we get Figure \\ref{fig:Shortest:path}. The shorted path has length $1+0+0+1+0+1+1=4$. \n\\begin{figure}[H]\n\t\\begin{center}\n\t\t\\caption{Shortest path}\\label{fig:Shortest:path}\n\t\t\\includegraphics[width=0.6\\textwidth]{q2-path}\n\t\\end{center}\n\\end{figure}\n\n% glossary\n%\\printglossaries\n\n% bibliography go here\n\n%\\bibliographystyle{unsrt}\n%\\addcontentsline{toc}{section}{Bibliography}\n%\\bibliography{origins,wikipedia}\n\n\\begin{appendices}\n\t\\section{Reference Implementation for Algorithm \\ref{alg:find:maximum:independent:set}}\\label{sect:ref:implementation}\n\tOne way to implement Algorithm \\ref{alg:find:maximum:independent:set} is with the following code.\n\t\\lstinputlisting[language=Python]{mis.py}\n\\end{appendices}\n\n\n\\end{document}\n", "meta": {"hexsha": "bee5ca810a2303c73460ed9330ec0ed4b533467b", "size": 5697, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "computations/exam2.tex", "max_stars_repo_name": "weka511/fractals", "max_stars_repo_head_hexsha": "fa4e39677ea3ed7713e40a55b9453b2826f11a6c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-07-22T01:39:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-28T04:36:22.000Z", "max_issues_repo_path": "computations/exam2.tex", "max_issues_repo_name": "weka511/fractals", "max_issues_repo_head_hexsha": "fa4e39677ea3ed7713e40a55b9453b2826f11a6c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2019-06-20T03:20:12.000Z", "max_issues_repo_issues_event_max_datetime": "2020-07-20T03:07:55.000Z", "max_forks_repo_path": "computations/exam2.tex", "max_forks_repo_name": "weka511/complexity", "max_forks_repo_head_hexsha": "435ffab978e4499aea7c2c83788533867cc9b062", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.7551020408, "max_line_length": 332, "alphanum_fraction": 0.7312620678, "num_tokens": 1815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.752012562644147, "lm_q2_score": 0.8354835350552604, "lm_q1q2_score": 0.6282941142438974}} {"text": "\\providecommand{\\main}{../..}\n\\documentclass[\\main/thesis.tex]{subfiles}\n\\begin{document}\n\n\\section{Systems with an Upper Bound}\\label{bounded}\n\nWe say that a system is \\textit{bounded} if there \\textit{exists} a maximum in\nsuch system. In Agda, an existential proposition like this is expressed with\na \\textit{dependent sum type}, which is essentially a \\textit{pair} where the\ntype of the second term is dependent on the first.\n\n\\begin{lstlisting}\ndata Σ (A : Set) (B : A → Set) : Set where\n _,_ : (x : A) → B x → Σ A B\n\nproj₁ : ∀ {A B} → Σ A B → A\nproj₂ : ∀ {A B} → (pair : Σ A B) → B (proj₁ pair)\n\\end{lstlisting}\n\nFor example, to prove that there exists a right identity for addition on natural numbers;\nwe place the ``evidence'' on the left and its justification on the right.\n\n\\begin{lstlisting}\nprop : ∀ n → Σ ℕ (λ e → n + e ≡ n)\nprop n = 0 , +-right-identity n\n\\end{lstlisting}\n\nWith a little syntax sugar, we can rewrite the proposition as\n{\\lstinline|Σ[ x ∈ A ] B|} instead of {\\lstinline|Σ A (λ x → B)|}.\n\n\\begin{lstlisting}\nBounded : ∀ b d o → Set\nBounded b d o = Σ[ xs ∈ Numeral b d o ] Maximum xs\n\\end{lstlisting}\n\nThe proposition above can be read as follows:\n\\textit{There exists a numeral {\\lstinline|xs|} of type\n{\\lstinline|Numeral b d o|} such that {\\lstinline|Maximum xs|} holds.}\nTo prove that a system is bounded, we have to place the numeral that we consider\nto be a maximum on the left of the pair, and a proof to justify it on the right.\n\n\\subsection{Properties of each Category}\n\n\\paragraph{NullBase}\n\nWe have proven that any numeral with the greatest digit as its LSD is a maximum.\n\n\\begin{lstlisting}\nBounded-NullBase : ∀ d o → Bounded 0 (suc d) o\nBounded-NullBase d o =\n (greatest-digit d ∙) ,\n (Maximum-NullBase-Greatest\n (greatest-digit d ∙)\n (greatest-digit-is-the-Greatest d))\n\\end{lstlisting}\n\n\\paragraph{NoDigits}\n\nSince {\\lstinline|¬ (Bounded b 0 o)|} reduces to {\\lstinline|Bounded b 0 o → ⊥|},\nwe are given an addition argument, which is a proof claiming that the system is\nbounded.\nPattern match on this argument yields a pair with a numeral of\n{\\lstinline|Numeral b 0 o|} on the left.\nAt this point, we do not actually care whether the numeral is a maximum\nbecause it should not have existed in the first place.\nHand the numeral to {\\lstinline|NoDigits-explode|} and we are done.\n\n\\begin{lstlisting}\nBounded-NoDigits : ∀ b o → ¬ (Bounded b 0 o)\nBounded-NoDigits b o (xs , claim) = NoDigits-explode xs\n\\end{lstlisting}\n\n\\paragraph{AllZeros}\n\nSimilar to that of {\\lstinline|Bounded-NullBase|}:\n\n\\begin{lstlisting}\nBounded-AllZeros : ∀ b → Bounded (suc b) 1 0\nBounded-AllZeros b = (z ∙) , Maximum-AllZeros (z ∙)\n\\end{lstlisting}\n\n\\paragraph{Proper}\n\nThis proposition is proven by contradicting the fact that\na proper numeral system has no maximum.\n\n\\begin{lstlisting}\nBounded-Proper : ∀ b d o → (proper : 2 ≤ suc (d + o))\n → ¬ (Bounded (suc b) (suc d) o)\nBounded-Proper b d o proper (xs , claim) =\n contradiction claim (Maximum-Proper xs proper)\n\\end{lstlisting}\n\n\\subsection{The Decidable Predicate}\n\nWe can determine whether a system is bounded by delegating the job to the\nhelper functions we have defined above.\n\n\\begin{lstlisting}\nBounded? : ∀ b d o → Dec (Bounded b d o)\nBounded? b d o with numView b d o\nBounded? _ _ _ | NullBase d o\n = yes (Bounded-NullBase d o)\nBounded? _ _ _ | NoDigits b o\n = no (Bounded-NoDigits b o)\nBounded? _ _ _ | AllZeros b\n = yes (Bounded-AllZeros b)\nBounded? _ _ _ | Proper b d o proper\n = no (Bounded-Proper b d o proper)\n\\end{lstlisting}\n\n\\paragraph{Summary}\n\n\\begin{center}\n \\begin{adjustbox}{max width=\\textwidth}\n \\begin{tabular}{ | l | c | c | c | c | }\n \\textbf{Properties} & \\textbf{NullBase} & \\textbf{NoDigits} & \\textbf{AllZeros} & \\textbf{Proper} \\\\\n \\hline\n has an maximum & yes & no & yes & no \\\\\n has an upper bound & yes & no & yes & no \\\\\n \\end{tabular}\n \\end{adjustbox}\n\\end{center}\n\n\\end{document}\n", "meta": {"hexsha": "d1a0c5f1efbad004d7f92adfcf7a65f223ed088a", "size": 3961, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/tex/constructions/bounded.tex", "max_stars_repo_name": "banacorn/numeral", "max_stars_repo_head_hexsha": "aae093cc9bf21f11064e7f7b12049448cd6449f1", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-04-23T15:58:28.000Z", "max_stars_repo_stars_event_max_datetime": "2015-04-23T15:58:28.000Z", "max_issues_repo_path": "Thesis/tex/constructions/bounded.tex", "max_issues_repo_name": "banacorn/numeral", "max_issues_repo_head_hexsha": "aae093cc9bf21f11064e7f7b12049448cd6449f1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/tex/constructions/bounded.tex", "max_forks_repo_name": "banacorn/numeral", "max_forks_repo_head_hexsha": "aae093cc9bf21f11064e7f7b12049448cd6449f1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2015-05-30T05:50:50.000Z", "max_forks_repo_forks_event_max_datetime": "2015-05-30T05:50:50.000Z", "avg_line_length": 31.4365079365, "max_line_length": 104, "alphanum_fraction": 0.6935117395, "num_tokens": 1237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.766293653760418, "lm_q1q2_score": 0.6282790600710013}} {"text": "\n% This LaTeX was auto-generated from MATLAB code.\n% To make changes, update the MATLAB code and republish this document.\n\n\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{color}\n\n\\sloppy\n\\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n \\input{preamble} \\begin{document} \\maketitle \\tableofcontents \n\n \n \n{}\n\n\n \n\n\n\\section{Problem 20.2-1}\n\n\\begin{par}\n\nSolute A is diffusing at unsteady state into a semi-infinite medium of\npure B and undergoes a first-order reaction with B. Solute A is\ndilute. Calculate the concentration $c_A$ at points z = 0, 4, and\n$10~\\textup{mm}$ from the surface for $t=1\\times 10^5~\\textup{s}$.\nPhysical property data are $D_{AB}=1\\times\n10^{-9}~\\textup{m}^2/\\textup{s}$, $k'=1\\times 10^{-4}~\\textup{s}^{-1}$,\n$c_{A0} = 1.0~\\textup{kg mol/m}^3$. Also calculate the\n$\\textup{kg mol absorbed/m}^2$.\nThe concentration of A at point z is given by \\vref{conc_empirical_1}\n(corrected version),\n\\begin{equation} \\label{conc_empirical_1}\n\\begin{split}\n\\frac{C_A}{C_{A0}}=&\\frac{1}{2}\\exp\\left(-z\\sqrt{k'/D_{AB}}\\right)\n * \\textup{erfc}\\left(\\frac{z}{2\\sqrt{tD_{AB}}}-\\sqrt{k't}\\right) \\\\\n &+ \\frac{1}{2}\\exp\\left(z\\sqrt{k'/D_{AB}}\\right)\n * \\textup{erfc}\\left(\\frac{z}{2\\sqrt{tD_{AB}}}+\\sqrt{k't}\\right)\n\\end{split}\n\\end{equation}\n\\Vref{conc_empirical_1} was incorrect in the textbook, and has\nbeen modified according to the instructor's email.\n\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nu = symunit;\nt = 1e5 * u.s;\nD_AB = 1e-9 * u.m^2 / u.s;\nk_prime = 1e-4 * u.s^-1;\nC_A0 = 1.0 * (u.kg * u.mol) / (u.m^3);\nC_A = @(z) C_A0 * ...\n (0.5*exp(-z*sqrt(k_prime/D_AB))*erfc((z/(2*sqrt(t*D_AB))) - sqrt(k_prime*t)) + ...\n 0.5*(exp(z*sqrt(k_prime/D_AB)))*erfc((z/(2*sqrt(t*D_AB))) + sqrt(k_prime*t)));\ndisp(unitString(C_A(0), 'C_A (0 mm)'))\ndisp(unitString(C_A(0.004 * u.m), 'C_A (4 mm)'))\ndisp(unitString(C_A(0.010 * u.m), 'C_A (10 mm)'))\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}C_A (0 mm): 1 (kg*mol)/m^3\nC_A (4 mm): 0.28226 (kg*mol)/m^3\nC_A (10 mm): 0.042328 (kg*mol)/m^3\n\\end{verbatim} \\color{black}\n \\begin{par}\n\nThe amount of A absorbed per square meter is given by\n\\vref{conc_empirical_2} (corrected version),\n\\begin{equation} \\label{conc_empirical_2}\nQ=C_{A0}\\sqrt{D_{AB}/k'}\\left[\\left(k't+\\frac{1}{2}\\right)\\textup{erf}\\sqrt{k't}+\\sqrt{k't/\\pi} e^{-k't}\\right]\n\\end{equation}\n\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nQ = C_A0*sqrt(D_AB/k_prime)*((k_prime*t+0.5)*erf(sqrt(k_prime*t) + ...\n sqrt((k_prime*t)/pi)*exp(-k_prime*t)));\ndisp(unitString(Q))\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}Q: 0.033204 (kg*mol)/m^2\n\\end{verbatim} \\color{black}\n \n\n\\section{Problem 21.1-2}\n\n\\begin{par}\nProve or show the following relationships, starting with the flux equations:\n\\end{par} \\vspace{1em}\n\\begin{par}\n\n\\subsection{Part a}\nConvert $k_c'$ to $k_y$ and $k_G$.\n\n\\end{par} \\vspace{1em}\n\\begin{par}\nThe flux equations involving these coefficients are, $$ N_A = k_c(C_{A1}-C_{A2}) = k_G(p_{A1}-p_{A2}) = k_y(y_{A1}-y_{A2}) $$ the conversion between $k_c'$ and $k_c$ is given in the flux equation for A diffusing through stagnant, non-diffusing B,\n\\end{par} \\vspace{1em}\n\\begin{par}\n\n\\[N_A = \\frac{k_c'}{x_{BM}}\\left(C_{A1}-C_{A2}\\right) =\nk_c\\left(C_{A1}-C_{A2}\\right)\\]\n\\[ k_c = \\frac{k_c'}{x_{BM}} = \\frac{k_c' P}{p_{BM}} \\]\n\\[ k_y = \\frac{k_c' P}{p_{BM}}\\cdot\\frac{C_{A1}-C_{A2}}{y_{A1}-y_{A2}} = \\frac{k_c' P}{RT\\cdot y_{BM}} \\]\n\\[ k_G = \\frac{k_c' P}{p_{BM}}\\cdot\\frac{C_{A1}-C_{A2}}{p_{A1}-p_{A2}} = \\frac{k_c' P}{RT\\cdot p_{BM}} \\]\n\n\\end{par} \\vspace{1em}\n\\begin{par}\n\n\\subsection{Part b}\nConvert $k_L$ to $k_x$ and $k_x'$.\n\n\\end{par} \\vspace{1em}\n\\begin{par}\nThe flux equations involving these coefficients are, $$ N_A = k_L(c_{A1}-c_{A2}) = k_x(x_{A1}-x_{A2}) $$ A relationship for $k_x$ can be found by substituting $x=c_{A}/c$, $$ k_x = k_L\\frac{c_{A1}-c_{A2}}{x_{A1}-x_{A2}} = k_L\\frac{c_{A1}-c_{A2}}{\\frac{c_{A1}}{c}-\\frac{c_{A2}}{c}} = k_L\\cdot c $$ $$ k_x' = k_x\\cdot x_{BM} = k_L \\cdot c \\cdot x_{BM} $$\n\\end{par} \\vspace{1em}\n\\begin{par}\n\n\\subsection{Part c}\nConvert $k_G$ to $k_y$ and $k_c$.\n\n\\end{par} \\vspace{1em}\n\\begin{par}\nThe flux equations involving these coefficients are, $$ N_A = k_G(p_{A1}-p_{A2}) = k_y(y_{A1}-y_{A2}) = k_c(C_{A1}-C_{A2}) $$ By substituting $p_A=y_A P$, $$ k_y = k_G\\frac{p_{A1}-p_{A2}}{y_{A1}-y_{A2}} = k_G\\frac{y_{A1}P-y_{A2}P}{y_{A1}-y_{A2}} = k_G\\cdot P $$ Finally, substituting $C=P/RT$, $$ k_c = k_G\\frac{p_{A1}-p_{A2}}{C_{A1}-C_{A2}} = k_G\\frac{p_{A1}-p_{A2}}{(p_{A1}-p_{A2})/RT} = k_G \\cdot RT $$\n\\end{par} \\vspace{1em}\n\n\n\\section{Problem 21.1-3}\n\n\\begin{par}\nIn a wetted-wall tower an air- $\\textup{H}_2\\textup{S}$ mixture is flowing by a film of water that is flowing as a thin film down a vertical plate. The $\\textup{H}_2\\textup{S}$ is being absorbed from the air to the water at a total pressure of $1.50~\\textup{atm abs}$ and $30~\\textup{C}$. A value for $k_c'$ of $9.567\\times 10^{-4}~\\textup{m/s}$ has been predicted for the gas-phase mass-transfer coefficient. At a given point, the mole fraction of $\\textup{H}_2\\textup{S}$ in the liquid at the liquid-gas interface is $2.0(10^{-5})$ and $p_A$ of $\\textup{H}_2\\textup{S}$ in the gas is $0.05~\\textup{atm}$. The Henry's law equilibrium relation is $p_A(\\textup{atm}) = 609x_A$ (mole fraction in liquid). Calculate the rate of absorption of $\\textup{H}_2\\textup{S}$. (\\textit{Hint}: Call point 1 the interface and point 2 the gas phase. Then, calculate $p_{A1}$ from Henry's law and the given $x_A$. The value of $p_{A2}$ is $0.05~\\textup{atm}$.)\n\\end{par} \\vspace{1em}\n\\begin{par}\nThe mass transfer coefficient given is $k_c'$, which can be converted to $k_G$ for a flux equation based on partial pressures. $$ k_G = \\frac{k_c' P}{RT\\cdot p_{BM}} $$ $$ N_A = k_G(p_{A1} - p_{A2}) $$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nu = symunit;\nP = 1.5 * u.atm;\nT = rewrite(30 * u.Celsius, u.K, 'Temperature', 'absolute');\nk_c_prime = 9.567e-4 * u.m / u.s;\nx_A1 = 2.0e-5;\np_A1 = 609 * x_A1 * u.atm;\np_A2 = 0.05 * u.atm;\np1 = separateUnits(p_A1); p2 = separateUnits(p_A2); p = separateUnits(P);\np_BM = ((p-p1)-(p-p2))/log((p-p1)/(p-p2)) * u.atm;\nR = 8.2057338e-5 * (u.m^3 * u.atm) / (u.mol * u.K);\nk_G = (k_c_prime * P) / (R * T * p_BM);\nN_A = k_G * (p_A1 - p_A2) * 1e-3 * u.kg; % convert to kg-mol\ndisp(unitString(N_A))\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}N_A: -1.4854e-06 (kg*mol)/(m^2*s)\n\\end{verbatim} \\color{black}\n \n\n\\section{Problem 21.2-1}\n\n\\begin{par}\nA fluid is flowing in a vertical pipe and mass tranfer is occurring from the pipe wall to the fluid. Relate the convective mass-transfer coefficient $k_c'$ to the variables $D$, $\\rho$, $\\mu$, $v$, $D_{AB}$, $g$, and $\\Delta{\\rho}$, where $D$ is pipe diameter, $L$ is pipe length, and $\\Delta{\\rho}$ is the density difference.\n\\end{par} \\vspace{1em}\n\\begin{par}\nAccording to Buckingham's pi theorem, given nine independent variables in three physical dimensions, there are six $\\pi-$ groups to construct.\n\\end{par} \\vspace{1em}\n\\begin{par}\nVector order: [Length Mass time]\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nD = [1 0 0];\nL = [1 0 0];\nrho = [-3 1 0];\nmu = [-1 1 -1];\nv = [1 0 -1];\nD_AB = [2 0 -1];\ng = [1 0 -2];\ndelta_rho = [-3 1 0];\nk_c_prime = [1 0 -1];\n\npi_1 = ([D' rho' mu']\\-k_c_prime')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_1 =\n\n 1 1 -1\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_1 = \\frac{D \\rho k_c'}{\\mu}$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\npi_2 = ([D' rho' mu']\\-v')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_2 =\n\n 1 1 -1\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_2 = \\frac{D \\rho v}{\\mu}$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\npi_3 = ([D' rho' mu']\\-D_AB')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_3 =\n\n 0 1 -1\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_3 = \\frac{\\rho D_{AB}}{\\mu}$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\npi_4 = ([D' rho' mu']\\-g')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_4 =\n\n 3 2 -2\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_4 = \\frac{D^3 \\rho^2 g}{\\mu^2}$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\npi_5 = ([D' rho' mu']\\-delta_rho')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_5 =\n\n 0 -1 0\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_5 = \\frac{\\Delta\\rho}{\\rho}$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\npi_6 = ([D' rho' mu']\\-L')'\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\npi_6 =\n\n -1 0 0\n\n\\end{verbatim} \\color{black}\n \\begin{par}\n$\\pi_6 = \\frac{L}{D}$\n\\end{par} \\vspace{1em}\n\\begin{par}\nThese dimensionless groups can be combined to describe the system. This combination was done pencil-and-paper with a lot of frustration, and the results are shown here.\n\\end{par} \\vspace{1em}\n\\begin{par}\n\n\\[ \\frac{\\pi_1}{\\pi_3} = f\\left(\\pi_4 \\pi_5 \\pi_6^3, \\pi_2,\n\\pi_3^{-1}\\right) \\]\n\\[ \\frac{k_c'D}{D_{AB}}=f\\left(\\frac{gL^3\\rho\\Delta\\rho}{\\mu^2},\n\\frac{Dv\\rho}{\\mu},\\frac{\\mu}{\\rho D_{AB}}\\right) \\]\n\n\\end{par} \\vspace{1em}\n\n\n\\section{Problem 21.3-1}\n\n\\begin{par}\nUsing the data and physical properties of Example 21.3-2, calculate the flux for a water velocity of $0.152~\\textup{m/s}$ and a plate length of $L = 0.137~\\textup{m}$. Do not assume that $x_{BM}=1.0$ but actually calculate its value.\n\\end{par} \\vspace{1em}\n\\begin{par}\nThe Schmidt and Reynolds numbers can be calculated from the given quantities,\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nu = symunit;\nT = rewrite(26.1 * u.Celsius, u.K, 'Temperature', 'absolute');\nL = 0.137 * u.m;\nv = 0.152 * u.m / u.s;\nsolubility = 0.02948 * (u.kg * u.mol) / (u.m^3);\nD_AB = 1.245e-9 * u.m^2 / u.s;\nmu = 8.71e-4 * u.Pa * u.s;\nrho = 996 * u.kg / u.m^3;\nN_Sc = simplify(mu / (rho * D_AB));\nN_Re = simplify((L * v * rho)/mu);\ndisp(unitString(N_Sc))\ndisp(unitString(N_Re))\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}N_Sc: 702.408 1\nN_Re: 23812.5189 1\n\\end{verbatim} \\color{black}\n \\begin{par}\nThis value for the Reynolds number corresponds to this equation for mass flux, $$ J_D = 0.99N_{Re,L}^{-0.5} = \\frac{k_c'}{v}\\left(N_{Sc}\\right)^{2/3} $$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nsyms k_c_prime;\nk_c_prime = solve(0.99*N_Re^-0.5 == (k_c_prime/v)*(N_Sc^(2/3)));\nFW_water = 18.02* u.kg / (u.kg * u.mol);\nc = rho / FW_water;\nx_A1 = 0;\nx_A2 = solubility / (solubility + c);\nx_BM = (1 - (1 - x_A2))/log(1/(1 - x_A2));\nk_c = k_c_prime / x_BM;\nN_A = k_c * solubility;\ndisp(unitString(N_A))\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}N_A: 3.6391e-07 (kg*mol)/(m^2*s)\n\\end{verbatim} \\color{black}\n \n\n\n \\section{Referenced Functions} \\subsection{unitString.m} \\begin{verbatim} function displayString = unitString(quantity, name)\n%UNITSTRING Display a 1x2 sym with symbolic units\n% USAGE: unitString(some_quantity, name)\n% OUTPUT:\n% - displayString: char vector containing name, scalar, and units\nif nargin < 2\n n = inputname(1);\nelse\n n = name;\nend\n\n[s, U] = separateUnits(quantity);\nformatSpec = '%s: %s %s';\ndisplayString = sprintf(formatSpec, n, num2str(double(s)), symunit2str(U));\nend \\end{verbatim} \\end{document}", "meta": {"hexsha": "453095d7c981e3f15e43a95c57b6ba3712cf19ff", "size": 11152, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "examples/homework_12/html/report.tex", "max_stars_repo_name": "ahoetker/autoreport", "max_stars_repo_head_hexsha": "26fb9e563e2c30ac9422d98e2567223dba77ab4a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-11-09T20:29:04.000Z", "max_stars_repo_stars_event_max_datetime": "2018-11-09T20:29:04.000Z", "max_issues_repo_path": "examples/homework_12/html/report.tex", "max_issues_repo_name": "ahoetker/autoreport", "max_issues_repo_head_hexsha": "26fb9e563e2c30ac9422d98e2567223dba77ab4a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "examples/homework_12/html/report.tex", "max_forks_repo_name": "ahoetker/autoreport", "max_forks_repo_head_hexsha": "26fb9e563e2c30ac9422d98e2567223dba77ab4a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.7939393939, "max_line_length": 944, "alphanum_fraction": 0.6425753228, "num_tokens": 4405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.766293653760418, "lm_q1q2_score": 0.6282790600710013}} {"text": "\\chapter{Fit of histograms}\n\\label{ch:fit}\n\nThis chapter illustrates how the fits were done in \\autoref{ch:anal}. They will\nturn out to be almost exactly a standard least squares fit of a histogram, but\nwe will give a Bayesian interpretation to all the procedures. If you do not\nfavor Bayesian methods, just do not think about it. To actually do the fits, we\nused the Python package \\texttt{lsqfit} \\cite{lepage2021}, which we warmly\nrecommend.\n\n\\section{Bayesian least squares}\n\nBayesian inference is based on applying Bayes' theorem to get the probability\ndistribution of the parameters conditional on the observed data. Call $\\theta$\nthe parameters and $y$ the data, then it reads\n%\n\\begin{equation}\n p(\\theta|y) = \\frac {p(y|\\theta) p(\\theta)} {p(y)}.\n\\end{equation}\n\nThe term $p(y|\\theta)$ is the likelihood, how the data depends on the\nparameters. $p(y)$ is the unconditional probability of the data; whatever that\nmeans, we can compute it with $\\int\\mathrm d\\theta\\,p(y|\\theta)p(\\theta)$.\n$p(\\theta)$ is the unconditional probability distribution of the parameters. It\nis called \\emph{prior} because it encodes what one knows about the parameters\nbefore having the data. By the same logic, $p(\\theta|y)$ is called\n\\emph{posterior}.\n\nLeast squares is a general procedure to produce an estimator, but a standard\ninterpretation is also as a maximum likelihood fit when the likelihood is\nGaussian. We can do the same in the Bayesian case if the prior is Gaussian too.\nWe switch to logarithms and drop addends which do not depend on the parameters:\n%\n\\begin{align}\n \\log p(\\mathbf y|\\boldsymbol\\theta) &=\n -\\sum_{i=1}^n \\log\\sigma_i\n -\\frac 12 \\sum_{i=1}^n\n \\left( \\frac {y_i - \\mu_i(\\boldsymbol\\theta)} {\\sigma_i} \\right)^2, \\\\\n %\n \\log p(\\boldsymbol\\theta) &=\n -\\sum_{j=1}^k \\log\\sigma^\\theta_j\n -\\frac 12 \\sum_{j=1}^k\n \\left( \\frac {\\mu^\\theta_j - \\theta_j} {\\sigma^\\theta_j} \\right)^2.\n\\end{align}\n\nNow these two terms have to be summed to obtain the log posterior. If we extend\nthe $\\mathbf y$, $\\boldsymbol\\mu$ and~$\\boldsymbol\\sigma$ vectors, we can see\nthe two sums together as a single sum of squares:\n%\n\\begin{align}\n y_{n+j} &\\equiv \\mu^\\theta_j, \\\\\n %\n \\mu_{n+j}(\\boldsymbol\\theta) &\\equiv \\theta_j, \\\\\n %\n \\sigma_{n+j} &\\equiv \\sigma^\\theta_j, \\qquad j = 1, \\ldots, k, \\\\\n %\n \\log p(\\boldsymbol\\theta|\\mathbf y) &=\n -\\sum_{i=1}^{n+k} \\log\\sigma_i\n - \\frac 12 \\sum_{i=1}^{n+k}\n \\left( \\frac {y_i - \\mu_i(\\boldsymbol\\theta)} {\\sigma_i} \\right)^2 \\equiv \\\\\n %\n &\\equiv -\\sum_{i=1}^{n+k} \\log\\sigma_i - \\frac12 \\chi^2. \\label{eq:chi2}\n\\end{align}\n\nSo formally the prior is equivalent to additional datapoints, one for each\nparameter. We have not dropped the sum with the $\\log\\sigma$ although it does\nnot depend on $\\boldsymbol\\theta$; it will be needed later. Note that the\ndegrees of freedom of the $\\chi^2$ now are $(n+k)-k = n$, instead of the usual\n$n-k$ without prior.\n\nIn the analysis we said we put uniform priors over the parameters defined in\nthe interval $(0,1)$. This is implemented by fitting a transformed parameter\nwith a standard Gaussian prior with zero mean and unitary variance, and\napplying as inverse transformation the Gaussian cumulative density function\n(cdf). Say, if we want to put a uniform prior on $\\theta_j$, we fit the\ntransformed parameter $\\theta_j'$, with\n%\n\\begin{align}\n \\mu^{\\theta'}_j &= 0, \\\\\n %\n \\sigma^{\\theta'}_j &= 1, \\\\\n %\n \\theta_j(\\theta'_j) &= \\Phi(\\theta'_j)\n \\equiv \\int_{-\\infty}^{\\theta'_j} \\mathrm du\\,\n \\frac {e^{-u^2/2}} {\\sqrt{2\\pi}}.\n\\end{align}\n\nBayesianly, this is equivalent to omitting the squared term for the prior and\nputting bounds on the parameter value, without transforming. However it is\nconvenient for the fitting routine. By the same logic, when we use the\nlogarithm to map a positive parameter to the real line, the prior on the\nuntransformed parameter is a log-Gaussian distribution.\n\nNow the least squares estimator would be obtained by minimizing the $\\chi^2$,\nwhile a Bayesian cares about the posterior as a distribution. If we approximate\nthe posterior as a Gaussian, we can do formally the same calculations. To\ncompute the Gaussian approximation we expand the logarithm of the distribution\nto second order around the maximum:\n%\n\\begin{align}\n \\log p(\\boldsymbol\\theta|\\mathbf y) &\\approx\n -\\frac 12 (\\boldsymbol\\theta-\\boldsymbol\\theta_0)^\\top\n V^{-1} (\\boldsymbol\\theta-\\boldsymbol\\theta_0), \\\\\n %\n \\boldsymbol\\theta_0\n &= \\operatorname*{arg\\,min}_{\\boldsymbol\\theta} \\chi^2, \\\\\n %\n V^{-1}_{jk} &= \\frac12 \\left.\n \\frac {\\partial^2 \\chi^2} {\\partial\\theta_j \\partial\\theta_k}\n \\right|_{\\boldsymbol\\theta_0}. \\label{eq:lsqpm}\n\\end{align}\n\nNow let us carry further the calculation of the precision matrix, to discuss\nan additional issue:\n%\n\\begin{align}\n \\frac{\\partial\\chi^2}{\\partial\\theta_j} &=\n 2 \\sum_i \\frac {y_i - \\mu_i(\\boldsymbol\\theta)} {\\sigma_i^2}\n \\frac {\\partial\\mu_i} {\\partial\\theta_j}, \\\\\n %\n \\frac {\\partial^2 \\chi^2} {\\partial\\theta_j \\partial\\theta_k} &=\n 2 \\sum_i \\frac1{\\sigma_i^2} \\left(\n -\\frac {\\partial\\mu_i} {\\partial\\theta_j}\n \\frac {\\partial\\mu_i} {\\partial\\theta_k}\n + (y_i - \\mu_i(\\boldsymbol\\theta))\n \\frac {\\partial^2 \\mu_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right). \\label{eq:lsqhess}\n\\end{align}\n\nIn standard least squares, the precision matrix of the estimator is estimated\nwith the above expression, but with the term with the second derivative\ndropped. The reason is for convenience, but also because the expectation of the\nresiduals $y_i - \\mu_i(\\boldsymbol\\theta_0(\\mathbf y))$ is approximately zero.\nA frequentist would like to compute this with the true value of the parameters,\nbut has to content herself with the least squares estimate, getting something\nwhich on average is the true precision matrix. So any term which is zero on\naverage can be removed.\n\nFor a Bayesian, instead, there is not some ideal error corresponding to an\nideal parameter value; there is only the covariance of the posterior obtained\nwith the particular values at hand. Different data, different posterior.\nHowever in our fits the residuals term is missing, as we said just because this\nis convenient for the implementation. In practice, the difference should be\nsmall.\n\n\\section{Least squares fit of a histogram}\n\nThe standard model of a histogram is Poisson distributions for the bin counts.\nLet $c_i$ be the count in bin~$i$, $f_i(\\boldsymbol\\theta)$ the model\nnormalized distribution of the bins, and $M$ a parameter for the mean of the\ntotal (Poisson) number of samples. Then the likelihood is\n%\n\\begin{equation}\n P(\\mathbf c|M,\\boldsymbol\\theta) =\n \\prod_i P(c_i|M,\\boldsymbol\\theta) =\n \\prod_i \\operatorname{poisson}(c_i;Mf_i(\\boldsymbol\\theta)),\n\\end{equation}\n%\nwhere $\\operatorname{poisson}(\\cdot;\\mu)$ is a Poisson probability mass\nfunction (pmf) with mean~$\\mu$.\n\nWe do not care about $M$, only about $\\boldsymbol\\theta$, so we marginalize\nthe posterior over $M$:\n%\n\\begin{align}\n p(\\boldsymbol\\theta|\\mathbf c)\n &= \\int \\mathrm dM\\, p(M,\\boldsymbol\\theta|\\mathbf c) = \\\\\n %\n &= \\int \\mathrm dM\\,\n \\frac {P(\\mathbf c|M,\\boldsymbol\\theta) p(M,\\boldsymbol\\theta)}\n {P(\\mathbf c)}.\n\\end{align}\n\nNow we assume that a priori $M$ is independent from $\\boldsymbol\\theta$, i.e., \n$p(M,\\boldsymbol\\theta) = p(M)p(\\boldsymbol\\theta)$. Then, continuing the\ncalculation:\n%\n\\begin{align}\n p(\\boldsymbol\\theta|\\mathbf c)\n &= \\frac {p(\\boldsymbol\\theta)} {P(\\mathbf c)}\n \\int \\mathrm dM\\, p(M)\n \\prod_i \\operatorname{poisson}(c_i;Mf_i(\\boldsymbol\\theta)) = \\\\\n %\n &= \\frac {p(\\boldsymbol\\theta)} {P(\\mathbf c)}\n \\int \\mathrm dM\\, p(M)\n \\prod_i\n e^{-Mf_i(\\boldsymbol\\theta)}\n \\frac {(Mf_i(\\boldsymbol\\theta))^{c_i}} {c_i!} = \\\\\n %\n &= \\frac {p(\\boldsymbol\\theta)} {P(\\mathbf c)}\n \\int \\mathrm dM\\, p(M)\n e^{-M \\sum_i f_i(\\boldsymbol\\theta)}\n M^{\\sum_i c_i}\n \\prod_i\n \\frac {f_i(\\boldsymbol\\theta)^{c_i}} {c_i!}\n\\end{align}\n%\nSince $f_i$ is normalized the sum over it yields~1, while we call $N$ the total\ncount, so\n%\n\\begin{align}\n p(\\boldsymbol\\theta|\\mathbf c)\n &= \\frac {p(\\boldsymbol\\theta)} {P(\\mathbf c)}\n \\prod_i\n \\frac {f_i(\\boldsymbol\\theta)^c_i} {c_i!}\n \\int \\mathrm dM\\, p(M) e^{-M} M^N.\n\\end{align}\n%\nThe integral does not depend on $\\boldsymbol\\theta$, so we call it $I(N)$ and\ndo not care about its value. Now we take the logarithm:\n%\n\\begin{align}\n \\log p(\\boldsymbol\\theta|\\mathbf c) &=\n -\\log P(\\mathbf c) + \\log I(N) - \\sum_i \\log c_i!\n + \\log p(\\boldsymbol\\theta) + \\sum_i c_i \\log f_i(\\boldsymbol\\theta).\n\\end{align}\n\nThis is the exact expression for the posterior, which we want to compare to\nleast squares, which yields\n%\n\\begin{align}\n Q(\\boldsymbol\\theta) &=\n \\log p(\\boldsymbol\\theta)\n -\\frac 12\n \\sum_i \\frac {(c_i - N f_i(\\boldsymbol\\theta))^2} {c_i}.\n\\end{align}\n\nNote the particular flavor of chisquare we have picked: there is no parameter\nfor the normalization, which is fixed to the actual total count $N$, and the\ndenominator is the observed count $c_i$.\n\nTo do the comparison, we expand both expressions to second order, ignoring the\nshared prior term which is already matched:\n%\n\\begin{align}\n L(\\boldsymbol\\theta) &\\equiv \\log p(\\boldsymbol\\theta|\\mathbf c), \\\\\n %\n \\frac {\\partial L} {\\partial\\theta_j} &=\n \\sum_i \\frac {c_i} {f_i(\\boldsymbol\\theta)}\n \\frac {\\partial f_i} {\\partial \\theta_j}, \\\\\n %\n \\frac {\\partial^2 L} {\\partial\\theta_j \\partial\\theta_k} &=\n \\sum_i c_i \\left[\n -\\frac 1 {f_i(\\boldsymbol\\theta)^2}\n \\frac {\\partial f_i} {\\partial \\theta_j}\n \\frac {\\partial f_i} {\\partial \\theta_k}\n + \\frac 1 {f_i(\\boldsymbol\\theta)}\n \\frac {\\partial^2 f_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right].\n\\end{align}\n\nFor $Q$ we start from \\eqref{eq:lsqhess}, obtaining\n%\n\\begin{align}\n \\frac {\\partial^2 Q} {\\partial\\theta_j \\partial\\theta_k} &=\n -N^2 \\sum_i\n \\frac 1 {c_i} \\left[\n \\frac {\\partial f_i} {\\partial \\theta_j}\n \\frac {\\partial f_i} {\\partial \\theta_k}\n - \\left( \\frac{c_i} N - f_i(\\boldsymbol\\theta) \\right)\n \\frac {\\partial^2 f_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right]. \\label{eq:Qjk}\n\\end{align}\n\nWe do not discuss the different results for the maximum because it is well\nknown that when there at least five counts per bin the least squares\napproximation is good enough. The covariance instead needs a discussion. Now we\nmanipulate $\\partial_{jk} L$ to show it is similar to $\\partial_{jk} Q$. First,\nwe approximate the observed counts with the model, $c_i \\approx N\nf_i(\\boldsymbol\\theta)$:\n%\n\\begin{align}\n \\frac {\\partial^2 L} {\\partial\\theta_j \\partial\\theta_k} &=\n -N^2 \\sum_i \\frac {c_i} {N^2 f_i(\\boldsymbol\\theta)^2} \\left[\n \\frac {\\partial f_i} {\\partial \\theta_j}\n \\frac {\\partial f_i} {\\partial \\theta_k}\n - f_i(\\boldsymbol\\theta)\n \\frac {\\partial^2 f_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right] \\approx \\\\\n %\n &\\approx\n -N^2 \\sum_i \\frac 1 {c_i} \\left[\n \\frac {\\partial f_i} {\\partial \\theta_j}\n \\frac {\\partial f_i} {\\partial \\theta_k}\n - f_i(\\boldsymbol\\theta)\n \\frac {\\partial^2 f_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right].\n\\end{align}\n\nNext, we observe that since $f$ is normalized, $\\sum_i \\partial_{jk} f_i = 0$,\nso we can add the missing term to obtain the residuals:\n%\n\\begin{align}\n \\frac {\\partial^2 L} {\\partial\\theta_j \\partial\\theta_k} &\\approx\n -N^2 \\sum_i \\frac 1 {c_i} \\left[\n \\frac {\\partial f_i} {\\partial \\theta_j}\n \\frac {\\partial f_i} {\\partial \\theta_k}\n + \\left( \\frac {c_i} N - f_i(\\boldsymbol\\theta) \\right)\n \\frac {\\partial^2 f_i} {\\partial\\theta_j \\partial\\theta_k}\n \\right]. \\label{eq:Ljk}\n\\end{align}\n\nComparing \\eqref{eq:Qjk} with \\eqref{eq:Ljk}, we see that the difference is the\nsign of the residuals term. As we explained above, in the least squares fit\nthat term is dropped altogether, so notwithstanding the Poisson model, we end\nup doing the same kind of approximation we did starting from Gaussians.\n\n\\section{The $\\chi^2/\\mathrm{dof}$ correction}\n\nStarting from the Gaussian model posterior \\eqref{eq:chi2}, we add a parameter\n$s$ that rescales uniformly all the errors:\n%\n\\begin{align}\n \\log p(\\boldsymbol\\theta|\\mathbf y)\n &= -\\sum_{i=1}^{n+k} \\log\\sigma_i - \\frac12 \\chi^2 \\mapsto \\\\\n %\n \\mapsto \\log p(\\boldsymbol\\theta,s|\\mathbf y)\n &= \\log p(s) -\\sum_{i=1}^{n+k} \\log(s \\sigma_i)\n - \\frac 12 \\sum_{i=1}^{n+k}\n \\left( \\frac {y_i - \\mu_i(\\boldsymbol\\theta)} {s \\sigma_i} \\right)^2 = \\\\\n %\n &= \\log p(s) - (n+k) \\log s - \\sum_{i=1}^{n+k} \\log\\sigma_i\n - \\frac 12 \\frac{\\chi^2}{s^2}.\n\\end{align}\n\nSince $s$ is a scale parameter, it makes sense to use a uniform prior over\n$\\log s$, i.e., $p(s) = 1/s$, because it is scale-invariant. Now that we have\nlet errors vary, we marginalize away $s$, freely dropping normalizations as\nusual:\n%\n\\begin{align}\n p(\\boldsymbol\\theta|\\mathbf y)\n &= \\int \\mathrm ds\\,\n p(\\boldsymbol\\theta,s|\\mathbf y) = \\\\\n %\n &= \\int \\frac {\\mathrm ds} {s^{n+k+1}}\n \\exp\\left(-\\frac12 \\frac {\\chi^2} {s^2}\\right) = \\\\\n %\n &= \\frac 1 {(\\chi^2)^{(n+k)/2}}.\n\\end{align}\n\nNow again we take the logarithm and expand to second order:\n%\n\\begin{align}\n P(\\boldsymbol\\theta) &\\equiv\n \\log p(\\boldsymbol\\theta|\\mathbf y)\n = -\\frac{n+k}2 \\log \\chi^2, \\\\\n %\n \\frac {\\partial P} {\\partial \\theta_j}\n &= -\\frac{n+k}2 \\frac 1{\\chi^2}\n \\frac {\\partial \\chi^2} {\\partial \\theta_j}, \\\\\n %\n \\frac {\\partial^2 P} {\\partial\\theta_j \\partial\\theta_k}\n &= -\\frac{n+k}2 \\left[\n -\\frac 1 {(\\chi^2)^2}\n \\frac {\\partial \\chi^2} {\\partial \\theta_j}\n \\frac {\\partial \\chi^2} {\\partial \\theta_k}\n + \\frac 1{\\chi^2}\n \\frac {\\partial^2 \\chi^2} {\\partial\\theta_j \\partial\\theta_k}\n \\right]. \\label{eq:Pjk}\n\\end{align}\n\nIn the maximum $\\partial_j P = 0$, so the first term in \\eqref{eq:Pjk}\nvanishes, leaving\n%\n\\begin{align}\n \\frac {\\partial^2 P} {\\partial\\theta_j \\partial\\theta_k}\n &= -\\frac12 \\frac 1 {\\chi^2/(n+k)}\n \\frac {\\partial^2 \\chi^2} {\\partial\\theta_j \\partial\\theta_k}.\n \\label{eq:chi2dof}\n\\end{align}\n%\nThis is to be compared with the precision matrix in \\eqref{eq:lsqpm}. The\nonly difference is that the matrix has been divided by $\\chi^2/(n+k)$. Note\nthat $\\mathrm{dof} = n$, so we obtained a different prescription than the\nconventional one. Nevertheless, in the analysis we used $\\chi^2/\\mathrm{dof}$\njust because it is what people expect; compared to \\eqref{eq:chi2dof}, this\ngives errors up to \\SI{25}\\% larger on our fits, see \\autoref{tab:ct}.\n\nSimilarly, we used the pvalue; since we always obtained a stark contrast\nbetween good and bad fits, we did not bother giving a precise Bayesian\nequivalent.\n", "meta": {"hexsha": "fbddfb4e96ca74be6430ae439e68a4e2a5f9f79a", "size": 14850, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "fit.tex", "max_stars_repo_name": "Gattocrucco/thesis", "max_stars_repo_head_hexsha": "0b8672b7dc6a4ec54c5c76ecfbb83edcf34729f7", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-07-15T07:16:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-15T07:16:46.000Z", "max_issues_repo_path": "fit.tex", "max_issues_repo_name": "Gattocrucco/thesis", "max_issues_repo_head_hexsha": "0b8672b7dc6a4ec54c5c76ecfbb83edcf34729f7", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "fit.tex", "max_forks_repo_name": "Gattocrucco/thesis", "max_forks_repo_head_hexsha": "0b8672b7dc6a4ec54c5c76ecfbb83edcf34729f7", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.772845953, "max_line_length": 80, "alphanum_fraction": 0.6783164983, "num_tokens": 4676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8198933425148214, "lm_q1q2_score": 0.6282790563776071}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage{array}\n\\usepackage{clrscode3e}\n\\usepackage{amsmath}\n\\usepackage{kbordermatrix}\n\n\\setlength{\\parindent}{0em}\n\\setlength{\\parskip}{1em}\n\n\\begin{document}\n\n\\title{Efficient Calculation of Interaction Features on Sparse Matrices}\n\\author{Andrew Nystrom}\n\\date{}\n\n\\maketitle\n\n\\begin{abstract}% <- trailing '%' for backward compatibility of .sty file\nFILL THIS IN\n\\end{abstract}\n\n\\section{Introduction}\n\nIntroduction\nInteraction features are a way of capturing correlations between features in a machine \nlearning setting. A feature vector $\\vec{x}$ of dimensionality $D$ has second degree interaction features \n$\\{x_i \\cdot x_j : i, j \\in \\{0,1,..., D-1\\} \\land i < j\\}$, \nso a $D$ dimensional vector has $\\binom{D}{2} = \\frac{D^2-D}{2}$ second degree interaction features. A naive\napproach to calculating these features is to simply iterate through the combinations of the column indices.\nFor a sparse vector, many of the resulting interaction features would be zero, and could therefore be ignored.\nThis work describes a method to efficiently calculate second degree interaction features for a sparse matrix \nthat has time and space complexities that decrease quadratically with the density of the input matrix with respect to the naive approach.\n\n\\section{Approach}\nLet the list of nonzero columns for a given row $\\vec{x}$ be denoted by $N_{zc}$. The nonzero second degree \ninteraction features are simply the products of all combinations of two elements whose \ncolumns are in $N_{zc}$. However, to properly place an interaction feature into the correct column, a mapping from the column \nindex pairs of $N_{zc}$ into the columns of the interaction matrix is needed. The mapping is \nfrom pairs $(a, b)$, where $a$ and $b$ are in $1,2,..., D$, and $a < b$, to $1,2,..., \\frac{D^2-D}{2}$, Such a mapping essentially consists of mapping the indices of entries in the upper triangle of a matrix to indices in a flat \nlist. We now describe the construction of such a mapping. \n\n\\subsection{Mapping construction}\nWe seek a map from matrix indices $(i, j)$ (with $i < j$ and $0 \\le i < D$) to numbers $f(i, j)$ with $0 \\le f(i, j) < \\frac{D(D-1)}{2}$, one that follows the pattern indicated by \n\\begin{align}\n\\begin{bmatrix}\nx & 0 & 1 & 2 \\\\\nx & x & 3 & 4 \\\\\nx & x & x & 5 \\\\\nx & x & x & x\n\\end{bmatrix}\n\\end{align}\nIt's considerably easier, however, to consider the same indices, but subtracted from $6$ (or more generally, from $\\frac{D(D-1)}{2}$; that gives the pattern\n\\begin{align}\n\\label{eq:4x4mat}\n\\begin{bmatrix}\nx & 6 & 5 & 4 \\\\\nx & x & 3 & 2 \\\\\nx & x & x & 1 \\\\\nx & x & x & x\n\\end{bmatrix}\n\\end{align}\nWe'll call the function defined by this example $(i, j) \\mapsto g(i, j)$, and then observe that \n\\begin{align}\nf(i, j) = \\frac{D(D-1)}{2} - g(i, j)\n\\end{align}\n\nTo simplify slightly, we introduce a notation for the $n$th triangular number, \n\\begin{align}\n\\begin{bmatrix}\nT_2(n) = \\frac{n(n+1)}{2}\n\\end{bmatrix}\n\\end{align}\n\\noindent\nThe subscript $2$ is there to indicate that these are triangles in two dimensions; we'll use $T_3(n)$ to indicate the $n$th tetrahedral number, and so on for higher dimensions. \n\nThe codomain of $g$ is now numbers from $1$ to $T_2(D-1)$, inclusive. \n\nObserve that in Equation~\\ref{eq:4x4mat}, each entry in row $i$ lies in the range\n\\begin{align}\nT_2(D-i-1) &< e \\le T_2(D-i).\n\\end{align}\n\\noindent\nFor instance, in row $2$ in our example, where $D = 4$, the entries range from $2$ to $3$, while $T_2(D-i-1) = T_2(1) = 1$ and $T_2(D-i) = T_2(2) = 3$. (Note that row indices start at zero.) Unfortunately, the numbers increase from right to left. The entry in column $j$ is just $T_2(D-i-1) + D-j$, which adds one for the rightmost column (because $D - (D-1) = 1$). \nThus, the formula for $g$ is simply\n\\begin{align}\ng(i, j) \n&= T_2(D-i-1) + (D-j) \\\\\n&= \\frac{(D-i-1)(D-i)}{2} + D-j \\\\\n&= \\frac{(D^2 - (2i)D - D - i^2 - i) + 2D - 2j}{2} \\\\\n&= \\frac{D^2 - (2i)D + D - i^2 - i - 2j}{2}\n\\end{align}\nand hence \n\\begin{align}\nf(i, j) \n&= \\frac{D(D-1)}{2} - g(i,j) \\\\\n&= \\frac{D^2-D}{2} - \\frac{D^2 - (2i)D + D - i^2 - i - 2j}{2}\\\\\n&= \\frac{D^2-D - D^2 + (2i)D - D + i^2 + i + 2j}{2}\\\\\n&= \\frac{(2i)D - 2D + i^2 + i + 2j}{2}\n\\end{align}\n[WRONG]\nCorrect result:\n\\begin{align}\nR = \\frac{ 2in - i^2 + 2j - 3i - 2}{2}.\n\\end{align}\n\\subsubsection{Other indices}\nWith one-based indexing, the formula above becomes\n\\begin{align}\nf_1(i, j) &= ....\n\\end{align}\n\n\n\n\n\n column \n\n\nINSERT JOHN'S PROOF HERE\n\nWith this mapping, an algorithm for generating second degree interaction features on a \nmatrix $A$ can be formulated as follows:\n\n\\begin{codebox}\n\\Procname{$\\proc{Sparse Interaction}(A)$}\n \\zi $\\func{map}(a, b) = \\frac{2Da-a^2+2b-3a-2}{2}$\n \\zi $N \\gets$ row count of $A$\n \\zi $D \\gets$ column count of $A$\n \\zi $B$ $\\gets$ Compressed Sparse Row Matrix of size $N \\times \\frac{D^2-D}{2}$\n \\zi \\For $\\id{row}$ in $A$ \\Do\n \\zi $N_{zc} \\gets$ nonzero columns of $row$\n \\zi \\For $i \\gets 0 \\To |N_{zc}|-1$ \\Do\n \\zi \\For $j \\gets i+1 \\To |N_{zc}|$ \\Do\n \\zi $k \\gets \\func{map}(i, j)$\n \\zi $r \\gets$ index of $\\id{row}$\n \\zi $B[r, k] \\gets \\id{row}[i] \\cdot \\id{row}[j]$\n \\End\n \\End\n \t\\End\n\\end{codebox}\n\n\\section{Complexity Analysis}\nAssume that A is a matrix with sparsity $0 < d < 1$, $N$ rows, and $D$ columns. Finding \ninteraction features with the proposed algorithm has time and space complexity \n$\\func{O}(d N D^2)$, \nwhereas a naive approach of using non-sparse matrices and multiplying all column \ncombinations has time and space complexity $\\func{O}(N D^2)$. The algorithm is therefore an \nimprovement by a factor of the density factor of $A$.\n\nThis can represent a large gain in speed and time. For example, the 20 Newsgroups dataset \nhas density $d$ of 0.12 when its unigrams are represented in a vector space model. This \nmeans the proposed approach would take less than $\\frac{1}{8}$ time and memory.\n\nThe real benefit of this method is revealed when the average complexity is analysed. The \nnumber of interactions calculated for a given row are $\\binom{|N_{zc}|}{2}$. If the matrix has \ndensity $d$, then on average, $N_{zc} = D d$, so the number of interaction features \ncalculated in total is \n\n\\begin{align*}\nN \\binom{d D}{2} &= N \\frac{(Dd)!}{2!(Dd-2)!}\\\\\n \\\\\n &= N \\frac{(D^2d^2-Dd)}{2}\n\\end{align*}\n\nThis means that the average complexity decreases quadratically with the density.\n\n\\section{Future Work}\nThe approach for generating second degree interaction features required a mapping from \ncombinations of two to the space $1,2,...,\\frac{D^2-D}{2}$, which is isomorphic to a mapping from \nthe indices of an upper triangular matrix to the indices of a flat list of the same size. \nTo generate third degree interaction features, a mapping from combinations of three \n$(a,b,c)$ to the space $1,2,...\\frac{D^3-3D^2+2D}{6}$ (which is $\\binom{D}{3}$), or the upper $3$-simplex of a tensor to a flat \nlist of the same size $\\frac{D^3-3D^2+2D}{6}$ would be required. In general, for interaction \nfeatures of degree k, the upper $k$-simplex of a $k$-dimensional tensor must be mapped to the \nspace $1,2,...\\frac{D!}{k!(D-k)!}$. A similar approach for finding these mappings could be taken \nas the one used here for $k=2$. \n\nMotivation for deriving mapping functions for higher orders\nof interaction features is that the average complexity of generating degree $k$ interaction\nfeatures is $N \\binom{Dd}{k}$, which decreases polynomially with respect to k compared to\ngenerating the features naively.\n\n \n\\vskip 0.2in\n\\bibliography{sample}\n\n\\end{document}", "meta": {"hexsha": "a3794a8639ceabfec1479aefde03e57764452e40", "size": 7660, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/spares_interaction.tex", "max_stars_repo_name": "AWNystrom/SparseInteraction", "max_stars_repo_head_hexsha": "68ac222d7a826a344675d0e5196d82cb1711a69a", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2016-01-08T17:17:55.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-16T15:35:26.000Z", "max_issues_repo_path": "paper/spares_interaction.tex", "max_issues_repo_name": "AWNystrom/SparseInteraction", "max_issues_repo_head_hexsha": "68ac222d7a826a344675d0e5196d82cb1711a69a", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/spares_interaction.tex", "max_forks_repo_name": "AWNystrom/SparseInteraction", "max_forks_repo_head_hexsha": "68ac222d7a826a344675d0e5196d82cb1711a69a", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-01-08T17:28:39.000Z", "max_forks_repo_forks_event_max_datetime": "2020-02-08T17:36:00.000Z", "avg_line_length": 40.9625668449, "max_line_length": 366, "alphanum_fraction": 0.6778067885, "num_tokens": 2453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8198933403143929, "lm_q1q2_score": 0.6282790546914326}} {"text": "\\documentclass[letterpaper]{article}\n\n\\usepackage[utf8]{luainputenc}\n\\usepackage{fullpage}\n\\usepackage{nopageno}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\n\\title{Notes}\n\\date{2 mars, 2015}\n\\maketitle\n\nwhen does taylor series converge?\n\nbetter question is when is $f(x)=\\sum\\limits{f^{(n)}(a)/n!\\cdot (x-a)^n}$\n\n\n\\section*{taylor's thm}\n$f\\in C^{\\infty}[a,b]$ and $f^{(n+1)}$ is defined on $[A,B]$ with $|f^{(n+1)}(x)|\\le M$ for $x\\in [A,B]$ then $R_n(x)=f(x)-\\sum{\\frac{f^{k}}{k!}(x-a)^k}$ satisfies $|R_n(x)|\\le M|x-a|^{n+1}/(n+1)!$\n\\begin{enumerate}\n\\item\n$f(x)=\\lim\\sum\\frac{f^{(k)}}{k!}(x-a)^k$ if and only if $\\lim R_n(x)=0$\n\nthe basic issue is that $M$ needs to not get too big to fast\n\\item\nif $f\\in C^\\infty[A,B]$ then these hypotheses happen automatically (it's infinitely differentiable), although there is no guarantee that the taylor series converges to $f$.\n\\end{enumerate}\n\nwe want to use induction (what are we inducting on?)\n\nwe will show that $|R_n^{(n-k)}(x)|\\le\\frac{M|x-a|^{k+1}}{(k+1)!}$.\n\nbase case is $k=0$. $R_n(x)=f(x)-\\sum{\\frac{f^{(k)}}{k!}(x-a)^k}=f^{(n)}-\\frac{n!}{n!}f^{(n)}x$ and so $|R_n(x)|=|f^{(n)}(x)-f^{(n)}(a)|$. by MVT we know $M|x-a|\\ge f^{(n+1)}(c)|x-a|=|R_n(x)$. So base case is done\n\nassume $|R_n^{(n-k)}(x)|\\le M|x-a|^{k+1}/(k+1)!$. consider $|R_n^{(n-(k+1))}(x)|=|R_n^{(n-(k+1))}(a)-\\int_a^x{R^{(n-k)}(t)\\;\\mathrm{d}t}|$\n\n\\subsection*{example}\n$f(x)=\\sin x$ and $a=\\frac{\\pi}{2}$.\n\n$P_n(x)=\\sum\\limits_{k=0}^n{\\frac{f^{(k)}(\\frac{\\pi}{2}(x-\\frac{\\pi}{2})^k}{k!}}$ and then we have $P_n=\\sum\\limits_{k=0}^n{(-1)^k(x-\\frac{\\pi}{2})^{2k}}/(2k)!$. Now $M_n=1$ (it is bounded by 1). Now then $R_n(x)\\le \\frac{1\\cdot |x-a|^{n+1}}{(n+1)!}$ and because factorials are bigger than powers, the limit is $R=0$.\n\nand so the power series gives the same value as the function all the time.\n\n$f(x)=\\log x$ (natural log). taylor series at 1. note that zero is a problem. one is nice because it's symmetric and easy to compute.\n\n$f'(x)=\\frac{1}{x}, f''(x)=-\\frac{1}{x^2}, f'''(x)=\\frac{2}{x^3},\\dots$\n\n$P_n=\\sum\\limits_{k=1}^n{(-1)^{k+1}\\frac{(k-1)!}{k!}(x-1)^k}$ now use ratio test\n\n$\\frac{\\frac{(-1)^{k+2}}{k+1}}{\\frac{(-1)^{k+1}}{k}}\\to$ radius of convergent of 1, centered at zero, so interval is $(0,2)$\n\\end{document}\n", "meta": {"hexsha": "3bef4a4e2b11fc1128d0406cfeb364bbf0dff5ec", "size": 2341, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ra2/ra2-notes-2015-03-02.tex", "max_stars_repo_name": "ylixir/school", "max_stars_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ra2/ra2-notes-2015-03-02.tex", "max_issues_repo_name": "ylixir/school", "max_issues_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ra2/ra2-notes-2015-03-02.tex", "max_forks_repo_name": "ylixir/school", "max_forks_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.8035714286, "max_line_length": 318, "alphanum_fraction": 0.6065783853, "num_tokens": 996, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8198933359135361, "lm_q1q2_score": 0.6282790513190841}} {"text": "\\section{Linkages and Frameworks}\n--izmestiev def of framework\n--freedom in machines def of linkage\n--used to describe configurations and freedoms \n\n\n\\section{Geometric Configuration Space}\n\n --pictures of flexing. \n\\begin{figure}[ht]\n %\\includegraphics[scale=0.1, angle=0]{tetris.eps}\n\\caption{Different geometric configurations of an octahedron intermediate.}\n\\label{fig:OctaGCS}\n\\end{figure}\n--embedding\n\n\\begin{mydef}\nThe \\textbf{constraint space} corresponding to a constraint function $c: \\mathbbm{R}^n \\to \\mathbbm{R}^{n-m}$ is its zero-set,\n\\begin{align}\n\\Omega \\doteq \\{z \\in \\mathbbm{R}^n : c(z) = 0\\}.\n\\end{align}\n\\end{mydef}\nIf $c$ is composed of polynomial functions, then the constraint space is an algebraic variety. If this variety has no singularities, then it is also a manifold. \n\nUsing the notion of a constraint space, we can define the space of different geometric configurations that a given intermediate can take. Since we are treating each intermediate as collection of rigid polygons attached to each other along hinged edges, there are two basic types of constraint we must enforce. First, we must ensure that the each individual face is rigid and has the correct geometric shape. Additionally, each edge to edge connection must remain fixed with only hinge-like motion allowed. \n \n\\begin{mydef}\nThe \\textbf{geometric configuration space} of a building game intermediate is a constraint space that enforces the rigidity of individual faces and hinged motion along connected edges.\n\\end{mydef}\n\nTo mathematically describe a particular configuration, we must specify the locations of the vertices of each face. Thus, for an intermediate $[x]$, its building game configuration space can be represented as a subset of the ambient space $\\mathbb{R}^{3\\times N_x}$ where $N_x \\doteq \\sum_{f\\in x} s_f$ where $s_f$ is the number of sides (and verticies) of face $f$. Using this representation, we must then identify the corresponding constaint equations that give rise to the combinatorial configuration space as a function of points in ambient space. \n\nIt is worth noting that while we represent each face with $3\\times s_f$ coordinates, only 6 are required to speciafy a face's position and orientation if they are chosen carefully. With this in mind, we will typically use a function of $6$ of a face's vertex coordinates to constrain each of the face's remaining vertex coordinates. \n\n\nNotationally, we refer to the $k$th vertex of the $j$th face of $x$ as $v^{jk} = \\left(v^{jk}_x,v^{jk}_y,v^{jk}_z\\right)$.\nSince the context of a constraint space requires a function $c: \\mathbbm{R}^n \\to \\mathbbm{R}^{n-m}$, we flatten the matrix of vertex coordinates into a vector of lenghth $n = 3N_x$. \n\\begin{align}\nz = \\begin{bmatrix} v^{1,1} \\\\ \\vdots \\\\ v^{1,s_{f_1}} \\\\ \\vdots \\\\ v^{|x|,1} \\\\ \\vdots \\\\ v^{|x|,s_{f_{|x|}}} \\end{bmatrix} \\in \\mathbbm{R}^n\n\\end{align}\n\nThere are four fundamental types of constraint equations: edge length constraints, angle constraints, and $2D$ face constraint to enforce the rigid structure of each face as well as vertex identification constraints to enforce the hinged connections.\n\nEdge length constraints enforce that the lengths of the edges of each face in an intermediate cannot change. If the $k$th edge is definted to be that between the $(k-1)$st and $k$th vertices, we use the following function to constrain its lengths to a known value $\\ell^{jk}$.\n\\begin{align}\nc_{edge}^{j,k}\\left(z\\right)& = \\left|v^{j,k} - v^{j,k-1}\\right|^2 - (\\ell_{j,k})^2 \\\\\n& = \\left(v_1^{j,k} - v_1^{j,k-1}\\right)^2 +\\left(v_2^{j,k} - v_2^{j,k-1}\\right)^2 +\\left(v_3^{j,k} - v_3^{j,k-1}\\right)^2 - (\\ell^{j,k})^2 \n\\end{align} \nThis uses the notational convention that $v^{j,0} \\doteq v^{j,s_j}$. For reasons that will be explained, we only explicitly enforce the lengths of two edges ($k=1,2$) per face, so there are a total of $2|x|$ edge length constraints. \n\nAngle constraints ensure that each face's polygonal angles are conserved. Using the dot product formula for angles, we can write this constraint as a polynomial.\n\\begin{align}\nc_{ang}^{j,k}\\left(z\\right) &= (v^{j,k-1} - v^{j,k})\\cdot(v^{j,k+1} - v^{j,k}) - \\ell^{j,k}\\ell^{j,k+1}\\cos(\\theta^{j,k})\n\\end{align} \nHere, $\\theta^{j,k}$ is the angle at $v^{j,k}$ between $k$th and $(k+1)st$ edges which is a constant that is known a priori. In practice, we only enforce that the $k=1$st angle constraint for each face.\n\nNow, between the two edge constraints and one angle constraints, we have described $3$ constraints as a function of $9$ vertex coordinates per face. Thus for any choice of the $6$ coordinates, $v^{j,1}_1, v^{j,1}_2, v^{j,1}_3, v^{j,2}_1, v^{j,2}_2, v^{j,0}_1$, the remaining $3$ coordinates, $v^{j,2}_3, v^{j,0}_2, v^{j,0}_3$, are specified by the $3$ constraint equations. Similarly, since each face's location and rotational orientation can be defined by this choice of $6$ coordinates, we have enough information to specify the coordinates of the remaining verticies. \n\nSince the positions of the first three verticies dictate the locations of the remaining verticies, the $2D$ face constraints use a map from the known vertex coordinates to the yet unknown locations. Using a template for what the ideal polygonal stucture for each face should be, we use a rotation matrix to specify these remaining verticies. If this template has vertices $\\hat{v}^{j,0}, \\hat{v}^{j,1}, \\hat{v}^{j,2}, \\dots, \\hat{v}^{j,k}, \\dots$ and the locations for $v^{j,0}, v^{j,1}, \\text{and } v^{j,2}$ are known, we can identify the location of $v^{j,k}$ for $k>2$. Using this template, we can define the following length and angle constants.\n\\begin{align}\n\\ell^{j,k_1, k_2} &\\doteq |\\hat{v}^{j,k_1} - \\hat{v}^{j,k_2}| \\\\ \n\\phi^{j,k_1,k_2,k_3} &\\doteq \\cos^{-1}\\left(\\frac{\\left(\\hat{v}^{j,k_1} - \\hat{v}^{j,k_2}\\right)\\cdot\\left(\\hat{v}^{j,k_3} - \\hat{v}^{j,k_2}\\right)}{(\\ell^{j,k_1, k_2})(\\ell^{j,k_3, k_2})}\\right) \n\\end{align}\n\n\\begin{figure}[ht]\n %\\includegraphics[scale=0.1, angle=0]{tetris.eps}\n\\caption{Face template and 2d face constraints.}\n\\label{fig:2DFC}\n\\end{figure}\n\nOur basic stratigy is to first place a point $\\bar{v}^{j,k}$ in the span of $v^{j,0} - v^{j,1}$ at a distance of $|\\bar{v}^{j,k} - v^{j,1}| = \\ell^{j,k,1}$. The choice $\\bar{v}^{j,k} = v^{j,1} + \\frac{\\ell^{j,k,1}}{\\ell^{j,0,1}}(v^{j,0} - v^{j,1})$ will work, since \n\\begin{align}\n |\\bar{v}^{j,k} - v^{j,1}| &= |\\frac{\\ell^{j,k,1}}{\\ell^{j,0,1}}(v^{j,0} - v^{j,1})| \\\\\n &= \\frac{\\ell^{j,k,1}}{\\ell^{j,0,1}}|v^{j,0} - v^{j,1}| \\\\\n &= \\ell^{j,k,1}.\n\\end{align}\nThen, a rotation matrix is used to rotate $\\bar{v}^{j,k}$ by the correct angle into its position $v^{j,k}$.\nThe rotation matrix is centered at $v^{j,1}$ and its axis of rotation is defined by $u = \\frac{1}{\\ell^{j,0,1}\\ell^{j,2,1}}(v^{j,0} - v^{j,1})\\times(v^{j,2} - v^{j,1})$. Similarly, the angle of rotation $\\phi^{j,0,1,k}$ is the angle created by the two line segments in the template $(\\hat{v}^{j,0},\\hat{v}^{j,1})$ and $(\\hat{v}^{j,2},\\hat{v}^{j,1})$. Thus, using $R = R(\\phi^{j,0,1,k}, u)$ and our equation for $v^{j,k}$ is \n\\begin{align}\nv^{j,k} &= v^{1,k} + R(\\bar{v}^{j,k} - v^{j,1})\\\\\n&= v^{1,k} + \\ell^{j,k,1}R(v^{j,0} - v^{j,1})\\\\\n\\end{align}\nSince $R$ is polynomial in $v^{j,0}, v^{j,1}, v^{j,2}$, we get the following polynomial $2$D face constraint for each $k > 2$.\n\\begin{align}\nc_{2D}^{j,k}(z) \\doteq v^{1,k} + \\ell^{j,k,1}R(v^{j,0} - v^{j,1}) - v^{j,k}\n\\end{align} \n\nThe final constraint type, vertex identification, is used to enforce that the connection between edges of two faces has the mobility of a hinge. To do this, we simply need to ensure that that corresponding vertices on each edge share identical locations. This results in the relatively simple constraints:\n\\begin{align}\nc_{ident}^{j_1,k_1,j_1,k_1,d}(z) \\doteq v^{j_1,k_1}_d - v^{j_2,k_2}_d \\\\\n\\end{align}\nwhere $v^{j_1,k_1}$ and $v^{j_2,k_2}$ are corresponding vertices from the faces $j_1$ and $j_2$ meeting at a hinged edge. If there are $|E_x|$ of these hinged connections in a building game state $x$, there must be $6|E_x|$ correcponding vertex identification constraints.\n\nWith all four constraint types explicitly defined, the aggregate constraint function for $x$ will have $n-m = 2|F_x|+ |F_x| + (N_x - 3|F_x|) + 6|E_x| = N_x + 6|E_x|$ total constraints. \n\n\n%To mathematically represent each of the possible geometric configurations, we parameterize the embedding using its vertex locations. If a building game state is composed of the faces $\\{f_1,\\dots, f_i\\}$, then any embedding of the intermediate can be described by a point in $\\mathbbm{R}^n$, where $n$ is the three times sum $3\\sum_{j=1}^i \\#\\text{vertices}(f_j)$, of the number of vertices in each face. Simply put, the point \n% which is the concatenation of all of the vertex locations $v_1, v_2, \\dots, v_{n/3} \\in \\mathbbm{R}^3$ for each verted on each face of an intermediate is used to represent a particular embedding. \n\n\\subsection{Special Case of Triangular Faces}\n\nIn the case where all of the faces of the polyhedron we consider are triangles (tetrahedron, octahedron, icosahedron, etc.) we notice that \n\n\\begin{mycon}\nThe geometric configuration space for each intermediate composed of equilateral triangular faces is a manifold.\n\\end{mycon}\n\n%\\begin{mycon}\n%\n%\\end{mycon}\n\n\n\\section{Degrees of Freedom}\nRoughly speeking, the degrees of freedom of a system are the different independent motions the system is able to exercise. Since the concept of degrees of freedom exists in many diverse scientific fields, such as mechanical engineering and statistical physics, many different formal definitions of degrees of freedom are used in the literature. McCarthy defines degrees of freedom of a mechanical system as follows. CITE \n\\begin{quote}\nWe derive formulas for the number of parameters needed to specify the configuration of a mechanism, in terms of the number of links and joints and the freedom of movement allowed at each joint. This number is the \\textit{degrees of freedom} or \\textit{mobility} of the mechanism. Changing the values of these parameters changes the configuration of the mechanism. Thus, if we view the set of all configuration available to a mechanism as a manifold, then the mobility of the mechanism is the dimension of this manifold. \n\\end{quote}\n\nSince our geometric configuration space is an algebraic variety and not a manifold in general, this definition must be modified to make degrees of freedom a statistic of each individual configuration rather than a global statistic of the geometric configuration space. \n\n\\begin{mydef}\nThe number of \\textbf{degrees of freedom} a building game intermediate at configuration $z$ is the dimension of the geometric configuration space algebraic variety at $z$. If $z$ is a singularity of the algebraic variety, then the degrees of freedom is undefined.\n\\end{mydef}\n\nIn most cases we consider, rigid rotation and rigid translation will preserve the value of the constraint function since they do not move the verticies relative to one another. In other words, if $c(z) = 0$ then we also have $c(Rz + T) = 0$ where $R \\in \\mathbbm{R}^{n}\\times n$ rotates each vertex in the configuration by some $\\hat{R} \\in SO(3)$ and translates each vertex by $\\hat{T} \\in \\mathbbm{R}^3$. With this definition, $R$ is the block diagonal matrix with each of the $n/3$ block being $\\hat{R} \\in SO(3)$ and $T \\in \\mathbbm{R}^n$ is composed of $n/3$ copies of $\\hat{T} \\in \\mathbbm{R}^3$ stacked upon each other. These rigid body rotations account for $6$ of the configuration's degrees of freedom: $3$ rotational degrees of freedom and $3$ translational.\n\\begin{mydef}\nThe \\textbf{trivial degrees of freedom} a building game intermediate at configuration $z$ are the $6$ degrees of freedom corresponding to rigid body rotations and translations.\n\\end{mydef}\nThus, the degrees of freedom we are most interested in are those that do represent movement of the verticies and faces relative to each other. \n\\begin{mydef}\nThe \\textbf{internal degrees of freedom} a building game intermediate at configuration $z$ are the degrees of freedom that are not trival.\n\\end{mydef}\n\n\n\\begin{mydef}\n\\textbf{cannonical configuraion}\n\\end{mydef}\n\n\\subsection{Computing Degrees of Freedom}\n--First for a general configuration space\n--Jacobian equations\n--rank of C\n--extension to non triangular polyhedra\n--implementing on computer\n\n\\[\n\t\\frac{\\partial\\psi_{edge}^{j,k}}{\\partial v} =\n \t\\begin{cases}\n \t2\\left(v^{j,k}_\\bullet-v^{j,k-1}_\\bullet\\right) \t& \\text{if } v = v^{j,k}_\\bullet \\\\\n \t\t-2\\left(v^{j,k}_\\bullet-v^{j,k-1}_\\bullet\\right) \t& \\text{if } v = v^{j,k-1}_\\bullet \\\\\n \t\t0 & \\text{else} \n \t\\end{cases}\n\\]\n\n\n\\subsection{Non-Manifold Example}\nFrom this definition, it is worth noting that if the geometric configuration space is indeed a manifold, then the degrees of freedom is the dimension of that manifold which is a global property. Here we present an interesting example in which this is not the case and, depending on the specific configuration, a different number of degrees of freedom exist.\n\n\\begin{figure}[ht]\n %\\includegraphics[scale=0.1, angle=0]{tetris.eps}\n\\caption{A linkage of 6 squares with configurations of different degrees of freedom.}\n\\label{fig:SixSq}\n\\end{figure}\n\n\n\\subsection{Results}\n\n--Pictures!!\n\n\\subsection{Explicit Removal of Trivial Degrees of Freedom}\n\n--fixe center of mass.\n\n%We consider the linkage of rigid 2-dimensional polygons in $\\mathbb{R}^3$ by means of ideal hinges located at the edges of the polygons. In this way, by treating intermediates of our various models as such a linkage, we can compute the non-trivial degrees of freedom (DoF) each intermediate has. \n\n%Every linkage of this type has 6 trivial degrees of freedom: 3 corresponding to translational movement and 3 corresponding to rotational freedom. While these freedoms do change the orientation of the intermediate, they do not cause the faces to move relative to each other. For this reason, we label them as trivial degrees of freedom and focus on the calculation the remaining, non-trivial degrees of freedom. \n\n%For our purposes, we define the \\textbf{degrees of freedom} of a configuration to be the difference between the number of parameters needed to specify a configuration in an ambient parameter space $\\Omega$ and the number of independent constraints imposed upon the configuration as a function of these parameters. Consequently, the non-trivial degrees of freedom is the number of degrees of freedom minus the six trivial degrees of freedom. If the ambiant parameter space is $\\mathbb{R}^N$ and there are $M$ constraint equations given by $\\varphi : \\mathbb{R}^N \\to \\mathbb{R}^M$, we find the degrees of freedom to be $N - \\operatorname{rank}(\\varphi)$ and the non-trivial degrees of freedom to be $N - \\operatorname{rank}(\\varphi) - 6$. Here we need $\\varphi$ to be a smooth map so that the rank is well defined. Furthermore, we define the \\textbf{configuration space} to be the subset of parameter space $\\{\\omega \\in \\Omega : \\varphi(\\omega) = 0\\}$ upon which the contraints are satisfied.\n\n%In our case, the location and orientation of each polygon in the linkage can be represented by a total of 6 parameters: 3 for location and 3 for roational orientation. Thus, the entire configuration of an intermediate $x$ can be represented in the parameter space $\\mathbb{R}^{F_x\\times6}$. With constraints corresponding to hinged connections we construct a function $\\varphi$ which is zero if and only if the constraints are satisfied. Thus the configuration space is $\\{z \\in \\mathbb{R}^{F_x\\times6} : \\varphi(z) =0\\}$. As we will show below, the constraint equation $\\varphi$ can be constructed as a polynomial which means that the configuration space (as the zero set of this polynomial) is an algebraic variety. \n\n% PARAGRAPH ON DIFFERENT DOF VALS AT DIFFERENT CONFIGS OF SAME INTERMEDIATE We refer to the embedded configuration of $x$ such that $x$ is a subset of the original polyhedron as the \\textbf{canonical configuration}.\n\n\n%The base constaints are put in place to fix one of the intermediate's faces in space to remove the 6 trivial degrees of freedom from the calculation. If $f_b$ is the face we wish to designate as the base we have the following $3\\times s_b$ constraint equations\n%\\begin{align}\n%\\psi_{base}^{k,x}\\left(\\mathbf{v}\\right)& = v^{b,k}_x - c^{b,k}_x \\\\\n%\\psi_{base}^{k,y}\\left(\\mathbf{v}\\right)& = v^{b,k}_y - c^{b,k}_y \\\\\n%\\psi_{base}^{k,z}\\left(\\mathbf{v}\\right)& = v^{b,k}_z - c^{b,k}_z\n%\\end{align} \n%for $k = 1,\\dots,s_b$ and with $c^{b,k}_\\bullet$ a known constant. The corresponding Jacobian calculation for each constraint equations is:\n%\\[\n%\\frac{\\partial\\psi_{base}^{k,\\bullet}}{\\partial v} =\n% \\begin{cases}\n% 1 & \\text{if } v = v^{b,k}_\\bullet \\\\\n% 0 & \\text{else} \n% \\end{cases}\n%\\]\n\n%Edge length constraints enforce that the lengths of the edges of each face in an intermediate cannot change. Since there are $s_j$ edges for each face $f_j$ of intermediate $x$, there are $N_x$ corresponsing edge constraints. \n%\\begin{align}\n%\\psi_{edge}^{j,k}\\left(\\mathbf{v}\\right)& = \\left|v^{j,k} - v^{j,k-1}\\right|^2 - \\ell_{j,k}^2 \\\\\n%& = \\left(v_x^{j,k} - v_x^{j,k-1}\\right)^2 +\\left(v_y^{j,k} - v_y^{j,k-1}\\right)^2 +\\left(v_z^{j,k} - v_z^{j,k-1}\\right)^2 - \\ell_{j,k}^2 \n%\\end{align} \n%for all $j: f_j \\subset x$ and with the convention that $v^{j,0} \\doteq v^{j,s_j}$ and $\\ell_{j,k}$ is a known constant. The resulting partial derivatives are\n%\\[\n%\t\\frac{\\partial\\psi_{edge}^{j,k}}{\\partial v} =\n% \t\\begin{cases}\n% \t2\\left(v^{j,k}_\\bullet-v^{j,k-1}_\\bullet\\right) \t& \\text{if } v = v^{j,k}_\\bullet \\\\\n% \t\t-2\\left(v^{j,k}_\\bullet-v^{j,k-1}_\\bullet\\right) \t& \\text{if } v = v^{j,k-1}_\\bullet \\\\\n% \t\t0 & \\text{else} \n% \t\\end{cases}\n%\\]\n%--definiteion of 'base configuration'\n%--representation of config space definition as system of constraint equations\n%--solution of the constraint equations is a manifold except perhaps at some 'sp0ecial' configurations\n%--define DoF as 6N - rank of the jacobian of the CEqs evaluated at the base config\n\n%--go through computation of jacobian \n\n\n\n\n\\section{Cyclohexane Application}\n??Include??\n%\\subsection{Sachse Model}\n%\n%Around the turn of the century, it was thought that cyclohexane's carbon atoms must lie in a plane. A young German assistant, Hermann Sachse, had the idea that allowing the carbons to lie outside the plan could alleviate the angle strain. Inspired by polyhedral geometry, he templates and outlined methods for creating 3D models of the chair and boat configurations his new theory conceptualized. Figure~\\ref{fig:sachse} shows a construction of these two models. Despite his best efforts, Sachse's ideas were not accepted by that chemistry community until after his death. \n%\n%\n%\n%\\begin{figure}\n% %\\centering\n% %\\begin{subfigure}[b]{0.24\\textwidth}\n% % \\includegraphics[width=\\textwidth]{chair_model_2.png}\n% % \\caption{Chair (top)}\n% % \\label{fig:CM1}\n% %\\end{subfigure}%\n% %~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc.\n% % %(or a blank line to force the subfigure onto a new line)\n% %\\begin{subfigure}[b]{0.24\\textwidth}\n% % \\includegraphics[width=\\textwidth]{chair_model_1.png}\n% % \\caption{Chair (side)}\n% % \\label{fig:CM2}\n% %\\end{subfigure}%\n% %~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc.\n% % %(or a blank line to force the subfigure onto a new line)\n% %\\begin{subfigure}[b]{0.24\\textwidth}\n% % \\includegraphics[width=\\textwidth]{boat_model_1.png}\n% % \\caption{Boat (top)}\n% % \\label{fig:BM1}\n% %\\end{subfigure}%\n% %~ %add desired spacing between images, e. g. ~, \\quad, \\qquad etc.\n% % %(or a blank line to force the subfigure onto a new line)\n% %\\begin{subfigure}[b]{0.24\\textwidth}\n% % \\includegraphics[width=\\textwidth]{boat_model_2.png}\n% % \\caption{Boat (side)}\n% % \\label{fig:BM2}\n% %\\end{subfigure}%\n% %\\caption{Sachse Models}\\label{fig:sachse}\n%\\end{figure}\n%\n% \n%\\subsection{Idealized Constraint Model}\\label{building_game}\n%\n%Withe the eclipsing and angle strains in mind, we heuristically define an idealized model of cyclohexane by imposing geometric constraints. Each configuration is represented by the 3-dimensional locations of the center of its carbon atoms and we explicitly parameterize these locations as $v_1, v_2, \\dots, v_6$ where $v_k \\in \\mathbb{R}^3$. For ease of exposition, $v_0 \\doteq v_6$ and $v_{-1} \\doteq v_5$ are notationally identified. \n%\n%First, we require that any two atoms sharing a bond have a known and fixed distance $\\ell$ from each other. Since we can re-scale our coordinate system, we assume $\\ell \\doteq 1$. Additionally, we assume that the connectivity of the cyclohexane molecule is such that $v_k$ is bonded to $v_{k-1}$ for $k = 1,\\dots,6$. This gives our first six constraint equations.\n%$$0 = \\varphi_{len}^k\\left(v_{k-1},v_{k}\\right) = \\|v_k-v_{k-1}\\| - 1 $$\n%for $k = 1,\\dots,6$. \n%\n%Additionally, we impose constraints representing the angle strain. Since the carbon atoms have the lowest energy when their bonds are at tetrahedral angles, we fix the angle of each set of three adjacent carbons to be at the tetrahedral angle. This is equivalent to the six angle constraint equations \n%$$0 = \\phi_{ang}^k\\left(v_{k-2},v_{k-1},v_{k}\\right) = \\left(v_k-v_{k-1}\\right)\\cdot \\left(v_{k-2}-v_{k-1}\\right) + \\frac{1}{3}$$\n%for $k = 1,\\dots,6$.\n%\n%\\subsection{Degrees of Freedom in Ideal Model}\n%\n%The first step in answering these questions is determining whether there is any degrees of freedom to each configuration or whether they are rigid. This can be determined by using established theory on rigidity of linkages and degrees of freedom. By computing the Jacobian matrix $J(v)$ of the system of constraint equations, we have the following equation for the degrees of freedom for a constraint satisfying set of coordinates $v$.\n%$$DoF(v) = 18 - rank(J(v))$$\n%When the Jacobian is of full rank 18, none of the constraints are dependent on each other and thus there is no freedom in the linkage. When testing the chair, we found there to be $0$ degrees of freedom, meaning that it is impossible to make a transition from the chair to another configuration without breaking one or more of the constraint equations. For the boat, however, one degree of freedom was found. This result shows that it is possible to deform the boat continuously while satisfying the constraints, but it is not informative as to which configurations it can deform to. In particular, we are interested in finding a path to he twist boat or another boat configuration. \n%\n\n\\section{Folding Configuration Space}\n???include??\n%\n%The number of degrees of freedom measures the rigidity of an intermediate, but some intermediates with the same number of degrees of freedom may have varying degrees of mobility. We seek to quantify an intermediate's mobility by the relative amount of movement the intermediate has for a small movement in its configuration space. Intermediates with the most mobility will move less than the less mobile intermediates for the same amount of movement in configuration space. \n%\n%\\subsection{Methods}\n%% What is the configuration space?\n%Each octahedron intermediate is composed of 8 equilateral triangles connected to each other along edges. We make the assumption that these triangles are rigid and cannot be deformed. Furthermore, when two triangles meet at an edge, they move relative to each other as if connected by an ideal hinge. Every configuration has 6 trivial degrees of freedom corresponding to the 3 translation degrees of freedom and the 3 rotational degrees of freedom. Since we are interested in the motion of an intermediate's faces relative to each other, we remove these trivial degrees of freedom by picking a single face to fix in space. Depending on the connectivity of the triangle's edges, an intermediate \n%may still have degrees of freedom. The Octahedron intermediate (83), being a convex polyhedron, is rigid and has no non-trivial degrees of freedom as given by Cauchy's Theorem. However, this is not the case for most of the intermediates.\n%\n%We formalize the concept of the configuration space, by defining it to be the subset of an ambient parameter space that satisfies constraint equations that correspond to our assumptions. Since the are 8 faces in each intermediate and each face has 3 vertices each with 3 spacial coordinates (x,y,z), we use $\\mathbb{R}^{8\\times 3 \\times 3} = \\mathbb{R}^{72}$ as our ambient space. We then have 3 types of constraint equations: base face constraints, rigid face constraints, and hinge constraints. Every admissible configuration of an intermediate will be a point in $\\mathbb{R}^{72}$ that satisfies the constraint equations and any admissible movement of a configuration (if possible) is a continuous movement in the subset of $\\mathbb{R}^{72}$ where these constraints are satisfied. \n%\n%Since our constraint equations can all be expressed as polynomials, the configuration space which is the corresponding solution set is an Algebraic variety. The number of degrees of freedom of a configuration is the local dimension of this algebraic variety as a subset of $\\mathbb{R}^{72}$. In the case of octahedron intermediates, the number of degrees of freedom is not an informative way of classifying which intermediates are dominant as intermediates $1-11$ have $7$ degrees of freedom, $12-33$ have $5$, $34-65$ have $3$, $66-82$ have $1$ and the Octahedron (83) and Boat (84) have $0$ degrees of freedom. Thus, a different measure the mobility of a configuration is required to differentiate intermediates. It is important to note that degrees of freedom of in intermediate can theoretically change based on the region of configuration space. However, in practice we have not observed an intermediate's degrees of freedom change in this way. \n%\n%We define the \\textit{canonical configuration} of an intermediate, to be that in which each hinge forms a $180^\\circ$ angle when the hinge does not have a vertex connection at either end and in the case of a closed vertex, the hinge's angle corresponds to the Octahedral angle. In cases where the intermediate cannot form an octahedron, we can use the Boat configuration's angles instead.\n%\n%% How do we move in the configuration space?\n%Given a particular configuration $X$, if we wish to make a small move in configuration space, we first find the null space of the Jacobian matrix of the constraint equations. Since this null space corresponds to the directions in which the constraint equations are not changing, the constraints will remain satisfied. The null space can be represented by an orthonormal basis $N \\in \\mathbb{R}^{72\\times d}$, where $d$ is the number of degrees of freedom of the configuration. Then, by taking a small step of size $\\epsilon$ in each of the $d$ directions we get the configurations $X + \\epsilon N_{\\cdot,k}$. \n%% How do we measure the magnitude of a movement in configuration space?\n%\n%We define the \\textit{mobility} of a configuration to be the $L^2$ norm of the gradient of the configuration space. To approximate this, we measure the mean squared distance between each point of $X$ and $X \\pm \\epsilon N_{\\cdot,k}$ in each of the $d$ directions of $N$, take the norm, and divide by $\\epsilon$. This is given by Equation~\\ref{eq:r1} where $\\triangle_f$ refers to the $f$th face of the intermediate and $r\\left(X\\right)$ is the point we are integrating over and $r\\left(X+\\sigma \\epsilon N_{\\cdot,k}\\right)$ is the corresponding point in the altered configuration. \n%\n%\n%To explicitly define the rotation and translation of each face by the movement in configuration space, we use the algorithm given by Arun et al~\\cite{Arunetal1987}. Given two sets of points $P,P' \\in \\mathbb{R}^{3\\times n}$ in 3D space, the algorithm finds the rotation matrix $R$ and translation vector $b$ minimizing the least squares error of $P' \\approx R P + b$. By using the three vertices of face $f$ as $P$ and their perturbed values as $P'$, we use this algorithm to find the rotation matrix $R_f$ and translation vector $b_f$ to describe the rigid movement of face $f$ in the configuration space. This leads to the formulation in Equation~\\ref{eq:r3}. \n%\\begin{figure}[h]\n% %\\centering\n% %\\includegraphics[width=0.61\\textwidth]{triangle_fig.png}\n% %\\caption{Coordinates of rotated triangle for integration.}\n% \\label{fig:tri}\n%\\end{figure}\n%\n%To make this integral easier to solve, we introduce a change of variables that enables us to integrate in the $x-y$ plane. If the the original triangle $f$ has vertices of $a', b',$ and $c'$, we consider the triangle with vertices at the coordinates $a = (0,0,0), b = (|a-b|,0,0),$ and $c = (\\alpha,\\beta,0)$ as seen in Figure~\\ref{fig:tri}. Here, the choices $\\alpha = \\frac{|a-b|^2 + |a-c|^2 - |b-c|^2 }{2|a-b|}$ and $\\beta = \\sqrt{|a-c|^2-\\alpha^2}$ yield a triangle congruent to the original. Using the Arun et al~\\cite{Arunetal1987} algorithm again, we find the rotation matrix $S_f$ and translation vector $c_f$ that gives $[a',b',c'] = S_f[a,b,c] + c_f$. Using this we perform a change of variables and integrate over $s$ and $t$. This results in an simple double integral over a quadratic function of $s$ and $t$ with constants $u,v,w \\in \\mathbb{R}^3$ as seen in Equation~\\ref{eq:r5}.\n%{\\tiny\n%\\begin{align}\n%\\mathcal{R}\\left(x\\right) &= \\frac{1}{\\epsilon}\\left(\\sum_{k=1}^d \\sum_{f=1}^8 \\int_{\\triangle_f}\\left|\\frac{r\\left(x+\\epsilon N_{\\cdot,k}\\right) - r\\left(x\\right)}{2\\sqrt{3}}\\right|^2\\right)^{\\frac{1}{2}} \\label{eq:r1} \\\\\n%&= \\frac{1}{2\\sqrt{3}\\epsilon}\\left( \\sum_{k=1}^d \\sum_{f=1}^8 \\int_{\\triangle_f}\\left|R_fr+b_f - r\\right|^2\\right)^{\\frac{1}{2}} \\label{eq:r2} \\\\\n%&= \\frac{1}{2\\sqrt{3}\\epsilon}\\left( \\sum_{k=1}^d \\sum_{f=1}^8 \\int_{\\triangle_f}\\left|\\left(R_f-I\\right)r+b_f\\right|^2\\right)^{\\frac{1}{2}} \\label{eq:r3} \\\\\n%&= \\frac{1}{2\\sqrt{3}\\epsilon}\\left( \\sum_{k=1}^d \\sum_{f=1}^8 \\int_{0}^{\\beta}\\int_{\\frac{\\alpha}{\\beta}t}^{|a-b| + \\frac{\\alpha- |a-b|}{\\beta}t}\\left|\\left(R_f-I\\right)\\left(S_f\\begin{bmatrix} s \\\\[0.3em] t \\\\[0.3em] 0 \\end{bmatrix} + c_f\\right)+b_f\\right|^2ds dt\\right)^{\\frac{1}{2}} \\label{eq:r4} \\\\\n%&= \\frac{1}{2\\sqrt{3}\\epsilon}\\left( \\sum_{k=1}^d \\sum_{f=1}^8 \\int_{0}^{\\beta}\\int_{\\frac{\\alpha}{\\beta}t}^{|a-b| + \\frac{\\alpha- |a-b|}{\\beta}t}\\left|u + sv + tw\\right|^2ds dt\\right)^{\\frac{1}{2}} \\label{eq:r5} \n%\\end{align} \n%} \n%% Why does the choice of the base constraint matter and how do we deal with it?\n%\n%Unfortunately, the particular choice of base face affects the mobility. To rectify this bias, we compute the mobility with each of the eight faces as the base and take the average. This gives the final mobility in Equation~\\ref{eq:r6}. \n%{\\tiny\n%\\begin{align}\n%\\mathcal{R} &= \\frac{1}{8}\\sum_{g=1}^8\\mathcal{R}_g \\\\\n%&= \\frac{1}{16\\sqrt{3}\\epsilon}\\sum_{g=1}^8\\left( \\sum_{k=1}^d \\sum_{f=1}^8 \\int_{0}^{\\beta}\\int_{\\frac{\\alpha}{\\beta}t}^{|a-b| + \\frac{\\alpha- |a-b|}{\\beta}t}\\left|u + sv + tw\\right|^2ds dt\\right)^{\\frac{1}{2}} \\label{eq:r6} \n%\\end{align} \n%}\n%\n%\\section{Results}\n%\n%Mobility was computed for all octahedral intermediates in their canonical configuration. The perturbation parameter $\\epsilon = 10^{-6}$ was selected, but the mobility showed to be robust to both smaller and larger values. Figure~\\ref{fig:t1} outlines the results. From these calculations, it is clear that while mobility is highly correlated to degrees of freedom, mobility gives additional information about an intermediate's configuration space. Interestingly, all nets had the same mobility. Similarly, the symmetric pairs (14,16), (15,18), (19,20), (34,38), and (36,39) have matching mobility values. Of the intermediates with 5 degrees of freedom, 19 and 20 had the lowest mobility and 17 had the highest. As for intermediates with 3 degrees of freedom, 35 was the least mobile and 37 was the most mobile. \n%\n%\\begin{figure}[h!]\n%\\label{fig:t1}\n%%\\centering\n%\\scalebox{.9}{\n%\\begin{tabular}{ l | c }\n% Intermediate & Mobility \\\\\n% \\hline\\hline\n%1 & 0.20518 \\\\\n%2 & 0.20518 \\\\\n%3 & 0.20518 \\\\\n%4 & 0.20518 \\\\\n%5 & 0.20518 \\\\\n%6 & 0.20518 \\\\\n%7 & 0.20518 \\\\\n%8 & 0.20518 \\\\\n%9 & 0.20518 \\\\\n%10 & 0.20518 \\\\\n%11 & 0.20518 \\\\\\hline\n%12 & 0.18376 \\\\\n%13 & 0.18313 \\\\\n%14 & 0.18249 \\\\\n%15 & 0.18171 \\\\\n%16 & 0.18249 \\\\\n%17 & 0.18634 \\\\\n%18 & 0.18171 \\\\\n%19 & 0.18102 \\\\\n%20 & 0.18102 \\\\\n%21 & 0.18255 \\\\\n%22 & 0.18485 \\\\\\hline\n%34 & 0.14487 \\\\\n%35 & 0.14350 \\\\\n%36 & 0.14530 \\\\\n%37 & 0.14973 \\\\\n%38 & 0.14487 \\\\\n%39 & 0.14530 \\\\\\hline\n%66 & 0.07954 \\\\\\hline\n%83 & 0.00000 \\\\\n%\\end{tabular}\n%}\n%%\\caption{A table with the mobility of octahedral intermediates.}\n%\\end{figure}\n", "meta": {"hexsha": "e5205876511548de2d172508e8c850a820780a81", "size": 32971, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/constraints_backup.tex", "max_stars_repo_name": "Danie1Johnson/thesis", "max_stars_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "content/constraints_backup.tex", "max_issues_repo_name": "Danie1Johnson/thesis", "max_issues_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/constraints_backup.tex", "max_forks_repo_name": "Danie1Johnson/thesis", "max_forks_repo_head_hexsha": "cc1137b2ab121771a6937a903f835268a5ec313e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 86.9947229551, "max_line_length": 993, "alphanum_fraction": 0.7188741621, "num_tokens": 9635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8198933359135361, "lm_q2_score": 0.7662936377487305, "lm_q1q2_score": 0.6282790469431254}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage{enumitem}\n\\usepackage{physics}\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{1em}\n\\relpenalty=10000\n\\binoppenalty=10000\n\n\\begin{document}\n\n\\section*{Graph theory}\n\\subsection*{Terminology}\n\\begin{enumerate}\n\t\\item \n\tGraph $G(V,E)$ is set of \\emph{vertices} $V$ and \\emph{edges} $E$ (pairs of vertices). If there is an edge $(x,y) \\in E$, then $x$ and $y$ are said to be \\emph{connected}.\n\t\\item \n\tA \\emph{proper graph} is a graph with no more than one edge between each pair of vertices and in which no vertex is connected to itself.\n\t\\item \n\tProper graph in which each pair of vertices is connected is called a \\emph{complete graph}.\n\t\\item\n\tA \\emph{complement} graph $\\overline{G}$ has vertices equal to vertices of $G$ and its vertices $x$ and $y$ are connected if and only if $x$ and $y$ are not connected in $G$.\n\t\n\t\\item \n\tAn \\emph{oriented graph} is one in which the pairs in the graph are ordered.\n\t\\item\n\tThe vertices of a \\emph{k-partite graph} can be partitioned into $k$ non-empty disjoint sets in such way that there are no connections between vertices in each set.\n\t\\item \n\tThe \\emph{degree} of a vertex $x$ is the number of times $x$ is the endpoint of an edge. Then\n\t$$\\sum_{x \\in V} d(x) = 2 \\abs{E} $$\n\t\\item \n\tA \\emph{trajectory (or path)} is a sequence of vertices for which subsequent vertices are connected.\n\t\\item \n\tA \\emph{circuit} is a path that ends and starts in the same vertex.\n\t\\item \n\tA \\emph{cycle} is a circuit in which no vertex appears more than once (except the initial/final vertex).\n\t\\item\n\tIn a \\emph{connected graph} there exists a path between any two points of the graph.\n\t\\item \n\t\\emph{Tree} is a connected graph with no circuits. A connected graph with $n$ vertices is a tree if and only if it has $n-1$ edges.\n\t\\item \n\t\\emph{Euler path} is a path in which every edge of the graph appears exactly once. Likewise \\emph{Euler circuit} is a circuit in which every edge appears exactly once.\n\t\\item \n\tIf each vertex in a connected graph has even degree, then the graph contains an Euler circuit.\n\t\\item \n\tIf a connected graph has exactly two vertices with odd degree, it contains an Euler path.\n\t\\item \n\tA \\emph{Hamilton circuit} is a circuit in which each vertex appears exactly once.\n\t\\item \n\tA \\emph{planar graph} can be embedded in a plane with edges corresponding to non-intersecting lines (not necessarily straight). A planar graph with $n$ vertices has at most $3n-6$ edges.\n\t\n\\end{enumerate}\n\n\\newpage\n\\subsection*{Theorems}\n\t\\begin{enumerate}\n\t\t\\item \\textbf{Ramsey's theorem: } For positive integers $r_1,\\dots,r_k$, there exists a finite Ramsey number. \n\t\t\\begin{enumerate}\n\t\t \t\\item case $k=2$ \\\\\n\t\t \tRamsey number $R(r_1,r_2)$ is a minimal number of people at the party such that there are either $r_1$ people who know each other or $r_2$ people who do not.\n\t\t \t\\item General case for $k$ \\\\\n\t\t \tRamsey's number of $R(r_1,r_2,\\dots,r_k)$ is the minimal number of points in a complete graph whose edges are coloured with $k$ different colours, such that there exists a colour $i$ for which we can find $r_i$ vertices that are all connected with edges of colour $i$.\n\t\t\\end{enumerate}\n\t\t\n\t\t\\item \\textbf{Kuratowski's theorem} % http://www.math.caltech.edu/~2014-15/2term/ma006b/10%20Planar3.pdf\n\t\t\\begin{enumerate}\n\t\t\t\\item $K_n$ is a complete graph with $n$ vertices and $K_{n,m}$ is a complete bipartite graph with $n$ and $m$ vertices in the subsets.\n\t\t\t\\item Graphs $K_5$ and $K_{3,3}$ are non-planar.\n\t\t\t\\item Every non-planar graph contains a subgraph which can be obtained from one of these graphs by a subdivision of its edges.\n\t\t\n\t\t\\end{enumerate}\n\t\t\\item \\textbf{Hall's marriage theorem} % http://www.math.cornell.edu/~levine/18.312/alg-comb-lecture-16.pdf\n\t\t\\begin{enumerate}\n\t\t\t\\item In a set of $n$ women and $n$ men, a woman $a_i$ is happy to be married to any man in set $A_i$ and a man is happy to marry a woman who wants to marry him. Hall's marriage theorem states that each person can be happily married if and only if the sets $A_1,A_2,\\dots,A_n$ meet the marriage condition.\n\t\t\t\\item The marriage condition states that for any subset of women $I$, the number of men whom at least one woman would be happy to marry $\\abs{\\bigcup_{i \\in I}A_i} \\geq \\abs{I}$\n\t\t\\end{enumerate}\n\t\\end{enumerate}\n\n\\subsection*{Problems}\n\\begin{enumerate}\n\t\\item\n\tProve that at least one of $G$ and $\\overline{G}$ is connected.\n\t\n\t\\item \n\t\\emph{(Dirac's theorem).} Prove that a graph with $n$ vertices contains a Hamilton cycle if the degree of each vertex is at least $n/2$. \t\n\t\n\t\\item \n\t\\emph{(Euler's formula).} A convex polyhedron has $E$ edges, $F$ faces and $V$ vertices. Prove that $E+2=F+V$.\n\n\t\\item % http://www.math.olympiaadid.ut.ee/arhiiv/baltitee/bt1997/bw97est.pdf\n\tIn a forest each of n animals ($n \\geq 3$) lives in its own cave, and there is exactly one separate path\n\tbetween any two of these caves. Before the election for King of the Forest some of the animals\n\tmake an election campaign. Each campaign-making animal visits each of the other caves exactly\n\tonce, uses only the paths for moving from cave to cave, never turns from one path to another\n\tbetween the caves and returns to its own cave in the end of its campaign. It is also known that\n\tno path between two caves is used by more than one campaign-making animal.\n\t\\begin{enumerate}\n\t\t\\item Prove that for any prime $n$, the maximum possible number of campaign-making animals is $\\tfrac{n-1}{2}$\n\t\t\\item Find the maximum number of campaign-making animals for $n = 9$.\t\n\t\\end{enumerate}\n\t\n\t\\item % https://mks.mff.cuni.cz/kalva/usa/usoln/usol991.html\n\tCertain squares of an $n \\times n$ board are coloured black and the rest white. Every white square shares a side with a black square. Every pair of black squares can be joined by chain of black squares, so that consecutive members of the chain share a side. Show that there are at least $\\tfrac{n^2-2}{3}$ black squares.\n\t\n\t\n\t\\item % https://cms.math.ca/Competitions/OMC/archive/sol2006.pdf\n\tConsider a round-robin tournament with $2n + 1$ teams, where each team plays each other team exactly once. We say\n\tthat three teams $X$, $Y$ and $Z$, form a \\emph{cycle triplet} if $X$ beats $Y$ , $Y$ beats $Z$, and $Z$ beats $X$. There are no ties.\n\t\\begin{enumerate}\n\t\t\\item Determine the minimum number of cycle triplets possible.\n\t\t\\item Determine the maximum number of cycle triplets possible.\n\t\\end{enumerate}\n\t \n\t\\item % https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Shortlist_Problems/C3\n\tDefine a $k$-clique to be a set of $k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.\t\n\t\n\t\\item % Latin squares\n\tFor integers $r,n$ where $r 0, b > 1)\n \\]\n\n \\begin{equation*}\n\t\\left.\\begin{aligned}\n\t f(n) \\\\\n\t af(\\frac{n}{b}) \\\\\n\t a^2f(\\frac{n}{b^2}) \\\\\n\t \\vdots \\\\\n\t a^{\\log_b^n}f(1) = n^{\\log_b^a}\n\t\\end{aligned}\\right\\}\n\t\\pause\n\t\\sum = \\left\\{\\begin{aligned}\n\t n^{\\log_b^a}\t\\quad & q > 1 \\\\\n\t n^{\\log_b^a}\\log n \\quad & q = 1 \\\\\n\t f(n)\t\\quad & q < 1 \\quad \\\\\n\t\\end{aligned}\\right.\n\t\\pause\n\t\\quad \\left.\\begin{aligned}\n\t f(n) &= O(n^{E-\\epsilon}) \\\\\n f(n) &= \\Theta(n^E) \\\\\n f(n) &= \\Omega(n^{E+\\epsilon}) \n\t\\end{aligned}\\right.\n \\end{equation*}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Solving recurrences (Problem 1.2.13, 1.2.16)}\n \\begin{columns}\n\t\\column{0.40\\textwidth}\n\t \\begin{enumerate}\n\t\t\\item $\\Theta(n^{\\log_3^2})$\n\t\t\\item \\textcolor{red}{$\\Theta(\\log^2 n)$}\n\t\t\\item $\\Theta(n)$\n\t\t\\item $\\Theta(n \\log n)$\n\t\t\\item \\textcolor{red}{$\\Theta(n \\log^2 n)$}\n\t\t\\item $\\Theta(n^2)$\n\t\t\\item $\\Theta(n^{\\frac{3}{2}}\\log n)$\n\t\t\\item $\\Theta(n)$\n\t\t\\item $\\Theta(n^{c+1})$\n\t\t\\item $\\Theta(c^{n+1})$\n\t\t\\item \\textcolor{blue}{$\\Theta(n)$}\n\t \\end{enumerate}\n\t\\column{0.60\\textwidth}\n\t \\[\n\t\tT(n) = T(n/2) + \\log n\n\t \\]\n\n\t \\[\n\t\tT(n) = 2T(n/2) + n\\log n\n\t \\]\n\n\t\\pause\n\t\\begin{alertblock}{Reference}\n\t \\[\n\t\tf(n) = \\Theta(n^{\\log_b^a} \\lg^{k}n) \\Rightarrow \\Theta(n^{\\log_b^a} \\lg^{k+1}n)\n\t \\]\n\t\\end{alertblock}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Solving recurrences (Problem 1.2.13, 1.2.16)}\n \\[\n\tT(n) = T(n/2) + T(n/4) + T(n/8) + n\n \\]\n\n \\pause\n \\vspace{0.30cm}\n \\centerline{By recursion-tree.}\n\n \\pause\n \\vspace{0.50cm}\n \n \\centerline{Exercise: Prove it by Mathematical Induction.}\n\n \\pause\n \\begin{alertblock}{Reference}\n\t``On the Solution of Linear Recurrence Equations'' by Akra \\& Bazzi, 1996.\n\n\t\\[\n\t T(n) = \\sum_{i=1}^{k} a_i T(n/b_i) + f(n)\n\t\\]\n \\end{alertblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Gaps (Problem 1.2.16)}\n \\[\n\tT(n) = 2T(n/2) + \\frac{n}{\\log n} \\pause = \\Theta(n \\log \\log n)\n \\]\n\n \\pause\n\n The regularity condition in Case 3:\n \\[\n\tbf(n/c) \\le cf(n), \\text{ for some } c < 1 \\text{ and sufficiently large } n\n \\]\n\n \\[\n\tT(n) = T(n/2) + n(2 - \\cos n)\n \\]\n\n \\[\n\tn^{E} = n^0 \\quad f(n) = n(2 - \\cos n) = \\Omega(n^{0 + \\epsilon})\n \\]\n\n \\pause\n \\[\n\tn = 2\\pi k (k \\text{ odd}) \\Rightarrow c \\ge \\frac{3}{2}\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{(Problem 1.2.15)}\n \\begin{align*} \n\t\\text{T}(n) &= \\sqrt{n}\\ \\text{T}(\\sqrt{n})+n \\\\\n\t\t&= n^{\\frac{1}{2}}\\ \\text{T}\\left(n^{\\frac{1}{2}} \\right )+n \\\\\n\t\t&= n^{\\frac{1}{2}}\\left( n^{\\frac{1}{2^2}}\\ \\text{T}\\left(n^{\\frac{1}{2^2}} \\right )+n^{\\frac{1}{2}} \\right )+n\\\\\n\t\t&= n^{\\frac{1}{2}+\\frac{1}{2^2}}\\ \\text{T}\\left(n^{\\frac{1}{2^2}}\\right ) +2n\\\\\n\t\t&= n^{\\frac{1}{2}+\\frac{1}{2^2}}\\left(n^{\\frac{1}{2^3}}\\ \\text{T}\\left(n^{\\frac{1}{2^3}}\\right ) +n^{\\frac{1}{2^2}} \\right )+2n\\\\\n\t\t&= n^{\\frac{1}{2}+\\frac{1}{2^2}+\\frac{1}{2^3}}\\ \\text{T}\\left(n^{\\frac{1}{2^3}}\\right ) +3n\\\\ \n\t\t&= \\cdots \\\\ \n\t\t&= n^{\\sum_{i=1}^{k}\\frac{1}{2^i}}\\ \\text{T}\\left(n^{\\frac{1}{2^k}}\\right ) +kn\\\\ \n \\end{align*}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{(Problem 1.2.15)}\n \\[\n\tn^{\\frac{1}{2^k}} = 2 \\Rightarrow k = \\log\\log n\n \\]\n\n \\pause\n\n \\begin{align*} \n\t\\text{T}(n) &= n^{\\sum_{i=1}^{k}\\frac{1}{2^i}}\\ \\text{T}\\left(n^{\\frac{1}{2^k}}\\right ) +kn\\\\ \n\t\t&=n^{\\sum_{i=1}^{\\log\\log n}\\frac{1}{2^i}}\\ \\text{T}(2) + n \\log\\log n\\\\ \n \\end{align*}\n\n \\pause\n \\[\n\t\\sum_{i=1}^{\\log_2\\log_2(n)}\\frac{1}{2^i} < 1 \\Rightarrow T(n) = \\Theta(n \\log \\log n)\n \\]\n\n \\pause\n \\vspace{0.30cm}\n \\centerline{Exercise: Prove it by Mathematical Induction.}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{(Problem 1.2.15)}\n \\[\n\tT(n) = \\sqrt{n} T(\\sqrt{n}) + n\n \\]\n\n \\[\n\tn = 2^k \\quad \\sqrt{n} = 2^{k/2} \\quad k = \\log n\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "25c04e62c378b6908fc57b0daea1bd2162b2a27b", "size": 3896, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-math-dc-20170329/sections/recurrences.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-math-dc-20170329/sections/recurrences.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-math-dc-20170329/sections/recurrences.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 24.049382716, "max_line_length": 131, "alphanum_fraction": 0.4969199179, "num_tokens": 1761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.82893881677331, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6281651740161535}} {"text": "% Copyright (C) He Guanyuming 2020\n% The file is licensed under the MIT license.\n\n\\section{Set Theory}\n\n\\subsection{Fundamentals}\n\\declareexercise{3.1.1}\n\\begin{proof}\nReflexive: $\\forall x \\in S, x \\in S$.\n\nSymmetric: \n\\begin{align*}\nX = Y\n&\\Longleftrightarrow \\\\\n\\forall x \\in X, x \\in Y \\wedge \\forall x \\in Y, x \\in X\n&\\Longleftrightarrow \\\\\nY = X\n\\end{align*}\n\nTransitive:\n$X=Y \\Longrightarrow \\forall x \\in X, x \\in Y$. Because $x \\in Y$ and $Y = Z$, we can conclude that \n$\\forall x \\in X, x \\in Z$. Conduct the process from inversely, we can get $\\forall x \\in Z, x \\in X$. \nTherefore, $X=Z$.\n\\end{proof}\n\nThe reason for the content beneath Axiom 3.2 is clearly demonstrated in the proof of Lemma 3.1.6.\n\nIn Remarks 3.1.9, there are three ``Why''s. The reason can be concluded as: Because of the ``if and only \nif'' in Axiom 3.3, or more precisely, ``only if'', if $x$ is a element in one of such sets, $x$ must \n$=a$ or $b$. And because of the ``if'', $x$ is thus in another set. So the two sets are equal according \nto Definition 3.1.4.\n\n\\declareexercise{3.1.2}\n\\begin{proof}\nAccording to Axiom 3.2, $\\varnothing$ exists, and is thus an object as stated by Axiom 3.1. Therefore, \nby Axiom 3.3, $\\{\\varnothing\\}$ also exists. $\\varnothing$ is an element of $\\{\\varnothing\\}$, but it \nis not an element of $\\varnothing$ because any object $\\notin$ $\\varnothing$.\n\nFor the same reason, any set that contains element(s) is not the same set as $\\varnothing$. Furthermore, \nthere exists an object $\\{\\varnothing\\}$ (Axiom 3.3 and 3.1), which is an element of \n$\\{\\varnothing, \\{\\varnothing\\}\\}$, but which is not an element of $\\{\\varnothing\\}$. So the two sets \nare not equal.\n\\end{proof}\n\n\\paragraph{Remarks 3.1.12}\n\\begin{proof}\nLet $x \\in A'\\cup B$. $x \\in A' \\Longrightarrow x \\in A$ And if $x \\notin A'$, $x \\in B$. So either way \n$x \\in A\\cup B$ and vice versa.\n\\end{proof}\n\n\\declareexercise{3.1.3}\n\\begin{proof}\n(1)\n\\[\nx \\in A \\cup B \\equiv (x \\in A \\vee x \\in B)\n\\]\n\\[\nx \\in A \\Longrightarrow x \\in B \\cup A\n\\]\n\\[\nx \\in B \\Longrightarrow x \\in B \\cup A\n\\]\nSo $x \\in A \\cup B \\Longrightarrow x \\in B \\cup A$. And vice versa.\n\n(2)\n$x \\in A \\Rightarrow x \\in A \\cup A$ and $x \\in A \\cup A \\Rightarrow x \\in A$.\n\n(3)\n\\begin{align*}\nx \\in A \\cup \\varnothing \n&\\Longrightarrow \\\\\nx \\in A \\vee x \\in \\varnothing\n&\\Longrightarrow \\\\\nx \\in A \\tag{$\\forall a, a \\notin \\varnothing$}\n\\end{align*}\n\nAnd obviously $x \\in A \\Rightarrow x \\in A \\cup \\varnothing$. So $A \\cup \\varnothing = A$.\n\nBy transitivity of equality, and commutativity of pairwise union, we can conclude the others.\n\\end{proof}\n\n\\paragraph{Examples 3.1.17}\n\\begin{proof}\n\\[\n\\forall x(x\\in A \\Longrightarrow x \\in A)\n\\]\n\nAnd\n\\[\n\\forall x(x \\in \\varnothing \\Longrightarrow x \\in A)\n\\]\nis vacuously true.\n\\end{proof}\n\n\\declareexercise{3.1.4}\n\\begin{proof}\n(1) \nOn one hand,\n\\[\nA \\subseteq B \\equiv \\forall x(x \\in A \\Longrightarrow x \\in B)\n\\]. \nOn the other hand,\n\\[\nB \\subseteq A \\equiv \\forall x(x \\in B \\Longrightarrow x \\in A)\n\\].\nThus $A=B$.\n\n(2)\nFirst, we prove that $A \\subsetneq B \\Longrightarrow \\exists x(x \\in B \\wedge x \\notin A)$. Suppose the \ncontradiction, that is, $\\forall x(x\\in B \\Longrightarrow x \\in A)$, which is impossible since \n$(A \\subseteq B \\equiv \\forall x(x\\in A \\Longrightarrow x \\in B))\\wedge A \\neq B$.\n\nAccording to what's proven in the book, \n$A \\subsetneq B \\wedge B \\subsetneq C \\Longrightarrow A \\subseteq C$.\n\nNow we prove that $\\exists x(x \\in C \\wedge x \\notin A)$. Since $x \\in A \\Longrightarrow x \\in B$, \n$x \\notin B \\Longrightarrow x \\notin A$. Because $B \\subsetneq C$, \n$\\exists x(x \\in C \\wedge x \\notin B)$, and thus for such $x$, $ x \\notin A$. Then $A \\neq C$.\n\nSo $A \\subsetneq C$. \n\\end{proof}\n\n\\paragraph{Axiom 3.5}\n(1) Because $x \\in \\{x \\in A : P(x)\\} \\Rightarrow x \\in A$.\n\n(2) Because both $\\in$ and $P(x)$ obey the axiom of substitution.\n\n\\declareexercise{3.1.5}\n\\begin{proof}\nFirst we prove that $A \\subseteq B \\equiv A \\cup B = B$. On one hand,\n\\begin{align*}\nA \\subseteq B \n&\\equiv \\\\\n\\forall x(x \\in A \\Longrightarrow x \\in B)\n&\\Longrightarrow \\\\\n\\forall x((x \\in A \\vee x \\in B) \\Longrightarrow x \\in B)\n&\\equiv \\\\\nA \\cup B = B\n\\end{align*}.\n\nOn the other hand,\n\\[\n\\forall x((x \\in A \\vee x \\in B) \\Longrightarrow x \\in B) \\Longrightarrow\n\\forall x(x \\in A \\Longrightarrow x \\in B)\n\\]. The statement is therefore proven.\n\nThen we prove that $A \\subseteq B \\equiv A \\cap B = A$. On one hand, \n\\begin{align*}\n(A \\cap B = A \\equiv \\forall x(x \\in A \\wedge x \\in B \\equiv x \\in A)) \n&\\Longrightarrow \\\\\n(\\forall x(x \\in A \\Rightarrow x \\in B) \\equiv (A \\subseteq B))\n\\end{align*}.\n\nOn the other hand, \n\\[\n\\forall x(x\\in A\\wedge x \\in B \\Longrightarrow x \\in A)\n\\]\nis always true (Vacuously true if $x \\notin B$).\n\nLogical equality is transitive, and thus all of the three statements are equal.\n\\end{proof}\n\n\\paragraph{Proposition 3.1.28} (Exercise 3.1.6) \\label{exercise3.1.6}\n\\begin{proof}\n(a) The two are identical to \n\\[\n\\forall x(x \\in A \\vee x \\in \\varnothing \\equiv x \\in A)\n\\], \nand \n\\[\n\\nexists x(x \\in A \\wedge x \\in \\varnothing)\n\\], \nwhich are all true since $\\forall x(x \\notin \\varnothing)$.\n\n(b) We have $A \\subseteq X$. According to what we have proven in \n\\hyperref[exercise3.1.5]{Exercise 3.1.5}, the two statements are all true.\n\n(c) Obvious since \n\\[\n\\forall x(x \\in A \\vee x \\in A \\equiv x \\in A)\n\\]\nand \n\\[\n\\forall x(x \\in A \\wedge x \\in A \\equiv x \\in A)\n\\]\n\n(d) All true since \\emph{logical or} and \\emph{logical and} are commutative.\n\n(e) See Lemma 3.1.13. I believe that this can be concluded by the fact that \\emph{logical or} and \n\\emph{logical and} are also associative.\n\n(f) \nFirst we prove the latter. On one hand, suppose \n\\[\nx \\in A \\cup (B \\cap C)\n\\] is ture.\n\nIf $x \\in A$, then $x \\in$ both $A \\cup B$ and $A \\cup C$, and thus $\\in$ \n$(A \\cup B)\\cap(A \\cup C)$.\n\nIf $x \\notin A$, then $x \\in B \\cap C$, then $x \\in$ both $A \\cup B$ and $A \\cup C$, and thus $\\in$ \n$(A \\cup B)\\cap(A \\cup C)$.\n\nOn the other hand, \nsuppose \n\\[\nx \\in (A \\cup B)\\cap(A \\cup C)\n\\] is true.\n\nIf $x \\in A$, obviously $x \\in A \\cup (B \\cap C)$.\n\nIf $x \\notin A$, then $x$ must $\\in B \\cap C$, and thus also $\\in A \\cup (B \\cap C)$.\n\nNow we prove the former. On one hand, suppose\n\\[\nx \\in A \\cap (B \\cup C)\n\\] is true.\n\nIf $x \\in A \\wedge x \\in B$, then $x \\in A \\cap B$, and thus $\\in (A \\cap B)\\cup(A \\cap C)$.\n\nIf $x \\notin A \\vee x \\notin B$, then\n\\begin{enumerate}\n\\item if $x \\notin A$, this is impossible.\n\\item if $x \\in A$, then $x \\notin B$. But $x \\in B \\cup C$, so $x \\in C$. And thus \n$x \\in A \\cap C \\Rightarrow x \\in (A \\cap B)\\cup(A \\cap C)$.\n\\end{enumerate}\n\nOn the other hand, suppose that \n\\[\nx \\in (A \\cap B)\\cup(A \\cap C)\n\\] is true.\n\nFirst we can see that $x \\in A$. \n\nIf $x \\in B$, then $x \\in B \\cup C$, and thus $\\in A \\cap (B \\cup C)$.\n\nIf $x \\notin B$, then $x \\in C$. So $x \\in B \\cup C$, and thus $\\in A \\cap (B \\cup C)$.\n\n(g)\nNow we prove the former: On one hand, suppose that \n\\[\nx \\in A \\cup (X-A)\n\\].\n\nIf $x \\in A$, then $x \\in X$ since $A \\subseteq X$.\n\nIf $x \\notin A$, then $x \\in X-A$, and thus also $\\in X$.\n\nOn the other hand, suppose that \n\\[\nx \\in X\n\\]\n\nIf $x \\in A$, then $x \\in A \\cup (X-A)$.\n\nIf $x \\notin A$, then $x \\in X-A$, and thus $\\in A \\cup (X-A)$.\n\n(h)\n$x \\in X - A$ requires $x \\notin A$. So $\\forall x(x \\in A \\cap (X-A))$ is always false. \nThus\n\\[\n\\forall x(x \\in A \\cap (X-A) \\Longleftrightarrow x \\in \\varnothing)\n\\] (vacuously true).\n\\end{proof}\n\n\\declareexercise{3.1.7}\n\\begin{proof}\n(1)\n$\\forall x(x \\in A \\cap B \\Longrightarrow x \\in A)$. Similarly, we can prove that \n$A \\cap B \\subseteq B$. (This can also be achieved via the commutativity).\n\n(2) \nOn one hand, suppose that \n\\[\nC \\subseteq A \\wedge C \\subseteq B\n\\] is true.\nThen, \n\\[\n\\forall x(x \\in C \\Longrightarrow x \\in A \\wedge x \\in B \\Longrightarrow x \\in A \\cap B)\n\\].\n\nOn the other hand, suppose that\n\\[\nC \\subseteq A \\cap B\n\\] is true.\nThen, \n\\[\n\\forall x(x \\in C \\Longrightarrow x \\in A \\wedge x \\in B)\n\\].\nThat is, $C \\subseteq A \\wedge C \\subseteq B$.\n\n(3) It is immediately given by \n\\[\n\\forall x(x \\in A \\Longrightarrow x \\in A \\cup B)\n\\]. \nSince $\\cup$ is commutative, the latter case is proven.\n\n(4) On one hand, suppose that $A \\subseteq C \\wedge B \\subseteq C$ and let $x \\in A \\cup B$.\n\nIf $x \\in A$, then $x \\in C$.\n\nIf $x \\notin A$, then $x \\in B$, and thus $x \\in C$.\n\nOn the other hand, suppose that $A \\cup B \\subseteq C$. Then, \n\\[\n\\forall x(x \\in A \\Longrightarrow x \\in A \\cup B \\Longrightarrow x \\in C)\n\\]\n\\[\n\\forall x(x \\in B \\Longrightarrow x \\in A \\cup B \\Longrightarrow x \\in C) \\qedhere\n\\].\n\\end{proof}\n\n\\declareexercise{3.1.8}\n\\begin{proof}\nThe former: On one hand, \nSuppose that \n\\[\nx \\in A \\cap (A \\cup B)\n\\].\n\nIf $x \\in A$, then $x \\in A$.\n\nIf $x \\notin A$, this is impossible.\n\nOn the other hand, suppose that $x \\in A$.\nThen $x \\in A \\wedge x \\in (A \\cup B)$, so $x \\in A \\cap (A \\cup B)$.\n\nThe latter: On one hand, suppose that $x \\in A \\cup (A \\cap B)$.\n\\[\nx \\in A \\Longrightarrow x \\in A.\n\\]\n\\[\nx \\notin A \\Longrightarrow x \\in (A \\cap B) \\Longrightarrow x \\in A\n\\].\n\nOn the other hand, Suppose that $x \\in A$, then $x \\in A \\cup (A \\cap B)$.\n\\end{proof}\n\n\\declareexercise{3.1.9}\n\\begin{proof}\n\\begin{lem} \\label{lem10}\n\\[\n\\nexists x\\forall B\\forall A(x \\in A \\wedge x \\in B \\wedge A \\cap B = \\varnothing)\n\\]\n\\end{lem}\n\\begin{proof}\nSuppose the negation: $x \\in A \\wedge x \\in B \\wedge A \\cap B = \\varnothing$, then \n$x \\in A \\cap B$, and thus $\\in \\varnothing$, which is impossible.\n\\end{proof}\n\nThe former: On one hand, suppose that $x \\in A$. Then $x \\notin B$ by Lemma \\ref{lem10}. And\n\\[\nx \\in A \\Longrightarrow x \\in A \\cup B \\Longrightarrow x \\in X\n\\].\nSo $x \\in (X-B)$.\n\nOn the other hand, suppose that $x \\in (X-B)$, then $x \\in A \\cup B$. But $x \\notin B$, so $x \\in A$ by \nLemma \\ref{lem10}.\n\nThe latter is immediately proven since $\\cap,\\cup$ are commutative.\n\\end{proof}\n\n\\paragraph{Exercise 3.1.10} \\label{exercise3.1.10}\n\\begin{proof}\nFirstly we prove that $(A-B)\\cap (A\\cap B) = \\varnothing$.\n\n$x \\in (A\\cap B)$ gives $x \\in B$, but $x \\in (A-B)$ gives $x \\notin B$. So the two statements can not \nbe true simultaneously. Which means \n\\[\nx\\in (A-B)\\cap (A\\cap B) \\Longrightarrow x \\in \\varnothing\n\\]\n\nAnd obviously \n\\[\nx\\in (A-B)\\cap (A\\cap B) \\Longleftarrow x \\in \\varnothing\n\\].\n\nSimilarly we can conclude all of the three sets are disjoint by the fact that $\\nexists x \\in$ either \ntwo of the three sets.\n\nNow we are showing that their union is $A \\cup B$.\n\nOn one hand, suppose that \n\\[\nx \\in (A-B)\\cap(A\\cap B)\\cap(B-A)\n\\].\n$x$ can at most be in one of these sets since they are disjoint.\nIf $x \\in A$, then $x \\in A \\cup B$.\n\nIf $x \\notin A$, then $x \\in (B-A)$, and thus $x \\in B$. So $x \\in A \\cup B$.\n\nOn the other hand, suppose that $x \\in A \\cup B$.\nThen $x$ either\n\\begin{enumerate}\n\\item $\\in A$, but $\\notin B$, or\n\\item $\\in B$, but $\\notin A$, or\n\\item $\\in$ both $A,B$.\n\\end{enumerate}\n\nIf (1), then $x \\in (A-B)$.\n\nIf (2), then $x \\in (B-A)$.\n\nIf (3), then $x \\in A\\cap B$.\n\nIn conclusion, we can see that $x \\in (A-B)\\cap(A\\cap B)\\cap(B-A)$.\n\\end{proof}\n\n\\paragraph{Exercise 3.1.11} \\label{exercise3.1.11}\n\\begin{proof}\nLet $S$ be a set.\nLet $P(x,y)$ be a property pertaining to $x \\in S$ and any object $y$, and is true iff \n$Q(x) \\wedge y = x$, where $Q(x)$ is a property pertaining to $x \\in S$.\n\nAccording to Axiom 3.6, there exists a set $Z$, such that \n$y \\in Z \\equiv x \\in S \\wedge P(x,y)$, which means $y \\in Z \\equiv x \\in S \\wedge Q(x) \\wedge x = y$. \nSo is the axiom of specification proven.\n\\end{proof}\n\n\\subsection{Russell's paradox}\nI think one major reason for building such a ``cumbersome'' axiom system is to restrict the way to \nconstruct sets. We can not construct just any set we want, there only exist certain kinds of sets.\n\n\\declareexercise{3.2.1}\n\\begin{proof}\n(Axiom 3.2) To prove the existence of the empty set, simply choose a property that is false for all \nobjects.\n\n(Axiom 3.3) To prove the existence of a \\emph{pair set}, say $\\{a,b\\}$, let $P(x)$ be a property \npertaining to any object $x$, and is true iff $x = a \\vee x = b$.\n\n(Axiom 3.4) Let the property be $P(x): x \\in A \\vee x \\in B$.\n\n(Axiom 3.5) Let the property be $Q(x): x \\in A \\wedge P(x)$, where $P(x)$ is a property pertaining to \nelements of $A$.\n\n(Axiom 3.6) Let the property be $Q(y): P(x,y)$ is true for some $x \\in A$.\n\\end{proof}\n\n\\declareexercise{3.2.2}\n\\begin{proof}\n(1)\nSuppose the negation: $\\exists A(A \\in A)$. Then by Axiom 3.3, construct a set $B:= \\{A\\}$. $A$ is \nthe only element in $B$. $A$ is a set. $A$ is not disjoint from $B$, for $A \\in A \\wedge A \\in B$. \n\n(2)\nSuppose the negation: $A \\in B \\wedge B \\in A$. Construct a set $S: \\{A,B\\}$. $A$ is an element of \n$S$. $A$ is a set. $A$ is not disjoint from $S$, for $B \\in A \\wedge B \\in S$.\n\\end{proof}\n\n\\declareexercise{3.2.3}\nOn one hand, if Axiom 3.8 is true, we can choose a property $P(x)$ which is true for all objects. Thus \nwe have $\\Omega$.\n\nOn the other hand, if there exists such a set as $\\Omega$, we can use Axiom 3.5 to construct any set \nwe want from it. (e.g. If we want a set to have these elements: $a,b,\\dots$, we can let \n$P(x):= x = a \\vee x = b, \\vee \\dots$.)\n\n\\subsection{Functions}\nIn Example 3.3.3, Tao asked why $x'=x \\Rightarrow f(x')=f(x)$. The reason is, the property $P(x,y)$ \nobeys the axiom of substitution. Thus, $P(x,y)\\equiv P(x',y)$. According to definition, since \n$x' \\in X$, $y$ is unique.\n\nIn Example 3.3.9, Tao asked why all functions whose domain is $\\varnothing$ and whose range is the same \nare equal. The reason is $x \\in \\varnothing \\Longrightarrow f(x) = g(x)$ is vacuously true.\n\n\\declareexercise{3.3.1}\n\\begin{proof}\nThe properties of equality are all true since in definition, we only use $f(x) = g(x)$, in which the $=$ \nobeys these rules, plus the fact that the output is unique.\n\nThen the substitution:\n\\begin{align*}\nf = \\overset{\\sim}{f} &\\Longrightarrow \\\\\nf(x) = \\overset{\\sim}{f}(x) &\\Longrightarrow \\\\\ng(f(x)) = g(\\overset{\\sim}{f}(x))\n\\end{align*}. \nAnd then $\\overset{\\sim}{g}(\\overset{\\sim}{f}(x)) = g(\\overset{\\sim}{f}(x)) = g(x)$.\n\\end{proof}\n\n\\declareexercise{3.3.2}\n\\begin{proof}\nThe former: Suppose the negation:\n\\[\n\\exists x \\exists x'(g(f(x)) = g(f(x')) \\wedge x \\neq x')\n\\]\nThen, \n\\begin{align*}\ng(f(x)) = g(f(x')) &\\Longrightarrow \\\\\nf(x)= f(x') \\tag{$g$ is injective} &\\Longrightarrow \\\\\nx = x' \\tag{$f$ is injective}\n\\end{align*}, \nwhich is impossible.\n\nThe latter: Suppose the negation:\n\\[\n\\exists z \\forall x(z \\in Z \\wedge g \\circ f(x) \\neq z) \n\\]\nThen, we can conclude that $\\exists y \\forall x(y \\in Y \\wedge y \\neq f(x))$, since $g$ is surjective. \nThis is impossible as $f$ is surjective.\n\\end{proof}\n\n\\declareexercise{3.3.3}\n\\begin{proof}\n\\begin{large}\n\\textbf{Attention:}\n\\end{large}\nDifferent interpretations for injectivity may result in different conclusions. \nI have asked a question at \n\\href{https://math.stackexchange.com/questions/3800240/how-to-interpret-the-definition-of-injectivity}{Stack Exchange} regarding this problem.\n\nLet the range be $Y$, and the function be $f$.\nInjectivity:\n\\[\n\\forall x'\\forall x((x \\in \\varnothing \\wedge x' \\in \\varnothing) \\Longrightarrow\n(x \\neq x' \\Longrightarrow f(x) \\neq f(x')))\n\\], \nwhich is always vacuously true.\n\nSurjectivity:\n\\[\n\\forall y(y \\in Y \\Longrightarrow \\exists x(x \\in \\varnothing \\wedge f(x) = y))\n\\], \nwhich is false if $Y \\neq \\varnothing$, and which is vacuously true if $Y = \\varnothing$.\n\nBijective: True if $Y = \\varnothing$.\n\\end{proof}\n\n\\declareexercise{3.3.4}\n\\begin{proof}\nThe former: $f,\\overset{\\sim}{f}$ have the same range and domain. \n\\[\n\\forall x(g \\circ f = g \\circ \\overset{\\sim}{f} \\Longrightarrow g(f(x)) = g(\\overset{\\sim}{f}(x)))\n\\]\nWe know that $g$ is injective, so $\\forall x \\in X, f(x) = \\overset{\\sim}{f}(x)$. Thus \n$f = \\overset{\\sim}{f}$.\n\nIt is not true if $g$ is not injective. Consider an extreme condition, where $g$ is constant. So \nwhatever $f,\\overset{\\sim}{f}$ are, $g \\circ f = g \\circ \\overset{\\sim}{f}$ are always equal.\n\nThe latter: Suppose the negation:$g \\neq \\overset{\\sim}{g}$.\n$g,\\overset{\\sim}{g}$ have the same range and domain. But they are not equal, so \n$\\exists y(y \\in Y \\wedge g(y) \\neq \\overset{\\sim}{g}(y))$. Because $f$ is surjective, \n$\\exists x(x \\in X \\wedge f(x) = y)$. However, $g \\circ f(x) = \\overset{\\sim}{g} \\circ f(x)$, so \nthis is impossible.\n\nIt is not true if $f$ is not surjective. We can make $g(y) = \\overset{\\sim}{g}(y)$ when $y=f(x)$, but \nas well make $g(y') \\neq \\overset{\\sim}{g}(y')$ if $\\nexists x(y'=f(x))$.\n\\end{proof}\n\n\\declareexercise{3.3.5}\n\\begin{proof}\nInjectivity:\nSuppose the negation, that \n\\[\n\\exists x \\exists x'(x \\neq x' \\wedge f(x) = f(x'))\n\\], which immediately gives \n\\[\ng(f(x)) = g(f(x'))\n\\], and thus is impossible.\n\n$g$ has not to be also injective, because $f$ being so ensures that an unique input $x$ gives an unique \ninput to $g$.\n\nSurjectivity:\nIf $g$ is not surjective, then $\\exists z \\forall y(z \\in Z \\wedge y \\in Y \\wedge z \\neq g(y))$\nAnd whatever $x$ is, $f(x) \\in Y$, so $g(f(x)) \\neq z$, which is a contradiction.\n\n$f$ has not to be surjective as long as its ``real'' domain is large enough to form the set $Z$ through \n$g$. For example (Informal), let $g$ be $z = |y|, \\mathbb{R} \\rightarrow \\mathbb{R}^{+}\\cup\\{0\\}$, and \nlet $f$ be $y = x, \\mathbb{R}^{+}\\cup\\{0\\} \\rightarrow \\mathbb{R}$.\n\\end{proof}\n\n\\declareexercise{3.3.6}\n\\begin{proof}\nThe latter:\nBy definition, $P(y,x)$ of $x = f^{-1}(y)$ is $f(x)=y$. Substitute $x$ with $f^{-1}(y)$, and here we \nhave $f(f^{-1}(y)) = y$, where $y \\in Y$.\n\nThe former: Let $y = f(x)$. According to what we have proven, \\\\\n$f(f^{-1}(y)) = y$. Substitute $y$ with \n$f(x)$, we have $f(f^{-1}(f(x))) = f(x)$. Since that $f(x)$ is injective, we have $f^{-1}(f(x)) = x$.\n\nNow we need to show that $f^{-1}$ is bijective. Assume that it is not injective, thus \n$\\exists x \\exists x'(x \\in Y \\wedge x' \\in Y \\Longrightarrow(x\\neq x' \\Longrightarrow f^{-1}(x) = \nf^{-1}(x')))$.\nHowever, according to the latter conclusion, $f^{-1}(x) = f^{-1}(x') \\Longrightarrow x=x'$, a \ncontradiction, so $f^{-1}$ must be injective.\n\nAnd it is also surjective. $\\forall x \\in X$, $\\exists y \\in Y, f^{-1}(y)=x$. According to the former \nconclusion, $y$ is $f(x)$.\n\nSo now $f^{-1}$ is bijective, and thus has its inverse. By definition, $P(x,y)$ of \n$y = (f^{-1})^{-1}(x)$ is $f^{-1}(y) = x$, where $x \\in X$. According to the former conclusion, \n$f^{-1}(f(x)) = x$. Thus\n\\[\nf^{-1}(y) = f^{-1}(f(x)) \\Longrightarrow y = f(x) \\Longrightarrow (f^{-1})^{-1}(x) = f(x)\n\\], which is true $\\forall x \\in X$. And since they have the same domain and range, $(f^{-1})^{-1} = f$.\n\\end{proof}\n\n\\declareexercise{3.3.7}\n\\begin{proof}\nInjectivity: \n\\[\ng \\circ f(x) = g \\circ f(x') \\Longrightarrow f(x) = f(x') \\Longrightarrow x = x'\n\\]\n\nSurjectivity:\nFor each $z \\in Z$, we need to find $x \\in X$ such that $g \\circ f(x) = z$. By the surjectivity of $g$, \nwe can find $y \\in Y$ such that $g(y) = z$. We can also find $a \\in X$ such that $f(a) = y$ as $f$ is \nsurjective. So $a$ is our desired $x$.\n\nThe $P(z,x)$ of $x = (g \\circ f)^{-1}(z)$ is $z = g \\circ f(x)$. Consider the following expression:\n\\begin{align*}\nf^{-1} \\circ g^{-1} (z)\n&= f^{-1} \\circ g^{-1} (g \\circ f(x)) \\\\\n&= f^{-1}(\\textcolor{red}{g^{-1}(g(}f(x)\\textcolor{red}{))}) \\\\\n&= f^{-1}(f(x)) \\\\\n&= x\n\\end{align*}\nSo $(g \\circ f)^{-1} = f^{-1} \\circ g^{-1}$. Therefore, they are equal as they have the same domain and \nrange.\n\\end{proof}\n\n\\declareexercise{3.3.8}\n\\begin{proof}\n(a) First they have the same domain and range. Finally, \n\\[\n\\forall x(x\\in X \\Longrightarrow x=x \\Longrightarrow \\iota_{Y \\rightarrow Z} \\circ \n\\iota_{X \\rightarrow Y} = \\iota_{X \\rightarrow Z})\n\\]\n\n(b) On one hand, they have the same domain and range.\n\nOn the other hand, \n\\begin{align*}\nf \\circ \\iota_{A \\rightarrow A}(x) \n&= f(\\iota_{A \\rightarrow A}(x)) \\\\\n&= f(x) \\\\\n&= \\iota_{B \\rightarrow B}(f(x)) \\\\\n&= \\iota_{B \\rightarrow B} \\circ f (x)\n\\end{align*}\n\n(c) It is easy to see that they have the same domain and range. \n\n\\[\nf \\circ f^{-1} (b) = b = \\iota_{B \\rightarrow B}\n\\]\n\\[\nf \\circ f^{-1} (a) = a = \\iota_{A \\rightarrow A}\n\\]\n\n(d) It is easy to see that they have the same domain and range.\n\nLet $h$ be \n$h(x) = f(x)$, if $x \\in X$, $h(x) = g(x)$, if $x \\in Y$. \n\nFor each $x \\in X$, $\\iota_{X \\rightarrow X \\cup Y}(x) = x$, so \n$h(\\iota_{X \\rightarrow X \\cup Y}(x)) =f(x)$.\n\nSimilarly we can prove $h(\\iota_{Y \\rightarrow X \\cup Y}(x)) = g(x)$ for each $x \\in Y$.\n\\end{proof}\n\n\\subsection{Images and inverse Images}\n\\paragraph{Definition 3.4.1}\nTo prove that $f(S)$ is well-defined by using the axiom of specification, we need to apply it to set \n$Y$, not $X$. Let $P(y)$ be a property pertaining to each $y \\in Y$, which is true iff \n$\\exists x(x \\in S \\wedge f(x) = y)$. According to the axiom of specification, there exists a set that \ncontains every $y \\in Y$ such that $P(y)$ is true.\n\nIn some places where Tao asked ``(Why?)'', the reason is obvious, so I don't write them here. \n\n\\paragraph{Example 3.4.6}\nThis is because \n\\[\nf^{-1}(f(\\{-1,0,1,2\\})) = f^{-1}(\\{1,0,4\\}) = \\{-1,1,0,2,-2\\}\n\\].\n\nMore generally, if $f$ (whose domain is $X$, and whose range is $Y$) is not injective, then \n\\[\n\\exists x \\exists x'((x \\in X \\wedge x' \\in X) \\wedge (x\\neq x' \\wedge f(x) = f(x')))\n\\]. \nLet $D \\subseteq X$ such that $x \\in D \\wedge x' \\notin D$. Then $f(x) = f(x') \\in f(D)$. And thus \n$x,x' \\in f^{-1}(f(D)) \\Longrightarrow f^{-1}(f(D)) \\neq D$. \n\n\\declareexercise{3.4.1}\n\\begin{proof} \n$f^{-1}(V)$ may be interpreted in two different ways:\n\n(1) Interpret $f^{-1}(V)$ as an inverse image, that is,\n\\[\n(\\forall x \\in X)(x \\in f^{-1}(V) \\equiv f(x) \\in V)\n\\]\n\\[\n(\\forall x \\notin X)(x \\notin f^{-1}(V))\n\\]\n\n(2) Interpret $f^{-1}(V)$ as an image, where we regard $f^{-1}$ as a function. So, \n\\[\n\\forall x(\\exists y(y \\in V \\wedge x = f^{-1}(y)) \\equiv x \\in f^{-1}(V))\n\\]\n\nWe need to show that if the two statements are well-defined($x \\in X$) , they are logically equivalent.\n\nLet $S_1$ be the set defined in form (1), $S_2$ be the set defined in form (2). For every $x \\in S_1$, \n$f(x) \\in V$. Let $y' = f(x), x' = f^{-1}(y')$, then by definition (2) we have $x' \\in S_2$. But \n$x' = x$, so $\\forall x(x \\in S_1 \\Longrightarrow x \\in S_2)$.\n\nOn the other hand, for every $a \\in S_2$, $\\exists b \\in V$, such that $a = f^{-1}(b)$. Then $a \\in X$. \n$f(a) = f(f^{-1}(b)) = b \\in V$, so $a \\in S_1$. Thus, $S_1 = S_2$.\n\\end{proof}\n\n\\declareexercise{3.4.2}\n(1) Generally we can say $S \\subseteq f^{-1}(f(S))$ but we cannot say that they are equal; (2) we can \nsay $f(f^{-1}(U)) \\subseteq U$ but we cannot say that they are equal.\n\\begin{proof}\n(1) $x \\in S \\Longrightarrow f(x) \\in f(S) \\Longrightarrow x \\in f^{-1}(f(S))$. However, it is possible \nthat $\\exists x(x \\in X \\wedge x \\notin S \\wedge f(x) \\in f(S))$\n\n(2) \n\\[\nx \\in f^{-1}(U) \\Longrightarrow f(x) \\in U \\Longrightarrow (y \\in f(f^{-1}(U)) \\Longrightarrow y \\in U)\n\\]\nHowever, it is still possible that \n\\[\n\\exists y(y \\in U \\wedge \\forall x(x \\in X \\Longrightarrow f(x) \\neq y))\n\\]\n\\end{proof}\n\n\\declareexercise{3.4.3}\n\\begin{proof}\n(1)\n\\[\nx \\in A \\cap B \\Longrightarrow f(x) \\in f(A) \\wedge f(x) \\in f(B) \\Longrightarrow \nf(x) \\in f(A) \\cap f(B)\n\\]\n\\[\ny \\in f(A \\cap B) \\equiv \\exists x(x \\in A \\cap B \\wedge f(x) = y)\n\\]\nSo $y \\in f(A) \\cap f(B)$, thus $f(A \\cap B) \\subseteq f(A) \\cap f(B)$.\n\n(2)\n\\[\nx \\in A \\setminus B \\Longrightarrow f(x) \\in f(A\\setminus B)\n\\]\n\\[\ny \\in f(A)\\setminus f(B) \\Longrightarrow \\exists x(x \\in A \\wedge x \\notin B \\wedge f(x) = y)\n\\]\nSo $y \\in f(A\\setminus B)$, thus $f(A)\\setminus f(B) \\subseteq f(A\\setminus B)$.\n\n(3)\nOn one hand, \n\\[\ny \\in f(A \\cup B) \\equiv \\exists x (x \\in A \\cup B \\wedge f(x) = y)\n\\]\n\\begin{align*}\nx \\in A \\cup B \\Longrightarrow x \\in A \\vee x \\in B \\Longrightarrow \\\\\nf(x) \\in f(A) \\vee f(x) \\in f(B) \\Longrightarrow f(x) \\in f(A)\\cup f(B)\n\\end{align*}\n\nOn the other hand, \n\\[\ny \\in f(A) \\cup f(B) \\Longrightarrow \\exists x((x \\in A \\vee x \\in B) \\wedge f(x) = y) \n\\]\n\\[\nx \\in A \\vee x \\in B \\Longrightarrow x \\in A \\cup B \\Longrightarrow f(x) \\in f(A\\cup B)\n\\]\n\\end{proof}\n\n(1) $\\subseteq$ can not be improved. Since it is possible that \n\\[\n\\exists x \\exists x'(x \\in A \\wedge x' \\in B \\wedge x \\neq x' \\wedge f(x) = f(x'))\n\\]\n\n(2) $\\subseteq$ can not be improved. Since it is possible that\n\\[\n\\exists x \\exists x'(x \\in A \\setminus B \\wedge x' \\in B \\wedge f(x) = f(x'))\n\\]\n\n\\declareexercise{3.4.4}\n\\begin{proof}\n(1) \n\\begin{align*}\nx \\in f^{-1}(U \\cup V) \\equiv (x \\in X \\wedge f(x) \\in U \\cup V) \\equiv \\\\\n(x \\in X \\wedge (f(x) \\in U \\vee f(x) \\in V))\n\\end{align*}\n\\begin{align*}\nx \\in f^{-1}(U) \\cup f^{-1}(V) \\equiv (x \\in X \\wedge f(x) \\in U) \\vee (x \\in X \\wedge f(x) \\in V) \\\\\n\\equiv (x \\in X \\wedge (f(x) \\in U \\vee f(x) \\in V))\n\\end{align*}\n\n(2) and (3) can be proven in similar manners. \n\\end{proof}\n\n\\declareexercise{3.4.5}\n\\begin{proof}\n(1) On one hand, if $f(f^{-1}(S)) = S$ for every $S \\subseteq Y$, then \\\\\n$f(f^{-1}(Y)) = Y$. That means, \n$y \\in Y \\Longrightarrow \\exists x (x \\in f^{-1}(Y) \\wedge f(x) \\in Y)$. So even $f^{-1}(Y)$ \nis enough for $f$ to be surjective. And $f^{-1}(Y) \\subseteq X$, so $f$ is surjective.\n\nOn the other hand, if $f$ is surjective, then for each $S \\subseteq Y$, \n\\[\ny \\in S \\Longrightarrow \\exists x(x \\in X \\wedge f(x) = y)\n\\]\nSuch $x$ are elements of $f^{-1}(S)$ of course, so $f(f^{-1}(S)) = S$.\n\n(2) On one hand, we show that $\\forall S(S \\subseteq X \\Longrightarrow f^{-1}(f(S)) = S)$ implies that \n$f$ is injective. Suppose the negation, that when \n\\[\n\\forall S(S \\subseteq X \\Longrightarrow f^{-1}(f(S)) = S)\n\\], but $f$ is not injective. Since $f$ is not injective, \n\\[\n\\exists x \\exists x'(x \\in X \\wedge x' \\in X \\wedge x \\neq x' \\wedge f(x) = f(x'))\n\\]\nLet $S \\subseteq X$ and $x \\in S \\wedge x' \\notin S$. There is always such a set $S$.\nFor example, we can let $S = X\\setminus \\{x'\\}$. So we have $f^{-1}(f(S)) \\neq S$ because $x' \\in$ it.\n\nOn the other hand, if $f$ is injective, then for every $S \\subseteq X$, and for every $x \\in S$, we have \n$f(x) \\in f(S)$. And $f(x)$ are the only elements in $f(S)$, that is, \n$y \\in f(S) \\Longrightarrow y = f(x)$ for some $x \\in S$. So now we know that \n$S \\subseteq f^{-1}(f(S))$. Moreover, for every $x' \\in f^{-1}(f(S))$, $f(x') \\in f(S)$. We can let \n$f(x') = y = f(x)$. As $f$ is injective, $x = x'$, so $x' \\in S$. That means $f^{-1}(f(S)) \\subseteq S$. \nSo $f^{-1}(f(S)) = S$.\n\\end{proof}\n\n\\declareexercise{3.4.6}\n\\begin{proof}\nMy own proof: According to Axiom 3.10, we can construct the set $X^X$. Apply the axiom \nof replacement to each element of $X^X$, we construct a set $Z$ such that\n\\[\n\\forall x(x \\in Z \\equiv \\exists f(f \\in X^X \\wedge x = f(X)))\n\\]\n\nLet $Y = \\{\\varnothing\\} \\cup Z$.\n\nNow we prove that $Y$ is the set we want. On one hand, for any $S \\subseteq X$, \nif $S = \\varnothing$, then $S \\in Y$, as $Y = \\{\\varnothing\\} \\cup Z$.\n\nIf $S \\neq \\varnothing$, there exists a surjective function $g: X \\rightarrow S$. $g\\in X^X$, and \n$g(X) = S$, so $S \\in Z$, and thus $S \\in Y$. (To show the existence of $g$, for example, let $x \\in X, \ng(x) = x$ if $x \\in S$, and for $x \\in X \\wedge x \\notin S$, $g(x)$ can be any element of $X$.)\n\n\nOn the other hand, for any $S' \\nsubseteq X$, $\\exists a(a \\in S' \\wedge a \\notin X)$. To prove that \n$S' \\notin Y$, we need to show that $\\nexists f(f \\in X^X \\wedge S' = f(X))$. We know that \n$\\nexists x(x \\in X \\wedge f(x) = a)$, so $a \\notin f(X)$. Therefore $S' \\neq f(X)$, so $S' \\notin Y$.\n\n$Y$ is the set we want.\n\nI posted a question \\href{https://math.stackexchange.com/questions/3803487/is-this-proof-to-the-existence-of-a-set-that-contains-all-subsets-of-another-set}{here} for verification for this proof. \nThanks to answers of people at Stack Exchange so that my proof can be refined.\n\nProof By Tao's Hint:\nFor each $S \\subseteq X$, let a function $f_S$ be $f_S(x) = 1$ if $x \\in S$, and $f_S(x)=0$ if \n$x \\in X \\wedge x \\notin S$. Then $f^{-1}_S(\\{1\\})$ gives $S$. \n\nNow we show that any element in \n$\\{0,1\\}^X$ is some $f_S$. Let $g \\in \\{0,1\\}^X$. Then if $\\forall x \\in X, g(x) = 0$, then \n$g = f_{\\varnothing}$. Otherwise, there exists a set that contains all $x$ such that $g(x) = 0$ by \naxiom of specification, namely $R$. Then $g = f_R$.\n\nOn the other hand, each $f_S$ is obviously an element of $\\{0,1\\}^X$.\n\nUse the axiom of replacement, we construct a set $Y$ such that\n\\[\n\\forall x (x \\in Y \\equiv \\exists f(f \\in \\{0,1\\}^X \\wedge x = f^{-1}(\\{1\\})))\n\\]\n\nAccording to what we have proven, $Y$ is the set we want.\n\\end{proof}\n\n\\declareexercise{3.4.7}\n\\begin{proof}\nAs stated by the previous exercise, there exists a set $\\mathbb{X}$ whose elements are all subsets of \n$X$, and a set $\\mathbb{Y}$ whose elements are all subsets of $Y$.\n\nFor every element $x \\in \\mathbb{X}$, apply the axiom of replacement to $\\mathbb{Y}$, to obtain a set \n$S_x := \\{y^x\\}$ for every element $y \\in \\mathbb{Y}$. \n\nAccording to the axiom of union, using $\\mathbb{X}$ as the index set, we have the set\n\\[\nZ = \\bigcup_{x \\in \\mathbb{X}} S_x\n\\]\n\nApply again the axiom of union to $Z$ to obtain $R$, which contains all elements of elements of $Z$. Now \nwe show that $R$ is the set we want.\n\nOn one hand, let $f$ be an arbitrary function with the domain of $X' \\subseteq X$, and the range of $Y' \n\\subseteq Y$. We can see that $f \\in {Y'}^{X'} \\in S_{X'}$. ${Y'}^{X'}$ becomes an element of $Z$. And \nthus $f$ becomes an element in $R$. \n\nOn the other hand, from the construction of $R$, we can see that $R$ contains only these elements.\n\n\\end{proof}\n\n\\declareexercise{3.4.8}\n\\begin{proof}\nLet $A,B$ be two arbitrary sets. They are also objects as stated by Axiom 3.1. So according to \nAxiom 3.3, there exists a set $S=\\{A,B\\}$. By Axiom 3.11, we have a set $Z$ such that \n\\[\n\\forall x(x \\in Z \\equiv \\exists X(X \\in S \\wedge x \\in X))\n\\]\n\nNow we show that $Z$ is the set we want. If $x \\in A \\vee x \\in B$, then \n$\\exists X(X \\in S \\wedge x \\in X)$ is true. So $x \\in Z$.\n\nIf $x \\notin A \\wedge x \\notin B$, then $\\forall X(X \\in S \\Longrightarrow x \\notin X)$, that is, \n$\\exists X(X \\in S \\wedge x \\in X)$ is false. So $x \\notin Z$.\n\n$Z$ is therefore the set we want. \n\\end{proof}\n\n\\paragraph{Example 3.4.11}\nIn (3.3), why do Tao choose some element $\\beta$ of $I$? This is because we need to apply the axiom of \nspecification to $A_\\beta$ with the restriction $x \\in A_\\alpha$ for all $\\alpha \\in I$.\n\n\\declareexercise{3.4.9}\n\\begin{proof}\nThis is quiet easy to prove. Let the left-handed side set be $S$, the RHS set be $S'$. For any \n$x \\in S$, $x \\in A_\\alpha$ for all $\\alpha \\in I$. So $x \\in A_{\\beta'}$. And $x \\in A_\\alpha$ for all \n$\\alpha \\in I$. Therefore $x \\in S'$. \n\nIt is nearly the same the prove $x \\in S' \\Longrightarrow x \\in S$.\n\\end{proof}\n\n\\paragraph{Exercise 3.4.10} \\label{exercise3.4.10}\n\\begin{proof}\nFor the sake of convenience, let \n$(\\bigcup_{\\alpha \\in I} A_{\\alpha})\\cup(\\bigcup_{\\alpha \\in J}A_{\\alpha})$ be $S$, \\\\\n$\\bigcup_{\\alpha \\in I \\cup J} A_{\\alpha}$ be $S'$, \n$(\\bigcap_{\\alpha \\in I} A_{\\alpha})\\cap(\\bigcap_{\\alpha \\in J}A_{\\alpha})$ be $Z$,\n$\\bigcap_{\\alpha \\in I \\cup J} A_{\\alpha}$ be $Z'$.\n\n(1) When $I,J \\neq \\varnothing$: \nOn one hand, \n\\[\nx \\in S \\Longrightarrow (x \\in \\bigcup_{\\alpha \\in I} A_{\\alpha} \\vee \nx \\in \\bigcup_{\\alpha \\in J}A_{\\alpha})\n\\]\nIf $x \\in \\bigcup_{\\alpha \\in I} A_{\\alpha}$, then $x \\in \\bigcup_{\\alpha \\in I \\cup J} A_{\\alpha}$.\nIf $x \\in \\bigcup_{\\alpha \\in J} A_{\\alpha}$, then $x \\in \\bigcup_{\\alpha \\in I \\cup J} A_{\\alpha}$.\n\nOn the other hand, if $x \\in S'$, then there exists an object $a \\in I \\cup J$ such that $x \\in A_a$.\nIf $a \\in I$ then $x \\in x \\in \\bigcup_{\\alpha \\in I} A_{\\alpha} \\Longrightarrow x \\in S$.\nIf $a \\in J$ then $x \\in x \\in \\bigcup_{\\alpha \\in J} A_{\\alpha} \\Longrightarrow x \\in S$.\n\nWhen $I,J$ are both empty, $S,S'$ are all empty.\n\nWhen there is only one of $I,J$ is empty, say it is $I$, then \n$S = \\varnothing \\cup \\bigcup_{\\alpha \\in J} = \\bigcup_{\\alpha \\in J}$. And \n$S' = \\bigcup_{\\alpha \\in \\varnothing \\cup J} A_{\\alpha} = \\bigcup_{\\alpha \\in J}$.\n\n(2)\n\\[\nx \\in Z \\equiv \\forall a(a \\in I \\Longrightarrow x \\in A_a) \\wedge \n\\forall b(b \\in J \\Longrightarrow x \\in A_b)\n\\], which is equal to $\\forall a(a \\in I \\cup J \\Longrightarrow x \\in A_a) \\equiv x \\in Z'$.\n\\end{proof}\n\n\\paragraph{Exercise 3.4.11} \\label{exercise3.4.11}\n\\begin{proof}\n(1) Let the LHS be $S$, the RHS be $S'$. \n\\begin{align*}\nx \\in S &\\equiv \\\\\nx \\in X \\wedge x \\notin \\bigcup_{\\alpha \\in I} A_{\\alpha} &\\equiv \\\\\nx \\in X \\wedge \\forall a(a \\in I \\Longrightarrow x \\notin A_{a})\n\\end{align*}\n\\begin{align*}\nx \\in S' &\\equiv \\\\\n\\forall a(a \\in I \\Longrightarrow x \\in X \\setminus A_a) &\\equiv \\\\\nx \\in X \\wedge \\forall a(a \\in I \\Longrightarrow x \\notin A_a)\n\\end{align*}.\n\nSo $S=S'$.\n\n(2) Let the LHS be $Z$, the RHS be $Z'$. \n\\begin{align*}\nx \\in Z &\\equiv \\\\\nx \\in X \\wedge x \\notin \\bigcap_{\\alpha \\in I} A_{\\alpha} &\\equiv \\\\\nx \\in X \\wedge \\neg(\\forall a(a \\in I \\Longrightarrow x \\in A_a)) &\\equiv \\\\\nx \\in X \\wedge \\exists a(a \\in I \\Longrightarrow x \\notin A_a)\n\\end{align*}\n\\begin{align*}\nx \\in Z' &\\equiv \\\\\nx \\in X \\wedge \\bigvee_{\\alpha \\in I}(x \\notin A_{\\alpha}) &\\equiv \\\\\nx \\in X \\wedge \\exists a(a \\in I \\Longrightarrow x \\notin A_a)\n\\end{align*}\n\nThus, $Z = Z'$\n\\end{proof}\n\n\\subsection{Cartesian products}\n\\declareexercise{3.5.1}\n\\begin{proof}\nFirst we show that $(x,y) = \\{\\{x\\},\\{x,y\\}\\}$ is a good definition. \nLet $S_1$ denote $(x_1,y_1) = \\{\\{x_1\\},\\{x_1,y_1\\}\\}$, $S_2$ denote \n$(x_2,y_2) = \\{\\{x_2\\},\\{x_2,y_2\\}\\}$.\n\nOn one hand, if $x_1=x_2\\wedge y_1=y_2$, then obviously $S_1=S_2$ for they have the same elements.\n\nOn the other hand, if $S_1 = S_2$, then \n\\[\n\\{x_1\\} \\in S_2 \\wedge \\{x_1,y_1\\} \\in S_2 \\wedge\n\\{x_2\\} \\in S_1 \\wedge \\{x_2,y_2\\} \\in S_1 \n\\].\nWe have that \n\\begin{align*}\n\\{x_1\\} \\in S_2 &\\equiv \\{x_1\\} = \\{x_2\\} \\vee \\{x_1\\} = \\{x_2,y_2\\} \\\\\n&\\equiv x_1=x_2 \\vee x_1=x_2=y_2 \\\\\n&\\Longrightarrow x_1=x_2\n\\end{align*}\n\\begin{align*}\n\\{x_1,y_1\\} \\in S_2 &\\equiv \\{x_1,y_1\\} = \\{x_2\\} \\vee \\{x_1,y_1\\} = \\{x_2,y_2\\} \\\\\n&\\equiv x_1=x_2=y_1 \\\\\n&\\vee \\textcolor{red}{((x_1=x_2\\wedge y_1=y_2)\\vee(x_1=y_2\\wedge y_1=x_2))} \n\\end{align*}\nSimilarly we have that\n\\begin{align*}\n\\{x_2,y_2\\} \\in S_1 \n&\\equiv x_2=x_1=y_2 \\\\\n&\\vee \\textcolor{red}{((x_2=x_1\\wedge y_2=y_1)\\vee(x_2=y_1\\wedge y_2=x_1))}\n\\end{align*}\nWe may notice that the red-colored text are two same statements. Thus from $\\{x_1,y_1\\} \\in S_2$ and \n$\\{x_2,y_2\\} \\in S_1$ we can always conclude that $y_1=y_2$. Therefore, \n$S_1 = S_2 \\Longrightarrow x_1=x_2\\wedge y_1=y_2$.\n\nThen we show that if $X,Y$ are two sets, then $X \\times Y$ is also a set. For each element $x \\in X$, \nconstruct a set $S_x$, where we replace each element $y \\in Y$ with $(x,y)$. Then construct the set \n$\\bigcup_{x \\in X}S_x$.\n\\end{proof}\n\n\\declareexercise{3.5.2}\n\\begin{proof}\nSince $x,y$ are two functions, they are equal means that $\\forall 1\\leq i \\leq n$, $x(i) = y(i)$. That \nis, $x_i = y_i, 1\\leq i \\leq n$.\n\nNow we show that $\\displaystyle \\prod_{1\\leq i\\leq n}X_i$ is a set. Let set $F$ be the set that contains \nall partial functions from $N = \\{i \\in \\mathbb{N}:1\\leq i\\leq n\\}$ to \n$\\displaystyle X = \\bigcup_{1\\leq i\\leq n}X_i$ (Exercise 3.4.7). Use the axiom of specification, select \nsuch elements $f$ from $F$ that:\n\\begin{enumerate}\n\\item the element is surjective, and\n\\item its domain is $N$, and \n\\item $f(i) \\in X_i$\n\\end{enumerate}, \nand use all of them to construct a set $Z$, which is the set we want.\n\\end{proof}\n\n\\declareexercise{3.5.3}\n\\begin{proof}\nThe definition is entirely based on the equality of objects (e.g. $x = x'$). The proof is immediately \ndone since this equality is reflective ($x = x$), symmetric ($x = x' \\equiv x' = x$), and transitive \n($x_0 = x_1 \\wedge x_1 = x_2 \\Longrightarrow x_0 = x_2$).\n\\end{proof}\n\n\\declareexercise{3.5.4}\n\\begin{proof}\n(1)\n\\begin{align*}\n(x,y) \\in A \\times (B \\cup C) &\\equiv x \\in A \\wedge y \\in (B \\cup C) \\\\\n&\\equiv x \\in A \\wedge (y \\in B \\vee y \\in C) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\vee (x \\in A \\wedge y \\in C) \\\\\n&\\equiv ((x,y) \\in A \\times B) \\vee ((x,y) \\in A \\times C) \\\\\n&\\equiv (x,y) \\in (A \\times B) \\cup (A \\times C)\n\\end{align*}\n\n(2)\n\\begin{align*}\n(x,y) \\in A \\times (B \\cap C) &\\equiv x \\in A \\wedge y \\in (B \\cap C) \\\\\n&\\equiv x \\in A \\wedge (y \\in B \\wedge y \\in C) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\wedge (x \\in A \\wedge y \\in C) \\\\\n&\\equiv ((x,y) \\in A \\times B) \\wedge ((x,y) \\in A \\times C) \\\\\n&\\equiv (x,y) \\in (A \\times B) \\cap (A \\times C)\n\\end{align*}\n\n(3)\n\\begin{align*}\n(x,y) \\in A \\times (B \\setminus C) &\\equiv x \\in A \\wedge y \\in (B \\setminus C) \\\\\n&\\equiv x \\in A \\wedge (y \\in B \\wedge \\neg (y \\in C)) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\wedge \\neg (x \\in A \\wedge y \\in C) \\\\\n\\tag{The statement $x \\in A$ implies $\\neg (x \\in A \\wedge y \\in C) \n\\Longrightarrow \\neg (y \\in C)$}\\\\\n&\\equiv ((x,y) \\in A \\times B) \\wedge \\neg((x,y) \\in A \\times C) \\\\\n&\\equiv (x,y) \\in (A \\times B) \\setminus (A \\times C)\n\\end{align*}\n\\end{proof}\n\n\\declareexercise{3.5.5}\n\\begin{proof}\n(1)\n\\begin{align*}\n(x,y) \\in (A \\times B) \\cap (C \\times D) \n&\\equiv (x,y) \\in (A \\times B) \\wedge (x,y) \\in (C \\times D) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\wedge (x \\in C \\wedge y \\in D) \\\\\n&\\equiv (x \\in A \\wedge x \\in C) \\wedge (y \\in B \\wedge y \\in D) \\\\\n&\\equiv x \\in A \\cap C \\wedge y \\in B \\cap D \\\\\n&\\equiv (x,y) \\in (A \\cap C) \\times (B \\cap D)\n\\end{align*}\n\n(2) It is not true since \n\\begin{align*}\n(x,y) \\in (A \\times B) \\cup (C \\times D) \n&\\equiv (x,y) \\in (A \\times B) \\vee (x,y) \\in (C \\times D) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\vee (x \\in C \\wedge y \\in D) \\\\\n&\\nLeftrightarrow (x \\in A \\vee x \\in C) \\wedge (y \\in B \\vee y \\in D)\n\\end{align*}\nGenerally \n\\[\n(x \\in A \\wedge y \\in B) \\vee (x \\in C \\wedge y \\in D) \\Longrightarrow \n(x \\in A \\vee x \\in C) \\wedge (y \\in B \\vee y \\in D)\n\\], \nbut\n\\[\n(x \\in A \\vee x \\in C) \\wedge (y \\in B \\vee y \\in D) \\nRightarrow\n(x \\in A \\wedge y \\in B) \\vee (x \\in C \\wedge y \\in D)\n\\].\n\n(3) It is not true since\n\\begin{align*}\n(x,y) \\in (A \\times B) \\setminus (C \\times D) \n&\\equiv (x,y) \\in (A \\times B) \\wedge (x,y) \\notin (C \\times D) \\\\\n&\\equiv (x \\in A \\wedge y \\in B) \\wedge (x \\notin C \\vee y \\notin D) \\\\\n&\\nLeftrightarrow (x \\in A \\wedge x \\notin C) \\wedge (y \\in B \\wedge y \\notin D)\n\\end{align*}\n\\end{proof}\n\n\\declareexercise{3.5.6}\n\\begin{proof}\n(1) On one hand, if $A \\subseteq C$ and $B \\subseteq D$, then \n\\begin{align*}\n(x,y) \\in A \\times B &\\equiv x \\in A \\wedge y \\in B \\\\\n&\\Longrightarrow x \\in C \\wedge y \\in D \\\\\n&\\Longrightarrow (x,y) \\in C \\times D\n\\end{align*}, \nwhich means $A \\times B \\subseteq C \\times D$.\n\nOn the other hand, if $A \\times B \\subseteq C \\times D$, but we suppose that \n\\[\n\\neg(A \\subseteq C \\wedge B \\subseteq D)\n\\]. \nWe only consider that $A \\nsubseteq C$, the other situations are similar. Then \n$\\exists x(x \\in A \\wedge x \\notin C)$. Let $p = (x,y)$, where $y \\in B$, then $p \\in A \\times B$. \nBut $x \\notin C$, so $p \\notin C \\times D$, a contradiction. Therefore, \n\\[\nA \\times B \\subseteq C \\times D \\Longrightarrow A \\subseteq C \\wedge B \\subseteq D\n\\]\n\n(2) On one hand, if $A = C \\wedge B = D$, then\n\\begin{align*}\n(x,y) \\in A \\times B &\\equiv x \\in A \\wedge y \\in B \\\\\n&\\equiv x \\in C \\wedge y \\in D \\\\\n&\\equiv (x,y) \\in C \\times D\n\\end{align*}.\n\nOn the other hand, if $A \\times B = C \\times D$, but we suppose that $\\neg(A = C \\wedge B = D)$. \nWe only consider that $A \\neq C$, the other situations are similar. Then we only consider \n$\\exists x(x \\in A \\wedge x \\notin C)$, for the other situations are similar. \n\n(3) It is easy to prove that $X \\times \\varnothing = \\varnothing$ and \n$\\varnothing \\times X = \\varnothing$. Let $A = \\varnothing$, we can see that even if $B \\nsubseteq D$, \n$A \\times B \\subseteq C \\times D$. \n\nLet $A = D = \\varnothing$, then even if $A \\neq C$, $A \\times B = C \\times D$.\n\\end{proof}\n\n\\declareexercise{3.5.7}\n\\begin{proof}\nExistence: Let $h(t):=(f(t),y(t))$. It is easy to verify that $h(t) \\in X \\times Y$, and that given a \n$t \\in Z$, $h(t)$ is unique. Therefore, $h$ is a function. And it is obvious that \n$\\pi_{X\\times Y \\rightarrow X} \\circ h = f$ and that $\\pi_{X\\times Y \\rightarrow Y} \\circ h = g$.\n\nUniqueness: $\\pi_{X\\times Y \\rightarrow X} \\circ h = f$ and $\\pi_{X\\times Y \\rightarrow Y} \\circ h = g$ \nimply that if there is another function $h'$ that satisfies the requirements, then $h'(t) = h(t)$. So $h$ \nis unique. \n\\end{proof}\n\n\\declareexercise{3.5.8}\n\\begin{proof}\nOn one hand, if for some $i, X_i = \\varnothing$, then \n\\[\n\\forall (x_i)_{1\\leq i \\leq n}(\\bigwedge^{n}_{i =1}x_i \\in X_i \\equiv (x_i)_{1\\leq i \\leq n} \\in \n\\varnothing)\n\\], \nwhich means that $\\varnothing = \\prod_{i=1}^{n}X_i$.\n\nOn the other hand, if $\\prod_{i=1}^{n}X_i = \\varnothing$ but we suppose that $X_i \\neq \\varnothing$. Then \nfor each $i$, $\\exists x_i \\in X_i$. We thus have a tuple $(x_i)_{1\\leq i \\leq n}$, which should be an \nelement of $\\prod_{i=1}^{n}X_i$. Therefore we have a contradiction.\n\\end{proof}\n\n\\declareexercise{3.5.9}\n\\begin{proof}\nOn one hand, let $x \\in (\\bigcup_{\\alpha \\in I}A_{\\alpha})\\cap(\\bigcup_{\\beta \\in J}B_{\\beta})$. Then \n\\[\n\\exists a(a \\in I \\wedge x \\in A_a) \\wedge \\exists b(b \\in J \\wedge x \\in B_b)\n\\]\nIt is obvious that $x \\in A_a \\cap B_b$ and that $(a,b) \\in I \\times J$. Therefore \n\\[\nx \\in \\bigcup_{(\\alpha,\\beta) \\in I \\times J}(A_\\alpha \\cap B_\\beta)\n\\].\n\nOn the other hand, let $x \\in \\bigcup_{(\\alpha,\\beta) \\in I \\times J}(A_\\alpha \\cap B_\\beta)$. Then \n\\begin{align*}\n\\exists (a,b) \\in I \\times J(x \\in A_a \\cap B_b) \n&\\Longrightarrow x \\in A_a \\wedge x \\in B_b \\\\\n&\\Longrightarrow x \\in \\bigcup_{\\alpha \\in I}A_{\\alpha} \\wedge x \\in \\bigcup_{\\beta \\in J}B_{\\beta} \\\\\n&\\Longrightarrow x \\in (\\bigcup_{\\alpha \\in I}A_{\\alpha})\\cap(\\bigcup_{\\beta \\in J}B_{\\beta})\n\\end{align*}\n\\end{proof}\n\n\\paragraph{Exercise 3.5.10} \\label{exercise3.5.10}\n\\begin{proof}\nWe denote $\\overset{\\sim}{f}$ as $f'$, the graph of $f$ as $G$, and the graph of $f'$ as $G'$ for the \nsake of simplification.\n\n(1) On one hand, if $f = f'$, then for every $(x,f(x)) \\in G$, we can find $(x,f'(x)) \\in G'$, and \nobviously $(x,f(x)) = (x,f'(x))$, and vice versa.\n\nOn the other hand, if $G = G'$, then for each $(x,f(x)) \\in G$, $(x,f(x)) \\in G'$. Note that each \nelement of $G'$ obeys the form $(x,f'(x))$, so $f(x) = f'(x)$ for every $x \\in X$, that is, $f=f'$.\n\n(2) Existence: Let $f(x)$ be such a value that $(x,f(x)) \\in G$. Thus the value is unique, so $f$ is a \nfunction. According to its definition, the graph of $f$ is $G$.\n\nUniqueness: As proven in (1), if $f,f'$ have the same graph, then they are equal.\n\\end{proof}\n\n\\paragraph{Exercise 3.5.11} \\label{exercise3.5.11}\nI think this exercise is meaningless. Lemma 3.4.6 is proven by the fact that $X^Y$ exists, which depends \non Axiom 3.10. Then the exercise asks us to prove Axiom 3.10 using Lemma 3.4.6. So I looked up some books \nabout set theory and found out that the power set axiom is essentially Lemma 3.4.6, not Axiom 3.10.\n\nNevertheless, here is the proof:\n\\begin{proof}\nLet set $Z$ contains all subsets of $X\\times Y$. The specify such element in $Z$ that obey the vertical \nline test, and let them form the set $S$. According to the previous exercise, for each element in $S$, \nthere exists an unique function whose graph is the element. Then we replace all elements in $S$ with \nthese functions to construct the set $F$. Obviously, each element in $F$ is a function with the domain \n$X$ and the range $Y$.\n\nNow we show that every function $f$ from $X$ to $Y$ is in $F$. Denote the graph of $f$ as $G$. We know \nthat $G$ obeys the vertical line test and $G \\subseteq X \\times Y$, so $G \\in S$. Since $G$ is the graph \nof $f$, $f \\in F$.\n\\end{proof}\n\n\\paragraph{Exercise 3.5.12} \\label{exercise3.5.12}\nI am confused by this exercise. It seems that simply applying induction to $a$ can solve the \nproblem, just like what we did in Proposition 2.1.16. What is wrong?\n\nBy the way, according to the \\href{https://terrytao.wordpress.com/books/analysis-i/}{corrections}, \nedit the exercise as the following:\n\\begin{quotation}\nLet $X$ be an arbitrary set containing at least an element $c$ and obeys the Peano axioms. Let $f$ be a \nfunction from $N \\times X$ to $X$. ... \n\nShow that there exists an unique function $a$ from $X$ to $X$ such that \n\\[\na(0) = c\n\\] \nand \n\\[\na(n++) = f(n,a(n)), \\forall n \\in X\n\\]\n...\n\nsuch that $a_N(0) = c$ and $a_{N}(n++) = f(n,a_N(n))$ ... \n\\end{quotation}\n\nNote that all properties (e.g. orders, addition) in section 2 are deduced from the Peano axioms and their \ndefinitions. Since $X$ obeys these rules, we use such properties on elements of $X$ without proof.\n\nThe proof is now reserved for further research.\n\\begin{proof}\n\\end{proof}\n\n\\paragraph{Exercise 3.5.13} \\label{exercise3.5.13}\n\\begin{proof}\nUse induction.\n\nExistence: We need to prove that for all $n \\in \\mathbb{N}$, $f(n)$ is defined. Use induction:\n$f(0) = 0'$ is define. And the definition is unique for $0$ is not the successor of any natural number. \nNow suppose that $f(n) = n'$ is defined, then $f(S(n)) = S'(f(n)) = S'(n')$ is also defined. The \ndefinition is also unique. So we know that $f$ exists.\n\nInjectivity: We need to prove that $f(m) = f(n) \\Longrightarrow m = n$. If $f(m) = f(n)$, then $m' = n'$, \nand thus $m=n$. \n\nSurjectivity: Use induction: \nThe basic case is, for $0' \\in \\mathbb{N}'$, $f(0) = 0'$. \n\nNow suppose that for $n' \\in \\mathbb{N}'$, we can find $n \\in \\mathbb{N}$ such that $f(n) = n'$, then \nfor $S'(n')$, we have $f(S(n)) = S'(n')$. We can close the induction now.\n\\end{proof}\n\n\\subsection{Cardinality of Sets}\n\n\\declareexercise{3.6.1}\n\\begin{proof}\nReflexivity: Let $f(x):= x, X \\rightarrow X$. $f$ is bijective since $f^{-1}(x) = x$ exists.\n\nSymmetry: If $X,Y$ have the same cardinality, then $\\exists f:X\\rightarrow Y$ which is bijective. So \n$f^{-1}$ exists, and is also a bijection. Thus $Y,X$ have the same cardinality. Since then, we can say \nthat two sets have the same cardinality without caring about the order.\n\nTransitivity: If $X,Y$ have the same cardinality, and $Y,Z$ also have the same cardinality, then there \nexist two bijections: $f:X \\rightarrow Y$ and $g:Y \\rightarrow Z$. It is easy to verify that $g \\circ f$ \nis also a bijection and is from $X$ to $Z$ (See \\exerciseref{3.3.7}).\n\\end{proof}\n\n\\paragraph{Remark 3.6.6}\nIt is $f(n) := S(n)$. We are now proving something stronger\n\\begin{lem} \\label{lem3.6.6}\nFor any natural number $m,n$, $\\{i \\in \\mathbb{N}:0\\leq i\\leq n\\}$ and \n$\\{i \\in \\mathbb{N}:m\\leq i\\leq n+m\\}$ have the same cardinality.\n\\end{lem}\n\\begin{proof}\nUse induction on $m$. When $m=0$, the statement is obviously true. Simply give the function $f(n):=n$.\n\nSuppose that for some $m$, we have proven the statement. Then there exists a bijection: \n\\[\nf:\\{i \\in \\mathbb{N}:0\\leq i\\leq n\\} \\rightarrow \\{i \\in \\mathbb{N}:m\\leq i\\leq n+m\\}\n\\].\nLet $g$ be a function from $\\{i \\in \\mathbb{N}:0\\leq i\\leq n\\}$ to $\\mathbb{N}$ such that \n$g(x) = S(f(x))$. We prove that $g$ is a bijection from $\\{i \\in \\mathbb{N}:0\\leq i\\leq n\\}$ to \n$\\{i \\in \\mathbb{N}:S(m)\\leq i\\leq n+S(m)\\}$.\n\nFirst we prove that $g(n)$ always in $\\{i \\in \\mathbb{N}:S(m)\\leq i\\leq n+S(m)\\}$, which is immediately \ngiven by the fact that addition preserves order. \n\nSurjectivity: For any $a \\in \\{i \\in \\mathbb{N}:S(m)\\leq i\\leq n+S(m)\\}$, $a$ is positive. Then $a$ is \nalways some number's successor, that is $a = S(b) = b+1$ for some natural number $b$. Since addition \npreserves order, $b \\in \\{i \\in \\mathbb{N}:m\\leq i\\leq n+m\\}$. $f$ being surjective implies that there is \nsome $x$ in the domain such that $f(x) = b$, and $g(x) = f(x) + 1 = a$.\n\nInjectivity: By cancellation law, $f(x) + 1 \\neq f(x') + 1 \\equiv f(x) \\neq f(x') \\equiv x \\neq x'$.\n\nWe can now close the induction.\n\\end{proof}\n\n\\paragraph{Lemma 3.6.9}\nEmpty functions are not injective when the range is not empty (See \\exerciseref{3.3.3}). \n\nNow we show that $g$ is bijective:\n\\begin{proof}\nInjectivity: $f$ being injective implies that \n\\[\n\\forall x \\forall x'((x \\in X \\wedge x' \\in X) \\Longrightarrow (f(x) = f(x') \\Rightarrow x = x'))\n\\]\nFor $a,a' \\in X - \\{x\\}$, they also $\\in X$. If $g(a) = g(a')$, then either directly $f(a) = f(a')$ or \n$f(a) - 1 = f(a') - 1$, which gives $f(a) = f(a')$. Thus $a = a'$. (Note that subtraction is not defined \nyet, see the footnote about this in the book).\n\nSurjectivity: The surjectivity of $f$ gives \n\\[\n(\\forall 1 \\leq i \\leq n)(\\exists a(a \\in X \\wedge f(a) = i))\n\\].\n\nIf $f(x) = n$, then $g(a) = f(a)$ for all meaningful $a$. Then for $1 \\leq i \\leq n-1$, we can find $a$ \nsuch that $a \\in X \\wedge a \\neq x$, that is, $x \\in X - \\{x\\}$. So $g(a)$ is meaningful, then $g$ is \nsurjective.\n\nIf $f(x) \\neq n$, then $f(x) < n$. For those $1 \\leq i < f(x)$, $g$ is obviously surjective. For \n$n-1 \\geq i \\geq f(x)$, since $S(i) \\leq n$, $\\exists a(a \\in X \\wedge f(a) = S(i))$. And we know that \n$S(i) \\neq f(x)$, then $a \\in X - \\{x\\}$. So $g(a) = f(a) - 1 = i$.\n\\end{proof}\n\n\\declareexercise{3.6.2}\n\\begin{proof}\nOn one hand, if $X$ is empty, then we know that the empty function whose range is also empty is injective, \n(See \\exerciseref{3.3.3}) so its cardinality is $0$. \n\nOn the other hand, if $\\# X = 0$ but $X \\neq \\varnothing$, then there exists an bijection \n$f:X \\rightarrow \\varnothing$, which is impossible.\n\\end{proof}\n\n\\declareexercise{3.6.3}\n\\begin{proof}\nWhen $n = 0$, this is vacuously true. The base case then becomes $n=1$. We simply let $M = f(1)$.\n\nSuppose that the statement for $n$ is true. And for $1\\leq i\\leq n$ we have the number $M$. Then $f(S(n))$ \neither $\\geq$ or $<$ $M$. On the former case, let $f(S(n))$ be $M'$, and on the latter case, let $M' = M$. \nIt is east to verify that $M'$ is the number we want.\n\\end{proof}\n\nFrom now on we will denote $\\{i\\in \\mathbb{N}:1\\leq i \\leq n\\}$ as $\\mathbb{N}_n$\n\n\\declareexercise{3.6.4}\n\\begin{proof}\n(a) \nLet $n = \\#X$. There is an injective $f$ from $X$ to $\\{i\\in \\mathbb{N}:1\\leq i\\leq n\\}$. Let $g$ be a \nfunction from $X \\cup \\{x\\}$ to $\\{i\\in \\mathbb{N}:1\\leq i\\leq n+1\\}$ such that $g(a) = f(a)$ if \n$a\\neq x$, and $g(x) = n+1$. Now we show that $g$ is bijective.\n\nInjectivity: We know that $\\forall x \\in X$, $g$ is already injective. Since that $g(x) = n+1 \\neq g(a)$ \nfor all $a \\in X$, so $g$ is injective on $X \\cup \\{x\\}$.\n\nSurjectivity: We know that $\\forall i \\in \\{i\\in \\mathbb{N}:1\\leq i\\leq n\\}$, we can find \n$a \\in X \\cup \\{x\\}$ such that $g(a) = i$. And we have $g(x) = n+1$, so \n$\\forall a \\in \\{i\\in \\mathbb{N}:1\\leq i\\leq n+1\\}$, we can find $a \\in X \\cup \\{x\\}$ such that \n$g(a) = i$. \n\n(b)\nFirst we prove that if $X,Y$ are disjoint, then $\\#X + \\#Y = \\#(X\\cup Y)$. Let $f$ be a bijection from $X$ \nto $\\mathbb{N}_{\\#X}$, and $g$ be a bijection from $Y$ to $\\mathbb{N}_{\\#Y}$. According to \n\\hyperref[lem3.6.6]{this Lemma}, there exists a bijection $h$ from $\\mathbb{N}_{\\#Y}$ to \n$\\{i\\in \\mathbb{N}:\\#X+1\\leq i \\leq \\#X+\\#Y\\}$. Thus $h \\circ g$ is also a bijection. Let $u$ be a \nfunction from $X \\cup Y$ to $\\mathbb{N}_{\\#X} \\cup\\{i\\in \\mathbb{N}:\\#X+1\\leq i \\leq \\#X+\\#Y\\}$. Now we \nshow that $u$ is bijective.\n\nInjectivity: For $x \\neq x'$ in the domain. If $x,x'$ are both in $X$ or $Y$, then $f(x)\\neq f(x')$ is \nimmediately given by the injectivity of $f$ and $h \\circ g$. If one of them is in $X$, and the other is \nin $Y$, then they can also never be equal because the ranges of the two functions are disjoint. \n\nSurjectivity: It is easy to verify that the range is equal to $\\mathbb{N}_{\\#X + \\#Y}$. For any $y$ in the \nrange, if $y \\in$ the range of $f$, then $u$ is surjective since $f$ is, and if $y \\in$ the range of $h \n\\circ g$, $u$ is surjective for the same reason. The range consists of only this two sets, so $u$ is \nsurjective on the whole range.\n\nThe proof is over. This also implies that $X \\cup Y$ is finite. Now we need only to show that \n$\\#(X \\cup Y) < \\#X + \\#Y$ when $X,Y$ are not disjoint. It is easy to see that\n\\begin{align*}\n\\#A + \\#B \n&= \\#(A - A \\cap B) + \\#(A \\cap B) + \\#(B - A \\cap B) + \\#(A \\cap B) \\\\\n&= (\\#(A - A \\cap B) + \\#(A \\cap B) + \\#(B - A \\cap B)) + \\#(A \\cap B) \\\\\n&= \\#(A \\cup B) + \\#(A \\cap B) \\\\\n&> \\#(A \\cup B)\n\\end{align*}\n\n(c)\nIf $X \\subseteq Y \\wedge X \\neq Y$, then $\\#(Y \\setminus X) \\neq 0$. \n\\[\n\\#Y = \\#X + \\#(Y \\setminus X) > \\#X\n\\].\n\nIf $X = Y$, then $\\#(Y \\setminus X) = 0$, and $\\#Y$ becomes $\\#X$.\n\n(d)\n$f: X \\rightarrow f(X)$ is always surjective. If $f$ is also injective, then $f$ is bijective. On this \noccasion, $\\#f(X) = \\#X$. If $f$ is not injective, we can select a set $X' \\subseteq X \\wedge X' \\neq X$, \non which $f$ is bijective. Then $\\#X' = \\#f(X') = \\#f(X)$. According to (c), $\\#X' < \\#X$, so \n$\\#f(X) < \\#X$.\n\n(e)\nSuppose that $\\#Y = n$. Use induction on $n$. \n\nWhen $n=0$, $Y$ is empty, then $\\#(X \\times Y) = 0 = \\#X \\times 0$. Here we additionally prove that \nwhen $n=1$, this is also true for further usage. When $n=1$, let $Y = \\{a\\}$. Then the bijection is \n$f(x):=(x,a), X \\rightarrow X \\times \\{a\\}$.\n\nSuppose that we have proven for some $n$, $\\#(X \\times Y) = \\#X \\times \\#Y$. Then when $\\#Y = S(n)$, \nlet $Y = Y\\setminus\\{x\\}\\cup\\{x\\}$, where $x \\in Y$. Lemma 3.6.9 tells us that \n$\\#(Y\\setminus\\{x\\}) = S(n)-1 = n$. And \\exerciseref{3.5.4} tells us that \n$X \\times Y = X \\times (Y\\setminus\\{x\\}) \\cup X \\times \\{x\\}$. \n\\begin{align*}\n\\#(X \\times Y) \n&= \\#(X \\times (Y\\setminus\\{x\\}) \\cup X \\times \\{x\\}) \\\\\n&= \\#(X \\times (Y\\setminus\\{x\\})) + \\#(X \\times \\{x\\}) \\\\\n&= \\#X \\times n + \\#X \\\\\n&= \\#X \\times S(n)\n\\end{align*}\n\nWe can now close the induction.\n\n(f)\nWe should first define $m^n$ for natural numbers $m,n$. It has not been done yet. Exponentiation is \ndefined for rational numbers at Definition 4.3.9.\n\\begin{definition}\n\\begin{itemize}\n\\item $m^0=1$,\n\\item $m^{S(n)} = m^n \\times m$\n\\end{itemize}\n\\end{definition}\n\nSuppose that $\\#Y = m,\\#X = n$. Use induction on $n$. \n\nWhen $n=0$, $X$ is empty, then $Y^X$ has one function $f:\\varnothing \\rightarrow Y$.\n\nSuppose that we have proven the statement for some $n$. Before we proceed the proof, we need some lemmas.\n\\begin{lem}\nIf $X$ is not empty, \n\\[\n\\#Y^{X\\setminus\\{x'\\}\\cup\\{x'\\}} = \\#Y^{X\\setminus\\{x'\\}} \\times \\#Y\n\\], \nwhere $x'$ is an element of $X$.\n\\end{lem}\n\\begin{proof}\nBy (e) we know that \n\\[\n\\#Y^{X\\setminus\\{x'\\}} \\times \\#Y = \\#(Y^{X\\setminus\\{x'\\}} \\times Y)\n\\].\n\nTry to build a bijection between $Y^{X\\setminus\\{x'\\}} \\times Y$ and $Y^X$. Let $f' \\in Y^X$.\n\nLet $h$ be a function from $Y^X$ to $Y^{X\\setminus\\{x'\\}} \\times Y$ such that\n\\[\nh(f') = (f,f'(x')), \n\\]\nwhere $f(x):=f'(x)$ when $x \\neq x'$. Now we show that $h$ is bijective.\n\nInjectivity: \nIf ${f_1}' \\neq {f_2}'$, then \n\\[\n{f_1}'(x') \\neq {f_2}'(x') \\vee \\exists x(x \\neq x' \\wedge {f_1}'(x) \\neq {f_2}'(x))\n\\]\nThat is, \n\\[\n{f_1}'(x') \\neq {f_2}'(x') \\vee f_1 \\neq f_2,\n\\]\nwhich means \n\\[\n(f_1,{f_1}'(x')) \\neq (f_2,{f_2}'(x')).\n\\]\n\nSurjectivity:\nFor any $(f,a) \\in Y^{X\\setminus\\{x'\\}} \\times Y$, let $f'$ be $f$ if $x\\neq x'$, and $f'(x') = a$. Then \n$f' \\in Y^X$ and $h(f') = (f,a)$.\n\nSo, \n\\[\n\\#Y^X = \\#(Y^{X\\setminus\\{x'\\}} \\times Y)\n\\], which gives the lemma.\n\\end{proof}\n\nNow we proceed the proof. Suppose that $\\#X = n+1$, then $\\#(X \\setminus\\{x'\\}) = n$. By induction \nhypothesis, $\\#(Y^{X \\setminus\\{x'\\}}) = m^n$. \n\nBy the lemma, \n\\[\n\\#Y^X = \\#Y^{X\\setminus\\{x'\\}\\cup\\{x'\\}} = \\#Y^{X\\setminus\\{x'\\}} \\times \\#Y,\n\\]\nwhich equals to $m^n \\times m$.\n\nNow we can close the induction.\n\nWe have proven that the cardinality of power sets obeys the definition of power. This ensures the \nexercise.\n\\end{proof}\n\n\\declareexercise{3.6.5}\n\\begin{proof}\nLet $f((x,y)):= (y,x), A\\times B \\rightarrow B \\times A$. The bijectivity is obvious. \n\nNow we are using set theory to prove the commutativity of multiplication of natural number. For any \nnatural number $m,n$, construct two sets: $M = \\mathbb{N}_{m}, N = \\mathbb{N}_{n}$. According to (e) in \nProposition 3.6.14, we have that $\\#(M \\times N) = \\#M \\times \\#N$. Then by what we have just proven, \n\\[\n\\#(M \\times N) = \\#(N \\times M) \\Longrightarrow \\#M \\times \\#N = \\#N \\times \\#M \\Longrightarrow mn = nm\n\\]\n\\end{proof}\n\n\\declareexercise{3.6.6}\n\\begin{proof}\nLet $c \\in C$, $f \\in (A^B)^C$. Then $f(c)$ is a function $B\\rightarrow A$. Let \n$b \\in B, h \\in A^{B\\times C}$. Let\n\\[\ng:A^{B\\times C} \\rightarrow (A^B)^{C}\n\\]\nbe such a function that for all $b,c$,\n\\[\ng(h) = f \\equiv h(b,c) = (f(c))\\,(b)\n\\]\n. Now we show that $g$ is bijective.\n\nInjectivity: \nIf $h \\neq h'$, then $\\exists b_0,c_0(h(b_0,c_0) \\neq h'(b_0,c_0))$. Let $g(h) =f, g(h') = f'$. Then we \nknow that $(f(c_0))\\,(b_0) \\neq (f'(c_0))\\,(b_0)$, so $f(c_0) \\neq f'(c_0) \\Longrightarrow f \\neq f'$. \nThat means, $g(h) \\neq g(h')$.\n\nSurjectivity:\nFor any $f \\in (A^B)^{C}$, let $h$ be such a function $\\in A^{B\\times C}$ that for all $b \\in B,c \\in C$, \n$h(b,c) := (f(c))\\,(b)$. It is easy to see that $h$ is well-defined. So $g(h) = f$.\n\nNote that by Proposition 3.6.14 we have $\\#M^N = m^n$ and $\\#(M \\times N) = mn$, where $\\#M =m, \\#N =n$. \nSuppose that $\\#A = a, \\#B = b, \\#C = c$, then\n\\[\n\\#(A^B)^C = (\\#A^B)^{\\#C} = (a^b)^c\n\\]\n\\[\n\\#A^{B\\times C} = \\#A^{\\#(B \\times C)} = a^{bc}\n\\]\nSo we have proven that $(a^b)^c = a^{bc}$.\n\nNow we try to prove $a^b \\times a^c = a^{b+c}$. Let $B,C$ be disjoint sets with the cardinality $b,c$ \nrespectively. What we need to show is that\n\\[\n\\#(A^B \\times A^C) = \\#(A^{B \\cup C}).\n\\]\n\nSimilarly, let \n\\[\nf: (A^{B \\cup C}) \\rightarrow (A^B \\times A^C)\n\\]\nbe such a function that \n\\[\nf(g) = (u,v) \\equiv \\forall x(x \\in B \\Rightarrow g(x) = u(x) \\wedge x \\in C \\Rightarrow g(x) = v(x)),\n\\]\nwhere $g\\in A^{B \\cup C}, (u,v) \\in A^B \\times A^C$.\n\nWe can verify the bijectivity of $f$ nearly in the same way as way did previously. So I won't write it \ndown here.\n\nThen, we know $B \\cap C = \\varnothing \\Rightarrow \\#(B \\cup C) = \\#B + \\#C$. So we can conclude that \n\\[\na^b \\times a^c = a^{b+c}\n\\]\n\\end{proof}\n\n\\declareexercise{3.6.7}\n\\begin{proof}\nOn one hand, if $\\#A = a \\leq \\#B = b$, we show that $A$ has lesser or equal cardinalty to $B$. Let $f$ \nbe a bijection from $A$ to $\\mathbb{N}_{a}$, $g$ be a bijection from $B$ to $\\mathbb{N}_b$. Let \n$\\iota(x):=x, \\mathbb{N}_{a} \\rightarrow \\mathbb{N}_b$. Then $g^{-1} \\circ \\iota \\circ f$ is an injection \nfrom $A$ to $B$.\n\nOn the other hand, suppose that there is an injection $f$ from $A$ to $B$. We know that \n$f:A\\rightarrow f(A)$ is bijective. So $\\#A = \\#f(A)$. Since $f(A) \\subseteq B$, $\\#f(A) \\leq B$ (See (c) \nin Proposition 3.6.14). That is, $\\#A \\leq \\#B$\n\\end{proof}\n\n\\declareexercise{3.6.8}\n\\begin{proof}\n$f:A \\rightarrow f(A)$ is bijective. So $f^{-1}: f(A) \\rightarrow A$ is surjective. Let $g$ be defined as:\n\\begin{itemize}\n\\item $b \\in f(A) \\Longrightarrow g(b) = f^{-1}(b)$\n\\item $b \\in B\\setminus f(A) \\Longrightarrow g(b)$ is any element of $A$.\n\\end{itemize}\n\nThen $g$ is surjective.\n\\end{proof}\n\n\\declareexercise{3.6.9}\n\\begin{proof}\n\\begin{align*}\n\\#A + \\#B \n&= \\#(A - A \\cap B) + \\#(A \\cap B) + \\#(B - A \\cap B) + \\#(A \\cap B) \\\\\n&= (\\#(A - A \\cap B) + \\#(A \\cap B) + \\#(B - A \\cap B)) + \\#(A \\cap B) \\\\\n&= \\#(A \\cup B) + \\#(A \\cap B)\n\\end{align*}\n\\end{proof}\n\n\\paragraph{Exercise 3.6.10} \\label{exercise3.6.10}\n\\begin{proof}\nPresume the negation:\n\\[\n\\forall i(i \\in \\{1,\\dots,n\\} \\Longrightarrow \\#(A_i) < 2)\n\\]\n\nUse mathematical induction for (b) in Proposition 3.6.14, we can easily get:\n\\[\n\\#\\bigcup_{i \\in \\{1,\\dots,n\\}}A_i \\leq \\sum_{i \\in \\{1,\\dots,n\\}} \\#A_i\n\\]\nWe can also use mathematical induction to furthermore enhance what we proved while dealing with natural \nnumbers to:\n\\[\n\\bigwedge_{i} a_{i} \\leq b_{i} \\Longrightarrow \\sum_{i} a_{i} \\leq \\sum_{i} b_i\n\\].\n\nThen because $\\# A_i \\leq 1$, so \n\\[\n\\sum_{i \\in \\{1,\\dots,n\\}} A_i \\leq (\\sum_{i \\in \\{1,\\dots,n\\}} 1 = n)\n\\], \nwhich is impossible.\n\\end{proof}", "meta": {"hexsha": "65b10fdd1befd8a29df1c751aa6b06468fc8ce2e", "size": 59789, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Set Theory.tex", "max_stars_repo_name": "Little-He-Guan/Notebook-for-Analysis-of-Tao", "max_stars_repo_head_hexsha": "e040260e4346ae65ce28af11dbd2bb5d9d5ac96b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Set Theory.tex", "max_issues_repo_name": "Little-He-Guan/Notebook-for-Analysis-of-Tao", "max_issues_repo_head_hexsha": "e040260e4346ae65ce28af11dbd2bb5d9d5ac96b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Set Theory.tex", "max_forks_repo_name": "Little-He-Guan/Notebook-for-Analysis-of-Tao", "max_forks_repo_head_hexsha": "e040260e4346ae65ce28af11dbd2bb5d9d5ac96b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.9308894231, "max_line_length": 196, "alphanum_fraction": 0.61880948, "num_tokens": 22767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746407, "lm_q2_score": 0.8289388146603365, "lm_q1q2_score": 0.6281651678806756}} {"text": "\\subsection{Quotient spaces}\n\\begin{defn}\nLet $X$ be a topological space and $\\sim$ an equivalence relation\non $X$. ($a\\sim a$, $a\\sim b\\Leftrightarrow b\\sim a,a\\sim b,b\\sim c,a\\sim c$.)\nLet $Y=X/\\sim$ be the set of equivalence classes and define\n\n\\begin{align*}\n\\pi:X\\to Y=X/\\sim\\\\\n\\pi(x)=[x]=\\text{equiv class of \\ensuremath{X}under \\ensuremath{\\sim}}\n\\end{align*}\n\n\nthen the quotient space $Y=X/\\sim$ has a \\emph{quotient topology}\nsuch that $V\\subseteq X/\\sim$ is open iff $\\pi^{-1}(V)$ is open\nin $X$.\n\nNote: A continuous map $f:X\\to Y$ is a \\emph{quotient map} if $f$\nis onto and $U\\subset Y$ is open $\\Leftrightarrow$ $f^{-1}(U)$\nis open in $X$.\n\\end{defn}\n\n\\begin{thm}\n$g:X/\\sim\\to Z$ is continuous $\\Leftrightarrow$ $g\\circ\\pi:X\\to Z$\nis continuous.\n\n\\begin{align*}\n\\xymatrix{X\\ar[r]^{g\\circ\\pi}\\ar[d]_{\\pi} & Z\\\\\nX/\\sim\\ar[ur]_{g}\n}\n\\end{align*}\n\\end{thm}\n\\begin{cor}\nLet $f:X\\to Z$ be a continuous map which is constant of equivalence\nclasses (if $u\\sim v$ then $f(u)=f(v)$), then $f$ induces a unique\ncontinuous map $\\bar{f}:X/\\sim\\to Z$ such that $\\bar{f}\\circ\\pi=f$.\n\\end{cor}\n\\begin{align*}\n\\xymatrix{X\\ar[r]^{f}\\ar[d]_{\\pi} & Z\\\\\nX/\\sim\\ar[ur]_{\\bar{f}}\n}\n\\end{align*}\n\n\nNote: $\\pi:X\\to X/\\sim$ is continuous, so if $X$ is compact or connected\nor path connected, then so is $X/\\sim$.\n\\begin{thm}\nIf $f:X\\to Y$ continuous and onto, and\\end{thm}\n\\begin{itemize}\n\\item $f$ is an open map, or\n\\item $f$ is a closed map\n\\end{itemize}\nthen $Y\\approx X/\\sim$ where $x\\sim y\\Leftrightarrow f(x)=f(y)\\forall x,y\\in X$.\n\\begin{defn}\nLet $X,Y$ be topological spaces. Two continuous maps $f,g:X\\to Y$\nare \\emph{homotopic} if there exists a continuous map $F:X\\times I\\to Y$\nsuch that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for all $x\\in X$. We\nwrite $f\\simeq g$ if $F$ exists, and say that $F$ is a homotopy\nfrom $f$ to $g$.\\end{defn}\n\\begin{thm}\nHomotopy is an equivalence relation on the set of all continuous maps\n$\\{\\text{cts maps}:X\\to Y\\}$ between spaces $X,Y$. I.e.\\end{thm}\n\\begin{itemize}\n\\item $f\\simeq f$\n\\item If $f\\simeq g\\implies g\\simeq f$\n\\item If $f\\simeq g$, $g\\simeq h$, $\\implies f\\simeq h$.\n\\end{itemize}\nNote: Let $X=A\\cup B$, $A,B$ subspaces of $X$. Suppose $f:A\\to Y$,\n$g:B\\to Y$ are continuous maps such that $f(x)=g(x)$ $\\forall x\\in A\\cap B$.\nThen define $h:A\\cup B\\to Y$ by \n\n\\begin{align*}\nh(x) & =\\begin{cases}\nf(x) & x\\in A\\\\\ng(x) & x\\in B\n\\end{cases}\n\\end{align*}\n\n\nthen $h$ is continuous if either\n\\begin{itemize}\n\\item $A,B$ are closed subsets\n\\item $A,B$ are open subsets\\end{itemize}\n\\begin{defn}\nLet $f,g:X\\to Y$ be continuous maps which agree on $A$. Then $f\\simeq g$\n\\emph{relative to $A$ }if there is a homotopy $F:f\\to g$ which fixes\nall points in $A$. I.e. we have $F:X\\times I\\to Y$ with $F(x,0)=f(x)$\n$F(x,1)=g(x)$ and $\\forall a\\in A$ $F(a,t)=f(a)=g(a)$ $\\forall t$.\n\\end{defn}\n\n\\begin{defn}\nA continuous map $f:X\\to Y$ is a \\emph{homotopy equivalence} if there\nexists a continuous map $g:Y\\to X$ such that\n\n\\begin{align*}\ng\\circ f\\simeq1_{X}:X\\to X\\\\\nf\\circ g\\simeq1_{y}:Y\\to Y\n\\end{align*}\n\n\nwe say that $X$ and $Y$ are homotopy equivalent of there exists\nsuch a map, and write $X\\simeq Y$.\n\\end{defn}\n\n\\begin{defn}\nA topological space $X$ is \\emph{contractible} if the identity map\n$\\text{id}_{X}:X\\to X$ is homotopic to a constant map $c:X\\to X,c(x)=x_{0}$\nfor all $x\\in X$.\n\nNote/Exercise: $X$ contractible $\\Leftrightarrow$ $X\\simeq\\{\\text{pt}\\}$.\n\\end{defn}\n\n\\begin{defn}\nGiven $A$ a subspace of $X$, a \\emph{(strong) deformation retraction}\nof $X$ onto $A$ is a homotopy rel $A$, $H:X\\times I\\to X$ such\nthat $H_{0}=1_{X}$, $H_{1}:X\\to A$. Then we have $A\\simeq X$.\n\\end{defn}\n\n\\begin{defn}\nLet $\\pi_{1}(X,x_{0})$ be a set of homotopy classes (rel endpoints)\nof loops in $X$ based at $x_{0}$ (with a suitably defined product.)\nThis is the \\emph{fundamental group of $X$ at $x_{0}$.}\\end{defn}\n\\begin{thm}\nIf $X$ is path connected, then $\\pi_{1}(X,x_{0})\\cong\\pi_{1}(X,x_{j})\\forall x_{j}\\in X$.\n\\marginpar{Maybe add in the picture?}\n\\end{thm}\n\n\n\\textbf{Effects of continuous maps}\n\nLet $f:X\\to Y$ be a continuous map. Then\n\\begin{itemize}\n\\item $\\alpha$ a path in $X$ $\\implies$ $f\\circ\\alpha=f\\alpha$ is a\npath in $Y$\n\\item $\\alpha\\sim\\alpha'$ in $X\\implies f\\circ\\alpha\\sim f\\circ\\alpha'$\nin $Y$.\n\\item $\\alpha$ a loop at $x_{0}$ $\\implies f\\circ\\alpha$ is a loop at\n$f(x_{0})$. So there is a well-defined function $f_{*}:\\pi_{1}(X,x_{0})\\to\\pi_{1}(F,f(x_{0}))$\ngiven by $f_{*}([\\alpha])=[f\\circ\\alpha]$.\\end{itemize}\n\\begin{thm}\n~\\end{thm}\n\\begin{enumerate}\n\\item $f_{*}$ is a group homomorphism.\n\\item If $f=\\text{identity}=1_{X}:X\\to X$ then the induced map is the identity\nmap $f_{*}:\\pi_{1}(X,x_{0})\\to\\pi_{1}(X,x_{0})$.\n\\item $(f\\circ g)_{*}=f_{*}\\circ g_{*}$.\\end{enumerate}\n\\begin{cor}\nIf $f:X\\to Y$ is a homeomorphism, then $f_{*}:\\pi_{1}(X,x_{0})\\to\\pi_{1}(Y,f(x_{0}))$\nis a group isomorphism. So the fundamental group is a topological\ninvariant.\\end{cor}\n\\begin{thm}\nLet $f,g:X\\to Y$ be homotopic maps via a homotopy $F:X\\times I\\to Y$.\nLet $\\gamma:I\\to Y$ be the path $\\gamma(t):F(x_{0},t)$ traced out\nby the basepoint $x_{0}$ under $F$ and let $y_{0}:y(0)=F(x_{0},0)$,\n$y_{1}=y(1)=F(x_{0},1)$, then $g_{*}=\\gamma_{*}\\circ f_{*}$ where\n$\\gamma_{*}([\\alpha])=[\\gamma^{-1}*\\alpha*\\gamma]=\\text{change of base point isomorphism}$,\ni.e. the diagram\n\n\\begin{align*}\n\\xymatrix{\\pi_{1}(X,x_{0})\\ar[dr]_{g_{*}}\\ar[r]^{f_{*}} & \\pi_{1}(Y,y_{0})\\ar[d]_{\\gamma_{*}\\cong}\\\\\n & \\pi_{1}(Y,y_{1})\n}\n\\end{align*}\n\n\ncommutes.\n\nNote: If $F$ does not move the basepoint $x_{0}$ then $\\gamma_{*}=\\text{id}$\nand $f_{*}=g_{*}$.\\end{thm}\n\\begin{lem}\nLet $G:I\\times I\\to Y$ be continuous and let $a,b:I\\to I\\times I$\nbe paths such that $a(0)=b(0)$ and $a(1)=b(1)$. Then $G\\circ a\\cong G\\circ b$\nrel endpoints.\n\\end{lem}\nApplication: If $f:X\\to Y$ is a homotopy-equivalence then $f_{*}:\\pi_{1}(X,x_{0})\\to\\pi_{1}(Y,f(x_{0}))$\nis an ismorphism for all $x_{0}\\in X$. So homotopically-equivalence\nspaces has isomorphic fundamental groups.\n\\begin{defn}\n$X$ is simply-connected if $X$ is path connected and $\\pi_{1}(X,x_{0})\\cong\\{1\\}$\nfor all $x_{0}\\in X$.\\end{defn}\n\\begin{lem}\nIf $X\\simeq Y$ then $X$ is path connected iff $Y$ is path connected.\\end{lem}\n\\begin{thm}\nLet $p_{1}:X\\times Y\\to X$ and $p_{2}:X\\times Y\\to Y$ be projection\nmaps. Then $(p_{1*})\\times(p_{2*}):\\pi_{1}(X\\times Y,(x_{0},y_{0}))\\to\\pi_{1}(X,x_{0})\\times\\pi_{1}(Y,y_{0})$\nis an isomorphism. Explicitly\n\n\\begin{align*}\np_{1*}\\times p_{2*}([\\alpha]) & =([p_{1}\\circ\\alpha],[p_{2}\\circ\\alpha])\n\\end{align*}\n\\end{thm}\n\\begin{lem}\n(Homotopy lemma) If $\\alpha_{0},\\alpha_{1}:I\\to S^{1}$ are paths\nsuch that $\\alpha_{0}\\simeq\\alpha_{1}$ rel endpoints, then $\\deg(\\alpha_{0})=\\deg(\\alpha_{1})$.\n(Where we defined the degree to be that ``distance moved upwards'',\ni.e. $\\tilde{\\alpha}(1)$, where $\\tilde{\\alpha}$ is the lifted path. \\end{lem}\n\\begin{thm}\n(Fundamental theorem of algebra) Every non-constant polynomial with\ncomplex coefficients has a complex zero.\n\nIdea:\\end{thm}\n\\begin{enumerate}\n\\item Let $f:B\\to\\mathbb{C}$ be continuous and defined on a closed 2-ball\n$B\\approx B^{2}$. Assume $f(z)\\neq0$ on the boundary, i.e. $\\forall z\\in\\partial B\\approx S^{1}$.\nDefine $g:\\partial B\\to S^{1}$ by $g(z)=f(z)/|f(z)|$ (continuous).\nThen If $g_{*}:\\pi_{1}(\\partial B)\\cong\\mathbb{Z}\\to\\pi_{1}(S^{1})\\cong\\mathbb{Z}$\nis non-trivial $f$ has a zero inside $B$. (To prove this, try the\ncontrapositive.)\n\\item To prove FTA: Apply this result to $f(z)=z^{n}+a_{n-1}z^{n-1}+\\cdots$\nand $B$ a sufficiently large ball of radius $r$ around the origin.\\end{enumerate}\n\\begin{thm}\nAny homeomorphism $B^{2}\\to B^{2}$ takes $\\partial B^{2}=S^{1}$\nto $S^{1}$. (Idea: Remove various points.)\n\\end{thm}\n\n", "meta": {"hexsha": "9a0619b2185f245b49472415773c22277f1eb03b", "size": 7692, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "quotient_spaces.tex", "max_stars_repo_name": "silky/alg-top-notes", "max_stars_repo_head_hexsha": "2e47a522a31a93487df6ab523ed555aaa2ca017c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-11-17T08:28:55.000Z", "max_stars_repo_stars_event_max_datetime": "2017-11-17T08:28:55.000Z", "max_issues_repo_path": "quotient_spaces.tex", "max_issues_repo_name": "silky/alg-top-notes", "max_issues_repo_head_hexsha": "2e47a522a31a93487df6ab523ed555aaa2ca017c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "quotient_spaces.tex", "max_forks_repo_name": "silky/alg-top-notes", "max_forks_repo_head_hexsha": "2e47a522a31a93487df6ab523ed555aaa2ca017c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.776744186, "max_line_length": 109, "alphanum_fraction": 0.6484659386, "num_tokens": 3083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8289388146603365, "lm_q1q2_score": 0.6281651678806754}} {"text": "\\documentclass[]{article}\n\n\\usepackage{amsmath}\n\\usepackage[backend=bibtex]{biblatex}\n\\addbibresource{main.bib}\n\n%opening\n\\title{Multiple Disjunctively Constrained Knapsack Problem}\n\\author{Marco Favorito}\n\\date{}\n\n\\begin{document}\n\n\\maketitle\n\n\\begin{abstract}\nIn this document, we will present a problem, \\emph{Multiple Disjunctively Constrained Knapsack Problem} (MDCKP), that is a combination of well-known problems: \\emph{Multiple 0-1 Knapsack Problem} (MKP) and \\emph{Disjunctively Constrained Knapsack Problem} (DCKP). Linear programming formulation is provided, as well as correlations with other Knapsack problem variants. Finally, we describe a (naive) algorithm.\n\\end{abstract}\n\n\\section{Introduction}\n\nMultiple Disjunctively Constrained Knapsack Problem (MDCKP) is a combination of two Knapsack-like problems, namely: \n\n\\begin{itemize}\n\t\\item Multiple 0-1 Knapsack Problem (MKP) \\cite{Lagoudakis96the0-1, Martello:1990:KPA:98124};\n\t\\item Disjunctively Constrained Knapsack Problem (DCKP) \\cite{Senisuka_reductionand}\n\\end{itemize}\n\nInformally, it is about how to assign $n$ items to $m$ knapsacks, each of them with limited capacity, in order to maximize the total profit of assigned items, where:\n\\begin{itemize}\n\t\\item each item has a weight;\n\t\\item there are some items that cannot be assigned to the same knapsack, due to intrinsic incompatibility.\n\\end{itemize}\n\n\\section{Linear Programming}\nMore formally, we have $n$ items, each of them with $p_1, \\dots, p_n$ profits and $w_1, \\dots, w_n$ weights. Each of them can be assigned to only one of the $m$ knapsacks, which have capacities $c_1, \\dots, c_m$. Let $E$ be the set of incompatible pairs, such that $(i, j)\\in E$ iff items $i$ and $j$ are incompatible. To avoid duplicates, we define $E \\subset \\{(i, j) | 1 \\le i \\neq j \\le n\\}$ and assume $E$ to be a reflective relation.\n\nNow follow a LP formulation of the problem:\n\n\\begin{align*}\n\\text{maximize } & \\sum_j^m \\sum_i^n p_i x_{ij} \t&\\\\\n\\text{subject to } & \\sum_i^n w_i x_{ij} \\le c_j \t& j \\in M = \\{1, \\dots m\\}\\\\\n\t\t\t\t & \\sum_j^m x_{ij} \\le 1 \t\t\t& i \\in N = \\{1, \\dots n\\}\\\\\n\t\t\t\t & x_{kj} + x_{hj} \\le 1\t\t\t& (k, h)\\in E, j\\in M\\\\\n\t\t\t\t & x_{ij} \\in \\{0, 1\\} \t\t\t& i \\in N, j \\in M\\\\\n\\end{align*}\nwhere:\n\n\\begin{equation*}\nx_{ij} = \\begin{cases}\n\t1 & \\text{if item $i$ is assigned to knapsack $j$} \\\\\n\t0 & \\text{otherwise}\n\\end{cases}\n\\end{equation*}\n\nNotice: with $E=\\emptyset$ and $m=1$, the problem reduces to the classic 0-1 Knapsack problem. Hence, MDCKP is $\\mathcal{NP}$-hard, since KP is $\\mathcal{NP}$-hard too.\n\n\\section{Naive Approximation Algorithm}\nThe simplest approximation algorithm one can think of is the following:\n\\begin{itemize}\n\t\\item sort the items by non-increasing profit per weight, that is:\n\t\\begin{equation*}\n\t\tp_1/w_1 \\ge p_2/w_2 \\ge \\dots \\ge p_n/w_n\n\t\\end{equation*}\n\t\\item start to fill up the knapsacks and, if there is not enough capacity or there is some incompatibility with other items, then go to the next knapsack. If no knapsack is available, then leave the item out.\n\t\\item perform some local search algorithm by defining the neighborhood of a solution as the following:\n\t\n\t\\paragraph{} Look for a pair of items, one inside and the outer outside the knapsacks, and see if it is possible to exchange those items. If the exchange yields a better solution, swap the items.\n\t\n\\end{itemize}\n\n\\section{Implementation}\nWe provided an implementation of the approximation algorithm in C++. \n\nThe input file should have this format:\n\\begin{align*}\n&n\\ m\t\t\t\t\t\t\t\t \\\\\n&c_1 \\; c_2 \\; \\dots \\; c_m\t\t\t \\\\\n&p_1 \\; w_1\t\t\t\t\t\t\t \\\\\n& \\vdots\t\t\t\t\t\t\t \\\\\n&p_n \\; w_n\t\t\t\t\t\t\t \\\\\n& i \\; j \\; \\;\t(\\forall (i,j)\\in E)\t\\\\\n\\end{align*}\n\nWhereas the output file:\n\n\\begin{align*}\n&\\sum_j^m \\sum_i^n p_i x_{ij}\t\t\t\\\\\n&x_{11} \\; x_{21} \\; \\dots \\; x_{n1} \\\\\n\\vdots\t\t\t\t\t\t\t\t\t\\\\\n&x_{1m} \\; x_{2m} \\; \\dots \\; x_{nm} \\\\\n\\end{align*}\n\n\\subsection{Example}\n\nTo build the solution:\n\\begin{verbatim}\nmkdir build && cd build\ncmake ..\nmake -j4\n\\end{verbatim}\n\nTo run the algorithm:\n\\begin{verbatim}\nbuild/apps/mdckp-cli/mdckp-cli < input_file\n\\end{verbatim}\n\n\n\\printbibliography\n\n\\end{document}\n", "meta": {"hexsha": "13a56c2087c8a288d7c7cbf8bd24d447a8672a08", "size": 4144, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/main.tex", "max_stars_repo_name": "MarcoFavorito/MDCKP", "max_stars_repo_head_hexsha": "db85a015ef23d37a1a0a14e73eb3b5f103af442f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-12-01T21:22:17.000Z", "max_stars_repo_stars_event_max_datetime": "2019-12-01T21:22:17.000Z", "max_issues_repo_path": "docs/main.tex", "max_issues_repo_name": "MarcoFavorito/MDCKP", "max_issues_repo_head_hexsha": "db85a015ef23d37a1a0a14e73eb3b5f103af442f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/main.tex", "max_forks_repo_name": "MarcoFavorito/MDCKP", "max_forks_repo_head_hexsha": "db85a015ef23d37a1a0a14e73eb3b5f103af442f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-11-29T09:19:28.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-29T09:19:28.000Z", "avg_line_length": 37.0, "max_line_length": 440, "alphanum_fraction": 0.6993243243, "num_tokens": 1338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8289388125473629, "lm_q1q2_score": 0.6281651662794759}} {"text": "\\documentclass{amsart}\n\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\n\\title{Probability: Quiz 1}\n\\author{Mark Ditsworth}\n\n\\begin{document}\n\t\\maketitle\n\t\\section{Problem 1}\n\tIndicate \\textit{True} or \\textit{False} for each claim.\n\t\\subsection{Part A}\n\tIf $\\mathbf{P}(A) \\leq \\mathbf{P}(B)$, then $A \\subseteq B$.\\\\\n\t\\textbf{False}. Let $A$ and $B$ be independent, but $\\mathbf{P}(A) \\leq \\mathbf{P}(B)$.\n\t\\\\\n\t\\subsection{Part B}\n\tIf $\\mathbf{P}(B) > 0$, $\\mathbf{P}(A|B)$ is at least as large as $\\mathbf{P}(A)$.\\\\\n\t\\textbf{False}. $\\mathbf{P}(A)$ can decrease if it is dependent on $B$. For example, the probability of rain today ($\\mathbf{P}(A)$) may historically be 35\\%. But the probability of rain today given clear skies ($\\mathbf{P}(A|B)$) is lower than $\\mathbf{P}(A)$.\\\\\n\t\\subsection{Part C}\n\tIf $\\mathbf{E}[X] > \\mathbf{E}[Y]$, then $\\mathbf{E}[X^2] > \\mathbf{E}[Y^2]$.\\\\\n\t\\textbf{False}. Let $X$ take the values 0 and 1 with equal probability, and $Y$ take the values -4 and -2 with equal probability. $\\mathbf{E}[X]=0.5 > \\mathbf{E}[Y] = -3$. But $\\mathbf{E}[X^2]=0.5 < \\mathbf{E}[Y^2] = 10$.\\\\\n\t\\subsection{Part D}\n\tSuppose $\\mathbf{P}(A) > 0$. Then $\\mathbf{E}[X] = \\mathbf{E}[X|A]+ \\mathbf{E}[X|A^C]$.\\\\\n\t\\textbf{False}. Assume $X$ is independent of $A$. Then $\\mathbf{E}[X]=\\mathbf{E}[X|A]= \\mathbf{E}[X|A^C]$, so $\\mathbf{E}[X|A]+ \\mathbf{E}[X|A^C] = 2\\mathbf{E}[X]$.\\\\\n\t\\subsection{Part E}\n\tIf $X$ and $Y$ are independent and $\\mathbf{P}(C)>0$, then $p_{X,Y|C}(x,y) = p_{X|C}(x)p_{Y|C}(y)$.\\\\\n\t\\textbf{False}. Independence between $X$ and $Y$ does not guarantee conditional independence.\\\\\n\t\\subsection{Part F}\n\tIf for some constant $c$, we have $\\mathbf{P}(\\{X>c\\})=\\frac{1}{2}$, then $\\mathbf{E}[X]>\\frac{c}{2}$.\\\\\n\t\\textbf{False}. Let $X$ take the values between -1 and 1, uniformly. $\\mathbf{P}(\\{X>0\\})=\\frac{1}{2}$, and $\\mathbf{E}[X] = 0 \\ngtr \\frac{0}{2}$.\\\\\n\t\\subsection{Part G}\n\tIn a game involving the flip of a fair coin, you win a dollar for each head flipped. The maximum number of flips is 10, but the game ends when you flip a tail. The expected gain from this game is 1.\\\\\n\t\\textbf{False}. $\\mathbf{E}[g(X)] = \\sum g(x) p_X(x) = \n\t\\sum_{k=0}^{9}\\left(k(0.5)^{k+1}\\right) + (0.5)^{10} = 0.99902$.\\\\\n\t\\subsection{Part H}\n\tLet $X$ be a uniformly distributed continuous random variable over some interval $[a,b]$. We can uniquely describe $f_X(x)$ from knowing its mean and variance.\\\\\n\t\\textbf{True}. For a uniform distribution, $\\mathbf{E}[X]=\\frac{a+b}{2}$ and $\\sigma_X^2 = \\frac{(b-a)^2}{12}$. Two equations and two unknowns result in a solvable system of equations for $a$ and $b$.\\\\\n\t\\subsection{Part I}\n\tLet $X$ be an exponentially distributed random variable with a probability density function $f_X(x)= e^{-x}$. $P\\left(\\{0 \\leq X \\leq 3\\} \\cup \\{2 \\leq X\\leq 4\\}\\right) = 1-e^{-4}$.\\\\\n\t\\textbf{True}. $P\\left(\\{0 \\leq X \\leq 3\\} \\cup \\{2 \\leq X\\leq 4\\}\\right) = \\int_{0}^{4}e^{-x} dx = 1-e^{-4}$.\\\\\n\t\\subsection{Part J}\n\tLet $X$ be a normal random variable with mean 1 and variance 2. Let $Y$ be a normal random variable with mean 1 and variance 1. $\\mathbf{P}(X<0) < \\mathbf{P}(Y<0)$.\\\\\n\t\\textbf{False}. Since $X$ has a larger variance than $Y$ its tails are thicker, thus the integral over the left tail will be greater for $X$ and $Y$.\n\t\\pagebreak\n\t\\section{Problem 2}\n\tMarketing Team A has found that 1/4 of their customers are low-frequency buyers. But out of their low-frequency buyers, 1/3 are high spenders, whereas 1/10 of their high-frequency buyers are high spenders. Assume that each customer is either low-frequency or high frequency, and low spender of high spender.\\\\\n\t\\subsection{Part A}\n\tCompute the probability that a randomly chosen customer is a high spender.\\\\\n\t$\\mathbf{P}(HS)=\\mathbf{P}(HFB)\\mathbf{P}(HS|HFB)+\\mathbf{P}(LFB)\\mathbf{P}(HS|LFB)=\\frac{1}{4}\\frac{1}{3}+\\frac{3}{4}\\frac{1}{10} = \\frac{19}{120}$.\\\\\n\t\\subsection{Part B}\n\tCompute the probability that a randomly chosen customer is a high-frequency buyer given that they are a low spender.\\\\\n\t$\\mathbf{P}(HFB|LS) = \\frac{\\mathbf{P}(HFB)\\mathbf{P}(LS|HFB)}{\\mathbf{P}(LS)} = \\frac{\\frac{3}{4}\\frac{9}{10}}{1-\\frac{19}{120}}=\\frac{81}{101}$\\\\\\\\\n\tYou are told that the only products sold are books (B), CDs (CD), and DVDs (DVD). Marketing team B has found three customer groups $C_1$, $C_2$, $C_3$. These groups are collectively exhaustive and mutually exclusive. Each customer is equally likely to be in any group, and customers are \\textit{i.i.d.}. Each customer only buys one item per day. Marketing team B has determined the following purhcasing conditional probabilities:\\\\\n\t\\begin{table}[h!]\n\t\t\\centering\n\t\t\\begin{tabular}{rcl}\n\t\t\t$\\mathbf{P}(B|C_1)$&=&$\\frac{1}{2}$\\\\\n\t\t\t$\\mathbf{P}(CD|C_1)$&=&$\\frac{1}{4}$\\\\\n\t\t\t$\\mathbf{P}(DVD|C_1)$&=&$\\frac{1}{4}$\\\\\n\t\t\t$\\mathbf{P}(B|C_2)$&=&$\\frac{1}{2}$\\\\\n\t\t\t$\\mathbf{P}(CD|C_2)$&=&$0$\\\\\n\t\t\t$\\mathbf{P}(DVD|C_2)$&=&$\\frac{1}{2}$\\\\\n\t\t\t$\\mathbf{P}(B|C_3)$&=&$\\frac{1}{3}$\\\\\n\t\t\t$\\mathbf{P}(CD|C_3)$&=&$\\frac{1}{3}$\\\\\n\t\t\t$\\mathbf{P}(DVD|C_3)$&=&$\\frac{1}{3}$\n\t\t\\end{tabular}\n\t\\end{table}\n\t\\subsection{Part C}\n\tCompute the probability that a customer purchases a book or CD.\\\\\n\tSince $C_1$, $C_2$, and $C_3$ are mutually exclusive and customers are \\textit{i.i.d.},\\\\\n\t$\\mathbf{P}(B\\cup CD) = \\mathbf{P}(B) + \\mathbf{P}(CD)$\\\\\n\t$\\mathbf{P}(B) = \\sum_{i=1}^{3} \\mathbf{P}(C_i)\\mathbf{P}(B|C_i) = \\frac{1}{6} + \\frac{1}{6} + \\frac{1}{9} = \\frac{4}{9}$\\\\\n\t$\\mathbf{P}(CD) = \\sum_{i=1}^{3} \\mathbf{P}(C_i)\\mathbf{P}(CD|C_i) = \\frac{1}{12} + 0 + \\frac{1}{9} = \\frac{7}{36}$\\\\\n\t$\\mathbf{P}(B\\cup CD) = \\frac{4}{9} + \\frac{7}{36} = \\frac{23}{36} \\approx 0.6389$\\\\\n\t\\subsection{Part D}\n\tCompute the probability that a customer is in group $C_2$ or $C_3$ given they purchased a book.\\\\\n\t$\\mathbf{P}(C_2 \\cup C_3 | B) = \\frac{\\mathbf{P}(C_2 \\cup C_3) \\mathbf{P}(B|C_2 \\cup C_3)}{\\mathbf{P}(B)}=\n\t\\frac{\\mathbf{P}(B \\cap (C_2 \\cup C_3))}{\\mathbf{P}(B)} = \\frac{\\mathbf{P}(B \\cap C_2) + \\mathbf{P}(B \\cap C_3)}{\\mathbf{P}(B)} = \\frac{5}{8}$\\\\\n\t\\\\\n\tEach book costs \\$15, each CD costs \\$10, and each DVD costs \\$15.\n\t\\subsection{Part E}\n\tCompute the PDF, expected value, and variance of the revenue ($R$) collected by a single customer's purchase.\n\t\\\\\n\t$\\mathbf{P}(R=\\text{\\$10})=\\mathbf{P}(CD) = \\frac{7}{36}$\\\\\n\t$\\mathbf{P}(R=\\text{\\$15})=\\mathbf{P}(B)+\\mathbf{P}(DVD) = \\frac{4}{9}+\\frac{13}{36}$\\\\\n\t\\[\n\tp_R(r) = \n\t\\begin{cases}\n\t\t\\frac{7}{36} & r = \\$10\\\\\n\t\t\\frac{29}{36} & r = \\$15\\\\\n\t\t0 & \\text{otherwise}\n\t\\end{cases}\n\t\\]\\\\\n\t$\\mathbf{E}[R] = \\sum r \\cdot p_R(r) = 10(\\frac{7}{36}) + 15(\\frac{29}{36}) = \\$14.03$\\\\\\\\\n\t$\\text{Var}(R) = \\sum (r-\\textbf{E}[R])^2 \\cdot p_R(r) \\approx 3.9159$\\\\\n\t\\subsection{Part F}\n\tIf the store gets $n$ customers in a day, what is the total expected revenue and variance?\\\\\n\t$\\text{Total Revenue}~ TR = nR$\\\\\n\t$\\mathbf{E}[TR] = \\mathbf{E}[nR] = n\\mathbf{R} = n\\cdot\\$14.03$\\\\\\\\\n\t$\\text{Var}(nR) = n^2\\text{Var}(R) = n^2\\cdot \\$3.9159$\\\\\n\t\\subsection{Part G}\n\tSkip is an abnormal customer. When he goes to the store, he flips a fair coin until he gets his seconds tails. The number of heads that have been flipped is the number of DVDs he buys. What is the expected dollar amount he spends and the variance?\\\\\n\t\\\\\n\tThis process follows the $k$th arrival time process, modeled by the pascal distribution. Each time Skip flips his coin, he has a 50\\% probability of getting tails. We want to know the expected number of flips it takes to get his second tails. The number of heads is 2 less than this number since 2 of them are the 2 arrivals of tails.\n\t\\\\\\\\\n\t$\\text{For pascal},~\\mathbf{E}[Y_k] = \\frac{k}{p}$\\\\\n\t\\\\\n\t$\\mathbf{E}[Y_2] = \\frac{2}{0.5} = 4$\\\\\n\tThus, the expected number of heads is 2, resulting in an expected revenue of \\$30.\n\t\\\\\\\\\n\t$\\text{For pascal},~\\text{Var}[Y_k] = \\frac{k(1-p)}{p^2}$\\\\\n\t\\\\\n\t$\\text{Var}(Y_2) = \\frac{2(1-0.5)}{0.5^2}=4$\\\\\n\t$\\text{Var}(R) = \\text{Var}(\\$15\\cdot Y_2) = \\$15^2\\text{Var}(Y_2) = \\$900$\n\t\\pagebreak\n\t\\section{Problem 3}\n\tWe have $s$ urns and $n$ balls ($n \\geq s$). Each ball is placed in an urn at random. Each ball has equal probability of being placed in any urn, and it's placement is independent of the other ball placements. Each urn can fit any number of balls.\\\\\n\tDefine the following random variables:\\\\\\\\\n\tLet $X_i$ be the number of balls in urn $i$.\\\\\n\tLet $Y_k$ be the number of urns that have exactly $k$ balls.\\\\\n\t\\subsection{Part A}\n\tAre the $X_i$'s independent? Why or why not?\\\\\n\t\\textbf{They are not independent.} Without loss of generality, if $X_1=j$, then $X_2$ through $X_s$ can be at most $n-j$.\\\\\n\t\\subsection{Part B}\n\tFind the PMF, mean, and variance of $X_i$.\\\\\n\t\\\\\n\tIf there are $k$ balls in urn $i$, there must have been $k$ occurrences of successful placement, and $n-k$ occurrences of unsuccessful placement: $p^k(1-p)^{(n-k)}$. Since the order in which the $k$ balls are placed in urn $i$ does not matter, there are $\\binom{n}{k}$ possible combinations of this placement. Since urn placement is equally distributed, the probability of placement in urn $i$ is $\\frac{1}{s}$. Thus, the PMF is:\n\t\\[\n\t\tp_{X_{i}}(k) = \\left(\\frac{1}{s}\\right)^k \\left(1-\\frac{1}{s}\\right)^{(n-k)}\\binom{n}{k}\n\t\\]\\\\\n\t\\\\\n\tThis PMF is the binomial distribution. Thus, the mean is:\\\\\n\t\\[\n\t\t\\textbf{E}(X_i) = np = \\frac{n}{s}\n\t\\]\\\\\n\tand the variance is:\\\\\n\t\\[\n\t\t\\text{Var}(X_i) = np(1-p) = \\frac{n}{s}\\left(1-\\frac{1}{s}\\right)\n\t\\]\\\\\n\t\\\\\n\t\\subsection{Part C}\n\tAssume $n=10$ and $s=3$. Find the probability that the first urn has 3 balls, the second has 2, and the third has 5.\\\\\n\t\\\\\n\tThe number of combinations of dividing the 10 balls into partitions of 3, 2, and 5 is $\\binom{10}{3,2,5}$. The total number of ways to divide 10 balls among the 3 urns is $3^{10}$. Thus, the probability is $\\frac{\\binom{10}{3,2,5}}{3^{10}} \\approx 0.0427$.\\\\\n\t\\\\\n\t\\subsection{Part D}\n\tCompute \\textbf{E}[$Y_k$].\\\\\n\t\\\\\n\tRecall that $Y_k$ is the number of urns that have exactly $k$ balls. Define:\\\\\n\t\\[\n\t\tI_i=\n\t\t\\begin{cases}\n\t\t\t1 & \\text{urn } i \\text{ has } k \\text{ balls}\\\\\n\t\t\t0 & \\text{otherwise}\n\t\t\\end{cases}\n\t\\]\n\tNow we can write $Y_k = \\sum_{i=1}^{s} I_i$. Due to linearity, $\\textbf{E}[Y_k] = \\textbf{E}\\left[\\sum_{i=1}^{s}\\right] =\\sum_{i=1}^{s} \\textbf{E}[I_i]$.\\\\\n\t\\\\\n\t$\\textbf{E}[I_i]$ is just the probability that urn $i$ has $k$ balls, $p_{X_i}(k)$, which is known from Part B. Thus,\n\t\\[\n\t\t\\textbf{E}[Y_k] = \\sum_{i=1}^{s} \\left(\\frac{1}{s}\\right)^k\\left(1-\\frac{1}{s}\\right)^{(n-k)}\\binom{n}{k} = s \\left(\\frac{1}{s}\\right)^k\\left(1-\\frac{1}{s}\\right)^{(n-k)}\\binom{n}{k}\n\t\\]\n\t\\\\\n\t\\subsection{Part E}\n\tCompute Var($Y_k$).\\\\\n\t\\\\\n\tVar($Y_k$)$= \\textbf{E}[Y_k^2] - \\textbf{E}[Y_k]^2$. The second term is known from Part D. Thus, we need to find $\\textbf{E}[Y_k^2]$.\\\\\n\t\\[\n\t\t\\textbf{E}[Y_k^2] = \\textbf{E}\\left[\\left( \\sum_{i=1}^{s}I_i\\right)^2\\right]\n\t\t=\n\t\t\\textbf{E}\\left[\n\t\t\\sum_{i=1}^{s}I_i^2\\right] + 2\\textbf{E}\\left[ \\sum_{i=1}^{s-1}\\sum_{j=i+1}^{s}I_iI_j\n\t\t\\right]\n\t\\]\\\\\n\tSince $I_i$ is either 0 or 1, $I_i^2 = I_i$. Thus, \n\t$\\textbf{E}\\left[\\sum_{i=1}^{s}I_i^2\\right] = \\textbf{E}\\left[\\sum_{i=1}^{s}I_i\\right] = \\textbf{E}[Y_k]$, which is known from Part D. We now need to find $\\textbf{E}\\left[\\sum_{i=1}^{s-1}\\sum_{j=i+1}^{s}I_iI_j\n\t\\right]$ to be done.\\\\\n\t\\\\\n\t$\\textbf{E}\\left[\\sum_{i=1}^{s-1}\\sum_{j=i+1}^{s}I_iI_j\n\t\\right] = \\sum_{i=1}^{s-1}\\sum_{j=i+1}^{s}\\textbf{E}\\left[I_iI_j\n\t\\right]$\\\\\\\\\n\t$\\textbf{E}\\left[I_iI_j\\right]$ is the probability that $X_i$ and $X_j$ both equal $k$. This is $2k$ successful placements and $n-2k$ unsuccessful placements. The number of combinations of these placements is the number of ways to partition the $n$ balls into 2 sets of size $k$ and a third set of size $n-2k$. Thus this probability is $\\binom{n}{k,k,n-2k}\\left(\\frac{1}{s}\\right)^{2k}\\left(1-\\frac{1}{s}\\right)^{n-2k}$.\\\\\n\t\\\\\n\tPutting it all together:\n\t\\[\n\t\\text{Var}(Y_k) = s \\left(\\frac{1}{s}\\right)^k\\left(1-\\frac{1}{s}\\right)^{(n-k)}\\binom{n}{k} + \n\t(s)(s-1)\\binom{n}{k,k,n-2k}\\left(\\frac{1}{s}\\right)^{2k}\\left(1-\\frac{1}{s}\\right)^{n-2k}\\]\n\t\\[-\\left(s \\left(\\frac{1}{s}\\right)^k\\left(1-\\frac{1}{s}\\right)^{(n-k)}\\binom{n}{k}\n\t\\right)^2\n\t\\]\n\t\n\t\n\t\t\n\\end{document}", "meta": {"hexsha": "1c7043cee277d7b45d4f2900208f53e1cb6f5cf3", "size": 12114, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Exams/Quiz1/quiz1.tex", "max_stars_repo_name": "markditsworth/psa", "max_stars_repo_head_hexsha": "af38e92111464c0d4ac7e833d9bb235d22c0a4ca", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Exams/Quiz1/quiz1.tex", "max_issues_repo_name": "markditsworth/psa", "max_issues_repo_head_hexsha": "af38e92111464c0d4ac7e833d9bb235d22c0a4ca", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Exams/Quiz1/quiz1.tex", "max_forks_repo_name": "markditsworth/psa", "max_forks_repo_head_hexsha": "af38e92111464c0d4ac7e833d9bb235d22c0a4ca", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.1230769231, "max_line_length": 432, "alphanum_fraction": 0.6315832921, "num_tokens": 4857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461390043208003, "lm_q2_score": 0.8418256412990658, "lm_q1q2_score": 0.6281189458106041}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\n\\title{The Factorization Problem\\\\In\\\\Computational Group Theory}\n\\author{Spencer T. Parkin}\n\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{definition}{Definition}[section]\n\\newtheorem{corollary}{Corollary}[section]\n\\newtheorem{identity}{Identity}[section]\n\\newtheorem{lemma}{Lemma}[section]\n\\newtheorem{result}{Result}[section]\n\n\\newcommand{\\stab}{\\mbox{stab}}\n\n\\begin{document}\n\\maketitle\n\n\\section*{The Problem And Solution}\n\nFor any given element $g$ of a finite permutation group $G$ generated\nby a set of permutation $S$, we wish to factor $g^{-1}$ in terms of $S$.\nSince $g\\in G=\\langle S\\rangle$, it is clear that for some positive integer $k$, there exists\na sequence of $k$ elements $\\{s_i\\}_{i=1}^k$ taken from $S$ such that\n\\begin{equation*}\ng = s_1s_2\\dots s_k,\n\\end{equation*}\nand therefore,\n\\begin{equation*}\ng^{-1} = s_k^{-1}s_{k-1}^{-1}\\dots s_1^{-1}.\n\\end{equation*}\nIndeed, there may be many such sequences.\nHow we might come up with such a sequence, however, is not immediately clear.\nOur goal is to come up with a systematic way of producing such a sequence, even if it\nis not of minimal length.\n\nAssociated with our group $G$ is a set $\\Omega$ over which each $g\\in G$ is defined.\nThat each $g$ is a permutation of this set $\\Omega$ is to say that $g$ is a one-to-one and onto\nmapping from $\\Omega$ to $\\Omega$.\nFor a given $\\omega\\in\\Omega$,\nwe define the stabilizer subgroup of $G$ with respect to $\\omega$ as\n\\begin{equation*}\n\\stab_{\\omega}(G) = \\{g\\in G|g(\\omega)=\\omega\\}.\n\\end{equation*}\nThis is the set of all permutation of $g$ that stabilize $\\omega$. It is not hard to show\nthat this is a subgroup of $G$. Furthermore, it is a normal subgroup of $G$. \nFor any $h\\in\\stab_{\\omega}(G)$,\nand any $g\\in G$, we have\n\\begin{equation*}\ng(h(g^{-1}(\\omega))) = g(g^{-1}(\\omega)) = \\omega.\n\\end{equation*}\nEDIT: WRONG!!!!!! You moron!!!\n\nBeing normal, we can consider the factor group\n\\begin{equation*}\nG/H = \\{gH|g\\in G\\},\n\\end{equation*}\nwhere $H=\\stab_{\\omega}(G)$. We now make the observation that since $S$\ngenerates $G$, we must have $SH=\\{sH|s\\in S\\}$ generating $G/H$. To see this, we write\nthe typical element of $G/H$ as\n\\begin{equation*}\ngH = \\left(\\prod_{i=1}^k s_i\\right)H = \\prod_{i=1}^k (s_iH).\n\\end{equation*}\n\nWe now let $R$ be a set of coset representatives for $G/H$, and $[\\cdot]:G\\to R$ a function\nmapping any element of $g$ to the coset representative in $R$ representing the coset in $G/H$ that contains $g$.\nThat is, $|R|=|G|/|H|=|G/H|$, and $G/H=\\{rH|r\\in R\\}$, and for all $g\\in G$, $gH=[g]H$ with $[g]\\in R$.\nComputationally, coming up with $R$ is really just the same problem as generating $G/H$ as we would simply use\ncoset representatives to represent each coset of $G/H$ anyway. An implementation of $[g]$ might,\nfor each $r\\in R$, check whether $g^{-1}r\\in H$.\n\nArmed with all this, and letting $g$ be any element of $G$, we're now going to begin to address our original\nquestion by finding an element $q$ in terms of the generators of $S$ such that $gq\\in H$. Clearly, a procedure\nfor such a thing is recursively applicable, for $H$ is simply a permutation group associated with $\\Omega-\\{\\omega\\}$,\nbringing us one step closer to the trivial group $\\{e\\}$. The trick is that at each step, we must keep track of all new\ngenerators in terms of the original generators in $S$.\n\nSo how do we go about finding $q$? Well, if $g\\in H$ (i.e., $g(\\omega)=\\omega$), then $q=e=ss^{-1}$ for any $s\\in S$,\nand we're done. If not, consider $q=[g]^{-1}$. Clearly $g[g]^{-1}\\in H$, since $gH=[g]H$, and furthermore,\nwe know the factorization of $[g]^{-1}$ in terms of the generators in $S$, because we know such a factorization\nfor $[g]\\in R$ by virtue of how we generated $R$ from $S$.\n\nWe are now almost there! All that remains is our ability to continue this process in $H$ with $gq\\in H$ and a stabilizer\nsubgroup of $H$ with respect to some element in $\\Omega-\\{\\omega\\}$. For that, we need a generating set for $H$,\nand this is where Schreier's Lemma comes into play. According to Schreier, this set of generators is given by\n\\begin{equation*}\n\\{rs[rs]^{-1}|r\\in R, s\\in S\\},\n\\end{equation*}\nprovided $e\\in R$.\nNote that computationally, while we might boil each generator down to its permutation $g\\in G$, we would need\nto keep its expression (or ``word'') in terms of our original generators in $S$.\n\nSo there we have it! In closing, it might be worth considering why the algorithm thus described does\nbetter than the most obvious, naive and impractical approach to solving the factorization problem, which is to\nsimply generate all of $G$ from $S$, then search for $g^{-1}$ in $G$. To begin, $|G|$ may be\nextremely large, making its complete generation take too long and require too much memory.\nThat being said, the stabilizer-chain approach may have its own, similar problems. For a group $G$\nof large order, how big is the order of a factor group likely to get? How big is a set of generators\nlikely to get? I'll have to give it a try. I believe it can also be shown that this method does not\nproduce solution sequences anywhere near minimal length.\n\n\\section*{Generating Groups}\n\nFor a finite group $G=\\langle S\\rangle$, how do we go about generating all elements of $G$ using that of $S$?\nLetting $G=\\{e\\}$, we proceed until $S$ is empty. Remove $s$ from $S$ and then, for each $g\\in G$,\nadd $gs$ or $sg$ to $S$ if they're not found in $G$ or $S$, then add $s$ to $G$.\n\nClearly this algorithm will terminate, but is it correct? As elements are added to $G$, keep them in order so\nthat in the end, we have $G$ as an array of elements showing the order that they were added. Then for any $g\\in G$,\nwe know that $G$ also contains $rg$ and $gr$ for every $r\\in G$ left of $g$. We also know that this is true for\nall $r\\in G$ right of $g$. In fine, we should have closure in $G$. Now since $G$ is a finite subset of a larger group (since\nthe original $S$ was a subset of that larger group),\nit must be a subgroup of that group. Lastly, can we claim that $G$ is the smallest group containing all original elements of $S$?\nWell, since we never added any elements to $G$ unnecessarily, I would have to think so.\n\nThis line of reasoning is not terribly rigorous, but it is the best I can come up with for now.\n\n% Maybe note a recursive algorithm for generating a group in the case that it is abelian.\n% In that case, all subgroups are normal. If H is a subgroup of such a G, and g is in G-H,\n% Then is the smallest subgroup of G containing g and H.\n\n\\end{document}", "meta": {"hexsha": "fdfe7013d7b9fb101d298a7823d5edcc65b6e6ba", "size": 6643, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Paper/TheFactorizationProblemInComputationalGroupTheory.tex", "max_stars_repo_name": "spencerparkin/PermutationGroup", "max_stars_repo_head_hexsha": "9932737ee77348869f8e2e715bdad3c84401b214", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-04-01T06:21:06.000Z", "max_stars_repo_stars_event_max_datetime": "2017-04-01T06:21:06.000Z", "max_issues_repo_path": "Paper/TheFactorizationProblemInComputationalGroupTheory.tex", "max_issues_repo_name": "spencerparkin/PermutationGroup", "max_issues_repo_head_hexsha": "9932737ee77348869f8e2e715bdad3c84401b214", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Paper/TheFactorizationProblemInComputationalGroupTheory.tex", "max_forks_repo_name": "spencerparkin/PermutationGroup", "max_forks_repo_head_hexsha": "9932737ee77348869f8e2e715bdad3c84401b214", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.144, "max_line_length": 130, "alphanum_fraction": 0.7084148728, "num_tokens": 2025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256432832333, "lm_q2_score": 0.7461389986757757, "lm_q1q2_score": 0.6281189425389424}} {"text": "\\documentclass[12pt,a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}\n\n\\title{Mathematical software - homework 1}\n\\author{Sebastiano Tronto}\n\n\\newtheorem{thm}{Theorem}\n\\newtheorem{prop}[thm]{Proposition}\n\n\\theoremstyle{definition}\n\\newtheorem{ex}{Exercise}\n\n\\theoremstyle{definition}\n\\newtheorem*{remark}{Remark}\n\n\\begin{document}\n\n\\noindent\\hrulefill\n\n\\begin{center}\n\\Huge{\\textbf{Mathematical Software - Homework 1}}\n\\end{center}\n\n\\noindent\\hrulefill\n\\begin{center}\n\\begin{tabular}{lcr}\n\\texttt{sebastiano.tronto@uni.lu} & \\qquad \\qquad \\qquad \\qquad &\n\\textbf{Deadline: Sunday, March 28th}\n\\end{tabular}\n\\end{center}\n\n\\vspace{1cm}\n\n\\begin{center}\n \\emph{\\large\n For each of the following exercises submit a .tex and a .pdf file.\n }\n\\end{center}\n\n\\vspace{1cm}\n\n\\begin{ex}\n Write a short Latex document that contains the following theorem-like\n environments using the \\texttt{\\textbackslash newtheorem} command of the\n \\texttt{amsthm} package (the box around the text is not needed):\n \\begin{center}\n \\fbox{\\parbox{0.95\\textwidth}{\n \\begin{prop}[Fundamental Theorem of Algebra]\n \\label{prop:fta}\n Let \\(p(x)\\) be a non-constant polynomial with coefficients in\n $\\mathbb C$. Then there is \\(z\\in\\mathbb C\\) such that $p(z)=0$.\n \\end{prop}\n\n \\begin{remark}\n Proposition \\ref{prop:fta} is not true for polynomials with\n coefficients in $\\mathbb R$. For example\n \\begin{align}\n p(x) = x^2+1\n \\end{align}\n does not have real roots.\n \\end{remark}\n\n \\begin{thm}\n If $X$ and $Y$ are $\\sigma$-finite measure spaces and $f:X\\times Y\\to\n \\mathbb R$ is measurable and such that\n \\begin{align*}\n \\int_{X\\times Y}|f(x,y)|\\mathrm d(x,y) < \\infty\n \\end{align*}\n then\n \\begin{align}\n \\label{eq:fubini}\n \\int_X\\left(\\int_Yf(x,y)\\mathrm d y\\right)\\mathrm d x =\n \\int_Y\\left(\\int_Xf(x,y)\\mathrm d x\\right)\\mathrm d y =\n \\int_{X\\times Y} f(x,y)\\mathrm d(x,y)\\,.\n \\end{align}\n \\end{thm}\n\n \\begin{remark}\n In practice, equation \\eqref{eq:fubini} means that we can switch the\n order of integration in a double integral.\n \\end{remark}\n }}\n \\end{center}\n Notice that Propositions, Theorems and some of the equations are numbered,\n and some of them are referred to in the Remarks. This numbering should change\n accordingly if more numbered Theorems and equations are added before this\n part of the text.\n\\end{ex}\n\n\\vspace{0.8cm}\n\n\\begin{ex}\n[This is basically a ``do whatever you want, just do something'' exercise]\nWrite a Mathematical document in Latex, where ``Mathematical'' means any text\nthat contains both inline and displaystyle symbols and formulas, and where most\nof the sentences contain a Mathematical symbol.\nFor example, you can submit (part of) your homework or notes for another course\nthat you are following at the moment.\n\nYou must follow the following rules:\n\\begin{itemize}\n \\item Your document must be at least 2 pages long.\n \\item Use the following in your preamble:\n \\begin{verbatim}\n\\documentaclass[12pt,a4paper]{article}\n\\usepackage[left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm]{geometry}\n \\end{verbatim}\n\\end{itemize}\nThe document itself can be about any area of Mathematics, Physics or similar.\nThe correctness of the Mathematics itself will not be judged for this course.\n\\end{ex}\n\n\\section*{Grading}\n\nThis homework assignment is worth 25\\% of your final grade.\n\n\\vspace{0.3cm}\n\\textbf{Exercise 1 (10 points).}\n\\begin{itemize}\n \\item A correct use of the \\texttt{\\textbackslash newtheorem} command is\n worth 4 out of 10 points.\n \\item A correct use of the labelling and reference system is worth 4 points.\n \\item Reproducing correctly the mathematical formulas is worth 2 points.\n\\end{itemize}\n\n\\textbf{Exercise 2 (10 points).}\n\\begin{itemize}\n \\item A resulting pdf file that looks readable and correct is worth 6 out of\n 10 points.\n \\item The correct use of latex commands and environments is worth 4 out of\n 10 points. \\emph{Examples of mistakes in this sense: adjusting manually\n the size of parenthesis instead of using \\texttt{\\textbackslash left}\n and \\texttt{\\textbackslash right}; using \\texttt{\\textbackslash Sigma}\n when you actually want to write a sum symbol.}\n\\end{itemize}\n\n\n\\end{document}", "meta": {"hexsha": "961a80d755c08eca3b14a2c6453b35f7e2d00cd5", "size": 4536, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/Homework1/Homework1.tex", "max_stars_repo_name": "sebastianotronto/mathsoftware", "max_stars_repo_head_hexsha": "e995905df49fefdef231aeb16e7f4afd7994c9fe", "max_stars_repo_licenses": ["AAL"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/Homework1/Homework1.tex", "max_issues_repo_name": "sebastianotronto/mathsoftware", "max_issues_repo_head_hexsha": "e995905df49fefdef231aeb16e7f4afd7994c9fe", "max_issues_repo_licenses": ["AAL"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/Homework1/Homework1.tex", "max_forks_repo_name": "sebastianotronto/mathsoftware", "max_forks_repo_head_hexsha": "e995905df49fefdef231aeb16e7f4afd7994c9fe", "max_forks_repo_licenses": ["AAL"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.7202797203, "max_line_length": 79, "alphanum_fraction": 0.7003968254, "num_tokens": 1307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389930307512, "lm_q2_score": 0.8418256432832333, "lm_q1q2_score": 0.628118937786816}} {"text": "\\documentclass[letter,10pt]{article}\n\n\\usepackage{amsmath}\n%\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n%\\usepackage[ngerman,english]{babel}\n%\\usepackage{hyperref}\n%\\usepackage{graphicx}\n%\\usepackage{listings}\n%\\usepackage{adjustbox}\n%\\usepackage[affil-it]{authblk}\n%\\usepackage[T1]{fontenc}\n%\\usepackage{lmodern}\n%\\usepackage[space]{grffile}\n\\usepackage{hyperref}\n\n\\begin{document}\n\n\n\n\\section*{Note on Analog-to-Digital Converter (ADC)}\n\n``What does the ADC output mean? What are those big numbers?''\n\n\\subsection*{Resolution}\nAn ADC linearly maps an input voltage to a range of numbers. The ADC we use has a 5V reference and is a 10-bit ADC, so it maps an input voltage between 0 and 5V to a number between 0 to $2^{10} - 1$:\n\\begin{equation}\nADC = V_{input} / V_{ref} * 2^{10} = V_{input} / 5V * 1024\n\\end{equation}\n(Reality is a bit more complicated than that. Divide by 1023 or 1024, which is correct? Or does it matter?)\\\\\n\\\\\nFor example, if the ADC outputs 365, then the voltage of the input is\n\\begin{equation}\n\\frac{365}{1024}*5V = 1.78V\n\\end{equation}\nYou should verify this with a voltmeter. Also try to verify that the reference voltage is indeed (not exactly) 5V. How does that affect your measurement?\n\n\\end{document}\n", "meta": {"hexsha": "cb5f95dd5c5e54f3ebbf3765bd393aebfcb500c4", "size": 1245, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "S18/week8/notes on adc.tex", "max_stars_repo_name": "stanleylio/OCN318", "max_stars_repo_head_hexsha": "98f0329ea38ddd4ed2966f8f845a0bb29e71b923", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-03-05T23:33:59.000Z", "max_stars_repo_stars_event_max_datetime": "2018-03-05T23:33:59.000Z", "max_issues_repo_path": "S18/week8/notes on adc.tex", "max_issues_repo_name": "stanleylio/OCN318", "max_issues_repo_head_hexsha": "98f0329ea38ddd4ed2966f8f845a0bb29e71b923", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "S18/week8/notes on adc.tex", "max_forks_repo_name": "stanleylio/OCN318", "max_forks_repo_head_hexsha": "98f0329ea38ddd4ed2966f8f845a0bb29e71b923", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.9230769231, "max_line_length": 199, "alphanum_fraction": 0.7421686747, "num_tokens": 384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998714925404, "lm_q2_score": 0.8080672181749422, "lm_q1q2_score": 0.6281105448447172}} {"text": "\\part{Limits and Continuity}\n\n\\chapter{Basic Limits}\n\n\\section{Basic Limits}\n\nSuppose we have a function $ y = f(x) $. The limit of $ f(x) $ as $ x $ approaches the number $ a $, denoted $ \\lim\\limits_{x \\rightarrow a}f(x) $. \\\\\n\nFor example, suppose we wanted to determine $ \\lim\\limits_{x \\rightarrow 3}x^2 $. \\\\\n\n\\begin{table}[H]\n\t\\centering\n\t\\begin{tabular}{|c|c|} \\hline\n\t\t\\textbf{$ x $ approaches 3 from the left} & \\textbf{$ x $ approaches 3 from the right} \\\\ \\hline\n\t\t$ f(2.9) = 8.41 $ & $ f(3.1) = 9.61 $ \\\\ \\hline\n\t\t$ f(2.99) = 8.9401 $ & $ f(3.01) = 9.0601 $ \\\\ \\hline\n\t\t$ f(2.999) = 8.994001 $ & $ f(3.001) = 9.006001 $ \\\\ \\hline\n\t\\end{tabular}\n\\end{table}\n\nNo matter which way we approach from, as $ x $ gets close 3, we can see that $ x^2 $ gets very close to 9. \\\\\n\n\\begin{exercise}\\nonumber\n\tFind $ \\lim\\limits_{x \\rightarrow 0}f(x) $ where \\\\\n\t\\begin{align}\n\t\tf(x) = \\begin{cases}\n\t\t\tcos(x) & x \\ne 0 \\\\\n\t\t\t-1 & x = 0\n\t\t\\end{cases}\n\t\\end{align}\n\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\begin{tikzpicture}\n\t\t\t\\begin{axis}[\n\t\t\t\t\txlabel={$ x $}, ylabel={$ y $},\n\t\t\t\t\txmin=-2*pi, xmax=2*pi,\n\t\t\t\t\tymin=-1.5, ymax=1.5,\n\t\t\t\t\txtick={-6.28319, -3.14159, 0, 3.14159, 6.28319},\n\t\t\t\t\txticklabels={$-2\\pi$, $-\\pi$, $0$, $\\pi$, $2\\pi$},\n\t\t\t\t\tline width=1pt,\n\t\t\t\t\taxis lines=center,\n\t\t\t\t]\n\t\t\t\t\\addplot[smooth,domain=-2*pi:-0.15, red!70]{cos(deg(x))};\n\t\t\t\t\\addplot[smooth,domain=0.15:2*pi, red!70]{cos(deg(x))};\n\t\t\t\\end{axis}\n\t\t\t\\node at (3.43,4.72) [red,circle,inner sep=1pt]{$\\circ$};\n\t\t\\end{tikzpicture}\n\t\\end{figure}\n\n\t\\begin{gather*}\n\t\t\\lim\\limits_{x \\rightarrow 0}f(x) = 1 \\\\\n\t\tf(0) = -1\n\t\\end{gather*}\n\\end{exercise}\n\n\\section{Law of Limits}\n\n\\begin{theorem}[Law of Limits]\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow a}(f+g)(x) = \\lim\\limits_{x \\rightarrow a}f(x) + \\lim\\limits_{x \\rightarrow a}g(x) \\\\\n\t\t & \\lim\\limits_{x \\rightarrow a}(f-g)(x) = \\lim\\limits_{x \\rightarrow a}f(x) - \\lim\\limits_{x \\rightarrow a}g(x) \\\\\n\t\t & \\lim\\limits_{x \\rightarrow a}(fg)(x) = \\lim\\limits_{x \\rightarrow a}f(x) \\cdot \\lim\\limits_{x \\rightarrow a}g(x) \\\\\n\t\t & \\lim\\limits_{x \\rightarrow a}\\left({f \\over g}\\right)(x) = {\\lim\\limits_{x \\rightarrow a}f(x) \\over \\lim\\limits_{x \\rightarrow a}g(x)}\n\t\\end{align}\n\\end{theorem}\n\n\\begin{exercise}\\nonumber\n\tEvaluate $ \\lim\\limits_{x \\rightarrow 0}xsin\\left({1 \\over x}\\right) $. \\\\\n\t\\begin{gather*}\n\t\t-1 \\le sin\\left({1 \\over x}\\right) \\le 1 \\\\\n\t\t-x \\le xsin\\left({1 \\over x}\\right) \\le x \\\\\n\t\t\\lim\\limits_{x \\rightarrow 0}-x \\le \\lim\\limits_{x \\rightarrow 0}xsin\\left({1 \\over x}\\right) \\le \\lim\\limits_{x \\rightarrow 0}x \\\\\n\t\t\\therefore \\lim\\limits_{x \\rightarrow 0}xsin\\left({1 \\over x}\\right) = 0\n\t\\end{gather*}\n\\end{exercise}\n\n\\section{One-sided Limits}\n\n$ x $ can approach the value a in two ways: \\\\\n\n\\begin{itemize}\n\t\\item\n\t left-hand limit: $ \\lim\\limits_{x \\rightarrow a^-}f(x) $ \\\\\n\n\t\\item\n\t right-hand limit: $ \\lim\\limits_{x \\rightarrow a^+}f(x) $ \\\\\n\\end{itemize}\n\nWe can tighten our definition of the limit of a function $ f(x) $ as $ x $ approaches $ a $. If the left-hand and right-hand limits of $ f(x) $ are both equal to a number $ L $, then we say that $ \\lim\\limits_{x \\rightarrow a}f(x) $ exists and is equal to $ L $. \\\\\n\n\\begin{exercise}\\nonumber\n\tConsider the piecewise function given by \\\\\n\t\\begin{align}\n\t\tf(x) = \\begin{cases}\n\t\t\tx & x < 1 \\\\\n\t\t\t-1 & x = 1 \\\\\n\t\t\t\\sqrt{x-1} & x > 1\n\t\t\\end{cases}\n\t\\end{align}\n\n\t(a) $ \\lim\\limits_{x \\rightarrow 1^-}f(x) = 1 $ \\\\\n\n\t(b) $ \\lim\\limits_{x \\rightarrow 1^+}f(x) = 0 $ \\\\\n\n\t(c) $ f(1) = -1 $ \\\\\n\n\t(d) Does $ \\lim\\limits_{x \\rightarrow 1}f(x) $ exist? No.\n\\end{exercise}\n\n\\chapter{Continuity}\n\n\\section{Continuity}\n\nIt seems like sometimes we can evaluate a limit by just plugging in, but sometimes we can't! \\\\\n\n\\begin{exercise}\\nonumber\n\tEvaluate $ \\lim\\limits_{x \\rightarrow 1}\\left(ln(\\sqrt{x}) + {1 \\over x}\\right) $. \\\\\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow 1}\\left(ln(\\sqrt{x}) + {1 \\over x}\\right) \\\\\n\t\t & = ln\\left(\\sqrt{x} + {1 \\over x}\\right) \\\\\n\t\t & = ln\\left(\\sqrt{1}\\right) + {1 \\over 1} \\\\\n\t\t & = 0 + 1 \\\\\n\t\t & = 1\n\t\\end{align}\n\\end{exercise}\n\nA function is continuous at a point a if each of the following conditions hold: \\\\\n\n\\begin{enumerate}\n\t\\item\n\t $ \\lim\\limits_{x \\rightarrow a}f(x) $ exists \\\\\n\n\t\\item\n\t $ f(a) $ exists \\\\\n\n\t\\item\n\t $ \\lim\\limits_{x \\rightarrow a}f(x) = f(a) $ \\\\\n\\end{enumerate}\n\n\\section{Discontinuity}\n\nKinds of discontinuities: \\\\\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{tikzpicture}[scale=0.7]\n\t\t\\draw[->] (-5,0) -- (4,0) node[right] {$ x $};\n\t\t\\draw[->] (0,-4) -- (0,4) node[above] {$ y $};\n\t\t\\draw[very thick,color=red,domain=-4.5:-1.3] plot (\\x,{1 / (\\x + 1)});\n\t\t\\draw[very thick,color=red,domain=-0.7:3] plot (\\x,{1 / (\\x + 1)});\n\t\t\\draw[very thick,densely dashed,color=red] (-1,4)--(-1,-4);\n\t\\end{tikzpicture}\n\t\\caption{essential}\n\\end{figure}\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{tikzpicture}\n\t\t\\begin{axis}[\n\t\t\t\txlabel={$ x $}, ylabel={$ y $},\n\t\t\t\txmin=-2*pi, xmax=2*pi,\n\t\t\t\tymin=-1.5, ymax=1.5,\n\t\t\t\txtick={-6.28319, -3.14159, 0, 3.14159, 6.28319},\n\t\t\t\txticklabels={$-2\\pi$, $-\\pi$, $0$, $\\pi$, $2\\pi$},\n\t\t\t\tline width=1pt,\n\t\t\t\taxis lines=center,\n\t\t\t]\n\t\t\t\\addplot[smooth,domain=-2*pi:1.07, red!70]{sin(deg(x))};\n\t\t\t\\addplot[smooth,domain=1.2:2*pi, red!70]{sin(deg(x))};\n\t\t\t\\node at (742,240) [red,circle,inner sep=1pt]{$\\circ$};\n\t\t\\end{axis}\n\t\\end{tikzpicture}\n\t\\caption{removable}\n\\end{figure}\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{tikzpicture}[scale=0.5,yscale=0.3]\n\t\t\\draw[->] (-10,0) -- (6,0) node[right] {$ x $};\n\t\t\\draw[->] (0,-5) -- (0,20) node[above] {$ y $};\n\t\t\\draw[very thick,color=red,domain=-9:-2.1] plot (\\x,{0.5 * \\x + 3});\n\t\t\\draw[very thick,color=red,domain=-2:2] plot (\\x,{0});\n\t\t\\draw[very thick,color=red,domain=2.1:4] plot (\\x,{\\x^2 - 1});\n\t\t\\foreach \\point in {(-2,0),(2,0)} {\n\t\t\t\t\\node at \\point [red,circle,fill,inner sep=1.5pt]{};\n\t\t\t}\n\t\t\\foreach \\point in {(-2,2),(2,3)} {\n\t\t\t\t\\node at \\point [red,circle,inner sep=1.5pt]{$\\circ$};\n\t\t\t}\n\t\\end{tikzpicture}\n\t\\caption{jump}\n\\end{figure}\n\nSo far, we know that evaluating $ \\lim\\limits_{x \\rightarrow a}f(x) $ is easy if $ f(x) $ is continuous at $ a $. But what if it isn't? The limit may still exist. \\\\\n\n\\begin{exercise}\\nonumber\n\tEvaluate the following limits. \\\\\n\n\t(a) $ \\lim\\limits_{x \\rightarrow 3^+}{1 \\over 3-x} = -\\infty $ \\\\\n\n\t(b) $ \\lim\\limits_{x \\rightarrow 3^-}{1 \\over 3-x} = \\infty $ \\\\\n\n\t(c) $ \\lim\\limits_{x \\rightarrow 0^+}ln(x) = -\\infty $ \\\\\n\n\t(d) $ \\lim\\limits_{x \\rightarrow -{\\pi \\over 2}^+}sec(x) = \\infty $\n\\end{exercise}\n\nA function doesn't have to go to $ \\pm \\infty $ at a discontinuity though. Consider, for instance, evaluating $ \\lim\\limits_{x \\rightarrow 3}{x^2-x-6 \\over x-3} $. Plugging in $ x = 3 $ gives us \"$ 0 / 0 $\", which is known as an indeterminate form. \\\\\n\nA good first step is to factor where possible. Once we cancel the factor $ x - 3 $ on top and bottom, this function is just the function $ f(x) = x + 2 $, but with a hole at $ x = 3 $. The hole is there because we cannot ignore that the original form of the function had issues at $ x = 3 $. \\\\\n\n\\begin{exercise}\\nonumber\n\tEvaluate the following limits. \\\\\n\n\t(a)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow 3}{x^2-x-6 \\over x-3} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 3}{(x-3)(x+2) \\over x-3} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 3}{x+2} \\\\\n\t\t & = 5\n\t\\end{align}\n\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\begin{tikzpicture}[scale=0.3]\n\t\t\t\\draw[->] (-6,0) -- (6,0) node[right] {$ x $};\n\t\t\t\\draw[->] (0,-4) -- (0,8) node[above] {$ y $};\n\t\t\t\\draw[very thick,color=red,domain=-5:2.85] plot (\\x,{\\x + 2});\n\t\t\t\\draw[very thick,color=red,domain=3.2:5] plot (\\x,{\\x + 2});\n\t\t\t\\node at (3,5) [red,circle,inner sep=1.5pt]{$\\circ$};\n\t\t\\end{tikzpicture}\n\t\\end{figure}\n\n\t(b)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow -1}{x^3 - 11x^2 + 8x + 20 \\over x + 1} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow -1}{(x+1)(x^2-12x+20) \\over x + 1} \\\\\n\t\t & = (-1)^2 - 12(-1) + 20 \\\\\n\t\t & = 33\n\t\\end{align}\n\t\\\\\n\n\t(c)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow 4}{\\sqrt{x+5} - 3 \\over x-4} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 4}{({{\\sqrt{x+5}-3} \\over {x-4}})({{\\sqrt{x+5}+3} \\over \\sqrt{x+5}+3})} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 4}{x-4 \\over (x-4)(\\sqrt{x+5}+3)} \\\\\n\t\t & = {1 \\over \\sqrt{9} + 3} \\\\\n\t\t & = {1 \\over 6}\n\t\\end{align}\n\t\\\\\n\n\t(d)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow 2}{{1 \\over x-4} + {1 \\over 2} \\over x-2} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 2}{{{2+(x-4)} \\over {2(x-4)}} \\over x-2} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 2}{\\left({x-2 \\over 2x-4}\\right)\\left({1 \\over x-2}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 2}{1 \\over 2(x-4)} \\\\\n\t\t & = {1 \\over 2(-2)} \\\\\n\t\t & = -{1 \\over 4}\n\t\\end{align}\n\\end{exercise}\n\n\\chapter{The Fundamental Sine Limit}\n\n\\section{The Fundamental Sine Limit}\n\nConsider the following limit:\n\n$$\n\t\\lim\\limits_{x \\rightarrow 0}{sin(x) \\over x}\n$$\n\nWe can quickly see that this is a \"$ 0 / 0 $\" limit, and therefore is indeterminate. We can't factor anything or otherwise simplify the expression to get rid of the issue. \\\\\n\n\\begin{figure}[H]\n\t\\centering\n\t\\begin{tikzpicture}[scale=1]\n\t\t\\begin{axis}[\n\t\t\t\tgrid=both,\n\t\t\t\txmin=-4,\n\t\t\t\txmax=4,\n\t\t\t\tymin=-2,\n\t\t\t\tymax=2,\n\t\t\t\txlabel=$x$,\n\t\t\t\tylabel=$y$,\n\t\t\t\taxis lines=center,\n\t\t\t\t>=stealth\n\t\t\t]\n\t\t\t\\addplot[\n\t\t\t\tdomain=-4:4,\n\t\t\t\tred,\n\t\t\t\tultra thick,\n\t\t\t\tsamples=100,\n\t\t\t] plot[smooth] {sin(deg(x))/x};\n\t\t\\end{axis}\n\t\\end{tikzpicture}\n\t\\caption{$ sin(x) \\over x $}\n\\end{figure}\n\nIt is pretty clear from the graph that \\\\\n\n\\begin{theorem}[The Fundamental Sine Limit]\n\t\\begin{align}\n\t\t\\lim\\limits_{x \\rightarrow 0}{sin(x) \\over x} = \\lim\\limits_{x \\rightarrow 0}{x \\over sin(x)} = 1\n\t\\end{align}\n\\end{theorem}\n\n\\begin{exercise}\\nonumber\n\tEvaluate the following limits. \\\\\n\n\t(a)\n\t\\begin{align}\n\t\t & \\lim\\limits_{t \\rightarrow 0}{5sin(t) \\over 2t} \\\\\n\t\t & = \\lim\\limits_{t \\rightarrow 0}{{5 \\over 2} \\cdot {sin(t) \\over t}} \\\\\n\t\t & = {5 \\over 2}\n\t\\end{align}\n\t\\\\\n\n\t(b)\n\t\\begin{align}\n\t\t & \\lim\\limits_{\\theta \\rightarrow 0}{\\theta \\over sin(4\\theta)} \\\\\n\t\t & = \\lim\\limits_{\\theta \\rightarrow 0}{4\\theta \\over 4sin(4\\theta)} \\\\\n\t\t & = \\lim\\limits_{\\theta \\rightarrow 0}{{1 \\over 4}{4\\theta \\over sin(4\\theta)}} \\\\\n\t\t & = {1 \\over 4}\n\t\\end{align}\n\t\\\\\n\n\t(c)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow -1}{sin^2(x+1) \\over (x+1)^2} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow -1}{{sin(x+1) \\over x+1} \\cdot {sin(x+1) \\over x+1}} \\\\\n\t\t & = 1\n\t\\end{align}\n\t\\\\\n\n\t(d)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow 0}{sin(3x) \\over sin(2x)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow 0}{{sin(3x) \\over 3} \\cdot {2x \\over sin(2x)} \\cdot {3 \\over 2}} \\\\\n\t\t & = {3 \\over 2}\n\t\\end{align}\n\t\\\\\n\n\t(e)\n\t\\begin{align}\n\t\t & \\lim\\limits_{\\theta \\rightarrow 0}{100\\theta \\over tan(3\\theta)} \\\\\n\t\t & = \\lim\\limits_{\\theta \\rightarrow 0}{10\\theta \\cdot {cos(3\\theta) \\over sin(3\\theta)}} \\\\\n\t\t & = \\lim\\limits_{\\theta \\rightarrow 0}{{3\\theta \\over sin(3\\theta)} \\cdot {1 \\over 3} \\cdot 10 \\cdot cos(3\\theta)} \\\\\n\t\t & = 1 \\cdot {10 \\over 3} \\cdot 1 \\\\\n\t\t & = {10 \\over 3}\n\t\\end{align}\n\t\\\\\n\n\t(f)\n\t\\begin{align}\n\t\t & \\lim\\limits_{h \\rightarrow 0}{cos(h)-1 \\over h} \\\\\n\t\t & = \\lim\\limits_{h \\rightarrow 0}{\\left({cos(h)-1 \\over h}\\right)\\left({cos(h)+1 \\over cos(h)+1}\\right)} \\\\\n\t\t & = \\lim\\limits_{h \\rightarrow 0}{cos^2h - 1 \\over h(cos(h) + 1)} \\\\\n\t\t & = \\lim\\limits_{h \\rightarrow 0}{-sin^2h \\over h\\left(cos(h) + 1\\right)} \\\\\n\t\t & = \\lim\\limits_{h \\rightarrow 0}{-{sin(h) \\over h} \\cdot {sin(h) \\over cos(h) + 1}} \\\\\n\t\t & = -1 \\cdot {0 \\over 1+1} \\\\\n\t\t & = 0\n\t\\end{align}\n\\end{exercise}\n\n\\chapter{Limits Approaching $ \\pm \\infty $}\n\n\\section{Limits Approaching $ \\pm \\infty $}\n\nSometimes, we will be interested in determining what happens as $ x $ gets really big, either in the positive or negative direction. \\\\\n\n\\begin{exercise}\\nonumber\n\tEvaluate the following limits. \\\\\n\n\t(a) $ \\lim\\limits_{x \\rightarrow \\infty}{1 \\over x} = 0 $ \\\\\n\n\t(b) $ \\lim\\limits_{x \\rightarrow -\\infty}{1 \\over x} = 0 $ \\\\\n\n\t(c) $ \\lim\\limits_{x \\rightarrow -\\infty}{401 \\over x^{203}} = 0 $ \\\\\n\n\t(d) $ \\lim\\limits_{x \\rightarrow -\\infty}{-238 \\over 23x^{1/4}} = 0 $\n\\end{exercise}\n\nFor problems like $ \\lim\\limits_{x \\rightarrow \\infty}{2x^2 + 3x + 1 \\over x^2 - 10x + 100} $, a good first step is to divide the top and the bottom by the highest power of $ x $ appearing in the denominator. \\\\\n\n\\begin{exercise}\\nonumber\n\tEvaluate the following limits. \\\\\n\n\t(a)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow \\infty}{2x^2 + 3x + 1 \\over x^2 - 10x + 100} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{\\left({2x^2+3x+1 \\over x^2-10x+100}\\right)\\left({1/x^2 \\over 1/x^2}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{\\left({2x^2+3x+1 \\over x^2-10x+100}\\right)\\left({1/x^2 \\over 1/x^2}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{{2 + {3 \\over x} + {1 \\over x^2}} \\over {1 - {10 \\over x} + {100 \\over x^2}}} \\\\\n\t\t & = {2 \\over 1} \\\\\n\t\t & = 2\n\t\\end{align}\n\t\\\\\n\n\t(b)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow -\\infty}{3 - 2x^3 \\over 1 + x + x^2} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow -\\infty}{\\left({3-2x^3 \\over 1+x+x^2}\\right)\\left({1/x^2 \\over 1/x^2}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow -\\infty}{{3 \\over x^2} - 2x \\over {{1 \\over x^2} + {1 \\over x} + 1}} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow -\\infty}{-2x \\over 1} \\\\\n\t\t & = \\infty\n\t\\end{align}\n\t\\\\\n\n\t(c)\n\t\\begin{align}\n\t\t & \\lim\\limits_{x \\rightarrow \\infty}{\\sqrt{x^2 - x + 1}} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{\\left(\\sqrt{x^2-x+1}-x\\right)\\left({\\sqrt{x^2-x+1}+x \\over \\sqrt{x^2-x+1}+x}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{(x^2-x+1)-x^2 \\over \\sqrt{x^2-x+1}+x} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{x+1 \\over \\sqrt{x^2-x+1}+x} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{\\left({x+1 \\over \\sqrt{x^2-x+1}+x}\\right)\\left({1/x \\over 1/x}\\right)} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{-1 + {1 \\over x} \\over {\\sqrt{x^2-x+1} \\over x} + 1} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{-1 + {1 \\over x} \\over \\sqrt{{x^2-x+1} \\over x^2} + 1} \\\\\n\t\t & = \\lim\\limits_{x \\rightarrow \\infty}{-1 \\over \\sqrt{1} + 1} \\\\\n\t\t & = -{1 \\over 2}\n\t\\end{align}\n\t\\\\\n\n\t(d)\n\t\\begin{align}\n\t\t & \\lim\\limits_{t \\rightarrow \\infty}{\\sqrt{3t^2 + 2t + 5} + t} \\\\\n\t\t & = \\infty\n\t\\end{align}\n\\end{exercise}", "meta": {"hexsha": "31ea891bf2b665df6931a19dae4795939cf5db8c", "size": 15964, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Calculus_Solution/parts/part3.tex", "max_stars_repo_name": "xdai02/Calculus", "max_stars_repo_head_hexsha": "45e2774a19879e687e0ca9cc8ada28cf290ce2ec", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Calculus_Solution/parts/part3.tex", "max_issues_repo_name": "xdai02/Calculus", "max_issues_repo_head_hexsha": "45e2774a19879e687e0ca9cc8ada28cf290ce2ec", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Calculus_Solution/parts/part3.tex", "max_forks_repo_name": "xdai02/Calculus", "max_forks_repo_head_hexsha": "45e2774a19879e687e0ca9cc8ada28cf290ce2ec", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.0394431555, "max_line_length": 294, "alphanum_fraction": 0.5186670008, "num_tokens": 5842, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.8080672227971211, "lm_q1q2_score": 0.6281105401000203}} {"text": "% Start of header.\n\\documentclass{article}\n\\usepackage{amsthm}\n\\usepackage{caption}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\usepackage{lmodern}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n\\newcommand{\\I}[0]{\\mathbb{I}}\n\\newcommand{\\Q}[0]{\\mathbb{Q}}\n\\newcommand{\\Z}[0]{\\mathbb{Z}}\n\\newcommand{\\C}[0]{\\mathbb{C}}\n\\newcommand{\\F}[0]{\\mathbb{F}}\n\\newcommand{\\R}[0]{\\mathbb{R}}\n\\newcommand{\\N}[0]{\\mathbb{N}}\n\\newcommand{\\unif}[0]{\\mathcal{U}}\n\\newcommand{\\Tor}[0]{\\operatorname{Tor}}\n\\newcommand{\\ra}[0]{\\rightarrow}\n\\newcommand{\\rr}[0]{\\Rightarrow}\n\\newcommand{\\llrr}[0]{\\Leftrightarrow}\n\\newcommand{\\subq}[0]{\\subseteq}\n\\newcommand{\\sub}[0]{\\subset}\n\\newcommand{\\nsubq}[0]{\\not\\subseteq}\n\\newcommand{\\nsub}[0]{\\not\\subset}\n\\newcommand{\\sep}[0]{\\hbox{ }}\n\\newcommand{\\inv}[0]{^{\\raisebox{.2ex,$\\scriptscriptstyle-1$}}}\n\\newcommand{\\thnew}[0]{^{\\text{th}}}\n\\newcommand{\\ifnew}[0]{\\text{ if }}\n\\newcommand{\\elsenew}[0]{\\text{ else }}\n\\newcommand{\\for}[0]{\\text{ for }}\n\n\\setlength{\\jot}{8pt}\n% End of header\n% Start of body.\n\n\\begin{document}\n\n\\section{Expected value of kth element of an ordered sequence of elements from some interval}\n\n\\subsection{Question:}\n\nLet each ordered sequence of $n$ unique elements $X_1, \\dots, X_n$ from the interval $(a, b)$ be equally probable. What is the expected value of the element $X_k$?\n\n\\subsection{Answer:}\n\nConsider random variables $Y_1, \\dots, Y_n$ sampled from a uniform distribution over $(a, b)$ such that no two $Y_i$ are equal. First, we show that any ordered sequence made from $Y_1, \\dots, Y_n$ is equally probable.\n\nLet $X_1 = Y_{i_1}, \\dots, X_n = Y_{i_n}$ be the ordered sequence made from our random variables $Y_1, \\dots, Y_n$. Then we see the probability density function, $f$, at a specific random sequence is\n\n$$f_X([X_1, \\dots, X_n]) = \\sum_{Y_1, \\dots, Y_n \\in \\text{ Permuations}(X_1, \\dots, X_n)} f_{Y_1, \\dots, Y_n}(Y_1, \\dots, Y_n) $$\n$$= \\sum_{Y_1, \\dots, Y_n \\in \\text{ Permuations}(X_1, \\dots, X_n)} f_{Y_1}(Y_1)\\cdots f_{Y_n}(Y_n)$$\n\nNote that the value of a uniform probability density does not change if we introduce a finite number of holes (as the integral over the pdf does not change in value), thus:\n\n$$\nf_{Y_i}(Y_i) = \n\\begin{cases}\n\\frac{1}{b-a}, & \\text{if $Y_i \\in (a, b)$ and $Y_i \\neq Y_j$ for $j \\neq i$} \\\\\n0, & \\text{otherwise}\n\\end{cases}\n$$\n\n\\textbf{NOTE}: \\emph{I think I can rewrite it this way, because all I'm doing is reorganizing the points of the composite $Y$ pdf into groups that produce the same ordered sequence. So when I integrate over the pdf of the ordered sequences, it's integrating over the same region as the composite $Y$ pdf.} \\textbf{But the multiplier of $n!$ can't be right or the pdf for $f_X$ will integrate to greater than 1!}\n\nTherefore we can rewrite $f_X$:\n\n$$f_X([X_1, \\dots, X_n]) = \\sum_{Y_1, \\dots, Y_n \\in \\text{ Permuations}(X_1, \\dots, X_n)} \\frac{1}{(b-a)^n}$$\n$$=\\frac{n!}{(b-a)^n}$$\n\\end{document}\n% End of body.", "meta": {"hexsha": "f5a857c23cb7111c2a20bb62c4fc398c4b3057e6", "size": 2952, "ext": "tex", "lang": "TeX", "max_stars_repo_path": ".archive/notes/examples/proofs.tex", "max_stars_repo_name": "abhmul/projects-repo", "max_stars_repo_head_hexsha": "6d3a8110ce69219e769e0e56d14fbe394f6389c6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": ".archive/notes/examples/proofs.tex", "max_issues_repo_name": "abhmul/projects-repo", "max_issues_repo_head_hexsha": "6d3a8110ce69219e769e0e56d14fbe394f6389c6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": ".archive/notes/examples/proofs.tex", "max_forks_repo_name": "abhmul/projects-repo", "max_forks_repo_head_hexsha": "6d3a8110ce69219e769e0e56d14fbe394f6389c6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.0, "max_line_length": 411, "alphanum_fraction": 0.6893631436, "num_tokens": 1039, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7772998663336157, "lm_q1q2_score": 0.6281105370831401}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{graphicx}\n\\usepackage{amsmath}\n\\usepackage{tikz}\n\\usepackage{hyperref}\n\\usepackage{enumitem}\n\n\\newcommand{\\fromlectures}{{\\\\ \\color{blue} \\hspace*{\\fill}(from lecture slides)} \\\\}\n\\newcommand{\\bydefn}{{\\\\ \\color{blue} \\hspace*{\\fill}(by definition)} \\\\}\n\\newcommand{\\given}{{\\\\ \\color{blue} \\hspace*{\\fill}(given)} \\\\}\n\\newcommand{\\rtp}{{\\\\ \\color{blue} \\hspace*{\\fill}(required to prove)} \\\\}\n\n\\newcommand{\\f}[1]{o_{#1}x_{#1}y_{#1}z_{#1}}\n\\newcommand{\\rx}[1]{\\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & cos(#1) & -sin(#1) & 0 \\\\ 0 & sin(#1) & cos(#1) & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}}\n\\newcommand{\\rz}[1]{\\begin{bmatrix} cos(#1) & -sin(#1) & 0 & 0 \\\\ sin(#1) & cos(#1) & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}}\n\\newcommand{\\iden}{\\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}}\n\\newcommand{\\trans}[3]{\\begin{bmatrix} 1 & 0 & 0 & #1 \\\\ 0 & 1 & 0 & #2 \\\\ 0 & 0 & 1 & #3 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}}\n\n\\title{CSci 5551 - HW4}\n\\author{Yashasvi Sriram Patkuri\\\\patku001@umn.edu}\n\n\\begin{document}\n\\maketitle\n\\pagebreak\n\n\\section{}\nA three link three joint planar robot with link lengths $l_1, l_2, l_3$ respectively is considered.\n\\given\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.6\\textwidth]{q1.png}\n \\caption{Three link three joint robot}\n \\label{fig:q1.1}\n\\end{figure}\nAssigning frames according to DH convention we have,\n\\fromlectures\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[x=1cm, y=1cm, z=-0.6cm]\n % F0\n \\draw [->] (0, 0) -- (2, 0) node [right] {$x_0$};\n \\draw (0, 0) circle (0.15cm);\n \\draw [fill] (0, 0) circle (0.07cm) node [left] {$z_0$};\n % F1\n \\draw [->] (4, 2) -- (5.7888, 2.8944) node [right] {$x_1$};\n \\draw (4, 2) circle (0.15cm);\n \\draw [fill] (4, 2) circle (0.07cm) node [left] {$z_1$};\n % F2\n \\draw [->] (6, 5) -- (7.414, 6.414) node [right] {$x_2$};\n \\draw (6, 5) circle (0.15cm);\n \\draw [fill] (6, 5) circle (0.07cm) node [left] {$z_2$};\n % F3\n \\draw [->] (9, 5) -- (11, 5) node [right] {$x_3$};\n \\draw (9, 5) circle (0.15cm);\n \\draw [fill] (9, 5) circle (0.07cm) node [left] {$z_3$};\n \\end{tikzpicture}\n \\caption{DH frame assignment}\n \\label{fig:q1.2}\n\\end{figure}\n\n\\begin{enumerate}[nolistsep]\n \\item The z-axes are chosen along the axes of rotation and out of paper for the end-effector.\n \\item The choice of $x_0$ is free so it is chosen to be parallel to ground for simplicity.\n \\item There is a unique common normal b/w $z_0$ and $z_1$, along which $x_1$ is chosen.\n \\item $x_2$ and $x_3$ are chosen in the same way.\n\\end{enumerate}\n\n\\subsubsection*{DH parameters}\nThe DH parameters for the frames shown in Figure \\ref{fig:q1.2} are as follows\n\\begin{center}\n\\begin{tabular}{ c | c c c c }\n \\hline\n $F_i \\to F_j$ & $\\theta$ & d & r & $\\alpha$ \\\\\n \\hline\n $0 \\to 1$ & $\\theta_1$ & 0 & $l_1$ & $0^{\\circ}$ \\\\\n $1 \\to 2$ & $\\theta_2$ & 0 & $l_2$ & $0^{\\circ}$ \\\\\n $2 \\to 3$ & $\\theta_3$ & 0 & $l_3$ & $0^{\\circ}$ \\\\\n \\hline\n\\end{tabular}\n\\end{center}\n\n\\subsubsection*{Finding transformation $T_{01}, T_{02}, T_{03}$}\nGiven frame i and frame j with DH parameters [ $\\theta$, d, r, $\\alpha$ ] the transformation matrix is\n\\[\n T_{i,j} \\equiv Rot_{z,\\theta} * Trans_{z, d} * Trans_{x, r} * Rot_{x, \\alpha}\n\\]\n\\fromlectures\nConsider $T_{01}$\n\\[\n T_{01} \\equiv Rot_{z,\\theta_1} * Trans_{z, 0} * Trans_{x, l_1} * Rot_{x, 0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\trans{0}{0}{0} * \\trans{l_1}{0}{0} * \\rx{0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\iden * \\trans{l_1}{0}{0} * \\iden\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\trans{l_1}{0}{0}\n\\]\n\\[\n T_{01} \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nWhere $c_1 \\equiv cos(\\theta_1), s_1 \\equiv sin(\\theta_1)$.\n$T_{12}, T_{23}$ have the same DH-parameter structure with changed variables, therefore we can use the final form of $T_{01}$ for them.\n\\[\n T_{12} \\equiv\n \\begin{bmatrix} c_2 & -s_2 & 0 & l_2c_2 \\\\ s_2 & c_2 & 0 & l_2s_2 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{23} \\equiv\n \\begin{bmatrix} c_3 & -s_3 & 0 & l_3c_3 \\\\ s_3 & c_3 & 0 & l_3s_3 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nBut\n\\[\n T_{ij} \\equiv T_{ik} * T_{kj}\n\\]\n\\fromlectures\nTherefore,\n\\[\n T_{02} \\equiv T_{01} * T_{12}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} c_2 & -s_2 & 0 & l_2c_2 \\\\ s_2 & c_2 & 0 & l_2s_2 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_1c_2 - s_1s_2 & -(s_1c_2 + c_1s_2) & 0 & l_1c_1 + l_2(c_1c_2 - s_1s_2) \\\\ s_1c_2 + c_1s_2 & c_1c_2 - s_1s_2 & 0 & l_1s_1 + l_2(s_1c_2 + c_1s_2) \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_{12} & -s_{12} & 0 & l_1c_1 + l_2c_{12} \\\\ s_{12} & c_{12} & 0 & l_1s_1 + l_2s_{12} \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nWhere $c_{12} \\equiv cos(\\theta_1 + \\theta_2), s_{12} \\equiv sin(\\theta_1 + \\theta_2)$.\nConsider $T_{03}$,\n\\[\n T_{03} \\equiv T_{02} * T_{23}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix} c_{12} & -s_{12} & 0 & l_1c_1 + l_2c_{12} \\\\ s_{12} & c_{12} & 0 & l_1s_1 + l_2s_{12} \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} c_3 & -s_3 & 0 & l_3c_3 \\\\ s_3 & c_3 & 0 & l_3s_3 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n c_{12}c_{3} - s_{12}s_{3} & -(s_{12}c_{3} + c_{12}s_{3}) & 0 & l_1c_1 + l_2c_{12} + l_3(c_{12}c_{3} - s_{12}s_{3})\\\\\n s_{12}c_{3} + c_{12}s_{3} & c_{12}c_{3} - s_{12}s_{3} & 0 & l_1s_1 + l_2s_{12} + l_3(s_{12}c_{3} + c_{12}s_{3})\\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n c_{123} & -s_{123} & 0 & l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n s_{123} & c_{123} & 0 & l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\n\\subsubsection*{Finding Jacobian}\nFor a 3R robot, the Jacobian of pose of end effector w.r.t frame 0 as a function of joint variables\n\\[\n J \\equiv\n \\begin{bmatrix} z_0^0 \\times a_{03}^0 & z_1^0 \\times a_{13}^0 & z_2^0 \\times a_{23}^0 \\\\ z_0^0 & z_1^0 & z_2^0 \\end{bmatrix}\n\\]\nwhere\n\\[\n a_{i3}^{0} \\equiv a_{3}^{0} - a_{i}^{0}\n\\]\nwhere $a_3^0$ is the origin of frame 3 in frame 0 and $a_i^0$ is origin of frame i in frame 0.\n$z_i^0$ can be obtained by the first three number of third column of $T_{0i}$.\n\\fromlectures\n\\[\n z_0^{0} \\equiv z_1^{0} \\equiv z_2^{0} \\equiv z_3^{0} \\equiv \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\nConsider $a_{3}^{0}$,\n\\[\n a_3^0 \\equiv T_{03} * O_3\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n c_{123} & -s_{123} & 0 & l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n s_{123} & c_{123} & 0 & l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\nConsider $a_{2}^{0}$,\n\\[\n a_2^0 \\equiv T_{02} * O_2\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix} c_{12} & -s_{12} & 0 & l_1c_1 + l_2c_{12} \\\\ s_{12} & c_{12} & 0 & l_1s_1 + l_2s_{12} \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12}\\\\\n l_1s_1 + l_2s_{12}\\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\nConsider $a_{1}^{0}$,\n\\[\n a_1^0 \\equiv T_{01} * O_1\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\nAnd,\n\\[\n a_0^0 \\equiv \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\nRemoving the homogeneous fourth coordinate we have,\n\\[\n a_0^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12}\\\\\n l_1s_1 + l_2s_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{03}^0$,\n\\[\n a_{03}^0 \\equiv a_3^0 - a_0^0\n\\]\n\\[\n a_{03}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n -\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}\n\\]\n\\[\n a_{03}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{13}^0$,\n\\[\n a_{13}^0 \\equiv a_3^0 - a_1^0\n\\]\n\\[\n a_{13}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_{13}^0 \\equiv\n \\begin{bmatrix}\n l_2c_{12} + l_3c_{123}\\\\\n l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{23}^0$,\n\\[\n a_{23}^0 \\equiv a_3^0 - a_2^0\n\\]\n\\[\n a_{23}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12}\\\\\n l_1s_1 + l_2s_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_{23}^0 \\equiv\n \\begin{bmatrix}\n l_3c_{123}\\\\\n l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider the cross product of the form,\n\\[\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times \\begin{bmatrix} a \\\\ b \\\\ 0 \\end{bmatrix}\n \\equiv\n \\begin{bmatrix} -b \\\\ a \\\\ 0 \\end{bmatrix}\n\\]\nFinally,\n\\[\n z_0^0 \\times a_{03}^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times\n \\begin{bmatrix}\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n l_1s_1 + l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + l_2s_{12} + l_3s_{123})\\\\\n l_1c_1 + l_2c_{12} + l_3c_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n z_1^0 \\times a_{13}^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times\n \\begin{bmatrix}\n l_2c_{12} + l_3c_{123}\\\\\n l_2s_{12} + l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_2s_{12} + l_3s_{123})\\\\\n l_2c_{12} + l_3c_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n z_2^0 \\times a_{23}^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times\n \\begin{bmatrix}\n l_3c_{123}\\\\\n l_3s_{123}\\\\\n 0\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_3s_{123})\\\\\n l_3c_{123}\\\\\n 0\n \\end{bmatrix}\n\\]\nUsing these to build the Jacobian,\n\\[\n J \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + l_2s_{12} + l_3s_{123}) & -(l_2s_{12} + l_3s_{123}) & -(l_3s_{123})\\\\\n l_1c_1 + l_2c_{12} + l_3c_{123} & l_2c_{12} + l_3c_{123} & l_3c_{123}\\\\\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 1 & 1 & 1 \\\\\n \\end{bmatrix}\n\\]\n\n\\subsubsection*{Finding $v$ and $\\omega$ of $O_c$}\nThe $v$ and $\\omega$ of $O_c$ are not dependent on the last joint.\nConsider a manipulator identical to the original one for the first one and half link.\nThe new manipulator will have its end-effector at $O_c$.\nThe required quantities can be found by repeating the same process of finding Jacobian as done for the original manipulator.\n\nAssigning frames for the new manipulator according to DH convention we have,\n\\fromlectures\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[x=1cm, y=1cm, z=-0.6cm]\n % F0\n \\draw [->] (0, 0) -- (2, 0) node [right] {$x_0$};\n \\draw (0, 0) circle (0.15cm);\n \\draw [fill] (0, 0) circle (0.07cm) node [left] {$z_0$};\n % F1\n \\draw [->] (4, 2) -- (5.7888, 2.8944) node [right] {$x_1$};\n \\draw (4, 2) circle (0.15cm);\n \\draw [fill] (4, 2) circle (0.07cm) node [left] {$z_1$};\n % F2\n \\draw [->] (5, 4) -- (6.414, 5.414) node [right] {$x_2$};\n \\draw (5, 4) circle (0.15cm);\n \\draw [fill] (5, 4) circle (0.07cm) node [left] {$z_2$};\n \\end{tikzpicture}\n \\caption{DH frame assignment}\n \\label{fig:q1.3}\n\\end{figure}\n\n\\begin{enumerate}[nolistsep]\n \\item The z-axes are chosen along the axes of rotation and out of paper for the end-effector.\n \\item The choice of $x_0$ is free so it is chosen to be parallel to ground for simplicity.\n \\item There is a unique common normal b/w $z_0$ and $z_1$, along which $x_1$ is chosen.\n \\item $x_2$ is chosen in the same way.\n\\end{enumerate}\n\n\\subsubsection*{DH parameters}\nThe DH parameters for the frames shown in Figure \\ref{fig:q1.3} are as follows\n\\begin{center}\n\\begin{tabular}{ c | c c c c }\n \\hline\n $F_i \\to F_j$ & $\\theta$ & d & r & $\\alpha$ \\\\\n \\hline\n $0 \\to 1$ & $\\theta_1$ & 0 & $l_1$ & $0^{\\circ}$ \\\\\n $1 \\to 2$ & $\\theta_2$ & 0 & $l_2 / 2$ & $0^{\\circ}$ \\\\\n \\hline\n\\end{tabular}\n\\end{center}\n\n\\subsubsection*{Finding transformation $T_{01}, T_{02}$}\nGiven frame i and frame j with DH parameters [ $\\theta$, d, r, $\\alpha$ ] the transformation matrix is\n\\[\n T_{i,j} \\equiv Rot_{z,\\theta} * Trans_{z, d} * Trans_{x, r} * Rot_{x, \\alpha}\n\\]\n\\fromlectures\nConsider $T_{01}$\n\\[\n T_{01} \\equiv Rot_{z,\\theta_1} * Trans_{z, 0} * Trans_{x, l_1} * Rot_{x, 0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\trans{0}{0}{0} * \\trans{l_1}{0}{0} * \\rx{0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\iden * \\trans{l_1}{0}{0} * \\iden\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\trans{l_1}{0}{0}\n\\]\n\\[\n T_{01} \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nWhere $c_1 \\equiv cos(\\theta_1), s_1 \\equiv sin(\\theta_1)$.\n$T_{12}$ has the same DH-parameter structure with changed variables, therefore we can use the final form of $T_{01}$ for it.\n\\[\n T_{12} \\equiv\n \\begin{bmatrix} c_2 & -s_2 & 0 & \\frac{l_2}{2}c_2 \\\\ s_2 & c_2 & 0 & \\frac{l_2}{2}s_2 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nBut\n\\[\n T_{ij} \\equiv T_{ik} * T_{kj}\n\\]\n\\fromlectures\nTherefore,\n\\[\n T_{02} \\equiv T_{01} * T_{12}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} c_2 & -s_2 & 0 & \\frac{l_2}{2}c_2 \\\\ s_2 & c_2 & 0 & \\frac{l_2}{2}s_2 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_1c_2 - s_1s_2 & -(s_1c_2 + c_1s_2) & 0 & l_1c_1 + \\frac{l_2}{2}(c_1c_2 - s_1s_2) \\\\ s_1c_2 + c_1s_2 & c_1c_2 - s_1s_2 & 0 & l_1s_1 + \\frac{l_2}{2}(s_1c_2 + c_1s_2) \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix} c_{12} & -s_{12} & 0 & l_1c_1 + \\frac{l_2}{2}c_{12} \\\\ s_{12} & c_{12} & 0 & l_1s_1 + \\frac{l_2}{2}s_{12} \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\nWhere $c_{12} \\equiv cos(\\theta_1 + \\theta_2), s_{12} \\equiv sin(\\theta_1 + \\theta_2)$.\n\n\\subsubsection*{Finding Jacobian}\nFor a 2R robot, the Jacobian of pose of end effector w.r.t frame 0 as a function of joint variables\n\\[\n J \\equiv\n \\begin{bmatrix} z_0^0 \\times a_{02}^0 & z_1^0 \\times a_{12}^0\\\\ z_0^0 & z_1^0 \\end{bmatrix}\n\\]\nwhere\n\\[\n a_{i2}^{0} \\equiv a_{2}^{0} - a_{i}^{0}\n\\]\nwhere $a_2^0$ is the origin of frame 2 in frame 0 and $a_i^0$ is origin of frame i in frame 0.\n$z_i^0$ can be obtained by the first three number of third column of $T_{0i}$.\n\\fromlectures\n\\[\n z_0^{0} \\equiv z_1^{0} \\equiv z_2^{0} \\equiv \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\nConsider $a_{2}^{0}$,\n\\[\n a_2^0 \\equiv T_{02} * O_2\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n c_{12} & -s_{12} & 0 & l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n s_{12} & c_{12} & 0 & l_1s_1 + \\frac{l_2}{2}s_{12} \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12} \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\nConsider $a_{1}^{0}$,\n\\[\n a_1^0 \\equiv T_{01} * O_1\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix} c_1 & -s_1 & 0 & l_1c_1 \\\\ s_1 & c_1 & 0 & l_1s_1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\nAnd,\n\\[\n a_0^0 \\equiv \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\nRemoving the homogeneous fourth coordinate we have,\n\\[\n a_0^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{02}^0$,\n\\[\n a_{02}^0 \\equiv a_2^0 - a_0^0\n\\]\n\\[\n a_{02}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n -\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}\n\\]\n\\[\n a_{02}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{12}^0$,\n\\[\n a_{12}^0 \\equiv a_2^0 - a_1^0\n\\]\n\\[\n a_{12}^0 \\equiv\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n l_1c_1\\\\\n l_1s_1\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_{12}^0 \\equiv\n \\begin{bmatrix}\n \\frac{l_2}{2}c_{12}\\\\\n \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\nConsider the cross product of the form,\n\\[\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times \\begin{bmatrix} a \\\\ b \\\\ 0 \\end{bmatrix}\n \\equiv\n \\begin{bmatrix} -b \\\\ a \\\\ 0 \\end{bmatrix}\n\\]\nFinally,\n\\[\n z_0^0 \\times a_{02}^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times\n \\begin{bmatrix}\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n l_1s_1 + \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + \\frac{l_2}{2}s_{12})\\\\\n l_1c_1 + \\frac{l_2}{2}c_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n z_1^0 \\times a_{12}^0 \\equiv\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times\n \\begin{bmatrix}\n \\frac{l_2}{2}c_{12}\\\\\n \\frac{l_2}{2}s_{12}\\\\\n 0\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(\\frac{l_2}{2}s_{12})\\\\\n \\frac{l_2}{2}c_{12}\\\\\n 0\n \\end{bmatrix}\n\\]\nUsing these to build the Jacobian,\n\\[\n J \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + \\frac{l_2}{2}s_{12}) & -(\\frac{l_2}{2}s_{12})\\\\\n l_1c_1 + \\frac{l_2}{2}c_{12} & \\frac{l_2}{2}c_{12}\\\\\n 0 & 0 \\\\\n 0 & 0 \\\\\n 0 & 0 \\\\\n 1 & 1 \\\\\n \\end{bmatrix}\n\\]\nFinally,\n\\[\n \\begin{bmatrix} v \\\\ \\omega \\end{bmatrix}\n \\equiv\n J *\n \\begin{bmatrix} \\frac{d\\theta_1}{dt} \\\\ \\frac{d\\theta_2}{dt} \\end{bmatrix}\n\\]\n\\[\n \\begin{bmatrix} v \\\\ \\omega \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + \\frac{l_2}{2}s_{12}) & -(\\frac{l_2}{2}s_{12})\\\\\n l_1c_1 + \\frac{l_2}{2}c_{12} & \\frac{l_2}{2}c_{12}\\\\\n 0 & 0 \\\\\n 0 & 0 \\\\\n 0 & 0 \\\\\n 1 & 1 \\\\\n \\end{bmatrix}\n *\n \\begin{bmatrix} \\frac{d\\theta_1}{dt} \\\\ \\frac{d\\theta_2}{dt} \\end{bmatrix}\n\\]\n\\[\n \\begin{bmatrix} v \\\\ \\omega \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + \\frac{l_2}{2}s_{12}) * \\frac{d\\theta_1}{dt} - (\\frac{l_2}{2}s_{12}) * \\frac{d\\theta_2}{dt} \\\\\n l_1c_1 + \\frac{l_2}{2}c_{12} * \\frac{d\\theta_1}{dt} + \\frac{l_2}{2}c_{12} * \\frac{d\\theta_2}{dt}\\\\\n 0 \\\\\n 0 \\\\\n 0 \\\\\n \\frac{d\\theta_1}{dt} + \\frac{d\\theta_2}{dt}\\\\\n \\end{bmatrix}\n\\]\nTherefore,\n\\[\n v\n \\equiv\n \\begin{bmatrix}\n -(l_1s_1 + \\frac{l_2}{2}s_{12}) * \\frac{d\\theta_1}{dt} - \\frac{l_2}{2}s_{12} * \\frac{d\\theta_2}{dt} \\\\\n (l_1c_1 + \\frac{l_2}{2}c_{12}) * \\frac{d\\theta_1}{dt} + \\frac{l_2}{2}c_{12} * \\frac{d\\theta_2}{dt}\\\\\n 0 \\\\\n \\end{bmatrix},\n \\omega\n \\equiv\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n \\frac{d\\theta_1}{dt} + \\frac{d\\theta_2}{dt}\\\\\n \\end{bmatrix}\n\\]\n\n\\pagebreak\n\n\\section{}\nA three link three joint robot with link lengths $l_1, l_2, l_3$ respectively is considered.\n\\given\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.6\\textwidth]{q2.png}\n \\caption{Three link three joint robot}\n \\label{fig:q2.1}\n\\end{figure}\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[x=1cm, y=1cm, z=-0.6cm]\n \\draw [dashed, red] (-3, 0) -- (-3, 3) node [right] {$l_1$};\n \\draw [dashed, red] (0, 5) -- (3, 5) node [right] {$l_2$};\n \\draw [dashed, red] (3, 4) -- (6, 4) node [right] {$l_3$};\n % F0\n \\draw [->] (0, 0) -- (0, 1) node [right] {$z_0$};\n \\draw (0, 0) circle (0.15cm);\n \\draw [fill] (0, 0) circle (0.07cm) node [left] {$x_0$};\n % F1\n \\draw [->] (0, 3) -- (1, 3) node [right] {$x_1$};\n \\draw (0, 3) circle (0.15cm);\n \\draw [fill] (0, 3) circle (0.07cm) node [left] {$z_1$};\n % F2\n \\draw [->] (3, 3) -- (4, 3) node [right] {$x_2$};\n \\draw (3, 3) circle (0.15cm);\n \\draw [fill] (3, 3) circle (0.07cm) node [left] {$z_2$};\n % F3\n \\draw [->] (6, 3) -- (7, 3) node [right] {$x_3$};\n \\draw (6, 3) circle (0.15cm);\n \\draw [fill] (6, 3) circle (0.07cm) node [left] {$z_3$};\n \\end{tikzpicture}\n \\caption{DH frame assignment}\n \\label{fig:q2.2}\n\\end{figure}\n\n\\subsubsection*{DH parameters}\nThe DH parameters for the frames shown in Figure \\ref{fig:q2.2} are as follows\n\\begin{center}\n\\begin{tabular}{ c | c c c c }\n \\hline\n $F_i \\to F_j$ & $\\theta$ & d & r & $\\alpha$ \\\\\n \\hline\n $0 \\to 1$ & $\\theta_1$ & $l_1$ & $ 0 $ & $90^{\\circ}$ \\\\\n $1 \\to 2$ & $\\theta_2$ & 0 & $l_2$ & $0^{\\circ}$ \\\\\n $2 \\to 3$ & $\\theta_3$ & 0 & $l_3$ & $0^{\\circ}$ \\\\\n \\hline\n\\end{tabular}\n\\end{center}\n\n\\subsubsection*{Finding transformation $T_{01}, T_{02}, T_{03}$}\nGiven frame i and frame j with DH parameters [ $\\theta$, d, r, $\\alpha$ ] the transformation matrix is\n\\[\n T_{i,j} \\equiv Rot_{z,\\theta} * Trans_{z, d} * Trans_{x, r} * Rot_{x, \\alpha}\n\\]\n\\fromlectures\nConsider $T_{01}$\n\\[\n T_{01} \\equiv Rot_{z,\\theta_1} * Trans_{z, l_1} * Trans_{x, 0} * Rot_{x, 90^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv Rot_{z,\\theta_1} * Trans_{z, l_1} * \\iden * Rot_{x, 90^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv Rot_{z,\\theta_1} * Trans_{z, l_1} * Rot_{x, 90^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{\\theta_1} * \\trans{0}{0}{l_1} * \\rx{90^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv\n \\begin{bmatrix}\n c_1 & -s_1 & 0 & 0 \\\\\n s_1 & c_1 & 0 & 0 \\\\\n 0 & 0 & 1 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix} * \\rx{90^{\\circ}}\n\\]\nwhere $c_i \\equiv cos(\\theta_i), s_i \\equiv sin(\\theta_i)$.\n\\[\n T_{01} \\equiv\n \\begin{bmatrix}\n c_1 & 0 & s_1 & 0 \\\\\n s_1 & 0 & -c_1 & 0 \\\\\n 0 & 1 & 0 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\nConsider $T_{12}$\n\\[\n T_{12} \\equiv Rot_{z,\\theta_2} * Trans_{z, 0} * Trans_{x, l_2} * Rot_{x, 0^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\rz{\\theta_2} * \\iden * \\trans{l_2}{0}{0} * \\rx{0^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\rz{\\theta_2} * \\iden * \\trans{l_2}{0}{0} * \\iden\n\\]\n\\[\n T_{12} \\equiv \\rz{\\theta_2} * \\trans{l_2}{0}{0}\n\\]\n\\[\n T_{12} \\equiv\n \\begin{bmatrix}\n c_2 & -s_2 & 0 & l_2c_2 \\\\\n s_2 & c_2 & 0 & l_2s_2 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n$T_{23}$ has same DH structure as $T_{12}$, therefore we can use the final form of the latter. Therefore,\n\\[\n T_{23} \\equiv\n \\begin{bmatrix}\n c_3 & -s_3 & 0 & l_3c_3 \\\\\n s_3 & c_3 & 0 & l_3s_3 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\nBut\n\\[\n T_{ij} \\equiv T_{ik} * T_{kj}\n\\]\n\\fromlectures\nTherefore,\n\\[\n T_{02} \\equiv T_{01} * T_{12}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix}\n c_1 & 0 & s_1 & 0 \\\\\n s_1 & 0 & -c_1 & 0 \\\\\n 0 & 1 & 0 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n c_2 & -s_2 & 0 & l_2c_2 \\\\\n s_2 & c_2 & 0 & l_2s_2 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix}\n c_1c_2 & -c_1s_2 & s_1 & l_2c_1c_2 \\\\\n s_1c_2 & -s_1s_2 & -c_1 & l_2s_1c_2 \\\\\n s_2 & c_2 & 0 & l_2s_2 + l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv T_{02} * T_{23}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n c_1c_2 & -c_1s_2 & s_1 & l_2c_1c_2 \\\\\n s_1c_2 & -s_1s_2 & -c_1 & l_2s_1c_2 \\\\\n s_2 & c_2 & 0 & l_2s_2 + l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n c_3 & -s_3 & 0 & l_3c_3 \\\\\n s_3 & c_3 & 0 & l_3s_3 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n c_1(c_2c_3 - s_2s_3) & -c_1(s_2c_3 + c_2s_3) & s_1 & c_1l_2c_2 + c_1l_3(c_2c_3 - s_2s_3) \\\\\n s_2(c_2c_3 - s_2s_3) & -s_1(s_2c_3 + c_2s_3) & -c_1 & s_1l_2c_2 + s_1l_3(c_2c_3 - s_2s_3) \\\\\n s_2c_3 + c_2s_3 & c_2c_3 - s_2s_3 & 0 & l_2s_2 + l_3(s_2c_3 + c_2s_3) + l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n c_1c_{23} & -c_1s_{23} & s_1 & l_2c_1c_2 + l_3c_1c_{23} \\\\\n s_1c_{23} & -s_1s_{23} & -c_1 & l_2s_1c_2 + l_3s_1c_{23} \\\\\n s_{23} & c_{23} & 0 & l_1 + l_2s_2 + l_3s_{23} \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\n\\subsubsection*{Finding Jacobian}\nFor a 3R robot, the Jacobian of pose of end effector w.r.t frame 0 as a function of joint variables\n\\[\n J \\equiv\n \\begin{bmatrix} z_0^0 \\times a_{03}^0 & z_1^0 \\times a_{13}^0 & z_2^0 \\times a_{23}^0 \\\\ z_0^0 & z_1^0 & z_2^0 \\end{bmatrix}\n\\]\nwhere\n\\[\n a_{i3}^{0} \\equiv a_{3}^{0} - a_{i}^{0}\n\\]\nwhere $a_3^0$ is the origin of frame 3 in frame 0 and $a_i^0$ is origin of frame i in frame 0.\n$z_i^0$ can be obtained by the first three number of third column of $T_{0i}$.\n\\fromlectures\n\\[\n z_0^{0} \\equiv \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n\\]\n\\[\n z_1^{0} \\equiv z_2^{0} \\equiv z_3^{0} \\equiv \\begin{bmatrix} s_1 \\\\ -c_1 \\\\ 0 \\end{bmatrix}\n\\]\nConsider $a_{3}^{0}$,\n\\[\n a_3^0 \\equiv T_{03} * O_3\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n c_1c_{23} & -c_1s_{23} & s_1 & l_2c_1c_2 + l_3c_1c_{23} \\\\\n s_1c_{23} & -s_1s_{23} & -c_1 & l_2s_1c_2 + l_3s_1c_{23} \\\\\n s_{23} & c_{23} & 0 & l_1 + l_2s_2 + l_3s_{23} \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23} \\\\\n 1\n \\end{bmatrix}\n\\]\nConsider $a_{2}^{0}$,\n\\[\n a_2^0 \\equiv T_{02} * O_2\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n c_1c_2 & -c_1s_2 & s_1 & l_2c_1c_2 \\\\\n s_1c_2 & -s_1s_2 & -c_1 & l_2s_1c_2 \\\\\n s_2 & c_2 & 0 & l_2s_2 + l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\n\\[\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 \\\\\n l_2s_1c_2 \\\\\n l_2s_2 + l_1 \\\\\n 1\n \\end{bmatrix}\n\\]\nConsider $a_{1}^{0}$,\n\\[\n a_1^0 \\equiv T_{01} * O_1\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n c_1 & 0 & s_1 & 0 \\\\\n s_1 & 0 & -c_1 & 0 \\\\\n 0 & 1 & 0 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\n\\[\n a_1^0 \\equiv\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n l_1 \\\\\n 1\n \\end{bmatrix}\n\\]\nRemoving homogeneous coordinate we have,\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n a_2^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 \\\\\n l_2s_1c_2 \\\\\n l_2s_2 + l_1\n \\end{bmatrix}\n a_1^0 \\equiv\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n l_1\n \\end{bmatrix}\n a_0^0 \\equiv\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_{03}^0$,\n\\[\n a_{03}^0 \\equiv a_3^0 - a_0^0\n\\]\n\\[\n a_{03}^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n \\end{bmatrix}\n\\]\n\\[\n a_{03}^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n\\]\nConsider $a_{13}^0$,\n\\[\n a_{13}^0 \\equiv a_3^0 - a_1^0\n\\]\n\\[\n a_{13}^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n l_1\n \\end{bmatrix}\n\\]\n\\[\n a_{13}^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n\\]\nConsider $a_{23}^0$,\n\\[\n a_{23}^0 \\equiv a_3^0 - a_2^0\n\\]\n\\[\n a_{23}^0 \\equiv\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n -\n \\begin{bmatrix}\n l_2c_1c_2 \\\\\n l_2s_1c_2 \\\\\n l_2s_2 + l_1\n \\end{bmatrix}\n\\]\n\\[\n a_{23}^0 \\equiv\n \\begin{bmatrix}\n l_3c_1c_{23} \\\\\n l_3s_1c_{23} \\\\\n l_3s_{23}\n \\end{bmatrix}\n\\]\nThe first 3x3 block of J is\n\\[\n J_{11} \\equiv\n \\begin{bmatrix} z_0^0 \\times a_{03}^0 & z_1^0 \\times a_{13}^0 & z_2^0 \\times a_{23}^0\\end{bmatrix}\n\\]\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_1 + l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n &\n \\begin{bmatrix} s_1 \\\\ -c_1 \\\\ 0 \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n &\n \\begin{bmatrix} s_1 \\\\ -c_1 \\\\ 0 \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n l_3c_1c_{23} \\\\\n l_3s_1c_{23} \\\\\n l_3s_{23}\n \\end{bmatrix}\n \\end{bmatrix}\n\\]\nConsider the cross product of the form,\n\\[\n \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\times \\begin{bmatrix} a \\\\ b \\\\ 0 \\end{bmatrix}\n \\equiv\n \\begin{bmatrix} -b \\\\ a \\\\ 0 \\end{bmatrix}\n\\]\nTherefore simplifying first column,\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n \\begin{bmatrix}\n -l_2s_1c_2 - l_3s_1c_{23} \\\\\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n 0\n \\end{bmatrix}\n &\n \\begin{bmatrix} s_1 \\\\ -c_1 \\\\ 0 \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n l_2s_1c_2 + l_3s_1c_{23} \\\\\n l_2s_2 + l_3s_{23}\n \\end{bmatrix}\n &\n \\begin{bmatrix} s_1 \\\\ -c_1 \\\\ 0 \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n l_3c_1c_{23} \\\\\n l_3s_1c_{23} \\\\\n l_3s_{23}\n \\end{bmatrix}\n \\end{bmatrix}\n\\]\nSimplifying second and third columns,\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n \\begin{bmatrix}\n -l_2s_1c_2 - l_3s_1c_{23} \\\\\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n 0\n \\end{bmatrix}\n &\n \\begin{bmatrix}\n -l_2c_1s_2 - l_3c_1s_{23} \\\\\n -l_2s_1s_2 - l_3s_1s_{23} \\\\\n l_2c_2 + l_3c_{23} (s_1^2 + c_1^2)\n \\end{bmatrix}\n &\n \\begin{bmatrix}\n -l_3c_1s_{23}\\\\\n -l_3s_1s_{23}\\\\\n (c_1^2 + s_1^2)l_3c_{23}\n \\end{bmatrix}\n \\end{bmatrix}\n\\]\nBut\n\\[\n c_1^2 + s_1^2 \\equiv 1\n\\]\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n \\begin{bmatrix}\n -l_2s_1c_2 - l_3s_1c_{23} \\\\\n l_2c_1c_2 + l_3c_1c_{23} \\\\\n 0\n \\end{bmatrix}\n &\n \\begin{bmatrix}\n -l_2c_1s_2 - l_3c_1s_{23} \\\\\n -l_2s_1s_2 - l_3s_1s_{23} \\\\\n l_2c_2 + l_3c_{23}\n \\end{bmatrix}\n &\n \\begin{bmatrix}\n -l_3c_1s_{23}\\\\\n -l_3s_1s_{23}\\\\\n l_3c_{23}\n \\end{bmatrix}\n \\end{bmatrix}\n\\]\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n -l_2s_1c_2 - l_3s_1c_{23} & -l_2c_1s_2 - l_3c_1s_{23} & -l_3c_1s_{23}\\\\\n l_2c_1c_2 + l_3c_1c_{23} & -l_2s_1s_2 - l_3s_1s_{23} & -l_3s_1s_{23}\\\\\n 0 & l_2c_2 + l_3c_{23} & l_3c_{23}\n \\end{bmatrix}\n\\]\nHence the required relation is proved (I used $l$'s to represent lengths instead of $a$'s).\n\n\\subsubsection*{If $\\theta_3 = 0^{\\circ}$}\nIf $\\theta_3 = 0^{\\circ}$, then $c_{23} = c_2, s_{23} = s_2$. Therefore,\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n -l_2s_1c_2 - l_3s_1c_{2} & -l_2c_1s_2 - l_3c_1s_{2} & -l_3c_1s_{2}\\\\\n l_2c_1c_2 + l_3c_1c_{2} & -l_2s_1s_2 - l_3s_1s_{2} & -l_3s_1s_{2}\\\\\n 0 & l_2c_2 + l_3c_{2} & l_3c_{2}\n \\end{bmatrix}\n\\]\nTaking common terms out,\n\\[\n J_{11} \\equiv\n \\begin{bmatrix}\n -(l_2 + l_3)s_1c_2 & (l_2 + l_3)(-c_1s_2) & l_3(-c_1s_{2})\\\\\n (l_2 + l_3)c_1c_2 & (l_2 + l_3)(-s_1s_2) & l_3(-s_1s_{2})\\\\\n 0 & (l_2 + l_3)c_2 & l_3c_{2}\n \\end{bmatrix}\n\\]\nObserve the second and third columns are linearly dependent (non-zero multiples of $[-c_1s_2, -s_1s_2, c_2]^T$). When two columns of a matrix are linearly dependent then its determinant $\\equiv 0$ or the matrix loses rank. Therefore $J_{11}$ loses rank i.e. $\\theta_3 = 0$ describes a family of singular configurations for $J_{11}$.\n\n\\pagebreak\n\n\\section{}\nTwo frames $F_0$ and $F_1$ are related by transformation H i.e.\n\\[\n p^0 \\equiv H * p^1\n\\]\n\\[\n H \\equiv\n \\begin{bmatrix} 0 & -1 & 0 & 1 \\\\ 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n\\]\n\\given\nLet\n\\[\n p^0 \\equiv (x^0, y^0, z^0), p^1 \\equiv (x^1, y^1, z^1)\n\\]\nThen\n\\[\n \\begin{bmatrix} x^0 \\\\ y^0 \\\\ z^0 \\\\ 1\\end{bmatrix} \\equiv\n \\begin{bmatrix} 0 & -1 & 0 & 1 \\\\ 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}\n *\n \\begin{bmatrix} x^1 \\\\ y^1 \\\\ z^1 \\\\ 1\\end{bmatrix}\n\\]\n\\[\n \\begin{bmatrix} x^0 \\\\ y^0 \\\\ z^0 \\\\ 1\\end{bmatrix} \\equiv\n \\begin{bmatrix} -y^1 + 1 \\\\ x^1 - 1 \\\\ z^1 \\\\ 1\\end{bmatrix}\n\\]\nTherefore\n\\[\n \\begin{bmatrix} x^0 \\\\ y^0 \\\\ z^0\\end{bmatrix} \\equiv\n \\begin{bmatrix} -y^1 + 1 \\\\ x^1 - 1 \\\\ z^1\\end{bmatrix}\n\\]\nTaking derivative w.r.t. time on both sides\n\\[\n \\frac{d}{dt}\\begin{bmatrix} x^0 \\\\ y^0 \\\\ z^0\\end{bmatrix} \\equiv\n \\frac{d}{dt}\\begin{bmatrix} -y^1 + 1 \\\\ x^1 - 1 \\\\ z^1\\end{bmatrix}\n\\]\nAs $\\frac{dx^0}{dt} \\equiv v_x^0$ and so on...,\n\\[\n \\begin{bmatrix} v_x^0 \\\\ v_y^0 \\\\ v_z^0\\end{bmatrix} \\equiv\n \\begin{bmatrix} -v_y^1 + 0 \\\\ v_x^1 - 0 \\\\ v_z^1\\end{bmatrix}\n\\]\n\\[\n \\begin{bmatrix} v_x^0 \\\\ v_y^0 \\\\ v_z^0\\end{bmatrix} \\equiv\n \\begin{bmatrix} -v_y^1 \\\\ v_x^1 \\\\ v_z^1\\end{bmatrix}\n\\]\nGiven $v^1(t) = (3, 1, 0)$, using the above identity we have\n\\[\n \\begin{bmatrix} v_x^0 \\\\ v_y^0 \\\\ v_z^0\\end{bmatrix} \\equiv\n \\begin{bmatrix} -1 \\\\ 3 \\\\ 0\\end{bmatrix}\n\\]\nTherefore the velocity w.r.t frame 0 $v^0(t) = (-1, 3, 0) \\frac{m}{s}$\n\n\\pagebreak\n\n\\section{}\nA three link cylindrical robot is considered.\n\\given\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.47\\textwidth]{q4.png}\n \\caption{Three link three joint robot}\n \\label{fig:q4.1}\n\\end{figure}\nIt's frame assignment is shown in Figure \\ref{fig:q4.2}. Let the length of link 1 be $l_1$ and the joint variables be $q_1, q_2, q_3$ respectively.\n\\begin{figure}[h]\n \\centering\n \\begin{tikzpicture}[x=1cm, y=1cm, z=-0.6cm]\n \\draw [dashed, red] (-3, 0) -- (-3, 3) node [right] {$l_1$};\n \\draw [dashed, red] (-2, 3) -- (-2, 6) node [right] {$q_2$};\n \\draw [dashed, red] (0, 5) -- (3, 5) node [right] {$q_3$};\n % F0\n \\draw [->] (0, 0) -- (0, 1) node [right] {$z_0$};\n \\draw (0, 0) circle (0.15cm);\n \\draw [fill] (0, 0) circle (0.07cm) node [left] {$x_0$};\n % F1\n \\draw [->] (0, 3) -- (0, 4) node [right] {$z_1$};\n \\draw (0, 3) circle (0.15cm);\n \\draw [fill] (0, 3) circle (0.07cm) node [left] {$x_1$};\n % F2\n \\draw [->] (0, 6) -- (1, 6) node [right] {$z_2$};\n \\draw (0, 6) circle (0.15cm);\n \\draw [fill] (0, 6) circle (0.07cm) node [left] {$x_2$};\n % F3\n \\draw [->] (3, 6) -- (4, 6) node [right] {$z_3$};\n \\draw (3, 6) circle (0.15cm);\n \\draw [fill] (3, 6) circle (0.07cm) node [left] {$x_3$};\n \\end{tikzpicture}\n \\caption{DH frame assignment}\n \\label{fig:q4.2}\n\\end{figure}\n\n\\subsubsection*{DH parameters}\nThe DH parameters for the frames shown in Figure \\ref{fig:q4.2} are as follows\n\\begin{center}\n\\begin{tabular}{ c | c c c c }\n \\hline\n $F_i \\to F_j$ & $\\theta$ & d & r & $\\alpha$ \\\\\n \\hline\n $0 \\to 1$ & $q_1$ & $l_1$ & $0$ & $0^{\\circ}$ \\\\\n $1 \\to 2$ & $0$ & $q_2$ & $0$ & $-90^{\\circ}$ \\\\\n $2 \\to 3$ & $0$ & $q_3$ & $0$ & $0^{\\circ}$ \\\\\n \\hline\n\\end{tabular}\n\\end{center}\n\n\\subsubsection*{Finding transformation $T_{01}, T_{02}, T_{03}$}\nGiven frame i and frame j with DH parameters [ $\\theta$, d, r, $\\alpha$ ] the transformation matrix is\n\\[\n T_{i,j} \\equiv Rot_{z,\\theta} * Trans_{z, d} * Trans_{x, r} * Rot_{x, \\alpha}\n\\]\n\\fromlectures\nConsider $T_{01}$\n\\[\n T_{01} \\equiv Rot_{z,q_1} * Trans_{z, l_1} * Trans_{x, 0} * Rot_{x, 0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{q_1} * \\trans{0}{0}{l_1} * \\trans{0}{0}{0} * \\rx{0^{\\circ}}\n\\]\n\\[\n T_{01} \\equiv \\rz{q_1} * \\trans{0}{0}{l_1} * \\iden * \\iden\n\\]\n\\[\n T_{01} \\equiv \\rz{q_1} * \\trans{0}{0}{l_1}\n\\]\n\\[\n T_{01} \\equiv\n \\begin{bmatrix}\n cq_1 & -sq_1 & 0 & 0 \\\\\n sq_1 & cq_1 & 0 & 0 \\\\\n 0 & 0 & 1 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\nWhere $cq_1 \\equiv cos(q_1), sq_1 \\equiv sin(q_1)$.\nConsider $T_{12}$\n\\[\n T_{12} \\equiv Rot_{z, 0} * Trans_{z, q_2} * Trans_{x, 0} * Rot_{x, -90^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\rz{0} * \\trans{0}{0}{q_2} * \\trans{0}{0}{0} * \\rx{-90^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\iden * \\trans{0}{0}{q_2} * \\trans{0}{0}{0} * \\rx{-90^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\trans{0}{0}{q_2} * \\rx{-90^{\\circ}}\n\\]\n\\[\n T_{12} \\equiv \\trans{0}{0}{q_2}\n *\n \\begin{bmatrix}\n 1 & 0 & 0 & 0 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & -1 & 0 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{12} \\equiv\n \\begin{bmatrix}\n 1 & 0 & 0 & 0 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & -1 & 0 & q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\nConsider $T_{23}$\n\\[\n T_{23} \\equiv Rot_{z, 0} * Trans_{z, q_3} * Trans_{x, 0} * Rot_{x, 0^{\\circ}}\n\\]\n\\[\n T_{23} \\equiv \\rz{0} * \\trans{0}{0}{q_3} * \\trans{0}{0}{0} * \\rx{0^{\\circ}}\n\\]\n\\[\n T_{23} \\equiv \\iden * \\trans{0}{0}{q_3} * \\trans{0}{0}{0} * \\iden\n\\]\n\\[\n T_{23} \\equiv \\trans{0}{0}{q_3}\n\\]\nBut\n\\[\n T_{ij} \\equiv T_{ik} * T_{kj}\n\\]\n\\fromlectures\nTherefore,\n\\[\n T_{02} \\equiv T_{01} * T_{12}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix}\n cq_1 & -sq_1 & 0 & 0 \\\\\n sq_1 & cq_1 & 0 & 0 \\\\\n 0 & 0 & 1 & l_1 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n 1 & 0 & 0 & 0 \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & -1 & 0 & q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{02} \\equiv\n \\begin{bmatrix}\n cq_1 & 0 & -sq_1 & 0 \\\\\n sq_1 & 0 & cq_1 & 0 \\\\\n 0 & -1 & 0 & l_1 + q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\\[\n T_{03} \\equiv T_{02} * T_{23}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n cq_1 & 0 & -sq_1 & 0 \\\\\n sq_1 & 0 & cq_1 & 0 \\\\\n 0 & -1 & 0 & l_1 + q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\trans{0}{0}{q_3}\n\\]\n\\[\n T_{03} \\equiv\n \\begin{bmatrix}\n cq_1 & 0 & -sq_1 & -q_3sq_1 \\\\\n sq_1 & 0 & cq_1 & q_3cq_1 \\\\\n 0 & -1 & 0 & l_1 + q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n\\]\n\n\\subsubsection*{Finding Jacobian}\nFor an RPP robot, the Jacobian of pose of end effector w.r.t frame 0 as a function of joint variables\n\\[\n J \\equiv\n \\begin{bmatrix}\n z_0^0 \\times a_{03}^0 & z_1^0 & z_2^0 \\\\\n z_0^0 & 0 & 0\n \\end{bmatrix}\n\\]\nwhere\n\\[\n a_{i3}^{0} \\equiv a_{3}^{0} - a_{i}^{0}\n\\]\nwhere $a_3^0$ is the origin of frame 3 in frame 0 and $a_i^0$ is origin of frame i in frame 0.\n$z_i^0$ can be obtained by the first three number of third column of $T_{0i}$.\n\\fromlectures\n\\[\n z_0^0 \\equiv z_1^0 \\equiv \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\n\\[\n z_2^0 \\equiv\n \\begin{bmatrix}\n -sq_1 \\\\\n cq_1 \\\\\n 0\n \\end{bmatrix}\n\\]\nConsider $a_3^0$,\n\\[\n a_3^0 \\equiv T_{03} * O_3\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n cq_1 & 0 & -sq_1 & -q_3sq_1 \\\\\n sq_1 & 0 & cq_1 & q_3cq_1 \\\\\n 0 & -1 & 0 & l_1 + q_2 \\\\\n 0 & 0 & 0 & 1\n \\end{bmatrix}\n *\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n\\]\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n -q_3sq_1 \\\\\n q_3cq_1 \\\\\n l_1 + q_2 \\\\\n 1\n \\end{bmatrix}\n\\]\nRemoving the homogeneous coordinate,\n\\[\n a_3^0 \\equiv\n \\begin{bmatrix}\n -q_3sq_1 \\\\\n q_3cq_1 \\\\\n l_1 + q_2\n \\end{bmatrix}\n\\]\nConsider $a_0^0$,\n\\[\n a_0^0 \\equiv \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n \\end{bmatrix}\n\\]\nTherefore,\n\\[\n a_{03}^0 \\equiv\n \\begin{bmatrix}\n -q_3sq_1 \\\\\n q_3cq_1 \\\\\n l_1 + q_2\n \\end{bmatrix}\n\\]\nConsider\n\\[\n z_0^0 \\times a_{03}^0 \\equiv\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 1\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n -q_3sq_1 \\\\\n q_3cq_1 \\\\\n l_1 + q_2\n \\end{bmatrix}\n \\equiv\n \\begin{bmatrix}\n -q_3cq_1 \\\\\n -q_3sq_1 \\\\\n 0\n \\end{bmatrix}\n\\]\nUsing these quantities to build the Jacobian,\n\\[\n J \\equiv\n \\begin{bmatrix}\n z_0^0 \\times a_{03}^0 & z_1^0 & z_2^0 \\\\\n z_0^0 & 0 & 0\n \\end{bmatrix}\n\\]\n\\[\n J \\equiv\n \\begin{bmatrix}\n -q_3cq_1 & 0 & -sq_1 \\\\\n -q_3sq_1 & 0 & cq_1 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 1 & 0 & 0\n \\end{bmatrix}\n\\]\n\n\\subsubsection*{Inspecting for singularities}\nConsider $J_\\omega$,\n\\[\n J_\\omega \\equiv\n \\begin{bmatrix}\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 1 & 0 & 0\n \\end{bmatrix}\n\\]\n$J_\\omega$ does not depend on configuration therefore it cannot lose rank with configuration.\nConsider $J_v$,\n\\[\n J_v \\equiv\n \\begin{bmatrix}\n -q_3cq_1 & 0 & -sq_1 \\\\\n -q_3sq_1 & 0 & cq_1 \\\\\n 0 & 1 & 0\n \\end{bmatrix}\n\\]\nConsider its determinant (using the third row),\n\\[\n det(J_v) \\equiv\n -(-q_3cq_1^2 - q_3sq_1^2)\n\\]\n\\[\n det(J_v) \\equiv\n q_3(cq_1^2 + sq_1^2)\n\\]\nBut $cos(\\theta)^2 + sin(\\theta)^2 \\equiv 1, \\forall \\theta \\in R$ therefore,\n\\[\n det(J_v) \\equiv q_3\n\\]\nAs (the length of third link) $q_3 > 0$ then determinant$(J_v) > 0 \\implies$ $J_v$ does not lose rank for any configuration.\n\nCombining above two results, there are no configurations where $J$ loses rank i.e. there are no singular configurations for the Jacobian $J$.\nHence proved.\n\n\\end{document}\n", "meta": {"hexsha": "c37c3df203ff4649ba6fbce55150aa0d63a2a2d2", "size": 41651, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw4/hw4.tex", "max_stars_repo_name": "yashorts/robotmath-hws", "max_stars_repo_head_hexsha": "c1ec2612504bd7095af5518a71363d5633d108a6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw4/hw4.tex", "max_issues_repo_name": "yashorts/robotmath-hws", "max_issues_repo_head_hexsha": "c1ec2612504bd7095af5518a71363d5633d108a6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw4/hw4.tex", "max_forks_repo_name": "yashorts/robotmath-hws", "max_forks_repo_head_hexsha": "c1ec2612504bd7095af5518a71363d5633d108a6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.3573099415, "max_line_length": 332, "alphanum_fraction": 0.5378022136, "num_tokens": 20537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.8080672158638527, "lm_q1q2_score": 0.6281105347107918}} {"text": "\\chapter{Einfügen von Quellcode}\n\\label{chap:code}\n\n\\section{Beispiel für einen Programmcode}\n\n\n\\if 1\\mintedinstall\n\\subsection{Beispiel Minted}\n\t\\begin{minted}[xleftmargin=20pt,linenos]{python}\n\timport numpy as np\n\t\n\tdef incmatrix(genl1,genl2):\n\t\tm = len(genl1)\n\t\tn = len(genl2)\n\t\tM = None #to become the incidence matrix\n\t\tVT = np.zeros((n*m,1), int) #dummy variable\n\t\n\t\t#compute the bitwise xor matrix\n\t\tM1 = bitxormatrix(genl1)\n\t\tM2 = np.triu(bitxormatrix(genl2),1) \n\t\n\t\tfor i in range(m-1):\n\t\t\tfor j in range(i+1, m):\n\t\t\t\t[r,c] = np.where(M2 == M1[i,j])\n\t\t\t\tfor k in range(len(r)):\n\t\t\t\t\tVT[(i)*n + r[k]] = 1;\n\t\t\t\t\tVT[(i)*n + c[k]] = 1;\n\t\t\t\t\tVT[(j)*n + r[k]] = 1;\n\t\t\t\t\tVT[(j)*n + c[k]] = 1;\n\t\t\t\t\t\n\t\t\t\t\tif M is None:\n\t\t\t\t\t\tM = np.copy(VT)\n\t\t\t\t\telse:\n\t\t\t\t\t\tM = np.concatenate((M, VT), 1)\n\t\t\t\t\t\n\t\t\t\t\tVT = np.zeros((n*m,1), int)\n\t\n\t\treturn M\n\t\\end{minted}\n\\fi\n\n\n\n\n\\subsection{Beispiel listings}\n\\begin{lstlisting}[language=Python]\nimport numpy as np\n\ndef incmatrix(genl1,genl2):\n\tm = len(genl1)\n\tn = len(genl2)\n\tM = None #to become the incidence matrix\n\tVT = np.zeros((n*m,1), int) #dummy variable\n\n\t#compute the bitwise xor matrix\n\tM1 = bitxormatrix(genl1)\n\tM2 = np.triu(bitxormatrix(genl2),1) \n\t\n\tfor i in range(m-1):\n\t\tfor j in range(i+1, m):\n\t\t\t[r,c] = np.where(M2 == M1[i,j])\n\t\t\tfor k in range(len(r)):\n\t\t\t\tVT[(i)*n + r[k]] = 1;\n\t\t\t\tVT[(i)*n + c[k]] = 1;\n\t\t\t\tVT[(j)*n + r[k]] = 1;\n\t\t\t\tVT[(j)*n + c[k]] = 1;\n\t\n\t\t\t\tif M is None:\n\t\t\t\t\tM = np.copy(VT)\n\t\t\t\telse:\n\t\t\t\t\tM = np.concatenate((M, VT), 1)\n\t\t\t\t\n\t\t\t\tVT = np.zeros((n*m,1), int)\n\t\t\t\t\n\treturn M\n\\end{lstlisting}\n\n", "meta": {"hexsha": "cfe1509920e39ad9d6d4828ac1e673c32e00fba8", "size": 1571, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "11_bsp.tex", "max_stars_repo_name": "silaskalmbach/Vorlage_Abschlussarbeiten_light", "max_stars_repo_head_hexsha": "37028edf2056af98646e447febe689fb8812e92f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "11_bsp.tex", "max_issues_repo_name": "silaskalmbach/Vorlage_Abschlussarbeiten_light", "max_issues_repo_head_hexsha": "37028edf2056af98646e447febe689fb8812e92f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "11_bsp.tex", "max_forks_repo_name": "silaskalmbach/Vorlage_Abschlussarbeiten_light", "max_forks_repo_head_hexsha": "37028edf2056af98646e447febe689fb8812e92f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 20.141025641, "max_line_length": 49, "alphanum_fraction": 0.570973902, "num_tokens": 600, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.6281105245768664}} {"text": "\\documentclass[12pt, a4paper]{article}\n\\usepackage[margin=1.25in]{geometry}\n\\usepackage{graphicx}\n\\usepackage{amsmath}\n\\usepackage{float}\n\\usepackage{listings}\n\\usepackage{caption}\n\\usepackage{physics}\n\\usepackage{mathrsfs}\n\\usepackage[shortlabels]{enumitem}\n\n\n\\setlength\\parindent{0pt}\n\\newcommand{\\code}{\\lstinline[basicstyle=\\small]}\n\\lstset{\n language=Python,\n basicstyle=\\scriptsize\n}\n\n\n\\title{EE2703: Applied Programming Lab \\\\ \\Large Assignment 8: The Digital Fourier Transform}\n\\author{Soham Roy \\\\ \\normalsize EE20B130}\n\\date{\\today}\n\\begin{document}\n\n\\maketitle % Insert the title, author and date\n\n\n\n\\section{Introduction}\nThe goal of this assignment is to analyse signals in the frequency domain. The Discrete Fourier Transform (DFT)\nis a powerful tool for this purpose, and has been calculated using the Fast Fourier Transform (FFT) Algorithm.\nThe Continuous Time Fourier Transform (CTFT) of a gaussian has also been approximated using the FFT. For this,\nwe use the \\code{numpy} library.\n\n\n\n\\section{Subquestions}\n\n\\subsection{Work through the Examples}\n\n\\subsubsection{Random Data}\nThe Fourier Transform of a random signal has been evaluated, and then the Inverse Fourier Transform of\nthe result is evaluated. The maximum absolute error of the result and the original signal varies, but usually\nhas an order of magnitude of -16.\n\\begin{lstlisting}\n x = np.random.rand(100)\n X = fft(x)\n y = ifft(X)\n np.c_[x, y]\n print(\"Maximum Absolute Error for Random Data: \", np.abs(x - y).max())\n\\end{lstlisting}\n\n\\subsubsection{Spectrum of $\\sin(5t)$}\nWe begin with the following rudimentary code:\n\\begin{lstlisting}\n x = np.linspace(0, 2 * np.pi, 128)\n y = np.sin(5 * x)\n Y = fft(y)\n\n plt.figure()\n plt.subplot(2, 1, 1)\n plt.title(\"Spectrum of $\\sin(5t)$ without Phase Wrapping\")\n plt.ylabel(\"$|Y|$\")\n plt.plot(np.abs(Y), lw=2)\n plt.grid(True)\n\n plt.subplot(2, 1, 2)\n plt.xlabel(\"$\\omega$\")\n plt.ylabel(\"Phase of $Y$\")\n plt.plot(np.unwrap(np.angle(Y)), lw=2)\n plt.grid(True)\n\\end{lstlisting}\nTo obtain:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.8]{eg1.png}\n\\end{figure}\nThe frequencies we sample must exclude $2\\pi$, as it is equivalent to 0. The phase plot must also be shifted\nto go from $-\\pi$ to $\\pi$. For this, a helper function \\code{plotter()} has also been written.\n\\begin{lstlisting}\n def plotter(w, Y, title, lim, out, xlabel, ylabels=(\"$|Y|$\", r\"$\\angle Y$\")):\n plt.figure(figsize=(8, 10))\n plt.subplot(2, 1, 1)\n plt.title(title, size=16)\n plt.ylabel(ylabels[0], size=14)\n plt.plot(w, abs(Y), lw=2)\n plt.xlim(-lim, lim)\n plt.grid(True)\n\n plt.subplot(2, 1, 2)\n plt.xlabel(xlabel, size=14)\n plt.ylabel(ylabels[1], size=14)\n ii = np.where(abs(Y) > 1e-3)\n plt.plot(w[ii], np.unwrap(np.angle(Y[ii])), \"go\", lw=2)\n plt.xlim(-lim, lim)\n plt.grid(True)\n\n plt.savefig(\"Assignment_08/LaTeX/\" + out)\n\n\n x = np.linspace(0, 2 * np.pi, 128, endpoint=False)\n w = np.linspace(-64, 64, 128, endpoint=False)\n\n Y = fftshift(fft(np.sin(5 * x))) / 128\n plotter(w, Y, \"Spectrum of $\\sin(5t)$\", 10, \"eg2\", \"$k$\")\n\\end{lstlisting}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{eg2.png}\n\\end{figure}\n\n\\subsubsection{Spectrum of Amplitude Modulated Wave}\nThe signal to be considered is:\n\\begin{equation*}\n f(t) = (1 + 0.1\\cos(t))\\cos(10t)\n\\end{equation*}\nThe same helper function has been used as such:\n\\begin{lstlisting}\n Y = fftshift(fft((1 + 0.1 * np.cos(x)) * np.cos(10 * x))) / 128\n plotter(w, Y, \"Spectrum of $(1 + 0.1\\cos(t))\\cdot\\cos(10t)$\", 15, \"eg3\")\n\\end{lstlisting}\nThis gives us:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{eg3.png}\n\\end{figure}\nThe number of samples used is not enough to resolve all the peaks, hence the time vector and the number of\nsamples have both been increased such that the sampling frequency remains constant.\n\\begin{lstlisting}\n x = np.linspace(-4 * np.pi, 4 * np.pi, 512, endpoint=False)\n w = np.linspace(-64, 64, 512, endpoint=False)\n\n Y = fftshift(fft((1 + 0.1 * np.cos(x)) * np.cos(10 * x))) / 512\n plotter(w, Y, \"Spectrum of $(1 + 0.1\\cos(t))\\cdot\\cos(10t)$\", 15, \"eg4\")\n\\end{lstlisting}\nThus, we obtain:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{eg4.png}\n\\end{figure}\n\n\n\\subsection{Generate the Spectrum of $\\sin^3(t)$ and $\\cos^3(t)$}\n\\subsubsection{$\\sin^3(t)$}\n$\\sin^3(t)$ can be expressed as a sum of sine waves as:\n\\begin{equation*}\n \\sin^3(t) = \\frac{3}{4}\\sin(t) - \\frac{1}{4}\\sin(3t)\n\\end{equation*}\nThus, 2 peaks are expected: at 1 and 3, with phases being of magnitude $= \\frac{\\pi}{2}$.\n\\begin{lstlisting}\n Y = fftshift(fft(np.sin(x) ** 3)) / 512\n plotter(w, Y, \"Spectrum of $\\sin^3(t)$\", 15, \"q2a\")\n\\end{lstlisting}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{q2a.png}\n\\end{figure}\n\n\\subsubsection{$\\cos^3(t)$}\n$\\cos^3(t)$ can be expressed as a sum of cosine waves as:\n\\begin{equation*}\n \\cos^3(t) = \\frac{3}{4}\\cos(t) + \\frac{1}{4}\\cos(3t)\n\\end{equation*}\nThus, 2 peaks are expected: at 1 and 3, with phases $= 0$.\n\\begin{lstlisting}\n Y = fftshift(fft(np.cos(x) ** 3)) / 512\n plotter(w, Y, \"Spectrum of $\\cos^3(t)$\", 15, \"q2b\")\n\\end{lstlisting}\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{q2b.png}\n\\end{figure}\n\n\n\\subsection{Frequency Modulated Wave: $\\cos(20t + 5\\cos(t))$}\nThe same helper function is invoked as follows:\n\\begin{lstlisting}\n Y = fftshift(fft(np.cos(20 * x + 5 * np.cos(x)))) / 512\n plotter(w, Y, \"Spectrum of $\\cos(20t + 5\\cos(t))$\", 30, \"q3\")\n\\end{lstlisting}\nThis gives us:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{q3.png}\n\\end{figure}\nIt can be seen that many more peaks are present, and that single peaks no longer carry a majority of the energy.\nThus, the signal is phase modulated.\n\n\n\\subsection{The Gaussian}\nWe know that the Fourier Transform of a signal is defined as:\n\\begin{equation*}\n X(\\omega) = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} x(t) e^{-j\\omega t} dt\n\\end{equation*}\nWe also know that a Gaussian tends to 0 for large magnitudes of $t$. Thus, for some window size $T$,\nwe can approximate its Transform as:\n\\begin{equation*}\n X(\\omega) = \\frac{1}{2\\pi} \\int_{-T/2}^{T/2} x(t) e^{-j\\omega t} dt\n\\end{equation*}\nApproximating the integral to a Reimann summation of $N$ terms, we get:\n\\begin{equation*}\n X(\\omega) \\approx \\frac{T}{2\\pi N} \\sum_{n = -N/2}^{N/2 - 1} x(nT/N) e^{-j\\omega nT / N}\n\\end{equation*}\nWhere $T / N$ is the time step. Then, let $\\omega = 2\\pi k / T$:\n\\begin{equation*}\n X(2\\pi k / T) \\approx \\frac{T}{2\\pi N} \\sum_{n = -N/2}^{N/2 - 1} x(nT/N) e^{-j2\\pi kn/N}\n\\end{equation*}\nWe observe that the summation is the Discrete Fourier Transform (DFT) of the signal. Thus, we get:\n\\begin{equation*}\n X(2\\pi k / T) \\approx \\frac{T}{2\\pi N} DFT\\{x(nT/N)\\}\n\\end{equation*}\nWe can improve the accuracy of our obtained approximation by choosing a larger window size while keeping the\nsampling frequency constant. We do this iteratively until our error is below a certain threshold.\n\nWe compare our approximation to the actual Continuous Time Fourier Transform (CTFT) of the signal:\n\\begin{equation*}\n \\mathscr{F}(e^{-\\frac{t^2}{2}}) = \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{\\omega^2}{2}}\n\\end{equation*}\n\nFor this, we use the following code:\n\\begin{lstlisting}\n T = 2 * np.pi\n N = 128\n iter_n = 0\n error = None\n threshold = 1e-6 # 6 decimals of precision\n\n while error is None or error > threshold:\n t = np.linspace(-T / 2, T / 2, N, endpoint=False)\n w = np.linspace(-np.pi, np.pi, N, endpoint=False) * N / T\n y = np.exp(-0.5 * t ** 2)\n Y = fftshift(fft(ifftshift(y))) * T / (2 * np.pi * N)\n\n Y_true = np.exp(-0.5 * w ** 2) / np.sqrt(2 * np.pi)\n error = np.sum(np.abs(Y - Y_true))\n\n T *= 2\n N *= 2\n iter_n += 1\n print(f\"Iteration {iter_n}: Total Error = {error:.2e}\")\n\n T /= 2\n N /= 2\n\n print(f\"Samples = {int(N)}, Time Period = {int(T / np.pi)} pi\")\n\n plotter(w, Y, \"Spectrum of Approximated Gaussian\", 5, \"q4a\")\n plotter(w, Y_true, \"Spectrum of True Gaussian\", 5, \"q4b\")\n\\end{lstlisting}\nThis gives us the graphs:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{q4a.png}\n\\end{figure}\nHere, we also have the following output:\n\\begin{lstlisting}\n Iteration 1: Total Error = 7.19e-03\n Iteration 2: Total Error = 5.35e-09\n Samples = 256, Time Period = 4 pi\n\\end{lstlisting}\nThus, we can say that our approximation converged to an accuracy of 6 decimals very fast.\n\\pagebreak\n\nTo compare, the Continuous Time Fourier Transform results in the following graph:\n\\begin{figure}[H]\n \\centering\n \\includegraphics[scale=0.6]{q4b.png}\n\\end{figure}\n\n\n\n\\section{Conclusion}\nWe have calculated the DFT of various signals using the FFT Algorithm. We started with a random signal,\nthen sinusoids, combinations of sinusoids, and finally a Gaussian.\n\n\n\n\\end{document}\n", "meta": {"hexsha": "4dda131ea0d8b892855480844caae0f2324dac42", "size": 9076, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment_08/LaTeX/Report.tex", "max_stars_repo_name": "sohamroy19/EE2703", "max_stars_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment_08/LaTeX/Report.tex", "max_issues_repo_name": "sohamroy19/EE2703", "max_issues_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment_08/LaTeX/Report.tex", "max_forks_repo_name": "sohamroy19/EE2703", "max_forks_repo_head_hexsha": "7ea141082815d80fe765344303d98f96f7a9a492", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.4142857143, "max_line_length": 112, "alphanum_fraction": 0.6549140591, "num_tokens": 2991, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7185944046238982, "lm_q2_score": 0.8740772417253255, "lm_q1q2_score": 0.6281070151129095}} {"text": "\\section*{Analytic Learning Theory}\nThis section covers basic definitions and concepts of statistical learning theory. we approach this from a functional analytic viewpoint.\nSupervised learning in its most abstract setting requires finding a function $f(x)$ given instances ${ (x_i ,f(x_i))}$. Typically ${x_i}$ is an independent and identially distribute (iid) sample from some unknown distribution. A loss function is a random variable\n\\[ L : Ran(f) \\times Ran(f) \\rightarrow \\dblr^+\\]\n defining the cost of misclassification. The risk associated with a candidate function $f'$ is defined to be the expectation of the loss over the sample space $\\Omega$,\n\\begin{equation*} R(f')=\\int L( f(\\omega), f'(\\omega)) d\\omega\\end{equation*}.\nStatistical learning theory is concerned with assessing the approximations to $f$ given by minimizing the empirical loss associated with a sample ${x_i ,f(x_i)}$.\n\nThe notion of a loss function goes back to the roots of modern probability theory and economics. The St. Petersburg paradox is an example of a random variable $S : \\dbln \\rightarrow \\dblr^=$ with infinite expectation limited utility. Let $W(k)$ be the winnings after k plays of from a game with outcome $S$ that pays $2^{i-1}$ with probability $p_i=1/2^i$. The expected payout is given by\n\\begin{equation*}\n\\lim_{k\\To\\infty} W(k)/k =E(S)=\\sum\\limits_{i=1}^{\\infty} p_i 2^{i-1}= \\infty\n\\end{equation*} The implication for a decision theory based only on expected value is that a rational player would pay an infinite amount of money to play this game. Bernoulli introduced the notion of expected utility which takes into account the fact that a payout of $2^i$ may not have twice the utility of a payout of $2^{i+1}$ when $i$ gets large. The utility $U$ is a random variable on the sample space representing preferences of an agent. Loss represents the aversion of an agent to the outcomes of the sample space, $L(\\omega)+U(\\omega) = \\alpha \\fall \\omega \\in \\Omega$ where $\\alpha$ is constant. Expected loss $R(f')$ is the risk associated with choosing the approximation $f'$. Restricting the class of functions to consider when minimizing the risk for a candidate approximation to $f$ is a key aspect of classifier design.\n\n(BBCREVISIT GP RKHS )Gaussian processes provide a class of models and learning algorithms for real world problems that have a long history and are well characterized. Learning algorithms are cast as minimization problems $min_\\mathcal(H) R() $ in a Hilbert space $ \\mathcal{H}$ with a dot product that encapsulates a model and sample data. Bayesian methods are often employed for estimation and inference with Gaussian processes. They allow an intuitive approach to incorporating prior knowledge in classification problems and the ability to obtain confidence intervals for predictions. Many common regression and classification algorithms can be cast as minimization problems in a Reproducing Kernel Hilbert Space (RKHS).\n\n\n\\section*{Learning With Kernels}\n\\cite{KLBurges98atutorial}, \\cite{KLKeerthi99improvementsto}, \\cite{KLProgramminglearningthe},\n\\cite{KLScholkopf00newsupport}, \\cite{KLScholkopf00statisticallearning}, \\cite{KLShevade99improvementsto},\n\\cite{KLTsang03distancemetric}, \\cite{KLWeston00featureselection}, \\cite{KLSchultz03learninga}\nKernel learning is a paradigm for classification and regression where prior belief is expressed in the construction of a similarity matrix of distances between points in a feature space $\\Omega$ by embedding via a non linear map $\\phi$ in a higher [often infinite] dimensional Hilbert space using the kernel as an inner product.\n\\begin{center}\\begin{eqnarray*}\n% \\nonumber to remove numbering (before each equation)\n K(x,x')= <\\phi(x),\\phi(x')> \\\\\n K \\succeq 0 \\\\\n SPD \\Rightarrow \\sum\\limits_{x \\in \\Omega}^{} \\sum\\limits_{x' \\in \\Omega}^{} f(x) K(x,x') f(x') \\geq 0 \\forall f\\ \\in \\ell^2(\\Omega)\n\\end{eqnarray*}\\end{center}\nRecall that infinitely divisible probability distributions aries as the sum of $iid$ random variables. Infinitely divisible kernels have the representation\n\\begin{center}\\begin{eqnarray*}\nK=K^{\\frac{1}{n}} \\ldots K^{\\frac{1}{n}} \\\\\nK= e^{\\beta H} \\\\\ne^{\\beta H} = \\lim\\limits_{n\\rightarrow \\infty} (1+ \\frac{\\beta H}{n} )^n\n\\end{eqnarray*}\\end{center}\nWe construct a mutli-resolution representation of the data with exponentiated kernels. The sequence of kernels $K(\\beta)$ represents a one parameter group associated with a diffusion on the graph of the data. A $\\beta \\rightarrow 0 \\infty$ the kernel moves from the identity to one that represents the clusters in the off diagonal components. The local structure of $\\Omega$ is preserved in $H$ while the global geometry of the data set is progressively revealed in $K(\\beta)$ as we push the diffusion forward with the one parameter group. We can construct exponentiated kernels over direct products of sets $\\Omega_1 \\bigotimes \\Omega_2$ that will allow for the class conditional representation [bbcrevisit term use multiclass]. Simply set $H = H_1 \\bigotimes I_{\\Omega_1} + H_2 \\bigotimes I_{\\Omega_2}$.\n\\begin{center}\\begin{eqnarray*}\nK(\\beta) = e^{\\beta H} = e^{(\\beta H_1 \\bigotimes I_{\\Omega_1} + H_2 \\bigotimes I_{\\Omega_2})} \\Rightarrow \\\\\n\\frac{d}{d \\beta} K(\\beta) = H (K_1(\\beta) \\bigotimes K_2(\\beta))\n\\end{eqnarray*}\\end{center}\n\nThe kernels thus constructed can be used to drive a diffusion on a graph by letting $H$ be the familiar graph Laplacian. Furthermore, the continuum limit of infinite data can be analyzed in within the framework of a discreet stochastic process much the way the convergence of finite element solutions of PDE's takes place.\n\n\n\\section*{Kernel Density Estimation}\nTo define the empirical distribution function of a sample of size $N$ - place mass $1/N$ at each member of the sample. This forms a nonparametric estimate of the marginal density $P(X)$. This is a singular form of kernel smoothing for density estimation. If $\\psi$ belongs to some nice class of function, and $\\int\\limits_{\\infty}^{-\\infty}\\psi(x) dx = 1$, we can form a parametric estimator for the pdf of a process from a sample\npopulation of size $N$ by calculating\n\\begin{equation*} p(x;\\theta)=\\frac{1}{N \\theta}\n\\sum\\limits_{i-1}^{N} \\phi( \\frac{x-X_n}{\\theta})\n\\end{equation*}\nIf $\\phi$ happens to be a density then $p(x,\\theta)$ is also a density. Letting $\\theta \\rightarrow 0$ for the right kernel, we get the empirical density of the sample population. The mean squared error of the estimator expressed as a bias term and a variance term is\n\\begin{equation*} Err_p(x;\\theta)= E[p(x;\\theta)-p(x)]^2 = E[ p(x;\\theta)-p(x)]^2 =\n(E[p(x;\\theta)]-p(x))^2 + Var[ p(x;\\theta)]\n\\end{equation*}\n\n\n\\section*{Distance Functions \\& The Affinity Matrix}\n\nThere are two key ingredients to forming the affinity matrix; a distance function, and a convention for which pairs to consider. If all or too many pairs are compared, sparse methods will not be possible. We can include $n$ nearest points, all points within $\\epsilon$ of $x$, or some other criteria. For instance when working with spatial data, the diffusion can be done on a graph formed by a tessellation of the locations. This is exactly how numerical solutions of heat diffusions are done.\n\nIf the feature vector is a histogram, the $L^2$ distance is not meaningful, in this case use the $\\chi^2$\n\n\\section*{The Graph Cut}\nSpectral clustering is a relaxation to an NP-hard problem of finding an optimal way to cut a graph. Here we describe some various cut criteria. Let\n\\begin{eqnarray*}\nX_1, \\hdots , X_n \\; \\in \\dblr^n \\\\\nW_{ij} = s(X_i,X_j) \\\\\nd_i = \\sum_{j} W_{i j}\\\\\n|A| = card (A) \\\\\nvol(A) = \\sum_{i \\in A} d_i\n\\end{eqnarray*}\n\n\\begin{defn}[Min Cut]\n\\begin{equation*} min cut(A,B) =\\sum_{i \\in A \\; j \\in B} W_ij \\end{equation*}\n\\end{defn}\n\n\\begin{defn}[Balanced Cut / Ratio Cut ]\n\\begin{equation*} min cut (A,B) \\frac{1}{|A|} \\frac{1}{|B|} \\end{equation*}\n\\end{defn}\n\\begin{defn}[n Cut ]\n\\begin{equation*} min cut (A,B) \\frac{1}{vol(A)} \\frac{1}{vol(B)} \\end{equation*}\n\\end{defn}\n\n\n\\section*{Graph Spectra}\nGraph spectral methods are some of the most successful heuristic approaches to partitioning algorithms in solving sparse linear systems, clustering and, ranking problems. Eigenvalues of the graph Laplacian are used to transform a combinatorial optimization problem to a continuous one, typically a SDP problem. Recent advances in SDP optimization techniques have opened new avenues of research in combinatorial optimization. For instance, isoperimetric properties of a graph are used to find efficient communication networks, and fast convergence of Markov Chains.\n\n\n\n\\section*{Matrix Factorization}\nMany forms of matrix factorization can be cast as an optimization problem that involves minimization of generalized Bregman divergences\\cite{BDAUnifiedViewMatrixFactorizationModels}. Factorization algorithms such as NNMF, Weighted SVD, Exponential Family PCA, , pLSI, Bregman co-clustering \\cite{CCBanerjee04ageneralized} can be cast in this framework. The approach uses an alternating projection algorithm for solving the optimization problem which allows for generalizations that include row, column, or relaxed cluster constraints. A brief description of the algorithm is given below. The description of a generalized Bregman divergence can be found in \\cite{BDGordon99approximatesolutions}.\n\n\n\n\\section*{PCA and its generalization to the Exponential Family}\nPCA finds linear combinations of the variables that correspond to directions of maximal variance in the data. Typically this is performed via a singular value decomposition (SVD) of the data matrix $A \\in R^{n,m}$, or via an eigenvalue decomposition if A is a covariance matrix in which case $A \\in R^{n,n}$. Representing the data in the directions of maximum variance allows for a dimension reduction that preserves information. Principal component directions are uncorrelated which can be useful. PCA has the disadvantage that components are usually linear combinations of all variables. Weights in the linear combination data elements are non-zero. Sparse PCA is an attempt to find a low dimension representation of the data that explainers most of the variance.\n\nHere we describe a generalization of Principal component analysis (PCA) to the Exponential Family of probability distributions. PCA is a popular dimensionality reduction technique that seeks to find a low-dimensional subspace passing close to a given set of points \\begin{equation*}\\{x_i\\} \\subset \\mathbb{R}^n\\end{equation*}. The procedure is to solve the optimization problem that minimizes the sum of squared differences of the data points to the projections on a subspace spanned by the empirical variance after centering the data to have mean $0$;\n\\begin{equation*}\n\\sum\\limits_{i=i}^{n} \\norm{x_i - \\theta_i}^2_{\\ell^2}\n\\end{equation*}. The choice of $\\ell^2$ norm here codifies the assumption of Gaussian data. An alternate interpretation of the algorithm is finding the parameters ${\\theta_i}$ that maximizes the log likelihood of the data which corresponds to \\begin{equation*}\n\\sum\\limits_{i=i}^{n} \\norm{x_i - \\theta_i}^2_{\\ell^2}\n\\end{equation*}. The goal of PCA is to find the the true low dimensional distribution of the data given the assumption that data is corrupted by Gaussian noise.\nBregman divergences\n \\begin{equation*}\n D_\\phi(A,B)=\\phi(A)-\\phi(B) - \\nabla \\phi(B) (A-B)\n\\end{equation*}\noffer a framework to extend PCA [and other spectral dimension reduction techniques] to the entire Exponential Family. Here $\\phi$ is a striclty convex function. The roles of\n\nLet $\\theta_i$ be the natural parameter for dimension $i$, with Exponential distribution $P_\\theta$. Then the conditional expectation is given by\n\\begin{equation*}\nlog P_\\theta (x| \\theta) = log P_0(x) + x \\theta - G(\\theta) \\sp : G \\ni \\int{ P_\\theta dx} =1\n\\end{equation*}\nWe can model multivariate data where the conditional distribution can vary along the feature space. The common feature of this PCA model and GLZ regression is the derivative of $G$ which is familiar link function and the loss function which is appropriate for $P_\\theta (x | \\theta)$. The non linear relationship in the GLZ regression model data is captured by the link function $h = \\frac{d}{d \\theta}G(\\theta)'$. This feature is also passed on to the generalized PCA. Instead of projecting on to a linear subspace, a Bregman divergence is used as the distortion measure. This gives a convex optimization problem to solve which can be shown to converge. In \\cite{BDAzoury99relativeloss} a dual function to $\\phi$ is defined by the relationship $\\phi(g(\\theta))+G(\\theta)=h(\\theta) \\theta$ which is used to write the log likelihood as a Bregman divergence\n\\begin{equation*}\n \\log P( x | \\theta ) = -log P_0(x) - \\phi(x) + D_\\phi (x,h(\\theta))\n\\end{equation*}. Typically $x$ is a vector but extending to matrices is straightforward.\n\n\nSparse PCA\n[CS Notes from http://ugcs.caltech.edu/~srbecker/\n\nClassic PCA is sensitive to outliers. Convex methods can be use to address this by attempting to factor the data $X$ as a sum of a low rank component $L$ and a sparse component $S$ by solving\n\\begin{equation*}\n\\underset{min}{S,L} \\norm{L}_* + \\lambda \\norm{S}_{\\ell_1} : X=L+S\n\\end{equation*}\nSee http://cvxr.com/tfocs/demos/rpca/ for a demo of this in action with the popular cvx software package.\n\n\\section*{Manifold Learning}\nThere are numerous machine learning techniques which accomplish some form of dimensionality reduction.\nManifold learning uses principal curves and manifolds to encode a natural geometric framework for nonlinear dimensionality reduction. These methods construct low-dimensional data representation using a cost function that retains local properties.\nContrasting methods such as MDS employ proximity data via a similarity or distance matrices. The important ISOMAP \\cite{MDS_ISOMAP}algorithm extends MDS by capturing geodesic measurements of non-local pairs on the data manifold $M$ via an multi-scale approximation. Non-local distances are approximated via a shortest path on a K nearest neighbor clustering of the data. Effectively a ball in data space is used to represent a cluster, and a graph is then constructed to encode the non-local information. The connectivity of the data points in the neighborhood graph are the nearest k Euclidean neighbors in the feature space. Dijkstra's algorithm for computing shortest paths with weights is used to construct the proximity matrix from the neighborhood graph. The top n eigenvectors encode the coordinates in the low dimensional Euclidean space. Choosing the correct number of neighbors is an essential component to an accurate representation.\nOther shortest path algorithms that may be employed to calculate the geodesic distances are listed below:\n\\begin{itemize}\n \\item Dijkstra's algorithm finds the single-pair, single-source, and single-destination shortest path.\n \\item Johnson's algorithm finds all pairs shortest paths\n \\item Bellman-Ford algorithm single source problem and allows negative edge weights.\n \\item Floyd-Warshall algorithm solves all pairs shortest paths.\n \\item A* search algorithm solves the single pair shortest path problem.\n\\end{itemize}\nIn \\cite{MDSBernstein00graphapproximations} a sampling condition is given which bounds the quality of the manifold embedding based on the quality of the neighborhood graph.\n\n\n\\section*{Graph Laplacian}\n\\cite{GLPBoydconvexoptimization}, \\cite{GLPChung93laplaciansof}, \\cite{GLPCrescenzi96toweight},\n\\cite{GLPGuatterygraphembeddings}, \\cite{GLPBoydconvexoptimization},\n\\cite{GLPChung93laplaciansof},\n\\cite{GLPSpectralAnalysisComplexLaplacianMatrices}, \\cite{GLPKellersignedgraph}\n\nLet $G$ be a connected simple graph with vertex set $V = {1, 2, ... , n}$ , edge set $E$ and let each edge be associated with a positive number, called the weight of the edge. The above graph is called a weighted graph. An unweighted graph is just a weighted graph with each of the edges bearing weight 1. The weight $w(i)$ of a vertex $V_i$ is the sum of the weights of the edges incident with it. There are a number of ways in which the Laplacian matrix $L$ is defined; the combinatorial Laplacian, the normalized Laplacian and the unsigned Laplacian. Spectra from graph matrix representations may be obtained from the adjacency matrix $A$ and the various Laplacian discretizations. Spectra can also be derived from the heat kernel matrix and path length distribution matrix.\n\nThe matrix representation of the graph Laplacian has a significant effect on the spectrum. Attributes may be accounted for by by a complex number that encodes the edge attributes. The node attributes may be encoded in the diagonal elements. The complex graph Laplacian matrix is Hermitian, and hence it has real eigenvalues and complex eigenvectors. Graph feature vectors can be embedded in a pattern space by PCA, MDS, and LDA( linear discriminant analysis). Attribute graphs may be characterized by the application of symmetric polynomials to the real and complex components of the eigenvectors. \\cite{GLPAnaFred} This gives rise to permutation invariants that can be used for pattern vectors. Partitioning a graph into three pieces, with two of them large and connected, and the third a small separator set can be accomplished using the second eigenvector [the Feidler Vector] of the graph Laplacian. In the case or sparse graphs, the first few eigenvectors can be efficiently computed using the Lanczos algorithm [see section below on ARPAC]. This graph partitioning algorithm can be extended to give a hierarchical subdivision of the graph.\n\n\n\\section*{Distance Metrics}\nA measure of similarity between data points is a vital component to clustering algorithms. The suitability of any given measure is dependent on the generative process providing the data.\n\n\n\\section*{Markov Chains}\nLet $X \\in \\mathcal{M}$, $P(x,y)$ the transition probability for an irreducible Markov chain. If $P$ is reversible relative to $\\pi$ then we have that $Q(x,y) = \\Pi(x) P(x,y) =\\Pi(y) P(y,x) \\forall x,y, \\in X$. This implies that $\\pi$ is the stationary distribution. Bounding the rate of convergence to the stationary distribution is related to $L_G$ and isoperimetric problems.\n\nDetailed balance is a property of a Markov chain where $\\pi(x) P(x,y) =\\pi(y) P(y,x) \\Rightarrow$ reversibility. This is stronger than the requirement of a stationary distribution. It is the property that implies for every closed cycle of states there is no flow of probability.\n\nThere are four aspects of Markov chain stability.\n\\begin{enumerate}\n\\item $\\pi$-irreducible\n\\item small set\n\\item Harris recurrence\n\\item Geometric Ergodicity\n\\end{enumerate}\n\nFor reversible Markov chains, the rate of convergence to $\\Pi$ for a finite state chain is determined by the second eigenvalue of $P$.\n\nThere are two types of bounds for the rates of convergence\n\\begin{enumerate}\n\\item graph theoretic Cheeger type - see Diaconis and Stook\n\\item Spectral\n\\end{enumerate}\nThe rate of convergence to $\\pi$ for finite state Markov chains is determined by the second eigenvalue $\\lambda_2$ of $P$.\n\nBBCREVISIT CLEAN UP LOTS\n\\begin{thm}[Perron Frobenius - $P \\in \\dblp^{n \\times n}$ ]\nThere is a simple largest eigenvalue, and it's eigenvector is positive.\n\n\\end{thm}\nThe growth of $P^k$ is determined by the Perron eigenvalue. Frobenious extended the theorem to include a class of non negative matrices. This is achieved through the concept of reducibility. $P$ is irreducible if\n\\begin{equation*}\n\\neg \\exists \\; \\Lambda \\st\\Lambda^\\dag L \\Lambda =\n \\left( \\begin{array}{cc}\nP_{11} & P_{12} \\\\\n0 & P_{22} \\end{array} \\right)\n\\end{equation*}\nwhere $\\Lambda$ is a projection $\\Lambda^2 = \\Lambda$\nThe irreducible matrices $L$ have a simple largest eigenvalue $\\lambda_n$, now called the Perron-Frobenius eigenvalue whose right eigenvector components are all positive. We can put this in Markov chain terminology; irreducibility is equivalent to the existence of a unique stationary distribution.\n\nBBC Make the Markov Chain section nice.\n\nHammerlsy Clifford\n\nThe Hammersley–Clifford theorem gives necessary and sufficient conditions for when a positive probability distribution can be represented as a Markov Random Field. It states that a probability distribution that has a positive mass or density satisfies one of the Markov properties with respect to an undirected graph $G$ if and only if it is a Gibbs random field. The density of the Gibbs random field can be factorized over the cliques (complete subgraphs) of $G$.\n\n\n\\section*{Spectral Graph Theory}\nModern spectral graph theory increasingly takes insights from geometry. Discrete analogues of isoperimetry results and heat flow on manifolds are just a few examples being put to use in modern applications. The normalized graph Laplacian is used to aid in consistency between spectral geometry and stochastic processes. We consider connected graphs $G = (E,V)$ in this work, in which case we can define the normalized graph Laplacian as $\\mathcal{L} = T^{\\frac{1}{2}} L T^{\\frac{-1}{2}} = I - T^{\\frac{1}{2}} A T^{-\\frac{1}{2}}$, where $A$ is the adjacency matrix, L is defined by\n\n\\begin{equation*}\nL(u,v) = \\Biggl\\{\n\\begin{array}{cc}\n d_v & :\\; u=v \\\\\n -1 & : u\\sim v \\\\\n 0 & : u \\nsim v \\end{array}\n\\end{equation*} and\n$T = diag\\{d_1, \\cdots , d_n\\}$ where $d_v$ is the degree of vertex $v$.\n\n$\\mathcal{L}$ is a difference operator :\n\n\\begin{equation*}\n \\mathcal{L} = \\frac{1}{\\sqrt{d_u}} \\sum\\limits_{}^{v : u \\sim v} ( \\frac{g(u)}{\\sqrt{d_u}} - \\frac{g(v)}{\\sqrt{d_v} } )\n\\end{equation*}\n\n\\begin{eqnarray}\nVol (G) = \\sum\\limits_{v \\in V}^{d_v} = Tr(T) \\\\\n\\sigma(\\mathcal{L}) \\in \\dblr^+ \\\\\nker ( \\mathcal{L} ) = span\\{ T^{\\frac{1}{2}} \\mathbb{1} \\}\n\\end{eqnarray}\n\n\n\n\n\n\\section*{Diffusion Maps}\n\\cite{DMBremerabstractdiffusion}, \\cite{DMCarnegieinformationdiffusion}, \\cite{DMCoifmandiffusionmaps},\n\\cite{DMKubota00reactiondiffusionsystems}, \\cite{DMLafferty05diffusionkernels},\n\\cite{DMNadler06diffusionmaps}.\n\nSpectral clustering involves constructing a Markov chain over a graph is constructed over the graph of the data and using the sign of the first non-constant eigenvector for graph cuts and cluster localization. This approach can be generalized to higher-order eigenvectors yielding a multi-resolution view of the data. Using multiple eigenvectors allows one to embed and parameterize the data in a lower dimensional space. Examples of this procedure include LLE, Laplacian \\& Hessian Eigenmaps. The common theme among these approaches is that eigenvectors of a Markov process can encode coordinates of the data set on a low dimensional manifold in a Euclidian space. The advantage over conventional methods is that the representation is non-linear and they preserve local structure. Kernel eigenmap embeddings can be generalized into a diffusion framework where a discrete Laplacian acts on a low dimensional representation space. This allows for a true multi-scale parametrization. Iterating a Markov process involves computing power of the transition matrix to run a random walk of the graph forward in time. By construction a one parameter map defining the diffusion and specifying boundary conditions the full power of diffusions on a smooth manifold may be brought to bear on parameterizing the geometry of the data. Different boundary conditions and diffusion operators give rise to a discrete approximations of familiar stochastic PDE's.\n\nLet $(X,\\mathcal{A},\\mu)$ be a measure space and $\\quad k: X \\times Y \\longrightarrow \\dblr $ a kernel function.\n \\begin{eqnarray}\n d(x) &=& \\int\\limits_{X} k(x,y) d \\mu(y) \\\\\n P(x,y) &=& \\frac{ k(x,y) }{ d(x) } \\\\\n (D_t (x,y))^2 &=& \\norm{P_t(x, \\cdot) - P_t(y,\\cdot )}_{ L^2(X,\\frac{d\\mu}{\\pi}) } \\\\\n \\pi(\\mu) &=& \\frac{d(y)}{z \\in Z^{d(z)} } \\\\\n \\pi(x) p(x,y) &=& \\pi(y) p(y,x)\n \\end{eqnarray}\n\n\n$D_t (x,y)$ is the functionally weighted $L^2$ distance between the 2 posteriors $\\mu \\rightarrow P_t(x,u) $ and $\\mu \\rightarrow P_t(y,u) $. This is related to isoperimetry. Think about what happens as the cardinality of paths connecting $x$ and $y$ is increased. $D_t$ can be computed using the eigenvalues of $P$.\n\n\\begin{equation*}\nD_t (x,y) = \\sqrt{ \\sum\\limits_{\\lambda \\geq 1 }{} \\lambda_{l} ( \\phi_l (x) - \\phi_l (y) )^2 }\n\\end{equation*}\n\nWe can define an embedding in Euclidian space via\n\n\\begin{equation*}\n\\Psi_t (x) = \\{ \\lambda_{1}^{t} \\phi_1(x) , ... \\lambda_{s(\\delta,t)}^{t} \\phi_{s(\\delta,t)} (x) \\}\n\\end{equation*}\n\n\n\n\n\\section*{Spectral Geometry}\nSpectral Geometry concerns itself with the relationships between a geometric structure and the spectra of a differential operator, typically the Laplacian. Inferring the geometry from the spectra is a type of inverse problem since two non isometric manifolds may share the same spectra. Going the other way, we encounter isoperimetric inequalities and spectral gap theorems. \"Can One Hear the Shape of a Drum?\" was the of an article by Mark Kac in the American Mathematical Monthly 1966. The frequencies at which a drum vibrate depends on its shape. The elliptic PDE $ \\nabla^2 A + k A = 0$ tells us the frequencies if we know the shape. These frequencies are the eigenvalues of the Laplacian in the region. Can the spectrum of the Laplacian tell us the shape if we know the frequencies? Hermann Weyl showed the eigenvalues of the Laplacian in the compact domain $\\Omega$ are distributed according to $ N(\\lambda) \\sim (2 \\pi)^{-d) \\omega_d \\lambda^{\\frac{d}{2}} vol(\\Omega}$\n\nThe Laplace Beltrami operator is the generalization of $\\nabla \\circ \\nabla = \\Delta$ to $\\mathcal{M}$\n\\begin{equation*}\n\\Delta f = tr(H(f))\n\\end{equation*}\nIn the exterior calculus we have $ \\Delta f = d^*d \\; f$.\n\n//BBCREVISIT - Fill this out and check\nThe Laplacian of a Gaussian has well known applications in image processing. Given $f(x,y)$, we get a scale space representation when we convolve by\n\\begin{equation*}\n g(x,y,t) = \\frac{e^{x^2+y^2}}{2 \\pi t}\n\\end{equation*}\n\n\\begin{equation*}\n L(x,y,t) =g(x,y,t) \\ast f(x,y)\n\\end{equation*}\nApplying $\\Delta$ to $L(x,y,t)$ gives response to blobs of extent $\\sqrt{t}$\n\nThere is a well known connection between diffusion processes and Schrodinger operators;\n\\begin{eqnarray*}\nH = \\nabla^2 + V(x) \\Phi \\in L^2(\\dblr^n) \\\\\nH \\Phi = E \\Phi \\\\\nE = \\sigma(H)\n\\end{eqnarray*}\n\n\n\n\\section*{Concentration of Measure}\n \\cite{MCArora04expanderflows}, \\cite{MCBartlett03convexity}, \\cite{MCBoucheron04concentrationinequalities},\n \\cite{MCFRIEDMAN96computingbetti}, \\cite{MCLedoux04spectralgap}, \\cite{MCMuyan_ablessing},\n \\cite{MCSinclair92improvedbounds}, \\cite{MCTalagrand95concentrationof}.\n\nFamiliar tools used when when dealing with additive functions of independent random variables are the CLT, LLN, and the inequalities of Markov, Chebychev, and Chernoff. When the differences are not independent we rely on the theory of martingales and use inequalities like Azuma's to provide concentration bounds.\n\nLet $(X,\\Sigma,\\mu)$ be a measure space and let $f$ be an real-valued measurable function defined on $X$. Then for any number $t > 0 \\in \\mathbb({R}$ we have\n\\begin{equation*}\n\\mu(x \\in X | f(x) \\geq t) < \\frac{1}{t} \\int_X |f(x)| d \\mu\n\\end{equation*} If we let $\\mu$ be a probability measure - $\\mu(X)=1$, then the above is equivalent to $P(|X| \\geq a) \\leq \\frac{E(|X|)}{a}$ commonly known as Markov's inequality.\n\nAnother familiar concentration inequality is the Chebychev inequality has it's origins as a measure theoretic inequality;\n\\begin{equation*}\n\\mu({x \\in X : \\abs{f(x)} \\geq t}) \\geq \\frac{1}{t^2}\nint_X \\abs{f}^2 d\\mu\n\\end{equation*} When $X$ has finite first moment $\\mu$ and non-zero second moment $\\sigma$, we have the more familiar\n\\begin{equation*}\nP( \\abs{X - \\mu} \\geq k \\sigma ) \\leq \\frac{1}{k^2} \\; \\forall k > 0\n\\end{equation*}\n\n\n%These notes generally follow Wiki on Isoperimetry\nThe isoperimetric inequality concerns the relationship between the length $l$ of a closed curve and the area $a$ of the planar region that it encloses. Specifically, $4 \\pi a \\leq l^2$. Equality holds in the case that the curve is a circle. The isoperimetric problem is to determine a plane figure of the largest area whose boundary has a given length. F\n\nFederer \\cite{federer1996geometric} is a good reference for a measure theoretic generalization to higher dimensions. We make a few remarks here that will be expanded on later.\n\\begin{prop}[Isoperimetric Inequality In $\\mathbb{R}^n$]\nLet $\\mu$ be Lebesgue measure in $\\mathbb{R}$ and $X \\ in \\mathbb{R} \\st \\mu(cl(X) < \\infty$, then\n\\begin{equation*}\n n \\omega^{\\frac{1}{n}}_{n} \\mu(cl(X))^{\\frac{n-1}{n}} \\leq M^{n-1}(\\partial X)\n\\end{equation*}\n\\end{prop}\n$M$ is the Minkowski content, which is the Hausdorff measure of $\\partial X$ for rectifiable $\\partial X$.\nThe proof relies on the Brunn–Minkowski theorem which states that $\\mu(A+B)^{\\frac{1}{n}} \\geq \\mu(A)^{\\frac{1}{n}} + \\mu(A)^{\\frac{1}{n}}$ where set addition in $\\mathbb{R}$ is in the sense of Minkowski. This addition behaves well with respect to the convex hull; $\\forall A,B \\in \\mathbb{R} \\; co(A + A) = co(A) + co(A)$\n\nFor smooth domains general Isoperimetry inequality is equivalent to a Sobolev inequality on $\\mathbb{R}$.\n\n\\begin{prop}[Sobolev inequality]\nLet $u$ be a continuously differentiable real-valued function on $\\mathbb{R}$ with compact support. Then for $1 \\le p < n$ there is a constant $C$ depending only on $n$ and $p$ such that\n\\begin{equation*}\n ||u||_{L^{\\frac{pn}{n-p}}} \\leq C || Du||_{L^p}\n\\end{equation*}\n\\end{prop}\n\nThe Sobolev embedding theorem relies on the\n\\begin{thm}[Hardy Littlewood Sobolev fractional integration theorem]\nLet $0 < \\alpha 0$, there exists a finite positive number $a(t)$ such\nthat, for all $f\\in L^1$ :\n\\begin{equation*}\\label{ult1}\n\\|T_tf\\|_{\\infty} \\leq a(t) \\|f\\|_1.\n\\end{equation*}\n\nAn equivalent formulation (by interpolation) of ultracontractivity is\nthat for any $t>0$, there exists a finite positive number $c(t)$ such\nthat, $\\forall f\\in L^2$,\n\\begin{equation*}\\label{ult2}\n\\|T_tf\\|_{\\infty} \\leq c(t) \\|f\\|_2\n\\end{equation*}\n Also by duality, the inequality (\\ref{ult2}) is equivalent to\n\\begin{equation*}\\label{ult3}\n\\|T_tf\\|_{2} \\leq c(t) \\|f\\|_1\n\\end{equation*}\nIt is known that, under the assumptions on the semigroup\n$(T_t)_{t\\geq 0}$, (\\ref{ult2}) implies (\\ref{ult1})\nwith $a(t)\\leq c^2(t/2)$\nand\n(\\ref{ult1}) implies (\\ref{ult2}) with $c(t) \\leq \\sqrt{a(t)}$.\n\\\\\n\nWe say that the generator $A$ satisfies LSIWP (logarithmic Sobolev inequality\nwith parameter) if there exist a monotonically decreasing continuous function\n${\\beta}: (0,+\\infty)\\rightarrow (0,+\\infty)$ such that\n\\begin{equation*}\\label{lsiwp}\n\\int f^2\\log f\\, d{\\mu} \\leq\n\\epsilon Q(f) +{\\beta}(\\epsilon) \\|f\\|^2_2 + \\|f\\|^2_2\\log \\|f\\|_2\n\\end{equation*}\nfor all $\\epsilon >0$ and $0\\leq f\\in \\mbox{Quad}(A)\\cap L^1\\cap\nL^{\\infty}$ where\n$\\mbox{Quad}(A)$ is the domain of $\\sqrt{A}$ in $L^2$ and\n$Q(f)=(\\sqrt{A}f,\\sqrt{A}f)$.\n\\\\\n\nThis inequality is modeled on the Gross inequality \\cite{}.\n\\\\\n\nIn \\cite{ds},\\cite{d}, the authors show that LSIWP implies\nultracontractivity property under an integrability condition on $\\beta$. This condition can be enlarged and be stated as follows:\n\n\\begin{thm}\nLet ${\\beta}(\\epsilon)$ be a monotonically decreasing continuous\nfunction of $\\epsilon$\nsuch that\n\\begin{equation*}\\label{vareps}\n\\int f^2\\log f \\, d{\\mu}\\leq\n\\epsilon Q(f) +{\\beta}(\\epsilon)\\, \\|f\\|^2_2 + \\|f\\|^2_2\\log \\|f\\|_2\n\\end{equation*}\nfor all $\\epsilon >0$ and $0\\leq f\\in \\mbox{Quad}(A)\\cap L^1\\cap\nL^{\\infty}$. Suppose that\nfor one ${\\eta}>-1$,\n\\begin{equation*}\\label{integral}\nM_{\\eta}(t)=({\\eta}+1)t^{-({\\eta}+1)})\\int_0^t\n{s}^{\\eta}{\\beta}\\left(\\frac{s}{\\eta+1}\\right)\n\\,ds\n \\end{equation*}\nis finite for all $t>0$. Then $e^{-At}$ is ultracontractive\nand\n\\begin{equation*}\\label{majo}\n\\| e^{-At} \\|_{\\infty,2}\\leq e^{M_{\\eta}(t)}\n\\end{equation*}\nfor all $0$ be the Bregman divergence. Let $\\chi = {X_i} \\in S \\in dblr^d$ be a random variable and take the encoding $X_i \\to S$ so the rate is zero and the code book is 1. The rate distortion is $E_\\nu [D_\\phi (\\chi,s)] = min \\s in S \\sum\\limits_{i=1}^{n} \\nu_i D_\\phi (x_i,x)$ which we call the Bregman information for the random variable $X$.\n\\begin{thm}\n$\\mu = E_\\nu [X]$ is the unique minimizer\n\\end{thm}\n\n\n\\section*{Sparse Representation}\nA Gaussian distribution is often an accurate density model for low dimensional data, but very rarely for high-dimensional data. High dimensional data is less likely to be Gaussian, because of the high degree of independence this demands. Recall the a Gaussian is a rotation of a distribution with completely independent coordinates. In a typical high dimensional application, one may be able to find a few features that are approximately independent, but generally as more features are added the dependencies between them will grow.\n\nDiaconis and Freedman showed that for \\textit{most} high dimensional point clouds, \\textit{most} low dimensional orthogonal projections are a mixture of normal spherically symmetric distributions.\n\n\\begin{lem}[Poincare Lemma]\nIf $\\sigma_n$ is uniform on $\\sqrt{n}S_{n-1} \\in \\dblr^n$, $d : \\Phi f* = f \\; f* \\succeq 0\n\\end{equation*}\nbbcrevisit\n\nMuch work has been done to determine when these two problems have the same solution, i.e. when is exact recovery possible. One sufficient condition for exact recovery is the RIP condition of Candes and Tao (see below).\n\n\\begin{defn}[RIP (Candes \\& Tao) ]\nLet $A \\in M_{m \\times n}(\\dblr)$ and $p \\in [1,n]$, then we say that $A$ has the restricted isometry property if\n\\begin{eqnarray*}\n \\exists \\delta_p \\st \\forall m,p A_s \\in A \\\\\n (1 - \\delta_p) \\norm{y}_{\\ell^2}^2 \\leq \\norm{A_s y}_{\\ell^2}^2 \\leq (1 + \\delta_p) \\norm{y}_{\\ell^2}^2 \\leq\n\\end{eqnarray*}\n\\end{defn}\nRIP is a property which classifies a matrix as being close to orthogonal. $\\delta_p$ is referred to as the RIC, and is NP-Hard to compute. Compressed sensing decoders are guaranteed to recover the sparsest solution to $Y = A x$ when $A$ is close to an isometry. Bounds on the RIC are available for some classes of random matrix ensembles.\n\n\n\\section*{Random Matrix Theory \\& OP}\n\nRMT concerns itself with the eigenvalue statistics of large matrices with random entries. We define the eigenvalue counting measure of a matrix $H$n as;\n\\begin{equation*}\n\\mu_H (A) = \\frac{| \\lambda_i \\in A |}{n} = N_{1_A, H}\n\\end{equation*}\n\nMore generally a eigenvalue statistic $N_{f,H} = \\frac{tr f(H)}{n}$\nFor many types of random matrices we have a CLT;\n\\begin{equation*}\n\\frac{N_{f,H}-\\int f(\\lambda) dN(\\lambda)}{\\sigma_{f,n}} \\rightarrow_d N(0,1)\n\\end{equation*}\n\n$GUE(n)$ Start with Gaussian measure\n\\begin{equation*}\n\\gamma^n(A) - \\frac{1}{(\\sqrt{2 \\pi} )^n} \\int_A e^ \\frac{-1}{2} \\norm{X}^2 d \\lambda^n\n\\end{equation*}\n\nWeiener space is the Hilbert space $L^{2,0}[0,1]$ of upon which a Gaussian measure can be defined. The inner product on Weiner space is\n\\begin{equation*}\n<\\sigma_i,\\sigma_j> = \\int\\limits_{0}^{1} <\\overset{\\cdot}{\\sigma_i},\\overset{\\cdot}{\\sigma_j}> dt\n\\end{equation*}\nwhere $\\overset{\\cdot}{\\sigma_i} = \\frac{d \\sigma)i}{dt}$\nThe Wiener measure is a Gaussian measure.\n\nLet $H$ be Hermitian, $Z_{GUE(n)} = 2^{\\frac{n}{2}} \\pi^{\\frac{n^2}{2}}$ We can write\n\\begin{equation*}\n \\gamma^n(A) - \\frac{1}{Z_{GUE(n)} } \\int_A e^ \\frac{-1}{2} tr(H^2) d \\mu\n\\end{equation*}\n\n\nThere are two domains of eigenvalue statistics, local and bulk. Locally we are concerned with level spacing ,$\\Delta \\lambda = \\lambda_i - \\lambda_{i-1}$, and edge statistics $P_{TW}(\\lambda_1), \\underset{n \\rightarrow \\infty}{lim} P_{TW}(\\lambda_n)$. The Tracy Widom law $P_{TW}$ gives edge statistics.\n\nThe empirical spectral measure of $H$ is\n\\begin{equation*} \\mu_H = \\frac{|eigs H \\in A |}{n} \\end{equation*}\n\nThe CDF of $H(n)$ is $N_n (\\lambda)$ and as $n \\rightarrow \\infty$ is $N_n (\\lambda) \\rightarrow W(\\lambda)$ where $W$ is the Winger Semi-Circle Law.\n\nBulk statistics look at the determinantal point process $E(\\lambda_0) = \\sum_j \\delta ( n \\rho(\\lambda_0 (\\lambda_j - \\lambda_0)$. The kernel of the process is the sine kernel $K(x,y) = \\frac{sin (\\pi (x-y) )}{\\pi (x-y)}$. $E(\\lambda)$ captures the statistics of the eigenvalues in the vicinity of $\\lambda_0$. Recall the joint densities of a determinantal point process with kernel $K$ are given by $\\rho_n(x_1, ...,x_n) = det(K(x_i,x_j)_{1 \\leq i, j \\leq j})$\n\n\n\\section*{The Hamburger Moment Problem - HMP}\n\nGiven a sequence ${m_i}$ the HMP seeks to find a measure that generates the moments.\n\\begin{equation*}\n \\exists \\mu : m_n = \\int\\limits^{+ \\infty}_{- \\infty} x^n d\\mu\n\\end{equation*}\n\nThe answer is affirmative when the Hankel kernel $A \\succ0 $, which\nis equivalent to $\\sigma(A) \\in \\mathbb{R}^{+}$\n\\begin{equation*}\nA =\\left(\n \\begin{array}{cccc}\n m_0 & m_1 & \\ldots & \\\\\n m_1 & m_2 & m_3 & \\\\\n m_2 & m_3 & m_4 & \\\\\n \\vdots & & & \\ddots \\\\\n \\end{array}\n\\right)\n\\end{equation*}\n\nRecall that A Hankel matrix is a square matrix with constant skew-diagonals; $A_{i,j} = A{i-1,j+1}$\nA Hankel matrix is an upside-down Toeplitz matrix.\n\nThe Hilbert matrix $ H{i,j} = \\frac{1}{i+j-1}$ is a special case of a Hankel matrix. The Hilbert matrices are canonical examples of ill-conditioned matrices.\nThey arise in the expression for the Grammian matrix of powers of $x$; $H_{ij} = \\int\\limits{0}{1} x^{i+j-2} dx$\nwhich shows up in the least squares approximation by polynomials.\n\nThe solutions to the HMP are either unique or infinite in number and form a convex set in $\\mathcal{H}$\n\nLet\n\\begin{equation*}\n\\Delta_n =\\left(\n \\begin{array}{cccc}\n m_0 & m_1 & \\ldots & m_n \\\\\n m_1 & m_2 & \\ldots & m_{n-1} \\\\\n & & \\ddots & \\\\\n m_n & m_{n+1} & & m_{2n}\\\\\n \\end{array}\n\\right)\n\\end{equation*}, then $A \\succ0 \\Rightarrow det(\\Delta_n) \\geq 0 \\forall n$\nIf $det(\\Delta_n)=0$ then $(\\mathcal(H), <,>)$ is finite dimensional and $T$ is self adjoint.\n\n$A$ gives us a sesquilinear form on $\\mathcal(H) = \\mathcal(l)^2$ via\n$ = \\hbar{x}^T A y$.\n\nLet $T$ be a shift operator. The HMP is closely related to OP in $\\mathbb{R}$.\nGram Schmidt gives basis ${\\phi_i}$ in which $T$ has tridiagonal Jacobi.\n\nThe Caley transform $Q(T) = (I-T)(I+T)^{-1}$ shows the connection to the Nevanlina class\nof functions (sub-harmonic).\n\n\n\n\\section*{Random Matrix Ensembles}\n\\cite{RMTAchlioptas04randommatrices}, \\cite{RMTAlon00bipartitesubgraphs}, \\cite{RMTAlon00onthe} ,\n\\cite{RMTCooper00onthe}, \\cite{RMTSoshnikov02anote}, \\cite{RMTTracy98correlationfunctions}\n\nThe classical ensembles of random matrix theory are GOE, GUE, GSE, Wishart, and MANOVA. These correspond to the weight functions of the equilibrium measure of the orthogonal polynomials Hermite, Laguerre,and Jacobi. The Jacobians of the well known matrix factorizations are used to compute the joint eigenvalue densities of these ensembles. The distribution of eigenvalues of the GOE ensemble follow the well know Winger Semi-circle distribution.\nThe joint densities up to a constant factor are listed below:\n\\begin{itemize}\n \\item Hermite \\item Laguerre \\item Jacobi\n\\end{itemize}\nWe generated histograms in Matlab for samples from the GOE, GUE, GSE, Wishart, and MANOVA ensembles.\nThe joint PDF of a generic Gaussian ramdom matrix is given by,\n\\begin{equation*}\nP(M)=G_\\beta(n,m)=\\frac{1}{2 \\pi^{\\frac{\\beta n m}{2} }} \\exp^{\\frac{-1}{2}\\norm{M}_F }\n\\end{equation*} where $\\beta$ encodes the dimension of the field. Note this leaves open the possibility to\ngeneralize to non integer $\\beta$.\n\nThe table below describes how to generate from the common ensembles starting from a sample $A \\in G_\\beta(n,n)$\n\\begin{eqnarray*}\n GOE \\{ M | M = \\frac{A+A^T}{2}, A \\in G_1(n,n)\\}\\\\ %[bbcrevisit is necessaria and sufficient?]\\\\\n GUE \\{ M | M = \\frac{A+A^\\dagger}{2}, A \\in G_2(n,n)\\}\\\\ %[bbcrevisit is necessaria and sufficient?]\\\\\n GSE \\{ M | M = \\frac{A+A^\\ddagger}{2}, A \\in G_4(n,n)\\} %[bbcrevisit is necessaria and sufficient?]\n\\end{eqnarray*}\n\n\nThe $CS$ decomposition is a matrix factorization equivalent to four $SVD$'s which correspond to rotation problems\n$\\left(\\begin{array}{cc}\n X \\rightarrow Y & X^\\perp \\rightarrow Y \\\\\n X \\rightarrow Y^\\perp & X^\\perp \\rightarrow Y^\\perp \\\\\n\\end{array}\\right)$\nWhich can be compactly written\n$[X | X^\\perp]^T [Y | Y^\\perp ]=\\left(\n \\begin{array}{cc}\n Q_{11} & Q_{12} \\\\\n Q_{21} & Q_{22} \\\\\n \\end{array}\n\\right)$\n $\\left(\n \\begin{array}{cc}\n Q_{11} & Q_{12} \\\\\n Q_{21} & Q_{22} \\\\\n \\end{array}\n\\right) = \\left(\n \\begin{array}{cc}\n U_1 & 0 \\\\\n 0 & U_2\n \\end{array}\n\\right) * \\left(\n \\begin{array}{cc}\n C & S \\\\\n -S & C\n \\end{array}\n\\right) * \\left(\n \\begin{array}{cc}\n V_1 & 0 \\\\\n 0 & V_2\n \\end{array}\n\\right)\n$\nWhere U, S are unary.\n\nThe Tracy-Widom law of order one is the limiting distribution of the largest eigenvalue of a Wishart matrix with identity covariance when properly scaled. This has some application to weighted directional graphs. The largest eigenvalue of the adjacency matrix of a random d-regular directed graph follows the Tracy-Widom law. The kernels of integrable operators describe the asymptotic eigenvalue distribution of self-adjoint random matrices from the unitary ensembles. Consider the discreet operator $K(n,m): \\l^2(N) \\rightarrow \\l^2(M)$ where $K(n,m) = \\frac{(}{m-n}$ the discrete Bessel kernel and kernels arising from the almost Mathieu equation. The celebrated paper of Tracy and Widom \\cite{RMTTracy98correlationfunctions} investigated integral kernels of the form\n\\begin{equation*}\nK(x,y)=\\frac{f(x)g(y)-f(y)g(x)}{x-y} : x \\neq y f(x), g(x) \\in L^2(0,\\infty)\n\\end{equation*}\n are solutions to the system of $ODE$'s\n\n\\begin{equation*}\n\\frac{d}{dx}\\left( \\begin{array}{c}\n f(x) \\\\\n g(x)\n \\end{array}\n\\right) = \\left(\n \\begin{array}{cc}\n \\alpha(x) & \\beta(x) \\\\\n -\\gamma(x) & -\\alpha(x)\n \\end{array}\n\\right) * \\left( \\begin{array}{c}\n f(x) \\\\\n g(x)\n \\end{array} \\right)\n\\end{equation*}\n\nLet $\\phi_i(x)$ be an orthogonal basis in a Hilbert Space $\\mathcal{H}$ where\n\\begin{equation*}\n\\Gamma_{\\phi}\n= \\{\\phi_(j+k-1)\\}_{j,k=1}^{\\infty}\n\\end{equation*}\nis the induced Hankel Matrix.\n\n\nLet $\\mathcal(L) : \\mathcal{H} \\rightarrow \\mathcal{H}$ be compact. Then $\\mathcal(L): f \\mapsto \\sum\\limits_{n=1}^{N} \\omega_n <\\phi_n, f> \\psi_n$ where $\\{\\phi_i\\}_{i=1}^{N}$\n\n\n\\section*{The Tracy Widom Law}\nThe Tracy-Widom distribution is related to to determinantal stochastic processes. A process following this law is distributed as the largest point of a point process on the real line where the kernel K is the so-called Airy kernel. In addition to describing the edge spectrum of random matrices, it arises in several place in combinatorial for instance the longest increasing subsequences of random permutations is described by the Tracy Widom law. In addition to the eigenvalues of random matrices, this type of point process is used in models such as fluctuations in first and last passage percolation, and the asymmetric exclusion process.\n\nThe path configuration of random viscous walkers is related to the Young tableaux. Statistical problems related to the Young tableaux include random growth, point processes, random permutation, and the random word problem. the asymptotic distribution of scaled variables from these models are described by the Tracy Widom distribution which is the limit distribution for the largest eigenvalue of $X \\in GUE$.\n\n\n\\section*{Nystr\\\"{o}m Method}\nThe Nystrom method is a technique to speed up large-scale learning applications by generating low-rank approximations to large matrices. The origins of the Nystrom method lie in the numerical solution of the integral eigenvalue problem\n\\begin{equation*}\n \\int p(y) k(x,y) \\phi(y) = \\lambda \\phi(y)\n\\end{equation*}\n(BBCREVISIT talk about the functional analysis and physics origin of this method for solving integral problems)\n\nProblems in computer vision, natural language processing, computational biology and other areas can involve data sets containing more training examples than can be held in memory. Low rank approximations to the kernel matrices found in kernel based machine learning algorithms such as spectral clustering, manifold learning,support vector machines offer the opportunity to train on large data sets. The performance of this technique relies on the ability of a matrix to be approximated by a subset of its columns. Coherence bounds from the field of compressed sensing and matrix completion can be used to bound the error of the Nystrom approximation. \\cite{Talwalkar_matrixcoherence}\n\nThe Nystrom method can be evaluated by comparing the approximation to a low rank approximation generated from the first $n$ singular vectors from the $SVD$. If the process generating the data is $N(\\mu,\\Sigma)$ random sampling can be used and exact error bounds can be calculated based on the number of columns sampled. (BBCREVISIT confirm and cite).\n\nChoosing the columns based on the underlying probability density $p(y)$ has the potential to improve the quality of the approximation \\cite{Zhang09density-weightednystrom}.\n\n\\section*{Statistical Leverage}\nMichael Mahoney - videolectures.net Statistical Leverage Given an m x n matrix A and a rank parameter k, define the leverage of the i-th row of A to be the i-th diagonal element of the projection matrix onto the span of the top k left singular vectors of A. In this case, \"high leverage\" rows have a disproportionately large amount of the \"mass\" in the top singular vectors. Historically, this statistical concept (and generalizations of it) has found extensive applications in diagnostic regression analysis. Recently, this concept has also been central in the development of improved randomized algorithms for several fundamental matrix problems that have broad applications in machine learning and data analysis. Two examples of the use of statistical leverage for improved worst-case analysis of matrix algorithms will be described. The first problem is the least squares approximation problem, in which there are n constraints and d variables. Classical algorithms, dating back to Gauss and Legendre, use O(nd2) time. We describe a randomized algorithm that uses only O(n d log d) time to compute a relative-error, i.e., 1+/-epsilon, approximation. The second problem is the problem of selecting a \"good\" set of exactly k columns from an m x n matrix, and the algorithm of Gu and Eisenstat provides the best previously existing result. We describe a two-stage algorithm that improves on their result. Recent applications of statistical leverage ideas in modern large-scale machine learning and data analysis will also be briefly described. This concept has proven to be particularly fruitful in large data applications where modeling decisions regarding what computations to perform are made for computational reasons, as opposed to having any realistic hope that the statistical assumptions implicit in those computations are satisfied by the data.\n\n\\section*{How are differential operators related to semi group}\nSobolev inequalities relate norms of Sobolev spaces and can be used to prove embedding of $W^{k,p}(\\Real^n) \\in W^{l,q}(\\Real^n)$ Logarithmic Sobolev inequalities. The Rellich–Kondrachov theorem makes this precise for the $L^p$ spaces.\n$A$ is an operator on a $L^p$ Banach space. We define this in an operator sense as the sum given by the Taylor expansion of $e^x = \\sum \\frac{x^n}{n!}$\n\\begin{equation*}\n e^{-t A} : L^p \\rightarrow L^q\n\\end{equation*} is contractive if this holds for $p=2$ and $q=4$.\n$e^{-t A}$ is a positive preserving contractive semi-group if\n\\begin{eqnarray*}\n e^{-t A} f &\\geq& 0 \\\\\n \\parallel e^{-t A} f \\parallel_p &\\leq& \\parallel f \\parallel_p\n\\end{eqnarray*}\nWhen $e^{-t A}$ maps $L^2$ to $L^\\infty$ we call the semi group untracontractive.\n\nCovering number\n\nUnion bound\n\nKashin representation\n\nCandes and Tao extend Uncertainty Principle to discrete domain\nFor $x \\in \\Complex^n$\n\\begin{equation*}\n |supp(x)| |supp(\\hat{x}) \\geq N\n\\end{equation*}\nUniform uncertainty principle is related to the restricted isometry property.\n\nThe $\\Lambda$ problem of Talingrad relates Fourier matrix to HUP question. Consider matrices which are euclidian projection of cube $Q^N = {x : \\parallel x \\parallel_\\infty \\leq 1}$, a problem form geometric functional analysis. Let $B^n = {x : \\parallel x \\parallel_2 \\leq 1}$\n\nKashin; $\\exists$ orthogonal projection $P$\n\\begin{equation*}\n P : Q^N \\rightarrow U \\in \\Complex^N \\st U \\approx B^n\n\\end{equation*}\nThis implies\n\\begin{equation*}\n \\exists A \\in M^{n,N} \\st B^n \\in \\frac{K}{\\sqrt{N}} A Q^N \\in K B^n where K = K(\\lambda) \\lambda=\\frac{N}{n}\n\\end{equation*}\n$\\lambda$ is the redundancy. This statement is essentially telling us that we can project the cube into the ball when N and n allow for it.\n\n\\section*{Learning with integral operators}\nMachine learning methods such as Principal Components Analysis (PCA), Laplacian-based spectral clustering, and manifold methods rely on estimating eigenvalues and eigenvectors of the underlying data-dependent matrices. These matrices can be interpreted as empirical versions of underlying integral operators or closely related objects, such as continuous Laplacian operators. Establishing a connection between the finite dimensional empirical operator represented by the data matrix and the continuous counterpart can yield insight, help with convergence theorems, and even provide the basis for new machine learning methods.\n\nTheoretical analysis of many problems, such as low-rank matrix recovery, covariance estimation and approximate matrix multiplication, is built upon exponential bounds for $P(\\| \\sum X_i \\| > t)$ where ${X_i}$ is a finite sequence of self-adjoint random matrices and k · k is the operator norm. Moment-generating function techniques can be used to produce generalizations of Chernoff, Bernstein and Friedman inequalities to the non-commutative case.\n\nConcentration inequalities extend classical exponential inequalities for sums of independent random variables to functions of independent random variables. In machine learning theory we are interested in functions such as the supremum of the deviations between the true risk and the empirical risk, the empirical VC-dimension, the empirical VC entropy, and the eigenvalues of the Gram matrix.\n$E[ \\hat{\\theta} - \\theta_0 ]$ ", "meta": {"hexsha": "bde325f59377bf0c7a677a48bb3ae373de8f3c92", "size": 60462, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "BookContent.tex", "max_stars_repo_name": "brucebcampbell/machine-learning-notes", "max_stars_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "BookContent.tex", "max_issues_repo_name": "brucebcampbell/machine-learning-notes", "max_issues_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "BookContent.tex", "max_forks_repo_name": "brucebcampbell/machine-learning-notes", "max_forks_repo_head_hexsha": "6c5229ef7b943455a4e890f0ec62764adf9a2c40", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 73.4653705954, "max_line_length": 1854, "alphanum_fraction": 0.7376699415, "num_tokens": 17273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519527944504227, "lm_q2_score": 0.7371581510799252, "lm_q1q2_score": 0.6280239467644492}} {"text": "\n\n\n\\index{Lucas Critique}%\n\\section{The Lucas Critique}\\label{sec:lucascritique}%\n We can use our chapter \\use{timeseries} tools concisely to describe reasoning that led Robert E. Lucas, Jr., (1976) to criticize a class of procedures that were widely used in applied work at the\n time that he wrote, and that occasionally still are.\nAssume a jointly stationary Gaussian stochastic process for $\\{x_t, y_t, d_t\\}_{t=0}^\\infty$ generated by a state-space system\n$$ \\EQNalign{ x_{t+1} & = A x_t + C w_{t+1} \\EQN lucas101;a \\cr\n y_t & = G x_t \\EQN lucas101;b \\cr\n d_t & = S x_t \\EQN lucas101;c \\cr\n S & = H G (I - \\delta A)^{-1} \\EQN lucas101;d \\cr} $$\nwhere $\\delta \\in (0,1)$ is a discount factor and $A$ is a stable matrix\n that in applications itself often depends on deeper parameters,\n and $x_0 \\sim {\\cal N}(\\mu, \\Sigma) $, where $\\mu, \\Sigma$ are the mean vector and covariance matrix, respectively, of a stationary distribution for the\n $\\{x_t\\}_{t=0}^\\infty$ process. As usual, $w_{t+1}$ is the time $t+1$ component of an i.i.d.~sequence of ${\\cal N}(0,I)$ random vectors.\n Here $d_t$ is a vector of decisions that an econometrician observes. Equations \\Ep{lucas101;c} and \\Ep{lucas101;d} assert that $d_t$ equals a fixed matrix $H$ times a vector of\nforecasts of discounted sums of current and future $y_t$ vectors, forecasts that equal appropriate conditional expectations\n (please see equation \\Ep{discount1}). To fix ideas with an example, take a univariate autoregression for $\\{y_t\\}$, namely,\n$$ y_{t+1} = \\alpha + \\rho_1 y_t + \\rho_2 y_{t-2} + \\sigma w_{t+1} \\EQN lucas102 $$\n that we can represent with equation \\Ep{lucas101;a}-\\Ep{lucas101;b}. We can think of\n$d_t$ as a time $t$ {\\it decision\\/} and $d_t = S x_t$ as a {\\it decision rule\\/} that tells how an economic agent or a collection of economic agents sets\n$d_t$ as a function of time $t$ information $x_t$ about future $y_{t+j}$'s, in particular, so that\n$d_t = H E_t \\sum_{j=0}^\\infty \\delta^j y_{t+j}$ where $E_t$ denotes a mathematical expectation conditional on time $t$ information.\nLucas (1976) offered three examples that can be fit into this framework. In those examples the decisions $d_t$ were\n choices either of a consumption rate or an investment rate \nor a labor supply (i.e., Lucas's Phillips curve example), and $y_t$ was a stochastic process embedding \ngovernment polices \n(e.g., a tax rate or inflation process), forecasts\nof which influenced time $t$ decisions $d_t$.\\NFootnote{A variety of other examples that we'll\nencounter later in this book also fit. Also see Hansen and Sargent (1980, 1981). In those examples, counterparts of the parameter $\\delta$\nin equation \\Ep{lucas101;d} are themselves functions of deeper parameters measuring costs of adjustment and other features of\ntechnologies and preferences.}\n\\auth{Lucas, Robert E., Jr.}%\n\n\nNote that by virtue of stationarity, $\\bmatrix{d_t \\cr x_t}$ is a multivariate normal vector with partitioned covariance matrix\n$\\bmatrix{ \\Sigma_{dd} & S\\Sigma_{dx} \\cr \\Sigma_{xd} S' & \\Sigma_{xx}}$, where $\\Sigma_{xx}$ satisfies the discrete Lyapunov equation $\\Sigma_{xx}\n= A \\Sigma_{xx} A' + C C'$ and $\\Sigma_{dx} = S \\Sigma_{xx}$. Application of the least squares formula \\Ep{leastsq12} confirms that $S$ equals a matrix of population regression coefficients of $d_t$ on $x_t$\n and that the $R^2$ for each component of $d_t$ is $1$.\\NFootnote{Apply formula \\Ep{leastsq12} for the matrix of regression coefficients\n and notice that $\\beta = S \\Sigma_{xx} \\Sigma_{xx}^{-1} = S$\n and that the conditional covariance matrix $\\hat \\Sigma_{dd} = 0$.}\n\n %\n%\n%\n%under the assumption\n%that the researcher does not understand the origin of the decision rule $S$ via restriction \\Ep{lucas101;d}. Instead,\n%the researcher has a large set of observations on $\\{d_t, x_t\\}$ and estimates $S$ by least squares regression of $d_t$ on $x_t$.\n\nLucas (1976) criticized a collection of studies that he said had used the following faulty logic.\n\n\\medskip\n\n\\item{a.} A researcher just wants to know the decision rule for $d_t$ and ignores that \n $S$ satisfies restriction \\Ep{lucas101;d}.\n\n\\medskip\n\n\\item{b.} The researcher gathers historical observations on $\\{d_t, x_t\\}_{t=0}^T$,\n correctly believes that $d_t = S x_t$ is a\nregression equation, and therefore estimates the matrix $S$ of regression coefficients by applying least squares equation-by-equation.\n\n\\medskip\n\n\\item{c.} Mistakenly as it will turn out, the researcher assumes that the decision rule $d_t = S x_t$ that he has estimated\n from historical data will remain invariant under\n arbitrary hypothetical alterations of the right-hand side variables $\\{x_t\\}$, including ones that would\n deviate from the historical law of motion\n \\Ep{lucas101;a}-\\Ep{lucas101;b}. He falsely believes that he can use that estimated decision rule to predict how\n variations in $\\{x_t\\}$ that systematically depart from system \\Ep{lucas101;a}-\\Ep{lucas101;b} will induce\n systematic variations in \n $\\{d_t\\}$.\\NFootnote{Of course, he could use the estimated decision rule to predict the consequences of variations in $\\{x_t\\}$ that\n {\\it are \\/} described by system \\Ep{lucas101;a}-\\Ep{lucas101;b}.}\n\n\\medskip\n\n\\item{d.} Holding the historically estimated decision rule $S $ fixed,\n the researcher simulates outcomes $\\{y_t, d_t\\}_{t=T+1}^{\\tilde T}$\nunder alternative rules for generating\n time paths for\n $\\{x_t\\}_{t=T+1}^{\\tilde T}$ and then evaluates the\nresulting $\\{y_t, d_t, x_t\\}_{t=T+1}^{\\tilde T}$\noutcomes according to a social welfare function that depends on these outcome paths. %, that are functions of the joint sequence $\\{d_t, x_t\\}$.\n\n\n\\medskip\n\n\\item{e.} After\nlearning how the welfare function varies with the hypothetical rule for setting the $\\{x_t\\}_{t=T+1}^{\\tilde T}$ sequence,\nthe researcher recommends a continuation $\\{x_t\\}_{t=T+1}^{\\tilde T}$\nsequence that yields the best social welfare. For example, the researcher holds $S$ fix and uses it as a component\nof a state-space system to which he applies a linear-quadratic optimal control theory to be described in chapter \\use{dplinear} to recommend a\nlaw of motion for $x_t$ that manifests itself as a new pair of matrices $\\tilde A, \\tilde G$ that in the future\nwill replace $A, G$ in the state-space representation \\Ep{lucas101}, while he holds $S$ fixed at the value estimated from historical data.\n\n\\medskip\n\n\\item{f.} When the decision makers who choose $d_t$ recognize that under the new government policy $\\tilde A, \\tilde G$ and not $A, G$ will describe future $y_t$'s, they adjust the decision rule $S$ to become \n $\\tilde S = H \\tilde G (I - \\delta \\tilde A)^{-1}$. By failing to understand that,\nthe researcher and policy maker will be disappointed to find that the regression that the researcher estimated on the historical data fails to describe data generated\nunder the regime $\\tilde G, \\tilde A$.\n\n\\medskip\n\nLucas (1976) said that the line of reasoning in points (a)-(e) characterized an influential 1940s-1960s econometric policy evaluation\nresearch method that applied optimal control techniques to Keynesian and monetarist\neconometric models to propose improved macroeconomic policies. \nIn point (f) Lucas said that the method was flawed because it ignored\n the key cross-equation restrictions \\Ep{lucas101;d}. Altering the policy-generating\nmatrices from the $G, A$ pair that prevailed historically to new pair $\\tilde G, \\tilde A$ that describes a hypothetical policy ``experiment'' that we want to study\n amounts to a positing a ``regime change'' that\n should be expected systematically to alter how private decision makers make $y_t$ depend on $x_t$, as\n described by\n % that would alter the $A,G$ pair %and the associated $x_t$\n%of matrices that appears\n%in\n system \\Ep{lucas101}. % and that governs the $\\{y_t, x_t\\}$ process.\n A\n change in the decision maker's environment $G, A$ alters the moment matrices\n$ \\Sigma_{xx}, \\Sigma_{dx}$ and therefore the population regression matrix of $d_t$ on $x_t$ that should be \nexpected to characterize outcomes\nunder the new regime. Thus, structure \\Ep{lucas101}, and in particular the cross-equation restrictions \\Ep{lucas101;d},\nimply that $S$ will vary {\\it systematically\\/} with hypothetical changes\nin $\\tilde G, \\tilde A$, belying the assumption that $S$ is invariant to the hypothetical policy interventions being investigated.\nA concise way to say this is that the joint probability distribution of the $\\{d_t, x_t\\}$ process and the implied\n probability distribution\nof $d_t$ conditional on $x_t$ process both depend on the $(A, C, G)$ triple of matrices.\n\nLucas (1976) summarized his argument as follows:\n\n\\epigraph{$\\ldots$ given that the structure of an econometric model consists of optimal decision rules of economic agents, and\nthat optimal decision rules vary systematically with changes in the structure of series relevant to the decision makers, it follows\nthat any change in policies will systematically affect the structure of econometric models.}{``Econometric Policy Evaluation: A Critique,'' 1976}\n\n\n Equation \\Ep{lucas101;d} is an effective way of capturing how ``optimal decision rules vary systematically with changes in the structure of\n series relevant to the decision makers.'' We already encountered an example of such restrictions on optimal decision rules in\n section \\use{sec:LQmodel} when we found that the LQ permanent income model of consumption smoothing implies an optimal decision rule\n for consumption of the form \\Ep{eqn:lucascritiquecons}. We shall encounter more examples in chapter \\use{dplinear}.\n We can read Lucas (1976) as calling for a rational expectations econometrics that incorporates the kinds of cross-equation\nrestrictions that system \\Ep{lucas101} encodes as an essential part of a better theory of econometric policy \nevaluation.\\NFootnote{See Sargent (1981) and the introductory essay in Lucas and Sargent (1981), a volume containing\n early papers on rational expectations econometrics. See Hansen and Sargent (1991a) for\nfurther\ntechnical contributions.}\n\\index{rational expectations!econometrics}\n\\auth{Lucas, Robert E., Jr.}%\n\\auth{Hansen, Lars P.}%\n\\auth{Sargent, Thomas J.}%\n\n\n\\section{Responses to the Lucas Critique}\\label{sec:lucascritique2}%\nSection \\use{sec:lucascritique} already mentioned a principal response to the Lucas Critique, namely,\nrational expectations econometrics, an estimation and interpretation strategy now widely used throughout applied economics.\nIn this section, we use a state-space system with coefficients that are functions of a finite state Markov process \nas a vehicle to describe\nanother response\n%We describe what amount to either extensions to, or alternatives to, or elaborations of rational expectations econometrics\nthat takes the perspective that \nchanges in government decision rules or ``regimes'' \nunfold in ways that private agents anticipate. %don't occur in a vacuum but were\n%understood during the historical\n%sample period being modeled econometrically.\n%\\NFootnote{Section 3 of Lucas (1976) suggested that longstanding evidence that coefficients in\n%macroeconomic models drifted over time reflected the workings of cross-equation restrictions \n%like \\Ep{lucas101;d}.} \nSargent and Wallace (1976) argued that\n\n\n\\epigraph{$\\ldots$ new rules are not adopted\nin a vacuum. Something would cause the change -- a change in adminstrations,\nnew appointments, and so on. Moreover, if rational agents live in a world in\nwhich rules can be and are changed, their behavior should take into account such\npossibilities and should depend on the process generating the rule changes. But\ninvoking this kind of complete rationality seems to rule out normative economics\ncompletely by, in effect, ruling out freedom for the policymaker. For in a model\nwith completely rational expectations, including a rich enough description\nof policy, it seems impossible to define a sense in which there is any scope for\ndiscussing the optimal design of policy rules. That is because the equilibrium\nvalues of the endogenous variables already reflect, in the proper way, the\nparameters\ndescribing the authorities' prospective subsequent behavior, including\nthe probability that this or that proposal for reforming policy will be adopted.}{``Rational Expectations and the Theory of Economic Policy,''\nSargent and Wallace (1976), p.~181}\n\n\n\\auth{Wallace, Neil}%\n\\auth{Sargent, Thomas J.}%\n\nThis view sees government decision rules as statistical processes\n that are known to the private agents living inside a rational expectations model. \n So those private agents properly anticipate changes in government decision rules.\n Sargent and Wallace's view that government policy rules are\ndetermined by a well understood process leaves econometricians and policy analysts with\n no ``free parameters'' to change,\ndisabling them from being able to offer policy advice.\nThis perspective was embraced by\n Christopher A. Sims (1982) and Sargent (1984).\\NFootnote{Musto and Yilmaz (2003) extended the approach \n to a political \n economy with complete financial markets.}\n \\auth{Musto, David K.}%\n \\auth{Yilmaz, Bilge}%\n \n\\auth{Sims, Christopher A.}%\n\\auth{Zha, Tao}%\n\nIn describing vector autoregressions with regime shifting coefficients and volatilities, Sims and Zha (2006) link\ntheir work to the Lucas Critique: %\\example\n\\epigraph\n{The model with time variation in coefficients in all equations might be expected to fit best if there \nwere policy regime changes, and the nonlinear effects of these changes on private sector dynamics, via\n changes in private sector forecasting behavior, were important. That this is possible was the main point \n of Robert E. Lucas (1972).\n$\\ldots$ % as Sims (1987) has explained at more length,\nonce we recognize that changes in policy must in principle themselves be modeled as \nstochastic, Lucas's argument can be seen as a claim that a certain sort of nonlinearity is important. Even if the public believes that policy is time-varying and tries to adjust its expectation formation accordingly, its behavior could be well approximated as linear and non-time-varying. As with any use of a linear approximation, it is an empirical matter whether the linear approximation is adequate for a particular sample or counterfactual analysis.}{``Were There Regime Switches in U.S. Monetary Policy?'' Sims and Zha (2006), p.~59}\n%\\endexample\n\n\n\\noindent Here we'll use chapter \\use{timeseries} tools to study these parts of Sims and Zha's analysis:\n\n\\medskip\n\\item{a.} How to represent ``nonlinearities'' in the form of dependencies of linear state-space model \ncoefficients that are functions of a discrete Markov state variable that\nstands in for a ``regime''.\n\n\\medskip\n\\item{b.} How agents who understand regime transition probabilities and who recognize the current regime shape their decisions.\n\n\\medskip\n\\item{c.} How a best-fitting fixed-coefficient linear model that ignores regime changes can approximate observed \noutcomes.\n\n\n\\medskip\n\n\\subsection{Markov switching linear state space model}\n\nWe can represent the regime-change-induced \nnonlinearities mentioned by Sims and Zha policy with a linear state space model with\n coefficients that are governed by a finite state Markov chain.\\NFootnote{Hamilton (1989) introduces this \n type of model of a stochastic process and labels it as ``non-stationary''. If we assume that \nchain $\\pi_0$ is a stationary distribution, then the model actually generates a stationary stochastic process,\n a property that we exploit below. Williams and Svensson (2008) and Svensson and Williams (2009) used linear quadratic dynamic programming problems with Markov-switching coefficients to model optimal\nmonetary policies. }\nLet $s_t \\in \\{1,2\\}$ be governed by a Markov chain with transition matrix $P$, where\n$P_{ij} = {\\rm Prob}(s_{t+1} = j| s_t = i) $, $\\pi_{0,i} = {\\rm Prob}(s_0 = i)$, and $\\pi_{t,i} = {\\rm Prob}(s_t = i)$.\nWe assume that $\\{x_{t}, y_t\\}_{t=0}^\\infty$ is governed by the linear state-space system with \n Markov-switching matrices \n$$ \\eqalign{ x_{t+1} & = A_{s_t} x_{t+1} + C_{s_t} w_{t+1} \\cr\n y_t & = G_{s_t} x_t \\cr } \\EQN MSLQ $$ %\nwhere $x_0 $ is a random initial vector drawn from density $\\phi_0$ and $w_{t+1} \\sim {\\cal N}(0,I)$ is an i.i.d.~sequence of random\nvectors.\nWe want to compute the conditional mathematical expectations\n $$ z_t = E ( \\sum_{j=0}^\\infty \\delta^t y_{t+j} ) \\Bigl| x_t, s_t $$\n and\n $$ \\tilde z_t = E ( \\sum_{j=0}^\\infty \\delta^t y_{t+j} ) \\Bigl| x_t = E z_t | x_t , $$\n where the last equality follows by an application of the \\idx{law of iterated expectations}.\n We also want to compute an associated time-invariant vector autoregression that is implied by \\Ep{MSLQ} and that takes the form\n $$ x_{t+1} = \\bar A x_t + v_{t+1} \\EQN VARtimeinv$$\n where $v_{t+1} $ is a serially uncorrelated shock with mean zero and contemporaneous covariance matrix $E v_{t+1} v_{t+1}' = \\bar \\Sigma$.\nThis is the object that Sims and Zha mention when they say of the time-varying system that ``its behavior could be well approximated as linear and non-time-varying''.\n\\auth{Hamilton, James D.}%\n\\auth{Williams, Noah}%\n\\auth{Svensson, Lars E. O.}%\n\\auth{Sims, Christopher A.}%\n\\auth{Zha, Tao}%\n\n\n\n\\subsection{Example}\nBefore computing these objects, let's take an example fashioned after a rational expectations version of Philip Cagan's (1956) model of\nhyperinflation. The rate of inflation is the first difference of the logarithm of the price level\nand the rate of growth of the money supply is the first difference of the logarithm of the money supply.\nLet $p_t$ be the rate of inflation and let $m_t$ be the rate of growth of the money supply. Assume\nthat\n$$ p_t = (1-\\lambda) m_t + \\lambda E_t p_{t+1}, \\quad \\lambda \\in (0,1) $$\nso that\n$$ p_t = (1-\\lambda) E_t \\sum_{j=0}^\\infty \\lambda^j m_{t+j} , $$\nwhere $E_t (\\cdot)$ is a mathematical expectation conditioned on time $t$ information, which will be either $x_t$ or $(x_t, s_t)$, where the pair $(x_t, s_t)$ is defined as follows.\nWe assume that $s_t \\in {1,2}$ is governed by a two-state Markov chain $P$ with initial distribution $\\pi_0$.\nSetting $x_t= \\bmatrix{1 & m_t & m_{t-1} }'$ lets us capture the following stochastic process for $\\{m_t\\}_{t=0}^\\infty$ with the state-space form \\Ep{MSLQ }:\n$$ m_{t+1} = \\rho_{1,s_t} m_t + \\rho_{2,s_t} m_{t-1} + c_{s_t}w_{t+1}, \\EQN second_order_AR $$\nwhere $w_{t+1}$ is an i.i.d.~univariate standardized normal random variable. In subsection \\use{sec:cond_dist_102},\nwe'll use this example\nas a vehicle for illustrating Sims and Zha's remark about how a linear time-invariant autoregression can approximate\na Markov-switching autoregression like \\Ep{second_order_AR}.\n\\auth{Cagan, Phillip}%\n\n\\subsection{Conditional distributions of geometric sums}\nWe begin by computing $z_t$, an expectation conditional on both $x_t$ and $s_t$.\nWe begin by guessing that\n$$ z_t = H_{s_t} x_t $$\nand substitute this guess into\n$$ z_t = y_t + \\delta E [ z_{t+1} | x_t, s_t ] $$\nto get\n$$ H_i = G_{i} + \\delta \\sum_{j} H_j A_i P_{ij}, $$\nwhich implies that\n$$ \\eqalign{ H_1 & = G_1 + \\delta [ H_1 A_1 P_{11} + H_2 A_1 P_{12} ] \\cr\n H_2 & = G_2 + \\delta [ H_1 A_2 P_{21} + H_2 A_1 P_{22} ]. }$$\nStacking these equations and solving for the matrices $H_1, H_2$, we compute\n$$ \\bmatrix{H_1 & H_2 } = \\bmatrix{G_1 & G_2} (I - \\delta \\bmatrix {A_1 P_{11} & A_2 P_{21} \\cr\n A_1 P_{12} & A_2 P_{22} } )^{-1}. $$\n\n\n\nNext we compute $ \\tilde z_t = E ( \\sum_{j=0}^\\infty \\delta^t y_{t+j} ) | x_t$.\n%We let $\\pi_t$ be the distribution over the finite Markov state $s_t$. Note that the\n We start knowing that the probability distribution of $x_{t+1}$ conditional on $(x_t, s_t)$ is multivariate normal:\n$$ x_{t+1} | x_t, s_t \\sim {\\cal N}( A_{s_t} x_t, C_{s_t}C_{s_t} ' ) . $$\nDistributions of $x_{t+1}, y_t$, and $z_t$ conditional on $x_t$ are {\\it mixtures of normals}, with $\\pi_{t,i}$ being the mixing probabilities over states $i$ at time $t$.\nDefining the matrix averages\n$$ \\eqalign{ \\bar A_t & = \\pi_{t,1} A_1 + \\pi_{t,2} A_2 \\cr\n \\bar G_t & =\\pi_{t,1} G_1 + \\pi_{t,2} G_2 \\cr\n \\bar H_t & = \\pi_{t,1} H_1 + \\pi_{t,2} H_2 ,}$$\nwe can deduce that means conditional on $x_t$ are \n$$ \\eqalign{ E x_{t+1} | x_t & = \\bar A_t x_t \\cr\n E y_t | x_t & = \\bar G_t x_t \\cr\n E z_t | x_t & \\equiv \\tilde z_t = \\bar H_t x_t . }$$\nThe appearance of time subscripts $t$ in the formulas for $\\bar A_t, \\bar G_t, \\bar H_t$ are a \nconsequence of $\\pi_t$ not necessarily\nbeing a stationary distribution for $s_t$. % for Markov chain $P$.\n\n\\subsection{More conditional distributions}\\label{sec:cond_dist_102}%\nHaving computed conditional means, let's compute covariances conditional on $x_t$.\nSince\n$$ x_{t+1} - \\bar A_t x_t = \\cases{ (A_1 - \\bar A_t) x_t + C_1 w_{t+1} & if $s_t =1$ \\cr\n (A_2 - \\bar A_t) x_t + C_2 w_{t+1} & if $s_t =2$ , } $$\nit follows that\n$$ \\EQNalign{ E (x_{t+1} - \\bar A_t x_t ) (x_{t+1} - \\bar A _tx_t ) ' | x_t & = \\sum_i \\pi_{t,i} [ (A_i - \\bar A_t) x_t x_t' (A_i - \\bar A_t)' + C_i C_i' ] \\cr\n & \\equiv \\Sigma_t(x_t) \\EQN Sigmatxt } $$\nwhere $i$ is present through $\\pi_{t,i}$ and $x_t$ is present because it appears on the right side of the first line.\nSince\n$$ y_t = \\cases{ G_1 x_t & if $s_t =1$ \\cr\n G_2 x_t & if $s_t =2 $ } $$\nand\n$$ y_t - \\bar G_t x_t = \\cases{ (G_1 - \\bar G_t) x_t & if $s_t =1$ \\cr\n (G_2 - \\bar G_t) x_t & if $s_t =2 $ , } $$\nit follows that\n$$ E (y_t - \\bar G_t x_t) (y_t - \\bar G_t x_t)' | x_t = \\sum_i \\pi_{t,i} \n(G_i - \\bar G_t) x_t x_t' (G_i - \\bar G_t)' \\equiv \\Sigma_{y,t}(x_t) $$\nand similarly that\n$$ E (z_t - \\bar H_t x_t) (z_t - \\bar H_t x_t)' | x_t = \\sum_i \\pi_{t,i} (H_i - \\bar H_t) x_t x_t'\n (H_i - \\bar H_t)' \\equiv \\Sigma_{z,t}(x_t), $$\n where the presence of $t$ subscripts in $ \\Sigma_{y,t}(x_t)$ and $\\Sigma_{z,t}(x_t)$ reflects that\n $\\pi_t$ might not be a stationary distribution, so that it and $\\bar G_t$ and $\\bar H_t$ all depend on $t$.\nWe can compute unconditional means and covariances of the $\\{x_t\\}_{t=0}^\\infty$ process as follows.\nDefine the unconditional moments\n$E x_t = \\mu_t$ and $\\Sigma_t = E (x_t - \\mu_t) (x_t - \\mu_t)'$.\nNote that\n$$ x_{t+1} - \\mu_{t+1} = \\cases{ A_1 (x_t - \\mu_t) + C_1 w_{t+1} & if $s_t =1$ \\cr\n A_2 (x_t - \\mu_t) + C_2 w_{t+1} & if $s_t =2$ .} $$\nIt follows that\n$$ \\mu_{t+1} = \\bar A_t \\mu_t$$\nand\n$$\\Sigma_{t+1} = \\pi_{t,1} (A_1 - \\bar A_t) \\Sigma_t (A_1 - \\bar A_t)' + \\pi_{t,2} (A_2 - \\bar A_t) \\Sigma_t (A_2 - \\bar A_t)' + (\\bar C \\bar C')_t $$\nwhere $(\\bar C \\bar C')_t = \\sum_i \\pi_{t,i} C_i C_i' $.\nWhen $\\pi_t \\rightarrow \\bar \\pi$ as $t$ goes to infinity, iterations on these two recursions converge, respectively, to\na stationary mean vector $\\mu$ and a stationary covariance matrix $\\Sigma$ that\nsatisfy\n$$ \\mu = \\bar A \\mu $$\nand\n$$\\bar \\Sigma = \\bar \\pi_1 (A_1 - \\bar A) \\bar \\Sigma (A_1 - \\bar A)' + \\bar \\pi_2 (A_2 - \\bar A) \\bar \\Sigma(A_2 - \\bar A)' + \\bar C \\bar C' $$\nwhere $\\bar \\pi_i = \\lim_{t \\rightarrow +\\infty} \\pi_{t,i}$ and $\\bar C \\bar C' = \\sum_i \\bar \\pi_i C_i C_i'$.\nA time-invariant vector autoregression of the form \\Ep{VARtimeinv}, namely,\n$$ x_{t+1} = \\bar A x_t + v_{t+1} .$$\nhas $E v_{t+1} v_{t+1}' = \\bar \\Sigma$.\nWhen $\\pi_t = \\bar \\pi_t$, formulas for $\\Sigma_y(x_t)$ and $\\Sigma_z(x_t)$ also simplify: \n$$ E (y_t - \\bar G x_t) (y_t - \\bar G x_t)' | x_t = \\sum_i \\pi_{i} \n(G_i - \\bar G) x_t x_t' (G_i - \\bar G)' \\equiv \\Sigma_{y}(x_t) $$\nand \n$$ E (z_t - \\bar H x_t) (z_t - \\bar H x_t)' | x_t = \\sum_i \\pi_{i} (H_i - \\bar H) x_t x_t'\n (H_i - \\bar H)' \\equiv \\Sigma_{z}(x_t), $$\n where $t$ subscripts have disappeared from $ \\Sigma_{t}(x_t)$ and $\\Sigma_{z}(x_t)$ because \n $\\pi = \\bar \\pi$ and also now $\\bar G$ and $\\bar H$ no longer depend on $t$.\n Under this stationary specification, we thus have the regression equations\n $$\\eqalign{ x_{t+1} & = \\bar A x_t + v_{t+1} \\cr\n y_t & = \\bar G x_t + \\epsilon_{y,t} \\cr \n z_t & = \\bar H x_t + \\epsilon_{z,t}, }\n$$\nwhere $\\bar H = \\bar G (I - \\delta \\bar A)^{-1}$, $E v_{t+1} v_{t+1}' = \\bar \\Sigma$,\n $E \\epsilon_{y,t} \\epsilon_{y,t}' = \\Sigma_y(x_t), \\ E \\epsilon_{z,t} \\epsilon_{z,t}' = \\Sigma_z(x_t) $.\n\n\nWe can use some of the formulas above and our example in equation \\Ep{second_order_AR} to revisit Sims and Zha's (2006)\ncomments about how a linear time invariant autoregression can approximate a Markov switching autoregression.\n We set $(\\rho_{1,1}, \\rho_{2,1}, c_{1}) =\n(1.2, -.3, .5)$ and $(\\rho_{1,2}, \\rho_{2,2}, c_{2}) =\n(.8, 0.0, .8)$, $P = \\bmatrix{.5 & .5 \\cr .5 & .5}$, and $\\pi_0 = \\bmatrix{.5 \\cr .5} $, so that the one step-ahead \n conditional variance in state $s_t=1$ is $.25$ and in state $s_t=2$ is $.64$. Applying the preceding formulas, we\ncan deduce the time-invariant second order autoregression\n$$ m_{t+1} = 1.0 m_t - .15 m_{t-1} + v_{t+1} $$\nwhere $E v_{t+1}^2 = \\bar \\Sigma = .46$, which is to be compared with the conditional variances $.25$ in state $s_t=1$ and\n$.64$ in state $s_t = 2$. \nFor this same example, it is instructive to set the discount factor $\\delta =.95$ and set $G_i =\n\\bmatrix{1 & 0 \\cr}'$ for $i = 1, 2$ so that $G_i$ does {\\it \\/} not depend on the Markov state $s_t$. \nIn this case, we obtain \n$$ \\bar H = \\bar G (I - \\delta \\bar A)^{-1} = XXXX\n $$\n\n% \\medskip\n% \\noindent{\\bf Request to Quentin:} Could you please add to your notebook code for computing the quantities\n% that required to compute $\\bar H$. The preceding can be computed \n% \"analytically\" using the above formulas, most of which you already have in the Jupyter notebook.\n% Another things that could be \"fun\" to compute would be the {\\it average\\/} value of $\\Sigma_z(x_t)$.\n% I would recommend doing this just by running a very very long simulation and taking a time series average \n% of $(H_i - \\bar H) x_t x_t'\n% (H_i - \\bar H)'$ as $s_t, x_t$ wander through time and states. \n\n\n\n\n\\auth{Sims, Christopher A.}%\n\\auth{Zha, Tao}%\n\n\n\n% \\noindent{\\bf Quentin R0:} Please go over the code in your notebook and reorganize to have all population computations at the front\n% of the notebook. By ``population'' I mean analytic with our formulas, not the simulations with the least squares regressions. Please put that material at the end of the notebook. Please make the notebook so that Zejin can understand the flow and approve of its readability.\n\n% \\noindent{\\bf Quentin R1:} Please compute and display conveniently for me the objects $C_1 C_1', C_2 C_2', \\bar \\Sigma, \\bar A$.\n\n\n% \\noindent{\\bf Quentin R2:} Please compute ``all of our objects'' for the above example with two $s_t$ dependent $\\rho_1, \\rho_2$ vectors, namely $1.2, -.3$ and $.95, 0$ with\n% a transition matrix $P = \\bmatrix{ .5 & .5 \\cr .5 & .5 } $ using as $\\pi_0 =\\pi_\\infty$ the stationary distribution $\\bmatrix{.5 & .5}'$\n% or with another set of parameter values that I can input easily. \n\n\n% \\medskip\n% \\noindent{\\bf Quentin R3:} Please check whether $\\bar H = \\bar G (I -\\delta \\bar A)^{-1}$ and display outcomes so that Zejin and I can readily\n% spot them.\n\n\n\\subsection{Time series averages}\nConsider the conditional covariance matrix\n$$\\Sigma(x_t) = \\sum_i \\pi_{t,i} [ (A_i - \\bar A) x_t x_t' (A_i - \\bar A)' + C_i C_i' ] .$$\nAssume that the pair $(P, \\pi_0)$ governing the Markov state $s_t$ is stationary. Then the joint process $\\{x_t, s_t\\}$ process is also stationary,\nand so is the process $\\{x_t\\}$. Say that $x_t$ has the unconditional probability distribution $\\Phi(x)$.\nThen a law of large numbers tells us that as $ {T \\rightarrow + \\infty}$\n$$ T^{-1} \\sum_{t=0}^T \\Sigma(x_t) \\rightarrow \\int \\Sigma(x) d \\Phi(x) \\equiv \\Sigma $$\n\n\n", "meta": {"hexsha": "e1b8d6fca766d11bbe9a087b6b842f129f11b7d4", "size": 28228, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/Lucas_Sims_notes.tex", "max_stars_repo_name": "QuantEcon/python-lecture-sandpit.myst", "max_stars_repo_head_hexsha": "8e6c8cb971ab06af12e5364a4a2cb9d0d07ab925", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lectures/Lucas_Sims_notes.tex", "max_issues_repo_name": "QuantEcon/python-lecture-sandpit.myst", "max_issues_repo_head_hexsha": "8e6c8cb971ab06af12e5364a4a2cb9d0d07ab925", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-12-14T07:16:37.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-05T00:06:51.000Z", "max_forks_repo_path": "lectures/Lucas_Sims_notes.tex", "max_forks_repo_name": "QuantEcon/python-lecture-sandpit.myst", "max_forks_repo_head_hexsha": "8e6c8cb971ab06af12e5364a4a2cb9d0d07ab925", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.5765765766, "max_line_length": 540, "alphanum_fraction": 0.7030253649, "num_tokens": 8521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8104789178257654, "lm_q2_score": 0.7745833841649232, "lm_q1q2_score": 0.6277835029638061}} {"text": "\\section{Proofs for Chapter 4}\n\\label{sec:a13chapter4}\n\n\\disableornamentsfornextheadingtrue\n\\subsection{Combinatorial Proof of the Combination Technique}\n\\label{sec:a131proofCombiTechnique}\n\n\\addtocontents{lop}{\\protect\\fi}\n\n\\begin{definition}[binomial coefficient for integer parameters]\n \\label{def:binomialCoefficient}\n The binomial coefficient $\\binom{n}{k}$ is defined for\n $n \\in \\natz$ and $k \\in \\integer$ as\n \\begin{equation}\n \\binom{n}{k}\n \\ceq\n \\begin{cases}\n \\frac{n (n - 1) \\dotsm (n - (k-1))}{k!},&0 < k < n,\\\\\n 1,&(k = 0) \\lor (k = n),\\\\\n 0,&(k < 0) \\lor (k > n).\n \\end{cases}\n \\end{equation}\n\\end{definition}\n\n\\begin{lemma}[inclusion-exclusion counting lemma]\n \\label{lemma:inclusionExclusionCountingLemma}\n For $a \\in \\natz$, $r \\ge a$, and $s \\in \\integer$, we have\n \\begin{equation}\n \\sum_{q=0}^a (-1)^q \\binom{a}{q} \\binom{r-q}{s}\n = \\binom{r-a}{s-a}.\n \\end{equation}\n\\end{lemma}\n\n\\begin{proof}\n We apply the upper negation formula\n (see Eq. (5.14) of \\cite{Graham94Concrete})\n to the second binomial of the \\lhs:\n \\begin{subequations}\n \\begin{align}\n \\sum_{q=0}^a (-1)^q \\binom{a}{q} \\binom{r-q}{s}\n &= (-1)^s \\sum_{q=0}^a \\binom{a}{0+q} \\binom{(s-r-1)+q}{s} (-1)^q.\\\\\n \\intertext{%\n This sum can be simplified using the identity\n in Eq.\\ (5.24) of \\cite{Graham94Concrete}\n (the sum has already been written in the same way as in\n \\cite{Graham94Concrete}):%\n }\n \\cdots\n &= (-1)^{s+a} \\binom{s-r-1}{s-a}.\\\\\n \\intertext{%\n Applying the upper negation formula again,%\n }\n \\cdots\n &= \\binom{r-a}{s-a},\n \\end{align}\n \\end{subequations}\n we obtain the desired quantity.\n\\end{proof}\n\n\\addtocontents{lop}{\\protect\\iffalse}\n\n\\propCombiTechniqueOne*\n\n\\begin{proof}\n Let $q = 0, \\dotsc, d - 1$ and $\\gp{\\*l,\\*i} \\in \\regsgset{n}{d}$, i.e.,\n $\\normone{\\*l} \\le n$ and $\\*i \\in \\hiset{\\*l}$.\n Note that for $\\*l' \\in \\natz^d$, we have\n $\\fgset{\\*l'} \\ni \\gp{\\*l,\\*i} \\iff \\*l' \\ge \\*l$.\n Hence,\n \\begin{subequations}\n \\begin{align}\n \\setsize{\n \\{\\*l' \\mid \\normone{\\*l'} = n - q,\\; \\fgset{\\*l'} \\ni \\gp{\\*l,\\*i}\\}\n }\n &= \\setsize{\n \\{\\*l' \\mid \\normone{\\*l'} = n - q,\\; \\*l' \\ge \\*l\\}\n }\\\\\n &= \\setsize{\n \\{\\*a \\in \\natz^d \\mid \\normone{\\*a} = n - q - \\normone{\\*l}\\}\n }\\\\\n \\intertext{%\n by mapping $\\*a \\ceq \\*l' - \\*l$.\n The size of the last set is\n known as the number of \\term{weak compositions}\n of $n - q - \\normone{\\*l}$ into $d$ parts\n and can be computed as%\n }\n \\cdots\n &= \\binom{n - q - \\normone{\\*l} + d - 1}{d - 1},\n \\end{align}\n \\end{subequations}\n see Theorem 2.2 of \\cite{Bona15Introduction}.\n Now, we can use \\cref{lemma:inclusionExclusionCountingLemma}\n with the values\n $a \\ceq s \\ceq d - 1$ and\n $r \\ceq n - \\normone{\\*l} + d - 1$\n to conclude that the \\lhs of the assertion\n \\eqref{eq:combiTechniqueOne} equals\n \\begin{equation}\n \\sum_{q=0}^{d-1} (-1)^q \\binom{d-1}{q} \\cdot\n \\binom{n - q - \\normone{\\*l} + d - 1}{d - 1}\n = \\binom{n - \\normone{\\*l}}{0}\n = 1,\n \\end{equation}\n proving the proposition.\n\\end{proof}\n\n\\addtocontents{lop}{\\protect\\fi}\n\n\\begin{shortlemma}[relation is equivalence relation]\n \\label{lemma:combiTechniqueEquivalenceRelation}\n $\\eq$ is an equivalence relation.\n\\end{shortlemma}\n\n\\begin{proof}\n We check reflexivity, symmetry, and transitivity of $\\eq$:\n \\begin{itemize}\n \\item\n \\emph{Reflexivity:}\n Using the same level $\\*l' = \\*l''$ implies\n $T_{\\*l',\\*l'} = \\{t \\mid l'_t < l_t\\}$.\n For all $t \\notin T_{\\*l',\\*l'}$, we have $l'_t \\ge l_t$.\n Consequently, $\\*l' \\eq \\*l'$.\n \n \\item\n \\emph{Symmetry:}\n We have\n $\\*l' \\eq \\*l'' \\iff \\*l'' \\eq \\*l'$, since\n $T_{\\*l',\\*l''} = T_{\\*l'',\\*l'}$ and\n $\\min\\{l'_t, l''_t\\} = \\min\\{l''_t, l'_t\\}$.\n \n \\item\n \\emph{Transitivity:}\n Let $\\*l' \\eq \\hat{\\*l}$, $\\hat{\\*l} \\eq \\*l''$, and\n $t \\notin T_{\\*l',\\*l''}$.\n From the definition of $T_{\\*l',\\*l''}$,\n it holds that either $l'_t \\not= l''_t$ or $l'_t = l''_t \\ge l_t$.\n As $l'_t = l''_t \\ge l_t$ already implies $\\min\\{l'_t, l''_t\\} \\ge l_t$,\n we assume that $l'_t \\not= l''_t$.\n Here, we have three cases:\n \\begin{itemize}\n \\item\n \\mbox{\\emph{Case 1:} ${l'_t \\not= \\hat{l}_t = l''_t}$.}\n $t \\notin T_{\\*l',\\hat{\\*l}}$ implies $l'_t \\ge l_t$ and\n $l''_t = \\hat{l}_t \\ge l_t$.\n Therefore, $\\min\\{l'_t, l''_t\\} \\ge l_t$.\n \n \\item\n \\mbox{\\emph{Case 2:} ${l'_t = \\hat{l}_t \\not= l''_t}$.}\n Analogously to the first case, we conclude $\\min\\{l'_t, l''_t\\} \\ge l_t$.\n \n \\item\n \\mbox{\\emph{Case 3:} ${l'_t \\not= \\hat{l}_t \\not= l''_t}$.}\n $t \\notin T_{\\*l',\\hat{\\*l}}$ implies $l'_t \\ge l_t$ and\n $t \\notin T_{\\*l'',\\hat{\\*l}}$ implies $l''_t \\ge l_t$.\n Hence, $\\min\\{l'_t, l''_t\\} \\ge l_t$.\n \\end{itemize}\n Therefore, it holds that $\\min\\{l'_t, l''_t\\} \\ge l_t$\n for all $t \\notin T_{\\*l',\\*l''}$, i.e., $\\*l' \\eq \\*l''$.\n \\end{itemize}\n This shows that $\\eq$ is an equivalence relation.\n\\end{proof}\n\n\\addtocontents{lop}{\\protect\\iffalse}\n\n\\lemmaCombiTechniqueIdenticalValues*\n\n\\begin{proof}\n First, we note that $T_{\\*l',\\*l''} \\not= \\emptyset$.\n Otherwise, for $T_{\\*l',\\*l''} = \\emptyset$,\n we have $\\min\\{l'_t, l''_t\\} \\ge l_t$\n for all $t = 1, \\dotsc, d$, which implies $\\*l' \\ge \\*l$, i.e.,\n $\\fgset{\\*l'} \\ni \\gp{\\*l,\\*i}$.\n This contradicts the fact that $\\*l' \\in L$, where $L$ is defined\n in \\eqref{eq:combiTechniqueSpecialLevelSet}\n (which holds as our equivalence relation is only defined on $L$).\n Therefore, $T_{\\*l',\\*l''} \\not= \\emptyset$ must hold.\n Without loss of generality,\n we assume that $T_{\\*l',\\*l''} = \\{1, \\dotsc, m\\}$\n for some $m \\in \\{1, \\dotsc, d\\}$.\n \n Let\n \\begin{equation}\n S \\ceq \\gp{\\*l,\\*i} + \\spn\\{\\stdbasis{1}, \\dotsc, \\stdbasis{m}\\}\n = \\{\\gp{\\*l,\\*i} + \\textstyle\\sum_{t=1}^m c_t \\stdbasis{t} \\mid\n c_1, \\dotsc, c_m \\in \\real\\}\n \\end{equation}\n be the $m$-dimensional affine subspace of $\\real^d$\n through $\\gp{\\*l, \\*i}$\n parallel to the dimensions $1, \\dotsc, m$,\n where $\\stdbasis{t}$ is the $t$-th standard basis vector.\n It holds that $S \\cap \\fgset{\\*l'} = S \\cap \\fgset{\\*l''}$ due to\n $l'_t = l''_t$ for $t \\le m$.%\n \\footnote{%\n In more detail:\n If we have an $\\gp{\\hat{\\*l},\\hat{\\*i}} \\in S \\cap \\fgset{\\*l'}$,\n then $\\fa{t \\le m}{\\hat{l}_t \\le l'_t = l''_t}$ and\n $\\fa{t > m}{\\hat{l}_t = l_t \\le l''_t}$, i.e.,\n $\\hat{\\*l} \\le \\*l''$ and therefore\n $\\gp{\\hat{\\*l},\\hat{\\*i}} \\in S \\cap \\fgset{\\*l''}$.%\n }\n \n On this $m$-dimensional grid $S \\cap \\fgset{\\*l'} = S \\cap \\fgset{\\*l''}$,\n the full grid interpolants $\\fgintp{\\*l'}$ and $\\fgintp{\\*l''}$\n coincide, as both interpolate the function values given by\n the objective function $\\objfun$:\n \\begin{equation}\n \\label{eq:proofCombiTechniqueIdenticalValues1}\n \\restrictfcn{\\fgintp{\\*l'}}{S \\cap \\fgset{\\*l'}}\n = \\restrictfcn{\\objfun}{S \\cap \\fgset{\\*l'}}\n = \\restrictfcn{\\fgintp{\\*l''}}{S \\cap \\fgset{\\*l'}}.\n \\end{equation}\n However, this does not suffice to conclude\n $\\fgintp{\\*l'}(\\gp{\\*l,\\*i}) = \\fgintp{\\*l''}(\\gp{\\*l,\\*i})$,\n since $\\gp{\\*l,\\*i} \\notin \\fgset{\\*l'}$.\n \n To this end, we recall from \\eqref{eq:interpFullGridMV} that\n \\begin{equation}\n \\fgintp{\\*l'}\n = \\sum_{\\*i'=\\*0}^{\\*2^{\\*l'}} \\interpcoeff{\\*l',\\*i'}\n \\basis{\\*l',\\*i'},\\quad\n \\interpcoeff{\\*l',\\*i'} \\in \\real.\n \\end{equation}\n This implies that the $m$-variate restricted interpolant\n $\\restrictfcn{\\fgintp{\\*l'}}{S \\cap \\clint{\\*0, \\*1}}$ can be written as\n \\begin{subequations}\n \\begin{align}\n (\\restrictfcn{\\fgintp{\\*l'}}{S \\cap \\clint{\\*0, \\*1}})\n (\\*x_{\\range{1}{m}})\n &\\mathrel{\\righthphantom{=}{\\ceq}}\n \\sum_{\\*i'_{\\range{1}{m}}=\\*0}^{\\*2^{\\*l'_{\\range{1}{m}}}}\n \\interpcoefftilde{\\*l'_{\\range{1}{m}},\\*i'_{\\range{1}{m}}}\n \\basis{\\*l'_{\\range{1}{m}},\\*i'_{\\range{1}{m}}}(\\*x_{\\range{1}{m}}),\\quad\n \\*x_{\\range{1}{m}} \\in \\clint{0, 1}^m,\\\\\n \\interpcoefftilde{\\*l'_{\\range{1}{m}},\\*i'_{\\range{1}{m}}}\n &\\ceq \\sum_{\\*i'_{\\range{m+1}{d}}=\\*0}^{\\*2^{\\*l'_{\\range{m+1}{d}}}}\n \\interpcoeff{\\*l',\\*i'}\n \\basis{\\*l'_{\\range{m+1}{d}},\\*i'_{\\range{m+1}{d}}}%\n (\\gp{\\*l_{\\range{m+1}{d}},\\*i_{\\range{m+1}{d}}})\n \\end{align}\n \\end{subequations}\n by factoring out tensor product factors corresponding to dimensions\n $m + 1, \\dotsc, d$.\n The subscripts $\\range{1}{m}$ and $\\range{m+1}{d}$ denote\n the entries with respect to the dimensions $1, \\dotsc, m$ and\n $m + 1, \\dotsc, d$, respectively.\n As a result,\n both $\\restrictfcn{\\fgintp{\\*l'}}{S \\cap \\clint{\\*0, \\*1}}$\n and, analogously,\n $\\restrictfcn{\\fgintp{\\*l''}}{S \\cap \\clint{\\*0, \\*1}}$\n are interpolants of $\\objfun$ in\n $\\ns{\\*l'_{\\range{1}{m}}} = \\ns{\\*l''_{\\range{1}{m}}}$.\n Due to \\thmref{lemma:tensorProductLinearIndependence},\n it follows from \\eqref{eq:proofCombiTechniqueIdenticalValues1}\n that they must be the same:\n \\begin{equation}\n \\restrictfcn{\\fgintp{\\*l'}}{S \\cap \\clint{\\*0, \\*1}}\n = \\restrictfcn{\\fgintp{\\*l''}}{S \\cap \\clint{\\*0, \\*1}}.\n \\end{equation}\n Consequently, $\\fgintp{\\*l'}(\\gp{\\*l,\\*i}) = \\fgintp{\\*l''}(\\gp{\\*l,\\*i})$\n as $\\gp{\\*l,\\*i} \\in S \\cap \\clint{\\*0, \\*1}$.\n\\end{proof}\n\n\\lemmaCombiTechniqueCharacterization*\n\n\\begin{proof}\n ``$\\subset$'':\n Let $\\*l' \\in L_0$.\n We have to prove that\n $\\fa{t \\in T_{L_0}}{l'_t = l^\\ast_t}$ and\n $\\fa{t \\notin T_{L_0}}{l'_t \\ge l_t}$.\n The first statement is clear by the definition of $T_{L_0}$.\n Therefore, let $t \\notin T_{L_0}$.\n By the definition of $T_{L_0}$,\n we have either\n $\\ex{\\hat{\\*l} \\in L_0}{l'_t \\not= \\hat{l}_t}$ or\n $\\fa{\\hat{\\*l} \\in L_0}{l'_t = \\hat{l}_t \\ge l_t}$.\n In the latter case, we obtain $l'_t \\ge l_t$\n (e.g., by setting $\\hat{\\*l}$ to $\\*l'$).\n In the former case, there is an $\\hat{\\*l} \\in L_0$ such that\n $l'_t \\not= \\hat{l}_t$.\n Due to $\\*l' \\eq \\hat{\\*l}$\n (since $\\*l'$ and $\\hat{\\*l}$ are both contained in the same\n equivalence class $L_0$) and $t \\notin T_{\\*l',\\hat{\\*l}}$,\n we have $\\min\\{l'_t, \\hat{l}_t\\} \\ge l_t$.\n This implies $\\fa{t \\notin T_{L_0}}{l'_t \\ge l_t}$, as desired.\n \n ``$\\supset$'':\n Let $\\*l' \\in L$ such that\n $\\fa{t \\in T_{L_0}}{l'_t = l^\\ast_t}$ and\n $\\fa{t \\notin T_{L_0}}{l'_t \\ge l_t}$.\n Furthermore,\n let $\\*l'' \\in L_0$ be an arbitrary representative of $L_0$.\n We prove that $\\*l' \\eq \\*l''$ (i.e., $\\*l' \\in L_0$).\n Note that $T_{L_0} \\subset T_{\\*l',\\*l''}$,\n as $t \\in T_{L_0}$ implies\n $l''_t = l^\\ast_t < l_t$, which can be combined with $l'_t = l^\\ast_t$\n to $l'_t = l''_t < l_t$, i.e., $t \\in T_{\\*l',\\*l''}$.\n \n To prove the equivalence of $\\*l'$ and $\\*l''$,\n let $t \\notin T_{\\*l',\\*l''}$,\n i.e., $t \\notin T_{L_0}$.\n By assumption on $\\*l'$, it holds $l'_t \\ge l_t$.\n Hence, it remains to show that $l''_t \\ge l_t$ as well.\n Again, by definition of $T_{L_0}$, we have either\n $\\ex{\\hat{\\*l} \\in L_0}{l''_t \\not= \\hat{l}_t}$ or\n $\\fa{\\hat{\\*l} \\in L_0}{l''_t = \\hat{l}_t \\ge l_t}$.\n In the second case, it holds $l''_t \\ge l_t$.\n In the first case, there is an $\\hat{\\*l} \\in L_0$ such that\n $l''_t \\not= \\hat{l}_t$.\n Due to $\\*l'' \\eq \\hat{\\*l}$\n (since $\\*l''$ and $\\hat{\\*l}$ are both contained in the same\n equivalence class $L_0$) and $t \\notin T_{\\*l'',\\hat{\\*l}}$,\n we have $\\min\\{l''_t, \\hat{l}_t\\} \\ge l_t$.\n In particular, $l''_t \\ge l_t$.\n In total, we have $\\min\\{l'_t, l''_t\\} \\ge l_t$ for all\n $t \\notin T_{\\*l',\\*l''}$, proving that $\\*l'$ and $\\*l''$ are equivalent,\n as asserted.\n\\end{proof}\n\n\\propCombiTechniqueZero*\n\n\\begin{proof}\n \\Cref{lemma:combiTechniqueIdenticalValues}\n implies that the summands $\\fgintp{\\*l'}(\\gp{\\*l,\\*i})$\n corresponding to levels $\\*l'$ of the same equivalence class\n $L_0 \\in \\eqclasses{L}{\\eq}$ are identical.\n Let $f_{L_0}$ denote the common function value.\n The sum in the \\lhs of the assertion can be reordered to combine\n levels of the equivalence classes $L_0 \\in \\eqclasses{L}{\\eq}$:\n \\begin{subequations}\n \\begin{align}\n &\\sum_{q=0}^{d-1} (-1)^q \\binom{d-1}{q} \\cdot\n \\sum_{\\substack{\\normone{\\*l'} = n - q\\\\\\fgset{\\*l'} \\notni \\gp{\\*l,\\*i}}}\n \\fgintp{\\*l'}(\\gp{\\*l,\\*i})\\\\\n \\label{eq:proofCombiTechniqueIdenticalValues2}\n &= \\sum_{L_0 \\in \\eqclasses{L}{\\eq}} f_{L_0} \\sum_{q=0}^{d-1}\n (-1)^q \\binom{d-1}{q} \\cdot\n \\setsize{\\{\\*l' \\in L_0 \\mid \\normone{\\*l'} = n - q\\}}.\n \\end{align}\n \\end{subequations}\n It now suffices to show that the inner sum vanishes\n for every equivalence class $L_0 \\in \\eqclasses{L}{\\eq}$.\n \n To this end, we have to calculate\n $\\setsize{\\{\\*l' \\in L_0 \\mid \\normone{\\*l'} = n - q\\}}$\n for a fixed equivalence class $L_0$.\n Without loss of generality,\n let $T_{L_0} = \\{1, \\dotsc, m\\}$ in the notation of\n \\thmref{lemma:combiTechniqueCharacterization}\n with $1 \\le m \\le d$.\n Note that the case $m = 0$ is impossible:\n Otherwise, $T_{L_0} = \\emptyset$ implies\n $\\fa{\\*l' \\in L_0}{\\*l' \\ge \\*l}$ by\n \\cref{lemma:combiTechniqueCharacterization}, and\n as equivalence classes are non-empty,\n there is at least one $\\*l' \\in L_0$ with $\\*l' \\ge \\*l$.\n However, this is equivalent to $\\fgset{\\*l'} \\ni \\gp{\\*l,\\*i}$,\n which contradicts $\\*l' \\in L$.\n Hence, we have $m > 0$.\n \n To enumerate all levels $\\*l' \\in L_0$ with $\\normone{\\*l'} = n - q$,\n we exploit the characterization of $L_0$ of\n \\cref{lemma:combiTechniqueCharacterization}.\n For notational convenience, we define the vector\n \\begin{equation}\n \\hat{\\*l}\n \\ceq (l^\\ast_1, \\dotsc, l^\\ast_m,\\; l_{m+1}, \\dotsc, l_d),\n \\end{equation}\n where $\\*l^\\ast$ is given as in \\cref{lemma:combiTechniqueCharacterization}.\n We show that $\\*a \\ceq (l'_t - l_t)_{t = m+1, \\dotsc, d}$\n constitutes a bijection between\n \\begin{equation}\n \\{\\*l' \\in L_0 \\mid \\normone{\\*l'} = n - q\\}\n \\quad\\text{and}\\quad\n \\{\\*a \\in \\natz^{d-m} \\mid \\normone{\\*a} = n - q - \\normone{\\hat{\\*l}}\\}\n \\colon\n \\end{equation}\n \\begin{itemize}\n \\item\n Let $\\*l' \\in L_0$ with $\\normone{\\*l'} = n - q$.\n Then, $\\fa{t=m+1,\\dotsc,d}{l'_t - l_t \\ge 0}$\n (by \\cref{lemma:combiTechniqueCharacterization}), i.e.,\n $\\*a \\in \\natz^{d-m}$, and\n \\begin{equation}\n \\normone{\\*a}\n = \\sum_{t=m+1}^d (l'_t - l_t)\n = \\paren*{\\normone{\\*l'} - \\sum_{t=1}^m l'_t} -\n \\sum_{t=m+1}^d \\hat{l}_t\n = n - q - \\normone{\\hat{\\*l}}.\n \\end{equation}\n \n \\item\n Conversely, let $\\*a \\in \\natz^{d-m}$ with\n $\\normone{\\*a} = n - q - \\normone{\\hat{\\*l}}$.\n If we define $\\*l'$ as\n \\begin{equation}\n \\*l'\n = (l^\\ast_1, \\dotsc, l^\\ast_m,\\;\n a_1 + l_{m+1}, \\dotsc, a_{d-m} + l_d),\n \\end{equation}\n then $\\fa{t=1,\\dotsc,m}{l'_t = l^\\ast_t < l_t}$ and\n $\\fa{t=m+1,\\dotsc,d}{l'_t \\ge l_t}$.\n By \\cref{lemma:combiTechniqueCharacterization},\n we obtain $\\*l' \\in L_0$ and\n \\begin{equation}\n \\normone{\\*l'}\n = \\normone{\\hat{\\*l}} + \\normone{\\*a}\n = n - q.\n \\end{equation}\n \\end{itemize}\n This bijection implies that\n \\begin{subequations}\n \\begin{align}\n \\setsize{\\{\\*l' \\in L_0 \\mid \\normone{\\*l'} = n - q\\}}\n &= \\setsize{\n \\{\\*a \\in \\natz^{d-m} \\mid \\normone{\\*a} = n - q - \\normone{\\hat{\\*l}}\\}\n }.\\\\\n \\intertext{%\n This is the number of weak decompositions of\n $n - q - \\normone{\\hat{\\*l}}$ into $d - m$ parts:%\n }\n \\cdots\n &= \\binom{n - q - \\normone{\\hat{\\*l}} + d - m - 1}{d - m - 1},\n \\end{align}\n \\end{subequations}\n see Theorem 2.2 of \\cite{Bona15Introduction}.\n We insert this quantity into the inner sum of\n \\eqref{eq:proofCombiTechniqueIdenticalValues2}:\n \\begin{subequations}\n \\label{eq:proofCombiTechniqueIdenticalValues3}\n \\begin{align}\n &\\sum_{q=0}^{d-1}\n (-1)^q \\binom{d-1}{q} \\cdot\n \\setsize{\\{\\*l' \\in L_0 \\mid \\normone{\\*l'} = n - q\\}}\\\\\n &= \\sum_{q=0}^{d-1} (-1)^q \\binom{d-1}{q} \\cdot\n \\binom{n - q - \\normone{\\hat{\\*l}} + d - m - 1}{d - m - 1}.\n \\end{align}\n \\end{subequations}\n Again, we apply \\thmref{lemma:inclusionExclusionCountingLemma}\n with the values $a \\ceq d - 1$,\n $r \\ceq n - \\normone{\\hat{\\*l}} + d - m - 1$, and\n $s \\ceq d - m - 1$ to infer that as claimed,\n \\eqref{eq:proofCombiTechniqueIdenticalValues3} is equal to\n \\begin{equation}\n \\label{eq:proofCombiTechniqueIdenticalValues4}\n \\binom{n - \\normone{\\hat{\\*l}} - m}{-m}\n = 0\n \\end{equation}\n by the convention for binomial coefficients in \\cref{def:binomialCoefficient}\n as $-m < 0$.\n \n Note that for the calculation in\n \\cref{%\n eq:proofCombiTechniqueIdenticalValues3,%\n eq:proofCombiTechniqueIdenticalValues4%\n }\n to be correct,\n we have to ensure that $n - \\normone{\\hat{\\*l}} - m \\ge 0$;\n otherwise, the binomial coefficients would not be well-defined.\n This is a direct consequence of the fact that $l^\\ast_t < l_t$\n for all $t = 1, \\dotsc, m$\n (see \\cref{lemma:combiTechniqueCharacterization}) as\n \\begin{equation}\n n - \\normone{\\hat{\\*l}} - m\n = n - \\sum_{t=1}^m l^\\ast_t - \\sum_{\\mathclap{t=m+1}}^d l_t - m\n \\ge n - \\sum_{t=1}^m (l_t - 1) - \\sum_{\\mathclap{t=m+1}}^d l_t - m\n = n - \\normone{\\*l}\n \\ge 0,\n \\end{equation}\n where we have used $l^\\ast_t \\le l_t - 1$ for $t = 1, \\dotsc, m$\n and $\\normone{\\*l} \\le n$.\n\\end{proof}\n\n\n\n\\fillsubsectionornament\n\\subsection{Correctness Proof of the Method of Residual Interpolation}\n\\label{sec:a132proofResidualInterpolation}\n\n\\propInvariantResidualInterpolation*\n\n\\begin{proof}\n \\newcommand*{\\eqwithref}[1]{%\n \\quad\\;%\n \\if\\relax\\detokenize{#1}\\relax%\n \\mathclap{=}%\n \\else%\n \\mathclap{\\overset{\\eqref{eq:propInvariantResidualInterpolation#1}}{=}}%\n \\fi%\n \\quad\\;%\n }%\n %\n We prove the assertion by induction over $j = 1, \\dotsc, m$.\n We will need the following two equations\n that directly follow from the algorithm\n (\\cref{%\n line:algResidualInterpolation3,%\n line:algResidualInterpolation2,%\n line:algResidualInterpolation4%\n }, respectively):%\n \\begin{subequations}\n \\begin{alignat}{3}\n \\label{eq:propInvariantResidualInterpolation5}\n r_{\\*l^{(j)}}^{(j-1)}(\\gp{\\*l,\\*i})\n &= r^{(j-1)}(\\gp{\\*l,\\*i}),\\quad\n &&\\*l \\le \\*l^{(j)},\\;\\;\n &&\\*i \\in \\hiset{\\*l},\\\\\n \\label{eq:propInvariantResidualInterpolation6}\n r^{(j)}(\\gp{\\*l,\\*i})\n &= r^{(j-1)}(\\gp{\\*l,\\*i}) - r_{\\*l^{(j)}}^{(j-1)}(\\gp{\\*l,\\*i}),\\quad\n &&\\*l \\in L,\\;\\;\n &&\\*i \\in \\hiset{\\*l}.\n \\end{alignat}\n \\end{subequations}\n \n \\vspace*{-2em}\n \\pagebreak\n \n \\noindent\n \\textbf{Induction base case:}\n For $j = 1$, there is nothing to show for\n \\eqref{eq:propInvariantResidualInterpolation1}.\n \\Cref{eq:propInvariantResidualInterpolation2}\n can be proven as follows:\n \\begin{equation}\n r^{(1)}(\\gp{\\*l,\\*i})\n \\eqwithref{6}\n r^{(0)}(\\gp{\\*l,\\*i}) - r_{\\*l^{(1)}}^{(0)}(\\gp{\\*l,\\*i})\n \\eqwithref{5} 0,\\qquad\n \\*l \\le \\*l^{(1)},\\; \\*i \\in \\hiset{\\*l}.\n \\end{equation}\n \\Cref{eq:propInvariantResidualInterpolation3}\n holds as $r_{\\*l^{(1)}}^{(0)} = f^{\\sparse,(1)}$\n (by \\cref{line:algResidualInterpolation2} in\n \\cref{alg:residualInterpolation}) and, therefore,\n \\begin{align}\n r^{(1)}(\\gp{\\*l,\\*i})\n \\eqwithref{6}\n r^{(0)}(\\gp{\\*l,\\*i}) - r_{\\*l^{(1)}}^{(0)}(\\gp{\\*l,\\*i})\n = \\fcnval{\\*l,\\*i} - f^{\\sparse,(1)}(\\gp{\\*l,\\*i}),\\quad\n \\*l \\in L,\\; \\*i \\in \\hiset{\\*l}.\n \\end{align}\n \n \\noindent\n \\textbf{Induction step case:}\n We show the three statements for the induction step $j \\to (j + 1)$.\n \\begin{itemize}\n \\item\n \\emph{Showing \\eqref{eq:propInvariantResidualInterpolation1} for $j + 1$:}\n Let $j' = 1, \\dotsc, j$, $\\*l \\le \\*l^{(j')}$,\n and $\\*i \\in \\hiset{\\*l}$.\n Due to the ordering of the levels $\\*l^{(1)}, \\dotsc, \\*l^{(m)}$,\n we can conclude from $j + 1 > j'$ that\n $\\normone{\\*l^{(j+1)}} \\le \\normone{\\*l^{(j')}}$.\n This implies that there must be a $t' \\in \\{1, \\dotsc, d\\}$\n such that $l_{t'}^{(j+1)} \\le l_{t'}^{(j')}$.\n Let $S$ be the line in $\\real^d$ defined by\n \\begin{equation}\n S\n \\ceq \\gp{\\*l,\\*i} + \\spn\\{\\stdbasis{t'}\\},\n \\end{equation}\n where $\\stdbasis{t'}$ is the $t'$-th standard basis vector.\n It holds that $S \\cap \\fgset{\\*l^{(j+1)}} \\subset \\fgset{\\*l^{(j')}}$.\n To show this, let $\\gp{\\*l',\\*i'} \\in S \\cap \\fgset{\\*l^{(j+1)}}$\n be arbitrary (with $\\*i' \\in \\hiset{\\*l'}$).\n Then, $\\fa{t \\not= t'}{l'_t = l_t \\le l_t^{(j')}}$\n (due to $\\gp{\\*l',\\*i'} \\in S$) and\n $l'_{t'} \\le l_{t'}^{(j+1)} \\le l_{t'}^{(j')}$\n (due to $\\gp{\\*l',\\*i'} \\in \\fgset{\\*l^{(j+1)}}$).\n This means that $\\*l' \\le \\*l^{(j')}$, which implies that\n $\\gp{\\*l',\\*i'} \\in \\fgset{\\*l^{(j')}}$.\n As $\\gp{\\*l',\\*i'}$ is arbitrary,\n this shows $S \\cap \\fgset{\\*l^{(j+1)}} \\subset \\fgset{\\*l^{(j')}}$.\n \n Thus, we infer\n \\begin{equation}\n \\label{eq:proofPropInvariantResidualInterpolation2}\n r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l',\\*i'})\n \\eqwithref{5}\n r^{(j)}(\\gp{\\*l',\\*i'})\n \\eqwithref{2}\n 0,\\quad\n \\gp{\\*l',\\*i'} \\in S \\cap \\fgset{\\*l^{(j+1)}}\n \\subset \\fgset{\\*l^{(j')}},\\;\n i' \\in \\hiset{\\*l'},\n \\end{equation}\n with the induction hypothesis\n \\eqref{eq:propInvariantResidualInterpolation2} for $j$.\n Unfortunately, this does not suffice to directly conclude that\n $r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i}) = 0$ as\n $\\gp{\\*l,\\*i}$ is in general not contained in $\\fgset{\\*l^{(j+1)}}$.\n \n As in the proof of \\cref{lemma:combiTechniqueIdenticalValues},\n we exploit the tensor product nature of the basis functions and\n restrict $r_{\\*l^{(j+1)}}^{(j)}$ to $S \\cap \\clint{0, 1}$:\n \\begin{subequations}\n \\begin{align}\n (\\restrictfcn{r_{\\*l^{(j+1)}}^{(j)}}{S \\cap \\clint{0, 1}})(x_{t'})\n &\\mathrel{\\righthphantom{=}{\\ceq}}\n \\sum_{l'_{t'}=0}^{l_{t'}^{(j+1)}}\n \\sum_{i'_{t'} \\in \\hiset{l'_{t'}}}\n \\surplustilde[(j+1)]{l'_{t'},i'_{t'}}\n \\basis{l'_{t'},i'_{t'}}(x_{t'}),\\quad\n x_{t'} \\in \\clint{0, 1},\\\\\n \\surplustilde[(j+1)]{l'_{t'},i'_{t'}}\n &\\ceq \\sum_{\\*l'_{-t'}=\\*0}^{\\*l^{(j+1)}_{-t'}}\n \\sum_{\\*i'_{-t'} \\in \\hiset{\\*l'_{-t'}}}\n \\surplus[(j+1)]{\\*l',\\*i'}\n \\basis{\\*l'_{-t'},\\*i'_{-t'}}(\\gp{\\*l_{-t'},\\*i_{-t'}}),\n \\end{align}\n \\end{subequations}\n where the subscript $-t'$ indicates all entries but the $t'$-th.\n As a consequence, this shows that\n $\\restrictfcn{r_{\\*l^{(j+1)}}^{(j)}}{S \\cap \\clint{0, 1}} \\in\n \\ns{l_{t'}^{(j+1)}}$ is an interpolant of the zero function\n (by \\eqref{eq:proofPropInvariantResidualInterpolation2}).\n Due to the linear independence of the univariate basis functions,\n we conclude\n \\begin{equation}\n \\restrictfcn{r_{\\*l^{(j+1)}}^{(j)}}{S \\cap \\clint{0, 1}}\n \\equiv 0.\n \\end{equation}\n Consequently, we obtain\n $r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i}) = 0$\n as $\\gp{\\*l,\\*i} \\in S \\cap \\clint{\\*0, \\*1}$.\n \n \\item\n \\emph{Showing \\eqref{eq:propInvariantResidualInterpolation2} for $j + 1$:}\n Let $j' = 1, \\dotsc, j + 1$, $\\*l \\le \\*l^{(j')}$,\n and $\\*i \\in \\hiset{\\*l}$.\n For the case $j' \\le j$, we obtain\n \\begin{equation}\n \\label{eq:proofPropInvariantResidualInterpolation1}\n r^{(j+1)}(\\gp{\\*l,\\*i})\n \\eqwithref{6}\n r^{(j)}(\\gp{\\*l,\\*i}) - r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i})\n = 0\n \\end{equation}\n due to $r^{(j)}(\\gp{\\*l,\\*i}) = 0$ by induction hypothesis\n (\\cref{eq:propInvariantResidualInterpolation2}) and\n $r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i}) = 0$ as shown above\n (\\cref{eq:propInvariantResidualInterpolation1} for $j + 1$).\n \n For the case $j' = j + 1$,\n \\cref{eq:proofPropInvariantResidualInterpolation1}\n still holds as the difference between\n $r^{(j)}(\\gp{\\*l,\\*i})$ and $r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i})$\n vanishes due to \\eqref{eq:propInvariantResidualInterpolation5}\n for $j + 1$\n (here, we need $\\*l \\le \\*l^{(j+1)}$).\n \n \\item\n \\emph{Showing \\eqref{eq:propInvariantResidualInterpolation3} for $j + 1$:}\n Let $\\*l \\in L$ and $\\*i \\in \\hiset{\\*l}$.\n Then,\n \\begin{subequations}\n \\begin{align}\n r^{(j+1)}(\\gp{\\*l,\\*i})\n &\\eqwithref{6}\n r^{(j)}(\\gp{\\*l,\\*i}) - r_{\\*l^{(j+1)}}^{(j)}(\\gp{\\*l,\\*i}).\\\\\n \\intertext{%\n The first term can replaced with the induction hypothesis\n (\\eqref{eq:propInvariantResidualInterpolation3} for $j$).\n For the second term, note that\n $r_{\\*l^{(j+1)}}^{(j)}\n = \\sum_{\\*l' \\in \\levelset} \\sum_{\\*i' \\in \\hiset{\\*l'}}\n \\surplus[(j+1)]{\\*l',\\*i'} \\basis{\\*l',\\*i'}\n = f^{\\sparse,(j+1)} - f^{\\sparse,(j)}$ by definition.\n Hence, we obtain as desired%\n }\n \\cdots\n &\\eqwithref{} (\\fcnval{\\*l,\\*i} - f^{\\sparse,(j)}(\\gp{\\*l,\\*i})) -\n (f^{\\sparse,(j+1)}(\\gp{\\*l,\\*i}) - f^{\\sparse,(j)}(\\gp{\\*l,\\*i}))\\\\\n &\\eqwithref{} \\fcnval{\\*l,\\*i} - f^{\\sparse,(j+1)}(\\gp{\\*l,\\*i}).\n \\end{align}\n \\end{subequations}\n \\end{itemize}\n This shows the validity of the statements in\n \\eqref{eq:propInvariantResidualInterpolationStatements}\n for $j + 1$.\n\\end{proof}\n\n\n\n\\subsection{Correctness Proof of Hierarchization with Breadth-First Search}\n\\label{sec:a133proofBFS}\n\n\\propInvariantBFS*\n\n\\begin{proof}\n We start with two observations:\n \n \\begin{itemize}\n \\item\n First, due to the breadth-first search nature of \\cref{alg:BFS} and the\n hierarchical relation \\eqref{eq:directAncestor},\n all grid points with level sum $< q$ are \\pop{}ped before\n the first point with level sum $\\ge q$ is \\pop{}ped.\n \n \\item\n Second, after \\pop{}ping all grid points with\n level sum $< q$, the output values of the grid points with\n level sum $\\le q$ remain unchanged for the rest of the algorithm:\n If \\cref{line:algBFS2} of the algorithm updates the output value of a point\n $(\\*l, \\*i)$ in the iteration of $(\\*l', \\*i') \\in \\liset$ with\n $\\normone{\\*l'} \\ge q$, then \\cref{line:algBFS1} implies\n $\\*l \\ge \\*l'$ and thus, $\\normone{\\*l} \\ge \\normone{\\*l'} \\ge q$.\n However, $\\normone{\\*l} = q$ is not possible as\n this would imply that $\\normone{\\*l} = \\normone{\\*l'}\n \\implies (\\*l, \\*i) = (\\*l', \\*i')$ by \\cref{line:algBFS1},\n but $(\\*l, \\*i) = (\\*l', \\*i')$ is explicitly excluded in the\n \\texttt{\\algorithmicfor} loop of \\cref{line:algBFS1}.\n Therefore, we must have $\\normone{\\*l} > q$.\n Hence, if a point $(\\*l, \\*i)$ with level sum $\\ge q$ has been \\pop{}ped,\n only surpluses of points with level sum $> q$ may be updated.\n \\end{itemize}\n \n \\noindent\n Now, we prove the asserted claim by induction over $q$.\n \n \\noindent\n \\textbf{Induction base case:}\n For $q = 0$, \\cref{alg:BFS} sets $\\linout{\\*l,\\*i}$ to\n $\\fcnval{\\*l,\\*i}$ in \\cref{line:algBFS3}.\n As the sum in \\eqref{eq:propInvariantBFS} is empty,\n the claim is correct for $q = 0$.\n \n \\noindent\n \\textbf{Induction step case:}\n Let $\\linout[(q)]{\\*l',\\*i'}$ and\n $\\linout[(q+1)]{\\*l',\\*i'}$\n be the surpluses after \\pop{}ping all\n grid points with level sum $< q$ and $< q + 1$, respectively.\n We show the induction step $q \\to (q + 1)$, i.e.,\n we assume that the assertion is true for $q$\n and prove that after \\pop{}ping all grid points with level sum $< q + 1$,\n it holds\n \\begin{equation}\n \\label{eq:proofPropInvariantBFS1}\n \\linout[(q+1)]{\\*l,\\*i}\n = \\fcnval{\\*l,\\*i} -\n \\largesum{\\normone{\\*l'} < q+1} \\linout[(q+1)]{\\*l',\\*i'}\n \\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i}),\\quad\n (\\*l, \\*i) \\in \\liset,\\;\\;\n \\normone{\\*l} = q+1.\n \\end{equation}\n Therefore, let $(\\*l, \\*i) \\in \\liset$ fulfill $\\normone{\\*l} = q+1$.\n The update in \\cref{line:algBFS2} can safely be applied\n with all grid points $(\\*l', \\*i')$ with level sum $q$.\n The grid points $(\\*l', \\*i')$ that do not satisfy the relation in the set in\n \\cref{line:algBFS1} do not contribute as\n $\\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i}) = 0$\n due to the necessary condition \\eqref{eq:fundamentalPropertyImplicationMV}.\n By summing all updates from \\cref{line:algBFS2}, we obtain\n \\begin{subequations}\n \\begin{align}\n \\linout[(q+1)]{\\*l,\\*i}\n &= \\linout[(q)]{\\*l,\\*i} -\n \\largesum{\\normone{\\*l'} = q} \\linout[(q)]{\\*l',\\*i'}\n \\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i}).\\\\\n \\intertext{%\n After inserting the induction hypothesis\n for the first $\\linout[(q)]{\\*l,\\*i}$, we have%\n }\n \\cdots\n &= \\paren*{\n \\fcnval{\\*l,\\*i} -\n \\largesum{\\normone{\\*l'} < q} \\linout[(q)]{\\*l',\\*i'}\n \\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i})\n } -\n \\largesum{\\normone{\\*l'} = q} \\linout[(q)]{\\*l',\\*i'}\n \\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i})\\\\\n &= \\hphantom{\\Biggl(}\\fcnval{\\*l,\\*i} -\n \\largesum{\\normone{\\*l'} < q + 1} \\linout[(q)]{\\*l',\\*i'}\n \\fundbasis{\\*l',\\*i'}(\\gp{\\*l,\\*i}).\n \\end{align}\n \\end{subequations}\n As noted above, we have\n $\\fa{(\\*l', \\*i'),\\, \\normone{\\*l'} < q + 1}{\n \\linout[(q)]{\\*l',\\*i'} = \\linout[(q+1)]{\\*l',\\*i'}\n }$\n (the values of points with level sum $< q + 1$\n do not change after \\pop{}ping all points with level sum $< q$).\n This shows the induction claim \\eqref{eq:proofPropInvariantBFS1}.\n\\end{proof}\n\n\n\n\\disableornamentsfornextheadingtrue\n\\subsection{%\n Proof for the Correctness of the Unidirectional Principle on\n Spatially Adaptive Sparse Grids%\n}\n\\label{sec:a134proofCorrectnessUnidirectionalPrincipleSASG}\n\n\\lemmaChainExistenceSufficient*\n\n\\begin{proof}\n We prove the assertion by induction over $j = 0, \\dotsc, d$.\n \n For $j = 0$, the operator $\\upop{\\emptyset}$ is by definition\n \\eqref{eq:upopProduct} the identity operator $\\idop$.\n The assumption $(\\upop{\\emptyset})_{\\*k'',\\*k'} \\not= 0$\n implies that $\\*k' = \\*k''$, since the identity matrix is diagonal.\n Therefore, the chain $(\\chain{0})$ from $\\*k'$ to $\\*k''$ is given by\n $\\chain{0} = \\*k'$, which is contained in $\\liset$.\n \n For the induction step $j \\to (j+1)$, we split \\cref{eq:upopProduct},\n i.e.,\n \\begin{equation}\n \\upop{t_1,\\dotsc,t_{j+1}}\n = \\upop{t_{j+1}} \\upop{t_j} \\dotsm \\upop{t_1}\n = \\upop{t_{j+1}} \\upop{t_1,\\dotsc,t_j},\n \\end{equation}\n and infer by assumption\n \\begin{equation}\n \\label{eq:proofLemmaChainExistenceSufficient1}\n 0\n \\not= (\\upop{t_1,\\dotsc,t_{j+1}})_{\\*k'',\\*k'}\n = \\sum_{\\*k \\in \\liset} (\\upop{t_{j+1}})_{\\*k'',\\*k}\n (\\upop{t_1,\\dotsc,t_j})_{\\*k,\\*k'}.\n \\end{equation}\n Consequently,\n there is at least one summation index $\\*k$\n for which both factors do not vanish.\n The first factor $(\\upop{t_{j+1}})_{\\*k'',\\*k}$ can by definition\n only be non-zero if $\\*k \\samepole{t_{j+1}} \\*k''$.\n The second factor $(\\upop{t_1,\\dotsc,t_j})_{\\*k,\\*k'}$ being\n non-zero implies that by induction hypothesis,\n $\\liset$ contains the chain $(\\chain{0}, \\dotsc, \\chain{j})$\n from $\\*k'$ to $\\*k$ with respect to $(t_1, \\dotsc, t_j)$.\n The combination of both statements leads to\n the chain $(\\chain{0}, \\dotsc, \\chain{j}, \\chain{j+1})$ from $\\*k'$\n to $\\*k''$ with respect to $(t_1, \\dotsc, t_{j+1})$.\n All points of the chain are contained in $\\liset$.\n\\end{proof}\n\n\\lemmaChainExistenceNecessary*\n\n\\begin{proof}\n Again, we prove the claim by induction over $j = 0, \\dotsc, d$.\n \n For $j = 0$, the operator $\\upop{\\emptyset}$ is the identity operator.\n Therefore, the \\lhs of \\eqref{eq:lemmaChainExistenceNecessary} is one\n (due to $\\chain{0} = \\*k'$).\n The \\rhs is by convention also one, as it is an empty product.\n \n \\setlength{\\abovedisplayskip}{9pt}%\n \\setlength{\\belowdisplayskip}{9pt}%\n For the induction step $j \\to (j+1)$, we consider again\n \\begin{equation}\n \\label{eq:proofLemmaChainExistenceNecessary1}\n (\\upop{t_1,\\dotsc,t_{j+1}})_{\\chain{j+1},\\*k'}\n = \\sum_{\\*k \\in \\liset} (\\upop{t_{j+1}})_{\\chain{j+1},\\*k}\n (\\upop{t_1,\\dotsc,t_j})_{\\*k,\\*k'}\n \\end{equation}%\n similar to \\eqref{eq:proofLemmaChainExistenceSufficient1}.\n Recall that\n \\begin{subequations}\n \\begin{alignat}{4}\n \\chainuv{j}_t\n &= k''_t\n &&\\;\\;\\text{and}\\;\\;\n &\\chainuv{j+1}_t\n &= k''_t\n &&\\quad\\text{for } t \\in \\{t_1, \\dotsc, t_j\\},\\\\\n \\label{eq:prooflemmaChainExistenceNecessary}\n \\chainuv{j}_t\n &= k'_t\n &&\\;\\;\\text{and}\\;\\;\n &\\chainuv{j+1}_t\n &= k''_t\n &&\\quad\\text{for } t = t_{j+1},\\\\\n \\chainuv{j}_t\n &= k'_t\n &&\\;\\;\\text{and}\\;\\;\n &\\chainuv{j+1}_t\n &= k'_t\n &&\\quad\\text{for } t \\notin \\{t_1, \\dotsc, t_j, t_{j+1}\\}.\n \\end{alignat}\n \\end{subequations}\n We now argue that all summands of\n \\eqref{eq:proofLemmaChainExistenceNecessary1} vanish,\n except the summand with index $\\chain{j}$.\n There are two cases for the summation index $\\*k$,\n if we assume $\\*k \\not= \\chain{j}$:\n \\begin{itemize}\n \\item\n If there is a $t \\in \\{t_1, \\dotsc, t_j\\}$ with $k_t \\not= k''_t$,\n then we have $k_t \\not= k''_t = \\chainuv{j+1}_t$.\n Consequently, $\\chain{j+1} \\not\\samepole{t_{j+1}} \\*k$ and\n $(\\upop{t_{j+1}})_{\\chain{j+1},\\*k} = 0$ due to \\eqref{eq:upopEntries},\n i.e., the first factor of the $\\*k$-th summand in\n \\eqref{eq:proofLemmaChainExistenceNecessary1} vanishes.\n \n \\item\n If there is a $t \\notin \\{t_1, \\dotsc, t_j\\}$ with $k_t \\not= k'_t$,\n then the second factor $(\\upop{t_1,\\dotsc,t_j})_{\\*k,\\*k'}$\n of the $\\*k$-th summand in\n \\eqref{eq:proofLemmaChainExistenceNecessary1} vanishes.\n Indeed, if we assume the contrary,\n then \\cref{lemma:chainExistenceSufficient} implies that there is\n a chain from $\\*k'$ to $\\*k$ with respect to $(t_1, \\dotsc, t_j)$.\n However, by definition of the chain, this means that\n $\\*k'$ and $\\*k$ coincide in all other dimensions\n (which are not in $\\{t_1, \\dotsc, t_j\\}$).\n This contradicts $k_t \\not= k'_t$ and therefore\n $(\\upop{t_1,\\dotsc,t_j})_{\\*k,\\*k'}$ must vanish.\n \\end{itemize}\n We infer that only the summand $\\*k = \\chain{j}$ remains in\n \\eqref{eq:proofLemmaChainExistenceNecessary1}:\n \\begin{equation}\n (\\upop{t_1,\\dotsc,t_{j+1}})_{\\chain{j+1},\\*k'}\n = (\\upop{t_{j+1}})_{\\chain{j+1},\\chain{j}}\n (\\upop{t_1,\\dotsc,t_j})_{\\chain{j},\\*k'}.\n \\end{equation}\n The first factor equals\n \\begin{equation}\n (\\upop{t_{j+1}})_{\\chain{j+1},\\chain{j}}\n = (\\upopuv{t_{j+1}}{\\eqclass{\\chain{j+1}}{\\samepole{t_{j+1}}}})%\n _{\\chainuv{j+1}_{t_{j+1}},\\chainuv{j}_{t_{j+1}}}\n = (\\upopuv{t_{j+1}}{\\eqclass{\\chain{j+1}}{\\samepole{t_{j+1}}}})%\n _{k''_{t_{j+1}},k'_{t_{j+1}}}\n \\end{equation}\n by \\cref{eq:upopEntries,eq:prooflemmaChainExistenceNecessary}\n (and due to $\\chain{j} \\samepole{t_{j+1}} \\chain{j+1}$).\n The second factor equals\n \\begin{equation}\n (\\upop{t_1,\\dotsc,t_j})_{\\chain{j},\\*k'}\n =\n (\\upopuv{t_1}{\\eqclass{\\chain{1}}{\\samepole{t_1}}})_{k''_{t_1},k'_{t_1}}\n \\dotsm\n (\\upopuv{t_j}{\\eqclass{\\chain{j}}{\\samepole{t_j}}})_{k''_{t_j},k'_{t_j}}\n \\end{equation}\n by induction hypothesis.\n Hence, the product of both factors is\n \\begin{equation}\n (\\upop{t_1,\\dotsc,t_{j+1}})_{\\chain{j+1},\\*k'}\n =\n (\\upopuv{t_1}{\\eqclass{\\chain{1}}{\\samepole{t_1}}})_{k''_{t_1},k'_{t_1}}\n \\dotsm\n (\\upopuv{t_{j+1}}{\\eqclass{\\chain{j+1}}{\\samepole{t_{j+1}}}})%\n _{k''_{t_{j+1}},k'_{t_{j+1}}}.\n \\vspace*{-2.5em}\n \\end{equation}\n\\end{proof}\n\n\\vspace*{1em}\n\\pagebreak\n\n\\propCorrectnessUPCharacterization*\n\n\\begin{proof}\n ``$\\implies$'':\n Let the \\up be correct for $\\linop$ and $(t_1, \\dotsc, t_d)$\n and $\\*k', \\*k'' \\in \\liset$ with $(\\linop)_{\\*k'',\\*k'} \\not= 0$.\n Then, we obtain\n \\begin{equation}\n (\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'}\n = (\\linop)_{\\*k'',\\*k'}\n \\not= 0.\n \\end{equation}\n By \\cref{lemma:chainExistenceSufficient},\n this implies that $\\liset$ contains the chain from $\\*k'$ to $\\*k''$\n with respect to $(t_1, \\dotsc, t_d)$.\n \n ``$\\impliedby$'':\n For the converse direction, we assume that there are chains\n from $\\*k'$ to $\\*k''$ with respect to $(t_1, \\dotsc, t_d)$\n for all $\\*k', \\*k'' \\in \\liset$ with $(\\linop)_{\\*k'',\\*k'} \\not= 0$.\n Let $\\*k', \\*k'' \\in \\liset$ be arbitrary.\n There are two cases:\n \\begin{itemize}\n \\item\n ${(\\linop)_{\\*k'',\\*k'} \\not= 0}$:\n By assumption, $\\liset$ contains the chain from $\\*k'$ to $\\*k''$\n with respect to $(t_1, \\dotsc, t_d)$.\n We apply \\cref{lemma:chainExistenceNecessary} with $j = d$\n to infer\n \\begin{equation}\n (\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'}\n =\n (\\upopuv{t_1}{\\eqclass{\\chain{1}}{\\samepole{t_1}}})_{k''_{t_1},k'_{t_1}}\n \\dotsm\n (\\upopuv{t_d}{\\eqclass{\\chain{d}}{\\samepole{t_d}}})_{k''_{t_d},k'_{t_d}}\n = (\\linop)_{\\*k'',\\*k'}\n \\end{equation}\n by the assumption \\eqref{eq:tensorProductOperator} on\n the tensor product structure of $\\linop$.\n \n \\item\n ${(\\linop)_{\\*k'',\\*k'} = 0}$:\n In this case, $(\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'}$ must vanish as well.\n Indeed, if we assume the contrary\n $(\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'} \\not= 0$,\n then we can apply \\cref{lemma:chainExistenceSufficient}\n to obtain that $\\liset$ contains the chain from $\\*k'$ to $\\*k''$\n with respect to $(t_1, \\dotsc, t_d)$.\n We conclude with \\cref{lemma:chainExistenceNecessary} as in the first case\n that $(\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'} = (\\linop)_{\\*k'',\\*k'} = 0$,\n which is a contradiction.\n \\end{itemize}\n In any case, we obtain\n $(\\upop{t_1,\\dotsc,t_d})_{\\*k'',\\*k'} = (\\linop)_{\\*k'',\\*k'}$,\n from which follows the correctness of the \\up\\punctfix{,}\n as $\\*k'$ and $\\*k''$ are arbitrary.\n\\end{proof}\n\n\n\n\\breakpagebeforenextheadingtrue\n\\subsection{Correctness Proof of Hermite Hierarchization}\n\\label{sec:a135proofHermiteHierarchization}\n\n\\propInvariantHermiteHierarchization*\n\n\\begin{proof}\n We prove the assertion by induction over $l = 0, \\dotsc, n$.\n \n For the induction base case $l = 0$ and $i \\in \\{0, 1\\}$, we have\n \\begin{equation}\n \\sum_{i'=0}^1\n \\linout{0,i'} \\deriv[q]{x}{\\bspl[\\wfs]{0,i'}{p}}(\\gp{0,i})\n = \\kronecker{q}{0} \\cdot \\fcnval{0,i} +\n \\kronecker{q}{1} \\cdot (\\fcnval{0,1} - \\fcnval{0,0})\n = \\deriv[q]{x}{\\fgintp{0}}(\\gp{0,i})\n \\end{equation}\n for $q = 0, \\dotsc, \\frac{p-1}{2}$ by\n \\cref{%\n line:algHermiteHierarchization1,%\n line:algHermiteHierarchization3%\n }\n of \\cref{alg:hermiteHierarchization}.\n \n For the induction step case $(l-1) \\to l$,\n it suffices to show that\n \\begin{equation}\n \\label{eq:proofPropInvariantHermiteHierarchization1}\n \\deriv[q]{x}{\\fgintp{l-1}}(\\gp{l,i})\n \\overset{!}{=} \\sum_{l'=0}^{l-1} \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\deriv[q]{x}{\\bspl[\\wfs]{l',i'}{p}}(\\gp{l,i}),\\quad\n i = 0, \\dotsc, 2^l,\\;\\;\n q = 0, \\dotsc, \\frac{p-1}{2}.\n \\end{equation}\n Indeed, if \\eqref{eq:proofPropInvariantHermiteHierarchization1} holds,\n then we obtain by\n \\cref{line:algHermiteHierarchization6,line:algHermiteHierarchization8}\n of \\cref{alg:hermiteHierarchization}\n \\begin{subequations}\n \\begin{align}\n \\deriv[q]{x}{\\fgintp{l}}(\\gp{l,i})\n &= \\deriv[q]{x}{\\fgintp{l-1}}(\\gp{l,i}) +\n \\deriv[q]{x}{r^{(l)}_l}(\\gp{l,i})\\\\\n &= \\sum_{l'=0}^{l-1} \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\deriv[q]{x}{\\bspl[\\wfs]{l',i'}{p}}(\\gp{l,i}) +\n \\sum_{i' \\in \\hiset{l}}\n \\linout{l,i'} \\deriv[q]{x}{\\bspl[\\wfs]{l,i'}{p}}(\\gp{l,i})\\\\\n &= \\sum_{l'=0}^l \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\deriv[q]{x}{\\bspl[\\wfs]{l',i'}{p}}(\\gp{l,i}),\n \\end{align}\n \\end{subequations}\n which is the desired relation\n \\eqref{eq:propInvariantHermiteHierarchization}\n for level $l$.\n \n To prove \\eqref{eq:proofPropInvariantHermiteHierarchization1},\n we separate two cases:\n \\begin{itemize}\n \\item\n \\mbox{${i \\notin \\hiset{l}}$: In this case,}\n the ``true'' level of $\\gp{l,i}$ is actually $\\le l - 1$.\n Therefore, we can apply the induction hypothesis\n for \\cref{eq:propInvariantHermiteHierarchization} to\n obtain \\eqref{eq:proofPropInvariantHermiteHierarchization1}.\n \n \\item\n \\mbox{${i \\in \\hiset{l}}$: In this case,}\n we cannot directly apply the induction hypothesis,\n as it only holds for grid points of levels $\\le l - 1$.\n However, we note that in\n \\eqref{eq:proofPropInvariantHermiteHierarchization1},\n the term $\\deriv[q]{x}{\\fgintp{l-1}}(\\gp{l,i})$\n is the $q$-th derivative of the Hermite interpolant of\n the data $\\deriv[q']{x}{\\fgintp{l-1}}(\\gp{l,i\\pm1})$\n ($q' = 0, \\dotsc, \\frac{p-1}{2}$),\n as determined in \\cref{line:algHermiteHierarchization4}\n of \\cref{alg:hermiteHierarchization}.\n The ``true'' level of the grid points $\\gp{l,i\\pm1}$ is\n actually $\\le l - 1$ due to $i \\in \\hiset{l}$.\n Hence, we can apply the induction hypothesis\n for \\cref{eq:propInvariantHermiteHierarchization}\n to conclude that the interpolated data of\n $\\deriv[q]{x}{\\fgintp{l-1}}(\\gp{l,i})$ are given by\n \\begin{equation}\n \\deriv[q']{x}{\\fgintp{l-1}}(\\gp{l,i\\pm1})\n = \\deriv[q']{x}{\n \\bracket*{\n \\sum_{l'=0}^{l-1} \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\bspl[\\wfs]{l',i'}{p}\n }\n }(\\gp{l,i\\pm1}),\\quad\n q' = 0, \\dotsc, \\frac{p-1}{2}.\n \\end{equation}\n The linear combination in square brackets\n is a polynomial of degree $\\le p$ on the interval\n $\\clint{\\gp{l,i-1}, \\gp{l,i+1}}$\n by construction of the hierarchical basis functions\n $\\bspl[\\wfs]{l',i'}{p}$ ($l' = 0, \\dotsc, l - 1$, $i' \\in \\hiset{l'}$).\n Due to the uniqueness of Hermite interpolation\n (\\cref{lemma:hermiteInterpolation}),\n the interpolation polynomial of the data must\n coincide on $\\clint{\\gp{l,i-1}, \\gp{l,i+1}}$\n with the term in square brackets.\n In particular, as $\\gp{l,i} \\in \\clint{\\gp{l,i-1}, \\gp{l,i+1}}$,\n we obtain the claim \\eqref{eq:proofPropInvariantHermiteHierarchization1}.\n \\end{itemize}\n In both cases, we obtain the desired relation\n \\eqref{eq:proofPropInvariantHermiteHierarchization1}.\n\\end{proof}\n\n\\corAlgHermiteHierarchizationCorrectness*\n\n\\begin{proof}\n By \\cref{prop:invariantHermiteHierarchization} ($q = 0$), we have\n \\begin{equation}\n \\sum_{l'=0}^n \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\bspl[\\wfs]{l',i'}{p}(\\gp{l,i})\n = \\sum_{l'=0}^l \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\bspl[\\wfs]{l',i'}{p}(\\gp{l,i})\n = \\fgintp{l}(\\gp{l,i}),\\quad\n l \\le n,\\; i \\in \\hiset{l},\n \\end{equation}\n as $\\bspl[\\wfs]{l',i'}{p}(\\gp{l,i}) = 0$ if $l' > l$\n (weakly fundamental property \\eqref{eq:weaklyFundamentalProperty}).\n \\Cref{line:algHermiteHierarchization8} of\n \\cref{alg:hermiteHierarchization} implies\n $\\fgintp{l}(\\gp{l,i})\n = \\fgintp{l-1}(\\gp{l,i}) + r^{(l)}_l(\\gp{l,i})$, and by\n \\cref{line:algHermiteHierarchization5,line:algHermiteHierarchization7},\n the second summand $r^{(l)}_l(\\gp{l,i})$ equals\n $\\fcnval{l,i} - \\fgintp{l-1}(\\gp{l,i})$,\n which cancels out the first summand, resulting in\n $\\fgintp{l}(\\gp{l,i}) = \\fcnval{l,i}$.\n Combining these statements, we obtain\n \\begin{equation}\n \\sgintp(\\gp{l,i})\n = \\fcnval{l,i},\\quad\n l \\le n,\\; i \\in \\hiset{l},\n \\quad\\text{where}\\quad\n \\sgintp\n \\ceq \\sum_{l'=0}^n \\sum_{i' \\in \\hiset{l'}}\n \\linout{l',i'} \\bspl[\\wfs]{l',i'}{p}.\n \\end{equation}\n This means that $\\sgintp$ is the correct hierarchical interpolant\n of the given function values\n (see \\cref{eq:hierarchizationInterpolant}).\n Due to the uniqueness of hierarchical surpluses,\n the coefficients $\\linout{l,i}$\n (which are the output of \\cref{alg:hermiteHierarchization})\n must coincide with the surpluses $\\surplus{l,i}$.\n\\end{proof}\n", "meta": {"hexsha": "579ec9f115333d449830a070e3143e8707c34ef9", "size": 43710, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/document/a13chapter4.tex", "max_stars_repo_name": "valentjn/thesis-arxiv", 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"max_line_length": 80, "alphanum_fraction": 0.5723861817, "num_tokens": 17689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7745833893685269, "lm_q1q2_score": 0.6277835000895631}} {"text": "%!TEX root = ../TTK18-Summary.tex\n\\section{Linear matrix inequalities for piecewise affine systems}\n\n\\subsection{Piecewise affine systems}\nA piecewise affine system is a system where the state space $X$ is divided into non-overlapping partitions $X_i$ with distinct models in each partition:\n%\n\\begin{itemize}\n \\item Partitions containing the origin are linear:\n \\begin{equation}\n \\begin{split}\n x &= A_i x \\\\\n x_{k+1} &= A_i x_k\n \\end{split}\n \\end{equation}\n \\item Partitions not containing the origin are affine:\n \\begin{equation}\n \\begin{split}\n x &= A_i x + a_i \\\\\n x_{k+1} &= A_i x_k + a_i\n \\end{split}\n \\end{equation}\n\\end{itemize}\n%\nThe partitions are indexed, with an index set\n%\n\\begin{equation}\n I = I_0 \\cup I_1\n\\end{equation}\n%\nwhere $I_0$ are the parititions containing the origin, and $I_1$ are the partitions not containing the origin.\n\n\\subsection{Lyapunov stability}\nSometimes you can find a Lyapunov function for the whole PWA system:\n%\n\\begin{itemize}\n \\item If $a_i = 0 \\forall i$ and there exists a $P = P^T > 0$ such that\n \\begin{equation}\n A_i^T P + P A_i < 0 \\quad \\forall i \\in I,\n \\end{equation}\n then the origin is exponentially stable.\n \\item If $a_i = 0 \\forall i$ and there exists an $R_i > 0 \\forall i \\in I$ such that\n \\begin{equation}\n \\sum_{i \\in I} (A_i^T R_i + R_i A_i) > 0,\n \\end{equation}\n then a common Lyapunov function can be found.\n\\end{itemize}\n%\nWhen a common function cannot be found, we must look for one that depends on the partition.\n\n\\subsubsection{Notation}\n\\begin{equation}\n \\bar{A}_i = \\begin{bmatrix} A_i & a_i \\\\ 0 & 0 \\end{bmatrix}\n\\end{equation}\n%\nEach partition is a polyhedron, so we can write\n%\n\\begin{equation}\n \\bar{E}_i = \\begin{bmatrix} E_i & e_i \\end{bmatrix}, \\quad\n \\bar{F}_i = \\begin{bmatrix} F_i & f_i \\end{bmatrix}\n\\end{equation}\n%\nwhere $e_i = 0$ and $f_i = 0$ for all $i \\in I_0$ such that\n%\n\\begin{gather}\n \\bar{E}_i \\begin{bmatrix} x \\\\ 1 \\end{bmatrix} \\geq 0 \\quad \\forall x \\in X_i \\forall i \\in I \\\\\n \\bar{F}_i \\begin{bmatrix} x \\\\ 1 \\end{bmatrix} = \\bar{F}_j \\begin{bmatrix} x \\\\ 1 \\end{bmatrix} \\quad \\forall\\; x \\in X_i \\cap X_j, \\forall i, j \\in I\n\\end{gather}\n%\nThat is, $E_i x + e_i \\geq 0$ for all values of $x$, and $F_i x + f_i = F_j x + f_j$ for all $x$ along borders between partitions.\n\n\\subsubsection{Lyapunov function}\n\\begin{equation}\n V(x) =\n \\begin{cases}\n x\\tp P_i x, &x \\in X_i, i \\in I_0 \\\\\n \\bmat{x\\\\1}\\tp \\bar{P}_i \\bmat{x\\\\1}, &x \\in X_i, i \\in I_1\n \\end{cases}\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\\label{eq:pwa-lyapunov-matrix}\n P_i = F_i^T T F_i,\\quad \\bar{P}_i = \\bar{F}_i^T T \\bar{F}_i\n\\end{equation}\n%\nand $T$ is some symmetric matrix with positive elements.\n\n\\subsubsection{Relaxing the LF}\nWhile $V(x) > 0$ is required for all $x$, we only need $P_i > 0$ for $x \\in X_i$. The S-procedure can be used to alter the LF to reflect this:\n%\n\\begin{equation}\n V(x) =\n \\begin{cases}\n x\\tp E_i\\tp U_i E_i x, &x \\in X_i, i \\in I_0 \\\\\n \\begin{bmatrix} x \\\\ 1 \\end{bmatrix}\\tp\n \\bar{E}_i\\tp U_i \\bar{E}_i\\tp\n \\begin{bmatrix} x \\\\ 1 \\end{bmatrix}, &x \\in X_i, i \\in I_1\n \\end{cases}\n\\end{equation}\n%\nwhere $U_i$ has only non-negative elements. The condition $\\dot{V}(x)<0$ only needs to hold within each partition, as well.\n\n\\subsubsection{Overall LF design}\nFind symmetric matrices $T$, $U_i$, $W_i$ ($U_i$ and $W_i$ with no negative elements), such that\n%\n\\begin{equation}\n \\begin{split}\n \\begin{rcases}\n A_i\\tp P_i + P_i A_i + E_i\\tp U_i E_i < 0 \\\\\n P_i - E_i\\tp W_i E_i > 0\n \\end{rcases} &i \\in I_0 \\\\\n \\begin{rcases}\n \\bar{A}_i\\tp \\bar{P}_i + \\bar{P}_i \\bar{A}_i + \\bar{E}_i\\tp U_i \\bar{E}_i < 0 \\\\\n \\bar{P}_i - \\bar{E}_i\\tp W_i \\bar{E}_i > 0\n \\end{rcases} &i \\in I_1\n \\end{split}\n\\end{equation}\n%\nwith \\eqref{eq:pwa-lyapunov-matrix} still holding:\n%\n\\begin{equation}\n P_i = F_i^T T F_i,\\quad \\bar{P}_i = \\bar{F}_i^T T \\bar{F}_i\n\\end{equation}\n\n$\\bar{E}_i$ and $\\bar{F}_i$ can be determined e.g. with explicit formulas found in papers by Johansson and Ratzer.\n\n\\subsection{Discrete-time Lyapunov stability}\nWith discrete systems\n%\n\\begin{itemize}\n \\item we don't need a continuous LF,\n \\item we use LF difference rather than LF differential,\n \\item we must keep track of possible next-timestep partitions.\n\\end{itemize}\n\nIf there exists matrices $R_i > 0$ such that\n%\n\\begin{equation}\n \\sum_{i \\in I} (A_i\\tp R_i A_i - R_I) > 0\n\\end{equation}\n%\nthen no common LF can be found for the PWL system.\n\n\\subsubsection{Stability criteria}\n$f(X_i)$ is a quadratic function positive on $X_i$, such as\n%\n\\begin{equation}\n f(X_i) =\n \\begin{cases}\n x\\tp E_i\\tp U_i E_i x, & x \\in I_0 \\\\\n \\bsmat{x \\\\ 1}\\tp \\bar{E}_i\\tp U_i \\bar{E}_i \\bsmat{x \\\\ 1}, & x \\in I_1\n \\end{cases}\n\\end{equation}\n\nRewrite the system\n%\n\\begin{equation}\n \\bar{x} = \\bmat{x \\\\ 1}, \\quad \\bar{x}_{k+1} = \\bar{A}_i \\bar{x}_k\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\n \\bar{A}_i =\n \\begin{cases}\n \\bmat{A_i & a_i \\\\ 0 & 1},&\n \\underbrace{x_k \\in X_i, i \\in I_1, x_{k+1} \\in X_j, j \\in I_1}\n _{x \\text{ moving between } I_1 \\text{ partitions}} \\\\\n \\bmat{A_i & a_i \\\\ 0 & 0},&\n \\underbrace{x_k \\in X_i, i \\in I_1, x_{k+1} \\in X_j, j \\in I_0}\n _{x \\text{ moving from } I_1 \\text{ to } I_0 \\text{ partition}} \\\\\n \\bmat{A_i & 0 \\\\ 0 & 0},&\n \\underbrace{x_k \\in X_i, i \\in I_0, x_{k+1} \\in X_j, j \\in I_0}\n _{x \\text{ moving between } I_0 \\text{ partitions}}.\n \\end{cases}\n\\end{equation}\n%\nThis formulation requires that $\\cup_{i \\in I_0} X_i$ is positively invariant.\\footnote{All trajectories in the set never leave it.} The system is then stable if\n%\n\\begin{equation}\n \\begin{split}\n P_i - f(X_i) > 0,\\quad & \\forall i \\\\\n P_i - \\bar{A}_i\\tp P_j \\bar{A}_i - f(X_{ij}) > 0,\\quad & \\forall (i,j) \\mbox{ with nonempty } X_{ij}\n \\end{split}\n\\end{equation}\n%\nwhere $X_{ij} \\subset X_i$ is the region where the state in the next timestep will move to $X_j$.\n\n\\subsubsection{Relaxations}\n$R$ is a bounded polyhedron and $V(R)$ is its vertice set. Then this function is quadratic and positive over $R$:\n\\begin{equation}\n f(R) = \\bsmat{x \\\\ 1}\\tp H \\bsmat{x \\\\ 1}\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\n \\begin{split}\n H &= \\bar{H} + C \\\\\n \\bar{H} &< 0 \\\\\n C &= \\bsmat{0 & 0 \\\\ 0 & c} \\\\\n c &> 0 \\\\\n \\bsmat{v_i \\\\ 1}\\tp (\\bar{H} + C) \\bsmat{v_i \\\\ 1} &> 0,\\quad \\forall v_i \\in V(R)\n \\end{split}\n\\end{equation}\n", "meta": {"hexsha": "3e11d3963860e45bf96c6a3fe419932dfdd5efb6", "size": 6483, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TTK18 Optimaliseringsbasert reguleringsdesign og analyse/tex/sec-lmis-for-pwa.tex", "max_stars_repo_name": "jakoblover/ntnu-course-summaries", "max_stars_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-05-30T09:19:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-11T02:42:40.000Z", "max_issues_repo_path": "TTK18 Optimaliseringsbasert reguleringsdesign og analyse/tex/sec-lmis-for-pwa.tex", "max_issues_repo_name": "jakoblover/ntnu-course-summaries", "max_issues_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "TTK18 Optimaliseringsbasert reguleringsdesign og analyse/tex/sec-lmis-for-pwa.tex", "max_forks_repo_name": "jakoblover/ntnu-course-summaries", "max_forks_repo_head_hexsha": "8ba859de2349b93c5079ca10a4cf2ec49c1f5dc0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.4708737864, "max_line_length": 161, "alphanum_fraction": 0.6399814901, "num_tokens": 2480, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789040926009, "lm_q2_score": 0.7745833841649232, "lm_q1q2_score": 0.627783492326325}} {"text": "\\documentstyle[11pt,reduce]{article}\n\\title{{\\tt TRIGSIMP}\\\\\nA REDUCE Package for the Simplification and Factorization of Trigonometric\nand Hyperbolic Functions}\n\\date{}\n\\author{Wolfram Koepf\\\\\n Andreas Bernig\\\\\n Herbert Melenk\\\\\n ZIB Berlin \\\\\n email: {\\tt Koepf@ZIB-Berlin.de}}\n\\begin{document}\n\\maketitle\n\\section{Introduction}\n\nThe REDUCE package TRIGSIMP is a useful tool for all kinds of trigonometric and\nhyperbolic simplification and factorization. There are three\nprocedures included in TRIGSIMP: trigsimp, trigfactorize and triggcd. \nThe first is for finding simplifications of trigonometric or \nhyperbolic expressions with many options, the second for factorizing \nthem and the third \nfor finding the greatest common divisor of two trigonometric or \nhyperbolic polynomials.\n\nTo start the package it must be loaded by:\n{\\small\n\\begin{verbatim}\n1: load trigsimp;\n\\end{verbatim}\n}\\noindent\n\n\\section{\\REDUCE{} operator {\\tt trigsimp}}\n\nAs there is no normal form for trigonometric and hyperbolic functions, the same \nfunction can convert in many different directions, e.g. \n$\\sin(2x) \\leftrightarrow 2\\sin(x)\\cos(x)$. \nThe user has the possibility to give several parameters to the \nprocedure {\\tt trigsimp} in order to influence the direction of transformations. \nThe decision whether a rational expression in trigonometric \nand hyperbolic functions vanishes or not is possible.\n\nTo simplify a function {\\tt f}, one uses {\\tt trigsimp(f[,options])}. Example:\n{\\small\n\\begin{verbatim}\n2: trigsimp(sin(x)^2+cos(x)^2);\n\n1\n\\end{verbatim}\n}\\noindent\n\nPossible options are (* denotes the default):\n\\begin{enumerate}\n\\item {\\tt sin} (*) or {\\tt cos}\n\\item {\\tt sinh} (*) or {\\tt cosh}\n\\item {\\tt expand} (*) or {\\tt combine} or {\\tt compact}\n\\item {\\tt hyp} or {\\tt trig} or {\\tt expon}\n\\item {\\tt keepalltrig}\n\\end{enumerate}\n\nFrom each group one can use at most one option, otherwise an error \nmessage will occur. The first group fixes the preference used while \ntransforming a trigonometric expression:\n{\\small\n\\begin{verbatim}\n3: trigsimp(sin(x)^2);\n\n 2\nsin(x)\n\n4: trigsimp(sin(x)^2,cos);\n\n 2\n - cos(x) + 1\n\\end{verbatim}\n}\\noindent\n\nThe second group is the equivalent for the hyperbolic functions. \nThe third group determines the type of transformations. With \nthe default {\\tt expand}, an expression is written in a form only using \nsingle arguments and no sums of arguments:\n{\\small\n\\begin{verbatim}\n5: trigsimp(sin(2x+y));\n\n 2\n2*cos(x)*cos(y)*sin(x) - 2*sin(x) *sin(y) + sin(y)\n\\end{verbatim}\n}\\noindent\n\nWith {\\tt combine}, products of trigonometric functions are transformed to \ntrigonometric functions involving sums of arguments:\n{\\small\n\\begin{verbatim}\n6: trigsimp(sin(x)*cos(y),combine);\n\n \n sin(x - y) + sin(x + y)\n-------------------------\n 2\n\\end{verbatim}\n}\\noindent\n\nWith {\\tt compact}, the REDUCE operator {\\tt compact} \\cite{hearns}\nis applied to {\\tt f}. \nThis leads often to a simple form, but in contrast to {\\tt expand} one \ndoesn't get a normal form. Example for {\\tt compact}:\n{\\small\n\\begin{verbatim}\n7: trigsimp((1-sin(x)**2)**20*(1-cos(x)**2)**20,compact);\n\n 40 40\ncos(x) *sin(x)\n\\end{verbatim}\n}\\noindent\n\nWith the fourth group each expression is transformed to a\ntrigonometric, hyperbolic or exponential form:\n{\\small\n\\begin{verbatim}\n8: trigsimp(sin(x),hyp);\n\n - sinh(i*x)*i\n\n9: trigsimp(sinh(x),expon);\n \n 2*x\n e - 1\n----------\n x\n 2*e\n\n10: trigsimp(e^x,trig);\n\n x x\ncos(---) + sin(---)*i\n i i\n\\end{verbatim}\n}\\noindent \n\nUsually, {\\tt tan}, {\\tt cot}, {\\tt sec}, {\\tt csc} are expressed in terms of\n{\\tt sin} and {\\tt cos}. It can \nbe sometimes useful to avoid this, which is handled by the option \n{\\tt keepalltrig}:\n{\\small\n\\begin{verbatim}\n11: trigsimp(tan(x+y),keepalltrig);\n\n - (tan(x) + tan(y))\n----------------------\n tan(x)*tan(y) - 1\n\\end{verbatim}\n}\\noindent\n\nIt is possible to use the options of different groups simultaneously:\n{\\small\n\\begin{verbatim}\n12: trigsimp(sin(x)**4,cos,combine);\n\n cos(4*x) - 4*cos(2*x) + 3\n---------------------------\n 8\n\\end{verbatim}\n}\\noindent\n\nSometimes, it is necessary to handle an expression in different steps:\n{\\small\n\\begin{verbatim}\n13: trigsimp((sinh(x)+cosh(x))**n+(cosh(x)-sinh(x))**n,expon);\n\n 2*n*x\n e + 1\n------------\n n*x\n e \n \n14: trigsimp(ws,hyp);\n\n2*cosh(n*x)\n\n15: trigsimp((cosh(a*n)*sinh(a)*sinh(p)+cosh(a)*sinh(a*n)*sinh(p)+ \n sinh(a - p)*sinh(a*n))/sinh(a));\n\ncosh(a*n)*sinh(p) + cosh(p)*sinh(a*n)\n\n16: trigsimp(ws,combine);\n\nsinh(a*n + p)\n\\end{verbatim}\n}\\noindent\n\n\\section{\\REDUCE{} operator {\\tt trigfactorize}}\n\nWith {\\tt trigfactorize(p,x)} one can factorize the trigonometric or \nhyperbolic polynomial {\\tt p} with respect to the argument x. Example:\n{\\small\n\\begin{verbatim}\n17: trigfactorize(sin(x),x/2); \n\n x x\n{2,cos(---),sin(---)}\n 2 2\n\\end{verbatim}\n}\\noindent\n\nIf the polynomial is not coordinated or balanced \\cite{art}, \nthe output will equal the input.\nIn this case, changing the value for x can help to find a factorization:\n{\\small\n\\begin{verbatim}\n18: trigfactorize(1+cos(x),x);\n\n{cos(x) + 1}\n\n19: trigfactorize(1+cos(x),x/2); \n\n x x\n{2,cos(---),cos(---)}\n 2 2\n\\end{verbatim}\n}\\noindent\n\nThe polynomial can consist of both trigonometric and hyperbolic functions:\n{\\small\n\\begin{verbatim}\n20: trigfactorize(sin(2x)*sinh(2x),x);\n\n{4, cos(x), sin(x), cosh(x), sinh(x)}\n\\end{verbatim}\n}\\noindent\n\n\\section{\\REDUCE{} operator {\\tt triggcd}}\n\nThe operator {\\tt triggcd} is an application of {\\tt trigfactorize}. \nWith its help the user can find the greatest common divisor of two \ntrigonometric or hyperbolic polynomials. It uses the method described \nin \\cite{art}. The syntax is: {\\tt triggcd(p,q,x)}, where p and q \nare the polynomials and x is the smallest unit to use. Example:\n\n{\\small\n\\begin{verbatim}\n21: triggcd(sin(x),1+cos(x),x/2);\n\n x\ncos(---)\n 2 \n\n22: triggcd(sin(x),1+cos(x),x);\n\n1\n\\end{verbatim}\n}\\noindent\n\nThe polynomials p and q can consist of both trigonometric and hyperbolic \nfunctions:\n{\\small\n\\begin{verbatim}\n23: triggcd(sin(2x)*sinh(2x),(1-cos(2x))*(1+cosh(2x)),x);\n\ncosh(x)*sin(x)\n\\end{verbatim}\n}\\noindent\n\n\n\\section{Further Examples}\n\nWith the help of the package the user can create identities:\n{\\small\n\\begin{verbatim}\n24: trigsimp(tan(x)*tan(y));\n\n sin(x)*sin(y)\n---------------\n cos(x)*cos(y)\n\n25: trigsimp(ws,combine);\n\n cos(x - y) - cos(x + y)\n-------------------------\n cos(x - y) + cos(x + y)\n\n26: trigsimp((sin(x-a)+sin(x+a))/(cos(x-a)+cos(x+a)));\n\n sin(x)\n--------\n cos(x)\n\n27: trigsimp(cosh(n*acosh(x))-cos(n*acos(x)),trig);\n\n0\n\n28: trigsimp(sec(a-b),keepalltrig);\n\n csc(a)*csc(b)*sec(a)*sec(b)\n-------------------------------\n csc(a)*csc(b) + sec(a)*sec(b)\n\n29: trigsimp(tan(a+b),keepalltrig);\n\n - (tan(a) + tan(b))\n----------------------\n tan(a)*tan(b) - 1\n\n30: trigsimp(ws,keepalltrig,combine);\n\ntan(a + b)\n\\end{verbatim}\n}\\noindent\n\nSome difficult expressions can be simplified:\n{\\small\n\\begin{verbatim}\n\n31: df(sqrt(1+cos(x)),x,4);\n\n 4 2 2 2\n(sqrt(cos(x) + 1)*( - 4*cos(x) - 20*cos(x) *sin(x) + 12*cos(x)\n\n 2 4 2\n - 4*cos(x)*sin(x) + 8*cos(x) - 15*sin(x) + 16*sin(x) ))/(16\n\n 4 3 2\n *(cos(x) + 4*cos(x) + 6*cos(x) + 4*cos(x) + 1))\n\n32: trigsimp(ws);\n\n sqrt(cos(x) + 1)\n------------------\n 16\n\n33: load taylor; \n\n34: taylor(sin(x+a)*cos(x+b),x,0,4);\n\ncos(b)*sin(a) + (cos(a)*cos(b) - sin(a)*sin(b))*x\n\n 2\n - (cos(a)*sin(b) + cos(b)*sin(a))*x\n\n 2*( - cos(a)*cos(b) + sin(a)*sin(b)) 3\n + --------------------------------------*x\n 3\n\n cos(a)*sin(b) + cos(b)*sin(a) 4 5\n + -------------------------------*x + O(x )\n 3\n\n35: trigsimp(ws,combine);\n\nsin(a - b) + sin(a + b) 2 2*cos(a + b) 3\n------------------------- + cos(a + b)*x - sin(a + b)*x - --------------*x\n 2 3\n\n sin(a + b) 4 5\n + ------------*x + O(x )\n 3\n\\end{verbatim}\n}\\noindent\nCertain integrals whose calculation was not possible in REDUCE \n(without preprocessing), are now computable:\n{\\small\n\\begin{verbatim}\n36: int(trigsimp(sin(x+y)*cos(x-y)*tan(x)),x);\n\n 2 2\n cos(x) *x - cos(x)*sin(x) - 2*cos(y)*log(cos(x))*sin(y) + sin(x) *x\n---------------------------------------------------------------------\n 2\n\n37: int(trigsimp(sin(x+y)*cos(x-y)/tan(x)),x);\n\n x 2\n(cos(x)*sin(x) - 2*cos(y)*log(tan(---) + 1)*sin(y)\n 2\n\n x\n + 2*cos(y)*log(tan(---))*sin(y) + x)/2\n 2\n\\end{verbatim}\n}\\noindent\n\nWithout the package, the integration fails, in the second case one doesn't \nreceive an answer for many hours.\n{\\small\n\\begin{verbatim}\n\n38: trigfactorize(sin(2x)*cos(y)**2,y/2);\n\n{2*cos(x)*sin(x),\n\n y y\n cos(---) + sin(---),\n 2 2\n\n y y\n cos(---) + sin(---),\n 2 2\n\n y y\n cos(---) - sin(---),\n 2 2\n\n y y\n cos(---) - sin(---)}\n 2 2\n\n39: trigfactorize(sin(y)**4-x**2,y);\n\n 2 2\n{ - sin(y) + x, - (sin(y) + x)}\n\n40: trigfactorize(sin(x)*sinh(x),x/2);\n\n x x x x\n{4,cos(---),sin(---),cosh(---),sinh(---)}\n 2 2 2 2\n\n41: triggcd(-5+cos(2x)-6sin(x),-7+cos(2x)-8sin(x),x/2);\n\n x x\n2*cos(---)*sin(---) + 1\n 2 2\n\n42: triggcd(1-2cosh(x)+cosh(2x),1+2cosh(x)+cosh(2x),x/2);\n\n x 2\n2*sinh(---) + 1\n 2\n\\end{verbatim}\n}\n\n\\begin{thebibliography}{99}\n\n\\bibitem{art}\nRoach, Kelly: Difficulties with Trigonometrics. Notes of a talk.\n\n\\bibitem{hearns}\nHearn, A.C.: COMPACT User Manual. \n\\end{thebibliography}\n\\end{document}\n", "meta": {"hexsha": "53e25803c42c00592d2fdac88b7848fb5056161b", "size": 10163, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "packages/trigsimp/otrgsimp.tex", "max_stars_repo_name": "arthurcnorman/general", "max_stars_repo_head_hexsha": "5e8fef0cc7999fa8ab75d8fdf79ad5488047282b", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "packages/trigsimp/otrgsimp.tex", "max_issues_repo_name": "arthurcnorman/general", "max_issues_repo_head_hexsha": "5e8fef0cc7999fa8ab75d8fdf79ad5488047282b", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "packages/trigsimp/otrgsimp.tex", "max_forks_repo_name": "arthurcnorman/general", "max_forks_repo_head_hexsha": "5e8fef0cc7999fa8ab75d8fdf79ad5488047282b", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.1503416856, "max_line_length": 81, "alphanum_fraction": 0.5682377251, "num_tokens": 3174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104788995148792, "lm_q2_score": 0.7745833789613197, "lm_q1q2_score": 0.627783484563087}} {"text": "\\documentclass[jou]{apa6}\n\n\\usepackage[american]{babel}\n\n\\usepackage{csquotes}\n\\usepackage[style=apa,sortcites=true,sorting=nyt,backend=biber]{biblatex}\n\\DeclareLanguageMapping{american}{american-apa}\n\\addbibresource{bibliography.bib}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%% Discrete Structures\n%% The start of RBS stuff\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Working internal and external links in PDF\n\\usepackage{hyperref}\n% Extra math symbols in LaTeX\n\\usepackage{amsmath}\n\\usepackage{gensymb}\n\\usepackage{amssymb}\n% Enumerations with (a), (b), etc.\n\\usepackage{enumerate}\n\n\\let\\OLDitemize\\itemize\n\\renewcommand\\itemize{\\OLDitemize\\addtolength{\\itemsep}{-6pt}}\n\n\\usepackage{etoolbox}\n\\makeatletter\n\\preto{\\@verbatim}{\\topsep=3pt \\partopsep=3pt }\n\\makeatother\n\n% These sizes redefine APA for A4 paper size\n\\oddsidemargin 0.0in\n\\evensidemargin 0.0in\n\\textwidth 6.27in\n\\headheight 1.0in\n\\topmargin -24pt\n\\headheight 12pt\n\\headsep 12pt\n\\textheight 9.19in\n\n\n\n\\title{Sample Quiz 4}\n\\author{Discrete Structures, Fall 2020}\n\\affiliation{RBS}\n\n\\leftheader{Discrete Sample Quiz 4}\n\n\\abstract{%\n}\n\n%\\keywords{}\n\n\\begin{document}\n\n%\\thispagestyle{empty}\n\n\\section{Quiz 4: Sets}\n\n\\vspace{10pt}\n{\\bf Question 1.} Define the universe $U$ to be all possible remainders \nwhen we divide by $360$: $\\{ 0, 1, 2, \\ldots, 359 \\}$. \nAlso define $3$ subsets in this universe: \n$$\\left\\{ \\begin{array}{rcl}\nK_2 & = & \\{ x \\in U \\,\\mid\\, x\\;\\text{divisible by}\\;2 \\},\\\\\nK_3 & = & \\{ x \\in U \\,\\mid\\, x\\;\\text{divisible by}\\;3 \\},\\\\\nK_5 & = & \\{ x \\in U \\,\\mid\\, x\\;\\text{divisible by}\\;5 \\},\\\\\n\\end{array} \\right.$$\n\nDenote by $\\Phi$ the subset of $U$ containing all numbers\nthat are mutually prime with $360$ (no common divisors greater than $1$):\n$\\Phi = \\{1,7,11,13,\\ldots,359\\}$.\nWhich set equality is valid regarding the subset $\\Phi$:\n\n\\noindent\n{\\bf (A)} $\\Phi = \\left( K_2 \\cup K_3 \\cup K_5 \\right)$\\\\\n{\\bf (B)} $\\Phi = \\left( K_2 \\cap K_3 \\cap K_5 \\right)$\\\\\n{\\bf (C)} $\\Phi = \\left( \\overline{K_2} \\cup \\overline{K_3} \\cup \\overline{K_5} \\right)$\\\\\n{\\bf (D)} $\\Phi = \\left( \\overline{K_2} \\cap \\overline{K_3} \\cap \\overline{K_5} \\right)$\\\\\n{\\bf (E)} $\\Phi = \\left( \\overline{K_2 \\cap K_3} \\cup \\overline{K_2 \\cap K_5} \\cup \\overline{K_3 \\cap K_5} \\right)$\n\n\nPick your answer as a single letter like this: {\\tt G}\n\n\\vspace{6pt}\n{\\bf Question 2.}\nFind the size of the set you constructed in the previous example. \n\nWrite your answer as a single non-negative integer like this: {\\tt 17}\n\n\n\\vspace{6pt}\n{\\bf Question 3.} We have the following sets:\\\\\n$A$ is the set of all finite sequences of even positive positive numbers (such as $(6,22,10,14,2,6)$, and so on)\\\\\n$B$ is the set of all infinite nondecreasing lists of even positive numbers (such as $(40 \\leq 40 \\leq 42 \\leq 46 \\leq \\ldots)$, and so on)\\\\\n$C$ is the set of all infinite nonincreasing lists of even positive numbers (such as $(64 \\geq 58 \\geq 58 \\geq 54 \\geq \\ldots)$, and so on).\\\\\nClearly, all three sets are infinite. Determine their cardinalities - which list \nof cardinalities is equal to the list $(|A|,|B|,|C|)$?\n\n\n\\noindent\n{\\bf (A)} $\\left( \\left| \\mathbb{N} \\right|, \\left| \\mathbb{N} \\right|, \\left| \\mathbb{N} \\right| \\right)$. \n{\\bf (B)} $\\left( \\left| \\mathbb{N} \\right|, \\left| \\mathbb{R} \\right|, \\left| \\mathbb{N} \\right| \\right)$.\n{\\bf (C)} $\\left( \\left| \\mathbb{N} \\right|, \\left| \\mathbb{N} \\right|, \\left| \\mathbb{R} \\right| \\right)$.\n{\\bf (D)} $\\left( \\left| \\mathbb{N} \\right|, \\left| \\mathbb{R} \\right|, \\left| \\mathbb{R} \\right| \\right)$.\n{\\bf (E)} $\\left( \\left| \\mathbb{R} \\right|, \\left| \\mathbb{R} \\right|, \\left| \\mathbb{R} \\right| \\right)$.\n\n\nPick your answer as a single letter like this: {\\tt G}\n\n\\vspace{6pt}\n{\\bf Question 4.}\nLet ${\\displaystyle f(x) = (x^2)\\;\\mathbf{mod}\\;11}$. Find the set $f(S)$ if $S = \\{ 0,1,2,3,4,5,6,7,8,9,10 \\}$. \n\nWrite the list of elements of $f(S)$ as a sorted list like this: {\\tt 1,2,3}\n\n\n\n\n\\vspace{6pt}\n{\\bf Question 5.}\nHow many 2-element sets are there in the powerset $\\mathcal{P}\\left( \\{ \\{ \\mathtt{A}, \\mathtt{B} \\}, \\mathtt{C}, \\mathtt{D}, \\mathtt{E} \\} \\right)$? \n\nWrite your answer as a non-negative integer like this: {\\tt 17}\n\n\n\\vspace{6pt}\n{\\bf Question 6.}\nGiven two sets $A = \\{ x, y \\}$ and $B = \\{x, \\{x \\}\\}$, check, if statements are true or false:\\\\\n{\\bf (A)} $x \\subseteq B$.\\\\\n{\\bf (B)} $\\emptyset \\in \\mathcal{P}(B)$.\\\\\n{\\bf (C)} $\\{x\\} \\subseteq A - B$.\\\\\n{\\bf (D)} $|\\mathcal{P}(A)| = 4$.\n\nWrite your answer as a sorted list of letters (which are true) like this: {\\tt A,B,C,D}\n\n\\vspace{6pt}\n{\\bf Question 7.}\nWe define functions $g\\,:\\, A \\rightarrow A$ and $f : A \\rightarrow A$, where $A \\{1, 2, 3, 4\\}$ by \nlisting all argument-value pairs: \n$f = \\{(1, 2), (2, 3), (3, 4), (4, 1)\\}$, $g = \\{(1, 3), (2, 1), (3, 4), (4, 2)\\}$.\nFind the value pairs for the function $(f \\circ g)^{-1}$. \n\nWrite your answer as a comma-separated list like this: {\\tt (1,1),(2,2),(3,3),(4,4)}\n\n\n\\vspace{6pt}\n{\\bf Question 8.} Find the value of this infinite sum:\n$1 - 1/3 + 1/9 - 1/27 + 1/81 - \\ldots$. \n\nWrite your answer as a simple fraction: {\\tt P/Q}\n\n\n\\vspace{6pt}\n{\\bf Question 9.} It is known that the function $f(n) = n^3 +88n^2 +3$ is in $O(n^3)$ \\textendash{}\nits asymptotic growth is as fast as the growth of the function $g(n) = n^3$. \n$\\exists C \\in \\mathbb{Z}^{+}\\;\\exists n_0 \\in \\mathbb{Z}^{+}\\;\\forall n \\in \\mathbb{Z}^{+},$\\\\\n$(n > n_0 \\rightarrow |f(n)| \\leq C\\cdot{}|g(n)|)$\nFind the smallest positive integer $C$ that would satisfy the above definition, \nand for your $C$ find the smallest possible $n_0$.\n\nWrite your answer $(C,n_0)$ as a pair of two numbers like this: {\\tt 17,17}\n\n\n\\vspace{6pt}\n{\\bf Question 10.}\n\"Big O notation\" allows to arrange functions according to the their\ngrowth rate for large $n$. \nIdentify, which list of functions is such that \nthe first element of this list is in the big-O of the \nnext element of that list and so on. (Intuitively, the first element\nin the list is the slowest growing function, the last element is the\nfastest growing one.)\n\n{\\bf (1)} $\\log (n^{10})$, {\\bf (2)} $(\\log n)^2$, {\\bf (3)} $\\log \\log n$,\n{\\bf (4)} $n\\log n$, {\\bf (5)} $\\log(n!)$, {\\bf (6)} $\\log 2^n$.\n\nWrite your answer as a comma-separated list like this: {\\tt 1,2,3,4,5,6}\n\n\n\\vspace{6pt}\n{\\bf Question 11.} Digits of all rational numbers $P/Q$ in $(0;1)$\nare eventually periodic: they infinitely repeat some group of digits (the period)\nstarting from some place. For example,\nthe fraction $11/205 = 0.05(36585)$ has period of 5 digits and a pre-period \"05\"\nof just two digits.\nFind the predicate logic expression that tells\nthat sequence of digits $d(1),d(2),d(3),\\ldots$ is eventually periodic (it may have\npre-period of any length, including length zero).\n\n\\noindent\n{\\bf (A)} $\\exists N \\in \\mathbb{Z}^{+}\\;\\exists T \\in \\mathbb{Z}^{+}\\;\\forall n \\in \\mathbb{Z}^{+},$\\\\\n$\\left(n \\geq N - 1 \\rightarrow d(n) = d(n+T)\\right)$.\\\\\n{\\bf (B)} $\\exists N \\in \\mathbb{Z}^{+}\\;\\forall n \\in \\mathbb{Z}^{+}\\;\\exists T \\in \\mathbb{Z}^{+},$\\\\\n$\\left(n \\geq N - 1 \\rightarrow d(n) = d(n+T)\\right)$.\\\\\n{\\bf (C)} $\\forall n \\in \\mathbb{Z}^{+}\\;\\exists N \\in \\mathbb{Z}^{+}\\;\\exists T \\in \\mathbb{Z}^{+},$\\\\\n$\\left(n \\geq N - 1 \\rightarrow d(n) = d(n+T)\\right)$.\\\\\n{\\bf (D)} $\\forall n \\in \\mathbb{Z}^{+}\\;\\forall N \\in \\mathbb{Z}^{+}\\;\\exists T \\in \\mathbb{Z}^{+},$\\\\\n$\\left(n \\geq N - 1 \\rightarrow d(n) = d(n+T)\\right)$. \n\nPick your answer as a single letter like this: {\\tt G}\n\n\n\n\n\\newpage\n\n\\subsection{Answers}\n\n\\vspace{6pt}\n{\\bf Question 1.} Answer {\\bf (D)}.\\\\\nAny number that is mutual prime with $360 = 2^3\\cdot{}3^2\\cdot{}5$\nis not divisible by any of the primes $2,3,5$. And also vice versa. \nThis set is expressed as intersection of the complements:\\\\\n$\\Phi = \\left( \\overline{K_2} \\cap \\overline{K_3} \\cap \\overline{K_5} \\right)$\n\n\n\\vspace{6pt}\n{\\bf Question 2.} Answer: {\\tt 96}.\\\\\n$${\\displaystyle |\\Phi| = 360 \\cdot \\left(1 - \\frac{1}{2}\\right) \\cdot \\left(1 - \\frac{1}{3}\\right) \\cdot \\left(1 - \\frac{1}{5}\\right) =}$$\n$${\\displaystyle = 360 \\cdot \\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{4}{5} = 96}.$$\nIn the above formula we start with all $360$ elements; then we throw\nout one half (all that are divisible by 2); then from the remaining ones \nwe throw out one third (all that are divisible by 3); finally from the remaining numbers we throw\nout one fifth (all that are divisible by 5). Since divisibility by $2$ does not \naffect divisibility by $3$ and $5$ (they are independent), \nall the ratios can be multiplied.\n\nAnother solution: Since we know the sizes of each \nset of numbers divisible by $2,3,5$:\n$$|K_2| = 180,\\; |K_3| = 120,\\; |K_5| = 72.$$\nWe can express their union by {\\em inclusion-exclusion principle}:\n$$|K_2 \\cup K_3 \\cup K_5| \\;=\\; |K_2| \\;+\\; |K_3| \\;+\\; |K_5| \\;-$$\n$$-\\;|K_2 \\cap K_3|\\;-\\;|K_2 \\cap K_5|\\;-\\;|K_3 \\cap K_5| \\;+\\; |K_2 \\cap K_3 \\cap K_5| \\;=$$\n$$=\\; 180 + 120 + 72 - 60 - 36 - 24 +12 = 264.$$\nWe then apply De Morgan's law to find the count of all elements that are {\\em outside}\nthat union of $K_2 \\cup K_3 \\cup K_5$: \n$$ \\left| \\overline{K_2} \\cap \\overline{K_3} \\cap \\overline{K_5} \\right| = \n\\left| \\overline{K_2 \\cup K_3 \\cup K_5} \\right| =360 - 264 = 96.$$ \n\n\\vspace{6pt}\n{\\bf Question 3.} Answer: {\\tt B}.\n\\begin{itemize} \n\\item $A$ (the set of all finite sequences of even natural numbers can be enumerated\nwith numbers from $\\mathbb{N}$). You can encode \nevery such sequence in a finite alphabet of $13$ symbols, \nusing just digits, commas and parentheses. For example, {\\tt (6, 22, 10, 14, 2, 6)}. \nThe shortest encoding is {\\tt (1)} \\textendash it consists of just three symbols: \ntwo parentheses and a digit. There can be only finite number of such lists of \nlength $3$; we sort the all lexicographically (i.e. in some alphabetical order), and\nassign them numbers.\\\\\nAfter that we enumerate all lists writeable with 4 symbols (sorted lexicographically) \nand so on. Eventually all the sequences will be sorted.\n\\item $B$ (the set of all infinite nondecreasing sequences of even numbers) \nhas cardinality $\\mathbb{R}$. You can repeat the diagonalization argument: \nAssume from the contrary that the elements from $B$ can be enumerated: we get\ninfinitely many infinite sequences $b_1,b_2,\\ldots$. \nThen take the first element from $b_1$ (and pick some even number that is bigger than that); \nthen take the second element from $b_2$ (and pick some even number that is bigger than \nthat; plus it is bigger than all the previously picked numbers), and so on.\\\\\nYou can also encode any subset $A \\subseteq \\mathbb{N}$ as such sequence (simply arrange all \nthe elements in increasing order and multiply them by $2$ to get even numbers). \nWe get that $B$ has at least as many elements as $\\mathcal{P}(\\mathbb{N})$. \n\\item $C$ (the set of all nonincreasing infinite sequences can be \nenumerated). Since the sequence is non-increasing, it can have only finitely many \nplaces where it actually decreases; since natural numbers cannot decrease infinitely. \nWe can encode all the ``constant runs'' of the sequence as pairs:\\\\\n$$(64,58,58,54,50,50,50,50,2,2,\\ldots) \\rightarrow $$\n$$\\rightarrow \\mathtt{((64,1),(58,2),(54,1),(50,4),(2,\\infty))}.$$\nAs we saw before, all the finite sequences that are encoded in an alfabet\nof $14$ symbols (10 digits, 2 parentheses, commas and infinity) can be enumerated.\n\\end{itemize}\n\n\n\\vspace{6pt}\n{\\bf Question 4.} Answer: {\\tt 0,1,3,4,5,9}.\\\\\nWe can square each number, compute the remainder and sort the \nresults (and eliminate duplicates).\n\n\n\\vspace{6pt}\n{\\bf Question 5.} Answer: {\\tt 6}.\\\\\nThe set $\\{ \\{ \\mathtt{A}, \\mathtt{B} \\}, \\mathtt{C}, \\mathtt{D}, \\mathtt{E} \\}$\nhas $4$ elements ({\\tt A}, {\\tt B} are always glued together). \nThere are $6$ ways to select two out of four elements. \n(Can be computed as a binomial coefficient $C_4^2 = \\frac{4!}{2!2!}$ or simply \nby listing all the $6$ pairs. \n\n\n\\vspace{6pt}\n{\\bf Question 6.} Answer: {\\tt B,D}.\\\\\n$x$ cannot be a subset of $A$ (since it is not a set itself). \n$\\{ x \\}$ is not a subset of $A - B = \\{ y \\}$.\n\n\\vspace{6pt}\n{\\bf Question 7.}\\\\ Answer: {\\tt (1,3),(2,2),(3,4),(4,1)}.\\\\\nWe first compute $f \\circ g$ (to get $(f \\circ g)(x) = f(g(x))$\nwe first apply $g$, then $f$): \n$$f \\circ g = \\{(1,4),(2,2),(3,1),(4,3)\\}.$$\nThe inverse happens, if we switch the order in all these pairs\n($4$ maps back to $1$ etc.)\\\\\n$$(f \\circ g)^{-1} = \\{(1,3),(2,2),(3,4),(4,1)\\}.$$\n\n\n\\vspace{6pt}\n{\\bf Question 8.} Answer: {\\tt 3/4}.\\\\\nThe sum of the infinite geometrical progression is $b_1/(1 - q)$. \nIn our case:\n$$\\frac{1}{1 - (-1/3)} = \\frac{1}{4/3} = \\frac{3}{4}.$$\n\n\\vspace{6pt}\n{\\bf Question 9.} Answer: {\\tt 2,88}. \nClearly, $f(n) = |n^3 +88n^2 +3|$ cannot be smaller than $C\\cdot{}|n^3|$, \nif $C=1$, because $88n^2$ is always positive and makes $f(n)$ larger\nthan simply $n^3$.\\\\\nIf we take $C = 2$, then the inequality starts to hold for all $n>88$. \nIt is possible to prove that for such $n$: \n$$n^3 +88n^2 +3 = n^2(n + 88) + 3 =$$\n$$ = n^2(n+n) + n^2(88 -n) + 3 = 2n^3 + n^2(88-n) + 3 \\leq 2n^3.$$\nThe last inequality is true, since $n^2(88-n) + 3 < 0$ for any \n$n > 88$.\n\n\n\\vspace{6pt}\n{\\bf Question 10.} Answer: {\\tt 3,1,2,6,4,5} (or {\\tt 3,1,2,6,5,4}).\\\\\nLogarithm of a logarithm is a very slowly growing function; \n$\\log n^{10}$ is just equal to $10$ times $\\log n$. \n$(\\log n)^2 = \\log^2 n$ is slightly faster than a logarithm.\\\\\nFinally $\\log 2^n$ is simply $n$; but both $\\log (n!)$ and\n$n \\log n$ grow slightly faster than $n$; they are \"Big-O\" of each other:\n$$n \\log n \\;\\text{is in}\\;O(\\log n!);$$\n$$\\log n! \\;\\text{is in}\\;O(n \\log n).$$\nIt does not matter, in which order we list them. \n\nTo verify all these claims, you need to prove various limits: \n$$\\lim_{n \\rightarrow \\infty} \\frac{\\log\\log n}{\\log n} = 0,$$\n$$\\lim_{n \\rightarrow \\infty} \\frac{\\log n}{(\\log n)^2} = 0,$$\nand so on. Most of these limits are easy to find (L'Hospital's Rule and so on). \nWith $\\log n!$ you might need to use integrals to estimate \nthe sum of $\\log 1 + \\log 2 + \\ldots + \\log n$. \n\n{\\em Note.} Unless noted otherwise, all logarithms in our course are base $2$. \n\n\n\n\\vspace{6pt}\n{\\bf Question 11.} Answer: {\\tt A}. \nClearly the $N$ and $T$ should not depend on $n$; so they are the first quantifiers.\\\\\n{\\bf (B)} describes a sequence of digits where some digit repeats itself infinitely \noften (which is true for any sequence of digits).\\\\\n{\\bf (C)} describes the set of all sequences; one can always pick $N$ that is larger than $n$, \nthen the condition is trivially true.\\\\ \n{\\bf (D)} describes a sequence where each digit appears infinitely often. \n\n\n\n\\end{document}\n\n", "meta": {"hexsha": "853cd3bdf17625dd71a37288109276e1c02c3d99", "size": 14595, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/site/discrete-spring2020/questionbase/quiz-04.tex", "max_stars_repo_name": "kapsitis/math", "max_stars_repo_head_hexsha": "f21b172d4a58ec8ba25003626de02bfdda946cdc", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/site/discrete-spring2020/questionbase/quiz-04.tex", "max_issues_repo_name": "kapsitis/math", "max_issues_repo_head_hexsha": "f21b172d4a58ec8ba25003626de02bfdda946cdc", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2020-07-20T03:40:52.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-10T21:50:18.000Z", "max_forks_repo_path": "src/site/discrete-spring2020/questionbase/quiz-04.tex", "max_forks_repo_name": "kapsitis/math", "max_forks_repo_head_hexsha": "f21b172d4a58ec8ba25003626de02bfdda946cdc", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.4293628809, "max_line_length": 150, "alphanum_fraction": 0.6465912984, "num_tokens": 5265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154240079185319, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.6277679544250367}} {"text": "% !TEX root = ../zeth-protocol-specification.tex\n\n\\chapter{Double spend attack on equivalent class}\\label{appendix:dbspd}\n\nThe primary inputs of our zk-SNARK are elements of $\\FFx{\\rCURVE}$ and they can be written over $\\fieldBitLen$ bits. Note that the projection of $\\BB^\\fieldBitLen$ onto $\\FFx{\\rCURVE}$ formed by interpreting elements in $\\BB^\\fieldBitLen$ as $\\fieldBitLen$-bit numbers and reducing modulo $\\rCURVE$, is surjective.\n\nWhen we pass the primary inputs to the $\\mixer$ contract, they are interpreted as elements of $\\BB^\\ethWordLen$, and $\\BB^\\fieldBitLen \\subset \\BB^\\ethWordLen$. As previously noted, this means that there exist pairs of elements in $\\BB^\\ethWordLen$ with the same projection in $\\FFx{\\rCURVE}$. An adversary could make use of this to perform a double spend attack.\n\nIndeed, to check that a note is not double spent, the contract stores the nullifiers of spent notes (as elements of $\\BB^\\ethWordLen$) and verifies that the nullifier of the note to be spent is not stored. The adversary could thus modify the nullifier to a different value with the same projection. As the SNARK verification operates in $\\FFx{\\rCURVE}$, the proof would still be valid. However, the value stored for this nullifier would be different from the adversarial one. Hence, the nullifier would be validated, the transaction would succeed and the note would be double spent.\nIn practice, the adversary can perform the attack by simply adding $\\rCURVE$ to one of the elements representing the nullifier.\n\nTo prevent this attack, the contract checks that all primary inputs are elements of $\\FFx{\\rCURVE}$, that is to say that they are smaller than $\\rCURVE$.\nAs one may see, the attack described above is not due to the packing of hash digests into field elements but to the contract storage of field elements as $\\ethereum$ words.\n", "meta": {"hexsha": "5b9ceb252e413ff3030416ebe461ebe264f51f3c", "size": 1843, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "appendices/appendix02-dbspd.tex", "max_stars_repo_name": "clearmatics/zeth-specifications", "max_stars_repo_head_hexsha": "ba29c67587395f5c7b26b52ee7ab9cba12f1cc6b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-04-29T18:22:00.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-29T18:22:00.000Z", "max_issues_repo_path": "appendices/appendix02-dbspd.tex", "max_issues_repo_name": "clearmatics/zeth-specifications", "max_issues_repo_head_hexsha": "ba29c67587395f5c7b26b52ee7ab9cba12f1cc6b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 13, "max_issues_repo_issues_event_min_datetime": "2020-10-27T10:41:50.000Z", "max_issues_repo_issues_event_max_datetime": "2021-04-16T10:57:05.000Z", "max_forks_repo_path": "appendices/appendix02-dbspd.tex", "max_forks_repo_name": "clearmatics/zeth-specifications", "max_forks_repo_head_hexsha": "ba29c67587395f5c7b26b52ee7ab9cba12f1cc6b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-07-26T04:51:29.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-26T04:51:29.000Z", "avg_line_length": 131.6428571429, "max_line_length": 582, "alphanum_fraction": 0.7759088443, "num_tokens": 440, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.877476793890012, "lm_q2_score": 0.7154239897159439, "lm_q1q2_score": 0.6277679487679474}} {"text": "\\documentclass{article}\n%\\usepackage{fullpage}\n%\\usepackage{nopageno} \n\\usepackage[margin=1.5in]{geometry}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage[normalem]{ulem}\n\\usepackage{fancyhdr}\n\\usepackage{cancel}\n%\\renewcommand\\headheight{12pt}\n\\pagestyle{fancy}\n\\lhead{April 16, 2014}\n\\rhead{Jon Allen}\n\\allowdisplaybreaks\n\n\\newcommand{\\abs}[1]{\\left\\lvert #1 \\right\\rvert}\n\n\\begin{document}\nRead Sections 7.2 and 7.3 and do the following problems from Chapter 7: \\#16, 17, 18, 19, 22, 24(a)(b)(c) (be sure to write your answer in closed form), 26, 28, 50 (Grad: 49).\n\\begin{enumerate}\n\\setcounter{enumi}{15}\n\\item\nFormulate a combinatorial problem for which the generating function is\n\\[(1+x+x^2)(1+x^2+x^4+x^6)(1+x^2+x^4+\\cdots)(x+x^2+x^3+\\cdots)\\]\n\nHow many combinations of red, yellow, green and blue colored pencils are there in which there are at most 2 red pencils, an even number of at most six yellow pencils, an even number of green pencils and at least one blue pencil?\n\\item\nDetermine the generating function for the number $h_n$ of bags of fruit of apples, oranges, bananas, and pears in which there are an even number of apples, at most two oranges, a multiple of three number of bananas, and at most one pear. Then find a formula for $h_n$ from the generating function.\n\n\\begin{align*}\n g(x)&=\\underbrace{\\left(\\sum\\limits_{n=0}^\\infty{x^{2n}}\\right)}_{\\text{even apples}}\\underbrace{\\left(\\sum\\limits_{n=0}^2{x^n}\\right)}_{\\text{oranges}\\le 2}\\underbrace{\\left(\\sum\\limits_{n=0}^\\infty{x^{3n}}\\right)}_{3\\mid \\text{bananas}}\\underbrace{\\left(\\sum\\limits_{n=0}^1{x^{n}}\\right)}_{\\text{pears}\\le 1}\\\\\n &=\\frac{1}{1-x^2}\\left[\\left(\\sum\\limits_{n=0}^\\infty{x^n}\\right)-\\left(\\sum\\limits_{n=3}^\\infty{x^n}\\right)\\right]\\frac{1}{1-x^3}\\left[\\left(\\sum\\limits_{n=0}^\\infty{x^n}\\right)-\\left(\\sum\\limits_{n=2}^\\infty{x^n}\\right)\\right]\\\\\n &=\\frac{1}{1-x^2}\\left[\\left(1-x^3\\right)\\left(\\sum\\limits_{n=0}^\\infty{x^n}\\right)\\right]\\frac{1}{1-x^3}\\left[\\left(1-x^2\\right)\\left(\\sum\\limits_{n=0}^\\infty{x^n}\\right)\\right]\\\\\n &=\\left(\\frac{1}{1-x^2}\\right)\\left(\\frac{1-x^3}{1-x}\\right)\\left(\\frac{1}{1-x^3}\\right)\\left(\\frac{1-x^2}{1-x}\\right)\\\\\n &=\\frac{1}{(1-x)^2}\n =\\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{1}{1-x}\\right)\n =\\frac{\\mathrm{d}}{\\mathrm{d}x}\\sum\\limits_{n=0}^\\infty{x^n}\\\\\n &=\\sum\\limits_{n=0}^\\infty{nx^{n-1}}\n =\\sum\\limits_{n=-1}^\\infty{(n+1)x^{n}}\n =0+\\sum\\limits_{n=0}^\\infty{(n+1)x^{n}}\\\\\n h_n&=n+1\n\\end{align*}\n\\item\nDetermine the generating function for the number $h_n$ of nonnegative integral solutions of\n\\[2e_1+5e_2+e_3+7e_4=n\\]\nThis is equivalent to saying how many ways can you pick $n$ items from the multiset $\\{\\infty\\cdot e_1,\\infty\\cdot e_2,\\infty\\cdot e_3,\\infty\\cdot e_4\\}$ where you have an even number of $e_1$'s, a multiple of 5 $e_2$'s and a multiple of 7 $e_4$'s.\n\\begin{align*}\n g(x)&=\\left(\\sum\\limits_{n=0}^\\infty{x^{2n}}\\right)\\left(\\sum\\limits_{n=0}^\\infty{x^{5n}}\\right)\\left(\\sum\\limits_{n=0}^\\infty{x^{n}}\\right)\\left(\\sum\\limits_{n=0}^\\infty{x^{7n}}\\right)\\\\\n &=\\left(\\frac{1}{1-x^2}\\right)\\left(\\frac{1}{1-x^5}\\right)\\left(\\frac{1}{1-x}\\right)\\left(\\frac{1}{1-x^7}\\right)\n\\end{align*}\n\\item\nLet $h_0,h_1,h_2,\\dots,h_n,\\dots$ be the sequence defined by $h_n=\\binom{n}{2},(n\\ge0)$. Determine the generating function for the sequence.\n\\begin{align*}\n g(x)&=\\sum\\limits_{n=0}^\\infty{\\binom{n}{2}x^n}\\\\\n &=\\sum\\limits_{n=0}^\\infty{\\frac{n!}{2!(n-2)!}x^n}\\\\\n &=\\frac{1}{2}\\sum\\limits_{n=0}^\\infty{n(n-1)x^n}\\\\\n h_n&=\\frac{1}{2}(n^2-n)\n\\end{align*}\n\\setcounter{enumi}{21}\n\\item\nDetermine the exponential generating function for the sequence of factorials: $0!,1!,2!,3!,\\dots,n!,\\dots$\n\\begin{align*}\n g^{(e)}(x)&=\\sum\\limits_{n=0}^\\infty{n!\\cdot\\frac{x^n}{n!}}\\\\\n &=\\sum\\limits_{n=0}^\\infty{x^n}\\\\\n &=\\frac{1}{1-x}\n\\end{align*}\n\\setcounter{enumi}{23}\n\\item\nLet $S$ denote the multiset $\\{\\infty\\cdot e_1,\\infty\\cdot e_2,\\dots,\\infty\\cdot e_k\\}$. Determine the exponential generating function for the sequence $h_0,h_1,h_2,\\dots,h_n,\\dots$, where $h_0=1$ and, for $n\\ge 1$\n\\begin{enumerate}\n\\item\n$h_n$ equals the number of $n$-permutations of $S$ in which each object occurs an odd number of times.\n\\begin{align*}\n g^{(e)}(x)&=\\left(x+\\frac{x^3}{3!}+\\frac{x^5}{5!}+\\dots\\right)^k\\\\\n &=\\left[\\left(1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\dots\\right)+\\left(-1+x-\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\dots\\right)\\right]^k\\frac{1}{2^k}\\\\\n &=\\left(\\frac{1}{2}\\sum\\limits_{n=0}^\\infty{\\frac{x^n+(-x)^n}{n!}}\\right)^k\\\\\n &=\\left(\\frac{e^x+e^{-x}}{2}\\right)^k\n\\end{align*}\n\\item\n$h_n$ equals the number of $n$-permutations of $S$ in which each object occurs at least four times.\n\\begin{align*}\n g^{(e)}(x)&=\\left(\\frac{x^4}{4!}+\\frac{x^5}{5!}+\\frac{x^6}{6!}+\\dots\\right)^k\\\\\n &=\\left(\\sum\\limits_{n=4}^\\infty{\\frac{x^n}{n!}}\\right)^k=\\left(-\\frac{x^3}{3!}+\\sum\\limits_{n=3}^\\infty{\\frac{x^n}{n!}}\\right)^k\\\\\n &=\\left(-1-x-\\frac{x^2}{2!}-\\frac{x^3}{3!}+\\sum\\limits_{n=3}^\\infty{\\frac{x^n}{n!}}\\right)^k\\\\\n &=\\left(e^x-1-x-\\frac{x^2}{2!}-\\frac{x^3}{3!}\\right)^k\n\\end{align*}\n\\item\n$h_n$ equals the number of $n$-permutations of $S$ in which $e_1$ occurs at least once, $e_2$ occurs at least twice, $\\dots,e_k$ occurs at least $k$ times.\n\\begin{align*}\n g^{(e)}(x)&=\\left(x+\\frac{x^2}{2!}+\\dots\\right)\\left(\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\dots\\right)\\dots\\left(\\frac{x^{k-1}}{(k-1)!}+\\frac{x^k}{k!}+\\dots\\right)\\\\\n &=\\prod_{i=1}^k{\\left(\\sum\\limits_{n=i}^\\infty{\\frac{x^n}{n!}}\\right)}\\\\\n &=\\prod_{i=1}^k{\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^n}{n!}}-\\sum\\limits_{n=0}^{i-1}{\\frac{x^n}{n!}}\\right)}\\\\\n &=\\prod_{i=1}^k{\\left(e^x-\\sum\\limits_{n=0}^{i-1}{\\frac{x^n}{n!}}\\right)}\\\\\n\\end{align*}\n\\end{enumerate}\n\\setcounter{enumi}{25}\n\\item\nDetermine the number of ways to color the squares of a 1-by-$n$ chessboard, using the colors red, blue, green, and orange if an even number of squares is to be colored red and an even number is to be colored green.\n\\begin{align*}\n g^{(e)}(x)&=\\underbrace{\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^{2n}}{(2n)!}}\\right)^2}_{\\text{red and green are even}}\\cdot\\overbrace{\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^{n}}{n!}}\\right)^2}^{\\text{blue and orange are unrestricted}}\\\\\n &=\\left(\\frac{1}{2}\\sum\\limits_{n=0}^\\infty{\\frac{x^n+(-x)^n}{n!}}\\right)\\left(e^x\\right)^2\\\\\n &=\\frac{e^{2x}}{2}\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^n}{n!}}+\\sum\\limits_{n=0}^\\infty{\\frac{(-x)^n}{n!}}\\right)\\\\\n &=\\frac{e^{2x}}{2}\\left(e^x+e^{-x}\\right)=\\frac{e^{3x}+e^x}{2}\\\\\n &=\\frac{1}{2}\\sum\\limits_{n=0}^\\infty{\\frac{(3x)^n+x^n}{n!}}=\\sum\\limits_{n=0}^\\infty{\\frac{3^n+1}{2}\\cdot\\frac{x^n}{n!}}\n\\end{align*}\nSo we can color the board $\\frac{3^n+1}{2}$ ways\n\\setcounter{enumi}{27}\n\\item\nDetermine the number of $n$-digit numbers with all digits at least 4, such that 4 and 6 each occur an even number of times, and 5 and 7 each occur at least once, there being no restriction on the digits 8 and 9.\n\\begin{align*}\n g^{(e)}(x)&=\\underbrace{\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^{2n}}{(2n)!}}\\right)^2}_{\\text{4,6 are even}}\\cdot\\overbrace{\\left(\\sum\\limits_{n=1}^\\infty{\\frac{x^n}{n!}}\\right)^2}^{\\text{5,7 occur at least once}}\\cdot\\underbrace{\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^n}{n!}}\\right)^2}_{\\text{8,9 are unrestricted}}\\\\\n &=\\left(\\frac{1}{2}\\sum\\limits_{n=0}^\\infty{\\frac{x^n+(-x)^n}{n!}}\\right)^2\\left(-1+\\sum\\limits_{n=0}^\\infty{\\frac{x^n}{n!}}\\right)^2\\left(\\sum\\limits_{n=0}^\\infty{\\frac{x^n}{n!}}\\right)^2\\\\\n &=\\frac{e^{2x}}{4}\\left(e^x+e^{-x}\\right)^2(e^x-1)^2\\\\\n &=\\frac{e^{2x}}{4}\\left(e^{2x}+2+e^{-2x}\\right)(e^{2x}-2e^x+1)\\\\\n &=\\frac{e^{2x}}{4}\\left(e^{4x}-2e^{3x}+e^{2x}+2e^{2x}-4e^x+2+1-2e^{-x}+e^{-2x}\\right)\\\\\n &=\\frac{e^{2x}}{4}\\left(e^{4x}-2e^{3x}+3e^{2x}-4e^x+3-2e^{-x}+e^{-2x}\\right)\\\\\n &=\\frac{1}{4}\\left(e^{6x}-2e^{5x}+3e^{4x}-4e^{3x}+3e^{2x}-2e^{x}+1\\right)\\\\\n &=\\frac{1}{4}+\\frac{1}{4}\\sum\\limits_{n=0}^\\infty{\\frac{(6x)^n-2(5x)^n+3(4x)^n-4(3x)^n+3(2x)^n-2x^n}{n!}}\\\\\n &=\\frac{1}{4}+\\frac{1}{4}\\sum\\limits_{n=0}^\\infty{\\left[6^n-2(5^n)+3(4^n)-4(3^n)+3(2^n)-2\\right]\\frac{x^n}{n!}}\n\\end{align*}\nSo the number of $n$-digit numbers that meet the criteria is 0 when $n=0$ and $\\frac{1}{4}\\left[6^n-2(5^n)+3(4^n)-4(3^n)+3(2^n)-2\\right]$ when $n\\ge1$\n\\setcounter{enumi}{49}\n\\item\nCall a subset of $S$ of the integers $\\{1,2,\\dots,n\\}$ \\emph{extraordinary} provided its smallest integer equals its size:\n\\[\\min\\{x:x\\in S\\}=\\lvert S\\rvert.\\]\nFor example, $S=\\{3,7,8\\}$ is extraordinary. Let $g_n$ be the number of extraordinary subsets of $\\{1,2,\\dots,n\\}$. Prove that\n\\[g_n=g_{n-1}+g_{n-2},\\quad(n\\ge3),\\]\nwith $g_1=1$ and $g_2=1$.\n\\subsubsection*{proof}\nThe number of extraordinary subsets of $\\{1,2,\\dots,n\\}$ is the sum of the number of extraordinary subsets of each size from 1 to $n$. A $k$ sized extraordinary subset of $\\{1,2,\\dots,n\\}$ contains $k$ and $k-1$ elements of the set $\\{k+1,k+2,\\dots,n\\}$. We can choose these $k-1$ elements in $\\binom{n-(k+1)+1}{k-1}$ ways. Lets simplify a little:\n\\begin{align*}\n \\binom{n-(k+1)+1}{k-1}&=\\binom{n-k}{k-1}\\\\\n g_n&=\\sum\\limits_{k=1}^n{\\binom{n-k}{k-1}}\\\\\n \\binom{n+1}{k+1}&=\\binom{n}{k}+\\binom{n}{k+1}\\\\\n g_n&=\\sum\\limits_{k=1}^n{\\binom{n-k-1}{k-1-1}}+\\sum\\limits_{k=1}^n{\\binom{n-k-1}{k-1}}\\\\\n &=\\binom{n-n-1}{n-2}+\\sum\\limits_{k=1}^{n-1}{\\binom{n-k-1}{k-1-1}}+\\binom{n-n-1}{n-1}+\\sum\\limits_{k=1}^{n-1}{\\binom{n-k-1}{k-1}}\\\\\n &=\\frac{(n-n-1)!}{(n-n-1-(n-2))!(n-2)!}+\\frac{(n-n-1)!}{(n-n-1-(n-1))!(n-1)!}\\\\&\\quad+\\sum\\limits_{k=1}^{n-1}{\\binom{n-k-1}{k-1-1}}+\\sum\\limits_{k=1}^{n-1}{\\binom{(n-1)-k}{k-1}}\\\\\n &=\\frac{(-1)!}{(1-n)!(n-2)!}+\\frac{(-1)!}{(-n)!(n-1)!}\\\\&\\quad+\\sum\\limits_{k=1}^{n-1}{\\binom{n-(k-1)}{(k-1)-1}}+g_{n-1}\\\\\n &=\\frac{(-1)!(n-1)}{(1-n)!(n-1)!}+\\frac{(-1)!(1-n)}{(1-n)!(n-1)!}\\\\&\\quad+\\sum\\limits_{k=0}^{n-2}{\\binom{n-k}{k-1}}+g_{n-1}\\\\\n &=\\frac{(-1)!n-(-1)!+(-1)!-(-1)!n}{(1-n)!(n-1)!}\\\\&\\quad+\\binom{n}{-1}+\\sum\\limits_{k=1}^{n-2}{\\binom{n-k}{k-1}}+g_{n-1}\\\\\n &=0+\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}-\\binom{n-(n-1)}{(n-1)-1}-\\binom{n-n}{n-1}+g_{n-1}\\\\\n &=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}-\\binom{1}{n-2}-\\binom{0}{n-1}+g_{n-1}\\\\\n% &=\\sum\\limits_{k=1}^{n-1}{\\binom{n-k}{k-1}}+\\binom{n-n}{n-1}\n\\end{align*}\n%Now if we have $g_n$, then what is $g_{n-1}$? We have \n%\\begin{align*}\n% g_{n-1}&=\\sum\\limits_{k=1}^{n-1}{\\binom{n-k-1}{k-1}}\\\\\n%\\end{align*}\n%Now lets see what happens if we add $g_{n}$ and $g_{n+1}$\n%\\begin{align*}\n% \\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n+1}{\\binom{(n+1)-k}{k-1}}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\binom{(n+1)-k}{k-1}}+\\binom{(n+1)-(n+1)}{(n+1)-1}\n%\\end{align*}\n%lets make an assumption and see if it holds\n%\\begin{align*}\n% \\sum\\limits_{k=1}^{n+2}{\\binom{(n+2)-k}{k-1}}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\binom{(n+1)-k}{k-1}}+\\binom{0}{n}\\\\\n% \\sum\\limits_{k=1}^{n}{\\binom{(n+2)-k}{k-1}}+\\binom{0}{n+1}+\\binom{1}{n}-\\sum\\limits_{k=1}^{n}{\\binom{(n+1)-k}{k-1}}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\binom{(n+2)-k}{k-1}}-\\sum\\limits_{k=1}^{n}{\\binom{(n+1)-k}{k-1}}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\frac{((n+2)-k)!}{(k-1)!((n+2)-k-(k-1))!}}\\qquad&\\\\\n% -\\sum\\limits_{k=1}^{n}{\\frac{((n+1)-k)!}{(k-1)!((n+1)-k-(k-1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-k(k-1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\frac{(n-k+2)!}{(k-1)!(n-2k+3)!}}\\qquad&\\\\\n% -\\sum\\limits_{k=1}^{n}{\\frac{(n-k+1)!}{(k-1)!(n-2k+2))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\frac{(n-k+2)(n-k+1)(n-k)!}{(k-1)!(n-2k+3)(n-2k+2)(n-2k+1)!}}\\qquad&\\\\\n% -\\sum\\limits_{k=1}^{n}{\\frac{(n-k+1)(n-k)!}{(k-1)!(n-2k+2)(n-2k+1)!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\begin{aligned}\\frac{(n-k+2)(n-k+1)}{(n-2k+3)(n-2k+2)}\\\\-\\frac{(n-k+1)}{(n-2k+2)}\\end{aligned}\\right]\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{\\begin{aligned}(n-k+2)(n-k+1)\\\\-(n-k+1)(n-2k+3)\\end{aligned}}{(n-2k+3)(n-2k+2)}\\right]\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{(n-k+1)(n-k+2-n+2k-3)}{(n-2k+3)(n-2k+2)}\\right]\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{(n-k+1)(k-1)}{(n-2k+3)(n-2k+2)}\\right]\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{1}{(n-2k+3)(n-2k+2)}\\right]\\frac{(n-k+1)!}{k!(n-2k+1))!}}&=\\sum\\limits_{k=1}^{n}{\\frac{(n-k)!}{(k-1)!(n-2k+1))!}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{1}{(n-2k+3)(n-2k+2)}\\right]\\binom{n-k+1}{k}}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% \\sum\\limits_{k=2}^{n+1}{\\left[\\frac{\\binom{(n-1)-(k-1)+1}{k-1}}{((n-1)-2(k-1)+3)((n-1)-2(k-1)+2)}\\right]}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% \\sum\\limits_{k=2}^{n+1}{\\left[\\frac{\\binom{n-k+1}{k-1}}{(n-2k+4)(n-2k+3)}\\right]}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% \\sum\\limits_{k=1}^{n}{\\left[\\frac{\\frac{n-k+1}{k}\\binom{n-k}{k-1}}{(n-2k+3)(n-2k+2)}\\right]}&=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}\\\\\n% &=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\frac{(n+1-k)!}{(k-1)!(n-k+1-k+1)!}}+\\binom{0}{n}\\\\\n% &=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\frac{(n+1-k)(n-k)!}{(k-1)!(n-k+1-k+1)(n-k+1-k)!}}\\\\\n% &=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\frac{(n+1-k)}{(n+2-2k)}\\binom{n-k}{k-1}}\\\\\n% &=\\sum\\limits_{k=1}^{n}{\\binom{n-k}{k-1}}+\\sum\\limits_{k=1}^{n}{\\frac{(n-k+1)}{(n-2k+2)}\\binom{n-k}{n-k+1}}\\\\\n%\\end{align*}\n\\end{enumerate}\n\\end{document}\n", "meta": {"hexsha": "775fb9b6f90045361d1598086a53ff1afc9a2c60", "size": 13668, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "combinatorics/combinatorics-hw-2014-04-16.tex", "max_stars_repo_name": "ylixir/school", "max_stars_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "combinatorics/combinatorics-hw-2014-04-16.tex", "max_issues_repo_name": "ylixir/school", "max_issues_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "combinatorics/combinatorics-hw-2014-04-16.tex", "max_forks_repo_name": "ylixir/school", "max_forks_repo_head_hexsha": "66d433f2090b6396c8dd2a53a733c25dbe7bc90f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 75.9333333333, "max_line_length": 347, "alphanum_fraction": 0.5847966052, "num_tokens": 6617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891305219504, "lm_q2_score": 0.7634837743174788, "lm_q1q2_score": 0.6277280605737049}} {"text": "% \\cleardoublepage\n% \\chapter*{Introduction}\n% \\markboth{Introduction}{Introduction}\n% \\addcontentsline{toc}{chapter}{Introduction}\n\n\\chapter{Introduction}\n\n\\section{Problem and Motivation}\n\nSuppose we are given samples of data say $\\mathbf{x}$ and $\\mathbf{y}$, s.t.\n\n\\begin{align*}\n \\mathbf{x} = x_1, ..., x_n \\\\\n \\mathbf{y} = y_1, ..., y_n \n\\end{align*}\n\nFor example, we may be measuring the blood pressure and heart rate of Alice at time $k$, \nsay $x_k$ and $y_k$ respectively, with $x_k, y_k \\in \\R$. Further, suppose we are unaware of her context, for example,\nBob hacked into Alice's Apple watch and so can only read $\\mathbf{x}$ and $\\mathbf{y}$ --- he has no idea of \nanything she might be up to.\n\nBob then observes the following trend:\n\n\\begin{figure}[htb]\n \\centering\n \\begin{tikzpicture}\n \\node (img) {\\includegraphics[width=.45\\textwidth]{correlated_hr.png}};\n \\node[below=of img, node distance=0cm, yshift=1cm,font=\\color{black}] {time};\n \\node[left=of img, node distance=0cm, rotate=90, anchor=center,yshift=-0.7cm,font=\\color{black}] {bpm/mmHg};\n \\end{tikzpicture}\n \\caption{Heart rate (HR) and Blood pressure (BP) measured in bbp and mmHg resp. Day 1.}\n\\end{figure}\n\nBob, having studied data science, is well aware of the fallacy of the law of small numbers\\footnote{\n The law of small numbers is the error of concluding too much from too few data (many such examples \n can be found in the book \"fooled by randomness\" by \\cite{taleb}).}. \n He therefore checks again the data the next day at a slightly different time. He again observes \na similar trend, and is now more confident in the existence of a causal relation and --- having neglected biology --- \nmakes the conjecture that either blood pressure causes heart rate, or \nperhaps the other way around. \n\nGiven this strong correlation, Bob asserts that he may either model $x$ as a function of $y$ or the other way \naround. He proceeds to find some $f$ s.t. $f(x) \\approx y$. The next day, to his dismay, he notices that his\nmodel has terrible performance when evaluated on new data. He then proceeds to see what is going on, and \nobserves the following:\n\n\\begin{figure}[htb]\n \\centering\n \\begin{tikzpicture}\n \\node (img) {\\includegraphics[width=.45\\textwidth]{uncorrelated_hr.png}};\n \\node[below=of img, node distance=0cm, yshift=1cm,font=\\color{black}] {time};\n \\node[left=of img, node distance=0cm, rotate=90, anchor=center,yshift=-0.7cm,font=\\color{black}] {bpm/mmHg};\n \\end{tikzpicture}\n \\caption{Heart rate (HR) and Blood pressure (BP) measured in bbp and mmHg resp. Day 3.}\n\\end{figure}\n\nAs it turns out, in the last few days, Alice was working hard on finishing her thesis and the deadline had been \nthe previous day. But how, Bob wondered, could this have changed the relationship between BP and HR? \nFinally, admitting to himself that curve fitting alone is not enough to understand the world; \nBob spends some time learning about the heart. It turns out, that fear triggers a \"flight or fight\"\nresponse that increases both the heart rate and blood pressure; Interestingly your heart rate and blood pressure \nwon’t always rise and fall in sync.\n\nSo what did Bob learn\\footnote{\n Note that heart rate and blood pressure are intimately linked, and the story between them is more complicated.\n The plots were randomly generated using a Gaussian process, however they do resemble some real examples that \n can be found in the internet.\n}?\n\n\\begin{enumerate}\n \\item When we train a model with some data, and then use it on some newly acquired data, we might\n face a \\textbf{covariate-shift}\\footnote{Another way to describe the covariate-shift is as follows, \n suppose we observe some samples, but\n these samples only represent a small portion of the data space, then when we acquire new data, this new data might be very \n different from our original acquisition. } --- that is, the distribution might change due to the context changing.\n \\item When we see correlation it might be spurious due to a \\textbf{confounder} --- fear was the \\textbf{confounder} \n of the heart rate and blood pressure.\n\\end{enumerate}\n\nIf Bob was able to incorporate these notions into his machine learning models, then it might have been more\nrobust to the covariate-shift. To give a more concrete example, there is a \"neural net tank urban legend\"\\footnote{\n More about this story here: https://www.gwern.net/Tanks.\n}\n, where a neural network accurately predicts if there is a tank or not in an image, but it turns out it uses the \nweather as a predictor. From this it is clear that the model will perform badly under covariate shift, and indeed\nit makes the case that incorporating causality to a model should make it more robust as\n\\cite{scholkopf2019causality} argues. Note that this is in effect the issue with generalization in machine learning:\nhow can we ensure that we learn \\textit{meaningful} representations (features about the tanks) rather than just\ncorrelations (the weather) useful for training accuracy. \n\nAs for confounders, it is impossible to say anything in general\\footnote{\n For most of the 20th century, a huge debate took place to determine the question of whether or not \n smoking caused cancer. A clever argument against a causal relation was that there existed a gene that \n made a person both want to smoke and more prone to cancer; even the father of modern statistics\n himself thought this explanation more plausible (For a good read on how science is and was used \n for wrong see the excellent book of \\cite{NaomiMerchants}).\n \n}. We must therefore specify a causal model, and\nthen see what guarantees we can give under what assumptions. Even in the absence of confounders it is highly \nnon trivial to determine causality.\n\nAs this simple example illustrates, causality is related to many interesting questions; perhaps, one of the most \nsimple questions we can ask --- and the one that we will explore --- is, given that either X causes Y, or Y causes X\n(we assume no confounders) then, when \ncan we predict the direction of causality? If yes, how? \n\nIn the figure bellow (figure \\ref{fig:simple_bivariate_example}) can you tell if $X$ causes $Y$? Or perhaps\nit is the other way around? The right answer is that $X$ causes $Y$, and we will show algorithms that \ncan accurately predict causality in such settings with as few as 75 samples. \n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=1\\textwidth]{bivariate_causal_example.png}\n \\caption{75 samples of data $X$, $Y$. The samples are generated independently as follows:\n $y_i = f(x_i) + n_i$ where $x_i$ is drawn from an exponential distribution and $n_i$ is drawn \n independently from a Gaussian one and $f(x) = 10 \\tanh(x) + 4\\sin(x) + x + x^2$}\n \\label{fig:simple_bivariate_example}\n\\end{figure}\n\n\\section{Causality}\n\n\\subsection{Causal models: FCM}\n\nWe can model any causal model by using a \\textbf{Function Causal Model (FCM)} which can be constructed as follows\n(\\cite{Peters2017}):\n\nWe \\textit{generate} a random vector\\footnote{That is, each $X_i$ is a random variable.}\n$X = (X_1, ..., X_d)$ by using a graph $\\mathcal{G}$ (encoding the relationships),\na set of functions $f = (f_1, ..., f_d)$ (encoding the type of relationship) and a noise distribution $\\mathcal{E}$\n(the randomness generator).\n\nFor each $i = 1, ..., d$\n\n$$\n X_i \\leftarrow f_i(X_{\\text{Pa}(i)}, E_i), \\quad E_i \\thicksim \\mathcal{E}\n$$\n\nwhere $\\text{Pa}(i)$ is the set of parents of $i$; and so $X_{\\text{Pa}(i)}$ is the set of random variables\nfor the parents of $X_i$. For example in Figure \\ref{fig:fcm_1}, $\\text{Pa}(4) = \\{1, 2\\}$, \nand so $X_{\\text{Pa}(4)} = \\{X_1, X_2\\}$.\n\n\\begin{figure}[!h]\n \\centering\n\n \\begin{tikzpicture}[\n > = stealth, % arrow head style\n shorten > = 1pt, % don't touch arrow head to node\n auto,\n node distance = 2.5cm, % distance between nodes\n thick % line style\n ]\n \n \\tikzstyle{every state}=[\n draw = black,\n thick,\n fill = white,\n minimum size = 4mm\n ]\n \n \\node[state] (a) {$X_1$};\n \\node[state] (b) [right of=a] {$X_2$};\n \\node[state] (c) [below left of=a] {$X_3$};\n \\node[state] (d) [below right of=a] {$X_4$};\n \n \\path[->] (a) edge node {} (c);\n \\path[->] (a) edge node {} (d);\n \\path[->] (b) edge node {} (d);\n \n \\end{tikzpicture}\n \\caption{ Example for FCM with $X = X_1, ... X_4$, with $E_i \\thicksim \\mathcal{E}$, \n with $X_1 = f_1(E_1)$, $X_2 = f_2(E_2)$, $X_3 = f_3(X_1, E_3)$, and \n $X_4 = f_4(X_1, X_2, E_4)$}\n \\label{fig:fcm_1}\n\n\\end{figure}\n\n\nNote that causal relations can also be \\textit{cyclic}, i.e. $X$ causes $Y$ which in turn causes \n$X$ ad infinitum; while this deserves consideration as many systems have feedback loops we will\nnot look at such settings for simplicity. \n\n\\subsection{Interventions}\n\nTo make precise the meaning of causality, suppose that we are given \ntwo random variables $X$, $Y$ with joint distribution $p_{x, y}$. Intuitively we \nwould say that $X$ causes $Y$, or $X \\rightarrow Y$, if we intervene on $X$ and then see an effect on $Y$. In particular\nwe will denote $\\operatorname{do}(x)$ --- short for $\\operatorname{do}(X = x)$ --- as an intervention\nthat forces the variable $X$ to have the value $x$, and leaves the rest of the system untouched. \nFollowing the convention inspired by \\cite{pearl2000causality}, \nwe define the resulting distribution as $p_{y|do(x)}$.\n\nThis motivates the following definition:\n\n\\begin{definition}\n We say that $X$ \\textbf{causes} $Y$ if $p_{y|do(x)} \\neq p_{y|do(x^\\prime)}$ for some\n $x \\neq x^\\prime$\n\\end{definition}\n\nWhen we talk about $p_{y|x}$, we often say, \"The chance of $y$ given that $x$ happened\". This sounds similar\nto $p_{y|do(x)}$; note however that \"$x$ happened\" and \"force $X = x$\" are very different. Imagine that \nthere indeed was a gene that made people both prone to smoking and cancer; then if we forced someone at \nrandom to smoke, he would on average be less likely to have cancer than someone who smoked because he wanted to.\nThis also illustrates \none of the limitations of causality: some interventions are not possible due to ethical issues.\n\nYou might have heard about randomized trials or A/B testing, these are both common ways to estimate $p_{y|do(x)}$.\nFor example, when developing cures, the idea of a random trial is to give experimental drugs to participants\nat random. When designing new UIs to maximize user participation in apps, developers implement A/B testing, \nthey assign new versions to people at random to estimate engagement. Note that in both of these, we are able \nto avoid a potential confounder by picking $x$'s at random, and \"forcing\" them to \"do$(x)$\".\n\nSince we will restrict ourselves to the observational setting, we will not be able to perform any interventions,\nwhich would allow us to estimate $p_{y|do(x)}$. In this setting however, in order to perform any meaningful\ninference, we will need to make concessions; in particular, we will make some assumptions about \nthe causal structure. If we are able to infer the causal structure in such a setting, we \nshall call it \\textbf{identifiable}.\n\n% f we have the random variables $X$, $Y$, and $Z$, then depending on how they are causaly linked, then \n% we will have different relationships between the marginals and the Pearl's \"do\" conditional. For example,\n% if $X \\rightarrow Z \\rightarrow Y$, then we will have that $p_X = p_{X|do(y)}$, but $p_Y \\neq p_{Y|do(x)}$. \n\n% Indeed, for all 6 different relations of $X$, $Y$, and $Z$ we get a different set of relationships between \n% the interventional and observational conditions. Then, assuming that we get both interventional and observational data it suffices to check to which of the 6 \n% possible structure the data corresponds to (at least in theory). \n\nWe restrict ourselves also to the bivariate case; one big difference worth noting is that in the\nmultivariate setting we can test conditional independence. Using conditional independence tests \nis a very powerful method for causal inference. Suppose \nwe have random variables $X$, $Y$ and $Z$, then if we can estimate that $X \\bigCI Y | Z$, then it must be \nthat all information between $X$ and $Y$ must flow through $Z$ (See Figure \\ref{fig:conditionalIndep}).\nThis test alone has has now greatly reduced the space of possible causal directions between these random variables.\n\nOne can in fact generalize the conditional independence such that $X$ and the other variables are \na collection of random variables, which gives a lot of flexibility to devise clever algorithms. \nThe theory comes from graphical models, which tries to understand the relationship between \ndistributions and their graphical counterparts, such as Figure \\ref{fig:conditionalIndep}.\nThe key difference is that in graphical models we do not care about the causal direction. \nSince we will not be using any of this theory, we will not go into any detail \n(a good overview can found in \\cite{bishop2006pattern}).\n\n\\begin{figure}[!h]\n \\centering\n\n \\begin{tikzpicture}[\n > = stealth, % arrow head style\n shorten > = 1pt, % don't touch arrow head to node\n auto,\n node distance = 2.5cm, % distance between nodes\n thick % line style\n ]\n \n \\tikzstyle{every state}=[\n draw = black,\n thick,\n fill = white,\n minimum size = 4mm\n ]\n \n \\node[state] (a) {$Z$};\n \\node[state] (b) [below left of=a] {$X$};\n \\node[state] (c) [below right of=a] {$Y$};\n \n \\path[-] (a) edge node {} (b);\n \\path[-] (a) edge node {} (c);\n \n \\end{tikzpicture}\n \\caption{An example of FCM with random variables $X$, $Z$ and $Y$; we leave it undirected}\n \\label{fig:conditionalIndep}\n\n\\end{figure}\n\nIn some sense the two variable case is hard because we cannot use conditional independence. As we will see, \nit is not possible to distinguish causality in the general bivariate setting when only observing \nobservational data; we will thus need to restrict \nthe class of such models. The underlying structure behind such causal models is what is known as \nStructural Equation Models (SEM) (\\cite{Peters2017}). Essentially \nit is a model specification; and the key insight is that it should not be reversible.\n\n\n\\section{Proposed Methods}\n\nWe propose a new type of inference method which exploits the i.i.d \nnoise assumption; the idea is to split the data in different intervals, and then to regress each \ninterval separately. In the causal direction one would expect that the residuals of each \ninterval will be more homogenous\\footnote{By homogenous residuals we mean that the residuals will be similar \nto each other, for example if we plot their respective histograms, these will asymptotically converge to the \nsame distribution} --- in the ANM settings, asymmetries introduced\nby inversion will break the i.i.d assumption. We will conclude by \nshowing that this intuition is indeed correct, by proving that the method is consistent --- \nassuming that causal discovery is indeed possible and that the additive noise is i.i.d. We will\nalso propose a second method, however while also consistent, it requires knowledge of the \nnoise distribution, which is not a good assumption in practice. \n\n\n\\section{Outline}\n\nWe begin by formalizing further the causal framework that we will explore; we will then present an overview of \npopular methods for inference on the bivariate causal setting. Next, we will overview some notions in statistical distance \n--- as these play a central role in causal inference --- such as maximum mean discrepancy (MMD) and f-divergences. \nFor the convenience of the reader we will also review some background material that will help gain a better \nintuition on the MMD. \n\nWe will end by going over the proposed methods, and showing both theoretical guarantees and experimental results. \n\n\n\n\n", "meta": {"hexsha": "9c6fc4406ab5c41de22fe5bdb2a8bc1f129835c9", "size": 16012, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main/ch1_introduction.tex", "max_stars_repo_name": "Nacho114/EPFL_thesis_template", "max_stars_repo_head_hexsha": "e92f8b0b2d14d0a514dce0fc4a83a358481a4d20", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "main/ch1_introduction.tex", "max_issues_repo_name": "Nacho114/EPFL_thesis_template", "max_issues_repo_head_hexsha": "e92f8b0b2d14d0a514dce0fc4a83a358481a4d20", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "main/ch1_introduction.tex", "max_forks_repo_name": "Nacho114/EPFL_thesis_template", "max_forks_repo_head_hexsha": "e92f8b0b2d14d0a514dce0fc4a83a358481a4d20", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.987012987, "max_line_length": 160, "alphanum_fraction": 0.7275168624, "num_tokens": 4158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.822189134878876, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6277280594754595}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\n\\markright{fmsin}\n\\section*{\\hspace*{-1.6cm} fmsin}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nSignal with sinusoidal frequency modulation.\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[y,iflaw] = fmsin(N)\n[y,iflaw] = fmsin(N,fmin)\n[y,iflaw] = fmsin(N,fmin,fmax)\n[y,iflaw] = fmsin(N,fmin,fmax,period)\n[y,iflaw] = fmsin(N,fmin,fmax,period,t0)\n[y,iflaw] = fmsin(N,fmin,fmax,period,t0,f0)\n[y,iflaw] = fmsin(N,fmin,fmax,period,t0,f0,pm1)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty fmsin} generates a sinusoidal frequency modulation, whose\nminimum frequency value is {\\ty fmin} and maximum is {\\ty fmax}. This\nsinusoidal modulation is designed such that the instantaneous frequency at\ntime {\\ty t0} is equal to {\\ty f0}, and the ambiguity between increasing or\ndecreasing frequency is solved by {\\ty pm1}.\\\\\n \n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8.5cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty N} & number of points\\\\\n {\\ty fmin} & smallest normalized frequency & {\\ty 0.05}\\\\\n {\\ty fmax} & highest normalized frequency & {\\ty 0.45}\\\\\n {\\ty period} & period of the sinusoidal frequency modulation & {\\ty N} \\\\ \n {\\ty t0} & time reference for the phase & {\\ty N/2} \\\\\n {\\ty f0} & normalized frequency at time {\\ty t0} & {\\ty 0.25}\\\\\n {\\ty pm1} & frequency direction at {\\ty t0} (-1 or +1)& {\\ty +1} \\\\\n \\hline {\\ty y} & signal\\\\\n {\\ty iflaw} & instantaneous frequency law \\\\\n \n\\hline\n\\end{tabular*}\n\n\\end{minipage}\n\\vspace*{1cm}\n\n\n{\\bf \\large \\sf Example}\n\\begin{verbatim}\n z=fmsin(140,0.05,0.45,100,20,0.3,-1.0);\n plot(real(z));\n\\end{verbatim}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\nfmconst, fmlin, fmodany, fmhyp, fmpar, fmpower.\n\\end{verbatim}\n\\end{minipage}\n \n\n\n\n", "meta": {"hexsha": "cc5d078897bdcf47fc3f36882570cba14f39afe2", "size": 2332, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/fmsin.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/fmsin.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/fmsin.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 26.5, "max_line_length": 85, "alphanum_fraction": 0.6252144082, "num_tokens": 840, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891392358015, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6277280583772138}} {"text": "\\section{Cycles}\n\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Cycle detection}\n \\begin{exampleblock}{Cycle detection (Problem 5.24--1)}\n\t\\begin{table}[ht]\n\t \\centering\n\t \\renewcommand{\\arraystretch}{1.1}\n\t \\begin{tabular}{c||c|c}\n\t \\hline\n\t\t& Digraph \t\t\t& Undirected graph \\\\ \\hline \\hline\n\t\tDFS & {\\uncover<2->{back edge $\\iff$ cycle}}\n\t\t& {\\uncover<3->{back edge $\\iff$ cycle}}\n\t\t\\\\ \\hline\n\t\tBFS & {\\uncover<5->{\\begin{tabular}[c]{@{}l@{}}back edge $\\implies$ cycle\\\\ cycle \\textcolor{red}{$\\centernot\\implies$} back edge \\end{tabular}}}\n\t\t& {\\uncover<4->{cross edge $\\iff$ cycle}}\n\t\t\\\\ \\hline\n\t \\end{tabular}\n\t\\end{table}\n \\end{exampleblock}\n\n \\begin{columns}\n\t\\column{0.50\\textwidth}\n\t \\uncover<6->{\\fignocaption{width = 0.30\\textwidth}{figs/bfs-digraph-cycle-without-back.png}}\n\t\\column{0.50\\textwidth}\n\t \\uncover<7->{%\n\t\t\\begin{alertblock}{Remark}\n\t\t How to identify back edges?\n\t\t\\end{alertblock}}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Evasiveness of acyclicity}\n \\begin{exampleblock}{Evasiveness of acyclicity (Problem 5.24--2)}\n\t\\[\n\t \\text{Evasiveness } \\triangleq \\text{ check } \\binom{n}{2} \\text{ edges (adjacency matrices)}\n\t\\]\n\n\t\\pause\n\t\\centerline{Is acyclicity evasive?}\n \\end{exampleblock}\n\n \\pause\n \\vspace{0.50cm}\n \\fignocaption{width = 0.20\\textwidth}{figs/adversary-logo.png}{\\centerline{\\pause Hint: Kruskal}}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Evasiveness of connectivity}\n \\begin{exampleblock}{Evasiveness of connectivity (Additional Problem)}\n\t\\[\n\t \\text{Evasiveness } \\triangleq \\text{ check } \\binom{n}{2} \\text{ edges}\n\t\\]\n\n\t\\centerline{Is connectivity evasive?}\n \\end{exampleblock}\n\n \\pause\n \\vspace{0.50cm}\n \\fignocaption{width = 0.20\\textwidth}{figs/adversary-logo.png}{\\centerline{Hint: Anti-Kruskal}}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Edge deletion}\n \\begin{exampleblock}{Edge deletion (Problem 5.20)}\n \\begin{itemize}\n \\item connected, undirected graph $G$\n \\item $\\exists? e \\in E: G \\setminus e$ is connected?\n \\item $O(|V|)$\n \\end{itemize}\n \\end{exampleblock}\n\n \\pause\n \\[\n\t\\exists \\text{ cycle} \\iff \\exists \\text{ such } e\n \\]\n\n \\pause\n \\[\n\tO(m + n)\n \\]\n\n \\pause\n \\[\n\t\\text{tree: } |E| = |V| - 1 \\implies \\text{ check } |E| \\ge |V|\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Orientation of undirected graph}\n \\begin{exampleblock}{Orientation of undirected graph (Problem 5.9)}\n\t\\begin{itemize}\n \\item undirected (connected) graph $G$ \n\t \\item edges oriented \\emph{s.t.} \n\t\t\\[\n\t\t \\forall v, \\text{in}[v] \\ge 1\n\t\t\\]\n\t\\end{itemize}\n \\end{exampleblock}\n\n \\pause\n \\[\n\t\\text{orientation} \\iff \\exists \\text{ cycle } C\n \\]\n\n \\pause\n \\[\n\t\\text{BFS/DFS from } v \\in C\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Shortest cycle of undirected graph}\n \\begin{exampleblock}{Shortest cycle of undirected graph (Problem 5.8)}\n\tShortest cycle of $G$:\n\t\\begin{itemize}\n\t \\item DFS on $G$\n\t \\item $\\forall v: \\text{level}[v]$\n\t \\item back edge $u \\to v: \\text{level}[u] - \\text{level}[v] + 1$\n\t\\end{itemize}\n \\end{exampleblock}\n\n \\pause\n \\vspace{0.50cm}\n \\begin{alertblock}{Question}\n\tWhat about digraphs?\n \\end{alertblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "a3f938d246f1b0ede0222c9d6f7e2f985adab1b6", "size": 3216, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-graph-decomposion-20170524/sections/cycle.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-graph-decomposion-20170524/sections/cycle.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-graph-decomposion-20170524/sections/cycle.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 25.5238095238, "max_line_length": 147, "alphanum_fraction": 0.6212686567, "num_tokens": 1147, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837527911057, "lm_q2_score": 0.8221891392358015, "lm_q1q2_score": 0.6277280495278387}} {"text": "\\section{Optimization objective}\\label{optimization-objective}\n\nIn support vector machine, we don't directly have $\\lambda$. Instead, C\nis used, which penalizes the training set instead of the $\\theta$. In\nreality, it's almost the same, as we can set a small value for C and\nthen we're giving more importance to the regularization term. We could\nthink of $C = \\frac{1}{\\lambda}$.\n\nInformally, the C parameter is a positive value that controls the\npenalty for misclassified training examples. A large C parameter tells\nthe SVM to try to classify all the examples correctly. \\smallskip\n\nA part from that, the objective function (minimizing cost function) is\nalmost the same as the logistic regression one:\n\n\\begin{equation} \nmin_{\\theta} C \\sum_{i=1}^m [y^{(i)}cost_1(\\theta^T x^{(i)}) + (1 - y^{(i)})cost_0(\\theta^T x^{(i)})] + \\frac{1}{2} \\sum_{j=1}^n \\theta_j^2\n\\end{equation}\n\nWhere $cost_0$ and $cost_1$ are functions that are applied when $y = 0$\nand $y = 1$, respectively. \\smallskip\n\nSVMs don't output probabilities, they output directly the class they\nclassificate into, using the following hyphotesis:\n\n\\begin{equation} \nh_\\theta(x) = \n\\begin{dcases}\n 1, & \\text{if\\,} \\theta^Tx \\geq 0 \\\\ \n 0, & \\text{otherwise}\n\\end{dcases} \n\\end{equation}\n\n\\section{Large Margin Intuition}\\label{large-margin-intuition}\n\nWe can see the SVM as a large margin classifier, meaning that if we have\nlinearly separable data, it will separate it optimizing the boundary\ndistance between the data (fig \\ref{fig:large_margin}).\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{img/large_margin.eps}\n\\caption{Example of linearly separable data correctly classified by an SVM.}\n\\label{fig:large_margin}\n\\end{figure}\n\n\\section{Kernels I}\\label{kernels-i}\n\nIn logistic regression, when we had a non-linear problem, we had to use\npolynomial features. Kernels are another way of getting more features\nout of the training sample. \\smallskip\n\nKernels are similarity functions. In the video, he selects 3 landmarks\nand uses a gaussian kernel. He then uses those 3 landmarks to compute 3\nfeatures, using the gaussian to know the similarity between x and each\nof the landmark.\n\nDoing that, we can create a decision boundary around the landmarks,\nwhere everything near them gets classified as 1 and everything that is\nfar from them gets classified as 0 (following a gaussian shape).\n\n\\section{Kernels II}\\label{kernels-ii}\n\n\\subsection{Choosing Landmarks}\\label{choosing-landmarks}\n\nFor every training example, get a landmark exactly at that location.\nThen, features will be based around the similarity of a point in our\ndata to other points. This way we have one feature per datapoint, so we\nhave m features. \\smallskip\n\nBasically, we just substitute the features we had and we use this ones.\n\nMathematical implemenation note:\n\n\\begin{equation} \n\\sum_{j=1}^n \\theta_j^2 = \\theta^t M \\theta\n\\end{equation}\n\nWhere M is a matrix that depends on the kernel. This is done to scale to\nbigger training sets.\n\n\\subsection{SVM Parameters}\\label{svm-parameters}\n\n\\begin{itemize}\n\\itemsep1pt\\parskip0pt\\parsep0pt\n\\item\n Large C: Lower bias, high variance (small $\\lambda$).\n\\item\n Small C: Higher bias, low variance (large $\\lambda$).\n\\end{itemize}\n\n\\section{Using an SVM}\\label{using-an-svm}\n\n\\begin{itemize}\n\\itemsep1pt\\parskip0pt\\parsep0pt\n\\item\n Use SVM software package to solve for paraemters $\\theta$.\n\\item\n You still need to specify the choice of parameter C and the kernel.\n\\item\n No kernel is sometimes called ``linear kernel''.\n\\item\n Do perfrom feature scaling before using the kernel.\n\\end{itemize}\n\nNot all similarity functions make valid kernels. They need to satisfy a\ntechnical condition called ``Mercer's Theorem''.\n\n\\subsection{Multiclass classification}\\label{multiclass-classification}\n\nMany SVM packages already have built-in multi-class classification\nfunctionality. Otherwise, use one-vs-all.\n\n\\subsection{Logistic regression vs\nSVMs}\\label{logistic-regression-vs-svms}\n\nLet n = number of features and m = number of training examples.\n\\smallskip\n\n\\begin{itemize}\n\\itemsep1pt\\parskip0pt\\parsep0pt\n\\item\n If n is large relative to m -\\textgreater{} Use logistic regression or\n SVM without kernel.\n\\item\n If n is small, m is intermediate -\\textgreater{} Use SVM with Gaussian\n kernel.\n\\item\n If n is small, m is large -\\textgreater{} Create/Add features and then\n use logistic regression or SVM without kernel.\n\\end{itemize}\n\nNeural networks is likely to work well for mos of these settings, but\nthey're way slower to train.\n\nSVMs don't have to worry about local optima, because they solve a convex\nproblem. NN may have problems with that, though.\n", "meta": {"hexsha": "6336dcc5d56a0e1a0905b680df64a5ddaf051997", "size": 4664, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "theory/07.01 - Support Vector Machines/07.01 - Support Vector Machines.tex", "max_stars_repo_name": "GMadorell/coursera-machine-learning", "max_stars_repo_head_hexsha": "7633ea83218e6f22f8ce4032f8f64d878732c2e9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-07-11T17:56:22.000Z", "max_stars_repo_stars_event_max_datetime": "2018-01-07T02:35:55.000Z", "max_issues_repo_path": "theory/07.01 - Support Vector Machines/07.01 - Support Vector Machines.tex", "max_issues_repo_name": "GMadorell/coursera-machine-learning", "max_issues_repo_head_hexsha": "7633ea83218e6f22f8ce4032f8f64d878732c2e9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "theory/07.01 - Support Vector Machines/07.01 - Support Vector Machines.tex", "max_forks_repo_name": "GMadorell/coursera-machine-learning", "max_forks_repo_head_hexsha": "7633ea83218e6f22f8ce4032f8f64d878732c2e9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.7971014493, "max_line_length": 139, "alphanum_fraction": 0.7647941681, "num_tokens": 1260, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837527911056, "lm_q2_score": 0.822189121808099, "lm_q1q2_score": 0.6277280362220709}} {"text": "\\documentclass[a4paper,11pt]{article}\n\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\usepackage{amsthm, mathtools}\n%\\usepackage{hyperref}\n\\usepackage{color}\n\\usepackage{jheppub}\n\\usepackage[T1]{fontenc} % if needed\n\n\\newcommand{\\be}{\\begin{equation}}\n\\newcommand{\\ee}{\\end{equation}}\n\\newcommand{\\bes}{\\begin{equation*}}\n\\newcommand{\\ees}{\\end{equation*}}\n\\newcommand{\\bea}{\\begin{flalign*}}\n\\newcommand{\\eea}{\\end{flalign*}}\n\n%\\linespread{1.0}\n%\\setlength{\\parindent}{0em}\n%\\setlength{\\parskip}{0.8em}\n\n\\title{\\textbf{Some Information about Gravitational Waves}}\n\\author{Aditya Vijaykumar}\n\\affiliation{International Centre for Theoretical Sciences, Bengaluru, India.}\n\\emailAdd{aditya.vijaykumar@icts.res.in}\n\\abstract{One needs to get accustomed with what gravitational waves are before jumping in and working on it. One hopes the two tomes by Maggiore will aid one's exploration.}\n\n\\begin{document}\n\\maketitle\n\n\\section{Gravitational Waves as Geometry}\n\n\\textit{Though Maggiore, the main reference for this tome, prefers not to use geometric units, I shall unapologetically use them.}\n\nThe gravitational action is a sum of the Einstein-Hilbert action and the matter action; $S = S_E + S_M$ where,\n\n$$S_E = \\frac{1}{16 \\pi} \\int d^4 x \\sqrt{-g}R$$\n\nThe \\textit{energy-momentum tensor} $T_{\\mu \\nu}$ is defined from the variation of the matter action under a metric change $g_{\\mu \\nu} \\rightarrow g_{\\mu \\nu} + \\delta g_{\\mu \\nu}$ as,\n\n$$\\delta S_M = \\frac{1}{2} \\int d^4 x \\sqrt{-g} \\ T^{\\mu \\nu} \\delta g_{\\mu \\nu}$$\n\nVariation of the total action with respect to $g_{\\mu \\nu}$ yields Einstein's equations,\n$$\\boxed{R_{\\mu \\nu} - \\frac{1}{2}g_{\\mu \\nu}R = 8 \\pi T_{\\mu \\nu}}$$\n\nGeneral relativity is invariant under a huge groups of transformations $x^\\mu \\rightarrow x'^\\mu(x)$, as long as $x'$ is invertible, differentiable, and has a differentiable inverse. Such transformations are called \\textit{diffeomorphisms}. Under diffeomorphisms, $g_{\\mu \\nu}$ transforms as,\n$$g_{\\mu \\nu}(x') = \\pdv{x^\\rho}{x'^\\mu} \\pdv{x^\\sigma}{x'^\\nu} g_{\\mu \\nu}(x)$$\n\n\\subsection{Expansions around flat space}\nLets imagine that our system is a perturbation around the flat space metric,\n$$g_{\\mu \\nu} = \\eta_{\\mu \\nu} + h_{\\mu \\nu}$$\nThe magnitude of $h_{\\mu \\nu}$ is infinitesmal, and we only consider equations upto first order in $h_{\\mu \\nu}$.\n\nLets stare at the above equation for a few seconds. We know that the numerical values of the tensor depends on the choice of reference frame. So on what basis are we calling $h_{\\mu \\nu}$ infinitesmal? Actually, what we merely wish to say is that for some choice of reference frame, the above equation will be valid and we can do all our calculations. While this choice helps us simplify our calculations, it should be noted that choosing a specific reference frame breaks the diffeomorphism invariance.\n\nEven after choosing the above transformation, we still have a residual gauge freedom left. Consider $$x'^\\mu \\rightarrow x^\\mu + \\xi^\\mu (x)$$ where the derivatives of $\\xi^\\mu$ as as inifinitesmal as $h_{\\mu \\nu}$. By using the diffeomorphism property,\n$$g'_{\\mu \\nu}(x') = \\pdv{x^\\rho}{x'^\\mu} \\pdv{x^\\sigma}{x'^\\nu} g_{\\rho \\sigma}(x)$$ and substituting for the partial derivatives, one gets,\n\\begin{flalign*}\n\tg'_{\\mu \\nu}(x') &= \\left(\\delta^\\rho_\\mu - \\pdv{\\xi^\\rho}{x'^\\mu}\\right) \\left(\\delta^\\sigma_\\nu - \\pdv{\\xi^\\sigma}{x'^\\nu}\\right) (\\eta_{\\rho \\sigma} + h_{\\rho \\sigma}(x)) \\\\\n\t\\eta_{\\mu \\nu} + h'_{\\mu \\nu}(x') &= \\left(\\delta^\\rho_\\mu \\delta^\\sigma_\\nu - \\pdv{\\xi^\\rho}{x'^\\mu}\\delta^\\sigma_\\nu - \\pdv{\\xi^\\sigma}{x'^\\nu} \\delta^\\rho_\\mu \\right) (\\eta_{\\rho \\sigma} + h_{\\rho \\sigma}(x))\\\\\n\th'_{\\mu \\nu}(x') &= h_{\\mu \\nu}(x) - \\pdv{\\xi_\\nu}{x'^\\mu} - \\pdv{\\xi_\\mu}{x'^\\nu} \n\\end{flalign*}\n\n\\end{document}\\grid\n", "meta": {"hexsha": "123990aa65d8ebcb856215ca3eaf82e484a6aaed", "size": 3814, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "selfstudy/gws/GWs.tex", "max_stars_repo_name": "adivijaykumar/courses", "max_stars_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "selfstudy/gws/GWs.tex", "max_issues_repo_name": "adivijaykumar/courses", "max_issues_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "selfstudy/gws/GWs.tex", "max_forks_repo_name": "adivijaykumar/courses", "max_forks_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 56.9253731343, "max_line_length": 503, "alphanum_fraction": 0.7026743576, "num_tokens": 1260, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677506936879, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6276439888993601}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{hyperref}\n\\usepackage{mathrsfs}\n\\usepackage{mathtools}\n\\usepackage{textcomp}\n\\usepackage{fullpage}\n\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{lemma}[theorem]{Lemma}\n\\theoremstyle{definition}\n\\newtheorem{definition}[theorem]{Definition}\n\n\\newcommand{\\defterm}[1]{\\textit{#1}}\n\n\\title{A Compressed Introduction to Compressed Sensing}\n\\author{Benjamin Peterson}\n\n\\begin{document}\n\\maketitle\n\n\\begin{abstract}\nWe attempt to convey a sense of compressed sensing. Specifically, we discuss how $\\ell_1$ minimization and the restricted isometry property for matrices can be used for sparse recovery of underdetermined linear systems even in the presence of noise.\n\\end{abstract}\n\n\\section{Introduction}\nSuppose $A \\in \\mathcal{M}_{m\\times n}(\\mathbb{R})$ is a matrix and we obtain a measurement $y \\in \\mathbb{R}^m$, which we know to be the image of some $x \\in \\mathbb{R}^n$ under $A$. If the system $Ax = y$ is underdetermined (i.e., $m < n$), then $x$ is not unique. Suppose however that we want a single solution and we know $x$ is ``small''. We might try to pick an $x$ which solves the equation with the least $\\ell_2$ norm by solving the convex program\n\\begin{equation}\\label{eq:l2-prog}\n\\min_{x \\in \\mathbb{R}^n}\\|x\\|_{\\ell_2}\\qquad Ax = y.\\tag{L}\n\\end{equation}\nThis program has the advantage that a unique solution is guaranteed. Unfortunately, the solution to \\eqref{eq:l2-prog} is often not the best one for applications.\n\nMany real world signals are known to be sparse (i.e.\\ having few nonzero entries) in some basis. Thus, we might alternatively seek a sparse solution for $Ax = y$. The sparsest solution to the linear equation is given by the program\n\\begin{equation}\\label{eq:l0-prog}\n\\min_{x\\in \\mathbb{R}^n} \\|x\\|_{\\ell_0}\\qquad Ax = y.\\tag{S}\n\\end{equation}\nHere the ``norm'' $\\|x\\|_{\\ell_0}$ counts the number of nonzero entries in a vector $x$. Solving \\eqref{eq:l0-prog} is a difficult NP-hard problem—the subset-sum problem may be reduced to it—in combinatorial optimization \\cite{sparse-np}. We will replace \\eqref{eq:l0-prog} with the convex program\n\\begin{equation}\\label{eq:l1-prog}\n\\min_{x \\in \\mathbb{R}^n}\\|x\\|_{\\ell_1}\\qquad Ax = y.\\tag{P}\n\\end{equation}\nSolving this program is often called ``basis pursuit'' and can be efficiently done with linear programming or specialized algorithms \\cite[ch. 15]{foucart-intro}. Relating the solutions of \\eqref{eq:l0-prog} and \\eqref{eq:l1-prog} is major goal in compressed sensing which we will explore in this paper.\n\nThere is an intuitive geometrical reason that we might expect the $\\ell_1$ norm to be better at finding sparse solutions than the $\\ell_2$ norm. The program~\\eqref{eq:l2-prog} finds smallest ball around the origin that intersects the solution subspace of $Ax = y$. The level sets of the $\\ell_1$ norm are polyhedra, which emphasize the axes. Therefore, it seems ``likely'' that the \\eqref{eq:l1-prog} finds a sparse solution in the solution subspace.\n\nWe now introduce a restriction on matrices that allows sparse solutions to be easily recovered. Recall that a vector is said to be $k$-sparse if it has at most $k$ nonzero entries.\n\\begin{definition}[\\cite{ct-decoding}]\nA matrix $A \\in \\mathcal{M}_{m\\times n}(\\mathbb{R})$ satisfies the \\defterm{restricted isometry property} (RIP) of order $k$ if there exists a $\\delta_k \\geq 0$ such that for all $k$-sparse vectors $x \\in \\mathbb{R}^n$, $$(1 - \\delta_k)\\|x\\|_{\\ell_2}^2 \\leq \\|Ax\\|_{\\ell_2}^2 \\leq (1 + \\delta_k)\\|x\\|_{\\ell_2}^2.$$\nThe smallest such $\\delta_k$ is called $A$'s \\defterm{restricted isometry constant} of order $k$.\n\\end{definition}\nNotice that $\\delta_1 \\leq \\delta_2 \\leq \\dotsb$. The RIP is related to vector space frames \\cite{ole-frames} and the Johnson-Lindenstrauss lemma \\cite{rip-jl}. Unfortunately, checking whether a matrix has the restricted isometry property is NP-Hard in general \\cite{rip-np}. On the other hand, many families of random matrices (e.g.\\ having Gaussian or Bernoulli entries) satisfy the RIP with high probability \\cite[\\S 1.3]{ctr-stable}.\n\n\\begin{theorem}[{\\cite[lemma 1.2]{ct-decoding}}]\\label{thm:unique-sparse}\nSuppose $A \\in \\mathcal{M}_{m\\times n}$ satisfies RIP with $\\delta_{2s} < 1$. Then the equation $Ax = y$ has an unique $s$-sparse solution given by \\eqref{eq:l0-prog}.\n\\end{theorem}\n\\begin{proof}\nSuppose $x, x\\in \\mathbb{R}^m$ are $s$-sparse and $Ax = Ax'$. By the RIP,\n$$\n0 \\leq (1 - \\delta_{2s})\\|x - x'\\|_{\\ell_2} \\leq \\|A(x - x')\\|_{\\ell_2} = 0.\n$$\nNecessarily, $x = x'$.\n\\end{proof}\n\nBefore stating the main theorem, we introduce the notion of error. In reality, we do not know $Ax$ exactly. Instead, we measure $y = Ax + z$, where $z \\in \\mathbb{R}^m$ is a small error vector satisfying $\\|z\\|_{\\ell_2} \\leq \\epsilon$. We would like our recovery of $x$ to be robust against error. To deal with this, we generalize \\eqref{eq:l1-prog} to the (still convex) program\n\\begin{equation}\\tag{P\\textquotesingle}\\label{eq:l1-approx}\n\\min_{x \\in \\mathbb{R}^n}\\|x\\|_{\\ell_1}\\qquad \\|Ax - y\\|_{\\ell_2} \\leq \\epsilon.\n\\end{equation}\n\nFinally, we can state the main theorem of this lecture. We denote the vector containing only the $s$ largest entries of $x$ by $x_s$.\n\\begin{theorem}[{\\cite[1.3]{candes-rip}}]\\label{thm:noisy-recovery}\nSuppose $A \\in \\mathcal{M}_{m\\times n}(\\mathbb{R})$ satisfies RIP with $\\delta_{2s} < \\sqrt{2} - 1$. Let $x^*$ denote the solution to \\eqref{eq:l1-approx}. Then, there are constants $C_0, C_1$ depending only on $\\delta_{2s}$ such that $$\\|x^* - x\\|_{\\ell_2} \\leq C_0s^{-1/2}\\|x - x_s\\|_{\\ell_1} + C_1\\epsilon.$$\n\\end{theorem}\n\nTheorem~\\ref{thm:noisy-recovery} tells us that when RIP holds, basis pursuit recovers solutions very close to sparse solutions of the equation.\n\n\\section{Proof of Theorem~\\ref{thm:noisy-recovery}}\nWe closely follow \\cite{candes-rip}.\n\nSet $x = x^* + h$. Our goal is to show $\\|h\\|_{\\ell_2}$ is small. The RIP for $A$ only gives us control over sparse vectors, so we start by breaking $h$ up into $s$-sparse vectors. Let $T_0$ be the indexes of the $s$ largest entires of $x$. Let $T_1$ be the indexes of the $k$ largest entries of $h_{T_0^c}$. Let $T_2$ be the indexes of the $s$ largest entries of $h_{(T_0\\cup T_1)^c}$ and so on. We observe that $h$ can be written as the sum of $s$-sparse vectors $h_{T_0} + h_{T_1} + h_{T_2} + \\dotsb$. Using the triangle inequality,\n\\begin{equation}\\label{eq:main-triang}\n\\|x - x^*\\|_{\\ell_2} = \\|h\\|_{\\ell_2} \\leq \\|h_{T_0\\cup T_1}\\|_{\\ell_2} + \\|h_{(T_0\\cup T_1)^c}\\|_{\\ell_2}.\n\\end{equation}\nWe will estimate the two terms of \\eqref{eq:main-triang} separately and then combine them to prove the theorem. We start by showing the $\\|h_{(T_0\\cup T_1)^c}\\|_{\\ell_2}$ term can be bounded in terms of the first term $\\|h_{T_0\\cup T_1}\\|_{\\ell_2}$. Several useful intermediate inequalities are obtained along the way.\n\n\\begin{lemma}[Tail estimates]\\label{lem:tail-estimate}\n\\begin{align}\\label{eq:small-h1}\n\\|h_{T_0^c}\\|_{\\ell_1} &\\leq \\|h_{T_0}\\|_{\\ell_1} + 2\\|x_{T_0^c}\\|_{\\ell_1}\n\\\\\n\\label{eq:tail-sum}\n\\sum_{j \\geq 2}\\|h_{T_j}\\|_{\\ell_2} &\\leq s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1} \\\\\n\\label{eq:small-h2}\n\\|h_{(T_0\\cup T_1)^c}\\|_{\\ell_2} &\\leq \\|h_{T_0}\\|_{\\ell_2} + 2s^{-1/2}\\|x - x_s\\|_{\\ell_1}\n\\end{align}\n\\end{lemma}\n\\begin{proof}\nSince $x$ is feasible for \\eqref{eq:l1-approx} and $x^*$ is a minimum,\n$$\n\\|x\\|_{\\ell_1} \\geq \\|x + h\\|_{\\ell_1} = \\sum_{i \\in T_0}|x_i + h_i| + \\sum_{i \\in T_0^c}|x_i + h_i| \\geq \\|x_{T_0}\\|_{\\ell_1} - \\|h_{T_0}\\|_{\\ell_1} + \\|h_{T_0^c}\\|_{\\ell_1} - \\|x_{T_0^c}\\|_{\\ell_1}.\n$$\nThe last step uses the triangle inequality twice. Rewriting and applying the reverse triangle inequality proves \\eqref{eq:small-h1}.\n$$\n\\|h_{T_0^c}\\|_{\\ell_1} \\leq \\|x\\|_{\\ell_1} - \\|x_{T_0}\\|_{\\ell_1} + \\|x_{T_0^c}\\|_{\\ell_1} + \\|h_{T_0}\\|_{\\ell_1} \\leq \\|h_{T_0}\\|_{\\ell_1} + 2\\|x_{T_0^c}\\|_{\\ell_1}\n$$\n\nIf $j \\geq 2$,\n$$\n\\|h_{T_j}\\|_{\\ell_2} \\leq s^{1/2}\\|h_{T_j}\\|_{\\ell_\\infty} \\leq s^{-1/2}\\|h_{T_{j - 1}}\\|_{\\ell_1}.\n$$\nSumming yields \\eqref{eq:tail-sum}.\n$$\n\\sum_{j \\geq 2}\\|h_{T_j}\\|_{\\ell_2} \\leq s^{-1/2}\\sum_{j \\geq 1}\\|h_{T_j}\\|_{\\ell_1}= s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1}\n$$\nFrom this, we immediately obtain\n\\begin{equation}\\label{eq:small-h3}\n\\|h_{(T_0\\cup T_1)^c}\\|_{\\ell_2} = \\left\\|\\sum_{j \\geq 2}h_{T_j}\\right\\|_{\\ell_2} \\leq \\sum_{j \\geq 2}\\|h_{T_j}\\|_{\\ell_2} \\leq s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1}.\n\\end{equation}\n\nBy the Cauchy-Schwarz inequality,\n\\begin{equation}\\label{eq:lem1-cs}\n\\|h_{T_0}\\|_{\\ell_1} \\leq s^{1/2}\\|h_{T_0}\\|_{\\ell_2}.\n\\end{equation}\nApplying \\eqref{eq:small-h3}, \\eqref{eq:small-h1}, and \\eqref{eq:lem1-cs} proves \\eqref{eq:small-h2}.\n$$\n\\|h_{(T_0\\cup T_1)^c}\\|_{\\ell_2} \\leq s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1} \\leq s^{-1/2}(\\|h_{T_0}\\|_{\\ell_1} + 2\\|x_{T_0^c}\\|_{\\ell_1}) \\leq \\|h_{T_0}\\|_{\\ell_2} + 2s^{-1/2}\\|x_{T_0^c}\\|_{\\ell_1}\n$$\n\\end{proof}\n\nNext, we want to bound the main part of the error term $\\|h_{T_0\\cup T_1}\\|_{\\ell_2}$. We begin with a lemma.\n\n\\begin{lemma}\\label{lem:innner-control}\nSuppose $x$ and $x'$ are $s$-sparse and $s'$-sparse respectively and supported on disjoint sets. Then\n$$\n|\\langle Ax, Ax'\\rangle| \\leq \\delta_{s + s'}\\|x\\|_{\\ell_2}\\|x'\\|_{\\ell_2}\n$$\n\\end{lemma}\n\\begin{proof}\nWe may assume without a loss of generality that $\\|x\\|_{\\ell_2} = \\|x'\\|_{\\ell_2} = 1$. The RIP tells us that\n$$\n2(1 - \\delta_{s + s'}) = (1 - \\delta_{s + s'})\\|x + x'\\|_{\\ell_2}^2 \\leq \\|Ax \\pm Ax'\\|_{\\ell_2}^2 \\leq (1 + \\delta_{s + s'})\\|x + x'\\|_{\\ell_2}^2 = 2(1 - \\delta_{s+s'}).\n$$\nBy the polarization identity,\n$$\n|\\langle Ax, Ax'\\rangle| \\leq \\frac{1}4\\left|\\|Ax + Ax'\\|_{\\ell_2}^2 - \\|Ax - Ax'\\|_{\\ell_2}^2\\right| \\leq \\delta_{s + s'}.\n$$\n\\end{proof}\n\n\\begin{lemma}[Main term estimate]\\label{lem:main-estimate}\n\\begin{equation}\\label{eq:main-term-est}\n\\|h_{T_0\\cup T_1}\\|_{\\ell_2} \\leq (1 - \\rho)^{-1}(\\alpha\\epsilon + 2\\rho s^{-1/2}\\|x - x_s\\|_{\\ell_1})\n\\end{equation}\nwhere\n$$\n\\alpha \\equiv \\frac{2\\sqrt{1 - \\delta_{2s}}}{1 - \\delta_{2s}},\\quad \\rho \\equiv \\frac{\\sqrt{2}\\delta_{2s}}{1 - \\delta_{2s}}.\n$$\n\\end{lemma}\n\\begin{proof}\nRIP allows us to control the size of $\\|h_{T_0\\cup T_1}\\|_{\\ell_2}$ with $\\|Ah_{T_0\\cup T_1}\\|_{\\ell_2}$, and so we start by bounding the latter. We break $\\|Ah_{T_0\\cup T_1}\\|_{\\ell_2}$ into parts using properties of the inner product.\n\\begin{equation}\\label{eq:main-bound}\n\\|Ah_{T_0\\cup T_1}\\|_{\\ell_2}^2 = \\langle A h_{T_0\\cup T_1}, Ah \\rangle - \\sum_{j \\geq 2}\\left(\\langle A h_{T_0}, Ah_{T_j}\\rangle + \\langle Ah_{T_1}, Ah_{T_j}\\rangle\\right)\n\\end{equation}\nFrom the triangle inequality and hypothesis,\n\\begin{equation}\\label{eq:Ah-bound}\n\\|Ah\\|_{\\ell_2} = \\|A(x - x^*)\\|_{\\ell_2} \\leq \\|Ax^* - y\\|_{\\ell_2} + \\|y - Ax\\|_{\\ell_2} \\leq 2\\epsilon.\n\\end{equation}\nThe first term of \\eqref{eq:main-bound} can be bounded using Cauchy-Schwarz, the RIP, and \\eqref{eq:Ah-bound}.\n\\begin{equation}\\label{eq:first-term-est}\n|\\langle Ah_{T_0\\cup T_1}, Ah\\rangle| \\leq \\|Ah_{T_0\\cup T_1}\\|_{\\ell_2}\\|Ah\\|_{\\ell_2} \\leq 2\\epsilon \\sqrt{1 + \\delta_{2s}}\\|h_{T_0\\cup T_1}\\|_{\\ell_2}\n\\end{equation}\nSince $T_0$ and $T_1$ are disjoint Cauchy-Schwarz gives\n$$\n\\|h_{T_0}\\|_{\\ell_2} + \\|h_{T_1}\\|_{\\ell_2} \\leq \\sqrt{2}\\|h_{T_0\\cup T_1}\\|_{\\ell_2}.\n$$\nTo estimate the sum term of \\eqref{eq:Ah-bound}, we apply lemma~\\ref{lem:innner-control} and \\eqref{eq:tail-sum}.\n\\begin{align}\n\\sum_{j \\geq 2}|\\langle Ah_{T_0}, Ah_{T_j}\\rangle| + |\\langle Ah_{T_1}, Th_{T_j}\\rangle| &\\leq \\delta_{2s}(\\|h_{T_0}\\|_{\\ell_2} + \\|h_{T_1}\\|_{\\ell_2})\\sum_{j \\geq 2}\\|h_{T_j}\\|_{\\ell_2}\n\\nonumber\\\\&\\leq \\delta_{2s}\\sqrt{2}s^{-1/2}\\|h_{T_0\\cup T_1}\\|_{\\ell_2}\\|h_{T_0^c}\\|_{\\ell_1} \\label{eq:second-term-est}\n\\end{align}\nWe now have\n$$\n(1 - \\delta_{2s})\\|h_{T_0\\cup T_1}\\|_{\\ell_2}^2 \\leq \\|Ah_{T_0\\cup T_1}\\|_{\\ell_2}^2\\leq \\|h_{T_0\\cup T_1}\\|_{\\ell_2}(2\\epsilon\\sqrt{1 +\\delta_{2s}} + \\sqrt{2}\\delta_{2s}s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1})\n$$\nfrom RIP, \\eqref{eq:first-term-est}, and \\eqref{eq:second-term-est}. We divide by $(1 - \\delta_{2s})\\|h_{T_0\\cup T_1}\\|_{\\ell_2}$.\n$$\n\\|h_{T_0\\cup T_1}\\|_{\\ell_2} \\leq \\alpha\\epsilon + \\rho s^{-1/2}\\|h_{T_0^c}\\|_{\\ell_1}\n$$\nFrom \\eqref{eq:small-h1} and \\eqref{eq:lem1-cs}, we obtain\n$$\n\\|h_{T_0 \\cup T_1}\\|_{\\ell_2} \\leq \\alpha\\epsilon + \\rho s^{-1/2}(\\|h_{T_0}\\|_{\\ell_1} + 2\\|x - x_s\\|_{\\ell_1}) \\leq \\alpha\\epsilon + \\rho \\|h_{T_0\\cup T_1}\\|_{\\ell_2} + 2\\rho s^{-1/2}\\|x - x_s\\|_{\\ell_1}.\n$$\nBy the hypothesis of theorem~\\ref{thm:noisy-recovery}, $\\rho < 1$, and we get\n$$\n\\|h_{T_0\\cup T_1}\\|_{\\ell_2} \\leq (1 - \\rho)^{-1}(\\alpha\\epsilon + 2\\rho s^{-1/2}\\|x - x_s\\|_{\\ell_1}),\n$$\nwhich completes the lemma's proof.\n\\end{proof}\n\nApplying our estimates \\eqref{eq:small-h2} and lemma~\\ref{lem:main-estimate} to the two terms of \\eqref{eq:main-triang}, we have\n$$\n\\|h\\|_{\\ell_2} \\leq 2\\|h_{(T_0\\cup T_1)}\\|_{\\ell_2} + 2s^{-1/2}\\|x - x_s\\|_{\\ell_1} \\leq 2(1 - \\rho)^{-1}(\\alpha\\epsilon + (1 + \\rho)s^{-1/2}\\|x - x_s\\|_{\\ell_1}).\n$$\nThis completes the proof of theorem~\\ref{thm:noisy-recovery}.\n\n\\section{Remarks}\nA huge amount of research in compressed sensing has appeared since the publication of the original papers \\cite{ctr-stable,donoho-cs} around 2004-2006. Among other things, researchers have investigated techniques for sparse recovery besides basis pursuit e.g.\\ matching pursuit \\cite{tg-matching}. There is also replacement condition for RIP called the nullspace property, which is a necessary and sufficient condition for \\eqref{eq:l0-prog} and \\eqref{eq:l1-prog} to have the same solutions \\cite{nullspace}.\n\nAs far as applications are concerned, compressed sensing is being used in areas as diverse as tomography, astronomy, machine linearing, linear coding, and experiment design \\cite{cs-science,dantzig-selector}. It is particularly useful in situations where minimizing the work done in sensors is important such as space probes. Gimmicks like single-pixel cameras \\cite{single-pixel} have captured the public imagination.\n\nFinally, for readers still curious, there are a lot of other (possibly more palatable) introductions to compressed sensing \\cite{cw-intro,cs-book,foucart-intro,users-guide,kutyniok-cs,qaisar-compressive,romberg-intro}.\n\n\\bibliographystyle{amsplain}\n\\bibliography{compressed_sensing}\n\n\\end{document}\n", "meta": {"hexsha": "e2d07a11ae2b20efda715a884f7156692b534f4b", "size": 14285, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compressed_sensing.tex", "max_stars_repo_name": "benjaminp/compressed-sensing-intro", "max_stars_repo_head_hexsha": "c284fec4ce921d41f208ab7be324bd51d253c27c", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2017-12-12T05:46:33.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T23:10:50.000Z", "max_issues_repo_path": "compressed_sensing.tex", "max_issues_repo_name": "benjaminp/compressed-sensing-intro", "max_issues_repo_head_hexsha": "c284fec4ce921d41f208ab7be324bd51d253c27c", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compressed_sensing.tex", "max_forks_repo_name": "benjaminp/compressed-sensing-intro", "max_forks_repo_head_hexsha": "c284fec4ce921d41f208ab7be324bd51d253c27c", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 65.8294930876, "max_line_length": 535, "alphanum_fraction": 0.6828841442, "num_tokens": 5579, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872243177518, "lm_q2_score": 0.8376199694135332, "lm_q1q2_score": 0.6274504179211038}} {"text": "\\section{Algorithms}\n\nIn this section we will describe the randomized algorithms in detail, provide the\ncorresponding computational complexity analysis, and state the main theoretical results\nthat guarantee the accuracy of the approximation.\n\nWe split this section in three main parts: Stage 1 \\ref{sec:stage1} and Stage 2\n\\ref{sec:stage2} that will study in depth the algorithmic details of both stages,\nand Full Algorithms \\ref{sec:full-algos} in which full algorithms will be presented\nfor a diversity of matrices with certain properties.\n\n\\subsection{Stage 1}\n\\label{sec:stage1}\nIn the introductory section, we provided some intuition on the randomized procedure\nand we developed a general Proto-Algorithm \\ref{alg:proto-algorithm} to find\nthe range-approximating matrix $\\mtx{Q}$. However, Proto-Algorithm \\ref{alg:proto-algorithm} is very general and can be\ntunned depending on the problem requirements.\nThe number $T_{\\rm basic}$ of flops required by Proto-Algorithm \\ref{alg:proto-algorithm} satisfies\n\\begin{equation}\n\\label{eq:cost_basic}\nT_{\\rm basic} \\sim \\ell n \\, T_{\\rm rand} + \\ell\\,T_{\\rm mult} + \\ell^{2}m %\\,T_{\\rm flop},\n\\end{equation}\nwhere $T_{\\rm rand}$ is the cost of generating a Gaussian random number\nand $T_{\\rm mult}$ is the cost of multiplying $\\mtx{A}$ by a vector.\nThe last term comes from the orthonormalization procedure of $\\mtx{Y}$.\n\nWe will now describe some specific realizations of Proto-Algorithm \\ref{alg:proto-algorithm} \nthat will be intended for problems with different requirements.\n\n%%%%%%%%%%%%%%%%%% RANDOMIZED RANGE FINDER %%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsubsection{Randomized Range Finder} \\label{alg:randomized-range-finder}\nThis is the most naive and simplest implementation of Proto-Algorithm \\ref{alg:proto-algorithm}.\nGiven an oversampling parameter $p$, the \\textit{Randomized Range Finder}\nperforms Proto-Algorithm \\ref{alg:proto-algorithm} with a gaussian test matrix \n$\\mtx{\\Omega}\\in\\Rspace{n\\times\\ell}$ with $\\ell=k+p$ and $k$ being a pre-specified\ntarget rank. Then, it orthogonalizes the rows of the resulting matrix $\\mtx{Y}$ by computing\na QR decomposition. A numerical issue arises when computing the orthogonalization\nprocedure due to the fact that the columns of $\\mtx{Y}$ are almost linearly\ndependent. The authors in ~\\cite{halko2011finding} found that\nusing the \\textit{double orthogonalization} \n~\\cite{bjorck1994numerics} was enough to guarantee stability\nof the procedure.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\fbox{\n\\begin{minipage}{.9\\textwidth}\n\\begin{center}\n\\textsc{Randomized Range Finder}\n\\end{center}\n\\begin{tabbing}\n\\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\kill\n\\anum{1} \\>Draw an $n\\times \\ell$ Gaussian random matrix $\\mtx{\\Omega}$.\\\\\n\\anum{2} \\>Form the $m\\times \\ell$ matrix $\\mtx{Y} = \\mtx{A}\\mtx{\\Omega}$.\\\\\n\\anum{3} \\>Construct an $m \\times \\ell$ matrix $\\mtx{Q}$ whose columns form an orthonormal\\\\\n \\> basis for the range of $\\mtx{Y}$, e.g., using the QR factorization $\\mtx{Y} = \\mtx{Q}\\mtx{R}$.\n\\end{tabbing}\n\\end{minipage}}\n\\end{center}\n\\end{figure}\n\nThe complexity analysis of the Algorithm \\ref{alg:randomized-range-finder}\ngives\n\n\\begin{equation}\\label{eq:analysis-rand-finder}\nT_{\\text{Randomized Range Finder}}\\sim \\mathcal{O}(mn\\ell)\n\\end{equation}\n\nthis is because generating a gaussian random number is $\\mathcal{O}(1)$\nand computing a matrix vector multiplication is $\\mathcal{O}(mn)$\n\\footnote{Note that parallel computation can be extremely helpful here\nto drastically reduce the effective complexity of the product $\\mtx{A}\\mtx{\\Omega}$.}. \nOne important observation is that the complexity $\\mathcal{O}(mn\\ell)$\ncan be prohibitive for massive datasets \\footnote{Obviously, we can't hope\nto do better than $\\mathcal{O}(mn)$ which is the size of the matrix\nwe want to approximate. The term that we want to address is the linear term\nin $\\ell$. Classical algorithms to compute the $\\ell$-SVD cost \n$\\mathcal{O}(mn\\ell)$, by first computing s rank-revealing QR factorization\n~\\cite{gu1996efficient} and then manipulating the terms to get to the desired\nfactorization.}. A variant of this procedure\nwill be studied in \\ref{alg:fast-randomized-range-finder}\nto get around this problem,\n\n%%%%%%%%%%%%%%%%%% ADAPTIVE RANDOMIZED RANGE FINDER %%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsubsection{Adaptive Randomized Range Finder}\n\\label{alg:adaptive-randomized-range-finder}\n\nOne important pitfall of the \\textit{Randomized Range Finder}\n\\ref{alg:randomized-range-finder} is that it\nrequires to know in advance the target rank $k$. However, if we intend\nto solve the \\textit{fixed-precision approximation problem}, we need a scheme to estimate\nthe error $\\|\\mtx{A}-\\mtx{Q}\\mtx{Q}^\\adj\\mtx{A}\\|$ during the algorithm\nin order to match the requited tolerance $\\epsilon$.\n\nThis scheme is possible and it is direct consequence of the following lemma.\n\\begin{lemma}\n\\label{thm:aposteriori}\nLet $\\mtx{B}$ be a real $m\\times n$ matrix.\nFix a positive integer $r$ and a real number $\\alpha > 1$.\nDraw an independent family $\\{ \\vct{\\omega}^{(i)} : i = 1, 2, \\dots, r \\}$\nof standard Gaussian vectors. Then\n\\begin{equation*}\n\\norm{\\mtx{B}}\n \\leq \\alpha \\sqrt{\\frac{2}{\\pi}} \\max_{i = 1, \\dots, r}\n \\smnorm{}{\\mtx{B}\\vct{\\omega}^{(i)} }\n\\end{equation*}\nexcept with probability $\\alpha^{-r}$.\n\\end{lemma}\n\nLemma \\ref{thm:aposteriori} says that we can bound the error with high\nprobability using inexpensive computations in an online manner. \nThe Lemma \\ref{thm:aposteriori} applied to our problem reads\n\n\\begin{equation}\n\\label{eq:errorest}\n\\norm{ (\\Id - \\mtx{Q}\\mtx{Q}^{\\adj})\\mtx{A}}\n \\leq 10 \\sqrt{\\frac{2}{\\pi}} \\max_{i = 1, \\dots, r}\n \\smnorm{}{ (\\Id - \\mtx{Q}\\mtx{Q}^{\\adj}) \\mtx{A}\\vct{\\omega}^{(i)} }\n\\end{equation}\nwith probability at least $1 - 10^{-r}$.\n\nThe high probability bound \\ref{eq:errorest} gives a simple online scheme\nto decide when we have a good enough $\\mtx{Q}$ that matches the pre-specified\ntolerance. The goal here is to find an integer $l$ and a $m\\times l$\northonormal matrix $\\mtx{Q}^{(l)}$ such that\n\\begin{equation} \\label{eqn:err_est_err_bd}\n\\smnorm{}{ \\big(\\Id - \\mtx{Q}^{(\\ell)} (\\mtx{Q}^{(\\ell)})^{\\adj} \\big)\\mtx{A} } \\leq \\eps.\n\\end{equation}\n\nWe call \\textit{Adaptive Randomized Range Finder}\n\\ref{alg:adaptive-randomized-range-finder} to the\nalgorithm derived from Lemma \\ref{thm:aposteriori} that solves this\nproblem.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\fbox{\n\\begin{minipage}{.9\\textwidth}\n\\begin{center}\n\\textsc{Adaptive Randomized Range Finder}\n\\end{center}\n\\begin{tabbing}\n\\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\kill\n\\anum {1} \\> Draw standard Gaussian vectors $\\vct{\\omega}^{(1)}, \\dots, \\vct{\\omega}^{(r)}$ of length $n$.\\\\\n\\anum{2} \\> For $i = 1,2,\\dots,r$, compute $\\vct{y}^{(i)} = \\mtx{A}\\vct{\\omega}^{(i)}$.\\\\\n\\anum{3} \\> $j=0$.\\\\\n\\anum{4} \\> $\\mtx{Q}^{(0)} = [\\ ]$, the $m\\times 0$ empty matrix. \\\\\n\\anum{5} \\> \\textbf{while} $\\displaystyle\n \\max\\left\\{\\smnorm{}{\\vct{y}^{(j+1)}},\\smnorm{}{\\vct{y}^{(j+2)}},\\dots,\\smnorm{}{\\vct{y}^{(j+r)}} \\right\\} >\n\\varepsilon/(10\\sqrt{2/\\pi})$,\\\\\n\\anum{6} \\> \\> $j = j + 1$.\\\\\n\\anum{7} \\> \\> Overwrite $\\vct{y}^{(j)}$ by $\\bigl(\\Id - \\mtx{Q}^{(j-1)}(\\mtx{Q}^{(j-1)})^{\\adj}\\bigr)\\vct{y}^{(j)}$.\\\\\n\\anum{8} \\> \\> $\\vct{q}^{(j)} = \\vct{y}^{(j)}/\\norm{\\vct{y}^{(j)}}$.\\\\\n\\anum{9} \\> \\> $\\mtx{Q}^{(j)} = [\\mtx{Q}^{(j-1)}\\ \\vct{q}^{(j)}]$.\\\\\n\\anum{10} \\> \\> Draw a standard Gaussian vector $\\vct{\\omega}^{(j+r)}$ of length $n$.\\\\\n\\anum{11} \\> \\> $\\vct{y}^{(j+r)} = \\left(\\Id - \\mtx{Q}^{(j)}(\\mtx{Q}^{(j)})^{\\adj}\\right)\\mtx{A}\\vct{\\omega}^{(j+r)}$.\\\\\n\\anum{12} \\> \\> \\textbf{for} $i = (j+1),(j+2),\\dots,(j+r-1)$,\\\\\n\\anum{13} \\> \\> \\> Overwrite $\\vct{y}^{(i)}$ by $\\vct{y}^{(i)} - \\vct{q}^{(j)}\\ip{\\vct{q}^{(j)}}{\\vct{y}^{(i)}}$.\\\\\n\\anum{14} \\> \\> \\textbf{end for}\\\\\n\\anum{15} \\> \\textbf{end while}\\\\\n\\anum{16} \\> $\\mtx{Q} = \\mtx{Q}^{(j)}$.\n\\end{tabbing}\n\\end{minipage}}\n\\end{center}\n\\end{figure}\n\nOne important question regarding Algorithm \\ref{alg:adaptive-randomized-range-finder}\n is how good the bound given by Lemma \\ref{thm:aposteriori}\nis in practice. If there is a significant gap between theory and practice\nthe optimal $l$ will be overestimated. This question will be addressed in the experimental section\n\\ref{sec:gaussian-matrices}.\n\n%%%%%%%%%%%%%%%%%% RANDOMIZED POWER ITERATION %%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsubsection{Randomized Power Iteration}\n\\label{alg:randomized-power-iteration}\n\nThe \\textit{Randomized Range Finder} \\ref{alg:randomized-range-finder}\nalgorithm assumes that the singular values\nof the matrix decay fast. This can be intuitively seen from equation \\ref{eq:iter-range},\nwhere the small singular\nvalues interfere with the calculation of the range. This intuition is made precise \nin Theorem \\ref{thm:avg-frob-error-gauss}, where the error of the approximation depends on\nthe $\\sigma_{k+1}$ and $\\sum_{j>k}\\sigma_j^2$.\n\nThe goal here is to reduce the weight of the\nsmall singular values by taking powers of the matrix whose range we want\nto approximate. Instead of applying the sampling scheme to $\\mtx{A}$, we will\napply it to $\\mtx{B}=(\\mtx{A}\\mtx{A}^\\adj)^q\\mtx{A}$ where $q>0$ is a small\ninteger.\n\nThe matrix $\\mtx{B}$ has the same singular vectors than $\\mtx{A}$ (hence, \nthe same range), but its singular values decay much more quickly.\n\n\\begin{equation}\\label{eq:sing-values-power}\n\\sigma_j(\\mtx{B}) = \\sigma_j(\\mtx{A})^{2q+1},\n\\hspace{0.5cm} j=1,2,3,\\ldots\n\\end{equation}\n\nThe \\textit{Randomized Power Iteration}\n\\ref{alg:randomized-power-iteration} algorithm is the same \nas the \\textit{Randomized Range Finder} \\ref{alg:randomized-range-finder}\n but replacing the formula\n$\\mtx{Y}=\\mtx{A}\\mtx{\\Omega}$ by $\\mtx{Y}=\\mtx{B}\\mtx{\\Omega}$.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\fbox{\n\\begin{minipage}{.9\\textwidth}\n\\begin{center}\n\\textsc{Randomized Power Iteration}\n\\end{center}\n\\begin{tabbing}\n\\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\kill\n\\anum{1} \\>Draw an $n\\times \\ell$ Gaussian random matrix $\\mtx{\\Omega}$.\\\\\n\\anum{2} \\>Form the $m\\times \\ell$ matrix $\\mtx{Y} = (\\mtx{A}\\mtx{A}^{\\adj})^{q}\\mtx{A}\\mtx{\\Omega}$ via alternating application\\\\\n \\>of $\\mtx{A}$ and $\\mtx{A}^{\\adj}$.\\\\\n\\anum{3} \\>Construct an $m \\times \\ell$ matrix $\\mtx{Q}$ whose columns form an orthonormal\\\\\n \\> basis for the range of $\\mtx{Y}$, e.g., via the QR factorization $\\mtx{Y} = \\mtx{Q}\\mtx{R}$.\n\\end{tabbing}\n\\end{minipage}}\n\\end{center}\n\\end{figure}\n\nThe computational complexity of the Algorithm \\ref{alg:randomized-power-iteration}\nis essentially the same because\nit only requires $2q+1$ as many matrix-multiplications as Algorithm\n\\ref{alg:randomized-range-finder}\nbut the number $q$ is in practice 2,3 or 4. This can be seen from \nCorollary \\ref{cor:power-method-spec-gauss},\nwhich shows that the power iteration drives the approximation\ngap to 1 exponentially fast as $q$ increases.\n\n\n%%%%%%%%%%%%%%%%%% FAST RANDOMIZED RANGE FINDER %%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsubsection{Fast Randomized Range Finder}\n\\label{alg:fast-randomized-range-finder}\n\nA simple inspection to equation \\ref{eq:cost_basic} reveals the computational\nbottleneck of the sampling procedure. This is the matrix multiplication\n$\\mtx{Y}=\\mtx{A}\\mtx{\\Omega}$ that takes $\\mathcal{O}(mn\\ell)$ operations\nfor dense matrices, which is the same as the $\\ell$-SVD.\n\nThe key idea is to use a \\textit{structured} random matrix that allows us \nto compute the product in $\\mathcal{O}(mn\\log(\\ell))$ operations.\n\nThe simplest structured random matrix that meets our goals is the so-called\n\\textit{subsampled random Fourier transform} (SRFT).\n\nAn SRFT is an $n \\times \\ell$ matrix of the form\n\\begin{equation}\n\\label{eq:def_srft}\n\\mtx{\\Omega} = \\sqrt{\\frac{n}{\\ell}} \\, \\mtx{DFR},\n\\end{equation}\nwhere\n\\lsp\n\\begin{itemize}\n\\item $\\mtx{D}$ is an $n \\times n$ diagonal matrix whose entries are\nindependent random variables uniformly distributed on the complex unit circle.\n\n\\item $\\mtx{F}$ is the $n \\times n$ unitary discrete Fourier transform (DFT),\nwhose entries take the values $f_{pq} = n^{-1/2} \\, \\econst^{-2\\pi\\iunit (p-1)(q-1)/n}$ for $p, q = 1, 2, \\dots, n$\n\n\\item $\\mtx{R}$ is an $n \\times \\ell$ matrix that samples $\\ell$ coordinates\nfrom $n$ uniformly at random, i.e., its $\\ell$ columns are drawn randomly\nwithout replacement from the columns of the $n \\times n$ identity matrix.\n\\end{itemize}\n\\lsp\n\nNow, via a subsampled FFT ~\\cite{woolfe2008fast}, we can compute the\nsample matrix $\\mtx{Y}=\\mtx{A}\\mtx{\\Omega}$ with\n$\\mathcal{O}(mn\\log(\\ell))$ operations.\n\nThe total number of operations required by this procedure is reduced to\n\\begin{equation}\n\\label{eq:cost_SRFT}\nT_{\\rm struct} \\sim mn \\log(\\ell) + \\ell^2 n\n\\end{equation}\n\nHence, the computational complexity of the approach is essentially\n$\\mathcal{O}(mn \\log(\\ell))$.\n\n% If $\\ell$ is substantially larger than the numerical rank $r(\\mtx{A})$\n% \\ref{eq:num-rank}, then the orthogonalization can be done in\n% $\\mathcal{O}(k\\ell n)$ instead of $\\mathcal{O}(\\ell^2n)$.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\fbox{\n\\begin{minipage}{.9\\textwidth}\n\\begin{center}\n\\textsc{Fast Randomized Range Finder}\n\\end{center}\n\\begin{tabbing}\n\\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\hspace{5mm} \\= \\kill\n\\anum{1} \\>Draw an $n\\times \\ell$ SRFT test matrix $\\mtx{\\Omega}$, as defined by \\eqref{eq:def_srft}. \\\\\n%\\pgnotate{Added reference.}\\\\\n\\anum{2} \\>Form the $m\\times \\ell$ matrix $\\mtx{Y} = \\mtx{A}\\mtx{\\Omega}$ using a (subsampled) FFT.\\\\\n\\anum{3} \\>Construct an $m \\times \\ell$ matrix $\\mtx{Q}$ whose columns form an orthonormal\\\\\n \\> basis for the range of $\\mtx{Y}$, e.g., using the QR factorization $\\mtx{Y} = \\mtx{Q}\\mtx{R}$.\n\\end{tabbing}\n\\end{minipage}}\n\\end{center}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% STAGE 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\n\\subsection{Stage 2} \\label{sec:stage2}\n The output of the Stage 1 \\ref{sec:stage1} produces an orthonormal matrix $\\mtx{Q}$ whose\n range captures the action of the matrix $\\mtx{A}$. The goal of Stage 2\n \\ref{sec:stage2} is \n to produce standard approximate matrix factorizations of $\\mtx{A}$\n using this $\\mtx{Q}$.\n\nThis subsection is divided into three parts; first, we will show how to compute\nstandard approximate matrix factorizations (SVD and QR) from a general \napproximate low-rank factorization. Recall that taking $\\mtx{B}=\\mtx{Q}$\nand $\\mtx{C}=\\mtx{Q}^\\adj\\mtx{A}$ we readily have a factorization that\nsatisfies $\\|\\mtx{A}-\\mtx{B}\\mtx{C}\\|\\leq\\varepsilon$.\nThen, we will describe in detail the \\textit{Direct SVD} algorithm, which will\nconsist in constructing an SVD from $\\mtx{B}$ and $\\mtx{C}$. Finally, we will\ncomment on other more involved methods that avoid computing the expensive\nproduct $\\mtx{C}=\\mtx{Q}^\\adj\\mtx{A}$.\n\n\\subsubsection{Compute standard QR and SVD from a general factorization}\nNow we will specify how we can compute the standards SVD and QR decompositions\nfrom a general low rank decomposition $\\|\\mtx{A} - \\mtx{B}\\mtx{C}\\|\\leq\\varepsilon$\nmaintaining the tolerance $\\epsilon$ from Stage 1 \\ref{sec:stage1}.\n\n\\begin{itemize}\n \\item \\label{itm:SVD-from-C} \\textit{SVD decomposition:} $\\|\\mtx{A}-\\mtx{U}\\mtx{\\Sigma}\\mtx{V}^\\adj\\|\\leq\\varepsilon$\n\\lsp\n\\begin{enumerate}\n\\item Compute a QR factorization of $\\mtx{B}$ so that $\\mtx{B} = \\mtx{Q}_{1}\\mtx{R}_{1}$.\n\\item Form the product $\\mtx{D} = \\mtx{R}_{1}\\mtx{C}$, and compute an SVD:\n $\\mtx{D} = \\mtx{U}_{2}\\mtx{\\Sigma}\\mtx{V}^{\\adj}$.\n\\item Form the product $\\mtx{U} = \\mtx{Q}_{1}\\mtx{U}_{2}$.\n\\end{enumerate}\n\\lsp\n% The result is a diagonal matrix $\\mtx{\\Sigma}$ and orthonormal matrices $\\mtx{U}$ and $\\mtx{V}$\n% such that $\\norm{\\mtx{A} - \\mtx{U}\\mtx{\\Sigma}\\mtx{V}^{\\adj}} \\leq \\varepsilon$.\n\n\\item \\label{itm:QR-from-C} \\textit{QR decomposition:} $\\|\\mtx{A}-\\mtx{Q}\\mtx{R}\\|\\leq\\varepsilon$\n\\lsp\n\\begin{enumerate}\n\\item Compute a QR factorization of $\\mtx{B}$ so that $\\mtx{B} = \\mtx{Q}_{1}\\mtx{R}_{1}$.\n\\item Form the product $\\mtx{D} = \\mtx{R}_{1}\\mtx{C}$, and compute a QR factorization: $\\mtx{D} = \\mtx{Q}_{2}\\mtx{R}$.\n\\item Form the product $\\mtx{Q} = \\mtx{Q}_{1}\\mtx{Q}_{2}$.\n\\end{enumerate}\n% The result is an orthonormal matrix $\\mtx{Q}$ and a weakly upper-triangular matrix $\\mtx{R}$ such\n% that $\\norm{\\mtx{A} - \\mtx{Q}\\mtx{R}} \\leq \\varepsilon$.\n\n\\end{itemize}\n\n\\subsubsection{Direct SVD} \\label{alg:direct-svd}\n\nThe procedure described in \\ref{itm:SVD-from-C} to compute the approximate \nSVD decomposition without sacrificing error, defines what we call\nthe \\textit{Direct SVD} \\ref{alg:direct-svd}.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\fbox{\\begin{minipage}{.9\\textwidth}\n\\begin{center}\n\\textsc{Direct SVD}\n\\end{center}\n\\anum{1} Form the matrix\n$\n\\mtx{B} = \\mtx{Q}^{\\adj}\\mtx{A}.\n%\\mtx{B} = \\bigl(\\mtx{A}^{\\adj}\\mtx{Q}\\bigr)^{\\adj}.\n$\n\n\\anum{2} Compute an SVD of the small matrix:\n$\n\\mtx{B} = \\widetilde{\\mtx{U}}\\mtx{\\Sigma}\\mtx{V}^{\\adj}.\n$\n\n\\noindent\n\\anum{3} Form the orthonormal matrix\n$\n\\mtx{U} = \\mtx{Q}\\widetilde{\\mtx{U}}.\n$\n\\end{minipage}}\n\\end{center}\n%\\caption{trash}\n%\\label{alg:Atranspose}\n\\end{figure}\n\nAlthough using \\textit{Direct SVD} \\ref{alg:direct-svd} algorithm for\nStage 2 \\ref{sec:stage2} does not incur additional errors, the computation\nof $\\mtx{C}=\\mtx{Q}^\\adj\\mtx{A}$ is in general too expensive for dense matrices.\n\nMore concretely, the product costs $\\mathcal{O}(mn\\ell)$, even more\nexpensive that the cost of Stage 1 \\ref{sec:stage1} when using\nthe accelerated Algorithm\n\\ref{alg:fast-randomized-range-finder}.\n\n\\subsubsection{Different Procedures} \\label{sec:dif-app}\n\n\\begin{itemize} \n \\item \\textit{Match the cost} $\\mathcal{O}(mn\\log(\\ell))$ \\textit{of accelerated\n Stage 1} \\ref{alg:fast-randomized-range-finder}.\n\n In order to match the complexity $\\mathcal{O}(mn\\log(\\ell))$ from \nStage 1 \\ref{sec:stage1} we must avoid the product $\\mtx{Q}^\\adj\\mtx{A}$.\n\nIn \\cite{halko2011finding}, the authors propose algorithms based\non row extraction of $\\mtx{Q}$ via its \\textit{Interpolative Decomposition}\n$\\mtx{Q}=\\mtx{X}\\mtx{Q}_{(J,:)}$ \\cite{cheng2005compression}.\nNow, $\\mtx{Q}_{(J,:)}$ is a $k\\times k$ matrix. The proposed\nalgorithm takes $\\mtx{Q}$ as input and constructs a rank-$k$ matrix \nfactorization\n\n\\begin{equation}\\label{eq:via-row-extraction}\n\\mtx{A}\\approx\\mtx{X}\\mtx{B}\n\\end{equation}\nwhere $\\mtx{B}$ is a $k \\times n$ matrix consisting of $k$ rows\nextracted from $\\mtx{A}$.\n\nThe key here is that \\ref{eq:via-row-extraction} can be produced without any large\nmatrix-matrix multiplication resulting in a total of $\\mathcal{O}(k^2(m+n))$\noperations. The drawback is that the initial error is larger than the one\nincurred by $\\mtx{Q}^\\adj\\mtx{Q}\\mtx{A}$\n\\footnote{In \\cite{halko2011finding}, they prove that the error is increased at \nmost by $\\left(\\sqrt{1+4k(n-k)}\\right)\\varepsilon$, which can be large\nwhen $\\varepsilon$ is not that small or $kn$ is large.}.\n\n\\item \\textit{Single-Pass algorithms}\n\nThe previously described algorithms require revisit the matrix $\\mtx{A}$\nmultiple times. In \\cite{halko2011finding} they propose single-pass\nalgorithms relying on the observation that all the information\nyou need to compute the decomposition is in the matrices $\\mtx{\\Omega},\\mtx{Y}$\nand $\\mtx{Q}$. They call the Algorithm \\textit{Eigenvalue Decomposition in\nOne-pass}. However, this algorithm too adds additional error to the Stage 1\n\\footnote{Authors in \\cite{halko2011finding} argue that this issue can be\naddressed with extra oversampling.}.\n\\end{itemize}\n\n\\subsection{Full Algorithms}\\label{sec:full-algos}\nLet's now propose a full algorithm depending on the input matrix properties.\n\n\\begin{itemize}\n\\item \\textit{General Matrices That Fit in Core Memory}\n\nIn this case, the appropiate method for Stage 1 is the accelerated\nAlgorithm \\ref{alg:fast-randomized-range-finder} that uses structured random\nmatrices. For Stage 2, use the \\textit{row-extraction technique}\ndescribed in \\ref{sec:dif-app}. The overall cost $T_{\\rm random}$ reduces to\n\n\\begin{equation} \\label{eq:cost-1}\n\\boxed{T_{\\rm random} \\sim mn \\log(k) + k^2 (m+n)}\n\\end{equation}\nThe overall approximation error satisfies\n\\footnote{This can be derived by combining the error analysis\nof range approximation with SRFT (which can be found in \\cite{halko2011finding})\nand the additional error incurred by the \\textit{row-extraction technique}.}\n\\begin{equation} \\label{eq:totalerror-1}\n\\norm{ \\mtx{A} - \\mtx{U\\Sigma V}^\\adj }\n \\lesssim n \\cdot \\sigma_{k+1},\n\\end{equation}\n\n\\item \\textit{Matrices for which Matrix-Vector Products are Cheap.}\n\nIn this set of matrices we include for example sparse and structured matrices for\nwhich the matrix-vector multiplication is substantially cheaper than the general case.\n\nThe appropiate method for Stage 1 is Algorithm \\ref{alg:randomized-range-finder}\nwith $p$ constant \\footnote{If we are dealing\nwith the \\textit{fixed-precision problem}, then we use \nAlgorithm \\ref{alg:adaptive-randomized-range-finder}.}, or more generally\nAlgorithm \\ref{alg:randomized-power-iteration} with $q>0$. For Stage 2, we apply\nthe \\textit{Direct SVD} \\ref{alg:direct-svd}. The total cost $T_{\\rm sparse}$ satisfies\n\\begin{equation}\n\\label{eq:sparsealg-cost}\n\\boxed{T_{\\rm sparse} = (2q+2)\\,(k + p) \\, T_{\\rm mult} + \\bigO(k^2 (m + n))}\n\\end{equation} \n\nThe overall approximation error satisfies (see Corollary \\ref{cor:power-method-spec-gauss}).\n\\begin{equation}\n\\label{eq:sparsealg-error2}\n\\norm{ \\mtx{A} - \\mtx{U\\Sigma V}^\\adj } \\lesssim (kn)^{1/2(2q+1)} \\cdot \\sigma_{k+1}.\n\\end{equation}\n\n\\end{itemize}\n\n\n\\newpage", "meta": {"hexsha": "57f30c917031213f9c88e42a01fde935b14fe6ec", "size": 21485, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/sections/algorithms.tex", "max_stars_repo_name": "alexnowakvila/ProbAlgosProj", "max_stars_repo_head_hexsha": "819251313cb2b9ca4748892ee62f40c602453ddc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-06-10T11:13:11.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-05T07:55:46.000Z", "max_issues_repo_path": "report/sections/algorithms.tex", "max_issues_repo_name": "alexnowakvila/ProbAlgosProj", "max_issues_repo_head_hexsha": "819251313cb2b9ca4748892ee62f40c602453ddc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/sections/algorithms.tex", "max_forks_repo_name": "alexnowakvila/ProbAlgosProj", "max_forks_repo_head_hexsha": "819251313cb2b9ca4748892ee62f40c602453ddc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-11-05T07:55:49.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-05T07:55:49.000Z", "avg_line_length": 43.668699187, "max_line_length": 130, "alphanum_fraction": 0.7006748895, "num_tokens": 6988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8376199754937772, "lm_q2_score": 0.7490872131147275, "lm_q1q2_score": 0.6274504130918599}} {"text": "\\section{Comparison with empirical portfolio returns}\n\\label{sec:comparison_returns}\n\nWith the definition of the four different distributions we proceed to compare\nthem with empirical data from 200 stocks traded in the S\\&P 500 stock market\nindex.\n\nAs described in Sect. \\ref{sec:exact_distributions}, we obtain normalized all\nthe time series in the market data, the we compute the covariance matrix and\nobtain the corresponding eigenvalues and eigenvectors. Using them we rotate and\nscale the returns, and finally, we aggregate the rotated and scaled returns\ninto a single univariate distribution.\n\nIn Fig. \\ref{fig:gg_dist} we plot four Gaussian-Gaussian distributions with\nfit parameter $N = 2, 3, 4, 5$. The best fit seems to be the distributions\nwith $N = 3$ and $N = 4$.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.7\\columnwidth]\n {figures/08_gg.png}\n \\caption{Gaussian-Gaussian probability density\n $\\left\\langle p \\right\\rangle_{GG'}^{\\left(k\\right)}$, in the\n Markovian case, normalized to unit standard deviation for\n different values of $N$, and aggregated distribution returns\n $\\left(\\tilde{r}\\right)$ for fixed covariance for $200$ companies\n selected from the S\\&P 500 dataset and $\\Delta t = 1d$. (left)\n semilog scale and (right) loglog scale.}\n \\label{fig:gg_dist}\n\\end{figure}\n\nFor the Gaussian-Algebraic distribution and Algebraic-Gaussian distribution,\nwe have to add an additional shape parameter $L$ and $l$, respectively. In\nFig. \\ref{fig:ga_dist} we plot four Gaussian-Algebraic distributions and in\nFig. \\ref{fig:ag_dist} we plot four Algebraic-Gaussian distributions with fit\nparameters $N = 2, 3, 4, 5$. In both cases, a good agreement when $N = 4$ can\nbe seen.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.7\\columnwidth]\n {figures/08_ga.png}\n \\caption{Gaussian-Algebraic probability density\n $\\left\\langle p \\right\\rangle_{GA'}^{\\left(k\\right)}$, in the\n Markovian case, normalized to unit standard deviation for\n different values of $N$, and aggregated distribution returns\n $\\left(\\tilde{r}\\right)$ for fixed covariance for $200$ companies\n selected from the S\\&P 500 dataset and $\\Delta t = 1d$. (left)\n semilog scale and (right) loglog scale.}\n \\label{fig:ga_dist}\n\\end{figure}\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.7\\columnwidth]\n {figures/08_ag.png}\n \\caption{Algebraic-Gaussian probability density\n $\\left\\langle p \\right\\rangle_{AG'}^{\\left(k\\right)}$, in the\n Markovian case, normalized to unit standard deviation for\n different values of $N$, and aggregated distribution returns\n $\\left(\\tilde{r}\\right)$ for fixed covariance for $200$ companies\n selected from the S\\&P 500 dataset and $\\Delta t = 1d$. (left)\n semilog scale and (right) loglog scale.}\n \\label{fig:ag_dist}\n\\end{figure}\n\nFinally, in Fig. \\ref{fig:aa_dist} we plot four Algebraic-Algebraic\ndistributions with fit parameters $N = 2, 3, 4, 5$. In this case we use the\nshape parameters $L$ and $l$. Again, a good agreement when $N = 4$ can be seen.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.7\\columnwidth]\n {figures/08_aa.png}\n \\caption{Algebraic-Algebraic probability density\n $\\left\\langle p \\right\\rangle_{AA'}^{\\left(k\\right)}$, in the\n Markovian case, normalized to unit standard deviation for\n different values of $N$, and aggregated distribution returns\n $\\left(\\tilde{r}\\right)$ for fixed covariance for $200$ companies\n selected from the S\\&P 500 dataset and $\\Delta t = 1d$. (left)\n semilog scale and (right) loglog scale.}\n \\label{fig:aa_dist}\n\\end{figure}\n", "meta": {"hexsha": "79c6aadd2c10454ca719f87fce30c897f06ffbc4", "size": 3888, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/exact_distributions_financial_paper/sections/08_empirical_results.tex", "max_stars_repo_name": "juanhenao21/exact_distributions_financial", "max_stars_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-20T18:24:22.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-15T07:25:50.000Z", "max_issues_repo_path": "paper/exact_distributions_financial_paper/sections/08_empirical_results.tex", "max_issues_repo_name": "juanhenao21/exact_distributions_financial", "max_issues_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/exact_distributions_financial_paper/sections/08_empirical_results.tex", "max_forks_repo_name": "juanhenao21/exact_distributions_financial", "max_forks_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.2857142857, "max_line_length": 79, "alphanum_fraction": 0.6867283951, "num_tokens": 1037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199471193039, "lm_q2_score": 0.7490872187162397, "lm_q1q2_score": 0.6274503965288432}} {"text": "\\chapter{Network Theory}\n\\label{ch:networktheory}\n\n\\begin{chapterabstract}\nThe theory underpinning complex networks is discussed, covering the representation of atomic systems as networks and the relationship of the dual network to ring structure.\nThe laws which govern the topological properties of physical networks are also introduced, namely Euler's law (mean node degree), \\lm's law (node degree distribution) and the \\aw{} law (node degree correlations).\n\\end{chapterabstract}\n\n\\section{Network Theory}\n\\label{sec:networktheory}\n\nThe scope of what constitutes a complex network is extremely broad, covering everything from the tangible (\\eg{} computational clusters) to the more abstract (\\eg{} social interactions). Yet part of the appeal and power of network science is the ability to quantify and relate these highly disparate systems with the same underlying theory.\nA network is simply a collection of components termed \\textit{nodes} and the connections between them termed \\textit{links}, an example of which is given in figure \\ref{fig:smallnet}.\nThere are then two fundamental classes of network based on the nature of the connections.\nNetworks in which the links between nodes are mutual are termed undirected, whereas those in which the links are one\\--way are termed directed \\cite{barabasi2016n}.\nAt the risk of dating this thesis, this is the difference between Facebook (an undirected social network of friends) and Twitter (a directed social network of followers).\nAll the networks considered in this work are undirected and all the theory assumes this property.\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=0.45\\textwidth]{./figures/methods/small_periodic_net.pdf}\n \\caption{Example of a periodic \\td{} network where nodes are represented by circles and links as lines. Nodes are coloured similarly according to their degree, whilst periodic images are faded to highlight the central repeating unit.}\n \\label{fig:smallnet}\n\\end{figure}\n\n\\subsection{Node Degree and Probability Distributions} \n\\label{s:theorynodeprob}\n\nA key concept in network science is the the node degree, defined as the number of links associated with each node.\nA node with $k$ links is then said simply to have degree $k$, where $k\\in\\mathbb{N}$.\nThis is illustrated in figure \\ref{fig:smallnet}, which consists of 5\\--, 6\\-- and 7\\-- degree nodes (coloured blue, grey and red respectively).\nThe occurrence and correlations of nodes of given degrees can then be described by a range of probability distributions.\n\nThe probability of a randomly selected node having degree $k$ is given by the node degree distribution, denoted $p_k$.\nThis is a normalised discrete distribution such that\n\\begin{equation}\n\t\\label{eq:pknorm}\n\t\\sumk p_k = 1.\n\\end{equation}\nThe $n$\\th{} moments of this distribution are then given by:\n\\begin{equation}\n\t\\label{eq:pkmoment}\n\t\\langle k^{n} \\rangle = \\sumk k^np_k.\n\\end{equation}\nAlternatively, one can also calculate the probability that a randomly selected link has a $k$\\--degree node at the end, denoted $q_k$.\nThis is not the same as the distribution above, as there is greater chance of selecting links which emanate from high degree nodes, in a manner which is proportional to the node degree.\nAs this distribution is normalised, this leads to the relations:\n\\begin{align}\n\t\\sumk q_k &= 1, \\label{eq:qknorm} \\\\\n\tq_k &= \\frac{kp_k}{\\langle k \\rangle} \\label{eq:qkpk}.\n\\end{align}\nIn addition, one can also evaluate the probability that a randomly chosen link has nodes of degree $j,k$ at either end.\nThis is the edge joint degree distribution, denoted $e_{jk}$. \nOnce again this is normalised and satisfies the following relationships:\n\\begin{align}\n\t\\sumjk e_{jk} &= 1, \\label{eq:ejknorm} \\\\\n\t\\sumk e_{jk} &= q_j, \\label{eq:ejkqk} \\\\\n\te_{jk} &= e_{kj} \\label{eq:ejkekj},\n\\end{align}\nwhere the final result arises from reciprocal nature of the links in an undirected network.\nAs an illustration of these different distributions, examples %for all three probability distributions \nare provided below for the network in figure \\ref{fig:smallnet}:%, where the adjacent numbers correspond to the node degrees:\n\\begin{align}\n\t\\mathbf{p} = \\frac{1}{16} \\, \\begin{blockarray}{*{1}{c} l}\n\t\\begin{block}{[*{1}{c}]>{$\\footnotesize}l<{$}}\n\t4 \\: \\bigstrut[t]& 5\\\\\n\t8 & 6 \\\\\n\t4 & 7 \\\\\n\t\\end{block}\n\t\\end{blockarray},\n\t\\qquad\n\t\\mathbf{q} = \\frac{1}{96} \\, \\begin{blockarray}{*{1}{c} l}\n\t\\begin{block}{[*{1}{c}]>{$\\footnotesize}l<{$}}\n\t20 \\: \\bigstrut[t]& 5\\\\\n\t48 & 6 \\\\\n\t28 & 7 \\\\\n\t\\end{block}\n\t\\end{blockarray},\n\t\\qquad\t\n\t\\mathbf{e} = \\frac{1}{96}\\: \\begin{blockarray}{*{3}{c} l}\n\t\\begin{block}{*{3}{>{$\\footnotesize}c<{$}} l}\n\t5 & 6 & 7 \\\\\n\t\\end{block}\n\t\\begin{block}{[*{3}{c}]>{$\\footnotesize}l<{$}}\n\t2 & 9 & 9 \\: \\bigstrut[t]& 5\\\\\n\t9 & 22 & 17 & 6 \\\\\n\t9 & 17 & 2 & 7\\\\\n\t\\end{block}\n\t\\end{blockarray}.\n%\t\\textbf{p} = \\frac{1}{96}\\: \\begin{blockarray}{*{5}{c} l}\n%\t\\begin{block}{*{5}{>{$\\footnotesize}c<{$}} l}\n%\t3 & 6 & 7 & 8 & 9 \\\\\n%\t\\end{block}\n%\t\\begin{block}{[*{5}{c}]>{$\\footnotesize}l<{$}}\n%\t0 & 0 & 5 & 10 & 1\\: \\bigstrut[t]& \\:3\\\\\n%\t0 & 0 & 1 & 1 & 0 & \\:6 \\\\\n%\t5 & 1 & 14 & 14 & 1 & \\:7\\\\\n%\t10 & 1 & 14 & 14 & 1 & \\:8\\\\\n%\t1 & 0 & 1 & 1 & 0 & \\:9\\\\\n%\t\\end{block}\n%\t\\end{blockarray}\n\\end{align}\n\n\\subsection{Atomic and Ring Networks}\n\\label{s:atomringnetworks}\n\nTo see how network theory relates to atomic materials, consider the amorphous graphene configuration in figure \\ref{fig:graphdualgraph}.\nIn this network the nodes represent carbon atoms and the links sp$^2$ bonds.\nThe node degree in the atomic network for all nodes is then equal to three, being equivalent to the atomic coordination number (which throughout this thesis will be denoted by $c$).\nThis is problematic, because whilst there is clear disorder in the system, it is not well captured by the atomic network.\nDue to the fact that the local environment around the atoms is identical, when examining the node degree distribution, any information about the glassy structure is lost.\nThis network is to at least first order indeterminable from a crystalline hexagonal lattice.\n\n\\begin{figure}[bt]\n \\centering\n \n \\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/graph.pdf}\n \\caption{Atomic network}\n \\label{fig:graphdualgraph}\n \\end{subfigure}\n \\hfill\n\t\\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/dual.pdf}\n \\caption{Ring network}\n \\label{fig:graphdualdual}\n \\end{subfigure}\n \\hfill\n \\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/graph_dual.pdf}\n \\caption{Dual relationship}\n \\label{}\n \\end{subfigure}\n \\hfill\n \n \\caption{Panel (a) gives an example of a 3\\--coordinate periodic atomic network with disordered ring structure. Nodes and links represent atoms and bonds respectively where rings are coloured by size. Panel (b) gives the corresponding ring network where nodes and links represent rings and their adjacencies, where nodes are coloured by degree. Panel (c) shows the dual relationship between the atomic and ring networks, where the node degree in the ring network is equal to the ring size in the atomic network.}\n \\label{fig:graphdual}\n\\end{figure}\n\nObserving figure \\ref{fig:graphdualgraph} one can see there is another level of structure in the network, namely that of the ring structure.\nA ring is strictly any closed path of sequentially linked nodes in a network, but this thesis will use the term in reference only to the primitive rings \\ie{} those which cannot be subdivided into two smaller rings \\cite{Yuan2002}.\nA ring of size $k$ (or $k$\\--ring) is then defined as a ring with $k$ constituent nodes.\nIt is clear that finding and counting the number of rings of each size, often termed calculating the ring statistics, is a method of quantifying the disorder in the system \\cite{Kumar2014}.\nThe ring statistics can be summarised by the normalised probability distribution, $p_k$.\n\nHowever, there is a more efficient way of representing and quantifying the ring structure in the system, and that is by constructing the dual network \\cite{Aboav1984}.\nThe dual is generated by placing a node at the centre of each ring and linking the nodes of adjacent (\\ie{} edge\\--sharing) rings, as in figure \\ref{fig:graphdualdual}.\nThis will be referred to as the ring network.\nThe ring network is a reciprocal lattice in which the node degree, $k$, is equivalent to the ring size in the atomic network.\nSimilarly, it consists solely of triangles, reflecting the 3\\--coordinate nature of the underlying atomic network.\nHence, the disorder is captured directly in the node properties of the ring network.\nThese characteristics make the ring network preferable for manipulating and analysing the systems in this thesis.\n\n\\section{Topological Laws}\n\\label{s:topolaws}\n\nThere are a number of laws which govern the topological properties of \\td{} network\\--forming materials.\nThese laws constrain the ring structure, influencing the network properties in a manner that makes physical networks unique in the field of network science.\nThese laws act on a number of ``levels'': Euler's law controls the overall mean ring size, \\lm's{} law the ring size distribution and the \\aw{} law the ring\\--ring correlations.\n\n\\subsection{Euler's Law}\n\\label{s:eulerslaw}\n\nEuler's law constrains the mean ring size, $\\ki$, in an atomic network, equivalent to the mean node degree of the ring network.\nThe atomic networks studied in this work are all \\td{}, connected (\\ie{} there is a path between any two nodes) and planar (\\ie{} they have no overlapping links) and so are subject to Euler's formula which states:\n\\begin{equation}\n\t\\label{eq:eulerformula}\n\tN + V - E = \\chi,\n\\end{equation}\nwhere $N$, $V$, $E$ are the number of rings, vertices and edges in the network and $\\chi$ in an integer termed the Euler characteristic, which is dependent on the global topology of the system.\nEach vertex represents an atom and the number of edges emanating from each vertex is then the coordination number.\n\nFor generality, consider an atomic network with atoms of assorted coordination numbers, $c$. \nIf the proportion of each coordination type is $x_c$, then the mean coordination number is given by $\\langle c \\rangle = \\sum\\limits_c cx_c$.\nThis allows the number of edges to be written in terms of the number of vertices as $E=\\frac{V}{2}\\langle c \\rangle$. \nIn turn the mean ring size is simply the total number of vertices per ring, allowing for multiple counting, such that $\\ki=\\frac{V}{N}\\langle c \\rangle$.\nSubstituting these two expressions into equation \\eqref{eq:eulerformula} leads to the expression:\n\\begin{equation}\n\t\\label{eq:avdegree}\n\t\\ki = \\frac{2\\langle c \\rangle\\left(1-\\chi/N\\right)}{\\langle c \\rangle - 2}.\n\\end{equation}\nHence the average node degree in the ring network (equivalent to the mean ring size of the physical network), is simply related to the average degree of the physical network (\\ie{} local coordination environment), the topology of the system and the number of rings.\n\nAlthough equation \\eqref{eq:avdegree} may appear simple, it is a very powerful constraint. \nTo demonstrate this consider a two\\--dimensional lattice with two possible coordination environments $c=3,4$. \nThe planar case with periodic boundary conditions (mimicking an infinite planar lattice) maps onto the torus with $\\chi=0$, and so:\n\\begin{equation}\n\t\\label{eq:2dplanarcases}\n\t\\ki = \\begin{cases}\n\t\t6, \\quad x_3 = 1 \\\\\n\t\t4, \\quad x_4 = 1 \\\\\n\t\t5, \\quad x_3 = 2/3,\\, x_4 =1/3\n\t\\end{cases}.\n\\end{equation}\nTo reiterate in plain terms, this means that if there is a material consisting of atoms all forming exactly three bonds (as for amorphous carbon), the mean ring size \\textit{must} be equal to six. \nSimilarly if all atoms form four bonds the mean ring size is four, and if there is a two\\--thirds to one\\--third mixture of coordination environments the mean ring size is five.\nThe simplest illustrations of these are the hexagonal, square and Cairo tilings, shown in figure \\ref{fig:lattices}, but this law holds equally well for amorphous configurations.\nFor aperiodic systems strictly $\\chi=1$, but as $N\\rightarrow \\infty$, the proportion of vertices with unsatisfied coordination on the sample perimeter become negligible overall, hence does the term in $\\chi$.\nTherefore in reality these relationships hold, and remain as applicable to amorphous graphene as the basalt columns in Fingal's Cave, and the Penrose tiling \\cite{Goehring2014,Ressouche2009}.\n\nThis analysis also extends to spherical topology where $\\chi=2$, and so:\n \\begin{equation}\n \t\\label{eq:2dsphericalcases}\n\t\\ki = \\begin{cases}\n\t\t\\frac{6N-12}{N}, \\quad x_3 = 1 \\\\\n\t\t\\frac{4N-8}{N}, \\quad x_4 = 1\n\t\\end{cases}\\,.\n\\end{equation}\nThese relationships are the origin of the 12 pentagon rule for 3-coordinate fullerenes (the ``football problem''), or equivalently an ``8 triangle rule'' in the 4\\--coordinate case, as this is the only way to satisfy these equations if the allowed ring sizes are limited to $k=5,6$ and $k=3,4$ respectively (as in figures \\ref{fig:latticesfull92} and \\ref{fig:latticesfull98}) \\cite{Fowler1996}.\nMuch of the richness in the structural behaviour of \\td{} physical networks stems from this fundamental constraint on the network average degree.\n\n\\begin{figure}[bt]\n \\centering\n \n \\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[height=2.4cm]{./figures/methods/hex.pdf}\n \\caption{Hexagonal}\n \\label{fig:latticeshex}\n \\end{subfigure}\n \\hfill\n \\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[height=2.4cm]{./figures/methods/cai.pdf}\n \\caption{Cairo}\n \\label{fig:latticescairo}\n \\end{subfigure}\n \\hfill\n\t\\begin{subfigure}[b]{0.3\\textwidth}\n \\centering\n \\includegraphics[height=2.4cm]{./figures/methods/sq.pdf}\n \\caption{Square}\n \\label{fig:latticessq}\n \\end{subfigure}\n \\hfill\n \\vspace{0.5cm}\n \n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[height=2.4cm]{./figures/methods/full92.pdf}\n \\caption{3\\--coordinate fullerene}\n \\label{fig:latticesfull92}\n \\end{subfigure}\n \\hspace{1cm}\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[height=2.4cm]{./figures/methods/full98.pdf}\n \\caption{4\\--coordinate fullerene}\n \\label{fig:latticesfull98}\n \\end{subfigure}\n \n \\caption{Panels (a)\\--(c) give planar tilings of 6\\--, 5\\-- and 4\\-- rings, where the ring size is related to the underlying atomic coordination. Panels (d) and (e) show the 3\\-- and 4\\-- coordinate tilings in spherical topology, where the mean ring size is reduced due to the change in the Euler characteristic.}\n \\label{fig:lattices}\n\\end{figure}\n\n\\subsection{\\lm's Law}\n\\label{s:lemaitre}\n\nKnowing that the mean node degree is fixed by Euler's law, the next level of available information is the form of the underlying degree distribution, $p_k$.\nInterestingly, the degree distributions found in physical ring networks seem relatively well defined.\nFor instance, it has been noted in models and realisations of \\td{} silica glass that the ring statistics looked to follow a lognormal distribution \\cite{Shackelford1981,Buchner2017}.\n\\lm{} \\etal{} demonstrated that the distribution in 3-coordinate networks systems can be well described by a maximum entropy distribution \\cite{Gervois1992}.\n\\lm's{} maximum entropy method is summarised here, trivially extended to arbitrary coordination.\n\nThe entropy of a probability distribution is defined as \n\\begin{equation}\n\t\\mathcal{S}=-\\sumk p_k\\log p_k. \n\\end{equation}\nIn addition, the degree distribution has the following constraints:\n\\begin{align}\n\t\t\\sumk p_k &=1, \\\\\n\t\t\\sumk kp_k&=\\ki, \\label{con:lm2}\\\\\n\t\t\\sumk \\frac{p_k}{k}&=\\text{constant} \\label{con:lm3},\n\\end{align}\nwhere the first two constraints correspond to the normalisation condition and the fixed mean ring size, and the final constraint will be discussed below.\nThe entropy can then be maximised using Lagrange's method of undetermined multipliers to yield the result:\n\\begin{equation}\n\t\\label{eq:mepk}\n\tp_k = \\frac{e^{-\\lambda_1 k - \\lambda_2 / k}}{\\sumk e^{-\\lambda_1 k - \\lambda_2 / k}},\n\\end{equation}\nwhich can be solved numerically by substitution into equations \\eqref{con:lm2} and \\eqref{con:lm3}. \nBy allowing the chosen constant to vary, a family of maximum entropy curves can be generated, as in figure \\ref{fig:lm1}.\nThe resulting distributions can be summarised by relating the variance, $\\mu_2=\\kii-\\ki^2$, to a single chosen node degree probability, leading to the plot known as \\lm's law, given in figure \\ref{fig:lm2}.\nIt is usually framed in the context of the proportion of hexagons in a system, $p_6$, for the precise reason that most networks have $\\ki=6$ and $p_6$ as the largest contribution.\nMany experimental and theoretical studies have shown good agreement to this law \\cite{Caer1993,Cerisier1996,Miklius2012}.\n\nSimple extensions of the classic law are however possible, by modifying the mean degree or the permitted degree range.\nFor instance, $k$ is usually taken in the interval $k\\geq3$ (as the triangle, $k=3$, is the smallest polygon), but there can be manifestations of physical systems where only certain degrees are accessible \\cite{Rivier1988}.\nGood examples of such systems are procrystalline lattices, which are explored in chapter \\ref{ch:procrystals}, in which the underlying lattice constraints may preclude the formation of odd membered rings or rings greater than a certain size. \n\\lm{} curves for a selection of $\\ki$ and $k$\\--range modifications are given in figure \\ref{fig:lm3}. \nThe maximum values of these curves, as indicated by circular points, are determined by removing constraint \\eqref{con:lm3}, equivalent to setting $\\lambda_2=0$ in equation \\eqref{eq:mepk}.\nOne can see that these modifications can lead to marked differences in the maximum entropy solutions.\nIn particular, removing the possibility of small rings ($k<\\ki$) forming reduces the maximum variance, as these are required to offset very large rings ($k\\gg\\ki$) in order to maintain the fixed mean ring size.\n\n\n\\begin{figure}[bt]\n \\centering\n \n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/lm_1.pdf}\n \\caption{Maximum entropy distributions}\n \\label{fig:lm1}\n \\end{subfigure}\n \\hfill\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/lm_2.pdf}\n \\caption{\\lm's law}\n \\label{fig:lm2}\n \\end{subfigure}\n \\hfill\n \n \\vspace{2mm}\n \\begin{subfigure}[b]{0.45\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{./figures/methods/lm_3.pdf}\n \\caption{Extensions to \\lm's law}\n \\label{fig:lm3}\n \\end{subfigure}\n \\hfill\n\n \n \\caption{Illustration of \\lm's maximum entropy method. Panel (a) gives examples of explicit maximum entropy distributions with different values of $p_6$. Panel (b) shows how these distributions can be summarised in a plot of $p_6$ \\vs{} $\\mu_2$ (\\lm's law). Panel (c) provides extensions to the law by modifying the underlying constraints of the mean ring size and allowable $k$\\--range, with circles indicating maximal values.}\n \\label{fig:lm}\n\\end{figure}\n\nThe only somewhat puzzling aspect of this successful theory is the choice of constraint \\eqref{con:lm3}.\nIt was originally rationalised on the basis that the areas of rings of a given size, $A_k$, can be well fit by an expression $A_k = ak+b+c/k$, where $a,b,c$ are constants.\nAs noted at the time, this is by no means true for all systems and in fact is contrary to the widely known Lewis law, which states that $A_k$ is linear in $k$ for many observable networks \\cite{Lewis1928,Fortes1995,Kim2014}.\nOverall, there does not appear to be a strong physical basis for this constraint, its justification relying on the perceived universality of the \\lm{} law, and hence it can be viewed as a tool that engineers the correct behaviour in the underlying distributions. \n\n%Despite this, the universality of the \\lm{} law suggests that there must be a physical basis to \\eqref{con:lm3}, and in the section \\ref{s:assortativity} it will be demonstrated that it can be regenerated by considering ring adjacencies.\n\n\\subsection{\\aw{} Law}\n\\label{s:awlaw}\n\nThe ring statistics given by \\lm's law are an important measure for physical networks, but they do not provide a complete characterisation of the ring structure, as they say nothing about the ring adjacencies. \nThis is important because whilst with the same ring statistics it is theoretically possible to organise the rings in many different arrangements, it is well known experimentally that only a subset of these are observed.\nThe vast majority of physical systems have a preference for small rings ($k<\\ki$) be adjacent to large rings ($k>\\ki$).\nThis effect was first noted in the grains of polycrystals by Aboav \\cite{Aboav1970}.\nAboav quantified these ring correlations by measuring the mean ring size about a $k$\\--ring, denoted $m_k$, and found empirically that $m_k \\approx 5 + 8/k$.\n\nIn an attempt to explain this observation, Weaire formulated the %following\nrelationship:\n\\begin{equation}\n\t\\label{eq:weairesumrule}\n\t\\sumk km_kp_k = \\sumk k^2p_k = \\mu_2 + \\ki^2,\n\\end{equation}\nknown as Weaire's sum rule \\cite{Weaire1974}.\nAlthough Weaire's original proof was not rigorous, the sum rule has since been shown to be exact, and formulations have also been constructed for use in higher dimensions \\cite{Chiu1994}.\nFrom this he suggested the modification of $m_k=5+\\left(6+\\mu_2\\right)/k$ which satisfied this rule.\nAboav's original equation then became a special case when $\\mu_2=2$, which is close to the expected value for a random collection of Voronoi polygons (see section \\ref{s:voronoiintro}).\nAboav proposed that if a generic form of $m_k = A + B/k$ was used in conjunction with Weaire's sum rule then\n\\begin{equation}\n\tm_k = A+\\frac{\\mu_2+\\ki^2-A\\ki}{k}.\n\\end{equation}\nThis is now more commonly expressed in the linear form \\cite{Chiu1995}:\n\\begin{equation}\n\t\\label{eq:aboavweaire}\n\tkm_k = \\mu_2+\\ki^2+\\ki\\left(1-\\alpha\\right)\\left(k-\\ki\\right).\n\\end{equation}\nEquation \\ref{eq:aboavweaire} is known as the \\aw{} law and relates the mean ring size about a given central ring to a single fitting parameter, $\\alpha$.\nThe value of $\\alpha$ describes the strength of the ring correlations, with a larger positive value indicating a greater tendency for small\\--large ring adjacencies.\nMore specifically, the random limit can be deduced by evaluating $\\frac{\\partial{m_k}}{\\partial{k}}=0$ as \\cite{Delannay1994}:\n\\begin{equation}\n\t\\label{eq:awrandlim}\n\t\\alpha=-\\frac{\\mu_2}{\\ki^2}.\n\\end{equation}\nHence all systems with $\\alpha>-\\mu_2/\\ki^2$ have more small\\--large ring adjacencies than would be expected from chance whilst conversely those with $\\alpha<-\\mu_2/\\ki^2$ have more small\\--small and large\\--large pairings.\n\n\\begin{figure}[tb]\n \\centering\n \\includegraphics[width=0.45\\textwidth]{./figures/methods/aw_demo.pdf}\n \\caption{Calculation of an \\aw{} fit for three configurations (shown in figures \\ref{fig:zach_high}\\--\\ref{fig:zach_low}). The value of the $\\alpha$ parameter quantifies the tendency of small rings to be adjacent to large rings, with a larger value indicating stronger small\\--large ring correlations.}\n \\label{fig:awdemo}\n\\end{figure}\n\nDespite the \\aw{} law being purely empirical and there being no topological requirement for $m_k$ to vary systematically with $k$, the law does seem to hold well for a diverse set of physical systems.\nThe law is used %for example \nin studies of materials, emulsions, biological tissues as well as in planetary science \\cite{LeRoux2013,Roy2018,Noever1992,Mombach1993,Pedro2008}.\nAs an example of the calculation% of the \\aw{} parameter\n, the plots of the fits for the systems in figure \\ref{fig:zach} are presented in figure \\ref{fig:awdemo}, along with the corresponding $\\alpha$ parameters.\nThis demonstrates two contrasting aspects of the \\aw{} law. \nFirstly the law holds very well, especially given the fact that these samples consist of just 20 rings each.\nHowever, it also demonstrates that the law is by no means exact and that some uncertainty is inevitably introduced during the linear regression.\nThis dichotomy will be discussed in detail in section \\ref{s:awdeficiencies}, where it will be shown that the \\aw{} law can be replaced by another measure from network theory, namely the assortativity.\n", "meta": {"hexsha": "f129737e835eb9e1d8670a0e1e57bac95cc61ed0", "size": 24945, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "text/theory.tex", "max_stars_repo_name": "dormrod/Thesis", "max_stars_repo_head_hexsha": "77ddd9fcb3b563a5dc93457682724046053e1137", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-05-14T11:17:18.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-14T11:17:18.000Z", "max_issues_repo_path": "text/theory.tex", "max_issues_repo_name": "dormrod/Thesis", "max_issues_repo_head_hexsha": "77ddd9fcb3b563a5dc93457682724046053e1137", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "text/theory.tex", "max_forks_repo_name": "dormrod/Thesis", "max_forks_repo_head_hexsha": "77ddd9fcb3b563a5dc93457682724046053e1137", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.7979539642, "max_line_length": 517, "alphanum_fraction": 0.7402685909, "num_tokens": 6658, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430604060731, "lm_q2_score": 0.7549149813536518, "lm_q1q2_score": 0.6274423479486677}} {"text": "\\section{Probability Distributions}\n\n \\subsection{What is a probability distribution?}\n A probability distribution is a method of displaying data that displays the probabilities of every possible outcome. We'll need to define a few types of distributions first though, which are uniform and non-uniform distributions.\n \\begin{definition}\n Uniform Distribution: A distribution is called uniform when all of the outcomes are equally likely, just like a single die roll.\n \\end{definition}\n \\begin{definition}\n Non-Uniform Distribution: A distribution is called non-uniform when not all of the outcomes are equally likely, just like rolling the sum of a pair of dice.\n \\end{definition}\n The variable we measure when calculating these, is called the random variable, and is denoted as $X$. The various possibilities for $X$ are the outcomes, denoted $x$. We'll typically find the probability distributions that rely on two different types of variables: discrete and continuous variables. \n \\begin{definition}\n Discrete Variables: These are values that are typically separate from each other, and have a finite number of possibilities.\n \\end{definition}\n \\begin{definition}\n Continuous Variables: These are values that have an infinite number of possibilities, such as a length of time.\n \\end{definition}\n \n \\subsection{Sigma Notation \\& Regular Probability Distributions}\n Sigma Notation, or $\\sum_{}^{}$, may look terrifying, but it simply means the sum of a lot of numbers. We use this when finding an expected value of something, which we'll denote as $E(x)$. To find this, we'll use the equation below.\n \\begin{equation*}\n E(x) = \\sum_{i=1}^{n}x_{i}P(x_{i})\n \\end{equation*}\n To break it down, the $i=1$ is the start, the $n$ is the stop, $x_{i}$ is the outcome, and $P(x_{i})$ is the possibility of that outcome.\n To show this in action, let's suppose that in a game, you'll get the amount of points you'll roll on a die.\n To find that, we'll need to use find $E(x)$.\n We know that the probability of rolling each number is $\\frac{1}{6}$, and we know all the outcomes, so let's fill out the equation. First, we'll need to set our limits though.\n \\begin{equation*}\n E(x) = \\sum_{i=1}^{6}\n \\end{equation*}\n This means that we must repeat the $x_{i}P(x_{i})$ starting at one, going to six, which would be six times.\n Now we'll expand that below.\n \\begin{equation*}\n E(x) = 1(\\frac{1}{6}) + 2(\\frac{1}{6}) + 3(\\frac{1}{6}) + 4(\\frac{1}{6}) + 5(\\frac{1}{6}) + 6(\\frac{1}{6})\n \\end{equation*}\n We can also simplify this, as we know it is a uniform distribution. Shown below is the updated equation for this uniform distribution.\n \\begin{equation*}\n E(x) = \\frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6)\n \\end{equation*}\n \n \\subsection{Binomial Distribution}\n Remember binomials from combinations? They look like this $\\binom{x}{y}$, and come in handy when working with distributions.\n We have to mention something called a Bernoulli Trial first.\n \\begin{definition}\n Bernoulli Trials: This refers to repeated, independent trials that are measured in terms of success, or failure (or other Boolean outcomes).\n \\end{definition}\n The probability distribution of the number of successes in Bernoulli Trials is called a binomial distribution, and is found using the equation below.\n \\begin{equation*}\n P(x) = \\binom{n}{x}p^{x}q^{n-x}\n \\end{equation*}\n The equation solves for the probability of exactly $x$ successful trials out of $n$ total trials. In this equation, $p$ represents the probability of success, and $q$ represents the probability of failure.\n To find the expected value for a binomial distribution, we'll resort to using the formula below.\n \\begin{equation*}\n E(x) = n\\cdot p\n \\end{equation*}\n\n \\subsection{Geometric Distributions}\n A geometric distribution is found when an experiment is repeated just \\textbf{until the first success}.\n It differs from binomial distributions, as the number of trials is not known at the start. \n Just before the success, there is something called a \"waiting period\". \n \\begin{definition}\n Waiting Period: This is the number of failures before the first and only success. \n \\end{definition}\n If you have 8 failures before your success, then you have a waiting period of 8.\n To solve for the probability of success after $x$ failures, we can use the formula below.\n Remember, $x$ represents the waiting period.\n \\begin{equation*}\n P(x) = q^{x}p^{1}\n \\end{equation*}\n In this equation, the variable $p$ \\textbf{must have a max of one} as you can never have more than one success in a Geometric Distribution. To find your expected waiting period, you must solve for $E(x)$, which you can do by using the formula below.\n \\begin{equation*}\n E(x) = \\frac{q}{p}\n \\end{equation*}\n \\subsection{Hyper\\textemdash Geometric Distribution}\n Hyper-Geometric Distribution are quite similar to Geometric Distributions, yet they are dependent, rather than independent.\n Repeated sampling without replacement exhibits this type of process.\n To find the probability of this, we must use the equation below, where $n$ represents the total number of choices,\n $r$ represents the total number of items chosen,\n $a$ represents the number of successes to choose from, and\n $x$ represents the successes chosen.\\\\\n \\begin{equation*}\n P(x) = \\frac{\\binom{a}{x}\\cdot \\binom{n-a}{r-x}}{\\binom{n}{r}}\n \\end{equation*}\n We can also find the number of expected successes chosen by using the formula below, using the same variables as in the equation above.\n \\begin{equation*}\n E(x) = r \\cdot \\frac{a}{n}\n \\end{equation*}", "meta": {"hexsha": "6f6cac04c426de792c52b872912dbbf2afb03609", "size": 5830, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "highschool-projects/MDM4UI/Distributions.tex", "max_stars_repo_name": "johnaoss/dead-projects", "max_stars_repo_head_hexsha": "f8a911a8d08dc34bf52a8d1afd8493a3fcb7f2ca", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "highschool-projects/MDM4UI/Distributions.tex", "max_issues_repo_name": "johnaoss/dead-projects", "max_issues_repo_head_hexsha": "f8a911a8d08dc34bf52a8d1afd8493a3fcb7f2ca", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "highschool-projects/MDM4UI/Distributions.tex", "max_forks_repo_name": "johnaoss/dead-projects", "max_forks_repo_head_hexsha": "f8a911a8d08dc34bf52a8d1afd8493a3fcb7f2ca", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.0114942529, "max_line_length": 304, "alphanum_fraction": 0.7063464837, "num_tokens": 1491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430562234877, "lm_q2_score": 0.7549149758396752, "lm_q1q2_score": 0.627442340208268}} {"text": "\\chapter{ODE Example Revisited}\\label{ext:ode-example}\nOne way by which modelers can improve the quality of parameter estimates is by collecting more data.\nWe consider what would happen if our measurement equipment were able to capture twice as many observations.\nFurthermore, we verify computationally that using more precise measurement equipment improves the precision of the MUD estimate.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Different Measurement Equipment}\nInstead of using equipment that operates at $100$Hz, we take $200$ measurements every second, resulting in 400 equispaced observations for $t \\in (1,3)$.\nAll other choices involved in the experiment (assumed equipment tolerance, number of trials, parameter samples), are kept the same.\nWe refer to this setup wit faster measurement equipment as the ``alternative'' design.\n\n\nWe show the resulting predictions for the signal using MUD points from twenty repeated trials in Figure~\\ref{fig:ode-alt-reference} using the first twenty (middle) and all of the measurements (bottom).\nThe top of the figure shows the solution from the original setup \\ref{fig:ode-reference} for visual comparison.\nThe true signal is well-recovered even with a small subset of the data collected.\nBy the time all measurements are used, the stability of the solutions\\---with respect to the noise that may have polluted them\\---is evidenced by the fact that the red lines representing solutions disappear against the backdrop of the true signal in black.\n\n\\begin{figure}[htbp]\n \\includegraphics[width=\\linewidth]{figures/ode/ode_20_reference_solution.png}\n \\includegraphics[width=\\linewidth]{figures/ode/ode-alt_20_reference_solution.png}\n \\includegraphics[width=\\linewidth]{figures/ode/ode-alt_400_reference_solution.png}\n \\caption{Gray lines are the initial parameter samples.\n The true signal (black) is well-recovered by the MUD estimates (red).\n (Top): First twenty measurements used to solve the original problem.\n (Middle): First twenty measurements used to solve the alternative problem.\n (Bottom): The entire value $\\param_i$ ($1\\leq i \\leq N$), in the sampled parameter set.\n }\n \\label{fig:ode-alt-reference}\n\\end{figure}\n\nTo quantify accuracy and stability of the MUD solutions, we solve the problem for the same choices of $S$ as the original problem (with the addition of $S=400$).\nWe show the resulting error plots for convergence in the right half of Figures~\\ref{fig:ode-convergence-alt}, juxtaposed against the original experimental design with $100$Hz equipment.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.475\\linewidth]{figures/ode/ode_convergence_mud_obs_mean_comp.png}\n \\includegraphics[width=0.425\\linewidth]{figures/ode/ode_convergence_mud_obs_var_comp.png}\n\n \\caption{Convergence of the MUD point given $N=1E3$ model evaluations for increasing numbers of observations for randomly placed sensors.\n Convergence rates are estimated using first-order linear regressions in $\\text{log}_{10}$-space.\n $100$Hz equipment demonstrates a reduction of uncertainty and improvement in precision as $S$ increases towards $200$.\n We observe the same rates of convergence for the alternative equipment and note the (slightly) lower overall error for equal numbers of measurements ($S=200$ corresponding to $t\\in (1,2)$ in this formulation).\n }\n \\label{fig:ode-convergence-alt}\n\\end{figure}\n\nThe convergence rates are similar (shown in the legend of \\ref{fig:ode-convergence}), and reduction in uncertainty is almost negligible at a given $S$.\nHowever, we note that in the alternative setup, for an equal number of measurements, the time elapsed is half of that in the original due to the different equipment being used.\nTo this end, we estimate convergence rates with respect to the time elapsed in the experiment rather than number of measurements used, and notice that the alternative setup (orange) exhibits much lower error at a given point of time.\nThis implies that we can achieve similar results with a shorter observational window by using equipment that allows for faster observations.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\FloatBarrier\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\subsection{Impact of Equipment Precision}\nTo achieve higher precision in the estimate of the MUD point, one can use more precise measurement equipment.\nWe expect that a method designed to address parameter estimation would see an improvement in accuracy if the data is collected with more precise instruments.\nHere we show that this is indeed the case for the time-series example introduced earlier by considering choices of $\\tau = 0.1, 0.05, 0.01, \\text{ and } 0.005$ for $\\mathbb{P}( \\abs{\\xi} < \\tau ) = 99\\%$ to select our $\\sigma$ in our normal additive noise model.\nWe sequentially incorporate $S=5, 10, 15, 20, 25, 50, 100, \\text{ and } 200$ measurements and study the error in our estimate of $\\paramref$.\n\n\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.475\\linewidth]{figures/ode/ode_convergence_mud_std_mean.png}\n \\includegraphics[width=0.425\\linewidth]{figures/ode/ode_convergence_mud_std_var.png}\n\n \\caption{Convergence of the MUD point given $N=1E3$ model evaluations incorporating measurements at a fixed point in time.\n As more precise measurements are incorporated, the accuracy and precision of the MUD solution improves.\n }\n \\label{fig:ode-convergence-std}\n\\end{figure}\n\nIn Figure~\\ref{fig:ode-convergence-std}, we study the absolute error's mean and variance as our measurement equipment gets more precise (lower tolerance), for both the $100$Hz (ode) and $200$Hz (ode-alt) variants of sensors we are simulating.\nIn the left half of the figure, we find that the convergence rates for the two designs are nearly identical but the equipment which records twice as many measurements has a persistent reduction in error.\nThe right half shows the convergence in variance of the absolute error, and the vertical displacement between the two designs is visually difficult to distinguish.\nHowever, as evidenced in the legend annotations of Fig~\\ref{fig:ode-convergence-std}, the alternative design (faster equipment) exhibits an increase in the rate of convergence from 1.87 to 1.92.\n\nWe have shown that the Data--Consistent approach to solving parameter identification problems manages to generalize to problems involving time-series data from a single Quantity of Interest.\nWe now turn our attention to an example where instead of temporal measurements, we incorporate spatial data to solve another 1-D parameter identification problem.\n", "meta": {"hexsha": "f27211dc37d55b56385676d8967d21f090e1c904", "size": 6648, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "extensions/decay_extension.tex", "max_stars_repo_name": "mathematicalmichael/thesis", "max_stars_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-04-24T08:05:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-28T20:34:29.000Z", "max_issues_repo_path": "extensions/decay_extension.tex", "max_issues_repo_name": "mathematicalmichael/thesis", "max_issues_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 59, "max_issues_repo_issues_event_min_datetime": "2019-12-27T23:15:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-24T17:52:57.000Z", "max_forks_repo_path": "extensions/decay_extension.tex", "max_forks_repo_name": "mathematicalmichael/thesis", "max_forks_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 79.1428571429, "max_line_length": 262, "alphanum_fraction": 0.7691034898, "num_tokens": 1457, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676284, "lm_q2_score": 0.8311430436757313, "lm_q1q2_score": 0.6274423399015854}} {"text": "\\chapter{Markov Chain Monte Carlo}\n\\section{Monte Carlo}\nMonte Carlo is a general term for computational techniques that use random\nnumbers.\nMonte Carlo can be used in classical and Bayesian statistics. A special kind\nof Monte Carlo\ncalled Markov Chain Monte Carlo (MCMC) was one of the main reasons\nfor the revival of Bayesian statistics in the second half of the 20th century.\nBefore MCMC became popular, one of the major drawbacks of the Bayesian approach\nwas that some of the calculations were too hard to do. MCMC enables us to solve\na wide range of Bayesian problems which cannot be solved using analytical\nmethods.\n\n\\subsection{Summaries}\nSo far, we have represented our probability distributions (prior and posterior)\nin a computer by using a vector of possible parameter values and a corresponding\nvector of probabilities. For example, suppose we have a single parameter $\\theta$\nand we have worked out the posterior distribution by using a Bayes' Box. This\nwill give us a vector {\\tt theta} of possible $\\theta$ values and a corresponding\nvector {\\tt post} containing the posterior probabilities. Well, one thing we could\ndo is plot the posterior distribution, resulting in a plot like the one in\nFigure~\\ref{fig:normal}.\n\\begin{verbatim}\nplot(theta, post, xlab=\"Theta\", ylab=\"Posterior Probability\")\n\\end{verbatim}\n\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.5]{Figures/normal.pdf}\n\\caption{\\it A posterior distribution can be represented in a computer by a discrete\nset of possible parameter values, and the corresponding probabilities.\\label{fig:normal}}\n\\end{center}\n\\end{figure}\n\nIf we want to obtain some summaries, we could do it like so:\n\\begin{verbatim}\npost_mean = sum(theta*post)\npost_sd = sqrt(sum(theta^2*post) - post_mean^2)\n\\end{verbatim}\n\nHowever,\nthere is an alternative way of representing this posterior distribution in a\ncomputer. It may not be immediately obvious why this is a good idea,\nbecause there is nothing wrong with the tried and true method we have used\nso far. But this second method has the advantage that it continues to\nwork well on much bigger problems, such as when we have more than one parameter.\nWith more than one parameter, the `vector of possible solutions'' approach\ncan fail very dramatically.\n\nOur new way of representing a probability distribution in a computer will be\nvia {\\it Monte Carlo} samples.\nInstead of having two vectors (one of $\\theta$ values and one of the\ncorresponding probabilities),\nimagine we had some method to compute a random sample of $\\theta$ values, drawn\nfrom the posterior distribution in Figure~\\ref{fig:normal}.\nThere would only be one vector {\\tt theta}. So how would we\nknow there is greater probability around $\\theta=1$? Well, {\\it more elements\nof the {\\tt theta} vector would be near 1}.\nInstead of carrying around a second vector of\nprobabilities, we understand that more probable regions will simply contain\nmore points. Say our vector of random\nsamples is called {\\tt theta}. Then we can look at the posterior distribution\nby plotting a histogram of samples:\n\\begin{verbatim}\nhist(theta, breaks=100)\n\\end{verbatim}\nThe histogram looks something like the one in Figure~\\ref{fig:normal2}.\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.5]{Figures/normal2.pdf}\n\\caption{\\it The posterior distribution for a parameter $\\theta$ can also be\nrepresented by a random sample of $\\theta$ values drawn from the posterior\ndistribution. Some $\\theta$ values are more probable than others, which is\nencoded by certain values appearing more frequently in the sample.\n\\label{fig:normal2}}\n\\end{center}\n\\end{figure}\nWe can also get our summaries, but the code looks different\n(it's actually easier than before!):\n\\begin{verbatim}\npost_mean = mean(theta)\npost_sd = sd(theta)\n\\end{verbatim}\nBecause of the randomness involved in generating the {\\tt theta} values,\nthe summaries aren't exact. For example, I know the actual posterior mean\nand standard deviation in this example were both 1, but the values obtained\nfrom the Monte Carlo samples were 0.9604 and 1.0008, respectively. However,\nthis doesn't matter much because the results indicate $\\theta$ is probably\nsomewhere around 1, with an uncertainty of about 1. The error introduced by\nusing random samples is much smaller than the amount of uncertainty inherent\nin the posterior distribution itself. For example, if I summarised the posterior\ndistribution by saying $\\theta = 0.9604 \\pm 1.0008$, for almost all practical\npurposes the conclusion is exactly the same as the true version of the summaries\n$\\theta = 1 \\pm 1$.\n\nIn this discussion, we haven't answered the question of how to actually generate\nrandom samples of $\\theta$ from the posterior distribution. This is the job\nof Markov Chain Monte Carlo.\n\n\\begin{framed}\n{\\bf\nThe purpose of Markov Chain Monte Carlo is to generate random samples of\nparameter values drawn from the posterior distribution. This makes it very easy\nto compute summaries even if you have more than one unknown parameter.}\n\\end{framed}\n\n\\section{Multiple Parameters}\nMCMC becomes extremely useful when we begin to look at Bayesian models involving\nmore than one unknown parameter. \nHaving posterior samples makes the process of {\\it marginalisation}\nmuch easier. Imagine we wanted to infer two parameters, called $a$ and $b$,\nfrom\ndata $x$. Bayes' rule (parameter estimation version) would give us the posterior\ndistribution:\n\\begin{eqnarray}\np(a, b | x) \\propto p(a, b)p(x|a,b)\n\\end{eqnarray}\n\nHowever, what if you didn't really care about the value of $b$ but\nonly really wanted to measure $a$? The terminology for this is that $b$ is a\n{\\it nuisance parameter}: you need it to define the model, but ultimately you\nare not really interested in knowing its value.\nWhat you need in this case is the\n{\\it marginal} posterior distribution for $a$ (that is, the posterior\ndistribution for $a$ on its own, not the joint distribution with $b$).\nThis can be obtained using the sum rule. The result is:\n\\begin{eqnarray}\np(a | x) &=& \\int_b p(a, b|x) \\, db\n\\end{eqnarray}\nor\n\\begin{eqnarray}\np(a | x) &=& \\sum_{b} p(a, b|x)\n\\end{eqnarray}\ndepending on whether the set of possible $b$ values is continuous or discrete.\nBefore MCMC, these integrals or sums usually couldn't be done without making\ncertain choices purely for mathematical convenience (e.g. choosing the prior\nto be a certain kind of distribution only because it will make the maths work\nout, rather than it being a good description of your prior beliefs).\n\nSamples of parameter values drawn from the posterior distribution\n(achieved using MCMC) make this hard problem much easier. We no longer need\nto worry about mathematical convenience.\nSee Figure~\\ref{fig:marginalisation} for an example showing how Monte Carlo\nsampling makes marginalisation trivial.\n\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.5]{Figures/marginalisation.pdf}\n\\caption{\\it An example of a posterior distribution for two parameters, $a$ and\n$b$. The left panels show the joint posterior distribution (which has a\ncorrelation) and the marginal posterior distribution for $a$, obtained by\nintegrating over all the possible $b$ values. The top right panel contains\nrandom samples\n(points) drawn from the posterior distribution for $a$ and $b$.\nThe only step needed to get the\nmarginal distribution for $a$ (lower right panel) is to ignore the $b$-values of\nthe points!\n\\label{fig:marginalisation}}\n\\end{center}\n\\end{figure}\n\n\n\\section{The Metropolis Algorithm}\nThe Metropolis algorithm is the most basic MCMC method.\nThe ideas behind it are fairly simple, yet Metropolis forms the basis of a large\nnumber of more advanced MCMC methods. In STATS 331 we will study the basic ideas\nbehind how the Metropolis algorithm works, which will involve a small amount\nof Markov chain theory. We will also look at a small amount of R code\nwhich implements the Metropolis algorithm, but for solving practical problems\nit is more convenient to use the JAGS program\\footnote{JAGS uses a number\nof MCMC methods internally, including Metropolis, ``Gibbs Sampling'', and\n``Slice Sampling'', which we will not study in this course.}.\n\nThe Metropolis algorithm was invented in the 1950s by physicists\n(including Nicholas Metropolis, for whom the algorithm is named), who used it\nto do calculations in the field of statistical mechanics. This intriguing field\nfocuses on calculating the macroscopic (large scale) properties of matter from\nknowledge of the small-scale properties. For example, if you know water is\ncomposed of H$_2$O molecules, you can use statistical mechanics to figure out\nwhat will happen if you have a lot\nof water molecules. For example, it will freeze at 0 degrees Celsius and boil at 100 degrees Celsius.\n\nIt took\nmany decades before people started to realise the Metropolis algorithm was\nuseful in Bayesian\nstatistics as well as statistical mechanics. The Bayesian approach\nseemed very elegant and\nuseful to many people, but it could always be criticised as unworkable,\nbecause you usually had to do difficult or\nimpossible integrals when solving practical problems\n(to summarise the posterior, or to get rid of nuisance\nparameters). MCMC changed all that, and is one of the reasons for the\nexplosion in the popularity of Bayesian statistics beginning in the 1990s.\n\nThe basic idea of MCMC is that we want a method which will travel between\ndifferent possible states (such as the possible hypotheses/parameter values in\na Bayesian analysis). We want the amount of time spent in\nany particular state to be proportional to the posterior probability of the state.\nThe computer ``explores'' the set of possible parameter values, spending a lot\nof time in the regions with high posterior probability, and only rarely visiting regions\nof low posterior probability. Figure~\\ref{fig:mcmc} shows an example of MCMC\napplied to a problem with only two possible hypotheses or parameter values.\nNobody would actually use MCMC on such a small problem, but it is helpful for\nexplaining how MCMC works.\n\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.65]{Figures/mcmc.pdf}\n\\caption{\\it An illustration of the basic idea behind MCMC. Imagine we had a\nBayes' Box with two possible hypotheses, and we knew the\n{\\tt prior $\\times$ likelihood} values. The amount of time the MCMC program will\nspend in each state is proportional to the posterior probability of the state.\nIn this example, the MCMC algorithm was in State 1 three times and in State 2\nseven times. Using this, we could estimate the posterior probability of State 2\nas being 0.7. This estimate would become more accurate if we ran the MCMC for\nmore iterations.\\label{fig:mcmc}}\n\\end{center}\n\\end{figure}\n\n\\subsection{Metropolis, Stated}\nThe Metropolis algorithm is given below. The first thing to do is start\n{\\it somewhere} in the ``parameter space'' (set of possible parameter values).\nYou then\n{\\it propose} to move somewhere else. There is an acceptance probability $\\alpha$\nthat determines whether to accept the proposal. If the proposal is better\n($h$, the prior times likelihood value is higher), then you accept it, and the\nproposed state becomes the new state of the algorithm.\nIf the\nproposal is worse, you can also accept it, but the probability of accepting is\ngiven by the ratio of the unnormalised posterior probabilities, i.e. $h'/h$.\nFor example, if the proposed point is 1/3 as good as the current point, the\nacceptance probability is 1/3. If the proposed point is rejected, the original\npoint remains the state of the algorithm, and gets counted again in the results.\n\nThe Metropolis algorithm works because it makes transitions towards low\nprobability\nstates rare, while transitions towards high probability states are common, because\nof the acceptance probability. It is hard to move into an improbable state, so\nnot much time will be spent there. Over time, the fraction of time spent in\nany given state is equal to the posterior probability of the corresponding\nhypothesis.\n\n\\begin{framed}\n\\begin{itemize}\n\\item Start in some state $\\theta$\n\\item Generate a ``proposal'' state $\\theta'$ from a {\\it proposal distribution} $q$ (assumed symmetric so that $q(\\theta'|\\theta)=q(\\theta|\\theta')$)\n\\item With probability $\\alpha = \\textnormal{min}(1, h'/h)$, replace the current\nstate with the proposed state\n\\item Repeat\n\\end{itemize}\n\\end{framed}\n\n\\section{A Two State Problem}\nWe will now study how the Metropolis algorithm works on a very simple\nexample, namely the two-ball problem from the beginning of the notes. There\nare two hypotheses and the posterior probabilities are $1/3$ and $2/3$.\nIt is important to note that MCMC is not actually needed for a problem this\nsimple, but it is a good test case to see precisely how an MCMC algorithm works.\nWhen we solve real data analysis problems with JAGS,\nwe won't have to think too\nmuch about how the MCMC works, but can concentrate instead on the Bayesian\nstatistics problem at hand.\n\nLet's\ncall the less probable hypothesis ``State 1'' and the more probable hypothesis\n``State 2'' for the purposes of this section.\nWhat we need is a Markov process that will spend 1/3 of the time in State 1 and\n2/3 of the time in State 2. The Metropolis algorithm described above will\ndo what we need. The main thing we need to compute is the acceptance probability\n$\\alpha_{ij}$ for a proposed transition from state $i$ to state $j$ where\n$i, j \\in \\{1, 2\\}$.\nThe acceptance probability $\\alpha$ for a proposed move from state $i$ to\nstate $j$ is given by\n\\begin{eqnarray}\n\\alpha_{ij} = \\textnormal{min}\\left(1, \\frac{h_j}{h_i}\\right)\n\\end{eqnarray}\nwhere $h_i$ and $h_j$ are proportional to the posterior probabilities of\nstates $i$ and $j$ respectively. If the proposal is to move\nto an equal or better (higher posterior probability)\nstate ($h_j \\geq h_i$) then the acceptance probability is 1.\nIf the proposal is to move to a less probable state then the acceptance probability\nis $h_j/h_i$, the ratio of the two posterior probabilities\\footnote{Note that\nthis algorithm can be used even if the marginal likelihood is unknown, because\nonly ratios of posterior probabilities are needed. This is useful because the\nmarginal likelihood is sometimes very hard to calculate in multi-parameter\nproblems.}.\n\nThe transition probability is the probability\nof being in state $j$ at the next iteration given that you are in state $i$\nat the current iteration. The transition probability is given by the product\nrule:\n\\begin{eqnarray}\np_{ij} = q_j \\alpha_{ij}\\label{eq:transition}\n\\end{eqnarray}\nfor $i \\neq j$.\nThe transition matrix of the Markov chain is a matrix with all the different\n$p_{ij}$ values in it:\n\\begin{eqnarray}\n\\mathbf{P} &=&\n\\left[\n\\begin{array}{cc}\np_{11} & p_{12}\\\\\np_{21} & p_{22}\n\\end{array}\n\\right]\n\\end{eqnarray}\nIn our particular case, we can work out the off-diagonal elements of\n$\\mathbf{P}$ using Equation~\\ref{eq:transition}:\n\\begin{eqnarray}\n\\mathbf{P}\n&=&\n\\left[\n\\begin{array}{cc}\n & \\frac{1}{2} \\times 1\\\\\n\\frac{1}{2}\\times\\frac{1}{2} & \n\\end{array}\n\\right]\n\\end{eqnarray}\nThe diagonal elements can be found by knowing the rows of $\\mathbf{P}$ must\nsum to 1. We must be in {\\it some} state at the next iteration. Therefore the\ntransition matrix of our Markov chain in this two-state problem is:\n\\begin{eqnarray}\n\\mathbf{P}\n&=&\n\\left[\n\\begin{array}{cc}\n\\frac{1}{2} & \\frac{1}{2}\\\\\n\\frac{1}{4} & \\frac{3}{4}\n\\end{array}\n\\right]\\label{eq:p_matrix}\n\\end{eqnarray}\n\nA Markov chain with a small number of possible states can be represented\ngraphically using a {\\it transition diagram}, as in\nFigure~\\ref{fig:transitions}.\n\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.65]{Figures/transitions.pdf}\n\\caption{\\it A transition diagram for a Markov Chain with two possible states.\nThe states (which correspond to hypotheses in Bayesian inference) are drawn\nas circles labelled ``1'' and ``2''. When the algorithm is in a particular state\n(e.g. State 1), and we apply one step of the Metropolis algorithm,\nthe probability it will be in State 1 is $p_{11}$ and the probability it\nwill be in State 2 is $p_{12}$.\nMCMC works by making it easy to move into states with high\nposterior probability, and hard to move out of them.\n\\label{fig:transitions}}\n\\end{center}\n\\end{figure}\n\n\\section{The Steady-State Distribution of a Markov Chain}\nOnce we have the transition matrix, we can work out the {\\it steady state\ndistribution} of the Markov chain. Imagine, instead of starting the MCMC\nfrom an arbitrary initial state, you have a probability distribution for the\ninitial state. After applying one iteration of the Metropolis algorithm, the\nprobability distribution for the updated state will usually be different from\nthe probability distribution for the initial state.\n\nHowever, one special probability distribution, called the {\\it steady state\ndistribution}, does not change after you apply an MCMC update. If your initial\npoint was drawn from the steady state distribution, then subsequent points will\nalso be drawn from the steady state distribution. In MCMC the steady state\ndistribution should be the same as the posterior distribution.\n\nImagine we are using MCMC on our two-state problem, and our initial position\nis State 1 with probability $v_1$ and State 2 with probability $v_2$. What\nis the probability of being in State 1 at the next iteration? There are two\nways for that to happen: by starting in State 1 and then making a transition\nfrom $1\\to 1$, or by starting in State 2 and making a transition from $2\\to 1$.\nThe total probability is then:\n\\begin{eqnarray}\nP(\\textnormal{State 1 after iteration}) &=& v_1p_{11} + v_2p_{21}.\n\\end{eqnarray}\nSimilarly for State 2:\n\\begin{eqnarray}\nP(\\textnormal{State 2 after iteration}) &=& v_1p_{12} + v_2p_{22}.\n\\end{eqnarray}\nIf $v_1$ and $v_2$ happened to be the steady state distribution, then these\nprobabilities would also be $v_1$ and $v_2$. This gives us two simultaneous\nequations\n\\begin{eqnarray}\nv_1 &=& v_1p_{11} + v_2p_{21}\\\\\nv_2 &=& v_1p_{12} + v_2p_{22}\n\\end{eqnarray}\nwhich can also be written in matrix form as $\\mathbf{v}\\mathbf{P} = \\mathbf{v}$\nwhere $\\mathbf{v} = \\left(v_1, v_2\\right)$ and $\\mathbf{P}$ is the transition\nmatrix\\footnote{Mathematics students might recognise this equation, which says\n$\\mathbf{v}$ is the left-eigenvector of $\\mathbf{P}$, with eigenvalue 1.\nAn alternative is to write the stationary distribution as a column vector and\nuse $\\mathbf{P}^T\\mathbf{v} = \\mathbf{v}$.}.\nThese two simultaneous equations can be solved for $v_1$ and $v_2$.\nHowever there is a third constraint, that is $v_1 + v_2 = 1$. Since there\nare three equations but two unknowns, it seems like the problem might be\n``over-determined'', but that is not actually the case because the transition\nmatrix rows are not all linearly independent.\nIt is left as an exercise for the reader to show the steady state\ndistribution for our Markov chain (Equation~\\ref{eq:p_matrix}) is in fact\nequal to the posterior distribution, so\n$\\mathbf{v} = \\left(\\frac{1}{3}, \\frac{2}{3}\\right)$.\n\n\\section{Tactile MCMC}\nIn class we will do ``Tactile MCMC'', which is an implementation of the\nMetropolis algorithm using coins and dice instead of the random number generators\nprovided in computer software such as R. This is a good way to get a feel for\nthe flow of MCMC algorithms.\n\nIn STATS 331, when we want to use MCMC in practice, we will use the JAGS program\nrather than using Metropolis directly. JAGS is a general purpose MCMC program\nwhich allows you to solve fairly complex problems without a large amount of\nprogramming.\n", "meta": {"hexsha": "e6a0e802970c4a1d788a550ec3fdbde39d7080ce", "size": 19548, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "mcmc.tex", "max_stars_repo_name": "xulinpan/stat331", "max_stars_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 55, "max_stars_repo_stars_event_min_datetime": "2015-03-09T18:03:23.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-25T03:36:54.000Z", "max_issues_repo_path": "mcmc.tex", "max_issues_repo_name": "xulinpan/stat331", "max_issues_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2015-07-07T05:00:32.000Z", "max_issues_repo_issues_event_max_datetime": "2015-07-10T08:48:27.000Z", "max_forks_repo_path": "mcmc.tex", "max_forks_repo_name": "xulinpan/stat331", "max_forks_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2015-07-29T14:34:51.000Z", "max_forks_repo_forks_event_max_datetime": "2020-06-04T20:04:47.000Z", "avg_line_length": 47.4466019417, "max_line_length": 150, "alphanum_fraction": 0.7776754655, "num_tokens": 4908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8311430436757313, "lm_q1q2_score": 0.6274423399015853}} {"text": "\\section{Time Complexity of MST Algorithms}\n\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Prim \\emph{vs.} Kruskal (Problem 6.4)}\n \\begin{exampleblock}{Prim \\emph{vs.} Kruskal (Problem 6.4)}\n\t\\begin{itemize}\n\t \\item Array \\emph{vs.} heap\n\t \\item $m = O(n)$ \\emph{vs.} $m = \\Omega(n^2)$\n\t\\end{itemize}\n\n\t\\[\n\t T(n,m) = O(n + n T(\\text{deleteMin}) + m T(\\text{decreaseKey}))\n\t\\]\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{MST on special graphs (Problem 6.3)}\n \\begin{exampleblock}{MST on special graphs (Problem 6.3)}\n\t\\begin{enumerate}\n\t \\item $K$-bounded degree graph\n\t \\item Planar graph\n\t\\end{enumerate}\n \\end{exampleblock}\n\n \\[\n\t(1)\\; m \\le \\frac{nk}{2}\n \\]\n\n \\[\n\t(2)\\; m \\le 3n - 6\n \\]\n\n \\begin{alertblock}{Reference}\n\t\\begin{itemize}\n\t \\item ``Finding Minimum Spanning Trees'' by David Cheriton and Robert Tarjan, 1976 (linear on planar graph).\n\t\\end{itemize}\n \\end{alertblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Prim on special graphs (Problem 6.1)}\n \\begin{exampleblock}{Prim on special graphs (Problem 6.1)}\n\t\\[\n\t E = \\set{v_1v_i \\mid i = 2 \\dots n}, W(v_1v_i) = 1\n\t\\]\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Prim on special graphs (Problem 6.2)}\n \\begin{exampleblock}{Prim on special graphs (Problem 6.2)}\n\t\\begin{enumerate}\n\t \\item $G = K_n$\n\t \\item $W(v_iv_j) = n + 1 -i, 1 \\le i < j \\le n$\n\t\\end{enumerate}\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "9bac5060755e56216d7b1c284d0ff4b106fea072", "size": 1446, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-mst-20170531/sections/mst-time.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-mst-20170531/sections/mst-time.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-mst-20170531/sections/mst-time.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 25.3684210526, "max_line_length": 111, "alphanum_fraction": 0.5961272476, "num_tokens": 528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430436757312, "lm_q2_score": 0.7549149758396752, "lm_q1q2_score": 0.6274423307357787}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{graphicx}\n\\usepackage{latexsym,algorithm,algorithmic}\n\\usepackage{times}\n\\usepackage{amsmath,amssymb,mathtools}\n\\usepackage{mathptmx} % use Times fonts if available on your TeX system\n\n\\setlength{\\textwidth}{7in}\n\\setlength{\\textheight}{9in}\n\\setlength{\\oddsidemargin}{-0.5in}\n\\setlength{\\evensidemargin}{-0.5in}\n\\setlength{\\topmargin}{-0.5in}\n\n\n\\newcommand{\\weightvec}{\\textbf{w}}\n\\newcommand{\\edge}[2]{{u\\rightarrow{}v}}\n\\newcommand{\\edgeuv}{{\\edge{u}{v}}}\n\n\\newcommand{\\vek}[1]{\\textbf{#1}}\n\\newcommand{\\ddw}{\\frac{\\partial}{\\partial\\vek{w}}}\n\\newcommand{\\M}{\\textrm{M}}\n\\newcommand{\\dM}{\\textrm{dM}}\n\\newcommand{\\df}{\\textbf{df}}\n\\newcommand{\\dt}{\\textbf{dt}}\n\\newcommand{\\vphi}{\\vec{\\phi}}\n\n\\begin{document}\n\n\\section{Derivation: PPR and its derivative}\n\nNotation: $\\vek{s}$ is seeds, $\\M$ is transition matrix, $\\vek{w}$ are\nparameters, and $\\vek{p}^\\infty$ is the PPR stationary distribution,\n$\\vek{p}^t$ is a point in the power-rule iteration for computating\nPPR, and $\\vek{d}^t$ is the partial derivative of $\\vek{p}^t$ wrt the\nparameters $\\vek{w}$.\n\n\\begin{eqnarray}\n\\vek{p}^{t+1} & \\equiv & \\alpha \\vek{s} + (1-\\alpha) \\M^\\top \\vek{p}^t \\\\\n\\vek{d}^t & \\equiv & \\ddw \\vek{p}^t\n\\end{eqnarray}\n\nNote $\\vek{p}^t$ and $\\vek{d}^t$ have different dimensions: while\n$\\vek{p}^t_u$ is a scalar score for $u$ under PPR, $\\vek{d}^t_u$ is a\nvector, giving the sensitivity of that score to each parameter in\n$\\vek{w}$.\n\n$\\M$ is defined as follows. There is a weight vector $\\vek{w}$, and\nfor an edge $\\edgeuv$, there is a feature vector $\\vphi_{uv}$, which is\nused to define a basic score $s_{uv}$ for the edge, which is passed\nthru a squashing function $f$, e.g., $f(x)\\equiv e^x$, and then\nnormalized to form $\\M$.\n\n\\begin{eqnarray}\n s_{uv} & \\equiv & \\vphi_{uv} \\cdot \\vek{w} \\\\\n t_u & \\equiv & \\sum_{v'} f(s_{uv'}) \\\\\n \\M_{u,v} & \\equiv & \\frac{ f(s_{uv}) }{ t_u }\n\\end{eqnarray}\n\nWe can define $\\vek{d}^t$ recursively:\n\n\n\\begin{eqnarray}\n\\vek{d}^{t+1} & = & \\ddw \\vek{p}^{t+1} \\\\\n & = & \\ddw \\left( \\alpha \\vek{s} + (1-\\alpha) \\M^\\top \\vek{p}^t \\right) \\\\\n & = & (1-\\alpha) \\ddw \\M^\\top \\vek{p}^t \\\\\n & = & (1-\\alpha) \\left( (\\ddw \\M^\\top) \\vek{p}^t + \\M^\\top \\ddw\\vek{p}^t \\right) \\\\\n & = & (1-\\alpha) \\left( (\\ddw \\M^\\top) \\vek{p}^t + \\M^\\top \\vek{d}^t \\right)\n\\end{eqnarray}\n\nNow let's look at $\\ddw\\M$, which I'll denote $\\dM$ below. Note that\neach $\\dM_{uv}$ is a vector, again giving the sensitivity of the\nweight $\\M_{uv}$ to each parameter in $\\vek{w}$.\n\n\\begin{eqnarray}\n\\dM_{uv} & = & \\ddw \\frac{ f(s_{uv}) }{ t_u } \\\\\n & = & \\frac{1}{t_u ^2} \\left( t_u \\ddw f(s_{uv}) - f(s_{uv}) \\ddw t_u \\right)\n\\end{eqnarray}\n\nTo continue this, define $\\df$ and $\\dt$ as the vectors\n\n\\begin{eqnarray}\n \\df_{uv} & \\equiv & \\ddw f(s_{uv}) = f'(s_{uv}) \\vphi_{uv} \\label{eqn:fprime} \\\\\n \\dt_{u} & \\equiv & \\ddw t_u = \\sum_{v'} \\df_{uv'}\n\\end{eqnarray}\n\nNote that $\\df_{uv}$ has no more non-zero components than\n$\\vphi_{uv}$, so it is sparse, and $\\dt_{u}$ is also sparse, but\nsomewhat less so.\n\nWe can continue the derivation as\n\n\\begin{eqnarray}\n\\dM_{uv} & = & \\frac{1}{t_u ^2} \\left( t_u \\ddw f(s_{uv}) - f(s_{uv}) \\ddw t_u \\right) \\\\\n & = & \\frac{1}{t_u ^2} \\left( t_u \\df_{uv} - f(s_{uv}) \\dt_u \\right)\n\\end{eqnarray}\n\nThis is a little different from the old algorithm: there is always an\n(implicit) reset with probability $\\alpha$, and that reset probability\ncan be increased by learning by weighting the features for the\n(explicit) reset links that are present already in the graph. No min\n$\\alpha$!\n\n\\section{Computation}\n\n\\begin{table} % Computing M and dM\n\\hrule\n\\begin{enumerate}\n\\item For each node/row $u$\n \\begin{enumerate}\n \\item $t_u = 0$\n \\item $\\dt_u = \\vek{0}$, an all-zeros vector\n \\item For each neighbor $v$ of $u$\n \\begin{enumerate} \n \\item $s_{uv} = \\vek{w} \\cdot \\vphi_{uv}$, a scalar.\n\n In detail: For $i\\in \\vphi_{uv}$: increment $s_{uv}$ by $\\vek{w}_i\\vphi_i$\n\n \\item $t_u +\\!\\!= f(s_{uv})$, a scalar\n \\item $\\df_{uv} = f'(s_{uv}) \\vphi_{uv}$, a vector, as sparse as $\\vphi_{uv}$\n\n In detail: For $i\\in \\vphi_{uv}$: set $\\df_{uv,i} = \\vphi_f * c$, where $c=f'(s_{uv})$\n\n \\item $\\dt_u +\\!\\!= \\df_{uv}$, a vector, as sparse as $\\sum_{v'} \\vphi_{uv'}$\n\n In detail: For $i\\in \\df_{uv}$: increment $\\dt_{u,i}$ by $\\df_{uv,i}$\n\n \\end{enumerate}\n Now $t_u = \\sum_{v'} f(s_{uv'})$ and $\\dt_{u} = \\sum_{v'} \\df_{uv'}$\n \\item For each neighbor $v$ of $u$ create the vector\n\n\\[ \\dM_{uv} = \\frac{1}{t_u ^2} \\left( t_u \\df_{uv} - f(s_{uv}) \\dt_u \\right) \n\\]\n\nOr in detail: For $i\\in\\dt_u$: $\\dM_{uv,i} = \\frac{1}{t_u ^2} \\left( t_u \\df_{uv,i} - f(s_{uv}) \\dt_u,i \\right)$. \n\nThere aren't any dimensions $i$ that are present in $\\df_{uv}$ but not\n$\\dt_u$, since $\\dt_u$ is a summation. Also create the scalar\n\n\\[ \\M_{uv} = \\frac{f(s_{uv})}{t_u} \n\\]\n\n You can now discard the intermediate values like $t_u$, $\\dt_u$, $\\df_{uv}$.\n\n \\end{enumerate}\n\\end{enumerate}\n\\caption{Computing $\\M$ and $\\dM$} \\label{alg:mat}\n\\hrule\n\\end{table}\n\nSince $\\dM$ and $M$ are reused many times, we should compute and store\n$M$ and $\\dM$ first. This will expand the size of the graph somewhat:\nin particular, we will need to store, not only the active edges $u,v$\nand their features $\\vphi_{uv}$, but also $\\dM_{uv}$, which includes\nweights for all features of vertexes $v'$ that are siblings of $v$\n(i.e., there is an edge $u,v'$). I'm not sure how bad this will be in\npractice: perhaps we should estimate it for some of our test cases.\nAfter this, operations should run in time linear in the size of the\nnew, less sparse graph encoded in $\\dM$. I believe that this scheme\nalso makes operations like the APR learning more modular.\n\nComputing $M$ and $\\dM$ is one pass over the graph, shown in Table\n\\ref{alg:mat}.\n\n\nThen you can start with $\\vek{p}^0 = \\vek{d}^0 = \\vek{0}$ and iterate\n\n\\begin{eqnarray}\n\\vek{p}^{t+1} & = & \\alpha \\vek{s} + (1-\\alpha) \\M^\\top \\vek{p}^t \\\\\n\\vek{d}^{t+1} & = & (1-\\alpha) \\left( \\dM^\\top \\vek{p}^t + \\M^\\top \\vek{d}^t \\right)\n\\end{eqnarray}\n\nIn more detail, the iteration for the updates on $\\vek{p}$ are shown\nin Table~\\ref{alg:updates}.\n\n\\begin{table} % updating p and d\n\\hrule\n\nUpdating $\\vek{p}$:\n\\begin{enumerate}\n\\item $\\vek{p}^{t+1} = \\vek{0}$\n\\item For each node $u$ \n \\begin{enumerate}\n \\item $\\vek{p}^{t+1}_u +\\!\\!= \\alpha \\vek{s}_u$\n \\item For each neighbor $v$ of $u$\n \\begin{enumerate}\n \\item $\\vek{p}^{t+1}_v +\\!\\!= (1-\\alpha) \\vek{p}^t_u \\M_{uv} $\n \\end{enumerate}\n \\end{enumerate}\n\\end{enumerate}\n\nUpdating $\\vek{d}$:\n\\begin{enumerate}\n\\item $\\vek{d}^{t+1} = \\langle \\vek{0}, \\ldots, \\vek{0} \\rangle$ --- i.e., for each node $u$ there is an all-zeros\n vector of weights.\n\\item For each node $u$ \n \\begin{enumerate}\n \\item For each neighbor $v$ of $u$\n \\begin{enumerate}\n \\item For each $i$ in $\\dM_{vu}$\n\n \\[ \\vek{d}^{t+1}_{v,i} +\\!\\!= (1-\\alpha) \\vek{p}^t_u \\dM_{uv,i} \\]\n \\item For each $i$ in $\\vek{d}^t_{v}$\n\n \\[ \\vek{d}^{t+1}_{v,i} +\\!\\!= (1-\\alpha) \\vek{d}^t_{u,i} \\M_{uv} \\]\n \\end{enumerate}\n \\end{enumerate}\n\\end{enumerate}\n\\caption{Updates for $\\vek{d}$ and $\\vek{p}$} \\label{alg:updates}\n\\hrule\n\\end{table}\n\n\\section{Loss functions and lazy regularization}\n\nFor SRW each example is a triple $(\\vek{s},P,N)$ where $\\vek{s}$ is\nthe seed distribution, $P=\\{a^1,\\ldots,a^I\\}$ are the positive (a-ok?)\nexamples, and $N=\\{b^1,\\ldots,b^J\\}$ are the negative (bad?) examples.\nI use $\\vek{p}$ for the PPR distribution starting at $\\vek{s}$, and\nwrite $\\vek{p}[u]$ for $\\vek{p}_u$ if I run out of space for\nsubscripts.\n\nThe loss function is\n\\begin{equation}\nL(\\vek{w}) \\equiv - \\Bigg (\\sum_{k=1}^I \\log \\vek{p}[a^k] + \\sum_{k=1}^J \\log (1 - \\vek{p}[b^k]) \\Bigg) + \\mu R(\\vek{w})\n\\end{equation}\nwhere $R(\\vek{w})$ is the regularization, eg\n$R(\\vek{w})\\equiv||\\vek{w}||^2_2$. This means loss increases as the objective function decreases, where the objective function is proportional to the probability of either hitting a positive solution or not-hitting a negative solution. Once we have $\\vek{d}$ it's easy\nto compute the gradient of this as\n\n\\begin{eqnarray}\n\\ddw L(\\vek{w}) & \\equiv & - \\Bigg (\\sum_{k=1}^I \\frac{1}{\\vek{p}[a^k]}\\vek{d}[a^k] \n - \\sum_{k=1}^J \\frac{1}{1 - \\vek{p}[b^k]}\\vek{d}[b^k] \\Bigg) + \\mu \\ddw R(\\vek{w})\n\\end{eqnarray}\n\nWe can split this into two parts: the empirical loss gradient, which is\n\n\\[\n-\\Bigg (\\sum_{k=1}^I \\frac{1}{\\vek{p}[a^k]}\\vek{d}[a^k] \n - \\sum_{k=1}^J \\frac{1}{1 - \\vek{p}[b^k]}\\vek{d}[b^k] \\Bigg)\n\\]\nand the regularization gradient, \n\\[\n \\mu \\ddw R(\\vek{w})\n\\]\nIf an example doesn't contain all features then the empirical gradient\nwill be a sparse vector, but the regularization gradient will be\ndense. So the following code might be more efficient for SGD than\njust computation of the full gradient and taking a step in that\ndirection.\n\n\\begin{enumerate}\n\\item Maintain a ``clock'' counter $m$ which is incremented when each\n example is processed. Also maintain a history $\\vek{h}_i$ which\n says, for each feature $i$, the last time $t$ an example containing\n $i$ was processed.\n\\item When a new example $(\\vek{s},P,N)$ arrives at time $t$\n \\begin{enumerate}\n \\item For each feature active in the example, initialize it, if\n necessary, and them perform the regularization-loss gradient\n update $t-\\vek{h}_i$ times.\n \\item Peform the empirical-loss update, using \n the new weights.\n \\end{enumerate}\n\\item When you finish learning at final time $T$, consider every\n feature $i$, and perform the regularization-loss gradient update\n $T-\\vek{h}_i$ times. Then write out the final parameters.\n\\end{enumerate}\n\n\\section{Inference: PPR and APR}\n\nInference using requires only $\\M$, which at theorem-proving time is\ncomputed on-the-fly. That is, whenever you need a row of $\\M$ (the\nset of values $\\M[u,v]$ for all $v$ near $u$) you simply compute the\nnormalized outlinks of $u$.\n\nThe power-iteration version of PPR, which in the codebase is called\nthe \\texttt{PPRProver}, simply iterates this step until convergence\n(or for a fixed number of iterations), starting with\n$\\vek{p}^0=\\vek{0}$.\n\n\\begin{eqnarray}\n\\vek{p}^{t+1} & \\equiv & \\alpha \\vek{s} + (1-\\alpha) \\M \\vek{p}^t \\\\\n\\end{eqnarray}\n\nBreaking this down, the one-step update is the following.\n\n\\begin{enumerate}\n\\item $\\vek{p}^{t+1} = \\vek{0}$\n\\item For each key $u$ with non-zero weight in $\\vek{s}$:\n \\begin{enumerate}\n \\item $\\vek{p}^{t+1}[u]$ += $\\alpha \\vek{s}[u]$\n \\end{enumerate}\n\\item For each key $u$ with non-zero weight in $\\vek{p}^t$:\n \\begin{enumerate}\n \\item For each node $v$ near $u$ in $\\M$:\n\\(\n \\vek{p}^{t+1}[v] \\mbox{~+=~} (1-\\alpha) \\M[u,v] \\vek{p}^t[u]\n\\)\n \\end{enumerate}\n\\end{enumerate}\n\nThe approximate PageRank is based on a more primitive \\texttt{push}\noperation. The full approximate PageRank algorithm also uses\n$\\vek{d}$, a vector of node degrees.\n\n\\begin{enumerate}\n\\item Let $\\vek{p}= \\vek{0}$ and $\\vek{r}= \\vek{s}$\n\\item While there is some vertex $u$ such that $r(u)\\geq{}\\epsilon\n \\vek{d}(u)$:\n \\begin{enumerate}\n \\item Perform the \\texttt{push}($u$) operation:\n \\begin{enumerate}\n \\item Save the current value of $u$ in $\\vek{r}$: $ru = \\vek{r}[u]$\n \\item $\\vek{p}[u]$ += $\\alpha \\vek{r}[u]$\n \\item $\\vek{r}[u]$ *= $(1-\\alpha)$\n \\item For each node $v$ near $u$ in $\\M$:\n\\(\n \\vek{r}[v] \\mbox{~+=~} (1-\\alpha) \\M[u,v] ru\n\\)\n \\end{enumerate}\n \\end{enumerate}\n\\end{enumerate}\n\nIn the old implementation, $\\vek{d}[u]$ was computed and cached each\ntime a new key $u$ was added to $\\vek{r}$. In the new implementation\nof the prover, this caching is done by the proof graph, so there's no\nneed to do it in the prover. \n\nThe old DPR prover implementation computed this approximation by\nfinding $u$'s with large $\\vek{r}[u]$ using a variation of the\nstandard depth-first prover: essentially, you traversed the tree\ndepth-first, and whenever you hit a node with $\\vek{r}[u]$ below\nthreshold, you stopped. Factoring this in the algorithm used the same\ninitial values of $\\vek{p}$ and $\\vek{r}$ but then called a recursive\nroutine \\texttt{proveState}$(u_0)$, where $u_0$ is the initial query\nnode (aka the initial state of the proof graph).\n\nWilliam Wang's got a new version which calls this iteratively with\nsmaller and smaller $\\epsilon$'s, which seems to work faster. That\nsuggests that it's helpful to do pushes on the nodes with larger\n$\\vek{r}$ values first. With that in mind, another thing we could\nconsider would be storing $u$'s in a heap/priority queue, ordered by\n$\\vek{r}$ values. I know this idea has been used in the past but I\ndon't know how much it helps. Obviously there's an overhead for the\nheap but my guess is that's not going to dominate.\n\n\\bigskip\n\n\\noindent Function \\texttt{proveState}$(u)$:\n\\begin{enumerate}\n\\item If $r(u)\\geq{}\\epsilon \\vek{d}(u)$:\n \\begin{enumerate}\n \\item Perform the \\texttt{push}$(u)$ operation\n \\item For each $v$ near $u$ (i.e., each subgoal of $u$): \\texttt{proveState}$(v)$\n \\end{enumerate}\n\\end{enumerate}\n\n\n\\section{Architectural Comments}\n\n\\emph{ [Katie's comments inline and signed with -k]}\n\n\\noindent\nThe routines that are suggested are:\n\\begin{itemize}\n\\item Loading: Compute $\\M$ and $\\dM$ from a graph and $\\vek{w}$. It\n seems necessary, altho a little non-modular, to initialize and\n create new features while the graph is read in. \\emph{ [Do this here, or in SGD? -k]} We should be able\n to estimate performance pretty well after this step.\n\nInputs: graph and current values of parameters $\\vek{w}$.\n\nOutputs: $\\M$ and $\\dM$, and a list of all active features $i$.\n\nParameters: $f$, $f'$; method to init new features $i$ for $f$, clock\ntime and lazy-regularization function.\n \n\\item Extended inference: compute $\\vek{p}$ and $\\vek{d}$.\n \nInputs: $\\vek{s}$, $\\vek{w}$, $\\M$ and $\\dM$.\n\nOutputs: $\\vek{p}$ and $\\vek{d}$.\n\n\\item SGD: Initialize new features $i$, use clock time and\n lazy-regularization function to update their weights, and update\n $\\vek{w}$ using the empirical (non-gradient) loss.\n\nInputs: $\\vek{p}$, $\\vek{d}$, and $P$, $N$.\n\nOutputs: Modified $\\vek{w}$.\n\n\\end{itemize}\n\n\\begin{table}\n\\hrule\n\\begin{verbatim}\nclass Graph {\n\n // names of features\n\n String[] featName;\n\n // space for labels for feature weights on edges, each is a\n // feature index and a weight\n\n int[] label_featIndex;\n double[] label_featWeight;\n\n // space for edges, each is a destination node and a\n // list of labels, which point into the label space\n\n int[] edge_dst;\n int[] edge_labels_lo;\n int[] edge_labels_hi;\n\n // space for list of neighbors of nodes, each \n // points into the edge space\n\n int [] node_near_lo;\n int [] node_near_hi;\n\n // list of nodes in graph\n int node_lo, node_hi;\n\n // space for weight derivatives, which are structurally \n // just like labels. these maybe should be vectors since \n // we don't know the size of this space in advance...?\n\n int[] deriv_featIndex;\n double[] deriv_featDeriv;\n\n // list of derivative features for M[u][euv]\n int[][] dM_lo;\n int[][] dM_hi;\n}\n\\end{verbatim}\n\\caption{Proposed data structure for graph} \\label{alg:data}\n\\hrule\n\\end{table}\n\n\\begin{table}\n\\hrule\n\\begin{verbatim}\nfor (int u = g.node_lo; u < g.node_hi; u++) {\n // euv is edge from u to v\n for (euv = g.node_near_lo[u]; euv < g.node_near_hi[u]; euv++) {\n int v = g.edge_dst[euv];\n // luvk is k-th label on edge euv\n for (luvk = g.edge_labels_lo[euv]; luvk = g.edge_labels_hi[euv]; luvk++) {\n int i = g.label_featIndex[luvk];\n double w = g.label_featWeight[luvk];\n ...\n }\n }\n}\nfor (int d = g.dM_lo[u][euv]; d < g.dM_hi[u][euv]; d++) {\n i = g.deriv_featIndex[d];\n double v = g.deriv_featDeriv[d];\n // v is partial/(partial feature i) if M[u][v]\n ...\n}\n\\end{verbatim}\n\\caption{Accessing the proposed data structure for graph} \\label{alg:access}\n\\hrule\n\\end{table}\n\nA proposed data structure which would be very efficient is shown in\nTable~\\ref{alg:data} and \\ref{alg:access}. Some things that aren't\nclear now are:\n\\begin{itemize}\n\\item Where does the feature-index $\\leftrightarrow$ feature-name\n symbol table go? Probably not in the graph as shown here.\n\\item Is $\\M$ and $\\dM$ part of the graph or a different structure?\n\\end{itemize}\nAlso a clarification: \\texttt{M} and \\texttt{dM\\_lo,dM\\_hi}, are not\ndense 2-matrixes, of size quadratic in the number of nodes, $n$.\nInstead \\texttt{M} is a length-$n$ array of variable-size arrays, and\n\\texttt{M[u]}, for node index \\texttt{u}, is an array of size $m_u$,\nwhere $m_u$ is the number of neighbors (edges away from) \\texttt{u}.\nAnd \\texttt{dM\\_lo,dM\\_hi} are parallel structures.\n\nA final note: the squashing function $f$ and its derivative $f'$ are\nonly used in computing $\\df$ in Equation~\\ref{eqn:fprime}. One future\nextension to ProPPR might be to have facts with weights defined by\nparameters: e.g., facts like \\texttt{sim(a,b)} which when used would\nlead to an edge with a computed weight $s_{uv}$ based on an internal\nset of parameters $\\vec{\\lambda}$ (think of a learned similarity\nsubroutine). A modular way of dealing with this might be worth\nthinking through. Conceptually, $\\vec{\\lambda}$ is a part of\n$\\vek{w}$: one could have an interface which given an edge $u,v$ with\nweight $s_{uv}$ returns $f'(s_{uv})\\ddw s_{uv}$ in terms of components\nof $\\vec{\\lambda}$. For training you would also need to pass in an\nupdate of the $\\vec{\\lambda}$ features to the learning subroutine....\n\n\\clearpage\n\\begin{appendix}\n\\section{Linear Algebra proof of imperative procedure for p and d updates}\n\n\\subsection{Computing the p update}\n\\begin{align*}\n\\vek{p}^{t+1} =& \\alpha \\vek{s} + (1-\\alpha)\\vek{p}^t \\M\\\\\n\\vek{p}^t \\M =& \\left[ \\begin{array}{ccccc} \np_0 & p_1 & p_2 & \\ldots & p_n \n\\end{array} \\right] \\left[ \\begin{array}{cccc}\n\\\\M_{00} & \\M_{01} & \\M_{02} & \\ldots \\\\\n\\\\M_{10} & \\M_{11} & \\M_{12} & \\ldots \\\\\n\\M_{20} & \\M_{21} & \\M_{22} & \\ldots \\\\\n\\vdots & \\vdots & \\vdots\n\\end{array} \\right]\\\\\n=& \\left[ \\begin{array}{cccc} p^{t+1}_0 & p^{t+1}_1 & p^{t+1}_2 & \\ldots \\end{array} \\right]\\\\\np^{t+1}_0 =& p_0 \\M_{00} + p_1 \\M_{10} + p_2 \\M_{20} + \\ldots\\\\\np^{t+1}_1 =& p_0 \\M_{01} + p_1 \\M_{11} + p_2 \\M_{21} + \\ldots\\\\\n\\end{align*}\n\nIf we set $p^{t+1}_u$, we must iterate across, over all terms of the first element, then all terms of the second element, etc:\n\n$$p^{t+1}_u += (1-\\alpha) p_v \\M_{vu}$$\n\nGiven our data structure for $\\M$ however, constructing $\\M_{vu}$ is difficult. To use $\\M_{uv}$ instead, we can set $p^{t+1}_v$ and iterate down, computing the first term of all elements, then the second term of all elements, etc:\n\n$$p^{t+1}_{v} += (1-\\alpha) p_u \\M_{uv}$$\n\n\\subsection{Computing the d update}\n\nComputing $d$ proceeds in much the same way:\n\n\\begin{align*}\n\\vek{d}^{t+1}_i =& (1-\\alpha)(\\vek{p}^t \\dM_i + \\vek{d}^t_i \\M)\n\\end{align*}\n\nWe'll start by deconstructing the first term only:\n\n\\begin{align*}\n\\vek{p}^t \\dM_i =& \\left[ \\begin{array}{cccc} p_0 & p_1 & p_2 & \\ldots \\end{array} \\right] \\\n\\left[ \\begin{array}{cccc}\n\\dM_{00,i} & \\dM_{01,i} & \\dM_{02,i} & \\ldots \\\\\n\\dM_{10,i} & \\dM_{11,i} & \\dM_{12,i} & \\ldots \\\\\n\\dM_{20,i} & \\dM_{21,i} & \\dM_{22,i} & \\ldots \\\\\n\\vdots & \\vdots & \\vdots\n\\end{array} \\right]\\\\\n=& \\left[ \\begin{array}{cccc} d^{t+1}_{0,i} & d^{t+1}_{1,i} & d^{t+1}_{2,i} & \\ldots \\end{array} \\right]\\\\\nd^{t+1}_{0,i} =& p_0 \\dM_{00,i} + p_1 \\dM_{10,i} + p_2 \\dM_{20,i} + \\ldots\\\\\np^{t+1}_{1,i} =& p_0 \\dM_{01,i} + p_1 \\dM_{11,i} + p_2 \\dM_{21,i} + \\ldots\\\\\n\\end{align*}\n\nWe have the same choice of setting $d^{t+1}_{u,i}$ or $d^{t+1}_{v,i}$, with the same dependence on using $\\dM_{vu,i}$ or $\\dM_{uv,i}$. Since we must use $\\dM_{uv,i}$ due to the way our data structure is constructed, we must update $d^{t+1}_{v,i}$:\n\n$$ d^{t+1}_{v,i} += (1-\\alpha) \\vek{p}^t_u \\dM_{uv,i} $$\n\nThe same principles hold for the second term:\n\n$$ d^{t+1}_{v,i} += (1-\\alpha) \\vek{d}^t_{u,i} \\M_{uv} $$\n\n\\end{appendix}\n\n\\end{document}\n\n", "meta": {"hexsha": "380494368cca1c23610b95beca26c1ff200e9c17", "size": 20097, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/proppr-math/deriv.tex", "max_stars_repo_name": "TeamCohen/ProPPR", "max_stars_repo_head_hexsha": "2e0d978f96c2377ef3d26b80c17f38d2804f496c", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 138, "max_stars_repo_stars_event_min_datetime": "2015-01-15T00:04:37.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-26T18:35:48.000Z", "max_issues_repo_path": "doc/proppr-math/deriv.tex", "max_issues_repo_name": "TeamCohen/ProPPR", "max_issues_repo_head_hexsha": "2e0d978f96c2377ef3d26b80c17f38d2804f496c", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2015-09-01T16:58:27.000Z", "max_issues_repo_issues_event_max_datetime": "2018-06-11T21:51:11.000Z", "max_forks_repo_path": "doc/proppr-math/deriv.tex", "max_forks_repo_name": "TeamCohen/ProPPR", "max_forks_repo_head_hexsha": "2e0d978f96c2377ef3d26b80c17f38d2804f496c", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 52, "max_forks_repo_forks_event_min_datetime": "2015-02-04T16:03:45.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-22T06:38:25.000Z", "avg_line_length": 36.0161290323, "max_line_length": 269, "alphanum_fraction": 0.6472110265, "num_tokens": 7218, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042767, "lm_q2_score": 0.7718434873426303, "lm_q1q2_score": 0.6274063523446788}} {"text": "\\documentclass{article}\n\\usepackage{amsmath, amsfonts, graphicx}\n\\begin{document}\n\\section{Problem Statement}\nWe consider a special type of piecewise affine system, that the boundaries of the hybrid modes intersect at the origin. Mathematically, the system has this form\n\\begin{align}\n\t\\dot{x} = f(x) = A_ix \\text{ if } P_ix\\le 0\n\\end{align}\nNotice that the righ-hand side of $P_ix\\le0$ is zero vector, which guarantees that all mode boundaries pass through the origin. This also implies that $f(kx) = kf(x)$ if $k>0$.\n\nWe want to find a Lyapunov function satisfying\n\\begin{align}\n\tV(x) > 0 \\forall x\\neq 0\\\\\n\t\\dot{V}(x) < 0 \\forall x\\neq 0\\\\\n\tV(x) = 0\n\\end{align}\n\n\\textbf{Goal}: We want to prove that if there exists a piecewise affine Lyapunov function in the form \n\\begin{align}\n\tV(x) = g_i^Tx + h_i \\text{ if } C_ix\\le d_i \\label{eq:piecewise_affine_lyapunov}\n\\end{align}\nNamely in the $i'th$ piece $\\mathcal{P}_i=\\{x | C_ix \\le d_i\\}$, the Lyapunov function is an affine function of state $x$. Then there has to exist a piecewise linear Lyapunov function in the form\n\\begin{align}\n\t\\bar{V}(x) = \\bar{g}_i^Tx \\text{ if } \\bar{C}_i x \\le 0\n\\end{align}\nNamely the piecewise linear Lyapunov function $\\bar{V}$ is linear inside each piece, and each piece is a conic region originating from the origin.\n\n\\section{Proof}\nThe first step is to prove that the origin is at the vertex of some pieces $\\mathcal{P}_i$ in Lyapunov function $V(x)$. To prove this, suppose that the origin lies in the $i'th$ piece $\\mathcal{P}_i$. Since the Lyapunov function is linear within each piece, and $V(x)$ obtains its global minimal at the origin (implied by the strict positivity of Lyapunov function), we know that the origin must be a vertex of the polyhedron region $\\mathcal{P}_i$ (This is because the minimal of an LP must be obtained at the vertex of the polyhedron region). Hence the origin is at the vertices of many neighbouring pieces, as shown in the plot \\ref{fig:global_convergence1}, it cannot be in the strict interior of any polyhedron piece $\\mathcal{P}_i$.\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.4\\textwidth]{./global_convergence1.eps}\n\t\\caption{The origin (black dot) has to be the shared vertex of the neighbouring pieces ($\\mathcal{P}_1, \\mathcal{P}_2, ..., \\mathcal{P}_6$ in the figure), it cannot be in the strict interior of any piece.}\n\t\\label{fig:global_convergence1}\n\\end{figure}\nAs $V(0)=0$, we know that $h_i = 0$ if $\\mathcal{P}_i$ has the origin as a vertex.\n\nNext we show that we can construct a piecewise linear Lyapunov function $\\bar{V}$ using the piecewise affine Lyapunov function $V$. The procedure is as follows\n\\begin{enumerate}\n\t\\item Select each of the piece $\\mathcal{P}_i$ that neighbours the origin.\n\t\\item For each of the piece $\\mathcal{P}_i$ in step 1, compute the cone of $\\mathcal{P}_i = \\{x | C_ix\\le d_i\\}$, by keeping the boundaries of $\\mathcal{P}_i$ that contain the origin, and remove the boundaries that don't contain the origin. We denote this conic region as $\\bar{\\mathcal{P}}_i=\\{x|\\bar{C}_i\\le 0\\}$.\n\t\\item Within each of the conic region $\\bar{\\mathcal{P}}_i$, the new Lyapunov function is defined as $\\bar{V}(x) = g_i^Tx \\text{ if } \\bar{C}_i \\le 0$. \n\\end{enumerate}\nThis procedure is shown in Fig.\\ref{fig:global_convergence2}.\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.4\\textwidth]{./global_convergence2.eps}\n\t\\caption{We construct the new Lyapunov function, by first forming the conic region of each piece that neighbours the origin. Hence we extend the region $\\mathcal{P}_1, ...\\mathcal{P}_6$ to infinity, and remove the pieces that don't neighbour the origin ($\\mathcal{P}_7, ...\\mathcal{P}_{11}$).}\n\t\\label{fig:global_convergence2}\n\\end{figure}\nNow we need to prove that $\\bar{V}$ is a valid Lyapunov function. Notice that in the polyhedron piece $\\mathcal{P}_i$, the Lyapunov function $V(x)$ is $V(x) = g_i^Tx$. Now for any state $x$ in $\\mathcal{P}_i$, we shoot a ray starting from the origin and passing $x$. We denote the state on this ray as $kx, k > 0$. Obviously this entire ray is in the conic piece $\\bar{\\mathcal{P}}_i$ (by the definition of the cone). We also know that $\\bar{V}(kx) = k\\bar{V}(x) = kV(x)>0$. The first equality is because the function $\\bar{V}$ is linear inside each piece, the second equality is because $\\bar{V}(x) = V(x) \\text{ if } x\\in\\mathcal{P}_i$. Finally, we know that $\\dot{\\bar{V}}(kx) = k\\dot{\\bar{V}}(x) = k\\dot{V}(x) < 0$. The first equality is because inside each conic piece $\\partial V/\\partial x$ is a constant, and the dynamics scales proportionally w.r.t $k$ ($f(kx) = kf(x)$).The second equality is because $\\bar{V}(x)$ is the same as $V(x)$ in the polyhedron region $\\mathcal{P}_i$. As a result, we prove that this new function $\\bar{V}(x)$ is a valid Lyapunov function. This piecewise linear function $\\bar{V}$ can be represented by neural network without bias term.\n\\end{document}\n", "meta": {"hexsha": "069eee36febe9f5b4a6f4ed6ae011e2f264e927d", "size": 4914, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/global_convergence_proof.tex", "max_stars_repo_name": "hongkai-dai/neural-network-lyapunov-1", "max_stars_repo_head_hexsha": "8843c13f69f7f39cbb939ab250413e76f61843f6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 58, "max_stars_repo_stars_event_min_datetime": "2021-06-21T08:59:52.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T14:35:23.000Z", "max_issues_repo_path": "doc/global_convergence_proof.tex", "max_issues_repo_name": "StanfordASL/neural-network-lyapunov", "max_issues_repo_head_hexsha": "9e5db1c7f91b42df729026c9aa8575bc126f66b6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2021-08-22T05:31:23.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-29T03:47:07.000Z", "max_forks_repo_path": "doc/global_convergence_proof.tex", "max_forks_repo_name": "StanfordASL/neural-network-lyapunov", "max_forks_repo_head_hexsha": "9e5db1c7f91b42df729026c9aa8575bc126f66b6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2021-06-21T04:29:59.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-30T05:54:43.000Z", "avg_line_length": 92.7169811321, "max_line_length": 1172, "alphanum_fraction": 0.7277167277, "num_tokens": 1534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.8128673110375458, "lm_q1q2_score": 0.6274063486302625}} {"text": "\\chapter{Prototype set classifier}\r\n\\label{ch_classifier}\r\n%\r\nIn this chapter, we introduce the prototype set classifier, consider how to fit it to data, test the fitting procedure on benchmark cases, and compare model performance to other algorithms.\r\n%\r\n\\section{Definition}\r\n\\label{sec_classifier_definition}\r\n%\r\nWe observe $N\\gg1$ i.i.d.\\ pairs $(X_n,Y_n)$ of random variables such that $X_n$ takes values in $\\R^D$, $D\\geq1$, and $Y_n$ is one of $K>1$ classes represented by the integers from 0 to $K-1$.\r\nOur goal is to estimate the conditional distribution of $Y_n$ given $X_n$.\r\nAs proset models are discriminative instead of generative, the distribution of $X_n$ is of no concern and we just deal with the observed realizations $x_n\\in\\R^D$ in practice.\r\nNominal or ordinal features can be included in the model via encoding as real vectors.\\par\r\n%\r\nA proset classifier is built from $B\\geq0$ ``batches'' of points selected from the available samples.\r\nEach batch is defined by a a nonempty subset $S_1,\\dots,S_B\\subset\\{1,\\dots,N\\}$ of the observation indices.\r\nThe collection of batches is denoted by\r\n%\r\n\\begin{equation}\r\n\\mathcal{S}:=\\{S_b:b\\in\\{1,\\dots,B\\}\\}\\label{eq_batches}\r\n\\end{equation}\r\n%\r\nor $\\mathcal{S}=\\emptyset$ if $B=0$.\r\nThe indices of each batch are $S_b=:\\{s_{b,1},\\dots,s_{b,J_b}\\}$ where $J_b:=|S_b|$.\r\nWe refer to the samples $(x_{s_{b,j}},y_{s_{b,j}})$ as `prototypes'.\r\nThe model treats each as a representative example for the distribution of $Y_n$ when $X_n$ is in a neighborhood of $x_{s_{b,j}}$.\r\nTo make this notion more precise, we require additional notation:\\par\r\n%\r\nThe empirical marginal probabilities of $Y$ are\r\n%\r\n\\begin{equation}\r\n\\hat{p}_{0,k}:=\\frac{1}{N}\\sum_{n=1}^N\\1_{\\{k\\}}(y_k)\\label{eq_p0k}\r\n\\end{equation}\r\n%\r\nwhere $\\1_A$ is the indicator function of a set $A$.\\par\r\n%\r\nThe unnormalized Gaussian kernel $G_v$ with feature weights (inverse bandwidths) $v\\in\\R^D$, $v_d\\geq0$, is\r\n%\r\n\\begin{equation}\r\nG_v:\\R^D\\rightarrow(0,1],\r\nz\\mapsto G_v(z):=\\exp\\left(-\\frac{1}{2}\\sum_{d=1}^D(v_dx_d)^2\\right)\\label{eq_kernel}\r\n\\end{equation}\r\n%\r\nWe associate each batch $b$ with a vector $v_b\\in\\R^D$, $v_{b,d}\\geq0$, and each prototype with a weight $w_{b,j}>0$ to estimate the conditional distribution as\r\n%\r\n\\begin{equation}\r\nP(Y=k|X=x)\\approx\\hat{p}_k(x):=\r\n\\frac{\\hat{p}_{0,k}+\\sum_{b=1}^B\\sum_{j=1}^{J_b}\\1_{\\{k\\}}(y_{s_{b,j}})w_{b,j}G_{v_b}(x-x_{s_{b,j}})}\r\n{1+\\sum_{b=1}^B\\sum_{j=1}^{J_b}w_{b,j}G_{v_b}(x-x_{s_{b,j}})}\r\n\\label{eq_pkx}\r\n\\end{equation}\r\n%\r\n\\begin{remark}\r\n\\begin{enumerate}\r\n\\item The sets $S_b$ are not required to be disjoint, so the same point can appear multiple times in (\\ref{eq_pkx}).\r\nAs each batch is associated with its own $v_b$, the impact on the model may be different every time a sample appears.\r\n%\r\n\\item We use unnormalized kernels, i.e., the integral over the kernel function is in general not equal to 1, since the scaling can be considered to be subsumed in the choice of $w_{b,j}$.\r\nThis avoids any complications related to the fact that the scaling for a Gaussian kernel depends on the number of features with nonzero coefficients.\r\n%\r\n\\item The kernels are parameterized in terms of inverse bandwidth to enable feature selection via an $L_1$ penalty on $v_{b,d}$.\r\nIf a weight is forced to zero by the penalty, the values of the corresponding feature in points of batch $b$ cease to affect the model.\r\nConversely, a large value of $v_{b,d}$ means that the model is very sensitive to variations in the feature.\r\nKernels are limited to a diagonal bandwidth structure (product kernel) instead of a full semi-positive definite matrix to be able to fit on large feature spaces with reasonable effort.\r\n%\r\n\\item The conditional probability (\\ref{eq_pkx}) is computed as a locally weighted average similar to the Nadaraya-Watson estimator (\\ref{eq_nadaraya_watson}) or the conditional distributions studied in \\cite{Hall_04}.\r\nHowever, there are two important differences:\r\n%\r\n\\begin{itemize}\r\n\\item The model uses a subset of the training samples with individual weights instead of the entire data with unit weights.\r\nThis is done to reduce the computational effort for training and scoring if data size is large.\r\nAlso, studying the prototypes selected for the model may help to understand its structure.\r\n%\r\n\\item The model adds the marginal probabilities to the contribution of the prototypes\r\nThis sets the scale for the weights $w_{b,j}$, which would otherwise be determined only up to a multiplicative constant.\r\nAdditionally, it defines a natural baseline for $B=0$ , i.e., the model that treats $Y_n$ as independent of the features.\r\n\\end{itemize}\r\n\\end{enumerate}\r\n\\end{remark}\r\n%\r\nFinding an optimal representation of the form (\\ref{eq_pkx}) for anything but a very small sample appears intractable.\r\nWe thus focus on providing a heuristic that results in models of good quality.\r\nThe general idea is to iteratively add batches of prototypes to a base model consisting of the marginal probabilities $p_{0,k}$.\r\nIn each iteration, the available samples are split into a set of candidates for prototypes and a remainder used for scoring.\r\nThe weights for the candidates and the feature weights for the new batch are chosen to maximize a modified log-likelihood function for the scoring data.\r\nModifications to the likelihood are\r\n%\r\n\\begin{enumerate}\r\n\\item reweighting the terms such that each class has the same overall weight as in the set of all samples.\r\nThis enables us to use a sampling scheme for candidates that does not draw proportionally from each class.\r\n%\r\n\\item adding elastic net penalties for both prototype and feature weights to suppress candidates and features with negligible impact on the model.\r\n\\end{enumerate}\r\n%\r\nThe method is greedy in so far that parameters selected during earlier iterations remain untouched.\r\nHyperparameters -- penalty weights and the number of batches -- are selected via cross-validation.\\par\r\n%\r\nTo state the modified likelihood and its derivatives, we make use of the following expression representing the unnormalized class probabilities truncated at batch $c\\geq0$:\r\n%\r\n\\begin{equation}\r\n\\hat{q}_{c,k}(x)=\\hat{p}_{0,k}\r\n+\\sum_{b=1}^c\\sum_{j=1}^{J_b}\\1_{\\{k\\}}(y_{s_{b,j}})w_{b,j}G_{v_b}(x-x_{s_{b,j}})\r\n\\label{eq_qckx}\r\n\\end{equation}\r\n%\r\nThese satisfy\r\n%\r\n\\begin{align}\r\n\\hat{p}_{k}(x)&=\\frac{\\hat{q}_{B,k}(x)}{\\sum_{l=0}^{K-1}\\hat{q}_{B,l}(x)}\\label{eq_q_properties}\\\\\r\n\\hat{q}_{0,k}(x)&=\\hat{p}_{0,k}\\notag\\\\\r\n\\forall c>0:\\hat{q}_{c,k}(x)&\r\n=\\hat{q}_{c-1,k}(x)+\\sum_{j=1}^{J_c}\\1_{\\{k\\}}(y_{s_{c,j}})w_{c,j}G_{v_c}(x-x_{s_{c,j}})\r\n\\notag\r\n\\end{align}\r\n%\r\nDuring the training phase, we do not yet know the final number of batches $B$ but grow the model iteratively.\r\nThe conditional probabilities using only batches up to $c$ are given by\r\n%\r\n\\begin{equation}\r\n\\hat{p}_{c,k}(x)=\\frac{\\hat{q}_{c,k}(x)}{\\sum_{l=0}^{K-1}\\hat{q}_{c,l}(x)}\\label{eq_pckx}\r\n\\end{equation}\r\n%\r\nwhich also satisfy $\\hat{p}_{0,k}(x)=\\hat{p}_{0,k}$.\\par\r\n%\r\nGiven the first $c-1$ batches, the parameters for batch $c>0$ are chosen to minimize the following function, which is the negative log-likelihood with regularization as per (\\ref{eq_regularization}) and reweighting as discussed above:\r\n%\r\n\\begin{align}\r\nf\\left(v_c,\\{w_{c,j}\\}_j|\\{x_n\\}_n\\right)\r\n&=-\\frac{1}{N}\\sum_{k=0}^{K-1}\\frac{N_k}{N_k-J_{c,k}}\\sum_{n\\in T_c}\r\n\\1_{\\{k\\}}(y_n)\\log(\\hat{p}_{c,k}(x_n))\\label{eq_log_likelihood}\\\\\r\n&+\\lambda_v\\left(\\frac{\\alpha_v}{2}\\sum_{d=1}^Dv_{c,d}^2\r\n+(1-\\alpha_v)\\sum_{d=1}^D|v_{c,d}|\\right)\\notag\\\\\r\n&+\\lambda_w\\left(\\frac{\\alpha_w}{2}\\sum_{j=1}^{J_c}w_{c,j}^2\r\n+(1-\\alpha_w)\\sum_{j=1}^{J_c}|w_{c,j}|\\right)\\notag\r\n\\end{align}\r\n%\r\nThe following expressions still need to be defined:\r\n%\r\n\\begin{itemize}\r\n\\item $N_k:=\\left|\\{n\\in\\{1,\\dots,N\\}:y_n=k\\}\\right|$ is the number of all samples that have class $k$.\r\n%\r\n\\item $J_{c,k}:=\\left|\\{j\\in\\{1,\\dots,J_c\\}:y_{s_{c,j}}=k\\}\\right|$\r\nis the number of all candidates in batch $c>0$ that have class $k$.\r\n%\r\n\\item $T_c:=\\{1,\\dots,N\\}\\setminus S_c$ is the set of samples not included as candidates for prototypes in batch $c>0$.\r\n%\r\n\\item $\\lambda_v\\geq0$ is the weight for the elastic net penalty applied to feature weights.\r\n%\r\n\\item $\\lambda_w\\geq0$ is the weight for the elastic net penalty applied to prototype weights.\r\n%\r\n\\item $\\alpha_v\\in[0,1]$ is the portion of $\\lambda_v$ assigned to the $L_2$ penalty for feature weights.\r\n%\r\n\\item $\\alpha_w\\in[0,1]$ is the portion of $\\lambda_w$ assigned to the $L_2$ penalty for prototype weights.\r\n\\end{itemize}\r\n%\r\nGiven a set of prototypes $S_c$, the objective function (\\ref{eq_log_likelihood}) is maximized subject to $v_{c,d}\\geq0$ and $w_{c,j}\\geq0$.\r\nNote that we permit $w_{c,j}=0$ as solution of the optimization problem which contradicts our earlier definition.\r\nHowever, since assigning zero weight to a point is equivalent to excluding it from the model, this does not cause any issues.\\par\r\n%\r\nAs all parameters are constrained to the first orthant, the fact that the $L_1$ penalty is not differentiable in zero does not pose a problem.\r\nIn fact, we can simply replace the absolute value with the identity function and solve the optimization task using a standard solver for continuous optimization with bounds like L-BFGS-B \\cite{Byrd_95}.\\par\r\n%\r\nTo gain the full advantage of using L-BFGS-B, we need to compute the gradient of (\\ref{eq_log_likelihood}) analytically.\r\nThe partial derivatives of $\\log(\\hat{p}_{c,k}(x))$, $c>0$, are:\r\n%\r\n\\begin{align}\r\n\\frac{\\partial}{\\partial v_{c,d}}\\log(\\hat{p}_{c,k}(x))&\r\n=\\frac{\\frac{\\partial}{\\partial v_{c,d}}\r\n\\sum_{j=1}^{J_c}\\1_{\\{k\\}}(y_{s_{c,j}})w_{c,j}G_{v_c}(x-x_{s_{c,j}})}\r\n{\\hat{q}_{c,k}(x)}\r\n\\label{eq_p_partial_v}\\\\\r\n&-\\frac{\\frac{\\partial}{\\partial v_{c,d}}\r\n\\sum_{j=1}^{J_c}w_{c,j}G_{v_c}(x-x_{s_{c,j}})}\r\n{\\sum_{l=0}^{K-1}\\hat{q}_{c,l}(x)}\\notag\\\\\r\n&=v_{c,d}\\left(\\frac{\\sum_{j=1}^{J_c}w_{c,j}\r\n(x_d-x_{s_{c,j},d})^2G_{v_c}(x-x_{s_{c,j}})}\r\n{\\sum_{l=0}^{K-1}\\hat{q}_{c,l}(x)}\\right.\\notag\\\\\r\n&\\left.-\\frac{\\sum_{j=1}^{J_c}\\1_{\\{k\\}}(y_{s_{c,j}})w_{c,j}\r\n(x_d-x_{s_{c,j},d})^2G_{v_c}(x-x_{s_{c,j}})}\r\n{\\hat{q}_{c,k}(x)}\\right)\\notag\\\\\r\n\\frac{\\partial}{\\partial w_{c,i}}\\log(\\hat{p}_{c,k}(x))&\r\n=\\frac{\\frac{\\partial}{\\partial w_{c,i}}\r\n\\sum_{j=1}^{J_c}\\1_{\\{k\\}}(y_{s_{c,j}})w_{c,j}G_{v_c}(x-x_{s_{c,j}})}\r\n{\\hat{q}_{c,k}(x)}\r\n\\label{eq_p_partial_w}\\\\\r\n&-\\frac{\\frac{\\partial}{\\partial w_{c,i}}\r\n\\sum_{j=1}^{J_c}w_{c,j}G_{v_c}(x-x_{s_{c,j}})}\r\n{\\sum_{l=0}^{K-1}\\hat{q}_{c,l}(x)}\\notag\\\\\r\n&=\\frac{\\1_{\\{k\\}}(y_{s_{c,i}})G_{v_c}(x-x_{s_{c,i}})}\r\n{\\hat{q}_{c,k}(x)}\r\n-\\frac{G_{v_c}(x-x_{s_{c,i}})}\r\n{\\sum_{l=0}^{K-1}\\hat{q}_{c,l}(x)}\\notag\r\n\\end{align}\r\n%\r\nThe partial derivatives of the objective function on the first orthant are\r\n%\r\n\\begin{align}\r\n\\frac{\\partial}{\\partial v_{c,d}}f(v_c,\\{w_{c,j}\\}_j|\\{x_n\\}_n)\r\n&=-\\frac{1}{N}\\sum_{k=0}^{K-1}\\frac{N_k}{N_k-J_{c,k}}\\sum_{n\\in T_c}\r\n\\1_{\\{k\\}}(y_n)\\frac{\\partial}{\\partial v_{c,d}}\\log(\\hat{p}_{c,k}(x_n))\r\n\\label{eq_l_partial_v}\\\\\r\n&+\\lambda(\\alpha_v v_{c,d}+(1-\\alpha_v))\\notag\\\\\r\n\\frac{\\partial}{\\partial w_{c,j}}f(v_c,\\{w_{c,j}\\}_j|\\{x_n\\}_n)\r\n&=-\\frac{1}{N}\\sum_{k=0}^{K-1}\\frac{N_k}{N_k-J_{c,k}}\\sum_{n\\in T_c}\r\n\\1_{\\{k\\}}(y_n)\\frac{\\partial}{\\partial w_{c,j}}\\log(\\hat{p}_{c,k}(x_n))\r\n\\label{eq_l_partial_w}\\\\\r\n&+\\lambda\\beta(\\alpha_w w_{c,j}+(1-\\alpha_w))\\notag\r\n\\end{align}\r\n%\r\nIt remains to consider how to choose the prototypes for each batch and the starting points for optimization.\r\nFor the former, we score the model for iteration $c-1$ on all samples to get a probability distribution for each $Y_n$.\r\nA sample is considered correctly classified iff the probability assigned to its true class is greater than that for the other classes.\r\nThis enables us to split the samples into $2K$ bins, i.e., the correctly and incorrectly classified samples for each class.\r\nWe now draw prototypes from these bins as evenly as possible, subject to the constraint that no bin is depleted, i.e., some samples from each bin should remain for scoring.\\par\r\n%\r\nTo make the above notion more precise, let $M>0$ be the total number of prototypes we want to consider for the new batch and $\\eta\\in(0,1)$ the maximum fraction of samples we want to take from one bin.\r\nAlso, let $g_1,\\dots, g_{2K}\\geq0$ be the number of samples actually available in each bin, which satisfy w.l.o.g.\\ $g_1\\leq g_2\\leq\\dots\\leq g_{2K}$.\r\nTo arrive at a number of samples $h_1,\\dots,h_{2K}$ to draw from each bin, we use the following algorithm:\r\n%\r\n\\begin{samepage}\r\n\\begin{algorithm}[Number of prototypes per bin]~\r\n\\label{alg_bins}\r\n%\r\n\\begin{description}\r\n\\item{[1]} Assign $i\\leftarrow1$ and $R\\leftarrow M$.\r\n%\r\n\\item{[2]} Compute $r:=\\frac{R}{2K+1-i}$.\r\n%\r\n\\item{[3]} If $r\\leq\\eta g_i$: assign $h_i,\\dots,h_{2K}\\leftarrow r$ and go to [5].\r\n%\r\n\\item{[4]} Assign $h_i\\leftarrow\\eta g_i$, $i\\leftarrow i+1$, $R\\leftarrow R-h_i$, and go to [2].\r\n%\r\n\\item{[5]} Round each $h_i$ to the closest integer number.\r\n\\end{description}\r\n\\end{algorithm}\r\n\\end{samepage}\r\n%\r\nThus, we draw the maximum number of samples from the smallest bin that has not been processed until the remaining bins are large enough to draw an equal amount from each.\r\nIn case $\\sum_{i=1}^{2K}g_i0.5(1-\\eta)^{-1}\\label{eq_min_cases_2}\r\n\\end{equation}\r\n%\r\nwhere the brackets indicate rounding up.\r\n\\end{lemma}\r\n%\r\n\\paragraph{Proof:} we show first that for any $k$ holds\r\n%\r\n\\begin{equation}\r\n\\forall n\\in\\{0,\\dots,N_k\\}:\\eta n\\geq0.5\\vee\\eta(N_k-n)\\geq0.5\\iff\\lceil0.5N_k\\rceil\\geq0.5\\eta^{-1}\\label{eq_condition_proof}\r\n\\end{equation}\r\n%\r\nThe left-hand side implies the right-hand if we choose $n=\\lceil0.5N_k\\rceil$.\\par\r\n%\r\nThe right-hand side implies the left-hand side since for any $n$, either $n$ or $N_k-n$ has to be greater or equal to $\\lceil0.5N_k\\rceil$.\\par\r\n%\r\nFor the remaining clauses, note that $\\eta n0.5$ so that we can use a similar argument to the one above.$\\quad\\Box$\\par\r\n%\r\nA check for condition (\\ref{eq_min_cases_2}) is implemented in the software to ensure that samples of each class are available both as prototypes and for scoring.\r\nHowever, it does not guarantee that the algorithm is able to find meaningful structure in the data.\r\nFor example, the recommended default $\\eta=0.5$ requires only that $N_k\\geq3$ for all $k$, which is inadequate for supervised learning.\\par\r\n%\r\nRegarding the starting values for optimization, we initialize all $v_{c,d}$ to $10D^{-1}$ and all $w_{c,j}$ to 1.\r\nThe former choice assumes that the features have a scale on the order of magnitude of 1, e.g., they have been scaled to unit variance.\r\nThus, setting the inverse bandwidth for a single feature to 10 means the kernels for the starting solution describe a structure that is more granular than the whole distribution.\r\nDividing the weight by the total number of features ensures that the sum of squares in the exponent of (\\ref{eq_kernel}) is always of a similar magnitude and the exponential function does not vanish.\r\nA prototype weight of 1 is of the same order of magnitude as the marginal probabilities.\r\n%\r\n\\begin{remark}\r\nThe optimization problem does in general have multiple stationary points, including the trivial solution where all weights are equal to zero.\r\nProvided the features are scaled, the proposed starting value for the feature weights appears to work well in practice.\\par\r\n%\r\nScaling the features is also recommended for a different reason.\r\nThe size of the elastic net penalty can only be interpreted relative to the scale of the partial derivatives.\r\nThus, to penalize all features equally, they need to be equally scaled.\r\n\\end{remark}\r\n%\r\nDespite regularization, there is one situation in which the selection of prototypes is not parsimonious.\r\nIf multiple candidate points have the same target and feature values for all features with positive weights, the $L_2$-penalty causes the algorithm to assign equal weight to all candidates.\r\nIf this weight is positive, the model retains multiple copies of what is effectively the same prototype (the original samples can differ in the values of excluded features).\r\nTo simplify the model representation, we add a clean-up stage to the algorithm:\\par\r\n%\r\nEach prototype in a batch is represented by the combined vector of the active features and target.\r\nTwo prototypes are considered equivalent if the maximum norm of the difference of their associated vectors does not exceed some small tolerance.\r\nThis notion is generalized from pairs to larger groups by finding all connected components in the graph where each node represents a prototype and an edge indicates pairwise equivalence.\r\nWe then replace all prototypes belonging to the same component by a single prototype using the combined weight, as well as the features of the sample appearing first in the training data.\r\n%\r\n\\section{Fit strategy}\r\n\\label{sec_classifier_fit}\r\n%\r\nFitting a proset classifier is controlled by seven hyperparameters:\r\n%\r\n\\begin{itemize}\r\n\\item The number of batches $B$.\r\n%\r\n\\item The number $M$ of candidates for prototypes evaluated per batch.\r\n%\r\n\\item The maximum fraction $\\eta$ of candidates drawn from one bin using Algorithm \\ref{alg_bins}.\r\n%\r\n\\item The penalty term $\\lambda_v$ for the feature weights.\r\n%\r\n\\item The penalty term $\\lambda_w$ for the prototype weights.\r\n%\r\n\\item The ratio $\\alpha_v$ of $\\lambda_v$ assigned as $L_2$ penalty term.\r\n%\r\n\\item The ratio $\\alpha_w$ of $\\lambda_w$ assigned as $L_2$ penalty term.\r\n\\end{itemize}\r\n%\r\nOne goal of the case studies in this section is to identify good default values and indicate which parameters are worthwhile to tune.\r\nSome of the key findings are summarized here:\r\n%\r\n\\begin{enumerate}\r\n\\item Fitting a few large batches is preferable to many small ones.\r\nThe first batch has the largest impact on the model score and a large first batch results in a better overall score.\r\nWe believe the reason for this is that model quality depends on finding good constellations of prototypes, not just individual prototypes.\r\nSuitable constellations are more likely to occur in larger batches.\\par\r\n%\r\nAdding too many batches to a model leads to saturation, not overfitting.\r\nThe number of prototypes chosen per batch decreases, possibly down to zero.\r\nThe corresponding model scores fluctuate slightly around a common level.\r\nThus, using a small number of batches is mostly a question of reducing computational effort.\\par\r\n%\r\nWe recommend $M=1,000$ candidates per batch and a single batch ($B=1$) as default.\r\nIf optimizing with respect to the number of batches, $B=10$ is a reasonable upper limit.\r\n%\r\n\\item If the number of samples in a bin is too small for drawing the desired number of candidates, a compromise is to use half of the samples as candidates and the others as reference points for computing the likelihood.\r\nThus, we fix $\\eta=0.5$.\r\n%\r\n\\item The penalty $\\lambda_v$ for the feature weights controls the smoothness of the model.\r\nIt is the main safeguard against overfitting.\\par\r\n%\r\nIn the case study for the classifier, good values lie in the range from $10^{-6}$ to $10^{-1}$.\r\nChoosing $10^{-3}$ as default works for all cases.\r\nHowever, if an experimenter wants to tune only one parameter, it should be $\\lambda_v$.\r\n%\r\n\\item The penalty $\\lambda_w$ for the prototype weights controls how much weight can be assigned to a single prototype.\r\nWe observe that increasing $\\lambda_w$ within a certain range can actually lead to more candidate points being included in the model with nonzero weight.\r\nFurther increases gradually lead to underfitting.\r\nHowever, reducing $\\lambda_w$ does not lead to appreciable overfitting.\r\nIt appears that values below a certain threshold mostly control the preference for few prototypes with large weights versus more prototypes with smaller weights.\r\nAbove the threshold, the estimated distribution is shrunk towards the marginal probabilities.\r\nThis is desirable to some degree for avoiding overfitting.\\par\r\n%\r\nIn the case study for the classifier, good values lie in the range from $10^{-9}$ to $10^{-4}$.\r\nChoosing $10^{-8}$ as default works for all cases.\r\n%\r\n\\item A dominant $L_2$-penalty ($\\alpha_v$ and $\\alpha_w$ close to 1.0) yields slightly better results in the case study than either a dominant $L_1$- or balanced penalty.\r\nThe default values recommended for the algorithm are $\\alpha_v=\\alpha_w=0.95$.\r\n\\end{enumerate}\r\n%\r\nThe case study uses the following procedure for hyperparameter search, subject to small variations described later:\r\n%\r\n\\begin{algorithm}[Hyperparameter selection]~\r\n\\label{alg_hyperparameters}\r\n%\r\n\\begin{enumerate}\r\n\\item Choose $M$, $\\eta$, $\\alpha_v$, and $\\alpha_w$ based on the above recommendations.\r\nChoose a range for $\\lambda_v$, a range for $\\lambda_w$, and a set of candidates for $B$ (e.g., the numbers from 0 to 10).\r\n%\r\n\\item Split the data into a training set (70 \\%) and test set (30 \\%), stratified by class.\r\n%\r\n\\item\\textbf{Stage 1}\r\n%\r\n\\begin{enumerate}\r\n\\item Randomly generate 50 pairs $(\\lambda_v,\\lambda_w)$.\r\nEach $\\lambda$ is sampled uniformly on the log-scale from the chosen range.\r\n%\r\n\\item Perform five-fold cross-validation on the training set using $B=1$.\r\n%\r\n\\item Compute the mean and standard deviation of log-loss for each pair of penalties over the left-out folds.\r\n%\r\n\\item Determine a threshold for model quality by taking the minimal mean log-loss and adding the corresponding standard deviation.\r\n%\r\n\\item Among all pairs whose mean log-loss is less than or equal to the threshold, choose the one maximizing the geometric mean $\\sqrt{\\lambda_v\\lambda_w}$.\r\n\\end{enumerate}\r\n%\r\n\\item\\textbf{Stage 2}\r\n\\begin{enumerate}\r\n\\item Perform five-fold cross-validation on the training set using the parameters from stage 1 and the maximal number of batches.\r\n%\r\n\\item For each value of $B$ up to the maximum, evaluate the models on the left-out fold and compute mean and standard deviation of log-loss.\r\n%\r\n\\item Determine a threshold for model quality by taking the minimal mean log-loss and adding the corresponding standard deviation.\r\n%\r\n\\item Among all candidates whose mean log-loss is less than or equal to the threshold, choose the smallest $B$.\r\n\\end{enumerate}\r\n%\r\n\\item Refit the model with parameters selected in stages 1 and 2 on all training data.\r\n%\r\n\\item Score the final model on the test data.\r\n\\end{enumerate}\r\n\\end{algorithm}\r\n%\r\nThe purpose of stage 1 is to identify good values for the most important parameters $\\lambda_v$ and $\\lambda_w$.\r\nBased on the observation that the first batch has the most impact, we fix $B=1$ during this stage to save computation time.\r\nControlling $B$ can be treated as a secondary concern, since it appears to be impossible to overfit by increasing $B$.\r\nIt is still necessary to test larger values as more complex problems may be underfitted with $B=1$.\\par\r\n%\r\nNote that stage 2 only fits five models up to the maximum number of batches as these can also be evaluated for smaller choices of $B$.\r\nThis introduces a dependency between the means and standard deviations as estimates reuse the same initial batches.\r\nHowever, as keeping $B$ small is a secondary concern, we consider this time-saving approach acceptable.\\par\r\n%\r\nThe parameters selected in each stage are not necessarily those that minimize mean cross-validation log-loss.\r\nInstead, we consider all sets of parameters that are `equivalent' to the optimizer in the sense that their mean log-loss is within one standard error of the optimum.\r\nFrom these, we pick the parameters that yield the sparsest model.\r\nThis `1 SE rule' is a recommendation from R package \\texttt{glmnet} \\cite{Friedman_10}.\r\nThe authors observe that a model using the optimal parameters tends to overfit the cross-validation sample, which is mitigated by the rule.\r\nThe multiplier of 1 is of course arbitrary but a common `rule of thumb' in statistics.\r\nIn case multiple parameters control sparseness, we need to decide which set we consider `sparsest'.\r\nFor stage 1, maximizing the geometric mean gives equal importance to both penalties.\\par\r\n%\r\nAs discussed in the introduction, we use the threshold obtained via the `1 SE rule' also to compare models derived via different fit strategies or classification algorithms.\r\nFor the classifier that performs best on testing data, the threshold found during stage 2 is used as upper bound on log-loss to determine which of the other classifiers are still considered `equivalent'.\r\n%\r\n\\section{Benchmarks for hyperparameter selection}\r\n\\label{sec_classifier_benchmarks}\r\n%\r\nIn this section, we test variations of the fit strategy outlined above on different benchmark cases.\r\nWe use four small data sets that come as `toy' examples with Python package \\texttt{sklearn} \\cite{Pedregosa_11}, plus two slightly larger artificial data sets:\r\n%\r\n\\begin{enumerate}\r\n\\item\\textbf{Iris 2f:} this data set consists of the first two features of Fisher's famous iris data set \\cite{Fisher_36}.\r\nWe limit the analysis to two of four features (sepal length and width) as this allows us to visualize the decision surface of the classifier as a 2d plot.\r\nThe data set comprises 150 samples for three species of iris flower, with 50 samples per class.\r\nOne class is linearly separable from the others, but measurements for the remaining two overlap.\r\n%\r\n\\item\\textbf{Wine:} this data set from the UCI Machine Learning Repository \\cite{Dua_19} consists of chemical analysis data for wines from three different cultivators.\r\nIt consists of 178 samples with between 48 and 71 samples per class.\r\nThe data is known to be separable \\cite{Aeberhard_92}.\r\n%\r\n\\item\\textbf{Cancer:} this data set from the UCI Machine Learning Repository \\cite{Dua_19} consists of medical analysis data from breast tissue samples.\r\nThe 569 samples are classified as either malignant (212 samples) or benign (357 samples).\r\n%\r\n\\item\\textbf{Digits:} this data set from the UCI Machine Learning Repository \\cite{Dua_19} consists of monochrome images of handwritten digits downsampled to an eight-by-eight grid.\r\nThe 1,797 samples are approximately balanced among the 10 digits.\r\n%\r\n\\item\\textbf{Checker:} for this artificial data set, we sample two features uniformly on the unit square and assign class labels deterministically to create an eight-by-eight checkerboard.\r\nThe pattern defeats methods that rely on global properties of the data like correlation, e.g., logistic regression.\r\nIt can be recovered successfully by methods that model local structure, e.g., a k-nearest neighbor classifier or decision tree.\r\nThe total number of samples is 6,400, so each square of the pattern contains approximately 100 data points.\r\n%\r\n\\item\\textbf{XOR 6f:} for this artificial data set, we sample six features independently and uniformly on the interval $[-1.0, 1.0]$.\r\nThe class label is assigned deterministically based on the sign of the product of features: a positive (or zero) sign is class 1, a negative sign is class 0.\r\nThis is similar to the `continuous XOR' problem found in the \\texttt{mlbench} library for R \\cite{Leisch_21}, except that we only distinguish two classes.\r\nDespite being deterministic, this problem appears to be a hard even for classifiers that model local structure.\r\nThe total number of samples is 6,400, so each orthant of the feature space contains approximately 100 data points.\r\n\\end{enumerate}\r\n%\r\nThe first three experiments consider the impact of $\\alpha_v$ and $\\alpha_w$ on model behavior.\r\nProset classifiers are fitted to all six data sets with fixed $M=1,000$ and $\\eta=0.5$.\r\nPenalty weights are sampled from the ranges $\\lambda_v\\in(10^{-6},10^{-1})$ and $\\lambda_w\\in(10^{-9},10^{-4})$, while the number of batches is allowed to vary between 0 and 10.\r\nThe first experiment uses a dominant $L_1$-penalty ($\\alpha_v=\\alpha_w=0.05$), the second uses balanced penalties ($\\alpha_v=\\alpha_w=0.50$), and the third a dominant $L_2$-penalty ($\\alpha_v=\\alpha_w=0.95$).\\par\r\n%\r\n\\begin{figure}\r\n\\caption{[E1] Parameter search for Iris 2F data}\r\n\\label{fig_parameter_search}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{Stage 1}]{\\includegraphics[height=0.4\\textheight]{figures/iris_2f_dominant_l1_2d_search.pdf}}\\\\\r\n\\subfloat[\\textbf{Both stages}]{\\includegraphics[width=0.99\\textwidth]{figures/iris_2f_dominant_l1_1d_search.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nFigure \\ref{fig_parameter_search} shows parameter search results for the first experiment using iris data.\r\nIn the lower plot, the log-loss for stage 1 as a function of either $\\lambda_v$ or $\\lambda_w$ appears highly variable.\r\nHowever, this is mostly due to changes in the other parameter, as evidenced by the surface plot.\r\nComplete results for all data sets are presented in tables \\ref{tab_e1}, \\ref{tab_e2}, and \\ref{tab_e3}.\r\nA comparison of the first six experiments is found in table \\ref{tab_e1_to_e6}.\r\nThe description of individual experiments comprises the following information:\r\n%\r\n\\begin{description}\r\n\\item[Data:] the number of classes, as well as the size of the whole data set and train-test split.\r\n%\r\n\\item[Candidates:] the approximate number of candidates for prototypes used to build the final model.\r\nWhile $M=1,000$ candidates are specified for each model, the effective maximum for small data sets is around 35 \\% of the available samples: training data is 70 \\% of all samples and $\\eta=0.5$ allows at most 50 \\% of data in one bin to be used as prototypes.\r\n%\r\n\\item[Stage 1:] results for selecting $\\lambda_v$ and $\\lambda_w$ using a single batch.\r\nLists the optimal and chosen parameters according to the `1 SE rule' (see algorithm \\ref{alg_hyperparameters}), together with the achieved mean log-loss from cross-validation.\r\nThe given threshold is the one for the `1 SE rule', i.e., the sum of the minimal mean log-loss and corresponding standard deviation.\r\n%\r\n\\item[Stage 2:] results for selecting $B$ using the penalty weights chosen in stage 1.\r\n%\r\n\\item[Final model:] information about the final model fitted on all training data with parameters determined in stages 1 and 2.\r\nThe number of features and prototypes, as well as the scores achieved for the test data.\r\nApart from log-loss, the measures used for evaluation are ROC-AUC and balanced accuracy.\\par\r\n%\r\nFor multi-class problems, the stated ROC-AUC values is the unweighted average for all pairwise comparisons of two classes.\r\nThis generalization to more than two classes is recommended in \\cite{Hand_01} as being robust to class imbalance.\\par\r\n%\r\nTo compute balanced accuracy, we use the `naive' rule that assigns each sample the class label with the highest estimated probability.\r\nThe reported score is the unweighted average of the correct classification rates for each class.\r\n\\end{description}\r\n%\r\n\\begin{table}\r\n\\caption{[E1] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.05$)}\r\n\\label{tab_e1}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$1.1\\times10^{-2}$&$6.9\\times10^{-4}$&$5.5\\times10^{-3}$&$1.2\\times10^{-5}$&$4.5\\times10^{-3}$&$1.0\\times10^{-2}$\\\\\r\nSelected $\\lambda_v$&$1.1\\times10^{-2}$&$3.5\\times10^{-3}$&$1.1\\times10^{-2}$&$1.8\\times10^{-3}$&$5.5\\times10^{-3}$&$1.1\\times10^{-2}$\\\\\r\nOptimal $\\lambda_w$&$1.0\\times10^{-5}$&$1.1\\times10^{-6}$&$8.3\\times10^{-8}$&$3.4\\times10^{-9}$&$1.3\\times10^{-8}$&$4.6\\times10^{-7}$\\\\\r\nSelected $\\lambda_w$&$1.8\\times10^{-5}$&$3.1\\times10^{-5}$&$1.8\\times10^{-5}$&$7.7\\times10^{-8}$&$8.3\\times10^{-8}$&$1.0\\times10^{-5}$\\\\\r\nOptimal log-loss&0.47&0.12&0.09&0.16&0.17&0.53\\\\\r\nThreshold&0.60&0.17&0.14&0.19&0.18&0.55\\\\\r\nSelected log-loss&0.54&0.12&0.09&0.19&0.18&0.55\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&2&5&1&9&1&1\\\\\r\nSelected batches&2&2&1&1&1&1\\\\\r\nOptimal log-loss&0.46&0.12&0.10&0.18&0.19&0.55\\\\\r\nThreshold&0.54&0.16&0.14&0.22&0.20&0.56\\\\\r\nSelected log-loss&0.46&0.13&0.10&0.19&0.19&0.55\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&6&3&19&2&6\\\\\r\nPrototypes&34&43&24&546&257&309\\\\\r\nLog-loss&0.62&0.14&0.13&0.15&0.17&0.55\\\\\r\nROC-AUC&0.86&0.99&0.99&1.00&0.99&0.81\\\\\r\nBalanced acc.&0.76&0.97&0.94&0.97&0.95&0.71\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E2] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.50$)}\r\n\\label{tab_e2}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$5.6\\times10^{-3}$&$1.0\\times10^{-2}$&$5.5\\times10^{-3}$&$1.2\\times10^{-5}$&$9.5\\times10^{-4}$&$5.5\\times10^{-3}$\\\\\r\nSelected $\\lambda_v$&$1.1\\times10^{-2}$&$3.5\\times10^{-3}$&$1.1\\times10^{-2}$&$4.5\\times10^{-3}$&$9.5\\times10^{-4}$&$1.0\\times10^{-2}$\\\\\r\nOptimal $\\lambda_w$&$1.7\\times10^{-5}$&$4.6\\times10^{-7}$&$8.3\\times10^{-8}$&$3.4\\times10^{-9}$&$5.6\\times10^{-9}$&$8.3\\times10^{-8}$\\\\\r\nSelected $\\lambda_w$&$1.0\\times10^{-5}$&$3.1\\times10^{-5}$&$1.8\\times10^{-5}$&$1.3\\times10^{-8}$&$5.6\\times10^{-9}$&$4.6\\times10^{-7}$\\\\\r\nOptimal log-loss&0.45&0.10&0.09&0.17&0.17&0.53\\\\\r\nThreshold&0.51&0.17&0.14&0.20&0.17&0.54\\\\\r\nSelected log-loss&0.46&0.13&0.09&0.19&0.17&0.54\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&4&10&1&3&1&1\\\\\r\nSelected batches&3&2&1&1&1&1\\\\\r\nOptimal log-loss&0.46&0.13&0.10&0.16&0.18&0.54\\\\\r\nThreshold&0.49&0.16&0.14&0.19&0.19&0.55\\\\\r\nSelected log-loss&0.47&0.14&0.10&0.18&0.18&0.54\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&7&3&18&2&6\\\\\r\nPrototypes&30&69&54&533&281&360\\\\\r\nLog-loss&0.49&0.20&0.13&0.18&0.16&0.53\\\\\r\nROC-AUC&0.90&0.98&0.99&1.00&0.99&0.82\\\\\r\nBalanced acc.&0.73&0.93&0.95&0.97&0.95&0.72\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E3] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.95$)}\r\n\\label{tab_e3}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$5.6\\times10^{-3}$&$3.8\\times10^{-5}$&$5.5\\times10^{-3}$&$7.8\\times10^{-6}$&$3.1\\times10^{-5}$&$5.5\\times10^{-3}$\\\\\r\nSelected $\\lambda_v$&$1.1\\times10^{-2}$&$1.1\\times10^{-2}$&$1.1\\times10^{-2}$&$9.5\\times10^{-4}$&$2.9\\times10^{-4}$&$5.5\\times10^{-3}$\\\\\r\nOptimal $\\lambda_w$&$1.7\\times10^{-5}$&$6.9\\times10^{-9}$&$8.3\\times10^{-8}$&$1.6\\times10^{-9}$&$4.3\\times10^{-9}$&$8.3\\times10^{-8}$\\\\\r\nSelected $\\lambda_w$&$1.0\\times10^{-5}$&$1.8\\times10^{-5}$&$1.8\\times10^{-5}$&$5.6\\times10^{-9}$&$6.9\\times10^{-8}$&$8.3\\times10^{-8}$\\\\\r\nOptimal log-loss&0.44&0.11&0.09&0.17&0.18&0.53\\\\\r\nThreshold&0.49&0.18&0.14&0.19&0.20&0.54\\\\\r\nSelected log-loss&0.46&0.17&0.11&0.18&0.19&0.53\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&5&5&1&8&1&10\\\\\r\nSelected batches&3&2&1&2&1&1\\\\\r\nOptimal log-loss&0.51&0.14&0.10&0.14&0.19&0.51\\\\\r\nThreshold&0.57&0.17&0.14&0.17&0.20&0.53\\\\\r\nSelected log-loss&0.56&0.15&0.10&0.16&0.19&0.53\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&7&4&30&2&6\\\\\r\nPrototypes&49&52&56&861&328&413\\\\\r\nLog-loss&0.43&0.14&0.13&0.15&0.18&0.52\\\\\r\nROC-AUC&0.91&1.00&0.99&1.00&0.99&0.82\\\\\r\nBalanced acc.&0.71&0.98&0.98&0.97&0.95&0.72\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\clearpage\r\n%\r\nIn the experiments, a dominant $L_1$-penalty yields slightly worse results than balanced penalties or a dominant $L_2$-penalty.\r\nAlso, we have occasionally observed (not in the results reported here) that a dominant $L_1$-penalty causes too few features to be selected for the final model.\r\nThis happens even though models created during cross-validation do not underfit.\r\nThe likely reason is that random candidate selection for small sample size can result in a set of candidates hat does not fully represent the data.\r\nSince a dominant $L_2$-penalty slightly outperforms the balanced case in the trials, we recommend $\\alpha_v=\\alpha_w=0.95$ as default values.\\par\r\n%\r\n\\begin{table}\r\n\\caption{[E4] Use optimal hyperparameters ($\\alpha_v=\\alpha_w=0.95$)}\r\n\\label{tab_e4}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$5.6\\times10^{-3}$&$3.8\\times10^{-5}$&$5.5\\times10^{-3}$&$7.8\\times10^{-6}$&$3.1\\times10^{-5}$&$5.5\\times10^{-3}$\\\\\r\nOptimal $\\lambda_w$&$1.7\\times10^{-5}$&$6.9\\times10^{-9}$&$8.3\\times10^{-8}$&$1.6\\times10^{-9}$&$4.3\\times10^{-9}$&$8.3\\times10^{-8}$\\\\\r\nOptimal log-loss&0.44&0.11&0.09&0.17&0.18&0.53\\\\\r\nThreshold&0.49&0.18&0.14&0.19&0.20&0.54\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&5&5&1&8&1&10\\\\\r\nOptimal log-loss&0.51&0.14&0.10&0.14&0.19&0.51\\\\\r\nThreshold&0.57&0.17&0.14&0.17&0.20&0.53\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&11&3&41&2&6\\\\\r\nPrototypes&65&204&93&3,053&595&403\\\\\r\nLog-loss&0.48&0.05&0.11&0.13&0.16&0.52\\\\\r\nROC-AUC&0.87&1.00&0.99&1.00&0.99&0.83\\\\\r\nBalanced acc.&0.69&0.98&0.95&0.97&0.94&0.74\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nThe fourth experiment shows the effect of using the optimal parameters from cross-validation instead of the equivalent sparser solution (see table \\ref{tab_e4}).\r\nIn half of the cases, results actually outperform those of the first three experiments.\r\nHowever, the log-loss scores for [E3] are still below the equivalence thresholds for [E4] in these cases.\r\nThis means the metrics for the benchmark cases do not indicate that the `1 SE rule' is necessary to prevent overfitting, but we can use it to obtain a sparser parameterization with negligible loss in quality.\r\n%\r\n\\clearpage\r\n%\r\n\\begin{figure}\r\n\\caption{Proset decision surfaces for case Iris 2f}\r\n\\label{fig_proset_decision_iris_2f}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{[E1] $\\lambda_v=\\lambda_w=0.05$, `1 SE rule'}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_dominant_l1_surf_test.pdf}}\r\n\\subfloat[\\textbf{[E2] $\\lambda_v=\\lambda_w=0.50$, `1 SE rule'}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_balanced_surf_test.pdf}}\\\\\r\n\\subfloat[\\textbf{[E3] $\\lambda_v=\\lambda_w=0.95$, `1 SE rule'}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_dominant_l2_surf_test.pdf}}\r\n\\subfloat[\\textbf{[E4] $\\lambda_v=\\lambda_w=0.95$, optimal hyperparameters}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_dominant_l2_surf_opt_test.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nIf we study the decision surfaces for the Iris 2f benchmark, it does look like [E4] has a slight tendency to overfit.\r\nFigure \\ref{fig_proset_decision_iris_2f} shows the surface plots for experiments [E1] to [E4].\r\nThe colors indicates the class with the highest estimated probability in that region of the feature space.\r\nOnly the result for [E3] appears fully convincing to us.\r\nIt separates the feature space near the training data into three contiguous regions with smooth boundaries.\r\nThe surfaces for [E1] and [E4] appear overly complex and somewhat arbitrary, which we take as a visual indicator of overfitting.\r\n[E2] oddly prefers class `virginica' in the lower left, although there is no training data in this region.\r\nThis is due to the layering of kernels with different bandwidths in a model with multiple batches.\\par\r\n%\r\n\\clearpage\r\n%\r\n\\begin{table}\r\n\\caption{[E5] Grid search for $\\lambda_v$ ($\\lambda_w=10^{-8}$ and $\\alpha_v=\\alpha_w=0.95$)}\r\n\\label{tab_e5}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$1.0\\times10^{-2}$&$3.2\\times10^{-6}$&$1.0\\times10^{-2}$&$1.0\\times10^{-4}$&$3.2\\times10^{-4}$&$1.0\\times10^{-2}$\\\\\r\nSelected $\\lambda_v$&$1.0\\times10^{-2}$&$3.2\\times10^{-2}$&$3.2\\times10^{-2}$&$3.2\\times10^{-2}$&$3.2\\times10^{-4}$&$1.0\\times10^{-2}$\\\\\r\nOptimal log-loss&0.51&0.11&0.08&0.20&0.17&0.52\\\\\r\nThreshold&0.63&0.16&0.12&0.22&0.18&0.53\\\\\r\nSelected log-loss&0.51&0.14&0.10&0.21&0.17&0.52\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&10&1&1&2&1&1\\\\\r\nSelected batches&1&1&1&1&1&1\\\\\r\nOptimal log-loss&0.60&0.11&0.10&0.21&0.17&0.52\\\\\r\nThreshold&0.71&0.21&0.15&0.26&0.19&0.53\\\\\r\nSelected log-loss&0.67&0.11&0.10&0.21&0.17&0.52\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&4&3&18&2&6\\\\\r\nPrototypes&25&25&69&410&409&421\\\\\r\nLog-loss&0.47&0.12&0.13&0.18&0.17&0.53\\\\\r\nROC-AUC&0.90&1.00&0.99&1.00&0.99&0.81\\\\\r\nBalanced acc.&0.73&0.97&0.96&0.97&0.95&0.71\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nThe fifth experiment shows what happens if we fix $\\lambda_w$ and perform cross-validation only with respect to $\\lambda_v$ and $B$.\r\nBased on the previous experiments, $\\lambda_w=10^{-8}$ appears to be a suitable choice.\r\nAs stage 1 now deals with a single parameter, we replace the random search with a grid search using 11 points equidistantly spaced on the log-scale.\r\nThe results in table \\ref{tab_e5} show that fixing the penalty on prototype weights has only a small impact on model quality.\r\nAll log-loss scores except that for the digits case are below the equivalence threshold for the best model found.\r\nTherefore, $\\lambda_w=10^{-8}$ is the recommended default for the proset classifier.\\par\r\n%\r\n\\clearpage\r\n%\r\n\\begin{table}\r\n\\caption{[E6] Stage 2 only ($\\lambda_v=10^{-3}$, $\\lambda_w=10^{-8}$, and $\\alpha_v=\\alpha_w=0.95$)}\r\n\\label{tab_e6}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\textbf{Candidates}&$\\sim50$&$\\sim60$&$\\sim200$&$\\sim630$&1,000&1,000\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&9&8&9&7&1&10\\\\\r\nSelected batches&4&2&1&4&1&3\\\\\r\nOptimal log-loss&0.54&0.10&0.10&0.15&0.19&0.50\\\\\r\nThreshold&0.72&0.17&0.15&0.17&0.20&0.52\\\\\r\nSelected log-loss&0.71&0.15&0.12&0.16&0.19&0.51\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&7&4&34&2&6\\\\\r\nPrototypes&82&72&149&1,270&252&1,261\\\\\r\nLog-loss&0.45&0.16&0.13&0.12&0.18&0.49\\\\\r\nROC-AUC&0.89&0.99&0.99&1.00&0.99&0.84\\\\\r\nBalanced acc.&0.76&0.95&0.95&0.98&0.95&0.75\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nThe sixth experiment considers fixing both $\\lambda_w$ and $\\lambda_v$ (see table \\ref{tab_e6}).\r\nThe former is again set to $\\lambda_w=10^{-8}$, the latter to $\\lambda_w=10^{-3}$, which is close to the parameters selected in previous experiments.\r\nThe mean scores for the digits and XOR 6f data are the best for any of the experiments.\r\nAll other results are equivalent to the best model.\r\nBased on this finding, we recommend $\\lambda_w=10^{-3}$ as default for the classifier.\\par\r\n%\r\n\\begin{table}\r\n\\caption{[E1--E6] Comparison of results (best log-loss bold)}\r\n\\label{tab_e1_to_e6}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E1] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.05$)}}\\\\\r\nActive features&2&6&3&19&2&6\\\\\r\nPrototypes&34&43&24&546&257&309\\\\\r\nLog-loss&0.62&0.14&0.13&0.15&0.17&0.55\\\\\r\nThreshold stage 2&0.54&0.16&0.14&0.22&0.20&0.56\\\\\r\nROC-AUC&0.86&0.99&0.99&1.00&0.99&0.81\\\\\r\nBalanced acc.&0.76&0.97&0.94&0.97&0.95&0.71\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E2] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.50$)}}\\\\\r\nActive features&2&7&3&18&2&6\\\\\r\nPrototypes&30&69&54&533&281&360\\\\\r\nLog-loss&0.49&0.20&0.13&0.18&\\textbf{0.16}&0.53\\\\\r\nThreshold stage 2&0.49&0.16&0.14&0.19&0.19&0.55\\\\\r\nROC-AUC&0.90&0.98&0.99&1.00&0.99&0.82\\\\\r\nBalanced acc.&0.73&0.93&0.95&0.97&0.95&0.72\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E3] Randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&7&4&30&2&6\\\\\r\nPrototypes&49&52&56&861&328&413\\\\\r\nLog-loss&\\textbf{0.43}&0.14&0.13&0.15&0.18&0.52\\\\\r\nThreshold stage 2&0.57&0.17&0.14&0.17&0.20&0.53\\\\\r\nROC-AUC&0.91&1.00&0.99&1.00&0.99&0.82\\\\\r\nBalanced acc.&0.71&0.98&0.98&0.97&0.95&0.72\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E4] Use optimal hyperparameters ($\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&11&3&41&2&6\\\\\r\nPrototypes&65&204&93&3,053&595&403\\\\\r\nLog-loss&0.48&\\textbf{0.05}&\\textbf{0.11}&0.13&\\textbf{0.16}&0.52\\\\\r\nThreshold stage 2&0.57&0.17&0.14&0.17&0.20&0.53\\\\\r\nROC-AUC&0.87&1.00&0.99&1.00&0.99&0.83\\\\\r\nBalanced acc.&0.69&0.98&0.95&0.97&0.94&0.74\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E5] Grid search for $\\lambda_v$ ($\\lambda_w=10^{-8}$ and $\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&4&3&18&2&6\\\\\r\nPrototypes&25&25&69&410&409&421\\\\\r\nLog-loss&0.47&0.12&0.13&0.18&0.17&0.53\\\\\r\nThreshold stage 2&0.71&0.21&0.15&0.26&0.19&0.53\\\\\r\nROC-AUC&0.90&1.00&0.99&1.00&0.99&0.81\\\\\r\nBalanced acc.&0.73&0.97&0.96&0.97&0.95&0.71\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E6] Stage 2 only ($\\lambda_v=10^{-3}$, $\\lambda_w=10^{-8}$, and $\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&7&4&34&2&6\\\\\r\nPrototypes&82&72&149&1,270&2527&1,261\\\\\r\nLog-loss&0.45&0.16&0.13&\\textbf{0.12}&0.18&\\textbf{0.49}\\\\\r\nThreshold stage 2&0.72&0.17&0.15&0.17&0.20&0.52\\\\\r\nROC-AUC&0.89&0.99&0.99&1.00&0.99&0.84\\\\\r\nBalanced acc.&0.76&0.95&0.95&0.98&0.95&0.75\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\clearpage\r\n%\r\n\\begin{table}\r\n\\caption{[E7] Stage 2, vary candidates ($\\lambda_v=10^{-3}$, $\\lambda_w=10^{-8}$, and $\\alpha_v=\\alpha_w=0.95$)}\r\n\\label{tab_e7}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Checker}&\\textbf{Checker}&\\textbf{Checker}&\\textbf{XOR 6f}&\\textbf{XOR 6f}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&2&2&2&2&2&2\\\\\r\nFeatures&2&2&2&6&6&6\\\\\r\nSamples&6,400&6,400&6,400&6,400&6,400&6,400\\\\\r\nTrain samples&4,480&4,480&4,480&4,480&4,480&4,480\\\\\r\nTest samples&1,920&1,920&1,920&1,920&1,920&1,920\\\\\r\n\\textbf{Candidates}&100&300&1,500&100&300&1,500\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&3&1&1&10&9&10\\\\\r\nSelected batches&2&1&1&7&4&4\\\\\r\nOptimal log-loss&0.41&0.29&0.18&0.57&0.53&0.50\\\\\r\nThreshold&0.42&0.31&0.19&0.58&0.55&0.51\\\\\r\nSelected log-loss&0.41&0.29&0.18&0.57&0.54&0.51\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&2&2&6&6&6\\\\\r\nPrototypes&112&125&300&365&487&2,188\\\\\r\nLog-loss&0.43&0.28&0.17&0.57&0.54&0.48\\\\\r\nROC-AUC&0.90&0.97&0.99&0.78&0.80&0.85\\\\\r\nBalanced acc.&0.79&0.90&0.96&0.70&0.72&0.76\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nThe seventh experiment shows the impact of different batch sizes $M$ (see table \\ref{tab_e7}).\r\nIt deals only with the checkerboard and XOR data, as the maximum batch size for the first four cases is limited by their small sample size.\r\nFor each of the two large data sets, we try using 100, 300, and 1,500 samples instead of the default $M=1,000$.\r\nThe model metrics for 100 and 300 samples are worse than before, while the results for 1,500 candidates are comparable.\r\nClearly, there is an advantage to using large batches as long as the sample size permits.\r\n%\r\n\\section{Comparison with other classifiers}\r\n\\label{sec_classifier_comparison}\r\n%\r\nIn this section, we compare the proset classifier to the $k$-nearest neighbor (kNN) and XGBoost methods.\r\nThe former is conceptually similar to proset, as it scores new points based on their proximity to training samples.\r\nHowever, there are at least three major differences between proset and kNN:\r\n%\r\n\\begin{enumerate}\r\n\\item kNN uses all training samples for scoring and gives equal importance to each sample.\r\nProset selects prototypes and assigns individual weights.\r\n%\r\n\\item kNN relies on the user's choice of features and distance metric.\r\nIrrelevant features or a badly scaled metric can degrade performance.\r\nProset adapts the metric to the problem and removes features that contribute little or nothing.\r\n%\r\n\\item kNN has no notion of absolute distance.\r\nThe nearest neighbors have the same impact on classification no matter how far away they are from the sample being scored.\r\nIf that sample is far away from the training data, estimates become arbitrary.\r\nFor proset, remote prototypes have negligible impact on classification.\r\nAs the distance of a sample to the set of prototype increases, the estimated probabilities converge to the marginals of the training data.\r\n\\end{enumerate}\r\n%\r\nXGBoost is a particular implementation of gradient boosting for decision trees (we do not consider other base learners in this study) \\cite{Chen_16}.\r\nIt has become a kind of industry standard for supervised learning outside the domain of deep learning.\r\nIn terms of the three points stated above, XGBoost relates to proset and kNN as follows:\r\n%\r\n\\begin{enumerate}\r\n\\item XGBoost does not retain any training samples.\r\nIt generates a set of rules in the form of decision trees and performs classification via weighted voting.\r\n%\r\n\\item XGBoost relies only on the ordering of features, not on any kind of distance metric.\r\nWhen building trees, the algorithm selects features using a greedy heuristic.\r\nIt is considered good practice to limit the number of features evaluated at each stage by subsampling \\cite{Chen_16}.\r\nThis results in a more diverse set of trees with less tendency to overfit.\r\nEven with this approach, features that have negligible impact on performance are unlikely to be selected.\r\n%\r\n\\item Using XGBoost to score a sample that is far away from the training data gives arbitrary results.\r\nThe decision trees have not been validated in that part of the feature space.\r\n\\end{enumerate}\r\n%\r\nTo fit kNN and XGBoost models, we follow the same general strategy as for proset.\r\nWe determine optimal hyperparameters using cross-validation and then choose an equivalent set of parameters that is less likely to overfit via the `1 SE rule' (see algorithm \\ref{alg_hyperparameters}).\r\nThis is easy to implement for kNN, since the sole hyperparameter is the number of neighbors $k$.\r\nNote that we choose the equivalent solution with the \\textit{largest} $k$ to achieve a high degree of smoothing.\\par\r\n%\r\nFor XGBoost, tuning is more complex as the number of hyperparameters that can be used to control the model fit is quite large.\r\nBased on prior experience, we focus on the following five only, leaving all others at their recommended defaults:\r\n%\r\n\\begin{enumerate}\r\n\\item Learning rate $\\eta$.\r\nControls the impact of each additional tree on the estimator.\r\nChoosing a smaller $\\eta$ can increase model performance slightly but requires more boosting iterations.\r\n%\r\n\\item Number of boosting iterations.\r\nHas to be sufficiently large for the model to capture all of the structure in the data.\r\nIncreasing it further leads to saturation instead of overfitting, i.e., the model quality on test data fluctuates around a common level.\r\n%\r\n\\item Maximum tree depth.\r\nControls the complexity of each tree.\r\nOur experience is that XGBoost performs best if the individual trees underfit.\r\n%\r\n\\item Fraction of features evaluated per split.\r\nUsing a random subset of the features in each split results in a more diverse set of trees.\r\nThis can result in a more favorable trade-off between bias and variance.\r\n%\r\n\\item Fraction of records used for training each tree.\r\nUsing a random subset of the training samples to build each tree has a similar effect as randomizing the features.\r\n\\end{enumerate}\r\n%\r\nWe follow a similar strategy as for proset (see algorithm \\ref{alg_hyperparameters}) to fit XGBoost classifiers.\r\nIn the first stage, we fix $\\eta=0.1$ and use 100 boosting iterations to determine suitable values for maximum tree depth, fraction of features per split, and fraction of records per tree.\r\nHyperparameter values are sampled randomly to generated 100 trial combinations:\r\n%\r\n\\begin{itemize}\r\n\\item Maximum tree depth is sampled from 0 (constant model) to 9 with equal probability.\r\nIn case the maximum was selected or model performance looked inadequate, we increased the maximum parameter value.\r\n%\r\n\\item Fraction of features per split is sampled uniformly between 0.1 and 0.9.\r\nNote that for a model with two features, any value of 0.5 or above just means to test both, while values below 0.5 means to select one at random.\r\n%\r\n\\item Fraction of records per tree is sampled uniformly between 0.1 and 0.9.\r\n\\end{itemize}\r\n%\r\nAfter five-fold cross-validation, we determine the parameter combination that minimizes log-loss, compute a threshold, and choose an equivalent combination.\r\nAmong all candidates, we use the one that minimizes the depth of the tree first, then the fraction of features per split, and finally the fraction of records per tree.\\par\r\n%\r\nIn the second stage, we fix $\\eta=0.01$ and use five-fold cross-validation to determine the number of boosting iterations up to a maximum of 10,000.\r\nWe again apply the `1 SE rule' to find an equivalent smaller number of iterations.\r\nThe model is then re-fitted to all training data and scored on test data.\\par\r\n%\r\n\\begin{table}\r\n\\caption{[E8] $k$-nearest neighbor classifier with grid search for $k$}\r\n\\label{tab_e8}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Cross-validation}}\\\\\r\nOptimal $k$&12&4&29&8&10&14\\\\\r\nSelected $k$&29&13&51&43&11&20\\\\\r\nOptimal log-loss&0.47&0.10&0.15&0.16&0.23&0.54\\\\\r\nThreshold&0.54&0.15&0.17&0.27&0.24&0.56\\\\\r\nSelected log-loss&0.53&0.15&0.17&0.27&0.24&0.55\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nLog-loss&0.48&0.11&0.14&0.22&0.21&0.54\\\\\r\nROC-AUC&0.93&1.00&1.00&1.00&0.98&0.83\\\\\r\nBalanced acc.&0.82&0.97&0.95&0.93&0.92&0.73\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E9] XGBoost classifier with randomized parameter search}\r\n\\label{tab_e9}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Data}}\\\\\r\nClasses&3&3&2&10&2&2\\\\\r\nFeatures&2&13&30&64&2&6\\\\\r\nSamples&150&178&569&1,797&6,400&6,400\\\\\r\nTrain samples&105&124&398&1,257&4,480&4,480\\\\\r\nTest samples&45&54&171&540&1,920&1,920\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal max.\\ depth&2&3&4&4&18&1\\\\\r\nSelected max.\\ depth&1&1&1&3&15&1\\\\\r\nOptimal colsample&0.43&0.48&0.38&0.27&0.79&0.81\\\\\r\nSelected colsample&0.11&0.27&0.11&0.24&0.39&0.81\\\\\r\nOptimal subsample&0.58&0.67&0.45&0.77&0.77&0.25\\\\\r\nSelected subsample&0.46&0.86&0.46&0.72&0.68&0.25\\\\\r\nOptimal log-loss&0.50&0.07&0.10&0.13&0.10&0.70\\\\\r\nThreshold&0.62&0.11&0.14&0.15&0.11&0.70\\\\\r\nSelected log-loss&0.52&0.09&0.12&0.15&0.11&0.70\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal iterations&1,586&2,344&5,579&6,187&9,973&16\\\\\r\nSelected iterations&337&887&668&1,168&1,519&1\\\\\r\nOptimal log-loss&0.50&0.08&0.10&0.11&0.07&0.69\\\\\r\nThreshold&0.62&0.11&0.14&0.14&0.09&0.69\\\\\r\nSelected log-loss&0.62&0.11&0.14&0.14&0.09&0.69\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&13&29&53&2&1\\\\\r\nLog-loss&0.63&0.11&0.12&0.11&0.06&0.69\\\\\r\nROC-AUC&0.89&1.00&0.99&1.00&1.00&0.50\\\\\r\nBalanced acc.&0.69&0.97&0.96&0.98&0.98&0.50\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E3, E8, E9] Comparison of results (best log-loss bold)}\r\n\\label{tab_e3_e8_e9}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrrr|}\r\n\\hline\r\n&\\multicolumn{6}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Iris 2f}&\\textbf{Wine}&\\textbf{Cancer}&\\textbf{Digits}&\\textbf{Checker}&\\textbf{XOR 6f}\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E3] Proset with randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&7&4&30&2&6\\\\\r\nLog-loss&\\textbf{0.43}&0.14&0.13&0.15&0.18&\\textbf{0.52}\\\\\r\nThreshold stage 2&0.57&0.17&0.14&0.17&0.20&0.53\\\\\r\nROC-AUC&0.91&1.00&0.99&1.00&0.99&0.82\\\\\r\nBalanced acc.&0.71&0.98&0.98&0.97&0.95&0.72\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E8] $k$-nearest neighbor classifier with grid search for $k$}}\\\\\r\nActive features&2&13&30&64&2&6\\\\\r\nLog-loss&0.48&\\textbf{0.11}&0.14&0.22&0.21&0.54\\\\\r\nThreshold CV&0.54&0.15&0.17&0.27&0.24&0.56\\\\\r\nROC-AUC&0.93&1.00&1.00&1.00&0.98&0.83\\\\\r\nBalanced acc.&0.82&0.97&0.95&0.93&0.92&0.73\\\\\r\n\\multicolumn{7}{|l|}{\\textbf{[E9] XGBoost classifier with randomized parameter search}}\\\\\r\nActive features&2&13&29&53&2&1\\\\\r\nLog-loss&0.63&\\textbf{0.11}&\\textbf{0.12}&\\textbf{0.11}&\\textbf{0.06}&0.69\\\\\r\nThreshold stage 2&0.62&0.11&0.14&0.14&0.09&0.69\\\\\r\nROC-AUC&0.89&1.00&0.99&1.00&1.00&0.50\\\\\r\nBalanced acc.&0.69&0.97&0.96&0.98&0.98&0.50\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\nResults for fitting kNN and XGBoost classifiers to the six examples used for the proset parameter study are given in tables \\ref{tab_e8} and \\ref{tab_e9}.\r\nThe structure of the tables and reported metrics are essentially the same as for proset in the previous section.\r\nFor XGBoost, the number of active features reported is the number of features with positive importance score using the method's built-in scoring.\\par\r\n%\r\nTable \\ref{tab_e3_e8_e9} summarizes the performance of all three models.\r\nFor proset, reference values are taken from the full parameters search using $\\alpha_v=\\alpha_w=0.95$ (see table \\ref{tab_e3}).\\par\r\n%\r\n\\begin{enumerate}\r\n\\item\\textbf{Iris 2f:} proset is best with kNN equivalent and XGBoost worse.\r\nHowever, the number of test samples is so small that these results are unreliable.\r\nChanging the random seed for the train-test split can have a large impact on the outcome.\\par\r\n%\r\nFigure \\ref{fig_comparison_decision_iris_2f} shows the decision surfaces resulting from the three algorithms.\r\nThe different modeling strategies are apparent from the plots: the surface for proset is composed of many localized kernels, the one for kNN resembles a Voronoi tesselation, and for XGBoost, it is possible to see the splits made by the decision trees perpendicular to the coordinate axes.\r\n%\r\n\\item\\textbf{Wine:} XGBoost and kNN achieve the same log-loss.\r\nThe score for for proset is above the threshold for XGBoost but not for kNN.\r\nIn this case, we consider the lower threshold for the overall ranking in Table \\ref{tab_classifier_comparison}.\r\nProset does use only 7 of 13 features, while XGBoost uses the full set (kNN can perform no selection).\r\n%\r\n\\item\\textbf{Cancer:} XGBoost is best but both other models are considered `equivalent'.\r\nProset reduces the number of features considerably from 30 to 4, while XGBoost uses 29.\r\n%\r\n\\item\\textbf{Digits}: there is a strict ranking with XGBoost being best and proset better than kNN.\r\nHowever, proset is close to XGBoost in terms of balanced accuracy while kNN is markedly worse.\r\nProset reduces the number of features from 64 to 30, XGBoost uses 54.\r\n%\r\n\\item\\textbf{Checker:} there is a strict ranking with XGBoost being best and proset better than kNN.\r\nXGBoost is considerably better than both other models on this data set.\r\nNote that the maximum tree depth for XGBoost was increased to 19 as the fit with a limit of 9 selected the upper bound.\r\n%\r\n\\item\\textbf{XOR 6f:} proset is best, kNN worse, while XGBoost fails to find any patterns in the data and returns a constant estimator.\r\nIncreasing the maximum tree depth for XGBoost to 99 does nothing to improve model quality.\r\n\\end{enumerate}\r\n%\r\n\\begin{figure}\r\n\\caption{Comparison of decision surfaces for case Iris 2f}\r\n\\label{fig_comparison_decision_iris_2f}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{[E3] proset}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_dominant_l2_surf_test.pdf}}\r\n\\subfloat[\\textbf{[E8] kNN}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_knn_surf_test.pdf}}\\\\\r\n\\subfloat[\\textbf{[E9] XGBoost}]{\\includegraphics[width=0.49\\textwidth]{figures/iris_2f_xgb_surf_test.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nWe draw the following conclusions from this study:\r\n%\r\n\\begin{itemize}\r\n\\item In terms of achieving a low log-loss, proset appears to rank in-between XGBoost and kNN.\r\n%\r\n\\item Proset achieves a greater reduction in the number of features than XGBoost with a preference for small decision trees.\r\nWhile the fit strategy for XGBoost may not always select the smallest possible feature set due to feature subsampling, the difference is marked.\r\n%\r\n\\item The extremely good performance of XGBoost on the checker data is possibly due to the fact that the approximating function is of the same class as the target.\r\nBoth are step functions with edges parallel to the main coordinate axes.\r\n%\r\n\\item XOR 6f appears to defeat tree ensembles, which choose features one at a time based on a greedy criterion.\r\nNo matter how the first split is made, the expected number of cases per class in each resulting subspace is 50 \\%, same as for the whole space.\r\nThus, the first split is always made in response to random fluctuations in the data.\r\nThis is also true for the checkerboard with an even number of squares per side, but the first split is almost sure to break the symmetry for this case.\r\nIn contrast, for XOR 6f, at least five splits are required to expose the underlying structure.\r\nThe partitions created by five random splits apparently do not contain sufficient information to build a meaningful decision tree.\r\n\\end{itemize}\r\n%\r\nBased on these observations, we define five additional test cases to further explore the differences between the three algorithms:\r\n%\r\n\\begin{enumerate}\r\n\\item\\textbf{Checker rot:} to determine how much of the good performance of XGBoost on the checker data is due to the axis-parallel steps, we rotate the checkerboard pattern by 45\\textdegree.\r\n%\r\n\\item\\textbf{XOR 3f, XOR 4f, XOR 5f:} in order to find the point at which XGBoost fails on the `continuous XOR' class of problems, we generate instances with three, four, and five features.\r\nAs for XOR 6f, these each have an average of 100 samples per orthant.\r\n%\r\n\\item\\textbf{XOR 6+6f:} to understand how easy it is to confuse the kNN classifier with irrelevant data, we add six more features to the XOR 6f problem.\r\nThese are drawn from the same distribution as the first six but have no impact on the target.\r\n\\end{enumerate}\r\n%\r\n\\begin{table}\r\n\\caption{[E10] New examples -- results for proset classifier}\r\n\\label{tab_e10}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrr|}\r\n\\hline\r\n&\\multicolumn{5}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Checker rot}&\\textbf{XOR 3f}&\\textbf{XOR 4f}&\\textbf{XOR 5f}&\\textbf{XOR 6+6f}\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Data}}\\\\\r\nClasses&2&2&2&2&2\\\\\r\nFeatures&2&3&4&5&12\\\\\r\nSamples&6,400&800&1,600&3,200&6,400\\\\\r\nTrain samples&4,480&560&1,120&2,240&4,480\\\\\r\nTest samples&1,920&240&480&960&1,920\\\\\r\n\\textbf{Candidates}&1,000&$\\sim280$&$\\sim560$&$1000$&1,000\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal $\\lambda_v$&$3.1\\times10^{-5}$&$4.5\\times10^{-2}$&$1.8\\times10^{-3}$&$5.5\\times10^{-3}$&$2.2\\times10^{-4}$\\\\\r\nSelected $\\lambda_v$&$2.9\\times10^{-4}$&$4.2\\times10^{-3}$&$1.0\\times10^{-2}$&$5.5\\times10^{-3}$&$1.9\\times10^{-3}$\\\\\r\nOptimal $\\lambda_w$&$4.3\\times10^{-9}$&$1.3\\times10^{-8}$&$7.7\\times10^{-8}$&$8.3\\times10^{-8}$&$5.7\\times10^{-8}$\\\\\r\nSelected $\\lambda_w$&$6.9\\times10^{-8}$&$1.5\\times10^{-6}$&$4.6\\times10^{-7}$&$8.3\\times10^{-8}$&$2.7\\times10^{-6}$\\\\\r\nOptimal log-loss&0.18&0.19&0.30&0.40&0.55\\\\\r\nThreshold&0.19&0.21&0.34&0.42&0.56\\\\\r\nSelected log-loss&0.19&0.20&0.32&0.40&0.56\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal batches&1&6&6&5&1\\\\\r\nSelected batches&1&1&1&2&1\\\\\r\nOptimal log-loss&0.19&0.19&0.32&0.39&0.57\\\\\r\nThreshold&0.20&0.23&0.33&0.40&0.57\\\\\r\nSelected log-loss&0.19&0.23&0.33&0.39&0.57\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&3&4&5&6\\\\\r\nPrototypes&320&89&124&572&425\\\\\r\nLog-loss&0.18&0.17&0.28&0.39&0.56\\\\\r\nROC-AUC&0.99&0.99&0.97&0.91&0.80\\\\\r\nBalanced acc.&0.95&0.96&0.91&0.83&0.72\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E11] New examples -- results for $k$-nearest neighbor classifier}\r\n\\label{tab_e11}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrr|}\r\n\\hline\r\n&\\multicolumn{5}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Checker rot}&\\textbf{XOR 3f}&\\textbf{XOR 4f}&\\textbf{XOR 5f}&\\textbf{XOR 6+6f}\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Data}}\\\\\r\nClasses&2&2&2&2&2\\\\\r\nFeatures&2&3&4&5&12\\\\\r\nSamples&6,400&800&1,600&3,200&6,400\\\\\r\nTrain samples&4,480&560&1,120&2,240&4,480\\\\\r\nTest samples&1,920&240&480&960&1,920\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Cross-validation}}\\\\\r\nOptimal $k$&10&7&8&9&100\\\\\r\nSelected $k$&11&16&9&13&100\\\\\r\nOptimal log-loss&0.23&0.22&0.34&0.43&0.70\\\\\r\nThreshold&0.24&0.27&0.35&0.44&0.71\\\\\r\nSelected log-loss&0.24&0.27&0.34&0.44&0.70\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nLog-loss&0.21&0.25&0.31&0.45&0.71\\\\\r\nROC-AUC&0.98&0.98&0.95&0.88&0.46\\\\\r\nBalanced acc.&0.92&0.91&0.86&0.80&0.48\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E12] New examples -- results for XGBoost classifier}\r\n\\label{tab_e12}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrr|}\r\n\\hline\r\n&\\multicolumn{5}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Checker rot}&\\textbf{XOR 3f}&\\textbf{XOR 4f}&\\textbf{XOR 5f}&\\textbf{XOR 6+6f}\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Data}}\\\\\r\nClasses&2&2&2&2&2\\\\\r\nFeatures&2&3&4&5&12\\\\\r\nSamples&6,400&800&1,600&3,200&6,400\\\\\r\nTrain samples&4,480&560&1,120&2,240&4,480\\\\\r\nTest samples&1,920&240&480&960&1,920\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Stage 1}}\\\\\r\nOptimal max.\\ depth&21&7&18&17&1\\\\\r\nSelected max.\\ depth&12&6&11&12&1\\\\\r\nOptimal colsample&0.77&0.81&0.79&0.72&0.81\\\\\r\nSelected colsample&0.26&0.88&0.88&0.81&0.81\\\\\r\nOptimal subsample&0.43&0.84&0.77&0.69&0.25\\\\\r\nSelected subsample&0.54&0.81&0.88&0.49&0.25\\\\\r\nOptimal log-loss&0.18&0.10&0.30&0.64&0.70\\\\\r\nThreshold&0.19&0.11&0.32&0.67&0.70\\\\\r\nSelected log-loss&0.19&0.11&0.31&0.66&0.70\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Stage 2}}\\\\\r\nOptimal iterations&8,455&2,677&2,926&895&2\\\\\r\nSelected iterations&3,364&690&1,154&96&1\\\\\r\nOptimal log-loss&0.13&0.10&0.28&0.65&0.69\\\\\r\nThreshold&0.14&0.15&0.31&0.68&0.69\\\\\r\nSelected log-loss&0.14&0.15&0.31&0.68&0.69\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{Final model, scores for test data}}\\\\\r\nActive features&2&3&4&5&1\\\\\r\nLog-loss&0.12&0.11&0.23&0.67&0.69\\\\\r\nROC-AUC&0.99&1.00&0.98&0.65&0.50\\\\\r\nBalanced acc.&0.95&0.97&0.94&0.60&0.50\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\begin{table}\r\n\\caption{[E10, E11, E12] Comparison of results (best log-loss bold)}\r\n\\label{tab_e10_e11_e12}\r\n%\r\n\\begin{center}\r\n\\small\r\n\\begin{tabular}{|lrrrrr|}\r\n\\hline\r\n&\\multicolumn{5}{c|}{\\textbf{\\hrulefill\\ Data set \\hrulefill}}\\\\\r\n&\\textbf{Checker rot}&\\textbf{XOR 3f}&\\textbf{XOR 4f}&\\textbf{XOR 5f}&\\textbf{XOR 6+6f}\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{[E10] Proset with randomized search for $\\lambda_v$ and $\\lambda_w$ ($\\alpha_v=\\alpha_w=0.95$)}}\\\\\r\nActive features&2&3&4&5&6\\\\\r\nLog-loss&0.18&0.17&0.28&\\textbf{0.39}&\\textbf{0.56}\\\\\r\nThreshold stage 2&0.20&0.23&0.33&0.40&0.57\\\\\r\nROC-AUC&0.99&0.99&0.97&0.91&0.80\\\\\r\nBalanced acc.&0.95&0.96&0.91&0.83&0.72\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{[E11] $k$-nearest neighbor classifier with grid search for $k$}}\\\\\r\nActive features&2&3&4&5&12\\\\\r\nLog-loss&0.21&0.25&0.31&0.45&0.71\\\\\r\nThreshold CV&0.24&0.27&0.35&0.44&0.71\\\\\r\nROC-AUC&0.98&0.98&0.95&0.88&0.46\\\\\r\nBalanced acc.&0.92&0.91&0.86&0.80&0.48\\\\\r\n\\multicolumn{6}{|l|}{\\textbf{[E12] XGBoost classifier with randomized parameter search}}\\\\\r\nActive features&2&3&4&5&1\\\\\r\nLog-loss&\\textbf{0.12}&\\textbf{0.11}&\\textbf{0.23}&0.67&0.69\\\\\r\nThreshold stage 2&0.14&0.15&0.31&0.68&0.69\\\\\r\nROC-AUC&0.99&1.00&0.98&0.65&0.50\\\\\r\nBalanced acc.&0.95&0.97&0.94&0.60&0.50\\\\\r\n\\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\\end{table}\r\n%\r\n\\clearpage\r\n%\r\nResults for proset, kNN, and XGBoost are summarized in tables \\ref{tab_e10}, \\ref{tab_e11}, and \\ref{tab_e12}.\r\nA comparison of the three algorithms is shown in table \\ref{tab_e10_e11_e12}:\r\n%\r\n\\begin{enumerate}\r\n\\item\\textbf{Checker rot:} XGBoost still yields the best model on this pattern, but the log-loss is higher than for the original checkerboard.\r\nThe largest value for maximum tree depth used in cross-validation was set to 29 as smaller bounds meant the bound was selected.\r\nThe metrics for the other two models are very similar to the original case.\r\n%\r\n\\item\\textbf{XOR 3f, XOR 4f, XOR 5f:} with three features, XGBoost performs better than other models, with four it is still best but the others are equivalent, and with five features, proset is better than the other two models, i.e., they are not affected by the orientation of the pattern.\r\n%\r\n\\item\\textbf{XOR 6+6f:} proset correctly selects the six relevant features and produces a model that is slightly worse than for the original XOR 6f.\r\nNeither kNN nor XGBoost are able to find any structure and return constant estimators.\r\n\\end{enumerate}\r\n%\r\nOverall, we believe that these results show that proset is a worthwhile addition to the supervised learning toolbox.\r\nAs intended, it performs feature selection as an integral part of model fitting and is able to identify a nonlinear relationship between the features and target.\r\n%\r\n\\endinput\r\n", "meta": {"hexsha": "fbe9fcc44a99e0b7db922fe06d53deb11ee2d063", "size": 74160, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/classifier.tex", "max_stars_repo_name": "NRuf77/proset", "max_stars_repo_head_hexsha": "101d491e05c2423faddca31029232982f46d8831", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/classifier.tex", "max_issues_repo_name": "NRuf77/proset", "max_issues_repo_head_hexsha": "101d491e05c2423faddca31029232982f46d8831", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/classifier.tex", "max_forks_repo_name": "NRuf77/proset", "max_forks_repo_head_hexsha": 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YES\n2. YES", "lm_q1_score": 0.9099070109242131, "lm_q2_score": 0.6893056231680122, "lm_q1q2_score": 0.6272040191900581}} {"text": "\\lab{Applications}{Data Visualization}{Data Visualization} \n\\objective{Use data visualizations to explore data and communicate it to others.}\n\\label{lab:DataVis}\n\n\\section*{What is data visualization?} \nData visualizations (or graphs) are used to understand and explore data as well as communicate results to others. \nJust as a picture is worth a thousand words, a graph is much easier than a thousand data points for the human mind to interpret. \nThough anyone working with data benefits from graphing it, data visualization is also an independent field that studies the interplay between mathematics, statistics, and the visual arts.\n\n\\section*{Types of visualizations}\nThere are many standard ways to visualize data, and some consistently reveal and communicate the most information about certain kinds of data sets. \nHere are some standard plots and some examples of data they are commonly used to visualize.\n\n\\begin{enumerate}\n\\item A \\emph{scatter plot} graphs $(x,y)$ tuples as points. \nA scatter plot can reveal correlation (or lack thereof) between $x$ and $y$. \nUse a scatter plot when your data is not ordered. \nYou can create a scatter plot in matplotlib with the command \\li{plt.scatter()}.\n\n\\item A \\emph{line plot} graphs $(x,y)$ tuples as points and then connects them with a line. \nYou should use a line plot when there is a natural order to your tuples. \nFor example, use a line plot to look for trends over time. \nRecall that you can create a line plot in matplotlib with the command \\li{plt.plot()}.\n\n\\item A \\emph{histogram} (or bar graph) depicts $(x,y)$ tuples as rectangles whose length is determined by $y$ (see Figure \\ref{fig:healthcare}). \nLike line plots, histograms can be used to look for trends over time. \nAnother use of a histograms is to investigate statistical distributions. \nYou can create a histogram in matplotlib with the command \\li{plt.hist()}. \n%TODO: Add a problem that requires them to build a histogram\n\n\\item A \\emph{pie chart} depicts parts of a whole as slices of a circle. \nOften, a pie chart is best replaced with a histogram. \nThis is because it is difficult for us to see how similar slices of a pie chart differ in size. \nBy contrast, we can easily detect even small differences in the lengths of rectangles in a bar chart. \nHowever, pie charts are ideal for some purposes, especially when the exact numbers from the data are not important. \nSee \\cite{piecharts} for some situations where data is best represented by a pie chart. \nYou can create a pie chart in matplotlib with the command \\li{plt.pie()}.\n\n\\item A \\emph{sparkline} is a word-sized graphic that is often embedded inline with the text. \nCreated by Edward Tufte, an innovator and recognized leader in the field of data visualizations, sparklines can be used to quickly communicate large-scale trends in a data set. They are also ideal for binary data sets. \nTwo sources for creating sparkplots in Python are \\cite{sparkplots} and \\cite{sparklineshtml}.\n%TODO: Add an example of a sparkline\n\n\\item A \\emph{pseudocolor plot} or heat map uses color to display a third dimension on a two-dimensional page (see Figure \\ref{fig:heatmap}). \nOne common use is to illustrate the temperature of an object. \nYou can create a pseudocolor plot in matplotlib with the \\li{plt.pcolormesh()} command.\n\n\\item \\emph{Small multiples} are several minature graphics of the same type in a single visualization. \nLike sparklines, they were pioneered by Edward Tufte. Any of the plots discussed so far may be assembled into a small multiple. \nSmall multiples are useful for viewing trends over time, when the data associated to each time is more complex than a single number. \nMore generally, they can be used to compare many sets of similar data (\\cite{tufte1990} p. 67). \nFigure \\ref{fig:log_plots} could be called a small multiple, though frequently they contain more subplots. \nRecall that this kind of plot can be created in matplotlib with the command \\li{plt.subplot()}.\n\\end{enumerate}\n\n\n\n\\section*{Exploring data with visualizations}\nWhen you visualize data you may notice trends that are not apparent from the numbers. \nThe following problem is an example of this.\n\n\\begin{problem}\\label{prob:anscombe}\nThe data sets I-IV in Table \\ref{table:anscombe} are known as Anscombe's quartet. \nEach dat set has identical statistical properties. \nIn each case,\n\\begin{itemize}\n\\item The mean of $x$ is 9 and the mean of $y$ is $7.5$.\n\\item The variance of $x$ is 11 and the variance of $y$ is 4.127.\n\\item The correlation between $x$ and $y$ is .816.\n\\item The linear regression line is $y=3+5x$.\n\\end{itemize}\nPlot each data set. What do you notice?\n\n\\begin{table}[H]\n\\begin{tabular}{l l | l l | l l | l l }\nI & & II & & III & & IV\\\\\nx & y & x & y & x & y & x & y \\\\\n\\hline\n10.0 & 8.04 & 10.0 & 9.14 & 10.0 & 7.46 & 8.0 & 6.58 \\\\\n8.0 & 6.95 & 8.0 & 8.14 & 8.0 & 6.77 & 8.0 & 5.76 \\\\\n13.0 & 7.58 & 13.0 & 8.74 & 13.0 & 12.74 & 8.0 & 7.71 \\\\\n9.0 & 8.81 & 9.0 & 8.77 & 9.0 & 7.11 & 8.0 & 8.84 \\\\\n11.0 & 8.33 & 11.0 & 9.26 & 11.0 & 7.81 & 8.0 & 8.47 \\\\\n14.0 & 9.96 & 14.0 & 8.10 & 14.0 & 8.84 & 8.0 & 7.04 \\\\\n6.0 & 7.24 & 6.0 & 6.13 & 6.0 & 6.08 & 8.0 & 5.25 \\\\\n4.0 & 4.26 & 4.0 & 3.10 & 4.0 & 5.39 & 19.0 & 12.50 \\\\\n12.0 & 10.84 & 12.0 & 9.13 & 12.0 & 8.15 & 8.0 & 5.56 \\\\\n7.0 & 4.82 & 7.0 & 7.26 & 7.0 & 6.42 & 8.0 & 7.91 \\\\\n5.0 & 5.68 & 5.0 & 4.74 & 5.0 & 5.73 & 8.0 & 6.89 \\\\\n\\end{tabular}\n\\caption{These four sets of data are known as Anscombe's quartet.}\n\\label{table:anscombe}\n\\end{table}\n\\end{problem}\n\nAs Problem \\ref{prob:anscombe} demonstrates, a picture can reveal a data set in ways numbers can't. \nLearning from a visualization of a data set is often a recursive process. \nYou must visualize the data, make observations, and then zoom, filter, or otherwise modify your visualization to further explore the data.\n\nOne issue to consider is the range. \nBy choosing to plot only a portion of your data, you may be able to focus your attention on the most relevant data. \nOn the other hand, you may also miss some interesting behavior in the data you didn't graph.\n\nAnother issue to consider is scale. \nFor example, suppose the number of dogs kept as pets in a certain city increases from 50,000 to 55,000 over 4 years. \nThis increase can appear small or large, depending on where we start the y-axis and how long we stretch out the x-axis (see Figure \\ref{fig:dog_plots}).\n\nThere are many software packages that facilitate the visual exploration of data. \nOne Python library is Glue (see \\cite{glue}).\n\n\n\\subsection*{Log plots}\n\nOne good way to analyze data is to plot it on a different scale. \nLog plots in particular can reveal important structure in a data set. \nYou can create a log-log plot by taking the logarithm of both the $x$- and $y$-values. \nAnalogously, you can create a log-lin plot (also called a log plot) by taking the logarithm of only the $y$-values of your data set, or a lin-log plot by taking the logarithm of only the $x$-values.\n\nAs an example, let us look at the cost of health care claims. \nOur data set is bogus but created to closely resemble real-life data. \nAn initial plotting of the data produces the histogram on the left in Figure \\ref{fig:healthcare}. \nFrom this picture, it is hard to see the nature of the data---the plot just looks like a single bar. \nWe may think that all health care claims are cheap. \nHowever, when we take the logarithm of the claims prices, we get the histogram on the right in Figure \\ref{fig:healthcare}. \nGraphed on this scale, the data has a bell-shaped distribution similar to a normal distribution. \nMoreover, we see that there are some very expensive claims being submitted, though they are few. \n\\begin{figure}\n\\centering\n\\begin{subfigure}{.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{healthcare_linscale.pdf}\n\\end{subfigure}%\n\\begin{subfigure}{.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{healthcare_logscale.pdf}\n\\end{subfigure}\n\\caption{The same data (meant to resemble the prices of health care claims) is plotted in both histograms above. \nThe plot at left is a lin-lin scale, whereas the plot at right is a log-lin scale. \nIn this case, changing scales revealed important information about the data.}\n\\label{fig:healthcare}\n\\end{figure}\n\nAs a general rule, log-lin plots are useful when the range of the $y$-values is orders of magnitude larger than the range of the $x$-values. \nOur health care data was an example of this. \nSimilarly, lin-log plots are useful when the range of the $x$-values is much bigger than the range of the $y$-values. \nIn any case, you can try applying a log scale to one or both of your axes as a way to explore your data.\n\nLet us analyze the mathematics behind log plots. \nSuppose we have some data that roughly follows the line $y=cx^a$ where $c$ and $a$ are constants. \nTaking the logarithm of both sides yields\n\\[\n\\log(y) = a\\log(x) + \\log(c).\n\\]\nIf we set new variables $X = \\log(x)$ and $Y = \\log(y)$, then this equation becomes $Y = aX + \\log(c)$, which is the equation of a straight line. \nSo on a log-log plot, polynomial-shaped data looks linear. \nSimilarly, taking the log of $y = a^x$ yields\n\\[\n\\log(y) = \\log(a)x,\n\\]\nso exponential data looks linear on a log-lin plot. \nFinally, data of the form $y = a\\log(x)+b$ will look linear on a lin-log plot. \nSee Figure \\ref{fig:log_plots}.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{log_plots.pdf}\n\\caption{Here are some functions plotted on various logarithmic and linear scales. \nThe dark blue line is $y=x$, the green line is $y=\\log(x)$, the red line is $y=e^x$, and the light blue line is $y=x^3$. \nWe have used different ranges on the $y$-axes to highlight appropriate parts of the graphs.}\n\\label{fig:log_plots}\n\\end{figure}\n\nBe warned that many different functions can look linear on log plots. \nThus, if you graph your data on a log-log plot and you see a line, you CANNOT conclude that your data must follow a polynomial. \nIn general, you need more information.\n\nIn matplotlib, you can create plots on a logarithmic scale directly by using the \\li{plt.loglog()}, \\li{plt.semilogy()}, and \\li{plt.semilogx()} commands. \nTheir syntax is identical to that of\\li{plt.plot()}. \nAlternatively, you can modify the scale of an existing plot using \\li{plt.yscale()} or \\li{plt.xscale()}.\n\n\n\\begin{problem}\nGraph your data from Problem 1 in Lab \\ref{lab:complexity} (Matrices and Complexity) on a (a) log-log scale, (b) log-lin scale, and (c) lin-log scale. \nIn your opinion, which plot makes the data easiest to understand?\n\\end{problem}\n\n\n\n\n\\section*{Communicating data with visualizations}\n\nAs the saying has it, ``a picture is worth a thousand words.'' \nCertainly, a data visualization is far more valuable than a thousand data points to a colleague who wants to understand the results of your analysis. \nData visualizations are critical for communication in both business and research.\n\n\\subsection*{Choosing the right visualization}\nAt this point, you have already used visualization to explore your data and draw conclusions, and you are ready to tell your results to someone else. \nThe first step in creating a graphic to do this is to choose the right type of visualization. \nAs we saw in the section ``Types of Visualizations,'' there are many possibilities, each with different strengths. \nYou should carefully consider which one best suits your data set and communication needs.\n\n\\subsection*{Simplify}\nOnce you have chosen the visualization best suited to your data, you should design the details so that all elements contribute to the communication of data. \nEdward Tufte offers two principles for simplifying graphics: (1) erase ink that does not communicate data, and (2) erase ink that communicates data redundantly. \nBoth of these principles should be applied within reason (\\cite{tufte2001} pp.96-100). \n\nAccording to these principles, decorative backgrounds, fancy lettering, and cute graphics in the corner of you plots should all be deleted.\n\nOther opportunities for simplification are harder to notice and implement. As an example, let us examine the plot on the right of Figure \\ref{fig:healthcare}. \nThis plot was created with the following code.\n\n\\begin{lstlisting}\nimport numpy as np\nimport scipy as sp\nfrom matplotlib import pyplot as plt\n\nm = 2.07\ns = 0.63\nnum_samples = 10000\nsamples = []\n\nfor i in xrange(num_samples):\n samples.append(sp.random.lognormal(m, s)) \n\nsp_samples = sp.array(samples)\n\n# Plot the histogram\nplt.hist(sp_samples, 100)\n\\end{lstlisting}\n\nFor purposes of this example, the only important part of the above code is the line \\li{plt.hist(sp_samples, 100)} that plots the histogram. \nWhat ink in this plot can we erase because it communicates no data?\n\nFirst, let's get rid of the vertical black lines that separate the bars of the histograph. \nThese are meaningless for our application. \nWe can do this by modifying our call to \\li{plt.hist()} as follows.\n\n\\begin{lstlisting}\nplt.hist(sp_samples, 100, histtype=\"stepfilled\")\n\\end{lstlisting}\n\nNext, let us turn off the top and right lines that box in the graph. \nTo do this, we need to access the ``axis'' object associated with the figure. \nWe can access the ``axis'' object with the command \\li{plt.gca()} (get current axis).\n\n\\begin{lstlisting}\n# Get current axis instance\naxis = plt.gca()\n\n# Hide top and right spines\naxis.spines['right'].set_visible(False)\naxis.spines['top'].set_visible(False)\n\\end{lstlisting}\n\nThese commands only turn off the sides of the box, not the tick marks. \nTo turn off the tick marks we run the following commands. \n\n\\begin{lstlisting}\n# Only show bottom and left tick marks\naxis.yaxis.set_ticks_position('left')\naxis.xaxis.set_ticks_position('bottom')\n\\end{lstlisting}\n\nFinally, we do not need so many tick marks on the $x$- and $y$-axes. \nWe adjust those along with specifying the range on each axis.\n\n\\begin{lstlisting}\n# Fix x and y ranges\nplt.xlim(0,70)\nplt.ylim(0, 800)\n\n# Use fewer axis ticks\nplt.xticks(np.arange(0, 71, 35))\nplt.yticks(np.arange(0, 801, 200))\n\\end{lstlisting}\n\nThe final graph is shown in Figure \\ref{fig:simplify}. \nNote how much cleaner this looks than the original.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{simplify.pdf}\n\\caption{A simplified version of the histogram in Figure \\ref{fig:healthcare}.}\n\\label{fig:simplify}\n\\end{figure}\n\n\n\\subsection*{Color}\n%TODO: explain how you can make a colorbar, either in this lab or the matplotlib lab\nColor should be added to your graphic thoughtfully. \nThe default settings in matplotlib should generally not be used. \nSeveral bright colors in a single plot often jar the viewer, so you should limit the number and intensity of colors and use them only to emphasize parts of your data. \nIf a grayscale plot contains just as much information as a colored plot, consider using grayscale.\n\nThe default color settings for pseudocolor plots in matplotlib are particularly problematic. \nThe default color gradient is a rainbow, which usually has no relation to the data. \nMost viewers have trouble remembering which colors are ``high\" and which are ``low\". \nInstead, you should set the gradient to cover one or two colors, and when possible these colors should be related to the data you are communicating. \nYou can change the color gradient used by \\li{plt.pcolormesh()} with the keyword argument \\li{cmap} (short for ``colormap''). \nA list of predefined colormaps is available at \\url{http://matplotlib.org/examples/color/colormaps_reference.html}, but you can also define your own. \nSee Figure \\ref{fig:heatmap} for an example of a good and bad pseudocolor plot.\n\n\n\\begin{figure}\n\\centering\n\\begin{subfigure}{.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{heatmap_color.png}\n\\end{subfigure}%\n\\begin{subfigure}{.5\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{heatmap_gray.png}\n\\end{subfigure}\n%\\includegraphics[width=\\textwidth]{heatmap.png}\n\\caption{Both of these pseudocolor plots depict the function $z = sin(x)sin(y)$ on the domain $[-1,4] \\times [-4,1]$. \nThe plot at left uses a rainbow gradient which has no relationship to the data. \nThe plot at right uses a greyscale, which has two benefits: the colors are not irritating, and ``black'' is associated with ``low.''}\n\\label{fig:heatmap}\n\\end{figure}\n\n%TODO: Rewrite this problem to use a specific image with a specific context\n\\begin{problem}\nChoose a graph you previously created that you think could use simplification. \nApply at least 3 of the following to simplify your plot:\n\\begin{enumerate}\n\\item Turn off some of the spines.\n\\item Fix the x or y range to better fit your data.\n\\item Change the number of x or y ticks.\n\\item Choose a better color scheme.\n\\end{enumerate}\nSimplify your image in any other ways you can think of.\n\\end{problem}\n\n\\subsection*{Tell the truth}\n\nJust as graphics can tell your reader about your data, they can also give your viewer false impressions. \nYou should never use data visualization to mislead your reader. \nThere are many ways pictures can be used to lie, and some of them can be unintentional.\n\nFor example, if we see differences in shapes by area, not width or height. \nSuppose you are creating a pictogram to demonstrate the recent increase in housing prices, which have doubled in the last $n$ years. \nSo, you draw two houses, one that is 1'' wide and one that is 2'' wide. \nTo a reader, it will appear that housing prices have increased by a factor of $2^2=4$. \nHence, you should make pictures proportional in \\emph{area} to the numbers they represent.\n\nScale is also important when you are drawing a line graph or a histogram. \nDifferences between data points can appear small if you zoom out, or large if you zoom in (see Figure \\ref{fig:dog_plots}). \nAlso, changing the scale partway along the $x$- or $y$-axis can distort data.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{dog_plots.pdf}\n\\caption{The graph of some data can look very different depending on the scale. \nHere are two graphs of the ``same'' line over the same range (1 to 4). \nThe only difference is the $y$-range. Notice easy it is to manipulate the y-scale to imply either a large or small increase in $y$. \nThese figures are especially misleading when we leave the range markers off of the $y$-axes.}\n\\label{fig:dog_plots}\n\\end{figure}\n\nThree-dimensional special effects also tend to be misleading. \nA 3-D pie chart, for example, can be used to distort the sizes of its slices. \nSimilarly, a 3-D histogram can use perspective to distort the relative differences between bars.\n\nAs in any form of communication, integrity is important in data visualization. \nYou will win more respect for yourself and those you represent if you avoid misleading diagrams.\n\n\\printbibliography", "meta": {"hexsha": "cfaa1440f1a571aeb4dbd6c4b0b59afd20a28f6c", "size": 18747, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Labs/ScientificVisualization/DataVisualization.tex", "max_stars_repo_name": "rachelwebb/numerical_computing", "max_stars_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Labs/ScientificVisualization/DataVisualization.tex", "max_issues_repo_name": "rachelwebb/numerical_computing", "max_issues_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Labs/ScientificVisualization/DataVisualization.tex", "max_forks_repo_name": "rachelwebb/numerical_computing", "max_forks_repo_head_hexsha": "e7416b43b97976060f6875fa46c7dca20a9f635f", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.0817438692, "max_line_length": 219, "alphanum_fraction": 0.7534005441, "num_tokens": 4928, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.72487026428967, "lm_q2_score": 0.8652240964782012, "lm_q1q2_score": 0.6271752194839447}} {"text": "\n\\section{Lower bound of finite element error estimates}\n%{Notation and basic results}\nIn this section, we will indicate that the convergence rate is optimal by showing the lower bound of finite element error estimates. \n%\\subsection{Lower bound of finite element approximation in any dimension}\n%In this subsection, \nWe will provide lower bound for finite element error estimates that are obtained by Lin, Xie and Xu~\\cite{lin2014lower}.\n\\begin{theorem}\\label{Lower_Bound_linear}\nAssume $u\\in H^{3}(\\Omega)$ and $u$ is not linear %$u\\notin V_h$ \nand for all $v_h\\in V_h$ being the linear finite element function.\nThe following lower bound of the approximation error holds when the family of mesh $\\mathcal{T}_h$\nis quasi-uniform %the mesh size is $h$ is sufficiently small:\n\\begin{equation}\\label{lower_convergence_0_1}\n\\inf_{v_h\\in V_h}\\|u-v_h\\|_{r}\\geq Ch^{2-r}\\approxeq N^{-\\frac{2-r}{d}}\\quad r=0, 1,\n\\end{equation}\nwhere $C$ is dependent on $u$.\n\\end{theorem}\n\\begin{proof}\nLet $\\Pi_2$ is the nodal interpolation to quadratic finite element, and denote $\\displaystyle |\\cdot|_{2,h}=\n\\left(\\sum_{\\tau \\in \\mathcal T_h} |\\cdot|^2_{2,\\tau}\\right)^{\\frac12}$. For any $v_h\\in V_h$, since $v_h$ is\nlinear, by the approximation property of $\\Pi_2$ and inverse inequality, we have \n\\begin{equation}\n\\begin{aligned}\n|u|_{2,\\Omega}&=|u-v_h|_{2,h}\\le |u-\\Pi_2 u|_{2,h}+|\\Pi_2u-v_h|_{2,h}\\\\\n& \\le C_1h |u|_{3,\\Omega}+C_2h^{r-2}\\|\\Pi_2u-v_h\\|_{r,\\Omega}\\\\\n& \\le C_1h |u|_{3,\\Omega}+C_2h^{r-2}\\|\\Pi_2u-u\\|_{r,\\Omega} +C_2h^{r-2}\\|u-v_h\\|_{r,\\Omega}\\\\\n& \\le C_1h |u|_{3,\\Omega}+C_3h^{r-2}h^{3-r}|u|_{3,\\Omega} +C_2h^{r-2}\\|u-v_h\\|_{r,\\Omega}\\\\\n\\end{aligned}\n\\end{equation}\nnamely \n\\begin{equation}\n\\begin{aligned}\n|u|_{2,\\Omega}-(C_1+C_3)h|u|_{3,\\Omega}\\le C_2h^{r-2}\\|u-v_h\\|_{r,\\Omega}.\n\\end{aligned}\n\\end{equation}\nLet $h_0(u)=\\frac{|u|_{2,\\Omega}}{2 (C_1+C_3)|u|_{3,\\Omega}}$ and we have\n$$\n|u|_{2,\\Omega}-(C_1+C_3)h_0|u|_{3,\\Omega}=\\frac12|u|_{2,\\Omega}.\n$$\nNoting that $v_h\\in V_h$ is arbitrary, then if $h\\le h_0$, we have \n\\begin{equation}\n\\inf_{v_h\\in V_h}\\|u-v_h\\|_{r,\\Omega}\\geq Ch^{2-r}.\n\\end{equation}\nIf $\\mathcal{T}_h$ is a triangulation with mesh size $h> h_0$, then by the newest vertex bisection strategy shown in \\cite{stevenson2008completion}, we \ncan refine $\\mathcal{T}_h$ such that the new triangulation \n$\\mathcal{T}_{h_0}$ with mesh \nsize decreasing to the order of $h_0$ and we have $V_h\\subset V_{h_0}$, which implies \n\\begin{equation}\n\\inf_{v_h\\in V_h}\\|u-v_h\\|_{r,\\Omega}\\geq \\inf_{v_{h_0}\\in V_{h_0}}\\|u-v_{h_0}\\|_{r,\\Omega} \\geq Ch_0^{2-r}.\n\\end{equation}\nTherefore \\eqref{lower_convergence_0_1} is desired. \n\\end{proof}\n\n%\\begin{conjecture}\n%\tLet $V_N$ be linear finite element spaces associated with \n%\tshape-regular simplicial grids $\\mathcal T_N$ of $N$-elements (or\n%\t$N$-grid points) in a polyhedral domain $\\Omega\\subset \\mathbb R^d$,\n%\tsay $\\Omega=(0,1)^d$. Then, for any reasonable function (which is, say,\n%\tnot locally linear, $|u|_{2}\\neq 0$), then the following lower bound holds:\n%\t\\begin{equation}\n%\t\\label{optimalFEMerror}\n%\t\\inf_{\\# \\mathcal T_N=\\mathcal O(N)} \\inf_{v_N\\in\n%\t\tV_N}\\|u-v_N\\|_{0,\\Omega}\\ge c(u) N^{-2/d} \n%\t%\\le C(u) N^{-2/d}\n%\t\\end{equation}\n%\\end{conjecture}\n%\n%Questions:\n%\\begin{enumerate}\n%\t\\item Is the above conjecture correct in some way? If yes, what are\n%\tthe more rigorous statements for such results?\n%\t\\item What are the most relevant references that contain such results?\n%\t\\item If the conjecture is incorrect, is there a counter example?\n%\\end{enumerate}\n\n\n\\begin{conjecture}\\label{conjecture1}\nFor any function $u\\in L^2(\\Omega)$ % is not smooth (for example $u\\in H^{1+\\epsilon}$ for any $\\epsilon>0$)\n that is not locally linear (namely $u$ is not\nlinear in any open subset), we have \n\\begin{equation}\\label{lower_convergence_0_2}\n\\inf_{v_h\\in V_N}\\|u-v_h\\| \\gtrsim N^{-\\frac{2}{d}}\n\\end{equation}\nfor any finite element space $V_N$ of dimension $N$ on a shape-regular grid.\n\\end{conjecture}\n%Questions:\n%\\begin{enumerate}\n%\t\\item Is the above conjecture correct in some way? If yes, what are\n%\tthe more rigorous statements for such results?\n%\t\\item What are the most relevant references that contain such results?\n%\t\\item If the conjecture is incorrect, is there a counter example?\n%\\end{enumerate}\n%\\begin{conjecture}\\label{con2}\n%Let $\\mathcal T_N$ be a grid of $\\Omega\\subset \\mathbb R^d$ with $N$ elements and $V_h$ be the finite element space on $\\mathcal T_N$. \n%If $u$ is a smooth function (for example $u\\in H^3(\\Omega)$) and the optimal partition $\\mathcal T^*_N$ such that \n%\\begin{equation}\\label{lower_convergence_0_3}\n%\\inf_{\\# \\mathcal T_N=N}\\inf_{v_h\\in V_h}\\|u-v_h\\|_{r}=\\inf_{v_h\\in V^*_h}\\|u-v_h\\|_{r}\\geq CN^{-\\frac{2-r}{d}}\\quad r=0, 1\n%\\end{equation}\n%where $V_h^*$ is the finite element space on $\\mathcal T^*_N$, then the partition $\\mathcal T^*_N$ can be a quasi-uniform partition.\n%\\end{conjecture}\n\n\\begin{conjecture}\\label{con2}\nLet %with $N$ elements \n $V_N$ be the linear finite element space on a grid $\\mathcal T_N$ of $\\Omega\\subset \\mathbb R^d$. \nIf $u$ is a smooth function (for example $u\\in H^2(\\Omega)$) and $u$ is not a linear function,\n\\begin{equation}\\label{lower_convergence_0_3}\n\\mathop{\\inf_{\\# \\mathcal T_N=N}}_{\\mathcal T_N~\\hbox{is shape regular}} \\inf_{v_h\\in V_N}\\|u-v_h\\|\\gtrsim \\mathop{\\inf_{\\# \\mathcal T_N=N}}_{\\mathcal T_N~\\hbox{is quasi-uniform}} \\inf_{v_h\\in V_N}\\|u-v_h\\|\\gtrsim N^{-\\frac{2}{d}}.\n\\end{equation}\n%where $V_N^*$ is the finite element space on $\\mathcal T^*_N$, then the partition $\\mathcal T^*_N$ can be a quasi-uniform partition.\n\\end{conjecture}\nQuestions:\n\\begin{enumerate}\n\t\\item Is the above Conjecture \\ref{conjecture1} or Conjecture \\ref{con2} correct in some way? If yes, what are\n\tthe more rigorous statements for such results?\n\t\\item What are the most relevant references that contain such results?\n\t\\item If the Conjecture \\ref{conjecture1} or Conjecture \\ref{con2} is incorrect, is there a counter example?\n\\end{enumerate}\n\n\n", "meta": {"hexsha": "5b0c4f977f604b2d8bc4155f5cc19cd74a4fe36f", "size": 6005, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/LowerBound.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/LowerBound.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/LowerBound.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.0416666667, "max_line_length": 231, "alphanum_fraction": 0.7025811823, "num_tokens": 2172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702642896702, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.6271752094074087}} {"text": "% !TEX root = ../main.tex\n\\newpage\n\\section{\\theory Mean Field Reductions} \\label{sec:MFRs}\nThe \\textsl{mean-field reduction} (\\MFR) is a theory that can predict the dynamics of the order parameter \\eqref{eq:orderparameter}. In \\cite{OttAntonsen2008, OttAntonsen2009, OttAntonsen2010} such a method was published for fully connected networks of indistinguishable oscillators with harmonic coupling. In \\cite{Restrepo2014} the authors extended their work to include networks with arbitrary degree distributions, applied to the Kuramoto model. Later this analysis was extended to networks of the Theta Neuron model \\cite{OttAntonsen2017}. We will now consider the limit $N \\gg 1$ and formulate an exact \\MFR for different types of networks, following the method in \\cite{OttAntonsen2017}. \n\n\\subsection{The Ott-Antonsen manifold}\nTo simplify notation, the authors incorporate the network size in to $P$ so that $\\sum_{\\k \\in \\K} P(\\k) = N$. To specify the probability of a link from a node of degree $\\kacc$ to one of degree $\\k$ we can define an assortativity function:\n\\begin{align}\na\\left(\\k_j \\rightarrow \\k_i \\right) = 0 \\leq \\frac{k^{\\rm{out}^{\\prime}}_j k^{\\rm{in}}_i}{N \\kmean} \\leq 1 \\label{eq:assortativityfunction}\n\\end{align}\nwhere we have chosen a neutral assortativity \\cite{OttAntonsen2017}. \\eqref{eq:assortativityfunction} is constrained so that the number of links in the network, $N \\kmean$, remains constant \\cite{Restrepo2014}:\n\\begin{align*}\n\\sum_{\\kacc \\in \\K} \\sum_{\\k \\in \\K} P\\left( \\kacc \\right) \\: a\\left(\\kacc \\rightarrow \\k\\right) P\\left(\\k\\right) = N \\kmean\n\\end{align*}\nWe can now assume that the state of all neurons can be represented by a probability density function $f(\\theta, \\eta | \\k, t)$. Hence, the marginal distribution \n\\begin{align*}\n \\int_{\\R} \\int_{\\mathbb{I}} f(\\vartheta, \\eta^{\\prime} | \\k, t) \\mathop{d \\vartheta} \\mathop{d \\eta^{\\prime}} \n\\end{align*}\ngives the fraction of nodes of degree $\\k$ with a phase in $\\mathbb{I} \\subset \\T$ at time $t$. Also, we assume $\\eta_i$ do not change over time, so that $\\int_{\\T} f(\\vartheta, \\eta^{\\prime} | \\k, t) \\: \\mathop{d \\vartheta}$ yields the excitability distribution $g(\\eta | \\k)$. \\\\\n\nTo describe the global synchronisation of the network of theta neurons \\eqref{eq:thetaneuronnetwork} we have introduced the order parameter $Z(t)$ in \\eqref{eq:orderparameter}. It is now hypothesized that $Z(t)$ can be approximated by a mean-field order parameter, defined by the continuum limit:\n\\begin{align}\n\\bar{Z}(t)= \\frac{1}{N} \\sum_{\\k \\in \\K}P(\\k) \\int_{\\R} \\int_{\\T} f\\left(\\vartheta, \\eta^{\\prime} | \\k, t\\right) e^{\\ic \\vartheta} \\mathop{d \\vartheta} \\mathop{d \\eta^{\\prime}} \\label{eq:meanfieldorderparameter}\n\\end{align}\nHere, $f$ is constrained by a continuity equation, as the number of oscillators is conserved:\n\\begin{align}\n\\frac{\\partial f}{\\partial t}+\\frac{\\partial}{\\partial \\theta}\\left(v_{\\theta} f\\right) = 0 \\label{eq:transportequation}\n\\end{align}\nwith $v_{\\theta}$ a continuum version of \\eqref{eq:thetaneuronnetwork}:\n\\begin{align*}\nv_{\\theta} &= (1-\\cos \\theta)+(1+\\cos \\theta) \\left[\\eta + I(\\k,t) \\right]\\\\\nI(\\k,t) &= \\frac{\\kappa}{\\kmean}\\sum_{\\kacc \\in \\K} P(\\kacc ) \\:a\\left(\\kacc \\rightarrow \\k \\right) \\times \\left[ \\int_{\\R} \\int_{\\T} f\\left(\\vartheta, \\eta^{\\prime} | \\kacc, t\\right) \\: a_{2} \\left(1-\\cos \\vartheta\\right)^{2} \\mathop{d \\vartheta} \\mathop{d \\eta^{\\prime}} \\right] &\n\\end{align*}\nIn \\cite{OttAntonsen2008} it is shown that there exists a manifold of invariant probability densities for the continuity equation. The exact \\MFR is obtained by expanding $f$ as a Fourier series, and expanding the pulse $\\mathcal{P}_s$ using the binomial theorem. When assuming $\\eta_i$ is distributed according to a Lorenz distribution:\n\\begin{align}\ng(\\eta |\\k)=\\frac{1}{\\pi} \\frac{\\sigma(\\k)}{(\\eta-\\eta_{0}(\\k))^{2}+\\sigma(\\k)^{2}} \\label{eq:Lorentzpdf}\n\\end{align}\nthe set of reduced equations then takes a particularly simple form, as \\eqref{eq:meanfieldorderparameter} can be evaluated at the poles of $g$ using the Cauchy residue theorem for the integration of complex variables and we find a closed form expression. We can now capture the dynamics by $z(\\k,t)$, the mean-field variable for nodes of degree $\\k$:\n\\begin{align}\n\\frac{\\partial z(\\k, t)}{\\partial t} &= -\\ic \\frac{(z(\\k, t)-1)^{2}}{2} + \\frac{(z(\\k, t)+1)^{2}}{2} \\cdot I(\\k, t) \\qquad \\qquad z \\in \\C^{M_\\k} \\nonumber \\\\\nI(\\k, t) &= -\\sigma(\\k) + \\ic \\eta_{0}(\\k) + \\ic H_2(\\k,t) \\label{eq:OttAntonsenSystemFull} \\\\\nH_2(\\k,t) &= \\frac{\\kappa}{\\kmean} \\sum_{\\kacc \\in \\K} P\\left(\\kacc\\right) \\: a\\left(\\kacc \\rightarrow \\k\\right) \\left( 1 + \\frac{z(\\kacc, t)^2 + (z(\\kacc, t)^c )^2}{6} - \\frac{4}{3} \\Re(z(\\kacc, t)) \\right) \\nonumber\n\\end{align}\nwith $z^c$ the complex conjugate. $H$ is a legacy term and has been computed in \\cite{Martens2020}.\nThe mean-field order parameter can now be expressed in terms of $z(\\k,t)$: using the constraints on $f$ and $g$ we can solve \\eqref{eq:meanfieldorderparameter} as:\n\\begin{align}\n\\bar{Z}(t) &= \\frac{1}{N} \\sum_{\\k \\in \\K} P(\\k) z(\\k, t) \\qquad \\bar{Z} \\in \\C \\label{eq:OttAntonsenMeanField}\n\\end{align}\nwhich clearly reflects the network architecture through $P(\\k)$, as in the limit, this is the number of neurons of degree $\\k$ that are present in the network. The mean-field dynamics of the whole network are thus equal to a weighed average of the degree dynamics of each node with unique degree $\\k \\in \\K$. \n\nWe have now formulated the evolution on the invariant manifold by a reduced set of ordinary differential equations. The \\MFR is computationally efficient, and in \\cite{OttAntonsen2017} many methods for improving this efficiency further are treated.\n\n\n\\subsection{Simplifications for fixed-degree networks}\nIn the case of a fixed-degree network, every node has $\\deg (\\theta_i) = (\\kmean, \\kmean)$ so:\n\\begin{align*}\n \\frac{1}{\\kmean} \\sum_{\\kacc \\in \\K} P\\left(\\kacc\\right) \\: a\\left(\\kacc \\rightarrow \\k\\right) = \\frac{1}{\\kmean} N \\left( \\frac{\\kmean \\kmean}{N \\kmean} \\right) = 1\n\\end{align*}\nFor any fixed-degree network, \\cref{eq:OttAntonsenSystemFull,eq:OttAntonsenMeanField} reduce to a single complex differential equation:\n\\begin{align}\n\\dot{Z}(t)= -\\ic \\frac{(Z-1)^2}{2}+\\frac{(Z+1)^2}{2} \\left(-\\sigma+ \\ic\\eta_{0}\n+ \\ic \\kappa \\left(1+\\frac{Z^{2} + (Z^c)^{2} }{6} - \\frac{4}{3} \\Re(Z)\\right)\\right) \\label{eq:MeanField}\n\\end{align}\nThis is an identical formulation as in \\cite{Luke2013} and \\cite{Martens2020}. For any fixed-degree network, the reduced system is thus a complex (two-dimensional) system with three bifurcation parameters $\\eta_0, \\sigma$ and $\\kappa$, \\cite{Luke2013, Martens2020}. We will start our analysis with \\eqref{eq:MeanField}.\\\\\n\nThree distinct macroscopic states can be identified. In the partially synchronous rest state (\\PSR) we can observe in Figure \\ref{fig:MFRPSR}, $Z$ settles onto a stable node. Most neurons can be found in a resting state $\\eta_0 + \\sigma \\lesssim 0$, and inhibit one another through $\\kappa < 0$. Most neurons are therefore inactive, though some spiking neurons from the tail of $g$ are present but have a neglegible effect. \\\\\n\nIn Figure \\ref{fig:MFRPSS} we can observe the partially synchronous spiking (\\PSS) state, where we can see how $Z$ settles onto a stable focus. This happens predominantly when $\\eta_0 - \\sigma \\gtrsim 0$ and most neurons inherently spike, with the coupling being either excitatory or weakly inhibitory. Although most neurons are active, the network is partially synchronous and organized such that phase cancellation occurs by continuous spiking among the neurons.\\\\\n\nLastly, in the collective periodic wave state (\\CPW) we can observe a limit cycle of the mean field, in Figure \\ref{fig:MFRCPW}. Most neurons are active and inhibitory: $\\eta_0 > 0$ and $\\kappa < 0$. The collective oscillation emerges from the interplay between the neurons’ inherent tendency to spike and the strong suppressive network interaction. \\CPW states are mediated through Hopf bifurcations and homoclinic bifurcations of $Z$. We can also see the occurrence of a saddle-node bifurcation in the lower hand corner, for low $\\sigma$. We will continue to study the \\CPW due to its interesting properties.\\\\\n\nA more detailed discussion of the different regimes and bifurcations can be found in \\cite{Luke2013}.\n\n% \\textcolor{red}{QUESTION}: \\textsl{I have worked on the analysis presented in \\cite{Luke2013} to make my own figures on the bifurcations. However, I cannot seem to get them plotted. Should I continue on this?}\n\n\\begin{figure}[H]\n\\centering\n\\begin{subfigure}[b]{0.32\\linewidth}\n \\centering\n \\includegraphics[width=\\linewidth]{../Figures/PhaseSpace/MFRPSR.pdf}\n \\caption{PSR state for $\\eta_0 = -0.9, \\sigma = 0.8$ and $\\kappa= -2$. The mean field settles onto a stable node.}\n \\label{fig:MFRPSR} \n\\end{subfigure} \\hfill\n\\begin{subfigure}[b]{0.32\\linewidth}\n \\centering\n \\includegraphics[width=\\linewidth]{../Figures/PhaseSpace/MFRPSS.pdf}\n \\caption{PSS state for $\\eta_0 = 0.5, \\sigma = 0.7$ and $\\kappa= 2$. The mean field settles onto a stable focus.}\n \\label{fig:MFRPSS}\n\\end{subfigure} \\hfill\n\\begin{subfigure}[b]{0.32\\linewidth}\n \\centering\n \\includegraphics[width=\\linewidth]{../Figures/PhaseSpace/MFRCPW.pdf}\n \\caption{CPW state for $\\eta_0 = 10.75, \\sigma = 0.5$ and $\\kappa= -9$. The mean field settles onto a stable limit cycle.}\n \\label{fig:MFRCPW}\n\\end{subfigure}\n \\caption{Three macroscopic states observed in the \\MFR inside the imaginary unit circle $|Z(t)| \\leqslant 1$. Green arrows mark the phase space vector field and blue trails mark solution curves. Red points indicate equilibrium points, with black arrows marking the direction of the eigenvectors in that point, scaled according to the magnitude of the corresponding eigenvalues. This information is found from the Jacobian, which we will discuss in Chapter \\ref{sec:MFRSUndirected}.}\n \\label{fig:macroscopicstatesfixeddegree}\n\\end{figure}\n\n\n\\subsection{Implications and challenges of the \\MFR}\nThe advantages of using the \\MFR can be found in the number of equations we now have left to investigate. As there are $M_{\\k}$ equations in \\eqref{eq:OttAntonsenSystemFull}, instead of $N$ equations for $N$ neurons, the reduction becomes more and more efficient for larger networks. As we have seen in \\eqref{eq:MeanField} this yields a single equation for a fixed-degree network, as all neurons have the same degree.\\\\\n\nWhile the \\MFR gives us the opportunity to use any arbitray univariate distribution $P(k)$ for undirected, symmetric networks or any bivariate distribution $P(\\k)$ for directed, asymmetric networks, none of the publications on the \\MFR have treated directed networks. The challenge there is that the support $\\K$ is a much larger set, as $\\K = \\koutb \\times \\kinb$. For example, the scale-free distribution \\eqref{eq:scalefreepdf} has $M_{\\k} = \\kmax - \\kmin$ number of degrees in its support. An example would be $M_{\\k} = $ 1250, \\cite{OttAntonsen2017}. For 10.000 neurons, that is a reduction of 87,5\\%. When we wish to extend \\eqref{eq:scalefreepdf} to a bivariate distribution, $M_{\\k}$ grows to $\\left( \\kmax - \\kmin \\right)^2$. A bivariate distribution would therefore need about $1.56 \\times 10^6$ equations for 10.000 neurons. It is not feasible to solve this many equations at once. Even though it is reported that solving only 10\\% of the equations and then interpolating across $z$ and $t$ yields a very good approximation of the whole system, \\cite{OttAntonsen2017}, the sheer number of equations remains too high.\n\n", "meta": {"hexsha": "495d5b472416824c6388ed2755054c8b0ce46e35", "size": 11587, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Writing/Mainmatter/Mean Field Reductions.tex", "max_stars_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_stars_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Writing/Mainmatter/Mean Field Reductions.tex", "max_issues_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_issues_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Writing/Mainmatter/Mean Field Reductions.tex", "max_forks_repo_name": "simonaertssen/AdaptiveNeuronalNetworks", "max_forks_repo_head_hexsha": "506a4e8aba392330f8a6ecc6b229e2c8322b8e83", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 109.3113207547, "max_line_length": 1127, "alphanum_fraction": 0.7307327177, "num_tokens": 3566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951143326726, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.6271558611971542}} {"text": "\n\n\\chapter{Multiscale Methods on the Sphere}\n\\label{ch_mms}\n\nIn this chapter, many multiscale decompositions will be built \nbased on the spherical harmonics and/or the HEALPix representation.\n\n\\section{Orthogonal Haar Wavelets on the Sphere}\n\\label{mrs_haar}\n\\index{wavelet!Haar on sphere}\n\\index{sphere!Haar}\n\nThe Haar wavelet transform on the sphere \\citep{wave:sweldens95a} at each resolution $j$ and pixel $\\bk=(k_x,k_y)$ on the sphere \nis based on a scaling function $\\phi_{j,\\bk}$ ($\\phi_{j,\\bk}(\\bx) = \\phi\\parenth{2^{-j}(\\bx-\\bk)}$, where $\\bx$ is the vector of \nCartesian coordinates on the sphere, and $\\phi$ is the Haar scaling function) and three Haar wavelet functions $\\psi^{d}_{j,\\bk}$ (see~\\eqref{def_whaar1})\nwith $d \\in\\{1,2,3\\}$. It uses the idea that a given pixel on the sphere at a given resolution $j$ in the HEALPix representation\nis directly related to four pixels at the next resolution $j-1$. \n% For Healpix pixelisation, resolution of the map is related \n% to the $N_{side}$ parameter, \n% which corresponds to the square root of number of pixels in each face.\n% $ N_{side} = 2^{j-1}$. At a given resolution $j$ we have $n_j=12 \\times 4^{j-1}$ pixels with a surface size $\\mu_j$. \n\nNoting, $\\bk_0$, the pixel. The scaling function $\\phi$ and three wavelet functions $\\psi$ are defined by:\n\\begin{eqnarray}\n\\label{def_whaar1}\n\\phi_{j,\\bk}(\\bx) & = & \\left\\{ \\begin{array}{ll}\n1 & \\textrm{if $t \\in S_{j,\\bk}$ }\\\\\n0 & \\textrm{otherwise}\n \\end{array} \\right. \\nonumber \\\\\n\\psi_{1,j+1,\\bk} & = & \\frac{\\phi_{j,\\bk_0} +\\phi_{j,\\bk_2}-\\phi_{j,\\bk_1}-\\phi_{j,\\bk_3} }{4} \\nonumber \\\\\n\\psi_{2,j+1,\\bk} & = & \\frac{\\phi_{j,\\bk_0} +\\phi_{j,\\bk_1}-\\phi_{j,\\bk_2}-\\phi_{j,\\bk_3} }{4} \\nonumber \\\\\n\\psi_{3,j+1,\\bk} & = & \\frac{\\phi_{j,\\bk_0} +\\phi_{j,\\bk_3}-\\phi_{j,\\bk_1}-\\phi_{j,\\bk_2} }{4}\n\\end{eqnarray}\n\n% \\begin{figure}[htb]\n%\\centering\n%\\includegraphics[width=10cm]{fig_haar_decomp.pdf}\n%\\label{fig_haar_decomp}\n%\\caption{On resolution to next using Haar.}\n%\\end{figure}\n\n% \\begin{figure*}\n% \\vbox{\n% \\centerline{\n% \\hbox{\n% \\includegraphics[angle=180,width=7cm]{haar0.pdf}\n% \\includegraphics[angle=180,width=7cm]{haar1.pdf}\n% }}\\centerline{\n% \\hbox{\n% \\includegraphics[angle=180,width=7cm]{haar2.pdf}\n% \\includegraphics[angle=180,width=7cm]{haar3.pdf}\n% }}\n% \\centerline{\n% \\hbox{\n% \\includegraphics[angle=180,width=7cm]{haar4.pdf}\n% \\includegraphics[angle=180,width=7cm]{haar5.pdf}\n% }}\n% }\n% \\label{fig_haar_decomp2}\n% \\caption{Haar wavelet coefficents.}\n% \\end{figure*}\n\nDenoting $\\bk_0,\\bk_1,\\bk_2,\\bk_3$ the four pixels at scale $j$, hierarchically related to the pixel $\\bk$ at scale $j+1$, \nscaling coefficients $c_{j+1,\\bk}$ at scale $j+1$ are derived from those at scale $j$ by:\n\\begin{eqnarray}\n\\label{def_haar1}\nc_{j+1}[\\bk] &=& \\frac{1}{4} \\sum_{d=0}^3 c_{j} [\\bk_d],\n\\end{eqnarray}\nand wavelet coefficients at scale $j+1$ from coefficients at scale $j$ by:\n\\begin{eqnarray}\n\\label{def_haar2}\nw^{1}_{j+1} [\\bk] &=& \\frac{1}{4} (c_{j}[\\bk_0]+c_{j} [\\bk_2] - c_{j} [\\bk_1]-c_{j}[\\bk_3]) \\nonumber \\\\\nw^{2}_{j+1} [\\bk] &=& \\frac{1}{4} (c_{j}[\\bk_0]+c_{j} [\\bk_1] - c_{j} [\\bk_2]-c_{j}[\\bk_3]) \\nonumber \\\\\nw^{3}_{j+1} [\\bk] &=& \\frac{1}{4} (c_{j}[\\bk_0]+c_{j} [\\bk_3] - c_{j} [\\bk_1]-c_{j}[\\bk_2]) .\n\\end{eqnarray}\n\n% Fig. \\ref{fig_haar_decomp} shows the calculation scheme of haar coefficients, from one resolution to the next. \n% Haar wavelet coefficients for the astronomical WMAP data set are shown in Fig. \\ref{fig_haar_decomp2}.\n\nThe Haar wavelet transform on the sphere is orthogonal and its reconstruction is exact. The inverse transformation is obtained by:\n\\begin{eqnarray}\n\\label{def_recons_haar}\nc_0[\\bx] = \\sum_{\\bk} c_{J}[\\bk] \\phi_{J,\\bk}(\\bx) + \\sum_{j=1}^{J} \\sum_{d=1}^3 \\sum_{\\bk} w^{d}_{j}[\\bk] \\psi^d_{j}(\\bx) .\n\\end{eqnarray}\n\nThis transform is very fast but its interest is relatively limited. Indeed, it is not rotation invariant, and more importantly \nthe Haar wavelet shape is not well adapted for most applications, because of the non-regular shape of the wavelet function.\n\n\n% ===============================================\n\n\\section{Continuous Wavelets on the Sphere}\n\n\\subsection{Stereoscopic Projection}\n\\label{chap_mex_hat}\n\n\\index{wavelet!axisymmetrical wavelets}\n\\index{wavelet!stereoscopic projection}\n\n\\begin{figure}[htb]\n\\centering\n\\includegraphics[width=15cm]{proj_stereo_inverse.pdf}\n\\caption{Inverse stereographic projections of a radial function from plane to the sphere.}\n\\label{figprojstereo_direct}\n\\end{figure}\n\nIn order to have more choice to design the wavelet function, we may want to use wavelets defined for regular 2D images to the sphere.\nThis is possible by using inverse stereographic projections of radial wavelet functions such the Mexican hat \\citep{wave:cayon01}.\nDefining the stereographic projection operator, $\\bR: {\\bf t} \\mapsto \\ {\\boldsymbol \\omega}$, with ${\\boldsymbol \\omega} = (\\theta(r),\\vartheta)$, \n$\\theta(r) = 2 \\arctan(r/2)$, the radial wavelets $\\psi_{\\mathrm{plane}}$ can be projected on the sphere by a unique rotation, \n${\\boldsymbol \\omega}_0 =(\\theta_0,\\vartheta_0)$, respectively around the two axes $O_y$ and $O_z$. Fig.~\\ref{figprojstereo_direct} shows \nthe projection of radial functions from the plane to the sphere. \n \nThe convolution on the sphere between a radial wavelet function $\\psi(\\theta)$ and a function $f({\\boldsymbol \\omega})$ is:\n \\begin{eqnarray}\n\\label{convol_sph}\n(\\psi \\ast f)(\\theta,\\vartheta)= \\int_{S^2} \\psi_{\\mathrm{plane}}^* (\\bR^{-1} {\\boldsymbol \\omega}) f({\\boldsymbol \\omega}) d{\\boldsymbol \\omega} .\n\\end{eqnarray}\n\nSuch wavelets are axisymmetric by construction. This property can be used to derive fast transformation algorithms using spherical harmonics.\nIndeed spherical harmonics coefficients $\\hat{\\psi}[l,m]$ of the wavelet function $\\psi$ on the sphere are equal to zero when $m \\neq 0$, \nand by the Funk-Hecke theorem, the convolution can be written using spherical harmonics by:\n\\begin{eqnarray}\n\\label{convol_sph_harm}\n(\\psi \\ast f)(\\theta,\\vartheta) = \\sum_{l=0}^{\\infty} \\sum_{m=-l}^{l} \\sqrt{\\frac{2l+1}{4\\pi} }\\hat{f}[l,m] \\hat{ \\psi}[l,0] Y_{lm}(\\theta,\\vartheta) ~.\n\\end{eqnarray}\nwhere $\\hat{f}[l,m]$ are the spherical harmonics coefficients of the function $f$,\\\\ \ni.e. $f = \\sum_{l=0}^{\\infty}\\sum_{m=-l}^{l} \\hat{f}[l,m] Y_{lm}$ and similary for $\\hat{\\psi}$.\n\nClassical wavelet dilations can also be derived on the sphere using the dilation operator $\\mathscr{D}_a$ by a factor $a > 0$ \\citep{wiaux07}:\n\\begin{eqnarray}\n\\label{proj_stereo}\n\\mathscr{D}_a(f)({\\boldsymbol \\omega}) = \\chi_a^{1/2} (a,\\theta) f(D^{-1}_a {\\boldsymbol \\omega}) ,\n\\end{eqnarray}\nwhere $D_a(\\theta,\\vartheta)=(\\theta_a(\\theta),\\vartheta)$ with the linear relation $\\tan\\theta_a(\\theta)/2=a\\tan\\theta/2$, \nand $D_a$ is the dilation operator that maps a sphere without its South pole on itself. $\\chi_a^{1/2}(a,\\theta)$ is a norm \npreservation term (i.e. $\\mathscr{D}_a$ is unitary):\n\\begin{eqnarray}\n\\label{def_lambda}\n\\chi_a^{1/2}(a,\\theta) = a^{-1}[1+\\tan^2(\\theta/2)]/[1+a^{-2}\\tan^2(\\theta/2)] ~.\n\\end{eqnarray}\n\n% \\begin{figure*}\n% \\centering\n% \\includegraphics[width=9cm]{proj_stereo.pdf}\n% \\caption{Projection st\\'er\\'eographique inverse sur la sphere.}\n% \\label{figprojstereo}\n% \\end{figure*}\n\n\\subsection{Mexican Hat Wavelet} \n\n\\index{sphere!mexican hat}\n\\index{Mexican hat!on sphere}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.9cm]{wav_mex_hat_0.pdf}\n\\includegraphics[angle=180,width=7.9cm]{wav_mex_hat_1.pdf}\n}}\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.9cm]{wav_mex_hat_2.pdf}\n\\includegraphics[angle=180,width=7.9cm]{wav_mex_hat_3.pdf}\n}}\n}\n\\caption{Mexican hat on the sphere for the dilation parameter equal to $a = \\{1,2,4,8\\}$.}\n\\label{fig_wav_mex_hat}\n\\end{figure}\n\n% \\begin{figure}[htb]\n% \\vbox{\n% \\centerline{\n% \\hbox{\n% \\includegraphics[angle=180,width=6.5cm]{mex_hat_0.pdf}\n% \\includegraphics[angle=180,width=6.5cm]{mex_hat_1.pdf}\n% }}\\centerline{\n% \\hbox{\n% \\includegraphics[angle=180,width=6.5cm]{mex_hat_8.pdf}\n% \\includegraphics[angle=180,width=6.5cm]{mex_hat_20.pdf}\n% }}\n% }\n% \\caption{Cosmic Microwave Background (CMB) simulated map and its mexican hat wavelet transform\n% with dilation parameter $a = \\{1,8,20\\}$.}\n% \\label{fig_trans_cmb_wav_mex_hat}\n% \\end{figure}\n\nThe 2D Mexican hat wavelet transform is the second derivative of a Gaussian:\n\\begin{equation}\n\\label{rad_mexhat}\n\\psi(r) = \\frac{1}{\\sqrt{2 \\pi}} \\frac{1}{a} \\Big( 2 - \\Big( \\frac{r}{a} \\Big)^2 \\Big) e^{- \\frac{r^2}{2a^2}}\n\\end{equation}\nwhere $a$ is a scale factor parameter and $r$ the distance to the wavelet center.\n\nUsing the inverse stereographic projection, it is possible to extend the Mexican hat wavelet\non the sphere \\citep{wave:antoine99,wave:tenerio99,wave:cayon01,wave:holschneider96,wave:vielva04}:\n\\begin{equation}\n\\psi_a (r) = \\frac{1}{\\sqrt{2 \\pi} C_a} \\Big( 1 + \\Big( \\frac{r}{2} \\Big)^2 \\Big)^2 \\Big( 2 - \\Big( \\frac{r}{a} \\Big)^2 \\Big) e^{- \\frac{r^2}{2a^2}}\n\\end{equation}\nwhere $a$ is a scale factor, $C_a$ is a normalization term $C_a = a \\Big( 1 + \\frac{a^2}{2} + \\frac{a^4}{4} \\Big)^{\\frac{1}{2}}$, \nand $r$ is the distance on the tangent plane, which is related to the polar angle $\\theta$ through $r = 2 \\textrm{tan} \\frac{\\theta}{2}$. \nThis transform may be useful to analyze the data, but it does not have a reconstruction operator, and can therefore not be used for \nrestoration applications.\n\nFig.~\\ref{fig_wav_mex_hat} shows the Mexican hat wavelet on the sphere for four different scales.\n% and Fig ~\\ref{fig_trans_cmb_wa_mex_hat} shows four scales for the a simulated CMB map.\n\n\n\\subsection{Directional Wavelets}\n\\label{dirwavelet}\n\\index{wavelet!directional wavelets}\n\n% \\index{sphere!directionnal wavelet}\n\n\\begin{figure}[htb]\n\\centering\n\\includegraphics[width=15cm]{projection_stereo_ondelette_directionnelle.pdf}\n\\caption{Inverse stereographic projections of a directional wavelet on the sphere.}\n\\label{figprojstereo_direct2}\n\\end{figure}\n\nTo study anisotropic structures, the previously described continuous wavelet transform can be extended to directional wavelets \\citep{wave:antoine01,vielva06,McEwen08}.\nFig.~\\ref{figprojstereo_direct2} shows the projection of an elliptic function from the plane to the sphere. \n\n\\subsubsection{Elongated Mexican Hat Wavelet}\n\\index{wavelet!mexican hat}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_0.pdf}\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_1.pdf}\n}}\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_2.pdf}\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_3.pdf}\n}}\n\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_4.pdf}\n\\includegraphics[angle=180,width=7.7cm]{dir_mex_hat_5.pdf}\n}}\n}\n\\caption{Elongated Mexican hat on the sphere for the dilation parameter equal to $a_x= 1$ and $a_y = \\{0.5,1.,1.25,1.5,2,4\\}$.}\n\\label{fig_dir_mex_hat}\n\\end{figure}\n\nThe elongated Mexican hat wavelet can be written as:\n\\begin{multline}\n\\label{directional_hat}\n\\psi_{a_x, a_y}(\\theta, \\vartheta) = \\sqrt{\\frac{2}{\\pi}} C(a_x,a_y) \\parenth{1+ \\tan^2 \\frac{\\theta}{2}} \\bigg( 1- \\frac{4 \\tan^2 \\theta/2 }{a_x^2+a_y^2} \\Big(\\frac{a_y^2}{a_x^2} \\cos^2 \\vartheta \\\\\n+ \\frac{a_x^2}{a_y^2} \\sin^2 \\vartheta\\Big) \\bigg) e^{-2 \\tan \\frac{\\theta}{2}(\\cos^2 \\vartheta / a^2_x + \\sin^2 \\vartheta / a^2_y )} ~\n\\end{multline}\nwhere $a_x$ and $a_y$ are the dilation factors along the two axes $O_x$ and $O_y$, $C(a_x,a_y)$ is a normalization constant defined as\n\\begin{eqnarray}\n\\label{directional_hat_norm}\nC(a_x,a_y)& =& (a_x^2+a_y^2) \\parenth{a_x a_y (3a_x^4+ 3a_y^4+2a_xa_y)}^{-1/2} ~\n\\end{eqnarray}\n\nFig.~\\ref{fig_dir_mex_hat} shows the wavelet functions for different dilation parameters $a_x$ and $a_y$.\n\n\n\\subsubsection{Morlet Wavelet}\n\\index{wavelet!Morlet}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.9cm]{dir_morlet_0.pdf}\n\\includegraphics[angle=180,width=7.9cm]{dir_morlet_1.pdf}\n}}\\centerline{\n\\hbox{\n\\includegraphics[angle=180,width=7.9cm]{dir_morlet_2.pdf}\n\\includegraphics[angle=180,width=7.9cm]{dir_morlet_3.pdf}\n}}\n}\n\\caption{Morlet wavelets on the sphere for the parameter $\\bk$ equal to (2,0), (4,0), (6,6) et (9,1).}\n\\label{fig_dir_morlet}\n\\end{figure}\n\nThe Morlet wavelet on the sphere, derived from the stereographic projection of the 2D function on the plane, is:\n\\begin{equation}\n\\label{real_morlet}\n\\psi_{a_x, a_y, {\\bf k}}(\\theta, \\vartheta) = \\sqrt{\\frac{2}{\\pi}} C({\\bf k}) (1+ \\tan^2 \\frac{\\theta}{2}) \\bigg(\\cos \\frac{ {\\bf k} \\cdotp R^{-1} {\\bf x} }{\\sqrt{2}} - e^{- |{\\bf k}|^2/4} \\bigg) e^{-2 \\tan^2(\\theta/2)} \n\\end{equation}\nwith $R^{-1} {\\bf x} = (2 \\tan (\\theta/2) \\cos \\vartheta,2 \\tan (\\theta/2) \\sin \\vartheta)$, ${\\bf k}=(k_x,k_y)$, $|{\\bf k}|^2 = k_x^2 + k_y^2$, \nand $C({\\bf k}) = (1+3e^{-|{\\bf k}|^2/2} -4e^{-3 |{\\bf k}|^2/8})^{-1/2} $. ${\\bf k}$ allows us to control the oscillations of the wavelet functions.\\\\ \\\\\nFig.~\\ref{fig_dir_morlet} shows the Morlet wavelet for ${\\bf k}$ equal respectively to $(2,0)$, $(4,0)$, $(6,6)$ and $(9,1)$.\\\\\n\nContinuous wavelet transforms have been intensively used in astrophysics, mainly to analyze the Cosmic Microwave Background \\citep{wave:vielva04}.\nDirectional wavelets based on steerable filters were also proposed in \\citep{wiaux06,McEwen08}. We present in the following a set of multiscale \ndecompositions on the sphere which have a fast exact inverse transform, and are therefore suitable for many applications such as restoration.\n \n\n\n\n%===============================================================\n\n\\section{Redundant Wavelet Transform on the Sphere with Exact Reconstruction}\n\\label{sect_wts}\n%We show in this section that we can build wavelet transforms on the sphere, based on the spherical harmonics transform, ewhich have an exact reconstruction and can therefore be easily used in different applications.\n\n\n\\subsection{Isotropic Undecimated Wavelet Transform on the Sphere}\nHere an undecimated isotropic transform (UWTS) is described which is similar in many respects to the undecimated isotropic transform \n\\citep{starck2010,starck:sta06}, and will therefore be a good candidate for restoration applications. Its isotropy is a favorable \nproperty when analyzing isotropic features. This isotropic transform is obtained using a scaling function ${\\phi}_{l_c}(\\theta, \\vartheta)$ \nwith cut-off frequency $l_c$ and azimuthal symmetry, meaning that ${\\phi}_{l_c}$ does not depend on the azimuth $\\vartheta$. \nHence the spherical harmonics coefficients $\\hat {\\phi}_{l_c} [l,m]$ of ${\\phi}_{l_c}$ vanish when $m \\ne 0$ so that:\n\\begin{eqnarray}\n{\\phi}_{l_c}(\\theta, \\vartheta)= {\\phi}_{l_c}(\\theta) = \\sum_{l = 0}^{l = l_c} \\hat {\\phi}_{l_c} [l,0] Y_{l0}(\\theta, \\vartheta) .\n\\end{eqnarray}\nThen, convolving a function $f(\\theta, \\vartheta) \\in L_2(S^2)$ with ${\\phi}_{l_c}$ is greatly simplified \nand the spherical harmonics coefficients of the resulting map $c_0$ are readily given by\n\\begin{eqnarray}\n \\hat c_{0}[l,m] = \\widehat{{\\phi}_{l_c} * f} [l,m] = \\sqrt{\\frac{2l+1}{4\\pi} } \\hat {\\phi}_{l_c} [l,0] \\hat f[l,m] .\n\\end{eqnarray}\n\\index{sphere!undecimated wavelet}\n\n\\subsubsection{From One Resolution to the Next}\n\nA sequence of smoother approximations of $f$ on a dyadic resolution scale can be obtained using the scaling function ${\\phi}_{l_c}$ as follows:\n\\begin{eqnarray}\n\\begin{split}\nc_0 & = & {\\phi}_{ l_{c} } \\ast f \\\\\nc_1 & = & {\\phi}_{2^{-1} l_{c} } \\ast f \t \\\\\n&\\ldots&\\\\ \nc_j &=& {\\phi}_{2^{-j} l_{c} } \\ast f ,\n\\end{split}\n\\end{eqnarray}\nwhere ${\\phi}_{2^{-j} l_{c} }$ is a rescaled version of ${\\phi}_{l_{c}}$. The above multiresolution sequence can actually be obtained recursively. \n\nDefine a low pass filter $h_{j}$ for each scale $j$ by:\n\\begin{eqnarray}\n \\widehat{H}_{j}[l,m] & = & \\sqrt{\\frac{4\\pi}{2l+1} } \\hat h_{j}[l,m] \\nonumber \\\\\n & = & \n \\begin{cases}\n \\frac { \\hat \\phi_{\\frac{l_{c}}{2^{j+1}} }[l,m] } { \\hat \\phi_{ \\frac{l_{c}}{2^{j}} }[l,m] } & \\mbox{if } l < \\frac{ l_{c}} {2^{j+1}} \\quad \\textrm{and}\\quad m = 0 , \\\\\n 0 & \\mbox{otherwise } . \n \\end{cases}\n\\end{eqnarray}\nIt is then easily shown that $c_{j+1}$ derives from $c_j$ by convolution on the sphere with $h_j$: $c_{j+1} = c_{j} \\ast h_j$.\n\n\n\\subsubsection{The Wavelet Coefficients}\n\nGiven an axisymmetric wavelet function $\\psi_{l_c}$, we can derive in the same way a \nhigh pass filter $g_j$ on each scale~$j$:\n\\begin{eqnarray}\n \\widehat{G}_{j}[l,m] & = & \\sqrt{\\frac{4\\pi}{2l+1} } \\hat{g}_{j}[l,m] \\nonumber \\\\\n & = & \n \\begin{cases}\n \\frac { \\hat \\psi_{\\frac{l_{c}}{2^{j+1}} }[l,m] } { \\hat \\phi_{ \\frac{l_{c}}{2^{j}} }[l,m] } & \\mbox{if } l < \\frac{ l_{c}} {2^{j+1}} \\quad \\textrm{and}\\quad m = 0 ,\\\\\n1 &\\mbox{if } l \\ge \\frac{ l_{c}} {2^{j+1}} \\quad \\textrm{and}\\quad m = 0 ,\\\\ \n0& \\mbox{otherwise } .\n \\end{cases}\n\\end{eqnarray}\nFrom this definition, the wavelet coefficients $w_{j+1} $ at scale $j+1$ are obtained from the previous scaling coefficients $c_j$ \nby a simple convolution on the sphere with $g_j$: $w_{j+1} = c_{j} \\ast g_j$.\n\nAs in the standard 2D isotropic undecimated transform algorithm, the wavelet coefficients can be defined as the difference between two consecutive resolutions, \n$w_{j+1}(\\theta, \\vartheta) = c_{j}(\\theta, \\vartheta) - c_{j+1}(\\theta, \\vartheta)$. This defines a zonal wavelet function $\\psi_{l_c}$ as:\n\\begin{eqnarray}\\label{wavelet}\n\\hat \\psi_{\\frac{l_c}{2^{j}}}[l,m] = \\hat \\phi_{\\frac{l_c}{2^{j-1}}} [l,m] - \\hat \\phi_{\\frac{l_c}{2^{j}}}[l,m] .\n\\end{eqnarray}\nThe high pass filters $g_j$ associated with this wavelet are expressed as: \n\\begin{eqnarray}\n\\begin{split}\n\\widehat{G}_{j}[l,m] & = \\sqrt{\\frac{4\\pi}{2l+1} } \\hat{g}_{j}[l,m] \\\\\n & = 1 - \\sqrt{\\frac{4\\pi}{2l+1} } \\hat{h}_j[l,m] = 1 - \\widehat{H}_j[l,m] .\n\\end{split}\n\\end{eqnarray}\nObviously other wavelet functions could be used just as well. For example, we can define the wavelet function as the difference between\nthe squares of the scaling functions \\cite{starck:book98}, i.e. $\\hat \\psi^2_{\\frac{l_c}{2^{j}}}[l,m] = \\hat \\phi^2_{\\frac{l_c}{2^{j-1}}} [l,m] - \\hat \\phi^2_{\\frac{l_c}{2^{j}}}[l,m]$. \n\n\\newpage\n\\subsubsection{Choice of the Scaling Function}\n\\index{wavelet!needlet}\n\\index{wavelet!Meyer}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n\\includegraphics[width=15cm,height=5cm]{fig_sphere_filterbank1.pdf}\n% \\includegraphics[width=14cm,height=5cm]{ch1_diff_uv_phi_psi.pdf}\n}}}\n\\caption{On the left, spherical harmonics coefficients $\\hat{{\\phi}}[l,0]$ of the the scaling function ${{\\phi}}$ and, on the right, those of the wavelet function ${\\psi}$.}\n\\label{fig_diff_uv_phi_psi}\n\\end{figure}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm]{fig_splinew.png}%[width=6.5cm,height=3.9cm]\n \\includegraphics[width=7.5cm]{fig_needletw.png}\n}}\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm, trim= 2cm 14cm 2cm 2cm, clip=]{fig_splinew_versus_needletw.pdf}\n \\includegraphics[width=7.5cm, trim= 2cm 14cm 2cm 2cm, clip=]{fig_abssplinew_versus_absneedletw.pdf}\n}}\n\\centerline{\n\\hbox{\n\\includegraphics[width=7.5cm, trim= 2cm 14cm 2cm 2cm, clip=]{fig_meyerw_versus_needletw.pdf}\n\\includegraphics[width=7.5cm, trim= 2cm 14cm 2cm 2cm, clip=]{fig_absmeyerw_versus_absneedletw.pdf}\n}}\n}\n\\caption{Comparaison between spline, needlet, and Meyer wavelet functions on the sphere.}\n\\label{Figure:cmp_needlet}\n\\end{figure}\n\nAny function with a cut-off frequency is a possible candidate. The \nB-spline function of order 3 \\citep{starck2010,starck:sta06} leads to :\n\\begin{eqnarray}\n{\\hat{\\phi}}_{l_c} [l,m] = \\frac{3}{2} B_{3} \\left( \\frac{2l}{l_{c}} \\right)\n\\end{eqnarray}\nwhere $B_3(t)$ is the scaling function:\n\\begin{equation}\nB_3(t) = \\frac{1}{12}({\\mid{t-2}\\mid}^3 - 4 {\\mid{t-1}\\mid}^3 + 6 {\\mid{t}\\mid}^3 - 4 {\\mid{t+1}\\mid}^3 + {\\mid{t+2}\\mid}^3)\n\\end{equation}\n\n\nIn Fig.~\\ref{fig_diff_uv_phi_psi} the spherical harmonics coefficients of the scaling function derived from a B$_3$-spline, \nand those of the associated wavelet function \\eqref{wavelet}, are plotted as a function of $l$. Other functions such as \nMeyer wavelets or the needlet function \\citep{marinucci08} can be used as well. \n\n\nMeyer and needlet wavelet functions have both a much better frequency localization than the wavelet function derived from the B$_3$-spline,\nand, as nothing is perfect, the price to pay is more oscillations in the direct space. To illustrate this, we \nshow in Fig.~\\ref{Figure:cmp_needlet} different wavelet functions. Top left and right shows the respectively the spline and the needet wavelet function at a given scale. Fig.~\\ref{Figure:cmp_needlet}~middle left shows a cut of the healpix face with contains the previous functions. Middle right is the same functions, but we have plotted the absolute value in order to better visualize their respective ringing.\nAs it can be seen, for wavelet functions with the same main lob, the needlet wavelet oscillate much more than the spline wavelet. Fig.~\\ref{Figure:cmp_needlet}~bottom left and right compares Meyer and needlet functions. They are relatively close and share\nthe same property of good frequency localization, but with more oscillations than the spline wavelet function.\nHence, the best wavelet choice certainly depends on the final applications. For statistical analysis, detection or restoration applications, \nwe may prefer to use a wavelet which does not oscillate too much and with a smaller support, and the spline wavelet is clearly the \ncorrect choice. For spectral or bispectral analysis, where the frequency localization is fundamental, then Meyer or needlet shoud be \npreferred to the spline wavelet.\n\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm]{fig_mars.png}%[width=6.5cm,height=3.9cm]\n \\includegraphics[width=7.5cm]{fig_mars_scale1.png}\n}}\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm]{fig_mars_scale2.png}\n \\includegraphics[width=7.5cm]{fig_mars_scale3.png}\n}}\n\\centerline{\n\\hbox{\n\\includegraphics[width=7.5cm]{fig_mars_scale4.png}\n\\includegraphics[width=7.5cm]{fig_mars_scale5.png}\n}}\n}\n\\caption{Mars topographic map and its UWTS (four wavelet detail scales and the scaling (smooth) band).}\n\\label{Figure:UWTS}\n\\index{data!Mars topography}\n\\end{figure}\n\nThe steps of the UWT on the sphere of a discrete image $X$ sampled from $f$ are summarized in Algorithm~\\ref{algo_uwts}. \nIf the wavelet function corresponds to the choice \\eqref{wavelet}, Step 3 in this UWTS algorithm reduces to $w_{j+1} = c_{j} - c_{j+1}$.\n\n% Their corresponding conjugate low pass and high pass filters $h$ and $g$ are plotted in Fig.~\\ref{fig_diff_uv_ht_gt}. \n\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{The Undecimated Wavelet Transform on the Sphere.}\n\\label{algo_uwts}\n\\noindent{\\bf Task:} Compute the UWTS of a discrete $X$.\\\\\n\\noindent{\\bf Parameters:} Data samples $X$ and number of of wavelet scales $J$.\\\\ \n\\noindent{\\bf Initialization:} \n\\begin{itemize}\n\\item $c_0=X$.\n\\item Compute the B$_3$-spline scaling function and derive $\\hat{\\psi}$, $\\widehat{H}$ and $\\widehat{G}$ numerically.\n\\item Compute the corresponding spherical harmonics transform of $c_0$.\n\\end{itemize}\n\\For{$j=0$ to $J-1$} {\n\\begin{enumerate}[1.]\n\\item Compute the spherical harmonics transform of the scaling coefficients: $\\hat{c}_{j+1}=\\hat{c}_j\\widehat{H}_{j}$.\n\\item Compute the inverse spherical harmonics transform of $\\hat{c}_{j+1}$ to get $c_{j+1}$.\n\\item Compute the spherical harmonics transform of the wavelet coefficients: $\\hat{w}_{j+1}=\\hat{c}_j\\widehat{G}_{j}$.\n\\item Compute the inverse spherical harmonics transform of $\\hat{w}_{j+1}$ to get $w_{j+1}$.\n\\end{enumerate}\n}\n\\noindent{\\bf Output:} ${\\cal W}=\\{w_1, w_2, \\dots, w_{J}, c_{J}\\}$ the UWTS of $X$.\n\\end{algorithm}\n}\n\nFig.~\\ref{Figure:UWTS} shows the Mars topographic map (top left)\n\\footnote{The Mars Orbiter Laser Altimeter (MOLA) generated altimetry profiles used to create global topographic maps. The MOLA instrument stopped acquiring altimetry data on June 30, 2001, and after that operated in passive radiometry mode until the end of the Mars Global Surveyor mission. MOLA data sets are produced by the MOLA Science Team and archived by the PDS Geosciences Node.} and its wavelet transform, using five scales (four wavelet scales + coarse scale). The sum of the five scales reproduces exactly the original image.\n\\index{Mars Orbiter Laser Altimeter}\n\n\\subsubsection{Inverse Transform}\n\nIf the wavelet is the difference between two resolutions, a straightforward reconstruction of an image from its wavelet coefficients ${\\cal W} = \\{w_1,\\dots, w_{J}, c_{J}\\}$ is: \n\n\\begin{eqnarray}\n c_{0}(\\theta, \\vartheta) = c_{J}(\\theta, \\vartheta) + \\sum_{j=1}^J w_j(\\theta, \\vartheta) .\n\\end{eqnarray}\nThis reconstruction formula is the same as with the 2D undecimated isotropic wavelet algorithm (i.e. with \\`a trous agorithm).\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n\\includegraphics[width = 15cm, height = 5cm]{fig_sphere_filterbank2.pdf} \n% \\includegraphics[width=14.5cm,height=5cm]{ch1_diff_uv_ht_gt.pdf}\n}}}\n\\caption{On the left, the filter $\\hat{\\tilde{h}}$, and on the right the \nfilter $\\hat{\\tilde{g}}$.}\n\\label{fig_diff_uv_ht_gt}\n\\end{figure}\n\nBut since the transform is redundant there is actually no unique way to reconstruct an image from its coefficients see~\\citep{starck:sta06}. Indeed, using the relations:\n\\begin{eqnarray}\n\\begin{split}\n\\hat c_{j+1}[l,m] = \\widehat H_{j} [l,m] \\hat c_{j} [l,m] \\\\\n\\hat w_{j+1}[l,m] = \\widehat G_{j} [l,m] \\hat c_{j} [l,m] \n\\end{split}\n\\end{eqnarray}\na least-squares estimate of $c_j$ from $c_{j+1}$ and $w_{j+1}$ gives:\n\\begin{eqnarray}\n\\hat{c}_{j} = \\hat{c}_{j+1} {\\widehat {\\tilde H}}_{j} + \\hat{w}_{j+1} {\\widehat {\\tilde G}}_{j} ~,\n\\end{eqnarray}\nwhere the dual filters $\\tilde h$ and $\\tilde g$ satisfy:\n\\begin{eqnarray}\n\\label{eqnht} \n\\begin{split}\n{\\widehat {\\tilde H}}_j = \\sqrt{\\frac{4\\pi}{2l+1} } {\\hat {\\tilde h}}_j & = {\\widehat H}_{j}^* /\n\\parenth{\\big|{\\widehat H}_{j}\\big|^2 + \\big|{\\widehat G}_j\\big|^2} \\\\\n{\\widehat {\\tilde G}}_j = \\sqrt{\\frac{4\\pi}{2l+1} } {\\hat {\\tilde g}}_j & = {\\widehat G}_{j}^* /\n\\parenth{\\big|{\\widehat H}_j\\big|^2 + \\big|{\\widehat G}_j\\big|^2} .\n\\end{split}\n\\end{eqnarray}\nFor the scaling function which is a B$_3$-spline function and a wavelet taken as the difference between two resolutions,\nthe corresponding conjugate low pass and high pass filters $\\widehat {\\tilde H}$ and $\\widehat {\\tilde G}$ are plotted in Fig.~\\ref{fig_diff_uv_ht_gt}. \nThe reconstruction algorithm is given in Algorithm~\\ref{algo_iuwts}.\n\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{Inverse UWT on the sphere.}\n\\label{algo_iuwts}\n\\noindent{\\bf Task:} Reconstruct an image from its UWTS coefficients.\\\\\n\\noindent{\\bf Parameters:} UWTS coefficients ${\\cal W}=\\{w_1, w_2, \\dots, w_{J}, c_{J}\\}$.\\\\\n\\noindent{\\bf Initialization:}\n\\begin{itemize}\n\\item Compute the B$_3$-spline scaling function and derive $\\hat{\\psi}$, $\\widehat{H}$, $\\widehat{G}$, $\\widehat{\\tilde H}$ and $\\widehat{\\tilde G}$ numerically.\n\\item Compute the spherical harmonics transform of $c_J$ to get ${\\hat c}_J$.\n\\end{itemize}\n\\For{$j=J-1$ to $0$, with step $=-1$} {\n\\begin{enumerate}[1.]\n\\item Compute the spherical harmonics transform of the wavelet coefficients $w_{j+1}$ to get $\\hat{w}_{j+1}$.\n\\item Multiply $\\hat{c}_{j+1}$ by ${\\widehat {\\tilde H}}_{j}$.\n\\item Multiply $\\hat{w}_{j+1}$ by ${\\widehat {\\tilde G}}_{j}$.\n\\item Get the spherical harmonics of $\\hat{c}_j=\\hat{c}_{j+1}+\\hat{w}_{j+1}$.\n\\end{enumerate}\n}\nCompute The inverse Spherical Harmonics transform of $\\hat c_0$.\\\\\n\\noindent{\\bf Output:} $c_0$ is the inverse UWT on the sphere.\n\\end{algorithm}\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\\includegraphics[width = 12cm,angle = 90]{fig_backwt_sphere.pdf}\n}}\n\\caption{Reconstruction from a single wavelet coefficient at different scales. Each map is obtained by setting all wavelet coefficients to zero but one, and by applying an inverse UWTS. Depending on the position and scale of the non-zero coefficient, the reconstructed map shows an isotropic feature \nat different scales and positions.}\n\\label{Figure:back_wt}\n\\end{figure}\n}\nFig.~\\ref{Figure:back_wt} shows the reconstruction by setting all wavelet coefficients but one at different scales and positions. \nDepending on the position and scale of the non-zero coefficient, the reconstructed map shows an isotropic feature at different scales and positions.\n \n\n%-------------------------\n\n\\subsection{Isotropic Pyramidal Wavelet Transform on the Sphere}\n\\index{wavelet!pyramidal transform}\n\\index{sphere!pyramidal wavelet}\n\n\\subsubsection{Forward Transform}\nIn the previous algorithm, no down-sampling is performed and each scale of the wavelet decomposition has the same number of pixels \nas the original data set. Therefore the number of pixels in the decomposition is equal to the number of pixels in the data multiplied \nby the number of scales. For some applications, we may prefer to introduce a decimation in the decomposition so as to reduce the \nrequired memory size and the computation time. This can be done easily by using a specific property of the chosen scaling function.\nIndeed, since we are considering here a scaling function with an initial cut-off $l_c$ in spherical harmonic multipole number $l$, \nand since the actual cut-off is reduced by a factor of two at each step, the number of significant spherical harmonics coefficients \nis then reduced by a factor of four after each convolution with the low pass filter $h$. Therefore, we need less pixels in the direct \nspace when we compute the inverse spherical harmonics transform. Using the HEALPix pixelization scheme \\citep{pixel:healpix}, \nthis can be done easily by dividing by 2 the $N_{\\mathrm{side}}$ parameter when calling the inverse spherical harmonics transform routine.\nThe pyramidal wavelet transform on the sphere algorithm is given in Algorithm~\\ref{algo_pwts}.\n\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{Pyramidal wavelet transform on the sphere.}\n\\label{algo_pwts}\n\\noindent{\\bf Task:} Compute the pyramidal WT on the sphere of a discrete image $X$.\\\\\n\\noindent{\\bf Parameters:} Data $X$ and number of of wavelet scales $J$.\\\\\n\\noindent{\\bf Initialization:} \n\\begin{itemize}\n\\item $c_0=X$.\n\\item Compute the B$_3$-spline scaling function and derive $\\hat{\\psi}$, $\\widehat{H}$ and $\\widehat{G}$ numerically.\n\\item Compute the corresponding spherical harmonics transform of $c_0$.\n\\end{itemize}\n\\For{$j=0$ to $J-1$} {\n\\begin{enumerate}[1.]\n\\item Compute the spherical harmonics transform of the scaling coefficients: $\\hat{c}_{j+1}=\\hat{c}_j\\widehat{H}_{j}$.\n\\item Compute the inverse spherical harmonics transform of $\\hat{c}_{j+1}$ to get $c_{j+1}$.\n\\item Down-sample $c_{j+1}$, since its support in the spherical harmonic domain has been divided by two.\n\\item Compute the spherical harmonics transform of the wavelet coefficients: $\\hat{w}_{j+1}=\\hat{c}_j\\widehat{G}_{j}$.\n\\item Compute the inverse spherical harmonics transform of $\\hat{w}_{j+1}$ to get $w_{j+1}$.\n\\end{enumerate}\n}\n\\noindent{\\bf Output:} ${\\cal W}=\\{w_1, w_2, \\dots, w_{J}, c_{J}\\}$ the Pyramidal WT on sphere of $X$.\n\\end{algorithm}\n}\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm]{fig_earth_elevation.png}%[width=6.5cm,height=3.9cm]\n \\includegraphics[width=7.5cm]{fig_earth_scale_1.png}\n}}\n\\centerline{\n\\hbox{\n \\includegraphics[width=7.5cm]{fig_earth_scale_2.png}\n \\includegraphics[width=7.5cm]{fig_earth_scale_3.png}\n}}\n\\centerline{\n\\hbox{\n\\includegraphics[width=7.5cm]{fig_earth_scale_4.png}\n\\includegraphics[width=7.5cm]{fig_earth_scale_5.png}\n}}\n}\n\\caption{Pyramidal wavelet transform on the sphere.}.\n\\label{Figure:PWTS}\n\\end{figure}\n\nFig.~\\ref{Figure:PWTS} shows an Earth image and its pyramidal wavelet transform (PWTS) using five scales.\nAs the scale number increases (i.e.\\ the resolution decreases), the pixel size becomes larger. The data are \nland and sea-floor elevations obtained from the ETOPO5 5-minute gridded elevation data set. A thorough explanation \nof the data set is provided at \\texttt{www.ngdc.noaa.gov} \\footnote{The ETOPO5 data are credited to ``Data Announcement 88-MGG-02, \nDigital relief of the Surface of the Earth. NOAA, National Geophysical Data Center, Boulder, Colorado, 1988''. The HEALPix image \nis available at \\texttt{http://astro.ic.ac.uk/$\\sim$pdineen/earth/index.html\\#earthmap}.}.\n\n\n%\\newpage\n\\subsubsection{Inverse Transform}\n \nThis reconstruction is not as straightforward as in the undecimated case, since the different scales do not have the same resolution.\nFor each resolution level, we have to up-sample the scaling band before co-adding it to the wavelet coefficients. Algorithm \\ref{algo_ipwts} describes this.\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{Inverse Pyramidal Wavelet Transform on the sphere.}\n\\label{algo_ipwts}\n{\\bf Task:} Reconstruct an image from its pyramidal WT on the sphere.\\\\\n\\noindent{\\bf Parameters:} Pyramidal WT coefficients ${\\cal W}=\\{w_1, w_2, \\dots, w_{J}, c_{J}\\}$.\\\\\n{\\bf Initialization:}\n\\begin{itemize}\n\\item Compute the B$_3$-spline scaling function and derive $\\hat{\\psi}$, $\\widehat{H}$, $\\widehat{G}$, $\\widehat{\\tilde H}$ and $\\widehat{\\tilde G}$ numerically.\n\\item Compute the spherical harmonics transform of $c_J$ to get ${\\hat c}_J$.\n\\end{itemize}\n\\For{$j=J-1$ to $0$} {\n\\begin{enumerate}[1.]\n\\item Upsample $c_{j+1}$ to the resolution of $c_j$.\n\\item Compute the spherical harmonics transform of the wavelet coefficients $w_{j+1}$ to get $\\hat{w}_{j+1}$.\n\\item Multiply $\\hat{c}_{j+1}$ by ${\\widehat {\\tilde H}}_{j}$.\n\\item Multiply $\\hat{w}_{j+1}$ by ${\\widehat {\\tilde G}}_{j}$.\n\\item Get the spherical harmonics of $\\hat{c}_j=\\hat{c}_{j+1}+\\hat{w}_{j+1}$.\n\\end{enumerate}\n}\nCompute The inverse spherical harmonics transform of $\\hat c_0$.\\\\\n\\noindent{\\bf Output:} $c_0$ is the inverse pyramidal WT on the sphere.\n\\end{algorithm}\n}\n \nThe wavelet transform on the sphere and its pyramidal version have both a reconstruction operator, \nso they are very well designed for any restoration application when the data contains isotropic features. \nIn the following, we present other transforms on the sphere more adapted to the analysis of anisotropic features.\n\n\n\n\n\n\\section{Ridgelet and Curvelet Transform on the Sphere (CTS) }\n\\label{sect_cur}\n\\subsection{Introduction.}\n\\index{curvelet}\n\\index{curvelet!sphere}\n\\index{sphere!curvelet}\n\nThe 2D curvelet transform, proposed in \\citep{cur:donoho99,starck:sta01_3,starck:sta02_3}, enables the directional analysis of an image \nin different scales. The fundamental property of the curvelet transform is to analyze the data with functions of length about $2^{-j/2}$ \nfor the $j^{\\textrm{th}}$ sub-band $[2^j, 2^{j+1}]$ of the two dimensional wavelet transform. Following the implementation described \nin \\citep{starck:sta01_3,starck:sta02_3}, the data first undergoes an Isotropic Undecimated Wavelet Transform (i.e. \\og{}\\`a trous \\fg{} algorithm). \nEach scale $j$ is then decomposed into smoothly overlapping blocks of side-length $B_j$ pixels in such a way that the overlap between two\nvertically adjacent blocks is a rectangular array of size $B_j \\times B_j/2$. And finally, the ridgelet transform \\citep{cur:candes99_1} is \napplied on each individual block. Recall that the ridgelet transform precisely amounts to applying a 1-dimensional wavelet transform to the\nslices of the Radon transform. More details on the implementation of the digital curvelet transform can be found in \\citep{starck:sta01_3,starck:sta02_3}.\nIt has been shown that the curvelet transform could be very useful for the detection and the discrimination of non-Gaussianity in CMB \\citep{starck:sta02_4}.\nThe curvelet transform is also redundant, with a redundancy factor of $16J+1$ whenever $J$ scales are employed. Its complexity scales like \nthat of the ridgelet transform that is as $O(n^2 \\log_2n)$. The curvelet transform was shown to sparsely represent anisotropic structures and smooth curves and edges of different lengths.\n\n\\subsection{Ridgelets and Curvelets on the Sphere.}\n\\index{curvelet transform}\nThe Curvelet transform on the sphere (CTS) can be similar to the 2D digital curvelet transform, but replacing the \\og{}\\`a trous \\fg{} algorithm \nby the Isotropic Wavelet Transform on the Sphere previously described. The CTS algorithm consists in the following three steps which \nwe describe in more details next.\n\\begin{itemize}\n\\item {\\it Isotropic Wavelet Transform on the Sphere.} \n\\item {\\it Partitioning.} Each scale is decomposed into blocks of an appropriate scale (of side-length $\\sim2^{-s}$),using the HEALPix pixelization.\n\\item {\\it Ridgelet Analysis.} Each square is analyzed via the discrete ridgelet transform.\n\\end{itemize}\nWe now describe these three steps.\n\n\\subsubsection{Partitioning Using the HEALPix Representation}\n\n \\begin{figure}\n\\centerline{\n\\hbox{\n\\includegraphics[width=15cm]{fig_flowgraph_ridgelet_sphere.pdf}\n% \\includegraphics[width=13.6cm,height=8cm]{fig_flowgraph_ridgelet_sphere.pdf}\n}}\n\\caption{Flowgraph of the ridgelet transform on the sphere.}\n\\label{Figure:rid_sphere}\n\\end{figure}\n\nThe HEALPix representation is a curvilinear hierarchical partition of the sphere into quadrilateral pixels \nof exactly equal area but with varying shape. The base resolution divides the sphere into 12 quadrilateral \nfaces of equal area placed on three rings around the poles and equator. Each face is subsequently divided \ninto $N_{\\mathrm{side}}^{2}$ pixels following a quadrilateral multiscale tree structure (see Fig.~\\ref{pixelhealpix}). \nThe pixel centers are located on iso-latitude rings, and pixels from the same ring are equispaced in azimuth. \nThis is critical for computational speed of all operations involving the evaluation of spherical harmonics transforms, \nincluding standard numerical analysis operations such as convolution, and power spectrum estimation. \n\nAn important geometrical feature of the HEALPix sampling grid is the hierarchical quadrilateral tree structure. \nThis defines a natural one-to-one mapping of the sphere sampled according to the HEALPix grid, into twelve \nflat images, on all scales. It is then easy to partition a spherical map using HEALPix into quadrilateral blocks \nof a specified size. One first extracts the twelve base-resolution faces, and each face is then decomposed into \noverlapping blocks of the specified size. This decomposition into blocks is an essential step of the traditional \nflat 2D curvelet transform. Based on the reversible warping of the sphere into a set of flat images made possible \nby the HEALPix sampling grid, the ridgelet and curvelet transforms can be extended to the sphere. \n\nWith the decomposition into blocks described above, there is no overlap between neighboring blocks belonging \nto different base-resolution faces. This may result for instance in blocking effects in denoising experiments \nvia nonlinear filtering. It is possible to overcome this difficulty in some sense by working simultaneously \nwith various rotations of the data with respect to the sampling grid. This will average out undesirable effects \nat edges between base resolution faces. \n\n\\subsubsection*{Ridgelet transform}\n\\index{ridgelet}\n\\index{ridgelet!sphere}\n\\index{sphere!ridgelet}\n\\index{Radon transform}\n\nOnce the partitioning is performed, the standard 2D ridgelet transform described in \\citep{starck:sta02_3} is applied in each individual block :\n\\begin{enumerate}\n\\item Compute the 2D Fourier transform.\n\\item Extract lines going through the origin in the frequency plane.\n\\item Compute the 1D inverse Fourier transform of each line. We get the Radon transform.\n\\item Compute the 1D wavelet transform of the lines of the Radon transform.\n\\end{enumerate}\nThe first three steps correspond to a Radon transform method called the {\\it linogram}. Other implementations of the Radon transform, such as \nthe {\\it Slant Stack Radon Transform} \\citep{cur:donoho_02}, can be used as well, as long as they offer an exact reconstruction.\n \nFig.~\\ref{Figure:rid_sphere} shows the flowgraph of the ridgelet transform \non the sphere and Fig.~\\ref{Figure:back_rid} shows the reconstruction from a single ridgelet \ncoefficient at different scales and orientations.\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{\n\\includegraphics[width=15cm]{fig_ridssr.pdf}\n% \\includegraphics[width=9.5cm,height=6cm]{fig_ridssr.pdf}\n}}\n\\caption{Ridgelet atoms on the sphere obtained by reconstruction from a few ridgelet coefficient at different scales and orientations.}\n\\label{Figure:back_rid}\n\\end{figure}\n \n \\subsection{Curvelet Transform Algorithm}\n\nThe curvelet transform algorithm on the sphere is described in \nAlgorithm \\ref{algo_curts}.\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{Curvelet Transform on the sphere.}\n\\label{algo_curts}\n{\\bf Task:} Compute the curvelet transform on the sphere of a discrete image $X$.\\\\\n{\\bf Parameters:} Image $X$ and number of scales $J$.\\\\\n{\\bf Initialization:} \n\\begin{itemize}\n\\item $B_1 = B_{\\min}$.\n\\item Compute the isotropic UWTS of $X$ with $J$ scales, get $\\{w_1,\\dots,w_J,c_J\\}$.\n\\end{itemize}\n\\For{$j=0$ to $J-2$} {\n\\begin{enumerate}[1.]\n\\item Partition the wavelet subband $w_j$ with a block size $B_j$.\n\\item Apply the digital ridgelet transform to each block; get the curvelet coefficients at scale $j$.\n\\end{enumerate}\n\\lIf{$j \\mbox{ modulo } 2 = 1$} $B_{j+1} = 2 B_{j}$, else $B_{j+1} = B_{j}$.\n}\n{\\bf Output:} The curvelet transform on the sphere of $X$.\n\\end{algorithm}\n}\n\nThe sidelength of the localizing windows is doubled {\\em at every\nother} dyadic subband, hence maintaining the fundamental property of\nthe curvelet transform which says that elements of length about\n$2^{-j/2}$ serve for the analysis and synthesis of the $j^{\\textrm{th}}$ subband\n$[2^j, 2^{j+1}]$. We used the default value $B_{\\min} = 16$\npixels in our implementation. Fig.~\\ref{Figure:cur_sphere}\ngives an overview of the organization of the algorithm.\n\n\\begin{figure}[htb]\n\\vbox{\n\\centerline{\n\\hbox{\n% \\includegraphics[width=9.cm,height=12cm]{fig_flowgraph_curvelet_sphere.pdf}\n\\includegraphics[width=15cm]{fig_flowgraph_curvelet_sphere.pdf}\n}}}\n\\caption{Flowgraph of the curvelet transform on the sphere.}\n\\label{Figure:cur_sphere}\n\\end{figure}\n\n\\begin{figure}\n\\vbox{\n\\centerline{\n\\hbox{\n% \\includegraphics[width=9.cm,height=12cm]{fig_back_cur_sphere.pdf}\n\\includegraphics[angle=90,width=\\textwidth]{fig_back_cur_sphere.pdf}\n}}}\n\\caption{Reconstruction from a single curvelet coefficient at different scales and orientations.}\n\\label{Figure:back_cur}\n\\end{figure}\nFig.~\\ref{Figure:back_cur} shows the backprojection of curvelet coefficients at \ndifferent scales and orientations.\n\n\\subsection{Pyramidal Curvelet Transform on the Sphere (PCTS)}\nThe CTS is very redundant, which may be a problem for handling huge data sets such as \nPlanck data (see Section \\ref{section:cmb} below). \nThe redundancy can be reduced by substituting, in the \ncurvelet transform algorithm, the pyramidal wavelet transform with the undecimated wavelet transform.\nThe second step which consists of applying the ridgelet transform on the wavelet scale is unchanged.\nThe pyramidal curvelet transform (PCTS) algorithm is summarized in Algorithm~\\ref{algo_pcurts}.\n\\index{sphere!pyramidal curvelet}\n\n{\\linespread{1}\n\\begin{algorithm}[h]\n\\caption{Pyramidal Curvelet Transform on the sphere.}\n\\label{algo_pcurts}\n{\\bf Task:} Compute the pyramidal curvelet transform on the sphere of a discrete image $X$.\\\\\n{\\bf Parameters:} Image $X$ and number of scales $J$.\\\\\n{\\bf Initialization:} \n\\begin{itemize}\n\\item $B_1 = B_{\\min}$.\n\\item Compute the pyramidal wavelet transform of $X$ with $J$ scales, get $\\{w_1,\\dots,w_J,c_J\\}$.\n\\end{itemize}\n\\For{$j=0$ to $J-2$} {\n\\begin{enumerate}[1.]\n\\item Partition the wavelet subband $w_j$ with a block size $B_j$.\n\\item Apply the digital ridgelet transform to each block; get the curvelet coefficients at scale $j$.\n\\end{enumerate}\n\\lIf{$j \\mbox{ modulo } 2 = 1$} $B_{j+1} = 2 B_{j}$, else $B_{j+1} = B_{j}$.\n}\n{\\bf Output:} The pyramidal curvelet transform on the sphere of $X$.\n\\end{algorithm}\n}\n\n% In the next section, it is shown how the pyramidal curvelet transform can be used for image filtering.\n ", "meta": {"hexsha": "61d3664c088c8a70b7996b58c64e1f9032cef647", "size": 44313, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_isap/multiscale.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_isap/multiscale.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_isap/multiscale.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.4129692833, "max_line_length": 536, "alphanum_fraction": 0.7289057387, "num_tokens": 13990, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438950986284992, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.6271558495263148}} {"text": "%% -*- coding:utf-8 -*-\n\\chapter{Generalities on algebraic extensions}\nWe introduce the basic notions such as a field extension, algebraic\nelement, minimal polynomial, finite extension, and study their very\nbasic properties such as the multiplicativity of degree in towers. \n\n\\section{Field extensions: examples}\n\n\\subsection{K-algebra}\n\\begin{definition}[K-algebra]\n \\label{def:kalgebra}\n Let $K$ is a \\mynameref{def:ring} and $A$ is a\n \\mynameref{def:module} over $K$ \n \\footnote{\n The definition in the lectures is more common and assumes that $K$\n is a ring but later $K$ will be mostly a \\mynameref{def:field}. Therefore\n $A$ has to be a \\mynameref{def:vectorspace} over $K$ in the case. \n }\n equipped with \n an additional binary operation $A \\times A \\rightarrow A$ that we\n denote as $\\cdot$ here. The $A$ is an algebra over $K$ if the\n following identities hold $\\forall x,y,z \\in A$ and for every\n elements (often called as scalar) $a, b \\in K$\n \\begin{itemize}\n \\item Right distributivity:\n $(x + y) \\cdot z = x \\cdot z + y \\cdot z$\n \\item Left distributivity:\n $z \\cdot (x + y) = z \\cdot x + z \\cdot y$\n \\item Compatibility with scalars:\n $(ax) \\cdot (by) = (ab) (x \\cdot y)$\n \\end{itemize}\n\\end{definition}\n\n\\begin{example}[Field of complex numbers $\\mathbb{C}$]\n The field of complex numbers $\\mathbb{C}$ can be considered as a\n K-algebra over the field of real numbers $\\mathbb{R}$.\n \\label{ex:complexnumbers}\n\\end{example}\n\n\\subsection{Field extension}\n\n\\begin{definition}[Field extension]\n Let $K$ and $L$ are fields.\n $L$ is an extension of $K$ if $L \\supset K$\n \\label{def:fextension1}\n\\end{definition}\nand another definition\n\\begin{definition}[Field extension]\n Let $K$ is a field then\n $L$ is an extension of $K$ if $L$ is a K-algebra\n \\footnote{\n $L$ in the definition is not the same object with $L$ from\n definition \\ref{def:fextension1}. Because $L$ in the definition is\n a $K$-algebra i.e. a ring but $L$ in the definition\n \\ref{def:fextension1} is a field. \n }\n \\label{def:fextension2}\n\\end{definition}\nWhy the 2 definitions are equivalent?\n\n\\begin{lemma}[K-algebra and homomorphism]\n Given a K-algebra is the same as\n having \\mynameref{def:homomorphism} $f: K \\rightarrow A$ of rings.\n \\begin{proof}\n Really if I have a K-algebra I can define the\n \\mynameref{def:homomorphism} $f(k) = k \\cdot 1_A$, where $1_A$ is an\n identity element of $A$. Thus $k \\cdot 1_A \\in A$.\n\n And conversely if I have the \\mynameref{def:homomorphism}\n $f: K \\rightarrow A$\n I can define the K-algebra structure by setting\n $k a = f(k) a$ because $f(k), a \\in A$ and there is a multiplication\n defined on $A$. As result I have a rule for multiplication a vector\n ($f(k) \\in A$) on a vector ($a \\in A$) that is required for\n \\mynameref{def:kalgebra} setup.\n \\end{proof}\n \\label{lem:lec1_homkalgebra}\n\\end{lemma}\n\n\\begin{lemma}[About homomorphism of fields]\n Any \\mynameref{def:fieldhomomorphism} is\n \\mynameref{def:injection}.\n \\footnote{\n But the statement is not valid for groups. In that case the\n homomorphism is injection if and only if the \n kernel is trivial and consists of only one element - identity.\n See theorem \\ref{thm:grouphomomorphsim}.\n }\n \\label{lem:lec1_homomorphism_is_injection}\n \\begin{proof}\n Lets proof by contradiction.\n Really if $f(x) = f(y)$ and $x \\ne y$ then\n \\begin{eqnarray}\n f(x) - f(y) = 0_A,\n \\nonumber \\\\\n f(x - y) = 0_A,\n \\nonumber \\\\\n f(x - y) f(\\left(x - y\\right)^{-1}) =\n f\\left(\\frac{x - y}{x - y}\\right) = f(1_K) = 1_A = 0_A\n \\nonumber\n \\end{eqnarray}\n that is impossible.\n \\end{proof} \n\\end{lemma}\n\nThere are some comments on the results. We have got that a\n\\mynameref{def:homomorphism} can be set between field $K$ and its\nK-algebra.\nThe \\mynameref{def:homomorphism} is \\mynameref{def:injection}\ntherefore we can allocate a sub-field $A' \\subset A$ for that we will\nhave the \\mynameref{def:homomorphism} is a \\mynameref{def:surjection} and\ntherefore we have an \\mynameref{def:isomorphism} between original field $K$\nand a sub-field $A'$. This means that we can say that the original\nfield $K$ is a sub-field for the K-algebra.\n\n\\begin{example}[Field extensions]\n $\\mathbb{C}$ is a field extension for $\\mathbb{R}$.\n $\\mathbb{R}$ is a field extension for $\\mathbb{Q}$\n \\label{ex:fieldextension}\n\\end{example}\n\n\\begin{example}[$K$-algebra is not a field]\n In the example\n \\footnote{the example was not present in the lectures}\n I will show\n that $K$ algebra is not a field. Consider\n $K = \\mathbb{R}$. \\mynameref{def:vectorspace} $A = \\mathbb{R}^2$ i.e.\n $A$ consists of vectors of the following form\n \\[\n x = \\left(\n \\begin{array}{c}\n x_1 \\\\\n x_2\n \\end{array}\n \\right),\n \\]\n where $x_1, x_2 \\in \\mathbb{R}$.\n I will define the multiplication for $L$ (our $K$ algebra) as\n follows\n \\[\n \\left(\n \\begin{array}{c}\n x_1 \\\\\n x_2\n \\end{array}\n \\right)\n \\cdot\n \\left(\n \\begin{array}{c}\n y_1 \\\\\n y_2\n \\end{array}\n \\right) =\n \\left(\n \\begin{array}{c}\n x_1 \\cdot y_1 \\\\\n x_2 \\cdot y_2\n \\end{array}\n \\right)\n \\]\n\n It can be seen that all requirements of \\mynameref{def:kalgebra} are\n satisfied\n \\begin{eqnarray}\n (x + y) \\cdot z =\n \\left(\n \\left(\n \\begin{array}{c}\n x_1 \\\\\n x_2\n \\end{array}\n \\right)\n +\n \\left(\n \\begin{array}{c}\n y_1 \\\\\n y_2\n \\end{array}\n \\right)\n \\right)\n \\left(\n \\begin{array}{c}\n z_1 \\\\\n z_2\n \\end{array}\n \\right) =\n \\nonumber \\\\\n =\n \\left(\n \\begin{array}{c}\n (x_1 + y_1)z_1 \\\\\n (x_2 + y_2)z_2\n \\end{array}\n \\right) =\n \\left(\n \\begin{array}{c}\n x_1 z_1 + y_1 z_1 \\\\\n x_2 z_2 + y_2 z_2\n \\end{array}\n \\right) =\n x \\cdot z + y \\cdot z\n \\nonumber \\\\\n z \\cdot (x + y) = z \\cdot x + z \\cdot y\n \\nonumber \\\\\n (ax) \\cdot (by) =\n \\left(\n \\begin{array}{c}\n a x_1 \\\\\n a x_2\n \\end{array}\n \\right)\n \\left(\n \\begin{array}{c}\n b y_1 \\\\\n b y_2\n \\end{array}\n \\right) =\n \\left(\n \\begin{array}{c}\n ab x_1 y_1 \\\\\n ab x_2 y_2\n \\end{array}\n \\right)\n (ab) (x \\cdot y)\n \\nonumber\n \\end{eqnarray}\n The multiplication identity element of $L$ is the following\n \\[\n 1_L = \n \\left(\n \\begin{array}{c}\n 1 \\\\\n 1\n \\end{array}\n \\right)\n \\]\n The zero is the standard one from vector space\n \\[\n 0_L = \n \\left(\n \\begin{array}{c}\n 0 \\\\\n 0\n \\end{array}\n \\right)\n \\]\n\n We can see that\n \\[\n \\left(\n \\begin{array}{c}\n 1 \\\\\n 0\n \\end{array}\n \\right)\n \\left(\n \\begin{array}{c}\n 0 \\\\\n 1\n \\end{array}\n \\right) = 0_L\n \\]\n i.e. we have 2 divisor of zero which are not zero itself. The\n elements do not have invert ones and as result the $L$ is\n not a field.\n\n From other side if we define $L' \\subset L$ as follows\n $L' = \\left\\{\n \\left(\n \\begin{array}{c}\n r \\\\\n r\n \\end{array}\n \\right)\n \\right\\}$, where $r \\in \\mathbb{R}$,\n then we will have that $L'$ is a field and $L' \\cong \\mathbb{R}$.\n \\label{ex:kalgebranotfield}\n\\end{example}\n\n\\begin{definition}[Characteristic]\n \\label{def:fieldcharacteristic}\n If $L$ is a field there are 2 possibilities\n \\begin{enumerate}\n \\item $1 + 1 + \\dots \\ne 0$. In this case\n $\\mathbb{Z} \\subset L$ but $\\mathbb{Z}$ is not a field therefore $L$\n is an extension of $\\mathbb{Q}$. In the case $char L = 0$\n \\item $1 + 1 + \\dots + 1 = \\sum_{i = 1}^m 1 = 0$ for some $m \\in\n \\mathbb{Z}$. The first time when it happens is for a prime number\n i.e. minimal $m$ with the property is prime. In this case $char L\n = p$, where $p = \\min m$ - the minimal $m$ (prime) with the\n property. In this \n case $\\mathbb{Z}/p\\mathbb{Z} \\subset L$. The\n $\\mathbb{Z}/p\\mathbb{Z}$ is a field denoted by $\\mathbb{F}_p$. The\n $L$ is an extension of $\\mathbb{F}_p$.\n \\end{enumerate}\n No other possibilities exist. The $\\mathbb{Q}$ and $\\mathbb{F}_p$ are\n the prime fields. Any field is an extension of one of those.\n\\end{definition}\n\n\\begin{definition}[Prime field]\n \\label{def:primefield}\n If $L$ is a field $\\mathbb{Q} \\subset L$, i.e. $char L = 0$ then\n $\\mathbb{Q}$ is the prime field for $L$. Otherwise, if $char L = p > 0$\n then $\\mathbb{F}_p$ is the prime field.\n\\end{definition}\n\n\\begin{claim}\n\\label{claim:lec1_sec14}\nLet $K\\left[X\\right]$ \\mynameref{def:ring} of polynomials.\nThe $P \\in K\\left[X\\right]$ is an\n\\mynameref{def:irreducible}. $\\left(P\\right)$ is an \n\\mynameref{def:ideal} formed by the polynomial.\n\\footnote{\n I.e.\n \\(\n \\left(P\\right) = \\left\\{ Q = G P \\right\\} \n \\)\n where $G \\in K\\left[X\\right]$\n}\nThe set of residues by\nthe polynomial forms a field that denoted by\n$K\\left[X\\right]/\\left(P\\right)$.\n\\begin{proof}\nHow we can see it?\nIf $Q \\in K\\left[X\\right]$ is a polynomial that $Q \\notin\n\\left(P\\right)$ when $Q$ is prime to $P$.\n\\footnote{\n As soon as $P$ is irreducible in $K\\left[X\\right]$ then there is only\n one possibility for $Q$ and $P$ to have common divisors: if $Q = G P$\n where $G \\in K\\left[X\\right]$ but this is in contradiction with\n $Q \\notin \\left(P\\right)$\n}\nThen with\n\\mynameref{lem:bezout} we can get $\\exists A, B \\in K\\left[X\\right]$\nsuch that\n\\[\nA P + B Q = 1,\n\\]\nor\n\\[\nB Q \\equiv 1 \\mod P,\n\\]\nthus $B$ is $Q^{-1}$ in $K\\left[X\\right]/\\left(P\\right)$.\n\\end{proof}\n\\end{claim}\n\n\\begin{claim}\n \\footnote{\n The claim is not a part of the lectures but it's very usefully in\n future lectures.\n }\n Let $K$ is a field and $a \\in K$ then $K\\left[X\\right]/\\left(X -\n a\\right)$ is also a field and there exists an\n \\mynameref{def:isomorphism} between the field and $K$ i.e.\n \\begin{equation}\n K\\left[X\\right]/\\left(X - a\\right) \\cong K\n \\label{eq:lec1_ex1_1}\n \\end{equation}\n \\label{claim:lec1_fieldquotionisomorphism}\n \\begin{proof}\n The $K\\left[X\\right]/\\left(X - a\\right)$ is a field just because\n $X-a$ is\n \\mynameref{def:irreducible} (see example \\ref{ex:irreducible} and\n claim \\ref{claim:lec1_sec14}).\n\n For the proof the main statement (\\ref{eq:lec1_ex1_1}) lets consider a\n polynomial $P \\in K\\left[X\\right]$ and \n define the following \\mynameref{def:fieldhomomorphism}:\n \\begin{equation}\n \\phi: K\\left[X\\right]/\\left(X - a\\right)\n \\xrightarrow[P(X) \\to P(a)]{} K\n \\label{eq:lec1_ex1_2}\n \\end{equation}\n\n The $\\phi$ defined by (\\ref{eq:lec1_ex1_2}) is\n \\mynameref{def:fieldhomomorphism}. For the proof of the claim lets take\n $P_1, P_2 \\in K\\left[X\\right]/\\left(X - a\\right)$. Clear that\n \\[\n \\phi\\left(P_1 + P_2\\right) = P_1\\left(a\\right) + P_2\\left(a\\right)\n = \\phi\\left(P_1\\right) + \\phi\\left(P_1\\right).\n \\]\n The same holds with the multiplication:\n \\[\n \\phi\\left(P_1 \\cdot P_2\\right) = P_1\\left(a\\right) \\cdot P_2\\left(a\\right)\n = \\phi\\left(P_1\\right) \\cdot \\phi\\left(P_1\\right).\n \\]\n %% For the inverse element image lets consider $P \\in\n %% K\\left[X\\right]/\\left(X - a\\right)$ such that $P \\ne 0$ and as\n %% result $\\phi\\left(P\\right) = P\\left(a\\right) \\ne 0$.\n %% There exists $P^{-1} \\in K\\left[X\\right]/\\left(X - a\\right)$ as\n %% soon as $K\\left[X\\right]/\\left(X - a\\right)$ is a field. Therefore\n %% we have\n %% \\begin{eqnarray}\n %% \\phi(1) = \\phi\\left(P P^{-1}\\right) =\n %% \\phi\\left(P\\right)\\phi\\left(P^{-1}\\right) =\n %% \\nonumber \\\\\n %% = P\\left(a\\right) P^{-1}\\left(a\\right) = 1,\n %% \\nonumber\n %% \\end{eqnarray}\n %% i.e.\n %% \\[\n %% \\phi\\left(P^{-1}\\right) = \\phi\\left(P\\right)^{-1}.\n %% \\]\n That completes the proof of the \\mynameref{def:fieldhomomorphism}.\n %% Division is more complex but also can be\n %% shown: if $P_2 \\ne 0$ when there exists\n %% $P_2^{-1} \\in K\\left[X\\right]/\\left(X - a\\right)$ as soon as\n %% $K\\left[X\\right]/\\left(X - a\\right)$ is the field then\n %% with $\\phi\\left(P_2^{-1}\\right) = P_2^{-1}\\left(a\\right) =\n %% \\frac{1}{\\phi\\left(P_2\\right)}$ one can get\n %% \\[\n %% \\phi\\left(\\frac{P_1}{P_2}\\right) =\n %% \\phi\\left(P_1 P_2^{-1}\\right) =\n %% \\phi\\left(P_1\\right) \\phi\\left(P_2^{-1}\\right) =\n %% \\frac{\\phi\\left(P_1\\right)}{\\phi\\left(P_2\\right)}\n %% \\]\n\n %% We have $\\ker{\\phi} = \\left(X - a\\right)$ because for any polynomial\n %% $P$ that is in the ideal $\\left(X - a\\right)$ has $P(a) = 0$ i.e. in\n %% the kernel of $\\phi$.\n\n Lemma \\mynameref{lem:lec1_homomorphism_is_injection} says that any\n \\mynameref{def:fieldhomomorphism} is \\mynameref{def:injection},\n i.e. $\\phi$ is an injection.\n \n Next we should show that $\\phi$ is \\mynameref{def:surjection} it's\n easy because $\\forall k \\in K$ we can consider constant polynomial\n $P = k$ from $K\\left[X\\right]$. For the polynomial we will have\n $\\phi\\left(k\\right) = k$.\n\n As result $\\phi$ is \\mynameref{def:injection} as well as\n \\mynameref{def:surjection} i.e. $\\phi$ is\n \\mynameref{def:bijection}. With the fact that $\\phi$ is\n \\mynameref{def:homomorphism} it will gives us that $\\phi$ is\n \\mynameref{def:isomorphism}. \n %% Now (\\ref{eq:lec1_ex1_1}) follows from the\n %% \\mynameref{thm:firstisomorphism}. \n \\end{proof}\n\\end{claim}\n\n\\section{Algebraic elements. Minimal polynomial}\n\n\\subsection{$K\\left[X\\right]/\\left(P\\right)$ field}\n\nAlternative proof of claim \\ref{claim:lec1_sec14} i.e. the fact that\n$K\\left[X\\right]/\\left(P\\right)$ is the \n\\mynameref{def:field} as follows.\nThe $\\left(P\\right)$ is a \\mynameref{def:maxideal}\n\\footnote{\n To prove that $\\left(P\\right)$ is a \\mynameref{def:maxideal} we have\n to use \\mynameref{lem:bezout}.\n}\nbut a quotient by a \\mynameref{def:maxideal} is a \\mynameref{def:field}\n(see \\mynameref{thm:maxideal}).\n\nWe also can say that $K\\left[X\\right]/\\left(P\\right)$ is an extension\nof $K$ because it's \n\\mynameref{def:kalgebra}.\n\n\\begin{example}[$K = \\mathbb{F}_2/\\left(X^2+X+1\\right)$]\n Lets consider the following field\n $\\mathbb{F}_2 = \\mathbb{Z}/2\\mathbb{Z} = \\{0,1\\}$ in the\n field polynomial $X^2+X+1$ is irreducible. It's very easy to verify\n it because $\\mathbb{F}_2$ has only 2 elements that can be (possible)\n a root:\n \\[\n 0^2+0+1 = 1\\ne 0\n \\]\n and\n \\[\n 1^2+1+1 = 1\\ne 0\n \\]\n \n The polynomial has the following\n residues: $\\bar{X} = X + \\left(X^2+X+1\\right)$ and\n $\\overline{X + 1} = X + 1 + \\left(X^2+X+1\\right)$. Thus the field\n $\\mathbb{F}_2/\\left(X^2+X+1\\right)$ consists of 4 elements:\n $\\{0, 1, \\bar{X}, \\overline{X+1}\\}$.\n\n It's easy to see that the third element ($\\bar{X}$) is a root of\n $P(X) = X^2+X+1$:\n \\[\n \\bar{X}^2 + \\bar{X} + 1 =\n P(X) + \\left(P(X)\\right) = \\left(P(X)\\right) \\equiv 0 \\mod P.\n \\]\n \n \\[\n \\bar{X}^2 + \\bar{X} + 1 = \\bar{0},\n \\]\n therefore \n \\[\n \\bar{X}^2 = - \\bar{X} - 1 = \\bar{X} + 1 = \\overline{X+1}.\n \\]\n This is because we are in field $\\mathbb{F}_2$ where\n \\[\n 2 \\left(X + 1\\right) \\mod 2 = 0 \n \\]\n and thus\n \\[\n - \\bar{X} - 1 = \\bar{X} + 1 = \\overline{X + 1}\n \\]\n \n Also\n \\[\n \\overline{X+1}^2 = \\bar{X},\n \\]\n and they are inverse each other\n \\[\n \\overline{X+1} \\bar{X} = 1,\n \\]\n\n So this is the structure of a field of four elements. The\n cardinality of $K=\\mathbb{F}_2/\\left(X^2+X+1\\right)$ is 4, one\n writes then $K=F_4$. Well, this might be \n strange at the first sight, because we only know that $K$ has four\n elements and if you write $F_4$ you somehow mean that there is only\n one field of four elements. Well, it is true, there is only one\n field of four elements. In fact, all finite fields of the same\n cardinality are isomorphic, and we will see it very shortly\n (see theorem \\ref{thm:lec3_1}). \n \\label{ex:F2overP}\n\\end{example}\n\nArguments for the \\autoref{ex:F2overP} can also be found in the\nfollowing remark (not part of the lectures)\n\\begin{remark}[About quotient field]\nConsider an arbitrary field $L$ ($L=\\mathbb{F}_2$ at the\n\\autoref{ex:F2overP}) and $P\\left(X\\right) = X^n + a_{n-1}X^{n-1} +\na_{n-2} X^{n-2} + \\dots + a_0 \\in L\\left[X\\right]$ is an\nirreducible polynomial over $L$. Let $\\alpha$ is a root of the\npolynomial then \n\\[\n\\alpha^n + a_{n-1} \\alpha^{n-1} + \\dots + a_0 = 0,\n\\]\nor\n\\[\n\\alpha^n = - a_{n-1} \\alpha^{n-1} - \\dots - a_0.\n\\]\nTherefore $\\forall l \\ge n, \\exists l_i \\in L$ such that \n\\[\n\\alpha^k = \\sum_{i=0}^{n-1}l_i \\alpha^{i}\n\\]\nand as result $\\forall x \\in L/\\left(P\\right)$ exists $x_i \\in L$\nsuch that \n\\[\nx = \\sum_{i=0}^{n-1}x_i \\alpha^{i}.\n\\]\nWe can say that we have the following basis: $\\{\\{L\\}, \\alpha, \\dots,\n\\alpha^{n-1}\\}$ where $\\{L\\}$ - L basis. For \\autoref{ex:F2overP} we\nhave\n$\\{L\\}= \\{\\mathbb{F}_2\\}= \\{0,1\\}$, $\\alpha = \\bar{X}$ and as result\n$K = \\mathbb{F}_2/\\left(X^2+X+1\\right)$ will have the following basis\n$\\{0,1,\\bar{X}\\}$.\n\\end{remark}\n\n\\subsection{Algebraic elements}\n\n\\begin{definition}[Algebraic element]\n Let $K \\subset L$ and $\\alpha \\in L$. $\\alpha$ is an algebraic\n element if $\\exists P \\in K\\left[X\\right]$ such that\n $P\\left(\\alpha\\right) = 0$. Otherwise the $\\alpha$ is called\n transcendental.\n \\label{def:algebraicelement}\n\\end{definition}\n\n\\subsection{Minimal polynomial}\n\n\\begin{lemma}[About minimal polynomial existence]\n If $\\alpha$ is \\mynameref{def:algebraicelement} then\n $\\exists!$ unitary polynomial $P$ of minimal degree such that\n $P\\left(\\alpha\\right) = 0$. It is irreducible. $\\forall Q$ such that\n $Q\\left(\\alpha\\right) = 0$ is divisible by $P$\n \\footnote{\n see also \\mynameref{thm:irreduciblediv}\n }\n \\begin{proof}\n We know that $K\\left[X\\right]$ is a \\mynameref{def:pid} and a\n polynomial $Q\\left(\\alpha\\right) = 0$ forms an\n \\mynameref{def:ideal}: $I \\left\\{Q \\in K\\left[X\\right] \\mid\n Q\\left(\\alpha\\right) = 0 \\right\\}$, so the ideal is generated by\n one element: $I = \\left(P\\right)$. This is an unique (up to\n constant) polynomial minimal degree in $I$.\n \n Lets prove that $P$ is irreducible. If $P$ is not irreducible then\n $\\exists Q,R \\in I$ such that $P = Q \n R$, $Q(\\alpha) = 0$ or $R(\\alpha) = 0$ and\n $deg R,Q < deg P$ that is in contradiction with the definition\n that $P$ is a polynomial of minimal degree. \n \\end{proof}\n \\label{lem:minpolynomial}\n\\end{lemma}\n\n\\begin{definition}[Minimal polynomial]\n If $\\alpha$ is \\mynameref{def:algebraicelement} then\n the unitary polynomial $P$ of minimal degree such that\n $P\\left(\\alpha\\right) = 0$ is called minimal polynomial and denoted\n by $P_{min}\\left(\\alpha, K\\right)$.\n \\label{def:minpolynomial}\n\\end{definition}\n\n\\begin{gapexample}[Minimal polynomial]\n Consider the following minimal polynomial\n $P_{min}\\left(\\alpha, \\mathbb{Q}\\right)$, where $\\alpha =\n \\sqrt{2}$. In GAP \\cite{url:gap} we have\n\\begin{verbatim}\n gap> x:=Indeterminate(Rationals,\"x\");;\n gap> alpha:=Sqrt(2);;\n gap> MinimalPolynomial(Rationals, alpha);\n x^2-2\n\\end{verbatim}\nI.e. $P_{min}\\left(\\alpha, \\mathbb{Q}\\right) = X^2 -2$.\n\\label{ex:minpolynomial}\n\\end{gapexample}\n\n\\section{Algebraic elements. Algebraic extensions}\n\n\\begin{definition}\n Let $K \\subset L$, $\\alpha \\in L$. The smallest sub-field contained\n $K$ and $\\alpha$ denoted by $K\\left(\\alpha\\right)$. The smallest\n sub-ring (or \\mynameref{def:kalgebra}) contained $K$ and $\\alpha$ denoted by $K\\left[\\alpha\\right]$.\n\\end{definition}\n\nAs soon as $K\\left[\\alpha\\right]$ is a \\mynameref{def:kalgebra} it is a\n\\mynameref{def:vectorspace} over $K$ generated by\n\\[\n1, \\alpha, \\alpha^2, \\dots,\n\\alpha^n, \\dots.\n\\]\n\n\\begin{example}[$\\mathbb{C}$]\n \\[\n \\mathbb{C} = \\mathbb{R}\\left(i\\right) = \\mathbb{R}\\left[i\\right]\n \\]\n $\\mathbb{C}$ is also a \\mynameref{def:vectorspace} generated by $1$\n and $i$: $\\forall z \\in \\mathbb{Z}$ it holds $z = x + i y$ where\n $x,y \\in \\mathbb{R}$.\n\\end{example}\n\n\\begin{proposition}\n The following assignment are equivalent\n \\begin{enumerate}\n \\item $\\alpha$ is algebraic over $K$\n \\item $K\\left[\\alpha\\right]$ is a finite dimensional\n \\mynameref{def:vectorspace} over $K$\n \\item $K\\left[\\alpha\\right] = K\\left(\\alpha\\right)$\n \\footnote{\n Contrary to the example \\ref{ex:kalgebranotfield} we see\n that $K$-algebra is a field there.\n }\n \\end{enumerate}\n \\begin{proof}\n Lets proof that 1 implies 2. If $\\alpha$ is algebraic over $K$\n then using lemma \\mynameref{def:minpolynomial} $\\exists\n P_{min}\\left(\\alpha, K\\right)$:\n \\[\n P_{min}\\left(\\alpha, K\\right) = \\alpha^d + a_{d-1}\\alpha^{d-1} +\n a_1 \\alpha + a_0 = 0,\n \\]\n where $a_k \\in K$. Then\n \\[\n \\alpha^d = - a_{d-1}\\alpha^{d-1} -\n a_1 \\alpha - a_0\n \\]\n this means that any $\\alpha^n$ can be represented as a linear\n combination of finite number of powers of $\\alpha$ i.e.\n $K\\left[\\alpha\\right]$ generated by $1, \\alpha, \\dots,\n \\alpha^{d-1}$ is a finite dimensional \\mynameref{def:vectorspace}.\n\n Lets proof that 2 implies 3. Its enough to prove that\n $K\\left[\\alpha\\right]$ is a field because\n $K\\left[\\alpha\\right] \\subset K\\left(\\alpha\\right)$.\n \n Let $x \\ne 0 \\in\n K\\left[\\alpha\\right]$ then lets look at an operation\n $x \\cdot K\\left[\\alpha\\right] \\rightarrow\n K\\left[\\alpha\\right]$. This is \\mynameref{def:injection}.\n \\footnote{\n If $y, z \\in K\\left[\\alpha\\right]$ and\n $\\dim K\\left[\\alpha\\right] = d < \\infty$ where\n $d = \\deg P_{min}\\left(\\alpha, K\\right)$. Then\n $y = \\sum_{i=0}^{d-1} y_i \\alpha^i$ and\n $z = \\sum_{i=0}^{d-1} z_i \\alpha^i$ where $y_i, z_i \\in K$.\n We have $y - z = \\sum_{i=0}^{d-1} \\left(y_i - z_i\\right)\n \\alpha^i \\ne 0$ if $y \\ne z$ (i.e.\n $\\exists i: y_i \\ne z_i$)\n because $y -z$ can be considered as a polynomial of degree\n $ \\le d - 1 < \\deg P_{min}\\left(\\alpha, K\\right)$ and cannot be equal\n to 0 by minimal polynomial definition.\n Continue we have $x \\cdot \\left(y - z\\right) \\ne 0$ because it\n also can be considered as a product of 2 polynomial of degree\n $< d$. \n Thus\n \\[\n x \\cdot y \\ne x \\cdot z\n \\]\n i.e. \\mynameref{def:injection} property is satisfied.\n }\n But the $K\\left[\\alpha\\right]$ is finite dimensional\n \\mynameref{def:vectorspace} and a \\mynameref{def:homomorphism} between\n 2 vector spaces with the same dimension is\n \\mynameref{def:surjection}\n \\footnote{\n Two vector spaces with same dimension are isomorphic each other\n (see lemma \\ref{lem:vsisomorphism})\n }\n thus $\\exists y \\in K\\left[\\alpha\\right]$\n such that $x \\cdot y = 1_{K\\left[\\alpha\\right]}$. Therefore $x$ is\n invertable and $K\\left[\\alpha\\right]$ is a\n \\mynameref{def:field}.\n\n Lets proof that 3 implies 1. Let $K\\left[\\alpha\\right]$ is a\n \\mynameref{def:field} but $\\alpha$ is not algebraic. Thus $\\forall P\n \\in K\\left[X\\right]$ $P(\\alpha) \\ne 0$. The we have an\n \\mynameref{def:injection} \\mynameref{def:homomorphism} $i$ :\n $K\\left[X\\right] \\to K\\left[\\alpha\\right] = K\\left(\\alpha\\right)$\n which sends $P\\left(X\\right)$ to $P\\left(\\alpha\\right)$.\n \\footnote{\n And if $P\\left(X\\right) \\ne 0$ then\n $P\\left(\\alpha\\right) \\ne 0$\n }\n But $K\\left[X\\right]$\n is not a field thus $K\\left[\\alpha\\right]$ should not be a field\n too that is in contradiction with the initial conditions.\n \\footnote{\n Alternative prove is the following.\n Let $x \\ne 0 \\in K\\left[X\\right]$ and $K\\left[\\alpha\\right]$ is a\n field then $i(x)$ is invertable i.e. $\\exists y \\in\n K\\left[X\\right]: i(x) i(y) = 1$ or \n $i(xy) = 1$ or finally $x$ - is invertable and $K\\left[X\\right]$\n is a field but $K\\left[X\\right]$ is a ring and therefore we just\n got a contradiction. \n }\n \\end{proof}\n \\label{prop:lec1_1}\n\\end{proposition}\n\n\\begin{definition}[Algebraic extension]\n $L$ an extension of $K$ is called algebraic over $K$ if $\\forall\n \\alpha \\in L$ - $\\alpha$ is algebraic over $K$. \n \\label{def:algebraicextension}\n\\end{definition}\n\n\\begin{proposition}\n If $L$ is algebraic over $K$ then any K-subalgebra of $L$ is a\n \\mynameref{def:field}.\n \\begin{proof}\n Let $L' \\subset L$ is a subalgebra and let $\\alpha \\in L'$. We want\n to show that $\\alpha$ is invertable. $\\alpha$ is algebraic\n therefore $\\alpha \\in K\\left[\\alpha\\right] \\subset L' \\subset L$\n and it's invertable.\n \\footnote{\n As soon as $K\\left[\\alpha\\right] = K\\left(\\alpha\\right)$ is a\n field then its any element (especially $\\alpha$) is invertable. \n }\n \\end{proof}\n \\label{prop:lec1_algebraicsubalgebra}\n\\end{proposition}\n\n\\begin{proposition}\n Let $K \\subset L \\subset M$. $\\alpha \\in M$ - algebraic over $K$\n then $\\alpha$ algebraic over $L$ and\n $P_{min}\\left(\\alpha, L\\right)$ divides $P_{min}\\left(\\alpha,\n K\\right)$. \n \\begin{proof}\n It is clear because $P_{min}\\left(\\alpha,K\\right) \\in\n L\\left[X\\right]$.\n \\footnote{\n Thus $\\exists P_L \\in L\\left[X\\right]$ such that\n $P_L\\left(\\alpha\\right) = 0$ i.e. $\\alpha$ is algebraic over $L$.\n \n As soon as $P_{min}\\left(\\alpha,K\\right) \\in L\\left[X\\right]$ then\n using \\mynameref{lem:minpolynomial} one can get that\n $P_{min}\\left(\\alpha,L\\right)$ divides\n $P_{min}\\left(\\alpha,K\\right)$.\n }\n \\end{proof}\n \\label{prop:lec1_algebraic}\n\\end{proposition}\n\nWe can consider the following example as an illustration for\nproposition \\ref{prop:lec1_algebraic}: \n\\begin{myexample}\n $K = \\mathbb{R}, L=M=\\mathbb{C}$. $\\alpha = i \\in M$ is algebraic\n over $K = \\mathbb{R}$ and therefore using the proposition\n \\ref{prop:lec1_algebraic} it is\n algebraic over $L = \\mathbb{C}$. Moreover $P_{min}(\\alpha, L) = X\n -i$ and it divides $P_{min}(\\alpha, K) = X^2 + 1$.\n\\end{myexample}\n\n\\section{Finite extensions. Algebraicity and finiteness}\n\n\\begin{definition}[Finite extension]\n $L$ is a finite extension of $K$ if $dim_k L < \\infty$. $dim_k L$ is\n called as degree of $L$ over $K$ and is denoted by\n $\\left[L:K\\right]$\n \\label{def:finiteextension}\n\\end{definition}\n\n\\begin{theorem}[The multiplicativity formula for degrees]\n Let $K \\subset L \\subset M$. Then $M$ is\n \\mynameref{def:finiteextension} over $K$ if and only if\n $M$ is \\mynameref{def:finiteextension} over $L$ and\n $L$ is \\mynameref{def:finiteextension} over $K$. In this case\n \\[\n \\left[M:K\\right] = \\left[M:L\\right] \\left[L:K\\right].\n \\]\n \\begin{proof}\n Let $\\left[M:K\\right] < \\infty$ but any linear independent set of\n vectors $\\left\\{m_1, m_2, \\dots, m_n\\right\\}$ over $L$ is also\n linear independent over $K$ thus\n \\[\n \\left[M:K\\right] < \\infty \\Rightarrow \\left[M:L\\right] < \\infty\n \\]\n also $L$ is a vector sub space of $M$ thus if\n $\\left[M:K\\right] < \\infty$ then $\\left[L:K\\right] < \\infty$.\n\n Let $\\left[M:L\\right] < \\infty$ and $\\left[L:K\\right] < \\infty$\n then we have the following basises\n \\begin{itemize}\n \\item $L$-basis over $M$: $\\left(e_1, e_2, \\dots, e_n\\right)$\n \\item $K$-basis over $L$:\n $\\left(\\varepsilon_1, \\varepsilon_2, \\dots, \\varepsilon_d\\right)$\n \\end{itemize}\n Lets proof that $e_i\\varepsilon_j$ forms a $K$-basis over $M$.\n $\\forall x \\in M$:\n \\[\n x = \\sum_{i=1}^n a_i e_i, \n \\]\n where $a_i \\in L$ and can be also written as\n \\[\n a_i = \\sum_{j=1}^d b_{ij} \\varepsilon_j,\n \\]\n where $b_{ij} \\in K$.\n Thus\n \\[\n x = \\sum_{i=1}^n \\sum_{j=1}^d b_{ij} \\varepsilon_j e_i, \n \\]\n therefore $\\varepsilon_j e_i = e_i \\varepsilon_j$ generates $M$\n over $K$. From the other side we should check that $\\varepsilon_j\n e_i$ linear independent system of vectors. Lets\n \\[\n \\sum_{i,j} c_{ij} \\varepsilon_j e_i =\n \\sum_{i=1}^n \\left( \\sum_{j=1}^d c_{ij} \\varepsilon_j \\right) e_i,\n \\]\n then $\\forall i$:\n \\[\n \\sum_{j=1}^d c_{ij} \\varepsilon_j = 0.\n \\]\n Thus $\\forall i,j: c_{ik} = 0$ that finishes the proof the linear\n independence. \n The number of linear independent vectors is $n \\times d$ i.e.\n \\[\n \\left[M:K\\right] = \\left[M:L\\right] \\left[L:K\\right].\n \\]\n \\end{proof}\n \\label{thm:mulformuladegrees}\n\\end{theorem}\n\n\\begin{definition}[$K\\left(\\alpha_1, \\dots, \\alpha_n\\right)$]\n $K\\left(\\alpha_1, \\dots, \\alpha_n\\right) \\subset L$ generated by\n $\\alpha_1, \\dots, \\alpha_n$ is the smallest sub field of $L$\n contained $K$ and $\\alpha_i \\in L$.\n\\end{definition}\n\n\\begin{theorem}[About towers]\n $L$ is finite over $K$ if and only if $L$ is generated by a finite\n number of algebraic elements over $K$.\n \\begin{proof}\n If $L$ is finite then $\\alpha_1, \\dots, \\alpha_d$ is a basis. In\n this case\n $L = K\\left[\\alpha_1, \\dots, \\alpha_d\\right] = K\\left(\\alpha_1,\n \\dots, \\alpha_d\\right)$. Moreover each\n $K\\left[\\alpha_i\\right]$ is finite dimensional thus by\n proposition \\ref{prop:lec1_1} $\\alpha_i$ is algebraic.\n\n From other side if we have a finite set of algebraic elements\n $\\alpha_1, \\dots, \\alpha_d$ then\n $K\\left[\\alpha_1\\right]$ is a finite dimensional\n \\mynameref{def:vectorspace} over $K$,\n $K\\left[\\alpha_1, \\alpha_2\\right]$ is a finite dimensional\n \\mynameref{def:vectorspace} over $K\\left[\\alpha_1\\right]$ and\n so on $K\\left[\\alpha_1, \\dots, \\alpha_d\\right]$ is a finite dimensional\n \\mynameref{def:vectorspace} over\n $K\\left[\\alpha_1, \\dots, \\alpha_{d-1}\\right]$. All elements\n are algebraic thus\n \\[\n K\\left[\\alpha_1, \\dots, \\alpha_i\\right] =\n K\\left(\\alpha_1, \\dots, \\alpha_i\\right)\n \\]\n Then using theorem \\ref{thm:mulformuladegrees} we can conclude\n that $K\\left(\\alpha_1, \\dots, \\alpha_d\\right)$ has finite\n dimension. \n \\end{proof}\n \\label{thm:lec1_2}\n\\end{theorem}\n\n\\section{Algebraicity in towers. An example}\n\n\\begin{theorem}\n $K \\subset L \\subset M$ then $M$ \\mynameref{def:algebraicextension}\n over $K$ if and only if $M$ algebraic over $L$ and $L$ algebraic\n over $K$. \n \\begin{proof}\n If $\\alpha \\in M$ is an \\mynameref{def:algebraicelement} over $K$ then\n $\\exists P \\in K\\left[X\\right]$ such that\n $P\\left(\\alpha\\right) = 0$ but the\n polynomial $P \\in K\\left[X\\right] \\subset L\\left[X\\right]$\n thus $\\alpha$ is algebraic over $L$.\n If $\\alpha \\in L \\subset M$ then $\\alpha$ is algebraic over $K$\n thus $L$ is algebraic over $K$.\n\n Let $M$ algebraic over $L$ and $L$ algebraic over $K$ and let\n $\\alpha \\in M$. We want to prove that $\\alpha$ is algebraic over\n $K$. Lets consider $P_{min}\\left(\\alpha, L\\right)$ the polynomial\n coefficients are from $L$ and they (as soon as they count is a\n finite) generate a finite extension $E$ over $K$ thus\n $E\\left(\\alpha\\right)$ is finite over $E$ (exists a relation\n between powers of $\\alpha$) is finite over $K$ thus $\\alpha$ is\n algebraic over $K$.\n \\footnote{\n $P_{min}\\left(\\alpha, L\\right) = \\sum_{i = 0}^{d-1} l_i\n \\alpha^i$ where $l_i \\in L$ and each $l_i$ is algebraic over $K$\n by algebraic extension definition\n \\ref{def:algebraicextension}. By theorem\n \\ref{thm:lec1_2} there are finite number of $l_i$\n and they forms an algebraic extension\n $E = K\\left(l_0, l_1, \\dots, l_{d-1}\\right)$. The\n $E\\left(\\alpha\\right)$ is finite over $E$ and therefore finite\n over $K$. As soon as $E\\left(\\alpha\\right)$ has a finite\n dimension over $K$ thus there exists a relation for powers of\n $\\alpha$ such that $\\sum_{i=0}^n k_i \\alpha^i = 0$ i.e. \n $\\alpha$ is algebraic. \n }\n \\end{proof}\n\\end{theorem}\n\\begin{example}[$\\mathbb{Q}$ extension]\n $\\mathbb{Q}\\left( \\sqrt[3]{2}, \\sqrt{3}\\right)$ algebraic and finite\n over $\\mathbb{Q}$:\n \\[\n \\mathbb{Q} \\subset \\mathbb{Q}\\left( \\sqrt[3]{2}\\right)\n \\subset \\mathbb{Q}\\left( \\sqrt[3]{2}, \\sqrt{3}\\right)\n \\]\n\n Minimal polynomial\n \\[\n P_{min}\\left(\\sqrt[3]{2}, \\mathbb{Q}\\right) = x^3 - 2.\n \\]\n\n $\\mathbb{Q}\\left( \\sqrt[3]{2}\\right)$ is generated over $\\mathbb{Q}$\n by $1, \\sqrt[3]{2}, \\sqrt[3]{4}$ thus\n $\\left[\\mathbb{Q}\\left( \\sqrt[3]{2}\\right): \\mathbb{Q}\\right] = 3$.\n\n But $\\sqrt{3} \\notin \\mathbb{Q}\\left( \\sqrt[3]{2}\\right)$ because\n otherwise $\\left[\\mathbb{Q}\\left( \\sqrt{3}\\right): \\mathbb{Q}\\right]\n = 2$ must devide \n $\\left[\\mathbb{Q}\\left( \\sqrt[3]{2}\\right): \\mathbb{Q}\\right] = 3$\n that is impossible.\n\n Therefore $x^2 - 3$ is irreducible over\n $\\mathbb{Q}\\left( \\sqrt[3]{2}\\right)$ and\n \\[\n P_{min}\\left(\\sqrt{3}, \\mathbb{Q}\\left( \\sqrt[3]{2}\\right)\\right) =\n x^2 - 3.\n \\]\n\n \\[\n \\left[\\mathbb{Q}\\left( \\sqrt[3]{2}, \\sqrt{3}\\right):\n \\mathbb{Q}\\right] = 3 \\cdot 2 = 6.\n \\]\n\\end{example}\n\n\\begin{proposition}[On dimension of extension]\n \\[\n \\left[K\\left(\\alpha\\right) : K\\right] =\n \\deg\\left(P_{min}\\left(\\alpha, K\\right)\\right),\n \\]\n if $\\alpha$ is algebraic.\n \\begin{proof}\n If $\\deg\\left(P_{min}\\left(\\alpha, K\\right)\\right) = d$ then $1, \\alpha,\n \\cdots, \\alpha^{d-1}$ - $d$ independent vectors and dimension\n $K\\left(\\alpha\\right)$ is $d$.\n \\end{proof}\n \\label{prop:dimextension}\n\\end{proposition}\n\n\\begin{proposition}[About algebraic closure]\n If $K \\subset L$ ($L$ extension of $K$). Consider\n \\[\n L' = \\left\\{\n \\alpha \\in L \\mid \\alpha \\mbox{ algebraic over } K\n \\right\\},\n \\]\n then $L'$ sub-field of $L$ and is called as algebraic closure of $K$\n in $L$.\n \\begin{proof}\n We have to prove that if $\\alpha, \\beta$ are algebraic then\n $\\alpha + \\beta$ and $\\alpha \\cdot \\beta$ are also algebraic. This\n is trivial because\n \\[\n \\alpha + \\beta, \\alpha \\cdot \\beta \\in K\\left[\\alpha, \\beta\\right]\n \\]\n \\footnote{\n We also have that $K\\left[\\alpha, \\beta\\right]$ is a field:\n $K\\left[\\alpha, \\beta\\right]\n = K\\left(\\alpha, \\beta\\right)$.\n Really\n $K\\left[\\alpha\\right] =\n K\\left(\\alpha\\right)$ (see proposition\n \\ref{prop:lec1_1}). $\\beta$ is algebraic over $K$ and \n therefore over $K\\left(\\alpha\\right)$ thus we can construct\n $K\\left(\\alpha\\right)\\left[\\beta\\right] = K\\left(\\alpha, \\beta\\right)$\n by proposition \\ref{prop:lec1_1}\n }\n \\end{proof}\n\\end{proposition}\n\n\n\\section{A digression: Gauss lemma, Eisenstein criterion}\n\nWhat we have seen so far:\n\n\\begin{itemize}\n\\item $K$ is a field, $\\alpha$ is an \\mynameref{def:algebraicelement}\n over $K$ if it is a root of a polynomial $P \\in K\\left[X\\right]$.\n\\item $L$ is an \\mynameref{def:algebraicextension} over $K$ if\n $\\forall \\alpha \\in L$: $\\alpha$ is an algebraic over $K$\n\\item $L$ is a \\mynameref{def:finiteextension} over $K$ if $dim_K L <\n \\infty$.\n\\item If an extension is finite then it is algebraic\n\\item An extension is finite if and only if it is algebraic and\n generated by a finite number of algebraic elements (see theorem\n \\ref{thm:lec1_2})\n\\item $\\left[K\\left[\\alpha\\right]:K\\right] =\n deg P_{min}\\left(\\alpha, K\\right)$ (see proposition\n \\ref{prop:dimextension}). \n\\end{itemize}\n\nHow to decide that a polynomial $P$ is irreducible over $K$?\nAbout polynomial $x^3 - 2$ it is easy to decide that it's irreducible\nover $\\mathbb{Q}$, but what's about $x^{100}-2$?\n\n\\begin{lemma}[Gauss]\n Let $P \\in \\mathbb{Z}\\left[X\\right]$, i.e. a polynomial with integer\n coefficients, then if $P$ decomposes over $\\mathbb{Q}$ ($P = Q\\cdot\n R, deg Q,R < deg P$) then it also decomposes over $\\mathbb{Z}$.\n \\begin{proof}\n Let $P = Q R$ over $\\mathbb{Q}$. Then\n \\begin{eqnarray}\n Q = m Q_1, Q_1 \\in \\mathbb{Z}\\left[X\\right],\n \\nonumber \\\\\n R = n R_1, R_1 \\in \\mathbb{Z}\\left[X\\right],\n \\nonumber\n \\end{eqnarray}\n thus\n \\[\n n m P = Q_1 R_1. \n \\]\n There exists $p$ that divides $mn$: $p \\mid mn$ thus in modulo $p$\n we have\n \\[\n 0 = \\overline{Q_1}\\overline{R_1}\n \\]\n but $p$ is prime and the equation is in the field $\\mathbb{F}_p$\n thus either $\\overline{Q_1} = 0$ or $\\overline{R_1} = 0$. Let\n $\\overline{Q_1} = 0$ thus $p$ divides all coefficients in $Q_1$\n and we can take $\\frac{Q_1}{p} = Q_2 \\in\n \\mathbb{Z}\\left[X\\right]$. Continue for all primes in $mn$ we can\n get that\n \\[\n P = Q_s R_t,\n \\]\n where $Q_s, R_t \\in \\mathbb{Z}\\left[X\\right]$.\n \\end{proof}\n \\label{lem:gauss}\n\\end{lemma}\n\n\\begin{example}[Eisenstein criterion]\n Lets consider the following polynomial $x^{100} -2$. It's\n irreducible. Lets prove it. If it reducible then\n $\\exists Q, R \\in \\mathbb{Z}\\left[X\\right]$ such that\n \\begin{equation}\n x^{100} -2 = Q R\n \\label{eq:ex_eisenstein}\n \\end{equation}\n Lets consider (\\ref{eq:ex_eisenstein}) modulo 2. In the case we will\n have\n \\[\n Q R \\equiv x^{100} \\mod 2,\n \\]\n therefore\n \\begin{eqnarray}\n Q \\equiv x^k \\mod 2,\n \\nonumber \\\\\n R \\equiv x^l \\mod 2,\n \\nonumber\n \\end{eqnarray}\n or\n \\[\n Q = x^k + \\dots + 2 \\cdot m\n \\]\n and\n \\[\n R = x^l + \\dots + 2 \\cdot n\n \\]\n thus\n \\[\n QR = x^{100} + 4 \\cdot nm\n \\]\n that is impossible because $n,m \\in \\mathbb{Z}$ and $nm \\ne\n -\\frac{1}{2}$. \n \\label{ex:eisenstein}\n\\end{example}\n\n\\begin{lemma}[Eisenstein criterion]\n Lets $P \\in \\mathbb{Z}\\left[X\\right]$ and\n $P = a_n X^n + a_{n-1} X^{n-1} + a_1 X + a_0$. If $\\exists p$ -\n prime such that $p \\nmid a_n$, $p \\mid a_i \\forall i < n$ and\n $p^2 \\nmid a_0$, then $P \\in \\mathbb{Z}\\left[X\\right]$ is\n irreducible. \n \\begin{proof}\n the same as for example \\ref{ex:eisenstein}.\n \\end{proof}\n \\label{lem:eisenstein}\n\\end{lemma}\n\nNote: that both: \\mynameref{lem:gauss} and \\mynameref{lem:eisenstein} are\nvalid by replacing $\\mathbb{Z}$ with an \\mynameref{def:ufd} $R$ and\n$\\mathbb{Q}$ by its factorization field.\n", "meta": {"hexsha": "ae11e2ee0f5fe025eaef31843d407b23892cc49c", "size": 37285, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture1.tex", "max_stars_repo_name": "JiuziLau/courseragalois", "max_stars_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2016-06-21T07:34:33.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-23T00:14:44.000Z", "max_issues_repo_path": "lecture1.tex", "max_issues_repo_name": "JiuziLau/courseragalois", "max_issues_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2016-06-28T21:26:22.000Z", "max_issues_repo_issues_event_max_datetime": "2016-06-30T05:36:37.000Z", "max_forks_repo_path": "lecture1.tex", "max_forks_repo_name": "JiuziLau/courseragalois", "max_forks_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-08-08T07:47:20.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-25T06:16:07.000Z", "avg_line_length": 33.0248007086, "max_line_length": 102, "alphanum_fraction": 0.6275177685, "num_tokens": 13565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124812, "lm_q2_score": 0.8438951064805861, "lm_q1q2_score": 0.6271558457609949}} {"text": "\\documentclass[twoside]{MATH77}\n\\usepackage{multicol}\n\\usepackage[fleqn,reqno,centertags]{amsmath}\n\\begin{document}\n\\begmath 17.3 Double Precision Complex Computation\n\n\\silentfootnote{$^\\copyright$1997 Calif. Inst. of Technology, \\thisyear \\ Math \\`a la Carte, Inc.}\n\n\\subsection{Purpose}\n\nThese subprograms do the following computations for double precision complex\ndata: sum, difference, product, quotient, square root, and absolute value.\nThe Fortran~77 standard does not support a double precision complex data\ntype. Using the convention of storing the real and imaginary parts of a\ncomplex number as an adjacent pair of double precision numbers, this set of\nsubprograms provides the above operations while staying within the standard.\n\n\\subsection{Usage}\n\n\\subsubsection{Program Prototype, Double Precision}\n\n\\begin{description}\n\n\\item[DOUBLE PRECISION] \\ {\\bf A}(2){\\bf , B}(2){\\bf , RESULT}(2){\\bf ,\nABSVAL, DZABS}\n\n\\end{description}\n\nAssign values to A(), or to A() and B(), and access the appropriate\nsubprogram. The results will be stored in RESULT() or ABSVAL.\n\\begin{center}\n\\fbox{\\begin{tabular}{@{\\bf \\ }ll@{\\ }}\nCALL ZSUM(A, B, RESULT) & Result $= a + b$\\\\\nCALL ZDIF(A, B, RESULT) & Result $= a - b$\\\\\nCALL ZPRO(A, B, RESULT) & Result $= a \\times b$\\\\\nCALL ZQUO(A, B, RESULT) & Result $= a\\ /\\ b$\\\\\nCALL ZSQRTX(A, RESULT) & Result $= \\sqrt a$\\\\\nABSVAL = DZABS(A) & Absval $= |a|$\\\\\n\\end{tabular}}\n\\end{center}\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\n\\item[A()] \\ [in] Contains the double precision complex value, $a$, with the real part in A(1) and\nthe imaginary part in A(2).\n\n\\item[B()] \\ [in] Contains the double precision complex value, $b$, with the real part in B(1)\nand the imaginary part in B(2).\n\n\\item[RESULT()] \\ [out] On return contains the double precision complex result with the real part\nin RESULT(1) and the imaginary part in RESULT(2). It is allowable for the\narray RESULT() to occupy the same storage locations as the arrays A()\nand/or B(). For example the statement, CALL ZPRO(Z, Z, Z), is\npermissible.\n\nIn the case of the complex square root, which is double valued, ZSQRTX\nreturns the root with nonnegative real part. If the real part of the root is\nzero it returns the root whose imaginary part is nonnegative. The other root\nis always the negative of the returned root.\n\n\\item[DZABS] \\ [out] Returns the double precision absolute value of the complex\nnumber represented by the pair, (A(1),A(2)).\n\n\\end{description}\n\n\\subsection{Examples and Remarks}\n\nThe program, DRZCOMP, with its output, ODZCOMP, illustrates the use of these\nsubprograms. The example begins with three complex constants, $a$, $b$, and $u$%\n. It computes $v = u+a$, $w = v\\times b$, and $z = \\sqrt w$. It then inverts this\nsequence of computations by computing $w_2 = z\\times z$, $v_2 = w_2/b$, and $u_2 =\nv_2-a$. Mathematically this should result in $w_2 = w$, $v_2 =\nv$, and $u_2 = u$. We test the last of these relations by computing TEST\\ $=\nu_2-u$. Additionally TEST2 is a measure of the error in an\napplication of DZABS. Both TEST and TEST2 are seen to be acceptably\nsmall given that the computation was done with double precision IEEE\narithmetic.\n\n\\subsection{Functional Description}\n\n\\subsubsection{Method}\n\nTo compute $(u,v)$ as the square root of $(x,y)$ the basic algorithm is\n\\begin{gather*}\nr = \\sqrt{x^2 + y^2}\\\\\nu = \\sqrt{(r+|x|)/2}\\\\\nv = |y|/(2u)\\\\\n\\text{if } x < 0 \\text{ swap }u\\text{ and }v\\\\\n\\text{if } y < 0 \\text{ set }v = -v\n\\end{gather*}\nSubroutine ZSQRTX contains special treatment for the cases of $x = 0$ or $y =\n0$, and uses scaling to avoid unnecessary overflow that could result from\ncomputing $x^2$ or $y^2.$\n\nSubprograms ZQUO and DZABS use scaling to avoid unnecessary overflow.\nIn the five subroutines the implementations permit the output array to\noccupy the same storage locations as the input arrays.\n\n\\subsubsection{Accuracy tests}\n\nThese subprograms have been tested using arguments on the four principal\naxes and interior to the eight octants. Results were consistent with the\nmachine precision.\n\n\\subsection{Error Procedures and Restrictions}\n\nIf B($1) = 0$ and B($2) = 0$ in ZQUO, the subroutine will execute a division\nby zero. We assume the host system will produce a runtime error diagnostic\nin this case.\n\nIf the function DZABS is used its name must be typed as double precision in\nthe referencing program.\n\n\\subsection{Supporting Information}\n\nThe source language is ANSI Fortran~77.\n\nDesigned by C. L. Lawson, JPL, May~1986.\n\nProgrammed by C. L. Lawson and S. Y. Chiu, JPL, May~1986, Feb.~1987,\nNov.~1987.\n\n\nEach of the subprograms, DZABS, ZDIF, ZPRO, ZQUO, ZSQRTX, and ZSUM is\ncontained in a program file of the same name.\n\n\\begcode\n\n\\medskip\\\n\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRZCOMP}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{zcomp}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODZCOMP}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{zcomp}}\n\\end{document}\n", "meta": {"hexsha": "c0a586656cef829a330c3c670a21141a9c4960dd", "size": 5013, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doctex/ch17-03.tex", "max_stars_repo_name": "jacobwilliams/math77", "max_stars_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2016-01-04T03:17:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-25T19:17:42.000Z", "max_issues_repo_path": "doc/doctex/ch17-03.tex", "max_issues_repo_name": "jacobwilliams/math77", "max_issues_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2019-01-17T02:48:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T16:04:58.000Z", "max_forks_repo_path": "doc/doctex/ch17-03.tex", "max_forks_repo_name": "jacobwilliams/math77", "max_forks_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-01-07T09:26:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-11-25T05:32:54.000Z", "avg_line_length": 35.3028169014, "max_line_length": 98, "alphanum_fraction": 0.7338918811, "num_tokens": 1454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8244619306896955, "lm_q1q2_score": 0.6271275056866941}} {"text": "\\section{Coherent state}\n\nLet\n\\begin{equation}\n\\widehat{b}^- = \\widehat{a}^- - \\alpha,\\qquad\n\\widehat{b}^+ = \\rbr{\\widehat{b}^-}^\\dagger = \\widehat{a}^+ - \\alpha^*.\n\\label{eq:displacement-b}\n\\end{equation}\nOne seeks the state $\\Ket{\\alpha}$ satisfying\n\\begin{equation}\n\\widehat{b}^-\\Ket{\\alpha} = 0.\n\\label{eq:coherent-state-def-1}\n\\end{equation}\nExpanding $\\widehat{b}^-$ in terms of $\\widehat{x}$ and $\\widehat{p}$,\n\\begin{equation}\n\\widehat{b}^- = \\sqrt{\\frac{m\\Omega}{2}}\\rbr{\\widehat{x}-x_\\alpha}\n+ \\frac{\\ii}{\\sqrt{2m\\Omega}}\\rbr{\\widehat{p}-p_\\alpha},\n\\end{equation}\nwhere\n\\begin{equation}\nx_\\alpha = \\sqrt{\\frac{2}{m\\Omega}}\\Re\\alpha,\\qquad\np_\\alpha = \\sqrt{2m\\Omega}\\Im\\alpha.\n\\end{equation}\n\nOne may define\n\\begin{align}\n\\widehat{y} & \\coloneqq \\widehat{x} - x_\\alpha \\widehat{1}\n= \\widehat{x} - \\ii x_\\alpha\\sbr{\\widehat{p},\\widehat{x}}_-\n= \\ee^{-\\ii x_\\alpha \\widehat{p}} \\widehat{x} \\ee^{+\\ii x_\\alpha \\widehat{p}},\n\\label{eq:coh-y}\\\\\n\\widehat{k} & \\coloneqq \\widehat{p} - p_\\alpha \\widehat{1}\n= \\widehat{p} + \\ii p_\\alpha\\sbr{\\widehat{x},\\widehat{p}}_-\n= \\ee^{+\\ii p_\\alpha \\widehat{x}} \\widehat{p} \\ee^{-\\ii p_\\alpha \\widehat{x}},\n\\label{eq:coh-k}\n\\end{align}\nwhere in the last steps, Lemma \\ref{lem:bch-sandwich} has been applied.\nOne can verify\n\\begin{equation}\n\\sbr{\\widehat{y}, \\widehat{k}}_- = \\ii \\widehat{1}.\n\\end{equation}\n\n\\subsection{Displacement operator}\nInspired by \\cref{eq:coh-y,eq:coh-k}, one attempts\n\\begin{equation}\n\\rfun{\\widehat{D}}{\\alpha}\\widehat{a}^-\\rfun{\\widehat{D}^{-1}}{\\alpha},\n\\end{equation}\nwhere\n\\begin{equation}\n\\rfun{\\widehat{D}}{\\alpha} \\coloneqq\n\\ee^{+\\alpha\\widehat{a}^+ - \\alpha^*\\widehat{a}^-}\n\\equiv \\ee^{\\ii\\rbr{-x_\\alpha\\widehat{p}+p_\\alpha\\widehat{x}}}\n\\end{equation}\nis the \\emph{displacement operator}. Note that\n\\begin{equation}\n\\rfun{\\widehat{D}^{-1}}{\\alpha} = \\rfun{\\widehat{D}}{-\\alpha}\n= \\rfun{\\widehat{D}^\\dagger}{\\alpha}.\n\\end{equation}\n\n\\begin{equation}\nb^- = a^- - \\alpha = a^- + \\alpha\\sbr{a^+, a^-}_-\n= a^- + \\sbr{\\alpha a^+ - \\alpha^* a^-, a^-}_-.\n\\end{equation}\nNote that\n\\begin{equation}\n\\ad_{\\alpha a^+ - \\alpha^* a^-}^{\\rbr{n}}a^- \\equiv 0,\\qquad\\forall n \\ge 2.\n\\end{equation}\nBy Lemma \\ref{lem:bch-sandwich},\n\\begin{equation}\nb^- = \\ee^{+\\alpha a^+ - \\alpha^* a^-}a^-\\ee^{-\\alpha a^+ + \\alpha^* a^-}\n= \\rfun{D}{\\alpha} a^- \\rfun{D}{\\alpha}^{-1}.\n\\end{equation}\nNow\n\\begin{equation}\na^-\\Ket{0} \\coloneqq 0 \\eqqcolon \\rfun{D^\\dagger}{\\alpha} b^- \\Ket{0}\n= a^- \\rfun{D^\\dagger}{\\alpha} \\Ket{\\alpha}.\n\\end{equation}\nWith $\\rfun{D}{\\alpha}$ unitary in mind, one finds\n\\begin{equation}\n\\Ket{\\alpha} = \\ee^{\\ii\\theta}\\rfun{D}{\\alpha}\\Ket{0}.\n\\end{equation}\nOne can fix the phase $\\theta$ to be zero.\n\n\\subsection{Time evolution}\n\nCompute\n\\begin{equation}\n\\rfun{\\widehat{\\mscrU}}{t} \\rfun{\\widehat{D}}{\\alpha}\n\\rfun{\\widehat{\\mscrU}^\\dagger}{t}.\n\\label{eq:tim-evo-disp-1}\n\\end{equation}\n\n\\Cref{eq:tim-evo-disp-1} simplifies to\n\\begin{equation}\n\\ee^{-\\ii \\Omega t a^+ a^-} \\rfun{D}{\\alpha}\n\\ee^{+\\ii \\Omega t a^+ a^-}\n\\label{eq:tim-evo-disp-2}\n\\end{equation}\n\nOne wishes to interchange $\\ee^{-\\ii \\Omega t a^+ a^-}$ and\n$\\rfun{D}{\\alpha}$. By Theorem \\ref{thm:bch-brading}, one computes\n\\begin{align}\n\\ad_{-\\ii \\Omega t a^+ a^-}^{(1)}\\rbr{\\alpha a^+ - \\alpha^* a^-}\n&= \\sbr{-\\ii \\Omega t a^+ a^- , \\alpha a^+ - \\alpha^* a^-}_-\n\\nonumber\\\\\n&= -\\ii\\Omega t \\rbr{\\alpha a^+\\sbr{a^-, a^+}_- - \\alpha^*\\sbr{a^+, a^-}_- a^-}\n\\nonumber\\\\\n&= -\\ii\\Omega t \\rbr{+\\alpha a^+ + \\alpha^* a^-} \\nonumber\\\\\n&=\t+ \\alpha\\rbr{-\\ii\\Omega t}^1 a^+\n\t- \\alpha\\rbr{+\\ii\\Omega t}^1 a^-;\n\\end{align}\n\\begin{align}\n\\ad_{-\\ii \\Omega t a^+ a^-}^{(2)}\\rbr{\\alpha a^+ - \\alpha^* a^-}\n&= \\sbr{-\\ii \\Omega t a^+ a^-,\n\t-\\ii\\Omega t \\rbr{+\\alpha a^+ + \\alpha^* a^-}}_-\n\\nonumber\\\\\n&= \\Omega^2 t^2\\rbr{-\\alpha a^+ + \\alpha^* a^-}\n\\nonumber\\\\\n&= + \\alpha\\rbr{-\\ii\\Omega t}^2 a^+\n- \\alpha\\rbr{+\\ii\\Omega t}^2 a^-.\n\\end{align}\nBy induction one can prove\n\\begin{equation}\n\\ad_{-\\ii \\Omega t a^+ a^-}^{(n)}\\rbr{\\alpha a^+ - \\alpha^* a^-}\n= + \\alpha\\rbr{-\\ii\\Omega t}^n a^+\n- \\alpha\\rbr{+\\ii\\Omega t}^n a^-,\n\\end{equation}\nso that \\cref{eq:tim-evo-disp-1} transforms to\n\\begin{align}\n&\\phantom{{}={}}\\rfun{\\exp}{+\\alpha\\ee^{-\\ii\\Omega t}a^+\n\t-\\alpha^*\\ee^{+\\ii\\Omega t}a^-}\n\t\\ee^{-\\ii \\Omega t a^+ a^-}\n\t\\ee^{+\\ii \\Omega t a^+ a^-} \\\\\n&= \\rfun{\\exp}{+\\alpha\\ee^{-\\ii\\Omega t}a^+\n-\\alpha^*\\ee^{+\\ii\\Omega t}a^-} \\\\\n&\\equiv \\rfun{D}{\\alpha^t}.\n\\end{align}\nby Theorem \\ref{thm:bch-brading}, where\n\\begin{equation}\n\\alpha^t\\coloneqq \\alpha\\ee^{-\\ii\\Omega t}.\n\\end{equation}\n\nAs a result, the time evolution of the coherent ground state is\n\\begin{equation}\n\\rfun{\\mscrU}{t} \\Ket{\\alpha} = \\rfun{\\mscrU}{t} \\rfun{D}{\\alpha} \\Ket{0}\n= \\rfun{D}{\\alpha^t} \\rfun{\\mscrU}{t} \\Ket{0}\n= \\ee^{-\\ii\\Omega t/2}\\Ket{\\alpha^t},\n\\end{equation}\nin terms of the coherent parameter $\\alpha$.\n\n\\subsection{Particle numbers}\n\n\\begin{namedthm}{Lemma}\n\\begin{equation}\n\\rfun{\\widehat{D}}{\\alpha} = \\ee^{-\\vbr{\\alpha}^2/2}\n\\ee^{+\\alpha\\widehat{a}^+} \\ee^{-\\alpha^*\\widehat{a}^-}.\n\\end{equation}\n\\end{namedthm}\n\\begin{proof}\nUse Corollary \\ref{thm:bch-merging}.\n\\end{proof}\n\n\\begin{namedthm}{Particle number representation of $\\Ket{\\alpha}$}\n\\begin{equation}\n\\Braket{n | \\alpha} = \\ee^{-\\vbr{\\alpha}^2/2}\n\\frac{\\alpha^n}{\\sqrt{n!}}.\n\\end{equation}\n\\end{namedthm}\n\n\n\\begin{namedthm}{$\\widehat{a}$-particle number in $\\Ket{\\alpha}$}\n\\begin{equation}\n\\Braket{\\alpha | \\widehat{a}^+ \\widehat{a}^- | \\alpha} = \\vbr{\\alpha}^2.\n\\end{equation}\n\\end{namedthm}\n\\begin{proof}\n\\begin{equation}\n\\rfun{\\widehat{D}^\\dagger}{\\alpha}\\widehat{a}^-\\rfun{\\widehat{D}}{\\alpha}\n= \\widehat{a}^- + \\sbr{-\\alpha \\widehat{a}^+, \\widehat{a}^-}_-\n= \\widehat{a}^- + \\alpha;\n\\end{equation}\n\\begin{equation}\n\\rfun{\\widehat{D}^\\dagger}{\\alpha}\\widehat{a}^+\n\\widehat{a}^-\\rfun{\\widehat{D}}{\\alpha}\n=\n\\rfun{\\widehat{D}^\\dagger}{\\alpha}\\widehat{a}^+\\rfun{\\widehat{D}}{\\alpha}\n\\rfun{\\widehat{D}^\\dagger}{\\alpha}\\widehat{a}^-\\rfun{\\widehat{D}}{\\alpha}\n=\n\\rbr{\\widehat{a}^+ + \\alpha^*}\\rbr{\\widehat{a}^- + \\alpha}.\n\\end{equation}\n\\end{proof}\n\n\\begin{namedthm}{$\\widehat{b}$-particle number in $\\widehat{a}$-ground state}\n\\begin{equation}\n\\Braket{0 | \\widehat{b}^+ \\widehat{b}^- | 0} = \\vbr{\\alpha}^2\n\\end{equation}\n\\end{namedthm}\n\n\\begin{proof}\n\\begin{equation}\n\\widehat{b}^+ \\widehat{b}^- =\n\\rbr{\\widehat{a}^+ - \\alpha^*}\\rbr{\\widehat{a}^- - \\alpha}\n\\end{equation}\naccording to \\ref{eq:displacement-b}.\n\\end{proof}\n\n\\begin{namedrem}{Remark}\nSince $\\Ket{0}$ is stationary, $\\Braket{0 | \\widehat{b}^+ \\widehat{b}^- | 0}$\nis a constant of motion.\n\n\\end{namedrem}\n\n\\subsection{Wave function}\nSince $\\sbr{+\\ii p_\\alpha\\widehat{x}, -\\ii x_\\alpha\\widehat{p}}_- =\n\\ii p_\\alpha x_\\alpha \\widehat{1}$ is central, one has\n\\begin{align}\n\\Bra{x}\\rfun{\\widehat{D}}{\\alpha}\n&= \\Bra{x}\\ee^{\\ii\\rbr{+p_\\alpha\\widehat{x}-x_\\alpha\\widehat{p}}}\n\\nonumber\\\\\n&= \\Bra{x}\\ee^{+\\ii p_\\alpha\\widehat{x}}\\ee^{-\\ii x_\\alpha\\widehat{p}}\n\\ee^{-\\sbr{+\\ii p_\\alpha\\widehat{x}, -\\ii x_\\alpha\\widehat{p}}_-/2}\n\\nonumber \\\\\n&= \\ee^{-\\ii p_\\alpha x_\\alpha/2} \\ee^{+\\ii p_\\alpha x}\n\\Bra{x} \\ee^{-\\ii x_\\alpha\\widehat{p}}\n\\nonumber \\\\\n&= \\ee^{-\\ii p_\\alpha x_\\alpha/2} \\ee^{+\\ii p_\\alpha x}\n\\Bra{x-x_\\alpha},\n\\label{eq:displacement-x-bra}\n\\end{align}\nusing \\cref{eq:displace-x-bra}. Similarly\n\\begin{align}\n\\Bra{x}\\rfun{\\widehat{D}}{\\alpha}\n&= \\ee^{-\\sbr{-\\ii x_\\alpha\\widehat{p}, +\\ii p_\\alpha\\widehat{x}}_-/2}\n\\ee^{-\\ii x_\\alpha\\widehat{p}}\\ee^{+\\ii p_\\alpha\\widehat{x}} \\Ket{x}\n\\nonumber \\\\\n&= \\ee^{+\\ii p_\\alpha x_\\alpha/2} \\ee^{+\\ii p_\\alpha x} \\Ket{x+x_\\alpha}.\n\\end{align}\nNote the extra phase factors compared with\n\\cref{eq:displace-x-ket,eq:displace-x-bra}.\n\n\\Cref{eq:displacement-x-bra} solves the wave function of coherent ground state\n\\begin{align}\n\\Braket{x | \\alpha} &= \\Braket{x | \\rfun{D}{\\alpha} | 0}\n\\nonumber \\\\\n&= \\rbr{\\frac{m\\Omega}{\\pp}}^{1/4} \\cfun{\\exp}%\n{-\\frac{1}{2}m\\Omega\\rbr{x-x_\\alpha}^2}\\cfun{\\expi}{p_\\alpha x - \\frac{1}{2} \np_\\alpha x_\\alpha},\n\\end{align}\nas well as its time evolution\n\\begin{align}\n&\\phantom{{}={}}\\rbr{\\frac{m\\Omega}{\\pp}}^{-1/4}\\Braket{x | \\rfun{\\mscrU}{t} | \n\\alpha} \\nonumber \\\\\n&=\\ee^{-\\ii\\Omega t/2}\n\\Braket{x | \\rfun{D}{\\alpha\\ee^{-\\ii\\Omega t}} | 0}\n\\nonumber \\\\\n&=\\cfun{\\exp}{-\\frac{1}{2}m\\Omega\\rbr{x-x_\\alpha^t}^2} \n\\cfun{\\expi}{-\\frac{1}{2}\\Omega t + x p_\\alpha^t\n- \\frac{1}{2} p_\\alpha^t x_\\alpha^t},\n\\end{align}\nwhere\n\\begin{equation}\nx_\\alpha^t \\coloneqq x_{\\alpha^t},\\qquad\np_\\alpha^t \\coloneqq p_{\\alpha^t},\n\\end{equation}\nor\n\\begin{equation}\n\\begin{pmatrix} \\rbr{m\\Omega}^{+1/2} x_\\alpha^t \\\\\n\\rbr{m\\Omega}^{-1/2} p_\\alpha^t \\end{pmatrix} \\coloneqq\n\\begin{pmatrix}\n\\cos\\Omega t & \\sin\\Omega t \\\\ -\\sin\\Omega t & \\cos\\Omega t\n\\end{pmatrix}\n\\begin{pmatrix} \\rbr{m\\Omega}^{+1/2} x_\\alpha \\\\\n\\rbr{m\\Omega}^{-1/2} p_\\alpha\\end{pmatrix}\n\\end{equation}\n\n\\subsection{Density matrix and Wigner function}\n\n\\begin{nameddef}{Density matrix in the particle number representation}\n\\begin{equation}\n\\Braket{n_1 | \\rfun{\\widehat{\\rho}}{t} | n_2} = \\ee^{-\\vbr{\\alpha}^2}\n\\frac{\\rbr{\\alpha^t}^{n_1} \\rbr{\\alpha^t{}^*}^{n_2}}{\\sqrt{n_1!n_2!}}\n= \\ee^{-\\vbr{\\alpha}^2} \\frac{\\alpha^{n_1} \n\\rbr{\\alpha^*}^{n_2}}{\\sqrt{n_1!n_2!}}\\ee^{-\\ii\\Omega\\rbr{n_1-n_2}t}.\n\\end{equation}\n\n\\end{nameddef} % Density matrix in the particle number representation\n\n\n\n\\begin{nameddef}{Density matrix in the position representation}\n\\begin{align}\n&\\phantom{{}={}} \\Braket{x_1 | \\rfun{\\widehat{\\rho}}{t} | x_2} \\nonumber \\\\\n&= \\rbr{\\frac{m\\Omega}{\\pp}}^{1/2} \\cfun{\\exp}{-\\frac{m\\Omega}{2}\\rbr{\\rbr{\nx_1-x_\\alpha^t}^2 + \\rbr{x_1-x_\\alpha^t}^2}}\\,\n\\ee^{+\\ii p_\\alpha^t\\rbr{x_1-x_2}}\n\\end{align}\n\\end{nameddef} % Density matrix in the position representation\n\n\\begin{nameddef}{Wigner function}\n\\begin{equation}\n\\rfun{W}{x, p; t} = \\frac{1}{\\pp} \\cfun{\\exp}{-m\\Omega \\rbr{x-x_\\alpha^t}^2\n-\\frac{1}{m\\Omega}\\rbr{p-p_\\alpha^t}^2}.\n\\end{equation}\n\\end{nameddef} % Wigner function\n\n", "meta": {"hexsha": "803ce38a7e218df74efbcb25a2c413774b73f60d", "size": 9827, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/gen-coh-squ/sections/coherent.tex", "max_stars_repo_name": "cmp0xff/Notes", "max_stars_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "old/gen-coh-squ/sections/coherent.tex", "max_issues_repo_name": "cmp0xff/Notes", "max_issues_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/gen-coh-squ/sections/coherent.tex", "max_forks_repo_name": "cmp0xff/Notes", "max_forks_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.2196721311, "max_line_length": 79, "alphanum_fraction": 0.6276584919, "num_tokens": 4259, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619177503205, "lm_q2_score": 0.7606506526772884, "lm_q1q2_score": 0.6271274958443501}} {"text": "\\documentclass[10pt]{article}\n\n\\usepackage{amsmath,amscd}\n\\usepackage{amssymb,array}\n\\usepackage{amsfonts,latexsym}\n\\usepackage[mathscr]{euscript}\n\\usepackage{graphicx,subfig,wrapfig}\n\\usepackage{times}\n\\usepackage{psfrag,epsfig}\n\\usepackage{verbatim}\n\\usepackage{tabularx}\n\\usepackage[toc,page]{appendix}\n\n\\newcommand{\\matlab}[1]{\\texttt{#1}}\n\\newcommand{\\setname}[1]{\\textsl{#1}}\n\\newcommand{\\Ce}{\\mathbb{C}}\n\\newcommand{\\Ree}{\\mathbb{R}}\n\\newcommand{\\p}{\\begin{pmatrix}}\n\\newcommand{\\pp}{\\end{pmatrix}}\n\\newcommand{\\bm}{\\begin{bmatrix}}\n\\newcommand{\\bb}{\\end{bmatrix}}\n\\newcommand{\\eul}[1]{e^{#1}}\n\n\\begin{document}\n\n\\title{ \\vspace{-30mm}Systems Bioengineering 3\\\\Homework 12}\n\\author{Greg Kiar}\n\n\\maketitle\n\\begin{enumerate}\n\\item\n%q1\n\\begin{enumerate}\n\\item For the following evaluation of average time, we make use of the fact that the transcript follows an exponential trend.\n\\begin{align*}\nE(T) = t_{avg} &= \\int_0^{\\infty} t e^{-\\alpha t} dt \\\\\n&= \\alpha \\begin{bmatrix} \\frac{-t}{\\alpha} - \\int_0^{\\infty} \\frac{1}{\\alpha} e^{-\\alpha t} dt \\end{bmatrix} \\\\\n&= \\alpha \\begin{bmatrix} \\frac{-t}{\\alpha}e^{-\\alpha t} + \\frac{-1}{\\alpha^2} e^{-\\alpha t}\\end{bmatrix}_{0,t} \\\\\nt_{avg} &= \\frac{1}{\\alpha} = 10 \\text{min}\n\\end{align*}\nAt equilibrium, $\\beta = \\alpha n$, $\\therefore \\mu_n = 10$ at equilibrium. The variance is $10$ since for independent stochastic events $\\sigma_n^2 = \\mu_n$.\n\\item We can see that the ODE of this system is: $\\dot{X}(t) = \\beta - \\alpha X(t)$. Solving this model using the equation: \\begin{align*} \\dot{X} &= \\frac{d}{dX} ln(X(t)) \\\\ 0 &= \\frac{d}{dX}\\begin{pmatrix} ln(\\frac{\\beta}{\\alpha} - n) \\end{pmatrix} \\\\\nn &= \\frac{\\beta}{\\alpha} = 10 \\end{align*}\n\\item Source code for this part was provided by the TAs. It has been modified slightly and is attached in an appendix.\n\\item When starting the simulation at $n=10$ and running it for $10,000$ min, the mean and variance were $\\mu = 10.1287$ and $\\sigma^2 = 9.8604$, respectively. These values are near but not exactly expected values. This makes sense because the discrete stochastic system has a finitely small timestep. If the timesteps were infinitely small or the trial infinitely long, the values would approach the theoretical values.\n\\end{enumerate}\n\n\\item \\begin{enumerate}\n\\item Solving $\\dot{X}(t) = \\beta - \\alpha X(t)$ for the case prior to equilibrium allows us to use the Laplace transform. The solution of this system becomes \\begin{align*} X(t) = \\frac{\\beta}{\\alpha} (1 - e^{-\\alpha t}) \\end{align*}\n\\item Shown below is the solution for $1000$ randomly generated trajectories each for $100 min$. \\\\ \\includegraphics[scale=0.2]{sbe3hw12q2b.png}\n\\item It can be seen by the figure below, a plot of the variance, that the variance follows the same exponential trend as the random variable $n$. From this, we can set an approximate equation for the variance to be $\\sigma^2 = \\frac{\\beta}{\\alpha} (1 - e^{-\\alpha t}) $. From this we can see that the variance is at a maximum as $t \\to \\infty$. \\includegraphics[scale=0.2]{sbe3hw12q2c.png}\n\\end{enumerate}\n\n\\item \\begin{enumerate}\n\\item Similarly to the equation found in 2a), we can solve $ \\dot{X}(t) = \\frac{\\beta}{b} - \\alpha X(t) $, where $b=5$ here. The result for $X(t)$ is as follows: \\begin{align*} X(t) &= \\frac{\\beta}{b \\alpha} (1 - e^{-\\alpha t})\\end{align*}\nWe can see from this result that the burst size, $b$, affects the forward rate of the system. The forward rate$ = \\frac{\\beta}{b}$.\n\\item Simulating the stochastic and ODE models of this system for $10,000 min$ with the inital condition $X(0) = n(0) = 2$, the equilibrium value, the discrete model produced the values shown below. \\begin{align*} \\text{Discrete} && \\mu &= 12.0428 && \\sigma^2 = 30.3180 \\\\ \\end{align*}\n\\item For the stochastic system with larger burst sizes, the mean and variance both increase. When testing this for different burst sizes, an approximately linear relationship is observed for both the mean and variance values. This relationship does not change as $\\beta$ or $\\alpha$ increase and decrease simultaneously to one another. The approximate expressions are as follows: \\begin{align*} \\mu &= \\frac{\\beta}{\\alpha}\\\\ \\sigma^2 &= \\frac{2 \\beta}{b \\alpha} \\end{align*}\n\\end{enumerate}\n\\end{enumerate}\n\\newpage\n\\begin{appendices}\n\\section{Trajectories}\n\\input{get_trajectory}\n\\section{Example main script}\n\\input{Prob2_forStudents}\n\\end{appendices}\n\\end{document}\n", "meta": {"hexsha": "cefd8a7605ed8f3e7a94e4b283856f255497711e", "size": 4418, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sbe3/Homework/HW12/sb3_hw12.tex", "max_stars_repo_name": "gkiar/sb2", "max_stars_repo_head_hexsha": "55eafc3078aaf248ea53ba5ccbc221aae9dab093", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2016-12-07T15:51:49.000Z", "max_stars_repo_stars_event_max_datetime": "2019-05-21T09:35:04.000Z", "max_issues_repo_path": "sbe3/Homework/HW12/sb3_hw12.tex", "max_issues_repo_name": "gkiar/sb2", "max_issues_repo_head_hexsha": "55eafc3078aaf248ea53ba5ccbc221aae9dab093", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2015-03-30T18:13:30.000Z", "max_issues_repo_issues_event_max_datetime": "2015-03-30T23:37:57.000Z", "max_forks_repo_path": "sbe3/Homework/HW12/sb3_hw12.tex", "max_forks_repo_name": "gkiar/jhu-bme-mse", "max_forks_repo_head_hexsha": "55eafc3078aaf248ea53ba5ccbc221aae9dab093", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-05-26T14:05:56.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-01T03:55:41.000Z", "avg_line_length": 64.0289855072, "max_line_length": 475, "alphanum_fraction": 0.713218651, "num_tokens": 1383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585786300049, "lm_q2_score": 0.8577681013541613, "lm_q1q2_score": 0.6270787289701311}} {"text": "\n\\subsection{Hodge stars}\n\nThe Hodge star operator is a generalisation of cross product.\nIn 3d space if we have a plane, we can get a vector perpendicular and visa versa.\nGenerally, we are in \\(n\\)-dimensional space and we input \\(k\\) vectors and get out \\(n-k\\) vectors.\n\n\\subsection{Hodge duals}\n\n", "meta": {"hexsha": "ba56cbd3fe3eb7fe71b28352f36d5553c69e3a9f", "size": 299, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/analysis/multiCalculus/04-03-hodge.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/analysis/multiCalculus/04-03-hodge.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/analysis/multiCalculus/04-03-hodge.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.9, "max_line_length": 100, "alphanum_fraction": 0.745819398, "num_tokens": 78, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8577681049901037, "lm_q2_score": 0.7310585727705126, "lm_q1q2_score": 0.6270787266021324}} {"text": "% file: problems/euclid.tex\n\n\\section{Extra Problem: $\\euclid(m, n)$} \\label{section:euclid}\n\nProve the recursive Euclid algorithm as shown in Algorithm~\\ref{alg:euclid} for \ncomputing the greatest common divisor (gcd) of two natural numbers are totally correct.\n\n\\begin{marginfigure}%\n \\includegraphics[width=0.60\\linewidth]{figs/euclid}\n \\label{fig:euclid}\n\\end{marginfigure}\n\n\\input{algs/euclid}\n\n\\subsection{Proof}\n\n\\marginnote{Also pay attention to the way how to write a mathematical induction proof.}\n\nWe prove the partial correctness of \\euclid{} by strong mathematical induction on $n$,\nwith $m$ any fixed natural number.\n\n\\begin{description}\n \\item[Basis:] $n = 0$. We have that\n \\marginnote{Make sure you understand each of these three ``='''s:\n \\begin{enumerate}[(1)]\n\t\\item By $n = 0$;\n\t\\item By the property of $\\gcd$;\n\t\\item By the \\euclid{} algorithm.\n \\end{enumerate}\n }\n \\[\n \\gcd(m,n) = \\gcd(m,0) = m = \\euclid(m,0).\n \\]\n \\item[Inductive Hypothesis:]\n Suppose that $n \\ge 1$ and\n \\[\n \\gcd(m, k) = \\euclid(m, k), \\;\\forall 0 \\le k \\le n - 1.\n \\]\n \\item[Inductive Step:]\n We need to prove that ($n \\ge 1$)\n \\[\n \\gcd(m, n) = \\euclid(m, n).\n \\]\n\n According to \\euclid{}, we have\n \\[\n \\euclid(m,n) = \\euclid(n, m \\bmod n).\n \\]\n Since $(m \\bmod n) < n$, by the inductive hypothesis, we have\n \\[\n \\euclid(n, m \\bmod n) = \\gcd(n, m \\bmod n).\n \\]\n Therefore, it suffices to prove that\n \\[\n \\boxed{\\gcd(m,n) = \\gcd(n, m \\bmod n).}\n \\]\n For notational convenience, we denote\n \\[\n d = \\gcd(m,n), \\quad d' = \\gcd(n, m \\bmod n).\n \\]\n Because $d, d' \\ge 0$, it is sufficient to obtain $d = d'$\n by showing that $d \\mid d'$ and $d' \\mid d$:\n\n \\begin{itemize}\n \\item $d \\mid d'$.\n \\item $d' \\mid d$.\n \\end{itemize}\n\\end{description}\n", "meta": {"hexsha": "a5316f47851a892fadfed69a7bdadd734a57e29e", "size": 1873, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2017/2017-4th/4-0-sample/problems/euclid.tex", "max_stars_repo_name": "courses-at-nju-by-junma/problem-solving-class-problems", "max_stars_repo_head_hexsha": "79de740506000972b2bec91cc6042fa639cd2e55", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 13, "max_stars_repo_stars_event_min_datetime": "2018-03-16T04:33:03.000Z", "max_stars_repo_stars_event_max_datetime": "2020-02-11T14:50:38.000Z", "max_issues_repo_path": "2017/2017-4th/4-0-sample/problems/euclid.tex", "max_issues_repo_name": "courses-at-nju-by-junma/problem-solving-class-problems", "max_issues_repo_head_hexsha": "79de740506000972b2bec91cc6042fa639cd2e55", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 23, "max_issues_repo_issues_event_min_datetime": "2018-03-19T10:36:11.000Z", "max_issues_repo_issues_event_max_datetime": "2019-12-03T04:58:39.000Z", "max_forks_repo_path": "2017/2017-4th/4-0-sample/problems/euclid.tex", "max_forks_repo_name": "courses-at-nju-by-junma/problem-solving-class-problems", "max_forks_repo_head_hexsha": "79de740506000972b2bec91cc6042fa639cd2e55", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 14, "max_forks_repo_forks_event_min_datetime": "2018-03-16T04:26:21.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-11T11:42:48.000Z", "avg_line_length": 27.1449275362, "max_line_length": 87, "alphanum_fraction": 0.5969033636, "num_tokens": 633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217431943271999, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.6270698078963841}} {"text": "\\paragraph{Adversarial accuracy:} check if data point in test set is classified correctly \\emph{and} network is robust in region around point (e.g., using PGD in $\\epsilon$ $L_\\infty$-ball). Often have tradeoff with standard accuracy. \n\\paragraph{Opt. Prob.:} $\\displaystyle \\argmin_\\theta \\E_{(x, y) \\sim D} \\left[ \\max_{x' \\in S(x)} \\loss(\\theta; x', y)\\right]$\n\\subsection*{PGD training}\n\\begin{enumerate}\n \\item Select mini-batch $B$\n \\item $\\displaystyle B_\\text{max} \\gets \\{\\argmax_{x' \\in S(x)} \\loss(\\theta; x', y) \\lvert x \\in B\\}$\n \\item $\\displaystyle \\theta \\gets \\theta - \\frac{1}{\\lvert B_\\text{max} \\rvert} \\sum_{(x, y) \\in B_\\text{max}} \\nabla_\\theta \\loss(\\theta; x, y)$\n\\end{enumerate}\n\\subsection*{TRADES}\n$$ \\argmin_\\theta \\E_{(x, y) \\sim D} \\left[ \\loss(\\theta; x, y) + \\lambda \\cdot \\max_{x' \\in S(x)} \\loss(\\theta; x', f_\\theta(x))\\right]$$", "meta": {"hexsha": "ad6521df51ad9144d0d53cb6de1f522ebfc626b3", "size": 873, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "adversarial-defenses.tex", "max_stars_repo_name": "cknabs/RIAI-summary-HS2020", "max_stars_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-01-20T21:27:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-24T20:28:56.000Z", "max_issues_repo_path": "adversarial-defenses.tex", "max_issues_repo_name": "cknabs/RIAI-summary-HS2020", "max_issues_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-01-25T09:29:16.000Z", "max_issues_repo_issues_event_max_datetime": "2021-01-25T10:50:09.000Z", "max_forks_repo_path": "adversarial-defenses.tex", "max_forks_repo_name": "cknabs/RIAI-summary-HS2020", "max_forks_repo_head_hexsha": "42a1ee3cc2e51c52188f842c78923792bd0f3edf", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 87.3, "max_line_length": 235, "alphanum_fraction": 0.6563573883, "num_tokens": 318, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267626522814, "lm_q2_score": 0.7217431943271998, "lm_q1q2_score": 0.6270698029936175}} {"text": "% !TeX root = ../diss.tex\r\n\r\n\\chapter{Preparation}\r\n\r\nThere were two main areas of preparation for my project. One was to research the Smith-Waterman (SW) algorithm, and the ways it has been modified and implemented by others.\r\nThe other area was researching the different architectures, programming environment and tools to use.\r\n\r\n\\section{Overview of the Smith-Waterman algorithm}\r\n\\label{sec:SW_Overview}\r\nSequence alignment algorithms are used to find similar regions between two strings.\r\nThis involves predicting where symbols in the two strings correlate with each other (both matching and mismatching symbols), or where one sequence has some symbols that have been inserted into that sequence or deleted from the other.\r\nThis can be represented as follows, where symbols in the same column correspond to one another, and dashes represent insertions or deletions (gaps):\r\n\\begin{center}\r\n\\begin{tabular}{c}\r\n\\begin{lstlisting}[basicstyle=\\ttfamily\\linespread{0.8}]\r\nAC-T\r\nACGT\r\n\\end{lstlisting}\r\n\\end{tabular}\r\n\\end{center}\r\n\r\nThe Smith-Waterman algorithm \\cite{SW_Original} is a local alignment algorithm; it can produce alignments that span a subset of the input sequences.\r\nThis is useful when aligning long segments of DNA which contain non-coding regions or multiple different genes (which occurs when aligning against an entire chromosome).\r\nWhen aligning a single gene against a chromosome from a different species, a local alignment algorithm can find the relevant region whilst excluding the rest of the longer sequence in the alignment.\r\n\r\n\\subsection{The Dynamic Programming approach}\r\n\\label{sec:SW_DP}\r\n\r\nThe Smith-Waterman algorithm \\cite{SW_Original} finds alignments using dynamic programming.\r\nFor sequences $A=a_1 a_2 \\ldots a_N$ and $B=b_1 b_2 \\ldots b_M$, a scoring matrix $H$ is constructed where cell $H_{i,j}$ gives the score of the best local alignment that ends with $a_i$ and $b_j$.\r\nUsing a similarity function $s(a,b)$ which gives the similarity of two sequence symbols, and gap score list $W_k$ which gives the penalty for a deletion of length $k$, $H_{i,j}$ is defined as follows:\r\n$$ H_{i,j} = \\max \\begin{cases}\r\nH_{i-1,j-1} + s(a_i, b_j) & \\text{(Including both symbols in alignment)} \\\\\r\n\\max_{k \\geq 1}(H_{i-k,j} - W_k) & \\text{(Skipping $k$ symbols from $a$)} \\\\\r\n\\max_{l \\geq 1}(H_{i,j-l} - W_l) & \\text{(Skipping $k$ symbols from $b$)} \\\\\r\n0 & \\text{(Starting alignment here)}\r\n\\end{cases}$$\r\n\r\nThe grid is initialised using $\\forall i\\leq N . H_{i,0}=0$ and $\\forall j \\leq M . H_{0,j}=0$.\r\n\r\nFor example, using sequences {\\ttfamily GACT} and {\\ttfamily ACGT} and the following definitions of $s(a,b)$ and $W_k$:\r\n$$s(a, b)= \\begin{cases}\r\n5 & a = b \\\\\r\n-4 & a \\neq b\r\n\\end{cases}\r\n\\qquad \\qquad \\qquad W_k = -k\r\n$$\r\n\r\n\\begin{center}\r\n$H = $ \\begin{tabular}{|c|ccccc|} \\hline\r\n& & {\\ttfamily A} & {\\ttfamily C} & {\\ttfamily G} & {\\ttfamily T} \\\\ \\hline\r\n& $0$ & $0$ & $0$ & $0$ & $0$ \\\\\r\n{\\ttfamily G} & $0$ & $0$ & $0$ & $3$ & $2$ \\\\\r\n{\\ttfamily A} & $0$ & $3$ & $2$ & $2$ & ${\\color{darkorange} 1}$ \\\\\r\n{\\ttfamily C} & $0$ & $2$ & ${\\color{darkblue}6}$ & $5$ & $4$ \\\\\r\n{\\ttfamily T} & $0$ & $1$ & $5$ & $4$ & $8$ \\\\ \\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\r\nTaking two examples from this:\r\n\\begin{align*}\r\n{\\color{darkblue}H_{3,2}} = 6 &= \\max \\begin{cases}\r\n 3 + s(\\text{\\ttfamily C}, \\text{\\ttfamily C}) =6\\\\\r\n \\max(2-1, 0 - 2, 0-3) =1 \\\\\r\n \\max(2-1, 0 - 2) =1\\\\\r\n 0\r\n\\end{cases} \\\\\r\n{\\color{darkorange} H_{2,4}} = 1 &= \\max \\begin{cases}\r\n 3 + s(\\text{\\ttfamily A}, \\text{\\ttfamily T}) =1 \\\\\r\n \\max(2-1, 0 - 2) =1 \\\\\r\n \\max(2-1, 2 - 2, 3-3, 0-4) =1 \\\\\r\n 0\r\n\\end{cases}\r\n\\end{align*}\r\nNote that in the case of $H_{2,4}$ (and others) there are multiple ways of getting to the maximum value (from above, the left, or diagonally in this case), and this leads to multiple alignments that have the same optimal score.\r\n\r\n\\subsection{Back-tracing}\r\n\\label{sec:SW_Back_tracing}\r\nStored with each grid cell is a ``pointer'' which locates the previous cell in the alignment ending with that cell, where pointer $p \\in \\{Above_k, Left_k, Diagonal, Nil\\}$. $Nil$ represents this cell not being part of an alignment, because all paths to that cell are worse than the threshold $0$. $Left_k$ represents skipping $k$ symbols from $a$, likewise for $Above_k$ skipping $k$ symbols from $b$. In the general case, this matrix of pointers $P$ is found using:\r\n$$ P_{i,j} = \\begin{cases}\r\n Diagonal, & H_{i,j} = H_{i-1,j-1} + s(a_i, b_j) \\\\\r\n Left_k, & H_{i,j} = \\max_{k \\geq 1}(H_{i-k,j} - W_k) \\\\\r\n Above_k, & H_{i,j} = \\max_{l \\geq 1}(H_{i,j-l} - W_l) \\\\\r\n Nil, & H_{i,j} = 0\r\n\\end{cases}$$\r\n\\pagebreak\r\n\r\nAn example of $P_{i,j}$ is as follows (where $\\cdot$ represents $Nil$):\r\n\\begin{center}\r\n$P = $ \\begin{tabular}{|p{0.035\\textwidth}|p{0.035\\textwidth}p{0.035\\textwidth}p{0.035\\textwidth}p{0.035\\textwidth}p{0.035\\textwidth}|} \\hline\r\n& & \\hfil {\\ttfamily A} & \\hfil {\\ttfamily C} & \\hfil {\\ttfamily G} & \\hfil {\\ttfamily T} \\\\ \\hline\r\n& \\hfil $\\cdot$ & \\hfil $\\cdot$ & \\hfil $\\cdot$ & \\hfil $\\cdot$ & \\hfil $\\cdot$ \\\\\r\n\\hfil {\\ttfamily G} & \\hfil ${\\color{darkorange} \\cdot}$ & \\hfil $\\cdot$ & \\hfil $\\cdot$ & \\hfil $\\nwarrow$ & \\hfil $\\leftarrow_1$ \\\\\r\n\\hfil {\\ttfamily A} & \\hfil $\\cdot$ & \\hfil ${\\color{darkorange} \\nwarrow}$ & \\hfil $\\leftarrow_1$ & \\hfil $\\uparrow_1$ & $\\leftarrow_1$ \\\\\r\n\\hfil {\\ttfamily C} & \\hfil $\\cdot$ & \\hfil $\\uparrow_1$ & \\hfil ${\\color{darkorange} \\nwarrow}$ & \\hfil ${\\color{darkorange} \\leftarrow_1}$ & $\\leftarrow_2$ \\\\\r\n\\hfil {\\ttfamily T} & \\hfil $\\cdot$ & \\hfil $\\uparrow_2$ & \\hfil $\\nwarrow$ & \\hfil $\\leftarrow_1$ & \\hfil ${\\color{darkorange} \\nwarrow}$ \\\\ \\hline\r\n\\end{tabular}\r\n\\end{center}\r\n\r\nA running maximum score is maintained during the production of the grid, and the cell $H_{i,j}$ with the highest score corresponds to the highest-scoring alignment, which finishes with $a_i$ and $b_j$.\r\nStarting here, the alignment can be found from end to start by following the pointers in the grid. The best path through $P_{i,j}$ is highlighted in {\\color{darkorange}orange}, and gives alignment:\r\n\\begin{center}\r\n\\begin{tabular}{c}\r\n\\begin{lstlisting}[basicstyle=\\ttfamily\\linespread{0.8}]\r\nAC-T\r\nACGT\r\n\\end{lstlisting}\r\n\\end{tabular}\r\n\\end{center}\r\n\r\n\\subsection{Algorithmic complexity}\r\n\\label{sec:SW_Complexity}\r\nFor sequences of lengths $N$ and $M$, this algorithm requires $O(NM)$ space to store the two matrices, each with $(N+1)\\times(M+1)$ elements.\r\nThe runtime is $O(N^2 M)$, where without loss of generality $N\\geq M$.\r\nEach cell needs to consider the cell to its diagonal (constant time), and all of the cells above it (of order $M$ cells), and all cells to its left (of order $N$ cells).\r\nThis gives a runtime for evaluating a cell as $O(N+M)$, which is $O(N)$ given $N\\geq M$.\r\nThere are $NM$ cells to be evaluated, so producing the grid takes $O(N^2 M)$ time.\r\n\r\nIf the gap evaluation scheme did not maximise over a row, and instead performed constant time operations, then the complexity is $O(NM)$.\r\nThis is discussed in \\cref{sec:SW_gaps}.\r\n\r\nEach back-tracing step requires constant time (a matrix lookup, and some pointer comparisons) to produce each element in the alignment, and the length of alignment is bounded by $N+M$, where the alignment is wholly made up of gaps.\r\nTherefore, this step requires $O(N)$ space and time, where $N\\geq M$.\r\n\r\nOverall, the space requirements of the algorithm as presented is $O(NM)$, and the time requirements are either $O(NM)$ or $O(N^2 M)$ depending on how gaps are evaluated.\r\nImprovements to this are discussed in \\cref{sec:SW_Linear_Prep}.\r\n\r\n\\subsection{Scoring alignments}\r\n\\label{sec:SW_Scoring}\r\n\r\nThe general approach to scoring alignments is given in the paper written by Smith and Waterman \\cite{SW_Original}, but the exact details are not specified.\r\nThere are two areas that need definition: how to compare a pair of sequence symbols, and how to score gaps in the alignment.\r\n\r\n\\subsubsection{Sequence similarity functions}\r\n\\label{sec:SW_similarity_functions}\r\nThe original paper requires similarity function $s(a,b)$ which gives the similarity of two sequence symbols. One simple implementation might be:\r\n$$s(a, b) = \\begin{cases}\r\nm, & a = b \\\\\r\n\\mu, & a \\neq b\r\n\\end{cases}$$\r\n\r\nFor two constants $m$ and $\\mu$ where typically $m>0$ and $\\mu<0$.\r\nThe implementation Smith-Waterman in the European Molecular Biology Open Software Suite (EMBOSS) \\cite{EMBOSS} uses an approach equivalent to using $m=5$ and $\\mu=-4$ for DNA alignments by default.\r\n\r\nHowever, the biological processes that leads to one symbol being substituted for another cannot be entirely expressed in these two constants, especially when some symbols are more likely than others to replace a given symbol.\r\nThis is particularly problematic in proteins, which naturally are made up of 22 different types of amino acids.\r\n\r\nA more granular approach is to base $s(a,b)$ on a substitution matrix, which encodes a score for each different pairing of symbols.\r\nThis has performance penalties (\\cref{sec:C_scoring_eval,sec:CUDA_scoring_eval,sec:SV_expected_v_actual,sec:FPGA_utilisation,sec:SV_Fmax}), because the first approach can use direct equality testing and hard-coded constants, whereas using a substitution matrix for proteins requires a lookup from a $22\\times22$ matrix.\r\n\r\nDiscussions of which substitution matrices are best for which applications are beyond the scope of this project, and they are all implemented in the same way.\r\nI chose to base investigations into the performance implications into this on the BLOSUM50 matrix for proteins \\cite{BLOSUM}, which is used for alignments between sequences that are not closely related.\r\n\r\n\\subsubsection{Scoring gaps}\r\n\\label{sec:SW_gaps}\r\nThe original paper \\cite{SW_Original} references a gap weight vector $W_k$, which defines the score penalty of a gap of length $k$, where gaps jump either $k$ elements to the left, or above.\r\nHowever, it makes the algorithm run in $O(N^2 M)$ time for $N\\geq M$ (\\cref{sec:SW_Complexity}), so other approaches are used instead.\r\n\r\nOne solution to this problem was suggested by Waterman and Smith \\cite{SW_Metrics} in an earlier paper about global alignment: to define $W_k=kw_1$ for some gap penalty $w_1$.\r\nUsing the recursive definition of $H_{i,j}$ the score using a $Left$ gap is:\r\n$$\\max_{k \\geq 1}(H_{i-k,j} -W_k) = H_{i-1,j} - w_1$$\r\nGotoh proposed a different way of solving this problem \\cite{Gotoh}.\r\nThe constant gap penalty models the cost of inserting or deleting symbols in a sequence by making this penalty proportional to the length of the insertion or deletion.\r\nThis is not ideal, because as Gotoh observes, ``long gap[s] can be produced by a single mutational event''.\r\nInstead he proposes what is later called the affine-gap method, where $W_k=uk+v$ for some gap penalties $u$ and $v$.\r\nThe penalty of starting to insert or delete a sequence is given by $u$, and $v$ accounts for the different lengths of such a sequence.\r\n\r\nThis approach requires more space than the original Smith-Waterman algorithm because choosing whether to open a gap has an impact on scores later in the alignment.\r\nWhen evaluating a mismatched cell in the dynamic-programming grid it is impossible to know whether to pay the mismatch penalty because a sequence of matches immediately follow, or if a long sequence of mismatches follow and it would be less expensive to open up a gap to cover them instead.\r\nTherefore, extra space is needed to keep track of scores when the decision to open a gap is and also is not taken.\r\n\r\nTo achieve this, three grids are constructed instead of one.\r\nThis triples the memory requirements in practice but does not change the asymptotic memory requirements.\r\nThe main score grid $H_{i,j}$ is calculated using a similar approach to before, representing the score of the best alignment using sequences $a_1\\ldots a_i$ and $b_1 \\ldots b_j$.\r\nAdditionally, there is grid $F_{i,j}$ which represents the score of the best alignment between $a_1\\ldots a_i$ and $b_1 \\ldots b_j$ provided it ended with an open left-gap.\r\n$G_{i,j}$ is similar for above gaps.\r\nThis gives grid definitions:\r\n\\begin{align*}\r\nH_{i,j} &= \\max(H_{i-1,j-1} + s(a_i, b_j), F_{i,j}, G_{i,j}, 0) \\\\\r\nF_{i,j} &= \\max(H_{i-1,j} + u + v, F_{i-1,j} + v) \\\\\r\nG_{i,j} &= \\max(H_{i,j-1} + u + v, G_{i,j-1} + v)\r\n\\end{align*}\r\n\r\nThis is modified from the original paper somewhat for notational consistency with the rest of the dissertation.\r\n\r\nPointer grids need to be stored for all three grids, where the back-tracer will start in grid $H$ (and this is the only grid where the running maximum is kept), but it will trace through $F$ and $G$ as well to determine where a gap started.\r\n\r\n\\subsection{Linear space modification to Smith-Waterman algorithm}\r\n\\label{sec:SW_Linear_Prep}\r\n\\subsubsection{Overview}\r\n\\label{sec:SW_Linear_Overview}\r\n\r\nA common approach to improving the runtime of the Smith-Waterman algorithm is to reduce the amount of space used by increasing the amount of time spent in doing arithmetic.\r\nHirschberg’s algorithm \\cite{Hirschberg} pioneered this approach for the Longest Common Subsequence problem.\r\nMyers and Miller apply this approach to local alignments \\cite{MyersMiller}, which is summarised below.\r\n\r\nThe purpose of this algorithm is to correctly and efficiently split the alignment into small chunks, manageable for the quadratic space implementation which is where the back-tracing is done.\r\nIt recursively divides a large grid into two smaller grids.\r\n\\Cref{fig:Linear_Space_Structure} shows this, where the optimal alignment is shown in orange.\r\nThe grid has been divided into two smaller grids by the first pass of the algorithm, a grid with blocks $1$ and $2$, and the grid with blocks $3$ and $4$.\r\nThe grey segments do not contain the alignment and can be ignored.\r\nThe black arrows represent the direction the blocks are evaluated in, discussed further on.\r\n\r\n\\Cref{lst:pseudo_sw} is a pseudo-code representation of the quadratic space Smith-Waterman algorithm, to be used in this linear space algorithm.\r\nParameter \\lstinline{fixedTop} specifies whether the alignment can start from anywhere in the grid, or just the top left corner (and is implemented by including or removing the options of $0$ and $Nil$ from the maximising functions).\r\nThis is necessary because the optimal alignment inside a subproblem may not start at the top-left, but the overall optimal alignment may pass through multiple sub-grids where the top-left of each grid must be connected to the bottom-right of the next grid.\r\nParameter \\lstinline{fixedBottom} is similar, specifying whether to start back-tracing at the best cell in the grid, or the bottom-right cell.\r\n\r\n\\begin{figure}\r\n \\centering\r\n \\includegraphics[width=(0.4\\textwidth)]{figs/linear_space_structure.pdf}\r\n \\caption{The divided dynamic programming matrix when using the linear space approach, after the first iteration}\r\n \\label{fig:Linear_Space_Structure}\r\n\\end{figure}\r\n\r\n\\begin{lstlisting}[basicstyle=\\linespread{0.9}\\ttfamily\\footnotesize, label={lst:pseudo_sw},captionpos=b,caption={Pseudo-code representation of quadratic space implementation of SW algorithm}]\r\nfun sw(seq1, seq2, fixedTop, fixedBottom):\r\n Dynamic programming approach using quadratic space\r\n if fixedTop:\r\n Only allow alignments to start at the top-left\r\n else:\r\n Allow alignments to start anywhere\r\n if fixedBottom:\r\n Return the alignment ending at the bottom-right grid cell\r\n else:\r\n Return the alignment ending at the best cell in the grid\r\n\\end{lstlisting}\r\n\r\nTo split a large grid into two smaller grids, the scores of the middle column of cells are found by starting from the top-left and working towards the middle, and also by starting from the bottom right and working towards the middle.\r\n\\Cref{lst:pseudo_sw_col} is a representation of this.\r\nIn \\cref{fig:Linear_Space_Structure}, blocks $1$ and $3$ are evaluated from their top-lefts to their middle columns, and blocks $2$ and $4$ from the bottom-right, and the direction of evaluation represented by the black arrows.\r\nThese scores are found using a linear space.\r\nThese two columns are added together, and each cell of the column gives the score for the best alignment that passes through that cell.\r\nThe optimal alignment might cross the middle column or be in one of these half-grids.\r\n\r\nThe middle columns are found using $O(N)$ space for sequences of lengths $N$ and $M$, and maintains $O(NM)$ time complexity, albeit with a greater constant factor.\r\nThis uses the fact that only the immediate neighbours are needed to calculate the score of a cell, so the values of a previous column can be discarded once the values in the next column have been computed. This data dependency is shown in \\cref{fig:Linear_Space_Dependencies}.\r\nHowever, discarding this information leads to recomputation of the cells in the smaller subproblems that are recursively evaluated.\r\n\r\nMaxima are calculated whilst computing each half, and compared to the maximum cell in the summed columns.\r\nIf the overall maximum is in one of the half-grids, restart the algorithm between that cell and the top-left or bottom-right, if on the left or right sides respectively.\r\nOtherwise, the algorithm is rerun between the end cells and the maximum middle cell.\r\nThis recursively breaks up the alignment until it is small enough to tackle directly using the quadratic space algorithm (\\cref{sec:SW_DP,sec:SW_Back_tracing,sec:SW_Complexity}), and then joins them back together.\r\n\\begin{figure}[h]\r\n \\centering\r\n \\includegraphics[width=(0.4\\textwidth)]{figs/linear_space_deps.pdf}\r\n \\caption{Data dependencies when using the linear space approach}\r\n \\label{fig:Linear_Space_Dependencies}\r\n\\end{figure}\r\n\\begin{lstlisting}[basicstyle=\\ttfamily\\linespread{0.9}\\footnotesize, label={lst:pseudo_sw_col},captionpos=b,caption={Pseudo-code representation of linear space algorithm which finds middle columns}]\r\n fun sw_col(seq1, seq2, direction, fixedStart):\r\n direction in {forwards, backwards}\r\n direction determines which half of grid is evaluated\r\n - Forwards: from top left, to column (seq2/2)\r\n - Backwards: from bottom right, to column (seq2/2)\r\n fixedStart similar to fixedTop of sw()\r\n Calculate grid column by column, returning:\r\n - Values of column (seq2/2)\r\n - Location of best cell, and score\r\n \\end{lstlisting}\r\n\\begin{minipage}{\\linewidth}\r\n\\begin{lstlisting}[basicstyle=\\linespread{0.9}\\ttfamily\\footnotesize, label={lst:pseudo_sw_linear},captionpos=b,caption={Pseudo-code representation of the linear space Smith-Waterman algorithm}]\r\nfun sw_linear(seq1, seq2, fixedTop, fixedBottom):\r\n if seq1 and seq2 are small:\r\n return sw(seq1, seq2, fixedTop, fixedBottom)\r\n else:\r\n (forwardCol, bestForwards) = sw_col(seq1, seq2, forwards, fixedTop)\r\n (backwardCol, bestBackwards) = sw_col(seq1, seq2,\r\n backwards, fixedBottom)\r\n middleCol = forwardCol + backwardCol # Using vector addition\r\n bestMiddle = max(middleCol)\r\n\r\n # Using list splicing: seq1[x:] = [seq1[x], seq1[x+1], ... seq1[n]]\r\n if best path is bottom-right to bestBackwards and not fixedTop:\r\n return sw_linear(seq1[bestForwards.i:], seq2[bestForwards.j:],\r\n fixedTop, fixedBottom)\r\n else if best path is top-left to bestForwards and not fixedBottom:\r\n return sw_linear(seq1[0:bestForwards.i], seq2[0:bestForwards.j],\r\n fixedTop, fixedBottom)\r\n else use middle cell:\r\n alignStart = sw_linear(seq1[0:bestMiddle.i],seq2[0:len(seq2/2)],\r\n fixedTop, true)\r\n alignEnd = sw_linear(seq1[bestMiddle.i:],seq2[len(seq2/2):],\r\n true, fixedBottom)\r\n return alignStart @ alignEnd\r\n\\end{lstlisting}\r\n\\end{minipage}\r\n\r\n\\subsubsection{Back-tracing}\r\n\\label{sec:SW_Linear_Back_tracing}\r\n\r\nThe process of building the final alignment, back-tracing, is done in parts and later joined, in accordance with the divide-and-conquer approach of this algorithm.\r\nWhen the large grid has been recursively divided to small enough grids, the quadratic space algorithm (\\cref{sec:SW_DP,sec:SW_Back_tracing}) is used to find an alignment.\r\nThis is because producing an alignment requires a grid of pointers taking $O(NM)$ space, so when the grid is small enough for a full grid of pointers to be made, a full grid of scores can also be made.\r\nThe resulting aligned sequences from each of these small alignments can be joined together to make the overall alignment.\r\n\r\n\\subsubsection{Runtime complexity}\r\n\\label{sec:SW_Linear_Complexity}\r\nThis additional arithmetic does not impact the runtime complexity.\r\nFor sequences of lengths $N$ and $M$ the linear space approach evaluates $2NM$ grid cells (each taking $O(1)$ time).\r\n\r\nThe amount of work done in each iteration halves. The first step evaluates every grid cell, taking $NM$ evaluations (over two halves of size $\\frac{N}{2} \\times M$).\r\nIf the alignment is in one half, then the next area to evaluate has at least halved.\r\nIf the alignment is split across the middle column, some point at position $m$ will be chosen on that column.\r\nThen two grids will be evaluated of area $\\frac{N}{2} \\times m$ and $\\frac{N}{2} \\times (M-m)$ which sum to $\\frac{NM}{2}$ for all possible $m$.\r\nUsing this repeated halving of work, the total number of grid evaluations required is:\r\n$$ \\text{Cells evaluated } = NM + \\frac{NM}{2} + \\frac{NM}{4} + \\cdots \\leq NM \\sum_{i=0}^\\infty 2^{-i} = 2NM $$\r\n\r\n\\section{Overview of architectures and programming languages used}\r\n\\label{sec:Architecture_prep}\r\nThe objective of my project was to implement the Smith-Waterman algorithm using different hardware architectures (general purpose CPUs, GPUs, and FPGAs).\r\nAs part of my preparation I learnt about these architectures so I could produce implementations that were best suited for these different architectures.\r\n\r\n\\subsection{Central Processing Units and C}\r\n\\label{sec:CPU_prep}\r\nModern general-purpose Central Processing Units (CPUs) can have very high clock speeds, multiple processing cores, and have rich Instruction Set Architectures (ISAs).\r\nThe CPU in my laptop, used by this project, is an Intel Core i7-8750H.\r\nThis has six processing cores, and has a maximum clock frequency of $\\SI{4.1}{\\giga\\hertz}$ \\cite{i7-8750H}.\r\nThis clock frequency is significantly higher than the other platforms I used, however other implementations will be able to do more work per clock cycle, using increased parallelism or specialisation.\r\n\r\nI chose to write my programs for the CPU in C. The major reason I chose to do so was that the GPU APIs available (CUDA and OpenCL) are APIs for C/C++, so for the sake of interoperability it made sense to implement my project in either C or C++.\r\nThis would allow me to later reuse segments of code for the GPU implementation, especially the control logic and testing framework.\r\nAlso, I had previous experience with C from Part IB and a summer internship, so I chose to use C.\r\n\r\nThe focus of my project is exploring parallelism in Smith-Waterman, and with a multicore CPU I could implement a solution that was multi-threaded.\r\nThis is not a native feature in the C language, so I used the Native POSIX Thread Library, part of the GNU C Library.\r\nThis allows multiple threads to be run on multiple processing cores, where all of the scheduling is handled by the library and the host operating system.\r\n\r\n\\subsection{Graphics Processing Units and CUDA}\r\n\\label{sec:GPU_prep}\r\nGraphics Processing Units (GPUs) are designed to exploit vector-level parallelism in programs. The GPU in my laptop, used by this project, is an Nvidia GTX 1050 Ti.\r\nIt has $768$ CUDA cores in it \\cite{1050-Ti}; it can compute up to $768$ different things in parallel.\r\nThis is much more parallelised than the CPU, though the device has a significantly lower clock speed of $\\SI{1392}{\\mega\\hertz}$.\r\n\r\nI chose to use CUDA, Nvidia’s proprietary API for programming their GPUs.\r\nAn alternative was OpenCL, but I chose CUDA because it has a more helpful profiling tool, which suggests possible sources of poor performance, and I used this to guide development.\r\n\r\nThe highly parallel nature of CUDA causes its execution model to be different to a single core of a CPU.\r\nIn CUDA, a body of work is a grid, which is a group of blocks, where a block is a group of threads.\r\nThe GPU has a collection of streaming multiprocessors (SMs, in my case $24$) which execute different blocks, and each of these SMs execute $32$ threads at once.\r\nThese groups of threads are called warps and are subsets of blocks.\r\nThreads in a warp execute in lockstep, with a single instruction for all $32$ threads in the warp.\r\nThe overall number of CUDA cores is $24 \\times 32=768$.\r\n\r\nUnits of code to be run on the GPU are called kernels, which are similar to C functions. When a kernel is called the number of blocks and threads in a block are specified.\r\nThere is a hard limit to the size of a block of $1024$ threads \\cite{CUDA_Guide}, therefore programs implemented in CUDA cannot require synchronisation over an arbitrary number of threads.\r\n\r\nAll of the threads are given the same arguments and execute the same blocks of code (though different paths may take different branches through the code).\r\nAll that distinguishes them are two variables, their \\lstinline{threadId} and \\lstinline{blockId}.\r\nAll threads within a block can be synchronised using the execution barrier \\lstinline{__syncthreads()}.\r\nHowever, there is no synchronisation between blocks, even when executing the same kernel, and they can execute in parallel to each other.\r\n\r\n\\subsection{Field Programmable Gate Arrays and SystemVerilog}\r\n\\label{sec:FPGA_prep}\r\nAn extension to my project was to implement the Smith-Waterman algorithm on an FPGA using a hardware description language (HDL).\r\nI chose to use SystemVerilog as my HDL, because I had studied it in Part IB.\r\nField Programmable Gate Arrays are programmable circuits, principally made up of programmable lookup tables, flip-flops, integrated memory blocks and programmable routing between the features on the chip.\r\nThese devices can be programmed to make custom circuitry on the FPGA chip.\r\n\r\nThis is appealing because it allows me to define a custom accelerator optimised just for this problem, able to evaluate a cell in the dynamic programming grid in just one clock cycle.\r\nMy other implementations require numerous clock cycles to do this same work.\r\nThe device is clocked significantly slower than the CPU or GPU, with the default clock frequency of my device being $\\SI{50}{\\mega\\hertz}$, and the theoretical maximum of $\\SI{550}{\\mega\\hertz}$.\r\nHowever, being able to do more work per clock cycle can reduce the impact of a slower clock frequency.\r\n\r\nAlthough an implementation on an FPGA will require much more power and be significantly slower than the same logic in an ASIC, working with an FPGA was possible for this project and manufacturing an ASIC would cost far too much money.\r\n\r\nI used a Terasic DE1-SoC board \\cite{DE1-SoC} loaned from the Computer Architecture Group.\r\nThis has an Altera Cyclone V SE FPGA SoC \\cite{CycloneV}, which has $\\SI{85000}{}$ logic elements; a fairly small FPGA, but also not the smallest produced. This device was significantly older than the CPU and GPU it was compared against. This FPGA uses a 28nm process and was made in 2012 \\cite{CycloneV}, whereas the CPU and GPU are from 2018 and use a 14nm process \\cite{i7-8750H} \\cite{1050-Ti}.\r\nA smaller process node might improve the maximum frequency of a design implemented on an FPGA, and there would be more resources on the device at a similar price, which could be used to do more work in parallel.\r\n\r\nThe FPGA SoC also has a dual-core ARM Cortex-A9 on the same package as the FPGA \\cite{CycloneV}, referred to as the Hard Processing System, HPS.\r\nThe ARM core can control the accelerator using an interconnect between the HPS and FPGA.\r\nIn particular, it provides a memory-mapped Avalon streaming interface, where the HPS requests to read or write to the FPGA as if it were memory and the FPGA’s responses to these requests can be defined in HDL.\r\n\r\nA major design consideration for my work on the FPGA was memory capacity, due to the quadratic memory requirement of the Smith-Waterman algorithm.\r\nThere are several different types of memory on the FPGA \\cite{CycloneV} and development board \\cite{DE1-SoC} I was using, suited for different things.\r\nThis is tabulated in \\cref{tab:FPGA_Memories}.\r\nFor the SDRAM memories, the throughput given (marked \\dag) is the peak throughput, but average throughput will be lower as selecting rows takes time.\r\nThe ways that these memories were used was discussed in \\cref{sec:Memory_in_SV}.\r\n\r\n\\begin{table}[h]\r\n \\centering\r\n \\begin{tabular}{|p{0.2\\textwidth}|p{0.4\\textwidth}p{0.1\\textwidth}p{0.2\\textwidth}|} \\hline\r\n Name & Description & Capacity & Throughput \\\\ \\hline\r\n ALM Registers & On-chip; all registers are directly accessible by routing elements & $\\SI{15.67}{\\kibi\\byte}$ & $\\SI{15.67}{\\kibi\\byte\\cycle}$ \\\\ \\hline\r\n M10K Blocks & On-chip; made up of $\\SI{10}{\\kibi\\bit}$ blocks units where $\\SI{40}{\\bits}$ can be accessed in a cycle, which are pipelined with one delay-slot between request and action. & $\\SI{49.63}{\\kibi\\byte}$ & $\\SI{1.94}{\\kibi\\byte\\cycle}$ \\\\ \\hline\r\n FPGA SDRAM & Off-chip; requires additional hardware to interface with it, and multiple delay slots as rows and columns are selected. & $\\SI{64}{\\mebi\\byte}$\r\n & $\\SI{2}{\\byte\\cycle}^\\dag$ \\\\ \\hline\r\n HPS DDR3 & Off-chip; requires additional hardware to interface with it, and multiple delay slots as rows and columns are selected. & $\\SI{1}{\\gibi\\byte}$ & $\\SI{8}{\\byte\\cycle}^\\dag$ \\\\ \\hline\r\n \\end{tabular}\r\n\r\n \\caption{Types of memory on the Terasic DE1-SoC development board}\r\n \\label{tab:FPGA_Memories}\r\n\\end{table}\r\n\r\n\\FloatBarrier\r\n\r\n\\subsection{Licencing}\r\n\\label{sec:Licencing_prep}\r\n\r\nThe majority of tools and libraries I used for my project are open source, and the rest are available for private use.\r\n\\begin{itemize}\r\n\\item GNU Compiler Collection; particularly the C compiler---GPLv3\r\n\\item GNU C Library, particularly Native POSIX Thread Library---LGPL\r\n\\item GNU Debugger; for C and CUDA code---GPLv3\r\n\\item Gprof; C profiler---GPLv3\r\n\\item JetBrains CLion; C IDE---student license\r\n\\item Visual Studio Code; general source-code editor---MIT License\r\n\\item CUDA Toolkit; including a compiler, debugger, memory-checking suite and profiler---proprietary, freely distributed\r\n\\item Quartus Prime Lite Edition; for simulating and synthesising HDL designs---proprietary, freely distributed\r\n\\item Git; for version control of all source code and dissertation text---GPLv2\r\n\\item LaTeX; for typesetting this dissertation---LPPL\r\n\\item BibTeX; for references in this dissertation---GPLv3\r\n\\end{itemize}\r\n\r\n\\section{Development methodology}\r\n\\label{sec:Methodology_prep}\r\n\r\nThe plan for the project was outlined in the project proposal (\\cref{sec:Proposal}), sequentially implementing designs for each platform.\r\nBy doing this, if any unexpected difficulty arose in the later stages of the project, then there would still be completed implementations from earlier in the project to evaluate.\r\nThe implementations were ordered in ascending difficulty and my descending confidence, to this end, with the first three required for the core success criteria.\r\nIn order these were:\r\n\\begin{enumerate}\r\n\\item Implementing the Smith-Waterman algorithm in single-threaded C code.\r\n\\item Modifying the C code to make it multi-threaded.\r\n\\item Implementing the Smith-Waterman algorithm in CUDA.\r\n\\item Implement the Smith-Waterman algorithm in SystemVerilog and simulate design.\r\n\\item Instantiate Verilog implementation on an FPGA.\r\n\\end{enumerate}\r\n\r\nI tested throughout development and had two main testing areas: correctness and performance. Any errors were likely to propagate between implementations where similar approaches were used, especially between C and CUDA, so it was important to catch errors as they arose instead of letting them propagate.\r\nI used an external implementation \\cite{EMBOSS} to check my first C programs.\r\nI tested the rest of my work using this implementation, because the reference implementation would only give one correct solution when there might be more than one for a given alignment problem, whereas my programs could decide whether a given alignment was valid and optimal.\r\n\r\n\\section{Starting Point}\r\n\\label{sec:Starting_Point}\r\n\r\nI started with reasonable experience with C, having studied it in Part IB and writing some utility programs in C during a summer job. I had no experience with CUDA, or GPU programming in general, and had written a small amount of SystemVerilog during Part IB, but never having attempted a project of this scale in it before.\r\nTherefore, in my project plan I assigned more time for the GPU and FPGA implementations compared to the CPU implementation, based on my relative levels of experience.\r\n\r\nThe only code used that was not my own was SystemVerilog used for administrative purposes.\r\nI used a template from the Part IB ECAD+Arch course for instantiating a SystemVerilog design on the development board I was using, which correctly arranged IO pins.\r\nMy supervisor, Peter Rugg, gave me the Verilog he had used to build an Avalon streaming interface between the HPS and the FPGA, to transfer data between my hardware design and the OS on the HPS.\r\nI based my streaming interface on this, and I am grateful for his assistance.\r\n\r\nSequence alignment and Smith-Waterman in particular are covered in detail by the first few lectures of the Part II Bioinformatics course, and is presented in \\cref{sec:SW_Overview}.\r\nThis course runs in Michaelmas term.\r\nThe linear space approach (\\cref{sec:SW_Linear_Prep}) is described for global alignment in that course, and I read a paper \\cite{MyersMiller} for its application to local alignments.\r\n", "meta": {"hexsha": "316e4b1977033b2215fe158f9d81d8d4fa21fac7", "size": 34274, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diss/chapters/prep.tex", "max_stars_repo_name": "benjimarshall/SequenceAlignment", "max_stars_repo_head_hexsha": "23fcfbf11567486afe03b9d671ce8640a83ca557", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "diss/chapters/prep.tex", "max_issues_repo_name": "benjimarshall/SequenceAlignment", "max_issues_repo_head_hexsha": "23fcfbf11567486afe03b9d671ce8640a83ca557", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "diss/chapters/prep.tex", "max_forks_repo_name": "benjimarshall/SequenceAlignment", "max_forks_repo_head_hexsha": "23fcfbf11567486afe03b9d671ce8640a83ca557", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 76.3340757238, "max_line_length": 468, "alphanum_fraction": 0.7437124351, "num_tokens": 8610, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324983301568, "lm_q2_score": 0.7690802370707283, "lm_q1q2_score": 0.6269792030835191}} {"text": "% Created 2020-04-15 mié 17:46\n% Intended LaTeX compiler: pdflatex\n\\documentclass[presentation,aspectratio=1610]{beamer}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{graphicx}\n\\usepackage{grffile}\n\\usepackage{longtable}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{amssymb}\n\\usepackage{capt-of}\n\\usepackage{hyperref}\n\\usepackage{khpreamble}\n\\usepackage{pgfplots}\n\\usepackage{pdfpages}\n\\usepackage{circuitikz}\n\\usepgfplotslibrary{groupplots}\n\\usetikzlibrary{positioning}\n\\renewcommand*{\\not}[1]{\\ensuremath{\\bar{#1}}}\n\\renewcommand*{\\not}[1]{\\ensuremath{\\overline{#1}}}\n\\usetheme{default}\n\\author{Kjartan Halvorsen}\n\\date{\\today}\n\\title{Boolean algebra, logic diagrams and truth tables}\n\\hypersetup{\n pdfauthor={Kjartan Halvorsen},\n pdftitle={Boolean algebra, logic diagrams and truth tables},\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 26.3 (Org mode 9.3.6)}, \n pdflang={English}}\n\\begin{document}\n\n\\maketitle\n\n\\section{Boolean algebra, minterms and maxterms}\n\\label{sec:org605a59f}\n\\begin{frame}[label={sec:orgb501097}]{AND and OR}\n\\(a, b \\in \\{0,1\\}\\)\n\\begin{columns}\n\\begin{column}{0.5\\columnwidth}\n\\begin{block}{AND}\n\\begin{center}\n\\begin{tabular}{|cc|c|}\n\\hline\n\\(a\\) & \\(b\\) & \\(a\\) AND \\(b\\), \\(ab\\)\\\\\n\\hline\n0 & 0 & 0\\\\\n0 & 1 & 0\\\\\n1 & 0 & 0\\\\\n1 & 1 & 1\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\begin{center}\n\\begin{tikzpicture}\n \\draw (0,0) to[switch, label=$a$, o-] (2,0) to[switch, label=$b$, -o] (4, 0);\n\\end{tikzpicture}\n\\end{center}\n\\alert{Closed circuit \\(\\Leftrightarrow\\) 1}\n\n\\alert{Open circuit \\(\\Leftrightarrow\\) 0}\n\n\\begin{center}\n\\includegraphics[width=0.5\\linewidth]{../../figures/and-gate.pdf}\n\\end{center}\n\\end{block}\n\\end{column}\n\n\\begin{column}{0.5\\columnwidth}\n\\begin{block}{OR}\n\\begin{center}\n\\begin{tabular}{|cc|c|}\n\\hline\n\\(a\\) & \\(b\\) & \\(a\\) OR \\(b\\), \\(a+b\\)\\\\\n\\hline\n0 & 0 & 0\\\\\n0 & 1 & 1\\\\\n1 & 0 & 1\\\\\n1 & 1 & 1\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\begin{center}\n\\begin{tikzpicture}\n \\draw (0,0) to[switch, label=$a$, o-o] (4,0);\n \\draw (1,0) to[short] (1,-1) to[switch, l_=$b$, ] (3, -1) to[short] (3, 0);\n\\end{tikzpicture}\n\\end{center}\n\n\\begin{center}\n\\includegraphics[width=0.5\\linewidth]{../../figures/or-gate.pdf}\n\\end{center}\n\\end{block}\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:org1573679}]{NAND and NOR}\n\\(a, b \\in \\{0,1\\}\\)\n\\begin{columns}\n\\begin{column}{0.5\\columnwidth}\n\\begin{block}{NAND}\n\\begin{center}\n\\begin{tabular}{|cc|c|}\n\\(a\\) & \\(b\\) & \\(a\\) NAND \\(b\\), \\(\\overline{a\\cdot{}b}\\)\\\\\n\\hline\n0 & 0 & 1\\\\\n0 & 1 & 1\\\\\n1 & 0 & 1\\\\\n1 & 1 & 0\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\begin{center}\n\\includegraphics[width=0.5\\linewidth]{../../figures/nand-gate.pdf}\n\\end{center}\n\\end{block}\n\\end{column}\n\n\\begin{column}{0.5\\columnwidth}\n\\begin{block}{NOR}\n\\begin{center}\n\\begin{tabular}{|cc|c|}\n\\(a\\) & \\(b\\) & \\(a\\) NOR \\(b\\), \\(\\overline{a+b}\\)\\\\\n\\hline\n0 & 0 & \\\\\n0 & 1 & \\\\\n1 & 0 & \\\\\n1 & 1 & \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\begin{center}\n\\includegraphics[width=0.5\\linewidth]{../../figures/nor-gate.pdf}\n\\end{center}\n\\end{block}\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n\\begin{frame}[label={sec:org2d0b9ef}]{Boolean algebra, contd}\n\\(x, y, z \\in \\{0,1\\}\\)\n\n\\begin{center}\n\\begin{tabular}{r|c|c|}\n & Property & Dual\\\\\n\\hline\nProperties of 0 and 1 & \\(x+0=x\\) & \\(x\\cdot 0=0\\)\\\\\n & \\(x+1=1\\) & \\(x \\cdot 1 = x\\)\\\\\nIdempotency & \\(x+x=x\\) & \\(x\\cdot x = x\\)\\\\\nComplementarity & \\(x+\\not{x}=1\\) & \\(x\\cdot \\not{x}=0\\)\\\\\nInvolution & \\(\\not{\\not{x}}=x\\) & \\\\\nCommutative & \\(x+y=y+x\\) & \\(x\\cdot y = y\\cdot x\\)\\\\\nAssociative & \\((x+y) + z = x + (y+z)\\) & \\((xy)z = z(yz)\\)\\\\\nDistributive & \\(x\\cdot (y+z) = xy + xz\\) & \\(x+yz=(x+y)(x+z)\\)\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}[label={sec:orge5065a4}]{Boolean algebra, contd}\n\\(x, y \\in \\{0,1\\}\\)\n\n\\begin{center}\n\\begin{tabular}{r|c|c|}\n & Theorem & Dual\\\\\n\\hline\nAbsorption & \\(x+xy=x(1+y)=x\\) & \\(x(x+y)=x\\)\\\\\nLogic adjacency & \\(xy + x\\not{y} = x(y+\\not{y}) =x\\) & \\((x+y)(x+\\not{y}) = x\\)\\\\\nDe Morgan's & \\(\\not{x+y}=\\not{x}\\cdot{}\\not{y}\\) & \\(\\not{xy} = \\not{x} + \\not{y}\\)\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}[label={sec:orgd14be63}]{DeMorgan's theorem}\n\\begin{center}\n\\includegraphics[width=0.4\\linewidth]{../../figures/Demorganlaws.png} From wikipedia\n\\end{center}\n\\end{frame}\n\\begin{frame}[label={sec:orgbb8fe29}]{Simplify functions}\n\\begin{enumerate}\n\\item \\(f = (a+b)(a+c)\\)\n\\item \\(f = a + \\not{a}b\\)\n\\end{enumerate}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:orgd9a80c9}]{Logic diagram \\(\\rightarrow\\) function}\nDetermine the function represented by the logic diagrams\n\\begin{center}\n\\includegraphics[width=0.7\\linewidth]{../../figures/exercise-gate-4.pdf}\n\\end{center}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:org42b2888}]{Function \\(\\rightarrow\\) logic diagram}\nDraw the diagram corresponding to the boolean function\n\\begin{enumerate}\n\\item \\(f = (a+b)(a+c)\\)\n\\item \\(f = a + \\not{a}b\\)\n\\end{enumerate}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:org98e40c3}]{Group exercise}\n\\begin{enumerate}\n\\item Enter breakout room\n\\item One of you downloads and shares this presentation\n\\item Work together on the problems in the previous three slides\n\\begin{enumerate}\n\\item Simplify functions\n\\item Determine function from logic diagram\n\\item Draw logic diagram from function\n\\end{enumerate}\n\\end{enumerate}\n\\end{frame}\n\\end{document}", "meta": {"hexsha": "477775b4d68aaaa224cea20b69fe4f5a8278e193", "size": 5415, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "modules/logic-control/logic-control.tex", "max_stars_repo_name": "kjartan-at-tec/mr2015", "max_stars_repo_head_hexsha": "1134f3a99ef72e4a17d44edb4d288daad84f3e70", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "modules/logic-control/logic-control.tex", "max_issues_repo_name": "kjartan-at-tec/mr2015", "max_issues_repo_head_hexsha": "1134f3a99ef72e4a17d44edb4d288daad84f3e70", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "modules/logic-control/logic-control.tex", "max_forks_repo_name": "kjartan-at-tec/mr2015", "max_forks_repo_head_hexsha": "1134f3a99ef72e4a17d44edb4d288daad84f3e70", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.75, "max_line_length": 86, "alphanum_fraction": 0.6506001847, "num_tokens": 2118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.8152324960856175, "lm_q1q2_score": 0.6269792013572882}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% CS630: Database Management Systems\n% Copyright 2014 Pejman Ghorbanzade \n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nFor questions 1 to 3, you are given the following relational schema.\n\n\\begin{terminal}\nStudents (@*\\underline{sid}*@:integer, sname:string, age:integer)\nGrades (@*\\underline{sid}*@:integer, @*\\underline{cid}*@:integer, grade:integer)\nCourses (@*\\underline{cid}*@:integer, cname:string, credits:integer)\n\\end{terminal}\n\nThe meaning of attributes is as follows.\n\n\\begin{enumerate}[topsep=0pt,itemsep=0ex,partopsep=5pt,parsep=0pt,label=-]\n\\item \\texttt{sid} : unique student identifier, primary key in table \\texttt{Students}\n\\item \\texttt{cid} : unique course identifier, primary key in table \\texttt{Courses}\n\\item \\texttt{sname} : student name\n\\item \\texttt{age} : student age\n\\item \\texttt{cname} : course name\n\\item \\texttt{credits} : number of credits for a course\n\\item \\texttt{grade} : the grade obtained by student identified by \\texttt{sid} for course identified by \\texttt{cid}; \\texttt{sid} and \\texttt{cid} are foreign keys referring to the \\texttt{sid} and \\texttt{cid} fields in the \\texttt{Students} and \\texttt{Courses} tables, respectively.\n\\end{enumerate}\n\n\\section*{Question 1}\n\nWrite \\textbf{relational algebra} expressions for the following queries given the schema above:\n\n\\begin{enumerate}[label=(\\alph*)]\n\n\\item Find the grades that students of age 20 obtained in courses with 4 credits.\n\n\\textbf{Solution:}\n\n$$ \\pi_{G.grade}(((\\sigma_{S.age=20}S)\\Join G)\\Join (\\sigma_{C.credits=4}C)) $$\n\n\\item Find the names of students who took a course named \\textit{Calculus} and did not get a \\textit{C} grade in any course.\n\n\\textbf{Solution:}\n\n$$ \\rho(S1,\\pi_{G.sid}((\\sigma_{C.cname='Calculus'}C)\\Join G)) $$\n$$ \\rho(S2,\\pi_{G.sid}(\\sigma_{G.grade='C'}G)) $$\n$$ \\pi_{S.sname}((S1-S2)\\Join S) $$\n\n\\item Find the ages of students who got an \\textit{A} in some course with 3 credits or who got a \\textit{B} in any course.\n\n\\textbf{Solution:}\n\n$$\\rho(S1,\\pi_{S.sid}((\\sigma_{G.grade='A'}((\\sigma_{C.credits=3}C)\\Join G))\\Join S)) $$\n$$\\rho(S2,\\pi_{G.sid}(\\sigma_{G.grade='B'}G)) $$\n$$ \\pi_{S.age} ((S1\\cup S2)\\Join S) $$\n\n\\item Find the maximum age among students who took \\textit{Calculus}.\n\n\\textbf{Solution:}\n\n$$ \\rho(S1,(((\\sigma_{C.cname='Calculus'}Courses)\\Join Grades)\\Join Students) ) $$\n$$ \\rho(S2(1\\rightarrow sid2,2\\rightarrow sname2, 3\\rightarrow age2),S1) $$\n$$ \\rho(S3(1\\rightarrow sid), \\pi_{S1.sid}(S1\\Join_{S1.age 1\n ) \\land (\n \\forall m_0 \\in \\mathbb{N},\n \\exists n \\in \\mathbb{N},\n c^{(f)}_{\\min}(n) \\geq m_0\n )\n \\end{equation}\n \\end{theorem}\n\n In other words, this function partitions the natural number set into infinitely number of finite subsets where the cardinality of them are unbounded.\n Each cycle is associated with a subset.\n\n\n \\section{Proof}\\label{sec:proof}\n\n A simple construction satisfies those properties is as follows:\n\n Suppose we found partition on $\\mathbb{N}$ of ($1^*$) infinitely many finite subsets where ($1^{**}$) each of them has the cardinality of at least 2, ($2^*$) the cardinality of these subsets is unbounded.\n\n \\begin{equation}\n \\label{eq:partition}\n \\mathbb{P} = \\{ P_i \\}_{i=1}^\\infty = \\{ P_1, P_2, P_3, \\dots \\}\n \\end{equation}\n\n Where we order all elements in each $P_i$, so that for every $P_i$, we have a minimum element, a maximum element and a function to yield the successor element if the input is not the maximum element namely\n\n \\begin{equation}\n \\label{eq:succ_func}\n succ_i : P_i \\backslash \\{ \\max{P_i} \\} \\to P_i \\backslash \\{ \\min{P_i} \\}\n \\end{equation}\n\n Define a function $f_i: P_i \\to P_i$ that returns minimum element of $P_i$ if the input is the maximum element of $P_i$, otherwise return its successor.\n\n \\begin{equation}\n \\label{eq:func_sub}\n f_i(n)=\n \\begin{cases}\n \\min P_i &\\text{if $n = \\max P_i$}.\\\\\n succ_i(n) &\\text{otherwise}.\n \\end{cases}\n \\end{equation}\n\n This function has a cycle number of the cardinality of $P_i$\n Since $\\mathbb{P}$ is a partition, these $P_i$ are disjoint and their union is $\\mathbb{N}$.\n We define the function $f: \\mathbb{N} \\to \\mathbb{N}$\n\n \\begin{equation}\n \\label{eq:func}\n f(n)= f_i(n) \\text{ if } n \\in P_i\n \\end{equation}\n\n ($1^* \\land 1^{**} \\to 1$) For every input $n$, the cycle number is $|P_i| \\geq 2$ where $P_i$ is the associated subset. ($2^* \\to 2$) Since these subsets are unbounded in size, the set of all cycle numbers of $f$ is also unbounded.\n\n In order to finish the proof, we will show a partition on $\\mathbb{N}$ that satisfies ($1^*$), ($1^{**}$) and ($2^*$).\n\n Let $S_i$ be the set of all natural numbers in the range $[2^i, 2^{i+1})$ for $i=0, 1, 2, \\dots$.\n e.g. $S_0 = \\{ 1 \\}$, $S_1 = \\{ 2, 3 \\}$, $S_2 = \\{ 4 , 5, 6, 7 \\}$, $S_3 = \\{ 8, 9, \\dots, 15 \\}$, etc.\n\n Our partition is in the form\n\n \\begin{equation}\n \\label{eq:partition_concrete}\n \\mathbb{P} = \\{\\{ S_0 \\cup S_1 \\}\\} \\cup \\{ S_i \\}_{i=2}^\\infty\n \\end{equation}\n\n\n\\end{document}\n", "meta": {"hexsha": "b0613910098ad4919c0cb620387b2c80073f8053", "size": 4990, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "blog/cycle/tex/main.tex", "max_stars_repo_name": "khanhhhh/khanhhhh.github.io", "max_stars_repo_head_hexsha": "f6313f0228a480b37e5f4ba9550e3cf011e60ba8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2021-12-14T08:27:45.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-14T14:51:34.000Z", "max_issues_repo_path": "blog/cycle/tex/main.tex", "max_issues_repo_name": "khanh-nguyen-code/khanh-nguyen-code.github.io", "max_issues_repo_head_hexsha": "4c047a941720e46b004d0bb922507c3b2bc7132f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "blog/cycle/tex/main.tex", "max_forks_repo_name": "khanh-nguyen-code/khanh-nguyen-code.github.io", "max_forks_repo_head_hexsha": "4c047a941720e46b004d0bb922507c3b2bc7132f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-07-30T05:30:49.000Z", "max_forks_repo_forks_event_max_datetime": "2020-07-30T05:30:49.000Z", "avg_line_length": 34.1780821918, "max_line_length": 236, "alphanum_fraction": 0.5991983968, "num_tokens": 1594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996144, "lm_q2_score": 0.7931059511841119, "lm_q1q2_score": 0.6269742925003878}} {"text": "\\documentclass{standalone}\n\\begin{document}\n\t\\chapter{Classification of numbers}\n\t\n\t\\begin{itemize}\n\t\t\\item{Natural numbers: $\\mathbb{N}$; (1, 2, 3, 4 \\ldots)}\n\t\t\\subitem{This set includes every number which is both positive and whole.}\n\t\t\\item{Integer numbers: $\\mathbb{Z}$; (-2, -1, 0, 1, 2, \\ldots)}\n\t\t\\subitem{The integer number set includes every negative and positive whole numbers, similarly to $\\mathbb{N}$}\n\t\t\\item{Rational numbers: $\\mathbb{Q}$; (-1, 2 , $\\frac{1}{2}$)}\n\t\t\\subitem{A number is s.t.b rational if expressed in the form $\\frac{p}{q};p, q \\in \\mathbb{Z}$.}\n\t\t\\item{Irrational numbers: $\\mathbb{Q'}$; ($\\pi, e, \\sqrt{2}, \\sqrt{5}, \\ldots$))}\n\t\t\\subitem{If a number is not classified as any of the above, it is referred to as irrational.}\n\t\t\\item{Real numbers: $\\mathbb{R}$}\n\t\t\\subitem{Anything mentioned above inclusively represent the set of Real numbers}\\\\\n\t\\end{itemize}\n\t\n\t\n\t\\emph{We can additionally refer to positive or negative numbers in any set by using the notation:}\\\\\n\t\\begin{center}\n\t\t$\\mathbb{R}^+$ and $\\mathbb{R}^-$\n\t\\end{center}\n\t%TODO: insert image of sets\n\t\n\t\n\t\\newpage\t\n\t\\end{document}", "meta": {"hexsha": "c6acb14d3e8aea1ca66ab695049fd5718a71209d", "size": 1127, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Pure Mathematics/Class_of_Numbers.tex", "max_stars_repo_name": "Girogio/My-LaTeX", "max_stars_repo_head_hexsha": "706ec7cb4d62af1b5a9ad7547589889240c755bb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-01-12T11:45:45.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-30T21:47:25.000Z", "max_issues_repo_path": "Pure Mathematics/Class_of_Numbers.tex", "max_issues_repo_name": "Girogio/My-LaTeX", "max_issues_repo_head_hexsha": "706ec7cb4d62af1b5a9ad7547589889240c755bb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Pure Mathematics/Class_of_Numbers.tex", "max_forks_repo_name": "Girogio/My-LaTeX", "max_forks_repo_head_hexsha": "706ec7cb4d62af1b5a9ad7547589889240c755bb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.7407407407, "max_line_length": 112, "alphanum_fraction": 0.6796805679, "num_tokens": 390, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059511841119, "lm_q2_score": 0.7905303087996143, "lm_q1q2_score": 0.6269742925003877}} {"text": "\\documentclass[../../../thesis.tex]{subfiles}\n\n\\begin{document} \n \n \\section{Localization}\n \n We will briefly explain the localization of a category at a set. We will look at local objects and localizations by adjoint functors. This section is based on \\cite{Krause21}.\n\n \\subsection{Definition and Local Objects}\n\n The localization of a category at some set of morphisms is a universal category where these morphisms have been inverted. In this manner, we will think of the localization as adding formal inverses to the morphisms in this set to the category.\n\n \\begin{definition}\n Let $\\mathcal{C}$ be a category and $S \\subseteq Mor(\\mathcal{C})$ a set of morphisms in $\\mathcal{C}$. We define the localization of $\\mathcal{C}$ at $S$ as a category $\\mathcal{C}[S^{-1}]$ together with a functor $Q : \\mathcal{C} \\rightarrow \\mathcal{C}[S^{-1}]$ satisfying the properties:\n \\begin{enumerate}\n \\item For any morphism $s\\in S$, $Qs$ is an invertible morphism.\n \\item Any functor $F : \\mathcal{C} \\rightarrow \\mathcal{D}$ satisfying $Fs$ is invertible if $s\\in S$ factors through $Q$. Thus there exists a unique functor $\\bar{F} : \\mathcal{C}[S]^{-1} \\rightarrow \\mathcal{D}$ and a natural isomorphism $\\alpha : F \\implies \\bar{F} \\circ Q$ relating the functors.\n \\end{enumerate}\n\n \\begin{center}\n \\begin{tikzcd}\n \\mathcal{C} \\ar{rd}{Q} \\ar{rr}{F} & & \\mathcal{D} \\\\\n & \\mathcal{C}[S^{-1}] \\ar[dashed]{ur}{\\bar{F}}\n \\end{tikzcd}\n \\end{center}\n \\end{definition}\n\n Informally, the category $\\mathcal{C}[S^{-1}]$ may be constructed as the category with the same objects as $\\mathcal{C}$. The morphisms are constructed with paths of morphisms in $\\mathcal{C}$, where we also allow formal inverse paths $s^{-1}$ for any morphism in $S$. The constant paths would be the new identity morphisms, while every path from an object $X$ to $Y$ represents a morphism.\n\n \\begin{remark}\n Beware whenever localizing a locally small category. Adding more morphisms to this category may make some set of morphisms between objects big.\n \\end{remark}\n\n By the universality of the functor $Q$ we have the following lemma.\n\n \\begin{lemma}\n Let $\\mathcal{D}$ be a category, then the pre-composition functor is fully faithful.\n \\begin{align*}\n \\_ \\circ Q : Fun(\\mathcal{C}[S^{-1}],\\mathcal{D}) & \\rightarrow Fun(\\mathcal{C}, \\mathcal{D}) \\\\\n F & \\mapsto F \\circ Q\n \\end{align*}\n Moreover, we may identify $Fun(\\mathcal{C}[S^{-1}],\\mathcal{D})$ as the full subcategory of functors in $Fun(\\mathcal{C}, \\mathcal{D})$ which sends every morphism in $S$ to an isomorphism.\n \\end{lemma}\n\n We proceed to define $S$-local objects, and informally, these are all the objects which does not change after the localization process.\n\n \\begin{definition}\n An object $Y\\in \\mathcal{C}$ is called $S$-local ($S$-closed or $S$-orthogonal) if for any $s\\in S$, the map $\\mathcal{C}(s, Y)$ is a bijection. Define $S\\perp$ to be the full subcategory of $S$-local objects.\n \\end{definition}\n\n The following lemma gives an equivalent definition for \"localness\".\n\n \\begin{lemma}\n An object $Y\\in \\mathcal{C}$ is $S$-local if and only if the canonical map $q$ is a natural isomorphism.\n \\begin{align*}\n q_{X,Y} : \\mathcal{C}(X,Y) \\rightarrow \\mathcal{C}[S^{-1}](X,Y)\n \\end{align*}\n \\end{lemma}\n\n \\subsection{Localizing at Adjoint Functors}\n\n Given two functors $F : \\mathcal{C} \\rightarrow \\mathcal{D}$ and $G : \\mathcal{D} \\rightarrow \\mathcal{C}$ such that they form an adjoint pair $F \\dashv G$, we want to see how they react with localizations.\n\n \\begin{proposition}\n Let $F \\dashv G$ be an adjoint pair as above, and define $S = $\\startset{$s\\in Mor\\mathcal{C}$ $\\mid$ $Fs$ is invertible}. Draw out the following diagram, with the relations $F = \\bar{F}\\circ Q$.\n \\begin{center}\n \\begin{tikzcd}\n & \\mathcal{C} \\ar{ld}[description]{Q} \\ar[bend right = 20]{dd}[description]{F} \\\\\n \\mathcal{C}[S^{-1}] \\ar{rd}[description]{\\bar{F}} \\\\\n & \\mathcal{D} \\ar[bend right = 20]{uu}[description]{G}\n \\end{tikzcd}\n \\end{center}\n The following are equivalent:\n \\begin{enumerate}\n \\item $G$ is fully faithful.\n \\item The counit $\\varepsilon : FG \\implies Id_\\mathcal{D}$ is invertible for any object $X\\in \\mathcal{D}$.\n \\item The functor $F$ induces an equivalence $\\bar{F}:\\mathcal{C}[S^{-1}] \\rightarrow \\mathcal{D}$.\n \\end{enumerate}\n \\end{proposition}\n \n In the light of the last proposition, we may define \"short exact sequences\" of pre-additive categories.\n\n \\begin{definition}\n Suppose there is a diagram of additive functors like below.\n \\begin{center}\n \\begin{tikzcd}[column sep = large]\n \\mathcal{A} \\ar[bend left = 30, tail]{r}{E} \\ar[phantom]{r}{\\bot} & \\mathcal{B} \\ar[bend left = 30, two heads]{r}{F} \\ar[bend left = 30, two heads]{l}{E_\\rho} \\ar[phantom]{r}{\\bot} & \\mathcal{C} \\ar[bend left = 30, tail]{l}{F_\\rho} \n \\end{tikzcd}\n \\end{center}\n It is called a localization sequence whenever the following holds:\n \\begin{enumerate}\n \\item $(E,E_\\rho)$ and $(F,F_\\rho)$ are adjoint pairs.\n \\item $E$ and $F_\\rho$ are fully faithful.\n \\item $ImE \\simeq KerF$.\n \\end{enumerate}\n \\end{definition}\n\n\\end{document}", "meta": {"hexsha": "2a570727b040842d658be8b1b158b602510bb5db", "size": 6187, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "appendices/localization/localization.tex", "max_stars_repo_name": "CelestialCry/Strongly-Homotopic-Quasi-iso", "max_stars_repo_head_hexsha": "a01fabe6f916696b7f3ad94af6014333117da208", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "appendices/localization/localization.tex", "max_issues_repo_name": "CelestialCry/Strongly-Homotopic-Quasi-iso", "max_issues_repo_head_hexsha": "a01fabe6f916696b7f3ad94af6014333117da208", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "appendices/localization/localization.tex", "max_forks_repo_name": "CelestialCry/Strongly-Homotopic-Quasi-iso", "max_forks_repo_head_hexsha": "a01fabe6f916696b7f3ad94af6014333117da208", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.1326530612, "max_line_length": 402, "alphanum_fraction": 0.5737837401, "num_tokens": 1651, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.793105941403651, "lm_q2_score": 0.7905303112671294, "lm_q1q2_score": 0.6269742867256378}} {"text": "\\chapter{Classification}\nLinear regression discussed in the previous chapter assumes a quantitative response value Y but often classification is needed to solve a problem. Consider the binary classification problem: Predicting whether a tumor is malignant or not based on the size of the tumor. Linear regression could be used by converting the wanted qualitative response variable into a quantitative response. The quantitative response can be converted into a qualitative response by saying that $p(maligant = true | tumorsize) > 0.5$. The model would then classify a tumor as malignant if the probability is higher than 0.5. \n\nA problem though arises from this, since the model will make estimates outside the interval $[0,1]$. Because some predictions are outside of this interval, then the prediction can not be thought of as probabilities. For example measuring a tumor size of 0 would predict a negative value. Another problem with linear regression is that it can not be extended to a classification problem with more than two classes. It can only be used for binary classification. The reason for this is that there might not be a natural ordering between the different classes and the gap between some classes might be bigger than others. This would make it difficult to make an accurate model using a linear regression \\citep[pp. 129-130]{ISLR}. Instead this chapter presents different methods that are more suited for classification.\n\n\\section{Logistic Regression}\n\\label{chp:logreg}\nLogistic regression is a classification method. It avoids the problem with values outside the interval of $[0,1]$. In logistic regression a logistic function is used and can be seen in (3.1).\n\n\\begin{equation}\n\tp(X) = \\dfrac{e^{\\beta_0 + \\beta_1 X}}{1 + e^{\\beta_0 + \\beta_1 X}}\n\\end{equation}\n\nBy taking the exponential of $\\beta_0 + \\beta_1 X$ the result will always be equal to or higher than 0. By dividing by the same equation + 1 the prediction will never be higher than 1. In the equation $\\beta_0$ and $\\beta_1$ is similar to what is used in linear regression. $\\beta_0$ describes the probability when the predictor is 0 and $\\beta_1$ is the regression coefficient that is multiplied with the predictor. The equation can be extended to multiple logistic regression using multiple predictors. Instead the equation looks as shown in (3.2).\n\n%\\begin{center}\n%\t$p(X) = \\dfrac{e^{\\beta_0 + \\beta_1 X_1 + ... + \\beta_p X_p}}{1 + e^{\\beta_0 + \\beta_1 X_1 + ... + \\beta_p X_p}}$ \n%\\end{center}\n\n\\begin{equation}\n\tp(X) = \\dfrac{e^{\\beta_0 + \\beta_1 X_1 + ... + \\beta_p X_p}}{1 + e^{\\beta_0 + \\beta_1 X_1 + ... + \\beta_p X_p}}\n\\end{equation}\n\n%\\todo{måske indsæt en figur der viser grafer med linear og logistic}\n\n\\subsection{Lab 4.6.1 - The Stock Market Data}\nIn this lab the correlation between the predictors and todays return value is examined. Todays return value describes the direction. If the number is negative the stock went down and if it is positive the stock went up. The available predictors are 5 lagging indicators \\emph{Lag1, Lag2, Lag3, Lag4} and \\emph{Lag5} and a \\emph{Volume} which describes the number of shares traded in billions.\n\nWith the python code shown in Listing \\ref{lst:correlation} a matrix with the pairwise correlation among the predictors and todays return is printed. Numpy is used for calculating the correlation matrix. Pandas is used for creating a table based on the calculated correlation matrix.\n\n\\begin{lstlisting}[language=Python, label=lst:correlation, caption=Printing correlation matrix]\ncor = np.corrcoef(preparedData)\nlabels = ['Year', 'Lag1', 'Lag2', 'Lag3', 'Lag4', 'Lag5', 'Volume', \n'Today']\ndf = pd.DataFrame(cor, columns=labels, index=labels)\nprint df\n\\end{lstlisting}\n\nThe printed correlation matrix is illustrated in figure \\ref{fig:lab461}. Here it is shown that there is very little correlation between the different Lag variables and todays return. This also makes sense because the variables are lagging indicators and therefore it only changes after the economy has already changed. For example if \\emph{Lag1} describes how the market went up yesterday then it is less likely to go up today. An interesting correlation can be seen between \\emph{Year} and \\emph{Volume}. This indicates that each year the average number of traded stocks increases. \n\n\\myFigure{461.png}{Pairwise correlation matrix between predictors and todays return}{fig:lab461}{1} \n\\FloatBarrier\n\n\\myFigure{461_plot.png}{Correlation between Volume and Year. Higher index describes forward in time}{fig:lab461Plot}{0.45}\n \n\\emph{Volume} is plotted in figure \\ref{fig:lab461Plot}. Here the increase in traded stock is obvious. When choosing predictors in our logistic regression model it must be noted that a correlation matrix not necessarily describes the best predictors, at least not if multiple predictors are used. When choosing predictors it is important to choose the predictors that actually are associated with the response variable. The following lab will therefore examine the data to find good predictors.\n\n\\subsection{Lab 4.6.2 - Logistic Regression}\n\\label{sec:lab462}\nIn this lab a logistic regression model is made based on the stock market data. The predictors were examined in Lab 4.6.1 and some of the predictors will be chosen to predict the direction of the stock market. The direction can either be \\emph{up} or \\emph{down} and hence this is a binary classification problem.\n\nIn the exercise the p-values will first be examined. This is done by fitting a logistic model with the predictors \\emph{Lag1, Lag2, Lag3, Lag4, Lag5} and \\emph{Volume} to predict the \\emph{direction}. The statsmodels python library is used to print a summary of the fitted model. This is shown in listing \\ref{lst:summary}.\n\n\\begin{lstlisting}[language=Python, label=lst:summary, caption=Printing summary]\nformula = 'Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5+Volume'\nmodel = smf.glm(formula=formula, data=data, family=sm.families.Binomial())\nprint(model.fit().summary())\n\\end{lstlisting}\n\n\\myFigure{Lab462Pvalue.PNG}{Summary of data with associated p-value describing good predictors}{fig:lab462Pvalue}{0.6} \n\\FloatBarrier\n\nThe printed summary is illustrated in figure \\ref{fig:lab462Pvalue}. Here it is seen that the lowest p-value is \\emph{Lag1} but it still has a value of 0.145, which is still quite high and it presents no evidence of any association between \\emph{Lag1} and \\emph{Direction}. But note here that only using the predictors with the lowest p-value could yield the most accurate result when predicting the \\emph{Direction}. The reason for this is that some predictors have so little association with the \\emph{Direction} that they will only make the prediction less accurate. This will also be shown in the following examples.\n\n\nIn Listing \\ref{lst:AllPredictors} the training data is divided into a training set and a test set. Then all the data from 2004 and before is used to train our logistic model and then it is tested on the data from 2005. In this implementation all predictors are used to fit the logistic model.\n\n\\begin{lstlisting}[language=Python, label=lst:AllPredictors, caption=Logistic regression using all predictors]\nX_train = data[:'2004'][['Lag1', 'Lag2', 'Lag3', 'Lag4', 'Lag5', 'Volume']]\nY_train = data[:'2004']['Direction']\n\nX_test = data['2005':][['Lag1', 'Lag2', 'Lag3', 'Lag4', 'Lag5', 'Volume']]\nY_test = data['2005':]['Direction']\n\nlr = linear_model.LogisticRegression()\nlr.fit(X_train,Y_train)\n\\end{lstlisting}\n\nBased on the test data it is examined how well our logistic model predicts the \\emph{Direction}. In Listing \\ref{lst:LRscore} a score method is called for the logistic regression model. This function returns the mean accuracy of the model. The score shows that the model only predicts correctly $48.8\\%$ of the time. Based on this result the model needs to be approved upon.\n\\begin{lstlisting}[language=Python, label=lst:LRscore, caption=Printing mean accuracy]\nprint(lr.score(X_test, Y_test))\n\\end{lstlisting}\n\nIn Listing \\ref{lst:OnlyLag1Lag2} only the predictors \\emph{Lag1} and \\emph{Lag2} are now used. \\emph{Lag1} is included because it has the lowest p-value. \\emph{Lag2} is also used as predictor because it has a very low correlation with \\emph{Lag1} and also the third lowest p-value.\n%It has a very low correlation with \\emph{Lag1} and also the third lowest p-value.\nNote though that the p-value of \\emph{Lag2} is only 0.006 higher than the second lowest p-value, which describes \\emph{Volume}. \\emph{Volume} also has almost twice as high correlation with \\emph{Lag1} and this might lead to a better result using \\emph{Lag1} and \\emph{Lag2} for prediction.\n\n\\begin{lstlisting}[language=Python, label=lst:OnlyLag1Lag2, caption=Logistic regression using only Lag1 and Lag2 as predictors]\nX_train = data[:'2004'][['Lag1', 'Lag2']]\nY_train = data[:'2004']['Direction']\n\nX_test = data['2005':][['Lag1', 'Lag2']]\nY_test = data['2005':]['Direction']\n\nlr = linear_model.LogisticRegression()\nlr.fit(X_train,Y_train)\n\\end{lstlisting}\n\nUsing only \\emph{Lag1} and \\emph{Lag2} as predictors gave a better result. Running the score function in Listing \\ref{lst:LRscore} showed a mean accuracy of $56\\%$. By only using \\emph{Lag1} and \\emph{Lag2} the model has been improved. This combination of predictors also showed the most accurate for predicting the \\emph{Direction}. Using p-values for choosing predictors is important and provides information about the association when multiple predictors are used.\n\n\\section{Linear Discriminant Analysis}\n\\label{chp:lindisana}\n\nIn chapter \\ref{chp:logreg} it was described how logistic regression could be used for binary classification. It was also stated how it could be extended to a classification problem with more than two classes. In practice logistic regression is though rarely used for classification problems with more than two classes \\citep[pp. 137]{ISLR}. Furthermore a problem with logistic regression is that it does not perform very stable under some circumstances. One big problem with logistic regression is that it might not converge if the classes are well-separated \\citep{convergencefailure}. An example could be the data in the table below. Here we see that the classes are well-separated as $X<0$ then $Y=0$ and if $X>0$ then $Y=1$. Because of this perfect separation the algorithm will never converge because it can keep optimizing.\n%\\begin{center} \n%\t\\begin{tabular}{rll}\n%\t\t\\multicolumn{1}{c}{\\textbf{Y}} &\n%\t\t\\multicolumn{1}{c}{\\textbf{X}} \\\\ \\hline\n%\t\t0 & -2 \\\\[0.05cm] \n%\t\t0 & -1 \\\\[0.05cm] \n%\t\t0 & -5 \\\\[0.05cm] \n%\t\t1 & 3 \\\\[0.05cm] \n%\t\t1 & 1 \\\\[0.05cm] \n%\t\t1 & 6 \\\\[0.05cm] \n%\t\\end{tabular}\n%\\end{center}\n\n\\begin{center} \n\t\\begin{tabular}{l|llllll}\n%\t\t\\multicolumn{1}{c}{\\textbf{Y}} &\n%\t\t\\multicolumn{1}{c}{\\textbf{X}} \\\\ \\hline\n\t\tX & -2 & -1 & -5 & 3 & 1 & 6 \\\\[0.05cm] \n\t\tY & 0 & 0 & 0 & 1 & 1 & 1 \\\\[0.05cm] \n\t\\end{tabular}\n\\end{center}\n\nInstead a method called Linear Discriminant Analysis \\emph{LDA} can be used. In logistic regression we directly model $Pr(Y|X)$. With LDA $P(X|Y)$ is modeled and prediction is done by using Bayes theorem to achieve $P(Y|X)$ and find the most probable class given the predictors. The following variables are used:\n\n\\begin{itemize}\n\t\\item Let $\\pi_k$ represent the prior probability $P(Y=k)$ for K classes.\n\t\\item Let $f_k(x)$ represent the probability $P(X=x|Y=k)$. So $f_k(x)$ is the probability of the predictors when the class Y is given.\n\\end{itemize} \n\nBy inserting these in Bayes formula and normalizing to get $P(Y=k|X=x)$ the following can be written.\n\n\\begin{equation}\nP(Y=k|X = x) = \\dfrac{\\pi_k f_k(x)}{\\sum_{l=1}^{K} \\pi_l f_l(x)}\n\\end{equation}\n\nFor the equation the two variables $\\pi_k$ and $f_k(x)$ still needs to be estimated. $\\pi_k$ can be estimated by computing the fraction of the training data that belongs to the k'th class, while $f_k(x)$ can be estimated as a Gaussian. This means that the equation differs when using only 1 predictor to using multiple. When only using one predictor and taking log of the equation it can be rewritten as the following\n\n\\begin{equation}\n\\delta_k(x) = x \\dfrac{\\mu_k}{\\sigma^2} - \\dfrac{\\mu_k^2}{2\\sigma^2} + log(\\pi_k)\n\\end{equation}\n\nThen the equation is calculated for each class k and is classified for the k that results in the highest probability. The equation for multiple predictors instead uses a mean vector and a covariance matrix and can be written as follows:\n\n\\begin{equation}\n\\delta_k(x) = x^T \\Sigma^{-1} \\mu_k - \\frac{1}{2} \\mu_k^{T} \\Sigma^{-1} \\mu_k + log (\\pi_k)\n\\end{equation}\n\nThe means of the Gaussian for each class are simply calculated by summing all values with a specific class and dividing by the total number. The covariance matrix is calculated as a common covariance for all classes.\n\nNote here that LDA assumes the observation to have a Gaussian distribution. If this is the case then LDA is likely to outperform Logistic regression, while if it is not, then Logistic regression is likely to outperform LDA \\citep{comparelogandlda}.\n\n\\subsection{Lab 4.6.3 - Linear Discriminant Analysis}\nFirst a linear discriminant analysis model is created with the python library sklearn. The model is created based on the two predictors \\emph{Lag1} and \\emph{Lag2}. \n\n\n\\begin{lstlisting}[language=Python, label=lst:ldaModel, caption=Creating linear discriminant analysis model sklearn]\nX_train = data[:'2004'][['Lag1', 'Lag2']]\nY_train = data[:'2004']['Direction']\n\nX_test = data['2005':][['Lag1', 'Lag2']]\nY_test = data['2005':]['Direction']\n\nlda = LinearDiscriminantAnalysis()\nfit = lda.fit(X_train,Y_train)\n\\end{lstlisting}\n\nAfter creating the model the calculated prior probability, means and coefficients can be read. The prior probability for \\emph{Down} is $\\pi_1=0.492$ while the prior probability for \\emph{Up} is $\\pi_2=0.508$. So based on the data there are more days where the stock market went up than down. \n\nThe means $\\mu_k$ are described by a matrix. The matrix describes a separate mean for both \\emph{Lag1} and \\emph{Lag2} and for each predictor a mean for \\emph{Up} and \\emph{Down}. The mean value for \\emph{Lag1} and \\emph{Lag2} has a tendency to be positive on days when the stock went up while they are negative when the stock went down. \n\n\\begin{center}\n\t$\\begin{matrix} \t & Lag1 & Lag2\t\\\\ \t Down & 0.0428 & 0.0339 \\\\ \t Up & -0.0395 & -0.0313 \\end{matrix} $\n\\end{center}\n\n\nThe coefficients are estimated to $-0.0554$ and $-0.0443$. These are the multipliers for \\emph{Lag1} and \\emph{Lag2}.\n\n%\\todo{Er det vigtigt at vise hvordan vi printer disse værdier?}\n%\\begin{lstlisting}[language=Python, label=lst:ldaProperties, %caption=printing lda prior mean and coefficients]\n%print \"Prior probabilities:\"\n%print lda.priors_\n%\n%print \"Mean:\"\n%print lda.means_\n%\n%print \"Coefficients:\"\n%print lda.coef_\n%\\end{lstlisting}\n\nIn listing \\ref{lst:ldaScore} the score is printed. The score show that the LDA model predicted correctly in $56\\%$ of the predictions on the test data. \n\\begin{lstlisting}[language=Python, label=lst:ldaScore, caption=Printing LDA score]\nprint(lda.score(X_test, Y_test))\n\\end{lstlisting}\n\nThis shows how linear discriminant analysis performs very similar to logistic regression on the stock market data.\n\n\\section{Quadratic Discriminant Analysis}\nSimilar to LDA the Quadratic discriminant analysis \\emph{QDA} also assumes that the observations are drawn from a Gaussian distribution. The difference lies in the fact that LDA assumes a shared covariance for the observations for all classes K, while QDA assumes a different covariance for the observations for each class.\n\n\\myFigure{lda_vs_qda.PNG}{Comparison of LDA and QDA}{fig:qdaCompare}{0.6}\n\nIn figure \\ref{fig:qdaCompare} the difference between LDA and QDA is illustrated. It shows the linear decision boundary of LDA, while the QDA can learn quadratic boundaries.\n\n\\subsection{Lab 4.6.4 - Quadratic Discriminant Analysis}\nAgain the stock market data is used for prediction with \\emph{Lag1} and \\emph{Lag2} as predictors. The QDA model is estimated with sklearn. This is shown in listing \\ref{lst:qdaModel}.\n\n\\begin{lstlisting}[language=Python, label=lst:qdaModel, caption=Creating QDA model]\nX_train = data[:'2004'][['Lag1', 'Lag2']]\nY_train = data[:'2004']['Direction']\n\nX_test = data['2005':][['Lag1', 'Lag2']]\nY_test = data['2005':]['Direction']\n\nqda = QuadraticDiscriminantAnalysis()\nfit = qda.fit(X_train,Y_train)\n\\end{lstlisting}\n\nIn listing \\ref{lst:qdaScore} the score for the QDA model is printed. The print shows that on the test data the prediction was correct in $59.9\\%$ of the cases. With the stock market data it actually performed better than both logistic regression and linear discriminant analysis. This might indicate that the relationship for the data set is better captures with a quadratic form.\n\\begin{lstlisting}[language=Python, label=lst:qdaScore, caption=Printing qda score]\nprint(qda.score(X_test, Y_test))\n\\end{lstlisting}\n\n\\section{K-Nearest Neighbors}\nK-nearest neighbors or \\emph{KNN} is a classification technique that classifies observations depending on distance to k other observations. \n\nSay you have a data set with two classes of observations A and B, with 10 observations each. Then a test observation is desired to be classified with $k=5$, which means the five nearest observations will be examined. If three of the examined observations are A and the other two are B then the estimated probabilities are $3/5$ for class A and $2/5$ for class B. \n\n\\subsection{Lab 4.6.5 - K-Nearest Neighbors}\n\nIn this lab exercise the method K-nearest neighbors is used. To perform this method, sklearn has a function called \\emph{KNeighborsClassifier}. This takes one parameter, which is number of k. After this the function fit is called, that fits the data.\nThe code is shown in listing \\ref{lst:kneighbor}.\n\n\\begin{lstlisting}[caption={Python KNN function, where k is set to 1}, label=lst:kneighbor, mathescape=true]\nX_train = data[:'2004'][['Lag1', 'Lag2']]\nY_train = data[:'2004']['Direction']\n\nX_test = data['2005':][['Lag1', 'Lag2']]\nY_test = data['2005':]['Direction']\n\n\nknn = KNeighborsClassifier(n_neighbors=1)\nfit = knn.fit(X_train, Y_train)\n\\end{lstlisting}\n\nThe function is used to predict the market's movement from the dates in 2005. If the value k is set to 1, 50\\% of the predictions are correctly predicted which is not very good.\nIf k is change to 3, the predictions turn out to be correct in 53\\% of the cases. So increasing the number of k from 1 to 3 made the model a bit better. All kinds of different values can be chosen for k, but it is important not to change the number too often because the model then will be customized to the training data which is not good.\n\nIf the results are compared with the results that QDA produces, the QDA provides better results than KNN.", "meta": {"hexsha": "d34b4f6ae21b65a364d0123343c02ead49235bcb", "size": 18841, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/chapter/logistic_regression.tex", "max_stars_repo_name": "Rotvig/F17Q4---Decision-Support-Systems", "max_stars_repo_head_hexsha": "62c3bb41a8fb4df75f4f038dea7af521cddafa4a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Report/chapter/logistic_regression.tex", "max_issues_repo_name": "Rotvig/F17Q4---Decision-Support-Systems", "max_issues_repo_head_hexsha": "62c3bb41a8fb4df75f4f038dea7af521cddafa4a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Report/chapter/logistic_regression.tex", "max_forks_repo_name": "Rotvig/F17Q4---Decision-Support-Systems", "max_forks_repo_head_hexsha": "62c3bb41a8fb4df75f4f038dea7af521cddafa4a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 74.7658730159, "max_line_length": 832, "alphanum_fraction": 0.759566902, "num_tokens": 5075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.7879311981328135, "lm_q1q2_score": 0.6269238400868391}} {"text": "\\subsection{Category equivalences}\\label{subsec:category_equivalences}\n\n\\begin{remark}\\label{rem:category_similarity}\n We have the following notions for expressing that two categories \\( \\cat{C} \\) and \\( \\cat{D} \\) are similar:\n\n \\begin{thmenum}\n \\thmitem{rem:category_similarity/equality} Obviously, if \\( \\cat{C} \\) and \\( \\cat{D} \\) are equal, they are similar.\n\n \\thmitem{rem:category_similarity/isomorphism} A slightly less obvious notion is \\term{isomorphism of categories}. This is an isomorphism, in the sense of \\fullref{def:morphism_invertibility/isomorphism}, in the category \\hyperref[def:category_of_small_categories]{\\( \\ucat{Cat} \\)} of small categories for a suitable \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\). That is, \\( \\cat{C} \\) and \\( \\cat{D} \\) are isomorphic if there exists an invertible functor between them.\n\n We rarely distinguish between objects and arrows of isomorphic categories, even if we do not have strict equality in the sense of the \\hyperref[def:zfc/extensionality]{axiom of extensionality} in \\hyperref[def:zfc]{\\logic{ZFC}}.\n\n Examples of isomorphic categories include \\fullref{thm:order_category_isomorphism} and \\fullref{ex:isomorphism_of_quiver_categories}.\n\n \\thmitem{rem:category_similarity/equivalence} A weaker but very useful notion is \\term{category equivalence} defined in \\fullref{def:category_equivalence}.\n \\end{thmenum}\n\\end{remark}\n\n\\begin{example}\\label{ex:set_discr_cat_isomorphism}\n There is an isomorphism between the category \\( \\ucat{Set} \\) of small sets and \\( \\ucat{DiscrCat} \\) of all small \\hyperref[def:discrete_category]{discrete categories}.\n\n Consider the pair of functors\n \\begin{equation*}\n U: \\ucat{DiscrCat} \\to \\ucat{Set},\n \\end{equation*}\n which for any small category \\( \\cat{C} \\) gives us its set of objects \\( \\obj(\\cat{C}) \\) and\n \\begin{equation*}\n D: \\ucat{Set} \\to \\ucat{DiscrCat},\n \\end{equation*}\n which for any small set \\( A \\) gives us the \\hyperref[def:discrete_category]{discrete category} whose set of objects is \\( A \\).\n\n These were discussed in \\fullref{ex:discrete_category_adjunction}, although with \\( \\cat{Cat} \\) rather than \\( \\ucat{DiscrCat} \\).\n\n It is obvious that for any small category \\( \\cat{C} \\), the functor \\( D \\bincirc U \\) is bijective on objects. We need to verify that it is bijective on morphisms, however, in order to prove that \\( D \\bincirc U \\) is the identity functor on \\( \\ucat{DiscrCat} \\).\n\n But any functor \\( F: \\cat{C} \\to \\cat{D} \\) is completely determined by how \\( F \\) acts on the objects of \\( \\cat{C} \\). Indeed, the only morphisms in \\( \\cat{C} \\) are the identity morphisms and, by \\ref{def:functor/CF1}, every object \\( X \\) in \\( \\cat{C} \\) determines how the identity \\( \\id_X \\in \\cat{C}(X) \\) is mapped by \\( F \\).\n\n Therefore, \\( D \\) is a left inverse of \\( U \\). It is also a right inverse --- for any function \\( f: A \\to B \\) between small sets,\n \\begin{equation*}\n [U \\bincirc D](f) = D(f)\\restr_{A} = f.\n \\end{equation*}\n\n Therefore, the forgetful functor \\( U \\) is invertible, and its inverse is \\( D \\).\n\\end{example}\n\n\\begin{definition}\\label{def:category_equivalence}\\mcite[def. 1.3.15]{Leinster2016Basic}\n An \\term{equivalence} between the \\hyperref[def:category]{categories} \\( \\cat{C} \\) and \\( \\cat{D} \\) is a quadruple\n \\begin{equation}\\label{eq:def:category_equivalence/signature}\n \\begin{aligned}\n F &: \\cat{C} \\to \\cat{D}, \\\\\n G &: \\cat{D} \\to \\cat{C}, \\\\\n \\eta &: \\id_{\\cat{C}} \\Rightarrow G \\bincirc F, \\\\\n \\varepsilon &: F \\bincirc G \\Rightarrow \\id_{\\cat{D}},\n \\end{aligned}\n \\end{equation}\n where \\( \\eta \\) and \\( \\varepsilon \\) are \\hyperref[thm:natural_isomorphism]{natural isomorphisms}.\n\n We call \\( \\eta \\) the \\term{unit} of the equivalence and \\( \\varepsilon \\) the \\term{counit}.\n\n If \\( (F, G, \\eta, \\varepsilon) \\) is an equivalence, we say that \\( \\cat{C} \\) and \\( \\cat{D} \\) are equivalent categories. This is justified because equivalence of categories is an equivalence relation --- see \\fullref{thm:category_equivalence_is_equivalence_relation}.\n\n Note that an equivalence is not necessarily an \\hyperref[def:category_adjunction]{adjunction}, they simply have a common setup. This is discussed in \\fullref{thm:adjoint_equivalence}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:category_equivalence_is_equivalence_relation}\n For every \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\), \\hyperref[def:category_equivalence]{category equivalence} is an \\hyperref[def:equivalence_relation]{equivalence relation} on the set \\( \\obj(\\ucat{Cat}) \\)\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf[def:binary_relation/reflexive]{reflexivity} Clearly \\( (\\id_{\\cat{C}}, \\id_{\\cat{C}}, \\id_{\\id_{\\cat{C}}}, \\id_{\\id_{\\cat{C}}}]) \\) is a self-equivalence for the category \\( \\cat{C} \\).\n\n \\SubProofOf[def:binary_relation/symmetric]{symmetry} If \\( (F, G, \\eta, \\varepsilon) \\) is an \\hyperref[def:category_equivalence]{equivalence} between the categories \\( \\cat{C} \\) and \\( \\cat{D} \\), then \\( (G, F, \\varepsilon^{-1}, \\eta^{-1}) \\) is an equivalence between \\( \\cat{D} \\) and \\( \\cat{C} \\).\n\n \\SubProofOf[def:binary_relation/transitive]{transitivity} By \\fullref{thm:def:morphism_invertibility/invertible_composition}, the composition of invertible natural isomorphisms is again a natural isomorphism, hence equivalence in \\( \\obj(\\ucat{Cat}) \\) is a transitive relation.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:discrete_category_equivalence}\n Let \\( \\cat{C} \\) and \\( \\cat{D} \\) be \\hyperref[def:discrete_category]{discrete categories}. Then \\( \\cat{C} \\) and \\( \\cat{D} \\) are \\hyperref[def:category_equivalence]{equivalent} if and only if the underlying sets \\( \\obj(\\cat{C}) \\) and \\( \\obj(\\cat{D}) \\) are \\hyperref[def:equinumerosity]{equinumerous}.\n\\end{proposition}\n\\begin{proof}\n \\SufficiencySubProof Suppose that \\( (F, G, \\eta, \\varepsilon) \\) be a category equivalence.\n\n The unit natural transformation \\( \\eta: \\id_{\\cat{C}} \\Rightarrow G \\bincirc F \\) consists of a morphism\n \\begin{equation*}\n \\eta_A: A \\to [G \\bincirc F](A)\n \\end{equation*}\n for every object \\( A \\) of \\( \\cat{C} \\). Since the only morphisms in \\( \\cat{C} \\) are the identities, it follows that \\( \\eta_A = \\id_A \\) and hence \\( [G \\bincirc F](A) = A \\). In particular, this implies that \\( \\eta \\) is the \\hyperref[eq:def:functor_category/identity]{identity natural transformation} on \\( \\id_{\\cat{C}} \\) and that the restriction \\( G\\restr_{\\obj(D)} \\) is a left inverse of \\( F\\restr_{\\obj(\\cat{C})} \\).\n\n Similarly, for the counit \\( \\varepsilon: F \\bincirc G \\Rightarrow \\id_{\\cat{D}} \\), for every object \\( X \\) in \\( \\cat{D} \\) we have \\( \\varepsilon_X = \\id_X \\) and hence \\( [F \\bincirc G](X) = X \\). Thus, \\( \\eta \\) is the identity natural transformation on \\( \\id_{\\cat{D}} \\) and \\( G\\restr_{\\obj(D)} \\) is a right inverse of \\( F\\restr_{\\obj(\\cat{C})} \\).\n\n Therefore, the sets \\( \\obj(\\cat{C}) \\) and \\( \\obj(D) \\) are equinumerous.\n\n \\NecessitySubProof Suppose that \\( F: \\obj(\\cat{C}) \\to \\obj(D) \\) is a bijective function. Then it is an isomorphism in the category \\( \\ucat{Cat} \\) for an appropriate universe \\( \\mscrU \\), hence it induces an equivalence between \\( \\cat{C} \\) and \\( \\cat{D} \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:opposite_of_category_equivalence}\n The \\hyperref[def:opposite_category]{opposite} of \\hyperref[def:category_equivalence]{equivalent categories} are equivalent.\n\n More precisely, if \\( (F, G, \\eta, \\varepsilon) \\) is an \\hyperref[def:category_equivalence]{equivalence} between the categories \\( \\cat{C} \\) and \\( \\cat{D} \\), then\n \\begin{equation*}\n \\begin{aligned}\n G^{\\opcat} &: \\cat{D}^{\\opcat} \\to \\cat{C}^{\\opcat}, \\\\\n F^{\\opcat} &: \\cat{C}^{\\opcat} \\to \\cat{D}^{\\opcat}, \\\\\n \\varepsilon^{\\opcat} &: \\id_{\\cat{D}} \\Rightarrow [F \\bincirc G]^{\\opcat}, \\\\\n \\eta^{\\opcat} &: \\underbrace{[G \\bincirc F]^{\\opcat}}_{G^{\\opcat} \\bincirc F^{\\opcat}} \\Rightarrow \\id_{\\cat{C}^{\\opcat}},\n \\end{aligned}\n \\end{equation*}\n is an equivalence between \\( \\cat{C}^{\\opcat} \\) and \\( \\cat{D}^{\\opcat} \\).\n\n This is part of the duality principles listed in \\fullref{thm:categorical_principle_of_duality}.\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor}\n In any \\hyperref[def:category_equivalence]{category equivalence} \\( (F, G, \\eta, \\varepsilon) \\), the functor \\( F \\) is \\hyperref[def:functor_invertibility/fully_faithful]{fully faithful} and \\hyperref[def:functor_invertibility/surjective_on_objects]{essentially surjective on objects}.\n\n The converse of this statement is \\fullref{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf[def:functor_invertibility/surjective_on_objects]{essential surjectivity} For any object \\( X \\) in \\( \\cat{D} \\), \\( A \\coloneqq G(X) \\) is an object in \\( \\cat{C} \\).\n\n By definition of category equivalence, the morphism\n \\begin{equation*}\n \\varepsilon_X: \\underbrace{[F \\bincirc G](X)}_{F(A)} \\to X\n \\end{equation*}\n is an isomorphism.\n\n Therefore, for every object \\( X \\) in \\( \\cat{D} \\), there exists some object \\( A \\) in \\( \\cat{C} \\) such that \\( F(A) \\cong X \\). Thus, \\( F \\) is essentially surjective.\n\n \\SubProofOf[def:functor_invertibility/faithful]{faithfulness} Fix some objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\). Let \\( f_1: A \\to B \\) and \\( f_2: A \\to B \\) be morphisms such that \\( F(f_1) = F(f_2) \\).\n\n From the naturality of \\( \\eta \\) it follows that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor/faithfullness}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__equivalence_induces_fully_faithful_and_essentially_surjective_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore, \\( \\eta_B \\bincirc f_1 = \\eta_B \\bincirc f_2 \\) and, since \\( \\eta_B \\) is left-cancellative, \\( f_1 = f_2 \\).\n\n \\SubProofOf[def:functor_invertibility/full]{fullness} Fix some objects \\( A \\) and \\( B \\) in \\( \\cat{C} \\). Let \\( g: F(A) \\to F(B) \\) be an arbitrary morphism.\n\n We can define a morphism \\( f: A \\to B \\) via the composition\n \\begin{equation}\\label{eq:thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor/fullness/def}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__equivalence_induces_fully_faithful_and_essentially_surjective_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n By naturality of \\( \\eta \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor/fullness/eta_nat}\n \\begin{aligned}\n \\includegraphics[page=3]{output/thm__equivalence_induces_fully_faithful_and_essentially_surjective_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore,\n \\begin{equation*}\n G(g) = [G \\bincirc F](f).\n \\end{equation*}\n\n We have already shown that \\( F \\) is faithful and, by \\fullref{thm:category_equivalence_is_equivalence_relation}, \\( G \\) is also faithful. Since \\( g \\) and \\( F(f) \\) are parallel, it follows that they are equal.\n\n Therefore, \\( F \\) is full.\n\\end{proof}\n\n\\begin{remark}\\label{rem:adjoint_equivalence_induces_fully_faithful_and_essentially_surjective_functor}\n In the fullness proof of \\fullref{thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor}, we can use another argument if \\( (F, G, \\eta, \\varepsilon) \\) is an \\hyperref[def:adjoint_equivalence]{adjoint equivalence}.\n\n If the triangle diagram \\eqref{eq:def:category_adjunction/d_triangle} commutes, the dashed lines in the following diagram also commute:\n \\begin{equation}\\label{eq:thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor/fullness/triangles}\n \\begin{aligned}\n \\includegraphics[page=1]{output/rem__adjoint_equivalence_induces_fully_faithful_and_essentially_surjective_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\\end{remark}\n\n\\begin{theorem}[Fully faithful and essentially surjective functor induces equivalence]\\label{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}\n Every \\hyperref[def:functor_invertibility/fully_faithful]{fully faithful} and \\hyperref[def:functor_invertibility/surjective_on_objects]{essentially surjective on objects} functor induces a \\hyperref[def:category_equivalence]{category equivalence}.\n\n More precisely, given a functor \\( F: \\cat{C} \\to \\cat{D} \\) that is fully faithful and essentially surjective on objects, there exists a functor \\( G: \\cat{D} \\to \\cat{C} \\) and \\hyperref[def:natural_transformation]{natural transformations}\n \\begin{align*}\n \\eta &: \\id_{\\cat{C}} \\Rightarrow G \\bincirc F, \\\\\n \\varepsilon &: F \\bincirc G \\Rightarrow \\id_{\\cat{D}},\n \\end{align*}\n such that the quadruple \\( (F, G, \\eta, \\varepsilon) \\) is a \\hyperref[def:category_equivalence]{category equivalence}.\n\n In \\hyperref[def:zfc]{\\logic{ZF}}, this theorem is equivalent to the \\hyperref[def:zfc/choice]{axiom of choice} --- see \\fullref{thm:axiom_of_choice_equivalences/fully_faithful_essentially_surjective}.\n\n We prove the converse of this statement separately in \\fullref{thm:equivalence_induces_fully_faithful_and_essentially_surjective_functor}.\n\\end{theorem}\n\\begin{proof}\n \\ImplicationSubProof[def:zfc/choice]{the axiom of choice}[thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence]{functors induce equivalences} Suppose that the axiom of choice holds and let \\( F \\) be a fully faithful functor that is surjective on objects.\n\n From essential surjectivity of \\( F \\), it follows that for every object \\( X \\) in \\( \\cat{D} \\), the preimage of \\( X \\) under \\( F \\) is nonempty. The preimage of \\( X \\) is the set \\( \\mscrA_X \\) of objects in \\( \\cat{C} \\) such that \\( A \\in \\mscrA_X \\) if and only if \\( F(A) \\cong X \\). We use the axiom of choice on the family \\( \\set{ \\mscrA_X }_{X \\in \\cat{D}} \\) to select a single preimage for every \\( X \\), which we denote by \\( G(X) \\).\n\n Again using the axiom of choice, we pick an isomorphism \\( \\varepsilon_X: F(G(X)) \\to X \\) for every \\( X \\).\n\n We have defined a function \\( G \\) from \\( \\obj(\\cat{D}) \\) to \\( \\obj(\\cat{C}) \\). In order to \\( G \\) to become a functor, we must extend it to morphisms. Let \\( X \\) and \\( Y \\) be objects in \\( \\cat{D} \\) and \\( g: X \\to Y \\) be any morphism.\n\n Consider the morphism\n \\begin{equation*}\n \\varepsilon_Z^{-1} \\bincirc f \\bincirc \\varepsilon_X: [F \\bincirc G](X) \\to [F \\bincirc G](Y).\n \\end{equation*}\n\n Since \\( F \\) is fully faithful, there exists a unique morphism \\( g \\) in \\( \\cat{D}(G(X), G(Y)) \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/inverse_morphism_definition}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n We define \\( G(g) \\coloneqq g \\).\n\n In order to prove that \\( G \\) is a functor, we need to show that \\ref{def:functor/CF1} and \\ref{def:functor/CF2} hold.\n\n For \\ref{def:functor/CF1}, note that the following diagram commutes for any object \\( X \\) in \\( \\cat{D} \\):\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/identity}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n Note that \\eqref{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/inverse_morphism_definition} also commutes if we replace \\( [F \\bincirc G](\\id_X) \\) with \\( F(\\id_{G(X)}) \\). Since \\( F \\) is fully faithful, this morphism is unique and it follows that\n \\begin{equation*}\n [F \\bincirc G](\\id_X) = F(\\id_{G(X)}).\n \\end{equation*}\n\n For \\ref{def:functor/CF2}, analogously, given morphisms \\( g: X \\to Y \\) and \\( f: Y \\to Z \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/composition}\n \\begin{aligned}\n \\includegraphics[page=3]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n We have implicitly used \\fullref{rem:inverting_isomorphisms_may_preserve_commutativity} above.\n\n By the same uniqueness argument used for \\ref{def:functor/CF1}, we conclude that\n \\begin{equation*}\n G(f \\bincirc q) = G(f) \\bincirc G(g).\n \\end{equation*}\n\n We have shown that \\( G \\) is a functor. Furthermore, \\( \\varepsilon \\) is a natural transformation since, for any morphism \\( g: X \\to Y \\) in \\( \\cat{D} \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/varepsilon}\n \\begin{aligned}\n \\includegraphics[page=4]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n To show that \\( F \\) induces an equivalence, it now only remains to define a unit natural transformation \\( \\eta: \\id_{\\cat{C}} \\to G \\bincirc F \\). For every object \\( A \\) in \\( \\cat{C} \\) we have an isomorphism\n \\begin{equation*}\n \\varepsilon_{F(A)}^{-1}: F(A) \\to [F \\bincirc G \\bincirc F](A).\n \\end{equation*}\n\n Using \\( G(\\varepsilon_{F(A)}^{-1}) \\) will get us nowhere. Fortunately, \\( F \\) is fully faithful, so there is a bijective function\n \\begin{equation*}\n \\varphi: \\cat{D}(F(A), [F \\bincirc G \\bincirc F](A)) \\to \\cat{C}(A, [F \\bincirc G](A)).\n \\end{equation*}\n\n Hence, we can define\n \\begin{equation*}\n \\begin{aligned}\n &\\eta: \\id_{\\cat{C}} \\Rightarrow F \\bincirc G \\\\\n &\\eta_A \\coloneqq \\varphi(\\varepsilon_{F(A)}^{-1})\n \\end{aligned}\n \\end{equation*}\n so that \\( F(\\eta_A) = \\varepsilon_{F(A)}^{-1} \\). By \\fullref{thm:def:functor_invertibility/fully_faithful_reflects_invertible}, since \\( \\varepsilon_{F(A)} \\) is an isomorphism, \\( \\eta_A \\) is also an isomorphism.\n\n By naturality of \\( F(\\eta_A) \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/varepsilon_image_nat}\n \\begin{aligned}\n \\includegraphics[page=5]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n Hence, by \\fullref{thm:commutative_diagrams_preserved_and_reflected}, the following diagram also commutes:\n \\begin{equation}\\label{eq:thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence/varepsilon_source_nat}\n \\begin{aligned}\n \\includegraphics[page=6]{output/thm__fully_faithful_and_essentially_surjective_functor_induces_equivalence.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore, the quadruple \\( (F, G, \\eta, \\varepsilon) \\) is an equivalence of categories.\n\n \\ImplicationSubProof[thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence]{functors induce equivalences}[def:zfc/choice]{the axiom of choice} Let \\( \\mscrA \\) be a family of nonempty sets. Let \\( \\cat{D} \\) be the \\hyperref[def:discrete_category]{discrete category} induced by \\( \\mscrA \\).\n\n Define the category \\( \\cat{C} \\) as follows:\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(\\cat{C}) \\) is the \\hyperref[def:disjoint_union]{disjoint union} \\( \\coprod_{A \\in \\mscrA} A \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{C}((A, x), (B, y)) \\) has a single morphism if \\( A = B \\) and no morphisms otherwise. This single morphism can be encoded as the triple \\( (A, x, y) \\).\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms} \\( (A, x, y) \\) and \\( (A, y, z) \\) is the morphism \\( (A, x, z) \\).\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the object \\( (A, x) \\in \\cat{C} \\) is \\( (A, x, x) \\).\n \\end{itemize}\n\n Define the functor\n \\begin{equation*}\n \\begin{aligned}\n &F: \\cat{C} \\to \\cat{D} \\\\\n &F(A, x) \\coloneqq A \\\\\n &F(A, x, y) \\coloneqq \\id_A\n \\end{aligned}\n \\end{equation*}\n that maps each point \\( x \\in A \\in \\mscrA \\) into the set \\( A \\) it belongs to. We have taken the disjoint union of \\( \\mscrA \\) since otherwise there may not be a canonical choice of set \\( A \\) for \\( F \\) to send \\( x \\) to. Thus, the functor is surjective on objects (not essentially surjective but actually surjective).\n\n Note that \\( \\cat{D}(F(A, x), F(B, y)) \\) has a single morphism if \\( A = B \\) and is empty otherwise. From this it follows that \\( F \\) is fully faithful.\n\n Therefore, \\( F \\) induces a \\hyperref[def:category_equivalence]{category equivalence} \\( (F, G, \\eta, \\varepsilon) \\). The functor \\( G \\) chooses an object \\( (A, x) \\) of \\( \\cat{C} \\) for each object \\( A \\) of \\( \\cat{D} \\). This induces a \\hyperref[def:choice_function]{choice function} on \\( \\mscrA \\).\n\n We have shown that the axiom of choice holds.\n\\end{proof}\n\n\\begin{definition}\\label{def:groupoid}\n A \\term{groupoid} is a category whose only morphisms are \\hyperref[def:morphism_invertibility/isomorphism]{isomorphisms}.\n\\end{definition}\n\n\\begin{definition}\\label{def:monoid_delooping}\\mcite[def. 1.1.7]{Perrone2019}\n Let \\( (M, \\cdot, e) \\) be a \\hyperref[def:monoid]{monoid}. The \\term{delooping} \\( \\cat{B}_M \\) of \\( M \\) is the following category:\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(\\cat{B}_M) \\) is the singleton set \\( \\set{ \\Anon } \\), where \\( \\Anon \\) is any set not in the set of all morphisms \\( M \\).\n\n \\item The only \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{B}_M(\\Anon) \\) is the underlying set \\( M \\) of the monoid.\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms} \\( x \\) and \\( y \\) is the multiplication:\n \\begin{equation*}\n y \\bincirc x \\coloneqq y \\cdot x.\n \\end{equation*}\n\n Note how we write composition in the same order as multiplication. This may seem to contradict the general convention, however it is consistent with groups being regarded as sets of invertible transformations.\n\n \\item The \\hyperref[def:category/identity]{identity morphism} is \\( e \\).\n \\end{itemize}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:delooping_of_group}\n The \\hyperref[def:monoid_delooping]{delooping} of a \\hyperref[def:group]{group} \\( G \\) is a \\hyperref[def:groupoid]{groupoid}.\n\n There is a restricted form of a converse --- see \\fullref{thm:connected_delooping}.\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{definition}\\label{def:connected_category}\\mcite[exer. 1.5.20]{Perrone2019}\n A \\hyperref[def:category]{category} is \\term{connected} if there exists a morphism between any two objects. That is, for a connected category \\( \\cat{C} \\), given two objects \\( X \\) and \\( Y \\), either \\( \\cat{C}(X, Y) \\) or \\( \\cat{C}(Y, X) \\) is nonempty.\n\n Connected categories are used in \\fullref{thm:connected_delooping} and \\fullref{thm:order_category_isomorphism/totally_ordered}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:connected_delooping}\n For every \\hyperref[def:connected_category]{connected} \\hyperref[def:groupoid]{groupoid} \\( \\cat{G} \\) there exists a \\hyperref[def:group]{group} \\( G \\) such that \\( \\cat{G} \\) is \\hyperref[def:category_equivalence]{equivalent} to the \\hyperref[def:monoid_delooping]{delooping} \\( \\cat{B}_G \\). Furthermore, if \\( \\cat{G} \\) has only one object, then this equivalence is an \\hyperref[rem:category_similarity/isomorphism]{isomorphism}.\n\n See \\fullref{thm:delooping_of_group} for a much simpler converse.\n\\end{proposition}\n\\begin{proof}\n Define the group \\( G \\) as follows:\n \\begin{itemize}\n \\item Let the underlying set of \\( G \\) be the set of all morphisms of \\( \\cat{G} \\).\n \\item Define \\( x \\cdot y = z \\) to hold whenever the following diagram commutes:\n \\begin{equation}\\label{eq:thm:connected_delooping/mult}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__connected_delooping.pdf}\n \\end{aligned}\n \\end{equation}\n \\item Pick the identity of \\( G \\) out of the identity morphisms in \\( G \\).\n \\item Let the inverse of \\( x \\) be the inverse morphism \\( x^{-1} \\).\n \\end{itemize}\n\n Finally, define the functor \\( F: \\cat{G} \\to \\cat{B}_G \\), which sends every object to \\( \\anon \\) and every morphism to itself.\n\n By definition, for every morphism \\( x: \\anon \\to \\anon \\) in the delooping, there exists a unique \\( f: X \\to Y \\) in \\( \\cat{G} \\) such that \\( F(f) = x \\). Hence, \\( f \\) is surjective on morphisms, and by \\fullref{thm:def:functor_invertibility/surjective}, full and surjective on objects.\n\n Furthermore, for any objects \\( X \\) and \\( Y \\) in \\( \\cat{G} \\), we have\n \\begin{equation*}\n \\cat{G}(X, Y) \\subseteq \\cat{B}_G(F(X), F(Y)).\n \\end{equation*}\n\n Hence, \\( G \\) is faithful. Therefore, by \\fullref{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}, the groupoid \\( \\cat{G} \\) is equivalent to the delooping \\( \\cat{B}_G \\).\n\n Furthermore, if \\( \\cat{G} \\) has only one object, it is also injective on objects and, by \\fullref{thm:def:functor_invertibility/injective}, injective on morphisms. In this case, we have an isomorphism due to \\fullref{thm:def:functor_invertibility/isomorphism}.\n\\end{proof}\n\n\\begin{definition}\\label{def:skeletal_category}\\mcite[91]{MacLane1994}\n The category \\( \\cat{S} \\) is called \\term{skeletal} if the only isomorphisms in \\( \\cat{S} \\) are the identity morphisms.\n\n If \\( \\cat{S} \\) is a subcategory of \\( \\cat{C} \\) and if they are \\hyperref[def:category_equivalence]{equivalent}, we say that \\( \\cat{S} \\) is a \\term{skeleton} of \\( \\cat{C} \\).\n\\end{definition}\n\n\\begin{example}\\label{ex:skeleton_of_set}\n Fix a Grothendieck universe \\( \\mscrU \\), the \\hyperref[def:cardinal_number]{cardinal numbers} in \\( \\mscrU \\) are a \\hyperref[def:skeletal_category]{skeletal subcategory} of \\( \\ucat{Set} \\).\n\\end{example}\n\n\\begin{theorem}[Category skeleton existence]\\label{thm:category_skeleton_existence}\n Every \\hyperref[def:category]{category} has a \\hyperref[def:skeletal_category]{skeleton}.\n\n In \\hyperref[def:zfc]{\\logic{ZF}}, this theorem is equivalent to the \\hyperref[def:zfc/choice]{axiom of choice} --- see \\fullref{thm:axiom_of_choice_equivalences/skeletons}.\n\\end{theorem}\n\\begin{proof}\n \\ImplicationSubProof[def:zfc/choice]{the axiom of choice}[thm:category_skeleton_existence]{skeleton existence} Suppose that the axiom of choice holds and let \\( F \\) be a fully faithful functor that is surjective on objects.\n\n Fix a category \\( \\cat{C} \\). We will build a subcategory \\( \\cat{S} \\) of \\( \\cat{C} \\) whose inclusion functor \\( \\Iota: \\cat{S} \\to \\cat{C} \\) is essentially surjective and fully faithful. By \\fullref{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}, this is sufficient for \\( \\cat{S} \\) and \\( \\cat{C} \\) to be equivalent.\n\n In order for \\( \\Iota \\) to be a full functor, \\( \\cat{S} \\) must be a full subcategory. Therefore, when building \\( \\cat{S} \\), we can only remove objects and must preserve the morphism sets for the remaining objects.\n\n Denote by \\( \\obj(\\cat{C}) / \\cong \\) the quotient of \\( \\cat{C} \\) by the isomorphism relation. Using the axiom of choice, we can obtain a \\hyperref[def:choice_function]{choice function} \\( c: (\\obj(\\cat{C}) / \\cong) \\to \\obj(\\cat{C}) \\).\n\n Define \\( \\cat{S} \\) as the subcategory induced by the image \\( c[\\obj(\\cat{C}) / \\cong] \\).\n\n Now consider the \\hyperref[def:subcategory]{inclusion functor} \\( \\Iota: \\cat{S} \\to \\cat{C} \\). For every pair \\( X \\) and \\( Y \\) of objects in \\( \\cat{S} \\), clearly\n \\begin{equation*}\n \\cat{S}(X, Y) = \\cat{C}(\\Iota(X), \\Iota(Y)).\n \\end{equation*}\n\n Hence, \\( \\Iota \\) is fully faithful.\n\n Now let \\( X \\) and \\( Y \\) be objects of \\( \\cat{C} \\). Since the objects of \\( \\cat{S} \\) were chosen from isomorphism classes of \\( \\cat{C} \\), there exist objects \\( X' \\) and \\( Y' \\) in \\( \\cat{S} \\) that are isomorphic to \\( X \\) and \\( Y \\), correspondingly. Hence, \\( \\Iota \\) is essentially surjective.\n\n Therefore, \\( F \\) satisfies \\fullref{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}, from which is follows that \\( \\cat{C} \\) and \\( \\cat{S} \\) are equivalent.\n\n \\ImplicationSubProof[thm:category_skeleton_existence]{skeleton existence}[def:zfc/choice]{the axiom of choice} Suppose that every category has a skeleton.\n\n Let \\( \\mscrA \\) be a family of nonempty sets. Construct a category \\( \\cat{C} \\)from the \\hyperref[def:disjoint_union]{disjoint union} \\( \\coprod_{A \\in \\mscrA} A \\), where a morphism exists only between members of the same set. This construction is performed in detail in the proof of \\fullref{thm:fully_faithful_and_essentially_surjective_functor_induces_equivalence}.\n\n Then \\( \\cat{C} \\) has a skeleton \\( \\cat{S} \\). All morphisms in \\( \\cat{C} \\) are isomorphisms, hence the set \\( \\obj(\\cat{S}) \\) contains exactly one representative for each set in the family \\( \\mscrA \\).\n\n More precisely, define the set\n \\begin{equation*}\n S \\coloneqq \\set{ x \\given (A, x) \\in \\obj(\\cat{S}) }.\n \\end{equation*}\n\n Then \\( S \\) satisfies \\fullref{thm:axiom_of_choice_equivalences/choice_sets}.\n\n Since the family \\( \\mscrA \\) is arbitrary, we conclude that the axiom of choice holds.\n\\end{proof}\n\n\\begin{remark}\\label{rem:skeletons_and_thin_categories}\n Rather than defining representatives of equivalence classes, as in \\fullref{thm:category_skeleton_existence}, we can define morphisms between the equivalence classes themselves, as in \\fullref{thm:preorder_to_partial_order}.\n\n This does not require the \\hyperref[def:zfc/choice]{axiom of choice}, but is rarely applicable, unfortunately. One case where it is applicable is in \\hyperref[def:thin_category]{thin categories} --- see \\fullref{thm:order_category_isomorphism}.\n\\end{remark}\n\n\\begin{definition}\\label{def:thin_category}\\mcite{nLab:thin_category}\n A \\hyperref[def:category]{category} is \\term{thin} if any two parallel morphisms are equal.\n\n This is equivalent to saying that the function for every two objects \\( A \\) and \\( B \\) in \\( \\cat{P} \\), whenever the set \\( \\cat{P}(A, B) \\) is at most a singleton.\n\n As shown in \\fullref{ex:preorder_nonuniqueness} and discussed in \\fullref{thm:order_category_isomorphism}, a thin category may not be \\hyperref[def:skeletal_category]{skeletal}.\n\n Thin categories are often conflated with preordered sets due to \\fullref{thm:order_category_isomorphism/preordered}.\n\\end{definition}\n\n\\begin{theorem}[Ordered sets as categories]\\label{thm:order_category_isomorphism}\n Regarding ordered sets, we have the following \\hyperref[rem:category_similarity/isomorphism]{category isomorphisms}:\n \\begin{thmenum}\n \\thmitem{thm:order_category_isomorphism/preordered} The categories \\hyperref[def:preordered_set/category]{\\( \\cat{PreOrd} \\)} and \\( \\cat{Thin} \\), where\n \\begin{itemize}\n \\item \\( \\cat{PreOrd} \\) is the category of small \\hyperref[def:preordered_set]{preordered sets} and (nonstrict) \\hyperref[def:preordered_set/homomorphism]{monotone maps}.\n\n \\item \\( \\cat{Thin} \\) is the subcategory of \\hyperref[def:category_of_small_categories]{\\( \\cat{Cat} \\)} \\hyperref[def:subcategory]{induced} by \\hyperref[def:thin_category]{thin categories}, i.e. the category of small thin categories.\n \\end{itemize}\n\n \\thmitem{thm:order_category_isomorphism/partially_ordered} The categories \\hyperref[def:partially_ordered_set/category]{\\( \\cat{Pos} \\)} and \\( \\cat{SkelThin} \\), where\n \\begin{itemize}\n \\item \\( \\cat{Pos} \\) is the subcategory of \\( \\cat{PreOrd} \\) induced by (nonstrict) \\hyperref[def:partially_ordered_set]{partially ordered sets}.\n\n \\item \\( \\cat{SkelThin} \\) is the subcategory of \\( \\cat{Thin} \\) induced by \\hyperref[def:skeletal_category]{skeletal categories}, i.e. the category of small thin skeletal categories.\n \\end{itemize}\n\n \\thmitem{thm:order_category_isomorphism/totally_ordered} The categories \\hyperref[def:totally_ordered_set]{\\( \\cat{Tos} \\)} and \\( \\cat{ConnSkelThin} \\), where\n \\begin{itemize}\n \\item \\( \\cat{Tos} \\) is the subcategory of \\( \\cat{Pos} \\) induced by (nonstrict) \\hyperref[def:totally_ordered_set]{totally ordered sets}.\n\n \\item \\( \\cat{ConnSkelThin} \\) is the subcategory of \\( \\cat{SkelThin} \\) induced by \\hyperref[def:connected_category]{connected categories}, i.e. the category of small thin skeletal connected categories.\n \\end{itemize}\n \\end{thmenum}\n\\end{theorem}\n\\begin{proof}\n \\SubProofOf{thm:order_category_isomorphism/preordered}\n \\SubProof*{Proof that preorders induce thin categories} Let \\( (P, \\leq) \\) be a small preordered set. Let \\( \\cat{P} \\) be the \\hyperref[def:quiver_free_category]{free category} obtained by regarding \\( (P, \\leq) \\) as a \\hyperref[def:quiver]{quiver}. Explicitly, the category \\( \\cat{P} \\) is built as follows:\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(\\cat{P}) \\) is simply \\( P \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{P}(x, y) \\) consists of the tuple \\( (x, y) \\) if \\( x \\leq y \\) and is empty otherwise.\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms} \\( (x, y) \\) and \\( (y, z) \\) is simply \\( (x, y) \\). This is well-defined because of the \\hyperref[def:binary_relation/transitive]{transitivity} of \\( \\leq \\).\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the object \\( x \\in \\cat{C} \\) is \\( (x, x) \\). This is well-defined because of the \\hyperref[def:binary_relation/reflexive]{reflexivity} of \\( \\leq \\).\n \\end{itemize}\n\n The category is clearly thin because \\( \\leq \\) is a binary relation and ordered tuples with the same elements are equal.\n\n Now let \\( f: P \\to Q \\) be a nonstrict monotone map. It induces the functor\n \\begin{equation*}\n \\begin{aligned}\n &F: \\cat{P} \\to \\cat{Q} \\\\\n &F(x) \\coloneqq f(x) \\\\\n &F(x, y) \\coloneqq (f(x), f(y)).\n \\end{aligned}\n \\end{equation*}\n\n \\ref{def:functor/CF1} is immediate and \\ref{def:functor/CF2} follows from \\eqref{eq:def:preordered_set/homomorphism}, hence \\( F \\) is indeed a functor.\n\n \\SubProof*{Proof that thin categories induce preorders} Let \\( \\cat{P} \\) be a small category. Define the binary relation\n \\begin{equation*}\n X \\leq Y \\T{if and only if} \\cat{P}(X, Y) \\neq \\varnothing.\n \\end{equation*}\n\n This is a binary relation over the set \\( P \\coloneqq \\obj(\\cat{P}) \\). It is reflexive because of the existence of identity morphisms in \\( \\cat{P} \\) and transitive because of the requirement that the composition of compatible morphisms exists.\n\n Therefore, \\( (P, \\leq) \\) is a preordered set.\n\n Given a functor \\( F: \\cat{P} \\to \\cat{Q} \\), the restriction \\( F\\restr_{\\obj(\\cat{P})} \\) is a monotone map from \\( (P, \\leq_P) \\) to \\( (Q, \\leq_Q) \\).\n\n Indeed, if \\( X \\leq Y \\) for \\( X, Y \\in P \\), then \\( \\cat{P}(X, Y) \\neq \\varnothing \\). Hence, \\( \\cat{P}(F(X), F(Y)) \\neq \\varnothing \\) and \\( F(X) \\leq F(Y) \\).\n\n \\SubProof*{Proof of isomorphism} We have implicitly defined a functor from \\( \\cat{PreOrd} \\) to \\( \\cat{Thin} \\) and vice versa. Isomorphism of \\( \\cat{PreOrd} \\) and \\( \\cat{Thin} \\) requires that these functors are mutually inverse.\n\n We need to prove that the induced preordered set for the induced thin category of a preordered set is the same as the original. In the other direction, we need to prove that the induced thin category for the induced preordered set of a thin category is the same as the original.\n\n Both of these proofs are trivial but nevertheless the fact that we need to perform such a check is important.\n\n \\SubProofOf{thm:order_category_isomorphism/partially_ordered} We have already shown in \\fullref{thm:order_category_isomorphism/preordered} the isomorphism between \\( \\cat{PreOrd} \\) and \\( \\cat{Thin} \\).\n\n For a thin category, there is an isomorphism between \\( x \\) and \\( y \\) if and only if there is both a morphism from \\( x \\) to \\( y \\) and one from \\( y \\) to \\( x \\). This isomorphism may not be unique as a consequence of \\fullref{ex:preorder_nonuniqueness}. Uniqueness requires \\( x = y \\) to hold in this case, which is in turn equivalent to partial order \\hyperref[def:binary_relation/antisymmetric]{antisymmetry}.\n\n \\SubProofOf{thm:order_category_isomorphism/totally_ordered} We have already shown in \\fullref{thm:order_category_isomorphism/partially_ordered} the isomorphism between \\( \\cat{Pos} \\) and \\( \\cat{SkelThin} \\).\n\n \\hyperref[def:connected_category]{Connectedness} of a category \\( \\cat{P} \\) is then equivalent to \\hyperref[def:binary_relation/total]{totality} of \\( (P, \\leq) \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:order_category_isomorphism_properties}\n Let \\( (P, \\leq) \\) be a \\hyperref[def:partially_ordered_set]{partially ordered set} and let \\( \\cat{P} \\) be its corresponding category induced by \\fullref{thm:order_category_isomorphism}.\n\n \\begin{thmenum}\n \\thmitem{thm:order_category_isomorphism_properties/opposite} The \\hyperref[def:opposite_category]{opposite category} \\( \\cat{P}^{\\opcat} \\) corresponds to the \\hyperref[def:partially_ordered_set/opposite]{opposite partially ordered set} \\( (P, \\geq) \\).\n\n \\thmitem{thm:order_category_isomorphism_properties/universal} If \\( \\cat{P} \\) has an \\hyperref[def:universal_objects/initial]{initial object}, since it is a skeletal category, this initial object is unique.\n\n An object \\( I \\) is an initial object if and only if it is the \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{bottom} of \\( P \\).\n\n \\hyperref[thm:categorical_principle_of_duality]{Dually}, \\( T \\) is a \\hyperref[def:universal_objects/terminal]{terminal object} if and only if it is the \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{top} of \\( P \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n", "meta": {"hexsha": "6c575969472f68c637acd8fe54f7ec8080651695", "size": 38365, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/category_equivalences.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/category_equivalences.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/category_equivalences.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.7545454545, "max_line_length": 502, "alphanum_fraction": 0.7069985664, "num_tokens": 12066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.795658090372256, "lm_q2_score": 0.787931190663057, "lm_q1q2_score": 0.6269238265077058}} {"text": "\\section{Properties of determinants}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Use the determinant of a square matrix to decide whether the\n matrix is invertible.\n \\item From the determinants of two matrices, calculate the\n determinant of their product.\n \\item From the determinant of a matrix, calculate the determinant of\n its inverse.\n \\item From the determinant of a matrix, calculate the determinant of\n its transpose.\n \\item Calculate the determinant of $kA$, if the determinant of $A$\n is known.\n \\item Without calculation, find the determinant of a matrix\n containing a row or column of zeros, or a matrix containing a row\n (or column) that is a scalar multiple of another row (or column).\n \\item Use algebraic properties to reason about determinants.\n \\end{enumerate}\n\\end{outcome}\n\nOne reason that the determinant is such an important quantity is that\nit permits us to tell whether a square matrix is invertible.\n\n\\begin{theorem}{Determinants and invertible matrices}{determinant-invertible}\n Let $A$ be an $n\\times n$-matrix. Then $A$ is invertible%\n \\index{determinant!and invertibility}%\n \\index{matrix!determinant!and invertibility}%\n \\index{invertible matrix!and determinant}%\n \\index{matrix!invertible!and determinant} if and\n only if $\\det(A) \\neq 0$.\n\\end{theorem}\n\n\\begin{proof}\n We know that every matrix $A$ can be converted to {\\ef} by\n elementary row operations. We also know from\n Theorem~\\ref{thm:determinant-row-operations} that no elementary row\n operation changes whether the determinant is zero or not. Let $R$\n be an {\\ef} of $A$. Because $R$ is an {\\ef}, it is also an upper\n triangular matrix. Case 1: $A$ is invertible. In that case, the rank\n of $R$ is $n$, and every diagonal entry of $R$ is a pivot entry\n (therefore non-zero). It follows that $\\det(R)\\neq 0$, which implies\n $\\det(A)\\neq 0$. Case 2: $A$ is not invertible. In that case, the\n triangular matrix $R$ contains a row of zeros. It follows that\n $\\det(R)=0$, and therefore $\\det(A)=0$.\n\\end{proof}\n\n\\begin{example}{Determinants and invertible matrices}{determinant-invertible}\n Determine which of the following matrices are invertible by\n computing their determinants.\n \\begin{equation*}\n A = \\begin{mymatrix}{rr}\n 3 & 6 \\\\\n 2 & 4 \\\\\n \\end{mymatrix},\\quad\n B = \\begin{mymatrix}{rr}\n 2 & 3 \\\\\n 5 & 1 \\\\\n \\end{mymatrix},\\quad\n C = \\begin{mymatrix}{rrr}\n 1 & 2 & -5 \\\\\n 2 & 0 & 2 \\\\\n 3 & 1 & 0 \\\\\n \\end{mymatrix}.\n \\end{equation*}\n\\end{example}\n\n\\begin{solution}\n We have $\\det(A) = 3\\cdot 4 - 2\\cdot 6 = 0$ and $\\det(B) = 2\\cdot\n 1-5\\cdot 3 = -13$. Therefore, $B$ is invertible and $A$ is not\n invertible. A quick way to compute the determinant of $C$ is to\n expand it along the third row. We have\n \\begin{equation*}\n \\det(C)\n = 3\\begin{absmatrix}{rr}\n 2 & -5 \\\\\n 0 & 2 \\\\\n \\end{absmatrix}\n - 1\\begin{absmatrix}{rr}\n 1 & -5 \\\\\n 2 & 2 \\\\\n \\end{absmatrix}\n = 3\\cdot 4 - 1\\cdot 12 = 0.\n \\end{equation*}\n Therefore, $C$ is not invertible.\n\\end{solution}\n\nAs an application of Theorem~\\ref{thm:determinant-invertible}, we note\nthat the determinant of an $n\\times n$-matrix can be used to predict\nwhether a homogeneous system of equations has non-trivial solutions.\n\n\\begin{corollary}{Determinants and homogeneous systems}{determinant-homogeneous}\n Let $A$ be an $n\\times n$-matrix. Then the homogeneous system\n $A\\vect{v}=\\vect{0}$ has non-trivial solutions if and only if $\\det(A)=0$.\n\\end{corollary}\n\n\\begin{proof}\n We know from Theorem~\\ref{thm:rank-homogeneous-solutions} that the\n homogeneous system has a non-trivial solution if and only if\n $\\rank(A)$, $<$ and $=$}\n\n\\starttabulate[|l|l|l|l|l|l|]\n\\NC $\\leq$ \\NC \\type{\\leq}\n\\NC $\\geq$ \\NC \\type{\\geq}\n\\NC $\\equiv$ \\NC \\type{\\equiv}\n\\NC\\NR\n\\NC $\\prec$ \\NC \\type{\\prec}\n\\NC $\\succ$ \\NC \\type{\\succ}\n\\NC $\\sim$ \\NC \\type{\\sim}\n\\NC\\NR\n\\NC $\\preceq$ \\NC \\type{\\preceq}\n\\NC $\\succeq$ \\NC \\type{\\succeq}\n\\NC $\\simeq$ \\NC \\type{\\simeq}\n\\NC\\NR\n\\NC $\\ll$ \\NC \\type{\\ll}\n\\NC $\\gg$ \\NC \\type{\\gg}\n\\NC $\\asymp$ \\NC \\type{\\asymp}\n\\NC\\NR\n\\NC $\\subset$ \\NC \\type{\\subset}\n\\NC $\\supset$ \\NC \\type{\\supset}\n\\NC $\\approx$ \\NC \\type{\\approx}\n\\NC\\NR\n\\NC $\\subseteq$ \\NC \\type{\\subseteq}\n\\NC $\\supseteq$ \\NC \\type{\\supseteq}\n\\NC $\\cong$ \\NC \\type{\\cong}\n\\NC\\NR\n\\NC $\\sqsubseteq$ \\NC \\type{\\sqsubseteq}\n\\NC $\\sqsupseteq$ \\NC \\type{\\sqsupseteq}\n\\NC $\\bowtie$ \\NC \\type{\\bowtie}\n\\NC\\NR\n\\NC $\\in$ \\NC \\type{\\in}\n\\NC $\\ni$ \\NC \\type{\\ni}\n\\NC $\\propto$ \\NC \\type{\\propto}\n\\NC\\NR\n\\NC $\\vdash$ \\NC \\type{\\vdash}\n\\NC $\\dashv$ \\NC \\type{\\dashv}\n\\NC $\\models$ \\NC \\type{\\models}\n\\NC\\NR\n\\NC $\\smile$ \\NC \\type{\\smile}\n\\NC $\\mid$ \\NC \\type{\\mid}\n\\NC $\\doteq$ \\NC \\type{\\doteq}\n\\NC\\NR\n\\NC $\\frown$ \\NC \\type{\\frown}\n\\NC $\\parallel$ \\NC \\type{\\parallel}\n\\NC $\\perp$ \\NC \\type{\\perp}\n\\NC\\NR\n\\stoptabulate\n\n\\section{Negated relations}\n\n\\starttabulate[|l|l|l|l|l|l|]\n\\NC $\\not<$ \\NC \\type{\\not}\\type{<}\n\\NC $\\not>$ \\NC \\type{\\not}\\type{>}\n\\NC $\\not=$ \\NC \\type{\\not}\\type{=}\n\\NC\\NR\n\\NC $\\not\\leq$ \\NC \\type{\\not}\\type{\\leq}\n\\NC $\\not\\geq$ \\NC \\type{\\not}\\type{\\geq}\n\\NC $\\not\\equiv$ \\NC \\type{\\not}\\type{\\equiv}\n\\NC\\NR\n\\NC $\\not\\prec$ \\NC \\type{\\not}\\type{\\prec}\n\\NC $\\not\\succ$ \\NC \\type{\\not}\\type{\\succ}\n\\NC $\\not\\sim$ \\NC \\type{\\not}\\type{\\sim}\n\\NC\\NR\n\\NC $\\not\\preceq$ \\NC \\type{\\not}\\type{\\preceq}\n\\NC $\\not\\succeq$ \\NC \\type{\\not}\\type{\\succeq}\n\\NC $\\not\\simeq$ \\NC \\type{\\not}\\type{\\simeq}\n\\NC\\NR\n\\NC $\\not\\subset$ \\NC \\type{\\not}\\type{\\subset}\n\\NC $\\not\\supset$ \\NC \\type{\\not}\\type{\\supset}\n\\NC $\\not\\approx$ \\NC \\type{\\not}\\type{\\approx}\n\\NC\\NR\n\\NC $\\not\\subseteq$ \\NC \\type{\\not}\\type{\\subseteq}\n\\NC $\\not\\supseteq$ \\NC \\type{\\not}\\type{\\supseteq}\n\\NC $\\not\\cong$ \\NC \\type{\\not}\\type{\\cong}\n\\NC\\NR\n\\NC $\\not\\sqsubseteq$ \\NC \\type{\\not}\\type{\\sqsubseteq}\n\\NC $\\not\\sqsupseteq$ \\NC \\type{\\not}\\type{\\sqsupseteq}\n\\NC $\\not\\asymp$ \\NC \\type{\\not}\\type{\\asymp}\n\\NC\\NR\n\\stoptabulate\n\n\\section{Some arrows}\n\n\\starttabulate[|l|l|l|l|l|l|]\n\\NC $\\leftarrow$ \\NC \\type{\\leftarrow}\n\\NC $\\longleftarrow$ \\NC \\type{\\longleftarrow}\n\\NC $\\uparrow$ \\NC \\type{\\uparrow}\n\\NC\\NR\n\\NC $\\Leftarrow$ \\NC \\type{\\Leftarrow}\n\\NC $\\Longleftarrow$ \\NC \\type{\\Longleftarrow}\n\\NC $\\Uparrow$ \\NC \\type{\\Uparrow}\n\\NC\\NR\n\\NC $\\rightarrow$ \\NC \\type{\\rightarrow}\n\\NC $\\longrightarrow$ \\NC \\type{\\Rightarrow}\n\\NC $\\downarrow$ \\NC \\type{\\downarrow}\n\\NC\\NR\n\\NC $\\Rightarrow$ \\NC \\type{\\Rightarrow}\n\\NC $\\Longrightarrow$ \\NC \\type{\\Longrightarrow}\n\\NC $\\Downarrow$ \\NC \\type{\\Downarrow}\n\\NC\\NR\n\\NC $\\leftrightarrow$ \\NC \\type{\\leftrightarrow}\n\\NC $\\longleftrightarrow$ \\NC \\type{\\longleftrightarrow}\n\\NC $\\updownarrow$ \\NC \\type{\\updownarrow}\n\\NC\\NR\n\\NC $\\Leftrightarrow$ \\NC \\type{\\Leftrightarrow}\n\\NC $\\Longleftrightarrow$ \\NC \\type{\\Longleftrightarrow}\n\\NC $\\Updownarrow$ \\NC \\type{\\Updownarrow}\n\\NC\\NR\n\\NC $\\mapsto$ \\NC \\type{\\mapsto}\n\\NC $\\longmapsto$ \\NC \\type{\\longmapsto}\n\\NC $\\nearrow$ \\NC \\type{\\nearrow}\n\\NC\\NR\n\\NC $\\searrow$ \\NC \\type{\\searrow}\n\\NC $\\swarrow$ \\NC \\type{\\swarrow}\n\\NC $\\nwarrow$ \\NC \\type{\\nwarrow}\n\\NC\\NR\n\\NC $\\hookleftarrow$ \\NC \\type{\\hookleftarrow}\n\\NC $\\hookrightarrow$ \\NC \\type{\\hookrightarrow}\n\\NC ~ \\NC ~\n\\NC\\NR\n\\stoptabulate\n\n\\section{Alternative commands}\n\n\\starttabulate[|l|l|l|l|l|l|l|l|l|l|]\n\\NC $\\ne$ \\NC \\type{\\ne}\n\\NC $\\{$ \\NC {\\tttf\\leftargument}\n\\NC $\\land$ \\NC \\type{\\land}\n\\NC $\\to$ \\NC \\type{\\to}\n\\NC $\\vert$ \\NC \\type{\\vert}\n\\NC\\NR\n\\NC $\\le$ \\NC \\type{\\le}\n\\NC $\\}$ \\NC {\\tttf\\rightargument}\n\\NC $\\lor$ \\NC \\type{\\lor}\n\\NC $\\gets$ \\NC \\type{\\gets}\n\\NC $\\Vert$ \\NC \\type{\\Vert}\n\\NC\\NR\n\\NC $\\ge$ \\NC \\type{\\ge}\n\\NC $\\owns$ \\NC \\type{\\owns}\n\\NC $\\lnot$ \\NC \\type{\\lnot}\n\\NC \\NC\n\\NC \\NC\n\\NC\\NR\n\\stoptabulate\n\n\\stopcomponent\n", "meta": {"hexsha": "b1477a5407e8ce2dc640bbda6e4958d8c5e8d763", "size": 9769, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "contextman/context-beginners/en/ma-cb-en-mathcharacters.tex", "max_stars_repo_name": "marcpaterno/texmf", "max_stars_repo_head_hexsha": "aa7ad70e0102492ff89b7967b16b499cbd6c7f19", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "contextman/context-beginners/en/ma-cb-en-mathcharacters.tex", "max_issues_repo_name": "marcpaterno/texmf", "max_issues_repo_head_hexsha": "aa7ad70e0102492ff89b7967b16b499cbd6c7f19", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "contextman/context-beginners/en/ma-cb-en-mathcharacters.tex", "max_forks_repo_name": "marcpaterno/texmf", "max_forks_repo_head_hexsha": "aa7ad70e0102492ff89b7967b16b499cbd6c7f19", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.5129032258, "max_line_length": 60, "alphanum_fraction": 0.5195004606, "num_tokens": 3790, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.8006919949619793, "lm_q1q2_score": 0.6266715816547157}} {"text": "Common Lagrangian in introductory courses of classical mechanics depends of $q$\nand $\\dot{q}$. We refer to them as \\emph{first-order Lagrangian} because the\nhighest derivative term is $\\dot{q}$, i.e.\\ the first derivative of $q(t)$ with\nrespect to time. If in the Lagrangian appears also $\\ddot{q}$ it will be a\n\\emph{second-order Lagrangian} and so on. \\\\\n\nIn the following, the mathematical machinery develop in Section~\\ref{section:\nlagrangin and hamiltonian formalism}, will be used to study two apparently\nsimilar Lagrangian and introduce the core problem of HD systems.\n\n\\subsection{First-order vs Second-order Lagrangian}\nThe simplest non-trivial Lagrangians are those in polynomial form like\n\\begin{equation*}\n L_1(q, \\dot{q}) = \\frac{\\dot{q}^2}{2} - \\alpha(q) \\qquad\n L_2(q, \\ddot{q}) = \\frac{\\ddot{q}^2}{2} - \\alpha(q)\n\\end{equation*}\nwhere $\\alpha$ is a smooth function of $q$ only~\\cite{Chen13}. No Greek indices\nappear on $q^{(i)}$ because we limit ourself to one-dimensional systems. $L_1$\nis a first-order Lagrangian while $L_2$ it's a second-order one.\n\n\\paragraph{First-order Lagrangian} In order to get the equation of motion we\nused the Euler-Lagrange equation~\\eqref{eq:euler-lagrange}\n\n\\begin{equation} \\label{eq: first-order_motion_eq_lagr}\n \\frac{\\partial L_1}{\\partial q} -\n \\frac{d}{dt}\\frac{\\partial L_1}{\\partial \\dot{q}} +\n \\underbrace{\n \\frac{d^2}{dt^2}\\frac{\\partial L_1}{\\partial \\ddot{q}} - \\ldots }_0 = 0\n \\qquad \\Rightarrow \\qquad\n \\ddot{q} = - \\frac{d\\alpha(q)}{dq}\n\\end{equation}\n\nIf $q$ is the position coordinate of a unit mass point, the function $\\alpha(q)$\ncan be interpreted as the potential energy and~\\eqref{eq:\nfirst-order_motion_eq_lagr} is nothing less than Newton's second law. To get the\nHamiltonian description apply the recipe in Section~\\ref{subsection:\nhamiltonian_formalism}\n\n\\begin{equation*}\n \\begin{cases}\n Q_1 := q \\\\\n P_1 := \\frac{\\delta L_1}{\\delta \\dot{q}} = \\dot{q}\n \\end{cases}\n \\qquad\n \\tilde{H}_1(q, \\dot{q}) :=\n \\frac{\\delta L_1}{\\delta \\dot{q}} \\dot{q} - L_1(q, \\dot{q})\n\\end{equation*}\n\n\\begin{equation} \\label{eq: first-order_motion_eq_ham}\n H_1(Q_1, P_1) = \\frac{P_1^2}{2} + \\alpha(Q_1) \\qquad\n \\begin{cases}\n \\dot{Q_1} = \\frac{\\partial H_1}{\\partial P_1} = P_1 \\\\\n \\dot{P_1} = - \\frac{\\partial H_1}{\\partial Q_1} = - \\frac{d\\alpha(Q_1)}{dQ_1}\n \\end{cases}\n\\end{equation}\n\nSince $L_1$ does not explicitly depend on time (only through $q$ and $\\dot{q}$),\nthe Hamiltonian $H_1$ can be interpreted as the total energy of the system (sum\nof kinetic and potential energy). Combining the two equation in~\\eqref{eq:\nfirst-order_motion_eq_ham} we restore~\\eqref{eq: first-order_motion_eq_lagr}\nproving the equivalence between Lagrangian and Hamiltonian mechanics for simple\nsystems describe by $L_1$.\n\n\n\\paragraph{Second-order Lagrangian} Proceeding as before, equation of motion\nfollow from Euler-Lagrange equation~\\eqref{eq:euler-lagrange}\n\n\\begin{equation} \\label{eq: second-order_motion_eq_lagr}\n \\frac{\\partial L_2}{\\partial q} -\n \\frac{d}{dt}\\frac{\\partial L_2}{\\partial \\dot{q}} +\n \\frac{d^2}{dt^2}\\frac{\\partial L_2}{\\partial \\ddot{q}} -\n \\underbrace{\n \\frac{d^3}{dt^3}\\frac{\\partial L_2}{\\partial q^{(3)}} + \\ldots\n }_0 = 0\n \\quad \\Rightarrow \\quad\n q^{(4)} = - \\frac{d\\alpha(q)}{dq}\n\\end{equation}\n\nEquation~\\eqref{eq: second-order_motion_eq_lagr} never appear in classical\nmechanics, but can it describe some kind of system? The study of the Hamiltonian\ncan give us some clues about the total energy of the system.\n\n\\begin{equation*}\n \\begin{cases}\n Q_1 := q \\\\\n P_1 := \\frac{\\delta L_2}{\\delta \\dot{q}}\n = \\frac{\\partial L_2}{\\partial \\dot{q}} -\n \\frac{d}{dt} \\left( \\frac{\\partial L_2}{\\partial \\ddot{q}} \\right)\n = 0 - \\frac{d}{dt} \\left( \\ddot{q} \\right) = -q^{(3)}\n \\end{cases}\n \\begin{cases}\n Q_2 := \\dot{q} \\\\\n P_2 := \\frac{\\delta L_2}{\\delta \\ddot{q}}\n = \\frac{\\partial L_2}{\\partial \\ddot{q}} = \\ddot{q}\n \\end{cases}\n\\end{equation*}\n\nTherefore\n\\begin{equation}\\label{eq:canonical_coordinates_second_order_lagrangian}\n \\begin{cases}\n q = Q_1 \\\\\n \\dot{q} = Q_2 \\\\\n \\ddot{q} = P_2\n \\end{cases}\n \\qquad\n \\begin{cases}\n \\frac{\\delta L_2}{\\delta \\dot{q}} = P_1 \\\\\n \\frac{\\delta L_2}{\\delta \\ddot{q}} = P_2\n \\end{cases}\n\\end{equation}\n\n\\begin{equation*}\n \\tilde{H}_2(q, \\dot{q}) :=\n \\frac{\\delta L_2}{\\delta \\dot{q}} \\dot{q} +\n \\frac{\\delta L_2}{\\delta \\ddot{q}} \\ddot{q} -\n L_2(q, \\ddot{q})\n = \\frac{\\delta L_2}{\\delta \\dot{q}} \\dot{q} +\n \\frac{\\delta L_2}{\\delta \\ddot{q}} \\ddot{q} -\n \\frac{\\ddot{q}^2}{2} + \\alpha(q)\n\\end{equation*}\n\nSubstituting relations~\\eqref{eq:canonical_coordinates_second_order_lagrangian}\nin the expression for $\\tilde{H}_2$ we get the Hamiltonian\n\n\\begin{align} \\label{eq: second-order_motion_eq_ham}\n H_2(Q_1, Q_2, P_1, P_2)\n &= P_1 Q_2 + P_2 P_2 - \\frac{P_2^2}{2} + \\alpha(Q_1) \\notag \\\\\n &= P_1 Q_2 + \\frac{P_2^2}{2} + \\alpha(Q_1)\n\\end{align}\n\nThere is a significant difference between the two spectra of $H_1$ and $H_2$:\nthe first is bounded from below while the latter is not (and both are not\nbounded from above).\n\n\n\\subsection{Linear Ostrogradskian instability}~\\label{subsection:\nlinear_ostrogradskian_instability}\nConsider an \\emph{isolated} system describe by the Hamiltonian $H_2$; if the\nenergy is conserve even though the spectra is not bounded the energy stay\nconstant. Things start to going wrong when ones consider \\emph{interacting}\nsystem (e.g. $H_1$-system interacting with $H_2$-system). The phase space of\n$H_2$-system infinitely extend where the Hamiltonian is negative. In the example\nthis is due to the \\emph{linear} term $P_1$. When two systems interact the\n$H_1$-system tends to occupy higher and higher energies states while the\n$H_2$-system fall lower and lower in negative energies; this behavior entails\nthe conservation of energy~\\cite{Kallosh08, Eliezer89}.\nThis is the so called \\emph{linear Ostrogradskian instability}. \\\\\n\nOne can encounter the same kind of instability in quantum mechanics when try to\ncanonically quantized higher order Hamiltonian (e.g. $H_2$). This process lead\nto \\emph{negative norm states} (or \\emph{negative energy states}) which are\noften called \\emph{``ghosts''}. Systems (even classical ones) involving this\ntype of ghosts are called \\emph{ghost-like} or \\emph{ghost-ridden} systems. \\\\\n\nOne may wonder when these ghosts are summoned; the answer is contains in the\nOstrogradsky theorem\n\n\\begin{theorem}[Ostrogradsky]\\label{th:ostrogradsky_classical}\n If in the Lagrangian~\\eqref{eq:general_lagrangian} $n \\geq 2$ and the\n canonical momentum $\\bm{P}_n$ does not vanish, the\n Hamiltonian~\\eqref{eq:general_hamiltonian} may acquire an arbitrary real\n value.\n\\end{theorem}\n\\begin{proof}\n The proof is given for a one-dimensional system (i.e. $\\mu=1$) but it is\n generalize straight to higher dimensions. Consider the\n Hamiltonian~\\eqref{eq:general_hamiltonian} in terms of conjugate coordinates\n \\begin{equation} \\label{eq:general_hamiltonian_1dim}\n H = P_n h + P_{n-1} Q_n + \\cdots + P_1 Q_2 - L(Q_1, Q_2, \\ldots, h)\n \\end{equation}\n The function $h$ express $q^{(n)}$ in terms of conjugate coordinates under\n the assumption of regular Lagrangian, i.e.\n \\begin{equation*}\n q^{(n)} = h(Q_1, \\ldots, Q_n, P_n)\n \\end{equation*}\n Observe that $h$ does not depend on $P_1, \\ldots, P_{n-1}$ so the only\n contribution of these momenta to the Hamiltonian is in the linear factor\n \\begin{equation*}\n P_{n-1} Q_n + \\cdots + P_1 Q_2\n \\end{equation*}\n which is linear in momenta and hence can acquire any real value and so does\n the Hamiltonian~\\eqref{eq:general_hamiltonian_1dim}.\n\\end{proof}\n\nFor many years, due to consequences of Theorem~\\ref{th:ostrogradsky_classical},\nHD theories are consider intrinsically sick and not worthy of further study.\nHowever a deeper investigation reveals that Ostrogradsky instability can be cure\nand HD theories should not be discard in the first instance. One way to\nexorcising Ostrogradsky's ghost is through imposition of\nconstraints~\\cite{Chen13}.\n\n\n\\subsection{Removing Ostrogradskian instability with constraints}\n\nIn Hamiltonian mechanics, \\emph{constraints} are relation between coordinates\nand conjugate momenta. Constraints are classify in two group: \\emph{first class\nconstraints} and \\emph{second class constraints}. Second class constraints can\nbe thought as ``physical'' (e.g.\\ train on a railway, particle on a plane) so\nthat the solutions of equations of motion are different with or without them.\nFirst-class are instead related to a gauge freedom: the solutions of equations\nof motion differ by a function of time and hence represent the same system.\nMoreover the Poisson Bracket of a first class constraint with all the other\nconstraints vanishes on the constraints surface in the phase space (i.e.\n$\\{\\phi, \\tilde{\\phi}\\} \\approx 0$, see later for the notation used).\n\n\nAnother independent classification can be made for constraints (and not confused\nwith the previous one): \\emph{primary constraints} and \\emph{secondary\nconstrains}. Primary constraints are relation between $Q$s and $P$s which are\ngiven at the beginning.\n\n\\begin{equation} \\label{eq:constraint}\n \\phi_1(Q, P) = 0\n\\end{equation}\n\nAs the name suggests, secondary constraints are derived from the primary ones\nimposing the conservations of the constraints during the evolution of the\nsystem. This generates a series of constraints where the constraints relations\nare called \\emph{consistency relations}.\n\n\\begin{equation} \\label{eq:consistency_relations}\n \\phi_1 \\approx 0 \\quad \\Rightarrow \\quad\n \\left\\{ \\phi_1 , H \\right\\} =: \\phi_2 \\approx 0 \\quad \\Rightarrow \\quad\n \\left\\{ \\phi_2 , H \\right\\} =: \\phi_3 \\approx 0 \\quad \\Rightarrow \\quad\n \\ldots\n\\end{equation}\n\nThe weak equality symbol ``$\\approx$'' has been used to highlight the fact that\nthese relations vanish only on the hypersurface where all constraints are\nsatisfied. When we will be said that a constrain $\\phi_i$ \\emph{vanishes} it is\nto be intended in this weak sense.\n\n\\subsubsection{Lagrangian and Hamiltonian with $m$ auxiliary variables}\nConstraints in Hamiltonian mechanics can be imposed with the help of auxiliary\nvariables $\\lambda_i$ in the Lagrangian:\n\n\\begin{equation*}\n L = L(\n q, \\dot{q}, \\ddot{q}, \\ldots, q^{(n)},\n \\lambda_1, \\lambda_2, \\ldots, \\lambda_m\n )\n\\end{equation*}\n\nSo the Lagrangian, besides the equations of motion~\\eqref{eq:euler-lagrange},\nneed to satisfy the constraints equation\n\\begin{equation}\n \\frac{\\partial L}{\\partial \\lambda_i} = 0 \\qquad i=1, 2, \\ldots, m\n\\end{equation}\n\nCanonical coordinates have to be given also for $\\lambda_i$ and the following\nchoice can be done\n\\begin{equation} \\label{eq:def_canonical_coordinates_lambda}\n \\Lambda_{i}:= \\lambda_{i}\n \\quad \\leftrightarrow \\quad\n \\Pi_{i} := \\frac{\\delta L}{\\delta \\dot{\\lambda_{i}}} = 0\n \\qquad i = 1, 2, \\ldots, m\n\\end{equation}\nso the primary constraints in~\\eqref{eq:constraint} simply become\n$\\phi_{1, i}: \\Pi_i = 0$. To the Hamiltonian~\\eqref{eq:Ham_in_q} are now added\nthe auxiliary variables terms and it becomes\n\\begin{equation} \\label{eq:Ham_constraints_in_q}\n \\tilde{H} :=\n \\sum_{j=1}^{n} \\frac{\\delta L}{\\delta q^{(j)}} q^{(j)} +\n \\sum_{i=1}^{m} \\frac{\\delta L}{\\delta \\dot{\\lambda_i}} \\dot{\\lambda_i} -\n L(q, \\ldots, q^{(n)}, \\lambda_1, \\ldots, \\lambda_m)\n\\end{equation}\n\nThe $\\dot{\\lambda_i}$ can be obtain by the consistency\nrelations~\\eqref{eq:consistency_relations} but there is no need to explicitly\ncalculate them. $\\dot{\\lambda_i}$ can be written as functions of canonical\ncoordinates: $\\dot{\\lambda_i} = u_i(Q_1, \\ldots, Q_n, P_1, \\ldots, P_n)$.\nExpressing~\\eqref{eq:Ham_constraints_in_q} using canonical\ncoordinates we obtain the Hamiltonian\n\\begin{align} \\label{eq:Ham_constraints}\n H =\\ & P_n h + P_{n-1} Q_n + \\cdots + P_1 Q_2 \\nonumber \\\\\n & + \\phi_m u_m + \\phi_{m-1} u_{m-1} + \\cdots + \\phi_1 u_1 \\nonumber \\\\\n & - L ( Q_1, Q_2, \\ldots, h, \\Lambda_1, \\ldots, \\Lambda_m)\n\\end{align}\n\n\nAfter calculating the secondary constraints\nwith~\\eqref{eq:consistency_relations}\n\n\\begin{equation*}\n \\phi_{2, i} := \\left\\{ \\phi_{1,i}, H \\right\\} =\n \\frac{\\partial \\phi_{1,i}}{\\partial \\Pi_i}\n \\frac{\\partial H}{\\partial \\Lambda_i} =\n \\frac{\\partial L}{\\partial \\lambda_i} \\Big|_{\\lambda_i = \\Lambda_i}\n \\approx 0\n\\end{equation*}\ntwo cases are possible:\n\n\\begin{enumerate}\n \\item $\\{\\phi_{1,i}, \\phi_{2,j}\\} \\not\\approx 0$: $\\phi_{1,i}$, $\\phi_{2,j}$ are\n both second-class constraints thus there are no further constraints\n derivable from consistency relations.\n \\item $\\{\\phi_{1,i}, \\phi_{2,j}\\} \\approx 0$: $\\phi_{1,i}$, $\\phi_{2,j}$ are\n not second-class constraints then further constraints can be found employing\n the chain~\\eqref{eq:consistency_relations} until either \\emph{case 1} is\n reached or the new constraint can be expressed using those found previously.\n\\end{enumerate}\n\nAfter all constraints have been found and the gauge symmetries are fixed, one\ncan use these equations to write the auxiliary canonical variables in terms of\nthe ``usual'' canonical variables\n\\begin{equation} \\label{eq:constraints_in_terms_of_canonical_vars}\n \\begin{cases}\n \\Lambda_i =\\ f_i(Q_1, \\ldots, Q_n, P_n) \\\\\n \\Pi_i =\\ 0\n \\end{cases}\n \\qquad i = 1, 2, \\ldots, m \\\\\n\\end{equation}\nand then substitute in the Hamiltonian~\\eqref{eq:Ham_constraints}\n\\begin{equation*}\n H =\\ P_n h + P_{n-1} Q_n + \\cdots + P_1 Q_2\n - L ( Q_1, Q_2, \\ldots, h, f_1, \\ldots, f_m)\n\\end{equation*}\n\nTo recap, we started form a phase space of dimension $2n$ (n$Q$ + n$P$), then we\nenlarged it by adding $2m$ auxiliary variables (m$\\Lambda$ + m$\\Pi$) so the\ntotal phase space dimension is $2(n+m)$. Then we associated with every pair of\ncanonical auxiliary variables two constraints (\\emph{case 1}) or more\n(\\emph{case 2}) have been found. During the last substitution the dimension of\nphase space is bring down to it's original dimensions (\\emph{case 1}: $2(n+m) -\n2m = 2n$) or further reduced (\\emph{case 2}: e.g. $2(n+m) - 3m < 2n$).\n\n\\subsubsection{Example of non-stable system with constraints}\nWe introduce here a widely studied HD Lagrangian: the\n\\emph{Pais-Uhlenbeck oscillator} (PU).\n\\begin{equation} \\label{eq:lagrangian_PU}\n L_{PU} = \\frac{1}{2} \\left[\n \\ddot{q}^2 - (\\omega_1^2 + \\omega_2^2) \\dot{q}^2 + \\omega_1^2 \\omega_2^2 q^2\n \\right]\n\\end{equation}\n\nAccording to Ostrogradsky Theorem~\\ref{th:ostrogradsky_classical} this system\nshows a problematic instability. Then we can try to cure that by introducing a\nconstraint such as $\\ddot{q}^2 - \\dot{q}^2 = 0$ and see if the Hamiltonian is\nstill two-side unbounded. Using the auxiliary variable $\\lambda$ one can write\nthe constraint version of~\\eqref{eq:lagrangian_PU}:\n\\begin{equation} \\label{eq:lagrangian_PUC}\n L_{PUC} = L_{PU} + \\frac{\\lambda}{2} \\left(\\ddot{q}^2 - \\dot{q}^2\\right)\n\\end{equation}\nand from it we derive the Hamiltonian following the given prescription.\n\\begin{align*}\n & \\begin{cases}\n Q_1 := q \\\\\n P_1 := \\frac{\\delta L_{PUC}}{\\delta \\dot{q}} =\n - (\\lambda + 1) q^{(3)} - \\dot{\\lambda}\\ddot{q}\n - (\\lambda + \\omega_1^2 + \\omega_2^2)\\dot{q}\n \\end{cases} \\\\\n & \\begin{cases}\n Q_2 := \\dot{q} \\\\\n P_2 := \\frac{\\delta L_{PUC}}{\\delta \\ddot{q}} =\n (\\lambda + 1) \\ddot{q}\n \\quad \\Rightarrow \\quad \\ddot{q} = \\frac{P_2}{(\\Lambda + 1)}\n \\end{cases} \\\\\n & \\begin{cases}\n \\Lambda := \\lambda \\\\\n \\Pi := \\frac{\\delta L_{PUC}}{\\delta \\lambda} = 0\n \\quad \\Rightarrow \\quad \\phi_1: \\Pi \\approx 0\n \\end{cases}\n\\end{align*}\n\\begin{equation} \\label{eq:ham_PUC_with_aux_vars}\n H_{PUC} =\\ P_1 Q_2 + \\frac{P_2^2}{2(\\Lambda + 1)} + \\phi_1 u_1\n + \\frac{Q_2^2}{2} \\left(\\Lambda + \\omega_1^2 + \\omega_2^2 \\right)\n - \\frac{Q_1^2}{2} \\omega_1^2 \\omega_2^2\n\\end{equation}\n\n\\begin{equation} \\label{eq:secondary_constraint_PU}\n \\phi_2 :=\n \\left\\{ \\phi_1, H_{PUC} \\right\\} =\n \\frac{1}{2} \\left[ \\frac{P_2}{\\Lambda + 1} - Q_2 \\right]\n \\left[ \\frac{P_2}{\\Lambda + 1} + Q_2 \\right]\n \\approx 0\n\\end{equation}\n\nThe secondary constraint equation~\\eqref{eq:secondary_constraint_PU}\nvanish if one of the factors in square brackets vanish. The two square bracket\nrepresent two hypersurfaces and one solution is chosen instead of the other\nbased on initial conditions. In the following, we choose the bracket with the\nminus sign. Moreover note that $\\phi_1$ and $\\phi_2$ are already second-class\nconstraints, indeed\n\\begin{equation*}\n \\begin{cases}\n \\phi_1: \\Pi = 0 \\\\\n \\phi_2: \\frac{P_2}{\\Lambda + 1} - Q_2 \\approx 0\n \\end{cases}\n \\quad \\Rightarrow \\quad\n \\left\\{ \\phi_1, \\phi_2 \\right\\} = -\n \\frac{\\partial\\phi_1}{\\partial\\Pi} \\frac{\\partial\\phi_2}{\\partial\\Lambda} =\n \\frac{P_2}{{(\\Lambda + 1)}^2} \\not\\approx 0\n\\end{equation*}\nso no further constraints are needed (\\emph{case 1}). The equation for\n$\\phi_1$ (trivial) and for $\\phi_2$ can be locally\ninverted~\\eqref{eq:constraints_in_terms_of_canonical_vars} obtaining that\n$\\Pi = 0$ and $\\Lambda = P_2/Q_2 - 1$. Substituting them\nin~\\eqref{eq:ham_PUC_with_aux_vars} we get the final Hamiltonian of the system\nbut unfortunately the Ostrogradsky instability persists.\n\\begin{equation} \\label{eq:ham_PUC_instable}\n H_{PUC} =\\ P_1 Q_2 + P_2 Q_2\n + \\frac{Q_2^2}{2} \\left(\\omega_1^2 + \\omega_2^2 - 1 \\right)\n - \\frac{Q_1^2}{2} \\omega_1^2 \\omega_2^2\n\\end{equation}\n\n\n\\subsubsection{Example of stable system with constraints}\nNow consider the following constrained Pais-Uhlenbeck oscillator~\\cite{Chen13}\n\\begin{equation}\n L_{PUC} = L_{PU} +\n 4 \\omega_1^2\\omega_2^2 q^2 \\lambda (1+ \\lambda) +\n 2 \\sqrt{2} \\omega_1\\omega_2 \\lambda q \\ddot{q}\n\\end{equation}\n\nAnalogously to the previous example, canonical variable and hamiltonian\n$H_{PUC}$ can be derived in similar fashion\n\\begin{align*}\n & \\begin{cases}\n Q_1 := q \\\\\n P_1 := - q^{(3)} -\n \\left( 2\\sqrt{2} \\omega_1\\omega_2\\lambda + \\omega_1^2 + \\omega_1^2 \\right)\n \\dot{q} - 2\\sqrt{2}\\omega_1\\omega_2\\dot{\\lambda} q\n \\end{cases} \\\\\n & \\begin{cases}\n Q_2 := \\dot{q} \\\\\n P_2 := \\ddot{q} + 2\\sqrt{2} \\omega_1\\omega_2\\lambda q\n \\quad \\Rightarrow \\quad\n \\ddot{q} = P_2 - 2\\sqrt{2} \\omega_1\\omega_2\\Lambda Q_1\n \\end{cases} \\\\\n & \\begin{cases}\n \\Lambda := \\lambda \\\\\n \\Pi := 0\n \\quad \\Rightarrow \\quad \\phi_1: \\Pi \\approx 0\n \\end{cases}\n\\end{align*}\n\\begin{align} \\label{eq:ham_PUC_with_aux_vars_stable}\n H_{PUC} =\\ & P_1 Q_2 + \\frac{P_2^2}{2} -\n \\omega_1^2\\omega_2^2\\frac{Q_1^2}{2} +\n \\left( \\omega_1^2 + \\omega_2^2 \\right) \\frac{Q_2^2}{2} \\\\\n & - 4 \\omega_1^2\\omega_2^2\\Lambda Q_1^2 -\n 2\\sqrt{2}\\omega_1\\omega_2\\Lambda P_2 Q_1 + \\phi_1 u_1\n\\end{align}\n\nSecondary constraints relations carry useful information up to the forth order\n\\footnote{\n To obtain $\\phi_4$, information from $\\phi_1$ and $\\phi_2$ have been used in\n order to write $\\phi_4$ in terms of $\\Lambda$.\n}\n\\begin{align*}\n & \\phi_1 : \\Pi = 0 \\\\\n & \\phi_2 := \\{\\phi_1, H_{PUC}\\} =\n P_2 + \\sqrt{2} \\omega_1\\omega_2 Q_1 \\approx 0 \\\\\n & \\phi_3 := \\{\\phi_2, H_{PUC}\\} =\n P_1 + \\left(\\omega_1^2 + \\omega_2^2 - \\sqrt{2}\\omega_1\\omega_2 \\right) Q_2\n \\approx 0 \\\\\n & \\phi_4 := \\{\\phi_3, H_{PUC}\\} =\n \\omega_1 \\omega_2 (3+8\\Lambda) - \\sqrt{2}\n \\left( \\omega_1^2 + \\omega_2^2 \\right) \\left(1 + 2\\Lambda \\right) \\approx 0\n\\end{align*}\n\nThese consistency relations can be inverted so that $P_2$, $Q_2$ and $\\Lambda$\nare express in terms of $P_1$ and $Q_1$. Substituting them\ninto~\\eqref{eq:ham_PUC_with_aux_vars_stable} obtaining\n\n\\begin{equation} \\label{eq:ham_PUC_stable}\n H_{PUC} =\\ \\frac{\\omega_1^2\\omega_2^2}{2} Q_1^2 +\n \\frac{\\omega_1\\omega_2}{\\sqrt{2} {\\left(\\sqrt{2} \\omega_1\\omega_2 -\n \\omega_1^2 - \\omega_2^2 \\right)}^2} P_1^2\n\\end{equation}\n\nThe Hamiltonian~\\eqref{eq:ham_PUC_stable} is bounded from below (and not from\nabove, just like an harmonic oscillator) and hence this system does not suffer\nof Ostrogradsky instability unlike the previous example. \\\\\n\nThe main difference between the two is the role played by constraints. In the\nfirst example the dimensions of phase space of the unconstrained and constrained\nHamiltonian are both four (i.e.\\ four variable appear in the reduced\nHamiltonian~\\eqref{eq:ham_PUC_instable}: $Q_1$, $P_1$, $Q_2$ and $P_2$). In the\nsecond example the constraints reduce the dimensionality of the phase space from\nthe original four to two (indeed just $Q_1$ and $P_1$ appear\nin~\\eqref{eq:ham_PUC_stable}). As shown in~\\cite{Chen13} a reduction of the\nphase space, due to imposed constraints, can cure the typical Ostrogradsky\ninstability of HD theories.\n", "meta": {"hexsha": "cd2a47c0f9f032a7cb009aa6432efc4e9695a700", "size": 20600, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "document/classical-higher-derivative-systems.tex", "max_stars_repo_name": "S1M0N38/thesis", "max_stars_repo_head_hexsha": "2dee034051a4b5ee017dae80759d849e99edb4e5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "document/classical-higher-derivative-systems.tex", "max_issues_repo_name": "S1M0N38/thesis", "max_issues_repo_head_hexsha": "2dee034051a4b5ee017dae80759d849e99edb4e5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "document/classical-higher-derivative-systems.tex", "max_forks_repo_name": "S1M0N38/thesis", "max_forks_repo_head_hexsha": "2dee034051a4b5ee017dae80759d849e99edb4e5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.5619834711, "max_line_length": 82, "alphanum_fraction": 0.7022330097, "num_tokens": 7085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245994514084, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6266705721793333}} {"text": "\\gotosection{2}{3}\n\\subsection{Matrix inverses and elementary matrices}\n\n\\begin{exercise}{2}\n\\begin{enumerate}\n\\item $\\xmat{1/6&5/54\\\\-1/6&1/54}$\n\n\\item Inverse doesn't exist, as $[A|I]$ is reduced to $\\xmat{1&3&1&0\\\\0&0&-3&1}$.\n\n\\item $\\xmat{2&-1&-3\\\\-2&2&3\\\\1&-1&-1}$\n\n\\item Inverse doesn't exist, as this is not a square matrix.\n\n\\item $\\xmat{1/7 & -1/2 & 1/14 \\\\ 4/21 & 5/6 & -1/14 \\\\ -4/21 & 1/6 & 1/14}$\n\n\\item $\\xmat{0&1&-1 \\\\ 1&-2& 1&-1&1}$\n\n\\item $\\xmat{4 & -6 & 4 & 1 \\\\ -6 & 14 & -11 & 3 \\\\ 4 & -11 & 10 & -3 \\\\ -1 & 3 & -3 & 1}$\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{3}\n\\begin{enumerate}\n\\item Suppose an matrix $A \\in M_{m,n}(\\mathbb{R})$. If $A$ has an inverse matrix, then $T_A$ must be bijective, or for any $\\Vect{b} \\in \\mathbb{R}^n$, $A\\Vect{x}=\\Vect{b}$ has a unique solution. If $m < n$, then $\\widetilde{A}$ has at least one column without a pivotal 1, so there exist infinitely many solutions, namely $T_A$ is not injective. If $m > n$, there must exist a row-reduced vector $\\widetilde{\\mathbf{b}}$ that contains a pivotal 1---since row operations are invertible, we can always find a $\\Vect{b}$ for any given $\\widetilde{\\mathbf{b}}$---there exists no solution, namely $T_A$ is not surjective. Only when $m = n$ could $A$ be reduced to an identity matrix and therefore has a unique solution for every possible $\\widetilde{\\mathbf{b}}$. Hence only squre matrices have inverses. \\rQED\n\n\\item $A = \\xmat{1&0&0\\\\0&1&0}, B = \\xmat{1&0\\\\0&1\\\\0&0}$.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{5}\nLabel them as $A, B, C$. $A+C$ and $B+2C$ gives $4x+4z=2, 4x+3y=3$, therefore, $x = (1-2z)/2$. Substituting $x$ with $z$ gives $y = (1+4z)/3$. Substituting $x$ and $y$ with $z$ into $A$ gives $z = 1/8$, then $x = 3/8$, $y = 1/2$.\n\n\\begin{enumerate}\n\\item Echelon form of this linear system is\n$\\xmat{1 & 0 & 0 & 3/8\\\\\n 0 & 1 & 0 & 1/2\\\\\n 0 & 0 & 1 & 1/8}$. Therefore, $x = 3/8$, $y = 1/2$, $z=1/8$.\n\n\\item $A^{-1} = \\xmat{3/16 & 1/4 & -1/16\\\\\n -1/4 & 0 & 3/4 \\\\\n 1/16 & -1/4 & 5/16}$. $A^{-1}\\xmat{1\\\\1\\\\1} = \\xmat{3/8\\\\1/2\\\\1/8}$.\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{8}\n\\begin{enumerate}\n\\item \\begin{enumerate}\n \\item Multiply the second row by 3.\n \\item Exchange the second and the third row.\n \\item Add a multiple 2 of the first row to the third row.\n \\end{enumerate}\n\n\\item \\begin{enumerate}\n \\item $\\xmat{1&0&-1\\\\6&3&3\\\\0&1&2}$.\n \\item $\\xmat{1&0&-1\\\\0&1&2\\\\2&1&1}$.\n \\item $\\xmat{1&0&-1\\\\2&1&1\\\\2&1&0}$.\n \\end{enumerate}\n\n\\item \\begin{enumerate}\n \\item Multiply the second column by 3.\n \\item Exchange the second and the third column.\n \\item Add a multiple 2 of the first column to the third column.\n \\end{enumerate}\n\n\\item \\begin{enumerate}\n \\item $\\xmat{1&0&-1\\\\2&3&1\\\\0&3&2}$.\n \\item $\\xmat{1&-1&0\\\\2&1&1\\\\0&2&1}$.\n \\item $\\xmat{1&0&1\\\\2&1&5\\\\0&1&2}$.\n \\end{enumerate}\n\\end{enumerate}\n\\end{exercise}\n\n\\begin{exercise}{11}\nColumn operations applied to a matrix $A$ can be seen as the transpose of the matrix generated by applying the row operations with the same parameters to $A^{\\top}$. Here ``the same parameters\" means the same index of row/column to multiply, the same multiplication constant, the same indexes of the two row/columns to exchange, or the same multiple of the same index of row/column to be added to another same index of row/column.\n\nGiven any elementary matrix $E$, $AE^{\\top} = (EA^{\\top})^{\\top}$. Since $E^{\\top}$ is also an elementary matrix, column operations can be achieved by multiplication on the right by elementary matrices. Note that if $E$ is of type 1 or 3, its transpose is the same as itself; but if $E = E_2(i,j,x)$ for some $i, j, x$, its transpose will be $E_2(j,i,x)$, which means that $AE_2(i,j,x)$ will add the multiple $x$ of the $i$th column of $A$ to its $j$th column, rather than adding that multiple of the $j$th column of $A$ to its $i$th column. \\rQED\n\\end{exercise}\n\n\\begin{exercise}{12}\n\\begin{enumerate}\n\\item Show $E_1(i,x)E_1(i,1/x) = E_1(i,1/x)E_1(i,x) = I$.\n\n\\Proof\\ For $E_1(i,x)E_1(i,1/x)$, the $i$th-row of $E_1(i,1/x)$ will be multiplied by $x$, then its $i,i$ entry will become $x \\cdot (1/x) = 1$, therefore the result is an identity matrix. For $E_1(i,1/x)E_1(i,x)$, the $i$th-row of $E_1(i,x)$ will be multiplied by $1/x$, then its $i,i$ entry will become $(1/x) \\cdot x = 1$, therefore the result is an identity matrix. This shows that $E_1(i,1/x)$ is the inverse matrix of $E_1(i,x)$, and that $E_1(i,x)$ is invertible. \\rQED\n\n\\item Show $E_2(i,j,x)E_2(i,j,-x) = E_2(i,j,-x)E_2(i,j,x) = I$.\n\n\\Proof\\ Consider $E_2(i,j,-x)E_2(i,j,x)I$, which can be seen as applying two row operations to the identity matrix. Let $r_k$ be the $k$-th row of $I$. After the first operation $E_2(i,j,x)$, the $i$th row becomes $xr_j + r_i$; after the second operation $E_2(i,j,-x)$, the $i$th row of the identity matrix becomes $xr_j + r_i - xr_j = r_i$. Since all other rows of the identity matrix keeps invariant during this two operations, after the two operations the identity itself keeps invariant, namely $E_2(i,j,-x)E_2(i,j,x)I = I$, or $E_2(i,j,-x)E_2(i,j,x)=I$. From what we just proved we can get $E_2(i,j,x)E_2(i,j,-x) = I$ by seeing $-x$ as $y$ and $x$ as $-y$. Then, $E_2(i,j,-x)$ is both the left and right inverse matrix of $E_2(i,j,x)$, namely $E_2(i,j,-x)$ is the inverse of $E_2(i,j,x)$ and the later is invertible. \\rQED\n\n\\item Show $E3(i,j)E3(i,j) = I$.\n\n\\Proof\\ Since $E3(i,j)I = E3(i,j)$, $E3(i,j)$ is just the identity matrix with its $i,j$ rows exchanged. $E3(i,j)E3(i,j)$ exchanges these two rows back, so the product matrix is just the identity matrix. This shows that $E3(i,j)E3(i,j) = I$ and that $E3(i,j)$ is the inverse of itself. \\rQED\n\\end{enumerate}\n\\end{exercise}", "meta": {"hexsha": "00e5994e12c4d2d27a0698652b74a3ecb5edf1ff", "size": 5859, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW4/sec0203.tex", "max_stars_repo_name": "notcome/fa15-linear-algebra", "max_stars_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "HW4/sec0203.tex", "max_issues_repo_name": "notcome/fa15-linear-algebra", "max_issues_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW4/sec0203.tex", "max_forks_repo_name": "notcome/fa15-linear-algebra", "max_forks_repo_head_hexsha": "0ee2fbe81d901271d747e5b314101378a1633852", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.0, "max_line_length": 827, "alphanum_fraction": 0.6338965694, "num_tokens": 2198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8333245911726382, "lm_q1q2_score": 0.6266705705850665}} {"text": "\\documentclass[../main]{subfiles}\n\\pagestyle{fancy}\n\n\\begin{document}\n\n\\chapter{Large Cardinals}\n\\label{apx.large-cardinals}\n\\thispagestyle{fancy}\n\nSince large cardinals came into existence in the beginning of the 20th century, a vast zoo of different types of such have appeared. The aim of this appendix is to act as a reference for the definitions of these as well as the relations between them.\n\n\\section{Inaccessibles}\n\n\\defi{\n A cardinal $\\kappa$ is \\textbf{regular} if $\\cof\\kappa=\\kappa$; i.e. that there are no $\\gamma<\\kappa$ with a cofinal function $f\\colon\\gamma\\to\\kappa$. $\\kappa$ is a \\textbf{strong limit} if $2^\\lambda<\\kappa$ for all cardinals $\\lambda<\\kappa$. If $\\kappa$ is both regular and a strong limit then we say that it is (strongly) \\textbf{inaccessible}.\n}\n\n\\qprop[\\cite{Kanamori} Proposition 1.2]{\n If $\\kappa$ is inaccessible then $(V_\\kappa, \\in)\\models\\zfc$.\n}\n\nG\\\"odel's Second Incompleteness Theorem from \\cite{godel-incompleteness} then immediately implies the following corollary.\n\n\\qcoro{\n \\zfc\\ can not prove the existence of any inaccessible cardinals. Indeed, not even the \\textit{consistency} of the existence of any inaccessible cardinals can be proven in \\zfc.\n}\n\n\\section{Weakly compacts}\n\n\\defi{\n For any function $f\\colon A\\to B$, a subset $H\\subset A$ is \\textbf{homogeneous for $f$} if $f\\restr H$ is a constant function.\n}\n\n\\defi{\n Let $\\kappa$ and $\\lambda$ be infinite cardinals, $\\gamma$ an ordinal and $n<\\omega$. Then the partition relation $\\kappa\\to(\\lambda)^n_\\gamma$ holds if to every function $f\\colon[\\kappa]^n\\to\\gamma$ there exists a subset $H\\subset[\\kappa]^n$ of size $\\lambda$ which is homogeneous for $f$. If $\\gamma=2$ then we usually leave it out and simply write $\\kappa\\to(\\lambda)^n$.\n}\n\n\\defi{\n An uncountable cardinal $\\kappa$ is \\textbf{weakly compact} if $\\kappa\\to(\\kappa)^2$.\n}\n\n\\qtheo[\\cite{Jech} Lemma 9.9]{\n Every weakly compact cardinal is a limit of inaccessible cardinals.\n}\n\n\\section{Ineffables and completely ineffables}\n\n\\defi{\n An uncountable cardinal $\\kappa$ is \\textbf{ineffable} if to any function $f\\colon[\\kappa]^2\\to 2$ there exists a \\textit{stationary} $H\\subset[\\kappa]^2$ which is homogeneous for $f$.\n}\n\nIneffable cardinals are weakly compact by definition, and the following theorem from \\cite{Friedman} shows that they are strictly stronger.\n\n\\qtheo[Friedman]{\n Ineffable cardinals are weakly compact limits of weakly compacts.\n}\n\nA way of improving ineffability is to ``close under homogeneity'', in the sense that if $H$ is homogeneous for $f\\colon[\\kappa]^2\\to 2$ and $g\\colon[H]^2\\to 2$ is any function, then there is a subset of $H$ which is homogeneous for $g$. To formalise this notion we use the concept of a \\textit{stationary class}.\n\n\\xdefi{\n For $X$ any set, a collection $\\R\\subset\\p(X)$ is a \\textbf{stationary class} if\n \\begin{itemize}\n \\item $\\R\\neq\\emptyset$;\n \\item Every $A\\in\\R$ is a stationary subset of $X$;\n \\item If $A\\in\\R$ and $B\\supset A$ then $B\\in\\R$.$\\hfill\\circ$\n \\end{itemize}\n}\n\n\\defi{\n An uncountable cardinal $\\kappa$ is \\textbf{completely ineffable} if there is a stationary class $\\R\\subset\\p(\\kappa)$ such that for every $A\\in\\R$ and $f\\colon[A]^2\\to 2$ there exists a $H\\in\\R$ which is homogeneous for $f$.\n}\n\nAs suspected, these completely ineffable cardinals are indeed strictly stronger than the ineffables, as the following theorem from \\cite{Abramson} shows.\n\n\\qtheo[Abramson et al]{\n Completely ineffable cardinals are ineffable limits of ineffable cardinals.\n}\n\n\\section{Measurables, strongs and supercompacts}\n\n\\defi{\n For two first-order structures $\\M$ and $\\N$ with underlying sets $M$ and $N$, an \\textbf{elementary embedding} $j\\colon\\M\\to\\N$ between them is a function $j\\colon M\\to N$ such that, for any first-order formula $\\varphi(v_1,\\dots,v_n)$ and sets $x_1,\\dots,x_n\\in\\M$ it holds that $\\M\\models\\varphi[x_1,\\dots,x_n]$ iff $\\N\\models\\varphi[j(x_1),\\dots,j(x_n)]$.\n}\n\nAs elementary embeddings in particular preserve equality, they are always injective. Identity embeddings are of course always elementary, so we say that an elementary embedding is \\textbf{non-trivial} if it is not the identity. The following then shows that in most situations these non-trivial embeddings can be associated to a unique ordinal.\n\n\\qprop[\\cite{Kanamori} Propostion 5.1]{\n If $j\\colon(\\M,\\in)\\to(\\N,\\in)$ is an elementary embedding such that $\\M$ is transitive and \\textit{either} $\\N\\subset\\M$ or $\\M\\models\\zfc$, then there exists an ordinal $\\alpha\\kappa$ exists a transitive class $\\M_\\theta$ satisfying that $H_\\theta\\subset\\M_\\theta$, and an elementary $j_\\theta\\colon(V,\\in)\\to(\\M_\\theta,\\in)$ with critical point $\\kappa$. We say that $\\kappa$ is \\textbf{$\\theta$-strong} if the property holds for a specific $\\theta$.\n}\n\n\\qprop[\\cite{Kanamori} 26.6]{\n Strong cardinals are measurable limits of measurable cardinals.\n}\n\n\\defi[\\gbc]{\n An uncountable cardinal $\\kappa$ is \\textbf{supercompact} if there to every cardinal $\\theta>\\kappa$ exists a transitive class $\\M_\\theta$ satisfying that $^{<\\theta}\\M_\\theta\\subset\\M_\\theta$, and an elementary $j_\\theta\\colon(V,\\in)\\to(\\M_\\theta,\\in)$ with critical point $\\kappa$.\n}\n\n\\prop{\n If $\\kappa$ is supercompact then\n \\eq{\n V_\\kappa\\models\\godel{\\text{There exists a proper class of strong cardinals}}.\\tag*{$(1)$}\n }\n}\n\\proof{\n (Sketch) By noting that the restrictions of the supercompact embedding is an element of the target model by supercompactness, $\\kappa$ is strong in the target model, so that a reflection argument shows $(1)$.\n}\n\n\\section{Woodins and Vop\\v enkas}\n\n\\defi[][defi.A-strong]{\n Let $A$ be any set. An uncountable cardinal $\\kappa$ is \\textbf{$A$-strong} if there to every cardinal $\\theta>\\kappa$ exists a transitive class $\\M_\\theta$ satisfying that $H_\\theta\\subset\\M_\\theta$, and an elementary $j_\\theta\\colon(V,\\in)\\to(\\M_\\theta,\\in)$ with critical point $\\kappa$, such that $A\\cap H_\\theta = j(A)\\cap H_\\theta$.\n}\n\n\\defi{\n An uncountable cardinal $\\delta$ is a \\textbf{Woodin cardinal} if there to every subset $A\\subset H_\\delta$ exists $\\kappa<\\delta$ such that $(H_\\delta, \\in, A)\\models\\godel{\\text{$\\kappa$ is $A$-strong}}$.\n}\n\n\\theo[\\cite{Kanamori} Theorem 26.14]{\n The following are equivalent for an uncountable cardinal $\\kappa$.\n \\begin{enumerate}\n \\item $\\kappa$ is a Woodin cardinal;\n \\item For any $f\\colon\\kappa\\to\\kappa$ there exists $\\alpha<\\kappa$ such that $f[\\alpha]\\subset\\alpha$, a transitive $\\M$ with $V_{j(f)(\\alpha)}\\subset\\M$ and an elementary embedding $j\\colon(V,\\in)\\to(\\M,\\in)$ with $\\crit j=\\kappa$.\n \\end{enumerate}\n}\n\n\\defi[\\gbc]{\n \\textbf{Vop\\v enka's Principle (\\vp)} postulates that to any first-order language $\\mathcal L$ and proper class $\\C$ of $\\mathcal L$-structures, there exist distinct $\\M,\\N\\in\\C$ and an elementary embedding $j\\colon\\M\\to\\N$.\n}\n\n\\defi{\n An uncountable cardinal $\\delta$ is \\textbf{Vop\\v enka} if $(V_\\delta,\\in;V_{\\delta+1})\\models\\vp$.\n}\n\nThe following theorem is from \\cite{Perlmutter}.\n\n\\qtheo[Perlmutter]{\n Vop\\v enka cardinals are equivalent to cardinals that are ``Woodin for supercompactness'', meaning a cardinal $\\delta$ such that to any subset $A\\subset H_\\delta$ there is a cardinal $\\kappa<\\delta$ such that $(H_\\delta,\\in,A)\\models\\godel{\\text{$\\kappa$ is $A$-supercompact}}$.\\footnote{Here $\\kappa$ is, in analogy with Definition \\ref{defi.A-strong}, \\textbf{$A$-supercompact} if there to every cardinal $\\theta>\\kappa$ exists a transitive class $\\M_\\theta$, closed under ${<}\\theta$-sequences, and an elementary $j_\\theta\\colon(V,\\in)\\to(\\M_\\theta,\\in)$ with critical point $\\kappa$, such that $A\\cap H_\\theta=j(A)\\cap H_\\theta$.}\n}\n\n\\section{Reinhardts and Kunen inconsistency}\n\n\\defi[\\gbc]{\n An uncountable cardinal $\\kappa$ is a \\textbf{Reinhardt cardinal} if there exists an elementary embedding $j\\colon(V,\\in)\\to (V,\\in)$ with $\\crit j=\\kappa$.\n}\n\n\\theo[Kunen inconsistency, \\gbc, \\cite{Kanamori} Theorem 23.12]{\n There are no Reinhardt cardinals. Even more, there is no non-trivial elementary $j\\colon(V_{\\lambda+2},\\in)\\to(V_{\\lambda+2}, \\in)$ for any uncountable cardinal $\\lambda$.\n}\n\nWhen we're dealing with the \\textit{virtual} large cardinals in Chapter \\ref{chapter.virtual-large-cardinals} we show that the property $j(\\kappa)>\\theta$ is a highly non-trivial assumption. However, when we're not in the virtual world then this is simply automatic. \n\n\\qprop[\\cite{Kanamori} 26.7]{\n If $j\\colon V\\to\\M_\\theta$ witnesses that $\\kappa:=\\crit j$ is a $\\theta$-strong cardinal then $j(\\kappa)>\\theta$.\n}\n\nNote that a crucial part of the proof of the above is Corollary 23.14 in \\cite{Kanamori}, which relies \\textit{heavily} on the Kunen inconsistency. There is also the following even stronger version of the Reinhardts.\n\n\\defi{\n An uncountable cardinal $\\kappa$ is \\textbf{super Reinhardt} if for all ordinals $\\lambda$ there exists an elementary embedding $j\\colon (V,\\in)\\to(V,\\in)$ with $\\crit j=\\kappa$ and $j(\\kappa)>\\lambda$.\n}\n\n\\section{Berkeleys}\n\n\\defi[\\gb]{\n An uncountable cardinal $\\delta$ is a \\textbf{proto-Berkeley cardinal} if to every transitive \\textit{set} $\\M$ such that $\\delta\\subset\\M$ there exists an elementary embedding $j\\colon(\\M,\\in)\\to(\\M,\\in)$ with $\\crit j<\\delta$.\n}\n\nNote that if $\\kappa$ is a proto-Berkeley cardinal then every $\\lambda>\\kappa$ is also proto-Berkeley, which makes it quite an uninteresting notion. But we can isolate the interesting cases, leading to the definition of a Berkeley cardinal. The following is Theorem 2.1.14 in \\cite{Cutolo}.\n\n\\qtheo[Cutolo]{\n If $\\delta_0$ is the least proto-Berkeley cardinal then we can choose the critical point of the embedding to be arbitrarily large below $\\delta_0$.\n}\n\nAs this property is clearly not preserved upwards, this makes for a good candidate for the large cardinal notion.\n\n\\defi[\\gb]{\n A proto-Berkeley cardinal $\\delta$ is \\textbf{Berkeley} if we can choose the critical point of the embedding to be arbitrarily large below $\\delta$. If we furthermore can choose the critical point as an element of any club $C\\subset\\delta$ then we say that $\\delta$ is \\textbf{club Berkeley}.\n}\n\nIn \\cite{Cutolo}, they furthermore mention that, among the above-mentioned cardinals, the non-trivial relative consistency implications currently known are the following, being Theorem 2.2.1 and 2.2.2 in \\cite{Cutolo}, respectively.\n\n\\qtheo[Cutolo]{\n Berkeley cardinals are consistency-wise strictly stronger than Reinhardt cardinals.\n}\n\n\\qtheo[Cutolo]{\n Club Berkeley cardinals are consistency-wise strictly stronger than super Reinhardt cardinals.\n}\n\n\\end{document}\n", "meta": {"hexsha": "c0beaff98e2fde0e6bc2582757fec15d94395f29", "size": 11925, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "frontbackmatter/apx-large-cardinals.tex", "max_stars_repo_name": "saattrupdan/phd", "max_stars_repo_head_hexsha": "21481596be517c874e311797f5a70829e0cba7d3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "frontbackmatter/apx-large-cardinals.tex", "max_issues_repo_name": "saattrupdan/phd", "max_issues_repo_head_hexsha": "21481596be517c874e311797f5a70829e0cba7d3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "frontbackmatter/apx-large-cardinals.tex", "max_forks_repo_name": "saattrupdan/phd", "max_forks_repo_head_hexsha": "21481596be517c874e311797f5a70829e0cba7d3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 55.4651162791, "max_line_length": 636, "alphanum_fraction": 0.7300628931, "num_tokens": 3604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8333245911726381, "lm_q1q2_score": 0.6266705705850664}} {"text": "\\section{Bimatrix Games and Labels}\n\\label{labels-sect}\n\nLet $(A,B)$ be bimatrix game, and let $X=\\Delta^1$ and\n$Y=\\Delta^2$ be the mixed strategy\nsimplices of player 1 and 2, respectively, that is,\n\\begin{equation}\n\\arraycolsep.2em\n\\label{mixed-strat-simplices}\n\\begin{array}{rcll}\nX&=&\\{\\, x\\in\\reals^m&\\mid x\\geq\\0,\\ \\1\\T x = 1 \\},\\\\\nY&=&\\{\\, y\\in\\reals^n&\\mid y\\geq\\0,\\ \\1\\T y = 1 \\}.\n\\end{array}\n\\end{equation}\nA {\\em labeling} of the sets $X$ and $Y$ assigns to each\n$x\\in X$ and $y\\in Y$ a set of {\\em labels}, which is a\nsubset of $[m+n]$, as follows:\n\\begin{enumerate}\n\\item the $m$ pure strategies of player 1 are denoted as $i\\in [m]$;\n\\item the $n$ pure strategies of player 2 are denoted as $m+j$ with $j\\in [n]$;\n\\item each mixed strategy $x\\in X$ of player 1 has\n \\begin{itemize}\n \\item label $i$ for each $i\\in [m]$ such that $x_i = 0$,\n \\item label $m + j$ for each $j\\in [n]$ such that the $j$-th pure strategy\n of player 2 is a best response to $x$;\n \\end{itemize}\n\\pagebreak[2]\n\\item each mixed strategy $y\\in Y$ of player 2 has\n \\begin{itemize}\n \\item label $m + j$ for each $j\\in [n]$ such that $y_j = 0$,\n \\item label $i$ for each $i\\in [m]$ such that the $i$-th pure strategy\n of player 1 is a best response to $y$.\n \\end{itemize}\n\\end{enumerate}\nThis labeling can be used to characterize the Nash equilibria of the game:\nA pair $(x,y)$ is called {\\em completely labeled} if each\npossible label in $[m+n]$ is a label of $x$ or of~$y$.\n\n\\begin{theorem}{\\rm (Shapley \\cite{shapley})}\\label{comp-label-thm}\nLet $(x,y)\\in X\\times Y$. Then $(x,y)$ is a Nash equilibrium of the bimatrix\ngame $(A,B)$ if and only if $(x,y)$ is completely labeled.\n\\end{theorem}\n\n\\begin{proof}\nThe mixed strategy $x\\in X$ has label $m + j$ for some $j\\in [n]$ if and\nonly if the $j$-th pure strategy of player 2 is a best response to $x$. By\nProposition \\ref{br-played-thm}, this is a necessary and sufficient condition\nfor\nplayer 2 to play his $j$-th strategy at every equilibrium where player 1\nchooses $x$. The analogous holds for the strategies $y\\in Y$ and player 1\nplaying her $i$-th strategy in response to player 2 choosing $y$.\nTherefore, at an equilibrium $(x,y)$ all labels $m + j$, with $j\\in [n]$,\nappear either as labels of $x$ or of $y$.\nConversely, if $(x,y)$ is not completely labeled, then some\nlabel does not appear as a label of $x$ or $y$. This label\nrepresents a pure strategy that is played with positive\nprobability but is not a best response, which contradicts\nthe equilibrium property.\n\\end{proof}\n\nA useful graphical representation of labels on the simplices $X$ and $Y$\nis done by labeling the outside of each simplex according to the player's own\npure strategies that are not played, and by subdividing its interior\ninto closed polyhedral sets, called {\\em best response\nregions}, that correspond to the other player's pure\nbest responses.\nWe give an example of this construction.\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\n\\label{br-game-ex}\nConsider the $3\\times 3$ game $(A,B)$ with\n\\begin{equation}\n\\label{AB}\nA = \\left(\\begin{matrix}1&0&0\\\\ 0&1&0\\\\\n0&0&1\\end{matrix}\\right),\n\\qquad\nB = \\left(\\begin{matrix}0&2&4\\\\ 3&2&0\\\\\n0&2&0\\end{matrix}\\right).\n\\end{equation}\nFigure~\\ref{br-regions-fig} shows the mixed strategy simplices $X$ and $Y$:\nthe exterior facets are labeled with the pure strategy that is played\n% In general, do not use a semicolon where period works.\nat the opposite vertex of the simplex. The interior is covered\nby the best response regions, labeled by the other player's pure best\nresponse strategies.\nFor example, the best-response region in $Y$ with label~$1$\nis the set of all the $(y_1,y_2,y_3)\\in\\reals^3$ such that $y_1\\ge y_2$ and\n$y_1\\ge y_3$.\nThere is only one pair $(x,y)$\nthat is completely labeled, namely $x=(\\frac13,\\frac23,0)$\nwith labels $3,4,5$, and $y=(\\frac12,\\frac12,0)$ with labels\n$1,2,6$. This is the only Nash equilibrium of the game.\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Xnondeg.pdf}%\n\\hfill\n%\\vrule height 52mm ~ \\vrule height 27ex\n\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Yimit.pdf}%\n\\hfill\\strut\n\\caption[The labeled mixed strategy simplices of a game]{%\nThe labeled best response regions of the mixed strategy simplices\nof player 1 (left) and player 2 (right) in game (\\ref{AB}).\n}\n\\label{br-regions-fig}\n\\end{figure}\n\\end{example}\n\nThe representation of a game and its Nash equilibria in terms of best\nresponse regions can be translated to an equivalent construction on\npolytopes.\nThe first step is to notice that the best-response regions can be obtained as\nprojections on $X$ and $Y$ of the {\\em best-response facets} of\nthe polyhedra\n\\begin{equation}\n\\label{br-polyhedron}\n\\arraycolsep.2em\n\\begin{array}{rcll}\n\\overline{P}&=&\\{ (x,v)\\in X\\times\\reals& \\mid B\\T x\\leq\\1 v \\},\\\\\n\\overline{Q}&=&\\{ (y,u)\\in Y\\times\\reals& \\mid A y\\leq\\1 u \\}. %\n\\end{array}\n\\end{equation}\nIn $\\overline{P}$, these facets are the points $(x,v)\\in X\\times\\reals$\nsuch that $(B\\T x)_j = v$, which in turn correspond to the strategies\n$x\\in X$ of player 1 that give exactly payoff $v$ to player~2\nwhen he plays strategy $j$; this payoff is the\nbest-response payoff by the definition of $P$.\nThe projection of the facet defined by $(B\\T x)_j = v$ to\n$X$ then has label~$j$.\nAnalogously, the facet of $\\overline{Q}$ given by the points\n$(y,u)\\in Y\\times\\reals$ such that $(A y)_i = u$ projects to the\nbest-response region of $Y$ with label $i$.\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\n\\label{br-polyhedra-ex}\nIn Example \\ref{br-game-ex}, the inequalities $B\\T x\\le\\1v$ are\n\\[\n\\arraycolsep.2em\n\\begin{array}{rcl}\n3x_2 & \\leq&v \\\\\n2x_1+ 2x_2+ 2x_3 & \\leq&v \\\\\n4x_1 & \\leq&v.\n\\end{array}\n\\]\nFigure~\\ref{br-polyhedron-fig} shows the best-response facets of\n$\\overline{P}$ and their projection to $X$ by ignoring the payoff\nvariable~$v$, which gives the subdivision of $X$ into best-response\nregions of Figure \\ref{br-regions-fig}.\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=40ex]{chapter-2/fig-labels/upenv.pdf}%\n\\hfill\\strut\n\\caption[The best response polyhedron of a game]{%\nThe best response polyhedron of player 1 in game (\\ref{AB}).}\n\\label{br-polyhedron-fig}\n\\end{figure}\n\\end{example}\n\nGiven the assumptions on non-negativity of $A$ and $B\\T$, we can\nchange coordinates to $x_i / v$ and $y_j / u$ and replace $\\overline{P}$ and\n$\\overline{Q}$ with the {\\em best-response polytopes}\n\\begin{equation}\n\\label{br-polytopes}\n\\arraycolsep.2em\n\\begin{array}{rcll}\nP&=&\\{\\, x\\in\\reals^m& \\mid\\,x\\geq\\0,\\ B\\T x\\leq\\1 \\},\\\\\nQ&=&\\{\\, y\\in\\reals^n& \\mid\\,A y\\leq\\1,~ y\\geq\\0 \\}.\n\\end{array}\n\\end{equation}\nIn (\\ref{br-polytopes}), both $P$ and $Q$ are defined by\n$m+n$ inequalities that correspond, in the listed order, to the\npure strategies of the two players with labels in $[m+n]$.\nThe polytope $P$ is the intersection of the $m+n$ half-spaces\nthat correspond to either player 1 not playing her $i$-th pure\nstrategy or to a best response $j$ of player~2,\nwhere $i\\in [m]$ and $j\\in [n]$.\nThe analogous statement holds for $Q$.\nFormally, a point $x\\in P$ has label $k$ if and only if either\n$x_k = 0$ for $k\\in [m]$ or $(B\\T x)_j = 1$ for $k=m+j$ with $j\\in [n]$, and\na point in $Q$ has label $k$ if and only if either\n$(A y)_k=1$ for $k\\in [m]$ or $y_j = 0$ for $k=m+j$ with $j\\in [n]$.\n\nHence, a point $(x,y)\\in P\\times Q$ is completely labeled if and only if\nit satisfies the {\\em complementarity condition} that states\n% the quantifiers should be outside to avoid confusion\nthat for all $i\\in[m]$ and all $j\\in[n]$,\n\\begin{equation}\n%\\label{}\n\\begin{array}{ccrll}\nx_i = 0 & \\text{ or }&(Ay)_i= 1~,\\\\\ny_j = 0 & \\text{ or }&(B\\T x)_j=1~.\n\\end{array}\n\\end{equation}\nTherefore, if $(x,y)\\in P\\times Q$ is completely labeled either the\ncorresponding point in $\\overline{P}\\times\\overline{Q}$\nis a Nash equilibrium or $(x,y)=(\\0,\\0)$; we refer to the latter case\nas {\\em artificial equilibrium}.\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\nKeeping on with Example \\ref{br-game-ex} and \\ref{br-polyhedra-ex},\nthe best response\npolyhedron $\\overline{P}$ of Figure \\ref{br-polyhedron-fig} becomes\nthe best response polytope of Figure \\ref{br-polytopes-fig}.\nNotice that the vertex $(x,y)=(\\0,\\0)$ is completely labeled, since it is labeled\nby the labels $1,2,3$ in $P$ and $4,5,6$ in $Q$.\n\\begin{figure}[htb]\n\\strut\\hfill\n\\includegraphics[width=40ex]{chapter-2/fig-labels/polytope.pdf}%\n\\hfill\\strut\n\\caption[The best response polytope of a game]{%\nThe best response polytope of player 1 in game (\\ref{AB}).\n}\n\\label{br-polytopes-fig}\n\\end{figure}\n\\end{example}\n\nWe now consider some special cases of games and how they are related to each\nother in terms of computational complexity.\nFirst of all, we note that any bimatrix game can be ``symmetrized''.\nThis result is due to Gale, Kuhn and Tucker \\cite{gale-kuhn-tucker}\nfor zero-sum games, while its extension to non-zero-sum games is a folklore\nresult.\n\n\\begin{proposition}\\label{symmetrize-c}\nLet $(A,B)$ be a bimatrix game and let $(x,y)$ be one of its Nash equilibria.\nThen $(z,z)$, where $z=(x\\alpha,y\\beta)$ for suitable\npositive scalars $\\alpha$ and $\\beta$, is a Nash equilibrium of the symmetric game\n$(C,C\\T)$, where\n\\begin{equation}\nC = \\left(\n \\begin{array}{cc}\n 0 & A \\\\\n B\\T & 0\n \\end{array}\n \\right).\n\\end{equation}\n\\end{proposition}\n\nMcLennan and Tourky \\cite{mclennan-tourky} have proven a result in the\nopposite direction of Proposition \\ref{symmetrize-c}: any symmetric game\ncan be translated into a {\\em imitiation game}, where the payoff matrix of\nplayer 1 is the identity matrix $I$. In any Nash equilibrium of $(I,B)$,\nthe mixed strategy $x$\nof player 1 corresponds exactly to the symmetric equilibrium $(x,x)$ in\nthe symmetric game defined by the payoff matrix of player 2.\nSince it takes polynomial time in the size of a matrix to calculate its\ntranspose, an algorithm that finds a Nash equilibrium of a bimatrix game\ncan be used to find a symmetric Nash equilibrium of a symmetric game.\n\n\\begin{theorem}{\\rm (McLennan and Tourky \\cite{mclennan-tourky})}\n\\label{imitation-thm}\nThe pair $(x,x)$ is a symmetric Nash equilibrium of the symmetric bimatrix\ngame $(C,C\\T)$ if and only if there is some $y$ such that $(x,y)$ is a\nNash equilibrium of the imitation game $(I,B)$ with $B=C\\T$.\n\\end{theorem}\n\nNotice that Theorem \\ref{imitation-thm} applies to the symmetric\nequilibria of the symmetric game but not to all its Nash equilibria; there\ncould be non-symmetric equilibria of $(C,C\\T)$ that are not found\nthrough the imitation game, as shown in the following example.\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\nAs an example, consider the symmetric game $(C,C\\T)$ with\n\\begin{equation}\n\\label{C}\nC = \\left(\\begin{matrix}0&3&0\\\\ 2&2&2\\\\\n4&0&0\\end{matrix}\\right),\n\\qquad\nC\\T = \\left(\\begin{matrix}0&2&4\\\\ 3&2&0\\\\\n0&2&0\\end{matrix}\\right).\n\\end{equation}\nThe corresponding imitation game is $(I,C\\T)=(A,B)$, seen in\nExample \\ref{br-game-ex}.\nFigure~\\ref{imit-fig} shows the labeled mixed-strategy simplices $X$ and $Y$\nfor the game~(\\ref{C}); since the game is symmetric, only the labels\nare different.\nIn addition to the symmetric equilibrium $(x,x)$ where\n$x=(\\frac13,\\frac23,0)$, the game has two non-symmetric equilibria in $(a,b)$\nand $(b,a)$ with $a=(\\frac12,\\frac12,0)$ and $b=(0,\\frac23,\\frac13)$.\nThe imitation game $(A,B)$, on the other hand, has only one equilibrium\n$(x,y)$, corresponding to $(x,x)$, with $y=(\\frac{1}{2},\\frac{1}{2},0)$.\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Xnondeg.pdf}%\n\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Ynondeg.pdf}%\n\\hfill\\strut\n\\caption[The best response regions of a symmetric game]{%\nThe best response regions of the symmetric game (\\ref{C}).\n}\n\\label{imit-fig}\n\\end{figure}\n\\end{example}\n\nThe characterization of Nash equilibria as completely labeled pairs $(x,y)$\nholds for arbitrary bimatrix games. From now on, we impose a\nfurther condition: all points in $P$ have at most $m$ labels,\nand all points in $Q$ have at most $n$ labels. These games are called\n{\\em nondegenerate}.\nThis condition is required for the Lemke-Howson algorithm (see\nsection \\ref{lh-sect}).\n% This is a loss of generality\nThe algorithm can also be applied to degenerate games by\nlexicographic perturbation, as shown in von Stengel \\cite{vs-agt}.\nIn an equilibrium $(x,y)$ of a nondegenerate game each label appears\nexactly once. This also means that the number of pure best response\nstrategies against a mixed strategy is never larger than the size of\nthe support of that mixed strategy.\nGeometrically, this means that no point of the best response polytope $P$\nlies on more than $m$ facets and no point of the best response polytope $Q$\nlies on more than $n$ facets, so both $P$ and $Q$ are simple.\nFurthermore, a point of\n$P$ has exactly $m$ labels if and only if it is a vertex, and a point\nof $Q$ has exactly $n$ labels if and only if it is a vertex. Therefore,\nall completely labeled points $(x,y)$ are vertices of the best response\npolytopes, and Nash equilibria are isolated points.\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\nAn example of degenerate game is given by $(C,C\\T)$ with\n\\begin{equation}\n\\label{Cdeg}\nC = \\left(\\begin{matrix}0&4&0\\\\ 2&2&2\\\\\n4&0&0\\end{matrix}\\right),\n\\qquad\nC\\T = \\left(\\begin{matrix}0&2&4\\\\ 4&2&0\\\\\n0&2&0\\end{matrix}\\right).\n\\end{equation}\nAs shown in Figure \\ref{deg-fig}, the mixed strategy\n$x=(\\frac12,\\frac12,0)$, that also defines the\nunique symmetric equilibrium $(x,x)$ of the game, has three pure best\nresponses. The Nash equilibria $(x,y)$ of the imitation game\n$(I,C\\T)$ are not unique, since any convex\ncombination of $(\\frac12,\\frac12,0)$ and $(\\frac13,\\frac13,\\frac13)$\ncan be chosen for~$y$.\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Xsymmdeg.pdf}%\n\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/Yimitdeg.pdf}%\n\\hfill\\strut\n\\caption[A degenerate symmetric game]{%\nLeft: The best-response regions of the degenerate symmetric\ngame (\\ref{Cdeg}). \\\\\nRight: The best-response regions for player 2 in the corresponding\nimitation game $(I,C\\T)$.\n}\n\\label{deg-fig}\n\\end{figure}\n\\end{example}\n\nA generalization of imitation games is the class of\n{\\em unit vector games}, introduced by Balthasar \\cite{balthasar}.\nThese are defined as bimatrix games of the form $(U,B)$ where the\ncolumns of the matrix $U$ are unit vectors.\nBy the results above, finding a Nash equilibrium of a bimatrix game is\nat least as hard as finding a Nash equilibrium of a unit vector game.\nSavani and von Stengel \\cite{uvg} have shown that\nthe problem of finding a completely labeled vertex of the product of the\nbest response polytopes $P\\times Q$ can be simplified for unit vector games:\nit is enough to find a completely labeled vertex of a single\npolytope $P^l$, for which the last $n$ facets are labeled following a\nthe labeling in $[m]$ that also encodes the matrix $U$.\n\n\\begin{theorem}{\\rm (Savani and von Stengel \\cite{uvg})}\n\\label{unit-vector-thm}\nLet $l:[n]\\to [m]$ be a function, and let $(U,B)$ be the unit vector game with\n$U=(e_{l(1)}\\ \\cdots\\ e_{l(n)})$.\n\nLet $N_i = \\{ j\\in [n]\\ \\mid\\ l(j)=i \\}$ for\n$i\\in [m]$, and define $P^l$ and $Q^l$ as\n\\begin{equation}\\label{p-l}\n\\arraycolsep.2em\n\\begin{array}{rcll}\nP^l&=&\\{ x\\in\\reals^m& \\mid\\,x\\geq\\0,\\ B\\T x\\leq\\1 \\}, \\\\\nQ^l&=&\\{ y\\in\\reals^n& \\mid\\,y\\geq\\0,\\\n\\sum_{j\\in N_i } y_j\\leq 1 \\hbox{ for } i\\in [m] \\}.\n\\end{array}\n\\end{equation}\nLet $l_f$ be the labeling of the facets of $P^l$ with labels\nin $[m]$ defined as follows:\n\\begin{equation}\n\\label{facet-labeling-unitv}\n\\begin{array}{rll}\nx_i\\geq 0 & \\text{ has label }i & \\text{ for }i\\in [m], \\\\\n(B\\T x)_j \\leq 1 & \\text{ has label }l(j) & \\text{ for }j\\in [n].\n\\end{array}\n\\end{equation}\nThen $x\\in P^l$ is a completely labeled vertex of\n$P^l\\setminus\\{\\0\\}$ (that is, has all labels in $[m]$)\nif and only if there is some $y\\in Q^l$ such that, after scaling,\nthe pair $(x,y)$ is a Nash equilibrium\nof $(U,B)$.\n\\end{theorem}\n\n\\begin{proof}\nLet $P$ and $Q$ be the best response polytopes of $(U,B)$ as\nin (\\ref{br-polytopes}),\nand let $(x,y)\\in P\\times Q\\setminus\\{ (\\0,\\0) \\}$ be a Nash equilibrium of\n$(U,B)$. Then $(x,y)$ is completely labeled with labels in $[m + n]$. If\n$x_i=0$, then $x$ has label $i\\in m$.\nIf $x_i > 0$, then $y$ has label $i$, so $(Uy)_i = 1$.\nTherefore for some $j\\in [n]$ we have $y_j > 0$ and $U_j = e_i$; that is,\nwe have $y_j > 0$ and $l(j)=i$ for some $j\\in [n]$.\nSince $y_j > 0$, $x\\in P$ has label $m+j$; then, $(B\\T x)_j = 1$;\ntherefore $x\\in P^l$ has label $l(j) = i$.\nHence, $x$ is a completely labeled vertex of $P^l$.\n\nConversely, let $x\\in P^l\\setminus \\{ \\0 \\}$ be completely labeled.\nIf $x_i > 0$, then there is $j\\in [n]$ such that $(B\\T x) = j$ and\n$l(j) = i$; that is, $j\\in N_i$. For all $i\\in [m]$ such that $x_i > 0$,\ndefine $y$ as follows: $y_j = 1$, and\n$y_h = 0$ for all $h\\in N_i\\setminus \\{ j \\}$.\nThen $(x,y)\\in P\\times Q$ is completely labeled.\n\\end{proof}\n\n\\begin{example}({\\rm Savani and von Stengel \\cite{uvg}})\nThe game in Example \\ref{br-game-ex} is a unit vector game with $l(i)=i$.\nIn the polytope $P^l$ of Figure~\\ref{p-l-fig} the labels 4, 5 and 6 of the\nbest response polytope $P$ of Figure \\ref{br-polytopes-fig}\nare replaced by 1, 2 and 3, since the corresponding columns of $A$\nare the unit vectors $e_1,e_2,e_3$.\nThe only completely labeled point of $P^l$ are the origin $\\0$, corresponding\nto the ``artificial'' equilibrium, and $x$, corresponding to the unique\nNash equilibrium of the unit vector game (\\ref{AB}).\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=30ex]{chapter-2/fig-labels/p-l-color.pdf}%\n\\hfill\\strut\n\\caption[The polytope $P^l$ of a unit vector game]{%\nThe polytope $P^l$ of the unit vector game (\\ref{AB}).\n}\n\\label{p-l-fig}\n\\end{figure}\n\\end{example}\n\nWe now move on the dual version of Theorem \\ref{unit-vector-thm}\ngiven by Balthasar \\cite{balthasar}.\nWe can translate the polytope $P^l$ of (\\ref{p-l}) to\n$P = \\{ x - \\1\\ |\\ x\\in P^l \\}$,\npossibly multiplying all payoffs in $B$ by a\nconstant so that $\\0$ is in the interior of $P$, which holds\nif $\\1$ in the interior of $P^l$, that is, all columns $b_j$\nof the matrix $B$ fulfill $\\1\\T b_j<1$, for $j\\in[n]$. Then\n\\begin{align*}\nP & = \\{ x + \\1\\geq\\0,\\ (x + \\1)\\T B\\leq\\1 \\} = \\\\\n & = \\{ x\\in\\reals^m\\ |\\ -x_i\\leq 1\\text{ for }i\\in [m],\\\n x\\T (b_j/(1 - \\1\\T b_j)) \\leq 1\\text{ for }j\\in [n] \\}.\n\\end{align*}\nThe polar of $P$ is then\n\\begin{equation}\\label{p-l-dual}\nP^\\Delta = \\conv(\\{-e_i\\ |\\ i\\in [m] \\}\\ \\cup\\ \\{ c_j\\ |\\ j\\in [n] \\})\n\\end{equation}\nwhere $c_j=b_j/(1 - \\1\\T b_j)$.\nSince $P$ and $P^{\\Delta}$ have $\\0$ in their interior,\n$P^{\\Delta\\Delta}=P$. Furthermore,\n$P^\\Delta $ is simplicial and its facets correspond to the vertices of $P$\nand vice versa.\nWe label the vertices\nof $P^\\Delta $ as the corresponding facets in $P^l$, so the completely\nlabeled facets of $P^\\Delta $ correspond to the completely\nlabeled vertices of $P^l$.\nIn particular, the facet corresponding to $\\0$ is\n\\begin{equation}\n\\label{f-0}\n\\arraycolsep.3em\n\\begin{array}{rcl}\nF_0 & = & \\{ x\\in P^\\Delta\\ \\mid\\ -\\1\\T x = 1 \\}\\\\\n & = & \\conv\\{-e_i \\mid i\\in [m] \\}.\n\\end{array}\n\\end{equation}\nTheorem \\ref{unit-vector-thm} then translates to the following.\n\n\\begin{theorem}{\\rm (Balthasar \\cite{balthasar})}\n\\label{unit-vector-dual-thm}\nLet $P^\\Delta $ be a labeled $m$-dimensional simplicial polytope with \\0 in\nits interior and vertices $e_1,\\ldots,e_m,c_1,\\ldots,c_n$ such that\n$F_0$ in $(\\ref{f-0})$ is a facet of $P^\\Delta $.\n\nLet $(U,B)$ be a unit vector game, with\n$U=(e_{l(1)}\\ \\cdots\\ e_{l(n)})$\nfor a labeling $l:[n]\\to [m]$ and $B = [\\,b_1\\,\\cdots\\,b_n\\,]$, where\n$b_j = c_j/(1 + \\1\\T c_j)$ for $j\\in [n]$.\n\nLet $l_v$ be the labeling\nof the vertices of $P^\\Delta $ given by\n\\begin{equation}\n\\label{vert-labeling-unitv}\n\\arraycolsep.2em\n\\begin{array}{rcll}\nl_v(-e_i)&=&i\\quad & \\text{ for }i\\in [m],\\\\\nl_v(c_j)&=&l(j)\\quad & \\text{ for }j\\in [n].\n\\end{array}\n\\end{equation}\nThen a facet $F\\neq F_0$ of $P^\\Delta $ with normal vector $v$ is completely\nlabeled if and only if $(x,y)$ is a Nash equilibrium of $(U,B)$, where\n$x = (v + \\1) / (\\1\\T (v + \\1))$, so that\n$x_i = 0$ if and only if $−e_i\\in F$ for $i\\in [m]$\nand the mixed strategy $y$ is the uniform distribution on the set of\nthe pure best replies to $x$, which in turn correspond to all $j\\in [n]$\nsuch that $c_j$ is a vertex of $F$.\n\\end{theorem}\n\nTheorem \\ref{unit-vector-thm} gives a correspondence between\ncompletely labeled vertices of $P^l$ and equilibria of the unit vector game\n$(U,B)$ with the ``artificial'' equilibrium corresponding to the vertex $\\0$.\nTheorem \\ref{unit-vector-dual-thm} gives a correspondence between\ncompletely labeled facets of $P^\\Delta $\nand equilibria of $(U,B)$ with the ``artificial'' equilibrium\ncorresponding to the facet $F_0$ in (\\ref{f-0}).\n\n% ... here you come back to the old Q!\nGiven a bimatrix game $(A,B)$, it takes polynomial time\nto write and solve the linear equations defining its best response polyhedra\n$\\overline{P},\\overline{Q}$ and its best response polytopes $P,Q$.\nIt also takes polynomial time to label $\\overline{P},\\overline{Q}$ and $P,Q$.\nAnalogously, given a unit vector game $(U,B)$, it takes polynomial time\nto construct and label the polytope $P^l$ and its polar.\nTherefore, Theorem \\ref{unit-vector-thm} gives a polynomial time\nreduction from the problem 2-{\\sc Nash} to the\nproblem {\\sc Another Completely Labeled Vertex} of Table \\ref{aclv} and\nTheorem \\ref{unit-vector-dual-thm} gives a dual reduction to the problem\n{\\sc Another Completely Labeled Facet} of Table \\ref{aclf}.\n\n\\begin{problem}\n{Another Completely Labeled Vertex}\n{An $m$-dimensional simple polytope $P$ with $m+n$ facets;\na labeling $l_f:[m+n]\\to [n]$;\na vertex $v_0$ of $P$ that is completely labeled by~$l_f$.}\n{A vertex $v\\neq v_0$ of $P$ that is completely labeled by $l_f$.}\n\\label{aclv}\n\\end{problem}\n\n\\begin{problem}\n{Another Completely Labeled Facet}\n{A simplicial $m$-dimensional polytope $P^\\Delta$ with $m+n$ vertices;\na labeling $l_v:[m+n]\\to [n]$;\na facet $F_0$ of $P^\\Delta$ that is completely labeled by $l_v$.}\n{A facet $F\\neq F_0$ of $P^\\Delta$ that is completely labeled by $l_v$.}\n\\label{aclf}\n\\end{problem}\n\n\n\\begin{proposition}\n\\label{nash-to-acl}\n{\\sc 2-Nash} reduces in polynomial time to\n{\\sc Another Completely Labeled Vertex} and\n{\\sc Another Completely Labeled Facet}.\n\\end{proposition}\n", "meta": {"hexsha": "436b72e08457d54abd4f99243fc952c87a365956", "size": 22551, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/chapter-2/labels.tex", "max_stars_repo_name": "mmcasetti/mphil-thesis", "max_stars_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_stars_repo_licenses": ["OLDAP-2.4"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "thesis/chapter-2/labels.tex", "max_issues_repo_name": "mmcasetti/mphil-thesis", "max_issues_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_issues_repo_licenses": ["OLDAP-2.4"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/chapter-2/labels.tex", "max_forks_repo_name": "mmcasetti/mphil-thesis", "max_forks_repo_head_hexsha": "6d9902c4f813cf3239d1b312e9453b3ea690c3db", "max_forks_repo_licenses": ["OLDAP-2.4"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.0765027322, "max_line_length": 82, "alphanum_fraction": 0.6983725777, "num_tokens": 7709, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.752012562644147, "lm_q2_score": 0.8333245891029456, "lm_q1q2_score": 0.6266705597656869}} {"text": "\\subsection{Real series}\\label{subsec:real_series}\n\n\\begin{proposition}\\label{thm:almost_all_terms_positive_implies_absolute_convergent}\n If only finitely many coefficients in a real \\hyperref[def:convergent_series]{convergent} series are negative, then the series converges absolutely.\n\\end{proposition}\n\\begin{proof}\n Let \\( N \\) be the index of the last negative coefficient in \\eqref{def:convergent_series/series}. Then the series\n \\begin{equation*}\n \\sum_{k={N+1}}^\\infty a_k\n \\end{equation*}\n is absolutely convergent since every coefficient is positive. Then\n \\begin{equation*}\n \\sum_{k=0}^\\infty \\abs{a_k} = \\sum_{k=0}^N \\abs{a_k} + \\sum_{k=N+1}^\\infty \\abs{a_k}\n \\end{equation*}\n is convergent since the first term on the right side is a finite sum and the second is a convergent series. Hence, the series \\eqref{def:convergent_series/series} converges absolutely.\n\\end{proof}\n\n\\begin{corollary}\\label{thm:almost_all_terms_negative_implies_absolute_convergent}\n If only finitely many coefficients in a real \\hyperref[def:convergent_series]{convergent} series are positive, then the series converges absolutely.\n\\end{corollary}\n\n\\begin{theorem}[Riemann's series permutation theorem]\\label{thm:riemanns_series_permutation_theorem}\\mcite[\\textnumero 247]{Фихтенгольц1968Том2}\n If the real series\n \\begin{equation*}\n \\sum_{k=0}^\\infty a_k\n \\end{equation*}\n is \\hyperref[def:convergent_series]{convergent}, but not absolutely convergent, then for any extended real number \\( x \\in \\BbbR \\cup \\{ -\\infty, +\\infty \\} \\) there exists a \\hyperref[def:symmetric_group]{permutation} \\( p \\) of the coefficients \\( a_0, a_1, a_2 \\)\n such that\n \\begin{equation*}\n \\sum_{k=0}^\\infty p(a_k) = x.\n \\end{equation*}\n\\end{theorem}\n\\begin{proof}\n If the series is not absolutely convergent, then there exist both infinitely many positive and infinitely many negative coefficients.\n\n First, assume that \\( x \\) is finite.\n\n Define the permuted series\n \\begin{equation*}\n \\sum_{k=0}^\\infty b_k\n \\end{equation*}\n as follows:\n \\begin{thmenum}\n \\thmitem{thm:riemanns_series_theorem/positive} Assign to \\( b_n \\) only nonnegative elements of the sequence \\( \\{ a_k \\}_{k=0}^\\infty \\) until \\( \\sum_{k=0}^n b_k \\geq x \\). Then go to \\ref{thm:riemanns_series_theorem/negative}.\n \\thmitem{thm:riemanns_series_theorem/negative} Assign to \\( b_n \\) only negative elements of the sequence \\( \\{ a_k \\}_{k=0}^\\infty \\) until \\( \\sum_{k=0}^n b_k \\geq x \\). Then go to \\ref{thm:riemanns_series_theorem/positive}.\n \\end{thmenum}\n\n This mutual recursion builds a series that converges to \\( x \\) because the coefficients \\( \\{ a_k \\}_{k=0}^\\infty \\) get arbitrarily close to each other.\n\n If \\( x = +\\infty \\), we can add positive coefficients until \\( \\sum_{k=0}^n b_k \\geq 1 \\), then add a single negative coefficient, then continue adding positive coefficients until \\( \\sum_{k=0}^n b_k \\geq 2 \\) and, so on.\n\n If \\( x = -\\infty \\), we use the same process, but with milestones of \\( -1, -2, -3, \\ldots \\).\n\\end{proof}\n\n\\begin{example}\\label{ex:riemanns_series_permutation_theorem/alternating_harmonic_series}\\cite[\\textnumero 247]{Фихтенгольц1968Том2}\n Consider the alternating harmonic series \\eqref{eq:ex:harmonic_series/alternating}. Denote its sum by \\( a \\).\n\n We can rearrange this series by repeating two negative terms and a single positive term as follows:\n \\begin{equation}\\label{ex:riemanns_series_permutation_theorem/alternating_harmonic_series/rearranged}\n 1 - \\frac 1 2 - \\frac 1 4 + \\frac 1 3 - \\frac 1 6 - \\frac 1 8 + \\cdots\n =\n \\sum_{m=1}^\\infty \\left( \\frac 1 {2m - 1} - \\frac 1 {4m - 2} - \\frac 1 {4m} \\right).\n \\end{equation}\n\n Note that \\fullref{ex:riemanns_series_permutation_theorem/alternating_harmonic_series/rearranged} is equivalent to\n \\begin{equation*}\n \\sum_{m=1}^\\infty \\left( \\frac 1 {2m - 1} - \\frac 1 {4m - 2} - \\frac 1 {4m} \\right)\n =\n \\sum_{m=1}^\\infty \\left( \\frac 1 {4m - 2} - \\frac 1 {4m} \\right)\n =\n \\frac 1 2 \\sum_{m=1}^\\infty \\left( \\frac 1 {2m - 1} - \\frac 1 {42} \\right)\n =\n \\frac a 2.\n \\end{equation*}\n\\end{example}\n\n\\begin{proposition}\\label{thm:positive_series_comparison}\\mcite[\\textnumero 237]{Фихтенгольц1968Том2}\n Fix two nonnegative series\n \\begin{equation}\\label{def:positive_series_comparison/a}\n \\sum_{k=0}^\\infty a_k\n \\end{equation}\n and\n \\begin{equation}\\label{def:positive_series_comparison/b}\n \\sum_{k=0}^\\infty b_k\n \\end{equation}\n that is, series with nonnegative real coefficients. Assume that there exists an index \\( K \\) such that\n \\begin{equation*}\n a_k \\leq b_k \\quad\\forall k \\geq K.\n \\end{equation*}\n\n We say that the series \\fullref{def:positive_series_comparison/b} \\term{dominates} the series \\fullref{def:positive_series_comparison/a}.\n\n Then\n \\begin{thmenum}\n \\thmitem{thm:positive_series_comparison/b_converges} If \\fullref{def:positive_series_comparison/b} converges, so does \\fullref{def:positive_series_comparison/a}.\n\n \\thmitem{thm:positive_series_comparison/a_diverges} If \\fullref{def:positive_series_comparison/a} diverges, so does \\fullref{def:positive_series_comparison/b}.\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:positive_series_comparison/b_converges} Suppose that \\fullref{def:positive_series_comparison/b} converges. Then by \\fullref{thm:real_monotone_sequence_converges_iff_bounded}, the sequence of partial sums is bounded. Therefore, the sequence of partial sums of \\fullref{def:positive_series_comparison/a} is also bounded and, by \\fullref{thm:real_monotone_sequence_converges_iff_bounded} again, the series is convergent.\n\n \\SubProofOf{thm:positive_series_comparison/a_diverges} Analogous to \\fullref{thm:positive_series_comparison/b_converges}, but using the negation of \\fullref{thm:real_monotone_sequence_converges_iff_bounded}.\n\\end{proof}\n\n\\begin{proposition}[Cauchy's root test]\\label{thm:cauchys_root_test}\\mcite[thm. 3.33]{Rudin1976Principles}\n Consider the nonnegative series \\fullref{def:positive_series_comparison/a}. Put\n \\begin{equation*}\n q \\coloneqq \\limsup_{k \\to \\infty} \\sqrt[k]{a_k},\n \\end{equation*}\n where \\( q = \\infty \\) if the limit does not exist. Then\n \\begin{itemize}\n \\item If \\( q < 1 \\), the series converges.\n \\item If \\( q > 1 \\), the series diverges.\n \\item If the limit does not exist (e.g. if \\( a_k = k^k \\)), the series diverges.\n \\item If \\( q = 1 \\), the series may either converge or diverge.\n \\end{itemize}\n\\end{proposition}\n\\begin{proof}\n The case when the limit \\( q \\) does not exist is obvious by the contraposition to \\fullref{thm:convergent_series_terms_vanish}.\n\n Suppose that the limit exists. Therefore, there exists an index \\( K \\) such that\n \\begin{equation*}\n \\sqrt[k]{a_k} \\leq q \\quad\\forall k \\geq K.\n \\end{equation*}\n\n Thus, we have the inequality\n \\begin{equation*}\n a_k \\leq q^k \\quad\\forall k \\geq K.\n \\end{equation*}\n\n The statement of the theorem now follows from comparison (\\fullref{thm:positive_series_comparison}) with the geometric series \\eqref{eq:def:geometric_progression/series}.\n\\end{proof}\n\n\\begin{proposition}[d'Alambert's ratio test]\\label{thm:dalamberts_ratio_test}\\mcite[thm. 3.33]{Rudin1976Principles}\n Consider the nonnegative series \\fullref{def:positive_series_comparison/a}. Put\n \\begin{equation*}\n q \\coloneqq \\limsup_{k \\to \\infty} \\frac {a_{k+1}} {a_k},\n \\end{equation*}\n where \\( q = \\infty \\) if the limit does not exist. Then\n \\begin{itemize}\n \\item If \\( q < 1 \\), the series converges.\n \\item If there exists an index \\( k_0 \\) such that \\( \\frac {a_{k+1}} {a_k} \\geq 1 \\) for all \\( k \\geq k_0 \\), the series diverges.\n \\item If the limit does not exist (e.g. if \\( a_k = k! \\)), the series diverges.\n \\end{itemize}\n\\end{proposition}\n\\begin{proof}\n All cases except for \\( q < 1 \\) are obvious by the contraposition to \\fullref{thm:convergent_series_terms_vanish}.\n\n Suppose that the limit exists. Therefore, there exists an index \\( k_0 \\) such that\n \\begin{equation*}\n a_{k+1} \\leq q a_k \\quad\\forall k \\geq k_0.\n \\end{equation*}\n\n Thus,\n \\begin{equation*}\n a_{k_0 + m} \\leq q^m a_{k_0} \\quad\\forall m \\geq \\BbbZ^{\\geq 0}.\n \\end{equation*}\n\n Convergence now follows from comparison (\\fullref{thm:positive_series_comparison}) of the geometric series \\eqref{eq:def:geometric_progression/series} with the subseries of \\fullref{def:positive_series_comparison/a} obtained by trimming the first \\( k_0 \\) elements.\n\\end{proof}\n\n\\begin{proposition}\\label{rem:nonnegative_series_convergence_test_equivalence}\n The values of \\( q \\) in \\fullref{thm:cauchys_root_test} and in \\fullref{thm:dalamberts_ratio_test} are equal.\n\\end{proposition}\n\\begin{proof}\n If we assume that they are not equal, then the same series would have to be convergent and divergent simultaneously in some region.\n\\end{proof}\n\n\\begin{definition}\\label{def:alternating_series}\n Series of the form\n \\begin{equation}\\label{def:alternating_series/series}\n \\pm \\sum_{k=0}^\\infty (-1)^k a_k,\n \\end{equation}\n where all \\( a_k, k = 0, 1, \\ldots \\) are nonnegative, are called \\term{alternating}.\n\\end{definition}\n\n\\begin{proposition}[Leibniz' alternating series test]\\label{thm:leibniz_alternating_series_test}\n Consider the alternating series \\fullref{def:alternating_series}. If the sequence of terms \\( \\{ a_k \\}_{k=0}^\\infty \\) decreases monotonically and if \\( \\lim_{k \\to \\infty} a_k = 0 \\), then the series converges.\n\\end{proposition}\n\n\\begin{theorem}\\label{thm:weierstrass_series_criterion_nessessity}\\mcite[\\textnumero 268]{Фихтенгольц1968Том2}\n \\Fullref{thm:weierstrass_series_criterion} is a necessary condition for nonnegative real functions.\n\\end{theorem}\n", "meta": {"hexsha": "d040b5fb3e2a8642840fb97bc761b395356e822b", "size": 9708, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/real_series.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/real_series.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/real_series.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.4756756757, "max_line_length": 435, "alphanum_fraction": 0.7301194891, "num_tokens": 3189, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581626286834, "lm_q2_score": 0.849971175657575, "lm_q1q2_score": 0.6265631901350799}} {"text": "\\section{Mixed States}\nIn many situations in physics we don't know what the wave function is,\nbut we do know that there is some probability of the wave function being\nsomething.\nSo to compute things we need to take averages.\nThere are two averages.\nOne is quantum mechanical, but one is classical.\nProbabilities in quantum mechanics don't add,\nthey might subtract.\nAmplitudes add,\nbut there are cross terms.\nThe second average is a classical average,\nwhich is due to my ignorance.\n\nIf one is to describe it with QM,\ninstead of having kets,\nI'm going to have an operator called the density operator.\nIt's a bit different,\nbecause even the state is an operator.\nIf I don't know which state I'm in,\nsay there's a probability $p_n$ of being $\\ket{n}$ ,\nthen the density operator is\n\\begin{align}\n \\rho &= \\sum_{n} p_n \\ket{n}\\bra{n}\n\\end{align}\nThen the expectation value of $A$ is\n\\begin{align}\n \\bar{A} = \\Tr(\\rho A)\n\\end{align}\nThere are special states though.\nIf I know 100\\% that my system is in state $\\ket{\\psi}$,\nthen in that sum,\nthe probability is zero in every state except for $\\ket{\\psi}$,\nin which case the density operator is\n\\begin{align}\n \\rho = \\ket{\\psi}\\bra{\\psi}.\n\\end{align}\nAnd so the expectation value is\n\\begin{align}\n \\Tr(\\rho A) &=\n \\Tr(\\ket{\\psi}\\bra{\\psi} A)\\\\\n &= \\sum_n \\braket{n}{\\psi} \\bra{\\psi}A\\ket{n}\\\\\n &= \\sum_n \\bra{A}A\\ket{n}\\braket{n}{\\psi}\\\\\n &= \\bra{\\psi}A\\ket{\\psi}\\\\\n &= \\bar{A}\n\\end{align}\nwhere we made a sandwich of whatever basis $\\ket{n}$ we wanted\nand used the resolution of the identity.\nSo this reduces to the usual rule for pure states.\n\nBut what happens when we don't have a pure state,\nbut a mixed state,\nin which case the $\\Tr(\\rho A)$ rule applies more generally.\n\nIt's not obvious, but there is a way to test if a density operator is a pure\nstate.\nIn fact,\n$\\rho$\nis a pure state if and only if\n$\\rho^2 = \\rho$.\n\nAnd we also required that it be positive definite\n\\begin{align}\n \\bra{\\phi}\\rho \\ket{\\rho} \\ge 0.\n\\end{align}\nAlso,\n$\\Tr \\rho = 1$.\n\n\nExample: Thermodynamic ensemble\nSuppose we have a system in thermal equilibrium.\nWe don't know where the microscopic particles are,\nso there are certain probabilities of certain wave functions.\nTurns out if you take most systems and let them evolve,\nthe wave function keeps changing,\nbut everything behaves as if we were in a basis state.\nI'm going to point out the following here.\nIf I wait some time,\nand make macroscopic measurements of energy, pressure, etc.\nEverything behaves as if it were a density operator of the following form.\n\\begin{align}\n \\rho &= \\frac{e^{-\\beta H}}{\\Tr e^{-\\beta H}}\n\\end{align}\nwhere $H$ is Hermitian so $e^{-\\beta H} = 1 - \\beta H + \\cdots$\nis also Hermitian.\nYou can check that $H^2$ is hermitian with\n$(H^2)^\\dagger = (HH)^\\dagger = H^\\dagger H^\\dagger = H^2$.\nDon't be confused if you have an $i$ factor instead,\nwhere $e^{itH}$ is unitary,\nbut $e^{-\\beta H}$ is hermitian.\nWe have to divide by the real number $\\Tr e^{-\\beta H}$\nto make the trace 1,\nwhich is obvious,\nbecause the trace of the numerator cancels the denominator.\n\nBut is it positive definite?\nIt's not obvious,\nyou can work it out yourself,\nbut the eigenvalues of $H$ are real,\nand when you exponentiate a real number,\nyou get a positive number,\nso the eigenvalues of $\\rho$ must be positive.\n\nThis is a perfectly good density matrix.\nIt's sometimes written in a slightly different way.\nI could write $\\rho$ in the eigenbasis of energy\nwith matrix elements $\\bra{n}\\rho\\ket{m}$.\nIn other words $H\\ket{n} = E_n \\ket{n}$.\n\\begin{align}\n \\bra{n}\\rho\\ket{m} &=\n \\frac{\\bra{n} e^{-\\beta H}\\ket{m}}{\\Tr e^{-\\beta H}}\\\\\n &= \\frac{e^{-\\beta E_m} \\delta_{nm}}{\n \\sum_n \\bra{n} e^{-\\beta E_n}\\ket{n}\n }\\\\\n &= \\frac{e^{-\\beta E_m} \\delta_{nm}}{\n \\sum_n e^{-\\beta E_n}\n }.\n\\end{align}\nI'm sure you're familiar with the top\n$e^{-\\beta E_n}$\nwhich is just the Boltzmann factor.\nStates with large energy are unlikely,\nthe states that are more likely have smaller energy.\nLarge and small compared to what?\n\nRemember that this $\\beta E = \\frac{E}{k_B T}$.\nSo for the exponential to be small,\nwe need $E$ to be large.\nSo higher temperatures are more likely to have more likely higher energy states.\nWe'll see this a lot later on.\n\n\nExample: Photo polarization\nI can have light linearly polarized in $x$ and $y$ directions,\nand various superpositions like circular, left, right, etc.\nBut I can also have mixed states.\n\\begin{itemize}\n \\item 50\\% linearly polarized along $\\hat{x}$ direction and\n 50\\% linearly polarized along $\\ket{y}$ direction\n can be written as the state\n \\begin{align}\n \\rho =\n \\frac{1}{2}\\ket{\\hat{x}}\\bra{\\hat{x}}\n + \\frac{1}{2}\\ket{\\hat{y}}\\bra{\\hat{y}}\n \\end{align}\n and this is just a specific case of the more general formula\n \\begin{align}\n \\rho = \\sum_n p_n \\ket{n}\\bra{n}\n \\end{align}\n \\item Now consider the state with\n 50\\% probability of being left circularly polarized\n and 50\\% probability of being right circularly polarized.\n Let's write down the density matrix for that.\n \\begin{align}\n \\rho &=\n \\frac{1}{2} \\ket{L}\\bra{L}\n + \\frac{1}{2} \\ket{R}\\bra{R}\\\\\n &=\n \\frac{1}{2}\n \\underbrace{\n \\frac{\n \\ket{\\hat{x}} + i\\ket{\\hat{y}}\n }{\\sqrt{2}}\n }_{\\ket{L}}\n \\underbrace{\n \\frac{\n \\bra{\\hat{x}} - i\\bra{\\hat{y}}\n }{\\sqrt{2}}\n }_{\\bra{L}}\n + \\frac{1}{2}\n \\underbrace{\n \\frac{\n \\ket{\\hat{x}} - i\\ket{\\hat{y}}\n }{\\sqrt{2}}\n }_{\\ket{R}}\n \\underbrace{\n \\frac{\n \\bra{\\hat{x}} + i\\bra{\\hat{y}}\n }{\\sqrt{2}}\n }_{\\bra{R}}\\\\\n &= \\frac{1}{2}\\ket{\\hat{x}}\\bra{\\hat{x}}\n + \\frac{1}{2}\\ket{\\hat{y}}\\bra{\\hat{y}}\n \\end{align}\n so although the description in words they sound different,\n they are in fact the same state,\n and there is no experiment in the world that will allow you to\n differentiate between the two.\n\\end{itemize}\nSo the description in words in different,\nbut in reality they are different.\n\n\\subsection{Entropy (max)minimization}\nLet me point something about entropy because people talk about it so I should\npoint it out as soon as possible.\n\nYou can talk about the entropy of a mixed state.\nThere are two concepts of entropy closely related.\nWe can define entropy the following way.\nSuppose your system is defined by a density matrix $\\rho$\nthat might be a mixed state.\nYou can compute the following quantity\n\\begin{align}\n S &= - \\Tr \\left( \\rho \\ln \\rho \\right)\n\\end{align}\nBut first,\ndo you know how to define the log of an operator?\nTaylor expansion is one way?\nInverse of the exponential operator?\nHowever,\nif you can diagonalize the operator,\nyou can take any function you want,\nbecause you can take the function of any diagonal matrix,\nyou just take the function of each eigenvalue on the diagonal,\nno different from taking the exponential of an operator.\nWhy should we call this entropy?\nEntropy is something you learn in high school.\n\nFor that,\nlet me write this in a slightly different way.\nI'm going to take a basis and I'm going to write\n\\begin{align}\n S &=\n - \\sum_n\n \\bra{n} \\rho \\ln\\rho \\ket{n}\n\\end{align}\nwhere $\\ket{n}$ is an eigenbasis of $\\rho$.\nTo make things a little better,\nI can insert a complete set of states here.\n\\begin{align}\n S &=\n - \\sum_{n,n'}\n \\bra{n} \\rho \\ket{n'}\\bra{n'}\\ln\\rho \\ket{n}\n\\end{align}\nand so you notice that because $\\rho$ is diagonal in this basis,\n\\begin{align}\n S &= \\sum_{n,n'} p_{n'}\\delta_{nn'} \\delta_{nn'}\\ln p_n\\\\\n &= - \\sum_n p_n \\ln p_n\n\\end{align}\nThis expression you may have seen before in thermodynamics.\nWhy is this important?\nImagine if $\\rhO$ is a pure state.\nIt's a pure state when I write $\\rho$ as a sum like\n\\begin{align}\n \\rho &= \\sum_n p_n \\ket{n}\\bra{n} = \\ket{\\psi}\\bra{\\psi}\n\\end{align}\nso all the $p_n=0$ except for one $n=n_0$ where $p_{n_0}=1$.\nSo what does that mean?\nThere are two kinds of terms in the sum,\nterms where $p_n=0$,\nand only one term where $p_n=1$,\nso then\n\\begin{align}\n S &= - 0 \\ln 0 - 1\\ln 1 = 0\n\\end{align}\nSo if I have a pure state\nand I compute the entropy,\nthe entropy is zero.\n\n\\begin{question}\n What about superpositions?\n Do we get non-zero entropy?\n\\end{question}\nNo no no no!\nLet's take the harmonic oscillator\nand say we have an excited state like\n\\begin{align}\n \\rho &=\n \\underbrace{\\left( \\frac{3}{5}\\ket{0} + \\frac{4}{5}\\ket{1} \\right)}_{%\n \\ket{\\psi}\n }\n \\underbrace{\\left( \\frac{3}{5}\\bra{0} + \\frac{4}{5}\\ket{1} \\right)}_{%\n \\bra{\\psi}\n }\\\\\n &\\ne \\left( \\frac{3}{5} \\right)^2 \\ket{0}\\bra{0}\n + \\left( \\frac{4}{5} \\right)^2 \\ket{1}\\bra{1}\n\\end{align}\nbecause of the cross terms.\n\nAnyway,\nlet us consider the uniform distribution\n$p_n=1/N$\nand the dimension $N$ of the Hilbert space is finite.\nThen the entropy is\n\\begin{align}\n S &= -\\sum_n p_n\\ln p_n\n = -\\frac{N}{N}\\ln \\frac{1}{N} = \\ln N\n\\end{align}\nand it turns out that this is the maximum value the entropy can be.\nIt represents complete ignorance.\n\n\\begin{question}\n This is just the Shannon entropy of each hermitian operator viewed as a\n random variable?\n\\end{question}\nYes, I can't parse that precisely but sure.\n\n\\begin{question}\n Is $N$ the same thing as the number of basis states?\n\\end{question}\nYes, that's the definition of dimension.\n\n\\subsection{Subsystems and mixed states}\nThe idea is the following.\nImagine I have a system made up of two subsystems.\nIf the system is made up of two subsystems,\nthen the total space is the tensor product.\nThe generic state of a composite system can be written like this\n\\begin{align}\n \\ket{\\psi} &=\n \\sum_{n,m} \\psi_{n,m}\n \\ket{n}\\otimes\\ket{m}\n\\end{align}\nwhere $\\ket{n}\\otimes\\ket{m}$ forms the basis for the composite system\n$\\mathcal{H}$,\n$\\ket{n}$ forms a basis for $\\mathcal{H}_1$ and\n$\\ket{m}$ forms a basis for $\\mathcal{H}_2$.\n\nNow imagine if you have an operator $A$,\nbut I only want to measure one subsystem.\nThe operator $A\\otimes \\mathbf{1}$ does this.\n\\begin{align}\n A\\otimes \\mathbf{1} \\ket{\\phi}\\otimes\\ket{\\chi}\n &=\n A \\ket{\\phi} \\otimes \\ket{\\chi}\n\\end{align}\nIt does nothing to subsystem 2.\n\nSuppose I want to find an expectation value like this.\nBut I'm going to assume the system is in what is called an entangled state,\nsomething like this\n\\begin{align}\n \\ket{\\psi} =\n \\frac{%\n \\ket{\\phi_1}\\otimes\\ket{\\chi_1}\n + \\ket{\\phi_2}\\otimes\\ket{\\chi_2}\n }{\\sqrt{2}}\n\\end{align}\nThere is a correlation between particles 1 and 2.\nThis is called \\emph{entanglement}.\nIt's a perfectly good state,\nso let's see what happens.\n\\begin{align}\n \\bra{\\psi} A\\otimes\\mathbf{1} \\ket{\\psi} &=\n \\frac{%\n \\ket{\\phi_1}\\otimes\\ket{\\chi_1}\n + \\ket{\\phi_2}\\otimes\\ket{\\chi_2}\n }{\\sqrt{2}}\n A\\otimes\\mathbf{1}\n \\frac{%\n \\bra{\\phi_1}\\otimes\\bra{\\chi_1}\n + \\bra{\\phi_2}\\otimes\\bra{\\chi_2}\n }{\\sqrt{2}}\\\\\n &=\n \\frac{1}{2}\n \\bra{\\phi_1}A\\ket{\\phi_1}\n \\bra{\\chi_1}1\\ket{\\chi_1}\n +\n \\frac{1}{2}\n \\bra{\\phi_1}A\\ket{\\phi_2}\n \\bra{\\chi_1}1\\ket{\\chi_2}\\\\\\nonumber\n &\\qquad+\n \\frac{1}{2}\n \\bra{\\phi_2}A\\ket{\\phi_1}\n \\bra{\\chi_2}1\\ket{\\chi_1}\n +\n \\frac{1}{2}\n \\bra{\\phi_1}A\\ket{\\phi_1}\n \\bra{\\chi_2}1\\ket{\\chi_2}\\\\\n &=\n \\frac{1}{2}\\bra{\\phi_1}A\\ket{\\phi_1}\n + \\frac{1}{2}\\bra{\\phi_2}A\\ket{\\phi_2}\n\\end{align}\nBut notice that we have seen this before and it is the same as an incoherent sum\n\\begin{align}\n \\frac{1}{2}\\bra{\\phi_1}A\\ket{\\phi_1}\n + \\frac{1}{2}\\bra{\\phi_2}A\\ket{\\phi_2}\n =\\Tr\\left[\n \\underbrace{\\left(\n \\frac{1}{2}\\ket{\\phi_1}\\bra{\\phi_1}\n + \\frac{1}{2}\\ket{\\phi_2}\\bra{\\phi_2}\n \\right)}_{\\rho_1} A\n \\right]\n\\end{align}\nThe situation with a pure state over 2 systems is indistinguishable from having\na classical mixed state in system 1.\nIt's like you destroyed the interference!\nThis density matrix $\\rho_1$ you have here is only for subsystem 1\nIf I had no access to subsystem 2,\nlike if I moved it to another galaxy,\nall I would see is a classical mixed state in subsystem 1.\n\n[break]\n\nLet me repeat this same calculation in a slightly more general setting so you\nunderstand the general rule.\nSuppose you want to calculate the expectation value of a pure state of a\ncomposite system.\nThe most general composite state looks like this\n\\begin{align}\n \\sum_{n'm'}\\ket{n'}\\otimes\\ket{m'}\n\\end{align}\nThis is a little tricky,\nso I'm going to do it slowly.\n\\begin{align}\n \\bra{\\psi}A\\otimes\\mathbf{1}\\ket{\\psi}\n &=\n \\sum_{n',m',n,m}\n \\bra{n}\\otimes\\bra{m}\n \\psi_{nm}^*\n A\\otimes\\mathmbf{1}\n \\psi_{n'm'}\\ket{n'}\\otimes\\ket{m'}\\\\\n &=\n \\sum_{n',m',n,m}\n \\psi_{nm}^* \\psi_{n'm'}\n \\bra{n}A\\ket{n'}\n \\underbrace{\\bra{m}\\mathbf{1}\\ket{m'}}_{\\delta_{mm'}}\\\\\n &=\n \\sum_{nn'}\\bra{n}A\\ket{n'}\n \\sum_m\\psi_{nm}^* \\psi_{n'm}\\\\\n &= \\Tr\\left[\n \\underbrace{\\sum_{n,m,n'} \\psi_{nm}^* \\psi_{n'm}\\ket{n}\\bra{n'}}_{%\n \\rho_1\n }\n A\n \\right]\n\\end{align}\nNote that the proper way to compute a trace is to sandwich a complete basis on\neach side and sum over.\nUsing the cyclic property of the trace is not a valid way of doing it,\nbut it just happens to work in this case.\n\nTensor products look like a mouthful on paper,\nbut it's completely trivial to manipulate.\n\nAnyway,\nthe point of this is that $\\rho_1$ may be a mixed state of only subsystem 1.\nThat's the price you pay for not knowing about subsystem 2.\n\n\nLet's doe something exciting.\nShooting particles through holes and interference.\nPeople say that if you somehow have a way of looking which way the particle goes\nthrough,\njust by having the possibility of knowing,\nthe interference pattern gets destroyed.\nIt's a bit mysterious because of the following.\nIf you know the particle is left or right,\nyou make a measurement,\nand you collapse the system.\n\nIf it was a superposition,\nbecause of measurement,\nyou collapse it.\n\\begin{align}\n \\frac{\\ket{L} + \\ket{R}}{\\sqrt{2}}\n \\to\n \\ket{L}\n\\end{align}\nBut what does it even mean to make a measurement.\nSuppose you had a machine that automatically writes left or right,\nbut nobody looks at that piece of paper,\nis that enough to destroy the interference or not.\nLet me not use the collapse of the wave function,\nbut using the knowledge we just learnt.\n\nSuppose my state is a superposition\n\\begin{align}\n \\ket{\\psi_0} &=\n \\frac{\\ket{L} + \\ket{R}}{\\sqrt{2}}\n\\end{align}\nbut suppose I have a piece of equipment that spies on the particle.\nBut now I have both the particle and the spy\nas a composite system.\nMy spy system $\\ket{\\text{spy}}$ lives in a Hilbert space\n$\\mathcal{H}_{\\text{spy}}$,\nwhich could be quite complicated\nbecause it's a complicated device.\n\n\\begin{align}\n \\ket{\\psi_0} &=\n \\frac{\\ket{L} + \\ket{R}}{\\sqrt{2}}\n \\otimes\n \\ket{\\text{spy}}\n\\end{align}\nBut then it evolves according to a Hamiltonian $H$ and we get\n\\begin{align}\n \\ket{\\psi} &=\n \\frac{\\ket{L} + \\ket{R}}{\\sqrt{2}}\n \\otimes\n \\ket{\\text{spy}'}\n\\end{align}\nBut now it's evolved in a way completely independent of the particle,\nbut that's not useful because it's not spying!\nSo what we really want is for the spy to distinguish between left and right,\nso we something like this.\n\\begin{align}\n \\ket{\\psi} &=\n \\frac{1}{\\sqrt{2}}\n \\ket{L}\n \\otimes\n \\ket{\\text{spy,left}}\n +\n \\frac{1}{\\sqrt{2}}\n \\ket{R}\n \\otimes\n \\ket{\\text{spy,right}}\n\\end{align}\nThat is a 100\\% precise measurement by the spy device.\nNow I'm going to ask the following question.\nThe spy device going left or right is writen on the piece of paper,\nbut no one is going to look at it.\nIt's not really a measurement.\nNow suppose the paper is burnt,\ndestroyed.\n\nThe particles are still going to hit the screen,\nso what I'm going to be interested in doing is measure things,\nsay $A$,\nthat refer only to the particle,\nso the operator is $A\\otimes\\mathbf{1}$.\nIf I only measure things like this,\nthen I know everything that I can possibly measure about the system\nis contained in a density matrix obtained the same as before.\n\nTake the density matrix intially which is a pure state\n\\begin{align}\n \\rho &= \\ket{\\psi}\\bra{\\psi}\\\\\n &=\n \\sum_{n,m,n',m'} \\psi_{n'm'}\\ket{n'}\\otimes\\ket{m'}\\\\\n \\bra{n}\\otimes\\bra{m}\\psi_{nm}^*\\\\\n &\\to\n \\sum_{n,m,n'} \\psi_{nm}^* \\psi_{n'm}\\ket{n}\\bra{n'}\n = \\rho_1\n\\end{align}\nBut how did I get there?\nI had to take a trace over system 2.\n\\begin{align}\n \\sum_{m''}\\bra{m''}\\rho\\ket{m''} &=\n \\sum_{m''}\\bra{m''}\\ket{\\psi}\\bra{\\psi}\\ket{m''}\\\\\n &=\n \\sum_{m''}\\bra{m''}\\sum_{n,m,n',m'} \\psi_{n'm'}\\ket{n'}\\otimes\\ket{m'}\n \\bra{n}\\otimes\\bra{m}\\psi_{nm}^*\\ket{m''}\\\\\n &\\to\n \\sum_{n,m,n'} \\psi_{nm}^* \\psi_{n'm}\\ket{n}\\bra{n'}\n = \\rho_1\n\\end{align}\n\nThis procedure here,\nthere's a name of it,\nit's called taking the trace of subsystem 2.\n\\begin{align}\n \\Tr_2 \\rho = \\sum_{m''}\\bra{m''} \\rho \\ket{m''}\n\\end{align}\nwhere $\\ket{m''}$ form a basis for subsystem 2.\nYou will be left with a possibly mixed state $\\rho_1$ for only subsystem 1.\nThis is sometimes called the \\emph{partial trace},\nan operation that is very common\nand you should know this.\n\nLet's go back to the problem.\nSo we have the state for subsystem 1\n\\begin{align}\n \\rho_1 &=\n \\Tr_2 \\rho\\\\\n &=\n \\Tr_2\n \\left(\n \\frac{1}{\\sqrt{2}}\\ket{L}\\otimes\\ket{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\ket{R}\\otimes\\ket{\\text{spy,r}}\n \\right)\n \\left(\n \\frac{1}{\\sqrt{2}}\\bra{L}\\otimes\\bra{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\bra{R}\\otimes\\bra{\\text{spy,r}}\n \\right)\\\\\n &= \n \\bra{\\text{spy.l}}\n \\left(\n \\frac{1}{\\sqrt{2}}\\ket{L}\\otimes\\ket{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\ket{R}\\otimes\\ket{\\text{spy,r}}\n \\right)\n \\left(\n \\frac{1}{\\sqrt{2}}\\bra{L}\\otimes\\bra{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\bra{R}\\otimes\\bra{\\text{spy,r}}\n \\right)\n \\ket{\\text{spy.l}}\\\\\\nonumber\n &\\qquad+\n \\bra{\\text{spy.r}}\n \\left(\n \\frac{1}{\\sqrt{2}}\\ket{L}\\otimes\\ket{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\ket{R}\\otimes\\ket{\\text{spy,r}}\n \\right)\n \\left(\n \\frac{1}{\\sqrt{2}}\\bra{L}\\otimes\\bra{\\text{spy,l}}\n + \\frac{1}{\\sqrt{2}}\\bra{R}\\otimes\\bra{\\text{spy,r}}\n \\right)\n \\ket{\\text{spy.r}}\\\\\n &= \\frac{1}{2}\\ket{L}\\bra{L} + \\frac{1}{2}\\ket{R}\\bra{R}\n\\end{align}\nThis gives the density matrix of the particle\nand you've seen this before.\nIt describes the situation where you don't know where the particle is,\nand it could left or right.\nIn particular,\nIf I try to compute expectation values of some of the particles,\nI just get the usual thing of incoherent sum, no intereference.\n\\begin{align}\n \\bar{A} &=\n \\Tr \\rho_A\\\\\n &= \\frac{1}{2}\\bra{L}A\\ket{L}\n + \\frac{1}{2}\\bra{R}A\\ket{R}\n\\end{align}\nbut I'm now going to look at this piece of paper and it doesn't matter.\nThe interference pattern is gone.\nIt works whether it's written by the piece of paper or not.\nIt could be a single photon going through the particle,\nand that reveals the position of the particle,\nbecause the future state of the photon depends where the particle is.\nThat entanglement means I end up with a situation indistinguishable from a\ncoherent sum.\nA single photon going through the experiment\ncorrelating with the particle destroys the interference.\n\nWhat counts as a measurmenet?\none thing you can tell from our calculation.\nIt's enough to have anything entangling with the particle and you won't be able\nto see the intereference.\nit doesn't have to be a monkey, a grad student, or whatever,\na single photon correlation is enough.\n\n\\begin{question}\n You call it incoherent because it has no inteference terms?\n\\end{question}\nYes that's right, some lingo I introduced before.\n\n\\begin{question}\n What if we're not aware of the entanglement?\n Will it stil affect the system?\n\\end{question}\nIt doesn't matter what Paulo sees,\nany system entangling with the particle will destroy interference.\nDoesn't matter if I know about it or not.\n\n\nLet me make a comment.\nwhat this suggests is that to see intference effects is tricky in pratice.\nA single straight photon will destroy it.\nIt's not hard with light.\nBut if you tried doint it with electrons,\nit's hard.\nPhotons don't interact with each other,\nbut there's always light to interact with electrons.\nSo to do experiments with electrons is difficult.\n\nThere's great work recently with quantum devices of all kinds\nQuantum computers being the most famous ones.\nIt's why quantum computers are so hard to make.\nYour quantum coputer needs a hilbert space of some dimension,\nbut all the coherence will disappear if your qubits get entangled with the\noutside world,\nand that could be anything,\neven the metal frame that's holding it.\nYou have to make sure the whole thing that is supporting your device doesn't get\nentangled with it.\ncertainly you have to make sure it's dark,\nbecause a little photon can interact with it and it can entangle,\nand you get a mixed state.\nIt's a tremendous technological challange.\nIt's not hard to get this in the lab,\nit's how they discovered QM.\nBut instead of an atom,\nbut a bigger system,\ncertainly a macroscopic thing,\nit's hard.\nThere's awhole fued nowadays in physics,\nquantum registers or other systems,\ncould they keep a superposition.\nI won't be surprised if half of you spend half your life trying to do this\nstuff.\nAny other comments?\n\n\\begin{question}\n Future of quantum computing?\n\\end{question}\nI cannot predict the future,\nbut I can tell you\nI have not invested in quantum computing companies.\nI can tell you I was on the ground on Amazon,\nand I did not buy Amazon stock either.\nI was like this is stupid,\nwho wants to buy books,\nbooks are heavy.\nI did not buy Amazon stock,\nI'm not buying quantum computing stock.\n\nSomeone's comment is that the effect of cosmic rays in classical computers is a\nthing.\nIt may happen, it does happen.\nIt happens a lot in satellites though.\nCosmic rays are like a hammer,\nit cna destroy annything,\nincluding classical informaiton.\nWhat I'm saying is way worse than this.\nClassical computers don't even use error correcting codes.\nThey're so precise they don't hvae to.\nQuantum computers are a non-starter without quanutm error correction.\nI'm not an experimentalist,\nI don't know,\nbut I'm excitedly waiting for a quantum computer.\nThe difficulty is that you don't need a measurement to collapse a wave funciton,\nall you need is a bit of entangelmetn.\n\nThere are many other things that can entangle.\n\nWe have 9 minutes left.\nI have more stuff to talk about,\nbut about a little bit of a different topic.\n\n\\begin{question}\n Before yo urevisited the double slit expreiment,\n we were looking at the expectation of $A$.\n If we have the same $\\psi$ on both sides,\n why do we need separate indices on both sides $m'$ and $n'$.\n\\end{question}\nEvery time there's an issue with sums and indices,\nthe best thing to do is do an example.\nLet's say $n,m,\\ldots, = 1,2$.\nSay your state is\n\\begin{align}\n \\ket{\\psi} = \\frac{\\ket{1} + \\ket{2}}{\\sqrt{2}}\n\\end{align}\nThen we have\n\\begin{align}\n \\bra{\\psi}A\\ket{\\psi} &=\n \\frac{\\bra{1} + \\bra{2}}{\\sqrt{2}} A\n \\frac{\\ket{1} + \\ket{2}}{\\sqrt{2}}\\\\\n &=\n \\frac{1}{2} \\bra{1} A\\ket{1}\n + \\frac{1}{2} \\bra{2} A\\ket{2}\n + \\frac{1}{2} \\bra{1} A\\ket{2}\n + \\frac{1}{2} \\bra{2} A\\ket{1}\n\\end{align}\nGreat, let's write things in terms of indices now?\nIs this the same thing as\n\\begin{align}\n \\sum_{n=1}2\\bra{n}A\\ket{m}?\n\\end{align}\nClearly not,\nbut it is the same as\n\\begin{align}\n \\sum_{n,m=1}2\\bra{n}A\\ket{m}?\n\\end{align}\nYou can settle these issues by doing an example that's trivial,\nand it's immediately completely obvious.\n\n\\begin{question}\n If I think of that line conceptually as no intference,\n we have a particle and it got entangled with another system,\n and we lost informaitno about the other system,\n and is that why we don't have inteference.\n\\end{question}\nI agree with everything we said,\nbut it doesn't mean one expalins another.\nI find it completely mind-boggling\nthat I start with a pure state,\nand if I look at just a subsystem,\nand I get a mixed state.\nIf somehow saying those words makes you happy,\nthen everything is alrgith.\nYou're smarter than me.\nYeah, I don't know what to say.\nI don't find anything at all.\nI just got used to it.\n\nA week ago,\nif I told you this,\nyou'd say oh yeah correlaing wiht another particle,\nnitference should disapear.\n\nYou've heard the double slit experiment a million times,\nand yo've been told if you look at it it will disappear.\nDid you know that entangling is a measurment devices and they destroy\ninteference.\nWhen I was a sutdent this wasn't explained.\nEverything you said was right,\nthat's the only thing I can tell you.\n\n\\begin{question}\n Does the uniform distirbution have maximum entropy?\n But what's the distribution that maximises entropy for the haronic\n osciollator?\n\\end{question}\nYes.\nThere is no upper bound for the entropy of an infinite dimensinoal system.\nEntropy increases with temperature,\nand there is no bound for temperature.\nThe entropy does have a maximum for finite-state systems.\n\nTentative new office hours Friday 1pm.\n", "meta": {"hexsha": "3967134e8062d8279b102f2652a1be29d1385753", "size": 25621, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "phys612/lecture11.tex", "max_stars_repo_name": "ehua7365/umdphysnotes", "max_stars_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-12-11T12:53:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-11T12:53:46.000Z", "max_issues_repo_path": "phys612/lecture11.tex", "max_issues_repo_name": "ehua7365/umdphysnotes", "max_issues_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "phys612/lecture11.tex", "max_forks_repo_name": "ehua7365/umdphysnotes", "max_forks_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.0663329161, "max_line_length": 80, "alphanum_fraction": 0.6733148589, "num_tokens": 7662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.875787001374006, "lm_q2_score": 0.7154239897159438, "lm_q1q2_score": 0.6265590306643541}} {"text": "%!TEX root = ../OGUSAdoc.tex\n\nThree markets must clear in \\ogindia---the labor market, the capital market, and the goods market. By Walras' Law, we only need to use two of those market clearing conditions because the third one is redundant. In the model, we choose to use the labor market clearing condition and the capital market clearing condition, and to ignore the goods market clearing condition. But we present all three market clearing conditions here. Further, the redundant goods market clearing condition---sometimes referred to as the resource constraint---makes for a nice check on the solution method to see if everything worked.\n\nWe also characterize here the law of motion for total bequests $BQ_t$. Although it is not technically a market clearing condition, one could think of the bequests law of motion as the bequests market clearing condition.\n\n\\section{Market Clearing Conditions}\\label{SecMarkClrMktClr}\n\n Labor market clearing \\eqref{EqMarkClrLab} requires that aggregate labor demand $L_t$ measured in efficiency units equal the sum of household efficiency labor supplied $e_{j,s}n_{j,s,t}$.\n \\begin{equation}\\label{EqMarkClrLab}\n L_t = \\sum_{s=E+1}^{E+S}\\sum_{j=1}^{J} \\omega_{s,t}\\lambda_j e_{j,s}n_{j,s,t} \\quad \\forall t\n \\end{equation}\n Capital market clearing \\eqref{EqMarkClrCap} requires that aggregate capital demand from firms $K_t$ and from the government $D_t$ equal the sum of capital savings and investment by households $b_{j,s,t}$.\n \\begin{equation}\\label{EqMarkClrCap}\n K_t + D_t = \\sum_{s=E+2}^{E+S+1}\\sum_{j=1}^{J}\\Bigl(\\omega_{s-1,t-1}\\lambda_j b_{j,s,t} + i_s\\omega_{s,t-1}\\lambda_j b_{j,s,t}\\Bigr) \\quad \\forall t\n \\end{equation}\n Note that the capital demand side of the capital market clearing equation \\eqref{EqMarkClrCap} includes both capital demand by firms $K_t$ and capital demand by government $D_t$. It is here that we can see the potential of government deficits to crowd out investment.\n\n Aggregate consumption $C_t$ is defined as the sum of all household consumptions, and aggregate investment is defined by the resource constraint $Y_t = C_t + I_t + G_t$ as shown in \\eqref{EqMarkClrGoods}.\n \\begin{equation}\\label{EqMarkClrGoods}\n \\begin{split}\n Y_t &= C_t + K_{t+1} - \\biggl(\\sum_{s=E+2}^{E+S+1}\\sum_{j=1}^{J}i_s\\omega_{s,t}\\lambda_j b_{j,s,t+1}\\biggr) - (1-\\delta)K_t + G_t \\quad\\forall t \\\\\n &\\quad\\text{where}\\quad C_t \\equiv \\sum_{s=E+1}^{E+S}\\sum_{j=1}^{J}\\omega_{s,t}\\lambda_j c_{j,s,t}\n \\end{split}\n \\end{equation}\n\n Note that the extra terms with the immigration rate $i_s$ in the capital market clearing equation \\eqref{EqMarkClrCap} and the resource constraint \\eqref{EqMarkClrGoods} accounts for the assumption that age-$s$ immigrants in period $t$ bring with them (or take with them in the case of out-migration) the same amount of capital as their domestic counterparts of the same age. Note also that the term in parentheses with immigration rates $i_s$ in the sum acts is equivalent to a net exports term in the standard equation $Y=C+I+G+NX$. That is, if immigration rates are positive, then immigrants are bringing capital into the country and the term in parentheses has a negative sign in front of it. Negative exports are imports.\n\n\n\\section{Total Bequests Law of Motion}\\label{SecMarkClrBQ}\n\n Total bequests $BQ_t$ are the collection of savings of household from the previous period who died at the end of the period. These savings are augmented by the interest rate because they are returned after being invested in the production process.\n \\begin{equation}\\label{EqMarkClrBQ}\n BQ_{t} = (1+r_{t})\\left(\\sum_{s=E+2}^{E+S+1}\\sum_{j=1}^J\\rho_{s-1}\\lambda_j\\omega_{s-1,t-1}b_{j,s,t}\\right) \\quad\\forall t\n \\end{equation}\n Because the form of the period utility function in \\eqref{EqHHPerUtil} ensures that $b_{j,s,t}>0$ for all $j$, $s$, and $t$, total bequests will always be positive $BQ_{j,t}>0$ for all $j$ and $t$.\n", "meta": {"hexsha": "7a6c0ba18a512e64d3eada84d69dcf24376f0e17", "size": 3940, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/LaTeXsource/Chapters/Chap_MarkClr.tex", "max_stars_repo_name": "keshavchoudhary87/OG-India", "max_stars_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-17T19:49:22.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-17T19:49:22.000Z", "max_issues_repo_path": "docs/LaTeXsource/Chapters/Chap_MarkClr.tex", "max_issues_repo_name": "keshavchoudhary87/OG-India", "max_issues_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2019-08-16T15:40:52.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-16T07:07:15.000Z", "max_forks_repo_path": "docs/LaTeXsource/Chapters/Chap_MarkClr.tex", "max_forks_repo_name": "keshavchoudhary87/OG-India", "max_forks_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 44, "max_forks_repo_forks_event_min_datetime": "2019-08-16T15:10:39.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-08T07:03:26.000Z", "avg_line_length": 106.4864864865, "max_line_length": 728, "alphanum_fraction": 0.7494923858, "num_tokens": 1116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.839733983715524, "lm_q2_score": 0.7461389873857264, "lm_q1q2_score": 0.6265582642828831}} {"text": " \\documentclass[a4paper,11pt]{article}\n\n\\usepackage{fancyhdr}\n\\usepackage[dvips]{graphicx} % for eps and MatLab Diagrams\n\\usepackage{amsmath}\n\\usepackage{psfrag,color}\n\\usepackage[framed]{/Applications/TeX/mcode}\n\n\\pagestyle{fancy}\n\n\\begin{document} % Begin Document\n\n% Title\n\\title{\\textsc{DERIVEST}}\n\n% Authors and Contact Information\n\\author{\\textbf{John R. D'Errico}\\\\\nEmail: woodchips@rochester.rr.com}\n\n\\maketitle\n\n\\section{Introduction - Derivative Estimation}\n\nThe general problem of differentiation of a function typically pops up in\nthree ways in Matlab.\n\n\\begin{itemize}\n \\item The symbolic derivative of a function.\n \\item Compute numerical derivatives of a function defined only by a sequence of data points.\n \\item Compute numerical derivatives of a analytically supplied function.\n\\end{itemize}\n\nClearly the first member of this list is the domain of the symbolic toolbox, or some\nset of symbolic tools. Numerical differentiation of a function defined by data points\ncan be achieved with the function gradient, or perhaps by differentiation of a curve fit\nto the data, perhaps to an interpolating spline or a least squares spline fit.\n\nThe third class of differentiation problems is where \\mcode{DERIVEST} is valuable. This\ndocument will describe the methods used in \\mcode{DERIVEST}.\n\n\\bigskip\n\n\\section{Numerical differentiation of a general function of one variable}\n\nSurely you recall the traditional definition of a derivative, in terms of a limit.\n\n\\begin{equation} \\tag{1}\n f'(x) = \\lim_{\\delta \\to 0}{\\frac{f(x+\\delta) - f(x)}{\\delta}}\n\\end{equation}\n\nFor small $\\delta$, the limit approaches $f'(x)$. This is a one-sided approximation for\nthe derivative. For a fixed value of $\\delta$, this is also known as a finite difference\napproximation (a forward difference.) Other approximations for the derivative are also\navailable. We will see the origin of these approximations in the Taylor series expansion\nof a function $f(x)$ around some point $x_0$.\n\n\\begin{multline} \\tag{2}\n f(x) = f(x_0) + (x - x_0)f'(x_0) + \\frac{(x - x_0)^2}{2} f''(x_0) + \\\\\n \\frac{(x - x_0)^3}{6} f^{(3)}(x_0) + \\frac{(x - x_0)^4}{24} f^{(4)}(x_0) + \\\\\n \\frac{(x - x_0)^5}{120} f^{(5)}(x_0) + \\frac{(x - x_0)^6}{720} f^{(6)}(x_0) +...\n\\end{multline}\n\nTruncate the series in (2) to the first three terms, then form the forward difference\napproximation (1), where $x = x_0 + \\delta$.\n\n\\begin{equation} \\tag{3}\n f'(x_0) = \\frac{f(x_0+\\delta) - f(x_0)}{\\delta} - \\frac{\\delta}{2} f''(x_0) - \\frac{\\delta^2}{6} f'''(x_0) + ...\n\\end{equation}\n\nWhen $\\delta$ is small, $\\delta^2$ and any higher powers are vanishingly small. So we tend\nto ignore those higher powers, and describe the approximation in (3) as a \"first\" order\napproximation since the error in this approximation approaches zero at the same rate as the first power of $\\delta$. \\footnote{We would normally write these additional terms using O() notation, where all that matters is that the error term is $O(\\delta)$ or perhaps $O(\\delta^2)$, but explicit understanding of these error terms will be useful in the Romberg extrapolation step later on.} The values of $f''(x_0)$ and $f'''(x_0)$, while unknown to us, are fixed constants as $\\delta$ varies.\n\nHigher order approximations arise in the same fashion. The central difference (4) is a second\norder approximation.\n\n\\begin{equation} \\tag{4}\n f'(x_0) = \\frac{f(x_0+\\delta) - f(x_0-\\delta)}{2\\delta} - \\frac{\\delta^2}{3} f'''(x_0) + ...\n\\end{equation}\n\n\n\\bigskip\n\n\\section{Unequally spaced finite difference rules}\n\nWhile most finite difference rules used to differentiate a function will use equally spaced points,\nthis fails to be appropriate when one does not know the final spacing. Adaptive quadrature\nrules can succeed by subdividing each sub-interval as necessary. But an adaptive\ndifferentiation scheme must work differently, since differentiation is a point estimate.\n\\mcode{DERIVEST} generates a sequence of sample points that follow a log spacing away\nfrom the point in question, then it uses a single rule (generated on the fly) to estimate the\ndesired derivative. Because the points are log spaced, the same rule applies at any scale,\nwith only a scale factor applied.\n\n\n\\bigskip\n\n\\section{Odd and even transformations of a function}\n\nReturning to the Taylor series expansion of $f(x)$ around some point $x_0$, an even function \\footnote{An even function is one which expresses an even symmetry around a given point. An even symmetry has the property that $f(x) = f(-x)$. Likewise, an odd function expresses an odd symmetry, wherein $f(x) = -f(-x)$.} around $x_0$ must have all the odd order derivatives vanish at $x_0$. An odd function has all its even derivatives vanish from its expansion. Consider the derived functions $f_{odd}(x)$ and $f_{even}(x)$.\n\n\\begin{equation} \\tag{5}\nf_{odd}(x) = \\frac{f(x - x_0) - f(-x - x_0)}{2} \n\\end{equation}\n\nThe Taylor series expansion of $f_{odd}(x)$ has the useful property that we have killed off any even order terms, but the odd order terms are identical to $f(x)$, as expanded around $x_0$.\n\n\\begin{multline} \\tag{6}\nf_{odd}(x) = (x - x_0)f'(x_0) + \\frac{(x - x_0)^3}{6} f^{(3)}(x_0) + \\\\\n\\frac{(x - x_0)^5}{120} f^{(5)}(x_0) + \\frac{(x - x_0)^7}{5040} f^{(7)}(x_0) +...\n\\end{multline}\n\nLikewise, $f_{even}(x)$ has no odd order terms or a constant term, but other even order terms that are\nidentical to $f(x)$.\n\n\\begin{equation} \\tag{7}\nf_{even}(x) = \\frac{f(-x-x_0) - 2f(x_0) + f(x-x_0)}{2}\n\\end{equation}\n\n\\begin{multline} \\tag{8}\nf_{even}(x) = \\frac{(x - x_0)^2}{2} f^{(2)}(x_0) + \\frac{(x - x_0)^4}{24} f^{(4)}(x_0) + \\\\\n \\frac{(x - x_0)^6}{720} f^{(6)}(x_0) + \\frac{(x - x_0)^8}{40320} f^{(8)}(x_0) + ...\n\\end{multline}\n\nThe point of these transformations is we can rather simply generate a higher order approximation\nfor any odd order derivatives of $f(x)$ by working with $f_{odd}(x)$. Even order derivatives of $f(x)$ are similarly generated from $f_{even}(x)$. For example, a second order approximation for $f'(x_0)$ is trivially written in (9) as a function of $\\delta$.\n\n\\begin{equation} \\tag{9}\n f'(x_0; \\delta) = \\frac{f_{odd}(x_0 + \\delta)}{\\delta} - \\frac{\\delta^2}{6} f^{(3)}(x_0)\n\\end{equation}\n\nWe can do better rather simply, so why not? (10) shows a fourth order approximation for $f'(x_0)$.\n\n\\begin{equation} \\tag{10}\n f'(x_0; \\delta) = \\frac{8 f_{odd}(x_0+\\delta)-f_{odd}(x_0+2\\delta)}{6\\delta} + \\frac{\\delta^4}{30} f^{(5)}(x_0)\n\\end{equation}\n\nAgain, the next non-zero term (11) in that expansion has a higher power of $\\delta$ on it, so we\nwould normally ignore it since the lowest order neglected term should dominate the behavior\nfor small $\\delta$.\n\n\\begin{equation} \\tag{11}\n \\frac{\\delta^6}{252} f^{(7)}(x_0)\n\\end{equation}\n\n\\mcode{DERIVEST} uses similar approximations for all derivatives of $f$ up to the fourth order.\nOf course, its not always possible for evaluation of a function on both sides of a point, as central difference rules will require. In these cases, you can specify forward or backward difference rules\nas appropriate.\n\n\n\\bigskip\n\n\\section{Romberg extrapolation methodology applied to derivative estimation}\n\nSome individuals might suggest that the above set of approximations are entirely adequate for\nany sane person. Can we do better?\n\nSuppose we were to generate several different estimates of the approximation in (3) for\ndifferent values of $\\delta$ at a fixed $x_0$. Thus, choose a single $\\delta$, estimate a\ncorresponding resulting approximation to $f'(x_0)$, then do the same for $\\delta/2$.\nIf we assume that the error drops off linearly as $\\delta \\to 0$, then it is a simple matter\nto extrapolate this process to a zero step size. Our lack of knowledge of $f''(x_0)$ is\nirrelevant. All that matters is $\\delta$ is small enough that the linear term dominates so we\ncan ignore the quadratic term, therefore the error is purely linear.\n\n\\begin{equation} \\tag{12}\n f'(x_0) = \\frac{f(x_0+\\delta) - f(x_0)}{\\delta} - \\frac{\\delta}{2} f''(x_0)\n\\end{equation}\n\nThe linear extrapolant for this interval halving scheme as $\\delta \\to 0$ is given by (13).\n\n\\begin{equation} \\tag{13}\n f'_0 = 2f'_\\delta - f'_{\\delta/2}\n\\end{equation}\n\nSince I've always been a big fan of convincing myself that something will work before I\nproceed too far, lets try this out in Matlab. Consider the function $e^x$. Generate a pair of \napproximations to $f'(0)$, once at $\\delta$ of 0.1, and the second approximation at $1/2$\nthat value. Recall that $\\frac{d(e^x)}{dx} = e^x$, so at x = 0, the derivative should be\nexactly 1. How well will we do?\n\n\\begin{lstlisting}\n>> format long g\n\n>> f = @(x) exp(x);\n>> del = 0.1;\n\n>> df1 = (f(del) - f(0))/del\ndf1 =\n 1.05170918075648\n\n>> df2 = (f(del/2) - f(0))/(del/2)\ndf2 =\n 1.02542192752048\n\n>> 2*df2 - df1\nans =\n 0.999134674284488\n\\end{lstlisting}\n\nIn fact, this worked very nicely, reducing the error to roughly 1 percent of our initial estimates.\nShould we be surprised at this reduction? Not if we recall that last term in (3). We saw there that\nthe next term in the expansion was $O(\\delta^2)$. Since $\\delta$ was 0.1 in our experiment, that\n1 percent number makes perfect sense.\n\nThe Romberg extrapolant in (13) assumed a linear process, with a specific reduction in $\\delta$\nby a factor of 2. Assume the two term (linear + quadratic) residual term in (3), evaluating our approximation there with a third value of $\\delta$. Again, assume the step size is cut in half again.\nThe three term Romberg extrapolant is given by (14).\n\n\\begin{equation} \\tag{14}\n f'_0 = \\frac{1}{3}f'_\\delta - 2f'_{\\delta/2} + \\frac{8}{3}f'_{\\delta/4}\n\\end{equation}\n\nA quick test in matlab yields much better results yet.\n\n\\begin{lstlisting}\n>> format long g\n>> f = @(x) exp(x);\n>> del = 0.1;\n\n>> df1 = (f(del) - f(0))/del\ndf1 =\n 1.05170918075648\n\n>> df2 = (f(del/2) - f(0))/(del/2)\ndf2 =\n 1.02542192752048\n\n>> df3 = (f(del/4) - f(0))/(del/4)\ndf3 =\n 1.01260482097715\n\n>> 1/3*df1 - 2*df2 + 8/3*df3\nans =\n 1.00000539448361\n\\end{lstlisting} \n\nAgain, \\mcode{DERIVEST} uses the appropriate multiple term Romberg extrapolants for all derivatives\nof $f$ up to the fourth order. This, combined with the use of high order approximations for the derivatives, allows the use of quite large step sizes.\n\n\\bigskip\n\n\\section{Uncertainty estimates for DERIVEST}\n\nWe can view the Romberg extrapolation step as a polynomial curve fit in the step size parameter\n$\\delta$. Our desired extrapolated value is seen as simply the constant term coefficient in that polynomial model. Remember though, this polynomial model (see (10) and (11)) has only a few\nterms in it with known non-zero coefficients. That is, we will expect a constant term $a_0$, a term\nof the form $a_1 \\delta^4$, and a third term $a_2 \\delta^6$. \n\nA neat trick to compute the \"statistical\" uncertainty in the estimate of our desired derivative is to\nuse statistical methodology for that error estimate. While I do appreciate that there is nothing\ntruly statistical or stochastic in this estimate, the approach still works nicely, providing a very reasonable estimate in practice. A three term Romberg-like extrapolant, then evaluated at four distinct values for $\\delta$, will yield an estimate of the standard error of the constant term, with one spare degree of freedom. The uncertainty is then derived by multiplying that standard error by the appropriate percentile from the Students-t distribution.\n\n\\begin{lstlisting}\n>> tcdf(12.7062047361747,1)\nans =\n 0.975\n\\end{lstlisting}\n\nThis critical level will yield a two-sided confidence interval of 95 percent.\n\nThese error estimates are also of value in a difference sense. Since they are efficiently generated\nat all the different scales, the particular spacing which yields the minimum predicted error is chosen\nas the best derivative estimate. This has been shown to work consistently well. A spacing too large\ntends to have large errors of approximation due to the finite difference schemes used. But a too\nsmall spacing is bad also, in that we see a significant amplification of least significant fit errors\nin the approximation. A middle value generally seems to yield quite good results. For example,\n\\mcode{DERIVEST} will estimate the derivative of $e^x$ automatically. As we see, the final overall\nspacing used was 0.1953125.\n\n\\begin{lstlisting}\n>> [d,e,del]=derivest(@(x) exp(x),1)\nd =\n 2.71828182845904\ne =\n 1.02015503167879e-14\ndel =\n 0.1953125\n\\end{lstlisting}\n\nHowever, if we force the step size to be artificially large, then approximation error takes over.\n\n\\begin{lstlisting}\n>> [d,e,del]=derivest(@(x) exp(x),1,'FixedStep',10)\nd =\n 2.3854987890005\ne =\n 3.90016042034995\ndel =\n 10\n\\end{lstlisting}\n\nAnd if the step size is forced to be too small, then we see noise dominate the problem.\n\n\\begin{lstlisting}\n>> [d,e,del]=derivest(@(x) exp(x),1,'FixedStep',.0000000001)\nd =\n 2.71826406220403\ne =\n 0.000327191484277048\ndel =\n 1e-10\n\\end{lstlisting}\n\n\\mcode{DERIVEST}, like Goldilocks in the fairy tale bearing her name, stays comfortably in\nthe middle ground.\n\n\n\\bigskip\n\n\\section{DERIVEST in action}\n\nHow does \\mcode{DERIVEST} work in action? A simple nonlinear function with a well\nknown derivative is $e^x$. At $x = 0$, the derivative should be 1.\n\n\\begin{lstlisting}\n>> [d,err] = derivest(@(x) exp(x),0)\nd =\n 0.999999999999997\n\nerr =\n 2.22066469352214e-14\n\\end{lstlisting}\n\nA second simple example comes from trig functions. The first four derivatives of the sine\nfunction, evaluated at $x = 0$, should be respectively $[cos(0), -sin(0), -cos(0), sin(0)]$,\nor $[1,0,-1,0]$.\n\n\\begin{lstlisting}\n>> d = derivest(@(x) sin(x),0,1)\nd =\n 0.999999999999999\n\n>> d = derivest(@(x) sin(x),0,2)\nd =\n 0\n\n>> d = derivest(@(x) sin(x),0,3)\nd =\n -1.00000000000046\n\n>> d = derivest(@(x) sin(x),0,4)\nd =\n 0\n\\end{lstlisting}\n\n\n\\bigskip\n\n\\section{Gradient (\\mcode{GRADEST}) and Hessian (\\mcode{HESSIAN}) estimation}\n\nEstimation of the gradient vector (\\mcode{GRADEST}) of a function of multiple variables is a\nsimple task, requiring merely repeated calls to \\mcode{DERIVEST}. Likewise, the diagonal\nelements of the hessian matrix are merely pure second partial derivatives of a function.\n\\mcode{HESSDIAG} accomplishes this task, again calling \\mcode{DERIVEST} multiple\ntimes. Efficient computation of the off-diagonal (mixed partial derivative) elements of the\nHessian matrix uses a scheme much like that of \\mcode{DERIVEST}, wherein\n\\mcode{DERIVEST} is called to determine an initial step size, then Romberg extrapolation\nis used to improve a set of second order finite difference estimates of those mixed partials.\n\n\\bigskip\n\n\\section{Conclusion}\n\n\\mcode{DERIVEST} is an a adaptive scheme that can compute the derivative of arbitrary\n(well behaved) functions. It is reasonably fast as an adaptive method. Many options have\nbeen provided for the user who wishes the ultimate amount of control over the estimation.\n\n\n\\bigskip\n\n\\section{Acknowledgments}\n\nMy thanks are due to Shaun Simmons for convincing me to learn enough\nLaTeX to write this document.\n\n\\bigskip\n\n\n\\begin{thebibliography}{3}\n\n\\bibitem{LM66} Lyness, J. M., Moler, C. B. (1966).\n\\newblock Vandermonde Systems and Numerical Differentiation.\n\\newblock \\emph{Numerische Mathematik}.\n\n\\bibitem{LM69} Lyness, J. M., Moler, C. B. (1969).\n\\newblock Generalized Romberg Methods for Integrals of Derivatives.\n\\newblock \\emph{Numerische Mathematik}.\n\n\\bibitem{LM66} \\emph{NAG Library}.\n\\newblock NAG Fortran Library Document: D04AAF\n\n\\end{thebibliography}\n\n\n\\end{document}\n", "meta": {"hexsha": "663edaf0f732168c9c9d1ca29565c3639cf8b109", "size": 15690, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/src/DERIVEST.tex", "max_stars_repo_name": "rparini/numdifftools", "max_stars_repo_head_hexsha": "2c88878df732c9c6629febea56e7a91fd898398d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 181, "max_stars_repo_stars_event_min_datetime": "2015-01-21T07:16:45.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-14T20:35:10.000Z", "max_issues_repo_path": "docs/src/DERIVEST.tex", "max_issues_repo_name": "rparini/numdifftools", "max_issues_repo_head_hexsha": "2c88878df732c9c6629febea56e7a91fd898398d", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 58, "max_issues_repo_issues_event_min_datetime": "2015-01-21T11:46:55.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T21:22:31.000Z", "max_forks_repo_path": "docs/src/DERIVEST.tex", "max_forks_repo_name": "rparini/numdifftools", "max_forks_repo_head_hexsha": "2c88878df732c9c6629febea56e7a91fd898398d", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 43, "max_forks_repo_forks_event_min_datetime": "2015-02-22T18:03:23.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-21T15:48:42.000Z", "avg_line_length": 39.5214105793, "max_line_length": 520, "alphanum_fraction": 0.7205226259, "num_tokens": 4614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461390043208003, "lm_q2_score": 0.8397339596505965, "lm_q1q2_score": 0.6265582605480592}} {"text": "\\subsection{Classification of Riemann Surfaces}\r\nWe roughly classified all Riemann surfaces in the preceding section by viewing them as quotients of their universal covers, which can only be one of $\\mathbb C_\\infty,\\mathbb C,\\mathbb D$.\r\nWe say the surface is uniformised by its universal cover.\r\nFor some of these cases, we can do something better.\r\n\\begin{proposition}\r\n If a Riemann surface $R$ is uniformised by $\\mathbb C_\\infty$, then $R$ is conformally equivalent to $\\mathbb C_\\infty$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Suppose $R=G\\backslash\\mathbb C_\\infty$, then we know that $G$ acts by conformal equivalences $\\mathbb C_\\infty\\to\\mathbb C_\\infty$.\r\n But this is just M\\\"obius transformations (from Example Sheet).\r\n However, any M\\\"obius transformation has at least one fixed point, but $G$ should act freely, hence necessarily $G$ is trivial and hence $R\\cong\\mathbb C_\\infty$.\r\n\\end{proof}\r\nWhat about $\\mathbb C$?\r\n\\begin{proposition}\r\n If a Riemann surface $R$ is uniformised by $\\mathbb C$, i.e. $R\\cong G\\backslash\\mathbb C$, then one of the following holds:\\\\\r\n (i) $G$ is trivial and $R\\cong\\mathbb C$.\\\\\r\n (ii) $G\\cong\\mathbb Z$ and $R\\cong\\mathbb C_\\star$.\\\\\r\n (iii) $G\\cong\\mathbb Z^2$ and $R\\cong\\mathbb C/\\Lambda$ for a lattice $\\Lambda$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n The conformal automorphisms of $\\mathbb C$ are simply the (nonconstant) linear maps $\\{z\\mapsto az+b:a\\in\\mathbb C_\\star,b\\in\\mathbb C\\}$ (again from Example Sheet).\r\n Which of them can $G$ act by?\r\n Note that if $a\\neq 1$, then $z\\mapsto az+b$ has a fixed point, therefore $G$ can only consist of translations.\r\n But then (Example Sheet again!) $G$ (identified by the values of $b$) can only be one of:\\\\\r\n (i) Trivial.\\\\\r\n (ii) $\\langle\\omega\\rangle\\cong\\mathbb Z$ for some $\\omega\\neq 0$.\\\\\r\n (iii) a lattice $\\Lambda\\cong\\mathbb Z^2$.\\\\\r\n And these corresponds to the three situations as stated.\r\n\\end{proof}\r\nOh, by the way, a surface cannot be uniformised by more than one of $\\mathbb C_\\infty,\\mathbb C,\\mathbb D$.\r\n\\begin{lemma}[Lifting Lemma]\r\n Let $f:R\\to S$ be an analytic map of Riemann surfaces.\r\n Suppose $R$ is simply connected, and let $\\pi:\\tilde{S}\\to S$ be the uniformising map of $S$, then there is an analytic map $F:R\\to\\tilde{S}$ such that $f=\\pi\\circ F$.\r\n\\end{lemma}\r\nSo we have the commutative diagram\r\n\\[\r\n \\begin{tikzcd}\r\n &\\tilde{S}\\arrow{d}{\\pi}\\\\\r\n R\\arrow[dashed]{ur}{F}\\arrow[swap]{r}{f}&S\r\n \\end{tikzcd}\r\n\\]\r\n\\begin{proof}\r\n Example Sheet.\r\n\\end{proof}\r\n\\begin{proposition}\r\n A Riemann surface $R$ is uniformised by at most one of $\\mathbb C_\\infty,\\mathbb C,\\mathbb D$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n By the previous discussion, we already know everything that is uniformised by $\\mathbb C$ or $\\mathbb C_\\infty$ and they are distinct.\r\n Now suppose $R$ is uniformised by $\\mathbb D$ and $\\tilde{R}$ which is either $\\mathbb C$ or $\\mathbb C_\\infty$.\r\n Let $\\pi,f$ be the respective uniformisation maps, then by the preceding lemma, there is $F:\\tilde{R}\\to\\mathbb D$ such that\r\n \\[\r\n \\begin{tikzcd}\r\n &\\mathbb D\\arrow{d}{\\pi}\\\\\r\n \\tilde{R}\\arrow[swap]{r}{f}\\arrow{ur}{F}&R\r\n \\end{tikzcd}\r\n \\]\r\n commutes.\r\n But then $F$ has to be constant by Liouville's Theorem, hence $f$ is also constant, contradiction.\r\n\\end{proof}\r\nSo any other Riemann surface must be uniformised by $\\mathbb D$.\r\n\\begin{proposition}\r\n Any conformal automorphisms of $\\mathbb D$ is in the form\r\n $$z\\mapsto e^{i\\theta}\\frac{z-a}{1-\\bar{a}z},a\\in\\mathbb C,\\theta\\in\\mathbb R$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n Complex Analysis.\r\n\\end{proof}\r\nThis is perhaps easier to picture if we send $\\mathbb D$ to the open upper half plane $\\mathbb H$ (by the M\\\"obius transformation $z\\mapsto (1+iz)/(1-iz)$), which has automorphisms in the form $z\\mapsto (az+b)/(cz+d)$ for $a,b,c,d\\in\\mathbb R,ad-bc=1$.\r\n\\begin{definition}\r\n A subgroup of $\\operatorname{PSL}(\\mathbb R)$ that acts properly discontinuously on $\\mathbb H$ is called a Fuchsian group.\r\n\\end{definition}", "meta": {"hexsha": "f3ba2dc5db6500e7c999d3e6aa0a2cf581775c87", "size": 4126, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "16/class.tex", "max_stars_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_stars_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": 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YES\n2. YES", "lm_q1_score": 0.766293653760418, "lm_q2_score": 0.817574478416099, "lm_q1q2_score": 0.6265021342867405}} {"text": "\\subsection{Abelian groups}\\label{subsec:abelian_groups}\n\n\\begin{definition}\\label{def:abelian_group}\n A \\hyperref[def:magma/commutative]{commutative} \\hyperref[def:group]{group} is usually called an \\term{abelian group}. We denote by \\( \\cat{Ab} \\) the category of abelian groups.\n\n By \\fullref{thm:ring_is_integer_algebra}, the abelian groups are precisely the \\hyperref[def:algebra_over_ring]{rings} over \\( \\BbbZ \\), and we have an \\hyperref[rem:category_similarity/isomorphism]{isomorphism of categories} \\( \\cat{Ab} \\cong \\cat{Mod}_\\BbbZ \\).\n\\end{definition}\n\n\\begin{remark}\\label{rem:additive_magma}\n General groups often arise as \\hyperref[def:automorphism_group]{automorphism groups}, which are, for the most part, non-commutative, while abelian groups are usually used as the main building block for \\hyperref[def:ring]{rings} and \\hyperref[def:module]{modules}.\n\n To make a further distinction, if the operation is denoted by \\( \\cdot \\) or juxtaposition, we say that the group is a \\term{multiplicative group}, and if the operation is denoted by \\( + \\), we say that the group is an \\term{additive group}. This terminology usually, but not necessarily, coincides with the group (or, more generally, the \\hyperref[def:magma]{magma}) being \\hyperref[def:magma/commutative]{commutative}.\n\n To make things explicit, a \\term{multiplicative magma} is any magma as defined in \\fullref{def:magma}. Compare this to \\term{additive magmas}, where\n \\begin{thmenum}\n \\thmitem{rem:additive_magma/addition} The magma operation is denoted by \\( + \\) and called \\term{addition}.\n\n \\thmitem{rem:additive_magma/multiplication} The magma \\hyperref[def:magma/exponentiation]{exponentiation operation} \\( x^n \\) is denoted by \\( n \\cdot x \\) or juxtaposition and called \\term{multiplication}. Thus, multiplication is not defined for two elements of the magma, but defined for a positive integer and an element of the magma. That is,\n \\begin{equation}\\label{eq:rem:additive_magma/multiplication}\n \\begin{aligned}\n &\\cdot: \\cdot: \\BbbN \\times R \\to R \\\\\n &n \\cdot x \\coloneqq \\begin{cases}\n 0_M, &n = 0, \\T{initial condition if} M \\T{is a monoid} \\\\\n x, &n = 1, \\T{initial condition if} M \\T{is not a monoid} \\\\\n n \\cdot x + x, &n > 1 \\\\\n -(n \\cdot x), &n < 0, \\\\\n \\end{cases}\n \\end{aligned}\n \\end{equation}\n\n In the case of a \\hyperref[def:magma/commutative]{commutative} \\hyperref[def:monoid]{monoid}, if multiplication is extended to two elements of the monoid, we instead talk about \\hyperref[def:semiring]{semirings}.\n\n \\thmitem{rem:additive_magma/identity} The \\hyperref[def:monoid]{identity} is usually denoted by \\( 0 \\).\n\n \\thmitem{rem:additive_magma/inverse} If an \\hyperref[def:monoid_inverse]{inverse} of \\( x \\) exists, it is denoted by \\( -x \\) rather than \\( x^{-1} \\).\n \\end{thmenum}\n\\end{remark}\n\n\\begin{proposition}\\label{thm:abelian_outer_automorphism_group}\n In an \\hyperref[def:abelian_group]{abelian group}, the full \\hyperref[def:automorphism_group]{automorphism group} \\( \\aut(G) \\) is isomorphic to the \\hyperref[def:inner_and_outer_automorphisms]{outer automorphism group} \\( \\op{out}(G) \\).\n\\end{proposition}\n\\begin{proof}\n If the group operation is \\hyperref[def:magma/commutative]{commutative}, then \\( xyx^{-1} = yxx^{-1} = y \\), which makes the \\hyperref[def:inner_and_outer_automorphisms]{conjugation action} trivial. Thus, the \\hyperref[def:inner_and_outer_automorphisms]{inner automorphism group} \\( \\op{int}(G) \\) is trivial, and hence \\( \\aut(G) \\cong \\op{out}(G) \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:abelian_normal_subgroups}\n All subgroups of an abelian group are \\hyperref[thm:normal_subgroup_equivalences]{normal}.\n\\end{proposition}\n\\begin{proof}\n Let \\( G \\) be abelian and \\( H \\) be a subgroup of \\( G \\). Then \\( x H x^{-1} = xx^{-1} H = H \\) for any \\( x \\in G \\) and thus \\( H \\) is normal.\n\\end{proof}\n\n\\begin{definition}\\label{def:congruence_modulo_normal_subgroup}\n Given a \\hyperref[thm:normal_subgroup_equivalences]{normal subgroup} \\( N \\) of an \\hyperref[def:abelian_group]{abelian group} \\( G \\), we say that two elements \\( x \\) and \\( y \\) of \\( G \\) are \\term{congruent modulo} \\( N \\) and write \\( x \\cong y \\pmod N \\) if \\( x - y \\in N \\).\n\n If \\( N = \\braket{ z } \\), this implies that \\( x \\cong y \\pmod z \\) if and only if \\( x - y \\in \\braket{ z } \\).\n\n This concept also extends to \\hyperref[def:semiring_ideal]{ring ideals} rather than normal subgroups, in which case \\( \\braket{ z } \\) is the \\hyperref[def:semiring_ideal/generated]{ideal generated} by \\( z \\) rather than the \\hyperref[def:cyclic_group]{cyclic subgroup} of \\( z \\).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:group_of_integers_modulo}\n The \\hyperref[def:set_of_integers]{integers} \\( \\BbbZ \\) form an abelian group under addition. For every positive integer \\( n \\), we define the group\n \\begin{equation*}\n \\BbbZ_n \\coloneqq \\set{ 0, 1, \\ldots, n - 1 }\n \\end{equation*}\n with the operation\n \\begin{equation*}\n x \\oplus y \\coloneqq \\rem(x + y, n)\n \\end{equation*}\n so that\n \\begin{equation*}\n x \\oplus y \\cong x + y \\pmod n.\n \\end{equation*}\n\n The group \\( \\BbbZ_n \\) is called the \\term{group of integers modulo} \\( n \\). Compare this result with \\fullref{thm:ring_of_integers_modulo}.\n\\end{proposition}\n\\begin{proof}\n We will prove that \\( \\BbbZ_n \\) is an abelian group.\n\n \\SubProofOf[def:magma/associative]{associativity} Addition in \\( \\BbbZ_n \\) is associative since\n \\begin{balign*}\n (x \\oplus y) \\oplus z\n &=\n \\rem((x \\oplus y) + z, n)\n = \\\\ &=\n \\rem(\\rem(x + y, n) + z, n)\n = \\\\ &=\n \\rem(x + y - n \\quot(x + y, n) + z, n)\n = \\\\ &=\n \\rem(x + y + z, n)\n = \\\\ &=\n \\ldots\n = \\\\ &=\n x \\oplus (y \\oplus z).\n \\end{balign*}\n\n \\SubProofOf[def:monoid]{identity} The zero is the identity.\n\n \\SubProofOf[def:monoid_inverse]{inverse} Fix \\( x \\in \\BbbZ_n \\). If \\( x = 0 \\), its inverse is \\( 0 \\). If \\( x > 0 \\), its inverse is \\( n - x \\) since \\( n - x \\in \\BbbZ_n \\) and\n \\begin{equation*}\n x \\oplus (n - x) = x + (n - x) - n = 0.\n \\end{equation*}\n\n \\SubProofOf[def:magma/commutative]{commutativity} Follows from\n \\begin{equation*}\n x \\oplus y\n =\n \\rem(x + y, n)\n =\n \\rem(y + x, n)\n =\n y \\oplus x.\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:integers_modulo_isomorphic_to_quotient_group}\n The group \\( \\BbbZ_n \\) of \\hyperref[thm:group_of_integers_modulo]{integers modulo \\( n \\)} is isomorphic to the quotient of \\( \\BbbZ \\) by \\( n\\BbbZ = \\set{ nz \\given z \\in \\BbbZ } \\). That is,\n \\begin{equation*}\n \\BbbZ_n \\cong \\BbbZ / n\\BbbZ.\n \\end{equation*}\n\\end{proposition}\n\\begin{proof}\n Define the function\n \\begin{align*}\n &\\varphi: \\BbbZ_n \\to \\BbbZ / n\\BbbZ \\\\\n &\\varphi(x) \\coloneqq x + n\\BbbZ.\n \\end{align*}\n\n It is a homomorphism because\n \\begin{balign*}\n \\varphi(x \\oplus y)\n &=\n \\varphi(\\rem(x + y, n))\n = \\\\ &=\n \\varphi(x + y - n \\quot(x + y, n))\n = \\\\ &=\n x + y - n \\quot(x + y, n) + n\\BbbZ\n = \\\\ &=\n x + y + n\\BbbZ\n = \\\\ &=\n (x + n\\BbbZ) + (y + n\\BbbZ)\n = \\\\ &=\n \\varphi(x) + \\varphi(y).\n \\end{balign*}\n\n Furthermore, this shows that \\( \\varphi \\) is also an isomorphism.\n\\end{proof}\n\n\\begin{example}\\label{ex:lagranges_theorem_for_groups/direct_product_zn}\n \\Fullref{thm:lagranges_theorem_for_groups} and \\fullref{thm:integers_modulo_isomorphic_to_quotient_group} imply that, for any positive integer \\( n \\), \\( (nm, k) \\mapsto nm + k \\) is a bijection between \\( n \\BbbZ \\times \\BbbZ_n \\) and \\( \\BbbZ \\). This bijection, however, is not necessarily a group isomorphism because \\eqref{eq:def:magma/homomorphism} may not hold.\n\n Consider the tuples \\( (nm_1, k_1) \\) and \\( (nm_2, k_2) \\) in \\( n \\BbbZ \\times \\BbbZ_n \\). We have\n \\begin{equation*}\n (nm_1, k_1) + (nm_2, k_2) = (nm_1 + nm_2, \\rem(k_1 + k_2, n)).\n \\end{equation*}\n\n Therefore, if \\( k_1 + k_2 \\geq n \\),\n \\begin{equation*}\n nm_1 + nm_2 + \\rem(k_1 + k_2, n) < (nm_1 + k_1) + (nm_2 + k_2).\n \\end{equation*}\n\\end{example}\n\n\\begin{proposition}\\label{thm:cyclic_group_isomorphic_to_integers_modulo_n}\n The \\hyperref[def:cyclic_group]{cyclic group} \\( C_n \\) is isomorphic to the group \\hyperref[thm:group_of_integers_modulo]{\\( \\BbbZ_n \\)} of integers modulo \\( n \\).\n\\end{proposition}\n\\begin{proof}\n The homomorphism\n \\begin{equation*}\n \\begin{aligned}\n &\\varphi: \\BbbZ_n \\to C_n \\\\\n &\\varphi(k) \\coloneqq a^k,\n \\end{aligned}\n \\end{equation*}\n and the analogous homomorphism for the infinite group, are isomorphisms.\n\\end{proof}\n\n\\begin{definition}\\label{def:monoid_grothendieck_completion}\\mcite{nLab:grothendieck_group_of_a_commutative_monoid}\n Let \\( M \\) be a \\hyperref[def:magma/commutative]{commutative} \\hyperref[def:monoid]{monoid}. Define the \\hyperref[def:equivalence_relation]{equivalence relation} \\( \\sim \\) on tuples of members of \\( M \\) to hold for \\( (a, b) \\sim (a', b') \\) if there exists an element \\( u \\) of \\( M \\) such that\n \\begin{equation*}\n a + b' + u = a' + b + u.\n \\end{equation*}\n\n Define addition on the \\hyperref[thm:equivalence_partition]{equivalence partition} \\( G \\coloneqq (M \\times M) / {\\sim }\\) componentwise as\n \\begin{equation*}\n [(a, b)] \\oplus [(c, d)] \\coloneqq [(a + c, b + d)]\n \\end{equation*}\n and fix a canonical embedding\n \\begin{equation*}\n \\begin{aligned}\n &\\iota_M: M \\to G \\\\\n &\\iota_M(m) \\coloneqq [(m, 0)].\n \\end{aligned}\n \\end{equation*}\n\n We call the obtained \\hyperref[def:abelian_group]{abelian group} \\( (G, \\oplus) \\) the \\term{Grothendieck completion} of \\( M \\).\n\\end{definition}\n\\begin{defproof}\n \\SubProof{Proof that \\( \\sim \\) is an equivalence relation}\n \\SubProofOf*[def:binary_relation/reflexive]{reflexivity}\n \\begin{equation*}\n (a, b) \\sim (a, b) \\T{if and only if} a + b + 0 = a + b + 0\n \\end{equation*}\n\n \\SubProofOf*[def:binary_relation/symmetric]{symmetry} By commutativity, if \\( (a, b) \\sim (a', b') \\), then there exists \\( u \\) such that\n \\begin{equation*}\n a + b' + u = a' + b + u\n =\n a' + b + u = a + b' + u,\n \\end{equation*}\n hence \\( (a', b') \\sim (a, b) \\).\n\n \\SubProofOf*[def:binary_relation/transitive]{transitivity} Suppose that \\( (a, b) \\sim (a', b') \\) and \\( (a', b') \\sim (a^\\dprime, b^\\dprime) \\). Thus, there exist elements \\( u \\) and \\( b \\) of \\( M \\) such that\n \\begin{align*}\n a + b' + u &= a' + b + u, \\\\\n a' + b^\\dprime + v &= a^\\dprime + b' + v.\n \\end{align*}\n\n Summing both sides, we obtain\n \\begin{equation*}\n (a + b' + u) + (a' + b^\\dprime + v) = (a' + b + u) + (a^\\dprime + b' + v)\n \\end{equation*}\n\n We reorder both sides to obtain\n \\begin{equation*}\n (a + b^\\dprime) + (a' + b' + u + v) = (a^\\dprime + b) + (a' + b' + u + v),\n \\end{equation*}\n which implies \\( (a, a^\\dprime) \\sim (b, b^\\dprime) \\).\n\n \\SubProof{Proof that \\( (G, \\oplus) \\) is an abelian group}\n\n \\SubProof*{Proof that \\( \\oplus \\) is well-defined} The addition operation on \\( G \\) does not depend on the representative of the equivalence class. Indeed, let \\( (a, b) \\sim (a', b') \\) and \\( (c, d) \\sim (c', d') \\). Then there exist \\( u \\) and \\( b \\) such that\n \\begin{align*}\n a + b' + u &= a' + b + u, \\\\\n c + d' + v &= c' + d + v.\n \\end{align*}\n\n When added combined, these give\n \\begin{equation*}\n (a + c) + (b' + d') + (u + v)\n =\n (a' + c') + (b + d) + (u + v),\n \\end{equation*}\n which implies that\n \\begin{equation*}\n (a + c, b + d) \\sim (a' + c', b' + d').\n \\end{equation*}\n\n \\SubProofOf*[def:magma/associative]{associativity} Associativity of multiplication in \\( G \\) is inherited from multiplication in \\( M \\).\n\n \\SubProofOf*[def:monoid]{identity} The equivalence class \\( [(0, 0)] \\) is an identity in \\( G \\) and contains the pairs \\( (x, x) \\) of identical elements.\n\n \\SubProofOf*[def:monoid_inverse]{inverse} For each member \\( (a, b) \\in M \\times M \\), its inverse is \\( (b, a) \\) because\n \\begin{equation*}\n [(a, b)] \\oplus [(b, a)] = [(a + b, b + c)],\n \\end{equation*}\n which, by commutativity, belongs to \\( [(0, 0)] \\).\n\n \\SubProofOf*[def:magma/commutative]{commutativity} Commutativity of the group operation \\( \\oplus \\) is also inherited from the monoid operation \\( + \\).\n\\end{defproof}\n\n\\begin{theorem}[Grothendieck monoid completion universal property]\\label{thm:grothendieck_monoid_completion_universal_property}\\mcite{nLab:grothendieck_group_of_a_commutative_monoid}\n The \\hyperref[def:monoid_grothendieck_completion]{Grothendieck completion} \\( \\overline{M} \\) of a commutative monoid \\( M \\) satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every abelian group \\( G \\) and every monoid homomorphism \\( \\varphi: M \\to G \\), there exists a unique group homomorphism \\( \\widetilde{\\varphi}: \\overline{M} \\to G \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:grothendieck_monoid_completion_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__monoid_grothendieck_completion_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Via \\fullref{rem:universal_mapping_property}, \\( \\overline{\\anon} \\) becomes \\hyperref[def:category_adjunction]{left adjoint} to the \\hyperref[def:concrete_category]{forgetful functor}\n \\begin{equation*}\n U: \\cat{Ab} \\to \\cat{CMon}.\n \\end{equation*}\n\n Compare this result to \\fullref{thm:grothendieck_semiring_completion_universal_property}.\n\\end{theorem}\n\\begin{proof}\n Let \\( \\varphi: M \\to G \\) be a monoid homomorphism into an abelian group \\( G \\). We want to define a homomorphism \\( \\overline{\\varphi} \\) such that\n \\begin{equation*}\n \\overline{\\varphi}(\\iota_M(a)) = \\overline{\\varphi}([(a, 0)]) = \\varphi(a).\n \\end{equation*}\n\n Each equivalence class \\( C \\) in \\( G \\) has a unique member \\( a \\) such that \\( (a, 0) \\in C \\), hence the above condition is well-posed.\n\n Fix pairs \\( (a, b) \\) and \\( (a', b') \\) from \\( M \\times M \\). Suppose that \\( (a, b) \\sim (a', b') \\). Then there exists \\( u \\in M \\) such that\n \\begin{equation*}\n a + b' + u = a' + b + u.\n \\end{equation*}\n\n An additional restriction on \\( \\overline{\\varphi} \\) is then\n \\begin{equation*}\n \\overline{\\varphi}\\parens[\\Big]{ [(a, b)] }\n =\n \\overline{\\varphi}\\parens[\\Big]{ [(a', b')] }.\n \\end{equation*}\n\n We need to cancel out \\( u \\). This uniquely determines \\( \\overline{\\varphi} \\) as\n \\begin{equation*}\n \\overline{\\varphi}([(a, b)]) \\coloneqq \\varphi(a) - \\varphi(b).\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:group_commutator}\\mcite[313]{Knapp2016BasicAlgebra}\n Let \\( G \\) be an arbitrary group. We define the \\term{commutator} of the elements \\( x \\) and \\( y \\) as\n \\begin{equation*}\n [x, y] \\coloneqq xyx^{-1}y^{-1}.\n \\end{equation*}\n\n The \\term{commutator subgroup} \\( [G, G] \\) of \\( G \\) is the subgroup \\hyperref[def:group/submodel]{generated} by all the commutators in \\( G \\).\n\\end{definition}\n\n\\begin{theorem}[Group abelianization universal property]\\label{thm:group_abelianization_universal_property}\\mcite[prop. 7.4]{Knapp2016BasicAlgebra}\n The commutator group \\( [G, G] \\) of any group \\( G \\) is \\hyperref[thm:normal_subgroup_equivalences]{normal} and the quotient \\( G / [G, G] \\) is an abelian group, which we call the \\term{abelianization} of \\( G \\), satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every abelian group \\( H \\), every group homomorphism \\( \\varphi: G \\to H \\) \\hyperref[def:factors_through]{uniquely factors through} \\( G / [G, G] \\). That is, there exists a unique group homomorphism \\( \\widetilde{\\varphi}: G / [G, G] \\to H \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:group_abelianization_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__group_abelianization_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Via \\fullref{rem:universal_mapping_property}, the abelianization functor becomes \\hyperref[def:category_adjunction]{left adjoint} to the \\hyperref[def:concrete_category]{forgetful functor}\n \\begin{equation*}\n U: \\cat{Ab} \\to \\cat{Grp}.\n \\end{equation*}\n\n This result extends to \\fullref{thm:ring_abelianization_universal_property}.\n\\end{theorem}\n\\begin{proof}\n Let \\( C \\coloneqq [G, G] \\).\n\n \\SubProof{Proof that \\( G / C \\) is abelian} Normality of \\( G / C \\) easily follows from\n \\begin{equation*}\n a xyx^{-1}y^{-1} a^{-1}\n =\n (a x a^{-1}) (a y a^{-1}) (a x a^{-1})^{-1} (a y a^{-1})^{-1}.\n \\end{equation*}\n\n Then for the cosets \\( a C \\) and \\( b C \\), we have\n \\begin{equation*}\n a C \\cdot b C\n =\n a b C\n =\n a b (b^{-1} a^{-1} b a) C\n =\n b a C.\n \\end{equation*}\n\n Therefore, the quotient group \\( G / C \\) is abelian.\n\n \\SubProof{Proof of universal mapping property} Let \\( H \\) be an abelian group and let \\( \\varphi: G \\to H \\) be a group homomorphism.\n\n Observe that \\( \\varphi(C) = e_H \\). Indeed, since \\( H \\) is abelian, for \\( [x, y] = xyx^{-1}y^{-1} \\in C \\) we have\n \\begin{equation*}\n \\varphi([x, y]) = \\varphi(x) \\varphi(y) \\varphi(x^{-1}) \\varphi(y^{-1}) = \\varphi(x) \\varphi(x^{-1}) \\varphi(y) \\varphi(y^{-1}).\n \\end{equation*}\n\n We want \\( \\overline{\\varphi}: G / C \\to H \\) to satisfy\n \\begin{equation*}\n \\overline{\\varphi}(\\underbrace{\\pi_G(x)}_{xC}) = \\varphi(x).\n \\end{equation*}\n\n This suggests the definition\n \\begin{equation*}\n \\overline{\\varphi}(xC) \\coloneqq \\varphi(x).\n \\end{equation*}\n\n It is well-defined because if \\( xC = yC \\), we have\n \\begin{equation*}\n \\varphi(x)\n =\n \\varphi(x) e_H\n =\n \\varphi(x) \\varphi(C)\n =\n \\varphi(x C)\n =\n \\varphi(y C)\n =\n \\ldots\n =\n \\varphi(y).\n \\end{equation*}\n\\end{proof}\n", "meta": {"hexsha": "0c98e32035c42df0c000625a05a5fbd04f5fda4a", "size": 18014, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/abelian_groups.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/abelian_groups.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/abelian_groups.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.3084832905, "max_line_length": 423, "alphanum_fraction": 0.6455534584, "num_tokens": 6266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744850834649, "lm_q2_score": 0.7662936377487304, "lm_q1q2_score": 0.6265021263051535}} {"text": "\\chapter{DC Circuit Analysis}\n\nIn the most basic circuit, you have only a battery and a resistor:\n\n\\includegraphics[width=0.8\\textwidth]{DC_Circuit_Diagram.png}\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$6V$] ++(0,3)\nto ++(3,0)\nto [R=$3\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i=2A] ++(0,-3) -- (0,0);\n\\end{circuitikz}\n\nIn this case, you only need Ohm's Law: $V = I R$. In this case, $6V = 3\\Omega \\times 2A$.\n% ADD: Define Ohm's Law\n\\begin{Exercise}[title={Ohm's Law}, label=ohms_check]\n\n How many amps are going around the circuit?\n \n \\vspace{1cm}\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$24V$] ++(0,3)\nto ++(3,0)\nto [R=$6\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i={? A}] ++(0,-3) -- (0,0);\n\\end{circuitikz}\n\n \n\\end{Exercise}\n\\begin{Answer}[ref=ohms_check]\n\n $V = I R$ so $I = \\frac{V}{R} = \\frac{24V}{6\\Omega} = 4A$.\n \n\\end{Answer}\n% KA: https://youtu.be/F_vLWkkOETI\n\n\\section{Resistors in Series}\n\nWhen you have two resistors wired together in a long line, we say they\nare ``in series''. If you have two resistors $R_1$ and $R_2$ wired in\nseries, the total resistance is $R_1 + R_2$.\n\nIn this diagram, for example, the total resistance is $5\\Omega$.\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$10V$] ++(0,5)\nto ++(3,0)\nto [R=$3\\Omega$] ++(0,-2.5)\nto [R=$2\\Omega$] ++(0,-2.5) -- (0,0);\n\\end{circuitikz}\n\nThe current flowing through the circuit, then, is $10/4 = 2A$.\n\nBy Ohm's law, the voltage drop across the upper resistor is $I R = 2A \\times 3\\Omega = 6V$.\n\nThe voltage drop across the lower resistor is $I R = 2A \\times 2\\Omega = 4V$.\n\nNotice that the battery pumps the voltage up to $10V$, then the two\nresistors drop it by exactly $10V$. This is known as ``Kirchhoff's\nVoltage Law'':\n% KA: https://youtu.be/4rsswT_Rv1M\n\n\\begin{mdframed}[style=important, frametitle={Kirchhoff's Voltage Law}]\\index{Kirchhoff's voltage law}\nAs you make a loop around a circuit, the sum of the voltage increase\nmust equal the sum of the voltage decrease.\n\\end{mdframed}\n\nThe negative end of the battery as connected to ``ground'' (\nit has zero voltage), then we can draw a diagram with the\nvoltages(That symbol in the lower right represents a connection to ground).\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$6V$] ++(0,5) \nto [-*] ++(3,0) node[anchor=west] {10V}\nto [R=$3\\Omega$,-*] ++(0,-2.5) node[anchor=west] {4V}\nto [R=$2\\Omega$,-*] ++(0,-2.5) node[anchor=west]{0V} node[ground]{} --(0,0);\n\\end{circuitikz}\n\n\n\\begin{Exercise}[title={Resistors In Series}, label=series_resistor]\n\n What is the current going around the circuit?\n \n What is the voltage drop across each resistor?\n \n \\vspace{1cm}\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$16V$] ++(0,5) \nto [-*] ++(3,0) node[anchor=west] {16V}\nto [R=$5\\Omega$,-*] ++(0,-2.5) node[anchor=west] {?}\nto [R=$3\\Omega$,-*] ++(0,-2.5) node[anchor=west]{0V} node[ground]{} --(0,0);\n\\end{circuitikz}\n\n\n\\end{Exercise}\n\\begin{Answer}[ref=series_resistors]\n\n There is a total resistance of $8\\Omega$, so your 16V will push 2A\n of current around the circuit.\n\n 2A going through a $5\\Omega$ resistor represents a 10V drop.\n\n 2A going through a $3\\Omega$ resitor represents a 6V drop.\n \n\\end{Answer}\n\n\n\\section{Resistors in Parallel}\n\nLook at this circuit. Note that the current can go two different paths.\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$12V$] ++(0,3)\nto ++(3,0)\nto [R=$2\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm] ++(0,-3) -- (0,0);\n\\draw (3,3) -- (5,3)\nto [R=$3\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm] ++(0,-3) -- (3,0);\n\\end{circuitikz}\n\nThere is 12 volts pushing current through both resistors. So 6A will\ngo through the 2$\\Omega$ resistor and 4A will go through the 3$\\Omega$\nresistor.\n\n\\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$12V$] ++(0,3)\nto ++(3,0)\nto [R=$2\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i=6A] ++(0,-3) -- (0,0);\n\\draw (3,3) -- (5,3)\nto [R=$3\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i=4A] ++(0,-3) -- (3,0);\n\\end{circuitikz}\n\nThus, a total of 10 A will be going through the battery.\n\nImagine you are a battery. You can't see that you have two resistors.\nWhat does it feel like to you? $\\frac{V}{I} = R$, and $V= 12$ and $I =\n10$. So the effective resistance of the two resistors in parallel is\n$\\frac{12}{10}$ or $\\frac{6}{5} \\Omega$.\n\n\\begin{mdframed}[style=important, frametitle={Resistance in Parallel}]\\index{resistance!in parallel}\nIf you have several resistances $R_1, R_2, \\ldots, R_n$ wired in\nparallel, their effective resistance $R_t$ is given by\n\n$$\\frac{1}{R_t} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\ldots + \\frac{1}{R_n}$$\n\n\\end{mdframed}\n\nIn our example:\n\n$$\\frac{1}{R_t} = \\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6}$$\n\nThus $R_t = \\frac{6}{5}\\Omega$.\n\n\\begin{Exercise}[title={Resistors In Parallel}, label=parallel_resistors]\n\n What is the current going through the battery?\n What is the drop over the $4\\Omega$ resistor?\n What is the current in each branch?\n\n \\vspace{1cm}\n\n \\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$12V$] ++(0,3)\nto [R=$4\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm] ++(3,0) node [yshift=0.3cm] {? V}\nto [R=$6\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i={? A}] ++(0,-3) node[ground]{} -- (0,0);\n\\draw (3,3) -- (5,3)\nto [R=$3\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i={? A}] ++(0,-3) -- (3,0);\n\\end{circuitikz}\n\n\\end{Exercise}\n\\begin{Answer}[ref=parallel_resistors]\n The effective resistance of the $6\\Omega$ and the $3\\Omega$ is $2\\Omega$ because \n\n $$\\frac{1}{R_T} = \\frac{1}{6} + \\frac{1}{3} == \\frac{1}{2}$$\n\n So the battery experiences a resistance of $4\\Omega + 2\\Omega =\n 6\\Omega$. A $12V$ will push 2A through a resistance of $6\\Omega$.\n\n The voltage drop across the $4\\Omega$ resistor is $2A \\times 4\\Omega\n = 8V$. Thus there will be a 4V drop across the two resistors in\n parallel. So 2/3 A will flow through the $6\\Omega$ resistor. 4/3 A\n will flow through the $3\\Omega$ resistor.\n\n \\begin{circuitikz}\n\\draw (0,0) to[battery1,invert,l=$12V$] ++(0,3)\nto [R=$4\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm] ++(3,0) node [yshift=0.3cm] {8 V}\nto [R=$6\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i={2/3 A}] ++(0,-3) node[ground]{}-- (0,0);\n\\draw (3,3) -- (5,3)\nto [R=$3\\Omega$, /tikz/circuitikz/bipoles/length=1.0cm,i={4/3 A}] ++(0,-3) -- (3,0);\n\\end{circuitikz}\n \n\\end{Answer}\n\n", "meta": {"hexsha": "2c323cfec732145f8f9962c36b0e50f47c2a6e5c", "size": 6347, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Modules/MatterEnergy/dc_circuits-en_US.tex", "max_stars_repo_name": "hillegass/sequence", "max_stars_repo_head_hexsha": "b7b4896d804c49cbc93fe86a0d2fce531afbcc1f", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2020-06-13T17:19:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-05T00:43:44.000Z", "max_issues_repo_path": "Modules/MatterEnergy/dc_circuits-en_US.tex", "max_issues_repo_name": "hillegass/sequence", "max_issues_repo_head_hexsha": "b7b4896d804c49cbc93fe86a0d2fce531afbcc1f", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Modules/MatterEnergy/dc_circuits-en_US.tex", "max_forks_repo_name": "hillegass/sequence", "max_forks_repo_head_hexsha": "b7b4896d804c49cbc93fe86a0d2fce531afbcc1f", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-05T00:43:58.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-05T00:43:58.000Z", "avg_line_length": 33.0572916667, "max_line_length": 102, "alphanum_fraction": 0.6623601702, "num_tokens": 2487, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.817574471748733, "lm_q2_score": 0.7662936430859597, "lm_q1q2_score": 0.6265021204504156}} {"text": "\\documentclass{report}\n\\def\\MLine#1{\\par\\hspace*{-\\leftmargin}\\parbox{\\textwidth}{\\[#1\\]}}\n\\usepackage{amsmath}\n\\begin{document}\n\nBook: Basic Mathematics\nAuthor: Serge Lang\n\n\\tableofcontents{}\n\n\\chapter{Numbers}\n\n\\section{The Integers}\nNo exercises in this section.\n\n\\section{Rules for Addition}\nJustify each step, using commutativity and associativity in proving the following identities.\n\n\\begin{enumerate}\n \\item (a + b) + (c + d) = (a + d) + (b + c)\n \\item (a + b) + (c + d) = (a + c) + (b + d)\n \\item (a - b) + (c - d) = (a + c) + (- b - d)\n \\item (a - b) + (c - d) = (a + c) - (b + d)\n \\item (a - b) + (c - d) = (a + d) - (c - b)\n \\item (a - b) + (c - d) = -(b + d) + (a + c)\n \\item (a - b) + (c - d) = -(b + d) - (-a - c)\n \\item ((x + y) + z) + w = (x + z) + (y + w)\n \\item (x - y) - (z - w) = (x + w) - y + z\n \\item (x - y) - (z - w) = (x - z) - (w - y)\n \\item Show that - (a + b + c) = -a + (-b) + (-c).\n \\item Show that -(a -b - c) = -a + b + c.\n \\item Show that -(a - b) = b - a.\n \\item[] Solve for x in the following equations.\n \\item -2 + x = 4\n \\item 2 - x = 5\n \\item x - 3 = 7\n \\item -x + 4 = 1\n \\item 4 - x = 8\n \\item -5 - x = -2\n \\item -7 + x = -10\n \\item -3 + x = 4\n \\item Prove the \\textbf{cancellation law for addition:}\n\t \\MLine{\\textnormal{If } a + b = a + c, \\textnormal{then } b = c}\n \\item Prove: If a + b = a, then b = 0\n\\end{enumerate}\n\n\\section{Rules for Multiplication}\n\n\\begin{enumerate}\n\t\\item Express each of the following expressions in the form $2^m3^na^rb^s$, where $m$, $n$, $r$, $s$ are positve integers.\n\t\t\\begin{enumerate}\n\t\t\t\\item $8a^2b^3(27a^4)(2^5ab)$\n\t\t\t\\item $16b^3a^2(6ab^4)(ab)^3$\n\t\t\t\\item $3^2(2ab)^3(16a^2b^5)(24b^2a)$\n\t\t\t\\item $24a^3(1ab^2)^3(3ab)^2$\n\t\t\t\\item $(3ab)^2(27a^3b)(16ab^5)$\n\t\t\t\\item $32a^4b^5a^3b^2(6ab^3)^4$\n\t\t\\end{enumerate}\n\t\\item Prove:\n\t\t\\MLine{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}\n\t\t\\MLine{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}\n\t\\item Obtain expansion for $(a + b)^4$ and $(a - b)^4$ similar to the expansions for $(a + b)^3$ and $(a - b)^3$ of the preceding exercise.\n\t\\item[] Expand the following expressions as sums of powers of $x$ multiplied by integers.\n\t\\item $(2 - 4x)^2$\n\t\\item $(1 - 2x)^2$\n\t\\item $(2x + 5)^2$\n\t\\item $(x-1)^2$\n\t\\item $(x + 1)(x-1)$\n\t\\item $(2x + 1)(x + 5)$\n\t\\item $(x^2 + 1)(x^2 - 1)$\n\t\\item $(1 + x^3)(1 - x^3)$\n\t\\item $(x^2 + 1)^2$\n\t\\item $(x^2 - 1)^2$\n\t\\item $(x^2 + 2)^2$\n\t\\item $(x^2 - 2)^2$\n\t\\item $(x^2 - 4)^2$\n\t\\item $(x^3 - 4)(x^3 + 4)$\n\t\\item $(2x^2 + 1)(2x^2 - 1)$\n\t\\item $(-2 + 3x)(-2 - 3x)$\n\t\\item $(x + 1)(2x + 5)(x-2)$\n\t\\item $(2x + 1)(1 - x)(3x + 2)$\n\t\\item $(3x -1)(2x + 1)(x + 4)$\n\t\\item $(-1 -x)(-2 + x)(1 - 2x)$\n\t\\item $(-4x + 1)(2 - x)(3 + x)$\n\t\\item $(1 - x)(1 + x)(2 - x)$\n\t\\item $(x - 1)^2(3 - x)$\n\t\\item $(1 - x)^2(2-x)$\n\t\\item $(1 - 2x)^2(3 + 4x)$\n\t\\item $(2x + 1)^2(2 - 3x)$\n\t\\item The population of a city in 1910 was 50,000, and it doubles every 10 years. What will it be (a) in 1970 (b) in 1990 (c) in 2000?\n\t\\item The population of a city in 1905 was 100,00 and it doubles every 25 years. What will it be after (a) 50 years (b) 100 years (c) 150 years?\n\t\\item The population of a city was 200 thousand in 1915, and it triples every 50 years. What will be the population:\n\t\t\\begin{enumerate}\n\t\t\t\\item in the year 2215?\n\t\t\t\\item in the year 2165?\n\t\t\\end{enumerate}\n\t\\item The population of a city was 25,000 in 1870, and it triples every 40 years. What will it be:\n\t\t\\begin{enumerate}\n\t\t\t\\item in 1990?\n\t\t\t\\item in 2030?\n\t\t\\end{enumerate}\n\\end{enumerate}\n\n\\section{Even and Odd Integers; Divisibility}\n\\begin{enumerate}\n\t\\item Give the proofs for the cases of Theorem 1 which were not proved in the text.\n\t\\item Prove: If $a$ is even and $b$ is any positive integer, then $ab$ is even.\n\t\\item Prove: If $a$ is even, then $a^3$ is even.\n\t\\item Prove: If $a$ is odd, then $a^3$ is odd.\n\t\\item Prove: if $n$ is even, then $(-1)^n = -1$.\n\t\\item Prove: if $n$ is odd, then $(-1)^n = -1$.\n\t\\item Prove: if $m$, $n$ are odd, then the product $mn$ is odd.\n\t\\item[] Find the largest power of 2 which divides the following integers.\n\t\\item 16\n\t\\item 24 \n\t\\item 32\n\t\\item 20\n\t\\item 50\n\t\\item 64\n\t\\item 100\n\t\\item 36\n\t\\item[] Find the largest power of 3 which divides the following integers.\n\t\\item 30 \n\t\\item 27\n\t\\item 63\n\t\\item 99\n\t\\item 60\n\t\\item 50\n\t\\item 42\n\t\\item 25\n\t\\item Let $a$, $b$ be integers. Define $a \\equiv b (\\textnormal{mod } 5)$, which we read\n\t\t``$a$ is \\textbf{congruent to} $b$ \\textbf{modulo} 5'', to mean that\n\t\t$a - b$ is divisible by 5. Prove: if $a \\equiv b (\\textnormal{mod } 5)$ and $x \\equiv y (\\textnormal{mod } 5)$, then\n\t\t\\MLine{a + x \\equiv b + y (\\textnormal{mod } 5)}\n\t\tand\n\t\t\\MLine{ax \\equiv by (\\textnormal{mod } 5)}\n\t\\item Let $d$ be a positive integer. Let $a$, $b$ be integers. Define\n\t\t\\MLine{a \\equiv b (\\textnormal{mod } d)}\n\t\tto mean that $a - b$ is divisible by $d$. Prove that if $a \\equiv b (\\textnormal{mod } d)$ and $x \\equiv y (\\textnormal{mod } d)$, then\n\t\t\\MLine{a + x \\equiv b + y (\\textnormal{mod } d)}\n\t\tand\n\t\t\\MLine{ax \\equiv by (\\textnormal{mod } d)}\n\t\\item Assume that every positive integer can be written in one of the forms $3k$, $3k + 1$, $3k + 2$ for some integer $k$.\n\t\tShow that if the square of a positive integer is divisible by 3, then so is the integer.\n\\end{enumerate}\n\n\\section{Rational Numbers}\n\\begin{enumerate}\n\t\\item Solve for $a$ in the following equations.\n\t\t\\begin{enumerate}\n\t\t\t\\item $2a = \\dfrac{3}{4}$\n\t\t\t\\item $\\dfrac{3a}{5} = -7$\n\t\t\t\\item $\\dfrac{-5a}{2} = \\dfrac{3}{8}$\n\t\t\\end{enumerate}\n\t\\item Solve for $x$ in the following equations.\n\t\t\\begin{enumerate}\n\t\t\t\\item $3x - 5 = 0$\n\t\t\t\\item $-2x + 6 = 1$\n\t\t\t\\item $-7x = 2$\n\t\t\\end{enumerate}\n\t\\item Put the following fractions in the lowest form.\n\t\t\\begin{enumerate}\n\t\t\t\\item $ \\dfrac{10}{25} $\n\t\t\t\\item $\\dfrac{3}{9}$\n\t\t\t\\item $\\dfrac{30}{25}$\n\t\t\t\\item $\\dfrac{50}{15}$\n\t\t\t\\item $\\dfrac{45}{9}$\n\t\t\t\\item $\\dfrac{62}{4}$\n\t\t\t\\item $\\dfrac{23}{46}$\n\t\t\t\\item $\\dfrac{16}{40}$\n\t\t\\end{enumerate}\n\\end{enumerate}\n\n\\end{document}\n\n\n\n", "meta": {"hexsha": "050fa236e9dc55723ab61882d1a9ad4f62d192d8", "size": 6049, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "basic_mathematics_serge_lang/exercises.tex", "max_stars_repo_name": "llulai/math_book_solutions", "max_stars_repo_head_hexsha": "c052b46031ca7cb8f8a34078f67a3eaa9c67298c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-12-05T20:00:45.000Z", "max_stars_repo_stars_event_max_datetime": "2018-12-05T20:00:45.000Z", "max_issues_repo_path": "basic_mathematics_serge_lang/exercises.tex", "max_issues_repo_name": "llulai/math_book_solutions", "max_issues_repo_head_hexsha": "c052b46031ca7cb8f8a34078f67a3eaa9c67298c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "basic_mathematics_serge_lang/exercises.tex", "max_forks_repo_name": "llulai/math_book_solutions", "max_forks_repo_head_hexsha": "c052b46031ca7cb8f8a34078f67a3eaa9c67298c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.9831460674, "max_line_length": 145, "alphanum_fraction": 0.5714994214, "num_tokens": 2585, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.817574471748733, "lm_q1q2_score": 0.6265021204504156}} {"text": "Clustering is a basic tool of exploratory data analysis. There are a variety of efficient algorithms---including $k$-means, EM for Gaussian mixtures, and hierarchical agglomerative schemes---that are widely used for discovering ``natural'' groups in data. Unfortunately, they don't always find a grouping that suits the user's needs.\n\nThis is inevitable. In any moderately complex data set, there are many different plausible grouping criteria. Should a collection of rocks be grouped according to value, or shininess, or geological properties? Should animal pictures be grouped according to the Linnaean taxonomy, or cuteness? Different users have different priorities, and an unsupervised algorithm has no way to magically guess these.\n\nAs a result, a rich body of work on {\\it constrained} clustering has emerged.\nIn this setting, a user supplies guidance, typically in the form of ``must-link'' or ``cannot-link'' constraints, pairs of points that must be placed together or apart. Introduced by \\citet{Wagstaff2001}, these constraints have since been incorporated into many different {\\it flat} clustering procedures~\\citep{Wagstaff2000,Bansal2004,Basu2004,Kulis2009,Biswas2014}.\n\nIn this chapter, we introduce constraints to {\\it hierarchical clustering}, the recursive partitioning of a data set into successively smaller clusters to form a tree. A hierarchy has several advantages over a flat clustering. First, there is no need to specify the number of clusters in advance. Second, the tree captures cluster structure at multiple levels of granularity, simultaneously. As such, trees are particularly well-suited for exploratory data analysis and the discovery of natural groups.\n\nThere are several well-established methods for hierarchical clustering, the most prominent among which are the bottom-up agglomerative methods such as average linkage (see, for instance, Chapter 14 of \\citet{Hastie2009}). But they suffer from the same problem of under-specification that is the scourge of unsupervised learning in general. And, despite the rich literature on incorporating additional guidance into flat clustering, there has been relatively little work on the hierarchical case.\n\nWhat form might the user's guidance take? The usual must-link and cannot-link constraints make little sense when data has hierarchical structure. Among living creatures, for instance, should {\\tt elephant} and {\\tt tiger} be linked? At some level, yes, but at a finer level, no. A more straightforward assertion is that {\\tt elephant} and {\\tt tiger} should be linked in a cluster that does not include {\\tt snake}. We can write this as a {\\it triplet} $(\\{\\mbox{\\tt elephant}, \\mbox{\\tt tiger}\\}, \\mbox{\\tt snake})$. We could also assert $(\\{\\mbox{\\tt tiger}, \\mbox{\\tt leopard}\\}, \\mbox{\\tt elephant})$. Formally, $(\\{a,b\\},c)$ stipulates that the hierarchy contains a subtree (that is, a cluster) containing $a$ and $b$ but not $c$.\n\nA wealth of research addresses learning taxonomies from triplets {\\it alone}, mostly in the field of phylogenetics: see \\citet{Felsenstein2004} for an overview, and \\citet{Aho1981} for a central algorithmic result. Let's say there are $\\numdata$ data items to be clustered, and that the user seeks a particular hierarchy $\\tree^*$ on these items. This $\\tree^*$ embodies at most ${\\numdata \\choose 3}$ triplet constraints, possibly less if it is not binary. It was pointed out in \\citet{Tamuz2011} that roughly $\\numdata \\log \\numdata$ carefully-chosen triplets are enough to fully specify $\\tree^*$ if it is balanced. This is also a lower bound: there are $\\numdata^{\\Omega(\\numdata)}$ different labeled rooted trees, so each tree requires $\\Omega(\\numdata \\log \\numdata)$ bits, on average, to write down---and each triple provides $O(1)$ bits of information, since there are just three possible outcomes for each set of points $a,b,c$. Although $\\numdata \\log \\numdata$ is a big improvement over $\\numdata^3$, it is impractical for a user to provide this much guidance when the number of points is large. In such cases, a hierarchical clustering cannot be obtained on the basis of constraints alone; the geometry of the data must play a role.\n\nWe consider an interactive process during which a user incrementally adds constraints.\n\\begin{itemize}\n\\item Starting with a pool of data $\\data \\subseteq \\R^\\d$, the machine builds a candidate hierarchy $\\tree$.\n\\item The set of constraints $C$ is initially empty.\n\\item Repeat:\n\\begin{itemize}\n\\item The machine presents the user with a small portion of $\\tree$: specifically, its restriction to $O(1)$ leaves $S \\subset X$. We denote this $\\tree|_S$.\n\\item The user either accepts $\\tree|_S$, or provides a triplet constraint $(\\{a,b\\},c)$ that is violated by it.\n\\item If a triplet is provided, the machine adds it to $C$ and modifies the tree $T$ accordingly.\n\\end{itemize}\n\\end{itemize}\nIn realizing this scheme, a suitable clustering algorithm and querying strategy must be designed. Similar issues have been confronted in flat clustering---with must-link and cannot-link constraints---but the solutions are unsuitable for hierarchies, and thus a fresh treatment is warranted.\n\n\\textbf{The clustering algorithm:}\nWhat is a method of hierarchical clustering that takes into account the geometry of the data points as well as user-imposed constraints? \n\nWe adopt an \\emph{interactive Bayesian} approach. The learning procedure is uncertain about the intended tree and this uncertainty is captured in the form of a distribution over all possible trees. Initially, this distribution is informed solely by the geometry of the data but once interaction begins, it is also shaped by the growing set of constraints.\n\nThe nonparametric Bayes literature contains a variety of different distributional models for hierarchical clustering. We describe a general methodology for extending these to incorporate user-specified constraints. For concreteness, we focus on the Dirichlet diffusion tree~\\citep{Neal2003}, which has enjoyed empirical success. We show that triplet constraints are quite easily accommodated: when using a Metropolis-Hastings sampler, they can efficiently be enforced, and the state space remains strongly connected, assuring convergence to the unique stationary distribution.\n\n\\textbf{The querying strategy:}\nWhat is a good way to select the subsets $S$? A simple option is to pick them at random from $\\data$. We show that this strategy leads to convergence to the target tree $\\tree^*$. Along the way, we define a suitable distance function for measuring how close $\\tree$ is to $\\tree^*$. \n\n%This allows us to give a simple rule for when to stop querying if we seek a tree $\\epsilon$-close to the target.\n\nWe might hope, however, that a more careful choice of $S$ would lead to faster convergence, in much the same way that intelligent querying is often superior to random querying in active learning. In order to do this, we show how the Bayesian framework allows us to quantify which portions of the tree are the most uncertain, and thereby to pick $S$ that focuses on these regions.\n\nQuerying based on uncertainty sounds promising, but is dangerous because it is heavily influenced by the choice of prior, which is ultimately quite arbitrary. Indeed, if only such queries were used, the interactive learning process could easily converge to the wrong tree. We show how to avoid this situation by interleaving the two types of queries.\n\nFinally, we present a series of experiments that illustrate how a little interaction leads to significantly better hierarchical clusterings.\n\n\n\\subsection*{Other related work}\n\nA related problem that has been studied in more detail~\\citep{Zoller2000,Eriksson2011,Krishnamurthy2012} is that of building a hierarchical clustering where the only information available is pairwise similarities between points, but these are initially hidden and must be individually queried.\n\nIn another variant of interactive flat clustering~\\citep{Balcan2008,Awasthi2010,Awasthi2013}, the user is allowed to specify that individual clusters be merged or split. A succession of such operations can always lead to a target clustering, and a question of interest is how quickly this convergence can be achieved.\n\nFinally, it is worth mentioning the use of triplet constraints in learning other structures, such as Euclidean embeddings~\\citep{Borg1997}.\n\n\\iffalse\n\\section{Bayesian hierarchical clustering}\n%\\subsection{Priors on trees}\n\n\\fi\n\\section{Adding interaction}\n\nImpressive as Bayesian nonparametric hierarchical clustering is, there is no reason to suppose that it will magically find a tree that suits the user's needs. But a little interaction can be helpful in improving the outcome.\n\nLet $\\tree^*$ denote the target hierarchical clustering. It is not necessarily the case that the user would be able to write this down explicitly, but this is the tree that captures the distinctions he/she is able to make, or wants to make. Figure~\\ref{fig:refinement} (left) shows an example, for a small data set of 5 points. In this case, the user does not wish to distinguish between points $1,2,3$, but does wish to place them in a cluster that excludes point $4$.\n\nWe could posit our goal as exactly recovering $\\tree^*$. But in many cases, it is good enough to find a tree that captures all the distinctions within $\\tree^*$ but also possibly has some extraneous distinctions, as in the right-hand side of Figure~\\ref{fig:refinement}.\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=3in]{img/ibhc/refinement.pdf}\n \\caption{Target tree $T^*$ (left) and a refinement of it.}\n \\label{fig:refinement}\n\\end{figure}\n\nFormally, given data set $\\data$, we say $S \\subseteq \\data$ is a {\\it cluster} of tree $\\tree$ if there is some node of $\\tree$ whose descendant leaves are exactly $S$. We say $\\tree$ is a {\\it refinement} of $\\tree^*$ if they have the same set of leaves, and moreover every cluster of $\\tree^*$ is also a cluster of $\\tree$. This, then, is our goal: to find a refinement of the target clustering $\\tree^*$.\n\n\\subsection{Triplets}\n\nThe user provides feedback in the form of triplets. The constraint $(\\{a,b\\},c)$ means that the tree should have a cluster containing $a$ and $b$ but not $c$. Put differently, the lowest common ancestor of $a,b$ should be a strict descendant of the lowest common ancestor of $a,b,c$.\n\nLet $\\Delta(\\tree)$ denote the set of all proper triplet constraints embodied in tree $\\tree$. If $\\tree$ has $n$ nodes, then $|\\Delta(\\tree)| \\leq {n \\choose 3}$. For non-binary trees, it will be smaller than this number. Figure~\\ref{fig:refinement} (left), for instance, has no triplet involving $1,2,3$. \n\nRefinement can be characterized in terms of triplets.\n\\begin{restatable}{lemma}{treerefinement}\n\\label{thm:treerefinement}\nTree $\\tree$ is a refinement of tree $\\tree'$ if and only if $\\Delta(\\tree') \\subseteq \\Delta(\\tree)$.\n\\end{restatable}\n\\begin{proof}\nSee \\autoref{chap:ibhc-appendix}.\n\\end{proof}\nIn particular, {\\it any triplet-querying scheme that converges to the full set of triplets of the target tree $\\tree^*$ is also guaranteed to produce trees that converge to a refinement of $\\tree^*$.}\n\nWith this lemma in mind, it is natural to measure how close a tree $\\tree$ is to the target $\\tree^*$ with the following (asymmetric) distance function, which we call {\\it triplet distance} (TD):\n\n%\\subsection{Triplet Distance}\n%To better understand the performance of clustering algorithm\n%and the various triplet querying schemes,\n%we first define a distance function between\n%a candidate tree and the desired tree.\n\n%$T$ is defined to be a \\emph{refinement} of $T^*$ if\n%if every cluster of $T^*$ is in $T$.\n%In the general setting, $T^*$ may be a $k$-ary tree\n%and our algorithm only produces a binary tree. \n%For example, in a $K$-way classification problem, the desired\n%tree $T^*$ would be a root node with $K$ children corresponding to %each class. \n%Each of these $K$ children would have all data belonging to its class %as\n%a child. See \\autoref{fig:kway} for an example. \n\n%If a binary tree $T$ is a refinement of $T^*$ in the $K$-way\n%classification problem,\n%there exist nodes in $T$ that contain\n%solely the data for each class and this is the closest binary tree we\n%can obtain.\n%Thus, we desire a distance function such that if $T$ is a refinement\n%of $T^*$, their distance should be 0.\n\n%\\begin{figure}[H]\n% \\centering\n% \\includegraphics[width=0.5\\textwidth]{img/ibhc/KWayTree}\n% \\caption{A 4-way classification tree on the left\n% and a refinement of it on the right. Nodes\n% with matching colors represent matching clusters.}\n% \\label{fig:kway}\n%\\end{figure}\n\n%Let $\\mathcal{C}(T)$ be the enumeration of all triplets\n%satisfied by tree $T$.\n%For a given desired \\emph{master tree} $T^*$\n%the triplet distance (TD) between $T^*$ and a candidate tree $T$\n%is\n\n\\begin{align}\n \\mathrm{TD}(\\tree^*, \\tree) = \\frac{\\sum_{c \\in \\Delta(\\tree^*)} \\mathbb{I}(c \\notin \\Delta(\\tree)) }{|\\Delta(\\tree^*)|}\n\\end{align}\nwhere $\\mathbb{I}$ is the indicator function.\n% Lemma\n%If $T$ is a refinement of $T^*$, their triplet distance is zero\n%as $\\mathcal{C}(T^*) \\subseteq \\mathcal{C}(T)$.\nThis distance is zero exactly when $\\tree$ is a refinement of $\\tree^*$, in which case we have reached our goal. \n\n%User guidance can assist in under-specified\n%unsupervised learning problems. We now\n%provide a concrete means of obtaining \n%information about hierarchical structure in data\n%from a user, in the form of \\emph{triplet feedback}.\n\n%For hierarchies, a triplet is a set of data\n%of the form $(\\{a, b\\}, c)$, such that\n%$a$ and $b$ should belong in a subtree\n%together without $c$. Let $\\text{lca}(u, v)$ be\n%the lowest common ancestor of nodes $u$ and $v$. \n%Formally, we can verify that a tree satisfies a triplet\n%$(\\{a, b\\}, c)$ if $\\text{lca}(a, b)$ is a descendant\n%of both $\\text{lca}(a, c)$ and $\\text{lca}(b, c)$.\n\nA simple strategy for obtaining triplets \nwould be to present the user with three randomly\nchosen data points and have the user pick the odd one out.\nThis strategy has several drawbacks.\nFirst, some sets of three points have no triplet constraint \n(for instance, points $1,2,3$ in \\autoref{fig:refinement}).\nSecond, the chosen set of points might correspond to a triplet\nthat has already been specified, or is {\\it implied} by \nspecified triplets. For example, knowledge of $(\\{a, b\\}, c)$ and\n$(\\{b, c\\}, d)$ implies $(\\{a, c\\}, d)$.\nEnumerating the set of implied\ntriplets is non-trivial for $n > 3$ triplets~\\citep{Bryant1995},\nmaking it difficult to avoid these implied triplets\nin the first place.\n\nWe thus consider another strategy---rather\nthan the user arranging three data points into\na triplet, the user observes\nthe hierarchy induced over some $O(1)$-sized\nsubset $S$ of the data and corrects\nan error in the tree by supplying a triplet.\nWe call this is a \\emph{subtree query}.\nFinally, we note that in this work\nwe only consider the \\emph{realizable} case\nwhere the triplets obtained\nfrom a user do not contain contradictory information\nand that there is a tree that satisfies\nall of them.\n\n%\\subsection{Triplet feedback on subtrees}\n%The interactive process integrates\n%user feedback into a hierarchy\n%after querying the user for a triplet.\n%We perform this query by\n%showing a user the subtree\n%over a particular subset of data\n%and ask to either\n%accept the subtree as correct, or\n%report a triplet that is violated by\n%the subtree.\n%\n%Formally, given a candidate hierarchy $T$\n%that satisfies a set of triplets $C$ ,\n%we first obtain some $O(1)$ size\n%subset of the dataset $S \\subset X$,\n%and find the induced subtree\n%$T|_S$.\n%$T|_S$ is shown to a user whereupon\n%the user can either accept the tree,\n%acknowledging that there are no errors in its\n%structure, or report a triplet\n%$(\\{a, b\\}, c)$ that is violated by this subtree.\n\n%This approach confers several benefits over\n%the simple approach of selecting three points\n%at random. Subtree queries have the benefit\n%of never returning a triplet \n%implied by $C$. A tree that satisfies $C$\n%will by definition will satisfy all of $C$'s\n%implied triplets. \n%In addition,\n%the triplet reported by the user\n%is likely to be more informative\n%as the triplet corresponds to an \n%error in a candidate hierarchy.\n%However, this method assumes\n%the existence of an algorithm\n%that can produce candidate trees\n%that satisfy $C$.\n\n\\subsection{Finding a tree consistent with constraints}\n\\label{sec:aho}\nWe start with a randomly initialized hierarchy $\\tree$\nover our data\nand show an induced subtree $\\tree|_S$ to the user, obtaining the\nfirst triplet. The next step is constructing \na new tree that satisfies the triplet.\nThis begins the feedback cycle; a user provides a triplet\ngiven a subtree and the triplet is incorporated into a clustering algorithm,\nproducing a new candidate tree.\nA starting point is \nan algorithm that returns \na tree consistent with a set of triplets.\n\nThe simplest algorithm to solve this problem is\nthe \\texttt{BUILD} algorithm, introduced in \\citet{Aho1981}.\nGiven a set of triplets $C$, \\texttt{BUILD}\nwill either return a tree that satisfies $C$, or error \nif no such tree exists.\nIn \\texttt{BUILD}, we first construct the {\\it Aho graph}\n$G_C$, which has a vertex for each data point and an \nundirected edge $\\{a,b\\}$ for each triplet constraint $(\\{a,b\\},c)$.\nIf $G_C$ is connected, there is no tree that satisfies all\ntriplets. Otherwise, the top split of the tree is a partition of \nthe connected components of $G_C$: any split is fine as long as \npoints in the same component stay together. Satisfied triplets are discarded, and \\texttt{BUILD} then continues recursively on the\nleft and right subtrees.\n\n\\texttt{BUILD} satisfies triplet constraints but ignores \nthe geometry of the data, whereas we wish to take both into\naccount. By incorporating triplets into the posterior DDT \nsampler, we obtain high likelihood trees that still satisfy $C$.\n\n\n\\subsection{Incorporating triplets into the sampler}\n\nIn this section, we present an algorithm\nto sample candidate trees \nfrom the posterior DDT, constrained\nby a triplet set $C$.\nIt is based on the subtree prune and regraft move.\n%The Metropolis-Hastings algorithm for the DDT\n%alternates sampling tree structure\n%via SPR moves and latent divergence locations\n%via Gibbs Sampling.\n%This process produces tree samples\n%that converge to the posterior\n%distribution of the DDT\n%given data.\n\n\\begin{figure*}[htp!]\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/ConstrainedSPRMove}\n \\caption{\n Visualized is a constrained-SPR move.\n Pruning is identical but\n a regraft location is selected from the valid regraft locations\n limited by triplets.\n In this image, we are constrained by the sole triplet $(\\{2, 3\\}, 4)$.\n }\n \\label{fig:constrainedsprmove}\n\\end{figure*}\n\nThe SPR move is of particular interest\nbecause we can efficiently\nenforce triplets to form a \\emph{constrained-SPR move},\nresulting in a sampler\nthat only produces trees that satisfy a set of triplets.\nA \\emph{constrained-SPR move} is defined as\nan SPR move that assigns zero probability to\nany resulting trees that would violate a set of triplets.\nRestricting the neighborhood of an SPR move\nruns the risk of partitioning the state space,\nlosing the convergence\nguarantees of the Metropolis-Hastings algorithm.\nFortunately, a constrained-SPR move does not compromise\nstrong connectivity.\nFor any realizable triplet set $C$,\nwe prove \nthe constrained-SPR move Markov chain's aperiodicity and irreducibility.\n\nConsider the Markov chain on the state space \nof rooted binary trees that is induced by the constrained sampler.\n%\\subsection{Proof of \\autoref{thm:apr}}\n\\begin{restatable}{lemma}{aperiodic}\nThe constrained-SPR Markov chain is aperiodic.\n\\end{restatable}\n\\begin{proof}\nA sufficient condition for aperiodicity\nis the existent of a ``self-loop'' in the transition matrix: a non-zero probability of a state transitioning to itself.\nSupposed we have pruned a subtree already.\nWhen regrafting, the ordinary SPR move\nhas a non-zero probability of choosing any branch,\nand a constrained-SPR move cannot regraft\nto branches that would violate triplets.\nSince the current tree in the Markov chain\nsatisfies triplet set $C$, there is a non-zero probability\nof regrafting to the same location. \nWe thus have an aperiodic Markov chain.\n\\end{proof}\n\n\n\n%\\subsection{Proof of \\autoref{thm:irr}}\n%\\label{app:irr}\n%\\irreducible*\n\\begin{restatable}{lemma}{irreducible}\n\\label{thm:irr}\nA constrained-SPR Markov chain is irreducible.\n\\end{restatable}\n\n\\begin{proof}\n(sketch) To show irreducibility,\nwe show that a tree $\\tree$ has an non-zero\nprobability of reaching an arbitrary tree\n$\\tree'$ via constrained-SPR moves where\nboth $\\tree$ and $\\tree'$ satisfy a set of triplets $C$.\nOur proof strategy\nis to construct a canonical tree \n$\\tree_C$, and show that there exists\na non-zero probability path from $\\tree$ to $\\tree_C$,\nand therefore from $\\tree'$ to $\\tree_C$.\nWe then show that for a given constrained-SPR move,\nthe reverse move has a non-zero probability.\nThus, there exists a path from $\\tree$ to $\\tree_C$ to\n$\\tree'$, satisfying irreducibility.\n\nRecall that the split at a node in a binary\ntree that satisfies triplet set $C$\ncorresponds to a binary partition\nof the Aho graph at the node (see Section \\ref{sec:aho}).\n$\\tree_C$ is a tree such that\nevery node in $\\tree_C$ is in \\emph{canonical form}.\nA node is in canonical form if it is a leaf node,\nor, the partition of the Aho graph\nat that node can be written as $(l, r)$.\n$l$ is the single connected component\ncontaining the point with the minimum data index,\nand $r$ is the rest of the components.\n\nTo convert a particular node $s$ into canonical form,\nwe first perform ``grouping'',\nwhich puts $l$ into a single descendant of $s$\nvia constrained-SPR moves.\nWe then make two constrained-SPR moves to convert\nthe partition at $s$ into the form $(l, r)$ (see \\autoref{fig:canonical}).\nWe convert all nodes into canonical form recursively, turning \nan arbitrary tree $\\tree$ into $\\tree_C$.\n\nFinally, the reverse constrained-SPR move has a non-zero\nprobability. Suppose\nwe perform a constrained-SPR move on tree $\\tree_1$, converting it into $\\tree_2$ by \ndetaching subtree $s$\nand attaching it to branch $(u, v)$.\nA constrained-SPR move on $\\tree_2$ can select\n$s$ for pruning \nand can regraft it to form $\\tree_1$ with a non-zero\nprobability since\n$\\tree_1$ satisfies the same constraint set as $C$.\nFor a full proof, please refer to \\autoref{chap:ibhc-appendix}.\n\\end{proof}\n\n\\begin{figure*}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/CanonicalTree}\n \\caption{The process of converting $s$ into canonical form.\n We first group nodes from $l$ into their own isolated subtree, then perform\n two constrained-SPR moves to put $s$ into canonical form.}\n \\label{fig:canonical}\n\\end{figure*}\n\n%\\begin{restatable}{lemma}{aperiodic}\n%\\label{thm:apr}\n%\\end{restatable}\n%\\begin{proof} See \\ref{app:apr}.\n%\\end{proof}\n%\n%\\begin{restatable}{lemma}{irreducible}\n%\\label{thm:irr}\n%A constrained-SPR Markov chain is irreducible.\n%\\end{restatable}\n%\\begin{proof} See \\ref{app:irr}.\n%\\end{proof}\n\n\nThe simplest possible scheme for a constrained-SPR move \nwould be rejection sampling. The Metropolis-Hastings\nalgorithm for the DDT would be the same as in the\nunconstrained case,\nbut any trees violating $C$ would have accept\nprobability $0$. Although this procedure is correct,\nit is impractical. As the number of triplets\ngrows larger, more trees will be rejected\nand the sampler will slow down over time.\n\nTo efficiently sample a tree that satisfies a set of triplets $C$, \nwe modify the regraft in the ordinary SPR move. \nThe constrained-SPR move must assign zero probability\nto any regraft branches that would result in a tree\nthat violates $C$.\nThis is accomplished by generating the path from the root\nin the same manner as sampling a branch,\nbut avoiding paths that would resulted in violated triplets.\n\n\n\\textbf{Description of constrained-SPR sampler}\nRecall that in the DDT's sampling procedure for regraft branches,\na particle at a node picks a branch, and either diverges from that\nbranch or recursively samples the node's child.\nLet $s$ be the root of the subtree we are currently grafting \nback onto tree $\\tree$, let $C$ be the\ntriplet set, and let $\\texttt{leaves}(u)$\ndenote the descendant-leaves of node $u$.\nSuppose we are are currently at node $u$,\ndeciding whether to diverge\nat the branch $(u, v)$ or to recursively sample $v$.\nConsider any triplet $(\\{a, b\\}, c) \\in C$. If all---or none---of $a,b,c$ are in $\\texttt{leaves}(s)$, then the triplet is \nunaffected by the graft, and can be ignored. Otherwise, \nsome checks are needed:\n\\begin{enumerate}\n\\item $c \\in \\texttt{leaves}(s)$\n\nThen we know $a,b \\not\\in \\texttt{leaves}(s)$. If $a$ and $b$ are split across $v$'s children,\nwe are banned from sampling $v$.\n\n\\item $a \\in \\texttt{leaves}(s)$ but $b,c \\not\\in \\texttt{leaves}(s)$\n\nIf both $b$ and $c$ are in $\\texttt{leaves}(v)$, we are\nrequired to sample $v$.\nIf just $c$ is in $\\texttt{leaves}(v)$, we are banned\nfrom both diverging at $(u, v)$ and sampling $v$.\nOtherwise we can either diverge at $(u, v)$ or sample $v$.\n\n\\end{enumerate}\n\n%\\begin{enumerate}\n%\\item[1.] A node $u$ is \\emph{path-banned} if for\n%any triplet $(\\{a, b\\}, c) \\in C$, \n%$(a \\in \\texttt{leaves}(s) \\land b \\notin \\texttt{leaves}(u) \\land c \\in \\texttt{leaves}(u)) \\lor (b \\in \\texttt{leaves}(s) \\land a \\notin \\texttt{leaves}(u) \\land c \\in \\texttt{leaves}(u))$.\n%Let $p$ be $u$'s parent.\n%Intuitively, if $u$ is \\emph{path-banned}, we cannot\n%graft to the branch $(p, u)$, as we would automatically\n%violate a triplet. This is because there is some triplet$(\\{a, b\\}, c)$\n%in $C$ such that attaching $s$ to $(p, u)$ would result in a subtree\n%that contains either $a$ and $c$ but not $b$ or $b$ and $c$ but not $a$.\n%\n%\\item[2.] A node $u$ is \\emph{sample-banned} if \n%for\n%any triplet $(\\{a, b\\}, c) \\in C$, \n%$(c \\in \\texttt{leaves}(s)) \\land ((a \\in \\texttt{leaves}(v) \\land b \\in \\texttt{leaves}(w)) \\lor (a \\in \\texttt{leaves}(w) \\land b \\in \\texttt{leaves}(v))$\n%where $v$ and $w$ are the children of $u$. \n%Intuitively,\n%this means that there exist some $a$ and $b$ in our triplet set\n%that are split between $u$'s children. Attaching a subtree containing\n%$c$ to a any branch below $u$ would violate a triplet.\n%\n%\\item[3.] A node $u$ is \\emph{path-required} if for\n%any triplet $(\\{a, b\\}, c) \\in C$, \n%$(a \\in \\texttt{leaves}(s) \\land b \\in \\texttt{leaves}(u)) \\lor (b \\in \\texttt{leaves}(s) \\land a \\in \\texttt{leaves}(u))$. \n%If $u$ is \\emph{path-required}, the particle must enter its branch,\n%as for some triplet $(\\{a, b\\}, c)$, if $a$ is in our\n%pruned subtree, then $b$ is in $u$ and $c$ is in $u$'s sibling.\n%Choosing the branch to $u$'s sibling would violate this triplet.\n%\n%\\item[4.] A node $u$ is \\emph{sample-required} if for\n%any triplet $(\\{a, b\\}, c) \\in C$, \n%$(a \\in \\texttt{leaves}(s) \\land b \\in \\texttt{leaves}(u) \\land c \\in \\texttt{leaves}(u)) \\lor (b \\in \\texttt{leaves}(s) \\land a \\in \\texttt{leaves}(u) \\land c \\in \\texttt{leaves}(u))$.\n%If $u$ being \\emph{sample-required} \n%means that there exists some $a$ and $c$ or $b$ and $c$\n%in our triplet set that are already in a subtree in $T$ together.\n%We must attach our subtree such that this triplet is not violated.\n%\n%\\end{enumerate}\n(The case where $b \\in \\texttt{leaves}(s)$ is\nsymmetric to case 2.)\nIf we choose to sample $v$, we remove\nconstraints from our current set $C$ that are now satisfied,\nand continue recursively.\nThis defines a procedure by which we can sample\na divergence branch that does not violate constraints.\n\nWhile the constrained-SPR sampler can produce\na set of trees given a set of static constraints,\nthe \\texttt{BUILD} algorithm is useful in\nadding new triplets into the sampler.\nSuppose we have been sampling trees with constrained-SPR moves\nwith satisfying triplet set $C$ \nand we obtain a new triplet $u = (\\{a, b\\}, c)$ from a user query. \nWe take the current tree $\\tree$ and find the least common ancestor \n(call it $z$) of $a$ and $b$. We then call \n\\texttt{BUILD}$(C + \\{u\\})$ on just the nodes in \n$\\texttt{leaves}(z)$, and \nwe substitute the resulting subtree at position $z$ in tree $\\tree$.\n\n\\subsection{Intelligent subset queries}\nWe now have a method to sample a constrained\ndistribution over candidate trees.\nGiven a particular candidate tree $\\tree$,\nour first strategy for subtree querying\nis to pick a random subset $S$ of the leaves\nof constant size, and show the user\nthe induced subtree over the subset, $\\tree|_S$.\nWe call this \\emph{random subtree querying}.\nBut can we use a set of trees produced by the sampler\nto make better subtree queries?\nIf tree structure is ambiguous in a particular region of data,\ni.e. there are several hierarchies that could explain\na particular configuration of data, \nthe MH algorithm will sample over\nthese different configurations. A query over points\nin these ambiguous regions may help our algorithm\nconverge to a better tree faster. By looking for\nthese regions in our samples, we can choose query\nsubsets $S$ for which the tree structure is highly variable, \nand hopefully the resulting triplet from the user will \nreduce the ambiguity.\n\nMore precisely, we desire a notion of tree variance.\nGiven a set of trees $\\trees$, what is the variance\n%TODO: check this\nover a given subset of the data $S$?\nWe propose using the notion of tree distance\nas a starting point. \nFor a given tree $\\tree$, the tree distance between two nodes $a$ and $b$,\ndenoted $\\texttt{treedist}_T(a, b)$,\nis the number of edges of $\\trees$ needed to get from $a$ to $b$.\nConsider two leaves $u$ and $v$. If the tree structure around\nthem is static, we expect the tree distance\nbetween $u$ and $v$ to change very little, as the surrounding tree\nwill not change. However, if there is\nambiguity in the surrounding structure, the tree distance will\nbe more variable.\nGiven a subset of data $S$ and a set of trees $\\trees$,\nthe tree distance variance (TDV) of the trees over the subset is defined as:\n\\begin{align}\n \\text{TDV}(\\trees, S) = \\max_{u, v \\in S}\\mathrm{Var}_{\\tree \\in \\trees}[\\texttt{treedist}_{\\tree|_S}(u, v)]\n\\end{align}\n\nThis measure of variance is\nthe \\emph{max} of the variance of tree distance between\nany two points in the subset. Computing this requires\n$O(|\\trees||S|^2 \\log|S|)$ time, and since since $|S|$ is constant,\nit is not prohibitively expensive.\n\nGiven a set of trees from the sampler $\\trees$,\nwe now select a high-variance subtree\nby instantiating $L$ random subsets of constant size, $S_1, \\ldots, S_L$\nand picking $\\text{argmax}_l \\text{TDV}(\\trees, S_l)$.\nWe call this \\emph{active subtree querying}.\nAlthough using tree variance will help reconcile ambiguity in the tree structure,\nif a set of samples from a tree all violate the same triplet,\nit is unlikely that active querying will recover that triplet.\nThus, interleaving random querying and active querying\nwill hopefully help the algorithm converge quickly, while avoiding\nlocal optima.\n\n\\section{Experiments}\n\n\\begin{figure}\n \\centering\n \\begin{subfigure}[]{0.6\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/Fisher-Iris-result.png} \\caption{Fisher Iris}\n \\label{fig:iris-result}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[]{0.6\\textwidth}\n \\centering\n \\includegraphics[width=\\textwidth]{img/ibhc/MNIST-result.png}\n \\caption{MNIST}\n \\label{fig:mnist-result}\n \\end{subfigure}\n \\caption{The average of four runs of constrained-SPR samplers\n for the Fisher Iris dataset and the MNIST dataset, using 5 different querying schemes. A query was made every 100 iterations.}\n \\label{fig:main-results}\n\\end{figure}\n\nWe evaluated the convergence properties of five different querying schemes.\nIn a ``simple query'', a user is presented with three random data\nand picks an odd one out.\nIn a ``smart query'', a user is unrealistically shown the entire candidate tree and reports a violated triplet.\nIn a ``random query'', the user is shown the induced candidate tree over\na random subset of the data.\nIn an ``active query'', the user is shown a high variance \nsubtree using tree-distance variance.\nFinally, in an ``interleaved query'', the user is alternatively shown a random\nsubtree and a high variance subtree.\nIn each experiment, $\\tree^*$ was known, so user queries were simulated\nby picking a triplet\nviolated by the root split of the queried tree, \nand if no such triplet existed, recursing on a child.\n%\\begin{enumerate}\n%\\item Simple querying: a user is presented with three random data and\n%reports a triplet of the data that should be satisfied.\n%\\item Smart querying (unrealistic): a user is presented with the entire tree\n%and reports a triplet that is violated.\n%\\item Random querying: the user is shown the induced subtree\n%over a random subset of the data of constant size\n%and reports a violated triplet.\n%\\item Active querying: the user is shown a high variance\n%induced subtree, chosen from a set of random subsets of data $\\{S_1, \\ldots , S_L\\}$,\n%with variance defined as tree distance variance. The user then reports\n%a violated triple from this induced subtree.\n%\\item Interleaved querying: we alternate performing a random query\n%and an active query.\n%\\end{enumerate}\nEach scheme was evaluated on four different datasets.\nThe first dataset, MNIST \\citep{Lecun1998}, is an 10-way image classification\ndataset where the data are 28 x 28 images of digits.\nThe target tree $\\tree^*$ is simply the $K$-way classification \ntree over the data.\nThe second dataset is Fisher Iris, a 3-way flower\nclassification problem, where each of 150 flowers has\nfive features.\nThe third dataset, Zoo \\citep{Lichman2013}, is a set of 93 animals\nand 15 binary morphological features for each of animals,\nthe target tree being the induced binary tree \nfrom the Open Tree of Life \\citep{Hinchliff2015}.\nThe fourth dataset is 20 Newsgroups \\citep{Joachims1997}, a\ncorpus of text articles on 20 different subjects. We use\nthe first 10 principal components as features\nin this classification problem.\nAll datasets were modeled with DDT's with\nacquisition function $a(t) = 1/(1 - t)$\nand Brownian motion parameter $\\sigma^2$ estimated from data.\nTo better visualize the different convergence rates\nof the querying schemes, MNIST and 20 Newsgroups were subsampled\nto 150 random points.\n\nFor each dataset and querying scheme, we instantiated a SPR\nsampler with no constraints. Every one hundred iterations of the \nsampler, we performed a query.\nIn subtree queries, we used subsets of size $|S| = 10$ \nand in active querying, the highest-variance subset was chosen from $L = 20$ different random subsets.\nAs baselines,\nwe measured the triplet distance of the vanilla DDT\nand the average linkage tree.\nFinally, results were averaged over four runs of each sampler.\nThe triplet distances for Fisher Iris and MNIST can be seen in \\autoref{fig:main-results}. Results for the other datasets\ncan be found in \\autoref{chap:ibhc-appendix}.\nAlthough unrealistic due to the size of the tree shown to the user, \nthe smart query performed the best, achieving minimum error\nwith the least amount of queries. Interleaved followed next,\nfollowed by active, random, and simple. In general, the vanilla\nDDT performed the worst, and the average linkage\nscore varied on each dataset, but in all cases, the\nsubtree querying schemes performed better than both the vanilla DDT\nand average linkage.\n\nIn three datasets (MNIST, Fisher Iris and Zoo), \ninteractive methods\nachieve higher data likelihood than the vanilla DDT.\nInitially, the sampler is often restructuring the tree\nwith new triplets\nand data likelihood is unlikely to rise. However, over time\nas less triplets are reported,\nthe data likelihood increases rapidly.\nWe thus conjecture that triplet constraints \nmay help the MH algorithm find better optima.\n\n\\section{Future Work}\nWe are interested in studying the non-realizable case, i.e.\nwhen there does not exist a tree that satisfies triplet set $C$. We would also like to better understand the effect of constraints on searching\nfor optima using MCMC methods.\n\n\\ibhcack\n", "meta": {"hexsha": "55d20017ebc3e72182e1f0bec0212f5898a995a6", "size": 36372, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "writeup/content/structure/ibhc.tex", "max_stars_repo_name": "sharadmv/thesis", "max_stars_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-04-30T01:28:54.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-30T01:28:54.000Z", "max_issues_repo_path": "writeup/content/structure/ibhc.tex", "max_issues_repo_name": "sharadmv/thesis", "max_issues_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "writeup/content/structure/ibhc.tex", "max_forks_repo_name": "sharadmv/thesis", "max_forks_repo_head_hexsha": "5fbf70c0645e44b2992f3cb4d7c2fbbbf7592d7f", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.409221902, "max_line_length": 1244, "alphanum_fraction": 0.755526229, "num_tokens": 9414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.7662936324115011, "lm_q1q2_score": 0.6265021151293575}} {"text": "% !TEX root = hott_intro.tex\n\n\\section{Contractible types and contractible maps}\n\\sectionmark{Contractible types and maps}\n\n\\index{contractible type|(} \nA contractible type is a type which has, up to identification, only one term. In other words, a contractible type is a type that comes equipped with a point, and an identification of this point with any point.\n\nWe may think of contractible types as singletons up to homotopy, and indeed we show that the unit type is an example of a contractible type. Moreover, we show that contractible types satisfy an induction principle that is very similar to the induction principle of the unit type, provided that we formulate the computation rule using the identity type rather than postulating a judgmental computation rule.\n\nAnother case of an inductive type with a single constructor is the type of identifications $p:a=x$ with a fixed starting point $a:A$. To specify such an identification, we have to give its end point $x:A$ as well as the identification $p:a=x$, and the path induction principle asserts that in order to show something about all such identifications, it suffices to show that thing in the case where the end point is $a$, and the path is $\\refl{a}$. This suggests that the total space\n\\begin{equation*}\n \\sm{x:A}a=x\n\\end{equation*}\nof all paths with starting point $a:A$ is contractible. This important fact will be shown in \\cref{thm:total_path}, and it is the basis for the fundamental theorem of identity types (\\cref{chap:fundamental}).\n\nIn the remainder of this section we will show that for any equivalence $e:A\\simeq B$ and any $b:B$, the type of all $a:A$ equipped with a path $p:e(a)=b$ is contractible. In other words, if a map $f:A\\to B$ is an equivalence, then the `preimage'\n\\begin{equation*}\n \\sm{a:A}f(a)=b\n\\end{equation*}\nis contractible for each $b:B$. The preimage of a map $f:A\\to B$ at a point $b:B$ is called the fiber of $f$ at $b$, and we say that a map is contractible if all its fibers are contractible. This condition is of course analogous to the set theoretic notion of bijective map, or $1$-to-$1$-correspondence. We will see that a map is contractible if and only if it is an equivalence.\n\n\\subsection{Contractible types}\n\n\\begin{defn}\n We say that a type $A$ is \\define{contractible} if it comes equipped with a term of type\\index{is-cont(A)r@{$\\iscontr(A)$}|see {contractible type}}\n \\begin{equation*}\n \\iscontr(A) \\defeq \\sm{c:A}\\prd{x:A}c=x.\n \\end{equation*}\n Given a term $(c,C):\\iscontr(A)$, we call $c:A$ the \\define{center of contraction}\\index{center of contraction}\\index{contractible type!center of contraction} of $A$, and we call $C:\\prd{x:A}c=x$ the \\define{contraction}\\index{contraction}\\index{contractible type!contraction} of $A$.\n\\end{defn}\n\n\\begin{rmk}\nSuppose $A$ is a contractible type with center of contraction $c$ and contraction $C$. Then the type of $C$ is (judgmentally) equal to the type\n\\begin{equation*}\n\\const_c\\htpy\\idfunc[A].\n\\end{equation*}\nIn other words, the contraction $C$ is a \\emph{homotopy} from the constant function to the identity function.\n\\end{rmk}\n\n\\begin{eg}\n The unit type is easily seen to be contractible.\\index{unit type!is contractible}\\index{is contractible!unit type} For the center of contraction we take $\\ttt:\\unit$. Then we define a contraction $\\prd{x:\\unit}\\ttt=x$ by the induction principle of $\\unit$. Applying the induction principle, it suffices to construct a term of type $\\ttt = \\ttt$, for which we just take $\\refl{\\ttt}$.\n\\end{eg}\n\n\\begin{defn}\n Suppose $A$ comes equipped with a term $a:A$. Then we say that $A$ satisfies \\define{singleton induction}\\index{singleton induction}\\index{induction principle!singleton induction} if for every type family $B$ over $A$, the map\\index{ev-pt@{$\\evpt$}}\n \\begin{equation*}\n \\evpt:\\Big(\\prd{x:A}B(x)\\Big)\\to B(a)\n \\end{equation*}\n defined by $\\evpt(f)\\defeq f(a)$ has a section. In other words, if $A$ satisfies singleton induction we have a function and a homotopy\\index{sing-ind@{$\\singind$}}\\index{sing-comp@{$\\singcomp$}}\n \\begin{align*}\n \\singind_{a} & : B(a)\\to \\prd{x:A}B(x) \\\\\n \\singcomp_{a} & : \\evpt\\circ \\singind_{a} \\htpy \\idfunc\n \\end{align*}\n for any type family $B$ over $A$.\n\\end{defn}\n\n\\begin{eg}\n Note that the singleton induction principle is almost the same as the induction principle for the unit type, the difference being that the \"computation rule\" in the singleton induction for $A$ is stated using an \\emph{identification} rather than as a judgmental equality. The unit type\\index{unit type!singleton induction} $\\unit$ comes equipped with a function\n \\begin{equation*}\n \\indunit:B(\\ttt)\\to \\prd{x:\\unit}B(x)\n \\end{equation*}\n for every type family $B$ over $\\unit$, satisfying the judgmental equality $\\indunit(b,\\ttt)\\jdeq b$ for every $b:B(\\ttt)$ by the computation rule. Thus we easily obtain the homotopy\n \\begin{equation*}\n \\lam{b}\\refl{b}:\\evpt\\circ\\indunit \\htpy\\idfunc,\n \\end{equation*}\n and we conclude that the unit type satisfies singleton induction. \n\\end{eg}\n\n\\begin{thm}\\label{thm:contractible}\nLet $A$ be a type. The following are equivalent:\\index{is contractible!iff singleton induction}\\index{singleton induction!iff contractible}\n\\begin{enumerate}\n\\item The type $A$ is contractible.\n\\item The type $A$ comes equipped with a term $a:A$, and satisfies singleton induction.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nSuppose $A$ is contractible with center of contraction $a$ and contraction $C$. \nFirst we observe that, without loss of generality, we may assume that $C$ comes equipped with an identification $p:C(a)=\\refl{a}$.\nTo see this, note that we can always define a new contraction $C'$ by\n\\begin{equation*}\nC'(x)\\defeq\\ct{C(a)^{-1}}{C(x)},\n\\end{equation*}\nwhich satisfies the requirement by the left inverse law, constructed in \\cref{defn:id_invlaw}.\n\nTo show that $A$ satisfies singleton induction let $B$ be a type family over $A$, and suppose we have $b:B(a)$. Our goal is to define\n\\begin{equation*}\n \\indsing_a(b):\\prd{x:A}B(x).\n\\end{equation*}\nLet $x:A$. Since we have an identification $C(x):a=x$, and a term $b$ in $B(a)$, we may transport $b$ along the path $C(x)$ to obtain\n\\begin{equation*}\n \\indsing_a(b,x)\\defeq \\tr_B(C(x),b):B(x).\n\\end{equation*}\nTherefore, the function $\\indsing_a(b)$ is defined to be the dependent function $\\lam{x}\\tr_B(C(x),b)$. Now we have to show that $\\indsing_a(b,a)=b$. Then we have the identifications\n\\begin{equation*}\n\\begin{tikzcd}\n\\tr_B(C(a),b) \\arrow[r,equals,\"\\ap{\\lam{\\omega}\\tr_B(\\omega,b)}{p}\"] &[4em] \\tr_B(\\refl{a},b) \\arrow[r,equals,\"\\refl{b}\"] & b.\n\\end{tikzcd}\n\\end{equation*}\nThis shows that the computation rule is satisfied, which completes the proof that $A$ satisfies singleton induction.\n\nFor the converse, suppose that $a:A$ and that $A$ satisfies singleton induction. Our goal is to show that $A$ is contractible. For the center of contraction we take the term $a:A$. By singleton induction applied to $B(x)\\defeq a=x$ we have the map \n\\begin{equation*}\n\\indsing_{a} : a=a \\to \\prd{x:A}a=x.\n\\end{equation*}\nTherefore $\\indsing_{a}(\\refl{a})$ is a contraction.\n\\end{proof}\n\n\\begin{thm}\\label{thm:total_path}\nFor any $a:A$, the type\n\\begin{equation*}\n\\sm{x:A}a=x\n\\end{equation*}\nis contractible.\\index{identity type!total space is contractible}\\index{is contractible!total space of identity type}\n\\end{thm}\n\n\\begin{proof}\n We will prove the statement by showing that $\\sm{y:A}x=y$ satisfies singleton induction, and then use \\cref{thm:contractible} to conclude that $\\sm{x:A}a=x$ is contractible. We will use the term $(a,\\refl{a}):\\sm{x:A}a=x$ as the center of contraction.\n\n Now let $P$ be a type family over $\\sm{x:A}a=x$. Note that we have a commuting triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=0]\n \\prd{t:\\sm{x:A}a=x}P(t) \\arrow[dr,swap,\"\\evpt\"] \\arrow[rr,\"\\evpair\"] & & \\prd{x:A}\\prd{p:a=x}P(x,p) \\arrow[dl,\"\\evrefl\"] \\\\\n \\phantom{\\prd{x:A}\\prd{p:a=x}P(x,p)} & P(a,\\refl{a})\n \\end{tikzcd}\n \\end{equation*}\n where the maps $\\evpair$ and $\\evrefl$ are defined as\n \\begin{align*}\n f & \\mapsto \\lam{x}\\lam{p}f(x,p) \\\\\n g & \\mapsto g(a,\\refl{a}),\n \\end{align*}\n respectively. By the induction principle for $\\Sigma$-types it follows that $\\evpair$ has a section, and by path induction it follows that $\\evrefl$ has a section. Therefore it follows from \\cref{ex:3_for_2} that the composite $\\evpt$ has a section.\n\\end{proof}\n\n\\subsection{Contractible maps}\n\n\\index{contractible map|(}\n\\begin{defn}\n Let $f:A\\to B$ be a function, and let $b:B$. The \\define{fiber}\\index{fiber}\\index{homotopy fiber|see {fiber}} of $f$ at $b$ is defined to be the type\\index{fib f b@{$\\fib{f}{b}$}}\n \\begin{equation*}\n \\fib{f}{b}\\defeq\\sm{a:A}f(a)=b.\n \\end{equation*}\n\\end{defn}\n\nIn other words, the fiber of $f$ at $b$ is the type of $a:A$ that get mapped by $f$ to $b$.\nOne may think of the fiber as a type theoretic version of the pre-image\\index{pre-image|see {fiber}} of a point.\n\n\\index{fiber!characterization of identity type|(}\n\\index{characterization of identity type!fiber|(}\n\\index{identity type!of a fiber|(}\nIt will be useful to have a characterization of the identity type of a fiber, so we will make such a characterization immediately.\n\n\\begin{defn}\n Let $f:A \\to B$ be a map, and let $(x,p),(x',p'):\\fib{f}{y}$ for some $y:B$.\n Then we define\\index{Eq fib@{$\\Eqfib$}}\\index{fiber!Eq fib@{$\\Eqfib$}}\n \\begin{equation*}\n \\Eqfib_f((x,p),(x',p'))\\defeq \\sm{\\alpha:x=x'}p=\\ct{\\ap{f}{\\alpha}}{p'}\n \\end{equation*}\n The relation $\\Eqfib_f:\\fib{f}{y}\\to\\fib{f}{y}\\to\\UU$ is a reflexive relation, since we have\n \\begin{equation*}\n \\lam{(x,p)}(\\refl{x},\\refl{p}):\\prd{(x,p):\\fib{f}{y}}\\Eqfib_f((x,p),(x,p)).\n \\end{equation*}\n\\end{defn}\n\n\\begin{lem}\n Consider a map $f:A\\to B$ and let $y:B$. The canonical map\n \\begin{equation*}\n ((x,p)=(x',p'))\\to\\Eqfib_f((x,p),(x',p'))\n \\end{equation*}\n induced by the reflexivity of $\\Eqfib_f$ is an equivalence for any $(x,p),(x',p'):\\fib{f}{y}$.\n\\end{lem}\n\n\\begin{proof}\n The converse map\n \\begin{equation*}\n \\Eqfib_f((x,p),(x',p'))\\to ((x,p)=(x',p'))\n \\end{equation*}\n is easily defined by $\\Sigma$-induction, and then path induction twice. The homotopies witnessing that this converse map is indeed a right inverse as well as a left inverse is similarly constructed by induction.\n\\end{proof}\n\n\\index{fiber!characterization of identity type|)}\n\\index{characterization of identity type!fiber|)}\n\\index{identity type!of a fiber|)}\n\nNow we arrive at the notion of contractible map.\n\n\\begin{defn}\nWe say that a function $f:A\\to B$ is \\define{contractible} if there is a term of type\\index{is-contr(f)@{$\\iscontr(f)$}|see {contractible map}}\\index{is a contractible map}\n\\begin{equation*}\n\\iscontr(f)\\defeq\\prd{b:B}\\iscontr(\\fib{f}{b}).\n\\end{equation*}\n\\end{defn}\n\n\\begin{thm}\\label{thm:equiv_contr}\nAny contractible map is an equivalence.\\index{contractible map!is an equivalence}\\index{is an equivalence!contractible map}\n\\end{thm}\n\n\\begin{proof}\nLet $f:A\\to B$ be a contractible map. Using the center of contraction of each $\\fib{f}{y}$, we obtain a term of type\n\\begin{align*}\n\\lam{y}\\pairr{g(y),G(y)}:\\prd{y:B}\\fib{f}{y}.\n\\end{align*}\nThus, we get map $g:B\\to A$, and a homotopy $G:\\prd{y:B} f(g(y))=y$. In other words, we get a section of $f$.\n\nIt remains to construct a retraction of $f$. Taking $g$ as our retraction, we have to show that $\\prd{x:A} g(f(x))=x$. Note that we get an identification $p:f(g(f(x)))=f(x)$ since $g$ is a section of $f$. It follows that $(g(f(x)),p):\\fib{f}{f(x)}$. Moreover, since $\\fib{f}{f(x)}$ is contractible we get an identification $q:\\pairr{g(f(x)),p}=\\pairr{x,\\refl{f(x)}}$. The base path $\\ap{\\proj 1}{q}$ of this identification is an identification of type $g(f(x))=x$, as desired.\n\\end{proof}\n\n\\subsection{Equivalences are contractible maps}\n\nIn \\cref{thm:contr_equiv} we will show the converse to \\cref{thm:equiv_contr}, i.e., we will show that any equivalence is a contractible map. We will do this in two steps.\n\nFirst we introduce a new notion of \\emph{coherently invertible map}, for which we can easily show that such maps have contractible fibers. Then we show that any equivalence is a coherently invertible map.\n\n Recall that an invertible map is a map $f:A\\to B$ equipped with $g:B\\to A$ and homotopies\n \\begin{equation*}\n G : f\\circ g \\htpy \\idfunc\\qquad\\text{and}\\qquad H:g\\circ f\\htpy \\idfunc.\n \\end{equation*}\n Then we observe that both $G \\cdot f$ and $f \\cdot H$ are homotopies of the same type\n \\begin{equation*}\n f\\circ g\\circ f \\htpy f.\n \\end{equation*}\n A coherently invertible map is an invertible map for which there is a further homotopy $G \\cdot f\\htpy f\\cdot H$.\n\n \\begin{defn}\n Consider a map $f:A\\to B$. We say that $f$ is \\define{coherently invertible}\\index{coherently invertible map} if it comes equipped with\n \\begin{align*}\n g & : B \\to A \\\\\n G & : f \\circ g \\htpy \\idfunc \\\\\n H & : g \\circ f \\htpy \\idfunc \\\\\n K & : G \\cdot f \\htpy f \\cdot H.\n \\end{align*}\n We will write $\\iscohinvertible(f)$\\index{is-coh-invertible(f)@{$\\iscohinvertible(f)$}} for the type of quadruples $(g,G,H,K)$.\n \\end{defn}\n\n Although we will encounter the notion of coherently invertible map on some further occasions, the following lemma is our main motivation for considering it.\n\n \\begin{lem}\\label{lem:contr-inv}\n Any coherently invertible map has contractible fibers.\\index{coherently invertible map!is a contractible map}\n \\end{lem}\n\n \\begin{proof}\n Consider a map $f:A\\to B$ equipped with\n \\begin{align*}\n g & : B \\to A \\\\\n G & : f \\circ g \\htpy \\idfunc \\\\\n H & : g \\circ f \\htpy \\idfunc \\\\\n K & : G \\cdot f \\htpy f \\cdot H,\n \\end{align*}\n and let $y:B$. Our goal is to show that $\\fib{f}{y}$ is contractible. For the center of contraction we take $(g(y),G(y))$. In order to construct a contraction, it suffices to construct a term of type\n \\begin{equation*}\n \\prd{x:A}\\prd{p:f(x)=y}\\Eqfib_f((g(y),G(y)),(x,p)).\n \\end{equation*}\n By path induction on $p:f(x)=y$ it suffices to construct a term of type\n \\begin{equation*}\n \\prd{x:A}\\Eqfib_f((g(f(x)),G(f(x))),(x,\\refl{f(x)})).\n \\end{equation*}\n By definition of $\\Eqfib_f$, we have to construct a term of type\n \\begin{equation*}\n \\prd{x:A}\\sm{\\alpha:g(f(x))=x}G(f(x))=\\ct{\\ap{f}{\\alpha}}{\\refl{f(x)}}.\n \\end{equation*}\n Such a term is constructed as $\\lam{x}(H(x),K'(x))$, where the homotopy $H:g\\circ f\\htpy \\idfunc$ is given by assumption, and the homotopy\n \\begin{align*}\n K' & : \\prd{x:A}G(f(x))=\\ct{\\ap{f}{H(x)}}{\\refl{f(x)}}\n \\end{align*}\n is defined as\n \\begin{equation*}\n K'\\defeq \\ct{K}{\\rightunithtpy(f\\cdot H)^{-1}}.\\qedhere\n \\end{equation*}\n \\end{proof}\n\n Our next goal is to show that for any map $f:A\\to B$ equipped with\n \\begin{equation*}\n g:B\\to A,\\qquad G:f\\circ g \\htpy \\idfunc,\\qquad\\text{and}\\qquad H:g\\circ f\\htpy \\idfunc,\n \\end{equation*}\n we can improve the homotopy $G$ to a new homotopy $G':f\\circ g\\htpy \\idfunc$ for which there is a further homotopy\n \\begin{equation*}\n f\\cdot H\\htpy G'\\cdot f.\n \\end{equation*}\n Note that this situation is analogous to the situation in the proof of \\cref{thm:contractible}, where we improved the contraction $C$ so that it satisfied $C(c)=\\refl{}$. The extra coherence $f\\cdot H\\htpy G'\\cdot f$ is then used in the proof that the fibers of an equivalence are contractible.\n\n\\begin{defn}\\label{defn:htpy_nat}\\index{homotopy!naturality}\nLet $f,g:A\\to B$ be functions, and consider $H:f\\htpy g$ and $p:x=y$ in $A$. We define the identification\\index{nat-nat@{$\\nathtpy$}}\\index{homotopy!nat-htpy@{$\\nathtpy$}}\n\\begin{equation*}\n\\nathtpy(H,p) \\defeq \\ct{\\ap{f}{p}}{H(y)}=\\ct{H(x)}{\\ap{g}{p}}\n\\end{equation*}\nwitnessing that the square\n\\begin{equation*}\n\\begin{tikzcd}\nf(x) \\arrow[r,equals,\"H(x)\"] \\arrow[d,equals,swap,\"\\ap{f}{p}\"] & g(x) \\arrow[d,equals,\"\\ap{g}{p}\"] \\\\\nf(y) \\arrow[r,equals,swap,\"H(y)\"] & g(y)\n\\end{tikzcd}\n\\end{equation*}\ncommutes. This square is also called the \\define{naturality square}\\index{naturality square of homotopies} of the homotopy $H$ at $p$.\n\\end{defn}\n\n\\begin{constr}\n By path induction on $p$ it suffices to construct an identification\n \\begin{equation*}\n \\ct{\\ap{f}{\\refl{x}}}{H(x)}=\\ct{H(x)}{\\ap{g}{\\refl{x}}}\n \\end{equation*}\n since $\\ap{f}{\\refl{x}}\\jdeq \\refl{f(x)}$ and $\\ap{g}{\\refl{x}}\\jdeq\\refl{g(x)}$, and since $\\ct{\\refl{f(x)}}{H(x)}\\jdeq H(x)$, we see that the path $\\rightunit(H(x))^{-1}$ is of the asserted type.\n\\end{constr}\n\n\\begin{defn}\\label{defn:retraction_swap}\nConsider $f:A\\to A$ and $H: f\\htpy \\idfunc[A]$. We construct an identification $H(f(x))=\\ap{f}{H(x)}$, for any $x:A$.\n\\end{defn}\n\n\\begin{constr}\nBy the naturality of homotopies with respect to identifications the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large]\nff(x) \\arrow[d,swap,equals,\"\\ap{f}{H(x)}\"] \\arrow[r,equals,\"H(f(x))\"] & f(x) \\arrow[d,equals,\"H(x)\"] \\\\\nf(x) \\arrow[r,swap,equals,\"H(x)\"] & x\n\\end{tikzcd}\n\\end{equation*}\ncommutes. This gives the desired identification $H(f(x))=\\ap{f}{H(x)}$.\n\\end{constr}\n\n\\begin{lem}\\label{lem:coherently-invertible}\n Let $f:A\\to B$ be a map, and consider $(g,G,H):\\hasinverse(f)$. Then there is a homotopy $G':f\\circ g\\htpy \\idfunc$ equipped with a further homotopy\n \\begin{equation*}\n K : G'\\cdot f \\htpy f\\cdot H.\n \\end{equation*}\n Thus we obtain a map $\\hasinverse(f)\\to\\iscohinvertible(f)$.\n\\end{lem}\n\n\\begin{proof}\n For each $y:B$, we construct the identification $G'(y)$ as the concatenation\n \\begin{equation*}\n \\begin{tikzcd}\n fg(y) \\arrow[r,equals,\"{G(fg(y))}^{-1}\"] &[2.5em] fgfg(y) \\arrow[r,equals,\"\\ap{f}{H(g(y))}\"] &[2.5em] fg(y) \\arrow[r,equals,\"G(y)\"] & y.\n\\end{tikzcd}\n \\end{equation*}\n In order to construct a homotopy $G'\\cdot f\\htpy f\\cdot H$, it suffices to show that the square\n \\begin{equation*}\n \\begin{tikzcd}[column sep=8em]\n fgfgf(x) \\arrow[r,equals,\"{G(fgf(x))}\"] \\arrow[d,equals,swap,\"\\ap{f}{H(gf(x))}\"] & fgf(x) \\arrow[d,equals,\"\\ap{f}{H(x)}\"] \\\\\n fgf(x) \\arrow[r,equals,swap,\"G(f(x))\"] & f(x)\n \\end{tikzcd}\n \\end{equation*}\n commutes for every $x:A$.\n Recall from \\cref{defn:retraction_swap} that we have $H(gf(x))=\\ap{gf}{H(x)}$. Using this identification, we see that it suffices to show that the square\n \\begin{equation*}\n \\begin{tikzcd}[column sep=8em]\n fgfgf(x) \\arrow[r,equals,\"G(fgf(x))\"] \\arrow[d,equals,swap,\"\\ap{fgf}{H(x)}\"] & fgf(x) \\arrow[d,equals,\"\\ap{f}{H(x)}\"] \\\\\n fgf(x) \\arrow[r,equals,swap,\"G(f(x))\"] & f(x)\n \\end{tikzcd}\n \\end{equation*}\n commutes. Now we observe that this is just a naturality square the homotopy $Gf:fgf\\htpy f$, which commutes by \\cref{defn:htpy_nat}.\n\\end{proof}\n\nNow we put the pieces together to conclude that any equivalence has contractible fibers.\n\n\\begin{thm}\\label{thm:contr_equiv}\nAny equivalence is a contractible map.\\index{equivalence!is a contractible map}\\index{is a contractible map!equivalence}\\index{is contractible!fiber of an equivalence}\n\\end{thm}\n\n\\begin{proof}\n We have seen in \\cref{lem:contr-inv} that any coherently invertible map is a contractible map. Moreover, any equivalence has the structure of an invertible map by \\cref{lem:inv_equiv}, and any invertible map is coherently invertible by \\cref{lem:coherently-invertible}.\n\\end{proof}\n\n\\begin{cor}\\label{cor:contr_path}\nLet $A$ be a type, and let $a:A$. Then the type\\index{is contractible!total space of opposite identity type}\n\\begin{equation*}\n\\sm{x:A}x=a\n\\end{equation*}\nis contractible.\n\\end{cor}\n\n\\begin{proof}\nBy \\cref{thm:id_equiv}, the identity function is an equivalence. Therefore, the fibers of the identity function are contractible by \\cref{thm:contr_equiv}. Note that $\\sm{x:A}x=a$ is exactly the fiber of $\\idfunc[A]$ at $a:A$.\n\\end{proof}\n\\index{contractible map|)}\n\n\\begin{exercises}\n\\exercise \\label{ex:prop_contr}Show that if $A$ is contractible, then for any $x,y:A$ the identity type $x=y$ is also contractible.\\index{contractible type!identity types of}\\index{is contractible!identity type of contractible type}\\index{identity type!of contractible type}\\index{characterization of identity type!contractible type}\n\\exercise \\label{ex:contr_retr}Suppose that $A$ is a retract of $B$. Show that\\index{contractible type!closed under retracts}\n \\begin{equation*}\n \\iscontr(B)\\to\\iscontr(A).\n \\end{equation*}\n\\exercise \\label{ex:contr_equiv}\n \\begin{subexenum}\n \\item Show that for any type $A$, the map $\\const_\\ttt : A\\to \\unit$ is an equivalence if and only if $A$ is contractible.\\index{contractible type!is equivalent to 1@{is equivalent to $\\unit$}}\n \\item Apply \\cref{ex:3_for_2} to show that for any map $f:A\\to B$, if any two of the three assertions\\index{contractible type!3-for-2 property}\\index{3-for-2 property!of contractible types}\n \\begin{enumerate}\n \\item $A$ is contractible\n \\item $B$ is contractible\n \\item $f$ is an equivalence\n \\end{enumerate}\n hold, then so does the third.\n \\end{subexenum}\n\\exercise Show that for any two types $A$ and $B$, the following are equivalent:\n \\index{contractible type!closed under cartesian product}\n \\index{is contractible!factor of contractible cartesian product}\n \\begin{enumerate}\n \\item Both $A$ and $B$ are contractible.\n \\item The type $A\\times B$ is contractible.\n \\end{enumerate}\n\\exercise \\label{ex:contr_in_sigma} Let $A$ be a contractible type with center of contraction $a:A$. Furthermore, let $B$ be a type family over $A$. Show that the map $y\\mapsto\\pairr{a,y}:B(a)\\to\\sm{x:A}B(x)$ is an equivalence.\\index{left unit law!of Sigma-types@{of $\\Sigma$-types}}\\index{dependent pair type!left unit law}\n\\exercise \\label{ex:proj_fiber}Let $B$ be a family of types over $A$, and consider the projection map \n \\begin{equation*}\n \\proj 1 : \\big(\\sm{x:A}B(x)\\big)\\to A.\n \\end{equation*}\n Show that for any $a:A$, the map\n \\begin{equation*}\n \\lam{((x,y),p)} \\tr_B(p,y) : \\fib{\\proj 1}{a} \\to B(a),\n \\end{equation*}\n is an equivalence. Conclude that $\\proj 1$ is an equivalence if and only if each $B(a)$ is contractible.\n \\index{family!fibers of projection map}\n \\index{pr 1@{$\\proj 1$}!of contractible family is an equivalence}\n \\index{is an equivalence!pr 1 of contractible family@{$\\proj 1$ of contractible family}}\n\\exercise \\label{ex:fib_replacement}Construct for any map $f:A\\to B$ an equivalence $e:\\eqv{A}{\\sm{y:B}\\fib{f}{y}}$ and a homotopy $H:f\\htpy \\proj 1\\circ e$ witnessing that the triangle\n \\begin{equation*}\n \\begin{tikzcd}[column sep=0em]\n A \\arrow[rr,\"e\"] \\arrow[dr,swap,\"f\"] & & \\sm{y:B}\\fib{f}{y} \\arrow[dl,\"\\proj 1\"] \\\\\n \\phantom{\\sm{y:B}\\fib{f}{y}} & B\n \\end{tikzcd}\n \\end{equation*}\n commutes. The projection $\\proj 1 : (\\sm{y:B}\\fib{f}{y})\\to B$ is sometimes also called the \\define{fibrant replacement}\\index{fibrant replacement} of $f$, because first projection maps are fibrations in the homotopy interpretation of type theory.\n\\end{exercises}\n\\index{contractible type|)}\n", "meta": {"hexsha": "e904237bce6b990576de5585901832ebffab433d", "size": 23032, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/contractible.tex", "max_stars_repo_name": "hemangandhi/HoTT-Intro", "max_stars_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 333, "max_stars_repo_stars_event_min_datetime": "2018-09-26T08:33:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T23:50:15.000Z", "max_issues_repo_path": "Book/contractible.tex", "max_issues_repo_name": "hemangandhi/HoTT-Intro", "max_issues_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2019-06-18T04:16:04.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-16T15:27:01.000Z", "max_forks_repo_path": "Book/contractible.tex", "max_forks_repo_name": "hemangandhi/HoTT-Intro", "max_forks_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 30, "max_forks_repo_forks_event_min_datetime": "2018-09-26T09:08:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-16T00:33:50.000Z", "avg_line_length": 54.4491725768, "max_line_length": 482, "alphanum_fraction": 0.6934699548, "num_tokens": 7600, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8267117983401364, "lm_q1q2_score": 0.6264775564926846}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n%%% \\documentclass{article}\n%%% \\usepackage{graphicx}\n%%% \\usepackage{color}\n\n%%% \\sloppy\n%%% \\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n%%% \\begin{document}\n\n \n \n\\subsection*{Spurious Free Dynamic Range by means of Fast fourier transform}\n\n\\begin{par}\nExample for algorithm SFDR.\n\\end{par} \\vspace{1em}\n\\begin{par}\nCalculates SFDR by calculating FFT spectrum.\n\\end{par} \\vspace{1em}\n\n\\subsubsection*{Contents}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item Generate sample data\n \\item Call algorithm\n \\item Display results\n\\end{itemize}\n\n\n\\subsubsection*{Generate sample data}\n\n\\begin{par}\nFirst quantity \\lstinline{y} representing 1 second of signal containing spurious component is prepared. Main signal component has nominal frequency 1 kHz, nominal amplitude 2 V, nominal phase 1 rad and offset 1 V sampled at sampling frequency 10 kHz.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDI = [];\nfsnom = 1e4; Anom = 4; fnom = 100; phnom = 1; Onom = 0.2;\nt = [0:1/fsnom:1-1/fsnom];\nDI.y.v = Anom*sin(2*pi*fnom*t + phnom);\n\\end{lstlisting}\n\\begin{par}\nA spurious component with amplitude at 1/100 of main carrier frequency is added. Thus by definition the SFDR in dBc has to be 40.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDI.y.v = DI.y.v + Anom./100*sin(2*pi*fnom*3.5*t + phnom);\n\\end{lstlisting}\n\n\n\\subsubsection*{Call algorithm}\n\n\\begin{par}\nUse QWTB to apply algorithm \\lstinline{SFDR} to data \\lstinline{DI}.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDO = qwtb('SFDR', DI);\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\nQWTB: no uncertainty calculation\n\\end{lstlisting} \\color{black}\n \n\n\\subsubsection*{Display results}\n\n\\begin{par}\nResult is the SFDR (dBc).\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nSFDR = DO.SFDRdBc.v\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\n\nSFDR =\n\n 40.0000\n\n\\end{lstlisting} \\color{black}\n \n\n\n%%% \\end{document}\n \n", "meta": {"hexsha": "2df64dbdd009c4de6cc14f0dd5a194714f5eb054", "size": 2095, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/algs_examples_published/doc_SFDR.tex", "max_stars_repo_name": "qwtb/qwtb", "max_stars_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-12-09T13:18:54.000Z", "max_stars_repo_stars_event_max_datetime": "2015-12-09T13:18:54.000Z", "max_issues_repo_path": "doc/algs_examples_published/doc_SFDR.tex", "max_issues_repo_name": "qwtb/qwtb", "max_issues_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 18, "max_issues_repo_issues_event_min_datetime": "2015-12-09T13:08:38.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-13T11:33:41.000Z", "max_forks_repo_path": "doc/algs_examples_published/doc_SFDR.tex", "max_forks_repo_name": "qwtb/qwtb", "max_forks_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-11-11T02:12:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-09-17T12:59:18.000Z", "avg_line_length": 23.2777777778, "max_line_length": 250, "alphanum_fraction": 0.7107398568, "num_tokens": 668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117983401363, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6264775519705877}} {"text": "\\documentclass[11pt]{article}\n\n\\input{preamble.tex}\n\n\\begin{document}\n \n % \\maketitle\n \n \\hypertarget{smooth-manifolds-made-concrete-for-dumbasses}{%\n\\section{Smooth Manifolds made Concrete for Dumbasses}\\label{smooth-manifolds-made-concrete-for-dumbasses}}\n\nThe intention behind this document is to provide an extended and comprehensive example of a smooth\nmanifold for dumbasses like me who can't always grok the heavily abstract without a more concrete\nexample, or as the OO framework types would say, a reification. The extended example is the\nprototypical/canonical stereographic projection considering the sphere \\(S^2\\) as a manifold.\nExamples cover deriving the charts, finding the inverses functions between manifolds including\nderiving an ellipsoid stereographic manifold, and tangent spaces including both equivalence classes\nof curves and derivation based approaches, illustrating proofs of vector space closure and change\nof basis of tangent vectors. The examples are based on using the open source\n\\href{https://www.sagemath.org/}{SageMath} Computer Algebra System (CAS) to create an interactive\nnotebook. Sage is available on most Linux distributions and the web site also has a Windows\ninstaller. The notebook has also been converted to a Latex/SageTex document which hides most of\nthe Sage specific manipulation and shows just the mathematics for those who aren't interested in\nSage (although you lose the ability to interact with the content including the graphs).\nNote Sage does have \n\\href{http://doc.sagemath.org/html/en/reference/manifolds/sage/manifolds/differentiable/manifold.html}{built in Manifold}\nsupport, but this document won't be using it as the intention is to provide examples from first principles.\nErrors or misunderstandings on my part or also more than likely as I am a computer scientist who dabbles in mathematics,\nfor Computer Vision, Machine Learning and Neural Networks, therefore any error corrections or suggestions \nfor improvements would also be welcome.\n\n\\begin{sagesilent}\nclear_vars()\n\ndef stereographic(pole):\n o_n = vector(SR, pole)\n var('x', 'y', 'z', 'u', 'v', 'l')\n assume(x, 'real', y, 'real', z, 'real', u, 'real', v, 'real', l, 'real')\n# p = vector(SR, (x, y, sqrt(x^2 + y^2)))\n p = vector(SR, (x, y, z))\n s = vector(SR, (u, v, 0))\n lhs = o_n - p\n rhs = l*(o_n - s)\n eq1 = lhs[0] == rhs[0]\n eq2 = lhs[1] == rhs[1]\n eq3 = lhs[2] == rhs[2]\n l_s = solve(eq3, l)[0].rhs()\n return (solve(eq1.substitute(l=l_s), u)[0], solve(eq2.substitute(l=l_s), v)[0])\n\ndef the_solution(sols, submap):\n for sol in sols:\n if sol.rhs().substitute(submap) != 0:\n return sol\n return None\n\no_n = vector(SR, (0,0,1))\nvar('x', 'y', 'z', 'u', 'v', 'l')\nassume(x, 'real', y, 'real', z, 'real', u, 'real', v, 'real', l, 'real')\np = vector(SR, (x, y, z))\ns = vector(SR, (u, v, 0))\nlhs = o_n - p\nrhs = l*(o_n - s)\neq1 = lhs[0] == rhs[0]\neq2 = lhs[1] == rhs[1]\neq3 = lhs[2] == rhs[2] \n\\end{sagesilent}\n\n\\hypertarget{the-stereographic-projection-atlas}{%\n\\subsection{The Stereographic Projection\nAtlas}\\label{the-stereographic-projection-atlas}}\n\nThe sterographic projection is the canonical example for a\ntopological/smooth manifold. The example shown here will map \\(S^2\\) to\n\\(R^2\\), although it can be extended to \\(S^n \\rightarrow R^n\\).\n\nThe sphere \\(S^2\\) is seen as a manifold having an atlas with two open\nsets and associated charts. The first chart is defined on the open set\ndefined by the full sphere excluding the north (top) pole \\((0,0,1)\\),\nwhile the second chart is defined on a set that excludes only the south\n(bottom) \\((0,0,1)\\) pole. Both charts project a point on the sphere\nsurface to the plane at \\(z=0\\) by extending a line from the excluded\npole through the point on the manifold (sphere) being charted onto the\nplane.\n\nTo derive the chart equation let \\(\\vec{o_n} = (0, 0, 1)\\),\n\\(\\vec{p_n} = (x, y, z)\\) be a point on the sphere and\n\\(\\vec{s_n} = (u, v, 0)\\) be the projection then (see the figure below):\n\n\\(\\vec{o_n} - \\vec{p_n} = l(\\vec{o_n} - \\vec{s_n}\\))\nwhich evaluates to: $\\sage{lhs} = \\sage{rhs}$\n \n Solving for the third component of the vector equation and substituting\ninto the first two:\n\n\\begin{sagesilent}\nsol1 = solve(eq3, l)[0]\nsol2 = solve(eq1, u)[0], solve(eq2, v)[0]\n\\end{sagesilent}\n\\begin{align*} \n \\sage{sol1} \\\\\n \\sage{sol2}\n\\end{align*} \n \n\\begin{sagesilent}\nOn = (0, 0, 1)\nOs = (0, 0, -1)\nCn = stereographic(On)\nCs = stereographic(Os)\ncn_0 = Cn[0].rhs()\ncn_1 = Cn[1].rhs()\ncs_0 = Cs[0].rhs()\ncs_1 = Cs[1].rhs()\n\\end{sagesilent}\n\n \\(\\therefore\\) the charts with chart coordinates given by \\(u\\) and \\(v\\) are given by.\n\nNorth pole:\\\\\n\n$(\\sage{cn_0}, \\sage{cn_1})$ \n \n South pole:\\\\\n\n $(\\sage{cn_0}, \\sage{cn_1})$ \n \n \\hypertarget{visualization}{%\n\\subsection{Visualization}\\label{visualization}}\n\nTo visualize the manifold and charts see Figure \\ref{fig:stereo1} which demonstrates manifold points\n\\((\\frac{1}{2}, \\frac{1}{2},\\frac{1}{\\sqrt{2}})\\) and \\((\\frac{1}{2}, \\frac{1}{2},-\\frac{1}{\\sqrt{2}})\\) \nprojected to the plane z = 0 (it does look a lot better in the original Jupyter notebook):\n\n\\begin{sagesilent}\nfrom sage.plot.plot3d.shapes2 import Line\ndef to_float(v): # Temp fix to convert to float from Sage type (see https://trac.sagemath.org/ticket/28949) for Sage 9\n vv = [None] * len(v)\n for i in range(0, len(v)):\n vv[i] = float(v[i])\n return vv \ntwosq = 1/1.4142\nPn = (0.5, 0.5, twosq)\nPs = (0.5, 0.5, -twosq)\nSn = sphere(center=(0, 0, 0),size=1, color='green', aspect_ratio=[1,1,1], opacity=5/10)\nFn = plot3d(lambda x, y: 0, (-2,2), (-2,2))\non = point(On, rgbcolor=(1,0,0), size=20)\nos = point(Os, rgbcolor=(0,0,0), size=20)\npn = point(Pn, rgbcolor=(1,0,0), size=20)\ntpn = text3d(' (%.2f,%.2f,%.2f)'%(Pn[0], Pn[1], Pn[2]), (Pn[0], Pn[1], Pn[2]+0.2), horizontal_alignment='left',color='red', fonsize='x-small')\nps = point(Ps, rgbcolor=(0,0,0), size=20)\ntps = text3d(' (%.2f,%.2f,%.2f)'%(Ps[0], Ps[1], Ps[2]), (Ps[0], Ps[1], Ps[2]-0.2), horizontal_alignment='left',color='black', fonsize='x-small')\nBn = (Cn[0].substitute({x:Pn[0], z:Pn[2]}).rhs(), Cn[1].substitute({y:Pn[1], z:Pn[2]}).rhs(), 0)\nbn = point(Bn, rgbcolor=(0,0,0), size=20)\ntbn = text3d(' (%.2f,%.2f)'%(Bn[0], Bn[1]), (Bn[0], Bn[1]+0.6, 0.2), horizontal_alignment='left',color='red', fonsize='x-small')\nBs = (Cs[0].substitute({x:Ps[0], z:Ps[2]}).rhs(), Cs[1].substitute({y:Ps[1], z:Ps[2]}).rhs(), 0)\nbs = point(Bs, rgbcolor=(0,0,0), size=20)\n#Ln = Line([On, Pn, Bn], color='red')\n#Ls = Line([Os, Ps, Bs], color='black')\nLn = Line([to_float(On), to_float(Pn), to_float(Bn)], color='red')\nLs = Line([to_float(Os), to_float(Ps), to_float(Bs)], color='black')\n#show(Sn+Fn+on+pn+tpn+bn+tbn+os+ps+tps+bs+Ln+Ls, figsize=8)\nsave(Sn+Fn+on+pn+tpn+bn+tbn+os+ps+tps+bs+Line([On, Pn, Bn], color='red')+Line([Os, Ps, Bs], color='black'), \"stereo1.png\")\n\\end{sagesilent}\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=0.95\\textwidth,keepaspectratio]{stereo1.png}\n \\caption{Visualization of stereographic projection.}\n \\label{fig:stereo1}\n \\end{figure}\n \n \\hypertarget{chart-inverses}{%\n\\subsection{Chart Inverses}\\label{chart-inverses}}\n\\begin{sagesilent}\nvar('x', 'y', 'z', 'u', 'v', 'l')\nassume(x, 'real', y, 'real', z, 'real', u, 'real', v, 'real', l, 'real')\no_n = vector(SR, On)\nuv = vector(SR, (u, v, 0))\nd = (o_n - uv).normalized()\nLn = o_n + d*l\n\\end{sagesilent}\n\nThe above charts are bijective and continuous within the domain\n\\(S^2 - (0, 0, 1), S^2 - (0, 0, -1)\\) and codomain \\((u, v) \\in R^2\\),\nhowever to be valid charts they must also be bicontinuous, ie the\ninverse must be continuous too, that is they must be homeomorphic to\n\\(R^2\\).\n\nTo find the inverse first find the unit vector in the direction of the\nline joining the pole to the projected coordinates, for example\n\\(\\vec{d} = \\frac{((0, 0, 1) - (u, v, 0))}{|(0, 0, 1) - (u, v, 0))|}\\) = $\\sage{d}$.\n \n Now express the line from the pole to the projection as a vector\nequation:\\\\ \n\\((0, 0, 1) + l \\vec{d}\\) = $\\sage{Ln}$\n\n\\begin{sagesilent}\nc_o = vector(SR, (0, 0, 0))\neq4 = (Ln - c_o).dot_product(Ln - c_o) == 1\nsols = solve(eq4, l)\nsolution = the_solution(sols, {u:1, v:1})\nIn = (Ln - c_o).substitute(l=solution.rhs())\npretty_print(In)\ncn_ix = In[0]\ncn_iy = In[1]\ncn_iz = In[2]\n\\end{sagesilent}\n \n The intersection of this line with the sphere occurs where\n\\(|(0, 0, 1) + l\\vec{d} - (0, 0, 0)| = 1\\), or more generally\n\\(|(0, 0, 1) + l\\vec{d} - \\vec{c}| = r\\) where \\(\\vec{c}\\) is the sphere\ncentre and r is the radius. This can also be expressed as\n\\([(0, 0, 1) + l\\vec{d} - (0, 0, 0)] \\cdot [(0, 0, 1) + l\\vec{d} - (0, 0, 0)] = r\\):\\\\\n$\\sage{eq4}$\n \n This quadratic will have one solution for \\(l = 0\\) at the pole. The\nother solution will be the required \\((x, y, z)\\) on the sphere\nmanifold. Letting Sage do the PT:\\\\\n% \\begin{tcolorbox}\n% sols = solve(eq4, l)\\\\\n% \\end{tcolorbox}\n$\\sage{sols}$\n\n and substituting the appropriate \\(l\\) back into the line equation:\\\\\n $\\sage{In}$ \n \n See also the\n\\href{https://en.wikipedia.org/wiki/Line\\%E2\\%80\\%93sphere_intersection}{Wikipeadia}\npage on sphere-line intersection (note their notation swops \\(\\vec{d}\\)\nand \\(l\\) as used in the above).\n\n\\begin{sagesilent}\no_s = vector(SR, Os)\nd = (o_s - uv).normalized()\nLs = o_s + d*l\neq5 = (Ls - c_o).dot_product(Ls - c_o) == 1\nsols = solve(eq5, l)\nsolution = the_solution(sols, {u:1, v:1}) \nIs = (Ls - c_o).substitute(l=solution.rhs())\ncs_ix = Is[0]\ncs_iy = Is[1]\ncs_iz = Is[2] \n\\end{sagesilent}\n \n Similarly the inverse for the south pole chart is given by (it could\nprobably also be derived by symmetry):\\\\\n$\\sage{Is}$\n \n \\hypertarget{verifying-the-charts}{%\n\\subsection{Verifying the Charts}\\label{verifying-the-charts}}\n\\begin{sagesilent}\n uu = cs_0.substitute({x: cn_ix, z: cn_iz}).full_simplify()\n vv = cs_1.substitute({y: cn_iy, z: cn_iz}).full_simplify()\n pretty_print(\"(\", uu, \", \", vv, \")\") \n\\end{sagesilent}\n\n The two charts described above are homeomorphic and are also\ndifferentiable with a differentiable inverse, therefore they are\ndiffeomorphic too. Assume the charts are named \\(c_n\\) and \\(c_s\\), then\nto verify the two charts define a compatible atlas it is required that\nthe intersection of their domain open sets be empty ie\n\\(S^2 - (0, 0, 1) \\cap S^2 - (0, 0, -1) = \\emptyset,\\) (definitely not\nthe case) or\n\\(c_s \\circ c_n^{-1} : c_n(S^2 - (0, 0, 1) \\cap S^2 - (0, 0, -1)) \\rightarrow c_s(S^2 - (0, 0, 1) \\cap S^2 - (0, 0, -1))\\)\nand the opposite (\\(c_n \\circ c_s^{-1}\\)) is diffeomorphic. After\nperforming the composition we get:\\\\\n$(\\sage{uu}, \\sage{vv})$\n\n which is homeomorphic and smooth over the specified domain.\n\n \\hypertarget{smooth-functions-between-manifolds}{%\n\\subsection{Smooth Functions between\nManifolds}\\label{smooth-functions-between-manifolds}}\n\nA smooth map \\(F: M \\rightarrow N\\) can be described \\parencite{lee} as\n\\say{if for every \\(p \\in M\\), there exist smooth charts\n\\((U, \\phi)\\) containing \\(p\\) and \\((V, \\psi)\\) containing \\(F(p)\\)\nsuch that \\(F(U) \\subseteq V\\) and the composite map\n\\(\\psi \\circ F \\circ \\phi^{-1}\\) is smooth from \\(\\phi(U)\\) to\n\\(\\psi(V)\\)}.\n\nWe will start by defining a manifold E on an ellipsoid to use as the\ndestination manifold corresponding to N above for the map. U will then\ncorrespond to either \\(S^2 - (0, 0, 1)\\) or \\(S^2 - (0, 0, -1)\\)\ndepending on the source chart selection. N will similarly correspond to\n\\(E - (0, 0, 1)\\) or \\(E - (0, 0, -1)\\). Because E uses $ (0, 0, 1)$\nand $(0, 0, -1)$ the a and b ellipse axes are arbitarily defined but\n\\(c = 1\\) (the z semi-axis) to ensure that the poles are at\n\\((0, 0, 1)\\) and \\((0, 0, -1)\\).\n\n\\hypertarget{the-ellipsoid-as-a-stereographical-manifold}{%\n\\subsubsection{The Ellipsoid as a Stereographical Manifold}\\label{the-ellipsoid-as-a-stereographical-manifold}}\n\\begin{sagesilent}\nen_0 = cn_0\nen_1 = cn_1\nes_0 = cs_0\nes_1 = cs_1\nvar('x', 'y', 'z', 'u', 'v', 'l', 'a', 'b', 'c')\nassume(x, 'real', y, 'real', z, 'real', u, 'real', v, 'real', l, 'real', a, 'real', b, 'real', c, 'real')\no_n = vector(SR, (0, 0, 1))\nuv = vector(SR, (u, v, 0))\nd = (o_n - uv).normalized()\nLn = o_n + d*l - vector(SR, (0, 0, 0))\n\\end{sagesilent}\n \nThe chart map from the ellipsoid to \\(R^2\\) can be derived similarly to\nthe sphere,resulting in \\(u=-\\frac{x}{z-1}, v=-\\frac{y}{z-1}\\) for the\nnorth chart and \\(u=\\frac{x}{z+1}, v=\\frac{y}{z+1}\\), but with \\(z\\)\ncalculated for the ellipse. The inverse is a little more problematic\nhowever. As with the sphere start by describing the line from the pole\nto the projection in terms of a unit vector pointing in the direction of\nthe line:\n\\begin{equation}\n\\label{eq:ellipsoid-line}\n\\scriptstyle\nL = (0, 0, 1) + l \\frac{(0, 0, 1) - (u, v, 0)}{|(0, 0, 1) - (u, v, 0)|} = \\sage{Ln}\n\\end{equation}\n \n Recall that a vector equation for an ellipsoid can be\n\\href{https://en.wikipedia.org/wiki/Ellipsoid\\#In_general_position}{defined}\nas \\((\\mathbf{x}-\\mathbf{o})^{\\mathrm{T}} A(\\mathbf{x}-\\mathbf{o})=1\\)\nwhere \\(0\\) is the origin coordinates and A is a positive definite\nmatrix which in the general case may include rotation, but in our case\ncontains only the ellipsoid axes:\\\\\n\\begin{sagesilent}\nA = matrix(SR, 3, 3, [1/a^2, 0, 0, 0, 1/b^2, 0, 0, 0, 1]) # c = 1 to make (0,0,1) possible\nel = Ln.row()*A*Ln\neq1 = el[0] == 1\nsols = solve(eq1, l)\n\\end{sagesilent}\n$\\sage{A}$\n \nThe ellipsoid equation is then \\(L^T A L\\):\\\\\n$\\sage{eq1}$\n \nwith two solutions again with one at 0 corresponding to the pole:\\\\\n$\\sage{sols}$\n \nSubstituting back into \\(L\\) (Equation \\ref{eq:ellipsoid-line}):\\\\\n\\begin{sagesilent}\nsolution = the_solution(sols, {u:1, v:1, a:1, b:1}) \nEIn = Ln.substitute(l=solution.rhs())\nen_ix = EIn[0]\nen_iy = EIn[1]\nen_iz = EIn[2]\n\no_s = vector(SR, (0, 0, -1))\nd = (o_s - uv).normalized()\nLs = o_s + d*l - vector(SR, (0, 0, 0))\nel = Ls.row()*A*Ls\neq1 = el[0] == 1\nsols = solve(eq1, l)\nsolution = the_solution(sols, {u:1, v:1, a:1, b:1}) \nEIs = Ls.substitute(l=solution.rhs())\nes_ix = EIs[0]\nes_iy = EIs[1]\nes_iz = EIs[2]\n\\end{sagesilent}\n$\\sage{EIn}$\n \nSimilarly for the south pole we get:\\\\\n$\\sage{EIs}$\n \n \\hypertarget{defining-the-inter-manifold-function}{%\n\\subsection{Defining the Inter-Manifold\nFunction}\\label{defining-the-inter-manifold-function}}\n\\begin{sagesilent}\nT = matrix(SR, 3, 3, [a, 0, 0, 0, b, 0, 0, 0, 1]) \nev = vector(SR, (x, y, sqrt(1 - x^2 - y^2)))\nF = T*ev\nFx = F[0]\nFy = F[1]\nFz = F[2] \nEllips = F.substitute({x: cn_ix, y: cn_iy})\n# pretty_print(Ellips)\nf1 = en_0.substitute({x:Ellips[0], z:Ellips[2]}).full_simplify()\nf2 = en_1.substitute({y:Ellips[1], z:Ellips[2]}).full_simplify()\n\\end{sagesilent}\n\nA linear transformation applied to a sphere results in an ellipse. In\nthis case we use a linear transformation matrix:\\\\\n$\\sage{T}$\n\nApplied to $\\sage{ev}$ resulting in \\(F: S \\rightarrow E\\) where S is the sphere manifold and E\nis the ellipsoid one:\\\\\n$F = \\sage{F}$\n \n Let the ellipse chart be \\((V, e_n): V \\subset E\\) (or\n\\((V, e_s): V \\subset E\\)) and the sphere chart be\n\\((U, s_n): U \\subset S\\) (or \\((U, s_s): U \\subset S\\)), then\n\n\\(e_n \\circ F \\circ s_n^{-1}[s_n(U)]\\)\n\nshould be a smooth function from \\(R^2\\) (the stereographic projection\ncoordinates for the sphere) to \\(R^2\\) (the stereographic projection\ncoordinates for the ellipse) (its probably best to take Sage's word for\nit unless you \\textbf{really} enjoy Algebra):\\\\\n$\\Bigg(\\sage{f1}, \\sage{f2}\\Bigg)$\n \n \\hypertarget{tangent-vectors-and-spaces}{%\n\\subsection{Tangent Vectors and\nSpaces}\\label{tangent-vectors-and-spaces}}\n\nConfusingly there are three different approaches to defining tangent\nspaces on manifolds \\parencite{montgomery}. These examples will\ncover the geometrically based curve approach and the more algebraic\nderivation approach. \n\n\\subsubsection{Tangent Spaces defined as Equivalence class of Curves}\n This approach starts with a curve with a domain on an\ninterval of \\(R^1\\) and the manifold \\(M\\) as the co-domain i.e\n\\(\\lambda: I \\subset R^1 \\rightarrow M\\). The simplest approach combines the\nco-domain of the curve with a chart map to define curves as being\ntangent at a point \\(p\\) if \\parencite[see][pp. 73]{isham}: \n\\begin{enumerate}\n \\item Values on the manifold correspond ie \\(\\lambda_{1}(t_0) = \\lambda_{2}(t_1) = p\\),\n \\item The velocities at the point in a chart are the same ie\n \\((c_n \\circ \\lambda_{1})^\\prime\\lvert_{t=t_0} = (c_n \\circ \\lambda_{2})^\\prime\\lvert_{t=t_1}\\)\n\\end{enumerate}\n\nThe curve used in the example is a spiral on \\(S^2\\) (from\n\\href{https://math.stackexchange.com/questions/140541/finding-parametric-curves-on-a-sphere}{math.stackexchange})\n\n\\(\\lambda_1 = (x, y, z): (\\sqrt{1-t^{2}} \\cos (a \\pi t), \\sqrt{1-t^{2}} \\sin (a \\pi t), t)\\)\n\nTo find the geometric tangent vector at a point we use the north pole\nchart to project a point from the manifold specified by the curve to the\n\\(R^2\\) Euclidean space of the stereographic projection:\\\\\n\\begin{sagesilent}\nvar('s', 't', 'p_4', 'a', 'pt')\na = 3\npt = 4/5\np_4 = pi/a\nK_1 = (sqrt(1-t*t)*cos(7*pi*t), sqrt(1-t*t)*sin(7*pi*t), t)\nT1 = vector(SR, (cn_0.substitute({x:K_1[0], z:K_1[2]}), cn_1.substitute({y:K_1[1], z:K_1[2]})))\nV = T1.diff(t)\n\\end{sagesilent}\nAs a reminder, the north chart: $(\\sage{cn_0}, \\sage{cn_1})$\\\\\n$(c_n \\circ \\lambda_1) = \\sage{T1}$\n \nNext we differentiate the above to get the tangent vector:\\\\\n$\\scriptstyle \\sage{V}$\n \nNext to find another tangent vector with the necessary properties\n(having the same value in the manifold at \\(t\\) and equal velocity\n(derivitive) in the chart), we find the inverse of the tangent line in\nthe chart which results in a circle in the manifold (\\(s\\) is the\nparametric variable for the tangent line/circle) \n$\\sage{T1.substitute(t=pt).n(digits=4)} + s \\sage{V.substitute(t=pt).n(digits=4)}$:\\\\\n\\begin{sagesilent}\nTn = T1.substitute(t=pt) + V.substitute(t=pt)*s # Tangent line in projection\nTm = vector(SR, (cn_ix.substitute(u=Tn[0], v=Tn[1]), cn_iy.substitute(u=Tn[0], v=Tn[1]), cn_iz.substitute(u=Tn[0], v=Tn[1]))).simplify_full()\n\\end{sagesilent}\n$\\lambda_2 = \\sage{Tm}$\n \nPlotting the results on the manifold (Figure \\ref{fig:stereo2}):\\\\\n\\begin{sagesilent}\nSn = sphere(center=(0, 0, 0),size=1, color='green', aspect_ratio=[1,1,1], opacity=0.7)\ncurv_1 = parametric_plot3d(K_1, (t,-1,1), color='red')\ncurv_2 = parametric_plot3d(Tm, (s,-1,1), color='blue')\nPn = (K_1[0].substitute(t=pt).n(), K_1[1].substitute(t=pt).n(), K_1[2].substitute(t=pt).n())\n#Pn = (K_2[0].substitute(t=iss).n(), K_2[1].substitute(t=iss).n(), K_2[2].substitute(t=iss).n())\npn = point(Pn, rgbcolor=(1,0,0), size=20)\nsave(Sn+curv_1+curv_2+pn, \"stereo2.png\")\n\\end{sagesilent}\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=0.95\\textwidth,keepaspectratio]{stereo2.png}\n \\caption{Visualization of tangent vector curves}\n \\label{fig:stereo2}\n \\end{figure}\n \n Verifying the two curves satisfy the equivalence class criteria:\\\\\n \\textbf{Value on Manifold}:\\\\\n $(\\sage{K_1[0].substitute(t=pt).n(digits=7)}, \\sage{K_1[1].substitute(t=pt).n(digits=7)}, \\sage{K_1[2].substitute(t=pt).n(digits=7)})$\\\\\n $(\\sage{Tm.substitute(s=0).n(digits=7)})$\\\\\n \\textbf{Velocity in chart}:\\\\\n $\\sage{V.substitute(t=pt).n(digits=7)}$\\\\\n $\\sage{Tn.diff(s).substitute(s=0).n(digits=7)}$\\\\\n \n \\hypertarget{derivation-based-approach}{%\n\\subsection{Derivation based Approach}\\label{derivation-based-approach}}\n\nA derivation at a point in a \\(C^\\infty\\) manifold is defined to be a\nlinear map \\(D(C^\\infty) \\rightarrow R\\),\n\n\\((D_{1}+D_{2})(f) = D_{1}(f)+D_{2}(f)\\)\n\n\\((\\lambda D)(f) = \\lambda D(f)\\) which also obeys the Leibnitz identity\n\\(D(f g)=D(f) \\cdot g(x)+f(x) \\cdot D(g)\\).\n\nComputer science types could see the derivation as defining an abstract\ninterface, with the actual derivitive being a concrete class\nimplementing the abstract interface.\n\nSee \\textcite{parzygnat} (lecture 12 part 3) for a description of\nderivations and associated associative algebra (in a Euclidean manifold\ncontext, but most of what is covered also applies to general manifolds,\nother than the manifold chart map directions are inverted).\n\nA tangent vector can then be applied to a \\(C^\\infty\\) function \\(f\\) given a curve \\(\\lambda\\) by\n\\(D(f \\circ \\lambda)[t]\\) where \\(\\lambda(t) = \\vec{p} \\in M\\) where \\(\\vec{p}\\) is the point in\nthe manifold where the tangent is taken. The derivation based velocities defined this way should\nform a vector space similarly to the previous section tangent vectors which can fairly easily be\nshown to form a vector space by adding/multiplying in chart space \\parencite[see][pp. 76]{isham} .\nProving a vector space using the \\(D(f \\circ \\lambda)[t]\\) approach is a little more difficult.\nAdditive closure is covered quite lengthily in lesson 9 of \\textcite{XylyXylyX} and more concisely\nby \\textcite{schuller} lecture 5, while Schuller also covers scalar multiplication.\n\nThe above proofs are regurgitated here for documentary purposes as I\ncouldn't find a book containing the proofs. Intermediate results in the\nproofs are also illustrated using the stereographic example.\n\nMultivariable differentation is denoted, as for derivations, with a \\(D\\) operator. For\nmultivariable functions and maps the differentation result will be a Jacobian, and in many cases\nwith tangent spaces mapping to \\(R^1\\) will be vectors, although transition maps between charts for\nexample will be matrices. In some cases the differention operator \\(D\\) will be subscripted with\nthe variables w.r.t. which the differentiation is taking place eg \\(D_t\\). This is in contrast with\nmany physics books which use the old fashioned multiple different chain rules to extensively\nutilize component indexing, instead of one chain rule using linear transformations with Jacobian\nmatrices applying in all cases (as for example in \\textcite{spivak} or more colorfully in\n\\textcites{ghrist}{ghrist-youtube}). Where applicable the comparable `standard' notation will also\nbe shown.\n\nWe will continue to use the curves defined in the previous section. We\nalso define \\(f(\\vec{p}) : S \\rightarrow R = 2x + 2y + 2z\\) (where z is\nconstrained to be on the manifold ie \\(z = \\sqrt{1 - x^2 - y^2}\\)). Now\n\\(D(f \\circ \\lambda_1)\\) and \\(D(f \\circ \\lambda_2)\\) are velocities:\\\\\n\\begin{sagesilent}\nvar('x', 'y', 'z')\nf = 2*x + 2*y + 2*z\nd1 = f.substitute(x = K_1[0], y = K_1[1], z = K_1[2]).diff(t)\nd2 = f.substitute(x = Tm[0], y = Tm[1], z = Tm[2]).diff(s) #.full_simplify()\n#print(latex(d2)) # Manually split the output\n\\end{sagesilent}\n\n$\\scriptstyle (f \\circ \\lambda_1)^\\prime = \\sage{d1}$\n\\begin{multline*}\n \\scriptstyle\n (f \\circ \\lambda_2)^\\prime = \n \\frac{4 \\, {({(3969 \\, \\pi^{2} + 625)} s + 225)}}{{(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90} - \n \\frac{4 \\, {({(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 72)} {({(3969 \\, \\pi^{2} + 625)} s + 225)}}{{({(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90)}^{2}} -\\\\\n \\scriptstyle\n \\frac{6 \\, {(63 \\, \\pi s {(\\sqrt{5} - 1)} - {(25 \\, s + 9)} \\sqrt{2 \\, \\sqrt{5} + 10})} {({(3969 \\, \\pi^{2} + 625)} s + 225)}}{{({(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90)}^{2}} - \n \\frac{6 \\, {(63 \\, \\pi s \\sqrt{2 \\, \\sqrt{5} + 10} + 25 \\, s {(\\sqrt{5} - 1)} + 9 \\, \\sqrt{5} - 9)} {({(3969 \\, \\pi^{2} + 625)} s + 225)}}{{({(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90)}^{2}} +\\\\ \n \\scriptstyle\n \\frac{3 \\, {(63 \\, \\pi {(\\sqrt{5} - 1)} - 25 \\, \\sqrt{2 \\, \\sqrt{5} + 10})}}{{(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90} +\n \\frac{3 \\, {(63 \\, \\pi \\sqrt{2 \\, \\sqrt{5} + 10} + 25 \\, \\sqrt{5} - 25)}}{{(3969 \\, \\pi^{2} + 625)} s^{2} + 450 \\, s + 90}\n\\end{multline*}\n \n To demonstrate the curves form a vector space (for this example anyway), before providing a proof (of closure under addition and scalar multiplication\n - the rest is left as an exercise for the reader \\smiley{}) we first set $\\lambda_3 = \\lambda_1 + \\lambda_2$ and then substituting the values for the \n intersection point specified by $s = 0$ and $t = \\frac{4}{5}$ into the three different velocities.\\\\\n\\begin{sagesilent}\nK_3 = vector(SR, ( (K_1[0] + Tm[0]).full_simplify(), (K_1[1] + Tm[1]), (K_1[2] + Tm[2]) ) ).simplify_full()\nd3 = f.substitute(x = K_3[0], y = K_3[1], z = K_3[2]).diff(s).full_simplify()\n\\end{sagesilent} \n$(D(f \\circ \\lambda_1)[\\sage{pt}] = \\sage{d1.substitute(t=pt).n(digits=7)}$\\\\\n$(D(f \\circ \\lambda_2)[0] = \\sage{d2.substitute(s=0).n(digits=7)}$\\\\\n$(D(f \\circ \\lambda_3)[\\sage{pt}] = \\sage{d3.substitute(s=0, t=pt).n(digits=7)}$\\\\\n% \\begin{multline*}\n% \\scriptscriptstyle\n% \\lambda_3 = -\\frac{750141 \\, \\sqrt{5} \\pi^{3} s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{750141 \\, \\pi^{3} s^{2} \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\\n% \\scriptscriptstyle \\frac{750141 \\, \\pi^{3} s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\\n% \\scriptscriptstyle \\frac{297675 \\, \\sqrt{5} \\pi^{2} s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{297675 \\, \\pi^{2} s^{2} \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{297675 \\, \\pi^{2} s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{214326 \\, \\sqrt{5} \\pi^{2} s}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{118125 \\, \\sqrt{5} \\pi s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{214326 \\, \\pi^{2} s \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{118125 \\, \\pi s^{2} \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\\n% \\scriptscriptstyle \\frac{500094 \\, \\pi^{2} s}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\\n% \\scriptscriptstyle \\frac{118125 \\, \\pi s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{46875 \\, \\sqrt{5} s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{46875 \\, s^{2} \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{46875 \\, s^{2}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{17010 \\, \\sqrt{5} \\pi}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{33750 \\, \\sqrt{5} s}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{17010 \\, \\pi \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{33750 \\, s \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{17010 \\, \\pi}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{78750 \\, s}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} -\\\\ \n% \\scriptscriptstyle \\frac{5400 \\, \\sqrt{5}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{5400 \\, \\sqrt{2 \\, \\sqrt{5} + 10}}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100} +\\\\ \n% \\scriptscriptstyle \\frac{21600}{15752961 \\, \\pi^{4} s^{4} + 4961250 \\, \\pi^{2} s^{4} + 3572100 \\, \\pi^{2} s^{3} + 714420 \\, \\pi^{2} s^{2} + 390625 \\, s^{4} + 562500 \\, s^{3} + 315000 \\, s^{2} + 81000 \\, s + 8100}\n% \\end{multline*}\n\n \\hypertarget{additive-closure-proof}{%\n\\subsubsection{Additive Closure Proof}\\label{additive-closure-proof}}\n\nThe proof starts with a construction, which given two existing curves\n\\(\\lambda_1,\\lambda_2\\) intersecting at point \\(\\vec{p}\\) at parameter\nvalues \\(t_1\\) and \\(t_2\\), constructs a third curve \\(\\lambda_3\\) also\nintersecting \\(\\vec{p}\\) at a parameter \\(t_3\\) ie\n\\(\\lambda_1(t_1) = \\lambda_2(t_2) = \\lambda_3(t_3) = \\vec{p}\\):\n\n\\(\\lambda_3(t) = c^{-1}\\{ (c \\circ \\lambda_1)[t_1 + t] + (c \\circ \\lambda_2)[t_2 + t] - (c \\circ \\lambda_1)[t_1]\\}\\)\n\nAt \\(t = 0\\) this evaluates to\n\\(c^{-1}\\{ c(\\vec{p}) + c(\\vec{p}) - c(\\vec{p}) \\} = c^{-1}(c(\\vec{p})) = \\vec{p}\\)\nso \\(t_3 = 0\\). Checking this construct results in the same values as\nthe other two curves:\\\\\n\\begin{sagesilent}\nt1 = vector(SR, (cn_0.substitute(x=K_1[0], z=K_1[2]).substitute(t=t+pt), cn_1.substitute(y=K_1[1], z=K_1[2]).substitute(t=t+pt)))\nt2 = vector(SR, (cn_0.substitute(x=Tm[0], z=Tm[2]), cn_1.substitute(y=Tm[1], z=Tm[2]))).substitute(s=t+0)\nt3 = vector(SR, (cn_0.substitute(x=K_1[0], z=K_1[2]), cn_1.substitute(y=K_1[1], z=K_1[2]))).substitute(t=pt)\ntt = t1 + t2 - t3\nK3 = vector(SR, (cn_ix.substitute(u=tt[0], v=tt[1]), cn_iy.substitute(u=tt[0], v=tt[1]), cn_iz.substitute(u=tt[0], v=tt[1]))).simplify_full()\n\\end{sagesilent}\n$\\textbf{Value on Manifold:}~~\" \\sage{K3.substitute(t=0).n(digits=7)}$\n \n Because the third curve parameter \\(t_3\\) evaluates to 0, we can\ntherefore write the velocity vector for \\(\\lambda_3\\) at \\(\\vec{p}\\) as\n\\((f \\circ \\lambda_3)^\\prime[t_3 = 0]\\).\n\nInserting a chart and its inverse and using associativity results in\n\n\\(\\{(f \\circ c^{-1}) \\circ (c \\circ \\lambda_3)\\}^\\prime[0]\\).\n\nUsing the chain rule with the \\(f \\circ c^{-1}\\) as the outer function:\n\n\\(D(f \\circ c^{-1})[(c \\circ \\lambda_3)[0])~D(c \\circ \\lambda_3)[0]\\)\n\nSubstituting the expression for \\(\\lambda_3\\) into the right side term:\n\n\\(D(c \\circ \\lambda_3)[0] = D(c \\circ c^{-1}\\{ (c \\circ \\lambda_1)[t_1 + t] + (c \\circ \\lambda_2)[t_2 + t] - (c \\circ \\lambda_1)[t_1]\\}) =\\)\n\n\\(D((c \\circ c^{-1} \\circ c \\circ \\lambda_1))[t_1 + t] + (c \\circ c^{-1} \\circ c \\circ \\lambda_2)[t_2 + t]) - (c \\circ c^{-1} \\circ c \\circ \\lambda_1))[t_1])\\)\n(distributivity, associativity)\n\nThe third term is not dependent on \\(t\\) so the derivitive would be\n\\(0 \\therefore\\) is will disappear and the inner \\(c^{-1} \\circ c\\) in\nthe other terms will cancel leaving:\n\n\\(D((c \\circ \\lambda_1))[t_1 + t] + (c \\circ \\lambda_2)[t_2 + t]))[0] = D(c \\circ \\lambda_1)[t_1] + D(c \\circ \\lambda_2)[t_2]\\)\n\n The left hand term can also be simplified:\n\n\\(D(f \\circ c^{-1})[(c \\circ \\lambda_3)[0]) \\equiv D(f \\circ c^{-1})[(c \\circ \\lambda_3[0])) = D(f \\circ c^{-1})[c(\\vec{p})]\\)\n\nas \\(\\lambda_3[0] = \\vec{p}\\) by construction.\n\nRecombining the simplified left and right sides:\n\n\\begin{equation}\n\\label{eq:addition-close} \n\\scriptstyle\nD(f \\circ c^{-1})[c(\\vec{p})] \\cdot (D(c \\circ \\lambda_1)[t_1] + D(c \\circ \\lambda_2)[t_2]) = D(f \\circ c^{-1})[c(\\vec{p})] D(c \\circ \\lambda_1)[t_1] + D(f \\circ c^{-1})[c(\\vec{p})] D(c \\circ \\lambda_2)[t_2])\n\\end{equation}\n\nUsing the chain rule to differentiate \\(D(f \\circ \\lambda_1)[t_1]\\) to show that it is the same as the first term above:\n\n\\(D(f \\circ \\lambda_1)[t_1] = D((f \\circ c^{-1}) \\circ (c \\circ \\lambda_1))[t_1] =\\)\n\n\\(D(f \\circ c^{-1})[(c \\circ \\lambda_1)[t_1]] \\cdot D(c \\circ \\lambda_1)[t_1] =\\)\n\n\\(D(f \\circ c^{-1})[(c(\\vec{p})] \\cdot D(c \\circ \\lambda_1)[t_1]\\)\n\nwhich is the first term in the previous expression (Equation \\ref{eq:addition-close}) (combining the\nsimplified left and right sides).\n\nSimilarly the second term\n\\(D(f \\circ c^{-1})[c(\\vec{p})] D(c \\circ \\lambda_2)[t_2])\\) can be\nshown by reversing the chain rule to be\n\\(D(f \\circ \\lambda_2)[t_2] \\therefore\\) the equation that shall remain\nunnumbered can be reduced to:\n\n\\(D(f \\circ \\lambda_1)[t_1] + D(f \\circ \\lambda_2)[t_2]\\)\n\nwhich proves closure under addition of velocity vectors in the tangent space. We can verify the above terms for our concrete example:\\\\\n\\begin{sagesilent}\nfl1 = f.substitute(x=K_1[0], y=K_1[1], z=K_1[2]).simplify_full()\nfl2 = f.substitute(x=Tm[0], y=Tm[1], z=Tm[2]).simplify_full()\nfl3 = f.substitute(x=K3[0], y=K3[1], z=K3[2]).simplify_full()\nvv1 = fl1.diff(t).substitute(t=pt).n(digits=7)\nvv2 = fl2.diff(s).substitute(s=0).n(digits=7)\nvv3 = fl3.diff(t).substitute(t=0).n(digits=7)\n\\end{sagesilent}\n$f \\circ \\lambda_{1,2,3} = \\sage{fl1.substitute(t=pt).n(digits=7)}, \\sage{fl2.substitute(s=0).n(digits=7)}, \\sage{fl3.substitute(t=0).n(digits=7)}$\\\\\n$D(f \\circ \\lambda_1) + D(f \\circ \\lambda_2) = \\sage{vv1} + \\sage{vv2} = \\sage{vv1+vv2} = D(f \\circ \\lambda_3) = \\sage{vv3}$\n \n\\hypertarget{scalar-multiplication-closure}{%\n\\subsubsection{Scalar Multiplication\nClosure}\\label{scalar-multiplication-closure}}\n\nProving closure under scalar multiplication is easier and does not\ninvolve any charts. Again start with a construction for the new curve\n\\(\\lambda_3\\) tangent at \\(\\vec{p}\\) ie\n\\(\\lambda_1(t_1) = \\lambda_3(t_3) = \\vec{p}\\) in terms of \\(\\lambda_1\\):\n\n\\(\\lambda_3(t) = \\lambda_1(a t + t_1) = (\\lambda_1 \\circ u_a)[t_3]~~a \\in R,~u_a(t) :R \\rightarrow R = a t + t_1\\)\n\nAt \\(t = 0\\) this evaluates to\n\\(\\lambda_3(0) = \\lambda_1(0 \\cdot a + t_1) = \\lambda_1(t_1) = \\vec{p}\\)\nso \\(t_3 = 0\\).\n\n\\(D(f \\circ \\lambda_3)[t_3] = D(f \\circ \\lambda_3)[0] = D((f \\circ \\lambda_1) \\circ u_a)[0]\\)\n\nApplying the chain rule:\n\n\\(D(f \\circ \\lambda_1)[u_a(0)] \\cdot D(u_a)[0] = D(f \\circ \\lambda_1)[t_1] \\cdot a = a D(f \\circ \\lambda_1)[t_1]\\)\n\nwhich proves closure under scalar multiplication.\n\n \\hypertarget{chart-realizations-for-tangent-space}{%\n\\subsection{Chart Realizations for Tangent\nSpace}\\label{chart-realizations-for-tangent-space}}\n\nWe start by introducing a chart \\(c\\) into a velocity vector applied to\nan arbitrary function \\(f\\) for a curve\n\\(\\lambda(t_1) = \\vec{p} \\in U \\subseteq M\\):\n\n\\(D((f \\circ c^{-1}) \\circ (c \\circ \\lambda))[t_1]\\)\n\nApplying the chain rule once again:\n\n\\(D_{u,v}(f \\circ c^{-1})[c \\circ \\lambda(t_1)] \\cdot D_t(c \\circ \\lambda)[t_1] = D_t(c \\circ \\lambda)[t_1] \\cdot D_{u,v}(f \\circ c^{-1})[c(\\vec{p})]\\)\n\nThe representation of derivation based tangent space vectors w.r.t\ncharts as used above in the proofs is quite complicated, so a simplified\nrepresentation, otherwise known as syntactic sugar in the computer\nscience community, is conventionally used, particularly in Physics,\nwhich is shown here along with an example.\n\nWe start by investigating what the left and right hand sides of the\nabove look like for a concrete example using the north chart of the\nstereographic projection and the spiral curve on the sphere. To do this\nwe take the Jacobian of both sides and then do matrix multiplication (or\ndot product as they are both vectors) on the jacobians:\\\\\n\\begin{sagesilent}\nt1 = vector(SR, (cn_0.substitute(x=K_1[0], z=K_1[2]), cn_1.substitute(y=K_1[1], z=K_1[2])))\nlhs = jacobian(t1, (t)).transpose()\n\n_pp = vector(SR, (cn_0.substitute(x=Pn[0], z=Pn[2]), cn_1.substitute(y=Pn[1], z=Pn[2])))\n_f = f.substitute(x=cn_ix, y=cn_iy, z=cn_iz).simplify_full()\nrhs = jacobian(_f, [u,v])\nres = (lhs.substitute(t=pt)*rhs.transpose().substitute({u:_pp[0], v:_pp[1]})).n(digits=6)\n\\end{sagesilent}\n\\begin{multline*}\n \\scriptstyle LHS = \n\\left(\\begin{array}{rr}\n \\scriptstyle \n \\frac{7 \\, \\pi \\sqrt{-t^{2} + 1} \\sin\\left(7 \\, \\pi t\\right)}{t - 1} + \\frac{t \\cos\\left(7 \\, \\pi t\\right)}{\\sqrt{-t^{2} + 1} {\\left(t - 1\\right)}} + \\frac{\\sqrt{-t^{2} + 1} \\cos\\left(7 \\, \\pi t\\right)}{{\\left(t - 1\\right)}^{2}} & -\\frac{7 \\, \\pi \\sqrt{-t^{2} + 1} \\cos\\left(7 \\, \\pi t\\right)}{t - 1} +\\\\ \n \\frac{t \\sin\\left(7 \\, \\pi t\\right)}{\\sqrt{-t^{2} + 1} {\\left(t - 1\\right)}} + \\frac{\\sqrt{-t^{2} + 1} \\sin\\left(7 \\, \\pi t\\right)}{{\\left(t - 1\\right)}^{2}}\n \\end{array}\\right)\n \\Big[t = \\sage{pt} \\Big]\n\\end{multline*}\n\n$\\scriptstyle RHS = \\sage{rhs.transpose()}\\Big[(u,v) = \\sage{_pp.n(digits=6)} \\Big]$\n \nThe right hand side of the expression is simplified as follows: \n\\begin{enumerate}[itemsep=1pt, topsep=1pt, partopsep=0pt]\n \\item As the derivative is being taken Leibnitz notation is used (odd - one would have thought physicists would prefer Newton), the D is replaced with a\n del ie \\(\\partial\\),\n \\item The function is used in the numerator and the chart is used in the denominator in the Leibnitz notation for differentiation \n i.e.~\\(\\frac{\\partial f}{\\partial c}\\),\n \\item The coordinate index of the component being differentiated in the vector is used as a superscript in the denominator (the use of vectors and\n matrices was implied in the Jacobian differentiation operator used in most of this document) \\(\\frac{\\partial f}{\\partial c^i}\\),\n \\item Finally the point on the manifold is given as a subscript i.e \\((\\frac{\\partial f}{\\partial c^i})_p\\).\n\\end{enumerate}\n \nThe left hand side is also simplified: \n\\begin{enumerate}[itemsep=1pt, topsep=1pt, partopsep=0pt]\n \\item As the curve is being differentiated with respect to a parameter which can be interpreted as time, Newton finally arrives on the scene, \n with the Newtonian notation with a dot on top of the curve being used i.e.~\\(\\dot \\lambda\\),\n \\item As with the RHS the component index is used as a superscript (which also ties in with using Einstein summation for the entire expression)\n i.e.~\\(\\dot \\lambda^i\\),\n \\item The curve is used as a subscript ie \\(\\dot \\lambda^i_c\\) 4. Finally the point in the manifold is added i.e.~\\(\\dot \\lambda^i_c(t_1)\\)\n\\end{enumerate} \n\nThe entire shorthand expression for a component is expressed as an operator which operates on\n\\(C^\\infty\\) functions: \\(\\dot \\lambda^i_c(t_1) (\\frac{\\partial }{\\partial c^i})_pf\\)\n\nThe \\((\\frac{\\partial }{\\partial c^i})_p\\) part (or more accurately its full expansion) forms a\nbasis for the vector space of derivations defining the tangent space.\n\\Textcite[see][pp.~82-86]{isham} provides a proof of this as well as the isomorphism between the\nderivation and the previously described geometric tangent space. This basis is known as a chart\ninduced basis.\n\nThe example shows the basis dependent on \\(u,v\\) and the components on\n\\(t\\):\\\\\n\\begin{multline*}\n \\left(\\begin{array}{r}\n \\scriptstyle\n 4 \\, {\\left(\\frac{7 \\, \\pi \\sqrt{-t^{2} + 1} \\sin\\left(7 \\, \\pi t\\right)}{t - 1} + \\frac{t \\cos\\left(7 \\, \\pi t\\right)}{\\sqrt{-t^{2} + 1} {\\left(t - 1\\right)}} + \\frac{\\sqrt{-t^{2} + 1} \\cos\\left(7 \\, \\pi t\\right)}{{\\left(t - 1\\right)}^{2}}\\right)} {\\left(\\frac{u + 1}{u^{2} + v^{2} + 1} - \\frac{{\\left(u^{2} + v^{2} + 2 \\, u + 2 \\, v - 1\\right)} u}{{\\left(u^{2} + v^{2} + 1\\right)}^{2}}\\right)} -\\\\\n \\scriptstyle\n 4 \\, {\\left(\\frac{7 \\, \\pi \\sqrt{-t^{2} + 1} \\cos\\left(7 \\, \\pi t\\right)}{t - 1} - \\frac{t \\sin\\left(7 \\, \\pi t\\right)}{\\sqrt{-t^{2} + 1} {\\left(t - 1\\right)}} - \\frac{\\sqrt{-t^{2} + 1} \\sin\\left(7 \\, \\pi t\\right)}{{\\left(t - 1\\right)}^{2}}\\right)} {\\left(\\frac{v + 1}{u^{2} + v^{2} + 1} - \\frac{{\\left(u^{2} + v^{2} + 2 \\, u + 2 \\, v - 1\\right)} v}{{\\left(u^{2} + v^{2} + 1\\right)}^{2}}\\right)}\n \\end{array}\\right)\n = \\sage{res[0]}\n\\end{multline*} \n \n\\hypertarget{change-of-basis-for-tangent-space-components}{%\n\\subsection{Change of basis for Tangent Space\nComponents}\\label{change-of-basis-for-tangent-space-components}}\n\nGiven two charts \\((U, c_1)\\) and \\((V, c_2)\\) where\n\\(U \\cap V \\neq \\emptyset\\), where the tangent vector to a point\n\\(\\vec{p} \\in U,V\\) is known relative to say \\(U\\), then the equivalent\nbasis vector for \\(V\\) can be derived:\n\n\\begin{enumerate}[itemsep=1pt, topsep=1pt, partopsep=0pt]\n\\item\n Recall the expression for the basic component from the previous\n section and once again apply the idiom of using the inverse of a map\n after the map, in this case the map being the chart from \\(V\\).\n \\((\\frac{\\partial f}{\\partial c_1^{i:1\\cdots m}})_p = D(f \\circ c_1^{-1})[c_1(\\vec{p})] = D((f \\circ c_2^{-1}) \\circ (c_2 \\circ c_1^{-1})[c_1(\\vec{p})]\\)\n\\item\n Utilise the chain rule again:\n \\(D_{u,v}(f \\circ c_2^{-1})[c_2 \\circ c_1^{-1}(c_1(\\vec{p})] \\cdot D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})] = D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})] \\cdot D_{u,v}(f \\circ c_2^{-1})[c_2(\\vec{p})]\\).\n\\item\n The full expression for the tangent vector in \\(U\\) is given by:\n \\(\\dot \\lambda^i_{c_1}(t_1) (\\frac{\\partial f}{\\partial c_1^{i:1\\cdots m}})_p = D(c_1 \\circ \\lambda)[t_1] \\cdot D(f \\circ c^{-1}_1)[c_1(\\vec{p})]\\)\n Substituting the expression from 2:\n \\(D(c_1 \\circ \\lambda)[t_1] \\cdot D(f \\circ c^{-1})[c(\\vec{p})] = D_t(c_1 \\circ \\lambda)[t_1] \\cdot D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})] \\cdot D_{u,v}(f \\circ c_2^{-1})[c_2(\\vec{p})]\\)\n\\item\n The full expression for the tangent vector in \\(V\\) is given by:\n \\(\\dot \\lambda^i_{c_2}(t_2) (\\frac{\\partial f}{\\partial c_2^{i:1\\cdots n}})_p = D(c_{2} \\circ \\lambda)[t_2] \\cdot D(f \\circ c^{-1}_{2})[c_2(\\vec{p})]\\)\n\\item\n As the curve on the manifold remains the same, the tangent vector is\n also the same, its just expressed differently in the two bases,\n therefore these expressions can be equated:\n \\(D_t(c_{2} \\circ \\lambda)[t_2] \\cdot D_{u,v}(f \\circ c^{-1}_{2})[c_2(\\vec{p})] = D_t(c_1 \\circ \\lambda)[t_1] \\cdot D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})] \\cdot D_{u,v}(f \\circ c_2^{-1})[c_2(\\vec{p})]\\)\n The components in V are thus\n \\(D_t(c_1 \\circ \\lambda)[t_1] \\cdot D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})]\\)\n\\end{enumerate}\n\nThe \\(D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})]\\) (Jacobian of the\noverlap function between \\(U\\) and \\(V\\)) is \\(R^m \\rightarrow R^n\\) so\nthe Jacobian will be a \\(n{\\times}m\\) matrix. This matrix is multiplied\nby a \\(m{\\times}1\\) vector of the map from the curve to chart\ncoordinates via the manifold.\n\nApplying this to the example using a change of basis from the north to\nthe south hemisphere of the stereographic projection (the point at which\nthe tangent are found is the northern hemisphere test point, and the\ncurve is the same as the one used in the previous tangent space\nexamples):\\\\\n\\begin{sagesilent}\npp1 = vector(SR, (cn_0.substitute(x=Pn[0], z=Pn[2]), cn_1.substitute(y=Pn[1], z=Pn[2])))\npp2 = vector(SR, (cs_0.substitute(x=Pn[0], z=Pn[2]), cs_1.substitute(y=Pn[1], z=Pn[2])))\nfc2 = f.substitute(x=cs_ix, y=cs_iy, z=cs_iz).simplify_full()\nc2c1 = vector(SR, (cs_0.substitute({x: cn_ix, z: cn_iz}).full_simplify(), cs_1.substitute({y: cn_iy, z: cn_iz}).full_simplify()))\nc1l = vector(SR, (cn_0.substitute(x=K_1[0], z=K_1[2]), cn_1.substitute(y=K_1[1], z=K_1[2])))\n\nDc2c1 = jacobian(c2c1, (u, v))\nDc1l = jacobian(c1l, (t))\npretty_print(Dc2c1, LatexExpr(\"\\\\cdot\"), Dc1l, LatexExpr(\"~=~\"), Dc2c1*Dc1l)\npretty_print(Dc2c1.substitute(u=pp1[0], v=pp1[1]).n(digits=7), LatexExpr(\"\\\\cdot\"), Dc1l.substitute(t=float(pt)).n(digits=7), LatexExpr(\"~=~\"), (Dc2c1.substitute(u=pp1[0], v=pp1[1])*Dc1l.substitute(t=float(pt))).n(digits=7)) \n\\end{sagesilent}\n% \\begin{multline*}\n% \\scriptstyle \n% \\sage{Dc2c1} \\cdot \\sage{Dc1l} =\\\\\n% \\sage{(Dc2c1.substitute(u=pp1[0], v=pp1[1])*Dc1l.substitute(t=float(pt))).n(digits=7)}\n% \\end{multline*}\n\n\\begin{multline*}\n \\hspace{-1.6cm}\n \\medmuskip=2mu\n \\scriptstyle \n \\sage{Dc2c1} \\cdot \\sage{Dc1l}[t=\\sage{float(pt)}] = \\\\\n \\scriptstyle \\sage{(Dc2c1*Dc1l.substitute(t=pt))}\n\\end{multline*}\n\n\\begin{multline*}\n \\hspace{-1.6cm}\n \\medmuskip=2mu\n \\scriptstyle \\sage{(Dc2c1*Dc1l.substitute(t=pt))}\\Big[\\begin{array}{c} u = \\sage{pp1[0].n(digits=5)} \\\\ v = \\sage{pp1[1].n(digits=5)}\\end{array}\\Big] =\\\\\n \\scriptstyle \\sage{(Dc2c1.substitute(u=pp1[0], v=pp1[1])*Dc1l.substitute(t=float(pt))).n(digits=7)}\n\\end{multline*}\n \n We can test the change of basis by calculating the component representation in the south chart directly:\\\\\n\\begin{sagesilent} \nlhss = vector(SR, (cs_0, cs_1))\nlhs = jacobian(vector(SR, (cs_0.substitute(x=K_1[0], z=K_1[2]), cs_1.substitute(y=K_1[1], z=K_1[2]))), (t)).transpose()\n\\end{sagesilent} \n\\begin{multline*}\n \\hspace{-1.6cm}\n \\medmuskip=2mu \n D_{x,y,z}\\sage{lhss}\\Big[\\begin{array}{c} x = \\sage{K_1[0]} \\\\ y = \\sage{K_1[1]} \\\\ z = \\sage{K_1[2]}\\end{array}\\Big] =\\\\ \n \\hspace{-1.6cm}\n \\scriptstyle \\sage{lhs}\\big[t=\\sage{float(pt)}\\big] =\\\\ \n \\scriptstyle \\sage{lhs.substitute(t=pt).n(digits=6)}\n\\end{multline*} \n \nMost (physics) textbooks use a shorthand notation for \\(D_{u,v}(c_2 \\circ c_1^{-1})[c_1(\\vec{p})]\n\\cdot D_t(c_1 \\circ \\lambda)[t_1]\\), namely \\(\\frac{\\partial c_2^j}{\\partial c_1^i} \\cdot\nX^i_{c_1}\\) where the \\(\\frac{\\partial c_2^j}{\\partial c_1^i}\\) refers to the Jacobian matrix with\nthe superscript indices corresponding to rows and columns from the matrix, while the \\(X^i_{c_1}\\)\nterm is a generic reference to the components in the convert-from chart minus any curve\ninformation. \n\n\\printbibliography \n\\end{document}\n", "meta": {"hexsha": "4e6effbb93dac5a9cc1392064212672ae80fe175", "size": 47304, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ConcreteManifolds.tex", "max_stars_repo_name": "donaldmunro/ConcreteManifolds", "max_stars_repo_head_hexsha": "dc7699ed029d6e5c87872654188d748b40249cca", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ConcreteManifolds.tex", "max_issues_repo_name": "donaldmunro/ConcreteManifolds", "max_issues_repo_head_hexsha": "dc7699ed029d6e5c87872654188d748b40249cca", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ConcreteManifolds.tex", "max_forks_repo_name": "donaldmunro/ConcreteManifolds", "max_forks_repo_head_hexsha": "dc7699ed029d6e5c87872654188d748b40249cca", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.9127516779, "max_line_length": 407, "alphanum_fraction": 0.6347666159, "num_tokens": 18165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8267117855317474, "lm_q1q2_score": 0.6264775467865593}} {"text": "\\documentclass[11pt]{article}\n\n\\usepackage[margin=1in, paperwidth=8.5in, paperheight=11in]{geometry}\n\\usepackage{amsfonts}\n\n\\begin{document}\n\\title{Mathematical}\n\\author{Mrinmoi Hossain}\n\\date{\\today}\n\\maketitle\n\n\\section{Mathematical Notation}\n\\begin{enumerate}\n\\item The set of nutural number is denoted by $\\mathbb{N}$.\n\\item The set of real number is denoted by $\\mathbb{R}$.\n\\end{enumerate}\n\n\\end{document}", "meta": {"hexsha": "16ac8e54c24e3a37b28b07a85a377959da3a5345", "size": 411, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MathematicalNotation/mathematicalSymbol.tex", "max_stars_repo_name": "MrinmoiHossain/HelloLateX", "max_stars_repo_head_hexsha": "ed8d424a21dc252d3d67356433053adf4b34ccae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-12-03T03:10:11.000Z", "max_stars_repo_stars_event_max_datetime": "2017-12-03T03:10:11.000Z", "max_issues_repo_path": "MathematicalNotation/mathematicalSymbol.tex", "max_issues_repo_name": "MrinmoiHossain/HelloLateX", "max_issues_repo_head_hexsha": "ed8d424a21dc252d3d67356433053adf4b34ccae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "MathematicalNotation/mathematicalSymbol.tex", "max_forks_repo_name": "MrinmoiHossain/HelloLateX", "max_forks_repo_head_hexsha": "ed8d424a21dc252d3d67356433053adf4b34ccae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 22.8333333333, "max_line_length": 69, "alphanum_fraction": 0.7615571776, "num_tokens": 135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.826711791935942, "lm_q1q2_score": 0.6264775425954286}} {"text": "% Copyright 2018-2021 Melvin Eloy Irizarry-Gelpí\n\\setcounter{chapter}{3}\n\\chapter{Faraday's Law of Induction}\n%\nIn this experiment you will learn about Faraday's law of induction and a relationship between electric and magnetic phenomena.\n%\n\\section{Preliminary}\n%\nFaraday's law of induction is a statement relating an \\textbf{electric} phenomenon to a \\textbf{magnetic} phenomenon. It states that a certain amount of \\textbf{voltage} $\\mathcal{E}$ is equal to the negative \\textbf{rate of change of the magnetic flux} $\\Phi_{B}$ with time:\n\\begin{equation}\n\t\\mathcal{E} = -\\frac{\\Delta \\Phi_{B}}{\\Delta t}\n\t\\label{eq.04.faradays.law}\n\\end{equation}\nHere $\\mathcal{E}$ is known as the \\textbf{electro-motive force} or EMF. Although ``force'' is part of its name, an EMF quantity has nothing to do with force (which is measured in newtons): an EMF quantity has units of volts.\n%\n\\subsection{Magnetic flux and its rate of change}\n%\nThe \\textbf{magnetic flux} is defined as the product of the amount of \\textbf{magnetic field} $B$ and the amount of \\textbf{area} $a$ in a surface where the magnetic field lines pass through:\n\\begin{equation}\n\t\\Phi_{B} = B a\n\\end{equation}\nThe \\textbf{SI unit} for magnetic field is the \\textbf{tesla} (T). The \\textbf{SI unit} for area is the square meter (m$^{2}$). Thus, magnetic flux is measured in units of T {\\textperiodcentered} m\\textsuperscript{2}.\n\nA non-zero EMF $\\mathcal{E}$ requires the magnetic flux $\\Phi_{B}$ to change with time. In this experiment you are going to keep the \\textbf{area fixed} and let the \\textbf{magnetic field change with time}. In this setting, the rate of change of the magnetic flux with time is proportional to the rate of change of the magnetic field with time:\n\\begin{align}\n\t\\Phi_{B} = B a && \\Longrightarrow && \\frac{\\Delta \\Phi_{B}}{\\Delta t} = \\left(\\frac{\\Delta B}{\\Delta t}\\right) a\n\t\\label{eq.04.flux.rate}\n\\end{align}\nThe rate of change of magnetic flux is measured in units of T {\\textperiodcentered} m\\textsuperscript{2}/s. But according to Faraday's law, these units are the same as the units for EMF (volts), so you must have the equivalence:\n\\begin{equation}\n\t1 \\ \\text{V} = 1 \\ \\text{T {\\textperiodcentered} m\\textsuperscript{2}/s}\n\\end{equation}\nThis allows you to express the tesla unit in terms of volts, meters, and seconds:\n\\begin{equation}\n\t1 \\ \\text{T} = 1 \\ \\text{V {\\textperiodcentered} s/m\\textsuperscript{2}}\n\\end{equation}\n%\n\\subsection{Producing the magnetic field}\n%\nWires with electric current produce magnetic fields. You are going to use a particular wire called a \\textbf{solenoid coil}. The \\textbf{magnetic field} $B$ inside the cylindrical space of a solenoid coil is (almost) uniform and the amount of field is directly proportional to the amount of current $I$ flowing through the coil:\n\\begin{equation}\n\tB = \\left(\\frac{\\mu_{0} N_{1}}{L}\\right) I\n\t\\label{eq.04.B.coil}\n\\end{equation}\nHere\n\\begin{itemize}\n\t\\item $B$ is the amount of \\textbf{magnetic field} (unit: T)\n\t\\item $\\mu_{0} = 4 \\pi \\times 10^{-7}$ T {\\textperiodcentered} m/A is a universal physical \\textbf{constant}\n\t\\item $N_{1}$ is the \\textbf{number of turns} in the (primary) coil (no units)\n\t\\item $L$ is the \\textbf{length} of the (primary) coil (unit: m)\n\t\\item $I$ is the amount of \\textbf{electric current} through the (primary) coil (unit: A)\n\\end{itemize}\nFor a given coil you are going to keep the \\textbf{number of turns fixed}, and the \\textbf{length fixed}, but you are going to use a \\textbf{current that changes with time}. In this case, the rate of change of the magnetic field with time is proportional to the rate of change of the current with time:\n\\begin{align}\n\tB = \\left(\\frac{\\mu_{0} N_{1}}{L}\\right) I && \\Longrightarrow && \\frac{\\Delta B}{\\Delta t} = \\left( \\frac{\\mu_{0} N_{1}}{L} \\right) \\frac{\\Delta I}{\\Delta t}\n\t\\label{eq.04.B.rate}\n\\end{align}\nUsing Equation (\\ref{eq.04.B.rate}) in Equation (\\ref{eq.04.flux.rate}) gives\n\\begin{equation}\n\t\\frac{\\Delta \\Phi_{B}}{\\Delta t} = \\left( \\frac{\\mu_{0} N_{1} a}{L} \\right) \\frac{\\Delta I}{\\Delta t}\n\\end{equation}\nThus, according to Faraday's law, for a solenoid coil you should have\n\\begin{equation}\n\t\\mathcal{E} = -\\left( \\frac{\\mu_{0} N_{1} a}{L} \\right) \\frac{\\Delta I}{\\Delta t}\n\t\\label{eq.04.emf.solenoid}\n\\end{equation}\nThat is, the EMF associated with the area $a$ is proportional to the rate of change of the electric current with respect to time.\n%\n\\section{Experiment}\n%\nThe overall goal of the experiment is to check that the EMF behaves like the rate of change of the electric current. To do this you need to measure the EMF voltage and the current. From the current you can estimate the rate of change and confirm that it agrees with the EMF data.\n\nYou are going to look at \\textbf{five ways for the current to change with time}:\n\\begin{enumerate}\n\t\\item \\textbf{Sinusoidal}: current will change according to a \\textbf{sine or cosine} profile.\n\t\\item \\textbf{Square}: current will stay almost \\textbf{constant}, but the \\textbf{sign changes}.\n\t\\item \\textbf{Triangular}: current will increase \\textbf{linearly}, then decrease \\textbf{linearly}.\n\t\\item \\textbf{Ramp-up}: current will \\textbf{increase linearly}, then \\textbf{abruptly drop} to its base value.\n\t\\item \\textbf{Ramp-down}: current will \\textbf{decrease linearly}, then \\textbf{abruptly rise} to its base value.\n\\end{enumerate}\nEach profile for the current leads to different results for the induced EMF.\n%\n\\subsection{Rates of Change}\n%\nHere is a quick review of rates of change.\n%\n\\subsubsection{Constant Quantity}\n%\nThe rate of change of a \\textbf{constant quantity} is \\textbf{zero}. That is, if the quantity is not changing, then no amount is being gained or lost.\n%\n\\subsubsection{Quantity Increasing Linearly}\n%\nThe rate of change of a \\textbf{quantity that increases linearly with time} is a \\textbf{constant}. Increasing linearly is the same as being proportional to time. This means that the slope is constant. Moreover, since the quantity is increasing, the \\textbf{slope is positive}.\n%\n\\subsubsection{Quantity Decreasing Linearly}\n%\nThe rate of change of a \\textbf{quantity that decreases linearly with time} is a \\textbf{constant}. Decreasing linearly is the same as being proportional to time. This means that the slope is constant. However, since the quantity is decreasing, the \\textbf{slope is negative}.\n%\n\\subsubsection{Sinusoidal Change}\n%\nThe rate of change of the \\textbf{sine function} is a cosine function:\n\\begin{equation}\n\t\\frac{\\Delta \\sin(t)}{\\Delta t} = \\cos(t)\n\\end{equation}\nSimilarly, the rate of change of the \\textbf{cosine function} is proportional to a sine function:\n\\begin{equation}\n\t\\frac{\\Delta \\cos(t)}{\\Delta t} = -\\sin(t)\n\\end{equation}\nHere the sign is not important. What is important is to remember that sine and cosine are very similar functions and that a cosine is the same as a sine function shifted by 90 deg (or $\\pi/2$ rad):\n\\begin{align}\n\t\\cos(t) = \\sin(t + 90\\text{\\textdegree}), && \\sin(t) = \\cos(t - 90\\text{\\textdegree})\n\\end{align}\nThus, if the quantity is sinusoidal, then the rate of change will also be sinusoidal but shifted by 90 deg. In particular, when $\\sin(t)$ is at a peak or valley, then $\\cos(t)$ is crossing the horizontal axis (i.e. close to zero), and when $\\sin(t)$ is crossing the horizontal axis (i.e. close to zero), then $\\cos(t)$ is either at a valley or a peak.\n%\n\\subsection{Solenoid Coils}\n%\nYou are going to use two solenoid coils. The smaller one fits inside the larger one. The smaller one is called the \\textbf{secondary coil}. This is the coil used to measure the EMF value. The larger coil is called the \\textbf{primary coil}. This coil is the one with the current and the one that produces the magnetic field.\n\nThe \\textbf{primary coil} has a length of 10 cm and 3300 turns. Thus,\n\\begin{align}\n\tL = 10 \\ \\text{cm} = 0.1 \\ \\text{m,} && N_{1} = 3300\n\\end{align}\nThe \\textbf{secondary coil} has 150 turns and an outer diameter of 1.8 cm. The inner diameter I measured it to be close to 1.1 cm. The average of these two diameters correspond to the diameter half-way between these two. This average is 1.45 cm. Since the wire was very thick, I think it is more accurate to use the average of the diameters to calculate the area of each loop. Thus,\n\\begin{align}\n\td = 1.45 \\ \\text{cm} = 0.0145 \\ \\text{m,} && N_{2} = 150\n\\end{align}\nThe area enclosed by a circular loop with diameter $d$ is given by\n\\begin{equation}\n\ta_{\\text{loop}} = \\frac{1}{4} \\pi d^{2} = \\frac{1}{4} \\pi \\left(0.0145 \\ \\text{m}\\right)^2 = 1.65 \\times 10^{-4} \\ \\text{m\\textsuperscript{2}}\n\\end{equation}\nThis is the amount of \\textbf{area-per-loop} in the secondary coil. The \\textbf{total amount of area} $a$ is the area-per-loop $a_{\\text{loop}}$ multiplied by the number of loops $N_{2}$ in the secondary coil:\n\\begin{equation}\n\ta = N_{2} a_{\\text{loop}} = 150 \\times \\left(1.65 \\times 10^{-4} \\ \\text{m\\textsuperscript{2}}\\right) = 2.48 \\times 10^{-2} \\ \\text{m\\textsuperscript{2}}\n\\end{equation}\nThis is the amount of area that you will use to measure the magnetic flux.\n%\n\\section{Analysis}\n%\nAccording to (\\ref{eq.04.emf.solenoid}), the amount of EMF is directly proportional to the amount of rate of change of the current with time. The constant of proportionality is\n\\begin{equation}\n\t\\frac{\\mu_{0} N_{1} a}{L} = 1.03 \\times 10^{-3} \\ \\text{V {\\textperiodcentered} s/A} = 1.03 \\ \\text{mV {\\textperiodcentered} s/A}\n\t\\label{eq.04.slope}\n\\end{equation}\nThis is the expected value. In principle, if you could chart $\\mathcal{E}$ versus $\\Delta I / \\Delta t$, the slope should correspond to the negative of this value. You can only do this for the triangular current profile.\n%\n\\subsection{Sinusoidal Current Profile}\n%\nFor the sinusoidal current profile, one way to check the validity of Faraday's law is by comparing the fit parameters for a sinusoidal fit for both current and voltage. A ``sine'' sinusoidal fit for the current is of the form\n\\begin{equation}\n\tI(t) = A \\sin(B t + C) + D\n\\end{equation}\nNote that\n\\begin{itemize}\n\t\\item The current amplitude $A$ has units of amps (A)\n\t\\item The current angular frequency $B$ has units of radians-per-second (rad/s)\n\t\\item The current angular shift $C$ has units of radians (rad)\n\t\\item The current shift $D$ has units of amps (A)\n\\end{itemize}\nA ``cosine'' sinusoidal fit for the EMF is of the form\n\\begin{equation}\n\t\\mathcal{E}(t) = W \\cos(X t + Y) + Z\n\\end{equation}\nNote that\n\\begin{itemize}\n\t\\item The EMF amplitude $W$ has units of volts (V)\n\t\\item The EMF angular frequency $X$ has units of radians-per-second (rad/s)\n\t\\item The EMF angular shift $Y$ has units of radians (rad)\n\t\\item The EMF shift $Z$ has units of volts (V)\n\\end{itemize}\nIf the sensors were zeroed correctly, you should find that both $D$ and $Z$ are close to \\textbf{zero}, and thus they can be neglected. There are three simple checks:\n\\begin{enumerate}\n\t\\item The value of $W$ should be very close to\n\t\\begin{equation}\n\t\tA B \\left(\\frac{\\mu_{0} N_{1} a}{L}\\right)\n\t\\end{equation}\n\t\\item The value of $B$ should be very close to the value of $X$.\n\t\\item The value of $\\vert C - Y \\vert$ should be very close to $\\pi$.\n\\end{enumerate}\nAnother check, similar to what you did for simple harmonic motion, is to make a \\textbf{phase space} chart, with EMF voltage in the vertical axis, and current in the horizontal axis. Just like for the mass on the spring, the phase space chart should have the \\textbf{shape of an ellipse}.\n%\n\\subsection{Square Current Profile}\n%\nSince the square current profile consists of segments where the current is almost constant, the rate of change of the current should be almost zero. This will result in an EMF value that is also almost zero. You can confirm this by examining a scatter chart with EMF voltage on the vertical axis and time in the horizontal axis.\n%\n\\subsection{Triangular Current Profile}\n%\nFor the triangular current, the chart of current versus time has regions of increasing current and regions of decreasing current. In the same time regions, the EMF voltage is approximately constant but either positive or negative.\n\nYou can isolate a time region of linear increase of current and find the slope. (In the spreadsheet use \\texttt{=SLOPE(Y,X)} with \\texttt{Y} the current and \\texttt{X} the time). This slope is a value for the rate of change of current with respect to time ($\\Delta I / \\Delta t$). The slope should be \\textbf{positive} since the current is increasing. In approximately the same time region the EMF voltage is constant (make sure you only include the region where the EMF is approximately flat and not suddenly rising or falling). You can calculate the time-average of the EMF in this time region. This time average value corresponds to the best direct estimate for the EMF ($\\mathcal{E}$).\n\nIn a similar way, you can analyze the time regions where the current decreases in a linear way. Now the rate of change of the current will be \\textbf{negative}. The EMF should be almost flat and \\textbf{positive}.\n\nAfter finding \\textbf{at least six pairs} of EMF and rate of change of current with respect to time, you can plot these quantities to test the relation in Equation (\\ref{eq.04.emf.solenoid}) and find the slope. Then you can compare this value with the expected slope in Equation (\\ref{eq.04.slope}). The chart should have two clusters of points. The best-fit line should be connecting these two clusters.\n%\n\\subsection{Ramp-Up and Ramp-Down Current Profiles}\n%\nFor the ramp-up current profile, the current only increases linearly. This means that the rate of change of the current should be a positive constant always. You can repeat the steps followed above with the triangular current and find the rate of change for the current and also the time-average EMF. However, you cannot plot $\\mathcal{E}$ and $dI/dt$ anymore, so instead just calculate the ratio\n\\begin{equation}\n\t\\text{ratio } = -\\frac{\\mathcal{E}}{\\Delta I/\\Delta t}\n\t\\label{eq.04.ratio}\n\\end{equation}\nThis ratio should be somewhat close to the expected value of the slope in Equation (\\ref{eq.04.slope}). A similar result holds for the ramp-down current profile.\n%\n\\section{My Data}\n%\nMy data consist of five runs:\n\\begin{itemize}\n\t\\item Run 1: Sinusoidal current profile\n\t\\item Run 2: Square current profile\n\t\\item Run 3: Triangular current profile\n\t\\item Run 4: Ramp-up current profile\n\t\\item Run 5: Ramp-down current profile\n\\end{itemize}\nHere are some comments from each run.\n%\n\\subsection{Run 1}\n%\nThe analysis for this run is qualitative. Figure \\ref{figure.04.run.1.I} contains the current data over time, and Figure \\ref{figure.04.run.1.V} contains the EMF voltage data over time. You can see that the EMF voltage behaves like the rate of change of the current because whenever the current is at a peak or valley, the EMF voltage is close to zero, and whenever the current is close to zero, the EMF voltage is at a peak or valley. The phase phase chart in Figure \\ref{figure.04.run.1.phase.space} also shows the expected elliptical shape.\n%\n\\subsection{Run 2}\n%\nThe analysis for this run is also qualitative. Figure \\ref{figure.04.run.2.I} contains the current data over time, and Figure \\ref{figure.04.run.2.V} contains the EMF voltage data over time. As expected, when the square current profile is used, the EMF voltage is zero whenever the current has a fixed value.\n%\n\\subsection{Run 3}\n%\nThe analysis for this run has a quantitative aspect. Figure \\ref{figure.04.run.3.I} contains the current data over time, and Figure \\ref{figure.04.run.3.V} contains the EMF voltage data over time. Since the current increases or decreases in a linear way, the EMF voltage is expected to be constant but switching signs. That is indeed the case. Furthermore, the slope for the current can be calculated. Table \\ref{table.04.run.3.I.V} contains the current slope and the time-average EMF voltage for a six segments. These six segments lead to two clusters in Figure \\ref{figure.04.run.3}, where the slope is close to the expected value. The results are in Table \\ref{table.04.run.3}.\n%\n\\subsection{Run 4}\n%\nThe analysis for this run has a quantitative aspect. Figure \\ref{figure.04.run.4.I} contains the current data over time, and Figure \\ref{figure.04.run.4.V} contains the EMF voltage data over time. Since the current only increases before abruptly taking the base value, you expect the EMF voltage to be constant and negative. That is indeed the case. The results are in Table \\ref{table.04.run.4}, where the last column contains the ratio in Equation (\\ref{eq.04.ratio}). These values are close to the expected slope.\n%\n\\subsection{Run 5}\n%\nThe analysis for this run has a quantitative aspect. Figure \\ref{figure.04.run.5.I} contains the current data over time, and Figure \\ref{figure.04.run.5.V} contains the EMF voltage data over time. Since the current only decreases before abruptly taking the base value, you expect the EMF voltage to be constant and positive. That is indeed the case. The results are in Table \\ref{table.04.run.5}, where the last column contains the ratio in Equation (\\ref{eq.04.ratio}). These values are also close to the expected slope.\n%\n\\section{Your Data}\n%\nYou should have five runs:\n\\begin{itemize}\n\t\\item Run 1: Sinusoidal current profile\n\t\\item Run 2: Square current profile\n\t\\item Run 3: Triangular current profile\n\t\\item Run 4: Ramp-up current profile\n\t\\item Run 5: Ramp-down current profile\n\\end{itemize}\nFor each run you should have three columns of data: time, voltage (electric potential), and current.\n% Note that the voltage column is in mV and should be converted to volts:\n% \\begin{equation}\n% \t1 \\text{ mV} = 0.001 \\ \\text{V}\n% \\end{equation}\n% Otherwise, agreement with the slope in Equation (\\ref{eq.04.slope}) will be off by some orders of magnitude.\n%\n% \\newpage\n% \\section{Your Lab Report}\n% %\n% For \\textbf{run 1}, your lab report should include:\n% \\begin{itemize}\n% \t\\item A scatter chart with \\textbf{current} in the vertical axis, and time in the horizontal axis.\n% \t\\item A scatter chart with \\textbf{voltage} in the vertical axis, and time in the horizontal axis.\n% \t\\item A brief argument discussing whether the voltage that you measured behaves \\textbf{qualitatively} like the rate of change of current with time.\n% \t\\item A scatter chart with \\textbf{voltage} in the vertical axis, and \\textbf{current} in the horizontal axis. Does this chart have the elliptical shape?\n% \\end{itemize}\n% For \\textbf{run 2}, your lab report should include:\n% \\begin{itemize}\n% \t\\item A scatter chart with \\textbf{current} in the vertical axis, and time in the horizontal axis.\n% \t\\item A scatter chart with \\textbf{voltage} in the vertical axis, and time in the horizontal axis.\n% \t\\item A brief argument discussing whether the voltage that you measured behaves \\textbf{qualitatively} like the rate of change of current with time.\n% \\end{itemize}\n% For \\textbf{run 3}, your lab report should include:\n% \\begin{itemize}\n% \t\\item A scatter chart with \\textbf{voltage} in the vertical axis, and time in the horizontal axis.\n% \t\\item A brief argument discussing whether the voltage that you measured behaves \\textbf{qualitatively} like the rate of change of current with time.\n% \t\\item A table like Table \\ref{table.04.run.3.I.V} with at least six values for the current slope, and the time-average EMF voltage.\n% \t\\item A scatter chart with EMF voltage in the vertical axis, and current slope in the horizontal axis. Include the best-fit line, and display the equation in the legend.\n% \t\\item A table like Table \\ref{table.04.run.3} with the expected slope, the experimental slope, and the percent difference.\n% \\end{itemize}\n% For \\textbf{run 4} and \\textbf{run 5}, your lab report should include:\n% \\begin{itemize}\n% \t\\item A scatter chart with \\textbf{voltage} in the vertical axis, and time in the horizontal axis.\n% \t\\item A brief argument discussing whether the voltage that you measured behaves \\textbf{qualitatively} like the rate of change of current with time.\n% \t\\item A table like Table \\ref{table.04.run.4} and Table \\ref{table.04.run.5} with at least four values for the current slope, the time-average EMF voltage, and the ratio (\\ref{eq.04.ratio}).\n% \\end{itemize}\n%\n\\newpage\n\\section{Tables}\n%\n\\begin{table}[ht!]\n\t\\centering\n\t\\begin{tabular}{r|r}\n\t\t\\textbf{Current Slope} (A/s) & \\textbf{Time-Average EMF Voltage} (mV) \\\\\n\t\t\\hline\n\t\t\\textminus 2.247 & 2.65 \\\\\n\t\t2.274 & \\textminus 2.67 \\\\\n\t\t\\textminus 2.266 & 2.66 \\\\\n\t\t2.243 & \\textminus 2.67 \\\\\n\t\t\\textminus 2.280 & 2.66 \\\\\n\t\t2.275 & \\textminus 2.68 \\\\\n\t\t\\hline\n\t\\end{tabular}\n\t\\caption{Current slope and EMF time-average results for run 3}\n\t\\label{table.04.run.3.I.V}\n\\end{table}\n%\n\\begin{table}[ht!]\n\t\\centering\n\t\\begin{tabular}{r|r|r}\n\t\t\\textbf{Expected Slope} (mV$\\cdot$s/A) & \\textbf{Observed Slope} (mV$\\cdot$s/A) & \\textbf{P.D.} (\\%) \\\\\n\t\t\\hline\n\t\t\\textminus 1.03 & \\textminus 1.18 & 14.55 \\\\\n\t\t\\hline\n\t\\end{tabular}\n\t\\caption{Results for run 3}\n\t\\label{table.04.run.3}\n\\end{table}\n%\n\\begin{table}[ht!]\n\t\\centering\n\t\\begin{tabular}{r|r|r}\n\t\t\\textbf{Current Slope} (A/s) & \\textbf{Time-Average EMF Voltage} (mV) & \\textbf{Ratio} (mV$\\cdot$s/A) \\\\\n\t\t\\hline\n\t\t0.507 & \\textminus 0.607 & \\textminus 1.20 \\\\\n\t\t0.516 & \\textminus 0.612 & \\textminus 1.19 \\\\\n\t\t0.505 & \\textminus 0.599 & \\textminus 1.19 \\\\\n\t\t0.505 & \\textminus 0.600 & \\textminus 1.19 \\\\\n\t\t\\hline\n\t\\end{tabular}\n\t\\caption{Results for run 4}\n\t\\label{table.04.run.4}\n\\end{table}\n%\n\\begin{table}[ht!]\n\t\\centering\n\t\\begin{tabular}{r|r|r}\n\t\t\\textbf{Current Slope} (A/s) & \\textbf{Time-Average EMF Voltage} (mV) & \\textbf{Ratio} (mV$\\cdot$s/A) \\\\\n\t\t\\hline\n\t\t\\textminus 0.507 & 0.596 & \\textminus 1.17 \\\\\n\t\t\\textminus 0.516 & 0.596 & \\textminus 1.17 \\\\\n\t\t\\textminus 0.505 & 0.600 & \\textminus 1.19 \\\\\n\t\t\\textminus 0.505 & 0.601 & \\textminus 1.18 \\\\\n\t\t\\hline\n\t\\end{tabular}\n\t\\caption{Results for run 5}\n\t\\label{table.04.run.5}\n\\end{table}\n%\n\\FloatBarrier\n\\newpage\n\\section{Figures}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-1-I.pdf}\n\t\\caption{Run 1 -- Current}\n\t\\label{figure.04.run.1.I}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-1-V.pdf}\n\t\\caption{Run 1 -- EMF Voltage}\n\t\\label{figure.04.run.1.V}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-1-phase-space.pdf}\n\t\\caption{Run 1 -- Phase Space}\n\t\\label{figure.04.run.1.phase.space}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-2-I.pdf}\n\t\\caption{Run 2 -- Current}\n\t\\label{figure.04.run.2.I}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-2-V.pdf}\n\t\\caption{Run 2 -- EMF Voltage}\n\t\\label{figure.04.run.2.V}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-3-I.pdf}\n\t\\caption{Run 3 -- Current}\n\t\\label{figure.04.run.3.I}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-3-V.pdf}\n\t\\caption{Run 3 -- EMF Voltage}\n\t\\label{figure.04.run.3.V}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-4-I.pdf}\n\t\\caption{Run 4 -- Current}\n\t\\label{figure.04.run.4.I}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-4-V.pdf}\n\t\\caption{Run 4 -- EMF Voltage}\n\t\\label{figure.04.run.4.V}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-5-I.pdf}\n\t\\caption{Run 5 -- Current}\n\t\\label{figure.04.run.5.I}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-5-V.pdf}\n\t\\caption{Run 5 -- EMF Voltage}\n\t\\label{figure.04.run.5.V}\n\\end{figure}\n%\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[scale=0.74]{image/04-faraday/run-3.pdf}\n\t\\caption{Run 3}\n\t\\label{figure.04.run.3}\n\\end{figure}", "meta": {"hexsha": "a71e19ade74ed48fc5d3c2770ccc0d27f03187a6", "size": 23869, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter/04-faraday.tex", "max_stars_repo_name": "meirizarrygelpi/phys-208L", "max_stars_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter/04-faraday.tex", "max_issues_repo_name": "meirizarrygelpi/phys-208L", "max_issues_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": 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YES\n2. YES\n\n", "lm_q1_score": 0.8031738057795403, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.6264698661781124}} {"text": "\\documentclass{enonce}\n\\selectlanguage{english}\n\\usepackage{palatino,mathpazo,color}\n\\usepackage{tutorial}\n\\setlength{\\overfullrule}{3pt}\n\\course{IFM 2010 - TLAPS tutorial}\n\\title{Peterson algorithm}\n\\author{D. Cousineau, S. Merz}\n%\\institute{INRIA Lorraine - MSR INRIA Joint Lab}\n\\date{}\n\n\\begin{document}\n\\maketitle\n\n\\noindent In this tutorial, we shall prove that Peterson's algorithm satisfies the mutual exclusion property, using the TLA Proof System and the Toolbox IDE.\n\n%\\tableofcontents\n\n\\section{Toolbox and TLAPS install}\nIf you have not done it yet, you can download the Toolbox from \n\\begin{center}\n\\url{http://www.tlaplus.net/tools/tla-toolbox/}\n\\end{center}\nand dowload the TLA Proof System from\n\\begin{center}\n\\url{http://www.msr-inria.inria.fr/~doligez/tlaps/}\n\\end{center}\n\n\\noindent Binaries are available for Windows, MacOS and Linux distributions. Notice that Cygwin (version 1.7 or higher) is required to install TLAPS on Windows.\n\n\\section{Peterson Algorithm}\nThis algorithm, formulated by Gary L. Peterson in 1981, is a concurrent programming algorithm for mutual exclusion that allows two processes to share a single-use resource without conflict, using only shared memory for communication.\n\n\\subsection{Description}\nThe algorithm allows two processes to share some resource with the three following properties:\n\\begin{itemize}\n\\item Mutual exclusion : the two processes cannot access the resource at the same time\n\\item Progress: a process cannot immediately re-access the resource if the other process wants also to access the resource.\n\\item Bounded waiting: there exists a limit on the number of times that the other process is allowed to access the resource after a process has made a request to access the resource.\n\\end{itemize}\n\n\\noindent We shall focus on the mutual exclusion property in the TLA+ specification we are to build.\n\n\\medskip\n\n\\noindent We call \\textit{critical section} the moment when some process is accessing the shared resource. The Mutual Exclusion property ensures that the two processes are not both in the critical section at the same time.\n\n\\medskip\n\n The idea of the algorithm is the following: we use two variables, a function called {\\tt flag}, which says for each of the processes, if it wants to access the resource or not. And a simple variable {\\tt turn} which holds the process whose turn it is to enter the critical section.\n\n\\noindent At the beginning of the algorithm, {\\tt flag[0]} and {\\tt flag[1]} are equal to {\\tt FALSE} (none of the processes wants to access the resource), and {\\tt turn} is equal to {\\tt 0} (process 0 has priority to access the resource).\n\n\\noindent Then, for making process {\\tt 0} (for example) access the shared resource, we do the following:\n\\begin{enumerate}\n\t\\item[({\\tt a1})] turn the value of {\\tt flag[0]} to {\\tt TRUE} \\\\(process {\\tt 0} does want to enter the critical section)\n\t\\item[({\\tt a2})] turn the value of {\\tt turn} to {\\tt 1} \\\\(give the priority to the other process)\n\t\\item[({\\tt a3})] wait for {\\tt flag[1]} to be equal to {\\tt FALSE} and {\\tt turn} to be equal to {\\tt 1} \\\\(process {\\tt 1} does not want to access the resource and process {\\tt 0} has priority to access it)\n\t\\item[({\\tt cs})] enter the critical section \\\\(access the resource)\n\t\\item[({\\tt a4})] turn the value of {\\tt flag[0]} to {\\tt FALSE} \\\\(process {\\tt 0} does not want to access the resource anymore)\n\\end{enumerate} \n\n\\medskip\n\n\n\\subsection{PlusCal version}\n\nPeterson's algorithm can be expressed in the PlusCal language as follows:\n\\smallskip\n\n\\begin{verbatim}\nNot(i) == IF i = 0 THEN 1 ELSE 0\n\n--algorithm Peterson {\n variables flag = [i \\in {0, 1} |-> FALSE], turn = 0;\n process (proc \\in {0,1}) {\n a0: while (TRUE) {\n a1: flag[self] := TRUE;\n a2: turn := Not(self);\n a3a: if (flag[Not(self)]) {goto a3b} else {goto cs} ;\n a3b: if (turn = Not(self)) {goto a3a} else {goto cs} ;\n cs: skip; \\* critical section\n a4: flag[self] := FALSE;\n } \\* end while\n } \\* end process\n }\n\\end{verbatim}\n\n\\noindent Notice that, for simplicity, the TLA+ specification we describe in the next section differs a little bit from the one that can be directly translated from the algorithm above, by the PlusCal translator.\n\n\\section{The specification}\nPeterson's algorithm for two processes can already be verified by the Toolbox model-checker. But we use it as an exercise to make you learn ho to use TLA Proof Sytem.\n\n\\medskip\n\n\\noindent Peterson's TLA+ specification can be downloaded at \n\\begin{center}\n\\url{http://www.msr-inria.inria.fr/~doligez/tlaps/IFM2010/Peterson.tla}\n\\end{center}\nLet us describe what you can find in this file.\n\n\\subsection{Variables}\nWe declare three variables, {\\tt flag}, {\\tt turn} and {\\tt pc} (which gives the current step of the algorithm).\nWe define {\\tt vars} as the tuple composed by those three variables, and {\\tt ProcSet} as the set $\\{0,1\\}$.\n\\begin{verbatim}\nVARIABLES flag, turn, pc\nvars == << flag, turn, pc >>\nProcSet == ({0,1})\n\\end{verbatim}\n\n\\subsection{The {\\tt Init} state}\nWe then express the initial state of the algorithm as follows:\n\\begin{verbatim}\nInit == /\\ flag = [i \\in {0, 1} |-> FALSE]\n /\\ turn = 0\n /\\ pc = [self \\in ProcSet |-> CASE self \\in {0,1} -> \"a0\"]\n\\end{verbatim}\n{\\tt flag[0]} and {\\tt flag[1]} are equal to {\\tt FALSE}.\n\\\\ {\\tt turn} is equal to {\\tt 0}.\n\\\\ Both processes are entering the while spin.\n\n\\subsection{The {\\tt Next} action}\nIn order to define the {\\tt Next} action of Peterson's algorithm, we have to describe, for each process, every possible step of the algorithm. For example the {\\tt a0} step describes the fact that the considered process ({\\tt self}) is going from line {\\tt a0} to line {\\tt a1} in the algorithm (in that case, values of variables {\\tt flag} and {\\tt turn} don't change).\n\n\\begin{verbatim}\na0(self) == /\\ pc[self] = \"a0\"\n /\\ pc' = [pc EXCEPT ![self] = \"a1\"]\n /\\ UNCHANGED << flag, turn >>\n\\end{verbatim}\n\n\\noindent The second line above expresses the fact that the next value of the function {\\tt pc} is the same as the current one, except that its value on {\\tt self} is now {\\tt \"a1\"}.\n\n\\medskip\n\n\\noindent We then do the same for each possible step of the algorithm:\n\\begin{verbatim}\na1(self) == /\\ pc[self] = \"a1\"\n /\\ flag' = [flag EXCEPT ![self] = TRUE]\n /\\ pc' = [pc EXCEPT ![self] = \"a2\"]\n /\\ UNCHANGED turn\n\na2(self) == /\\ pc[self] = \"a2\"\n /\\ turn' = Not(self)\n /\\ pc' = [pc EXCEPT ![self] = \"a3a\"]\n /\\ UNCHANGED flag\n\na3a(self) == /\\ pc[self] = \"a3a\"\n /\\ IF flag[Not(self)]\n THEN /\\ pc' = [pc EXCEPT ![self] = \"a3b\"]\n ELSE /\\ pc' = [pc EXCEPT ![self] = \"cs\"]\n /\\ UNCHANGED << flag, turn >>\n\n\\end{verbatim}\n\\newpage\n\\begin{verbatim} \na3b(self) == /\\ pc[self] = \"a3b\"\n /\\ IF turn = Not(self)\n THEN /\\ pc' = [pc EXCEPT ![self] = \"a3a\"]\n ELSE /\\ pc' = [pc EXCEPT ![self] = \"cs\"]\n /\\ UNCHANGED << flag, turn >>\n\ncs(self) == /\\ pc[self] = \"cs\"\n /\\ TRUE\n /\\ pc' = [pc EXCEPT ![self] = \"a4\"]\n /\\ UNCHANGED << flag, turn >>\n\na4(self) == /\\ pc[self] = \"a4\"\n /\\ flag' = [flag EXCEPT ![self] = FALSE]\n /\\ pc' = [pc EXCEPT ![self] = \"a0\"]\n /\\ UNCHANGED turn\n\\end{verbatim}\n\n\\noindent Now we can define {\\tt proc(self)} as the fact that one of the previous actions is being accomplished:\n\\begin{verbatim}\nproc(self) == \\/ a0(self) \\/ a1(self) \\/ a2(self) \\/ a3a(self) \n \\/ a3b(self) \\/ cs(self) \\/ a4(self)\n\\end{verbatim}\n \n\\noindent Finally, we define the {\\tt Next} action, as the fact that either {\\tt proc} is accomplished for one of the processes, or the algorithm has finished (to prevent deadlock on termination).\n\n\\begin{verbatim} \nNext == \\E self \\in {0,1}: proc(self)\n\\end{verbatim}\n\n\\medskip\n\\subsection{Specification}\nThe specification of the algorithm is given by the facts that the initial state is satisfied and that at every step either the action {\\tt Next} is satisfied or the variables in {\\tt vars} keep their values.\n\n\\begin{verbatim}\nSpec == Init /\\ [][Next]_vars\n\\end{verbatim}\n\n\\medskip\n\\subsection{Mutual Exclusion}\n%\\noindent The termination property is defined by the fact that there exists some state such that for that state and all the next ones, both processes are in step {\\tt \"Done\"}.\n% \n%\\begin{verbatim}\n%Termination == <>(\\A self \\in ProcSet: pc[self] = \"Done\")\n%\\end{verbatim}\n%\n%\\noindent The safeness property is defined by the fact that the two processes don't access the resource at the same time.\n%\n%\\begin{verbatim}\n%Safe == \\E i \\in {0, 1} : pc[i] # \"cs\"\n%\\end{verbatim}\n%\n%\\noindent The liveness property, is defined by the fact that both processes access the resource at some time.\n%\n%\\begin{verbatim}\n%Live == \\A i \\in {0, 1} : []<>(pc[i] = \"cs\")\n%\\end{verbatim}\n\nThe property we want the algorithm to satisfy can be defined as the fact that {\\tt pc[0]} and {\\tt pc[1]} have not both value {\\tt \"cs\"} at the same time.\n\n\\begin{verbatim}\nMutualExclusion == ~ (pc[0] = \"cs\" /\\ pc[1] = \"cs\")\n\\end{verbatim}\n\n\\medskip\n\\subsection{The invariant}\nLet us first define the property that ensures well-typedness of the variables .\n%\\textcolor{red}{sm: if you remove the second disjunct of Next, also remove that remark.}\n\n\\begin{verbatim}\nTypeOK == /\\ pc \\in [{0,1} -> {\"a0\",\"a1\",\"a2\",\"a3a\",\"a3b\",\"cs\",\"a4\"}]\n /\\ turn \\in {0,1}\n /\\ flag \\in [{0,1} -> BOOLEAN]\n\\end{verbatim}\n\n\\noindent Now we can define the invariant we want the algorithm to satisfy, in order to prove the \\linebreak {\\tt MutualExclusion} property. It is defined as the fact that for each process {\\tt i}, \n\\begin{itemize}\n\\item if it is in step {\\tt \"a2\"}, {\\tt \"a3a\"}, {\\tt \"a3b\"}, {\\tt \"cs\"} or {\\tt \"a4\"}, then its flag is equal to {\\tt TRUE} (i.e. it does want to access to the resource)\n\\item if it is in step {\\tt \"cs\"} or {\\tt \"a4\"}, then the other process is not in one of these steps, and if the other process is in step {\\tt \"a3a\"} or {\\tt \"a3b\"} then process {\\tt i} has priority to access the resource.\n\\end{itemize}\n\n\\smallskip\n \n\\begin{verbatim} \nI == \\A i \\in {0, 1} :\n /\\ (pc[i] \\in {\"a2\", \"a3a\", \"a3b\", \"cs\", \"a4\"} => flag[i])\n /\\ (pc[i] \\in {\"cs\", \"a4\"})\n => /\\ pc[Not(i)] \\notin {\"cs\", \"a4\"}\n /\\ (pc[Not(i)] \\in {\"a3a\", \"a3b\"}) => (turn = i)\n\\end{verbatim}\n\n\\noindent Finally, we can define the actual invariant as the conjunction of the previous one and the fact that {\\tt TypeOK} is satisfied. \n\n\\begin{verbatim}\nFullInv == TypeOK /\\ I\n\\end{verbatim}\n\n\n\\section{Checking a simple proof from the Toolbox}\n\\label{simple}\nAs usual when we want to prove that a property is always satisfied (here {\\tt MutualExclusion}), we reason by induction and define a invariant property (here {\\tt FullInv}) such that:\n\\begin{enumerate}\n\\item the initial state (here {\\tt Init}) satisfies the invariant,\n\\item the invariant is preserved by the next-state relation (here {\\tt [Next]\\_vars}),\n\\item the invariant ensures the correctness property ({\\tt MutualExclusion}). \n\\end{enumerate}\n\n\\noindent Proving the first fact is obvious. Let us check that by asking TLAPS to prove the following theorem (recall that we have to indicate explicitly which definitions we want to be usable in a proof):\n\n\\begin{verbatim}\nTHEOREM InitFullInv == Init => FullInv\n BY DEF Init, FullInv, TypeOK, I, ProcSet\n\\end{verbatim}\n\n%\\textcolor{red}{sm: this should be \\texttt{FullInvariant}, not \\texttt{MutualExclusion}}\n\n\\noindent Now, in the Toolbox, put the cursor on the first line of that theorem and then right-click on \"Prove step or module\". The Toolbox then colors that theorem in green, which means that it is proved (in this case, Zenon has succeeded in finding a proof).\n\n\\section{The main theorem}\n\n\\noindent Let us now focus fact number 2 above. For proving that fact it is easier to an intermediate lemma, which states that property {\\tt TypeOK} is preserved by the next-state relation.\n\n%consider two intermediate lemmas. The first one expresses that if {\\tt TypeOK} is satisfied, then any of the process can be in step {\\tt \"Done\"}. It comes directly from the definitions of {\\tt TypeOK} and {\\tt ProcSet}.\n%\\begin{verbatim}\n%LEMMA NeverDone == \n% TypeOK /\\ (\\A self \\in ProcSet: pc[self] = \"Done\") => FALSE\n% BY DEFS TypeOK, ProcSet\n%\\end{verbatim}\n%\n%\\textcolor{red}{sm: this lemma would disappear, and the proof of the subsequent one simplified}\n%\\noindent The second one states that property {\\tt TypeOK} is preserved by the next-state relation.\n\n\\begin{verbatim}\nLEMMA TypeCorrect == TypeOK /\\ Next => TypeOK'\n <1>1. ASSUME NEW i \\in {0,1}\n PROVE TypeOK /\\ proc(i) => TypeOK'\n BY DEFS TypeOK, proc, a0, a1, a2, a3a, a3b, cs, a4, Not\n <1>. QED\n BY <1>1, NeverDone DEF Next\n\\end{verbatim}\n\n\\noindent When asking TLAPS to prove the theorem above, you can notice that step {\\tt<1>1} gets first colored black. That means that Zenon has been trying to prove it for more than three seconds. Then it gets colored red. That means that Zenon has failed to find a proof with the default timeout (ten seconds). Finally, it gets colored green since Isabelle succeeds in proving it.\n\n\\bigskip\n\n\\noindent Finally, let us prove the second item of section \\ref{simple}.\n\n\\begin{verbatim}\nTHEOREM Inductive == TypeOK /\\ I /\\ Next => I'\n\\end{verbatim}\n\n\\noindent We first rephrase the sequent to be proved in a more perspicuous form:\n\n\\begin{verbatim}\n<1>1. ASSUME TypeOK,\n I,\n NEW i \\in {0,1},\n proc(i)\n PROVE I'\n\\end{verbatim}\n\n\\noindent We then assert {\\tt TypeOK'} (that will be useful in the following of the proof):\n\n\\begin{verbatim}\n <2>0. TypeOK'\n\\end{verbatim}\n\n\\noindent We then split the result we want to prove into three assertions:\n\n\\begin{verbatim}\n <2>1. ASSUME NEW j \\in {0,1},\n pc'[j] \\in {\"a2\", \"a3a\", \"a3b\", \"cs\", \"a4\"}\n PROVE flag'[j]\n \n <2>2. ASSUME NEW j \\in {0,1},\n pc'[j] \\in {\"cs\", \"a4\"}\n PROVE pc'[Not(j)] \\notin {\"cs\", \"a4\"}\n \n <2>3. ASSUME NEW j \\in {0,1},\n pc'[j] \\in {\"cs\", \"a4\"},\n pc'[Not(j)] \\in {\"a3a\", \"a3b\"}\n PROVE turn' = j\n\\end{verbatim}\n\n\\noindent Those three assertions can be proved by case on the step in which process {\\tt i} is.\nIf you ask TLAPS to prove that theorem, you can notice that all those proof-steps are colored yellow. That means that the proof is missing. It's up to you now to make them green...\n\n\\end{document}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\n\\begin{appendix}\n\\section{Prover options}\n\n\n\n\n\n\n\n\\section{Practical hints}\n\nThe TLA+ proof system is designed to check the validity of claims as\nindependently as possible of specific proof back-ends. We believe that users\nshould concentrate on writing proofs in terms of their particular applications,\nnot in terms of the capabilities of a particular proof system. In particular,\nTLAPS invokes its back-ends with some default setup for automatic proof, and we\ntry to make it hard for users to change this default setup. Expert users of\nback-end provers may be frustrated because they may have to develop proofs\nsomewhat further than what would be necessary with a fine-tuned tactic script.\nThe main payoff of limited access to the nitty-gritty details of provers is\ngreater clarity of the resulting proofs. They are also easier to maintain across\nminor changes of the specification or new releases of the TLA prover.\n\nOn some occasions users will encounter situations where the prover cannot prove\nan \"obvious\" proof obligation. Here are a few hints on what to try to make the\nproof go through. Your additions to this list are welcome.\n\n\n\\subsection{Control the size of formulas and expressions}\nOur provers are currently not good at making abstractions that humans understand\nimmediately. They are easily confused by moderately big proof obligations and\nare just as likely to work on a top-level conjunction as on a set construction\nburied deeply inside the formula. This can cause back-ends to become very slow\nor even unable to complete seemingly trivial steps. While we intend to improve\nthe back-ends in this respect, you can help them by using local definitions in\nproofs and hiding these definitions in order to keep expressions small. (Keep in\nmind that definitions introduced in a proof are usable by default and must be\nhidden explicitly, unlike definitions in specifications, which must be\nexplicitly $\\USE$d.)\n\n\\medskip\n\n\\noindent Here is a contrived example:\n\n\\begin{verbatim}\nLEMMA /\\ x \\in SomeVeryBigExpression\n /\\ y \\in AnotherBigExpression\n <=>\n /\\ y \\in AnotherBigExpression\n /\\ x \\in SomeVeryBigExpression\n \n<1> DEFINE S == SomeVeryBigExpression\n \\** here and in the following, you may use positional names to\n \\** avoid repeating the big expressions\n<1> DEFINE T == AnotherBigExpression\n<1>1. x \\in S <=> x \\in SomeVeryBigExpression\n OBVIOUS\n \\** The provers should not have any trouble proving that\n \\** a very large expression is equivalent to itself. If it\n \\** can't prove this, you've almost certainly made a mistake\n \\** in the definition of S.\n<1>2. y \\in T <=> y \\in AnotherBigExpression\n OBVIOUS\n<1> HIDE DEF S, T\n<1>3. /\\ x \\in S\n /\\ y \\in T\n <=> \n /\\ y \\in T\n /\\ x \\in S\n OBVIOUS\n<1>4. QED\n BY <1>1, <1>2, <1>3\n\\end{verbatim}\n\n\\medskip\n\nThis kind of problem typically arises when reasoning about LET expressions,\nwhich are \\linebreak silently expanded by the proof manager. In a proof, introduce local\ndefinitions corresponding to the LET (using copy and paste from the\nspecification or subexpression names), show that the overall expression equals the body of the LET\n(trivial by reflexivity), establish the necessary facts about these locally\ndefined operators, and HIDE the definitions afterwards.\n\n\\medskip\n\nIntroducing definitions to hide irrelevant expressions is a useful\nthing to do whenever you're having trouble proving something. It\nmakes the obligations whose proofs fail easier to read, making\nmistakes easier to find.\n\n\\subsection{Avoid \"circular\" (sets of) equations}\n\nRewriting is one effective way to reason about equations, and it underlies the\nautomatic proof methods used by the Isabelle back-end. The basic idea is to\norient equalities such that the expressions on the left-hand side are\nsystematically replaced by the right-hand sides. However, if the set of\nequations contains cycles as in\n\n\\begin{verbatim}\ns = f(t)\nt = g(s)\n\\end{verbatim}\n\nthen rewriting may never terminate. Isabelle employs some (incomplete)\nheuristics to detect such cycles and will refuse to rewrite equations that it\ndetermines to be circular. This usually leads to its inability to infer anything\nabout these equations. If circularity is not detected, it may cause Isabelle to\nenter an infinite loop. The suggested remedy is again to introduce local\ndefinitions that are hidden to break the loops.\n\n\\medskip\n\n\\noindent As a concrete example consider the following proof snippet:\n\n\\begin{verbatim}\n <4>17. foo.name = \"xyz\"\n <5>1. foo = [name |-> \"xyz\", value = foo.value]\n BY <2>2\n <5>2. QED\n BY <5>1 \\** may not work because <5>1 is a circular equation\n\\end{verbatim}\n\n\\noindent One possible workaround is as follows:\n\n\\begin{verbatim}\n <4>17. foo.name = \"xyz\"\n <5> DEFINE fooval == foo.value\n <5>1. foo = [name |-> \"xyz\", value = fooval]\n BY <2>2\n <5> HIDE DEF fooval\n <5>2. QED\n BY <5>1\n\\end{verbatim}\n%\\end{appendix}\n%\\end{document}\n\\subsection{Reasoning about CHOOSE expressions}\n\nConsider a definition such as\n\n\\begin{verbatim}\n foo == CHOOSE x \\in S : P(x)\n\\end{verbatim}\n\nIn order to prove a property Q(foo), you will typically prove the two following\nassertions:\n\n\\begin{verbatim}\n(a) \\E x \\in S : P(x)\n(b) \\A x \\in S : P(x) => Q(x)\n\\end{verbatim}\n\nIn some cases, assertion (b) can be trivial and need not be shown explicitly.\nReasoning about an unbounded CHOOSE expression is analogous.\n\n\\medskip\n\nRemember that CHOOSE always denotes some value, even if P(x) holds for no\n$x \\in S$ (in particular, if $S = \\{\\}$), in which case the CHOOSE expression is\nfixed, but arbitrary. In practice, CHOOSE expressions usually arise when\ncondition (a) is satisfied. Should you have designed your property to work even\nif the domain of the CHOOSE is empty, property Q must be trivial in that case,\nand you can structure your proof as follows:\n\n\\begin{verbatim}\n <3>5. Q(foo)\n <4>1. CASE \\E x \\in S : P(x)\n <5>1. \\A x \\in S : P(x) => Q(x)\n <5>2. QED\n BY <4>1, <5>1\n <4>2. CASE ~ \\E x \\in S : P(x)\n <5>1. \\A x : Q(x)\n <5>2. QED\n BY <5>1\n <4>3. QED\n BY <4>1, <4>2\n\\end{verbatim}\n\nA frequent TLA+ idiom is to define a \"null\" value by writing\n\n\\begin{verbatim}\nNoValue == CHOOSE x : x \\notin Value\n\\end{verbatim}\n\nThe laws of set theory ensure that no set is universal, hence there exists an x\nthat is not an element of set Value, ensuring condition (a) above. The theorem\nNoSetContainsEverything in the standard module TLAPS can be used to prove this\ncondition.\n\n\\subsection{Help the Prover When Reasoning About Records}\n\nIn one proof, we had\n\n\\begin{verbatim}\n mb == [type |-> \"1b\", bal |-> b, acc |-> self,\n mCBal |-> maxCBal[self], mCVal |-> maxCVal[self]]\n\\end{verbatim}\n\nand were trying to prove\n\n\\begin{verbatim}\n m1 # mb /\\ m2 # mb\n\\end{verbatim}\n\nfrom facts that included\n\n\\begin{verbatim}\n m1.type = \"2av\" /\\ m2.type = \"2av\"\n\\end{verbatim}\n\nZenon failed on the proof and Isabelle proved it only after a long\ntime. (In fact, we originally stopped the proof because it was taking\nso long.) However, Zenon proved it instantly when we added $mb.type =\n\"1b\"$ to the BY statement's list of facts. The provers are reluctant\nto try finding relations of the form record.field = value; they often\nneed help.\n\n\n\\subsection{Divide and Conquer}\n\nWhen the provers can't prove something that you think is obvious, it's\nusually because it isn't true. You can easily spend hours looking at\na proof obligation without noticing a tiny mistake. The best way to\nfind a mistake is by breaking the proof into simpler steps.\nContinuing to do this on the step or steps whose proof fails will\neventually lead you to discover the problem--usually a missing\nhypothesis or a mistake in a formula. When you correct the mistake in\nthe original proof step, the prover will usually be able to prove it.\n\n\\subsection{Don't Reinvent Mathematics}\n\nWe expect that most people who use TLAPS will do so because they want\nto verify properties of an algorithm or system. We have therefore not\ndevoted our limited resources to building libraries of mathematical\nresults. If you want to create such libraries, we would welcome your\nhelp. However, if you are concerned with an algorithm or system, you\nshould not be spending your time proving basic mathematical facts.\nInstead, you should assert the mathematical theorems you need as\nassumptions or theorems.\n\n\\medskip\n\nAsserting facts is dangerous, because it's easy to make a mistake and\nassert something false, making your entire proof unsound.\nFortunately, you can use the TLC model checker to avoid such mistakes.\nFor example, our example correctness proof of Euclid's algorithm uses\nthis assumption\n\n\\begin{verbatim}\n ASSUME GCDProperty3 == \n \\A p, q \\in Nat \\ {0}: (p < q) => GCD(p, q) = GCD(p, q-p)\n\\end{verbatim}\n\nTLC cannot check this assumption because it can't evaluate a\nquantification over an infinite set. However, you can tell TLC to\nreplace the definition of Nat with\n\n\\begin{verbatim}\n Nat == 0..50\n\\end{verbatim}\n\n(In the Toolbox, use the Definition Override section of the model's\nAdvanced Options page.) TLC quickly verifies this assumption. (TLC\nchecks each ASSUME; to add an assumption that you don't want TLC to check,\nmake it an AXIOM.) \n\nThis kind of checking is almost certain to catch an error in\nexpressing a fundamentally correct mathematical result--except when\nthe only counterexamples are infinite. Fortunately, this is rarely\nthe case when the result is needed for reasoning about an algorithm or\nsystem.\n\n\\subsection{It's Easier to Prove Something if it's True}\n\nBefore trying to prove a property of an algorithm or system, try to\ncheck it with TLC. Even if TLC cannot check a large enough model to\ncatch all errors, running it on a small model can still catch many\nsimple errors. You will save a lot of time if you let TLC find these\nerrors instead of discovering them while writing the proof.\n\n\n\\end{appendix}\n\n\\end{document}\n", "meta": {"hexsha": "e9475ac0c5803bb5695067237050fb34226c5007", "size": 25333, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/presentations/2010-ifm/tutoriel-Peterson/Peterson_IFM2010.tex", "max_stars_repo_name": "damiendoligez/tlapm", "max_stars_repo_head_hexsha": "13a1993263642092a521ac046c11e3cb5fbcbc8b", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 31, "max_stars_repo_stars_event_min_datetime": "2016-08-16T14:58:40.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-19T18:38:07.000Z", "max_issues_repo_path": "doc/presentations/2010-ifm/tutoriel-Peterson/Peterson_IFM2010.tex", "max_issues_repo_name": "damiendoligez/tlapm", "max_issues_repo_head_hexsha": "13a1993263642092a521ac046c11e3cb5fbcbc8b", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 49, "max_issues_repo_issues_event_min_datetime": "2020-03-04T18:13:13.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-07T17:43:24.000Z", "max_forks_repo_path": "doc/presentations/2010-ifm/tutoriel-Peterson/Peterson_IFM2010.tex", "max_forks_repo_name": "damiendoligez/tlapm", "max_forks_repo_head_hexsha": "13a1993263642092a521ac046c11e3cb5fbcbc8b", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 12, "max_forks_repo_forks_event_min_datetime": "2020-02-26T19:58:37.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-12T22:18:25.000Z", "avg_line_length": 38.7947932619, "max_line_length": 380, "alphanum_fraction": 0.682824774, "num_tokens": 6766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8031738034238807, "lm_q1q2_score": 0.6264698643407146}} {"text": "\\documentclass[10pt]{article}\n\\usepackage[usenames]{color} %used for font color\n\\usepackage{amssymb} %maths\n\\usepackage{amsmath} %maths\n\\usepackage[utf8]{inputenc} %useful to type directly\n\\usepackage{tikz}\n%diacritic characters\n\\begin{document}\n\\paragraph{Quotient Set Notation}\nThere are many ways denote the quotient set (check the \\emph{Book of Proof} for one way).\nA common method is to get a notation for an individual equivalence class first.\nSay $[x] = \\{ y \\in C: x \\sim y\\}$ which is the set of all bundles in $C$ that are indifferent to $x$.\nThen we might create a class of sets (set of sets) using the indexed collection of sets notation you covered in the prerequisite readings.\nFor example $\\mathcal{I} = \\{ [x] : x \\in C\\}$ where $[x]=\\{ y \\in C: x \\sim y\\}$.\nRemember that sets only contain distinct elements so even though we ``loop'' through all values $x$ any two values $x,y$ such that $x \\sim y$ will lead to $[x] = [y]$ and the distinct equivalence class will only show up once in the collection $\\mathcal{I}$.\n\n\\paragraph{Problem 10}\nLet $\\succsim$ be rational preference relation and define $\\sim$ as the indifference relation where $x\\sim y$ iff $[x\\succsim y \\wedge y\\succsim x]$.\nBecause $\\succsim$ is rational we know that it is complete, reflexive and transitive.\nTo show $(x,x) \\in \\sim, \\; \\forall x$ we can refer to completeness to know that either $x\\succsim x$ or $x\\succsim x$ and by reflexiveness we know both statements are true so $x\\sim x$.\n\nSymmetry would imply that for any $x,y\\in C$ if $(x,y)\\in \\sim$ then $(y,x) \\in \\sim$.\nAgain we now know that $\\sim$ is reflexive and we still know $\\succsim$ is complete, reflexive and transitive and we should use these facts to establish symmetry and transitivity of $\\sim$.\nLet $x,y\\in C$ and suppose $x \\sim y$.\nThen we know $[x\\succsim y \\wedge y\\succsim x]$ is true by definition of $\\sim$.\nBecause the order of arguments in the conjunction don't alter its truth value we could say\n$[y\\succsim x \\wedge x \\succsim y]$ which is the definition of $y\\sim x$.\n\nFor transitivity, suppose $x,y,z \\in C$ and $[x\\sim y \\wedge y \\sim z]$.\nThen by definition $[x\\succsim y \\wedge y\\succsim x]$ and $[y\\succsim z \\wedge z\\succsim y]$.\nHence we have $x \\succsim y \\succsim z$ which by transitivity of $\\succsim$ implies $x\\succsim z$ and we also have $z\\succsim y \\succsim x$ which by transitivity of $\\succsim$ implies $z\\succsim x$.\nSince we have $[x\\succsim z \\wedge z\\succsim x]$ we see that $x \\sim z$ and therefore $\\sim$ is transitive.\n\\end{document}", "meta": {"hexsha": "8e9fd4bf98683098b62d4c53d7ad73c25bfc4ee9", "size": 2532, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assets/pdfs/math_bootcamp/2016/HW1Prob10Suggest.tex", "max_stars_repo_name": "joepatten/joepatten.github.io", "max_stars_repo_head_hexsha": "4b9acc8720f3a33337368fee719902b54a6f2f68", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assets/pdfs/math_bootcamp/2016/HW1Prob10Suggest.tex", "max_issues_repo_name": "joepatten/joepatten.github.io", "max_issues_repo_head_hexsha": "4b9acc8720f3a33337368fee719902b54a6f2f68", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2020-08-09T16:28:31.000Z", "max_issues_repo_issues_event_max_datetime": "2020-08-10T14:48:57.000Z", "max_forks_repo_path": "assets/pdfs/math_bootcamp/2016/HW1Prob10Suggest.tex", "max_forks_repo_name": "joepatten/joepatten.github.io", "max_forks_repo_head_hexsha": "4b9acc8720f3a33337368fee719902b54a6f2f68", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 76.7272727273, "max_line_length": 257, "alphanum_fraction": 0.7235387046, "num_tokens": 774, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.8031737987125612, "lm_q1q2_score": 0.6264698606659188}} {"text": "\\subsection{Simple algebraic structures}\\label{subsec:simple_algebraic_structures}\n\nWe list here several basic algebraic structures, we will use mostly as building blocks for more complicated structures.\n\nAs discussed in \\fullref{rem:first_order_model_notation}, denoting all operations explicitly is cumbersome, and we will usually avoid it.\n\n\\begin{definition}\\label{def:pointed_set}\\mcite[26]{MacLane1994}\n The simplest algebraic structure is a \\term{pointed set}. It is simply a nonempty set \\( \\mscrX \\) equipped with a distinguished element \\( e \\). It is an algebraic structure because \\( e \\) can be regarded as the sole value of a nullary function \\( *: \\mscrX^0 \\to \\mscrX \\).\n\n We will call \\( e \\) the \\term{origin} of \\( \\mscrX \\) based on the terminology for \\hyperref[def:euclidean_plane_coordinate_system/origin]{affine coordinate systems}.\n\n \\begin{thmenum}\n \\thmitem{def:pointed_set/theory} Pointed sets can also be viewed as \\hyperref[def:first_order_semantics/satisfiability]{models} of an empty \\hyperref[def:first_order_theory]{theory} for a \\hyperref[def:first_order_language]{first-order logic language} with a constant symbol, i.e. a nullary \\hyperref[def:first_order_language/func]{functional symbol}.\n\n \\thmitem{def:pointed_set/homomorphism} A \\hyperref[def:first_order_homomorphism]{homomorphism} between the pointed sets \\( (\\mscrX, e_{\\mscrX}) \\) and \\( (\\mscrY, e_{\\mscrY}) \\) is, explicitly, a function \\( \\varphi: \\mscrX \\to \\mscrY \\) that satisfies\n \\begin{equation}\\label{eq:def:pointed_set/homomorphism}\n \\varphi(e_{\\mscrX}) = e_{\\mscrY}.\n \\end{equation}\n\n \\thmitem{def:pointed_set/submodel} The set \\( S \\subseteq \\mscrX \\) is a \\hyperref[def:first_order_substructure]{substructure} if \\( \\mscrX \\) if \\( e \\in S \\).\n\n \\thmitem{def:pointed_set/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} pointed set is, up to an isomorphism, the set \\( \\set{ e } \\).\n\n It is a \\hyperref[def:universal_objects/initial]{zero object} in \\( \\cat{Set_*} \\) as discussed in \\fullref{ex:def:universal_objects/grp}.\n\n \\thmitem{def:pointed_set/category} We denote the \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} for this theory by \\( \\ucat{Set}_* \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:pointed_set_kernel}\\mimprovised\n The \\term{kernel} \\( \\ker(\\varphi) \\) of a \\hyperref[def:pointed_set/homomorphism]{pointed set homomorphism} \\( \\varphi: \\mscrX \\to \\mscrY \\) is the \\hyperref[def:zero_locus]{zero locus} of \\( \\varphi \\). That is, \\hyperref[thm:def:function/properties/preimage]{preimage} \\( \\varphi^{-1}(e_\\mscrY) \\).\n\n These are the \\hyperref[def:zero_morphisms/kernel]{categorical kernels} in \\hyperref[def:pointed_set/category]{\\( \\cat{Set_*} \\)}. This is discussed in \\fullref{ex:zero_morphisms_in_grp/kernel} for \\hyperref[def:group]{groups}, where the argument only depends on groups being pointed sets.\n\\end{definition}\n\n\\begin{definition}\\label{def:set_with_involution}\\mimprovised\n A \\term{set with an involution} is a \\hyperref[def:set]{set} \\( \\mscrX \\) with a unary operation \\( (\\anon)^{-1} \\) such that\n \\begin{equation*}\n (x^{-1})^{-1} = x\n \\end{equation*}\n for every \\( x \\in \\mscrX \\).\n\n Such an operation is called, surprisingly, an \\term{involution}.\n\n \\begin{thmenum}\n \\thmitem{def:set_with_involution/theory} We define the theory of sets with involution as a theory over the language consisting of a single unary functional symbol \\( \\anon^{-1} \\) and the sole axiom\n \\begin{equation}\\label{eq:def:set_with_involution/theory/axiom}\n (\\xi^{-1})^{-1} \\doteq \\xi.\n \\end{equation}\n\n \\thmitem{def:set_with_involution/homomorphism} A \\hyperref[def:first_order_homomorphism]{homomorphism} between sets with involutions \\( \\mscrX \\) and \\( \\mscrY \\) is a function \\( \\varphi: \\mscrX \\to \\mscrY \\) satisfying\n \\begin{equation}\\label{eq:def:set_with_involution/homomorphism}\n \\varphi(x^{-1})\n =\n \\varphi(x)^{-1}.\n \\end{equation}\n\n \\thmitem{def:set_with_involution/submodel} Any subset of a set with involution is again a set with involution.\n\n \\thmitem{def:set_with_involution/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} set with involution is the empty set.\n\n \\thmitem{def:set_with_involution/category} We denote the \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} for this theory by \\( \\ucat{Inv} \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:magma}\n A \\term{magma} is a set \\( \\mscrM \\) equipped with a \\hyperref[def:multi_valued_function/arguments]{binary function} \\( \\cdot: \\mscrM \\times \\mscrM \\to \\mscrM \\), called the \\term{magma operation}. Unless specified otherwise, we denote this operation by juxtaposition as \\( xy \\) instead of \\( x \\cdot y \\).\n\n We often call the operation \\term{multiplication} or, in the case of \\hyperref[def:endomorphism_monoid]{endomorphism monoids} --- \\term{composition}. See also the notes in \\fullref{rem:additive_magma} regarding additive magmas and in \\fullref{def:monoid_delooping} regarding the order of operands.\n\n \\begin{thmenum}[series=def:magma]\n \\thmitem{def:magma/theory} In analogy to the \\hyperref[def:pointed_set/theory]{theory of pointed sets}, we can define the theory of magmas as an empty theory over a language with a single \\hyperref[rem:first_order_formula_conventions/infix]{infix} binary functional symbol.\n\n \\thmitem{def:magma/homomorphism} A \\hyperref[def:first_order_homomorphism]{homomorphism} between the magmas \\( (\\mscrM, \\cdot_{\\mscrM}) \\) and \\( (\\mscrN, \\cdot_{\\mscrN}) \\) is, explicitly, a function \\( \\varphi: \\mscrM \\to \\mscrN \\) such that\n \\begin{equation}\\label{eq:def:magma/homomorphism}\n \\varphi(x \\cdot_{\\mscrM} y) = \\varphi(x) \\cdot_{\\mscrN} \\varphi(y)\n \\end{equation}\n for all \\( x, y \\in \\mscrM \\).\n\n \\thmitem{def:magma/submodel} The set \\( S \\subseteq \\mscrM \\) is a \\hyperref[def:first_order_substructure]{first-order submodel} of \\( \\mscrM \\) if it is closed under the magma operation. That is, if \\( x, y \\in S \\) implies \\( xy \\in S \\).\n\n We call \\( S \\) a \\term{submagma} of \\( \\mscrM \\).\n\n \\thmitem{def:magma/category} We denote the \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} for the theory of magmas by \\( \\ucat{Mag} \\).\n\n \\thmitem{def:magma/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial magma} is the empty set with an empty operation. It is the unique \\hyperref[def:universal_objects/initial]{zero object} in \\( \\cat{Mag} \\).\n\n \\thmitem{def:magma/opposite} The \\term{opposite magma} of \\( (\\mscrM, \\cdot) \\), also called the \\term{dual magma}, is the magma \\( (\\mscrM, \\star) \\) with multiplication reversed:\n \\begin{equation*}\n x \\star y \\coloneqq y \\cdot x.\n \\end{equation*}\n\n We denote the opposite magma by \\( \\mscrM^{-1} \\).\n\n \\thmitem{def:magma/exponentiation} We define an additional \\term{exponentiation} operation for positive integers \\( n \\) inductively as\n \\begin{equation}\\label{eq:def:magma/exponentiation}\n x^n \\coloneqq \\begin{cases}\n x, & n = 1 \\\\\n x \\cdot x^{n-1}, & n > 1\n \\end{cases}\n \\end{equation}\n\n \\thmitem{def:magma/power_set} It is customary to perform magma operations with sets. That is, if \\( A \\) and \\( B \\) are sets in the magma \\( \\mscrM \\), it is customary to write\n \\begin{equation*}\n A \\cdot B \\coloneqq \\set{ a \\cdot b \\colon a \\in A, b \\in B }.\n \\end{equation*}\n\n This actually turns the power set \\( \\pow(\\mscrM) \\) into a magma, which we will call the \\term{power set magma} of \\( \\mscrM \\). This is especially useful with the convention \\fullref{rem:singleton_sets} since it allows us to write \\( aB \\) for \\( a \\in M \\) and \\( B \\subseteq M \\).\n\n See \\fullref{thm:power_set_magma_preservation}.\n \\end{thmenum}\n\n We list some additional restrictions that are often imposed on magmas.\n \\begin{thmenum}[resume=def:magma]\n \\thmitem{def:magma/associative} We can add the (\\hyperref[thm:implicit_universal_quantification]{universal closure} of) following axiom to the theory:\n \\begin{equation}\\label{eq:def:magma/associative}\n (x \\cdot y) \\cdot z = x \\cdot (y \\cdot z).\n \\end{equation}\n\n If \\eqref{eq:def:magma/associative} is satisfied, we say that the operation \\( \\cdot \\) and, by extension, the magma itself, are \\term{associative}. Associative magmas are usually called \\term{semigroups}. Associativity imposes no additional restrictions on the homomorphisms, hence semigroups are a \\hyperref[def:subcategory]{full subcategory} of \\( \\cat{Mag} \\).\n\n \\thmitem{def:magma/commutative} Another common axiom is \\term{commutativity}:\n \\begin{equation}\\label{eq:def:magma/commutative}\n x \\cdot y = y \\cdot x.\n \\end{equation}\n\n Commutative magmas also form a full subcategory. Obviously \\( \\mscrM = \\mscrM^{-1} \\) in a commutative magma.\n\n \\thmitem{def:magma/idempotent} We say that the operation \\( \\cdot \\) is \\term{idempotent} if\n \\begin{equation}\\label{eq:def:magma/idempotent}\n x \\cdot x = x.\n \\end{equation}\n\n \\thmitem{def:magma/cancellative} We say that \\( \\cdot \\) is \\term{left-cancellative} (resp. \\term{right-cancellative}) if\n \\begin{equation}\\label{eq:def:magma/cancellative}\n z \\cdot x = z \\cdot y \\T{for all} z \\in \\mscrM \\T{implies that} x = y\n \\quad\n (\\T{resp.} x \\cdot z = y \\cdot z).\n \\end{equation}\n\n The operation is \\term{cancellative} if it is both left and right cancellative. Cancellative magmas also form a full subcategory.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:def:magma}\n \\hfill\n \\begin{thmenum}\n \\thmitem{ex:def:magma/composition} The quintessential example of a non-\\hyperref[def:magma/commutative]{commutative} operation is \\hyperref[def:multi_valued_function/composition]{composition} in any set of functions or, more generally, \\hyperref[def:category/composition]{morphism composition} in any \\hyperref[def:category]{category}.\n\n Composition is \\hyperref[def:magma/associative]{associative}. \\hyperref[def:magma/cancellative]{Cancellation} with respect to composition is discussed in \\fullref{def:morphism_invertibility} and, for function composition, in \\fullref{thm:function_invertibility_categorical}.\n\n \\thmitem{ex:def:magma/midpoint} The operation\n \\begin{equation*}\n (x, y) \\mapsto \\dfrac {x + y} 2\n \\end{equation*}\n makes \\( \\BbbR \\) a commutative and cancellative magma, which is not associative.\n \\end{thmenum}\n\\end{example}\n\n\\begin{remark}\\label{rem:additive_magma}\n General groups often arise as \\hyperref[def:automorphism_group]{automorphism groups}, which are, for the most part, non-commutative, while abelian groups are usually used as the main building block for \\hyperref[def:semiring/ring]{rings} and \\hyperref[def:left_module]{modules}.\n\n To make a further distinction, if the operation is denoted by \\( \\cdot \\) or juxtaposition, we say that the group is a \\term{multiplicative group}, and if the operation is denoted by \\( + \\), we say that the group is an \\term{additive group}. This terminology usually, but not necessarily, coincides with the group (or, more generally, the \\hyperref[def:magma]{magma}) being \\hyperref[def:magma/commutative]{commutative}.\n\n To make things explicit, a \\term{multiplicative magma} is any magma as defined in \\fullref{def:magma}. Compare this to \\term{additive magmas}, where\n \\begin{thmenum}\n \\thmitem{rem:additive_magma/addition} The magma operation is denoted by \\( + \\) and called \\term{addition}.\n\n \\thmitem{rem:additive_magma/multiplication} The magma \\hyperref[def:magma/exponentiation]{exponentiation operation} is denoted by \\( \\cdot \\) or juxtaposition and called \\term{multiplication}. Thus, multiplication is not defined for two elements of the magma, but defined for a positive integer and an element of the magma. In the case of a \\hyperref[def:magma/commutative]{commutative} \\hyperref[def:unital_magma/monoid]{monoid}, if multiplication is extended to two elements of the monoid, we instead talk about \\hyperref[def:semiring]{semirings}.\n\n \\thmitem{rem:additive_magma/identity} The \\hyperref[def:magma_identity]{identity} is usually denoted by \\( 0 \\).\n\n \\thmitem{rem:additive_magma/inverse} If an \\hyperref[def:unital_magma_inverse_element]{inverse} of \\( x \\) exists, it is denoted by \\( -x \\) rather than \\( x^{-1} \\).\n \\end{thmenum}\n\\end{remark}\n\n\\begin{proposition}\\label{thm:power_set_magma_preservation}\n \\hyperref[def:magma/associative]{Associativity} and \\hyperref[def:magma/commutative]{commutativity} from a magma \\( \\mscrM \\) are preserved in \\( \\pow(\\mscrM) \\), unlike \\hyperref[def:magma/cancellative]{cancellation}.\n\\end{proposition}\n\\begin{proof}\n Associativity and commutativity are obviously preserved.\n\n To show that cancellation is not, consider the group \\hyperref[def:group_of_integers_modulo]{\\( \\BbbZ_2 \\)}. It is a cancellative magma by \\fullref{thm:def:group/properties/cancellative}. Define the sets \\( A \\coloneqq \\{ 0, 1 \\} \\) and \\( B \\coloneqq \\{ 0 \\} \\). Then\n \\begin{equation*}\n A + A = A = A + B,\n \\end{equation*}\n however we cannot cancel \\( A \\) from the left because \\( A \\neq B \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:magma_exponentiation_properties}\n Fix a magma \\( \\mscrM \\). \\hyperref[def:magma/exponentiation]{Magma exponentiation} in \\( \\mscrM \\) has the following basic properties:\n\n \\begin{thmenum}\n \\thmitem{thm:magma_exponentiation_properties/commutativity} We have the following \\hyperref[def:magma/commutative]{commutativity}-like property: for \\( x \\in M \\) and \\( n = 1, 2, \\ldots \\),\n \\begin{equation}\\label{eq:thm:magma_exponentiation_properties/commutativity}\n x^n = x x^{n-1} = x^{n-1} x.\n \\end{equation}\n\n \\thmitem{thm:magma_exponentiation_properties/distributivity} Exponentiation distributes over multiplication: for any member \\( x \\in M \\) and any two positive integers \\( n \\) and \\( m \\),\n \\begin{equation}\\label{eq:thm:magma_exponentiation_properties/multiplication}\n x^{n + m} = x^n x^m.\n \\end{equation}\n\n \\thmitem{thm:magma_exponentiation_properties/repeated} For any member \\( x \\in M \\) and any two positive integers \\( n \\) and \\( m \\),\n \\begin{equation}\\label{eq:thm:magma_exponentiation_properties/repeated}\n (x^n)^m = x^{nm}.\n \\end{equation}\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:magma_exponentiation_properties/commutativity} We use induction on \\( n \\). The cases \\( n = 1 \\) and \\( n = 2 \\) are obvious. For \\( n > 2 \\), we have\n \\begin{equation*}\n x^n\n \\reloset {\\eqref{eq:def:magma/exponentiation}} =\n x x^{n-1}\n \\reloset {\\T{ind.}} =\n x x^{n-2} x\n \\reloset {\\eqref{eq:def:magma/exponentiation}} =\n x^{n-1} x.\n \\end{equation*}\n\n \\SubProofOf{thm:magma_exponentiation_properties/distributivity} We use induction on \\( n \\). The case \\( n = 1 \\) follows directly from \\eqref{eq:def:magma/exponentiation}. The case \\( n > 1 \\) follows from\n \\begin{equation*}\n x^{n + m}\n \\reloset {\\eqref{eq:def:magma/exponentiation}} =\n x x^{n + (m - 1)}\n \\reloset {\\T{ind.}} =\n x x^{n-1} x^m\n \\reloset {\\eqref{eq:def:magma/exponentiation}} =\n x^n x^m.\n \\end{equation*}\n\n \\SubProofOf{thm:magma_exponentiation_properties/repeated} We use induction on \\( n \\). The case \\( n = 1 \\) is obvious and the rest follows from\n \\begin{equation*}\n (x^n)^m\n \\reloset {\\eqref{eq:def:magma/exponentiation}} =\n x^n (x^n)^{m-1}\n \\reloset {\\T{ind.}} =\n x^n x^{n (m - 1)}\n \\reloset {\\eqref{eq:thm:magma_exponentiation_properties/multiplication}} =\n =\n x^{nm}.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:preordered_magma}\n A \\term{preordered magma} is a magma \\( \\mscrM \\) equipped with a \\hyperref[def:preordered_set]{preorder} \\( \\leq \\) such that \\( x \\leq y \\) implies \\( xz \\leq yz \\) and \\( zx \\leq zy \\) for all \\( z \\in M \\).\n\n The category of small preordered magmas is both \\hyperref[def:magma/category]{\\( \\cat{Mag} \\)}-\\hyperref[def:concrete_category]{concrete} and \\hyperref[def:preordered_set/category]{\\( \\cat{PreOrd} \\)}-concrete.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:preordered_magma_max_distributivity}\n In a \\hyperref[def:preordered_magma]{preordered magma} \\( \\mscrM \\),\n \\begin{equation}\\label{eq:thm:preordered_magma_max_distributivity}\n \\max \\set{a b, c d} \\leq \\max \\set{a, c} \\cdot \\max \\set{b, d}.\n \\end{equation}\n\\end{proposition}\n\\begin{proof}\n Since \\( a \\leq \\max \\set{a, c} \\), then\n \\begin{equation*}\n ab\n \\leq\n \\max \\set{a, c} \\cdot b\n \\leq\n \\max \\set{a, c} \\cdot \\set{b, d}\n \\end{equation*}\n\n Analogously, \\( cd \\leq \\max \\set{a, c} \\cdot \\set{b, d} \\) and\n \\begin{equation*}\n \\max \\set{a b, c d} \\leq \\max \\set{a, c} \\cdot \\set{b, d}.\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:topological_magma}\n A \\term{topological magma} is a magma equipped with a \\hyperref[def:topological_space]{topology} such that the magma operation is continuous.\n\n The category of small topological magmas is both \\hyperref[def:magma/category]{\\( \\cat{Mag} \\)}-\\hyperref[def:concrete_category]{concrete} and \\hyperref[def:category_of_small_topological_spaces]{\\( \\cat{Top} \\)}-concrete.\n\\end{definition}\n\n\\begin{definition}\\label{def:magma_identity}\n An element \\( e \\) of a magma \\( \\mscrM \\) is called a \\term{left identity} (resp. \\term{right identity}) if \\( ex = x \\) for all \\( x \\in M \\) (resp. \\( xe = x \\) for all \\( x \\in M \\)).\n\n If \\( e \\) is simultaneously a left and right identity, we call a \\term{two-sided identity} or simply \\term{identity} of \\( \\mscrM \\).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:magma_identity_unique}\n A two-sided \\hyperref[def:magma_identity]{magma identity} \\( e \\), if it exists, is unique.\n\\end{proposition}\n\\begin{proof}\n If \\( f \\) is another identity, then \\( e = ef = f \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:unital_magma}\n A \\hyperref[def:magma]{magma} with an \\hyperref[def:magma_identity]{identity} is called \\term{unital}. This makes it a \\hyperref[def:pointed_set]{pointed set}. We can consider it as a pair \\( (\\mscrM, \\cdot) \\) rather than a triple \\( (\\mscrM, \\cdot, e) \\) because, by \\fullref{thm:magma_identity_unique}, a two-sided identity is uniquely determined by the magma operation.\n\n \\begin{thmenum}\n \\thmitem{def:unital_magma/theory} The theory of unital magmas consists of the single axiom\n \\begin{equation}\\label{eq:def:unital_magma/theory/identity}\n \\qforall \\xi (e \\cdot \\xi \\doteq \\xi \\wedge \\xi \\cdot e \\doteq \\xi)\n \\end{equation}\n over the combined language of \\hyperref[def:pointed_set/theory]{pointed sets} and \\hyperref[def:magma/theory]{magmas}.\n\n \\thmitem{def:unital_magma/homomorphism} A \\hyperref[def:first_order_homomorphism]{homomorphism} between unital magmas is a function that satisfies both \\eqref{eq:def:pointed_set/homomorphism} and \\eqref{eq:def:magma/homomorphism}.\n\n \\thmitem{def:unital_magma/category} The \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} \\( \\ucat{Mag}_* \\) of unital magmas is \\hyperref[def:concrete_category]{concrete} with respect to both \\hyperref[def:pointed_set/category]{\\( \\ucat{Set}_* \\)} and \\hyperref[def:magma/category]{\\( \\ucat{Mag} \\)}.\n\n \\thmitem{def:unital_magma/submodel} The set \\( S \\subseteq X \\) is a \\hyperref[def:first_order_substructure]{substructure} if \\( \\mscrX \\) if \\( e \\in S \\). This is equivalent to \\( S \\) being a pointed subset.\n\n We say that \\( S \\) is a \\term{unital submagma}.\n\n \\thmitem{def:unital_magma/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} unital magma is the \\hyperref[def:pointed_set/trivial]{trivial pointed set} \\( \\set{ e } \\).\n\n \\thmitem{def:unital_magma/monoid} An \\hyperref[eq:def:magma/associative]{associative} unital magma is called a \\term{monoid}. The category \\( \\cat{Mon} \\) of monoids is a full subcategory of \\( \\cat{Mag}_* \\).\n\n \\thmitem{def:unital_magma/exponentiation} We extend \\hyperref[def:magma/exponentiation]{magma exponentiation} to all nonnegative integers by defining\n \\begin{equation*}\n x^0 \\coloneqq e.\n \\end{equation*}\n\n \\thmitem{def:unital_magma/power_set} The \\hyperref[def:magma/power_set]{power set magma} \\( \\pow(\\mscrM) \\) of a unital magma \\( \\mscrM \\) with identity \\( e \\) is again a unital magma with identity \\( \\set{ e } \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:monoid_cancellation_not_preserved_by_homomorphism}\\mcite{MathSE:magma_cancellation_not_preserved}\n \\hyperref[def:unital_magma/homomorphism]{Monoid homomorphisms} may not preserve the \\hyperref[def:magma/cancellative]{cancellation property}. For example, the \\hyperref[def:set_of_natural_numbers]{natural numbers} \\( \\BbbN \\) are a cancellative monoid under addition, as shown in \\fullref{thm:def:natural_number_addition/properties}), but the magma homomorphism\n \\begin{equation*}\n \\begin{aligned}\n &h: (\\BbbN, +) \\to (\\hyperref[thm:galois_field_existence]{\\BbbF_2}, \\max) \\\\\n &h(n) \\coloneqq \\begin{cases}\n 0, &n = 0 \\\\\n 1, &n > 0\n \\end{cases}\n \\end{aligned}\n \\end{equation*}\n does not preserve the cancellative property.\n\n Indeed, \\( \\max\\set{ 0, 1} = \\max\\set{ 1, 1 } \\), but \\( 0 \\neq 1 \\).\n\\end{example}\n\n\\begin{proposition}\\label{thm:unital_magma_kernel_is_submagma}\n The \\hyperref[def:pointed_set_kernel]{kernel} of a unital magma homomorphism \\( \\varphi: \\mscrM \\to \\mscrN \\) is a \\hyperref[def:first_order_substructure]{unital submagma} of \\( \\mscrM \\).\n\\end{proposition}\n\\begin{proof}\n By \\eqref{eq:def:pointed_set/homomorphism}, \\( e_{\\mscrM} \\in \\ker(\\varphi) \\), therefore \\( \\ker(\\varphi) \\) inherits its unital magma structure from \\( \\mscrM \\). It remains only to show that it is closed under the magma operation. But this is trivial since, if \\( x, y \\in \\ker(\\varphi) \\), then\n \\begin{equation*}\n \\varphi(xy) = \\varphi(x) \\varphi(y) = e_{\\mscrN} e_{\\mscrN} = e_{\\mscrN}.\n \\end{equation*}\n\\end{proof}\n", "meta": {"hexsha": "96823136dc5d619a1d638dffaf08decb854cc96a", "size": 22186, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/algebraic_building_blocks.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/algebraic_building_blocks.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/algebraic_building_blocks.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "32b364d134182efedaf961d5b3b67e8d1a9d37cf", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.5702005731, "max_line_length": 553, "alphanum_fraction": 0.7099522221, "num_tokens": 7071, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737869342623, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.6264698514789293}} {"text": "\\chapter{Mathematical Background}\n\\section{One-Shot Games}\n\\subsection{Prisoner's Dilemma}\nThe Prisoner's Dilemma is the canonical game of game theory, finding its use in models for everything from share prices to the relationship between lions and small birds. Essentially, in any situation where there is the possibility of cooperation, the Prisoner's Dilemma can be a useful model in understanding the situation to a first approximation.\n\\subsubsection{The Story}\nThe motivating story goes that two accomplices are caught at a crime scene, arrested and placed in seperate cells with no contact. The police give them each a choice: they can confess their crimes to the authorities (defect) or they can stay silent (cooperate).\\\\\nThey learn that if they both cooperate, they only get one year in prison. If they both defect, they both get two years. However, if one of them cooperates and the other defects, the cooperator gets 3 years and the defector gets out of jail immediately. As they are in seperate cells, they cannot communicate.\\\\\nThe smallest combined time in prison for the couple is if they both cooperate and take 1 year in jail each. But even if they could communicate and agree to the appealing option of both cooperating, it would still make sense to defect from the `contract' and defect. But just 1 year in jail is clearly better than 2. How should they work through this? As with most things in life, they should get mathematical.\n\\subsection{Making this Mathematical}\n\\subsubsection{Key Concepts}\nThis story has all the features of a game-theoretic game. Let's go through it, pick out the key elements and name them.\\\\\nThe four essential elements of a game have a useful acronym: \\textit{PAPI} \\cite{rasmusen}. This stands for:\\\\\nthe \\textbf{P}layers of the game,\\\\\nthe \\textbf{A}ctions available to each player,\\\\\nthe \\textbf{P}ayoffs for each outcome\\footnote{An \\textit{outcome} defines an action for each player. It can be written $(C,C)$ for two players both playing the action $(C)$. Note, actions are usually represented as a single letter i.e. $(C)$ for cooperate and $(D)$ for defect. } and\\\\\nthe \\textbf{I}nformation available to each player.\\\\\n\nIn our story, there are two players: Lou and Avery. The actions are cooperate with the partner in crime (i.e. stay quiet) or defect (i.e. confess and betray partner). The payoffs are given by the years in jail, $-x$ for $x$ years in jail. As they cannot communicate, there is no information available about the other player's choice prior to choosing.\n\\subsubsection{Representing games}\n\\subsubsection{Payoff Matrix}\nA payoff matrix is a way of defining two-player simultaneous games completely i.e. it tells you about every element of \\textit{PAPI}. The payoff matrix for the Prisoner's Dilemma given in the story would be:\\\\\n\\setlength{\\extrarowheight}{2pt}\n\\begin{tabular}{cc|c|c|}\n\t& \\multicolumn{1}{c}{} & \\multicolumn{2}{c}{Avery}\\\\\n\t& \\multicolumn{1}{c}{} & \\multicolumn{1}{c}{$C$} & \\multicolumn{1}{c}{$D$} \\\\\\cline{3-4}\n\t\\multirow{2}*{Lou} & $C$ & $(-1,-1)$ & $(-3,0)$ \\\\\\cline{3-4}\n\t& $D$ & $(0,-3)$ & $(-2,-2)$ \\\\\\cline{3-4}\n\\end{tabular}\n\\\\\n\\\\\nThe players, actions and payoffs are defined. Also there is no information available about the other player's action. So this matrix completely describes the game. Note that the the ordered pair $(x,y)$ in each matrix entry gives the payoffs for each outcome. $x$ gives the payoff for the horizontal player and $y$ gives the vertical player's. So outcome $(D,C)$ has payoff $(0,-3)$ meaning a payoff of $0$ for Lou for defecting and a payoff of $-3$ for Avery for cooperating.\\\\\nThe payoff matrix is an extremely useful representation, allowing concise, complete descriptions of games. Consider, for example:\\\\\n\\setlength{\\extrarowheight}{2pt}\n\\begin{tabular}{cc|c|c|c|}\n\t& \\multicolumn{1}{c}{} & \\multicolumn{3}{c}{Avery} \\\\\n\t& \\multicolumn{1}{c}{} & \\multicolumn{1}{c}{$R$} & \\multicolumn{1}{c}{$P$} & \\multicolumn{1}{c}{$S$} \\\\\\cline{3-5}\n\t& $R$ & $(0,0)$ & $(-1,1)$ & $(1,-1)$ \\\\ \\cline{3-5}\n\tLou & $P$ & $(1,-1)$ & $(0,0)$ & $(-1,1)$ \\\\\\cline{3-5}\n\t& $S$ & $(-1,1)$ & $(1,-1)$ & $(0,0)$ \\\\\\cline{3-5}\n\\end{tabular}\n\\\\\n\\\\\nThis game might initially look foreign, but it is simply Rock, Paper, Scissors. You can create stories to motivate them but the payoff matrix completely defines all the relevant mathematical aspects of the game. This allows us to abstract from the messy details of the human world and find `solutions' to the game.\n\\subsubsection{Solution Concepts}\nA \\textit{strategy} is a set of rules for which actions to play at each point of the game. It can respond to the other player's actions.\n\\subsubsection{Strategies}\n\\paragraph{Socially Optimal Strategy}Lou is able to recognise the \\textit{socially optimal strategy} as the option of both cooperate so that they get only 1 year each. This is the outcome that leads to the highest joint payoff for all players.\n\\paragraph{Nash Equilibrium} However, the outcome with both defecting is the Nash Equilibrium. This is an outcome in which, even if all players knew what the rest of the players actions were, none of them would choose to change their action as it would not improve their payoff. In this game, both players defecting is the unique Nash Equilibrium.\n\\paragraph{Dominant Strategy}The outcome of $(D, D)$ actually satisfies a stronger condition as the action defect is a \\textit{dominant strategy} for both players. This means that no matter what action the opponent plays, the player can always increase their payoff by defecting. This is the way (defect, defect) `pulls' players in, despite it having a worse payoff for both players than cooperate cooperate. Game theory assumes that players are \\textit{rational} (i.e. the payoffs we have described actually match their desires and they are aware of how the game works). So if Lou and Avery are rational, they will always end up both defecting, betraying each other and being rewarded with a longer jail term. So it goes.\n\n\\section{Repeated Games}\n\\subsubsection{Introduction}\nThe previous games are called \\textit{one-shot games} meaning, after the first game is played, no further games are played. It happens once and never again, with no possible repurcussions that are not already described in the payoff matrix.\\\\\n\nUnlike one-shot games, repeated games have players play multiple games against each other. This means that it is harder or even impossible to `solve' these games to find the best actions and equilibria. However, this makes them more interesting, allowing for complex dynamics. For example, the prisoner's dilemma no longer has a best strategy when repeated.\n\\subsubsection{Repeated Prisoner's Dilemma Tournament}\nWe can create a tournament of different strategies all playing several hundred games against each other. The winner is the strategy with the highest payoff overall. This numerical experiment was famously first done by Robert Axelrod\\cite{axelrod} in the 1980s. In Axelrod's first tournament, the winner was `Tit for Tat', a strategy that always cooperates on the first round and then copies the opponents previous move forever after.\\\\\n(\nMy simulation of Prisoner's Dilemma tournament.\\\\\nInclude Tit for Tat, Grim Trigger, Random, always cooperate, always defect, tit for two tats etc.\\\\\nbrief analysis of my results- graph of results\n)\n\n\\subsubsection{Repeated Games with Evolution}\n[Include replicator equations at some point here.]\\\\\nIncorporating repetitions, allows the possibility of 'evolutionary' behaviour. For example, one can create simulations where higher payoffs are more likely to produce 'offspring' i.e. players that use the same strategy. For example, below is a simulation of strategies playing Rock, Paper Scissors defined by the payoff matrix as above.\\\\\n(\nMy simulation of Rock, Paper, Scissors\n)\\\\\nThe effect is a self-balancing system. If the strategy \\textit{rock} becomes more populous, \\textit{paper} will start to get higher payoffs on average. This in turn brings the population back towards $\\frac{1}{3}$ for each strategy.\\\\\nThe same scenario works with the admitedlly less well-known game Rock, Paper, Scissors, Spock\\cite{for game}\\cite{for code}.\\\\\n(\nMy simulation of Rock, Paper Scissors, Spock\n)\\\\\nHowever, now with the different rules we can have the extinction of strategies.\n\\section{Games on Graphs}\nIn real life, games are usually not isolated situations that happen as if in a laboratory. Instead, we take part in many at the same time with several different players. Sometimes the different games effect each other. To model this, we can use graphs.\n\\subsubsection{Graphs}\nA graph $G=(V,E)$ is a set of vertices $V$ and a set of edges $E$ which are pairs of vertices. For now we will consider only undirected ($(u,v)\\in E\\iff(v,u)\\in E$) graphs with no loops ($\\nexists v\\textnormal{ s.t. } (v,v)\\in E$). A graph looks something like this:\\\\\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{directedGraph.png}\n\t\t\\caption{Directed graph}\n\t\t\\label{fig:dir}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{loopGraph.png}\n\t\t\\caption{Graph with loops}\n\t\t\\label{fig:loop}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{graph.png}\n\t\t\\caption{Undirected graph without loops}\n\t\t\\label{fig:undirected}\n\t\\end{subfigure}\n\t\\caption{We will consider graphs of type \\ref{fig:undirected} and ignore the others}\n\t\\label{fig:complete graphs}\n\\end{figure}\nTwo vertices $u,v$ are \\textit{adjacent} iff $(u,v)\\in E$.\n\\subsubsection{Games on graphs we have already seen}\nWe can now consider games on graphs by allowing each vertex to represent a player and only allowing players to play a game against players they are connected to. In fact, we have sneakily been doing this the whole time but now we want to make this explicit.\\\\\nThe two person one-shot prisoner's dilemma in 1.1.2 was a game on a trivial graph $G=(\\{Lou,Avery\\},\\{(Lou,Avery)\\})$.\\\\\n\\begin{figure}[h]\n\t\\centering\n\t\\begin{tikzpicture}\n\t\\draw\n\t(1,1) node[anchor=east,circle,draw]{Lou}--\n\t(3,1) node[anchor=west,circle,draw]{Avery};\n\t\\end{tikzpicture}\n\t\\caption{The trivial graph underlying the one-shot Prisoner's Dilemma}\n\\end{figure}\nSimilarly, for the tournament if we call the players $p_1,...,p_n$, the tournament was a repeated game on a \\textit{complete graph}. This is the graph with each vertex connected to everyother vertex: $K_n=\\{\\{p_1,...,p_n\\},\\{(p_i,p_j):\\forall i\\neq j\\}\\}$\\\\\n\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.5\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{K5.pdf}\n\t\t\\caption{$K_5$}\n\t\t\\label{fig:K5}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.5\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{K16.pdf}\n\t\t\\caption{$K_{16}$}\n\t\t\\label{fig:K16}\n\t\\end{subfigure}\n\t\\caption{The Complete Graphs $K_5$ and $K_{16}$}\n\t\\label{fig:complete graphs}\n\\end{figure}\n\n\\subsubsection{Some new Graphs}\nNow we have made the graphs we are playing games on explicit we can try some new ones.\\\\\nA graph often used in models is the 2D lattice as it is both instructive and relatively easy to analyse.\\cite{eq_of_life}\\\\\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=.5\\linewidth]{2d_lattice.pdf}\n\t\\caption{A 2D lattice}\n\\end{figure}\nThe von Neumann neighbourhood of the 2D lattice (Figure \\ref{fig:vonneumann}) makes each point a neighbour of the $4$ points vertically and horizontally next to it.\\\\\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.5\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{vonNeumannNeighbourhood.pdf}\n\t\t\\caption{The von Neumann Neighbourhood}\n\t\t\\label{fig:vonneumann}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.5\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{mooreNeighbourhood.png}\n\t\t\\caption{The Moore neighbourhood}\n\t\t\\label{fig:moore}\n\t\\end{subfigure}\n\t\\caption{Different neighbourhoods on the 2D Lattice}\n\t\\label{fig: lattice neighbourhoods}\n\\end{figure}\nThe Moore neighbourhood of the 2D lattice (Figure \\ref{fig:vonneumann}) includes the nearest vertical and horizontal neighbours as well as the nearest diagonal points.\\\\\nTo avoid boundary effects when playing games on these lattices (i.e. the side players having fewer opponents), we can 'wrap' the 2D lattice around itself to create a torus. So each neighbour at the top of the lattice is joined to the neighbour in the same column displayed visually at the bottom of the lattice.\n\\subsection{Prisoner's Dilemma on a Torus}\n\\subsubsection{Setup}\nWe can extend the Prisoner's Dilemma to a repeated game on 2D Lattice.\\\\\nInitialisation:\\\\\nMake an $n \\times n$ grid with wrapped ends to avoid boundary effects, creating a torus. Then let each vertex be a player of the Prisoner's Dilemma. In the first round, each player plays $C$ with probability $p$ and $D$ with probability $(1-p)$. They play their strategy simultaneously against every player in their Moore neighbourhood (Figure \\ref{fig:moore}), using the same strategy against each of them.\\\\\nEvery stage after:\\\\\nAfter playing the previous round, each player looks at their neighbours scores. They then adopt the strategy of the highest scoring neighbour as their strategy for the next round. Again they play the same strategy against every player in their neighbourhood.\nThe payoff matrix retains characteristics of the original given Prisoner's Dilemma and is given by:\\\\\n\\setlength{\\extrarowheight}{2pt}\n\\begin{tabular}{cc|c|c|}\n\t& \\multicolumn{1}{c}{} & \\multicolumn{2}{c}{Lou}\\\\\n\t& \\multicolumn{1}{c}{} & \\multicolumn{1}{c}{$C$} & \\multicolumn{1}{c}{$D$} \\\\\\cline{3-4}\n\t\\multirow{2}*{Avery} & $C$ & $(1,1)$ & $(\\epsilon,b)$ \\\\\\cline{3-4}\n\t& $D$ & $(b,\\epsilon)$ & $(0,0)$ \\\\\\cline{3-4}\n\\end{tabular}\n\\\\\n\\\\\nwhere $\\epsilon<11.\\bar{6}$, the board eventually tends to an equilibrium with mostly defectors. Clearly the rewards of non-cooperation are too high to a sustain a more socially optimal situation.\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/1b=17.png}\n\t\t\\caption{Early}\n\t\t\\label{fig:dir}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/2b=17.png}\n\t\t\\caption{Developing}\n\t\t\\label{fig:loop}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/3b=17.png}\n\t\t\\caption{Equilibrium}\n\t\t\\label{fig:undirected}\n\t\\end{subfigure}\n\t\\caption{The simulation running with $b=1.7$}\n\t\\label{fig:complete graphs}\n\\end{figure}\nSimilarly, for $b<1.6$, the simulation tends towards a static equilibrium of mainly cooperators.\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/1b=15.png}\n\t\t\\caption{Early}\n\t\t\\label{fig:dir}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/2b=15.png}\n\t\t\\caption{Developing}\n\t\t\\label{fig:loop}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/3b=15.png}\n\t\t\\caption{Equilibrium}\n\t\t\\label{fig:undirected}\n\t\\end{subfigure}\n\t\\caption{The simulation running with $b=1.5$}\n\t\\label{fig:complete graphs}\n\\end{figure}\nBetween these two parameter regions, exists the third which exhibits the most interesting behaviour with dynamics between cooperation and defection.\n\\begin{figure}\n\t\\centering\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/1b=163.png}\n\t\t\\caption{Early}\n\t\t\\label{fig:dir}\n\t\\end{subfigure}%\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/2b=163.png}\n\t\t\\caption{Developing}\n\t\t\\label{fig:loop}\n\t\\end{subfigure}\n\t\\begin{subfigure}{.3\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=.9\\linewidth]{/mooreLattice/3b=163.png}\n\t\t\\caption{Dynamic equilibrium}\n\t\t\\label{fig:undirected}\n\t\\end{subfigure}\n\t\\caption{The simulation running with $b=1.63$}\n\t\\label{fig:complete graphs}\n\\end{figure}\n\\subsubsection{Quantative Analysis}\nlooking at individual squares\n%\\subsubsection{Prisoner's Dilemma on a Scale Free Graph}\n\n\n", "meta": {"hexsha": "8a7facc091a3f1ee56bef2d295c2dddb8ef45443", "size": 16524, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Writing/TeX_files/Old/chapter01.tex", "max_stars_repo_name": "joekroese/math-of-revolution", "max_stars_repo_head_hexsha": "c831ea3d5f6c56c3861522f71ec47e1a22f9ff2c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-12-07T18:16:46.000Z", "max_stars_repo_stars_event_max_datetime": "2020-06-16T10:54:20.000Z", "max_issues_repo_path": "Writing/TeX_files/Old/chapter01.tex", "max_issues_repo_name": "joekroese/math-of-revolution", "max_issues_repo_head_hexsha": "c831ea3d5f6c56c3861522f71ec47e1a22f9ff2c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Writing/TeX_files/Old/chapter01.tex", "max_forks_repo_name": "joekroese/math-of-revolution", "max_forks_repo_head_hexsha": "c831ea3d5f6c56c3861522f71ec47e1a22f9ff2c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.5909090909, "max_line_length": 722, "alphanum_fraction": 0.7529653837, "num_tokens": 4697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569014, "lm_q2_score": 0.7799928900257127, "lm_q1q2_score": 0.6264698506133428}} {"text": "\\section{On-policy Prediction with Approximation}\n\n\\subsection{Exercise 9.1}\n\\subsubsection*{Q}\nShow that tabular methods such as presented in Part I of this book are a special case of linear function approximation. What would the feature vectors be?\n\\subsubsection*{A}\nWrite $\\hat{V}(s, \\vec{w}) = w$ and we get that $\\grad_{\\vec{w}} \\hat{V}(s, \\vec{w}) = 1$ so we return to tabular TD learning. In this case the features are $x(s) = 1$ $\\forall s \\in \\mathcal{S}$.\n\n\\subsection{Exercise 9.2}\n\\subsubsection*{Q}\nWhy does (9.17) define $(n+1)^k$ distinct features for dimension $k$?\n\\subsubsection*{A}\nEach of the $k$ terms can independently have one of $n+1$ exponents, hence the total number of features is $(n+1)^k$.\n\n\\subsection{Exercise 9.3}\n\\subsubsection*{Q}\nWhat $n$ and $c_{i,j}$ produce the feature vectors $\\vec{x}(s) = (1, s_1, s_2, s_1s_2, s_1^2, s_2^2, s_1s_2^2, s_1^2 s_2, s_1^2 s_2^2)$?\n\\subsubsection*{A}\n$n=2$ and $c_i, j = C_{ij}$ where\n\\[\nC =\n\\begin{pmatrix}\n 0 & 0 \\\\\n 1 & 0 \\\\\n 0 & 1 \\\\\n 1 & 1 \\\\\n 2 & 0 \\\\\n 0 & 2 \\\\\n 1 & 2 \\\\\n 2 & 1 \\\\\n 2 & 2\n\\end{pmatrix}\n\\]\n\n\\subsection{Exercise 9.4}\n\\subsubsection*{Q}\nSuppose we believe that one of two state dimensions is more likely to have an effect on the value function than is the other, that generalization should be primarily across this dimension rather than along it. What kind of tilings could be used to take advantage of this prior knowledge?\n\\subsubsection*{A}\nTiles that are thin along the dimension of interest and long across it. Rectangles, for instance.\n\n\\subsection{Exercise 9.5}\n\\subsubsection*{Q}\nSuppose you are using tile coding to transform a seven-dimensional continuous state space into binary feature vectors to estimate a state value function $\\hat{v}(s, \\vec{w}) \\approx v_\\pi(s)$. You believe that the dimensions do not interact strongly, so you decide to use eight tilings of each dimension separately (stripe tilings), for $7 \\times 8 = 56$ tilings. In addition, in case there are some pairwise interactions between the dimensions, you also take all $\\binom{7}{2} = 21$ pairs of dimensions and tile each pair conjunctively with rectangular tiles. You make two tilings for each pair of dimensions, making a grand total of $21 \\times 2 + 56 = 98$ tilings. Given these feature vectors, you suspect that you still have to average out some noise, so you decide that you want learning to be gradual, taking about 10 presentations with the same feature vector before learning nears its asymptote. What step-size parameter $\\alpha$ should you use? Why?\n\n\\subsubsection*{A}\nEach tiling is a partition, so each tiling has exactly one tile activated per state. This means that in our case the number of features is 98. We consider each of these equally likely because we are uninformed. We therefore take\n\\[\n\\alpha = \\frac{1}{10 \\times 98} = \\frac{1}{980}.\n\\]\nSo that on average we see each feature 10 times before asymptote. [Note that this assumes a constant target.]\n", "meta": {"hexsha": "e1f106e24ac92517ce582e891818b31e8f22524b", "size": 2980, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "exercises/chapters/chapter9/chapter9_content.tex", "max_stars_repo_name": "ElliotMunro200/reinforcement_learning_an_introduction", "max_stars_repo_head_hexsha": "c4fccb46a4bb00955549be3505144ec49f0132e5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 234, "max_stars_repo_stars_event_min_datetime": "2018-09-01T00:26:29.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T03:55:50.000Z", "max_issues_repo_path": "exercises/chapters/chapter9/chapter9_content.tex", "max_issues_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_issues_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2018-11-29T21:04:36.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-13T17:11:50.000Z", "max_forks_repo_path": "exercises/chapters/chapter9/chapter9_content.tex", "max_forks_repo_name": "15779235038/reinforcement_learning_an_introduction", "max_forks_repo_head_hexsha": "a0ac9e5da6eaeae14d297a560c499d1a6e579c2a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 63, "max_forks_repo_forks_event_min_datetime": "2018-07-31T04:53:21.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T04:03:43.000Z", "avg_line_length": 58.431372549, "max_line_length": 958, "alphanum_fraction": 0.7265100671, "num_tokens": 870, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321843145405, "lm_q2_score": 0.7772998663336158, "lm_q1q2_score": 0.6262177891817513}} {"text": "\\hypertarget{classnumpp_1_1differentiation_1_1symbolic_1_1sinus}{}\\section{numpp\\+:\\+:differentiation\\+:\\+:symbolic\\+:\\+:sinus$<$ T $>$ Class Template Reference}\n\\label{classnumpp_1_1differentiation_1_1symbolic_1_1sinus}\\index{numpp\\+::differentiation\\+::symbolic\\+::sinus$<$ T $>$@{numpp\\+::differentiation\\+::symbolic\\+::sinus$<$ T $>$}}\n\\subsection*{Public Types}\n\\begin{DoxyCompactItemize}\n\\item \n\\mbox{\\Hypertarget{classnumpp_1_1differentiation_1_1symbolic_1_1sinus_a5bde5934631bcfb972ad6c15c81ccca5}\\label{classnumpp_1_1differentiation_1_1symbolic_1_1sinus_a5bde5934631bcfb972ad6c15c81ccca5}} \n{\\footnotesize template$<$std\\+::size\\+\\_\\+t Active$>$ }\\\\using {\\bfseries derivative} = simplify\\+\\_\\+multiplication$<$ \\hyperlink{classnumpp_1_1differentiation_1_1symbolic_1_1cosinus}{cosinus}$<$ T $>$, typename T\\+::template derivative$<$ Active $>$ $>$\n\\end{DoxyCompactItemize}\n\\subsection*{Static Public Member Functions}\n\\begin{DoxyCompactItemize}\n\\item \n\\mbox{\\Hypertarget{classnumpp_1_1differentiation_1_1symbolic_1_1sinus_a25872983c585075c533169136769caa7}\\label{classnumpp_1_1differentiation_1_1symbolic_1_1sinus_a25872983c585075c533169136769caa7}} \nstatic C\\+O\\+N\\+S\\+T\\+E\\+X\\+PR auto {\\bfseries calculate} (auto \\&\\&values)\n\\end{DoxyCompactItemize}\n\n\nThe documentation for this class was generated from the following file\\+:\\begin{DoxyCompactItemize}\n\\item \ndifferentiation/symbolic/trigonometric.\\+hpp\\end{DoxyCompactItemize}\n", "meta": {"hexsha": "ab33503349dc8351690d6475f4509af8c564faa2", "size": 1439, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1sinus.tex", "max_stars_repo_name": "szymonmaszke/numpp", "max_stars_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2018-06-06T01:51:17.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-02T15:17:00.000Z", "max_issues_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1sinus.tex", "max_issues_repo_name": "vyzyv/numpp", "max_issues_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-11-28T12:15:46.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-16T00:03:38.000Z", "max_forks_repo_path": "docs/classnumpp_1_1differentiation_1_1symbolic_1_1sinus.tex", "max_forks_repo_name": "szymonmaszke/numpp", "max_forks_repo_head_hexsha": "9149c9d81f70a6ce833fdd1d2f0f2b584e2ac4d9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-08-06T13:58:27.000Z", "max_forks_repo_forks_event_max_datetime": "2018-04-06T06:45:22.000Z", "avg_line_length": 71.95, "max_line_length": 256, "alphanum_fraction": 0.8047255038, "num_tokens": 538, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998663336157, "lm_q2_score": 0.8056321796478256, "lm_q1q2_score": 0.6262177855543143}} {"text": "%%=========================================================================\n%% Section 3.4: Boundedness of the Exponentially Weighted Hilbert Transform\n%%=========================================================================\n\n\\documentclass[../dissertation.tex]{subfiles}\n\n\\begin{document}\n\\section{Boundedness of the Exponentially Weighted Hilbert Transform}\\label{sec3:BndE}\n\nProving that the exponentially weighted Hilbert transform $E$ is bounded on \n$L^p(\\mathbb R)$ for $1< p < \\infty$ is a multistep process in which we first compute\nthe Fourier multiplier of $E$ (Lemma \\ref{lma:FMultE}), use the Fourier multiplier\nof $E$ to show that $E$ is strong type $(2,2)$ (Theorem \\ref{thm:Estrong}), prove $E$ is \nweak type $(1, 1)$ (Theorem \\ref{thm:Eweak}), and then use these results to ultimately\nprove $E$ is strong type $(p, p)$ for $p>1$ (Theorem \\ref{thm:EbndOp}). Note that an \noperator is said to be weak type $(p,q)$ if it is a bounded operator from $L^p$ to \n$weak$-$L^q$ (\\textit{i.e.} $L^{q, \\infty}$) and strong type $(p,q)$ if it \nis a bounded operator from $L^p$ to $L^q$. For a defintion of the space\n$L^{p,\\infty}$, please see the \\hyperref[app:HA]{``Harmonic Analysis Results''} \nappendix.\n\n\n\\begin{lma}\\label{lma:FMultE}\n\tThe Exponentially weighted Hilbert transform $E$ has Fourier multiplier\n\t$$m_E(\\xi) = \\frac{1}{\\pi} \\arctan(\\xi/\\pi).$$\\label{sym3:Emult}\n\\end{lma}\n\\begin{proof}\n\tIn order to compute the Fourier multiplier of $E$, it suffices to compute the Fourier\n\ttransform of the function $x^{-1} e^{-\\pi|x|}$, which is $2\\pi i$ times the \n\tconvolution kernel of $E$. Letting $\\theta$ denote the Heaviside function, we \n\twrite $x^{-1} e^{-\\pi|x|}$ as \n\t\\begin{align}\\label{eq:Ekern}\n\t\t\\frac{e^{\\pi |x|}}{x} \n\t\t\t=\n\t\t\t\\frac{e^{\\pi \\, x}}{x} \\theta(-x) + \\frac{e^{-\\pi \\, x}}{x} \\theta(x),\n\t\\end{align}\n\tand compute the Fourier transform of two terms on the right-hand side of \n\t\\eqref{eq:Ekern} separately. By further splitting $\\frac{e^{\\pi \\, x}}{x} \\theta(-x)$\n\tinto even and odd parts, we obtain\n\t\\begin{align}\n\t\t\\mc F\\left( \\frac{e^{\\pi (\\dotarg)}}{(\\dotarg)} \\theta(-\\dotarg) \\right)\n\t\t\t&= \\mc F^{(c)} \\left( \n\t\t\t\t\t\\frac{e^{\\pi (\\dotarg)}\\theta(-\\dotarg)-e^{-\\pi(\\dotarg)}\\theta(\\dotarg)}\n\t\t\t\t\t\t{2(\\dotarg)} \n\t\t\t\t\\right)(\\xi)\n\t\t\t\t\\\\\n\t\t\t&\\quad + i \\, \\mc F^{(s)} \\left(\n\t\t\t\t\t\\frac{e^{\\pi(\\dotarg)}\\theta(-\\dotarg)+e^{-\\pi(\\dotarg)}\\theta(\\dotarg)}\n\t\t\t\t\t\t{2(\\dotarg)} \n\t\t\t\t\\right)(\\xi),\n\t\t\t\t\\nonumber\n\t\\end{align}\n\twhere $\\mc F^{(c)}$ and $\\mc F^{(s)}$ respectively denote the Fourier cosine and\n\tFourier sine transformations. Direct computation yields \n\t\\begin{align*}\n\t\t\\mc F^{(c)} \\left( \n\t\t\t\\frac{e^{\\pi (\\dotarg)}\\theta(-\\dotarg)-e^{-\\pi(\\dotarg)}\\theta(\\dotarg)}\n\t\t\t\t{2(\\dotarg)} \n\t\t\\right)(\\xi)\n\t\t\t&= \\frac{1}{\\pi}\n\t\t\t\t\\int_0^\\infty \n\t\t\t\t\t\\frac{e^{\\pi \\,x}\\theta(-x)-e^{-\\pi\\,x}\\theta(x)}{2x}\n\t\t\t\t\\, \\cos(\\xi x) \\, \\mathrm{d}x \n\t\t\t\t\\\\\n\t\t\t&= \\frac{1}{2} \\big(\n\t\t\t\t\t\\log(\\xi^2 + \\pi^2) + 2 \\gamma\n\t\t\t\t\\big)\n\t\t\t\t\\nonumber\n\t\\end{align*}\n\tand \n\t\\begin{align*}\n\t\t\\mc F^{(s)} \\left( \n\t\t\t\\frac{e^{\\pi (\\dotarg)}\\theta(-\\dotarg)+e^{-\\pi(\\dotarg)}\\theta(\\dotarg)}\n\t\t\t\t{2(\\dotarg)} \n\t\t\\right)(\\xi)\n\t\t\t&= \\frac{1}{2\\pi} \n\t\t\t\t\\int_0^\\infty \n\t\t\t\t\t\\frac{e^{\\pi \\,x}\\theta(-x)+e^{-\\pi\\,x}\\theta(x)}{2x}\n\t\t\t\t\\, \\sin(\\xi x) \\, \\mathrm{d}x \n\t\t\t\t\\\\\n\t\t\t&= \\frac{1}{2} \\big(\n\t\t\t\t\t\\arctan(\\xi/\\pi)\n\t\t\t\t\\big),\n\t\t\t\t\\nonumber\n\t\\end{align*}\n\twhere $\\gamma$ denotes the Euler-Mascheroni constant, which implies\n\t\\begin{align}\n\t\t\\mc F \\left( \n\t\t\t \\frac{e^{\\pi (\\dotarg)}}{(\\dotarg)} \\theta(-\\dotarg)\n\t\t\\right)(\\xi)\n\t\t\t&=\n\t\t\t\t\\gamma\n\t\t\t\t+ \\frac{1}{2} \\log(\\xi^2 + \\pi^2) \n\t\t\t\t+ i \\, \\arctan(\\xi/\\pi).\n\t\\end{align}\n\tA similar computation also shows \n\t\\begin{align}\n\t\t\\mc F \\left( \n\t\t\t \\frac{e^{-\\pi (\\dotarg)}}{(\\dotarg)} \\theta(\\dotarg)\n\t\t\\right)(\\xi)\n\t\t\t&= \n\t\t\t\t- \\gamma\n\t\t\t\t- \\frac{1}{2} \\log(\\xi^2 + \\pi^2) \n\t\t\t\t+ i \\, \\arctan(\\xi/\\pi),\n\t\\end{align}\n\tfrom which the result follows.\n\\end{proof}\n\n\\begin{thm}\\label{thm:Estrong}\n\t$E$ is strong type $(2, 2)$.\n\\end{thm}\n\\begin{proof}\n\tThis result is an immediate consequence of Lemma \\ref{lma:FMultE}, \n\tPlancherel's Theorem, and the density of $\\mathscr S(\\mathbb R)$ \n\tin $L^2(\\mathbb R)$.\n\\end{proof}\n\n\n\\begin{thm}\\label{thm:Eweak}\n\t$E$ is weak type $(1, 1)$. That is, $E$ is a bounded operator from \n $L^1(\\mathbb R)$ into\n $L^{1, \\infty}(\\mathbb R)$, where $L^{1, \\infty}$ denotes weak $L^1$.\n \\label{sym3:weakLp}\n\\end{thm}\n\\begin{proof}\n Fix $\\Lambda > 0$, let $f$ be Schwartz class, and assume without loss of \n generality that $f \\in \\mathscr S(\\mathbb R)$ is real-valued and nonnegative \n (otherwise, we can decompose $f$ in the appropriate pieces).\n Let $\\{ I_j \\}$ be the sequence sequence of dyadic intervals in the \n Calder\\'on-Zygmund decomposition of \n $f$ at height $\\Lambda$. For $\\Omega := \\bigcup_j I_j$, define \n \\begin{align}\\label{eq:C-Zdecomp}\n g(x) := \n \\begin{cases}\n \\displaystyle f(x), & x \\notin \\Omega \\\\[0.5\\baselineskip]\n \\displaystyle \\frac{1}{|I_j|} \\int_{I_j} f, & x \\in I_j\n \\end{cases},\n \\qquad \\text{and} \\qquad\n b(x) := \\sum_j b_j(x),\n \\end{align}\n where \n \\[\n b_j(x) = \\left( f(x) - \\dfrac{1}{|I_j|} \\int_{I_j} f \\right) \\chi_{I_j}(x).\n \\]\n Note that $f = g+b$ and\n \\begin{align} \\label{eq:GoodPartBnd}\n g(x) \\leq 2 \\Lambda \\quad \\forall x \\in \\mathbb R.\n \\end{align}\n To show that $Eg$ and $Eb$ are well defined, it suffices\n to bound $d_{Eg}(\\Lambda)$ and $d_{Eb}(\\Lambda)$ in terms of only $\\Lambda$ and \n $\\|f\\|_1$, where\n the notation \n \\[\n d_h (\\Lambda) := \\big| \\big\\{ x \\in \\mathbb R \\, : \\, |f(x)| > \\Lambda \\big\\} \\big|\n \\]\n is used to denote the \\textit{distributional function} of a function $h$. Using\n Chebyshev's inequality, the $L^2$ boundedness of $E$ found in Theorem \\ref{thm:Eweak}\n and equations \\eqref{eq:GoodPartBnd} and \\eqref{eq:C-Zdecomp}, we find\n \\begin{align}\\label{eq:dEgBound}\n d_{Eg}(\\Lambda)\n &\\leq \\frac{1}{\\Lambda^2} \\int_{\\mathbb R} E g(x)^2 \\, \\mathrm{d}x \\\\\n &\\leq \\frac{C}{\\Lambda^2} \\int_{\\mathbb R} g(x)^2 \\, \\mathrm{d}x \\nonumber\\\\\n &\\leq \\frac{C}{\\Lambda^2} \\int_{\\mathbb R} g(x) \\, \\mathrm{d}x \\nonumber\\\\\n &= \\frac{C}{\\Lambda^2} \\left( \\int_{\\Omega} g \n + \\int_{\\mathbb{R} \\backslash \\Omega} g \\right) \\nonumber\\\\\n &\\leq \\frac{C}{\\Lambda} \\| f \\|_1. \\nonumber\n \\end{align}\n\n\n On the other hand, for $Eb$, let $\\Omega^* = \\bigcup_j 2 I_j$. Using the \n Calder\\'on-Zygmund Covering Lemma and Chebyshev's inequality we find\n \\begin{align} \\label{eq:dEb}\n d_{Eb}(\\Lambda) \n \\leq |\\Omega^*| + |\\{ x \\notin \\Omega^* \\, : \\, |Eb(x)| > \\Lambda \\}|\n \\leq \\frac{2}{\\Lambda} \\|f\\|_1 + \\frac{1}{\\Lambda} \n \\int_{\\mathbb R \\backslash \\Omega^*} |Eb(x)| \\, \\mathrm{d}x.\n \\end{align}\n So, to show that $Eb$ is well defined, we need to bound the integral on the \n right hand side of the above inequality by $\\|f\\|_1$. To that end, note that\n if $x \\notin \\Omega^*$, then for each $j$, $x \\notin 2 I_j$ and\n \\begin{align*}\n E b_j(x) \n = \\frac{1}{2\\pi} \n \tp.v. \\int_{\\mathbb R} \n \t\t\\frac{e^{-\\pi|x - x'|}}{x - x'} b_j(x') \n \t\\, \\mathrm{d}x'\n = \\frac{1}{2\\pi} \n \t\\int_{I_j} \n \t\t\\frac{e^{-\\pi|x - x'|}}{x - x'} b_j(x') \n \t\\, \\mathrm{d}x \n < \\infty,\n \\end{align*}\n as $\\text{supp} \\, b_j \\subseteq I_j$. Since $E$ is a tempered distribution\n (Lemma \\ref{lma:FMultE}) and\n $f \\in \\mathscr S(\\mathbb R)$ means that $Ef \\in L^2(\\mathbb R)$ and, hence\n $\\sum_j E b_j$ converges to $Eb$ in the $L^2$ norm, it follows that\n \\[\n |Eb(x)| \\leq \\sum_j |Eb_j(x)| \\quad a.e.\n \\]\n As such, proving $Eb$ is well defined reduces to showing\n \\begin{align}\n \\int_{\\mathbb R \\backslash \\Omega^*} \\sum_{j} |E b_j(x)| \\, \\mathrm{d}x \n \\leq C \\| f \\|_1.\n \\end{align}\n\n If we let $c_j$ denote the center of $I_j$, then, for $x \\notin \\Omega^*$, \n since $b_j$ has zero average\n \\begin{align*}\n \\left|\\int_{I_j} \\frac{e^{-\\pi|x - x'|}}{x - x'} b_j(x') \\, \\mathrm{d}x'\\right|\n &= \\left|\\int_{I_j} e^{-\\pi|x - x'|}\n \\left(\\frac{b_j(x')}{x - x'} - \\frac{b_j(x')}{x - c_j} \\right)\\, \\mathrm{d}x'\\right| \\\\\n &\\leq \\int_{I_j} e^{-\\pi|x - x'|}\n \\left|\\frac{b_j(x')(x' - c_j)}{(x - x')(x-c_j)}\\right| \\, \\mathrm{d}x' \\\\\n &\\leq \\int_{I_j} |b_j(x')| \\frac{|I_j|}{(x - c_j)^2} \\, \\mathrm{d}x'\n \\end{align*}\n as $|x - x'| \\geq |x - c_j|/2$ and $|x' - c_j| \\leq |I_j|/2$. Moreover, \n \\begin{align} \\label{eq:IjIntBnd}\n \\int_{\\mathbb R \\backslash \\Omega^*} \\frac{|I_j|}{(x - c_j)^2} \\, \\mathrm{d}x'\n \\leq \\int_{\\mathbb R \\backslash I_j} \\frac{|I_j|}{(x - c_j)^2} \\, \\mathrm{d}x'\n \\leq 4,\n \\end{align}\n and so, by Fubini's Theorem,\n \\begin{align} \\label{eq:sumEbj}\n \\sum_j \\int_{\\mathbb R \\backslash \\Omega^*} \\left| E b_j(x) \\right| \\, \\mathrm{d}x\n &\\leq \\frac{1}{2\\pi} \\sum_j\\int_{\\mathbb R \\backslash \\Omega^*} \n \\left| \\int_{I_j} e^{-\\pi|x-x'|} \\frac{b_j(x')}{x - x'} \\, \\mathrm{d}x' \\right| \\, \\mathrm{d}x \n \\\\\n &\\leq \\frac{1}{2\\pi} \\sum_j \\int_{\\mathbb R \\backslash \\Omega^*} \\int_{I_j}\n |b_j(x')| \\frac{|I_j|}{(x - c_j)^2} \\, \\mathrm{d}x' \\, \\mathrm{d}x \n \\nonumber \\\\\n &\\leq \\frac{2}{\\pi} \\sum_j \\int_{I_j} |b_j(x')| \\, \\mathrm{d}x' \n \\nonumber \\\\\n &\\leq \\frac{4}{\\pi} \\| f\\|_1. \\nonumber\n \\end{align}\n Putting everything together, we find\n \\begin{align} \\label{eq:E11bnd}\n d_{Ef}(\\Lambda) \n \\leq d_{Eg}\\left(\\frac{\\Lambda}{2}\\right) + d_{Eb}\\left(\\frac{\\Lambda}{2}\\right)\n \\leq \\frac{C}{\\Lambda} \\| f\\|_1,\n \\end{align}\n where $C>0$ is independent of $\\Lambda$ and $f$. Since $f$ is Schwartz class, we can \n therefore extend the inequality \\eqref{eq:E11bnd} to $L^1$ \\textit{via} density to\n conclude that $E$ is weak (1,1).\n\\end{proof}\n\n\n\\begin{thm}\\label{thm:EbndOp}\n\tThe operator $E$ is strong type $(p, p)$ for $p>1$.\n\\end{thm}\n\\begin{proof}\n\tTheorems \\ref{thm:Estrong} and \\ref{thm:Eweak} in conjunction with the \n\tMarcinkiewicz Interpolation Theorem \\cite[Theorem 1.3.2]{Grafakos} immediately imply that $E$ is strong\n\ttype $(p,p)$ for $p\\in (1, 2]$. As such, it remains only to show that\n\t$E$ is strong type $(p, p)$ for $p > 2$. \n\n Let $E'$ denote the adjoint of $E$. By density, we need only consider \n $f \\in \\mathscr S(\\mathbb R)$. Fix $p >2$ and let $q$ \n denote its H\\\"older Conjugate. Note that the map\n \\[\n g \\mapsto \\int_{\\mathbb R} E f \\bar{g} =: \\left\\langle Ef, g \\right\\rangle \n \\]\n is a linear functional on $L^q$ with norm $\\| E f \\|_p$. As such, \n we see by H\\\"older's inequality that\n \\begin{align*}\n \\| E f \\|_p\n &= \\sup_{\\|g\\|_q = 1} |\\langle Ef, g \\rangle| \\\\\n &= \\sup_{\\|g\\|_q = 1} |\\langle f, E'g \\rangle| \\\\\n &\\leq \\|f\\|_p \\|E\\|_q \\\\\n &\\leq C \\|f\\|_p,\n \\end{align*}\n as $p>2$ implies $1 < q < 2$, which means $E$ is strong type $(q, q)$, \n and Theorem 5 from Chapter VII.3 of \\cite{yosida} implies \n $\\|E\\|_q = \\|E'\\|_q$.\n\\end{proof}\n\n\\end{document}", "meta": {"hexsha": "a9412e6f7fbbedd29dd0511707594e350b3fa45c", "size": 11462, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chapter3-xContin/3.4-BndnessETransform.tex", "max_stars_repo_name": "ADGC/ilw-dsm-dissertation", "max_stars_repo_head_hexsha": "de0f27b6389ee55c24d155ff482743acbe6a35a1", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Chapter3-xContin/3.4-BndnessETransform.tex", "max_issues_repo_name": "ADGC/ilw-dsm-dissertation", "max_issues_repo_head_hexsha": "de0f27b6389ee55c24d155ff482743acbe6a35a1", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Chapter3-xContin/3.4-BndnessETransform.tex", "max_forks_repo_name": "ADGC/ilw-dsm-dissertation", "max_forks_repo_head_hexsha": "de0f27b6389ee55c24d155ff482743acbe6a35a1", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.5289855072, "max_line_length": 111, "alphanum_fraction": 0.5471121968, "num_tokens": 4404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.8056321796478255, "lm_q1q2_score": 0.6262177813981187}} {"text": "%---------------------------Shear and Size-----------------------------\n\\section{Shear and Size\\label{s:quad-shear-and-size}}\n\nLet $R$ be the relative size squared as defined in \\S\\ref{s:quad-rel-size-squared}\nand $H$ be the shear as defined in \\S\\ref{s:quad-shear}.\nThe shear and size metric is the product of these two numbers:\n\\[\nq = RH\n\\]\n\n\\quadmetrictable{shear and size}%\n{$1$}% Dimension\n{$[0.2,1]$}% Acceptable range\n{$[0,1]$}% Normal range\n{$[0,1]$}% Full range\n{Dependent on $\\overline{A}$}% Unit square\n{\\cite{knu:03}}% Citation\n{v\\_quad\\_shear\\_and\\_size}% Verdict function name\n\n", "meta": {"hexsha": "fb53233a9085241cbb73bd773384b89642ed9718", "size": 781, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadShearAndSize.tex", "max_stars_repo_name": "Lin1225/vtk_v5.10.0", "max_stars_repo_head_hexsha": "b54ac74f4716572862365fbff28cd0ecb8d08c3d", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-06-01T00:15:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T10:49:02.000Z", "max_issues_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadShearAndSize.tex", "max_issues_repo_name": "Armand0s/homemade_vtk", "max_issues_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2022-02-17T11:40:17.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-02T21:23:25.000Z", "max_forks_repo_path": "Utilities/verdict/docs/VerdictUserManual2007/QuadShearAndSize.tex", "max_forks_repo_name": "Armand0s/homemade_vtk", "max_forks_repo_head_hexsha": "6bc7b595a4a7f86e8fa969d067360450fa4e0a6a", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2015-03-23T21:13:19.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-03T11:15:39.000Z", "avg_line_length": 39.05, "max_line_length": 82, "alphanum_fraction": 0.4775928297, "num_tokens": 191, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8872045996818986, "lm_q2_score": 0.7057850278370112, "lm_q1q2_score": 0.6261757230836132}} {"text": "\\sec{Cauchy Euler equations}\n\\subsection{The first challenge} \\label{ssec:cauchyeulerhomo}\n\\begin{mdframed}[style=boxstyle, frametitle={The Setup}]\n\tAn ODE of the form\n\t\\begin{equation*} \n\t\tx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + \\cdots + a_{1}xy' + a_0y = 0\n\t\\end{equation*}\n\tis called a \\defin{Cauchy Euler equation}. (Each $a_i$ is a constant.)\n\\end{mdframed}\nWe shall be interested in solving the above equation only on $(0, \\infty).$\n\\begin{mdframed}[style=boxstyle, frametitle={The Solution}]\n\tWe plug the trial solution $y = x^m.$ This gives us the polynomial equation:\n\t\\begin{align} \n\t\tm(m - 1)\\cdots(m-(n-1)) + a_{n-1}m(m-1)\\cdots(m-(n-2)) + \\nonumber \\\\\n\t\t \\cdots + a_1m + a_0 = 0. \\label{eq:indic}\n\t\\end{align}\n\tNote that if $m$ is a root of the above, then $x^m$ is a solution of the ODE.\\\\\n\tThus, if the above equation has $n$ distinct roots $m_1, \\ldots, m_n,$ then the original ODE has $n$ linearly independent solutions $x^{m_1}, \\ldots, x^{m_n}$ and we are done.\n\\end{mdframed}\n\\begin{mdframed}[style=boxstyle, frametitle={Repeated roots}]\n\tSuppose $m_0$ is a repeated root of (\\ref{eq:indic}). Suppose that it is repeated $k$ times. Then, one can show that the following $k$ functions are solutions of the original ODE:\n\t\\begin{equation*} \n\t\tx^{m_0}, x^{m_0}\\ln x, \\ldots, x^{m_0}(\\ln x)^{k-1}.\n\t\\end{equation*}\n\tThese are linearly independent as well.\\\\\n\tThus, we are now done for any case as the total number of roots is always going to be $n$ when counted with multiplicity. (This is assuming that we work in $\\mathbb{C},$ which we shall do.)\n\\end{mdframed}\n\\begin{mdframed}[style=boxstyle, frametitle={Getting real}]\n\tSuppose $m = a+ib$ is a root. Then, we have\n\t\\begin{align*} \n\t\tx^m &:= \\exp(m \\ln x) & (\\text{By definition})\\\\\n\t\t&= \\exp(a \\ln x + ib\\ln x)\\\\\n\t\t&= \\exp(a \\ln x)\\cdot\\exp(ib \\ln x)\\\\\n\t\t&= \\exp(a \\ln x)\\cdot\\big(\\cos(b\\ln x) + i\\sin(b \\ln x)\\big)\\\\\n\t\t&=x^a(\\cos(b\\ln x) + i\\sin(b \\ln x))\n\t\\end{align*}\n\tAs before, if each $a_i$ is real, then the roots appear in the same multiplicity as their conjugates which will give us the real pair of solutions as:\n\t\\begin{equation*} \n\t\tx^a\\cos(b\\ln x), x^a\\sin(b\\ln x).\n\t\\end{equation*}\n\tIn case of repetition twice, we get the four solutions:\n\t\\begin{equation*} \n\t\tx^a\\cos(b\\ln x), x^a\\sin(b\\ln x),\\quad x^a(\\ln x)\\cos(b\\ln x), x^a(\\ln x)\\sin(b\\ln x).\n\t\\end{equation*}\n\tThe general case is (hopefully) clear.\n\\end{mdframed}\n\\subsection{Welcome back, annihilators}\nThis time, we consider annihilators of polynomial (polylogmial?) functions.\\\\\nAs before, we shall see that it suffices to consider the case of just the monomials first. The annihilator table is particularly simple this time:\n\\begin{mdframed}[style=boxstyle, frametitle={Annihilators of special functions}]\n\t\\begin{center}\n\t\\bgroup\n\t\\def\\arraystretch{1.25}\n\t\\begin{tabular}{|l|l|}\n\t\t\\hline\n\t\tFunction & Annihilator\\\\\n\t\t\\hline\n\t\t$x^n$ & $xD - n$\\\\\n\t\t$x^n(\\ln x)^k$ & $(xD - n)^{k+1}$\\\\\n\t\t\\hline\n\t\\end{tabular}\n\t\\egroup\n\t\\end{center}\n\\end{mdframed}\n\\begin{mdframed}[style=boxstyle, frametitle={A word on $xD$}]\n\tNote that $xD$ is the operator which acts on a function as:\n\t\\begin{equation*} \n\t\t(xD)(f) = xf'.\n\t\\end{equation*}\n\tOn the other hand, $Dx$ is an operator which acts as:\n\t\\begin{equation*} \n\t\t(Dx)(f) = D(xf) = xf' + f.\n\t\\end{equation*}\n\tIn particular, $xD \\neq Dx.$ This means that $(xD)^2 \\neq x^2D^2$ and so on.\n\\end{mdframed}\n\\emph{Note:} $xD$ is an operator but there is no $xP$ operator. \\hfill (A joke.)\n\\begin{mdframed}[style=boxstyle, frametitle={Some arithmetic}]\n\tFor ease of calculations, one may note the following useful identities:\n\t\\begin{align*} \n\t\tx^2D^2 &= xD(xD - 1)\\\\\n\t\tx^3D^3 &= xD(xD - 1)(xD - 2)\\\\\n\t\t&\\vdots\n\t\\end{align*}\n\\end{mdframed}\n\\subsection{The main problem}\n\\begin{mdframed}[style=boxstyle, frametitle={The Setup}]\n\tWe consider a linear ODE of the form\n\t\\begin{equation} \n\t\tx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + \\cdots + a_{1}xy' + a_0y = Q(x),\n\t\\end{equation}\n\twhere each $a_i$ is a constant and $Q(x)$ is a linear combination of functions of the form $x^k(\\ln x)^m$.\n\\end{mdframed}\n\\begin{mdframed}[style=boxstyle, frametitle={The Solution}]\n\tThe steps are now identical as the case of \\nameref{sec:undcoeff}.\\\\\n\tWe shall assume that $Q(x) = x^k(\\ln x)^m$ and not a linear combination. The general case follows as before by considering the linear combination of solutions.\n\t\\begin{enumerate}[leftmargin=*, label = \\Roman*.]\n\t\t\\item We first consider the equation\n\t\t\\begin{equation*} \n\t\t\tx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + \\cdots + a_{1}xy' + a_0y = 0.\n\t\t\\end{equation*}\n\t\tThis can be solved by the method we saw in \\S\\S\\ref{ssec:cauchyeulerhomo}.\n\t\t\\item We then apply the annihilator of $Q(x)$ from the table earlier.\\\\\n\t\t\\textbf{\\emph{This again gives us a Cauchy Euler equation.}}\\\\\n\t\tWe solve this again using the method of \\S\\S\\ref{ssec:cauchyeulerhomo}.\n\t\t\\item We follow the step III as in the case of \\nameref{sec:undcoeff} to determine the coefficients of the new functions we obtain in Step II.\n\t\\end{enumerate}\n\\end{mdframed}\nNote that $D^{11}$ is also an annihilator of $x^{10},$ however we do not use that.\\\\\nThe first reason is that the calculations would be a nightmare.\\\\\nSecondly, applying this annihilator to both sides of the equation wouldn't technically give us a Cauchy Euler equation again. \\\\\nAlso, note that the method we have used doesn't require $k$ to be an integer. This would work even if we wish to solve something like $xy'' + y = x^{1/2}.$ Another reason to prefer the annihilator $xD - 1/2.$", "meta": {"hexsha": "49e497ae6a66d6b29f2c8e469045233f8c11e357", "size": 5516, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tuts/ma-108/summary/cauchyeuler.tex", "max_stars_repo_name": "siddhanttripathy/siddhanttripathy.github.io", "max_stars_repo_head_hexsha": "aa22ff5a198e9ede5d07001384fda67a4606f8ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2020-12-17T20:34:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-10T09:58:57.000Z", "max_issues_repo_path": "tuts/ma-108/summary/cauchyeuler.tex", "max_issues_repo_name": "siddhanttripathy/siddhanttripathy.github.io", "max_issues_repo_head_hexsha": "aa22ff5a198e9ede5d07001384fda67a4606f8ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2020-03-04T10:23:16.000Z", "max_issues_repo_issues_event_max_datetime": "2020-05-28T10:29:01.000Z", "max_forks_repo_path": "tuts/ma-108/summary/cauchyeuler.tex", "max_forks_repo_name": "siddhanttripathy/siddhanttripathy.github.io", "max_forks_repo_head_hexsha": "aa22ff5a198e9ede5d07001384fda67a4606f8ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 26, "max_forks_repo_forks_event_min_datetime": "2019-08-30T18:19:52.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-17T12:09:09.000Z", "avg_line_length": 49.6936936937, "max_line_length": 208, "alphanum_fraction": 0.6829224075, "num_tokens": 1987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.705785040214066, "lm_q2_score": 0.8872045832787205, "lm_q1q2_score": 0.6261757224874754}} {"text": "\\section{Deep Learning} \\label{sec:bt/DNN}\n\nDeep learning is a powerful extension of the simple learning algorithms detailed above. In theory, any problem which can be modeled as the mapping of an input vector to an output vector can be solved by deep learning \\cite{goodfellow2016}. In other words, any function of any complexity can be approximated given sufficiently large models and sufficiently large labeled training examples. \n\n\\subsection{Deep Feedforward Neural Networks}\n\nFeedforward neural networks constitute a branch of deep learning covering models with a one-directional data flow from input to output. They are called \\textit{neural networks} because they are loosely inspired by the neural circuitry of the human brain. Extending this analogy, all human perception may be considered one vast deep feedforward neural network. It works by mapping sensory input from sight, touch, smell, taste, and hearing through countless layers of neurons to output what we perceive of the world. Artificial neural networks work similarly by mapping a vector of input data points to one or more output values.\n\nEven in their simplest form, deep feedforward neural networks overcome many challenges that linear models face. Linear regression, for instance, may never accurately approximate a non-linear function or capture interactions between two or more input variables. This limitation may only be overcome by introducing a nonlinearity to the input $\\bm{x}$. We encapsulate this nonlinearity as $\\phi(\\bm{x})$. Thus, the equivalent to equation \\ref{eq:bt_linReg} becomes equation \\ref{eq:bt/dnn/dfnn}.\n\n\\begin{equation}\n \\label{eq:bt/dnn/dfnn}\n y=f(\\bm{x};\\bm{\\theta},\\bm{x})=\\phi(\\bm{x};\\bm{\\theta})^T\\bm{w}\n\\end{equation}\n\nWe now have another set of trainable parameters $\\bm{\\theta}$, meaning that the final form of $\\phi$ is yet to be determined. By allowing the model to experience a given set of labeled training data, $\\bm{\\theta}$ will gradually converge towards a value that enables the full model to approximate the data optimally.\n% be trained from experiencing a given set of labeled training data. \nThe parameters defined by $\\bm{w}$ may subsequently map from $\\phi(\\bm{x})$ to the desired output. $\\phi$ is called a \\textit{hidden layer} because its outputs are unrelated to either the input or output vectors. It can instead be thought of as a transformation of the input layer. \n% We can think of $\\phi$ as a transformation of the input layer. \nIts output is hence a new representation or interpretation of the information provided by $\\bm{x}$.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.5\\textwidth]{figures/bt_dfnn.png}\n \\caption{Example of a simple \\acrshort{mlp}. Left representation depicts individual neurons of each layer and the edges between them. The right representation is more common, depicting each layer with vector notation. }\n \\source{\\cite{goodfellow2016}}\n \\label{fig:bt_dfnn}\n\\end{figure}\n\n\\subsubsection{\\acrlong{mlp}}\n\nA \\acrfull{mlp} is one form of feedforward neural network, the simplest implementation of which is illustrated in figure \\ref{fig:bt_dfnn}. Here we have two input neurons, one hidden layer with two neurons, and one output layer with one neuron. With vector notation, we say that the hidden layer $\\bm{h}$ is a function of the input layer $\\bm{x}$, with the relation $\\bm{h}=f^{(1)}(\\bm{x};\\bm{W}, \\bm{c})$. $\\bm{h}$ is then the input to a third layer (which in this case is also the output layer) with the relation $y=f^{(2)}(\\bm{h};\\bm{W}, \\bm{c})$. We can keep adding depth and width to the model this way, further increasing its complexity and potential hypothesis space. The entire model may now be given by equation \\ref{eq:bt/dnn/mlp1}.\n\n\\begin{equation}\n \\label{eq:bt/dnn/mlp1}\n f(\\bm{x};\\bm{W},\\bm{c},\\bm{w},b)=f^{(2)}(f^{(1)}(\\bm{x}))\n\\end{equation}\n\nTo provide the nonlinearity we want, we define $\\bm{h}=g(\\bm{W}^T\\bm{x}+\\bm{c})$. This functon is commonly known as the \\textit{activation function} of the layer. For deeper networks, it would be added to the output of every hidden layer. There are many activation functions to choose from, but the default recommendation in the machine learning community is to use the \\acrfull{relu} \\cite{agarap2018}. It is defined as $g(z)=max\\{0,z\\}$ and has been made popular for its simplicity without trading off on optimization with gradient-based backpropagation. The complete network of figure \\ref{fig:bt_dfnn} can thus be given by equation \\ref{eq:bt/dnn/mlp3}.\n\n\\begin{equation}\n \\label{eq:bt/dnn/mlp2}\n y=f(\\bm{x};\\bm{W},\\bm{c}, \\bm{w}, b)=\\bm{w}^Tg(\\bm{W}^T\\bm{x}+\\bm{c}) + b\n\\end{equation}\n\n\\begin{equation}\n \\label{eq:bt/dnn/mlp3}\n y=f(\\bm{x};\\bm{W},\\bm{c}, \\bm{w}, b)=\\bm{w}^Tmax\\{\\bm{W}^T\\bm{x}+\\bm{c}\\} + b\n\\end{equation}\n\n\\newpage\n\\subsection{\\acrlong{cnn}s} \\label{sec:bt/DNN/CNN}\n\nA \\acrfull{cnn} is a specialized kind of deep feedforward neural network that has shown remarkable results in computer vision and language processing. When first proposed by \\textcite{lecun1989} in 1989, it revolutionized machine learning problems where data had a grid-structured topology. Such data could be interpreted with more insight while allowing for much deeper and scaleable model architectures. Before its inception, most practitioners had little faith in the advent of neural networks for machine learning. The success of \\acrshort{cnn}s triggered a wave of interest, which laid the foundation for further research in neural networks and deep learning \\cite{goodfellow2016}.\n\n\\subsubsection{Convolution}\n\nTo understand \\acrshort{cnn}s, we need to understand what a \\textit{convolution} is. Used in many engineering disciplines, it is simply a mathematical operator which expresses how two functions overlap \\cite{weisstein2003}. In signal processing, for instance, convolution may be used to mathematically replicate how a given sound signal would behave in any conceivable environment. All one would need is the impulse response of this environment, which could be the recording of a clap or similar fast transient sound. Convolving these two signals would have the effect of \"placing\" the sound in this environment, artificially reproducing all reverberations. The operation is typically denoted with an asterisk and is defined as equations \\ref{eq:bt/dnn/conv1}-\\ref{eq:bt/dnn/conv2}. This particular version is known as discrete convolution, which is what we will be encountering for neural networks.\n\n\\begin{equation}\n \\label{eq:bt/dnn/conv1}\n s(t)=(x*k)(t)\n\\end{equation}\n\n\\begin{equation}\n \\label{eq:bt/dnn/conv2}\n s(t)=\\sum_{a=-\\infty}^{\\infty}x(a)k(t-a)\n\\end{equation}\n\nWith \\acrshort{cnn} terminology, $x$ is the input to a convolutional layer and $k$ is its \\textit{kernel}. Since the input to a \\acrshort{cnn} is often an image, time series, or other signals, $x$ and $k$ are usually multidimensional. Taking image classification as an example, the convolutional operations performed on all input pixels would look like equation \\ref{eq:bt/dnn/conv3}.\n\n\\begin{equation}\n \\label{eq:bt/dnn/conv3}\n S(i,j)=(K*X)(i,j)=\\sum_{m}^{}\\sum_{n}^{}X(i-m,j-n)K(m,n)\n\\end{equation}\n\nEach pixel is denoted by $i$ and $j$ for its spatial coordinate in the input image. The output from convolving a kernel $K$ with an image $X$ is called a \\textit{feature map} because it is essentially a mapping to a new \"image\" where features represented by the kernel are highlighted. A visual representation of how a 2-dimensional convolutional operation may be applied to an input is displayed in figure \\ref{fig:bt_conv}.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.8\\textwidth]{figures/bt_conv.png}\n \\caption{Example of 2-dimensional convolution operation. If this were image classification, values in the input matrix would represent individual pixel activations in the image. The final value of each output cell is calculated by the convolution operation given by equation \\ref{eq:bt/dnn/conv3}.}\n \\source{\\cite{goodfellow2016}}\n \\label{fig:bt_conv}\n\\end{figure}\n\nThe kernel $K$ is what defines all properties of a convolutional model. Instead of learning individual parameters for every edge between every neuron, as is the case for \\acrshort{mlp}s, \\acrshort{cnn}s learn the internal weights of all kernels in the network. As we will see, \\acrshort{cnn}s usually contain several kernels for every convolutional layer, each capturing one feature in the input to each layer.\n\n\\subsubsection{Properties of \\acrshort{cnn}s}\n\nWhat makes \\acrshort{cnn}s so efficient is based on three essential ideas. First, since parameters are only present in each kernel, as described above, the network does not need a single parameter for all neurons in every layer. This idea is known as \\textit{parameter sharing} and will reduce the storage requirements of large models. Another consequence of the convolutional layers is that each input neuron is only connected to a few outputs. As can be seen in figure \\ref{fig:bt_conv}, only four neurons of the input are required to calculate the value of a cell in the output. This idea is called \\textit{sparse interactions}. Contrary to an \\acrshort{mlp}, which requires connections from every input to every output, this feature drastically improves both computing efficiency and memory requirements during training.\n\nFinally, a concept known as \\textit{equivariant representations} allows for the detection of features that are independent of spatial (or temporal) position in the input. In other words, if there is an object in the input to a convolutional layer that causes high activations in a feature map, the same object will cause an equally high activation in the same feature map if it were in another position. This effect is another consequence of the way in which feature maps are created by \"moving\" the same kernel across the entire input. \\acrshort{mlp}s, in contrast, lack this property since each neuron is associated with an individual learned weight. \n% The kernel will eventually reach and generate the convolutional output for that respective feature regardless of its po\n\n\n% \\subsection{Terms} \\label{sec:bt/DNN/terms}\n\n% Much machine learning terminology is considered common knowledge in most literature and is often neglected to be explained in detail. Some such terms, many of which will be used frequently during the course of this thesis, are detailed below.\n\n% \\subsubsection{Batch Normalization}\n% \\subsubsection{Dropout}\n% % \\subsubsection{Perceptron}\n% % \\subsubsection{\\acrlong{relu}}\n% \\subsubsection{Max Pooling}\n% \\subsubsection{Softmax}\n% \\subsubsection{Cross Entropy Loss}\n% \\subsubsection{Precision, Accuracy and Recall}", "meta": {"hexsha": "5830adabfc471504d1d76a86d9c84d0b6766c290", "size": 10699, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/background/DNN.tex", "max_stars_repo_name": "JLysberg/thesis-NTNU", "max_stars_repo_head_hexsha": "c0a9631f89a0112b2ade27d05c22818745706fb8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/background/DNN.tex", "max_issues_repo_name": "JLysberg/thesis-NTNU", "max_issues_repo_head_hexsha": "c0a9631f89a0112b2ade27d05c22818745706fb8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/background/DNN.tex", "max_forks_repo_name": "JLysberg/thesis-NTNU", "max_forks_repo_head_hexsha": "c0a9631f89a0112b2ade27d05c22818745706fb8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 99.0648148148, "max_line_length": 899, "alphanum_fraction": 0.7705393027, "num_tokens": 2675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8872045847699186, "lm_q2_score": 0.7057850216484838, "lm_q1q2_score": 0.626175707068471}} {"text": "\\lab{Kalman Filter}{Kalman Filter}\n\\objective{Understand how to implement the standard Kalman Filter. Apply to the problem of projectile tracking.}\n\nMeasured observations are often prone to significant noise, due to restrictions on measurement accuracy.\nFor example, most commercial GPS devices can provide a good estimate of geolocation, but only within a dozen meters or so.\nA Kalman filter is an algorithm that takes a sequence of noisy observations made over time and attempts to get rid of the noise,\nproducing more accurate estimates than the original observations.\nTo do this, the algorithm needs information about the system being observed.\n\nConsider the problem of tracking a projectile as it travels through the air.\nShort-range projectiles approximately trace out parabolas, but a sensor that is recording measurements of the projectile's position over time will likely show a path that is much\nless smooth.\nBecause we know something about the laws of physics, we can filter out the noise in the measurements using basic Newtonian mechanics, recovering a more accurate estimate\nof the projectile's trajectory.\nIn this lab, we will simulate measurements of a projectile and implement a Kalman filter to estimate the complete trajectory of the projectile.\n\n\\section*{Linear Dynamical Systems}\nThe standard Kalman filter assumes that:\n(1) we have a linear dynamical system,\n(2) the state of the system evolves over time with some noise, and\n(3) we receive noisy measurements about the state of the system at each iteration.\nMore formally, letting $\\mathbf{x}_{k}$ denote the state of the system at time $k$, we have\n\\begin{equation}\n\\mathbf{x}_{k+1} = F_{k} \\mathbf{x}_{k} + G_{k}\\mathbf{u}_{k} + \\mathbf{w}_{k}\n\\label{eq:state}\n\\end{equation}\nwhere $F_{k}$ is a state-transition model, $G_{k}$ is a control-input model, $\\mathbf{u}_{k}$ is a control vector, and $\\mathbf{w}_{k}$ is the noise present in state $k$.\nThis noise is assumed to be drawn from a multivariate Gaussian distribution with zero mean and covariance matrix $Q_{k}$.\nThe control-input model and control vector allow the assumption that the state can be additionally influenced by some other factor than the linear state-transition model.\n\nWe further assume that the states are ``hidden,'' and we only get the noisy observations\n\\begin{equation}\n\\mathbf{z}_{k} = H_{k}\\mathbf{x}_{k} + \\mathbf{v}_{k}\n\\label{eq:obs}\n\\end{equation}\nwhere $H_{k}$ is the observation model mapping the state space to the observation space, and $\\mathbf{v}_{k}$ is the observation noise present at iteration $k$.\nAs with the aforementioned error, we assume that this noise is drawn from a multivariate Gaussian distribution with zero mean and covariance matrix $R_{k}$.\n\nThe dynamics stated above are all taken to be linear.\nThus, for our purposes, the operators $F_k$, $G_k$, and $H_k$ are all matrices, and $\\mathbf{x}_k$, $\\mathbf{u}_k$, $\\mathbf{z}_k$, and $\\mathbf{v}_k$ are all vectors.\n\nWe will assume that the transition and observation models, the control vector, and the noise covariances are constant, i.e. for each $k$, we will replace $F_{k}, H_k, \\mathbf{u}_{k}, Q_{k},$ and $R_{k}$ with $F, H, \\mathbf{u}, Q,$ and $R$.\nWe will also assume that $G = I$ is the identity matrix, so it can safely be ignored.\n\n\\begin{problem}\nBegin implementing a \\li{KalmanFilter} class by writing an initialization method that stores the transition and observation models, noise covariances, and control vector.\nWe provide an interface below:\n\\begin{lstlisting}\nclass KalmanFilter(object):\n def __init__(self,F,Q,H,R,u):\n \"\"\"\n Initialize the dynamical system models.\n\n Parameters\n ----------\n F : ndarray of shape (n,n)\n The state transition model.\n Q : ndarray of shape (n,n)\n The covariance matrix for the state noise.\n H : ndarray of shape (m,n)\n The observation model.\n R : ndarray of shape (m,m)\n The covariance matric for observation noise.\n u : ndarray of shape (n,)\n The control vector.\n \"\"\"\n pass\n\\end{lstlisting}\n\\end{problem}\n\nWe now derive the linear dynamical system parameters for a projectile traveling through $\\mathbb{R}^2$ undergoing a constant downward gravitational force of 9.8 $m/s^2$.\nThe relevant information needed to describe how the projectile moves through space is its position and velocity.\nThus, our state vector has the form\n\\[\n\\mathbf{x} = \\left( \\begin{array}{c} s_{x} \\\\ s_{y} \\\\ V_{x} \\\\ V_{y} \\end{array} \\right),\n\\]\nwhere $s_x$ and $s_y$ give the $x$ and $y$ coordinates of the position (in meters), and $V_x$ and $V_y$ give the horizontal and vertical components of the velocity (in meters per second), respectively.\n\nHow does the system evolve from one time step to the next?\nAssuming each time step is $0.1$ seconds, it is easy enough to calculate the new position:\n\\begin{align*}\ns_x' &= s_x + 0.1V_x\\\\\ns_y' &= s_y + 0.1V_y.\n\\end{align*}\nFurther, since the only force acting on the projectile is gravity (we are ignoring things like wind resistance), the horizontal velocity remains constant:\n\\[\nV_x' = V_x.\n\\]\nThe vertical velocity, however, does change due to the effects of gravity.\nFrom basic Newtonian mechanics, we have\n\\[\nV_y' = V_y - 0.1\\cdot9.8.\n\\]\nIn summary, over one time step, the state evolves from $\\mathbf{x}$ to $\\mathbf{x}'$, where\n\\[\n\\mathbf{x}' = \\left( \\begin{array}{c} s_{x} + 0.1V_x \\\\ s_{y} + 0.1V_y \\\\ V_{x} \\\\ V_{y} - 0.98 \\end{array} \\right).\n\\]\nFrom this equation, you can extract the state transition model $F$ and the control vector $u$.\n\nWe now turn our attention to the observation model.\nImagine that a radar sensor captures (noisy) measurements of the projectile's position as it travels through the air.\nAt each time step, the radar transmits the observation $z = (z_x, z_y)$ given by\n\\begin{align*}\nz_x &= s_x + v_x\\\\\nz_y &= s_y + v_y,\n\\end{align*}\nwhere $(v_x, v_y)$ is a noise vector assumed to be drawn from a multivariate Gaussian with mean zero and some known covariance.\nThese equations indicate the appropriate choice of observation model.\n\n\\begin{problem}\nWork out the transition and observation models $F$ and $H$, along with the control vector $\\mathbf{u}$, corresponding to the projectile.\nAssume that the noise covariances are given by\n\\begin{align*}\nQ &= 0.1 \\cdot I_4\\\\\nR &= 5000 \\cdot I_2.\n\\end{align*}\nInstantiate a \\li{KalmanFilter} object with these values.\n\\end{problem}\n\n\\begin{figure}\n \\centering\n \\begin{tabular}{cc}\n \\includegraphics[width=.45\\textwidth]{states_evolution} &\n \\includegraphics[width=.45\\textwidth]{obs_evolution}\n \\end{tabular}\n \\caption{State sequence (left) and sampling of observation sequence (right).}\n \\label{fig:evolution}\n\\end{figure}\n\nWe now wish to simulate a sequence of states and observations from the dynamical system.\nIn addition to the system parameters, we need an initial state $\\mathbf{x}_0$ to get started.\nComputing the subsequent states and observations is simply a matter of following equations \\ref{eq:state} and \\ref{eq:obs}.\n\n\\begin{problem}\nAdd a method to your \\li{KalmanFilter} class to generate a state and observation sequence by evolving the system from a given initial state (the function \\li{numpy.random.multivariate_normal} will be useful).\nTo do this, implement the following:\n\\begin{lstlisting}\ndef evolve(self,x0,N):\n \"\"\"\n Compute the first N states and observations generated by the Kalman system.\n\n Parameters\n ----------\n x0 : ndarray of shape (n,)\n The initial state.\n N : integer\n The number of time steps to evolve.\n\n Returns\n -------\n states : ndarray of shape (n,N)\n States 0 through N-1, given by each column.\n obs : ndarray of shape (m,N)\n Observations 0 through N-1, given by each column.\n \"\"\"\n pass\n\\end{lstlisting}\n\nSimulate the true and observed trajectory of a projectile with initial state\n\\[\n\\mathbf{x}_0 = \\left( \\begin{array}{c} 0\\\\ 0 \\\\ 300 \\\\ 600\\end{array} \\right).\n\\]\nApproximately 1250 time steps should be sufficient for the projectile to hit the ground (i.e. for the $y$ coordinate to return to 0).\nYour results should qualitatively match those given in Figure \\ref{fig:evolution}.\n\\label{prob:simulation}\n\\end{problem}\n\n\n\n\\section*{State Estimation with the Kalman Filter}\nThe Kalman filter is a recursive estimator that smooths out the noise in real time, estimating each current state based on the past state estimate and the current measurement.\nThis process is done by repeatedly invoking two steps: Predict and Update.\nThe predict step is used to estimate the current state based on the previous state.\nThe update step then combines this prediction with the current observation, yielding a more robust estimate of the current state.\n\nTo describe these steps in detail, we need additional notation. Let\n\\begin{itemize}\n \\item $\\widehat{\\mathbf{x}}_{n|m}$ be the state estimate at time $n$ given only measurements up through time $m$; and\n \\item $P_{n|m}$ be an error covariance matrix, measuring the estimated accuracy of the state at time $n$ given only measurements up through time $m$.\n\\end{itemize}\n\nThe elements $\\widehat{\\mathbf{x}}_{k|k}$ and $P_{k|k}$ represent the state of the filter at time $k$, giving the state estimate and the accuracy of the estimate.\n\n%\\begin{figure}\n% \\centering\n% \\includegraphics[width=.90\\textwidth]{norms}\n% \\caption{Norms of the estimated accuracy matrices as the Kalman filter progresses. }\n% \\label{fig:norms}\n%\\end{figure}\n\nWe evolve the filter recursively, as follows:\n\\begin{align*}\n\\textbf{Predict} & & \\widehat{\\mathbf{x}}_{k|k-1} & = F\\widehat{\\mathbf{x}}_{k-1|k-1} + \\mathbf{u} \\\\\n & & P_{k|k-1} & = FP_{k-1|k-1}F^{T} + Q \\\\\n\\textbf{Update} & & \\tilde{\\mathbf{y}}_{k} & = \\mathbf{z}_{k} - H\\widehat{\\mathbf{x}}_{k|k-1} \\\\\n & & S_{k} & = HP_{k|k-1}H^{T} + R \\\\\n & & K_{k} & = P_{k|k-1}H^{T}S_{k}^{-1} \\\\\n & & \\widehat{\\mathbf{x}}_{k|k} & = \\widehat{\\mathbf{x}}_{k|k-1} + K_{k}\\tilde{\\mathbf{y}}_{k} \\\\\n & & P_{k|k} & = (I - K_{k}H)P_{k|k-1}\n\\end{align*}\n\nThe more observations we have, the greater the accuracy of these estimates becomes (i.e the norm of the accuracy matrix converges to $0$).\n%This is reflected in Figure \\ref{fig:norms}.\n\n\\begin{problem}\nAdd code to your \\li{KalmanFilter} class to estimate a state sequence corresponding to a given observation sequence and initial state estimate.\nImplement the following class method:\n\\begin{lstlisting}\ndef estimate(self,x,P,z):\n \"\"\"\n Compute the state estimates using the Kalman filter.\n If x and P correspond to time step k, then z is a sequence of\n observations starting at time step k+1.\n\n Parameters\n ----------\n x : ndarray of shape (n,)\n The initial state estimate.\n P : ndarray of shape (n,n)\n The initial error covariance matrix.\n z : ndarray of shape(m,N)\n Sequence of N observations (each column is an observation).\n\n Returns\n -------\n out : ndarray of shape (n,N)\n Sequence of state estimates (each column is an estimate).\n \"\"\"\n pass\n\\end{lstlisting}\n\\end{problem}\n\n\\begin{figure}\n \\centering\n \\begin{tabular}{cc}\n \\includegraphics[width=.45\\textwidth]{estimate_macro} &\n \\includegraphics[width=.45\\textwidth]{estimate_micro}\n \\end{tabular}\n \\caption{State estimates together with observations and true state sequence (detailed view on the right).}\n \\label{fig:estimates}\n\\end{figure}\n\nReturning to the projectile example, we now assume that our radar sensor has taken observations from time steps $200$ through $800$ (take the corresponding slice of the\nobservations produced in Problem \\ref{prob:simulation}).\nUsing these observations, we seek to estimate the corresponding true states of the projectile.\nWe must first come up with a state estimate $\\widehat{\\mathbf{x}}_{200}$ for time step $200$, and then feed this into the Kalman filter\nto obtain estimates $\\widehat{\\mathbf{x}}_{201}, \\ldots, \\widehat{\\mathbf{x}}_{800}$.\n\\begin{problem}\nCalculate an initial state estimate $\\widehat{\\mathbf{x}}_{200}$ as follows:\nFor the horizontal and vertical positions, simply use the observed position at time $200$.\nFor the velocity, compute the average velocity between the observations $\\mathbf{z}_k$ and $\\mathbf{z}_{k+1}$ for\n$k = 200,\\ldots,208$, then average these 9 values and take this as the initial velocity estimate.\n(Hint: the NumPy function \\li{diff} is useful here.)\n\nUsing the initial state estimate, $P_{200} = 10^{6} \\cdot Q$, and your Kalman filter, compute the next $600$ state estimates,\ni.e. compute $\\widehat{\\mathbf{x}}_{201}, \\ldots, \\widehat{\\mathbf{x}}_{800}$.\nPlot these state estimates as a smooth green curve together with the radar observations (as red dots) and the entire true state sequence (as a blue curve).\nZoom in to see how well it follows the true path. Your plots should be similar to Figure \\ref{fig:estimates}.\n\\label{prob:state_estimate}\n\\end{problem}\n\nIn the absence of observations, we can still estimate some information about the state of the system at some future time.\nWe can do this by recognizing that the expected state noise $\\mathbb{E}\\left[\\boldsymbol{\\epsilon}_{k}\\right] = 0$ at any time $k$.\nThus, given a current state estimate $\\widehat{\\mathbf{x}}_{n|m}$ using only measurements up through time $m$, the expected state at time $n+1$ is\n\\begin{equation*}\n\\widehat{\\mathbf{x}}_{n+1|m} = F \\widehat{\\mathbf{x}}_{n|m} + \\mathbf{u}\n\\end{equation*}\n\n\\begin{problem}\nAdd a function to your class that predicts the next $k$ states given a current state estimate but in the absence of observations.\nDo so by implementing the following function:\n\\begin{lstlisting}\ndef predict(self,x,k):\n \"\"\"\n Predict the next k states in the absence of observations.\n\n Parameters\n ----------\n x : ndarray of shape (n,)\n The current state estimate.\n k : integer\n The number of states to predict.\n\n Returns\n -------\n out : ndarray of shape (n,k)\n The next k predicted states.\n \"\"\"\n pass\n\\end{lstlisting}\n\\end{problem}\n\n\\begin{figure}\n \\centering\n \\begin{tabular}{cc}\n \\includegraphics[width=.45\\textwidth]{impact_macro} &\n \\includegraphics[width=.45\\textwidth]{impact_micro}\n \\end{tabular}\n \\caption{Predicted vs. actual point of impact (detailed view on right).}\n \\label{fig:impact}\n\\end{figure}\n\nWe can use this prediction routine to estimate where the projectile will hit the surface.\n\\begin{problem}\nUsing the final state estimate $\\widehat{\\mathbf{x}}_{800}$ that you obtained in Problem \\ref{prob:state_estimate}, predict the future states of\nthe projectile until it hits the ground.\nPredicting approximately the next $450$ states should be sufficient.\n\nPlot the actual state sequence together with the predicted state sequence (as a yellow curve), and observe how near the prediction is to the actual point of impact.\nYour results should be similar to those shown in Figure \\ref{fig:impact}.\n\\end{problem}\n\nIn the absence of observations, we can also reverse the system and iterate backward in time to infer information about states of the system prior to measured observations.\nThe system is reversed by\n\\begin{equation*}\n\\mathbf{x}_{k} = F^{-1}(\\mathbf{x}_{k+1} - \\mathbf{u} - \\boldsymbol{\\epsilon}_{k+1}).\n\\end{equation*}\nConsidering again that $\\mathbb{E}\\left[\\boldsymbol{\\epsilon}_{k}\\right] = 0$ at any time $k$, we can ignore this term, simplifying the recursive estimation backward in time.\n\n\n\n\n\\begin{figure}[htb]\n \\centering\n \\begin{tabular}{cc}\n \\includegraphics[width=.45\\textwidth]{origin_macro} &\n \\includegraphics[width=.45\\textwidth]{origin_micro}\n \\end{tabular}\n \\caption{Predicted vs. actual point of origin (detailed view on right).}\n \\label{fig:origin}\n\\end{figure}\n\n\\begin{problem}\nAdd a function to you class that rewinds the system from a given state estimate, returning predictions for the previous states.\nDo so by implementing the following function:\n\\begin{lstlisting}\ndef rewind(self,x,k):\n \"\"\"\n Predict the k states preceding the current state estimate x.\n\n Parameters\n ----------\n x : ndarray of shape (n,)\n The current state estimate.\n k : integer\n The number of preceding states to predict.\n\n Returns\n -------\n out : ndarray of shape (n,k)\n The k preceding predicted states.\n \"\"\"\n pass\n\\end{lstlisting}\n\\end{problem}\n\nReturning to the projectile example, we can now predict the point of origin.\n\n\\begin{problem}\nUsing your state estimate $\\widehat{\\mathbf{x}}_{250}$, predict the point of origin of the projectile along with\nall states leading up to time step $250$. Note that you may have to take a few extra time steps to predict the point of origin.\n(The point of origin is the first point along the trajectory where the $y$ coordinate is $0$.)\nPlot these predicted states (in cyan) together with the original state sequence.\nZoom in to see how accurate your prediction is.\nYour plots should be similar to Figure \\ref{fig:origin}.\n\\label{prob:origin_pt}\n\nRepeat the prediction starting with $\\widehat{\\mathbf{x}}_{600}$. Compare to the previous results. Which is better? Why?\n\\end{problem}\n", "meta": {"hexsha": "c78895d6e6a00f04fb715b93179c400e8d659db4", "size": 17198, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Volume3/KalmanFilter/KalmanFilter.tex", "max_stars_repo_name": "chrismmuir/Labs-1", "max_stars_repo_head_hexsha": "13c23611b90d73b0c2c7d275bce9808f829009f2", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 190, "max_stars_repo_stars_event_min_datetime": "2015-07-17T01:57:06.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-15T19:16:19.000Z", "max_issues_repo_path": "Volume3/KalmanFilter/KalmanFilter.tex", "max_issues_repo_name": "chrismmuir/Labs-1", "max_issues_repo_head_hexsha": "13c23611b90d73b0c2c7d275bce9808f829009f2", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": 184, "max_issues_repo_issues_event_min_datetime": "2015-07-16T17:56:06.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-06T23:47:14.000Z", "max_forks_repo_path": "Volume3/KalmanFilter/KalmanFilter.tex", "max_forks_repo_name": "chrismmuir/Labs-1", "max_forks_repo_head_hexsha": "13c23611b90d73b0c2c7d275bce9808f829009f2", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 76, "max_forks_repo_forks_event_min_datetime": "2015-08-06T02:53:11.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-27T11:08:57.000Z", "avg_line_length": 46.4810810811, "max_line_length": 239, "alphanum_fraction": 0.7222932899, "num_tokens": 4470, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8791467548438124, "lm_q2_score": 0.712232184238947, "lm_q1q2_score": 0.6261566134689905}} {"text": "\\documentclass{article}\n\\usepackage{leonine,amsmath,amssymb,amsthm,graphicx}\n\\setkeys{Gin}{width=\\linewidth,totalheight=\\textheight,keepaspectratio}\n\\graphicspath{{graphics/}}\n% Prints a trailing space in a smart way.\n\\usepackage{xspace}\n% Inserts a blank page\n\\newcommand{\\blankpage}{\\newpage\\hbox{}\\thispagestyle{empty}\\newpage}\n% \\usepackage{units}\n% Typesets the font size, leading, and measure in the form of 10/12x26 pc.\n\\newcommand{\\measure}[3]{#1/#2$\\times$\\unit[#3]{pc}}\n\n\\theoremstyle{definition}\n\\newtheorem{pred}[thm]{Prediction}\n\n\\title{Askesis: Introduction} \\author{Eric Purdy}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Perceptron model of a neuron}\n\nA perceptron has two possible outputs, $1$ (representing firing) and\n$0$ (representing not firing). Let its inputs be denoted by $x_1,\n\\dots, x_n$. The rule is then\n$$y = \\begin{cases} 1 & \\sum_i w_i x_i > \\theta \\\\ -1 & \\mbox{otherwise} \n\\end{cases}$$\n\nThe $x_i$ can usually be taken to be either $0$ (if the input neuron\nfires) or $1$ (if the input neuron does not fire). Neurons that do not\nfire are thus not included in the sum. Occasionally, it is helpful to\nthink about $x_i$ between zero and one, in which case it can be\nencoded by the firing rate. So, if the neuron fires at its maximum\nrate, that is represented by a one, and if it doesn't fire at all,\nthat is represented by a zero. If it fires at a tenth its firing rate,\nthat is represented by $0.1$.\n\nThe $w_i$ are called the synapse weights; each is a summary of\nproperties of the synapse that occurs between the $i$-th input neuron\nand the output neuron. These properties include the physical distance\nbetween the two neurons, as well as the concentration of\nneurotransmitter receptors in the output neuron, and other things as\nwell. If $w_i > 0$, then we call the synapse ``excitatory'' - the\nfiring of the input neuron excites the output neuron. Most excitatory\nsynapses use the neurotransmitter glutamate. If $w_i < 0$, we call the\nsynapse ``inhibitory'' - the firing of the input neuron inhibits the\noutput neuron. Most inhibitory synapses use the neurotransmitter GABA\n(Gamma-Aminobutyric acid). In actual neurons, a single neuron releases\na single main neurotransmitter, and is therefore only ever excitatory\n(all synapses it outputs to are excitatory) or inhibitory (all\nsynapses it outputs to are inhibitory). The parameter $\\theta$\ncontrols how excitable the neuron is, and can be modified by\nlearning. This is called ``intrinsic plasticity''.\n\n\\footnote{I like to think of glutamate as the ``black ink'' and GABA as the\n ``red ink'' of the nervous system.}\n\nA perceptron is much simpler than a neuron, but is a reasonable\nfirst-pass approximation. Our model does not require anything more\ncomplicated than perceptrons.\n\nThe word ``perceptron'' actually refers to two things: the model\nabove, and a learning algorithm. We will describe the learning\nalgorithm here. First, we add an $x_0$ that is always $1$, and let\n$w_0 = - \\theta$. The perceptron then fires if $\\sum_{i=0}^n w_i x_i >\n0$. We then suppose that we see a sequence of inputs $x_i(t)$, each\nlabeled either $1$ or $-1$ by a desired output $\\widehat{y}(t)$. The\nlearning rule is then as follows: if the output $y$ of the neuron was\nthe same as $\\widehat{y}(t)$, we do nothing. If $y(t) \\ne\n\\widehat{y}(t)$, then we update the weights as follows:\n$$w_i(t+1) = w_i(t) + \\alpha (\\widehat{y}(t) - y(t)) x_i(t).$$ If we\nstart with all $w_i$ equal to zero, then the $w_i(t)$ (considered as a\nvector) will be equal to the sum of the $\\mathbf{x}(t)$ for which\n$\\widehat{y}(t)=1,y(t)=0$, minus the sum of the $\\mathbf{x}(t)$ for\nwhich $\\widehat{y}(t)=0, y(t)=1$. The parameter $\\alpha$ is called\nthe learning rate; high values of $\\alpha$ result in a system that\nchanges its parameters more quickly and thus learns more quickly. A\nlower learning rate can be desirable, since the system will be less\nlikely to lose the information it already has.\n\nThe perceptron algorithm is impossible to implement in a single\nneuron, since it requires the same synapse to be both excitatory and\ninhibitory, depending on what the training data looks like. The\nperceptron algorithm is still useful to know, since it can be\nimplemented by multiple neurons, and has long been thought to be a\nlarge part of how the cerebellum works.\n\n\n\\section{Basic overview of the cerebellum}\n\n\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{organization.png}\n\\caption{Inputs and outputs of the cerebellum. The two arrows marked\n with double lines represent two neural pathways that convey copies\n of the same information.}\n\\label{fig-organization}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/global.png}\n\\caption{How the cerebellum fits into the rest of the brain. Courtesy\n of\n http://what-when-how.com/wp-content/uploads/2012/04/tmp15F121.jpg}\n\\label{fig-global}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/pathways.png}\n\\caption{Pathways into and out of the cerebellum. The fastigial,\n interposed, and dentate nuclei (and some others) are collectively\n referred to as the deep nuclei, as they lie deep within the\n cerebellum. Courtesy of\n http://www.dizziness-and-balance.com/anatomy/brain/cerebellum.htm}\n\\label{fig-pathways}\n\\end{figure}\n\n\n\\section{Cell types}\n\nThe cerebellum contains a number of different types of cells:\n\\begin{description}\n\\item[Mossy fiber] Mossy fibers are the axons of neurons outside of\n the cerebellum. They constitute one of the two sources of input of\n the cerebellum. One of the major sources of mossy fibers is the\n pontine nuclei, which relay information from the cerebrum.\n\n All mossy fibers are excitatory. The mossy fibers excite granular\n cells and Golgi cells, and collaterals (side branches) of the mossy\n fibers excite deep nuclear cells.\n\n Mossy fibers participate in both the positive and negative pathways.\n\n\\item[Deep nuclear cell] These are the main source of output from the\n cerebellum. (Some Purkinje cells project to the vestibular nuclei,\n which seem to be pretty similar to the deep nuclei.)\n\n Some of the deep nuclei are: interposed nucleus, fastigial nucleus,\n dentate nucleus.\n\n There are two kinds of deep nuclear cell, large and small. The large\n cells can be excitatory or inhibitory, and project to premotor\n areas. The small cells are inhibitory and project to the inferior\n olive.\n\n Deep nuclear cells receive input from mossy fiber collaterals,\n climbing fiber collaterals, and Purkinje cells.\n\n Deep nuclear cells participate in both the positive and negative\n pathways.\n\n\\item[Inferior olive cell] These are the second source of input to the\n cerebellum. They receive an ``efference'' copy of the signals sent\n from the cerebrum to the spinal cord.\n\n The inferior olive cells are excitatory and send climbing fibers\n into the cerebellum to make contact with the Purkinje cells,\n stellate cells, and basket cells. A single climbing fiber impulse\n will cause a Purkinje cell to fire. \n\n Collaterals of the climbing fiber make contact with the deep nuclear\n cells.\n\n Inferior olive cells participate in both the positve and negative\n pathways.\n\n\\item[Purkinje cell] The Purkinje cells are inhibitory, and send their\n output to the deep nuclear cells. They receive input mainly from the\n parallel fibers, which are the output of the granular cells. They\n also receive some input directly from nearby granular cells.\n\n Purkinje cells fire at a certain base rate even when not receiving\n any input.\n\n Purkinje cells participate in the negative pathway.\n\n\\item[Basket/stellate cell] Basket and stellate cells inhibit the\n Purkinje cells (and thus disinhibit the deep nuclear cells by\n proxy).\n\n The stellate cells occupy the outermost portions of the cerebellar\n cortex, while basket cells are slightly more inward. There is also\n an intermediate type of cell that occurs between the basket cells\n and stellate cells and has an appearance halfway between the two.\n\n Basket and stellate cells participate in the negative pathway.\n\n\\item[Granular cell] Granular cells receive input from 4-5 mossy\n fibers. Granular cells are excitatory. The output of the granular\n cell is the parallel fiber, which sends input to the Purkinje cells,\n basket cells, and stellate cells.\n\n Granular cells are generally thought to fire only when multiple of\n their inputs are active.\n\n Granular cells are the most numerous neurons in the brain; about\n three-fourths of the neurons in the brain are granular cells in the\n cerebellum. (Granular cells are very small, so the cerebellum only\n takes up about 10\\% of the volume of the brain.)\n\n Granular cells participate in the negative pathway.\n\\end{description}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{cells.png}\n\\caption{Cells of the cerebellum. Excitatory cells are shown in black,\n while inhibitory cells are shown in red. Inputs to a cell are shown\n as an empty half circle, while outputs are shown as a full half\n circle. Cells of the inferior olive are shown in green, to\n emphasize that their output is used to train the targeted cells;\n they are excitatory. There are two kinds of deep nuclear cell: one\n excitatory and projecting either back into the cerebellum or to\n premotor areas; and one inhibitory and projecting to the inferior\n olive. Since the cells receive the same input, we have simplified\n the picture by identifying them together. It is also worth noting\n that state cells are actually represented twice here: they are both\n ``mossy fiber cells'' and ``deep nuclear cells''.}\n\\label{fig-cells}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/028.png}\n\\caption{Cells of the cerebellum. Courtesy of\n https://nanohub.org/resources/18948/watch?resid=19060}\n\\label{fig-physical-1}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/029.png}\n\\caption{Cells of the cerebellum. Courtesy of\n https://nanohub.org/resources/18948/watch?resid=19060}\n\\label{fig-physical-2}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/image004.png}\n\\caption{Cells of the cerebellum. Courtesy of http://www.scritub.com/limba/engleza/health/CEREBELLUM25854.php}\n\\label{fig-physical-3}\n\\end{figure}\n\n\\begin{figure}\n\\includegraphics[width=\\linewidth]{nanohub/nature.png}\n\\caption{Cells of the cerebellum. Courtesy of\n http://www.nature.com/nrn/journal/v9/n4/images/nrn2332-f3.jpg.\nPCN:pontine nuclei.\nMF: mossy fiber.\nLg: Lugaro cell.\n5-HT fibres: beaded fibers.\nUB: unipolar brush cells (found only in the vestibulocerebellum).\nGo: Golgi cell.\nGr: Granular cell.\nN-C: state cell pathway.\nCN/VN: deep nuclei and vestibular nuclei.\nPF: parallel fiber.\nLTD: long-term depression, the learning mechanism of the Purkinje cell.\nSC: stellate cell.\nBC: basket cell.\nCF: climbing fiber.\nN-O: training suppression cell pathway to inferior olive.\nIO: inferior olive.\nR-O: red nucleus to inferior olive pathway.\npRN: parvocellular (small cells) red nucleus. \nPd: unknown, seems to refer to another kind of beaded fiber.\n}\n\\label{fig-physical-4}\n\\end{figure}\n\n\n\\section{Learning}\n\nAs a brief aside to anyone not versed in machine learning or neuroscience,\n``learning'' is a term that describes changing the structure or parameters of a\nsystem in order to make that system better achieve some goal or perform some\npurpose correctly. The best understood mechanism for learning in the brain is\nchanging the strength of synapses, the gaps between two neurons, usually by\nchanging the number of receptors that respond to neurotransmitters.\n\n\\section{Thoughts}\n\n\\begin{itemize}\n\\item Discuss microzones. In particular, how big are they? What\n constraints do they impose?\n\\item Give sigmoid rule with probability.\n\n\\item Hebbian learning\n\n\\item Explain electrical synapses in the inferior olive?\n\\end{itemize}\n\n\\end{document}\n", "meta": {"hexsha": "791859ed065a58ce401642b71872c0a06370dea6", "size": 11860, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "askesis/neuro.tex", "max_stars_repo_name": "advancedresearch/ethicophysics", "max_stars_repo_head_hexsha": "52806b53d6d3ee92e1bd2a8c00f7728cebc9e684", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2018-04-26T17:10:58.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-19T01:56:36.000Z", "max_issues_repo_path": "askesis/neuro.tex", "max_issues_repo_name": "advancedresearch/ethicophysics", "max_issues_repo_head_hexsha": "52806b53d6d3ee92e1bd2a8c00f7728cebc9e684", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 46, "max_issues_repo_issues_event_min_datetime": "2018-04-26T16:25:59.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-15T10:13:05.000Z", "max_forks_repo_path": "askesis/neuro.tex", "max_forks_repo_name": "advancedresearch/ethicophysics", "max_forks_repo_head_hexsha": "52806b53d6d3ee92e1bd2a8c00f7728cebc9e684", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-05-13T17:37:48.000Z", "max_forks_repo_forks_event_max_datetime": "2018-05-13T17:37:48.000Z", "avg_line_length": 40.4778156997, "max_line_length": 110, "alphanum_fraction": 0.7699831366, "num_tokens": 3117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424489603726, "lm_q2_score": 0.7401743505760728, "lm_q1q2_score": 0.6261449027839765}} {"text": "\\documentclass{ut-thesis}\n\n%**************************** General libraries *******************************\n\\usepackage{mathtools}\n\\usepackage{cancel}\n\\usepackage[utf8]{inputenc} \n\\usepackage{amssymb}\n\\usepackage{ntheorem}\n\\usepackage{tensor}\n%\\usepackage{physics}\n\\usepackage[italicdiff]{physics}\n\\usepackage{calc}\n\\usepackage{caption}\n%\\usepackage{subcaption}\n\\usepackage{tcolorbox}\n\\usepackage{chngcntr}\n\\usepackage{titlesec}\n\\usepackage {graphicx,float} \n\\usepackage{subfig}\n\\usepackage{siunitx}\n\\usepackage{xcolor}\n\\usepackage{etoolbox} %ifthen\n\\usepackage[outline]{contour} % glow around text\n\n%**************************** Bibliography management *******************************\n\\usepackage{biblatex} %Imports biblatex package\n\\addbibresource{sample.bib} %Import the bibliography file\n\n\n\n%**************************** Tikz libraries *******************************\n\\usepackage{tikz}\n\\usetikzlibrary{shapes,arrows,arrows.meta,spy}\n\\usetikzlibrary{calc}% needed for BB\n\\usepackage{pgfplots}\n\\usetikzlibrary{math} % for \\tikzmath\n\\usetikzlibrary{angles,quotes} % for pic (angle labels)\n\\usetikzlibrary{decorations.pathmorphing,decorations.markings}\n\\usetikzlibrary{decorations.pathreplacing} % for curly braces\n\\usetikzlibrary {decorations.markings,shapes.arrows}\n\\usetikzlibrary{patterns}\n\\usepackage{tikz-3dplot}\n\n%\\usepackage[margin=0cm,nohead]{geometry}\n%\\usepackage[active,tightpage]{preview}\n%\\usepackage[pdf]{pstricks}\n\n%**************************** custom macro's *******************************\n\\newsavebox\\CBox\n\\newcommand\\hcancel[2][0.5pt]{%\n \\ifmmode\\sbox\\CBox{$#2$}\\else\\sbox\\CBox{#2}\\fi%\n \\makebox[0pt][l]{\\usebox\\CBox}% \n \\rule[0.5\\ht\\CBox-#1/2]{\\wd\\CBox}{#1}}\n%\\newcommand{\\edal}{\\end{align}}\n%\\newcommand{\\bgal}{\\begin{align}\\end{align}}\n\\newcommand{\\Lagr}{\\mathcal{L}}\n\\newcommand{\\questeq}{\\overset{?}{=}}\n%\\newtheorem{theorem}{Theorem}\n\\newtheorem{lemma}{Lemma}\n\\renewcommand\\thelemma{\\unskip}\n\\newcommand{\\christ}[3]{\\ensuremath{\\Gamma^{#1}_{#2#3}}}\n\\newcommand{\\half}{\\ensuremath{\\frac{1}{2}}}\n\\newcommand{\\kwart}{\\ensuremath{\\frac{1}{4}}}\n\\newcommand\\del{\\overline{\\boldsymbol{{\\triangledown}}}}\n\\newcommand\\maal{\\boldsymbol{\\times}}\n\\newcommand\\spatie{\\quad\\quad\\quad\\quad}\n\\newcommand\\RAr{\\quad\\Rightarrow\\quad}\n%\\newcommand{\\myabsdv}[3]{{\\frac{\\delta^2 \\ensuremath{{#1}}}{{\\delta %%\\ensuremath{{#2}}}{\\delta\\ensuremath{{#3}}}}}}\n\\newcommand\\myabsdv[3][1]{{\\frac{{\\delta^{2}}{#1}}{\\delta {#2} \\delta {#3}}}}\n\n%**************************** layout setings *******************************\n\\counterwithin*{equation}{chapter}\n\\counterwithin*{equation}{section}\n\\counterwithin*{equation}{subsection}\n\\renewcommand{\\theequation}{\\arabic{equation}}\n\\titleformat{\\chapter}{\\normalfont\\huge}{\\thechapter.}{20pt}{\\huge\\it}\n\\titleformat{\\chapter}[display]\n {\\normalfont\\bfseries}{}{0pt}{\\Huge}\n\\titlespacing*{\\chapter}{0pt}{50pt}{*2}\n\n%**************************** Tikz setings *******************************\n\\begin{comment}\n\\tikzset{\n circ/.style={draw, circle,inner sep=0pt,minimum size=8mm, font=\\scriptsize},\n triangle/.tip={Computer Modern Rightarrow[open,angle=120:3pt]}}\n \n\\tikzset{\n every point/.style = {circle, inner sep={.75\\pgflinewidth}, opacity=1, draw, solid, fill=white},\n point/.style={insert path={node[every point, #1]{}}}, point/.default={},\n colored point/.style = {point={fill=#1}},\n point name/.style = {insert path={coordinate (#1)}},\n inherit/.style = {point/.style={insert path={node[circle, inner sep={.75\\pgflinewidth}, draw, fill, #1]{}}}}\n}\n\\end{comment}\n\n%**************************** Where to find images *****************************\n\\graphicspath{ {D:/MathLatex/images/} }\n\n\n%**************************** Document itself *******************************\n\\author{Bernard Carrette}\n\\gradyear{2020}\n\\title{Tensor Calculus\\\\J.L. Synge and A.Schild (Dover Publication)\\\\ Solutions to exercises}\n\\begin{document}\n\\maketitle\n\n\\section*{Remarks and warnings}\nYou're welcome to use these notes, but they may contain errors, so proceed with caution. If you do find an error, however, I'd be happy to receive bug reports, suggestions, and the like through Github.\n\\section*{Some notation conventions}\n\\begin{center}\n\\begin{tabular}{ c c }\n$\\partial_r \\equiv \\pdv{}{x^r}$ & \\\\\\\\\n$\\Gamma^r_{mn} \\equiv \n\\begin{Bmatrix}\nr\\\\\nm n\\\\\n\\end{Bmatrix}$ & Christoffel symbol of the second kind\\\\\\\\\n\n\\end{tabular}\n\\end{center}\n\n\\tableofcontents\n\\listoffigures\n\\include{chapter1}\n\\include{chapter2}\n\\include{chapter3}\n\\include{chapter4}\n%\\include{chapter5}\n\\printbibliography %Prints bibliography\n\\end{document}", "meta": {"hexsha": "150fbb50846631c0a31e75ff8b2b20cf988537b4", "size": 4557, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main.tex", "max_stars_repo_name": "Niohori/Synge-Tensor-Calculus", "max_stars_repo_head_hexsha": "a11e45d9d8c4f78b9e9504391b532bb4f0d0587f", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "main.tex", "max_issues_repo_name": "Niohori/Synge-Tensor-Calculus", "max_issues_repo_head_hexsha": "a11e45d9d8c4f78b9e9504391b532bb4f0d0587f", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "main.tex", "max_forks_repo_name": "Niohori/Synge-Tensor-Calculus", "max_forks_repo_head_hexsha": "a11e45d9d8c4f78b9e9504391b532bb4f0d0587f", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.0538461538, "max_line_length": 201, "alphanum_fraction": 0.6570111916, "num_tokens": 1374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542925, "lm_q2_score": 0.8198933337131076, "lm_q1q2_score": 0.6259752481363589}} {"text": "\\subsection*{Stream Support}\n\nThe following stream processing functions are supported:\n\n\\begin{itemize}\n\\item \\lstinline{stream_tail(x)}: \\textit{primitive}, assumes that the tail (second component) of the\n pair \\lstinline{x} is a nullary function, and returns the result of\n applying that function.\\\\\n\\emph{Laziness:} Yes: \\lstinline{stream_tail} only forces the direct tail of a given\nstream,\nbut not the rest of the stream, i.e. not the tail of the tail, etc.\n\\item \\lstinline{stream(x1, x2,..., xn)}: \\textit{primitive}, returns a stream with $n$ elements. The\nfirst element is \\lstinline{x1}, the second \\lstinline{x2}, etc.\\\\\n\\emph{Laziness:} No: In this implementation, we generate first a\n complete list, and then a stream using \\lstinline{list_to_stream}.\n\\item \\lstinline{is_stream(x)}: Returns \\lstinline{true} if\n \\lstinline{x} is a stream as defined in the lectures, and\n \\lstinline{false} otherwise.\\\\\n\\emph{Laziness:} No: \\lstinline{is_stream} needs to force the given stream.\n\\item \\lstinline{list_to_stream(xs)}: transforms a given list to a stream.\\\\\n\\emph{Laziness:} Yes: \\lstinline{list_to_stream} goes down the list only when forced.\n\\item \\lstinline{stream_to_list(s)}: transforms a given stream to a list.\\\\\n\\emph{Laziness:} No: \\lstinline{stream_to_list} needs to force the whole stream.\n\\item \\lstinline{stream_length(s)}: Returns the length of the stream\n \\lstinline{s}.\\\\\n\\emph{Laziness:} No: The function needs to force the whole stream.\n\\item \\lstinline{stream_map(f, s)}: Returns a stream that results from stream\n \\lstinline{s} by element-wise application of \\lstinline{f}.\\\\\n\\emph{Laziness:} Yes: The argument stream is only explored as forced by\n the result stream.\n\\item \\lstinline{build_stream(n, f)}: Makes a stream with \\lstinline{n}\nelements by applying the unary function \\lstinline{f} to the numbers 0\nto \\lstinline{n - 1}.\\\\\n\\emph{Laziness:} Yes: The result stream forces the applications of fun\n for the next element.\n\\item \\lstinline{stream_for_each(f, s)}: Applies \\lstinline{f} to every\n element of the stream \\lstinline{s}, and then returns\n \\lstinline{true}.\\\\\n\\emph{Laziness:} No: \\lstinline{stream_for_each} forces the exploration of the entire stream.\n\\item \\lstinline{stream_reverse(s)}: Returns finite stream \\lstinline{s} in reverse\n order. Does not terminate for infinite streams.\\\\\n\\emph{Laziness:} No: \\lstinline{stream_reverse} forces the exploration of the entire stream.\n\\item \\lstinline{stream_append(xs, ys)}: Returns a stream that results from \nappending the stream \\lstinline{ys} to the stream \\lstinline{xs}.\\\\\n\\emph{Laziness:} Yes: Forcing the result stream activates the actual append operation.\n\\item \\lstinline{stream_member(x, s)}: Returns first postfix substream\nwhose head is equal to\n\\lstinline{x} (\\lstinline{===}); returns \\lstinline{null} if the\nelement does not occur in the stream.\\\\\n \\emph{Laziness:} Sort-of: \\lstinline{stream_member} forces the stream only until the element is found.\n\\item \\lstinline{stream_remove(x, s)}: Returns a stream that results from\ngiven stream \\lstinline{s} by removing the first item from \\lstinline{s} that\nis equal (\\lstinline{===}) to \\lstinline{x}.\nReturns the original list\nif there is no occurrence.\\\\\n\\emph{Laziness:} Yes: Forcing the result stream leads to construction of each next element.\n\\item \\lstinline{stream_remove_all(x, s)}: Returns a stream that results from\ngiven stream \\lstinline{s} by removing all items from \\lstinline{s} that\nare equal (\\lstinline{===}) to \\lstinline{x}.\\\\\n\\emph{Laziness:} Yes: The result stream forces the construction of each next element.\n\\item \\lstinline{stream_filter(pred, s)}: Returns a stream that contains\nonly those elements for which the one-argument function\n\\lstinline{pred}\nreturns \\lstinline{true}.\\\\\n\\emph{Laziness:} Yes: The result stream forces the construction of\n each next element. Of course, the construction\n of the next element needs to go down the stream\n until an element is found for which \\lstinline{pred} holds.\n\\item \\lstinline{enum_stream(start, end)}: Returns a stream that enumerates\nnumbers starting from \\lstinline{start} using a step size of 1, until\nthe number exceeds (\\lstinline{>}) \\lstinline{end}.\\\\\n \\emph{Laziness:} Yes: Forcing the result stream leads to the construction of\n each next element.\n\\item \\lstinline{integers_from(n)}: Constructs an infinite stream of integers\nstarting at a given number \\lstinline{n}.\\\\\n \\emph{Laziness:} Yes: Forcing the result stream leads to the construction of \n each next element.\n\\item \\lstinline{eval_stream(s, n)}: Constructs the list\nof the first \\lstinline{n} elements\nof a given stream \\lstinline{s}.\\\\\n\\emph{Laziness:} Sort-of: \\lstinline{eval_stream} only forces the computation of\n the first \\lstinline{n} elements, and leaves the rest of\n the stream untouched.\n\\item \\lstinline{stream_ref(s, n)}: Returns the element\nof stream \\lstinline{s} at position \\lstinline{n}, \nwhere the first element has index 0.\\\\\n \\emph{Laziness:} Sort-of: \\lstinline{stream_ref} only forces the computation of\n the first \\lstinline{n} elements, and leaves the rest of\n the stream untouched.\n\\end{itemize}\n", "meta": {"hexsha": "520fd7fba9491abe22b4d7188d5aa66fadec37a3", "size": 5299, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/specs/source_streams.tex", "max_stars_repo_name": "jaesimin/js-slang", "max_stars_repo_head_hexsha": "153596c436998e4aa182a61be455febf77eb5510", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 48, "max_stars_repo_stars_event_min_datetime": "2018-07-09T06:16:03.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T09:40:24.000Z", "max_issues_repo_path": "docs/specs/source_streams.tex", "max_issues_repo_name": "jaesimin/js-slang", "max_issues_repo_head_hexsha": "153596c436998e4aa182a61be455febf77eb5510", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1117, "max_issues_repo_issues_event_min_datetime": "2018-07-09T08:08:25.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-29T14:47:08.000Z", "max_forks_repo_path": "docs/specs/source_streams.tex", "max_forks_repo_name": "jaesimin/js-slang", "max_forks_repo_head_hexsha": "153596c436998e4aa182a61be455febf77eb5510", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 80, "max_forks_repo_forks_event_min_datetime": "2018-08-24T08:55:58.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-10T08:56:48.000Z", "avg_line_length": 58.2307692308, "max_line_length": 104, "alphanum_fraction": 0.7371202114, "num_tokens": 1468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527632, "lm_q2_score": 0.7745833841649232, "lm_q1q2_score": 0.625915436906419}} {"text": "\\documentclass{scrartcl}\n\\usepackage[a4paper,left=1in,right=1in,top=1.2in,bottom=1in]{geometry}\n\\usepackage{siunitx}\n\\usepackage{graphicx}\n\\usepackage{mathtools}\n\\setkomafont{disposition}{\\normalfont\\bfseries}\n\\newcommand*\\diff{\\mathop{}\\!\\mathrm{d}}\n\\newcommand*\\Diff[1]{\\mathop{}\\!\\mathrm{d^#1}}\n\\newcommand*\\colvec[3][]{\n \\begin{pmatrix}\\ifx\\relax#1\\relax\\else#1\\\\\\fi#2\\\\#3\\end{pmatrix}\n}\n\n%title\n\\title{Exercise 02:\\\\State space analysis}\n\\subtitle{Theoretical Neuroscience II}\n\\author{Johannes G\\\"atjen \\and Lorena Morton}\n\n%use these for structure/overview\n\\newcommand\\Question{%\n \\textbf{Question:}%\n}\n\\newcommand\\Answer{%\n \\textbf{Answer:}%\n}\n\n\\begin{document}\n\\maketitle\n\n\\section{Task A}\n\nWe investigate a linear recurrent network with 2 populations. The time-development is given by the equation\n\n\\begin{align*}\n\\frac{\\diff \\mathbf{v}}{\\diff t} = \\mathbf{A} \\cdot \\mathbf{v} + \\mathbf{b} \\qquad \\text{ with } \\qquad \\mathbf{A} = \\left( \\begin{array}{cc}\n-2 & 3 \\\\\n-3 & 2 \\end{array} \\right).\n\\end{align*}\n\nWe determine the input vector $\\mathbf{b}$ such that the steady state of the network is at\n\n\\begin{align*}\n\\mathbf{v}_{ss} = \\colvec{10}{10} \\qquad \\text{ by solving the equation } \\qquad \\mathbf{A} \\cdot \\mathbf{v}_{ss} + \\mathbf{b} = 0. \\\\\n\\shortintertext{Substituting our values we obtain:}\n\\left( \\begin{array}{cc}\n-2 & 3 \\\\\n-3 & 2 \\end{array} \\right) \\cdot \\colvec{10}{10} + \\mathbf{b} &= \\colvec{10}{-10} + \\mathbf{b} = 0 \\\\\n\\Leftrightarrow \\mathbf{b} &= \\colvec{-10}{10} \\\\\n\\shortintertext{With this we get the full dynamic equation:}\n\\colvec{\\frac{\\diff v_1}{\\diff t}}{\\frac{\\diff v_2}{\\diff t}} = \\left( \\begin{array}{cc}\n-2 & 3 \\\\\n-3 & 2 \\end{array} \\right) \\cdot \\colvec{v_1}{v_2} + \\colvec{-10}{10} \\\\\n\\shortintertext{And we get two line equations when we set the gradient for both populations to zero:}\n\\text{$v_1$-isocline:} & -2 v_1 + 3 v_2 - 10 = 0 \\\\\n\\Leftrightarrow & v_2 = \\frac{2}{3} v_1 + \\frac{10}{3} \\\\\n\\text{$v_2$-isocline:} & -3 v_1 + 2 v_2 + 10 = 0 \\\\\n\\Leftrightarrow & v_2 = \\frac{3}{2} v_1 - 5 \\\\\n\\shortintertext{Get the intersection point, by solving the system of linear equations for $v_1$ and $v_2$.}\n& \\frac{2}{3} v_1 + \\frac{10}{3} = \\frac{3}{2} v_1 - 5 \\\\\n\\Leftrightarrow & v_1 = 10 \\\\\n\\shortintertext{Replace the value in one of the original equations.}\nv_2 = \\frac{3 \\cdot 10}{2} - 5 = 10 \\\\\n\\shortintertext{And we have confirmed that the steady state is}\n\\mathbf{v}_{ss} = \\colvec{10}{10}.\n\\end{align*}\n\n\\section{Task B}\n\nWe obtain the eigenvalues of the connectivity matrix by solving the characteristic equation.\n\\begin{align*}\n&(-2 - \\lambda) \\cdot (2 - \\lambda) + 9 = 0 \\\\\n\\Leftrightarrow & -4 + 2\\lambda - 2\\lambda + \\lambda^2 + 9 = 0 \\\\\n\\Leftrightarrow & \\lambda^2 = -5 \\\\\n\\Leftrightarrow & \\lambda_1 = i\\sqrt{5} \\qquad \\lambda_2 = -i\\sqrt{5} \n\\end{align*}\nBecause $Im(\\lambda_1) = - Im(\\lambda_2)$ we expect a spiraling behavior from the dynamical system, and because $Re(\\lambda_1) = Re(\\lambda_2) = 0$ it is neutrally stable.\n\n\\begin{figure}\n\\centering\n\\includegraphics[trim = {1.3cm 0 0.5cm 0.2cm}, width=0.7\\textwidth, clip]{../pics/traj}\n\\caption{Three trajectories of the population activity in the state space for the linear recurrent network.}\n\\label{label}\n\\end{figure}\n\n\\section{Task C}\n\nAs $\\kappa$ increases, the frequency of the oscillations do not change, but the steady state value for the amplitude of the oscillations gets smaller, such that the trajectory is damped more strongly. The steady state also appears to change from neutrally stable to stable with higher $\\kappa$.\n\n\\begin{figure}\n\\centering\n\\includegraphics[trim = {0.8cm 0 0.5cm 0.2cm}, width=0.7\\textwidth, clip]{../pics/nonlin}\n\\caption{The evolution of the activity of the two populations over time for the non-linear recurrent network, with $\\kappa = 2$.}\n\\label{label}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\includegraphics[trim = {0.8cm 0 0.5cm 0.2cm}, width=0.7\\textwidth, clip]{../pics/nonlintraj}\n\\caption{The trajectory of the population activity in the state space for the non-linear recurrent network, with initial condition at $(7, 11)$.}\n\\label{label}\n\\end{figure}\n\n%\\operatorname{\\mathbf{v}}(t) = \n\n%include picture\n%\\begin{figure}\n%\\centering\n%\\includegraphics[trim = {1.3cm 0 2cm 0.9cm}, width=\\textwidth, clip]{../pics/picname}\n%\\caption{caption text}\n%\\label{label}\n%\\end{figure}\n\\end{document}", "meta": {"hexsha": "574f422d103f865d4b544498187e23a7c91ff48e", "size": 4365, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ex2/pdf/ex2.tex", "max_stars_repo_name": "gaetjen/TNSII_Exercises", "max_stars_repo_head_hexsha": "d82eb790132e9066c6ad41e7f90ba193145a2e8f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ex2/pdf/ex2.tex", "max_issues_repo_name": "gaetjen/TNSII_Exercises", "max_issues_repo_head_hexsha": "d82eb790132e9066c6ad41e7f90ba193145a2e8f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ex2/pdf/ex2.tex", "max_forks_repo_name": "gaetjen/TNSII_Exercises", "max_forks_repo_head_hexsha": "d82eb790132e9066c6ad41e7f90ba193145a2e8f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.3243243243, "max_line_length": 294, "alphanum_fraction": 0.6966781214, "num_tokens": 1537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.749087201911703, "lm_q2_score": 0.8354835452961425, "lm_q1q2_score": 0.6258500311891569}} {"text": "\n\\section{Coulomb Integrals with error function Operator}\n\nIn a lot of applications it's useful to modify the normal Coulomb operator with \nerror function($\\omega$ is some given constant):\n\\begin{equation}\\label{erf_coulomb_integral:1}\n \\frac{1}{r_{12}} = \\frac{erf(\\omega r_{12})}{r_{12}} + \\frac{erfc(\\omega r_{12})}{r_{12}}\n\\end{equation}\nwhere the error function is:\n\\begin{align}\\label{erf_coulomb_integral:2}\n erf(\\omega r_{12}) &= \\frac{2}{\\sqrt{\\pi}}\\int_{0}^{\\omega r_{12}} e^{- t^{2}} dt \\nonumber \\\\ \n erfc(\\omega r_{12}) &= \\frac{2}{\\sqrt{\\pi}}\\int_{\\omega r_{12}}^{+\\infty} e^{- t^{2}} dt\n\\end{align}\ntherefore $erf(\\omega r_{12}) + erfc(\\omega r_{12}) = 1$. Based on the decomposition of the Coulomb operator, \nnow the normal four body integral could be rewritten as:\n\\begin{equation}\\label{erf_coulomb_integral:3}\n I_{ijkl} = \\int \\chi_{i}(r_{1})\\chi_{j}(r_{1})\\frac{erf(\\omega r_{12})}{r_{12}}\n\\chi_{k}(r_{2})\\chi_{l}(r_{2}) dr_{1} dr_{2}\n\\end{equation}\n\nHow to integrate the $I_{ijkl}$? By following the idea in \\ref{os_eri}, we need to transform\nthe operator with erf or erfc function into another form, just like we use Laplace transformation\nfor the $r_{12}$ operator:\n\\begin{equation}\\label{erf_coulomb_integral:5}\n \\frac{1}{|r-r^{'}|} = \\frac{2}{\\pi^{1/2}}\\int^{\\infty}_{0}\ne^{(r-r^{'})^{2}u^{2}}du\n\\end{equation}\nin \\ref{os_eri}. However, here we have a better idea; just to rewrite the normal erf/erfc function\nin the following way:\n\\begin{align}\\label{erf_coulomb_integral:4}\n erf(\\omega r_{12}) \n &= \\frac{2}{\\sqrt{\\pi}}\\int_{0}^{\\omega r_{12}} e^{- t^{2}} dt \\quad \\text{set t=u*$\\omega r_{12}$}\n \\Rightarrow \\nonumber \\\\ \n &= \\frac{2}{\\sqrt{\\pi}}\\int_{0}^{\\omega r_{12}} e^{- (u\\omega r_{12})^{2}} d(u\\omega r_{12}) \\nonumber \\\\\n &= \\frac{2\\omega}{\\sqrt{\\pi}}\\int_{0}^{1} r_{12}e^{- u^{2}\\omega^{2}r_{12}^{2}} du \n\\end{align}\nhere the trick is to use variable $u$ to replace the variable $t$, so that to transform the \nintegral limit from $\\omega r_{12}$ to 1. By using this form of erf function, we can get:\n\\begin{equation}\\label{erf_coulomb_integral:6}\n \\frac{erf(\\omega r_{12})}{r_{12}} = \n \\frac{2\\omega}{\\sqrt{\\pi}}\\int_{0}^{1} e^{- u^{2}\\omega^{2}r_{12}^{2}} du \n\\end{equation}\nso the $r_{12}$ inside the integrate of \\ref{erf_coulomb_integral:4} is canceled with the \none in the denominator. It's interesting that the result in \\ref{erf_coulomb_integral:6}\nis same with operator transformation in \\ref{erf_coulomb_integral:5}, except that here\nthe integration goes from 0 to 1 and in \\ref{erf_coulomb_integral:5} it goes from 0 to \n$+\\infty$. Similarly, the operator with erfc function is:\n\\begin{equation}\\label{erf_coulomb_integral:7}\n \\frac{erfc(\\omega r_{12})}{r_{12}} = \n \\frac{2\\omega}{\\sqrt{\\pi}}\\int_{1}^{+\\infty} e^{- u^{2}\\omega^{2}r_{12}^{2}} du \n\\end{equation}\n\nSuch derivation leads to an important idea, that the integral with operator \n$\\dfrac{erf(\\omega r_{12})}{r_{12}}$ etc. can be used for the recurrence relation described\nin \\ref{os_eri} without any problem. For example, the result in \\ref{OS_ERI_eq:2}\ncan be rewritten as for operator with erf function:\n\\begin{equation}\n \\begin{split}\n (ab|\\frac{erf(\\omega r_{12})}{r_{12}}|cd) &= \\frac{2\\omega}{\\pi^{1/2}}\\int^{1}_{0} du \n\\int dr \\int dr^{'} \\chi_{a}(r)\\chi_{b}(r) e^{(r-r^{'})^{2}(u\\omega)^{2}}\n\\chi_{c}(r^{'})\\chi_{d}(r^{'}) \\\\\n&= \\frac{2\\omega}{\\pi^{1/2}}\\int^{1}_{0} du\n\\int dr^{'}\\chi_{c}(r^{'})\\chi_{d}(r^{'}) \n\\left( \\int dr \\chi_{a}(r)\\chi_{b}(r) e^{(r-r^{'})^{2}(u\\omega)^{2}}\\right) \\\\ \n&= \\frac{2\\omega}{\\pi^{1/2}}\\int^{1}_{0} du (ab|u\\omega|cd)\n \\end{split}\n\\end{equation}\nLet's name $u^{'} = u\\omega$. By repeating all of same procedure in \\ref{os_eri} in terms of\n$u^{'}$, we can get the same recurrence relation for ERI as shown in \\ref{OS_ERI_result}(here \nbelow we omit the operator of $\\dfrac{erf(\\omega r_{12})}{r_{12}}$):\n\\begin{equation}\n \\begin{split}\n((a+\\iota_{i})b|cd)^{(m)} &= (P_{i} - A_{i})(ab|cd)^{(m)} +\n\\left(W_{i} -P_{i}\\right)(ab|cd)^{(m+1)} \\\\\n&+\\frac{N_{i}(A)}{2\\epsilon}\\left(((a-\\iota_{i})b|cd)^{(m)}-\\frac{\\rho}{\n\\epsilon }((a-\\iota_{i})b|cd)^{(m+1)}\\right) \\\\\n&+\\frac{N_{i}(B)}{2\\epsilon}\\left((a(b-\\iota_{i})|cd)^{(m)}-\\frac{\\rho}{\n\\epsilon }(a(b-\\iota_{i})|cd)^{(m+1)}\\right) \\\\\n&+\\left(\\frac{N_{i}(C)}{2}\\right)\\frac{1}{\\epsilon+\\eta}\n(ab|(c-\\iota_{i})d)^{(m+1)} \\\\\n&+\\left(\\frac{N_{i}(D)}{2}\\right)\\frac{1}{\\epsilon+\\eta}\n(ab|c(d-\\iota_{i}))^{(m+1)}\n\\end{split}\n\\label{erf_coulomb_integral:8}\n\\end{equation}\nwhere the integral of $(ab|cd)^{(m)}$ in \\ref{OS_ERI_eq:28} is expressed as:\n\\begin{equation}\n\\label{erf_coulomb_integral:9}\n (ab|\\frac{erf(\\omega r_{12})}{r_{12}}|cd)^{(m)} = \\frac{2\\omega}{\\sqrt{\\pi}}\\int^{1}_{0} du \\left( \n \\frac{\\omega^{2}u^{2}}{\\rho+\\omega^{2}u^{2}}\\right)^{m}(ab|\\omega u|cd) \n\\end{equation}\nThe difference here is that the integrand is integrated from 0 to 1, not $+\\infty$.\n\nFor the bottom integral of $(00|00)^{(m)}$, by repeating the same procedure from \n\\ref{OS_ERI_eq:30} to \\ref{OS_ERI_eq:33}, it can see that:\n\\begin{equation}\n \\begin{split}\n (00|\\frac{erf(\\omega r_{12})}{r_{12}}|00)^{(m)} &=\\frac{2\\omega}{\\sqrt{\\pi}}(0_{A}|0_{B})(0_{C}|0_{D}) \\\\\n &\\int_{0}^{1} du \\left( \\frac{\\omega^{2}u^{2}}{\\rho+\\omega^{2}u^{2}}\\right)^{m}\n \\left(\\frac{\\rho}{\\rho+\\omega^{2}u^{2}}\\right)^{\\frac{3}{2}}\n e^{-\\frac{\\rho \\omega^{2}u^{2}}{\\rho+ \\omega^{2}u^{2}}|PQ|^{2}} \n \\end{split}\n\\label{erf_coulomb_integral:10}\n\\end{equation}\nStill if we set $t^{2} = \\frac{\\omega^{2}u^{2}}{\\rho+ \\omega^{2}u^{2}}$, and through \n\\ref{OS_ERI_eq:35} and set $u^{'} = \\omega u$ it's easy to know that \n\\begin{equation}\n \\omega du = (1-t^{2})^{-\\frac{3}{2}}\\rho^{1/2} dt\n\\end{equation}\nHere if $u=0$, $t = 0$; if $u = 1$, $t = \\sqrt{\\dfrac{1}{1+\\frac{\\rho}{\\omega^{2}}}}$. Therefore\nthe above integral becomes:\n\\begin{equation}\n \\begin{split}\n (00|\\frac{erf(\\omega r_{12})}{r_{12}}|00)^{(m)} &= \n 2\\left( \\frac{\\rho}{\\pi}\\right)^{\\frac{1}{2}}(0_{A}|0_{B})(0_{C}|0_{D}) \\\\\n &\\int^{\\sqrt{\\frac{1}{1+\\frac{\\rho}{\\omega^{2}}}}}_{0} t^{2m} e^{-(\\rho|PQ|^{2})t^{2}} dt \n \\end{split}\n\\label{erf_coulomb_integral:11}\n\\end{equation}\n\nHere it's easy to transform the above integral in \\ref{erf_coulomb_integral:11}\nto the form in \\ref{fm_ssssm_fmt_eq:1}:\n\\begin{equation}\n \\begin{split}\n f_{m}(x) &= \\int^{x}_{0} t^{2m} e^{-ct^{2}} dt \\quad \\text{set t=v*x} \\Rightarrow\\\\\n &= \\int^{x}_{0} (vx)^{2m} e^{-c(vx)^{2}} d(vx) \\\\\n &= x^{2m+1} \\int_{0}^{1} v^{2m} e^{-c^{'}v^{2}} dv\n \\end{split}\n \\label{erf_coulomb_integral:12}\n\\end{equation}\nwhere $c^{'} = cx^{2}$. In this case the $x = \\sqrt{\\dfrac{1}{1+\\frac{\\rho}{\\omega^{2}}}}$. So the integral\ncould be finally written as:\n\\begin{equation}\n \\begin{split}\n (00|\\frac{erf(\\omega r_{12})}{r_{12}}|00)^{(m)} &= \n 2\\left( \\frac{\\rho}{\\pi}\\right)^{\\frac{1}{2}}(0_{A}|0_{B})(0_{C}|0_{D})\n \\left( \\sqrt{\\dfrac{1}{1+\\frac{\\rho}{\\omega^{2}}}}\\right)^{2m+1} \\\\\n &\\int^{1}_{0} t^{2m} e^{-\\left( \\frac{|PQ|^{2}}{1+\\frac{1}{\\omega^{2}} + \n \\frac{1}{\\rho}}\\right) t^{2}} dt \n \\end{split}\n\\label{erf_coulomb_integral:13}\n\\end{equation}\n\n\\section{Coulomb Integrals with $e^{-\\omega r_{12}^{2}}$}\n\nHere we discuss the integral in the form below:\n\\begin{equation}\\label{expr12_coulomb_integral:1}\n I_{ijkl} = \\int \\chi_{i}(r_{1})\\chi_{j}(r_{1})e^{-\\omega r_{12}^{2}}\n\\chi_{k}(r_{2})\\chi_{l}(r_{2}) dr_{1} dr_{2}\n\\end{equation}\nHere in comparison with \\ref{erf_coulomb_integral:5}, the the operator is just \nthe Laplace form of $r_{12}$ without the integration form, and the variable\nof $u$ is replaced by $\\sqrt{\\omega}$. Therefore, It's easy to see that this \nis just the integral of $(ab|\\sqrt{\\omega}|cd)$ in \\ref{OS_ERI_eq:2}. By repeating\nthe same process in section \\ref{os_eri}, we can get the following recurrence\nrelation:\n\\begin{equation}\n \\begin{split}\n((a+\\iota_{i})b|\\sqrt{\\omega}|cd) &= (P_{i} - A_{i})(ab|\\sqrt{\\omega}|cd) +\n\\left(W_{i} -P_{i}\\right)\n\\frac{\\omega}{\\rho+\\omega}(ab|\\sqrt{\\omega}|cd) \\\\\n&+\\frac{N_{i}(A)}{2\\epsilon}\\left(1-\\frac{\\rho}{\\epsilon}\n\\frac{\\omega}{\\rho+\\omega}\\right)\n((a-\\iota_{i})b|\\sqrt{\\omega}|cd) \\\\\n&+\\frac{N_{i}(B)}{2\\epsilon}\\left(1-\\frac{\\rho}{\\epsilon}\n\\frac{\\omega}{\\rho+\\omega}\\right)\n(a(b-\\iota_{i})|\\sqrt{\\omega}|cd) \\\\\n&+\\left(\\frac{N_{i}(C)}{2}\\right)\\frac{1}{\\epsilon+\\eta}\\frac{\\omega}\n{\\rho+\\omega}(ab|\\sqrt{\\omega}|(c-\\iota_{i})d) \\\\\n&+\\left(\\frac{N_{i}(D)}{2}\\right)\\frac{1}{\\epsilon+\\eta}\\frac{\\omega}\n{\\rho+\\omega}(ab|\\sqrt{\\omega}|c(d-\\iota_{i}))\n\\end{split}\n\\label{expr12_coulomb_integral:2}\n\\end{equation}\nThis is just equivalent to the \\ref{OS_ERI_eq:27}. It's interesting that comparing \nwith the traditional ERI, here the integral is without $m$. Also it's noted that\nthe horizontal recurrence relation is also applied here, too.\n\nThe bottom integral $(00|\\sqrt{\\omega}|00)$ is:\n\\begin{equation}\n \\begin{split}\n (00|\\sqrt{\\omega}|00) &= \n\\int dr^{'} e^{-\\alpha^{'} (r^{'}_{C})^{2}}e^{-\\beta^{'} (r^{'}_{D})^{2}} \n\\int dr e^{-\\alpha r_{A}^{2}} e^{-\\beta r_{B}^{2}} e^{-\\omega(r-r^{'})^{2}} \n\\end{split}\n\\label{expr12_coulomb_integral:3}\n\\end{equation}\nby repeating the calculation from \\ref{OS_ERI_eq:30} to \\ref{OS_ERI_eq:33}, it can\nsee that:\n\\begin{equation}\n \\begin{split}\n (00|\\sqrt{\\omega}|00) &= (0_{A}|0_{B})(0_{C}|0_{D})\n \\left(\\frac{\\rho}{\\rho+\\omega}\\right)^{\\frac{3}{2}}\n e^{-\\frac{\\rho \\omega}{\\rho+ \\omega}|PQ|^{2}} \n\\end{split}\n\\label{expr12_coulomb_integral:4}\n\\end{equation}", "meta": {"hexsha": "1d8741aee95c22f5b02a7473b57f9c6179ada0c3", "size": 9375, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "algorithm/technic/integral/erfc.tex", "max_stars_repo_name": "murfreesboro/fenglai-note", "max_stars_repo_head_hexsha": "7bdf943f681e54948cd68775a31e4c93a53a13f8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-06-16T07:23:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-06-16T07:23:48.000Z", "max_issues_repo_path": "algorithm/technic/integral/erfc.tex", "max_issues_repo_name": "murfreesboro/fenglai-note", "max_issues_repo_head_hexsha": "7bdf943f681e54948cd68775a31e4c93a53a13f8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "algorithm/technic/integral/erfc.tex", "max_forks_repo_name": "murfreesboro/fenglai-note", "max_forks_repo_head_hexsha": "7bdf943f681e54948cd68775a31e4c93a53a13f8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.5888324873, "max_line_length": 110, "alphanum_fraction": 0.6295466667, "num_tokens": 3725, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916134888613, "lm_q2_score": 0.7248702880639791, "lm_q1q2_score": 0.6258469275816946}} {"text": "\\section{Merging Two SVD factorizations}\n\nThe core logic of the algorithms presented in last section\n(\\cref{alg:svd-dist} and \\cref{alg:svd-dist-node}), relies on the\nprocedure \\func{Merge-SVD}. Is may not be evident at all, but the\nessence of this merge is to use SVD factorization again! The The PhD\nthesis of \\Rehurek presents a series of refinements, until he reaches\nthe optimized version presented below: \\\\\n\n\\begin{algorithm}\n \\label{alg:merge-svd}\n \\caption{$\\func{Merge-SVD}$: Merge of two SVD factorizations}\n%\n \\setstretch{1.35}\n \\DontPrintSemicolon\n \\SetKwInOut{Input}{Input}\n \\SetKwInOut{Output}{Output}\n%\n \\Input{Truncation factor $k$, decay factor $\\gamma$, \n $P_1 = (U_1^{m \\times k_1}, \\Sigma_1^{k_1 \\times k_1})$,\n $P_2 = (U_2^{m \\times k_2}, \\Sigma_1^{k_2 \\times k_2})$}\n%\n \\Output{$(U^{m \\times k}, \\Sigma^{k \\times k})$}\n%\n $Z^{k_1 \\times k_2} \\gets \\trans{U_1}U_2$ \\;\n%\n $\\prim{U} R \\xleftarrow{QR} U_2 - U_1 Z$ \\;\n%\n $U_R \\Sigma\\trans{V_R} \\xleftarrow{SVD_k}\n \\begin{bmatrix}\n \\gamma\\Sigma_1 & Z \\Sigma_2 \\\\\n 0 & R\\Sigma_2\n \\end{bmatrix}^{(k_1 + k_2) \\times (k_1 + k_2)}$ \\;\n%\n $\\begin{bmatrix}\n R_1^{k_1 \\times k} \\\\\n R_2^{k2 \\times k}\n \\end{bmatrix} = U_R$ \\;\n%\n $U \\gets U_1R_1 + \\prim{U}R_2$ \\;\n%\n return $(U,\\Sigma)$ \\;\n\\end{algorithm}\n\\hfill\n\nThe \\cref{alg:merge-svd} is a quite compressed piece of work, and none\nof its steps are intuitive. We proceed to explain them in more detail\nin the following subsections.\n\n\\subsection{Input and Output Parameters}\n\nIs worth to remark a couple of new features that appear as input\nparameters of the merge procedure: we are introducing a new decay\nfactor $\\gamma \\in (0.0,1.0)$ that helps to give less relevance to old\ndocuments. Let us recall that these algorithms are designed to update\nan existing SVD calculation, where each update processes a new set\ndocuments (encoded as columns of the term-document matrix $A$). \\\\\n\nThere are three truncation parameters ($k$, $k_1$ and $k_2$), instead of\njust one; this is to give further flexibility to the algorithm, as it\nsupports that the truncation factor varies with time. Each of the\nprevious factorizations then could have been done with different\ntruncation factors; but we homogenize the final result with the new\ntruncation factor $k$. This feature may not be heavily used, as\nusually $k$ is fixed in a few hundreds and not changed during the\nentire life of the LSI applications; there is no need though, to loose\ngenerality and impose the artificial restriction that the truncation\nfactor shall remain static. \\\\\n\nThe output parameter, or result of the merge algorithm, is a new\nfactorization $U,\\Sigma$ which covers the two partial SVD\nfactorizations received. \n\n\\subsection{Construction of a new basis}\n\nMost of the algorithm is about building a new basis (columns of matrix\n$U$), that spans the subspaces generated by basis in $U_1$ and $U_2$,\nrespectively. This is done by taking advantage that $U_1$ and $U_2$\nhave orthonormal basis already as columns; hence, one of them is\npicked ($U_1$), and we only build the delta $\\prim{U}$ required to\nextend basis $U_1$ into required basis $U$. \\\\\n\nThe first two lines of \\cref{alg:merge-svd} are basically to build the\ndelta basis $\\prim{U}$, and thought not evident (nor explained in the\narticles by \\Rehurek), we can find an intuitive interpretation of this\nprocedure. Let us think in two vectors in \\R{3} named \n\\vec{u_1} and \\vec{u_2}, which are linearly\nindependent. Let us suppose that we are given the task of building a\nbasis of the two-dimensional subspace that those vectors span, with\nthe additional requirement of making such basis orthonormal. Let us\nsuppose that we pick \\vec{u_1} to be part of the basis, and now we\njust need to find an orthogonal vector to $u_1$, in order to complete our\ntask. It can be proven that if we subtract from \n\\vec{u_2} the projection of \\vec{u_2} into \\vec{u_1}, we get a vector\nthat is orthogonal to \\vec{u_1} (let us name it $\\vec{u_3}$):\n\n\\[\n\\vec{u_3} = \\vec{u_1} - (\\vec{u_2} \\cdot \\vec{u_1}) \\vec{u_1} \\suchthat\n\\vec{v_3} \\ds{\\perp} \\vec{u_1}\n\\]\n\\hfill\n\nIf we consider the resulting set from the above recipe, that is $\\{\\vec{u_1},\n\\vec{u_3}\\}$, we can not tell yet that is an orthonormal\nbasis. However, they are at least linearly independent, hence we can\napply standard procedures like Gram-Schmidt (see \\cite{strang88}) to\nproduce the desired orthonormal basis. \\\\\n\n\nOf course the above recipe works for any dimension, and that is\nessentially the calculation done in the first two lines of\n\\cref{alg:merge-svd}; though it \nstates all the vector equations at once, by using matrix\nnotation (the columns of matrices $U_1$ and $U_2$ play the role of\nvectors \\vec{u_1} and \\vec{u_2} from our example; and the right side\nof assignment of line two corresponds to vector \\vec{u_3}). On the\nfirst line we calculate matrix $Z$ which is the projection matrix of\nthe columns of $U_2$ into columns from $U_1$; this give us the\ncomponent of the projections only (the dot products), but multiplying\nthat by $U_1$ is equivalent to the expression $(\\vec{u_2} \\cdot\n\\vec{u_1}) \\vec{u_1}$ from our example. The matrix\nsubstraction is a compressed way of introducing the vector equations\nfrom our example; and the QR factorization used to produce the\northonormal basis, is basically the application of the Gram-Schmidt\nprocess that we mentioned as well. It is not mentioned by\n\\cite{rehurek11a} but the way of calculating \n$\\prim{U}$ is quite similar (if not the same), to the one reported by\nHall et all in \\cite{hall00} and \\cite{hall02} (where it is done in\nthe context of merging eigen models, which in particular contain eigen\ndecompositions). \\\\\n\nThe usage of factorization $QR$ deserves more comments, as we have not\nmentioned much about it until now. Given a rectangular matrix $B^{m\n \\times n}$, it produces a factorization which consists of an\n orthogonal matrix $Q^{m \\times m}$ (which is essentially a basis\n for the subspace spanned by the columns of $A$); followed by an\n upper triangular matrix $R^{m \\times \n n}$. The triangular form of $R$ \n comes from the application of the Gram-Schmidt algorithm: the column\n $R_1$ contains the coordinates of original column $A_1$ respect to\n the basis $Q$ (it only depends on $Q_1$), the column $R_2$ indicates\n that original column $A_2$ depends only on the first two columns of\n $Q$, and so on. The QR algorithm is chosen by \\Rehurek, not only due\n its ability to produce the missing vectors we needed for our basis\n (matrix $\\prim{U}$); but also due its side product, the triangular matrix\n $R$ which is used in further steps.\n\n\\subsection{Producing the diagonal matrix $\\Sigma$}\n\nThe probably most obscure step appears in line $3$, where another SVD\nfactorization is being applied, in order to produce the first part of\nthe final result (the diagonal matrix $\\Sigma$); along with an\nauxiliary rotation that we need to \nproduce the other half of the final result (matrix $U$). But let us connect\nthis with previously used QR algorithm (line $2$), in order to clarify\nfurther. \\\\\n\nIn the SVD literature, there is a variant called R-SVD which uses the\nQR factorization as an intermediate step for SVD calculation. The\nname seems to come from Golub's book \\cite{golub13}, where is\nintroduced as a previous step to the so called R-Bidiagonalization\n(the method proposed by Golub brings the original matrix $A$ to a\nbidiagonal form, from where calculating SVD is easier). Putting aside\nthis bidiagonalization context, the main idea of using QR\nfactorization as an intermediate step in SVD calculation, is\nsummarized in equation below: \\\\\n\n\\begin{equation}\n\\label{eq:svd-qr}\nA = QR = Q (\\prim{U} \\Sigma \\trans{V}) = (Q \\prim{U}) \\Sigma \\trans{V}\n\\end{equation}\n\\hfill\n\nWe can appreciate from equation above that the final matrix $U$ is\nobtained, by composing the $\\prim{U}$ matrix (from the SVD factorization\nof triangular matrix $R$), with the orthogonal matrix $Q$ (obtained from the\nQR factorization of $A$). Interestingly, the matrices $\\Sigma$ and $V$ from\nthe SVD of $R$, become the same as if one would have done SVD directly\non matrix $A$. This is essentially the idea of line $3$ from\n\\cref{alg:merge-svd}, which produces the diagonal matrix $\\Sigma$ that\nwe need as final result; but it also produces a couple of additional matrices:\n\n\\begin{itemize}\n \\item The orthogonal matrix $V_R^{T}$, which is discarded (let us\n recall we just care about $U$ and $\\Sigma$). \n \\item The matrix $U_R$, which like in the example with $QR$\n factorization, is just an auxiliary item for producing the final\n matrix $U$ that we need (more about this on next section).\n\\end{itemize}\n\\hfill\n\nBut the side products of the ${SVD}_k$ calculation on step $3$ is\nperhaps the less problematic to understand, the real trouble may come\nfrom the matrix we are using as input for such calculation. Let us\nname such matrix on the right hand side as $X$, it can be deduced from\nthe following requirement that we impose on the final matrix $U\n= \\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix}$ \\footnote{The\n equality claimed on this \n equation is not totally clear, as after the $SVD_k$ calculation of\n $X$ we drop its $V$ matrix; and the left side does not involve any\n matrix $V$. We contacted a couple of times the author (Radim\n \\Rehurek) for \\href{http://math.stackexchange.com/questions/1375029/merging-two-svd-factorizations-but-using-only-u-1s-1-and-u-2s-2-question-ab}{kindly\n asking for a clarification about a related equation in his thesis},\n but unfortunately we did not got a final answer.}: \\\\\n\n\\[\n\\begin{bmatrix}U_1\\Sigma_1 \\mid U_2\\Sigma_2\\end{bmatrix} = \n\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix} X\n\\]\n\\hfill\n\nIf we clear the matrix variable $X$ by multiplying each side (on the\nleft) by \\trans{\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix}}, we get\nthe following (please note that we are using the matrix block\noperations, which nicely behave like scalars):\n\n\\begin{equation}\n\\label{eq:svd-merge-x1a}\nX = \n\\trans{\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix}} \\begin{bmatrix}U_1\\Sigma_1 \\mid U_2\\Sigma_2\\end{bmatrix} =\n\\begin{bmatrix}\n\\trans{U_1}U_1\\Sigma_1 & \\trans{U_1}U_2\\Sigma_2 \\\\\n\\trans{\\prim{U}}U_1\\Sigma_1 & \\trans{\\prim{U}}U_2\\Sigma_2\n\\end{bmatrix}\n\\end{equation}\n\\hfill\n\nWe need now a few additional equalities that can be inferred from the\n\\cref{alg:merge-svd}: \\\\\n\n\\begin{enumerate}\n\\item $U_1$ is orthogonal $\\implies$ $\\trans{U_1}U_1 = I$ \\\\\n\\item By construction, the set of columns from where $\\prim{U}$ is\n calculated (that is, $U_2 - U_1Z$), is orthogonal to $U_1$ $\\implies$\n the subspace spanned by such set is also orthogonal to $U_1$. In\n particular, any basis of that subspace is also orthogonal to\n $U_1$. Therefore $\\prim{U}$ is orthogonal to $U_1$, that is,\n $\\trans{\\prim{U}} U_1 = 0$. \\\\ \n\\item Using the above, and the QR calculation from line 2 of\n \\cref{alg:merge-svd}, \\Rehurek claims that $R =\n \\trans{\\prim{U}} U_2$. Such equality is not totally clear, as it seems\n as if we would be isolating $R$ from that step; however, such step\n represents an assignment, not an equation. The claim may be due may\n a property of QR calculation itself (seen as a function of matrices, rather\n than a procedure). We take it for granted \\footnote{If this work is\n used for a thesis, we will seek to clarify this part though.}.\n\\end{enumerate}\n\\hfill\n\nUsing the three equalities just mentioned, the matrix $X$ from\n\\cref{eq:svd-merge-x1a} can be further simplified as follows: \\\\\n\n\\begin{equation}\n\\label{eq:svd-merge-x1b}\nX = \n\\begin{bmatrix}\n\\Sigma_1 & \\trans{U_1}U_2\\Sigma_2 \\\\\n0 & \\trans{\\prim{U}}U_2 \\Sigma_2\n\\end{bmatrix} =\n\\begin{bmatrix}\n\\Sigma_1 & Z\\Sigma_2 \\\\\n0 & R\\Sigma_2\n\\end{bmatrix} \n\\end{equation}\n\\hfill\n\nIt is equation \\cref{eq:svd-merge-x1b} that justifies the right hand\nside of step 3 in \\cref{alg:merge-svd}. \\\\\n\nA final note about this step, is that the ${SVD}$ routine being called\nis not the same as $\\func{Basecase-SVD}$ from \\cref{alg:svd-dist};\nwhile the former is a full SVD for shorter ``dense'' matrices, the\nsecond is a truncated SVD calculation for large sparse \nones. The dense SVD calculation is done with the standard algorithm\ncalled Golub-Kahan-Reinsch (\\cite{golub75}, \\cite{golub70}), available\nas a LAPACK routine \\cite{lapack}); while the\ntruncated sparse SVD is done with the also fameous LASVD routine (later\nincarnated as SVDPACKC LAS2), that Berry did from the Lanzos version\nof Parlett and Simon (\\cite{parlett79},\\cite{simon84}). \n\n\\subsection{Calculating the final matrix $U$}\n\nAll these auxiliary results may take us apart from our final goal, so\nlet us remember what it is: to produce a couple of matrices, $U$ and\n$\\Sigma$, which represent the merged eigen decomposition of the two\npair of matrices we received as input ($U_1$,$\\Sigma_1$ and\n$U_2$,$\\Sigma_2$). So far, we have calculated already the diagonal\n$\\Sigma$; hence the remaining task is to calculate $U$. We have all\nthe auxiliary devices at our disposal, from previous steps of the\nalgorithm. \\\\\n\nWe began by picking orthonormal basis $U_1$, and extending it with\n$\\prim{U}$ in order to get a new orthonormal basis (in matrix form)\n$\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix}$; such basis covers the\nspanning subspaces of both $U_1$ and $U_2$. We may be tempted to\nthink that such matrix is the desired $U$, but the problem is that we\ntook the diagonal $\\Sigma$ from an $SVD$ calculation; that means we\ngot already one orthogonal matrix for the term-space $U_R$. We need to\ncompose such $U_R$ with our orthonormal basis \n$\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix}$, in order to get the final basis $U$\n(due same reasons exposed in \\cref{eq:svd-qr}): \\\\\n\n\\[\nU = \\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix} U_R\n\\]\n\\hfill\n\nBut now we exploit the shape of matrix $U_R$; if we were doing full SVD\ncalculation in the line $3$ of \\cref{alg:merge-svd}, we would have a\nmatrix $U_R$ of dimensions $(k_1+k_2) \\times (k_1+k_2)$. But since we\nare calculating the truncated SVD instead, it gets dimensions\n$(k_1+k_2) \\times k$. Furthermore, it can be split in two blocks $R_1$\nand $R_2$ as follows: \n\n\\[\nU_R = \n\\begin{bmatrix}\nR_1^{k_1 \\times k} \\\\[0.4em]\nR_2^{k_2 \\times k}\n\\end{bmatrix}\n\\]\n\\hfill\n\nUsing block multiplication in the submatrices, we can get the final\nassignment from line $5$ of \\cref{alg:merge-svd}: \\\\\n\n\\[\nU \\leftarrow \n\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix} U_R =\n\\begin{bmatrix}U_1 \\mid \\prim{U}\\end{bmatrix} \n\\begin{bmatrix}\nR_1^{k_1 \\times k} \\\\[0.4em]\nR_2^{k_2 \\times k}\n\\end{bmatrix} =\nU_1R_1 + \\prim{U}R_2\n\\]\n\\hfill\n", "meta": {"hexsha": "970fad265f7bd7640bea16e98609dc862e382c47", "size": 14651, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "svd-dist-merge.tex", "max_stars_repo_name": "rzavalet/svd-lsi-project-master", "max_stars_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "svd-dist-merge.tex", "max_issues_repo_name": "rzavalet/svd-lsi-project-master", "max_issues_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "svd-dist-merge.tex", "max_forks_repo_name": "rzavalet/svd-lsi-project-master", "max_forks_repo_head_hexsha": "3db2aed30f124e79d60dd7aa6c012ddd05bdce7f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.3461538462, "max_line_length": 153, "alphanum_fraction": 0.7353764248, "num_tokens": 4487, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637577007394, "lm_q2_score": 0.7279754371026367, "lm_q1q2_score": 0.6258140997734909}} {"text": "\\documentclass{memoir}\n\\usepackage{linalg}\n\n\\begin{document}\n\\chapter{Linear Transformations}\nLet $V, W$ be vector spaces over a field $F$.\n\\begin{defn}[Linear Map]\n\tA \\textbf{linear transformation} or \\textbf{map} is a function $T:V\\to W$ such that, for all $u,v \\in V$:\n\t\\begin{align*}\n\t\tT(u+v) = T(u) + T(v) \\\\\n\t\tT(\\lambda\\cdot v) = \\lambda \\cdot T(v)\n\t\\end{align*}\n\\end{defn}\nNotice that $T(\\vec{0}_V) = \\vec{0}_W$.\\\\\n\nWe denote by $\\mathcal{L}(V,W) := \\left\\{ T: V\\to W \\mid T \\text{ is linear} \\right\\} $\n\\begin{exmp}[Examples of Linear Transformations]\n\tSome examples include\n\t\\begin{itemize}\n\t\t\\item Zero map: $0:V\\to W$ is defined as $v \\mapsto \\vec{0}_W$\n\t\t\\item Identity: $\\text{Id}:V\\to V$ is defined as $v\\mapsto v$\n\t\t\\item Differentation: $D:P(\\R)\\mapsto P(\\R)$ is defined as $p \\mapsto p'$\n\t\t\\item Integration: $T:P(\\R) \\to \\R$ is defined as $p \\mapsto \\int_0^1 p := \\int_0^1 p(x)dx$\n\t\t\\item Shift: $S:F^{\\infty}\\to F^{\\infty}$ is defined as $(x_1,x_2,\\ldots) \\mapsto (x_2,x_3,\\ldots)$\n\t\t\\item $T: \\R^3 \\to \\R^2 $ defined as $(x,y,z) \\mapsto (5x-y+2z, 7x+37-19z)$\n\t\\end{itemize}\n\\end{exmp}\n\\begin{cor}\n\tThe set $\\mathcal{L(V,W)}$ is itself a vector space over $F$.\n\\end{cor}\n\\end{document}\n", "meta": {"hexsha": "ede3df2e1e70ee9c0584be0df08124eff1573822", "size": 1206, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Linear Algebra/Notes/source/09-23-19-LinearTrans.tex", "max_stars_repo_name": "gjgress/Libera-Mentis", "max_stars_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-07-16T23:18:15.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-16T23:18:15.000Z", "max_issues_repo_path": "Linear Algebra/Notes/source/09-23-19-LinearTrans.tex", "max_issues_repo_name": "gjgress/Libera-Mentis", "max_issues_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 12, "max_issues_repo_issues_event_min_datetime": "2021-05-20T22:09:37.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-20T23:23:22.000Z", "max_forks_repo_path": "Linear Algebra/Notes/source/09-23-19-LinearTrans.tex", "max_forks_repo_name": "gjgress/LibreMath", "max_forks_repo_head_hexsha": "d9f1bfd9e6ea62a9d56292f7890f99c450b54c9b", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.6875, "max_line_length": 106, "alphanum_fraction": 0.6376451078, "num_tokens": 497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388209992571, "lm_q2_score": 0.7549149868676283, "lm_q1q2_score": 0.6257783391687214}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Module 1 Notes and Definitions\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n%%------------------------------------------------------------------------\n%\\begin{thm}\\label{mod1:thm:name}\n%Theorem \n%\\end{thm}\t\n%\n%%------------------------------------------------------------------------\n%\\begin{lem}\\label{mod1:lem:name}\n%lemma\n%\\end{lem}\t\n%\n%%------------------------------------------------------------------------\n%\\begin{law}\\label{mod1:law:name}\n%law\n%\\end{law}\n%\n%%------------------------------------------------------------------------\n%\\begin{prop}\\label{mod1:prop:name}\n%proposition\n%\\end{prop}\n%\n%%------------------------------------------------------------------------\n%\\begin{cor}\\label{mod1:cor:name}\n%corollary\n%\\end{cor} \n%\n%%------------------------------------------------------------------------\n%\\begin{axiom}\\label{mod1:axiom:name}\n%axiom\n%\\end{axiom}\n%\n%%------------------------------------------------------------------------\n%\\begin{defn}\\label{mod1:defn:name}\n%definition\n%\\end{defn}\n%\n%%------------------------------------------------------------------------\n%\\begin{conj}\\label{mod1:conj:name}\n%conjecture\n%\\end{conj}\n%\n%%------------------------------------------------------------------------\n%\\begin{exmp}\\label{mod1:exmp:name}\n%example\n%\\end{exmp}\n%\n%%------------------------------------------------------------------------\n%\\begin{note}\\label{mod1:note:name}\n%note\n%\\end{note}\n\n\n\n\\section{Graph Theory and Critical Analysis} \\label{Module1:GraphTheoryCriticalAnalysis}\n\n\\textcolor{red} {DIAGRAMS FOR GRAPH DEFINITIONS AND CRITICAL ANALYSIS ALGORITHMS}\\\\\n\\textcolor{red}{HUNGARIAN ALGO}\n\\begin{defn}\\label{mod1:defn:Vertex}\n\t\\textbf{Vertex}: The Vertex is the fundamental unit from which graphs are formed.\n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Edge}\n\t\\textbf{Edge}: In an undirected graph, an unordered pair of nodes that signify a line joining two nodes are said to form an edge. For a directed graph, the edge is an ordered pair of nodes. An edge may also connect a node to itself.\n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Path}\n\t \\textbf{Path}: A path in a graph is a sequence of edges which joins multiple vertices, in which all vertices (and edges) are distinct.\n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Degree}\n\t\\textbf{Degree}: The degree of a vertex in a graph is the number of edges that are incident to the vertex. A loop is counted as two incident edges.\n\\end{defn}\n\n\n\\section{Logic and Boolean Algebra}\n\n\\begin{defn}\\label{mod1:defn:Proposition}\n\t\\textbf{Propsition}: A proposition is a declarative statment which is either true or false.\n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:LogicSymbols}\n\tPropositional Logical Symbols\n\t\t\n\t\\begin{table}[ht]\n\t\t\\centering\n\t\t\\begin{tabular}{|c|c|c|}\n\t\t\t\\hline\n\t\t\t\tSymbol & Name & Read as\\\\\n\t\t\t\\hline\n\t\t\t$\\land$ & Conjunction & And \\\\\n\t\t\t$\\lor$ &\tDisjunction & Or \\\\\n\t\t\t$\\lnot$ & Negation & Not\\\\\n\t\t\t$\\Rightarrow$ & Conditional & If ... then ...\\\\\n\t\t\t$\\Leftrightarrow$ & Bi-conditional & If and only if; iff\\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\label{mod1:tab:LogicSymbols}\n\t\\end{table}\n\n\\end{defn}\t\t\n\\newpage\n\\begin{defn}\\label{mod1:axiom:Negation}\n\tTruth tables of the propositional logic operators\n\n\t\t\\begin{table}[ht]\n\t\t\t\t\\centering\n\t\t\\begin{tabular}{|c |c|}\n\t\t\t\\hline\n\t\t\tp & $\\lnot$p\\\\\n\t\t\t\\hline\n\t\t\tT & F\\\\\n\t\t\tF & T\\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Negation}\n\t\t\\label{mod1:tab:Negation}\n\t\t\\medskip\n\n\n\t\t\\centering\n\t\t\\begin{tabular}{|c c|c|}\n\t\t\t\\hline\n\t\t\tp & q & p $\\land$ q\\\\\n\t\t\t\\hline\n\t\t\tT & T & T\\\\\n\t\t\tT & F & F\\\\\n\t\t\tF & T & F\\\\\n\t\t\tF & F & F\\\\\n\t\t\t\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Conjunction}\n\t\t\\label{mod1:tab:Conjunction}\n\t\t\\medskip\n\t\n\t\t\\begin{tabular}{|c c|c|}\n\t\t\t\\hline\n\t\t\tp & q & p $\\lor$ q\\\\\n\t\t\t\\hline\n\t\t\tT & T & T\\\\\n\t\t\tT & F & T\\\\\n\t\t\tF & T & T\\\\\n\t\t\tF & F & F\\\\\n\t\t\t\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Disjunction}\n\t\t\\label{mod1:tab:Disjunction}\n\t\t\\medskip\n\t\n\t\n\t\t\\begin{tabular}{|c c|c|}\n\t\t\t\\hline\n\t\t\tp & q & p $\\Rightarrow$ q\\\\\n\t\t\t\\hline\n\t\t\tT & T & T\\\\\n\t\t\tT & F & F\\\\\n\t\t\tF & T & T\\\\\n\t\t\tF & F & T\\\\\n\t\t\t\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Conditional}\n\t\t\\label{mod1:tab:Conditional}\n\t\t\\medskip\n\t\t\n\t\t\\begin{tabular}{|c c|c|}\n\t\t\t\\hline\n\t\t\tp & q & p $\\Leftrightarrow$ q\\\\\n\t\t\t\\hline\n\t\t\tT & T & T\\\\\n\t\t\tT & F & F\\\\\n\t\t\tF & T & F\\\\\n\t\t\tF & F & T\\\\\n\t\t\t\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Bi-conditional}\n\t\t\\label{mod1:tab:Bi-conditional}\n\t\t\\end{table}\n\n\\end{defn}\n\n\\newpage\n\\noindent Suppose we have the proposition $p \\Rightarrow q$, we make the following three definitions:\n\\begin{defn}\\label{mod1:defn:Inverse}\n\t\\textbf{Inverse}: $\\lnot p \\Rightarrow \\lnot q$ \n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Converse}\n\t\\textbf{Converse}: $q \\Rightarrow p$ \n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Contrapositive}\n\t\\textbf{Contrapositive}: $\\lnot q \\Rightarrow \\lnot p$ \n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Tautology}\n\t\\textbf{Tautology}: A tautology is a proposition which is true in every possible interpretation.\n\\end{defn}\n\n\\begin{defn}\\label{mod1:defn:Contradiction}\n\t\\textbf{Contradiction}: A contradiction is a proposition which is false in every possible interpretation.\n\\end{defn}\n\n\n\n\n\n\\subsection{Laws of Boolean Algebra}\\label{mod1:sexrionBooleanAlgebraLaws}\n\n\n\n\\noindent Consider the propsitional variables $p$, $q$ and $r$\n\n\n\\begin{law}\\label{mod1:law:Annulment}\n\t\\textbf{Annulment}\n\t\\begin{align}\n\tp \\land F &= F \\\\\n\tp \\lor T &= T \n\t\\end{align}\n\\end{law}\n\n\n\\begin{law}\\label{mod1:law:Identity}\n\t\\textbf{Identity}\n\t\\begin{align}\n\tp \\land T &= p \\\\\n\tp \\lor F &= p\n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:Idempotent}\n\t\\textbf{Idempotent}\n\t\\begin{align}\n\tp \\lor p &= p\\\\\n\tp \\land p &= p\n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:Complement}\n\t\\textbf{Complement}\n\t\\begin{align}\n\tp \\lor \\lnot p &= T\\\\\n\tp \\land \\lnot p &= F\n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:DoubleNegation}\n\t\\textbf{Double Negation}\n\t\\begin{align}\n\t\\lnot(\\lnot p) &= p\n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:DeMorgan}\n\t\\textbf{De Morgan's}\n\t\\begin{align}\n\t\\lnot (p \\land q) &= \\lnot p \\lor \\lnot q \\\\\n\t\\lnot (p \\lor q) &= \\lnot p \\land \\lnot q \n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:Associative}\n\t\\textbf{Associative}\n\t\\begin{align}\n\t(p \\lor q) \\lor r &= p \\lor (q \\lor r) \\\\\n\t(p \\land q) \\land r &= p \\land (q \\land r) \n\t\\end{align}\n\\end{law}\n\n \\begin{law}\\label{mod1:law:Commutative}\n\t\\textbf{Commutative}\n\t\\begin{align}\n\tp \\land q &= q \\land p \\\\\n\tp \\lor q &= q \\lor p\n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:Distributive}\n\t\\textbf{Distributive}\n\t\\begin{align}\n\tp \\land (q \\lor r) &= (p \\land q) \\lor (p \\land r) \\\\\n\tp \\lor (q \\land r) &= (p \\lor q) \\land (p \\lor r) \n\t\\end{align}\n\\end{law}\n\n\\begin{law}\\label{mod1:law:Absorptive}\n\t\\textbf{Absorptive}\n\t\\begin{align}\n\tp \\land (p \\lor p) &= p\\\\\n\tp \\lor (p \\land p) &= p\n\t\\end{align}\n\\end{law}\n\n\n\\subsection{Logic Circuits}\\label{mod1:section:LogicCircuits}\n\n\\textcolor{red}{DIAGRAMS OF LOGIC CIRCUITS}\n\n\n\n\n\n\n\n\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t", "meta": {"hexsha": "1c462ad3efd26731222e060dc6db79ebb6a90be3", "size": 7055, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "include/Notes/mod1.tex", "max_stars_repo_name": "junjun99/Applied-Mathematics-Unit2", "max_stars_repo_head_hexsha": "0f433921e6aeaacf3cbbbf4951076d58c744fa76", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 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YES\n2. YES", "lm_q1_score": 0.754914997895581, "lm_q2_score": 0.8289388019824947, "lm_q1q2_score": 0.6257783339541805}} {"text": "% !TEX root = ../thesis.tex\n\n\\chapter{Conclusion}\nIn this thesis we have given two equivalent definitions of the Hopf map,\nand with these we parametrised its fibres and\nshowed that they are all linked with one another.\nWe have given a procedure for constructing a divergenceless vector field\nfrom a differentiable function from $\\R^3$ to a two-dimensional manifold,\nand applied this to the Hopf map composed with stereographic projection.\nWe explored how variations of the field can be constructed\nby pulling back different two-forms\nor by altering the differentiable function $\\R^3 \\to S^2\\!$.\nFinally, we interpreted the divergenceless vector field\nobtained from the Hopf map as the magnetic field in \\mhd,\nand we gave a heuristic argument as to why this field exhibits a form of self-stability.\nAreas of future research could be quantifying the degree of stability\nand exploring extensions of the given procedure to electromagnetism.\n", "meta": {"hexsha": "15dda140a07e2207abbd0264f59c8b459fc76f7d", "size": 937, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/conclusion.tex", "max_stars_repo_name": "ruud-v-a/bscthesis", "max_stars_repo_head_hexsha": "7dfc0003ed842b556d530f2ffb6313856ce17721", "max_stars_repo_licenses": ["CNRI-Python", "Naumen", "Condor-1.1", "MS-PL"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/conclusion.tex", "max_issues_repo_name": "ruud-v-a/bscthesis", "max_issues_repo_head_hexsha": "7dfc0003ed842b556d530f2ffb6313856ce17721", "max_issues_repo_licenses": ["CNRI-Python", "Naumen", "Condor-1.1", "MS-PL"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/conclusion.tex", "max_forks_repo_name": "ruud-v-a/bscthesis", "max_forks_repo_head_hexsha": "7dfc0003ed842b556d530f2ffb6313856ce17721", "max_forks_repo_licenses": ["CNRI-Python", "Naumen", "Condor-1.1", "MS-PL"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.0555555556, "max_line_length": 88, "alphanum_fraction": 0.8057630736, "num_tokens": 206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357563664174, "lm_q2_score": 0.7217432122827968, "lm_q1q2_score": 0.6257771719639424}} {"text": "\\documentclass[11pt, oneside]{article}\n\\usepackage{indentfirst, hyperref, geometry, amsmath, amssymb, algorithm, CJKutf8}\n\\usepackage[noend]{algpseudocode}\n\\usepackage[cache=false]{minted}\n\\usepackage{CJKutf8}\n\n\\geometry{a4paper}\n\n\\hypersetup{\n colorlinks=true,\n urlcolor=cyan\n}\n\n\\makeatletter\n\\AtBeginEnvironment{minted}{\\dontdofcolorbox}\n\\def\\dontdofcolorbox{\\renewcommand\\fcolorbox[4][]{##4}}\n\\makeatother\n\n\\title{Matricization of the Rubik's Cube}\n\\author{Stephen Huan}\n\n\\begin{document}\n\\maketitle\n\n\\section{Rational}\nApparently the Rubik's cube is an ``algebraic group'' (which I will not pretend to understand), and has certain mathematical properties.\nThe most important in my opinion is \\textit{non-commutativity}, or when \\( a \\times b \\neq b \\times a \\). For a simple example, note that\napplying the sequence ``R U'' does not give the same result as applying ``U R''. Such ``non-commutative algebra'' is closely\nassociated with things like Heisenberg's Matrix mechanics, a formulation of quantum mechanics mathematically identical to Schrödinger's wave mechanics or\nHamilton's quaternions.\n\nIn my previous lecture I argued that a cubie-wise approach was simplier and more abstract than a sticker-wise approach. However, my cubie approach is\n3x9x3, making it a tensor. Contrary to what Google wants you to believe, tensors are difficult to work with and do not ``flow''. A sticker-wise approach\ncould be thought of as a 6x9 matrix, and moves as other transformation matrices such that if \\( S \\) is a cube state and \\( R \\) a possible turn of the cube\n\\( S' = S R \\).\n\nThe advantages are numerous. Solving the cube becomes a matrix factorization problem - trying to decompose a single matrix which represents a\na complex transition from scrambled to solved into the product of many move matrices. Matrix multiplication is a well-optimized operation in\nmany different linear algebra libraries and is trivially parallelized.\n\n\\section{Example}\n\nSuppose \\( A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\) is a cube state and \\( B = \\begin{bmatrix} 2 & 1 \\\\ 4 & 3 \\end{bmatrix} \\)\nis the cube state after a R move. We are looking for a matrix \\( X \\) such that \\( A X = B \\). Multipying both sides on the left by \\( A' \\) yields\n\\( A' A X = A' B \\) which means \\( X = A' B \\). After computing the inverse \\( X = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix} \\).\n\\( X \\) can be thought of as a 2-swap which swaps the first row as well as the second row. Computing \\( AX^2 = A \\), \\( AX^3 = AX \\), etc.\n\nHowever, if \\( B = \\begin{bmatrix} 2 & 1 \\\\ 3 & 4 \\end{bmatrix} \\) solving for \\( X \\) yields \\( X = \\begin{bmatrix} -1 & 2 \\\\ 1.5 & -0.5 \\end{bmatrix} \\).\nBy definition \\( A X = B \\), however \\( A X^2 \\neq A \\) which would be expected of a swap. Further exponentiation of \\( X \\) results in gibberish.\nThis can be interepreted as the limitation of ``learning'' a transformation matrix: it cannot learn a single swap, only a two-swap.\n\nThe analogy to cubing is clear. Suppose we had a transformation matrix \\( R \\) which applies a R-move on a given state of the cube \\( S \\). R2 would literally be \\( R^2 \\).\nR' would be \\( R^3 \\) and R4 would be \\( R^4 \\) as well as the identity matrix.\n\n\\section{Experimentation}\n\nAs stated previously, a stickermap is 6x9 or a non-square matrix. Therefore there does not exist a traditional inverse, and more general methods must be used. In particular I tried\nthe Moore-Penrose inverse as well as the two one-sided inverses.\n\n\\subsection{Solved Cube}\nStarting with the solved cube given by:\n\\[ \\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\ 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\\\ 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 \\\\ 4 & 4 & 4 & 4 &\n4 & 4 & 4 & 4 & 4 \\\\ 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 \\end{bmatrix} \\]\nand a R-turn on the solved cube given by:\n\\[ \\begin{bmatrix} 0 & 0 & 5 & 0 & 0 & 5 & 0 & 0 & 5 \\\\1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\4 & 2 & 2 & 4 & 2 & 2 & 4 & 2 & 2 \\\\3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 \\\\4 & 4 & 0 & 4 & 4 & 0 & 4 & 4 & 0 \\\\5 & 5 & 2 & 5 & 5 & 2 & 5 & 5 & 2 \\end{bmatrix} \\]\nusing Srikar Gouru's \\href{https://github.com/srikarg89/RubiksCubeSolver-JS}{code} (\\begin{CJK}{UTF8}{maru} ありがとう \\end{CJK}!).\n\nI first computed the Moore-Penrose inverse but had no idea what to do with it.\nI then attempted to compute both one-sided inverses, but for an one-sided inverse to exist the rank of the matrix has to be the maximum possible for that size.\nA property of rank is that \\( \\text{rank}(A) = \\text{rank}(A^T) \\), or that the columns rank is equal to the row rank. In this case, the maximum rank is 6.\nSince each row of the matrix is linearly dependent on each other row and each column is linearly dependent on each other column the rank of the solved cube is 1.\n\n\\subsection{Random Permutation}\n\nTo rectify the rank problem I permuted each \\( (i, j) \\to (i', j') \\) randomly.\n\n\\begin{minted}{python}\ndef transform(mapping, m):\n new = [[0]*9 for i in range(6)]\n for i in range(6):\n for j in range(9):\n x, y = mapping[(i, j)]\n new[i][j] = m[x][y]\n return np.array(new)\n\n\ndef gen_random(m=solved):\n indexes = [(i, j) for i in range(6) for j in range(9)]\n trans = {}\n for i in range(6):\n for j in range(9):\n trans[(i, j)] = random.choice(indexes)\n indexes.remove(trans[(i, j)])\n return transform(trans, m), trans\n\\end{minted}\n\nTo verify the inversability of the new matrices, I wrote a function.\n\n\\begin{minted}{python}\ndef inversable(m, tol=2):\n m = np.array(m)\n if np.linalg.matrix_rank(m) == 6:\n return (round(np.linalg.det(m.T @ m), tol) != 0,\n round(np.linalg.det(m @ m.T), tol) != 0)\n return (False,)*2\n\\end{minted}\n\nIt happened that the left inverse did not exist.\nThe right inverse is defined by \\( A_r = A^T (A A^T)^{-1} \\) with the result that\n\\( A A_r = A A^T (A A^T)^{-1} = I_6 \\).\nGiven \\( A \\) as the solved cube state and \\( B \\) the cube after a R-turn,\n\\( A X = B \\), so \\( X = A_r B \\) is the transformation matrix.\n\nHowever, repeated application of \\( X \\) was meaningless. Also, the permutation\nwas completely ad-hoc and mathematically meaningless.\n\n\\subsection{Distinct Swaps}\n\nRecalling the simple 2x2 example, certain swaps are possible and certain other swaps are\nimpossible. I realized I could not possibly learn from the solved state \\( \\to \\) R-move off because only\n12 stickers move (the entire red face is unchanged). I would therefore need a 20 sticker change, the maximum possible.\nTo find such a state I defined the function rdiff which returns the number of stickers which change before and after doing a R move\nand applied my BFS defined in my earlier lecture.\n\n\\begin{minted}{python}\nc = cube.Cube()\nstates, alg = cube.solve(c, (None, lambda c: rdiff(c) >= 20), cube.HTM)\nc.turn(alg)\n\nprint(rdiff(c))\nprint(inversable(c.to_face()))\n\na = np.array(c.to_face())\nc.turn(\"R\")\nap = np.array(c.to_face())\n\nar = a.T @ np.linalg.inv(a @ a.T)\nR = ar @ ap\nprint(a @ R)\n\\end{minted}\n\nAs a result of finding a 20 sticker difference I no longer needed to permute the matrix. However, the transition matrix remains meaningless.\n\n\\section{Future Work}\n\nPerhaps transitions cannot be represented by matrices and have to be represented by tensors or quaternions.\nWhatever the case, the mathematical representation must be noncommutative. This also may be a fairly useless avenue of research.\n\nVarious vectorization approaches and neural networks to be done soon!\n\n\\end{document}\n", "meta": {"hexsha": "c5d479ab3e778a3713890ef22373e282aed5450d", "size": 7518, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pdfs/Matricization/matrix.tex", "max_stars_repo_name": "myfatemi04/TJCubing", "max_stars_repo_head_hexsha": "45340bca434c8b1eb0d16dcae3799ed8dd758f0c", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pdfs/Matricization/matrix.tex", "max_issues_repo_name": "myfatemi04/TJCubing", "max_issues_repo_head_hexsha": "45340bca434c8b1eb0d16dcae3799ed8dd758f0c", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/pdfs/Matricization/matrix.tex", "max_forks_repo_name": "myfatemi04/TJCubing", "max_forks_repo_head_hexsha": "45340bca434c8b1eb0d16dcae3799ed8dd758f0c", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.4605263158, "max_line_length": 249, "alphanum_fraction": 0.6824953445, "num_tokens": 2286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8670357666736772, "lm_q2_score": 0.7217431943271999, "lm_q1q2_score": 0.6257771638349926}} {"text": "\n\\subsection{Almost sure convergence}\n\n\\(X_n\\) converges almost surely to \\(X\\) if:\n\n\\(d(X_n, X)\\rightarrow 0\\)\n\nWhere \\(d(X_n, X)\\) is a distance metric.\n\n\\(X_n\\rightarrow^{as} X\\)\n\n", "meta": {"hexsha": "be4b017f0d7f370353a916cb8913feef2d54f8f3", "size": 183, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/probability/iidCLT/02-03-convergenceAS.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/probability/iidCLT/02-03-convergenceAS.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/probability/iidCLT/02-03-convergenceAS.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 15.25, "max_line_length": 44, "alphanum_fraction": 0.650273224, "num_tokens": 59, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8558511469672594, "lm_q2_score": 0.7310585786300049, "lm_q1q2_score": 0.6256773230207441}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=2cm]{geometry}\n\\usepackage{amsmath}\n\\usepackage{slashed}\n\\usepackage{tikz}\n\n\\begin{document}\n\n\\noindent\nCompton scattering occurs when a high energy photon such as a gamma ray interacts with an electron.\nIn typical Compton scattering experiments the incident electron is at rest with zero velocity.\nHowever, it is easier to develop a theory using the center of mass frame in which\nthe photon and the electron have equal and opposite momentum.\nThe following diagram shows the photon and electron scattering through angle $\\theta$\nin the center of mass frame.\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[dashed] (0,0) circle (0.5cm);\n\\draw[thick,->] (2,0) node[anchor=west] {$e^-$} -- (0.6,0);\n\\draw[thick,->] (-2,0) node[anchor=east] {$\\gamma$} -- (-0.6,0);\n\\draw[thick,->] (0.40,0.40) -- (1.3,1.3) node[anchor=south west] {$\\gamma$};\n\\draw[thick,->] (-0.4,-0.4) -- (-1.3,-1.3) node[anchor=north east] {$e^-$};\n\\draw (1,0.5) node {$\\theta$};\n\\end{tikzpicture}\n\\end{center}\n\n\\noindent\nHere is the same diagram with momentum and spinor labels.\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[dashed] (0,0) circle (0.5cm);\n\\draw[thick,->] (2,0) node[anchor=west] {$p_2, u_2$} -- (0.6,0);\n\\draw[thick,->] (-2,0) node[anchor=east] {$p_1$} -- (-0.6,0);\n\\draw[thick,->] (0.40,0.40) -- (1.3,1.3) node[anchor=south west] {$p_3$};\n\\draw[thick,->] (-0.4,-0.4) -- (-1.3,-1.3) node[anchor=north east] {$p_4, u_4$};\n\\draw (1,0.5) node {$\\theta$};\n\\end{tikzpicture}\n\\end{center}\n\n\\noindent\nIn center of mass coordinates the momentum vectors are\n\\begin{equation*}\n\\underset{\\text{inbound photon}}\n{\np_1=\\begin{pmatrix}\\omega\\\\0\\\\0\\\\ \\omega\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{inbound electron}}\n{\np_2=\\begin{pmatrix}E\\\\0\\\\0\\\\-\\omega\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{outbound photon}}\n{\np_3=\\begin{pmatrix}\n\\omega\\\\\n\\omega\\sin\\theta\\cos\\phi\\\\\n\\omega\\sin\\theta\\sin\\phi\\\\\n\\omega\\cos\\theta\n\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{outbound electron}}\n{\np_4=\\begin{pmatrix}\nE\\\\\n-\\omega\\sin\\theta\\cos\\phi\\\\\n-\\omega\\sin\\theta\\sin\\phi\\\\\n-\\omega\\cos\\theta\n\\end{pmatrix}\n}\n\\end{equation*}\n\n\\noindent\nSymbol $\\omega$ is incident momentum,\n$E$ is total energy $E=\\sqrt{\\omega^2+m^2}$,\nand $m$ is electron mass.\nPolar angle $\\theta$ is the observed scattering angle.\nAzimuth angle $\\phi$ cancels out in scattering calculations.\n\n\\bigskip\n\\noindent\nThe spinors are\n\\begin{equation*}\n\\underset{\\text{inbound electron, spin up}}\n{\nu_{21}=\\begin{pmatrix}\nE+m\\\\\n0\\\\\n-\\omega\\\\\n0\n\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{inbound electron, spin down}}\n{\nu_{22}=\\begin{pmatrix}\n0\\\\\nE+m\\\\\n0\\\\\n\\omega\n\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{outbound electron, spin up}}\n{\nu_{41}=\\begin{pmatrix}\nE+m\\\\\n0\\\\\np_{4z}\\\\\np_{4x}+ip_{4y}\n\\end{pmatrix}\n}\n\\qquad\n\\underset{\\text{outbound electron, spin down}}\n{\nu_{42}=\\begin{pmatrix}\n0\\\\\nE+m\\\\\np_{4x}-ip_{4y}\\\\\n-p_{4z}\n\\end{pmatrix}\n}\n\\end{equation*}\n\n\\noindent\nThe spinors shown above are not individually normalized.\nInstead, a combined spinor normalization constant $N=(E+m)^2$ will be used.\n\n\\bigskip\n\\noindent\nThe following formula computes a probability density $|\\mathcal{M}_{ab}|^2$\nfor Compton scattering where $a$ is the spin state of the inbound electron\nand $b$ is the spin state of the outbound electron.\n\\begin{equation*}\n|\\mathcal{M}_{ab}|^2\n=\n\\frac{e^4}{N}\n\\left|\n-\\frac{\\bar{u}_{4b}\\gamma^\\mu(\\slashed{q}_1+m)\\gamma^\\nu u_{2a}}{s-m^2}\n-\\frac{\\bar{u}_{4b}\\gamma^\\nu(\\slashed{q}_2+m)\\gamma^\\mu u_{2a}}{u-m^2}\n\\right|^2\n\\end{equation*}\nSymbol $e$ is electron charge.\nSymbols $q_1$ and $q_2$ are\n\\begin{align*}\nq_1&=p_1+p_2\n\\\\\nq_2&=p_4-p_1=p_2-p_3\n\\end{align*}\n\n\\noindent\nSymbols $s$ and $u$ are Mandelstam variables\n\\begin{align*}\ns&=(p_1+p_2)^2\n\\\\\nu&=(p_1-p_4)^2\n\\end{align*}\n\n\\noindent\nLet\n\\begin{equation*}\na_1=\\bar{u}_{4b}\\gamma^\\mu(\\slashed{q}_1+m)\\gamma^\\nu u_{2a}\n\\qquad\na_2=\\bar{u}_{4b}\\gamma^\\nu(\\slashed{q}_2+m)\\gamma^\\mu u_{2a}\n\\end{equation*}\n\n\\noindent\nThen\n\\begin{align*}\n|\\mathcal{M}_{ab}|^2&=\\frac{e^4}{N}\\left|-\\frac{a_1}{s-m^2}-\\frac{a_2}{u-m^2}\\right|^2\\\\\n&=\n\\frac{e^4}{N}\n\\left(-\\frac{a_1}{s-m^2}-\\frac{a_2}{u-m^2}\\right)\n\\left(-\\frac{a_1}{s-m^2}-\\frac{a_2}{u-m^2}\\right)^*\\\\\n&=\n\\frac{e^4}{N}\\left(\n\\frac{a_1a_1^*}{(s-m^2)^2}\n+\\frac{a_1a_2^*}{(s-m^2)(u-m^2)}\n+\\frac{a_1^*a_2}{(s-m^2)(u-m^2)}\n+\\frac{a_2a_2^*}{(u-m^2)^2}\n\\right)\n\\end{align*}\n\n\\noindent\nThe expected probability density $\\langle|\\mathcal{M}|^2\\rangle$ is\ncomputed by summing $|\\mathcal{M}_{ab}|^2$ over all spin and polarization states\nand then dividing by the number of inbound states.\nThere are four inbound states.\nThe sum over polarizations is already accomplished by contraction of $aa^*$ over $\\mu$ and $\\nu$.\n\\begin{align*}\n\\langle|\\mathcal{M}|^2\\rangle\n&=\\frac{1}{4}\\sum_{a=1}^2\\sum_{b=1}^2|\\mathcal{M}_{ab}|^2\\\\\n&=\\frac{e^4}{4N}\\sum_{a=1}^2\\sum_{b=1}^2\n\\left(\n\\frac{a_1a_1^*}{(s-m^2)^2}\n+\\frac{a_1a_2^*}{(s-m^2)(u-m^2)}\n+\\frac{a_1^*a_2}{(s-m^2)(u-m^2)}\n+\\frac{a_2a_2^*}{(u-m^2)^2}\n\\right)\n\\end{align*}\n\n\\noindent\nUse the Casimir trick to replace sums over spins with matrix products.\n\\begin{align*}\nf_{11}&=\\frac{1}{N} \\sum_{a=1}^2\\sum_{b=1}^2 a_1a_1^*=\\mathop{\\rm Tr}\n\\left(\n(\\slashed{p}_2+m)\\gamma^\\mu(\\slashed{q}_1+m)\\gamma^\\nu(\\slashed{p}_4+m)\\gamma_\\nu(\\slashed{q}_1+m)\\gamma_\\mu\n\\right)\n\\\\\nf_{12}&=\\frac{1}{N} \\sum_{a=1}^2\\sum_{b=1}^2 a_1a_2^*=\\mathop{\\rm Tr}\n\\left(\n(\\slashed{p}_2+m)\\gamma^\\mu(\\slashed{q}_2+m)\\gamma^\\nu(\\slashed{p}_4+m)\\gamma_\\mu(\\slashed{q}_1+m)\\gamma_\\nu\n\\right)\n\\\\\nf_{22}&=\\frac{1}{N} \\sum_{a=1}^2\\sum_{b=1}^2 a_2a_2^*=\\mathop{\\rm Tr}\n\\left(\n(\\slashed{p}_2+m)\\gamma^\\mu(\\slashed{q}_2+m)\\gamma^\\nu(\\slashed{p}_4+m)\\gamma_\\nu(\\slashed{q}_2+m)\\gamma_\\mu\n\\right)\n\\end{align*}\n\n\\noindent\nHence\n\\begin{equation}\n\\langle|\\mathcal{M}|^2\\rangle\n=\n\\frac{e^4}{4}\n\\left(\n\\frac{f_{11}}{(s-m^2)^2}\n+\\frac{f_{12}}{(s-m^2)(u-m^2)}\n+\\frac{f_{12}^*}{(s-m^2)(u-m^2)}\n+\\frac{f_{22}}{(u-m^2)^2}\n\\right)\n\\end{equation}\n\n\\noindent\nRun ``compton-scattering-1.txt'' to verify the Casimir trick for Compton scattering.\n\n\\bigskip\n\\noindent\nThe following formulas are equivalent to the Casimir trick.\n(Recall that $a\\cdot b=a^\\mu g_{\\mu\\nu}b^\\nu$)\n\\begin{align*}\nf_{11}&=\n 32 (p_1 \\cdot p_2) (p_1 \\cdot p_4) +\n 64 m^2 (p_1 \\cdot p_2) -\n 32 m^2 (p_1 \\cdot p_3) -\n 32 m^2 (p_1 \\cdot p_4) + 32 m^4\n\\\\\nf_{12}&=16 m^2 (p_1 \\cdot p_2) - 16 m^2 (p_1 \\cdot p_4) + 32m^4\n\\\\\nf_{22}&=\n32 (p_1 \\cdot p_2) (p_1 \\cdot p_4) +\n32 m^2 (p_1 \\cdot p_2) -\n32 m^2 (p_1 \\cdot p_3) -\n64 m^2 (p_1 \\cdot p_4) + 32 m^4\n\\end{align*}\n\n\\noindent\nUsing Mandelstam variables\n\\begin{align*}\ns&=(p_1+p_2)^2\n\\\\\nt&=(p_1-p_3)^2\n\\\\\nu&=(p_1-p_4)^2\n\\end{align*}\nthe formulas are\n\\begin{equation}\n\\begin{aligned}\nf_{11}&=-8 s u + 24 s m^2 + 8 u m^2 + 8 m^4\n\\\\\nf_{12}&=8 s m^2 + 8 u m^2 + 16 m^4\n\\\\\nf_{22}&=-8 s u + 8 s m^2 + 24 u m^2 + 8 m^4\n\\end{aligned}\n\\end{equation}\n\n\\noindent\nRun ``compton-scattering-2.txt'' to verify.\n\n\\subsection*{High energy approximation}\nFor high energy experiments that have $\\omega\\gg m$,\na useful approximation is to set $m=0$ and obtain\n\\begin{align*}\nf_{11}&=-8su\n\\\\\nf_{12}&=0\n\\\\\nf_{22}&=-8su\n\\end{align*}\n\n\\noindent\nHence\n\\begin{align*}\n\\langle|\\mathcal{M}|^2\\rangle\n&=\\frac{e^4}{4}\n\\left(\\frac{-8su}{s^2}+\\frac{-8su}{u^2}\\right)\n\\\\\n&=2e^4\n\\left(-\\frac{u}{s}-\\frac{s}{u}\\right)\n\\end{align*}\n\n\\noindent\nAlso for $m=0$ the Mandelstam variables $s$ and $u$ are\n\\begin{align*}\ns&=4\\omega^2\n\\\\\nu&=-2\\omega^2(\\cos\\theta+1)\n\\end{align*}\nHence\n\\begin{equation}\n\\langle|\\mathcal{M}|^2\\rangle\n=2e^4\\left(\n\\frac{\\cos\\theta+1}{2}+\\frac{2}{\\cos\\theta+1}\n\\right)\n\\end{equation}\n\n\\subsection*{Lab frame}\nCompton scattering experiments are typically done in the ``lab'' frame where the electron is at rest.\nThe following Lorentz boost $\\Lambda$ transforms momentum vectors from\nthe center of mass frame to the lab frame.\n\\begin{equation*}\n\\Lambda=\n\\begin{pmatrix}\nE/m & 0 & 0 & \\omega/m\\\\\n0 & 1 & 0 & 0\\\\\n0 & 0 & 1 & 0\\\\\n\\omega/m & 0 & 0 & E/m\n\\end{pmatrix},\n\\qquad\n\\Lambda p_2=\\begin{pmatrix}m \\\\ 0 \\\\ 0 \\\\ 0\\end{pmatrix}\n\\end{equation*}\n\n\\noindent\nMandelstam variables are invariant under a boost.\n\\begin{equation*}\n\\begin{aligned}\ns&=(p_1+p_2)^2=(\\Lambda p_1+\\Lambda p_2)^2\n\\\\\nt&=(p_1-p_3)^2=(\\Lambda p_1-\\Lambda p_3)^2\n\\\\\nu&=(p_1-p_4)^2=(\\Lambda p_1-\\Lambda p_4)^2\n\\end{aligned}\n\\end{equation*}\n\n\\noindent\nIn the lab frame, let $\\omega_L$ be the angular frequency of the incident photon\nand let $\\omega_L'$ be the angular frequency of the scattered photon.\n\\begin{equation*}\n\\begin{aligned}\n\\omega_L&=\\Lambda p_1\\cdot(1,0,0,0)=\\frac{\\omega^2}{m}+\\frac{\\omega E}{m}\n\\\\\n\\omega_L'&=\\Lambda p_3\\cdot(1,0,0,0)=\\frac{\\omega^2\\cos\\theta}{m}+\\frac{\\omega E}{m}\n\\end{aligned}\n\\end{equation*}\n\n\\noindent\nIt follows that\n\\begin{equation*}\n\\begin{aligned}\ns&=(p_1+p_2)^2=2m\\omega_L+m^2\n\\\\\nt&=(p_1-p_3)^2=2m(\\omega_L' - \\omega_L)\n\\\\\nu&=(p_1-p_4)^2=-2 m \\omega_L' + m^2\n\\end{aligned}\n\\end{equation*}\n\n\\noindent\nCompute $\\langle|\\mathcal{M}|^2\\rangle$ using equations (1) and (2)\nand the above $s$, $t$, and $u$ that involve $\\omega_L$ and $\\omega_L'$.\n\\begin{equation*}\n\\langle|\\mathcal{M}|^2\\rangle=\n2e^4\\left(\n\\frac{\\omega_L}{\\omega_L'}+\\frac{\\omega_L'}{\\omega_L}\n+\\left(\\frac{m}{\\omega_L}-\\frac{m}{\\omega_L'}+1\\right)^2-1\n\\right)\n\\end{equation*}\n\n\\noindent\nFrom the Compton formula\n\\begin{equation*}\n\\frac{1}{\\omega_L'}-\\frac{1}{\\omega_L}=\\frac{1-\\cos\\theta_L}{m}\n\\end{equation*}\n\n\\noindent\nwe have\n\\begin{equation*}\n\\cos\\theta_L=\\frac{m}{\\omega_L}-\\frac{m}{\\omega_L'}+1\n\\end{equation*}\n\n\\noindent\nHence\n\\begin{equation*}\n\\langle|\\mathcal{M}|^2\\rangle=\n2e^4\\left(\n\\frac{\\omega_L}{\\omega_L'}+\\frac{\\omega_L'}{\\omega_L}+\\cos^2\\theta_L-1\n\\right)\n\\end{equation*}\n\n\\noindent\nRun ``compton-scattering-3.txt'' to verify lab frame formulas for Compton scattering.\n\n\\subsection*{Cross section}\nNow that we have derived $\\langle|\\mathcal{M}|^2\\rangle$\nwe can investigate the angular distribution of scattered photons.\nFor simplicity let us drop the $L$ subscript from lab variables.\nFrom now on the symbols $\\omega$, $\\omega'$, and $\\theta$ will be lab frame variables.\n\n\\bigskip\n\\noindent\nThe differential cross section for Compton scattering is\n\\begin{align*}\n\\frac{d\\sigma}{d\\Omega}&=\\frac{\\hbar^2}{64\\pi^2m^2c^4}\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\\langle|\\mathcal{M}|^2\\rangle\n\\\\\n&=\\frac{e^4\\hbar^2}{32\\pi^2m^2c^4}\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\n\\left(\n\\frac{\\omega}{\\omega'}+\\frac{\\omega'}{\\omega}+\\cos^2\\theta-1\n\\right)\n\\end{align*}\n\n\\noindent\nSubstitute $e^4=16\\pi^2\\alpha^2$ to obtain\n\\begin{equation*}\n\\frac{d\\sigma}{d\\Omega}\n=\\frac{\\alpha^2\\hbar^2}{2m^2c^4}\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\n\\left(\n\\frac{\\omega}{\\omega'}+\\frac{\\omega'}{\\omega}+\\cos^2\\theta-1\n\\right)\n\\end{equation*}\n\n\\noindent\nThe scattered photon frequency $\\omega'$ is computed from the Compton equation.\n\\begin{equation*}\n\\omega'=\\frac{m\\omega}{m+\\omega(1-\\cos\\theta)}\n\\end{equation*}\n\n\\noindent\nWe can integrate $d\\sigma$ to obtain a cumulative distribution function.\nRecall that\n\\begin{equation*}\nd\\Omega=\\sin\\theta\\,d\\theta\\,d\\phi\n\\end{equation*}\nHence\n\\begin{equation*}\nd\\sigma\n=\\frac{\\alpha^2\\hbar^2}{2m^2c^4}\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\n\\left(\\frac{\\omega}{\\omega'}+\\frac{\\omega'}{\\omega}+\\cos^2\\theta-1\\right)\n\\sin\\theta\\,d\\theta\\,d\\phi\n\\end{equation*}\n\n\\bigskip\n\\noindent\nLet $I(\\theta)$ be the following integral of $d\\sigma$.\n\\begin{align*}\nI(\\theta)\n&=\\left(\\frac{2m^2c^4}{\\alpha^2\\hbar^2}\\right)\\frac{1}{2\\pi}\\int_0^{2\\pi}\\int d\\sigma\n\\\\\n&=\\int\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\n\\left(\\frac{\\omega}{\\omega'}+\\frac{\\omega'}{\\omega}+\\cos^2\\theta-1\\right)\n\\sin\\theta\\,d\\theta,\n\\quad0\\le\\theta\\le\\pi\n\\end{align*}\n\n\\noindent\nAssume that $I(\\theta)-I(0)$ is computable given $\\theta$ by either symbolic or numerical integration.\n\n\\bigskip\n\\noindent\nLet $C$ be the normalization constant\n\\begin{equation*}\nC=I(\\pi)-I(0)\n\\end{equation*}\n\n\\noindent\nThen the cumulative distribution function $F(\\theta)$ is\n\\begin{equation*}\nF(\\theta)=\\frac{I(\\theta)-I(0)}{C},\n\\quad0\\le\\theta\\le\\pi\n\\end{equation*}\n\n\\noindent\nThe probability of observing scattering events in the interval\n$\\theta_1$ to $\\theta_2$ can now be computed.\n\\begin{equation*}\nP(\\theta_1\\le\\theta\\le\\theta_2)=F(\\theta_2)-F(\\theta_1)\n\\end{equation*}\n\n\\noindent\nThe probability density function $f(\\theta)$ is the derivative of $F(\\theta)$.\n\\begin{equation*}\nf(\\theta)=\\frac{dF(\\theta)}{d\\theta}\n=\\frac{1}{C}\n\\left(\\frac{\\omega'}{\\omega}\\right)^2\n\\left(\\frac{\\omega}{\\omega'}+\\frac{\\omega'}{\\omega}+\\cos^2\\theta-1\\right)\n\\sin\\theta\n\\end{equation*}\n\n\\noindent\nRun ``compton-scattering-4.txt'' to draw a graph of $f(\\theta)$ for\n$\\omega=500\\,\\text{keV}=1.2\\times10^{20}\\,\\text{Hz}$.\n\n\\begin{center}\n\\includegraphics[scale=0.5]{compton-scattering.png}\n\\end{center}\n\n\\noindent\nProbability distribution for $45^\\circ$ bins\n($\\omega=500\\,\\text{keV}=1.2\\times10^{20}\\,\\text{Hz}$).\n\n\\begin{center}\n\\begin{tabular}{|c|c|c|}\n\\hline\n$\\theta_1$ & $\\theta_2$ & $P(\\theta_1\\le\\theta\\le\\theta_2)$\\\\\n\\hline\n$0^\\circ$ & $45^\\circ$ & 0.35 \\\\\n$45^\\circ$ & $90^\\circ$ & 0.34 \\\\\n$90^\\circ$ & $135^\\circ$ & 0.22 \\\\\n$135^\\circ$ & $180^\\circ$ & 0.09 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\subsection*{Thomson scattering}\nWhen $\\omega$ is much smaller than the electron mass $m$ we have\n\\begin{equation*}\n\\frac{m}{m+\\omega(1-\\cos\\theta)}\\approx1\n\\end{equation*}\n\n\\noindent\nHence for $\\omega\\ll m$ the differential cross section is approximately\n\\begin{equation*}\n\\frac{d\\sigma}{d\\Omega}=\\frac{\\alpha^2}{2m^2}(1+\\cos^2\\theta)\n\\end{equation*}\n\n\\noindent\nwhich is the formula for Thomson scattering.\n\n\\subsection*{Data from a CERN LEP experiment}\nSee ``Compton Scattering of Quasi-Real Virtual Photons at LEP,''\narxiv.org/abs/hep-ex/0504012.\n\n\\begin{center}\n\\begin{tabular}{|c|c|}\n\\hline\n$x$ & $y$\\\\\n\\hline\n$-0.74$ & $13380$\\\\\n$-0.60$ & $\\phantom{0}7720$\\\\\n$-0.47$ & $\\phantom{0}6360$\\\\\n$-0.34$ & $\\phantom{0}4600$\\\\\n$-0.20$ & $\\phantom{0}4310$\\\\\n$-0.07$ & $\\phantom{0}3700$\\\\\n$\\phantom{+}0.06$ & $\\phantom{0}3640$\\\\\n$\\phantom{+}0.20$ & $\\phantom{0}3340$\\\\\n$\\phantom{+}0.33$ & $\\phantom{0}3500$\\\\\n$\\phantom{+}0.46$ & $\\phantom{0}3010$\\\\\n$\\phantom{+}0.60$ & $\\phantom{0}3310$\\\\\n$\\phantom{+}0.73$ & $\\phantom{0}3330$\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent\nThe data are for the center of mass frame and have the following relationship with the differential cross section formula.\n\\begin{equation*}\nx=\\cos\\theta\\qquad y=\\frac{d\\sigma}{d\\cos\\theta}=2\\pi\\frac{d\\sigma}{d\\Omega}\n\\end{equation*}\n\n\\noindent\nFrom equation (3) we have for the center of mass frame\n\\begin{equation*}\n\\langle|\\mathcal{M}|^2\\rangle\n=\n2e^4\\left(\n\\frac{\\cos\\theta+1}{2}+\\frac{2}{\\cos\\theta+1}\n\\right)\n\\end{equation*}\n\n\\noindent\nThe corresponding cross section formula is\n\\begin{equation*}\n\\frac{d\\sigma}{d\\Omega}\n=\\frac{\\langle|\\mathcal{M}|^2\\rangle}{64\\pi^2s}\n=\\frac{e^4}{32\\pi^2s}\n\\left(\n\\frac{\\cos\\theta+1}{2}+\\frac{2}{\\cos\\theta+1}\n\\right),\\quad s\\gg m\n\\end{equation*}\n\n\\noindent\nSubstituting $e^4=16\\pi^2\\alpha^2$ yields\n\\begin{equation*}\n\\frac{d\\sigma}{d\\Omega}\n=\\frac{\\alpha^2}{2s}\n\\left(\n\\frac{\\cos\\theta+1}{2}+\\frac{2}{\\cos\\theta+1}\n\\right)\n\\end{equation*}\n\n\\noindent\nMultiply by $2\\pi$ to obtain\n\\begin{equation*}\n\\frac{d\\sigma}{d\\cos\\theta}\n=\\frac{\\pi\\alpha^2}{s}\\left(\n\\frac{\\cos\\theta+1}{2}+\\frac{2}{\\cos\\theta+1}\n\\right)\n\\end{equation*}\n\n\\noindent\nTo compute predicted values $\\hat{y}$ from the above formula,\nmultiply by $(\\hbar c)^2$ to convert to SI\nand multiply by $10^{40}$ to convert square meters to picobarns.\n\\begin{equation*}\n\\hat{y}\n=\n\\frac{\\pi\\alpha^2}{s}\n\\left(\n\\frac{x+1}{2}+\n\\frac{2}{x+1}\n\\right)\n\\times(\\hbar c)^2\n\\times10^{40}\n\\end{equation*}\n\n\\noindent\nThe following table shows $\\hat{y}$\nfor $s=40\\,\\text{GeV}^2$ (i.e., $\\omega=100\\,\\text{MeV}$).\n\n\\begin{center}\n\\begin{tabular}{|c|c|c|}\n\\hline\n$x$ & $y$ & $\\hat{y}$\\\\\n\\hline\n$-0.74$ & $13380$ & $12739$\\\\\n$-0.60$ & $\\phantom{0}7720$ & $\\phantom{0}8468$\\\\\n$-0.47$ & $\\phantom{0}6360$ & $\\phantom{0}6577$\\\\\n$-0.34$ & $\\phantom{0}4600$ & $\\phantom{0}5472$\\\\\n$-0.20$ & $\\phantom{0}4310$ & $\\phantom{0}4723$\\\\\n$-0.07$ & $\\phantom{0}3700$ & $\\phantom{0}4259$\\\\\n$\\phantom{+}0.06$ & $\\phantom{0}3640$ & $\\phantom{0}3936$\\\\\n$\\phantom{+}0.20$ & $\\phantom{0}3340$ & $\\phantom{0}3691$\\\\\n$\\phantom{+}0.33$ & $\\phantom{0}3500$ & $\\phantom{0}3532$\\\\\n$\\phantom{+}0.46$ & $\\phantom{0}3010$ & $\\phantom{0}3420$\\\\\n$\\phantom{+}0.60$ & $\\phantom{0}3310$ & $\\phantom{0}3338$\\\\\n$\\phantom{+}0.73$ & $\\phantom{0}3330$ & $\\phantom{0}3291$\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent\nThe coefficient of determination $R^2$ measures how well predicted values fit the real data.\n\\begin{equation*}\nR^2=1-\\frac{\\sum(y-\\hat{y})^2}{\\sum(y-\\bar{y})^2}=0.97\n\\end{equation*}\n\n\\noindent\nThe result indicates that the model $d\\sigma$ explains 97\\% of the variance in the data.\n\n\\subsection*{Notes}\nHere are a few notes regarding the Eigenmath scripts.\n\n\\bigskip\n\\noindent\nStart by writing out $a_1$ and $a_2$ in full component form.\n\\begin{equation*}\na_1^{\\mu\\nu}\n=\\bar{u}_{4\\alpha}\\gamma^{\\mu\\alpha}{}_\\beta(\\slashed{q}_1+m)^\\beta{}_\\rho\\gamma^{\\nu\\rho}{}_\\sigma u_2^\\sigma\n\\qquad\na_2^{\\nu\\mu}\n=\\bar{u}_{4\\alpha}\\gamma^{\\nu\\alpha}{}_\\beta(\\slashed{q}_2+m)^\\beta{}_\\rho\\gamma^{\\mu\\rho}{}_\\sigma u_2^\\sigma\n\\end{equation*}\n\n\\noindent\nTranspose $\\gamma$ tensors to form inner products over $\\alpha$ and $\\rho$.\n\\begin{equation*}\na_1^{\\mu\\nu}\n=\\bar{u}_{4\\alpha}\\gamma^{\\alpha\\mu}{}_\\beta(\\slashed{q}_1+m)^\\beta{}_\\rho\\gamma^{\\rho\\nu}{}_\\sigma u_2^\\sigma\n\\qquad\na_2^{\\nu\\mu}\n=\\bar{u}_{4\\alpha}\\gamma^{\\alpha\\nu}{}_\\beta(\\slashed{q}_2+m)^\\beta{}_\\rho\\gamma^{\\rho\\mu}{}_\\sigma u_2^\\sigma\n\\end{equation*}\n\n\\noindent\nConvert transposed $\\gamma$ to Eigenmath code.\n\\begin{equation*}\n\\gamma^{\\alpha\\mu}{}_\\beta\n\\quad\\rightarrow\\quad\n\\text{\\tt gammaT = transpose(gamma)}\n\\end{equation*}\n\n\\noindent\nThen to compute $a_1$ we have\n\\begin{multline*}\na_1=\\bar{u}_{4\\alpha}\\gamma^{\\alpha\\mu}{}_\\beta(\\slashed{q}_1+m)^\\beta{}_\\rho\\gamma^{\\rho\\nu}{}_\\sigma u_2^\\sigma\n\\\\\n\\rightarrow\\quad\n\\text{\\tt a1 = dot(u4bar[s4],gammaT,qslash1 + m I,gammaT,u2[s2])}\n\\end{multline*}\n\n\\noindent\nwhere $s_2$ and $s_4$ are spin indices.\nSimilarly for $a_2$ we have\n\\begin{multline*}\na_2=\\bar{u}_{4\\alpha}\\gamma^{\\alpha\\nu}{}_\\beta(\\slashed{q}_2+m)^\\beta{}_\\rho\\gamma^{\\rho\\mu}{}_\\sigma u_2^\\sigma\n\\\\\n\\rightarrow\\quad\n\\text{\\tt a2 = dot(u4bar[s4],gammaT,qslash2 + m I,gammaT,u2[s2])}\n\\end{multline*}\n\n\\noindent\nIn component notation the product $a_1a_1^*$ is\n\\begin{equation*}\na_1a_1^*=a_1^{\\mu\\nu}a_1^{*\\mu\\nu}\n\\end{equation*}\n\n\\noindent\nTo sum over $\\mu$ and $\\nu$ it is necessary to lower indices with the metric tensor.\nAlso, transpose $a_1^*$ to form an inner product with $\\nu$.\n\\begin{equation*}\na_1a_1^*=a_1^{\\mu\\nu}a_{1\\nu\\mu}^*\n\\end{equation*}\n\n\\noindent\nConvert to Eigenmath code.\nThe dot function sums over $\\nu$ and the contract function sums over $\\mu$.\n\\begin{equation*}\na_1a_1^*\n\\quad\\rightarrow\\quad\n\\text{\\tt a11 = contract(dot(a1,gmunu,transpose(conj(a1)),gmunu))}\n\\end{equation*}\n\n\\noindent\nSimilarly for $a_2a_2^*$ we have\n\\begin{equation*}\na_2a_2^*\n\\quad\\rightarrow\\quad\n\\text{\\tt a22 = contract(dot(a2,gmunu,transpose(conj(a2)),gmunu))}\n\\end{equation*}\n\n\\noindent\nThe product $a_1a_2^*$ does not require a transpose because $a_1a_2^*=a_1^{\\mu\\nu}a_2^{*\\nu\\mu}$.\n\\begin{equation*}\na_1a_2^*\n\\quad\\rightarrow\\quad\n\\text{\\tt a12 = contract(dot(a1,gmunu,conj(a2),gmunu))}\n\\end{equation*}\n\n\\noindent\nIn component notation, a trace operator becomes a sum over an index, in this case $\\alpha$.\n\\begin{align*}\nf_{11}\n&=\n\\mathop{\\rm Tr}\n\\left(\n(\\slashed{p}_2+m)\\gamma^\\mu(\\slashed{q}_1+m)\\gamma^\\nu(\\slashed{p}_4+m)\\gamma_\\nu(\\slashed{q}_1+m)\\gamma_\\mu\n\\right)\\\\\n&=\n(\\slashed{p}_2+m)^\\alpha{}_\\beta\n\\gamma^{\\mu\\beta}{}_\\rho\n(\\slashed{q}_1+m)^\\rho{}_\\sigma\n\\gamma^{\\nu\\sigma}{}_\\tau\n(\\slashed{p}_4+m)^\\tau{}_\\delta\n\\gamma_\\nu{}^\\delta{}_\\eta\n(\\slashed{q}_1+m)^\\eta{}_\\xi\n\\gamma_\\mu{}^\\xi{}_\\alpha\n\\end{align*}\n\n\\noindent\nAs before, transpose $\\gamma$ tensors to form inner products.\n\\begin{equation*}\nf_{11}=\n(\\slashed{p}_2+m)^\\alpha{}_\\beta\n\\gamma^{\\beta\\mu}{}_\\rho\n(\\slashed{q}_1+m)^\\rho{}_\\sigma\n\\gamma^{\\sigma\\nu}{}_\\tau\n(\\slashed{p}_4+m)^\\tau{}_\\delta\n\\gamma^\\delta{}_{\\nu\\eta}\n(\\slashed{q}_1+m)^\\eta{}_\\xi\n\\gamma^\\xi{}_{\\mu\\alpha}\n\\end{equation*}\n\n\\noindent\nTo convert to Eigenmath code, use an intermediate variable for the inner product.\n\\begin{equation*}\nT^{\\alpha\\mu\\nu}{}_{\\nu\\mu\\alpha}\n\\quad\\rightarrow\\quad\n\\text{\\tt T = dot(P2,gammaT,Q1,gammaT,P4,gammaL,Q1,gammaL)}\n\\end{equation*}\n\n\\noindent\nNow sum over the indices of $T$.\nThe innermost contract sums over $\\nu$ then the next contract sums over $\\mu$.\nFinally the outermost contract sums over $\\alpha$.\n\\begin{equation*}\nf_{11}\\quad\\rightarrow\\quad\n\\text{\\tt f11 = contract(contract(contract(T,3,4),2,3))}\n\\end{equation*}\n\n\\noindent\nFollow suit for $f_{22}$.\nFor $f_{12}$ the order of the rightmost $\\mu$ and $\\nu$ is reversed.\n\\begin{equation*}\nf_{12}=\\mathop{\\rm Tr}\n\\left(\n(\\slashed{p}_2+m)\\gamma^\\mu(\\slashed{q}_2+m)\\gamma^\\nu(\\slashed{p}_4+m)\\gamma_\\mu(\\slashed{q}_1+m)\\gamma_\\nu\n\\right)\n\\end{equation*}\n\n\\noindent\nThe resulting inner product is $T^{\\alpha\\mu\\nu}{}_{\\mu\\nu\\alpha}$\nso the contraction is different.\n\\begin{equation*}\nf_{12}\n\\quad\\rightarrow\\quad\n\\text{\\tt f12 = contract(contract(contract(T,3,5),2,3))}\n\\end{equation*}\n\n\\noindent\nThe innermost contract sums over $\\nu$ followed by sum over $\\mu$ then sum over $\\alpha$.\n\n\\end{document}\n", "meta": {"hexsha": "241f72c5303d1622a334d117893b5e6a10361890", "size": 21276, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compton-scattering.tex", "max_stars_repo_name": "georgeweigt/georgeweigt.github.io", "max_stars_repo_head_hexsha": "94fc6dfbc8dee95cca58c9822533699e8ed79a51", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "compton-scattering.tex", "max_issues_repo_name": "georgeweigt/georgeweigt.github.io", "max_issues_repo_head_hexsha": "94fc6dfbc8dee95cca58c9822533699e8ed79a51", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.8558511543206819, "lm_q2_score": 0.7310585669110202, "lm_q1q2_score": 0.6256773183668202}} {"text": "\\section{The Theory of Covering Spaces}\r\n\\begin{definition}\r\n Suppose $\\pi:\\tilde{X}\\to X$ is a covering map and $\\gamma:[0,1]\\to X$ is a path.\r\n A lift of $\\gamma$ along $\\pi$ is a path $\\tilde{\\gamma}:[0,1]\\to\\tilde{X}$ such that $\\pi\\circ\\tilde{\\gamma}=\\gamma$.\r\n\\end{definition}\r\nObviously, lifts are not usually unique.\r\n\\begin{example}\r\n The function $\\exp:\\mathbb C\\to\\mathbb C_\\star$ is a covering map.\r\n Consider the path $\\gamma:[0,1]\\to \\mathbb C$ representing the unit circle, i.e. $\\gamma(t)=e^{2\\pi it}$, then both $t\\mapsto 2\\pi it$ and $t\\mapsto 2\\pi i+2\\pi it$ are lifts of $\\gamma$.\r\n\\end{example}\r\nAn interesting observation is that although we exhibited two different lifts, they do start at different points.\r\nAnd it is indeed a correct intuition.\r\n\\begin{proposition}[Uniqueness of Lifts]\r\n Suppose $\\tilde{\\gamma}_1,\\tilde{\\gamma}_2$ are both lifts of $\\gamma$ along a covering $\\pi:\\tilde{X}\\to X$.\r\n If $\\tilde{\\gamma}_1(0)=\\tilde{\\gamma}_2(0)$ then $\\tilde{\\gamma}_1=\\tilde{\\gamma}_2$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Consider the set\r\n $$I=\\{t\\in[0,1]:\\tilde{\\gamma}_1(t)=\\tilde{\\gamma}_2(t)\\}$$\r\n We shall show that it is both open and closed, which shows the proposition as $[0,1]$ is connected.\r\n It is obviously closed as $\\tilde{\\gamma}_1,\\tilde{\\gamma}_2$ are continuous and $[0,1]^2$ is Hausdorff.\r\n So it remains to show it is open.\r\n Let $t\\in I$.\r\n As $\\pi$ is a covering map, $\\tilde{\\gamma}_1(t)=\\tilde{\\gamma}_2(t)$ has a neighbourhood $\\tilde{N}$ such that $\\pi|_{\\tilde{N}}$ is a homeomorphism.\r\n As $\\gamma$ is continuous, there is $\\delta>0$ such that $\\gamma(t-\\delta,t+\\delta)\\in N$.\r\n But for any $s$, $\\pi\\circ\\tilde{\\gamma}_1(s)=\\gamma(s)=\\pi\\circ\\tilde{\\gamma}_2(s)$.\r\n So pick any $s\\in (t-\\delta,t+\\delta)$, we have\r\n $$\\tilde{\\gamma}_1(s)=(\\pi|_{\\tilde{N}})^{-1}\\circ\\gamma(s)=\\tilde{\\gamma}_2(s)$$\r\n Therefore $s\\in I$.\r\n This shows that $I$ is open, as desired.\r\n\\end{proof}\r\nNow, even if $\\pi$ is surjective, lifts may not exist.\r\n\\begin{example}[Counterexample]\r\n Consider $X=\\mathbb C_\\star,\\tilde{X}=\\mathbb R+i(-\\pi,2\\pi)$ and $\\pi=\\exp|_D$, but then we cannot lift the path $t\\mapsto e^{2\\pi it}$.\r\n\\end{example}\r\nBut note that the $\\pi$ in the example above is not a regular covering map.\r\nIn fact, a lift does exist if $\\pi$ is regular.\r\n\\begin{proposition}[Path-Lifting Lemma]\r\n Let $\\pi:\\tilde{X}\\to X$ be a regular covering map and $\\gamma:[0,1]\\to X$ is a path.\r\n Suppose $\\pi(\\tilde{x})=\\gamma(0)$, then there is a unique lift $\\tilde\\gamma$ of $\\gamma$ such that $\\tilde{\\gamma}(0)=\\tilde{x}$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Suffices to show the existence.\r\n Let\r\n $$I=\\{t\\in [0,1]:\\gamma|_{[0,t]}\\text{ can be lifted to some }\\tilde\\gamma\\text{ with }\\tilde{\\gamma}(0)=\\tilde{x}\\}$$\r\n Again we will show that $I$ is both open and closed.\\\\\r\n To see $I$ is closed, consider a sequence $t_n\\to\\tau$ where $t_n\\in I$ for all $I$.\r\n We shall show that $\\tau\\in I$.\r\n As $\\pi$ is regular, there is some open $U\\ni\\gamma(\\tau)$ such that\r\n $$\\pi^{-1}(U)\\cong\\coprod_{\\delta\\in D}U_\\delta$$\r\n for some set $D$.\r\n Throwing away finitely many terms we can assume $\\gamma(t_n)\\in U$ for any $t$, consequently $\\tilde{\\gamma}(t_n)$ are all in the same $U_\\delta$.\r\n Set $\\tilde{\\gamma}(\\tau)=(\\pi|_{U_\\delta})^{-1}\\circ\\gamma(\\tau)$ extends $\\tilde{\\gamma}$ continuous to $\\tau$, so $\\tau\\in I$.\\\\\r\n To see $I$ is open, let $\\tau\\in I$ and choose open $U\\ni \\gamma(\\tau)$ such that\r\n $$\\pi^{-1}(U)\\cong\\coprod_{\\delta\\in D}U_\\delta$$\r\n for a set $D$.\r\n There is a unique $\\delta$ such that $\\tilde{\\gamma}(\\tau)\\in U_\\delta$.\r\n Choose $\\epsilon>0$ such that $|t-\\tau|<\\epsilon\\implies \\gamma(t)\\in U$.\r\n So we want to extend $\\tilde{\\gamma}$ via $\\tilde\\gamma(t)=(\\pi|_{U_\\delta})^{-1}\\circ\\gamma(t)$ for $|t-\\tau|<\\epsilon$, which works.\r\n Therefore $I$ is open, as required.\r\n\\end{proof}\r\n\\begin{definition}\r\n Let $X$ be a topological space and $\\alpha,\\beta:[0,1]\\to X$ paths with $\\alpha(0)=\\beta(0),\\alpha(1)=\\beta(1)$.\r\n We say $\\alpha,\\beta$ are homotopic (or $\\alpha\\simeq\\beta$) if there is a family of paths $(\\alpha_s)_{s\\in[0,1]}$ such that:\\\\\r\n 1. $\\alpha_0=\\alpha,\\alpha_s(1)=\\beta$.\\\\\r\n 2. $\\alpha_s(0)=\\alpha(0),\\alpha_s(1)=\\alpha(1)$ for any $s$.\\\\\r\n 3. The map $(t,s)\\mapsto\\alpha_s(t)$ is continuous.\r\n\\end{definition}\r\n\\begin{definition}\r\n A topological space $X$ is simply connected if:\\\\\r\n 1. $X$ is path-connected.\\\\\r\n 2. Every pair of paths $\\alpha,\\beta:[0,1]\\to X$ with the same endpoints are homotopic.\r\n\\end{definition}\r\n\\begin{remark}\r\n Let $D\\subset\\mathbb C$ be a convex domain, then the formula\r\n $$\\alpha_s(t)=(1-s)\\alpha(t)+s\\beta(t)$$\r\n gives a homotopy between any two paths $\\alpha,\\beta$ with same endpoints.\r\n\\end{remark}\r\n\\begin{example}\r\n $\\mathbb C$, the unit disk and half-plane are simply connected.\r\n\\end{example}\r\n\\begin{theorem}[Monodromy Theorem, aka Homotopy Lifting Lemma]\\label{monodromy}\r\n Let $\\pi:\\tilde{X}\\to X$ be a covering map and $\\alpha,\\beta$ in $X$ be such that:\\\\\r\n 1. $\\alpha\\simeq\\beta$ in $X$.\\\\\r\n 2. There are lifts $\\tilde\\alpha$ of $\\alpha$ and $\\tilde\\beta$ of $\\beta$ along the covering.\\\\\r\n 3. Every path $\\gamma$ in $X$ with $\\gamma(0)=\\alpha(0)=\\beta(0)$ has a lift $\\tilde{\\gamma}$ to $\\tilde{X}$ with $\\tilde{\\gamma}(0)\\tilde{\\alpha}(0)=\\tilde{\\beta}(0)$.\\\\\r\n Then $\\tilde{\\alpha}\\simeq\\tilde{\\beta}$.\r\n In particular, $\\tilde{\\alpha}\\simeq\\tilde{\\beta}$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n See Algebraic Topology.\r\n\\end{proof}\r\n\\begin{note}\r\n The requirements (ii) and (iii) are automatically satisfied if $\\pi$ is regular.\r\n\\end{note}", "meta": {"hexsha": "0c7f079491ed4037b607a077f2533fb71ec23157", "size": 5757, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "7/cover.tex", "max_stars_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_stars_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "7/cover.tex", "max_issues_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_issues_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "7/cover.tex", "max_forks_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_forks_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.6, "max_line_length": 192, "alphanum_fraction": 0.6381796074, "num_tokens": 1978, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585669110203, "lm_q2_score": 0.855851143290548, "lm_q1q2_score": 0.6256773103031463}} {"text": "\\section{First Order Homogeneous Linear Equations}\\label{sec:first order homogeneous linear}\nA simple, but important and useful, type of separable equation is the\n\\dfont{first order homogeneous linear equation}:\n\n\\begin{definition}{First Order Homogeneous Linear Equation}{First Order Homogeneous Linear Equation}\\label{First Order Homogeneous Linear Equation}\nA first order homogeneous linear differential equation\nis one of\nthe form $\\ds y' + p(t)y=0$\nor equivalently\n$\\ds y' = -p(t)y$.\n\\end{definition}\n\n``Homogeneous'' refers to the zero on the right side of the equation, provided that $y'$ and $y$ are on the left. ``Linear'' in this definition indicates that both $y'$ and $y$ appear independently and explicitly; we don't see $y'$ or $y$ to any power greater than 1, or multiplied by each other (i.e. $y'y$).\n\n\\begin{example}{Linear Examples}{Linear Examples}\\label{Linear Examples}\n The equation $\\ds y' = 2t(25-y)$ can be written\n$\\ds y' + 2ty= 50t$. This is linear, but not homogeneous. The\nequation $\\ds y'=ky$, or $\\ds y'-ky=0$ is linear and\nhomogeneous, with a particularly simple $p(t)=-k$.\nThe equation $y'+y^2=0$ is homogeneous, but not linear.\n\\end{example}\n\nSince first order homogeneous linear equations are separable, we can\nsolve them in the usual way:\n\\begin{eqnarray*}\ny' &=& -p(t)y\\cr\n\\int {1\\over y}\\,dy &=& \\int -p(t)\\,dt\\cr\n\\ln|y| &=& P(t)+C\\cr\ny&=&\\pm\\,e^{P(t)}\\cr\ny&=&Ae^{P(t)},\n\\end{eqnarray*}\nwhere $P(t)$ is an anti-derivative of $-p(t)$. As in previous\nexamples, if we allow $A=0$ we get the constant solution $y=0$.\n\n\\begin{example}{Solving an IVP}{Solving an IVP}\\label{Solving an IVP}\n Solve the initial value problem \n$$\\ds y' + y\\cos t =0,$$\nsubject to $y(0)=1/2$ and $y(2)=1/2$.\n\\end{example}\n\n\\begin{solution}\nWe start with\n$$P(t)=\\int -\\cos t\\,dt = -\\sin t,$$\nso the general solution to the differential equation is\n$$y=Ae^{-\\sin t}.$$\nTo compute $A$ we substitute:\n$$ {1\\over 2} = Ae^{-\\sin 0} = A,$$\nso the solutions is \n$$ y = {1\\over 2} e^{-\\sin t}.$$\nFor the second problem,\n\\begin{eqnarray*}\n{1\\over 2} &=& Ae^{-\\sin 2}\\cr\nA &=& {1\\over 2}e^{\\sin 2}\n\\end{eqnarray*}\nso the solution is \n$$ y = {1\\over 2}e^{\\sin 2}e^{-\\sin t}.$$\n\\vskip-15pt\\end{solution}\n\n\\begin{example}{}{}\\label{}\n Solve the initial value problem $ty'+3y=0$, $y(1)=2$,\nassuming $t>0$.\n\\end{example}\n\n\\begin{solution}\nWe write the equation in standard form: $y'+3y/t=0$. Then\n$$P(t)=\\int -{3\\over t}\\,dt=-3\\ln t$$\nand \n$$ y=Ae^{-3\\ln t}=At^{-3}.$$\nSubstituting to find $A$:\n$\\ds 2=A(1)^{-3}=A$, so the solution is $\\ds y=2t^{-3}$.\n\\end{solution}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\Opensolutionfile{solutions}[ex]\n\\section*{Exercises for \\ref{sec:first order homogeneous linear}}\n\n\\begin{enumialphparenastyle}\n\nFind the general solution of each equation in the following exercises.\n\n\\begin{multicols}{2}\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y'+5y=0$\n\\begin{sol}\n $\\ds y=Ae^{-5t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y'-2y=0$\n\\begin{sol}\n $\\ds y=Ae^{2t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y'+{y\\over 1+t^2}=0$\n\\begin{sol}\n $\\ds y=Ae^{-\\arctan t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y'+t^2y=0$\n\\begin{sol}\n $\\ds y=Ae^{-t^3/3}$\n\\end{sol}\n\\end{ex}\n\n\\end{multicols}\n\nIn the following exercises, solve the initial value problem.\n\n\\begin{multicols}{2}\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' + y=0$, $y(0)=4$\n\\begin{sol}\n $\\ds y=4e^{-t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' -3y=0$, $y(1)=-2$\n\\begin{sol}\n $\\ds y=-2e^{3t-3}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' + y\\sin t = 0$, $y(\\pi)=1$\n\\begin{sol}\n $\\ds y=e^{1+\\cos t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' +ye^t=0$, $y(0)=e$\n\\begin{sol}\n $\\ds y=e^2e^{-e^t}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' +y\\sqrt{1+t^4}=0$, $y(0)=0$\n\\begin{sol}\n $\\ds y=0$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds y' + y\\cos(e^t)=0$, $y(0)=0$\n\\begin{sol}\n $\\ds y=0$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds ty' - 2y = 0$, $y(1)=4$\n\\begin{sol}\n $\\ds y=4t^2$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds t^2y' + y = 0$, $y(1)=-2$, $t>0$\n\\begin{sol}\n $\\ds y=-2e^{(1/t)-1}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds t^3y' = 2y$, $y(1)=1$, $t>0$\n\\begin{sol}\n $\\ds y=e^{1-t^{-2}}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n $\\ds t^3y' = 2y$, $y(1)=0$, $t>0$\n\\begin{sol}\n $\\ds y=0$\n\\end{sol}\n\\end{ex}\n\n\\end{multicols}\n\n\n%%%%%%%%%%\n\\begin{ex}\n A function $y(t)$ is a solution of $\\ds y' +\nky=0$. Suppose that $y(0)=100$ and $y(2)=4$. Find $k$ and find $y(t)$.\n\\begin{sol}\n $k=\\ln 5$, $\\ds y=100e^{-t\\ln 5}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n A function $y(t)$ is a solution of $\\ds y' +\nt^ky=0$. Suppose that $y(0)=1$ and $y(1)=e^{-13}$. Find $k$ and find\n$y(t)$. \n\\begin{sol}\n $k=-12/13$, $\\ds y=\\exp(-13 t^{1/13})$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n A bacterial culture grows at a rate proportional to its\npopulation. If the population is one million at $t=0$ and 1.5\nmillion at $t=1$ hour, find the population as a function of time.\n\\begin{sol}\n $\\ds y=10^6e^{t\\ln(3/2)}$\n\\end{sol}\n\\end{ex}\n\n\n%%%%%%%%%%\n\\begin{ex}\n A radioactive element decays with a half-life of 6 years. If\na mass of the element weighs ten pounds at $t=0$, find the amount of\nthe element at time $t$.\n\\begin{sol}\n $\\ds y=10e^{-t\\ln(2)/6}$\n\\end{sol}\n\\end{ex}\n\n\\end{enumialphparenastyle}", "meta": {"hexsha": "54ad6fc37481b6f14979d295d1e6c103b5bcd4eb", "size": 5329, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "10-differential-equations/10-2-first-order-homo-linear.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "10-differential-equations/10-2-first-order-homo-linear.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "10-differential-equations/10-2-first-order-homo-linear.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 20.6550387597, "max_line_length": 309, "alphanum_fraction": 0.5937324076, "num_tokens": 2054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680199891789, "lm_q2_score": 0.8418256472515683, "lm_q1q2_score": 0.625617899444057}} {"text": "%%%%%%%%%%%%%%%%%%%%%definitions%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\n\r\n\\input{../../doc/related_pages/header.tex}\r\n\\input{../../doc/related_pages/newcommands.tex}\r\n\r\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%DOCUMENT%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\n\\begin{document}\r\n\r\n\\title{The heat diffusion project}\r\n\\author{ M.~Wiesenberger and M.~Held}\r\n\\maketitle\r\n\r\n\\begin{abstract}\r\n This is a program for the 3d heat diffusion in arbitrary magnetic field\r\n geometry.\r\n \\end{abstract}\r\n\r\n\\section{Equations}\r\nThe temperature $T$ follows a non-isotropic heat diffusion equation\r\n\\begin{align}\r\n\\frac{\\partial T}{\\partial t} = \\nu_\\perp \\Delta_\\perp T + \\nu_\\parallel\\Delta_\\parallel T\r\n\\label{eq:temperature}\r\n\\end{align}\r\nwith\r\n\\begin{align}\r\n\\Delta_\\parallel T &= \\nabla\\cdot ( \\bhat\\bhat\\cdot\\nabla T) \\\\\r\n\\Delta_\\perp T &= -\\nabla\\cdot( \\bhat\\times(\\bhat\\times \\nabla T)) \\equiv \\nabla\\cdot (h \\nabla T) \\\\\r\nh^{ij} &= g^{ij} - b^ib^j\r\n\\end{align}\r\n$\\bhat = \\bhat(\\vec x)$ is the prescribed magnetic field unit vector, $g$ is the\r\nmetric and $h$ is the projection\r\ntensor.\r\n$\\nu_\\parallel$ and $\\nu_\\perp$ are the respective conduction\r\ncoefficients parallel and perpendicular to this field.\r\n\r\n\\subsection{Simulation domain}\r\nWe define the simulation box as\r\n$[ R_{\\min}, R_{\\max}]\\times [Z_{\\min}, Z_{\\max}] \\times [0,2\\pi]$,\r\nwhere we define\r\n\\begin{align} \\label{eq:box}\r\n R_{\\min}&=R_0-\\varepsilon^{R-}a\\quad\r\n &&R_{\\max}=R_0+\\varepsilon^{R+}a\\nonumber\\\\\r\n Z_{\\min}&=-\\varepsilon^{Z-}ae\\quad\r\n &&Z_{\\max}=\\varepsilon^{Z+}ae\r\n\\end{align}\r\nwhere $a$ is the minor radius, $e$ is the elongation of the flux surfaces and\r\nthe $\\varepsilon$ are free parameters to be specified by the user.\r\n\r\n\\subsection{Initialization}\r\nInitialization of $T$ is a Gaussian\r\n\\begin{align} \\label{eq:initial}\r\n T(R,Z,\\varphi) = 1 + A\\exp\\left(\r\n -\\frac{(R-R_i)^2 + (Z-Z_i)^2}{2\\sigma^2} - \\frac{(\\varphi-\\pi)^2}{2\\sigma_\\varphi^2}\\right)\r\n\\end{align}\r\nwhere $R_i = p_x a$ and $Z_i = p_y a$ denote the initial position, $A$ is the amplitude and $\\sigma$ the\r\nradius of the blob.\r\n\\subsection{Diagnostics}\r\nIntegrating Eq.~\\eqref{eq:temperature} over the volume yields\r\n\\begin{align}\r\n \\frac{\\partial}{\\partial t} \\int T \\dV &=\r\n \\int \\dV ( \\nu_\\perp \\Delta_\\perp + \\nu_\\parallel \\Delta_\\parallel T) \\\\\r\n \\frac{\\partial}{\\partial t} \\int T^2/2 \\dV &=\r\n \\int \\dV ( \\nu_\\perp T\\Delta_\\perp T + \\nu_\\parallel T\\Delta_\\parallel T)\r\n\\label{}\r\n\\end{align}\r\n\\section{Numerical methods}\r\ndiscontinuous Galerkin on structured grid\r\n\\rowcolors{2}{gray!25}{white} %%% Use this line in front of longtable\r\n\\begin{longtable}{ll>{\\RaggedRight}p{7cm}}\r\n\\toprule\r\n\\rowcolor{gray!50}\\textbf{Term} & \\textbf{Method} & \\textbf{Description} \\\\ \\midrule\r\ncoordinate system & Cylindrical 3D & equidistant discretization of $[R_0-a,R_0+a] \\times [-a,a] \\times [0,2\\pi]$, equal number of Gaussian nodes in R and Z, one node in $\\varphi$ \\\\\r\n$\\Delta_\\parallel$ & FCI & forward-backward symmetric sym, See the documentation on DS\\\\\r\ntime & Adaptive embedded ARK-4-2-3 & $3rd$ order accurate in both\r\nexplicit and implicit parts\\\\\r\n\\bottomrule\r\n\\end{longtable}\r\n\r\n\\section{Compilation and useage}\r\nThere are two programs heat.cu and heat\\_hpc.cu . Compilation with\r\n\\begin{verbatim}\r\nmake device = \r\n\\end{verbatim}\r\nRun with\r\n\\begin{verbatim}\r\npath/to/feltor/src/heat/heat input.json geometry_params.json\r\npath/to/feltor/src/heat/heat_hpc input.json geometry_params.json output.nc\r\n\\end{verbatim}\r\nAll programs write performance informations to std::cout.\r\nThe first opens a terminal window with life simulation results\r\nthe\r\nother writes the results to disc. Both programs run on shared memory\r\nsystems.\r\nFor the hpc code, there is another mode:\r\n\\begin{verbatim}\r\npath/to/feltor/src/heat/heat_hpc input.json geometry_params.json output.nc input.nc\r\n\\end{verbatim}\r\nIn this case the temperature field of \\code{input.nc} at the last output\r\ntime is read\r\nand taken as a reference solution.\r\n\r\n\\subsection{Input file structure}\r\nInput file format: json\r\n\r\n%%This is a booktabs table\r\n\\begin{longtable}{llll>{\\RaggedRight}p{7cm}}\r\n\\toprule\r\n\\rowcolor{gray!50}\\textbf{Name} & \\textbf{Type} & \\textbf{Example} & \\textbf{Default} & \\textbf{Description} \\\\ \\midrule\r\n n & integer & 3 & - &\\# Gaussian nodes in x and y \\\\\r\n Nx & integer &20& - &\\# grid points in x \\\\\r\n Ny & integer &40& - &\\# grid points in y \\\\\r\n Nz & integer &60& - &\\# grid points in z \\\\\r\n dt & integer &0.001& - &initial time step\\\\\r\n cx & integer & 2 & 1 & output contains n*Nx/cx points in x,\r\n has to divde Nx evenly\\\\\r\n cy & integer & 2 & 1 & output contains n*Ny/cy points in y,\r\n has to divde Ny evenly\\\\\r\n itstp & integer &2 & - & steps between outputs \\\\\r\n maxout & integer &100& - & \\# outputs excluding first \\\\\r\n eps\\_time & float &1e-10 & - & accuracy of implicit solver in time-stepper \\\\\r\n rtol & float & 1e-6 & - & desired tolerance of adaptive time-stepper \\\\\r\n nu\\_parallel & float &100 & - & parallel viscosity $\\nu_\\parallel$ \\\\\r\n nu\\_perp & float &0 & - & perpendicular viscosity $\\nu_\\perp$ \\\\\r\n amplitude & float &0.1 & - & amplitude $A$ of the blob in Eq.~\\eqref{eq:initial}\\\\\r\n sigma & float &5 & - & blob radius $\\sigma$ in $R$ and $Z$ in Eq.~\\eqref{eq:initial} \\\\\r\n sigma\\_z & float &0.025 & - & variance in $\\varphi$ in units of $R_0$ in Eq.~\\eqref{eq:initial} \\\\\r\n posX & float &0.8 & - & blob $R$-position in units of $a$ in Eq.~\\eqref{eq:initial} \\\\\r\n posY & float &0 & - & blob $Z$-position in units of $a$ in Eq.~\\eqref{eq:initial} \\\\\r\n bcx & char & \"NEU\" & - & boundary condition in R (one of PER, DIR, NEU, DIR\\_NEU or NEU\\_DIR) \\\\\r\n bcy & char & \"NEU\" & - & boundary condition in Z (one of PER, DIR, NEU, DIR\\_NEU or NEU\\_DIR) \\\\\r\n boxscaleRp & float & 1.05, & - & $\\varepsilon^{R+}$ in Eq.~\\eqref{eq:box} \\\\\r\n boxscaleRm & float & 1.05, & - & $\\varepsilon^{R-}$ in Eq.~\\eqref{eq:box} \\\\\r\n boxscaleZp & float & 1.05, & - & $\\varepsilon^{Z+}$ in Eq.~\\eqref{eq:box} \\\\\r\n boxscaleZm & float & 1.15, & - & $\\varepsilon^{Z-}$ in Eq.~\\eqref{eq:box} \\\\\r\n mx & integer & 10 & - & refinement factor of DS in x \\\\\r\n my & integer & 10 & - & refinement factor of DS in y \\\\\r\n rk4eps & float & 0.01 & 1e-5 & Accuracy of fieldline integration in DS \\\\\r\n diff & char & non-adjoint & non-adjoint & adjoint, non-adjoint, elliptic (discretization for $\\Delta_\\parallel$) \\\\\r\n\\bottomrule\r\n\\end{longtable}\r\n\r\nThe default value is taken if the value name is not found in the input file. If there is no default and\r\nthe value is not found,\r\nthe program exits with an error message.\r\n\r\n\\subsection{Geometry file structure}\r\nFile format: json\r\n\r\n%%This is a booktabs table\r\n\\begin{longtable}{llll>{\\RaggedRight}p{7cm}}\r\n\\toprule\r\n\\rowcolor{gray!50}\\textbf{Name} & \\textbf{Type} & \\textbf{Example} & \\textbf{Default} & \\textbf{Description} \\\\ \\midrule\r\n A & float & 1 & - & Solovev parameter \\\\\r\n R\\_0 & float & - & - & Major radius in $\\rho_s$ \\\\\r\n C & float[12] & - & - & Solovev coefficients \\\\\r\n elongation & float & 1 & - & Elongation \\\\\r\n triangularity & float & 0 & - & Triangularity \\\\\r\n inverseaspectratio & float & 0.16667 & - & $a/R_0$ \\\\\r\n\\bottomrule\r\n\\end{longtable}\r\n\r\n\\subsection{Structure of output file}\r\nOutput file format: netcdf-4/hdf5\r\n%\r\n%Name | Type | Dimensionality | Description\r\n%---|---|---|---|\r\n\\begin{longtable}{lll>{\\RaggedRight}p{7cm}}\r\n\\toprule\r\n\\rowcolor{gray!50}\\textbf{Name} & \\textbf{Type} & \\textbf{Dimension} & \\textbf{Description} \\\\ \\midrule\r\ninputfile & text attribute & 1 & verbose input file as a string \\\\\r\ngeomfile & text attribute & 1 & verbose geometry input file as a string \\\\\r\nenergy\\_time & Dataset & 1 & timesteps at which 1d variables are written \\\\\r\ntime & Dataset & 1 & time at which temperature is written \\\\\r\nx & Dataset & 1 & x-coordinate \\\\\r\ny & Dataset & 1 & y-coordinate \\\\\r\nz & Dataset & 1 & z-coordinate \\\\\r\nT & Dataset & 4 (time, z, y, x) & temperature $T$ \\\\\r\nenergy & Dataset & 1 (energy\\_time) & total energy integral \\\\\r\nentropy & Dataset & 1 (energy\\_time) & total entropy integral \\\\\r\nmass & Dataset & 1 (energy\\_time) & mass integral \\\\\r\ndEdt & Dataset & 1 (energy\\_time) & change of energy per time \\\\\r\ndissipation & Dataset & 1 (energy\\_time) & diffusion integrals \\\\\r\naccuracy & Dataset & 1 (energy\\_time) & accuracy in time \\\\\r\nerror & Dataset & 1 (energy\\_time) & relative distance to temperature field at time 0\\\\\r\nrelerror & Dataset & 1 (energy\\_time) & relative difference to reference solution if one is provide, else 0 \\\\\r\n\\bottomrule\r\n\\end{longtable}\r\n\r\n%..................................................................\r\n\\bibliography{../../doc/related_pages/references}\r\n%..................................................................\r\n\r\n\r\n\\end{document}\r\n", "meta": {"hexsha": "2af553b885f989da695972661ca5cdb7832c607a", "size": 9071, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/heat/heat.tex", "max_stars_repo_name": "RaulGerru/FELTOR_Raul", "max_stars_repo_head_hexsha": "a566f8a9003ade437e093334877f839f3dfd0260", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 18, "max_stars_repo_stars_event_min_datetime": "2016-06-28T14:34:29.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-06T08:50:48.000Z", "max_issues_repo_path": "src/heat/heat.tex", "max_issues_repo_name": "RaulGerru/FELTOR_Raul", "max_issues_repo_head_hexsha": "a566f8a9003ade437e093334877f839f3dfd0260", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2017-01-18T16:06:15.000Z", "max_issues_repo_issues_event_max_datetime": "2021-02-22T19:11:42.000Z", "max_forks_repo_path": "src/heat/heat.tex", "max_forks_repo_name": "mrheld/feltor", "max_forks_repo_head_hexsha": "c70bc6bb43f39261f6236df88e16610d08cb98ca", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 12, "max_forks_repo_forks_event_min_datetime": "2016-06-27T13:18:11.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-03T08:12:25.000Z", "avg_line_length": 46.5179487179, "max_line_length": 182, "alphanum_fraction": 0.6252893838, "num_tokens": 2853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256591565729, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6256178987142795}} {"text": "\\documentclass[pre,aps,superscriptaddress,nofootinbib]{revtex4}\n\n\\usepackage{amsmath,amsfonts,amssymb,bm,graphicx,hyperref,listings,xcolor,float}\n\\setlength{\\parindent}{0pt}\n\n\\begin{document}\n\n\\title{2 RTPs on a ring}\n\\author{Yann-Edwin Keta}\n\\maketitle\n\n% \\maketitle\n\n\\nocite{Slowman2016, Nemoto2016, Chetrite2015}\n\n\\section{Model}\n\n\\subsection{Fokker-Planck equations}\n\nWe consider 2 run-and-tumble particles (RTPs), with swim speed $v_0$ and tumble rate $\\tau^{-1}$ on a linear ring of length $L$. With $r_1, r_2 \\in [0, L]$ the positions and $\\alpha_1, \\alpha_2 = \\pm$ the states of each particles, and $V(r_1, r_2) = V(|r_1 - r_2|)$ the potential of interactions between these particles, we have the equations of motion\n\\begin{equation}\n\\dot{r}_i = \\alpha_i v_0 - \\partial_{r_i} V(|r_1 - r_2|)\n\\label{EOM}\n\\end{equation}\nand the Fokker-Planck equations for the joint distribution of positions\n\\begin{equation}\n\\begin{aligned}\n\\dot{P}_{\\alpha_1, \\alpha_2}(r_1, r_2) =\n &- v_0(\\alpha_1 \\partial_{r_1} + \\alpha_2 \\partial_{r_2}) P_{\\alpha_1\\alpha_2}(r_1, r_2)\n &&\\to \\text{Propulsion}\\\\\n &+ \\partial_{r_1} (P_{\\alpha_1\\alpha_2}(r_1, r_2)\\partial_{r_1}V(|r_1 - r_2|))\n &&\\to \\text{Interaction}\\\\\n &+ \\partial_{r_2} (P_{\\alpha_1\\alpha_2}(r_1, r_2)\\partial_{r_2}V(|r_1 - r_2|))\n &&\\\\\n &+ \\tau^{-1} (P_{\\overline{\\alpha_1}\\alpha_2}(r_1, r_2) + P_{\\alpha_1\\overline{\\alpha_2}}(r_1, r_2) - 2 P_{\\alpha_1\\alpha_2}(r_1, r_2))\n &&\\to \\text{Tumble}\\\\\n =& \\, \\mathcal{L}^{\\dagger}_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r_1, r_2)\n\\end{aligned}\n\\label{FP0}\n\\end{equation}\nwith $\\mathcal{L}^{\\dagger}_{\\alpha_1\\alpha_2}$ the Fokker-Planck operator.\\\\\n\nWe introduce $r \\equiv r_2 - r_1 \\geq 0$, such that $\\partial_{r_1} \\equiv - \\partial_r$ and $\\partial_{r_2} = \\partial_r$, with which we can define the symmetries\n\\begin{eqnarray}\n\\label{evensymmetry}\nP_{++}(r) = P_{--}(r),\\\\\n\\label{oddsymmetry}\nP_{+-}(r) = P_{-+}(L - r),\n\\end{eqnarray}\nand the interaction potential\n\\begin{align*}\nV(r) = \\begin{cases} 0 &\\text{ if } r \\in ]0, L[ \\\\ \\infty &\\text{ otherwise} \\end{cases}\n\\end{align*}\nsuch that, with Eq. (\\ref{EOM}), we get for particles with opposite directions\n\\begin{equation}\n- \\partial_r V(0) = \\partial_r V(L) = v_0.\n\\end{equation}\nbecause of mutual hindrance.\\\\\n\nWe can solve the steady state problem with the normalisation condition\n\\begin{equation}\n\\int_0^L \\mathrm{d}r \\, \\sum_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r) = 1\n\\label{normalisation}\n\\end{equation}\nthe bulk equations\n\\begin{equation}\n\\dot{P}_{\\alpha_1\\alpha_2}(r \\in ]0, L[) = 0\n\\label{bulk}\n\\end{equation}\nthe left boundary equations\n\\begin{equation}\n\\int_{0^-}^{0^+} \\mathrm{d}r \\, \\dot{P}_{\\alpha_1\\alpha_2}(r) = 0\n\\label{leftboundary}\n\\end{equation}\nand the right boundary equations\n\\begin{equation}\n\\int_{L^-}^{L^+} \\mathrm{d}r \\, \\dot{P}_{\\alpha_1\\alpha_2}(r) = 0.\n\\label{rightboundary}\n\\end{equation}\n\n\\subsection{Biased ensemble}\n\nWe define our biasing quantity\n\\begin{equation}\n\\dot{Z} = \\sum_{i=1}^2 - \\partial_{r_i} V(r) \\alpha_i v_0 = v_0 (\\alpha_1 - \\alpha_2) \\partial_r V(r)\n\\end{equation}\nwhich is non-zero only at contact, and the tilted generator\n\\begin{equation}\n\\mathcal{W}^{\\dagger}_{\\alpha_1\\alpha_2} = \\mathcal{L}^{\\dagger}_{\\alpha_1\\alpha_2} - s \\dot{Z}\n\\end{equation}\nwith $s$ the biasing parameter, which eigenvalue problem\n\\begin{equation}\n\\psi P_{\\alpha_1\\alpha_2}(r) = \\mathcal{W}^{\\dagger}_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r)\n\\label{eigenproblem}\n\\end{equation}\nwe have to solve for $\\psi(s)$ its largest eigenvalue, which defines the dynamical free energy.\\\\\n\nWhile the symmetries (Eqs. (\\ref{evensymmetry}, \\ref{oddsymmetry})) and the normalisation condition (Eq. (\\ref{normalisation})) remains unchanged, we have that the bulk equations (Eq. (\\ref{bulk})), the left boundary equations (Eq. \\ref{leftboundary}), and the right boundary equations (Eq. \\ref{rightboundary}) become\n\\begin{eqnarray}\n\\label{bulk:biased}\n\\psi P_{\\alpha_1\\alpha_2}(r \\in ]0, L[) = \\mathcal{W}^{\\dagger}_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r \\in ]0, L[)\\\\\n\\label{leftboundary:biased}\n\\int_{0^-}^{0^+} \\mathrm{d}r \\, \\psi P_{\\alpha_1\\alpha_2}(r) = \\int_{0^-}^{0^+} \\mathrm{d}r \\, \\mathcal{W}^{\\dagger}_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r)\\\\\n\\label{rightboundary:biased}\n\\int_{L^-}^{L^+} \\mathrm{d}r \\, \\psi P_{\\alpha_1\\alpha_2}(r) = \\int_{L^-}^{L^+} \\mathrm{d}r \\, \\mathcal{W}^{\\dagger}_{\\alpha_1\\alpha_2} P_{\\alpha_1\\alpha_2}(r)\n\\end{eqnarray}\nrespectively.\n\n\\section{Unbiased steady state distribution}\n\nWe use the following ansatz for the unbiased ($s = 0$) steady state distribution\n\\begin{equation}\nP_{\\alpha_1\\alpha_2}(r) = a_{\\alpha_1\\alpha_2} + b_{\\alpha_1\\alpha_2} \\delta(r) + c_{\\alpha_1\\alpha_2} \\delta(L - r)\n\\end{equation}\nwhich satisfies the normalisation condition (Eq. (\\ref{normalisation}))\n\\begin{equation}\n\\sum_{\\alpha_1\\alpha_2} \\left(L a_{\\alpha_1\\alpha_2} + b_{\\alpha_1\\alpha_2} + c_{\\alpha_1\\alpha_2}\\right) = 1\n\\end{equation}\nthe bulk equations (Eq. (\\ref{bulk}))\n\\begin{eqnarray}\n\\tau^{-1} (a_{+-} + a_{-+} - 2 a_{++}) = 0\\\\\n\\tau^{-1} (a_{+-} + a_{-+} - 2 a_{--}) = 0\\\\\n\\tau^{-1} (a_{++} + a_{--} - 2 a_{+-}) = 0\\\\\n\\tau^{-1} (a_{++} + a_{--} - 2 a_{-+}) = 0\n\\end{eqnarray}\nthe left boundary equations (Eq. (\\ref{leftboundary}))\n\\begin{eqnarray}\n\\tau^{-1}(b_{+-} + b_{-+} - 2 b_{++}) = 0\\\\\n\\tau^{-1}(b_{+-} + b_{-+} - 2 b_{--}) = 0\\\\\n2 v_0 a + \\tau^{-1}(b_{++} + b_{--} - 2 b_{+-}) = 0\\\\\n- 2 v_0 a + \\tau^{-1}(b_{++} + b_{--} - 2 b_{-+}) = 0\n\\end{eqnarray}\nthe right boundary equations (Eq. (\\ref{rightboundary}))\n\\begin{eqnarray}\n\\tau^{-1}(c_{+-} + c_{-+} - 2 c_{++}) = 0\\\\\n\\tau^{-1}(c_{+-} + c_{-+} - 2 c_{--}) = 0\\\\\n- 2 v_0 a + \\tau^{-1}(c_{++} + c_{--} - 2 c_{+-}) = 0\\\\\n2 v_0 a + \\tau^{-1}(c_{++} + c_{--} - 2 c_{-+}) = 0\n\\end{eqnarray}\nand the symmetries (Eqs. (\\ref{evensymmetry}, \\ref{oddsymmetry})).\\\\\n\nWe get\n\\begin{eqnarray}\n\\label{s0++}\nP_{++}(r) = a + l a \\delta(r) + l a \\delta(L - r)\\\\\n\\label{s0--}\nP_{--}(r) = a + l a \\delta(r) + l a \\delta(L - r)\\\\\n\\label{s0+-}\nP_{+-}(r) = a + 2 l a \\delta(r)\\\\\n\\label{s0-+}\nP_{-+}(r) = a + 2 l a \\delta(L - r)\n\\end{eqnarray}\nwith\n\\begin{equation}\na = \\frac{1}{4(L + 2l)}\n\\label{smooth_unbiased}\n\\end{equation}\nand the persistence length $l = v_0 \\tau$.\n\n\\section{Biased steady state distribution}\n\\label{sec:biased-distribution}\n\n\\subsection{Exact solution}\n\nWe use the following ans\\\"atze for the steady state distribution\n\\begin{eqnarray}\nP_{++}(r) = P_{--}(r) = \\beta(r) + \\gamma_- \\delta(r) + \\gamma_+ \\delta(L - r)\\\\\nP_{+-}(r) = \\varepsilon(r) + \\zeta \\delta(r)\\\\\nP_{-+}(r) = \\theta(r) + \\zeta \\delta(L - r)\\\\\n\\end{eqnarray}\nwith $\\varepsilon(r) = \\theta(L - r)$ according to Eq. (\\ref{oddsymmetry}).\\\\\n\nWe write the eigenvalue equations (Eq. (\\ref{eigenproblem}))\n\\begin{eqnarray}\n\\label{FP1}\n\\psi P_{\\alpha\\alpha}(r) = 2 \\partial_r(P_{\\alpha\\alpha}(r) \\partial_r V(r)) + \\tau^{-1} (P_{+-}(r) + P_{-+}(r) - 2 P_{\\alpha\\alpha}(r))\\\\\n\\label{FP2}\n\\psi P_{+-}(r) = 2 v_0 \\partial_r P_{+-}(r) + 2 \\partial_r(P_{+-}(r) \\partial_r V(r)) + \\tau^{-1}(2 P_{\\alpha\\alpha}(r) - 2 P_{+-}(r)) - 2 s v_0 \\partial_r V(r) P_{+-}(r)\\\\\n\\label{FP3}\n\\psi P_{-+}(r) = - 2 v_0 \\partial_r P_{-+}(r) + 2 \\partial_r(P_{-+}(r) \\partial_r V(r)) + \\tau^{-1}(2 P_{\\alpha\\alpha}(r) - 2 P_{-+}(r)) + 2 s v_0 \\partial_r V(r) P_{-+}(r)\n\\end{eqnarray}\nand integrate from Eqs. (\\ref{FP1}, \\ref{FP2}, \\ref{FP3}) the sum of $\\psi(2 P_{\\alpha\\alpha} + P_{+-} + P_{-+})$ between $0^-$ and $L^+$ to get\n\\begin{equation}\n\\psi = 4 s v_0 \\partial_r V(L) \\zeta = 4 s v_0^2 \\zeta\n\\label{psiZeta}\n\\end{equation}\nlinking the dynamical free energy $\\psi$ and the sticking term $\\zeta$.\\\\\n\nWe have the bulk equations (Eq. (\\ref{bulk:biased}))\n\\begin{eqnarray}\n\\label{bulk:biased1}\n\\psi \\beta(r) = \\tau^{-1} (\\varepsilon(r) + \\theta(r) - 2 \\beta(r))\\\\\n\\label{bulk:biased2}\n\\psi \\epsilon(r) = 2 v_0 \\varepsilon^{\\prime}(r) + \\tau^{-1} (2 \\beta(r) - 2 \\varepsilon(r))\\\\\n\\label{bulk:biased3}\n\\psi \\theta(r) = -2 v_0 \\theta^{\\prime}(r) + \\tau^{-1} (2 \\beta(r) - 2 \\theta(r))\n\\end{eqnarray}\nsuch that Eq. (\\ref{bulk:biased1}) gives\n\\begin{equation}\n\\beta(r) = (2 + \\tau\\psi)^{-1} (\\epsilon(r) + \\theta(r))\n\\end{equation}\nand the difference and sum of Eqs. (\\ref{bulk:biased2}, \\ref{bulk:biased3}) give\n\\begin{eqnarray}\n2 l (\\varepsilon(r) + \\theta(r))^{\\prime} = (2 + \\tau \\psi)(\\varepsilon(r) - \\theta(r))\\\\\n\\begin{aligned}\n2 l (\\varepsilon(r) - \\theta(r))^{\\prime} &= -4 \\beta(r) + (\\tau\\psi + 2)(\\varepsilon(r) + \\theta(r))\\\\\n&= \\left((\\tau\\psi + 2) - \\frac{4}{\\tau\\psi + 2}\\right)(\\varepsilon(r) + \\theta(r))\n\\end{aligned}\n\\end{eqnarray}\nwhere we set $A(r) = \\varepsilon(r) + \\theta(r)$ and $B(r) = \\varepsilon(r) - \\theta(r)$, which on the one hand verify\n\\begin{equation}\nA^{\\prime\\prime}(r) - k^2 A(r) = 0\n\\end{equation}\nwhere\n\\begin{equation}\nk^2 l^2 = \\tau \\psi \\left(\\frac{\\tau \\psi}{4} + 1\\right)\n\\label{kPsi}\n\\end{equation}\nand which general solution is\n\\begin{equation}\nA(r) = A_+ e^{- k r} + A_- e^{-k (L - r)}\n\\end{equation}\nand on the other hand\n\\begin{equation}\nB(r) = l(1 + \\tau \\psi/2)^{-1} A^{\\prime}(r) = kl (1 + \\tau\\psi/2)^{-1} (A_- e^{-k(L - r)} - A_+ e^{-k r})\n\\end{equation}\nfrom which we infer\n\\begin{eqnarray}\n\\label{epsilon}\n\\begin{aligned}\n\\varepsilon(r) &= \\frac{1}{2} (A(r) + B(r))\\\\\n&= \\frac{1}{2}\\left(1 - k l (1 + \\psi \\tau/2)^{-1}\\right) A_+ e^{- k r} + \\frac{1}{2}\\left(1 + k l (1 + \\psi \\tau/2)^{-1}\\right) A_- e^{- k (L - r)}\n\\end{aligned}\n\\mbox{}\\\\\n\\label{theta}\n\\begin{aligned}\n\\theta(r) &= \\frac{1}{2} (A(r) - B(r))\\\\\n&= \\frac{1}{2}\\left(1 + k l (1 + \\psi \\tau/2)^{-1}\\right) A_+ e^{- k r} + \\frac{1}{2}\\left(1 - k l (1 + \\psi \\tau/2)^{-1}\\right) A_- e^{- k (L - r)}\n\\end{aligned}\n\\end{eqnarray}\nwhere we note that the symmetry condition of Eq. (\\ref{oddsymmetry}) implies that $A_- = A_+$, and\n\\begin{equation}\n\\begin{aligned}\n\\beta(r) &= (\\tau\\psi + 2)^{-1}(\\varepsilon(r) + \\theta(r)) = (\\tau \\psi + 2)^{-1} A(r)\\\\\n&= (1 + \\tau\\psi/2)^{-1}(A_+ e^{-k r} + A_- e^{-k (L - r)})\n\\end{aligned}\n\\label{beta}\n\\end{equation}\nwhere we need to determine $A_+ = A_-$.\\\\\n\nWe have the left and right boundary equations (Eqs. (\\ref{leftboundary:biased}, \\ref{rightboundary:biased})) for $P_{\\alpha\\alpha}$\n\\begin{eqnarray}\n\\psi \\gamma_- = \\tau^{-1} (\\zeta - 2 \\gamma_-)\\\\\n\\psi \\gamma_+ = \\tau^{-1} (\\zeta - 2 \\gamma_+)\n\\end{eqnarray}\nsuch that\n\\begin{equation}\n\\gamma = \\gamma_- = \\gamma_+ = (\\tau \\psi + 2)^{-1} \\zeta\n\\label{ratio_gamma}\n\\end{equation}\nwhich with Eq. (\\ref{psiZeta}) gives\n\\begin{equation}\n\\psi = 4 s v_0^2 (\\tau \\psi + 2) \\gamma\n\\label{gammaPsi}\n\\end{equation}\nwhere we need to determine $\\gamma$.\\\\\n\nWe have the left boundary equations (Eq. (\\ref{leftboundary:biased})) for $P_{\\alpha\\overline{\\alpha}}$\n\\begin{eqnarray}\n\\label{LBe1}\n\\psi \\zeta = 2 v_0 \\varepsilon(0^+) + \\tau^{-1} (2 \\gamma - 2 \\zeta) - 2 s v_0 \\partial_r V(0) \\zeta\\\\\n\\label{LBe2}\n0 = - 2 v_0 \\varepsilon(L^-) + \\tau^{-1} 2 \\gamma\n\\end{eqnarray}\nfrom which we get\n\\begin{eqnarray}\n\\label{S1}\n\\left[(\\tau\\psi + 2)(\\tau\\psi + 2 - 2 s l v_0) - 2\\right] \\gamma - l \\left(1 - kl (1 + \\tau\\psi/2)^{-1}\\right) A_+ - l \\left(1 + kl(1 + \\tau\\psi/2)^{-1}\\right) e^{-kL} A_- = 0\\\\\n\\label{S2}\n2 \\gamma - l \\left(1 - kl (1 + \\tau\\psi/2)^{-1}\\right) e^{-kL} A_+ - l \\left(1 + kl (1 + \\tau\\psi/2)^{-1}\\right) A_- = 0\n\\end{eqnarray}\nand the normalisation condition (Eq. (\\ref{normalisation}))\n\\begin{equation}\n(2 \\tau \\psi + 8) \\gamma + \\frac{1}{k} (1 - e^{-kL})(1 + (1 + \\tau\\psi/2)^{-1})(A_+ + A_-) = 1\n% 2((\\tau\\psi + 2) + 1) \\gamma + \\frac{1}{k} \\left(1 - e^{-kL}\\right)\\left(1 + (\\tau\\psi + 2)^{-1}\\right) A_+ + \\frac{1}{k} \\left(1 - e^{-kL}\\right)\\left(1 + (\\tau\\psi + 2)^{-1}\\right) A_- = 1\n\\label{S3}\n\\end{equation}\nso we can solve the system of Eqs. (\\ref{S1}, \\ref{S2}, \\ref{S3}) with respect to $\\gamma$, $A_+$, $A_-$, or equivalently with $A_+ = A_-$\n\\begin{eqnarray}\n\\label{S01}\n\\left[(\\tau\\psi + 2)(\\tau\\psi + 2 - 2 s l v_0) - 2\\right] \\gamma - l \\left[\\left(1 - kl (1 + \\tau\\psi/2)^{-1}\\right) + \\left(1 + kl(1 + \\tau\\psi/2)^{-1}\\right) e^{-kL}\\right] A_- = 0\\\\\n\\label{S02}\n% 2((\\tau\\psi + 2) + 1) \\gamma + \\frac{2}{k} \\left(1 - e^{-kL}\\right)\\left(1 + (\\tau\\psi + 2)^{-1}\\right) A_+ = 1\n(2 \\tau \\psi + 8) \\gamma + \\frac{2}{k} (1 - e^{-kL})(1 + (1 + \\tau\\psi/2)^{-1}) A_+ = 1\n\\end{eqnarray}\nwith respect to $A_+$ and $\\gamma$.\\\\\n\nWe use \\textsc{SageMath} to solve Eqs. (\\ref{S01}, \\ref{S02})\n\\begin{lstlisting}[backgroundcolor=\\color{lightgray!20!white}, language=Python, xleftmargin=0.5cm, xrightmargin=0.5cm, framexleftmargin = 0.5em, frame=tlbr,framesep=4pt]\n# variables\npsi = var('psi')\ntau = var('tau')\nk = var('k')\nL = var('L')\nl = var('l')\nv0 = var('v0')\ns = var('s')\n\n# system [gamma, A]\nsystem = Matrix([\n\n [(tau*psi + 2)*(tau*psi + 2 - 2*s*l*v0) - 2,\n -l*((1 - k*l/(1 + (tau*psi)/2)) + (1 + k*l/(1 + (tau*psi)/2))*exp(-k*L))],\n\n [2*tau*psi + 8,\n (2/k)*(1 - exp(-k*L))*(1 + 1/(tau*psi + 2))]\n\n])\n\n# solution\n[[gamma], [A]] = system \\ Matrix([[0], [1]])\n\\end{lstlisting}\nand get\n\\begin{eqnarray}\n\\gamma = -\\frac{1}{2 \\, {\\left(\\frac{{\\left(\\psi \\tau + 4\\right)} {\\left({\\left(\\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} e^{\\left(-L k\\right)} - \\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} l}{{\\left(2 \\, l s v_{0} - \\psi \\tau - 2\\right)} {\\left(\\psi \\tau + 2\\right)} + 2} + \\frac{{\\left(\\frac{1}{\\psi \\tau + 2} + 1\\right)} {\\left(e^{\\left(-L k\\right)} - 1\\right)}}{k}\\right)}}\\\\\nA_+ = A_- = \\frac{{\\left({\\left(\\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} e^{\\left(-L k\\right)} - \\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} l}{2 \\, {\\left({\\left(2 \\, l s v_{0} - \\psi \\tau - 2\\right)} {\\left(\\psi \\tau + 2\\right)} + 2\\right)} {\\left(\\frac{{\\left(\\psi \\tau + 4\\right)} {\\left({\\left(\\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} e^{\\left(-L k\\right)} - \\frac{2 \\, k l}{\\psi \\tau + 2} + 1\\right)} l}{{\\left(2 \\, l s v_{0} - \\psi \\tau - 2\\right)} {\\left(\\psi \\tau + 2\\right)} + 2} + \\frac{{\\left(\\frac{1}{\\psi \\tau + 2} + 1\\right)} {\\left(e^{\\left(-L k\\right)} - 1\\right)}}{k}\\right)}}.\n\\end{eqnarray}\n\n\\subsection{Scaling regime}\n\nWe assume there exists a scaling form\n\\begin{equation}\n\\psi(s) = L^{-2} \\varphi(s L)\n\\end{equation}\nso that for $s = \\mathcal{O}(L^{-1})$ we have $\\psi(s) = \\mathcal{O}(L^{-2})$ and $k = \\mathcal{O}(L^{-1})$ from Eq. (\\ref{kPsi}).\\\\\n\nWe give an expression of Eqs. (\\ref{S2}, \\ref{S3}) at the lowest order of $L^{-1}$ and using $A_+ = A_-$\n\\begin{eqnarray}\n\\label{SC1}\n2 \\gamma - l(1 + e^{-kL}) A_+ = 0\\\\\n\\label{SC2}\n8 \\gamma + \\frac{4}{k} (1 - e^{-kL}) A_+ = 1\n\\end{eqnarray}\nfrom which we infer\n\\begin{equation}\n\\label{SCgamma}\n\\gamma = \\frac{1}{8} k l \\frac{1 + e^{-kL}}{1 - e^{-kL}}\n\\end{equation}\nand an expression of Eq. (\\ref{gammaPsi}) at the lowest order of $L^{-1}$\n\\begin{equation}\n\\gamma = \\frac{\\psi}{8 s v_0^2}\n\\end{equation}\nwhich with Eq.(\\ref{kPsi}) linking $k$ and $\\psi$, also at the lowest order of $L^{-1}$,\n\\begin{equation}\n\\label{SCkl}\nk^2 l^2 = \\tau \\psi \\Rightarrow k = \\begin{cases} \\frac{1}{l} \\sqrt{\\tau\\psi} &\\text{ if } s > 0, \\\\ 0 &\\text{ if } s = 0, \\\\ \\mathrm{i} \\, \\frac{1}{l} \\sqrt{- \\tau\\psi} &\\text{ if } s < 0, \\end{cases}\n\\end{equation}\ngive\n\\begin{eqnarray}\n\\frac{2}{s L v_0} = \\frac{\\coth\\left(\\sqrt{\\frac{\\psi L^2}{4 l v_0}}\\right)}{\\sqrt{\\frac{\\psi L^2}{4 l v_0}}} \\Leftrightarrow \\frac{1}{\\Lambda} = \\frac{\\coth\\left(\\sqrt{\\Psi}\\right)}{\\sqrt{\\Psi}} &\\qquad\\qquad s > 0, \\psi > 0\\\\\n- \\frac{2}{s L v_0} = \\frac{\\cot\\left(\\sqrt{\\frac{|\\psi| L^2}{4 l v_0}}\\right)}{\\sqrt{\\frac{|\\psi| L^2}{4 l v_0}}} \\Leftrightarrow - \\frac{1}{\\Lambda} = \\frac{\\cot\\left(\\sqrt{|\\Psi|}\\right)}{\\sqrt{|\\Psi|}} &\\qquad\\qquad s < 0, \\psi < 0\n\\end{eqnarray}\nwhere we have set\n\\begin{eqnarray}\n\\label{Lambda}\n\\Lambda = \\frac{s L v_0}{2}\\\\\n\\label{Phi}\n\\Psi = \\frac{\\psi L^2}{4 l v_0} = \\frac{\\varphi}{4 l v_0}\n\\end{eqnarray}\nfor convenience, and introduce\n\\begin{equation}\n\\Gamma = \\frac{\\Psi}{\\Lambda} = \\frac{\\psi}{8 s v_0^2} \\frac{4 L}{l} = \\gamma \\frac{4 L}{l}\n\\label{eqGamma}\n\\end{equation}\nwhich we plot in Fig. \\ref{Gamma}.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.5\\textwidth]{gamma.eps}\n\\caption{Rescaled sticking term $\\Gamma = (4L/l) \\, \\gamma$ as a function of rescaled biasing parameter $\\Lambda = s L v_0/2$.}\n\\label{Gamma}\n\\end{figure}\n\nAt the lowest order in $L^{-1}$ we have from Eqs. (\\ref{epsilon}, \\ref{theta}, \\ref{beta})\n\\begin{equation}\n\\varepsilon(r) = \\theta(r) = \\beta(r) = \\frac{1}{2} A_+ (e^{-kr} + e^{-k(L-r)})\n\\end{equation}\nand from Eqs. (\\ref{SC1}, \\ref{SC2}, \\ref{SCgamma})\n\\begin{equation}\nA_+ = \\frac{2\\gamma}{l} \\frac{1}{1 + e^{-kL}} = \\frac{k}{4(1 - e^{-kL})}\n\\end{equation}\nso we can write with Eqs. (\\ref{SCkl}, \\ref{Lambda}, \\ref{Phi}, \\ref{eqGamma})\n\\begin{equation}\nL \\varepsilon(r) = \\frac{1}{4} \\Gamma \\begin{cases} \\frac{\\cosh\\left(\\sqrt{\\Psi}(1 - 2r/L)\\right)}{\\cosh\\left(\\sqrt{\\Psi}\\right)} &\\text{ if } s > 0, \\\\ 1 &\\text{ if } s = 0, \\\\ \\frac{\\cos\\left(\\sqrt{|\\Psi|}(1 - 2r/L)\\right)}{\\cos\\left(\\sqrt{|\\Psi|}\\right)} &\\text{ if } s < 0, \\end{cases}\n\\end{equation}\nwhich we plot in Fig. \\ref{fig:epsilon}.\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.5\\textwidth]{epsilon.eps}\n\\caption{Rescaled regular part $L \\varepsilon(r)$ as a function of rescaled distance $r/L$ for different values of the rescaled biasing parameter $\\Lambda = s L v_0/2$.}\n\\label{fig:epsilon}\n\\end{figure}\n\n\\section{Solution to the adjoint eigenproblem}\n\n\\subsection{General solution}\n\nWe have computed the left eigenvectors $P_{\\alpha_1\\alpha_2}$ of Eq.~\\ref{eigenproblem} corresponding to the SCGF $\\psi(\\lambda)$ in Sec.~\\ref{sec:biased-distribution}, and will now compute the right eigenvectors $Q_{\\alpha_1\\alpha_2}$.\nWe introduce $Y_{\\alpha_1\\alpha_2}(r) = P_{\\alpha_1\\alpha_2}(r) Q_{\\alpha_1\\alpha_2}(r)$ which has the same symmetries as $P_{\\alpha_1\\alpha_2}(r)$, so that $Q_{\\alpha_1\\alpha_2}(r)$ also has these properties.\\\\\n\nWe assume that the normalisation condition\n\\begin{equation}\n\\int_0^L \\sum_{\\alpha_1, \\alpha_2 = \\pm 1} Y_{\\alpha_1\\alpha_2}(r) \\, \\mathrm{d}r = 1\n\\label{normalisation_pave}\n\\end{equation}\nas well as the general form\n\\begin{eqnarray}\n\\label{gfY1}\nY_{\\alpha\\alpha}(r) = \\hat{\\varepsilon}_{\\alpha\\alpha}(r) + \\hat{\\gamma}_{\\alpha\\alpha} \\delta (r) + \\hat{\\gamma}_{\\alpha\\alpha} \\delta (L - r)\\\\\n\\label{gfY2}\nY_{+-}(r) = \\hat{\\varepsilon}_{+-}(r) + \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\delta (r)\\\\\n\\label{gfY3}\nY_{-+}(r) = \\hat{\\varepsilon}_{-+}(r) + \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\delta (L - r)\n\\end{eqnarray}\nalso hold for $Y_{\\alpha_1\\alpha_2}$.\\\\\n\nWe write the hermitian adjoints of Eqs.~\\ref{FP1}, \\ref{FP2}, \\ref{FP3} for the right eigenvectors\n\\begin{eqnarray}\n\\label{FPr1}\n\\psi Q_{\\alpha\\alpha}(r) = -2 \\partial_r V(r) \\partial_r Q_{\\alpha\\alpha}(r) + \\tau^{-1} (Q_{\\alpha\\overline{\\alpha}}(r) + Q_{\\overline{\\alpha}\\alpha}(r) - 2 Q_{\\alpha\\alpha}(r))\\\\\n\\label{FPr2}\n\\psi Q_{\\alpha\\overline{\\alpha}}(r) = - \\alpha 2 v_0 \\partial_r Q_{\\alpha\\overline{\\alpha}}(r) - 2 \\partial_r V(r) \\partial_r Q_{\\alpha\\overline{\\alpha}}(r) + \\tau^{-1} (2 Q_{\\alpha\\alpha}(r) - 2 Q_{\\alpha\\overline{\\alpha}}(r)) - \\alpha 2 s v_0 \\partial_r V(r) Q_{\\alpha\\overline{\\alpha}}(r)\n\\end{eqnarray}\nfollowing the symmetries we have introduced.\\\\\n\nWe notice that the bulk equations for Eqs.~\\ref{FPr1}, \\ref{FPr2} are the bulk equations \\eqref{bulk:biased1}, \\eqref{bulk:biased2}, \\eqref{bulk:biased3} under the transformation $\\alpha \\to \\overline{\\alpha}$ so we get on $]0;L[$\n\\begin{eqnarray}\n\\label{qaa}\nQ_{\\alpha\\alpha}(r) = A^{\\prime} (\\tau \\psi + 2)^{-1} (e^{-k r} + e^{-k (L - r)})\\\\\n\\label{qaoa}\n2 Q_{\\alpha\\overline{\\alpha}}(r) = \\left(1 + \\frac{\\alpha 2 k l}{\\tau \\psi + 2}\\right) A^{\\prime} e^{- k r} + \\left(1 - \\frac{\\alpha 2 k l}{\\tau \\psi + 2}\\right) A^{\\prime} e^{- k (L - r)}\n\\end{eqnarray}\nwhere $k$ is defined by Eq.~\\ref{kPsi} and $A^{\\prime}$ is a new constant.\n\n\\subsubsection{Complete determination method}\n\nWe write for $Y_{\\alpha_1\\alpha_2}$ the equations\n\\begin{eqnarray}\n\\label{FPrY1}\n\\begin{aligned}\n\\psi Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha}(r)}{P_{\\alpha\\overline{\\alpha}}(r)} = &\\left[- \\alpha 2 v_0 \\partial_r Q_{\\alpha\\overline{\\alpha}}(r) - 2 \\partial_r V(r) \\partial_r Q_{\\alpha\\overline{\\alpha}}(r)\\right] P_{\\alpha\\alpha}(r)\\\\\n&+ \\tau^{-1} \\left(2 Y_{\\alpha\\alpha}(r) - 2 Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha}(r)}{P_{\\alpha\\overline{\\alpha}}(r)}\\right) - \\alpha 2 s v_0 \\partial_r V(r) Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha}(r)}{P_{\\alpha\\overline{\\alpha}}(r)}\n\\end{aligned}\\\\\n% \\label{FPrY1}\n% \\psi Y_{\\alpha\\alpha}(r) = - 2 \\partial_r V(r) P_{\\alpha\\alpha}(r) \\partial_r Q_{\\alpha\\alpha}(r) + \\tau^{-1} \\left(Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha(r)}}{P_{\\alpha\\overline{\\alpha}}(r)} + Y_{\\overline{\\alpha}\\alpha}(r) \\frac{P_{\\alpha\\alpha(r)}}{P_{\\overline{\\alpha}\\alpha}(r)} - 2 Y_{\\alpha\\alpha}(r)\\right)\\\\\n% \\label{FPrY2}\n% \\psi Y_{\\alpha\\alpha}(r) = 2 Q_{\\alpha\\alpha}(r) \\partial_r (P_{\\alpha\\alpha}(r) \\partial_r V(r)) + \\tau^{-1} \\left(Y_{\\alpha\\alpha}(r) \\frac{P_{\\alpha\\overline{\\alpha}}(r)}{P_{\\alpha\\alpha}(r)} + Y_{\\alpha\\alpha}(r) \\frac{P_{\\overline{\\alpha}\\alpha}(r)}{P_{\\alpha\\alpha}(r)} - 2 Y_{\\alpha\\alpha}(r)\\right)\\\\\n\\label{FPrY3}\n\\begin{aligned}\n\\psi Y_{\\alpha\\overline{\\alpha}}(r) = &- \\alpha 2 v_0 P_{\\alpha\\overline{\\alpha}}(r) \\partial_r Q_{\\alpha\\overline{\\alpha}}(r) - 2 \\partial_r V(r) P_{\\alpha\\overline{\\alpha}}(r) \\partial_r Q_{\\alpha\\overline{\\alpha}}(r) + \\tau^{-1} \\left(2 Y_{\\alpha\\alpha}(r) \\frac{P_{\\alpha\\overline{\\alpha}}(r)}{P_{\\alpha\\alpha}(r)} - 2 Y_{\\alpha\\overline{\\alpha}}(r)\\right)\\\\\n&- \\alpha 2 s v_0 \\partial_r V(r) Y_{\\alpha\\overline{\\alpha}}(r)\n\\end{aligned}\\\\\n\\label{FPrY4}\n\\begin{aligned}\n\\psi Y_{\\alpha\\overline{\\alpha}}(r) = &\\alpha 2 v_0 Q_{\\alpha\\overline{\\alpha}}(r) \\partial_r P_{\\alpha\\overline{\\alpha}}(r) + 2 Q_{\\alpha\\overline{\\alpha}}(r) \\partial_r (Q_{\\alpha\\overline{\\alpha}}(r) \\partial_r V(r)) + \\tau^{-1} \\left(2 Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha}(r)}{P_{\\alpha\\overline{\\alpha}}(r)} - 2 Y_{\\alpha\\overline{\\alpha}}(r)\\right)\\\\\n&- \\alpha 2 s v_0 \\partial_r V(r) Y_{\\alpha\\overline{\\alpha}}(r)\n\\end{aligned}\n\\end{eqnarray}\nwhere \\eqref{FPrY1} is derived from the multiplication of \\eqref{FPr2} by $P_{\\alpha\\alpha}(r)$,\n% where \\eqref{FPrY1} is derived from the multiplication of \\eqref{FPr1} by $P_{\\alpha\\alpha}(r)$, \\eqref{FPrY2} is derived from the multiplication of \\eqref{FP1} by $Q_{\\alpha\\alpha}(r)$,\n\\eqref{FPrY3} is derived from the multiplication of \\eqref{FPr2} by $P_{\\alpha\\overline{\\alpha}}(r)$, \\eqref{FPrY4} is derived from the multiplication of (\\ref{FP2}, \\ref{FP3}) by $Q_{\\alpha\\overline{\\alpha}}(r)$\n, and finally substract\n%\n% We substract Eqs.~\\ref{FPrY1},~\\ref{FPrY2}\n% \\begin{equation}\n% 0 = - 2 \\partial_r (\\partial_r V(r) Y_{\\alpha\\alpha}(r)) + \\tau^{-1} \\left(Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha(r)}}{P_{\\alpha\\overline{\\alpha}}(r)} + Y_{\\overline{\\alpha}\\alpha}(r) \\frac{P_{\\alpha\\alpha(r)}}{P_{\\overline{\\alpha}\\alpha}(r)} - Y_{\\alpha\\alpha}(r) \\frac{P_{\\alpha\\overline{\\alpha}}(r)}{P_{\\alpha\\alpha}(r)} - Y_{\\alpha\\alpha}(r) \\frac{P_{\\overline{\\alpha}\\alpha}(r)}{P_{\\alpha\\alpha}(r)}\\right)\n% \\end{equation}\n% and also\nEqs.~\\ref{FPrY3},~\\ref{FPrY4}\n\\begin{equation}\n0 = - \\alpha 2 v_0 \\partial_r Y_{\\alpha\\overline{\\alpha}}(r) - 2 \\partial_r (Y_{\\alpha\\overline{\\alpha}}(r) \\partial_r V(r)) + \\tau^{-1} \\left(2 Y_{\\alpha\\alpha}(r) \\frac{P_{\\alpha\\overline{\\alpha}}(r)}{P_{\\alpha\\alpha}(r)} - 2 Y_{\\alpha\\overline{\\alpha}}(r) \\frac{P_{\\alpha\\alpha}(r)}{P_{\\alpha\\overline{\\alpha}}(r)}\\right)\n\\end{equation}\nto get equations where derivatives should have nice integrals.\\\\\n\nWe integrate Eqs.~\\ref{FPrY1},~\\ref{FPrY4} between $0^-$ and $0^+$\n\\begin{eqnarray}\n\\label{BCr1}\n\\psi \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}} = \\tau^{-1} \\left(2 \\hat{\\gamma}_{\\alpha\\alpha} - 2 \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}\\right) + 2 s v_0^2 \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}\\\\\n\\label{BCr2}\n0 = - v_0 \\hat{\\varepsilon}_{+-}(0^+) + \\tau^{-1} \\left(\\hat{\\gamma}_{\\alpha\\alpha} \\frac{\\gamma_{\\alpha\\overline{\\alpha}}}{\\gamma_{\\alpha\\alpha}} - \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}\\right)\n\\end{eqnarray}\nwhere we have discarded in \\eqref{BCr1} the integral of the first term of the right hand side of \\eqref{FPrY1} since $\\partial_r V(0) = -v_0$, and with\n\\begin{eqnarray}\n\\label{he+-}\n\\begin{aligned}\n\\hat{\\varepsilon}_{+-}(r) &= Q_{+-}(r) P_{+-}(r) \\qquad (r \\in ]0; L[)\\\\\n&= \\frac{A A^{\\prime}}{4} \\left[1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right](e^{-2 k r} + e^{-2 k (L - r)})\\\\\n&\\quad+ \\frac{A A^{\\prime}}{4} \\left[\\left(1 + \\frac{2 k l}{\\tau \\psi + 2}\\right)^2 + \\left(1 - \\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right] e^{-k L}\\\\\n&= \\frac{A A^{\\prime}}{4} \\left[1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right](e^{-2 k r} + e^{-2 k (L - r)}) + \\frac{A A^{\\prime}}{2} \\left[1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right] e^{-k L}\n\\end{aligned}\\\\\n\\end{eqnarray}\nwhere we will introduce\n\\begin{equation}\n\\Omega = \\frac{2 k l}{\\tau \\psi + 2}\n\\label{Omega}\n\\end{equation}\nto get\n\\begin{eqnarray}\n\\label{he+-lim}\n\\begin{aligned}\n\\hat{\\varepsilon}_{+-}(0^+) = \\lim_{r \\to 0^+} \\hat{\\varepsilon}_{+-}(r) &= \\frac{A A^{\\prime}}{4} \\left[1 - \\Omega^2\\right](1 + e^{-2 k L}) + \\frac{A A^{\\prime}}{2} \\left[1 + \\Omega^2\\right] e^{-k L}\\\\\n&= A A^{\\prime} e^{-k L} \\left[\\left(\\frac{e^{kL} + e^{-k L}}{2}\\right)^2 - \\Omega^2 \\left(\\frac{e^{kL} - e^{-k L}}{2}\\right)^2\\right]\n% &= \\frac{A A^{\\prime}}{2} e^{-k L}\\left[\\left(1 - \\left(\\frac{2 k l}{\\tau\\psi + 2}\\right)^2\\right) \\cosh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau\\psi + 2}\\right)^2\\right)\\right]\n\\end{aligned}\n\\end{eqnarray}\nfrom Eqs.~\\ref{epsilon},~\\ref{qaoa}, with $A = A_{\\pm}$ which appear in Eqs.~\\ref{epsilon}, \\ref{theta}, \\ref{beta}. We solve Eqs.~\\ref{BCr1},~\\ref{BCr2} for $\\hat{\\gamma}_{\\alpha\\alpha}$ and $\\hat{\\gamma}_{\\alpha\\overline{\\alpha}}$\n\\begin{eqnarray}\n\\label{hgaa}\n\\hat{\\gamma}_{\\alpha\\alpha} = \\frac{[(\\tau\\psi + 2) - 2 s l v_0] l \\hat{\\varepsilon}_{+-}(0^+)}{(\\tau \\psi + 2) [(\\tau\\psi + 2) - 2 s l v_0] - 2}\\\\\n% \\hat{\\gamma}_{\\alpha\\alpha} = \\frac{\\left[\\frac{1}{2} (\\tau \\psi + 2) - s l v_0\\right] \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}} l \\hat{\\varepsilon}_{+-}(0^+)}{\\frac{1}{2}(\\tau\\psi + 2) - s l v_0 - \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}}\\\\\n\\label{hgaoa}\n\\hat{\\gamma}_{\\alpha\\overline{\\alpha}} = \\frac{2 (\\tau\\psi + 2) l \\hat{\\varepsilon}_{+-}(0^+)}{(\\tau \\psi + 2) [(\\tau\\psi + 2) - 2 s l v_0] - 2}\n% \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} = \\frac{l \\hat{\\varepsilon}_{+-}(0^+)}{\\frac{1}{2}(\\tau\\psi + 2) - s l v_0 - \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}}\n\\end{eqnarray}\nas a function of $\\hat{\\varepsilon}_{+-}(0^+)$ and where we have used $\\gamma_{\\alpha\\alpha}/\\gamma_{\\alpha\\overline{\\alpha}} = (\\tau \\psi + 2)^{-1}$ \\eqref{ratio_gamma}.\\\\\n\nWe can finally write the normalisation condition \\eqref{normalisation_pave}\n\\begin{equation}\n\\begin{aligned}\n1 &= 4 \\hat{\\gamma}_{\\alpha\\alpha} + 2 \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) \\, \\mathrm{d}r\\\\\n&= 4 \\hat{\\gamma}_{\\alpha\\alpha} \\left(1 + \\frac{1}{2} \\frac{\\hat{\\gamma}_{\\alpha\\overline{\\alpha}}}{\\hat{\\gamma}_{\\alpha\\alpha}}\\right) + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) \\, \\mathrm{d}r\n\\label{normalisation_pave_full}\n\\end{aligned}\n\\end{equation}\nwhere we have\n\\begin{equation}\n\\hat{\\varepsilon}_{\\alpha\\alpha}(r) = P_{\\alpha\\alpha}(r \\in ]0; L[) Q_{\\alpha\\alpha}(r \\in ]0; L[) = \\frac{A A^{\\prime}}{(\\tau \\psi + 2)^2} \\left(e^{-2 k r} + e^{-2 k (L - r)} + 2 e^{-k L}\\right)\n\\label{heaa}\n\\end{equation}\naccording to Eqs.~\\ref{beta},~\\ref{qaa}. We compute from \\eqref{heaa}\n\\begin{equation}\n\\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r = A A^{\\prime} e^{-k L} \\frac{2}{(\\tau \\psi + 2)^2} \\left[\\frac{1}{k} \\left(\\frac{e^{k L} - e^{-k L}}{2}\\right) + L\\right]\n\\label{int_heaa}\n\\end{equation}\nand from \\eqref{he+-}\n\\begin{equation}\n\\int_0^L \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) \\, \\mathrm{d}r = A A^{\\prime} e^{-k L} \\frac{1}{2} \\left[\\left(1 -\\Omega^2\\right) \\frac{1}{k} \\left(\\frac{e^{k L} - e^{-k L}}{2}\\right) + \\left(1 + \\Omega^2\\right) L\\right]\n\\end{equation}\non the one hand, and use on the other hand \\eqref{he+-lim} to re-write \\eqref{normalisation_pave_full}\n\\begin{equation}\n\\begin{aligned}\n\\tilde{\\mathcal{N}} =~& (A A^{\\prime} e^{-k L} l)^{-1}\\\\\n=~& \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2)[(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left[1 + \\frac{\\tilde{\\psi} + 2}{(\\tilde{\\psi} + 2) - 2 \\tilde{s}}\\right] \\left[\\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)^2 - \\Omega^2 \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right)^2\\right]\\\\\n&+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - \\Omega^2)\\right] \\frac{1}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + \\Omega^2)\\right] \\tilde{L}\n\\end{aligned}\n\\label{normalisation_factor_pave}\n\\end{equation}\n% \\begin{equation}\n% \\begin{aligned}\n% 1 = &\\left[4 \\frac{\\left(\\frac{1}{2}(\\tau \\psi + 2) - s l v_0\\right) \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}} l}{\\frac{1}{2}(\\tau \\psi + 2) - s l v_0 - \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}} + 2 \\frac{l}{\\frac{1}{2}(\\tau \\psi + 2) - s l v_0 - \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}}\\right]\\\\\n% &\\qquad \\times \\frac{A A^{\\prime}}{2} e^{-k L} \\left[\\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\cosh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right)\\right]\\\\\n% &+ \\frac{4 A A^{\\prime}}{(\\tau \\psi + 2)^2} e^{-k L} \\left[\\frac{1}{k}\\sinh(kL) + L\\right]\\\\\n% &+ A A^{\\prime} e^{-k L} \\left[\\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\frac{1}{k} \\sinh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) L\\right]\\\\\n% = & A A^{\\prime} e^{-k L}\\\\\n% &\\times\\Bigg[\\frac{l}{\\frac{1}{2}(\\tau \\psi + 2) - s l v_0 - \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}}} \\left(2 \\left(\\frac{1}{2} (\\tau \\psi + 2) - s l v_0\\right) \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}} + 1\\right)\\\\\n% &\\qquad\\qquad \\times \\left(\\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\cosh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right)\\right)\\\\\n% &\\qquad + \\frac{4}{(\\tau \\psi + 2)^2} \\left(\\frac{1}{k}\\sinh(kL) + L\\right) + \\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\frac{1}{k} \\sinh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) L\\Bigg]\n% \\end{aligned}\n% \\end{equation}\n% and conclude\n% \\begin{equation}\n% \\begin{aligned}\n% A^{\\prime} =& \\frac{e^{k L}}{A}\\\\\n% &\\times\\Bigg[\\frac{l}{\\frac{1}{2}(\\tau \\psi + 2)^2 - s l v_0 (\\tau \\psi + 2) - 1} \\left(2 (\\tau \\psi + 2) - 2 s l v_0\\right)\\\\\n% &\\qquad\\qquad \\times \\left(\\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\cosh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right)\\right)\\\\\n% &\\qquad + \\frac{4}{(\\tau \\psi + 2)^2} \\left(\\frac{1}{k}\\sinh(kL) + L\\right) + \\left(1 - \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) \\frac{1}{k} \\sinh(k L) + \\left(1 + \\left(\\frac{2 k l}{\\tau \\psi + 2}\\right)^2\\right) L\\Bigg]^{-1}\n% \\end{aligned}\n% \\end{equation}\nwhere we have denoted the dimensionless quantities $\\tilde{\\psi} = \\tau \\psi$, $\\tilde{k} = k l$, $\\tilde{L} = L/l$, $\\tilde{s} = s l v_0$, $\\tilde{\\mathcal{N}} = \\mathcal{N}/l$.\n% We can easily check that $\\lim_{s \\to 0} \\tilde{\\mathcal{N}^{-1}} = 8 + 4 \\tilde{L}$, and\n% \\begin{eqnarray}\n% \\label{normalisation_factor_pave_sneg}\n% \\begin{aligned}\n% \\tilde{\\mathcal{N}}^{-1}(s < 0) =~& \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2)[(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left[1 + \\frac{\\tilde{\\psi} + 2}{(\\tilde{\\psi} + 2) - 2 \\tilde{s}}\\right]\\left(\\cos(|\\tilde{k}| \\tilde{L})^2 - |\\Omega|^2 \\sin(|\\tilde{k}| \\tilde{L})^2\\right)\\\\\n% &+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + |\\Omega|^2)\\right] \\frac{1}{|\\tilde{k}|} \\sin(|\\tilde{k}| \\tilde{L}) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - |\\Omega|^2)\\right] \\tilde{L}\n% \\end{aligned}\\\\\n% \\label{normalisation_factor_pave_spos}\n% \\begin{aligned}\n% \\tilde{\\mathcal{N}}^{-1}(s > 0) =~& \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2)[(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left[1 + \\frac{\\tilde{\\psi} + 2}{(\\tilde{\\psi} + 2) - 2 \\tilde{s}}\\right]\\left(\\cosh(\\tilde{k} \\tilde{L})^2 - \\Omega^2 \\sinh(\\tilde{k} \\tilde{L})^2\\right)\\\\\n% &+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - \\Omega^2)\\right] \\frac{1}{\\tilde{k}} \\sinh(\\tilde{k} \\tilde{L}) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + \\Omega^2)\\right] \\tilde{L}\n% \\end{aligned}\n% \\end{eqnarray}\n% depending on the sign of $s$.\\\\\n%\n% We define the polarisation\n% \\begin{equation}\n% \\nu = 4 \\hat{\\gamma}_{\\alpha\\alpha} + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r\n% \\label{polarisation_def}\n% \\end{equation}\n% which we can compute with (\\ref{hgaa}, \\ref{int_heaa}, \\ref{normalisation_factor_pave})\n% \\begin{equation}\n% \\begin{aligned}\n% \\nu =~& \\tilde{\\mathcal{N}} \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2) [(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left[\\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)^2 - \\Omega^2 \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right)^2\\right]\\\\\n% &+ \\tilde{\\mathcal{N}} \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left[\\frac{1}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right) + \\tilde{L}\\right]\n% \\end{aligned}\n% \\label{polarisation_value}\n% \\end{equation}\n% where we can check that $\\lim_{s \\to 0} \\nu = 1/2$ and write depending on the sign of $s$\n% \\begin{eqnarray}\n% \\label{polarisation_sneg}\n% \\nu(s < 0) = \\tilde{\\mathcal{N}} \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2) [(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left(\\cos(|\\tilde{k}| \\tilde{L})^2 - |\\Omega|^2 \\sin(|\\tilde{k}| \\tilde{L})^2\\right) + \\tilde{\\mathcal{N}} \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left[\\frac{1}{|\\tilde{k}|} \\sin(|\\tilde{k}| \\tilde{L}) + \\tilde{L}\\right]\\\\\n% \\label{polarisation_spos}\n% \\nu(s > 0) = \\tilde{\\mathcal{N}} \\frac{4 [(\\tilde{\\psi} + 2) - 2 \\tilde{s}]}{(\\tilde{\\psi} + 2) [(\\tilde{\\psi} + 2) - 2 \\tilde{s}] - 2} \\left(\\cosh(\\tilde{k} \\tilde{L})^2 - |\\Omega|^2 \\sinh(\\tilde{k} \\tilde{L})^2\\right) + \\tilde{\\mathcal{N}} \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left[\\frac{1}{\\tilde{k}} \\sinh(\\tilde{k} \\tilde{L}) + \\tilde{L}\\right]\n% \\end{eqnarray}\n% with (\\ref{normalisation_factor_pave_sneg}, \\ref{normalisation_factor_pave_spos}).\n\n\\subsubsection{Polarisation-oriented method}\n\nWe have that the form of Eqs.~\\ref{FPr1}, \\ref{FPr2} indicates that there cannot be $\\delta$-function terms in the right eigenvectors $Q_{\\alpha_1\\alpha_2}$, therefore these are completely described by Eqs.~\\ref{qaa}, \\ref{qaoa}. We can thus write\n\\begin{eqnarray}\n\\hat{\\varepsilon}_{\\alpha\\alpha}(r) = \\frac{A^{\\prime}}{A} \\varepsilon_{\\alpha\\alpha}(r)^2\\\\\n\\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) = \\frac{A^{\\prime}}{A} \\varepsilon_{\\alpha\\overline{\\alpha}}(r) \\varepsilon_{\\overline{\\alpha}\\alpha}(r)\\\\\n\\hat{\\gamma}_{\\alpha\\alpha} = \\gamma_{\\alpha\\alpha} \\frac{A^{\\prime}}{A} \\varepsilon_{\\alpha\\alpha}(0^+)\\\\\n\\hat{\\gamma}_{\\alpha\\overline{\\alpha}} = \\gamma_{\\alpha\\overline{\\alpha}} \\frac{A^{\\prime}}{A} \\varepsilon_{+-}(L^-)\n\\end{eqnarray}\nwith the notations of Eqs.~\\ref{gfY1}, \\ref{gfY2}, \\ref{gfY3}.\\\\\n\nWe want to compute the average polarisation\n\\begin{equation}\n\\nu = 4 \\hat{\\gamma}_{\\alpha\\alpha} + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r\n\\label{nu_pave}\n\\end{equation}\ncorresponding to the propensity of particles to be aligned, and use the normalisation condition \\eqref{normalisation_pave_full} to re-write it\n\\begin{equation}\n\\begin{aligned}\n\\nu &= \\frac{4 \\hat{\\gamma}_{\\alpha\\alpha} + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r}{4 \\hat{\\gamma}_{\\alpha\\alpha} + 2 \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\alpha}(r) \\, \\mathrm{d}r + 2 \\int_0^L \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) \\, \\mathrm{d}r}\\\\\n&= \\frac{4 e^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} \\frac{\\varepsilon_{\\alpha\\alpha}(0^+)}{A} + 2 e^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\alpha}(r)^2}{A^2} \\, \\mathrm{d}r}{4 e^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} \\frac{\\varepsilon_{\\alpha\\alpha}(0^+)}{A} + 2 e^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} (\\tau \\psi + 2) \\frac{\\varepsilon_{+-}(L^-)}{A} + 2 e^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\alpha}(r)^2}{A^2} \\, \\mathrm{d}r + 2 e^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\overline{\\alpha}}(r)\\varepsilon_{\\overline{\\alpha}\\alpha}(r)}{A^2} \\, \\mathrm{d}r}\n\\end{aligned}\n\\label{nu_pave_full}\n\\end{equation}\nwhere we have used $\\gamma_{\\alpha\\overline{\\alpha}} = (\\tau \\psi + 2) \\gamma_{\\alpha\\alpha}$ \\eqref{ratio_gamma} from the first to the second line.\\\\\n\nWe can compute from Eqs.~\\ref{epsilon}, \\ref{LBe2}\n\\begin{equation}\n\\frac{\\gamma_{\\alpha\\alpha}}{A} = l \\frac{\\varepsilon_{+-}(L^-)}{A} = \\frac{l}{2} \\left[(1 + e^{-k L}) + \\Omega (1 - e^{-k L})\\right]\n\\end{equation}\nfrom Eqs.~\\ref{epsilon}, \\ref{theta}\n\\begin{eqnarray}\n\\frac{\\varepsilon_{\\alpha\\alpha}(0^+)}{A} = \\frac{1}{\\tau\\psi + 2}(1 + e^{-k L})\\\\\n\\begin{aligned}\n\\frac{\\varepsilon_{+-}(L^-)}{A} &= \\frac{1}{2} \\left[(1 + e^{-k L}) + \\Omega (1 - e^{-k L})\\right]\n\\end{aligned}\n\\end{eqnarray}\nand from Eqs.~\\ref{heaa}, \\ref{he+-}\n\\begin{eqnarray}\n\\int_0^L \\frac{\\varepsilon_{\\alpha\\alpha}(r)^2}{A^2} \\, \\mathrm{d}r = e^{-k L} \\frac{2}{(\\tau \\psi + 2)^2} \\left[\\frac{1}{k} \\left(\\frac{e^{k L} - e^{- k L}}{2}\\right) + L\\right]\\\\\n\\int_0^L \\frac{\\varepsilon_{\\alpha\\overline{\\alpha}}(r)\\varepsilon_{\\overline{\\alpha}\\alpha}(r)}{A^2} \\, \\mathrm{d}r = e^{-k L} \\frac{1}{2} \\left[(1 - \\Omega^2) \\frac{1}{k} \\left(\\frac{e^{k L} - e^{-k L}}{2}\\right) + (1 + \\Omega^2) L\\right]\n\\end{eqnarray}\nso that we can write\n\\begin{eqnarray}\n\\label{gaa-full}\ne^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} \\frac{\\varepsilon_{\\alpha\\alpha}(0^+)}{A} = \\frac{1}{4} \\left[2 \\frac{\\Omega}{k} \\left(1 + \\left(\\frac{e^{k L} + e^{-k L}}{2}\\right)\\right) + 2 \\frac{\\Omega^2}{k} \\left(\\frac{e^{k L} - e^{-k L}}{2}\\right)\\right]\\\\\n\\label{gaba-full}\n\\begin{aligned}\ne^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} (\\tau \\psi + 2) \\frac{\\varepsilon_{+-}(L^-)}{A} = \\frac{1}{2} \\Bigg[l(\\tau \\psi + 2)(1 - \\Omega^2) + l(\\tau\\psi + 2)(1 + \\Omega^2)\\left(\\frac{e^{k L} + e^{-k L}}{2}\\right)\\\\\n+ (\\tau \\psi + 2)^2 \\frac{\\Omega^2}{k} \\left(\\frac{e^{k L} - e^{-k L}}{2}\\right)\\Bigg]\n\\end{aligned}\\\\\ne^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\alpha}(r)^2}{A^2} \\, \\mathrm{d}r = \\frac{1}{2} \\left[\\frac{4}{(\\tau\\psi + 2)^2} \\left(\\frac{1}{k}\\left(\\frac{e^{k L} - e^{- k L}}{2}\\right) + L\\right)\\right]\\\\\ne^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\overline{\\alpha}}(r)\\varepsilon_{\\overline{\\alpha}\\alpha}(r)}{A^2} \\, \\mathrm{d}r = \\frac{1}{2} \\left[(1 - \\Omega^2) \\frac{1}{k}\\left(\\frac{e^{k L} - e^{-k L}}{2}\\right) + (1 + \\Omega^2) L\\right]\n\\end{eqnarray}\nwhere we have used the quantity $\\Omega$ \\eqref{Omega}.\\\\\n\nWe can first check\n\\begin{eqnarray}\n\\lim_{s \\to 0} 4 e^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} \\frac{\\varepsilon_{\\alpha\\alpha}(0^+)}{A} = 4 l\\\\\n\\lim_{s \\to 0} 2 e^{k L} \\frac{\\gamma_{\\alpha\\alpha}}{A} (\\tau \\psi + 2) \\frac{\\varepsilon_{+-}(L^-)}{A} = 4 l\\\\\n\\lim_{s \\to 0} 2 e^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\alpha}(r)^2}{A^2} \\, \\mathrm{d}r = 2 L\\\\\n\\lim_{s \\to 0} 2 e^{k L} \\int_0^L \\frac{\\varepsilon_{\\alpha\\overline{\\alpha}}(r)\\varepsilon_{\\overline{\\alpha}\\alpha}(r)}{A^2} \\, \\mathrm{d}r = 2 L\n\\end{eqnarray}\nso that, replacing in \\eqref{nu_pave_full}, we get $\\lim_{s \\to 0} \\nu = 1/2$ as expected.\\\\\n\nWe finally give the full expressions for $s < 0$\n\\begin{equation}\n\\nu(s < 0) = \\frac{\n\\begin{aligned}\n2 \\frac{|\\Omega|}{|\\tilde{k}|} \\left(1 + \\cos(|\\tilde{k}| \\tilde{L})\\right) - 2 \\frac{|\\Omega|^2}{|\\tilde{k}|} \\sin(|\\tilde{k}| \\tilde{L}) + \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left(\\frac{1}{|\\tilde{k}|} \\sin(|\\tilde{k}| \\tilde{L}) + \\tilde{L}\\right)\n\\end{aligned}\n}{\n\\begin{aligned}\n2 \\frac{|\\Omega|}{|\\tilde{k}|} \\left(1 + \\cos(|\\tilde{k}| \\tilde{L})\\right) - (2 + (\\tilde{\\psi} + 2)^2) \\frac{|\\Omega|^2}{|\\tilde{k}|} \\sin(|\\tilde{k}| \\tilde{L}) + (\\tilde{\\psi} + 2)(1 + |\\Omega|^2) + (\\tilde{\\psi} + 2)(1 - |\\Omega|^2) \\cos(|\\tilde{k}| \\tilde{L})\\\\\n+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + |\\Omega|^2)\\right] \\frac{1}{|\\tilde{k}|} \\sin(|\\tilde{k}|\\tilde{L}) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - |\\Omega|^2)\\right] \\tilde{L}\n\\end{aligned}\n}\n\\end{equation}\nand for $s > 0$\n\\begin{equation}\n\\nu(s > 0) = \\frac{\n\\begin{aligned}\n2 \\frac{\\Omega}{\\tilde{k}} \\left(1 + \\cosh(\\tilde{k} \\tilde{L})\\right) + 2 \\frac{\\Omega^2}{\\tilde{k}} \\sinh(\\tilde{k} \\tilde{L}) + \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left(\\frac{1}{\\tilde{k}} \\sinh(\\tilde{k} \\tilde{L}) + \\tilde{L}\\right)\n\\end{aligned}\n}{\n\\begin{aligned}\n2 \\frac{\\Omega}{\\tilde{k}} \\left(1 + \\cosh(\\tilde{k} \\tilde{L})\\right) + (2 + (\\tilde{\\psi} + 2)^2) \\frac{\\Omega^2}{\\tilde{k}} \\sinh(\\tilde{k} \\tilde{L}) + (\\tilde{\\psi} + 2)(1 - \\Omega^2) + (\\tilde{\\psi} + 2)(1 + \\Omega^2) \\cosh(\\tilde{k} \\tilde{L})\\\\\n+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - \\Omega^2)\\right] \\frac{1}{\\tilde{k}} \\sinh(\\tilde{k}\\tilde{L}) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + \\Omega^2)\\right] \\tilde{L}\n\\end{aligned}\n}\n\\end{equation}\nwhere we have introduced the dimensionless quantities $\\tilde{\\psi} = \\tau \\psi$, $\\tilde{k} = k l$ and $\\tilde{L} = L/l$.\n\n\\subsubsection{Full expressions}\n\nWe summarise here our expressions for the full-time distributions.\n\nWe have the normalisation factor (Eq.~\\ref{normalisation_factor_pave})\n\\begin{equation}\n\\begin{aligned}\n\\tilde{\\mathcal{N}} =~&\\left(A A^{\\prime} e^{-\\tilde{k}\\tilde{L}} l\\right)^{-1}\\\\\n=~&2 \\frac{\\Omega}{\\tilde{k}} \\left(1 + \\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\right) + (2 + (\\tilde{\\psi} + 2)^2) \\frac{\\Omega^2}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\\\\n&+ (\\tilde{\\psi} + 2)(1 - \\Omega^2) + (\\tilde{\\psi} + 2)(1 + \\Omega^2) \\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\\\\n&+ \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 - \\Omega^2)\\right] \\frac{1}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right) + \\left[\\frac{4}{(\\tilde{\\psi} + 2)^2} + (1 + \\Omega^2)\\right] \\tilde{L}\n\\end{aligned}\n\\end{equation}\nand write from Eqs.~\\ref{Omega}, \\ref{he+-}, \\ref{heaa}, the regular part of the full-time distribution functions\n\\begin{align}\n\\tilde{\\mathcal{N}} l \\hat{\\varepsilon}_{\\alpha\\alpha}(r) &= \\frac{4}{(\\tilde{\\psi} + 2)^2} \\left(\\frac{e^{\\tilde{k} \\tilde{L} \\left(1/2 - \\tilde{r}/\\tilde{L}\\right)} + e^{-\\tilde{k} \\tilde{L} \\left(1/2 - \\tilde{r}/\\tilde{L}\\right)}}{2}\\right)^2\\\\\n\\tilde{\\mathcal{N}} l \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) &= \\frac{1}{2}\\left(1 - \\Omega^2\\right) \\left(\\frac{e^{2 \\tilde{k}\\tilde{L}\\left(1/2 - \\tilde{r}/\\tilde{L}\\right)} + e^{-2 \\tilde{k}\\tilde{L}\\left(1/2 - \\tilde{r}/\\tilde{L}\\right)}}{2}\\right) + \\frac{1}{2}\\left(1 + \\Omega^2\\right)\n\\end{align}\nwhere $\\tilde{r} = r/l$, such that we can check $\\tilde{\\mathcal{N}} \\hat{\\varepsilon}_{\\alpha\\alpha}(r) = \\tilde{\\mathcal{N}} \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r) = 1$ for $s = 0$, and from Eqs.~\\ref{gaa-full}, \\ref{gaba-full} the sticking terms\n\\begin{eqnarray}\n\\tilde{\\mathcal{N}} \\hat{\\gamma}_{\\alpha\\alpha} = \\frac{1}{2} \\left[\\frac{\\Omega}{\\tilde{k}} \\left(1 + \\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\right) + \\frac{\\Omega^2}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\right]\\\\\n\\begin{aligned}\n\\tilde{\\mathcal{N}} \\hat{\\gamma}_{\\alpha\\overline{\\alpha}} =~& \\frac{1}{2} \\Bigg[(\\tilde{\\psi} + 2)(1 - \\Omega^2) + (\\tilde{\\psi} + 2)(1 + \\Omega^2)\\left(\\frac{e^{\\tilde{k} \\tilde{L}} + e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\\\\n&+ (\\tilde{\\psi} + 2)^2 \\frac{\\Omega^2}{\\tilde{k}} \\left(\\frac{e^{\\tilde{k} \\tilde{L}} - e^{-\\tilde{k} \\tilde{L}}}{2}\\right)\\Bigg]\n\\end{aligned}\n\\end{eqnarray}\nwhich we plot in Fig.~\\ref{fig-full-expr}.\n\n\\subsection{Unbiased ensemble check}\n\nWe have in the unbiased ensemble $\\psi(s = 0) = 0$, therefore $k = 0$ according to \\eqref{kPsi}. It follows that\n\\begin{equation}\n\\hat{\\varepsilon}_{\\alpha\\alpha}(r, s=0) = \\hat{\\varepsilon}_{\\alpha\\overline{\\alpha}}(r, s=0) = A A^{\\prime}\n\\end{equation}\nfrom Eqs.~\\ref{he+-},~\\ref{heaa}. We know from Eqs.~\\ref{s0++}, \\ref{s0--}, \\ref{s0+-}, \\ref{s0-+} that\n\\begin{equation}\n\\left. \\frac{\\gamma_{\\alpha\\alpha}}{\\gamma_{\\alpha\\overline{\\alpha}}} \\right|_{s=0} = \\frac{1}{2}\n\\end{equation}\ntherefore we get\n\\begin{eqnarray}\n\\hat{\\gamma}_{\\alpha\\alpha}(s=0) = l A A^{\\prime}\\\\\n\\hat{\\gamma}_{\\alpha\\overline{\\alpha}}(s=0) = 2 l A A^{\\prime}\n\\end{eqnarray}\nfrom Eqs.~\\ref{hgaa},~\\ref{hgaoa}. We can then write the normalisation condition \\eqref{normalisation_pave_full}\n\\begin{equation}\n1 = 4 l A A^{\\prime} + 4 l A A^{\\prime} + 2 A A^{\\prime} L + 2 A A^{\\prime} L\n\\end{equation}\nsuch that\n\\begin{equation}\nA^{\\prime} = \\frac{1}{A( 8 l + 4 L)} = 1\n\\end{equation}\nwhere the second equality derives from Eq.~\\ref{smooth_unbiased}, and which consequence is that\n\\begin{equation}\nQ_{\\alpha\\alpha}(r, s=0) = Q_{\\alpha\\overline{\\alpha}}(r, s=0) = 1\n\\end{equation}\naccording to Eqs.~\\ref{qaa},~\\ref{qaoa}.\n\n%%%%%%%%%%%%%%\n% REFERENCES %\n%%%%%%%%%%%%%%\n\n\\bibliographystyle{alpha}\n{\\renewcommand{\\bibname}{References}\\bibliography{references}}\n\n\\end{document}\n", "meta": {"hexsha": "fb83c41619a155007ccf3de5f8b29f9c877f29a6", "size": 45942, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Summaries/RTPring/main.tex", "max_stars_repo_name": "yketa/DAMTP_2019_Wiki", "max_stars_repo_head_hexsha": "9995202a586d5f301f2bdffe868b1fcec6fbb990", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Summaries/RTPring/main.tex", "max_issues_repo_name": "yketa/DAMTP_2019_Wiki", "max_issues_repo_head_hexsha": "9995202a586d5f301f2bdffe868b1fcec6fbb990", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Summaries/RTPring/main.tex", "max_forks_repo_name": "yketa/DAMTP_2019_Wiki", "max_forks_repo_head_hexsha": "9995202a586d5f301f2bdffe868b1fcec6fbb990", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 59.2036082474, "max_line_length": 598, "alphanum_fraction": 0.6093335075, "num_tokens": 19957, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8418256393148982, "lm_q2_score": 0.743167997235783, "lm_q1q2_score": 0.6256178743913855}} {"text": "\n\\subsection{Random optimisation}\n\nThis is similar to random search, however we use a multivariate Gaussian distribution around our current point rather than a hypersphere.\n\n", "meta": {"hexsha": "cfe57226d115375269f971b25b560c990f396825", "size": 174, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/probability/optimisation/01-02-randomOptimisation.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/probability/optimisation/01-02-randomOptimisation.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/probability/optimisation/01-02-randomOptimisation.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.0, "max_line_length": 137, "alphanum_fraction": 0.8218390805, "num_tokens": 33, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8705972751232808, "lm_q2_score": 0.718594386544335, "lm_q1q2_score": 0.6256063148443837}} {"text": "\\chapter{Explainability}\r\n\\label{ch_explainability}\r\n%\r\nDuring evaluation of the benchmark study, we found that proset models are very open to inspection after fitting.\r\nThis is due to their built-in feature selection and geometric structure.\r\nIn this chapter, we review diagnostic plots and other techniques that provide a better understanding of model behavior.\r\n%\r\n\\section{Low-dimensional representations of the data}\r\n\\label{sec_low_dimensional}\r\n%\r\nVerification of the model structure is greatly simplified if the algorithm is able to reduce the inputs to a small set.\r\nWe can then conduct what amounts to post-fit exploratory data analysis.\\par\r\n%\r\nIn the benchmark study, the proset classifier achieves the largest reduction in the number of features for the cancer data set, where only 4 out of 30 features yield a model that is still equivalent to XGBoost in terms of log-loss (see Table \\ref{tab_e3_e8_e9}).\r\nWith only four inputs, we can visually inspect all scatter plots involving a combination of two features.\r\nFigure \\ref{fig_scatter_plots_cancer} shows the positions of the prototypes (large circles) and test samples (small circles).\r\nThe three misclassified cases are circled in black.\r\nThe density plots on the diagonal show the marginal distributions for each feature, both for the prototypes (solid curves / vertical lines) and test samples (dashed curves / dots).\r\nTwo interesting findings are that (a) any pair of features allows us to distinguish the two classes fairly well and (b) the prototypes are more clearly separated with lower variance than the whole population.\\par\r\n%\r\n\\begin{figure}\r\n\\caption{Scatter plots for active features of the cancer data model}\r\n\\label{fig_scatter_plots_cancer}\r\n%\r\n\\begin{center}\r\n\\includegraphics[width=0.95\\textheight, angle=90]{figures/cancer_scatter_plot_testing_data.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nA `small' model is also easier to review for domain experts.\r\nFor example, an oncologist should be able to provide a clear assessment whether the four features (in decreasing order of weights: worst radius, worst concave points, worst texture, worst smoothness) are a natural and sufficient choice of indicators.\r\nIf we have access to the original image data, we can ask experts to review the misclassified cases in detail and compare them to the most similar prototypes.\r\nThis can reveal labeling errors, important features excluded by the algorithm due to dearth of relevant training samples, or genuine `hard cases' that are difficult to classify even for humans.\\par\r\n%\r\nA technique that can work even if the number of features remains large is to create low-dimensional maps of the data that take model fit results into account.\r\nFor the visualization we call the `batch 1 map', we perform doubly weighted PCA on the prototypes of the first batch:\\par\r\n%\r\nLet $Z\\in\\R^{J_1\\times D}$ be the matrix obtained by rescaling the features for the prototypes of the first batch with the corresponding feature weights:\r\n%\r\n\\begin{equation}\r\nZ_{j,d}:=v_{1,d}x_{s_{1,j},d}\\label{eq_scaled_x}\r\n\\end{equation}\r\n%\r\nThis gives larger impact to features for which the algorithm is more sensitive and removes any deselected features from the analysis.\r\nNow let $\\overline{z}$ be the row mean of $Z$ weighted using the prototype weights:\r\n%\r\n\\begin{equation}\r\n\\overline{z}:=\\frac{\\sum_{j=1}^{J_1}w_{1,j}Z_{j,\\bullet}}\r\n{\\sum_{j=1}^{J_1}w_{1,j}}\\label{eq_weighted_mean}\r\n\\end{equation}\r\n%\r\nDenote by $\\overline{Z}$ the centered matrix\r\n%\r\n\\begin{equation}\r\n\\overline{Z}_{j,d}:=Z_{j,d}-\\overline{z}_d\\label{eq_centered_Z}\r\n\\end{equation}\r\n%\r\nLet $W\\in\\R^{J_1\\times J_1}$ be the diagonal matrix with diagonal elements $w_{1,j}$ and compute the matrix decomposition\r\n%\r\n\\begin{equation}\r\n\\overline{Z}^TW\\overline{Z}=:U\\Sigma U^T\\label{eq_weighted_svd}\r\n\\end{equation}\r\n%\r\nIn the above, $\\Sigma$ is the diagonal matrix of eigenvalues with row and column dimension equal to the rank of $\\overline{Z}^TW\\overline{Z}$, while $U$ is the corresponding orthogonal matrix of eigenvectors.\r\nIf $\\Sigma$ has at least rank two, the `batch 1 map' coordinates for the prototypes of the first batch are the first two columns of the product $\\overline{Z}U$.\r\nOther samples can be mapped via the same transform, i.e., scale with $v_1$, center with $\\overline{z}$, and multiply with $U$.\\par\r\n%\r\n\\begin{figure}\r\n\\caption{`Batch 1 map' of cancer data showing test samples}\r\n\\label{fig_batch_1_map_test}\r\n%\r\n\\begin{center}\r\n\\includegraphics[height=0.4\\textheight]{figures/cancer_batch_1_map_testing_data.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nThe `batch 1 map' for the prototypes and test set of the cancer data is shown in Figure \\ref{fig_batch_1_map_test}.\r\nIt appears that the two classes are almost linearly separable except for the three misclassified samples.\r\nTwo are close to the boundary between the classes, while the last is a `malignant' case surrounded by `benign' cases.\r\n%\r\n\\section{Assessing the familiarity of new samples}\r\n\\label{sec_familiarity}\r\n%\r\n\\begin{figure}\r\n\\caption{Model behavior outside the `area of competence'}\r\n\\label{fig_area_of_competence}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{kNN on the rotated checkerboard}]{\\includegraphics[width=0.49\\textwidth]{figures/checker_rot_knn_surf_test.pdf}}\r\n\\subfloat[\\textbf{XGBoost on the rotated checkerboard}]{\\includegraphics[width=0.49\\textwidth]{figures/checker_rot_xgb_surf_test.pdf}}\\\\\r\n\\subfloat[\\textbf{proset on the rotated checkerboard}]{\\includegraphics[width=0.49\\textwidth]{figures/checker_rot_surf_test.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nOne problem faced by all machine learning is that their behavior becomes arbitrary for samples that are far removed from the original training data.\r\nThe way a model extrapolates outside its `area of competence' depends entirely on the specific algorithm and does not reflect any empirical evidence.\r\nThis is illustrated in Figure \\ref{fig_area_of_competence} for the rotated checkerboard example.\r\nThe plots show the probability surfaces for each of the three classifiers studied in Chapter \\ref{ch_classifier}, i.e., the color scale represents the estimated probability for class 1 from green (0.0) to red (1.0).\r\nWhile kNN and XGBoost create idiosyncratic patterns in regions of the feature space with no training data, proset reverts to the marginal distribution.\r\nThe proset estimate is no more correct than that of the other two algorithms, but we consider it the most `honest' in the absence of information.\r\nFor balanced training data as in the example, proset does not assign a high probability to an arbitrary class as may happen with kNN or XGBoost.\r\nGiven that users tend to interpret the probability of the class with the highest probability estimate as the model's `confidence' in the result, this is an important advantage.\\par\r\n%\r\nThe proset algorithm can even provide an explicit indicator that a new sample is far away from the training data.\r\nFor any $x\\in\\R^D$, we define the familiarity of a proset model with $x$ as\r\n%\r\n\\begin{equation}\r\n\\mathfrak{f}(x):=\\sum_{b=1}^B\\sum_{j=1}^{J_b}w_{b,j}G_{v_b}(x-x_{s_{b,j}})\\label{eq_familiarity}\r\n\\end{equation}\r\n%\r\nThis is just the denominator of (\\ref{eq_pkx}) minus 1, which can be provided alongside the estimate with no additional effort.\r\nFamiliarity measures the total contribution of the weighted prototypes to the estimate relative to the marginal distribution represented by the 1.\\par\r\n%\r\nInstead of using the absolute size of $\\mathfrak{f}(x)$ to assess new samples, we find it more useful to express it as a quantile of the distribution of familiarity observed for test data.\r\nAs the model may overfit its training data slightly, it is better to use held-out data for this analysis.\r\nFigure \\ref{fig_area_of_competence} shows the resulting 1 \\% contour line of familiarity for the proset classifier.\r\nFor a new sample, familiarity expressed as a quantile of the previously observed distribution can be interpreted as p-value of a statistical test whether the estimate is inside the model's `area of competence'.\\par\r\n%\r\nFor a supervised learning problem with many features, the training data is typically close to a `thin' manifold in the high-dimensional feature space.\r\nUsing quantiles of familiarity as indicated above, we can measure proximity to that manifold via a scalar indicator.\r\nThis is a powerful tool for monitoring a machine learning model in production and to detect data drift.\r\nHowever, there is a trade-off between feature selection and the usefulness of familiarity.\r\nConsider the case where proset does not select a feature because of insufficient variation in the training data.\r\nA new sample that has a markedly different value in that one feature is still assigned a high familiarity score if the features actually used by the model lie in the expected range.\r\n%\r\n\\section{Explaining individual estimates}\r\n\\label{sec_individual_explanations}\r\n%\r\n\\subsection{Explanation report}\r\n\\label{sec_explanation_report}\r\n%\r\n\\begin{figure}\r\n\\caption{Example of explanation report for cancer data}\r\n\\label{fig_explanation_report}\r\n%\r\n\\begin{center}\r\n\\includegraphics[width=0.99\\textwidth]{figures/cancer_2d_95_model_explain.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nOur implementation of proset can export information about the model as tabular data.\r\nThe basic report contains information on all prototypes, their weights, and active features.\r\nA more interesting report relates a particular sample to the selected prototypes.\r\nFigure \\ref{fig_explanation_report} shows an example analysis for the proset classifier.\r\nIt deals with the one misclassified `malignant' case from the cancer data example that is surrounded by `benign' cases.\r\nThe report contains the following information:\r\n%\r\n\\begin{description}\r\n\\item[Batch:] the number of the batch to which each prototype belongs.\r\nThe model in the example has only a single batch.\r\nThe first three rows contain information on the misclassified sample and the marginal distribution, they are not associated with any batch.\r\n%\r\n\\item[Sample:] the zero-based index of the prototype with respect to the ordering of the training data after the train-test split.\r\n%\r\n\\item[Sample name:] the record for the test sample states its zero-based index within the set of test data, the class with the highest estimated probability, and familiarity expressed as a quantile of the distribution for the test data.\r\nThe remaining rows indicate whether they belong to the marginal distribution or individual prototypes.\r\n%\r\n\\item[Target:] the numeric representation of the class, either 0 (`benign') or 1 (`malignant').\r\nThe first rows contains the true target for the test sample.\r\n%\r\n\\item[Prototype weight:] the value $w_{b,j}$ for each prototype.\r\nThe test sample has no entry, records for the marginal distribution show the marginal probability for the corresponding class.\r\n%\r\n\\item[Similarity:] the value of the unnormalized Kernel $G_{v_b}$ at $x-x_{s_{b,j}}$.\r\nThis is a number in $(0,1]$, where $1$ indicates that the sample features $x$ are identical to those of the prototype.\r\nThe test sample has no entry, records for the marginal probabilities list a 1 as they affect all estimates equally.\r\n%\r\n\\item[Impact:] the product of prototype weight and similarity is the total impact of the prototype on the estimated probability (\\ref{eq_pkx}).\r\nBy default, the report lists the prototypes in order of descending impact.\r\n%\r\n\\item[Dominant set:] a binary indicator of observations that have the largest impact on the estimate.\r\nThe target values for prototypes outside the dominant set can be changed without affecting which class is assigned the highest probability.\r\nSee below for a precise definition of the dominant set.\r\n%\r\n\\item[p class 0:] the record for the test sample shows the estimated probability for class 0 (`benign').\r\nThe remaining rows contain the additive contribution of the marginal distribution and prototypes to this estimate.\r\nThe contribution is equal to the impact listed for the row divided by the total impact if the target is 0.\r\nOtherwise, the contribution is zero.\r\n%\r\n\\item[p class 1:] as above for class 1 (`malignant').\r\n\\end{description}\r\n%\r\nThe report provides a complete breakdown of the estimate for one sample into the contribution of the marginals and prototypes.\r\nThe multivariate structure of the feature space is replaced by a simple list that can be filtered and sorted by different criteria to identify major contributors.\r\nWhile the total number of prototypes can be much larger than in the example, most tend to have negligible impact on any particular case.\\par\r\n%\r\nFor the computation of the `dominant set' indicator, we use the following procedure:\r\n%\r\n\\begin{algorithm}[Dominant set]~\r\n\\label{alg_dominant_set}\r\n%\r\n\\begin{enumerate}\r\n\\item Sort the prototypes in descending order of impact and assign a rank to each.\r\nPrototypes with the same impact receive the same rank.\r\n%\r\n\\item For each rank $r$, compute $p_{r,k}$ as sum of additive contributions (columns `p class 0', etc.\\ in the report) to the probability estimate for class $k$ from the marginal distribution and prototypes up to rank $r$.\r\nThese satisfy $0<\\sum_{k=0}^{K_1}p_{r,k}\\leq 1$.\r\nThe remainder $t_r:=1-\\sum_{k=0}^{K-1}p_{r,k}$ is the total contribution of prototypes with rank greater than $r$.\r\n%\r\n\\item For each rank $r$, denote by $k_{1,r}$ and $k_{2,r}$ the classes with the largest and second largest value of $p_{r,k}$.\r\nDetermine the set $R:=\\{r:p_{r,k_{1_r}}-p_{r,k_{2,r}}>t_r\\}$.\r\n%\r\n\\item If $R$ is empty, the dominant set is empty.\r\nOtherwise, all prototypes with a rank less than or equal to the minimal element in $R$ form the dominant set.\r\n\\end{enumerate}\r\n\\end{algorithm}\r\n%\r\n\\begin{remark}\r\nBy constructions, if $R$ is not empty, it contains all ranks greater than or equal to its minimal element and all share the same value for $k_{1,r}$.\r\nThus, it contains all ranks where the remainder $t_r$ is too small to affect which class is assigned the highest probability.\r\n\\end{remark}\r\n%\r\nIn the example report, the prototypes of the dominant set together with the marginal distribution contribute 53 percentage points of the 99 \\% probability estimated for class 0.\r\nThus, class 0 would be assigned the highest probability even if all remaining prototypes belonged to class 1.\r\nThe purpose of the dominant set is to serve as a guideline which prototypes need to be reviewed for a better understanding of a particular result.\\par\r\n%\r\n\\begin{figure}\r\n\\caption{`Batch 1 map' of cancer data showing prototype impact}\r\n\\label{fig_batch_1_map_impact}\r\n%\r\n\\begin{center}\r\n\\includegraphics[height=0.4\\textheight]{figures/cancer_batch_1_map_impact.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nThe impact of prototypes on a particular estimate can also be visualized in the `batch 1 map'.\r\nFigure \\ref{fig_batch_1_map_impact} does this for the case from the example report.\r\nThe size of each circle is proportional to the impact of the prototype, numbers refer to the sample index.\r\n%\r\n\\subsection{Combining SHAP values and geometric analysis}\r\n\\label{sec_shap_values}\r\n%\r\nSo far, we have looked at model-specific explanations for proset estimators on their own.\r\nIn this section, we combine them with a model-agnostic approach to gain additional insight into the model structure.\r\nWe choose the SHAP (SHapley Additive exPlanations) method for assessing feature importance proposed by Lundberg and Lee \\cite{Lundberg_17}.\r\nSHAP provides explanations for individual estimates and is the only additive feature attribution method that has certain desirable properties.\r\nA downside is that the computational effort grows exponentially in the number of features, although an efficient algorithm exists for decision trees \\cite{Lundberg_18}.\r\nTo facilitate usage of SHAP, our implementation of proset allows `shrinking' the model to take only the features with nonzero weights as input.\\par\r\n%\r\nSHAP explains an estimate by comparing the sample of interest to a reference point.\r\nThe relative change in the estimator between the points is attributed to the relative change in the features.\r\nThere appears to be little research so far on the best choice of reference point, but Izzo et.\\ al.\\ \\cite{Izzo_20} consider the issue.\r\nThe authors point out that the common choice of using the origin of the feature space as baseline is not satisfactory.\r\nFor a binary classifier, they recommend to select a reference point on the decision boundary between the two classes.\r\nThey refer to this as a `neutral baseline', since the algorithm is indifferent which class to choose.\r\nWe cannot follow this approach, as we want to study the probability estimate itself without reference to a decision rule\\footnote{\r\nWe do use the `naive' decision rule that selects the class with the highest estimated probability to define bins for Algorithm \\ref{alg_bins} and to compute balanced accuracy as a secondary metric for the benchmark study.\r\nSee the discussion in the next paragraph why we do not consider this rule for further analysis.\r\n}.\r\nAlso, several of the benchmark cases have more than two classes and it is not certain that a point can be found that is `neutral' with regards to all of them at once.\\par\r\n%\r\nWe believe that classification should always be carried out in two stages by combining a stochastic model with a decision rule.\r\nThe purpose of the former is to accurately estimate the probability distribution with no regard to the practical problem we want to solve.\r\nThe purpose of the latter is to optimize the expected value of a problem-specific utility or loss function under the estimated distribution.\r\nAny algorithm that conflates the two stages implicitly applies an arbitrary decision rule that is likely to yield sub-optimal results.\r\nAs an example, consider a credit scoring algorithm that estimates the default probability for loans.\r\nIt is doubtful that the lending company wants to grant loans to all applicants with default probability less than 50 \\%.\r\nA more thorough discussion of this issue can be found, e.g., in a blog article by Harrell \\cite{Harrell_20}.\\par\r\n%\r\nFor the above reason, we want to find a suitable reference point for explaining the estimated probabilities themselves regardless of any decision rule that could be applied later.\r\nWe expect such a point to have the following three properties:\r\n%\r\n\\begin{enumerate}\r\n\\item The point itself represents a feasible combination of features.\r\n%\r\n\\item The point is not extreme in terms of the feature space.\r\nFor every feature, there are training samples that have either a higher or lower value.\r\n%\r\n\\item The point is not extreme in terms of the estimated probabilities.\r\nFor every class, there are training samples that have either a higher or lower estimate.\r\n\\end{enumerate}\r\n%\r\n\\begin{remark}\r\n\\begin{enumerate}\r\n\\item Property 1 acknowledges that the baseline is part of the explanation and should have no `unnatural' characteristics.\r\nThere is less risk of misinterpreting the feature importance if they are computed with respect to a relatable feature vector.\r\n%\r\n\\item Properties 2 and 3 mean that SHAP can explore changes of the input and output in all directions.\r\n%\r\n\\item Property 3 can be viewed as a weaker notion of the `neutral baseline' proposed in \\cite{Izzo_20}.\r\n%\r\n\\item Properties 1 and 2 are mutually exclusive for binary features.\r\nHere, we prefer property 1, i.e., the baseline should be a genuine case and not a hypothetical half-way point.\r\n\\end{enumerate}\r\n\\end{remark}\r\n%\r\nOne way to satisfy the first condition is to use an actually observed point as reference.\r\nThe first two conditions imply that a medoid\\footnote{\r\nA medoid is any point from a set for which the average absolute distances to all other points is minimal.\r\nIn one dimension, a medoid is also a median of the data.\r\n} of the feature vectors for the training data is a suitable choice.\r\nLikewise, condition 1 and 3 suggest to use a point whose estimated probabilities are a medoid of the probability vectors for the training data.\r\nWe thus propose the following procedure for choosing a baseline:\r\n%\r\n\\begin{algorithm}[SHAP baseline]~\r\n\\label{alg_shap_baseline}\r\n%\r\n\\begin{enumerate}\r\n\\item Rank all training samples in increasing order of average absolute distance to all other training samples in the feature space.\r\nUse only features selected by proset to compute distances.\r\n%\r\n\\item Rank all training samples in increasing order of average absolute distance to all other training samples in the space of estimated probability vectors.\r\n%\r\n\\item Combine both rankings using Borda's rule\\footnote{\r\nBorda's rule converts rankings to scores by assigning each candidate half a point for each other candidate ranked the same, plus a full point for each candidate ranked worse.\r\nScores are totaled across rankings and the winners are the candidates with the most points.\r\n} and use the winner as reference.\r\nIn case of a tie, choose the candidate whose probability vector has the highest entropy.\r\nIf still tied, use the point with the lowest sample index.\r\n\\end{enumerate}\r\n\\end{algorithm}\r\n%\r\n\\begin{remark}\r\n\\begin{enumerate}\r\n\\item We do not recommend to combine the active features and estimated probabilities into a single vector.\r\nThe medoid property depends on the scaling of of individual variables and there is no obvious common scale.\r\nAlso, if there are many more features than classes, the former might dominate the decision.\r\nComputing two separate rankings and using Borda's rule results in a choice that balances properties 2 and 3.\r\n%\r\n\\item The method requires computing pairwise distances for all training samples, which may be unattractive for large problems.\r\nIn this case, we recommend subsampling.\r\n\\end{enumerate}\r\n\\end{remark}\r\n%\r\n\\begin{figure}\r\n\\caption{`Batch 1 map' of cancer data showing reference point}\r\n\\label{fig_batch_1_map_reference}\r\n%\r\n\\begin{center}\r\n\\includegraphics[height=0.4\\textheight]{figures/cancer_batch_1_map_training_data_reference.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\n\\begin{figure}\r\n\\caption{SHAP feature importance}\r\n\\label{fig_feature_importance}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{Global importance}]{\\includegraphics[width=0.99\\textwidth]{figures/cancer_global_shap.pdf}}\\\\\r\n\\subfloat[\\textbf{Misclassified case}]{\\includegraphics[width=0.99\\textwidth]{figures/cancer_example_shap.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\n\\begin{figure}\r\n\\caption{Scatter plots for important features}\r\n\\label{fig_scatter_important}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{Global importance}]{\\includegraphics[height=0.4\\textheight]{figures/cancer_scatter_global_shap.pdf}}\\\\\r\n\\subfloat[\\textbf{Misclassified case}]{\\includegraphics[height=0.4\\textheight]{figures/cancer_scatter_example_shap.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\n\\begin{figure}\r\n\\caption{Probability surfaces for important features}\r\n\\label{fig_probability_important}\r\n%\r\n\\begin{center}\r\n\\subfloat[\\textbf{Global importance}]{\\includegraphics[height=0.4\\textheight]{figures/cancer_surface_global_shap.pdf}}\\\\\r\n\\subfloat[\\textbf{Misclassified case}]{\\includegraphics[height=0.4\\textheight]{figures/cancer_surface_example_shap.pdf}}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nIf we apply the above rule to the cancer data, it selects a baseline near the center of the point cloud of training samples.\r\nThe estimated probability for being `benign' is 92 \\%, which is far from the 50 \\% that would be achievable given the structure of the data.\r\nHowever, the data is unbalanced in favor of `benign' cases and we have seen for the misclassified case in the previous section that probabilities in excess of 92 \\% are achievable.\r\nFigure \\ref{fig_batch_1_map_reference} shows the reference point in a `batch 1 map' together with training data and the misclassified case we want to explain.\\par\r\n%\r\nFigure \\ref{fig_feature_importance} shows the results of SHAP analysis for the whole test data and the misclassified case.\r\nNote that the global values are just the averages of the absolute scores of each feature across all samples.\r\nOrdering the features by descending global importance yields the same sequence as for the feature weights, but we do not believe that this necessarily holds for every data set.\r\nThe two most important features globally are `worst radius' and `worst concave points'.\r\nFor the misclassified case, the estimate for class `benign' is shifted from 92 \\% (baseline) to 99 \\% because of the extremely low values for `worst concave points' and `worst smoothness'.\r\nFigure \\ref{fig_scatter_important} shows the bivariate scatter plots for both pairs of features, this time with training data as supplementary information.\r\nAs an alternative, we can study the probability surface resulting for the most important features if the other features are fixed at their values for the misclassified case.\r\nThis is shown in Figure \\ref{fig_probability_important}.\\par\r\n%\r\n\\clearpage\r\n%\r\n\\begin{figure}\r\n\\caption{`Batch 1 map' for digits data showing test samples}\r\n\\label{fig_batch_1_map_test_digits}\r\n%\r\n\\begin{center}\r\n\\includegraphics[width=0.95\\textheight, angle=90]{figures/digits_batch_1_map.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\n\\begin{figure}\r\n\\caption{Explanation combining prototypes and SHAP values}\r\n\\label{fig_impact_shap}\r\n%\r\n\\begin{center}\r\n\\includegraphics[width=0.95\\textheight, angle=90]{figures/digits_8_vs_1_explanation.pdf}\r\n\\end{center}\r\n\\end{figure}\r\n%\r\nAs a final example, we show how to create an explanation for the digits data set that relies both on prototypes with high impact and SHAP values.\r\nThis classification problem is more complex than the one for cancer data.\r\nFigure \\ref{fig_batch_1_map_test_digits} presents the `batch 1 map' including test data and a reference point selected using the ranking rule.\r\nSeveral classes are well differentiated in this plot, but a two-dimensional representation clearly cannot capture the whole structure.\\par\r\n%\r\nFigure \\ref{fig_impact_shap} is a custom plot for explaining a particular misclassified digit.\r\nIt shows the case itself -- an `8' for which the highest probability is estimated for class `1' -- as well as the five prototypes with the largest impact on the result.\r\nThe second row overlays the SHAP values for the probability estimate of the true class `8', the third does the same for class `1'.\r\nA red color indicates that the estimate is increased because of the grayscale value of the associated pixel, a blue color that it is decreased.\r\nNote that the estimated probabilities for the baseline are 61 \\% for class `8', 27 \\% for class `3', and negligible for the other digits.\r\nThus, the SHAP values indicate the importance of the features for a shift away from this distribution towards 60 \\% for class `1' and 32 \\% for class `8'.\r\n%\r\n\\endinput\r\n", "meta": {"hexsha": "f649d4562e0fda1542ff31e14fc87dbc6a6adf76", "size": 26719, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/explainability.tex", "max_stars_repo_name": "NRuf77/proset", "max_stars_repo_head_hexsha": "101d491e05c2423faddca31029232982f46d8831", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/explainability.tex", "max_issues_repo_name": "NRuf77/proset", "max_issues_repo_head_hexsha": "101d491e05c2423faddca31029232982f46d8831", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7718435030872968, "lm_q1q2_score": 0.625562880046471}} {"text": "\\subsection{Exponential function}\\label{subsec:exponential_function}\n\n\\begin{definition}\\label{def:exponential_function}\n We define the \\term{exponential function}\n \\begin{equation}\\label{def:exponential_function/series}\n \\exp(z) \\coloneqq \\sum_{k=0}^\\infty \\frac {z^k} {k!}\n \\end{equation}\n and \\term{Euler's number}\n \\begin{equation*}\n e \\coloneqq \\exp(1) = \\sum_{i=0}^k \\frac 1 {k!}.\n \\end{equation*}\n\n \\Fullref{thm:def:exponential_function/interpolates_power} justifies the notation \\( e^z = \\exp(z) \\).\n\\end{definition}\n\\begin{proof}\n We will show that \\( \\exp(z) \\) converges everywhere. By \\fullref{thm:power_series_radius_of_convergence}, the radius of convergence is\n \\begin{equation*}\n \\limsup_{k \\to \\infty} \\frac {k!} {(k-1)!}\n =\n \\limsup_{k \\to \\infty} k\n =\n +\\infty\n \\end{equation*}\n\n Hence, the radius of convergence of \\( \\exp(x) \\) is infinite.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:exponential_function}\n The exponential function \\( \\exp(z) \\) has the following basic properties (not that we do not use the notation \\( e^z \\) here in order to reduce confusion with yet-undefined power \\hyperref[def:power_function]{functions}):\n\n \\begin{thmenum}\n \\thmitem{thm:def:exponential_function/eulers_identity} (Euler's identity)\n \\begin{equation*}\n \\exp(i \\pi) = -1.\n \\end{equation*}\n\n \\thmitem{thm:def:exponential_function/derivative} \\( \\exp(z) \\) is its own derivative.\n\n \\thmitem{thm:def:exponential_function/homomorphism} \\( \\exp(x + y) = \\exp(x) \\exp(y) \\). Stated in another way, \\( \\exp \\) is a homomorphism from the additive group of \\( \\BbbC \\) to the multiplicative group.\n\n \\thmitem{thm:def:exponential_function/interpolates_power} The notation \\( \\exp(x) \\) is consistent with iterated multiplication as defined in \\fullref{def:semiring/identity}, that is, \\( \\exp(n) = \\underbrace{e \\cdot \\ldots \\cdot e}_{n \\text{times}} \\) and for positive integers \\( n \\), \\( \\exp(n) = \\) and \\( \\exp(-n) =\\tfrac 1 {\\exp(n)} \\).\n\n \\thmitem{thm:def:exponential_function/negative_power}\n \\begin{equation*}\n \\exp(z) = \\frac 1 {\\exp(-z)}.\n \\end{equation*}\n\n \\thmitem{thm:def:exponential_function/real_positive} For real \\( t \\), \\( e^t \\) is a positive real number.\n\n \\thmitem{thm:def:exponential_function/conjugate} \\( \\overline{\\exp(z)} = \\exp(\\overline{z}) \\).\n\n \\thmitem{thm:def:exponential_function/unit_circle} For any \\( c \\in \\BbbR \\), the function \\( t \\mapsto \\exp(it) \\) is a bijection between any half-open interval \\( [c, c + 2\\pi) \\) and the unit circle in \\( \\BbbC \\).\n\n \\thmitem{thm:def:exponential_function/real_bijective} \\( t \\mapsto \\exp(t) \\) is a bijection from \\( \\BbbR \\) to \\( [0, \\infty) \\).\n\n \\thmitem{thm:def:exponential_function/bijective} For any \\( c \\in \\BbbR \\), \\( \\exp(z) \\) is a bijection between the strip \\( S \\coloneqq \\{ a + bi \\colon c \\leq b < c + 2\\pi \\} \\) and the complex plane \\( \\BbbC \\setminus \\{ 0 \\} \\).\n\n \\thmitem{thm:def:exponential_function/periodic} \\( \\exp(z) \\) is \\( 2i\\pi \\)-\\hyperref[def:periodic_function]{periodic}.\n\n \\thmitem{thm:def:exponential_function/compound_interest} For nonnegative real \\( t \\geq 0 \\) we have\n \\begin{equation*}\n \\exp(t) = \\lim_{n \\to \\infty} \\left(1 + \\frac t n \\right)^n\n \\end{equation*}\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:exponential_function/eulers_identity} By \\fullref{eq:thm:trigonometric_function_basic_roots/pi} and \\fullref{thm:exponential_trigonometric_identities/eulers_formula}, we have\n \\begin{equation*}\n \\exp(i\\pi) = \\cos(\\pi) + i\\sin(\\pi) = -1.\n \\end{equation*}\n\n \\SubProofOf{thm:def:exponential_function/derivative} Follows from \\fullref{thm:power_series_are_locally_uniform_convergent} and \\fullref{thm:derivative_limit_exchange}.\n\n \\SubProofOf{thm:def:exponential_function/homomorphism} The Cauchy product of \\( \\exp(x) \\) and \\( \\exp(y) \\) is\n \\begin{balign*}\n \\exp(x) \\exp(y)\n & =\n \\left( \\sum_{k=0}^\\infty \\frac {x^k} {k!} \\right) \\left( \\sum_{k=0}^\\infty \\frac {y^k} {k!} \\right)\n = \\\\ &=\n \\sum_{k=0}^\\infty \\sum_{m=0}^k \\frac {x^m} {m!} \\frac {x^{k-m}} {(k-m)!}\n = \\\\ &=\n \\sum_{k=0}^\\infty \\frac 1 {k!} \\sum_{m=0}^k \\binom{k}{m} x^m y^{k-m}\n \\reloset {\\ref{thm:binomial_theorem}} = \\\\ &=\n \\sum_{k=0}^\\infty \\frac {(x + y)^k} {k!}\n =\n \\exp(x + y).\n \\end{balign*}\n\n \\SubProofOf{thm:def:exponential_function/interpolates_power} We use induction on \\( n \\) to prove \\( \\exp(n) = e^n \\). The case \\( \\exp(0) = 1 \\) is obvious. If we assume that \\( \\exp(n) = e^n \\), by \\fullref{thm:def:exponential_function/homomorphism}, we have\n \\begin{equation*}\n \\exp(n + 1)\n =\n \\exp(n) \\exp(1)\n =\n e^n \\cdot e\n =\n e^{n+1}.\n \\end{equation*}\n\n Note that this works for negative \\( n \\) too.\n\n \\SubProofOf{thm:def:exponential_function/negative_power} Note that\n \\begin{equation*}\n 1 = \\exp(0) = \\exp(z - z) = \\exp(z) \\exp(-z),\n \\end{equation*}\n hence\n \\begin{equation*}\n \\exp(-z) = \\frac 1 {\\exp(z)}.\n \\end{equation*}\n\n \\SubProofOf{thm:def:exponential_function/real_positive} For \\( t > 0 \\), the following\n \\begin{equation*}\n \\exp(t) = \\sum_{k=0}^\\infty \\frac {t^k} {k!}\n \\end{equation*}\n is a series of positive real numbers. To see its convergence, we apply \\fullref{thm:dalamberts_ratio_test}:\n \\begin{equation*}\n \\frac {t^k} {k!} \\cdot \\frac {(k-1)!} {t^{k-1}}\n =\n \\frac t k\n \\xrightarrow[k \\to \\infty]{} 0.\n \\end{equation*}\n\n Thus, \\( \\exp(t) \\) is a nonnegative real number. Furthermore, since the sequence of partial sums is monotone, \\( \\exp(t) \\) cannot be zero. Hence, for \\( t > 0 \\), we have \\( \\exp(t) > 0 \\).\n\n Notice that \\( \\exp(t) \\exp(-t) > 0 \\), hence if \\( \\exp(t) > 0 \\), then \\( \\exp(-t) > 0 \\).\n\n \\SubProofOf{thm:def:exponential_function/conjugate} By \\fullref{thm:exponential_trigonometric_identities/eulers_formula},\n \\begin{balign*}\n \\overline{\\exp(a + bi)}\n & \\reloset {\\ref{thm:def:exponential_function/homomorphism}} =\n \\overline{\\exp(a) \\exp(bi)}\n \\reloset {\\ref{thm:def:exponential_function/real_positive}} = \\\\ &=\n \\exp(a) \\overline{(\\cos(b) + i\\sin(b))}\n = \\\\ &=\n \\exp(a) (\\cos(b) - i\\sin(b))\n \\reloset {\\ref{thm:power_series_parity}} = \\\\ &=\n \\exp(a) (\\cos(-b) + i\\sin(-b))\n = \\\\ &=\n \\exp(a) \\exp(-bi)\n = \\\\ &=\n \\exp(a - bi)\n = \\\\ &=\n \\exp(\\overline{a + bi}).\n \\end{balign*}\n\n \\SubProofOf{thm:def:exponential_function/periodic} By \\fullref{thm:def:exponential_function/eulers_identity},\n \\begin{equation*}\n \\exp(x + 2i\\pi) = \\exp(x) \\exp(2i\\pi) = \\exp(x).\n \\end{equation*}\n\n Furthermore, this is also the minimal period. If we assume that \\( \\sin(x) \\) has another period, say \\( p \\in (0, 2\\pi) \\), we would have \\( \\sin(p) = \\sin(0) = 0 \\) and \\fullref{thm:trigonometric_identities/pythagorean_identity} would imply that \\( \\cos(p) \\in \\{ -1, 1 \\} \\). But then \\( \\cos(p) \\) would be an extreme point for \\( \\cos \\), which is not possible because \\( \\cos \\) is convex in \\( [0, 2\\pi] \\) and only has three extremal points --- \\( 0, \\pi, 2\\pi \\).\n\n \\SubProofOf{thm:def:exponential_function/unit_circle} For \\( c, t \\in \\BbbR \\) we have\n \\begin{equation*}\n \\abs{\\exp(it)}\n =\n \\abs{\\cos(t) + i\\sin(t)}\n =\n \\sqrt{\\cos(t)^2 + \\sin(t)^2}\n \\reloset {\\eqref{eq:thm:trigonometric_identities/pythagorean_identity}} =\n 1.\n \\end{equation*}\n\n Furthermore, if \\( r \\) is another real number,\n \\begin{equation}\n \\exp(ir)\n =\n \\exp(i(t + (r - t)))\n =\n \\exp(it) \\exp(i(r - t)).\n \\end{equation}\n\n It follows that \\( \\exp(ir) \\neq \\exp(it) \\) if and only if \\( \\exp(i(r - t)) \\neq 0 \\). If \\( t, r \\in [c, c + 2\\pi) \\) and \\( t \\neq r \\), this is satisfied.\n\n Hence, \\( t \\mapsto \\exp(it) \\) is indeed an injection of \\( [c, c + 2\\pi) \\) into the unit circle of \\( \\BbbC \\). It is also a surjection because of the intermediate value theorem.\n\n \\SubProofOf{thm:def:exponential_function/real_bijective} First, assume that \\( e^t \\) is not injective on \\( \\BbbR \\). Then there exist \\( t, r \\in \\BbbR \\), \\( t \\neq r \\), such that \\( e^t = e^r \\). By \\fullref{thm:def:exponential_function/real_positive}, both are positive real numbers. In particular, we can divide by \\( e^t \\) to obtain\n \\begin{equation*}\n 1\n =\n \\frac {e^r} {e^t}\n \\reloset {\\ref{thm:def:exponential_function/negative_power}} =\n =\n e^r e^{-t}\n \\reloset {\\ref{thm:def:exponential_function/homomorphism}} =\n e^{r - t}.\n \\end{equation*}\n\n We know that \\( e^0 = 1 \\) from \\fullref{thm:def:exponential_function/interpolates_power}. Thus, it is enough to show that \\( e^t = 1 \\) if and only if \\( t = 0 \\).\n\n Assume that \\( e^t = 1 \\) holds for some \\( t > 0 \\). The partial sums are monotonely increasing, so in order for them to converge to \\( 1 \\), for any fixed index \\( n \\) we must have\n \\begin{balign*}\n 0 & \\leq \\sum_{k=0}^n \\frac {t^k} {k!} = 1 + \\sum_{k=1}^n \\frac {t^k} {k!} \\leq 1, \\\\\n -1 & \\leq \\sum_{k=1}^n \\frac {t^k} {k!} \\leq 0.\n \\end{balign*}\n\n But \\( \\sum_{k=1}^n \\frac {t^k} {k!} > 0 \\) because \\( t > 0 \\). The obtained contradiction proves that \\( e^t \\neq 1 \\) for positive \\( t \\).\n\n For negative \\( t \\), note that\n \\begin{equation*}\n e^t e^{-t} = 1.\n \\end{equation*}\n\n Since \\( -t \\) is positive, \\( e^{-t} \\neq 1 \\) and hence \\( e^t \\neq 1 \\).\n\n Therefore, the function \\( t \\mapsto e^t \\) is injective on \\( \\BbbR \\). It is also surjective onto \\( \\BbbR^{>0} \\) because of the intermediate value theorem.\n\n \\SubProofOf{thm:def:exponential_function/bijective} Fix \\( a + bi \\in S_c \\), that is, \\( b \\in [c, c + 2\\pi) \\). By \\fullref{thm:def:exponential_function/homomorphism},\n \\begin{equation*}\n e^{a + bi} = e^a e^{bi}.\n \\end{equation*}\n\n By \\fullref{thm:def:exponential_function/unit_circle}, \\( b \\mapsto e^{bi} \\) is injective for \\( b \\in [c, c + 2\\pi) \\) and by \\fullref{thm:def:exponential_function/real_bijective}, \\( a \\mapsto e^a \\) is injective on \\( \\BbbR \\). It follows that their product is also injective.\n\n \\SubProofOf{thm:def:exponential_function/compound_interest}\\mcite[3.31]{Rudin1976Principles}By \\fullref{thm:binomial_theorem},\n \\begin{balign*}\n \\left(1 + \\frac t n \\right)^n\n & =\n \\sum_{k=0}^n \\binom{n}{k} \\left(\\frac t n\\right)^k 1^{n-k}\n = \\\\ &=\n \\sum_{k=0}^n \\frac {n!} {(n-k)! k!} \\frac {t^k} {n^k}\n = \\\\ &=\n \\sum_{k=0}^n \\frac {n!} {(n-k)! n^k} \\frac {t^k} {k!}\n = \\\\ &=\n \\sum_{k=0}^n \\left[ \\prod_{j=1}^k \\left(1 - \\frac {k+j} n \\right) \\right] \\frac {t^k} {k!}.\n \\end{balign*}\n\n Fix an index \\( m \\). Since the series is nonnegative, there exists an index \\( N \\) such that for \\( n \\geq N \\)\n \\begin{equation*}\n \\sum_{k=0}^m \\frac {t^k} {k!}\n \\leq\n \\sum_{k=0}^n \\left[ \\prod_{j=1}^k \\left(1 - \\frac {k+j} n \\right) \\right] \\frac {t^k} {k!}.\n \\end{equation*}\n\n Note that\n \\begin{equation*}\n \\left[ \\prod_{j=1}^k \\left(1 - \\frac {k+j} n \\right) \\right] \\frac {t^k} {k!}\n \\leq\n \\frac {t^k} {k!},\n \\end{equation*}\n hence\n \\begin{equation*}\n \\sum_{k=0}^m \\frac {t^k} {k!}\n \\leq\n \\sum_{k=0}^n \\left[ \\prod_{j=1}^k \\left(1 - \\frac {k+j} n \\right) \\right] \\frac {t^k} {k!}\n \\leq\n \\sum_{k=0}^n \\frac {t^k} {k!}.\n \\end{equation*}\n\n By \\fullref{thm:squeeze_lemma},\n \\begin{equation*}\n \\lim_{n \\to \\infty} \\left(1 + \\frac t n \\right)^n\n =\n \\lim_{n \\to \\infty} \\sum_{k=0}^n \\frac {t^k} {k!}\n =\n \\exp(t).\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:logarithm}\n Fix \\( c \\in \\BbbR \\). Unless specified otherwise, we assume \\( c = 0 \\).\n\n We define the \\term{natural logarithm} \\( \\log(x) \\) as \\hyperref[def:multi_valued_function/inverse]{inverse function} of \\( e^x \\) from \\( \\BbbC \\setminus \\{ 0 \\} \\) to the strip \\( S_c \\coloneqq \\{ a + bi \\colon c \\leq b < c + 2\\pi \\} \\).\n\n We also define the \\term{base \\( b \\) logarithm} \\( \\log_b(x) \\) for \\( b > 0 \\) over the same domain as\n \\begin{equation*}\n \\log_b(x) \\coloneqq \\frac {\\log(x)} {\\log(b)}.\n \\end{equation*}\n\\end{definition}\n\\begin{proof}\n The well-definedness follows from \\fullref{thm:def:exponential_function/bijective}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:logarithm}\n \\hfill\n \\begin{thmenum}\n \\thmitem{thm:def:logarithm/homomorphism} \\( \\log(xy) = \\log(x) \\log(y) \\)\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:logarithm/homomorphism} Follows from \\fullref{thm:def:exponential_function/homomorphism}.\n\\end{proof}\n\n\\begin{definition}\\label{def:power_function}\n For each positive real number \\( y > 0 \\), we define the \\term{power function}\n \\begin{equation*}\n x^y \\coloneqq e^{y \\ln x}\n \\end{equation*}\n as a function of \\( x \\).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:def:power_function}\n \\hfill\n \\begin{thmenum}\n \\thmitem{thm:def:power_function/composition} \\( (x^y)^z = x^{yz} \\).\n \\thmitem{thm:def:power_function/derivative} \\( D_x(x^y) = \\log(x) x^y \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:power_function/composition}\n \\begin{equation*}\n (x^y)^z\n =\n e^{z \\log(e^{y \\log(x)})}\n =\n e^{z y \\log(x)}\n =\n x^{yz}.\n \\end{equation*}\n\n \\SubProofOf{thm:def:power_function/derivative} Using the chain rule for differentiation, we obtain\n \\begin{equation*}\n D_x(x^y) = D_x(e^{\\log(y) x}) = \\log(x) e^{\\log(y) x} = \\log(x) x^y.\n \\end{equation*}\n\\end{proof}\n\\begin{proposition}\\label{thm:exponential-trigonometric_identities}\n We have the following exponential-trigonometric identities:\n \\thmitem{thm:exponential_trigonometric_identities/eulers_formula} (Euler's formula) For any \\( z \\in \\BbbC \\),\n \\begin{equation}\\label{thm:exponential_trigonometric_identities/eulers_formula/identity}\n e^{iz} = \\cos(z) + i \\sin(z).\n \\end{equation}\n\n \\thmitem{thm:exponential_trigonometric_identities/inverse_eulers_formula} (Inverse Euler's identities) For any \\( z \\in \\BbbC \\),\n \\begin{balign}\n \\sin(z) & = \\real(e^z) = \\frac {e^{iz} - e^{-iz}} {2i} \\label{thm:exponential_trigonometric_identities/inverse_eulers_formula/sin} \\\\\n \\cos(z) & = \\imag(e^z) = \\frac {e^{iz} + e^{-iz}} 2 \\label{thm:exponential_trigonometric_identities/inverse_eulers_formula/cos}\n \\end{balign}\n\n \\thmitem{thm:exponential_trigonometric_identities/de_moivre} (De Moivre's formula) For any complex number \\( z \\) and any nonnegative integer \\( n \\),\n \\begin{equation}\\label{thm:exponential_trigonometric_identities/de_moivre/identity}\n (\\cos(z) + i \\sin(z))^n = \\cos(nz) + i \\sin(nz).\n \\end{equation}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:exponential_trigonometric_identities/eulers_formula} Simply note that \\fullref{def:exponential_function} is a termwise sum of \\fullref{def:trigonometric_functions/sine} and \\fullref{def:trigonometric_functions/cosine}, therefore \\fullref{thm:exponential_trigonometric_identities/eulers_formula/identity} holds.\n\n \\SubProofOf{thm:exponential_trigonometric_identities/inverse_eulers_formula} Follows from \\fullref{thm:exponential_trigonometric_identities/eulers_formula}.\n\n \\SubProofOf{thm:exponential_trigonometric_identities/de_moivre} From \\fullref{thm:exponential_trigonometric_identities/eulers_formula},\n \\begin{equation*}\n (\\cos(z) + i \\sin(z))^n\n =\n {e^{iz}}^n\n \\reloset {\\ref{thm:def:power_function/composition}} {=}\n =\n e^{i(zn)}\n =\n \\cos(nz) + i \\sin(nz).\n \\end{equation*}\n\\end{proof}\n", "meta": {"hexsha": "46566d77da5c3336a09346edc06dcdbd5e4c4840", "size": 15778, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/exponential_function.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/exponential_function.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/exponential_function.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.9515669516, "max_line_length": 474, "alphanum_fraction": 0.6191532514, "num_tokens": 5720, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.810478913248044, "lm_q1q2_score": 0.6255628793261822}} {"text": "\\chapter*{Appendix C: Derivative Needed for the Minimization}\n\\label{annexB}\n\\addcontentsline{toc}{chapter}{Appendix C: Derivative Needed for the \nMinimization}\n\n\\subsection*{The error function}\n The error function is written erf(x) and the complementary error function\n erfc(x). Their definitions are:\n\\begin{eqnarray}\n \\mbox{erf(x)} = \\frac{2}{\\sqrt{\\pi}}\\int_{0}^{x} e^{-t^{2}} dt \n\\end{eqnarray}\n\n\\begin{eqnarray}\n \\mbox{erfc}(x) = 1-\\mbox{erf}(x) = \\frac{2}{\\sqrt{\\pi}}\\int_{x}^{\\infty} e^{-t^{2}} dt \n\\end{eqnarray}\n\nThese functions have the following limits and symmetries:\n\\begin{eqnarray}\n \\begin{tabular}{c|c} \n \\hline\n \\mbox{erf}(0) = 0 & $ \\mbox{erf}(\\infty) = 1 $ \\\\\n \\mbox{erfc}(0) = 1 & $ \\mbox{erfc}(\\infty) = 0 $ \\\\\n \\mbox{erf}(-x) = \\mbox{erf}(x) & \\mbox{erfc}(-x) = 2-\\mbox{erfc}(x) \\\\\n \\hline\n \\end{tabular}\n\\end{eqnarray}\n\n\\subsection*{N1-MSE}\nNow we compute the contribution of the wavelet coefficient $x$ to the noise information: \n\\begin{eqnarray}\nh_{n}(x) = \\frac{x^2}{2\\sigma^2}\\mbox{erfc}(\\frac{x}{\\sqrt{2}\\sigma})\n\\end{eqnarray}\n\n\\begin{eqnarray}\n\\frac{ d h_{n}(x)}{dx} = \\frac{x}{\\sigma^2}\\mbox{erfc}(\\frac{x}{\\sqrt{2}\\sigma}) + \\frac{x^2}{2\\sigma^2} \\frac{\\partial \\mbox{erfc}(\\frac{x}{\\sqrt{2}\\sigma})}{\\partial x} \n\\end{eqnarray}\nWe derive the function erfc:\n\\begin{eqnarray}\n\\frac{\\partial \\: \\mbox{erfc}(x)}{\\partial x} = - \\frac{2}{\\sqrt{\\pi}} e^{-x^{2}} \n\\end{eqnarray}\nthen \n\\begin{eqnarray}\n\\frac{ d h_{n}(x)}{dx} = \\frac{x}{\\sigma^2}\\mbox{erfc}(\\frac{x}{\\sqrt{2}\\sigma}) -\\frac{x^2}{\\sqrt{2\\pi}\\sigma^3}e^{-\\frac{x^{2}}{2\\sigma^2}} \n\\end{eqnarray}\n\nIn order to minimize the functional~\\ref{eqn_func1}, we may want to calculate\nthe derivative of $h_s(x-y)$, $h_s(x-y)$ measuring the amount of information\ncontained in the residual ($y$ being the data).\n\\begin{eqnarray}\nh_s(x-y) = \\frac{(y-x)^2}{2\\sigma^2}\\mbox{erf}(\\frac{y-x}{\\sqrt{2}\\sigma}) \n\\end{eqnarray}\nDenoting $z = y - x$, we have\n\\begin{eqnarray}\nh_s(z) = \\frac{z^2}{2\\sigma^2}\\mbox{erf}(\\frac{z}{\\sqrt{2}\\sigma}) \n\\end{eqnarray}\n\\begin{eqnarray}\n\\frac{ d h_{s}(z)}{dz} & = & \\frac{z}{\\sigma^2}\\mbox{erf}(\\frac{z}{\\sqrt{2}\\sigma}) + \\frac{z^2}{2\\sigma^2} \\frac{\\partial \\mbox{erf}(\\frac{z}{\\sqrt{2}\\sigma})}{\\partial z} \\\\\n & = & \\frac{z}{\\sigma^2}\\mbox{erf}(\\frac{z}{\\sqrt{2}\\sigma}) + \\frac{z^2}{\\sqrt{2\\pi}\\sigma^3} e^{-\\frac{z^{2}}{2\\sigma^2}}\n\\end{eqnarray}\nand\n\\begin{eqnarray}\n\\frac{ d h_{s}(x-y)}{dx} = - \\frac{ d h_{s}(z)}{dz}\n\\end{eqnarray}\n\n\n\\subsection*{N2-MSE}\nWe compute the contribution of the wavelet coefficient $x$ to the noise information: \n\\begin{eqnarray}\nh_{n}(x) = \n\\frac{1}{\\sigma^{2}}\\int_{0}^{x} t \\: \\mbox{erfc}( \\frac{x-t}{\\sqrt{2}\\sigma}) dt \n\\end{eqnarray}\n\n\\begin{eqnarray}\n\\frac{d h_{n}(x)}{dx} & = & h_{n}(x+dx) - h_{n}(x) \\\\\n & = & \\frac{1}{\\sigma^{2}} \\int_{0}^{x+dx} \n \\mbox{erfc}(\\frac{x+dx-t}{\\sqrt{2}\\sigma}) dt -\n\\frac{1}{\\sigma^{2}}\\int_{0}^{x} \\mbox{erfc} (\\frac{x-t}{\\sqrt{2}\\sigma}) dt \n\\end{eqnarray}\n\n\\begin{eqnarray}\n\\frac{d h_{n}(x)}{dx} = \n\\frac{x}{\\sigma^{2}} \\, \\mbox{erfc}( \\frac{x-x}{\\sqrt{2}\\sigma}) +\n\\frac{1}{\\sigma^{2}}\\int_{0}^{x} \n \\frac{ \\partial \\: t \\, \\mbox{erfc} (\\frac{x-t}{\\sqrt{2}\\sigma})}{ \\partial x} dt \n\\end{eqnarray}\n\nNow, because erfc(0) = 1 we have:\n\\begin{eqnarray}\n\\frac{d H_{n}(x)}{dx} = \n\\frac{x}{\\sigma^{2}} +\n\\frac{1}{\\sigma^{2}}\\int_{0}^{x} \n \\frac{ \\partial \\: t \\, \\mbox{erfc} (\\frac{x-t}{\\sqrt{2}\\sigma})}{ \\partial x} dt \n\\end{eqnarray}\n\nWe derive the function erfc:\n\\begin{eqnarray}\n\\frac{\\partial \\: \\mbox{erfc}(x)}{\\partial x} = - \\frac{2}{\\sqrt{\\pi}} e^{-x^{2}} \n\\end{eqnarray}\n\n\\begin{eqnarray}\n\\frac{ \\partial\\: \\mbox{erfc}( \\frac{(x-t)}{\\sqrt{2}\\sigma} )}{\\partial x} = \n-\\frac{2}{\\sqrt{\\pi}} \\frac{1}{\\sqrt{2}\\sigma} \n e^{-\\frac{(x-t)^{2}}{2\\sigma^{2}}} =\n -\\sqrt{\\frac{2}{\\pi}} \\frac{1}{\\sigma} e^{-\\frac{(x-t)^{2}}{2\\sigma^{2}}}\n\\end{eqnarray}\n\nNow we deduce for the derivative of $h_n$:\n\n\\begin{eqnarray}\n\\frac{d h_{n}(x)}{dx} = \n\\frac{x}{\\sigma^{2}} +\n\\frac{1}{\\sigma^{2}}\\int_{0}^{x} \n -\\sqrt{\\frac{2}{\\pi}}\\frac{1}{\\sigma}t \\, e^{-\\frac{(x-t)^{2}}{2\\sigma^{2}}}dt \n\\end{eqnarray}\n\n\\begin{eqnarray}\n\\frac{d h_{n}(x)}{dx} = \n\\frac{x}{\\sigma^{2}} +\n\\frac{1}{\\sigma^{3}}\\sqrt{\\frac{2}{\\pi}} \\int_{0}^{x} t \\, e^{-\\frac{(x-t)^{2}}{2\\sigma^{2}}}dt \n\\end{eqnarray}\n\nWe create the variable J\n\\begin{eqnarray}\nJ = \\int_{0}^{x} t \\, e^{-\\frac{(x-t)^{2}}{2\\sigma^{2}}}dt \n\\end{eqnarray}\n\nWe create the variable u\n\\begin{eqnarray}\n\\begin{array}{cc} \n u = \\frac{t-x}{\\sqrt{2}\\sigma} & t = x+u \\, \\sqrt{2}\\sigma \\\\\n & dt = \\sqrt{2}\\sigma du \\\\\n t=0 \\Rightarrow u = \\frac{-x}{\\sqrt{2}\\sigma} &\n t=x \\Rightarrow u = 0 \\\\\n \\end{array}\n\\end{eqnarray}\n\n\nThe variable J can be written with u\n\\begin{eqnarray}\n J = \\int_{\\frac{-x}{\\sqrt{2}\\sigma} }^{0} (x+u \\,\\sqrt{2}\\sigma) e^{-u^{2}}\\sqrt{2}\\sigma du\n\\end{eqnarray}\n\\begin{eqnarray}\n J = \\sqrt{2}\\sigma x \\int_{\\frac{-x}{\\sqrt{2}\\sigma} }^{0} \n e^{-u^{2}} du +\n 2 \\sigma^{2} \\int_{\\frac{-x}{\\sqrt{2}\\sigma} }^{0} u \\,\n e^{-u^{2}} du\n\\end{eqnarray}\n\nThe first part of J can be rewritten as:\n\\begin{eqnarray}\n J_{0} = \\sqrt{2}\\sigma x \\int_{0}^{\\frac{x}{\\sqrt{2}\\sigma} } \n e^{-u^{2}} du \n\\end{eqnarray}\n$ J_{0} $ can be expressed with the error function.\n\\begin{eqnarray}\n J_{0} = \\sqrt{2}\\sigma \\frac{\\sqrt{\\pi}}{2} x \\mbox{erf}(\\frac{x}{\\sqrt{2}\\sigma}) \n J_{0} = \\sigma \\sqrt{\\frac{\\pi}{2}} \\: x \\: \\mbox{erf}(\\frac{x}{\\sqrt{2}\\sigma}) \n\\end{eqnarray}\nNow the second part of J is obvious\n\\begin{eqnarray}\n J_{1} = 2 \\sigma^{2} \\int_{\\frac{-x}{\\sqrt{2}\\sigma} }^{0} u \\,\n e^{-u^{2}} du \n\\end{eqnarray}\n or\n\\begin{eqnarray}\n\\frac{d e^{-u^{2}}} {du} = -2 u e^{-u^{2}} \n\\end{eqnarray}\n\nWe replace\n\\begin{eqnarray}\n J_{1} = -\\sigma^{2} \\int_{\\frac{-x}{\\sqrt{2}\\sigma} }^{0} d(e^{-u^{2}}) \n\\end{eqnarray}\n\\begin{eqnarray}\n J_{1} = \\sigma^{2} [e^{-\\frac{x^{2}}{2\\sigma^{2}}} - 1] \n\\end{eqnarray}\n\nNow we can write J\n\\begin{eqnarray}\n J = J_{0} + J_{1}\n = \\sigma \\sqrt{\\frac{\\pi}{2}} \\: x \\: \\mbox{erf}(\\frac{x}{\\sqrt{2}\\sigma}) \n + \\sigma^{2} [e^{-\\frac{x^{2}}{2\\sigma^{2}}} - 1 ] \n\\end{eqnarray}\n\nNow we can write the derivative of $h_n$\n\\begin{eqnarray}\n\\frac{d h_{n}(x)}{dx} & = & \n\\frac{x}{\\sigma^{2}} -\n\\frac{1}{\\sigma^{3}} \\sqrt{\\frac{2}{\\pi} } \\: J \\\\\n & = & \\frac{x}{\\sigma^{2}} -\n \\frac{x}{\\sigma^{2}}\\: x \\: \\mbox{erf}(\\frac{x}{\\sqrt{2}\\sigma})\n + \\frac{1}{\\sigma} \\sqrt{\\frac{2}{\\pi} } \\:\n [1 - e^{-\\frac{x{2}}{2\\sigma{2}}}] \\\\\n & = & \\frac{x}{\\sigma^{2}}\\mbox{erfc}(\\frac{x}{\\sqrt{2}\\sigma})\n + \\frac{1}{\\sigma} \\sqrt{\\frac{2}{\\pi} } \\:\n [1 - e^{-\\frac{x^{2}}{2\\sigma^{2}}}] \n\\end{eqnarray}\n\nIn order to minimize the functional~\\ref{eqn_func1}, we may want to calculate\nthe derivative of $h_s(x \\mid y)$, $h_s(x \\mid \ny)$ measuring the amount of information\ncontained in the residual ($y$ being the data).\n\\begin{eqnarray}\nh_s(x \\mid y) = \\frac{1}{\\sigma^{2}} \\int_{0}^{y-x} t \\mbox{ erf}(\\frac{y-x-t}{\\sqrt{2}\\sigma}) dt\n\\end{eqnarray}\nDenoting $z = y - x$, we have\n\\begin{eqnarray}\nh_s(z) & = & \\frac{1}{\\sigma^{2}} \\int_{0}^{z} t \\mbox{ erf}(\\frac{z-t}{\\sqrt{2}\\sigma}) dt \\\\\n & = & \\frac{1}{\\sigma^{2}} \\int_{0}^{z} t dt - \\frac{1}{\\sigma^{2}} \\int_{0}^{z} t \\mbox{ erfc}(\\frac{z-t}{\\sqrt{2}\\sigma}) dt\n\\end{eqnarray} \nand\n\\begin{eqnarray}\n\\frac{d h_s(x)}{dx} = \\frac{\\partial h_s(z)}{\\partial z} \\frac{\\partial z}{\\partial x}\n\\end{eqnarray}\n\\begin{eqnarray}\n\\frac{\\partial h_s(z)}{\\partial z} & = \\frac{z}{\\sigma^2} - \\frac{\\partial h_n(z)}{\\partial z}\n \\end{eqnarray}\nthen\n\\begin{eqnarray}\n\\frac{d h_s(x)}{dx} & = & - \\frac{y-x}{\\sigma^2} + \\frac{y-x}{\\sigma^2}\\mbox{erfc}(\\frac{y-x}{\\sqrt{2}\\sigma}) + \\sqrt{\\frac{2}{\\pi}}\\frac{1}{\\sigma}[1 - e^{-\\frac{(y-x)^{2}}{2\\sigma^{2}}}] \\\\\n & = & - \\frac{y-x}{\\sigma^2} \\mbox{erf}(\\frac{y-x}{\\sqrt{2}\\sigma}) + \\sqrt{\\frac{2}{\\pi}}\\frac{1}{\\sigma}[1 - e^{-\\frac{(y-x)^{2}}{2\\sigma^{2}}}]\n\\end{eqnarray}\n", "meta": {"hexsha": "266235418de14986ff4012cff7ff81b095254d67", "size": 8048, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr2/annex_deriv1.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr2/annex_deriv1.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr2/annex_deriv1.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.2522522523, "max_line_length": 192, "alphanum_fraction": 0.5683399602, "num_tokens": 3532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104788995148792, "lm_q2_score": 0.7718435083355187, "lm_q1q2_score": 0.6255628772334747}} {"text": "\\subsubsection{Seasonal and Trend Decomposition using Loess}\n\\label{stl}\n\nA time series $y_t$ may exhibit different types of patterns; to fully capture\n each of them, the series must be decomposed.\nThen, each component is forecast with a distinct model.\nMost commonly, the components are the trend $t_t$, seasonality $s_t$, and\n remainder $r_t$.\nThey are themselves time series, where only $s_t$ exhibits a periodicity $k$.\nA decomposition may be additive (i.e., $y_t = s_t + t_t + r_t$) or\n multiplicative (i.e., $y_t = s_t * t_t * r_t$); the former assumes that\n the effect of the seasonal component is independent of the overall level\n of $y_t$ and vice versa.\nThe seasonal component is centered around $0$ in both cases such that its\n removal does not affect the level of $y_t$.\nOften, it is sufficient to only seasonally adjust the time series, and model\n the trend and remainder together, for example, as $a_t = y_t - s_t$ in the\n additive case.\n\nEarly approaches employed moving averages (cf., Sub-section \\ref{ets}) to\n calculate a trend component, and, after removing that from $y_t$, averaged\n all observations of the same seasonal lag to obtain the seasonal\n component.\nThe downsides of this are the subjectivity in choosing the window lengths for\n the moving average and the seasonal averaging, the incapability of the\n seasonal component to vary its amplitude over time, and the missing\n handling of outliers.\n\nThe X11 method developed at the U.S. Census Bureau and described in detail by\n \\cite{dagum2016} overcomes these disadvantages.\nHowever, due to its background in economics, it is designed primarily for\n quarterly or monthly data, and the change in amplitude over time cannot be\n controlled.\nVariants of this method are the SEATS decomposition by the Bank of Spain and\n the newer X13-SEATS-ARIMA method by the U.S. Census Bureau.\nTheir main advantages stem from the fact that the models calibrate themselves\n according to statistical criteria without manual work for a statistician\n and that the fitting process is robust to outliers.\n\n\\cite{cleveland1990} introduce a seasonal and trend decomposition using a\n repeated locally weighted regression - the so-called Loess procedure - to\n smoothen the trend and seasonal components, which can be viewed as a\n generalization of the methods above and is denoted by the acronym STL.\nIn contrast to the X11, X13, and SEATS methods, the STL supports seasonalities\n of any lag $k$ that must, however, be determined with additional\n statistical tests or set with out-of-band knowledge by the forecaster\n (e.g., hourly demand data implies $k = 24 * 7 = 168$ assuming customer\n behavior differs on each day of the week).\nMoreover, the seasonal component's rate of change, represented by the $ns$\n parameter and explained in detail with Figure \\ref{f:stl} in Section\n \\ref{decomp}, must be set by the forecaster as well, while the trend's\n smoothness may be controlled via setting a non-default window size.\nOutliers are handled by assignment to the remainder such that they do not\n affect the trend and seasonal components.\nIn particular, the manual input needed to calibrate the STL explains why only\n the X11, X13, and SEATS methods are widely used by practitioners.\nHowever, the widespread adoption of concepts like cross-validation (cf.,\n Sub-section \\ref{cv}) in recent years enables the usage of an automated\n grid search to optimize the parameters.\nThe STL's usage within a grid search is facilitated even further by its being\n computationally cheaper than the other methods discussed.\n", "meta": {"hexsha": "f832f0e802bdf67b7620d49779e49504baa21cff", "size": 3658, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/2_lit/2_class/4_stl.tex", "max_stars_repo_name": "webartifex/urban-meal-delivery-paper-demand-forecasting", "max_stars_repo_head_hexsha": "9ee3396a24ce20c9886b4cde5cfe2665fd5a8102", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-25T19:40:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-25T19:40:56.000Z", "max_issues_repo_path": "tex/2_lit/2_class/4_stl.tex", "max_issues_repo_name": "webartifex/urban-meal-delivery-demand-forecasting", "max_issues_repo_head_hexsha": "9ee3396a24ce20c9886b4cde5cfe2665fd5a8102", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/2_lit/2_class/4_stl.tex", "max_forks_repo_name": "webartifex/urban-meal-delivery-demand-forecasting", "max_forks_repo_head_hexsha": "9ee3396a24ce20c9886b4cde5cfe2665fd5a8102", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 59.0, "max_line_length": 78, "alphanum_fraction": 0.7635319847, "num_tokens": 844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104789086703225, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.6255628757928976}} {"text": "\\documentclass[a4paper]{article}\n\n% Language and Font Encoding\n\\usepackage[english]{babel}\n\\usepackage[utf8x]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{listings}\n\\usepackage{float}\n\\usepackage{caption} % For captions\n\\usepackage{fancyhdr} % For header and footer\n\\usepackage{rotating}\n\\usepackage{amssymb}\n\\usepackage{varwidth}\n\n\n% Page size and margins\n\\usepackage[a4paper,top=3cm,bottom=2cm,left=2cm,right=2cm,marginparwidth=1.75cm]{geometry}\n\n% Useful Packages\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{tikz}\n\\usetikzlibrary{automata, positioning}\n\\usepackage{amsthm}\n\n\\DeclareMathOperator*{\\argmax}{argmax}\n\\DeclareMathOperator*{\\argmin}{argmin}\n\\DeclareMathOperator*{\\softmax}{softmax}\n\n\\newcommand{\\squeezeup}{\\vspace{-2.5mm}}\n\\newcommand{\\norm}[1]{\\left\\lVert#1\\right\\rVert}\n\n\\usepackage[colorinlistoftodos]{todonotes}\n\\usepackage[colorlinks=true, allcolors=blue]{hyperref}\n\n\\pagestyle{fancy}\n\\lhead{Linear Algebra Notes}\n\\rhead{}\n\n\\begin{document}\n\n\\title{Linear Algebra: Vector Spaces, Subspaces, and Linear Maps}\n\\author{Ibrahim Akbar\\\\}\n\\maketitle\n\n%--------------------------------------------------------------------------------------------------------Vector Space--------------------------------------------------------------------------------------------------------------------\n\\section{Vector Space}\n\\textbf{Definition:} A collection, $\\mathcal{V}$, is known as a vector (linear) space over the field $\\mathbb{F}$ if it is equipped with the operations $+:\\mathcal{V}\\times\\mathcal{V} \\rightarrow \\mathcal{V}$ and $\\cdot:\\mathbb{F}\\times\\mathcal{V}\\rightarrow\\mathcal{V}$ such that the following properties hold.\\\\\n\n\\begin{itemize}\n\\item Closure under Vector Addition: $\\mathbf{u + v} \\in \\mathcal{V} \\quad \\forall \\mathbf{u,v} \\in \\mathcal{V}$\n\\item Closure under Scalar Multiplication: $\\alpha \\mathbf{v} \\in \\mathcal{V} \\quad \\forall \\mathbf{v} \\in \\mathcal{V}, \\alpha \\in \\mathbb{F}$\n\\item $(\\alpha + \\beta)\\mathbf{v} = \\alpha \\mathbf{v} + \\beta \\mathbf{v}$\n\\item $\\alpha(\\beta\\mathbf{v}) = (\\alpha\\beta)\\mathbf{v}$\n\\item $\\alpha(\\mathbf{u + v}) = \\alpha\\mathbf{u} + \\alpha\\mathbf{v} $\n\\item $\\exists \\{\\mathbf{0}\\} \\hspace{2mm} \\textnormal{s.t.} \\hspace{2mm} \\mathbf{v} +\\{\\mathbf{0}\\} = \\mathbf{v} \\quad \\forall\\mathbf{v}\\in\\mathcal{V}$\n\\item For each $\\mathbf{v}\\in\\mathcal{V}\\quad\\exists \\mathbf{(-v)}\\hspace{2mm} \\textnormal{s.t.} \\hspace{2mm} \\mathbf{v} +\\mathbf{(-v)} = \\{\\mathbf{0}\\}$\n\\item $1\\in\\mathbb{F},\\mathbf{v}\\in\\mathcal{V} \\Rightarrow \\mathbf{1v} = \\mathbf{v}$\n\\end{itemize}\n\n\\subsection{Examples}\nSome examples of vector spaces are: $\\mathbb{R}^{n}, \\mathbb{C}^{n}$\n\n%----------------------------------------------------------------------------------------------------------Sub Space --------------------------------------------------------------------------------------------------------------------\n\\section{Subspace}\n\\textbf{Definition:} A subspace, $\\mathcal{U}$ of a vector space, $\\mathcal{V}$, over the field, $\\mathbb{F}$, is a non-empty set such that,\n\n\\begin{itemize}\n\\item $\\mathbf{u,v}\\in\\mathcal{W}\\Rightarrow\\mathbf{u+v}\\in\\mathcal{W}$\n\\item $\\mathbf{v}\\in\\mathcal{W}\\Rightarrow\\alpha\\mathbf{w}\\in\\mathcal{W},\\forall\\alpha\\in\\mathbb{F}$\n\\item $\\exists \\{\\mathbf{0}\\} \\hspace{2mm} \\textnormal{s.t.} \\hspace{2mm} \\mathbf{w} +\\{\\mathbf{0}\\} = \\mathbf{w} \\quad \\forall\\mathbf{w}\\in\\mathcal{W}$\n\\end{itemize}\n\n\\subsection{Examples}\nSome examples of subspaces are:\n\n$$\nV := \\{x\\in\\mathbb{R}^{n}\\mid x^{2} \\geq 4x\\}\n$$\n\nNote that $V' := \\{x\\in\\mathbb{R}^{n}\\mid x^{2} > 4x\\}$ would not be a subspace nor would any polynomial that does not equal zero. Why?\\\\\nIn general, all spaces in $\\mathbb{R}^{n}$ that are defined by a homogeneous system of linear equations adhere to these properties. For a visual/geometric representation it is a point, line, plane, hyper-plane that goes through the origin $\\{\\mathbf{0}\\}$.\n\n%----------------------------------------------------------------------------------------------------Linear Independence-----------------------------------------------------------------------------------------------------------------\n\\section{Linearly Independent}\n\\textbf{Definition:} A set of vectors $\\{\\mathbf{x}_{1},\\mathbf{x}_{2},\\ldots,\\mathbf{x}_{n}\\}$ is said to be linearly independent if:\n$$\n\\sum_{i = 1}^{n}\\alpha_{i}\\mathbf{x}_{i} = 0 \\iff \\alpha_{i} = 0 \\hspace{2mm} \\forall i\n$$\n\nThe vectors cannot be written as a linear combination of each other and thus cannot describe each other. Therefore it is the same as saying there is no redundant information in the vectors.\n\n\\subsection{Example}\nLet $\\mathcal{S} = \\Bigg\\{\\begin{bmatrix}2\\\\1\\\\-1\\\\\\end{bmatrix},\\begin{bmatrix}3\\\\2\\\\-1\\\\\\end{bmatrix},\\begin{bmatrix}-4\\\\-2\\\\1\\\\\\end{bmatrix}\\Bigg\\}$, then we see that $-2v_{1} + 0v_{2} = v_{3}$ and therefore $\\mathcal{S}$ is not a linearly independent set.\\\\\n\n%------------------------------------------------------------------------------------------------------------Span------------------------------------------------------------------------------------------------------------------------\n\\section{Span}\n\\textbf{Definition:} The span of a set of vectors $\\mathcal{S} = \\{\\mathbf{x}_{1}, \\mathbf{x}_{2}, \\ldots, \\mathbf{x}_{n}\\}$ is the set of vectors, $\\mathcal{V}$, that is described by all possible linear combinations of $\\mathcal{S}$.\n$$\nV := span(\\mathcal{S}) = \\{y \\mid y=\\sum_{i = 1}^{n}\\alpha_{i}x_{i}\\hspace{2mm}\\forall\\alpha_{i}\\in\\mathbb{F}\\}\n$$ \n\n\\subsection{Examples}\nGiven the vector $v = \\begin{bmatrix}1\\\\0\\\\\\end{bmatrix}$ over the field of all reals it would span the space \n$$\n\\mathcal{S} = \\bigg\\{\\begin{bmatrix}\\alpha\\\\0\\\\\\end{bmatrix}\\mid \\alpha\\in\\mathbb{R}\\bigg\\}\n$$\nLets also consider $u = \\begin{bmatrix}0\\\\1\\\\\\end{bmatrix}$ in the same field. Together these would span\n$$\n\\mathcal{T} = \\bigg\\{\\begin{bmatrix}\\alpha\\\\\\beta\\\\\\end{bmatrix}\\mid\\alpha,\\beta\\in\\mathbb{R}\\bigg\\} = \\mathbb{R}^{2}\n$$\n\n%------------------------------------------------------------------------------------------------------------Basis-----------------------------------------------------------------------------------------------------------------------\n\\section{Basis}\n\\textbf{Definition:} A basis for a vector space, $\\mathcal{V}$, is a set of vectors, $\\mathcal{T}$, such that it is the maximal linear independent set and minimal spanning set of $\\mathcal{V}$.\n\n% Existence of a Basis and Uniqueness of a Basis\nIs a basis for a space unique?\\\\\nLet their be two bas\nDoes a basis for a space exist?\\\\\n\\subsection{Examples}\nA basis for $\\mathbb{R}^{2}$ is $\\begin{bmatrix} 0 & 1\\\\ 1 & 0\\\\ \\end{bmatrix}$.\\\\\n\nWe need to show that the set of vectors are linearly independent and span the space of $\\mathbb{R}^{2}$.\n$$\n\\alpha_{1}v_{1} + \\alpha_{2}v_{2} = 0 \\iff \\alpha_{i} = 0\\hspace{2mm}i = \\{1,2\\}\n$$\n$$\n\\alpha_{1}\\begin{bmatrix}0\\\\1\\\\\\end{bmatrix} + \\alpha_{2}\\begin{bmatrix}1\\\\0\\\\\\end{bmatrix} = 0\n$$\n$$\n\\Rightarrow\n\\begin{bmatrix}0\\\\\\alpha_{1}\\end{bmatrix} + \\begin{bmatrix}\\alpha_{2}\\\\0\\\\\\end{bmatrix} = 0 \n$$\n$$\n\\Rightarrow\n\\begin{bmatrix}\\alpha_{2}\\\\\\alpha_{1}\\\\\\end{bmatrix} = 0 \\Rightarrow \\alpha_{1} = \\alpha_{2} = 0\n$$\n\nTo see that it spans $\\mathbb{R}^{2}$ can be observed from the previous section.\\\\\n%----------------------------------------------------------------------------------------------------------Linear Maps-------------------------------------------------------------------------------------------------------------------\n\\section{Linear Maps}\n\\textbf{Definition:} Given vector spaces, $\\mathcal{U,V}$ over the field $\\mathbb{F}$ a linear map, $T: \\mathcal{U}\\rightarrow\\mathcal{V}$ is a transformation that preserves additivity and homogeneity.\n\n\\begin{itemize}\n\\item $T(\\mathbf{u + w}) = T(\\mathbf{u}) + T(\\mathbf{w})\\hspace{2mm} \\forall\\mathbf{u,w}\\in\\mathcal{U}$\n\\item $T(\\alpha\\mathbf{u}) = \\alpha T(\\mathbf{u})$\n\\end{itemize}\n\n\\subsection{Examples}\n\\begin{itemize}\n\\item $T(x): x\\rightarrow x^{2}$ is not linear.\n\\item $T(x): x\\rightarrow 0$ is linear.\n\\end{itemize}\n\nIf $\\mathcal{U,V}$ are finite dimensional and a basis exists for them, the linear maps can be defined as matrices.\n\n\\subsection{Range}\n\\textbf{Definition:} Let $T: \\mathcal{V}\\rightarrow\\mathcal{U}$ be a linear map and denote the Range(Image) of $T$ as,\n$$\n\\mathcal{R}(T) = \\{\\mathbf{u}\\mid T(v) = u,\\hspace{2mm} \\mathbf{u}\\in\\mathcal{U},\\hspace{1mm}\\mathbf{v}\\in\\mathcal{V}\\}\n$$\n\n%-Is the range space unique? Does the range space exist?\n\n\\subsection{Nullspace}\n\\textbf{Definition:} Let $T: \\mathcal{V}\\rightarrow\\mathcal{U}$ be a linear map, then denote the Nullspace of $T$ as,\n$$\n\\mathcal{N}(T) = \\{\\mathbf{v}\\mid T(v) = 0,\\hspace{2mm}\\mathbf{u}\\in\\mathcal{U}\\}\n$$\n\n% Is the nullspace unique?\n\n\\end{document}\n\n\n\n\n\n\n\n", "meta": {"hexsha": "2d552408b3bae4141e30155cf98fc8b4401549a8", "size": 8729, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "files/notes/tex/lin-alg-space-maps.tex", "max_stars_repo_name": "ibrahimkakbar/ibrahimkakbar.github.io", "max_stars_repo_head_hexsha": "5e6b0ea67f5e5f8f3a7bb4394095ea7f7992673a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "files/notes/tex/lin-alg-space-maps.tex", "max_issues_repo_name": "ibrahimkakbar/ibrahimkakbar.github.io", "max_issues_repo_head_hexsha": "5e6b0ea67f5e5f8f3a7bb4394095ea7f7992673a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "files/notes/tex/lin-alg-space-maps.tex", "max_forks_repo_name": "ibrahimkakbar/ibrahimkakbar.github.io", "max_forks_repo_head_hexsha": "5e6b0ea67f5e5f8f3a7bb4394095ea7f7992673a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.9615384615, "max_line_length": 313, "alphanum_fraction": 0.5844884867, "num_tokens": 2693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195269001831, "lm_q2_score": 0.8519528000888386, "lm_q1q2_score": 0.6254351865425044}} {"text": "\\subsection{Higgs mechanism and Electroweak symmetry breaking}\n\\label{symbreaking}\n\nAs shown in previous subsection, the Lagrangian $L_{gauge}$ does not involve any mass term due to the requirement of gauge invariance.\nSo all the W and B bosons should be massless. But experimental observations show that the gauge bosons are massive.\nTherefore, the gauge invariance must be broken spontaneously.\nThe Higgs field is introduced to break the $SU(2)_{L} \\times U(1)_{Y}$ symmetry and\ngauge bosons and fermions can interact with Higgs filed to acquire their masses.\nAnd this specific process is named \\textit{Higgs mechanism} in SM.\n\nThe Higgs field $\\phi$ is a doublet and can be written in a Hermitian basis as\n\\begin{equation}\n\t\\phi = \\binom{\\phi^{+}}{\\phi^{0}} = \\frac{1}{\\sqrt{2}} \\binom{\\phi_{1} - i\\phi_{2}}{\\phi_{3} - i\\phi_{4}}\n\\end{equation}\nwhere $\\phi_{i} = \\phi_{i}^{+}$ stand for four Hermitian field. \nIn this new basis, the Higgs potential in Eq.~\\ref{eq:Vhiggs} can be expressed as:\n\\begin{equation}\n\tV(\\phi) = \\frac{1}{2}\\mu^{2}\\left(\\sum_{i=1}^{4}\\phi_{i}^{2}\\right) + \\frac{1}{4}\\lambda\\left(\\sum_{i=1}^{4}\\phi_{i}^{2}\\right)^{2}\n\\end{equation}\nTo simplify the situation, the axis in this four-dimensional space can be chosen to satisfied\n~$\\left<0\\left| \\phi_{i} \\right|0\\right> = 0$ for $i = 1, 2, 4$, and $<0\\left| \\phi_{3} \\right|0> = v$. Thus,\n\\begin{equation}\n\tV(\\phi) \\rightarrow V(v) = \\frac{1}{2}\\mu^{2}v^{2} + \\frac{1}{4}\\lambda v^{4}\n\\end{equation}\nThe minimization of this potential depends on the sign of $\\mu^{2}$ as shown in figure~\\ref{fig:C2_Higgs_potential}.\nWhen $\\mu^{2} > 0$ the minimum occurs at $v = 0$, namely the vacuum is empty space and $SU(2)_{L} \\times U(1)_{Y}$ symmetry is unbroken.\nIn the case of $\\mu^{2} < 0$, the $v = 0$ symmetric point is no longer stable and the minimum occurs at nonzero value of \n$v = \\left( -\\mu^{2}/\\lambda\\right)^{1/2}$ which breaks the $SU(2)_{L} \\times U(1)_{Y}$ symmetry.\n\\begin{figure}[!htb]\n \\centering\n \\includegraphics[width=0.7\\textwidth]{figures/Theory/Vhiggs.png}\n \\caption{The Higgs potential $V(\\phi)$ with $\\mu^{2}>0$ (left) and $\\mu^{2}<0$ (right).}\n \\label{fig:C2_Higgs_potential}\n\\end{figure}\nThus, the classical vacuum $\\phi_{0}$ of Higgs doublet can be expressed by\n\\begin{equation}\n\t\\phi_{0} = \\frac{1}{\\sqrt{2}}\\binom{0}{v}\n\\end{equation}\nAnd to quantize around the classical vacuum in a general form:\n\\begin{equation}\n\t\\phi = \\frac{1}{\\sqrt{2}} \\binom{0}{v+H}\n\\end{equation}\nWhere H is a Hermitian field for physical Higgs scalar.\nIn this gauge, the Lagrangian $L_{Higgs}$ in Eq.~\\ref{eq:Lhiggs} takes a simple form\n\\begin{equation}\n\\begin{split} \\label{eq:Lhiggs2}\n\tL_{Higgs} & = \\left(D^{\\mu}\\phi\\right)^{\\dagger}D_{\\mu}\\phi - V(\\phi) \\\\\n\t& = M_{W}^{2}W^{\\mu+}W_{\\mu}^{-}\\left(1+\\frac{H}{\\nu}\\right)^{2} + \\frac{1}{2}M_{Z}^{2}Z^{\\mu}Z_{\\mu}\\left(1+\\frac{H}{\\nu}\\right)^{2} \\\\ \n & + \\frac{1}{2}\\left(\\partial_{\\mu}H\\right)^{2} - V(\\phi)\n\\end{split}\n\\end{equation}\nwhere the W and Z fields are\n\\begin{equation}\n\\begin{split}\n\t& W^{\\pm} = \\frac{1}{\\sqrt{2}} \\left(W^{1} \\mp iW^{2}\\right) \\\\\n\t& Z = - sin\\theta_{W}B + cos\\theta_{W}W^{3}\n\\end{split}\n\\end{equation}\nTherefore, in Eq.~\\ref{eq:Lhiggs2} spontaneous symmetry breaking brings out masses for the W and Z gauge bosons\n\\begin{equation}\n\\begin{split}\n\t& M_{W} = \\frac{gv}{2} \\\\\n\t& M_{Z} = \\sqrt{g^{2} + g'^{2}} \\frac{v}{2} = \\frac{M_{W}}{cos\\theta_{W}}\n\\end{split}\n\\end{equation}\nwhere $\\theta_{W}$ is the weak angle defined as\n\\begin{equation}\n\tsin\\theta_{W} = \\frac{g'}{\\sqrt{g^{2} + g'^{2}}} \\qquad cos\\theta_{W} = \\frac{g}{\\sqrt{g^{2} + g'^{2}}} \\qquad tan\\theta_{W} = \\frac{g'}{g}\n\\end{equation}\nThen another gauge boson photon remains massless with the field of\n\\begin{equation}\n\tA = cos\\theta_{W}B + sin\\theta_{W}W^{3}\n\\end{equation}\n\nAfter the symmetry breaking, the Higgs potential in unitary gauge can be written into\n\\begin{equation}\n\tV(\\phi) = -\\frac{\\mu^{4}}{4\\lambda} - \\mu^{4}H^{2} + \\lambda\\nu H^{3} + \\frac{\\lambda}{4}H^{4}\n\\end{equation}\nThe first term in $V$ is a constant, while the second term denotes a (tree-level) mass of Higgs boson\n\\begin{equation}\n\tM_{H} = \\sqrt{-2\\mu^{2}} = \\sqrt{2\\lambda}v\n\\end{equation}\nDue to the unknown of quartic Higgs coupling $\\lambda$, the Higgs mass is not predicted.\nThe third and fourth terms in the Higgs potential $V$ denote the induced cubic and quartic interactions of the Higgs scalar.\n\nThrough the Higgs mechanism, fermions can also acquire their masses.\nIn the unitary gauge, Yukawa Lagrangian ($L_{Yukawa}$) can be written as a simple form of \\cite{Pich:2015lkh}\n\\begin{equation}\n\tL_{Yukawa} = -\\left(1+\\frac{H}{v}\\right) \\left(m_{d}\\bar{d}d + m_{u}\\bar{u}u + m_{l}\\bar{l}l\\right)\n\\end{equation}\nin which $m_{f} = \\frac{y_{f}v}{\\sqrt{2}}$ for $f = d, u, l$.\n", "meta": {"hexsha": "ace8bfd9e508de454629b7f0acac19da44da7a23", "size": 4787, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/Theory/symbreaking.tex", "max_stars_repo_name": "zhuhel/PhDthesis", "max_stars_repo_head_hexsha": "55ec32affb5c105143798989d78043467c88da8e", "max_stars_repo_licenses": ["LPPL-1.3c"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/Theory/symbreaking.tex", "max_issues_repo_name": "zhuhel/PhDthesis", "max_issues_repo_head_hexsha": "55ec32affb5c105143798989d78043467c88da8e", "max_issues_repo_licenses": ["LPPL-1.3c"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/Theory/symbreaking.tex", "max_forks_repo_name": "zhuhel/PhDthesis", "max_forks_repo_head_hexsha": "55ec32affb5c105143798989d78043467c88da8e", "max_forks_repo_licenses": ["LPPL-1.3c"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.0326086957, "max_line_length": 140, "alphanum_fraction": 0.6747440986, "num_tokens": 1761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8519527944504227, "lm_q2_score": 0.734119521083126, "lm_q1q2_score": 0.6254351774473752}} {"text": "\\subsection{Convex functions}\\label{subsec:convex_functions}\n\nLet \\( X \\) be a Hausdorff \\hyperref[def:topological_vector_space]{topological vector space} and \\( D \\) be a \\hyperref[def:convex_set]{convex} subset of \\( X \\).\n\n\\begin{definition}\\label{def:convex_functions}\n A function \\( f: D \\to \\BbbR \\) is called \\term{convex} if any of the following equivalent conditions hold:\n\n \\begin{thmenum}\n \\thmitem{def:convex_functions/ineq} For any two points \\( x, y \\in D \\) and any \\( t \\in [0, 1] \\) we have\n \\begin{equation*}\n f(tx + (1-t)y) \\leq tf(x) + (1-t)f(y).\n \\end{equation*}\n\n \\thmitem{def:convex_functions/epi} The \\hyperref[def:epigraph]{epigraph}\n \\begin{equation*}\n \\epi f \\coloneqq \\{ (x, a) \\in X \\times \\BbbR \\colon f(x) \\leq a \\}\n \\end{equation*}\n is convex.\n \\end{thmenum}\n\n If \\( -g \\) is convex for some function \\( g: D \\to \\BbbR \\), we call \\( g \\) \\term{concave}.\n\n Note that definitions do not require any topological structure on \\( X \\). Most of their properties, however, require a topology.\n\\end{definition}\n\\begin{proof}\n Let \\( x, y \\in D \\) and let \\( t \\in [0, 1] \\).\n\n \\ImplicationSubProof{def:convex_functions/ineq}{def:convex_functions/epi} Let \\( \\epi f \\) be a convex set. Obviously \\( (x, f(x)) \\in D \\) and \\( (y, f(y)) \\in D \\). By the convexity of \\( \\epi f \\), we have\n \\begin{equation*}\n f(tx + (1-t)y) \\leq tf(x) + (1-t)f(y).\n \\end{equation*}\n\n Thus, \\( f \\) is a convex function.\n\n \\ImplicationSubProof{def:convex_functions/epi}{def:convex_functions/ineq} Let \\( f \\) be convex. Let \\( a \\geq f(x) \\) and \\( b \\geq f(y) \\), so that \\( (x, a) \\in \\epi f \\) and \\( (y, b) \\in \\epi f \\). Hence,\n \\begin{equation*}\n f(tx + (1-t)y) \\leq tf(x) + (1-t)f(y) \\leq ta + (1-t)b,\n \\end{equation*}\n which implies that\n \\begin{equation*}\n (tx + (1-t)y, ta + (1-t)b) \\in \\epi f.\n \\end{equation*}\n\n Thus, \\( \\epi f \\) is a convex set.\n\\end{proof}\n\n\\begin{definition}\\label{def:affine_functions_concave_and_convex}\n \\hyperref[def:affine_operator]{Affine functions} \\( f: X \\to \\BbbR \\) are simultaneously convex and concave.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:convex_subdifferential_is_convex_and_weak*_closed}\\mcite[exer. 1.10]{Phelps1993}\n For any convex function \\( f \\) and any \\( x \\in D \\), the set \\( \\partial f(x) \\) is convex and weak* closed.\n\\end{proposition}\n\\begin{proof}\n Fix \\( x \\in D \\). If \\( \\partial f(x) \\) is empty, then the theorem is trivially true.\n\n Suppose it is nonempty and \\( y^*, z^* \\in \\partial f(x) \\). For any \\( x \\in D \\) we then have\n \\begin{balign*}\n & \\inprod{y^*} {x - x} \\leq f(x) - f(x), \\\\\n & \\inprod{z^*} {x - x} \\leq f(x) - f(x).\n \\end{balign*}\n\n Fix \\( t \\in [0, 1] \\) and \\( x \\in D \\). It follows that\n \\begin{balign*}\n \\inprod{t y^* + (1-t) z^*} {x - x}\n & =\n t \\inprod{y^*} {x - x} + (1-t) \\inprod{z^*} {x - x}\n \\leq \\\\ &\\leq\n t [f(x) - f(x)] + (1-t) [f(x) - f(x)]\n = \\\\ &=\n f(x) - f(x),\n \\end{balign*}\n thus \\( t y^* + (1-t)z^* \\in \\partial f(x) \\) and hence \\( \\partial f(x) \\) is convex.\n\n To prove weak*-closedness, we consider the decomposition\n \\begin{balign*}\n \\partial f(x)\n & =\n \\{ x^* \\in E^* \\colon \\forall x \\in D, \\inprod {x^*} {x - x} \\leq f(x) - f(x) \\}\n = \\\\ &=\n \\bigcap_{x \\in D} \\{ x^* \\in E^* \\colon \\inprod {x^*} {x - x} \\leq f(x) - f(x) \\}\n = \\\\ &=\n \\bigcap_{x \\in D} L(x)^{-1} (-\\infty, f(x) - f(x)],\n \\end{balign*}\n where\n \\begin{balign*}\n & L: E \\to E^{**} \\\\\n & L(x)(x^*) = \\inprod {x^*} {x - x}.\n \\end{balign*}\n\n For each \\( x \\in E \\), the functionals \\( L(x) \\) are weak*-to-weak continuous because the image \\( L(E) \\subseteq E^{**} \\) is isometrically isomorphic to a translation of \\( E \\). Hence, the preimage \\( L(x)^{-1} (-\\infty, f(x) - f(x)] \\) is closed and \\( \\partial f(x) \\) is weak*-closed as the intersection of weak*-closed sets.\n\\end{proof}\n\n\\begin{lemma}\\label{thm:convex_difference_quotient_grows}\n For every point \\( x \\in X \\) and every direction \\( h \\in S_X \\) the difference quotient is a monotone function of \\( t > 0 \\), i.e. for \\( 0 < s < t \\)\n \\begin{balign*}\n \\frac {f(x + sh) - f(x)} s\n \\leq\n \\frac {f(x + th) - f(x)} t\n \\end{balign*}\n\\end{lemma}\n\\begin{proof}\n \\begin{balign*}\n \\frac {f(x + sh) - f(x)} s\n =\n \\frac t s \\frac {f(x + \\frac s t t h) - f(x)} t\n =\n \\frac t s \\frac {f\\left(\\frac s t (x + th) + (1 - \\frac s t) x \\right) - f(x)} t\n \\leq \\\\ \\leq\n \\frac t s \\frac {\\frac s t f(x + t h) + (1 - \\frac s t) f(x) - f(x)} t\n =\n \\frac t s \\frac s t \\frac {f(x + th) - f(x)} t\n =\n \\frac {f(x + th) - f(x)} t\n \\end{balign*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:convex_one_sided_derivatives_exist}\n For every point \\( x \\in X \\) and every direction \\( h \\in S_X \\) the one-sided derivative \\( f_+'(x)(h) \\) exists.\n\\end{proposition}\n\\begin{proof}\n We use the convexity of \\( f \\) to obtain\n \\begin{balign*}\n f(x) = f \\left(x + \\frac {th} 2 - \\frac {th} 2 \\right) \\leq \\frac {f(x + th) + f(x - th)} 2,\n \\\\\n 0 \\leq [f(x - th) - f(x)] + [f(x + th) - f(x)],\n \\\\\n -[f(x - th) - f(x)] \\leq [f(x + th) - f(x)],\n \\\\\n -\\frac {f(x + t(-h)) - f(x)} t \\leq \\frac {f(x + th) - f(x)} t,\n \\end{balign*}\n thus the difference quotient in \\( f_+'(x)(h) \\) is bounded below by the difference quotient for \\( -f_+'(x)(-h) \\).\n\n \\Fullref{thm:convex_difference_quotient_grows} implies that the right difference quotient is non-increasing, thus both limits exist and\n \\begin{equation*}\n -f_+'(x)(-h) \\leq f_+'(x)(h).\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:convex_one_sided_derivatives_sublinear}\n For every point \\( x \\in X \\) and every direction \\( h \\in S_X \\) the one-sided derivative \\( f_+'(x)(h) \\) is a \\hyperref[def:sublinear_functional]{sublinear functional}.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{def:sublinear_functional/subadditivity} It follows directly from\n \\begin{balign*}\n \\frac {f(x + t(a + b)) - f(x)} t\n & =\n \\frac {f(\\tfrac 1 2 (x + 2ta) + \\tfrac 1 2 (x + 2tb)) - f(x)} t\n \\leq \\\\ &\\leq\n \\frac {\\tfrac 1 2 f(x + 2ta) + \\tfrac 1 2 f(x + 2tb) - f(x)} t\n = \\\\ &=\n \\frac {f(x + 2ta) - f(x)} {2t} + \\frac {f(x + 2tb) - f(x)} {2t}.\n \\end{balign*}\n\n \\SubProofOf{def:sublinear_functional/positive_homogeneity} For \\( \\lambda > 0 \\) the equality \\( f_+'(x)(\\lambda h) = \\lambda f_+'(x)(h) \\) follows from\n \\begin{balign*}\n \\frac {f(x + t \\lambda h) - f(x)} t\n =\n \\lambda \\frac {f(x + t \\lambda h) - f(x)} {t \\lambda}\n \\end{balign*}\n\\end{proof}\n\n\\begin{corollary}\\label{thm:convex_one_sided_derivative_negative_inequality}\n \\begin{equation*}\n -f_+'(x)(-h) \\leq f_+'(x)(h)\n \\end{equation*}\n\\end{corollary}\n\\begin{proof}\n \\begin{equation*}\n 0 = f_+'(x)(h + (-h)) \\leq f_+'(x)(h) + f_+'(x)(-h)\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:convex_iff_subdifferential_nonempty}\n The continuous function \\( f: D \\to X \\) is convex if and only if its subdifferential \\( \\partial f(x) \\) (see \\fullref{def:subdifferentials/convex}) is nonempty for every \\( x \\) in \\( D \\).\n\\end{proposition}\n\\begin{proof}\n \\todo{Prove}.\n\\end{proof}\n\n\\begin{proposition}\n \\label{thm:convex_one_sided_derivative_is_max}\n For every direction \\( h \\in S_X \\), we have that\n \\begin{equation*}\n f_+'(x)(h) = \\max\\{ \\inprod {x^*} h \\colon x^* \\in \\partial f(x) \\}.\n \\end{equation*}\n\\end{proposition}\n\n\\begin{theorem}\\label{thm:singleton_subdifferential_implies_gateaux}\n If \\( f \\) is continuous and if the subdifferential \\( \\partial f(x) \\) at \\( x \\in X \\) is a singleton with element \\( x^* \\), then \\( f \\) is Gateaux differentiable at \\( x \\) and \\( f_G'(x) = x^* \\).\n\\end{theorem}\n\\begin{proof}\n Let \\( h \\in S_X \\) be arbitrary. \\Fullref{thm:convex_one_sided_derivatives_exist} implies that the one-sided derivatives \\( f_+'(x)(-h) \\) and \\( f_+'(x)(h) \\) exist and\n \\begin{equation*}\n -f_+'(x)(-h) \\leq f_+'(x)(h).\n \\end{equation*}\n\n Assume that \\( f \\) is not Gateaux differentiable at \\( x \\), i.e. for some \\( h_0 \\in X \\), we have a strict inequality. Then by \\fullref{thm:convex_one_sided_derivative_is_max}\n \\begin{balign*}\n \\min\\{ \\inprod {x^*} {h_0} \\colon x^* \\in \\partial f(x) \\}\n =\n -\\max\\{ \\inprod {x^*} {-h_0} \\colon x^* \\in \\partial f(x) \\}\n =\n -f_+'(x)(-h_0)\n < \\\\ <\n f_+'(x)(h_0)\n =\n \\max\\{ \\inprod {x^*} {h_0} \\colon x^* \\in \\partial f(x) \\},\n \\end{balign*}\n which implies that there is more that one functional \\( x^* \\in \\partial_C f(x) \\). This contradicts the assumption of the theorem.\n\n Thus, \\( f \\) is Gateaux differentiable at \\( x \\).\n\\end{proof}\n\n\\begin{theorem}\\label{thm:rn_continuous_convex_partial_derivatives_imply_gateaux}\\mcite[exer. 1.15(b]{Phelps1993})\n In \\( \\BbbR^n \\), the existence of the partial derivatives at \\( x \\) for a continuous convex function \\( f: D \\to \\BbbR \\) at a point \\( x \\in D \\) implies Gateaux differentiability.\n\\end{theorem}\n\\begin{proof}\n Let \\( D \\subseteq \\BbbR^n \\) be an open and convex set and let \\( f: D \\to \\BbbR \\) be continuous and convex. Then \\( f_+'(x) \\) exists everywhere by \\fullref{thm:convex_one_sided_derivatives_exist} and is a subdifferential functional by \\fullref{thm:convex_one_sided_derivatives_sublinear}.\n\n Let \\( e_1, \\ldots, e_n \\) be the canonical basis for \\( \\BbbR^n \\).\n\n The partial derivatives\n \\begin{balign*}\n \\frac {\\partial f} {\\partial x_i} (x)\n \\coloneqq\n \\lim_{t \\to 0} \\frac {f(x + t e_i) - f(x)} t\n =\n f_+'(x)(e_i)\n \\end{balign*}\n exist, hence the projections of \\( f_+'(x) \\) along the coordinate exes are linear.\n\n Define line linear functional\n \\begin{equation*}\n l(h) \\coloneqq \\sum_{i=1}^n h_i \\inprod{\\frac {\\partial f} {\\partial x_i} (x)} h,\n \\end{equation*}\n where \\( h_1, \\ldots, h_n \\) are the coordinates of \\( h \\) along \\( e_1, \\ldots, e_n \\).\n\n We will show that \\( l \\cong f_+' \\). Fix \\( h \\in S_X \\). We have\n \\begin{balign}\\label{thm:rn_continuous_convex_partial_derivatives_imply_gateaux/diff_dominated}\n f_+'(x)(h)\n & =\n f_+'(x)\\left(\\sum_{i=1}^n h_i e_i \\right)\n \\reloset {\\text{sublinearity}} \\leq \\nonumber \\\\ &\\leq\n \\sum_{i=1}^n f_+'(x)(h_i e_i)\n \\reloset {\\text{linearity along } e_i} = \\nonumber \\\\ &=\n \\sum_{i=1}^n h_i f_+'(x)(e_i)\n =\n \\sum_{i=1}^n h_i \\inprod{\\frac {\\partial f} {\\partial x_i} (x)} h.\n \\end{balign}\n\n Thus,\n \\begin{balign*}\n \\inprod l h\n =\n -\\inprod l {-h}\n \\reloset {\\ref{thm:rn_continuous_convex_partial_derivatives_imply_gateaux/diff_dominated}} \\leq\n -f_+'(x)(-h)\n \\reloset {\\text{\\ref{thm:convex_one_sided_derivative_negative_inequality}}} \\leq\n f_+'(x)(h)\n \\reloset {\\ref{thm:rn_continuous_convex_partial_derivatives_imply_gateaux/diff_dominated}} \\leq\n \\inprod l h,\n \\end{balign*}\n i.e. \\( f_+'(x)(h) = \\inprod l h \\) for all \\( h \\in S_X \\), hence \\( f_+'(x) \\) is a linear functional and \\( f \\) is Gateaux differentiable at \\( x \\).\n\\end{proof}\n\n\\begin{theorem}\\label{thm:rn_continuous_convex_gateaux_implies_frechet}\\mcite[exer. 1.15(a]{Phelps1993})\n In \\( \\BbbR^n \\), Gateaux differentiability of a continuous convex function \\( f: D \\to \\BbbR \\) at a point \\( x \\in D \\) implies Frechet differentiability.\n\\end{theorem}\n\\begin{proof}\n Since \\( f \\) is Gateaux differentiable (\\fullref{def:differentiability/gateaux}) at \\( x \\), the derivative \\( f'(x) = f_+'(x) \\) is linear.\n\n Because \\( f \\) is continuous and convex, it is locally Lipschitz with constant \\( L \\) in some \\( \\delta \\)-ball with center \\( x \\).\n\n Suppose that \\( f \\) is not Frechet differentiable at \\( x \\). Inverting the condition in \\fullref{def:differentiability/frechet}, we obtain that there exist \\( \\varepsilon > 0 \\) and a sequence \\( \\{ h_n \\}_n \\subseteq B(x, \\delta) \\setminus \\{ 0 \\} \\) such that \\( \\norm{h_n} \\to 0 \\) and yet for all \\( n \\in \\BbbZ_{>0} \\),\n \\begin{balign}\\label{thm:rn_continuous_convex_gateaux_implies_frechet/frechet_assumption}\n \\abs{f(x + h_n) - f(x) - \\inprod{f'(x)} {h_n}} > \\varepsilon \\norm{h_n}.\n \\end{balign}\n\n Define\n \\begin{balign*}\n t_n \\coloneqq \\norm{h_n}\n & &\n u_n \\coloneqq \\frac{h_n} {\\norm {h_n}}.\n \\end{balign*}\n\n Obviously \\( t_{n_k} \\downarrow 0 \\). The vectors \\( h_n \\) are linearly independent since otherwise \\( f \\) would not be Gateaux differentiable at \\( x \\), hence \\( u_n \\) are not all equal.\n\n Since \\( S_{\\BbbR^n} \\) is compact, by the Bolzano-Weierstrass theorem, there exists a convergent subsequence \\( \\{ u_{n_k} \\}_k \\underset {k \\to \\infty} \\to u_0 \\) of \\( \\{ u_n \\}_n \\). We have\n\n \\begin{balign}\\label{thm:rn_continuous_convex_gateaux_implies_frechet/frechet_estimate}\n & \\phantom= \\abs{\\frac {f(x + t_{n_k} u_{n_k}) - f(x)} {t_{n_k}} - \\inprod{f'(x)} {u_{n_k}}}\n \\leq \\nonumber\n \\abs{\\frac {f(x + t_{n_k} u_{n_k}) - f(x + t_{n_k} u_0)} {t_{n_k}}} + \\\\ &+ \\abs{\\frac {f(x + t_{n_k} u_0) - f(x)} {t_{n_k}} - \\inprod{f'(x)} {u_0}} + \\abs{\\inprod{f'(x)} {u_0 - u_{n_k}}}\n \\leq \\nonumber \\\\ &\\leq\n L \\norm{u_{n_k} - u_0} + \\abs{\\frac {f(x + t_{n_k} u_0) - f(x)} {t_{n_k}} - \\inprod{f'(x)} {u_0}} + \\norm{f'(x)} \\norm{u_0 - u_{n_k}}.\n \\end{balign}\n\n Fix \\( \\delta > 0 \\). Because of the Gateaux differentiable of \\( f \\) at \\( x \\), we can pick \\( k_0 \\) such that\n \\begin{equation*}\n \\abs{\\frac {f(x + t_{n_{k_0}} u_0) - f(x)} {t_{n_{k_0}}} - \\inprod{f'(x)} {u_0}} < \\delta.\n \\end{equation*}\n\n Because \\( \\{ u_{n_k} \\}_k \\) converges to \\( u_0 \\), we can choose \\( k_1 \\) such that\n \\begin{equation*}\n \\norm{u_0 - u_{n_{k_1}}} < \\delta.\n \\end{equation*}\n\n Thus, for \\( k > \\max \\{ k_0, k_1 \\} \\), \\fullref{thm:rn_continuous_convex_gateaux_implies_frechet/frechet_estimate} is bounded by\n \\begin{balign*}\n \\abs{\\frac {f(x + t_{n_k} u_{n_k}) - f(x)} {t_{n_k}} - \\inprod{f'(x)} {u_{n_k}}}\n \\leq\n (L + 1 + \\norm{f'(x)}) \\delta.\n \\end{balign*}\n\n It suffices to choose \\( \\delta > 0 \\), so that\n \\begin{equation*}\n \\delta < \\frac 1 {L + 1 + \\norm{f'(x)}}\n \\end{equation*}\n in order to have, for \\( k > \\max \\{ k_0, k_1 \\} \\),\n \\begin{equation*}\n \\abs{\\frac {f(x + t_{n_k} u_{n_k}) - f(x)} {t_{n_k}} - \\inprod{f'(x)} {u_{n_k}}} < \\varepsilon.\n \\end{equation*}\n\n But this contradicts \\fullref{thm:rn_continuous_convex_gateaux_implies_frechet/frechet_assumption}, hence \\( f \\) is Frechet differentiable at \\( x \\).\n\\end{proof}\n\n\\begin{corollary}\\label{thm:rn_continuous_convex_partial_derivatives_imply_frechet}\n In \\( \\BbbR^n \\), the existence of the partial derivatives at \\( x \\) for a continuous convex function \\( f: D \\to \\BbbR \\) at a point \\( x \\in D \\) is equivalent to Frechet differentiability.\n\\end{corollary}\n\\begin{proof}\n A direct consequence of and \\fullref{thm:rn_continuous_convex_partial_derivatives_imply_gateaux} and \\fullref{thm:rn_continuous_convex_gateaux_implies_frechet}.\n\\end{proof}\n\n\\begin{theorem}\\label{thm:rn_continuous_convex_frechet_almost_everywhere}\\mcite[exer. 1.17]{Phelps1993}\n In \\( \\BbbR^n \\), continuous convex functions \\( f: D \\to \\BbbR \\) are differentiable almost everywhere.\n\\end{theorem}\n\\begin{proof}\n For all \\( h \\in S_X \\) and small enough \\( t > 0 \\) we define\n \\begin{balign*}\n & \\varphi_h^t: D \\to \\BbbR\n & \\varphi_h^t(x) \\coloneqq \\frac {f(x + th) - f(x)} t\n \\end{balign*}\n and \\( \\varphi_h(x) \\coloneqq f_+'(x)(h) = \\lim_{t \\downarrow 0} \\varphi_h^t(x) \\).\n\n Considered as functions of \\( x \\), \\( \\varphi_h^t \\) are obviously continuous hence Borel measurable, and so \\( \\varphi_h \\) is also Borel measurable.\n\n Denote by\n \\begin{balign*}\n B_h\n \\coloneqq\n \\{ x \\in D \\colon -f_+'(x)(-h) < f_+'(x)(h) \\}\n =\n \\{ x \\in D \\colon -\\varphi_{-h}(x) - \\varphi_h(x) < 0 \\}\n \\end{balign*}\n the set of points \\( x \\in D \\) where the one-sided derivative \\( f_+'(x)(h) \\) is not linear, given a fixed direction \\( h \\in S_X \\). If \\( B_h \\) is nonempty, \\( f \\) is not differentiable at \\( x \\).\n\n The sets \\( B_h \\) are Borel sets since they are the preimages of \\( (-\\infty, 0) \\) under a Borel function. We will show that it is a null set for every direction \\( h \\).\n\n Fix \\( h \\in S_X \\). Denote by \\( \\delta_x \\coloneqq \\sup \\{ t > 0 \\colon x + th \\in D \\} \\).\n\n The function \\( t \\mapsto f(x + th) \\) is a convex function of one variable. By \\cite[theorem 1.16]{Phelps1993}, it is differentiable \\( \\mu_1 \\)-almost everywhere in \\( [0, \\delta_x) \\), where \\( \\mu_m \\) is the Lebesgue \\( m \\)-measure.\n\n Denote\n \\begin{balign*}\n & H \\coloneqq \\linspan\\{ h \\} \\cong \\BbbR^1,\n \\\\\n & H^\\perp \\cong \\BbbR^{n-1} \\text{ - the orthogonal complement of \\( H \\) in \\( \\BbbR^n \\)},\n \\\\\n & L_x \\coloneqq \\{ x + th, 0 \\leq t < \\delta_x \\} - half-open segments in D.\n \\end{balign*}\n\n THe whole domain \\( D \\) can be represented as \\( D = \\cup \\{ L_x \\colon x \\in H^\\perp \\} \\).\n\n We can now use Fubini's theorem to show that \\( B_h \\) is a null set:\n \\begin{balign*}\n \\mu_n(B_h)\n =\n \\int_{B_h} dz\n =\n \\int_{\\BbbR^n = H^\\perp \\oplus H} \\chi_{B_h} (z) dz\n =\n \\int_{H^\\perp} \\int_{L_x} \\chi_{B_h} (y) dy dx\n = \\\\ =\n \\int_{H^\\perp} \\mu_1(B_h \\cap L_x) dx\n =\n \\int_{H^\\perp} 0 dx\n =\n 0.\n \\end{balign*}\n\n Hence, for all \\( h \\in S_X \\), \\( -f_+'(x)(-h) = f_+'(x)(h) \\) for almost all \\( x \\in D \\).\n\n In particular, if \\( e_1, \\ldots, e_n \\) is the canonical basis of \\( \\BbbR^n \\), the \\( i \\)-th partial derivative \\( \\frac{\\partial f} {\\partial x_i} (x) \\) exists only in \\( D \\ B_{e_i} \\).\n\n The gradient\n \\begin{equation*}\n \\nabla f(x) = \\left( \\frac{\\partial f} {\\partial x_1} (x), \\ldots, \\frac{\\partial f} {\\partial x_n} (x) \\right)\n \\end{equation*}\n then exists in\n \\begin{equation*}\n \\hat D \\coloneqq (D \\ B_{e_1}) \\cap \\ldots \\cap (D \\ B_{e_n}) = D \\setminus \\left( \\bigcup_{i=1}^n B_{e_i} \\right).\n \\end{equation*}\n\n \\Fullref{thm:rn_continuous_convex_partial_derivatives_imply_frechet} then implies that \\( f \\) is Frechet differentiable in \\( \\hat D \\), i.e. almost everywhere in \\( D \\).\n\\end{proof}\n", "meta": {"hexsha": "bfe7a408f8d7ba52a87cc3f3141b6440bdc1c4bd", "size": 18144, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/convex_functions.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/convex_functions.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/convex_functions.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.5798525799, "max_line_length": 335, "alphanum_fraction": 0.5986552028, "num_tokens": 6950, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.822189134878876, "lm_q1q2_score": 0.625398702069792}} {"text": "%!TEX root = ../main.tex\n%-------------------------------------------------------------------------------\n\\subsection{Economic framework}\n%-------------------------------------------------------------------------------\nEKW models describe sequential decision-making under uncertainty \\citep{Gilboa.2009, Machina.2014}. At time $t = 1, \\hdots, T$ each individual observes the state of the economic environment $s_t\\in S$ and chooses an action $a_t$ from the set of admissible actions $\\mathcal{A}$. The decision has two consequences: an individual receives an immediate utility $u_t(s_t, a_t)$ and the economy evolves to a new state $s_{t + 1}$. The transition from $s_t$ to $s_{t + 1}$ is affected by the action but remains uncertain. Individuals are forward-looking. Thus they do not simply choose the alternative with the highest immediate utility. Instead, they take the future consequences of their current action into account.\n\nA policy $\\pi = (a^\\pi_1(s_1), \\hdots, a^\\pi_T(s_T))$ provides the individual with instructions for choosing an action in any possible future state. It is a sequence of decision rules $a^\\pi_t(s_t)$ that specify the action at a particular time $t$ for any possible state $s_t$ under $\\pi$. The implementation of a policy generates a sequence of utilities that depends on the objective transition probability distribution $p_t(s_t, a_t)$ for the evolution of state $s_t$ to $s_{t + 1}$ induced by the model. Individuals have rational expectations \\citep{Muth.1961} so their subjective beliefs about the future agree with the objective transition probabilities of the model.\n\n\\autoref{Timing} depicts the timing of events in the model for two generic periods. At the beginning of period $t$, an individual fully learns about the immediate utility of each alternative, chooses one of them, and receives its immediate utility. Then the state evolves from $s_t$ to $s_{t + 1}$ and the process is repeated in $t + 1$.\n%\n\\begin{figure}[b]\n\t%\\vspace{1.0cm}\n\t\\centering\n\t\\scalebox{0.755}{\\input{../material/fig-timing-events.tex}}\n\t\\caption{Timing of events}\n\t\\label{Timing}\n\\end{figure}\n%\n\nIndividuals face uncertainty and they seek to maximize the expected total discounted utilities. An exponential discount factor ${0 < \\delta < 1}$ parameterizes their time preference and captures a taste for immediate over future utilities.\n\nEquation~\\eqref{Objective Risk} provides the formal representation of the individual's objective. Given an initial state $s_1$, individuals implement the policy $\\pi$ from the set of all possible policies $\\Pi$ that maximizes the expected total discounted utilities over all $T$ decision periods given the information available in the first-period state, $s_1$:\n%\n\\begin{align}\\label{Objective Risk}\n\t\\max_{\\pi \\in\\Pi} \\E_{s_1}^\\pi\\left[\\left.\\sum^{T}_{t = 1} \\delta^{t - 1} u_t(s_t, a^\\pi_t(s_t))\\right]\\right..\n\\end{align}\n%\nThe superscript of the expectation emphasizes that each policy $\\pi$ induces a different probability distribution over the sequences of utilities.\n", "meta": {"hexsha": "f9e906f9fcdfe134aa7d7f6c982d7f41cec8d046", "size": 3031, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/sections/s-setup-economics.tex", "max_stars_repo_name": "OpenSourceEconomics/handout-eckstein-keane-wolpin-models", "max_stars_repo_head_hexsha": "68cc55540c8b8772a3b204b7ba063fb324b08fdb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/sections/s-setup-economics.tex", "max_issues_repo_name": "OpenSourceEconomics/handout-eckstein-keane-wolpin-models", "max_issues_repo_head_hexsha": "68cc55540c8b8772a3b204b7ba063fb324b08fdb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2020-03-05T07:53:32.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-30T11:59:58.000Z", "max_forks_repo_path": "paper/sections/s-setup-economics.tex", "max_forks_repo_name": "OpenSourceEconomics/handout-eckstein-keane-wolpin-models", "max_forks_repo_head_hexsha": "68cc55540c8b8772a3b204b7ba063fb324b08fdb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-09-17T17:09:27.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-15T19:27:05.000Z", "avg_line_length": 104.5172413793, "max_line_length": 712, "alphanum_fraction": 0.7268228307, "num_tokens": 721, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891392358015, "lm_q2_score": 0.7606506472514406, "lm_q1q2_score": 0.6253987009228172}} {"text": "\\section{Verification of Parallel Concatenation on Monoids}\\label{sec:parallelization}\nIn this section we define how to concatenate a list of monoids\ninto a monoid. \n%\nThen, we parallelize list concatenation \nand use LiquidHaskell to prove that the \nparallel and sequential versions are equivalent. \n\\NV{What does equivalence means here?}\n\n\\subsection{Concatenation of lists using monoids}\nIf @m@ is a monoid, then @mconcat@ reduces a list of monoids @xs@\nto one monoid. \n\\begin{code}\nmconcat :: List m -> m \nmconcat N = mempty\nmconcat (C x xs) = x mappend (mconcat xs)\n\\end{code}\n%\nConcatenation folds the elements of the list using @(mappend)@ \nand returns @mempty@ when the list is empty. \n\nNext, we define a parallel version of list concatenation \n%\n\\begin{code}\npmconcat :: Int -> List m -> m\npmconcat i xs | i <= 1 || llen xs <= i\n = mconcat xs \npmconcat i xs \n = pmconcat i $ withStrategy parStrategy $ \n map mconcat (chunk i xs)\n\\end{code}%$\nThe function @pmconcat i xs@ calls @mconcat xs@ in the base case, \notherwise it\n(1) chunks the list @xs@ in lists of size @i@, \n(2) runs in parallel @mconcat@ to each chunk, \n(3) recursively runs itself in the result list.\n%\nTermination of @pmconcat@ holds, as the length of @chunk i xs@\nis smaller than the length of @xs@, when @1 < i@. \n%\nConcretely, we define @chunk@ using the standard definition for \nlist @take@ and @drop@.\n\\begin{code}\nchunk :: i:Pos -> xs:List a \n -> {v:List (List a) | chunkRes i xs } \nchunk i xs \n | length xs <= i \n = C xs N \n | otherwise\n = C (take i xs) (chunk i (drop i xs))\n\nchunkRes i xs\n | length xs <= i = length v == 1 \n | i == 1 = length v == length xs\n | otherwise = length v < length xs\n\\end{code}\n\nWhere @take@ and @drop@ type as follows\n\\begin{code}\ndrop :: i:Nat -> xs:{List a | i <= length xs } \n -> {v:List a | length v == length xs - i }\ntake :: i:Nat -> xs:{List a | i <= length xs } \n -> {v:List a | length v == i }\n\\end{code}\n\n\n\\paragraph{Runtime Parallelism.}\nTo parallelize concatenation of the chunks we use parallel strategies \nfrom the Haskell library @Control.Parallel.Strategies@.\n%\nFor example, in our implementation @parStrategy@ is defined as\n\\begin{code}\nparStrategy :: Strategy (List a)\nparStrategy = parTraversable rdeepseq\n\\end{code}\n\nNote that parallelization of concatenation leads to runtime speedups\nonly when @(mappend)@ runs in constant time with respect to the monoid size. \n%\nOtherwise, at each recursive call of @pmconcat@,\nruntime will be dominated by the size of the biggest list element. \n\nEven though runtime of the parallel version depends on the implementation details \nof the monoids, correctness holds for every monoid. \n%\nIn the rest of this section we prove equivalence of the sequential and parallel concatenation operations. \n\n\n\\paragraph{Parallelism in the Logic.}\nTo prove equivalence we make the \\textit{assumption}\nthat the Haskell function @withStrategy s x@ \nreturns @x@ for every strategy.\n\\begin{code}\nassume withStrategy :: Strategy a -> x:a -> {v:a | v == x}\n\\end{code}\n%\nMoreover, to prove equivalence, we need to represent \nthe implementation of @pmconcat@ in logic.\n%\nWhich, in turn, requires representation of @withStrategy@. \n%\nLiquidHaskell represents @withStrategy@ in logic as a function that merely returns\nits second argument @withStrategy _ x = x@\nand does not reason about parallelism. \n\n\\subsection{Equivalence of Parallel and Sequential Concatenation}\n\n\\begin{theorem}[Monoid Concatenation Equivalence]\\label{theorem:equivalence:concat}\nIf @m@ is monoid, then the parallel and sequential concatenations are equivalent:\n\\begin{code}\n pmconcatEq :: i:Int -> is:List m -> {pmconcat i is == mconcat is }\n\\end{code}\n\\end{theorem}\n\n\\begin{proof}\nWe prove the theorem by providing a Haskell implementation of @pmconcatEq@\nthat is safe under LiquidHaskell. The details of the proof can be found in~\\cite{implementation}, \nhere we provide the sketch of the proof. \n\nFirst we prove that @mconcat@ distributes over list splitting\n\\begin{code}\nmconcatSplit \n :: i:Nat -> xs:{List m | i <= llen xs} \n -> { mconcat xs == mconcat (take i xs) \n <> mconcat (drop i xs) }\n\\end{code}\n%\nThe proofs proceeds by structural induction, using monoid left identity in the base case\nand associavity in the inductive step.\n\nWe generalize the above\nto prove that @mconcat@ distributes over list chunking\n\\begin{code}\nmconcatChunk \n :: i:Pos -> xs:List (Monoid a) \n -> { mconcat xs == mconcat (map mconcat (chunk i xs)) }\n\\end{code}\n%\nThe proofs proceeds by structural induction, using monoid left identity in the base case\nand lemma @mconcatSplit@ in the inductive step.\n\nLemma @mconcatChunk@ is sufficient to prove @pmconcatEq@ by structural induction, \nusing monoid left identity in the base case. \n\\cqed\\end{proof}\n\nNote that the proof uses only the monoid laws and not the implementations of the monoid\noperators, thus the theorem holds \\textit{for any} type that satisfies the monoid laws. \n%\nNext we define a monoid structure for string matching and prove it satisfies the monoid laws. \nThus we conclude that string matching concatenation can be parallelized. \n \n", "meta": {"hexsha": "6dee2e37c2e04ca7132c8118502dad1daa508e50", "size": 5167, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "text/stringmatcher/parallelization.tex", "max_stars_repo_name": "nikivazou/thesis", "max_stars_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 11, "max_stars_repo_stars_event_min_datetime": "2016-12-02T00:46:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-02-20T07:04:01.000Z", "max_issues_repo_path": "text/stringmatcher/parallelization.tex", "max_issues_repo_name": "nikivazou/thesis", "max_issues_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "text/stringmatcher/parallelization.tex", "max_forks_repo_name": "nikivazou/thesis", "max_forks_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2016-12-02T00:46:51.000Z", "max_forks_repo_forks_event_max_datetime": "2016-12-02T00:46:51.000Z", "avg_line_length": 33.9934210526, "max_line_length": 106, "alphanum_fraction": 0.7259531643, "num_tokens": 1385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506418255928, "lm_q2_score": 0.8221891261650247, "lm_q1q2_score": 0.6253986865194494}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath, amssymb}\n\\usepackage{amsthm}\n\\usepackage{enumitem}\n\\usepackage[legalpaper, margin=1in]{geometry}\n\\usepackage{amsfonts}\n\\usepackage[mathscr]{eucal}\n\\usepackage{hyperref}\n\\newtheorem{thm}{Theorem}\n%this makes it so that everything labelled thm is numbered by section\n\\newtheorem{claim}[thm]{Claim}\n\\newtheorem{cor}[thm]{Corollary}\n\\newtheorem{conclusion}{Conclusion}\n%\\newtheorem{remark}[thm]{Remark}\n\\newtheorem{lemma}[thm]{Lemma}\n\\newtheorem{prop}[thm]{Proposition}\n\\newtheorem{prob}[thm]{Problem}\n%\\newtheorem{exam}[thm]{Example}\n\\theoremstyle{definition}\n%this makes the following things non-italicized\n\\newtheorem{defn}[thm]{Definition}\n\\newtheorem{postulate}{Postulate}\n\\newtheorem{remark}[thm]{Remark}\n\\newtheorem{discrepency}[thm]{Discrepency}\n\\newtheorem{exam}[thm]{Example}\n\\newtheorem{innercustomthm}{Word Postulate}\n\\newenvironment{customthm}[1]\n{\\renewcommand\\theinnercustomthm{#1}\\innercustomthm}\n{\\endinnercustomthm}\n\n\\title{Time Series Analysis Notes}\n% \\author{Mozahid Haque}\n% \\date{October 2020}\n\n\\begin{document}\n\\maketitle\n\\section{Overview of Time Series Characteristics}\n\\begin{defn}\nA \\textbf{univariate time series} is a sequence of measurements of one variable at regular time steps.\n\\end{defn} \n\\noindent Such data need not be IID.\n\n\\subsection{Objectives of Analysis}\n\\begin{enumerate}\n \\item Describe important features of any time series pattern\n \\item Explain how past affects future or how two time series \"interact\"\n \\item Forecasting\n \\item Use time series as a control standard, e.g., measurements of the quality of some manufacturing product\n\\end{enumerate}\n\n\\subsection{Model Types and Considerations}\nThere are two basic types:\n\\begin{enumerate}\n \\item Relating time series values to past values and past prediction errors -- called Autoregressive Integrated Moving Average or ARIMA models.\n \\item Regular regression models with time indices as x-variables.\n\\end{enumerate}\n\n\\noindent Initial data exploration:\n\\begin{itemize}\n \\item Trend -- overall progression of measurements\n \\item Seasonality -- periodic behavior based around calendar intervals such as seasons, quarters, months, weeks, etc.\n \\item Outliers -- data away from original data or away from some manipulation of it\n \\item Long-run Cycle -- periodic behavior unrelated to seasonality type\n \\item Constant Variance -- variance changing or not\n \\item Abrupt Changes -- significant disturbances to series or variance or other things\n\\end{itemize}\n\n\\subsection{Autoregressive Models: Autocorrelation and Partial Autocorrelation}\n\\begin{defn}\nLet $\\{y_t\\}_{t=1}^n$ be a time series indexed by $t$. An \\textbf{autoregressive model} is one where time series values are regressed on previous values. For example, a first-order autoregression AR(1) would look like:\n\\[ y_t = \\beta_0 + \\beta_1 y_{t-1} + \\epsilon_t \\]\nand a second-order autoregression AR(2) would look like:\n\\[ y_t = \\beta_0 + \\beta_1 y_{t-1} + \\beta_2 y_{t-2} + \\epsilon_t. \\]\nIn general, a $k^\\text{th}$-order autoregression AR(k) is given by \n\\[ y_t = \\left( \\beta_0, \\beta_1, \\hdots, \\beta_n, 1 \\right) \\left( 1, y_{t-1}, \\hdots, \\beta_{t-n}, \\epsilon_t \\right)^T. \\]\nUsually the errors $\\epsilon_t \\overset{iid}{\\sim}N(0, \\sigma_{\\epsilon}^2)$ and independent of $y$.\n\\end{defn}\n\nThis begs the question:\nHow do we choose the order for a given time series?\nThere are two methods:\n\\begin{enumerate}\n \\item Autocorrelation function (ACF) and\n \\item Partial Autocorrelation function (PACF).\n\\end{enumerate}\n\\begin{defn}\nThe coefficient of correlation between two values in a time series given by:\n\\[ \\text{Corr}(y_t, y_{t+k}) = r_k = \\frac{c_k}{c_0} ~~~~\\text{where}~~~~\nc_k = \\frac{1}{n} \\sum_{t=1}^{n-k} (y_t - \\bar{y})(y_{t+k} - \\bar{y}) \\]\nis called the \\textbf{autocorrelation function (ACF)}.\nRecall what familiar term $c_0$ is. Notice the similarity to the covariance formula which is meant to exhibit the linear relationship between two variables. In this case, this is the covariance (or linear relationship) of lagged values. Here the \\textbf{lag} is given by \\textbf{k}.\n\\end{defn}\n\nThis is one method. This method includes in it the influence of other lags on $y_t$. The second method removes this influence of the other lags in between...\n\\begin{defn}\nHere the \\textbf{Partial Autocorrelation Function (PACF)} is given by\n\\[ f_k = \n\\begin{cases} \n r_1 = \\text{Corr}(y_t, y_0) & \\text{if}~ k=1; \\\\\n \\text{Corr}(y_t - y_t^{t-1}, y_0 - y_0^{k-1}) & \\text{if}~ k \\geq 2\n \\end{cases}.\n\\]\nEssentially, this has the effect of determining the linear correlation between $y_t$ and $y_{t+k}$ but removing the linear dependence of lags in between $t$ and $t+k$, namely $t_i$ such that $t < t_i < t+k$. One can also think of this as subtracting away the projection of $y_t$ on the linear subspace spanned by $y_{t+1}, \\hdots, y_{t+k-1}$.\n\\end{defn}\n\nWe can additionally complicate the whole thing by considering certain trend behavior within our original time series via decomposition.\n\n\\subsection{Complexifying ACF/PACF via Decomposition and/or Higher Order Trends}\nThe typical decomposition of a time series involves:\n\\begin{itemize}\n \\item overall trend, $m_t$\n \\item seasonality, $s_t$, and\n \\item error, $\\epsilon_t$\n\\end{itemize}\nso that\n\\[ y_t = m_t + s_t + \\epsilon_t. \\]\nTypical estimation involves first estimating the overall trend through linear filters. One example of this is a moving average given by some \"window\" size:\n\\[ \\hat{m_t} = \\sum_{k=-a}^a \\left( \\frac{1}{1+2a} \\right) y_{t+k}. \\]\nWe can experiment with the window size to get a feel for a good overall trend. Then once this is done, we can estimate the seasonality by looking at what remains:\n\\[ \\hat{s_t} = y_t - \\hat{m_t}. \\]\nNote that this $\\hat{s_t}$ depends on the window size. So we can take the average of the window sized seasonality estimates to fix a single $s_t$. Then this gives us a way to calculate the error:\n\\[ \\epsilon_t = x_t - \\hat{m_t} - \\hat{s_t}. \\]\nLuckily there are packages for such decomposition.\n\nAnother way to complexify our model for our time series is to add quadratic trends to the model by considering not just linear time factors $t$, but higher order factors and interactions such as $t^2$, $t^3$, etc.\n\\end{document}\n", "meta": {"hexsha": "896349253c15abae0ae9ceaabb54a1b80143fbae", "size": 6348, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main.tex", "max_stars_repo_name": "mh86/timeseries_learning", "max_stars_repo_head_hexsha": "ae12dcfcfd3edddf088e309911948ee373d21c30", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "main.tex", "max_issues_repo_name": "mh86/timeseries_learning", "max_issues_repo_head_hexsha": "ae12dcfcfd3edddf088e309911948ee373d21c30", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "main.tex", "max_forks_repo_name": "mh86/timeseries_learning", "max_forks_repo_head_hexsha": "ae12dcfcfd3edddf088e309911948ee373d21c30", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.59375, "max_line_length": 343, "alphanum_fraction": 0.7399180844, "num_tokens": 1857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.8128673223709252, "lm_q1q2_score": 0.6251601929960792}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{comment}\n \n\\usepackage[noend]{algpseudocode}\n\\usepackage{algorithm}\n\\usepackage{float}\n\\usepackage{graphicx}\n\\usepackage[margin=.75in]{geometry} \n\\usepackage{amsmath,amsthm,amssymb}\n\\usepackage{dsfont}\n\\usepackage{amsthm}\n\\usepackage{mathtools,amssymb}\n\\usepackage{wrapfig,caption,subcaption}\n\\allowdisplaybreaks\n\n\\newtheorem*{definition}{Definition}\n\\newtheorem*{question}{Question}\n\\newtheorem{theorem}{Theorem}\n\\newtheorem{proposition}[theorem]{Proposition}\n\\newtheorem{claim}[theorem]{Claim}\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{corollary}[theorem]{Corollary}\n\\newtheorem{conjecture}[theorem]{Conjecture}\n \n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\Rgz}{\\mathbb{R}_{\\ge 0}}\n\n\\newcommand{\\ip}[2]{\\left\\langle{#1},{#2}\\right\\rangle}\n\\newcommand{\\norm}[1]{\\left\\lVert{#1}\\right\\rVert}\n\\newcommand{\\sizeof}[1]{\\left\\lvert{#1}\\right\\rvert}\n\n\\newcommand{\\woloss}{without loss of generality }\n\n\\DeclareMathOperator*{\\argmin}{arg\\,min}\n\\DeclareMathOperator*{\\argmax}{arg\\,max}\n\\DeclareMathOperator*{\\cone}{cone}\n\\DeclareMathOperator*{\\hull}{hull}\n\\DeclareMathOperator*{\\indif}{indif}\n\n\\newcommand{\\1}[1]{\\mathds{1}[{#1}]}\n\\renewcommand{\\P}[1]{\\mathds{P}\\left[{#1}\\right]}\n\\newcommand{\\E}[1]{\\mathds{E}\\left[{#1}\\right]}\n\\newcommand{\\Var}[1]{\\mathrm{Var}[{#1}]}\n\n\\newcommand{\\unit}{\\mathds{1}}\n\\newcommand{\\lo}{\\succ}\n\n% \\renewcommand{\\thesubsection}{\\thesection.\\alph{subsection}}\n% \\renewcommand\\thesubsection{\\ \\ (\\alph{subsection})}\n\n\n\\begin{document}\n\n% \\renewcommand{\\qedsymbol}{\\filledbox}\n \n\\title{\n Dim Prefs\n}\n\\author{\n Clay Thomas \\\\\n claytont@princeton.edu\n}\n\\maketitle\n\n\\section{Lemmas}\n In this section, we set up the tools needed to reason about \n $d$-dimensional preferences.\n\n \\subsection{Lemmas based on the structure of $X$}\n These first few lemmas relate geometric properties of $X$ to\n limitations on the structure of $>_a$ for a specific, fixed $a$.\n\n The most simple way $X$ gives structure to preferences is if\n one outcome is better in all attributes.\n \\begin{definition}\n Let $x,y\\in X\\subseteq \\Rgz^d$.\n We say $x$ \\emph{dominates} $y$, denoted $x\\gg y$,\n if $x[k] > y[k]$ for each $k=1,\\ldots,d$.\n \\end{definition}\n \\begin{proposition}\n If $x \\gg y$, then $x >_a y$ for any nonzero $a\\in \\Rgz^d$.\n \\end{proposition}\n\n When comparing different outcomes, a useful tool is\n the familiar geometric notion of a convex hull.\n Intuitively, if a preference weight does not like any of a set of options,\n it will not like any outcome in the hull of those options either.\n Thus, a point ``dominated by the hull'' of a set of options\n cannot be preferred to all those options.\n \\begin{definition}\n For points $x_1,\\ldots,x_n \\in \\R^d$, let $\\hull(x_1,\\ldots,x_n)\n = \\{u_1x_1 + \\ldots + u_nx_n | 0\\le u_i\\le 1, \\sum_{i=1}^n u_i = 1\\}$\n denote the convex hull of $x_1,\\ldots,x_n$.\n \\end{definition}\n \\begin{lemma}\\label{lem:agreementHull}\n Let $z,x_1,\\ldots,x_k \\in \\Rgz^d$ and $a\\in \\Rgz^d \\setminus \\{0\\}$.\n If $z >_a x_i$ for $i=1,\\ldots,k$, then $z >_a w$\n for any $w\\in \\hull(x_1,\\ldots,x_k)$.\n \\end{lemma}\n \\begin{proof}\n We have $\\ip{a}{z} > \\ip{a}{x_i}$ for each $i=1,\\ldots,k$.\n If $w = u_1x_1+ \\ldots + u_nx_n$ and $\\sum_i u_i =1$, then\n $\\ip{a}{w} = u_1\\ip{a}{x_1}+\\ldots+u_n\\ip{a}{x_n}\n < u_1\\ip{a}{z} + \\ldots + u_n\\ip{a}{z} = \\ip{a}{z}$.\n \\end{proof}\n\n \\begin{proposition}\\label{prop:domHull}\n Let $U, V \\subseteq \\Rgz^d$ be finite sets of points.\n Suppose that there exists $u\\in \\hull U$ and $v\\in \\hull V$\n such that $u \\ll v$.\n Then no $a$ satisfies $u_i >_a v_i$ for each $i=1,\\ldots, k$.\n % Suppose that there exists $w\\in \\hull(x_1,\\ldots,x_k)$\n % such that $z \\gg w$.\n % Then no $a$ satisfies $x_i >_a z$ for each $i=1,\\ldots, k$.\n \\end{proposition}\n \\begin{proof}\n For contradiction, suppose such an $a$ exists.\n First, note $u_i >_a v \\in \\hull V$,\n then see that $\\hull U \\ni u >_a v$ as well.\n But $u\\ll v$, so this is a contradiction.\n \\end{proof}\n\n The converse of the last theorem also holds:\n \\begin{proposition}\n Suppose that no $a\\in \\R_{>0}^d$ satisfies $u_i \\ge_a v_j$ for each \n $u_i \\in U$ and $v_j\\in V$.\n Then there exist $u\\in\\hull(U)$ and $v\\in\\hull(V)$ with $u\\ll v$.\n NOTE: SHOULD USE THE ``SEMISTRICT`` DEFINITION OF << \n \\end{proposition}\n \\begin{proof}\n If no $a$ has this property, then in particular the following linear\n program is infeasible:\n \\begin{alignat*}{2}\n \\text{min: } &\n \\sum_{k=0}^d 0\\cdot a_k \\\\\n \\text{s.t. } &\n \\begin{aligned}[t]\n \\sum_{k=1}^d a_k (u^i_k - v^j_k)\n & \\ge 0\n \\quad \\forall i\\in U, j\\in V\n \\\\ a_k & \\ge 1 \\quad \\forall k\n \\end{aligned}\n \\end{alignat*}\n The dual of this linear program is\n \\begin{alignat*}{2}\n \\text{max: } &\n \\sum_{i\\in U, j\\in V} 0\\cdot b_{ij} + 1\\cdot c_k \\\\\n \\text{s.t. } &\n \\begin{aligned}[t]\n \\sum_{i\\in U, j\\in V} b_{ij}(u^i_k - v^j_k) + c_k\n & \\le 0\n \\quad \\forall k\n \\\\ b_{ij} & \\ge 0 \\quad \\forall i,j\n \\\\ c_k & \\ge 0 \\quad \\forall i,j\n \\end{aligned}\n \\end{alignat*}\n This dual program is always feasible (with the all zeros solution)\n so by strong duality it must be unbounded.\n Take some solution and some $k$ with $c_k > 1$.\n At this solution, $b_{ij}$ are not all zero.\n Thus, take\n $\\sum_{i,j} b_{ij}=B$, and consider:\n \\[ \\sum_{i\\in U} \\left(\\sum_{j\\in V}b_{ij}/B \\right) u^i_k\n \\le - \\frac {c_k} B + \\sum_{j\\in V} \\left(\\sum_{i\\in U}b_{ij}/B \\right) v^j_k\n < \\sum_{j\\in V} \\left(\\sum_{i\\in U}b_{ij} / B\\right) v^j_k\n \\]\n NOTE: THIS CANNOT BE CORRECT BECAUSE IT IMPLIES EG THAT ALL\n POINTS ARE MORE THAN 1 APART\n \\end{proof}\n\n\n This motivates the following definition:\n \\begin{definition}\n A set of points dominates, denoted $U \\ll V$,\n when DEFINITION\n \\end{definition}\n\\end{document}\n", "meta": {"hexsha": "4f7a1305b093e1c68f3af428609a268369e37693", "size": 6238, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/system.tex", "max_stars_repo_name": "ClathomasPrime/linear-prefs", "max_stars_repo_head_hexsha": "e700589a82667ca0f307459816e4f5b211fbd7d2", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/system.tex", "max_issues_repo_name": "ClathomasPrime/linear-prefs", "max_issues_repo_head_hexsha": "e700589a82667ca0f307459816e4f5b211fbd7d2", 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YES\n2. YES", "lm_q1_score": 0.8128673178375735, "lm_q2_score": 0.7690802370707281, "lm_q1q2_score": 0.625160189509568}} {"text": "%\n% 281\n%\n\\chapter{The Hypergeometric Function}\n\n\\Section{14}{1}{The hypergeometric series.}\n\nWe have already \\hardsubsectionref{2}{3}{8}) considered the hypergeometric series*\n\n1 + - ., ( + l) ( + l) a(a + l)(a+2)6(6+l)(6 + 2) 3 l.c 1.2.c(c + l)\n\"* 1.2.3.c(c + l)(c + 2) \"\n\nfrom the point of view of its convergence. It follows from \\hardsubsectionref{2}{3}{8} and \\hardsectionref{5}{3} that the series defines a function which is analytic when i 2 j\n< 1.\n\nIt will appear later \\hardsubsectionref{14}{5}{3}) that this function has a branch point\nat 2 = 1 and that if a cutf (i.e. an impassable barrier) is made from\n+ 1 to + along the real axis, the function is analytic and one-valued\nthroughout the cut plane. The function will be denoted by F a, b;c;z).\n\nMany important functions employed in Analysis can be expressed by\nmeans of hypergeometric functions. Thus|\n\n(1 +2)\" = F(-n, :;-z),\n\nlog l + z) = zF(l, l;2;-z),\n\ne'=\\ im F l,0;l;zl0).\n\nExample. Shew that d\n\nF a, b; c; z) = ~F a + \\, 6 + 1; c + 1; 2).\n\n1411. The valuel of F(a, b; c; I) when E(c-a-b)>0.\n\nThe reader will easily verify, by considering the coefficients of\n.2;'* in the\n\n* The name was given bj- Wallis in 1655 to the series whose Kth term\nis a + b a + 2b ... a + n-l)b . Euler used the term hypergeometric in\nthis sense, the modern use of the term being apparently due to Kummer,\nJournal fiir Math. xv. (1S36).\n\nt The plane of the variable z is said to be cut along a curve when it\nis convenient to consider only such variations in z which do not\ninvolve a passage across the curve in question; so that the cut may\nbe regarded as an impassable barrier.\n\nX It will be a good exercise for the reader to construct a rigorous\nproof of the third of these results.\n\n§ This analysis is due to Gauss. A method more easy to i-emember but\nmore difficult to justify is given in \\hardsectionref{14}{6} example 2.\n\n%\n% 282\n%\n\nvarious series, that if \\$ ir < 1, then c c-l- 2c-a-b-l)x]F (a, b : c\n; x) + (c - a) c - b) xF (a, b; c + I; x)\n\n= c(c-l) l-x)F(a,b;c-l;x)\n\n= C(C -l)jl+ (Un-Un-i)x'>'l,\n\nwhere Un is the coefficient of \" in i''(a, 6; c - 1; x).\n\nNow make x->l. By \\hardsubsectionref{3}{7}{1}, the right-hand side tends to zero if\n\n1+2 (un - tin-i) converges to zero, i.e. if ?i,j- >0, which is the\ncase when =i\n\nR c- a - 6) > 0.\n\nAlso, by \\hardsubsectionref{2}{3}{8} and \\hardsubsectionref{3}{7}{1}, the left-hand side tends to\n\nc(a+b-c)F a,b;c;l) + c-a)(c- b) F(a, b;c+l;l) under the same condition\n; and therefore\n\nRepeating this process, we see that\n\nF(a, b: c;l) = i Yl ) -- -, ( F(a, b:c + 7n; 1)\n\n ' '\\ n=o c + n) c-a-b + n)\\ ' ' '\n\n= ] lira n ) ., f [ lim F(a, b;c + m; 1),\n\nif these two limits exist.\n\nBut ( 12'13) the former limit is tt r-Fr, rz, if c is not a negative\n\n T c - a) 1 (c- 6) °\n\ninteger; and, if Un(ci, b, c) be the coefficient oi x in F (a, b; c;\nx), and\n\nm > I c I, we have\n\n00\n\n\\ F a, b] c + m; l) - l\\ \\ 2 \\ un(a, b, c + m)\\\n\nn = l\n\n00\n\n X Un (ja,, ib\\, m - \\ c )\n\nn=l\n\n' Q,b I \"\n\n<, 2 Mn (I a i -h 1, I 6 I + 1, 7?l -f- 1 - I C I).\n\nm-\\ c\\ n=o\n\nNow the last series converges, when ??i > ic| 4- a -f j l - 1, and is\na positive decreasing function of m; therefore, since [m - '\\ c\\ \\ }~\n->0, we have\n\nlim F(a, b: c + m; 1)= 1;\n\nand therefore, finally,\n$$\nTODO\n$$\n%\n% 283\n%\n\n\\Section{14}{2}{The differential equation satisfied by F (a, b; c: z).}\n\nThe reader will verify without difficulty, by the methods of § lO'S,\nthat the hypergeometric series is an integral valid near = of the\nhypergeometric\n\nequation*\n\n ~ d? + (c - (a + + 1) - - \\&' =;\n\nfrom \\hardsectionref{10}{3}, it is apparent that every point is an 'ordinary point' of\nthis equation, with the exception of 0, 1, x, and that these are '\nregular points.'\n\nExample. Shew that an integral of the equation\n\ni\n\nz< F a- a, 6 + a; a- + 1; z).\n\n\\Section{14}{3}{Solutions of Riemann's P-equation by hypergeometric functions.}\n\nIn \\hardsubsectionref{10}{7}{2} it was observed that Riemann's differential equation f\n\ndru j] -a -a' 1- -/3' l-y-y\\ du dz- \\ z- a z-b z - c ) dz\n\n\\ aa' (a-b)(a-c) jB' h - c) b - a ) \\ 77 (c-a)(c-6) z- a z - b z - c\n\nu\n\n z - a) z - b) z - c)\n\n= 0.\n\nby a suitable change of variables, could be reduced to a\nhypergeometric equation; and, carrying out the change, we see that a\nsolution of Riemann's equation is\n\nTODO\n\nprovided that a - a' is not a negative integer; for simplicity, we\nshall, throughout thi section, suppose that no one of the exponent\ndifferences a - a', /3 -, 7 - 7' is zero or an integer, as \\hardsubsectionref{10}{3}{2})\nin this exceptional case the general solution of the differential\nequation may involve logarithmic terms; the formulae in the\nexceptional case will be found in a memoir :J: by Lindelof, to which\nthe reader is referred.\n\nNow if a be interchanged with a', or 7 with 7', in this expression, it\nmust still satisfy Riemann's equation, since the latter is unaffected\nby this change.\n\n* This equation was given by Gauss.\n\nt The constants are subject to the condition a + a' + + ' + y + y' =\nl.\n\nJ Acta Soc. Sclent. Fennicae, xix. (1893). See also Klein's\nlithographed Lectures, Veher die hypergeometrische Funktion (Leipzig,\n1894).\n\n%\n% 284\n%\n\nWe thus obtain altogether four expressions, namely,\n\n'z-aX /z-c\\ y f .,, - .,,,\\ c-h) z-a)\\\n\nM, =\n\n(z-ay /z-c\\ y p, ry, -, > c-b) z-a)]\n\n ==(.- i) U- ft) J | + /3 + 7', o + +t; l+°-'''; (o-a)(.-6) j-\n\nwhich are all solutions of the differential equation.\n\nMoreover, the differential equation is unaltered if the triads (a, a',\na), (yS, y8', h), (y, y', c) are interchanged in any manner. If\ntherefore we make such changes in the above solutions, they will still\nbe solutions of the differential equation.\n\nThere are five such changes possible, for we may write\n\n b, c, a\\, c, a, b], a, c, b], [c, b, a], [b, a, c]\n\nin turn in place of [a, b, c], with corresponding changes of a, a, /S,\n(3', 7, 7'.\n\nWe thus obtain 4 x = 20 new expressions, which with the original four\nmake altogether twenty-four particular solutions of Riemann's\nequation, in terms of hypergeometric series.\n\nThe twenty new solutions may be written down as follows :\n\nu. =\n\nz-by fz-aY (,,, a-c)(z-b)\\ iP'j/8 + 7 + a, /3 + 7'+a; l + yS-/3'; )\n, (\n\nz-b\\ \\ ' [z-ay t q,, o', ', . 1, /D' o. (a-c)( -\\&)]\n\nUa =\n\n s=('- ) ('- Y Fy+y + oc, '+y'+a-l + /3'-/3-,. .\n\n\\ z - cj \\ z - cj [ \" a-b) z- c)\n\n(z-b fz - a-y \\ (, \\,,, (a - c)(z - b)\n\n\\ z - cl \\ z - c \\ (a -b) z - c)\n\ni /3+7+a, /3+7+a; l + /3'-/3; ~ '-\\\n\n\\ Z-C' \\ Z-C' y I ' r- I ) r- r- ' ( \\ J - c)\\\n\n(z-c\\ y (z-b\\ \\, o -., b-a) z- c)]\n\n\" \"=U- ) U- J |7 +,7+a-+;l + y-7; \\; \\ |\n\nU,,= l - )'( - r F\\ y a + ',y c '; I + 7-7'; j H \\ z-aj \\ z-al [' ' \"\n' ' ' (l,-c)(z-a)\\\n\n = (lllfy7l Yf f/'+o + ZS'. 7'+a+/3'; 1 + 7-7; P - \\ z-aj \\ z-a) ' \"\n' ' ' > (i,-c) z- a)\n\n%\n% 285\n%\n\nZ ~ Cj\n\na,.=\n\nMlQ -\n\nFL + y + ',a + y' + '; I + a - a; l' '\\ \\ ' ' ''\\ \\, [ b-a) z-c)\n\n/ -cy'/2-a\\ j ', /o', 1, ' a-b)(z-c))\n\n.T b) U- 6J |t+/3 +,7+ +< : 1+7-7; \\;( \\ |,\n\n\\ z - a J \\ z-a) [ ' (c-b)(z- a) '\n\n- / i c - o)(z - a))\n\n1 V - / \\ z-aj ( ' \" I r ' (c-6)( -a)J '\n\nBy writing 0, 1 - C,, B, 0, 6'- -5, a; for a, a', yg, S', y, y,\n\n/ - AU - respectively, we obtain 24 solutions of the hypergeoraetric\n\nequation satisfied by F A, B; G; x).\n\nThe existence of these 24 solutions was first shewn by Kuinmer*.\n\n\\Section{14}{4}{Relations betiueen particular solutions of the hyper geometric equation.}\n\nIt has just been shewn that 24 expressions involving hypergeometric\nseries are solutions of the hypergeometric equation; and, from the\ngeneral theory of linear differential equations of the second order,\nit follows that, if any three have a common domain of existence, there\nmust be a linear relation with constant coefficients connecting those\nthree solutions.\n\nIf Ave simplify n-, u.,, ih, ih] u, u g-, u i, W22 in the manner\nindicated at\n\n* Journal fUr Math. xv. (1836), pp. 39-83, 127-172. They are obtained\nin a different manner in Forsyth's Treatise on Differential Equations,\nChap. vi.\n\n' 22 =,\n\n\\ z - aj \\ z\n\n%\n% 286\n%\n\nthe end of \\hardsectionref{14}{3}, we obtain the following solutions of the\nhypergeometric equation with elements A, B, G, x:\n\ny, =F A,B;C-x),\n\ny = (- xy- 'FiA -C+\\,B-G+1;2-C;x),\n\ny,=(l-xr-''- F C-B,C-A;G;xl\n\ny = (\\ ccY-'' l - xf- - 'Fil -B,l-A;2-G;x),\n\ny,r = F A,B; A + B - G + 1; 1 - x),\n\ny = xf- -''F G-B, G-A; G-A-B-hl; l-x),\n\ny = (\\ xY 'FiA, A-G+\\;A-B + \\; x-%\n\ny = (- x)- F B,B-G + l]B-A + l; x-'). If 1 arg(l - a;) I < TT, it is\neasy to see from \\hardsubsectionref{2}{5}{3} that, when j .r | < 1, the relations\nconnecting y, y.2, y y must be y y-i, y2= rj, by considering the\nform of the expansions near c; = of the series involved.\n\nIn this manner we can group the functions Wj, ... z/24 into six sets\nof four*, viz. u-i, v-i, Mi3, Wis; Un, Hi, u, iiis; ii, u, M21, u\n', Wg, u, U22, M24 j '\"9? Wji, t ' 19 j Wjo, U12, u-18, '' ho,\nsuch that members of the same set are constant multiples of one\nanother throughout a suitably chosen domain.\n\nIn particular, we observe that Ui, u, u s, u are constant multiples\nof a function which (by §§ 5*4, 2\"o3) can be expanded in the form\n\n z-aY\\ l+ i en(2-a)4\n\nwhen \\ z - a\\ is sufficiently small; when arg z - a) is so restricted\nthat (z - a)* is one-valued, this solution of Riemann's equation is\nusually written P'\"*. And P'\"'; P' ', P' '; P* ', P* * are defined in\na similar manner when\n\nz - a\n\n\\ z - b\\, \\ z - c\\ respectively are sufficiently small.\n\nTo obtain the relations which connect three members of separate sets\nof solutions is much more difficult. The relations have been obtained\nby elaborate transformations of the double circuit integrals which\nwill be obtained later in \\hardsubsectionref{14}{6}{1}; but a more simple and singularly\nelegant method has recently been discovered by Barnes; of his\ninvestigation we shall give a brief account.\n\n\\Section{14}{5}{Barnes contour integrals for the hypergeometric function.}\n\nConsider r-. ~ -- p/, . - - '(-zyds,\n\n273-1 j\\ xj r(c + s)\n\nwhere |arg(- -z)] <7r, and the path of integration is curved (if\nnecessary) to ensure that the poles of r(a + 5)r(6 + 6'), viz. s = - a\n- n, -h-n n = 0, 1, 2, . . .),\n\n* The special formula\n\nFiA,l;C;.) F( C-A,l;C; ),\n\nwhich is derivable from the relation counectiug Mj with i3, was\ndiscovered in 1730 by Stirling, Methodus Differential is, prop. vii.\n\nt Piuc. London Math. Soc. (2), vi. (1908), pp. 141-177. References to\nprevious work on similar topics by Pincherle, Mellin and Barnes are\nthere given.\n\n%\n% 287\n%\n\nlie on the left of the path and the poles of r(- 5), viz. s = 0, 1, 2,\n..., lie on the right of the path *.\n\nFrom \\hardsectionref{13}{6} it follows that the integrand is\n\nOils j +''-c-i exp - arg - z) . I (s) - 7r\\ I s)\\ \\ ]\n\nas s- ►CO on the contour, and hence it is easily seen (| 5\"32) that\nthe integrand is an analytic function of z throughout the domain\ndefined by the inequality I arg 2\\ TT - S, where B is any positive\nnumber.\n\nNow, taking note of the relation F (- 5) F (1 + s) = - tt cosec s-rr,\nconsider\n\nT(a+s)r(b+s) iri-zy\n\nIf IttiJ c\n\nds,\n\n27rij(7 F(c + s) F(l +s) sinsTT\n\nwhere C is the semicircle of radius iV + - on the right of the\nimaginary axis\n\nwith centre at the origin, and N is an integer.\n\nNow, by \\hardsectionref{13}{6}, we have\n\nr(a+ )F(6 + ) -n-i-zy . a+,\\,\\,) (- y T(c + s)r l+s) sinsTT ' sin\nstt\n\nas N-> 00, the constant implied in the symbol being independent of\narg s when s is on the semicircle; and, if 6 = f j\\ \\ + j e''* and j\n| < 1, we have\n\n(- zy cosec sir = exp j ( iV + .j j cos 6 log | | - ( lY + j sin arg\n(- z)\n\n-(iV + )7r|sm ||]\n\nexp I i\\ \\ + . ) cos 6' log I I - (iV + ] 8 1 sin ( 1 11\n\nexp|2- iV +. )logj\n\nexp|-2- 8(i\\ r + i)|\n\n=\n\n I 6* k 7 TT,\n\n1 I zii 1\n\nHence if log | | is negative (i.e. \\ z\\ < 1), the integrand tends to\nzero sufficiently rapidly (for all values of 6 under consideration) to\nensure that\n\n/.\n\n'0 as iV- oo\n\nNow\n\nJ -ooi [J ~xi J C J (iV+i) i)\n\nby Cauchy's theorem, is equal to minus 2'Tri times the sum of the\nresidues of the integrand at the points s = 0, 1,2, ... N. Make N->cc\n, and the last\n\n* It is assumed that a and b are such that the contour can be drawn,\ni.e. that a and b are not negative integers (in which case the\nhypergeometric series is merely a polynomial).\n\n%\n% 288\n%\n\nthree integrals tend to zero when | arg (- ) | tt - S, and \\ z \\ < 1,\nand so, in these circumstances,\n\nthe general term in this summation being the residue of the integrand\nat\n\nS = 71.\n\nThus, an analytic function namely the integral under consideration)\nexists throughout the domain defined by the inequality j arg | < tt,\nand, when \\ z\\ < 1, this analytic function may he represented by the\nsei'ies\n\nV\n\nr(a + n)r(b + n) r c + n) .n !\n\nThe symbol F(a, h; c; z) will, in future, be used to denote this\nfunction divided by r( .)r(6)/r(c).\n\n\\Subsection{14}{5}{1}{The continuation of the hyper geometric series.}\n\nTo obtain a representation of the function F (a, b; c; z) in the form\nof series convergent when j 2 | > 1, we shall employ the integral\nobtained in \\hardsectionref{14}{5}. If -D be the semicircle of radius p on the left of\nthe imaginary axis with centre at the origin, it may be shewn* by the\nmethods of § 14'o that\n\nV a + s)V h + s)T -s)\n\nJd r c + s)\n\nas p- x, provided that j arg - z) \\ < tt, \\ z \\ > 1 and p-> 1,\n\n1 r r a + s)r(b + s)T -s)\n\n\\ V r(a+7i)r(l -c + a + n) sin ( c- a - n) tt /\\ x- -h ~ rZo r (1 + w)\nr (1 - 6 + ft + n) cos n-rr sin (b - ft - n) it\n\n T(b + n)r l-c + b + n) sin (c - 6 - ?i) tt f \\\n\nn=o r (1 + n) r (1 - ft + 6 + n) cos nir sin (ft - 6 - w) tt\n\nthe expressions in these summations being the residues of the\nintegrand at the points s = - a- n, s = - b - n respectively.\n\nIt then follows at once on simplifying these series that the analytic\n\n* In considering the asymptotic expansion of the integrand when | s [\nis large on the contour or on D, it is simplest to transform T (a +\ns), V (b + s), T c + s) by the relation of \\hardsubsectionref{12}{1}{4}.\n\n%\n% 289\n%\n\ncontinuation of the series, by which the hypergeometric function was\noriginally defined, is given by the equation\n\nr(c) \"\"' '- ' r(a-c) -(- r' - ( >i- +;i- + )\n\n1 0 - c) where arg (- z)\\ < tt.\n\nIt is readily seen that each of the three terms in this equation is a\nsolution of the hypergeometric equation (see \\hardsectionref{14}{4}).\n\nThis result has to be modified when a - 6 is an integer or zero, as\nsome of the poles of r a + s)r h + s) are double poles, and the\nright-hand side then may involve logarithmic terms, in accordance with\n1 43.\n\nCorollary. Putting h = c, we see that, if | arg - z)\\ < n,\n\nr (a) z)-\" = -. f \"' r a+s) r ( -s) ( - zyds, 'ZniJ \\ xt\n\nwhere (1 - )~<'- -l as 2-*.0, and so the value of [ arg(l - z) \\ which\nis less than tt always has to be taken in this equation, in virtue of\nthe cut (sec i:; 14'1) from to -f-oo caused by the inequality | arg( -\nz) \\ < n.\n\n14*52. Barnes' lemma that, if the 'path of integration is curved so\nthat the poles of T y - s)T b - s) lie on the right uf the path and\nthe poles of T (a + s) r (/3 + s) lie on the left*, then\n\nWrite / for the expression on the left.\n\nIf C be defined to be the semicircle of radius p on the right of the\nimaginary axis with centre at the origin, and if p-s-oo in such a way\nthat the lower bound of the distance of C from the poles of r (y - s)\nr (5 - s) is positive (not zero), it is readily seen that\n\nT a + s)T \\& + s)T y-s)T b-s) = - ''' '\n\n= 0[..' - +>+ - exp -2 1/( )l J, as ] s |- -oo on the imaginary axis\nor on C.\n\nHence the original integral converges; and / - -Oasp-a-oc, when (a +\n/3-|-y-|-S- 1)<0.\n\nThus, as in \\hardsectionref{14}{5}, the integral involved in 7 is - 2ni times the sum\nof the residues of the integrand at the poles oiT y - s)T b - s);\nevaluating these residues we gett\n\n/= i r( a + y+%)rO + y + ? ) TT I V a + b + n)T + b + n)\n\na=oV n + ) T + y-8 + 7l) sin(S-y)7r,,=o T (?i+l) r (1 -f-S-y + w)\nsin(y-S)7r*\n\n* It is supposed that a, /3, y, 5 are such that no pole of the first\nset coincides with any pole of the second set.\n\nt These two series converge \\hardsubsectionref{2}{3}{8}).\n\nW. M. A. 19\n\n%\n% 290\n%\n\nAnd so, using the result of 12\"14 freely, by \\hardsubsectionref{14}{1}{1} :\n\n\\ 7rr(l-a-/3-y -5) r T (a + d) T (3 + 5) r(a + y ) rO + y) )\n\nsin(y-8)7r |r(l-a-y)r(l-/3-y) T (1 - a -8)~r (1 - -8)J\n\nr(a+y)r( + y)r(a + g)r( + a) f r, ro,\n\n- r(a+/J + y + 8)sm(a + + y + 5).sin(y-8). \" ( + ) ' \" ('\n\n-sin (a + S) 77 sin 04-S)7r|.\n\nBut 2sin(fi+y)7rsiu(/3 + y)7r-2sin(a + S)7rsin (3 + S)7r\n\n= cos (a - /3) TT - cos (a + /3 + 2y) tt - cos (a - jS) tt + cos (a +\n+ 28) tt\n\n= 2sin(y-8)7r sin(a+/3 + y + S)7r.\n\nTherefore Vja + y)n ym-\\ + ) V )\n\n (a + + y + S)\n\nwhich is the required result; it has, however, only been prt)ved when\n\n (o + /3 + y + 8-l)<0;\n\nhut, )y the theory of analytic continuation, it is true throughout the\ndomain through which both sides of the equation are analytic functions\nof, say, a: and hence it is true for all values of a, /ii, y, 8 for\nwhich none of the poles of r (a + s) r (/3 + s), qua function of s,\ncoincide with any of the poles of r (y - s) r (8 - s).\n\nCorollary. Writing s + k., a - k, \\$-k, y + k, 8 + k in place of s, a,\n/3, y, 8, we see that the result is still true when the limits of\nintegration are -;, + cx) i, where k is any real constant.\n\n14 \"53. The connexion between hypergeometrie functions of z and of \\\n-z. We hav§ seen that, if | arg ( - s) | < tt,\n\nr(c) ' ' ' 2ni j \\ x,- T c + s) \" '\n\n= . \\ \\ r-.\\ T a t)Y h-irt)T s-t)V c-a-h-t)dt\\\n\n'' TTi J \\ x i l TTl J ~k-x i )\n\n.T c-a)T c-b) ' by Barnes' leinma.\n\nIf k be so chosen that the lower bound of the distance between the s\ncontour and the t contour is positive (not zero), it may be shewn that\nthe order of the integrations* may be interchanged.\n\nCarrying out the interchange, we see that if arg (1 -z) be given its\nprincipal value,\n\nT(C'-a)T c-h)T a)r b)F a, b; c; 2)/r (c)\n\n= s- / r a + t)r b+t)r c-a-b~t) -. r s-t)ri-s)(~zYds dt\n\n T lJ-k-Jii 'ZtvI J -jci\n\n1 f-k+oci\n\n= r-. I r a + t)r b+t)r c~a-b-t)r -t) i-zy dt.\n\n2ni J -A-K/\n\n* Methods similar to those of \\hardsubsectionref{4}{5}{1} may be used, or it may be proved\nwithout much difficulty that conditions established by Bromwich,\nInfinite Series, % 177, are satisfied.\n\n%\n% 291\n%\n\nNow, when I arg (1 -k) \\ < Stt and | I -z\\ < 1, this last integral may\nbe evaluated by the methods of Barnes' lemma \\hardsubsectionref{14}{5}{2}); and so we\ndeduce that\n\nr c-a)r c-b)T a)rib)F a,b;c;z)\n\n= r c)r a)r b)T c-a-b) F a, b; a + b-c + l; l-z)\n\n+ r c)r c - a) r (c-b) r a + b - c) I -zy-'\"-\" F c- a, c-b; c-a-b + l;\nz),\n\na result which shews the nature of the singularity of F a, b; c; z)\nat z=.\n\nThis result has to be modified if c - a - 6 is an integer or zero, as\nthen\n\nY a-- t) V b->r t) T c - a -b - t)T -t)\n\nhas double poles, and logarithmic terms may appear. With this\nexception, the result is valid when | arg -z)\\ < tt, | arg (1 - 2); <\ntt.\n\nTaking | s | < 1, we may make 2 tend to a real value, and we see that\nthe result still holds for real values of 2 such that < 2 < 1.\n\n\\Section{14}{6}{Solution of Riemann's equation by a contour integral.}\n\nWe next proceed to establish a result relating to the expression of\nthe liypergeometric function by means of contour integrals.\n\nLet the dependent variable u in Riemann's equation \\hardsectionref{10}{7}) be\nreplaced by a new dependent variable defined by the relation\n\nu = (z - aY z - bf (z - c)y I.\n\nThe differential equation satisfied by / is easily found to be\n\nd f 1 + a a' l+ -§' 1-f - 7) dl dz- \\ z - a z - h z - c ] dz\n\n( g + /3 + 7) (g + + ry + 1) + Sa (g + /?' + 7' - 1)1 z - a) z - 0) z\n- c)\n\nwhich can be written in the form\n\nQ(z) -[ X-2)Q' z)+R z)]f\n\n+ li ( - 2) (X - 1 ) (/' (z) - (X-DR (z)] 1 = 0,\n\nwhere / \\=l-ot- -j = a' + ' + y',\n\niQ(z) z-a)(z-h) z-c),\n\n(R(z) = X a' +/3 + ry) z-b) z-c).\n\nIt must be observed that the function / is not analytic at x, and\nconsequently the above differential equation in / is not a case of the\ngeneralised hypergeometric equation.\n\nWe shall noiu sheiu that this differential equation can be satisfied\nby an integral of the form\n\n1=1 t- ay'+ +y- (t - 6)-+ '+y-i (t - c)-+ -y -1 z - )- - -y dt, J c\n\njirovlded that C, the contour of integration, is suitably cJiosen.\n\n19-2\n\n%\n% 292\n%\n\nFor, if we substitute this value of / in the differential equation,\nthe con- dition* that the equation should be satisfied becomes\n\nj (t- ay+ +y-' t - by+ '+y-' t - cY+p+y'-' z - t)-''- -y--Kdt = 0,\n\nJ c where\n\nir = (\\ - 2) |q (z) + t- z) Q' (z) + l(t- zj Q\" z)\n\n+ (t-z) R(z) + (t-z)R'(z)]\n\n= ( 2) Q (t) -it- zy] -it-z) [R (t) -a- zy s ( ' + /3 + 7)1\n\n= - (1 + a + /3 + 7) ( - a) ( - 6) (i - c)\n\n+ S (a + /9 + 7) ( - ) ( - c) t - z).\n\nf dV It follows that the condition to be satisfied reduces to -7- dt =\n0, where\n\nJ c dt\n\nv=(t- ay+ +y (t - by+ '+y (t - cy+ +y (t - )-(i+-+p+v .\n\nThe integral / is therefore a solution of the differential equation,\nwhen C is such that V resumes its initial value after t has described\nC.\n\nNow\n\nV= t- ay'+ +y-' (t - by+ '+y-' (t - cy+ +y-' [z - t)-''- -y U,\n\nwhere U = (t- a) t -b) t- c) (z - ty\\\n\nNow Z7 is a one-valued function of t; hence, if C be a closed\ncontour, it must be such that the integrand in the integral / resumes\nits original value after t has described the contour.\n\nHence finally any integral of the type\n\n(z-ay(z-by(z - c)y \\ (t-ay+y+''-\\ t-b)y+''+ '-' t-cy+ y'-' z-ty''~ -y\ndt,\n\nJC\n\nwhere C is either a closed contour in the t-plane such that the\nintegrand resumes its initial value after t has described it, or else\nis a simple curve such that V lias the same value at its termini, is a\nsolution of the differential equation of the general hypergeometric\nfunction.\n\nThe reader is referred to the memoirs of Pochhammer, Math. An)i. xxxv.\n(1890), pp. 495-526, and Hob.son, Phil. Trans. 187 a (1896), pp.\n443-531, for an account of the methods by which integrals of this type\nare transformed so as to give rise to the relations of \\hardsubsectionref{14}{5}{1} and\n14-53.\n\nExample 1. To deduce a real definite integral which, in certain\ncircumstances, represents the hypergeometric series.\n\n The iliEferentiations imder the sign of integration are legitimate\n\\hardsectionref{4}{2}) if the path C does not depend on z a3d does not pass through\nthe points a, b, c, z; if C be an infinite contour or if C passes\nthrough the points a, b, c or z, further conditions are necessary.\n\n%\n% 293\n%\n\nThe hypergeometric series F a, b; c; z) is, as already shewn, a\nsolution of the differential equation defined by the scheme\n\nr 00 1 \\\n\nP a zy\n\n\\ \\ -c b c- a- b ] If in the integral\n\nwhich is a constant multiple of that just obtained, we make 6 -x\n(without paying attention to the validity of this process), we are led\nto consider\n\n/ t'''' t-\\ y- - t-zydt.\n\nNow the limiting form of V in question is\n\nand this tends to zero at t= 1 and = cc, provided R (c) > R(b)>0.\n\nWe accordingly consider / t\"~'= t - Vf-''-' t-z)~\" dt, where z is not*\npositive and greater tlum 1.\n\nIn this integral, write t = u~; the integral becomes\n\n/ ? ''-' (1-jt)' -''-' (l-M2)- 0??<. .'\n\nWe are therefore led to expect that this integral may be a solution of\nthe differential equation for the hypergeometric series.\n\nThe reader will easily see that if R (c) > R b)> 0, and if arg u = arg\n(1 - m) =0, while the branch of l-uz is specified by the fact that\n(1-m2)- 1 as m- 0, the integral just found is\n\nT b)r c-b)\n\n (, \\&; c;z .\n\nThis can be proved by expandingf (1 - s)-a in ascending powers of z\nwhen | 3 | <1 and using § 12 \"41.\n\nExample 2. Deduce the result of \\hardsubsectionref{14}{1}{1} from the preceding example.\n\n\\Subsection{14}{6}{1}{Determination of an integral which represents TODO}\n\nWe shall now shew how an integral which represents the particular\nsolution i ' \\hardsectionref{14}{3}) of the hypergeometric difterential equation can\nbe found.\n\nWe have seen \\hardsectionref{14}{6}) that the integral\n\nl= z- df [z~bf z - cy ( t-a)P+y+- '-\\ t-b)y+<'+P'- (t-c)'' +\n+y'-Ht-z)-<'~ -ydt\n\nsatisfies the difl:erential equation of the hypergeometric function,\nprovided C is a closed contour such that the integrand resumes its\ninitial value after t has described C. Now the singularities of this\nintegrand in the plane are the points a, b, c, z; and after describing\nthe double circuit contour \\hardsubsectionref{12}{4}{3}) symbolised by (6 +, c-h, 6 -, c\n- ) the integi-and returns to its original value.\n\n* This ensures that the point t - ljz is not on the path of\nintegration, t The justification of this process by \\hardsectionref{4}{7} is left to\nthe reader.\n\n%\n% 294\n%\n\nNow, if z lie iu a circle whose centre is, the circle not containing\neither of the points h and c, we can choose the path of integration so\nthat t is outside this circle, and so \\ z - a\\ < i\\ t - a\\ for all\npoints t on the path.\n\nNow choose arg(2 - a) to be numericall ' less than it and arg(i - 6),\nai*g(2-c) so that they reduce to* arg(a - 6), arg(a-c) when z- a\\ fix\narg( - a), arg (< - 6), arg (< - c) at the point iV at which the path\nof integration starts and ends; also choose arg t - z) to reduce to\narg t - a) when z- a.\n\nThen (z-bf = (a-bf |l +/3 (j ' ) + ...j-,\n\n(.-c)V=( -o)' l+ (i ?) + ...l, and since we can expand (t-z)~' ~°~'\ninto an absolutely and uuifornily convergent series\n\nwe may expand the integral into a series which converges absolutely.\n\nMultiplying up the absolutely convergent series, we get a series of\ninteger powers of z - a multiplied by (s - a) . Consequently we must\nhave\n\nWe can define P '''\\ P \\ P '\\ F y\\ P ' by double circuit integrals in\na similar manner.\n\n\\Section{14}{7}{Relations between contiguous hypergeometric functions.}\n\nLet P z) be a solution of Riemann's equation with argument z,\nsingularities a, h, c, and exponents a, a', /3, j3', y, 7'. Further\nlet P(z) be a constant multiple of one of the six functions P *,\nTODO.\nLet TODO denote the function\nwhich is obtained by replacing two of the exponents, I and m, in P 2)\nhy I + 1 and m - 1 respectively. Such functions P;+i, i\\ i (z) are\nsaid to be contiguous to P (z). There are 6 x 5 = -30 contiguous\nfunctions, since I and m may be any two of the six exponents.\n\nIt was first shewn by Riemannf that the function P(z) and any tivo of\nits contiguous functions are connected by a linear relation, the\ncoeffi,cients in which are polynomials in z.\n\nThere will clearly be 5 x 30 x 29 = 435 of these relations. To shew\nhow to obtain them, we shall take P z in the form\n\nF z) = z- ay (z - hY (z - c)y f t- ay+y+' '- (t - 6)v+' + -\n\nJ c\n\n(t - cY+ +y'-' (z - ty- -y dt,\n\nwhere C is a double circuit contour of the type considered in \\hardsubsectionref{14}{6}{1}.\n\n* The values of arg (a-b), arg (a - c) being fixed.\n\nt Abh. der k. Ges. der Wiss. zu Gottingen, 1857; Gauss had previously\nobtained 15 rehitions between contiguous hypergeometric functions.\n\n%\n% 295\n%\n\nFirst, since the integral round G of the differential of any function\nwhich resumes its initial value after i has described C is zero, we\nhave\n\n= [ I- (t - ay+ +y (t - by ' y-Ht - cY+ +y'-' t - zy- - -y] dt. J c dt\n\nOn performing the differentiation by differentiating each factor in\nturn,\n\nwe get\n\n(a' + + y)P + (a + ' + y-l) Pa-+,, '-, + (a + (3 + y'-l) Pa'+i,y-i\n\n\\ (a + /3 + 7) p\n\nConsiderations of symmetry shew that the right-hand side of this\nequation can be replaced by\n\nThese, together with the analogous formulae obtained by cyclical\ninter- change* of (a, a, a) with (6, /3, ') and (c, 7, 7'), are six\nlinear relations connecting the hypergeometric function P with the\ntwelve contiguous functions\n\nPa+], '-l, -I p+i y'-i, -/y+i,a\\ l, -ta-],v'-l5 -Lfi+i a'-l, y~\\, '~\\\ni -I a.'+l, '- 1) -I a'4-],Y'\\ i, i '4.),-y'\\ i, -t j3'+l, a'-l > ''\ny'+I, a'- 1 ) -*7'+l, '- 1-\n\nNext, writing t - a = (t- h) + (6 - ), and usingf Pa'\\ i to denote the\nresult of writing a' - 1 for a' in P, we have\n\nP = P..\\ :,,.+:- (6- )P.-:.\n\nS-imilarly P = P \\ i,y'+i + (c - a) Pa'-i-\n\nEliminating Pa-i from these equations, we have\n\n c-h)P + (a - c) Pa-\\ :, vi + (b- a) Pa'-.,y-,, = 0. This and the\nanalogous formulae are three more linear relations con- necting P with\nthe last six of the twelve contiguous functions written above.\n\nNext, writing ( - ) = ( - a) - ( - a), we readily find the relation\n\nP = P +:, v'-i - ( - )\" ' ( - f y ( - 0\n\nJc which gives the equations\n\n( \\ a)-i P - ( - 6)- P +,, Y--: = ( - h)- [P- z- c)- Py+, a'-i\n\n= Kz - c)-' [P- z- a)-' Pa+i, \\ i .\n\n* The interchange is to be made only in the integrands; the contour C\nis to remain unaltered.\n\nt Pa'-l is not a function of Riemann's type since the sum of its\nexponents at a, h, c is not unity.\n\n%\n% 296\n%\n\nThese are two more linear equations between P and the above twelve\ncontiguous functions. *\n\nWe have therefore now altogether found eleven linear relations between\nP and these twelve functions, the coefficients in these relations\nbeing rational functions of 2. Hence each of these functions can be\nexpressed linearly in terms of P and some selected one of them; that\nis, between P and any two of the above functions there exists a linear\nrelation. The coefficients in this relation will be rational functions\nof z, and therefore will become polynomials in z when the relation is\nmultiplied throughout by the least common multiple of their\ndenominators.\n\nThe theorem is therefore proved, so far as the above twelve contiguous\nfunctions are concerned. It can, without difficulty, be extended so as\nto be established for the rest of the thirty contiguous functions.\n\nCorollary. If functions be derived from P by replacing the exponents\na, a, /3, /3', y, y\n\nby a+p, a' + g, /3 + r, jS'+s, y + t, y' + x, where p, q, r, s, f, u\nare integers satisfying the\n\nrelation\n\np- -q + r + s + t + u=0,\n\nthen between P and any two such functions there exists a linear\nrelation, the coefficients in which are polynomials in z.\n\nThis result can be obtained by connecting P with the two functions by\na chain of intermediate contiguous functions, writing down the linear\nrelations which connect them with P and the two functions, and from\nthese relations eliminating the intermediate contiguous functions.\n\nMany theorems which will be established subsequently, e.g. the\nrecurrence-formulae for the Legendre functions \\hardsubsectionref{15}{2}{1}), are really\ncases of the theorem of this article.\n\nREFERENCES. C. F. Gauss, Ges. Werke, in. pp. 12.3-163, 207-229. E. E.\nKuMMER, Journal fur Math. xv. (1836), pp. 39-83, 127-172. G. F. B.\nRiemann, Ges. Math. Werke, pp. 67-84. E. Papperitz, Math. Ann. xxv.\n(1885), pp. 212-221. S. PiNCHERLE, Rend. Accad. Lincei (4), iv.\n(1888), pp. 694-700, 792-799. E. Y. Barxes, Proc. London Math. Soc.\n(2), vi. (1908), pp. 141-177. Hj. Mellin, Acta Soc. Fennicae, xx.\n(1895), No. 12.\n\nMiscellaneous Examples.\n\n1. Shew that\n\nF a, h + \\; c; z)-F a, b; c; z) = - F a + l, h + l; c + l; z).\n\n2. Shew that if o is a negative integer while /3 and y are not\nintegers, then the ratio F(a, IB; a + /3 + l -y; \\ - .v)-7-F a,; y;\nA') is independent of x, and find its value.\n\n%\n% 297\n%\n\ndP d P\n\n3. If P iz) be a hypergeometrie function, express its derivates -,-\nand -- linearly in\n\ndP\n\nterms of P and contiguous functions, and hence find the linear\nrelation between P, -j-,\n\nd'-P and -TV, i.e. verify that P satisfies the hypergeometrie\ndifterential equation.\n\n4. Shew that i j, j; 1; 4 (1 -z)) satisfies the hypergeometrie\nequation satisfied by jP(|, I; 1; z). Shew that, in the left-hand\nhalf of the lemniscate \\ z z) | = j, these two functions are equal;\nand in the right-hand half of the lemniscate, the former function is\nequal to F \\, |; 1; 1 -2).\n\n5. \\ i Fu, =F a +,h; c; .r), / \\ =i (a - 1, h; c; x), determine the\n15 linear relations with polynomial coefficients which connect F a, h;\nc; x) with pairs of the six functions Fa, Fa-, F,,, F,\\, F,, F,\\ .\n\\addexamplecitation{Gauss.}\n\n6. Shew that the hypergeometrie equation\n\nx(x-l/J - y-(a+ + l)x ' + a i/ =\n\nis satisfied by the two integrals (suppo.sed convergent)\n\n[\\ \\ -\\ l-z)y- -' l-xz)-''dz J\n\nand ['/-i(l-2)\"-i' l-(l- )2 ~\"rf2.\n\n.' \"\n\n7. Shew that, for values of x between and 1, the solution of the\nequation\n\nis AF a,i; i; \\ --Ixy-l + B l-2x) F h a + l), h fi + l); |; l-2xy-,\n\nwhere A, B are arbitrary constants and TODO represents the\nhypergeometric series.\n\n\\addexamplecitation{Math. Trip. 1896.}\n\n8. Shew that\n\nJim|F(,/3,y,..)- J (-),;r(y-a)r(y- )r( )r( ) ' J\n\n\\ r(y-a- )r(y ) (y-a) (y- )\n\nwhere h is the integer such that k R (a + - y)) F ia + b+n)T 2b+7i),,\n\n'' ' '' r 2a)T 2b)T a + b)n=o nl T c + n)T 2c- 1+n) *' '\n\n\\addexamplecitation{Clausen, Journal fiir Math, iii.}\n\n12. Shew that, if | | -r\\ j ..y V9; r(|)r('/i + f)\n\n(Heymann, Zeitschrift fur Math, und Phys. XLiv.)\n\n16. If il-xY+ -y F 2a,2;2y; x) = + Bx + C:'f- + Dx +..., shew that\n\ni (a, /3; y+; ) i (y-a, y-/3; y+l; x)\n\ny+l (7+l)(7 + i-) (y+*)(y + f)(7+|)\n\n(Cayley, PAz7. Mag. (4), xvi. (1858), pp. 356-357. See also Orr, Camb.\nPhil. Trans, xvii. (1899), pp. 1-15.)\n\n17. If the function F a, ji, /3', y; .*;, y) be defined by the\nequation\n\ni (a,,/3', y; x,y) = \\ - \\ \\ i' -\\ \\ ~xc)y-''-\\ \\ - ux)-\\ l-uy)- '\ndu,\n\nr (a) r (y - a) j\n\nthen shew that between F and any three of its eight contiguous\nfunctions\n\ni (a±l), F ii± ), F ii'±l), F y±\\ \\\n\nthere exists a homogeneous linear equation, whose coefficients are\npolynomials in x and y.\n\n\\addexamplecitation{Le Vavasseur.}\n\n%\n% 299\n%\n\n18. If y - a - iS < 0, shew that, a.s x - 1-0,\n\nand that, if -0-/3 = 0, the corresponding approximate formula is\n\n\\addexamplecitation{Math. Trip. 1893.}\n\n19. Shew that, when \\ x\\ < 1,\n\n/(2 +,0 +,a:-,0-) 1 1 fl\n\n= \\,r/' sin OTT sin (y - a) TT . ' - \" - \"V (a, /3; y; ),\n\nwhere c denotes a point on the straight line joining the points 0, x,\nthe initial arguments of v - x and of v are the same as that of .>;,\nand arg (1 - j/) -*-0 as v- 0.\n\n\\addexamplecitation{Pochhammer.}\n\n20. If, when \\ arg (1 -x)\\ < 2n,\n\nK x)=-r T -s)Ti; + s)\\ \\ ' xYds,\n\nLTTl J \\ y,,\n\nand, when | arg.*; | < 27r,\n\nl-lZl J \\ K,-\n\nby changing the variable s in the integral or otherwise, obtain the\nfollowing relations : K (.r) = A\" (1 - ), if I arg (1 - :f) I <\n\nK l-x) = K'(x), if|argj;|<7r.\n\nK x) = l - .r)-4 A' ( -, if i arg (1 - .r) \\ < n.\n\nK l-x) x-h K f\"--, if I arg.r | <7r.\n\nK' x) = x 2 / ' (Ijx), if I arg.r |'< tt.\n\nA\" (1 - a;) = (1 - X) - i A\" ( \\ J\\ - \\ \"), if j arg (1 - .x-) I < -\n\\addexamplecitation{Barnes.}\n\n21. With the notation of the preceding example, obtain the following\nresults :\n\n11=0 [ 'I- )\n\n2 K' x)= - l f lt- j'- - log -4log 2 + 4 (1 - 1 + -- .\n\nwhen \\ x\\ < l, | arg .i? | < tt; and\n\niT ( ) = + i ( - ) - 5 a: (l/.r) + ( - ) - 2 ' (1/:??),\n\nwhen j arg - x)\\ < ir, the ambiguous sign being the same as the .sign\nof / (x).\n\n\\addexamplecitation{Barnes.}\n\n%\n% 300\n%\n\n22. Hypergeometric series in two variables are defined by the\nequations F, a;, ';y; .r, y) = 2 \"Xn f\" '\"J/'S\n\nF, (a; /3, '; y, y;,r, y) = 2 '-, x y .\n\n-ts a,a,,; y; x, y)= 2, --,-\\ x 'y\"\n\n n( n Pm Pn,( ??l . n i y i,1\n\nFUa,;y,y';.,y) .XpV ' '\"'\" '' where a, = a(a + l). ..(a + m - 1), and\n2 means 2 2 .\n\n?n, n m=0 =0\n\nObtain the differential equations\n\nc\"F. d'F ?Ft ?F\n\n-(l-)8 +y(l-)5j;5 + r-( +e+l)- 5;j'-* -. /-,=0,\n\nd F d F dF;\n\n.il-o:)- +y Hy- a + + l)x]j -a F, = 0,\n\nand four similar equations, derived from these by interchanging x with\ny and a, /3, y with a', 3', y when a', ', y occur in the corresponding\nseries.\n\n\\addexamplecitation{Appell, Comptes Rendus, xc.} 23. If a is negative, and if\n\na= -V + a,\n\nwhere v is an integer and a is positive, shew that\n\nr (x) r (a)\n\nu D (-)\"( -l)(a-2) ... (a-%), where = - - - - ' - - \" ' G( -n),\n\n\" i (*/ = - . \\addexamplecitation{Hermite, Joiirnal mr Math, xcii.}\n\nx+n \\ 1 . /\n\n24. When a < 1, shew that\n\nT x)T a-x) I \\ l Rn\n\nT a) n=\\ X + n n=\\ X - a - n''\n\nwhere (-) a(a + l) ... (a + .-l)\n\n7i !\n\n25. When a > 1, and v and a are respectively the integral and\nfractional parts of a, shew that\n\nT x)T a-x) G x)p.,\\ - (?( ')p, r(a) =i .r + ?i =i . r-a - n\n\n-G x) + + ...+ f'-'X '\\ \\ X - a X-a-\\ .r-a-i + lj'\n\n%\n% 301\n%\n\nwhere (?( )=(l-? (l -- ) ... (l \\ . \\ )\n\nV a/ V a+l/ \\ a + v - lj\n\najjj \\ (-)\"a(a+l)... (a + -l)\n\nIt,;\n\n\\addexamplecitation{Hermite, Journal fiir Math, xcil.} 26. If\n\n/ (T, \\ r - '(jZ + + n-l) ., x x+l)(7/ + v + n-l)( + v + n)\n\n./ (.r, y, .)- 1 - C, - - -,) + C.3 3,( + 1) ( +,)( ., + i) - ...,\n\nwhere n is a positive integer and C'i, C2, ... are binomial\ncoefficients, shew that\n\nf (T y, r(y)rCy- -H/ )r(.r-n;)r(y+?0\n\n./H .y. ) Y y-x)T y + n)V :6)V x-irV n)'\n\n(Saalschiitz, Zeitschrift fur Math. xxxv.; a number of similar\nresults are given by Dougall, Proc. Edinburgh Math. Soc. xxv.)\n\nshew that, when liib + e -,a - ) > 0, then\n\ni (a,a-5+l, -c + l'; S, e; l)=2- ..A P W \" \" - ' ' ' ' ' r(8-\na)r(f-ia)r( + ia)r(S + e-a-l)\n\n(A. C. Dixon, P/'oc. London Math. Soc. xxxv.)\n\n28. Shew that, if R (a) < .if, then\n\n; f a(a+l)... (a-l-n- l)l,.,,, . r(l-3a)\n\n(Morley, Proc. London Math. Soc. xxxiv.) 2!). If\n\n[' / '....-I (1 \\ .,.)>-iy-i (1 \\ )A-. (1 -a,yy--J->'dxdi/ = B(i,J,\nX-, m), J i> J i>\n\nshew, by integrating with respect to jc, and also with respect to i\nthat B i,j, /, m) is a symmetric function of i+J,j + i\\ / + + in,\n/n+i.\n\nDeduce that\n\nF a,ld,y; fi, f; l) r (8) F ( ) T (6 + 6 - a -/3-y)\n\nis a symmetric function of 8, f, S + e - a - /i, 8 + e - - y, 8 + e -\ny - a.\n\n(A. ('. Dixon, Proc. London Math. Soc. (2), II. (1905), jjp. 8-16. For\na proof of a special case by Barnes' methods, see Barnes, Quarterly\nJournal, XLI. (1910), pp. 136-140.)\n\n30. If\n\nx ~ il-x f'\"- d\" I\n\n/ = i (-, a +;.; y, .r) = - - y- - W ( x) \\\n\nshew that, when n is a large positive integer, and 0< .r < 1,\n\n =, - (sin, )i-V(cos0)V- -icos (2u + a)( -i,r(2y-l) + o( 1 ),\n\nwhere .r = sin2 .\n\n(This result is contained in the great memoir by Darboux, \"Sur\nI'approxi- mation des fonctions de trfes grands nombres,\" Journal de\nMath. (3), iv. (1878), pp. 5-56, 377-416. For a systematic development\nof hyper- geometric functions in which one (or more) of the constants\nis large, see Camh. Phil. Trans, xxii. (1918), pp. 277-308.)\n", "meta": {"hexsha": "5a85fd946f95b98f74eba1ebae67a4cd7df20a0e", "size": 39125, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/tex/wandw-ch14.tex", "max_stars_repo_name": "CdLbB/Whittaker-and-Watson", "max_stars_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/tex/wandw-ch14.tex", "max_issues_repo_name": "CdLbB/Whittaker-and-Watson", "max_issues_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/tex/wandw-ch14.tex", "max_forks_repo_name": "CdLbB/Whittaker-and-Watson", "max_forks_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.026504298, "max_line_length": 176, "alphanum_fraction": 0.6225942492, "num_tokens": 14144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.8128673133042217, "lm_q1q2_score": 0.6251601860230567}} {"text": "\\section{Factorization in Polynomial Rings}\r\n\\begin{theorem}\\label{ufd_poly}\r\n Let $R$ be a UFD, then $R[X]$ is also a UFD.\r\n\\end{theorem}\r\n\\begin{remark}\r\n One can apply this recursively to show that the polynomial ring in $n$ variables over a UFD is a UFD.\r\n\\end{remark}\r\nIn particular, $\\mathbb Z[X]$ is a UFD and $\\mathbb C[X_1,\\ldots,X_n]$ is a UFD.\\\\\r\nIn this section, we shall always assume that $R$ is a UFD.\r\nIt is an integral domain, so it has a field of fraction $F$, so $R[X]\\subset F[X]$ which is a UFD since it is a ED.\r\n\\begin{definition}\r\n The content of a polynomial $f(X)=a_0+a_1X+\\cdots+a_nX^n$ for $a_i\\in R$ is $c(f)=\\gcd(a_0,\\ldots,a_n)$.\r\n We say $f$ is primitive if $c(f)$ is a unit, i.e. the coefficients are coprime.\r\n\\end{definition}\r\n\\begin{lemma}\\label{poly_content}\r\n 1. Any prime element in $R$ is also prime in $R[X]$.\\\\\r\n 2. If $f,g\\in R[X]$ are primitive polynomials, then $fg$ is also primitive.\\\\\r\n 3. If $f,g\\in R[X]$, then $c(fg)=c(f)c(g)$ up to associates.\r\n\\end{lemma}\r\n\\begin{proof}\r\n 1. Given a prime $p\\in R$, so $R/(p)$ is an integral domain.\r\n For $a\\in R$, let $\\tilde{a}\\in R/(p)$ be its image under the quotient map.\r\n Define $\\theta:R[X]\\to R/(p)[X]$ by\r\n $$a_0+a_1X+\\cdots+a_nX^n\\mapsto \\tilde{a}_0+\\tilde{a}_1X+\\cdots+\\tilde{a}_nX^n$$\r\n which is a homomorphism.\r\n So for $p|fg$ where $f,g\\in R[X]$, we have $\\theta(fg)=0$, so $\\theta(f)\\theta(g)=0$.\r\n But $R/(p)[X]$ is an integral domain since $R/(p)$ is.\r\n Hence WLOG $\\theta(f)=0$, so $p|f$.\\\\\r\n 2. If $f,g$ are primitive but $fg$ is not, then there is some irreducible $p\\in R$ that divides $fg$.\r\n Since $R$ is a UFD, $p$ is prime in $R$, hence is prime in $R[X]$ by 1, hence $p|f$ or $p|g$, contradiction to their primitivity.\\\\\r\n 3. Write $f=c(f)f_0$, so $f_0$ is primitive.\r\n Do the same to $g$ gives $g=c(g)g_0$ for a primitive $g_0$, so $fg=c(f)c(g)(f_0g_0)$.\r\n But $f_0g_0$ is primitive by 2, so $c(fg)=c(f)c(g)$ up to associates.\r\n\\end{proof}\r\n\\begin{remark}\r\n Take a polynomial $f$ in $F[X]$, then we can write $f=ab^{-1}f_0$ where $f_0\\in R[X]$ where $a,b\\in R,b\\neq 0$ and $f_0$ is primitive.\r\n We can just take $b$ to be a common multiple of the denominators (since $F$ is the field of fractions of $R$) and the rest follows.\r\n\\end{remark}\r\n\\begin{lemma}\\label{primitive_div_fof}\r\n Let $f,g$ be polynomials in $R[X]$ and $g$ is primitive, then if $g|f$ in $F[X]$, then $g|f$ in $R[X]$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n If $g|f$ in $F[X]$, then $f=gh$ with $h\\in F[X]$.\r\n Write $h=ab^{-1}h_0$ for $a,b\\in R,b\\neq 0$ and $h_0\\in R[X]$ is primitive.\r\n So $bf=agh_0$, then by taking content $bc(f)=a$, so $h=c(f)h_0\\in R[X]$ which is what we wanted.\r\n\\end{proof}\r\n\\begin{lemma}[Gauss's Lemma]\\label{gauss_poly}\r\n Let $R$ be a UFD with field of fraction $F$.\r\n Suppose $f\\in R[X]$ be primitive and irreducible in $R[X]$, then $f$ is irreducible in $F[X]$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Assume that $\\deg f>0$ since otherwise $f$ has to be a constant.\r\n Then in this case $f$ is not a unit in $F[X]$.\r\n Suppose we can write $f=gh$ for $g,h\\in F[X]$ with $\\deg g,\\deg h>0$.\r\n Replacing $g,h$ by $\\lambda g$ and $\\lambda^{-1}h$ for some $\\lambda\\in F^\\times$, we can assume WLOG that $g\\in R[X]$ and is primitive.\r\n So by the Lemma \\ref{primitive_div_fof}, $g|f$ in $R[X]$, so $h\\in R[X]$, contradiction.\r\n\\end{proof}\r\n\\begin{lemma}\\label{prime_fof}\r\n Let $g\\in R[X]$ be primitive and is prime in $F[X]$, then $g$ is prime in $R[X]$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Suppose $f_1,f_2\\in R[X]$ and $g|f_1f_2$, then $g|f_1$ or $g|f_2$ in $R[X]$, but by Lemma \\ref{primitive_div_fof} $g|f_1$ or $g|f_2$.\r\n\\end{proof}\r\n\\begin{proof}[Proof of Theorem \\ref{ufd_poly}]\r\n Let $f\\in R[X]$, write $f=c(f)f_0$ where $f_0$ is primitive.\r\n Since $R$ is a UFD, we can write $c(f)$ as a product of irreducibles.\r\n Also, $f_0$ can also be written as a product of irreducibles by induction on its degree.\r\n So it suffices to show that every irreducible in $R[X]$ is prime.\r\n Take $f\\in R[X]$ irreducible, then either $f$ is constant or $f$ must be primitive.\\\\\r\n For a constant $f$, then $f$ is prime in $R$, so it is prime in $R[X]$ by the first part of Lemma \\ref{poly_content}.\\\\\r\n For a primitive $f$, combining Lemma \\ref{gauss_poly} and Lemma \\ref{prime_fof} gives the result.\r\n\\end{proof}\r\n\\begin{theorem}[Eisenstein's Criterion]\r\n Let $R$ be a UFD and $f=a_0+\\cdots+a_nX^n\\in R[X]$ is primitive.\r\n Suppose $p\\in R$ is irreducible and $p\\nmid a_n$ but $p|a_i,0\\le i\\le n-1$ and $p^2\\nmid a_0$, then $f$ is irreducible.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Suppose $f=gh$ where $g,h\\in R[X]$ are not unit.\r\n Since $f$ is primitive, $g,h$ both have positive degree.\r\n Write $g(X)=r_0+\\cdots+r_kX^k$ and $h(X)=s_0+\\cdots+s_lX^l$ where $k+l=n$.\r\n Now $a_n=r_ks_l$, so $p\\nmid r_k,p\\nmid s_l$.\r\n Also $a_0=r_0s_0$, so exactly one of $r_0,s_0$ is divisible by $p$ since $R$ is a UFD.\r\n Assume WLOG $p|r_0$.\r\n Choose $j$ such that $p|r_0,\\cdots,p|r_{j-1}$ but $p\\nmid r_j$.\r\n Note that we can choose this since $p\\nmid r_k$, in particular $j\\le k0$}\\\\\r\n 1\\text{, otherwise}\r\n \\end{cases}$$\r\n So if $\\alpha+\\beta\\notin\\{0,1\\}$, we have $p_{11}^{(n)}\\to\\beta/(\\alpha+\\beta)$, similarly\r\n $$P^n\\to\\frac{1}{\\alpha+\\beta}\\begin{pmatrix}\r\n \\beta&\\alpha\\\\\r\n \\beta&\\alpha\r\n \\end{pmatrix},n\\to\\infty$$\r\n Hence $(\\beta/(\\alpha+\\beta),\\alpha/(\\alpha+\\beta))$ is an invariant distribution.\r\n\\end{example}\r\nUsually we cannot easily compute the entries of $P^n$ and take the limit to find an invariant distribution.\r\nHowever, there is an obvious other way to get one.\r\n\\begin{example}\r\n Consider\r\n $$P=\\begin{pmatrix}\r\n 0&1&0\\\\\r\n 0&1/2&1/2\\\\\r\n 1/2&0&1/2\r\n \\end{pmatrix}$$\r\n So if we want $\\pi P=\\pi$, it gives the set of linear equations\r\n $$\\begin{cases}\r\n \\pi_1=\\pi_3/2\\\\\r\n \\pi_2=\\pi_1+\\pi_2/2\\\\\r\n \\pi_3=\\pi_2/2+\\pi_3/2\r\n \\end{cases}$$\r\n which we can solve to get $\\pi_1=1/5,\\pi_2=\\pi_3=2/5$ which is indeed an invariant distribution.\r\n\\end{example}\r\n\\begin{definition}\r\n For each state $k\\in I$, let $\\gamma_i^k$ be the expected time spent in the state $i$ between two visits to $k$, so\r\n $$\\gamma_i^k=\\mathbb E_k\\sum_{n=0}^{T_k-1}1_{X_n=i}=\\mathbb E_k\\sum_{n=1}^{T_k}1_{X_n=i}$$\r\n\\end{definition}\r\n\\begin{theorem}\r\n Let $P$ be irreducible and recurrent, then:\\\\\r\n (a) $\\gamma_k^k=1$.\\\\\r\n (b) $\\gamma^k=(\\gamma_i^k)_{i\\in I}$ is an invariant measure.\\\\\r\n (c) $0<\\gamma_i^k<\\infty$ for all $i\\in I$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n (a) is obvious.\r\n For (b), since $P$ is recurrent, we know $\\mathbb P_k[T_k<\\infty]=1$, so for $j\\neq k$,\r\n \\begin{align*}\r\n \\gamma_j^k&=\\mathbb E_k\\sum_{n=1}^{T_k}1_{X_n=j}=\\mathbb E_k\\sum_{n=1}^\\infty 1_{X_n=j,n\\le T_k}\\\\\r\n &=\\sum_{n=1}^\\infty\\mathbb P_k[X_n=j,n\\le T_k]\\\\\r\n &=\\sum_{j\\in I}\\sum_{n=1}^\\infty\\mathbb P_k[X_{n-1}=i,X_n=j,n\\le T_k]\\\\\r\n &=\\sum_{j\\in I}\\sum_{n=1}^\\infty\\mathbb P_k[X_{n-1}=i,n\\le T_k]\\mathbb P[X_n=j|X_{n-1}=i]\\\\\r\n &=\\sum_{j\\in I}p_{ij}\\sum_{n=1}^\\infty\\mathbb P_k[X_{n-1}=i,n\\le T_k]\\\\\r\n &=\\sum_{j\\in I}p_{ij}\\sum_{n=1}^\\infty\\mathbb E_k[1_{X_{n-1}=i,n\\le T_k}]\\\\\r\n &=\\sum_{j\\in I}p_{ij}\\mathbb E_k\\sum_{n=0}^{T_k-1}1_{X_n=i}\\\\\r\n &=\\sum_{i\\in I}p_{ij}\\gamma_i^k=(\\gamma^kP)_j\r\n \\end{align*}\r\n For (c), as $P$ is irreducible, there is $n,m\\ge 0$ such that $p_{ik}^{(n)}>0,p_{ki}^{(m)}>0$.\r\n So by (b) and (a),\r\n $$\\gamma_i^k\\ge\\gamma_k^kp_{ki}^{(m)}=p_{ki}^{(m)}>0,1=\\gamma_k^k\\ge\\gamma_i^kp_{ik}^{(n)}\\implies \\gamma_i^k\\le\\frac{1}{p_{ik}^{(n)}}<\\infty$$\r\n As desired.\r\n\\end{proof}\r\nThis theorem has a partial inverse.\r\n\\begin{theorem}\\label{inv_measure_exp}\r\n Let $P$ be irreducible and $\\lambda$ be an invariant measure with $\\lambda_k=1$ for some $k$.\r\n Then $\\lambda_i\\ge\\gamma_i^k$ for every $i$.\\\\\r\n If in addition $P$ is recurrent, then $\\lambda=\\gamma^k$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Since $\\lambda$ is invariant, for $j\\neq k$,\r\n \\begin{align*}\r\n \\lambda_j&=\\sum_{i_1\\in I}\\lambda_{i_1}p_{i_1j}\\\\\r\n &=\\sum_{i_1\\neq k}\\lambda_{i_1}p_{i_1j}+p_{kj}\\\\\r\n &=\\sum_{i_1\\neq k}\\left( \\sum_{i_2\\neq k}\\lambda_{i_2}p_{i_2i_1}+p_{ki_1} \\right)p_{i_1j} +p_{kj}\\\\\r\n &=\\cdots\\\\\r\n &=\\sum_{i_1,\\ldots,i_n\\neq k}\\lambda_{i_n}p_{i_ni_{n-1}}\\cdots p_{i_1j}\\\\\r\n &\\quad+\\left( p_{kj}+\\sum_{i_1\\neq k}p_{ji_1}p_{i_1k}+\\cdots\\sum_{i_1,\\ldots,i_{n-1}\\neq k}p_{ki_{n-1}}\\cdots p_{i_2i_1}p_{i_1j} \\right)\\\\\r\n &\\ge p_{kj}+\\sum_{i_1\\neq k}p_{ji_1}p_{i_1k}+\\cdots\\sum_{i_1,\\ldots,i_{n-1}\\neq k}p_{ki_{n-1}}\\cdots p_{i_2i_1}p_{i_1j}\\\\\r\n &=\\mathbb P_k[X_1=j,T_k\\ge 1]+\\mathbb P_k[X_2=j,T_k\\ge 2]+\\cdots+\\mathbb P_k[X_n=j,T_k\\ge n]\\\\\r\n &=\\mathbb E_k\\left[ \\sum_{m=1}^{\\min\\{n,T_k\\}}1_{X_m=j} \\right]\\\\\r\n &=\\mathbb E_k\\left[ \\sum_{m=0}^{\\min\\{n,T_k-1\\}}1_{X_m=j} \\right]\\\\\r\n &\\to \\gamma_j^k,n\\to\\infty\r\n \\end{align*}\r\n which proves the first part of the theorem.\r\n Now define $\\mu=\\lambda-\\gamma^k$ which is obviously also an invariant measure.\r\n As $P$ is irreducible, for any $i$, there is some $n$ such that $p_{ik}^{(n)}>0$, therefore\r\n $$0=\\mu_k=\\sum_{j=I}\\mu_jp_{jk}^{(n)}\\ge\\mu_ip_{ik}^{(n)}\\implies \\mu_i=0$$\r\n Hence $\\mu=0$ which shows the second part.\r\n\\end{proof}\r\n\\begin{example}\r\n 1. The simple symmetric random walk on $\\mathbb Z$ is clearly irreducible and recurrent.\r\n The measure $\\pi_i=1$ for all $i\\in\\mathbb Z$ is invariant.\r\n By the theorem, every invariant measure are of the form $\\pi_i=a$ for all $i\\in\\mathbb Z$ for some fixed $a$.\r\n Consequently, there is no invariant distribution on this Markov chain.\\\\\r\n 2. (non-example) The simple symmetric random walk on $\\mathbb Z^3$ has an invariant measure, but is not recurrent.\r\n\\end{example}\r\nNote that a recurrent $i\\in I$ does not necessarily have finite expected return time $m_i=\\mathbb E_i[T_i]$.\r\n\\begin{definition}\r\n A recurrent state $i\\in I$ is positive recurrent if $m_i<\\infty$, and is null recurrent otherwise.\r\n\\end{definition}\r\n\\begin{theorem}\r\n Let $P$ be irreducible, then the following are equivalent:\\\\\r\n (a) Every state is positive recurrent.\\\\\r\n (b) Some state is positive recurrent.\\\\\r\n (c) $P$ admits an invariant distribution $\\pi$.\\\\\r\n Moreover, when (c) holds, then $m_i=1/\\pi_i$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n (a) clearly implies (b).\r\n Assuming (b) and choose positive recurrent $i\\in I$.\r\n Now $\\gamma^i$ is an invariant measure and\r\n $$\\sum_{j\\in I}\\gamma_j^i=m_i<\\infty$$\r\n Therefore $\\pi_j=\\gamma_j^i/m_i$ defines an invariant distribution.\\\\\r\n Assuming (c), then $\\forall k\\in I$, $\\pi_k=\\sum_{i\\in I}\\pi_ip_{ik}^{(n)}>0$ for some $n$ as $P$ is irreducible.\r\n Fix any $k$ and set $\\lambda_i=\\pi_i/\\pi_k$, then $\\lambda$ is an invariant measure with $\\lambda_k=1$, therefore $\\lambda\\ge\\gamma^k$ by Theorem \\ref{inv_measure_exp}.\r\n So\r\n $$m_k=\\sum_{i\\in I}\\gamma_i^k\\le\\sum_{i\\in I}\\frac{\\pi_i}{\\pi_k}=\\frac{1}{\\pi_k}<\\infty$$\r\n which means $k$ is positive recurrent.\r\n Also, this means that if (c) holds, then $P$ has to be recurrent and the inequality has to be equality, that is $m_k=1/\\pi_k$.\r\n\\end{proof}\r\n\\begin{example}\r\n There exists Markov chains with more than one linearly independent invariant measures.\r\n Consider the general random walk on $\\mathbb Z$ with $p_{i,i+1}=p,p_{i,i-1}=q=1-p$ where $p\\notin \\{0,1/2,1\\}$.\r\n Then the constant and $\\pi_i=(p/q)^i$ are both invariant measures but they are linearly independent.\r\n\\end{example}", "meta": {"hexsha": "fd51ea4e0bcf188098f5d8d48163ecbecefe1597", "size": 8710, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "7/inv.tex", "max_stars_repo_name": "david-bai-notes/IB-Markov-Chains", "max_stars_repo_head_hexsha": "cef4f20b59106a1deaed4de2f503e594e3ffc61d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "7/inv.tex", "max_issues_repo_name": "david-bai-notes/IB-Markov-Chains", "max_issues_repo_head_hexsha": "cef4f20b59106a1deaed4de2f503e594e3ffc61d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "7/inv.tex", "max_forks_repo_name": "david-bai-notes/IB-Markov-Chains", "max_forks_repo_head_hexsha": "cef4f20b59106a1deaed4de2f503e594e3ffc61d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.4355828221, "max_line_length": 177, "alphanum_fraction": 0.6136624569, "num_tokens": 3342, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125848754472, "lm_q2_score": 0.831143054132195, "lm_q1q2_score": 0.6250300365392257}} {"text": "\\chapter{Integer Partitions}\n\\label{chap-partitions}\n\nAn (integer) partition $p = p(1) \\cdots p(n)$ is a weakly decreasing sequence of positive integers, which are called the \\emph{parts} of $p$. The \\emph{size} of a partition, denoted $\\size{p}$, is the sum of its parts. For convenience, we say that for $i > n$ the $i\\th$ part of $p$ is $0$ and write $p(i) = 0$. We say that a partition $p = p(1) \\cdots p(m)$ is contained in another partition $q = q(1) \\cdots q(n)$ if $p(i) \\le q(i)$ for all $i$. This partial order on partitions is simply the one of Young's lattice, namely containment of Ferrers diagrams. The \\emph{Ferrers diagram} of a partition $p = p(1) \\cdots p(m)$ is a visual representation of $p$ consisting of an arrangement of $p(1)$ cells in the first (topmost) row, $p(2)$ cells in the second row, and so on. An example of partition containment displayed through Ferrers diagrams is presented in Figure~\\ref{fig-ptn-ferrers}.\n\n\\begin{figure}[ht]\n\\captionsetup{justification=centering}\n\t\\begin{tikzpicture}[scale={1/3}]\n\t\t\\ferrers{3,1}\n\t\t\\node at (5.5,-1) {$\\le$};\n\n\t\t\\begin{scope}[shift={(6,1)}]\n\t\t\t\\ferrers{4,2,1,1}\n\t\t\t\\ferrersfilled{3,1}\n\t\t\\end{scope}\n\n\t\t\\begin{scope}[shift={(5.5,-4)}]\n\t\t\t\\node at (-3,0) {$31$};\n\t\t\t\\node at ( 0,0) {$\\le$};\n\t\t\t\\node at (3.5,0) {$4211$};\n\t\t\\end{scope}\n\t\\end{tikzpicture}\n\\caption{An example partition containment presented by way of Ferrers diagrams.}\n\\label{fig-ptn-ferrers}\n\\end{figure}\n\nThe \\emph{conjugate} of a partition $p$ is the partition whose Ferrers diagram is the reflection of the Ferrers diagram of $p$ reflected along the anti-diagonal $y = -x$. An example of a partition and its conjugate is displayed in Figure~\\ref{fig-ptn-conjugate}.\n\\begin{figure}[ht]\n\\captionsetup{justification=centering,margin=1in}\n\t\\begin{tikzpicture}[scale={1/3}]\n\t\t\\ferrers{6,3,3,2,1}\n\t\t\\draw[dashed, gray] (1,0) -- ++(3.5,-3.5);\n\n\t\t\\begin{scope}[shift={(9,0.5)}]\n\t\t\t\\ferrers{5,4,3,1,1,1}\n\t\t\t\\draw[dashed, gray] (1,0) -- ++(3.5,-3.5);\n\t\t\\end{scope}\n\t\\end{tikzpicture}\n\\caption{The Ferrers diagram of a partition and its conjugate, both displayed with the anti-diagonal.}\n\\label{fig-ptn-conjugate}\n\\end{figure}\n\nGiven a finite set of partitions $S$, the unique smallest partition that contains each partition in $S$ is the partition whose $i\\th$ part is equal to the largest $i\\th$ part among all members of $S$. \n\\begin{observation}\n\\label{obs-ptn-join}\n\tIf $S$ is a finite set of partitions, then the smallest partition $p$ that is $S$-universal has $i\\th$ part given by\n\t\\[\n\t\tp(i) \n\t\t= \n\t\t\\max\\{q(i) \\st q \\in S\\}.\n\t\\]\n\\end{observation}\nThus, to construct the smallest $m$-universal partition for the set of all partitions, one must only observe that the largest $i\\th$ part among all partitions of size $m$ is $\\floor{m/i}$, meaning the unique smallest $m$-universal partition has $i\\th$ part equal to $\\floor{m/i}$ for all $i$.\n\n\\begin{theorem}\n\\label{thm-ptn-universal}\n\tThe unique smallest $m$-universal partition has size\n\t\\[\n\t\t\\phi(m)\n\t\t=\n\t\t\\floor{m/1} + \\floor{m/2} + \\cdots + \\floor{m/m}\n\t\t\\footnote{The sequence $\\phi$ appears as sequence \\OEISlink{A006218} in the OEIS~\\cite{sloane:the-on-line-enc:}}.\n\t\\]\n\\end{theorem}\nFigure~\\ref{fig-ptn-universal} displays the $24$-universal partition of size $\\phi(24) = 84$. In~\\cite{dirichlet:uber-die:}, Dirichlet shows that $\\phi(m) = m(\\log(m) + 2\\gamma-1) + \\Delta(m)$, where $\\gamma \\approx 0.5772$ is the Euler--Mascheroni constant and $\\Delta(m) = \\oO{\\sqrt{m}}$, and thus $\\phi(m) \\sim m \\log m$. The asymptotics of $\\Delta(m)$ are the subject of the \\emph{Dirichlet Divisor Problem}, which aims to find the smallest value $\\theta$ so that $\\Delta(m) = \\oO{m^\\theta}$. The best known bound is due to Huxley~\\cite{huxley:exponential-sums-and:}, who showed that $\\inf \\theta \\le 131/416 \\approx 0.3149$. \n\n\\begin{figure}[ht]\n\\captionsetup{justification=centering}\n\t\\begin{tikzpicture}[scale={24/120}]\n\t\t\\ferrers{24,12,8,6,4,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1}\n\t\\end{tikzpicture}\n\\caption{The unique $24$-universal partition of size $\\phi(24) = 84$.}\n\\label{fig-ptn-universal}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Proper Classes of Partitions}\n\\label{sec-ptn-classes}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nIn this section, we show that any proper class of partitions admits universal partitions of linear size. For reference, some known or computed values of minimum sizes for universal partitions for various classes are presented in Appendix~\\ref{appendix-partitions}.\n\nWe require one more definition to discuss proper classes of partitions. We say that an \\emph{potentially infinite partition} $p$ is an infinite weakly decreasing sequence over the infinite ordered alphabet $\\{0 < 1 < \\cdots < \\omega\\}$, where $\\omega$ represents an infinite part. If $p(n) = \\ell > 0$ for each $n > N$, then we abbreviate $p$ as $p(1) \\cdots p(N) \\ell^\\omega$, where $\\ell^\\omega$ representing an infinite number of parts equal to $\\ell$. If $p(n) = 0$ for each $n > N$, then we abbreviate $p$ as $p(1) \\cdots p(N)$. Infinite partitions give way to infinite Ferrers diagrams, and we say that the \\emph{age} of a potentially infinite partition $p$, denoted $\\Age(p)$, is the set of finite partitions whose Ferrers diagrams embed into the Ferrers diagram of $p$ (the term \\emph{age} dates to Fra{\\\"i}ss{\\'e}~\\cite{fraisse:sur-lextension-:}). An example of containment of a finite Ferrers diagram into an infinite one, together with the corresponding set containment, is presented in Figure~\\ref{fig-ptn-age}.\n\n\\begin{figure}[ht]\n\\captionsetup{justification=centering}\n\t\\begin{tikzpicture}[scale={1/3}]\n\t\t\\ferrers{4,2,1,1}\n\t\t\n\t\t\\node at (6.5,-2) {$\\le$};\n\t\t\\begin{scope}[shift={(7.5,0.5)}]\n\t\t\t\\foreach \\d in {0.25, 0.50, 0.75} {\n\t\t\t\t\\filldraw[black] ({6+\\d},-0.5) circle [radius=0.05cm];\n\t\t\t\t\\filldraw[black] ({6+\\d},-1.5) circle [radius=0.05cm];\n\t\t\t\t\\filldraw[black] (1.5,-{6+\\d}) circle [radius=0.05cm];\n\t\t\t\t% \\filldraw[black] (2.5,-{6+\\d}) circle [radius=0.05cm];\n\t\t\t}\n\t\t\t\\ferrers{5,5,4,1,1}\n\t\t\t\\ferrersfilled{4,2,1,1}\n\t\t\\end{scope}\n\t\t\n\t\t\\begin{scope}[shift={(6.5,-6.5)}]\n\t\t\t\\node at (-3.5,0) {$4211$};\n\t\t\t\\node at ( 0 ,0) {$\\in$};\n\t\t\t\\node at ( 5 ,0) {$\\Age(\\omega \\omega 4 1^\\omega)$};\n\t\t\\end{scope}\n\t\\end{tikzpicture}\n\\caption{An example of containment of a finite Ferrers digram into an infinite one.}\n\\label{fig-ptn-age}\n\\end{figure}\n\nWe show that every proper class of partitions admits linear-size universal partitions in two parts. First, we show that every proper class of partitions is contained in an age of the form $\\Age(\\omega^k \\ell^\\omega)$ in Proposition~\\ref{prop-ptn-age-containment}, and then that every such age admits linear-size universal partitions in Theorem~\\ref{thm-ptn-universal-proper}.\n\\begin{proposition}\n\\label{prop-ptn-age-containment}\n\tFor any proper partition class $\\P$, there are nonnegative integers $k, \\ell$ such that $\\P \\subseteq \\Age(\\omega^k \\ell^\\omega)$.\n\\end{proposition}\n\\begin{proof}\n\tLet $\\ell$ be the largest integer such that $\\P$ contains the partition $\\ell^n$ for all $n$ and let $k$ be the largest integer such that $\\P$ contains $(\\ell+1)^k$. We claim that $\\P \\subseteq \\Age(\\omega^k \\ell^\\omega)$. For any partition $p \\in \\P$, we must have $p(n) \\le \\ell$ for each $n > k$, as otherwise $p$ contains $(\\ell+1)^n$, and as $\\P$ is closed downward, we would have $(\\ell+1)^n \\in \\P$, a contradiction. The condition that $p(n) \\le \\ell$ for each $n > k$ is precisely the condition that defines containment in $\\Age(\\omega^k \\ell^\\omega)$, and thus $p \\in \\Age(\\omega^k \\ell^\\omega)$, completing the proof.\n\\end{proof}\n\n\\begin{theorem}\n\\label{thm-ptn-universal-proper}\n\tLet $\\P = \\Age(\\omega^k \\ell^\\omega)$ be a class of partitions. Then for $m \\ge k\\ell$, the smallest $\\P_m$-universal partitions have size\n\t\\[\n\t\t\\u_{\\P}(m)\n\t\t= \n\t\t\\sum_{i=1}^{k} \\floor{\\frac{m}{i}} + \\sum_{i=1}^{\\ell} \\floor{\\frac{m}{i}} - k\\ell\n\t\\]\n\tand thus $\\u_{\\P}(m) = \\oTheta{m}$.\n\\end{theorem}\n\nBefore proving Theorem~\\ref{thm-ptn-universal-proper}, observe that if $P \\subseteq Q$ are sets of partitions and $q$ is a $Q$-universal partition, then $q$ is $P$-universal as well, as $q$ necessarily contains each partition in $P$. Thus, by proving that each age of the form $\\Age(\\omega^k \\ell^\\omega)$ admits universal partitions of linear size, Proposition~\\ref{prop-ptn-age-containment} implies that every proper partition class does as well.\n\n\\newenvironment{proof-of-thm-ptn-universal-proper}{%\n\t\\medskip\\noindent {\\it Proof of Theorem~\\ref{thm-ptn-universal-proper}.\\/}%\n}{%\n\t\\qed\\bigskip%\n}\n\\begin{proof-of-thm-ptn-universal-proper}\n\tBy Observation~\\ref{obs-ptn-join}, the unique smallest $\\P_m$-universal partition $P$ has its $i\\th$ part equal to the largest $i\\th$ part among all partitions in $\\P_m$. The only restriction on partitions in $\\P$ is that their $i\\th$ part is at most $\\ell$ for $i > k$, so we have\n\t\\[\n\t\tP(i)\n\t\t=\n\t\t\\begin{cases}\n\t\t\t\\floor{\\dfrac{m}{i}} & \\text{if $1 \\le i \\le k$,} \\\\\n\t\t\t\\min\\left\\{\\floor{\\dfrac{m}{i}}, \\ell\\right\\} & \\text{if $k < i \\le m$.}\n\t\t\\end{cases}\n\t\\]\n\tAs $\\floor{\\dfrac{m}{i}} \\ge \\ell$ if and only if $\\floor{\\dfrac{m}{\\ell}} \\ge i$, this may be simplified to\n\t\\[\n\t\tP(i)\n\t\t=\n\t\t\\begin{cases}\n\t\t\t\\floor{\\dfrac{m}{i}} & \\text{if $1 \\le i \\le k$,} \\\\\n\t\t\t\\ell & \\text{if $k < i \\le \\floor{\\dfrac{m}{\\ell}}$,} \\\\\n\t\t\t\\floor{\\dfrac{m}{i}} & \\text{if $\\floor{\\dfrac{m}{\\ell}} < i \\le m$.}\n\t\t\\end{cases}\n\t\\]\n\tThus, for $m \\ge k\\ell$, the size of $P$ is\n\t\\[\n\t\t\\size{P} \n\t\t= \n\t\t\\underbrace{\\floor{\\dfrac{m}{1}} + \\floor{\\dfrac{m}{2}} + \\cdots + \\floor{\\dfrac{m}{k}} + \\ell + \\ell + \\cdots + \\ell}_{\\text{$\\floor{\\dfrac{m}{\\ell}}$ terms}} + \\floor{\\dfrac{m}{\\floor{\\tfrac{m}{\\ell}}+1}} + \\cdots + \\floor{\\dfrac{m}{m}}.\n\t\\]\n\n\tThe sum of the parts $P(k+1), \\cdots, P(m)$ is equal to $\\sum_{i = 1}^{\\ell} \\floor{\\frac{m}{i}} - k\\ell$, as these correspond to the first $\\ell$ parts of the conjugate of $P$, which are $\\floor{\\dfrac{m}{i}}$ for $1 \\le i \\le \\ell$, and we need to subtract those cells that overlap with the first $k$ parts of $P$. As $m \\ge k\\ell$, we have both $\\floor{\\frac{m}{k}} \\ge \\ell$ and $\\floor{\\frac{m}{\\ell}} \\ge k$, and thus\n\t\\[\n\t\t\\size{P}\n\t\t= \n\t\t\\sum_{i=1}^{k} \\floor{\\frac{m}{i}} + \\sum_{i=1}^{\\ell} \\floor{\\frac{m}{i}} - k\\ell.\n\t\\]\n\tFor all $n$ and $j$, the inequality\n\t\\[\n\t\t\\log(j+1)n-j\n\t\t\\le\n\t\t\\sum_{i = 1}^{j} \\floor{\\frac{n}{j}} \n\t\t\\le \n\t\t\\left(\\log(j)+1\\right)n,\n\t\\]\n\tholds, and thus $\\size{P} = \\oTheta{m}$, as desired.\n\\end{proof-of-thm-ptn-universal-proper}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Concluding Remarks}\n\\label{sec-ptn-conclusion}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nObservation~\\ref{obs-ptn-join} establishes that the smallest $m$-universal partition is unique, and Theorem~\\ref{thm-ptn-universal} establishes its precise size for all $m$. For any proper class of partitions $\\P$, there are proper $m$-universal partitions for $\\P$ for all $m$ if and only if $\\P$ satisfies the joint-embedding property, which is equivalent to $\\P$ being the age of some potentially infinite partition $u$, as the following result of Fra{\\\"i}ss{\\'e} (which we have specialized to our context here) shows:\n\\begin{theorem}[Fra{\\\"i}ss{\\'e}~\\cite{fraisse:sur-lextension-:}; see also Hodges~{\\cite[Section 7.1]{hodges:model-theory:}}] \n\tThe following are equivalent for a class $\\P$ of integer partitions or compositions:\n\t\\begin{enumerate}\n\t\t\\item $\\P$ cannot be expressed as the union of two proper subclasses,\n\t\t\\item $\\P$ satisfies the \\emph{joint embedding property}, meaning that for every $a, b \\in \\P$ there is some $c \\in \\P$ such that $a, b \\le c$, and \n\t\t\\item $\\P = \\Age(u)$ for some word $u \\in (\\mathbb{P} \\cup \\{n^\\omega \\st n \\in \\mathbb{P}\\} \\cup \\{\\omega, \\omega^\\omega\\})^\\ast$.\n\t\\end{enumerate}\n\\end{theorem}\n\n", "meta": {"hexsha": "db4f7714dafe5fe375dfa29da67938ad29f9e430", "size": 11917, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chap-partitions.tex", "max_stars_repo_name": "engenmt/doctoral-dissertation", "max_stars_repo_head_hexsha": "b5e2caee30ee40653be58ec190028fae9cfb2df6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chap-partitions.tex", "max_issues_repo_name": "engenmt/doctoral-dissertation", "max_issues_repo_head_hexsha": "b5e2caee30ee40653be58ec190028fae9cfb2df6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chap-partitions.tex", "max_forks_repo_name": "engenmt/doctoral-dissertation", "max_forks_repo_head_hexsha": "b5e2caee30ee40653be58ec190028fae9cfb2df6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.1868686869, "max_line_length": 1023, "alphanum_fraction": 0.6558697659, "num_tokens": 4063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8311430562234877, "lm_q1q2_score": 0.6250300288732089}} {"text": "\\section{Task A}\n\\label{sec:task-a}\n\nAccording to equation \\eqref{eq:free-energy} in \\cite{Bonet2008}, the free energy\nfor compressible Neo-Hookean material is defined as follows:\n\\begin{equation} \\tag{6.27}\n \\label{eq:free-energy}\n \\Psi = \\frac{\\mu}{2} \\left( I_{C_{e}} - 3 \\right) - \\mu \\ln J_{e} +\n \\frac{\\lambda}{2} \\left( \\ln J_{e} \\right)^{2}\n\\end{equation}\nwith \\(I_{C_{e}} = \\utilde{I} : \\utilde{C}_{e} \\) and the elastic Lame constants\n\\(\\mu = G = 0.5 E / (1+\\nu)\\), \\(\\lambda = E \\nu / (1 - 2\\nu) / (1+\\nu)\\).\n\nYield function of von Mises type:\n\\begin{equation}\n \\label{eq:yield_func}\n \\phi = \\sqrt{\\frac{3}{2}} \\left| \\utilde{\\overline{M}}_{dev} \\right|- \n \\left( \\tau_{y} - H \\overline{\\varepsilon}_{p} \\right)\n\\end{equation}\nwith \\(\\overline{\\varepsilon}_{p} = \\int_{0}^{t} \\dot{\\gamma} \\, dt\\).\n\nMandel stress is defined as\n\\begin{equation}\n \\label{eq:mandel}\n \\utilde{\\overline{M}} = \\utilde{C}_{e} \\cdot \\widetilde{\\utilde{S}}\n\\end{equation}\nwith intermediate 2nd Piola-Kirchhoff stress defined as a pull-back of the\nKirchhoff stress to the intermediate configuration as follows:\n\\begin{equation}\n \\label{eq:PK2int}\n \\widetilde{\\utilde{S}} = \\utilde{F}_{e}^{-1} \\cdot \\utilde{\\tau} \\cdot\n \\utilde{F}_{e}^{-T} = 2 \\frac{\\partial \\Psi}{\\partial \\utilde{C}_{e}}\n\\end{equation}\nand intermediate elastic deformation tensor\n\\begin{equation}\n \\label{eq:Cint}\n \\utilde{C}_{e} = \\utilde{F}_{p}^{-T} \\cdot \\utilde{C} \\cdot \\utilde{F}_{p}^{-1}\n\\end{equation}\nand plastic part of the deformation gradient\n\\begin{equation}\n \\label{eq:F}\n \\utilde{F} = \\utilde{F}_{e} \\cdot \\utilde{F}_{p}\n\\end{equation}\n\nLet us derive expression for \\(\\widetilde{\\utilde{S}}\\):\n\\begin{equation}\n \\label{eq:PK2intDeriv}\n \\widetilde{\\utilde{S}} = 2 \\frac{\\partial \\Psi}{\\partial \\utilde{C}_{e}} = \n 2 \\frac{\\partial \\Psi}{\\partial I_{C_{e}}} \\cdot\n \\frac{\\partial I_{C_{e}}}{\\partial \\utilde{C}_{e}} +\n 2 \\frac{\\partial \\Psi}{\\partial J_{e}} \\cdot\n \\frac{\\partial J_{e}}{\\partial \\utilde{C}_{e}}\n\\end{equation}\n\\begin{align}\n \\frac{\\partial \\Psi}{\\partial I_{C_{e}}} &= \\frac{\\mu}{2} \\\\\n \\frac{\\partial I_{C_{e}}}{\\partial \\utilde{C}_{e}} &= \\utilde{I} \\\\\n \\frac{\\partial \\Psi}{\\partial J_{e}} &= - \\frac{\\mu}{J_{e}} + \n \\frac{\\lambda}{J_{e}} \\ln J_{e} \\\\\n \\frac{\\partial J_{e}}{\\partial \\utilde{C}_{e}} &= \\frac{J_{e}}{2} \\utilde{C}_{e}^{-1}\n\\end{align}\nTherefore,\n\\begin{equation}\n \\label{eq:PK2intFinal}\n \\widetilde{\\utilde{S}} = \\mu \\utilde{I} + \\left( \\lambda \\ln J_{e} -\n \\mu \\right) \\utilde{C}_{e}^{-1}\n\\end{equation}\nAt this point we can assess the yield function. \nThe Kuhn-Tucker conditions are\n\\begin{equation}\n \\label{eq:kuhn-tucker}\n \\phi \\leq 0, \\quad \\dot{\\gamma} \\geq 0, \\quad \\dot{\\gamma} \\phi = 0 \n\\end{equation}\nThe evolution equation is of associative type:\n\\begin{equation}\n \\label{eq:evolution}\n \\overline{\\utilde{L}}_{p} = \\dot{\\gamma}\n \\frac{\\partial \\phi}{\\partial \\overline{\\utilde{M}}} = \n \\dot{\\gamma} \\sqrt{\\frac{3}{2}} \n \\frac{\\overline{\\utilde{M}}_{dev}}{\\left| \\overline{\\utilde{M}}_{dev} \\right|} =\n \\dot{\\gamma} \\utilde{\\nu}\n\\end{equation}\nApply backward Euler integration:\n\\begin{align}\n \\frac{\\prescript{n+1}{}{\\utilde{F}_{p} -\\prescript{n}{}{\\utilde{F}_{p}}}}{\\Delta t}\n \\prescript{n+1}{}{\\utilde{F}}_{p}^{-1} &= \\frac{\\Delta \\gamma}{\\Delta t} \n \\prescript{n+1}{}{\\utilde{\\nu}} \\\\\n \\utilde{I} - \\prescript{n}{}{\\utilde{F}}_{p} \\cdot\n \\prescript{n+1}{}{\\utilde{F}}_{p}^{-1} &\\approx\n \\Delta \\gamma \\prescript{n+1}{}{\\utilde{\\nu}}\n\\end{align}\nApply semi-explicit simplification \n\\(\\prescript{n+1}{}{\\utilde{\\nu}} \\approx \\prescript{n}{}{\\utilde{\\nu}}\\):\n\\begin{align}\n \\utilde{I} - \\prescript{n}{}{\\utilde{F}}_{p} \\cdot\n \\prescript{n+1}{}{\\utilde{F}}_{p}^{-1} &\\approx\n \\Delta \\gamma \\prescript{n}{}{\\utilde{\\nu}} \\\\\n \\prescript{n+1}{}{\\utilde{F}}_{p}^{-1} &\\approx\n \\prescript{n}{}{\\utilde{F}}_{p}^{-1} \\left( \\utilde{I} -\n \\Delta \\gamma \\prescript{n}{}{\\utilde{\\nu}} \\right)\n\\end{align}\nSolve the simplified local constitutive problem:\n\\begin{equation}\n \\label{eq:local}\n \\phi \\left( \\Delta \\gamma \\right) = \\sqrt{\\frac{3}{2}} \n \\left| \\overline{\\utilde{M}}_{dev} \\right| - \\left[ \\tau_{y} - \n H \\left( \\prescript{n}{}{\\overline{\\varepsilon}_{p}} + \\Delta \\gamma\\right) \\right]\n\\end{equation}\nwith \n\\begin{equation}\n \\label{eq:Mexplicit}\n \\overline{\\utilde{M}} = \\utilde{C}_{e} \\cdot \\widetilde{\\utilde{S}} = \n \\utilde{C}_{e} \\left( \\utilde{F}_{p}^{-1} \\left( \\Delta \\gamma \\right),\n \\utilde{C} \\right) \\cdot\n \\widetilde{\\utilde{S}} \\left( \\utilde{C}_{e}\n \\left( \\utilde{F}_{p}^{-1} \\left( \\Delta \\gamma \\right), \\utilde{C} \\right)\n \\right)\n\\end{equation}\nusing Newton--Raphson method:\n\\begin{align}\n \\phi \\left( \\Delta \\gamma_{n+1} \\right) &\\approx \n \\phi \\left( \\Delta \\gamma_{n} \\right) +\n \\frac{d \\, \\phi \\left( \n \\Delta \\gamma_{n} \\right)}{d \\,\n \\Delta \\gamma} \\blacktriangle \\gamma \n = 0 \\\\\n \\blacktriangle \\gamma &= - \\left[ \\frac{d \\, \\phi \\left( \n \\Delta \\gamma_{n} \\right)}{d \\, \\Delta \\gamma} \n \\right]^{-1} \\cdot \\phi \\left( \\Delta \\gamma_{n} \\right)\\\\\n \\Delta \\gamma_{n+1} &= \\Delta \\gamma_{n} + \\blacktriangle \\gamma\n\\end{align}\nwhere the derivative of the yield function can be expanded to\n\\begin{equation}\n \\label{eq:yield-deriv}\n \\frac{d \\phi}{d \\gamma} = \\frac{\\partial \\phi}{\\partial \\overline{\\utilde{M}}} :\n \\frac{\\partial \\overline{\\utilde{M}}}{\\partial \\utilde{C}_{e}} :\n \\frac{\\partial \\utilde{C}_{e}}{\\partial \\utilde{F}_{p}^{-1}} :\n \\frac{d \\utilde{F}_{p}^{-1}}{d \\Delta \\gamma} + \n \\frac{\\partial \\phi}{\\partial \\Delta \\gamma}\n\\end{equation}\nwith \\(\\dfrac{\\partial \\phi}{\\partial \\Delta \\gamma} = - H\\) and the rest of\nthe derivatives given by equations (34)--(37) of \\cite{Ekh2016}, in which\n\\begin{equation}\n \\label{eq:dSdCe}\n \\frac{\\partial \\widetilde{\\utilde{S}}}{\\partial \\utilde{C}_{e}} = \n \\frac{\\lambda}{2} \\utilde{C}_{e}^{-1} \\otimes \\utilde{C}_{e}^{-1} +\n \\left( \\mu - \\lambda \\ln J_{e} \\right) \\utilde{C}_{e}^{-1} \\overline{\\otimes}\n \\utilde{C}_{e}^{-1}\n\\end{equation}\nFinally, compute the 2nd Piola-Kirchhoff stress\n\\begin{equation}\n \\label{eq:PK2final}\n \\utilde{S} = \\utilde{F}_{p}^{-1} \\cdot \\widetilde{\\utilde{S}} \\cdot\n \\utilde{F}_{p}^{-T}\n\\end{equation}\nand the material stiffness\n\\begin{equation}\n \\label{eq:mat-stiff}\n \\uutilde{C} = 2 \\frac{d \\utilde{S}}{d \\utilde{C}}\n\\end{equation}\n\\textit{Note: if computed numerically, \\(\\uutilde{C}\\) must be symmetrised to get\ncorrect element stiffness.}\n\nThe material model was implemented in Matlab and can be found in\n\\texttt{neo\\_hooke\\_plast.m} (see section \\ref{app:matlab-code}).\n\nFigure~\\ref{fig:sigma12-eps12} shows the test of the constitutive driver,\nwhere component of Cauchy stress plotted against\nengineering strain for the situation of uniaxial strain control\n\\(\\utilde{F} = F_{12} \\ubar{\\bm{e}}_{1} \\otimes \\ubar{\\bm{E}}_{2}\\).\n\\begin{figure}[th]\n \\centering\n \\begin{tikzpicture}\n \\begin{axis}[\n width = 0.95\\textwidth,\n height=\\axisdefaultheight,\n tick label style={/pgf/number format/fixed},\n try min ticks=6,\n minor tick num=1,\n grid=both,\n xmin=0, xmax=0.2,\n xlabel = {\\( \\varepsilon_{12}\\), [-]},\n ylabel = {\\( \\sigma_{12} \\), [MPa]},\n ]\n \\addplot table[skip first n=1] {data/sigma12_eps12.dat};\n \\end{axis}\n \\end{tikzpicture} \n \\caption{Cauchy stress component \\(\\sigma_{12}\\) versus strain\n \\(\\varepsilon_{12} = F_{12} \\).}\n \\label{fig:sigma12-eps12}\n\\end{figure}\n\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: \"../main\"\n%%% End:\n", "meta": {"hexsha": "37d742a8a5fe52801c85344fa4814056b4a25ea4", "size": 7870, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/sec/task_a.tex", "max_stars_repo_name": "iamrosk/large_elast_plast_deform", "max_stars_repo_head_hexsha": "62986469f7df01d6874f4a27b5887b7116c8a38a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-09-14T00:15:08.000Z", "max_stars_repo_stars_event_max_datetime": "2020-09-14T00:15:08.000Z", "max_issues_repo_path": "doc/sec/task_a.tex", "max_issues_repo_name": "iamrosk/large_elast_plast_deform", "max_issues_repo_head_hexsha": "62986469f7df01d6874f4a27b5887b7116c8a38a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/sec/task_a.tex", "max_forks_repo_name": "iamrosk/large_elast_plast_deform", "max_forks_repo_head_hexsha": "62986469f7df01d6874f4a27b5887b7116c8a38a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-09-14T03:27:42.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-05T01:44:26.000Z", "avg_line_length": 39.9492385787, "max_line_length": 87, "alphanum_fraction": 0.6048284625, "num_tokens": 2926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6250300162917812}} {"text": "%!TEX root = ../Tibt.tex\n\n\\exercise{4.2}\n\n\\subsection*{(a)}\n\nThe linear discriminant functions are:\n\n\\begin{eqnarray*}\n \\hat{\\delta}_k(x) & = & x^T \\hat{\\Sigma}^{-1} \\hat{\\mu}_k - \\frac{1}{2}\\, \\hat{\\mu}_k ^T \\hat{\\Sigma}^{-1} \\hat{\\mu}_k + \\log\\, \\hat{\\pi}_k \\\\\n \\hat{\\pi}_k & = & N_k / N\n\\end{eqnarray*}\nSo (a) follows directly from $\\hat{\\delta}_2(x) > \\hat{\\delta}_1(x)$. \n\n\n\\subsection*{(b), (c), (d)}\nLet's denote $\\mathbf{e}_{(1)}$ and $\\mathbf{e}_{(2)}$ the one-hot variables associated\nto the labels $g$. The coefficients $\\hat{\\beta}$ of the OLS regression with the bias\nterm included can be obtained by regressing $\\mathbf{y}$ against the demeaned\nversion of the original predictors:\n\\begin{eqnarray*}\n \\hat{\\beta} & = & \\left(\\mathbf{X}_{dm}^T \\mathbf{X}_{dm}\\right)^{-1} \\,\n \\mathbf{X}_{dm}^T \\mathbf{y} \\\\\n \\mathbf{X}_{dm} & \\equiv & \\mathbf{X} - \\mathbf{e} \\, \\hat{\\mu}^T \\\\\n & = & \\mathbf{X} - (\\mathbf{e}_{(1)} + \\mathbf{e}_{(2)}) \\, \\hat{\\mu}^T \\\\\n \\hat{\\mu} & \\equiv & \\frac{1}{N} \\mathbf{X}^T \\mathbf{e}\n\\end{eqnarray*}\nNotice that $\\hat{\\mu}$ is the vector of predictor means, regardless of the class.\nOn the other hand, the LDA covariance matrix is obtained as:\n\\begin{eqnarray*}\n \\hat{\\Sigma} & = & \\frac{1}{N - 2} \\mathbf{X}_{r}^T \\mathbf{X}_r\n\\end{eqnarray*}\nwhere $\\mathbf{X}_{r}$ are the values of predictors centered around the class\naverages:\n\\begin{eqnarray*}\n \\mathbf{X}_{r} & = & \\mathbf{X} - \\mathbf{e}_{(1)}\\,\\hat{\\mu}_1^T - \\mathbf{e}_{(2)}\\,\\hat{\\mu}_2^T \\\\\n \\hat{\\mu}_1 & \\equiv & \\frac{1}{N_1} \\mathbf{X}^T \\mathbf{e}_{(1)}\\\\\n \\hat{\\mu}_2 & \\equiv & \\frac{1}{N_2} \\mathbf{X}^T \\mathbf{e}_{(2)}\n\\end{eqnarray*}\nWe have therefore:\n\\begin{equation*}\n\\mathbf{X}_{dm} = \\mathbf{X}_r + \\mathbf{e}_{(1)} \\left(\\hat{\\mu}_1 - \\hat{\\mu}\\right)^T\n + \\mathbf{e}_{(2)} \\left(\\hat{\\mu}_2 - \\hat{\\mu}\\right)^T\n\\end{equation*}\nBy construction, the residuals $\\mathbf{X}_r$ are orthogonal to $\\mathbf{e}_{(1)}$, \n$\\mathbf{e}_{(2)}$:\n\\begin{equation*}\n\\mathbf{X}_r^T \\,\\mathbf{e}_{(1)} = \\mathbf{X}_r^T \\,\\mathbf{e}_{(2)} = 0\n\\end{equation*}\nAlso:\n\\begin{eqnarray*}\n \\mathbf{e}_{(k)}^T\\, \\mathbf{e}_{(k)} & = & N_k \\\\\n \\mathbf{e}_{(1)}^T \\mathbf{e}_{(2)} & = & 0 \\\\\n \\hat{\\mu} & = & \\frac{N_1}{N} \\hat{\\mu}_1 + \\frac{N_2}{N} \\hat{\\mu}_2 \n\\end{eqnarray*}\nHence:\n\\begin{eqnarray*}\n \\mathbf{X}_{dm}^T \\mathbf{X}_{dm} & = & \\mathbf{X}_r^T \\mathbf{X}_r + \n N_1\\,(\\hat{\\mu}_1 - \\hat{\\mu}) (\\hat{\\mu}_1 - \\hat{\\mu})^T + N_2\\,(\\hat{\\mu}_2 - \\hat{\\mu}) (\\hat{\\mu}_2 - \\hat{\\mu})^T \\\\\n & = & (N - 2)\\, \\hat{\\Sigma} + N_1 \\, \\left(\\frac{N_2}{N} \\hat{\\mu}_1 - \\frac{N_2}{N} \\hat{\\mu}_2\\right) \\left(\\ldots\\right) + N_2 \\, \\left(\\frac{N_1}{N} \\hat{\\mu}_2 - \\frac{N_1}{N} \\hat{\\mu}_1\\right) \\left(\\ldots\\right) \\\\\n & = & (N - 2)\\, \\hat{\\Sigma} + \\frac{N_1\\,N_2}{N^2} (N_1 + N_2) \n \\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)\\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)^T \\\\\n & = & (N - 2)\\, \\hat{\\Sigma} + \\frac{N_1\\,N_2}{N} \n \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right)\\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right)^T\n\\end{eqnarray*}\nConsider the general target encoding:\n\\begin{equation*}\n\\mathbf{y} = y_1\\, \\mathbf{e}_{(1)} + y_2\\, \\mathbf{e}_{(2)}\n\\end{equation*}\nWe have:\n\\begin{eqnarray*}\n\\mathbf{X}_{dm}^T \\mathbf{y} & = & N_1\\, y_1\\, (\\hat{\\mu}_1 - \\hat{\\mu}) + N_2\\, y_2\\, (\\hat{\\mu}_2 - \\hat{\\mu}) \\\\\n& = & \\frac{N_1}{N}\\, y_1\\, (N\\,\\hat{\\mu}_1 - N_1 \\hat{\\mu}_1 - N_2 \\hat{\\mu}_2) + \\frac{N_2}{N}\\, y_2\\, (N\\,\\hat{\\mu}_2 - N_1 \\hat{\\mu}_1 - N_2 \\hat{\\mu}_2) \\\\\n& = & \\frac{N_1 \\, N_2}{N} \\left(y_2 - y_1\\right) \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right)\n\\end{eqnarray*}\nTherefore:\n\\begin{eqnarray*}\n \\hat{\\beta} & = & \\left((N - 2)\\, \\hat{\\Sigma} + \\frac{N_1\\,N_2}{N} \n \\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)\\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)^T\\right)^{-1} \\frac{N_1 \\, N_2}{N} \\left(y_2 - y_1\\right) \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right) \\\\\n & = & \\left((N - 2)\\, \\hat{\\Sigma} + \\frac{N_1\\,N_2}{N} \n \\hat{\\Sigma}_B\\right) ^{-1} \\frac{N_1 \\, N_2}{N} \\left(y_2 - y_1\\right) \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right) \\\\\n \\hat{\\Sigma}_B & \\equiv & \\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)\\left(\\hat{\\mu}_1 - \\hat{\\mu}_2\\right)^T\n\\end{eqnarray*}\nThis proves (d) and, in particular, substituting $y_1 = - N / N_1$, $y_2 = N/N_2$\nyields the equation in the text. (c) is a direct consequence of this result.\n\n\n\\subsection*{(e)}\n\nAfter predictor demeaning:\n\\begin{equation*}\n\\hat{f} = \\hat{\\beta}_0 + \\hat{\\beta}^T x_{dm}\n\\end{equation*}\nthe OLS value $\\hat{\\beta}_0$ is the sample average of the response, which\nis zero with the encoding indicated in the text. Therefore:\n\\begin{eqnarray*}\n\\hat{f} & = & \\hat{\\beta}^T \\hat{\\Sigma}^{-1}\\left(x - \\hat{\\mu}\\right) \\\\\n & \\propto & \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1\\right)^T \\hat{\\Sigma}^{-1}\\left(x - \\frac{N_1}{N} \\hat{\\mu}_1 - \\frac{N_2}{N} \\hat{\\mu}_2\\right) \\\\\n & = & x^T \\hat{\\Sigma}^{-1} \\left(\\hat{\\mu}_2 - \\hat{\\mu}_1 \\right) - \n \\frac{N_2}{N} \\hat{\\mu}_2^T \\hat{\\Sigma}^{-1} \\hat{\\mu}_2 +\n \\frac{N_1}{N} \\hat{\\mu}_1^T \\hat{\\Sigma}^{-1} \\hat{\\mu}_1 + \n \\frac{N_2 - N_1}{N} \\hat{\\mu}_1 ^T \\hat{\\Sigma}^{-1} \\hat{\\mu}_2\n\\end{eqnarray*}\nThis expression coincides with the one for LDA only when $N_1 = N_2$.\n", "meta": {"hexsha": "ec4cb92c999d09bc53222a0480140169922e7a17", "size": 5252, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/Chapter4/ex_4_02.tex", "max_stars_repo_name": "pinoeottavio/ESLEx", "max_stars_repo_head_hexsha": "9d203da5b46c8d66ade827c237c738a35928af48", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-01-16T22:33:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-16T22:33:48.000Z", "max_issues_repo_path": "notes/Chapter4/ex_4_02.tex", "max_issues_repo_name": "pinoeottavio/ESLEx", "max_issues_repo_head_hexsha": "9d203da5b46c8d66ade827c237c738a35928af48", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/Chapter4/ex_4_02.tex", "max_forks_repo_name": "pinoeottavio/ESLEx", "max_forks_repo_head_hexsha": "9d203da5b46c8d66ade827c237c738a35928af48", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.019047619, "max_line_length": 227, "alphanum_fraction": 0.5693069307, "num_tokens": 2367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6250300162917812}} {"text": "\\subsection{Categorical limits}\\label{subsec:categorical_limits}\n\n\\begin{definition}\\label{def:comma_category}\\mcite[def. 2.3.1]{Leinster2016Basic}\n Comma categories allow us to define morphisms between morphisms, which becomes useful in, for example, the concise definition of a limit in \\fullref{def:category_of_cones/limit}.\n\n \\begin{thmenum}\n \\thmitem{def:comma_category/variable} We first prove the most general construction. Let \\( F: \\cat{A} \\to \\cat{C} \\) and \\( G: \\cat{B} \\to \\cat{C} \\) be any two functors with a common codomain. We define their \\term{comma category} \\( (F \\downarrow G) \\) as follows:\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(F \\downarrow G) \\) is the set of triples \\( (A, s, B) \\), where \\( A \\in \\cat{A} \\), \\( B \\in \\cat{B} \\) and \\( s: F(A) \\to G(B) \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( [F \\downarrow G]((A, s, B), (A', s', B')) \\) is the set of pairs\n \\begin{equation*}\n (f: A \\to A', g: B \\to B'),\n \\end{equation*}\n such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:comma_category/variable}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__comma_category.pdf}\n \\end{aligned}\n \\end{equation}\n\n \\item The \\hyperref[def:category/composition]{composition of morphisms} is their pairwise composition, analogically to \\hyperref[def:product_category]{product categories}.\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the object \\( (A, s, B) \\) is the identity pair \\( (\\id_A, \\id_B) \\).\n \\end{itemize}\n\n \\thmitem{def:comma_category/fixed} It is often the case where either \\( F \\) or \\( G \\) are constant. If, instead of \\( G \\), we are given an object \\( X \\) in \\( \\cat{C} \\), we use the constant functor \\( \\Delta_X^{\\cat{1}}: \\cat{1} \\to \\cat{C} \\) with domain the \\hyperref[def:universal_categories]{terminal category} \\( \\cat{1} \\), in order to use the comma category \\( (F \\downarrow \\Delta_X^{\\cat{1}}) \\).\n\n We can thus simplify \\fullref{def:comma_category/variable} as follows:\n\n \\begin{minipage}[t]{0.43\\textwidth}\n \\begin{equation*}\n (F \\downarrow X) \\coloneqq (F \\downarrow \\Delta_X^{\\cat{1}}).\n \\end{equation*}\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(F \\downarrow X) \\) is\n \\begin{equation*}\n \\set{ (A, s) \\given s: F(A) \\to X }\n \\end{equation*}\n\n \\item The \\hyperref[def:category/morphisms]{hom-set}\n \\begin{equation*}\n [F \\downarrow X]\\parens[\\Big]{ (A, s), (A', s') }\n \\end{equation*}\n is the set of \\( f: A \\to A' \\), such that\n \\begin{equation}\\label{eq:def:comma_category/fixed/right}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__comma_category.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{itemize}\n \\end{minipage}\n \\begin{minipage}[t]{0.43\\textwidth}\n \\begin{equation*}\n (X \\downarrow G) \\coloneqq (\\Delta_X^{\\cat{1}} \\downarrow G)\n \\end{equation*}\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(X \\downarrow G) \\) is\n \\begin{equation*}\n \\set{ (s, B) \\given s: X \\to F(B) }\n \\end{equation*}\n\n \\item The \\hyperref[def:category/morphisms]{hom-set}\n \\begin{equation*}\n [X \\downarrow G]\\parens[\\Big]{ (s, B), (s', B') }\n \\end{equation*}\n is the set of \\( g: B \\to B' \\), such that\n \\begin{equation}\\label{eq:def:comma_category/fixed/left}\n \\begin{aligned}\n \\includegraphics[page=3]{output/def__comma_category.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{itemize}\n \\end{minipage}\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:factors_through}\\mcite[133]{Knapp2016BasicAlgebra}\n Consider a morphism \\( f: A \\to B \\). We say that \\( f \\) \\term{factors through} the object \\( X \\) if there exist morphisms \\( g: A \\to X \\) and \\( h: X \\to B \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:factors_through}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__factors_through.pdf}\n \\end{aligned}\n \\end{equation}\n\n If, given \\( g \\), \\( h \\) is unique or vice versa, we say that \\( f \\) \\term{uniquely factors through} \\( X \\). In some cases, for example in \\fullref{def:zero_morphisms/morphism}, \\( f \\) there exists unique \\( g \\) and \\( h \\) and neither of them must be given beforehand.\n\\end{definition}\n\n\\begin{definition}\\label{def:category_of_cones}\n Let \\( D: \\cat{I} \\to \\cat{C} \\) be a \\hyperref[def:categorical_diagram]{diagram}.\n\n \\begin{thmenum}\n \\thmitem{def:category_of_cones/category} We define \\term{category of cones} to \\( D \\) as the \\hyperref[def:comma_category/fixed]{constant-functor comma category}\n \\begin{equation*}\n \\cat{Cone}(D) \\coloneqq \\underbrace{ (\\Delta^{\\cat{I}} \\downarrow D) }_{(\\Delta^{\\cat{I}} \\downarrow \\Delta^{\\cat{1}}_{D})},\n \\end{equation*}\n where \\( \\Delta^{\\cat{I}}: \\cat{C} \\to [\\cat{I}, \\cat{C}] \\) is the \\( \\cat{I} \\)-shaped \\hyperref[def:diagonal_functor]{diagonal functor} on \\( \\cat{C} \\).\n\n \\thmitem{def:category_of_cones/cone} A \\( D \\)-\\term{cone} is simply a member of \\( \\cat{Cone}(D) \\).\n\n Explicitly, a cone with \\term{vertex} \\( A \\in \\cat{C} \\) is a pair \\( (\\Delta^{\\cat{I}}_A, \\alpha) \\), where \\( \\Delta^{\\cat{I}}_A \\) is the constant functor at \\( A \\) and \\( \\alpha \\) is a natural transformation from \\( \\Delta^{\\cat{I}}_A \\) to \\( D \\).\n\n Even more explicitly, a cone is a family of morphisms\n \\begin{equation}\\label{eq:def:category_of_cones/cone}\n \\seq{ \\alpha_k: A \\to D(k) }_{k \\in \\cat{I}}.\n \\end{equation}\n satisfying a simplified naturality condition (compared to \\eqref{eq:def:natural_transformation/diagram}). For every morphism \\( u: k \\to m \\) in \\( \\cat{I} \\), the following diagram must commute:\n \\begin{equation}\\label{eq:def:category_of_cones/cone_nat}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__category_of_cones.pdf}\n \\end{aligned}\n \\end{equation}\n\n Note that this diagram is very different from \\eqref{eq:def:comma_category/fixed/right}, they merely look similar.\n\n \\thmitem{def:category_of_cones/limit} A \\term{limit cone} of the diagram \\( D \\) is a \\hyperref[def:universal_objects/terminal]{terminal object} of the cone category \\( \\cat{Cone}(D) \\).\n\n Explicitly, \\( (L, \\lambda) \\) is limit cone if, for every cone \\( (A, \\alpha) \\), there exists a unique \\hyperref[eq:def:comma_category/fixed/right]{cone morphism} \\( l_A: (A, \\alpha) \\to (L, \\lambda) \\).\n\n Even more explicitly, \\( (L, \\lambda) \\) is a limit cone if it satisfies the following \\hyperref[rem:limit_universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every cone \\( (A, \\alpha) \\), there exists a unique morphism \\( l_A: A \\to L \\) such that following diagram commutes for every index morphism \\( u: k \\to m \\):\n \\begin{equation}\\label{eq:def:category_of_cones/limit}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__category_of_cones.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Thus, \\( \\alpha_k \\) \\hyperref[def:factors_through]{uniquely factors through} \\( L \\) for every \\( k \\in \\cat{I} \\). Furthermore, \\( l_A \\) is compatible with morphisms in \\( \\cat{I} \\) and does not depend on \\( k \\).\n\n From \\fullref{thm:def:universal_objects/terminal} and \\fullref{thm:universal_objects_as_adjunctions/terminal} it follows that a limit, if it exists, is unique up to a unique isomorphism.\n\n Without further context, we usually refer to \\( L \\) as the limit vertex and \\( (L, \\lambda) \\) as the limit cone. By \\enquote{the limit}, we usually mean the vertex \\( L \\).\n\n \\thmitem{def:category_of_cones/cocone} \\hyperref[thm:categorical_principle_of_duality]{Dually}, a \\( D \\)-\\term{cocone} with vertex \\( A \\) is a family of morphisms\n \\begin{equation}\\label{eq:def:category_of_cones/cocone}\n \\seq{ \\alpha_k: D(k) \\to A }_{k \\in \\cat{I}}.\n \\end{equation}\n satisfying the naturality condition that for every morphism \\( u: k \\to m \\) in \\( \\cat{I} \\), the following diagram must commute:\n \\begin{equation}\\label{eq:def:category_of_cones/cocone_nat}\n \\begin{aligned}\n \\includegraphics[page=3]{output/def__category_of_cones.pdf}\n \\end{aligned}\n \\end{equation}\n\n \\thmitem{def:category_of_cones/colimit} A \\term{colimit cocone} of the diagram \\( D \\) is an \\hyperref[def:universal_objects/initial]{initial object} of the cocone category \\( (D \\downarrow \\Delta) \\).\n\n There are two major differences compared to limits: a colimit cocone is an initial object, not a terminal object, and its underlying comma category is \\( (D \\downarrow \\Delta) \\), not \\( (\\Delta^{\\cat{I}} \\downarrow D) \\).\n\n The analogous diagram to \\eqref{eq:def:category_of_cones/limit} is exactly its \\hyperref[thm:categorical_principle_of_duality]{opposite}:\n \\begin{equation}\\label{eq:def:category_of_cones/colimit}\n \\begin{aligned}\n \\includegraphics[page=4]{output/def__category_of_cones.pdf}\n \\end{aligned}\n \\end{equation}\n\n In particular, \\( \\alpha_k \\) \\hyperref[def:factors_through]{uniquely factors through} \\( L \\) for every \\( k \\in \\cat{I} \\).\n\n Without further context, we usually refer to \\( L \\) as the colimit vertex and \\( (L, \\lambda) \\) as the colimit cocone. By \\enquote{the colimit}, we usually mean the vertex \\( L \\).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:categorical_limit_duality}\n For every \\hyperref[def:category_of_cones/cone]{cone} \\( (A, \\alpha) \\) of the \\hyperref[def:categorical_diagram]{diagram} \\( D \\) in \\( \\cat{C} \\), \\( (A, \\alpha^{\\opcat}) \\) is a \\hyperref[def:category_of_cones/cone]{cocone} of \\( D^{\\opcat} \\) in the \\hyperref[def:opposite_category]{opposite category} \\( \\cat{C}^{\\opcat} \\).\n\n Even more, for every \\hyperref[def:category_of_cones/limit]{limit} \\( (L, \\lambda) \\) of \\( D \\) in \\( \\cat{C} \\), \\( (L, \\lambda^{\\opcat}) \\) is a \\hyperref[def:category_of_cones/colimit]{colimit} of \\( D^{\\opcat} \\) in \\( \\cat{C}^{\\opcat} \\).\n\n This is part of the duality principles listed in \\fullref{thm:categorical_principle_of_duality}.\n\\end{proposition}\n\\begin{proof}\n Note that the defining diagrams \\eqref{eq:def:category_of_cones/cone}, \\eqref{eq:def:category_of_cones/cone_nat} and \\eqref{eq:def:category_of_cones/limit} are dual to \\eqref{eq:def:category_of_cones/cocone}, \\eqref{eq:def:category_of_cones/cocone_nat} and \\eqref{eq:def:category_of_cones/colimit}.\n\\end{proof}\n\n\\begin{lemma}\\label{thm:categorical_limit_uniqueness_lemma}\n Any two limits (resp. colimits) of a diagram are isomorphic.\n\n We prove a stronger result in \\fullref{thm:categorical_limit_uniqueness}.\n\\end{lemma}\n\\begin{proof}\n Let \\( (L', \\lambda') \\) and \\( (L^\\dprime, \\lambda^\\dprime) \\) be two limit cones over the diagram \\( D: \\cat{I} \\to \\cat{C} \\). The definition \\eqref{eq:def:category_of_cones/limit} of a limit implies that\n \\begin{equation}\\label{eq:thm:categorical_limit_uniqueness}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__categorical_limit_uniqueness.pdf}\n \\end{aligned}\n \\end{equation}\n commutes.\n\n Therefore, the limits \\( L' \\) and \\( L^\\dprime \\) are isomorphic.\n\n Now let \\( (L', \\lambda') \\) and \\( (L^\\dprime, \\lambda^\\dprime) \\) be colimit cocones. By \\fullref{thm:categorical_limit_duality}, \\( (L', \\lambda'^{\\opcat}) \\) and \\( (L^\\dprime, {\\lambda^\\dprime}^{\\opcat}) \\) are limits in the opposite category and are thus isomorphic. By \\fullref{thm:morphism_invertibility_duality}, the colimits are isomorphic.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:categorical_limit_is_adjoint}\n Suppose that, for a given category \\( \\cat{C} \\), the limits over all \\( \\cat{I} \\)-shaped diagrams exist. Denote by \\( \\lim(D) \\) the vertex of the limiting cone of the diagram \\( D \\).\n\n Given a natural transformation \\( \\alpha: D \\to E \\) between diagrams, the diagram \\eqref{eq:thm:categorical_limit_is_adjoint/c_triangle} defining a limit uniquely determines a morphism \\( \\lim(D) \\) to \\( \\lim(E) \\). Denote this morphism by \\( \\lim(\\alpha) \\).\n\n We have defined a functor\n \\begin{equation*}\n \\lim: [\\cat{I}, \\cat{C}] \\to \\cat{C}.\n \\end{equation*}\n\n This functor is \\hyperref[def:category_adjunction]{right adjoint} to the diagonal functor\n \\begin{equation*}\n \\Delta: \\cat{C} \\to [\\cat{I}, \\cat{C}]\n \\end{equation*}\n\n \\hyperref[thm:categorical_principle_of_duality]{Dually}, the colimit functor\n \\begin{equation*}\n \\co\\lim: [\\cat{I}, \\cat{C}] \\to \\cat{C}\n \\end{equation*}\n is left adjoint to \\( \\Delta \\).\n\\end{proposition}\n\\begin{proof}\n It is sufficient to prove this for limits since the statement for colimits follows from the duality principles \\fullref{thm:category_adjunction_duality} and \\fullref{thm:categorical_limit_duality}. The unit \\( \\eta: \\id_{\\cat{C}} \\to [{\\lim} \\bincirc \\Delta] \\) of the adjunction\n \\begin{equation*}\n \\Delta \\dashv \\lim\n \\end{equation*}\n is the unique morphism from an object \\( A \\) of \\( \\cat{C} \\) to the limit of its constant diagram \\( \\Delta_A: \\cat{I} \\to \\cat{C} \\) such that \\eqref{eq:def:category_of_cones/limit} commutes. The counit is more complicated because its components are themselves natural transformations:\n \\begin{equation*}\n \\begin{aligned}\n \\varepsilon: &[\\Delta \\bincirc \\lim] \\Rightarrow \\id_{[\\cat{I}, \\cat{C}]} \\\\\n \\varepsilon_D: &\\Delta(\\lim D) \\Rightarrow D \\\\\n \\varepsilon_{D,k}: &\\lim(D) \\to D(k),\n \\end{aligned}\n \\end{equation*}\n where \\( \\varepsilon_{D,k} \\) are the projections of the limit.\n\n For any diagram \\( D: \\cat{I} \\to \\cat{C} \\), the following triangle commutes:\n \\begin{equation}\\label{eq:thm:categorical_limit_is_adjoint/ic_triangle}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__categorical_limit_is_adjoint.pdf}\n \\end{aligned}\n \\end{equation}\n\n Indeed,\n \\begin{equation*}\n \\eta_{\\lim(D)}: \\lim(D) \\to \\smash{ \\overbrace{\\lim(\\Delta_{\\lim(D)})}^{[{\\lim} \\bincirc \\Delta \\bincirc {\\lim}](D)} }\n \\end{equation*}\n is the unique morphism such that \\eqref{eq:def:category_of_cones/limit} commutes, and (somewhat) similarly for\n \\begin{equation*}\n \\lim(\\varepsilon_D): [{\\lim} \\bincirc \\Delta \\bincirc {\\lim}](D) \\to \\lim(D).\n \\end{equation*}\n\n It follows that both \\( \\lim(D) \\) and \\( [{\\lim} \\bincirc \\Delta \\bincirc {\\lim}](D) \\) are limits over the same diagram. By \\fullref{thm:categorical_limit_is_adjoint}, they are isomorphic, and hence \\eqref{eq:thm:categorical_limit_is_adjoint/ic_triangle} commutes.\n\n Also, for any object \\( A \\) in \\( \\cat{C} \\), the following triangle also commutes:\n \\begin{equation}\\label{eq:thm:categorical_limit_is_adjoint/c_triangle}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__categorical_limit_is_adjoint.pdf}\n \\end{aligned}\n \\end{equation}\n\n Indeed,\n \\begin{equation*}\n \\Delta(\\eta_A) = \\seq{ \\eta_A: A \\to \\lim(\\Delta_A) }_{k \\in \\cat{I}}\n \\end{equation*}\n is the constant family consisting of \\( \\eta_A \\) and\n \\begin{equation*}\n \\varepsilon_{\\Delta(A)} = \\seq{ \\lambda_k^{\\Delta_A}: \\lim(\\Delta_A) \\to \\underbrace{\\Delta_A(k)}_{A} }_{k \\in \\cat{I}}.\n \\end{equation*}\n is a constant family of the single projection of the limit to \\( A \\). From the commutativity of \\eqref{eq:def:category_of_cones/limit} it follows that \\eqref{eq:thm:categorical_limit_is_adjoint/c_triangle} also commutes.\n\\end{proof}\n\n\\begin{corollary}\\label{thm:categorical_limit_uniqueness}\n A limit (resp. colimit) of a diagram, if it exists, is unique up to a unique isomorphism.\n\n This statement strengthens \\fullref{thm:categorical_limit_uniqueness_lemma}.\n\\end{corollary}\n\\begin{proof}\n Follows from \\fullref{thm:functor_adjoint_uniqueness} and \\fullref{thm:categorical_limit_is_adjoint}.\n\\end{proof}\n\n\\begin{remark}\\label{rem:limit_universal_mapping_property}\n The limit diagram \\eqref{eq:def:category_of_cones/limit} may seem unrelated to the universal mapping properties discussed in \\fullref{rem:universal_mapping_property}, however it is actually a special case.\n\n Suppose that, for a given category \\( \\cat{C} \\), the limits over all \\( \\cat{I} \\)-shaped diagrams exist and fix a diagram \\( D: \\cat{I} \\to \\cat{C} \\). Consider the functors\n \\begin{align*}\n \\lim: &[\\cat{I}, \\cat{C}] \\to \\cat{C} \\\\\n \\Delta: &\\cat{C} \\to [\\cat{I}, \\cat{C}]\n \\end{align*}\n discussed in \\fullref{thm:categorical_limit_is_adjoint}. We have established that \\( \\Delta \\dashv \\lim \\).\n\n For every diagram \\( D: \\cat{I} \\to \\cat{C} \\), there exist unique up to a unique isomorphism object \\( \\lim(D) \\) in \\( \\cat{C} \\) and canonical projection map \\( \\lambda: [\\Delta \\bincirc {\\lim}(D)] \\to D \\) satisfying the following universal mapping property:\n \\begin{displayquote}\n For every object \\( A \\) in \\( \\cat{C} \\) and every natural transformation \\( \\alpha: \\Delta(A) \\Rightarrow D \\), there exists a unique morphism \\( l: A \\to \\lim(D) \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:rem:limit_universal_mapping_property/ic_triangle}\n \\begin{aligned}\n \\includegraphics[page=1]{output/rem__limit_universal_mapping_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n For a fixed index object \\( k \\in \\cat{I} \\), this becomes:\n \\begin{equation}\\label{eq:rem:limit_universal_mapping_property/c_triangle_basic}\n \\begin{aligned}\n \\includegraphics[page=2]{output/rem__limit_universal_mapping_property.pdf}\n \\end{aligned}\n \\end{equation}\n\n The defining diagram \\eqref{eq:def:category_of_cones/limit} of a limit simply encodes the cone naturality condition \\eqref{eq:def:category_of_cones/cone_nat} into \\eqref{eq:rem:limit_universal_mapping_property/c_triangle_basic}.\n\n Except for being simpler to check, limits defined via universal mapping properties have the advantage (compared to adjoint functors) that limits can exist for some diagrams and not for others.\n\n The construction for colimits is dual.\n\\end{remark}\n\n\\begin{proposition}\\label{thm:limits_of_empty_diagram}\n The \\hyperref[def:category_of_cones/limit]{limits} of an empty diagram over \\( \\cat{C} \\) are the \\hyperref[def:universal_objects/terminal]{terminal objects} and the \\hyperref[def:category_of_cones/colimit]{colimits} are the \\hyperref[def:universal_objects/initial]{initial object}.\n\n Compare this result with \\fullref{thm:limits_of_identity_functor}.\n\\end{proposition}\n\\begin{proof}\n A cone of the empty diagram is simply an object of \\( \\cat{C} \\). A limit cone is then an object \\( L \\) such that every other object \\( A \\) has a unique morphism \\( l_A: A \\to L \\). This is precisely the definition of a terminal object.\n\n The statement for colimits follows by \\hyperref[thm:categorical_principle_of_duality]{duality}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:limits_of_identity_functor}\n Fix an arbitrary category \\( \\cat{C} \\).\n\n \\begin{thmenum}\n \\thmitem{thm:limits_of_identity_functor/initial_object_ks_limit} If \\( I \\) is an \\hyperref[def:universal_objects/initial]{initial object} of \\( \\cat{C} \\), \\( (I, \\xi) \\) is a limit cone of the \\hyperref[eq:def:category_of_small_categories/identity]{identity functor} \\( \\id_{\\cat{C}} \\), where\n \\begin{equation*}\n \\xi \\coloneqq \\seq{ \\xi_A: I \\to A }_{A \\in \\cat{C}}\n \\end{equation*}\n is the family of unique morphisms with domain \\( I \\).\n\n \\thmitem{thm:limits_of_identity_functor/limit_ks_knitial_object} Conversely, if \\( (L, \\lambda) \\) is a limit cone of the identity, then \\( L \\) is an initial object.\n \\end{thmenum}\n\n \\hyperref[thm:categorical_principle_of_duality]{Dually}, by \\fullref{thm:universal_object_duality} and \\fullref{thm:categorical_limit_duality}, the cocones (and colimits) of the identity functor are the terminal objects.\n\n Compare this result with \\fullref{thm:limits_of_empty_diagram}.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:limits_of_identity_functor/initial_object_ks_limit} For an initial object \\( I \\) with morphisms \\( \\xi \\), and for any morphism \\( f: B \\to C \\), from the uniqueness of the arrows in \\( \\xi \\) it follows that \\( \\xi_C = f \\bincirc \\xi_B \\). Thus, the following naturality diagram commutes:\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/initial_object_ks_limit/nat}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore, \\( (I, \\xi) \\) is a cone.\n\n Now let \\( (A, \\alpha) \\) be another cone. The naturality of \\( \\alpha \\) implies that the following diagram commutes for every object \\( B \\):\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/initial_object_ks_limit/half_limit}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n In particular, since \\( \\xi_I = \\id_I \\), for every morphism \\( g: A \\to I \\) we have\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/initial_object_ks_limit/cone_morphism_uniqueness}\n \\begin{aligned}\n \\includegraphics[page=3]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n which shows that \\( \\alpha_I = g \\) and thus \\( \\alpha_I \\) is the unique morphism from \\( A \\) to \\( I \\).\n\n Therefore, for any morphism \\( f: B \\to C \\), the following diagram commutes:\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/initial_object_ks_limit/limit}\n \\begin{aligned}\n \\includegraphics[page=4]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n This is precisely the defining diagram for a limit cone. Therefore, \\( (I, \\xi) \\) is a limit cone.\n\n \\SubProofOf{thm:limits_of_identity_functor/limit_ks_knitial_object} Let \\( (L, \\lambda) \\) be a limit cone of the identity. The component \\( \\lambda_B \\) of the natural transformation \\( \\lambda \\) is a morphism from \\( L \\) to \\( B \\). In order for \\( L \\) to be an initial object, these morphisms must be unique.\n\n Since \\( (L, \\lambda) \\) is a limit cone, for any cone \\( (A, \\alpha) \\) and any morphism \\( f: B \\to C \\), there exists a unique morphism \\( l_A: A \\to L \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/limit_ks_knitial_object/limit}\n \\begin{aligned}\n \\includegraphics[page=5]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n In particular, there exists a unique map \\( l_L: L \\to L \\) for the cone \\( (L, \\lambda) \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/limit_ks_knitial_object/endomorphism_uniqueness}\n \\begin{aligned}\n \\includegraphics[page=6]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n Since \\eqref{eq:def:thm:limits_of_identity_functor/limit_ks_knitial_object/endomorphism_uniqueness} commutes with \\( \\id_L \\) instead of \\( l_L \\), by uniqueness it follows that \\( l_L = \\id_L \\). Since, by the naturality of \\( \\lambda \\), \\( \\lambda_L \\) also satisfies this condition, \\( \\lambda_L = \\id_L \\).\n\n From the naturality of \\( \\lambda \\), for any map \\( f: L \\to C \\) it follows that the following diagram commutes:\n \\begin{equation}\\label{eq:def:thm:limits_of_identity_functor/limit_ks_knitial_object/morphism_uniqueness}\n \\begin{aligned}\n \\includegraphics[page=7]{output/thm__limits_of_identity_functor.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore, \\( f = \\lambda_B \\). Since \\( f \\) was an arbitrary morphism with domain \\( L \\), we conclude that \\( L \\) has a unique morphism to every object in \\( \\cat{C} \\). Hence, \\( L \\) is an initial object.\n\\end{proof}\n\n\\begin{example}\\label{ex:limits_of_partially_ordered_set}\n We will show that limits and colimits correspond to suprema and infima of partially ordered sets.\n\n Let \\( (P, \\leq) \\) be a \\hyperref[def:partially_ordered_set]{partially ordered set} and \\( \\cat{P} \\) be its corresponding \\hyperref[def:thin_category]{thin} \\hyperref[def:skeletal_category]{skeletal} category. The correspondence is discussed in \\fullref{thm:order_category_isomorphism/preordered}.\n\n The image of a diagram \\( D: \\cat{I} \\to \\cat{P} \\) is a pair \\( (A, R) \\), where \\( A \\) is the set of objects \\( D(\\obj(\\cat{I})) \\) in the image \\( D(\\cat{I}) \\) and \\( R \\) is a subrelation of \\( \\leq \\). This relation is reflexive, however it may not even be a preorder as shown in \\fullref{ex:functor_image_not_a_category}.\n\n Conversely, every subset \\( A \\subseteq P \\) with a corresponding category \\( \\cat{A} \\) is given by the diagram \\( \\Iota_A \\), where the \\hyperref[def:subcategory]{inclusion functor} \\( \\Iota_A: \\cat{A} \\to \\cat{P} \\).\n\n A cone of a diagram \\( D: \\cat{I} \\to \\cat{P} \\) is a \\hyperref[def:partially_ordered_set_extremal_points/upper_and_lower_bounds]{lower bound} of the set \\( D(\\obj(\\cat{I})) \\). The relation induced by \\( D \\) does not matter here.\n\n Indeed, the cone \\( (x, R) \\) in \\( \\cat{P} \\) consists of morphisms with domain \\( x \\). That is, \\( R \\) is a subrelation of \\( \\leq \\) whose first component is \\( x \\). Clearly then \\( x \\) is a lower bound of the set \\( D(\\obj(\\cat{I})) \\).\n\n For the limiting cone \\( (y, S) \\), it holds that \\( x \\leq y \\) for every other cone \\( (x, R) \\). Both \\( R \\) and \\( S \\) are subrelations of \\( \\leq \\) with the same second components. Thus, \\( y \\) is the greatest lower bound of \\( D(\\obj(\\cat{I})) \\).\n\n \\hyperref[thm:categorical_principle_of_duality]{Dually}, cocones are upper bounds and colimits are suprema.\n\n As mentioned, the relation induced by the diagram \\( D \\) does not actually matter. Therefore, we may choose, without loss of generality, \\( D \\) to be a \\hyperref[def:discrete_category]{discrete category}. It then follows that infima and suprema correspond to \\hyperref[def:discrete_category_limits]{products} and \\hyperref[def:discrete_category_limits]{coproduct}, however what we have shown here is more general.\n\\end{example}\n\n\\begin{definition}\\label{def:direct_and_inverse_limits}\n We will define limits and colimits of diagrams over infinite \\hyperref[def:partially_ordered_set_chain_and_antichain]{chains} of integers. These notions predate limits and colimits, which explains why their names may seem inconsistent with other limits and colimits.\n\n \\begin{thmenum}\n \\thmitem{def:direct_and_inverse_limits/direct}\\mcite[exmpl. 5.2.15]{Leinster2016Basic} Consider the category \\hyperref[thm:order_category_isomorphism]{induced} by the positive integers\n \\begin{equation}\\label{eq:def:direct_and_inverse_limits/direct}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n\n A \\hyperref[def:category_of_cones/colimit]{colimit} over a diagram of this shape is called a \\term{direct limit}. In this context, the diagram itself is sometimes called a \\term{direct system}.\n\n \\thmitem{def:direct_and_inverse_limits/inverse}\\mcite[exmpl. 5.1.21(d)]{Leinster2016Basic} Now consider the opposite category\n \\begin{equation}\\label{eq:def:direct_and_inverse_limits/inverse}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n\n Somewhat confusingly, a \\hyperref[def:category_of_cones/limit]{limit} of a diagram of this shape is called an \\term{inverse limit}. In this context, the diagram itself is sometimes called an \\term{inverse system}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:def:direct_and_inverse_limits}\n We will list several examples of \\hyperref[def:direct_and_inverse_limits/direct]{direct} and \\hyperref[def:direct_and_inverse_limits/inverse]{inverse} limits.\n\n \\begin{thmenum}\n \\thmitem{ex:def:direct_and_inverse_limits/vector_space_direct} Consider the chain of \\hyperref[def:module_of_tuples]{tuple vector spaces}\n \\begin{equation}\\label{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n where \\( \\iota_n^m: \\BbbK^n \\to \\BbbK^m \\) is simply the canonical inclusion map.\n\n Let \\( \\BbbK_0^\\infty \\) be the vector space of all sequences in \\( \\BbbK \\) with only finitely many nonzero elements. For each positive integer \\( n \\), denote by \\( \\iota_n^\\infty: \\BbbK^n \\to \\BbbK_0^\\infty \\) the canonical inclusion.\n\n Then \\( (\\BbbK_0^\\infty, \\iota) \\) is a \\hyperref[def:category_of_cones/cocone]{cocone} of \\eqref{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}. We will show that it is a colimit cocone, i.e. that \\( \\BbbK_0^\\infty \\) is a \\hyperref[def:direct_and_inverse_limits]{direct limit} of \\eqref{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}.\n\n Let \\( (A, \\alpha) \\) be another cocone. We want to define a linear map \\( l_A: \\BbbK_0^\\infty \\to A \\) so that the following diagram commutes:\n \\begin{equation}\\label{eq:ex:def:direct_and_inverse_limits/vector_space_direct/limit}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n\n For every vector \\( x \\in \\BbbK^m \\), we must have\n \\begin{equation*}\n \\alpha_n(x) = l_A(\\iota_n^\\infty(x)).\n \\end{equation*}\n\n This implies the obvious definition where, given a vector \\( x \\in \\BbbK_0^\\infty \\) whose greatest nonzero element has index \\( m \\), we define\n \\begin{equation*}\n l_A(x) \\coloneqq \\alpha_m(x).\n \\end{equation*}\n\n This map is well-defined because the compatibility with inclusion maps guarantees that \\( \\alpha_{m+1}(x) = \\iota_{m+1}(\\alpha^{m+1}_m(x)) \\), i.e. the result obtained by using \\( \\alpha_m \\) and \\( \\alpha_{m+1} \\) is the same.\n\n Therefore, \\( (\\BbbK_0^\\infty, \\iota) \\) is a direct limit of the diagram \\eqref{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}.\n\n \\thmitem{ex:def:direct_and_inverse_limits/vector_space_inverse} \\hyperref[thm:categorical_principle_of_duality]{Dually}, consider the chain\n \\begin{equation}\\label{eq:ex:def:direct_and_inverse_limits/vector_space_inverse/diagram}\n \\begin{aligned}\n \\includegraphics[page=3]{output/ex__def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n where\n \\begin{equation*}\n \\begin{aligned}\n &\\pi_n^m: \\BbbK^m \\to \\BbbK^n \\\\\n &\\pi_n^m(x_1, \\ldots, x_n, x_{k+1}, \\ldots, x_m) \\coloneqq (x_1, \\ldots, x_n).\n \\end{aligned}\n \\end{equation*}\n\n Let \\( \\BbbK^\\infty \\) be the vector space of all sequences in \\( \\BbbK \\) and, for each positive integer \\( n \\), define \\( \\pi_n^\\infty \\) as a truncation in the obvious way.\n\n Then \\( (\\BbbK^\\infty, \\pi^\\infty) \\) is a \\hyperref[def:category_of_cones/cone]{cone} of \\eqref{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}. We will show that it is a limit cone, i.e. that \\( \\BbbK^\\infty \\) is a \\hyperref[def:direct_and_inverse_limits]{direct limit} of \\eqref{eq:ex:def:direct_and_inverse_limits/vector_space_direct/diagram}.\n\n Let \\( (A, \\alpha) \\) be another cone. We want to define \\( l_A: A \\to \\BbbK^\\infty \\) so that the following diagram commutes:\n \\begin{equation}\\label{eq:ex:def:direct_and_inverse_limits/vector_space_inverse/limit}\n \\begin{aligned}\n \\includegraphics[page=4]{output/ex__def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n\n For every \\( a \\in A \\), we must have\n \\begin{equation*}\n \\alpha_n(a) = \\pi_n^\\infty(l_A(a)),\n \\end{equation*}\n which implies the obvious definition where the \\( n \\)-th coordinate of the vector \\( l_A(a) \\) is\n \\begin{equation*}\n [l_A(a)]_n \\coloneqq [\\alpha_n(a)]_n.\n \\end{equation*}\n\n That is, \\( l_A(a) \\) is a sequence whose \\( n \\)-th coordinate is the \\( n \\)-th coordinate \\( \\alpha_n(a) \\). This is well-defined because, for \\( m > n \\), we have\n \\begin{equation*}\n \\pi_m^n \\bincirc \\alpha_m = \\alpha_n.\n \\end{equation*}\n\n \\thmitem{ex:def:direct_and_inverse_limits/generalized_intersection} In \\hyperref[def:concrete_category]{concrete categories}, \\hyperref[def:direct_and_inverse_limits/inverse]{inverse limits} are generalizations of \\hyperref[thm:zfc_existence_theorems/arbitrary_intersection]{set intersections}.\n\n As an example, consider the chain of sets\n \\begin{equation}\\label{eq:ex:inverse_limit_as_intersection/diagram}\n \\begin{aligned}\n \\includegraphics[page=5]{output/ex__def__direct_and_inverse_limits.pdf}\n \\end{aligned}\n \\end{equation}\n where \\( A_{k+1} \\subseteq A_k \\) for every positive integer \\( k \\) and \\( \\iota_k^m: A_m \\to A_k \\) is simply the canonical inclusion map.\n\n Then the intersection \\( \\bigcap_{k=1}^\\infty A_k \\) along with its inclusion maps is a limit of \\eqref{eq:ex:inverse_limit_as_intersection/diagram}. This is a consequence of the discussion in \\fullref{ex:limits_of_partially_ordered_set}.\n\n We can replace the inclusion maps \\( \\iota_k^m \\) with other injective functions, or even non-injective functions. In this case, we would obtain a \\enquote{generalized intersection}.\n \\end{thmenum}\n\\end{example}\n\n\\begin{definition}\\label{def:discrete_category_limits}\n Fix an arbitrary category \\( \\cat{C} \\). A \\term{product} in \\( \\cat{C} \\) is a \\hyperref[def:category_of_cones/limit]{limit} over a diagram \\( D: \\cat{I} \\to \\cat{C} \\), whose domain \\( \\cat{I} \\) is a \\hyperref[def:discrete_category]{discrete category}. \\hyperref[thm:categorical_principle_of_duality]{Dually}, a \\term{coproduct} or \\term{sum} in \\( \\cat{C} \\) is a colimit of \\( D \\).\n\n It will be convenient for us to speak about the product of an \\hyperref[def:cartesian_product/indexed_family]{indexed family} \\( \\seq{ X_k }_{k \\in \\mscrK} \\) of objects in \\( \\cat{C} \\).\n\n \\begin{minipage}[t]{0.47\\textwidth}\n A cone with vertex \\( A \\) consists of morphisms with signatures\n \\begin{equation*}\n \\alpha = \\seq{ \\alpha_k: A \\to X_k }.\n \\end{equation*}\n \\end{minipage}\n \\hfill\n \\begin{minipage}[t]{0.47\\textwidth}\n A cocone with vertex \\( A \\) consists of morphisms with signatures\n \\begin{equation*}\n \\alpha = \\seq{ \\alpha_k: X_k \\to A }.\n \\end{equation*}\n \\end{minipage}\n\n \\begin{minipage}[t]{0.47\\textwidth}\n A product \\hyperref[def:category_of_cones/cone]{cone} \\( (\\prod_{k \\in \\mscrK} X_k, \\pi) \\) satisfies the following \\hyperref[rem:limit_universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every cone \\( (A, \\alpha) \\), there exists a unique morphism\n \\begin{equation*}\n l_A: A \\to \\prod_{k \\in \\mscrK} X_k,\n \\end{equation*}\n such that the following diagram commutes:\n \\end{displayquote}\n \\begin{equation}\\label{eq:def:discrete_category_limits/product}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__discrete_category_limits.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{minipage}\n \\hfill\n \\begin{minipage}[t]{0.47\\textwidth}\n A coproduct \\hyperref[def:category_of_cones/cocone]{cocone} \\( (\\coprod_{k \\in \\mscrK} X_k, \\iota) \\) satisfies the following \\hyperref[rem:limit_universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every cone \\( (A, \\alpha) \\), there exists a unique morphism\n \\begin{equation*}\n l_A: A \\to \\coprod_{k \\in \\mscrK} X_k,\n \\end{equation*}\n such that the following diagram commutes:\n \\end{displayquote}\n \\begin{equation}\\label{eq:def:discrete_category_limits/coproduct}\n \\begin{aligned}\n \\includegraphics[page=3]{output/def__discrete_category_limits.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{minipage}\n\n \\begin{minipage}[t]{0.47\\textwidth}\n We call the morphism \\( \\pi_k \\) the \\term{canonical projection} of the product onto \\( X_k \\), even though it may not be a surjective function.\n \\end{minipage}\n \\hfill\n \\begin{minipage}[t]{0.47\\textwidth}\n We call the morphism \\( \\iota_k \\) the \\term{canonical inclusion} of \\( X_k \\) into the coproduct, even though it may not be an injective function.\n \\end{minipage}\n \\medskip\n\n From \\fullref{thm:limits_of_empty_diagram} it follows that the product (resp. coproduct) of an empty family is a terminal (resp. initial) object of \\( \\cat{C} \\).\n\n As in the case of general limits, we call \\( \\prod_{k \\in \\mscrK} X_k \\) \\hi{the} product of the family \\( X \\).\n\n In the case of only two objects, their product is given by the following diagram:\n \\begin{equation}\\label{eq:def:discrete_category_limits/product/binary}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__discrete_category_limits.pdf}\n \\end{aligned}\n \\end{equation}\n and their coproduct by\n \\begin{equation}\\label{eq:def:discrete_category_limits/coproduct/binary}\n \\begin{aligned}\n \\includegraphics[page=4]{output/def__discrete_category_limits.pdf}\n \\end{aligned}\n \\end{equation}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:discrete_category_limits_in_set}\n The \\hyperref[def:discrete_category_limits]{product} in the category \\hyperref[def:category_of_small_sets]{\\( \\ucat{Set} \\)} of \\( \\mscrU \\)-small sets of a family \\( \\mscrA = \\seq{ A_k }_{k \\in \\mscrK} \\) is their \\hyperref[def:cartesian_product/product]{Cartesian product} \\( \\prod_{k \\in \\mscrK} A_k \\) and the \\hyperref[def:discrete_category_limits]{coproduct} is their \\hyperref[def:disjoint_union]{disjoint union} \\( \\coprod_{k \\in \\mscrK} A_k \\).\n\\end{proposition}\n\\begin{proof}\n \\SubProof{Proof for products} Consider the Cartesian product\n \\begin{equation*}\n L \\coloneqq \\prod_{k \\in \\mscrK} A_k = \\set*{ f: A \\to \\bigcup_{k \\in \\mscrK} A_k \\given* \\qforall {k \\in \\mscrK} k \\in A_k }.\n \\end{equation*}\n\n Since \\( \\mscrU \\) is a model of \\logic{ZFC} (with or without the axiom of infinity), hence the product is an object of \\( \\ucat{Set} \\). Define the projection morphisms\n \\begin{equation*}\n \\pi_k(f) \\coloneqq f(k).\n \\end{equation*}\n\n Then \\( (L, \\pi) \\) is a cone for (some diagram forming) \\( \\mscrA \\). Let \\( (B, \\beta) \\) also be a cone and consider the following diagram:\n \\begin{equation}\\label{eq:thm:discrete_category_limits_in_set/limit}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__discrete_category_limits_in_set.pdf}\n \\end{aligned}\n \\end{equation}\n\n We want to define a function \\( l_B: B \\to L \\) such that\n \\begin{equation*}\n \\underbrace{ \\pi_k(l_B(b)) }_{[l_B(b)](k)} = \\beta_k(b)\n \\end{equation*}\n for every \\( b \\in B \\) and \\( k \\in \\mscrK \\). This is uniquely determined by \\( b \\) and \\( k \\), hence\n \\begin{equation*}\n l_B(b) \\coloneqq \\seq{ \\beta_k(b) }_{k \\in \\mscrK}.\n \\end{equation*}\n\n The cone \\( (B, \\beta) \\) was chosen randomly, and we showed that there exists a unique morphism \\( l_B: B \\to L \\) such that \\eqref{eq:thm:discrete_category_limits_in_set/limit} commutes. Therefore, \\( (L, \\pi) \\) is a categorical product of the family \\( \\mscrA \\).\n\n \\SubProof{Proof for coproducts} Now consider the disjoint union\n \\begin{equation*}\n L \\coloneqq \\coprod_{k \\in \\mscrK} A_k = \\set{ (k, a) \\given k \\in \\mscrK \\T{and} a \\in A_k }\n \\end{equation*}\n and the inclusions\n \\begin{equation*}\n \\iota_k\\parens[\\Big]{ (k, a) } = a.\n \\end{equation*}\n\n We must show that, for any other cocone \\( (B, \\beta) \\), the following diagram commutes:\n \\begin{equation}\\label{eq:thm:discrete_category_limits_in_set/colimit}\n \\begin{aligned}\n \\includegraphics[page=2]{output/thm__discrete_category_limits_in_set.pdf}\n \\end{aligned}\n \\end{equation}\n\n Analogously to products, the condition\n \\begin{equation*}\n \\underbrace{ l_B(\\iota_k(a)) }_{l_B(k, a)} = \\beta_k(a)\n \\end{equation*}\n for every \\( b \\in B \\) and \\( k \\in \\mscrK \\) uniquely identifies the function\n \\begin{equation*}\n l_B(k, a) \\coloneqq \\beta_k(a).\n \\end{equation*}\n\\end{proof}\n\n\\begin{definition}\\label{def:equalizers}\\mcite[def. 5.1.11]{Leinster2016Basic}\n Consider the index category\n \\begin{equation}\\label{eq:def:equalizers/index}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n\n A diagram indexed by \\eqref{eq:def:equalizers/index} is sometimes called a \\term{fork}. A \\hyperref[def:category_of_cones/limit]{limit} of a fork is called an \\term{equalizer} and a \\hyperref[def:category_of_cones/colimit]{colimit} --- a \\term{coequalizer}.\n\n Note that in \\fullref{def:categorical_diagram} we defined commutativity only when at least one path is nontrivial. That is, \\eqref{eq:def:equalizers/index} cannot commute by definition since all of its paths have length either \\( 0 \\) or \\( 1 \\).\n\n We will describe equalizers in more detail. Fix a fork in \\( \\cat{C} \\) which we will denote by\n \\begin{equation}\\label{eq:def:equalizers/raw_diagram}\n \\begin{aligned}\n \\includegraphics[page=2]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n\n \\begin{minipage}[t]{0.47\\textwidth}\n A \\hyperref[def:category_of_cones/cone]{cone} with vertex \\( A \\) over this diagram has a single morphism \\( f: A \\to X \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:equalizers/cone}\n \\begin{aligned}\n \\includegraphics[page=3]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n\n This is equivalent to requiring\n \\begin{equation*}\n s \\bincirc f = t \\bincirc f.\n \\end{equation*}\n\n An equalizer limit cone \\( (L, \\iota) \\) satisfies the following \\hyperref[rem:limit_universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every cone \\( (A, f) \\), \\( f \\) \\hyperref[def:factors_through]{uniquely factors through} \\( L \\). That is, there exists a unique morphism\n \\begin{equation*}\n l_A: A \\to L\n \\end{equation*}\n such that the following diagram commutes:\n \\end{displayquote}\n \\begin{equation}\\label{eq:def:equalizers/equalizer}\n \\begin{aligned}\n \\includegraphics[page=4]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{minipage}\n \\hfill\n \\begin{minipage}[t]{0.47\\textwidth}\n A \\hyperref[def:category_of_cones/cocone]{cocone} with vertex \\( A \\) over this diagram has a single morphism \\( f: Y \\to A \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:def:equalizers/cocone}\n \\begin{aligned}\n \\includegraphics[page=5]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n\n This is equivalent to requiring\n \\begin{equation*}\n f \\bincirc s = f \\bincirc t.\n \\end{equation*}\n\n A coequalizer cocone \\( (L, \\pi) \\) satisfies the following \\hyperref[rem:limit_universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every cocone \\( (A, f) \\), \\( f \\) \\hyperref[def:factors_through]{uniquely factors through} \\( L \\). That is, there exists a unique morphism\n \\begin{equation*}\n l_A: L \\to A\n \\end{equation*}\n such that the following diagram commutes:\n \\end{displayquote}\n \\begin{equation}\\label{eq:def:equalizers/coequalizer}\n \\begin{aligned}\n \\includegraphics[page=6]{output/def__equalizers.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{minipage}\n\n The requirement that \\eqref{eq:def:equalizers/equalizer} commutes does not mean that \\( p = q \\). As discussed in \\fullref{def:categorical_diagram}, for commutative diagrams, we only consider a pair of paths if at least one of them is nontrivial. We made this requirement in order to allow parallel morphisms.\n\n Note how we interchanged the notation for the projections and inclusions compared to \\fullref{def:discrete_category_limits} --- \\( (L, \\iota) \\) is a \\hi{limit} of a fork, while \\( (L, \\pi) \\) is a \\hi{colimit}. The reason for this is that equalizers are usually canonical inclusions, while coequalizers are projections.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:equalizer_invertibility}\n If \\( (L, \\iota) \\) is an \\hyperref[def:equalizers]{equalizer} in the category \\( \\cat{C} \\), then \\( \\iota \\) is a \\hyperref[def:morphism_invertibility/left_cancellative]{monomorphism}.\n\n \\hyperref[thm:categorical_principle_of_duality]{Dually}, if \\( (L, \\pi) \\) is a \\hyperref[def:equalizers]{coequalizer}, then \\( \\pi \\) is an \\hyperref[def:morphism_invertibility/right_cancellative]{epimorphism}.\n\\end{proposition}\n\\begin{proof}\n Fix an equalizer cone \\( (L, \\iota) \\) of the fork \\eqref{eq:def:equalizers/raw_diagram}. Fix any object \\( A \\) in \\( \\cat{C} \\) and any two parallel morphisms \\( a_1, a_2: A \\to X \\) such that\n \\begin{equation*}\n \\iota \\bincirc a_1 = \\iota \\bincirc a_2.\n \\end{equation*}\n\n Then\n \\begin{equation*}\n g \\bincirc \\iota \\bincirc a_1 = h \\bincirc \\iota \\bincirc a_2,\n \\end{equation*}\n and thus \\( \\iota \\bincirc a_1 = \\iota \\bincirc a_2 \\) is the morphism of a cone. Hence, both \\( (A, \\iota \\bincirc a_1) \\) and \\( (A, \\iota \\bincirc a_2) \\) are cones and thus \\( a_1 \\) and \\( a_2 \\) are the unique maps such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:equalizer_invertibility/monomorphism}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__equalizer_invertibility.pdf}\n \\end{aligned}\n \\end{equation}\n\n Therefore, \\( a_1 = a_2 \\) and, since \\( a_1 \\) and \\( a_2 \\) were arbitrary, it follows that \\( \\iota \\) is a monomorphism.\n\n Dually, if \\( (L, \\pi) \\) is a coequalizer colimit cocone in \\( \\cat{C} \\), then by \\fullref{thm:categorical_limit_duality}, \\( (L, \\pi^{\\opcat}) \\) is an equalizer in \\( \\cat{C}^{\\opcat} \\). Hence, \\( \\pi^{\\opcat} \\) is a monomorphism, and by \\fullref{thm:morphism_invertibility_duality}, \\( \\pi \\) is an epimorphism.\n\\end{proof}\n\n\\begin{example}\\label{ex:equalizers_in_set}\n \\hfill\n \\begin{thmenum}\n \\thmitem{ex:equalizers_in_set/equalizer} For a pair of functions \\( s, t: X \\to Y \\) in \\( \\cat{Set} \\), define the set\n \\begin{equation*}\n E = \\set{ x \\in X \\given s(x) = t(x) }.\n \\end{equation*}\n\n With the inclusion map \\( \\iota: E \\to X \\), this is an \\hyperref[def:equalizers]{equalizer} cone for \\( s \\) and \\( t \\).\n\n We will now prove that it is a limit cone. The pair \\( (E, \\iota) \\) is obviously a cone. For any other cone \\( (A, f) \\), we must find a map \\( l_A: A \\to E \\) such that the following equalizer diagram commutes:\n \\begin{equation}\\label{eq:ex:equalizers_in_set/equalizer}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__equalizers_in_set.pdf}\n \\end{aligned}\n \\end{equation}\n\n In order for \\( (A, f) \\) to be a cone, the image \\( f[A] \\) must be a subset of \\( E \\). In order for \\eqref{eq:ex:equalizers_in_set/equalizer} to commute, it only makes sense to take \\( l_A \\) to be \\( f \\) with its codomain restricted to \\( E \\).\n\n It follows that \\( (E, \\iota) \\) is a limit cone.\n\n \\thmitem{ex:equalizers_in_set/coequalizer} The coequalizer of \\( s, t: X \\to Y \\) is more nuanced. Suppose that \\( (L, \\pi) \\) is a colimit cocone and \\( (A, f) \\) is any cocone for the coequalizer diagram\n \\begin{equation}\\label{eq:ex:equalizers_in_set/coequalizer}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__equalizers_in_set.pdf}\n \\end{aligned}\n \\end{equation}\n\n In order for \\( (A, f) \\) to be a cocone, for every \\( x \\in X \\), we must have \\( f(s(x)) = f(t(x)) \\). Thus, \\( A \\) must be a partition of \\( Y \\) in a way such that \\( s(x) \\) and \\( t(x) \\) belong to the same coset if and only if \\( f(s(x)) = f(t(x)) \\). Outside the images of \\( s \\) and \\( t \\), \\( f \\) is free to take any value.\n\n Let \\( {\\sim} \\) be the smallest equivalence relation on \\( Y \\) such that \\( s(x) \\sim t(x) \\) for every \\( x \\in X \\). Explicitly, this is the \\hyperref[thm:equivalence_closure]{equivalence closure} of the relation\n \\begin{equation*}\n \\set{ (s(x), t(x)) \\given x \\in X }.\n \\end{equation*}\n\n Consider the partition \\( Y / {\\sim} \\) with the projection map\n \\begin{equation*}\n \\begin{aligned}\n &\\pi: Y \\to Y / {\\sim} \\\\\n &\\pi(y) \\coloneqq [y]\n \\end{aligned}\n \\end{equation*}\n\n This is a cocone. Furthermore, it is a colimit cocone because, for any other cocone \\( (A, f) \\), we can define \\( l_A([y]) \\coloneqq f(y) \\) so that \\eqref{eq:ex:equalizers_in_set/coequalizer} commutes.\n\n Coequalizers mostly make sense in the context of \\hyperref[def:group/quotient]{quotient groups}, where partitions are especially well-behaved and admit a much simpler description. See \\fullref{def:group/quotient}.\n \\end{thmenum}\n\\end{example}\n", "meta": {"hexsha": "d3dfa486b4487a30366ca91c78390e03a2125d84", "size": 48841, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/categorical_limits.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/categorical_limits.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/categorical_limits.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.9155609168, "max_line_length": 455, "alphanum_fraction": 0.6909563686, "num_tokens": 15348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.7520125626441471, "lm_q1q2_score": 0.625030007053086}} {"text": "\\section{Subset of Supported Operations}\n\n\\setlongtables\n\\begin{longtable}{| p{4.8in} |}\n%\\hline\n\\endhead\n%\\hline\n\\endfoot\n%\n%\n\\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Negate( PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{$ A \\becomes - A $.} \n\\parameter{\\tt A}{descriptor of data to be updated}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Obj_set_diagonal( PLA_Obj sigma, PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\nSet the diagonal of $ A $ to $ \\sigma I $.\nAll other values in $ A $ are unaffected.\n} \n\\parameter{\\tt sigma, A}{descriptor for $ \\sigma $ and $ A $}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Shift_diagonal( PLA_Obj sigma, PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ A \\becomes A + \\sigma I $.\n} \n\\parameter{\\tt sigma, A}{descriptor for $ \\sigma $ and $ A $}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Symmetrize( PLA_Uplo uplo, PLA_Conj conj, PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ A \\becomes \\mbox{symm}(A) $ or \n$ A \\becomes \\mbox{herm}(A) $,\nwhere {\\tt uplo} indicates whether\n$ A $ is originally stored only in the upper or\nlower triangular part of {\\tt A}.\n} \n\\parameter{\\tt uplo}{\nindicates where $ A$ is stored.\n} \n\\parameter{\\tt conj}{\nindicates whether to create a symmetric or hermitian matrix.\n} \n\\parameter{\\tt A}{\ndescriptor for matrix $ A $.\n}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Triangularize( PLA_Uplo uplo, PLA_Diag diag, PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ A \\becomes \\mbox{lower}(A) $ or \n$ A \\becomes \\mbox{upper}(A) $.\n}\n\\parameter{\\tt uplo}{\nindicates whether whether $ A$ should be upper of\nlower triangular.\n}\n\\parameter{\\tt diag}{\nindicates whether set diagonal to zero or one.\n}\n\\parameter{\\tt A}{\ndescriptor for matrix $ A $.\n}\n% \\\\ \\hline \\hline\n% \\begin{minipage}{4.8in}\n% \\footnotesize\n% \\begin{verbatim}\n% PLA_LU( PLA_Obj A, PLA_Obj ipiv )\n% \\end{verbatim}\n% \\end{minipage}\n% \\purpose{Compute the\n% LU factorization with partial pivoting of\n% matrix $ A $, overwriting $ A $ with factors\n% $ L $ and $ U $. The pivot matrix $ P $ is\n% stored as a vector of indices, in {\\tt ipiv}.\n% } \n% \\parameter{\\tt A}{descriptor for $ A $} \n% \\parameter{\\tt ipiv}{descriptor for vector in which $ P $ is\n% stored} \n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Nrm1( PLA_Obj A, PLA_Obj alpha )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{$ \\alpha \\becomes \\| A \\|_1 $.\n} \n\\parameter{\\tt A, alpha}{descriptor for $ A $ and $ \\alpha $}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Nrm_inf( PLA_Obj A, PLA_Obj alpha )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ \\alpha \\becomes \\| A \\|_\\infty $.\n} \n\\parameter{\\tt A, alpha}{descriptor for $ A $ and $ \\alpha $}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Scal_x( PLA_Uplo uplo, PLA_Obj alpha, PLA_Obj A )\n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ A \\becomes \\alpha A $\nwhere {\\tt uplo} indicates whether $ A $ is stored in the upper\nor lower triangular part of {\\tt A}.\n} \n\\parameter{\\tt alpha, A}{descriptor for $ \\alpha $ and $ A $}\n\\\\ \\hline \\hline\n\\begin{minipage}{4.8in}\n\\footnotesize\n\\begin{verbatim}\nPLA_Swap_x( PLA_Trans trans, PLA_Obj A, PLA_Obj B ) \n\\end{verbatim}\n\\end{minipage}\n\\purpose{\n$ A, B \\becomes \\op_{trans}(A), \\op_{trans}(B) $.\n} \n\\parameter{\\tt trans}{indicates whether to transpose and/or conjugate the data} \n\\parameter{\\tt A, B}{descriptors for $ A $ and $ B $}\n\\\\ \\hline\n\\end{longtable}\n", "meta": {"hexsha": "676a5a362da44732e7f12e2f4169a4335d2c6928", "size": 3605, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/libflame/old/appendixB.tex", "max_stars_repo_name": "haampie/libflame", "max_stars_repo_head_hexsha": "a6b27af9b7ef91ec2724b52c7c09b681379a3470", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 199, "max_stars_repo_stars_event_min_datetime": "2015-02-06T06:05:32.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-18T05:20:33.000Z", "max_issues_repo_path": "docs/libflame/old/appendixB.tex", "max_issues_repo_name": "haampie/libflame", "max_issues_repo_head_hexsha": "a6b27af9b7ef91ec2724b52c7c09b681379a3470", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 44, "max_issues_repo_issues_event_min_datetime": "2015-05-10T18:14:52.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-22T08:22:10.000Z", "max_forks_repo_path": "docs/libflame/old/appendixB.tex", "max_forks_repo_name": "haampie/libflame", "max_forks_repo_head_hexsha": "a6b27af9b7ef91ec2724b52c7c09b681379a3470", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 70, "max_forks_repo_forks_event_min_datetime": "2015-02-07T04:53:03.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-18T05:20:36.000Z", "avg_line_length": 23.8741721854, "max_line_length": 80, "alphanum_fraction": 0.6882108183, "num_tokens": 1265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430394931456, "lm_q2_score": 0.752012562644147, "lm_q1q2_score": 0.6250300070530859}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amssymb}\n\n\\title{\\LaTeX{} Maths examples}\n\\author{Becky Nisbet}\n\\date{January 2021}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Symbols}\nInfinity: $\\infty$\\\\\nDouble bold maths (like real numbers): $\\mathbb{R}, \\mathbb{C}, \\mathbb{N}$ \\\\ %needs package amssymb\nSet notation: $\\in, \\notin, \\ni, \\subset, \\subseteq, \\supset, \\supseteq, \\cup, \\cap \\varnothing$\\\\\nLogic symbols: $\\forall, \\exists, \\rightarrow, \\Rightarrow, \\equiv \\iff, \\leftrightarrow, \\neg, \\sim$\n\n\\section{Calculus}\nDerivatives:\\\\\n$\\frac{dy}{dx} =x^2+3\\\\\n\\frac{d}{dx}y=x^2+3\\\\\n\\frac{\\partial y}{\\partial x}=xy+1\\\\\n\\frac{\\partial}{\\partial x}[xy]=y$\\\\\nIntegrals:\\\\\n$\\int x = \\frac{x^2}{2}+c\\\\\n\\int_0^2 x = 2$\n\n\\section{Vectors and linear algebra}\nA matrix;\\\\\n$M = \\left( %this has to go before the begin array if you want the brackets\n\\begin{array}{cc}\n \n 1 & 0 \\\\\n 0 & 1\n\\end{array}\n\\right)$\\\\\nA vector;\\\\\n$\\vec{v}=\\left(\n\\begin{array}{c}\n a \\\\\n b\n\\end{array}\n\\right)$ \\\\%note how similar this is to constructing tables!\nDeterminants;\\\\\n$\\det{A} =\n\\left|\n\\begin{array}{cc}\n a & b \\\\\n c & d\n\\end{array}\n\\right|$\n\n\\end{document}\n", "meta": {"hexsha": "3c4ae4f89763304f31c6b759bdf06abb742879cf", "size": 1173, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "maths-examples.tex", "max_stars_repo_name": "renisbet/LaTeX-resources", "max_stars_repo_head_hexsha": "36f80d0d5a246839cec71f901947b9e26fa15997", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "maths-examples.tex", "max_issues_repo_name": "renisbet/LaTeX-resources", "max_issues_repo_head_hexsha": "36f80d0d5a246839cec71f901947b9e26fa15997", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "maths-examples.tex", "max_forks_repo_name": "renisbet/LaTeX-resources", "max_forks_repo_head_hexsha": "36f80d0d5a246839cec71f901947b9e26fa15997", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.3272727273, "max_line_length": 101, "alphanum_fraction": 0.641943734, "num_tokens": 434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.831143031127974, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6250300053817195}} {"text": "Optimization problems can appear in almost all situations on life, but there are special important cases where those problems appears, specially on the industry, since if we can achieve an optimal solution, we will improve the efficiency of the industrial process and then get lower production cost with the same (or better) efficiency. This improvement of the costs have an effect on the competitiveness of the companies and on the final quality of their products, including that this is an important money-saving factor. \n \n The challenging part is when this kind of problems become really big, since then they have a attached really high computational complexity and cost, therefore we need some methods to face them. \n \n To do so, we have two approaches, the always-optimal methods that obtain the optimal solution but taking into account every possible combination of the problems' solutions, as for instance Integer Linear Programming does. By the other hand we have those methods that concerns about fair execution times and are looking for a trade-off between acceptable execution times and the quality of the solution.\n \n In this work we propose two approaches in order to solve the scheduling of nurses in a hospital taking into account several constraints. The first one is using Integer Linear Programming and the second one by using metaheuristics. In the metaheuristics part we are focusing specifically in the Greedy Randomized Adaptative Procedure (GRASP) and the Biased-Random Key Genetic Algorithm (BRKGA).\n \n Firstly we compare the efficiency in terms of solving time and wellness of ILP and metaheuristics over medium-sized problems. \n \n Lastly, since large problems are intractable for ILP due to combinatorial explosion reasons, we compare the metaheuristics among them to compare such kind of problems.\n \n \\subsection{Problem Statement}\n \n A public hospital needs to design the working schedule of their nurses. As a first approximation, we are asked to help in designing the schedule of a single day. We know, for each hour h, that at least demandh nurses should be working at the hospital. We have available a set of nNurses nurses and we need to determine at which hours each nurse should be working. However, there are some limitations that should be taken into account:\n \n \\begin{itemize}\n \\item Each nurse should work at least minHours hours.\n \\item Each nurse should work at most maxHours hours.\n \\item Each nurse should work at most maxConsec consecutive hours.\n \\item No nurse can stay at the hospital for more than maxPresence hours (e.g. if maxPresence is 7, it is OK that a nurse works at 2am and also at 8am, but it not possible that he/she works at 2am and also at 9am).\n \\item No nurse can rest for more than one consecutive hour (e.g. working at 8am, resting at 9am and 10am, and working again at 11am is not allowed, since there are two consecutive resting hours).\n \\end{itemize}\n\nThe goal of this project is to determine at which hours each nurse should be working in order to minimize the number of nurses required and satisfy all the aforementioned constraints.\n\n\\subsection{Document Structure}\n\nThe structure of this document is the following: The problem definition is done in chapter 2. Here it is explained the problem that is faced, which constraints is needed to take into account and what we want to optimize. At chapter 3 is explained the ILP model that has been developed, i.e. the decision variables and also the constraints with a mathematical nomenclature. After this chapter, at chapter 4 is explained how it has been used the heuristics approach in order to face with the problem, two meta-heuristics has been used: GRASP3 and BRKGA4 . After perform several executions for those approaches, a comparison in terms of time and quality of the result is done at chapter 5. Finally conclusions of the project are explained at chapter 6.\n\n \n\n\n\\pagebreak", "meta": {"hexsha": "4ca3dc9bab905dad1ba7d5cc4ae2ac0cfbfdcb83", "size": 3948, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documentation/tex/Introduction.tex", "max_stars_repo_name": "presmerats/Nurse-Scheduling-LP-and-Heuristics", "max_stars_repo_head_hexsha": "0b4796d082908f6644bd28ad4bfad9552879ea75", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-02-10T02:38:12.000Z", "max_stars_repo_stars_event_max_datetime": "2021-02-10T02:38:12.000Z", "max_issues_repo_path": "Documentation/tex/Introduction.tex", "max_issues_repo_name": "presmerats/Nurse-Scheduling-LP-and-Heuristics", "max_issues_repo_head_hexsha": "0b4796d082908f6644bd28ad4bfad9552879ea75", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Documentation/tex/Introduction.tex", "max_forks_repo_name": "presmerats/Nurse-Scheduling-LP-and-Heuristics", "max_forks_repo_head_hexsha": "0b4796d082908f6644bd28ad4bfad9552879ea75", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-02-10T02:38:13.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-10T02:38:13.000Z", "avg_line_length": 116.1176470588, "max_line_length": 749, "alphanum_fraction": 0.7938196555, "num_tokens": 830, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.746138993030751, "lm_q2_score": 0.8376199714402813, "lm_q1q2_score": 0.624980922032898}} {"text": "\\section{Algebraic view of systems of equations}\n\n\\begin{outcome}\n \\begin{enumerate}\n \\item Recognize the difference between a linear equation and a\n non-linear equation.\n \\item Determine whether a tuple of real numbers is a solution for a\n system of linear equations.\n \\item Understand what it means for a system of linear equations to\n be consistent or inconsistent.\n \\end{enumerate}\n\\end{outcome}\n\nWe have taken an in-depth look at graphical representations of systems\nof equations, as well as how to find possible solutions\ngraphically. Our attention now turns to working with systems\nalgebraically.\n\n\\begin{definition}{Linear equation}{linear-equation}\n A \\textbf{linear equation}%\n \\index{linear equation} is an equation of the form\n \\begin{equation*}\n a_1x_1 + a_2x_2 + \\ldots + a_nx_n = b.\n \\end{equation*}\n Here, $a_1,\\ldots,a_n$ are real numbers called the\n \\textbf{coefficients}%\n \\index{coefficient} of the equation, $b$ is a real number called the\n \\textbf{constant term}%\n \\index{constant term} of the equation, and $x_1,\\ldots,x_n$ are\n \\textbf{variables}%\n \\index{variable}.\n\\end{definition}\n\nReal numbers, such as the coefficients $a_1,\\ldots,a_n$, the constant\nterm $b$, or the values of the variables $x_1,\\ldots,x_n$, will also\nbe called \\textbf{scalars}%\n\\index{scalar}. For now, the word ``scalar'' is just a synonym for\n``real number''. Later, in Section~\\ref{sec:fields}, we will discover\nother kinds of scalars.\n\n\\begin{example}{Linear vs. non-linear equation}{linear-vs-non-linear}\n Which of the following equations are linear?\n \\begin{equation*}\n \\begin{array}{c}\n 2x+3y=5 \\\\\n 2x^2+3y=5 \\\\\n 2\\sqrt{x} + 3y = 5 \\\\\n (\\sqrt{2}) x + 3y = 5^2\n \\end{array}\n \\end{equation*}\n\\end{example}\n\n\\begin{solution}\n The equation $2x+3y=5$ is linear. The equation $2x^2+3y=5$ is not\n linear, because it contains the square of a variable instead of a\n variable. The equation $2\\sqrt{x} + 3y = 5$ is also not linear,\n because the square root is applied to one of the variables. On the\n other hand, the equation $(\\sqrt{2}) x + 3y = 5^2$ is linear,\n because $\\sqrt{2}$ and $5^2$ are real numbers, and can\n therefore be used as coefficients and constant terms.\n\\end{solution}\n\nWe also permit minor notational variants of linear equations. The\nequation $2x-3y=5$ is linear although\nDefinition~\\ref{def:linear-equation} does not mention subtraction,\nbecause it can be regarded as just another notation for\n$2x+(-3)y = 5$. Similarly, the equation $2x=5+3y$ can be regarded as\nlinear, because it can be easily rewritten as $2x-3y=5$ by bringing\nall the variables (and their coefficients) to the left-hand side.\nWhen we need to emphasize that some linear equation is literally of\nthe form $a_1x_1 + a_2x_2 + \\ldots + a_nx_n = b$, we say that the\nequation is in \\textbf{standard form}%\n\\index{linear equation!standard form}%\n\\index{standard form!of a linear equation}. Thus, the standard form of\nthe equation $2x=5+3y$ is $2x+(-3)y=5$.\n\nA \\textbf{solution}%\n\\index{linear equation!solution} to a linear equation is an assignment\nof real numbers to the variables, making the equation true. More\nprecisely, if $r_1,\\ldots,r_n$ are real numbers, the assignment\n$x_1=r_1$, \\ldots, $x_n=r_n$ is a solution to the equation in\nDefinition~\\ref{def:linear-equation} if the real number\n$a_1r_1 + a_2r_2 + \\ldots + a_nr_n$ is equal to the real number\n$b$. To save space, we often write solutions in \\textbf{tuple\n notation}%\n\\index{tuple}\\footnote{The terminology ``tuple'' arose as follows. A\n collection of two items is called a ``pair'', a collection of three\n items is called a ``triple'', followed by ``quadruple'',\n ``quintuple'', ``sextuple'', and so on. You have to know Latin to\n know what the next ones are called. To avoid these Latin terms,\n mathematicians started saying $4$-tuple, $5$-tuple, $6$-tuple and so\n on, and more generally, $n$-tuple for an ordered collection of $n$\n items. When $n$ doesn't matter or is clear from the context, we\n often just say ``tuple''.} as\n$(x_1,\\ldots,x_n) = (r_1,\\ldots,r_n)$. When there is no doubt about\nthe order of the variables, we also often simply write the solution as\n$(r_1,\\ldots,r_n)$.\n\n\\begin{example}{Solutions of a linear equation}{linear-equation-solutions}\n Consider the linear equation $2x+3y-4z=5$. Which of the following\n are solutions? (a) $(x,y,z)=(1,1,0)$, (b) $(x,y,z)=(0,3,1)$, (c)\n $(x,y,z)=(1,1,1)$.\n\\end{example}\n\n\\begin{solution}\n The assignment $(x,y,z) = (1,1,0)$ is a solution because\n $2(1)+3(1)-4(0) = 5$. The assignment $(x,y,z) = (0,3,1)$ is also a\n solution, because $2(0) + 3(3) - 4(1) = 5$. On the other hand,\n $(x,y,z) = (1,1,1)$ is not a solution, because\n $2(1) + 3(1)-4(1) = 1 \\neq 5$.\n\\end{solution}\n\nA system of linear equations is just several linear equations taken\ntogether.\n\n\\begin{definition}{System of linear equations}{system-of-linear-equations}\n A \\textbf{system of linear equations}%\n \\index{system of linear equations} is a list of equations\n \\begin{equation*}\n \\begin{array}{c@{~}c@{~}c}\n a_{11}x_1+a_{12}x_2+\\ldots+a_{1n}x_n&=&b_1 \\\\\n a_{21}x_1+a_{22}x_2+\\ldots+a_{2n}x_n&=&b_2 \\\\\n \\vdots \\\\\n a_{m1}x_1+a_{m2}x_2+\\ldots+a_{mn}x_n&=&b_m,\n \\end{array}\n \\end{equation*}\n where $a_{ij}$ and $b_i$ are scalars (i.e., real numbers). The above\n is a system of $m$ equations in the $n$ variables,\n $x_1,x_2\\ldots,x_n$. As before, the numbers $a_{ij}$ are called the\n \\textbf{coefficients}%\n \\index{coefficient} and the numbers $b_i$ are called the\n \\textbf{constant terms}%\n \\index{constant term} of the system of equations.\n\\end{definition}\n\nThe relative size of $m$ and $n$ is not important here. We may have\nmore variables than equations, more equations than variables, or an\nequal number of equations and variables.\n\nA \\textbf{solution}%\n\\index{system of linear equations!solution} to a system of linear\nequations is an assignment of real numbers to the variables that is a\nsolution to {\\em all} of the equations in the system.\n\n\\begin{example}{Solutions of a system of linear equations}{linear-system-solutions}\n Consider the system of linear equations\n \\begin{equation*}\n \\begin{array}{c@{~}c@{~}c@{~}c@{~}c@{~}c@{~}c}\n 2x&+&3y&-&4z &=& 5 \\\\\n -2x&+&y&+&2z &=& -1.\n \\end{array}\n \\end{equation*}\n Which of the following are solutions of the system? (a)\n $(x,y,z)=(1,1,0)$, (b) $(x,y,z)=(6,3,4)$, (c) $(x,y,z)=(0,3,1)$.\n\\end{example}\n\n\\begin{solution}\n The assignment $(x,y,z)=(1,1,0)$ is a solution of this system of\n equations, because it is a solution to the first equation and the\n second equation. Also, $(x,y,z)=(6,3,4)$ is another solution of this\n system of equations (check this!). On the other hand,\n $(x,y,z)=(0,3,1)$ is not a solution of the system, because although\n it is a solution to the first equation, it is not a solution to the\n second equation.\n\\end{solution}\n\nRecall from Section~\\ref{sec:systems-geometric} that a system of\nequations either has a unique solution, infinitely many solutions, or\nno solution. It is very important to us whether a system of equations\nhas solutions or not. For this reason, we introduce the following\nterminology:\n\n\\begin{definition}{Consistent and inconsistent systems}{consistent-and-inconsistent}\n A system of linear equations is called \\textbf{consistent}%\n \\index{consistent system}%\n \\index{system of linear equations!consistent} if there exists at\n least one solution. It is called \\textbf{inconsistent}%\n \\index{inconsistent system}%\n \\index{system of linear equations!inconsistent} if there is no\n solution.\n\\end{definition}\n\nIf we think of each equation as a condition that must be satisfied by\nthe variables, consistent means that there is some choice of values\nfor the variables which can satisfy \\textbf{all} of the\nconditions. Inconsistent means that there is no such choice of values\nfor the variables. In the following sections, you will learn a method\nfor determining whether a system of equations is consistent or not,\nand in case it is consistent, to find all of its solutions.\n\n", "meta": {"hexsha": "bc1e2ec7083c0c33c318fbdd8189eecf568a7cc4", "size": 8098, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "baseText/content/SystemsofEquations-Algebraic.tex", "max_stars_repo_name": "selinger/linear-algebra", "max_stars_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-03-21T06:37:13.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-30T16:23:10.000Z", "max_issues_repo_path": "baseText/content/SystemsofEquations-Algebraic.tex", "max_issues_repo_name": "selinger/linear-algebra", "max_issues_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "baseText/content/SystemsofEquations-Algebraic.tex", "max_forks_repo_name": "selinger/linear-algebra", "max_forks_repo_head_hexsha": "37ad955fd37bdbc6a9e855c3794e92eaaa2d8c02", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-09T11:12:03.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-30T16:23:12.000Z", "avg_line_length": 42.3979057592, "max_line_length": 84, "alphanum_fraction": 0.712398123, "num_tokens": 2515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389930307512, "lm_q2_score": 0.8376199694135332, "lm_q1q2_score": 0.6249809205206622}} {"text": "\\chapter{Integrating differential forms}\nWe now show how to integrate differential forms over cells,\nand state Stokes' theorem in this context.\nIn this chapter, all vector spaces are finite-dimensional and real.\n\n\\section{Motivation: line integrals}\nGiven a function $g : [a,b] \\to \\RR$,\nwe know by the fundamental theorem of calculus that\n\\[\n\t\\int_{[a,b]} g(t) \\; dt = f(b) - f(a)\n\\]\nwhere $f$ is a function such that $g = df/dt$.\n%(You might recognize this more readily\n%as the more customary $\\int_a^b g(t) \\; dt$.)\nEquivalently, for $f : [a,b] \\to \\RR$,\n\\[ \\int_{[a,b]} g \\; dt = \\int_{[a,b]} df = f(b) - f(a) \\]\nwhere $df$ is the exterior derivative we defined earlier.\n\nCool, so we can integrate over $[a,b]$.\nNow suppose more generally, we have $U$ an open subset of our real vector space $V$\nand a $1$-form $\\alpha : U \\to V^\\vee$.\nWe consider a \\vocab{parametrized curve}, which is a smooth function $c : [a,b] \\to U$.\nPicture:\n\\begin{center}\n\t\\begin{asy}\n\t\tsize(8.5cm);\n\t\tpair A = (-13,0);\n\t\tpair B = (-9,0);\n\t\tdraw(A--B, grey);\n\t\tdot(A, grey); dot(B, grey);\n\t\tlabel(\"$[a,b]$\", A--B, dir(90), grey);\n\t\tdot(\"$t$\", 0.3*A+0.7*B, dir(-90), grey);\n\n\t\tdraw( (-8,0) -- (-6,0) , EndArrow);\n\t\tlabel(\"$c$\", (-7,0), dir(90));\n\n\t\tbigblob(\"$U$\");\n\t\tpair a = (-2,-2);\n\t\tpair b = (3,0);\n\t\tpair p = (0,1);\n\t\tpair q = (2,0);\n\t\tlabel(\"$c$\", q, dir(45), red);\n\t\tdraw(a..p..q..b, red);\n\t\tdot(\"$p = c(t)$\", p, dir(90), blue);\n\t\tdraw(p--(p+1.5*dir(-10)), blue, EndArrow);\n\n\t\tdraw( (0,-4)--(0,-8), EndArrow );\n\t\tlabel(\"$\\alpha$\", (0,-6), dir(0));\n\t\tlabel(\"$\\alpha_p(v) \\in \\mathbb R$\", (0,-9), heavygreen);\n\n\t\tdraw( (-10,-1)--(-1,-8), EndArrow);\n\t\tlabel(\"$c^\\ast \\alpha$\", (-5.5,-4.5), dir(225));\n\t\\end{asy}\n\\end{center}\n\nWe want to define an $\\int_c \\alpha$ such that:\n\\begin{moral}\n\tThe integral $\\int_c \\alpha$ should add up all the $\\alpha$ along the curve $c$.\n\\end{moral}\nOur differential form $\\alpha$ first takes in a point $p$ to get $\\alpha_p \\in V^\\vee$.\nThen, it eats a tangent vector $v \\in V$ to the curve $c$ to finally give a real number $\\alpha_p(v) \\in \\RR$.\nWe would like to ``add all these numbers up'',\nusing only the notion of an integral over $[a,b]$.\n\n\\begin{exercise}\n\tTry to guess what the definition of the integral should be.\n\t(By type-checking, there's only one reasonable answer.)\n\\end{exercise}\n\nSo, the definition we give is\n\\[ \\int_c \\alpha \\defeq\n\t\\int_{[a,b]} \\alpha_{c(t)} \\left( c'(t) \\right) \\; dt. \\]\nHere, $c'(t)$ is shorthand for $(Dc)_{c(t)}(1)$.\nIt represents the \\emph{tangent vector} to the curve $c$ at the point $p=c(t)$,\nat time $t$.\n(Here we are taking advantage of the fact that $[a,b]$ is one-dimensional.)\n\nNow that definition was a pain to write, so we will define a differential\n$1$-form $c^\\ast \\alpha$ on $[a,b]$ to swallow that entire thing:\nspecifically, in this case we define $c^\\ast\\alpha$ to be\n\\[ \\left( c^\\ast \\alpha \\right)_t (\\eps) = \\alpha_{c(t)} \\cdot (Dc)_{t} (\\eps) \\]\n(here $\\eps$ is some displacement in time).\nThus, we can more succinctly write\n\\[ \\int_c \\alpha \\defeq \\int_{[a,b]} c^\\ast \\alpha. \\]\nThis is a special case of a \\emph{pullback}:\nroughly, if $\\phi : U \\to U'$ (where $U \\subseteq V$, $U' \\subseteq V'$),\nwe can change any differential $k$-form $\\alpha$ on $U'$\nto a $k$-form on $U$.\nIn particular, if $U = [a,b]$,\\footnote{OK,\n\tso $[a,b]$ isn't actually open, sorry.\n\tI ought to write $(a-\\eps, b+\\eps)$, or something.}\nwe can resort to our old definition of an integral.\nLet's now do this in full generality.\n\n\\section{Pullbacks}\nLet $V$ and $V'$ be finite dimensional real vector spaces (possibly different dimensions)\nand suppose $U$ and $U'$ are open subsets of each;\nnext, consider a $k$-form $\\alpha$ on $U'$.\n\nGiven a map $\\phi : U \\to U'$ we now want to define a pullback in\nmuch the same way as before.\nPicture:\n\\begin{center}\n\t\\begin{asy}\n\t\tsize(13cm);\n\t\tbigblob(\"$U$\");\n\t\tpair p = (-1,0);\n\t\tdot(\"$p$\", p, dir(90), red);\n\t\tpair p1 = p + 1.4*dir(150);\n\t\tpair p2 = p + 1.7*dir(-50);\n\t\tdraw(p--p1, red, EndArrow);\n\t\tdraw(p--p2, red, EndArrow);\n\n\t\tadd(scale(0.8)*shift(14*dir(180))*CC());\n\t\tbigblob(\"$U'$\");\n\t\tpair q = (-0.5,0.5);\n\t\tdot(\"$q = \\phi(p)$\", q, dir(90), blue);\n\t\tpair q1 = q + 1.8*dir(-100);\n\t\tpair q2 = q + 2.3*dir(-10);\n\t\tdraw(q--q1, blue, EndArrow);\n\t\tdraw(q--q2, blue, EndArrow);\n\n\t\tdraw((-9,0)--(-3,0), EndArrow);\n\t\tlabel(\"$\\phi$\", (-6,0), dir(90));\n\t\n\t\tdraw( (0,-4)--(0,-8), EndArrow );\n\t\tlabel(\"$\\alpha$\", (0,-6), dir(0));\n\t\tlabel(\"$\\alpha_q(\\dots) \\in \\mathbb R$\", (0,-9), heavygreen);\n\n\t\tdraw( (-11,-3)--(-1,-8), EndArrow);\n\t\tlabel(\"$\\phi^\\ast \\alpha$\", (-6,-6), dir(225));\n\t\\end{asy}\n\\end{center}\n\nWell, there's a total of about one thing we can do.\nSpecifically: $\\alpha$ accepts a point in $U'$ and $k$ tangent vectors in $V'$,\nand returns a real number.\nWe want $\\phi^\\ast \\alpha$ to accept a point in $p \\in U$\nand $k$ tangent vectors $v_1, \\dots, v_k$ in $V$,\nand feed the corresponding information to $\\alpha$.\n\nClearly we give the point $q = \\phi(p)$.\nAs for the tangent vectors, since we are interested in volume, we take the\nderivative of $\\phi$ at $p$, $(D\\phi)_p$, which will scale each of our vectors $v_i$\ninto some vector in the target $V'$.\nTo cut a long story short:\n\\begin{definition}\n\tGiven $\\phi : U \\to U'$ and $\\alpha$ a $k$-form, we define the \\vocab{pullback}\n\t\\[\n\t\t(\\phi^\\ast \\alpha)_p(v_1, \\dots, v_k)\n\t\t\\defeq \\alpha_{\\phi(p)}\n\t\t\\left( (D\\phi)_p(v_1), \\dots, (D\\phi)_p(v_k) \\right).\n\t\\]\n\\end{definition}\n\nThere is a more concrete way to define the pullback using bases.\nSuppose $w_1, \\dots, w_n$ is a basis of $V'$\nand $e_1, \\dots, e_m$ is a basis of $V$.\nThus, by the projection principle (\\Cref{thm:project_principle}) \nthe map $\\phi : V \\to V'$ can be thought of as\n\\[ \\phi(v) = \\phi_1(v) w_1 + \\dots \\phi_n(v) w_n \\]\nwhere each $\\phi_i$ takes in a $v \\in V$ and returns a real number.\nWe know also that $\\alpha$ can be written concretely as\n\\[ \\alpha = \\sum_{J \\subseteq \\{1, \\dots, n\\}} f_J w_J. \\]\nThen, we define\n\\[\n\t\\phi^\\ast\\alpha\n\t= \\sum_{I \\subseteq \\{1, \\dots, m\\}}\n\t(f_I \\circ \\phi) (D\\phi_{i_1} \\wedge \\dots \\wedge D\\phi_{i_k}).\n\\]\nA diligent reader can check these definitions are equivalent.\n\\begin{example}\n\t[Computation of a pullback]\n\tLet $V = \\RR^2$ with basis $\\ee_1$ and $\\ee_2$,\n\tand suppose $\\phi : V \\to V'$ is given by sending\n\t\\[ \\phi(a\\ee_1 + b\\ee_2) = (a^2+b^2)w_1 + \\log(a^2+1) w_2 + b^3 w_3 \\]\n\twhere $w_1$, $w_2$, $w_3$ is a basis for $V'$.\n\tConsider the form $\\alpha_q = f(q) w_1 \\wedge w_3$, where $f : V' \\to \\RR$.\n\tThen\n\t\\[ (\\phi^\\ast\\alpha)_p = f(\\phi(p)) \\cdot (2a \\ee_1^\\vee + 2b\\ee_2^\\vee) \\wedge (3b^2 \\ee_2^\\vee)\n\t\t= f(\\phi(p)) \\cdot 6ab^2 \\cdot \\ee_1^\\vee \\wedge \\ee_2^\\vee. \\]\n\\end{example}\n\nIt turns out that the pullback basically behaves nicely as possible, e.g.\\\n\\begin{itemize}\n\t\\ii $\\phi^\\ast(c\\alpha + \\beta) = c\\phi^\\ast \\alpha + \\phi^\\ast\\beta$ (linearity)\n\t\\ii $\\phi^\\ast(\\alpha\\wedge\\beta)\n\t= (\\phi^\\ast \\alpha)\\wedge(\\phi^\\ast \\beta)$\n\t\\ii $\\phi_1^\\ast(\\phi_2^\\ast(\\alpha)) \n\t= (\\phi_2 \\circ \\phi_1)^\\ast(\\alpha)$ (naturality)\n\\end{itemize}\nbut I won't take the time to check these here\n(one can verify them all by expanding with a basis).\n\t\n\\section{Cells}\n\\prototype{A disk in $\\RR^2$ can be thought of as the cell\n\t$[0,R]\\times[0,2\\pi] \\to \\RR^2$ by\n\t$(r,\\theta) \\mapsto (r\\cos\\theta)\\ee_1 + (r\\sin\\theta)\\ee_2$.}\nNow that we have the notion of a pullback,\nwe can define the notion of an integral for more general spaces.\nSpecifically, to generalize the notion of integrals we had before:\n\\begin{definition}\n\tA \\vocab{$k$-cell} is a smooth function $c : [a_1, b_1] \\times [a_2,b_2] \\times \\dots [a_k, b_k] \\to V$.\n\\end{definition}\n\\begin{example}\n\t[Examples of cells]\n\tLet $V = \\RR^2$ for convenience.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii A $0$-cell consists of a single point.\n\t\t\\ii As we saw, a $1$-cell is an arbitrary curve.\n\t\t\\ii A $2$-cell corresponds to a $2$-dimensional surface.\n\t\tFor example, the map $c : [0,R] \\times [0,2\\pi] \\to V$ by\n\t\t\\[ c : (r,\\theta) \\mapsto (r\\cos\\theta, r\\sin\\theta) \\]\n\t\tcan be thought of as a disk of radius $R$.\n\t\\end{enumerate}\n\\end{example}\nThen, to define an integral\n\\[ \\int_c \\alpha \\]\nfor a differential $k$-form $\\alpha$ and a $k$-cell $c : [0,1]^k \\to V$,\nwe simply take the pullback\n\\[ \\int_{[0,1]^k} c^\\ast \\alpha \\]\nSince $c^\\ast \\alpha$ is a $k$-form on the $k$-dimensional unit box,\nit can be written as $f(x_1, \\dots, x_n) \\; dx_1 \\wedge \\dots \\wedge dx_n$,\nso the above integral can be written as\n\\[ \\int_0^1 \\dots \\int_0^1 f(x_1, \\dots, x_n) \\; dx_1 \\wedge \\dots \\wedge dx_n \\]\n\n\\begin{example}[Area of a circle]\n\tConsider $V = \\RR^2$ and let $c : (r,\\theta) \\mapsto (r\\cos\\theta)\\ee_1 + (r\\sin\\theta)\\ee_2$\n\ton $[0,R] \\times [0,2\\pi]$ as before.\n\tTake the $2$-form $\\alpha$ which gives $\\alpha_p = \\ee_1^\\vee \\wedge \\ee_2^\\vee$ at every point $p$.\n\tThen\n\t\\begin{align*}\n\t\tc^\\ast\\alpha &= \n\t\t\\left( \\cos\\theta dr - r\\sin\\theta d\\theta \\right)\n\t\t\\wedge\n\t\t\\left( \\sin\\theta dr + r\\cos\\theta d\\theta \\right) \\\\\n\t\t&= r(\\cos^2\\theta+\\sin^2\\theta) (dr \\wedge d\\theta) \\\\\n\t\t&= r \\; dr \\wedge d\\theta\n\t\\end{align*}\n\tThus,\n\t\\[ \\int_c \\alpha\n\t\t= \\int_0^R \\int_0^{2\\pi} r \\; dr \\wedge d\\theta\n\t\t= \\pi R^2 \\]\n\twhich is the area of a circle.\n\\end{example}\n\nHere's some geometric intuition for what's happening.\nGiven a $k$-cell in $V$, a differential $k$-form $\\alpha$ accepts a point $p$ and some tangent vectors $v_1$, \\dots, $v_k$\nand spits out a number $\\alpha_p(v_1, \\dots, v_k)$,\nwhich as before we view as a signed hypervolume.\nThen the integral \\emph{adds up all these infinitesimals across the entire cell}.\nIn particular, if $V = \\RR^k$ and we take the form $\\alpha : p \\mapsto \\ee_1^\\vee \\wedge \\dots \\wedge \\ee_k^\\vee$,\nthen what these $\\alpha$'s give is the $k$th hypervolume of the cell.\nFor this reason, this $\\alpha$ is called the \\vocab{volume form} on $\\RR^k$.\n\nYou'll notice I'm starting to play loose with the term ``cell'':\nwhile the cell $c : [0,R] \\times [0,2\\pi] \\to \\RR^2$ is supposed to be a function\nI have been telling you to think of it as a unit disk (i.e.\\ in terms of its image).\nIn the same vein, a curve $[0,1] \\to V$ should be thought of as a curve in space,\nrather than a function on time.\n\nThis error turns out to be benign.\nLet $\\alpha$ be a $k$-form on $U$ and $c : [a_1, b_1] \\times \\dots \\times [a_k, b_k] \\to U$ a $k$-cell.\nSuppose $\\phi : [a_1', b_1'] \\times \\dots [a_k', b_k'] \\to [a_1, b_1] \\times \\dots \\times [a_k, b_k]$;\nit is a \\vocab{reparametrization} if $\\phi$ is bijective and $(D\\phi)_p$ is always invertible\n(think ``change of variables'');\nthus\n\\[ c \\circ \\phi : [a_1', b_1'] \\times \\dots \\times [a_k',b_k'] \\to U \\]\nis a $k$-cell as well.\nThen it is said to \\vocab{preserve orientation} if $\\det(D\\phi)_p > 0$ for all $p$\nand \\vocab{reverse orientation} if $\\det(D\\phi)_p < 0$ for all $p$.\n\\begin{exercise}\n\tWhy is it that exactly one of these cases must occur?\n\\end{exercise}\n\n\\begin{theorem}\n\t[Changing variables doesn't affect integrals]\n\tLet $c$ be a $k$-cell, $\\alpha$ a $k$-form, and $\\phi$ a reparametrization.\n\tThen\n\t\\[ \\int_{c \\circ \\phi} \\alpha\n\t\t=\n\t\t\\begin{cases}\n\t\t\t\\int_c \\alpha & \\phi \\text{ preserves orientation} \\\\\n\t\t\t- \\int_c \\alpha & \\phi \\text{ reverses orientation}.\n\t\t\\end{cases}\n\t\\]\n\\end{theorem}\n\\begin{proof}\n\tUse naturality of the pullback to reduce it to the corresponding\n\ttheorem in normal calculus.\n\\end{proof}\n\nSo for example, if we had parametrized the unit circle as $[0,1] \\times [0,1] \\to \\RR^2$\nby $(r,t) \\mapsto R\\cos(2\\pi t) \\ee_1 + R\\sin(2\\pi t) \\ee_2$, we would have arrived at the same result.\nSo we really can think of a $k$-cell just in terms of the points it specifies.\n\n\\section{Boundaries}\n\\prototype{The boundary of $[a,b]$ is $\\{b\\}-\\{a\\}$. The boundary of a square goes around its edge counterclockwise.}\nFirst, I introduce a technical term that lets us consider multiple cells at once.\n\\begin{definition}\n\tA \\vocab{$k$-chain} $U$ is a formal\n\tlinear combination of $k$-cells over $U$,\n\ti.e.\\ a sum of the form\n\t\\[ c = a_1 c_1 + \\dots + a_m c_m \\]\n\twhere each $a_i \\in \\RR$ and $c_i$ is a $k$-cell.\n\tWe define $\\int_c \\alpha = \\sum_i a_i \\int c_i$.\n\\end{definition}\nIn particular, a $0$-chain consists of several points, each with a given weight.\n\nNow, how do we define the boundary?\nFor a $1$-cell $[a,b] \\to U$, as I hinted earlier we want the answer to be the $0$-chain $\\{c(b)\\}-\\{c(a)\\}$.\nHere's how we do it in general.\n\\begin{definition}\n\tSuppose $c : [0,1]^k \\to U$ is a $k$-cell.\n\tThen the \\vocab{boundary} of $c$, denoted $\\partial c : [0,1]^{k-1} \\to U$,\n\tis the $(k-1)$-chain defined as follows.\n\tFor each $i = 1,\\dots,k$ define\n\t\\begin{align*}\n\t\tc_i^{\\text{start}}(t_1, \\dots, t_{k-1}) &\n\t\t= (t_1, \\dots, t_{i-1}, 0, t_i, \\dots, t_k) \\\\\n\t\tc_i^{\\text{stop}}(t_1, \\dots, t_{k-1}) &\n\t\t= (t_1, \\dots, t_{i-1}, 1, t_i, \\dots, t_k).\n\t\\end{align*}\n\tThen\n\t\\[ \\partial c \\defeq\n\t\\sum_{i=1}^k (-1)^{i+1} \\left( c_i^{\\text{stop}} - c_i^{\\text{start}} \\right). \\]\n\tFinally, the boundary of a chain is the sum of the boundaries of each cell (with the appropriate weights).\n\tThat is, $\\partial(\\sum a_ic_i) = \\sum a_i \\partial c_i$.\n\\end{definition}\n\\begin{ques}\n\tSatisfy yourself that one can extend this definition to\n\ta $k$-cell $c$ defined on $c : [a_1, b_1] \\times \\dots \\times [a_k, b_k] \\to V$\n\t(rather than from $[0,1]^k \\to V$).\n\\end{ques}\n\n\\begin{example}\n\t[Examples of boundaries]\n\tConsider the $2$-cell $c : [0,1]^2 \\to \\RR^2$ shown below.\n\t\\begin{center}\n\t\t\\begin{asy}\n\t\t\tsize(7cm);\n\t\t\tpen e1 = heavyred;\n\t\t\tpen e2 = orange;\n\t\t\tpen e3 = olive;\n\t\t\tpen e4 = heavymagenta;\n\t\t\tdraw((0,0)--(2,0), e1, EndArrow);\n\t\t\tdraw((2,0)--(2,2), e2, EndArrow);\n\t\t\tdraw((2,2)--(0,2), e3, EndArrow);\n\t\t\tdraw((0,2)--(0,0), e4, EndArrow);\n\t\t\tlabel(scale(0.8)*\"$[0,1]^2$\", (1,1));\n\t\t\tdraw( (3,1)--(6,1), EndArrow);\n\t\t\tlabel(\"$c$\", (4.5,1), dir(90));\n\t\t\tpair p1 = (7,-1);\n\t\t\tpair p2 = (12,-2);\n\t\t\tpair p3 = (11,3);\n\t\t\tpair p4 = (8,2);\n\t\t\tfill(p1--p2--p3--p4--cycle, palecyan);\n\t\t\tdraw(p1--p2, e1, EndArrow, Margins);\n\t\t\tdraw(p2--p3, e2, EndArrow, Margins);\n\t\t\tdraw(p3--p4, e3, EndArrow, Margins);\n\t\t\tdraw(p4--p1, e4, EndArrow, Margins);\n\t\t\tdot(\"$p_1$\", p1, dir(225), blue+4);\n\t\t\tdot(\"$p_2$\", p2, dir(315), blue+4);\n\t\t\tdot(\"$p_3$\", p3, dir( 45), blue+4);\n\t\t\tdot(\"$p_4$\", p4, dir(135), blue+4);\n\t\t\tlabel(\"$c$\", (p1+p2+p3+p4)/4);\n\t\t\\end{asy}\n\t\\end{center}\n\tHere $p_1$, $p_2$, $p_3$, $p_4$ are the images of $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$, respectively.\n\tThen we can think of $\\partial c$ as\n\t\\[ \\partial c = [p_1,p_2] + [p_2,p_3] + [p_3,p_4] + [p_4,p_1] \\]\nwhere each ``interval'' represents the $1$-cell shown by the reddish arrows on the right.\n\tWe can take the boundary of this as well, and obtain an empty chain as\n\t\\[ \\partial(\\partial c) = \\sum_{i=1}^4 \\partial([p_i, p_{i+1}]) = \\sum_{i=1}^4 \\{p_{i+1}\\}-\\{p_i\\} = 0. \\]\n\\end{example}\n\n\\begin{example}\n\t[Boundary of a unit disk]\n\tConsider the unit disk given by\n\t\\[ c : [0,1] \\times [0,2\\pi] \\to \\RR^2 \\quad\\text{by}\\quad\n\t(r,\\theta) \\mapsto s\\cos(2\\pi t)\\ee_1 + s\\sin(2\\pi t)\\ee_2. \\]\n\tThe four parts of the boundary are shown in the picture below:\n\t\\begin{center}\n\t\t\\begin{asy}\n\t\t\tsize(7cm);\n\t\t\tpen e1 = heavyred;\n\t\t\tpen e2 = orange;\n\t\t\tpen e3 = olive;\n\t\t\tpen e4 = heavymagenta;\n\t\t\tdraw((0,0)--(2,0), e1, EndArrow);\n\t\t\tdraw((2,0)--(2,2), e2, EndArrow);\n\t\t\tdraw((2,2)--(0,2), e3, EndArrow);\n\t\t\tdraw((0,2)--(0,0), e4, EndArrow);\n\t\t\tlabel(\"$r$\", (2,0), dir(-45));\n\t\t\tlabel(\"$\\theta$\", (0,2), dir(135));\n\n\t\t\tlabel(scale(0.8)*\"$[0,1]^2$\", (1,1));\n\t\t\tdraw( (3,1)--(6,1), EndArrow);\n\t\t\tlabel(\"$c$\", (4.5,1), dir(90));\n\n\t\t\tpair O = (9,1);\n\t\t\tpair P = O + 2*dir(0);\n\t\t\tfill(CP(O,P), palecyan);\n\t\t\treal eps = 0.3;\n\t\t\tdraw(shift(0,eps) * (O--P), e1, EndArrow, Margins);\n\t\t\tdraw(shift(0,-eps) * (P--O), e3, EndArrow, Margins);\n\t\t\tdraw(CP(O,P), e2, EndArrow, Margins);\n\t\t\tdot(O, e4+4);\n\t\t\\end{asy}\n\t\\end{center}\n\tNote that two of the arrows more or less cancel each other out when they are integrated.\n\tMoreover, we interestingly have a \\emph{degenerate} $1$-cell at the center of the circle;\n\tit is a constant function $[0,1] \\to \\RR^2$ which always gives the origin.\n\\end{example}\n\nObligatory theorem, analogous to $d^2=0$ and left as a problem.\n\\begin{theorem}[The boundary of the boundary is empty]\n\t$\\partial^2 = 0$, in the sense that for any $k$-chain $c$ we have $\\partial^2(c) = 0$.\n\\end{theorem}\n\n\\section{Stokes' theorem}\n\\prototype{$\\int_{[a,b]} dg = g(b) - g(a)$.}\n\nWe now have all the ingredients to state Stokes' theorem for cells.\n\\begin{theorem}\n\t[Stokes' theorem for cells]\n\tTake $U \\subseteq V$ as usual, let $c : [0,1]^k \\to U$ be a $k$-cell\n\tand let $\\alpha : U \\to \\Lambda^{k-1}(V^\\vee)$ be a $k-1$-form.\n\tThen\n\t\\[ \\int_c d\\alpha = \\int_{\\partial c} \\alpha. \\]\n\tIn particular, if $d\\alpha = 0$ then the left-hand side vanishes.\n\\end{theorem}\nFor example, if $c$ is the interval $[a,b]$ then $\\partial c = \\{b\\} - \\{a\\}$,\nand thus we obtain the fundamental theorem of calculus.\n\n\\section\\problemhead\n\\begin{dproblem}[Green's theorem]\n\tLet $f,g : \\RR^2 \\to \\RR$ be smooth functions.\n\tProve that\n\t\\[ \\int_c \\left( \\fpartial gx - \\fpartial fy \\right) \\; dx \\wedge dy\n\t= \\int_{\\partial c} (f \\; dx + g \\; dy). \\]\n\t\\begin{hint}\n\t\tDirect application of Stokes' theorem to $\\alpha = f \\; dx + g \\; dy$.\n\t\\end{hint}\n\\end{dproblem}\n\n\\begin{problem}\n\tShow that $\\partial^2 = 0$.\n\t\\label{prob:partial_zero}\n\t\\begin{hint}\n\t\tThis is just an exercises in sigma notation.\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}[Pullback and $d$ commute]\n\tLet $U$ and $U'$ be open sets of vector spaces $V$ and $V'$\n\tand let $\\phi : U \\to U'$ be a smooth map between them.\n\tProve that for any differential form $\\alpha$ on $U'$ we have\n\t\\[ \\phi^\\ast(d\\alpha) = d(\\phi^\\ast\\alpha). \\]\n\t\\begin{hint}\n\t\tThis is a straightforward (but annoying) computation.\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}[Arc length isn't a form]\n\tShow that there does \\emph{not} exist a $1$-form $\\alpha$ on $\\RR^2$ such that\n\tfor a curve $c : [0,1] \\to \\RR^2$,\n\tthe integral $\\int_c \\alpha$ gives the arc length of $c$.\n\t\\begin{hint}\n\t\tWe would want $\\alpha_p(v) = \\norm{v}$.\n\t\\end{hint}\n\\end{problem}\n\n\\begin{problem}\n\tAn \\vocab{exact} $k$-form $\\alpha$ is one satisfying $\\alpha = d\\beta$ for some $\\beta$.\n\tProve that\n\t\\[ \\int_{C_1} \\alpha = \\int_{C_2} \\alpha \\]\n\twhere $C_1$ and $C_2$ are any concentric circles in the plane\n\tand $\\alpha$ is some exact $1$-form.\n\t\\begin{hint}\n\t\tShow that $d^2=0$ implies $\\int_{\\partial c} \\alpha = 0$ for exact $\\alpha$.\n\t\tDraw an annulus.\n\t\\end{hint}\n\\end{problem}\n", "meta": {"hexsha": "7104bb7582d203fe12ddbf61ed403300f860b0fe", "size": 18440, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/diffgeo/stokes.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/diffgeo/stokes.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/diffgeo/stokes.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.0206185567, "max_line_length": 122, "alphanum_fraction": 0.6293926247, "num_tokens": 6966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389930307512, "lm_q2_score": 0.8376199673867852, "lm_q1q2_score": 0.6249809190084266}} {"text": "\\RequirePackage{luatex85}\n\\documentclass[addpoints]{exam}\n\\usepackage{multicol, amsmath}\n\\everymath{\\displaystyle}\n\\begin{document}\n\\begin{questions}\n\\section{Elementary algebra. Equations.}\n\\question Solve $2=-\\frac32+b$. \\qquad\\qquad\n\\begin{oneparchoices}\n\t\\correctchoice $\\frac83$\n\t\\choice $\\frac43$\n\t\\choice 3\n\t\\end{oneparchoices}\n\n\\section{Elementary algebra. Functions.}\n\\question Let $f(x) = x^2$. Find $f(5)$.\n\\begin{oneparchoices}\n\t\\correctchoice 25\n\t\\choice $f(25)$\n\t\\end{oneparchoices}\n\n\\section{Algebra. Functions. Composition.}\n\\question Let $f(x) = -2+\\sqrt{x+3}$ and let $g(x) = \\frac1x$. Find the composition $(f\\circ g)(x)$.\n\\begin{multicols}3\n\\begin{choices}\n\t\\choice $\\frac1{-2+\\sqrt{x+3}}$\n\t\\choice $-2+\\sqrt{x+3}$\n\t\\choice $-2+\\frac1{\\sqrt{x+3}}$\n\t\\choice $-2+\\frac1{\\sqrt x+3}$\n\t\\correctchoice $-2+\\sqrt{\\frac1x+3}$\n\t\\end{choices}\n\t\\end{multicols}\n\n\\section{Algebra. Functions. Domain.}\n\\question Find the domain of $-2+\\sqrt{3+1/x}$.\n\\begin{multicols}3\n\\begin{choices}\n\t\\choice $(-\\infty,0)\\cup(0,\\infty)$ %6/30\n\t\\choice $(-\\infty,\\infty)$ %5/30\n\t\\choice $[0,\\infty)$ %5/30\n\t\\choice $(1,\\infty)$ %3/30\n\t\\choice $(-\\infty,0)\\cup(0,1)$ %3/30\n\t\\correctchoice $(-\\infty,-\\tfrac13]\\cup(0,\\infty)$ %2/30\n\t\\end{choices}\n\t\\end{multicols}\n\\question Find the domain of $\\frac1{-2+\\sqrt{3+x}}$.\n\\begin{multicols}3\n\\begin{choices}\n\t\\correctchoice $[-3,1)\\cup(1,\\infty)$\n\t\\choice $(-\\infty,0)\\cup(0,\\infty)$\n\t\\choice $(-\\infty,\\infty)$\n\t\\choice $(-\\infty,1)\\cup(1,\\infty)$\n\t\\choice $(-\\infty,-3)\\cup(-3,\\infty)$\n\t\\end{choices}\n\t\\end{multicols}\n\n\\section{Algebra. Trig. Right triangle apps.}\n\\question The angle of elevation is $26^\\circ$. Which diagram is correctly labelled.\n\n\n\\section{Calculus I. Limits. Algebraic.}\n\\question Find $\\lim_{x\\to 25} \\frac{\\sqrt x-5}{x-25}$. \\qquad\\qquad\n\\begin{oneparchoices}\n\t\\correctchoice $\\frac1{10}$\n\t\\choice $\\frac1{50}$ % +reasoning/level; +student stats\n\t\\choice $\\frac1{60}$\n\t\\choice $\\frac15$\n\t\\choice 0\n\t\\choice DNE\n\t\\choice 0.10001\n\t\\end{oneparchoices}\n\\question Find $\\lim_{h\\to0} \\frac0h$. \\qquad\\qquad\n\\begin{oneparchoices}\n\t\\correctchoice 0\n\t\\choice 1\n\t\\choice $\\infty$\n\t\\choice DNE\n\t\\end{oneparchoices}\n\n\n\\section{Calculus Lab}\n\\question Write $x^{1/(1-x)}$ using computer notation.\n\\begin{oneparchoices}\n\t\\correctchoice $x\\wedge(1/(1-x))$\n\t\\choice $(x\\wedge 1)/(1-x))$\n\t\\choice $x\\wedge 1/(1-x)$\n\t\\end{oneparchoices}\n\n\\end{questions}\n\\end{document}", "meta": {"hexsha": "5b4cfcff231a80a59e98e9b94e46f5719eaa3f03", "size": 2410, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "distractopedia.tex", "max_stars_repo_name": "gatewaycat/latex", "max_stars_repo_head_hexsha": "ae8c701522875dcd6df5919378dff3550002e06a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-11-11T12:20:40.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-11T12:20:40.000Z", "max_issues_repo_path": "distractopedia.tex", "max_issues_repo_name": "gatewaycat/byzorn", "max_issues_repo_head_hexsha": "ae8c701522875dcd6df5919378dff3550002e06a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "distractopedia.tex", "max_forks_repo_name": "gatewaycat/byzorn", "max_forks_repo_head_hexsha": "ae8c701522875dcd6df5919378dff3550002e06a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.7777777778, "max_line_length": 100, "alphanum_fraction": 0.6751037344, "num_tokens": 956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199511728003, "lm_q2_score": 0.7461390043208003, "lm_q1q2_score": 0.6249809163673106}} {"text": "\\chapter{Assignment- Differential Equations}\n\\section{MCQ}\n\\begin{enumerate}\n\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\t\\int \\frac{y}{\\sqrt{1+y^{2}}} d y+\\int \\frac{x}{\\sqrt{1+x^{2}}} d x=c \\Rightarrow \\sqrt{1+y^{2}}+\\sqrt{1+x^{2}}=c\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (c)}\n\t\\end{answer}\n\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tx d y&=-\\cot y d x \\Rightarrow \\tan y d y=-\\frac{d x}{x}\\\\\n\t\t\\Rightarrow \\int \\tan y d y&=-\\int \\frac{d x}{x}+\\log A \\Rightarrow \\log \\sec y=-\\log x+\\log A \\Rightarrow \\log \\sec y=\\log \\frac{A}{x}\\\\\n\t\t\\Rightarrow x&=A \\cos y \\Rightarrow \\sqrt{2}=A \\cos \\frac{\\pi}{4} \\Rightarrow A=2 \\Rightarrow x=2 \\cos y\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\t\\end{answer}\n\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\frac{d y}{d x}&=-\\frac{x}{y+1} \\Rightarrow x d x+y d y+d y=0 \\Rightarrow \\frac{x^{2}}{2}+\\frac{y^{2}}{2}+y=C_{1} \\Rightarrow x^{2}+y^{2}+2 y=2 C_{1}\\\\\n\t\t\\Rightarrow(x-0)^{2}+(y+1)^{2}&=2 C_{1}+1=C\n\t\t\\intertext{which is family of circles with different radii.}\n\t\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\t\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\tx y \\frac{d y}{d x}&=3 y^{2}+x^{2} \\Rightarrow \\frac{d y}{d x}=\\frac{3 y^{2}+x^{2}}{x y}\\text{ which is a homogeneous equation.}\\\\\n\t\\text{Put }y&=v x \\Rightarrow \\frac{d y}{d x}=v+x \\frac{d v}{d x} \\Rightarrow v+x \\frac{d v}{d x}=\\frac{3 v^{2} x^{2}+x^{2}}{v x^{2}}=\\frac{3 v^{2}+1}{v}\\\\\n\t\\Rightarrow x \\frac{d v}{d x}&=\\frac{3 v^{2}+1}{v}-v=\\frac{2 v^{2}+1}{v} \\Rightarrow \\frac{v}{2 v^{2}+1} d v=\\frac{d x}{x}\\\\\n\t\\Rightarrow \\frac{1}{4} \\ln \\left(2 v^{2}+1\\right)&=\\ln x+\\ln c^{\\prime} \\Rightarrow \\ln \\left(2 v^{2}+1\\right)=\\ln x^{4}+\\ln c \\Rightarrow\\left(2 v^{2}+1\\right)=c x^{4}\\\\\n\t\\Rightarrow\\left(2 \\frac{y^{2}}{x^{2}}+1\\right)&=c x^{4}\\\\\n\ty=2\\text{ when }x&=1 \\Rightarrow\\left(2 \\frac{4}{1}+1\\right)=c .1 \\Rightarrow c=9 \\Rightarrow 2 y^{2}+x^{2}=9 x^{6}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\frac{d y}{d x}=\\frac{x+2 y-1}{x+2 y+1}\\text{. Let }x+2 y&=z \\Rightarrow 1+2 \\frac{d y}{d x}=\\frac{d z}{d x} \\Rightarrow \\frac{d z}{d x}=\\frac{3 z-1}{z+1}\\\\\n\t\\Rightarrow \\int\\left(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{3 z-1}\\right) d z&=\\int d x+c \\Rightarrow \\frac{z}{3}+\\frac{4}{9} \\log (3 z-1)=x+c\\\\\n\t\\Rightarrow 3 x-3 y+a&=2 \\log (3 x+6 y-1)\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (b)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\text{I.F. }=e^{\\int 3 d x}&=e^{3 x} \\Rightarrow y \\times e^{3 x}=\\int e^{2 x} \\times e^{3 x} d x+c \\Rightarrow y \\times e^{3 x}=\\frac{e^{5 x}}{5}+c \\Rightarrow y=c_{1} e^{2 x}+c_{2} e^{-3 x}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (c)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\frac{d y}{d x}+\\frac{1}{x} y&=x^{3} \\Rightarrow \\mathrm{I} . F .=e^{\\int \\frac{1}{x} d x}=e^{\\ln x}=x\\\\\n\t\\Rightarrow y \\times x&=\\int x^{3} \\times x d x+c \\Rightarrow y \\times x=\\frac{x^{5}}{5}+c \\Rightarrow y=\\frac{x^{4}}{5}+\\frac{c}{x}\\\\\n\\text{\tSince }y&=1\\text{ at }x=1 \\Rightarrow 1=\\frac{1}{5}+\\frac{c}{1} \\Rightarrow c=\\frac{4}{5} \\Rightarrow y=\\frac{x^{4}}{5}+\\frac{4}{5 x}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (d)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\frac{d x}{d t}+\\frac{K_{2}}{K_{1}} x&=\\frac{K_{3}}{K_{1}} \\Rightarrow \\frac{d x}{d t}+A x=B \\Rightarrow I . F .=e^{\\int A d t}=e^{A t} \\Rightarrow x \\times e^{A t}=\\int\\left(B \\times e^{A t}\\right) d t+c\\\\\n\t\\Rightarrow x \\times e^{A t}&=B \\times \\frac{e^{A t}}{A}+c \\Rightarrow x=c_{1}+c e^{-A t}\\\\\n\t\\text{Since }x&=0\\text{ at }t=0 \\Rightarrow c_{1}+c=0 \\Rightarrow c_{1}=-c \\Rightarrow x=c\\left(1-e^{-A t}\\right)\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\text{Divide by }y t \\Rightarrow \\frac{1}{y} \\frac{d y}{d t}+\\frac{1}{t} \\log y&=e^{t}\\text{, put }\\log y=z \\Rightarrow \\frac{1}{y} \\frac{d y}{d t}=\\frac{d z}{d t}\\\\\n\t\\Rightarrow \\frac{d z}{d t}+\\frac{1}{t} \\cdot z&=e^{t} \\Rightarrow I \\cdot F=e^{\\int_{t}^{1} d t}=e^{\\log t}=t\\\\\n\t\\Rightarrow z t&=\\int t e^{t} d t+c \\Rightarrow t \\log y=t e^{t}-e^{t}+c\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (d)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(1+y^{2}\\right) d x&=\\left(\\tan ^{-1} y-x\\right) d y \\Rightarrow \\frac{d x}{d y}=\\frac{\\tan ^{-1} y-x}{1+y^{2}}\\\\\n\t\\Rightarrow \\frac{d x}{d y}+\\frac{x}{1+y^{2}}&=\\frac{\\tan ^{-1} y}{1+y^{2}}.\\text{ This is a linear differential equation.}\\\\\n\tI . F .&=e^{\\int \\frac{1}{1+y^{2}} d y}=e^{\\tan ^{-1} y}\\\\\n\t\\text{\tIts solution is }x . e^{\\tan ^{-1} y}&=\\int e^{\\tan ^{-1} y} \\frac{\\tan ^{-1} y}{1+y^{2}} d y+c\\\\\n\t\\text{Put }\\tan ^{-1} y&=t\\text{ on R.H.S. so that }\\frac{1}{1+y^{2}} d y=d t\\\\\n\tx . e^{\\tan ^{-1} y}&=\\int e^{t} t d t+C=t . e^{t}-e^{t}+C=e^{\\tan ^{-1} y}\\left(\\tan ^{-1} y-1\\right)+C\\\\\n\t\\Rightarrow x&=\\left(\\tan ^{-1} y-1\\right)+c e^{-\\tan ^{-1} y}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (d)}\n\\end{answer}\n\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\ty \\log y \\frac{d x}{d y}+x-\\log y&=0 \\Rightarrow \\frac{d x}{d y}+\\frac{x}{y \\log y}=\\frac{1}{y}\\\\\n\tI . F .&=e^{\\int \\frac{1}{y \\log y} d y}=e^{\\log (\\log y)}=\\log y\\\\\n\t\\text{Its solution is }x \\cdot \\log y&=\\int \\frac{1}{y}(\\log y) d y+c \\Rightarrow x \\cdot \\log y=\\frac{1}{2}(\\log y)^{2}+c\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\tM&=\\left(e^{y}+2\\right) \\sin x, N=-e^{y} \\cos x \\Rightarrow \\frac{\\partial M}{\\partial y}=e^{y} \\sin x, \\frac{\\partial N}{\\partial x}=e^{y} \\sin x\\\\\n\t\\Rightarrow \\frac{\\partial M}{\\partial y}&=\\frac{\\partial N}{\\partial x} \\Rightarrow \\int\\left(e^{y}+2\\right) \\sin x d x+0=c^{\\prime} \\Rightarrow\\left(e^{y}+2\\right) \\cos x=-c^{\\prime}=c\n\t\\end{align*}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\tM&=\\left(y^{4}+2 y\\right), N=\\left(x y^{3}+2 y^{4}-4 x\\right)\\\\\n\t\\Rightarrow \\frac{\\partial M}{\\partial y}&=4 y^{3}+2, \\frac{\\partial N}{\\partial x}=y^{3}-4 \\Rightarrow \\frac{\\frac{\\partial N}{\\partial x}-\\frac{\\partial M}{\\partial y}}{M}=-\\frac{3}{y}=f(y)\\\\\n\\text{\tthen I.F. }&=e^{\\int \\frac{3}{y} d y}=e^{-3 \\log y}=\\frac{1}{y^{3}}\\\\\n\\text{Multiplying by }&\\frac{1}{y^{3}}\\text{ we get }\\frac{1}{y^{3}}\\left(y^{4}+2 y\\right) d x+\\frac{1}{y^{3}}\\left(x y^{3}+2 y^{4}-4 x\\right) d y=0\\text{ which is an exact}\n\t\\end{align*}\n\tSo the correct answer is \\textbf{Option (c)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(D^{2}-3 D+2\\right)&=0 \\Rightarrow D=1,2 \\Rightarrow C . F .=c_{1} e^{t}+c_{2} e^{2 t}\\\\\n\tP . I .&=\\frac{1}{\\left(D^{2}-3 D+2\\right)} e^{3 t}=\\frac{1}{\\left(3^{2}-3 \\times 3+2\\right)} e^{3 t}=\\frac{1}{2} e^{3 t}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(D^{2}+2 D+101\\right)&=0 \\Rightarrow D=\\frac{-2 \\pm \\sqrt{4-4 \\times 101}}{2}=\\frac{-2 \\pm 20 i}{2}\\\\&=-1 \\pm 10 iC.F. =e^{-x}(A \\cos 10 x+B \\sin 10 x)\\\\\n\tP.I. &=\\frac{1}{\\left(D^{2}+2 D+101\\right)} 10.4 e^{x}=\\frac{1}{(1+2 \\times 1+101)} 10.4 e^{x}=\\frac{10.4}{104} e^{x}=0.1 e^{x}\\\\\n\t\\Rightarrow y&=e^{-x}(A \\cos 10 x+B \\sin 10 x)+0.1 e^{x}\\\\\n\t\\because y(0)&=1.1 \\Rightarrow 0=1(A+0)+0.1=1.1 \\Rightarrow A=1 .\\\\\n\t\\Rightarrow y&=e^{-x}(\\cos 10 x+B \\sin 10 x)+0.1 e^{x}\\\\\n\t\\Rightarrow \\frac{d y}{d x}&=e^{-x}(-10 \\sin 10 x+10 B \\cos 10 x)-e^{-x}(\\cos 10 x+B \\sin 10 x)+0.1 e^{x}\\\\\n\t\\left.\\because \\frac{d y}{d x}\\right|_{x=0}&=-0.9 \\Rightarrow-0.9=1(-0+10 B)-1(1+0)+0.1 \\Rightarrow 10 B=1-0.1-0.9 \\Rightarrow B=0 .\\\\\n\t\\Rightarrow y&=e^{-x} \\cos 10 x+0.1 e^{x}\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (a)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(D^{2}+4\\right)=0 \\Rightarrow D=+2 i,-2 i \\Rightarrow C . F .=A \\cos 2 t+B \\sin 2 t\n\t\\end{align*}\n\tSo the correct answer is \\textbf{Option (c)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(D^{2}-1\\right)&=0 \\Rightarrow D=+1,-1 \\Rightarrow C . F .=c_{1} e^{t}+c_{2} e^{-t}\\\\\n\tP.I. &=\\frac{1}{\\left(D^{2}-1\\right)} 2\\left(\\frac{e^{t}+e^{-t}}{2}\\right)=\\frac{1}{\\left(D^{2}-1\\right)}\\left(e^{t}+e^{-t}\\right)=t \\frac{1}{2 D}\\left(e^{t}+e^{-t}\\right)=\\frac{t}{2}\\left(e^{t}-e^{-t}\\right)\\\\\n\t\\Rightarrow y(t)&=c_{1} e^{t}+c_{2} e^{-t}+\\frac{t}{2}\\left(e^{t}-e^{-t}\\right)\\\\\n\ty(0)&=0 \\Rightarrow c_{1}+c_{2}=0 \\\\\n\t\\frac{d y}{d t}&=c_{1} e^{t}-c_{2} e^{-t}+\\frac{1}{2}\\left(e^{t}-e^{-t}\\right)+\\frac{t}{2}\\left(e^{t}+e^{-t}\\right),\\left.\\frac{d y}{d t}\\right|_{t=0}=0 \\Rightarrow c_{1}-c_{2}=0\\\\\n\\text{\tThus }c_{1}&=c_{2}=0 \\Rightarrow y(t)=\\frac{t}{2}\\left(e^{t}-e^{-t}\\right)=t \\sinh (t)\n\t\\end{align*}\n\tSo the correct answer is \\textbf{Option (d)}\n\\end{answer}\n\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\t\\left(D^{2}-1\\right)&=0 \\Rightarrow D=+1,-1 \\Rightarrow C . F .=C_{1} e^{t}+C_{2} e^{-t}\\\\\n\tP.I. &=\\frac{1}{\\left(D^{2}-1\\right)} 2\\left(\\frac{e^{t}-e^{-t}}{2}\\right)=\\frac{1}{\\left(D^{2}-1\\right)}\\left(e^{t}-e^{-t}\\right)=t \\frac{1}{2 D}\\left(e^{t}-e^{-t}\\right)=\\frac{t}{2}\\left(e^{t}+e^{-t}\\right)\\\\\n\t\\Rightarrow y(t)&=c_{1} e^{t}+c_{2} e^{-t}+\\frac{t}{2}\\left(e^{t}+e^{-t}\\right)\\\\\n\ty(0)&=0 \\Rightarrow c_{1}+c_{2}=0\\\\\n\t\\frac{d y}{d t}&=c_{1} e^{t}-c_{2} e^{-t}+\\frac{1}{2}\\left(e^{t}+e^{-t}\\right)+\\left.\\frac{t}{2}\\left(e^{t}-e^{-t}\\right) \\Rightarrow \\frac{d y}{d t}\\right|_{t=0}=0 \\Rightarrow c_{1}-c_{2}=0\\\\\n\\text{\tThus }c_{1}&=c_{2}=0 \\Rightarrow y(t)=\\frac{t}{2}\\left(e^{t}+e^{-t}\\right)=t \\cosh (t)\n\t\\end{align*}\n\t\tSo the correct answer is \\textbf{Option (c)}\n\\end{answer}\n\\end{enumerate}\n\\section{NAT}\n\\begin{enumerate}\n\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\frac{d x}{d t}&=x^{2} \\Rightarrow \\int \\frac{d x}{x^{2}}=\\int d t \\Rightarrow \\frac{x^{-2+1}}{-2+1}=t+C \\Rightarrow \\frac{-1}{x}=t+C\\\\\n\t\t\\Rightarrow x(0)&=1 \\Rightarrow \\frac{-1}{1}=0+C \\Rightarrow C=-1 \\Rightarrow \\frac{-1}{x}=t-1 \\Rightarrow x=\\frac{1}{1-t}\\text{ as }t \\rightarrow 1, x\\text{ blows up.}\\\\\n\t\t\\end{align*}\n\t\t\tSo the correct answer is 1\n\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\tM&=\\left(x y^{2}+\\lambda x^{2} y\\right), N=(x+y) x^{2} d y \\Rightarrow \\frac{\\partial M}{\\partial y}=2 x y+\\lambda x^{2}, \\frac{\\partial N}{\\partial x}=3 x^{2}+2 x y\\\\\n\t\t\\Rightarrow \\frac{\\partial M}{\\partial y}&=\\frac{\\partial N}{\\partial x} \\Rightarrow \\lambda=3\n\t\t\\end{align*}\n\t\t\tSo the correct answer is 3\n\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\tP.I. &=\\frac{1}{D^{2}+D+1} \\cdot \\cos 2 x=\\frac{1}{-2^{2}+D+1} \\cdot \\cos 2 x=\\frac{1}{D-3} \\cdot \\cos 2 x\\\\\n\t\t\t\\Rightarrow P . I .&=\\frac{D+3}{D^{2}-9} \\cdot \\cos 2 x=\\frac{D+3}{-2^{2}-9} \\cdot \\cos 2 x=\\frac{1}{13}(2 \\sin 2 x-3 \\cos 2 x)\n\t\t\\end{align*}\n\t\tSo the correct answer is 13\n\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\tP \\cdot I .&=\\frac{1}{D^{2}+2 D+2} \\sin x=\\frac{1}{-1+2 D+2} \\sin x=\\frac{2 D-1}{4 D^{2}-1} \\sin x=-\\frac{1}{5}(2 D-1) \\sin x\\\\\n\t\\Rightarrow P \\cdot I .&=-\\frac{1}{5}(2 \\cos x-\\sin x)=\\frac{1}{5}(\\sin x-2 \\cos x)\n\t\\end{align*}\n\t\tSo the correct answer is 2\n\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\t\\begin{answer}\n\t\t\t\\begin{align*}\n\t\t\tP \\cdot I \\cdot&=\\frac{1}{D^{2}-5 D+6} e^{t} \\cos 2 t=e^{t} \\frac{1}{(D+1)^{2}-5(D+1)+6} \\cos 2 t\\\\\n\t\t\tP \\cdot I \\cdot&=e^{t} \\frac{1}{D^{2}-3 D+2} \\cos 2 t=e^{t} \\frac{1}{-4-3 D+2} \\cos 2 t\\\\\n\t\t\tP \\cdot I \\cdot&=-e^{t} \\frac{1}{3 D+2} \\cos 2 t=-e^{t} \\frac{3 D-2}{9 D^{2}-4} \\cos 2 t\\\\\n\t\t\tP \\cdot I \\cdot&=-e^{t} \\frac{3 D-2}{9 \\times-4-4} \\cos 2 t=\\frac{e^{t}}{40}(3 D-2) \\cos 2 t=-\\frac{e^{t}}{20}(3 \\sin 2 t+\\cos 2 t)\n\t\t\t\\end{align*}\n\t\t\t\tSo the correct answer is 3\n\t\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\\begin{answer}\n\t\\begin{align*}\n\tP . I .&=\\frac{1}{D^{2}-4 D+4} \\cdot x^{3} e^{2 x}=e^{2 x} \\frac{1}{(D+2)^{2}-4(D+2)+4} \\cdot x^{3}\\\\\n\t\\Rightarrow P . I .&=e^{2 x} \\frac{1}{D^{2}} \\cdot x^{3}=e^{2 x} \\frac{1}{D}\\left(\\frac{x^{4}}{4}\\right)=e^{2 x} \\frac{x^{5}}{20}\n\t\\end{align*}\n\t\tSo the correct answer is 5\n\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t P.I. &=\\left[D^{2}+5 D+4\\right]^{-1}(3-2 x)=\\frac{1}{4}\\left[1+\\frac{5}{4} D+\\frac{5}{4} D^{2}\\right]^{-1}(3-2 x)\\\\\n\t\t\\Rightarrow P.I. &=\\frac{1}{4}\\left[1-\\frac{5}{4} D-\\frac{5}{4} D^{2}\\right](3-2 x)=\\frac{1}{4}\\left[3-2 x-\\frac{5}{4} \\times-2\\right]=\\frac{1}{8}[11-4 x]\n\t\t\\end{align*}\n\t\tSo the correct answer is 11\n\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\left(D^{2}+1\\right)&=0 \\Rightarrow D=\\pm i \\Rightarrow y=c_{1} e^{i x}+c_{2} e^{-i x}=A \\cos x+B \\sin x\\\\\n\t\t\\because y(0)&=1 \\Rightarrow 1=A \\times 1+B \\times 0 \\Rightarrow A=1 \\Rightarrow \\dot{y}=-A \\sin x+B \\cos x \\\\\n\t\t\\Rightarrow \\dot{y}(0)&=1 \\Rightarrow 1=-A \\times 0+B \\times 1 \\Rightarrow B=1 \\Rightarrow y=\\cos x+\\sin x\\\\\n\t\t\\text{For maxima, }y^{\\prime}&=-\\sin x+\\cos x=0 \\Rightarrow \\sin x=\\cos x \\Rightarrow x=45^{\\circ}\\\\\n\t\ty^{\\prime \\prime}&=-\\cos x-\\sin x, \\quad y^{\\prime \\prime}<0 \\text { for } x=45^{\\circ} \\\\\n\t\t\\Rightarrow y(\\max )&=\\cos 45^{\\circ}+\\sin 45^{\\circ}=\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}\n\t\t\\end{align*}\n\t\t\tSo the correct answer is 1.41\n\t\\end{answer}\n\t\t\\item $\\left. \\right. $\n\t\t\\begin{answer}\n\t\t\t\\begin{align*}\n\t\t\t\t\\left(D^{2}+2 \\alpha D+1\\right)=0 \\Rightarrow m_{1}, m_{1}=\\frac{-2 \\alpha \\pm \\sqrt{4 \\alpha^{2}-4}}{2} \\quad \\because m_{1}=m_{1} \\Rightarrow \\alpha=1\n\t\t\t\\end{align*}\n\t\t\tSo the correct answer is 1\n\t\t\\end{answer}\n\\item $\\left. \\right. $\n\t\\begin{answer}\n\t\t\\begin{align*}\n\t\t\\left(D^{2}+3 D+2\\right)&=0 \\Rightarrow D=-1,-2 \\Rightarrow x(t)=C_{1} e^{-t}+C_{2} e^{-2 t}\\\\\n\t\t \\Rightarrow x(1)&=\\frac{10}{e}=C_{1} e^{-1}+C_{2} e^{-2} \\Rightarrow C_{1}+C_{2} e^{-1}=10\\text{ and }C_{1}+C_{2}=20\\\\\n\t\t\\Rightarrow C_{1}&=\\frac{10 e-20}{e-1} ; C_{2}=\\frac{10 e}{e-1} \\\\\n\t\tx(2)&=\\left(\\frac{10 e-20}{e-1}\\right) e^{-2}+\\left(\\frac{10 e}{e-1}\\right) e^{-4}=0.8566\n\t\t\\end{align*}\n\t\t\tSo the correct answer is 0.8566\n\t\\end{answer}\n\\end{enumerate}", "meta": {"hexsha": "02017243feae654c138210b5edb5edc9e442f141", "size": 14212, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "CSIR- Mathematical Physics/chapter/Assignments/Assignment Differential Equations solutions.tex", "max_stars_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_stars_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "CSIR- Mathematical Physics/chapter/Assignments/Assignment Differential Equations solutions.tex", "max_issues_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_issues_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "CSIR- Mathematical Physics/chapter/Assignments/Assignment Differential Equations solutions.tex", "max_forks_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_forks_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.4285714286, "max_line_length": 211, "alphanum_fraction": 0.5678300028, "num_tokens": 6575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8376199754937772, "lm_q2_score": 0.7461389817407016, "lm_q1q2_score": 0.6249809156005983}} {"text": "% !TEX root = Thesis.tex\n\\section{Relational Model}\n\tThe relational data model, in its most basic form, is built upon sets and tuples. Each of these sets consist of a set of finite values. Tuples are constructed from these sets to form relations.\n\t\n\t\\begin{defn}[Named Tuple]\n\t\\label{def:named-tuple}\n\t\tA named tuple \\(\\tuple\\) is an instance of a relation \\(\\relation\\), consisting of values corresponding to the attributes of \\(\\relation\\). For example,\n\t\\end{defn}\n\t\n\t\\begin{ex}\n\t\tGiven a tuple \\(\\tuple = \\{\\text{code}: \\text{``CDPS 101''}, \\text{title}: \\text{``Human-Mutant Relations''}, \\text{subject}: \\text{``CDPS''}\\}\\), we denote the attributes of \\(\\tuple\\) as \\(\\attributes{\\tuple} = \\{\\text{code}, \\text{title}, \\text{subject}\\}\\). The values are \\(\\tuple\\lbrack\\text{code}\\rbrack = \\text{``CDPS 101''}\\), \\(\\tuple\\lbrack\\text{title}\\rbrack = \\text{``Human-Mutant Relations''}\\), and \\(\\tuple\\lbrack\\text{subject}\\rbrack = \\text{``CDPS''}\\).\n\t\\end{ex}\n\t\n\t\\begin{defn}[Relation]\n\t\\label{def:relation}\n\t\tA relation \\(\\relation\\) is a set of named tuples, \\(\\relation = \\{\\tuple_1, \\tuple_2, \\dotsc, \\tuple_n\\}\\), such that all the named tuples share the same attributes.\n\t\t\\begin{equation}\n\t\t\t\\forall \\tuple, \\tuple' \\in \\relation, \\attributes{\\tuple} = \\attributes{\\tuple'}\n\t\t\\end{equation}\n\t\\end{defn}\n\t\n\t\\begin{ex}\n\t\\label{ex:relation}\n\t\tAn example Course relation, \\(\\relation\\), would be\n\t\t\\[\n\t\t\t\\relation = \\left\\{\n\t\t\t\t\\begin{array}{llllll}\n\t\t\t\t\t\\{\\text{code} &: \\text{``CDPS 101''}, &\\text{title} &: \\text{``Human-Mutant Relations''}, &\\text{subject} &: \\text{``CDPS''}\\}, \\\\\n\t\t\t\t\t\\{\\text{code} &: \\text{``CDPS 201''}, &\\text{title} &: \\text{``Humans and You''}, &\\text{subject} &: \\text{``CDPS''}\\}, \\\\\n\t\t\t\t\t\\{\\text{code} &: \\text{``MATH 360''}, &\\text{title} &: \\text{``Complex Analysis''}, &\\text{subject} &: \\text{``MATH''}\\}\n\t\t\t\t\\end{array}\n\t\t\t\\right\\}\n\t\t\\]\n\t\\end{ex}\n\t\t\n\tRelations are typically referred to, and represented, as tables. \\cref{ex:relation} is shown in its tabular form in \\cref{tbl:course-relation}.\n\t\n\t\\begin{table}\n\t\t\\centering\n\t\t\n\t\t\\begin{tabular}{lll}\n\t\t\t\\toprule\n\t\t\tcode & title & subject \\\\\n\t\t\t\\midrule\n\t\t\tCDPS 101 & Human-Mutant Relations & CDPS \\\\\n\t\t\tCDPS 201 & Humans and You & CDPS \\\\\n\t\t\tMATH 360 & Complex Analysis & MATH \\\\\n\t\t\t\\bottomrule\n\t\t\\end{tabular}\n\t\t\n\t\t\\caption{Tabular representation of the Course relation}\n\t\t\\label{tbl:course-relation}\n\t\\end{table}\n\t\n\t\\begin{defn}[Keys]\n\t\\label{def:keys}\n\t\tKeys are constraints imposed on relations.\tA key constraint \\(\\key\\) on a relation \\(\\relation\\) is a subset of \\(\\attributes{\\relation}\\) which must uniquely identify a tuple.\tFormally, we say \\(\\relation\\) satisfies the key constraint \\(\\key\\), denoted as \\(\\relation \\models \\key\\), subject to\n\t\t\\[\n\t\t\t\\forall \\tuple, \\tuple' \\in \\relation, \\tuple \\not= \\tuple' \\implies \\tuple\\lbrack\\key\\rbrack \\not= \\tuple'\\lbrack\\key\\rbrack\n\t\t\\]\n\t\t\n\t\tFor example, in \\cref{tbl:course-relation}, the relation satisfies the key constraint \\(\\{\\text{code}\\}\\) or \\(\\{\\text{title}\\}\\), but not \\(\\{\\text{subject}\\}\\).\n\t\\end{defn}\n\t\n\t\\begin{defn}[Foreign Keys]\n\t\\label{def:foreign-keys}\n\t\t\\glsreset{fk} % Want to spell it out in the definition\n\t\tA \\gls{fk} constraint applies to two relations, \\(\\relation_1, \\relation_2\\). It asserts that values of certain attributes of \\(\\relation_1\\) must appear as values of some corresponding attributes of \\(\\relation_2\\). A \\gls{fk} constraint is written as\n\t\t\\[\n\t\t\t\\theta = \\relation_1(\\attribute_{1,1}, \\attribute_{1,2}, \\dotsc, \\attribute_{1,k}) \\rightarrow \\relation_2(\\attribute_{2,1}, \\attribute_{2,2}, \\dotsc, \\attribute_{2,k})\n\t\t\\]\n\t\t\n\t\twhere \\(\\attribute_{1,i} \\subseteq \\attributes{\\relation_1}\\) and \\(\\attribute_{2,i} \\subseteq \\attributes{\\relation_2}\\). We say \\((\\relation_1, \\relation_2)\\) satisfies \\(\\theta\\), denoted as \\((\\relation_1, \\relation_2) \\models \\theta\\), if for every tuple \\(\\tuple \\in \\relation_1\\), there exists a tuple \\(\\tuple' \\in \\relation_2\\) such that \\(\\tuple\\lbrack\\attribute_{1,1}, \\attribute_{1,2}, \\dotsc, \\attribute_{1,k}\\rbrack = \\tuple'\\lbrack\\attribute_{2,1}, \\attribute_{2,2}, \\dotsc, \\attribute_{2,k}\\rbrack\\).\n\t\t\n\t\tWe say \\(\\relation_1\\) is the source, while \\(\\relation_2\\) is the target.\n\t\t\t\t\n\t\t\\begin{ex}\n\t\t\tSuppose we have a relation \\(\\rel{Course}{code, title, subject}\\).\t We impose a \\gls{fk} constraint of\n\t\t\t\\begin{equation}\n\t\t\t\t\\theta = \\rel{Course}{subject} \\rightarrow \\rel{Subject}{id}\n\t\t\t\\end{equation}\n\t\t\t\n\t\t\twhich asserts \\((\\text{Course}, \\text{Subject}) \\models \\theta\\). Therefore, if\n\t\t\t\\[\n\t\t\t\t\\tuple = \\{\\text{code}: \\text{``CDPS 101''}, \\text{title}: \\text{``Human-Mutant Relations''}, \\text{subject}: \\text{``CDPS''}\\}\n\t\t\t\\]\n\t\t\t\n\t\t\tthen \\(\\exists! \\tuple' \\in \\text{Subject}\\) such that \\(\\tuple'\\lbrack\\text{id}\\rbrack = \\text{``CDPS''}\\).\n\t\t\\end{ex}\n\t\\end{defn}\n\t\n\t\\begin{defn}[Relational Database]\n\t\\label{def:relational-database}\n\t\tA relational database, \\(\\db\\), is a named collection of relations (\\cref{def:relation}), keys (\\cref{def:keys}), and foreign key constraints (\\cref{def:foreign-keys}).\n\t\t\n\t\tWe use \\(\\name{\\db}\\) to denote the name of \\(\\db\\), \\(\\relations{\\db}\\) the list of relations in \\(\\db\\), \\(\\keys{\\db}\\) the list of key constraints of \\(\\db\\), and \\(\\fks{\\db}\\) the list of foreign key constraints of \\(\\db\\).\n\t\\end{defn}\n\t\n\t\\subsection{Schema Group}\n\t\t\\begin{defn}[Schema Graph]\n\t\t\\label{def:schema-graph}\n\t\t\tIf we view relations as vertices, and \\gls{fk} constraints as edges, a database \\(\\db\\) can be viewed as a schema graph, \\(\\sgraph{}\\), formally defined as\n\t\t\t\\begin{align}\n\t\t\t\t\\text{vertices} &: \\text{V}(\\sgraph) = \\relations{\\db} \\\\\n\t\t\t\t\\text{edges} &: \\text{E}(\\sgraph) = \\fks{\\db}\n\t\t\t\\end{align}\n\t\t\\end{defn}\n\t\t\n\t\t\\begin{ex}\n\t\t\tGiven the schema in \\cref{tbl:schema} and the \\gls{fk} constraints in \\cref{eqn:fk-constraints}, we produce the schema graph in \\cref{fig:schema-graph}.\n\t\t\t\n\t\t\t\\begin{table}\n\t\t\t\t\\centering\n\t\t\t\t\n\t\t\t\t\\begin{tabular}{ll}\n\t\t\t\t\t\\toprule\n\t\t\t\t\tRelation & Attributes \\\\\n\t\t\t\t\t\\midrule\n\t\t\t\t\tCourse & \\underline{code}, title, subject \\\\\n\t\t\t\t\tSection & \\underline{id}, actual, campus, capacity, credits, levels, registration\\_start, \\\\\n\t\t\t\t\t & registration\\_end, semester, sec\\_code, sec\\_number, year, course \\\\\n\t\t\t\t\tSchedule & \\underline{id}, date\\_start, date\\_end, day, schedtype, hour\\_start, hour\\_end, \\\\\n\t\t\t\t\t & min\\_start, min\\_end, classtype, location, section\\_id \\\\\n\t\t\t\t\tInstructor & \\underline{id}, name \\\\\n\t\t\t\t\tTeaches & \\underline{id}, schedule\\_id, instructor\\_id, position \\\\\n\t\t\t\t\t\\bottomrule\n\t\t\t\t\\end{tabular}\n\t\t\t\t\n\t\t\t\t\\caption{Subset of mycampus dataset schema}\n\t\t\t\t\\label{tbl:schema}\n\t\t\t\\end{table}\n\t\t\t\n\t\t\t\\begin{figure}\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\rel{Schedule}{section\\_id} &\\rightarrow \\rel{Section}{id} \\\\\n\t\t\t\t\t\\rel{Section}{course} &\\rightarrow \\rel{Course}{code} \\\\\n\t\t\t\t\t\\rel{Teaches}{schedule\\_id} &\\rightarrow \\rel{Schedule}{id} \\\\\n\t\t\t\t\t\\rel{Teaches}{instructor\\_id} &\\rightarrow \\rel{Instructor}{id} \\\\\n\t\t\t\t\\end{align*}\n\t\t\t\t\n\t\t\t\t\\caption{\\gls{fk} constraints on schema in \\cref{tbl:schema}}\n\t\t\t\t\\label{eqn:fk-constraints}\n\t\t\t\\end{figure}\n\t\t\t\n\t\t\t\\begin{figure}\n\t\t\t\t\\centering\n\t\t\t\t\n\t\t\t\t\\begin{dot2tex}[neato]\n\t\t\t\t\tdigraph G {\n\t\t\t\t\t\tnode [shape=plaintext]; Course; Section; Schedule; Instructor; Teaches;\n\t\t\t\t\t\t\n\t\t\t\t\t\tSchedule -> Section;\n\t\t\t\t\t\tSection -> Course;\n\t\t\t\t\t\tTeaches -> Schedule;\n\t\t\t\t\t\tTeaches -> Instructor;\n\t\t\t\t\t}\n\t\t\t\t\\end{dot2tex}\n\t\t\t\t\n\t\t\t\t\\caption{Graph representation of relations (\\cref{tbl:schema}) and \\gls{fk} (\\cref{eqn:fk-constraints})}\n\t\t\t\t\\label{fig:schema-graph}\n\t\t\t\\end{figure}\n\t\t\\end{ex}\n\t\t\n\t\tThe relational data model is particularly powerful for analytic queries.\n\t\t\n\t\\subsection{Entity Group}\n\t\t\\begin{defn}[Entity Group]\n\t\t\\label{def:entity-group}\n\t\t\tAn entity group is a forest, \\(\\egraph\\), of tuples interconnected by join conditions defined by the \\gls{fk} constraints in the schema graph \\(\\sgraph\\).\n\t\t\t\n\t\t\tGiven two vertices \\((\\tuple, \\tuple') \\in \\text{V}(\\egraph)\\), \\(\\exists (\\relation_1, \\relation_2) \\in \\relations{\\db}\\) such that \\(\\tuple \\in \\relation_1\\), \\(\\tuple'\\in \\relation_2\\), and \\((\\relation_1, \\relation_2)\\in \\sgraph\\). That is, \\(\\tuple\\) and \\(\\tuple'\\) belong to two relations that are directly connected by the schema graph.\n\t\t\\end{defn}\n\t\t\n\t\t\\begin{ex}\n\t\t\tLet \\(\\relation_1(\\attribute_{11}, \\attribute_{12}, \\dotsc, \\attribute_{1k}) \\to \\relation_2(\\attribute_{21}, \\attribute_{22}, \\dotsc, \\attribute_{2k})\\) be the \\gls{fk} that connects \\((\\relation_1, \\relation_2)\\). We assert that \\(\\tuple\\lbrack\\attribute_{11}, \\attribute_{12}, \\dotsc, \\attribute_{1k}\\rbrack = \\tuple'\\lbrack\\attribute_{21}, \\attribute_{22}, \\dotsc, \\attribute_{2k}\\rbrack\\).\n\t\t\\end{ex}\n\t\t\n\t\tEntity groups define complex, structured objects that include more information than individual tuples in the relations.\n\t\t\n\t\t\\begin{ex}\n\t\t\tThe information in \\cref{tbl:hmr-properties-rel} all relates to the Course titled Human-Mutant Relations, however no single tuple in the database has all of this information as a result of database normalization.\n\t\t\t\n\t\t\t\\begin{table}\n\t\t\t\t\\centering\n\t\t\t\t\n\t\t\t\t\\begin{tabular}{ll}\n\t\t\t\t\t\\toprule\n\t\t\t\t\tAttribute & Value \\\\\n\t\t\t\t\t\\midrule\n\t\t\t\t\tcode & CDPS 101 \\\\\n\t\t\t\t\ttitle & Human-Mutant Relations \\\\\n\t\t\t\t\tsubject & Community Development \\& Policy Studies \\\\\n\t\t\t\t\t\\bottomrule\n\t\t\t\t\\end{tabular}\n\t\t\t\t\n\t\t\t\t\\caption{Properties of the Course titled Human-Mutant Relations.}\n\t\t\t\t\\label{tbl:hmr-properties-rel}\n\t\t\t\\end{table}\n\t\t\t\n\t\t\tWe require an entity group to join together all pieces of information related to this course. An example of this is given in \\cref{fig:hmr-entity-group}, where an example entity group comprised of a Course and Subject is shown.\n\t\t\t\n\t\t\t\\begin{figure}\n\t\t\t\t\\centering\n\t\t\t\t\n\t\t\t\t\\begin{dot2tex}[dot]\ndigraph G {\n\tnode [shape=plaintext]; \"Human-Mutant Relations\"; \"Community Development & Policy Studies\";\n\t\n\t\"Human-Mutant Relations\" -> \"Community Development & Policy Studies\";\n}\n\t\t\t\t\\end{dot2tex}\n\t\t\t\t\n\t\t\t\t\\caption{Human-Mutant Relations entity group}\n\t\t\t\t\\label{fig:hmr-entity-group}\n\t\t\t\\end{figure}\n\t\t\\end{ex}\n\n\t\\subsection{Benefits and Drawbacks of the Relational Model}\n\t\tTo better understand the motivation behind this work, it is important to examine both the strong and weak points of the relational model.\n\t\t\n\t\t\\subsubsection{Benefits}\n\t\t\tThe enforcement of constraints is essential to the relational model. Types of constraints include uniqueness and \\glspl{fk}. The first constraint maintains uniqueness.\n\t\t\t\n\t\t\tThe Course relation (\\cref{tbl:course-relation}) has the attribute \\texttt{code} as its primary key.\tIn order for other relations to reference a specific named tuple, the \\texttt{code} attribute must be unique.\n\t\t\t\n\t\t\t\\begin{ex}[Unique Constraint]\n\t\t\t\\label{ex:unique-constraint}\n\t\t\t\tAttempt to insert another course with a \\texttt{code} of ``CDPS 101.''\n\t\t\t\t\n\t\t\t\t\\begin{singlespaced}\n\t\t\t\t\t\\begin{pygments}{sql}\nINSERT INTO course\nVALUES ('CDPS 101',\n 'Mutant-Human Relations',\n 'CDPS');\n\t\t\t\t\t\\end{pygments}\n\t\t\t\t\\end{singlespaced}\n\t\t\t\t\n\t\t\t\tThe \\gls{rdbms} enforces the primary key constraint on the \\texttt{code} attribute, rejecting the insertion.\n\t\t\t\t\n\t\t\t\t\\begin{verbatim}\nError: column code is not unique\n\t\t\t\t\\end{verbatim}\n\t\t\t\\end{ex}\n\t\t\t\n\t\t\tWith the uniqueness of named tuples guaranteed (as demonstrated in \\cref{ex:unique-constraint}), we must ensure that any named tuples that are referenced actually exist. If they do not, the database must not permit the operation to continue. Doing so leads to dangling references.\n\t\t\t\n\t\t\t\\begin{ex}[Referential Integrity]\n\t\t\t\tAttempt to insert the tuple (``CHEM 101'', ``Introductory Chemistry'', ``'CHEM'') in the Course relation.\n\t\t\t\t\n\t\t\t\t\\begin{singlespaced}\n\t\t\t\t\t\\begin{pygments}{sql}\nINSERT INTO course\nVALUES ('CHEM 101',\n 'Introductory Chemistry',\n 'CHEM');\n\t\t\t\t\t\\end{pygments}\n\t\t\t\t\\end{singlespaced}\n\t\t\t\t\n\t\t\t\tAgain we see the \\gls{rdbms} protecting the integrity of the data.\n\t\t\t\t\n\t\t\t\t\\begin{verbatim}\nError: foreign key constraint failed\n\t\t\t\t\\end{verbatim}\n\t\t\t\\end{ex}\n\t\t\t\n\t\t\tThe relational model, in addition to enforcing consistency, is capable of providing higher-level views of the data through aggregation.\n\t\t\t\n\t\t\t\\begin{ex}[Aggregation]\n\t\t\t\tFind the number of sections offered for the subject named ``Community Development \\& Policy Studies.''\n\t\t\t\t\n\t\t\t\t\\begin{singlespaced}\n\t\t\t\t\t\\begin{pygments}{sql}\nSELECT Count(*)\nFROM section\n JOIN course\n ON section.course = course.code\n JOIN subject\n ON subject.id = course.subject\nWHERE subject.name = 'Community Development & Policy Studies';\n\t\t\t\t\t\\end{pygments}\n\t\t\t\t\\end{singlespaced}\n\t\t\t\\end{ex}\n\t\t\t\n\t\t\tInformation stored within a properly designed database is normalized. That is, no information is repeated.\n\t\t\t\n\t\t\t\\begin{ex}[Normalization]\n\t\t\t\tFor example, suppose Emma Frost became headmistress and the subject named ``Community Development \\& Policy Studies'' was renamed to ``Community Destruction \\& Policy Studies.'' If this information were not normalized, each course in this subject needs to be updated. Since this information is normalized, the following query will suffice.\n\t\t\t\t\n\t\t\t\t\\begin{singlespaced}\n\t\t\t\t\t\\begin{pygments}{sql}\nUPDATE subject\nSET name = 'Community Destruction & Policy Studies'\nWHERE id = 'CDPS';\n\t\t\t\t\t\\end{pygments}\n\t\t\t\t\\end{singlespaced}\n\t\t\t\\end{ex}\n\t\t\t\n\t\t\tThe above examples are some of the most important reasons for choosing the relational model over others.\tUnfortunately, the relational model is not without its downsides.\n\t\t\n\t\t\\subsubsection{Drawbacks}\n\t\t\tWhile the relational model excels at ensuring data consistency, aggregation, and reporting; it is not suitable for every task.\tA user must be familiar with the schema to issue queries. This requires specific domain knowledge of the data.\n\t\t\t\n\t\t\tA casual user is unlikely to determine the correct join path, know the exact name of the tables and attributes, etc. This is in contrast to the document model where the data is semi-structured or unstructured, requiring minimal domain knowledge.\n\n\t\t\tThe relational model is also rigid in structure. If a relation is modified, every query referencing said relation may require a rewrite. Even a simple attribute being renamed (e.g.~\\(\\rho_{\\text{name/alias}}(\\text{Person})\\)) is capable of modifying the join paths. This rigidity places additional cognitive burden on users.\n\t\t\t\n\t\t\tIn addition to having a rigid structure, most relational database management systems lack flexible string matching options.\t Assuming basic SQL-92 compliance, a \\gls{rdbms} only supports the \\texttt{LIKE} predicate \\cite{sql-11}.\n\t\t\t\n\t\t\t\\begin{ex}[\\texttt{LIKE} Predicate]\n\t\t\t\tFind all courses with a title that contains ``man.''\n\t\t\t\t\n\t\t\t\t\\begin{singlespaced}\n\t\t\t\t\t\\begin{pygments}{sql}\nSELECT *\nFROM course\nWHERE title LIKE '%man%';\n\t\t\t\t\t\\end{pygments}\n\t\t\t\t\\end{singlespaced}\n\t\t\t\\end{ex}\n\t\t\t\n\t\t\tA couple of limitations to the \\texttt{LIKE} predicate exist. First, it only supports basic substring matching. If a user accidentally searches for all courses with a title containing ``men,'' nothing would be found.\n\t\t\t\n\t\t\tSecond, unless the predicate is applied to the end of the string and the column is indexed, performance will be poor. The \\gls{rdbms} must scan the entire relation to answer the query, resulting in performance of \\(\\mathcal{O}(n)\\), where \\(n\\) is the number of named tuples in the relation.", "meta": {"hexsha": "096d017424c9f65db158cecd593aa676ba19362f", "size": 15407, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/document/relational-model.tex", "max_stars_repo_name": "rdrake/Molly", "max_stars_repo_head_hexsha": "f074ba926e6459d0074d980297d81dcad7f736d4", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "thesis/document/relational-model.tex", "max_issues_repo_name": "rdrake/Molly", "max_issues_repo_head_hexsha": "f074ba926e6459d0074d980297d81dcad7f736d4", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-12-13T19:40:08.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-13T19:40:08.000Z", "max_forks_repo_path": "thesis/document/relational-model.tex", "max_forks_repo_name": "rdrake/Molly", "max_forks_repo_head_hexsha": "f074ba926e6459d0074d980297d81dcad7f736d4", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.5524691358, "max_line_length": 519, "alphanum_fraction": 0.6809891608, "num_tokens": 4564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696747, "lm_q2_score": 0.7905303162021596, "lm_q1q2_score": 0.624938182785332}} {"text": "% Copyright 2018-2021 Melvin Eloy Irizarry-Gelpí\n\\setcounter{chapter}{4}\n\\chapter{Capacitors}\n%\n...\n%\n\\section{Preliminary}\n%\nResistance is a property of the components in an electric circuit. You can use units of ohm to measure electric resistance. One ohm of electric resistance is equivalent to\n\\begin{equation}\n\t1 \\text{ ohm} = 1 \\text{ V/A}\n\\end{equation}\nAnother property is capacitance. You can measure capacitance in units of farads (F). One farad of electric capacitance is equivalent to\n\\begin{equation}\n\t1 \\text{ F} = 1 \\text{ C/V}\n\\end{equation}\nIn some ways, capacitance measures the amount of electric charge per unit voltage.\n\nA component in a circuit with an electric resistance is called a \\textbf{resistor}. Similarly, a component in a circuit with a capacitance is called a \\textbf{capacitor}. A circuit with both capacitors and resistors is known as an RC circuit. The simplest RC circuit has one resistor and one capacitor connected in series to a battery (a DC source).\n\nA particular quantity $\\tau_{C}$ is given by multiplying resistance by capacitance:\n\\begin{equation}\n\t\\tau_{C} = R C\n\t\\label{eq.05C.tauC}\n\\end{equation}\nThis quantity is known as the \\textbf{capacitive time constant}. If resistance is in ohms and capacitance is in farads, then $\\tau_{C}$ is in seconds. The capacitive time constant is important because it describes how quickly voltage and current change in circuits with resistors and capacitors. During a \\textbf{charging event}, the \\textbf{electric current} drops in an exponential way:\n\\begin{equation}\n\tI(t) = I_{0} \\exp{\\left(- \\frac{t}{\\tau_{C}} \\right)}\n\t\\label{eq.05C.It}\n\\end{equation}\nDuring a \\textbf{discharging event}, the \\textbf{voltage} drops in an exponential way:\n\\begin{equation}\n\tV(t) = V_{0} \\exp{\\left(- \\frac{t}{\\tau_{C}} \\right)}\n\t\\label{eq.05C.Vt}\n\\end{equation}\n%\n\\section{Experiment}\n%\nThere are two kinds of experiments: the charging event and the discharging event. In each experiment you measure the current flowing between the resistor and the capacitor, and the voltage across the capacitor.\n\nFor the \\textbf{charging event}, the data collection is started with the switch off (i.e. incomplete circuit) and shortly after turned on. You should analyze the \\textbf{current} data for such events, as it follows the behavior in (\\ref{eq.05C.It}).\n\nFor the \\textbf{discharging event}, the data collection is started with the switch on (i.e. complete circuit) and shortly after turned off. You should analyze the \\textbf{voltage} data for such events, as it follows the behavior in (\\ref{eq.05C.Vt}).\n\nThere are 9 runs of data:\n\\begin{enumerate}\n\t\\item Charging capacitor with $R = 10$ ohm and $C = 0.025$ F.\n\t\\item Discharging capacitor with $R = 10$ ohm and $C = 0.025$ F.\n\t\\item Charging capacitor with $R = 51$ ohm and $C = 0.025$ F.\n\t\\item Discharging capacitor with $R = 51$ ohm and $C = 0.025$ F.\n\t\\item Charging capacitor with $R = 68$ ohm and $C = 0.025$ F.\n\t\\item Discharging capacitor with $R = 68$ ohm and $C = 0.025$ F.\n\t\\item Discharging capacitor with $R = 22 \\times 10^{3}$ ohm and $C = 10^{-5}$ F.\n\t\\item Discharging capacitor with $R = 47 \\times 10^{3}$ ohm and $C = 10^{-5}$ F.\n\t\\item Discharging capacitor with $R = 100 \\times 10^{3}$ ohm and $C = 10^{-5}$ F.\n\\end{enumerate}\nEach time the resistance or the capacitance changes, the time constant $\\tau_{C}$ will also change.\n%\n\\section{Analysis}\n%\nThe analysis is pretty simple: you need to make a graph of the desired quantity versus time and then include an \\textbf{exponential} fit. The exponential fit is of the form:\n\\begin{equation}\n\ty = A e^{-Bx}\n\t\\label{eq.05C.exp.fit}\n\\end{equation}\nComparing (\\ref{eq.05C.exp.fit}) with (\\ref{eq.05C.It}) and also (\\ref{eq.05C.Vt}) leads to identifying the experimental estimate of $\\tau_{C}$ with $1 / B$.\n\nThe trouble is that it takes some effort to prepare the data in such a way that the exponential fit is appropriate and accurate. You need to do two things:\n\\begin{enumerate}\n\t\\item Truncate the time region before the switch is turned on/off.\n\t\\item Truncate the long time tail after the exponential drop.\n\\end{enumerate}\nFor the first truncation, just find the time value where the voltage (in discharging events) or the current (in charging events) suddenly jumps. Remove all the values before this time and perhaps one or two events right after the exponential drop begins.\n\nFor the second truncation, one good rule of thump is to use the expected value of $\\tau_{C}$. You should keep a time region that is about $5\\tau_{C}$ long in time, starting from the beginning of the exponential drop.\n%\n\\section{My Data}\n%\nIn the spreadsheet that I shared with you, there is a sheet called ``Results''. This sheet has a table with eight columns and nine rows. Each of the rows corresponds to one of the nine runs mentioned above. Here are the columns:\n\\begin{itemize}\n\t\\item Column 1: run label.\n\t\\item Column 2: value of resistance (in ohm) for the resistor used.\n\t\\item Column 3: value of the capacitance (in F) for the capacitor used.\n\t\\item Column 4: Expected value of $\\tau_{C}$ (in s). Use (\\ref{eq.05C.tauC}) with columns 2 and 3 to calculate these values.\n\t\\item Column 5: Expected value of $1 / \\tau_{C}$ (in 1/s). Just take the reciprocal of the values in column 4.\n\t\\item Column 6: Fit value of $1 / \\tau_{C}$ (in 1/s). This correspond to the $B$ value in the exponential fit. See (\\ref{eq.05C.exp.fit}).\n\t\\item Column 7: Fit value of $\\tau_{C}$ (in s). This is just the reciprocal of the values in column 6.\n\t\\item Column 8: The percent difference between the theoretical values in column 4 and the experimental values in column 7.\n\\end{itemize}\nFor the percent difference, use\n\\begin{equation}\n\t\\text{Percent Difference } = 100 \\times (\\text{experiment } - \\text{ theory}) / \\text{theory}\n\\end{equation}\nIn my case I found mixed agreement, with the largest disagreements not being larger than 20\\%.\n%\n\\section{Your Data}\n%\nYour data is structured in the same way as my data, but the time delay for turning on/off the switch is different.\n%\n\\section{Your Lab Report}\n%\nIn your lab report you should include:\n\\begin{enumerate}\n\t\\item One voltage versus time graph. Choose one from runs 2, 4, or 6. Include the exponential fit line, and also the equation.\n\t\\item One current versus time graph. Choose one from runs 1, 3, or 5. Include the exponential fit line, and also the equation.\n\t\\item Another voltage versus time graph. Choose one from runs 7, 8, or 9. Include the exponential fit line, and also the equation.\n\t\\item A table in your printed lab report similar to my table in the ``Results'' sheet from the shared spreadsheet.\n\\end{enumerate}\nYou should also include answers to the following questions:\n\\begin{enumerate}\n\t\\item Are the fit values for $\\tau_{C}$ for runs 1, 3, and 5 consistent with the corresponding values for runs 2, 4, and 6?\n\t\\item Note that some runs involve current measurements and the other runs involve voltage measurements. In general, which runs give fit values for $\\tau_{C}$ that are closer to the expected values: charging events (current) or discharging events (voltage)?\n\t\\item Why are the current measurements in runs 7, 8 and 9 so small?\n\\end{enumerate}", "meta": {"hexsha": "d1de0343340eeeb0dec64215dd178719fb9e1583", "size": 7191, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter/05-RC.tex", "max_stars_repo_name": "meirizarrygelpi/phys-208L", "max_stars_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter/05-RC.tex", "max_issues_repo_name": "meirizarrygelpi/phys-208L", "max_issues_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapter/05-RC.tex", "max_forks_repo_name": "meirizarrygelpi/phys-208L", "max_forks_repo_head_hexsha": "ce8b093f9e730c87a2b4c6208b4d4da518035267", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.0789473684, "max_line_length": 388, "alphanum_fraction": 0.7446808511, "num_tokens": 1985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996143, "lm_q2_score": 0.79053032607222, "lm_q1q2_score": 0.6249381827853319}} {"text": "\\subsection{Implementation of Ruder style transfer for video}\n\\label{sec:Ruder style transfer}\nAs a part of our project we have implemented some of the components described in the paper by Ruder et al. \\cite{Ruder:1}. Here we try to reduce the noise that appears when styling each image independently by introducing a temporal constraint in order to penalize the model for straying too far away from the previous image.\\newline\\newline\nThe temporal constraint we have implemented is an additional loss-component: \\newline\n\n\\begin{equation}\n\\mathcal{L}_{temporal}(x, \\omega, c) = \\frac{1}{D}\\sum_{k=1} c_k \\cdot (x_k - \\omega_k)^2\n\\end{equation}\nSince we are using the style and content-loss used in Gatys style transfer, our total loss-function with a shortterm temporal constraint is this:\\newline\n\\begin{equation}\n\\mathcal{L}_{shortterm}(p^{(i)}, a, x) = \\alpha \\mathcal{L}_{content} + \\beta \\mathcal{L}_{style} + \\gamma \\mathcal{L}_{temporal}(x^{(i)}, \\omega_{i-1}^i(x^{i-1}, c^{i-1, i}))\n\\end{equation}\nWe use different weights for different resolutions, generally increasing the style-loss and decreasing the temporal constraint with increasing resolution.\\newline\nAs well as using this shortterm temporal loss function, we also initialize every frame after the first stylized frame with the previously stylized frame \\textit{i-1} warped to the current image \\texit{i}. This is done by using the flow between the content image $i-1$ and $i$. \nTo calculate this flow we need a flow algorithm. In our implementation we have used cv2's DeepFlow function. The calculated flow between frames is also used to create our c-matrix using the inequalities described in section \\ref{sec:usage_of_optical_flow}. These inequalities result in c-matrix masks like these:\\newline \n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.5]{report/Method/images/shrek_flow.png}\n\\caption{A frame from the shrek video and the flow image calculated, indicating what parts of the image are moving and should be recalculated}\n\\label{fig:architecture}\n\\end{center}\n\\end{figure}\n\\newline\nTo calculate these values for each frame, we have primarily used the numpy library. We have also implemented the long-term temporal constraint described in section \\ref{sec:long_term_temporal_loss}. However, a lot of additional computation is needed for this. With our current access to hardware and the negligible results we got from testing our implementation, we decided to not use this in our results.\\newline\\newline\nWith the implemented temporal constraint, we are able to have some consistency between frames, resulting in less noise: \\newline\n\\begin{figure}[!ht]\n\\begin{center}\n\\includegraphics[scale=0.11]{report/Method/images/ruder.png}\n\\caption{Two consecutive frames produced by Ruders method using La Muse (picasso) as style refernce, source video is from Parasite}\n\\label{fig:architecture}\n\\end{center}\n\\end{figure}Here we can see the same artifacts staying in the same place in both adjacent frames, while the person to the left has moved. The results from this implementation can be seen in \\ref{seq:ruder_result}", "meta": {"hexsha": "906fc4553d7a786c9de8501178347e698edd1551", "size": 3102, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/Method/Ruder.tex", "max_stars_repo_name": "kjerand/video-style-transfer", "max_stars_repo_head_hexsha": "fe44a1b486e976725cddd6db4981b161dea13813", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/Method/Ruder.tex", "max_issues_repo_name": "kjerand/video-style-transfer", "max_issues_repo_head_hexsha": "fe44a1b486e976725cddd6db4981b161dea13813", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/Method/Ruder.tex", "max_forks_repo_name": "kjerand/video-style-transfer", "max_forks_repo_head_hexsha": "fe44a1b486e976725cddd6db4981b161dea13813", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 96.9375, "max_line_length": 421, "alphanum_fraction": 0.7885235332, "num_tokens": 754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.7905303137346446, "lm_q1q2_score": 0.6249381808346866}} {"text": "%\\subsection{Variational Inference in Sparse GPs Revisited}\n%\n%In this section we review inference in sparse GPs (as presented in Titsias).\n%The posterior in an augmented model is $p(\\f, \\u | \\y) = p(\\f | \\u, \\y) p(\\u | \\y)$.\n%The key property of such augmented GPs is the notion of \\textit{sufficient statistics}: given the inducing points $\\u$, the latent values $\\f$ are independent with any other set of latent values (e.g. the test set).\n%In the optimal setting when $\\u$ is the sufficient statistics of $\\f$, it should hold that $p(\\f | \\u, \\y) = p(\\f | \\u)$ as $\\y$ is only the noisy version of $\\f$.\n%This leads to choosing a variational approximation of the posterior which factorizes as $q(\\f, \\u | \\y) = p(\\f | \\u) q(\\u | \\y)$.\n%Since the conditional $p(\\f | \\u)$ is known, variational inference becomes learning an optimal posterior $q(\\u | \\y)$ only. \\\\\n%\nWe approximate the posterior over the latent variables $\\vec{g}, \\vec{h}, \\vec{u}, \\vec{v}$ given observations $\\vec{y}$ using variational inference \\citep{jordan-variational-99}.\nIn section \\ref{sec:variationalLowerBound} we derive a lower bound of the marginal likelihood which has the key property of factorizing over the data points and outputs. Section \\ref{sec:svi} takes advantage of this factorization to \nderive stochastic variational inference, allowing the model to scale to very large datasets.\nSection \\ref{sec:complexity} compares the complexity of the model with previous multi-output methods.\n\n\\subsection{VARIATIONAL LOWER BOUND \\label{sec:variationalLowerBound}}\n\\newcommand{\\ug}{\\u_g}\n\\newcommand{\\uh}{\\u^h}\n\\newcommand{\\mgj}{\\m_j}\n\\newcommand{\\mhi}{\\m^h_i}\n\\newcommand{\\Sgj}{\\S_j}\n\\newcommand{\\Shi}{\\S^h_i}\nIn variational inference, we find the ``closest'' approximate distribution to the true posterior in terms of the $\\KL$ divergence.\nWe first observe that the true posterior distribution can be written as:\n\\begin{align}\np(\\g, \\h, \\u, \\v | \\y) = p(\\g | \\u, \\y) p(\\h | \\v, \\y) p(\\u, \\v | \\y).\n\\end{align}\nHere we recall the modeling assumption that each set of inducing variables is the sufficient statistics of the corresponding latent process. \nThis motivates replacing the true posteriors over $\\g$ and $\\h$ with their conditional distributions given the inducing variables, leading to a distribution of the form:\n\\begin{align}\nq(\\g, \\h, \\u, \\v | \\y)\n= p(\\g|\\u) p(\\h|\\v) q(\\u,\\v),\n\\end{align}\nwith\n\\begin{align}\nq(\\u,\\v) &= \\prod_{j=1}^Q \\underbrace{\\Normal(\\u_j; \\mgj, \\Sgj)}_{q(\\u_j)} \\prod_{i=1}^P \\underbrace{\\Normal(\\v_i; \\mhi, \\Shi)}_{q(\\v_i)}.\n\\end{align}\nThis technique has been used by \\citet{titsias2009variational} and \\citet{hensmangaussian} \nto derive variational inference algorithms for the single output case.\nSince the conditionals $p(\\g | \\u)$ and $p(\\h | \\v)$ are known (Equations \\eqref{eq:gu} and \\eqref{eq:hv}), \nwe only need to \nlearn $q(\\u, \\v)$ so as to minimize the divergence between the approximate posterior and\nthe true posteriors.\nThe quality of approximation depends entirely on the posterior over the inducing variables, thus underlining their pivotal role in the model as previously discussed.\n\n\\noindent\nTo find the best $q(\\u, \\v)$, we optimize the evidence lower bound (ELBO) of the log marginal:\n\\begin{align}\n%\\nonumber\n%\\log p(\\y) \\ge& \\int q(\\u, \\v) \\log \\frac{p(\\y | \\u, \\v) p(\\u, \\v)}{q(\\u, \\v)} \\der \\u \\der \\v \\\\\n%\\nonumber\n%=& \\int q(\\u, \\v) \\log p(\\y | \\u, \\v) \\der \\u \\der \\v \n%+ \\int q(\\u, \\v) \\log \\frac{p(\\u, \\v)}{q(\\u, \\v)} \\der \\u \\der \\v \\\\\n\\nonumber\n&\\log p(\\y) \\ge \\int q(\\u, \\v) \\log p(\\y | \\u, \\v) \\der \\u \\der \\v \\\\\n&- \\sum_{j=1}^Q \\KL[q(\\u_j) || p(\\u_j)] - \\sum_{i=1}^P \\KL[q(\\v_i) || p(\\v_i)],\n\\label{eq:elbo}\n\\end{align}\nwhich is derived using Jensen's inequality and the fact that both of $q(\\u, \\v)$ and $p(\\u, \\v)$ fully factorize.\nSince $q(\\u_j), q(\\v_i), p(\\u_j), p(\\v_i)$ are all multivariate Gaussian distributions, the KL divergence terms are analytically tractable.\nTo compute the expected likelihood term in the ELBO we first see that\n\\begin{align}\n\\nonumber\n\\log \\text{ } &p(\\y | \\u, \\v)\n% &= \\log \\Eb{p(\\y | \\g, \\h)}_{p(\\g,\\h | \\u, \\v)} \\\\\n% \\nonumber\n\\ge \\Eb{\\log p(\\y | \\g, \\h)}_{p(\\g,\\h | \\u, \\v)} \\\\\n&= \\sum_{i=1}^P \\sum_{n=1}^N \\Eb{\\log p(y_{in} | \\g_n, h_{in}) }_{p(\\g | \\u) p(\\h_i | \\v_i)} \n\\end{align}\nwhere $\\g_n = \\{g_{jn} = (\\g_j)_n\\}_{j=1}^Q$.\nThe inequality is due to Jensen's inequality and the equality is due to %the fact that the likelihood fully factorizes.\nthe factorization of the likelihood. \n\\newcommand{\\Ahi}{\\A^h_i}\n\\newcommand{\\Zi}{\\Z_i}\n\nThe ELBO can be computed by first solving for the individual expectations \n%$l_{in} \\define \\Eb{\\log p(y_{in} | \\g_n, h_{in}) }_{p(\\g_n | \\u) p(\\h_in | \\v_i)}$ \n$ \\Eb{\\log p(y_{in} | \\g_n, h_{in}) }$ over ${p(\\g | \\u) p(\\h_i | \\v_i)}$ and \nthen substituting these into Equation \\eqref{eq:elbo} (see the supplementary material for details). \nHence the resulting lower bound is given by:\n\\begin{align}\n\\nonumber\n%&\\log p(\\y)\n & \\calL = \n\\sum_{i,n}\n\\bigg( \\log \\Normal(y_{in}; \\tilde{\\mu}_{in}, \\beta_i^{-1})\n - \\frac{1}{2} \\beta_i \\sum_{j=1}^Q w_{ij}^2 \\tilde{k}_{jnn} \\\\ \\nonumber\n &- \\frac{1}{2} \\beta_i \\tilde{k}^h_{inn}\n - \\frac{1}{2} \\beta_i \\sum_{j=1}^Q \\trace w_{ij}^2 \\S_j \\mat{\\Lambda}_{jn} - \\beta_i \\frac{1}{2} \\trace \\S^h_i \\mat{\\Lambda}_{in} \n\\bigg) \\\\\n\\nonumber\n& - \\sum_{j=1}^Q \\bigg( \\frac{1}{2} \\log |\\K_{jzz} \\S_j^{-1}| + \\frac{1}{2} \\trace \\K_{jzz}^{-1} \\left( \\m_j \\m_j^T + \\S_j \\right)\\bigg) \\\\\n\\nonumber\n& - \\sum_{i=1}^P \\bigg( \\frac{1}{2} \\log |\\K_{izz}(\\Shi)^{-1}| \\\\\n& \\quad \\qquad + \\frac{1}{2} \\trace \\K_{izz}^{-1} \\left( \\mhi (\\mhi)^T + \\Shi \\right) \\bigg) \\text{,}\n \\label{eq:L}\n\\end{align}\n%\nwhere $\\K_{jzz} = k(\\Zj, \\Zj)$, $\\K_{izz} = k(\\Zhi, \\Zhi)$, and:\n\\begin{align}\n\\tilde{\\mu}_{in}\n%&= \\sum_{j=1}^Q w_{ij} k(\\x_n, \\Zj)k(\\Zj,\\Zj)^{-1}\\m_j + k(\\x_n, \\Zhi)k(\\Zhi,\\Zhi)^{-1}\\mhi \\\\\n&= \\sum_{j=1}^Q w_{ij} \\A_j(n,:) \\m_j + \\Ahi(n,:) \\mhi \\text{,}\\\\\n\\mat{\\Lambda}_{jn}\n%&= k(\\Zj,\\Zj)^{-1} k(\\Zj, \\x_n) k(\\x_n, \\Zj) k(\\Zj,\\Zj)^{-1} \\\\\n&= \\A_j(n,:)^T \\A_j(n,:) \\text{,}\\\\\n\\mat{\\Lambda}_{in}\n%&= k(\\Zhi,\\Zhi)^{-1} k(\\Zhi, \\x_n) k(\\x_n, \\Zhi) k(\\Zhi,\\Zhi)^{-1}\n&= \\Ahi(n,:)^T \\Ahi(n,:) \\text{,}\n\\end{align}\nwith $\\tilde{k}_{jnn} = (\\tilde{\\K}_j)_{nn}$; \n$\\tilde{k}^h_{inn} = (\\tilde{\\K}^h_i)_{nn}$; \n$\\mu_{jn} = (\\Mu_j)_n$;\n $\\mu^h_{in} = (\\Mu^h_i)_n$; \nand we have defined the auxiliary matrices $\\A_j = k(\\X,\\Zj) \\K_{jzz}^{-1}$ and $\\Ahi = k(\\X_i,\\Zhi)\\K_{izz}^{-1}$ and \nused $\\A_j(n,:)$ to denote the $n$-th row vector of $\\A_j$. \nNotice that this ELBO generalizes the bound for standard GP regression derived in \\cite{hensmangaussian}, which can be recovered by setting $P = Q = 1$, $w_{ij} = 1$ and $h_i(\\x) = 0$.\n\nThe novelty of the variational lower bound in Equation \\eqref{eq:L} is that it decomposes across both inputs and outputs.\nThis enables the use of stochastic optimization methods, which \nallow the model to handle very large datasets for which existing GP-based multi-output models are simply \nimpractical. \n%\n\\subsection{STOCHASTIC VARIATIONAL INFERENCE \\label{sec:svi}}\n\\newcommand{\\oi}{\\vec{o}_i}\n% some notation\n\nSo far in the description of the model and inference we have\n implicitly assumed that every output has full observations at all inputs $\\X$.\nTo discern where learning occurs for each output, we make the missing data scenario more explicit.\nSpecifically, each output $i$ can have observations at a different set of inputs $\\X_i$. \nWe shall use $\\oi$ to denote the indices of $\\X_i$ (in the set $\\X$) and use the indexing operator $\\B(\\oi)$ to select the rows corresponding to $\\oi$ from any arbitrary matrix $\\B$.\nWe also overload $\\y_i$ as the observed targets of output $i$. \n%\n\\subsubsection{Learning the Parameters of the Variational Distribution}\n%------------------------------------------\n% derivatives of q(u_j)\n\\newcommand{\\Lgj}{\\calL^g_j}\n\\newcommand{\\ynoj}{\\y_i^{\\backslash j}}\n\\newcommand{\\Kjzz}{\\mat{K}_{jzz}}\n We can obtain the derivatives of the ELBO in Equation \\eqref{eq:L}\n wrt the variational parameters for optimization.\nThe derivatives of $\\calL$ wrt the parameters of $q(\\u_j)$ are given by:\n\\begin{align}\n\\deriv{\\calL}{\\m_j} \n=& \\sum_{i=1}^P \\beta_i w_{ij} \\A_j(\\oi)^T \\ynoj \\\\\n\\nonumber\n&- \\bigg[\\Kjzz^{-1} + \\sum_{i=1}^P \\beta_i w_{ij}^2 \\A_j(\\oi)^T \\A_j(\\oi) \\bigg] \\m_j \\text{,} \\\\\n\\deriv{\\calL}{\\S_j} \n=& \\frac{1}{2} \\S_j^{-1} - \\frac{1}{2} \\bigg[ \\Kjzz^{-1} + \\sum_{i=1}^P \\beta_i w_{ij}^2 \\A_j(\\oi)^T \\A_j(\\oi) \\bigg],\n\\end{align}\nwhere $\\y_i^{\\backslash j} = \\y_i - \\Ahi(\\oi) \\mhi - \\sum_{j' \\neq j} w_{ij'} \\A_{j'}(\\oi) \\m_{j'}$.\n%\\noindent As a function of the parameters of $q(\\u_j)$, the lowerbound $\\calL$ is:\n%\\begin{align}\n%\\nonumber\n%\\calL^g_j \\define&\n% \\sum_{i=1}^P \\log \\Normal(\\y_i; \\sum_{j=1}^Q w_{ij} \\A_j(\\oi,:) \\m_j + \\Ahi \\mhi, \\beta_i^{-1} \\I) \\\\\n% \\nonumber\n% &- \\frac{1}{2} \\sum_{i=1}^P \\bigg(\\beta_i \\trace w_{ij}^2 \\tilde{\\K}_j(\\oi,\\oi) \n% + \\beta_i \\trace w_{ij}^2 \\S_j \\A_j(\\oi,:)^T \\A_j(\\oi,:) \\bigg)\n% \\\\\n% &- \\frac{1}{2} \\log |k(\\Zj,\\Zj) \\S_j^{-1}| -\\frac{1}{2} \\trace k(\\Zj,\\Zj)^{-1} (\\m_j \\m_j^T + \\S_j) ,\n%\\end{align}\n%where $\\A_j = k(\\X,\\Zj)k(\\Zj,\\Zj)^{-1}$, which gives $\\A_j(\\oi,:) = k(\\X_i,\\Zj) k(\\Zj,\\Zj)^{-1}$, and \n%$\\Ahi = k(\\X_i,\\Zhi)k(\\Zhi,\\Zhi)^{-1}$. \\\\\n\n%-------------------------------------------\n% derivatives of q(v_i)\n\\newcommand{\\Lhi}{\\calL^h_i}\n\\newcommand{\\Kizz}{\\mat{K}_{izz}}\n\\noindent The derivatives of $\\calL$ wrt the parameters of $q(\\v_i)$ are given by:\n\\newcommand{\\ynoh}{\\y_i^{\\backslash h}}\n\\begin{align}\n\\nonumber\n\\deriv{\\calL}{\\mhi}\n= & \\beta_i \\Ahi(\\oi)^T \\ynoh \\\\\n&- \\bigg[\\Kizz^{-1} + \\beta_i \\Ahi(\\oi)^T \\Ahi(\\oi) \\bigg] \\m_i \\text{,} \\\\\n\\deriv{\\calL}{\\Shi} \n=& \\frac{1}{2} \\S_i^{-1} - \\frac{1}{2} \\bigg[ \\Kizz^{-1} + \\beta_i \\Ahi(\\oi)^T \\Ahi(\\oi) \\bigg] ,\n\\end{align}\nwhere $\\ynoh = \\y_i - \\sum_{j=1}^Q w_{ij} \\A_j(\\oi,:) \\m_j$.\n%\\noindent As a function of the parameters of $q(\\v_i)$, the lower bound $\\calL$ is:\n%\\begin{align}\n%\\nonumber\n%\\Lhi \\define&\n% \\log \\Normal(\\y_i; \\sum_{j=1}^Q w_{ij} \\A_j(\\oi,:) \\m_j + \\Ahi \\mhi, \\beta_i^{-1} \\I)\n% - \\frac{1}{2} \\beta_i \\trace \\tilde{\\K}^h_i(\\oi,\\oi)\n% - \\frac{1}{2} \\beta_i \\trace \\Shi (\\Ahi)^T \\Ahi\n% \\\\\n% &- \\frac{1}{2} \\log |k(\\Zhi,\\Zhi) (\\Shi)^{-1}| -\\frac{1}{2} \\trace k(\\Zhi,\\Zhi)^{-1} (\\mhi (\\mhi)^T + \\Shi) ,\n%\\end{align}\n\n% comment on computation\n\\noindent It can be seen that the derivatives of the parameters of $q(\\v_i)$ only involve the observations of the output $i$.\nThe derivatives of the parameters of $q(\\u_j)$ involve the observations of all outputs but is a sum of contributions from individual outputs.\nComputation of the derivatives can therefore be easily distributed or parallelized.\n%Therefore, computation of the derivatives can be distributed or parallelized easily.\n%This property further enhances the scalability of the model, allowing it to handle big data problems with a very large number of inputs and outputs.\n\nSince the optimal distributions $q(\\u_j)$ and $q(\\v_i)$ are in the exponential family, it is more convenient to use stochastic variational inference \\citep{hensman2012fast,hensmangaussian} to perform update of their canonical parameters.\nThis works by taking a step of length $l$ in the direction of the natural gradient approximated by mini-batches of the data.\nFor instance, consider $q(\\u_j)$ whose canonical parameters are $\\Phi_1 = \\S_j^{-1} \\m_j$ and $\\Phi_2 = -\\frac{1}{2}\\S_j^{-1}$.\nTheir stochastic update equations at time $t + 1$ are given by:\n\\begin{align}\n\\nonumber\n& \\Phi_{1(t+1)} = \\S_{j(t)}^{-1} \\m_{j(t)} \\\\\n & \\quad \\qquad + l \\bigg(\\sum_{i=1}^P \\beta_i w_{ij} \\A_j(\\oi)^T \\ynoj - \\S_{j(t)}^{-1} \\m_{j(t)} \\bigg) \\\\\n& \\Phi_{2(t+1)} = -\\frac{1}{2} \\S_{j(t)}^{-1} + l \\bigg(\\frac{1}{2} \\S_{j(t)}^{-1} - \\frac{1}{2} \\mat{\\Lambda} \\bigg),\n\\end{align}\n\\normalsize\nwhere $\\mat{\\Lambda} = \\Kjzz^{-1} + \\sum_{i=1}^P \\beta_i w_{ij}^2 \\A_j(\\oi)^T \\A_j(\\oi)$.\n%\n\\subsubsection{Inducing Inputs and Hyper-parameters \\label{sec:hyperparameters}}\nTo learn the hyperparameters, which in this model include the mixing weights, the covariance hyperparameters of the latent processes, and the noise precision of each output, we follow standard practice in GP inference.\nFor this model this involves taking derivatives of the ELBO and applying standard stochastic gradient descent in alternative steps with the variational parameters, much like a variational EM algorithm. \nThe derivatives are given in the supplementary material.\n\nLearning of the inducing inputs, which was not considered in the single output case in \\citet{hensmangaussian}, is also possible in our stochastic optimization approach.\nIn the supplementary material, we show that \n%with the help of vectorized operations, \nthe additional cost of computing the derivatives of the lower bound wrt the inducing inputs is not significantly \nhigher than the cost of updating the variational parameters.\n%can be computed with a cost \\emph{independent of the input dimensionality}.\nThis makes optimizing the inducing locations a practical option, which can be critical \nin high-dimensional problems. %where fixing the inducing inputs may hinder the performance of the model.\nIndeed, our experiments on a large scale multi-output problem show that automatic learning of the inducing inputs can lead to significant performance gain with little overhead in computation. \n%\n\\subsection{COMPLEXITY ANALYSIS \\label{sec:complexity}}\nIn this section we analyze the complexity of the model and compare it to existing multi-output approaches.\nFor consistency, we first unify common notations used for all models.\nWe use $P$ as the number of outputs; $N$ as the number of inputs; $Q$ as the number of shared latent processes; and $M_t$ as the \\emph{total} number of inducing inputs.\nIt is worth noting that $M_t = (P + Q) \\times M$ in our COGP model, assuming that each sparse process has equal number of inducing points. \nAlso, COGP has $P$ additional individual processes, one for each output.\n\nThe complexity of COGP can be read off by inspecting the ELBO in Equation \\eqref{eq:L}, with the key observation that it contains \na sum over the outputs as well as over the inputs.\nThis means a mini-batch containing a small subset of the inputs and outputs can be used for stochastic optimization.\nTechnically, the cost is $\\calO(M^3)$ or $\\calO(N_bM^2)$, where $N_b$ is the size of the mini-batches, depending on which is larger between $M$ and $N_b$. \nIn practice, we may use $N_b > M$ as a large batch size (e.g.~$N_b = 1000$) helps reduce stochasticity of the optimization.\nHowever, here we use $\\calO(M^3)$ for easier comparison with other models whose time and storage demands are \ngiven in Table \\ref{tab:complexity}. \nWe see that COGP has \na computational complexity that is independent of the size of the inputs and outputs, \nwhich makes it the only method capable of handling large scale problems.\n%\n\\setlength{\\tabcolsep}{4pt}\n\\begin{table}[t]\n\\caption{Comparison of the time and storage complexity of approximate inference of multi-output GP models. \nA\\&W, 2009 refers to \\citet{alvarez-lawrence-nips-08}. COGP is the only method with complexity \\emph{independent} of the\n number of inputs $N$ and outputs $P$, thus it can scale to very large datasets.}\n\\label{tab:complexity}\n\\begin{center}\n\\small\n\\begin{tabular}{lcc}\n\\toprule\n\\textbf{METHOD} & \\textbf{TIME} & \\textbf{STORAGE} \\\\ \\hline\nCOGP, this paper & $\\calO(M^3)$ & $\\calO(M^2)$ \\\\\nSLFM, \\citep{teh-et-al-aistats-05} & $\\calO(QNM_t^{2})$ & $\\calO(QNM_t)$ \\\\\nMTGP, \\citep{bonilla-et-al-nips-08} & $\\calO(PNM_t^2)$ & $\\calO(PNM_t)$\\\\ \nCGP-FITC (A\\&W, 2009)& $\\calO(PNM_t^2)$ & $\\calO(PNM_t)$ \\\\\nGPRN, \\citep{wilson-et-al-icml-12} & $\\calO(PQN^3)$ & $\\calO(PQN^2)$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n\\end{table}\n\n\\subsection{PREDICTION}\nThe predictive distribution of the $i$-th output for a test input $\\x_*$ is given by:\n\\begin{align}\np(\\fstar | \\y, \\x_*) = \\Normal(\\fstar; \\sum_{j=1}^Q w_{ij} \\mu_{j*} + \\mu^h_{i*}, w_{ij}^2 s_{j*} + s^h_{i*}), \n\\end{align}\nwhere $\\mu_{j*}$ and $s_{j*}$ are the mean and variance of the prediction for $g_{j*} = g_j(\\x_*)$, i.e. $p(g_{j*} | \\y, \\x_*) = \\Normal(g_{j*}; \\mu_{j*}, s_{j*})$.\nLikewise, $\\mu^h_{i*}$ and $s^h_{i*}$ are the mean and variance of the prediction for $h_{i*} = h_i(\\x_*)$, $p(h_{i*} | \\y, \\x_*) = \\Normal(h_{i*}; \\mu^h_{i*}, s^h_{i*})$.\nThese predictive means and variances are given by: \n\\begin{align}\n\\mu_{j*} &= \\vec{k}_{j*z} \\K_{jzz}^{-1} \\m_j \\text{,}\\\\ \ns_{j*} &= k_{j**} - \\vec{k}_{j*z} \\left(\\K_{jzz}^{-1} - \\K_{jzz}^{-1} \\S_j \\K_{jzz}^{-1} \\right) \\vec{k}_{j*z}^T \\text{,}\\\\\n\\mu^h_{i*} &= \\vec{k}_{i*z} \\K_{izz}^{-1} \\mhi \\text{,} \\\\\ns^h_{i*} &= k_{i**} - \\vec{k}_{i*z} \\left(\\K_{izz}^{-1} - \\K_{izz}^{-1} \\S_i \\K_{izz}^{-1} \\right) \\vec{k}_{i*z}^T \\text{,}\n\\end{align}\nwhere $k_{j**} = k_j(\\x_*, \\x_*)$, $k_{i**} = k^h_i(\\x_*, \\x_*)$, $\\vec{k}_{j*z}$ is the covariance between $\\x_*$ and $\\Zj$, and $\\vec{k}_{i*z}$ is the covariance between $\\x_*$ and $\\Zhi$.\n\n%\\noindent Also as a consequence of sufficient statistics, the approximate prediction for test targets $\\vfstar$ at test inputs $\\X_*$ is\n%\\begin{align}\n%\\nonumber\n%p(\\vfstar | \\y, \\X_*) &= \\int p(\\vfstar | \\f, \\u, \\X_*) q(\\f, \\u | \\y) \\der \\f \\der \\u \\\\\n%\\nonumber\n%&= \\int p(\\vfstar | \\u) q(\\u| \\y) p(\\f | \\u) \\der \\f \\der \\u \\\\ \\nonumber\n%&= \\int p(\\vfstar | \\u) q(\\u| \\y) \\der \\u \\\\\n%\\label{eq:sorprediction}\n%&= \\Normal(\\vfstar; \\bs{\\mu_*},\\vec{s_*})\n%\\end{align}\n%where,\n%\\begin{align}\n%\\nonumber\n%\\bs{\\mu_*} &= \\K_{*z} \\K_{zz}^{-1}\\m \\\\ \n%\\nonumber\n%\\S_* &= \\K_{**} - \\K_{*z} \\left(\\K_{zz}^{-1} - \\K_{zz}^{-1} \\S \\K_{zz}^{-1} \\right) \\K_{*z}^T.\n%\\end{align}\n% Here $\\K_{*z}$ is the covariance matrix between test and inducing inputs, $\\K_{zz}$ is the covariance matrix of the inducing inputs.\n \n%\\begin{align}\n%p(\\fstar | \\y, \\x_*) = \\int \\Normal(\\fstar; \\sum_{j=1}^Q w_j g_{j*} + h_{i*}, 0) p(\\g_* | \\y, \\x_*) p(h_{i*} | \\y, \\x_*) \\der \\g_* \\der h_{i*},\n%\\end{align}\n", "meta": {"hexsha": "2baac4c1f574cc3f0d6f2418a4f6138c8ef680f6", "size": 17977, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/inference.tex", "max_stars_repo_name": "fkopsaf/cogp", "max_stars_repo_head_hexsha": "3b07f621ff11838e89700cfb58d26ca39b119a35", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 15, "max_stars_repo_stars_event_min_datetime": "2015-05-28T13:46:13.000Z", "max_stars_repo_stars_event_max_datetime": "2020-10-10T11:02:08.000Z", "max_issues_repo_path": "paper/inference.tex", "max_issues_repo_name": "fkopsaf/cogp", "max_issues_repo_head_hexsha": "3b07f621ff11838e89700cfb58d26ca39b119a35", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-07-30T08:52:36.000Z", "max_issues_repo_issues_event_max_datetime": "2019-06-04T01:44:21.000Z", "max_forks_repo_path": "paper/inference.tex", "max_forks_repo_name": "trungngv/cogp", "max_forks_repo_head_hexsha": "3b07f621ff11838e89700cfb58d26ca39b119a35", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 14, "max_forks_repo_forks_event_min_datetime": "2016-04-03T03:18:18.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-23T13:28:55.000Z", "avg_line_length": 59.1348684211, "max_line_length": 237, "alphanum_fraction": 0.6548923625, "num_tokens": 6452, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303087996143, "lm_q2_score": 0.7905303137346446, "lm_q1q2_score": 0.6249381730321045}} {"text": "\\chapter{Multi-layer Kernel Machines}\n\\label{chap_mkm}\n%This chapter introduces the concept of MKMs proposed in \\cite{saul} et al., followed by an empirical study on some popular datasets cited extensively in deep learning literatures. This chapter is organized as follows; section \\ref{chap2_mkm} gives a brief introduction to MKMs and the multi-layer composition of few kernel functions, section \\ref{chap2_exp} contains the results of empirical study using MKMs, section \\ref{chap2_tsne} talks about the visualization technique used to plot the distribution of digits(from \\cite{mnist} dataset) in a low-dimensional space, section \\ref{chap2_mkm_mix} contains the results of empirical study on MKMs with mixed kernels and section \\ref{chap2_conc} concludes the chapter. \n\n\n%\\label{chap2_mkm}\nMulti-layer Kernel Machines (MKMs\\nomenclature{MKM}{Multi-layer Kernel Machines}) were first introduced by \\cite{saul} et al. In their framework, arc-cosine kernel which itself is having a layered architecture was used. The architecture of MKMs consists of $L$ layers and in each layer the data is subjected to unsupervised dimensionality reduction methods, viz. kernel PCA(\\cite{kpca} et al.), followed by a supervised feature selection. \n\nThe MKM machines use multilayer kernel function, whose description is given in the next section. The rest of the chapter is organized as follows: section \\ref{chap2_exp} contains the results of empirical study using MKMs, section \\ref{chap2_tsne} describes the visualization technique used to plot the distribution of digits (from \\cite{mnist} dataset) in a low-dimensional space, section \\ref{chap2_mkm_mix} contains the results of empirical study on MKMs with mixed kernels and concluding remarks are given in section \\ref{chap2_conc}. \n\n\\section{Multi-layer Kernel}\nCorresponding to each Reproducing kernel Hilbert spaces (RKHS) $\\mathcal{F}$ there exists a unique reproducing kernel \n$k: \\mathcal{X} \\times \\mathcal{X} \\rightarrow \\mathcal{F}$, where $ k(x,y) = \\phi(x) \\cdot \\phi(y), x, y \\in \\mathcal{X} \\; \\text{where} \\; \\mathcal{X} \\text{ is a normed space, } \\phi(x), \\phi(y) \\in \\mathcal{F}$ and $\\cdot$ is the inner product defined on $\\mathcal{F}$.\n\nIteratively applying the mapping $\\phi(\\cdot)$ on the inputs $x \\textrm{ and } y$ and then taking their inner product, we can obtain an L-layer kernel function as\n\\[ k^{(L)}(x,y) = \\underbrace{\\phi(\\phi(\\ldots \\phi(x)))}_{\\textrm{L times}} \\textrm{ } \\cdot \\textrm{ } \\underbrace{\\phi(\\phi(\\ldots \\phi(y)))}_{\\textrm{L times}} \\]\n\nThus a $L$ layer kernel machine consists of $L$ RKHS's.\n\n\\subsection{Multi-layer Composition of Polynomial and Gaussian Kernels}\n\\label{chap2_mlother}\nA polynomial kernel of degree d is defined as follows\n\\[k(x, y) = (x \\cdot y)^d\\]\nConsider a two layer composition of polynomial kernel which can be computed as\n\\begin{equation*}\n\\begin{aligned}\n\\phi(\\phi(x)) \\cdot \\phi(\\phi(y)) &= \\bigg(\\phi(x) \\cdot \\phi(y)\\bigg)^d \\\\\n& = {(x \\cdot y)^d}^d = {(x \\cdot y)^d}^2 \n\\end{aligned}\n\\end{equation*}\nThus the higher order compositions of polynomial kernels are simply polynomials of higher degree than the one from which it is constructed.\n\nThe two layer composition for gaussian kernels $k(x, y) = e^{-\\lambda\\norm{x-y}^2} $ is given by\n\\begin{equation*}\n\\begin{aligned}\n\\phi(\\phi(x)) \\cdot \\phi(\\phi(y)) &= e^{-\\lambda\\norm{\\phi(x)-\\phi(y)}^2} \\\\\n& = e^{-2\\lambda(1-k(x, y))}\n\\end{aligned}\n\\end{equation*}\n\n\\subsection{Multi-layer Composition of Arc-cosine Kernels}\n\\label{chap2_mlarc}\nLet x, y be two inputs in $\\mathbb{R}^d$. Define $\\theta$ as the angle between them.\n\\[ \\theta = cos^{-1}\\left ( \\frac{x\\cdot y}{\\left \\| x \\right \\| \\left \\| y \\right \\|} \\right ) \\]\nThen the kernel function computed by the arc-cosine kernel is\n\\begin{equation}\nk_n(x,y) = \\frac{1}{\\pi}\\left \\| x \\right \\|^n \\left \\| y \\right \\|^n J_n(\\theta)\n\\label{arc_cosine_kernel} \n\\end{equation}\nwhere n is called the \\textit{degree of the kernel} and\n\\[ J_n(\\theta) = (-1)^n(sin\\theta)^{2n+1} \\left ( \\frac{1}{sin\\theta} \\frac{\\partial}{\\partial \\theta} \\right )^n \\left ( \\frac{\\pi-\\theta}{sin\\theta} \\right ) \\]\n\n$J_n(\\theta)$ for n=0, 1, 2 is computed as shown below.\n\\[ J_0(\\theta) = \\pi-\\theta \\]\n\\[ J_1(\\theta) = sin\\theta + (\\pi-\\theta)cos\\theta \\]\n\\[ J_2(\\theta) = 3sin\\theta cos\\theta + (\\pi-\\theta)(1+2cos^2\\theta) \\]\nfor n=0, it takes the simple form\n\\[k_0(x,y) = 1- \\frac{1}{\\pi}cos^{-1}\\left ( \\frac{x\\cdot y}{\\left \\| x \\right \\| \\left \\| y \\right \\|} \\right ) \\]\nhence the name arc-cosine kernel is given.The kernel function computed by arc-cosine kernel is related to the computation of multi-layer threshold networks as shown below.\n\nConsider a single layer neural network with weights $W_{ij}$ that connects the $j^{th}$ input unit to the $i^{th}$ output unit(see figure \\ref{fig_singnet}). The network maps input $x$ to output $f(x)$ by applying a non-linear map \n\\[ f(x) = g(W \\cdot x) \\]\nwhere the non-linearity is described by the network`s activation function\n\\[g_n(z) = \\Theta(z)z^n \\]\nwith \n\\[ \\Theta(z) = \\frac{1}{2}(1+sign(z)) \\]\nthis activation function is called one-sided polynomial activation function, whose graph for different n values is also shown in figure \\ref{fig_singnet}.\n\n\\begin{figure}[H]\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[width=1.0\\linewidth, height=7cm]{figures/singnet}\n \\caption{Single layer network and thresholded activation functions(\\cite{saul} et al.)}\n \\label{fig_singnet}\n\\end{figure}\n\nLet $f(x)$ and $f(y)$ be the outputs corresponding to inputs $x$ and $y$. Then the inner product of $f(x)$ and $f(y)$ is\n\\[ f(x)\\cdot f(y) = \\sum_{i=1}^m \\Theta(w_i\\cdot x) \\Theta(w_i\\cdot y)(w_i\\cdot x)^n (w_i\\cdot y)^n\\]\nHere $w_i$ is the $i^{th}$ row of weight matrix $W$ and $m$ is the no of output units. Assume $W_{ij}$ are Gaussian distributed with zero mean and unit variance and the network has an infinite number of output units. Then\n\\[ lim_{m\\rightarrow \\infty} \\frac{2}{m}f(x)\\cdot f(y) = k_n(x,y) \\]\nwhere\n\\begin{equation}\nk_n(x,y) = 2\\int dw \\frac{e^{-\\frac{\\norm{w}^2}{2}}}{(2\\pi )^{d/2}} \\Theta (w\\cdot x) \\Theta (w\\cdot y) (w\\cdot x)^n (w\\cdot y)^n \n\\label{arc_cosine_integral}\n\\end{equation}\nThe kernel function obtained in equation \\ref{arc_cosine_integral} can be converted into an alternate form as in equation \\ref{arc_cosine_kernel} with the derivation shown in \\cite{saul} et al.\n\nThe multi-layer composition of arc-cosine kernel can be recursively computed as\n\\[ k^{(L+1)}_n(x,y) = \\frac{1}{\\Pi} \\Bigg[ k^{(L)}(x,x) \\textrm{ } k^{(L)}(y,y)\\Bigg]^{\\frac{n}{2}} J_n(\\theta_n^{(L)}) \\]\nwhere $\\theta_n^{(L)}$ is the angle between images of $x \\textrm{ and } y$ in the feature space after L layer composition\n\\[ \\theta_n^{(L)} = \\textrm{cos}^{-1}\\Bigg( k^{(L)}(x,y) \\Bigg[ k^{(L)}(x,x) \\textrm{ } k^{(L)}(y,y)\\Bigg]^{\\frac{-1}{2}} \\Bigg) \\]\nIn the above formulation, we have assumed that the arc-cosine kernels have the same degree $n$ at every layers of recursion. We can also use kernels of different degree at different layers. The intuition behind the multi-layer composition in case of arc-cosine kernels is, if the base kernel $k(x,y) = \\phi(x) \\cdot \\phi(y)$ can mimic the computation of a single-layer network, then the iterated mapping in $k^{(L)}(x,y)$ can mimic the computation of multi-layer network(\\cite{saul} et al.).\n\n\\subsection{MKM Architecture}\n The architecture of MKMs for solving classification tasks is similar to that of neural network based deep learning machines, with unsupervised feature extraction(using Kernel PCA, \\cite{kpca} et al.) followed by supervised feature selection in each layer. Figure \\ref{fig_mkm} shows the architecture of an MKM consisting of $L$ layers of non-linear transformations.\n\n\\begin{figure}[h]\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.5]{figures/mkm}\n \\caption{An MKM with L layers of transformations. Each layer consists of unsupervised feature extraction(using kernel PCA) followed by supervised feature selection.}\n \\label{fig_mkm}\n\\end{figure}\n\nThe supervised feature selection allows us to retain only those set of features that are most informative for the pattern recognition task. The number of features to be passed to the next layer determines the width of that layer. The optimal layer width is computed by ranking the features based on their importance and selecting the required set of feature with the help of a lightweight classifier (detailed procedure is given in the experiments section). This procedure determines the architecture of the network in a greedy, layer-by-layer fashion. In their implementation \\cite{saul} et al. used exhaustive search in the range 10 to 300 to determine the optimal set of features. The output from the final layer can be passed to any classifier. Though any kernel can be used for the kernel PCA based feature extraction, \\cite{saul} et al. emphasized on using arc-cosine kernels due to their similarity with deep learning architecture of neural networks and the inclusion of multiple layers is significant only in case of arc-cosine kernels.\n\n\\section{Experiments on Arc-cosine Kernel MKMs}\n\\label{chap2_exp}\nEmpirical study was conducted on four datasets, in which three were created from \\cite{mnist} dataset of handwritten digits by adding noise in the background and one was a binary classification problem on shape images. The detailed experimental set up and a short description about datasets used is given below. \n\\subsection{Experimental Set-Up}\nIn the training phase, for each layer we set apart 10000 datapoints for training the model and 2000 datapoints for cross validating kernel parameters. For the kernel PCA we chose 3000 datapoints from the training set of 10000 images randomly. In each layer after extracting features with KPCA, we train a lightweight classifier like kNN and test the performance on the held out 2000 images. The best kernel parameter was chosen based on this performance values. With that parameter value, we did feature extraction on the entire dataset and then passed it to feature selection module. The feature selection was done by using univariate feature selection method available in \\cite{scikit} library. This method produces a ranking of features with a univariate statistical test. Based on this rank we can choose the required number of important features. We chose top 5 percent features based on this rank, since empirically it was giving a consistant performance.\n\nIn the final classification stage SVMs with arc-cosine kernels are used. The metric used for comparing the performance is, percentage loss in test dataset. Percentage loss is estimated as\n\\[ \\textrm{loss in percentage} = \\Bigg(1 - \\frac{\\# \\textrm{correct classifications}}{\\# \\textrm{datapoints}} \\Bigg) \\times 100 \\]\n\n\\subsection{Mnist-back-rand Dataset}\nThe \\textit{mnist-back-rand} dataset was created by filling the image background with random pixel values. Each pixel value of the background was generated uniformly between 0 and 255. Image size is 28$\\times$28 and the dataset contains 12000 training and 50000 testing images.\n\\subsection{Mnist-back-image Dataset}\nThe \\textit{mnist-back-image} dataset was generated by filling the image background with random image patches. The patches were extracted randomly from a set of 20 images downloaded from the internet. The dataset contains contains 12000 training and 50000 testing images, each of size 28$\\times$28.\n\\subsection{Mnist-rot-back-image Dataset}\nThe \\textit{mnist-rot-back-image} is a rotated variant of \\textit{mnist-back-image} where the rotation angle is generated uniformly between $0$ and $2\\pi$. Image size and number of samples are also the same as \\textit{mnist-back-image} dataset.\n\\subsection{Rectangles-image Dataset}\nFor \\textit{rectangles-image} dataset, the classification task is to identify whether a rectangle contained in an image has larger width or length. The dataset is constructed by uniformly sampling the height and width of the rectangles. Then random image patches are added in the background, where image patches are extracted from one of the 20 images used by \\textit{mnist-back-image}. Each image is of size 28$\\times$28 and each dataset contains 12000 training images and 50000 testing images.\n\nFigure \\ref{samples} contains some sample images from first three dataset and Figure \\ref{chap2_rect} contains the samples from \\textit{rectangles-image} dataset.\n\\begin{figure}[h]\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.55]{figures/data_sample1}\n \\caption{sample images from \\textit{mnist-back-rand}(first row), \\textit{mnist-back-image}(second row) and \\textit{mnist-rot-back-image}(third row) datasets.}\n \\label{samples}\n\\end{figure}\n\n\\begin{figure}[h]\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.55]{figures/rect_image}\n \\caption{sample images from \\textit{rectangles-image} dataset.}\n \\label{chap2_rect}\n\\end{figure}\n\n\\subsection{Results and Analysis}\n\\renewcommand{\\arraystretch}{2.1}\n\\begin{table*}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Dataset}} & \\multicolumn{7}{ |c| }{\\textbf{Loss in Percentage}} \\\\\n \\cline{2-8}\n & $\\textrm{SVM}_{\\textrm{RBF}}$ & $\\textrm{SVM}_{\\textrm{Poly}}$ & NNet & DBN-3 & SAA-3 & DBN-1 & MKMs\\\\\n \\hline \n \\textit{back-rand} & 14.58 & 16.62 & 20.04 & \\textbf{6.73} & 11.28 & 9.80 & 10.55\\\\\n \\hline\n \\textit{back-image} & 22.61 & 24.01 & 27.41 & 16.31 & 23.00 & \\textbf{16.15} & 21.39\\\\\n \\hline\n \\textit{rot-back-image} & 55.18 & 56.41 & 62.16 & \\textbf{47.39} & 51.93 & 52.21 & 51.61\\\\\n \\hline\n \\textit{rect-image} & 24.04 & 24.05 & 33.20 & 23.69 & 24.05 & \\textbf{22.50} & 23.01\\\\\n \\hline\n\\end{tabular}\n\\caption{Experimental Results of MKMs with Arc-cosine Kernels}\n\\label{tab_results_mkm}\n\\end{table*}\n\\renewcommand{\\arraystretch}{1}\n\nTable \\ref{tab_results_mkm} also contains best results obtained from other models(\\cite{dbn} et al.) like SVM with RBF kernel($\\textrm{SVM}_{\\textrm{RBF}}$), SVM with polynomial kenel($\\textrm{SVM}_{\\textrm{Poly}}$), single hidden layer feed-forward neural network(NNet\\nomenclature{NNet}{Neural Network with one hidden layer}), Deep Belif Networks(DBN)\\nomenclature{DBN}{Deep Belif Networks} with 1 hidden layer(DBN-1), DBN with 3 hidden layer(DBN-3) and 3 hidden layer Stacked Autoassociator Network(SAA-3). The first three models comes under shallow architectures and the remaining are deep architectures. From the table, it can be observed that MKMs outperforms all the remaining models except Deep Belif Networks(DBN). Compared to DBN the architecture, parameter tuning and optimization are fairly simple in MKMs.\n\nThe change in the performance of the classifier when each adding layer to the model was also analyzed. Figures \\ref{mkm_rand} and \\ref{mkm_image} shows the variations in the classifier performance when model complexity was increased by adding more layers(on \\textit{mnist-back-rand} and \\textit{mnist-back-image} datasets respectively). The results indicates that in some cases better representation can be obtained even with a less complex model (eg: in case of \\textit{mnist-back-image} dataset, where the performance degrades when model complexity increases).\n\n\n\\begin{figure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.65]{figures/mkm_rand}\n \\caption{Change in classifier performance on \\textit{mnist-back-rand} dataset when increasing the number of layers.}\n \\label{mkm_rand}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.65]{figures/mkm_image}\n \\caption{Change in classifier performance on \\textit{mnist-back-image} dataset when increasing the number of layers.}\n \\label{mkm_image}\n\\end{figure}\n\n\\section{Visualizing Features using tSNE}\n\\label{chap2_tsne}\nIn order to visualize the features learned by the MKM, t-distributed Stochastic Neighbor Embedding(tSNE, proposed by \\cite{tsne} et al.) was used. tSNE\\nomenclature{tSNE}{t-distributed Stochastic Neighbur Embedding} is a tool for visualizing high-dimensional data. It converts the similarities between datapoints to joint probabilities and tries to minimize the Kullback-Leibler divergence(KL divergence) between the joint probabilities of the low-dimensional embedding and the high dimensional data(\\cite{tsne} et al.). tSNE is a variant of stochastic neighbor embedding \\nomenclature{SNE}, that is much easier to optimize and produce significantly better visualizations by reducing the tendency of the points to get accumulated at the center of the map.\n\nSuppose $X = \\{x_1, x_2, \\ldots, x_n\\}$ are the high-dimensional datapoints, and $Y = \\{y_1, y_2, \\ldots, y_n\\}$ the corresponding low-dimensional map points. In SNE, we first convert high-dimensional representation of datapoints into pairwise similarities that can be interpreted as conditional probabilities. Intuitively, the similarity of a datapoint $x_i$ to $x_j$ is interpreted as the probability of $x_i$ picking up $x_j$ as its neighbor with a Gaussian centered at $x_i$(denoted as $p_{j/i}$). Mathematically this is formulated as\n\\[ p_{j/i} = \\frac{exp(-\\norm{x_i-x_j}^2/2 \\sigma_i^2}{\\sum_{k\\neq i} exp(-\\norm{x_i - x_k}^2/2 \\sigma_i^2)} \\]\n\nwhere $\\sigma_i$ is the variance of the Gaussian centered at $x_i$. For every $x_i$ we calculate corresponding $\\sigma_i$ which will vary according to the density of datapoints around $x_i$(if the region around $x_i$ is dense $\\sigma_i$ will be low and vice versa).\n\nSimilarly for the low-dimensional representations $y_i$ and $y_j$ corresponding to $x_i$ and $x_j$ respectively, we compute the conditional probability(denoted as $q_{j/i}$) by setting the variance of the Gaussian to $\\frac{1}{\\sqrt{2}}$. Thus\n\\[ q_{j/i} = \\frac{exp(-\\norm{y_i-y_j}^2}{\\sum_{k\\neq i} exp(-\\norm{y_i - y_k}^2)} \\]\n\nSince we are only interested in modelling pairwise similarity, we set $p_{i/i} = q_{i/i} = 0$. The crux of SNE as stated by \\cite{tsne} et al. is ``if the map points $y_i$ and $y_j$ correctly model the similarity between high-dimensional datapoints $x_i$ and $x_j$ then the conditional probabilities $p_{j/i}$ and $q_{j/i}$ will be equal''. In this case, the cost function can be formulated to minimize the difference between conditional probabilities $p_{j/i}$ and $q_{j/i}$ using KL divergence. The cost function $C$ is defined as\n\\begin{equation}\nC = \\sum_{i}\\textrm{KL}(\\textrm{P}_i \\big|\\big| \\textrm{Q}_i) = \\sum_{i}\\sum_{j} p_{j/i} log \\frac{p_{j/i}}{q_{j/i}}\n\\label{cost_tsne}\n\\end{equation}\nwhere $\\textrm{P}_i$ is the conditional probability distribution over all other datapoints given datapoint $x_i$ and $\\textrm{Q}_i$ is the conditional probability distribution over all other map points given the map point $y_i$.\n\nThe minimization of the cost function in \\ref{cost_tsne} is performed by using gradient descent method. The gradient is computed as\n\\[ \\frac{\\partial C}{\\partial y_i} = 2\\sum_j(p_{j/i}-q_{j/i} + p_{i/j}-q_{i/j})(y_i - y_j) \\]\nThough SNE can produce good quality visualizations, it has the following limitations.\n\\begin{itemize}\n\\item Since the KL divergence is not symmetric, different types of error in the pairwise distances in low-dimensional map are not weighted equally(there is a large cost for using widely separated map points to represent nearby datapoints, but the cost is small for using nearby map points to represent widely separated datapoints).\n\\item crowding of map points in the center of the map.\n\\end{itemize}\n\ntSNE is a variant of SNE which improves the visualization quality by alleviating the above mentioned problems.\n\\begin{itemize}\n\\item It uses a symmetric version of the SNE cost function by employing a joint probability distribution instead of the conditional probability distribution used by SNE. This also results in simpler gradients.\n\\item It uses student-t distribution to compute the similarity between map points. tSNE employs a heavy-tailed distribution for the map points to alleviate both the crowding problem and optimization problems of SNE.\n\\end{itemize}\n\nIn the symmetric version of the tSNE the cost function is computed as shown below.\n\\begin{equation*}\nC = \\textrm{KL}(\\textrm{P} \\big|\\big| \\textrm{Q}) = \\sum_{i}\\sum_{j} p_{ij} log \\frac{p_{ij}}{q_{ij}}\n\\end{equation*}\nwhere $P$ and $Q$ are the joint probability distribution functions in the high-dimensional and low-dimensional space respectively. Here also we set $p_{ii}$ and $q_{ii}$ to zero. The symmetry in the cost function is achieved due to the fact that $p_{ij} = p_{ji}$ and $q_{ij} = q_{ji}$ $\\forall i,j$. The joint probability in high-dimensional space($p_{ij}$) and low-dimensional space($q_{ij}$) are computed as\n\\[ p_{ij} = \\frac{exp(-\\norm{x_i-x_j}^2/2 \\sigma^2}{\\sum_{k\\neq l} exp(-\\norm{x_k - x_l}^2/2 \\sigma^2)} \\]\n\\[ q_{ij} = \\frac{exp(-\\norm{y_i-y_j}^2}{\\sum_{k\\neq l} exp(-\\norm{y_k - y_l}^2)} \\]\nThe gradient of the symmetric SNE has a fairly simple form\n\\[ \\frac{\\partial C}{\\partial y_i} = 4\\sum_j(p_{ij}-q_{ij})(y_i - y_j) \\]\nSymmetric SNE still faces problems when a datapoint $x_i$ is an outlier; when $x_i$ is an outlier $\\norm{x_i-x_j}^2$ is large for all $x_j$ with $x_i$, hence $p_{ij}$ are extremely low, so the location of its low-dimensional map point $y_i$ has very little effect on the cost function. This problem is addressed in tSNE by defining $p_{ij} = \\frac{p_{j/i} + p_{i/j}}{2n}$, which ensures that $\\sum_j p_{ij} \\ge \\frac{1}{2n}$ $\\forall \\textrm{ } x_i$, hence each datapoint $x_i$ makes a significant contribution to the cost function. The crowding problem is addressed in tSNE by using a student t-distribution with one degree of freedom as the heavy-tailed distribution in the low-dimensional map. Thus the joint probability $q_{ij}$ is computed as\n\\[ q_{ij} = \\frac{(1+\\norm{y_i-y_j}^2)^-1}{\\sum_{k \\neq l} (1+\\norm{y_k-y_l}^2)^-1} \\] \nThen, finally the gradient of the cost function is obtained as\n\\[ \\frac{\\partial C}{\\partial y_i} = 4\\sum_j(p_{ij}-q_{ij})(y_i - y_j)(1+\\norm{y_i-y_j}^2)^-1) \\]\n\ntSNE was employed in our experiments to understand the changes in the data distribution before and after the feature learning process. Figure \\ref{tsne_rand_raw} shows the tSNE embedding of the raw data from \\textit{mnist-back-rand} dataset and figure \\ref{tsne_rand_mkm} shows the same embedding applied on the features produced by MKM with arc-cosine kernels. The tSNE embedding of features learned by MKMs are crowded near the center, hence it was difficult to interpret the separability.\n\n\\begin{sidewaysfigure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.45]{figures/tsne_rand_rawdata}\n \\caption{tSNE embedding of raw data from \\textit{mnist-back-rand} dataset.}\n \\label{tsne_rand_raw}\n\\end{sidewaysfigure}\n\n\n\\begin{sidewaysfigure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.45]{figures/tsne_rand_mkmbest}\n \\caption{tSNE embedding of features obtained by MKM from \\textit{mnist-back-rand} dataset.}\n \\label{tsne_rand_mkm}\n\\end{sidewaysfigure}\n\n\n\n\\section{MKMs with Mixed Kernels}\n\\label{chap2_mkm_mix}\nWe tried different kernel functions in different layers of MKMs (Gaussian, arc-cosine, polynomial etc.) while performing KPCA. It had been seen that mixing a layer of Gaussian or polynomial kernel in between successive layers of arc-cosine kernel improves the result. This further supported the belief that, with more similarity information (by using more kernels) in hand, we can build better representations. Table \\ref{tab_results_mix} shows the result obtained with the mixed kernel models. Because of the improved performance, we used Multi-layer Multiple Kernel Machines(ML-MKL) model for our analysis, whose discussion is given in the next chapter.\n\n\\renewcommand{\\arraystretch}{2.1}\n\\begin{table}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Dataset}} & \\multicolumn{7}{ |c| }{\\textbf{Loss in Percentage}} \\\\\n \\cline{2-8}\n & $\\textrm{SVM}_{\\textrm{RBF}}$ & $\\textrm{SVM}_{\\textrm{Poly}}$ & NNet & DBN-3 & SAA-3 & DBN-1 & $\\textrm{MKMs}_\\textrm{(mix)}$\\\\\n \\hline \n \\textit{back-rand} & 14.58 & 16.62 & 20.04 & \\textbf{6.73} & 11.28 & 9.80 & 9.54\\\\\n \\hline\n \\textit{back-image} & 22.61 & 24.01 & 27.41 & 16.31 & 23.00 & \\textbf{16.15} & 20.94\\\\\n \\hline\n \\textit{rot-back-image} & 55.18 & 56.41 & 62.16 & \\textbf{47.39} & 51.93 & 52.21 & 54.03\\\\\n \\hline\n \\textit{rect-image} & 24.04 & 24.05 & 33.20 & 23.69 & 24.05 & \\textbf{22.50} & 25.04\\\\\n \\hline\n\\end{tabular}\n\\caption{Experimental Results of MKMs with Mixed Kernels}\n\\label{tab_results_mix}\n\\end{table}\n\\renewcommand{\\arraystretch}{1}\n\nAs in MKMs with arc-cosine kernel, we chose top 5 percent features based on univariate test score for the mixed kernel MKMs. The classifier used in the output layer was an SVM\\nomenclature{SVM}{Support Vector Machines} with arc-cosine kernel. The results indicates that MKMs with mixed kernels improved the classifier performance for \\textit{mnist-back-rand} and \\textit{mnist-back-image} datasets, whereas the classification accuracy declined in the case of \\textit{mnist-rot-back-image} and \\textit{rectangles-image} dataset. The tSNE embedding of the features produced by MKMs with mixed kernels is shown in figure \\ref{tsne_rand_mkmmix} (for \\textit{mnist-back-rand} dataset). The best result in this dataset was obtained from a model consisting of three layers with a Gaussian kernel in the middle layer. The visualization indicates that, MKMs with mixed kernels has good separability between different classes.\n\\begin{sidewaysfigure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.45]{figures/tsne_rand_mkmmix}\n \\caption{tSNE embedding of features obtained by MKM with mixed kernels from \\textit{mnist-back-rand} dataset.}\n \\label{tsne_rand_mkmmix}\n\\end{sidewaysfigure}\n\n\n\\section{Conclusion}\n\\label{chap2_conc}\nIn this chapter we analysed the MKMs framework which has the characteristics of deep learning algorithms by employing kernel methods. The empirical study indicated that MKMs are performing comparable with that of popular deep learning algorithms like DBN, SAA etc. We also analysed the performnace of MKMs with mixed kernels. The mixing produces better results when other kernels are sandwiched between two arc-cosine kernels.\n\nThe visualization of features learned by MKMs are difficult to interpret due to crowding of datapoints near the center, but for MKMs with mixed kernels the visualization indicates that the separation is good enough between different classes, even though the samples from same class is distributed in different locations.\n", "meta": {"hexsha": "d5bfc994a99f4d7b006dd0ebef12014a3506817e", "size": 26831, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/chapter2.tex", "max_stars_repo_name": "akhilpm/Masters-Project", "max_stars_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Thesis/chapter2.tex", "max_issues_repo_name": "akhilpm/Masters-Project", "max_issues_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/chapter2.tex", "max_forks_repo_name": "akhilpm/Masters-Project", "max_forks_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 90.3400673401, "max_line_length": 1045, "alphanum_fraction": 0.7515560359, "num_tokens": 7652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.78793120560257, "lm_q2_score": 0.7931059536292271, "lm_q1q2_score": 0.6249129302136529}} {"text": "\\documentclass[a4paper,11pt]{article}\n\n\\usepackage{physics}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\usepackage{amsthm, mathtools}\n%\\usepackage{hyperref}\n\\usepackage{color}\n\\usepackage{jheppub}\n\\usepackage[T1]{fontenc} % if needed\n\n\\newcommand{\\be}{\\begin{equation}}\n\\newcommand{\\ee}{\\end{equation}}\n\\newcommand{\\bes}{\\begin{equation*}}\n\\newcommand{\\ees}{\\end{equation*}}\n\\newcommand{\\bea}{\\begin{flalign*}}\n\\newcommand{\\eea}{\\end{flalign*}}\n\n%\\linespread{1.0}\n%\\setlength{\\parindent}{0em}\n%\\setlength{\\parskip}{0.8em}\n\n\\title{\\textbf{Methods for Data}}\n\\author{Aditya Vijaykumar}\n\\affiliation{International Centre for Theoretical Sciences, Bengaluru, India.}\n\\emailAdd{aditya.vijaykumar@icts.res.in}\n\\abstract{}\n\n\\begin{document}\n\\maketitle\n\\section{A Few Basics}\n\nPhysicists believe in a \\textit{cause and effect} philosophy of the world. If we are given a fair coin, we can assign equal probabilities to \\textit{head} and \\textit{tail} and predict the outcomes given the number of times the coin is tossed. We could deal with unfair coins equally well, and given the properties of the unfair coin (\\textit{ie.} probabilities of landing heads/tails/neither) are specified, we can make deductions about the outcomes of our \\textit{experiment}.\n\nBut, generally, we have information about the observed effects, from which we are supposed to ascertain what models/scenarios these effects would have arisen from. Consider tossing a single coin ten times. The nature of the coin is not specified, but we find that all ten times the coin lands heads up. Are we to believe that the coin is \\textit{biased}? Not quite. The most one can do is make some inference based on the data and our prior knowledge (\\textit{ie}. we say that the coin is \\textit{most likely} biased), adding a caveat that we reserve the right to alter our result if new information pops up in the future (\\textit{ie.} we have the data from more tosses of the same coin).\n\nThis makes the whole problem of \\textit{data analysis} open-ended. Through this exploration, we hope to understand the mathematical rules governing the analyses.\n\n\\subsection{Operations on Probabilities}\nAll the definitions of probability made below are with respect to some background information denoted by $ I $.\n\\begin{itemize}\n\t\\item \\textbf{Sum Rule} - Probability of an X being true and that of it being false should add up to 1.\n\t\\begin{equation*}\n\tP(X | I) + P (\\overline{X} | I) = 1\n\t\\end{equation*}\n\t\\item \\textbf{Product Rule} - Probability of both $ X $ and $ Y $ being true should be the product of probability of $ Y $ being true and the probability of $ X $ being true given $ Y $ is true.\n\t\\begin{equation*}\n\tP(X,Y| I) = P(X|Y,I) \\cross P(Y|I) = P(Y|X,I) \\cross P(X|I)\n\t\\end{equation*}\n\tThe symmetry property of the AND operation on $ X $ and $ Y $ means that we can interchange $ X $ and $ Y $ on the LHS.\n\t\\item \\textbf{Bayes' Theorem} - The probability of $ X $ conditioned on $ Y $ is proportional to the probability of $ Y $ conditioned on $ X $.\n\t\\begin{equation*}\n\tP(X|Y,I) =\\dfrac{ P(Y|X,I) \\cross P(X|I)}{P(Y|I)}\n\t\\end{equation*}\n\tThe proof follows from the second and third expressions in the product rule. The magic of Bayes' theorem becomes evident when we replace $ X $ with \\textit{hypothesis} and $ Y $ with \\textit{data}.\n\t\\begin{equation*}\n\t\\underbrace{P(hypothesis|data,I)}_{\\text{posterior}} \\propto \\underbrace{ P(data|hypothesis,I)}_{\\text{likelihood}} \\cross \\underbrace{ P(data|hypothesis,I)}_{\\text{prior}}\n\t\\end{equation*}\n\tThe statement of the Bayes' theorem now becomes a powerful tool for inversion! Some definitions follow,\n\t\\begin{itemize}\n\t\t\\item \\textbf{Prior} - This encodes our degree of ignorance about the hypothesis before we have even touched the data.\n\t\t\\item \\textbf{Likelihood Function} - This encodes how well data agrees with a particular model (\\textit{Details Later}).\n\t\t\\item \\textbf{Posterior} - This gives us the degree of knowledge of the truth of the model in light of the data.\n\t\t\\item \\textbf{Evidence} - $ P(data| I) $ which we omitted while writing the expression of the Bayes' theorem. In most cases, it will remain a proportionality constant.\n\t\\end{itemize}\n\n\t\\item \\textbf{Marginalization} - The probability of $ X $ is the same as the probability of $ X $ given $ Y_k $, summed over all $ k $. Here, $ \\{Y_k\\} $ is the set of all possibilities in $ Y $.\n\t\\begin{equation*}\n\tP(X| I) = \\sum_k P(X|Y_k, I)\n\t\\end{equation*}\n\tA continuous, integral form of the equation can be written as follows,\n\t\\begin{equation*}\n\tP(X|I) = \\int_{-\\infty}^\\infty P(X,Y|I) \\dd{Y} \n\t\\end{equation*}\n\\end{itemize}\n\\end{document}", "meta": {"hexsha": "59c94cb4545b5bf43b35707cf9bc1028579025f0", "size": 4642, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "selfstudy/data/notes/data.tex", "max_stars_repo_name": "adivijaykumar/courses", "max_stars_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "selfstudy/data/notes/data.tex", "max_issues_repo_name": "adivijaykumar/courses", "max_issues_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "selfstudy/data/notes/data.tex", "max_forks_repo_name": "adivijaykumar/courses", "max_forks_repo_head_hexsha": "c0aebb67332ccf0b116a3348923ab2631b586dac", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.2857142857, "max_line_length": 688, "alphanum_fraction": 0.7361051271, "num_tokens": 1348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059609645724, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.6249129280943225}} {"text": "In game theory, the default assumption is that all players are perfectly rational and that this is common knowledge between all players. From this assumption, often without additional arguments, game theorists assume that \\emph{backward induction} can be used to determine outcomes and strategies of perfect-information games. Backward induction is an algorithm that computes the outcome of a game assuming that all players want to maximize their payoffs from all states of the game.\n\nIn this section we shall investigate what rationality is. We shall also examine why common knowledge of rationality implies correctness of backward induction.\n\n\\paragraph*{Backward induction}\nThe idea of backward induction is straightforward: When a game tree has been constructed and payoffs have been assigned to the leaves, use a bottom-up approach to determine the \\emph{inductive outcome}. This approach can also determine the strategies of the players. \n\nStarting in the last stages (the nodes before the leaves) of the game tree, determine the action which yields maximal payoff and save the payoffs in the current node, so that the maximal payoffs propagate upwards in the tree until the root is encountered.\n\n\\subsection{Definitions and Assumptions}\nIn order to talk about rationality in perfect-information games, some important definitions and assumptions must be presented. Note that the Liar's Dice example is not used in this section, as it is not a game of perfect information.\n\n\\paragraph*{Strategy} A player participating in a game chooses a strategy, as defined previously in \\secref{seq:perfect-information}. This involves that a player, when choosing a strategy and deciding what action to take at some node $v$, act as though $v$ has been reached.\n\n\\paragraph*{Rationality} Rationality can now be defined as \\textit{choosing an action which yields maximum payoff at any node $v$}. In other word, a player will never choose a strategy that results in less payoff than another strategy, independent of which node he/she resides in. \nIt is important to understand that, even if $v$ is not reached, a player must still act rationally in $v$ for backward induction to work. This is also linked to the definition of strategy, as the player acts as though $v$ has been reached.\n\n\\paragraph*{Common knowledge of Rationality} The simple definition of rationality is enough, to show that backward induction works. Therefore the notion of common knowledge in epistemic logic is used. \nEssentially common knowledge of rationality boils down to all players being rational and all players knowing this fact. This piece of information is also known to all players, which results in an infinite chain of knowledge, which defines common knowledge. \n\n\\subsection{Results}\nAs stated before, it's important to assume that common knowledge of rationality holds, in order for backward induction to work.\nIntuitively, this requirement makes perfect sense, which is illustrated in \\figref{prg:lec5}:\nAssume that the black nodes in the game tree is where the player Ann has to select an action. She could either end the game by taking action 1, and receive 3 utils as payoff, or continue the game by choosing action 2. By backward induction it is obvious that she should end the game by choosing action 1, because she knows that Bob is rational and will choose payoffs $A:2$, $E:4$. By choosing action 2, Ann would not optimize her payoff, because she would only receive 2 utils. Let's say Ann doesn't know that Bob will act rationally. If Ann considers it possible that Bob is not rational, Bob could possibly choose the payoffs $A:7$, $E:1$, which then would maximize Anns payoff. Therefore the common knowledge of rationality is needed.\n\n\n\\begin{figure}[htbp]\n\\centering\n\t\\begin{tikzpicture}[every node/.style={draw,circle}, align=center, node distance=1.9cm]\n\t%Nodes\n\t\\node[draw,fill=black] \t(q1) {};\n\t\\node[draw, below right=1cm of q1,style={draw=none}] (q2) {$_{E:2}^{A:3}$};\n\t\\node[draw, below left of=q1] (q3) {};\n\t\\node[draw,fill=none, below right of=q3,style={draw=none}] (q4) {$_{E:4}^{A:2}$};\n\t\\node[draw,fill=none, below left of=q3,style={draw=none}] (q5) {$_{E:1}^{A:7}$};\n\tíges\n\t\\draw [line] (q1) -- (q2) node [label={[shift={(-0.6,-0.5)}]2},very near start, right, draw=none, fill=none] (TextNode) {1};\n\t\\draw [line] (q1) -- (q3) node [very near start, left, draw=none, fill=none] (TextNode) {2};\n\t\\draw [line] (q3) -- (q4) node [very near start, right, draw=none, fill=none] (TextNode) {1};\n\t\\draw [line] (q3) -- (q5) node [very near start, left, draw=none, fill=none] (TextNode) {2};\n\t\\end{tikzpicture}\n\t\\caption{Game tree for a perfect-information game between Ann and Bob.}\n\t\\label{prg:lec5}\n\\end{figure}\n\n\nHere, it can be seen that common knowledge of rationality implies backward induction. To proof that this is actually the case, one must look into epistemic logic to try and model the notion of common knowledge of rationality. \n\nFirst step is to create a possible worlds model $M$, where $\\Omega$ is the set of all states of the world. \nLet $s$ be a function defined by\n$$\ns(\\omega) = \\times_i S_i\n$$\nwhere $S_i$ is the set of all strategies of player $i$. The function $s$ is therefore a function that maps a state $\\omega$ in the world model, to a tuple of the players strategies at $\\omega$.\nLet the set of states in $M$, where the strategies of $s(\\omega)$ are rational for all players in the game, be denoted $R$. Let the set of states where $R$ is common knowledge be denoted $CR$. At some states in the $M$, the strategy represented in the state leads to the inductive outcome. Let $I$ denote these states.\nIn \\cite{aumann1995a} a detailed proof is given to show that\n$$\nCR \\subset I,\n$$\nwhich means that common knowledge of rationality implies correctness of backward induction.\n\n\\subsection{Discussion}\nIt is worth to investigate a play of a game where some nodes are not reached. Previously we assumed that a player would act rationally at every point in the game, but let's assume that we only demand rationality at the nodes that are actually reached in the game tree.\nIf a node is never reached, then one could possibly say that any decision made in that node is actually rational, but it's obvious that this prevents backward induction from working correctly. Therefore it is important to assume rationality at all nodes.\n\nCommon knowledge of rationality is a quite rare scenario in practice, given that the players are human. Often human players are not rational at all or at least it is not common knowledge. In these cases it would be irrational to choose a strategy based on the result of the backwards induction, because a player do not know the reasons for his/hers opponents choices.\n\nFrom a human perspective, perfect-information games with common knowledge of rationality are not fun to play, as the course of the game is completely known before the first move is made. It would be a complete waste of time to play a game and reach an outcome that was determined beforehand.\n", "meta": {"hexsha": "e595a8538a3cb2698d5ad25b6fe66696852e3ff7", "size": 7035, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Report/rationality.tex", "max_stars_repo_name": "Andefar/LiarsDice", "max_stars_repo_head_hexsha": "f47123e80283a11500232d34102db2e894a4affb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Report/rationality.tex", "max_issues_repo_name": "Andefar/LiarsDice", "max_issues_repo_head_hexsha": "f47123e80283a11500232d34102db2e894a4affb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2016-11-24T13:56:28.000Z", "max_issues_repo_issues_event_max_datetime": "2016-12-01T17:21:52.000Z", "max_forks_repo_path": "Report/rationality.tex", "max_forks_repo_name": "Andefar/LiarsDice", "max_forks_repo_head_hexsha": "f47123e80283a11500232d34102db2e894a4affb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 101.9565217391, "max_line_length": 738, "alphanum_fraction": 0.7694385217, "num_tokens": 1700, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311906630568, "lm_q2_score": 0.7931059438487663, "lm_q1q2_score": 0.6249129106587059}} {"text": "\\section{Modular Arithmetic}\n\n% From prime numbers to simple criptography\n\n\\frame{\n{Part 4: Modular Arithmetic}\n\n\\tableofcontents[currentsection,hideallsubsections, firstsection=1, sections={1-5}]\n}\n\n% Modular arithmetic and congruence\n\n\\subsection{Congruence}\n\\begin{frame}{Modular Arithmetic and Congruence}\n\n Using \\structure{modular arithmetic}, we can make the Turing Algorithm stronger.\\bigskip\n\n Modular arithmetic is centred around the concept of \\structure{{\\bf Congruence}}:\n\n \\begin{equation*}\n a\\equiv b (\\text{modulo } n) \\iff n|(a-b)\n \\end{equation*}\\bigskip\n\n \\structure{Examples:}\n \\begin{itemize}\n\\item $30\\equiv12 (\\text{modulo } 9)$,\\hspace{4.3cm} because $9|(30-12)$\n\\item $66666663 \\equiv 788253 (\\text{modulo } 10)$,\\hspace{2cm} because $10|66666663 - 788253$\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}{Modular Arithmetic and Congruence}{Remainder Theorem}\n\n One important implication of the definition of equivalence is the remainder theorem:\n \\begin{equation*}\n a \\equiv b (\\text{mod } n) \\iff \\text{ rem}(a,n) = \\text{ rem}(b,n)\n \\end{equation*}\n\n \\begin{proof}\n Proof by two way implication. Let the remainder of $a$ and $b$ be $r_{a,b}$:\n\n \\begin{itemize}\n \\item Let $a = q_an + r_{a,n}$, and $b = q_bn + r_{b,n}$\n \\item (right side) {\\bf if} $r_{a,n} = r_{b,n}$ then $a-b = (q_a-q_b)n \\implies n|(a-b)$\n \\item (left side) {\\bf if} $n|(a-b)$ then $n|((q_a-q_b)n+(r_{a,n}-r_{b,n}))$\n \\begin{itemize}\n \\item but $0 \\leq r_{*,n} < n$ so $r_{a,n}-r_{b,n}$ must be 0\n \\end{itemize}\n \\end{itemize}\n \\end{proof}\n\\end{frame}\n\n% Congruence is preserved by addition and multiplication\n\\begin{frame}{Modular Arithmetic and Congruence}{Consequences of the Remainder Theorem}\n\n Remainder Theorem:\n \\begin{equation*}\n a \\equiv b (\\text{mod } n) \\iff \\text{ rem}(a,n) = \\text{ rem}(b,n)\n \\end{equation*}\n\n \\begin{itemize}\n \\item $a \\equiv b$ (mod n) {\\bf and} $b \\equiv c$ (mod n) {\\bf implies}\n $a \\equiv c$ (mod n)\n \\item $a \\equiv $ rem(a,n) (mod n) \\hspace{1cm} (\\alert{important!})\\bigskip\n \\item $a \\equiv b$ (mod n) {\\bf implies} $a+c \\equiv b+c$ (mod n)\n \\item $a \\equiv b$ (mod n) {\\bf implies} $ac \\equiv bc$ (mod n)\n \\item $a \\equiv b$ (mod n) {\\bf and} $c\\equiv d$ (mod n)\\\\\n \\hspace{2cm}{\\bf implies} $a+c \\equiv b+d$ (mod n) {\\bf and} $ac \\equiv bd$ (mod n)\n \\end{itemize}\\bigskip\n\n The last three consequences show that we can freely use \\structure{addition} and \\structure{multiplication} in modular arithmetic. Try to prove some of these consequences!\n\\end{frame}\n\n% Remainder Arithmetic\n\\begin{frame}{Using Modular Arithmetic}{General Principle of Modular Arithmetic}\n\n \\begin{block}{}\n You can simplify modulo operations composed of additions and multiplications by replacing integer operands by their remaiders.\n \\end{block}\n\n Example: What is $287^9 \\equiv$ ? (mod 4)\n\n \\begin{itemize}\n \\item Simplify $287^9$ (mod 4) to $3^9$ (mod 4), becase $r_{287,4} = 3$\n \\item $3^9 \\to 3^8\\times 3 \\to 9^4 \\times 3$\n \\item Simplify $9^4 \\times 3$ (mod 4) to $1^4 \\times 3$, because $r_{9,4} = 1$\n \\item $289^9 \\equiv 1^4 \\times 3 \\equiv 3$ (mod 4)\n \\end{itemize}\\bigskip\n\n We calculated a large exponent without actually calculating the exponentiation!\n\\end{frame}\n\n% Remainder Inverse\n% Remainder inverse and relatively prime numbers\n\n\\subsection{Modular Inverses}\n\n\\begin{frame}{Modular Arithmetic and Division}\n\n We so that modular arithmetic works for addition and multiplication, so what about \\structure{division}?\\bigskip\n\n \\begin{itemize}\n \\item $8\\times 2 \\equiv 3\\times 2$ (mod 10)\n \\item $8\\times \\not{2} \\equiv 3\\times \\not{2}$ (mod 10)\n \\item $8 \\equiv 3$ (mod 10) \\hspace{2cm} \\alert{\\bf FALSE!}\n \\end{itemize}\\bigskip\n\n We cannot cancel out multiplications arbitrarily!\n\\end{frame}\n\n\\begin{frame}{Modular Arithmetic and Division}{Modular Inverses}\n\n When we cancel the multiplication of a real number, it is equivalent of multiplying it by its {\\bf multiplicative inverse}:\n\n \\begin{equation*}\n an = bn \\to a\\not{n} = b\\not{n} \\to a(n\\times \\frac{1}{n}) = b(n\\times \\frac{1}{n})\n \\end{equation*}\n\n So a multiplicative inverse of $n$ is the number $k$ so that $n\\times k =1$.\\bigskip\n\n In modulo arithmetic, we have: $n\\times k \\equiv 1$ (mod m). But, it turns out that it is not possible to find a $k$ that solves this equation for every $n$ and $m$!\n\\end{frame}\n\n\n\\begin{frame}{Modular Inverses}{When does $n$ have an inverse, and how can we find it?}\n\n It turns out that $n$ only has a modular inverse (module $m$) if $gcd(n,m) = 1$.\\bigskip\n\n It is relatively easy to find the modular inverse:\n \\begin{itemize}\n \\item First we calculate gcd(n,m) using the pulverizer, and obtaining $s$ and $t$ so that $sm+tn=1$\n \\item The multiplicative inverse of $n$ modulo $m$ is given by: $r_{m,t}$\n \\item So $n\\times r_{m,t} \\equiv 1$ (mod $m$)\n \\end{itemize}\n\\end{frame}\n\n\\subsection{Turing Code 2.0}\n\n\\begin{frame}{Turing Code 2.0}\n Using Modular Arithmetic, we can solve some of the problems of the cryptography algorithm described in the last section:\n\n \\begin{enumerate}\n \\item The sender and the receiver agree on a large prime number $n$, which will be the module, as well as the secret key $k < n$\n \\item {\\bf Encription:} The message $m$ is a prime number ($0 < m < n$). The sender calculates the secret message $\\hat{m}$ using modulo multiplication:\n \\begin{equation*}\n \\hat{m} = m\\times k \\text{ mod }n\n \\end{equation*}\n \\item {\\bf Decription:} The receiver multiplies the encrypted message $\\hat{m}$ by the inverse of the key: $k^{-1}$ (mod $n$):\n \\begin{equation*}\n m = \\hat{m}0\\times k^{-1} \\text{ mod }n\n \\end{equation*}\n \\end{enumerate}\n We cannot find $k$ anymore by having two encrypted messages, so is this algorithm safe?\n\\end{frame}\n\n\\begin{frame}{Turing Code 2.0}{Breaking the Turing Code with a Plaintext Attack}\n\n The new algorithm can be broken by what is called a \\structure{{\\bf Plaintext Attack}}.\\bigskip\n\n Imagine that an attacker is able to acquire an encrypted message $\\hat{m}$ and its respective clear message $m$. To break this code:\\bigskip\n\n \\begin{itemize}\n \\item Calculate the multiplicative inverse of $m$ modulo $n$: $m^{-1}$, using the Pulverizer.\n \\item Multiply the result by the encrypted message:\n \\begin{itemize}\n \\item $m^{-1}\\times \\hat{m} = m^{-1}\\times (m \\times k) = (m^{-1} \\times m) \\times k = 1\\times k$ (mod $n$)\n \\end{itemize}\n \\item This allows us to recover the secret key $k$.\n \\end{itemize}\\bigskip\n\n So this algorithm is not so good after all! :-(\n\\end{frame}\n", "meta": {"hexsha": "394c638d6b6e2a2deb8a4803f93a90c5ac80de1f", "size": 6632, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week03/04_ModularArithmetic.tex", "max_stars_repo_name": "caranha/MathCS", "max_stars_repo_head_hexsha": "f3ce6705d09c55541f629cd542191bfd3e9adf34", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-09-13T18:59:55.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-16T02:14:56.000Z", "max_issues_repo_path": "week03/04_ModularArithmetic.tex", "max_issues_repo_name": "caranha/MathCS", "max_issues_repo_head_hexsha": "f3ce6705d09c55541f629cd542191bfd3e9adf34", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "week03/04_ModularArithmetic.tex", "max_forks_repo_name": "caranha/MathCS", "max_forks_repo_head_hexsha": "f3ce6705d09c55541f629cd542191bfd3e9adf34", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.5581395349, "max_line_length": 173, "alphanum_fraction": 0.6749095296, "num_tokens": 2172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.7956581073313276, "lm_q1q2_score": 0.6248371391402043}} {"text": "\\chapter{Massless Spin-1 Particles}\nWe approach the quantum description of photons via classical electromagnetism, since we know via the correspondence principle that Maxwell's equations describe their dynamics.\n\n\\section{Maxwell's equation and the classical potential}\\label{sec:maxwell}\nFor fields $\\vec{E}$ and $\\vec{B}$ in a vacuum (i.e.~not in material), Maxwell's equations are:\n\\begin{enumerate}\n\\item $\\div\\vec{E} = \\rho$, the charge density (Poisson's equation);\n\\item $\\div\\vec{B} = 0$, no magnetic monopoles;\n\\item $\\curl\\vec{E} = -\\pdv{\\vec{B}}{t}$, Faraday's law;\n\\item $\\curl\\vec{B} = \\vec{j} + \\pdv{\\vec{E}}{t}$ where $\\vec{j}$ is the charge current density (Amp{\\'e}re's law).\n\\end{enumerate}\n\nWe define the vector potential such that\n\\begin{equation}\\boxed{\n\\vec{B} = \\curl \\vec{A}\n}\\end{equation}\nso Maxwell~II is automatically satisfied. Substituting into Maxwell~III,\n\\begin{align}\n\\curl\\vec{E} &= -\\pdv{t} \\left( \\curl \\vec{A} \\right) \\nonumber \\\\\n&= -\\curl \\pdv{\\vec{A}}{t}\n\\end{align}\n\\begin{equation}\n\\Rightarrow \\quad \\curl \\left( \\vec{E} + \\pdv{\\vec{A}}{t} \\right) = 0.\n\\end{equation}\nThe solution to this equation defines the scalar potential,\n\\begin{equation}\n\\vec{E} + \\pdv{\\vec{A}}{t} = -\\grad \\phi.\n\\end{equation}\nSubstituting into Maxwell~I,\n\\begin{align}\n\\rho &= -\\laplacian{\\phi} - \\pdv{t} {\\div\\vec{A}} \\label{eq:MaxwellDensity}\n\\end{align}\nNow consider Maxwell~IV. The left-hand side is\n\\begin{align}\n\\curl \\vec{B} &= \\curl(\\curl{\\vec{A}}) \\\\\n&= \\grad(\\div\\vec{A}) - \\laplacian{\\vec{A}} \\label{eq:Maxwell4LHS}\n\\end{align}\nand the right-hand side is\n\\begin{align}\n\\vec{j} + \\pdv{\\vec{E}}{t} &= \\vec{j} - \\pdv{t}\\left(\\grad{\\phi} + \\pdv{\\vec{A}}{t}\\right) \\nonumber \\\\\n&= \\vec{j} - \\grad{\\pdv{\\phi}{t}} - \\pdv[2]{\\vec{A}}{t} \\label{eq:Maxwell4RHS}.\n\\end{align}\nEquating \\eqref{eq:Maxwell4LHS} and \\eqref{eq:Maxwell4RHS} gives an expression for the 3-current,\n\\begin{equation}\n\\vec{j} = \\pdv[2]{\\vec{A}}{t} - \\laplacian{\\vec{A}} + \\grad(\\pdv{\\phi}{t} + \\div\\vec{A}).\n\\end{equation}\nAlso, \\eqref{eq:MaxwellDensity} can be written\n\\begin{equation}\n\\rho = \\pdv[2]{\\phi}{t} - \\laplacian{\\phi} - \\pdv{t}(\\pdv{\\phi}{t} + \\div\\vec{A})\n\\end{equation}\nso we can write Maxwell's equations in covariant form:\n\\begin{equation}\\boxed{\nj^\\mu = \\partial^2 A^\\mu - \\partial^\\mu \\partial_\\nu A^\\nu \\label{eq:MaxwellCovariant}\n}\\end{equation}\nwhere the 4-current is $j^\\mu = \\mqty(\\rho \\\\ \\vec{j})$ and the 4-potential is $A^\\mu = \\mqty(\\phi \\\\ \\vec{A})$.\n\n\\section{Gauge transformations and the Lorenz condition}\nEquation \\eqref{eq:MaxwellCovariant} is invariant under gauge transformations that have the form\n\\begin{equation}\nA^\\mu \\rightarrow A^\\mu + \\partial^\\mu f\n\\end{equation}\nfor some scalar $f$; this is easily verified by direct substitution. For a global gauge transformation $f$ is constant but for local gauge transformations $f$ can be a function of the coordinates.\n\nThis degree of freedom may be frozen out by the choice of a particular `gauge'. The Lorenz\\footnote{The Lorenz gauge is named after Ludwig Lorenz and is Lorentz invariant, named after the work of Hendrik Lorentz.} gauge is the most natural choice with the condition\n\\begin{equation}\n\\partial_\\mu A^\\mu = 0 \\label{eq:LorentzGauge}\n\\end{equation}\nsuch that all electrodynamics (Maxwell's equations) are simply described by\n\\begin{equation}\\boxed{\n\\partial^2 A^\\mu = j^\\mu.\n}\\end{equation}\n\n\\subsection{Polarisation states of a free photon}\nSince the 4-potential $A^\\mu$ describes the photon, there are 4 possible orthogonal polarisations available to it. However, the Lorenz gauge removes one degree of freedom.\n\nFor a free photon, we have that the charge current $j^\\mu = 0$. Therefore,\n\\begin{equation}\n\\partial^2 A^\\mu = 0\n\\end{equation}\nwith polarised plane wave solutions $A^\\mu = \\epsilon^\\mu_i \\, e^{-ipx}$ where $\\epsilon^\\mu_i$ are the polarisation 4-vectors. Therefore, the Lorenz condition becomes\n\\begin{equation}\np_\\mu \\epsilon^\\mu_i = 0.\n\\end{equation}\nThis has removed one degree of freedom from the available polarisation states.\n\nNow consider the gauge transformation\n\\begin{equation}\nA^\\mu \\rightarrow A^\\mu + \\partial^\\mu\\left( i\\alpha e^{-ipx} \\right).\n\\end{equation}\nThen\n\\begin{equation}\n\\partial_\\mu \\partial^\\mu \\left( \\alpha e^{-ipx} \\right) = -i \\alpha p^2 e^{-ipx} = 0\n\\end{equation}\nsince $p^2 = 0$ for a free photon (it's massless). Therefore the photon is unchanged by such a gauge transformation, which can be written\n\\begin{equation}\n\\epsilon_i^\\mu \\rightarrow \\epsilon_i^\\mu + \\alpha p^\\mu.\n\\end{equation}\n\nWe are therefore free to make the gauge choice $\\epsilon_i^0 = \\epsilon_i^3 = 0$ (with the spatial axes chosen such that $p^0 = p^3 = E$) which corresponds to $\\alpha = -\\epsilon_\\mu^i p^\\mu / \\epsilon_\\nu^i \\epsilon_i^\\nu$.\n\nFor a free photon with 4-momentum $p = (E,\\,0,\\,0,\\,E)^T$, the surviving polarisation states are\n\\begin{equation*}\n\\epsilon_1^\\mu = \\mqty(0 \\\\ 1 \\\\ 0 \\\\ 0) \\quad \\text{and} \\quad \\epsilon_2^\\mu = \\mqty(0 \\\\ 0 \\\\ 1 \\\\ 0)\n\\end{equation*}\nwhich can be combined to give circular polarisations\n\\begin{equation*}\n\\epsilon_R^\\mu = \\frac{1}{\\sqrt{2}} (\\epsilon_1^\\mu + i\\epsilon_2^\\mu) \\quad \\text{and} \\quad \\epsilon_L^\\mu = \\frac{1}{\\sqrt{2}} (\\epsilon_1^\\mu - i\\epsilon_2^\\mu)\n\\end{equation*}\nwhich correspond to the $\\pm 1$ helicity photon.\n\nIn summary, a free spin-1 particle loses one degree of freedom under gauge fixing in electrodynamics. A second gauge fixing removes another degree of freedom, but this relies on the fact that the particle is both massless and free. Therefore, the free photon has only two possible transverse polarisations. In contrast, a virtual photon may also have timelike or longitudinal polarisation.\n\n\\section{Virtual photons and the photon propagator}\nWe saw when discussing the electrodynamics of scalars that the potential \\eqref{eq:potential}, $A^\\mu = -g^{\\mu\\nu}j_\\nu / q^2$, is a solution of Maxwell's equations in the Lorenz gauge, $\\partial^2 A^\\mu = j^\\mu$. There, this was verified by substitution but now it will be derived via the propagator approach.\n\nAs a functional of the spatial propagator, $G(x^\\prime; x)$, the potential is given by\n\\begin{equation}\nA^\\mu(x^\\prime) = \\int G(x^\\prime; x) \\, j^\\mu(x) \\, \\dd[4]{x}.\n\\end{equation}\nApplying the d'Alambertian differential operator, the left-hand side of the Lorenz condition may be expressed\n\\begin{align}\n\\partial^2 A^\\mu(x^\\prime) &= \\int \\left[ \\partial^2 G(x^\\prime; x) \\right] \\, j^\\mu(x) \\, \\dd[4]{x}\n\\end{align}\nwhere we have used that $\\partial_\\mu j^\\mu = 0$ since the 4-current should be conserved (not explicitly shown for spin-1). The right hand side is\n\\begin{equation}\nj^\\mu = \\int \\delta^{(4)}(x^\\prime - x) \\, j^\\mu(x) \\, \\dd[4]{x}.\n\\end{equation}\nTherefore, we have that\n\\begin{equation}\n\\partial^2 G(x^\\prime; x) = \\delta^{(4)}(x^\\prime - x).\n\\end{equation}\nNow Fourier transform into momentum-space using $q^\\mu = -i\\partial^\\mu$,\n\\begin{equation}\n\\frac{1}{(2\\pi)^4} \\int (iq)^2 \\, \\mathcal{G}(q) \\, e^{-iq(x^\\prime - x)} \\, \\dd[4]{q} = \\frac{1}{(2\\pi)^4} \\int e^{-iq(x^\\prime - x)} \\, \\dd[4]{q}\n\\end{equation}\nfrom which we see that the momentum-space propagator is $\\mathcal{G}(q) = -1/q^2$. Then the potential is\n\\begin{equation}\nA^\\mu(x) = \\frac{-j^\\mu(x)}{q^2} = \\frac{-g^{\\mu\\nu} j_\\nu(x)}{q^2}\n\\end{equation}\nBy convention, we actually define $\\boxed{\\mathcal{G}(q) = -i g^{\\mu\\nu}/q^2}$.\n\n\\subsection{Momentum-space propagators}\nGeneralising the above, the momentum-space propagator may be obtained by inverting the equation of motion for a free particle and multiplying by $-i$.\n\nThe Klein-Gordon equation gives $(\\partial^2 + m^2)\\psi = 0$ for free scalars. Therefore, the propagator is\n\\begin{align}\n\\frac{-i}{(iq)^2 + m^2} = \\frac{i}{q^2-m^2}.\n\\end{align}\n\nNow consider the Dirac equation for free spin-$\\frac{1}{2}$ particles. From \\eqref{eq:covDirac}, the equation of motion is\n\\begin{equation*}\n(i \\gamma^\\mu \\partial_\\mu - m)\\psi = 0.\n\\end{equation*}\nThen, using that $i\\partial_\\mu = p_\\mu$ (note the lower-index form), the propagator is\n\\begin{align}\n\\frac{-i}{\\gamma^\\mu q_\\mu - m} &= \\frac{i}{\\gamma^\\mu q_\\mu - m} \\nonumber \\\\\n&= \\frac{i\\,(\\fsl{q}+m)}{(\\fsl{q}-m)(\\fsl{q}-m)} \\nonumber \\\\\n&= \\frac{i \\, (\\fsl{q}+m)}{q^2 - m^2} \\nonumber \\\\\n&= i \\frac{\\sum u_S \\overline{u}_{S^\\prime}}{q^2 - m^2}\n\\end{align}\nwhere on the third line we have used that, from the properties of the $\\gamma$-matrices, $\\fsl{q}^2 = \\gamma^\\mu q_\\mu \\gamma^\\nu q_\\nu = q^2$ and the spinor completeness relation has been used to obtain the last line. The sum is over all possible spins, $S$ and $S^\\prime$.\n\n\\section{Longitudinal and timelike virtual photons}\nSince we cannot imply the Lorenz gauge condition, virtual photons have four available polarisation states: two transverse, longitudinal, and timelike.\n\n\\subsection{Coulomb's law}\n\nThe amplitude for photon exchange between two currents is given by\n\\begin{align}\nA &= j_\\mu^A \\left( \\frac{-ig^{\\mu\\nu}}{q^2} \\right) j^\\nu_B \\nonumber \\\\\n&= j_\\mu^A \\left( \\frac{-i}{q^2} \\right) j^\\mu_B.\n\\end{align}\nNow the momentum exchange is $q^\\mu$. But we know that the four-current is conserved by QED and therefore $q_\\mu j^\\mu = 0$.\n\nWe are free to choose that the photon travels along the $z$- or $t$-axes, so $q_1 j^1 = q_2 j^2 = 0$. Then the continuity equation becomes\n\\begin{equation}\nq_\\mu j^\\mu = q_0 j^0 - q_3 j^3 = 0.\n\\end{equation}\nSubstituting back into the amplitude,\n\\begin{align}\nA &= \\frac{-i}{q^2} \\left[j_0^A j^0_B - j_3^A j^3_B\\right] \\nonumber \\\\\n&= \\frac{-i}{q^2} \\left[j_0^A j^0_B - \\frac{q_0^2}{q_3^2}(j_0^A j^0_B)\\right] \\nonumber \\\\\n&= \\frac{-i}{q_0^2 - q_3^2} \\left[ \\frac{q_3^2 - q_0^2}{q_3^2} j_0^A j^0_B \\right] \\nonumber \\\\\n&= i \\, \\frac{j_0^A j^0_B}{q_3^2}.\n\\end{align}\nThis is Coulomb's law for electromagnetism in 3-momentum space.\n\n\\subsection{Completeness relation}\nFor free photons, the two allowed transverse polarisations give the completeness relation\n\\begin{equation}\n\\sum_{P, Q} \\epsilon^P_\\mu \\epsilon_Q^\\nu = \\mqty(0,\\,1,\\,0,\\,0) \\mqty(0\\\\1\\\\0\\\\0) + \\mqty(0,\\,0,\\,1,\\,0) \\mqty(0\\\\0\\\\1\\\\0) = \\mqty(1&0\\\\0&1) = I_2.\n\\end{equation}\n\nIn the case of virtual photons with four allowed polarisations, this becomes\n\\begin{equation}\n\\sum_{P, Q} \\epsilon^P_\\mu \\epsilon_Q^\\nu = \\mqty(\\dmat[0]{-1,1,1,1}) = -g^{\\mu\\nu}\n\\end{equation}\nwhich embeds the completeness relation for real photons.\n", "meta": {"hexsha": "987668208e56249ae87a6b37626efb467956b43f", "size": 10345, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/12_Massless_Spin-1_Particles.tex", "max_stars_repo_name": "adambozson/Standard-Model-I", "max_stars_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/12_Massless_Spin-1_Particles.tex", "max_issues_repo_name": "adambozson/Standard-Model-I", "max_issues_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/12_Massless_Spin-1_Particles.tex", "max_forks_repo_name": "adambozson/Standard-Model-I", "max_forks_repo_head_hexsha": "9ea0d388c93f21d1c636ee18c9210a1b91169308", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.9849246231, "max_line_length": 389, "alphanum_fraction": 0.6930884485, "num_tokens": 3587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085808877581, "lm_q2_score": 0.7956581073313275, "lm_q1q2_score": 0.6248371391402043}} {"text": "\\newcommand\\chapternumber{3}\n\\input{../tex_header/header.tex}\n\\usepackage{enumerate}\n\\usepackage{float}\n\n\\begin{document}\nCollaborators: Julia Zimmerman, Jordan Donovan, Sam Rosenblatt, Milo Trujillo, Phil Nguyen, Nicholas Vartanian,Connor Klopfer, Brett Meyer...\nWe might have discussed problems related to the assignments via Slack.\n\\section{Manual Gradient Descent}\n\\begin{align*}\n \\frac{\\partial \\ell_i}{\\partial \\theta_j} = -2 x_{ij} (y_i - \\sum_{j'=1}^{p} x_{ij'} \\theta_{j'}) \\\\\n\\end{align*}\n\\begin{align*}\n \\nabla_{\\theta} \\ell & = \\sum_{i=1}^{N} \\begin{bmatrix}\n \\frac{\\partial \\ell_i}{\\partial \\theta_1} \\\\\n \\frac{\\partial \\ell_i}{\\partial \\theta_2} \\\\\n \\vdots \\\\\n \\frac{\\partial \\ell_i}{\\partial \\theta_p} \\\\\n \\end{bmatrix} \\\\\\\\\n & = -2 x^T (y-x\\theta)\n\\end{align*}\n\nI used the dataset this \\href{https://www.kaggle.com/sulianova/cardiovascular-disease-dataset}{Cardiovascular Disease dataset}.\n\nThe are 70k records in this dataset.\nThe task is to predict whether a person has cardiovascular disease or not base on his age, height, etc.\nI split the dataset into 80\\% training and 20\\% testing.\n\nThe source code is on \\href{https://github.com/liusida/ds2/blob/main/assignment3/code/q1.py}{GitHub q1.py}.\n\nI discovered that larger batch size and larger learning rate will result in faster convergence,\nhowever, if learning rate is too large, the learning might diverge,\nand if batch size and learning rate is too large will cause overflow.\nSo I normalized the input to reduce the possibility of overflow, and choose a batch size of 7k, and a small learning rate.\nNow the learning process with 1k epochs takes 1.5 seconds. (If I use a batch size of 100, it will take 9.8 seconds.)\n\nThe final test accuracy is 0.643.\nI also report the final confusion matrix:\nSensitivity 0.574, Specificity 0.712,\nPrecision 0.667, Negative Predictive Value 0.625.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=.8\\textwidth]{./q1.pdf}\n \\caption{Learning curve and two test points.}\n\\end{figure}\n\nWhy would I choose to manually do gradient descent?\nBecause I want to learn how it works, and also, as David mentioned in class, sometimes we have constraints in practice, for example there's possibility that PyTorch is not supported.\n\n\\newpage\n\\section{2D Rosenbrock function}\n\\begin{figure}[h]\n \\includegraphics[width=.99\\textwidth]{./rosenbrock.pdf}\n \\caption{Visualization of 2D Rosenbrock function. $x \\in (-30,30)$, $y \\in (-300,800)$}\n\\end{figure}\n\nIt is hard for SGD because the low area is not a dot but a long belt.\nIf the learning rate is large, it will diverge.\nIf the learning rate is low, it will be very slow in finding the minima on that flat belt.\n\nAnalytically, we need to solve the equations: $\\frac{\\partial f}{\\partial x} = 0$ and $\\frac{\\partial f}{\\partial y} = 0$.\n\n\\begin{align*}[left = \\empheqlbrace]\n \\frac{\\partial}{\\partial x} (1-x)^2+100(y-x^2)^2 & = 0 \\\\\n \\frac{\\partial}{\\partial y} (1-x)^2+100(y-x^2)^2 & = 0 \\\\\n\\end{align*}\n\n\\begin{align*}[left = \\empheqlbrace]\n 400x^3-400xy+2x-2 & = 0 \\\\\n y & =x^2 \\\\\n\\end{align*}\n\n\\begin{align*}\n 400x^3-400x^3+2x-2 & = 0 \\\\\n x & =1 \\\\\n y & =1 \\\\\n\\end{align*}\n\nUsing routine gradient descent, start from $x_0=0, y_0=0$, learning rate $\\gamma=10^{-3}$, run for 10k steps.\nThe optimization successfully converged to the minimum.\n\nThis is because the initialization is quite close to the minimum,\nand because it is smooth in $x \\in (0,1)$ and $ y \\in (0,1)$.\nFigure \\ref{fig:routine_gd} shows the trajectory of the optimization.\n\n\\begin{figure}[h]\n \\includegraphics[width=.8\\textwidth]{./routine_gd.png}\n \\caption{Visualization of the optimization trajectory. $x \\in (0,1)$, $y \\in (0,1)$.}\n \\label{fig:routine_gd}\n\\end{figure}\n\nIf we are not that lucky and use a bad initialization, such as $x_0 = 3$, $y_0 = -2$,\nthen the routine gradient descent will be harder.\nThe learning rate need to be smaller, otherwise it will diverge.\nThen momentum is needed to speed the optimization up.\nI set momentum to 0.99, learning rate $\\gamma=10^{-5}$, and run for 10k steps.\nThe optimization successfully converged to the minimum.\nFigure \\ref{fig:gd_m} shows the trajectory.\n\n\\begin{figure}[h]\n \\includegraphics[width=.8\\textwidth]{./gd_m.png}\n \\caption{Visualization of the optimization trajectory with momentum. $x \\in (0,3)$, $y \\in (-2,5)$.}\n \\label{fig:gd_m}\n\\end{figure}\n\nThe source code is on \\href{https://github.com/liusida/ds2/blob/main/assignment3/code/q2.py}{GitHub q2.py}.\n\n\\newpage\n\\section{Math notations and plate notations}\n\\subsection{Basic Parametric Density Estimation}\nMathematically, let $l=\\text{loc}$, $s=\\text{inv scale}$, we have:\n\\begin{align*}\n & p(l, s, d_i) \\\\\n & \\propto p(l)p(s)\\prod_{i=1}^N p(d_i|l,s) \\\\\n & = \\text{Normal}(0,1) \\text{Inv-Gamma}(3,2) \\prod_{i=1}^N \\text{Normal}(l, s) \\\\\n & = \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}l^2} \\frac{2^3}{2!}s^{-3-1} e^{-\\frac{2}{s}} \\prod_{i=1}^N \\frac{1}{s\\sqrt{2\\pi}} e^{-\\frac{1}{2}(\\frac{d_i-l}{s})^2} \\\\\n & = 4 (\\frac{1}{\\sqrt{2\\pi}})^{N+1} s^{-4-N} e^{-\\frac{l^2}{2} -\\frac{2}{s} - \\frac{1}{2}\\sum_{i=1}^N (\\frac{d_i-l}{s})^2} \\\\\n\\end{align*}\nIt doesn't seem to have conjugate prior to me.\n\nPlate notation:\n\\begin{figure}[h]\n \\centering\n \\tikz{\n % nodes\n \\node[obs] (data) {$d_i$};%\n \\node[latent,above=of data] (loc) {loc};\n % \\node[latent,above=of loc,xshift=-1cm] (mu) {$\\mu$}; %\n % \\node[latent,above=of loc,xshift=1cm] (sigma) {$\\sigma$}; %\n \\node[latent,left=of data] (scale) {scale};\n \\node[latent,left=of scale] (inv_scale) {inv\\_scale};\n % \\node[latent,left=of inv_scale,yshift=1cm] (alpha) {$\\alpha$};\n % \\node[latent,left=of inv_scale,yshift=-1cm] (beta) {$\\beta$};\n % plate\n \\plate [inner sep=.3cm,xshift=.02cm,yshift=.2cm] {plate1} {(data)} {N}; %\n % edges\n % \\edge {sigma,mu} {loc}\n % \\edge {alpha,beta} {inv_scale}\n \\edge {inv_scale} {scale}\n \\edge {scale,loc} {data}\n }\n \\caption{Plate notation}\n\\end{figure}\n\n\\newpage\n\\subsection{Continuous, nonstationary hidden Markov}\n\nMathematically, let $l=\\text{loc}$, $s=e^{\\text{log scale}}$, $o=\\text{obs scale}$, we have:\n\\begin{align*}\n & p(l,s,o,x_{n,t},y_{n,t}) \\\\\n & =p(l)p(s)p(o)\\prod_{n=1}^N p(x_{n,0}|l,s) \\prod_{t=1}^T p(x_{n,t}|l,x_{n,t-1},s) p(y_{n,t}|x_{n,t}, o) \\\\\n & =\\text{Normal}(0,1) \\text{LogNormal}(0,1) \\text{Gamma}(2,2) \\prod_{n=1}^N \\text{Normal}(l,s) \\prod_{t=1}^T \\text{Normal}(l+x_{n,t-1},s) \\text{Normal}(x_{n,t},o) \\\\\n & =\\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{l^2}{2}} \\cdot\n \\frac{1}{s\\sqrt{2\\pi}}e^{-\\frac{\\log(s)^2}{2}} \\cdot\n \\frac{1}{2!2^2}o^{2-1}e^{-\\frac{o}{2}} \\cdot\n \\prod_{n=1}^N\n \\frac{1}{s\\sqrt{2\\pi}}e^{-\\frac{(x_{n,0}-l)^2}{2s^2}}\n \\prod_{t=1}^T\n \\frac{1}{s\\sqrt{2\\pi}}e^{-\\frac{(x_{n,t}-l-x_{n,t-1})^2}{2s^2}}\n \\frac{1}{o\\sqrt{2\\pi}}e^{-\\frac{(y_{n,t}-x_{n,t})^2}{2o^2}} \\\\\n\\end{align*}\n\nPlate notation (Reference: \\href{https://davidrushingdewhurst.com/blog/2020-07-28keep-using-plate-notation.html}{this blog post}):\n\\begin{figure}[h]\n \\centering\n \\tikz{\n % \\draw[lightgray,ultra thin] (-6,-6) grid (6,6);\n % nodes\n \\node[obs] (y_t) {$y_{n,t}$};%\n \\node[latent,left=of y_t] (x_t) {$x_{n,t}$};%\n \\node[latent,above=of x_t] (x_0) {$x_{n,0}$};%\n \\node[latent,above=of x_0,xshift=1.5cm] (loc) {loc};%\n \\node[latent,right=of loc] (scale) {scale};%\n \\node[latent,above=of scale] (log_scale) {log scale};%\n \\node[latent,right=of y_t] (obs_scale) {obs scale};%\n % plate\n \\plate [inner sep=.3cm,xshift=.02cm,yshift=.2cm] {episode} {(x_t)(y_t)} {T}; %\n \\plate [inner sep=.3cm,xshift=.02cm,yshift=.2cm] {plate1} {(x_0)(episode)} {N}; %\n % edges\n \\edge {obs_scale,x_t} {y_t}\n \\edge {loc,scale} {x_t}\n \\edge {loc,scale} {x_0}\n \\edge {log_scale} {scale}\n % hmm...\n \\draw[blue, thick, ->,shorten >=1pt] (x_0) to [out=-90,in=90] node[left,yshift=.2cm,xshift=-.2cm] {$t=1$} (x_t);\n \\Cycle[blue, ->]{x_t}{180}{7mm}[{node[anchor=0,pos=0.5]{$t=2,...,T (t-1\\rightarrow t)$}}]{7mm}\n }\n \\caption{Plate notation}\n\\end{figure}\n\n\\newpage\n\\subsection{Switching model}\nMathematically, let $s'=\\text{log scale}$, $s=e^{s'}$, $w=\\text{switch}$, $m_1=\\text{model1}$, $m_2=\\text{model2}$, we have:\n\\begin{align*}\n & p(s,z_0,z_t,p_t,m_{1,t},m_{2,t},w_{t,n},y_{t,n},x_{t,n}) \\\\\\\\\n & =p(s)p(z_0|s) \\times \\\\\n & \\prod_{t=1}^T p(z_t|z_{t-1}) p(p_t|z_t) \n \\prod_{n=1}^N p(w_{t,n}|p_t) \n p(m_{1,t}|t) p(m_{2,t}|t) p(y_{t,n}|w_{t,n},m_{1,t},m_{2,t}) \n p(x_{t,n},y_{t,n}) \\\\\\\\\n & =\\text{LogNormal}(0,1)\n \\text{Normal}(0,s) \\times \\\\\n & \\prod_{t=1}^T \\text{LogitNormal}(z_{t-1},1)\n \\prod_{n=1}^N \\text{Bernoulli}(p_t)\n p(m_{1,t}|t) p(m_{2,t}|t)\n p(y_{t,n}|w_{t,n},m_{1,t},m_{2,t})\n \\text{Poisson}(y_{t,n}) \\\\\\\\\n & =\\frac{1}{s\\sqrt{2\\pi}}e^{-\\frac{\\log(x)^2}{2}} \\cdot\n \\frac{1}{s\\sqrt{2\\pi}}e^{-\\frac{z_0^2}{2s^2}} \\cdot \\\\\n &\\prod_{t=1}^T \\frac{1}{\\sqrt{2\\pi}} e^{-\\frac{\\text{logit}(z_t-z_{t-1})^2}{2}} \\cdot\n \\prod_{n=1}^N p_t^{w_{t,n}} (1-p_t)^{1-w_{t,n}}\n p(m_{1,t}|t)^{w_{t,n}} p(m_{2,t}|t)^{1-w_{t,n}}\n \\frac{y_{t,n}^{x_{t,n}} e^{-y_{t,n}}}{x_{t,n}!}\n\\end{align*}\n\n\nPlate notation (next page):\n\\begin{figure}[!t]\n \\centering\n \\tikz{\n % \\draw[lightgray,ultra thin] (-6,-6) grid (6,6);\n % nodes\n \\node[obs] (x_t) {$x_{n,t}$};%\n \\node[latent, left=of x_t] (y_t) {$y_{n,t}$};\n \\node[latent, below=of y_t, xshift=-2cm] (model1) {model1$_t$};\n \\node[latent, below=of y_t] (model2) {model2$_t$};\n \\node[latent, left=of y_t, yshift=2cm] (switch) {switch$_{t,n}$};\n \\node[latent, left=of switch] (p) {$p_t$};\n \\node[latent, left=of p] (z) {$z_t$};\n \\node[latent, above=of z] (z0) {$z_0$};\n \\node[latent, above=of z0] (scale) {scale};\n \\node[latent, left=of scale] (log_scale) {log scale};\n % plate\n \\plate [inner sep=.3cm] {plate_switch} {(x_t)(y_t)(switch)} {N};\n \\plate [inner sep=.3cm] {episode} {(plate_switch)(model1)(model2)(p)(z)} {T};\n % edges\n \\edge {scale} {z0};\n \\edge {log_scale} {scale};\n \\edge {z} {p};\n \\edge {p} {switch};\n \\edge {switch, model1, model2} {y_t};\n \\edge {y_t} {x_t};\n \\draw[blue, thick, ->,shorten >=1pt] (z0) to [out=-90,in=90] node[left,yshift=.2cm,xshift=-.2cm] {$t=1$} (z);\n \\Cycle[blue, ->]{z}{180}{7mm}[{node[anchor=0,pos=0.5]{$t=2,...,T (t-1\\rightarrow t)$}}]{7mm}\n }\n \\caption{Plate notation}\n\\end{figure}\n\n\\newpage\n\\section{PPL via Pyro}\n\nThe source code is on GitHub:\n\\begin{enumerate}\n \\item \\href{https://github.com/liusida/ds2/blob/main/assignment3/code/q4.1.py}{q4.1.py}.\n \\item \\href{https://github.com/liusida/ds2/blob/main/assignment3/code/q4.2.py}{q4.2.py}.\n \\item \\href{https://github.com/liusida/ds2/blob/main/assignment3/code/q4.3.py}{q4.3.py}.\n\\end{enumerate}\n\n\\newpage\n\\section{Project Thoughts}\n\\subsection{Bayesian Learning Signal}\nI like Deep Learning.\n\nIn Supervised Learning, the learning signal comes from the difference between the truth label $y$ and the prediction $\\hat{y}$.\n\nTo analyze its assumption, I perform a derivation using MAP:\n\nSuppose we observe a image $\\cal{I}$, we want to know a vector representation of the image $h$ that can tell us $y=\\text{argmax}(h)$.\n\nAssuming the prior to be $h_0 \\sim \\textbf{Uniform}(a,b)$, here $a$ and $b$ are constants.\nThe observation $\\hat{h}$ is the representation produced by the network after seeing the image,\nAssuming $\\hat{h} \\sim \\textbf{Normal}(h, \\sigma)$, here we don't know $\\sigma$ yet.\nThis Normal distribution is saying the network can produce a prediction that is centered at $h$ but very noisy.\nThese two are quite strong assumptions.\n\nNow I can use MAP to compute the point estimate for $h_0$ given $\\hat{h}$:\n\\begin{align*}\n p(h_0|\\hat{h}) &\\propto p(\\hat{h}|h_0) p(h_0)\\\\\n &= \\textbf{Normal}(h,\\sigma) \\cdot \\textbf{Uniform}(a,b) \\\\\n &= \\frac{1}{\\sigma \\sqrt{2 \\pi}} e^{-\\frac{(h-\\hat{h})^2}{2\\sigma^2}} \\cdot 1\\\\\n\\end{align*}\n\nSolve $\\frac{d}{dh}p(h_0|\\hat{h})=0$ to find the maximum:\n\\begin{align*}\n \\frac{h_0-\\hat{h}}{\\sigma^3} e^{-\\cdots} &= 0\\\\\n h_0-\\hat{h} = 0\n\\end{align*}\n\nSo we want to update our belief from $h_0$ to $h_1$: $h_1 = \\hat{h}$.\n\nThis is how we find the update target, I think?\nAnd we apply L2 loss to it, it'll just like a normal supervised learning signal.\n\nNow if the agent has the chance to observe a very similar image $\\cal{I}'$, here we don't use the i.i.d. assumption,\nso a better prior should be $h_1$ after the update of the first step.\nThe problem here is that by using MAP, we only have a point estimator, \nwe don't have a distribution for $h_1$, how can I discover a good $h_1$?\n\nWhy I am doing this?\n\nImagine there is a robot with a camera on its face.\nIt will observe the world, frame by frame.\nI assume the current frame of the world is very much similar to the previous frame.\nIf I can derive a learning signal from this assumption,\nthe agent will be able to learn without human providing labels.\n\nOnce there is a good way to produce the representation $h$ of the image,\nwe can use whatever existing supervised learning or reinforcement learning to utilize this representation,\nand the learning will be faster, I hope.\n\nI read \\cite{wang_survey_2020}. It focus of this paper is to connect Bayesian Inference and Deep Learning at an application level, which is not my focus. \nBut I learned some basic concepts from it.\nThe so called Bayesian treatments might give the distribution of $h$.\n\n\\subsection{Thought two}\n\nWith a second thought, the previous idea turns out to be a Recurrent Neural Network.\n\nThe representation $h_t$ is just the hidden state.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=.8\\textwidth]{./rnn.pdf}\n \\caption{A RNN with Bayesian updated hidden states.}\n\\end{figure}\n\nThe purpose of the observation is to build a good $h_t$, \nso that if other modules want to use this representation,\nit is readily accessible.\n\nSo the job for the network module $N$ is to update $h_t$ properly using Bayesian Theorem.\n\nThis is different from the Bayesian Neural Networks (BNN).\nIn BNN, people tries to make the weights and biases to be r.v.'s.\n\nIn the Long Short Term Memory (LSTM) model, there is a cell state, which is similar to this $h$.\nMaybe replacing this gating and addition with a Bayesian update will give what I want.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=.8\\textwidth]{./LSTM3-C-line.png}\n \\caption{Cell States are updated by Gating and Addition.}\n\\end{figure}\n\nLater,\nI realized it is related to the World Model \\cite{ha_world_2018},\nwhere they use a variational autoencoder to get a latent $z$ (we call that $h$ earlier), and pass that into a RNN.\nHowever, the $h$ produced by the RNN in the World Model is not parameters of distributions, maybe I can improve this.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=.5\\textwidth]{./world_model.png}\n \\caption{In the World Model, the $z$ is produced by VAE, so $z$ is the parameters of some distributions.}\n\\end{figure}\n\n\n\n\\subsection{Automated Model Search?}\n\nAnother thought for the course project is from the idea of automated architecture search.\nThe architectures are analogous to the generative models in Bayesian Inference.\nAnd people use Evolutionary Algorithms to search for better architectures rather than design it manually \\cite{real_regularized_2019}.\n\nSo a typically case could be, the EA produces a population of DAGs, \nand BI get all the parameters, and we have a set of final performances,\nand EA produces a new population of DAGs by mutating the best DAGs,\nand BI get all the parameters again, and we get another set of the performances.\n\nBut I am still not familiar with the whole Bayesian Inference process, so I'm not sure if the idea of automated design can be applied to Bayesian Inference.\n\n\\bibliographystyle{apalike}\n\\bibliography{b}\n\n\\end{document}\n", "meta": {"hexsha": "e1268f36b8e9de9331221a98d47095edb15e2a3e", "size": 17412, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assignment3/main.tex", "max_stars_repo_name": "liusida/ds2", "max_stars_repo_head_hexsha": "1a4c6b3e0590d987c1e66d83bda1fb3382bf034e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assignment3/main.tex", "max_issues_repo_name": "liusida/ds2", "max_issues_repo_head_hexsha": "1a4c6b3e0590d987c1e66d83bda1fb3382bf034e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assignment3/main.tex", "max_forks_repo_name": "liusida/ds2", "max_forks_repo_head_hexsha": "1a4c6b3e0590d987c1e66d83bda1fb3382bf034e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.3085106383, "max_line_length": 182, "alphanum_fraction": 0.6049850678, "num_tokens": 5496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085808877581, "lm_q2_score": 0.7956581024858786, "lm_q1q2_score": 0.6248371353350318}} {"text": "\nThe scattering matrix, or S-matrix (also called the scattering function matrix, \\cite{tsang2000scattering}), embeds the scattering behavior of an object as a mapping between incident and scattered plane-waves. It transforms between the incident/scattered directions as well as incident/scattered polarizations. This chapter gives the basic definition of the S-matrix, its relation to radar cross sections, routines for the S-matrix of simple objects, and the derivation of an object S-matrix under the Born approximation. \n\n\\section{Definition}\n\nFrom \\cite{tsang2000scattering}, let an incident wave be a plane wave with wave vector $\\bb{k}_i$, the polarization of which is decomposed into two linearly independent vectors that are perpendicular to $\\hat{k}_i$, as\n\\begin{equation}\n\\bb{E}_i = \\left( E_{pi} \\hat{p}_i+ E_{qi}\\hat{q}_i\\right) e^{i\\bb{k}_i \\cdot \\br}\n\\end{equation}\n\n\\noindent where $\\bb{r}$ is the position vector, ${\\bb{k}}_i = k\\hat{{k}}_i$ and $\\hat{p}_i$, $\\hat{q}_i$ and $\\hat{{k}}_i$ form an orthonormal system. The scattered field is then defined \n\\begin{equation}\n\\bb{E}_s = \\left( E_{ps}\\hat{p}_s + E_{qs}\\hat{q}_s\\right) \\dfrac{ e^{i k r}}{r}\n\\end{equation}\n\n\\noindent where the scattered field polarization vectors $\\hat{p}_s$ and $\\hat{q}_s$ form a right-handed orthonormal system with the scattered field direction $\\hat{{k}}_s$. The scattered field components are linearly related to the incident components through the scattering matrix, or S-matrix, \n\\begin{equation}\n\\twobyone{E_{ps}(\\hat{k}_s)}{E_{qs}(\\hat{k}_s)} = \\twobytwo{S_{pp}(\\hat{k}_s,\\hat{k}_i) }{S_{pq}(\\hat{k}_s,\\hat{k}_i) }{S_{qp}(\\hat{k}_s,\\hat{k}_i) }{S_{qq}(\\hat{k}_s,\\hat{k}_i) } \\twobyone{E_{pi}(\\hat{k}_i)}{E_{qi}(\\hat{k}_i)} \n\\end{equation}\n \nThese definitions for the fields and S-matrix give them the following units: the incident field amplitudes $E_{pi}$ and $E_{qi}$ have units of electric field (V/m), the S-matrix elements have units of length (m), and the scattering field amplitudes $E_{ps}$ and $E_{qs}$ have units of (Vm/m = V) because the length dimension of the factor of $1/r$ needs to be included. This depends on convention, because some definitions of the S-matrix use $1/kr$ outside the S-matrix, which makes the S-matrix unitless. \n\nWe use the wave vector and polarization convention of \\cite{tsang2000scattering}. The polarizations are taken as $\\hat{p} = \\hat{v}$ and $\\hat{q} = \\hat{h}$ with wave vector directions\n\\ea{\\hat{k}_i &=& \\sin\\theta_i \\cos\\phi_i \\hat{x} + \\sin\\theta_i \\sin\\phi_i \\hat{y} + \\cos\\theta_i \\hat{z} \\label{khati} \\\\\n\\hat{v}_i &=& \\cos\\theta_i \\cos\\phi_i \\hat{x} + \\cos\\theta_i\\sin\\phi_i \\hat{y} - \\sin\\theta_i\\hat{z} \\\\\n\\hat{h}_i &=& -\\sin\\phi_i\\hat{x} + \\cos\\phi_i \\hat{y} \\\\\n\\hat{k}_s &=& \\sin\\theta_s \\cos\\phi_s \\hat{x} + \\sin\\theta_s \\sin\\phi_s \\hat{y} + \\cos\\theta_s \\hat{z} \\label{ss1} \\\\\n\\hat{v}_s &=& \\cos\\theta_s \\cos\\phi_s \\hat{x} + \\cos\\theta_s\\sin\\phi_s \\hat{y} - \\sin\\theta_s\\hat{z} \\\\\n\\hat{h}_s &=& -\\sin\\phi_s\\hat{x} + \\cos\\phi_s \\hat{y} \\label{hhats}}\n\n\\noindent where $\\theta$, $\\phi$ are spherical directions of the propagation waves. Here, the wave vectors are treated as radial vectors, and $\\hat{v}$ and $\\hat{h}$ are equivalent to the $\\hat{\\theta}$ and $\\hat{\\phi}$ spherical unit vectors. The polarizations for $\\hat{z}$ propagating waves ($\\theta = 0$ or $\\theta = \\pi$) are well-defined by using $\\phi=0$ as a reference\n \\ea{\\hat{k}(0,0) &=& \\hat{z} \\\\\n\\hat{v}(0,0) &=& \\hat{x} \\\\\n\\hat{h}(0,0) &=& \\hat{y} \\\\\n\\hat{k}(\\pi,0) &=& -\\hat{z} \\\\\n\\hat{v}(\\pi,0) &=& -\\hat{x} \\\\\n\\hat{h}(\\pi,0) &=& - \\hat{y} }\n\nThese special cases can be used as definitions of the $\\hat{\\theta}$ and $\\hat{\\phi}$ unit vectors at the poles.\n\n\\section{Radar Cross Sections from S-matrix}\n\nWe list different radar cross sections and their relations to the S-matrix. Most of these can be found in \\cite{tsang2000scattering}.\n\n\\paragraph{Radar Cross Section} The bistatic radar cross section is defined \n\\ea{\\sigma_{pq}(\\hat{k}_s,\\hat{k}_i) &=& \\lim_{r \\rightarrow \\infty} 4\\pi r^2 \\dfrac{\\left\\vert \\hat{p} \\cdot \\bb{E}_s \\right\\vert^2}{\\left\\vert \\hat{q} \\cdot \\bb{E}_i \\right\\vert^2} \\\\\n\\ &=& 4\\pi \\vert S_{pq} \\vert^2 \\label{rcsfromSpq} }\n\n\\noindent where $p$ and $q$ are any two orthogonal polarizations. The units of radar cross section are area or length-squared, showing again that $S_{pq}$ has units of length.\n\n\\paragraph{Scattering Cross Section} The total scattering cross section for incident direction $\\hat{k}_i$ and arbitrary incident polarization $\\hat{\\beta}$, such that $\\hat{\\beta} \\cdot \\hat{k}_i = 0$, is computed as the integral over scattered field directions $\\hat{k}_s$ as\n\\eq{\\sigma_{s\\beta}(\\hat{k}_i) = \\int \\left( \\vert S_{p\\beta}(\\hat{k}_s,\\hat{k}_i) \\vert^2 + \\vert S_{q\\beta}(\\hat{k}_s,\\hat{k}_i) \\vert^2 \\right) d\\Omega_s }\n\n\\noindent where\n\\eq{\n\\twobyone{S_{p\\beta} }{S_{q\\beta} } = \n\\twobytwo{S_{pp}}{S_{pq}}{S_{qp}}{S_{qq}} \\twobyone{\\hat{p}\\cdot \\hat{\\beta}}{\\hat{q}\\cdot \\hat{\\beta}} \\label{projectedsmatrix} \n}\n\n\\noindent and $\\beta$ is the angle of the incident polarization relative to the two orthogonal polarizations of the S-matrix, which is \n\\eq{\\hat{\\beta} = \\cos\\beta \\hat{p} + \\sin\\beta \\hat{q}}\n\n\\paragraph{Extinction Cross Section} The extinction cross section (or total cross section) for a given incident direction is given by the optical theorem, \\cite{zhang2019generalized}, as the imaginary part of the co-polarized scattered field in the forward direction\n\\eq{\\sigma_{ext,\\beta}(\\hat{k}_i) = \\dfrac{4\\pi}{k}\\textrm{Im}\\left[S_{\\beta\\beta}(\\hat{k}_i,\\hat{k}_i)\\right]}\n\n\\paragraph{Absorption Cross Section} The absorption cross section is defined as the power absorbed by the scattering object divided by the power scattered. It can be written in terms of the extinction and total cross sections as \n\\eq{\\sigma_a = \\sigma_{ext} - \\sigma_s}\n\nThis subtraction is sometimes numerically inaccurate and so the absorption cross section can also be defined directly in terms of the internal field, \\cite{yurkin2007discrete}.\n\n\\paragraph{Polarization and Orientation Averaged Scattering Cross Section}\n\nThe polarization and orientation averaged scattered cross section is given by the integral of the scattering cross section over all possible incident directions and polarizations including normalization:\n%\\eq{\\left< \\sigma \\right> = \\left< \\sigma_{s\\beta}(\\hat{k}_i) \\right>_{\\beta,\\hat{k}_i} = \\dfrac{1}{2\\pi} \\int_0^{2\\pi}\\left( \\dfrac{1}{4\\pi} \\int \\sigma_{s\\beta}(\\hat{k}_i) d\\Omega_i \\right) d\\beta }\n\\eq{\\left< \\sigma \\right> = \\dfrac{1}{2\\pi} \\int_0^{2\\pi}\\left( \\dfrac{1}{4\\pi} \\int \\sigma_{s\\beta}(\\hat{k}_i) d\\Omega_i \\right) d\\beta \\label{polorientavescs} }\n\nThis requires projecting a given incident polarization $\\beta$ onto $\\hat{p}$ and $\\hat{q}$, computing the projected S-matrix elements \\eqref{projectedsmatrix}, computing the scattering cross section for all incident directions, then integrating over incident direction and incident angle $\\beta$. Computing this does not require one to explicitly rotate or recompute the S-matrix once it is known.\n\n\n\\paragraph{Scattering Efficiencies} The quantities above can be reinterpreted as efficiencies by dividing by the cross sectional area of the target. For example, the scattering efficiency for a given incident direction is given by the scattering cross section divided by cross sectional area of the target, $A$, \n\n\\eq{Q_{sca} = \\dfrac{\\sigma_{s}}{A} }\n\nThe same can be applied to extinction and absorption cross sections. \n\n\\section{S-matrix Rotation}\n\nAn S-matrix will often be computed in a reference frame aligned naturally with the geometry of the object. If an object is rotated relative to a global frame, then the incident and scattered directions and polarization in the global frame will appear to rotate in the frame of the object. It can be efficient to transform the global wave vectors and polarizations such that the S-matrix is evaluated in the frame of the object, rather than try to obtain the S-matrix of the rotated object in the global frame. Both \\cite{tsang2000scattering} and \\cite{van2011synthetic} give equations and procedures for rotating an object with cylindrical symmetry. We give a general procedure next.\n\n%For the purposes of rotation, both the wave vectors and polarization vectors can be treated as Cartesian unit vectors with their tail at the origin. An extrinsic rotation is first applied to the incident and scattered directions to \n%\n%The polarization vectors are then decomposed into the local $\\hat{v}$ and $\\hat{h}$ polarizations of the object frame. Then the S-matrix is evaluated in its frame. In this procedure, there is no need to 'rotate back', because it is sufficient to have transformed both the incident and scattered quantities to the object frame. \n\nLet an object rotation be described by ZXZ Euler angles $(\\alpha,\\beta,\\gamma)$ with corresponding rotation matrix $\\bb{R}$. $\\bb{R}$ describes a forward rotation, i.e., a point rides with the rotated frame. The inverse rotation is $\\bb{R}^t$, where points are fixed in the global frame and we view them as though we ride with the rotating frame. The inverse rotation is first applied to the incident and scattered directions of the global frame so that they are viewed from the rotated object frame. Let these angles be $(\\theta_{os},\\phi_{os};\\theta_{oi},\\phi_{oi})$. The polarization vectors are then evaluated in the object frame at these points giving $(\\hat{v}_{os},\\hat{h}_{os};\\hat{v}_{oi},\\hat{h}_{oi})$. The object-frame polarization vectors now need to be viewed from the global frame. This is done by rotating the object-frame polarization vectors with an forward rotation, so that they appear as vectors in the global frame:\n\\eq{\\hat{v}_{o}' = \\bb{R}(\\hat{v}_o)}\n\nIn matrix notation, S-matrix of the rotated object, as viewed in the global frame, is given by \n\\ea{\n\\twobytwo{S_{vv}(\\theta_s,\\phi_s;\\theta_i,\\phi_i)}{S_{vh}(\\theta_s,\\phi_s;\\theta_i,\\phi_i)}{S_{hv}(\\theta_s,\\phi_s;\\theta_i,\\phi_i)}{S_{hh}(\\theta_s,\\phi_s;\\theta_i,\\phi_i)}\n&=&\n\\twobytwo{\\hat{v}_s \\cdot \\hat{v}_{os}' }{\\hat{v}_s \\cdot \\hat{h}_{os}'}{\\hat{h}_s \\cdot \\hat{v}_{os}'}{\\hat{h}_s \\cdot \\hat{h}_{os}'} \\cdot \\nonumber \\\\\n\\ & \\ & \n\\twobytwo{S_{vv}(\\theta_{os},\\phi_{os};\\theta_{oi},\\phi_{oi})}{S_{vh}(\\theta_{os},\\phi_{os};\\theta_{oi},\\phi_{oi})}{S_{hv}(\\theta_{os},\\phi_{os};\\theta_{oi},\\phi_{oi})}{S_{hh}(\\theta_{os},\\phi_{os};\\theta_{oi},\\phi_{oi})} \\cdot \\nonumber \\\\\n\\ & \\ & \n\\twobytwo{\\hat{v}_{oi}' \\cdot \\hat{v}_{i} }{\\hat{v}_{oi}' \\cdot \\hat{h}_{i} }{\\hat{h}_{oi}' \\cdot \\hat{v}_{i} }{\\hat{h}_{oi}' \\cdot \\hat{h}_{i} }\n} \n \nThe inner matrix is the object S-matrix evaluated in its native frame using points rotated from the global frame. The outer two matrices decompose and project the polarization vectors between the two frames. \n\n\n%\\subsection{Euler Rotation}\n\n%\\subsection{Cylindrical Symmetry}\n\n\\section{Object S-matrix}\n\n\\subsection{Thin Circular Cylinder}\n\nFrom \\cite{sarabandi1990low, stiles1996scattering} the far field scattering solution for a thin dielectric cylinder with cross sectional area much smaller than a wavelength and oriented along the $z$ axis is given by \n\\eq{\\bb{E}_s = -E_o\\dfrac{e^{ikr}}{4\\pi r} k^2 L A \\left[ \\hat{k}_s \\times \\hat{k}_s \\times (\\bb{P} \\cdot \\hat{e}_i ) \\right] \\sinc{(U)}}\n\\eq{U = \\dfrac{kL}{2} \\left(\\cos\\theta_s - \\cos\\theta_i\\right) }\n\n\\noindent where $L$ is the length, $A$ is the cross sectional area, $\\hat{e}_i$ is the incident polarization, $k$ is the background wavenumber, $\\theta_i$ and $\\theta_s$ are the incident and scattered angles from the $z$ axis, and $\\bb{P}$ is the polarization tensor. Also, $\\textrm{sinc}(x) = \\sin(x)/x$. The polarization tensor for a homogenous dielectric cylinder with circular cross section is diagonal and given by \n\\ea{P_{xx} &=& 2 \\dfrac{\\epsilon_r - 1}{\\epsilon_r + 1} \\\\\nP_{yy} &=& 2 \\dfrac{\\epsilon_r - 1}{\\epsilon_r + 1}\\\\\nP_{zz} &=& \\epsilon_r - 1}\n\n\\noindent where $\\epsilon_r$ is the relative permittivity. For a cylinder with radius $a$, the $\\hat{v}$ and $\\hat{h}$ S-matrix elements are \n\\ea{\\twobytwo{S_{vv}}{S_{vh}}{S_{hv}}{S_{hh}} &=& \\dfrac{k^2 L a^2 }{4} \\sinc{(U)} \\bb{C} }\n\n%\\ea{\\twobytwo{S_{vv}}{S_{vh}}{S_{hv}}{S_{hh}} &=& -\\dfrac{k^2 L a^2 }{4} \\sinc{(U)} \\onebytwo{\\hat{v}_s}{\\hat{h}_s} \\cdot \\twobyone{ \\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{v}_i \\right)}{ \\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{h}_i \\right) } \\\\\n%\\ &=& \\dfrac{k^2 L a^2 }{4} \\sinc{(U)} \\bb{C} }\n\n\\eq{\\bb{C} = \\twobytwo\n{(\\hat{v}_s)\\cdot(\\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{v}_i \\right))}\n{(\\hat{v}_s)\\cdot\\left(\\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{h}_i \\right) \\right)}\n{(\\hat{h}_s)\\cdot(\\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{v}_i \\right))}\n{(\\hat{h}_s)\\cdot\\left(\\hat{k}_s \\times \\hat{k}_s \\times \\left(\\bb{P} \\cdot \\hat{h}_i \\right) \\right)} \\label{matrixCforS}}\n\nUsing \\eqref{khati}-\\eqref{hhats} and simplifying the elements of $\\bb{C}$ are\n\\ea{C_{vv} &=& P_{zz}\\sin\\theta_s\\sin\\theta_i + P_{xx}\\cos\\theta_s\\cos\\phi_s\\cos\\theta_i\\cos\\phi_i + P_{yy}\\cos\\theta_s\\cos\\theta_i\\sin\\phi_s\\sin\\phi_i \\\\\nC_{vh} &=& \\cos\\theta_s(P_{yy}\\sin\\phi_s\\cos\\phi_i - P_{xx}\\cos\\phi_s\\sin\\phi_i ) \\\\\nC_{hv} &=& \\cos\\theta_i(P_{yy}\\cos\\phi_s\\sin\\phi_i - P_{xx}\\sin\\phi_s\\cos\\phi_i) \\\\\nC_{hh} &=& P_{yy}\\cos\\phi_s\\cos\\phi_i + P_{xx}\\sin\\phi_s\\sin\\phi_i}\n\nApplying symmetry $P_{xx} = P_{yy}$ and taking $\\phi_i = 0$, these can be simplified to just\n\\ea{C_{vv} &=& P_{zz}\\sin\\theta_s\\sin\\theta_i + P_{xx}\\cos\\theta_s\\cos\\phi_s\\cos\\theta_i \\\\\nC_{vh} &=& P_{xx} \\cos\\theta_s\\sin\\phi_s\\\\\nC_{hv} &=& -P_{xx} \\cos\\theta_i \\sin\\phi_s\\\\\nC_{hh} &=& P_{xx}\\cos\\phi_s}\n\nThis solution is valid for edge-on incidence, and the $hh$ and $vv$ fields become equal, which we expect. This solution is valid for $\\vert n \\vert ka \\ll 1$ and $a \\ll L$, where $n = \\sqrt{\\epsilon_r}$, \\cite{schiffer1979light}.\n\nThe routine \\texttt{smatrix\\char`_thin\\char`_circular\\char`_cylinder} returns the S-matrix of a thin circular cylinder. It takes the parameters of a single cylinder and computes the four block S-matrices for any number of incident and scattered directions. The block matrices have scattered directions in rows and incident directions in columns. The polarizations $\\hat{v}$ and $\\hat{h}$ are treated as spherical $\\hat{\\theta}$ and $\\hat{\\phi}$ in the forward scattering convention.\n\n{\\footnotesize\n\\VerbatimInput{\\code/Smatrix/smatrix_thin_circular_cylinder.m}\n}\n\n\\clearpage\n\\newpage\n\n\\subsection{Thin Circular Disk}\n\nScattering from both needles and disks is derived in \\cite{schiffer1979light} and the results for needles match those in the previous section. The far field scattered field for a thin disk is\n\\ea{\\bb{E}_s &=& -E_o\\dfrac{e^{ikr}}{2\\pi r} k^2 L A \\left[ \\hat{k}_s \\times \\hat{k}_s \\times (\\bb{P} \\cdot \\hat{e}_i ) \\right] \\textrm{jinc}(U)\\\\\nU &=& kL \\Omega \\\\\n\\Omega^2 &=& (\\sin\\theta_i \\cos\\phi_i- \\sin\\theta_s\\cos\\phi_s)^2 + (\\sin\\theta_i \\sin\\phi_i- \\sin\\theta_s\\sin\\phi_s )^2 }\n\n\\noindent where $L$ is the disk thickness, $A$ is cross sectional area of the disk, and $\\textrm{jinc}(x) = J_1(x)/x$ where $J_1(x)$ is the Bessel function of degree 1. The polarization tensor of the circular disk is \n\\ea{P_{xx} &=& \\epsilon_r - 1 \\\\\nP_{yy} &=& \\epsilon_r - 1 \\\\\nP_{zz} &=& \\dfrac{\\epsilon_r - 1}{\\epsilon_r }}\n \n%(In comparing \\cite{schiffer1979light} to \\cite{sarabandi1990low, stiles1996scattering}, the the factor of 2 between the cylinder and disk scattered field appears correct.)% For the disk, we take the small argument limit of the Bessel function (small radius) which kicks out a factor of 2, while for the cylinder, we take the small argument limit of the sinc (small length) which is equal to 1. The factors in the polarization tensor are otherwise the same (going from \\cite{schiffer1979light} to \\cite{sarabandi1990low, stiles1996scattering} the factor of $(\\epsilon_r-1)$ is brought into $\\bb{P}$ and we swap the $x$ and $z$ tensor components). \n \nThen the S-matrix elements are \n\\eq{\\twobytwo{S_{vv}}{S_{vh}}{S_{hv}}{S_{hh}} = \\dfrac{k^2 L a^2 }{2} \\dfrac{J_1\\left(U\\right)}{ U}\\bb{C} }\n\n\\noindent where $\\bb{C}$ is given by \\eqref{matrixCforS}. In comparing \\cite{schiffer1979light} to \\cite{sarabandi1990low, stiles1996scattering}, the factor of 2 appears correct between the cylinder and disk scattered field. This solution is valid for $\\vert n \\vert kL \\ll 1$ and $L \\ll a$, where $n = \\sqrt{\\epsilon_r}$, \\cite{schiffer1979light}.\n\nThe routine \\texttt{smatrix\\char`_thin\\char`_circular\\char`_disk} returns the S-matrix of a thin circular disk. It takes the parameters of a single disk and computes the four block S-matrices for any number of incident and scattered directions. The block matrices have scattered directions in rows and incident directions in columns. The polarizations $\\hat{v}$ and $\\hat{h}$ are treated as spherical $\\hat{\\theta}$ and $\\hat{\\phi}$ in the forward scattering convention.\n\n{\\footnotesize\n\\VerbatimInput{\\code/Smatrix/smatrix_thin_circular_disk.m}\n}\n\n\n\\section{S-matrix Under the Born Approximation}\n\\label{smatrixborn}\n\nThe far-field Born approximation is explained in Section \\ref{bornapprox}. Because the approximation is written in terms of plane waves, it can be cast as an S-matrix. Recalling \\eqref{baesca}, using $\\overline{\\bb{I}} - \\hat{r}\\hat{r} = \\hat{\\theta}_s\\hat{\\theta}_s + \\hat{\\phi}_s\\hat{\\phi}_s$, and equating the spherical unit vectors with $\\hat{v}$ and $\\hat{h}$, we obtain the S-matrix \n\\eq{\\twobytwo{S_{vv}}{S_{vh}}{S_{hv}}{S_{hh}} = \\dfrac{1}{4\\pi} \\twobytwo{\\hat{v}_{s} \\cdot \\hat{v}_{i} }{\\hat{v}_{s} \\cdot \\hat{h}_{i} }{\\hat{h}_{s} \\cdot \\hat{v}_{i} }{\\hat{h}_{s} \\cdot \\hat{h}_{i} } \\int O(\\br) e^{i (\\bb{k}_i-\\bb{k}_s) \\cdot \\br} dV \\label{eqbasmat1} }\n\n%\\eq{\\bb{E}_{sca}(\\br)= \\dfrac{e^{ikr}}{4\\pi r} \\left[ \\hat{\\theta}_s\\hat{\\theta}_s + \\hat{\\phi}_s\\hat{\\phi}_s \\right] \\cdot \\bb{E}_i \\int O(\\br') \\exp(i (\\bb{k}_i-\\bb{k}_s) \\cdot \\br') dV' }\n\n\\noindent where the object function is given by \\eqref{objectfunction} and the wave vectors and polarizations are given by \\eqref{khati}-\\eqref{hhats}. This shows again that under the Born approximation, the object does not influence the polarization. The depolarization is merely the projection between incident and scattered polarizations. \n\nWhen the object is homogenous \\eqref{eqbasmat1} becomes \n\\eq{\\twobytwo{S_{vv}}{S_{vh}}{S_{hv}}{S_{hh}} = \\dfrac{1 }{4\\pi}\\twobytwo{\\hat{v}_{s} \\cdot \\hat{v}_{i} }{\\hat{v}_{s} \\cdot \\hat{h}_{i} }{\\hat{h}_{s} \\cdot \\hat{v}_{i} }{\\hat{h}_{s} \\cdot \\hat{h}_{i} } k^2(\\epsilon_r - 1) I(\\theta_{s},\\phi_{s};\\theta_{i},\\phi_{i}) }\n\\eq{I(\\theta_{s},\\phi_{s};\\theta_{i},\\phi_{i}) = \\int e^{i (\\bb{k}_i-\\bb{k}_s) \\cdot \\br} dV \\label{volphaseint}}\n\n\\noindent where $k$ is the wavenumber of the background medium which is also used for the wave vectors, $\\epsilon_r$ is complex relative permittivity of the object, and $I$ is the volume phase integral. The volume phase integral is the 3D Fourier transform over the domain of the object. \n\n\\clearpage\n\n\\section{Volume Phase Integral}\n\\label{sec:volumephaseint}\nThe volume phase integral occurs in the S-matrix formulation of the Born approximation for homogeneous dielectric objects, Section \\ref{smatrixborn}. The volume phase integral for several simple objects is derived below and summarized in Table \\ref{tablevolphaseint}. These are just the 3D Fourier transform over the object domain. The volume phase integral can be expressed as a product of the object volume and a function that has maximum value of one that depends on the incident/scattered directions. To facilitate derivations, \\eqref{volphaseint} is written in terms of the wave vector difference, $\\bb{K}$, as \n\\eq{I(\\theta_{s},\\phi_{s};\\theta_{i},\\phi_{i}) = \\int e^{i \\bb{K} \\cdot \\br} dV \\label{volphaseint2}}\n\\noindent where\n\\ea{\\bb{K} &=& \\bb{k}_i-\\bb{k}_s \\label{Kvector} = k(\\hat{k}_i-\\hat{k}_s) \\\\\n \\hat{k}_i &=& \\sin\\theta_i \\cos\\phi_i \\hat{x} + \\sin\\theta_i \\sin\\phi_i \\hat{y} + \\cos\\theta_i \\hat{z} \\\\\n\\hat{k}_r &=& \\sin\\theta_r \\cos\\phi_r \\hat{x} + \\sin\\theta_r \\sin\\phi_r \\hat{y} + \\cos\\theta_r \\hat{z} \\\\\n\\bb{r} &=& x \\hat{x} + y \\hat{y} + z \\hat{z} }\n\\vspace{-7mm}\n\n\n\\begin{table}[H]\n\\caption{Table of Volume Phase Integrals}\n\\vspace{-6mm}\n\\begin{center}\n\\begin{tabular}{|p{1.4cm}|p{3cm}|c|c|p{2.7cm}|}\n\\hhline{|=====|} \n%\\multicolumn{5}{|c|}{ \\ } \\\\\n\\multicolumn{5}{|c|}{ $\\displaystyle I(\\theta_{s},\\phi_{s};\\theta_{i},\\phi_{i}) = \\int e^{i \\bb{K} \\cdot \\br} dV = V f(\\cdot) $} \\\\\n\\multicolumn{5}{|c|}{$\\displaystyle \\bb{K} = \\bb{k}_i-\\bb{k}_s$} \\\\\n%\\multicolumn{5}{|c|}{ \\ } \\\\\n\\hhline{|=====|}\nObject & Geometry & $V$ & $f(\\cdot)$ & Notes \\\\\n\\hline\n\nCuboid & \\parbox[c]{1em}{\\includegraphics[width=1.3in]{Smatrix/Figures/Cuboid}} & $L_xL_yL_z$ & $ \\textrm{sinc}\\left(\\dfrac{L_x K_x}{2}\\right)\\textrm{sinc}\\left(\\dfrac{L_y K_y}{2}\\right) \\textrm{sinc}\\left(\\dfrac{L_z K_z}{2}\\right)$ & $\\textrm{sinc}(x) = \\dfrac{\\sin(x)}{x}$ \\\\ \\hline\n\nCircular Cylinder & \\quad \\parbox[c]{1em}{\\includegraphics[width=0.9in]{Smatrix/Figures/Cylinder}} & $\\pi a^2 L_z$ & $ 2 \\dfrac{ J_1(K_{\\rho} a)}{ K_{\\rho} a} \\textrm{sinc}\\left(\\dfrac{L_z K_z}{2}\\right)$ & $K_{\\rho} = \\sqrt{K_x^2 + K_y^2} $ \\\\ \\hline\n\nSphere & \\parbox[c]{1em}{\\includegraphics[width=1.2in]{Smatrix/Figures/Sphere}} & $\\dfrac{4}{3}\\pi a^3$ & $ \\dfrac{3(\\sin(K a) - K a \\cos(K a))}{(Ka)^3} $ & $K = \\vert \\bb{K} \\vert $ \\\\ \\hline\n\nEllipsoid & \\parbox[c]{1em}{\\includegraphics[width=1.3in]{Smatrix/Figures/Ellipsoid}} & $\\dfrac{4}{3}\\pi abc$ & $ \\dfrac{3(\\sin(K') - K' \\cos(K'))}{K'^3} $ & $\\begin{array}{c} K' = \\vert \\bb{K'} \\vert \\\\ (K_x',K_y',K_z') = \\\\ (a K_x, b K_y, c K_z)\\end{array}$ \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\\label{tablevolphaseint}\n\\end{table}%\n\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{1}}\n\\subsection{Cuboid}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\nLet a cuboid volume be centered at the origin with side lengths $L_x$, $L_y$, $L_z$ and let it be aligned with the Cartesian axes. In Cartesian coordinates, the volume phase integral, \\eqref{volphaseint2}, becomes\n\\ea{\nI &=& \\int_{-L_x/2}^{L_x/2} \\int_{-L_y/2}^{L_y/2} \\int_{-L_z/2}^{L_z/2} e^{i(K_x x + K_y y + K_z z)} dx dy dz }\n%\\ &=& \\int_{-L_x/2}^{L_x/2} e^{i K_x x} dx \\int_{-L_y/2}^{L_y/2} e^{i K_y y} dy \\int_{-L_z/2}^{L_z/2} e^{i K_z z} dz \\\\\n\nThe integral separates, then using the fact that \n\\eq{\\int_{-a/2}^{a/2} e^{i b z} dz = \\dfrac{2}{b} \\sin\\left(\\dfrac{ab}{2}\\right) \\label{sincinz}}\n\n\\noindent and multiplying top and bottom by $L_xL_yL_z$ to convert the sines to sinc functions the volume phase intergral over a cuboid is \n%\\ea{I &=& \\dfrac{4}{K_x K_y} \\sin\\left(\\dfrac{L_x K_x}{2}\\right)\\sin\\left(\\dfrac{L_y K_y}{2}\\right) \\\\\n%\\ & = & L_x L_y \\textrm{sinc}\\left(\\dfrac{L_x K_x}{2}\\right)\\textrm{sinc}\\left(\\dfrac{L_y K_y}{2}\\right) }\n\n\\eq{I = L_x L_y L_z \\textrm{sinc}\\left(\\dfrac{L_x K_x}{2}\\right)\\textrm{sinc}\\left(\\dfrac{L_y K_y}{2}\\right) \\textrm{sinc}\\left(\\dfrac{L_z K_z}{2}\\right)}\n\n\\noindent where $\\textrm{sinc} = \\sin(x)/x$. This is equal to the volume of the cuboid times a product of directionally dependent $\\textrm{sinc}$ functions.\n\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{1}}\n\\subsection{Circular Cylinder}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\nLet a circular cylinder be centered at the origin and aligned with the $z$ axis with radius $a$ and length $L_z$. The volume phase integral, \\eqref{volphaseint2}, in cylindrical coordinates is written \n\\ea{I &=& \\int_{-L_z/2}^{L_z/2} \\int_0^{2\\pi} \\int_0^{a} e^{i (K_x \\rho \\cos\\phi + K_y \\rho \\sin\\phi + K_z z)} \\rho d\\rho d\\phi dz }\n\n\\noindent where we have used the position vector $\\br = \\rho \\cos\\phi \\hat{x} + \\rho \\sin\\phi \\hat{y} + \\hat{z}$. The integral separates as \n\\ea{I &=& \\int_0^{2\\pi} \\int_0^{a} e^{i (K_x \\rho \\cos\\phi + K_y \\rho \\sin\\phi)} \\rho d\\rho d\\phi \\int_{-L_z/2}^{L_z/2} e^{i K_z z} dz }\n\nThe last integral is a sinc function in $z$ given by \\eqref{sincinz}. Using the identity \n\\eq{\\int_0^{2\\pi} e^{ u \\cos t +v \\sin t} d t = 2 \\pi I_0 \\left( \\sqrt{u^2 + v^2} \\right)}\n\n\\noindent the integral over $\\phi$ evaluates to \n\\eq{\\int_0^{2\\pi} e^{i K_x \\rho \\cos\\phi + K_y \\rho \\sin\\phi} d\\phi = 2\\pi I_0 \\left( i K_{\\rho} \\rho \\right)}\n\n\\noindent where $K_{\\rho} = \\sqrt{K_x^2 + K_y^2}$. The quantity $K_{\\rho}$ is the component of the wave vector difference in the X-Y plane. Last, the integral in $\\rho$ is evaluated using\n\\eq{\\int_0^a I_0(i c x) x dx = \\dfrac{a J_1(a c)}{c}}\n\nPutting these together, the volume phase integral over a circular cylinder is\n\\ea{I &=& 2\\pi \\dfrac{a J_1(K_{\\rho} a)}{ K_{\\rho} } L_z \\textrm{sinc}\\left(\\dfrac{L_z K_z}{2}\\right) }\n\nUsing the volume of the cylinder, $V = \\pi a^2 L_z$, this can be written \n\\ea{I &=& V 2 \\dfrac{ J_1(K_{\\rho} a)}{ K_{\\rho} a} \\textrm{sinc}\\left(\\dfrac{L_z K_z}{2}\\right) }\n\nSimilar to the other shapes, this is equal to the volume of the cylinder multiplied by a directionally dependent term that has maximum value of 1. \n\n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{1}}\n\\subsection{Sphere}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\n\\label{sectionvolintsphere}\nLet a spherical volume with radius $a$ be centered at the origin. The volume phase integral, \\eqref{volphaseint2}, can be written in spherical coordinates as \n\\vspace{-2mm}\n\\eq{ I = \\int_0^{2\\pi} \\int_0^{\\pi} \\int_0^a e^{i \\bb{K} \\cdot \\br} r^2 \\sin\\theta dr d\\theta d\\phi }\n\nFrom symmetry, the dot product is evaluated relative to a common fixed axis such that $\\bb{K} \\cdot \\br = K r \\cos t$ where $t$ is the angle between $\\bb{K}$ and $\\br$. Using this, the integral becomes\n\\eq{ I = \\int_0^{2\\pi} \\int_0^{\\pi} \\int_0^a e^{i K r \\cos t} r^2 \\sin t dr dt d\\phi \\label{eqtemp21}}\n\\vspace{-2mm}\n\nNext we use the identity \n\\eq{j_n(z) = \\dfrac{(-i)^n}{2} \\int_0^{\\pi} e^{i z \\cos u} P_n(\\cos u) \\sin u du}\n\n\\noindent where $j_n(z)$ is the spherical Bessel function and $P_n(\\cos u)$ is the Legendre polynomial. Applying this and integrating in $\\phi$ \\eqref{eqtemp21} becomes\n\\eq{ I = 4\\pi \\int_0^a j_0(K r) r^2 dr \\label{eqtmp22}}\n\nNote that $j_0(x) = \\sin(x)/x $. This integral is given generally as\n\\eq{\\int_0^a j_0 (c x) x^2 dx = \\dfrac{\\sin(a c) - a c \\cos(a c)}{c^3}}\n\nThe volume phase integral over a sphere is given by\n\\eq{ I = 4\\pi \\dfrac{\\sin(K a) - K a \\cos(K a)}{K^3} \\label{volphasphere1} }\n\n\\noindent where $K$ is the magnitude of \\eqref{Kvector}. Using the volume of the sphere, $V = 4/3 \\pi a^3$, this can be written\n\\ea{ I &=& V \\dfrac{3(\\sin(K a) - K a \\cos(K a))}{(Ka)^3}}\n\n%\n%\n%The volume phase integral over a sphere is given by\n%\\eq{ I = 4\\pi \\dfrac{\\sin(K a) - K a \\cos(K a)}{K^3} \\label{volphasphere1} }\n%\n%\\noindent where $K$ is the magnitude of \\eqref{Kvector}. This can be written in terms of the volume of the sphere $V = 4/3 \\pi a^3$ as \n%\\ea{ I &=& V \\dfrac{3(\\sin(K a) - K a \\cos(K a))}{(Ka)^3}}\n\nThe multiplying function acts like a directionally dependent weighting function that has a maximum value of 1 when $K a \\rightarrow 0$. \n\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{1}}\n\\subsection{Ellipsoid}\n\\addtocontents{toc}{\\protect\\setcounter{tocdepth}{2}}\n\nLet an ellipsoid be centered on the origin with axes $(a, b, c)$ along XYZ, respectively. We define a new position vector, $\\br'$, in spherical coordinates using the change of variables \n\\ea{\\br' &=& r a \\sin\\theta\\cos\\phi \\hat{x} + r b \\sin\\theta\\sin\\phi \\hat{y} + r c\\cos\\theta \\hat{z} \\\\\ndV &=& abc r^2 \\sin\\theta dr d\\theta d\\phi}\n\nThe volume phase integral, \\eqref{volphaseint2}, can then be written as an integral over the unit sphere\n\\eq{ I = abc \\int_0^{2\\pi} \\int_0^{\\pi} \\int_0^1 e^{i \\bb{K} \\cdot \\br'} r^2 \\sin\\theta dr d\\theta d\\phi }\n\nNext, transfer the coefficients $(a,b,c)$ that are in $\\br'$ to a new wave vector difference, $\\bb{K}'$, so that, $(K_x',K_y',K_z') = (a K_x, b K_y, c K_z)$. The position vector returns to its unstretched form and we have\n\\eq{ I = abc \\int_0^{2\\pi} \\int_0^{\\pi} \\int_0^1 e^{i \\bb{K}' \\cdot \\br} r^2 \\sin\\theta dr d\\theta d\\phi }\n\nUsing the results of Section \\ref{sectionvolintsphere}, \\eqref{volphasphere1}, the volume phase integral over an ellipsoid is \n\\eq{ I = 4\\pi abc \\dfrac{\\sin(K') - K' \\cos(K')}{K'^3} }\n\n\\noindent where $K'$ is the magnitude of $\\bb{K}'$. The volume of an ellipsoid is $V = 4/3\\pi abc$, and this can be written\n\\eq{ I = V \\dfrac{3(\\sin(K') - K' \\cos(K'))}{K'^3} }\n\n\\clearpage\n\n\n\n\\clearpage\n\\newpage\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "ae6a7753bad0141a2504e3f47b884e4180a6fcef", "size": 28620, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Tex/Smatrix/Smatrix.tex", "max_stars_repo_name": "nasa-jpl/Waveport", "max_stars_repo_head_hexsha": "caeb9540693185e000e08d826bc2ccabb6aa82bd", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2021-08-29T13:29:21.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-18T20:09:47.000Z", "max_issues_repo_path": "Tex/Smatrix/Smatrix.tex", "max_issues_repo_name": "ruzakb/Waveport", "max_issues_repo_head_hexsha": "caeb9540693185e000e08d826bc2ccabb6aa82bd", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Tex/Smatrix/Smatrix.tex", "max_forks_repo_name": "ruzakb/Waveport", "max_forks_repo_head_hexsha": "caeb9540693185e000e08d826bc2ccabb6aa82bd", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2021-08-29T13:28:44.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-08T19:58:04.000Z", "avg_line_length": 77.5609756098, "max_line_length": 938, "alphanum_fraction": 0.6908805031, "num_tokens": 9932, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.795658090372256, "lm_q2_score": 0.7853085808877581, "lm_q1q2_score": 0.6248371258220999}} {"text": "\\section{Non-parametric DF Estimation}\\label{S:NonparamDFEstimation}\nSo far, we have been interested in some estimation problems involved in parametric experiments. In parametric experiments, the parameter space $\\BB{\\Theta}$ can have many dimensions, but these are finite. For example, in the $n$ IID $\\bernoulli(\\theta^*)$ and the $n$ IID $\\exponential(\\lambda^*)$ experiments:\n\\begin{eqnarray*}\nX_1,\\ldots,X_n \\overset{IID}{\\sim} \\bernoulli(\\theta^*), & \\qquad & \\theta^* \\in \\BB{\\Theta}=[0,1] \\subset \\Rz^1 \\enspace , \\\\\nX_1,\\ldots,X_n \\overset{IID}{\\sim} \\exponential(\\lambda^*), & \\qquad & \\lambda^* \\in \\BB{\\Lambda}=(0,\\infty) \\subset \\Rz^1 \\enspace ,\n\\end{eqnarray*}\nthe parameter spaces $\\BB{\\Theta}$ and $\\BB{\\Lambda}$ are of dimension $1$. Similarly, in the $n$ IID $\\normal(\\mu,{\\sigma}^2)$ and the $n$ IID $\\lognormal(\\lambda,\\zeta)$, experiments:\n\\begin{eqnarray*}\nX_1,\\ldots,X_n \\overset{IID}{\\sim} \\normal(\\mu,\\sigma^2), &\\qquad & (\\mu,\\sigma^2) \\in \\BB{\\Theta}=(-\\infty,+\\infty) \\times (0,+\\infty) \\subset \\Rz^2 \\\\\nX_1,\\ldots,X_n \\overset{IID}{\\sim} \\lognormal(\\lambda,\\zeta), &\\qquad & (\\lambda,\\zeta) \\in \\BB{\\Theta}=(0,+\\infty) \\times (0,+\\infty) \\subset \\Rz^2 \n\\end{eqnarray*}\nthe parameter space is of dimension $2$.\n\nAn experiment with an infinite dimensional parameter space $\\BB{\\Theta}$ is said to be {\\bf non-parametric} . Next we consider a non-parametric experiment in which $n$ IID samples are drawn according to some fixed and possibly unknown DF $F^*$ from the space of {\\bf All Distribution Functions}:\n\\[\n\\boxed{\nX_1,X_2,\\ldots,X_n \\overset{IID}{\\sim} F^*, \\qquad F^* \\in \\BB{\\Theta} = \\{ \\text{All DFs}\\} := \\{ \\, F(x; F) : F~is~a~DF \\, \\} \n}\n\\]\nwhere the DF $F(x; F)$ is indexed or parameterised by itself. Thus, the parameter space $\\BB{\\Theta}=\\{ \\text{All DFs}\\}$ is the {\\bf infinite dimensional} space of {\\bf All DFs}. In this section, we look at estimation problems in non-parametric experiments with an infinite dimensional parameter space. That is, we want to estimate the DF $F^*$ from which our IID data are drawn.\n\nThe next proposition is often referred to as the {\\bf fundamental theorem of statistics} and is at the heart of non-parametric inference, empirical processes, and computationally intensive bootstrap techniques. Recall \\hyperref[D:ECDF]{Definition~\\ref*{D:ECDF}} of the $n$-sample empirical distribution function (EDF or ECDF) $\\widehat{F}_n$ that assigns a probability mass of $1/n$ at each data point $x_i$:\n\\begin{eqnarray*} %\\label{E:ECDF}\n\\widehat{F}_n(x) = \\frac{ \\sum_{i=1}^n \\BB{1}(X_i \\leq x) }{n} \\ , & \\quad where \\qquad\n\\BB{1}(X_i \\leq x) :=\n\\begin{cases}\n& 1 \\quad \\text{if $x_i \\leq x$} \\\\\n& 0 \\quad \\text{if $x_i > x$} \n\\end{cases}\n\\end{eqnarray*}\n\n\\begin{figure}[htpb]\n\\caption{Plots of ten distinct ECDFs $\\widehat{F}_n$ based on $10$ sets of $n$ IID samples from $\\uniform(0,1)$ RV $X$, as $n$ increases from $10$ to $100$ to $1000$. The DF $F(x)=x$ over $[0,1]$ is shown in red. The script of \\hyperref[Mf:GilvenkoCantelliUnif01n10n100n100ECDFs]{Labwork \\ref*{Mf:GilvenkoCantelliUnif01n10n100n100ECDFs}} was used to generate this plot. \\label{F:GilvenkoCantelliUnif01n10n100n100ECDFs}}\n\\centering \\makebox{\\includegraphics[width=6.50in]{figures/GilvenkoCantelliUnif01n10n100n100ECDFs}}\n\\end{figure}\n\\begin{prop}[Gilvenko-Cantelli Theorem]\\label{P:Gilvenko-Cantelli}\nLet $X_1,X_2,\\ldots,X_n \\overset{\\IID}{\\sim} F^*$. Then:\n\\[\n\\sup_x { | \\widehat{F}_n(x) - F^*(x) | } \\overset{P}{\\longrightarrow} 0 \\ .\n\\]\n%{\\scriptsize\n%\\begin{proof}\n%Proof to be seen in STAT 318 or another advanced Statistics course.\n%\\end{proof}\n%}\n\\end{prop}\n{\\bf Heuristic Interpretation of the Gilvenko-Cantelli Theorem}: As the sample size $n$ increases, the empirical distribution function $\\widehat{F}_n$ converges to the true DF $F^*$ in probability, as shown in \\hyperref[F:GilvenkoCantelliUnif01n10n100n100ECDFs]{Figure~\\ref*{F:GilvenkoCantelliUnif01n10n100n100ECDFs}}.\n\n\\begin{prop}[The Dvoretzky-Kiefer-Wolfowitz (DKW) Inequality]\nLet $X_1,X_2,\\ldots,X_n \\overset{\\IID}{\\sim} F^*$. Then, for any $\\epsilon>0$:\n\\begin{equation}\\label{E:DKWNeq}\nP \\left( \\sup_x | \\widehat{F}_n(x) - F^*(x) | > \\epsilon \\right) \\leq 2 \\exp {(-2 n \\epsilon^2)}\n\\end{equation}\nRecall that $\\sup(A)$ or supremum of a set $A \\subset \\Rz$ is the least upper bound of every element in $A$.\n\\end{prop}\n\n\\subsection{Estimating DF}\\label{S:EstimDF}\nLet $X_1,X_2,\\ldots,X_n \\overset{\\IID}{\\sim} F^*$, where $F^*$ is some particular DF in the space of all possible DFs, i.e.~the experiment is non-parametric. Then, based on the data sequence $X_1,X_2,\\ldots,X_n$ we want to estimate $F^*$.\n\nFor any fixed value of $x$, the expectation and variance of the empirical DF \\eqref{E:ECDF} are:\n\\begin{eqnarray}\n\\E \\left( \\widehat{F}_n(x) \\right) &=& F^*(x) \\implies \\mathsf{bias}_n \\left( \\widehat{F}_n(x) \\right) = 0 \\label{E:FnhatUnbiased} \\\\\n\\V \\left( \\widehat{F}_n(x) \\right) &=& \\frac{F^*(x) (1-F^*(x))}{n} \\implies \\lim_{n \\to \\infty}\\mathsf{se}_n \\left( \\widehat{F}_n(x) \\right) = 0 \\label{E:Fnhatseto0} \n\\end{eqnarray}\nTherefore, by \\hyperref[P:AsympConsistencyUnbiasedSE0]{Proposition~\\ref* {P:AsympConsistencyUnbiasedSE0}}, the empirical DF evaluated at $x$, i.e.~$\\widehat{F}_n(x)$ is an asymptotically consistent estimator of the DF evaluated at $x$, i.e.~$F^*(x)$. More formally,\n\\eqref{E:FnhatUnbiased} and \\eqref{E:Fnhatseto0}, by \\hyperref[P:AsympConsistencyUnbiasedSE0]{Proposition~\\ref* {P:AsympConsistencyUnbiasedSE0}}, imply that for any fixed value of $x$:\n\\[\n\\widehat{F}_n(x) \\overset{P}{\\longrightarrow} F^*(x) \\ .\n\\]\nWe are interested in a point estimate of the entire DF $F^*$, i.e.~$F^*(x)$ over all $x$. A point estimator $T_n=T_n(X_1,X_2,\\ldots,X_n)$ of a fixed and possibly unknown $F \\in \\{ \\text{All DFs} \\}$ is the empirical DF $\\widehat{F}_n$. This estimator has an asymptotically desirable property: \n\\[\n\\sup_x { | \\widehat{F}_n(x) - F^*(x) | } \\overset{P}{\\longrightarrow} 0 \n\\]\nbecause of the Gilvenko-Cantelli theorem in Proposition~\\ref*{P:Gilvenko-Cantelli}. Thus, we can simply use $\\widehat{F}_n$, based on the realized data $(x_1,x_2,\\ldots,x_n)$, as a point estimate of $F^*$.\n\nOn the basis of the DKW inequality \\eqref{E:DKWNeq}, we can obtain a $1-\\alpha$ confidence set or {\\bf confidence band} $C_n(x) := [\\underline{C}_{\\, n}(x), \\overline{C}_{\\, n}(x)]$ about our point estimate of $F^*$: \n\\begin{eqnarray}\n\\underline{C}_{\\, n}(x) &=& \\max \\{ \\widehat{F}_n(x)-\\epsilon_n, 0 \\}, \\notag \\\\\n\\overline{C}_{\\, n}(x) &=& \\min \\{ \\widehat{F}_n(x)+\\epsilon_n, 1 \\}, \\notag \\\\\n\\epsilon_n &=& \\sqrt{ \\frac{1}{2n} \\log \\left( \\frac{2}{\\alpha}\\right)} \\ .\n\\end{eqnarray}\nIt follows from \\eqref{E:DKWNeq} that for any fixed and possibly unknown $F^*$:\n\\[\nP \\left( \\underline{C}_{\\, n}(x) \\leq F^*(x) \\leq \\overline{C}_{\\, n}(x) \\right) \\geq 1-\\alpha \\ .\n\\]\nLet us look at a simple example next.\n\\begin{labwork}[Estimating the DF of $\\uniform(0,1)$ RV]\nConsider the problem of estimating the DF of $\\uniform(0,1)$ RV $U$ on the basis of $n_=10$ samples. We use the function {\\tt ECDF} of Labwork~\\ref*{Mf:ECDF} and {\\sc Matlab}'s built-in function {\\tt stairs} to render the plots. Figure~\\ref*{F:UniformECDFsBands} was generated by {\\tt PlotUniformECDFsConfBands.m} given below.\n{\\VrbMf[label=PlotUniformECDFsConfBands.m]{scripts/PlotUniformECDFsConfBands.m}}\n\\begin{figure}[htpb]\n\\caption{The empirical DFs $\\widehat{F}^{(1)}_{n}$ from sample size $n=10, 100, 1000$ (black), is the point estimate of the fixed and known DF $F(x)=x, x \\in[0,1]$ of $\\uniform(0,1)$ RV (red). The $95\\%$ confidence band for each $\\widehat{F}_n$ are depicted by green lines. \\label{F:UniformECDFsBands}}\n\\centering \\makebox{\\includegraphics [width=5.5in]{figures/UniformECDFsBands}}\n\\end{figure}\n\\end{labwork}\nNext we look at a more interesting example involving real-world data.\n\\begin{labwork}[Non-parametric Estimation of the DF of Times Between Earth Quakes]\\label{LW:NZSIEQTimesECDFsConfBands}\nSuppose that the $6,128$ observed times between Earth quakes in NZ between 18-Jan-2008 02:23:44 and 18-Aug-2008 19:29:29 are:\n$$X_1,\\ldots,X_{6128} \\overset{IID}{\\sim} F^*, \\qquad F^* \\in \\{ \\text{all DFs} \\} \\enspace .$$\nThen the non-parametric point estimate of the unknown $F^*$ is $\\widehat{F}_{6128}$, the ECDF of the inter earth quake times. We plot the non-parametric point estimate as well as the 95\\% confidence bands for $F^*$ in \\hyperref[F:NZSIEQTimesECDFsConfBands]{Figure~\\ref*{F:NZSIEQTimesECDFsConfBands}}.\n\\begin{figure}[htpb]\n\\caption{The empirical DF $\\widehat{F}_{6128}$ for the inter earth quake times and the $95\\%$ confidence bands for the non-parametric experiment. \\label{F:NZSIEQTimesECDFsConfBands}}\n\\centering \\makebox{\\includegraphics[width=6.5in]{figures/NZSIEQTimesECDFsConfBands}}\n\\end{figure}\n\\VrbMf[label=NZSIEQTimesECDFsConfBands.m]{scripts/NZSIEQTimesECDFsConfBands.m}\n\\end{labwork}\n\nRecall the poor fit of the Exponential PDF at the MLE for the Orbiter waiting time data. We can attribute the poor fit to coarse resolution of the waiting time measurements in minutes and the rigid decaying form of the exponential PDFs. Let us revisit the Orbiter waiting time problem with our non-parametric estimator. \n\n\\begin{labwork}[Non-parametric Estimation of Orbiter Waiting Times DF]\\label{LW:OrbiterECDFsConfBands}\nSuppose that the waiting times at the Orbiter bus stop are:\n$$X_1,\\ldots,X_{132} \\overset{IID}{\\sim} F^*, \\qquad F^* \\in \\{ \\text{all DFs} \\} \\enspace .$$\nThen the non-parametric point estimate of $F^*$ is $\\widehat{F}_{132}$, the ECDF of the $132$ Orbiter waiting times.\n\\begin{figure}[htpb]\n\\caption{The empirical DF $\\widehat{F}_{132}$ for the Orbiter waiting times and the $95\\%$ confidence bands for the non-parametric experiment. \\label{F:OrbiterECDFsConfBands}}\n\\centering \\makebox{\\includegraphics[width=5.5in]{figures/OrbiterECDFsConfBands}}\n\\end{figure}\nWe compute and plot the non-parametric point estimate as well as the 95\\% confidence bands for the unknown DF $F^*$ beside the parametric estimate and $95\\%$ confidence bands from \\hyperref[LW:ExponentialMLECIOrbiter]{Labwork~\\ref*{LW:ExponentialMLECIOrbiter}}. Clearly, the non-parametric estimate is preferable to the parametric one for this example. Notice how the non-parametric confidence bands do not contain the parametric estimate of the DF. \n\\VrbMf[label=OrbiterECDFsConfBands.m]{scripts/OrbiterECDFsConfBands.m}\n\\end{labwork}\n\n\n\\begin{example}\nFirst take a look at \\hyperref[DA:WebLogs]{Data~\\ref*{DA:WebLogs}} to understand how the web login times to our Maths \\& Stats Department's web server (or requests to our WWW server) were generated. \\hyperref[F:WebLogTimesECDFs]{Figure~\\ref*{F:WebLogTimesECDFs}} shows the login times in units of seconds over a 24 hour period starting at 0357 hours and 30 seconds (just before 4:00AM) on October 1st, 2007 (red line) and on October 2nd, 2007 (magenta). \n\\begin{figure}[htpb]\n\\caption{The empirical DFs $\\widehat{F}^{(1)}_{n_1}$ with $n_1= 56485$, for the web log times starting October 1, and $\\widehat{F}^{(2)}_{n_2}$ with $n_2=53966 $, for the web log times starting October 2. Their $95\\%$ confidence bands are indicated by the green. \\label{F:WebLogTimesECDFs}}\n\\centering \\makebox{\\includegraphics [width=4.5in]{figures/WebLogTimesECDFsBands}}\n%WebLogTimesECDFs.eps}}\n\\end{figure}\nIf we assume that some fixed and unknown DF $F^{(1)}$ specifies the distribution of login times for October 1st data and another DF $F^{(2)}$ for October 2nd data, then the non-parametric point estimates of $F^{(1)}$ and $F^{(2)}$ are simply the empirical DFs $\\widehat{F}^{(1)}_{n_1}$ with $n_1= 56485$ and $\\widehat{F}^{(2)}_{n_2}$ with $n_2=53966 $, respectively, as depicted in \\hyperref[F:WebLogTimesECDFs]{Figure~\\ref*{F:WebLogTimesECDFs}}. See the script of {\\tt WebLogDataProc.m} in \\hyperref[DA:WebLogs]{Data~\\ref*{DA:WebLogs}} to appreciate how the ECDF plots in \\hyperref[F:WebLogTimesECDFs]{Figure~\\ref*{F:WebLogTimesECDFs}} were made.\n\\end{example}\n\n\\section{Plug-in Estimators of Statistical Functionals}\\label{S:PlugIn}\nRecall from \\hyperref[S:Statistics]{Chapter~\\ref*{S:Statistics}} that a {\\bf statistical functional} is simply any function of the DF $F$. For example, the median $T(F) = F^{[-1]}(1/2)$ is a statistical functional. Thus, $T(F): \\{ \\text{All DFs }\\} \\to \\Tz$, being a map or function from the space of DFs to its range $\\Tz$, is a functional. \nThe idea behind the plug-in estimator for a statistical functional is simple: just plug-in the point estimate $\\widehat{F}_n$ instead of the unknown DF $F^*$ to estimate the statistical functional of interest.\n\\begin{definition}[Plug-in estimator]\nSuppose, $X_1,\\ldots,X_n \\overset{IID}{\\sim} F^*$. The plug-in estimator of a statistical functional of interest, namely, $T(F^*)$, is defined by:\n\\[\n\\widehat{T}_n := \\widehat{T}_n (X_1,\\ldots,X_n) = T(\\widehat{F}_n) \\ .\n\\]\n\\end{definition}\n\n\\begin{definition}[Linear functional]\nIf $T(F) = \\int r(x) dF(x)$ for some function $r(x):\\Xz\\to \\Rz$, then $T$ is called a {\\bf linear functional}. Thus, $T$ is linear in its arguments:\n\\[\nT(a F + a' F') = a T(F) + a' T(F') \\enspace .\n\\]\n\\end{definition}\n\n\\begin{prop}[Plug-in Estimator of a linear functional]\nThe plug-in estimator for a linear functional $T = \\int r(x) dF(x)$ is:\n\\[\n\\boxed{\nT(\\widehat{F}_n) = \\int r(x) d \\widehat{F}_n(x)=\\frac{1}{n}\\sum_{i=1}^n r(X_i) \n} \\enspace .\n\\]\n\\end{prop}\n\nSome specific examples of statistical linear functionals we have already seen include:\n\\begin{enumerate}\n\\item The {\\bf mean} of RV $X \\sim F$ is a function of the DF $F$: \n\\[\nT(F) = \\E(X) = \\int x\\,dF(x) \\ .\n\\]\n\\item The {\\bf variance} of RV $X \\sim F$ is a function of the DF $F$: \n\\[\nT(F) = \\E(X-\\E(X))^2 = \\int (x-\\E(X))^2\\,dF(x) \\ .\n\\]\n\\item The {\\bf value of DF at a given $x \\in \\Rz$} of RV $X \\sim F$ is also a function of DF $F$:\n\\[\nT(F) = F(x) \\ .\n\\]\n\\item The {\\bf $q^{\\text{th}}$ quantile} of RV $X \\sim F$: \n\\[\nT(F) = F^{[-1]}(q) \\ \\text{ where } q \\in [0,1] \\ .\n\\]\n\\item The {\\bf first quartile} or the {\\bf $0.25^{\\text{th}}$ quantile} of the RV $X \\sim F$: \n\\[\nT(F) = F^{[-1]}(0.25) \\ .\n\\]\n\\item The {\\bf median} or the {\\bf second quartile} or the {\\bf $0.50^{\\text{th}}$ quantile} of the RV $X \\sim F$: \n\\[\nT(F) = F^{[-1]}(0.50) \\ .\n\\]\n\\item The {\\bf third quartile} or the {\\bf $0.75^{\\text{th}}$ quantile} of the RV $X \\sim F$: \n\\[\nT(F) = F^{[-1]}(0.75) \\ .\n\\]\n\\end{enumerate}\n\n\n\\begin{labwork}[Plug-in Estimate for Median of Web Login Data]\nCompute the plug-in estimates for the median for each of the data arrays:\n\\[\n\\text{{\\tt WebLogSeconds20071001035730} and {\\tt WebLogSeconds20071002035730} }\n\\]\nthat can be loaded into memory by following the commands in the first $13$ lines of the script file {\\tt WebLogDataProc.m} of \\hyperref[DA:WebLogs]{Data~\\ref*{DA:WebLogs}}.\n\\end{labwork}\n\n\\begin{labwork}[Plug-in Estimates of Times Between Earth Quakes]\\label{LW:PlugInEstimatesEarthQuakes}\nCompute the plug-in estimates for the median and mean time in minutes between earth quakes in NZ using the data in {\\tt earthquakes.csv}. \n\\VrbMf[label=NZSIEQTimesPlugInEstimates.m]{scripts/NZSIEQTimesPlugInEstimates.m}\n\\begin{VrbM}\n>> NZSIEQTimesPlugInEstimates\nPlugInMedianEstimate = 0.0177\nPlugInMedianEstimateMinutes = 25.5092\nPlugInMeanEstimate = 0.0349\nPlugInMeanEstimateMinutes = 50.2278\n\\end{VrbM}\n\\end{labwork}\n\nNote that any statistical functional can be estimated using the plug-in estimator. However, to produce a $1-\\alpha$ confidence set for the plug-in point estimate, we need bootstrap methods. The subject of next chapter.\n\n", "meta": {"hexsha": "3773284696efb359e022eb8e051eed55e2ba9a55", "size": 15591, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "matlab/csebook/NonParamEstimDFPlugIn.tex", "max_stars_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_stars_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-02-19T07:54:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-14T13:55:18.000Z", "max_issues_repo_path": "matlab/csebook/NonParamEstimDFPlugIn.tex", "max_issues_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_issues_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "matlab/csebook/NonParamEstimDFPlugIn.tex", "max_forks_repo_name": "raazesh-sainudiin/computational-statistical-experiments", "max_forks_repo_head_hexsha": "edb33db9a05b32645e8337c03729c0b8d02fa728", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2018-07-18T07:47:52.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-19T11:28:24.000Z", "avg_line_length": 73.1971830986, "max_line_length": 650, "alphanum_fraction": 0.706818036, "num_tokens": 5439, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145997, "lm_q2_score": 0.824461932846258, "lm_q1q2_score": 0.6247726210606849}} {"text": "\\section{Introduction to train a Neural Network} %Jianhong / b-j8497 / Jul 10\n\\subsection{PyTorch Basics} % Author: Jonathan Siegel\nIn this section, we will introduce PyTorch \\footnote{\\url{https://pytorch.org}} which is an open source machine learning framework that accelerates the path from research prototyping to production deployment. \n\nThe basic class underlying pytorch is the tensor class. This class simply represents an array of numeric values (usually floats or doubles) and behaves very similar to the corresponding class in numpy.\n\\begin{python}\nimport torch\nimport numpy as np\n\n# We can initialize a pytorch tensor from a python list.\nx = torch.tensor([[1,2],[3,4]])\nprint('x is', x)\n\n# Or from a numpy array.\ny = torch.tensor(np.array([[1,2],[3,4]]))\nprint('y is', y)\n\n# We can also initialize a zeroed tensor.\na = torch.zeros([2,2])\nb = torch.zeros([3,3], dtype=torch.int32)\n\n# Note that the default type of such a tensor is a 32-bit float.\nprint('Type of a:', a.dtype)\nprint('Type of b:', b.dtype)\n\\end{python}\nThe output will be\n\\begin{python}\nx is tensor([[1, 2],\n [3, 4]])\ny is tensor([[1, 2],\n [3, 4]])\nType of a: torch.float32\nType of b: torch.int32\n\\end{python}\nArithmetic operations on PyTorch tensors work as one would expect.\n\\begin{python}\nz = 2*x\nprint(z)\n\nz = y - z\nprint(z)\n\nz = torch.matmul(x,y)\nprint(z)\n\n# etc. see the documentation for more details.\n\\end{python}\nIt gives\n\\begin{python}\ntensor([[2, 4],\n [6, 8]])\ntensor([[-1, -2],\n [-3, -4]])\ntensor([[ 7, 10],\n [15, 22]])\n\\end{python}\n\n\nAt this point, there isn't much difference between Pytorch and Numpy. What distinguishes Pytorch and makes it useful for Deep Learning is the automatic differentiation supported by Pytorch (essentially, Pytorch can automatically apply the chain rule to calculate derivatives). We illustrate this below with a simple example.\n\\begin{python}\n# In this example we calculate the derivative of (x.y)^2 w.r.t x.\n\n# In order to calculate the gradient with respect to a tensor,\n# we must set the requires_grad flag to True.\nx = torch.tensor([1.0, 0.0], requires_grad=True)\ny = torch.tensor([1.0, 1.0])\n\n# By default, requires_grad is set to False if possible.\nprint('y.requires_grad:', y.requires_grad)\n\nz = torch.dot(x, y)\n\n# Because z depends on x, which requires a gradient, z.requires_grad\n# is automatically set to true. This is done because the gradient of z\n# is needed to calculate the gradient of x.\nprint('z.requires_grad:', z.requires_grad)\n\nout = z * z\n\n# Calculate the derivative of out w.r.t. x. Automatically applies\n# the chain rule as needed.\ngrad = torch.autograd.grad(outputs=out, inputs=x)\nprint(grad)\n\\end{python}\nThe output will be\n\\begin{python}\ny.requires_grad: False\nz.requires_grad: True\n(tensor([2., 2.]),)\n\\end{python}\n\nUsing what we have so far, we can automatically differentiate polynomials (since they are just compositions of elementary arithmetic operations). But what about more complicated differentiable functions? We can deal with those by subclassing torch.autograd.Function as we show in the next example.\n\\begin{python}\n# We illustrate this by implementing the pointwise ReLU function\n# (which is of course already implemented in Pytorch, but this\n# process can be used to define more complex custom functions as well).\n\n# We subclass torch.autograd.Function to define new functions\nclass MyReLU(torch.autograd.Function):\n \n # The forward method applies the function to the input argument.\n # The ctx argument can be used to cache information for the subsequent\n # gradient computation. You can cache an object using using the \n # ctx.save_for_backward method.\n @staticmethod\n def forward(ctx, input):\n # The input is needed later to calculate the gradient.\n ctx.save_for_backward(input)\n return input.clamp(min=0)\n \n # The backward method calculates the gradient of the input, given that\n # the gradient of the output is the output_grad argument.\n def backward(ctx, output_grad):\n # Recover the original input from ctx.\n input, = ctx.saved_tensors\n grad_input = output_grad.clone()\n # Zero out the output gradient where the input is negative to obtain\n # the input gradient.\n grad_input[input < 0] = 0\n return grad_input\n \n# Apply the new function to a tensor.\nprint(MyReLU.apply(x - 0.5))\n \n# Calculating derivatives utilizes the backward method.\nz = torch.mul(MyReLU.apply(x - 0.5), y)\n# In this case z is not a scalar, so we must provide the derivatives\n# of the outputs. What is actually computed is the gradient of \n# z dot output_grad with respect to x.\noutput_grad = torch.ones([2])\nprint(torch.autograd.grad(outputs=z, inputs=x, grad_outputs = output_grad))\n\\end{python}\n\\begin{python}\ntensor([0.5000, 0.0000], grad_fn=)\n(tensor([1., 0.]),)\n\\end{python}\nWe have seen how to define a custom function. Luckily, we won't have to do this much work very often. As we will see, many common functions in machine learning have already been implemented for us.\n\nNext we introduce the Variable class in torch.autograd. This class used to be necessary for autograd, but since regular tensors now allow for automatic differentiation, it is basically just a wrapper around the tensor class which introduces a few additional methods, most notably the .backward() method.\n\\begin{python}\nfrom torch.autograd import Variable\n\n# Variables can be copy constructed from tensors. When constructed, we can \n# also specify whether the gradient is required or not.\nx = Variable(x, requires_grad=True)\ny = Variable(y, requires_grad=False)\n\n# All functions which can be applied to tensors can also be applied to \n# Variables. In the code below, z will be a Variable since x and y are.\n# In general, if at least one input is a Variable, then the output will be as\n# well.\nz = torch.mul(MyReLU.apply(x - 0.5), y)\n\n# Notably, we can now call .backward() from z_variable, which will compute \n# the gradient with respect to z for all tensors it depends on (if their\n# requires_grad field is True). Note that since z isn't a scalar, we still\n# need to pass the gradients of each of its components.\nz.backward(gradient=output_grad)\n\n# The resulting gradients are stored in the .grad field of the corresponding\n# tensors. The gradients are only calculated if requires_grad = True.\nprint('x gradient:', x.grad)\nprint('y gradient:', y.grad)\n\\end{python}\n\\begin{python}\nx gradient: tensor([1., 0.])\ny gradient: None\n\\end{python}\nSince Variables contain all of the functionality as tensors, but contain additional methods (especially the very useful .backward() method) as well, we recommend using Variables for all non-constant quantities in your code. We illustrate some of the additional behaviors of the .backward() function below.\n\\begin{python}\n# A very important note is that gradients are accumulated, i.e. added to\n# whatever is already stored in the .grad field.\nz = torch.mul(MyReLU.apply(x - 0.5), y)\nz.backward(gradient=output_grad)\nprint('x.grad has now been doubled:', x.grad)\n\n# In order to avoid this, we must first clear the gradient of x. Note also\n# that in the above example we need to recompute z in terms of x and y \n# in order to recompute the gradient. We can avoid this by setting \n# retain_graph = True\nx.grad.data.zero_()\nz = torch.mul(MyReLU.apply(x - 0.5), y)\nz.backward(gradient=output_grad, retain_graph=True)\nprint('x gradient:', x.grad)\n\n# Since we set retain_graph=True during the last backward call, we can \n# recompute the gradients without recalculating z. This feature is not\n# commonly used and considered bad practice, though.\nx.grad.data.zero_()\nz.backward(gradient=output_grad)\nprint('x gradient:', x.grad)\n\\end{python}\n\\begin{python}\nx.grad has now been doubled: tensor([2., 0.])\nx gradient: tensor([1., 0.])\nx gradient: tensor([1., 0.])\n\\end{python}\n\n\\subsection{Building Neural Networks} % Author: Jonathan Siegel\nBuilding and training neural networks directly using Pytorch Variables and automatic differentiation would be very cumbersome. Luckily, Pytorch provides built-in functionality which makes it much easier to build and train neural networks. In this section, we will describe how to use this functionality to build (i.e. define) a network.\n\nThe most important library for building neural networks is the torch.nn library. This library allows us to build neural networks by concatenating different types of layers.\n\\begin{python}\nimport torch.nn as nn\n\n# Here we define a sample neural network. The class defining our\n# network should inherit from nn.Module.\nclass Net(nn.Module):\n def __init__(self):\n super(Net, self).__init__()\n # Here we define the layers of the network.\n # The nn.Sequential method constructs a model by concatenating\n # the layers which are input to it.\n self.apply = nn.Sequential( # Sequentially apply a\n nn.Linear(10,20), # Linear function R^10 -> R^20\n nn.ReLU(), # Pointwise ReLU\n nn.Linear(20,20), # Linear function R^20 -> R^20\n nn.ReLU(), # Pointwise ReLU\n nn.Linear(20,10)) # Linear function R^20 -> R^10\n\n # The following method must be overloaded. It specifies how to\n # evaluate the model given an input x.\n def forward(self, x):\n return self.apply(x) # In our instance we simply pass x through\n # the previously defined layers. This could\n # potentially contains something more complex.\n\\end{python}\nImplementing a Logistic Regression\nWe will now consider the problem of implementing a logistic regression (trained using stochastic gradient descent) using Pytorch. We will test this model on the MNIST handwritten data imageset.\n\\begin{python}\nimport torch\nimport torchvision\nimport torchvision.transforms as transforms\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n# Transforms images from [0,255] to [0,1] range.\ntransform = transforms.Compose(\n [transforms.ToTensor(),\n transforms.Normalize(mean=[0], std=[1])])\n\n# Load the set of training images.\ntrainset = torchvision.datasets.MNIST(root='./data', train=True, download=True, transform=transform)\ntrainloader = torch.utils.data.DataLoader(trainset, batch_size=4, shuffle=True, num_workers=2)\n\n# Load the set of test images.\ntestset = torchvision.datasets.MNIST(root='./data', train=False, download=True, transform=transform)\ntestloader = torch.utils.data.DataLoader(testset, batch_size=4, shuffle=False, num_workers=2)\n\n# Let's plot some of the images to see what we're dealing with.\ndef plot_images(imgs):\n for i in range(imgs.size()[0]):\n npimg = imgs.numpy()[i,0,:,:]\n plt.imshow(npimg, cmap='gray')\n plt.ion()\n plt.show()\n plt.pause(.05)\n\ndata = iter(testloader)\nimages, labels = data.next()\nprint(labels)\nplot_images(images)\n\\end{python}\nNow that we've loaded our dataset, let's build the logistic regression model. In the language of deep learning, the logistic regression is the same as a neural network with no hidden layers trained with a cross entropy loss (exercise: work this out yourself!).\n\\begin{python}\n# This library contains a lot of useful classes for constructing \n# neural networks and other machine learning models.\nimport torch.nn as nn\n\n# This model simply takes an input of dimension 784 and multiplies it \n# by a 784x10 matrix of parameters.\nclass LogisticRegression(nn.Module):\n def __init__(self):\n super(LogisticRegression, self).__init__()\n self.dense_linear = nn.Sequential(\n nn.Linear(28 * 28, 10))\n\n def forward(self, x):\n x = x.view(-1,28*28)\n x = self.dense_linear(x)\n return x\n \n# Construct an instance of the model.\nmodel = LogisticRegression()\n\\end{python}\n\nNow we need to fit the parameters of this logistic regression model (the 784x10 matrix). We do this by defining a loss function (the cross entropy) and using the stochastic gradient descent optimization algorithm. Luckily, both of these have already been implemented for you in Pytorch!\n\\begin{python}\n# This library contains implementations of a number of useful optimization algorithms.\nimport torch.optim as optim\n\n# The cross entropy loss is already implemented in Pytorch.\ncriterion = nn.CrossEntropyLoss()\n\n# The stochastic gradient descent algorithm with a step size of 0.1.\noptimizer = optim.SGD(model.parameters(), lr=0.1)\n\n# Write a loop to train the model using the given optimizer and loss functions.\nfor i in range(20):\n for data in trainloader:\n # extract the images and labels.\n inputs, labels = data\n \n # This must be called to zero out the accumulated gradients.\n optimizer.zero_grad()\n \n # Calculate the predictions made based on the model.\n outputs = model(inputs)\n \n # Evaluate the loss function based on the model predictions.\n loss = criterion(outputs, labels)\n \n # Calculate the gradient of the parameters with respect to the loss.\n loss.backward()\n \n # Take a optimization step.\n optimizer.step()\n \n print('Completed epoch %d' % i)\nprint('Completed training')\n\\end{python}\nNow that we've trained our model, we will test its accuracy on the test dataset.\n\\begin{python}\n# Calculate the total number of test samples and the number of correctly \n# classified test samples\ncorrect = 0\ntotal = 0\nfor data in testloader:\n images, labels = data\n outputs = model(images)\n # Take the most likely label as the predicted label.\n _, predicted = torch.max(outputs.data, 1)\n total += labels.size(0)\n correct += (predicted == labels).sum()\n\nprint('Out of %d samples, the model correctly classified %d' % (total, correct))\n\\end{python}\n\n\\subsection{Datasets}\nIn this section, we will talk about how to train a neural network. According to the course in the morning, there are a lot of pictures in a dataset, so, at the beginning, we will introduce some famous and popular datasets. \n\\begin{itemize}\n\\item {\\bf MNIST}\\footnote{\\url{http://yann.lecun.com/exdb/mnist/}} contains 60,000 pictures of the ten handwritten digits from 0 to 9. The size of each picture is $28\\times 28$ pixels. There are 60,000 pictures for training and 10,000 for testing.\n\\begin{figure}\n\\centering\n\\includegraphics[scale=0.5]{./figures/497Proj_mnist}\n\\caption{MNIST}\n\\end{figure}\n\\item {\\bf CIFAR10}\\footnote{\\url{https://www.cs.toronto.edu/~kriz/cifar.html}} is a datasets consists of $60,000$ color images in $10$ classes, with $6,000$ images per class. There are $50,000$ training data and $10,000$ testing data. Since there are colored pictures, there are three channels, red, green and blue, of the input data. The size is $32\\times 32$ pixels for each. It's a subclass of CIFAR100.\n\\begin{figure}\n\\centering\n\\includegraphics[scale=0.5]{./figures/497Proj_cifar10}\n\\caption{CIFAR10}\n\\end{figure}\n\\item {\\bf ImageNet}\\footnote{\\url{http://www.image-net.org}} is a much larger dataset containing 1.2 millions pictures. The size is $224\\times 224$ pixels. The size is much bigger and the quantity is also much larger. So you may get a good performance with a simple neural network on MNIST or CIFAR10, but it's much hard to get good accuracy on ImageNet. %So that why we need some optimization algorithms on such large datasets.\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.4]{./figures/497Proj_imagenet}\n\\caption{ImageNet}\n\\end{figure}\n\\end{itemize}\nThese are three common used datasets that we will use to train our neural networks. \n\n\\subsection{Train an image classifier}\nNow we move to a brief PyTorch \\footnote{see a quick tutorial \\url{https://pytorch.org/tutorials/beginner/deep_learning_60min_blitz.html}} tutorial. To train your first image classifier, these are basically five steps in order:\n\\begin{enumerate}\n\\item Load and normalizing the CIFAR10 training and test datasets using torchvision (Here we use CIFAR10 as an example).\n\\item Define a Convolutional Neural Network. (You can also use a Deep Neural Networks or Recurrent Neural Networks whichever you like.)\n\\item Define a loss function. (As we just defined before as the function $f$ that is the sum of the total losses.) \n\\item Train the network on the training data. (Given the loss function and training data, we can do the steepest descent, stochastic gradient descent or some other optimization algorithms to decrease the loss function on the training set.)\n\\item Test the network on the test data. (After you have got a good model on your neural network from the last step, then you test your neural network on the test set to see how well it performs. So the final criterion that judges your neural network is the test accuracy because the test set doesn't involve the training process. So it is very important to get a good test accuracy.)\n\\end{enumerate}\nThese are all the steps you need to train a neural network. Now let start from the first step.\n\n\\subsubsection{Load a dataset}\nTo load the dataset, we need the libraries, like torch and some others. So we need to import the libraries first.\n\\begin{python}\nimport torch\nimport torchvision\nimport torchvision.transforms as transforms\n\\end{python}\nThis is the first step for all the other codes. And then here is some preprocess of the dataset. Because pictures can be various, you need to preprocess and normalize the pictures to make them have better properties, like the zero mean and small variance, for better interpretation.\n\\begin{python}\ntransform = transforms.Compose(\n [transforms.ToTensor(), # Transform the picture to the torch tensor\n transforms.Normalize((0.5, 0.5, 0.5), (0.5, 0.5, 0.5))]) # and then normalize it\n\\end{python}\nThen we define a training set using the \\emph{`CIFAR10'} function since CIFAR10 is a default dataset in PyTorch. The \\emph{`root'} is the folder to which we want to store the data. Since we are using the training set, set \\emph{`train'} is true. And because we haven't download it before, set \\emph{`download'} is true. \\emph{`transform=transform'} means we want to transform these pictures following the \\emph{`transform'} function that we defined in the preprocess part.\n\\begin{python}\ntrainset = torchvision.datasets.CIFAR10(root='./data', train=True, download=True, transform=transform)\n\\end{python}\nNow we have already defined our training set. But there are still some problems. The dataset is too large to load them all to compute the gradient on the full batch. That is why we choose the stochastic gradient descent. So we choose a small batch (mini-batch) to calculate the corresponding gradient. Here we set the batch size to be 4. So we can compute the loss and the gradient very conveniently on each step. \n\\begin{python}\ntrainloader = torch.utils.data.DataLoader(trainset, batch_size=4, shuffle=True, num_workers=2)\n\\end{python}\nIt is the same for the test set except that the \\emph{`train'} and \\emph{`shuffle'} are false. \n\\begin{python}\ntestset = torchvision.datasets.CIFAR10(root='./data', train=False,download=True, transform=transform)\ntestloader = torch.utils.data.DataLoader(testset, batch_size=4,shuffle=False, num_workers=2)\n\\end{python}\n\nHere are just some parts to show the images. If you are interested about what the images look like, you can use the \\emph{`matplotlib'} library to plot them. If not, just skip this part.\n\\begin{python}\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n# functions to show an image\n\ndef imshow(img):\n img = img / 2 + 0.5 # unnormalize\n npimg = img.numpy()\n plt.imshow(np.transpose(npimg, (1, 2, 0)))\n plt.show()\n\n# get some random training images\ndataiter = iter(trainloader)\nimages, labels = dataiter.next()\n\n# show images\nimshow(torchvision.utils.make_grid(images))\n# print labels\nprint(' '.join('%5s' % classes[labels[j]] for j in range(4)))\n\\end{python}\n\nThat's basically how you load a dataset. Let's go to the second step.\n\\subsubsection{Define a Convolution Neural Network}\nSince we are dealing with the image classification, we prefer to using the Convolution Neural Network. And we will introduce more details the part later. Now we just go through the codes for the simple example line by line.\nThe first part is to initialize the class. Then we define how we compute a forward process by given an input as \\emph{`x'} in second part. First, the input will be convoluted with the 2-d convolution operation, then use the ReLU function as the nonlinear activation function, after that we make a pooling. This is what a convolution layer composed, first a convolution operator, then a nonlinear activation, and then a pooling. We define two convolution layers this way. After that, the function \\emph{`x.view'} is used to flat the picture, because the output of the convolution layer is still a two dimensional picture. Here, we use the \\emph{`view'} function to flat the two dimensional picture to be one dimensional of length 400. Then we make a fully connect operator and a ReLU function. After that, we have the final classification layer to output the prediction outcome. So that is basically what the CNN composed.\n\\begin{python}\nimport torch.nn as nn\nimport torch.nn.functional as F\n\nclass Net(nn.Module):\n def __init__(self): # Initialize the class\n super(Net, self).__init__()\n # the construction of the neural network\n self.conv1 = nn.Conv2d(3, 6, 5) # a 2-d convolution layer named as `conv1'\n self.pool = nn.MaxPool2d(2, 2) # a max pooling layer named `pool'\n self.conv2 = nn.Conv2d(6, 16, 5)\n self.fc1 = nn.Linear(16 * 5 * 5, 120) # linear layer\n self.fc2 = nn.Linear(120, 84)\n self.fc3 = nn.Linear(84, 10)\n\n def forward(self, x): \n x = self.pool(F.relu(self.conv1(x))) \n x = self.pool(F.relu(self.conv2(x))) \n x = x.view(-1, 16 * 5 * 5)\n x = F.relu(self.fc1(x))\n x = F.relu(self.fc2(x))\n x = self.fc3(x)\n return x\n\n\nnet = Net()\n\\end{python}\nDifferent choices of these hyper-parameters may lead to different results. We usually need to tune these hyper-parameters to get a better model. \nSince the input of each layer in the neural network is the output of its previous layer, there are two parts of hyper-parameters of an operator, like \\emph{`Conv2d'}, \\emph{`Linear'}. One part depends on the input, which is the output of the previous layer, of this operator. The other can be arbitrarily choose to get a good model, and this part may decide a part of parameter of the next operator/layer. For example, `Linear(input dimension, output dimension)', the dimension of output of \\emph{`fc1'} is 120, then the first parameter of \\emph{`fc2'} should be same as it, 120. \n\nAnother important hyper-parameters of convolution layers is channel. For \\emph{`Conv2d (number of input channels, number of output channels, kernel size)'}, we need keep the number of input channel same as the number of output channel of the previous layer. \n\nTo choose parameters (called hyper-parameters in neural network) of the torch functions, we recommend to read the corresponding documents in PyTorch website for more precise details.\n\nHere we just give a simple working example. Later on, we will talk about more details and how to modify them.\n\n\\subsubsection{Define a loss function and optimization}\nThere are actually a lot of possible loss functions. For image classification, the common choice is the cross-entropy. You can also use the square loss or any reasonable functions based on the problems. Actually, the common loss functions have already been defined in the \\emph{`torch.nn'} library. The way to use them is like the \\emph{`criterion'} in the following code. And the \\emph{`optimizer'} is the corresponding optimization algorithm used to decrease the loss function. The common algorithms can also be found in the \\emph{`torch.optim'} library. In the example, we choose stochastic gradient descent (SGD). And there are a few optional parameters, like learning rate \\emph{`lr'}, momentum. For large learning rate, you may behave very widely, but for small learning rate, you may zig-zaging around very small range. The learning rate is one of the most important hyper-parameter that we need to tune to get a good model.\n\\begin{python}\nimport torch.optim as optim\n\ncriterion = nn.CrossEntropyLoss() #use the cross-entropy loss defined in torch library\noptimizer = optim.SGD(net.parameters(), lr=0.001, momentum=0.9)\n\\end{python} \n\nNow we have almost defined everything. Then we can start to train our networks. \n\\subsubsection{Train the network}\nFor the training, we first define what the epoch means. Because we have like 60,000 pictures in this training set, and we divided it into a few small batches (mini-batch). Whenever you go over all the pictures in the training set, then it means you finish an epoch.\nAnd when we do one optimization step (calculate gradient and update the parameters) in one mini-batch, we say it an iteration.\n\\begin{python}\nfor epoch in range(2): # loop over the dataset multiple times\n\n running_loss = 0.0 \n for i, data in enumerate(trainloader, 0): # i is the index and data are pictures in the corresponding mini-batch\n # get the inputs; data is a list of [inputs, labels]\n inputs, labels = data\n\n # zero the parameter gradients\n optimizer.zero_grad()\n\n # forward + backward + optimize\n outputs = net(inputs) # forward propagation to get the output\n loss = criterion(outputs, labels) # compute the loss, see if it satisfy the terminal criterion\n loss.backward() # backward propagation to get the gradient\n optimizer.step() # update the parameters by optimization algorithm (like SGD) to decrease the loss\n\n # print statistics\n running_loss += loss.item()\n if i % 2000 == 1999: # print every 2000 mini-batches\n print('[%d, %5d] loss: %.3f' %\n (epoch + 1, i + 1, running_loss / 2000))\n running_loss = 0.0\n\nprint('Finished Training') # After 2 epochs, we finish the training.\n\\end{python}\n\nNow, after training process, we want to see how well the model behave. So let's start to test the model we got on the test set, which is independent to the training.\n\\subsubsection{Test the network on the test data}\n\\begin{python}\ncorrect = 0 \ntotal = 0\nwith torch.no_grad(): # We don't need compute the gradients in the test process since we don't need optimization\n for data in testloader:\n images, labels = data\n outputs = net(images)\n _, predicted = torch.max(outputs.data, 1) # get the predicted class by neural network\n total += labels.size(0)\n correct += (predicted == labels).sum().item() # compare the predicted and the real label\n\nprint('Accuracy of the network on the 10000 test images: %d %%' % (100 * correct / total))\n\\end{python}\nNow we finish training a simple neural network here.\n\n\n\n", "meta": {"hexsha": "f8ca2e0495f8a0d0be795491fd1850fd037da388", "size": 26715, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/497Proj_introNN.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/497Proj_introNN.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/497Proj_introNN.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.0803059273, "max_line_length": 931, "alphanum_fraction": 0.7406700356, "num_tokens": 6466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7826624789529375, "lm_q1q2_score": 0.6247108421331493}} {"text": "\\subsection{Ordinals}\\label{subsec:ordinals}\n\n\\begin{remark}\\label{rem:ordinal_definition}\n Ordinals are generalizations of \\hyperref[def:set_of_natural_numbers]{natural numbers}. We will find characterizing properties of the natural numbers (defined as members of \\hyperref[thm:smallest_inductive_set_existence]{\\( \\omega \\)}), so that it is clear what we want to generalize.\n\n Every natural number is defined as a set of other natural numbers:\n \\begin{align*}\n &0 = \\varnothing \\\\\n &1 = \\set{ \\varnothing } = \\set{ 0 } \\\\\n &2 = \\set{ \\varnothing, \\set{ \\varnothing } } = \\set{ 0, 1 } \\\\\n &3 = \\set{ \\varnothing, \\set{ \\varnothing }, \\set{ \\varnothing, \\set{ \\varnothing } } } = \\set{ 0, 1, 2 }\n \\end{align*}\n\n It just, so happens that each natural number \\( n \\) is the set of natural numbers that are smaller with respect to the strict order relation \\( < \\) defined in \\eqref{eq:def:natural_number_ordering/strict_predicate}.\n\n Therefore, \\( \\in \\) and \\( < \\) are equivalent on the set \\( \\omega \\). It follows from \\fullref{thm:natural_numbers_are_well_ordered} that \\( \\in \\) is a \\hyperref[def:totally_ordered_set]{strict total order} on \\( \\omega \\). It is even \\hyperref[def:well_ordered_set]{well-ordered} by \\( \\in \\), but the latter condition is redundant due to \\fullref{thm:set_membership_is_well_founded}.\n\n For an arbitrary set \\( A \\), set membership is not even a \\hyperref[def:partially_ordered_set/strict]{strict partial order} --- irreflexivity is implied by \\fullref{thm:set_membership_is_well_founded}, but transitivity of \\( \\in \\) as a binary relation on \\( A \\) fails to hold in general, not speaking about trichotomy.\n\n A very simple counterexample for transitivity of \\( \\in \\) is the set \\( \\set{ \\varnothing, \\set{ \\varnothing }, \\set{ \\set{ \\varnothing } } } \\). Clearly \\( \\varnothing \\in \\set{ \\varnothing } \\) and \\( \\set{ \\varnothing } \\in \\set{ \\set{ \\varnothing } } \\), but \\( \\varnothing \\not\\in \\set{ \\set{ \\varnothing } } \\).\n\n In order for a set \\( A \\) to be a member of \\( \\omega \\), it is not sufficient for \\( \\in \\) to be a strict total ordered on \\( A \\). Except for the members of \\( \\omega \\), another set that is totally ordered by \\( \\in \\) is \\( A = \\set{ 0, 2, 4 } \\).\n\n If we require \\( A = \\set{ 0, 2, 4 } \\) to be a \\hyperref[def:transitive_set]{transitive set}, however, it will be a natural number. Indeed, since \\( 4 \\) is a member of \\( A \\) and \\( 1 \\) and \\( 3 \\) are members of \\( 4 \\), then by adding \\( 1 \\) and \\( 3 \\) to \\( A \\) we obtain the set \\( \\set{ 0, 1, 2, 3, 4 } \\), which by our definition of natural numbers is \\( 5 \\).\n\n note that transitivity of the relation \\( \\in \\) on \\( A \\) and transitivity of the set \\( A \\) itself are entirely different concepts, although we will use both. Every member of \\( \\omega \\) is a transitive set by \\fullref{thm:omega_is_transitive} and the relation \\( \\in \\) is a strict total order by \\fullref{thm:natural_numbers_are_well_ordered}.\n\n This is the reasoning behind our definition of an ordinal --- \\fullref{def:ordinal}. From this definition it will follow that the ordinals are unique representatives of order-isomorphisms classes of well-ordered sets.\n\n As a final note, the above two conditions are not sufficient for \\( A \\) to be a member of \\( \\omega \\) (they are too general), but if we additionally require \\( A \\) to be a \\hyperref[def:set_finiteness]{finite set}, then \\( A \\) will be a member of \\( \\omega \\). We have yet to define finiteness, however.\n\\end{remark}\n\n\\begin{definition}\\label{def:ordinal}\\mcite[def. 63.12]{OpenLogicFull}\n An \\term{ordinal number} or simply \\term{ordinal} is a \\hyperref[def:transitive_set]{transitive set} \\( A \\) such that set membership (as a binary relation on \\( A \\)) \\hyperref[def:well_ordered_set]{well-orders} \\( A \\). By tradition, ordinals are denoted by initial small Greek letters like \\( \\alpha \\) and \\( \\beta \\).\n\n Because of \\fullref{thm:set_membership_is_well_founded}, it is sufficient for set membership to be a \\hyperref[def:totally_ordered_set]{strict total order} on \\( A \\). Since well-foundedness also implies \\hyperref[def:binary_relation/irreflexive]{irreflexivity}, it follows that set membership must only be \\hyperref[def:binary_relation/transitive]{transitive} and \\hyperref[def:binary_relation/trichotomic]{trichotomic} on \\( A \\).\n\n In the absence of the \\hyperref[def:zfc/foundation]{axiom of foundation}, we additionally require set membership to be a \\hyperref[def:well_founded_relation]{well-founded relation} on \\( A \\), so that \\( A \\) is well-ordered.\n\n See \\fullref{rem:ordinal_definition} for a further discussion of the definition, especially the different notions of transitivity.\n\n We introduce the notation \\( \\alpha < \\beta \\) for \\( \\alpha \\in \\beta \\) in analogy with natural numbers. This is not a binary relation since there is no set of all ordinals by \\fullref{thm:burali_forti_paradox}, however it does satisfy the properties of a well-order due to \\fullref{thm:ordinals_are_well_ordered/trichotomy}.\n\n Finally, we introduce the following \\hyperref[rem:predicate_formula]{predicate formula}\n \\begin{equation*}\\taglabel[\\op{IsOrdinal}]{eq:def:ordinal/predicate}\n \\begin{aligned}\n \\ref{eq:def:ordinal/predicate}[\\tau] \\coloneqq\n &\\ref{eq:def:transitive_set/predicate}[\\tau]\n \\wedge \\\\ \\wedge&\n \\parens[\\Big]\n {\n \\qforall {\\xi \\in \\tau}\n \\qforall {\\eta \\in \\tau}\n \\parens[\\Big]\n {\n \\eta \\in \\xi \\vee \\eta \\doteq \\xi \\vee \\xi \\in \\eta\n }\n }\n \\wedge \\\\ \\wedge&\n \\parens[\\Big]\n {\n \\qforall {\\xi \\in \\tau}\n \\qforall {\\eta \\in \\tau}\n \\qforall {\\zeta \\in \\tau}\n \\parens[\\Big]\n {\n (\\xi \\in \\eta \\wedge \\eta \\in \\zeta) \\rightarrow \\xi \\in \\zeta\n }\n }.\n \\end{aligned}\n \\end{equation*}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:omega_is_an_ordinal}\n The \\hyperref[thm:smallest_inductive_set_existence]{smallest inductive set} \\( \\omega \\) is an \\hyperref[def:ordinal]{ordinal}.\n\\end{proposition}\n\\begin{proof}\n From \\fullref{thm:omega_is_transitive} it follows that \\( \\omega \\) is a transitive set.\n\n Also, as discussed in \\fullref{rem:ordinal_definition}, from \\fullref{thm:natural_numbers_are_well_ordered} it follows that set membership is a strict total order on \\( \\omega \\).\n\n Therefore, \\( \\omega \\) is an \\hyperref[def:ordinal]{ordinal}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:member_of_ordinal_is_ordinal}\\mcite[lemma 63.13]{OpenLogicFull}\n Every member of an ordinal is an ordinal.\n\\end{proposition}\n\\begin{proof}\n Let \\( \\alpha \\) be an ordinal and let \\( \\beta \\in \\alpha \\). We will show that \\( \\beta \\) is an ordinal.\n\n By transitivity of \\( \\alpha \\), we have \\( \\beta \\subseteq \\alpha \\) and by \\fullref{def:partially_ordered_set/submodel}, \\( (\\beta, \\in) \\) is a (strictly) totally ordered set as a \\hyperref[def:first_order_substructure]{substructure} of \\( (\\alpha, \\in) \\).\n\n It remains to show that \\( \\beta \\) is itself transitive. Let \\( x \\in \\beta \\). We have that \\( \\beta \\subseteq \\alpha \\) since \\( \\alpha \\) is transitive, hence \\( x \\in \\alpha \\).\n\n Fix \\( y \\in x \\). Again from the transitivity of \\( \\alpha \\) it follows that \\( y \\in \\alpha \\). Also, \\( \\in \\) is a total order on \\( \\alpha \\) and hence from \\( y \\in x \\) and \\( x \\in \\beta \\) it follows that \\( y \\in \\beta \\).\n\n Since \\( y \\in x \\) was chosen arbitrarily, it follows that \\( x \\subseteq \\beta \\). Since \\( x \\) was chosen arbitrarily, it follows that \\( \\beta \\) is transitive.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:initial_segment_of_ordinal}\n Let \\( \\alpha \\) be an ordinal. For any \\( \\beta \\in \\alpha \\), the \\hyperref[def:partially_ordered_set_interval/ray]{open initial segment} \\( \\alpha_\\beta \\) equals \\( \\beta \\).\n\n This is the bounded version of \\fullref{thm:ordinal_is_set_of_smaller_ordinals}.\n\\end{proposition}\n\\begin{proof}\n Let \\( \\beta \\in \\alpha \\). Consider the initial segment\n \\begin{equation*}\n \\alpha_{<\\beta} = \\set{ \\gamma \\in \\alpha \\given \\gamma \\in \\beta }.\n \\end{equation*}\n\n Clearly \\( \\alpha_{<\\beta} = \\alpha \\cap \\beta \\). Given that \\( \\alpha \\) is a transitive set, however, we have \\( \\beta \\subseteq \\alpha \\) and thus \\( \\alpha \\cap \\beta = \\beta \\).\n\n Therefore, \\( \\alpha_{<\\beta} = \\beta \\).\n\\end{proof}\n\n\\begin{corollary}\\label{thm:natural_numbers_are_ordinals}\n The natural numbers (as members of \\hyperref[thm:smallest_inductive_set_existence]{\\( \\omega \\)}) are ordinals.\n\\end{corollary}\n\\begin{proof}\n Follows from \\fullref{thm:omega_is_an_ordinal} and \\fullref{thm:member_of_ordinal_is_ordinal}.\n\\end{proof}\n\n\\begin{definition}\\label{def:transfinite_sequence}\n For any ordinal \\( \\alpha \\) we call any function with \\( \\alpha \\) as its domain a \\( \\alpha \\)-indexed \\term{transfinite sequence}.\n\n In particular, the case \\( \\alpha = \\omega \\) corresponds to the standard \\hyperref[def:sequence]{natural number sequences}.\n\\end{definition}\n\n\\begin{theorem}[Bounded transfinite induction]\\label{thm:bounded_transfinite_induction}\n For every formula \\( \\varphi \\) in the language of set theory not containing \\( \\tau \\), \\( \\eta \\) nor \\( \\zeta \\) as free variables, the following is a theorem of \\logic{ZFC}:\n \\small\n \\begin{equation*}\n \\qforall \\tau\n \\parens[\\Bigg]\n {\n \\ref{eq:def:ordinal/predicate}[\\tau]\n \\rightarrow\n \\parens[\\Big]\n {\n \\qforall {\\eta \\in \\tau}\n \\overbrace\n {\n \\underbrace{ \\parens[\\Big]{ \\qforall {\\zeta \\in \\eta} \\varphi[\\xi \\mapsto \\zeta] } }_{\\mathclap{\\substack{\\T{inductive} \\\\ \\T{hypothesis}}}}\n \\rightarrow\n \\underbrace{ \\varphi[\\xi \\mapsto \\eta] }_{\\mathclap{\\substack{\\T{inductive step} \\\\ \\T{conclusion}}}}\n }^{\\T{inductive step}}\n }\n \\rightarrow\n \\underbrace{ \\qforall {\\eta \\in \\tau} \\varphi[\\xi \\mapsto \\eta] }_{\\T{conclusion}}\n }.\n \\end{equation*}\n \\normalsize\n\n See the comments in \\fullref{def:peano_arithmetic/PA3} regarding variables and quantification in axiom schemas and \\fullref{rem:induction} for a general discussion of induction.\n\n See \\fullref{rem:transfinite_induction} about a reformulation that is often useful in practice.\n\\end{theorem}\n\\begin{proof}\n This theorem is a special case of \\fullref{thm:epsilon_induction} with the formula \\( \\xi \\in \\tau \\rightarrow \\varphi \\) that is explicitly universally quantified by the parameter \\( \\tau \\) which ranges over all ordinals.\n\n Note that it is unnecessary to verify that \\( \\eta \\) and \\( \\zeta \\) are ordinals because \\fullref{thm:member_of_ordinal_is_ordinal} ensures that \\( \\eta \\) is only quantified over ordinals.\n\\end{proof}\n\n\\begin{theorem}[Bounded transfinite recursion]\\label{thm:bounded_transfinite_recursion}\\mcite[177]{Enderton1977Sets}\n Fix an \\hyperref[def:ordinal]{ordinal} \\( \\alpha \\) and a nonempty set \\( A \\). Suppose that we are given some transformation \\( T: \\pow(\\alpha \\times A) \\to A \\) whose value for any relation between \\( \\alpha \\). Then there exists a unique \\( \\alpha \\)-indexed \\hyperref[def:transfinite_sequence]{transfinite sequence} \\( f: \\alpha \\to A \\) such that for any \\( \\beta \\in \\alpha \\) we have \\( f(\\beta) = T(f\\restr_\\beta) \\).\n\n This is a vast generalization of \\fullref{thm:omega_recursion} from \\hyperref[def:sequence]{natural number sequences} to \\hyperref[def:transfinite_sequence]{transfinite sequences}.\n\n See \\fullref{rem:transfinite_induction} about a reformulation that is often useful in practice.\n\\end{theorem}\n\\begin{proof}\n The proof is analogous to that of \\fullref{thm:omega_recursion}, but we will give it anyway to highlight the difference between using \\fullref{thm:omega_induction} and \\fullref{thm:bounded_transfinite_induction}.\n\n Let \\( G \\subseteq \\pow(\\alpha \\times A) \\) be the set of all \\hyperref[def:partial_function]{partial single-valued functions} \\( g: \\alpha \\to A \\) such that\n \\begin{itemize}\n \\item There exists some \\( \\beta_g \\in \\alpha \\) such that \\( g \\) is defined only in the \\hyperref[def:partially_ordered_set_interval/ray]{open initial segment} \\( \\alpha_{< \\beta_g} \\). That is, \\( g \\) is defined for all \\( \\beta \\) up to not including \\( \\beta_g \\).\n\n \\item \\( g(\\beta) = T(g\\restr_\\beta) \\) for all \\( \\beta < \\beta_g \\).\n \\end{itemize}\n\n Clearly \\( G \\) is nonempty because the function \\( \\set{ (\\varnothing, T(\\varnothing)) } \\) belongs to \\( G \\).\n\n Define \\( f \\coloneqq \\bigcup G \\). At this point \\( f \\) is a \\hyperref[def:multi_valued_function]{multi-valued function}. We must now show that \\( f \\) has all the properties that we want.\n\n \\SubProofOf[def:multi_valued_function/total]{totality} First, we will use \\fullref{thm:bounded_transfinite_induction} to show that \\( f \\) is total.\n\n Fix \\( \\beta \\in \\dom f \\). Then there exists a function \\( g \\in G \\) defined for all \\( \\gamma < \\beta \\).\n\n \\begin{itemize}\n \\item If \\( g \\) is also defined at \\( \\beta \\) also, this directly proves that \\( \\beta \\in \\dom f \\).\n \\item If \\( g \\) is not defined at \\( \\beta \\), consider\n \\begin{equation*}\n \\widehat g \\coloneqq g \\cup \\set{ (\\beta, T(g\\restr_\\beta) }.\n \\end{equation*}\n\n The function \\( \\widehat g \\) is again a single-valued partial function and thus it belongs to \\( G \\), hence \\( \\beta \\in \\dom f \\).\n \\end{itemize}\n\n Therefore, \\fullref{thm:bounded_transfinite_induction} allows us to conclude that \\( f: \\alpha \\multto A \\) is a total multi-valued function.\n\n \\SubProofOf[def:function]{single-valuedness} Now that we know that \\( f \\) is total, we will prove that it is single-valued and thus is a function in the usual sense of the term.\n\n Fix \\( \\beta \\in \\alpha \\). Suppose that \\( f \\) is single-valued for all \\( \\gamma < \\beta \\). Since \\( f \\) is total, there exist at least one partial function \\( g \\) in \\( G \\) that is defined at \\( \\beta \\). Let \\( g \\) and \\( h \\) both be such (single-valued partial) functions.\n\n Then\n \\begin{equation*}\n g(\\beta) = T(g\\restr_\\beta) = T(f\\restr_\\beta) = T(h\\restr_\\beta) = h(\\beta),\n \\end{equation*}\n hence \\( g \\) and \\( h \\) coincide at \\( \\beta \\), which in turn implies that \\( f \\) is single-valued at \\( \\beta \\).\n\n Therefore, \\fullref{thm:omega_induction} allows us to conclude that \\( f \\) is a single-valued total function.\n\n \\SubProofOf[def:function]{uniqueness} Now that it is clear that \\( f \\) satisfies the theorem, we must verify that it is unique.\n\n Suppose that \\( f_1 \\) and \\( f_2 \\) both satisfy the theorem. Fix some \\( \\beta \\in \\alpha \\) and suppose that for every \\( \\gamma < \\beta \\), we have \\( f_1(\\gamma) = f_2(\\gamma) \\). Then\n \\begin{equation*}\n f_2(\\beta) = T(f_1\\restr_\\beta)) = T(f_2\\restr_\\beta) = f_2(\\beta).\n \\end{equation*}\n\n Therefore, \\fullref{thm:omega_induction} allows us to conclude that \\( f_1 = f_2 \\). So there is at most one function that satisfies the theorem and we have already shown that \\( f \\) is such a function.\n\\end{proof}\n\n\\begin{theorem}[Transfinite induction]\\label{thm:transfinite_induction}\\mcite[thm. 63.15]{OpenLogicFull}\n It turns out that \\fullref{thm:bounded_transfinite_recursion} is valid for all ordinals simultaneously.\n\n For every formula \\( \\varphi \\) in the language of set theory not containing \\( \\eta \\) nor \\( \\zeta \\) as free variables, the following is a theorem of \\logic{ZFC}:\n \\small\n \\begin{equation*}\n \\qforall \\eta\n \\parens[\\Bigg]\n {\n \\ref{eq:def:ordinal/predicate}[\\eta]\n \\rightarrow\n \\parens[\\Big]\n {\n \\overbrace\n {\n \\underbrace{ \\parens[\\Big] { \\qforall {\\zeta \\in \\eta} \\varphi[\\xi \\mapsto \\zeta] } }_{\\mathclap{\\substack{\\T{inductive} \\\\ \\T{hypothesis}}}}\n \\rightarrow\n \\underbrace{ \\varphi[\\xi \\mapsto \\eta] }_{\\mathclap{\\substack{\\T{inductive step} \\\\ \\T{conclusion}}}}\n }^{\\T{inductive step}}\n }\n }\n \\rightarrow\n \\qforall \\eta \\parens[\\Bigg]\n {\n \\ref{eq:def:ordinal/predicate}[\\eta]\n \\rightarrow\n \\underbrace{ \\varphi[\\xi \\mapsto \\eta] }_{\\T{conclusion}}\n }.\n \\end{equation*}\n \\normalsize\n\n This theorem could be a special case of \\fullref{thm:bounded_transfinite_recursion}, but there exists no set of all ordinals due to \\fullref{thm:burali_forti_paradox}.\n\n See the comments in \\fullref{def:peano_arithmetic/PA3} regarding variables and quantification in axiom schemas and \\fullref{rem:induction} for a general discussion of induction.\n\n See \\fullref{rem:transfinite_induction} about a reformulation that is often useful in practice.\n\\end{theorem}\n\\begin{proof}\n The proof is similar to the proof of \\fullref{thm:well_founded_induction}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ordinals_are_well_ordered}\n The ordinals are \\hyperref[def:well_ordered_set]{well-ordered}. Since there exists no set of all ordinals due to \\fullref{thm:burali_forti_paradox}, we cannot say that the ordinals form a well-ordered set. We will instead state a more concrete result.\n\n \\begin{thmenum}\n \\thmitem{thm:ordinals_are_well_ordered/transitivity} For any three ordinals \\( \\alpha \\), \\( \\beta \\) and \\( \\gamma \\) such that \\( \\alpha < \\beta < \\gamma \\) we have \\( \\alpha < \\gamma \\).\n\n \\thmitem{thm:ordinals_are_well_ordered/trichotomy} For any two ordinals \\( \\alpha \\) and \\( \\beta \\), exactly one of \\( \\alpha = \\beta \\), \\( \\alpha < \\beta \\) or \\( \\alpha > \\beta \\) holds.\n \\end{thmenum}\n\n As discussed in \\fullref{def:ordinal}, irreflexivity and well-foundedness hold by \\fullref{thm:set_membership_is_well_founded}.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:ordinals_are_well_ordered/transitivity} Let \\( \\alpha \\), \\( \\beta \\) and \\( \\gamma \\) be ordinals and let \\( \\alpha \\in \\beta \\in \\gamma \\). Since \\( \\gamma \\) is a transitive set, it follows that \\( \\beta \\subseteq \\gamma \\) and thus \\( \\alpha \\in \\gamma \\).\n\n Therefore, we have used the fact that \\( \\gamma \\) is a transitive set to prove that set membership is a transitive relation, thus obtaining a connection between two distinct concepts both named \\enquote{transitivity}.\n\n \\SubProofOf{thm:ordinals_are_well_ordered/trichotomy} Let \\( \\alpha \\) and \\( \\beta \\) be ordinals.\n\n Due to \\fullref{thm:simple_foundation_theorems/member_of_itself}, it is not possible for both \\( \\alpha \\in \\beta \\) and \\( \\alpha = \\beta \\) to hold simultaneously. Due to \\fullref{thm:simple_foundation_theorems/mutual_membership}, it is not possible for both \\( \\alpha \\in \\beta \\) and \\( \\beta \\in \\alpha \\) to hold simultaneously.\n\n Thus, at most one of \\( \\alpha = \\beta \\), \\( \\alpha \\in \\beta \\) or \\( \\beta \\in \\alpha \\) holds.\n\n We will use \\fullref{thm:transfinite_induction} on \\( \\beta \\) to show that at least one of them holds. Fix an ordinal \\( \\beta_0 \\). Our inductive hypothesis is that for every ordinal \\( \\alpha \\) and every \\( \\gamma \\in \\beta_0 \\) at least one of \\( \\alpha = \\gamma \\), \\( \\alpha \\in \\gamma \\) or \\( \\gamma \\in \\alpha \\) holds.\n\n Fix some ordinal \\( \\alpha_0 \\). We will show that at least one of \\( \\alpha_0 = \\beta_0 \\), \\( \\alpha_0 \\in \\beta_0 \\) or \\( \\beta_0 \\in \\alpha_0 \\) holds. Since the case \\( \\beta_0 = \\alpha_0 \\) is trivial, we exclude it from consideration.\n\n \\begin{itemize}\n \\item If there exists \\( \\gamma \\in \\beta_0 \\) such that \\( \\gamma = \\alpha_0 \\), clearly \\( \\alpha_0 \\in \\beta_0 \\).\n \\item If there exists \\( \\gamma \\in \\beta_0 \\) such that \\( \\alpha_0 \\in \\gamma \\), then by transitivity \\( \\alpha_0 \\in \\beta_0 \\).\n \\item If for every \\( \\gamma \\in \\beta_0 \\) we have \\( \\gamma \\in \\alpha_0 \\), then \\( \\beta_0 \\subsetneq \\alpha_0 \\). Let \\( \\gamma_0 \\) be the smallest member of \\( \\alpha_0 \\setminus \\beta_0 \\). We will show that \\( \\gamma_0 = \\beta_0 \\).\n\n Our first goal is to show that \\( \\gamma_0 \\subseteq \\beta_0 \\). Aiming at a contradiction, suppose that there exists some \\( \\lambda \\in \\gamma_0 \\setminus \\beta \\). Since \\( \\gamma_0 \\) is a transitive set, we have \\( \\lambda \\in \\alpha \\). Thus, \\( \\lambda \\in \\alpha \\setminus \\beta \\) and \\( \\lambda \\in \\gamma_0 \\), contradicting the minimality of \\( \\gamma_0 \\). Therefore, \\( \\gamma_0 \\subseteq \\beta \\).\n\n Now we will use the existing inductive hypothesis for \\( \\alpha = \\gamma_0 \\) to show that \\( \\beta_0 \\subseteq \\gamma_0 \\).\n\n \\begin{itemize}\n \\item If there exists \\( \\lambda \\in \\beta_0 \\) such that \\( \\lambda = \\gamma_0 \\), clearly \\( \\gamma_0 \\in \\beta_0 \\). But that contradicts our choice of \\( \\gamma_0 \\) as a member of \\( \\alpha_0 \\setminus \\beta_0 \\).\n\n \\item If there exists \\( \\lambda \\in \\beta_0 \\) such that \\( \\gamma_0 \\in \\lambda \\), then by transitivity \\( \\gamma_0 \\in \\beta_0 \\), which again contradicts our choice of \\( \\gamma_0 \\).\n\n \\item Finally, if for every \\( \\gamma \\in \\beta_0 \\) we have \\( \\gamma \\in \\gamma_0 \\), then \\( \\beta_0 \\subseteq \\gamma_0 \\).\n \\end{itemize}\n\n Thus, both \\( \\gamma_0 \\subseteq \\beta_0 \\) and \\( \\beta_0 \\subseteq \\gamma_0 \\), giving us the equality \\( \\beta_0 = \\gamma_0 \\). This allows us to conclude that \\( \\beta_0 \\in \\alpha_0 \\).\n \\end{itemize}\n\n We have shown by transfinite induction that for a fixed ordinal \\( \\beta_0 \\), for every other ordinal \\( \\alpha \\) at least one of \\( \\beta_0 = \\alpha \\), \\( \\beta_0 \\in \\alpha \\) or \\( \\alpha \\in \\beta_0 \\) holds. We have already shown that at most one of the three holds, hence exactly one of the three holds.\n\n Since \\( \\beta_0 \\) is itself arbitrary, we can conclude that trichotomy holds for any two ordinals \\( \\alpha \\) and \\( \\beta \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ordinal_ordering_via_subsets}\n For any two ordinals \\( \\alpha \\) and \\( \\beta \\) we have \\( \\beta \\in \\alpha \\) if and only if \\( \\beta \\subsetneq \\alpha \\).\n\\end{proposition}\n\\begin{proof}\n \\NecessitySubProof Since \\( \\alpha \\) is a transitive set, from \\( \\beta \\in \\alpha \\) it follows that \\( \\beta \\subseteq \\alpha \\).\n\n We cannot have \\( \\beta = \\alpha \\) due to \\fullref{thm:simple_foundation_theorems/member_of_itself}, hence \\( \\beta \\subsetneq \\alpha \\).\n\n \\SufficiencySubProof Suppose that \\( \\beta \\subsetneq \\alpha \\).\n\n By \\fullref{thm:ordinals_are_well_ordered/trichotomy}, the ordinals \\( \\beta \\) and \\( \\alpha \\) must be related by set membership.\n \\begin{itemize}\n \\item If \\( \\alpha = \\beta \\), this directly contradicts our assumption that \\( \\beta \\subsetneq \\alpha \\).\n \\item If \\( \\alpha \\in \\beta \\), then \\( \\alpha \\in \\alpha \\) which contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n \\item It remains for \\( \\beta \\in \\alpha \\) to hold.\n \\end{itemize}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ordinal_successor_strictly_monotone_on_ordinals}\n The ordinal successor operation is strictly monotone on ordinals. That is, if \\( \\alpha < \\beta \\), then \\( \\op{succ}(\\alpha) < \\op{succ}(\\beta) \\).\n\\end{proposition}\n\\begin{proof}\n Let \\( \\alpha \\in \\beta \\) and let \\( \\gamma \\in \\op{succ}(\\alpha) \\).\n \\begin{itemize}\n \\item If \\( \\gamma \\in \\alpha \\), clearly \\( \\gamma \\in \\beta \\) because \\( \\beta \\) is a transitive set.\n\n \\item If \\( \\gamma = \\alpha \\), then \\( \\gamma = \\alpha \\in \\beta \\).\n \\end{itemize}\n\n We have shown that \\( \\op{succ}(\\alpha) \\subseteq \\beta \\). Thus, we either have \\( \\op{succ}(\\alpha) = \\beta \\in \\op{succ}(\\beta) \\) or else by \\fullref{thm:ordinal_ordering_via_subsets} we have \\( \\op{succ}(\\alpha) \\in \\beta \\in \\op{succ}(\\beta) \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:transitive_set_of_transitive_sets}\\mcite{MathSE:transitive_set_of_transitive_sets}\n A \\hyperref[def:transitive_set]{transitive set} whose members are transitive sets is an ordinal.\n\n In particular, a transitive set of ordinals is an ordinal. If a set of ordinals is not transitive, we can instead take its \\hyperref[def:transitive_closure_of_a_set]{transitive closure}.\n\\end{proposition}\n\\begin{proof}\n Let \\( A \\) be a set whose members are themselves transitive sets.\n\n We will first show that set membership is transitive on \\( A \\). If \\( x \\), \\( y \\) and \\( z \\) are members of \\( A \\) such that \\( x \\in y \\in z \\), then since \\( z \\) is transitive we have \\( y \\subseteq z \\) and thus \\( x \\in z \\).\n\n Therefore, we can conclude that set membership is a \\hyperref[def:partially_ordered_set/strict]{strict partial order} on \\( A \\). Now define the set\n \\begin{equation*}\n B \\coloneqq \\set{ x \\in A \\given \\qexists {y \\in A} x \\not\\in y \\wedge y \\not\\in x }\n \\end{equation*}\n of all members of \\( A \\) which are not related to at least one other member. If \\( B \\) is empty, then set membership is trichotomous on \\( A \\).\n\n If \\( B \\) is nonempty, let \\( b \\) be a minimal element of \\( B \\). A minimal element must exist by \\fullref{thm:infinite_descent_partial_order} because set membership is a well-founded partial order on \\( A \\). We have chosen \\( b \\), so that every member of \\( b \\) is related to every other member of \\( A \\).\n\n Define the set\n \\begin{equation*}\n C \\coloneqq \\set{ x \\in A \\given x \\neq b \\wedge x \\not\\in b \\wedge b \\not\\in x }\n \\end{equation*}\n of all members of \\( A \\) which are not related to \\( b \\) and let \\( c \\) be a minimal element of \\( C \\). We will now show that \\( b = c \\), which is a contradictions with our choice of \\( c \\).\n\n Let \\( x \\in b \\). As we have already mentioned, \\( x \\) is related to every other member of \\( A \\), including \\( c \\).\n \\begin{itemize}\n \\item If \\( c = x \\), then \\( c \\in b \\), which contradicts our choice of \\( c \\).\n\n \\item Suppose that \\( c \\in x \\). We have chosen \\( x \\) to be a member of \\( b \\) and we thus have \\( c \\in x \\in b \\). The set \\( b \\) is transitive as a member of \\( A \\), hence \\( x \\) is also a member of \\( A \\). Since set membership is a transitive relation on \\( A \\), it follows that \\( c \\in b \\), which contradicts our choice of \\( c \\).\n\n \\item It remains for \\( x \\in c \\) to hold.\n \\end{itemize}\n\n Therefore, \\( b \\subseteq c \\). The converse inclusion \\( c \\subseteq b \\) can be obtained analogously by noting that \\( c \\) is a minimal element of \\( C \\) and hence every \\( x \\in c \\) is related to \\( b \\). Thus, we obtain \\( b = c \\), which contradicts our choice of \\( c \\) as a member of \\( C \\).\n\n The obtained contradiction shows that the set \\( B \\) is empty and thus every member of \\( A \\) is related to every other member, proving trichotomy of set membership on \\( A \\). This allows us to conclude that \\( A \\) is an ordinal because it is a transitive set and we have already shown in the beginning of the proof that set membership is a transitive relation on \\( A \\).\n\\end{proof}\n\n\\begin{definition}\\label{def:transitive_closure_of_a_set}\\mcite[def. 64.8]{OpenLogicFull}\n To show that every set has a rank we must introduce additional definitions. We use \\hyperref[rem:natural_number_recursion]{natural number recursion} to define the \\term{transitive closure} of a set \\( A \\) as\n \\begin{equation*}\n \\cl^T(A) \\coloneqq \\bigcup \\set{ \\cl_n^T(A) \\given n \\in \\omega },\n \\end{equation*}\n where\n \\begin{equation*}\n \\cl_n^T(A) \\coloneqq \\begin{cases}\n A, &n = 0 \\\\\n \\bigcup \\cl_{n-1}^T(A), &n > 0\n \\end{cases}\n \\end{equation*}\n\n Note that this is different from the transitive closure of a relation defined in \\fullref{def:relation_closures/transitive}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:transitive_closure_of_a_set}\\mcite[thm. 64.9]{OpenLogicFull}\n The \\hyperref[def:transitive_closure_of_a_set]{transitive closure} \\( \\cl^T(A) \\) of the set \\( A \\) is the smallest transitive set containing \\( A \\).\n\\end{proposition}\n\\begin{proof}\n It is clear that \\( A = \\cl_0^T(A) \\) is a subset of \\( \\cl^T(A) \\).\n\n If \\( x \\in \\cl^T(A) \\), then there exists some natural number \\( n \\) for which \\( x \\in \\cl_n^T(A) \\). Therefore, \\( x \\subseteq \\bigcup \\cl_n^T(A) = \\cl_{n+1}^T(A) \\) and thus \\( x \\subseteq \\cl^T(A) \\).\n\n Now suppose that \\( B \\) is a transitive subset of \\( \\cl^T(A) \\) which contains \\( A \\). Let \\( x_0 \\in \\cl^T(A) \\).\n\n Suppose that \\( x_0 \\not\\in B \\). Then there must exist a smallest nonzero number \\( n \\) such that \\( x_0 \\in \\cl_n^T(A) \\). Then \\( x_0 \\) belongs to some member \\( x_1 \\) of \\( \\cl_{n-1}^T(A) \\). If \\( x_1 \\in A \\), then \\( x_0 \\) must belong to \\( B \\) since it is transitive. But this contradicts out choice of \\( x_0 \\). Then \\( x_1 \\not\\in A \\), in which case there exists some \\( x_2 \\in \\cl_{n-2}^T(A) \\) such that \\( x_1 \\in x_2 \\). If \\( x_2 \\in A \\), then again \\( x_0 \\in B \\), which contradicts our choice of \\( x_0 \\). We can thus recursively construct a sequence \\( \\set{ x_k }_{k=0}^\\infty \\) such that for every \\( k \\geq 0 \\) both \\( x_k \\in x_{k+1} \\) and \\( x_k \\not\\in A \\) hold. The existence of such a sequence contradicts \\fullref{thm:set_membership_is_well_founded}.\n\n Therefore, \\( \\cl^T(A) \\subseteq B \\), hence \\( \\cl^T(A) \\) is the smallest transitive set containing \\( A \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:ordinal_is_set_of_smaller_ordinals}\n Every ordinal equals the set of all smaller ordinals.\n\n This is the unbounded version of \\fullref{thm:initial_segment_of_ordinal}.\n\\end{proposition}\n\\begin{proof}\n Let \\( \\alpha \\) be an ordinal and let \\( A \\) be the set of all ordinals smaller than \\( \\alpha \\). We will show that \\( A = \\alpha \\). We will first show that \\( A \\) is a transitive set. Let \\( \\beta \\in A \\) and \\( \\gamma \\in \\beta \\). Since \\( \\alpha \\) is a transitive set that contains \\( \\beta \\), we have \\( \\gamma \\in \\alpha \\). Thus, \\( \\gamma \\) is smaller than \\( \\alpha \\) and hence it belongs to \\( A \\). Therefore, \\( A \\) is a transitive set of ordinals and by \\fullref{thm:transitive_set_of_transitive_sets}, it is itself an ordinal.\n\n \\Fullref{thm:ordinals_are_well_ordered} implies that \\( \\alpha \\) and \\( A \\) are either equal or related by set membership.\n \\begin{itemize}\n \\item If \\( \\alpha \\in A \\), then \\( \\alpha \\) is smaller than itself, which contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n\n \\item If \\( A \\in \\alpha \\), then \\( A \\) is smaller than itself, which again contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n\n \\item It remains for \\( A \\) to be equal to \\( \\alpha \\).\n \\end{itemize}\n\\end{proof}\n\n\\begin{theorem}[Burali-Forti paradox]\\label{thm:burali_forti_paradox}\\mcite[thm. 63.19]{OpenLogicFull}\n Assuming \\logic{ZFC}, there is no set of all ordinals.\n\\end{theorem}\n\\begin{proof}\n Aiming at a contradiction, suppose that \\( A \\) is a containing all ordinals. If \\( \\alpha \\in A \\) and \\( \\beta \\in \\alpha \\), transitivity \\( \\beta \\) implies \\( \\beta \\in A \\). Thus, \\( A \\) is a transitive set of ordinals, which \\fullref{thm:transitive_set_of_transitive_sets} is itself an ordinal. Hence, \\( A \\in A \\).\n\n But this contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}. Hence, there is no set of all ordinals.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:union_of_set_of_ordinals}\\mcite{MathSE:union_of_set_of_ordinals}\n Let \\( A \\) be a set of ordinals and denote \\( \\alpha \\coloneqq \\bigcup A \\).\n\n \\begin{thmenum}\n \\thmitem{thm:union_of_set_of_ordinals/is_ordinal} The union \\( \\alpha \\) is itself an ordinal.\n\n \\thmitem{thm:union_of_set_of_ordinals/supremum} The union \\( \\alpha \\), which is the \\hyperref[def:partially_ordered_set_extremal_points/supremum_and_infimum]{supremum} of \\( A \\) with respect to set inclusion, is also the supremum of \\( A \\) with respect to ordinal ordering.\n\n That is, either \\( \\alpha = A \\) or \\( \\alpha \\) is the smallest ordinal that is larger than every member of \\( A \\).\n\n \\thmitem{thm:union_of_set_of_ordinals/ordinal} If \\( A \\) is an ordinal, then \\( \\alpha \\leq A \\). Furthermore, in the case \\( \\alpha < A \\), there is no ordinal between \\( A \\) and \\( \\alpha \\). That is, if \\( \\alpha < A \\), then \\( A \\) is the smallest ordinal strictly larger than \\( \\alpha \\).\n\n See \\fullref{def:successor_and_limit_ordinal/union} for a further distinction between \\( \\alpha = A \\) and \\( \\alpha < A \\).\n \\end{thmenum}\n\n For recursive definitions like \\fullref{def:ordinal_arithmetic/addition} this proposition justifies using \\( \\sup A \\) instead of the more confusing \\( \\bigcup A \\).\n\n Compare this result to \\fullref{thm:union_of_set_of_cardinals}.\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:union_of_set_of_ordinals/is_ordinal} Let \\( A \\) be a set of ordinals. Denote its union by \\( \\alpha \\coloneqq \\bigcup A \\).\n\n We will show that \\( \\alpha \\) is a transitive set. Due to \\fullref{thm:transitive_set_of_transitive_sets}, this is sufficient for \\( \\alpha \\) to be an ordinal.\n\n Let \\( \\beta \\in \\alpha \\). Then there exists an ordinal \\( \\gamma \\) in \\( A \\) such that \\( \\beta \\in \\gamma \\). Since \\( \\gamma \\) is itself a transitive set, we have \\( \\beta \\subseteq \\gamma \\). But \\( \\gamma \\subseteq \\alpha \\), hence \\( \\beta \\subseteq \\alpha \\).\n\n Therefore, \\( \\alpha = \\bigcup A \\) is a transitive set and thus an ordinal.\n\n \\SubProofOf{thm:union_of_set_of_ordinals/supremum} For every \\( \\beta \\in A \\) we have \\( \\beta \\subseteq \\alpha \\), which by \\fullref{thm:ordinal_ordering_via_subsets} corresponds to \\( \\beta \\leq \\alpha \\) with respect to ordinal ordering.\n\n We will show that \\( \\alpha \\) is the smallest ordinal with this property. Let \\( \\lambda_0 \\) be any other ordinal such that \\( \\gamma \\leq \\lambda_0 \\) for any \\( \\gamma \\in A \\). Fix some \\( \\gamma_0 \\in \\alpha \\). Then there exists an ordinal \\( \\beta_0 \\in A \\) such that \\( \\gamma_0 \\in \\beta_0 \\). Since \\( \\beta_0 \\in A \\) and \\( A \\subseteq \\lambda_0 \\), since \\( \\lambda_0 \\) itself is a transitive set we have \\( \\gamma_0 \\in \\lambda_0 \\).\n\n Therefore, \\( \\alpha \\) is the least upper bound of \\( A \\) with respect to ordinal ordering.\n\n \\SubProofOf{thm:union_of_set_of_ordinals/ordinal} Assume that \\( A \\) is an ordinal.\n\n If \\( \\gamma \\in \\alpha \\), by transitivity of \\( \\alpha \\) we have \\( \\gamma \\subseteq \\alpha \\). Thus, \\( \\gamma \\in A \\). Hence, \\( \\alpha \\subseteq A \\).\n\n Now suppose that there exists some \\( \\gamma \\in A \\) such that \\( \\alpha \\in \\gamma \\). Then\n \\begin{equation*}\n \\gamma \\subseteq \\bigcup A = \\alpha \\in \\gamma,\n \\end{equation*}\n which contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n\n Hence, there is not ordinal between \\( \\alpha = \\bigcup A \\) and \\( A \\).\n\\end{proof}\n\n\\begin{proposition}\\label{thm:successor_of_ordinal}\n The \\hyperref[def:ordinal_successor]{successor} \\( \\alpha \\coloneqq \\op{succ}(\\beta) \\) of an ordinal \\( \\beta \\) is the smallest ordinal larger than \\( \\alpha \\).\n\\end{proposition}\n\\begin{proof}\n We must show that \\( \\alpha \\) is a transitive set and thus by \\fullref{thm:transitive_set_of_transitive_sets} an ordinal. Note that \\( \\alpha = \\op{succ}(\\beta) = \\beta \\cup \\set{ \\beta } \\).\n\n Let \\( \\gamma \\in \\alpha \\).\n \\begin{itemize}\n \\item If \\( \\gamma \\in \\beta \\), then since \\( \\beta \\) is a transitive set, we have \\( \\gamma \\subseteq \\beta \\). Furthermore, since \\( \\beta \\subseteq \\alpha \\), by transitivity of set inclusion \\( \\gamma \\subseteq \\alpha \\).\n\n \\item If \\( \\gamma = \\beta \\), then \\( \\gamma \\subseteq \\alpha \\) by definition of successor.\n \\end{itemize}\n\n Therefore, \\( \\alpha \\) is a transitive set and thus an ordinal.\n\n Now suppose that there is another ordinal \\( \\gamma \\) such that \\( \\beta \\in \\gamma \\in \\alpha = \\beta \\cup \\set{ \\beta } \\).\n \\begin{itemize}\n \\item If \\( \\gamma \\in \\beta \\), this would contradict \\fullref{thm:simple_foundation_theorems/mutual_membership}.\n\n \\item If \\( \\gamma = \\beta \\), this would contradict \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n \\end{itemize}\n\n The obtained contradictions show that there is no ordinal between \\( \\alpha \\) and \\( \\beta \\).\n\\end{proof}\n\n\\begin{remark}\\label{rem:def:ordinal_successor}\n It follows from \\fullref{thm:ordinal_is_set_of_smaller_ordinals} and \\fullref{thm:successor_of_ordinal} and that for any ordinal \\( \\beta \\) we have\n \\begin{align*}\n \\beta &= \\set{ \\gamma \\given \\gamma \\T{is an ordinal and} \\gamma < \\beta }, \\\\\n \\op{succ}(\\beta) &= \\set{ \\gamma \\given \\gamma \\T{is an ordinal and} \\gamma \\leq \\beta }.\n \\end{align*}\n\n This shows that the \\hyperref[def:ordinal_successor]{ordinal successor} operation is very natural in the context of ordinals.\n\\end{remark}\n\n\\begin{definition}\\label{def:successor_and_limit_ordinal}\n We say that the ordinal \\( \\alpha \\) is a \\term{successor ordinal} if any of the following equivalent conditions hold:\n\n \\begin{thmenum}\n \\thmitem{def:successor_and_limit_ordinal/successor} The ordinal \\( \\alpha \\) is the successor of another ordinal. That is, there exists another ordinal \\( \\beta \\) such that \\( \\alpha = \\op{succ}(\\beta) \\).\n\n \\thmitem{def:successor_and_limit_ordinal/smaller_successor} There exists some \\( \\beta \\in \\alpha \\) such that \\( \\op{succ}(\\beta) \\) does not belong to \\( \\alpha \\).\n\n \\thmitem{def:successor_and_limit_ordinal/union} We have \\( \\bigcup \\alpha \\in \\alpha \\).\n \\end{thmenum}\n\n If \\( \\alpha \\) is neither zero nor a successor ordinal, we call it a \\term{limit ordinal}. See \\fullref{thm:limit_ordinal_order_topology} for a more involved equivalent condition.\n\n These notions should not be confused with \\hyperref[def:successor_and_limit_cardinal]{successor and weak/strong limit cardinals}.\n\\end{definition}\n\\begin{proof}\n \\ImplicationSubProof{def:successor_and_limit_ordinal/successor}{def:successor_and_limit_ordinal/smaller_successor} If \\( \\alpha = \\op{succ}(\\beta) \\), then \\( \\beta \\) satisfies \\fullref{def:successor_and_limit_ordinal/smaller_successor}.\n\n \\ImplicationSubProof{def:successor_and_limit_ordinal/smaller_successor}{def:successor_and_limit_ordinal/successor} Let \\( \\beta \\in \\alpha \\) be such that \\( \\op{succ}(\\beta) \\not\\in \\alpha \\). Then by trichotomy we have that either \\( \\op{succ}(\\beta) = \\alpha \\) or \\( \\op{succ}(\\beta) > \\alpha \\).\n\n If \\( \\op{succ}(\\beta) > \\alpha \\), then either \\( \\alpha = \\beta \\), which would contradict \\fullref{thm:simple_foundation_theorems/member_of_itself}, or \\( \\alpha \\in \\beta \\), which would contradict \\fullref{thm:simple_foundation_theorems/mutual_membership}.\n\n Thus, it remains for \\( \\op{succ}(\\beta) \\) to be equal to \\( \\alpha \\).\n\n \\ImplicationSubProof{def:successor_and_limit_ordinal/successor}{def:successor_and_limit_ordinal/union} Suppose that \\( \\alpha = \\op{succ}(\\beta) \\).\n\n We have\n \\begin{balign*}\n \\bigcup \\alpha\n &=\n \\bigcup (\\beta \\cup \\set{ \\beta })\n = \\\\ &=\n \\set{ \\gamma \\given \\qexists \\delta (\\delta \\in \\beta \\T{or} \\delta = \\beta) \\T{and} \\gamma \\in \\delta }\n \\reloset {\\eqref{eq:def:semilattice/distributive_lattice/finite/meet_over_join}} = \\\\ &=\n \\set{ \\gamma \\given \\qexists \\delta (\\delta \\in \\beta \\T{and} \\gamma \\in \\delta) \\T{or} (\\delta = \\beta \\T{and} \\gamma \\in \\delta }\n = \\\\ &=\n \\set*{ \\gamma \\given* \\parens[\\Big]{ \\qexists {\\delta \\in \\beta} \\gamma \\in \\delta } \\T{or} \\gamma \\in \\beta }\n = \\\\ &=\n \\parens*{ \\bigcup \\beta } \\cup \\beta\n \\reloset{ \\bigcup \\beta \\subseteq \\beta } = \\\\ &=\n \\beta.\n \\end{balign*}\n\n Thus, \\( \\beta = \\bigcup \\alpha \\in \\alpha \\).\n\n \\ImplicationSubProof{def:successor_and_limit_ordinal/union}{def:successor_and_limit_ordinal/successor} Let \\( \\bigcup \\alpha \\in \\alpha \\). From \\fullref{rem:def:ordinal_successor} we have that\n \\begin{equation*}\n \\op{succ}\\parens*{ \\bigcup \\alpha }\n =\n \\set*{ \\gamma \\given* \\gamma \\subseteq \\bigcup \\alpha }\n =\n \\set{ \\gamma \\given \\gamma \\subsetneq \\alpha }\n =\n \\alpha.\n \\end{equation*}\n\\end{proof}\n\n\\begin{remark}\\label{rem:transfinite_induction}\n It is sometimes efficient to reformulate transfinite induction and recursion. Although analogous principles hold for \\fullref{thm:bounded_transfinite_induction} and \\fullref{thm:bounded_transfinite_recursion}, we will only demonstrate them for \\fullref{thm:transfinite_induction}. The original statement is that in order to prove that some formula is satisfied for all sets, it is sufficient to only prove one inductive step.\n\n More precisely, let \\( \\varphi \\) be a formula in the language of set theory, let \\( \\mscrV = (V, I) \\) be a \\hyperref[rem:standard_model_of_set_theory]{standard} \\hyperref[rem:transitive_model_of_set_theory]{transitive} model of set theory and let \\( v: \\boldop{Var} \\to V \\) be some variable assignment. This allows us to fix any parameters that would otherwise be present in the induction schema. We will say that the set \\( A \\) satisfies \\( \\varphi \\) if \\( \\Bracks{v_{\\xi \\mapsto A}} = T \\).\n\n \\Fullref{thm:transfinite_induction} states that in order to prove that \\( \\varphi \\) holds for any ordinal, the following is sufficient:\n \\begin{thmenum}[series=rem:transfinite_induction]\n \\thmitem{rem:transfinite_induction/single_inductive_step} For every ordinal \\( \\alpha \\), by assuming that every smaller ordinal satisfies \\( \\varphi \\), we must prove that \\( \\alpha \\) does.\n \\end{thmenum}\n\n We have just defined in \\fullref{def:successor_and_limit_ordinal} three mutually exclusive types of ordinals. We can now restate the principles of transfinite induction as follows:\n \\begin{thmenum}[resume=rem:transfinite_induction]\n \\thmitem{rem:transfinite_induction/base_case} In the \\term{base case}, we must prove that \\( 0 \\) satisfies \\( \\varphi \\).\n\n \\thmitem{rem:transfinite_induction/successor_case} In the \\term{successor case}, by assuming that \\( \\alpha \\) satisfies \\( \\varphi \\), we must prove that \\( \\op{succ}(\\alpha) \\) does. It is in line with \\fullref{thm:transfinite_induction} assume that every ordinal smaller that or equal to \\( \\alpha \\) satisfies \\( \\varphi \\), however it is often enough to do, so only for \\( \\alpha \\) itself.\n\n \\thmitem{rem:transfinite_induction/limit_case} In the \\term{limit case}, if \\( \\lambda \\) is a limit ordinal, by assuming that every smaller ordinal satisfies \\( \\varphi \\) we must prove that \\( \\lambda \\) satisfies \\( \\varphi \\).\n \\end{thmenum}\n\n We must note that \\fullref{rem:transfinite_induction/limit_case} is essentially the same as the single inductive step \\fullref{rem:transfinite_induction/single_inductive_step} except that it is restricted to limit ordinals. The reasoning for this is that the proofs for zero and every successor ordinal can be different from those for limit ordinals.\n\n See \\fullref{thm:ordinal_bijection_and_choice} and \\fullref{def:cumulative_hierarchy} for concrete examples.\n\\end{remark}\n\n\\begin{theorem}[Structural recursion]\\label{thm:structural_recursion}\\mimprovised\n The most general recursion principle we will consider is \\term{structural recursion}. It is sometimes also called \\term{mutual recursion}. See \\fullref{thm:well_founded_induction} for its corresponding induction principle.\n\n Let \\( X \\) be a set and let \\( T: \\pow(X) \\to \\pow(X) \\) be a transformation. Then \\( T \\) has a \\hyperref[def:partially_ordered_set_extremal_points/maximum_and_minimum]{smallest} \\hyperref[def:fixed_point]{fixed point}. That is, there exists a unique subset \\( A_0 \\subseteq X \\) such that \\( T(A_0) = A_0 \\) and no proper subset of \\( A_0 \\) has this property.\n\n See \\fullref{rem:structural_recursion_and_induction} for how this theorem is used.\n\\end{theorem}\n\\begin{proof}\n The result follows by applying \\fullref{thm:knaster_tarski_theorem} to the \\hyperref[thm:boolean_algebra_of_subsets]{Boolean algebra of all subsets} of \\( X \\) with \\( R(A) \\coloneqq A \\cup T(A) \\) as the monotone operator.\n\\end{proof}\n\n\\begin{remark}\\label{rem:structural_recursion_and_induction}\\mimprovised\n We will now show the connection between \\fullref{thm:well_founded_induction} and \\fullref{thm:structural_recursion}.\n\n Let \\( \\mscrF \\) be a set of functions, where each function \\( f \\) has a signature \\( f: X^{\\#f} \\to X \\) for some nonnegative integer \\( \\#f \\). Define the operator\n \\begin{equation*}\n \\begin{aligned}\n &T: \\pow(X) \\to \\pow(X) \\\\\n &T(A) \\coloneqq \\set*{ x \\in X \\given[\\Big] \\qexists{f \\in \\mscrF} \\qexists{t_1, \\ldots, t_{\\#f} \\in A_0} f(t_1, \\ldots, t_{\\#f}) = x }.\n \\end{aligned}\n \\end{equation*}\n\n Now we can use \\fullref{thm:structural_recursion} to obtain the smallest fixed point \\( A_0 \\) of \\( T \\). The set \\( A_0 \\) is closed under all the functions in \\( \\mscrF \\), i.e. \\( f\\bracks*{ A_0^{\\#f} } \\subseteq A_0 \\) for any \\( f \\in \\mscrF \\).\n\n Define a binary relation \\( \\to \\) on \\( A_0 \\) by declaring that, for every function \\( f \\) in \\( \\mscrF \\), every sequence \\( x_1, \\ldots, x_{\\#f} \\) and every index \\( k = 1, \\ldots, \\#f \\), we have\n \\begin{equation*}\n x_k \\to f(x_1, \\ldots, x_{\\#f}).\n \\end{equation*}\n\n This relation can be defined by taking unions of smaller relations rather than via recursion.\n\n If the \\hyperref[def:quiver]{quiver} happens to be \\hyperref[def:well_founded_relation]{well-founded}, we can use \\fullref{thm:well_founded_induction} to prove universal statements about \\( A_0 \\).\n\n In \\fullref{subsec:first_order_satisfiability}, for example, we use structural recursion to define first-order substitution in \\fullref{def:first_order_substitution} and then use structural induction to prove certain semantic equivalences like \\fullref{thm:renaming_assignment_compatibility} and \\fullref{thm:first_order_substitution_equivalence}.\n\\end{remark}\n\n\\begin{remark}\\label{rem:induction}\n \\term{Mathematical induction} is a very valuable proof technique for universal statements. The proof of \\fullref{thm:nonzero_natural_numbers_have_predecessors} contains remarks regarding its usage and its difference from deduction principles that are formalized via \\hyperref[def:deductive_system]{deductive systems}. Although it is a logical tool, this remark belongs to this section because it contains several induction and recursion principles.\n\n More generally, given a \\hyperref[def:first_order_syntax/formula]{first-order formula} \\( \\varphi \\) over some \\hyperref[def:first_order_syntax]{first-order language}, certain \\hyperref[def:first_order_theory]{logical theories} allow us to prove indirectly \\( \\qforall \\xi \\varphi[\\xi \\mapsto \\eta] \\) by proving simpler statements (\\fullref{def:peano_arithmetic/PA3} contains very important remarks regarding the free variables of \\( \\varphi \\)). This can be done in cases where every \\hyperref[def:first_order_semantics/satisfiability]{model} \\( X = (X, I) \\) of the theory allows us to exhaust its universe \\( X \\) in a small finite number of steps. We can sometimes use the same steps to instead build objects. The latter principles is called \\term{recursion}.\n\n It should be noted that induction and recursion are used interchangeably in the literature, especially regarding \\hyperref[rem:structural_recursion_and_induction]{structural induction}, however we will aim to distinguish between the two.\n\n Not much more can be said at this level of generality, so we list several induction principles and give examples of their usage:\n \\begin{thmenum}\n \\thmitem{rem:induction/peano_arithmetic} The most basic induction principles is the (weak) natural number induction. It is best described via the axiom schema \\eqref{eq:def:peano_arithmetic/PA3}. \\Fullref{thm:nonzero_natural_numbers_have_predecessors} contains detailed commentary regarding its usage and most of the proofs in \\fullref{subsec:natural_numbers} are performed inductively.\n\n It its set-theoretic form \\fullref{thm:omega_induction} it is important as a tool for introducing a model of Peano arithmetic. It is used directly for proving \\fullref{thm:omega_is_transitive} and \\fullref{thm:omega_recursion}. It is essentially the same as \\fullref{rem:transfinite_induction} without \\fullref{rem:transfinite_induction/limit_case}.\n\n \\Fullref{thm:omega_recursion} is an important standalone tool that allows us to perform recursive definitions for natural numbers. The latter is used, often implicitly, in a great variety of places, from the ability to define natural number operations in \\fullref{def:omega_operations} to the definition of magma exponentiation in \\fullref{def:magma/exponentiation}. \\Fullref{rem:natural_number_recursion} contains notes regarding its practical usage.\n\n \\thmitem{rem:induction/well_founded} A vast generalization of natural number induction is \\fullref{thm:well_founded_induction}. It is stated in a very general setting, but is not frequently used. It can be used to prove \\fullref{thm:epsilon_induction}, which however is even less frequently used. We do not use neither in practice, however the special case where \\( X = \\BbbN \\) is called strong induction on natural numbers.\n\n The usual (weak) natural number induction which is performed by proving the statement for \\( 0 \\) and then proceeding to prove it for \\( n + 1 \\) by assuming that it holds for \\( n \\). Strong induction instead has no base cases and is performed by proving a statement for \\( n \\) by assuming that it holds for all natural numbers strictly smaller than \\( n \\).\n\n Well-founded induction and epsilon-induction have no corresponding recursion principle.\n\n \\thmitem{rem:induction/transfinite} Another vast generalization of natural number induction is \\fullref{thm:bounded_transfinite_induction}, which is further generalized by \\fullref{thm:transfinite_induction}.\n\n Both principles are used to prove fundamental properties of the ordinals. Outside of set theory, transfinite induction is immensely useful, but it is rarely used directly. Instead, it is usually combined with the \\hyperref[def:zfc/choice]{axiom of choice} via \\fullref{thm:zorns_lemma}. \\Fullref{rem:transfinite_induction} contains notes on how it is used directly.\n\n Only bounded transfinite induction has a corresponding recursion principle --- \\fullref{thm:bounded_transfinite_recursion}. Unbounded transfinite induction cannot define such a principle because that would easily lead to \\fullref{thm:burali_forti_paradox}. \\Fullref{rem:unbounded_transfinite_recursion} shows how to circumvent this, however.\n\n Transfinite recursion is used to construct the cumulative hierarchy in \\fullref{def:cumulative_hierarchy}.\n\n \\Fullref{rem:cardinal_transfinite_recursion_and_induction} provides alternative transfinite recursion and induction principles for cardinals rather than for ordinals.\n\n \\thmitem{rem:induction/structural} A very general recursion principle is \\fullref{thm:structural_recursion}. Inductive proofs can be performed via \\fullref{thm:well_founded_induction} sometimes --- this is discussed in \\fullref{rem:structural_recursion_and_induction}.\n\n Within this document, we use \\fullref{thm:structural_induction_on_unambiguous_grammars} in certain special cases, for example \\hyperref[def:propositional_syntax/formula]{propositional} and \\hyperref[def:first_order_syntax/formula]{first-order formulas}.\n \\end{thmenum}\n\\end{remark}\n\n\\begin{proposition}\\label{thm:ordinal_isomorphism}\n Two ordinals are equal if and only if they are order-isomorphic.\n\\end{proposition}\n\\begin{proof}\n \\SufficiencySubProof Trivial.\n\n \\NecessitySubProof Let \\( \\alpha \\) and \\( \\beta \\) be two ordinals. The case \\( \\alpha = \\beta \\) is clear. Without loss of generality, suppose that \\( \\beta < \\alpha \\). \\Fullref{thm:ordinal_ordering_via_subsets} implies that \\( \\beta \\subsetneq \\alpha \\).\n\n Let \\( f: \\alpha \\to \\beta \\) be an order isomorphism. Let \\( \\gamma_0 \\) be the smallest value in \\( \\alpha \\setminus \\beta \\).\n\n From \\fullref{thm:well_ordered_embedding_inflationary} it follows that \\( \\gamma_0 \\leq f(\\gamma_0) \\). But \\( f(\\gamma_0) \\in \\beta \\), hence \\( \\gamma_0 < \\beta \\). But this contradicts our choice of \\( \\gamma_0 \\) as a member of \\( \\alpha \\setminus \\beta \\).\n\n Therefore, \\( \\gamma_0 \\in \\beta \\). Since \\( \\gamma_0 \\) was arbitrary, we conclude that \\( \\beta < \\alpha \\) leads to a contradiction.\n\n We can obtain an analogous contradiction for \\( \\alpha < \\beta \\), hence it only remains possible for \\( \\alpha \\) and \\( \\beta \\) to be equal.\n\\end{proof}\n\n\\begin{theorem}[Hartogs' lemma]\\label{thm:hartogs_lemma}\n For every set \\( A \\) there exists a smallest ordinal \\( \\alpha \\) such that no function from \\( \\alpha \\) to \\( A \\) is injective.\n\\end{theorem}\n\\begin{proof}\n Define the set\n \\begin{equation*}\n W \\coloneqq \\set{ (P, \\leq) \\given P \\subseteq A \\T*{and} \\leq \\T*{well-orders} P }.\n \\end{equation*}\n\n Let \\( \\beta \\) be an ordinal for which there exists an injective function \\( f: \\beta \\to A \\). At least one such pair of a ordinal and function exists because \\( \\varnothing: 0 \\to A \\) is an injective function.\n\n The image of \\( f \\) can be well-ordered by\n \\begin{equation*}\n f(\\delta) < f(\\gamma) \\T{if and only if} \\delta \\in \\gamma,\n \\end{equation*}\n where \\( \\delta \\) and \\( \\gamma \\) are members of \\( \\beta \\). Thus, if we restrict the range of \\( f \\) to its image, it would be an explicit order isomorphism of \\( (\\beta, \\in) \\) and \\( (\\img f, <) \\).\n\n We have shown that every ordinal \\( \\beta \\) and every injective function \\( f: \\beta \\to A \\) induces a well-ordered set which belongs to \\( W \\). Furthermore, if \\( f_1: \\beta_1 \\to A \\) and \\( f_2: \\beta_2 \\to A \\) are two such injective functions and if the induced well-ordered sets \\( (\\img f_1, \\leq_1) \\) and \\( (\\img f_2, \\leq_2) \\) are order-isomorphic, then \\( \\beta_1 \\) and \\( \\beta_2 \\) are themselves order-isomorphic and thus \\( \\beta_1 = \\beta_2 \\) by \\fullref{thm:ordinal_isomorphism}.\n\n Therefore, for any well-ordered set in \\( W \\) there exists at most one ordinal that induces it via some function. Let \\( W' \\) be the subset of all well-ordered sets in \\( W \\) induced by exactly one ordinal.\n\n We have that \\( W' \\) is a set and for each member of \\( W' \\) there corresponds exactly one ordinal. It follows from the \\hyperref[def:zfc/replacement]{axiom schema of replacement} that these ordinals form a set. Denote this set by \\( B \\).\n\n There is no set of all ordinals by \\fullref{thm:burali_forti_paradox}, hence there must exist an ordinal not in \\( B \\). But every ordinal that has an injective function into \\( A \\) belongs to \\( B \\). Hence, there exists some ordinal \\( \\alpha \\) such that no function from \\( \\alpha \\) to \\( A \\) is injective.\n\n If \\( \\alpha \\) is not the smallest ordinal with this property, we can now easily take the smallest member of \\( \\alpha \\) with this property.\n\\end{proof}\n\n\\begin{lemma}\\label{thm:ordinal_bijection_and_choice}\\mcite{MathSE:order_type_existence}\n Let \\( A \\) be a nonempty set and let \\( c \\) be a \\hyperref[def:choice_function]{choice function} on \\( A \\). Then there exists an ordinal \\( \\alpha \\) and a bijection between \\( A \\) and \\( \\alpha \\).\n\\end{lemma}\n\\begin{proof}\n We will explicitly build the desired ordinal. We will use \\fullref{thm:bounded_transfinite_recursion} in its structured form presented in \\fullref{rem:transfinite_induction} to build a \\hyperref[def:transfinite_sequence]{transfinite sequence} of injective maps from ordinals into \\( A \\).\n\n Let \\( \\delta \\) be the smallest ordinal such that no function from \\( \\delta \\) to \\( A \\) is injective. Such an ordinal exists by \\fullref{thm:hartogs_lemma}. Note that \\( \\delta \\) cannot be zero because the empty function is always injective.\n\n Any smaller than \\( \\delta \\) ordinal has at least one injective function into \\( A \\). We will use recursion bounded by \\( \\delta \\) to construct a \\hyperref[def:transfinite_sequence]{transfinite sequence} \\( \\seq{ \\iota_\\beta }_{\\beta < \\delta} \\) satisfying the invariant that for any \\( \\beta \\in \\delta \\), the function \\( \\iota_\\beta \\) is injective and strictly contains \\( \\iota_\\gamma \\) for any \\( \\gamma < \\beta \\).\n\n \\begin{itemize}\n \\item The zero ordinal has only one possible injective function into \\( A \\), namely the empty function \\( \\varnothing: 0 \\to P \\).\n\n \\item Now let \\( \\iota_\\beta: \\beta \\to A \\) be an injective function.\n\n If \\( \\iota_\\beta \\) is surjective, then it is the desired bijection and the rest of the construction is irrelevant.\n\n If \\( \\iota_\\beta \\) is not surjective, define\n \\begin{equation*}\n \\begin{aligned}\n &\\iota_{\\op{succ}(\\beta)}: \\op{succ}(\\beta) \\to P \\\\\n &\\iota_{\\op{succ}(\\beta)}(\\gamma) \\coloneqq \\begin{cases}\n \\iota_\\beta(\\gamma), &\\gamma < \\beta \\\\\n c(A \\setminus \\img(\\iota_\\beta)), &\\gamma = \\beta.\n \\end{cases}\n \\end{aligned}\n \\end{equation*}\n\n This function is clearly injective.\n\n \\item Let \\( \\lambda \\) be a limit ordinal and let \\( \\iota_\\beta \\) be an injective function for any \\( \\beta < \\lambda \\). Note that the domain of \\( \\iota_\\beta \\) is not necessarily \\( \\beta \\) --- it may be a smaller ordinal \\( \\gamma \\) in case \\( \\iota_\\gamma \\) is surjective (and thus the desired bijection).\n\n In any case, we have a family \\( \\set{ \\iota_\\beta }_{\\beta < \\lambda} \\) of functions such that \\( \\iota_\\gamma \\subseteq \\iota_\\beta \\) if and only if \\( \\gamma < \\beta \\). We simply take their union\n \\begin{equation*}\n \\iota_\\lambda \\coloneqq \\bigcup \\set{ \\iota_\\beta \\given \\beta < \\lambda }.\n \\end{equation*}\n\n This function is injective because each of the constituent functions it injective.\n \\end{itemize}\n\n We will now thoroughly build the order isomorphism itself.\n\n Let \\( \\alpha \\leq \\delta \\) be the (nonstrict) initial segment of \\( \\delta \\) such that \\( \\iota_\\beta \\) is fully defined on \\( \\beta \\) for every \\( \\beta \\in \\alpha \\). More precisely, let\n \\begin{equation*}\n \\alpha \\coloneqq \\set{ \\beta \\in \\delta \\given \\dom(\\iota_\\beta) = \\beta }.\n \\end{equation*}\n\n We will show that \\( \\alpha \\) is a successor ordinal. Note that \\( \\alpha \\) cannot be zero because \\( \\dom(\\iota_0) = \\varnothing = 0 \\). Aiming at a contradiction, assume that \\( \\alpha \\) is a limit ordinal.\n\n Since for every \\( \\beta \\in \\alpha \\) the value \\( \\iota_{\\op{succ}(\\beta)}(\\beta) \\) is defined, we can conclude that the union\n \\begin{equation*}\n \\bigcup\\set{ \\iota_\\beta \\given \\beta \\in \\alpha }\n \\end{equation*}\n is an injective function from \\( \\alpha \\) to \\( A \\).\n\n If \\( \\alpha = \\delta \\), this contradicts our choice of \\( \\delta \\). If \\( \\alpha \\in \\delta \\), this means that the function \\( \\iota_\\alpha \\) is equal to \\( \\iota_\\gamma \\) for some \\( \\gamma < \\alpha \\). But then \\( \\iota_{\\op{succ}(\\gamma)} \\) is not defined on the entirely of \\( \\gamma \\). This implies that \\( \\gamma \\geq \\alpha \\), which contradicts our assumption that \\( \\gamma < \\alpha \\).\n\n Therefore, \\( \\alpha \\) cannot be a limit ordinal. It remains for \\( \\alpha \\) to be a successor ordinal. Then there exists \\( \\beta \\) such that \\( \\alpha = \\op{succ}(\\beta) \\).\n\n Suppose that \\( \\iota_\\beta \\) is not surjective. We can then construct \\( \\iota_{\\op{succ}(\\beta)} \\) as in the corresponding recursion step. This will be an injective function from \\( \\alpha = \\op{succ}(\\beta) \\) to \\( A \\). But then we would have \\( \\alpha \\in \\alpha \\), which contradicts \\fullref{thm:simple_foundation_theorems/member_of_itself}.\n\n Therefore, \\( \\iota_\\beta \\) is a surjective function.\n\\end{proof}\n\n\\begin{theorem}\\label{thm:well_ordered_order_type_existence}\n Any \\hyperref[def:well_ordered_set]{well-ordered set} \\( (P, \\leq) \\) is \\hyperref[def:partially_ordered_set/homomorphism]{order-isomorphic} to a unique ordinal. This ordinal is called the \\term{order type} of \\( (P, \\leq) \\) and is denoted by \\( \\ord(P, \\leq) \\) or simply \\( \\ord(P) \\).\n\n Furthermore, this isomorphism is unique.\n\\end{theorem}\n\\begin{proof}\n For \\( P = \\varnothing \\), then the empty function \\( \\iota: 0 \\to P \\) is the desired isomorphism.\n\n We use \\fullref{thm:ordinal_bijection_and_choice} on \\( P \\) with the choice function\n \\begin{equation*}\n \\begin{aligned}\n &c: \\pow(P) \\setminus \\set{ \\varnothing } \\to P \\\\\n &c(B) \\coloneqq \\min B\n \\end{aligned}\n \\end{equation*}\n to obtain an ordinal \\( \\alpha \\) and a bijection \\( \\iota: \\alpha \\to P \\).\n\n The function \\( \\iota \\) is directly verified to be an order embedding in each of the recursion cases. Therefore, \\( \\iota \\) is a bijective order homomorphism and it follows from \\fullref{thm:totally_ordered_strict_isomorphisms} that \\( \\iota \\) is an order isomorphism.\n\n Uniqueness of \\( \\iota \\) follows from \\fullref{thm:well_ordered_isomorphism_is_unique}.\n\\end{proof}\n\n\\begin{theorem}[Zermelo's well-ordering theorem]\\label{thm:well_ordering_theorem}\\mcite[thm. 69.6]{OpenLogicFull}\n Any \\hyperref[def:set]{set} can be \\hyperref[def:well_ordered_set]{well-ordered}.\n\n Within \\hyperref[def:zfc]{\\logic{ZF}}, this theorem is equivalent to the \\hyperref[def:zfc/choice]{axiom of choice} --- see \\fullref{thm:axiom_of_choice_equivalences/well_ordering}.\n\\end{theorem}\n\\begin{proof}\n \\ImplicationSubProof[def:zfc/choice]{the axiom of choice}[thm:well_ordering_theorem]{well-ordering theorem} The empty set is trivially well-ordered.\n\n Let \\( A \\) be any nonempty set. By the axiom of choice, there exists a \\hyperref[def:choice_function]{choice function} \\( c \\) for \\( A \\). We use \\fullref{thm:ordinal_bijection_and_choice} on \\( A \\) and \\( c \\) to obtain an ordinal \\( \\alpha \\) and a bijection \\( \\iota: \\alpha \\to A \\). Since \\( \\alpha \\) itself is well-ordered, then the relation\n \\begin{equation*}\n x <_A y \\T{if and only if} \\iota^{-1}(x) <_\\alpha \\iota^{-1}(y)\n \\end{equation*}\n defines a well-order on \\( A \\).\n\n \\ImplicationSubProof[thm:well_ordering_theorem]{well-ordering theorem}[def:zfc/choice]{axiom of choice} Let \\( A \\) be any set and suppose that \\( < \\) well-orders \\( A \\).\n\n Define the function\n \\begin{equation*}\n \\begin{aligned}\n &c: \\pow(A) \\setminus \\set{ \\varnothing } \\to A \\\\\n &c(B) \\coloneqq \\min B.\n \\end{aligned}\n \\end{equation*}\n\n In the case where \\( A \\) is the empty set, \\( c \\) is the empty function.\n\n It is clear that \\( c(B) \\in B \\) for every subset \\( B \\) of \\( A \\). Therefore, \\( c \\) is a choice function for \\( A \\).\n\\end{proof}\n", "meta": {"hexsha": "35af7adf56ce41e2fa156c5acb1ed9f84f92873b", "size": 62119, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/ordinals.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/ordinals.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/ordinals.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 74.1276849642, "max_line_length": 794, "alphanum_fraction": 0.6918334165, "num_tokens": 18529, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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limits\\color{black}\\_\\{\\}\n\\item \\color{nounibaredI}\\textbackslash dots\\color{black}\n\\item \\color{nounibaredI}\\textbackslash cdot\\color{black}\n\\item \\color{nounibaredI}\\_\\color{black}\n\\item \\color{nounibaredI}\\^~\\color{black}\n\\end{itemize}\n\\end{multicols}\n\\end{block}\n\n\\end{frame}\n\n%-------------------------------------------------------------------------------\n\\begin{frame}\n\\frametitle{Formulas}\n\\framesubtitle{\\ldots ~A Marvel Of Beauty In \\LaTeX !}\n\n\\begin{columns}\n\\begin{column}{.3\\textwidth}\n{\\huge $2 \\sqrt{\\frac{\\pi ^2}{3}\\cdot c_{2}}$}\n\\end{column}\n\n\\begin{column}{.7\\textwidth}\n\t$\\underbrace{\n\t\\color{unibayellowI}\\text{\\$}\n\t\\color{black}2\n\t\\color{nounibaredI}\\backslash \\text{sqrt}\n\t\\color{black}\\{\n\t\\color{nounibaredI}\\backslash \\text{frac}\n\t\\color{black}\\{\n\t\\color{nounibaredI}\\backslash \\text{pi}\\color{nounibaredI}\n\t~\\hat{}~\\color{black}2\\}\\{3\\color{black}\\}\n\t\\color{nounibaredI}\\backslash\n\t\\color{nounibaredI}\\text{cdot}~\n\t\\color{black} \\text{c}\n\t\\color{nounibaredI}\\_\n\t\\color{black}2\\}\n\t\\color{unibayellowI}\\text{\\$}\n}$\n\\color{black}\n\nThe formula-environment begins and ends with \\color{unibayellowI}\\$ \\color{black} .\n\n\\medskip\n\t$\\underbrace{\n\t\\color{nounibaredI}\\backslash \\text{sqrt}\n\t\\color{black}\\{\n\t\\color{nounibaredI}\\backslash \\text{frac}\n\t\\color{black}\\{\n\t\\color{nounibaredI}\\backslash \\text{pi}\\color{nounibaredI}\n\t~\\hat{}~\\color{black}2\\}\\{3\\color{black}\\}\n\t\\color{nounibaredI}\\backslash\n\t\\color{nounibaredI}\\text{cdot}~\n\t\\color{black} \\text{c}\n\t\\color{nounibaredI}\\_\n\t\\color{black}2\\}\n}$\n\\color{black}\n\n\nThis whole part is the radical.\n\n\\bigskip\n$\\underbrace{\n\t\\color{nounibaredI}\\backslash \\text{frac}\n\t\\color{black}\\{\n\t\\color{nounibaredI}\\backslash \\text{pi}\\color{nounibaredI}\n\t~\\hat{}~\\color{black}2\\}\\{3\\color{black}\\}\n}$\n\nA fraction has always a numerator and a denominator.\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n%-------------------------------------------------------------------------------\n\n\\begin{frame}\n\\frametitle{Formulas}\n\\framesubtitle{\\ldots ~A Marvel Of Beauty In \\LaTeX !}\n\\begin{columns}\n\t\\begin{column}{.4\\textwidth}\n\t\t\\flushright\n\t\t$\\int_0^\\infty$\n\t\\end{column}\n\t\\begin{column}{.6\\textwidth}\n\t\t\\flushleft\n\t\t{\\ttfamily\\color{unibayellowI}\\$\\color{nounibaredI}\\textbackslash\\color{nounibaredI}int\\_\\color{black}0\\color{nounibaredI}\\textasciicircum \\textbackslash infty\\color{unibayellowI}\\$}\n\t\\end{column}\n\\end{columns}\n\\begin{columns}\n\t\\begin{column}{.4\\textwidth}\n\t\t\\flushright\n\t\t$\\sum_{i=1}^n$\n\t\\end{column}\n\t\\begin{column}{.6\\textwidth}\n\t\t\\flushleft\n\t\t{\\ttfamily \\color{unibayellowI}\\$\\color{nounibaredI}\\textbackslash\n\t\t\t\\color{nounibaredI}sum\\_\\color{black}\\{i=1\\}\\color{nounibaredI}\\textasciicircum\n\t\t\t\\color{black}n\\color{unibayellowI}\\$}\n\t\\end{column}\n\\end{columns}\n\n\\begin{columns}\n\t\\begin{column}{.4\\textwidth}\n\t\t\\flushright\n\t\t$\\lim_{n \\rightarrow \\infty}$\n\t\\end{column}\n\t\\begin{column}{.6\\textwidth}\n\t\t\\flushleft\n\t\t{\\ttfamily \\color{unibayellowI}\\$\\color{nounibaredI}\\textbackslash\n\t\t\t\\color{nounibaredI}lim\\_\\color{black}\\{n \\color{nounibaredI}\\textbackslash\n\t\t\t\\color{nounibaredI}rightarrow \\color{nounibaredI}\\textbackslash infty\\color{black}\\}\\color{unibayellowI}\\$}\n\t\\end{column}\n\\end{columns}\n\n\\begin{columns}\n\t\\begin{column}{.4\\textwidth}\n\t\t\\flushright\n\t\t$\\prod\\limits_{i=1}^{n+1}i = 1 \\cdot 2 \\cdot \\ldots \\cdot n \\cdot (n+1)$\n\t\\end{column}\n\t\\begin{column}{.6\\textwidth}\n\t\t\\flushleft\n\t\t{\\ttfamily \\color{unibayellowI}\\$%\n\t\t\t\\color{nounibaredI}\\textbackslash\\color{nounibaredI}prod\\textbackslash limits\\_\\color{black}\\{i=1\\}\\color{nounibaredI}\\^{}\\color{black}\\{n+1\\} i = 1 \\color{nounibaredI}\\textbackslash \\color{nounibaredI}cdot \\color{black}2 \\color{nounibaredI}\\textbackslash \\color{nounibaredI}cdot \\color{nounibaredI}\\textbackslash \\color{nounibaredI}ldots \\color{nounibaredI}\\textbackslash\\color{nounibaredI}cdot \\color{black}n \\color{nounibaredI}\\textbackslash \\color{nounibaredI}cdot \\color{black}(n+1)\\color{unibayellowI}\\$}\n\t\\end{column}\n\\end{columns}\n\\bigskip\nThe American Mathematical Society has a wonderful guide for the {\\ttfamily amsmath}-package.\\footnote{ftp://ftp.ams.org/pub/tex/doc/amsmath/amsldoc.pdf}\n\\end{frame}", "meta": {"hexsha": "6cfb40060899f75c8c7ec2190925393651977668", "size": 4899, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "200+ beamer 模板合集/LaTeX-Tutorial-master(培训课件)/beamer/content/formulas.tex", "max_stars_repo_name": "lemoxiao/Awesome-Beamer-Collection", "max_stars_repo_head_hexsha": "3ab28a23fb60cb0a97fcec883847e2d8728b98c0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 13, "max_stars_repo_stars_event_min_datetime": "2019-07-30T04:09:54.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-24T09:27:26.000Z", "max_issues_repo_path": "200+ beamer 模板合集/LaTeX-Tutorial-master(培训课件)/beamer/content/formulas.tex", "max_issues_repo_name": "lemoxiao/Awesome-Beamer-Collection", "max_issues_repo_head_hexsha": "3ab28a23fb60cb0a97fcec883847e2d8728b98c0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "200+ beamer 模板合集/LaTeX-Tutorial-master(培训课件)/beamer/content/formulas.tex", "max_forks_repo_name": "lemoxiao/Awesome-Beamer-Collection", "max_forks_repo_head_hexsha": "3ab28a23fb60cb0a97fcec883847e2d8728b98c0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2019-11-02T03:10:26.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-12T04:13:23.000Z", "avg_line_length": 31.8116883117, "max_line_length": 514, "alphanum_fraction": 0.7109614207, "num_tokens": 1804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.7981867705385762, "lm_q1q2_score": 0.6247108364971615}} {"text": "\\chapter{ELEC 360 CheatSheet}\nSome keywords used can be found in the glossary section including:\n\\begin{itemize}\n\t\\item \\gls{partFrac}, \\gls{openLoop}, \\gls{SISO}, \\gls{MIMO}, and \\gls{LTI}\n\t\\item \\gls{conSys}, \\gls{closedLoop}, \\gls{DC Motors}, and \\gls{Op Amps}\n\\end{itemize}\n\n\\begin{multicols}{3}\n\n%\\includegraphics[width=1.77165in,height=0.95669in]{media/image1.png}\n\n\\section{LAPLACE TRANSFORMS}\n\nFinal Value Theorem\n\nIn Control Engineering, the Final Value Theorem is used most frequently to determine the steady-state value of a system. The real part of the poles of the function must be < 0.\n\\begin{align}\n& \\lim\\limits_{t \\rightarrow \\infty} x(t) = \\lim\\limits_{t \\rightarrow 0} sX(s)\n\\end{align}\n%\\includegraphics[width=1.77165in,height=0.92126in]{media/image2.png}\n\n%\\includegraphics[width=1.77165in,height=1.14961in]{media/image3.png}\n\nInitial Value Theorem\n\n\\begin{align}\n& \\lim\\limits_{t \\rightarrow 0} x(t) = \\lim\\limits_{t \\rightarrow \\infty} sX(s)\n\\end{align}\n%\\includegraphics[width=1.77165in,height=0.90551in]{media/image4.png}\n\n%\\includegraphics[width=1.77165in,height=0.77953in]{media/image5.png}\n\n\\section{SOLUTION OF LINEAR DIFFERENTIAL EQUATION}\n\n\\[A\\ddot{y} + B\\dot{y} + Cy = u\\left( t \\right)\\]\n\n\\[\\text{with initial conditions }\\dot{y}\\left( 0 \\right)\\ and \\ y(0)\\]\n\n\\[\\text{is solved by constructing the equation}\\]\n\n%\\[A\\left\\lbrack s^{2}Y\\left( s \\right) - sy\\left( 0 \\right) - \\dot{y}\\left( 0 \\right) \\right\\rbrack + B\\left\\lbrack \\text{sY}\\left( s \\right) - y\\left( 0 \\right) \\right\\rbrack + CY(s)\\]\n\n\\begin{align*}\n &A\\left\\lbrack s^{2}Y\\left( s \\right) - sy\\left( 0 \\right) - \\dot{y}\\left( 0 \\right) \\right\\rbrack + B\\left\\lbrack \\text{sY}\\left( s \\right) - y\\left( 0 \\right) \\right\\rbrack \\\\\n & + CY(s)\n\\end{align*}\n\\[that\\ is\\ to\\ say:\\]\n\n\\[\\mathcal{L}\\left\\{ \\ddot{y} \\right\\} = s^{2}Y\\left( s \\right) - sy\\left( 0 \\right) - \\dot{y}(0)\\]\n\n\\[\\text{and}\\]\n\n\\[\\mathcal{L}\\left\\{ \\dot{y} \\right\\} = sY\\left( s \\right) - y\\left( 0 \\right)\\]\n\n\\section{STATESPACE REPRESENTATIONS}\n\nTo generate from a differential equation:\n\nUse the \\textbf{closed loop transfer function}\n\n\\[\\frac{Y\\left( s \\right)}{U\\left( s \\right)} = \\frac{s + A}{s^{3} + \\text{Bs}^{2} + s + A}\\]\n\nSeparate and take the inverse Laplace transform\n\n\\begin{align*} &s^{3}Y\\left( s \\right) + \\text{Bs}^{2}Y\\left( s \\right) + sY\\left( s \\right) + AY\\left( s \\right) \\\\\n &= sU\\left( s \\right) + AU\\left( s \\right)\n \\end{align*}\n\n\\[= \\dddot{y} + B\\ddot{y} + \\dot{y} + Ay = \\dot{u} + Au\\]\n\nThen define state variables:\n\n\\(x_{1} = y\\) \\(\\dot{x_{1}} = \\dot{y} = x_{2}\\)\n\n\\(x_{2} = \\dot{y}\\) \\(\\dot{x_{2}} = \\ddot{y} = x_{3}\\)\n\n\\(x_{3} = \\ddot{y}\\)\n\\(\\dot{x_{3}} = \\dddot{y} = \\dot{u} + Au - B\\ddot{y} - \\dot{y} - Ay\\)\n\nThen construct the state space matrix\n\n\\[\\begin{bmatrix}\n\\dot{x_{1}} \\\\\n\\dot{x_{2}} \\\\\n\\dot{x_{3}} \\\\\n\\end{bmatrix} = \\begin{bmatrix}\n0 & 1 & 0 \\\\\n0 & 0 & 1 \\\\\n - A & - 1 & - B \\\\\n\\end{bmatrix}\\begin{bmatrix}\nx_{1} \\\\\nx_{2} \\\\\nx_{3} \\\\\n\\end{bmatrix} + \\begin{bmatrix}\n\\beta_{1} \\\\\n\\beta_{2} \\\\\n\\beta_{3} \\\\\n\\end{bmatrix}u\\]\n\nand\n\n\\[y = \\begin{bmatrix}\n1 & 0 & 0 \\\\\n\\end{bmatrix}\\begin{bmatrix}\nx_{1} \\\\\nx_{2} \\\\\nx_{3} \\\\\n\\end{bmatrix} + \\beta_{0}u\\]\n\n\\(\\beta\\) values can be calculated as follows:\n\n\\[\\beta_{0} = b_{0}\\]\n\n\\[\\beta_{1} = b_{1} - a_{1}\\beta_{0}\\]\n\n\\[\\beta_{2} = b_{2} - a_{1}\\beta_{1} - a_{2}\\beta_{0}\\]\n\n\\[\\beta_{3} = b_{3} - a_{1}\\beta_{2} - a_{2}\\beta_{1} - a_{3}\\beta_{0}\\]\n\nWhere the values for \\(a_{x}\\text{\\ and\\ }b_{x}\\)come from:\n\n\\[\\dddot{y} + a_{1}\\ddot{y} + a_{2}\\dot{y} + a_{3}y = b_{0}\\dddot{u} + b_{1}\\ddot{u} + b_{2}\\dot{u} + b_{3}u\\]\n\n\n\\[\\ddot{x}=(F-c\\dot{x_{1}}-kx)/m\\]\nwhere\n\tx=position,$\\dot{x}$=speed/velocity,$\\ddot{x}$=acceleration\n\tc = damping constant,\n\tm = mass,\n\tF = force\n\tk = spring constant,\n\nTo perform inverse (find transfer function from statespace model)\n\n\\[G\\left( s \\right) = d + c{(sI - A)}^{- 1}b\\]\n\n\nwhere\n\n\\[\\begin{bmatrix}\n\\dot{x_{1}} \\\\\n\\dot{x_{2}} \\\\\n\\dot{x_{3}} \\\\\n\\end{bmatrix} = \\begin{bmatrix}\nA_{11} & A_{12} & A_{13} \\\\\nA_{21} & A_{22} & A_{23} \\\\\nA_{31} & A_{32} & A_{33} \\\\\n\\end{bmatrix}\\begin{bmatrix}\nx_{1} \\\\\nx_{2} \\\\\nx_{3} \\\\\n\\end{bmatrix} + \\begin{bmatrix}\nb_{1} \\\\\nb_{2} \\\\\nb_{3} \\\\\n\\end{bmatrix}u\\]\n\nAnd\n\n\\[y = \\begin{bmatrix}\nc_{1} & c_{2} & c_{3} \\\\\n\\end{bmatrix}\\begin{bmatrix}\nx_{1} \\\\\nx_{2} \\\\\nx_{3} \\\\\n\\end{bmatrix} + \\lbrack d\\rbrack u\\]\n\nThe inverse of a square 2x2 matrix is found by:\n\n\\[\\begin{bmatrix}\na & b \\\\\nc & d \\\\\n\\end{bmatrix}^{- 1} = \\frac{1}{ad - bc}\\begin{bmatrix}\nd & - b \\\\\n - c & a \\\\\n\\end{bmatrix}\\]\n\nThe inverse of a square 3x3 matrix is found by:\n%\\includegraphics[width=2.52147in,height=1.02367in]{media/image8.png}\n\nFor a 3×3 matrix\n$$A=\\begin{bmatrix}\n\ta_{11} & a_{12} & a_{13} \\\\\n\ta_{21} & a_{22} & a_{23}\\\\\n\ta_{31} & a_{32} & a_{33}\n\\end{bmatrix}$$\t\nthe matrix inverse is:\n\n$$A^{-1}= \\frac{1}{|A|}\\begin{bmatrix}\n\\begin{bmatrix} a_{22} & a_{23} \\\\ a_{32} & a_{33}\\end{bmatrix} & \\begin{bmatrix} a_{13} & a_{12} \\\\ a_{33} & a_{32}\\end{bmatrix} & \\begin{bmatrix} a_{12} & a_{13} \\\\ a_{22} & a_{23}\\end{bmatrix} \\\\\n\\begin{bmatrix} a_{23} & a_{21} \\\\ a_{33} & a_{31}\\end{bmatrix}& \\begin{bmatrix} a_{11} & a_{13} \\\\ a_{31} & a_{33}\\end{bmatrix} & \\begin{bmatrix} a_{13} & a_{11} \\\\ a_{23} & a_{21}\\end{bmatrix} \\\\\n\\begin{bmatrix} a_{21} & a_{22} \\\\ a_{31} & a_{32}\\end{bmatrix} & \\begin{bmatrix} a_{12} & a_{11} \\\\ a_{32} & a_{31}\\end{bmatrix} & \\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22}\\end{bmatrix}\t\n\\end{bmatrix}$$\nSolving a three order polynomial, without a fancy calculator\n\\begin{align*}\n& x = \\left[q + \\left[ q^2 + (r-p^2)^3\\right ]^{1/2} \\right]^{1 /3} \\\\\n& \\left[q - \\left[ q^2 + (r-p^2)^3\\right ]^{1/2} \\right]^{1 /3} + p\n\\end{align*}\nWhere \n\\begin{align*}\n& p = \\frac{-b}{3a}, \\ \\ q =p^3 + \\frac{bc-3ad}{3a^2}, \\ \\ r =\\frac{c}{3a}\n\\end{align*}\n\\section{SECOND ORDER SYSTEMS}\n\n\\[G\\left( s \\right) = \\frac{C(s)}{R(s)} = \\frac{\\omega_{n}^{2}}{s^{2} + 2\\zeta\\omega_{n}s + \\omega_{n}^{2}}\\]\n\n\\[K = \\omega_{n}^{2};\\ \\ \\ \\ T = 2\\zeta\\omega_{n} = 2\\sigma;\\ \\ \\ \\ \\ \\zeta = \\frac{T}{2\\sqrt{K}};\\ \\ \\ \\ \\omega_{d} = \\omega_{n}\\sqrt{1 - \\zeta^{2}}\\]\n\n\\[\\zeta = damping\\ ratio; \\sigma = real\\ part\\ of\\ root;\\]\n\n\\(\\ \\ \\ \\omega_{d} = damped\\ natural\\ frequency\\)\n\n\\( omega_{n} = undamped\\ natural\\ frequency \\)\n%\\includegraphics[width=1.96736in,height=1.37361in]{media/image9.png}\n\n\\[ Undamped:\\ \\zeta = 0;\\] \n\\[ Critically\\ Damped: \\zeta = 1 \\] \n\\[ Over\\ Damped:\\ \\zeta > 1\\]\n\nImaginary axis:\n\nFrequency of oscillations\n\nReal axis:\n\nDecay time\n\n\\textbf{UNIT STEP RESPONSE OF A 2\\textsuperscript{ND} ORDER UNDAMPED\nSYSTEM}\n\n$ t_{d}$ = delay time - to reach 50\\% of $c\\left( \\infty \\right) $\\text{for the first time}.\n$ t_{r}$ = rise time - time to reach 100 \\% of $c\\left( \\infty \\right)$ \\text{for first time} . \n$ t_{p}$ = peak time - time to reach first peak. \n$ t_{s}$ = settling time - time to reach \\& stay within 2\\% or 5\\% \n$ M_{p}$ = maximum overshoot. \n%\\[{t_{d} = delay\\ time - to\\ reach\\ 50\\%\\ of\\ c\\left( \\infty \\right)\\text{for\\ the\\ first\\ time}\\backslash n}{t_{r} = rise\\ time - time\\ to\\ reach\\ 100\\%\\ of\\ c\\left( \\infty \\right)\\text{for\\ first\\ time}\\backslash n}{t_{p} = peak\\ time - time\\ to\\ reach\\ first\\ peak\\backslash n}{t_{s} = settling\\ time - time\\ to\\ reach\\ \\&\\ stay\\ within\\ 2\\%\\ or\\ 5\\%\\backslash n}{M_{p} = maximum\\ overshoot\\ \\left( \\% \\right)\\backslash n}\\]\n\\[{t_{r} = \\frac{1}{\\omega_{d}}\\operatorname{}\\left( - \\frac{\\omega_{d}}{\\sigma} \\right);\\ \\ \\ \\ t_{p} = \\frac{\\pi}{\\omega_{d}}\\backslash n}\\]\n\n\\[{M_{p} = e^{- \\frac{\\zeta\\omega_{n}\\pi}{\\omega_{d}}} = e^{- \\frac{\\eta \\pi}{\\sqrt{1 - \\zeta^{2}}}} = e^{- \\frac{\\sigma \\pi}{\\omega_{d}}}}\\]\n\n\\[t_{s} = \\frac{4}{\\sigma} = \\frac{4}{\\zeta\\omega_{n}}\\ \\left( 2\\%\\ band \\right)\\]\n\\[t_{s} = \\frac{3}{\\sigma} = \\frac{3}{\\zeta\\omega_{n}}\\ \\left( 5\\%\\ band \\right)\\]\n\nDominant poles are the ones closest to the imaginary axis\n\n\\section{ROUTH-HURWITZ STABILITY TEST}\n\\[a_0s^n + a_1s^{n-1}+ \\cdots + a{n-1} s + a_n = 0\\]\n\\[ \\begin{array}{lllll}\n\\mbox{row n} & a_0 & a_2 & a_4 & \\cdots \\\\\n\\mbox{row n-1} & a_1 & a_3 & a_5 & \\cdots \\\\\n\\mbox{row n-2} & b_1 & b_2 & b_3 & \\cdots \\\\\n\\mbox{row n-3} & c_1 & c_2 & c_3 & \\cdots \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\n\\mbox{row 2} & * & * & & \\cdots \\\\\n\\mbox{row 1} & * & & & \\cdots \\\\\n\\mbox{row 0} & * & & & \\cdots \\end{array} \\]\n\n\\[ b_1=-\\frac{det\\left[\\begin{array}{cc}a_0&a_2\\\\a_1&a_3\\end{array}\\right]}{a_1}\n=\\frac{a_1a_2-a_0a_3}{a_1} \\]\n\\[ b_2=-\\frac{det\\left[\\begin{array}{cc}a_0&a_4\\\\a_1&a_5\\end{array}\\right]}{a_1}\n=\\frac{a_1a_4-a_0a_5}{a_1} \\]\n\\[ b_3=-\\frac{det\\left[\\begin{array}{cc}a_0&a_6\\\\a_1&a_7\\end{array}\\right]}{a_1}\n=\\frac{a_1a_6-a_0a_7}{a_1} \\]\n\n\\[ c_1=-\\frac{det\\left[\\begin{array}{cc}a_1&a_3\\\\b_1&b_2\\end{array}\\right]}{b_1}\n=\\frac{b_1a_3-a_1b_2}{b_1} \\]\n\\[ c_2=-\\frac{det\\left[\\begin{array}{cc}a_1&a_5\\\\b_1&b_3\\end{array}\\right]}{b_1}\n=\\frac{b_1a_5-a_1b_3}{b_1} \\]\n\\[ c_3=-\\frac{det\\left[\\begin{array}{cc}a_1&a_7\\\\b_1&b_4\\end{array}\\right]}{b_1}\n=\\frac{b_1a_7-a_1b_4}{b_1} \\]\n\\section{STEADY STATE ERROR ANALYSIS}\n\n\\[K_{p} = \\operatorname{}{G\\left( s \\right)H(s)}\\]\n\n\\[K_{v} = \\operatorname{}{\\text{sG}\\left( s \\right);\\ \\ \\ \\ K_{v} = \\operatorname{}{s\\left( \\text{KG}\\left( s \\right) \\right)}}\\]\n\n\\[K_{a} = \\operatorname{}{s^{2}G\\left( s \\right);\\ \\ \\ \\ K_{a} = \\operatorname{}{s^{2}\\left( \\text{KG}\\left( s \\right) \\right)}}\\]\n\nThe type of system is determined by the number of poles at the origin.\nFor example:\n\n%\\includegraphics[width=2.60736in,height=0.52328in]{media/image10.png}\n\n\\section{ROOT LOCUS}\n\nRoot Locus presents the poles of the closed loop system when the gain K\nchanges from zero to infinity.\n\n\\textbf{Construction of the Root Locus}\n\nOpen loop transfer function\n\n\\[\\text{KH}\\left( s \\right)G\\left( s \\right) = K\\frac{B(s)}{A(s)}\\]\n\nm: the order of the \\textbf{open-loop} numerator polynomial\n\nn: the order of the \\textbf{open-loop} denominator polynomial\n\n\\textbf{Rule 1:} number of branches equals the number of poles of the\nopen-loop transfer function\n\n\\textbf{Rule 2:} If the total number of poles and zeros of the open-loop\nsystem to the right of the s-point on the real axis is odd, then this\npoint lies on the locus.\n\n\\textbf{Rule 3:} The locus starting point (K=0) are at the open-loop\npoles and the locus ending points (K=$\\infty$) are at the open loop zeros and\nn-m branches terminate at infinity.\n\n\\textbf{Rule 4:} Slope of asymptotes of root locus as `s' approaches\ninfinity\n\n\\textbf{Rule 5:} Abscissa of the intersection between asymptotes of root\nlocus and real-axis.\n\n%\\includegraphics[width=2.13497in,height=0.89069in]{media/image11.png}\n\n\\textbf{Rule 6:} Break-away and break-in points. From the characteristic\nequation\n\n\\[f\\left( s \\right) = A\\left( s \\right) + KB\\left( s \\right) = 0\\ \\ \\ \\ and\\ \\ \\ \\ K = - \\frac{A\\left( s \\right)}{B\\left( s \\right)}\\]\n\nThe break-away and break-in points can be found from\n\n\\[\\frac{\\text{dK}}{\\text{ds}} = - \\frac{A^{'}\\left( s \\right)B\\left( s \\right) - A\\left( s \\right)B^{'}\\left( s \\right)}{B^{2}\\left( s \\right)} = 0\\]\n\n\\textbf{Rule 7:} Angle of departure from complex poles or zeros.\nSubtract from 180° the sum of all angles from all other zeros and poles\nof the open-loop system to the complex pole (or zero) with appropriate\nsigns.\n\n\\textbf{Rule 8:} Imaginary-axis crossing points. Use Ruth-Hurwitz table\nto find value of K where system becomes unstable.\n\n\\section{BODE DIAGRAMS}\n\n%\\includegraphics[width=2.08800in,height=1.27279in]{media/image12.png}\n\n\\subsection{1. Gain Factor K:} Horizontal straight line at magnitude:\n\\(20\\log{(K)}\\text{dB}\\)\n\nPhase is zero.\n\n\\subsection{2. Integral or derivative factors}\n\\(\\mathbf{(j\\omega)}^{\\mathbf{\\pm 1}}\\)\n\n\\[\\left( \\text{j}\\omega \\right)^{- 1}\\ \\rightarrow 20\\log{\\left| \\frac{1}{\\text{j}\\omega} \\right| = - 20\\log\\omega}\\]\n\nMagnitude: strait line with slope -20 dB/decade\n\nPhase: -90°\n\n\\[\\left( \\text{j}\\omega \\right) \\rightarrow 20\\log\\left| \\text{j}\\omega \\right| = 20\\log\\omega\\]\n\nMagnitude: straight line with slope 20dB/decade\n\nPhase: +90°\n\n\\subsection{3. First Order Factors}\n\\(\\left( \\mathbf{1 + j\\omega T} \\right)^{\\mathbf{\\pm 1}}\\)\n\n\\begin{align}\n& \\left( 1 + j\\omega T \\right)^{- 1} \\rightarrow 20\\log\\left| \\frac{1}{1 + j\\omega T} \\right| \\\\ \\notag\n&= - 20\\log\\sqrt{1 + \\omega^{2}T^{2}}\\ \\lbrack dB\\rbrack\n\\end{align}\n\nApproximation for Magnitude:\n\n\\[For\\ \\omega\\ between\\ 0\\ and\\ \\frac{1}{T}\\ \\rightarrow 0dB\\]\n\\[For\\ \\omega\\ \\gg \\frac{1}{T} \\ \\rightarrow - 20dB/decade\\]\n\nPhase:\n\n\\[\\omega = 0\\ \\rightarrow \\varphi = 0\\]\n\\[\\omega = \\frac{1}{T} \\rightarrow \\varphi = - 45\\]\n\\[\\omega = \\infty \\rightarrow \\varphi = - 90 \\]\n\n\\[\\left( \\mathbf{1 + j\\omega T} \\right)^{\\mathbf{+ 1}}\\]\n\n\\subsection{4. Quadratic Factors}\n\n\\[G\\left( \\text{j}\\omega \\right) = \\frac{1}{1 + 2\\zeta\\left( \\frac{\\omega}{\\omega_{n}} \\right) + \\left( \\frac{\\text{j}\\omega}{\\omega_{n}} \\right)^{2}}\\ \\ ;\\ \\ 0 < \\zeta < 1\\]\n\nApproximation for magnitude:\n\n\\[\\omega \\ll \\omega_{n} \\rightarrow 0dB\\]\n\n\\[\\omega \\gg \\omega_{n} \\rightarrow - 20\\log\\left( \\frac{\\omega^{2}}{\\omega_{n}^{2}} \\right) = - 40\\log{\\left( \\frac{\\omega}{\\omega_{n}} \\right)\\text{dB}}\\]\n\n\\[Phase:\\]\n\n\\[\\omega = 0\\ \\rightarrow \\varphi = 0\\]\n\\[\\frac{\\omega}{\\omega_{n}} = 1 \\rightarrow \\varphi = - 90\\backslash \\]\n\\[\\omega = \\infty \\rightarrow \\varphi = - 180 \\]\n\n\\[Resonant\\ Frequency:\\]\n\t\\[\\omega_{r} = \\omega_{n}\\sqrt{1 - 2\\zeta^{2}}\\ \\ for\\ 0 < \\zeta < 0.707\\]\n\n\\[Resonant\\ Peak\\ Value:\\]\n\\[M_{r} = \\left| G\\left( \\text{j}\\omega \\right) \\right|_{\\max} = \\frac{1}{2\\zeta\\sqrt{1 - \\zeta^{2}}} \\ for\\ 0 < \\zeta < 0.707 \\]\n\nConsider\n\n\\[ G_{1}\\left( s \\right) = \\frac{1}{1 + Ts} \\ G_{2}\\left( s \\right) = \\frac{1}{1 - Ts} \\ G_{3}\\left( s \\right) = \\frac{1}{Ts - 1} \\]\n\nThen\\ldots{}\n\n\\[\\left| G_{1}(j\\omega) \\right| = \\left| G_{2}(j\\omega) \\right| = \\left| G_{3}(j\\omega) \\right|\\]\n\nAnd\\ldots{}\n\n\\[\\angle G_{2}\\left( \\text{j}\\omega \\right) = - \\angle G_{1}\\left( \\text{j}\\omega \\right) and \\]\n\\[ \\angle G_{3}\\left( \\text{j}\\omega \\right) = 180 - \\angle G_{1}\\left( \\text{j}\\omega \\right)\\]\n\n\\[+ 90\\ and\\ \\angle G_{3}\\left( \\text{j}\\omega \\right)\\ goes\\ from - 180\\ to - 90\\]\n\n\nGenerate based on the Bode Plot.\n\n\\textbf{The Nyquist Stability Criterion:} relates the stability of the\nclosed loop system to the frequency response of the open loop system.\n\n\\[Z = N + P\\]\n\n\\textbf{Z:} Number of zeros of $(1+H(S)G(s))$ in the right half plane =\nnumber of unstable poles of the closed-loop system.\n\n\\textbf{N:} Number of clockwise encirclements of the point $-1+j0$.\n\n\\textbf{P:} Number of poles of $G(s)H(s)$ in the right half plane.\n\nIf the plot makes a counter-clockwise encirclement of the $-1+j0$ point\nthen N becomes -1.\n\nIf Z = 0 the closed loop system is stable. If Z \\textgreater{} 0 the\nclosed loop system has Z unstable poles. If Z \\textless{} 0 a mistake\nhas been made and the calculations need to be rechecked.\n\n\\section{PHASE AND GAIN MARGINS}\n\nA measure for relative stability of the closed-loop system is how close\n\\(G(j\\omega)\\), the frequency response of the open-loop system, comes to\nthe point $-1+j0$. This is represented by the phase and gain margins.\n\n\\textbf{Phase Margin:} The amount of additional phase lag at the Gain\nCrossover Frequency \\(\\omega_{0}\\) required to bring the system to the\nverge of instability.\n\nGain crossover frequency:\n\\(\\omega_{0}\\text{\\ for\\ which\\ }\\left| G\\left( j\\omega_{0} \\right) \\right| = 1\\)\n\nPhase margin:\n\\(\\gamma = 180 + \\angle G\\left( j\\omega_{0} \\right) = 180 + \\phi\\)\n\n\\textbf{Gain Margin:} The reciprocal of the magnitude\n\\(\\left| G(j\\omega_{1}) \\right|\\) at the Phase crossover frequency\n\\(\\omega_{1}\\) required to bring the system to the verge of instability.\n\nPhase crossover frequency:\n\\(\\omega_{1}\\ where\\ \\angle G\\left( j\\omega_{1} \\right) = - 180\\)\n\nGain margin:\n\n\\[K_{g} = \\frac{1}{\\left| G(j\\omega_{1}) \\right|}\\]\n\n\\[K_{g} = - 20\\log\\left| G\\left( j\\omega_{1} \\right) \\right|\\]\n\n\\[K_{g}\\ in\\ dB > 0 = stable\\] for minimum phase systems. \n\t\\[K_{g}in\\ dB < 0\\] = unstable for minimum phase systems. \n\n\\textbf{Minimum phase systems:} all poles and zeros are in the left half\nplane.\n\nIf the open-loop system is minimum phase and has both phase and gain\nmargins positive then the closed-loop system is stable.\n\nFor good relative stability both margins are required to be positive.\n\nGood values for minimum phase system are:\n\nPhase Margin: 30°-60°\n\nGain Margin: above 6dB\n\n\n\\end{multicols}", "meta": {"hexsha": "190311f319af5ebc981373a2878b862fef1c1024", "size": 16307, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "public/CheatSheets/ELEC360Cheat.tex", "max_stars_repo_name": "FriendlyUser/PortfolioWebsite", "max_stars_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "public/CheatSheets/ELEC360Cheat.tex", "max_issues_repo_name": "FriendlyUser/PortfolioWebsite", "max_issues_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "public/CheatSheets/ELEC360Cheat.tex", "max_forks_repo_name": "FriendlyUser/PortfolioWebsite", "max_forks_repo_head_hexsha": "82843816c07239c457d2d820bb50577333a75855", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.5534979424, "max_line_length": 428, "alphanum_fraction": 0.6294842706, "num_tokens": 6446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936537604181, "lm_q2_score": 0.8152324826183822, "lm_q1q2_score": 0.6247074777698167}} {"text": "\\iffalse\nIt's unfair to say that mathematicians aren't real doctors, we perform surgeries all the time. In this class we'll introduce the notion of a topological manifold via simplicial (delta) complexes. Spend a day or two doing examples and go over several notions like orientation, cobordism and of course surgery.\n\nKeywords: simplicial complex, manifold, orientation, cobordism, surgery\n\nType: Lecture\nHomework: Recommended\nPrereqs: None\n\\fi\n\n\n\n\\input{../preamble}\n\n\\rhead{\\scshape Mathcamp 2017 : All things Manifoldy}\n\n\\begin{document}\n\\title{Surfaces}\n\\author{Apurva Nakade}\n\\thispagestyle{fancy}\n\\maketitle\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Real projective space $\\R\\P^2$}\nOne way to define the projective plane is by gluing the antipodal points on the disc. In this section we'll see how this corresponds to lines in $\\R^3$.\n\nConsider the upper hemisphere\n\\begin{align*}\n\tS^2_+ = \\{ (x,y,z) : x^2 + y^2 + z^2 = 1 , z\\ge 0\\}\n\\end{align*}\nEvery line in $\\R^3$ intersects $S^2_+$ in exactly one point, unless the line is along the $x-y$ plane in which case it intersects $S^2_+$ boundary in antipodal points. Thus the set of lines in $\\R^3$ is in 1-1 correspondence with the points on $S^2_+$ but with the antipodal points on the boundary circle glued, but this is precisely $\\R\\P^2$!\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.25\\linewidth]{images/ProjectivePlane}\n\t\\caption{Real projective plane}\n\\end{figure}\n\n\n\\begin{exercise}\n\tDescribe the space $\\R\\P^1$, the space of lines in $\\R^2$.\n\\end{exercise}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Poincar\\'e conjecture}\n\nPoincar\\'e conjecture is perhaps the most celebrated conjecture in the theory of manifolds. It's solution(s) led to creation of a lot of amazing mathematics. For 3 dimensional manifolds the conjecture is as follows.\n\n\\begin{thm}[Poincar\\'e conjecture]\n\tLet $M$ be a compact connected 3 dimensional manifold (without boundary). Suppose every loop on $M$ can be shrunk to a point then $M$ is homeomorphic to the 3 dimensional sphere $S^3$.\n\\end{thm}\n\nThere is a generalization of this conjecture to higher dimensions which requires a little more language to state. Surprisingly the conjecture was first proven to be true for dimensions $\\ge 5$ by Smale, next for dimension 4 by Freedman and finally for dimension 3 by Perelman. The conjecture is \\textbf{false} in the `category' of smooth manifolds i.e. if we replace homeomorphism by something stronger like diffeomorphism. The fact that the conjecture is true in the topological `category' but false in the smooth `category' implies the existence of \\textbf{exotic manifolds}. Exotic spheres were first discovered by Milnor and only exist in dimensions 7 and higher. Freedman proved that $\\R^4$ is the only Euclidean space which supports an exotic structure.\n\nThe Poincar\\'e conjecture in dimension 2 follows from the classification theorem for surfaces.\n\n\\begin{thm}[Classification of surfaces]\n\tEvery compact connected 2-dimensional manifold (without boundary) is homeomorphic to one of the following:\n\t\\begin{enumerate}\n\t\t\\item $S^2$\n\t\t\\item $T^{\\#k}$ for some $k$\n\t\t\\item $(\\R\\P^2)^{\\#k}$ for some $k$\n\t\\end{enumerate}\n\\end{thm}\n\nThe Poincare conjecture for 2 dimensions is an easy consequence of this theorem.\n\n\\begin{exercise}\n\tFind a loop on each of the following spaces $T$, $\\R\\P^2$ and $K$ which cannot be shrunk to a point.\n\\end{exercise}\n\n\\begin{exercise}\n\tShow that for any compact connected manifold $M$ there is a loop in each of the spaces $M \\# T$ and $M \\# \\R\\P^2$ which cannot be shrunk to a point.\n\\end{exercise}\n\n\\begin{exercise}\n\tFor a compact connected 2 dimensional manifold $M$ (without boundary) show that if every loop on $M$ can be shrunk to a point then $M$ is homeomorphic to $S^2$.\n\\end{exercise}\n\n\n\n\n\n\n\n\n\n\n\n\n\\iffalse\nA two dimensional manifold is called a surface. The surface of the earth is an example of a two dimensional manifold. By the above definition a surface is set $S$ such every point $s \\in S$ has a neighborhood homeomorphism to a 2 dimensional disk.\n\n\\begin{figure}[H]\n\t\\centering \\includegraphics[width=0.85\\linewidth]{images/GegusGSurfaces}\n\t\\caption{Surfaces}\n\\end{figure}\n\n\\subsection{Connected sums}\nIt is not hard to see that we can make sense of `addition' of two surfaces, after all when we think of a genus $g$ surface we think of $g$ tori glued to each other.\n\nIf $M_1$ and $M_2$ are two surfaces of genus $g_1$ and $g_2$ then we can `add' $M_1$ and $M_2$ to get a surface of genus $g_1+ g_2$. This is done by removing a small disc from $M_1$ and $M_2$ each and gluing $M_1$ and $M_2$ along the holes. The resulting manifold is denoted by $M_1 \\# M_2$ and is called their \\textbf{connected sum}.\n\nWe will denote a torus by $T$ and a genus $g$ surface by $T \\# \\cdots \\# T$, $g$ times or simply as $T^{\\#g}$.\n\n\\begin{figure}[h]\n\t\\centering \\includegraphics[width=0.40\\linewidth]{images/ConnectedSum}\n\t\\caption{Connected sum}\n\\end{figure}\n\nThus it is possible to add two surfaces but not subtract them. There is also an \\emph{identity element} for this connected sum operation (what is it?). This is more formally expressed by saying that the set of surfaces (up to homeomorphism) is a \\textbf{monoid}.\n\n\\begin{exercise}\n\tDefine connected sum for arbitrary manifolds.\n\\end{exercise}\n\n\\begin{exercise}\n\tIn the definition of connected sum we removed arbitrary discs from the manifolds $M_1$ and $M_2$. Draw pictures to show that if any two such choices can be deformed into each other and hence up to homeomorphism the manifold $M_1 \\# M_2$ is well defined. Can you come up with a rigorous proof for this?\n\\end{exercise}\n\n\\begin{ques}\n\tIs every surface a genus $g$ surface for some integer $g$?\n\\end{ques}\nThe answer is no. The most famous example is perhaps the Klein bottle. The Klein bottle is an example of a non-orientable surface (a surface with only 1 side) and it cannot be embedded in $\\R^3$.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.15\\linewidth]{images/KleinBottle}\n\t\\caption{Klein bottle}\n\\end{figure}\n\nThere is an amazing theorem that holds for 2 dimensional manifold but which does not hold in any higher dimension.\n\n\\begin{thm}[Classification of surfaces]\n\tEvery compact connected 2 dimensional manifold is homeomorphic to one of the following\n\t\\begin{enumerate}\n\t\t\\item $S^2$\n\t\t\\item $T^{\\#g}$\n\t\t\\item $(\\R\\P^2)^{\\#g}$\n\t\\end{enumerate}\n\\end{thm}\nWe'll define the surface $\\R\\P^2$, called the \\textbf{real projective plane} below. If we think of a 2 dimensional manifold sitting inside some $\\R^n$ then \\textbf{compact connected} means that as a set the manifold is \\textbf{closed} and \\textbf{bounded}. If you do not know what these words mean then intuitively compact connected means that the manifold has no punctures in it and it does not extend to infinity.\n\nHow might one try to prove such a theorem? We'll use a technique very popular in topology called cutting and pasting!\n\n\\subsection{Gluing diagrams}\n\\begin{ques}\n\tWhat does a connected sum of a Klein bottle and a torus give us? What is the connected sum of two Klein bottles?\n\\end{ques}\n\nTo answer the above question we try to covert the problem into combinatorics by cutting open the surface. If cut a torus once we get a cylinder and we cut the cylinder again we get a square. On the square we label the sides to remember which side should be glued to which to get back the torus. Such a diagram is called a \\textbf{gluing diagram}.\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.20\\linewidth]{images/GluingDiagramTorus}\n\t\\caption{Gluing diagram for a torus}\n\\end{figure}\n\nCan we find gluing diagrams for other surfaces? The connected sum comes to the rescue. We can use connected sums to construct gluing diagrams using $2g$ sided polygons to construct $g$ genus surfaces, as in figure \\eqref{ConnectedSumGluingDiagram}.\n\n\\begin{figure}[h]\n\t\\centering\n\t\\includegraphics[width=0.50\\linewidth]{images/GluingDiagramConnectedSum1}\n\t\\includegraphics[width=0.60\\linewidth]{images/GluingDiagramConnectedSum2}\n\t\\caption{Connected sum and gluing diagrams}\n\t\\label{ConnectedSumGluingDiagram}\n\\end{figure}\n\nThis is an interesting development. Perhaps we can even find a gluing diagram for the Klein bottle. What happens when you cut open the Klein bottle?\n\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.20\\linewidth]{images/GluingDiagramKleinBottle}\n\t\\caption{Gluing diagram for a Klein bottle}\n\\end{figure}\n\n\\begin{ques}\n\tWhat happens if we do not put arrows on all the edges?\n\\end{ques}\n\n\\begin{ques}\n\tWhat do open sets and line segments look like on the gluing diagrams?\n\\end{ques}\n\n\\begin{ques}\n\tWhen does the gluing diagram give us a manifold?\n\\end{ques}\n\n\\begin{ques}\n\tCan you come up with a gluing diagram for a sphere?\n\\end{ques}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Classifying surfaces}\nNotice that if we flip the directions of both the arrows labelled $a$ on the gluing diagram for a torus we get the torus back, but if we flip only one of the $a$ while keeping the other fixed we get a Klein bottle. This begs the question,\n\n\\begin{ques}\n\tWhat happens when we flip both the arrows on the torus gluing diagram?\n\\end{ques}\nThis gives us the \\textbf{Projective plane} denoted by $\\R\\P^2$. This is the space that we encountered in the classification theorem.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=0.20\\linewidth]{images/GluingDiagramProjectivePlane}\n\t\\caption{Gluing diagram for the Projective Plane $\\R\\P^2$}\n\\end{figure}\n\nThe proof of the classification theorem has two main steps. First comes from point set topology. First we show that any compact connected surface gives rise to a gluing diagram. This is a highly non-trivial fact that we will assume. The proof of this requires some knowledge of the fundamental group.\n\nThe second step is the showing that any gluing diagram is of one of three types listed in the classification theorem.\n\n\\subsection{The algebra of gluing diagrams}\nWe saw how connected sums allowed us to construct gluing diagrams for genus $g$ surfaces. A connected sum is a cutting and pasting operation. We can do more interesting cutting and pasting operations and produce some non-trivial results about gluing diagrams.\n\n\\begin{figure}[h]\n\t\\centering\n\t\\includegraphics[width=0.75\\linewidth]{images/GluingDiagramKleinBottleMobiusStrip}\n\t\\caption{Klein bottle is the connected sum of two Mobius strips}\n\\end{figure}\n\\fi\n\\end{document}\n\n\n\n\n\n\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Modular origami}\n\n\\subsection{Orientation of surfaces}\nLet us start with surfaces. Roughly speaking a surface is called orientable if has two sides. A sphere or a torus are easily seen to be orientable. In fact any surface that can be \\textit{embedded} in $\\R^3$ is orientable. A Mobius strip on the other hand (which is not a surface as it has a boundary) is not orientable. With a little more mental effort one might be able to see that a Klein bottle or a Projective space are not orientable. We want an equivalent definition which can be used on gluing diagrams.\n\nWe begin by defining orientation of a polygon. An \\textbf{orientation of a polygon} is simply a cyclic ordering of it's vertices. For example this is one of the two possible ways to orient a triangle with vertices $(0,1,2)$.\n\n\\begin{center}\n\t\\begin{tabular}{c}\n\t\t\\centering \\includegraphics[height=3cm]{../noImageAvailable}\n\t\\end{tabular}\n\\end{center}\n\n\\begin{exercise}\n\tThis is related to the fact that the polygon has two sides via the right hand rule in physics. Do you see the connection?\n\\end{exercise}\n\nAn \\textbf{orientation} of (a gluing diagram of) a surface is a compatible choice of orientation for each triangle, where the orientations of two adjacent triangles need to be compatible in the following way,\n\\begin{center}\n\t\\begin{tabular}{c c c}\n\t\t\\centering \\includegraphics[height=3cm]{../noImageAvailable} & \\: & \\centering \\includegraphics[height=3cm]{../noImageAvailable}\n\t\\end{tabular}\n\\end{center}\nThis forces an edge to be directed in two different directions in adjacent triangles.\n\n\\begin{exercise}\n\tAdd the diagonals to the standard gluing diagrams to obtain triangulations and try to find orientations for them. Conclude that $S^2,T$ are orientable and $\\R\\P^2,K$ are not.\n\\end{exercise}\n\n\\begin{exercise}\n\tUse gluing diagrams to show that $T\\#T$ is orientable. Generalize this to argue that $T ^{\\# n}$ are orientable.\n\\end{exercise}\n\nThis same method also allows us to understand manifolds in higher dimensions as well.\n\n\n\n\n\n\n\n\\subsection{Simplices}\nAs we go to higher dimensions polygons (or is it polytopes?) become whacky are themselves quite hard to understand. So instead of looking at arbitrary polygons we look at the simplest polygons: triangles. We'll upgrade the definition of a triangle to a simplex which can live in arbitrary dimensions.\n\n\\begin{definition}\n\tAn $n$ dimensional \\textbf{simplex} $\\Delta^n$ is any set which is homeomorphic to the following region in $\\R^{n}$.\n\t\\begin{align}\n\t\t\\{ (x_0, x_1, \\cdots, x_n) : x_0 + \\cdots + x_n \\le 1 \\mbox{ and each } x_i \\ge 0 \\}\n\t\\end{align}\n\\end{definition}\nA 1 dimensional simplex is a segment and a 2 dimensional simplex is a triangle.\nSimplices are topologists best friend.\n\nNote that by adding extra diagonal lines we could have made the gluing diagrams entirely out of triangles.\nWe represent a simplex by its set of vertices.\n\n\\begin{figure}[h]\n\t\\centering \\includegraphics{../noImageAvailable}\n\t\\caption{Gluing diagram with labeled simplices}\n\t\\label{}\n\\end{figure}\n\nAs with origami you can glue these simplices any way you want and get relaly interesting objects. We're going to glue them to create manifolds.\n\n\\subsection{Manifolds from simplices}\nA delta complex is a collection of simplices glued along the faces.\n\n\\begin{definition}\n\tFaces of a simplex followed by examples.\n\\end{definition}\n\n\\begin{definition}\n\tDefinition of a delta complex followed by several examples.\n\\end{definition}\n\nTalk about spheres, tori and projective spaces in 3 dimensions.\n\n\\begin{ques}\n\tWhen is a delta complex a manifold?\n\\end{ques}\n\nExamples of manifolds and non-manifolds.\n\n\n\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{More on manifolds}\nThe last day is a little more on the expository side.\n\nWe saw how a Dehn surgery can be used to simplify a knot on a torus. We also saw how a connected sum can be used to build higher genus surfaces out of simpler ones.\n\nIs it possible to simply go the other way? Can one reverse the connected sum operation and go from a more complicated manifold to a simple one?\n\n\\subsection{Orientation of manifolds}\n\nShow that $\\R\\P^1$ and $\\R \\P^3$ are orientable but $\\R\\P^2$ is not using Delta complexes.\n\n\\subsection{Cutting $S^3$ into tori}\n\n\\subsection{Using Dehn twists to create 3 manifolds}\n\n\n\n\n\\end{document}\n", "meta": {"hexsha": "2b889162e5c8ed35efe795ef3f3c4b3dabf02fd5", "size": 15174, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "01 All things manifoldy/03 - Surfaces.tex", "max_stars_repo_name": "apurvnakade/mc2017", "max_stars_repo_head_hexsha": "ebec59bce5ee1979872e0f37208da6abd91dbb75", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "01 All things manifoldy/03 - Surfaces.tex", "max_issues_repo_name": "apurvnakade/mc2017", "max_issues_repo_head_hexsha": "ebec59bce5ee1979872e0f37208da6abd91dbb75", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "01 All things manifoldy/03 - Surfaces.tex", "max_forks_repo_name": "apurvnakade/mc2017", "max_forks_repo_head_hexsha": "ebec59bce5ee1979872e0f37208da6abd91dbb75", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.1104651163, "max_line_length": 759, "alphanum_fraction": 0.7346777382, "num_tokens": 3953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8152324938410783, "lm_q1q2_score": 0.6247074776675321}} {"text": "\\section{Properties of a Random Sample}\n\n\\begin{definition}[Random Sample]\n A collection of random variables $X_1, \\dots, X_n$ is a \\emph{random sample of size n from population $f(x)$} if they are iid with pdf/pmf f(x).\n\\end{definition}\n\n\\begin{definition}[Statistic]\n $Y = T(X_1, \\dots, X_n)$ is a \\emph{statistic} if the domain of $T$ contains the sample space of $(X_1, \\dots, X_n)$. The distribution of $Y$ is the \\emph{sampling distribution of $Y$}.\n\\end{definition}\n\n\\begin{remark}\n A statistic is any function of the data. The only restriction is that the statistic is not also a function of some other parameters.\n\\end{remark}\n\n\\begin{definition}[Sample Mean and Sample Variance]\n The \\emph{sample mean} $\\bar{X}$ and \\emph{sample variance} $S^2$ of a random sample $X_1, \\dots, X_n$ are, respectively,\n \\begin{itemize}\n \\item $\\bar{X} = \\frac1n \\sum_{i=1}^n X_i$\n \\item $S^2 = \\frac{1}{n-1}\\sum_{i=1}^n (X_i - \\bar{X})^2$.\n \\end{itemize}\n The \\emph{sample standard deviation} is $S = \\sqrt{S^2}$.\n\\end{definition}\n\n\\begin{theorem}[Some facts]\n Let $X_1, \\dots, X_n$ be a random sample from a population with mean $\\mu$ and finite variance $\\sigma^2$, then\n \\begin{enumerate}[a.]\n \\item $\\E{}[\\bar{X}] = \\mu$\n \\item $\\Var{}[\\bar{X}] = \\frac{\\sigma^2}{n}$\n \\item $\\E{}[S^2] = \\sigma ^2$.\n \\end{enumerate}\n\\end{theorem}\n\n\\begin{theorem}\n Take $x_1, \\dots, x_n \\in \\R{}$ and let $\\bar{x}$ be their mean. Then,\n \\begin{enumerate}[a.]\n \\item $\\min_a \\sum_{i=1}^n (x_i - a)^2 = \\sum_{i=1}^n (x_i - \\bar{x})^2$\n \\item $(n-1)s^2 = \\sum_{i=1}^n(x_i - \\bar{x})^2 = \\sum_{i=1}^n x_i^2 - n\\bar{x}^2$.\n \\end{enumerate}\n\\end{theorem}\n\n\\begin{theorem}\n Let $X_1, \\dots, X_n$ be a random sample from population $f(x\\vert{}\\vec{\\theta})$ belonging to an exponential family\n \\[\n f(x \\vert{} \\vec{\\theta}) = h(x)c(\\vec{\\theta}) \\exp\\left(\\sum_{i=1}^k w_i(\\vec{\\theta})t_i(x)\\right).\n \\]\n Define the statistics \n \\[\n T_i(X_1, \\dots, X_n) = \\sum_{j=1}^n t_i(X_j), \\quad i=1, \\dots, k.\n \\]\n Then if the set $\\{(w_1(\\vec{\\theta}), \\dots, w_n(\\vec{\\theta}), \\, \\theta \\in \\Theta\\}$ contains and open subset of $\\R{}^k$ then the distribution of $(X_1, \\dots, X_n)$ is an exponential family of the form\n \\[\n f(u_1, \\dots, u_n \\vert{} \\vec{\\theta}) = H(u_1, \\dots, u_n) c(\\vec{\\theta})^n \\exp\\left(\\sum_{i=1}^kw_i(\\vec{\\theta})u_i\\right).\n \\]\n\\end{theorem}\n\n\\begin{remark}\n The open condition eliminates curved exponential families from this result.\n\\end{remark}\n\n\\subsection{Sampling from the Normal Distribution}\n\n\\begin{theorem}\n Let $X_1, \\dots, X_n$ be a random sample from a $\\n(\\mu, \\sigma)$ distribution. Then,\n \\begin{itemize}\n \\item $\\bar{X}$ and $S^2$ are independent\n \\item $\\bar{X} \\sim \\n(\\mu, \\sigma^2/n)$\n \\item $(n-1)S^2/\\sigma^2 \\sim \\chi^2_{n-1}$.\n \\end{itemize}\n\\end{theorem}\n\n\\begin{lemma}[Covariance and independence]\n In the case of samples from a multivariate normal\n \\begin{itemize}\n \\item Independence $\\iff$ vanishing covariance\n \\item Pairwise independence $\\iff$ independence\n \\end{itemize}\n\\end{lemma}\n\n\n\\subsection{Convergence Concepts}\n\n\\subsubsection{Convergence in Probability}\n\\begin{definition}\n A sequence of random variables $\\{X_i: i \\in \\N \\}$ \\emph{converges in probability} to a random variable $X$ if $ \\forall \\epsilon > 0$\n \\[\n \\lim_{n\\to\\infty}\\P{}(\\abs{X_n - X} \\geq \\epsilon) = 0\n \\]\n (or equivalently if $\\lim_{n\\to\\infty}\\P{}(\\abs{X_n - X} < \\epsilon) = 1$).\n\\end{definition}\n\n\\begin{theorem}[Weak Law of Large Numbers]\n Let $X_1, X_2, \\dots$ be iid RVs with mean $\\mu$ and $\\E{}[\\abs{X_i}] < \\infty$. Define $\\bar{X}_n = \\frac1n \\sum_{i=1}^n$. Then the sequence $\\bar{X}_n$ converges in probability to $\\mu$. That is, $\\forall \\epsilon > 0$\n \\[\n \\lim_{n\\to\\infty} \\P{}(\\abs{X_n - \\mu} < \\epsilon) = 1\n \\]\n\\end{theorem}\n\nProof is using Chebychev's inequality.\n\n\\begin{theorem}\n If $X_1, X_2, \\dots$ converges in probability to $X$ and $h$ is a continuous function, then $h(X_1), h(X_2), \\dots$ converges in probability to $h(X)$.\n\\end{theorem}\n\n\\subsubsection{Almost Sure Convergence}\n\n\\begin{definition}\n A sequence of random variables $X_1, X_2, \\dots$ \\emph{converges almost surely} to a random variable $X$ if $ \\forall \\epsilon > 0$\n \\[\n \\P{}\\left(\\lim_{n\\to\\infty} \\abs{\\bar{X}_n - X} < \\epsilon\\right) = 1.\n \\]\n\\end{definition}\n\n\\begin{comments}\n \\mbox{}\n \\begin{itemize}[+]\n \\item Almost sure convergence is much stronger than convergence in probability. Convergence in probability states that the sequence of measures of the sets on which the sequence has finite difference from its the limit converges to 0. Almost sure convergence states that any place where the sequence has finite difference from its limit must have measure 0. It's like a sequence of integrals converging vs. whether the integrands converge.\n \\item Almost sure convergence implies convergence in probability but not the other way around.\n \\end{itemize}\n\\end{comments}\n\n\\begin{theorem}\n If a sequence converges in probability then it is possible to find a subsequence that converges almost surely.\n\\end{theorem}\n\n\\begin{theorem}[Strong Law of Large Numbers]\n Let $X_1, X_2, \\dots$ be iid RVs with mean $\\mu$ and $\\E{}[\\abs{X_i}] < \\infty$. Define $\\bar{X}_n = \\frac1n \\sum_{i=1}^n$. Then the sequence $\\bar{X}_n$ converges almost surely to $\\mu$. That is, $\\forall \\epsilon > 0$\n \\[\n \\P{}\\left(\\lim_{n\\to\\infty}\\abs{\\bar{X}_n - \\mu} < \\epsilon\\right) = 1.\n \\]\n\\end{theorem}\n \n\\subsubsection{Convergence in Distribution}\n\n\\begin{definition}\n A sequence of random variables $X_1, X_2, \\dots$ \\emph{converges in distribution} to a random variable $X$ if\n \\[\n \\lim_{n \\to \\infty} F_{X_n}(x) = F_X(x)\n \\]\n at all points where $F_X$ is continuous.\n\\end{definition}\n\n\\begin{remark}\n Here it is really the cdfs that converge, rather than the random variables. In this way convergence in distribution differs from the previous two concepts.\n\\end{remark}\n\n\\begin{theorem}\n Convergence in probability implies convergence in distribution\n\\end{theorem}\n\n\n\\begin{theorem}[Central Limit Theorem]\n Let $X_1, X_2, \\dots$ be a sequence of iid random variables with $\\E{}[X_i] = \\mu$ and finite variance $\\Var{}[X_i] = \\sigma^2 < \\infty$. Define $\\bar{X}_n = \\frac1n \\sum_{i=1}^n X_i$. Let $G_n(x)$ denote the cdf of $\\sqrt{n}(\\bar{X}_n - \\mu)/\\sigma$. Then $\\forall x \\in \\R$,\n \\[\n \\lim_{n\\to\\infty} G_n(x) = \\frac{1}{\\sqrt{2\\pi}} \\int_{-\\infty}^x e^{-y^2/2}\\d{}y.\n \\]\n That is, $\\sqrt{n}(\\bar{X}_n - \\mu)/\\sigma$ converges in distribution to the standard normal.\n\\end{theorem}\n\n\\begin{theorem}[Slutsky's Theorem]\n If $X_n \\to X$ in distribution and $Y_n \\to a$ in probability with $a$ constant, then\n \\begin{enumerate}[a.]\n \\item $Y_nX_n \\to aX$ in distribution\n \\item $X_n + Y_n \\to X + a$ in distribution\n \\end{enumerate}\n\\end{theorem}\n\n\\begin{remark}\n This tells us, for instance, that \n \\[\n \\frac{\\sqrt{n}(\\bar{X}_n - \\mu)}{S_n} \\to \\n(0, 1)\n \\]\n in distribution, since we know that $S_n \\to \\sigma$ in probability.\n\\end{remark}\n\n\\subsubsection{The Delta Method}\nIf we are interested in the convergence of some function of a sequence of RVs, rather than the RVs themselves, then we can use the Delta Method (follows from an application of Taylor's theorem and Slutsky's theorem).\n\n\\begin{theorem}[Delta Method]\n Let $Y_n$ be a sequence of random variables that satisfies $\\sqrt{n}(Y_n - \\theta) \\to \\n(0, \\sigma^2)$ in distribution. For a given function $g$ and a specific value of $\\theta$, suppose that $g'(\\theta)$ exists and is non-zero. Then\n \\[\n \\sqrt{n}[g(Y_n) - g(\\theta)] \\to \\n(0, \\sigma^2g'(\\theta)^2)\n \\]\n in distribution.\n\\end{theorem}\n\n\n\\begin{remark}\n There exists a corresponding multivariate result.\n\\end{remark}\n\nIf $g'(\\theta) = 0$ then we take the next term in the Taylor series.\n\n\\begin{theorem}[Second Order Delta Method]\n Let $Y_n$ be a sequence of random variables that satisfies $\\sqrt{n}(Y_n - \\theta) \\to \\n(0, \\sigma^2)$ in distribution. For a given function $g$ and a specific value of $\\theta$, suppose that $g'(\\theta) = 0$ and $g''(\\theta)$ exists and is non-zero. Then\n \\[\n \\sqrt{n}[g(Y_n) - g(\\theta)] \\to \\n(0, \\sigma^2g'(\\theta)^2)\n \\]\n in distribution.\n\\end{theorem}\n\n\n\\subsection{Generating A Random Sample}\n\n\\begin{definition}[Direct Method]\n A \\emph{Direct Method} of generating a random sample uses the probability integral transform to map draws from a $\\text{uniform}(0, 1)$ random variable to draws from the distribution of interest.\\\\\n \n The Probability Integral Transform states that if $X$ has continuous cdf $F_X(x)$ then\n \\[\n F_X(X) \\sim \\text{uniform}(0, 1).\n \\]\n\\end{definition}\n\n\\begin{definition}[Accept-Reject Algorithm]\n Let $Y \\sim f_Y(y)$ and $V \\sim f_V(v)$ where $f_Y$ and $f_V$ have common support with\n \\[\n M = \\sup_y f_Y(y) / f_V(y) < \\infty.\n \\]\n To generate a random variable $Y \\sim f_Y$:\n \\begin{enumerate}[a.]\n \\item Generate $U \\sim \\text{uniform}(0, 1)$, $V \\sim f_V$ independent.\n \\item If $U < \\frac 1M f_Y(V)/f_V(V)$, return $V$ as a sample of $Y$; otherwise go back to (a.).\n \\end{enumerate}\n\\end{definition}\n\n\\begin{remark}\n \\mbox{}\n \\begin{itemize}\n \\item It is typical to call $V$ the \\emph{candidate density} and $Y$ the \\emph{target density}.\n \\item One would normally try to choose a candidate density with heavier tails than the target density (e.g. Cauchy and normal) to ensure that the tails of the target are well represented. If the target has heavy tails, however, it can be hard to find a candidate that results in finite $M$. In this case people turn to MCMC methods.\n \\item Note that $\\P{}(\\texttt{terminate}) = 1/M$. The number of trials to generate one sample of $Y$ is therefore $\\text{geometric}(1/M)$, with $M$ the expected number of trials.\n \\item The intuition behind this algorithm is that if we consider placing the density of a random variable $Y$ in a box (2d for simplicity) with coordinates $(v,u)$, we express the cdf of $Y$ using $V, U \\sim \\text{uniform}(0, 1)$\n \\[\n \\P{}(Y \\leq y) = \\P{}(V \\leq y \\vert{} U \\leq \\frac1c f_Y(V))\n \\]\n where $c = \\sup_y f_Y(y)$. In the actual algorithm we take $U \\sim \\text{uniform}(0,1)$ and $V$ to be an RV that has common support with $Y$.\n \\end{itemize}\n\\end{remark}\n\n\n", "meta": {"hexsha": "a3c2c925d923321995aa3be798979f5ec5720afc", "size": 10658, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/chapters/chapter5/content.tex", "max_stars_repo_name": "brynhayder/statistical_inference", "max_stars_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-01-25T05:52:48.000Z", "max_stars_repo_stars_event_max_datetime": "2020-10-12T07:20:16.000Z", "max_issues_repo_path": "notes/chapters/chapter5/content.tex", "max_issues_repo_name": "brynhayder/statistical_inference", "max_issues_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-10-17T15:44:01.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-24T09:31:29.000Z", "max_forks_repo_path": "notes/chapters/chapter5/content.tex", "max_forks_repo_name": "brynhayder/statistical_inference", "max_forks_repo_head_hexsha": "fc3e770650e9c145aa9d45e604c9e67624c2a013", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-04-29T11:11:32.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-17T09:47:07.000Z", "avg_line_length": 44.4083333333, "max_line_length": 447, "alphanum_fraction": 0.6488084068, "num_tokens": 3431, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8152324915965392, "lm_q1q2_score": 0.624707475947556}} {"text": "%!TEX root = ../../master.tex\n\n\\subsection{Results of the dispersion method in three dimensions}\n\n\\begin{figure}[!htb]\n\t\\centering\n \\scalebox{0.8}{\\input{figure/ere-contact-fitted_a-inv_+0.0_zeta_dispersion_projector_a1g_n-eigs_200.pgf}}\n \\caption{The same as \\Figref{unimproved spherical}, but tuned and subsequently analyzed using the appropriate latticized \\Luscher function, matching the cutoff on the sum to the lattice scale and accounting for the dispersion relation.\n We emphasize the results are on a log scale, and the tuning was to $-1/a_3 = 0$.\n }\n \\label{fig:unimproved dispersion}\n\\end{figure}\n\nIn this section, we again attempt to tune our contact interaction to unitarity by matching the first zero of the zeta function.\nHowever, the difference is that at each lattice spacing we tune to that spacing's respective $S^{\\dispersion}_3$, leveraging the dispersion relation for that derivative.\nThen, when we extract finite-volume and finite-spacing energy levels, we put them through the dispersion equation \\eqref{dispersion-zeta-contact} using the same $S^{\\dispersion}$ function.\nThe numerical results of said procedure are shown in \\Figref{unimproved dispersion}.\nNote that the results for $p\\cot\\delta$ are now flat across the spectrum, matching the known result for a contact interaction.\nMoreover, comparing the scale to that in, for example, \\Figref{unimproved spherical}, there the deviations were of order~1, while here the results remain within $10^{-8}$ of zero, with the value entirely reflecting how well the contact interaction was tuned.\nPut another way, we have verified that the dispersion zeta function provides exact finite-spacing energy levels for our contact interaction \\eqref{p space hamiltonian}, just as one would hope for a contact interaction in the continuum.\n\n\\begin{figure}[!hbt]\n\t\\centering\n \\scalebox{0.7}{\\input{figure/contact-scaling-contact-fitted_a-inv_+0.0_zeta_dispersion_projector_a1g_n-eigs_200.pgf}}\n \\caption{\n Scaling of the contact interaction strength $C_R^{(n_s)}(\\epsilon)$ fitted using the dispersion method at unitarity.\n Data points are values of the contact interaction fitted to the first intersection of the phase shifts with the dispersion zeta function.\n The solid lines are analytic scaling predictions following \\eqref{dispersion-renormalization} and the dashed line corresponds to the spherical counter term $ \\mathcal{L}^{\\spherical}_3 = 2 \\pi$.\n Bar diagrams below present the absolute error between prediction and extracted value.\n }\n \\label{fig:dispersion running of strength}\n\\end{figure}\n\nIn \\Figref{dispersion running of strength} we show how the strength of the contact interaction runs with the lattice scale according to the analytic expectation \\eqref{dispersion-renormalization}.\nNote that the lines are not fits to the data; though the difference is down at $10^{-12}$ or better.\nAgain, this difference depends on the accuracy of the tuning.\n\nWe note that\n\\begin{enumerate}\n\t\\item when matching the contact interaction parameter using spherical \\Luscher data and finite spacing eigenvalues, the data points did not exactly match the analytic spherical contact scaling.\n\tThe error at the smallest lattice spacing had a relative error on the percent scale and it got worse for larger lattice spacings.\n\t\\item even in the limit of $n_s \\to \\infty$ the dispersion counter term $\\mathcal{L}^\\dispersion_3$ will not match the spherical counter term $\\mathcal{L}^{\\spherical}_3$.\nAt any finite $N$ the spherical integral and cartesian integrals differ---if the radius of the sphere is $N/2$, the corners of the lattice's Brillouin zone are absent; the cartesian integral matches the Brillouin zone correctly, critical for any finite-$N$ result.\n\\end{enumerate}\n\n\n\n\\subsection{Momentum-induced terms of \\texorpdfstring{$S^\\bigcirc_3(x^\\dispersion)$}{S3-spherical} due to discretization\\label{sec:3d induced momenta}}\nThe zeta function in L\\\"uscher's formula, $S^\\bigcirc_3(x)$, is derived in the continuum.\nAs such, it requires continuum energies $x$ for its argument.\nIf one instead feeds discretized energies $x^\\dispersion$ through $S^\\bigcirc_3(x)$ then momentum-dependent terms are subsequently induced.\n\n\\begin{figure}[htb]\n \\scalebox{0.8}{\\input{figure/3dtuned.pgf}}\n \\caption{\n Here we show a contact interaction in three dimensions with the ground state tuned to the first zero of the spherical zeta function $S^{\\spherical}_3$ on cubic lattices with $N=10,$ 20, 40, 80 (squares, diamonds, hexagons, and circles, respectively), with the resulting spectrum analyzed with $S^{\\spherical}_3$ (colored points) and the $N$-appropriate $S^{\\dispersion}_3$ (black points).\n The gray dashed line is $S^\\spherical_3$ and the thin vertical lines are at the non-interacting $x$s where it diverges.\n The colored lines are the second-order analytic prediction for the difference between the dispersion and spherical analysis as a function of $x$.\n For clarity of the continuum limit we show, in the bottom panel, a limited range in $x$ and $pL\\cot\\delta_{30}$, where it is clear that each $N$ hits the zero of $S^\\spherical_3$ but that the flat behavior at any finite $N$ is away from an infinite scattering length when analyzed with $S^\\dispersion_3$.\n }\n \\label{fig:3d-corrections}\n\\end{figure}\n\nThis is particularly evident for the contact interaction as was observed, for example, in \\Ref{Endres:2012cw}.\nTo understand the source of these terms, consider\n\\begin{multline}\n\\frac{1}{\\pi L}S^\\bigcirc_3(x^\\dispersion)=\\frac{1}{\\pi L}\\left(S^{\\dispersion}_3(x^\\dispersion)+\\left(S^\\bigcirc_3(x^\\dispersion)-S^{\\dispersion}_3(x^\\dispersion)\\right)\\right)=\\frac{-1}{a_3}+\\frac{1}{\\pi L}\\left(S^\\bigcirc_3(x^\\dispersion)-S^{\\dispersion}_3(x^\\dispersion)\\right)\\\\\n=\\frac{-1}{a_3}+\\lim_{\\eta \\to\\infty}\\frac{1}{\\pi L}\\left(\\sum_{\\bm n\\notin \\mathrm{B.Z.}}^{|\\bm n|<\\eta / 2} \\frac{1}{\\bm n^{2}-x^\\dispersion}-\\mathcal{L}^\\bigcirc_3\\frac{\\eta}{2}+\\mathcal{L}^\\dispersion_3\\frac{N}{2}\\right)\\ .\n\\end{multline}\nIn the first line we added and subtracted $S^\\dispersion_3$ and used the dispersion results~\\eqref{dispersion-zeta-form} and~\\eqref{dispersion-zeta-contact} to introduce the scattering length in the case of a contact interaction.\nFor convenience we assume $n_s=\\infty$\\footnote{\nThe logic of the following derivation remains them same also for $n_s < \\infty$, but, in this case, the expressions $\\bm n^2$ must be replaced with the proper dispersion $\\tilde K^{(n_s)}_{\\bm n \\bm n}$ \\eqref{normalized-kinetic-hamitlonian}, which makes it difficult to obtained closed expressions. Also, within the Brillouin zone the different dispersion relations cause the two sums differ by \\order{x/N^2} term-by-term.}.\nIn the second line, since $\\bm n$ is now restricted to be \\emph{outside} the Brillouin zone, we can assume that $\\bm n^2\\gg x^\\dispersion$ and expand in small $x^\\dispersion$ under the summation,\n\\begin{align}\n\t\\label{eqn:observable}\n\tS^\\bigcirc_3(x^\\dispersion)\n\t&=\n\t\\frac{-\\pi L}{a_3}\n\t+\\mathcal{L}^\\dispersion_3\\frac{N}{2}\n\t+\\lim_{\\eta \\to\\infty}\\left(\\sum_{\\bm n\\notin \\mathrm{B.Z.}}^{|\\bm n|<\\eta / 2} \\frac{1}{\\bm n^{2}}\n\t-\\mathcal{L}^\\bigcirc_3\\frac{\\eta}{2}\\right)\n\t+x^\\dispersion\\lim_{\\eta \\to\\infty}\\sum_{\\bm n\\notin \\mathrm{B.Z.}}^{|\\bm n|<\\eta / 2} \\frac{1}{\\bm n^{4}}\n\t+(x^\\dispersion)^2\\lim_{\\eta \\to\\infty}\\sum_{\\bm n\\notin \\mathrm{B.Z.}}^{|\\bm n|<\\eta / 2} \\frac{1}{\\bm n^{6}}\n\t+ \\ldots\n\t\\\\\n\t\\label{eq:small x}\n\t&\\equiv \\frac{-\\pi L}{a_3}+\\alpha_1(N)+\\alpha_2(N)x^\\dispersion+\\alpha_3(N)(x^\\dispersion)^2\n\t+\\ldots\n\\end{align}\nThe last line above shows explicitly the induced momentum-dependence in $x^\\dispersion$ and defines the coefficients $\\alpha_i(N)$ in terms of particular lattice summations similar to those of the three-dimensional zeta function. The dependence of these coefficients on $N$ comes from the \\emph{exclusion} of momentum modes within the Brillouin zone in the summation. The fact that these coefficients do \\emph{not} depend on $L$ is a unique feature of the contact interaction. The numerical values of the coefficients $\\alpha_i(N)$ can be determined using standard acceleration techniques (see, for example, Appendix B of~\\Ref{Luu:2011ep}). We provide values for select cases of $N$ in \\Tabref{slopes}.\n\\begin{table}\n\\caption{Coefficients $\\alpha_i(N)$ as a function of $N$ in 3-D.\\label{tab:slopes}}\n\\center\n\\begin{tabular}{S[table-format=1.0]S[table-format=-1.15,table-auto-round=false]S[table-format=-1.15,table-auto-round=false]S[table-format=-1.15,table-auto-round=false]}\n{$N$} & {$\\alpha_1$} & {$\\alpha_2$} & {$\\alpha_3$} \\\\ \\midrule\n10 & 0.34622847019345 &2.1088361299026 &0.02096728133239\\\\\n20 & 0.17384029798483 &1.0470052482673 &0.00253774588732\\\\\n40 & 0.08701147975728 &0.5225652776531 &0.00031456311910\\\\\n50 & 0.06961796407968 &0.4179620004936 &0.00016089237674\\\\\n80 & 0.04351717442702 &0.2611651268184 &0.00003923695720\\\\\n100 &0.03481483765136 &0.2089208128674 &0.00002008418957\n\\end{tabular}\n\\end{table}\n\nIn \\Figref{3d-corrections} we show the result of tuning a finite-spacing contact interaction to the first zero of the continuum zeta function $S^\\spherical_3$.\nAt each spacing the spectrum is fed through the continuum zeta for analysis, resulting in an apparent spacing-dependent momentum dependence that matches the small-$x$ expansion~\\eqref{small x} discussed in the next section.\nThe same spectrum is also fed through the spacing-appropriate dispersion zeta $S^\\dispersion$, resulting in the flat black lines.\nShown in detail in the bottom panel, it's clear that the continuum limit taken this way results in any finite spacing having a nonzero scattering length that vanishes with the continuum limit.\nIn contrast, tuning to the dispersion function directly, as in \\Figref{unimproved dispersion}, is flat and nearly zero at each individual lattice spacing.\nWe expect that this difference explains the induced momentum dependence of, for example, \\Refs{Endres:2011er,Endres:2012cw}.\n", "meta": {"hexsha": "a2aaf1789bb7532835a3f544dbdd5955ad1f188b", "size": 10034, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/luescher-nd/section/three-dimensions/dispersion.tex", "max_stars_repo_name": "ckoerber/luescher-nd", "max_stars_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-04-12T22:19:38.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-26T14:06:49.000Z", "max_issues_repo_path": "paper/luescher-nd/section/three-dimensions/dispersion.tex", "max_issues_repo_name": "ckoerber/luescher-nd", "max_issues_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2019-12-16T19:49:19.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-02T00:50:31.000Z", "max_forks_repo_path": "paper/luescher-nd/section/three-dimensions/dispersion.tex", "max_forks_repo_name": "ckoerber/luescher-nd", "max_forks_repo_head_hexsha": "d1bc6bff0c6ee9f4dc0d1d0bb4bcfa842c44cceb", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 90.3963963964, "max_line_length": 707, "alphanum_fraction": 0.7566274666, "num_tokens": 2815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324848629214, "lm_q2_score": 0.7662936377487305, "lm_q1q2_score": 0.624707466436545}} {"text": "\\documentclass[a4paper,12pt]{article}\n\\usepackage{amsmath,amssymb,wasysym}\n\\usepackage[margin=2cm]{geometry}\n\n\\usepackage{comment}\n\\usepackage{longtable}\n\\usepackage{tabularx,booktabs}\n\\usepackage{listings}\n\\usepackage{graphicx}\n\\usepackage{hyperref}\n\\lstset{basicstyle=\\ttfamily}\n\\newcommand{\\py}{\\lstinline}\n\n% Math notations\n\\newcommand{\\Z}{{\\mathbb{Z}}} % set of integers\n\n% MoA notations\n\\newcommand{\\minus}{{}^{\\boldsymbol{\\mbox{-}}\\!}} % scalar negation operator\n\\newcommand{\\rop}[1]{\\,\\mathrm{#1}^*\\,} % relational operation\n\\newcommand{\\op}[1]{\\,\\mathrm{#1}\\,} % binary operation\n\\newcommand{\\uop}[1]{\\mathrm{#1}\\,} % unary operation\n\\newcommand{\\hop}[2]{{{}_{#1}\\!\\Omega_{#2}}\\,} % higher order operation\n\\newcommand{\\id}[1]{\\mathrm{id}(\\op{#1})} % identity of operations\n\\newcommand{\\dims}{\\delta\\,} % array dimension operator\n\\newcommand{\\shape}{\\rho\\,} % array shape operator\n\\newcommand{\\size}{\\tau\\,} % array size operator\n\\newcommand{\\reshape}{\\,\\widehat{\\rho}\\,} % reshape operator\n\\newcommand{\\drop}{\\,\\nabla\\,} % drop operator\n\\newcommand{\\take}{\\,\\Delta\\,} % take operator\n\\newcommand{\\product}{\\pi\\,} % product operator\n\\DeclareMathOperator{\\rav}{rav}\n\\newcommand{\\ravel}{\\rav\\,} % ravel operator\n\\newcommand{\\range}{\\iota\\,} % range operator\n\\newcommand{\\transpose}{\\bigcirc\\!\\!\\!\\!\\!\\backslash\\;} % transpose operator, need a better symbol\n\\newcommand{\\vc}[1]{<#1>} % vector with one component\n\\newcommand{\\vcc}[2]{<#1\\;#2>} % vector with two components\n\\newcommand{\\vccc}[3]{<#1\\;#2\\;#3>} % vector with four components\n\\newcommand{\\outerprod}[1]{\\,\\bullet_{#1}\\,} % outer product opetation\n\\newcommand{\\innerprod}[2]{\\,{}_{#1}\\!\\!\\bullet_{#2}\\,} % inner product opetation\n\\DeclareMathOperator{\\red}{red}\n\\newcommand{\\reduce}[1]{{}_{#1}\\!\\red\\,} % reduce operator\n\\newcommand{\\getitem}[2]{{#2}\\,\\psi\\,{#1}} % psi operator\n\n\n\\newcommand{\\scan}[1]{{}_{\\op{#1}\\!}\\mathrm{scan}\\,}\n\\newcommand{\\kron}{\\bigcirc\\,\\!\\!\\!\\!\\!\\!\\times\\;}\n\\newcommand{\\cat}{+\\!\\!\\!+}\n\\newcommand{\\gu}{\\mathrm{gu}\\,}\n\\newcommand{\\gd}{\\mathrm{gd}\\,}\n\\newcommand{\\compress}{\\,\\notslash\\,}\n\\newcommand{\\expand}{\\,\\notbackslash\\,}\n\\newcommand{\\reverse}{\\phi\\,}\n\\newcommand{\\rotate}[1]{{#1}\\theta\\,}\n\n\\title{MoA formalism and arrays in software}\n\\author{Pearu Peterson and Hameer and Saul and Lenore and Travis\\\\\nQuansight, Labs}\n\n\\begin{document}\n\\maketitle\n\\section{Introduction}\n\nMoA - Mathematics of Arrays - defines a formalism (based on APL) to\ndescribe multidimensional arrays and operations on these. On the other\nhand, contemporary array processing software (NumPy, Xnd, Apache\nArrow, etc) implement multidimensional array container objects and\ndefine UI to interact with the array objects. The aim of this document\nis to provide a mapping between MoA formalism and UI-s of widely used\narray processing software.\n\n\\section{Basic definitions}\n\nThe MoA formalism is originally defined in ``A Mathematics of Arrays''\nby Lenore Mullin, PhD thesis, 1988\\cite{mul00}. We use NumPy UI as a\nrepresentative of a software implementing UI for multidimensional\narrays. For brevity, we assume \\verb+from numpy import *+.\n\n\\noindent\nIn general, \\emph{an array is an object that has a mapping to get\n array items} \\footnote{A mapping $f$ is a set of argument and value\n pairs,$(i,v)$, and is represented as $f(i)=v$. We consider only\n single-valued mappings.} . The arguments to the mapping are called\n\\emph{indices}, all valid indices define the so-called \\emph{index\n set} of an array. The items can be arbitrary objects.\nTwo arrays are \\emph{equal} iff their get items mappings are equal: their shapes and components are equivalent.\n\\emph{Element-wise operations} of two arrays are defined iff the index\nsets of the two arrays are equal.\nArrays can be classified according to how the mappings are defined:\n\\begin{description}\n\\item[Multidimensional arrays:] The index set of\n an $N$-dimensional array is an $N$-dimensional hyperrectangle in\n $\\Z^N$: $[0,\\ldots,s_0-1]\\times\\cdots\\times[0,\\ldots,s_{N-1}-1]$\n where non-negative integers $(s_0,\\ldots,s_{N-1})$ define the shape\n of an array.\n\\item[Ragged arrays:] The index set of an\n $N$-dimensional ragged array is a connected subset of an\n $N$-dimensional hyperrectangle in $\\Z^N$. The concrete\n representation of the index set may have different forms.\n\\item[Sparse arrays:] The index set of an sparse array is a possibly\n disconnected subset of $N$-dimensional hyperrectangle in $\\Z^N$ and\n is extended to the full hyperrectangle by defining the default\n value.\n\\end{description}\n\nThe MoA deals with multidimensional arrays only. The NumPy implements\narray object for multidimensional arrays only. There is one-to-one\nrelation between MoA formalism and NumPy UI. Xnd implements a more\ngeneral container object than NumPy ndarray.\n\nThe goal is to extend MoA formalism to ragged arrays and then use this\nto define an UI as well as optimized algorithms to the array object\nimplementation software such as Xnd. \n\nThe current working hypothesis is that the ragged arrays can be\nrepresented as arrays of arrays where the subarrays are\nmultidimensional arrays.\n\n\\section{Optimization of Arrays in NumPy: Dense to Start}\nWe show how the inclusion of MoA and the Psi Calculus for dense arrays\nreduces to a normal form. This is a powerful concept in that programmers may design an algorithm differently, one may be efficient, another may not, even if the answer is the same. \n\\noindent\nWe show that there will not be performance issues for that choice because all algorithms using the MoA algebra will reduce to the same normal form. Another important concept is that all intermediate arrays are eliminated. Performance analysis and graphs will show this.\n\n\\subsection{An example}\n\nIn the following we demonstrate how to reduce a NumPy array expression\nto its normal form using MoA.\n\nConsider the following NumPy arrays\n\\begin{verbatim}\n>>> A = np.arange(30).reshape((2,3,5))\n>>> B = 2 + A\n>>> A\narray([[[ 0, 1, 2, 3, 4],\n [ 5, 6, 7, 8, 9],\n [10, 11, 12, 13, 14]],\n\n [[15, 16, 17, 18, 19],\n [20, 21, 22, 23, 24],\n [25, 26, 27, 28, 29]]])\n>>> B\narray([[[ 2, 3, 4, 5, 6],\n [ 7, 8, 9, 10, 11],\n [12, 13, 14, 15, 16]],\n\n [[17, 18, 19, 20, 21],\n [22, 23, 24, 25, 26],\n [27, 28, 29, 30, 31]]])\n\\end{verbatim}\nthat in MoA formalism are represented as\n\\begin{align}\n A^3 &\\equiv \\vccc235\\reshape(\\range30), \\\\\n B^3 &\\equiv 2+A^3,\n\\end{align}\nand the following array expression as an example case\n\\begin{verbatim}\n>>> r = np.inner(A[1,0,:], np.outer(A[1,0,:], B[0,1,:])[2,:])\n>>> r\n13175\n\\end{verbatim}\nthat in MoA formalism reads\n\\begin{equation}\nr \\equiv (\\vcc10\\psi A^3) \\innerprod+\\times (\\vc2\\psi((\\vcc10\\psi A^3) \\outerprod\\times (\\vcc01\\psi B^3))).\n\\end{equation}\nNote that when evaluating the array expression directly, the\ncomputation involves 30 multiplication and 4 addition\noperations. Using MoA, we reduce the above array expression to its\nnormal form where the number of operations is minimal.\n\nFirst, the items of the left operand of the inner product are\n\\begin{align*}\n \\getitem{(\\vcc10\\psi A^3)}{\\vc i} = \\getitem{A^3}{\\vccc10i}\n\\end{align*}\nThe items of the right operand of the inner product are\n\\begin{align}\n&\\getitem{\n (\\getitem{((\\getitem{A^3}{\\vcc10}) \\outerprod\\times (\\getitem{B^3}{\\vcc01}))}{\\vc2})\n}{\\vc i} \\\\\n&\\equiv \\getitem{((\\getitem{A^3}{\\vcc10}) \\outerprod\\times (\\getitem{B^3}{\\vcc01})))}{\\vcc2i} \\\\\n&\\equiv \\getitem{(\\getitem{A^3}{\\vcc10})}{\\vc2} \\times \\getitem{(\\getitem{B^3}{\\vcc01})}{\\vc i}\\\\\n&\\equiv \\getitem{A^3}{\\vccc102} \\times \\getitem{B^3}{\\vccc01i}\n\\end{align}\nAs a result, we have\n\\begin{align*}\n r &\\equiv ((\\vcc10\\psi A^3) \\innerprod+\\times \\getitem{B^3}{\\vcc01})\\times\\getitem{A^3}{\\vccc102}\n\\end{align*}\nthat in NumPy syntax reads\n\\begin{verbatim}\n>>> r_nf = np.inner(A[1,0,:], B[0,1,:]) * A[1,0,2]\n\\end{verbatim}\nwhich contains 6 multiplications and 4 additions.\n\n\\subsection{Performance analysis}\n\nNow let us benchmark the original form of this expression against it's reduced normal form.\nFor $n$ in ${\\{1, 10, 20, ..., 200\\}}$ where $\\shape A \\equiv \\vccc2 3 5 \\times n$ we measured the\nexecution of the original expression and the reduced expression.\n\n\\begin{figure}[h]\n \\includegraphics[width=1\\textwidth]{benchmarks/visualization-ratio.png}\n \\caption{The ratio of the mean execution time of evaluating the reduced and the original expression comparing decreases as the shape increases.}\n \\label{fig:plot}\n\\end{figure}\n\n\\begin{figure}[h]\n \\includegraphics[width=1\\textwidth]{benchmarks/visualization-ratio.png}\n \\caption{The ratio of the mean execution time of evaluating the reduced and the original expression comparing decreases as the shape increases.}\n \\label{fig:plot2}\n\\end{figure}\n\nAs seen in Figure \\ref{fig:plot}, the reduced normal form version is takes a third of the time when $n=1$ and a hundredth of the time when\n~$n=80$. So as $n$ increases, so does the proportional time saving from reducing the expression.\n\nTo reproduce the benchmarks and for all the benchmark data as well as machine information, please go to \\href{https://github.com/Quansight-Labs/uarray-docs/tree/fcb6baf8bb7284ac7223de73a70cf47063df9e4b/benchmarks}{github.com/Quansight-Labs/uarray-docs}.\n\n\\section*{Appendix}\n\n\\subsection*{Multidimensional arrays: MoA and NumPy}\n\n\\begin{comment}\n\\newcommand{\\MoADefinition}[1]{\n \\begin{itemize}\n #1\n \\end{itemize}\n}\n\\newcommand{\\MoAItem}[3]{\\item {#1}\n \\begin{description}\n \\item[MoA:] {#2}\n \\item[NumPy:] {#3}\n \\end{description}\n} \n\\end{comment}\n%\\begin{comment}\n\\newcommand{\\MoADefinition}[1]{\n\\renewcommand{\\arraystretch}{2}\n\\begin{longtable}{| p{.5\\textwidth} | p{.5\\textwidth} |}\n#1\n\\end{longtable}\n\\renewcommand{\\arraystretch}{1}\n}\n\\newcommand{\\MoAItem}[3]{\n\\multicolumn2{l}{\n \\begin{minipage}[l]{0.9\\linewidth} \\vspace{2ex}\n $\\bullet$ #1\\\\[0.1ex]\n \\end{minipage}\n}\\\\\\hline\n\\begin{minipage}[l]{0.95\\linewidth}\n{#2} \n\\end{minipage}\n&\n\\begin{minipage}[l]{0.95\\linewidth}\n{#3} \n\\end{minipage}\\\\\\hline\n}\n%\\end{comment}\n\n\n\n\nWe have the following correspondence table for basic notions:\n\\MoADefinition{\n \\MoAItem{a multidimensional array}{$\\xi$}{\\py'x=array(...)'}\n \\MoAItem{a one dimensional array, a vector}{$\\vec v\\equiv$}{\\py'v=array([v_0, v_1, ...])'}\n \\MoAItem{total number of array items}{$\\size\\xi$}{\\py'x.size'}\n \\MoAItem{the shape of an array is a vector of non-negative integers}{$\\shape\\xi\\equiv$}{\\py`x.shape==(s_0,...,s_n1)`}\n \\MoAItem{the dimensionality of an array is non-negative integer}{$\\dims\\xi\\equiv n\\equiv\\size\\shape\\xi$}{\\py'x.ndim==n==len(x.shape)'}\n \\MoAItem{an array item}{$\\getitem\\xi{\\vec i}\\equiv\\xi[i_0;\\ldots;i_{n-1}]$}{\\py'x[i_0,...,i_n1]'}\n \\MoAItem{negative of $k$}{$\\minus k$}{\\py'-k'}\n \\MoAItem{$\\gamma$ function is used to define the mapping between the indices of an array and its ravelled version}{\n \\begin{eqnarray*}\n\\gamma(<>;<>)&\\equiv&0\\\\\n\\gamma(\\vec a;\\vec b)&\\equiv&a_{-1} + b_{-1}\\gamma({\\minus1}\\drop{\\vec a}; {\\minus1}\\drop{\\vec b})\n \\end{eqnarray*}\n}{}\n \\MoAItem{inverse of $\\gamma$ with respect to the first argument}{$\\gamma'\\ldots \\gamma(\\gamma'(n;\\vec x); \\vec x)=n$}{}\n}\n\nMoA defines a number of operators on multidimensional arrays:\n\\MoADefinition{\n \\MoAItem{collapse an array to one dimension (row ordering is assumed)}{$\\ravel \\xi$}{\\py'x.ravel()'}\n \\MoAItem{product of vector items}{$\\product\\vec v$}{\\py'v.prod()'}\n\n}\n\nMoA defines array construction operators:\n\\MoADefinition{\n \\MoAItem{join a sequences of vectors}{$\\vec v,\\vec w,\\ldots$}{\\py'concatenate((v, w, ...))'}\n \\MoAItem{a range of integers}{$\\range n\\equiv 0,1+\\range(n-1)$}{\\py'arange(n)'}\n \\MoAItem{compress}{${\\vec v}_l\\compress{\\vec v}_r$}{\\py''}\n \\MoAItem{}{${\\vec v}_l\\expand{\\vec v}_r$}{\\py''}\n % \\MoAItem{}{$$}{\\py''}\n % \\MoAItem{}{$$}{\\py''}\n}\n\n\nMoA defines various slicing operations:\n\\MoADefinition{\n \\MoAItem{take first $k$ items; take last $k$ items}{$k\\take{\\vec v}\\,;\\,{\\minus k}\\take{\\vec v} $}{\\py'v[:k] ; v[-k:]'}\n \\MoAItem{drop first $k$ items; drop last $k$ items}{$k\\drop{\\vec v}\\,;\\,{\\minus k}\\drop{\\vec v} $}{\\py'v[k:] ; v[:-k]'}\n \\MoAItem{}{$\\sigma\\take\\xi$}{\\py''}\n \\MoAItem{}{$\\sigma\\drop\\xi$}{\\py''}\n \\MoAItem{}{${\\vec v}\\take\\xi$}{\\py''}\n \\MoAItem{}{${\\vec v}\\drop\\xi$}{\\py''}\n}\n\nMoA defines the following relational operators,$(\\rop{R}, \\text{\\py'rop'}) \\in \\{(\\rop{>}, \\text{\\py'gt'}), (\\rop{<}, \\text{\\py'lt'}), (\\rop{=}, \\text{\\py'eq'}), \\ldots\\}$ :\n\\MoADefinition{\n \\MoAItem{element-wise relational operators}{$\\xi_l\\rop{R}\\xi_r$}{\\py'x_l.____(x_r).all()'}\n \\MoAItem{scalar extension of relational operators}{$\\xi\\rop{R}a$}{\\py'x.____(a*ones(x.shape)).all()'}\n \\MoAItem{}{$a\\,R^{*}\\xi$}{\\py''}\n}\n\nMoA defines the following binary and unary operators, $(\\op{op},\\text{\\py'op'})\\in\\{\n(\\op+, \\text{\\py'add'}), (\\op-, \\text{\\py'sub'}), (\\op\\times, \\text{\\py'mul'}), \\ldots\n\\}$:\n\\MoADefinition{\n \\MoAItem{element-wise binary operators}{$\\xi_l\\op{op}\\xi_r$}{\\py'x_l.____(x_r)'}\n \\MoAItem{scalar extension of binary operators}{$\\xi\\op{op}a\\quad;\\quad a\\op{op}\\xi$}{\\py'x.__op__(a*ones(x.shape))'}\n \\MoAItem{scalar extension of unary operators}{$\\uop{uop}\\xi$}{\\py'x.____()'}\n}\n\nMoA defines various array transformation operators:\n\\MoADefinition{\n \\MoAItem{reshape a vector}{$\\vec s\\reshape\\vec v$}{\\py'v.reshape(s)'}\n \\MoAItem{reverse vector items}{$\\reverse \\vec v$}{\\py'v[::-1]'}\n \\MoAItem{}{$\\rotate{\\sigma}\\xi$}{\\py''}\n \\MoAItem{transpose of an array}{$\\transpose{\\vec v}\\xi$}{\\py''}\n \\MoAItem{reduction}{$\\red{op}\\xi\\equiv(\\getitem\\xi{<0>})\\op{op}\\red{op}(1\\drop\\xi)$}{}\n}\n\nOperations with arrays\n\\MoADefinition{\n \\MoAItem{higher order operation}{\n $\\getitem{\\hop{f}{\\vec d}}{\\vec i} \\equiv f(\\getitem\\xi{\\vec i})$\n }{}\n \\MoAItem{outer product $\\xi_l\\outerprod{\\op{op}}\\xi_r$}{\n \\begin{align*}\n \\getitem{\\xi_l\\outerprod{\\op{op}}\\xi_r}{(\\vec i\\cat\\vec j)} \\equiv\n (\\getitem{\\xi_l}{\\vec i}) \\op{op} (\\getitem{\\xi_r}{\\vec j})\n \\end{align*}\n }\n {\\py|outer(x_l, x_r) FIXME|}\n \\MoAItem{inner product $\\xi_l\\innerprod{\\op{op_0}}{\\op{op_1}}\\xi_r$}{\n \\begin{align*}\n \\getitem{\\xi_l\\innerprod{\\op{op_0}}{\\op{op_1}}\\xi_r}{(\\vec i\\cat \\vec j)}\n \\equiv\\\\ \n \\reduce{\\op{op_0}} (\\getitem{\\xi_l}{\\vec i}) \\op{op_1} (\\getitem{\\xi_r}{(\\range q\\hop{\\cat}{\\vcc01}\\vec j)})\n \\end{align*}\n}{\\py|inner(x_l, x_r) FIXME|}\n}\n\nTODO:\\\\\n\\begin{tabularx}{\\textwidth}{lX}\n\\begin{tabular}[m]{m{0.2\\linewidth}|l|m{0.45\\linewidth}|}\n MoA & Python & Description \\\\\\hline\n$\\vec i\\psi\\xi;\\qquad \\size\\vec i <\\size\\shape\\xi$ & \\verb|x[i+(Ellipsis,)]| & slice of an array\\\\\n$\\xi_l\\psi\\xi_r$ &&\\\\\n\\multicolumn2{l|}{$\\id{op}\\ldots \\sigma\\op{op}\\id{op}\\equiv\\sigma$}& identity of an operator $\\op{op}$\\\\\n$\\scan{op}$ && \\\\\n$\\xi_l,\\xi_r$ & & \\\\\n$\\xi,\\sigma\\equiv \\xi, (1,(1\\drop{\\shape\\xi}))\\shape\\sigma$ &&\\\\\n$\\vec v[\\vec x]$ &&\\\\\n$\\gu\\vec v\\,;\\,\\gd\\vec v$ &&\\\\\n$\\hop{f}{\\vec a}\\xi$&&\\\\\n$\\xi_l(\\hop{f}{\\vec a})\\xi_r$&&\\\\\n\\end{tabular} \n\\end{tabularx}\n\n\nReferencing Travis \\cite{travis0}, and Mullin \\cite{mul00}.\n\\bibliography{paper}\n\\bibliographystyle{plain}\n\\end{document}\n", "meta": {"hexsha": "d677c7c06b75a6fde9b818bc9adec1b891f36c1a", "size": 15495, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notation.tex", "max_stars_repo_name": "Quansight-Labs/uarray-docs", "max_stars_repo_head_hexsha": "8c9f9d766649d34922e3cf670d91da8746fd67aa", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2018-09-21T17:38:58.000Z", "max_stars_repo_stars_event_max_datetime": "2020-06-30T00:37:25.000Z", "max_issues_repo_path": "notation.tex", "max_issues_repo_name": "Quansight-Labs/uarray-docs", "max_issues_repo_head_hexsha": "8c9f9d766649d34922e3cf670d91da8746fd67aa", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-09-27T10:55:22.000Z", "max_issues_repo_issues_event_max_datetime": "2018-12-08T19:02:27.000Z", "max_forks_repo_path": "notation.tex", "max_forks_repo_name": "Quansight-Labs/uarray-docs", "max_forks_repo_head_hexsha": "8c9f9d766649d34922e3cf670d91da8746fd67aa", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2018-11-28T01:47:49.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-24T15:55:45.000Z", "avg_line_length": 41.4304812834, "max_line_length": 269, "alphanum_fraction": 0.6673765731, "num_tokens": 5255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324803738429, "lm_q2_score": 0.7662936324115011, "lm_q1q2_score": 0.6247074586455098}} {"text": "\n\\subsection{Greedy search}\n\nFind absolute distance from goal for each node. choose node with shortest distance.\n\nCost of each is \\(f(n)=h(n)\\), where \\(h(n)\\) is the heuristic cost of node \\(n\\).\n\n", "meta": {"hexsha": "0278978bfc95e6f05d0c8b632ed1dd524e5dc797", "size": 198, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/computer/treeHeuristic/01-01-greedy.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/computer/treeHeuristic/01-01-greedy.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/computer/treeHeuristic/01-01-greedy.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 24.75, "max_line_length": 83, "alphanum_fraction": 0.702020202, "num_tokens": 51, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8933094032139576, "lm_q2_score": 0.6992544335934766, "lm_q1q2_score": 0.6246505607681025}} {"text": "\\section{Mixture Models for Jet Tagging}\n\\label{sec:tagging}\n\n\\subsection{Gaussian Mixture Models}\n\nAs a first step we pick the boosted top tagging problem. From Figure \\ref{fig:efp-histogram}, it is apparent that the jets are points N-dimentional space (here 1000 dimentional, we have limited to EFPs of degree <= 7) which originate from several multivariate distributions. The obvious choice to model this data is to fit a Bayesian Gaussian Mixture Model, each distribution being a multivariate Gaussian:\n\\begin{equation}\n f_{pdf}(X) = \\frac{1}{\\sqrt{(2\\pi)^k \\vert \\Sigma \\vert}} e^{-\\frac{1}{2} (x - \\mu)^T \\Sigma^{-1} (x - \\mu)}\n\\end{equation}\nWe train the model using Expectation Maximization algorithm.\nThe performance of the algorithm is at par with that of LDA (89.2\\% on the top tagging dataset).\n\n\n\\subsection{Maxwellian Mixture Models}\n\nThe performance of the GMM is marred by the fact that the distributions of EFPs are actually maxwellian. The skew of the distributions is high for the top quarks even at low degrees and increases as the degree of the EFP increases. Therefore it behooves us to model the data using the Maxwellian distribution, which follows.\n\n\nWe model the multivariate form of the Maxwell-Boltzmann Distribution as follows ~\\cite{maxwellian_multivariate}.\n\\begin{equation}\n f_{pdf}(X) = \\frac{b^{1 + \\frac{n}{2}} \\vert B \\vert^\\frac{n}{2} \\Gamma(\\frac{n}{2})}{\\pi^{\\frac{n}{2}} \\Gamma(1 + \\frac{n}{2})} [XBX^T] e^{-b(XBX^T)}\n\\end{equation}\nThe parameters being learnt will be the $n \\times n$ sized matrix $B$. $X$ is the input vector to the model of shape $n \\times 1$ and the the constant $b$ is the determinant of the matrix $B$, which together with the Gamma functions serves the task of normalization.\n\n\\textcolor{red}{This distribution will be trained on using the Expectation Maximization algorithm, details of the E- step and M- step will be added here, together with any improvements in performance. ~\\cite{maxwellian_plasma}}\n\n\n\\subsection{Performance of the Models}\n\nFollowing is the performance of the algorithms on the Top Tagging dataset \\cite{data_toptagging}.\n\\begin{table}\n \\caption{Performance on Top Tagging \\cite{tagging_review}}\n \\label{tab:1}\n \\begin{tabular}{lll}\n \\hline\\noalign{\\smallskip}\n Model Name & Accuracy & ROC AUC \\\\\n \\noalign{\\smallskip}\\hline\\noalign{\\smallskip}\n \\textbf{Bayessian Maxwellian Mixture Model with EFP} & UNK & UNK \\\\\n \\textbf{Bayesian Gaussian Mixture Model with EFP} & 89.224\\% & UNK \\\\\n Latent Dirichlet Allocation & 89.2\\% & 0.955 \\\\\n \\noalign{\\smallskip}\\hline\n Linear Discriminant Analysis with EFP & 93.2\\% & 0.980 \\\\\n ParticleNet (Graph Neural Net on Point Cloud) & 93.8\\% & 0.985 \\\\\n \\noalign{\\smallskip}\\hline\n \\end{tabular}\n\\end{table}\n\n\\textcolor{red}{Add accuracy figures on the QCD tagging dataset \\cite{data_qcdtagging}}\n\n\\subsection{Features of probabilistic tagging}\n\nWhile boosted top tagging is solved to a very high accuracy, the following attributes are still sought for:\n\\begin{itemize}\n \\item To know what confidence we have tagged a single jet with? - As seen from the histograms, some jets are very clearly in the Top or QCD domain, the overlap of the two is the set of jets where most algorithms fail. When tagging, a model should be able to output both the class label and the confidence figure.\n \\item Resilliance to unseen input, and ability to fail gracefully if the input jet is not unlike what the model has seen before? This may be due to errors by the clustering algorithm which has clustered multiple jets together or because of an unknown decay type.\n\\end{itemize}\nA probabilistic Bayesian generative model is the best attempt at modelling the probability with which we are tagging a jet.\n\n\\textcolor{red}{Show experimental proof that we are indeed resilliant against mutliple-jet in one jet image and against unknown jet types}.\n\\textcolor{yellow}{We need to explore what set of features help models like ParticleNet \\cite{particle_net} perform better in the overlap zone, and it's performance relative to our confidence, as in the Orange comparative scatters.}", "meta": {"hexsha": "0393a9eff19aa637a812316020fcab593c152b2c", "size": 4292, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/tex/tagging.tex", "max_stars_repo_name": "Blizzard57/jet-tagging", "max_stars_repo_head_hexsha": "b338b0d55380a6d21e2f78d7098650b0806f7cfc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/tex/tagging.tex", "max_issues_repo_name": "Blizzard57/jet-tagging", "max_issues_repo_head_hexsha": "b338b0d55380a6d21e2f78d7098650b0806f7cfc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/tex/tagging.tex", "max_forks_repo_name": "Blizzard57/jet-tagging", "max_forks_repo_head_hexsha": "b338b0d55380a6d21e2f78d7098650b0806f7cfc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.5333333333, "max_line_length": 406, "alphanum_fraction": 0.7269338304, "num_tokens": 1098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8933093946927838, "lm_q2_score": 0.6992544273261175, "lm_q1q2_score": 0.6246505492109432}} {"text": "%!TEX root = ../OGUSAdoc.tex\n\nThis appendix contains derivations from the theory in the body of this book.\n\n\n\\section{Properties of the CES Production Function}\\label{SecAppDerivCES}\n\n The constant elasticity of substitution (CES) production function of capital and labor was introduced by \\citet{Solow:1956} and further extended to a consumption aggregator by \\citet{Armington:1969}. The CES production function of aggregate capital $K_t$ and aggregate labor $L_t$ we use in Chapter \\ref{Chap_Firms} is the following,\n \\begin{equation}\\tag{\\ref{EqFirmsCESprodfun}}\n Y_t = F(K_t, L_t) \\equiv Z_t\\biggl[(\\gamma)^\\frac{1}{\\ve}(K_t)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}(e^{g_y t}L_t)^\\frac{\\ve-1}{\\ve}\\biggr]^\\frac{\\ve}{\\ve-1} \\quad\\forall t\n \\end{equation}\n where $Y_t$ is aggregate output (GDP), $Z_t$ is total factor productivity, $\\gamma$ is a share parameter that represents the capital share of income in the Cobb-Douglas case ($\\ve=1$), and $\\ve$ is the elasticity of substitution between capital and labor. The stationary version of this production function is given in Chapter \\ref{Chap_Stnrz}. We drop the $t$ subscripts, the ``$\\:\\hat{\\,}\\:$''stationary notation, and use the stationarized version of the production function \\eqref{EqStnrzCESprodfun} for simplicity.\n \\begin{equation}\\tag{\\ref{EqStnrzCESprodfun}}\n Y = Z\\biggl[(\\gamma)^\\frac{1}{\\ve}(K)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}(L)^\\frac{\\ve-1}{\\ve}\\biggr]^\\frac{\\ve}{\\ve-1}\n \\end{equation}\n The Cobb-Douglas production function is a nested case of the general CES production function with unit elasticity $\\ve=1$.\n \\begin{equation}\\label{EqAppDerivCES_CobbDoug}\n Y = Z(K)^\\gamma(L)^{1-\\gamma}\n \\end{equation}\n\n\n\n \\subsection{Wages as a function of interest rates}\\label{SecAppDerivCESwr}\n\n An important property of the CES production function for the solution method of \\ogindia is that the interest rate $r_t$ and wage $w_t$ are functions of the capital labor ratio. This property implies that the wage every period $w_t$ is just a function of the interest rate $r_t$, and vice versa. The first step is to show that the output-capital ratio ($Y/K$) and the output-labor ($Y/L$) ratio are both functions of the capital-labor ratio ($K/L$).\n \\begin{equation}\\label{EqAppDerivCES_YL}\n \\begin{split}\n Y &= Z\\biggl[(\\gamma)^\\frac{1}{\\ve}(K)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}(L)^\\frac{\\ve-1}{\\ve}\\biggr]^\\frac{\\ve}{\\ve-1} \\\\\n &= Z\\left[(\\gamma)^\\frac{1}{\\ve}(K)^\\frac{\\ve-1}{\\ve}\\left(\\frac{L^\\frac{\\ve-1}{\\ve}}{L^\\frac{\\ve-1}{\\ve}}\\right) + (1-\\gamma)^\\frac{1}{\\ve}(L)^\\frac{\\ve-1}{\\ve}\\right]^\\frac{\\ve}{\\ve-1} \\\\\n &= ZL\\left[(\\gamma)^\\frac{1}{\\ve}\\left(\\frac{K}{L}\\right)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\right]^\\frac{\\ve}{\\ve-1} \\\\\n \\Rightarrow\\quad \\frac{Y}{L} &= Z\\left[(\\gamma)^\\frac{1}{\\ve}\\left(\\frac{K}{L}\\right)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\right]^\\frac{\\ve}{\\ve-1}\n \\end{split}\n \\end{equation}\n\n \\begin{equation}\\label{EqAppDerivCES_YK}\n \\begin{split}\n Y &= Z\\biggl[(\\gamma)^\\frac{1}{\\ve}(K)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}(L)^\\frac{\\ve-1}{\\ve}\\biggr]^\\frac{\\ve}{\\ve-1} \\\\\n &= Z\\left[(\\gamma)^\\frac{1}{\\ve}(K)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}(L)^\\frac{\\ve-1}{\\ve}\\left(\\frac{K^\\frac{\\ve-1}{\\ve}}{K^\\frac{\\ve-1}{\\ve}}\\right)\\right]^\\frac{\\ve}{\\ve-1} \\\\\n &= ZK\\left[(\\gamma)^\\frac{1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\left(\\frac{L}{K}\\right)^\\frac{\\ve-1}{\\ve}\\right]^\\frac{\\ve}{\\ve-1} \\\\\n \\Rightarrow\\quad \\frac{Y}{K} &= Z\\left[(\\gamma)^\\frac{1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\left(\\frac{L}{K}\\right)^\\frac{\\ve-1}{\\ve}\\right]^\\frac{\\ve}{\\ve-1}\n \\end{split}\n \\end{equation}\n\n Solving for the firm's first order conditions for capital and labor demand from profit maximization \\eqref{EqStnrzProfit} gives the following equations in their respective stationarized forms from Chapter \\ref{Chap_Stnrz}.\n \\begin{align}\n w &= (Z)^\\frac{\\ve-1}{\\ve}\\left[(1-\\gamma)\\left(\\frac{Y}{L}\\right)\\right]^\\frac{1}{\\ve} \\tag{\\ref{EqStnrzFOC_L}} \\\\\n r &= (1 - \\tau^{corp})(Z)^\\frac{\\ve-1}{\\ve}\\left[\\gamma\\left(\\frac{Y}{K}\\right)\\right]^\\frac{1}{\\ve} - \\delta + \\tau^{corp}\\delta^\\tau \\tag{\\ref{EqFirmFOC_K}}\n \\end{align}\n As can be seen from \\eqref{EqStnrzFOC_L} and \\eqref{EqFirmFOC_K}, the wage $w$ and interest rate $r$ are functions of $Y/L$ and $Y/K$, respectively. Equations \\eqref{EqAppDerivCES_YL} and \\eqref{EqAppDerivCES_YK} show that both $Y/L$ and $Y/K$ are functions of the capital-labor ratio $K/L$. Substituting \\eqref{EqAppDerivCES_YL} and \\eqref{EqAppDerivCES_YK} into \\eqref{EqStnrzFOC_L} and \\eqref{EqFirmFOC_K}, respectively, gives expressions of the wage $w$ and interest rate $r$ in terms of the capital-labor ratio $K/L$.\n \\begin{align}\n w &= (1-\\gamma)^\\frac{1}{\\ve}Z\\left[(\\gamma)^\\frac{1}{\\ve}\\left(\\frac{K}{L}\\right)^\\frac{\\ve-1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\right]^\\frac{1}{\\ve-1} \\label{EqAppDerivCES_FOCL} \\\\\n r &= (1 - \\tau^{corp})(\\gamma)^\\frac{1}{\\ve}Z\\left[(\\gamma)^\\frac{1}{\\ve} + (1-\\gamma)^\\frac{1}{\\ve}\\left(\\frac{L}{K}\\right)^\\frac{\\ve-1}{\\ve}\\right]^\\frac{1}{\\ve-1} - \\delta + \\tau^{corp}\\delta^\\tau \\label{EqAppDerivCES_FOCK}\n \\end{align}\n In the Cobb-Douglas unit elasticity case ($\\ve=1$) of the CES production function, the first order conditions are more easily expressed in terms of the capital-labor ratio.\n \\begin{align}\n \\text{if}\\:\\:\\,\\ve=1:\\quad w &= (1-\\gamma)Z\\left(\\frac{K}{L}\\right)^\\gamma \\label{EqAppDerivCES_CDFOCL} \\\\\n \\text{if}\\:\\:\\:\\ve=1:\\quad r &= (1 - \\tau^{corp})\\gamma Z\\left(\\frac{L}{K}\\right)^{1-\\gamma} - \\delta + \\tau^{corp}\\delta^\\tau \\label{EqAppDerivCES_CDFOCK}\n \\end{align}\n\n With $w$ and $r$ expressed in terms of the capital-labor ratio $K/L$ in \\eqref{EqAppDerivCES_FOCL} and \\eqref{EqAppDerivCES_FOCK}, we can write the expressions for the capital-labor ratio as a function of the interest rate and, therefore, the wage as a function of the interest rate. We first solve equation \\eqref{EqAppDerivCES_FOCK} for the capital-labor ratio to get the expression for $K/L$ as a function of the interest rate $r$.\n \\begin{equation}\\label{EqAppDerivCES_KLr}\n \\frac{K}{L} = \\left(\\frac{(1-\\gamma)^\\frac{1}{\\ve}}{\\left[\\frac{r + \\delta - \\tau^{corp}\\delta^\\tau}{(1 - \\tau^{corp})\\gamma^\\frac{1}{\\ve}Z}\\right]^{\\ve-1} - \\gamma^\\frac{1}{\\ve}}\\right)^\\frac{\\ve}{\\ve-1}\n \\end{equation}\n In the Cobb-Douglas unit elasticity case ($\\ve=1$), we solve equation \\eqref{EqAppDerivCES_CDFOCK} for the capital-labor ration to get the expression for $K/L$ as a function of the interest rate $r$.\n \\begin{equation}\\label{EqAppDerivCES_CDKLr}\n \\text{if}\\:\\:\\:\\ve=1:\\quad \\frac{K}{L} = \\left[\\frac{(1 - \\tau^{corp})\\gamma Z}{r + \\delta - \\tau^{corp}\\delta^\\tau}\\right]^\\frac{1}{1-\\gamma}\n \\end{equation}\n\n Substituting \\eqref{EqAppDerivCES_KLr} into \\eqref{EqAppDerivCES_FOCL} gives the expression for the wage $w$ as a function of the interest rate $r$ in the general CES case.\n \\begin{equation}\\label{EqAppDerivCES_wr}\n w = (1-\\gamma)^\\frac{1}{\\ve}Z\\left[(\\gamma)^\\frac{1}{\\ve}\\left(\\frac{(1-\\gamma)^\\frac{1}{\\ve}}{\\left[\\frac{r + \\delta - \\tau^{corp}\\delta^\\tau}{(1 - \\tau^{corp})\\gamma^\\frac{1}{\\ve}Z}\\right]^{\\ve-1} - \\gamma^\\frac{1}{\\ve}}\\right) + (1-\\gamma)^\\frac{1}{\\ve}\\right]^\\frac{1}{\\ve-1}\n \\end{equation}\n In the Cobb-Douglas unit elasticity case ($\\ve=1$), we substitute \\eqref{EqAppDerivCES_CDKLr} into \\eqref{EqAppDerivCES_CDFOCL} gives the expression for the wage $w$ as a function of the interest rate $r$.\n \\begin{equation}\\label{EqAppDerivCES_CDwr}\n \\text{if}\\:\\:\\:\\ve=1:\\quad w = (1-\\gamma)Z\\left[\\frac{(1 - \\tau^{corp})\\gamma Z}{r + \\delta - \\tau^{corp}\\delta^\\tau}\\right]^\\frac{\\gamma}{1-\\gamma}\n \\end{equation}\n", "meta": {"hexsha": "64c832db73c0bf8bda79762e1303788396997433", "size": 7833, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/LaTeXsource/Chapters/Chap_Deriv.tex", "max_stars_repo_name": "keshavchoudhary87/OG-India", "max_stars_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2018-05-31T11:19:15.000Z", "max_stars_repo_stars_event_max_datetime": "2018-05-31T11:19:15.000Z", "max_issues_repo_path": "docs/LaTeXsource/Chapters/Chap_Deriv.tex", "max_issues_repo_name": "keshavchoudhary87/OG-India", "max_issues_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": 11, "max_issues_repo_issues_event_min_datetime": "2019-08-16T15:40:52.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-16T07:07:15.000Z", "max_forks_repo_path": "docs/LaTeXsource/Chapters/Chap_Deriv.tex", "max_forks_repo_name": "keshavchoudhary87/OG-India", "max_forks_repo_head_hexsha": "269ee172b837882c826ee7f99507d93f9643128e", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 44, "max_forks_repo_forks_event_min_datetime": "2019-08-16T15:10:39.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-08T07:03:26.000Z", "avg_line_length": 101.7272727273, "max_line_length": 526, "alphanum_fraction": 0.6565811311, "num_tokens": 2875, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951143326726, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6246295278790287}} {"text": "\\section{Introduction}\\label{introduction}\n%=========================================\n\nThe aim of this paper is to introduce an approach to 2D and 3D\ncomputational topology and geometry using a \\emph{GraphBLAS} sparse matrix\nrepresentation of \\emph{chain complexes} with linear operators $\\partial_p$ and $\\delta_{p-1} = \\partial_p^\\top$ between linear \\emph{chain} spaces $C_p$:\n\\[ \nC_\\bullet = (C_p, \\partial_p) := \nC_3 \\ \n\\substack{\n\\delta_2 \\\\\n\\longleftarrow \\\\[-1mm]\n\\longrightarrow \\\\\n\\partial_3 \n}\n\\ C_2 \\ \n\\substack{\n\\delta_1 \\\\\n\\longleftarrow \\\\[-1mm]\n\\longrightarrow \\\\\n\\partial_2 \n}\n\\ C_1 \\ \n\\substack{\n\\delta_0 \\\\\n\\longleftarrow \\\\[-1mm]\n\\longrightarrow \\\\\n\\partial_1 \n}\n\\ C_0 .\n\\] \nLet us to remark preliminary that when $p\\in \\{0,1\\}$, a chain complex is a representation of a \\emph{graph}, with $[\\partial_1]$ the \\emph{incidence matrix} between 1-chains (edges) in $C_1$ and 0-chains (vertices) in $C_0$. Also, the multiplication $[\\partial_1][\\delta_0]=[\\partial_1][\\partial_1]^t$ gives the \\emph{adjacency} matrix of the graph, while the diagonal entries provide the degrees of vertices, defined by the number of incident edges on each vertex. The \\emph{GraphBLAS} standard~\\cite{GraphBLAS:standard} provides a small set of matrix primitives to compute graph properties that, combined together, allow for easy implementation of fast algorithms on large graphs.\n\nIn several application of computational geometry and topology---including\ngeo-mapping, building information modeling (BIM), medical imaging, CAD and\nsolid modeling, virtual and augmented reality, finite element modeling\nand simulation (FEM, FEA), etc.---the amount and detail of 2D and/or 3D data\ncontinue to grow. Analogously, the need for an unified approach to\ngraph algorithms and for unified and simplified interfaces, has been\nwell intercepted by the GraphBLAS initiative and the related \nstandardization effort~\\href{http://graphblas.org}{\\texttt{graphblas.org}}. \nThis paper aim is to show that the domaincovering of this standard library on graphs can be greatly extended, to cover the representation of more general \\emph{cellular $d$-complexes} ($1\\leq d\\leq 3$) and their $p$-skeletons ($0\\leq p\\leq 3$).\n\n(Co)chain complexes, as well (co)boundary operators, are well-known\nbasic tools of algebraic topology and homological algebra. In\nparticular, a \\emph{chain complex} is a graded sequence of such linear\noperators between graded linear spaces of ``chains''. A $p$-chain can be seen as a subset of a finite set $\\texttt{V}$ of Euclidean points whose affine hull has dimension $p$.\nChain spaces and sparse matrices are the components of the Linear Algebraic\nRepresentation (\\texttt{LAR})~\\cite{Dicarlo:2014:TNL:2543138.2543294}, that is being used for\nboundary, decompositive, and enumerative\nrepresentations~\\cite{Requicha:1980:RRS:356827.356833} of models of rigid solid objects.\n\nSome numerical methods aiming to integrate domain modeling, differential\ntopology, mathematical modeling, and physical simulations were based on chains and\ncochains, starting with~\\cite{PALMER1995733,Palmer1993}. In particular,\nDiscrete Exterior Calculus (DEC) with simplicial complexes was\nintroduced by \\cite{Hirani:2003:DEC:959640} and made popular by\n\\cite{Desbrun:2006:DDF:1185657.1185665,Elcott:2006:BYO:1185657.1185666}.\nFinite Element Exterior Calculus \n(FEEC) is an advance in the mathematics of finite element \nmethods~\\cite{arnold_falk_winther_2006,Arnold:2010,Arnold:2018} that employs differential\ncomplexes to construct stable numerical schemes. The Cell\nMethod~(CM) is a purely algebraic computational method\nfor modeling and\nsimulation~\\cite{Tonti:1975,Tonti:2013,Ferretti:2014}\nbased on boundary/coboundary maps and a direct discrete formulation of\nfield laws. Our own research in geometrical and physical modeling with\nchain and cochain complexes was introduced in\n\\cite{DiCarlo:2009:DPU:1629255.1629273,ieee-tase,Dicarlo:2014:TNL:2543138.2543294}.\n\nMore recently, we provided---using sparse\nmatrices---an algorithmic pipeline~\\cite{TSAS:19} to compute the \\emph{arrangement} of the\nEuclidean \\(d\\)-space (\\(d=2,3\\)), i.e.~the partition of\nit into a cellular \\(d\\)-complex, starting from a collection of (possibly intersecting) cellular\n(\\(d\\)-1)-complexes embedded in \\(\\E^d\\). In~\\cite{paoluzzi2019finite} we\nhave shown that the \\emph{atoms} of the Boolean algebra generated by\nsuch space partition correspond one-to-one to the columns of the $[\\partial_d]$ matrix\nof the boundary operator \\(\\partial_d : C_d \\to C_{d-1}\\). This allows for fast\nnative reconstruction of every solid expression in the \\emph{solid algebra} of\n\\(d\\)-space generated by the input terms, usually called Constructive Solid Geometry (CSG)\nin solid modeling~\\cite{Requicha:1980:RRS:356827.356833}.\n\nThe topological background of linear chain spaces and chain complexes is summarized in \nSection~\\ref{sec:background}, together with the concepts of boundary and coboundary linear maps between chain spaces.\nIn Section~\\ref{chain-adjacencies-and-incidensies} an operational definition of topological queries through composition of operators corresponding to products of their sparse matrices is discussed.\nIn Section~\\ref{boolean-algebras} an interpretation as a Boolean algebra of the partition of the Euclidean $d$-space produced by a collection of geometric objects is provided.\nThe implementation of such concepts using GraphBLAS is given in Section~\\ref{graphblas-implementation}. Some simple examples of topology computation using linear algebra are shown in Section~\\ref{examples}.\nIn the Conclusion Section we briefly summarize our main points, state our findings, and propose some possible extension of this approach.\n", "meta": {"hexsha": "45e0df8a2b840817a64e002060035b1e109a87d4", "size": 5696, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/introduction.tex", "max_stars_repo_name": "cvdlab/Chain-BLAS", "max_stars_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/introduction.tex", "max_issues_repo_name": "cvdlab/Chain-BLAS", "max_issues_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/introduction.tex", "max_forks_repo_name": "cvdlab/Chain-BLAS", "max_forks_repo_head_hexsha": "38a2413ccefd1bc47ae404215e3616d21b16a89e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.0, "max_line_length": 683, "alphanum_fraction": 0.7808988764, "num_tokens": 1518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951104066293, "lm_q2_score": 0.7401743620390162, "lm_q1q2_score": 0.624629524973072}} {"text": "\\section{Transaction Costs Problem}\n\\label{sec:83problem}\n\n\\minitoc[-6mm]{70mm}{4}\n\n\\vspace{-1.5em}\n\n\\paragraph{Description}\n\nIn the \\term{transaction costs problem,}\nthe individual can invest their money risk-free in bonds\n(with a fixed interest rate similar to a bank account)\nor in $m_{\\vstock} \\in \\nat$ different risk-affected stocks\n\\cite{Schober18Solving}.\nEvery stock transaction,\ni.e., buy $\\buy_{t,j}$ or sell $\\sell_{t,j}$,\ninflicts transaction costs $\\tac \\buysell_{t,j}$ ($\\tac \\in \\nonnegreal$)\nproportional to the amount $\\buysell_{t,j}$ bought or sold\n($j = 1, \\dotsc, m_{\\vstock}$).\nThe individual only wants to invest a fixed\namount $\\wealth_0$ in stocks, i.e., we omit the individual's income.\n\n\n\n\\subsection{Unnormalized Problem}\n\\label{sec:831unnormalized}\n\n\\paragraph{Consumption and state transition}\n\nIn the following,\n$\\stock_{t,j}$ denotes the fraction of the total wealth $\\wealth_t$\nthat is invested in the $j$-th stock.\nWe combine these \\term{stock fractions} $\\stock_{t,j}$\nin a vector $\\vstock_t \\ceq (\\stock_{t,1}, \\dotsc, \\stock_{t,m_{\\vstock}})$;\nsimilarly, $\\vbuysell_t \\ceq (\\buysell_{t,1}, \\dotsc, \\buysell_{t,m_{\\vstock}})$\ncombines buy and sell amounts.\nThen, the consumption can be computed as a residual variable\n(i.e., a variable that can be fully computed from $\\state$ and $\\policy$\nand is thus omitted from $\\policy$),\nwhich is given by\n\\begin{equation}\n \\consume_t\n \\ceq (1 - \\sumfcn(\\vstock_t)) \\wealth_t - \\bond_t -\n (1 + \\tac) \\sumfcn(\\vbuy_t) + (1 - \\tac) \\sumfcn(\\vsell_t),\n\\end{equation}\nwhere $\\sumfcn(\\*a) \\ceq \\tr{\\*1} \\*a$\nis the sum of all entries of $\\*a$.\nThe state transition is computed by adding the returns of bonds and stocks:%\n\\begin{equation}\n \\wealth_{t+1}\n \\ceq \\bond_t \\bondreturn_t +\n %\\sum_{j=1}^{m_{\\vstock}}\n %(\\stock_{t,j} \\wealth_t + \\buy_{t,j} - \\sell_{t,j}) \\stockreturn_{t,j},\\\\\n \\tr{(\\vstock_t \\wealth_t + \\vbuy_t - \\vsell_t)} \\vstockreturn_t,\\qquad\n \\vstock_{t+1}\n \\ceq \\frac{\n (\\vstock_t \\wealth_t + \\vbuy_t - \\vsell_t) \\compmult \\vstockreturn_t\n }{\\wealth_{t+1}},\n\\end{equation}\nwhere $\\bondreturn_t \\in \\real$ is the bond interest rate,\n$\n \\vstockreturn_t\n = (\\stockreturn_{t,1}, \\dotsc, \\stockreturn_{t,m_{\\vstock}})\n \\in \\real^{m_{\\vstock}}\n$\nis the vector of (stochastic) stock return rates, and\n$\\compmult$ is component-wise multiplication.\n\n\n\n\\subsection{Normalization}\n\\label{sec:832normalized}\n\n\\paragraph{State transition}\n\nThe above equations can be normalized with respect to the wealth\n$\\wealth_t$:\nBy setting\n$\\normconsume_t \\ceq \\consume_t/\\wealth_t$,\n$\\normbond_t \\ceq \\bond_t /\\wealth_t$, and\n$\\vnormbuysell_t \\ceq \\vbuysell_t/\\wealth_t$, we obtain\n\\begin{subequations}\n \\label{eq:normalizedTCPStateTransition}\n \\begin{align}\n \\normconsume_t\n &\\mathrel{\\righthphantom{=}{\\ceq}}\n (1 - \\sumfcn(\\vstock_t)) - \\normbond_t -\n (1 + \\tac) \\sumfcn(\\vnormbuy_t) + (1 - \\tac) \\sumfcn(\\vnormsell_t),\\\\\n \\wealthratio_{t+1}\n &\\ceq \\normbond_t \\bondreturn_t +\n \\tr{(\\vstock_t + \\vnormbuy_t - \\vnormsell_t)} \\vstockreturn_t,\n \\qquad(= \\wealth_{t+1}/\\wealth_t)\\\\\n \\vstock_{t+1}\n &\\mathrel{\\righthphantom{=}{\\ceq}}\n \\frac{\n (\\vstock_t + \\vnormbuy_t - \\vnormsell_t) \\compmult \\vstockreturn_t\n }{\\wealthratio_{t+1}},\n \\end{align}\n\\end{subequations}\nwhere $\\normconsume_t$ and $\\wealthratio_{t+1}$ are residual\nvariables that specify \\term{normalized consumption} and\n\\term{wealth ratio,} respectively.\nAll in all, the resulting dynamic portfolio choice model has\nthe following variables:\n\\begin{itemize}\n \\item\n $\\centerhphantom{d}{m_{\\stochastic}} = m_{\\vstock}$\n state variables $\\normstate_t$:\n Stock fractions $\\stock_{t,1}, \\dotsc, \\stock_{t,m_{\\vstock}}$\n \n \\item\n $\\centerhphantom{m_{\\policy}}{m_{\\stochastic}} = 2m_{\\vstock} + 1$\n policy variables $\\normpolicy_t$:\n Normalized bonds $\\normbond_t$,\n normalized buy amounts $\\normbuy_{t,1}, \\dotsc, \\normbuy_{t,m_{\\vstock}}$ and\n normalized sell amounts $\\normsell_{t,1}, \\dotsc, \\normsell_{t,m_{\\vstock}}$\n \n \\item\n $m_{\\stochastic} = m_{\\vstock}$\n stochastic variables $\\stochastic_t$:\n Stock return rates $\\stockreturn_{t,1}, \\dotsc, \\stockreturn_{t,m_{\\vstock}}$\n\\end{itemize}\nThe state space and policy space constraints are given by\n\\begin{subequations}\n \\label{eq:normalizedTCPConstraints}\n \\newcommand*{\\centereqline}[1]{%\n \\mathclap{\\hphantom{\\mathrm{(8.99a)}}#1}%\n }%\n \\begin{gather}\n \\label{eq:normalizedTCPConstraintsShort}\n \\centereqline{\n \\vstock_t \\ge \\*0,\\qquad\n \\sumfcn(\\vstock_t) \\le 1,\\qquad\n \\normbond_t \\ge 0,\\qquad\n \\vnormbuysell_t \\ge \\*0,\\qquad\n \\vnormsell_t \\le \\vstock_t,\\qquad\n \\wealthratio_{t+1} \\ge 0,\n }\\\\\n \\label{eq:normalizedTCPConstraintsLong}\n \\centereqline{\n \\normconsume_{\\min} + \\normbond_t +\n (1 + \\tac) \\sumfcn(\\vnormbuy_t) - (1 - \\tac) \\sumfcn(\\vnormsell_t)\n \\le 1 - \\sumfcn(\\vstock_t),\n }\n \\end{gather}\n\\end{subequations}\nwhere $\\normconsume_{\\min} \\in \\nonnegreal$ is some minimal consumption\nthat must be maintained.\n\n\\paragraph{Bellman equation}\n\nConsequently, the Bellman equation \\eqref{eq:gridBellmanCET}\nafter the certainty-equiva\\-lent transformation has to be\nnormalized as well.\nBy setting $\\normcetvalueintp_t(\\state_t^{(k)})\n\\ceq \\cetvalueintp_t(\\state_t^{(k)})/\\wealth_t$, we obtain\n\\begin{subequations}\n \\begin{align}\n %\\normcetvalueintp_t(\\state_t^{(k)})\n %&= \\frac{\\cetvalueintp_t(\\state_t^{(k)})}{\\wealth_t}\\\\\n %&= \\max_{\\policy_t} \\left(\n % \\left(\n % \\left(\n % \\frac{\\consume_t(\\state_t^{(k)}, \\policy_t)}{\\wealth_t}\n % \\right)^{1-\\riskav} +\n % \\patience \\expectation[t]{\n % \\left(\n % \\frac{\n % \\cetvalueintp_{t+1}(\n % \\statefcn_t(\\state_t^{(k)}, \\policy_t, \\stochastic_t)\n % )\n % }{\n % \\wealth_t\n % }\n % \\right)^{1-\\riskav}\n % }\n % \\right)^{1/(1-\\riskav)}\n %\\right)\\\\\n &\\hphantom{=}\\hspace{0.6em} \\normcetvalueintp_t(\\state_t^{(k)})\n = \\wealth_t^{-1} \\cetvalueintp_t(\\state_t^{(k)})\\\\\n &= \\max_{\\policy_t} \\left(\n \\left(\n \\left(\n \\wealth_t^{-1} \\consume_t(\\state_t^{(k)}, \\policy_t)\n \\right)^{1-\\riskav} +\n \\patience \\expectation[t]{\n \\left(\n \\wealth_t^{-1} \\cetvalueintp_{t+1}(\n \\statefcn_t(\\state_t^{(k)}, \\policy_t, \\stochastic_t)\n )\n \\right)^{1-\\riskav}\n }\n \\right)^{1/(1-\\riskav)}\n \\right)\\\\\n \\label{eq:normalizedTCPBellmanEquation}\n &= \\max_{\\normpolicy_t} \\left(\n \\left(\n \\normconsume_t(\\state_t^{(k)}, \\normpolicy_t)^{1-\\riskav} +\n \\patience \\expectation[t]{\n \\bigl(\n \\wealthratio_{t+1} \\normcetvalueintp_{t+1}(\n \\normstatefcn_t(\\state_t^{(k)}, \\normpolicy_t, \\stochastic_t)\n )\n \\bigr)^{1-\\riskav}\n }\n \\right)^{1/(1-\\riskav)}\n \\right).\n \\end{align}\n\\end{subequations}\nThis means that compared with %the unnormalized Bellman equation\n\\eqref{eq:gridBellmanCET},\nthe value function in the expectation has to be multiplied by\nthe wealth ratio $\\wealthratio_{t+1}$ introduced above in\n\\eqref{eq:normalizedTCPStateTransition}.\nSince there is no inheritance, the optimal terminal solution\nis to sell all stocks and consume everything:\n\\begin{equation}\n \\normcetvalueintp_t(\\state_T^{(k)})\n = 1 - \\tac \\sumfcn(\\vstock_T^{(k)}),\\quad\n \\normbond_T^{\\opt}(\\state_T^{(k)})\n = 0,\\quad\n \\vnormbuy[\\opt]_T(\\state_T^{(k)})\n = \\*0,\\quad\n \\vnormsell[\\opt]_T(\\state_T^{(k)})\n = \\vstock_T^{(k)}.\n\\end{equation}\n\n\n\n\\subsection{State Space Cropping}\n\\label{sec:833cropping}\n\n\\paragraph{Sparse grids on non-rectangular domains}\n\nUnfortunately, the constraint $\\sumfcn(\\vstock_t) \\le 1$\nfrom \\cref{eq:normalizedTCPConstraints} limits the feasible state space\nregion to a proper subset (which is the unit simplex)\nof the unit hypercube $\\clint{\\*0, \\*1}$,\nwhich impedes the direct application of sparse grids.\nThere are three possible remedies:\ntransforming the unit hypercube to the feasible state space,\napplying extrapolation techniques as discussed in\n\\cref{sec:825interpolation}, or\nchoosing a model-tailored approach to obtain\nfunction values outside the feasible state space.\n\n\\paragraph{Virtual selling of stocks}\n\nWe choose the third remedy and \\term{virtually sell,}\nif $\\sumfcn(\\vstock_t) > 1$,\nas many stocks as needed to meet the constraint $\\sumfcn(\\vstock_t) \\le 1$.\nWe already might need to sell stocks\neven if $\\sumfcn(\\vstock_t)$ is smaller but close to one\nin order to satisfy the minimum consumption requirement\n\\eqref{eq:normalizedTCPConstraintsLong}.\nIn detail, we replace $\\vstock_t$ by $\\normcropfactor \\vstock_t$\nwhenever $\\normcropfactor < 1$,\nwhere $\\normcropfactor \\in \\posreal$ is a \\term{cropping factor}\nthat is determined by\n\\begin{equation}\n \\label{eq:virtualSelling}\n \\Bigl[\n 1 - \\tac\\, \\bigl(\n \\sumfcn(\\vstock_t) - \\sumfcn(\\normcropfactor\\vstock_t)\n \\bigr)\n \\Bigr]\n \\cdot \\bigl(1 - \\sumfcn(\\normcropfactor\\vstock_t)\\bigr)\n = \\normconsume_{\\min}.\n\\end{equation}\nHere, $\\bigl(\\sumfcn(\\vstock_t) - \\sumfcn(\\normcropfactor\\vstock_t)\\bigr)$\nis the amount of virtually sold stocks.\nHence, the term in square brackets is the fraction of wealth\nthat is still available after deducting the induced transaction costs.\nThe product of this term with\n$\\bigl(1 - \\sumfcn(\\normcropfactor\\vstock_t)\\bigr)$\nis the fraction of wealth that can be consumed after the virtual selling,\nwhich needs to be at least $\\normconsume_{\\min}$.\nSolving \\cref{eq:virtualSelling} for $\\normcropfactor$ and\nchoosing the positive solution, we finally obtain\n\\begin{equation}\n \\newcommand*{\\sumX}{\\sumfcn(\\vstock_t)}\n \\newcommand*{\\cMin}{\\normconsume_{\\min}}\n \\normcropfactor\n \\ceq \\frac{\n \\tac\\, \\bigl(1 + \\sumX\\bigr) - 1 +\n \\sqrt{\n \\tac^2\\, \\bigl(1 - \\sumX\\bigr)^2\n - 2 \\tac\\, \\bigl(2 \\cMin - 1 + \\sumX\\bigr) + 1\n }\n }{\n 2 \\tac \\sumX\n }.\n\\end{equation}\n\n\n\n\\subsection{Euler Equation Errors}\n\\label{sec:834eulerErrors}\n\n\\paragraph{Motivation}\n\nDue to the curse of dimensionality,\nreasonably accurate full grid reference solutions\nof the transaction costs problem can only be computed\nif the number $m_{\\vstock}$ of stocks is small.\nMainly (but not only) in higher-dimensional settings,\na different means of assessing the\nquality of sparse grid solutions is desirable.\nWe use Euler equation errors to measure the deviation in\nthe first-order optimality conditions.\n\n\\paragraph{Derivation}\n\nIn the following, we fix the state $\\normstate_t \\in \\clint{\\*0, \\*1}$\nfor which we want to compute the Euler equation error.\nWe abbreviate\nthe\n%normalized and certainty-equivalent-transformed\nvalue function interpolant\n$\n \\normcetvalueintp_t\n \\ceq \\normcetvalueintp_t(\\normstate_t)\n$,\nthe state transition function\n$\n \\normstatefcn_t\n \\ceq \\normstatefcn_t(\\normstate_t, \\normpolicy_t, \\stochastic_t)\n$,\nthe wealth ratio\n$\n \\wealthratio_{t+1}\n \\ceq \\wealthratio_{t+1}(\n \\normstate_t, \\normpolicy_t, \\stochastic_t\n )\n$, and\nthe consumption\n$\n \\normconsume_t\n \\ceq \\normconsume_t(\\normstate_t, \\normpolicy_t)\n$.\nThe Lagrangian of the optimization problem corresponding\nto the Bellman equation \\eqref{eq:normalizedTCPBellmanEquation}\nof the normalized transaction costs problem\nwith respect to the problem's constraints\n\\eqref{eq:normalizedTCPConstraints} is given by\n{%\n \\setlength{\\abovedisplayskip}{9pt}%\n \\setlength{\\belowdisplayskip}{9pt}%\n \\begin{equation}\n \\begin{split}\n \\lagrangian_t(\\normstate_t, \\normpolicy_t, \\*\\multiplier)\n &\\ceq \\left(\n (\\normconsume_t)^{1-\\riskav} +\n \\patience \\expectation[t]{\n \\bigl(\n \\wealthratio_{t+1}\\;\n \\normcetvalueintp_{t+1}(\\normstatefcn_t)\n \\bigr)^{1-\\riskav}\n }\n \\right)^{1/(1-\\riskav)}\\\\\n &\\hspace*{6mm} {}\n - \\multiplier_1 \\normbond_t\n - \\tr{\\*\\multiplier_2} \\vnormbuy_t\n - \\tr{\\*\\multiplier_3} \\vnormsell_t\n + \\tr{\\*\\multiplier_4}\\; (\\vnormsell_t - \\vstock_t)\n + \\multiplier_5\\; (\\normconsume_{\\min} - \\normconsume_t)\n \\end{split}\n \\end{equation}%\n}%\nwith $\n\\*\\multiplier \\ceq (\n \\multiplier_1,\n \\*\\multiplier_2,\n \\*\\multiplier_3,\n \\*\\multiplier_4,\n \\multiplier_5\n)$,\n$\\multiplier_1, \\multiplier_5 \\in \\real$, and\n$\\*\\multiplier_2, \\*\\multiplier_3, \\*\\multiplier_4 \\in \\real^{m_{\\vstock}}$.\nAccording to the first-order conditions\n\\term{(Karush--Kuhn--Tucker (KKT) conditions),}\nthe partial derivative\n$\n \\partialderiv{\\partialdiff \\normbond_t}{\\lagrangian_t}(\n \\normstate_t, \\normpolicy_t, \\*\\multiplier\n )\n$\nwith respect to $\\normbond_t$\nvanishes in the exact optimum\n$\\normpolicy_t = \\optnormpolicyfcn_t \\ceq \\optnormpolicyfcn_t(\\normstate_t)$,\ni.e.,\n{%\n \\setlength{\\abovedisplayskip}{9pt}%\n \\setlength{\\belowdisplayskip}{9pt}%\n \\begin{equation}\n \\label{eq:eulerErrorFirstOrderCondition}\n \\partialderiv{\\partialdiff \\normbond_t}{\n \\left(\n (\\normconsume_t^{\\opt})^{1-\\riskav} +\n \\patience \\expectation[t]{\n \\bigl(\n \\wealthratio_{t+1}^{\\opt}\\;\n \\normcetvalueintp_{t+1}(\\normstatefcn[\\opt]_t)\n \\bigr)^{1-\\riskav}\n }\n \\right)^{1/(1-\\riskav)}\n }\n - \\multiplier_1\n - \\multiplier_5\n \\partialderiv{\\partialdiff \\normbond_t}{\\normconsume_t^{\\opt}}\n = 0,\n \\end{equation}%\n}%\nwhere\n$\n \\normstatefcn[\\opt]_t\n \\ceq \\normstatefcn_t(\\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t)\n$,\n$\n \\wealthratio_{t+1}^{\\opt}\n \\ceq \\wealthratio_{t+1}(\n \\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t\n )\n$, and\n$\n \\normconsume_t^{\\opt}\n \\ceq \\normconsume_t(\\normstate_t, \\optnormpolicyfcn_t)\n$.\nWe now neglect binding constraints,\ni.e., we assume that $\\multiplier_1 = \\multiplier_5 = 0$,\notherwise we cannot compute the error.\nAfter calculating the derivatives,\n\\cref{eq:eulerErrorFirstOrderCondition} becomes\n{%\n \\setlength{\\abovedisplayskip}{9pt}%\n \\setlength{\\belowdisplayskip}{9pt}%\n \\begin{equation}\n \\patience \\bondreturn_t\n \\cdot \\expectationsign[t]\\Bigl[\n \\bigl(\n \\normcetvalueintp_t\n - \\tr{(\\gradient{\\normstate_t}{\\normcetvalueintp_t})}\n \\normstatefcn[\\opt]_t\n \\bigr) \\cdot\n \\bigl(\n \\wealthratio_{t+1}^{\\opt}\\; \\normcetvalueintp_t\n \\bigr)^{-\\riskav}\n \\Bigr]\n = (\\normconsume_t^{\\opt})^{-\\riskav}.\n \\end{equation}%\n}%\nThis equation can be used as an error measure by substituting\n$\\optnormpolicyfcn_t$ for the interpolated optimum\n$\\optnormpolicyintp_t = \\optnormpolicyintp_t(\\normstate_t)$.\nBy multiplying the resulting equation by\n$\n (\\normconsume_t^{\\opt,\\sparse})^{\\riskav}\n \\ceq (\\normconsume_t(\\normstate_t, \\optnormpolicyintp_t))^{\\riskav}\n$, we obtain the\n\\term{unit-free Euler equation errors $\\eulererror_t(\\normstate_t)$}\nwith respect to $\\normbond_t$:\n\\begin{equation}\n \\eulererror_t(\\normstate_t)\n \\ceq \\Bigl|\n 1 - \\Bigl(\n \\patience \\bondreturn_t (\\normconsume_t^{\\opt,\\sparse})^{\\riskav}\n \\cdot \\expectationsign[t]\\Bigl[\n \\bigl(\n \\normcetvalueintp_t\n - \\tr{(\\gradient{\\normstate_t}{\\normcetvalueintp_t})}\n \\normstatefcn[\\opt,\\sparse]_t\n \\bigr) \\cdot\n \\bigl(\n \\wealthratio_{t+1}^{\\opt,\\sparse}\\;\n \\normcetvalueintp_t\n \\bigr)^{-\\riskav}\n \\Bigr]\n \\Bigr)^{-1/\\riskav}\n \\Bigr|\n\\end{equation}\nwith\n$\n \\normstatefcn[\\opt,\\sparse]_t\n \\ceq \\normstatefcn_t(\\normstate_t, \\optnormpolicyintp_t, \\stochastic_t)\n$ and\n$\n \\wealthratio_{t+1}^{\\opt,\\sparse}\n \\ceq \\wealthratio_{t+1}(\n \\normstate_t, \\optnormpolicyintp_t, \\stochastic_t\n )\n$.\n\n%\\begin{equation}\n% \\begin{split}\n% \\patience \\bondreturn_t\n% \\cdot \\expectationsign[t]\\Bigl[\n% &\\bigl(\n% \\normcetvalueintp_t%(\\normstate_t)\n% - \\tr{\n% (\n% \\gradient{\\normstate_t}{\\normcetvalueintp_t}%(\\normstate_t)\n% )\n% }\n% \\statefcn_t%(\\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t)\n% \\bigr)\\\\[-1mm]\n% &{} \\cdot\n% \\bigl(\n% \\wealthratio_{t+1}%(\n% % \\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t\n% %)\n% \\;\n% \\normcetvalueintp_t%(\\normstate_t)\n% \\bigr)^{-\\riskav}\n% \\Bigr]\n% = \\normconsume_t%(\\normstate_t, \\optnormpolicyfcn_t)\n% ^{-\\riskav}\n% \\end{split}\n%\\end{equation}\n%\n%\\begin{equation}\n% \\begin{split}\n% \\eulererror_t(\\normstate_t)\n% \\ceq \\Bigl|\n% 1 - \\Bigl(\n% \\patience \\bondreturn_t\n% \\cdot \\normconsume_t%(\\normstate_t, \\optnormpolicyfcn_t)\n% ^{\\riskav}\n% \\cdot \\expectationsign[t]\\Bigl[\n% &\\bigl(\n% \\normcetvalueintp_t%(\\normstate_t)\n% - \\tr{\n% (\n% \\gradient{\\normstate_t}{\\normcetvalueintp_t}%(\\normstate_t)\n% )\n% }\n% \\statefcn_t%(\\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t)\n% \\bigr)\\\\[-2mm]\n% &{} \\cdot\n% \\bigl(\n% \\wealthratio_{t+1}%(\n% % \\normstate_t, \\optnormpolicyfcn_t, \\stochastic_t\n% %)\n% \\;\n% \\normcetvalueintp_t%(\\normstate_t)\n% \\bigr)^{-\\riskav}\n% \\Bigr]\n% \\Bigr)^{-1/\\riskav}\n% \\Bigr|,\n% \\end{split}\n%\\end{equation}\n\n\\paragraph{Weighted Euler equation errors}\n\nHowever, the state space cropping as introduced above\ndistorts Euler equation errors:\nThe error $\\eulererror_t(\\normstate_t)$ does not vanish\neven for the exact solution and even inside the feasible state space.\nThis is because the cropping already occurs for large stock holdings\n$\\sumfcn(\\normstate_t)$ that are less than one,\nas stocks have to be sold to maintain\nminimum consumption $\\normconsume_{\\min}$.\nNumerical experiments show that due to this issue,\nthe error attains large values in the region near the hyperplane\n$\\sumfcn(\\normstate_t) = 1$.\nEconomically, this region is not significant\nas such large stock fractions are highly unusual,\nwhich is confirmed by Monte Carlo simulations.\nWe therefore use the \\term{weighted Euler equation error}\n\\begin{equation}\n \\weightedeulererror_t(\\normstate_t)\n \\ceq \\bigl(1 - \\sumfcn(\\normstate_t)\\bigr) \\cdot\n \\eulererror_t(\\normstate_t)\n\\end{equation}\ninstead of $\\eulererror_t$,\nalthough other strategies exist\nsuch as restricting the state domain where the error is computed or\nweighting the error with the probability that a given state\noccurs in Monte Carlo simulations.\n", "meta": {"hexsha": "e579fe0115a1b47e4dbc339e2f7d1a66125c98d5", "size": 18416, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/document/83problem.tex", "max_stars_repo_name": "valentjn/thesis", "max_stars_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2022-01-15T19:50:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-15T20:16:10.000Z", "max_issues_repo_path": "tex/document/83problem.tex", "max_issues_repo_name": "valentjn/thesis", "max_issues_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/document/83problem.tex", "max_forks_repo_name": "valentjn/thesis", "max_forks_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.8065630397, "max_line_length": 80, "alphanum_fraction": 0.6653453519, "num_tokens": 6050, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951025545426, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6246295191611588}} {"text": "\\documentclass{article}\r\n\\usepackage[utf8]{inputenc}\r\n\\usepackage[letterpaper, margin=1in]{geometry}\r\n\\usepackage{amsmath}\r\n\\usepackage{tikz}\r\n\\usepackage{pgfplots}\r\n\\usepackage{amssymb }\r\n\r\n\\title{CS5780 HW4}\r\n\\author{Renhao Lu, NetID: rl839}\r\n\t\r\n\\begin{document}\r\n\t\\maketitle\r\n\t\r\n\t\\section{Problem: Linear Regression}\r\n\t\\subsection{Compute the closed form solution for $\\textbf{w}$}\r\n\tUsing the formula in class:\r\n\t\\[\r\n\t\\textbf{w} = \r\n\t\t(\\textbf{X} \\textbf{X}^T) ^{-1}\r\n\t\t\\textbf{X}\r\n\t\t\t\\textbf{y}^T\r\n\t\\]\r\n\twhere $\\textbf{X}=\\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\\\ -1 & 0 & 1 & 2 & 3\\end{pmatrix}$ and $\\textbf{y}=\\begin{pmatrix} 4 & 3 & -4& 3 & 7 \\end{pmatrix}$\\\\\r\n\tHence,\r\n\t\\[\\begin{split}\r\n\t\t\\textbf{w}&=(\\textbf{X}\\textbf{X}^T)^{-1}\\textbf{X}\\textbf{y}^T\\\\\r\n\t\t&=\\begin{pmatrix} 5 & 5 \\\\ 5 & 15\\end{pmatrix}^{-1}\\textbf{X}\\textbf{y}^T\\\\\r\n\t\t&=\\begin{pmatrix} 0.3 & -0.1 \\\\ -0.1 & 0.1\\end{pmatrix}\\textbf{X}\\textbf{y}^T\\\\\r\n\t\t&=\\begin{pmatrix} 0.4 & 0.3 & 0.2 & 0.1 & 0 \\\\ -0.2 & -0.1 & 0 & 0.1 & 0.2\\end{pmatrix}\\textbf{y}^T\\\\\r\n\t\t&=\\begin{pmatrix} 2 \\\\ 0.6\\end{pmatrix}\r\n\t\\end{split}\\]\r\n\t\r\n\t\\subsection{Calculate the training loss}\r\n\t\\[\r\n\t\t\\begin{split}\r\n\t\t\tl(\\textbf{w})&=\\sum (y_i-\\textbf{w}^T\\phi(x_i)^2)\\\\\r\n\t\t\t&={||\\textbf{y}-\\textbf{w}^textbf{X}||}_2^2\\\\\r\n\t\t\t&={||\\begin{pmatrix} 2.6 & 1 & -6.6 & -0.2 & 3.2 \\end{pmatrix}||}_2^2\\\\\r\n\t\t\t&=61.6\r\n\t\t\\end{split}\r\n\t\\]\r\n\t\r\n\t\\section{Problem: Linearity of Gaussian Naive Bayes}\r\n\t\\subsection{Show that the decision rule}\r\n\tAccording to the law of total probability:\r\n\t\\[\r\n\t\t\\begin{split}\r\n\t\t\tp(\\textbf{x}) &= p(y=1)p(\\textbf{x}|y=1)+p(y=0)p(\\textbf{x}|y=0)\\\\\r\n\t\t\t&=p(y=1)\\prod_{\\alpha=1}^d {p([\\textbf{x}]_{\\alpha}|y=1)} +p(y=0)\\prod_{\\alpha=1}^d {p([\\textbf{x}]_{\\alpha}|y=0)}\r\n\t\t\\end{split}\r\n\t\\]\r\n\tHence,\r\n\t\\[\r\n\t\t\\begin{split}\r\n\t\t\tp(y=1|\\textbf{x}) &=\\frac{\r\n\t\t\t\tp(y=1) \r\n\t\t\t\t\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t} \r\n\t\t\t}{\r\n\t\t\t\tp(\\textbf{x})\r\n\t\t\t} \\\\\r\n\t\t\t&=\\frac{\r\n\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t}\r\n\t\t\t}{\r\n\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t}\r\n\t\t\t\t+p(y=0)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=0)\r\n\t\t\t\t}\r\n\t\t\t}\r\n\t\t\\end{split}\r\n\t\\]\r\n\t\r\n\t\\subsection{Show how to rewrite}\r\n\tFrom the formula we got from 1:\r\n\t\\[\r\n\t\t\\begin{split}\r\n\t\t\tp(y=1|\\textbf{x}) \r\n\t\t\t&=\\frac{\r\n\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t}\r\n\t\t\t}{\r\n\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t}\r\n\t\t\t\t+p(y=0)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=0)\r\n\t\t\t\t}\r\n\t\t\t}\\\\\r\n\t\t\t&=\\frac{1}{\r\n\t\t\t\t1 + \\frac{\r\n\t\t\t\t\tp(y=0)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=0)\r\n\t\t\t\t\t}\r\n\t\t\t\t}{\r\n\t\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t\t}\r\n\t\t\t\t}\r\n\t\t\t}\\\\\r\n\t\t\t&=\\frac{1}{\r\n\t\t\t\t1 + exp(\r\n\t\t\t\t\t-log(\r\n\t\t\t\t\t\t\\frac{\r\n\t\t\t\t\t\t\tp(y=1)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=1)\r\n\t\t\t\t\t\t\t}\r\n\t\t\t\t\t\t}{\r\n\t\t\t\t\t\t\tp(y=0)\\prod_{\\alpha=1}^d {\r\n\t\t\t\t\t\t\t\tp([\\textbf{x}]_{\\alpha}|y=0)\r\n\t\t\t\t\t\t\t}\r\n\t\t\t\t\t\t}\r\n\t\t\t\t\t)\r\n\t\t\t\t)\r\n\t\t\t}\\\\\r\n\t\t\\end{split}\r\n\t\\]\r\n\t\\subsection{Show that Naive Bayes is a linear model}\r\n\tBecause\r\n\t\\[\r\n\t\tp([\\textbf{x}]_{\\alpha}|y=1)=\r\n\t\t\\frac{1}{\\sqrt{2\\pi[\\sigma_{\\alpha}]}}\r\n\t\t\\exp\\left({\r\n\t\t\t\\frac{\r\n\t\t\t-([\\textbf{x}]_{\\alpha}-[\\mu_1]_{\\alpha})^2\r\n\t\t\t}{\r\n\t\t\t\t2[\\sigma]_{\\alpha}\r\n\t\t\t}\r\n\t\t}\\right)\r\n\t\\]\r\n\t\\[\r\n\t\tp([\\textbf{x}]_{\\alpha}|y=0)=\r\n\t\t\\frac{1}{\\sqrt{2\\pi[\\sigma_{\\alpha}]}}\r\n\t\t\\exp\\left({\r\n\t\t\t\\frac{\r\n\t\t\t-([\\textbf{x}]_{\\alpha}-[\\mu_0]_{\\alpha})^2\r\n\t\t\t}{\r\n\t\t\t\t2[\\sigma]_{\\alpha}\r\n\t\t\t}\r\n\t\t}\\right)\r\n\t\\]\r\n\tHence,\r\n\t\\[\r\n\t\t\\begin{split}\r\n\t\t\tp(y=1|\\textbf{x})&=\\frac{1}{1+\\frac{p(y=0)}{p(y=1)}\\exp{\\left(\\sum_{\\alpha=1}^d{-\\log{\\frac{p([\\textbf{x}]_\\alpha|y=1)}{p([\\textbf{x}]_\\alpha|y=0)}}}\\right)}}\\\\\r\n\t\t\t&=\\frac{1}{1+\\frac{p(y=0)}{p(y=1)}\\exp{\\left(\\sum_{\\alpha=1}^d{\\frac{([\\textbf{x}]_{\\alpha}-[\\mu_0]_{\\alpha})^2-([\\textbf{x}]_{\\alpha}-[\\mu_1]_{\\alpha})^2}{2[\\sigma]_{\\alpha}}}\\right)}}\\\\\r\n\t\t\t&=\\frac{1}{1+\\frac{p(y=0)}{p(y=1)}\\exp{\\left(\\sum_{\\alpha=1}^d{\\frac{2([\\mu_1]_{\\alpha}- [\\mu_0]_{\\alpha})[\\textbf{x}]_{\\alpha}+([\\mu_0]_{\\alpha}^2-[\\mu_1]_{\\alpha}^2)}{2[\\sigma]_{\\alpha}}}\\right)}}\r\n\t\t\\end{split}\r\n\t\\]\r\n\tWe assume:\r\n\t\\[\\textbf{w}=[w_1, w_2,...,w_b], w_\\alpha=\\frac{[\\mu_1]_{\\alpha}- [\\mu_0]_{\\alpha}}{[\\sigma]_{\\alpha}}\\]\r\n\t\\[b=\\sum{b_\\alpha}+b_0, b_0=\\log{\\frac{p(y=0)}{p(y=1)}}, b_\\alpha=\\frac{[\\mu_0]_{\\alpha}^2-[\\mu_1]_{\\alpha}^2}{2[\\sigma]_{\\alpha}}\\]\r\n\tHence,\r\n\t\\[\\begin{split}\r\n\t\tp(y=1|\\textbf{x})&=\\frac{1}{1+\\exp{\\left(b_0+\\sum_{\\alpha=1}^d {(w_\\alpha[\\textbf{x}]_\\alpha+b_\\alpha)}\\right)}}\\\\\r\n\t\t&=\\frac{1}{1+\\exp{\\left(\\textbf{w}^T\\textbf{x}+b\\right)}}\r\n\t\\end{split}\\]\r\n\tFrom the formula above, whether $p(y=1|\\textbf{x}) > p(y=0|\\textbf{x})$ is decided by whether $\\textbf{w}^T\\textbf{x}+b>0$, hence this Naive Bayes is a linear model.\r\n\t\r\n\t\\section{Problem: Gradient for Logistic Regression}\r\n\t\\subsection{Show the sigmoid function property}\r\n\t\\[\\begin{split}\r\n\t\t\\sigma(s)+\\sigma(-s)&=\\frac{1}{1+e^{-s}}+\\frac{1}{1+e^{s}}\\\\\r\n\t\t&=\\frac{e^s}{1+e^{s}}+\\frac{1}{1+e^{s}}=1\r\n\t\\end{split}\\]\r\n\tHence, $\\sigma(-s)=1-\\sigma(s)$\r\n\t\r\n\t\\subsection{Compute the gradient of the log likelihood function}\r\n\t\\subsubsection{first derivative}\r\n\t\\[\\sigma'(s)=-\\frac{1}{(1+e^{-s})^2}(-e^{-s})=\\frac{1}{1+e^{-s}}\\frac{1}{1+e^s}=\\sigma(s)\\sigma(-s)=\\sigma(s)(1-\\sigma(s))\\]\r\n\r\n\t\\subsubsection{Show the gradient of the log likelihood function}\r\n\t\\[\r\n\t\\begin{split}\r\n\t\t\\nabla_w\\log(\\textbf{y}|X, \\textbf{w})&=\\nabla_w\\sum_{i=1}^n{\\log\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i))}\\\\\r\n\t\t&=\\sum_{i=1}^n {[\\frac{\\partial}{\\partial w_1}\\log\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i))...\\frac{\\partial}{\\partial w_d}\\log\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i))]}\r\n\t\\end{split}\r\n\t\\]\r\n\tConsidering that,\r\n\t\\[\r\n\t\t\\frac{\\partial}{\\partial w_\\alpha}\\log\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i)) = \\frac{1}{\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i))}\\sigma'(y_i(\\textbf{w}^T\\textbf{x}_i))y_i\\frac{\\partial}{\\partial w_\\alpha}\\textbf{w}^T\\textbf{x}_i=(1-\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i)))y_i\\textbf{x}_{i\\alpha}\r\n\t\\]\r\n\tHence,\r\n\t\\[\\nabla_w\\log(\\textbf{y}|X, \\textbf{w})=\\sum_{i=1}^n {(1-\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i)))y_i\\textbf{x}_{i}}\\]\r\n\t\r\n\t\\subsubsection{the weights are in the span of the feature vectors}\r\n\tAssume in $k^{th}$ iteration, the step size is $t_k$\\\\\r\n\tProof by induction:\\par\r\n\tBase case: in the $0^{th}$ iteration, $\\textbf{w}_0=\\textbf{0}=\\sum_{i=1}^n0\\times\\textbf{x}_i$\\par\r\n\tInduction assumption: assume in $k-1^{th}$ iteration, $\\textbf{w}_{k-1}=\\sum_{i=1}^nc_{(k-1)i}\\textbf{x}_i$, then $\\textbf{w}_{k}=\\textbf{w}_{k-1}+t_k\\nabla_w\\log(\\textbf{y}|X, \\textbf{w})=\\textbf{w}_{k-1}+t_k\\sum_{i=1}^n {(1-\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i)))y_i\\textbf{x}_{i}}$. We note $c_{ki}=c_{(k-1)i}+t_k(1-\\sigma(y_i(\\textbf{w}^T\\textbf{x}_i)))y_i$, then $\\textbf{w}_{k}=\\sum_{i=1}^nc_{ki}\\textbf{x}_i$\\par\r\n\tHence, at every iteration of gradient descent, the weights are in the span of the feature vectors.\\par\r\n\t\r\n\t\\section{Problem: Optimization with Gradient Descent}\r\n\t\\subsection{Perform 3 steps of gradient descent}\r\n\t$f'(w)=6(w-3)$, hence $w_k=w_{k-1}-\\frac 1 \\alpha f'(w_{k-1})=w_{k-1}-\\frac 1 2 (w_{k-1}-3)$.\r\n\t\\[\\begin{split}\r\n\t\t&w_0=5, f(w_0)=3\\times2^2=12\\\\\r\n\t\t&w_1=5-\\frac 1 2\\times 2=4, f(w_1)=3\\times1^2=3\\\\\r\n\t\t&w_2=4-\\frac 1 2\\times 1=3.5, f(w_2)=3\\times0.5^2=0.75\\\\\r\n\t\t&w_3=3.5-\\frac 1 2\\times 0.5=3.25, f(w_3)=3\\times0.25^2=0.1875\r\n\t\\end{split}\\]\r\n\t\\begin{center}\r\n\t\t\\begin{tikzpicture}\r\n\t\t\t\\begin{axis}[\r\n\t\t\t\txmin = 0, xmax = 6,\r\n\t\t\t\tymin = -2, ymax = 24,\r\n\t\t\t\txtick distance = 1,\r\n\t\t\t\tytick distance = 3,\r\n\t\t\t\tgrid = both,\r\n\t\t\t\tminor tick num = 1,\r\n\t\t\t\tmajor grid style = {lightgray},\r\n\t\t\t\tminor grid style = {lightgray!25},\r\n\t\t\t\twidth = .5\\textwidth,\r\n\t\t\t\theight = .5\\textwidth,\r\n\t\t\t\txlabel = {$x$},\r\n\t\t\t\tylabel = {$y$},\r\n\t\t\t\tlegend cell align = {left},\r\n\t\t\t]\r\n\t\t\t\t\\addplot[\r\n\t\t\t\t\tdomain = 0:6,\r\n\t\t\t\t\tsamples = 200,\r\n\t\t\t\t\tsmooth,\r\n\t\t\t\t\tthick,\r\n\t\t\t\t\tblue,\r\n\t\t\t\t] {3*(x-3)^2};\r\n\t\t\t\t\\addplot [only marks, mark=o,mark size=2.9pt] table {\r\n\t\t\t\t\t5 12\r\n\t\t\t\t};\r\n\t\t\t\t\\addplot [only marks, mark=halfcircle*,mark size=2.9pt] table {\r\n\t\t\t\t\t4 3\r\n\t\t\t\t};\r\n\t\t\t\t\\addplot [only marks, mark=square,mark size=2.9pt] table {\r\n\t\t\t\t\t3.5 0.75\r\n\t\t\t\t};\r\n\t\t\t\t\\addplot [only marks, mark=x,mark size=2.9pt] table {\r\n\t\t\t\t\t3.25 0.1825\r\n\t\t\t\t};\r\n\t\t\t\t\\legend{$f(w) = 3(w-3)^2$,$w_0$,$w_1$,$w_2$,$w_3$}\r\n\t\t\t\\end{axis}\r\n\t\t\\end{tikzpicture}\r\n\t\t\\end{center}\r\n\t\r\n\t\\subsection{Gradient descent can sometimes fail to converge}\r\n\t$f'(w)=20(w-11)^3$, hence $w_k=w_{k-1}-\\frac 1 \\alpha f'(w_{k-1})=w_{k-1}-\\frac 1 2 (w_{k-1}-11)^3$.\r\n\t\\[\\begin{split}\r\n\t\t&w_0=13, f(w_0)=5\\times2^4=80\\\\\r\n\t\t&w_1=13-\\frac 1 2\\times 2^3=9, f(w_1)=5\\times2^4=80\\\\\r\n\t\t&w_2=9-\\frac 1 2\\times (-2)^3=13, f(w_2)=5\\times2^4=80\\\\\r\n\t\t&w_3=13-\\frac 1 2\\times 2^3=9, f(w_3)=5\\times2^4=80\\\\\r\n\t\t&...\r\n\t\\end{split}\\]\r\n\tThe gradient descent is bouncing between $9$ and $13$ and will never converge.\r\n\t\\begin{center}\r\n\t\t\\begin{tikzpicture}\r\n\t\t\t\\begin{axis}[\r\n\t\t\t\txmin = 8, xmax = 14,\r\n\t\t\t\tymin = -1, ymax = 116,\r\n\t\t\t\txtick distance = 1,\r\n\t\t\t\tytick distance = 9,\r\n\t\t\t\tgrid = both,\r\n\t\t\t\tminor tick num = 1,\r\n\t\t\t\tmajor grid style = {lightgray},\r\n\t\t\t\tminor grid style = {lightgray!25},\r\n\t\t\t\twidth = .5\\textwidth,\r\n\t\t\t\theight = .5\\textwidth,\r\n\t\t\t\txlabel = {$x$},\r\n\t\t\t\tylabel = {$y$},\r\n\t\t\t\tlegend cell align = {left}\r\n\t\t\t]\r\n\t\t\t\t\\addplot[\r\n\t\t\t\t\tdomain = 8:14,\r\n\t\t\t\t\tsamples = 200,\r\n\t\t\t\t\tsmooth,\r\n\t\t\t\t\tthick,\r\n\t\t\t\t\tblue,\r\n\t\t\t\t] {5*(x-11)^4};\r\n\t\t\t\t\\addplot [only marks, mark=o,mark size=2.9pt] table {\r\n\t\t\t\t\t13 80\r\n\t\t\t\t};\r\n\t\t\t\t\\addplot [only marks, mark=halfcircle*,mark size=2.9pt] table {\r\n\t\t\t\t\t9 80\r\n\t\t\t\t};\r\n\t\t\t\t\\legend{$f(w) = 5(w - 11)^4$, ${w_0,w_2,...}$,${w_1,w_3,...}$}\r\n\t\t\t\\end{axis}\r\n\t\t\\end{tikzpicture}\r\n\t\t\\end{center}\r\n\t\r\n\t\\section{Problem: Derivation for Hard-margin Linear SVMs}\r\n\t\\subsection{Prove the optimal solution of formulation A is a feasible solution for formulation B}\r\n\tNote the optimal solution of frmulation A is $\\textbf{w}_A$. Then $\\forall_iy_i(\\textbf{w}_A^T\\textbf{x}_i+b)\\geq 0$ and $min_i|\\textbf{w}_A^T\\textbf{x}_i+b|=1$. Because $y_i=1or-1$, $min_iy_i(\\textbf{w}_A^T\\textbf{x}_i+b)=1$. Hence, $\\forall_iy_i(\\textbf{w}_A^T\\textbf{x}_i+b)\\geq 1$. $\\textbf{w}_A$ is a feasible solution for B.$\\blacksquare$\r\n\t\\subsection{Prove the optimal solution of formulation B is a feasible solution for formulation A}\r\n\tNote the optimal solution of frmulation B is $\\textbf{w}_B$, then $\\forall_i,y_i(\\textbf{w}_B^T\\textbf{x}_i+b)\\geq1$. Then obviously $\\forall_iy_i(\\textbf{w}_B^T\\textbf{x}_i+b)\\geq0$. Now we need to consider $min_i|\\textbf{w}_B^T\\textbf{x}_i+b|$\\par\r\n\tWe can prove $min_i|\\textbf{w}_B^T\\textbf{x}_i+b|=1$ by contradiction. We assume $min_i|\\textbf{w}_B^T\\textbf{x}_i+b|=p>1$, then we use a new weight vector $\\textbf{w}_B'=\\frac{\\textbf{w}_B}{p}$. Because $p>1$, $\\textbf{w}_B'^T\\textbf{w}_B'=\\frac{\\textbf{w}_B^T\\textbf{w}_B}{p^2}<\\textbf{w}_B^T\\textbf{w}_B$ And $\\forall_i,y_i(\\textbf{w}_B'^T\\textbf{x}_i+b)\\geq\\frac p p=1$. Then $\\textbf{w}_B$ cannot be optimal because $\\textbf{w}_B'$ is better. $\\blacksquare$\r\n\t\r\n\t\\subsection{Prove that for the optimal solution in formulation A is the optimal solution for formulation B and vice versa}\r\n\tWe note $\\textbf{w}_A$ as the optimal solution in A, and $\\textbf{w}_B$ as the optimal solution in B. From the previous questions, we know that $\\textbf{w}_A$ is a feasible solution for B, so $\\textbf{w}_A\\leq\\textbf{w}_B$. Similiarly, $\\textbf{w}_B$ is also a feasible solution for A, so $\\textbf{w}_B\\leq\\textbf{w}_A$. Hence $\\textbf{w}_B=\\textbf{w}_A\\blacksquare$\r\n\\end{document}", "meta": {"hexsha": "f6dd680a4a261e759e42bd06f58bbd4e65953cec", "size": 11518, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "CS5780 HW4.tex", "max_stars_repo_name": "lurenhaothu/CS5780", "max_stars_repo_head_hexsha": "2599c6348f76dede29282a8551c18f04e19414a5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "CS5780 HW4.tex", "max_issues_repo_name": "lurenhaothu/CS5780", "max_issues_repo_head_hexsha": "2599c6348f76dede29282a8551c18f04e19414a5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "CS5780 HW4.tex", "max_forks_repo_name": "lurenhaothu/CS5780", "max_forks_repo_head_hexsha": "2599c6348f76dede29282a8551c18f04e19414a5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.0132013201, "max_line_length": 464, "alphanum_fraction": 0.5737107137, "num_tokens": 5008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929104825007, "lm_q2_score": 0.8006920068519376, "lm_q1q2_score": 0.6245340888245172}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\t\\section{Lattices and crystal structure} \\label{sec:lattice}\n\t\\subsection{Theory} \\label{sec:lattice_theory}\n\tThere are three parts to the description of a crystal structure. First is the lattice. This is the mathematical ``framework'' upon which the physical part of the crystal lies. It can be defined in a number of ways, however the one we will use here is standard and can be found in for example \\cite{simon} or \\cite{kittel}.\n\t\n\t\\textbf{A lattice is defined as the infinite set of points produced by a linear combination of independent \\emph{primitive lattice vectors}, with integer coefficients.}\n\t\n\tThroughout this thesis, the primitive lattice vectors are labelled $ \\V{a}_i $, and the coefficients $ n_i $, so for a $ d $-dimensional lattice, the lattice points $ \\V{R} $ are given by\n\t\\begin{equation}\\label{eq:lattice_points}\n\tR = \\sum_{i = 1}^{d} n_i \\V{a}_i.\n\t\\end{equation}\n\tFor this thesis we will mainly focus on the cases of $ d = 3 $ (visualization of lattices, families of lattice planes and scattering) and $ d = 2 $ (visualization of the band structure).\n\t\n\tA thing to note is that the choice of primitive lattice vectors is not unique. A new set of primitive lattice vectors can be created by taking a linear combination of the original primitive lattice vectors, with integer coefficients. If the original set is ordered in a matrix $ A = \\begin{pmatrix*} \\V{a}_1 & \\V{a}_2 & \\cdots & \\V{a}_n \\end{pmatrix*}$, and the new set in a matrix $ B = \\begin{pmatrix*} \\V{b}_1 & \\V{b}_2 & \\cdots & \\V{b}_n \\end{pmatrix*} $, then $ B = MA $, where $ M $ is the matrix containing the coefficients. This matrix must have integer entries, and its inverse likewise. The integer entries of the direct coefficient matrix $ M $ makes sure that any lattice point generated with the new set of primitive lattice vectors will also have integer coefficients when expressed in the old set of primitive lattice vectors. The integer entries of the inverse matrix $ M\\inverse $ then makes sure that this process will also happen in reverse.\n\t\n\tThe second part is the unit cell. This is the building block of the lattice. It is a region of space which, when stacked will completely tile the space. Like with the choice of primitive lattice vectors, the choice of unit cell is not unique. In particular we distinguish between two type of unit cells: the smallest possible unit cell and everything else. The smallest possible unit cell is called a \\textit{primitive} unit cell, and must contain only one lattice point (it cannot contain zero lattice points, as then it would not recreate the lattice when tiled). Any unit cell containing more than one lattice point is called a \\textit{conventional unit cell}. Usually a conventional unit cell is chosen for ease of calculation (as we will see in the scattering simulation), where the primitive lattice vectors constitute an orthogonal set.\n\t\n\tThe third part of the crystal structure is the basis. This is a description of the physical objects that make up the structure, and their positions in relation to the lattice. In our case the objects are atoms. The basis is specified as a list of vectors that are to be added to the lattice points, specifying the position of the atoms in the crystal.\n\t\n\tThe user can create any type of crystal they want by specifying any set of primitive lattice vector and supplying any desired basis. However a small selection of crystals will be available as presets. These include the 14 Bravais lattices with a 1 atom basis [at $ (0,0,0) $], named in table \\ref{tab:bravais}. Each of these will specify the primitive lattice vectors for a corresponding primitive unit cell. Furthermore five other crystal presets will be available, named in table \\ref{tab:presets}. Specifications of all of these presets are available in the appendix \\ref{app:lattice}.\n\t\\begin{table}[H]\n\t\t\\centering\n\t\t\\begin{tabular}{|l|l|}\n\t\t\t\\hline\n\t\t\tSimple cubic & base centred cubic (bcc) \\\\\n\t\t\t\\hline\n\t\t\tface centred cubic (fcc) & tetragonal \\\\\n\t\t\t\\hline\n\t\t\tbody centred tetragonal & orthorhombic \\\\\n\t\t\t\\hline\n\t\t\tbody centred orthorhombic & face centred orthorhombic \\\\\n\t\t\t\\hline\n\t\t\tbase centred orthorhombic & simple monoclinic \\\\\n\t\t\t\\hline\n\t\t\tbase centred monoclinic & hexaonal \\\\\n\t\t\t\\hline\n\t\t\ttriclinic & rhombohedral \\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{The 14 Bravais lattices}\n\t\t\\label{tab:bravais}\n\t\\end{table}\n\t\\begin{table}[H]\n\t\t\\centering\n\t\t\\begin{tabular}{|l|l|}\n\t\t\t\\hline\n\t\t\tfcc, conventional & bcc, conventional \\\\\n\t\t\t\\hline\n\t\t\tzincblende & wurtzite \\\\\n\t\t\t\\hline\n\t\t\tdiamond (zincblende with 1 atom) & \\\\\n\t\t\t\\hline\n\t\t\\end{tabular}\n\t\t\\caption{Other available crystal presets}\n\t\t\\label{tab:presets}\n\t\\end{table}\n\t\n\t\\subsection{Implementation}\t\n\tMathematically speaking, a lattice is infinite. A physical crystal, of course, is not. However, even though a real crystal is finite, plotting all of the atoms would be infeasible, both due to the number of atoms, and the fact that each atom would be way too small. So for the purposes of this program, only a couple of unit cells will be plotted. A good amount seems to be 8 unit cells: 2 in each of the directions specified by the primitive lattice vectors. This keeps the size of the plot relatively small, whilst still showing the important parts of the crystal structure. As such, in the following we assume that $ n_i \\in \\{0, 1, 2\\} $ for all $ n_i $.\n\t\n\tIn creating a program that plots these structures, the thought should always be on how the end product looks. Mainly we want the plot to be as clear and instructive as possible. This constitutes plotting only full unit cells: no unit cell can have missing atoms. There is a caveat to this however. Say we define the unit cell to have atoms on the boundary, then these atoms will be in multiple unit cells. One could say no matter how many unit cells we plot, we would always get some amount of unfinished unit cells. As such we will plot the 8 unit cells along with any atoms on the border of these.\n\t\\begin{wrapfigure}{r}{2in}\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=\\linewidth]{figures/lattice_unfinished_1.pdf}\n\t\t\\end{center}\n\t\t\\caption{One conventional unit cell for a bcc lattice. If we naively plot all atoms associated with a lattice point, we will end up with atoms outside the unit cells we want to plot.}\n\t\t\\label{fig:lattice_unfinished_1}\n\t\\end{wrapfigure}\n\t\n\tEven with the above considerations we need to remember that the screen is still just a two dimensional projection of the actual three dimensional phenomena. As such without some form of depth perception, the crystal will just look like a weird two dimensional pattern. This is fixed by including grid lines. They, along with the ability to interactively rotate the crystal, give the required depth perception to allow the user to comprehend the crystal as a three dimensional structure, and not as a two dimensional sheet of dots. \n\t\n\tFor the crystals that can be expressed as a lattice with orthogonal primitive lattice vectors and a basis (cubic, tetragonal and orthorhombic), we usually want to plot orthogonal grid lines, whilst for other crystals plotting grid lines along the lattice vectors will be more useful.\n\t\n\tIn the case of plotting primitive unit cells, the ones furthest away from the origin (say with $ n_1 = n_2 = n_3 = 2 $) may not be filled. Say the basis consists of only one atom. Then this atom will just be placed on the lattice points, and there is no problem, as this is the edge of the unit cell. However, if the basis consists of two atoms (or more), where one is placed at $ (0,0,0) $ and the other has some positive coordinates, then this second atom will be in a unit cell which is not supposed to be plotted (it would necessitate plotting the lattice points corresponding to $ n_i = 3 $). An example of this is figure \\ref{fig:lattice_unfinished_1}, which shows one unit cell of a simple cubic lattice, with a two atom basis (corresponding to a conventional bcc unit cell). A way to correct for this is to write each plotted atoms position as a linear combination of the primitive lattice vectors with coefficients $ n'_i $ (where the coefficients here are real numbers, as they do not necessarily align with the lattice points). If the coefficients are $ 0 \\leq n'_i \\leq 2 $ for all $ n'_i $, then we plot the point.\n\t\n\tFor plotting conventional unit cells when the user inputs primitive lattice vectors (for cubic, tetragonal and orthorhombic lattices), we need something similar, otherwise a situation like in figure \\ref{fig:lattice_unfinished_2} may occur. To fix this we calculate the side lengths of the cuboid plot box that just contains the parallelepiped arising from plotting the lattice points with the specified coefficients ($ n_i \\in \\{0, 1, 2\\}$), and fill this plot box completely with atoms.\n\t\n\tTo calculate this, the program creates the 8 possible vectors arising from linear combinations of the primitive lattice vectors, with coefficients from the minimum and maximum coefficients and taking the limits of the plot box as the minimum and maximum values for these 8 vectors. For the specified lattice points, these 8 vectors are:\n\t\\begin{equation}\n\t\t\\begin{array}{llll}\n\t\t\t\\V{v}_1 = \\V{0}, & \\V{v}_2 = 2\\V{a}_1, & \\V{v}_3 = 2\\V{a}_2, & \\V{v}_4 = 2\\V{a}_3, \\\\\n\t\t\t\\V{v}_5 = 2(\\V{a}_1+\\V{a}_2), & \\V{v}_6 = 2(\\V{a}_1+\\V{a}_3), & \\V{v}_7 = 2(\\V{a}_2+\\V{a}_3), &\\V{v}_8 = 2(\\V{a}_1+\\V{a}_2+ \\V{a}_3).\n\t\t\\end{array}\n\t\\end{equation}\n\tThese are the vertices of the aforementioned parallelepiped. The side lengths of the plot box are then taken as the maximum coordinates for these vectors, minus the minimum values (the side length in $ x $ might be $ v_{8,x}-v_{0,x} $ for example). Then we plot atoms for a larger range of coefficients to completely fill out this plot box. If some atoms fall outside of the box we hide them.\n\t\n\t\\begin{wrapfigure}{r}{2in}\n\t\t\\begin{center}\n\t\t\t\\includegraphics[width=\\linewidth]{figures/lattice_unfinished_2.pdf}\n\t\t\\end{center}\n\t\t\\caption{An attempt to plot eight conventional unit cells for an FCC lattice, by using primitive lattice vectors with coefficients in the set $ \\{0, 1, 2\\} $. This does not fill out all eight unit cells, so we need to plot more lattice points than this. In practise we need $ \\{-2, -1, 0, 1, 2, 3, 4\\} $}\n\t\t\\label{fig:lattice_unfinished_2}\n\t\\end{wrapfigure}\n\t\n\t\n\t\\subsection{Step-by-step}\n\tWhen the program is run it will either load the chosen crystal preset or plot the user's manually input crystal. If a crystal preset is chosen, the program loads it in such a way as to make the resulting plot as informative as possible (eg. place lattice vectors along cardinal axes for an orthogonal lattice, to make plotting grid lines easier). If the user manually specifies the lattice and basis, the program attempts to classify the lattice according to the specifications in the appendix \\ref{app:lattice} (except for zincblende, diamond and wurtzite). Next the program checks whether or not the crystal should be rotated to make plotting prettier.\n\t\n\tIn general the rotation algorithm tries to align one lattice vector with the $ x $-axis. $ \\V{a}_1 $ is preferred, but is only chosen to lie along the $ x $-axis if it forms an orthogonal pair with at least one other primitive lattice vector. The second primitive lattice vector of the pair ($\\V{a}_2$ being preferred) is then aligned along the $ y $-axis. If the three primitive lattice vectors form an orthogonal set, then the last vector of the set will now be aligned along the $ z $-axis.\n\t\n\tThe actual rotation is done by rotating the whole crystal (each primitive lattice vector and all vectors in the basis) around the cross product between the initial vector and the destination vector. Rotating the crystal such that $ \\V{a}_1 $ lies along the $ x $-axis is done by rotating along $ \\V{a}_1 \\times \\U{x} $, with an angle of $ \\sin \\theta = |\\V{a}_1 \\times \\U{x}| / |\\V{a}_1| $. However, this might rotate the crystal the wrong way, depending on the orientation between the two vectors. Because of this the program checks whether or not the rotated initial vector and the destination vector are parallel (that is, if the rotated $ \\V{a}_1 $ is parallel to $ \\U{x} $). If this is not the case, the whole crystal is rotated about the same vector, with an angle $ -2\\theta $.\n\t\n\tFive of the Bravais lattices have specialised rotation functions: hexagonal, base centred monoclinic and the three face centred lattices.\n\t\n\tFor the hexagonal lattice, the program detects which primitive lattice vectors constitute the triangular lattice and orients them such that one is along the $ x $-axis and the other is in the $ xy $-plane (easily done by rotating the third primitive lattice vector such that it is parallel with the $ z $-axis). The same approach is used for the base centred monoclinic, but here the program always aligns $ \\V{a}_1 $ along the $ x $-axis, and $ \\V{a}_2 $ in the $ xy $-plane. Here however, the easy option of rotating $ \\V{a}_3 $ is not available. Instead the program uses the fact that the vector rejection of $ \\V{a}_2 $ with $ \\V{a}_1 $ is orthogonal to $ \\V{a}_1 $ (the vector rejection of $ \\V{a}_2 $ with $ \\V{a}_1 $ being $ \\V{a}_2 $ minus the projection of $ \\V{a}_2 $ along $ \\V{a}_1 $):\n\t\n\t\\begin{equation}\n\t\t\\V{a}_{2,rej} = \\V{a}_2 - \\V{a}_{2, proj} = \\V{a}_2 - \\frac{\\V{a}_2 \\D \\V{a}_1}{|\\V{a}_1|} \\V{a}_1.\n\t\\end{equation}\n\tThis vector rejection is then in the $ yz $-plane, and the crystal can be rotated around $ \\V{a}_1 $ with the angle the rejection makes with $ \\U{y} $: $ \\cos \\theta = \\V{a}_{2,rej} \\D \\U{y} / |\\V{a}_{2,rej}|$.\n\t\n\tFor the face centred lattices, the ideal, rotated lattice is created from the magnitudes of the primitive lattice vectors: if $ |\\V{a}_1| = a, |\\V{a}_2| = b, |\\V{a}_3| = c $, then $ \\V{a}'_1 = (a/2, b/2, 0), \\V{a}'_2 = (a/2, 0, c/2), \\V{a}'_3 = (0, b/2, c/2) $. First the crystal is rotated such that $ \\V{a}_1 $ aligns with $ \\V{a}'_1 $, by using their cross product. Then the crystal is rotated such that the now rotated $ \\V{a}_2 $ aligns with $ \\V{a}'_2 $, via the vector rejection of $ \\V{a}_2 $ with $ \\V{a}'_2 $. A thing to note is that these 5 rotation functions necessitates primitive lattice vectors of the form specified in the appendix. If the user specifies a lattice unlike those in the appendix, the program will not be able to identify it, and as such may rotate it ``incorrectly''. This could lead to the program not being able to impose grid lines along the cardinal axes for an fcc lattice, for example.\n\t\n\tWith the lattice and basis rotated, the program finds the limits of the plot box as written above, after which the crystal is generated. This is done by looping over the three ranges specified by the minimum and maximum coefficients, creating each lattice point $ \\V{R} $ by Eq. \\eqref{eq:lattice_points}. For each lattice point the program calculates the positions of the $ n $ atoms in the basis, by adding one of the $ n $ vectors in the basis to the lattice point, i.e. $ \\V{r}_{atom, i} = \\V{R} + \\V{r}_{basis, i} $. Lists of the colours and sizes associated with each atom are also created at this point. After creating all the atoms, the program deletes any that may lie outside the limits of the plot box.\n\t\n\tThe only thing missing now is to create the grid lines and plot everything. The program has two ways of creating grid lines: along the primitive lattice vectors and along the cardinal axes.\n\t\n\tCreating grid lines along the primitive lattice vectors works by taking each lattice point $ \\V{R}_n $, and finding the lattice point $ \\V{R}_m $ the furthest away from it, in the (positive) direction of these lattice vectors, such that $ \\V{R}_m = \\V{R}_n + \\alpha \\V{a}_i $, where $ \\alpha > 0 $, for all $ i \\in \\{1,2,3\\}$, and creating lines between $ \\V{R}_n $ and $ \\V{R}_m $. This does create duplicate grid lines, but these do not show on the final plot, since they are all plotted with the same width and colour.\n\t\n\tCreating grid lines along the cardinal axes works by finding the minimum spacing between lattice points on these axes (called $ a_x, a_y$ and $ a_z $), and using these as the spacing between grid lines. The program then finds the maximum and minimum coordinates of lattice points along the cardinal axes (called $x_{min}, x_{max}$, etc.). This is used to create ranges corresponding to each lattice points on the cardinal axes: The range for the $ x $-axis starting at $ x_{min} $ and ending at $ x_{max} $ with steps of $ a_x $.\n\t\n\tThese ranges then specify a grid of lattice points on the $ xy$, $ xz $ and $ yz $-planes. The program then creates lines orthogonal to these planes, stretching from $ z_{min} $ to $ z_{max} $ for the points in the $ xy $-plane, and similarly for the other two planes.\n\t\n\tWith everything calculated a blank figure is created with limits as calculated above. The atoms are plotted with colours and sizes specified by the user. The grid lines are plotted with a uniform size and colour and lastly the primitive lattice vectors are plotted with corresponding labels.\n\t\n\t\\subsection{Examples}\n\tAs an example, we plot a simple cubic lattice with a two atom basis (corresponding to the basis of a bcc lattice with conventional unit cells), where the atoms on the lattice points are grey and the body centred atoms are blue (figure \\ref{fig:lattice_demo_1}), we write the following:\n\\begin{lstlisting}\nLattice(lattice_name=\"conventional bcc\",\n\t\tcolors=[\"xkcd:cement\", \"b\"])\n\\end{lstlisting}\n\tOr we could plot a hexagonal lattice with a one atom basis (figure \\ref{fig:lattice_demo_2}):\n\\begin{lstlisting}\nLattice(lattice_name=\"hexagonal\")\n\\end{lstlisting}\n\tWhen plotted using the program, these plots are interactive and can be rotated.\n\n\n\t\\begin{figure}[h]\n\t\t\\centering\n\t\t\\begin{minipage}{.4\\textwidth}\n\t\t\t\\centering\n\t\t\t\\includegraphics[width=2in]{figures/lattice_demo_1.pdf}\n\t\t\t\\captionof{figure}{A simple cubic lattice with a two atom basis: One atom on the lattice points and another in the middle of each unit cell.}\n\t\t\t\\label{fig:lattice_demo_1}\n\t\t\\end{minipage}%\n\t\t\\hfil\n\t\t\\begin{minipage}{.4\\textwidth}\n\t\t\t\\centering\n\t\t\t\\includegraphics[width=2in]{figures/lattice_demo_2.pdf}\n\t\t\t\\captionof{figure}{A hexagonal lattice with a one atom basis. The lattice consists of a series of triangular lattices stacked on top of each other.}\n\t\t\t\\label{fig:lattice_demo_2}\n\t\t\\end{minipage}\n\t\\end{figure}\n\t\n\\end{document}", "meta": {"hexsha": "87c5219157fe5f037eeefdd6334b2dff2814310c", "size": 18438, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/lattices.tex", "max_stars_repo_name": "NikolaiNielsen/Bachelor", "max_stars_repo_head_hexsha": "e26f3cee6dcfc858b606b5d3112f553836dd3990", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-05-17T02:07:14.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-06T09:18:38.000Z", "max_issues_repo_path": "thesis/lattices.tex", "max_issues_repo_name": "NikolaiNielsen/Bachelor", "max_issues_repo_head_hexsha": "e26f3cee6dcfc858b606b5d3112f553836dd3990", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/lattices.tex", "max_forks_repo_name": "NikolaiNielsen/Bachelor", "max_forks_repo_head_hexsha": "e26f3cee6dcfc858b606b5d3112f553836dd3990", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-07-19T05:12:31.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-24T08:32:13.000Z", "avg_line_length": 117.4394904459, "max_line_length": 1128, "alphanum_fraction": 0.747315327, "num_tokens": 4905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929053683038, "lm_q2_score": 0.800691997339971, "lm_q1q2_score": 0.6245340773103542}} {"text": "\\documentclass[a4paper,12pt]{article}\n% decent example of doing mathematics and proofs in LaTeX.\n% An Incredible degree of information can be found at\n% http://en.wikibooks.org/wiki/LaTeX/Mathematics\n\n% Use wide margins, but not quite so wide as fullpage.sty\n\\marginparwidth 0.1in \n\\oddsidemargin 0.05in \n\\evensidemargin 0.05in \n\\marginparsep 0.05in\n\\topmargin 0.05in \n\\textwidth 6in \\textheight 8 in\n% That's about enough definitions\n\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amssymb}\n\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n\n\\makeatletter\n\\renewenvironment{proof}[1][\\proofname] {\\par\\pushQED{\\qed}\\normalfont\\topsep6\\p@\\@plus6\\p@\\relax\\trivlist\\item[\\hskip\\labelsep\\bfseries#1\\@addpunct{.}]\\ignorespaces}{\\popQED\\endtrivlist\\@endpefalse}\n\\makeatother\n\n\\newtheoremstyle{break}\n {\\topsep}{\\topsep}%\n {\\itshape}{}%\n {\\bfseries}{}%\n {\\newline}{}%\n\\theoremstyle{break}\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{corollary}{Corollary}[theorem]\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{definition}[theorem]{Definition}\n\n\n\\begin{document}\n\n\\title{Linear Algebra Theorems}\n\\author{Hyeonsu Lyu, Department of Electrical Engineering, UNIST}\n\\date{June 29, 2018}\n\\maketitle\n\n\\section{Linear Transformation}\n\\begin{definition} [Linear Transformation from $\\mathbb{R}^n$ to $\\mathbb{R}^m$]\n A \\textit{linear transformation} $T : \\mathbb{R}^n\\to\\mathbb{R}^m$ is a mapping such that for all scarars $\\textit{a}$ and all $\\overrightarrow{v}, \\overrightarrow{w} \\in \\mathbb{R}^n,$\n\\begin{align}\n T(\\overrightarrow{v}+\\overrightarrow{w}=T(\\overrightarrow{v}+\\overrightarrow{w})\\ and\\ T(a\\overrightarrow{v})=aT(\\overrightarrow{v})\n\\end{align}\n\\end{definition}\n\n\\begin{theorem} [Matrices and Linear transformations]\\label{thm:mlt}\n 1. Any $m\\times n$ matrix A defines a linear transformation $T:\\mathbb{R}^n\\to\\mathbb{R}^m$ by matrix multiplication:\n \\begin{align}\n T(\\overrightarrow{v})=A\\overrightarrow{v}.\n \\end{align}\n 2. Every linear transformation $T:\\mathbb{R}^n\\to\\mathbb{R}^m$ is given by multiplication by the $m\\times n$ matrix $[T]$:\n \\begin{align}\n T(\\overrightarrow{v})=[T]\\overrightarrow{v}\n \\end{align}\n ,where the \\textit{i}th column of $[T]$ is $T(\\overrightarrow{e_i}).$\n\\end{theorem}\n\n\\begin{proof}[Proof of theorem \\ref{thm:mlt}] \\hfill \\\\\n1. By definition of \\textit{linear transformation}, mapping (2) is \\textit{linear}.\\\\\n2. For any vector $\\overrightarrow{v}\\in \\mathbb{R}^n$, $v$ can be represented as basis of $\\mathbb{R}^n$, $\\{\\overrightarrow{e_1},\\overrightarrow{e_2}, \\dots, \\overrightarrow{e_n}\\}$.\n\\begin{align}\n \\overrightarrow{v} = v_1\\overrightarrow{e_1}+v_2\\overrightarrow{e_2}+\\dots+v_n\\overrightarrow{e_n}\n\\end{align}\nSince the linearity of mapping T,\n\\begin{align*}\n T(\\overrightarrow{v})&=v_1T(\\overrightarrow{e_1})+v_2T(\\overrightarrow{e_2})+\\dots+v_nT(\\overrightarrow{e_n})\\\\\n &=[T(\\overrightarrow{e_1})|T(\\overrightarrow{e_2})|\\dots|T(\\overrightarrow{e_n})]\\begin{bmatrix}v_1 \\\\ v_2 \\\\ \\vdots \\\\ v_n\\end{bmatrix}\\\\\n &= [T]\\overrightarrow{v}\n\\end{align*}\n\\end{proof}\n\n\\begin{definition} [Linear Transformation is onto]\n Let $A$ be $m\\times n$ matrix. The followings are logically equivalent: \\\\\n \\indent1. The function $Ax=b$ is onto $\\mathbb{R}^m$.\\\\\n \\indent2. For each $b\\in \\mathbb{R^m}$, the equation Ax=b has a solution.\\\\\n \\indent3. $b\\in \\mathbb{R^m}$ is a linear combination of the columns of $A$.\\\\\n \\indent4. The columns of $A$ span $\\mathbb{R^m}$.\\\\\n \\indent5. $A$ has a pivot position in every row.\\\\\n \\indent6. $A$ has rank m\\\\\n\\end{definition}\n\n\\begin{definition} [Linear Transformation is 1-1]\n Let $A$ be an $m\\times n$ matrix. The followings are logically equivalent:\\\\\n \\indent1. The function $Ax=b$ is $1-1$.\\\\\n \\indent2. The equation $Ax=b$ has at most on solution for every b.\\\\\n \\indent3. The equation $Ax=0$ has only the trivial solution.\\\\\n \\indent4. The columns of A are linearly independent.\\\\\n \\indent5. A has a pivot posiition in every column.\\\\\n \\indent6. A has rank n.\\\\\n\\end{definition}\n\nRelated theroems : Inverse function theorem, Implicit function theorem, Tangent space on implicit function.\n\n\\section{Another Section}\n\n\\end{document}\n\n", "meta": {"hexsha": "9bbc70c6240ec5fa4ec234335e6be1e02e098200", "size": 4247, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Mathematics/LinearAlgebra.tex", "max_stars_repo_name": "lhs55349780/Latex-Suite", "max_stars_repo_head_hexsha": "3ccd4c123c25571ac21ad2ccbaa5600a7f4c5568", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Mathematics/LinearAlgebra.tex", "max_issues_repo_name": "lhs55349780/Latex-Suite", "max_issues_repo_head_hexsha": "3ccd4c123c25571ac21ad2ccbaa5600a7f4c5568", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Mathematics/LinearAlgebra.tex", "max_forks_repo_name": "lhs55349780/Latex-Suite", "max_forks_repo_head_hexsha": "3ccd4c123c25571ac21ad2ccbaa5600a7f4c5568", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.8365384615, "max_line_length": 199, "alphanum_fraction": 0.7077937368, "num_tokens": 1442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799928900257126, "lm_q2_score": 0.8006920092299293, "lm_q1q2_score": 0.6245340742997472}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% CS671: Machine Learning\n% Copyright 2015 Pejman Ghorbanzade \n% Creative Commons Attribution-ShareAlike 4.0 International License\n% More info: https://github.com/ghorbanzade/beacon\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section*{Question 2}\n\nConsider the set of points in $\\mathbb{R}^2$ shown in Figure \\ref{fig1}.\nPositive examples are shown as black circles, while negative examples are shown as gray circles.\n\\begin{enumerate}[label=(\\alph*)]\n\\item Find classifiers for the sample using at least four kernels available in the package of $R$ software.\n\\item Graph the total time required by these kernels as a function of the error rate obtained on the training set.\n\\end{enumerate}\n\n\\begin{figure}[H]\\centering\n\\tikzstyle{black}=[circle, draw, fill=black]\n\\tikzstyle{white}=[circle, draw, fill=gray!20]\n\\begin{tikzpicture}\n \\draw [thick, color=gray, step=0.5cm, dashed] (0,0) grid (5,5);\n \\draw [<->,thick] (0,5.5) node (yaxis) [left] {$y$} |- (5.5,0) node (xaxis) [below] {$x$};\n \\foreach \\Point in { (4.5,0.5), (4,1), (3.5,1.5), (3,2), (4,2), (2.5,2.5), (3.5,2.5), (3,3), (4,3), (3.5,3.5), (4,4), (4.5,4.5) } {\n\t\\fill[black] \\Point circle (3pt);\n }\n \\foreach \\Point in {(1,0.5), (1,1.5), (1.5,1.5), (1.5,2.5), (1,3), (2,3.5), (2.5,4), (1,4.5), (3, 2.5)} {\n \\fill[white] \\Point circle (3pt);\n }\n\\end{tikzpicture}\n\\caption{Training Sample}\\label{fig1}\n\\end{figure}\n\n\\vfill\n\n\\subsection*{Solution}\n\n\\begin{table}[H]\\centering\n\\begin{tabular}{r l c c}\nKernel & Function Name & Training Error & Execution Time\\\\\n\\hline\nLinear & \\texttt{vanilladot} & 4.761905 & 0.01\\\\\nRadial Basis & \\texttt{rbfdot} & 4.761905 & 0.02\\\\\nPolynomial & \\texttt{polydot} & 4.761905 & 0.01\\\\\nANOVA RBF & \\texttt{anovadot} & 4.761905 & 0.02\\\\\nBessel & \\texttt{besseldot} & 4.761905 & 0.00\\\\\nLaplacian & \\texttt{laplacedot} & 4.761905 & 0.02\\\\\nSpline & \\texttt{splinedot} & 9.52381 & 0.01\\\\\nHyperbolic Tangent & \\texttt{tanhdot} & 19.04762 & 0.00\\\\\n\\hline\n\\end{tabular}\n\\caption{Performance of Different Kernel Functions on Training Sample given in Fig \\ref{fig1}}\\label{tab1}\n\\end{table}\n\nThe training dataset shown in Figure \\ref{fig1} is constructed in comma-separated value format.\nUsing R programming-language, the training dataset is imported to workspace and is used to construct classifiers using kernels available in the \\texttt{kernlab} package.\nTable \\ref{tab1} provides the list of kernels used for SVM-based classification as well as their training error and execution time for their construction over given training dataset.\n\nFig \\ref{fig2}, generated by R, presents the training error resulting construction of classifiers using different kernel functions.\nAs can be seen, \\texttt{splinedot} and \\texttt{tanhdot} kernels have failed to classify the training dataset accurately.\nThe similar training error achieved by other kernel functions can be justified by noting the limited amount of samples in the training dataset as well as the way they are formed which makes it fairly easy to classify.\n\n\\begin{figure}\\centering\n\\includegraphics[width=0.8\\textwidth]{\\imgDirectory/umb-cs671-2015s-f01-02.png}\n\\caption{Training Error for Different Kernels}\\label{fig2}\n\\end{figure}\n\nAlso, plot of execution time of different kernel functions in constructing classifier for the training dataset is given in Figure \\ref{fig3}.\n\n\\begin{figure}\\centering\n\\includegraphics[width=0.8\\textwidth]{\\imgDirectory/umb-cs671-2015s-f01-03.png}\n\\caption{Execution Time for Different Kernel Functions}\\label{fig3}\n\\end{figure}\n\nFigure \\ref{fig4} depicts the execution time based on training error for different kernel functions.\n\n\\begin{figure}\\centering\n\\includegraphics[width=0.8\\textwidth]{\\imgDirectory/umb-cs671-2015s-f01-04.png}\n\\caption{Execution-Time based on Error-Rate for Different Kernels}\\label{fig4}\n\\end{figure}\n\nIt can be inferred that using kernel functions that promise better performance in terms of training error might necessitate higher computational costs which lead to longer execution times.\nHowever, this particular example does not prove this point as can be seen in Fig \\ref{fig4}.\nIn fact, rerunning the same script would give different results each time.\nThis is due to the limited amount of the training dataset that will make the errors in measurement of the execution time quite considerable.\nTo prove our claim, the same code in R programming language has been applied to the larger database given for letter recognitions, results of which are included in the files attached to this document.\n\nThe code in R data-analysis software to construct and compare classifiers over the training dataset given in \\ref{fig1} has been given as appendix to this report.\n", "meta": {"hexsha": "9a40fc08a7b8d939be7157e71b699c35002edcd8", "size": 4799, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "umb-cs671-2015s/src/tex/f01/f01q02.tex", "max_stars_repo_name": "ghorbanzade/beacon", "max_stars_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-11-13T20:00:10.000Z", "max_stars_repo_stars_event_max_datetime": "2020-01-01T11:16:51.000Z", "max_issues_repo_path": "umb-cs671-2015s/src/tex/f01/f01q02.tex", "max_issues_repo_name": "ghorbanzade/beacon", "max_issues_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "umb-cs671-2015s/src/tex/f01/f01q02.tex", "max_forks_repo_name": "ghorbanzade/beacon", "max_forks_repo_head_hexsha": "c36e3d1909b9e1e47b1ad3cda81f7f33b713adc4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-09-20T05:58:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T17:18:05.000Z", "avg_line_length": 54.5340909091, "max_line_length": 218, "alphanum_fraction": 0.7284851011, "num_tokens": 1397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.800691997339971, "lm_q1q2_score": 0.6245340732154576}} {"text": "%\n% CMPT 310: Artificial Intelligence - A Course Overview\n% Section: Search Algorithms\n%\n% Author: Jeffrey Leung\n%\n\n\\section{Search Algorithms}\n\t\\label{sec:search-algorithms}\n\\subsection{Uninformed Algorithms}\n\t\\label{subsec:uninformed-algorithms}\n\\begin{easylist}\n\n& \\textbf{Breadth-first search:} Utilizes a queue\n\t&& Time and space complexity: Bounded by the number of nodes the same distance away from the root as the goal node\n\t&& Complete if branching factor is finite\n\n& \\textbf{Uniform-cost search:} Utilizes a priority queue to follow the path with least cost\n\t&& Time and space complexity: Bounded by the branching factor to the power of the average action cost\n\t&& Complete if branching factor is finite and step costs $\\leq \\epsilon$\n\n& \\textbf{Depth-first search:} Utilizes a stack\n\t&& Time complexity: Bounded by the maximum depth of the tree\n\t&& Space complexity: Bounded by the length of the maximum-length path, and its remaining unexpanded child nodes for each node on the path\n\t&& Not complete\n\t&& More space-efficient than breadth-first search\n\n& \\textbf{Iterative deepening (depth-first) search:} Depth-first search with maximum limited depth, which increases until the goal is found\n\t&& Complete if branching factor is finite\n\t&& Time complexity: Bounded by the number of nodes the same distance away from the root as the goal node\n\t&& Space complexity: Bounded by the length of the goal path, and its remaining unexpanded child nodes for each node on the path\n\n\\end{easylist}\n\\subsection{Informed Algorithms}\n\t\\label{subsec:informed-algorithms}\n\\begin{easylist}\n\n& \\textbf{Admissible heuristic:} Underestimation of the true cost from a given node to the goal\n\t&& Notation for a function evaluating a heuristic: $h(n)$\n\t&& Will provide improved time and space complexity over the worst time\n\t&& To create an admissible heuristic, relax a rule/constraint and find the shortest solution given\n\t&& \\textbf{Manhattan distance:} Number of unrestricted 4-directional moves from the goal state\n\t&& \\textbf{Dominance:} Characteristic of an admissible heuristic which has a result greater than or equal to another admissible heuristic for all $n$\n\t\t&& Given admissible heuristics $h_a(n)$ and $h_b(n)$, $h(n) = \\textrm{max}\\left(h_a(n), h_b(n)\\right)$ is also admissible and dominates $h_a(n)$ and $h_b(n)$\n\n& \\textbf{Greedy search:} Best-first search algorithm which chooses subsequent nodes by minimizing direct cost to the goal\n\t&& Complete: Only if state is checked for repeats\n\t&& Time and space complexity: At worst, bounded by all nodes\n\n& \\textbf{A* search:} Best-first search algorithm which expands only the paths with less cost than the optimal cost to the goal\n\t&& Evaluation function:\n\n\t\\end{easylist}\n\t\\begin{align*}\n\t\tf(n) & = g(n) + h(n) \\\\\n\t\t\\textrm{where }\n\t\t& f(n) = \\textrm{ estimated total cost of the path through } n \\textrm{ to the goal} \\\\\n\t\t& g(n) = \\textrm{ cost so far to reach } n \\\\\n\t\t& h(n) = \\textrm{ heuristic-estimated cost from } n \\textrm{ to the goal}\n\t\\end{align*}\n\t\\begin{easylist}\n\n\t&& Complete: Yes (unless there are infinitely many nodes with $f \\leq f(G)$)\n\n\\end{easylist}\n\\subsection{Complexity of Algorithms}\n\t\\label{subsec:complexity-of-algorithms}\n\n\\begin{figure}[!htb]\n\t\\caption{Complexity of Algorithms}\n\t\\label{fig:algos-complexity}\n\t\\center\n\t\\begin{tabular}{ l | c c p{4cm} }\n\t\t& Time & Space & Is the Goal Path Optimal? \\\\\n\t\t\\hline\n\t\tBreadth-first search & $O(b^{d+1})$ & $O(b^d)$ & Only when $d = 1$ or when paths have no cost \\\\\n\t\tUniform-cost search & $O(b^{\\left\\lfloor \\frac{C^*}{\\epsilon} \\right\\rfloor})$ & $O(b^{\\left\\lfloor \\frac{C^*}{\\epsilon} \\right\\rfloor})$ & Yes \\\\\n\t\tDepth-first search & $O(b^m)$ & $O(b \\cdot m)$ & No \\\\\n\t\tIterative deepening search & $O(b^d)$ & $O(b \\cdot d)$ & Yes \\\\\n\t\t\\hline\n\t\tGreedy search & $O(b^m)$ & $O(b^m)$ & No \\\\\n\t\tA\\textsuperscript{*} search & $O(b^m)$ & $O(b^m)$ & Yes \\\\\n\t\t\\hline\n\t\tMinimax algorithm & $O(b^m)$ (with $\\alpha / \\beta$, $O(b^\\frac{m}{2})$) & $O(b \\cdot m)$ & Yes (with rational adversary) \\\\\n\t\tBacktracking search & & & Yes \\\\\n\t\tHill climbing search & & & No \\\\\n\t\tSimulated annealing search & & & Yes\n\t\\end{tabular}\n\n\t\\begin{align*}\n\t\t\\textrm{where }\n\t\tb &= \\textrm{ branching factor} \\\\\n\t\td &= \\textrm{ depth of the goal} \\\\\n\t\tC^* &= \\textrm{ cost of the optimal solution} \\\\\n\t\t\\epsilon &= \\textrm{ the minimum cost of an action on the goal path} \\\\\n\t\tm &= \\textrm{ maximum depth}\n\t\\end{align*}\n\\end{figure}\n\n\\clearpage\n", "meta": {"hexsha": "51be775460ea8dae340c523b9977c8d8a021a57e", "size": 4449, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/search-algorithms.tex", "max_stars_repo_name": "AmirNaghibi/notes", "max_stars_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2019-08-11T08:45:10.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T02:37:39.000Z", "max_issues_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/search-algorithms.tex", "max_issues_repo_name": "AmirNaghibi/notes", "max_issues_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cmpt-310-artificial-intelligence-survey/course-overview/tex/search-algorithms.tex", "max_forks_repo_name": "AmirNaghibi/notes", "max_forks_repo_head_hexsha": "c4640bbcb65c94b8756ccc3e4c1bbc7d5c3f8e92", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2020-11-18T09:17:46.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-27T21:44:56.000Z", "avg_line_length": 44.0495049505, "max_line_length": 159, "alphanum_fraction": 0.7064508878, "num_tokens": 1325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006919925839875, "lm_q2_score": 0.7799929002541068, "lm_q1q2_score": 0.6245340695058242}} {"text": "\\section{Edge discretization}\\label{edge_discretization}\nWe calculate the location $l$ of the boundary between a significant and nonsignificant portion of an edge analytically.\nFor example, the parabolic MAT edge generated from the outline vertex $(0,1)$ and an outline segment aligned with the X-axis follows $y(x) = \\nicefrac12 x^2$ and $R(x) = y(x)$.\nWe can determine the significant portion $[-x_\\text{bound}, x_\\text{bound}]$ by evaluating $\\frac{\\partial R}{\\partial x} > \\cos(\\alpha_\\text{max} / 2)$, which is $ | x_\\text{bound} | = (\\tan(\\alpha_\\text{max} / 2))^{-1}$.\nSimilarly, a MAT edge generated from two vertices at $(0,0)$ and $(0,1)$ follows $y(x) = \\nicefrac12$ and $R(x) = \\sqrt{\\nicefrac14 + x^2}$.\nThe boundaries of significance are given by $ | x_\\text{bound} | = \\nicefrac12 (\\tan(\\alpha_\\text{max} / 2))^{-1}$.\nFrom these we can derive the locations $l = (\\pm x_\\text{bound}, y(x_\\text{bound}))$.\nThese specific cases can easily be transformed into all possible cases using scaling and rotation operations.", "meta": {"hexsha": "a7b1fa3666c897fa84fffdaea013df038e98b47a", "size": 1029, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "19_edge_discretization.tex", "max_stars_repo_name": "BagelOrb/variable_width_paper", "max_stars_repo_head_hexsha": "8b8b7a2b9b913e56267a578d2a15ed7e97fa3503", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-23T10:22:42.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-23T10:22:42.000Z", "max_issues_repo_path": "19_edge_discretization.tex", "max_issues_repo_name": "BagelOrb/variable_width_paper", "max_issues_repo_head_hexsha": "8b8b7a2b9b913e56267a578d2a15ed7e97fa3503", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "19_edge_discretization.tex", "max_forks_repo_name": "BagelOrb/variable_width_paper", "max_forks_repo_head_hexsha": "8b8b7a2b9b913e56267a578d2a15ed7e97fa3503", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 128.625, "max_line_length": 222, "alphanum_fraction": 0.7123420797, "num_tokens": 299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9059898178450964, "lm_q2_score": 0.6893056040203135, "lm_q1q2_score": 0.6245038586259679}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\n\\title{Math Proofs}\n\\author{Andrew Archibald}\n\\date{July 2017}\n\n\\begin{document}\n\n\\begin{titlepage}\n\\maketitle\n\\end{titlepage}\n\n\\section{Pi by the sum of inverse squares}\n\n$$\\frac{\\pi^2}{6} = \\sum_{n=1}^{\\infty}\\frac{1}{n^2}$$\n\n\\section{Pi by inverse tangent Taylor series}\n\n\\begin{align*}\n\\arctan x &= x - \\frac{x^3}{3} + \\frac{x^5}{5} - \\frac{x^7}{7} + \\frac{x^9}{9} - \\cdots \\\\\n &= \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n + 1}}{2n + 1}\n\\end{align*}\n$$x \\in [-1, 1]$$\n$$\\frac{\\pi}{4} = \\arctan 1\n= \\sum_{n=0}^{\\infty} \\frac{(-1)^n}{2n + 1}$$\n\n\\section{Pi by inscribed and circumscribed polygons}\n\n\\subsection{Inscribed}\n\n\\subsubsection{Law of sines}\n\n$$\\frac{\\sin(\\frac{\\pi}{2}-\\frac{\\pi}{n})}{r} = \\frac{\\sin\\frac{2\\pi}{n}}{l}$$\n$$= \\frac{\\cos\\frac{\\pi}{n}}{r} = \\frac{2\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}}{l}$$\n$$l = 2r\\sin\\frac{\\pi}{n}$$\n$$P = ln\n= 2rn\\sin\\frac{\\pi}{n}$$\n\n\\subsubsection{Law of cosines}\n\n$$l^2 = 2r^2 - 2r^2cos\\frac{2\\pi}{n}\n= 2r^2\\left(1 - cos\\frac{2\\pi}{n}\\right)\n= 4r^2\\left(\\frac{1 - cos\\frac{2\\pi}{n}}{2}\\right)\n= 4r^2\\sin^2{\\frac{\\pi}{n}}$$\n$$l = 2r\\sin\\frac{\\pi}{n} = r\\sqrt{2 - 2\\cos\\frac{2\\pi}{n}}$$\n$$P = 2n\\sin\\frac{\\pi}{n}$$\n\n\\subsubsection{Pythagorean theorem}\n\n$$r^2 = \\left(\\frac{l}{2}\\right)^2 + h^2$$\n$$h^2 = r^2 - \\frac{1}{4}l^2\n= r^2 - r^2\\sin^2\\frac{\\pi}{n}\n= r^2\\left(1 - \\sin^2\\frac{\\pi}{n}\\right)\n= r^2\\cos^2\\frac{\\pi}{n}$$\n$$h = r\\cos\\frac{\\pi}{n}$$\n$$A = \\frac{1}{2}lhn\n= \\frac{1}{2}\\left(2r\\sin\\frac{\\pi}{n}\\right)\\left(rcos\\frac{\\pi}{n}\\right)n\n= nr^2\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}\n= \\frac{1}{2}nr^2\\sin\\frac{2\\pi}{n}$$\n\n\\subsubsection{Results}\n\nPerimeter\n\n$$2\\pi{r} = \\lim_{n\\to\\infty} P\n= \\lim_{n\\to\\infty} 2rn\\sin\\frac{\\pi}{n}$$\n\nArea\n\n$$\\pi{r^2} = \\lim_{n\\to\\infty} \\frac{1}{2}nr^2\\sin\\frac{2\\pi}{n}$$\n$$\\pi = \\lim_{n\\to\\infty} \\frac{1}{2}n\\sin\\frac{2\\pi}{n}\n= \\lim_{n\\to\\infty} n\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}$$\n\n\\subsection{Circumscribed}\n\n\\subsubsection{Law of sines}\n\n$$\\frac{\\sin\\frac{\\pi}{n}}{L} = \\frac{\\sin\\left(\\pi - \\frac{2\\pi}{n}\\right)}{l}$$\n$$\\sin\\left(\\pi - \\frac{2\\pi}{n}\\right) = \\sin\\left[2\\left(\\frac{\\pi}{2} - \\frac{\\pi}{n}\\right)\\right]\n= 2\\sin\\left(\\frac{\\pi}{2} - \\frac{\\pi}{n}\\right)\\cos\\left(\\frac{\\pi}{2} - \\frac{\\pi}{n}\\right)\n= 2\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}$$\n$$L = \\frac{l\\sin\\frac{\\pi}{n}}{2\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}}\n= \\frac{1}{2}l\\sec\\frac{\\pi}{n}\n= \\frac{1}{2}\\left(2r\\sin\\frac{\\pi}{n}\\right)\\sec\\frac{\\pi}{n}\n= r\\tan\\frac{\\pi}{n}$$\n\n\\subsubsection{Law of cosines}\n\n$$l^2 = 2L^2 - 2L^2cos\\frac{\\pi - 2\\pi}{n}\n= L^2\\left(2 - 2\\cos\\left(\\pi - \\frac{2\\pi}{n}\\right)\\right)$$\n$$\\cos\\left(\\pi - \\frac{2\\pi}{n}\\right) = \\cos\\left(2\\left(\\frac{\\pi}{2} - \\frac{\\pi}{n}\\right)\\right)\n= 1 - 2\\sin^2\\left(\\frac{\\pi}{2} - \\frac{\\pi}{n}\\right)\n= 1 - 2\\cos^2\\frac{\\pi}{2}$$\n$$l^2 = L^2\\left[2 - 2\\left(1 - 2\\cos^2\\frac{\\pi}{n}\\right)\\right]\n= 4L^2\\cos^2\\frac{\\pi}{n}$$\n$$l = 2L\\cos\\frac{\\pi}{n}$$\n$$2r\\sin\\frac{\\pi}{n} = 2L\\cos{\\pi}{n}$$\n$$L = r\\tan\\frac{\\pi}{n}$$\n\n\\subsubsection{Pythagorean theorem}\n\n$$L^2 = \\left(\\frac{l}{2}\\right)^2 + H^2$$\n$$H^2 = r^2tan^2\\frac{\\pi}{n} - r^2\\sin^2\\frac{\\pi}{n}\n= r^2\\left(\\tan^2\\frac{\\pi}{n} - \\sin^2\\frac{\\pi}{n}\\right)$$\n$$tan^2\\frac{\\pi}{n} - \\sin^2\\frac{\\pi}{n} = tan^2\\frac{\\pi}{n} - \\cos^2\\frac{\\pi}{2}\\tan\\frac{\\pi}{2}\n= \\tan^2\\frac{\\pi}{n}\\left(1 - \\cos^2\\frac{\\pi}{n}\\right)\n= \\sin^2\\frac{\\pi}{n}tan^2\\frac{\\pi}{n}$$\n$$H^2 = R^2sin^2\\frac{\\pi}{n}\\tan^2{\\pi}{n}$$\n$$H = r\\sin\\frac{\\pi}{n}\\tan\\frac{\\pi}{n}$$\n\n\\subsubsection{Results}\n\nPerimeter\n\n$$P = 2nL = 2nr\\tan\\frac{\\pi}{n}$$\n$$2\\pi{r} = \\lim_{n\\to\\infty} 2nr\\tan\\frac{\\pi}{n}$$\n$$\\pi = \\lim_{n\\to\\infty} n\\tan\\frac{\\pi}{n}$$\n\nArea\n\n$$A = nH\\frac{L}{2} + A_l$$\n$$ = n\\left(r\\sin\\frac{\\pi}{n}\\tan\\frac{\\pi}{n}\\right)\\left(r\\sin\\frac{\\pi}{n}\\right) + A_l$$\n$$ = nr^2\\sin^2\\frac{\\pi}{n}\\tan\\frac{\\pi}{n} + A_l$$\n$$ = nr^2\\sin^2\\frac{\\pi}{n}\\tan\\frac{\\pi}{n} + nr^2\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}$$\n$$ = nr^2\\sin\\frac{\\pi}{n}\\left(\\sin\\frac{\\pi}{n}\\tan\\frac{\\pi}{n} + \\cos\\frac{\\pi}{n}\\right)$$\n$$ = nr^2\\tan\\frac{\\pi}{n}\\left(\\sin^2\\frac{\\pi}{n} + \\cos^2\\frac{\\pi}{n}\\right)$$\n$$ = nr^2\\tan\\frac{\\pi}{n}$$\n$$\\pi{r^2} = \\lim_{n\\to\\infty} nr^2\\tan\\frac{\\pi}{n}$$\n$$\\pi = \\lim_{n\\to\\infty} n\\tan\\frac{\\pi}{n}$$\n\n\\subsection{Simplified results}\n\n$$\\pi = \\lim_{n\\to\\infty} n\\sin\\frac{\\pi}{n}$$\n$$\\pi = \\lim_{n\\to\\infty} n\\sin\\frac{\\pi}{n}\\cos\\frac{\\pi}{n}$$\n$$\\pi = \\lim_{n\\to\\infty} n\\tan\\frac{\\pi}{n}$$\n\n\\section{Multi-dimensional spheres}\n\n$$V_1 = \\int\\limits_{-R}^{R} dx = 2R$$\n$$V_2\n= \\int\\limits_{-R}^{R} \\int_{-\\sqrt{R^2-x^2}}^{\\sqrt{R^2-x^2}} dydx\n= \\int\\limits_{-R}^{R} 2\\sqrt{R^2-x^2} dx\n= \\int\\limits_{-R}^{R} 2R\\sin\\arccos\\frac{x}{R} dx$$\n$$V_2\n= \\int\\limits_0^{2\\pi} \\int\\limits_0^{R} r drd\\theta\n= \\int\\limits_0^{2\\pi} \\frac{1}{2}R^2 d\\theta\n= \\pi R^2$$\n\n\\section{Trigonometric integrals}\n\n\\begin{align*}\n\\int \\sin^2{x} dx\n &= \\int \\frac{1-\\cos{2x}}{2} dx \\\\\n &= \\frac{1}{2}(x - \\sin{x}\\cos{x}) + C\n\\\\\\\\\n\\int \\cos^2{x} dx\n &= \\int \\frac{1+\\cos{2x}}{2} dx \\\\\n &= \\frac{1}{2}(x + \\sin{x}\\cos{x}) + C\n\\end{align*}\n\n\\begin{align*}\n\\int \\sin^3{x} dx\n &= \\int (1 - \\cos^2{x})\\sin{x} dx \\\\\n &= \\int (\\sin{x} - \\cos^2{x}\\sin{x}) dx \\\\\n &= -\\cos{x} + \\frac{1}{3}\\cos^3{x} + C\n\\\\\\\\\n\\int \\cos^3{x} dx\n &= \\int (1 - \\sin^2{x})\\cos{x} dx \\\\\n &= \\int (\\cos{x} - \\sin^2{x}\\cos{x}) dx \\\\\n &= \\sin{x} - \\frac{1}{3}\\sin^3{x} + C\n\\end{align*}\n\n\\begin{align*}\n\\int \\sin^4{x} dx\n &= \\int \\left(\\frac{1 -\\cos{2x}}{2}\\right)^2 dx \\\\\n &= \\int \\frac{1}{4}(1 - 2\\cos{2x} + \\cos^2{2x}) dx \\\\\n &= \\int \\frac{1}{4}\\left(1 - 2\\cos{2x} + \\frac{1 + \\cos{4x}}{2}\\right) dx \\\\\n &= \\int \\frac{1}{8}(3 - 4\\cos{2x} + \\cos{4x}) dx \\\\\n &= \\frac{1}{8}\\left(3x - 2\\sin{2x} + \\frac{1}{4}\\sin{4x}\\right) + C\n\\\\\\\\\n\\int \\cos^4{x} dx\n &= \\int \\left(\\frac{1 +\\cos{2x}}{2}\\right)^2 dx \\\\\n &= \\int \\frac{1}{4}(1 + 2\\cos{2x} + \\cos^2{2x}) dx \\\\\n &= \\int \\frac{1}{4}\\left(1 + 2\\cos{2x} + \\frac{1 + \\cos{4x}}{2}\\right) dx \\\\\n &= \\int \\frac{1}{8}(3 + 4\\cos{2x} + \\cos{4x}) dx \\\\\n &= \\frac{1}{8}\\left(3x + 2\\sin{2x} + \\frac{1}{4}\\sin{4x}\\right) + C\n\\end{align*}\n\n\\begin{align*}\n\\int \\sin^5{x} dx\n &= \\int (1 - \\cos^2{x})^2\\sin{x} dx \\\\\n &= \\int (\\sin{x} - 2\\cos^2{x}\\sin{x} + \\cos^4{x}\\sin{x}) dx \\\\\n &= -\\cos{x} + \\frac{2}{3}\\cos^3{x} - \\frac{1}{5}\\cos^5{x} + C\n\\\\\\\\\n\\int \\cos^5{x} dx\n &= \\int (1 - \\sin^2{x})^2\\cos{x} dx \\\\\n &= \\int (\\cos{x} - 2\\sin^2{x}\\cos{x} + \\sin^4{x}\\cos{x}) dx \\\\\n &= \\sin{x} - \\frac{2}{3}\\sin^3{x} + \\frac{1}{5}\\sin^5{x} + C\n\\end{align*}\n\n\\end{document}\n", "meta": {"hexsha": "f7b1b787de65384fbe100a3112ed80fc4ebfba8c", "size": 6517, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "math.tex", "max_stars_repo_name": "andrewarchi/math-proofs", "max_stars_repo_head_hexsha": "b259e422258d658e1333d5ecf5745483846c23e0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "math.tex", "max_issues_repo_name": "andrewarchi/math-proofs", "max_issues_repo_head_hexsha": "b259e422258d658e1333d5ecf5745483846c23e0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "math.tex", "max_forks_repo_name": "andrewarchi/math-proofs", "max_forks_repo_head_hexsha": "b259e422258d658e1333d5ecf5745483846c23e0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.1034482759, "max_line_length": 102, "alphanum_fraction": 0.5347552555, "num_tokens": 3307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8807970873650401, "lm_q2_score": 0.7090191399336402, "lm_q1q2_score": 0.6245019933396161}} {"text": "% Section with some maths\n\n\\section{Ein Abschnitt mit Matheformeln}\n\\blindmathpaper\n\n\n\\subsection{Formeln inkl. Subequations and Labels}\n\\label{sec:maths-incl-subeq}\n\\blindtext\n\\newcommand{\\iu}{{i\\mkern1mu}}\n\\vspace{-.4cm}\n\\begin{subequations}\n \\begin{align}\n \\iu \\hbar \\partial_t \\Psi &= - \\iu \\hbar c \\sum_{i=1}^3 \\alpha_i \\partial_i \\Psi + \\beta m c^2 \\Psi\\\\ &= \\left( c \\vec \\alpha \\cdot \\vec p + \\beta m c^2 \\right) \\Psi \\\\\n \\Longrightarrow \\quad 0 &= \\left( \\iu \\hbar \\gamma^\\mu \\partial_\\mu - mc \\right) \\Psi\n \\end{align}\n\\end{subequations}\n\\blindtext\n\\begin{align}\n F(x,y)=0 ~~\\mbox{and}~~\n \\left| \\begin{array}{ccc}\n F''_{xx} & F''_{xy} & F'_x \\\\\n F''_{yx} & F''_{yy} & F'_y \\\\\n F'_x & F'_y & 0\n \\end{array}\\right| = 0\n\\label{eq:1}\n\\end{align}\n\\blindtext\n\n\n% End of file\n", "meta": {"hexsha": "717a52fe18e830525d27b51aab3abc5e581e6ee0", "size": 838, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/maths.tex", "max_stars_repo_name": "kzoch/kLabCourse-template", "max_stars_repo_head_hexsha": "efaffbde3bea7ea826357398414b35bf54b934c2", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/maths.tex", "max_issues_repo_name": "kzoch/kLabCourse-template", "max_issues_repo_head_hexsha": "efaffbde3bea7ea826357398414b35bf54b934c2", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/maths.tex", "max_forks_repo_name": "kzoch/kLabCourse-template", "max_forks_repo_head_hexsha": "efaffbde3bea7ea826357398414b35bf54b934c2", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.1875, "max_line_length": 172, "alphanum_fraction": 0.5942720764, "num_tokens": 323, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9149009480320035, "lm_q2_score": 0.6825737214979745, "lm_q1q2_score": 0.6244873449002296}} {"text": "% !TEX root = ../../main.tex\n\\subsection{xokker-Planck equation in population genetics}\n\nIn the last section after we performed a Taylor expansion of the master\nequation we ended up with an equally complicated partial differential equation\nof infinite order. So in order to make any progress towards simplifying the\ntreatment of the equation what we will do is trade accuracy for simplicity.\nMore concretely if our assumptions on the shape of the transition probability\n$\\phi_t(x; r)$ being sharply peaked are to be taken seriously, we can then\ntruncate the Kramers-Moyal expansion to include only up to second order\nderivatives. This truncated equation would locally approximate the time\nevolution of the probability distribution $P(x, t)$. \\eref{eq_km_expansion}\nthen is of the form\n\\begin{equation}\n \\ddt{P(x, t)} = - {\\partial \\over \\partial x}\n \\left[\n a^{(1)}(x, t) P(x, t)\n \\right] +\n {1 \\over 2}{\\partial^2 \\over \\partial f^2}\n \\left[\n a^{(2)}(x, t) P(x, t)\n \\right].\n \\label{eq_fokker_planck}\n\\end{equation}\nIn the physics literature this is known as the Fokker-Planck equation, while in\nthe mathematics literature is known as the Kolmogorov forward equation. One\nintriguing aspect of how to arrive to this equation is the seemingly arbitrary\nchoice of truncating up to the second moment. The argument that is often thrown\naround is that the ``art'' of these truncations is to stop at the first\nnon-vanishing moment. But for the specific case of the Kramers-Moyal expansion\nof the master equation there is a theorem - Pawula theorem - that shows that\nfor the solutions of the Kramers-Moyal expansion to be interpreted as\nprobability densities the expansion must either contain one, two or an infinite\nnumber of moments. So two sounds much better than infinite, doesn't it?\n\\mrm{Need to include appendix with the proof of Pawula theorem.}\n\n\\subsubsection{Determination of the Fokker-Planck coefficients}\n\nThe two jump moments in \\eref{eq_fokker_planck}, $a^{(1)}$ and $a^{(2)}$\ndefined by \\eref{eq_jump_mom} have a specific interpretation in population\ngenetics. Here is where a little bit of a terminology conflict between physics\nand evolutionary biology comes into play. The directional term, i.e. the one\nwith the first order derivative in \\eref{eq_fokker_planck} is known in the\nphysics literature as the drift term on a diffusion-like equation. This is\nconfusing because in evolutionary theory the diffusive term, i.e. the one with\nthe second derivative in \\eref{eq_fokker_planck} is known as the genetic drift\nterm. I will try to be as consistent as possible on these notes using the terms\ndirectional and diffusive to avoid confusion.\n\nComing back to the directional term, we define $M(x, t)$ to be\n\\begin{equation}\n M(x, t) \\equiv \\ee{r(t)} = \\int_{-\\infty}^{\\infty} dr \\; \\phi_t(x; r) r,\n\\end{equation}\ni.e. the mean of the jump distribution. As we will see in coming sections this\nterm captures the effect of directional evolutionary forces such as selection,\nmutation and migration. For the diffusive term we define $V(x, t)$ as the\nsecond moment of the jump distribution\n\\begin{equation}\n V(x, t) \\equiv \\ee{r^2(t)} = \\int_{-\\infty}^{\\infty} dr \\; \\phi_t(x; r) r^2.\n\\end{equation}\nThis term captures the random sampling of alleles, also known as genetic drift.\nIn all of the population genetics literature I have encounter so far, this term\n$V(x, t)$ is treated as the \\textbf{variance} rather than the second moment.\nThis is partly because computing the specific functional form of the variance\nfor different models of reproduction is much simpler. The variance\n$\\sigma_r^2(t)$ is defined as\n\\begin{equation}\n \\sigma_r^2(t) = \\ee{\\left( r - \\ee{r} \\right)^2} = \\ee{r^2} - \\ee{r}^2.\n\\end{equation}\nWe can therefore work with this more convenient quantity if we assume that\n$\\ee{r}^2 \\approx 0$. This is a reasonable assumption given that for the\nFokker-Planck equation to be accurate we assumed a tight distribution for the\njumpt size $\\phi_t(x; r)$. The peaked nature of this distribution must imply\nthat $\\ee{r} \\ll 1$, but more importantly, we will assume that\n$\\ee{r}^2 \\ll \\ee{r^2}$. So upon using these two definitions we arrive to one\nof the main results in population genetics, the Kimura diffusion equation\n\\begin{equation}\n \\ddt{P(x, t)} = - {\\partial \\over \\partial x}\n \\left[\n M(x, t) P(x, t)\n \\right] +\n {1 \\over 2}{\\partial^2 \\over \\partial f^2}\n \\left[\n V(x, t) P(x, t)\n \\right].\n \\label{eq_kimura_diffusion}\n\\end{equation}\nThe power of diffusion theory is that in these two terms $M(x, t)$ and $V(x,\nt)$ we can include all evolutionary forces acting simultaneously. The\nfunctional forms of these specific terms depend on the reproduction model used.\nWe will explore that more specifically later on.\n\n\\subsubsection{Equilibrium distribution}\n\nIn the limit when $t \\rightarrow \\infty$ we expect the distribution of allele\nfrequencies to reach a steady-state $P_{ss}(x)$. For the 1D case we have\nstudied so far, i.e. two alleles with frequencies $x$ and $1 - x$ this steady\nstate is equivalent to an equilibrium distribution since detailed balance has\nto be satisfied. To emphasize this point let us rewrite\n\\eref{eq_kimura_diffusion} as a statement of conservation of probability. This\nis\n\\begin{equation}\n \\ddt{P(x, t)} = - {\\partial J(x, t) \\over \\partial x},\n\\end{equation}\nwhere $J(x, t)$ is the probability flux at point $x$. If we set the time\nderivative to zero there are only two options (in reality there is only one\noption for 1D systems):\n\\begin{enumerate}\n \\item ${\\partial J \\over \\partial x} = 0; \\; J \\neq 0 \\Rightarrow$ Steady\n state on a rotating or non-conservative field.\n \\item ${\\partial J \\over \\partial x} = 0; \\; J = 0 \\Rightarrow$ Equilibrium\n distribution that satisfies detailed balance.\n\\end{enumerate}\nFor our one-locus two-alleles case the second of these cases must be true.\nWhat this implies is that at steady state the flux $J_{ss}(x)$ takes the form\n\\begin{equation}\n J_{ss}(x) = - M(x) P_{eq}(x) + {\\partial \\over \\partial x}\n \\left[ V(x) P_{eq}(x) \\right] = 0,\n \\label{eq_flux_eq}\n\\end{equation}\nwhere we use $P_{eq}(x)$ to define that this is not only a steady-state\ndistribution, but an equilibrium distribution satisfying detailed balance.\n\\eref{eq_flux_eq} is a first order homogeneous ordinary differential equation.\nWe can solve it using the integration factor method. For this we define\n$G(x) \\equiv V(x)P_{eq}(x)$. Substituting this into \\eref{eq_flux_eq} gives\n\\begin{equation}\n {- M(x) \\over V(x)} G(x) + {\\partial \\over \\partial x} G(x) = 0.\n \\label{eq_ode_ss}\n\\end{equation}\nIn this form we define the integration factor to be\n\\begin{equation}\n h(x) = \\exp \\left( \\int_0^x -{M(x') \\over V(x')} \\; dx' \\right).\n\\end{equation}\nNotice that we chose the limits of integration to be $[0, x]$. This is because\nthe fundamental theorem of calculus states that for any function $f(x)$ defined\nin $[a, b]$, the antiderivative $F(x)$ is defined as\n\\begin{equation}\n F(x) = \\int_a^x f(x') \\; dx',\n\\end{equation}\nregardless of the lower limit of integration. Multiplying both sides of\n\\eref{eq_ode_ss} by the integration factor $h(x)$ results in\n\\begin{equation}\n {- M(x) \\over V(x)} G(x)\n \\exp \\left( \\int_0^x -{M(x') \\over V(x')} \\; dx' \\right)\n + {\\partial \\over \\partial x} G(x)\n \\exp \\left( \\int_0^x -{M(x') \\over V(x')} \\; dx' \\right)\n = 0.\n \\label{eq_ode_int_fact}\n\\end{equation}\nThe specific form of the integration factor was chosen such that we could\nrewrite \\eref{eq_ode_int_fact} as\n\\begin{equation}\n {d \\over dx}\n \\left[\n \\exp \\left(\n - \\int_0^x {M(x') \\over V(x')}\n \\right)G(x) \\; dx'\n \\right] = 0.\n\\end{equation}\nWritten in this form we can simply integrate both sides with respect to $x$ as\n\\begin{equation}\n \\int_0^x {d \\over dx''}\\left[\n \\exp \\left(\n - \\int_0^{x''} {M(x') \\over V(x')}\n \\right) G(x'')\\; dx'\n \\right] =\n \\int_0^x 0 \\; dx''\n\\end{equation}\nEvaluating these integrals results in\n\\begin{equation}\n \\exp \\left(\n - \\int_0^{x} {M(x') \\over V(x')}\n \\right) G(x)\\; dx' = C,\n\\end{equation}\nwhere $C$ is an integration constant. Notice that for a specific interval\n$[a, b] \\in \\mathbb{R}$ the integral of zero is\n\\begin{equation}\n \\int_a^b 0 dt = 0,\n\\end{equation}\nbut when we set the upper integration limit as an independent variable, what we\nare asking for is the antiderivative of zero which is a constant $C$.\nSubstituting the definition of $G(x) = V(x) P_{eq}(x)$ gives\n\\begin{equation}\n \\exp \\left(\n - \\int_0^x dx' \\; {M(x') \\over V(x')}\\right)\n V(x) P_{eq}(x) = C.\n\\end{equation}\nWe can then solve for the equilibrium allele distribution $P_{eq}(x)$ obtaining\nthe result we were aiming for\n\\begin{equation}\n P_{eq}(x) = {C \\over V(x)} \\exp \\left(\n \\int_0^x dx' \\; {M(x) \\over V(x)}\n \\right).\n\\end{equation}\nThis is a Boltzmann-like distribution! The analogy with statistical mechanics\nbecomes even more clear when we substitute specific functional forms for the\ndirectional term $M(x)$ and the diffusive term $V(x)$.\n\nIn the next section we will explore how to obtain the coefficients for our\nFokker-Planck equation given the Langevin dynamics that we defined in\n\\secref{sec_langevin_intro}.", "meta": {"hexsha": "df0a47e51a5da83cb9f31edd4fe90f1c886bd31b", "size": 9153, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/book_draft/chapters/classic_diffusion/04_fokker_planck.tex", "max_stars_repo_name": "mrazomej/stat_gen", "max_stars_repo_head_hexsha": "abafd9ecc63ae8a804c8df5b9658e47cabf951fa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/book_draft/chapters/classic_diffusion/04_fokker_planck.tex", "max_issues_repo_name": "mrazomej/stat_gen", "max_issues_repo_head_hexsha": "abafd9ecc63ae8a804c8df5b9658e47cabf951fa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-03-05T00:17:26.000Z", "max_issues_repo_issues_event_max_datetime": "2019-03-05T00:17:26.000Z", "max_forks_repo_path": "doc/book_draft/chapters/classic_diffusion/04_fokker_planck.tex", "max_forks_repo_name": "mrazomej/pop_gen", "max_forks_repo_head_hexsha": "abafd9ecc63ae8a804c8df5b9658e47cabf951fa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 45.5373134328, "max_line_length": 79, "alphanum_fraction": 0.7263192396, "num_tokens": 2711, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240791017535, "lm_q2_score": 0.7217432182679956, "lm_q1q2_score": 0.6244696113738624}} {"text": "\n\\subsection{Endomorphisms as group actions}\n\nWe can view each member of the group \\(g\\) as a homomorphim on \\(s\\).\n\nWhere \\(s\\) is a vector space \\(V\\), the representation on each group member is an invertible square matrix.\n\nIf the set we use is the vector space \\(V\\), then we can represent each group element with a square matrix acting on \\(V\\).\n\nFaithful means \\(a\\ne b\\) holds for repesentation too.\n\nRepresentation theory. groups defined by \\(ab=c\\). if we can match each eleemnt to amatrix where this holds we have represented the matrix.\n\n", "meta": {"hexsha": "4149d1732961f462413c1c158d17a73f45158589", "size": 549, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/geometry/endomorphisms/04-02-action.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/geometry/endomorphisms/04-02-action.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/geometry/endomorphisms/04-02-action.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.2142857143, "max_line_length": 139, "alphanum_fraction": 0.7395264117, "num_tokens": 138, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8652240964782012, "lm_q2_score": 0.7217431943271999, "lm_q1q2_score": 0.6244696032010423}} {"text": "\\input{../header_function}\r\n\r\n%---------- start document ---------- %\r\n \\section{cubic\\_root -- cubic root, residue, and so on}\\linkedzero{cubic\\_root}\r\n%\r\n \\subsection{c\\_root\\_p -- cubic root mod p}\\linkedone{cubic\\_root}{c\\_root\\_p}\r\n \\func{c\\_root\\_p}{\\hiki{a}{integer},\\ \\hiki{p}{integer}}{\\out{list}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the cubic root of \\param{a} modulo prime \\param{p}. (i.e. solutions of the equation $x^3 = \\param{a} \\pmod{\\param{p}}$).\\\\\r\n \\spacing\r\n % added document\r\n %\\spacing\r\n % input, output document\r\n \\quad \\param{p} must be a prime integer.\\\\\r\n This function returns the list of all cubic roots of \\param{a}.\\\\\r\n%\r\n \\subsection{c\\_residue -- cubic residue mod p}\\linkedone{cubic\\_root}{c\\_residue}\r\n \\func{c\\_residue}{\\hiki{a}{integer},\\ \\hiki{p}{integer}}{\\out{integer}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Check whether the rational integer \\param{a} is cubic residue modulo prime \\param{p}.\\\\\r\n \\spacing\r\n % added document\r\n \\quad If $\\param{p}\\ |\\ \\param{a}$, then this function returns $0$,\r\n elif \\param{a} is cubic residue modulo \\param{p}, then it returns $1$,\r\n otherwise (i.e. cubic non-residue), it returns $-1$.\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{p} must be a prime integer.\\\\\r\n%\r\n \\subsection{c\\_symbol -- cubic residue symbol for Eisenstein-integers}\\linkedone{cubic\\_root}{c\\_symbol}\r\n \\func{c\\_symbol}{\\hiki{a1}{integer},\\ \\hiki{a2}{integer},\\ \\hiki{b1}{integer},\\ \\hiki{b2}{integer}}{\\out{integer}}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the (Jacobi) cubic residue symbol of two Eisenstein-integers $\\left(\\frac{\\param{a1}+\\param{a2}\\omega}{\\param{b1}+\\param{b2}\\omega}\\right)_3$,\r\nwhere $\\omega$ is a primitive cubic root of unity.\\\\\r\n \\spacing\r\n % added document\r\n If $\\param{b1}+\\param{b2}\\omega$ is a prime in $\\mathbb{Z}[\\omega]$, it shows $\\param{a1}+\\param{a2}\\omega$ is cubic residue or not.\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad We assume that $\\param{b1}+\\param{b2}\\omega$ is not divisible $1-\\omega$.\\\\\r\n%\r\n \\subsection{decomposite\\_p -- decomposition to Eisenstein-integers}\\linkedone{cubic\\_root}{decomposite\\_p}\r\n \\func{decomposite\\_p}{\\hiki{p}{integer}}{(\\out{integer},\\ \\out{integer})}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return one of prime factors of \\param{p} in $\\mathbb{Z}[\\omega]$.\\\\\r\n \\spacing\r\n % added document\r\n \\quad If the output is (\\param{a},\\ \\param{b}), then $\\frac{\\param{p}}{\\param{a}+\\param{b}\\omega}$ is a prime in $\\mathbb{Z}[\\omega]$.\r\n In other words, \\param{p} decomposes into two prime factors $\\param{a}+\\param{b}\\omega$ and $\\param{p}/(\\param{a}+\\param{b}\\omega)$ in $\\mathbb{Z}[\\omega]$.\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{p} must be a prime rational integer.\r\n We assume that $\\param{p}\\equiv 1 \\pmod 3$.\\\\\r\n%\r\n \\subsection{cornacchia -- solve $x^2+dy^2=p$}\\linkedone{cubic\\_root}{cornacchia}\r\n \\func{cornacchia}{\\hiki{d}{integer},\\ \\hiki{p}{integer}}{(\\out{integer},\\ \\out{integer})}\\\\\r\n \\spacing\r\n % document of basic document\r\n \\quad Return the solution of $x^2 + \\param{d}y^2 = \\param{p}$.\\\\\r\n \\spacing\r\n % added document\r\n \\quad This function uses Cornacchia's algorithm. See \\cite{Cohen1}.\\\\\r\n \\spacing\r\n % input, output document\r\n \\quad \\param{p} must be prime rational integer.\r\n \\param{d} must be satisfied with the condition $0<\\param{d}<\\param{p}$.\r\n This function returns (\\param{x},\\ \\param{y}) as one of solutions of the equation $x^2 + \\param{d} y^2 = \\param{p}$.\\\\\r\n%\r\n\\begin{ex}\r\n>>> cubic_root.c_root_p(1, 13)\r\n[1, 3, 9]\r\n>>> cubic_root.c_residue(2, 7)\r\n-1\r\n>>> cubic_root.c_symbol(3, 6, 5, 6)\r\n1\r\n>>> cubic_root.decomposite_p(19)\r\n(2, 5)\r\n>>> cubic_root.cornacchia(5, 29)\r\n(3, 2)\r\n\\end{ex}%Don't indent!(indent causes an error.)\r\n\\C\r\n\r\n%---------- end document ---------- %\r\n\r\n\\input{../footer}\r\n", "meta": {"hexsha": "1eb068260b3d74d36d4c4d94e6a98ea63799c2ba", "size": 3956, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "manual/en/cubic_root.tex", "max_stars_repo_name": "turkeydonkey/nzmath3", "max_stars_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-05-26T19:22:17.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-26T19:22:17.000Z", "max_issues_repo_path": "manual/en/cubic_root.tex", "max_issues_repo_name": "turkeydonkey/nzmath3", "max_issues_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "manual/en/cubic_root.tex", "max_forks_repo_name": "turkeydonkey/nzmath3", "max_forks_repo_head_hexsha": "a48ae9efcf0d9ad1485c2e9863c948a7f1b20311", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.9555555556, "max_line_length": 162, "alphanum_fraction": 0.6385237614, "num_tokens": 1313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8577681158979307, "lm_q2_score": 0.7279754548076478, "lm_q1q2_score": 0.6244341342902953}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\\chapter{Asymptotic Analysis}\n\n\\epigraph{In algorithms, as in life, persistence usually pays off.}{Steven S.\\ Skiena}\n\n\\minitoc\n\n\\section{Introduction}\n\nWhat is an algorithm?\n\n\\begin{defn}[Algorithm]\n\tA recipe/description of a process that accomplishes a goal\n\\end{defn}\n\nWhy study algorithms? First, you will gain tools to \\textit{compare} algorithms. Second, studying algorithms will help you to creatively make \\textit{new} algorithms.\n\nWhy study algorithms in discrete math? Computers are discrete. Computers run programs, which are based on algorithms. Hence, algorithms must be discrete.\n\n\\section{\\(\\Oh\\), \\(\\Omega\\), and \\(\\Theta\\) Notations}\n\nOur motivation here is to describe the \\textit{growth-rate} of functions. The growth-rate class gives us key insights to what the actual function does on large inputs.\n\n\\begin{defn}[Big-Oh]\n\tAn upper-bound on the growth-rate of a function. Formally: \\[f(n) \\in \\Oh(g(n)) \\Leftrightarrow (\\exists n_0 \\in \\N, \\exists c \\in \\R^+, \\forall n \\in \\N^{\\geq n_0})[f(n) \\leq c \\cdot g(n)]\\]\n\\end{defn}\n\n\\begin{defn}[Big-Omega]\n\tA lower-bound on the growth-rate of a function. Formally: \\[f(n) \\in \\Omega(g(n)) \\Leftrightarrow (\\exists n_0 \\in \\N, \\exists c \\in \\R^+, \\forall n \\in \\N^{\\geq n_0})[f(n) \\geq c \\cdot g(n)]\\]\n\\end{defn}\n\n\\begin{defn}[Big-Theta]\n\tAn exact growth-rate of a function. Formally: \\[f(n) \\in \\Theta(g(n)) \\Leftrightarrow f(n) \\in \\Oh(g(n)) \\land f(n) \\in \\Omega(g(n))\\]\n\\end{defn}\n\nInformally, \\(f(n) \\in \\Oh(g(n))\\) just means that \\textbf{eventually} (for big enough \\(n\\)) we have that (some constant-multiple of) \\(g\\) overtakes \\(f\\).\nFor example, \\(2^n \\in \\Oh(n!)\\) since \\(2^n < n!\\) (\\(n!\\) overtakes \\(2^n\\)) for \\(n \\geq 4\\) (\\(n \\in \\N\\)).\nSo the growth-rate of \\(2^n\\) is bounded \\textit{above} by \\(n!\\).\nSimilarly, \\(n! \\in \\Omega(2^n)\\) since \\(n! > 2^n\\) for \\(n \\geq 4\\).\nSo \\(n!\\) is bounded \\textit{below} by \\(2^n\\).\nOftentimes in algorithm analysis we aim to show an algorithm has exact run-time (\\(\\Theta\\)), and we do this by showing it has both \\(\\Oh\\) and \\(\\Omega\\) run-time.\nSometimes, though, this is hard to do.\n\n\\begin{rem}\n\tYou may see \\(\\Oh,\\Omega,\\Theta\\) defined using \\(\\forall n \\in \\N^{> n_0}\\).\n\tIt does not actually matter.\n\tJust consult whichever definition your instructor uses.\n\\end{rem}\n\n\\begin{rem}\n\tYou may see \\(f(n) = \\Oh(g(n))\\) instead of \\(f(n) \\in \\Oh(g(n))\\).\n\tBoth notations represent the same thing.\n\\end{rem}\n\n\\begin{rem}\n\tWe need not use \\(n\\) for our function variable.\n\\end{rem}\n\n\\exproof{\n\tShow that \\(f(x) = x^2 + 2x - 4 = \\Oh(x^2)\\).\n}{\n\tTo show \\(\\Oh\\), it suffices to find a constant \\(c\\) and starting point \\(x_0\\).\n\tA typical way to do this is by selecting a starting point that might help us down the line.\n\tConsider \\(x_0 = 1\\).\n\tThen, \\(x \\geq x_0 = 1\\).\n\tThen \\(x \\leq x^2\\) so \\(2x \\leq 2x^2\\) (this is where our starting point choice helps us).\n\tThus \\(x^2 + 2x - 4 \\leq x^2 + 2x \\leq x^2 + 2x^2 = 3x^2\\).\n\tSo we let \\(c = 3\\) and we are done.\n}\n\n\\exproof{\n\tShow that \\(f(x) = x^2 + 2x - 4 = \\Omega(x^2)\\).\n}{\n\tWe take a similar approach as in the previous example.\n\tLet \\(x_0 = 4\\).\n\tThen \\(x \\geq x_0 = 4\\).\n\tThus \\(x \\geq 4 = \\frac{2}{1-\\frac{1}{2}}\\) so \\((1-\\frac{1}{2}x \\geq 2\\) so \\(x - \\frac{1}{2}x \\geq 2\\) thus \\(x - 2 \\geq \\frac{1}{2}x\\).\n\tSince \\(x+2 \\geq x-2\\) then \\(x+2 \\geq \\frac{1}{2}x\\).\n\tThus \\((x+2)(x-2) \\geq (\\frac{1}{2}x)^2 = \\frac{1}{4}x^2\\).\n\tSince \\((x+2)(x-2) = x^2 - 4\\) and \\(x \\geq 4 > 0\\) so \\(2x > 0\\), then \\(x^2+2x-4 \\geq x^2-4 \\geq \\frac{1}{4}x^2\\).\n\tSo we let \\(c = \\frac{1}{4}\\) and we are done.\n}\n\n\\begin{rem}\n\tWe can conclude that our \\(f(x)\\) defined above is in \\(\\Theta(x^2)\\) since it satisfies both \\(\\Oh\\) and \\(\\Omega\\).\n\tNote that the \\(c,x_0\\) used in the \\(\\Oh\\) proof need not be the same as those in the \\(\\Omega\\) proof (and vice versa).\n\tThey only need to exist.\n\\end{rem}\n\nA picture, to see what is going on.\n\n%\\pgfmathdeclarefunction{bound}{1}{%\n%\t\\pgfmathparse{#1*x^2}%\n%}\n\\begin{boxx}\n\t\\begin{center}\n\t\t\\begin{tikzpicture}\n\t\t\t\\begin{axis}[\n\t\t\t\tmark=none,smooth,\n\t\t\t\taxis x line=bottom, % no box around the plot, only x and y axis\n\t\t\t\taxis y line=left,\n\t\t\t\tevery fill between plot/.append style={fill=gray,fill opacity=0.1},\n\t\t\t\tsamples=50]\n\t\t\t\t\\addplot [name path=up,color=red,domain=4:6.5] {(3)*x^2} node[pos=1] (up) {};\n\t\t\t\t\\addplot [name path=lo,color=blue,domain=4:6.5] {(1/4)*x^2} node[pos=1] (lo) {};\n\t\t\t\t\\addplot [gray] fill between[of=up and lo];\n\t\t\t\t\\addplot [color=black,domain=0:6.5] {x^2+2*x-4} node[pos=1] (func) {};\n\t\t\t\t\\addplot [color=red,domain=0:6.5] {(3)*x^2} node[pos=1] (upper) {};\n\t\t\t\t\\addplot [color=blue,domain=0:6.5] {(1/4)*x^2} node[pos=1] (lower) {};\n\t\t\t\t\\node [label=135:{\\(x_0\\)},circle,fill,inner sep=1pt] at (axis cs:1,3) {};\n\t\t\t\t\\node [label=135:{\\(x_0\\)},circle,fill,inner sep=1pt] at (axis cs:4,4) {};\n\t\t\t\\end{axis}\n\t\t\t\\node [above, color=black] at (func) {\\(f(x)\\)};\n\t\t\t\\node [above, color=red] at (upper) {\\(\\Oh\\)};\n\t\t\t\\node [above, color=blue] at (lower) {\\(\\Omega\\)};\n\t\t\\end{tikzpicture}\n\t\\end{center}\n\\end{boxx}\n\nIt should be easy to see from this picture that we can make tighter bounds by changing \\(c\\).\nIf you change your \\(c\\), then your \\(x_0\\) usually will also change.\nIt should also be easy to see from this picture that for a given \\(c\\), we may be able to find better \\(x_0\\).\nIn this case, we could move our \\(\\Omega\\) \\(x_0\\) closer to 2.\nNote that our function \\(f\\) is still bounded by our \\(\\Oh\\) and \\(\\Omega\\) functions.\nAs long as \\(c,x_0\\) \\textit{exist}, we are good to go.\n\nIt might be confusing seeing these proofs, as we somewhat made everything appear out of nowhere.\nOften, we solve these proofs using scratch work, then cleaning it up for presentation.\n\n\\begin{example}\n\tFor \\(\\Omega\\), we started at the goal of \\(x^2 + 2x - 4 \\stackrel{?}{\\geq} cx^2\\).\n\tNote that \\(x^2 + 2x - 4 \\geq x^2-4 = (x+2)(x-2)\\).\n\tSince \\(x+2 > x-2\\) then if we can find \\(d > 0\\) such that \\(x-2 \\geq dx\\) then we can conclude that \\((x+2)(x-2) \\geq d^2x^2\\).\n\tThen we let \\(c = d^2\\).\n\t\n\tSo now our goal is to solve \\(x-2 \\stackrel{?}{\\geq} dx\\).\n\tRearranging yields \\(x \\stackrel{?}{\\geq} \\frac{2}{(1-d)}\\).\n\tBy letting \\(d = \\frac{1}{2}\\), then we get \\(x \\geq 4\\) and we can thus work backwards for the actual proof to find \\(c = \\frac{1}{4}\\) with \\(x_0 = 4\\).\n\\end{example}\n\n% todo -- limit definitions\n\n\\section{Analyzing Algorithms}\n\nWe can express algorithms in terms of functions. The function input is typically \\(n\\), the number of elements inputted into, or input size for, the algorithm, and the function output is typically time. Usually \\textit{time} will be the number of certain operations performed during the algorithm.\n\n\\exsol{\n\tCome up with a function \\(T(n)\\) that describes the run-time of the following algorithm. Define run-time as the number of conditional evaluations. In the following algorithm, \\(L\\) is a list of integers of length \\(n\\).\n\t\\begin{algorithmic}[1]\n\t\t\\Function{GetMax}{\\(L\\), \\(n\\)\\@}\n\t\t\t\\State \\(M \\gets -\\infty\\)\n\t\t\t\\State \\(i \\gets 0\\)\n\t\t\t\\While{\\(i < n\\)}\n\t\t\t\t\\If{\\(L[i] > M\\)}\n\t\t\t\t\t\\State \\(M \\gets L[i]\\)\n\t\t\t\t\\EndIf\n\t\t\t\t\\State \\(i \\gets i+1\\)\n\t\t\t\\EndWhile\n\t\t\t\\State \\Return \\(M\\)\n\t\t\\EndFunction\n\t\\end{algorithmic}\n}{\n\tWe evaluate the if-statement in line 5 exactly \\(n\\) times, and we evaluate the while-loop condition exactly \\(n+1\\) times (the last time is when we break out of the loop). Thus, \\(T(n) = n + n + 1 = 2n + 1\\)\n}\n\n\\exsol{\n\tNow come up with a function \\(U(n)\\) that outputs the number of \\(\\gets\\) operations. Compare \\(T(n)\\) to \\(U(n)\\).\n}{\n\tWe have two initial stores from lines 1 and 2. Each while-loop iteration has a store command in line 8. The while-loop runs exactly \\(n\\) times, so this gives us \\(n\\) stores. Line 6 has a store command which is run \\(\\leq n\\) times. We can model this using a natural number \\(c \\leq n\\) where \\(c\\) is the number of times line 5 evaluates to true. Our function is thus \\(U(n) = n + c + 2\\)\n\t\n\tSince \\(c \\leq n\\) we have that \\(U(n) \\leq T(n)\\) for all \\(n > 1\\). Note that when \\(n = 1\\) we have \\(T(n) = 3\\) and \\(U(n) = 4\\) (since \\(c = 1\\)). In any case, we have that the growth-rates of both functions are solely dependent on \\(n\\) and are hence the same. Both functions are in \\(\\Theta(n)\\)\n}\n\n\\begin{defn}[The Loop-Heuristic]\n\t(algorithm analysis) For each nested loop in an algorithm, multiply \\(n\\) by the number of nested levels. This can be helpful in an initial analysis, however it will only take you so far.\n\\end{defn}\n\n\\begin{rem}\n\tWe state the loop-heuristic as a definition, however it can be rephrased as a provable theorem.\n\\end{rem}\n\nIt is important to think about what is going on during the algorithm instead of blindly following the loop-heuristic. We present a few related examples.\n\n\\exsol{\n\tAnalyze the following algorithm:\n\t\\begin{algorithmic}[1]\n\t\t\\Function{AnalyzeList}{\\(L\\), \\(n\\)\\@}\n\t\t\t\\State \\(M \\gets 0\\)\n\t\t\t\\For{\\(i \\gets 0\\); \\(i < n\\); \\(i \\gets i + 1\\)}\n\t\t\t\t\\State \\(k \\gets L[i]\\)\n\t\t\t\t\\For{\\(j \\gets i\\); \\(j < n\\); \\(j \\gets j + 1\\)}\n\t\t\t\t\t\\State \\(k \\gets k * L[j]\\)\n\t\t\t\t\\EndFor\n\t\t\t\t\\State \\(M \\gets M + k\\)\n\t\t\t\\EndFor\n\t\t\t\\State \\Return \\(M\\)\n\t\t\\EndFunction\n\t\\end{algorithmic}\n}{\n\tThe algorithm contains 2 nested loops, so by the loop-heuristic our run-time is \\(n^2\\). We examine further. The outer loop goes through the list \\(n\\) times. For each iteration \\(i\\), we have a loop that goes through the list \\(n - i\\) times. The actual \\textit{work} is done in the inner loop. Thus, the amount of work equals \\(1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2} = \\frac{1}{2}n^2 + \\frac{1}{2}n\\) by the Gaussian sum. This has a growth rate of \\(n^2\\), so we see that the loop-heuristic worked and the algorithm runs in \\(\\Oh(n^2)\\). In fact, the amount of work does not change with respect to different input ordering, so the algorithm runs in \\(\\Theta(n^2)\\)\n}\n\n\\exsol{\n\tExamine the following algorithm:\n\t\\begin{algorithmic}[1]\n\t\t\\Function{AnalyzeGraph}{\\(G = (V,E)\\)\\@}\n\t\t\t\\State \\(c \\gets 0\\)\n\t\t\t\\For{\\(v \\in V\\)}\n\t\t\t\t\\For{\\(w \\in \\text{UnvisitedNeighbors}(v)\\)}\n\t\t\t\t\t\\State Visit \\(w\\)\n\t\t\t\t\t\\State \\(c \\gets c + 1\\)\n\t\t\t\t\\EndFor\n\t\t\t\\EndFor\n\t\t\t\\State \\Return \\(c\\)\n\t\t\\EndFunction\n\t\\end{algorithmic}\n}{\n\t\\(G\\) represents a graph, where \\(V\\) is the set of vertices and \\(E\\) is the set of edges. We usually denote \\(|V| = n\\) and \\(|E| = m\\).\n\t\n\tLet us first think about what the algorithm is \\textit{doing}. We instantiate \\(c=0\\), then go through each unvisited vertex we increment \\(c\\). We consider the amount of work in the inner loop \\(\\Theta(1)\\), and we do this exactly \\(|V| = n\\) times. So our algorithm runs in \\(\\Theta(V)\\) (or \\(\\Theta(n)\\)). % todo : undirected or directed? define neighbors?\n}\n\n\\begin{rem}\n\tHow would this change had we instead iterated through the \\textit{neighbors}, instead of \\textit{unvisited neighbors}? We will come back to this % todo\n\\end{rem}\n\nOkay, so how does one determine whether an algorithm is \\textit{good}? Well, it depends on how you define \\textit{good}. Typically we define \\textit{good} as having a low run-time. Unfortunately, an algorithm's run-time can change depending on the input. Remember: when writing algorithms, you cannot predetermine the input! This heeds way to the idea of \\textit{best-case}, \\textit{worst-case}, and \\textit{average-case} analysis.\n\n\\begin{defn}[Best-case Analysis]\n\tAnalyzing an algorithm based on an optimal input that minimizes its run-time.\n\\end{defn}\n\nBest-case analysis is akin to bounding an algorithm's run-time function below, which is just determining the tightest Big-Omega bound. So, \\(\\Omega\\) essentially tells you how fast the algorithm will run in the best case.\n\n\\begin{defn}[Worst-case Analysis]\n\tAnalyzing an algorithm based on a pathological input that maximizes its run-time.\n\\end{defn}\n\nBy contrast, worst-case analysis is akin to bounding an algorithm's run-time function above, which is just determining the tightest Big-Oh bound. So, \\(\\Oh\\) essentially tells you how fast the algorithm will run in the worst case.\n\n\\begin{defn}[Average-case Analysis]\n\tAnalyzing an algorithm's run-time based on typical, expected, input.\n\\end{defn}\n\n\\begin{rem}\n\tYou may be tempted to think that, as the analogy would imply, average-case analysis is akin to \\(\\Theta\\). This is not true. \\(\\Theta\\) tells you the \\textit{exact} run-time, which is when \\(\\Oh\\) and \\(\\Omega\\) agree. If you know your algorithm run-time has a \\(\\Theta\\) bound, then the average-case analysis will be this bound. The converse is not true though. If you have the average-case bound, this does not necessarily mean the worst-case or best-case is the same. An important example of this is the Quicksort algorithm.\n\\end{rem}\n\nWe present a pathological example to explain why these different analysis types are important.\n\n\\exsol{\n\tExamine the following algorithm:\n\t\\begin{algorithmic}[1]\n\t\t\\Function{Pathological}{\\(L\\), \\(n\\)\\@}\n\t\t\\If{\\(n \\equiv 0 \\Mod{2}\\)}\n\t\t\t\\State \n\t\t\\Else\n\t\t\t\\State \\Return \\(L[0]\\)\n\t\t\\EndIf\n\t\t\\State \\(c \\gets 0\\)\n\t\t\\For{\\(v \\in V\\)}\n\t\t\\For{\\(w \\in \\text{UnvisitedNeighbors}(v)\\)}\n\t\t\\State Visit \\(w\\)\n\t\t\\State \\(c \\gets c + 1\\)\n\t\t\\EndFor\n\t\t\\EndFor\n\t\t\\State \\Return \\(c\\)\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\t% todo function that takes exponential on if-branch, and constant on else\n}{\n\t\n}\n\n% maybe another that is less pathological\n\n\\subsection{Solving Summations}\n\n% todo\n\n% we did a similar thing during the number theory chapter.\n% i guess make it more explicit here? idk\n% we use the sum of c, i and i^2 here\n\n% recall addition is linear, so blah blah blah\n\n\\subsection{Solving Recurrences}\n\n% todo\n\n% tree table descending example\n\n\\begin{thm}[The Master Theorem]\n\t% todo\n\\end{thm}\n\nProof omitted.\n\n% another example, but use master thm\n\n\\section{Common Complexities}\n\nThere are a handful of common run-time complexities that you should be familiar with. In this book, we give them to you in ascending order of growth rate in Big Theta notation.\n\n\\begin{defn}[Constant]\n\t\\[\\Theta(1)\\]\n\\end{defn}\n\n\\begin{defn}[Logarithmic]\n\t\\[\\Theta(\\log n)\\]\n\\end{defn}\n\n\\begin{defn}[Linear]\n\t\\[\\Theta(n)\\]\n\\end{defn}\n\n\\begin{defn}[Polynomial]\n\t\\[\\Theta(n^c),\\ c > 1\\]\n\tWe call \\(\\Theta(n^2)\\) Quadratic\n\\end{defn}\n\n\\begin{defn}[Exponential]\n\t\\[\\Theta(c^n),\\ c > 1\\]\n\\end{defn}\n\n\\section{P vs NP}\n\nThere exist many YouTube videos discussing the famous P vs.\\ NP problem (\\href{https://www.youtube.com/watch?v=EHp4FPyajKQ}{ex 1}, \\href{https://www.youtube.com/watch?v=YX40hbAHx3s}{ex 2}). Moreover, if you solve it, the Clay Mathematics Institute has \\href{https://www.claymath.org/millennium-problems/p-vs-np-problem}{\\$1,000,000} waiting for you. So what is it?\n\n\\begin{defn}[P]\n\tThe class of computational problems that can be solved in \\textbf{P}olynomial time.\n\\end{defn}\n\n\\begin{defn}[NP]\n\tThe class of computational problems where a given solution can be \\textit{verified} in polynomial time.\n\\end{defn}\n\n\\begin{rem}\n\tNP stands for \\textit{Nondeterministic Polynomial}. It does \\textbf{not} stand for \\textit{non-polynomial} as some people think.\n\\end{rem}\n\n\n\n\\begin{defn}[Reductions]\n\t\n\\end{defn}\n\n\\begin{defn}[NP-Complete]\n\tA computational problem is NP-Complete if it satisfies two conditions:\n\t\\begin{enumerate}\n\t\t\\item The problem is in NP\n\t\t\\item Every problem in NP is reducible to it in polynomial time\n\t\\end{enumerate}\n\\end{defn}\n\n\\begin{rem}\n\tProblems that satisfy the second condition above, and not necessarily the first, are called \\textbf{NP-Hard}.\n\\end{rem}\n\nThe second condition is tricky.\nAs of now, the set of NP-Complete problems is empty.\nSo we need some form of \\textit{starter problem} to which all reductions can be made.\n\n\\begin{thm}[The Cook-Levin Theorem]\n\tThe Boolean satisfiability problem is NP-Complete.\n\\end{thm}\n\n\\begin{rem}\n\tWhile at this point in the course you likely can understand the proof, we omit it for scope.\n\\end{rem}\n\nWe leave this discussion with one final remark.\n\\(\\text{P}=\\text{NP}\\) will render most (all) cryptographic algorithms and services broken.\nThis is because most modern cryptography is built on the observation that factoring very large primes is \\textit{hard}.\nIf you are interested, look into RSA and/or the Diffie-Hellman key exchange.\n\n\\section{The Halting Problem}\n\nInfinite loops are a software developer's worst nightmare. They are hard to find, debug, and (sometimes) fix. If only there were a piece of software that could examine your code and tell you whether there are any infinite loops. More abstractly, if it could tell you whether your program continues forever or eventually stops.\n\nWell, we are sorry to break the news to you, but this piece of software can never exist; sort-of.\n\nThe halting problem is most commonly used as an introduction to computability theory. What problems can we compute? What does it even mean to be computable? These questions are out of scope of this book, but if the following problem interests you, consider taking a course on theory of computation.\n\n\\begin{defn}[The Halting Problem]\n\tA computational problem of determining whether an arbitrary computer program, with input, will stop or run forever.\n\\end{defn}\n\nThink to yourself whether you can come up with a solution program. It is a difficult problem. Alan Turing proved in 1936 that no such algorithm could exist \\textit{for all} pairs of programs and inputs. This proof led to (created) the Turing machine, and arguably all of modern computer science. The proof sketch follows. We call it a sketch because the actual proof is significantly more in-depth.\n\n\\begin{proof}\n\tAssume that such an algorithm (code description, function, method, whatever you want to call it) \\(h\\) exists. The function \\(h : P \\times I \\rightarrow \\mathbb{B}\\) given by \\(h : (p,i) \\mapsto b\\) tells us whether a given input program \\(p \\in P\\) (the set of all programs) with input \\(i \\in I\\) (the set of all program inputs) will halt (stop) or not: \\(b \\in \\mathbb{B} = \\{\\text{True}, \\text{False}\\}\\).\n\t\n\tThen the following procedure \\(\\textsc{g} \\in P\\):\n\t\\begin{algorithmic}[1]\n\t\t\\Function{g}{\\(\\cdot\\)\\@}\n\t\t\t\\If{\\(h(\\textsc{g},\\cdot)\\)}\n\t\t\t\t\\While{True}\n\t\t\t\t\\Comment{loop forever}\n\t\t\t\t\\EndWhile\n\t\t\t\\EndIf\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\t\n\tWe have two cases: \\(h(\\textsc{g},\\cdot)\\) returns True, or \\(h(\\textsc{g},\\cdot)\\) returns False.\n\t\n\tCase 1: \\(h(\\textsc{g},\\cdot) = \\text{True}\\). Then \\textsc{g} stops. But since \\(h\\) returned True, the body of the if-statement will be executed. Then \\textsc{g} will loop forever. But this contradicts the output of \\(h\\).\n\t\n\tCase 2: \\(h(\\textsc{g},\\cdot) = \\text{False}\\). Then \\textsc{g} does not stop. But since \\(h\\) returned False, the body of the if-statement is \\textit{not} executed, so \\textsc{g} will stop. But this contradicts the output of \\(h\\).\n\t\n\tThus, the halting problem is undecidable (impossible to answer for all inputs).\n\\end{proof}\n\n% todo\n\n\\section{Summary}\n\n\\begin{itemize}\n\t\\item \n\t\\item \n\t\\item \n\\end{itemize}\n\n\\section{Practice}\n\n\\begin{enumerate}\n\t\\item Analyze the following algorithm:\n\t\\begin{algorithmic}[1]\n\t\t\\Function{SomeFunc}{\\(n\\)\\@}\n\t\t\t\\State something\n\t\t\\EndFunction\n\t\\end{algorithmic}\n\t\\item Order the following complexities: \\(\\Theta(n \\log n)\\), \\(\\Theta(n \\log (n^2))\\), \\(\\Theta(n)\\), \\(\\Theta(n^2)\\). Explain why your ordering is correct.\n\t\\item Order the following complexities: \\(\\Theta(\\log (3n))\\), \\(\\Theta(n^n)\\), \\(\\Theta(n!)\\), \\(\\Theta(\\log (\\log n))\\).\n\\end{enumerate}\n\n%\\section{Solutions}\n%\n%\\begin{enumerate}\n%\t\\item \n%\t\\item Since \\(n \\log (n^2) = 2n \\log n\\) we have that \\(n \\log (n^2) \\in \\Theta(n \\log n)\\). So the growth-rate of those two functions are the same. It suffices to order \\(\\Theta(n \\log n)\\), \\(\\Theta(n)\\), \\(\\Theta(n^2)\\). We know \\(n \\in \\Oh(n^2)\\). It should be clear that \\(n \\in \\Oh(n \\log n)\\) (consider \\(1 < \\log_b n \\Rightarrow n < n \\log_b n\\) for all \\(n > b\\)). Similar logic can be applied to show that \\(n \\log n \\in \\Oh(n^2)\\). Thus our ordering is \\(n < n \\log n < n \\log (n^2) < n^2\\) where \\(a

\\max\\{a_1, \\ldots, a_e \\}$, the exponent of $X_e$ is simply the base $k$ expression with digits $a_e, a_1, \\ldots, a_{e-1}$. Now write $t$ as a linear combination of monomials $X_1^{a_1} \\cdots X_e^{a_e}$ and expand them in terms of $t_1, \\ldots, t_{e-1}, X_e$. From the observation above, different $(a_1, \\ldots, a_e)$ contributes a different exponent of $X_e$ whenever $k \\gg 0$. Adjusting $t$ by $\\Bbbk^\\times$, we get the asserted property.\n\\end{proof}\n\n\\begin{lemma}\\label{prop:normalization-aux}\n\tLet $\\mathfrak{a}$ be a nonzero ideal of a domain $R$, then $\\dim(R/\\mathfrak{a}) + 1 \\leq \\dim R$.\n\\end{lemma}\n\\begin{proof}\n\tAny prime chain $\\mathfrak{p}_0 \\supsetneq \\cdots \\supsetneq \\mathfrak{p}_n$ in $R$ with $\\mathfrak{p}_n \\supset \\mathfrak{a}$ can be extended to a prime chain length $n+1$, namely by adjoining $\\mathfrak{p}_{n+1} := \\{0\\}$.\n\\end{proof}\n\n\\begin{theorem}[E.\\ Noether, M.\\ Nagata]\\label{prop:Noether-normalization}\\index{Noether normalization}\n\tLet $B$ be a finitely generated $\\Bbbk$-algebra of dimension $n$. Consider a chain of proper ideals $I_1 \\subsetneq \\cdots \\subsetneq I_m$, with $\\dim(B/I_j) = d_j$ and $d_1 > \\cdots > d_m \\geq 0$. Then there exist a $\\Bbbk$-subalgebra $A \\subset B$ together with an isomorphism $A \\simeq \\Bbbk[X_1, \\ldots, X_n]$, satisfying\n\t\\begin{compactitem}\n\t\t\\item $B$ is a finitely generated $A$-module, in particular $B$ is integral over $A$;\n\t\t\\item $I_j \\cap A \\simeq (X_{d_j+1}, \\ldots, X_n)$ under the isomorphism above, for each $1 \\leq j \\leq m$.\n\t\\end{compactitem}\n\\end{theorem}\n\nNote that the assumption $I_j \\subsetneq I_{j+1}$ is merely for convenience. Allowing $I_j = I_{j+1}$ and $d_j = d_{j+1}$ for some $j$ is surely possible..\n\n\\begin{proof}\n\tWrite $B = \\Bbbk[Y_1, \\ldots, Y_r]/J$, where $r \\geq n$ (Corollary \\ref{prop:fg-dimension-bound}), and denote the preimage of $I_j$ in $\\Bbbk[Y_1, \\ldots, Y_r]$ by $\\tilde{I}_j$. The first step is to reduce to the case $J=\\{0\\}$. To see this, we adjoin $\\tilde{I}_0 = \\{0\\}$ (with $d_0 = n$) into the ideal chain; it may happen that $\\tilde{I}_0 = \\tilde{I}_1$, but that's harmless. Suppose we can find $\\tilde{A} \\subset \\Bbbk[Y_1, \\ldots, Y_r]$, $\\tilde{A} \\simeq \\Bbbk[X_1, \\ldots, X_r]$ with the required properties relative to $\\tilde{I}_\\bullet$. Taking quotient by $J$, we obtain the corresponding properties for $I_\\bullet$. Indeed, the passage from $\\tilde{A}$ to $A := \\tilde{A}/(\\tilde{A} \\cap J)$ truncates the variables $X_{n + 1}, \\ldots, X_r$, whereas\n\t\\[ I_j \\cap A = \\frac{\\tilde{I}_j \\cap (\\tilde{A} + J)}{J} = \\frac{(\\tilde{I}_j \\cap \\tilde{A}) + J}{J} \\simeq \\frac{\\tilde{I}_j \\cap \\tilde{A}}{J \\cap \\tilde{A}}; \\]\n\ttry to convince yourself of the middle equality.\n\n\tSecondly, having reduced to the case $B = \\Bbbk[Y_1, \\ldots, Y_r]$ (thus $r = n$), it suffices to pick $x_1, \\ldots, x_n \\in B$ such that\n\t\\begin{enumerate}[(a)]\n\t\t\\item $B$ is finitely generated as a module over $A := \\Bbbk[x_1, \\ldots, x_n]$, and\n\t\t\\item $I_j \\cap A \\supset (x_{d_j+1}, \\ldots, x_n)$ for all $j$.\n\t\\end{enumerate}\n\tIndeed, (a) implies $\\mathrm{Frac}(A)$ and $\\mathrm{Frac}(B)$ have the same transcendence degree $n$ over $\\Bbbk$, therefore $x_1, \\ldots, x_n$ must be algebraically independent over $\\Bbbk$. On the other hand, $\\dim(B/I_j) = \\dim(A/I_j \\cap A)$ by (a) and Proposition \\ref{prop:dim-integral}; but if $I_j \\cap A \\supsetneq (x_{d_j+1}, \\ldots, x_n)$ then $A/I_j \\cap A$ is a proper quotient of the domain $\\Bbbk[x_1, \\ldots, x_{d_j}]$, therefore would have dimension $< d_j$ by Lemma \\ref{prop:normalization-aux}.\n\t\n\tWe shall construct $x_1, \\ldots, x_n$ step by step. Suppose $0 \\leq e \\leq n$ and that we have produced elements $x'_1, \\ldots, x'_e$ and $x_{e+1}, \\ldots, x_n$ in $B$ satisfying\n\t\\begin{enumerate}[(i)]\n\t\t\\item $B$ is finitely generated as a module over $\\Bbbk[x'_1, \\ldots, x'_e, x_{e+1}, \\ldots, x_n] =: S_e$;\n\t\t\\item $I_j \\cap S_e \\supset (x_{e+1}, \\ldots, x_n)$ when $d_j \\leq e$;\n\t\t\\item $I_j \\cap S_e \\supset (x_{d_j + 1}, \\ldots, x_n)$ when $d_j \\geq e$ (with $j = 1, \\ldots, m$).\n\t\\end{enumerate}\n\n\tFor the initial case $e=n$, simply take $x'_i := Y_i$. Our aim is $e=0$. Let us explain the induction step from $e \\geq 1$ to $e-1$. The prior argument based on transcendence degrees implies the algebraic independence among\n\t\\[ x'_1, \\ldots, x'_e, x_{e+1}, \\ldots, x_n. \\]\n\tIf $e \\leq d_j$ for all $j$, we are done. Otherwise set $j := \\min\\{j': e > d_{j'} \\}$, we contend that\n\t\\[ I_j \\cap \\Bbbk[x'_1, \\ldots, x'_e] \\neq \\{0\\}. \\]\n\tIf not, we would have $I_j \\cap S_e = (x_{e+1}, \\ldots, x_n)$ since $I_j \\cap S_e \\supset (x_{e+1}, \\ldots, x_n)$ by (ii). We have $\\dim(S_e/I_j \\cap S_e) = \\dim(B/I_j) = d_j$ by integrality, whilst $\\dim(S_e/(x_{e+1}, \\ldots, x_n)) = e$. Contradiction.\n\t\n\tNow take $x_e \\in I_j \\cap \\Bbbk[x'_1, \\ldots, x'_e] \\smallsetminus \\{0\\}$. Note that $x_e \\notin \\Bbbk$ as $I_j \\neq B$. By Lemma \\ref{prop:normalization-Nagata}, we may choose $x''_1, \\ldots, x''_{e-1} \\in \\Bbbk[x'_1, \\ldots, x'_e]$ such that $\\Bbbk[x'_1, \\ldots, x'_e]$ is finitely generated over $\\Bbbk[x''_1, \\ldots, x''_{e-1}, x_e]$ as a module. It remains to verify that the new sequence\n\t\\[ x''_1, \\ldots, x''_{e-1}, x_e, \\ldots, x_n \\]\n\tsatisfies (i)---(iii) above with $e-1$ replacing $e$.\n\t\n\tFirst, $S_e$ is a finitely generated module over its subalgebra $\\Bbbk[x''_1, \\ldots, x_n]$ by construction, hence so is $B$ and (i) follows. Next, let $1 \\leq j' \\leq m$. If $d_{j'} > e-1$, the procedure above does not affect $x_{d_j + 1}, \\ldots, x_n$, so they belong to $I_{j'} \\cap \\Bbbk[x''_1, \\ldots, x_n]$. If $d_{j'} < e$, setting $j := \\min\\{j'' : e > d_{j''} \\}$ we have $j' \\geq j$ and\n\t\\[ I_{j'} \\cap \\Bbbk[x''_1, \\ldots, x_n] \\supset I_j \\cap \\Bbbk[x''_1, \\ldots, x_n] \\supset (x_{e+1}, \\ldots, x_n) + (x_e). \\]\n\tAll in all, we obtain (ii) and (iii).\n\\end{proof}\n\n\\begin{corollary}[Dimension formula]\\label{prop:fg-dim-formula}\\index{dimension formula}\n\tLet $B$ be a finitely generated algebra over a field $\\Bbbk$. Suppose $B$ is a domain, then for all $\\mathfrak{q} \\in \\Spec(B)$ we have\n\t\\[ \\dim(B/\\mathfrak{q}) + \\mathrm{ht}(\\mathfrak{q}) = \\dim B. \\]\n\\end{corollary}\n\\begin{proof}\n\tChoose a subalgebra $A$ of $B$ as in Theorem \\ref{prop:Noether-normalization} and put $\\mathfrak{p} = \\mathfrak{q} \\cap A$. Since $B$ is a domain and $A$ is normal, the Cohen--Seidenberg Theorem \\ref{prop:Cohen-Seidenberg} asserts the going-down property for $A \\hookrightarrow B$. Proposition \\ref{prop:dim-integral} implies that $\\dim A = \\dim B$, $\\dim A/\\mathfrak{p} = \\dim B/\\mathfrak{q}$ and $\\text{ht}(\\mathfrak{p}) = \\text{ht}(\\mathfrak{q})$, so we are again reduced to the case $B = \\Bbbk[X_1, \\ldots, X_n]$ and $\\mathfrak{q} = (X_{d+1}, \\ldots, X_n)$. This is known by Corollary \\ref{prop:poly-alg-ht}.\n\\end{proof}\n\nThe dimension formula allows us to compute $\\dim B$ by choosing any prime $\\mathfrak{q}$, a prime chain of longest length below $\\mathfrak{q}$ (i.e. in $B_{\\mathfrak{q}}$) and another one above $\\mathfrak{q}$ (i.e. in $B/\\mathfrak{q}$); their concatenation will then be a prime chain in $B$ with maximal length $\\dim B$. By applying the dimension formula to $\\mathfrak{q}, \\mathfrak{q}' \\subset B$ and to $\\mathfrak{q}'/\\mathfrak{q} \\subset B/\\mathfrak{q}$, it follows that $\\text{ht}(\\mathfrak{q}') = \\text{ht}(\\mathfrak{q}) + \\text{ht}(\\mathfrak{q}'/\\mathfrak{q})$ for all $\\mathfrak{q}' \\supsetneq \\mathfrak{q}$ in $B$.\n\nFor a general Noetherian domain $B$, we call a prime chain \\emph{maximal} if it is not properly contained in any prime chain. A priori, a maximal prime chain does not necessarily have length equal to $\\dim B$. If\n\\[ \\forall \\mathfrak{q}' \\supset \\mathfrak{q}: \\text{primes}, \\quad \\text{ht}(\\mathfrak{q}') = \\text{ht}(\\mathfrak{q}) + \\text{ht}(\\mathfrak{q}'/\\mathfrak{q}), \\]\nwe say $B$ is a \\emph{catenary} domain\\index{catenary}. This means that $\\text{ht}(\\mathfrak{q}'/\\mathfrak{q}) = \\dim B_{\\mathfrak{q}'}/\\mathfrak{q}B_{\\mathfrak{q}'}$ is the common length of all maximal prime chains between $\\mathfrak{q}'$ and $\\mathfrak{q}$. In particular, maximal prime chains in $B_{\\mathfrak{q}'}$ are automatically longest. As shown above, finitely generated domains over a field are catenary. For a finer analysis of catenary and universally catenary rings, we refer to \\cite[\\S 14]{Mat80}.\n\n\\begin{corollary}\n\tSuppose $B$ is a domain finitely generated over a field $\\Bbbk$. Set $L := \\mathrm{Frac}(B)$. Then $\\dim B = \\mathrm{tr.deg}_\\Bbbk(L)$ and it is the common length of maximal prime chains.\n\\end{corollary}\n\\begin{proof}\n\tChoose a subalgebra $A$ of $B$ as in Theorem \\ref{prop:Noether-normalization}. Since the field $\\text{Frac}(B)$ is a finite extension of $\\text{Frac}(A)$ and $\\dim A = \\dim B$ by Proposition \\ref{prop:dim-integral}, the first assertion reduces immediately to the case $B = \\Bbbk[X_1, \\ldots, X_n]$, which is obvious. As to the second assertion, consider a maximal prime chain $\\mathfrak{q}_0 \\supsetneq \\cdots \\supsetneq \\mathfrak{q}_n = \\{0\\}$ in $B$. Maximality implies $\\text{ht}(\\mathfrak{q}_i/\\mathfrak{q}_{i+1}) = 1$ for all $i$ and $\\dim B/\\mathfrak{q}_0 = 0$. Applying Corollary \\ref{prop:fg-dim-formula} repeatedly, we see\n\t\\begin{multline*}\n\t\t\\dim B = \\dim B/\\mathfrak{q}_n = \\dim B/\\mathfrak{q}_{n-1} + \\text{ht}(\\mathfrak{q}_{n-1}/\\mathfrak{q}_n) = \\\\\n\t\t\\dim B/\\mathfrak{q}_{n-2} + \\text{ht}(\\mathfrak{q}_{n-1}/\\mathfrak{q}_{n-2}) + \\text{ht}(\\mathfrak{q}_{n-1}/\\mathfrak{q}_n) = \\cdots = \\dim B/\\mathfrak{q}_0 + \\sum_{i=0}^{n-1} \\text{ht}(\\mathfrak{q}_i/\\mathfrak{q}_{i+1})\n\t\\end{multline*}\n\twhich equals $n$.\n\\end{proof}\n\n\\begin{remark}\n\tRecall that finitely generated domains over an algebraically closed field $\\Bbbk$ are objects ``opposite'' to the irreducible affine $\\Bbbk$-varieties. It is instructive to make a comparison with the analytic theory when $\\Bbbk=\\CC$. Let $\\mathcal{X}$ be a compact connected complex manifold of (complex) dimension $n$. Denote by $\\mathcal{M}(\\mathcal{X})$ the field of meromorphic functions on $\\mathcal{X}$. Siegel proved that $\\text{tr.deg}_{\\CC}(\\mathcal{M}(\\mathcal{X})) \\leq n$. When $\\mathcal{X}$ is a projective algebraic $\\CC$-variety, equality holds and we have $\\mathcal{M}(\\mathcal{X}) = \\text{Frac}(A)$ if $\\Spec(A)$ is any open dense affine subscheme in $\\mathcal{X}$. In general, the abundance of meromorphic functions on $\\mathcal{X}$ is a subtle issue, cf. the case of Riemann surfaces ($n=1$). Compact connected complex manifolds with $\\text{tr.deg}_{\\CC}(\\mathcal{M}(\\mathcal{X})) = \\dim_{\\CC} \\mathcal{X}$ are called \\emph{Moishezon manifolds}.\\index{Moishezon manifolds}\n\t\n\tNon-algebraic Moishezon manifolds do exist, and Moishezon proved that a Moishezon manifold is projective, hence algebraic, if and only if it is Kähler. Following M.\\ Artin and D.\\ Knutson, one can enlarge the category of $\\CC$-schemes into that of \\emph{algebraic spaces} over $\\CC$, and there is an analytification functor $\\mathcal{X} \\mapsto \\mathcal{X}_{\\text{an}}$ that sends algebraic spaces of finite type over $\\CC$ to complex analytic varieties. M. Artin \\cite[\\S 7]{Ar70} showed that the analytification establishes an equivalence between the category of smooth proper algebraic spaces of finite type over $\\CC$ and that of Moishezon manifolds. Therefore, such manifolds still retain an algebraic flavor: they are quotients of certain $\\CC$-schemes by étale equivalence relations.\n\\end{remark}\n", "meta": {"hexsha": "594e2385d66fa9b18eb89edc08fcd2c144e5f31a", "size": 26676, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "YAlg3-6.tex", "max_stars_repo_name": "wenweili/Yanqi-Algebra-3", "max_stars_repo_head_hexsha": "4223e9973c97342ecb09b444b9fc3c30ffd53aa3", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 26, "max_stars_repo_stars_event_min_datetime": "2019-07-09T06:22:14.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-05T14:44:14.000Z", "max_issues_repo_path": "YAlg3-6.tex", "max_issues_repo_name": "wenweili/Yanqi-Algebra-3", "max_issues_repo_head_hexsha": "4223e9973c97342ecb09b444b9fc3c30ffd53aa3", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "YAlg3-6.tex", "max_forks_repo_name": "wenweili/Yanqi-Algebra-3", "max_forks_repo_head_hexsha": "4223e9973c97342ecb09b444b9fc3c30ffd53aa3", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-07-10T23:47:58.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-21T03:32:08.000Z", "avg_line_length": 116.4890829694, "max_line_length": 992, "alphanum_fraction": 0.6703778677, "num_tokens": 10247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085808877581, "lm_q2_score": 0.7853085758631158, "lm_q1q2_score": 0.6167095632700498}} {"text": "\\chapter{Learning coupled dictionaries}\n\\label{chap_dictionarylearning} \nThis chapter describes an alternative approach to learn the relation between large and small scales of turbulence based on \\textit{dictionary learning} methods. The approach finds coupled dictionaries to represent the low and high resolution turbulent fields and to reconstruct the missing small-scale information. This method has been applied successfully in image processing, and remains a very active research subject. It potentially outperforms regression models, which are sometimes too simplistic to represent the underlying phenomenon in turbulence. Indeed learned dictionaries could encode part of the physics of the flow. \n\nThis chapter is organized as follows. First, different data representations are discussed, from predefined bases such as Fourier or wavelets to learned bases such as principal component analysis (PCA). Second, dictionary learning, a generalization of PCA, is discussed for the first time as a representation for turbulent fields. This section reviews also two algorithms to learn the dictionaries. Third, different approaches to learn coupled dictionaries of low and high resolution fields are presented. Last, the approach is tested on the DNS database of an isotropic turbulence presented in section \\ref{sec:data_isotropic}. \n\n\\section{From bases to dictionaries}\nThis section briefly reviews conventional representations of turbulent signals. Given a vector $ \\x_t $, the main idea of all representation methods is to find a ``\\textit{dictionary}'' $ \\dict $, a set of basis functions or the so-called ``\\textit{atoms}'' from which $ \\x_t $ can be represented via a linear combination. This representation by coefficients $ \\adictco_t $ is called a ``\\textit{code}''. The representation can be exact:\n\\begin{equation}\n\\x_t = \\dict \\adictco_t\n\\end{equation}\nor approximate:\n\\begin{equation}\n \\x_t = \\dict \\adictco_t +\\n_t\n\\end{equation}\nwhere $ \\n_t $ is an estimation noise term. Seeking the representation of the signal is an optimization problem to estimate the dictionary $ \\dict $ and the code $ \\adictco_t $. The dictionary can be mathematically predefined or learned from the data. The estimation of the coefficients $ \\adictco_t $ can be very simple with very fast algorithms for predefined or orthogonal bases, or more complex with redundant dictionaries. \n\n\\subsection{Predefined dictionaries}\nVarious types of dictionaries are studied to represent the data. They can be predefined \\textit{a priori} such as the well-known Fourier transform, the group of \\textit{wavelets} or \\textit{curvelets} \\citep{mallat1989theory,mallat1999wavelet}. These predefined transforms are easy to compute thanks to fast algorithms. Fourier transform aims at describing the signal via its frequency content by decomposing signals as an infinite series of sines and cosines. This transformation is localized in frequency but not in space/time. The constraint is overcome using wavelet transforms, replacing the sines and cosines by wavelet functions localized both in space/time and frequency. This property permits wavelets to better represent the signal with discontinuities or sharp spikes. In such cases, the representations are much more compact with wavelet functions than sines and cosines. This makes wavelets standard in signal compression, for example JPEG 2000 for images, while Fourier transform is mainly used for spectral analysis.\n\nRepresentations using predefined bases suffer from some limitations. Fourier transform is limited to spectral analysis due to its localization in frequency. Wavelet transforms overcome this constraint by its carefully designed functions, which permit the localization both in space/time and frequency domain. The aim is to optimize the ``\\textit{sparsity}'' of the representation, which is to minimize the number of basis functions to accurately represent the data. However, despite the large family of wavelets, all kinds of signals cannot be represented with high level of sparsity. Analogously for turbulence data, different flows or a single flow at various positions with respect to a wall can have significantly different physics, leading to signals with completely different properties.\n\n\\subsection{Adaptive dictionaries}\nTo further optimizing the representation, it is more appealing to learn dictionaries adaptively from the data \\citep{bengio2013representation}. The first and most common method is the \\textit{principal component analysis} (PCA). This approach learns a basis that maximizes the variance of the projected values. This is interesting for turbulence studies, since large scales of higher variances are of interest. This representation imposes orthogonal bases, which implies that the number of atoms is limited by the dimension of the input vectors. PCA is potentially subject to certain limitations due to this constraint. This chapter discusses the generalized version of PCA, which permits to learn a ``\\textit{redundant}'' dictionary. The method is called \\textit{dictionary learning}, which is widely used and remains an active research topic in image processing. It will be discussed further in this section before applying to the problem of turbulent field reconstruction. \n\nAdaptive dictionaries are useful in many applications \\citep{tovsic2011dictionary}. The first one is \\textit{dimension reduction} \\citep{burges2010dimension}, which is useful to build models or visualize large amount of data. For modeling, high dimensional input variables potentially causes the problem of high variance as discussed in chapter \\ref{chap_linearregression}, or intractable computation complexity. For visualization, it is a real constraint since the maximum number of dimensions one can observe and analyze is probably no more than three. The core idea is to project data onto a set of atoms in a dictionary. This projection is lossy, meaning that a certain amount of less informative variance is discarded. Another important application is to solve inverse problems such as super-resolution or denoising. These problems are very ill-posed and can not be solved directly by least-squares methods. A good representation of the data can play the role as a regularization term, making the problem easier to solve. The choice of a dictionary defines the space in which we search for solutions.\n\n\\section{Proper Orthogonal Decomposition (POD) as a representation of turbulent fields}\nPCA \\citep{jolliffe2002principal}, known also as \\textit{Karhunen–Loève} transform, is commonly used in many fields of signal and image processing. It was first introduced in turbulence studies by \\citet{lumley1967} under the name ``\\textit{proper orthogonal decomposition}'' (POD). It is then used as a standard approach for dimensionality reduction, which is largely beneficial in large scales reconstruction, flow control and coherent structure studies. PCA decomposes a sequence of snapshots into a dictionary $ \\dict $ to represent spatial structures, and a coefficient matrix $ \\dictco $, to capture temporal dynamics. The works on coherent structures and flow information extraction focus on the dictionary $ \\dict $ \\citep{bonnet1994stochastic, gordeyev2000coherent}. Each atom $ \\adict_i $, as ranked by its variance (an equivalent measure of the kinetic energy), represents the most energetic structures of the flow. Flow control and modeling works focus more on the use of projection coefficients $ \\dictco $. After learning the fixed dictionary $ \\dict $, independent of time, from given training fields, the temporal dynamics of the flow is presented in the projection coefficients $ \\dictco $ only. The efforts to model and control the flow are reduced tremendously by considering only several coefficients of high-variance atoms. This property is beneficial for reduced-order modeling works in flow control and reconstruction of large-scale velocity fields \\citep{ravindran2000reduced, ly2001modeling, taylor2004towards}.\n\nTo derive POD or PCA, let first denote a data matrix $ \\X $ of size $ \\dimsh \\times \\dimtl $ containing a sequence of $ \\dimsh- $dimensional input vectors $ \\x_t, t=1,2,...,\\dimtl $,\n\\begin{equation}\n\\X \\mydef \\begin{bmatrix} \\x_1, \\x_2, ..., \\x_\\dimtl \\end{bmatrix},\n\\end{equation} \nassumed centered \\textit{a priori}:\n\\begin{equation}\n\\sum\\limits_{t=1}^{\\dimtl}\\x_t = \\mybold{0}\n\\end{equation}\nFrom $ \\X $, a dictionary $ \\dict $ of size $ \\dimsh \\times \\dimsh $ is estimated, which contains $ \\dimsh $ atoms $ \\adict_i \\in \\R ^\\dimsh, i=1,2,...,\\dimsh $. The objective is to project $ \\x_t $ onto a reduced-order subspace of dimension $ \\dimtrunc \\leq \\dimsh $ while maximizing the amount of variance. This idea, when applying to turbulent fields where variance represents the kinetic energy, is interpreted as determining the most energetic structures among a sequence of snapshots. \n\nThe first step is to find the most dominant basis function $ \\adict_1 \\in \\R^\\dimsh$, assuming $ \\normtwo{\\adict_1} = 1 $. Each input vector is projected onto this direction as $ \\adict_1^\\mytrans \\x_t $. The variance of this projection is:\n\\begin{equation}\ns_1 = \\frac{1}{\\dimtl} \\sum\\limits_{t=1}^{\\dimtl}\\left( \\adict_1^\\mytrans \\x_t\\right)^2 = \\adict_1^\\mytrans \\Sigma \\adict_1\n\\end{equation}\nwhere $ \\Sigma $ is the covariance matrix:\n\\begin{equation}\n\\Sigma = \\frac{1}{\\dimtl} X X^\\mytrans\n\\end{equation}\nThe aim now is to find $ \\adict_1 $ such that $ s_1 $ is maximum, with the constraint that $ \\normtwo{\\adict_1} = 1$:\n\\begin{equation}\n\\adict_1 = \\argmax_{\\adict_1}{ \\left\\lbrace \\adict_1^\\mytrans \\Sigma \\adict_1 \\right\\rbrace } \\:\\:\\:\\:\\:\\: s.t \\:\\:\\:\\:\\:\\: \\normtwo{\\adict_1} = 1\n\\end{equation}\nThis optimization problem can be rewritten in an unconstrained manner by introducing a Lagrange multiplier $ \\lambda_1 $:\n\\begin{equation}\n\\adict_1 = \\argmax_{\\adict_1}{ \\left\\lbrace \\adict_1^\\mytrans \\Sigma \\adict_1 + \\lambda_1 \\left(1-\\adict_1^\\mytrans \\adict_1\\right) \\right\\rbrace } \n\\end{equation}\nSetting the derivative with respect to $ \\adict_1 $ to zero, one obtains:\n\\begin{equation}\n\\Sigma \\adict_1 = \\lambda_1 \\adict_1\n\\end{equation}\nwhich implies that $ \\adict_1 $ is an eigenvector of $ \\Sigma $, and $ \\lambda_1 $ is the equivalent eigenvalue. Also, multiplying both side with $ \\adict_1^\\mytrans $, using $ \\adict_1^\\mytrans\\adict_1=1 $, one has:\n\\begin{equation}\n\\adict_1^\\mytrans \\Sigma \\adict_1 = \\lambda_1\n\\end{equation}\nThe above expression implies that $ \\lambda_1 $ is the variance of the projection onto $ \\adict_1 $. This variance is maximized by selecting the maximum eigenvalue of the covariance matrix $ \\Sigma $. This process is repeated to find the second basis of the dictionary $ \\adict_2 $ such that it is orthogonal to the first one, i.e. $ \\adict_1^{\\mytrans}\\adict_2=0 $. This is identical to find the second eigenvector corresponding to the second largest eigenvalue $ \\lambda_2 $. The procedure of seeking all $ \\dimsh $ bases is identical to eigenvalue decomposition and SVD of the covariance matrix $ \\Sigma $. Sorting $ \\dimsh $ eigenvalues in a descending order as $ \\lambda_1 \\geq \\lambda_2 \\geq ... \\geq \\lambda_\\dimsh \\geq 0 $, one obtains the dictionary, or codebook, learned from the training samples $ \\X $ by putting the equivalent eigenvectors together as $ \\dict \\mydef \\left[ \\adict_1, \\adict_2, ..., \\adict_\\dimsh \\right] $. The projection matrix $ \\dictco $ is found such that:\n\\begin{equation}\n\\X = \\dict \\dictco\n\\end{equation}\n\n\n\\section{Dictionary learning as a new representation}\nDue to the orthogonality and variance maximization properties, PCA gains its success in many problems such as dimensionality reduction, low-order modeling and lossy compression. However, in solving different inverse problems, PCA is subject to several limitations. The number of atoms is at most the dimension of input vectors, leading to the ``limited expressiveness'' property \\citep{tovsic2011dictionary}. The representation learned by PCA can be efficient for training data, but a good generalization is not guaranteed. It is desirable to ignore the constraint on the number of basis functions and learn a redundant dictionary. Due to the redundancy, the sparsity constraint can be imposed and play the role of the regularization to solve ill-posed inverse problems.\n\n\\subsection{Redundant dictionary and sparse representation}\nDictionary learning (DL) ignores the constraint on the number of basis functions, or atoms, permitting to find a \\textit{redundant} (or \\textit{overcomplete}) dictionary $ \\dict \\in \\R^{\\dimsh \\times \\dimdict} $ to represent the data. $ K $ is the number of atoms, and redundancy means that $ \\dimdict $ is potentially larger than $ \\dimsh $. These vectors are therefore not necessarily orthogonal. The companion of redundancy is sparsity, where the linear transformation matrix $ \\dictco $ contains only a few nonzero coefficients. The representation then relies on the duality between redundancy and sparsity, which will be discussed further in this section. With more atoms, the representation using dictionary learning is expected to be more adaptive to the signal and to better represent new data. This is the reason why the approach often gives the state-of-the-art results in most inverse problems in image processing \\citep{elad2010on,yang2010image}.\n\nThe problem of representing data $ \\X $ using the redundant dictionary $ \\dict $ and sparse coefficients $ \\dictco $ includes two alternating optimization problems. The first one is to estimate the projection coefficients $ \\dictco $. PCA, with the orthogonality among atoms, simply estimates $ \\dictco $ by dot products. Dictionaries, which are not necessarily orthogonal, can have more atoms than the dimension ($ \\dimdict > \\dimsh $). There are potentially many matrices $ \\dictco $ such that $ \\X = \\dict \\dictco $. The sparsity constraint, meaning that $ \\dictco $ has a minimal number of nonzero coefficients, is imposed to make the solution unique. The representation of $ \\X $ becomes approximate, i.e. $ \\X \\approx \\dict \\dictco $. Finding $ \\dictco $ becomes an optimization problem of the form:\n\\begin{equation}\n\\dictco = \\argmin_{\\dictco}{\\normp{\\dictco}} \\quad s.t. \\quad \\X = \\dict \\dictco\n\\end{equation}\nor re-arranged in the form of a regularized cost function as in chapter \\ref{chap_linearregression}:\n\\begin{equation}\n\\dictco = \\argmin_{\\dictco}{\\normtwo{\\X - \\dict \\dictco} + \\lambda \\normp{\\dictco}}\n\\end{equation}\nThe common \\textit{$ \\ell^p $ norm} $ \\normp{.} $ is with $ 0 \\leq p \\leq 1 $. Solving the problem with the $ \\ell^0 $ norm, i.e. counting the number of nonzero coefficients, leads to orthogonal matching pursuit (OMP) \\citep{tropp2007signal}, while solving with the $ \\ell^1 $ norm, summing the absolute values of the coefficients, leads to LASSO as discussed in chapter \\ref{chap_linearregression}. Least Angle Regression (LARS) \\citep{efron2004least} is sometimes used as another efficient algorithm to solve $ \\ell^1 $ penalty problems and gives results very similar to LASSO. These regularizations help selecting a limited number of atoms that best approximate the input $ \\X $. The second problem is the choice of the dictionary, which is solved by efficient algorithms discussed in the next section.\n\n\\subsection{Dictionary learning methods}\nMany algorithms have been proposed to learn the dictionary from data, starting with gradient descend method \\citep{olshausen1996emergence}, then K-SVD \\citep{aharon2006ksvd}, feature-sign approach \\citep{lee2006efficient} and online dictionary learning \\citep{mairal2010online}. This section reviews some of those that are used later in this chapter. \n\n\\subsubsection{Alternate optimization}\nDictionary learning is an alternate optimization problem to search the redundant dictionary $ \\dict $ and the sparse matrix $ \\dictco $. The approach has started by using a gradient descent approach \\citep{olshausen1996emergence}. It has recently gained popularity thanks to the progress in solving regularized optimization problems with $ \\ell^p $ penalty \\citep{tibshirani1996regression, donoho2003optimally, efron2004least, donoho2012sparse}. One of the first efficient dictionary learning approach is the \\textit{method of optimal directions} (MOP) \\citep{engan1999method} where $ \\dict $ and $ \\dictco $ are estimated by solving:\n\\begin{equation}\n(\\dict, \\dictco) = \\argmin_{\\dict, \\dictco} \\normtwo{\\X - \\dict \\dictco} \\subjectto \\normzero{\\adictco_t} < \\dimtrunc \\quad \\forall t\n\\end{equation}\n$ \\dimtrunc $ is the sparsity constraint- the maximum number of nonzero coefficients. This optimization problem is combinatorial and highly non-convex. The search for a local minimum is done by alternating two steps. With a fixed dictionary $ \\dict $, the \\textit{sparse coding} step finds the representation of the input vectors by solving:\n\\begin{equation}\n\\dictco = \\argmin_{\\dictco} \\left\\lbrace \\normtwo{\\X - \\dict\\dictco} + \\lambda \\normp{\\dictco}\\right\\rbrace\n\\end{equation}\nThis step uses OMP for $ \\ell^0 $ penalty and LASSO or LARS for $ \\ell^1 $ penalty. The \\textit{dictionary update} step re-estimates $ \\dict $ with fixed $ \\dictco $ via the Moore-Penrose pseudo-inverse:\n\\begin{equation}\n\\dict = \\X \\dictco^\\mypseudo = \\X \\dictco^\\mytrans (\\dictco \\dictco^\\mytrans)^{-1}\n\\end{equation}\nThis scheme converges rapidly after some iterations, but requires complex matrix inversion, which is not efficient in most cases \\citep{aharon2006ksvd}. \n\\subsubsection{K-SVD}\nThe \\textit{K-SVD} algorithm was proposed later by \\citet{aharon2006ksvd} and rapidly gained its popularity. The \\textit{dictionary update} is done by a block-relaxation approach. Instead of inversing the matrix, the algorithm updates each atom in an efficient way by generalizing the \\textit{k-means} clustering method \\citep{bishop2006pattern}. The $ k- $th atom is estimated by minimizing a quadratic error:\n\\begin{equation}\n\t\\left\\lbrace \\adict_k, \\adictco_k \\right\\rbrace = \\argmin_{\\adict_k, \\adictco_k} \\normtwo{E_k \\: - \\: \\adict_k \\adictco_k^\\mytrans}\n\\end{equation} \nwhere $ E_k $ is defined as the residual matrix:\n\\begin{equation}\nE_k \\mydef \\X_k \\: - \\: \\sum_{j\\neq k} \\left(\\adict_j \\adictco_j^\\mytrans \\: - \\: \\adict_k \\adictco_k^\\mytrans\\right)\n\\end{equation}\nwhere $ \\X_k $ gathers examples $ \\x_t $ using $ \\adict_k $ in their representation. The update of both $ \\adict_k $ and $ \\adictco_k $ is done simultaneously via a rank-1 approximation, i.e. considering only the first eigenvector of $ E_k $ after performing a SVD.\n\n\\subsubsection{Online dictionary learning (ODL)}\nK-SVD algorithm has certain constraints, mainly with a potential local minimum while iteratively learning the dictionary \\citep{rubinstein2010dictionaries}. The algorithm is also limited to small numbers of samples due to its computational cost. Online dictionary learning (ODL) has been proposed by \\citet{mairal2010online} to learn $ \\dict $ from massive datasets. The use of stochastic gradient descent to update the dictionary for each example (or a group of them), and LARS algorithm for sparse coding, permits the learning over millions of samples. Algorithm \\ref{algo_ODL} summarizes this approach.\n\n\\begin{algorithm}[t]\n\\caption{Online dictionary learning algorithm by \\citet{mairal2010online}} \\label{algo_ODL}\n\\begin{algorithmic}[1]\n\t\\State Input: \n\t\\begin{itemize}\n\t\t\\item a set of $ \\mathcal{S}(\\x) $ of $\\dimtl $ input variable $ \\x_t \\in \\R^\\dimsh$;\n\t\t\\item $\\lambda \\in \\R $ : regularization parameter ($ \\lambda \\geq 0 $)\n\t\t\\item $ \\dict_0 \\in \\R^{\\dimsh \\times K}$: initial dictionary\n\t\\end{itemize}\n\t\\State $ \\mathbf{G}_0 \\gets 0$, $ \\mathbf{H}_0 \\gets 0 $\n\t\\For {$ t = 1, 2, ..., \\dimtl$}\n\t\t\\State Draw $ \\x_t $ from $\\mathcal{S}(\\x)$\n\t\t\\State Sparse coding:\n\t\t\\begin{equation}\n\t\t\t\\adictco_t = \\argmin_{\\adictco \\in \\R^K} \\left\\lbrace \\x_t - \\dict_{t-1}\\adictco +\\lambda \\normone{\\adictco}\\right\\rbrace\n\t\t\\end{equation}\n\t\t\\State $ \\mathbf{G}_t \\gets \\mathbf{G}_{t-1}+\\adictco_t \\adictco_t^\\mytrans$\n\t\t\\State $ \\mathbf{H}_t \\gets \\mathbf{H}_{t-1}+\\adictco_t \\x_t^\\mytrans$\n\t\t\\State Update $ \\dict_t $:\n\t\t\\begin{equation}\n\t\t\t\\begin{split}\n\t\t\t\\dict_t = & \\argmin_{\\dict \\in \\mathcal{C}} \\left\\lbrace \\frac{1}{t} \\sum\\limits_{i=1}^{t}\\frac{1}{2} \\normtwo{\\x_i - \\dict\\adictco_i} + \\lambda \\normone{\\adictco_i} \\right\\rbrace \\\\\n\t\t\t=& \\argmin_{\\dict \\in \\mathcal{C}} \\left\\lbrace \\frac{1}{t} \\sum\\limits_{i=1}^{t}\\frac{1}{2} \\mytrace{\\dict^\\mytrans\\dict \\mathbf{G}_t} - \\mytrace{\\dict^\\mytrans \\mathbf{H}_t} \\right\\rbrace\n\t\t\t\\end{split}\n\t\t\\end{equation}\n\t\\EndFor\n\t\\State Return $ \\dict_\\dimtl $\n\\end{algorithmic}\n\\end{algorithm}\n\n\\section{Learning coupled dictionaries}\n\\label{joint_learning}\nIn this section, DL is used to solve the problem of reconstructing HR fields from LR ones. The approach is inspired by the \\textit{single image super-resolution} application \\citep{yang2010image,yang2012coupled,zeyde2012single}, i.e. estimating HR images with finer details from LR ones. The approach jointly learns coupled dictionaries at LR and HR from given training samples. These dictionaries are forced to represent the data using the same coefficients. In the prediction phase, the projection coefficients are estimated from the LR images and then combined with the learned HR dictionary to reconstruct the HR images. \n\n\\subsection{Patch-based approach}\n\\label{subsec:patch_based_estimation}\n\\begin{figure}\n\\centering\n\t\\includegraphics[width=0.9\\textwidth]{./images/DL/ScSR.png}\n\t\\caption{\\label{fig:ScSR} An example of single image super-resolution using coupled dictionary learning and patch-wise approach. Photo credit: \\citet{yang2010image}.}\n\\end{figure}\n\nDL requires a sparse coding step, i.e. solving the $ l^1 $-penalty optimization problem. The computation complexity is $ \\mathcal{O}(\\dimsh^3+\\dimtl\\dimsh^2) $ \\citep{efron2004least}, where $ \\dimsh $ is the dimension of the fields and $ \\dimtl $ is the number of samples. Since $ \\dimsh \\gg \\dimtl$ in most cases, the complexity is $ \\mathcal{O}(\\dimsh^3) $, which is a very heavy procedure. The computation is therefore intractable with the whole field. To overcome this difficulty, the ``\\textit{patch-wise}'' approach uses small patches instead of the whole fields. These patches are extracted from the original fields at all positions by moving pixel-by-pixel in both horizontal and vertical directions. Let $ \\dimptl $ be the total number of patches, and $ \\dimpsh $ be their dimension ($\\dimptl \\gg \\dimpsh $), the computation complexity becomes $ \\mathcal{O}(\\dimptl\\dimpsh^2) $, which is tractable when $ \\dimpsh $ is small. In the reconstruction phase, since one pixel belongs to several intersecting patches, the estimated field is reconstructed by aggregating all possible estimates of each pixel after putting them back into the whole scene.\n\nThe path-wise approach comes with several advantages. First, it localizes the information. A shared dictionary can be learned to represent all type of images with different contents. Second, it reduces tremendously the computation. Standard images are of dimension few millions, while typical patch sizes are around $ 8 \\times 8 $, which is a few orders of magnitude smaller. Last, since the learning is on small patches, the number of samples required for the algorithm to converge is much smaller. From a single image, millions of small patches can be extracted for learning. Working with the original image, the number of training samples is necessary at least one order of magnitude higher than its dimension, which is already millions of pixels. \n\nTo formulate the patch-based approach, let $ \\z_t \\in \\R^{\\dimsh} $ be a snapshot of a high-resolution field as a column vector. The operator $ \\extracthigh{k}:\\R^{\\dimsh} \\mapsto \\R^{\\dimpsh} $ is to extract the 2D patch and put them in a lexicographical order to form a column vector $ \\patchhigh{k} \\mydef \\extracthigh{k} \\z_t \\in \\R^\\dimpsh $, where $ \\dimpsh $ is the size of 2D patches $ \\patchhigh{k} $. If the overlapping is such that all patches are translated pixel-by-pixel, the total number of patches is $ \\left( \\sqrt{\\dimsh} - \\sqrt{\\dimpsh} + 1 \\right)^2 $. The reconstruction of the whole image is done as:\n\\begin{equation}\n\\hat{\\z}_t= \\left[ \\sum\\limits_{k} \\left(\\extracthigh{k}\\right)^{\\mytrans}\\extracthigh{k}\\right]^{-1}\\sum\\limits_{k} \\left(\\extracthigh{k}\\right)^{\\mytrans}\\hat{\\mybold{p}}_h^k\n\\end{equation}\nwhere $ \\hat{\\mybold{p}}_h^k $ and $ \\hat{\\z}_t $ are the estimates of the reference $ \\patchhigh{k} $ and $ \\z_t $ respectively. The term $ \\left(\\extracthigh{k}\\right)^{\\mytrans}: \\R^{n} \\mapsto \\R^{\\dimsh}$ puts the equivalent patch into the global 2D scene and zero-pad elsewhere. The term $ \\left(\\extracthigh{k}\\right)^{\\mytrans}\\extracthigh{k} $ just counts the number of estimates for each pixels. Its inversion plays the role of a normalization factor. In practice, the whole process is done by estimating each pixel as its mean or median of all estimates from all patches that it belongs to. \n\n\\subsection{The approach}\n\\label{sec:chap3_theapproach}\nLet assume that small patches from the field can be represented as a linear combination of several atoms from the learned dictionary. This assumption is the so-called ``\\textit{Sparse-Land}'' prior \\citep{zeyde2012single}. The idea is to learn coupled dictionaries by imposing that the representation coefficients are the same at low and high resolution. The following section recalls the main ideas. More details can be found in \\citet{zeyde2012single}.\n\nSuppose that $ \\z_t \\in \\R^{\\dimsh}$ is the true high resolution velocity field. The corresponding low-resolution field $ \\y_t \\in \\R^{\\dimsl} $ $(\\dimsl < \\dimsh) $ is obtained as:\n\\begin{equation}\n\\y_t = \\Sub_s\\LPF_s\\z_t + \\mybold{v}_t\n\\label{eq:DL_approach1}\n\\end{equation}\nwhere $ \\mybold{v}_t $ is a random noise, $ \\LPF_s $ is an anti-aliasing low-pass filter, and $ \\Sub_s $ is a subsampling operator. The presence of $ \\LPF_s $ is to avoid the problem of aliasing when subsampling the field. In the case of direct subsampling, this filter is omitted from the above model. The goal is to find a HR estimate $ \\hat{\\z}_t $ containing both large and small scales. The most naive and simple idea is to interpolate from $ \\y $, i.e. $ \\hat{\\z}_t = \\Interp_s\\y_t \\in \\R^{\\dimsh}$. However, it will give no access to small-scale information above the cutoff frequency defined by the low-resolution grid. The coupled dictionary approach permits to learn small scales from training data and uses it for the reconstruction. \n\nDL uses the patch-based approach, where the couples of LR and HR patches are extracted from the LR and HR fields respectively. Let $ \\patchhigh{k} = \\mathcal{R}_h^k \\z_t \\in \\R^\\dimpsh $ be the $ k-${th} HR patch. By assuming the Sparse-Land model for high and low resolution training fields, each patch can be estimated as a linear combination of atoms in a dictionary:\n\\begin{equation}\n\\patchhigh{k} = \\dict_h \\adictco^k + \\mybold{\\epsilon}^k\n\\label{eq:DL_approach2}\n\\end{equation} \nwhere $ \\dict_h \\in \\R^{\\dimpsh \\times K}$ is the HR dictionary, $ \\adictco^k \\in \\R^K$ is the coefficient vector and $ \\mybold{\\epsilon}^k $ is the reconstruction error. The dictionary is redundant ($ K > \\dimpsh $) and the coefficients are sparse ($ \\Arrowvert \\adictco^k \\Arrowvert_0 < \\dimtrunc $), with the sparsity constraint $ \\dimtrunc \\ll K $. The corresponding LR patch $ \\patchlow{k} \\in \\R^\\dimpsl $ is: \n\\begin{equation}\n\t\\patchlow{k} = \\extractlow{k} \\y_t = \\extractlow{k} \\Sub_s\\LPF_s\\z_t + \\extractlow{k}\\mybold{v}_t\n\\label{eq:DL_approach3}\n\\end{equation}\t\n$ \\extractlow{k} $ is just an extraction operator, and $ \\Sub_s\\LPF_s $ is a transformation operator going from HR to LR fields. Since $ \\Sub_s $ and $ \\LPF_s $ are spatially independent operators, there exit local $ \\Sub_s^{loc} $ and $ \\LPF_s^{loc} $ that transform HR to LR patches:\n\\begin{equation}\n\t\\patchlow{k} = \\Sub_s^{loc}\\LPF_s^{loc}\\patchhigh{k} + \\mybold{v}_\\ell^k\n\t\\label{eq:DL_approach4}\n\\end{equation} \nwhere $ \\mybold{v}_\\ell^k $ is a random noise of the same dimension as the LR patches. From equations \\ref{eq:DL_approach2} and \\ref{eq:DL_approach4}, one can write:\n\\begin{equation}\n\\begin{split}\n\t\\patchlow{k} &= \\Sub_s^{loc}\\LPF_s^{loc}\\dicthigh \\adictco^k + \\Sub_s^{loc}\\LPF_s^{loc}\\mybold{\\epsilon}^k + \\mybold{v}_\\ell^k \\\\\n\t\t\t\t &= \\Sub_s^{loc}\\LPF_s^{loc}\\dicthigh \\adictco^k + \\tilde{\\mybold{v}}_\\ell^k\n\\end{split}\n\\label{eq:DL_approach5}\t\n\\end{equation}\nwhere $ \\tilde{\\mybold{v}}_\\ell^k \\mydef \\Sub_s^{loc}\\LPF_s^{loc}\\mybold{\\epsilon}^k + \\mybold{v}_\\ell^k$ is also a random noise term. Denoting $ \\dictlow \\mydef \\Sub_s^{loc}\\LPF_s^{loc}\\dicthigh $, the above equation illustrates that there exists also a Sparse-Land model for LR patches. These models for LR and HR patches also share the same sparse coefficient $ \\adictco^k $. The LR dictionary is also a downsampled version of the HR one.\n\n\\subsection{Joint learning methods}\n\\label{sec:joint_learning_methods}\nIn practice, $ \\Sub_s\\LPF_s $ or $ \\Sub_s^{loc}\\LPF_s^{loc} $ are usually not given. \\citet{yang2010image} has addressed this problem by proposing the approach to jointly learn coupled dictionaries. \\citet{zeyde2012single} follow the main idea with some modifications. All approaches contains two main steps: \\textit{learning phase} and \\textit{reconstruction phase}. The learning phase can be done \\textit{offline}, i.e. training \\textit{a priori} from the data. With learned dictionaries, the reconstruction phase can be \\textit{online}.\n\nGiven the training HR velocity fields $ \\z_t \\in \\R^{\\dimsh} $, corresponding LR fields are virtually extracted as $ \\y_t=\\Sub_s\\LPF_s\\z_t \\in \\R^{\\dimsl}$. Couples of LR and HR patches $ \\left\\lbrace \\patchlow{k},\\patchhigh{k} \\right\\rbrace $ are extracted from the training fields as $ \\patchlow{k}= \\extractlow{k} \\y_t\\in \\R^\\dimpsl $ and $ \\patchhigh{k}= \\extracthigh{k} \\z_t\\in \\R^\\dimpsh $, where $ \\dimpsl $ and $ \\dimpsh $ are the size of LR and HR patches respectively, and $ \\dimpsh/\\dimpsl = \\dimsh / \\dimsl $. Let $ \\mathbf{P}_\\ell $ and $ \\mathbf{P}_h $ denote matrices of all LR and HR patches:\n\\begin{equation}\n\t\\begin{split}\n\t\t\\mathbf{P}_\\ell&=\\left[\\patchlow{1} \\:\\: \\patchlow{2} \\:\\: ... \\:\\: \\patchlow{\\dimptl}\\right]_{\\dimpsl \\times \\dimptl} \\\\\n\t\t\\mathbf{P}_h&=\\left[\\mybold{p}_h^1 \\:\\: \\mybold{p}_h^2 \\:\\: ... \\:\\: \\mybold{p}_h^{\\dimptl}\\right]_{\\dimpsh \\times \\dimptl}\n\t\\end{split}\n\t\\label{eq:jointlearning1}\n\\end{equation}\nwhere $ \\dimptl $ is the total number of patches, and $ \\dimptl = \\left(\\sqrt{\\dimsl} - \\sqrt{\\dimpsl} + 1\\right)^2 \\times \\dimtl$ at most, where $ \\dimtl $ is the number of training planes. The LR patches are collected by one-pixel overlapping, while HR ones are obtained by overlapping $ \\dimpsh/\\dimpsl $ pixels to ensure the same number of LR and HR patches. Coupled dictionaries $ \\left\\lbrace \\dicthigh ,\\dictlow \\right\\rbrace $ are learned from the coupled training patches $ \\left\\lbrace \\mathbf{P}_h ,\\mathbf{P}_\\ell \\right\\rbrace $ imposing to share the same sparse representation:\n\\begin{equation}\n\\begin{cases}\n\t\\mathbf{P}_h \\approx \\dicthigh \\dictco\\\\\n\t\\mathbf{P}_\\ell \\approx \\dictlow \\dictco\\\\\n\\end{cases}\n\\end{equation} \nThe first learning approach is proposed by \\cite{yang2010image}, which aims at solving an optimization problem:\n\\begin{equation}\n\\left\\lbrace \\dicthigh ,\\dictlow, \\dictco \\right\\rbrace = \\argmin_{\\dicthigh ,\\dictlow ,A} \\left\\lbrace \\frac{1}{\\dimpsh} \\Vert \\mathbf{P}_h-\\dicthigh \\dictco\\Vert^2_2 + \\frac{1}{\\dimpsl} \\Vert \\mathbf{P}_\\ell-\\dictlow \\dictco\\Vert^2_2 + \\lambda_1 \\left( \\frac{1}{\\dimpsh}+\\frac{1}{\\dimpsl} \\right) \\Vert \\dictco \\Vert_1 \\right\\rbrace\n\\label{eq:jointlearning2}\n\\end{equation}\n$ \\dictco $ is the shared sparse coefficients, which ensures a compromise between small reconstruction errors of patches and sparsity constraint. The two normalization terms $ 1/\\dimpsl $ and $ 1/\\dimpsh $ are to balance the two mean-square error terms. The above cost function can be re-written as:\n\\begin{equation}\n\\dict = \\argmin_{\\dict,\\dictco} \\left\\lbrace \\Vert \\mathbf{P}-\\dict \\dictco\\Vert^2_2 + \\lambda_1 \\Vert \\dictco \\Vert_1 \\right \\rbrace\n\\label{eq:jointlearning3}\n\\end{equation}\nwhere\n\\begin{equation}\n\\mathbf{P}= \\begin{bmatrix} \\frac{1}{\\sqrt{\\dimpsh}}\\mathbf{P}_h \\\\ \\frac{1}{\\sqrt{\\dimpsl}}\\mathbf{P}_\\ell \\end{bmatrix}, \\:\\: \\dict= \\begin{bmatrix} \\frac{1}{\\sqrt{\\dimpsh}}\\dicthigh \\\\ \\frac{1}{\\sqrt{\\dimpsl}}\\dictlow \\end{bmatrix}\n\\label{eq:jointlearning4}\n\\end{equation}\nThis problem can be solved using the standard DL algorithms.\n\nThe second approach is proposed by \\citet{zeyde2012single}, which contains a direct model to estimate the HR dictionary. The LR dictionary is first learned from LR patches:\n\\begin{equation}\n\\left\\lbrace \\dictlow ,\\dictco \\right\\rbrace =\\argmin_{\\dictlow, \\dictco} \\left\\lbrace \\normtwo{\\mathbf{P}_\\ell-\\dictlow \\dictco} + \\lambda_1\\Vert \\dictco \\Vert_1 \\right \\rbrace\n\\label{eq:jointlearning5}\n\\end{equation}\nAssuming that the representation of HR patches $ \\mathbf{P}_h $ via the HR dictionary $ \\dicthigh $ will share the same sparse code $ \\dictco $, HR dictionary is estimated as:\n\\begin{equation}\n\\dicthigh = \\argmin_{\\dicthigh } \\left\\lbrace \\Vert \\mathbf{P}_h-\\dicthigh \\dictco\\Vert^2_2 \\right \\rbrace\n\\label{eq:jointlearning6}\n\\end{equation}\nfor which the solution is easy to obtain by a pseudo-inverse:\n\\begin{equation}\n\\dicthigh =\\mathbf{P}_h \\dictco^{\\dagger}=\\mathbf{P}_h\\dictco^{\\mytrans}(\\dictco\\dictco^{\\mytrans})^{-1}\n\\label{eq:jointlearning7}\n\\end{equation}\n\nAs originally proposed for image super-resolution where sharp edges are important, \\citet{yang2010image,yang2012coupled,zeyde2012single} use different pre-processing techniques before learning coupled dictionaries. Interpolated images $ \\Interp_s\\y_t $ of the same dimension as HR ones are considered as LR images. \\citet{yang2010image,yang2012coupled} couple HR patches, i.e. $ \\patchhigh{k}=\\extracthigh{k}\\z_t$, with the derivatives of interpolated LR images $ \\varmathbb{F} \\Conv \\Interp_s\\y_t $. The operator $ \\Conv $ is the convolution operator. These derivatives are obtained from the convolution of four different 1D kernels $ \\varmathbb{F} $ of first and second order:\n\\begin{equation}\n\t\\left[ \\begin{matrix} -1 & 0 & 1 \\end{matrix} \\right] , \\:\\: \\left[ \\begin{matrix} -1 & 0 & 1 \\end{matrix} \\right]^{\\mytrans} , \\:\\: \\left[ \\begin{matrix} 1 & 0 & -2 & 0 & 1 \\end{matrix} \\right] , \\:\\: \\left[ \\begin{matrix} 1 & 0 & -2 & 0 & 1 \\end{matrix} \\right]^{\\mytrans}\n\\label{eq:jointlearning8}\n\\end{equation}\nLR patches are then extracted as $\\patchlow{k}= \\extracthigh{k}(\\varmathbb{F} \\Conv \\Interp_s\\y_t) $. \\citet{zeyde2012single} couples the same features $ \\varmathbb{F} \\Conv \\Interp_s\\y $, but with the residual between HR and interpolated images, i.e. $ \\patchhigh{k}=\\extracthigh{k}(\\z_t-\\Interp_s\\y_t) $. Moreover, the dimension is reduced using PCA, which corresponds to an adaptive low-pass filter. This step in practice is important, since applying the four filters to low resolution fields bring redundant and superfluous information and lead to unnecessary heavy computation. \n\n\\begin{table}\n\t\\caption{\\label{tab:DLapproaches}\n\tNotations for three different methods of coupled dictionaries learning as proposed by \\citet{yang2010image,zeyde2012single}. The dimension of LR and HR patches are $ \\dimpsl $ and $ \\dimpsh $ respectively.}\n\t\\vspace{.5cm}\n\t\\centering\n\t\\begin{tabular}{ccccccc} \n\t\t\\toprule \\multirow{2}{*}{Notation}&\\multicolumn{1}{c}{}&\\multicolumn{2}{c}{Patch extraction}&\\multicolumn{1}{c}{}&\\multicolumn{2}{c}{Patch dimension}\\\\\n\t\t\\cmidrule{3-4} \\cmidrule{6-7}\n\t & & {LR} & {HR} & & {LR} & {HR} \\\\\n\t\t\\midrule \n\t\tSR1 && $ \\extractlow{k} \\Sub_s\\LPF_s\\z_t $ & $ \\extractlow{k} \\z_t $ && $ \\dimpsl $ & $ \\dimpsh $\\\\\n\t\tSR2 && $ \\extractlow{k} \\Interp_s\\Sub_s\\LPF_s\\z_t $ & $ \\extractlow{k} \\z_t $ && $ \\dimpsh $ & $ \\dimpsh $\\\\\n\t\tSR3 && $ \\extractlow{k} \\extracthigh{k}(\\varmathbb{F} \\Conv \\Interp_s\\y_t) $ & $ \\extractlow{k} \\left\\lbrace \\z_t - \\Interp_s\\y_t\\right\\rbrace$ && $ 4\\dimpsh $ & $ \\dimpsh $ \\\\\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\nBased on the above works by \\citet{yang2010image,zeyde2012single}, we study three different approaches, namely ``\\textit{SR1}'',``\\textit{SR2}'' and ``\\textit{SR3}'', to learn coupled dictionaries. Descriptions of LR and HR patches with their sizes are summarized in table \\ref{tab:DLapproaches}. The dimension of LR patches when using derivatives is $ 4 \\dimpsh $; However in practice, it is reduced significantly via PCA while retaining $ 99.9 \\% $ of energy content.\n\n\\subsection{Reconstruction using learned dictionaries}\nHaving the coupled dictionaries at hand and given a LR field $ \\y^\\ext $, the objective is to reconstruct the HR $ \\z^\\ext $. The superscript ``$ \\ext $'' stands for ``\\textit{external}'', meaning outside of the training planes. First, all LR patches $ \\mathbf{P}^\\ext_\\ell$ are extracted from $ \\y^\\ext $ with the overlapping of one pixel. $ \\mathbf{P}^\\ext_\\ell$ is then centered by removing the mean of each patch $ m_l $. Next, the sparse code $ \\dictco $ are estimated by solving the regularized least-squares problem:\n\\begin{equation}\n\\dictco^\\ext =\\argmin_{\\dictco^\\ext} \\left\\lbrace \\Vert \\mathbf{P}_\\ell-\\dictlow \\dictco^\\ext\\Vert^2_2 + \\lambda_2 \\Vert \\dictco^\\ext \\Vert_1 \\right\\rbrace \n\\label{eq:DLrec1}\n\\end{equation}\nHR patches are estimated using this shared code and putting back the mean $ m_l $:\n\\begin{equation}\n\\hat{\\mathbf{P}}^\\ext_h = \\sqrt{\\frac{\\dimpsh}{\\dimpsl}} \\dicthigh \\dictco^\\ext + m_l\n\\label{eq:DLrec2}\n\\end{equation}\nFinally, the HR field is reconstructed by solving an optimization problem:\n\\begin{equation}\n\\hat{\\z}^\\ext = \\argmin_{\\z^\\ext} \\left\\lbrace\\sum\\limits_{k} \\Arrowvert \\extracthigh{k}\\hat{\\z}^\\ext- \\hat{\\mybold{p}}_h^k\\Arrowvert^2_2 \\right\\rbrace\n\\label{eq:DLrec3}\n\\end{equation}\nThis problem aims to find the best compromise between all estimates, and the closed-form least-squares solution is:\n\\begin{equation}\n\\hat{\\z}^\\ext= \\left[ \\sum\\limits_{k} \\left(\\extracthigh{k}\\right)^{\\mytrans}\\extracthigh{k}\\right]^{-1}\\sum\\limits_{k} \\left(\\extracthigh{k}\\right)^{\\mytrans}\\hat{\\mybold{p}}_h^k\n\\label{eq:DLrec4}\n\\end{equation}\nThis is the overlapping procedure as discussed in section \\ref{subsec:patch_based_estimation}. A pseudo algorithm of the whole process including the learning and reconstruction phases is summarized in algorithm \\ref{algo_SR}.\n\n\\begin{algorithm}\n\\caption{Coupled dictionary learning for reconstruction of high-resolution fields from low-resolution ones} \\label{algo_SR}\n\\begin{algorithmic}[1]\n\t\\State \\textbf{Input}:\n\t\\begin{itemize}\n\t\t\\item training high-resolution fields $ \\left\\lbrace \\z_t \\right\\rbrace, t=1,2,...,\\dimtl$\n\t\t\\item testing low-resolution field $ \\y^\\ext $\n\t\\end{itemize}\n\t\\State \\textbf{Step 1}: \\textit{Learning phase} (offline)\n\t\\begin{itemize}\n\t\t\\item Extract virtual low-resolution fields the same way as $ \\y^\\ext $ is recorded: \n\t\t\\begin{equation}\n\t\t\t\\label{eq:algo_SR_1}\n\t\t\t\\y_t \\mydef \\Sub_s\\LPF_s\\z_t + \\mybold{v}_t\n\t\t\\end{equation}\n\t\t\\item Extract and join coupled patches $ \\left\\lbrace \\mathbf{P}_l,\\mathbf{P}_h \\right\\rbrace $ from the fields $ \\left\\lbrace \\y_t, \\z_t \\right\\rbrace, t=1,2,...,\\dimtl $:\n\t\t\\begin{equation}\n\t\t\t\\label{eq:algo_SR_2}\t\t\n\t\t\t\\mathbf{P} \\mydef \\left[ \\frac{1}{\\sqrt{\\dimpsh}}\\mathbf{P}_h ; \\frac{1}{\\sqrt{\\dimpsl}}\\mathbf{P}_\\ell \\right]\n\t\t\\end{equation}\n\t\t\\item Learn the joint dictionary $ \\dict \\mydef \\left[\\frac{1}{\\sqrt{\\dimpsh}}\\dicthigh; \\frac{1}{\\sqrt{\\dimpsl}}\\dictlow \\right] $ as:\n\t\t\\begin{equation}\n\t\t\t\\label{eq:algo_SR_3}\t\t\n\t\t\t\\left(\\dict, \\dictco\\right) = \\argmin_{\\dict,\\dictco} \\left\\lbrace \\Vert \\mathbf{P}-\\dict \\dictco\\Vert^2_2 + \\lambda_1 \\Vert \\dictco \\Vert_1 \\right\\rbrace\n\t\t\\end{equation}\n\t\\end{itemize}\t\n\t\\State \\textbf{Step 2}: \\textit{Reconstruction phase} (online)\n\t\\begin{itemize}\n\t\t\\item Extract LR patches: $ \\mathbf{P}_\\ell ^\\ext \\in \\R^{\\dimpsl \\times \\dimptl}$ from $ \\y^\\ext $\n\t\t\\item Estimate the sparse code:\n\t\t\\begin{equation}\n\t\t\t\\label{eq:algo_SR_4}\t\t\n\t\t\t\\dictco^\\ext = \\argmin_{\\dictco^\\ext} \\left\\lbrace \\Vert \\mathbf{P}_\\ell^\\ext-\\dictlow \\dictco^\\ext\\Vert^2_2 + \\lambda_2 \\Vert \\dictco^\\ext \\Vert_1 \\right\\rbrace \n\t\t\\end{equation}\n\t\t\\item Reconstruct HR patches $ \\hat{\\mathbf{P}}_h^\\ext = \\dicthigh\\dictco^\\ext \\in \\R^{\\dimpsh \\times \\dimptl} $\n\t\\end{itemize}\n\t\\State \\textbf{Output:} Reconstruct HR field $ \\z^\\ext $ from $ \\hat{\\mathbf{P}}_h^\\ext $ by overlapping.\n\\end{algorithmic}\n\\end{algorithm}\n \n\\section{Dictionary learning for isotropic turbulence fields}\nThe section applies dictionary learning approach to the DNS database of the isotropic turbulence discussed in chapter \\ref{sec:data_isotropic}. This data is chosen because the fields are periodic, isotropic and homogeneous. These properties will facilitate all computations and the handling of boundary conditions. Two main problems are addressed. First, the efficiency of the representation using dictionary learning will be studied, comparing with other approaches such as wavelet transform or PCA. Second, the coupled dictionaries approach is discussed to solve the problem of reconstructing HR velocity fields from LR measurements. \n\nTo compare with other approaches, we consider a similar configuration as in chapter \\ref{chap_linearregression} to study regression models. The reference HTHS data is subsampled to obtain the measurements of HTLS $ \\{\\y_t\\} $ and LTHS $ \\{\\x_t\\} $ (see figure \\ref{fig:space-time_measurements}). The subsampling ratios are $ \\dimsh/\\dimsl = 4 \\times 4$ in space and $ \\dimth/\\dimtl = 6 $ in time, corresponding to a moderate amount of energy losses in every directions. We will use $ \\{\\x_t\\} $ only for the training, then uses $ \\{\\y_t\\} $ to predict $ \\{\\z_t\\} $.\n\n\\subsection{On the choice of parameters}\n\\label{subsec:DL_choice_params}\n\\begin{table} \n\t\\caption{\\label{tab:DLparams}\n\tSet of parameters for coupled dictionaries approach (notations are consistent with Algorithm \\ref{algo_SR}): sparsity constrains $ \\lambda_1 $ for learning and $ \\lambda_2 $ for reconstruction; dimension of LR patches $ \\dimpsl $ and HR patches $ \\dimpsh $; number of atoms $ \\dimdict $; number of training patches $ \\dimptl $ }\n\t\\vspace{.5cm}\n\t\\centering\n\t\\begin{tabular}{cccccc} \n\t\t\\toprule\n\t\t{$\\lambda_1$} & {$\\lambda_2$} & {$\\sqrt{\\dimpsl}$} & {$\\sqrt{\\dimpsh}$} & {$\\dimdict$} & {$ \\dimptl $} \\\\ \n\t\t\\midrule \n\t\t$ 0.1 \\sim 0.2 $ & $ 10^{-5} $ & 4 & 16 & $ 2(\\dimpsl \\times \\dimpsh) $ & $~ 10^5 $ \\\\ %\\addlinespace\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\nResults in the following sections are obtained with a consistent set of parameters presented in table \\ref{tab:DLparams}. Notations are consistent with the algorithm \\ref{algo_SR} to learn coupled dictionaries. This section briefly discusses these choices.\n\nThe first set of parameters are sparsity constraints $ \\lambda_1 $ and $ \\lambda_2 $. For the learning phase, $ \\lambda_1 $ is about $ 0.1 \\sim 0.2 $ to learn either the coupled dictionaries in equation \\ref{eq:jointlearning2} or the LR dictionary in equation \\ref{eq:jointlearning5}. This constraint imposes that the reconstruction of training patches uses only $ 10\\sim15 $ nonzero coefficients in average, corresponding to the sparsity level of $ 0.90\\sim0.95 $ (only $ 5 \\sim 10\\% $ coefficients are nonzero). As will be seen latter in figure \\ref{fig:Sparsity_vs_NRMSE}, this is a strong constraint. However, decreasing $ \\lambda_1 $, i.e. reducing the sparsity level, downgrades reconstruction results of coupled dictionaries. This is probably the \\textit{overfitting} problem discussed in chapter \\ref{chap_linearregression}. With small $ \\lambda_1 $, the dictionaries represent well the training data but with a poor generalization capability. For reconstruction phase, $ \\lambda_2 $ is chosen to be close to zero. In such case, the sparsity level of $ \\dictco^\\ext $ in equation \\ref{eq:algo_SR_4} is very low, i.e. the maximum number of atoms is used to reconstruct $ \\mathbf{P}^\\ext_\\ell $.\n\nThe second set of parameters are to describe the training patches. The sizes of the LR and HR patches are $ \\dimpsl = 4 \\times 4 $ and $ \\dimpsh = 16 \\times 16 $, respectively. The LR patch size is small compared to the whole LR field of size $ \\dimsl = 16 \\times 16 $, permitting to localize the information. Numerical experiments show a slight improvement when increasing this size, but it leads to a high computation requirements and a large number of training planes. Coupled patches are extracted from $ 37 \\times 16 $ LTHS planes for training. From each snapshot, a total of $ (16-4+1) \\times (16-4+1) =169$ LR patches can be extracted using one-pixel overlapping. The same number of corresponding HR patches are also extracted from HR training fields. Around $ m=10^5 $ couples of LR-HR patches are chosen randomly from the whole set of $ 37 \\times 16 \\times 169 $ patches in total and used to train the coupled dictionaries. The objective is to reduce the computational cost, since many patches are very similar. \n\nThe last parameter is the number of atoms. There is no general rule, but new findings using non-parametric approaches of learning dictionaries with an adaptive number of atoms \\citep{dang2016towards} show that the needed level of redundancy maybe not very high. Numerical experiments in this work show that a good size is about two times the dimension of input variables. For the case of coupled dictionary learning, the choice is $ K = 2(\\dimpsh+\\dimpsl) $, where $ (\\dimpsh+\\dimpsl) $ is the dimension of the coupled LR and HR patches.\n\n\\subsection{Efficiency of representations: a comparative study} \nSince the sparsity may play a key role, this section compares the efficiency of different representations. ``\\textit{Efficiency}'' means the quality of the approximation with respect to different sparsity levels. Mathematical predefined representation using wavelets, learned bases using PCA and dictionary learning methods are compared. Both KSVD \\citep{aharon2006ksvd} and ODL \\citep{mairal2010online} are investigated. The representations are learned from the LTHS planes $ \\left\\lbrace \\x_t \\right\\rbrace $ in the cases of PCA and DL. These representations are then tested on random subsampled fields from HTHS planes $ \\z_t $ that are different from the set of $ \\{\\x_t\\} $ used for learning. Reconstruction errors are estimated by comparing with reference fields. \n\n\\begin{figure}[t]\n\\centering\n\t\\includegraphics[width=0.45\\textwidth]{./images/DL/DLstat/PCA_patchsize04.eps}\n\t\\includegraphics[width=0.45\\textwidth]{./images/DL/DLstat/D_HR_lambda005.eps}\n\t\\caption{\\label{fig:PODvsDL} Dictionaries leaned by PCA and ODL from the set of HR patches of size $ \\dimpsh = 16 \\times 16 $. With ODL, only 256 atoms are chosen from 512 atoms.}\n\\end{figure}\n\nFigure \\ref{fig:PODvsDL} shows the adaptive dictionaries learned by PCA (left) and ODL (right) as ranked by their energy contents. The redundant dictionary by ODL has two times more atoms compared to PCA, which contains exactly $ \\dimpsh = 16 \\times 16 $ atoms as the dimension of the input patches. Only 256 over 512 atoms obtained by ODL are shown to be comparable with the PCA dictionary. The atoms are completely different in the two dictionaries. PCA dictionary has a sharp decline of variance content. Also, since the number of training patches is sufficiently large, the atoms look similar to discrete cosine transform (DCT) basis functions, which have modulated sine-wave patterns. The redundant dictionary from ODL contains more ``patterns'' for each level of scale, which is expected to be more adaptive to the data for sparse representation. \n\nUsing a wavelet transform, there is no learning since the bases are predefined. The fields are decomposed into approximation and detail coefficients (horizontal, vertical and diagonal). The common Daubechies wavelet is used for its compact support and fast computation. The transform is performed on full fields. To test the sparsity effects, different thresholds are used. Detail coefficients larger than each threshold are retained while setting others to zeros, while approximation coefficients are kept unchanged. Inverse wavelet transform reconstructs the fields using these unchanged approximation and filtered detail coefficients. The sparsity is defined as the ratio of nonzero coefficients $ \\dimtrunc $ (including both approximation and retained detail coefficients after filtering) and the dimension of the field $ \\dimsh $. The NRMSE between reconstructed fields $ \\hat{\\z}_t $ and reference ones $ \\z_t $ is estimated as in equation \\ref{eq:NRMSE} to qualify the reconstruction for each level of sparsity $ (1- \\dimtrunc/\\dimsh )$.\n\nWith PCA, the dictionary $ \\dict $ is learned from training patches $ \\mathbf{P}_h $ extracted from all $ \\left\\lbrace \\x_t \\right\\rbrace, t=1,...,\\dimtl $. Its atoms $ \\left\\lbrace \\adicthigh{i} \\right\\rbrace $ are ranked by their variances $ \\lambda_i $, i.e. $ \\lambda_1 > \\lambda_2 > ... > \\lambda_{\\dimsh} $. The dimensionality is reduced such that the retained information from only the first $ \\dimtrunc $ principal components ($ \\dimtrunc < \\dimpsh $) is maximized. For a new field $ \\z_t $, $ \\mathbf{P}_h $ are extracted and projected onto the first $ \\dimtrunc $ vectors:\n\\begin{equation}\n\t\\adictco_i = \\frac{\\mathbf{P}_h \\mydot \\adicthigh{i}}{\\normtwo{\\adicthigh{i}}} \\:, i=1,2,...,\\dimtrunc\n\t\\label{eq:DL_efficiency3}\t\n\\end{equation}\nThe filtered patches are estimated by combining the projected coefficients with the corresponding functions:\n\\begin{equation}\n\t\\hat{\\mathbf{P}}_h = \\sum\\limits_{i=1}^{\\dimtrunc} {\\adictco_i \\adicthigh{i}}\n\t\\label{eq:DL_efficiency4}\n\\end{equation}\nFinally, the reconstructed field $ \\hat{\\z}_t $ is estimated by one-pixel overlapping as discussed in section \\ref{subsec:patch_based_estimation}. The sparsity level is defined as $ 1-\\dimtrunc/\\dimpsh $, where the patch size at HR is $ \\dimpsh = 16 \\times 16 $. NRMSEs are estimated using equation \\ref{eq:NRMSE}. \n\nKSVD and ODL learn $ \\dict $ from $ \\mathbf{P}_h $ \\textit{a priori} with a high sparsity level ($ \\lambda =0.2$, corresponding to about $ 15 $ non-zero coefficients). In the reconstruction step, all possible patches $ \\mathbf{P}_h $ are extracted from each field $ \\z_t $. The sparse code is estimated using LARS algorithm to solve:\n\\begin{equation}\n\t\\dictco = \\argmin_{\\dictco} \\left\\lbrace \\normtwo{\\mathbf{P}_h - \\dict \\dictco_t} + \\lambda \\normone{\\dictco} \\right\\rbrace\n\t\\label{eq:DL_efficiency2}\t\n\\end{equation}\nThe efficiency of the learned dictionary $ \\dict $ is studied by varying $ \\lambda $ in this step. For each $ \\lambda $, $ \\dictco $ is estimated and then used to re-estimate the patches $ \\hat{\\mathbf{P}}_h = \\dict \\dictco$ before reconstructing the global scene $ \\hat{\\z}_t $. Sparsity is measured as the average of $ 1- \\normzero{\\adictco_t}/\\dimpsh $, where $ \\adictco_t $ is the $ t- $th row of $ \\dictco $. NRMSEs are estimated the same way as of wavelet transform or PCA.\n\n\\begin{figure}[t]\n\\centering\n\t\\includegraphics[width=\\textwidth]{./images/DL/DLstat/Sparsity_vs_NRMSE_PCA_ODL_KSVD_WL_Dau_patchsize04.eps}\n\t\\caption{\\label{fig:Sparsity_vs_NRMSE} Sparsity vs error outside (left) and zoom in the region of low sparsity at semi-log scale (right) for different representations: wavelet (Daubechies), PCA, dictionary learning (ODL or KSVD). Wavelet transform is for the whole fields of size $ 96 \\times 96$, while PCA and DL are for patches of size $ \\dimpsh = 16 \\times 16 $.}\n\\end{figure}\n\nFigure \\ref{fig:Sparsity_vs_NRMSE} shows the curves of NRMSE as functions of sparsity levels. These errors are estimated from five testing planes equally far from neighboring LTHS snapshots. All curves behave similarly as reducing the error when sparsity decreases. At high levels (larger than 0.5), there is a clear benefit of using DL. With the same number of non-zero coefficients, both ODL and KSVD give lower errors than PCA and wavelet transform. Few atoms from DL better represent the data than high variance principal components of PCA or predefined ones. Comparing the two DL methods, ODL is better than KSVD. When using more non-zero coefficients, errors by DL methods saturate at nonzero values. Since DL provides only an approximate solution. Wavelet transform and PCA give zero NRMSEs when using all coefficients because the transforms are exact.\n\nThe above comparisons demonstrate the advantages of representations using learned dictionaries over predefined ones. Comparing redundant and orthogonal representations, i.e. DL versus PCA, redundant dictionaries represent the fields more efficiently. They achieve the same level of error using less atoms. This suggests also that the sparsity and redundancy priors can be good candidates to help solving the ill-posed inverse problem of reconstructing HR fields from LR ones. Comparing the two common DL methods, ODL shows clear advantages both in term of efficient representation and computation effort and will be used in the rest of this chapter. \n \t\n\\subsection{Reconstruction of high resolution fields- subsampling cases}\nTo compare to other methods, LR fields are first subsampled from HR ones by a factor of $ 4 \\times 4 $, i.e. $ \\y_t = \\Sub_s \\z_t $, where $ \\Sub_s: \\R^\\dimsh \\mapsto \\R^\\dimsl $, $ \\dimsh/\\dimsl = 4 \\times 4 $. Coupled LR and HR patches are extracted to train the dictionaries. Due to direct subsampling, the aliasing, which has not been addressed in previous works, will play an important role. This section investigates the ability of dictionary learning approaches to handle this aliasing problem.\n\n\\subsubsection*{Learning step}\n\\begin{figure}[t]\n\\centering\n\t\\includegraphics[width=\\textwidth]{./images/DL/SR_sspacing04/subsampling/coupleddictionary_HRLR_lambda010.eps}\n\t\\caption{\\label{fig:D_HR_LR} Coupled dictionary of high and low resolution patches, with LR patches are of size $ 4\\times 4 $, directly subsampled from their equivalent HR patches of size $ 16 \\times 16 $. The regularization parameter $ \\lambda $ for join learning are chosen such that about 16 non-zero coefficients are retained for reconstructing the joint patches of the training.}\n\\end{figure}\n\n\\begin{figure}[t]\n\\centering\n\t\\includegraphics[width=\\textwidth]{./images/DL/SR_sspacing04/subsampling/coupleddictionary_HRHRinterp_lambda010.eps}\n\t\\caption{\\label{fig:D_HR_HRinterp} Coupled dictionary of high and low resolution patches, where LR patches are of size $ 4\\times 4 $, directly subsampled from their equivalent HR patches of size $ 16 \\times 16 $. The regularization parameter $ \\lambda $ for join learning are chosen such that about 16 non-zero coefficients are retained for reconstructing the joint patches of the training.}\n\\end{figure}\n\nODL \\citep{mairal2010online} is used to learn a joint dictionary of HR and LR from the LTHS planes. The subsampling ratio in space is $ \\dimsh/\\dimsl = 4 \\times 4 $, while the training samples are taken every $ \\dimth/\\dimtl = 6 $ snapshots in streamwise direction. From a total of $ 37 \\times 16 $ training planes, small patches of size $ 4 \\times 4 $ at LR are extracted via the extraction operator $ \\extractlow{k} $, coupled with HR patches of size $ 16 \\times 16 $ by $ \\extracthigh{k} $. The choice of parameters is discussed in section \\ref{subsec:DL_choice_params}.\n\nThree approaches \\textit{SR1}, \\textit{SR2} and \\textit{SR3} (table \\ref{tab:DLapproaches}) are investigated. The interpolation step in \\textit{SR2} plays only the role of transforming the field from LR to the same dimension as the HR one. The content should be rather the same, since the interpolation does not introduce any small-scale information. However, starting from the interpolated fields can bring the benefit of having a good large-scale information \\textit{a priori}. The model will focus on small scales only. This approach have shown to be beneficial in many image processing problems.\n\nFigure \\ref{fig:D_HR_LR} shows dictionaries for HR and LR patches and figure \\ref{fig:D_HR_HRinterp} shows dictionaries for HR and interpolated LR patches. In both figures, atoms are sorted according to their variances $ \\normtwo{\\adictco_i} $ when learning the dictionary (from the top-left). The most significant atoms contain mostly large scales, while less significant ones represent high-frequency contents. Coupled atoms show similar patterns. The relation between LR and HR patches are now encoded in the relation between LR and HR dictionaries. The assumption of coupled representations in this case also implies that LR atoms are approximately subsampled from the HR ones as shown in equation \\ref{eq:DL_approach5}. \n\n\\subsubsection*{Reconstruction step}\nHaving learned dictionaries $ \\{\\dicthigh,\\dictlow \\} $ at hand, from given $ \\y^\\ext $ different from the training data, the HR field $ \\z^\\ext $ is reconstructed by using algorithm \\ref{algo_SR}. $ \\y^\\ext $ is also assumed to be directly subsampled from $ \\z^\\ext $ the same way as in the training data, i.e. $ \\y^\\ext = \\Sub_s \\z^\\ext $. From $ \\y^\\ext $, all LR patches $ \\mathbf{P}_l^\\ext $ with one-pixel overlapping are extracted. The sparse code is estimated by solving the optimization problem in equation \\ref{eq:algo_SR_4}. $ \\dictco^\\ext $ is then used to reconstruct HR patches as $ \\hat{\\mathbf{P}}_h^\\ext = \\dicthigh \\dictco^\\ext $ and put back into the global scene of $ \\hat{\\z}^\\ext $. \n\nLR and HR patches of sizes $ \\dimpsl = 4 \\times 4 $ and $ \\dimpsh = 16 \\times 16$ respectively are extracted from all training planes. The sparse code $ \\dictco^\\ext $ is estimated from LR patches. Each row has at most $ \\normzero{\\dictco^\\ext} = \\dimpsl = 16 $ nonzero coefficients. It means that HR patches are reconstructed from maximum $ 16 $ atoms within $ \\dicthigh $, a strong constraint on the reconstruction accuracy (see figure \\ref{fig:Sparsity_vs_NRMSE}). The nature of the data also affects the accuracy in the sense how good is the Sparse-Land prior. Also, this constraint addresses the problem of designing more efficient representations of the data to reduce the error when using the same number of atoms. \n\nUsing all three approaches presented in table \\ref{tab:DLapproaches}, the models can reconstruct the fields in the same accuracy as the spline interpolation but not better. This is due to the severe situation where the presence of aliasing makes equation \\ref{eq:DL_approach5} a very crude assumption. The subsampling of a small atoms brings a very strong aliasing effect that coupled dictionaries could not efficiently handle. The next section will study the capability of the present approach when the aliasing problem is absent from the LR data.\n\n\\subsection{Reconstruction of high resolution fields- the downsampling case}\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.6\\textwidth]{./images/DL/SR_sspacing04/downsampling/dictionary_couplefeatures_patchesHR.eps}\n\t\\caption{Dictionary of the residual between HR and interpolated LR.}\n\t\\label{fig:SR_dictionary_residual}\n\\end{figure}\n\nThis section investigates the possibility of the current approach to estimate the HR fields given the LR ones in the case of downsampling, i.e. with anti-aliasing prefiltering:\n\\begin{equation}\n\\y_t = \\Sub_s \\LPF_s \\z_t\n\\end{equation}\nwhere $ \\Sub_s $ and $ \\LPF_s $ are the spatial subsampling and low pass filter respectively. Bicubic filter in Matlab built-in function \\textit{imresize} is used for the prefiltering and interpolation step. It was shown also in section \\ref{sec:chap3_theapproach} that the assumptions Sparse-Land model at HR also lead to the relation $ \\mathbf{P}_\\ell = \\Sub_s^\\ext \\LPF_s^\\ext \\mathbf{P}_h$ between LR and HR dictionaries. $ \\Sub_s^\\ext $ and $ \\LPF_s^\\ext $ are local versions of $ \\Sub_s $ and $ \\LPF_s $ applying to patches. \n\nWe compare the three methods of coupled dictionary learning, the so-called \\textit{SR1}, \\textit{SR2} or \\textit{SR3}, presented in table \\ref{tab:DLapproaches}. To recall \\textit{SR1} and \\textit{SR2} couple either LR or interpolated LR patches with HR ones, while \\textit{SR3} couples the residuals with the features of derivatives. The procedure follows the previous section of subsampled fields, with learning parameters in table \\ref{tab:DLparams}. HR patches $ \\mathbf{P}_h$ are extracted from LTHS fields and coupled with LR patches $ \\mathbf{P}_\\ell$. The dictionaries are trained \\textit{offline} from $ \\{\\mathbf{P}_h, \\mathbf{P}_\\ell\\}$, and used in \\textit{online} reconstruction stage for all HTLS measurements to reconstruct HTHS fields. \n\nWe use three quantities for comparisons: the average NRMSE between reconstructed and reference velocity fields estimated using equation \\ref{eq:NRMSE}, the 2D energy spectra of the velocity fields and of the errors. These three quantities give a complete view to qualify different approaches. Only the most difficult planes, which are equally far from LTHS measurements, are used to estimate the errors. Error estimated using these planes will better represent the generalization capability of the approaches.\n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.9\\textwidth]{./images/DL/SR_sspacing04/downsampling/NRMSE_compare_all_spacespacing_04.eps}\n\t\\caption{Means and standard deviations of NRMSEs estimated between reference and reconstructed fields of all middle planes (at the center of blocks bounded by the two LTHS planes). The reconstructions are by spline interpolation and super-resolution using three different methods:: \\textit{SR1}, \\textit{SR2} and \\textit{SR3} (see table \\ref{tab:DLapproaches}). NRMSEs are $ 0.267 \\pm 0.021 $, $ 0.235 \\pm 0.019 $, $ 0.233 \\pm 0.018 $ and $ 0.230 \\pm 0.019 $ respectively. The NRSME of spline interpolation in the equivalent subsampling case is 0.276 (dashed black line).}\n\t\\label{fig:NRMSE_compare_all_spacespacing_04}\n\\end{figure}\n\nFigure \\ref{fig:NRMSE_compare_all_spacespacing_04} shows the average NRMSEs of different reconstructions, either interpolation or reconstruction by coupled dictionaries using \\textit{SR1}, \\textit{SR2} or \\textit{SR3} approaches. The three DL models reduce NRMSEs by $ 11.99 \\% $, $ 12.73 \\% $ and $ 13.86 \\% $ respectively compared to the simple interpolation. \\textit{SR3} gives the most accurate reconstructions by coupling the residuals, essentially contain only small scales, with derivatives of large-scale structures.\n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=0.75\\textwidth]{./images/DL/SR_sspacing04/downsampling/spectra2d_spacespacing_04.eps}\n\t\\caption{2D spectra of all planes used to computed the NRMSEs in figure \\ref{fig:NRMSE_compare_all_spacespacing_04}, from reference, interpolation and SR by three different methods: \\textit{SR1}, \\textit{SR2} and \\textit{SR3} (see table \\ref{tab:DLapproaches}). For scales from $ 0.5k_c $ to $ 1.5 k_c$, energy losses of SR fields compared to reference ones is $ 24\\% $, $ 22\\% $ and $ 21\\% $ respectively, while that of interpolation is $85 \\% $.}\n\t\\label{fig:spectra2d_spacespacing_04}\n\\end{figure}\n\nTo further understand the quality of the reconstructed fields at different scales, figure \\ref{fig:spectra2d_spacespacing_04} shows the 2D spectra of reference fields and different reconstructed ones. All methods capture good large scales till about $ 0.5 k_c $, where $ k_c $ is the cutoff wave number defined by the subsampling ratio. The interpolation, starting from the downsampled fields with prefiltering step to avoid aliasing, loses already energy at large scales and capture almost no small scales. By coupling the dictionaries, the reconstructed fields recover the large-scale information with some small scales. If considering only the scales between $ 0.5k_c $ and $ 1.5k_c $, the energy loss of interpolated fields is $ 85\\% $, while those are $ 24\\% $, $ 22\\% $ and $ 21\\% $ for \\textit{SR1}, \\textit{SR2} and \\textit{SR3} respectively. The benefit of SR is significant in this most interesting range of scales. Larger than $ 1.5k_c $, all reconstructions are not reliable. Smaller than $ 0.5 k_c $, spectra of all reconstructed fields are already very accurate. \n\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[width=\\textwidth]{./images/DL/SR_sspacing04/downsampling/errspectra2d_nonnormalized_normalized_timespacing_06_spacespacing_04.eps}\n\t\\caption{(Left) 2D spectra of errors, which are the different between the reference and reconstruction by interpolation and three different SR methods as described in \\ref{sec:joint_learning_methods}. (Right) 2D spectra of errors normalized by the energy spectrum of the reference (the black curve in figure \\ref{fig:spectra2d_spacespacing_04}).}\n\t\\label{fig:errspectra2d_nonnormalized_normalized_timespacing_06_spacespacing_04}\n\\end{figure}\n\nFigure \\ref{fig:errspectra2d_nonnormalized_normalized_timespacing_06_spacespacing_04} (left) shows the spectra of the errors, which are the differences between the reference and reconstructed fields by all methods. Errors are very low at small wave numbers, reach a maximum around $ k_c $ and reduce at higher wave numbers. Integrals of these curves give the mean-square errors of each reconstruction method. To better qualify the error at each scale, the curves are normalized by the energy spectrum of the reference (the black curve in Figure \\ref{fig:spectra2d_spacespacing_04}). The normalized spectra of errors are interpreted as the percentage of error at each scale. The interpolation gives very small relative errors at large scales, but grow rapidly near $ k_c $. Error spectra of all SR methods collapse at $ 0.5k_c $ and reach a maximum of $ 100 \\% $ error at around $ 1.5k_c $. Smaller than $ 0.5k_c $, \\textit{SR1} and \\textit{SR2} gives slightly higher errors compared to \\textit{SR3}. \\textit{SR3} is more accurate since it starts from the interpolated large scales, while those information are re-estimated by \\textit{SR1} and \\textit{SR2}. Also, \\textit{SR3} focuses more on the missing information of small scales by using derivatives as features and following the patterns of larger ones. \n\nFrom the above comparisons, coupled dictionary approaches demonstrate clear benefits compared to the simple interpolation, with errors reduced by about $ 15\\% $. From the energy spectra or spectra of errors, the benefits mostly come from the range of scales between $ 0.5k_c $ and $ 1.5k_c $, where $ k_c $ is the cutoff defined by the grid of LR measurements. The loss of energy in this range is reduced from $ 80\\% $ with interpolation to $ 20\\% $ by DL approaches. The spectra of the errors also show that most of benefits come from this range of scales, before reaching $ 100\\% $ at around $ 1.5k_c $.\n\n\\section{Concluding remarks}\nThis chapter has discussed the possibilities of applying dictionary learning, a successful approach in the field of signal and image processing, to turbulence studies. The method finds a representation for the data by generalizing principal component analysis to sparse representation in a redundant dictionary. \\textit{Redundancy} means that the number of atoms can be larger than the dimension of input vectors, ignoring the orthogonality constraint of PCA. This implies the \\textit{sparsity}, i.e. the representation is composed by a linear combination of only a few atoms. These properties make the learned dictionary a more adaptive representation of the data. Sparsity can also play the role of a prior about the system when solving the inverse problem of HR field reconstruction. \n\nTo investigate the efficiency of DL in representing the data, the learned dictionaries by this approach have been compared against PCA and predefined wavelet dictionaries. Reconstruction errors as functions of sparsity, i.e. the number of atoms used, are shown as the measure of efficiency. Adaptive dictionaries show some superiority compared to the predefined ones. The benefits are mainly from the high sparsity levels, when less than half the number of coefficients are non-zero. \n\nDL is then used to reconstruct HR velocity fields from LR measurements. The approach is called \\textit{coupled dictionary learning}, inspired from the single image super-resolution application \\citet{yang2010image,zeyde2012single}. By learning coupled representations of LR and HR fields from the data, a nonlinear relation is established and generalized to perform the reconstruction task. With the same idea, different preprocessing techniques are tested. The coupling can be between HR and LR fields or their interpolation. Another approach focuses more on the small-scale information by coupling the small scales with the derivatives of interpolated fields. \n\nThe first attempt is for the configuration where LR fields are directly subsampled from HR ones without any anti-aliasing prefiltering step. This is \\textit{a priori} not a favorable case due to the presence of aliasing terms. However, it is worth studying since this setup mimics what would happen in a real experiment. It is interesting also to see how well learned dictionaries can handle aliasing. Results show that DL is inoperative to recover some small scales on top of the interpolated large-scale information. \n\nTo understand whether the failure comes from the approach or from the aliasing, another case is investigated where LR fields are downsampled from the HR ones with a prefiltering step. The same approach with identical parameters is used. Results show significant improvements of reconstruction accuracy, with about $ 15 \\% $ reduction of NRMSEs compared to interpolation. Spectral analyses also show that most benefits are at the frequency range of $ 0.5 k_c$ to $ 1.5k_c $, where $ k_c $ is the cutoff wave number corresponding to the downsampling ratio. In term of energy, simple interpolation loses $ 85 \\% $ in this range scales, while this loss is reduced to about $ 20 \\% $ with DL approaches.\n\nThe above results demonstrate the capability and limitations of DL approaches in solving the reconstruction problem in turbulence. The corresponding prior, which is the duality between sparsity and redundancy, is robust. The first attempt has not succeeded to reconstruct HR fields from direct subsampled LR ones due to the aliasing problem. This attempt addresses also the question on designing a good sensing system or post-processing techniques to deal with aliasing terms if exist.\n\nThis chapter has presented a similar configuration as regression models to learn the mapping function between large and small scales. However, the learning is localized thanks to the patch-wise approach. This property of dictionary learning approach can be beneficial when applying to other configurations where the training and testing samples are of different scenes. For example, the training HR fields could be a small region of the whole field, while LR one can be larger. In such case, dictionary approach is the only candidate among all methods presented in this thesis. \n\n\n", "meta": {"hexsha": "0a4268e84adf3a0d8bd7575c8a0828d991e40639", "size": 70924, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/manuscripts/final_ver1/corps/dictionarylearning.tex", "max_stars_repo_name": "linhvannguyen/PhDworks", "max_stars_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-01-01T14:41:09.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-12T07:08:06.000Z", "max_issues_repo_path": "thesis/manuscripts/final_ver1/corps/dictionarylearning.tex", "max_issues_repo_name": "vanlinhnguyen/PhDworks", "max_issues_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/manuscripts/final_ver1/corps/dictionarylearning.tex", "max_forks_repo_name": "vanlinhnguyen/PhDworks", "max_forks_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 138.7945205479, "max_line_length": 1536, "alphanum_fraction": 0.7657492527, "num_tokens": 19026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.7853085708384736, "lm_q1q2_score": 0.616709559324155}} {"text": "\\Lecture{Jayalal Sarma}{Oct 21, 2020}{17}{Generating Functions(continued)}{Pragnya}{$\\alpha$}{JS}\n\n\\section{Introduction}\nIn this section we'll see some examples of ordinary generating functions and get introduced to exponential generating functions.\n\n\\subsection{Example 3 : }\nTo show two combinatorial qualities are equal it sufficies to show that they have same generating functions. Consider the following,\n$$ B_n(m) = \\{(x_1, x_2, \\dots x_n) ~|~ \\forall i ~ x_i \\in \\Z , \\sum |x_i| \\leq m \\}$$\nlet $b_{n,m} = |B_n(m)|$. Let's see properties of $b_{n,m}$ :\n\\begin{enumerate}\n \\item $b_{n, m} = \\sum_{k=0}^n {n \\choose k}{m \\choose k} 2^k$.\n \\item $b_{n, m} = b_{m, n}$. This can also be proved using bijection.\n \\item $b_{n, m} = d_{m, n}$. \n\\end{enumerate}\nWe'll prove property $3$ by showing they have same generating functions.\n\\begin{align*}\n B_{x,y} &= \\sum_{n,m \\geq 0} b_{n, m} x^n y^m\\\\\n &= \\sum_{n,m \\geq 0} (\\sum_{k=0}^n {n \\choose k}{m \\choose k} 2^k) x^n y^m \\\\\n &= \\sum_{n,m,k \\geq 0}{n \\choose k}{m \\choose k} 2^k x^n y^m \\\\\n &= \\sum_{k \\geq 0} 2^k \\sum_{n,m \\geq 0} {n \\choose k}{m \\choose k} x^n y^m \\\\\n &= \\sum_{k \\geq 0} 2^k (\\sum_{n \\geq 0} {n \\choose k} x^n)(\\sum_{m \\geq 0} {m \\choose k} y^m) \\\\\nB_{x, y}&= \\sum_{k \\geq 0} 2^k (x^k \\sum_{n \\geq 0} {n \\choose k} x^{n-k})(y^k \\sum_{m \\geq 0} {m \\choose k} y^{m-k})\n\\end{align*}\nConsider $\\frac{1}{(1-x)^{k+1}}$ : \n$$\\frac{1}{(1-x)^{k+1}} = \\frac{1}{(1-x)}.\\frac{1}{(1-x)}. \\dots \\frac{1}{(1-x)} ~(k+1 ~times)$$\nCoefficient of $x^{n-k}$ in $\\frac{1}{(1-x)^{k+1}}$ is equivalent to no.of solutions of $a_1 + a_2 + \\dots + a_{k+1} = n-k$ which is $ = {(n-k) + (k+1) -1 \\choose n-k} = {n \\choose k}$.\nHence\n$$ \\sum_{n \\geq 0} {n \\choose k} x^{n-k} = \\frac{1}{(1-x)^{k+1}} $$\nSimilarly \n$$ \\sum_{m \\geq 0} {m \\choose k} y^{m-k} = \\frac{1}{(1-y)^{k+1}} $$\nSubstituting them in the above derived $B_{x, y}$ - \n\\begin{align*}\n B_{x, y} &= \\sum_{k \\geq 0} 2^k x^k y^k \\frac{1}{(1-x)^{k+1}} \\frac{1}{(1-y)^{k+1}} \\\\\n &= \\sum_{k \\geq 0} (2xy)^k \\frac{1}{(1-x)^{k+1}} \\frac{1}{(1-y)^{k+1}} \\\\\n &= \\frac{1}{(1-x)(1-y)} \\sum_{k \\geq 0} \\frac{(2xy)^k}{(1-x)^k(1-y)^k} \\\\\n &= \\frac{1}{(1-x)(1-y)} \\sum_{k \\geq 0} (\\frac{(2xy)}{(1-x)(1-y)}) ^k \\\\\n &= \\frac{1}{(1-x)(1-y)} \\frac{1}{1- \\frac{2xy}{(1-x)(1-y)}} \\\\\n &= \\frac{1}{(1-x)(1-y) - 2xy} \\\\\n B(x, y) &= \\frac{1}{1 - x - y - xy} = D(x, y)\n\\end{align*}\nSince $b_{n, m}$ and $d_{n, m}$ have same generating functions, $b_{n, m} = d_{n, m}$. Hence $b_{n, m}$ also satisfies recurrence relation of $d_{n, m}$ - \n$$ b_{n, m} = b_{n-1, m} + b_{n, m-1} + b_{n-1, m-1}$$\n\n\n\\subsection{Example 4 : Stirling number of second kind} \nAs discussed in previous lectures, number of ways to partition set $\\{1, 2, 3 \\dots n \\}$ into $k$ non-empty parts is called stirling number of second kind. Let's represent by $S_{n,k}$. It's recurrence relation is given by - \n$$S_{n, k} = S_{n-1, k-1} + k S_{n-1, k} $$\nLHS : number of ways to partition set $\\{1, 2, 3 \\dots n \\}$ into $k$ non-empty parts $ = S_{n, k}$ \\\\\nRHS : \\begin{parts}\n\\item If element $1$ occurs in a singleton set. No.of ways to partition remaining $n-1$ elements to $k-1$ sets $= S_{n-1, k-1}$.\n\\item If element $1$ doesn't occur in a singleton set. Then we can partition remaining $n-1$ elements to $k$ sets and add element $1$ to one of these $k$ sets $= k S_{n-1, k}$\n\\end{parts}\nWe can also see that $S_{0,0} = 1 ,~ S_{n, 0} = 0 ,~ S_{0, k} = 0$.\n%$S_{x, y} = {\\sum_{n, k \\geq 0} ^ {\\infty} S_{n,k} x^n y^k}$ \\\\\n%\\begin{equation}\n\\begin{align*}\n S(x, y) &= {\\sum_{n, k \\geq 0} ^ {\\infty} S_{n,k} x^n y^k} \\\\\n&= S_{0,0}~x^0y^0 + {\\sum_{n = 0, k \\geq 1}^ {\\infty} S_{0,k}~x^0y^k} + {\\sum_{n \\geq 1, k = 0}^ {\\infty} S_{n,0}~x^n y^0} + {\\sum_{n \\geq 1, k \\geq 1}^ {\\infty} S_{n,k}~x^n y^k} \\\\\n &= 1 + \\sum_{n \\geq 1 , k \\geq 1} ^ {\\infty} S_{n,k} ~ x^n y^k \\\\\n &= 1 + \\sum_{n \\geq 1 , k \\geq 1} S_{n-1, k-1}~ x^n y^k + \\sum_{n \\geq 1, k \\geq 1} k S_{n-1, k}~ x^n y^k \\\\\n &= 1 + xy \\sum_{n \\geq 1, k \\geq 1} S_{n-1, k-1} x^{n-1} y^{k-1} + x\\sum_{n \\geq 1, k \\geq 1} k S_{n-1, k}~ x^{n-1} y^k \\\\\n &= 1 + xy~ S(x, y) + x \\sum_{n \\geq 0, k \\geq 1} k S_{n, k}~ x^{n} y^k \\\\\n &= 1 + xy~ S(x, y) + \\frac{\\partial}{\\partial y} S(x,y)\n\\end{align*}\nNote : $\\frac{\\partial}{\\partial y} S(x, y) = \\sum_{n \\geq 0, k \\geq 1} k S_{n,k}~x^n y^{k-1}$ \n\n\nConsider $y^k$ coefficients on both sides : \n\\begin{equation}\n \\begin{split}\n LHS &= \\sum_{n \\geq 0} S_{n, k} x^n \\\\\n RHS &= x\\sum_{n \\geq 0} S_{n, k-1} x^n + xk \\sum_{n \\geq 0} S_{n, k}x^n\n\\end{split} \n\\end{equation}\nEquating LHS and RHS :\n\\begin{align*}\n\\sum_{n \\geq 0} S_{n, k}~ x^n &= x\\sum_{n \\geq 0} S_{n, k-1} ~x^n + xk \\sum_{n \\geq 0} S_{n, k}~x^n \\\\\n\\sum_{n \\geq 0} S_{n, k} ~x^n &= \\frac{x}{1-xk} \\sum_{n \\geq 0} S_{n, k-1}~ x^n \\\\\n&= \\frac{x}{1-xk} \\frac{x}{1-x(k-1)} \\sum_{n \\geq 0} S_{n, k-2}~ x^n \\\\\n&= \\frac{x}{1-xk} \\frac{x}{1-x(k-1)} \\dots \\frac{x}{1-x(k-(k-1))}\n\\sum_{n \\geq 0} S_{n, 0}~ x^n \\\\\n&= \\frac{x^k}{(1-x)(1-2x)\\dots (1-kx)} \\times 1 ~( Note : S_{0,0} = 1, S_{n, 0} = 0) \\\\\n\\sum_{n \\geq 0} S_{n, k}~ x^n &= x^k \\times (\\frac{A_1}{1-x} + \\frac{A_2}{1-2x} + \\dots + \\frac{A_k}{1-kx})\n\\end{align*}\nSolving for $A_1, A_2, \\dots A_k $ we'll get $A_r = (-1)^{k-r} \\frac{r^{k-1}}{(r-1)! (k-r)!}$. \\\\\n$S_{n,k}$ is the coefficient of $x^n$ in RHS. i.e., \n\\begin{align*}\nS_{n,k} &= coeff~ of ~x^n ~in~ x^k \\times (\\frac{A_1}{1-x} + \\frac{A_2}{1-2x} + \\dots + \\frac{A_k}{1-kx}) \\\\\n&= coeff ~ of ~ x^{n-k} ~ in ~ \\sum_{r=1}^k \\frac{A_r}{1-rx}\n\\end{align*}\nCoefficient of $x^{p}$ in $\\frac{1}{1-rx} = r^p$. hence,\n\\begin{align*}\nS_{n,k} &= \\sum_{r=1}^k A_r r^{n-k} \\\\\n&= \\sum_{r=1}^k (-1)^{k-r} \\frac{r^{k-1}}{(r-1)! (k-r)!} ~ r^{n-k}\\\\\nS_{n,k} &= \\sum_{r=1}^k (-1)^{k-r} \\frac{r^{n}}{(r-1)! (k-r)!}\n\\end{align*}\nThe above expression is Stirling number of second kind\n%\\end{equation}\n%$$ = S_{0,0}~x^0y^0 + {\\sum_{n = 0, k \\geq 1}^ {\\infty} S_{0,k}~x^0y^k} + {\\sum_{n \\geq 1, k = 0}^ {\\infty} S_{n,0}~x^ny^0} + {\\sum_{n \\geq 1, k \\geq 1}^ {\\infty} S_{n,k}~x^ny^k} $$ \\\\\n%$&= 1 + \\sum_{\\substack{n \\geq 1 \\\\ k \\geq 1}} ^ {\\infty} S_{n,k} ~ x^n y^k$\n\n\\section{Exponential generating functions}\nIn ordinary generating functions we associate sequence, ${(a_n)}_{n \\geq 0}$ with $G(x) = \\sum_{n \\geq 0} a_n x^n$. In $G(x)$ we chose basis $\\{ 1, x, x^2, x^3 \\dots\\}$ for set of all polynomials in one variable. But there are many other basis for set of polynomials, like $\\{1, x, x(x-1), x(x-1)(x-2), \\dots\\}$. We chose basis $\\{ 1, x, x^2, x^3 \\dots\\}$ because it has combinatorial meaning. Other such meaning full basis are $\\{\\frac{x^n}{n!}\\}_{n \\in \\N}$, $\\{e^{-x} \\frac{x^n}{n!}\\}_{n \\in \\N}$ and $\\{\\frac{1}{n^x}\\}_{n \\in \\N}$. In this lecture we'll explore exponential generating functions which use basis $\\{\\frac{x^n}{n!}\\}_{n \\in \\N}$ . \\\\\nSo ${(a_n)}_{n \\geq 0}$ is associated with $E(x) = \\sum_{n \\geq 0} a_n \\frac{x^n}{n!}$ . Let's see a few examples - \n\\begin{align*}\n(1, 1, 1, \\dots ) &\\xrightarrow[generating function]{exponential} \\sum_{n \\geq 0} \\frac{x^n}{n!} = e^x \\\\\n&\\xrightarrow[generating function]{ordinary} \\sum_{n \\geq 0} x^n = \\frac{1}{1-x} \\\\\n(1!, 2!, 3!, \\dots) &\\xrightarrow[generating function]{exponential} \\sum_{n \\geq 0} n! \\frac{x^n}{n!} = \\sum_{n \\geq 0} x^n = \\frac{1}{1-x}\n\\end{align*}\n$\\frac{1}{1-x}$ is ordinary generating function(ogf) of $(1, 1, 1, \\dots )$ and exponential generating function(egf) of $(1!, 2!, 3!, \\dots )$.\\\\\n\\\\\n\\textbf{\\Large {Operations of EGF}}\n\\begin{parts}\n\\item{\\textbf{Addition :} } It's similar to ogf.\n \\begin{align*}\n \\{a_n\\}_{n \\geq 0} &\\xrightarrow[]{egf} E(x) \\\\\n \\{b_n\\}_{n \\geq 0} &\\xrightarrow[]{egf} F(x) \\\\\n \\{a_n + b_n \\}_{n \\geq 0} &\\xrightarrow[]{egf} E(x) + F(x) \\\\\n \\end{align*}\n\\item {\\textbf{Shifting :} } Multiplying ogf by $x$ shifts the sequence to left as seen in earlier lectures. \n \\begin{align*}\n \\{a_0, a_1, a_2 \\dots \\} &\\xrightarrow[]{egf} E(x) \\\\\n \\{0, a_0, a_1, a_2 \\dots \\} &\\xrightarrow[]{egf} xE(x) \\\\\n \\end{align*}\nDifferentiating egf function will shift the sequence to right.\n\\begin{gather*}\n \\{a_0, a_1, a_2 \\dots \\} \\xrightarrow[]{egf} ~~ E(x) \\endline\n \\{a_1, a_2, a_3 \\dots \\} \\xrightarrow[]{egf} \\frac{d}{dx} E(x) \\\\\n \\frac{d}{dx} E(x) = \\sum_{n \\geq 1} a_n~ \\frac{n. x^{n-1}}{n!} = \\sum_{n \\geq 1} a_n~ \\frac{x^{n-1}}{(n-1)!} = \\sum_{n \\geq 0} a_{n+1}~ \\frac{x^{n}}{n!}\n\\end{gather*}\n\\item {\\textbf{Multiplication :} } EGFs are used if the sequence counts labelled structures like permutations, derangements and partitions. Let $(a_n)_{n \\geq 0}$, $(b_n)_{n \\geq 0}$ count arrangements of type $A$ and type $B$ respectively using $n$ labelled objects. If we want to count type $C$ arrangements, that can be obtained by a unique split of $n$ objects into two sets and then arranging first set according to type $A$ and second set according to type $B$ - \\\\\n No.of arrangements of type $C$ of size $n, c_n = \\sum_{k=0}^n {n \\choose k} a_k b_{n-k}$.\\\\\n Now Let's see how multiplication of $A(x)$(egf of $A$) and $B(x)$(egf of $B$) is useful \n \\begin{align*}\n A(x).B(x) &= (\\sum_{n=0}^{\\infty} a_n \\frac{x^n}{n!})(\\sum_{n=0}^{\\infty} b_n \\frac{x^n}{n!}) \\\\\n &= \\sum_{n=0}^{\\infty}(\\sum_{k=0}^{n} \\frac{a_k}{k!}.\\frac{b_{n-k}}{(n-k)!}) x^n \\\\\n &= \\sum_{n=0}^{\\infty}(\\sum_{k=0}^{n} \\frac{n!}{k!(n-k)!} a_k b_{n-k}) \\frac{x^n}{n!} \\\\\n &= \\sum_{n=0}^{\\infty}(\\sum_{k=0}^{n} {n \\choose k} a_k b_{n-k}) \\frac{x^n}{n!} \\\\\n &= \\sum_{n=0}^{\\infty} c_n \\frac{x^n}{n!} \\\\\n A(x).B(x) &= C(x)\n \\end{align*}\n \n Now Let's see few examples of egf \n \\subsection{Derangements } Recall that we've discussed derangements in PIE and recurrence relations. Now let's derive it using egf. Let $D_n$ represent set of derangements of $n$ objects and let $d_n = |D_n|$. we can see that $d_0 = 1 ,~ d_1 = 0,~ d_2 = 1$. Recall the recurrence relation : $$ d_{n+2} = (n+1)(d_{n+1} + d_n)$$\n \\begin{align*}\n D(x) &= \\sum_{n=0}^{\\infty} d_n \\frac{x^n}{n!} \\\\\n D'(x) &= \\sum_{n=0}^{\\infty} d_{n+1} \\frac{x^n}{n!} ~(by ~shifting ~operation)\\\\\n &= \\sum_{n=1}^{\\infty} n(d_n + d_{n-1}) \\frac{x^n}{n!} \\\\\n &= \\sum_{n=1}^{\\infty} nd_n \\frac{x^n}{n!} + \\sum_{n=1}^{\\infty} nd_{n-1} \\frac{x^n}{n!} \\\\\n &= x \\sum_{n=1}^{\\infty} d_n \\frac{x^{n-1}}{(n-1)!} + x \\sum_{n=1}^{\\infty} d_{n-1} \\frac{x^{n-1}}{(n-1)!} \\\\\n &= x \\sum_{n=0}^{\\infty} d_{n+1} \\frac{x^n}{n!} + x \\sum_{n=0}^{\\infty} d_n \\frac{x^n}{n!} \\\\\n D'(x) &= xD'(x) + xD(x) \\\\\n (1-x)D'(x) &= xD(x) \\\\\n \\frac{D'(x)}{D(x)} &= \\frac{x}{1-x} = \\frac{1}{1-x} - 1\n \\end{align*}\nIntegrating on both sides\n $$\\ln{D(x)} = \\ln{(1-x)} - x + c$$\nSince $D(0) = d_0 = 1 \\implies c=0$.\n $$\\ln{D(x)} = \\ln{(1-x)} - x$$\n $$D(x) = \\frac{e^{-x}}{1-x} = \\sum_{n=0}^{\\infty} d_n \\frac{x^n}{n!}$$\nTo get $d_n$ we need coefficient of $\\frac{x^n}{n!}$ in LHS.\n$$e^{-x} \\frac{1}{1-x} = (\\sum_{n=0}^{\\infty} (-1)^n \\frac{x^n}{n!})(\\sum_{n=0}^{\\infty} n! \\frac{x^n}{n!})$$\nCoefficient of $\\frac{x^n}{n!}$ by multiplication property = $\\sum_{k=0}^{n} {n \\choose k} a_k b_{n-k}$\n\\begin{align*}\n d_n = \\sum_{k=0}^{n} {n \\choose k} (-1)^{n-k} k! \n = \\sum_{k=0}^{n} (-1)^{n-k} k! {n \\choose k}\n\\end{align*}\n$d_n$ is count of derangements of $n$ objects.\n\n\\subsection{Bell Numbers}\nLet $S_{n,k}$ represent number of ways of partitioning \\{$1, 2 ,3 \\dots n $\\} into $k$ non empty blocks and $B_n$ represent number of ways of partitioning \\{$1, 2 ,3 \\dots n $\\} ($B_0 = 1$). By definitions, \n$$B_n = \\sum_{k=0}^n S_{n,k}$$\nEquivalent intrepretation : Consider a number whose prime factorization is square free i.e., $k \\in \\N$ such that $k = p_1p_2 \\dots p_n$ where $\\{p_1, p_2, \\dots p_n\\}$ are distinct primes. Number of ways of writing $k$ as product of natural numbers $\\geq 2 = $ number of ways of partitioning $\\{p_1, p_2, \\dots p_n\\}$ $= B_n$. \\\\\nRecurrence Relation : \n $$ B_n = \\sum_{k=0}^{n-1} {{n-1} \\choose k} B_k $$ \n% LHS : Number ways of partitioning \\{$1, 2 ,3 \\dots n $\\}. \\\\\n% RHS : \\begin{parts}\n% \\item If $1$ occurs in singleton set. No.of such partitions = $B_{n-1}$\n% \\item If $1$ occurs in set with two elements. No.of ways element can be chosen $ = n-1$.No.of such partitions = ${{n-1} \\choose 1}B_{n-2}$.\n% \\item $1$ occurs in set with $k+1$ elements. No.of ways elements in this set can be chosen $ = {n-1 \\choose k}$. No.of such partitions = ${{n-1} \\choose k}B_{n-(k+1)}$.\n \\end{parts}\n% Hence total $= \\sum_{k=0}^{n-1} {{n-1} \\choose k} B_{n-k-1} = \\sum_{k=0}^{n-1} {{n-1} \\choose n-k-1} B_{n-k-1} = \\sum_{n-k-1=0}^{n-1} {{n-1} \\choose k} B_{k} = \\sum_{k=0}^{n-1} {{n-1} \\choose k} B_{k} $ \\\\\n Let's derive closed form expression for $B_n$ :\n \\begin{align*}\n B(x) &= \\sum_{n=0}^{\\infty} B_n \\frac{x^n}{n!} \\\\\n B'(x) &= \\sum_{n=0}^{\\infty} B_{n+1} \\frac{x^n}{n!} ~(by ~ shifting ~rule) \\\\\n &= \\sum_{n=0}^{\\infty} (\\sum_{k=0}^{\\n} {n \\choose k} B_k) \\frac{x^n}{n!} \\\\\n &= \\sum_{n=0}^{\\infty} (\\sum_{k=0}^{\\n} {n \\choose k}. ~1 B_k) \\frac{x^n}{n!} \\\\\n &= (\\sum_{n=0}^{\\infty} B_n \\frac{x^n}{n!})(\\sum_{n=0}^{\\infty} 1 \\frac{x^n}{n!}) ~(by ~ multiplication ~rule) \\\\\n B'(x) &= B(x) . e^x \\\\\n \\frac{B'(x)}{B(x)} &= e^x\n \\end{align*}\n Integrating on both sides \n $$\\ln{B(x)} = e^x + c$$\n Since $B(0) = B_0 = 1 \\implies c = -1$. Hence,\n %$$ B(x) = e^{e^x - 1} = \\frac{e^{e^x}}{e}$$\n \\begin{align*}\n B(x) &= e^{e^x - 1} \\\\\n &= \\frac{e^{e^x}}{e} \\\\\n &= \\frac{1}{e} (\\sum_{k=0}^{\\infty} \\frac{(e^x)^k}{k!}) \\\\\n &= \\frac{1}{e} (\\sum_{k=0}^{\\infty} \\frac{e^{kx}}{k!}) \\\\\n &= \\frac{1}{e} (~\\sum_{k=0}^{\\infty} \\frac{1}{k!}~ (\\sum_{n=0}^{\\infty} \\frac{(kx)^n}{n!})~) \\\\\n &= \\frac{1}{e} (~\\sum_{n=0}^{\\infty} \\frac{x^n}{n!}~ (\\sum_{k=0}^{\\infty} \\frac{k^n}{k!})~)\\\\\nB(x)&= \\sum_{n=0}^{\\infty} \\frac{1}{e} (\\sum_{k=0}^{\\infty} \\frac{k^n}{k!}) \\frac{x^n}{n!} \\\\\n B_n &= \\frac{1}{e} (\\sum_{k=0}^{\\infty} \\frac{k^n}{k!})\n \\end{align*}\n We've derived closed form expression for bell number. Above expression for $B_n$ is also called as Dobinski's formula.\n \n\\end{parts}\n", "meta": {"hexsha": "d4fd87d0b74cd2fdd231cba4a22a3bda5231c3f3", "size": 14054, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week06_part2.tex", "max_stars_repo_name": "pot8ohead/theory-toolkit", "max_stars_repo_head_hexsha": "177249454691f7e264d9ee7d5a354e180a54e095", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "week06_part2.tex", "max_issues_repo_name": "pot8ohead/theory-toolkit", "max_issues_repo_head_hexsha": "177249454691f7e264d9ee7d5a354e180a54e095", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "week06_part2.tex", "max_forks_repo_name": "pot8ohead/theory-toolkit", "max_forks_repo_head_hexsha": "177249454691f7e264d9ee7d5a354e180a54e095", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 66.9238095238, "max_line_length": 651, "alphanum_fraction": 0.5472463356, "num_tokens": 6389, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.782662489091802, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.6166841908149476}} {"text": "\\chapter{MIP Heuristics}\nMixed Integer Program Heuristics can be applied in each MIP problem independently from the context. These do not consider the specific formulation of the TSP problem however they could have good applicability. In this report two MIP Heuristics will be discussed and compared with the metaheuristics.\n\n\\section{Hard-fixing}\nA simple idea to reduce the complexity of the problem is to get an initial solution like an incumbent, fix some active edges and resolve the subproblem with CPLEX. This is exactly the idea behind hard fixing approach.\\\\\nThe implementation proposed is applied in sTSP problem, with the optimization of the general callback (\\texttt{subtour\\_callback\\_general}).\nThe algorithm can be divided in steps:\n\\begin{enumerate} \\label{hard-fix-step}\n\t\\item Calculate an initial solution: in our implementation this is done by CPLEX with \\texttt{subtour\\_callba \\\\ ck\\_general} and \\texttt{CPX\\_PARAM\\_INTSOLLIM} set to 1. When the first incumbent is available, the optimization terminate and a solution is obtained.\n\t\\item Fix a percentage of edges (fixing rate $ f_r $): edges are fixed using \\texttt{CPXchgbds} method that change the upper and lower bound of the decision variables. The fixing percentage is an important parameter: fixing too much edges leads to a fast resolution of the subproblem, however increase the risk of obtaining the same solution and fall in a loop. Fixing too low edges involves slower resolution. In the first iteration $ f_r = 0.9 $.\n\t\\item CPLEX optimization with time limit: After the fixing phase, the problem is optimized with \\texttt{CPX\\_mipopt} and a short time limit is set (in our implementation $50 $s).\n\t\\item Fixing rate update: after the last phase, if the returned solution is improved better than a fixed gap ($ good\\_gap $), it is considered a good solution and the $ f_r $ increases by a constant ($ incr\\_f_r $) until a max ($ max\\_f_r $), otherwise if the new solution is not increased enough (less than $ optimal\\_gap $), then $ f_r $ decreases of a constant ($ decr\\_f_r $) until a min ($ min\\_f_r $). Note that $ good\\_gap $ and $ optimal\\_gap $ are expressed as fraction of the current best lower bound.\n\t\\item Check end condition: if the time limit is reached or $ f_r = 0.0 $ and the solution is not improved in the last iteration, then the solution is returned. In the second case, the best solution is found. If no ending condition is satisfied, the algorithm continue with point 3.\n\\end{enumerate}\n\n\\section{Local-Branching}\nThe Local Branching is a relatively recent algorithm developed by Matteo Fischetti and Andrea Lodi, who wanted to propose an alternative method to Hard Fixing. The Hard Fixing idea, as stated above, fix randomly a percentage of edges and optimize the problem, Local Branching, instead, let the MIP model to fix the percentage of edges automatically and than optimize. \\\\\nConsider $ x^H $ as non optimal tour and $ \\chi $ the fraction of edges to fix, Local Branching add\n\\begin{equation}\n \\sum_{ e\\in E: x_e^H = 1 } x_e \\ge \\chi n\n\\end{equation}\nto automatically fix the desired number of edges. \\\\\nIn this paper, the structure of the algorithm is very similar to that of Hard Fixing. Initialization is analogous to Point 1 (\\ref{hard-fix-step})\nIn the CPLEX optimization with time limit (Point 3), the time limit is a little higher and is set to 300 according to the choice of the fixing rate update. Thus, $ \\chi n $ assume values in $ \\{3, 5, 10, 15, 20\\}$. It starts from the first value of 3 which is kept until better solutions are found. It is updated to the next value as soon as the algorithm returns a non-improving solution. This choice is motivated by the idea that keeping a low fixing rate allows you to find a solution in a short time and therefore not reach the time limit, which instead is set with a higher value to allow a longer search when the value increase. If the last value is reached and there is still no improvement solution, then the value is doubled until either a better solution is found or the number of nodes of the problem is reached. In the latter case, it is assigned the entire remaining time to find the best tour. In fig. \\ref{fig:local_branching_a280} there is an example of the lower bound trend executed on \\textit{a280.tsp} instance.\\\\\n\\begin{figure}[!h]\n\t\\centering\n\t\\includegraphics[width=0.5\\columnwidth]{../res/local_branching_a280.png}\n\t\\caption{Solution cost profile.}\n\t\\label{fig:local_branching_a280}\n\\end{figure}\n\nThe exit conditions evaluate both the time limit and the gap calculated by CPLEX, obtainable through the \\texttt{generic\\_callback} using \\texttt{CPX\\_CALLBACKCONTEXT\\_GLOBAL\\_PROGRESS} as \\texttt{Context}. \nFor a detailed explanation of the reasons for this algorithm refer to \\cite{article}.\n\n\n\\begin{figure}[h]\n\t\\centering\n\t\\begin{subfigure}{\\columnwidth}\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lmip_meta_LA_time.png}\n\t\t\\caption{}\n\t\t\\label{fig:Lmip_meta_LA_time}\n\t\\end{subfigure}\n\t\\begin{subfigure}{\\columnwidth}\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lmip_meta_LA_lb.png}\n\t\t\\caption{}\n\t\t\\label{fig:Lmip_meta_LA_lb}\n\t\\end{subfigure}\n\\caption{Comparison of mip heuristics and meta heuristics. \\texttt{n\\_greedy\\_best\\_two\\_opt} is the first part of \\texttt{vns} therefore it is faster, however the last one introduce a little improvement in the solutions cost. Most of the times \\texttt{hard\\_fixing} can find the best solution of sTSP, indeed it find the shorter tours of the analyzed heuristics. }\n\\label{fig:Lmip_meta_LA}\n\\end{figure}\n\n%\\section{RINS}\n%\\section{Feasibility Pump}\n%\\section{Proximity Search}\n%\\section{Polishing}\n", "meta": {"hexsha": "86f97317dafcd8badff55b0bcceef84d157b451f", "size": 5639, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Latex/mip-heuristics.tex", "max_stars_repo_name": "Fisher4537/OR2", "max_stars_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Latex/mip-heuristics.tex", "max_issues_repo_name": "Fisher4537/OR2", "max_issues_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Latex/mip-heuristics.tex", "max_forks_repo_name": "Fisher4537/OR2", "max_forks_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 100.6964285714, "max_line_length": 1033, "alphanum_fraction": 0.7767334634, "num_tokens": 1427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6166841868205838}} {"text": "\\paragraph{Question 3.}\n\nTry to simplify the regular expressions\n\\(\\lparen \\epsilon \\, \\disjM{} \\, a\\kleeneM{} \\, \\disjM{} \\,\n b\\kleeneM{} \\, \\disjM{} \\, a \\, \\disjM{} \\, b \\rparen\\kleeneM\\) and\n\\(a \\lparen a \\, \\disjM{} \\, b \\rparen\\kleeneM{} b \n \\, \\disjM{} \\, \\lparen a b \\rparen\\kleeneM{} \\, \\disjM{} \\, \\lparen\n b a \\rparen\\kleeneM\\).\n", "meta": {"hexsha": "756c69484a69a20da7d45e8c67a7aa60196e1a39", "size": 343, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "regexp_question_03.tex", "max_stars_repo_name": "rinderknecht/Book", "max_stars_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "regexp_question_03.tex", "max_issues_repo_name": "rinderknecht/Book", "max_issues_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "regexp_question_03.tex", "max_forks_repo_name": "rinderknecht/Book", "max_forks_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.1111111111, "max_line_length": 69, "alphanum_fraction": 0.5801749271, "num_tokens": 156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6166841868205838}} {"text": "\\vsssub\n\\subsubsection{~$S_{ln}$: Cavaleri and Malanotte-Rizzoli 1981} \\label{sec:LN1}\n\\vsssub\n\n\\opthead{LN1}{Pre-WAM}{H. L. Tolman}\n\n\\noindent \nA linear input source term is useful to allow for the consistent spin-up of a\nmodel from quiescent conditions, and to improve initial wave growth\nbehavior. The parameterization of \\cite{art:CMR81} is available in \\ws, with a\nfilter for low-frequency energy as introduced by \\cite{tol:JPO92}. The input\nterm can be expressed as\n\\begin{equation}\n\\cS_{lin}(k,\\theta) = 80 \\left( \\frac{\\rho_a}{\\rho_w} \\right ) ^2\n g^{-2} k^{-1} \\max \\left [ 0 , u_* \\cos (\\theta - \\theta_w) \\right ]^4 \\: G\n \\:\\:\\: , \\label{eq:CMR81}\n\\end{equation}\n\n\\noindent\nwhere $\\rho_a$ and $\\rho_w$ are the densities of air and water, respectively,\nand where $G$ is the filter function\n\n\\begin{equation}\nG = \\exp \\left [ - \\left ( \\frac{f}{f_{filt}} \\right ) ^{-4} \\right ]\n\\:\\:\\: . \\label{eq:GSln}\n\\end{equation}\n\n\\noindent\nIn \\cite {tol:JPO92} the filter frequency $f_{filt}$ was given as the\nPierson-Moskowitz frequency $f_{PM}$, which in turn was estimated as in\nEq.~(\\ref{eq:f_PM}). In the present implementation, the filter can be related\nto both $f_{PM}$ and the cut-off frequency of the prognostic part of the\nspectrum $f_{hf}$ as defined in Eq.~(\\ref{eq:tail_E_f})\n\n\\begin{equation}\nf_{filt} = \\max \\left [ \\alpha_{PM} f_{PM} , \\alpha_{hf} f_{hf} \\right ]\n\\:\\:\\: ,\n\\end{equation}\n\n\\noindent\nwhere the constants $\\alpha_{PM}$ and $\\alpha_{hf}$ are user-defined. Default\nvalues of these constants are set to $\\alpha_{PM} = 1$ and $\\alpha_{hf} =\n0.5$. Addition of the dependency on $f_{hf}$ assures consistent growth\nbehavior at all fetches, without the possibility of low-frequency linear\ngrowth to dominate at extremely short fetches.\n\n", "meta": {"hexsha": "8c9a3bb9adeea1b7d94f7aba654d60ecbae99376", "size": 1764, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "WW3/manual/eqs/LN1.tex", "max_stars_repo_name": "minsukji/ci-debug", "max_stars_repo_head_hexsha": "3e8bbbe6652b702b61d2896612f6aa8e4aa6c803", "max_stars_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "WW3/manual/eqs/LN1.tex", "max_issues_repo_name": "minsukji/ci-debug", "max_issues_repo_head_hexsha": "3e8bbbe6652b702b61d2896612f6aa8e4aa6c803", "max_issues_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2021-05-31T15:49:26.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-04T14:17:45.000Z", "max_forks_repo_path": "WW3/manual/eqs/LN1.tex", "max_forks_repo_name": "minsukji/ci-debug", "max_forks_repo_head_hexsha": "3e8bbbe6652b702b61d2896612f6aa8e4aa6c803", "max_forks_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-06-01T09:29:46.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-01T09:29:46.000Z", "avg_line_length": 37.5319148936, "max_line_length": 78, "alphanum_fraction": 0.701814059, "num_tokens": 581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8902942173896131, "lm_q2_score": 0.6926419894793248, "lm_q1q2_score": 0.6166551579546801}} {"text": "\\section{Sets and Functions}\r\n\\subsection{Naive Set Theory}\r\n\\begin{definition}[Naive definition of sets]\r\n A set is any\r\n \\footnote{Not quite, see later.}\r\n collection of (mathematical) objects.\r\n\\end{definition}\r\n\\begin{example}\r\n $\\mathbb N,\\mathbb Q,\\mathbb R,\\mathbb C,\\mathbb Z,[0,1],\\{1,2,3\\},\\ldots$\r\n\\end{example}\r\nA set is determined by its members: $(a\\in A\\iff a\\in B)\\iff(A=B)$.\r\nHence, the set is not ordered per se: $\\{1,3,7\\}=\\{7,1,3\\},\\{1,3,3,7\\}=\\{1,3,7\\}$.\r\nFrom the second example, we also know that there is no multiple membership.\\\\\r\nOf course, we would want to make new sets from old.\r\n\\begin{definition}\r\n A set $A$ is a subset of another set $B$ iff $x\\in A\\implies x\\in B$.\r\n We write it as $A\\subset B$ or $A\\subseteq B$.\r\n\\end{definition}\r\nSo $A=B$ is equivalent to $A\\subset B$ and $B\\subset A$.\r\n\\begin{example}\r\n $\\{1,7\\}\\subset \\{1,3,7\\}\\subset\\mathbb N\\subset\\mathbb Z\\subset \\mathbb Q\\subset\\ldots, [0,1]\\subset\\mathbb R$.\r\n\\end{example}\r\n\\begin{definition}\r\n Given a set $A$ and a property $P$ on $A$, we can form the set of all elements in $A$ having property $P$, $\\{x\\in A:P(x)\\}$\r\n\\end{definition}\r\n\\begin{example}\r\n We can construct the primes by $\\{n\\in\\mathbb N:\\text{$n$ is prime}\\}$.\r\n\\end{example}\r\n\\begin{definition}\r\n Given sets $A,B,U$ such that $A,B\\subset U$, we can construct their union by\r\n $$A\\cup B=\\{x\\in U:x\\in A\\lor x\\in B\\}$$\r\n and intersection by\r\n $$A\\cap B=\\{x\\in U:x\\in A\\land x\\in B\\}$$\r\n We say $A,B$ are disjoint if $A\\cap B=\\varnothing$.\\\\\r\n We also have the minus action\r\n $$A\\setminus B=\\{x\\in A:x\\notin B\\}$$\r\n\\end{definition}\r\nWe can view $A\\cap B$ as a subset selection, since $A\\cap B=\\{x\\in A:x\\in B\\}$.\r\nUnions and intersections are commutative and associatives.\r\nAlso, the union is distributive over the intersection: $A\\cup (B\\cap C)=(A\\cup B)\\cap (A\\cup C)$.\r\nThe intersection is distributive over the union as well: $A\\cap (B\\cup C)=(A\\cap B)\\cup (A\\cap C)$.\r\nOne can easily check them.\\\\\r\nWe can also have arbitrary unions and intersections.\r\n\\begin{definition}\r\n For $A_i\\subset U$ where $i\\in I$ for some index set $I$, the union of all $A_i$'s can be defined by\r\n $$\\bigcup_{i\\in I}A_i=\\{x\\in U:\\exists i\\in I,x\\in A_i\\}$$\r\n the intersection by\r\n $$\\bigcap_{i\\in I}A_i=\\{x\\in U:\\forall i\\in I,x\\in A_i\\}$$\r\n\\end{definition}\r\nIt coincides with our previous definition in the finite case due to the associativity and commutativity of unions and intersections.\r\n\\begin{definition}\r\n In $\\mathbb R$, let $A_n=[1-1/n,1+1/n],n\\in\\mathbb N$, so $\\bigcup_{n\\in\\mathbb N}A_n=[0,2],\\bigcap_{n\\in N}A_n=\\{1\\}$, note that there is no limiting operation going on here.\\\\\r\n Let let $A_n=(1-1/n,1+1/n),n\\in\\mathbb N$, so $\\bigcup_{n\\in\\mathbb N}A_n=(0,2),\\bigcap_{n\\in N}A_n=\\{1\\}$\r\n\\end{definition}\r\n\r\n\\begin{definition}\r\n Given two objects $a,b$, you can form the ordered pair $(a,b)$,\r\n And that $(a,b)=(c,d)$ iff $a=c,b=d$.\r\n For sets $A,B$, one can form the collection $A\\times B=\\{(a,b):a\\in A,b\\in B\\}$, which is called the product, or Cartesian product of $A,B$.\r\n\\end{definition}\r\n\\begin{example}\r\n The plane $\\mathbb R^2$ can be viewed as $\\mathbb R\\times\\mathbb R$.\r\n\\end{example}\r\nSimilarly, we can construct things like $\\mathbb R^n$ by recognising the collection of all ordered $n$-tuples.\\\\\r\nNote that if we wish to do it, we could define $(a,b)$ by the set $\\{\\{a\\},\\{a,b\\}\\}$.\r\nOne can check that $(a,b)=(c,d)\\iff a=c,b=d$ under this notion.\r\n\\footnote{Quoting the lecturer, ``You're completely nuts if you really think of this in that way'}\r\n\\begin{definition}\r\n For any set $A$, we can form the power set of $A$, written as $\\mathbb P(A)$ or $2^A$,\r\n \\footnote{Different from the lecturer, the author of this set of notes prefers the latter notation.}\r\n which is set of all subsets of $A$.\r\n\\end{definition}\r\n\\begin{example}\r\n Let $X=\\{1,2\\}$, then $2^X=\\{\\varnothing, \\{1\\},\\{2\\},\\{1,2\\}\\}$.\r\n\\end{example}\r\n\\begin{remark}\r\n Warning: Given a set $A$, we know we can form $\\{x\\in A:P(x)\\}$, but you should not form $\\{x:P(x)\\}$.\r\n Suppose we could form $X=\\{x:x\\notin x\\}$, then do we have $X\\in X$?\r\n Indeed, this gives a contradiction.\\\\\r\n This is called the Russel's Paradox.\r\n\\end{remark}\r\nSimilarly, there is not an universal set $U$ such that $\\forall x,x\\in V$, because it would induce the Russel's Paradox.\r\nWe can only guarantee that a given set exists if it is obtained, in some way, from known sets.\r\n\\subsection{Finite Sets and Their Sizes}\r\n\\begin{definition}\r\n A set $A$ has size $n$ where $n\\in\\mathbb N_0$ if we can write $A=\\{a_1,a_2,\\ldots,a_n\\}$ such that $a_i$'s are distinct.\r\n\\end{definition}\r\n\\begin{example}\r\n 1. $\\{1,3,7\\}$ has size $3$.\\\\\r\n 2. $\\varnothing$ has size $0$.\r\n\\end{example}\r\n\\begin{proposition}\r\n If a set has size $n$ and size $m$, then $m=n$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n It's trivial but whatever.\\\\\r\n Indeed, if $A$ has size $n$ and size $m$ with $n>m$, then if $m=0$ it is trivial, otherwise, removing an element from $A$ gives a set of sizes $n-1,m-1$, so it is done by induction on $m$.\r\n\\end{proof}\r\n\\begin{proposition}\\label{2power}\r\n If $A$ has size $n$, then $2^A$ has size $2^n$.\r\n\\end{proposition}\r\nIt is obvious when $n=0$, so we assume henceforth it is not.\r\n\\begin{proof}\r\n We can relabel the element in $A$ by $\\{1,2,\\ldots,n\\}$.\r\n So to specify a subset $S$, we must specify if $1\\in S$, $2\\in S$, and so on, so the size of $2^A$ is $2\\times 2\\times 2\\times\\cdots\\times 2$ $n$ times, which is $2^n$.\r\n\\end{proof}\r\n\\begin{proof}[Alternative proof]\r\n Induction on $n$.\r\n\\end{proof}\r\nThe alternative proof can be viewed as a more formal version of the first proof.\\\\\r\nWe are tired of saying ``the size of blah blah blah'' so we write $|A|$ to denote the size of $A$.\\\\\r\nA set of size $n$ is sometimes called an $n$-set.\r\n\\begin{definition}[Binomial Coefficients]\r\n Let $A=\\{1,2,\\ldots,n\\}$ for $n\\ge 1$, then the binomial coefficient is defined as\r\n $$\\binom{n}{k}=\\{S\\subset A:|S|=k\\}$$\r\n is the number of ways to choose a $k$-set from an $n$-set.\r\n\\end{definition}\r\n\\begin{example}\r\n $$\\binom{4}{2}=6$$\r\n by listing.\r\n\\end{example}\r\nWe always have $\\binom{n}{n}=\\binom{n}{0}=1,\\binom{n}{1}=n$ and $\\binom{n}{k}=\\binom{n}{n-k}$.\r\nAlso, by Proposition \\ref{2power}, we instantly have\r\n$$\\binom{n}{0}+\\binom{n}{1}+\\cdots+\\binom{n}{n}=2^n$$\r\nIn addition,\r\n$$\\binom{n}{k}=\\binom{n-1}{k-1}+\\binom{n-1}{k}$$\r\nsince both size counted the number of $k$-sets in an $n$-set, that is, the number of $k$-sets which include some specified element and the number of $k$ sets which do not include.\r\nHence we have the Pascal's triangle.\r\n\\begin{proposition}\r\n $$\\binom{n}{k}=\\frac{n(n-1)(n-2)\\cdots (n-k+1)}{k!}$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n We first count the number of ordered $k$-sets is $n(n-1)(n-2)\\cdots (n-k+1)$.\r\n But here we have overcounted each $k$-sets by $k(k-1)(k-2)\\cdots 1=k!$, therefore the formula.\r\n\\end{proof}\r\n\\begin{corollary}\r\n When $n$ is big, $\\binom{n}{k}\\sim n^k/k!$.\r\n\\end{corollary}\r\nAn appplication of the binomial coefficient is the binomial theorem.\r\n\\begin{theorem}[Binomial Theorem]\r\n $$(a+b)^n=\\sum_{k=0}^n\\binom{n}{k}a^{n-k}b^k$$\r\n\\end{theorem}\r\n\\begin{proof}\r\n When we expand $(a+b)^n=(a+b)(a+b)\\cdots (a+b)$ where there are altogether $n$ brackets, we obtain terms of the form $a^{n-k}b^{k}$ where $0\\le k\\le n$.\r\n But the number of terms that we get in each is the number of ways to choose $k$ (or equivalently, $n-k$) brackets from the $n$ brackets, so it is $\\binom{n}{k}$, thus the theorem.\r\n\\end{proof}\r\nIn particular, $(1+x)^n=1+nx+\\binom{n}{2}x^2+\\cdots +x^n$, hence for given $n$, $(1+x)^n\\sim 1+nx$ when $x$ is small, and we can get better approximations when we take more term(s).\\\\\r\nNow, how do sizes of unions and intersections (for finite sets) relate to each other?\r\n\\begin{example}\r\n We have $|A\\cup B|=|A|+|B|-|A\\cap B|$, and $|A\\cup B\\cup C|=|A|+|B|+|C|-|A\\cap B|-|B\\cap C|-|C\\cap A|+|A\\cap B\\cap C|$.\r\n\\end{example}\r\n\\begin{theorem}[Inclusion/Exclusion Principle]\r\n For finite sets $(S_i)_{i=1}^n$, $|S_1\\cup S_2\\cup\\cdots\\cup S_n|$ equals\r\n $$\\sum_i|S_i|-\\sum_{i|B|$, $f$ cannot be injective.\r\n If $|A|<|B|$, $f$ cannot be surjective.\\\\\r\n If $|A|=|B|$, then surjectivity, injectivity and bijectivity are equivalent.\r\n Thus any $f:A\\to A$ cannot biject $A$ to a proper subset $A'\\subsetneq A$.\r\n But this is not necessarily true for infinite (i.e. not finite) sets, for example, the ``adding-$1$'' function is a bijection $\\mathbb N\\to\\mathbb N\\setminus\\{1\\}$ (assuming $\\mathbb N$ starts with $1$), this also shows that injectivity does not imply surjective.\r\n Also the function $\\mathbb N\\to\\mathbb N$ by $1\\mapsto 1$ and $n\\mapsto n-1$ for $n\\ge 2$ is surjective but not injective.\r\n\\end{remark}\r\nWe have more examples of functions.\r\n\\begin{example}\r\n 1. For any set $A$, we have the identity function $A\\to A$ which sends each element to itself\r\n It is a bijection.\\\\\r\n 2. Given a set $X$ and a subset $A\\subset X$, the function $\\chi_A:X\\to\\{0,1\\}$ by $\\chi_A(x)=1$ if $x\\in S$ and $\\chi_A(x)=0$ otherwise is a function.\r\n This is called the indicator (or characteristic) function of $A$.\\\\\r\n 3. A sequence on a set $X$ is a function $\\mathbb N\\to X$.\\\\\r\n 4. The addition and multiplication, say on $\\mathbb N$, are functions $\\mathbb N\\times\\mathbb N\\to\\mathbb N$.\\\\\r\n 5. A finite set $A$ has set $m$ if and only if there is a bijection $\\{1,2,3,\\ldots,m\\}\\to A$.\r\n\\end{example}\r\n\\begin{definition}\r\n Let $f:A\\to B,g:B\\to C$, then the composition $g\\circ f$ is a function $A\\to C$ defined by $(g\\circ f)(a)=g(f(a))$.\r\n\\end{definition}\r\nNote that function compositions are not necessarily commutative (if $A=C$).\r\n\\begin{example}\r\n If $A=B=C=\\mathbb R$ and $f: x\\mapsto 2x,g: x\\mapsto x+1$, then $f\\circ g: x\\mapsto 2x+2,g\\circ f:x\\mapsto 2x+1$, which are indeed different.\r\n\\end{example}\r\nHowever, function compositions are associative.\r\n\\begin{proposition}\r\n If $f:A\\to B,g:B\\to C,h:C\\to D$, then $h\\circ (g\\circ f)=(h\\circ g)\\circ f$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n $\\forall x\\in A, (h\\circ (g\\circ f))(x)=h(g(f(x)))=((h\\circ g)\\circ f)(x)$.\r\n\\end{proof}\r\n\\begin{definition}\r\n $f:A\\to B$ is invertible if and only if there is some other function $g:B\\to A$ such that $f\\circ g=\\operatorname{id}_B,g\\circ f=\\operatorname{id}_{A}$.\r\n We say $g$ is the inverse of $f$.\r\n Note that $g$ is invertible as well with inverse $f$.\r\n\\end{definition}\r\n\\begin{example}\r\n Consider $f:\\mathbb R\\to\\mathbb R$ by $f(x)=2x+1$, consider $g:\\mathbb R\\to\\mathbb R$ by $g(x)=(x-1)/2$, then $\\forall x\\in\\mathbb R,(f\\circ g)(x)=(g\\circ f)(x)=x$, thus $f$ is invertible and $g$ is the inverse of $f$.\r\n\\end{example}\r\n\\begin{remark}\r\n Note that we have to check both sides.\r\n Consider $f_0:\\mathbb N\\to\\mathbb N$ by $f_0(x)=x+1$ and $f_1:\\mathbb N\\to\\mathbb N$ by $f_1(x)=x-1$ for $x>1$ and $f_1(1)=1$, then $f_1\\circ f_0=\\operatorname{id}_{\\mathbb N}$ but $f_0\\circ f_1\\neq\\operatorname{id}_{\\mathbb N}$.\r\n\\end{remark}\r\n\\begin{proposition}\r\n A function is invertible if and only if it is a bijection.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\n\\subsection{Equivalence Relations}\r\n\\begin{definition}\r\n Let $A$ be a set, then a relation $R$ is a subset of $A\\times A$.\r\n We say $x,y\\in A$ are related, or $xRy$, if $(x,y)\\in R$.\r\n\\end{definition}\r\n\\begin{example}\r\n 1. On $\\mathbb N$, the relation $aRb\\iff a\\equiv b\\pmod{7}$ is a relation.\\\\\r\n 2. On $\\mathbb N$, $aRb\\iff a|b$ is a relation.\\\\\r\n 3. On any set, $aRb\\iff a\\neq b$ is a relation.\\\\\r\n 4. On $\\mathbb N$, $aRb\\iff a=b\\pm 1$.\\\\\r\n 5. On $\\mathbb N$, $aRb\\iff |a-b|\\le 2$.\\\\\r\n 6. On $\\mathbb N$, $aRb\\iff (a,b<5\\lor a,b\\ge 5)$.\r\n\\end{example}\r\n\\begin{definition}\r\n A relation $R$ is reflexive iff $\\forall x\\in A, xRx$.\r\n\\end{definition}\r\nThe first, second, fifth and sixth examples above are reflexive relations.\r\n\\begin{definition}\r\n A relation $R$ is symmetric iff $\\forall x,y\\in A, xRy\\implies yRx$\r\n\\end{definition}\r\nThe first, third, fourth, fifth and sixth examples above are symmetric relations.\r\n\\begin{definition}\r\n A relation $R$ is transitive iff $\\forall x,y,z\\in A, (xRy\\land yRz)\\implies xRz$\r\n\\end{definition}\r\nThe first, second and sixth exampels above are transitive relations.\r\n\\begin{definition}\r\n We say $R$ is an equivalence relation if it is reflexive, symmetric and transitive.\r\n\\end{definition}\r\nThe first and sixth relations above are equivalence relations.\r\n\\begin{example}\\label{partition}\r\n Let $X$ be a set, then consider a partition $\\{C_i\\}_{i\\in I}$ of $X$.\r\n That is, $C_i\\neq \\varnothing$, $i\\neq j\\implies C_i\\cap C_j=\\varnothing$ and $X=\\bigcup_{i\\in I}C_i$.\r\n Then $xRy\\iff \\exists i\\in I, x,y\\in C_i$ is a equivalence relation.\r\n\\end{example}\r\n\\begin{proposition}\\label{eq_class}\r\n All equivalence relations on $X$ can be written in the form of Example \\ref{partition}.\r\n\\end{proposition}\r\n\\begin{definition}\r\n For an equivalence relation $R$ on a set $X$ and $x\\in X$, the equivalence class containing $x$, written as $C_x$ or $[x]$, is $\\{y\\in X:xRy\\}=\\{y\\in X:yRx\\}$.\r\n\\end{definition}\r\n\\begin{example}\r\n In the first example, $[2]=6+7\\mathbb Z=[16]=[23]=\\ldots$.\r\n Note that we have $7$ equivalence classes in total, namely $i+7\\mathbb Z,i\\in\\{0,1,2,3,4,5,6\\}$.\r\n\\end{example}\r\n\\begin{proof}[Proof of Proposition \\ref{eq_class}]\r\n let $X$ be the set with the equivalence relation $R$.\r\n We shall show that the equivalence classes partitions $X$.\r\n Note that $x\\in [x]$, so their union is $X$, so it remains to show that different equivalence classes are disjoint.\r\n If $y\\in [x]$ and $y\\in [z]$, then for any $w\\in [x]$, we know $wRx,xRy$, so $wRy$, but $yRz$, so $wRz$, therefore $w\\in [z]$, hence $[x]\\subset [z]$.\r\n Similarly $[z]\\subset [x]\\implies [x]=[z]$.\r\n\\end{proof}\r\nHence equivalence relation is just a partition of the set.\r\n\\begin{definition}\r\n The collection $X/R=\\{[x]:x\\in X\\}$ is called the quotient of $X$ with respect to an equivalence relation $R$.\r\n\\end{definition}\r\n\\begin{example}\r\n 1. $xRy\\iff x\\equiv y\\pmod{7}$ is an equivalence relation on $\\mathbb Z$.\r\n It gives the quotient $\\mathbb Z/R=\\{i+7\\mathbb Z:i\\in \\{0,1,2,3,4,5,6\\}\\}$.\\\\\r\n 2. $(a,b)R(c,d)\\iff ad=bc$ is an equivalence relation on $\\mathbb Z\\times\\mathbb N$.\r\n So we could construct $\\mathbb Q$ by $\\mathbb Q=(\\mathbb Z\\times\\mathbb N)/R$.\r\n The equivalence classes are like, for example, $[(1,2)]=\\{(1,2),(2,4),(4,8),\\ldots\\}$.\r\n In that way we can recognize $a/b=[(a,b)]$.\r\n\\end{example}\r\n\\begin{definition}\r\n The quotient map (or projection map) $q:X\\to X/R$ is defined by $q(x)=[x]$.\r\n\\end{definition}\r\nThis is well-defined since equivalence classes partitions the set $X$.\r\n", "meta": {"hexsha": "d15097efa272cc4b7f49ee467726814e39189f90", "size": 18229, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6/set.tex", "max_stars_repo_name": "david-bai-notes/Numbers-and-Sets", "max_stars_repo_head_hexsha": "2c8ca0c4983c1d575b5f55f2a91d34d6ef534845", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T21:17:29.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T21:17:29.000Z", "max_issues_repo_path": "6/set.tex", "max_issues_repo_name": "david-bai-notes/Numbers-and-Sets", "max_issues_repo_head_hexsha": "2c8ca0c4983c1d575b5f55f2a91d34d6ef534845", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6/set.tex", "max_forks_repo_name": "david-bai-notes/Numbers-and-Sets", "max_forks_repo_head_hexsha": "2c8ca0c4983c1d575b5f55f2a91d34d6ef534845", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.5047318612, "max_line_length": 268, "alphanum_fraction": 0.6491305063, "num_tokens": 6490, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.7905303211371898, "lm_q1q2_score": 0.6166080379226071}} {"text": "\\lab{Applications}{Wave Phenomena}{Wave Phenomena}\n\\label{lab:waveeqn}\n\n\\section*{Advection Equation}\nThe advection equation (or transport equation) is given by $u_t + s u_x = 0$, where $s$ is a nonzero constant.\nConsider the Cauchy problem\n\\begin{align*}\n\t& u_t + su_x = 0, \\quad -\\infty < x < \\infty,\\\\\\\n\t& u(x,0) = f(x).\n\\end{align*}\nThe function $f(x)$ may be thought of as an initial wave or signal.\nThe general solution of this initial boundary value problem is $u(x,t) = f(x-st)$ (check this!).\nThe solution $u(x,t)$ is a travelling wave that takes the signal $f(x)$ and moves it along at a constant speed $s$ - to the right if $s > 0$, and to the left if $s < 0$.\n\n\\section*{Wave Equation}\nMany different wave phenomena can be described using a hyperbolic PDE called the wave equation.\nThese wave phenomena occur in fields such as electromagnetics, fluid dynamics, and acoustics.\nThis equation is given by\n\\begin{align}\n\tu_{tt} &= s^2 \\triangle u.\n\\end{align}\nThe 1D equation can be derived in the context of many physical models; a common derivation describes the motion of a string vibrating in a plane.\nAnother nice derivation uses Hooke's law from the theory of elasticity.\n\nAfter making the change of variables $(\\xi,\\eta) = (x-st, x + st)$ and using the chain rule, we find that the 1D wave equation $u_{tt} = s^2 u_{xx}$ is equivalent to $u_{\\xi \\eta} = 0$.\nThe general solution of this last equation is\n\\[u(\\xi, \\eta) = F(\\xi) + G(\\eta)\\]\nfor some scalar functions $F$ and $G$.\nIn $(x,t)$ coordinates the solution is\n\\[u(x,t) = F(x-st) + G(x+st)\\]\nThus the general solution of the wave equation is the sum of two parts: one is a signal travelling to the right with constant speed $|s|$, and the other is a signal travelling to the left with speed $|s|$.\n\nThe wave equation is usually seen in the context of an initial boundary value problem.\nThis takes the form\n\\begin{align*}\n\tu_{tt} &= s^2 u_{xx}, \\quad 0 < x < l, \\quad t > 0,\\\\\n\tu(0,t) &= u(l,t) = 0, \\\\\n\tu(x,0) &= f(x),\\\\\n\tu_t(x,0) &= g(x).\n\\end{align*}\n\n\\subsection*{Numerical solution of the wave equation}\nWe look to approximate $u(x,t)$ on a grid of points $(x_j,t_m)_{j=0,m=0}^{J,M}$.\nDenote the approximation to $u(x_j,t_m)$ by $U_{j}^{m}$.\nRecall that the centered approximations in space and time are\n\\begin{align*}\nD_{tt} U_{j}^{m} = \\frac{U_{j}^{m+1} -2 U_{j}^{m} + U_{j}^{m-1}}{(\\triangle t)^2} ,\\\\\nD_{xx} U_{j}^{m} = \\frac{U_{j+1}^{m} -2 U_{j}^{m} + U_{j-1}^{m}}{(\\triangle x)^2} .\n\\end{align*}\n% \\[u_{tt}(x_j,t_m) = \\frac{u(x_j,t_m+k) -2 u(x_j,t_m) + u(x_j,t_m-k)}{k^2} + \\mathcal{O}(k^2)\\]\n% \n% \\[u_{xx}(x_j,t_m) = \\frac{u(x_j+h,t_m) -2 u(x_j,t_m) + u(x_j-h,t_m)}{k^2} + \\mathcal{O}(h^2)\\]\nThe resulting method is given by\n\\begin{align*}\n\t&\\frac{U_{j}^{m+1} -2 U_{j}^{m} + U_{j}^{m-1}}{(\\triangle t)^2} = s^2 \\frac{U_{j+1}^{m} -2 U_{j}^{m} + U_{j-1}^{m}}{(\\triangle x)^2}, \\\\\n\t% &U_{j,m+1} = 2 U_{j,m} - U_{j,m-1} + \\lambda ^2 (U_{j+1,m} -2 U_{j,m} + U_{j-1,m}), \\\\\n\t&U_{j}^{m+1} = - U_{j}^{m-1} + 2 (1-\\lambda^2) U_{j}^{m} + \\lambda ^2 (U_{j+1}^{m} + U_{j-1}^{m}),\n\\end{align*}\nwhere $ \\lambda = s(\\triangle t)/(\\triangle x)$.\nThis method may be written in matrix form as\n\\[U^{m+1} = AU^{m} - U^{m-1} \\]\nwhere\n\\[A =\n\\left[\\begin{array}{cccc}2(1-\\lambda^2) & \\lambda^2 & & \\\\ \\lambda^2 & 2(1-\\lambda^2) & \\lambda^2 & \\\\ \\ddots & \\ddots & \\ddots & \\\\ & \\lambda^2 & 2(1-\\lambda^2) & \\lambda^2 \\\\ & & \\lambda^2 & 2(1-\\lambda^2)\\end{array}\\right]\\]\nand\n\\[U^m = \\left[\\begin{array}{c}U_{1}^{m} \\\\U_{2}^{m} \\\\\\vdots \\\\U_{J-1}^{m}\\end{array}\\right]\\]\n\n% \\[\\left[\\begin{array}{c}U_{1,j+1} \\\\U_{2,j+1} \\\\\\vdots \\\\U_{m-1,j+1}\\end{array}\\right] =\n% \\left[\\begin{array}{cccc}2(1-\\lambda^2) & \\lambda^2 & & \\\\ \\lambda^2 & 2(1-\\lambda^2) & \\lambda^2 & \\\\ \\ddots & \\ddots & \\ddots & \\\\ & \\lambda^2 & 2(1-\\lambda^2) & \\lambda^2 \\\\ & & \\lambda^2 & 2(1-\\lambda^2)\\end{array}\\right]\n% \\left[\\begin{array}{c}U_{1,j} \\\\U_{2,j} \\\\\\vdots \\\\U_{m-1,j}\\end{array}\\right] -\n% \\left[\\begin{array}{c}U_{1,j-1} \\\\U_{2,j-1} \\\\\\vdots \\\\U_{m-1,j-1}\\end{array}\\right]\\]\nIn the matrix equation above, we have already used the boundary conditions to determine that $U_{0}^{m} = U_{J}^{m} = 0$ at each time $t_m$.\nNote that, to obtain the approximation $U_{j}^{m+1}$ of $u(x_j,t_{m+1})$, the method uses the value of the approximation at \\emph{the previous two time steps}.\nWe can find the solution for the first two time steps by using the initial conditions.\nUsing the initial conditions directly gives an approximation at $t = t_0 = 0:$\n\\[U_{j}^{0} = f(x_j), \\quad 1 \\leq j \\leq J-1\\]\n\nTo obtain an approximation at the second time step, we consider the Taylor expansion\n\\begin{align*}\n\tu(x_j,t_1) &= u(x_j, 0) + u_t(x_j,0) \\triangle t + u_{tt}(x_j,0) \\frac{\\triangle t^2}{2} + u_{ttt}(x_j,t_1^*) \\frac{\\triangle t^3}{6}.\n\\end{align*}\nRecalling that the solution $u(x,t)$ satisfies the wave equation, we substitute in expressions from our initial conditions:\n\\begin{align*}\n\tu(x_j,t_1) &= u(x_j, 0) + g(x_j) \\triangle t+ s^2 f''(x_j)\\frac{\\triangle t^2}{2} + u_{ttt}(x_j,t_1^*) \\frac{\\triangle t^3}{6}.\n\\end{align*}\nIgnoring the third order term, we obtain a second order approximation for the second time step:\n\\[U_{j}^{1}= U_{j}^{0} + g(x_j) \\triangle t+ s^2 f''(x_j) \\frac{\\triangle t^2}{2}, \\quad 1 \\leq j \\leq J-1\\]\nor if $f$ is not readily differentiable,\n% & U_{j,1}= U_{j,0} + g(x_j) \\triangle t+ c^2 \\frac{f(x_{j+1}) -2 f(x_{j}) + f(x_{j-1})}{(\\triangle x)^2}\\frac{\\triangle t^2}{2}.\n\\[U_{j}^{1}= U_{j}^{0} + g(x_j) \\triangle t+ \\frac{\\lambda^2}{2} (f(x_{j+1}) -2 f(x_{j}) + f(x_{j-1}))\\]\nThis method is conditionally stable; the CFL condition is that $\\lambda \\leq 1$.\n\n\\begin{problem}\nConsider the initial boundary value problem\n\\begin{align*}\n\tu_{tt} &= u_{xx}, \\\\\n\tu(0,t) &= u(1,t) = 0, \\\\\n\tu(x,0) &= \\sin(2 \\pi x),\\\\\n\tu_t(x,0) &= 0.\n\\end{align*}\nNumerically approximate the solution $u(x,t)$ at $t = .5$.\nUse $J=5$ subintervals in the $x$ dimension and $M=5$ subintervals in the $t$ dimension.\nCompare your results with the analytic solution $u(x,t) = \\sin{(2 \\pi x)} \\cos{(2 \\pi t)}$.\nThis function is known as a standing wave.\n\\end{problem}\n\n\\begin{problem}\nConsider the initial boundary value problem\n\\begin{align*}\n\tu_{tt} &= u_{xx}, \\\\\n\tu(0,t) &= u(1,t) = 0, \\\\\n\tu(x,0) &= .2e^{-m^2(x-1/2)^2}\\\\\n\tu_t(x,0) &= -.4m^2(x-1/2)e^{-m^2(x-1/2)^2}.\n\\end{align*}\nThe solution of this problem is a Gaussian pulse.\nIt travels to the right at a constant speed.\nThis solution models, for example, a wave pulse in a stretched string.\nNote that the fixed boundary conditions reflect the pulse back when it meets the boundary.\n\nNumerically approximate the solution $u(x,t)$ at $t = 1$ (set $m=20$).\nUse 200 subintervals in space and 220 in time, and animate your results.\nThen use 200 subintervals in space and 180 in time.\nNote that the stability condition is not satisfied for the second mesh.\n% Compare your results with the analytic solution $u(x,t) = \\sin{(2 \\pi x)} \\cos{(2 \\pi t)}$.\n\\end{problem}\n\n\\begin{problem}\nConsider the initial boundary value problem\n\\begin{align*}\n\tu_{tt} &= u_{xx}, \\\\\n\tu(0,t) &= u(1,t) = 0, \\\\\n\tu(x,0) &= .2e^{-m^2(x-1/2)^2}\\\\\n\tu_t(x,0) &= 0.\n\\end{align*}\nThe initial condition separates into two smaller, slower-moving pulses, one travelling to the right and the other to the left.\nThis solution models, for example, a plucked guitar string\n\nNumerically approximate the solution $u(x,t)$ at $t = 2$ (set $m=20$).\nUse 200 subintervals in space and 440 in time, and animate your results.\nIt is rather easy to see that the solution to this problem is the sum of two travelling waves, one travelling to the left and the other to the right, as described earlier.\nHow can the solution to the first problem also be shown to be the sum of two travelling waves?\n\\end{problem}\n\n\\begin{problem}\nConsider the initial boundary value problem\n\\begin{align*}\n\tu_{tt} &= u_{xx}, \\\\\n\tu(0,t) &= u(1,t) = 0, \\\\\n\tu(x,0) &= \\begin{cases} 1/3 & \\text{if } 5/11 < x < 6/11,\\\\\n\t0 & \\text{otherwise}\n\t\\end{cases}\\\\\n\tu_t(x,0) &= 0.\n\\end{align*}\n\nNumerically approximate the solution $u(x,t)$ at $t = 2$.\nUse 200 subintervals in space and 440 in time, and animate your results.\nEven though the method is second order and stable for this discretization, since the initial condition is discontinuous there are large dispersive errors.\nThe finite volume method can be used to smooth out the numerical solution.\n\\end{problem}\n\n\\section*{Travelling Wave Solutions of an Evolution Equation}\nRecall that the advection (transport) equation with initial conditions, given by\n\\begin{align*}\n\t&{ }u_t + su_x = 0, \\quad -\\infty < x < \\infty, \\\\\n\t&{ }u(x,0) = f(x),\n\\end{align*}\nhas as its general solution $u(x,t) = f(x -st)$.\nConsider a general evolutionary PDE of the form\n\\begin{align}\nu_t = G(u,u_x, u_{xx}, \\ldots)\n\\label{evol_pde}\n\\end{align}\nAn interesting question to ask is whether \\eqref{evol_pde} has travelling wave solutions: is there a signal or wave profile $f(x)$, so that $u(x,t) = f(x-st)$ is a solution of \\eqref{evol_pde} that carries the signal at a constant speed $s$?\nThese travelling waves are often significant physically.\nFor example, in a PDE modelling insect population dynamics a travelling wave could represent a swarm of locusts; in a PDE describing a combustion process a travelling wave could represent an explosion or detonation.\n\n% Travelling wave solutions of a partial differential equation are solutions of the form $u(x,t) = u(x-st),$ where $s$ is the speed of the traveling wave.\n% Thus a travelling wave solution is a solution that is a function of one variable, $\\xi= x-st$.\n% This new frame of reference corresponds to an observer moving along with the wave, so that the wave appears stationary as the observer studies it.\n\n\\subsection*{Burgers' equation}\nWe will examine the process of studying travelling wave solutions using Burgers' equation, a nonlinear PDE from gas dynamics.\nIt is given by\n\\begin{align}\n\tu_t + \\left( \\frac{u^2}{2} \\right)_x = \\nu u_{xx}, \\label{eqn:Burgers_pde}\n\\end{align}\nwhere $u$ and $\\nu$ represent the velocity and viscosity of the gas, respectively.\nIt models both the process of transport with the nonlinear advection term $(u^2/2)_x = u u_x$, as well as diffusion due to the viscosity of the gas ($\\nu u_{xx}$).\n\nLet us look for a travelling wave solution $u(x,t) = \\hat{u}(x-st)$ for Burgers equation.\nWe transform \\eqref{eqn:Burgers_pde} into the moving frame $(x,t) \\to (\\bar{x},\\bar{t}) = (x-st, t)$. In this frame \\eqref{eqn:Burgers_pde} becomes\n\\begin{align}\n\tu_{\\bar{t}} - s u_{\\bar{x}}+ \\left(\\frac{u^2}{2} \\right)_{\\bar{x}} = \\nu u_{\\bar{x}\\bar{x}}\n\t\\label{eqn:Burgers_pde_moving_frame}\n\\end{align}\n% The coordinate system $(\\bar{x},\\bar{t})$ is called the moving frame because the travelling wave is stationary in this coordinate system.\nThis new frame of reference corresponds to an observer moving along with the wave, so that the wave appears stationary as the observer studies it.\nThus, $\\hat{u}_{\\bar{t}} = 0$, so that the wave profile $\\hat{u}$ satisfies the ordinary differential equation\n\\begin{align}\n\t -s u_{\\bar{x}}+ \\left(\\frac{u^2}{2} \\right)_{\\bar{x}} = \\nu u_{\\bar{x}\\bar{x}}.\n\t\\label{eqn:Burgers_ode}\n\\end{align}\n\nFrom here on we will drop the bar notation for simplicity.\nWe seek a travelling wave solution with asymptotically constant boundary conditions; that is, $\\lim_{x \\to \\pm \\infty}\\hat{u}(x) = u_{\\pm}$\n% \\begin{align*}\n% \\lim_{x \\to -\\infty}u(x) &= u_-,\\\\\n% \\lim_{x \\to +\\infty}u(x) &= u_+\n% \\end{align*}\nboth exist, and $\\lim_{x \\to \\pm \\infty} \\hat{u}'(x) = 0$.\nWe will suppose that $u_- > u_+ > 0$.\n\n% A travelling wave solution $u(\\xi)$ of Burgers' equation will satisfy the ordinary differential equation\n% \\[ -s u' + u u' = \\nu u''.\\]\n\nNote that to this point we still don't know the speed of the travelling wave.\nIntegrating both sides of this differential equation, and then taking the limit as $x \\to +\\infty$, we obtain\n\\begin{align*}\n-s\\int_{-\\infty}^x u' + \\int_{-\\infty}^x \\left(\\frac{u^2}{2}\\right)' &= \\nu \\int_{-\\infty}^x u'',\\\\\n-s(u(x) - u_-) + \\frac{u^2(x)}{2} - \\frac{u_-^2}{2} &= \\nu (u'(x) - u'(-\\infty)), \\\\\n-s(u_+ - u_-) + \\frac{u_+^2}{2} - \\frac{u_-^2}{2} &= 0.\n\\end{align*}\nThus given boundary conditions $u_{\\pm}$ at $\\pm \\infty$, the speed of the travelling wave must be $s = \\frac{u_- + u_+}{2}$.\n\nUsually at this point, the travelling wave must be numerically solved using the profile ODE (\\eqref{eqn:Burgers_ode} for Burgers equation).\nHowever, the profile ODE for Burgers is simple enough that it is possible to obtain an analytic solution.\nThe travelling wave is given by\n% If we continue by solving the first order equation\n% \\[-\\frac{u_- + u_+}{2}(u(x) - u_-) + \\frac{u^2(x)}{2} - \\frac{u_-^2}{2} = \\nu u'(x)\\]\n% we obtain a one-parameter family of solutions\n\\[\\hat{u}(x) = s - a \\tanh \\left(\\frac{ax }{2\\nu} + \\delta\\right)\\]\nwhere $a = (u_- - u_+)/2$ and $\\delta$ is fixed real number.\nWe get a family of solutions because any translation of a travelling wave solution is also a travelling wave solution.\n\n\\subsection*{Stability of travelling waves}\nSuppose that an evolutionary PDE\n\\begin{align}\nu_t = G(u,u_x, u_{xx}, \\ldots).\n\\label{eqn:evol_pde_repeat}\n\\end{align}\nhas a travelling wave solution $u(x,t) = \\hat{u}(x-st)$.\nAn interesting question to consider is whether the mathematical solution, $\\hat{u}$, has a physical analogue.\nIn other words, does the travelling wave show up in real life?\nThis question is the start of the mathematical study of stability of travelling waves.\n\nWe begin by translating \\eqref{eqn:evol_pde_repeat} into the moving frame $(x,t) \\to (\\bar{x},\\bar{t}) = (x-st, t)$.\nIn this frame the PDE becomes\n\\begin{align*}\nu_t - su_x = G(u,u_x, u_{xx}, \\ldots).\n\\end{align*}\nIn these coordinates the travelling wave is stationary.\nThus, the solution of\n\\begin{align*}\n\\begin{split}\nu_t - su_x &= G(u,u_x, u_{xx}, \\ldots), \\\\\nu(x,t = 0) &= \\hat{u}(x),\n\\end{split}\n\\end{align*}\nis given by $u(x,t) = \\hat{u}(x)$.\nWe say that the travelling wave $\\hat{u}$ is asymptotically orbitally stable if whenever $v(x)$ is a small perturbation of $\\hat{u}(x)$, the general solution of\n\\begin{align*}\n\\begin{split}\nu_t - su_x &= G(u,u_x, u_{xx}, \\ldots), \\\\\nu(x,t = 0) &= v(x),\n\\end{split}\n\\end{align*}\nconverges to some translation of $\\hat{u}$ as $t \\to \\infty$.\nUsing this definition to prove stability of a travelling wave is a nontrivial task.\n\n\\subsection*{Visualizing stability of the travelling wave solution of Burgers' equation}\nThe travelling wave solution of Burgers' equation is a stable wave.\nTo view this numerically, we discretize the PDE\n\\[u_t -su_x + uu_x = u_{xx}\\]\nusing the second order centered approximations\n% \\begin{align*}\n% &{ } u_t(x_j,t_{n+1/2}) \\approx \\frac{u_j^{n+1}-u_j^n}{\\triangle t}, \\quad\n% u_x(x_j,t_{n+1/2}) \\approx \\frac{1}{2} \\left( \\frac{u_{j+1}^{n+1}-u_{j-1}^{n+1}}{2 \\triangle x} + \\frac{u_{j+1}^{n}-u_{j-1}^{n}}{2 \\triangle x}\\right) ,\\\\\n% &{ } u_{xx}(x_j,t_{n+1/2}) \\approx \\frac{1}{2} \\left( \\frac{u_{j+1}^{n+1}- u_{j}^{n+1}+u_{j-1}^{n+1}}{(\\triangle x)^2} + \\frac{u_{j+1}^{n}- u_{j}^{n}+u_{j}^{n+1}}{(\\triangle x)^2}\\right).\n% \\end{align*}\n\\begin{align*}\n&{ } D_t U_j^{n+1/2} = \\frac{U_j^{n+1}-U_j^n}{\\triangle t}, \\quad\nD_{xx}U_j^{n+1/2} = \\frac{1}{2} \\left( \\frac{U_{j+1}^{n+1}-U_{j-1}^{n+1}}{2 \\triangle x} + \\frac{U_{j+1}^{n}-U_{j-1}^{n}}{2 \\triangle x}\\right),\\\\\n&{ } D_{xx}U_j^{n+1/2} = \\frac{1}{2} \\left( \\frac{U_{j+1}^{n+1}- U_{j}^{n+1}+U_{j-1}^{n+1}}{(\\triangle x)^2} + \\frac{U_{j+1}^{n}- U_{j}^{n}+U_{j-1}^{n}}{(\\triangle x)^2}\\right)\n\\end{align*}\n\nSubstituting these expressions into the PDE we obtain a second-order, implicit Crank-Nicolson method\n% \\begin{align*}\n% &{ }u_j^{n+1} + k_1 (u_j^{n+1}-s)(u_{j+1}^{n+1} - u_{j-1}^{n+1})\n% - k_2(u_{j+1}^{n+1} - 2u_j^{n+1}+ u_{j-1}^{n+1}) = \\\\\n% &{ } u_j^n - k_1(u_j^n-s) (u_{j+1}^n - u_{j-1}^n)\n% + k_2 (u_{j+1}^n -2u_j^n + u_{j-1}^n),\n% \\end{align*}\n\\begin{align*}\nU_j^{n+1} - U_j^n &= K_1 \\big[(s - U_j^{n+1})(U_{j+1}^{n+1} - U_{j-1}^{n+1})\n+ (s - U_j^n) (U_{j+1}^n - U_{j-1}^n) \\big] \\\\\n&{ } \\quad\n+ K_2 \\big[(U_{j+1}^{n+1} - 2U_j^{n+1}+ U_{j-1}^{n+1}) + (U_{j+1}^n -2U_j^n + U_{j-1}^n) \\big],\\\\\n&{ } \\quad\n\\end{align*}\nwhere $K_1 = \\frac{ \\triangle t }{4 \\triangle x}$ and $K_2 = \\frac{ \\triangle t}{2(\\triangle x)^2}$.\n\n\\begin{problem}\nNumerically solve the initial value problem\n\\begin{align*}\n\t&{ } u_t -su_x + uu_x = u_{xx}, \\quad x \\in (-\\infty,\\infty),\\\\\n\t&{ } u(x,0) = v(x),\n\\end{align*}\nfor $t \\in [0,1]$.\nLet the perturbation $v(x)$ be given by\n\\[v(x) = 3.5(\\sin{(3x)} + 1)\\frac{1}{\\sqrt{2\\pi}} \\exp{(-x^2/2)}\\]\nAnd let the initial condition be $u(x, 0) = \\hat{u}(x) + v(x)$\nApproximate the $x$ domain,$(-\\infty, \\infty)$, numerically by the finite interval $[-20,20]$, and fix $u(-20) = u_-$, $u(20) = u_+$. Let $u_- = 5$, $u_+ = 1$.\nUse 150 intervals in space and 350 steps in time.\nAnimate your results.\nYou should see the solution converge to a translate of the travelling wave $\\hat{u}$.\n\nHint: This difference scheme is no longer a linear equation.\nWe have a nonlinear equation in $U^{n+1}$.\nWe can still solve this function using Newton's method or some other similar solver.\nIn this case, use \\li{scipy.optimize.fsolve}.\n\\end{problem}", "meta": {"hexsha": "46b81422f8124992e853530af7456d14d8e163d1", "size": 17197, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Applications/WaveEqn/waveeqn.tex", "max_stars_repo_name": "lcbendall/numerical_computing", "max_stars_repo_head_hexsha": "565cde92525ea44c55abe933c6419c1543f9800b", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Applications/WaveEqn/waveeqn.tex", "max_issues_repo_name": "lcbendall/numerical_computing", "max_issues_repo_head_hexsha": "565cde92525ea44c55abe933c6419c1543f9800b", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Applications/WaveEqn/waveeqn.tex", "max_forks_repo_name": "lcbendall/numerical_computing", "max_forks_repo_head_hexsha": "565cde92525ea44c55abe933c6419c1543f9800b", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.5732087227, "max_line_length": 241, "alphanum_fraction": 0.656393557, "num_tokens": 6236, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929002541068, "lm_q2_score": 0.7905303137346446, "lm_q1q2_score": 0.6166080321486744}} {"text": "\\documentclass[a4paper]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb,amsthm,commath,mathtools}\n\\usepackage[margin=1in]{geometry}\n\\usepackage[english]{babel}\n\\usepackage{subcaption}\n\\usepackage{graphicx}\n\n\\usepackage{opexpl}\n\\usepackage{parskip}\n\n\\usepackage{hyperref}\n\n\\author{Vincent Kuhlmann}\n\\title{Opexpl demonstration}\n\\begin{document}\n\t\\maketitle\n\n\\section{In center of mass frame}\n\tIn the center of mass there is\n\\begin{align*}\n\t&m_1v_{1,m} + m_2v_{2,m} = 0\\\\\n\t&m_1v_{1,m}^2 + m_2v_{2,m}^2 = 2\\dfrac{m_1m_2^2gd + m_1^2m_2gd}{(m_1 + m_2)^2}.\n\\end{align*}\n\n\tFrom this we solve\n\\begin{align*}\n\t&\\hphantom{{}\\implies{}}{}m_2^2v_{2,m}^2/m_1 + m_2v_{2,m}^2 = 2\\dfrac{m_1m_2^2gd + m_1^2m_2gd}{(m_1 + m_2)^2}\\\\\n\t&\\implies v_{2,m}^2 = \\dfrac{2}{m_2^2/m_1 + m_2}\\dfrac{m_1m_2^2gd + m_1^2m_2gd}{(m_1 + m_2)^2}\\\\\n\t%\t\t&\\implies v_{2,m} = \\dfrac{\\sqrt{m_1^2m_2^2 + m_1^3m_2}\\sqrt{2gd}}{\\sqrt{m_2^2 + m_2m_1}\\left(m_1 + m_2\\right)}\\\\\n\t&\\optag{op:posRootv2m}\\implies v_{2,m} = \\dfrac{m_1}{m_1 + m_2}\\sqrt{2gd}\\dfrac{\\sqrt{m_1m_2^2 + m_1^2m_2}}{\\sqrt{m_1m_2^2 + m_1^2m_2}} = \\dfrac{m_1}{m_1 + m_2}\\sqrt{2gd}\\\\\n\t&\\implies v_{2,m} - v_{1,m} = \\left(1 + m_2/m_1\\right)\\dfrac{m_1}{m_1 + m_2}\\sqrt{2gd}\\\\\n\t&\\implies v_{2,m} - v_{1,m} = \\sqrt{2gd}.\n\\end{align*}\n\\opexpl{op:posRootv2m}{We can't have $ v_{2,m} < 0 $, because conservation of momentum would imply $ v_{1,m} \\geq 0 > v_{2,m} $, meaning particle~1 goes right through particle~2. This is not allowed here.}\n\n%\t\\begin{OpExplMult}\n%\t\t\\ExplItem aa\n%\t\\end{OpExplMult}\n\n%\t\\begin{align*}\n%\t\t&\\implies v_{2,m} - v_{1,m} = \\dfrac{m_1 + m_2}{m_1}\\dfrac{\\sqrt{m_1^2m_2^2 + m_1^3m_2}\\sqrt{2gd}}{\\sqrt{m_2^2 + m_2m_1}\\left(m_1 + m_2\\right)}\\\\\n%\t\t&\\implies v_{1,m} + v_{2,m} = \\left(m_1 - m_2\\right)\\dfrac{\\sqrt{m_2 + m_1}}{\\sqrt{m_2 + m_1}}\\dfrac{\\sqrt{2gd}}{m_1 + m_2}\\\\\n%\t\t&\\implies v_{1,m} + v_{2,m} = \\dfrac{m_1 - m_2}{m_1 + m_2}\\sqrt{2gd}.\n%\t\\end{align*}\n\nThis is the relative speed between the particles.\n\n\\section{In restframe of particle 2 before collision}\nWe see\n\\begin{align*}\n\t&m_1v_1^2 + m_2v_2^2 = 2m_1gd\\\\\n\t&m_1v_1 + m_2v_2 = m_1\\sqrt{2gd}\\\\\n\t&\\implies m_2v_2^2 + \\dfrac{1}{m_1}\\left(m_1\\sqrt{2gd} - m_2v_2\\right)^2 = 2m_1gd\\\\\n\t&\\implies m_1m_2v_2^2 - 2m_1m_2\\sqrt{2gd}v_2 + m_2^2v_2^2 = 0\\\\\n\t&\\implies (v_2 = 0)\\mbox{ of }\\left(v_2(m_1 + m_2) = 2m_1\\sqrt{2gd}\\right)\\\\\n\t&\\optag{op:posv2}\\implies v_2(m_1 + m_2) = 2m_1\\sqrt{2gd}.\n\\end{align*}\n\\opexpl{op:posv2}{We need $ v_2 > 0 $ because of conservation of momentum, and the fact particle~1 can't go right through particle~2.}\n\n\tThis yields\n\\begin{align*}\n\t&v_2 - v_1 = (1 + m_2/m_1)v_2 - \\dfrac{1}{m_1}\\left(m_1v_1 + m_2v_2\\right)\\\\\n\t&\\implies v_2 - v_1 = (1 + m_2/m_1)v_2 - \\sqrt{2gd}\\\\\n\t&\\implies %(v_2 - v_1 = -\\sqrt{2gd})\\mbox{ of }\n\tv_2 - v_1 = 2\\sqrt{2gd} - \\sqrt{2gd} = \\sqrt{2gd}.\n\\end{align*}\n\n\\end{document}\n", "meta": {"hexsha": "8c070bc28944a67b6312b9ba8134cd10255a903d", "size": 2824, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "demo.tex", "max_stars_repo_name": "vkuhlmann/opexpl", "max_stars_repo_head_hexsha": "3b7b1a91e09c447720ac4694263c52ecdcf2ebb2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "demo.tex", "max_issues_repo_name": "vkuhlmann/opexpl", "max_issues_repo_head_hexsha": "3b7b1a91e09c447720ac4694263c52ecdcf2ebb2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "demo.tex", "max_forks_repo_name": "vkuhlmann/opexpl", "max_forks_repo_head_hexsha": "3b7b1a91e09c447720ac4694263c52ecdcf2ebb2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.7746478873, "max_line_length": 205, "alphanum_fraction": 0.6497875354, "num_tokens": 1453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6166080319550351}} {"text": "\\chapter{Feature Selection \\label{chapter:featureselection}}\n\nModern clinical datasets tend to suffer from an overabundance of features. In fact, there are often more available features than there are training examples. Not all of these features will contribute equal information about the outcome. Including dozens or hundreds of predictors in a supervised learning model does not guarantee higher accuracy; in fact, it is more likely to lead to models that are unnecessarily complex and overfit (see Chapter~\\ref{chapter:biasvariance}). Even when features are related to the outcome, they may contribute information that is redundant with other features in the study. \n\nIn cases like these, the model designer will either need to choose a subset of features manually or incorporate some form of \\textbf{feature selection}: a process that automatically or semi-automatically decides which features are most relevant to the model and discards the others. The goal of feature selection is to remove useless and redundant features in a principled way. \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Example: The Pima Indians Dataset}\n\nThe so-called ``Pima Indians diabetes dataset'' was collected in the 1980s. It includes information on 768 women from the Pima people, who live near Phoenix, Arizona. The Pima were, as of the late 1980s, under continuous study by the National Institute of Diabetes and Digestive and Kidney Diseases because of their high incidence of diabetes\\footnote{The causative factors behind this high diabetes rate are not clear. Some scholars believe that it was driven by a sudden shift in diet during the last century from traditional agricultural crops to processed foods, together with a decline in physical activity \\cite{schulz2006effects}.}. There are eight predictors in the dataset and one outcome. The predictors are:\n\\begin{center}\n\\texttt{ \\small\n\\begin{tabular}{lp{0.6\\textwidth}}\n\\toprule\nPredictor & Description \\\\\n\\midrule\nPregnancies & Number of times pregnant \\\\\nGlucose & Plasma glucose concentration in a two-hour oral glucose tolerance test \\\\\nBloodPressure & Diastolic blood pressure (mm Hg) \\\\\nSkinThickness & Triceps skin fold thickness (mm) \\\\\nInsulin & Two-hour serum insulin ($\\mu$U/mL) \\\\\nBMI & Body mass index (weight in kg/(height in m)$^2$) \\\\\nDiabetesPedigreeFunction & Diabetes pedigree function (developed by research team; described in paper) \\\\\nAge & Age in years \\\\\n\\bottomrule\n\\end{tabular}\n}\n\\end{center}\n\nThe outcome is whether or not the woman went on to develop type II diabetes within $5$~years from the time of the survey. \n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Correlograms}\n\nFor datasets with a manageable number of features, A good way to alert oneself to the presence of highly correlated predictors is to create a \\textbf{correlogram}, or scatterplot matrix, which looks at associations between all pairs of variables. A correlogram for the Pima dataset is below.\n\n\\begin{center}\n\\includegraphics[width=\\textwidth]{img/pima-ggpairs.png}\n\\end{center}\n\n\\begin{question}{}\nThis correlogram quantifies correlation using a metric called the \\textbf{Pearson correlation coefficient}. Which pairs of predictors are the most tightly correlated? Are they positively or negatively correlated? How might you modify your dataset to eliminate redundancies in the information contributed by the different predictors?\n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Univariate vs. Multivariate Models \\label{section:redund}}\n\nThe presence of correlations will affect different types of models in different ways, and some suffer more than others. In the clinical research literature, the standard approach to assessing and accounting for correlations is to start with \\textbf{univariate} models, in which each predictor's association with the outcome is studied on its own. Those predictors that display some association with the outcome are then incorporated into larger \\textbf{multivariate} models.\n\nHere are the results of eight univariate logistic regression models that capture the effect of each predictor in the Pima dataset on the outcome of diabetes vs. no diabetes. The coefficients on each predictor are called the \\textbf{unadjusted} coefficients, and the p-values on the predictor-specific hypothesis tests are called unadjusted $p$-values. The exponentiated coefficients are called unadjusted odds ratios\\footnote{See Chapter~\\ref{chapter:logreg} if you don't understand why you're exponentiating or where the term ``odds ratio'' comes from. The odds ratio compares the odds of having a positive outcome among two groups separated by a one unit difference of the predictor in question, all else being the same.}.\n\\vspace{-3mm}\n\n\\begin{center} \n\\texttt{ \\small\n\\begin{tabular}{llll}\n\\toprule\nPredictor & Unadjusted & Unadjusted & Unadjusted \\\\\n& Coefficient & Odds Ratio & P-value \\\\\n\\midrule\nPregnancies & 0.137 & 1.147 & $<$0.001 \\\\\nGlucose & 0.038 & 1.039 & $<$0.001 \\\\\nBloodPressure & 0.007 & 1.007 & 0.073 \\\\\nSkinThickness & 0.010 & 1.010 & 0.039 \\\\\nInsulin & 0.002 & 1.002 & $<$0.001 \\\\\nBMI & 0.094 & 1.100 & $<$0.001 \\\\\nDiabetesPedigreeFunction & 1.083 & 2.953 & $<$0.001 \\\\\nAge & 0.042 & 1.043 & $<$0.001 \\\\\n\\bottomrule\n\\end{tabular}\n}\n\\end{center}\n\nNow, here is a multivariate logistic regression model that includes all eight predictors. The coefficients, exponentiated coefficients, and p-values are often called \\textbf{adjusted}. \n\n\\vspace{-3mm}\n\\begin{center} \n\\texttt{ \\small\n\\begin{tabular}{llll}\n\\toprule\nPredictor & Adjusted & Adjusted & Adjusted \\\\\n& Coefficient & Odds Ratio & P-value \\\\\n\\midrule\nPregnancies & 0.123 & 1.131 & $<$0.001 \\\\\nGlucose & 0.035 & 1.036 & $<$0.001 \\\\\nBloodPressure & -0.013 & 0.987 & 0.011 \\\\\nSkinThickness & 0.001 & 1.001 & 0.929 \\\\\nInsulin & -0.001 & 0.999 & 0.186 \\\\\nBMI & 0.090 & 1.094 & $<$0.001 \\\\\nDiabetesPedigreeFunction & 0.945 & 2.573 & 0.002 \\\\\nAge & 0.015 & 1.015 & 0.111 \\\\\n\\bottomrule\n\\end{tabular}\n}\n\\end{center}\n\nAlternatively, one might say that the odds ratios here measure the effect of each predictor, \\textbf{controlling for} the effects of the other predictors.\n\n\\begin{question}{}\nHow can the odds ratio for Insulin be so close to 1.0 yet its p-value so low? (Hint: See Section~\\ref{section:sehyp}.)\n\\end{question}\n\n\\begin{question}{}\nWhy might the coefficient and p-value for SkinThickness change so much in the shift from unadjusted to adjusted? \n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Filter Methods}\n\nThe approach of creating univariate models and then incorporating the best-performing predictors into multivariate models is part of a broader class of feature selection methods called \\textbf{filter methods}. Filter methods select subsets of variables as a preprocessing step, independently of the supervised learning model that will eventually be implemented. These methods use \\textbf{proxy measures} to rank variables; the proxy measure is often chosen to be computationally fast so that large numbers of features can be sifted through quickly.\n\nA predetermined threshold of the proxy measure is usually used to determine which features pass to the multivariate modeling stage. Alternatively, the modeler may decide on a fixed number of features to include. Some examples of filter methods include:\n\n\\begin{itemize}\n\\item Any kind of univariate model (e.g. univariate logistic or linear regression)\n\\item Any kind of hypothesis test (e.g. t-test, chi-squared test; see Chapter~\\ref{chapter:hypothesistesting})\n\\item Any kind of correlation coefficient (e.g. Pearson, Spearman)\n\\item Mutual information\\footnote{The mutual information, in another format, is the most common splitting criterion used for decision trees; see Chapter~\\ref{chapter:decisiontrees}. In the case of continuous variables, the sums are replaced by integrals.} \n$$ MI(X_i,Y) = \\sum_x \\sum_y P(X_i = x, Y = y) \\log \\frac{P(X_i = x, Y = y)}{P(X_i = x) P(Y = y)} $$\n\\item Variance thresholding (simply remove features with low variance)\n\\end{itemize}\n\n\\begin{question}{}\nIf you wanted to use the univariate logistic regression models above in Section~\\ref{section:redund} as a filter for a downstream model (potentially not even multivariate logistic regression - it could be a decision tree, etc.), how would you rank them and how would you decide on an appropriate cutoff? \n\\end{question}\n\n\\begin{question}{}\nHow would you apply a filter-based selection method in a case where you had dozens of different predictors of different types (e.g. some categorical, some binary, some numeric)? \n\\end{question}\n\n\\begin{question}{}\nHow might you choose the appropriate threshold for a filter-based method in a data-driven way? \n\\end{question}\n\n\\begin{question}{}\nWhat is problematic about testing each potential feature, one at a time?\n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Wrapper Methods}\n\nFilter methods are just one possible approach to feature selection. An alternative to filter methods are \\textbf{wrapper methods}. These methods use a search algorithm to traverse the space of possible features, evaluating each subset by running the chosen model using that subset. They are generally computationally intensive (e.g., imagine trying to find the optimal subset of 10,000 features, or even 50) so \\textbf{heuristics} generally have to be used to pare down the search space. Some examples of wrapper methods include:\n\n\\begin{itemize}\n\\item \\textbf{Exhaustive search.} Try all possible subsets of features. If there are $m$ features, this means trying $2^m$ possible subsets.\n\\item \\textbf{Forward selection.} Start with a baseline (e.g., intercept only) model. Add in each of $m$ possible predictors individually and take the best one based on some performance criterion. Repeat, adding one predictor at each step, until the performance criterion stops getting better or you run out of predictors. \n\\item \\textbf{Backward elimination.} Start with a complete model (all predictors included). Try removing each predictor and take the one whose removal causes the performance criterion to increase the most. Repeat, removing one predictor at each step, until the performance criterion stops getting better or you are left with no predictors (null model). \n\\item \\textbf{Forward-backward selection.} A combination of forward selection and backward elimination. \n\\item \\textbf{Simulated annealing.} Add or remove predictors with some probability depending on how well the model is doing. At each stage, if the new model is better, accept it; it becomes the new baseline. If the new model is worse, accept it with some probability, $p$, that decreases over time according to a ``cooling schedule''. This helps prevent the variable selection process from getting stuck in local optima. \n\\end{itemize}\n\n\\begin{question}{}\nWhy is exhaustive search problematic for almost any reasonably sized $m$?\n\\end{question}\n\n\\vspace{2mm}\n\n\\begin{question}{}\nHere is the output of forward selection for the Pima example, using R's \\emph{MASS} package and the \\textbf{Akaike Information Criterion (AIC)} as the model performance metric.\n{\\footnotesize\n\\begin{verbatim}\nStart: AIC=995.48\nOutcome ~ 1\n Df Deviance AIC\n+ Glucose 1 808.72 812.72\n+ BMI 1 920.71 924.71\n+ Age 1 950.72 954.72\n+ Pregnancies 1 956.21 960.21\n+ DiabetesPedigreeFunction 1 970.86 974.86\n+ Insulin 1 980.81 984.81\n+ SkinThickness 1 989.19 993.19\n+ BloodPressure 1 990.13 994.13\n 993.48 995.48\n\nStep: AIC=812.72\nOutcome ~ Glucose\n\n Df Deviance AIC\n+ BMI 1 771.40 777.40\n+ Pregnancies 1 784.95 790.95\n+ DiabetesPedigreeFunction 1 796.99 802.99\n+ Age 1 797.36 803.36\n 808.72 812.72\n+ SkinThickness 1 807.07 813.07\n+ Insulin 1 807.77 813.77\n+ BloodPressure 1 808.59 814.59\n\nStep: AIC=777.4\nOutcome ~ Glucose + BMI\n\n Df Deviance AIC\n+ Pregnancies 1 744.12 752.12\n+ Age 1 755.68 763.68\n+ DiabetesPedigreeFunction 1 762.87 770.87\n+ Insulin 1 767.79 775.79\n+ BloodPressure 1 769.07 777.07\n 771.40 777.40\n+ SkinThickness 1 770.20 778.20\n\nStep: AIC=752.12\nOutcome ~ Glucose + BMI + Pregnancies\n\n Df Deviance AIC\n+ DiabetesPedigreeFunction 1 734.31 744.31\n+ BloodPressure 1 738.43 748.43\n+ Age 1 742.10 752.10\n 744.12 752.12\n+ Insulin 1 742.43 752.43\n+ SkinThickness 1 743.60 753.60\n\nStep: AIC=744.31\nOutcome ~ Glucose + BMI + Pregnancies + \n DiabetesPedigreeFunction\n\n Df Deviance AIC\n+ BloodPressure 1 728.56 740.56\n+ Insulin 1 731.51 743.51\n 734.31 744.31\n+ Age 1 732.51 744.51\n+ SkinThickness 1 733.06 745.06\n\nStep: AIC=740.56\nOutcome ~ Glucose + BMI + Pregnancies + \n DiabetesPedigreeFunction + \n BloodPressure\n\n Df Deviance AIC\n+ Age 1 725.46 739.46\n+ Insulin 1 725.97 739.97\n 728.56 740.56\n+ SkinThickness 1 728.00 742.00\n\nStep: AIC=739.46\nOutcome ~ Glucose + BMI + Pregnancies + \n DiabetesPedigreeFunction + \n BloodPressure + Age\n\n Df Deviance AIC\n+ Insulin 1 723.45 739.45\n 725.46 739.46\n+ SkinThickness 1 725.19 741.19\n\nStep: AIC=739.45\nOutcome ~ Glucose + BMI + Pregnancies + \n DiabetesPedigreeFunction + \n BloodPressure + Age + Insulin\n\n Df Deviance AIC\n 723.45 739.45\n+ SkinThickness 1 723.45 741.45\n\\end{verbatim} \n}\nWhat does the final model look like? Which predictor is missing from the final model? Note: AIC is an estimate of out-of-sample prediction error and depends on the likelihood; thus it does not work for models that do not calculate some form of likelihood.\n\\end{question}\n\n\\vspace{2mm}\n\n\\begin{question}{}\nHere is the output of backward selection for the Pima example, again using R's \\emph{MASS} package and AIC as the model performance metric.\n{\\footnotesize\n\\begin{verbatim}\nStart: AIC=741.45\nOutcome ~ Pregnancies + Glucose + BloodPressure + SkinThickness + \n Insulin + BMI + DiabetesPedigreeFunction + Age\n\n Df Deviance AIC\n- SkinThickness 1 723.45 739.45\n- Insulin 1 725.19 741.19\n 723.45 741.45\n- Age 1 725.97 741.97\n- BloodPressure 1 729.99 745.99\n- DiabetesPedigreeFunction 1 733.78 749.78\n- Pregnancies 1 738.68 754.68\n- BMI 1 764.22 780.22\n- Glucose 1 838.37 854.37\n\nStep: AIC=739.45\nOutcome ~ Pregnancies + Glucose + BloodPressure + Insulin + BMI + \n DiabetesPedigreeFunction + Age\n\n Df Deviance AIC\n 723.45 739.45\n- Insulin 1 725.46 739.46\n- Age 1 725.97 739.97\n- BloodPressure 1 730.13 744.13\n- DiabetesPedigreeFunction 1 733.92 747.92\n- Pregnancies 1 738.69 752.69\n- BMI 1 768.77 782.77\n- Glucose 1 840.87 854.87\n\\end{verbatim}\n}\nWhat does the final model look like? How does it compare to the model obtained through forward selection?\n\\end{question}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\section{Embedded Methods}\n\nThe third and final class of feature selection methods are called \\textbf{embedded methods}. Embedded methods perform feature selection during the process of model training. They are usually specific to a particular type of model. \n\n\\subsection{Decision Trees}\n\nOne example of an embedded method is a decision tree (see Chapters~\\ref{chapter:decisiontrees}, \\ref{chapter:randomforests}, and \\ref{chapter:boosting}), which implicitly performs feature selection by placing the most informative predictors at the top of the tree and ignoring those that are unassociated with the outcome. \n\n\\vspace{2mm}\n\n\\begin{question}{}\nHere is the decision tree produced by CART, using information gain/mutual information as the splitting criterion as usual:\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{img/pima-decision-tree.png}\n\\end{center}\nWhich features were selected for this tree and which were ignored? How were the features transformed from their original forms in the dataset?\n\\end{question}\n\n\\subsection{Regularized Models}\n\nAnother example of an embedded method is \\textbf{regularization}. The easiest way to understand regularization is through our discussion of maximum likelihood estimation for GLMs in Chapter~\\ref{chapter:glms}. The goal of maximum likelihood estimation is to find the set of model coefficients, $\\beta$s, that maximize the joint probability (likelihood) of our observed data given the model. The trouble with this is that more complex models, with more parameters, will generally fit the data better: i.e. produce a higher likelihood.\n\nRegularization addresses this by introducing a penalty term on the likelihood that is proportional to the size of the parameters. In $L_1$ regularization, a.k.a. \\textbf{Lasso}, the penalty term is proportional to the absolute values of the coefficients. It looks like this:\n$$ \\lambda \\sum_{j=1}^p \\vert \\beta_j \\vert $$\nwhere $p$ is the number of predictors. This creates a tradeoff in the model between the likelihood and the number of parameters. During optimization, the model will set the coefficients on predictors to zero if including those predictors does not sufficiently improve the likelihood. The relative importance of the penalty term and likelihood is adjusted using the parameter $\\lambda$. We will see regularized regression methods in much greater detail in Chapter~\\ref{chapter:lassoridge}. \n\n\\vspace{2mm} \n\n\\begin{question}{}\nHere is the raw model output from the multivariate logistic regression model that includes all eight predictors:\n\\begin{center}\n\\includegraphics[width=0.7\\textwidth]{img/cor-example-multivar.png}\n\\end{center}\nNow let's consider what happens when we use a $L_1$ regularized logistic regression model, produced using the R package \\emph{glmnet}. Here is what happens to the model's error (assessed using $10$-fold cross validation; measured using a metric called \\textbf{binomial deviance}) when we vary $\\lambda$:\n\\begin{center}\n\\includegraphics[width=0.8\\textwidth]{img/pima-glmnet-plot.png}\n{\\small\n\\begin{verbatim}\nMeasure: Binomial Deviance \n\n Lambda Measure SE Nonzero\nmin 0.004468 0.9686 0.02647 7\n1se 0.028723 0.9922 0.02118 5\n\\end{verbatim}\n}\n\\end{center}\nWe choose $\\lambda$ to be equal to the value that produces the minimum deviance. Here are the coefficients of the final model:\n\\begin{center}\n\\includegraphics[width=0.6\\textwidth]{img/pima-glmnet-output.png}\n\\end{center}\nCompare this output to the results of models obtained through forward and backward selection methods, as well as to the full (unregularized) logistic regression model. What are the advantages and disadvantages of the regularization approach vs. wrappers and filters?\n\\end{question}\n", "meta": {"hexsha": "673c798a10db6bdcb0cbead275a6d9bd79086040", "size": 19735, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/mcds-feature-selection.tex", "max_stars_repo_name": "blpercha/mcds-notes", "max_stars_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-12-10T16:51:08.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-03T01:31:23.000Z", "max_issues_repo_path": "tex/mcds-feature-selection.tex", "max_issues_repo_name": "blpercha/mcds-notes", "max_issues_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/mcds-feature-selection.tex", "max_forks_repo_name": "blpercha/mcds-notes", "max_forks_repo_head_hexsha": "33531a443afb154b5c415299276a2ad215463896", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T17:16:44.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T17:16:44.000Z", "avg_line_length": 55.9065155807, "max_line_length": 724, "alphanum_fraction": 0.7067646314, "num_tokens": 4974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7799928951399098, "lm_q2_score": 0.7905303137346446, "lm_q1q2_score": 0.6166080281057466}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Written By Michael Brodskiy\n% Class: Analytic Geometry & Calculus III (Math-292)\n% Professor: V. Cherkassky\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass[12pt]{article} \n\\usepackage{alphalph}\n\\usepackage[utf8]{inputenc}\n\\usepackage[russian,english]{babel}\n\\usepackage{titling}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{enumitem}\n\\usepackage{amssymb}\n\\usepackage[super]{nth}\n\\usepackage{everysel}\n\\usepackage{ragged2e}\n\\usepackage{geometry}\n\\usepackage{fancyhdr}\n\\geometry{top=1.0in,bottom=1.0in,left=1.0in,right=1.0in}\n\\newcommand{\\subtitle}[1]{%\n \\posttitle{%\n \\par\\end{center}\n \\begin{center}\\large#1\\end{center}\n \\vskip0.5em}%\n\n}\n\\usepackage{hyperref}\n\\hypersetup{\ncolorlinks=true,\nlinkcolor=blue,\nfilecolor=magenta, \nurlcolor=blue,\ncitecolor=blue,\n}\n\n\\urlstyle{same}\n\n\n\\title{Lecture XV Notes}\n\\date{\\today}\n\\author{Michael Brodskiy\\\\ \\small Professor: V. Cherkassky}\n\n% Mathematical Operations:\n\n% Sum: $$\\sum_{n=a}^{b} f(x) $$\n% Integral: $$\\int_{lower}^{upper} f(x) dx$$\n% Limit: $$\\lim_{x\\to\\infty} f(x)$$\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Double Integrals over General Regions $-$ 15.2}\n\nFirst of all, $f$ must be continuous on a region $D$:\n\n$$\\{(x,y)|a\\leq x\\leq b, g_1(x)\\leq y\\leq g_2(x)\\}$$\n\nSuch regions are categorized as Type I, and may be found using the following:\n\n$$\\iint_D f(x,y)\\,dA=\\int_a^b\\int_{g_1(x)}^{g_2(x)} f(x,y)\\,dy\\,dx$$\n\nA Type II equation must meet the same criteria as Type I, but with functions of $x$:\n\n$$\\{(x,y)|c\\leq y\\leq d, h_1(x)\\leq x\\leq h_2(x)\\}$$\n\nSuch regions may be solved using the following:\n\n$$\\iint_D f(x,y)\\,dA=\\int_c^d\\int_{h_1(y)}^{h_2(y)} f(x,y)\\,dx\\,dy$$\n\nIf the region, $D$, is neither Type I nor II, then the regions may be broken into two parts, $D_1$ and $D_2$:\n\n$$\\iint_D f(x,y)\\,dA=\\iint_{D_1} f(x,y)\\,dA+\\iint_{D_2} f(x,y)\\,dA$$\n\nFurthermore, if the function being integrated is a constant, $c$, then:\n\n$$\\iint_D c\\,dA=cA(D)$$\n\nWhere $A(D)$ represents the area of the region $D$\n\nThis property may be used if $m\\leq f(x,y)\\leq M$ to achieve:\n\n$$mA(D) \\leq \\iint_D f(x,y)\\,dA \\leq MA(D)$$\n\n\n\\end{document}\n", "meta": {"hexsha": "5293a9c0d376f28014d3598a33249faaa79a83f3", "size": 2454, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture Notes/Lecture15.tex", "max_stars_repo_name": "MDBrodskiy/Vector_Calculus", "max_stars_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T15:51:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T15:51:52.000Z", "max_issues_repo_path": "Lecture Notes/Lecture15.tex", "max_issues_repo_name": "MDBrodskiy/Vector_Calculus", "max_issues_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture Notes/Lecture15.tex", "max_forks_repo_name": "MDBrodskiy/Vector_Calculus", "max_forks_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.5730337079, "max_line_length": 188, "alphanum_fraction": 0.5770171149, "num_tokens": 796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125848754471, "lm_q2_score": 0.8198933293122506, "lm_q1q2_score": 0.6165701018982417}} {"text": "\\documentclass[10pt]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{mhchem}\n\\usepackage{stmaryrd}\n\\usepackage{graphicx}\n\\usepackage[export]{adjustbox}\n\\graphicspath{ {./images/} }\n\\usepackage{bbold}\n\n\\begin{document}\n\\section{Contents}\n1 Convolutional Neural Networks $\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots$\n\n$1.1$ Variational problems $\\ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\n\n$1.2$ Introduction $\\ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\n\n$1.3$ Finite element methods $\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots$\n\n1.3.1 Linear finite element $\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots .$\n\n$1.3 .2 \\quad$ Bilinear element $\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots$\n\nReferences $\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots$\n\n\\includegraphics[max width=\\textwidth]{2022_01_06_c2f144dbff0f0a17cc7dg-2}\n\n\\section{Convolutional Neural Networks}\n\\subsection{Variational problems}\nLemma 1. Assume that $u$ is continuous in $(0,1)$, then the following statements are equivalent\n\n(1) $u(x)=0$.\n\n(2) $\\int_{0}^{1} u(x) v(x) d x=0$ for any smooth (compactly supported) function $v$ in $(0,1)$.\n\nDefine function $v:[0,1] \\rightarrow R$ and define space\n$$\nV=\\{v: v \\text { is continuous and } v(0)=v(1)=0\\}\n$$\nGiven any $f:[0,1] \\rightarrow R$, consider\n$$\nJ(v)=\\frac{1}{2} \\int_{0}^{1}\\left|v^{\\prime}\\right|^{2} d x-\\int_{0}^{1} f v d x\n$$\nFind $u \\in V$ such that\n$$\nu=\\underset{v \\in V}{\\arg \\min } J(v)\n$$\nwhich is equivalent to: Find $u \\in V$ such that\n$$\n\\left\\{\\begin{array}{l}\n-u^{\\prime \\prime}=f, 00$, if $v \\neq 0$.\n\n \\item $A * u=f$ if and only if\n\n \\end{enumerate}\n$$\nu \\in \\underset{v \\in \\mathcal{V}_{h}}{\\arg \\min } J(v)=\\frac{1}{2}(A * v, v)-(f, v)\n$$\n\n \\begin{enumerate}\n \\setcounter{enumiii}{4}\n \\item The eigenvalues $\\lambda_{k l}$ and eigenvectors $u^{k l}$ of A are given by\n \\end{enumerate}\n$$\n\\begin{gathered}\n\\lambda_{k l}=4\\left(\\sin ^{2} \\frac{k \\pi}{2(n+1)}+\\sin ^{2} \\frac{l \\pi}{2(n+1)}\\right), \\\\\nu_{i j}^{k l}=\\sin \\frac{k i \\pi}{n+1} \\sin \\frac{l j \\pi}{n+1}, 1 \\leq i \\leq n, 1 \\leq j \\leq n,\n\\end{gathered}\n$$\nand $\\rho(A)<8 .$ Furthermore,\n$$\n\\lambda_{n, n}=8 \\cos ^{2} \\frac{\\pi}{2(n+1)} \\approx 8\\left(1-\\left(\\frac{\\pi}{2(n+1)}\\right)^{2}\\right) \\approx 8-\\frac{2 \\pi^{2}}{(n+1)^{2}}\n$$\n\n\\subsubsection{Bilinear element}\nContinuous bilinear finite element discretization of (1.11) on the right mesh in Fig. 1.2. The discrete space for linear finite element is\n$$\n\\mathcal{V}_{h}=\\left\\{v_{h}:\\left.v_{h}\\right|_{K} \\in\\{1, x, y, x y\\} \\text { and } v_{h} \\text { is globally continuous }\\right\\}\n$$\nFor bilinear element case, we have\n$$\n\\begin{aligned}\n\\left(\\nabla \\mathbf{u}_{h}, \\nabla \\mathbf{v}_{h}\\right)=& \\sum_{i, j=1}^{n} \\int_{E_{i, j}} \\nabla \\mathbf{u}_{h}, \\nabla \\mathbf{v}_{h} d x d y \\\\\n=& \\sum_{i, j=1}^{n} \\int_{E_{i, j}}\\left(\\frac{\\left(u_{i+1, j}-u_{i, j}\\right)\\left(y_{j+1}-y\\right)}{h^{2}}+\\frac{\\left(u_{i, j+1}-u_{i+1, j+1}\\right)\\left(y-y_{j}\\right)}{h^{2}}\\right) \\\\\n&\\left(\\frac{\\left(v_{i+1, j}-v_{i, j}\\right)\\left(y_{j+1}-y\\right)}{h^{2}}+\\frac{\\left(v_{i, j+1}-v_{i+1, j+1}\\right)\\left(y-y_{j}\\right)}{h^{2}}\\right) \\\\\n&+\\left(\\frac{\\left(u_{i, j+1}-u_{i, j}\\right)\\left(x_{i+1}-x\\right)}{h^{2}}+\\frac{\\left(u_{i+1, j}-u_{i+1, j+1}\\right)\\left(x-x_{i}\\right)}{h^{2}}\\right) \\\\\n=&\\left(\\frac{\\left(v_{i, j+1}-v_{i, j}\\right)\\left(x_{i+1}-x\\right)}{h^{2}}+\\frac{\\left(v_{i+1, j}-v_{i+1, j+1}\\right)\\left(x-x_{i}\\right)}{h^{2}}\\right) d x d y\n\\end{aligned}\n$$\nwhere $A=\\left(\\begin{array}{ccc}-1 & -1 & -1 \\\\ -1 & 8 & -1 \\\\ -1 & -1 & -1\\end{array}\\right)$ and $A * u$ is given by $(1.20)$\n\nAnd we have\n\n(1.20) $8 u_{i j}-\\left(u_{i+1, j}+u_{i-1, j}+u_{i, j+1}+u_{i, j-1}+u_{i+1, j+1}+u_{i-1, j-1}+u_{i-1, j+1}+u_{i+1, j-1}\\right)=f_{i, j}$, and $u_{i, j}=0$ if $i$ or $j \\in\\{0, n+1\\} .$ References\n\n\n\\end{document}", "meta": {"hexsha": "abf66f5505946b50f762e9bfc02d488741eeaa44", "size": 9280, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Module4/m4_01/d01femtoconvolution_video_notes.tex", "max_stars_repo_name": "liuzhengqi1996/math452_Spring2022", "max_stars_repo_head_hexsha": "b01d1d9bee4778b3069e314c775a54f16dd44053", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Module4/m4_01/d01femtoconvolution_video_notes.tex", "max_issues_repo_name": "liuzhengqi1996/math452_Spring2022", "max_issues_repo_head_hexsha": "b01d1d9bee4778b3069e314c775a54f16dd44053", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Module4/m4_01/d01femtoconvolution_video_notes.tex", "max_forks_repo_name": "liuzhengqi1996/math452_Spring2022", "max_forks_repo_head_hexsha": "b01d1d9bee4778b3069e314c775a54f16dd44053", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.9915966387, "max_line_length": 380, "alphanum_fraction": 0.6060344828, "num_tokens": 3905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.819893340314393, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6165700965015202}} {"text": "\n\\subsection{Finding neighbours}\n\nSay we have a distance function and a sample. How can we identify the \\(k\\)-nearest neighbours?\n\nWe can find the distance for all points, sort this and take the top \\(k\\) observations.\n", "meta": {"hexsha": "9c1ae63432b78fde713b6c5bdf8b189e4b5f4646", "size": 219, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/distance/04-01-neighbours.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/distance/04-01-neighbours.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/distance/04-01-neighbours.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.2857142857, "max_line_length": 95, "alphanum_fraction": 0.7579908676, "num_tokens": 49, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8198933183101077, "lm_q2_score": 0.7520125793176223, "lm_q1q2_score": 0.6165700890676684}} {"text": "% Copyright 2018 Melvin Eloy Irizarry-Gelpí\n\\chapter{Hyper Perplex Numbers}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nAn exo-2 split binion has the form\n\\begin{equation}\n a_{0} + a_{1} S + a_{2} W + a_{3} SW + a_{4} X + a_{5} SX + a_{6} WX + a_{7} SWX\n\\end{equation}\nThese follow from a parabolic Cayley-Dickson construct on the exo-1 split binions.\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Arithmetic Operations}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Multiplication}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Conjugate Operations}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Asterisk Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Quadrance}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Cloak Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Dagger Conjugation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Hodge Star}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Differential Operators}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n...", "meta": {"hexsha": "f3a733e50011f6ccbb7a89213010c5444964ba05", "size": 2153, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter/Q.tex", "max_stars_repo_name": "meirizarrygelpi/plexifications", "max_stars_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter/Q.tex", "max_issues_repo_name": "meirizarrygelpi/plexifications", "max_issues_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapter/Q.tex", "max_forks_repo_name": "meirizarrygelpi/plexifications", "max_forks_repo_head_hexsha": "0976772cc56b7f4b6cb9856c6d2bb052fed0b525", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.9318181818, "max_line_length": 84, "alphanum_fraction": 0.2052949373, "num_tokens": 255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9032942119105696, "lm_q2_score": 0.6825737279551494, "lm_q1q2_score": 0.6165648976641062}} {"text": "% !TeX root = main.tex\n\n\\chapter{Prefix Sum and Histogram}\n\\glsresetall\n\\label{chapter:prefixsum}\n\n\\section{Prefix Sum}\n\\label{sec:prefixSum}\n\nPrefix sum is a common kernel used in many applications, e.g., recurrence relations, \ncompaction problems, string comparison, polynomial evaluation, histogram, radix sort, and quick sort \\cite{blelloch1990prefix}. Prefix sum requires restructuring in order to create an efficient FPGA design.\n\nThe prefix sum is the cumulative sum of a sequence of numbers. Given a sequence of inputs $in_n$, the prefix sum $out_n$ is the summation of the first $n$ inputs, namely $out_n = in_0 + in_1 + in_2 + \\cdots + in_{n-1} + in_n$. The following shows the computation for the first four elements of the output sequence $out$.\n\\begin{align*} \nout_0 & = in_0 &\\\\\nout_1 & = in_0 + in_1 &\\\\\nout_2 & = in_0 + in_1 + in_2 \\\\\nout_3 & = in_0 + in_1 + in_2 + in_3 \\\\\n\\cdots\n\\end{align*}\n\nOf course, in practice we don't want to store and recompute the sum of all of the previous inputs, so the prefix sum is often computed by the recurrence equation:\n\\begin{equation}\nout_n = out_{n-1} + in_n\n\\end{equation}\n\nThe disadvantage of the recurrence equation is that we must compute $out_{n-1}$ before computing $out_n$, which fundamentally limits the parallelism and throughput that this computation can be performed. In contrast, the original equations have obvious parallelism where each output can be computed independently at the expense of a significant amount of redundant computation. C code implementing the recurrence equation is shown in Figure \\ref{fig:prefixsumSW}. Ideally, we'd like to achieve $II=1$ for the loop in the code, but this can be challenging even for such simple code. Implementing this code with \\VHLS results in behavior like that shown in Figure \\ref{fig:prefixsumSW}.\n\\begin{figure}\n\\begin{minipage}{.5\\textwidth}\n\\lstinputlisting{examples/prefixsumBO.cpp}\n\\end{minipage}\n\\begin{minipage}{.5\\textwidth}\n\\centering\n\\includesvg{prefixsumBO_behavior}\n\\end{minipage}\n\\caption{ Code for implementing prefix sum, and its accompanying behavior. }\n\\label{fig:prefixsumSW}\n\\end{figure}\n\nThe way this code is written, each output is written into the output memory \\lstinline|out[]| and then in the next iteration is read back out of the memory again. Since the memory read is has a latency of one, data read from memory cannot be processed until the following clock cycle. As a result, such a design can only achieve a loop II of 2. In this case there is a relatively easy way to rewrite the code: we can simply perform the accumulation on a separate local variable, rather than reading the previous value back from the array. Avoiding extra memory accesses in favor of register storage is often advantageous in processor code, but in HLS designs it is often more significant since other operations are rarely a performance bottleneck. Code that does this is shown in Figure \\ref{fig:prefixsum_optimized}.\n\n\\begin{figure}\n\\begin{minipage}{.5\\textwidth}\n\\lstinputlisting{examples/prefixsum_optimized.cpp}\n\\end{minipage}\n\\begin{minipage}{.5\\textwidth}\n\\centering\n\\includesvg{prefixsum_optimized_behavior}\n\\end{minipage}\n\\caption{ Code for implementing an optimized prefix sum, and its accompanying behavior. }\n\\label{fig:prefixsum_optimized}\n\\end{figure}\n\nYou might ask why the compiler is not able to optimize the memory loads and stores automatically in order to improve the II of the design. It turns out that \\VHLS is capable of optimizing loads and stores to array, but only for reads and writes within the scope of a single basic block. You can see this if we unroll the loop, as shown in Figure \\ref{fig:prefixsum_unrolled}. Note that we also have to add appropriate \\lstinline{array_partition} s in order to be able to read and write multiple values at the interfaces. In this case, \\VHLS is able to eliminate most of the read operations of the \\lstinline{out[]} array within the body of the loop, but we still only achieve a loop II of 2. In this case the first load in the body of the loop is still not able to be removed. We can, however, rewrite the code manually to use a local variable rather than read from the \\lstinline{out[]} array.\n\n\\begin{figure}\n\\begin{minipage}{.5\\textwidth}\n\\lstinputlisting{examples/prefixsum_unrolled.cpp}\n\\end{minipage}\n\\begin{minipage}{.5\\textwidth}\n\\raggedleft\n\\includesvg{prefixsum_unrolled_behavior}\n\\end{minipage}\n\\caption{ Optimizing the prefixsum code using \\lstinline{unroll}, \\lstinline{pipeline}, and \\lstinline{array_partition} directives. }\n\\label{fig:prefixsum_unrolled}\n\\end{figure}\n\n%results in the following message:\n%\\lstinputlisting[basicstyle=\\ttfamily\\footnotesize]{examples/prefixsumBO.message}\n\n%\\begin{figure}\n%\\lstinputlisting{examples/prefixsumBO.cpp}\n%\\caption{ The initial code for implementing prefix sum. }\n%\\label{fig:prefixsumSW}\n%\\end{figure}\n\nIdeally, when we unroll the inner loop, we the perform more operations per clock and reduce the interval to compute the function. If we unroll by a factor of two, then the performance doubles. A factor of four would increase the performance by factor four, i.e., the performance scales in a linear manner as it is unrolled. While this is mostly the case, as we unroll the inner loop there are often some aspects of the design that don't change. Under most circumstances, such as when the iteration space of loops execute for a long time, these aspects represent a small fixed overhead which doesn't contribute significantly to the performance of the overall function. However, as the number of loop iterations decreases, these fixed portions of the design have more impact. The largest fixed component of a pipelined loop is the depth of the pipeline itself. The control logic generate by \\VHLS for a pipelined loop requires the pipeline to completely flush before code after the loop can execute.\n\n\\begin{exercise}\nUnroll the \\lstinline{for} loop corresponding to the prefix sum code in Figure \\ref{fig:prefixsum_optimized} by different factors in combination with array partitioning to achieve a loop II of 1. How does the \\lstinline{prefixsum} function latency change? What are the trends in the resource usages? Why do you think you are seeing these trends? What happens when the loop becomes fully unrolled?\n\\end{exercise}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width= .7\\textwidth]{images/architectures_prefixsum}\n\\caption{ Part a) displays an architecture corresponding to the code in Figure~\\ref{fig:prefixsumSW}. The dependence on the \\lstinline{out[]} array can prevent achieving a loop II of 1. Computing the recurrence with a local variable, as shown in the code in Figure~\\ref{fig:prefixsum_optimized}, is able to reduce the latency in the recurrence and achieve a loop II of 1.}\n\\label{fig:architecture_prefixsum}\n\\end{figure}\n\nFigure~\\ref{fig:architecture_prefixsum} shows the hardware architecture resulting from synthesizing the code from Figure \\ref{fig:prefixsumSW} and Figure~\\ref{fig:prefixsum_optimized}. In part a), we can see that the `loop' in the circuit includes the output memory that stores the \\lstinline{out[]} array, whereas in part b), the loop in the circuit includes only a register that stores the accumulated value and the output memory is only written. Simplifying recurrences and eliminating unnecessary memory accesses is a common practice in optimizing HLS code.\n\n%This \\lstinline{A} variable maps to a hardware register. A register is helpful because a it can be read to and written from on every cycle. Unlike a memory, which has limited number of read and write ports, the register value can be read from and subsequently sent to a large number of places that need that data on every cycle with very limited penalty. The only major issues is wire routing, which in the grand scheme of things is not much additional overhead in terms of resources. Nor does it typically incur a large penalty in terms of performance. \n\n\\note{Could add a whole part about doing reduction. Probably not necessary.}\n\\note{Should have something here about what to do with floating point accumulation. This is fundamentally more problematic than what's above (which is relatively easily handled by improving store->load optimization.}\n\nThe goal of this section is to show that even a small changes in the code can sometimes have a significant effect on the hardware design. Some changes may not necessarily be intuitive, but can be identified with feedback from the tool.\n\n\\section{Histogram}\n\\label{sec:histogram}\n\nA Histogram models the probability distribution of a discrete signal. Given a sequence of discrete input values, the histogram counts the number of times each value appears in the sequence. When normalized by the total number of input values, the histogram becomes the probability distribution function of the sequence. Creating a histogram is a common function used in image processing, signal processing, database processing, and many other domains. In many cases, it is common to quantize high-precision input data into a smaller number of intervals or \\term{bins} as part of the histogram computation. For the purpose of this section, we will skip the actual process by which this is done and focus on what happens with the binned data.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width= .7\\textwidth]{images/histogram_introd}\n\\caption{ An example of a histogram. }\n\\label{fig:histogram_introd}\n\\end{figure}\n\nFigure~\\ref{fig:histogram_introd} provides a simple example of a histogram. The data set consists of a sequence of binned values, in this case represented by an integer in $[0,4]$. The corresponding histogram, consisting of a count for each bin, is shown below along with a graphical representation of the histogram, where the height of each bar corresponding to the count of each separate value. Figure~\\ref{fig:histogramSW} shows baseline code for the \\lstinline{histogram} function.\n\n\\begin{figure}\n\\lstinputlisting{examples/histogramSW.cpp}\n\\caption{ Original code for calculating the histogram. The \\lstinline{for} loop iterates across the input array and increments the corresponding element of the \\lstinline{hist} array. }\n\\label{fig:histogramSW}\n\\end{figure}\n\nThe code ends up looking very similar to the prefix sum in the previous section. The difference is that the prefix sum is essentially only performing one accumulation, while in the \\lstinline|histogram| function we compute one accumulation for each bin. The other difference is that in the prefix sum we added the input value each time, in this case we only add 1. When pipelining the inner loops using the \\lstinline|pipeline| directive, we return to the same problem as with the code in Figure \\ref{fig:prefixsumSW}, where we can only achieve a loop II of 2 due to the recurrence through the memory. This is due to the fact that we are reading from the \\lstinline{hist} array and writing to the same array in every iteration of the loop. \nFigure~\\ref{fig:architecture_histogram} shows the hardware architecture for the code in Figure~\\ref{fig:histogramSW}. You can see that the \\lstinline{hist} array has a read and write operation. The \\lstinline{val} variable is used as the index into the \\lstinline{hist} array, and the variable at that index is read out, incremented, and written back into the same location.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width= .5\\textwidth]{images/architectures_histogram}\n\\caption{ An architecture resulting from the code in Figure \\ref{fig:histogramSW}. The \\lstinline{val} data from the \\lstinline{in} array is used to index into the \\lstinline{hist} array. This data is incremented and stored back into the same location.}\n\\label{fig:architecture_histogram}\n\\end{figure}\n\n\\section{Histogram Optimization and False Dependencies}\nLet's look deeper at the recurrence here. In the first iteration of the loop, we read the \\lstinline{hist} array at some location $x_0$ and write back to the same location $x_0$. The read operation has a latency of one clock cycle, so the write has to happen in the following clock. Then in the next iteration of the loop, we read at another location $x_1$. Both $x_0$ and $x_1$ are dependent on the input and could take any value, so we consider the worst case when generating the circuit. In this case, if $x_0 == x_1$, then the read at location $x_1$ cannot begin until the previous write has completed. As a result, we must alternate between reads and writes. \n\nIt turns out that we must alternate between reads and writes as long as $x_0$ and $x_1$ are independent. What if they are {\\em not} actually independent? For instance, we might know that the source of data never produces two consecutive pieces of data that actually have the same bin. What do we do now? If we could give this extra information to the HLS tool, then it would be able to read at location $x_1$ while writing at location $x_0$ because it could guarantee that they are different addresses. In \\VHLS, this is done using the \\lstinline|dependence| directive.\n\nThe modified code is shown in Figure \\ref{fig:histogram_dependence}. Here we've explicitly documented (informally) that the function has some preconditions. In the code, we've added an \\lstinline|assert()| call which checks the second precondition.\nin \\VHLS, this assertion is enabled during simulation to ensure that the simulation testvectors meet the required precondition. The \\lstinline|dependence| directive captures the effect of this precondition on the circuit, generated by the tool. Namely, it indicates to \\VHLS that reads and writes to the \\lstinline|hist| array are dependent only in a particular way. In this case, \\lstinline|inter|-iteration dependencies consisting of a read operation after a write operation (RAW) have a distance of 2. In this case a distance of $n$ would indicate that read operations in iteration $i+n$ only depend on write operations in iteration $i$. In this case, we assert that \\lstinline|in[i+1] != in[i]|, but it could be the case that \\lstinline|in[i+2] == in[i]| so the correct distance is 2.\n\n\\begin{figure}\n\\lstinputlisting{examples/histogram_dependence.cpp}\n\\caption{ An alternative function for computing a histogram. By restricting the inputs and indicating this restriction to \\VHLS via the \\lstinline|dependence| directive, II=1 can be achieved without significantly altering the code. }\n\\label{fig:histogram_dependence}\n\\end{figure}\n\n\\begin{exercise}\nIn Figure \\ref{fig:histogram_dependence}, we added a precondition to the code, checked it using an assertion, and indicated the effect of the precondition to the tool using the \\lstinline|dependence| directive. What happens if your testbench violates this precondition? What happens if you remove the \\lstinline|assert()| call? Does \\VHLS still check the precondition? What happens if the precondition is not consistent with the \\lstinline|dependence| directive?\n\\end{exercise}\n\nUnfortunately, the \\lstinline{dependence} directive doesn't really help us if we are unwilling to accept the additional precondition. It's also clear that we can't directly apply same optimization as with the \\lstinline|prefixsum| function, since we might need to use all of the values stored in the \\lstinline|hist| array. Another alternative is implement the \\lstinline|hist| array with a different technology, for instance we could partition the \\lstinline|hist| array completely resulting in the array being implemented with \\gls{ff} resources. Since the data written into a \\gls{ff} on one clock cycle is available immediately on the next clock cycle, this solves the recurrence problem and can be a good solution when a small number of bins are involved. The architecture resulting from such a design is shown in Figure \\ref{fig:histogram_partitioned}. However, it tends to be a poor solution when a large number of bins are required. Commonly histograms are constructed with hundreds to thousands of bins and for large data sets can require many bits of precision to count all of the inputs. This results in a large number of FF resources and a large mux, which also requires logic resources. Storing large histograms in \\gls{bram} is usually a much better solution.\n\n\\begin{figure}\n\\centering\n\\note{fixme!}\n\\includegraphics[width= .5\\textwidth]{images/architectures_histogram}\n\\caption{ An architecture resulting from the code in Figure \\ref{fig:histogramSW} when the \\lstinline|hist| array is completely partitioned.}\n\\label{fig:histogram_partitioned}\n\\end{figure}\n\nReturning to the code in Figure~\\ref{fig:histogramSW}, we see that there are really two separate cases that the architecture must be able to handle. One case is when the input contains consecutive values in the same bin. In this case, we'd like to use a simple register to perform the accumulation with a minimal amount of delay. The second case is when the input does not contain consecutive values in the same bin, in which case we need to read, modify, and write back the result to the memory. In this case, we can guarantee that the read operation of the \\lstinline|hist| array can not be affected by the previous write operation. We've seen that both of these cases can be implemented separately, perhaps we can combine them into a single design. The code to accomplish this is shown in Figure~\\ref{fig:histogramOpt1}. This code uses a local variable \\lstinline|old| to store the bin from the previous iteration and another local variable \\lstinline{accu} to store the count for that bin. Each time through the loop we check to see if we are looking at the same bin as the previous iteration. If so, then we can simply increment \\lstinline|accu|. If not, then we need to store the value in \\lstinline|accu| in the \\lstinline|hist| array and increment the correct value in the \\lstinline|hist| array instead. In either case, we update \\lstinline|old| and \\lstinline|accu| to contain the current correct values. The architecture corresponding to this code is shown in Figure \\ref{fig:architecture_histogram_restructured}. \n\nIn this code, we still need a \\lstinline|dependence| directive, just as in Figure \\ref{fig:histogram_dependence}, however the form is slightly different. In this case the read and write accesses are to two different addresses in the same loop iteration. Both of these addresses are dependent on the input data and so could point to any individual element of the \\lstinline|hist| array. Because of this, \\VHLS assumes that both of these accesses could access the same location and as a result schedules the read and write operations to the array in alternating cycles, resulting in a loop II of 2. However, looking at the code we can readily see that \\lstinline|hist[old]| and \\lstinline|hist[val]| can never access the same location because they are in the \\lstinline|else| branch of the conditional \\lstinline|if(old == val)|. Within one iteration (an \\lstinline|intra|-dependence) a read operation after a write operation (\\lstinline|RAW|) can never occur and hence is a \\lstinline|false| dependence. In this case we are not using the \\lstinline|dependence| directive to inform the tool about a precondition of the function, but instead about a property of the code itself. \n\n\\begin{figure}\n\\lstinputlisting{examples/histogram_opt1.cpp}\n\\caption{ Removing the read after write dependency from the \\lstinline{for} loop. This requires an \\lstinline{if/else} structure that may seem like it is adding unnecessary complexity to the design. However, it allows for more effective pipelining despite the fact that the datapath is more complicated. }\n\\label{fig:histogramOpt1}\n\\end{figure}\n\n\\begin{exercise}\nSynthesize the code from Figure \\ref{fig:histogramSW} and Figure \\ref{fig:histogramOpt1}. What is the initiation interval (II) in each case? What happens when you remove the \\lstinline{dependence} directive from the code in Figure \\ref{fig:histogramOpt1}? How does the loop interval change in both cases? What about the resource usage?\n\\end{exercise}\n\n\\begin{aside}\nFor the code in Figure \\ref{fig:histogramOpt1}, you might question why a tool like \\VHLS cannot determine this property. In fact, while in some simple cases like this one better code analysis could propagate the \\lstinline|if| condition property into each branch, we must accept that there are some pieces of code where properties of memory accesses are actually undecidable. The highest performance in such cases will only be achieved in a static schedule with the addition of user information. Several recent research works have looked to improve this by introducing some dynamic control logic into the design\\cite{winterstein13dynamic, liu17elasticflow, dai17dynamic}.\n\\end{aside}\n\nA pictorial description of the restructured code from Figure \\ref{fig:histogramOpt1} is shown in Figure \\ref{fig:architecture_histogram_restructured}. Not all of the operations are shown here, but the major idea of the function is there. You can see the two separate \\lstinline{if} and \\lstinline{else} regions (denoted by dotted lines). The \\lstinline{acc} variable is replicated twice in order to make the drawing more readable; the actual design will only have one register for that variable. The figure shows the two separate datapaths for the \\lstinline{if} and the \\lstinline{else} clause with the computation corresponding to the \\lstinline{if} clause on the top and the \\lstinline{else} clause datapath on the bottom. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width= .6\\textwidth]{images/architectures_histogram_restructured}\n\\caption{ A depiction of the datapath corresponding to the code in Figure \\ref{fig:histogramOpt1}. There are two separate portions corresponding to the \\lstinline{if} and \\lstinline{else} clauses. The figure shows the important portions of the computation, and leaves out some minor details. }\n\\label{fig:architecture_histogram_restructured}\n\\end{figure}\n\n\\section{Increasing Histogram Performance}\n\nWith some effort, we've achieved a design with a loop II of 1. Previously we have seen how further reducing the execution time of a design can be achieved by partial unrolling of the inner loop. However, with the \\lstinline|histogram| function this is somewhat difficult for several reasons. One reason is the challenging recurrence, unless we can break up the input data in some fashion, the computation of one iteration of the loop must be completed with the computation of the next iteration of the loop. A second reason is that with a loop II of 1, the circuit performs a read and a write of the \\lstinline|hist| array each clock cycle, occupying both ports of a \\gls{bram} resource in the FPGA. Previously we have considered \\gls{arraypartitioning} to increase the number of memory ports for accessing an array, but there's not a particularly obvious way to partition the \\lstinline|hist| array since the access order depends on the input data.\n\nAll is not lost, however, as there is a way we can expose more parallelism by decomposing the histogram computation into two stages. In the first stage, we divide the input data into a number of separate partitions. The histogram for each partition can be computed independently using a separate instance, often called a \\gls{pe}, of the histogram solution we've developed previously. In the second stage, the individual histograms are combined to generate the histogram of the complete data sets. This partitioning (or mapping) and merging (or reducing) process is very similar to that adopted by the MapReduce framework \\cite{dean08mapreduce} and is a common pattern for parallel computation. The map-reduce pattern is applicable whenever there is recurrence which includes a commutative and associative operation, such as addition in this case. This idea is shown in Figure~\\ref{fig:architecture_histogram_parallel}. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width= .9\\textwidth]{images/architectures_histogram_parallel}\n\\caption{The histogram computation implemented using a map-reduce pattern. The processing element (PE) in Part a) is the same architecture as shown in Figure \\ref{fig:architecture_histogram_restructured}. The \\lstinline|in| array is partitioned and each partition is processed by a separate \\gls{pe}. The merge block combines the individual histograms to create the final histogram. }\n\\label{fig:architecture_histogram_parallel}\n\\end{figure}\n\nThe code for implementing this architecture is shown in Figure \\ref{fig:histogram_parallel}. The \\lstinline{histogram_map} function implements the `map' portion of the map-reduce pattern and will be instantiated multiple times. The code is very similar to the code in Figure \\ref{fig:histogramOpt1}. The main difference is that we have added the additional code to initialize the \\lstinline|hist| array. The \\lstinline{histogram_map} function takes an input array \\lstinline{in} which will contain a partition of the data being processed and computes the histogram of that partition in the \\lstinline|hist| array. The \\lstinline{histogram_reduce} function implements the `reduce' portion of the pattern. It takes as input a number of partial histograms and combines them into complete histogram by adding together the count for each histogram bin. In our code example in Figure \\ref{fig:histogram_parallel}, we have only two processing elements, thus the merge has two input arrays \\lstinline{hist1} and \\lstinline{hist2}. This can easily be extended to handle more processing elements.\n\nThe new \\lstinline{histogram} function takes as an input two partitions of the input data, stored in the \\lstinline{inputA} and \\lstinline{inputB} arrays. It computes the histogram of each partition using the \\lstinline {histogram_map} function, which are then stored in the \\lstinline{hist1} and \\lstinline{hist2} arrays. These are feed into the \\lstinline{histogram_reduce} function which combines them and stores the result in the \\lstinline{hist} array, which is the final output of the top level function \\lstinline{histogram}. \n\n\\begin{figure}\n\\lstinputlisting{examples/histogram_parallel.cpp}\n\\caption{ Another implementation of histogram that uses task level parallelism and pipelining. The histogram operation is split into two sub tasks, which are executed in the two \\lstinline{histogram_map} functions. These results are combined in the final histogram result using the \\lstinline{histogram_reduce} function. The \\lstinline{histogram} function is the top level function that connects these three functions together. }\n\\label{fig:histogram_parallel}\n\\end{figure}\n\n\\begin{exercise}\nModify the code in Figure \\ref{fig:histogram_parallel} to support a parameterizable number \\lstinline|NUM_PE| of \\glspl{pe}? Hint: You'll need to combine some of the arrays into a single array that is correctly partitioned and add some loops that depend on \\lstinline|NUM_PE|. What happens to the throughput and task interval as you vary the number of \\glspl{pe}? \n\\end{exercise}\n\nWe use the \\lstinline{dataflow} directive in the \\lstinline{histogram} function in order to enable a design with \\gls{taskpipelining}. In this case there are three processes: two instances of the \\lstinline{histogram_map} function and one instance of the \\lstinline{histogram_reduce} function. Within a single task, the two \\lstinline{histogram_map} processes can execute concurrently since they work on independent data, while the \\lstinline{histogram_reduce} function must execute after since it uses the results from the \\lstinline{histogram_map} processes. Thus, the \\lstinline{dataflow} directive essentially creates a two stage task pipeline with the \\lstinline{histogram_map} functions in the first stage and the \\lstinline{histogram_reduce} function in the second stage. As with any dataflow design, the interval of the entire \\lstinline{histogram} function depends upon the maximum initiation interval of the two stages. The two \\lstinline{histogram_map} functions in the first stage are the same and will have the same interval ($II_\\mathrm{histogram\\_map}$). The \\lstinline{histogram_reduce} function will have another interval ($II_\\mathrm{histogram\\_reduce}$). The interval of the toplevel \\lstinline{histogram} function $II_\\mathrm{histogram}$ is then $\\max (II_\\mathrm{histogram\\_map}, II_\\mathrm{histogram\\_reduce})$.\n\n\\begin{exercise}\nWhat happens when you add or change the locations of the \\lstinline{pipeline} directives? For example, is it beneficial to add a \\lstinline{pipeline} directive to the \\lstinline{for} loop in the \\lstinline{histogram_reduce} function? What is the result of moving the \\lstinline{pipeline} directive into the \\lstinline{histogram_map} function, i.e., hoisting it outside of the \\lstinline{for} loop where it currently resides? \n\\end{exercise}\n\nThe goal of this section was to walk through the optimization the histogram computation, another small but important kernel of many applications. The key takeaway is that there are often limits to what tools can understand about our programs. In some cases we must take care in how we write the code and in other cases we must actually give the tool more information about the code or the environment that the code is executing in. In particular, properties about memory access patterns often critically affect the ability of HLS to generate correct and efficient hardware. In \\VHLS, these properties can be expressed using the \\lstinline|dependence| directive. Sometimes these optimizations might even be counter-intuitive, such as the addition of the \\lstinline{if/else} control structure in \\ref{fig:histogramOpt1}. In other cases optimizations might require some creativity, as in applying the map-reduce pattern in Figures \\ref{fig:architecture_histogram_parallel} and \\ref{fig:histogram_parallel}. \n\n\\section{Conclusion}\nIn this section, we've looked at the prefix sum and histogram kernels. Although these functions seem different, they both contain recurrences through a memory access. These recurrences can limit throughput if the memory access is not pipelined. In both cases, by rewriting the code we can remove the recurrence. In the case of the prefix sum, this is much easier since the access patterns are deterministic. In the case of the histogram we must rewrite the code to address the recurrence or ensure that recurrence never happens in practice. In either case we needed a way to describe to \\VHLS information about the environment or about the code itself that the tool was unable to determine for itself. This information is captured in the \\lstinline{dependence} directive. Lastly, we looked at ways of parallelizing both algorithms yet further, so that they could process a number of data samples each clock cycle.", "meta": {"hexsha": "33bdd3574be8bedc24306c41d385fa605ec893ac", "size": 30720, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "prefix_histogram.tex", "max_stars_repo_name": "mithro/pp4fpgas", "max_stars_repo_head_hexsha": "ddede5bd337f4fa33915d7e4ca98f97a7b31413a", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 418, "max_stars_repo_stars_event_min_datetime": "2018-05-09T17:28:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T05:51:12.000Z", "max_issues_repo_path": "prefix_histogram.tex", "max_issues_repo_name": "jmuuu/pp4fpgas", "max_issues_repo_head_hexsha": "f604b68289b0a9998ace596c57ea7209f2d60fd5", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 19, "max_issues_repo_issues_event_min_datetime": "2018-05-13T16:26:23.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-06T06:06:57.000Z", "max_forks_repo_path": "prefix_histogram.tex", "max_forks_repo_name": "jmuuu/pp4fpgas", "max_forks_repo_head_hexsha": "f604b68289b0a9998ace596c57ea7209f2d60fd5", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 107, "max_forks_repo_forks_event_min_datetime": "2018-05-12T16:43:35.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-23T22:59:42.000Z", "avg_line_length": 139.6363636364, "max_line_length": 1538, "alphanum_fraction": 0.7965494792, "num_tokens": 7142, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8104789109591831, "lm_q1q2_score": 0.6164913126022805}} {"text": "\\documentclass[11pt,a4paper]{report}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{gensymb}\n\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\usetikzlibrary{positioning}\n\\usepackage{geometry}\n\\geometry{\n left=2cm,\n right=0.64cm,\n top=0.64cm,\n bottom=2cm\n}\n\\usepackage{multicol}\n\\setlength{\\columnsep}{1cm}\n\\graphicspath{ {images/} }\n\n\\begin{document}\n\n\\chapter{Semester 2 Examination 2015-2016\\\\CZ4041 Machine Learning}\n\n\\begin{multicols*}{2}\n\n\\section{Question 1}\n\\noindent \\textbf{Question 1a}\n\\noindent Information given in the question:\n\\begin{itemize}\n \\item $P(D) = 0.001$\n \\item $P(\\sim D) = 0.999$\n \\item $P(T_1|D) = 0.9$\n \\item $P(T_2|D) = 0.95$\n \\item $P(T_1|\\sim D) = 0.01$\n \\item $P(T_2|\\sim D) = 0.1$\n\\end{itemize}\n\n\\begin{equation*}\n\\begin{split}\nP(D|T_1 T_2) &= \\frac{P(T_1 | D) P(T_2 | D) P(D)}{P(T_1 T_2)}\\\\\n&= \\frac{0.9 \\times 0.95 \\times 0.001}{P(T_1 T_2)}\\\\\n&= \\frac{0.855 \\times 10^{-3}}{P(T_1 T_2)}\n\\end{split}\n\\end{equation*}\n\n\\begin{equation*}\n\\begin{split}\nP(\\sim D|T_1 T_2) &= \\frac{P(T_1 | \\sim D) P(T_2 | \\sim D) P(\\sim D)}{P(T_1 T_2)}\\\\\n&= \\frac{0.01 \\times 0.1 \\times 0.999}{P(T_1 T_2)}\\\\\n&= \\frac{0.999 \\times 10^{-3}}{P(T_1 T_2)}\n\\end{split}\n\\end{equation*}\n\n\\noindent Since $P(\\sim D|T_1 T_2) > P(D|T_1 T_2)$, we predict that the patient does not have the disease.\\\\\n\n\\noindent \\textbf{Question 1b}\n\n\\noindent Let action $a_1$ be the action predicting the patient to have the disease, and action $a_2$ be the action predicting the patient to not have the disease.\n\n$$R(a_1|T_1 T_2) = 1 - P( D | T_1 T_2)$$\n$$R(a_2|T_1 T_2) = 1 - P( \\sim D | T_1 T_2)$$\n\n\\noindent We choose the action with minimum risk. Since $P(\\sim D|T_1 T_2) > P(D|T_1 T_2)$, thus $R(a_2|T_1 T_2) < R(a_1|T_1 T_2)$, so the action $a_2$ has lower risk. As a result, we predict that the patient does not have the disease.\\\\\n\n\\noindent \\textbf{Question 1c}\n\n\\noindent Information given in the question:\n\\begin{itemize}\n \\item $\\lambda_{12} = 0.05$\n \\item $\\lambda_{21} = 1$\n\\end{itemize}\n\n\\noindent Where $\\lambda_{12}$ is the lost occur when the action is predict the patient to have disease $a_1$, but the ground truth is the patient do not have the disease $\\sim D$. The lost $\\lambda_{21}$ is defined in similar way. The risks are:\n\n\\begin{equation*}\n\\begin{split}\nR(a_1|T_1 T_2) &= \\lambda_{12} P(\\sim D | T_1 T_2)\\\\\n&= 0.05 \\times \\frac{0.999 \\times 10^{-3}}{P(T_1 T_2)}\\\\\n&= \\frac{0.4995 \\times 10^{-3}}{P(T_1 T_2)}\n\\end{split}\n\\end{equation*}\n\n\\begin{equation*}\n\\begin{split}\nR(a_2|T_1 T_2) &= \\lambda_{21} P( D | T_1 T_2)\\\\\n&= 1 \\times \\frac{0.855 \\times 10^{-3}}{P(T_1 T_2)}\\\\\n&= \\frac{0.855 \\times 10^{-3}}{P(T_1 T_2)}\\\\\n\\end{split}\n\\end{equation*}\n\n\\noindent Since $R(a_1|T_1 T_2) < R(a_2|T_1 T_2)$, we predict that the patient has the disease.\n\n\\section{Question 2}\n\\noindent \\textbf{Question 2a}\n\n\\noindent We define the inputs to Hidden Layer as:\n\n$$X = \\begin{bmatrix} X_1 \\\\ X_2 \\end{bmatrix}$$\n\n\\noindent The weight from Input Layer to Hidden Layer is:\n\n$$W =\n\\begin{bmatrix}\nw_{13} & w_{23} \\\\\nw_{14} & w_{24}\n\\end{bmatrix}$$\n\n\\noindent The activation function is:\n\n$$\nf(u) =\n\\begin{cases}\n1 & u \\ge 0\\\\\n-1 & u < 0\n\\end{cases}\n$$\n\n\\noindent The output of the Hidden Layer (which is also the input to Output Layer) is:\n\n$$Z = f(W X)$$\n\n\\noindent The weight from Hidden Layer to Output Layer is:\n\n$$V =\n\\begin{bmatrix}\nw_{35} & w_{45}\n\\end{bmatrix}$$\n\n\\noindent The output of Output Layer is:\n\n$$y = f(V Z)$$\n\n\\noindent For each $X$, we calculate $y$:\n\n\\begin{center}\n\\begin{tabular}{|c | c | c | c |}\n\\hline\n$X_1$ & $X_2$ & $y$ & Predict \\\\ \\hline\n2 & -0.5 & 1 & + \\\\\n1 & 1 & 1 & - \\\\\n3 & 1 & 1 & - \\\\\n2 & -2 & -1 & + \\\\\n1.5 & 2 & 1 & - \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent The error rate is $80\\%$.\\\\\n\n\\noindent \\textbf{Question 2b}\n\n\\noindent We define $X$ as:\n\n$$X =\n\\begin{bmatrix}\n2 & -0.5 \\\\\n1 & 1 \\\\\n3 & 1 \\\\\n2 & -2 \\\\\n1.5 & 2\n\\end{bmatrix}$$\n\n\\noindent The pairwise inner products between the five data points is:\n\n$$X X^{T} =\n\\begin{bmatrix}\n4.25 & 1.50 & 5.50 & 5.00 & 2.00\\\\\n1.50 & 2.00 & 4.00 & 0.00 & 3.50\\\\\n5.50 & 4.00 & 10.00 & 4.00 & 6.50\\\\\n5.00 & 0.00 & 4.00 & 8.00 & -1.00\\\\\n2.00 & 3.50 & 6.50 & -1.00 & 6.25\n\\end{bmatrix}$$\n\n\\noindent \\textbf{Question 2c}\n\n\\begin{tikzpicture}\n\\begin{axis}[\n enlargelimits=false,\n axis equal\n]\n\\addplot+[\n nodes near coords,\n only marks,\n point meta=explicit symbolic\n]\ntable[meta=label] {\n x y label\n 2 -0.5 ID1\n 1 1 ID2\n 3 1 ID3\n 2 -2 ID4\n 1.5 2 ID5\n 7 4 ID6\n 5 0.2 Q2d\n};\n\\addplot [\n domain=0:7,\n samples=2,\n color=red,\n]\n{0.25};\n\\end{axis}\n\\end{tikzpicture}\n\n\\noindent The decision boundary is $y=0.25$.\\\\\n\n\\noindent \\textbf{Question 2d}\n\n\\noindent Predict the test data point to be + class.\n\n\\section{Question 3}\n\n\\noindent \\textbf{Question 3a}\n\n\\noindent If we discard the last eigenvector to reduce the dimension to 2:\n\n$$\\text{POV} = \\frac{9.78 + 2.11}{9.78 + 2.11 + 0.11} = 0.991$$\n\n\\noindent If we discard the last eigenvector to reduce the dimension to 1:\n\n$$\\text{POV} = \\frac{9.78}{9.78 + 2.11 + 0.11} = 0.815$$\n\n\\noindent \\textbf{Question 3b}\n\n$$z = U_{\\text{reduce}} x^{T}$$\n$$U_{\\text{reduce}} = \\begin{bmatrix}\n0.78 & -0.54 & 0.32 \\\\\n0.04 & 0.56 & 0.83\n\\end{bmatrix}$$\n\n\\begin{center}\n\\begin{tabular}{| c | c | c |}\n\\hline\nData & Original & Reduced\\\\ \\hline\n$x_1$ & $(2,0,2)$ & $(2.20,1.74)$ \\\\\n$x_2$ & $(-1,0,-1)$ & $(-1.10,0.87)$ \\\\\n$x_3$ & $(3,-3,0)$ & $(3.96,-1.56)$ \\\\\n$x_4$ & $(-3,2,-2)$ & $(-4.06,-0.66)$ \\\\\n$x_5$ & $(-1,1,1)$ & $(-1.00,1.35)$ \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent \\textbf{Question 3c}\n\n\\begin{center}\n\\begin{tabular}{| c | c c c c c |} \\hline\n & $x_1$ & $x_2$ & $x_3$ & $x_4$ & $x_5$ \\\\ \\hline\n$x_1$ & 0 & 3.4128 & 3.7400 & 6.7043 & 3.2237 \\\\\n$x_2$ & 3.4128 & 0 & 5.6132 & 3.3320 & 0.49031 \\\\\n$x_3$ & 3.7400 & 5.6132 & 0 & 8.0703 & 5.7506 \\\\\n$x_4$ & 6.7043 & 3.3320 & 8.0703 & 0 & 3.6611 \\\\\n$x_5$ & 3.2237 & 0.49031 & 5.7506 & 3.6611 & 0 \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent We merge $x_2$ and $x_5$:\n\n\\begin{center}\n\\begin{tabular}{| c | c c c c |} \\hline\n & $x_2 \\cap x_5$ & $x_1$ & $x_3$ & $x_4$ \\\\ \\hline\n$x_2 \\cap x_5$ & 0 & 3.2237 & 5.6132 & 3.3320 \\\\\n$x_1$ & 3.2237 & 0 & 3.7400 & 6.7043 \\\\\n$x_3$ & 5.6132 & 3.7400 & 0 & 8.0703 \\\\\n$x_4$ & 3.3320 & 6.7043 & 8.0703 & 0 \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent We merge $x_2 \\cap x_5$ with $x_1$:\n\n\\begin{center}\n\\begin{tabular}{| c | c c c |} \\hline\n & $x_2 \\cap x_5 \\cap x_1$ & $x_3$ & $x_4$ \\\\ \\hline\n$x_2 \\cap x_5 \\cap x_1$ & 0 & 3.7400 & 3.3320 \\\\\n$x_3$ & 3.7400 & 0 & 8.0703 \\\\\n$x_4$ & 3.3320 & 8.0703 & 0 \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent We merge $x_2 \\cap x_5 \\cap x_1$ with $x_4$:\n\n\n\\begin{center}\n\\begin{tabular}{| c | c c |} \\hline\n & $x_2 \\cap x_5 \\cap x_1 \\cap x_4$ & $x_3$ \\\\ \\hline\n$x_2 \\cap x_5 \\cap x_1 \\cap x_4$ & 0 & 3.7400 \\\\\n$x_3$ & 3.7400 & 0 \\\\ \\hline\n\\end{tabular}\n\\end{center}\n\n\\noindent Finally we merge $x_3$ to $x_2 \\cap x_5 \\cap x_1 \\cap x_4$.\n\n\\begin{tikzpicture}[sloped]\n\\node (a) at (0,0) {a};\n\\node (b) at (1,0) {b};\n\\node (c) at (2,0) {c};\n\\node (d) at (3,0) {d};\n\\node (e) at (4,0) {e};\n\\node (dummy) at (3.5,5) {};\n\n\\node (ab) at (0.5,0.4903) {};\n\\node (abc) at (1.5,3.00) {}; % change the value abit from 3.22\n\\node (abcd) at (2.5,3.30) {};\n\\node (abcde) at (3.5,3.74) {};\n\n\\draw (a) |- (ab.center);\n\\draw (b) |- (ab.center);\n\\draw (ab) |- (abc.center);\n\\draw (c) |- (abc.center);\n\\draw (abc) |- (abcd.center);\n\\draw (d) |- (abcd.center);\n\\draw (abcd) |- (abcde.center);\n\\draw (e) |- (abcde.center);\n\\draw (abcde) |- (dummy.center);\n\n\\end{tikzpicture}\n\n\\noindent \\textbf{Question 3d}\n\\noindent [Help wanted!]\\\\\n\n\\section{Question 4}\n\n\\noindent \\textbf{Question 4a}\n\\noindent [Copied and pasted from lecture notes] Steps:\n\\begin{itemize}\n\\item Train C1 with dataset D1\n\\item Test C1 with dataset D2\n\\item Find error samples and an equal amount of correct samples and form D3\n\\item Train C2 with dataset D3\n\\item Find error samples and equal amount of correct samples and form D4\n\\item Train C1and C2 with dataset D4\n\\item Find errors and equal number of correct samples and form D5\n\\item Train C3 with D5...\n\\end{itemize}\n\n\\noindent \\textbf{Question 4b}\n\\noindent [Help wanted!]\\\\\n\n\\noindent \\textbf{Question 4c i}\n\n\\begin{equation*}\n\\begin{split}\nP(A) &= \\sum_{B,E} P(A,B,E)\\\\\n &= \\sum_{B,E} P(A | B,E) P(B,E)\\\\\n &= \\sum_{B,E} P(A | B,E) P(B) P(E)\\\\\n &= P(A | B, E) P( B) P( E) +\\\\\n &\\ \\ \\ \\ P(A | B,\\sim E) P( B) P(\\sim E) +\\\\\n &\\ \\ \\ \\ P(A | \\sim B, E) P(\\sim B) P( E) +\\\\\n &\\ \\ \\ \\ P(A | \\sim B,\\sim E) P(\\sim B) P(\\sim E)\\\\\n &= 0.95 \\times 0.01 \\times 0.02 +\\\\\n &\\ \\ \\ \\ 0.90 \\times 0.01 \\times 0.98 +\\\\\n &\\ \\ \\ \\ 0.25 \\times 0.99 \\times 0.02 +\\\\\n &\\ \\ \\ \\ 0.01 \\times 0.99 \\times 0.98 \\\\\n &= 0.02366\\\\\nP(\\sim A) &= 0.9763\n\\end{split}\n\\end{equation*}\n\n\\noindent \\textbf{Question 4c ii}\n\n\\begin{equation*}\n\\begin{split}\nP(S) &= \\sum_{A} P(S,A)\\\\\n &= \\sum_{A} P(S|A)P(A)\\\\\n &= P(S| A)P( A) +\\\\\n &\\ \\ \\ \\ P(S|\\sim A)P(\\sim A)\\\\\n &= 0.70 \\times 0.02366 + 0.05 \\times 0.9763\\\\\n &= 0.06538\n\\end{split}\n\\end{equation*}\n\n\\noindent \\textbf{Question 4c iii}\n\n\\begin{equation*}\n\\begin{split}\nP(S|B) &= \\sum_{A} P(S,A|B)\\\\\n &= \\sum_{A} \\frac{P(S,A,B)}{P(B)}\\\\\n &= \\sum_{A} \\frac{P(S|A,B)P(A,B)}{P(B)}\\\\\n &= \\sum_{A} \\frac{P(S|A,B)P(A|B)P(B)}{P(B)}\\\\\n &= \\sum_{A} P(S|A,B)P(A|B)\\\\\n &= \\sum_{A} P(S|A)P(A|B)\\\\\n &= P(S| A)P( A|B)+\\\\\n &\\ \\ \\ \\ P(S|\\sim A)P(\\sim A|B)\\\\\n &= 0.70 \\times P(A|B) + 0.05 \\times P(\\sim A|B)\\\\\n\\end{split}\n\\end{equation*}\n\n\\begin{equation*}\n\\begin{split}\nP(A|B) &= \\sum_{E} P(A,E|B)\\\\\n &= \\sum_{E} \\frac{P(A,E,B)}{P(B)}\\\\\n &= \\sum_{E} \\frac{P(A|E,B)P(E,B)}{P(B)}\\\\\n &= \\sum_{E} \\frac{P(A|E,B)P(E)P(B)}{P(B)}\\\\\n &= \\sum_{E} P(A|E,B)P(E)\\\\\n &= P(A|E,B)P(E) + P(A|\\sim E,B)P(\\sim E)\\\\\n &= 0.95 \\times 0.02 + 0.90 \\times 0.98\\\\\n &= 0.901\\\\\nP(\\sim A|B) &= 0.099\\\\\nP(S|B) &= 0.70 \\times 0.901 + 0.05 \\times 0.099\\\\\n &= 0.6357\n\\end{split}\n\\end{equation*}\n\n\\begin{equation*}\n\\begin{split}\nP(B|S) &= \\frac{P(S|B)P(B)}{P(S)}\\\\\n &= \\frac{0.6357 \\times 0.01}{0.06538}\\\\\n &= 0.09723\n\\end{split}\n\\end{equation*}\n\n\n\\end{multicols*}\n\\end{document}", "meta": {"hexsha": "1aa180f5dcf71b0e52a6a39663f374b77ae225f0", "size": 10970, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cz4041/cz4041-15-16-2.tex", "max_stars_repo_name": "Andyccs/ntucs-pyp", "max_stars_repo_head_hexsha": "b0bcf074b86da28e3b6af08d618652487e7e880c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, 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"alphanum_fraction": 0.5424794895, "num_tokens": 4820, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8104789018037399, "lm_q1q2_score": 0.6164913056381867}} {"text": "\\documentclass[a4paper]{scrartcl}\n\\usepackage{amssymb, amsmath} % needed for math\n\\usepackage[utf8]{inputenc} % this is needed for umlauts\n\\usepackage[ngerman]{babel} % this is needed for umlauts\n\\usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf\n\\usepackage[margin=2.5cm]{geometry} %layout\n\\usepackage{hyperref} % links im text\n\\usepackage{parskip} % no indentation on new paragraphs\n\\usepackage{color}\n\\usepackage{framed}\n\\usepackage{enumerate} % for advanced numbering of lists\n\\usepackage{algorithm,algpseudocode}\n\\usepackage{braket} % needed for \\Set\n\\clubpenalty = 10000 % Schusterjungen verhindern\n\\widowpenalty = 10000 % Hurenkinder verhindern\n\n\\hypersetup{\n pdfauthor = {Martin Thoma},\n pdfkeywords = {Google Code Jam, Round 1C 2013, Pogo},\n pdftitle = {Proof of correctness for an algorithm for pogo}\n}\n\n% From http://www.matthewflickinger.com/blog/archives/2005/02/20/latex_mod_spacing.asp\n% Thanks!\n\\makeatletter\n\\def\\imod#1{\\allowbreak\\mkern10mu({\\operator@font mod}\\,\\,#1)}\n\\makeatother\n\n\\renewcommand{\\algorithmicrequire}{\\textbf{Input: }}\n\\renewcommand{\\algorithmicensure}{\\textbf{Output: }}\n\n\\newenvironment{myindentpar}[1]%\n {\\begin{list}{}%\n {\\setlength{\\leftmargin}{#1}}%\n \\item[]%\n }\n {\\end{list}}\n\n\\begin{document}\n\\section{The Problem}\nYou're on a two-dimensional grid $\\mathbb{Z} \\times \\mathbb{Z}$ and\nhave to find a way to get to one coordinate $(x,y) \\in \\mathbb{Z} \\times \\mathbb{Z}$. You start at\n$(0, 0)$.\n\nIn your\n$i$-th step you move either $\\underbrace{(+i,0)}_{=: E}$,\n$\\underbrace{(-i,0)}_{=: W}$, $\\underbrace{(0,+i)}_{=: N}$ or\n$\\underbrace{(0,-i)}_{=: S}$.\n\n\\section{The algorithm}\n\\begin{algorithm}\n \\begin{algorithmic}\n \\Function{calculateSteps}{$x \\in \\mathbb{Z}$, $y \\in \\mathbb{Z}$}\n \\State $s \\gets 0$\n \\State $dist \\gets |x| + |y|$\n \\\\\n \\While{$\\overbrace{\\frac{s^2 + s}{2} < dist}^\\text{condition 1}$ or $\\overbrace{\\frac{s^2 + s}{2} \\not\\equiv dist \\imod{2}}^\\text{condition 2}$}\n \\State $s \\gets s + 1$\n \\EndWhile\n \\\\\n \\State \\Return $s$\n \\EndFunction\n \\end{algorithmic}\n \\caption{Algorithm to calculate the minimum amount of steps}\n \\label{alg:calculateSteps}\n\\end{algorithm}\n\n\\clearpage\n\n\\begin{algorithm}[ht!]\n \\begin{algorithmic}[ht!]\n \\Function{solvePogo}{$x \\in \\mathbb{Z}$, $y \\in \\mathbb{Z}$}\n \\State $s_{\\min} \\gets$ \\Call{calculateSteps}{$x, y$}\n \\\\\n \\State $solution \\gets \\varepsilon$\n \\For{$i$ in $s_{\\min}, \\dots, 1$}\n \\If{$|x| > |y|$}\n \\If{$x > 0$}\n \\State $solution \\gets solution + E$\n \\State $x \\gets x - i$\n \\Else\n \\State $solution \\gets solution + W$\n \\State $x \\gets x + i$\n \\EndIf\n \\Else\n \\If{$y > 0$}\n \\State $solution \\gets solution + N$\n \\State $y \\gets y - i$\n \\Else\n \\State $solution \\gets solution + S$\n \\State $y \\gets y + i$\n \\EndIf\n \\EndIf\n \\EndFor\n \\\\\n \\State \\Return \\Call{reverse}{$solution$}\n \\EndFunction\n \\end{algorithmic}\n \\caption{Algorithm to solve the pogo problem}\n \\label{alg:solvePogo}\n\\end{algorithm}\n\n\\section{Correctness}\n\\subsection{calculateSteps}\nLet $x,y \\in \\mathbb{Z}$ and $s := \\Call{calculateSteps}{x, y}$.\n\nLet $s_{\\min}$ be the minimum amount of necessary steps to get from $(0,0)$\nto $(x,y)$ when you move $i$ units in your $i$'th step.\n\n\\textbf{Theorem: } $s = s_{\\min}$\n\nIt's enough to proof $s \\geq s_{\\min}$ and $s \\leq s_{\\min}$.\n\n\\begin{myindentpar}{1cm}\n\\textbf{Theorem: } $s \\leq s_{\\min}$ (we don't make too many steps)\n\n\\textbf{Proof: }\n\\begin{myindentpar}{1cm}\nWe have to get from $(0,0)$ to $(x, y)$. As we may only move in\ntaxicab geometry we have to use the taxicab distance measure $d_1$:\n\\[d_1 \\left (p, q \\right ) := \\sum_{i=1}^2 |p_i -q_i|\\]\n\nSo in our scenario:\n\\[d_1 \\left ((0,0), (x,y) \\right ) = |x| + |y|\\]\n\nThis means we have to move at least $|x| + |y|$ units to get\nfrom $(0,0)$ to $(x, y)$. As we move $i$ units in the $i$'th step,\nwe have to solve the following equations for $s_{\\min1}$:\n\\begin{align}\n \\sum_{i=1}^{s_{\\min1}} i &\\geq |x| + |y| &&\\text{ and } &|x| + |y| &> \\sum_{i=1}^{s_{\\min1} - 1} i\\\\\n \\frac{s_{\\min1}^2 + s_{\\min1}}{2} &\\geq |x| + |y| && & &> \\sum_{i=1}^{s_{\\min1} - 1} i &\n\\end{align}\n\nThis is what algorithm \\ref{alg:calculateSteps} check with \\texttt{condition 1}.\nAs the algorithm increases $s$ only by one in each loop, it makes\nsure that $\\sum_{i=1}^{s_{\\min1} - 1} i$ is bigger than $|x| + |y|$.\n\nYou can undo moves by going back. But this will always make an even\nnumber undone. When you go $(+i, 0)$ and later $(-j, 0)$ it is the\nsame as if you've been going $(i-j, 0)$. So $2\\cdot i$ steps got undone.\nBut $2\\cdot i$ is an even number. You will never be able to undo\nan odd number of moved units. This means, the parity of the minimum\nnumber of units you would have to move if you would move one unit per\nstep has to be the same as the parity of the moves you actually do.\nThis is exactly what \\texttt{condition 2} makes sure.\n\nSo we need at least $s$ steps $\\Rightarrow s \\leq s_{\\min} \\square$\n\\end{myindentpar}\n\n\\textbf{Theorem: } $s \\geq s_{\\min}$ (we make enough steps)\n\n\\textbf{Proof: }\n\\begin{myindentpar}{1cm}\nWe chose $s$ in a way that \\texttt{condition 1} is true.\nAs we have to go $i \\in 1,\\dots,s$, we can get every possible sum $\\Sigma \\in \\Set{-\\frac{s^2+s}{2}, \\dots +\\frac{s^2+s}{2}}$\nwith a subset of $\\Set{1, \\dots, s}$\\footnote{This can easily be proved by induction over $\\Sigma$.}.\nThis means we can make a partition $(A, \\underbrace{\\Set{1, \\dots, s} \\setminus A}_{=: B})$\nsuch that $|\\sum_{i \\in A} i| = |x|$ and $|\\sum_{i \\in B} i|-2\\cdot j = |y|$.\nThis means, we can reach $(x,y)$ from $(0,0)$.\n\\end{myindentpar}\n\\end{myindentpar}\n\n\\subsection{solvePogo}\n\\textbf{Theorem: } \\Call{solvePogo}{$x,y$} returns a valid, minimal sequence of steps to get from $(0, 0)$ to $(x,y)$\n\n\\textbf{Proof: }\n\\begin{myindentpar}{1cm}\nAs $s_{\\min}$ is the minimum amount of steps you need to get from\n$(0,0)$ to $(x,y)$, \\Call{solvePogo}{$x,y$} will return a minimal\nsequence of steps to get from $(0, 0)$ to $(x,y)$ (see proof above).\n\nWe only have to prove that the sequence of steps that \\Call{solvePogo}{$x,y$}\nis valid, i.e. that you will get from $(0,0)$ to $(x,y)$ with the given\nsequence.\n\nTODO.\n\\end{myindentpar}\n\n\\end{document}\n", "meta": {"hexsha": "3e63f61b5e6303f8badac27bab20ce6148731824", "size": 6782, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documents/proof-of-correctness-pogo/proof-of-correctness-pogo.tex", "max_stars_repo_name": "RalfGuder/LaTeX-examples", "max_stars_repo_head_hexsha": "a1bf9fe422969be1ca4674394ebd2170c07f7693", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1231, "max_stars_repo_stars_event_min_datetime": "2015-01-07T04:04:25.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T17:43:29.000Z", "max_issues_repo_path": "documents/proof-of-correctness-pogo/proof-of-correctness-pogo.tex", "max_issues_repo_name": "DoubleL61/LaTeX-examples", "max_issues_repo_head_hexsha": "cd0d97f85fadb59b7c6e9062b37a8bf7d725ba0c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2015-05-10T13:10:47.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-02T21:28:49.000Z", "max_forks_repo_path": "documents/proof-of-correctness-pogo/proof-of-correctness-pogo.tex", "max_forks_repo_name": "DoubleL61/LaTeX-examples", "max_forks_repo_head_hexsha": "cd0d97f85fadb59b7c6e9062b37a8bf7d725ba0c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 400, "max_forks_repo_forks_event_min_datetime": "2015-01-05T06:22:18.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-19T04:07:59.000Z", "avg_line_length": 37.6777777778, "max_line_length": 156, "alphanum_fraction": 0.602034798, "num_tokens": 2251, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506418255928, "lm_q2_score": 0.8104789063814616, "lm_q1q2_score": 0.6164913003251633}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[colorlinks]{hyperref}\n\\usepackage{amsmath,esint,bm}\n\\title{Divergence Theorem}\n\\author{Naitree Zhu}\n\\date{Last modified: \\today}\n\\begin{document}\n\\maketitle\nIn vector calculus, the Divergence theorem\\footnote{For more information:\\url{http://en.wikipedia.org/wiki/Divergence_theorem}}, also known as Gauss's theorem or Ostrogradsky's theorem,is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface.\n\nMore precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that \\emph{the sum of all sources minus the sum of all sinks gives the net flow out of a region.}\n\nIn physics and engineering, the divergence theorem is usually applied in three dimensions. However, it generalizes to any number of dimensions. In one dimension, it is equivalent to the fundamental theorem of calculus. In two dimensions, it is equivalent to Green's theorem.\n\nThe theorem is a special case of the more general Stokes' theorem.\n\n\\part{Mathematical statement}\nSuppose \\textit{V} is a subset of $R^{n}$ (in the case of n = 3, \\textit{V} represents a volume in 3D space) which is compact and has a piecewise smooth boundary \\textit{S}. If $\\boldsymbol{F}$ is a continuously differentiable vector field defined on a neighborhood of \\textit{V}, then we have:\n\\begin{equation}\n\\iiint_V \\nabla\\cdot\\boldsymbol{F}\\mathrm{d}V = \\oiint_S \\boldsymbol{F}\\cdot\\mathrm{d}\\boldsymbol{S}\n\\end{equation}\nThe closed manifold $\\partial V$ is quite generally the boundary of \\textit{V} oriented by outward-pointing normals.\n\\section*{Corollaries}\nBy applying the divergence theorem in various contexts, other useful identities can be derived:\n\\begin{itemize}\n\\item Applying the divergence theorem to the product of a scalar function \\textit{g} and a vector field $\\boldsymbol{F}$, the result is\n\\begin{equation}\n\\iiint_V \\left[\\boldsymbol{F}\\cdot\\left(\\nabla g\\right) + g\\left(\\nabla\\cdot\\boldsymbol{F}\\right)\\right]\\mathrm{d}V=\\oiint_S g\\boldsymbol{F}\\cdot\\mathrm{d}\\boldsymbol{S}\n\\end{equation}\nA special case of this is $\\mathbf{F}=\\nabla f$, in which case the theorem is the basis for Green's identities.\n\\item Applying the divergence theorem to the product of a scalar function, \\textit{f}, and a non-zero constant vector, the following theorem can be proven(Proof is \\hyperlink{proof1}{here}):\n\\begin{equation}\n\\iiint_V \\nabla f \\mathrm{d}V=\\oiint_S f\\mathrm{d}\\boldsymbol{S}\n\\end{equation}\n\\item Similarly, applying the divergence theorem to the cross-product of a vector field $\\mathbf{F}$ and a non-zero constant vector, the following theorem can be proven(Proof is \\hyperlink{proof2}{here}):\n\\begin{equation}\n\\iiint_V \\nabla\\times\\boldsymbol{F}\\mathrm{d}V=\\oiint_S \\mathrm{d}\\boldsymbol{S}\\times\\boldsymbol{F}\n\\end{equation}\n\\end{itemize}\n\\part{Generalizations}\n\\section{Multiple dimensions}\nOne can use the \\emph{general} Stokes' Theorem to equate the n-dimensional volume integral of the divergence of a vector field \\textbf{F} over a region \\textit{U} to the (n-1)-dimensional surface integral of \\textbf{F} over the boundary of \\textit{U}:\n\\begin{equation}\n\\int_U \\nabla\\cdot\\boldsymbol{F}\\mathrm{d}V_{n}=\\oint_{\\partial U} \\boldsymbol{F}\\cdot\\hat{n}\\mathrm{d}S_{n-1}\n\\end{equation}\nThis equation is also known as the Divergence theorem.\n\nWhen n = 2, this is equivalent to Green's theorem.\n\nWhen n = 1, it reduces to the Fundamental theorem of calculus.\n\\part{History}\nThe theorem was first discovered by Lagrange in 1762, then later independently rediscovered by Gauss in 1813, by Ostrogradsky, who also gave the first proof of the general theorem, in 1826, by Green in 1828, etc. Subsequently, variations on the divergence theorem are correctly called Ostrogradsky's theorem, but also commonly Gauss's theorem, or Green's theorem.\n\\part{2-dimensional version}\n\\begin{equation}\n\\iint_S \\nabla\\cdot\\bm{F}\\mathrm{d}S=\\oint_L \\bm{F}\\cdot\\hat{n}\\mathrm{d}l\n\\end{equation}\nwhere $\\hat{n}$ is the outward-pointing unit normal vector, and the direction of line integral observes right-hand rule.\n\nBy expansion in Cartesian coordinates, it's easily seen that Green theorem is essentially an identical formula with a different form.\n\\part{Proofs}\n\\section{Proof to eq.~(3)}\n\\[\n\\iiint_V \\nabla f \\mathrm{d}V=\\oiint_S f\\mathrm{d}\\boldsymbol{S}\n\\]\n\\hypertarget{proof1}{Proof}:\n\nIf the vector field $\\mathbf{F}=f\\mathbf{C}$, where $\\mathbf{C}$ is an arbitrary nonzero constant vector, then \n\\[\n\\nabla\\cdot\\boldsymbol{F}=\\boldsymbol{C}\\cdot\\nabla f\n\\]\nApply divergence theorem\n\\begin{align*}\n\\int_V \\nabla\\cdot\\boldsymbol{F}\\mathrm{d}V\n&=\\boldsymbol{C}\\cdot\\oint_S f\\mathrm{d}\\boldsymbol{S}\\\\\n&=\\boldsymbol{C}\\cdot\\int_V \\nabla f\\mathrm{d}V\n\\end{align*}\nSince \\textbf{C} is nonzero and arbitrary, it cannot always be perpendicular to $\\oint_S f\\mathrm{d}\\boldsymbol{S}$ and $\\int_V \\nabla f\\mathrm{d}V$. So it's only logical that\n\\[\n\\iiint_V \\nabla f \\mathrm{d}V=\\oiint_S f\\mathrm{d}\\boldsymbol{S}\n\\]\n\\section{Proof to eq.~(4)}\n\\hypertarget{proof2}{Proof}:\nBy setting $\\mathbf{G}=\\mathbf{C}\\times\\mathbf{F}$, where \\textbf{C} is arbitrary nonzero vector, eq.~(4) can be proved, similar to the proof above.\n\nUsing:\n\\[\n\\nabla\\cdot\\left(\\boldsymbol{a}\\times\\boldsymbol{b}\\right)=\\boldsymbol{b}\\cdot\\left(\\nabla\\times\\boldsymbol{a}\\right)-\\boldsymbol{a}\\cdot\\left(\\nabla\\times\\boldsymbol{b}\\right)\n\\]\nthen,\n\\begin{align*}\n\\int_V \\boldsymbol{G}\\cdot\\mathrm{d}V&=-\\boldsymbol{C}\\cdot\\int_V \\nabla\\times\\boldsymbol{F}\\mathrm{d}V\\\\&=\\int_S \\left(\\boldsymbol{C}\\times\\boldsymbol{F}\\right)\\cdot\\mathrm{d}\\boldsymbol{S}\\\\&=\\boldsymbol{C}\\cdot\\int_S \\boldsymbol{F}\\times\\mathrm{d}\\boldsymbol{S}\n\\end{align*}\nWith the similar logic applied with \\textbf{C} in the proof above, we get\n\\[\n\\iiint_V \\nabla\\times\\boldsymbol{F}\\mathrm{d}V=\\oiint_S \\mathrm{d}\\boldsymbol{S}\\times\\boldsymbol{F}\n\\]\n\\end{document}", "meta": {"hexsha": "b7031244390faf99628a4b2b1806588780ce9e7e", "size": 6036, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "math/Divergence theorem/Divergence theorem.tex", "max_stars_repo_name": "Naitreey/notes-and-knowledge", "max_stars_repo_head_hexsha": "48603b2ad11c16d9430eb0293d845364ed40321c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2018-05-16T06:06:45.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-12T08:46:18.000Z", "max_issues_repo_path": "math/Divergence theorem/Divergence theorem.tex", "max_issues_repo_name": "Naitreey/notes-and-knowledge", "max_issues_repo_head_hexsha": "48603b2ad11c16d9430eb0293d845364ed40321c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-04-06T01:46:22.000Z", "max_issues_repo_issues_event_max_datetime": "2019-02-13T03:11:33.000Z", "max_forks_repo_path": "math/Divergence theorem/Divergence theorem.tex", "max_forks_repo_name": "Naitreey/notes-and-knowledge", "max_forks_repo_head_hexsha": "48603b2ad11c16d9430eb0293d845364ed40321c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-04-11T11:02:32.000Z", "max_forks_repo_forks_event_max_datetime": "2020-06-27T11:59:09.000Z", "avg_line_length": 62.2268041237, "max_line_length": 363, "alphanum_fraction": 0.7614314115, "num_tokens": 1842, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998714925403, "lm_q2_score": 0.7931059585194573, "lm_q1q2_score": 0.6164811596371422}} {"text": "\\documentclass[a4paper,10pt]{article}\n%\\documentclass[a4paper,10pt]{scrartcl}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage{lineno}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage{times}\n\\usepackage[T1]{fontenc} % to get the appropriate fonts for icelandic thorn\n\\usepackage{textcomp}\n\\usepackage{natbib}\n\\usepackage{placeins}\n\\usepackage{listings}\n\\usepackage{xcolor}\n\n\\definecolor{codegreen}{rgb}{0,0.6,0}\n\\definecolor{codered}{rgb}{1.0,0.0,0.0}\n\\definecolor{codeblue}{rgb}{0.0,0,1.0}\n\\definecolor{codeyellow}{rgb}{1.0,0.7,0.0}\n\\definecolor{backcolour}{rgb}{0.95,0.95,0.92}\n\n\\lstdefinestyle{mystyle}{\n\tbackgroundcolor=\\color{backcolour}, \n\tcommentstyle=\\color{codered},\n\tkeywordstyle=\\color{codeblue},\n\tnumberstyle=\\tiny\\color{codegreen},\n\tstringstyle=\\color{codeyellow},\n\tbasicstyle=\\ttfamily\\footnotesize,\n\tbreakatwhitespace=false, \n\tbreaklines=true, \n\tcaptionpos=b, \n\tkeepspaces=true, \n\tnumbers=left, \n\tnumbersep=5pt, \n\tshowspaces=false, \n\tshowstringspaces=false,\n\tshowtabs=false, \n\ttabsize=2\n}\n\n\\lstset{style=mystyle}\n\\DeclareMathOperator{\\atantwo}{arctan2}\n\n\\title{Lomb-Scargle Algorithm Documentation}\n\\author{Matt James}\n\\date{}\n\n\\pdfinfo{%\n /Title ()\n /Author ()\n /Creator ()\n /Producer ()\n /Subject ()\n /Keywords ()\n}\n\n\\begin{document}\n\\maketitle\n\t\n\\section{Description of Lomb-Scargle Periodogram}\n\nEverything in this document is pieced together from \\citet{Lomb1976,Scargle1983,Hocke1998}. This algorithm will calculate the power, $P$, amplitude, $A$, and phase, $\\phi$, of a wave with a given frequency, $f$, within an irregularly sampled time series of length $n$ samples.\n\nLet us consider sinusoidal waves of the form \n\\begin{equation}\n\ty_f(t_i) = a \\cos{\\omega(t_i - \\tau)} + b \\sin{\\omega(t_i - \\tau)}, \\label{EqWave}\n\\end{equation}\nwhere $\\omega = 2 \\pi f$ and $1 \\le i \\le n$.\n\n$\\tau$ is defined in \\citet{Scargle1983} as\n\\begin{equation}\n\t\\tan{2 \\omega \\tau} = \\frac{\\sum_{i=1}^{n} \\sin{2 \\omega t_i}}{\\sum_{i=1}^{n} \\cos{2 \\omega t_i}},\n\\end{equation}\nwhich can easily be rewritten as\n\\begin{equation}\n\t\\tau = \\frac{\\atantwo{\\left( \\sum_{i=1}^{n} \\sin{2 \\omega t_i}, \\sum_{i=1}^{n} \\cos{2 \\omega t_i} \\right)}}{2 \\omega}\n\\end{equation}\n\nThe constants $a$ and $b$ are defined as\n\\begin{equation}\n\ta = \\frac{\\sqrt{\\frac{2}{n}}\\sum_{i=1}^{n}y_i \\cos{\\omega (t+i - \\tau)}}{\\left( \\sum_{i=1}^{n} \\cos^2{\\omega (t+i - \\tau)} \\right)^{\\frac{1}{2}}}\n\\end{equation}\nand\n\\begin{equation}\nb = \\frac{\\sqrt{\\frac{2}{n}}\\sum_{i=1}^{n}y_i \\sin{\\omega (t+i - \\tau)}}{\\left( \\sum_{i=1}^{n} \\sin^2{\\omega (t+i - \\tau)} \\right)^{\\frac{1}{2}}}.\n\\end{equation}\n\nThe periodogram is calculated using \n\\begin{equation}\n\tP(\\omega) = \\frac{1}{2\\sigma^2}\\frac{n}{2} (a^2 + b^2),\n\\end{equation}\nwhere $\\sigma = \\frac{1}{n-1}\\sum_{i=1}^{n} y_i^2$ is the variance (assuming that the mean has been subtracted from the data already).\n\nThe amplitude is calculated using \n\\begin{equation}\n\tA(\\omega) = \\sqrt{\\frac{4\\sigma^2}{n}P(\\omega)}\n\\end{equation}\nor, equivalently,\n\\begin{equation}\n\tA(\\omega) = \\sqrt{a^2 + b^2}.\n\\end{equation}\n\nEquation \\ref{EqWave} can be expressed differently:\n\\begin{equation}\n\ty_f(t_i) = A(\\omega) \\cos{\\left[\\omega (t_i - \\tau) + \\phi \\right]},\n\\end{equation}\nwhere the wave phase is given by,\n\\begin{equation}\n\t\\phi = -\\atantwo(b,a).\n\\end{equation}\n\n\\section{Example Python Code}\nThe code used in this module is based on the following Python code.\n\t\t\t\t\t\t\n\t\t\t\t\t\t\n\\begin{lstlisting}[language=Python]\ndef LombScargle(t,x,f):\n\t'''\n\tCalculates the Lomb-Scargle periodogram using the method defined in\n\tHocke 1998. This method assumes that any mean is removed from the\n\tdata.\n\t'''\n\t\n\t#preformat the input variables\n\tt = np.array([t],dtype='float64').flatten()\n\tx = np.array([x],dtype='float64').flatten()\n\tf = np.array([f],dtype='float64').flatten()\n\t\n\t#get array sizes\n\tnf = np.size(f)\n\tn = np.size(t)\n\t\n\t#create output arrays (Power, amplitude, phase, a, b)\n\tP = np.zeros((nf,),dtype='float64')\n\tA = np.zeros((nf,),dtype='float64')\n\tphi = np.zeros((nf,),dtype='float64')\n\ta = np.zeros((nf,),dtype='float64')\n\tb = np.zeros((nf,),dtype='float64')\n\t\n\t#convert f to omega\n\tw = 2*np.pi*f\n\t\n\t#calculate variance (sigma**2)\n\to2 = np.sum(x**2)/(n-1)\n\t\n\t#loop through each frequency\n\tfor i in range(0,nf):\n\t\t#calculate sums to get Tau\n\t\tss2w = np.sum(np.sin(2*w[i]*t))\n\t\tsc2w = np.sum(np.cos(2*w[i]*t))\n\t\n\t\t#calculate Tau\n\t\tTau = np.arctan2(ss2w,sc2w)/(2*w[i])\n\t\t\n\t\t#calculate w(t - Tau)\n\t\twtT = w[i]*(t - Tau)\n\t\t\n\t\t#calculate some more sums\n\t\tsyc = np.sum(x*np.cos(wtT))\n\t\tsys = np.sum(x*np.sin(wtT))\n\t\tsc2 = np.sum(np.cos(wtT)**2)\n\t\tss2 = np.sum(np.sin(wtT)**2)\n\t\n\t\t#get a and b\n\t\trt2n = np.sqrt(2/n)\n\t\ta[i] = rt2n*syc/np.sqrt(sc2)\n\t\tb[i] = rt2n*sys/np.sqrt(ss2)\n\t\n\t#calculate the periodogram\n\ta2b2 = a**2 + b**2\n\tP[:] = (n/(4*o2))*a2b2\n\t\n\t#calculate amplitude\n\tA[:] = np.sqrt(a2b2)\n\t\n\t#calculate phase\n\tphi[:] = -np.arctan2(b,a)\n\t\n\treturn P,A,phi,a,b\n\\end{lstlisting}\n\\bibliographystyle{agu08}\n\\bibliography{references}\n\\end{document}\n", "meta": {"hexsha": "a080b6169918b37cbdd95ac104dd64fdc603df22", "size": 5132, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "wavespec/__data/liblombscargle/doc/lsdoc.tex", "max_stars_repo_name": "mattkjames7/wavespec", "max_stars_repo_head_hexsha": "2928ac3f4fb921a21caba3663b0d7c22e43692fa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-04T15:08:57.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-04T15:08:57.000Z", "max_issues_repo_path": "wavespec/__data/liblombscargle/doc/lsdoc.tex", "max_issues_repo_name": "mattkjames7/wavespec", "max_issues_repo_head_hexsha": "2928ac3f4fb921a21caba3663b0d7c22e43692fa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "wavespec/__data/liblombscargle/doc/lsdoc.tex", "max_forks_repo_name": "mattkjames7/wavespec", "max_forks_repo_head_hexsha": "2928ac3f4fb921a21caba3663b0d7c22e43692fa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.5913978495, "max_line_length": 276, "alphanum_fraction": 0.6549103663, "num_tokens": 1803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.7931059536292271, "lm_q1q2_score": 0.6164811476528191}} {"text": "\\section{Introduction}\n\nA skeleton is a lower dimensional entity which represents shape of it's parent object. It being simpler than the parent object, operations like pattern recognition, approximation, similarity estimation, collision detection, animation, matching and deformation can be performed efficiently on it than on the parent object. \n\nSkeletons can be computed via various mathematical formulations such as Medial Axis Transform (MAT), Chordal Axis Transform (CAT), Thinning etc. Table \\ref{Medials} briefly summarizes these methods of Midcurves creation and their strengths-weaknesses.\n\n\\begin{table}\n\\caption{Current Medial Computation Methods}\n\\begin{tabular}[htbp]{@{} p{0.14\\linewidth} p{0.22\\linewidth} p{0.22\\linewidth} p{0.23\\linewidth} @{}} \\toprule\n{\\bf Method } & {\\bf Medial } & {\\bf Description} & {\\bf Comments}\\\\\n\\midrule\n%------------------------------------------------------------------------------------------------------------------------------------\n\\raisebox{-.9\\height}{MAT \\cite{Ramanathan2004}} &\n\\raisebox{-.9\\height}{\\includegraphics[scale=0.3]{..//Common/images/MAT.png} }&\nLocii of centers of maximal disk traversing within boundary &\nComputable for any shape. But has unwanted branches. Sensitive to boundary perturbations. \\\\\n\n\n%------------------------------------------------------------------------------------------------------------------------------------\n\\raisebox{-.9\\height}{CAT \\cite{Quadros2008}}&\n\\raisebox{-.9\\height}{\\includegraphics[scale=0.45]{..//Common/images/CAT.png}}&\nCreates triangulation first then joins midpoints of sides &\nGaps at end. Expensive triangulation. \\\\\n\n\n%------------------------------------------------------------------------------------------------------------------------------------\n{Straight Skeleton \\cite{Henrik2004}} &\n\\raisebox{-.9\\height}{\\includegraphics[scale=0.3]{..//Common/images/Straight.png}} &\nGoes on thinning from boundary. &\nBisectors mot equidistant. Has unnecessary branches.\\\\\n\n\\bottomrule\n\\end{tabular}\n\\label{Medials}\n\\end{table}\n\n\nIn this paper we focus on 2D planar sketch profiles. Even in 2D profiles, shapes vary enormously. At the first level of simplification, we would deal with 2D polygons only(with an assumption that curved shapes can be converted to polygonal shape by faceting). Divide-and-Conquer is one of the widely used strategy for dealing with complex models. Shape decomposition partitions given shape into sub-shapes and then skeletonization can be performed on simpler sub-polygons. \n\nA polygon can be decomposed into convex regions by dividing at all reflex(concave) vertices. Generally criterion for decomposition is to produce a minimum number of convex components or to minimize the total length of the boundary of these components. Within the minimum component criterion methods further classification could be based on whether or not Steiner points (brand new, non polygonal vertices) are allowed. \n\n", "meta": {"hexsha": "d74d9cf75b703a2af83ebebee9e864400dc5b932", "size": 2943, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "TalksPublications/MidcurveNN_Vixra/Introduction_Midcurves_etes_paper.tex", "max_stars_repo_name": "yogeshhk/MidcurveNN", "max_stars_repo_head_hexsha": "98e11abdac1c6ab5c76e59dcbdd9ce8c08549a67", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 22, "max_stars_repo_stars_event_min_datetime": "2019-06-08T03:25:06.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-15T19:35:38.000Z", "max_issues_repo_path": "TalksPublications/MidcurveNN_Vixra/Introduction_Midcurves_etes_paper.tex", "max_issues_repo_name": "yogeshhk/MidcurveNN", "max_issues_repo_head_hexsha": "98e11abdac1c6ab5c76e59dcbdd9ce8c08549a67", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2019-10-06T08:10:24.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-08T01:33:44.000Z", "max_forks_repo_path": "TalksPublications/MidcurveNN_Vixra/Introduction_Midcurves_etes_paper.tex", "max_forks_repo_name": "yogeshhk/MidcurveNN", "max_forks_repo_head_hexsha": "98e11abdac1c6ab5c76e59dcbdd9ce8c08549a67", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 7, "max_forks_repo_forks_event_min_datetime": "2019-07-06T05:16:32.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-09T09:35:03.000Z", "avg_line_length": 70.0714285714, "max_line_length": 474, "alphanum_fraction": 0.6755011893, "num_tokens": 636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.7931059462938814, "lm_q1q2_score": 0.6164811419510561}} {"text": "\\chapter{Abelian categories}\nIn this chapter I'll translate some more familiar concepts into categorical language;\nthis will require some additional assumptions about our category,\nculminating in the definition of a so-called ``abelian category''.\nOnce that's done, I'll be able to tell you what this ``diagram chasing'' thing is all about.\n\nThroughout this chapter, ``$\\injto$'' will be used for monic maps and ``$\\surjto$'' for epic maps.\n\n\\section{Zero objects, kernels, cokernels, and images}\n\\prototype{In $\\catname{Grp}$, the trivial group and homomorphism\nare the zero objects and morphisms.\nIf $G$, $H$ are abelian then the cokernel\nof $\\phi : G \\to H$ is $H/\\img \\phi$.}\n\nA \\vocab{zero object} of a category is an object $0$ which is both initial and terminal;\nof course, it's unique up to unique isomorphism.\nFor example, in $\\catname{Grp}$ the zero object is the trivial group, \nin $\\catname{Vect}_k$ it's the zero-dimensional vector space consisting of one point, and so on.\n\\begin{ques}\n\tShow that $\\catname{Set}$ and $\\catname{Top}$ don't have zero objects.\n\\end{ques}\nFor the rest of this chapter, all categories will have zero objects.\n\nIn a category $\\AA$ with zero objects, any two objects $A$ and $B$ thus have a distinguished morphism\n\\[ A \\to 0 \\to B \\]\nwhich is called the \\vocab{zero morphism} and also denoted $0$.\nFor example, in $\\catname{Grp}$ this is the trivial homomorphism.\n\nWe can now define:\n\\begin{definition}\n\tConsider a map $A \\taking f B$.\n\tThe \\vocab{kernel} is defined as the equalizer of this map and the map $A \\taking 0 B$.\n\tThus, it's a map $\\ker f : \\Ker f \\injto A$ such that\n\t\\begin{diagram}\n\t\t\\Ker f && \\\\\n\t\t\\dInj^{\\ker f} & \\rdDashed^0 & \\\\\n\t\tA & \\rTo_f & B\n\t\\end{diagram}\n\tcommutes, and moreover any other map with the same property factors uniquely through $\\Ker A$\n\t(so it is universal with this property).\n\tBy \\Cref{prob:equalizer_monic}, $\\ker f$ is a monic morphism, \n\twhich justifies the use of ``$\\injto$''.\n\\end{definition}\nNotice that we're using $\\ker f$ to represent the map and $\\Ker f$ to represent the object\nSimilarly, we define the cokernel, the dual notion:\n\\begin{definition}\n\tConsider a map $A \\taking f B$.\n\tThe \\vocab{cokernel} of $f$ is a map $\\coker f: B \\surjto \\Coker f$ such that\n\t\\begin{diagram}\n\t\tA & \\rTo^f & B \\\\\n\t\t& \\rdDashed_0 & \\dSurj_{\\coker f} \\\\\n\t\t&& \\Coker f\n\t\\end{diagram}\n\tcommutes, and moreover any other map with the same property factors\n\tuniquely through $\\Coker f$ (so it is universal with this property).\n\tThus it is the ``coequalizer'' of this map and the map $A \\taking 0 B$.\n\tBy the dual of \\Cref{prob:equalizer_monic}, $\\coker f$ is an epic morphism,\n\twhich justifies the use of ``$\\surjto$''.\n\\end{definition}\nThink of the cokernel of a map $A \\taking f B$ as ``$B$ modulo the image of $f$'', e.g.\n\\begin{example}\n\t[Cokernels]\n\tConsider the map $\\Zc6 \\to D_{12} = \\left\\langle r,s \\mid r^6=s^2=1, rs=sr\\inv\\right\\rangle$.\n\tThen the cokernel of this map in $\\catname{Grp}$ is $D_{12} / \\left\\langle r \\right\\rangle \\cong \\Zc 2$.\n\\end{example}\nThis doesn't always work out quite the way we want since in general the image of\na homomorphism need not be normal in the codomain.\nNonetheless, we can use this to define:\n\\begin{definition}\n\tThe \\vocab{image} of $A \\taking f B$ is the kernel of $\\coker f$.\n\tWe denote $\\Img f = \\Ker(\\coker f)$.\n\tThis gives a unique map $\\img f : A \\to \\Img f$.\n\\end{definition}\nWhen it exists, this coincides with our concrete notion of ``image''.\nPicture:\n\\begin{diagram}\n\tA & \\rTo^f & B \\\\\n\t& \\rdTo(1,2)~{\\exists!} \\rdDashed(3,2)^{0\\qquad\\qquad} \\ruInj(1,2) & & \\rdSurj(1,2)^{\\coker f} \\\\\n\t& \\Img f & \\rDashed^0 & \\Coker f\n\\end{diagram}\nNote that by universality of $\\Img f$,\nwe find that there is a unique map $\\img f : A \\to \\Img f$ that makes the entire diagram commute.\n\n\\section{Additive and abelian categories}\n\\prototype{$\\catname{Ab}$, $\\catname{Vect}_k$, or more generally $\\catname{Mod}_R$.}\nWe can now define the notion of an additive and abelian category,\nwhich are the types of categories where this notion is most useful.\n\n\\begin{definition}\n\tAn \\vocab{additive category} $\\AA$ is one such that:\n\t\\begin{itemize}\n\t\t\\ii $\\AA$ has a zero object, and any two objects have a product.\n\t\t\\ii More importantly: every $\\Hom_\\AA(A, B)$ forms an \\emph{abelian group} (written additively)\n\t\tsuch that composition distributes over addition:\n\t\t\\[ (g+h)\\circ f = g\\circ f + h\\circ f\n\t\t\t\\quad\\text{and}\\quad\n\t\t\tf\\circ(g+h) = f\\circ g + f \\circ h. \\]\n\t\tThe zero map serves as the identity element for each group.\n\\end{itemize}\n\\end{definition}\n\\begin{definition}\n\tAn \\vocab{abelian category} $\\AA$ is one with the additional properties that\n\tfor any morphism $A \\taking f B$,\n\t\\begin{itemize}\n\t\t\\ii The kernel and cokernel exist, and\n\t\t\\ii The morphism factors through the image so that $\\img(f)$ is epic.\n\t\\end{itemize}\n\tSo, this yields a diagram\n\t\\begin{diagram}\n\t\t\\Ker(f) & \\rInj^{\\quad\\ker(f)} & A & \\rSurj^{\\img(f)\\quad} & \\Img(f) & \\rInj & B & \\rSurj^{\\coker(f)} \\Coker(f).\n\t\\end{diagram}\n\\end{definition}\n\n\\begin{example}[Examples of abelian categories]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii $\\catname{Vect}_k$, $\\catname{Ab}$ are abelian categories,\n\t\twhere $f+g$ takes its usual meaning.\n\t\t\\ii Generalizing this, the category $\\catname{Mod}_R$ of $R$-modules is abelian.\n\t\t\\ii $\\catname{Grp}$ is not even additive, because there is no way to assign\n\t\ta commutative addition to pairs of morphisms.\n\t\\end{enumerate}\n\\end{example}\n\nIn general, once you assume a category is abelian, all the properties you would want\nof these kernels, cokernels, \\dots\\ that you would guess hold true.\nFor example,\n\\begin{proposition}[Monic $\\iff$ trival kernel]\n\tA map $A \\taking f B$ is monic if and only if its kernel is $0 \\to A$.\n\tDually, $A \\taking f B$ is epic if and only if its cokernel is $B \\to 0$.\n\\end{proposition}\n\\begin{proof}\n\tThe easy direction is:\n\t\\begin{exercise}\n\t\tShow that if $A \\taking f B$ is monic, then $0 \\to A$ is a kernel.\n\t\t(This holds even in non-abelian categories.)\n\t\\end{exercise}\n\tOf course, since kernels are unique up to isomorphism, monic $\\implies$ $0$ kernel.\n\tOn the other hand, assume that $0 \\to A$ is a kernel of $A \\taking f B$.\n\tFor this we can exploit the group structure of the underlying homomorphisms now.\n\tAssume the diagram\n\t\\begin{diagram}\n\t\tZ & \\pile{\\rTo^g \\\\ \\rTo_h} & A & \\rTo^f & B\n\t\\end{diagram}\n\tcommutes.\n\tThen $(g - h) \\circ f = g \\circ f - h \\circ f = 0$, and we've arrived at a commutative diagram.\n\t\\begin{diagram}\n\t\tZ && \\\\\n\t\t\\dTo^{g-h} & \\rdDotted^0 & \\\\\n\t\tA & \\rTo_f & B\n\t\\end{diagram}\n\tBut since $0 \\to A$ is a kernel it follows that $g-h$ factors through $0$,\n\tso $g-h = 0 \\implies g = h$, which is to say that $f$ is monic.\n\\end{proof}\n\\begin{proposition}[Isomorphism $\\iff$ monic and epic]\n\tIn an abelian category,\n\ta map is an isomorphism if and only if it is monic and epic.\n\\end{proposition}\n\\begin{proof}\n\tOmitted. (The Mitchell embedding theorem\n\tpresented later implies this anyways for\n\tmost situations we care about,\n\tby looking at a small sub-category.)\n\\end{proof}\n\n\\section{Exact sequences}\n\\prototype{$0 \\to G \\to G \\times H \\to H \\to 0$ is exact.}\nExact sequences will seem exceedingly unmotivated until you learn about homology groups,\nwhich is one of the most natural places that exact sequences appear.\nIn light of this, it might be worth trying to read the chapter on homology groups\nsimultaneously with this one.\n\nFirst, let me state the definition for groups, to motivate the general categorical definition.\nA sequence of groups\n\\[ G_0 \\taking{f_1} G_1 \\taking{f_2} G_2 \\taking{f_3} \\dots \\taking{f_n} G_n \\]\nis \\emph{exact} at $G_k$ if the image of $f_k$ is the kernel of $f_{k+1}$.\nWe say the entire sequence is exact if it's exact at $k=1,\\dots,n-1$.\n\\begin{example}\n\t[Exact sequences]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii The sequence\n\t\t\\[ 0 \\to \\Zc 3\n\t\t\t\\overset{\\times 5}{\\injto} \\Zc{15}\n\t\t\t\\surjto \\Zc{5}\n\t\t\t\\to 0 \\]\n\t\tis exact.\n\t\tActually, $0 \\to G \\injto G \\times H \\surjto H \\to 0$ is exact in general.\n\t\t(Here $0$ denotes the trivial group.)\n\t\t\\ii For groups, the map $0 \\to A \\to B$ is exact if and only if $A \\to B$ is injective.\n\t\t\\ii For groups, the map $A \\to B \\to 0$ is exact if and only if $A \\to B$ is surjective.\n\t\\end{enumerate}\n\\end{example}\n\nNow, we want to mimic this definition in a general \\emph{abelian} category $\\AA$.\nSo, let's write down a criterion for when $A \\taking f B \\taking g C$ is exact.\nFirst, we had better have that $g \\circ f = 0$,\nwhich encodes the fact that $\\img(f) \\subseteq \\ker(g)$.\nAdding in all the relevant objects, we get the commutative diagram below.\n\\begin{diagram}\n\tA && \\rDashed^0 && C \\\\\n\t\\dSurj^{\\img f} & \\rdTo(2,1)^f & B & \\ruTo(2,1)^g & \\uDashed_0 \\\\\n\t\\Img f & \\ruInj(2,1)_{\\iota} & \\rDotted_{\\exists!} & \\luInj(2,1) & \\Ker g \\\\\n\\end{diagram}\nHere the map $A \\surjto \\Img f$ is epic since we are assuming $\\AA$ is an abelian category.\nSo, we have that\n\\[ 0 = (g \\circ \\iota) \\circ \\img f = g \\circ (\\iota \\circ \\img f) = g \\circ f = 0 \\]\nbut since $\\img f$ is epic, this means that $g \\circ \\iota = 0$.\nSo there is a \\emph{unique} map $\\Img f \\to \\Ker g$, and we require that this diagram commutes.\nIn short,\n\\begin{definition}\n\tLet $\\AA$ be an abelian category. The sequence\n\t\\[ \\dots \\to A_{n-1} \\taking{f_n} A_n \\taking{f_{n+1}} A_{n+1} \\to \\dots \\]\n\tis \\vocab{exact} at $A_n$ if $f_n \\circ f_{n+1} = 0$ and\n\tthe canonical map $\\Img f_n \\to \\Ker f_{n+1}$ is an isomorphism.\n\tThe entire sequence is exact if it is exact at each $A_i$.\n\t(For finite sequences we don't impose condition on the very first and very last object.)\n\\end{definition}\n\n\\begin{exercise}\n\tShow that, as before, $0 \\to A \\to B$ is exact $\\iff$ $A \\to B$ is monic.\n\\end{exercise}\n\n\\section{The Freyd-Mitchell embedding theorem}\nWe now introduce the Freyd-Mitchell embedding theorem,\nwhich essentially says that any abelian category can be realized as a concrete one.\n\n\\begin{definition}\n\tA category is \\vocab{small} if $\\obj(\\AA)$ is a set (as opposed to a class),\n\ti.e.\\ there is a ``set of all objects in $\\AA$''.\n\tFor example, $\\catname{Set}$ is not small because there is no set of all sets.\n\\end{definition}\n\n\\begin{theorem}\n\t[Freyd-Mitchell embedding theorem]\n\tLet $\\AA$ be a small abelian category.\n\tThen there exists a ring $R$ (with $1$ but possibly non-commutative)\n\tand a full, faithful, exact functor onto the category of left $R$-modules.\n\\end{theorem}\nHere a functor is \\vocab{exact} if it preserves exact sequences.\nThis theorem is good because it means\n\\begin{moral}\n\tYou can basically forget about all the weird definitions\n\tthat work in any abelian category.\n\\end{moral}\nAny time you're faced with a statement about an abelian category,\nit suffices to just prove it for a ``concrete'' category\nwhere injective/surjective/kernel/image/exact/etc.\\\nagree with your previous notions.\nA proof by this means is sometimes called \\emph{diagram chasing}.\n\n\\begin{remark}\n\tThe ``small'' condition is a technical obstruction\n\tthat requires the objects $\\AA$ to actually form a set.\n\tI'll ignore this distinction,\n\tbecause one can almost always work around it\n\tby doing enough set-theoretic technicalities.\n\\end{remark}\n\nFor example, let's prove:\n\\begin{lemma}\n\t[Short five lemma]\n\tIn an abelian category, consider the commutative diagram\n\t\\begin{diagram}\n\t\t0 & \\rTo & A & \\rInj^p & B & \\rSurj^q & C & \\rTo & 0 \\\\\n\t\t&& \\dTo^\\cong_\\alpha && \\dTo_\\beta && \\dTo^\\cong_\\gamma && \\\\\n\t\t0 & \\rTo & A' & \\rInj^{p'} & B' & \\rSurj^{q'} & C' & \\rTo & 0\n\t\\end{diagram}\n\tand assume the top and bottom rows are exact.\n\tIf $\\alpha$ and $\\gamma$ are isomorphisms, then so is $\\beta$.\n\\end{lemma}\n\n\\begin{proof}\n\tWe prove that $\\beta$ is epic (with a similar proof to get monic).\n%\tOne can show that it's possible to take a small subcategory of $\\AA$\n%\tcontaining the $10$ elements and $13$ arrows above, as well as all necessary\n%\tkernels, cokernels, et cetera.\n%\t(Essentially, let $\\BB_0$ be the diagram above and let $\\BB_{i+1}$ add in any needed objects;\n%\tthen $\\bigcup \\BB_i$ is a set-sized category).\n\tBy the embedding theorem we can treat the category as $R$-modules over some $R$.\n\tThis lets us do a so-called ``diagram chase'' where we move elements around the picture,\n\tusing the concrete interpretation of our category as $R$-modules.\n\n\tLet $b'$ be an element of $B'$.\n\tThen $q'(b') \\in C'$, and since $\\gamma$ is surjective, we have a $c$ such that $\\gamma(c) = b'$,\n\tand finally a $b \\in B$ such that $q(b) = c$.\n\tPicture:\n\t\\begin{diagram}\n\t\tb \\in B & \\rMapsto^q & c \\in C \\\\\n\t\t\\dDashed_\\beta && \\dMapsto^\\cong_\\gamma \\\\\n\t\tb' \\in B' & \\rMapsto^{q'} & c' \\in C'\n\t\\end{diagram}\n\tNow, it is not necessarily the case that $\\beta(b) = b'$.\n\tHowever, since the diagram commutes we at least have that\n\t\\[ q'(b') = q'(\\beta(b)) \\]\n\tso $b' - \\beta(b) \\in \\Ker q' = \\Img p'$, and there is an $a' \\in A'$ such that\n\t$p'(a') = b' - \\beta(b)$;\n\tuse $\\alpha$ now to lift it to $a \\in A$.\n\tPicture:\n\t\\begin{diagram}\n\t\ta \\in A && b \\in B && \\\\\n\t\t\\dMapsto && && \\\\\n\t\ta' \\in A' & \\rMapsto & b'-\\beta(b) \\in B' & \\rMapsto & 0 \\in C'\n\t\\end{diagram}\n\tThen, we have\n\t\\[\n\t\t\\beta(b + q(a)) = \\beta b + \\beta p a\n\t\t= \\beta b + p' \\alpha a\n\t\t= \\beta b + (b' - \\beta b)\n\t\t= b'\n\t\\]\n\tso $b' \\in \\Img \\beta$ which completes the proof that $\\beta'$ is surjective.\n\\end{proof}\n\n\\section{Breaking long exact sequences}\n\\prototype{First isomorphism theorem.}\n\nIn fact, it turns out that any exact sequence breaks into short exact sequences.\nThis relies on:\n\\begin{proposition}[``First isomorphism theorem'' in abelian categories]\n\t\\label{prop:break_exact}\n\tLet $A \\taking f B$ be an arrow of an abelian category.\n\tThen there is an exact sequence\n\t\\[ 0 \\to \\Ker f \\taking{\\ker f} A \\taking{\\img f} \\Img f \\to 0. \\]\n\\end{proposition}\n\n\\begin{example}\nLet's analyze this theorem in our two examples of abelian categories:\n\\begin{enumerate}[(a)]\n\t\\ii In the category of abelian groups, \n\tthis is basically the first isomorphism theorem.\n\t\\ii In the category $\\catname{Vect}_k$,\n\tthis amounts to the rank-nullity theorem, \\Cref{thm:rank_nullity}.\n\\end{enumerate}\n\\end{example}\nThus, any exact sequence can be broken into short exact sequences, as\n\\begin{diagram}\n\t&& 0 & & 0 && 0 && 0 \\\\\n\t&&& \\rdDotted(1,1) C_n \\ruDotted(1,1) &&&& \\rdDotted(1,1) C_{n+2} \\ruDotted(1,1) & \\\\\n\t\\dots & \\rTo & A_{n-1} \\ruDotted(1,1) & \\rTo^{f_{n-1}} & \\rdDotted(1,1) A_n & \\rTo^{f_n} &\n\tA_{n+1} \\ruDotted(1,1) & \\rTo^{f_{n+1}} & \\rdDotted(1,1) \\dots \\\\\n\t& \\rdDotted(1,1) C_{n-1} \\ruDotted(1,1) &&&& \\rdDotted(1,1) C_{n+1} \\ruDotted(1,1) &&& \\\\\n\t0 \\ruDotted(1,1) && \\rdDotted(1,1) 0 && 0 \\ruDotted(1,1) && \\rdDotted(1,1) 0 &&\n\\end{diagram}\nwhere $C_k = \\img f_{k-1} = \\ker f_k$ for every $k$.\n\n\\section\\problemhead\n\n\\begin{problem}\n\t[Four lemma]\n\tIn an abelian category, consider the commutative diagram\n\t\\begin{diagram}\n\t\tA & \\rTo^p & B & \\rTo^q & C & \\rTo^r & D \\\\\n\t\t\\dSurj_\\alpha && \\dInj_\\beta && \\dTo_\\gamma && \\dInj_\\delta \\\\\n\t\tA' & \\rTo_{p'} & B' & \\rTo_{q'} & C' & \\rTo_{r'} & D'\n\t\\end{diagram}\n\twhere the first and second rows are exact.\n\tProve that if $\\alpha$ is epic, and $\\beta$ and $\\delta$ are monic,\n\tthen $\\gamma$ is epic.\n\t\\begin{soln}\n\t\tLet $c \\in C$ with $\\gamma(c) = 0$.\n\t\tWe show $c = 0$.\n\t\tThis proceeds in a diagram chase:\n\t\t\\begin{itemize}\n\t\t\t\\ii Note that $0 = r'(\\gamma(c)) = \\delta(r(c))$, and since $\\delta$\n\t\t\tis injective, it follows that $r(c) = 0$.\n\t\t\t\\ii Since the top row is exact,\n\t\t\tit follows $c = q(b)$ for some $b \\in B$.\n\t\t\t\\ii Then $q'(\\beta(b)) = 0$,\n\t\t\tso if we let $b' = \\beta(b)$,\n\t\t\tthen $b' \\in \\ker(q')$.\n\t\t\tAs the bottom row is exact, there exists $a'$ with $p'(a') = b'$.\n\t\t\t\\ii Since $\\alpha$ is injective,\n\t\t\tthere is $a \\in A$ with $\\alpha(a') = a$.\n\t\t\t\\ii Since $\\beta$ is injective,\n\t\t\tit follows that $p(a) = b$.\n\t\t\t\\ii Since the top row is exact, and $b$ is in the image of $p$,\n\t\t\tit follows that $0 = q(b) = c$ as needed.\n\t\t\\end{itemize}\n\n\n\t\\end{soln}<++>\n\\end{problem}\n\n\\begin{problem}\n\t[Five lemma]\n\t\\gim\n\tIn an abelian category, consider the commutative diagram\n\t\\begin{diagram}\n\t\tA & \\rTo^p & B & \\rTo^q & C & \\rTo^r & D & \\rTo^s & E \\\\\n\t\t\\dSurj_\\alpha && \\dTo^\\cong_\\beta && \\dTo_\\gamma && \\dTo^\\cong_\\delta && \\dInj_\\eps \\\\\n\t\tA' & \\rTo_{p'} & B' & \\rTo_{q'} & C' & \\rTo_{r'} & D' & \\rTo_{s'} & E'\n\t\\end{diagram}\n\twhere the first and second rows are exact.\n\tProve that if $\\alpha$ is epic, $\\eps$ is monic, and $\\beta$, $\\delta$ are isomorphisms,\n\tthen $\\gamma$ is an isomorphism as well.\n\tThus this is a stronger version of the short five lemma.\n\\end{problem}\n\n\\begin{sproblem}\n\t[Snake lemma]\n\t\\yod\n\tIn an abelian category, consider the diagram\n\t\\begin{diagram}\n\t\t&& A & \\rTo^f & B & \\rSurj^g & C & \\rTo & 0 \\\\\n\t\t&& \\dTo_a && \\dTo_b && \\dTo_c && \\\\\n\t\t0 & \\rTo & A' & \\rInj_{f'} & B' & \\rTo_{g'} & C' && \n\t\\end{diagram}\n\twhere the first and second rows are exact sequences.\n\tProve that there is an exact sequence\n\t\\[ \\Ker a \\to \\Ker b \\to \\Ker c \\to \\Coker a \\to \\Coker b \\to \\Coker c. \\]\n\\end{sproblem}\n", "meta": {"hexsha": "69e18d2b9a0f0f54f1c52c5fec0eae4132d5e103", "size": 16964, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/cats/abelian.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.7931059511841119, "lm_q2_score": 0.7772998560157665, "lm_q1q2_score": 0.6164811416606577}} {"text": "\\chapter{angular momentum solutions}\n\\begin{abox}\n\tPractice set 1 solutions\n\t\\end{abox}\n\\begin{enumerate}\n\t\\begin{minipage}{\\textwidth}\n\t\t\\item The Hamiltonian of an electron in a constant magnetic field $\\vec{B}$ is given by $H=\\mu \\vec{\\sigma} \\cdot \\vec{B}$. where $\\mu$ is a positive constant and $\\vec{\\sigma}=\\left(\\sigma_{1}, \\sigma_{2}, \\sigma_{3}\\right)$ denotes the Pauli matrices. Let $\\omega=\\mu B / \\hbar$ and $I$ be the $2 \\times 2$ unit matrix. Then the operator $e^{i H t / \\hbar}$ simplifies to\n\t\t\\exyear{NET JUNE 2011}\n\t\\end{minipage}\n\t\\begin{tasks}(2)\n\t\t\\task[\\textbf{A.}] $I \\cos \\frac{\\omega t}{2}+\\frac{i \\vec{\\sigma} \\cdot \\vec{B}}{B} \\sin \\frac{\\omega t}{2}$\n\t\t\\task[\\textbf{B.}]$I \\cos \\omega t+\\frac{i \\vec{\\sigma} \\cdot \\vec{B}}{B} \\sin \\omega t$\n\t\t\\task[\\textbf{C.}]$I \\sin \\omega t+\\frac{i \\vec{\\sigma} \\cdot \\vec{B}}{B} \\cos \\omega t$\n\t\t\\task[\\textbf{D.}]$I \\sin 2 \\omega t+\\frac{i \\vec{\\sigma} \\cdot \\vec{B}}{B} \\cos 2 \\omega t$\n\t\\end{tasks}\n\\begin{answer}\n\t$H=\\mu \\vec{\\sigma} \\vec{B}$ where $\\vec{\\sigma}=\\left(\\sigma_{1}, \\sigma_{2}, \\sigma_{3}\\right)$ are pauli spin matrices and $\\vec{B}$ are constant magnetic field. $\\vec{\\sigma}=\\left(\\sigma_{1} \\hat{i}, \\sigma_{2} \\hat{j}, \\sigma_{3} \\hat{k}\\right), \\vec{B}=\\left(B_{x} \\hat{i}+B_{y} \\hat{j}+B_{z} \\hat{k}\\right)$ and Hamiltonion $H=\\mu \\vec{\\sigma} \\cdot \\vec{B}$ in matrices form is given by\n\t$$\n\tH=\\mu\\left(\\begin{array}{cc}\n\tB_{z} & B_{x}-i B_{y} \\\\\n\tB_{x}+i B_{y} & -B_{z}\n\t\\end{array}\\right) .\n\t$$\n\tEigenvalue of given matrices are given by $+\\mu \\mathrm{B}$ and $-\\mu \\mathrm{B} . H$ matrices are not diagonals so $e^{i H t / \\hbar}$ is equivalent to\n\t$$\n\tS^{-1}\\left(\\begin{array}{cc}\n\te^{\\frac{i \\mu B t}{\\hbar}} & 0 \\\\\n\t0 & e^{\\frac{-i \\mu B t}{\\hbar}}\n\t\\end{array}\\right) S\n\t$$\n\twhere $S$ is unitary matrices\\\\\n\tand $\\quad S^{-1}=S=\\left(\\begin{array}{cc}\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}}\\end{array}\\right)$.\\\\\n\t$S^{-1}\\left(\\begin{array}{cc}e^{\\frac{i \\mu B t}{\\hbar}} & 0 \\\\ 0 & e^{\\frac{-i \\mu B t}{\\hbar}}\\end{array}\\right) S=\\left(\\begin{array}{cc}\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}}\\end{array}\\right)\\left(\\begin{array}{cc}e^{\\frac{i \\mu B t}{\\hbar}} & 0 \\\\ 0 & e^{\\frac{-i \\mu B t}{\\hbar}}\\end{array}\\right)\\left(\\begin{array}{cc}\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}}\\end{array}\\right)$,\\\\ where $\\omega=\\mu \\mathrm{B} / \\hbar$.\\\\\n\t$e^{i H t / \\hbar}=\\left(\\begin{array}{cc}\\cos \\omega t & i \\sin \\omega t \\\\ i \\sin \\omega t & \\cos \\omega t\\end{array}\\right)$,\\\\\n\t which is equivalent to $I \\cos \\omega t+i \\sigma_{x} \\sin \\omega t$ can be written\n\tas $I \\cos \\omega t+\\frac{i \\vec{\\sigma} \\cdot \\vec{B}}{B} \\sin \\omega t$, where $\\sigma_{x}=\\frac{i \\vec{\\sigma} \\cdot B}{B}$\\\\\n\tThe correct option is \\textbf{(b)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item In a system consisting of two spin $\\frac{1}{2}$ particles labeled 1 and 2, let $\\vec{S}^{(1)}=\\frac{\\hbar}{2} \\vec{\\sigma}^{(1)}$ and $\\vec{S}^{(2)}=\\frac{\\hbar}{2} \\vec{\\sigma}^{(2)}$ denote the corresponding spin operators. Here $\\vec{\\sigma} \\equiv\\left(\\sigma_{x}, \\sigma_{y}, \\sigma_{z}\\right)$ and $\\sigma_{x}, \\sigma_{y}, \\sigma_{z}$ are the three Pauli matrices.\\\\\n\t$\\text { In the standard basis the matrices for the operators } S_{x}^{(1)} S_{y}^{(2)} \\text { and } S_{y}^{(1)} S_{x}^{(2)} \\text { are respectively, }$\n\t\\exyear{NET JUNE 2011}\n\\end{minipage}\n\\begin{tasks}(1)\n\t\\task[\\textbf{A.}]$\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cc}\n\t\t1 & 0 \\\\\n\t\t0 & -1\n\t\\end{array}\\right), \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{rr}\n\t\t-1 & 0 \\\\\n\t\t0 & 1\n\t\\end{array}\\right)$\n\t\\task[\\textbf{B.}]$\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cc}\n\t\ti & 0 \\\\\n\t\t0 & -i\n\t\\end{array}\\right), \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{rr}\n\t\t-i & 0 \\\\\n\t\t0 & i\n\t\\end{array}\\right)$\n\t\\task[\\textbf{C.}]$\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t\t0 & 0 & 0 & -i \\\\\n\t\t0 & 0 & i & 0 \\\\\n\t\t0 & -i & 0 & 0 \\\\\n\t\ti & 0 & 0 & 0\n\t\\end{array}\\right), \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t\t0 & 0 & 0 & -i \\\\\n\t\t0 & 0 & -i & 0 \\\\\n\t\t0 & i & 0 & 0 \\\\\n\t\ti & 0 & 0 & 0\n\t\\end{array}\\right)$\n\t\\task[\\textbf{D.}]$\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t\t0 & 1 & 0 & 0 \\\\\n\t\t1 & 0 & 0 & 0 \\\\\n\t\t0 & 0 & 0 & -i \\\\\n\t\t0 & 0 & i & 0\n\t\\end{array}\\right), \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t\t0 & -i & 0 & 0 \\\\\n\t\ti & 0 & 0 & 0 \\\\\n\t\t0 & 0 & 0 & 1 \\\\\n\t\t0 & 0 & 1 & 0\n\t\\end{array}\\right)$\n\\end{tasks}\n\\begin{answer}\n$\\mathrm{S}_{\\mathrm{x}}^{(1)} \\mathrm{S}_{\\mathrm{y}}^{(2)}=\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{ll}\n\t0 & 1 \\\\\n\t1 & 0\n\\end{array}\\right) \\otimes\\left(\\begin{array}{cc}\n\t0 & -\\mathrm{i} \\\\\n\t\\mathrm{i} & 0\n\\end{array}\\right) \\Rightarrow \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t0 & 0 & 0 & -\\mathrm{i} \\\\\n\t0 & 0 & \\mathrm{i} & 0 \\\\\n\t0 & -\\mathrm{i} & 0 & 0 \\\\\n\t\\mathrm{i} & 0 & 0 & 0\n\\end{array}\\right)$\\\\\n$S_{y}^{(1)} S_{x}^{(2)}=\\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cc}\n\t0 & -i \\\\\n\ti & 0\n\\end{array}\\right) \\otimes\\left(\\begin{array}{cc}\n\t0 & 1 \\\\\n\t1 & 0\n\\end{array}\\right) \\Rightarrow \\frac{\\hbar^{2}}{4}\\left(\\begin{array}{cccc}\n\t0 & 0 & 0 & -i \\\\\n\t0 & 0 & -i & 0 \\\\\n\t0 & i & 0 & 0 \\\\\n\ti & 0 & 0 & 0\n\\end{array}\\right)$\t\\\\\nThe correct option is \\textbf{c}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item $\\text { These two operators of above QUESTION satisfy the relation }$\n\t\\exyear{NET JUNE 2011}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\left\\{S_{x}^{(1)} S_{y}^{(2)}, S_{y}^{(1)} S_{x}^{(2)}\\right\\}=S_{z}^{(1)} S_{z}^{(2)}$\n\t\\task[\\textbf{B.}]$\\left\\{S_{x}^{(1)} S_{y}^{(2)}, S_{y}^{(1)} S_{x}^{(2)}\\right\\}=0$\n\t\\task[\\textbf{C.}]$\\left[S_{x}^{(1)} S_{y}^{(2)}, S_{y}^{(1)} S_{x}^{(2)}\\right]=i S_{z}^{(1)} S_{z}^{(2)}$\n\t\\task[\\textbf{D.}] $\\left[S_{x}^{(1)} S_{y}^{(2)}, S_{y}^{(1)} S_{x}^{(2)}\\right]=0$\n\\end{tasks}\n\\begin{answer}\n We have matrix $S_{x}^{(1)} S_{y}^{(2)}$ and $S_{y}^{(1)} S_{x}^{(2)}$ from question 6(A) so commutation is given by $\\left[S_{x}^{(1)} S_{y}^{(2)}, S_{y}^{(1)} S_{x}^{(2)}\\right]=0 .$\\\\\n The correct option is \\textbf{(d)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item The component along an arbitrary direction $\\hat{n}$, with direction $\\operatorname{cosines}\\left(n_{x}, n_{y}, n_{z}\\right)$, of the spin of a spin $-\\frac{1}{2}$ particle is measured. The result is\n\t\\exyear{NET JUNE 2012}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] 0\n\t\\task[\\textbf{B.}]$\\pm \\frac{\\hbar}{2} n_{z}$\n\t\\task[\\textbf{C.}]$\\pm \\frac{\\hbar}{2}\\left(n_{x}+n_{y}+n_{z}\\right)$\n\t\\task[\\textbf{D.}]$\\pm \\frac{\\hbar}{2}$\n\\end{tasks}\n\\begin{answer}\n\t$S_{x}=\\frac{\\hbar}{2}\\left(\\begin{array}{ll}\n\t\t0 & 1 \\\\\n\t\t1 & 0\n\t\\end{array}\\right), S_{y}=\\frac{\\hbar}{2}\\left(\\begin{array}{ll}\n\t\t0 & -i \\\\\n\t\ti & 0\n\t\\end{array}\\right), S_{z}=\\frac{\\hbar}{2}\\left(\\begin{array}{ll}\n\t\t1 & 0 \\\\\n\t\t0 & -1\n\t\\end{array}\\right)$\\\\\n\t$\\overrightarrow{\\mathrm{n}}=\\mathrm{n}_{\\mathrm{x}} \\hat{\\mathrm{i}}+\\mathrm{n}_{\\mathrm{y}} \\hat{\\mathrm{j}}+\\mathrm{n}_{\\mathrm{z}} \\hat{\\mathrm{k}} \\text { and } n_{x}^{2}+n_{y}^{2}+n_{z}^{2}=1, \\vec{S}=S_{x} \\hat{i}+S_{y} \\hat{j}+S_{z} \\hat{k}$\\\\\n\t$\\vec{n} \\cdot \\vec{S}=n_{x}\\left(\\begin{array}{cc}\n\t\t0 & \\frac{\\hbar}{2} \\\\\n\t\t\\frac{\\hbar}{2}\n\t\\end{array}\\right)+n_{y}\\left(\\begin{array}{cc}\n\t\t0 & -\\frac{i \\hbar}{2} \\\\\n\t\t\\frac{i \\hbar}{2} & 0\n\t\\end{array}\\right)+n_{z}\\left(\\begin{array}{cc}\n\t\t\\frac{\\hbar}{2} & 0 \\\\\n\t\t0 & \\frac{-\\hbar}{2}\n\t\\end{array}\\right)$\\\\\n\t$\\vec{n} \\cdot \\vec{S}=\\left(\\begin{array}{ll}\n\t\tn_{z} \\frac{\\hbar}{2} & \\frac{\\hbar}{2}\\left(n_{x}-i n_{y}\\right) \\\\\n\t\t\\frac{\\hbar}{2}\\left(n_{x}+i n_{y}\\right) & -n_{z} \\frac{\\hbar}{2}\n\t\\end{array}\\right)$\\\\\n\tLet $\\lambda$ is eigen value of $\\vec{n} \\cdot \\vec{S}$\n\t$$\n\t\\left|\\begin{array}{ll}\n\tn_{z} \\frac{\\hbar}{2}-\\lambda & \\frac{\\hbar}{2}\\left(n_{x}-i n_{y}\\right. \\\\\n\t\\frac{\\hbar}{2}\\left(n_{x}+i n_{y}\\right) & -n_{z} \\frac{\\hbar}{2}-\\lambda\n\t\\end{array}\\right|=0\n\t$$\n\t$\\Rightarrow-\\left(\\frac{\\mathrm{n}_{\\mathrm{z}} \\hbar}{2}-\\lambda\\right)\\left(\\frac{\\mathrm{n}_{\\mathrm{z}} \\hbar}{2}+\\lambda\\right)-\\frac{\\hbar^{2}}{4}\\left(\\mathrm{n}_{\\mathrm{x}}^{2}+\\mathrm{n}_{\\mathrm{y}}^{2}\\right)=0 \\Rightarrow-\\left(\\frac{n_{z}^{2} \\hbar^{2}}{4}-\\lambda^{2}\\right)-\\frac{\\hbar^{2}}{4}\\left(n_{x}^{2}+n_{y}^{2}\\right)=0$\\\\\n\t$\\Rightarrow-\\frac{\\hbar^{2}}{4}\\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\\right)+\\lambda^{2}=0 \\Rightarrow \\lambda=\\pm \\frac{\\hbar}{2}$\\\\\n\tThe correct option is \\textbf{(d)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item In a basis in which the $z$ - component $S_{z}$ of the spin is diagonal, an electron is in a spin state $\\psi=\\left(\\begin{array}{c}(1+i) / \\sqrt{6} \\\\ \\sqrt{2 / 3}\\end{array}\\right) .$ The probabilities that a measurement of $S_{2}$ will yield the values $\\hbar / 2$ and $-\\hbar / 2$ are, respectively,\n\t\\exyear{NET JUNE 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $1 / 2$ and $1 / 2$\n\t\\task[\\textbf{B.}]$2 / 3$ and $1 / 3$\n\t\\task[\\textbf{C.}]$1 / 4$ and $3 / 4$\n\t\\task[\\textbf{D.}]$1 / 3$ and $2 / 3$\n\\end{tasks}\n\\begin{answer}\nEigen state of $S_{z}$ is $\\left|\\phi_{1}\\right\\rangle=\\left(\\begin{array}{l}1 \\\\ 0\\end{array}\\right)$ and $\\left|\\phi_{2}\\right\\rangle=\\left(\\begin{array}{l}0 \\\\ 1\\end{array}\\right)$ corresponds to Eigen value $\\frac{\\hbar}{2}$ and $-\\frac{\\hbar}{2}$ respectively.\n$$\nP\\left(\\frac{\\hbar}{2}\\right)=\\frac{\\left|\\left\\langle\\phi_{1} \\mid \\psi\\right\\rangle\\right|^{2}}{\\langle\\psi \\mid \\psi\\rangle}=\\left|\\frac{1+i}{\\sqrt{6}}\\right|^{2}=\\frac{2}{6}=\\frac{1}{3}, \\quad P\\left(-\\frac{\\hbar}{2}\\right)=\\frac{\\left|\\left\\langle\\phi_{2} \\mid \\psi\\right\\rangle\\right|^{2}}{\\langle\\psi \\mid \\psi\\rangle}=\\frac{2}{3}\n$$\nThe correct option is \\textbf{(d)}\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item A spin $-\\frac{1}{2}$ particle is in the state $\\chi=\\frac{1}{\\sqrt{11}}\\left(\\begin{array}{c}1+i \\\\ 3\\end{array}\\right)$ in the eigenbasis of $S^{2}$ and $S_{2}$. If we measure $S_{z}$, the probabilities of getting $+\\frac{h}{2}$ and $-\\frac{h}{2}$, respectively are\n\t\\exyear{NET DEC 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{1}{2}$ and $\\frac{1}{2}$\n\t\\task[\\textbf{B.}]$\\frac{2}{11}$ and $\\frac{9}{11}$\n\t\\task[\\textbf{C.}] 0 and 1\n\t\\task[\\textbf{D.}]$\\frac{1}{11}$ and $\\frac{3}{11}$\n\\end{tasks}\n\\begin{answer}\n\t$P\\left(\\frac{\\hbar}{2}\\right)=\\left|\\frac{1}{\\sqrt{11}}(10)\\left(\\begin{array}{c}1+i \\\\ 3\\end{array}\\right)\\right|^{2}=\\frac{1}{11} \\times 2=\\frac{2}{11} \\quad \\because\\langle\\psi \\mid \\psi\\rangle=1$\\\\ $P\\left(-\\frac{\\hbar}{2}\\right)=\\left|\\frac{1}{\\sqrt{11}}(01)\\left(\\begin{array}{c}1+i \\\\ 3\\end{array}\\right)\\right|^{2}=\\frac{9}{11}$\\\\\n\ti.e. probability of $S_{z}$ getting $\\left(\\frac{\\hbar}{2}\\right)$ and $\\left(-\\frac{\\hbar}{2}\\right)$\\\\\n\tThe correct option is \\textbf{b}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\\item Let $\\vec{\\sigma}=\\left(\\sigma_{1}, \\sigma_{2}, \\sigma_{3}\\right)$, where $\\sigma_{1}, \\sigma_{2}, \\sigma_{3}$ are the Pauli matrices. If $\\vec{a}$ and $\\vec{b}$ are two arbitrary constant vectors in three dimensions, the commutator $[\\vec{a} \\cdot \\vec{\\sigma}, \\vec{b} \\cdot \\vec{\\sigma}]$ is equal to (in the following $I$ is the identity matrix)\n\t\\exyear{NET DEC 2014}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $(\\vec{a} \\cdot \\vec{b})\\left(\\sigma_{1}+\\sigma_{2}+\\sigma_{3}\\right)$\n\t\\task[\\textbf{B.}]$2 i(\\vec{a} \\times \\vec{b}) \\cdot \\vec{\\sigma}$\n\t\\task[\\textbf{C.}]$(\\vec{a} \\cdot \\vec{b}) I$\n\t\\task[\\textbf{D.}]$|\\vec{a}||\\vec{b}| I$\n\\end{tasks}\n\\begin{answer}\n\t$\\vec{a}=a_{1} \\hat{i}+a_{2} \\hat{j}+a_{3} \\hat{k}, \\vec{b}=b_{1} \\hat{i}+b_{2} \\hat{j}+b_{3} \\hat{k}, \\sigma=\\sigma_{x} \\hat{i}+\\sigma_{y} \\hat{j}+\\sigma_{z} \\hat{k}$\n\t\n\t\\begin{align*}\n\t&{[\\vec{a} \\cdot \\vec{\\sigma}, \\vec{b} \\cdot \\vec{\\sigma}]=\\left[a_{1} \\sigma_{x}+a_{2} \\sigma_{y}+a_{3} \\sigma_{z}, b_{1} \\sigma_{x}+b_{2} \\sigma_{y}+b_{3} \\sigma_{z}\\right]} \\\\\n\t&{[\\vec{a} \\cdot \\vec{\\sigma}, \\vec{b} \\cdot \\vec{\\sigma}]=a_{1} b_{1}\\left[\\sigma_{x}, \\sigma_{x}\\right]+a_{1} b_{2}\\left[\\sigma_{x}, \\sigma_{y}\\right]+a_{1} b_{3}\\left[\\sigma_{x}, \\sigma_{z}\\right]+a_{2} b_{1}\\left[\\sigma_{y}, \\sigma_{x}\\right]+a_{2} b_{2}\\left[\\sigma_{y}, \\sigma_{y}\\right]} \\\\\n\t&+a_{2} b_{3}\\left[\\sigma_{y}, \\sigma_{z}\\right]+a_{3} b_{1}\\left[\\sigma_{z}, \\sigma_{x}\\right]+a_{3} b_{2}\\left[\\sigma_{z}, \\sigma_{y}\\right]+a_{3} b_{3}\\left[\\sigma_{z}, \\sigma_{z}\\right] \\\\\n\t&=a_{1} b_{1} \\cdot 0+a_{1} b_{2} \\cdot 2 i \\sigma_{z}-2 i a_{1} b_{3} \\sigma_{y}-a_{2} b_{1} \\cdot 2 i \\sigma_{z}+0+a_{2} b_{3} \\cdot 2 i \\sigma_{x}+a_{3} b_{1} \\cdot 2 i \\sigma_{y}-a_{3} b_{2} \\cdot 2 i \\sigma_{x}+0 \\\\\n\t&\\Rightarrow[\\vec{a} \\cdot \\vec{\\sigma}, \\vec{b} \\cdot \\vec{\\sigma}]==2 i(\\vec{a} \\times \\vec{b}) \\cdot \\vec{\\sigma}\n\t\\end{align*}\n\tThe correct option is \\textbf{(b)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $L_{i}$ are the components of the angular momentum operator $\\vec{L}$, then the operator $\\sum_{i=1,2,3}\\left[\\vec{L}, L_{i}\\right]$ equals\n\t\\exyear{NET JUNE 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\vec{L}$\n\t\\task[\\textbf{B.}]$2 \\vec{L}$\n\t\\task[\\textbf{C.}]$3 \\vec{L}$\n\t\\task[\\textbf{D.}]$-\\vec{L}$\n\\end{tasks}\n\\begin{answer}\n\t Let $\\vec{L}=L_{x} \\hat{i}+L_{y} \\hat{j}+L_{z} \\hat{k}$\n\t\\begin{align*}\n\t&x=1, y=2, z=3 \\\\\n\t&{\\left[\\vec{L}, L_{x}\\right]=\\left[L_{y}, L_{x}\\right] j+\\left[L_{z}, L_{x}\\right] \\hat{k}=-i \\hbar L_{z} \\hat{j}+L_{y} \\hat{k} i \\hbar} \\\\\n\t&{\\left[\\left[\\vec{L}, L_{x}\\right], L_{x}\\right]=i \\hbar\\left[-L_{z}, L_{x}\\right] \\hat{j}+\\left[L_{y}, L_{x}\\right] i \\hbar-i \\hbar . i \\hbar L_{y} \\hat{j}-(i \\hbar) L_{z}(i \\hbar) L_{z}(i \\hbar) \\cdot \\hat{k}=\\hbar^{2}\\left[L_{y} \\hat{j}+L_{z} \\hat{k}\\right]} \\\\\n\t&\\text { similarly, }\\left[\\left[\\vec{L}, L_{y}\\right] L_{y}\\right]=\\hbar^{2}\\left[L_{x} \\hat{i}+L_{z} \\hat{k}\\right] \\\\\n\t&{\\left[\\left[\\vec{L}, L_{z}\\right] L_{z}\\right]=\\hbar^{2}\\left[L_{x} \\hat{i}+L_{y} \\hat{j}\\right]} \\\\\n\t&\\sum_{i=1,2,3}[[L, L i] L i]=2 \\hbar^{2}\\left[L_{x} \\hat{i}+L_{y} \\hat{j}+L_{z} \\hat{k}\\right]=2 \\vec{L} \\quad \\text { put } \\hbar=1\n\t\\end{align*}\n\tThe correct option is \\textbf{(b)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item The Hamiltonian for a spin- $\\frac{1}{2}$ particle at rest is given by $H=E_{0}\\left(\\sigma_{z}+\\alpha \\sigma_{x}\\right)$, where $\\sigma_{x}$ and $\\sigma_{z}$ are Pauli spin matrices and $E_{0}$ and $\\alpha$ are constants. The eigenvalues of this Hamiltonian are\n\t\\exyear{NET DEC 2015}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\pm E_{0} \\sqrt{1+\\alpha^{2}}$\n\t\\task[\\textbf{B.}]$\\pm E_{0} \\sqrt{1-\\alpha^{2}}$\n\t\\task[\\textbf{C.}]$E_{0}$ (doubly degenerate)\n\t\\task[\\textbf{D.}]$E_{0}\\left(1 \\pm \\frac{1}{2} \\alpha^{2}\\right)$\n\\end{tasks}\n\\begin{answer}\n\t$ H=E_{0}\\left(\\dot{\\sigma}_{z}+\\alpha \\sigma_{x}\\right)=E_{0}\\left[\\left(\\begin{array}{cc}1 & 0 \\\\ 0 & -1\\end{array}\\right)+\\alpha\\left(\\begin{array}{ll}0 & 1 \\\\ 1 & 0\\end{array}\\right)\\right] \\Rightarrow H=E_{0}\\left(\\begin{array}{cc}1 & \\alpha \\\\ \\alpha & -1\\end{array}\\right)$\n\tif $\\lambda$ is eigen value, then\n\t$$\n\tH-\\lambda I=0 \\Rightarrow E_{0}\\left(\\begin{array}{cc}\n\t(1-\\lambda) & \\alpha \\\\\n\t\\alpha & -(1+\\lambda)\n\t\\end{array}\\right)=0, \\quad \\lambda=\\pm E_{0} \\sqrt{1+\\alpha^{2}}\n\t$$\n\tThe correct option is \\textbf{(a)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $\\hat{L}_{x}, \\hat{L}_{y}, \\hat{L}_{z}$ are the components of the angular momentum operator in three dimensions the commutator $\\left[\\hat{L}_{x}, \\hat{L}_{x} \\hat{L}_{y} \\hat{L}_{z}\\right]$ may be simplified to\n\t\\exyear{NET JUNE 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $i \\hbar L_{x}\\left(\\hat{L}_{z}^{2}-\\hat{L}_{y}^{2}\\right)$\n\t\\task[\\textbf{B.}]$i \\hbar \\hat{L}_{z} \\hat{L}_{y} \\hat{L}_{x}$\n\t\\task[\\textbf{C.}] $i \\hbar L_{x}\\left(2 \\hat{L}_{z}^{2}-\\hat{L}_{y}^{2}\\right)$\n\t\\task[\\textbf{D.}]0\n\\end{tasks}\n\\begin{answer}\n\t\\begin{align*}\n\t\t&:\\left[L_{x}, L_{x} L_{y} L_{z}\\right]=L_{x}\\left[L_{x}, L_{y} L_{z}\\right]+\\left[L_{x}, L_{x}\\right] L_{y} L_{z} \\\\\n\t\t&=L_{x}\\left[L_{x}, L_{y}\\right] L_{z}+L_{x} L_{y}\\left[L_{x}, L_{z}\\right]+0=L_{x}\\left[i \\hbar L_{z}\\right] L_{z}+L_{x} L_{y}\\left(-i \\hbar L_{y}\\right) \\\\\n\t\t&=i \\hbar L_{x} L_{z}^{2}-i \\hbar L_{x} L_{y}^{2}=i \\hbar L_{x}\\left(L_{z}^{2}-L_{y}^{2}\\right)\n\t\\end{align*}\n\tThe correct option is \\textbf{(a)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item The Hamiltonian of a spin $\\frac{1}{2}$ particle in a magnetic field $\\vec{B}$ is given by $H=-\\mu \\cdot \\vec{B} \\cdot \\vec{\\sigma}$, where $\\mu$ is a real constant and $\\vec{\\sigma}=\\left(\\sigma_{x}, \\sigma_{y}, \\sigma_{z}\\right)$ are the Pauli spin matrices. If $\\vec{B}=\\left(B_{0}, B_{0}, 0\\right)$ and the spin state at time $t=0$ is an eigenstate of $\\sigma_{x}$, then of the expectation values $\\left\\langle\\sigma_{x}\\right\\rangle,\\left\\langle\\sigma_{y}\\right\\rangle$ and $\\left\\langle\\sigma_{z}\\right\\rangle$\n\t\\exyear{NET JUNE 2018}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] only $\\left\\langle\\sigma_{x}\\right\\rangle$ changes with time\n\t\\task[\\textbf{B.}] only $\\left\\langle\\sigma_{y}\\right\\rangle$ changes with time\n\t\\task[\\textbf{C.}]only $\\left\\langle\\sigma_{z}\\right\\rangle$ changes with time\n\t\\task[\\textbf{D.}]all three change with time\n\\end{tasks}\n\\begin{answer}\n $\\left\\langle\\sigma_{x}\\right\\rangle,\\left\\langle\\sigma_{y}\\right\\rangle$ and $\\left\\langle\\sigma_{z}\\right\\rangle$ will changes with time because Eigen state of $\\sigma_{x}$ ie $\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{l}1 \\\\ 1\\end{array}\\right)$ and $\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{c}1 \\\\ -1\\end{array}\\right)$ and can be written in basis of eigen state of $H=-\\mu \\cdot \\vec{B} \\cdot \\vec{\\sigma}=-B_{0}\\left(\\begin{array}{cc}0 & 1-i \\\\ 1+i & 0\\end{array}\\right)$\\\\\n THe correct option is \\textbf{(d)}\n\\end{answer}\n\\end{enumerate}\n\n\n\n\n\n\n\\newpage\n\\begin{abox}\n\tPractice set 2 solutions\n\t\\end{abox}\n\\begin{enumerate}\n\\begin{minipage}{\\textwidth}\n\t\\item $\\text { For a spin-s particle, in the eigen basis of } \\vec{S}^{2}, S_{x} \\text { the expectation value }\\left\\langle s m\\left|S_{x}^{2}\\right| s m\\right\\rangle \\text { is }$\n\t\\exyear{GATE 2010}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\frac{\\hbar^{2}\\left\\{s(s+1)-m^{2}\\right\\}}{2}$\n\t\\task[\\textbf{B.}] $\\hbar^{2}\\left\\{s(s+1)-2 m^{2}\\right\\}$\n\t\\task[\\textbf{C.}]$\\hbar^{2}\\left\\{s(s+1)-m^{2}\\right\\}$\n\t\\task[\\textbf{D.}]$\\hbar^{2} m^{2}$\n\\end{tasks}\n\\begin{answer}\n\\begin{align*}\n\t&\\left\\langle s m\\left|S_{x}^{2}\\right| s m\\right\\rangle=\\frac{1}{4}\\left\\langle s m\\left|\\left(S_{+}+S_{-}\\right)^{2}\\right| s m\\right\\rangle=\\frac{1}{4}\\left\\langle s m\\left|S_{+}^{2}+S_{-}^{2}+S_{+} S_{-}+S_{-} S_{+}\\right| s m\\right\\rangle \\\\\n\t&=\\frac{1}{4}\\left\\langle s m\\left|S_{+} S_{-}+S_{-} S_{+}\\right| s m\\right\\rangle=\\frac{\\hbar^{2}}{2}\\left[s(s+1)-m^{2}\\right] \\quad\\left[\\because S_{+} S_{-}+S_{-} S_{+}=2\\left(S^{2}-S_{z}^{2}\\right)\\right]\n\\end{align*}\nThe correct option is \\textbf{(a)}\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $L_{x}, L_{y}$ and $L_{z}$ are respectively the $x, y$ and $z$ components of angular momentum operator $L$. The commutator $\\left[L_{x} L_{y}, L_{z}\\right]$ is equal to\n\t\\exyear{GATE 2011}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $i \\hbar\\left(L_{x}^{2}+L_{y}^{2}\\right)$\n\t\\task[\\textbf{B.}]$2 i \\hbar L_{z}$\n\t\\task[\\textbf{C.}]$i \\hbar\\left(L_{x}^{2}-L_{y}^{2}\\right)$\n\t\\task[\\textbf{D.}]0\n\\end{tasks}\n\\begin{answer}\n\t$\\left\\lfloor L_{x} L_{y}, L_{z}\\right\\rfloor=L_{x}\\left[L_{y} L_{z}\\right]+\\left[L_{x}, L_{z}\\right] L_{y}=i \\hbar\\left(L_{x}^{2}-L_{y}^{2}\\right)$\\\\\n\tThe correct option is \\textbf{(c)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item Which one of the following commutation relations is NOT CORRECT? Here, symbols have their usual meanings.\n\t\\exyear{GATE 2013}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $\\left[L^{2}, L_{z}\\right]=0$\n\t\\task[\\textbf{B.}]$\\left\\lfloor L_{x}, L_{y}\\right\\rfloor=i \\hbar L_{z}$\n\t\\task[\\textbf{C.}]$\\left[L_{z}, L_{+}\\right]=\\hbar L_{+}$\n\t\\task[\\textbf{D.}]$\\left[L_{z}, L_{-}\\right]=\\hbar L_{-}$\n\\end{tasks}\n\\begin{answer}\nThe correct option is \\textbf{(d)}\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item A spin-half particle is in a linear superposition $0.8|\\uparrow\\rangle+0.6|\\downarrow\\rangle$ of its spin-up and spindown states. If $|\\uparrow\\rangle$ and $|\\downarrow\\rangle$ are the eigenstates of $\\sigma_{z}$, then what is the expectation value up to one decimal place, of the operator $10 \\sigma_{z}+5 \\sigma_{x}$ ? Here, symbols have their usual meanings.\n\t\\exyear{GATE 2013}\n\\end{minipage}\n\\begin{answer}\n\\begin{align*}\n\t&\\psi\\rangle=.8|\\uparrow\\rangle+.6|\\downarrow\\rangle=0.8\\left(\\begin{array}{l}\n\t\t1 \\\\\n\t\t0\n\t\\end{array}\\right)+0.6\\left(\\begin{array}{l}\n\t\t0 \\\\\n\t\t1\n\t\\end{array}\\right)=\\left(\\begin{array}{l}\n\t\t0.8 \\\\\n\t\t0.6\n\t\\end{array}\\right) \\\\\n\t&\\text { Operator } A=10 \\sigma_{z}+5 \\sigma_{x}=10\\left(\\begin{array}{cc}\n\t\t1 & 0 \\\\\n\t\t0 & -1\n\t\\end{array}\\right)+5\\left(\\begin{array}{ll}\n\t\t0 & 1 \\\\\n\t\t1 & 0\n\t\\end{array}\\right) \\Rightarrow A=\\left(\\begin{array}{cc}\n\t\t10 & 5 \\\\\n\t\t5 & -10\n\t\\end{array}\\right) \\\\\n\t&\\qquad\\langle A\\rangle=\\langle\\psi|A| \\psi\\rangle=\\left(\\begin{array}{ll}\n\t\t0.8 & 0.6\n\t\\end{array}\\right)\\left(\\begin{array}{cc}\n\t\t10 & 5 \\\\\n\t\t5 & -10\n\t\\end{array}\\right)\\left(\\begin{array}{l}\n\t\t0.8 \\\\\n\t\t0.6\n\t\\end{array}\\right)=(8.8-1.2)=7.6\n\\end{align*}\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $\\vec{L}$ is the orbital angular momentum and $\\bar{S}$ is the spin angular momentum, then $\\vec{L} \\cdot \\vec{S}$ does not commute with\n\t\\exyear{GATE 2014}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $S_{z}$ \n\t\\task[\\textbf{B.}]$L^{2}$\n\t\\task[\\textbf{C.}]$S^{2}$\n\t\\task[\\textbf{D.}]$(\\vec{L}+\\vec{S})^{2}$\n\\end{tasks}\n\\begin{answer}\nThe correct option is \\textbf{(d)}\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $L_{+}$and $L_{-}$are the angular momentum ladder operators then the expectation value of $\\left(L_{+} L_{-}+L_{-} L_{+}\\right)$in the state $|l=1, m=1\\rangle$ of an atom is $\\hbar^{2}$\n\t\\exyear{GATE 2014}\n\\end{minipage}\n\\begin{answer}\n\t$\\left(L_{+} L_{-}+L_{-} L_{+}\\right)=2\\left(L^{2}-L_{z}^{2}\\right)=2\\left(l .(l+1)-m^{2}\\right) \\hbar^{2}=2 \\hbar^{2}$\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item The Pauli matrices for three spin $-\\frac{1}{2}$ particles are $\\vec{\\sigma}_{1}, \\vec{\\sigma}_{2}$ and $\\vec{\\sigma}_{3}$, respectively. The dimension of the Hilbert space required to define an operator $\\hat{O}=\\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2} \\times \\vec{\\sigma}_{3}$ is\n\t\\exyear{GATE 2015}\n\\end{minipage}\n\\begin{answer}\n\t$\\sigma_{2} \\times \\sigma_{3} \\text { has dimension of } 4 \\text { and } \\sigma_{1} . \\sigma_{2} \\times \\sigma_{3} \\text { has dimension of } 2 \\times 4=8$\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item Let the Hamiltonian for two spin-1/2 particles of equal masses $m$, momenta $\\vec{p}_{1}$ and $\\vec{p}_{2}$ and positions $\\vec{r}_{1}$ and $\\vec{r}_{2}$ be $H=\\frac{1}{2 m} p_{1}^{2}+\\frac{1}{2 m} p_{2}^{2}+\\frac{1}{2} m \\omega^{2}\\left(r_{1}^{2}+r_{2}^{2}\\right)+k \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2}$, where $\\vec{\\sigma}_{1}$ and $\\vec{\\sigma}_{2}$ denote the corresponding Pauli matrices, $\\hbar \\omega=0.1 \\mathrm{eV}$ and $k=0.2 \\mathrm{eV}$. If the ground state has net spin zero, then the energy (in $\\mathrm{eV}$ ) is\n\t\\exyear{GATE 2015}\n\\end{minipage}\n\\begin{answer}\n$H=\\frac{1}{2 m} p_{1}^{2}+\\frac{1}{2 m} p_{2}^{2}+\\frac{1}{2} m \\omega^{2}\\left(r_{1}^{2}+r_{2}^{2}\\right)+k \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2}$\n\\begin{align*}\n&\\vec{\\sigma}=\\overrightarrow{\\sigma_{1}}+\\vec{\\sigma}_{2} \\Rightarrow \\vec{\\sigma}^{2}=\\sigma_{1}^{2}+\\sigma_{2}^{2}+2 \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2} \\Rightarrow 2 \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2}=\\vec{\\sigma}^{2}-\\sigma_{1}^{2}-\\sigma_{2}^{2} \\\\\n&\\Rightarrow 2 \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2}=0-3 I-3 I=-6 I \\Rightarrow \\vec{\\sigma}_{1} \\cdot \\vec{\\sigma}_{2}=-3\n\\end{align*}\nNow energy $E=2 \\times \\frac{3}{2} \\hbar \\omega+k(-3)=3 \\times(0.1)+(0.2)(-3)=-0.3 \\mathrm{eV}$\t\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item If $\\vec{s}_{1}$ and $\\vec{s}_{2}$ are the spin operators of the two electrons of a He atom, the value o $\\left\\langle\\vec{s}_{1} \\cdot \\vec{s}_{2}\\right\\rangle$ for the ground state is\n\t\\exyear{GATE 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $-\\frac{3}{2} \\hbar^{2}$\n\t\\task[\\textbf{B.}]$-\\frac{3}{4} \\hbar^{2}$\n\t\\task[\\textbf{C.}] 0\n\t\\task[\\textbf{D.}] $\\frac{1}{4} \\hbar^{2}$\n\\end{tasks}\n\\begin{answer}\n\t$\\vec{s}=\\vec{s}_{1}+\\vec{s}_{2}, s_{1}=\\frac{1}{2}, s_{1}=\\frac{1}{2}, s=0,1$\\\\ $$\\left\\langle\\vec{s}_{1} \\cdot \\vec{s}_{2}\\right\\rangle=\\frac{s(s+1) \\hbar^{2}-s_{1}\\left(s_{1}+1\\right) \\hbar^{2}-s_{2}\\left(s_{2}+1\\right) \\hbar^{2}}{2}$$\n\tFor $$s=1,\\left\\langle\\vec{s}_{1} \\cdot \\vec{s}_{2}\\right\\rangle=\\frac{2 \\hbar^{2}-\\frac{3}{4} \\hbar^{2}-\\frac{3}{4} \\hbar^{2}}{2}=\\frac{3}{4} \\hbar^{2}$$\n\t$$\n\ts=0,\\left\\langle\\vec{s}_{1} \\cdot \\vec{s}_{2}\\right\\rangle=\\frac{0 \\hbar^{2}-\\frac{3}{4} \\hbar^{2}-\\frac{3}{4} \\hbar^{2}}{2}=-\\frac{3}{4} \\hbar^{2}\n\t$$\n\tThe correct option is \\textbf{(b)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item $\\sigma_{x}, \\sigma_{y} \\text { and } \\sigma_{z} \\text { are the Pauli matrices. The expression } 2 \\sigma_{x} \\sigma_{y}+\\sigma_{y} \\sigma_{x} \\text { is equal to }$\n\t\\exyear{GATE 2016}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $-3 i \\sigma_{z}$\n\t\\task[\\textbf{B.}]$-i \\sigma_{z}$\n\t\\task[\\textbf{C.}]$i \\sigma_{z}$\n\t\\task[\\textbf{D.}]$3 i \\sigma_{z}$\n\\end{tasks}\n\\begin{answer}\n\t$2 \\sigma_{x} \\sigma_{y}+\\sigma_{y} \\sigma_{x} \\Rightarrow \\sigma_{x} \\sigma_{y}+\\sigma_{x} \\sigma_{y}+\\sigma_{y} \\sigma_{x} \\Rightarrow \\sigma_{x} \\sigma_{y}=i \\sigma_{z}$\\\\\n\tThe correct option is \\textbf{(c)}\n\\end{answer}\n\\begin{minipage}{\\textwidth}\n\t\\item For the Hamiltonian $H=a_{0} I+\\vec{b} \\cdot \\vec{\\sigma}$ where $a_{0} \\in R, \\vec{b}$ is a real vector, $I$ is the $2 \\times 2$ identity matrix, and $\\vec{\\sigma}$ are the Pauli matrices, the ground state energy is\n\t\\exyear{GATE 2017}\n\\end{minipage}\n\\begin{tasks}(2)\n\t\\task[\\textbf{A.}] $|b|$\n\t\\task[\\textbf{B.}]$2 a_{0}-|b|$\n\t\\task[\\textbf{C.}]$a_{0}-|b|$\n\t\\task[\\textbf{D.}]$a_{0}$\n\\end{tasks}\n\\begin{answer}\n$a_{0} I+\\vec{b} \\cdot \\vec{\\sigma}=a_{0}\\left(\\begin{array}{cc}\n\t1 & 0 \\\\\n\t0 & 1\n\\end{array}\\right)+b_{x}\\left(\\begin{array}{cc}\n\t0 & 1 \\\\\n\t1 & 0\n\\end{array}\\right)+b_{y}\\left(\\begin{array}{cc}\n\t0 & -i \\\\\n\ti & 0\n\\end{array}\\right)+b_{z}\\left(\\begin{array}{cc}\n\t1 & 0 \\\\\n\t0 & -1\n\\end{array}\\right)=\\left(\\begin{array}{cc}\n\ta_{0}+b_{z} & b_{x}-i b_{y} \\\\\n\tb_{x}+i b_{y} & a_{0}-b_{z}\n\\end{array}\\right)$\\\\\n$H=a_{0} I+\\vec{b} \\cdot \\vec{\\sigma}=\\left(\\begin{array}{cc}\n\ta_{0}+b_{z} & b_{x}-i b_{y} \\\\\n\tb_{x}+i b_{y} & a_{0}-b_{z}\n\\end{array}\\right)$\\\\\nFor eigen value $\\left(\\begin{array}{cc}a_{0}+b_{z}-\\lambda & b_{x}-i b_{y} \\\\ b_{x}+i b_{y} & a_{0}-b_{z}-\\lambda\\end{array}\\right)=0$\n$$\n\\begin{aligned}\n&\\left(a_{0}+b_{z}-\\lambda\\right)\\left(a_{0}-b_{z}-\\lambda\\right)-\\left(b_{x}^{2}+b_{y}^{2}\\right)=0 \\\\\n&\\lambda_{1}=a_{0}-|b|, \\lambda_{1}=a_{0}+|b|\n\\end{aligned}\n$$\nThe correct option is \\textbf{(c)}\n\\end{answer}\n\\end{enumerate}\n", "meta": {"hexsha": "5fecd5a4cbf52ddb4df50bc816c2555b3ea6ddc4", "size": 26994, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "QM -CSIR/chapter/angular momentum solutions.tex", "max_stars_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_stars_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "QM -CSIR/chapter/angular momentum solutions.tex", "max_issues_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_issues_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "QM -CSIR/chapter/angular momentum solutions.tex", "max_forks_repo_name": "archives-futuring/CSIR-Physics-Study-Material", "max_forks_repo_head_hexsha": "689cff91895fec36b4bb0add178f13a0f68648ab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.5595238095, "max_line_length": 538, "alphanum_fraction": 0.600133363, "num_tokens": 11794, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.7931059438487662, "lm_q1q2_score": 0.6164811400504683}} {"text": "\\documentclass[a4paper,11pt]{article}\n\\usepackage{amsmath, amsthm, amssymb, commath, graphicx, bbold, endnotes, graphicx, subfigure, multirow, setspace, siunitx}\n\\usepackage{amsmath}\n\\usepackage[font=footnotesize]{caption}\n\\usepackage[top=1.25in, bottom=1.25in, left=1.25in, right=1.25in]{geometry}\n\\linespread{1}\n\\setlength{\\parindent}{30pt}\n\n\\title{Cosmological Parameters and Conversions Memo}\n\\author{Ruby Byrne}\n\\date{October 2018}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Parallel to the Line-of-Sight}\n\nFrom Morales and Hewitt 2004, frequency $\\Delta f$ is related to the co-moving line-of-sight distance $\\Delta r_z$ by\n\\begin{equation}\n\\Delta f \\approx \\frac{f_{21} E(z)}{D_H (1+z)^2} \\Delta r_z\n\\end{equation}\nThe Fourier inverse of $\\Delta f$ is $\\eta$ and the Fourier inverse of $\\Delta r_z$ is $k_z$. These are related by\n\\begin{equation}\n\\eta \\approx \\frac{D_H (1+z)^2}{2 \\pi f_{21} E(z)} k_z\n\\end{equation}\n$f_{21}$ is the frequency of the 21-cm signal, so $f_{21} = \\frac{c}{0.21 \\si{m}}$. $D_H$ is the Hubble distance. From Hogg 2000,\n\\begin{equation}\nD_H = \\frac{c}{H_0} = 3000 \\frac{\\si{Mpc}}{h}\n\\end{equation}\nwhere $H_0$ is the Hubble constant in the present epoch and $h$ is a dimensionless quantity that is thought to be $0.6 < h < 0.9$ (the \\textit{IDL} function \\texttt{cosmology\\_measures} defaults $h=0.71$). $H_0$ is related to $h$ by\n\\begin{equation}\nH_0 = 100 \\frac{h\\si{km}}{\\si{s.Mpc}}\n\\end{equation}\nAgain from Morales and Hewitt 2004,\n\\begin{equation}\nE(z) = \\sqrt{\\Omega_M(1+z)^3+\\Omega_k(1+z)^2+\\Omega_{\\Lambda}}\n\\end{equation}\nwhere $\\Omega_M$ is the matter constant, $\\Omega_k$ is the curvature constant, and $\\Omega_{\\Lambda}$ is the ``lambda'' constant (whatever that is). \n\nThe \\textit{IDL} function \\texttt{cosmology\\_measures} defaults these parameters to $\\Omega_M=0.27$, $\\Omega_k=0$, and $\\Omega_{\\Lambda}=0.73$. Plugging these to $E(z)$ for $z=7$ in gives $E(7) \\approx 11.79$. Therefore at $z=7$, the conversion between $\\eta$ and $k_z$ is given by\n\\begin{equation}\nk_z \\approx \\left(5.51\\times10^5 \\frac{h}{\\si{s.Mpc}}\\right) \\eta\n\\end{equation}\n\n\n\\section{Perpendicular to the Line-of-Sight}\n\nFrom Morales and Hewitt 2004, sky positions $\\theta_x$ and $\\theta_y$ are related to co-moving distances $r_x$ and $r_y$ according to\n\\begin{equation}\n\\theta_x = \\frac{r_x}{D_M(z)}, \\quad \\theta_y = \\frac{r_y}{D_M(z)}\n\\end{equation}\nThe Fourier inverses of $\\theta_x$ and $\\theta_y$ are $u$ and $v$ and the Fourier inverses of $r_x$ and $r_y$ are $k_x$ and $k_y$. These are related by\n\\begin{equation}\nu = \\frac{k_x D_M(z)}{2 \\pi}, \\quad v = \\frac{k_y D_M(z)}{2 \\pi}\n\\end{equation}\n$D_M$ is the ``transverse co-moving distance.''\n\nFrom Hogg 2000,\n\\begin{equation}\n D_M=\\begin{cases}\n D_H \\frac{1}{\\sqrt{\\Omega_k}} \\text{sinh}(\\sqrt{\\Omega_k}D_C/D_H), & \\text{for $\\Omega_k>0$}\\\\\n D_C, & \\text{for $\\Omega_k=0$}\\\\\n D_H \\frac{1}{\\sqrt{\\Omega_k}} \\text{sin}(\\sqrt{\\Omega_k}D_C/D_H), & \\text{for $\\Omega_k<0$}\n \\end{cases}\n\\end{equation}\n$D_C$ is the co-moving distance and is defined as\n\\begin{equation}\nD_C = D_H \\int_0^z \\frac{dz'}{E(z')} = D_H \\int_0^z \\frac{dz'}{\\sqrt{\\Omega_M(1+z')^3+\\Omega_k(1+z')^2+\\Omega_{\\Lambda}}}\n\\end{equation}\n\nThis integral must be evaluated numerically. With $z=7$, $\\Omega_M=0.27$, $\\Omega_k=0$, and $\\Omega_{\\Lambda}=0.73$ as above and evaluating the integral with \n\\textit{Mathematica}, we get that\n\\begin{equation}\nD_C \\approx D_H(2.09) = 6.27 \\times 10^3 \\frac{\\si{Mpc}}{h}\n\\end{equation}\nThis integral is also evaluated by \\textit{IDL}'s \\texttt{cosmology\\_measures} function. Since we are assuming $\\Omega_k=0$, $D_M=D_C$ and\n\\begin{equation}\nD_M \\approx 6.27 \\times 10^3 \\frac{\\si{Mpc}}{h}\n\\end{equation}\nTo convert from $(u, v)$ coordinates to $(k_x,k_y)$ coordinates, we therefore get\n\\begin{equation}\nk_x \\approx \\left(1.00\\times10^{-3} \\frac{h}{\\si{Mpc}}\\right)u, \\quad k_y \\approx \\left(1.00\\times10^{-3} \\frac{h}{\\si{Mpc}}\\right)v\n\\end{equation}\n\n\\end{document}", "meta": {"hexsha": "8a4f2466a7e20282926c85052388b22b6f758f82", "size": 3958, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "002_Cosmological_Units/Cosmological_Units.tex", "max_stars_repo_name": "EoRImaging/Memos", "max_stars_repo_head_hexsha": "216dbda634c1686be25cda25bb258664067a3aad", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-08-05T08:38:01.000Z", "max_stars_repo_stars_event_max_datetime": "2020-08-05T08:38:01.000Z", "max_issues_repo_path": "002_Cosmological_Units/Cosmological_Units.tex", "max_issues_repo_name": "EoRImaging/Memos", "max_issues_repo_head_hexsha": "216dbda634c1686be25cda25bb258664067a3aad", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2019-03-07T22:45:46.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-20T20:19:43.000Z", "max_forks_repo_path": "002_Cosmological_Units/Cosmological_Units.tex", "max_forks_repo_name": "EoRImaging/Memos", "max_forks_repo_head_hexsha": "216dbda634c1686be25cda25bb258664067a3aad", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-03-07T01:12:56.000Z", "max_forks_repo_forks_event_max_datetime": "2019-03-07T01:12:56.000Z", "avg_line_length": 46.023255814, "max_line_length": 281, "alphanum_fraction": 0.6960586155, "num_tokens": 1468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998508568417, "lm_q2_score": 0.7931059414036511, "lm_q1q2_score": 0.6164811299667331}} {"text": "% ----------------------------------------------------------------------\n\\section{Incompressible Isotropic Elasticity with Infinitesimal Strain (Bathe)}\n\nIn this section we apply a similar approach to the one we use for the\nelasticity equation to the case of an incompressible material. We only\nconsider the quasistatic case (neglect inertia) without faults. As the\nbulk modulus ($K$) approaches infinity, the volumetric strain\n($\\Tr(\\epsilon)$) approaches zero and the pressure remains finite,\n$p = -K \\Tr(\\epsilon)$. We consider pressure $p$ as an independent\nvariable and decompose the stress into the pressure and deviatoric\ncomponents. As a result, we write the stress tensor in terms of both\nthe displacement and pressure fields,\n\\begin{equation}\n \\tensor{\\sigma}(\\vec{u},p) = \\tensor{\\sigma}^\\mathit{dev}(\\vec{u}) - p\\tensor{I}.\n\\end{equation}\n\nThe strong form is\n\\begin{gather}\n % Solution\n \\vec{s}^T = \\left( \\vec{u} \\quad \\ p \\right)^T, \\\\\n % Elasticity\n \\vec{f}(t) + \\tensor{\\nabla} \\cdot \\left(\\tensor{\\sigma}^\\mathit{dev}(\\vec{u}) - p\\tensor{I}\\right) = \\vec{0} \\text{ in }\\Omega, \\\\\n % Pressure\n \\vec{\\nabla} \\cdot \\vec{u} + \\frac{p}{K} = 0 \\text{ in }\\Omega, \\\\\n % Neumann\n \\tensor{\\sigma} \\cdot \\vec{n} = \\vec{\\tau} \\text{ on }\\Gamma_\\tau, \\\\\n % Dirichlet\n \\vec{u} = \\vec{u}_0 \\text{ on }\\Gamma_u, \\\\\n p = p_0 \\text{ on }\\Gamma_p.\n\\end{gather}\nWe place all terms for the elasticity and pressure equations on the\nleft-hand-side, consistent with PETSc TS implicit time stepping.\n\n\\begin{table}[htbp]\n \\caption{Mathematical notation for incompressible elasticity with\n infinitesimal strain.}\n \\label{tab:notation:incompressible:elasticity}\n \\begin{tabular}{lcp{3.5in}}\n \\toprule\n {\\bf Category} & {\\bf Symbol} & {\\bf Description} \\\\\n \\midrule\n Unknowns & $\\vec{u}$ & Displacement field \\\\\n & $p$ & Pressure field ($p>0$ corresponds to negative mean stress \\\\\n Derived quantities & $\\tensor{\\sigma}$ & Cauchy stress tensor \\\\\n & $\\tensor{\\epsilon}$ & Cauchy strain tensor \\\\\n Common constitutive parameters & $\\rho$ & Density \\\\\n & $\\mu$ & Shear modulus \\\\\n & $K$ & Bulk modulus \\\\\nSource terms & $\\vec{f}$ & Body force per unit volume, for example $\\rho \\vec{g}$ \\\\\n \\bottomrule\n \\end{tabular}\n\\end{table}\n\nUsing trial functions $\\trialvec[u]$ and $\\trialscalar[p]$ and\nincorporating the Neumann boundary conditions, we write the weak form\nas\n\\begin{gather}\n % Displacement\n \\int_\\Omega \\trialvec[u] \\cdot \\vec{f}(t) + \\nabla \\trialvec[u] : \\left(-\\tensor{\\sigma}^\\mathit{dev}(\\vec{u}) + p\\tensor{I}\n \\right)\\, d\\Omega + \\int_{\\Gamma_\\tau} \\trialvec[u] \\cdot \\vec{\\tau}(t) \\, d\\Gamma, = 0 \\\\\n % Pressure\n \\int_\\Omega \\trialscalar[p] \\cdot \\left(\\vec{\\nabla} \\cdot \\vec{u} + \\frac{p}{K} \\right) \\, d\\Omega = 0.\n\\end{gather}\n\n\\subsection{Residual Pointwise Functions}\n\nIdentifying $F(t,s,\\dot{s})$, we have\n\\begin{gather}\n \\label{eqn:incompressible:elasticity:displacement}\n F^u(t,s,\\dot{s}) = \\int_\\Omega \\trialvec[u] \\cdot \\eqnannotate{\\vec{f}(t)}{f_0^u} + \\nabla \\trialvec[u] :\n \\eqnannotate{\\left(-\\tensor{\\sigma}^\\mathit{dev}(\\vec{u}) + p\\tensor{I}\\right)}{f_1^u} \\, d\\Omega\n + \\int_{\\Gamma_\\tau} \\trialvec[u] \\cdot \\eqnannotate{\\vec{\\tau}(t)}{f_0^u} \\, d\\Gamma, \\\\\n%\n \\label{eqn:incompressible:elasticity:pressure}\n F^p(t,s,\\dot{s}) = \\int_\\Omega \\trialscalar[p] \\cdot \\eqnannotate{\\left(\\vec{\\nabla} \\cdot \\vec{u} + \n\\frac{p}{K} \\right)}{f_0^p} \\, d\\Omega.\n\\end{gather}\n\n\\subsection{Jacobians Pointwise Functions}\n\nWith two fields we have four Jacobian pointwise functions for the LHS:\n\\begin{align}\n % JF uu\n J_F^{uu} &= \\frac{\\partial F^u}{\\partial u} + s_\\mathit{tshift} \\frac{\\partial F^u}{\\partial \\dot{u}} =\n \\int_\\Omega \\nabla \\trialvec[u] : \\frac{\\partial}{\\partial u}(-\\tensor{\\sigma}^\\mathit{dev}) \\, d\\Omega \n = \\int_\\Omega \\trialscalar[u]_{i,k} \\, \\eqnannotate{\\left(-C^\\mathit{dev}_{ikjl}\\right)} {J_{f3}^{uu}} \\, \\basisscalar[u]_{j,l}\\, d\\Omega \\\\\n % JF up\n J_F^{up} &= \\frac{\\partial F^u}{\\partial p} + s_\\mathit{tshift} \\frac{\\partial F^u}{\\partial \\dot{p}} =\n \\int_\\Omega \\nabla\\trialvec[u] : \\tensor{I} \\basisscalar[p] \\, d\\Omega\n = \\int_\\Omega \\trialscalar[u]_{i,k} \\eqnannotate{\\delta_{ik}}{J_{f2}^{up}} \\, \\basisscalar[p] \\, d\\Omega \\\\\n % JF pu\n J_F^{pu} &= \\frac{\\partial F^p}{\\partial u} + s_\\mathit{tshift} \\frac{\\partial F^p}{\\partial \\dot{u}} =\n \\int_\\Omega \\trialscalar[p] \\left(\\vec{\\nabla} \\cdot \\basisvec[u]\\right) \\, d\\Omega\n = \\int_\\Omega \\trialscalar[p] \\eqnannotate{\\delta_{jl}}{J_{f1}^{pu}} \\basisscalar[u]_{j,l} \\, d\\Omega\\\\\n % JF pp\n J_F^{pp} &= \\frac{\\partial F^p}{\\partial p} + s_\\mathit{tshift} \\frac{\\partial F^p}{\\partial \\dot{p}} =\n \\int_\\Omega \\trialscalar[p] \\eqnannotate{\\frac{1} {K}}{J_{f0}^{pp}} \\basisscalar[p] \\, d\\Omega\n\\end{align}\n\nFor isotropic, linear incompressible elasticity, the deviatoric elastic constants are:\n\\begin{align}\n C_{1111} &= C_{2222} = C_{3333} = +\\frac{4}{3} \\mu \\\\\n C_{1122} &= C_{1133} = C_{2233} = -\\frac{2}{3} \\mu \\\\\n C_{1212} &= C_{1313} = C_{2323} = \\mu\n\\end{align}\n", "meta": {"hexsha": "1b72db5332590c4d2852b352743ff009b2aaef49", "size": 5144, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/userguide/governingeqns/incompressible_elasticity.tex", "max_stars_repo_name": "Grant-Block/pylith", "max_stars_repo_head_hexsha": "f6338261b17551eba879da998a5aaf2d91f5f658", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 93, "max_stars_repo_stars_event_min_datetime": "2015-01-08T16:41:22.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-25T13:40:02.000Z", "max_issues_repo_path": "doc/userguide/governingeqns/incompressible_elasticity.tex", "max_issues_repo_name": "Grant-Block/pylith", "max_issues_repo_head_hexsha": "f6338261b17551eba879da998a5aaf2d91f5f658", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 277, "max_issues_repo_issues_event_min_datetime": "2015-02-20T16:27:35.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-30T21:13:09.000Z", "max_forks_repo_path": "doc/userguide/governingeqns/incompressible_elasticity.tex", "max_forks_repo_name": "Grant-Block/pylith", "max_forks_repo_head_hexsha": "f6338261b17551eba879da998a5aaf2d91f5f658", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 71, "max_forks_repo_forks_event_min_datetime": "2015-03-24T12:11:08.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-03T04:26:02.000Z", "avg_line_length": 48.5283018868, "max_line_length": 154, "alphanum_fraction": 0.634525661, "num_tokens": 1801, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.839733983715524, "lm_q2_score": 0.7341195152660687, "lm_q1q2_score": 0.6164651050776853}} {"text": "\\documentclass[]{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n%opening\n\\title{Actor Critic Agent}\n\\author{Marco Marini}\n\n\\begin{document}\n\n\\maketitle\n\n\\begin{abstract}\nDescription of Actor Critic Agent and definitions of performance indicators\n\\end{abstract}\n\n\\tableofcontents\n\n\\section{Actor Critic Agent}\n\nThe Policy Actor Critic Agent is composed by an artificial neural network (ANN) that generates the differential state value of critic component and the policy parameters for action selection of each actor (the action space dimensions) from the state input signals.\nAt each iteration with the environment, the agent computes the average reward $r_\\pi$ and the differential TD error $ \\delta_t $\n\\begin{equation}\n\\delta_t = R_{t+1} - \\bar R + \\hat v(S_{t+1}) - \\hat v(S_t)\t\n\\end{equation}\n\nThe differential TD error indicate how different is the differential state value determined by the current reward from the estimated differential state value.\n\nIf the error is positive the agent changes the policy reinforcing the current strategy. It increments the probabilities of selected actions and at the same time reduces the probabilities of the other actions.\n\nIf the error is negative the agent changes the policy weakening the current strategy. It reduces the probabilities of selected actions and increments the probabilities of the other actions.\n\nThe magnitude of changes is proportional to the TD Error by a $ \\alpha_i $ parameter for each $ i $ actor that allows to control the convergence of the algorithm.\n\nThe TD error is used to determine the updated output values and train the AAN with these expected outputs.\n\n\\subsection{Discrete Policy Actor}\n\nThe policy actor computes the probabilities $ \\pi(a|s) $ of choose action $ a $ at status $ s $.\nThe probabilities are the softmax function of the action preferences $ h(s,a,\\vec \\theta) $ given by the ANN \n\\begin{align}\n\t\\pi(a|s, \\vec \\theta)\n\t&= \\frac{e^{h_a(s,a,\\vec{\\theta})}}{\\sum_{b \\in A} e^{h(s,b,\\vec{\\theta})}}\n\\end{align}\n\nTo train the ANN the agent updates the $ \\vec \\theta $ parameters at each step $ t $\n\\begin{align}\n\t\\vec \\theta_{t+1} =\n\t\\vec \\theta_t + \\alpha \\delta_t \\nabla \\ln \\pi(A_t|S_t, \\vec \\theta_t) \n\\end{align}\n\nThe policy gradient function is\n\\begin{align*}\n\\begin{split}\n\t\\nabla \\ln \\pi(A_t|S_t, \\vec \\theta)\n\t& = \\frac{1}{\\pi(A_t|S_t, \\vec \\theta)}\n\t\\nabla \\pi(A_t|S_t, \\vec \\theta)\n\t\\\\\n\t\\pi(a|s, \\vec \\theta)\n\t&= \\frac{e^{h_a(s,a,\\vec{\\theta})}}{D(s,\\vec \\theta)}\n\t\\\\\n\tD(s,\\vec \\theta)\n\t&= \\sum_{b \\in A} e^{h(s,b,\\vec{\\theta})}\n\t\\\\\n\t\\nabla \\pi(A_t|S_t, \\vec \\theta)\n\t& =\n\t\\frac{1}{D^2(S_t,\\vec \\theta)}\n\t\\left[\n\t\tD(S_t, \\vec \\theta) \\nabla e^{h(S_t,A_t,\\vec \\theta)} - e^{h(S_t,A_t,\\vec \\theta)} \\nabla D(S_t, \\vec \\theta)\n\t\\right] \n\t\\\\\n\t& =\t\\frac{1}{D^2(S_t,\\vec \\theta)}\n\t\\left[\n\t\tD(S_t, \\vec \\theta) e^{h(S_t,A_t,\\vec \\theta)} \\nabla h(S_t,A_t,\\vec \\theta)\n\t\t- e^{h(S_t,A_t,\\vec \\theta)} \\nabla D(S_t, \\vec \\theta)\n\t\\right] \n\t\\\\\n\t& =\n\t\\frac{\\pi(A_t|S_t,\\vec \\theta)}{D(S_t,\\vec \\theta)}\n\t\\left[\n\t\tD(S_t, \\vec \\theta) \\nabla h(S_t,A_t,\\vec \\theta)\n\t\t- \\nabla D(S_t, \\vec \\theta)\n\t\\right] \n\t\\\\\n\t\\nabla D(s,\\vec \\theta)\n\t&= \\sum_{b \\in A} \\nabla e^{h(s,b,\\vec \\theta)}\n\t\\\\\n\t&= \\sum_{b \\in A} e^{h(s,b,\\vec \\theta)}\t\\nabla h(s,b,\\vec \\theta)\n\t\\\\\n\t\\nabla \\pi(A_t|S_t, \\vec \\theta)\n\t& =\n\t\\frac{\\pi(A_t|S_t,\\vec \\theta)}{D(S_t,\\vec \\theta)}\n\t\\left[\n\t\tD(S_t, \\vec \\theta) \\nabla h(S_t,A_t,\\vec \\theta)\n\t\t- \\sum_{b \\in A} e^{h(s,b,\\vec \\theta)}\t\\nabla h(s,b,\\vec \\theta)\n\t\\right] \n\\end{split}\n\\end{align*}\n\nLet be\n\\begin{align*}\n\tI(a, b) = 1, \\Rightarrow a = b\n\t\\\\\n\tI(a,b) = 0 \\Rightarrow a \\ne b\n\\end{align*}\nthen\n\\begin{align*}\n\\begin{split}\n\t\\sum_{b \\in A} I(At, b) \\nabla h(St, b), \\vec \\theta)\n\t&= \\nabla h(St, At), \\vec \\theta)\n\t\\\\\n\t\\nabla \\pi(A_t|S_t, \\vec \\theta)\n\t&=\n\t\\frac{\\pi(A_t|S_t,\\vec \\theta)}{D(S_t,\\vec \\theta)}\n\t\\sum_{b \\in A}\n\t\\left[\n\t\tD(S_t, \\vec \\theta) I(A_t,b)\n\t\t- e^{h(S_t,b,\\vec \\theta)}\n\t\\right] \t\n\t\\nabla h(S_t,b,\\vec \\theta)\n\t\\\\\n\t\\nabla \\ln \\pi(A_t|S_t, \\vec \\theta)\n\t&=\n\t\\sum_{b \\in A}\n\t\\left[\n\t\tI(A_t,b)\n\t\t- \\pi(S_t,b,\\vec \\theta)\n\t\\right] \t\n\t\\nabla h(S_t,b,\\vec \\theta)\n\t\\\\\n\t\\delta_h(S_t, A_t, b, \\theta_t)\n\t&= \\delta_t\n\t\\left[\n\t\tI(A_t,b) - \\pi(S_t,b,\\vec \\theta)\n\t\\right] \n\t\\\\\n\t\\vec \\theta_{t+1}\n\t&=\n\t\\vec \\theta_t +\n\t\\sum_{b \\in A}\n\t\\alpha \\delta_h(S_t, A_t, b, \\theta_t) \\nabla h(S_t,b,\\vec \\theta_t)\n\\end{split}\n\\end{align*}\n\nThe same result is obtained by training the network with labels\n\\begin{align}\n\\begin{split}\n\th^*(S_t,A_t,b) = h(S_t,b,\\vec \\theta_t) + \n\t\\alpha \\delta_h(S_t, A_t, b, \\vec \\theta_t)\n\\end{split}\n\\end{align}\n\n\\subsection{Gaussian policy actor}\n\nThe Gaussian policy actor computes the probabilities $ \\pi(a|s) $ of choose a continuous action $ a $ at status $ s $ as a normal distributed function of two parameters $ \\mu(s, \\vec \\theta_\\mu) $ and $ \\sigma(s,\\vec \\theta_h) $.\n\n\\begin{align}\n\t\\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)\n\t&= \\frac{1}{\\sigma(s,\\vec \\theta_h) \\sqrt{2 \\pi}} e^{-\\frac{(a - \\mu(s, \\vec \\theta_\\mu))^2}{\\sigma(s, \\vec \\theta_h)^2}}\n\t\\\\\n\t\\sigma(s, \\vec \\theta_h)\n\t&\t= e^{h(s, \\vec \\theta_h)}\t\n\\end{align}\n\nTo train the ANN the agent updates the $ \\vec \\theta_\\mu, \\vec \\theta_h $ parameters at each step $ t $\n\\begin{align}\n\t\\vec \\theta_{\\mu t+1}\n\t&= \\vec \\theta_{\\mu t}\n\t+ \\alpha_\\mu \\delta_t \\nabla \\ln \\pi(A_t|S_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{h t})\n\t\\\\\n\t\\vec \\theta_{h+1}\n\t&= \\vec \\theta_{h t}\n\t+ \\alpha_h \\delta_t \\nabla \\ln \\pi(A_t|S_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{h t})\n\\end{align}\n\nThe policy gradient functions for $\\vec \\theta_\\mu $ is\n\\begin{align*}\n\t\\nabla \\ln \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)\n\t& = \\pi^{-1}(a|s,\\vec \\theta_\\mu, \\vec \\theta_h) \\nabla \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)\n\t\\\\\n\t& = \\pi^{-1}(a|s,\\vec \\theta_\\mu, \\vec \\theta_h) \\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)}{\\partial \\mu(s, \\vec \\theta_\\mu)} \\nabla \\mu(s, \\vec \\theta_\\mu)\n\t\\\\\n\t\\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)}{\\partial \\mu(s, \\vec \\theta_\\mu)}\n\t& = - \\frac{1}{\\sigma^3(s, \\vec \\theta_h) \\sqrt{2 \\pi} } e^{-\\frac{(a-\\mu(s, \\vec \\theta_\\mu))^2}{\\sigma^2(s, \\vec \\theta_h)}} \\frac{\\partial }{\\partial \\mu(s,\\vec \\theta_\\mu)} (a-\\mu(s, \\vec \\theta_\\mu))^2\n\t\\\\\n\t& = -\\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h) (-2) (a - \\mu(s, \\vec \\theta_\\mu)) \\sigma^{-2}(s, \\vec \\theta_h)\n\t\\\\\n\t\\nabla \\ln \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)\n\t& = \\frac{2 [a - \\mu(s, \\vec \\theta_\\mu)]}{\\sigma^2(s, \\vec \\theta_h)} \\nabla \\mu(s, \\vec \\theta_\\mu)\n\\end{align*}\nThe policy gradient functions for $\\vec \\theta_h $ is\n\\begin{align*}\n\t\\nabla \\ln \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )\n\t& = \\pi^{-1}(a|s,\\vec \\theta_\\mu, \\vec \\theta_h ) \\nabla \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)\n\t\\\\\n\t& = \\pi^{-1}(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )\n\t\\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )}{\\partial h(s, \\vec \\theta_h)} \\nabla h(s, \\vec \\theta_h)\n\t\\\\\t\n\t\\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )}{\\partial h(s, \\vec \\theta_h)}\n\t&= \\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )}{\\partial \\sigma(s, \\vec \\theta_h)}\n \t\\frac{\\partial \\sigma(s, \\vec \\theta_h)}{\\partial h(s, \\vec \\theta_h)}\n\t\\\\\n\t\\frac{\\partial \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )}{\\partial \\sigma(s, \\vec \\theta_h)}\n\t& = \\frac{\n\t\\left\\{\n\t\t\\sigma(s,\\vec \\theta_h) \\frac{\\partial}{\\partial \\sigma(s,\\vec \\theta_h)}\n\t\t\\left[\n\t\t\te^{-\\frac{(a-\\mu(s, \\vec \\theta_\\mu))^2}{\\sigma^2(s,\\vec \\theta_h)}}\n\t\t\\right]\n\t\t- e^{-\\frac{(a-\\mu(s, \\vec \\theta_\\mu))^2}{\\sigma^2(s, \\vec \\theta_h)}}\n\t\\right\\}}{\\sigma^2(s,\\vec \\theta_h) \\sqrt{2 \\pi}}\t\n\t\\\\\n\t& = \\frac{\n\te^{-\\frac{(a-\\mu(s, \\vec \\theta_\\mu))^2}{\\sigma^2(s,\\vec \\theta_h)}}\n\t\\left\\{\n\t\t-\\sigma(s,\\vec \\theta_h)\n\t\t(a-\\mu(s,\\vec \\theta_\\mu))^2\n\t\t\\frac{\\partial}{\\partial \\sigma(s,\\vec \\theta_h)}\n\t\t\\left[\n\t\t\t\\sigma^{-2}(s,\\vec \\theta_h)\n\t\t\\right]\n\t\t- 1\n\t\\right\\} }{\\sigma^2(s,\\vec \\theta_h) \\sqrt{2 \\pi}}\t\n\t\\\\\n\t& = \\frac{\\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h)}{\\sigma(s,\\vec \\theta_h)}\n\t\\left\\{\n\t\t2 \\frac{(a-\\mu(s,\\vec \\theta_\\mu))^2} {\\sigma^2(s,\\vec \\theta_h)} - 1\n\t\\right\\}\n\t\\\\\n\t\\frac{\\partial \\sigma(s,\\vec \\theta_h)}{\\partial h(s,\\vec \\theta_h)}\n\t&= e^{h(s, \\vec \\theta_h)}\n\t\\\\\n\t&= \\sigma(s, \\vec \\theta_h)\n\t\\\\\n\t\\nabla \\ln \\pi(a|s,\\vec \\theta_\\mu, \\vec \\theta_h )\n\t& = 2 \\frac{(a-\\mu(s,\\vec \\theta_\\mu))^2} {\\sigma^2(s,\\vec \\theta_h)} - 1\n\\end{align*}\n\nLet be\n\\begin{align*}\n\t\\delta_\\mu(s, a, \\vec \\theta_\\mu, \\vec \\theta_h )\n\t&= \\delta_t 2 \\frac{a - \\mu(s, \\vec \\theta_\\mu)}{\\sigma^2(s, \\vec \\theta_h)}\n\t\\\\\n\t\\delta_h(s, a, \\vec \\theta_\\mu, \\vec \\theta_h )\n\t&= \\delta_t \n\t\\left[\n\t\t2 \\frac{(a-\\mu(s,\\vec \\theta_\\mu))^2} {\\sigma^2(s,\\vec \\theta_h)} - 1\n\t\\right]\n\\end{align*}\n\nThe $ \\vec \\theta_\\mu, \\vec \\theta_h $ parameters are updated by\n\\begin{align*}\n\t\\vec \\theta_{\\mu t+1}\n\t&= \\vec \\theta_{\\mu t} + \\alpha_\\mu \\delta_\\mu(S_t, A_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{h t} ) \n\t\\\\\n\t\\vec \\theta_{h t+1}\n\t&= \\vec \\theta_{h t} + \\alpha_h \\delta_h(S_t, A_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{ht} ) \n\\end{align*}\n\nThe same result is obtained by training the network with labels\n\\begin{align}\n\t\\mu^*(S_t) = \\mu(S_t,\\vec \\theta_{\\mu t}) + \n\t\\alpha_\\mu \\delta_\\mu(S_t, A_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{ht})\n\t\\\\\n\th^*(S_t) = h(S_t,\\vec \\theta_{\\mu t}) + \n\\alpha_h \\delta_h(S_t, A_t, \\vec \\theta_{\\mu t}, \\vec \\theta_{ht})\n\\end{align}\n\n\\subsection{Learning process}\n\nInitially the agent do not have any experience and the differential state value generated by the ANN starts from a random initial estimation. The same is for the behavior generated by the actors where the preferences are random initialized.\n\nSince the very first interactions the agent generates wrong differential state value estimation with high value of TD Error. Consequently the agent easily changes the policy behavior, trying different policies with the exploration of both the policy and state space. It tries to improve the average reward, the value state estimation at the same time selecting better policies.\n\nDifferent factors influence the learning process.\nExternal agent factors like the number of states, actions and policies not yet explored tent to generate high error rate.\nThe error increases as the agent experiences new environment states or try new policies, but as the agent faces states already visited it can exploit the past experiences reducing the error and improving the state value estimation and the policy.\n\nInstead the agent internal factor, such as the ANN approximation, generates errors that can be reduced only to a low-bound limit value.\n\nThe ANN learning parameters can influence both in the quality of approximated functions and in the speed of adaptation.\n\nHigh value of step-size parameter $ \\eta $ increases the correction of weights increasing consequently the speed of adaptation, but can easily generate corrective value too big bringing the network to not converging behaviors.\nOn the other hand small value of step-size parameter weakens the correction of weights decreasing the speed of adaptation.\n\nAlso the actor parameters $ \\alpha_i $, which modulate the policy changes from the critic, influence the improvement capacity.\nHigh value of $ \\alpha_i $ determine strong policy changes even in the face of small errors, making it difficult a fine correction, the other way around very small value of $ \\alpha_a $ bring very little variations on policy without improvement.\n\nIf the actor parameters that govern the policy have a limited range of values we may choose an $ \\alpha_i = \\alpha^*_i $ so that the average of corrections is $ \\varepsilon_i $ a fraction of the limited range.\n\nFor the discrete policy actors we calculate the average of squared preferences variations $ \\Delta^2h_{i} $\n\\begin{align*}\n\th^*_{ia}(t) - h_{ia}(t)&=\t\\Delta h_{ia}(t)\n\t\\\\\n\t&=\t\\alpha_i \\delta_{h_{ia}}(t)\n\t\\\\\n\t\\Delta^2h_{i}(t)&=\t\t\t\\frac{1}{m} \\sum_a^m \\Delta h_{ia}(t)\n\t\\\\\n\t\\Delta^2h_{i}&=\t\t\t\\frac{1}{n} \\sum_t^n \\Delta^2 h_{i}(t)\n\t\\\\\n\t&=\t\t\t\\frac{1}{n}\\frac{1}{m}\\sum_t^n \\sum_a^m \\Delta^2 h_{ia}(t)\n\t\\\\\n\t&=\t\\alpha^2_i\\frac{1}{n}\\frac{1}{m}\\sum_t^n \\sum_a^m \\delta^2_{h_{ia}}(t)\n\t\\\\\n\t\\Delta^2h_i &=\t\\varepsilon^2_i\n\t\\\\\n\t\\alpha^*_i&=\t\\frac{\\varepsilon_i}{\\sqrt{\\frac{1}{n} \\frac{1}{m}\\sum_t^n \\sum_a^m \\delta^2 h_{ia}(t)}}\n\\end{align*}\nif $ \\alpha_i $ is the value used in the session then\n\\begin{align}\n\t\\alpha^*_i&=\t\\frac{\\varepsilon_i}{\\sqrt{\\frac{1}{n} \\frac{1}{m} \\sum_t^n \\sum_a^m \\Delta^2 h_{ia}(t)}} \\alpha_i\n\\end{align}\n\nFor the Gaussian policy actors we calculate the average of squared parameter variations $ \\Delta^2\\mu_i$ and $ \\Delta^2 h_{i\\sigma} $\n\\begin{align*}\n\t\\mu^*_i(t) - \\mu_i(t)&=\t\\Delta \\mu_i(t)\n\t\\\\\n\t&=\t\\alpha_{i\\mu} \\delta_{i\\mu}(t)\n\t\\\\\n\t\\Delta^2\\mu_{i}&=\t\t\t\\frac{1}{n} \\sum_t^n \\Delta^2 \\mu_i(t)\n\t\\\\\n\t&=\t\\alpha^2_{i\\mu}\\frac{1}{n}\\sum_t^n \\delta^2_{\\mu_{i}}(t)\n\t\\\\\n\t\\Delta^2\\mu_i &=\t\\varepsilon^2_{i\\mu}\n\t\\\\\n\t\\alpha^*_{i\\mu}&=\t\\frac{\\varepsilon_{i\\mu}}{\\sqrt{\\frac{1}{n} \\sum_t^n \\delta^2 \\mu_i(t)}}\n\\end{align*}\nif $ \\alpha_{i\\mu} $ is the value used in the session then\n\\begin{align}\n\t\\alpha^*_{i\\mu}&=\t\\frac{\\varepsilon_{i\\mu}}{\\sqrt{\\frac{1}{n} \\sum_t^n \\Delta^2 \\mu_i(t)}} \\alpha_{i\\mu}\n\\end{align}\n\n\\begin{align*}\n\th^*_{i\\sigma}(t) - h_{i\\sigma}(t)&=\t\\Delta h_{i\\sigma}(t)\n\t\\\\\n\t&=\t\\alpha_{i\\sigma} \\delta_{ih_\\sigma}(t)\n\t\\\\\n\t\\Delta^2h_{i\\sigma}&=\t\t\t\\frac{1}{n} \\sum_t^n \\Delta^2 h_{i\\sigma}(t)\n\t\\\\\n\t&=\t\\alpha^2_{i\\sigma}\\frac{1}{n}\\sum_t^n \\delta^2_{ih_\\sigma}(t)\n\t\\\\\n\t\\Delta^2h_{i\\sigma} &=\t\\varepsilon^2_{i\\sigma}\n\t\\\\\n\t\\alpha^*_{i\\sigma}&=\t\\frac{\\varepsilon_{i\\sigma}}{\\sqrt{\\frac{1}{n} \\sum_t^n \\delta^2 h_{i\\sigma}(t)}}\n\\end{align*}\nif $ \\alpha_{i\\sigma} $ is the value used in the session then\n\\begin{align}\n\t\\alpha^*_{i\\sigma}&=\t\\frac{\\varepsilon_{i\\sigma}}{\\sqrt{\\frac{1}{n} \\sum_t^n \\Delta^2 h_{i\\sigma}(t)}} \\alpha_{i\\sigma}\n\\end{align}\n\n\\subsection{Performance}\n\nThe average reward is the main indicators of learning quality, since the agent interacts with the environment the average reward grows up.\n\nBecause the process is adaptive the average reward can vary a lot during the interaction with the environment therefore it is necessary to evaluate the trend of the indicator. The logarithmic regression can be used to evaluate the performance\n\\begin{align}\n\tr_\\pi(t) & = \\ln (mt + q)\n\\end{align}\n\nAn increasing trend indicates the agent is selecting better policies and is getting higher and higher rewards.\nA decreasing trend indicates the agent is not selecting good policies, this may be caused by low $ \\alpha_a $ or $ \\eta $ parameters.\n\nAnother indicator of performance is the squared TD error, as for the average reward it is necessary to evaluated the trend.\nThe error has not negative values that should approach zero when the agent select the local best policy. The exponential regression can be used to evaluate the performance\n\\begin{align}\n\t\\delta^2(t) & = e^{(mt + q)}\n\\end{align}\n\nA decreasing trend indicates the estimation of value states is getting better.\nAn increasing trend indicates the estimation of value states is getting worse it may be caused by the ANN which cannot correct the estimation due to $ \\eta $ parameter too high.\n\nBoth the critic and actors approximate the value function and policy function with neural network.\nTo monitor the learning activity of ANN we compute the MSE of estimated functions.\n\n\nThe ANN is feed with normalized values label $ O^*_i $ so that the values are in the range $ -1, +1 $ by linear transformation\n\n\\begin{align*}\n\tO_i &= (O_C, O_{Ai})\n\t\\\\\n\tO_C &= m_c \\delta_t + q_c\n\t\\\\\n\tO_{Ai} &= m_{ai} \\delta_{hi} + q_{ai}\n\\end{align*}\n\nThe MSE of ANN is\n\\begin{align*}\n\tJ(t)&\t= \\sum_i [O^*_i(t) - O_i(t)]^2\n\\end{align*}\nand the MSE after the training is\n\\begin{align*}\n\tJ'(t)&\t= \\sum_i [O^*_i(t) - O'_i(t)]^2\n\\end{align*}\n\nThe ratio between the total MSE after and before the training activity indicates the quality of such activity.\n\\begin{align}\n\\begin{split}\n\tK(t)&\t= \\frac{J'(t)}{J(t)}\n\\end{split}\n\\end{align}\n\nA ratio $ K(t) \\ge 1 $ means the error after training gets worst due a step-size parameter $ \\alpha $ too high.\nA ratio $ K(t) = 1 $ means no change on error and therefore no improvement. This can be affected by a local minimum reached or a step-size parameter too low with very poor capacity of learning.\nA ratio $ K(t) < 1 $ means an improvement of neural network due to correct step-size parameter.\nA ratio $ K(t) = 0 $ means a perfect fit of neural network.\n\nWe can classify the steps in three class:\n\\begin{description}\n\n\t\\item[$ C_0 $]\n\tThe steps that created a bad approximation with an increased of MSE ($ J > \\varepsilon \\cup K > 1$)\n\n\t\\item[$ C_1 $]\n\tThe steps that create a trivial approximation with a small reduction of MSE ($ J > \\varepsilon \\cup K_0 \\le K \\le 1 $ with $ K_0 = 0.9$ )\n\n\t\\item[$ C_2 $]\n\tThe remaining steps that have a small MSE or that have reduced significantly the MSE\n\\end{description}\n\nThe ideal distribution should have \n\n\\begin{align}\n\\begin{split}\n\tC_0 & = 0\n\t\\\\\n\tC_1 & = 0\n\t\\\\\n\tC_2 & = 1\n\\end{split}\n\\end{align}\n\nA step parameter $ \\eta $ too high generates over correction increasing $ C_0 $ and reducing $ C_2 $.\n\nAn $ \\eta $ parameter too small generates under corrections increasing $ C_1 $ and reducing $ C_2 $.\n\nBecause it is difficult to determine the effects of step parameter changes on the result MSE, an empirical way to reduce or increment the parameter is applied, for example increasing or reducing by exponential factors ($ \\dots $, $ \\times 0.01 $, $ \\times 0.03 $, $ \\times 0.1$, $\\times 0.3 $, $ \\times 3 $, $ \\times 10 $, $ \\times 30$, $\\times 100 $, $ \\dots $).\n\n\\begin{figure}[ht]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{ann.png}\n\t\\caption{Step classification}\n\t\\label{fig:steps}\n\\end{figure}\n\n\\end{document}\n", "meta": {"hexsha": "d056dc4e855c0e28c8a1b1238f1ef44af26a9eec", "size": 17683, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/ActorCritiAgent.tex", "max_stars_repo_name": "m-marini/scalarl", "max_stars_repo_head_hexsha": "8883a672da955ebac2f4b536c3879f04b58c6c74", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-05-20T16:54:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-05-20T16:54:32.000Z", "max_issues_repo_path": "docs/ActorCritiAgent.tex", "max_issues_repo_name": "m-marini/scalarl", "max_issues_repo_head_hexsha": "8883a672da955ebac2f4b536c3879f04b58c6c74", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 68, "max_issues_repo_issues_event_min_datetime": "2019-01-30T20:30:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-09T12:41:45.000Z", "max_forks_repo_path": "docs/ActorCritiAgent.tex", "max_forks_repo_name": "m-marini/scalarl", "max_forks_repo_head_hexsha": "8883a672da955ebac2f4b536c3879f04b58c6c74", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.5592841163, "max_line_length": 377, "alphanum_fraction": 0.6682689589, "num_tokens": 6152, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8774767874818408, "lm_q2_score": 0.7025300636233415, "lm_q1q2_score": 0.6164538233376229}} {"text": "%\\section{Surface flux}\n{\\bf \\Large \n\\begin{tabular}{ccc}\n\\hline\n Corresponding author & : & Seiya Nishizawa\\\\\n\\hline\n\\end{tabular}\n}\n\n\n\\subsection{Monin-Obukhov similarity}\nFirst of all, we assume that in the boundary layer\n1. fluxes are constant, and 2. variables are horizontally uniform.\n\nRelations between flux and vertical gradient are:\n\\begin{align}\n \\frac{kz}{u_*} \\frac{\\partial u}{\\partial z} &= \\phi_m\\left(\\frac{z}{L}\\right), \\label{eq: flux-gradient u} \\\\\n \\frac{kz}{\\theta_*} \\frac{\\partial \\theta}{\\partial z} &= \\phi_h\\left(\\frac{z}{L}\\right), \\label{eq: flux-gradient t} \\\\\n \\frac{kz}{q_*} \\frac{\\partial q}{\\partial z} &= \\phi_q\\left(\\frac{z}{L}\\right),\n\\end{align}\nwhere $k$ is the Von Karman constant.\n$L$ is the Monin-Obukhov scale height:\n\\begin{equation}\n L = \\frac{\\theta u_*^2}{kg\\theta_*},\n\\end{equation}\nwhere $g$ is gravity.\nThe scaling velocity, $u_*$, temperature, $\\theta_*$,\nand water vapor, $q_*$, are defined from the vertical eddy fluxes of momentum, sensible heat, and water vapor:\n\\begin{align}\n \\overline{u'w'} &= -u_*u_*, \\\\\n \\overline{w'\\theta'} &= - u_*\\theta_*, \\\\\n \\overline{w'q'} &= - u_*q_*.\n\\end{align}\n\nThe integration between roughness length $z_0$ to height $z$ of the lowest model level, eqs. (\\ref{eq: flux-gradient u}) and (\\ref{eq: flux-gradient t}) becomes:\n\\begin{align}\n u(z) &= \\frac{u_*}{k} \\left\\{\\ln(z/z_0)-\\Phi_m(z/L)+\\Phi_m(z_0/L)\\right\\}, \\\\\n \\Delta\\theta &= R\\frac{\\theta_*}{k} \\left\\{\\ln(z/z_0)-\\Phi_h(z/L)+\\Phi_h(z_0/L)\\right\\},\n\\end{align}\nwhere $\\Delta\\theta = \\theta-\\theta_0$,\nand\n\\begin{align}\n \\Phi_m(z) = \\int^z \\frac{1-\\phi_m(z')}{z'} dz', \\\\\n \\Phi_h(z) = \\int^z \\frac{R-\\phi_h(z')}{R z'} dz'.\n\\end{align}\n\n\n\\subsection{Louis’s (1979) Model}\nLouis (1979) introduced a parametric model of vertical eddy fluxes.\n\nThe $L$ becomes:\n\\begin{equation}\n L = \\frac{\\theta u^2}{g\\Delta\\theta}\n \\frac{\\ln(z/z_0)-\\Phi_h(z/L)+\\Phi_h(z_0/L)}{\\left\\{\\ln(z/z_0)-\\Phi_m(z/L)+\\Phi_m(z/L)\\right\\}^2}.\n\\end{equation}\nThe bulk Richardson number for the layer $Ri_B$ is:\n\\begin{equation}\n Ri_B = \\frac{gz\\Delta\\theta}{\\theta u^2},\n\\end{equation}\nand its form implies a relationship with the Monin-Obukhov scale height $L$.\nThe fluxes can then be written as:\n\\begin{align}\n u_*^2 &= a^2 u^2 F_m\\left(\\frac{z}{z_0},Ri_B\\right), \\label{eq: u_*^2} \\\\\n u_*\\theta_* &= \\frac{a^2}{R} u \\Delta \\theta F_h\\left(\\frac{z}{z_0},Ri_B\\right), \\label{eq: u_*t_*}\n\\end{align}\nwhere\n$R$ is ratio of the drag coefficients for momentum and heat in the neutral limit(the turbulent Prandtl number),\nand\n\\begin{equation}\n a^2 = \\frac{k^2}{\\left\\{\\ln\\left(z/z_0\\right)\\right\\}^2}\n\\end{equation}\nis the drag coefficient under neutral conditions.\n\nFor unstable conditions ($Ri_B<0$),\n$F_i$s ($i=m,h$) could be:\n\\begin{equation}\n F_i = 1 - \\frac{b Ri_B}{1 + c_i \\sqrt{|Ri_B|}},\n \\label{eq: F_i unstable}\n\\end{equation}\nunder the consideration that\n$F_i$ must behave as $1/u$ (i.e., $\\sqrt{|Ri_B|}$) in the free convection limit ($u \\to 0$) and becomes $1$ under neutral conditions ($Ri_B \\to 0$).\nOn the other hand, under stable conditions ($Ri_b$), Louis (1979) adopted the following form for $F_i$:\n\\begin{equation}\n F_i = \\frac{1}{(1 + b' Ri_B)^2}.\n \\label{eq: F_i stable}\n\\end{equation}\n\nThe constants are estimated as $R=0.74$ by Businger et al. (1971),\nand $b=2b'=9.4$ by Louis (1979).\nBy dimensional analysis:\n\\begin{equation}\n c_i = C^*_i a^2 b \\sqrt{\\frac{z}{z_0}},\n\\end{equation}\nand $C^*_m = 7.4, C^*_h = 5.3$, which result best fits curves.\n\n\n\\subsection{Uno et al.’s (1995) Model}\nUno et al. (1995) extended the Louis Model,\nconsidering differences in roughness lengths related to momentum and temperature, i.e., $z_0$ and $z_t$, respectively.\n\nThe potential temperature difference between $z=z$ and $z=z_t$,\n$\\Delta\\theta_t$, is:\n\\begin{align}\n \\Delta\\theta_t\n &= R\\frac{\\theta_*}{k}\\left\\{\\ln(z_0/z_t) - \\Phi_h(z_0/L) + \\Phi_h(z_t/L)\\right\\} + \\Delta\\theta_0, \\nonumber \\\\\n &= R\\frac{\\theta_*}{k}{\\ln(z_0/z_t)} + \\Delta\\theta_0, \\nonumber \\\\\n &= \\Delta\\theta_0 \\left\\{\\frac{R\\ln(z_0/z_t)}{\\Psi_h} + 1\\right\\},\n\\end{align}\nwhere $\\Delta\\theta_0 = \\theta_z - \\theta_{z_0} (=\\Delta\\theta)$:\n\\begin{equation}\n \\Psi_h = \\int_{z_0}^z\\frac{\\phi_h}{z'}dz', \\label{eq: Psi_h}\n\\end{equation}\nand $\\phi_h$ is assumed to be $R$ in the range $z_t < z < z_0$.\nThus:\n\\begin{equation}\n \\Delta\\theta_0 = \\Delta\\theta_t \\left\\{\\frac{R\\ln(z_0/z_t)}{\\Psi_h}+1\\right\\}^{-1},\n \\label{eq: Delta t_0}\n\\end{equation}\nor equivalently,\n\\begin{equation}\n Ri_{B0} = Ri_{Bt} \\left\\{\\frac{R\\ln(z_0/z_t)}{\\Psi_h}+1\\right\\}^{-1}.\n \\label{eq: Ri_B0}\n\\end{equation}\n\nFrom eqs. (\\ref{eq: u_*^2}) and (\\ref{eq: u_*t_*}):\n\\begin{equation}\n \\Delta\\theta_0 = \\frac{R\\theta_*}{k}\\ln\\left(\\frac{z}{z_0}\\right)\\frac{\\sqrt{F_m}}{F_h},\n\\end{equation}\nwhile\n\\begin{equation}\n \\Delta\\theta_0 = \\frac{\\theta_*}{k}\\Psi_h,\n\\end{equation}\nfrom eqs. (\\ref{eq: flux-gradient t}) and (\\ref{eq: Psi_h}).\nTherefore:\n\\begin{equation}\n \\Psi_h = R\\ln\\left(\\frac{z}{z_0}\\right)\\frac{\\sqrt{F_m}}{F_h}.\n \\label{eq: Psi}\n\\end{equation}\n\nBecause $\\Psi_h$ depends on $Ri_{B0}$,\n$Ri_{B0}$ cannot be calculated from $Ri_{Bt}$ with eq. (\\ref{eq: Ri_B0})\ndirectly, so numerical iteration is required to obtain $Ri_{B0}$\n\\footnote{In the stable case, it can be solved analytically\nwith eq. (\\ref{eq: F_i stable}), but the solution is too complicated.}.\nStarting from $Ri_{Bt}$ as the first estimation of $Ri_{B0}$,\nthe second estimate by the Newton-Raphson iteration becomes:\n\\begin{equation}\n \\hat{Ri}_{B0} = Ri_{Bt} - \\frac{Ri_{Bt}R\\ln(z_0/z_t)}{\\ln(z_0/z_t) + \\hat{\\Psi}_h},\n \\label{eq: Ri_B0 estimation}\n\\end{equation}\nwhere $\\hat{\\Psi}_h$ is the estimate of $\\Psi_h$ using $Ri_{Bt}$ instead of $Ri_{B0}$.\nApproximate values for $F_m, F_h$, and $\\Psi_h$ are re-calculated\nbased on the $\\hat{Ri}_{B0}$, and then $\\Delta\\theta_0$, and the surface fluxes $u_*^2$ and $u_*\\theta_*$ are calculated from eqs. (\\ref{eq: Delta t_0}), (\\ref{eq: u_*^2}), and (\\ref{eq: u_*t_*}), respectively.\n\n\n\n\n\\subsection{Discretization}\n\n\\def\\half{\\frac{1}{2}}\n\nAll the fluxes are calculated based on the velocity at the first full-level (k=1)\n($z=\\Delta z/2$).\nThe absolute velocities $U$ are:\n\\begin{align}\n U_{i+\\half,j,1}^2 &=\n \\left\\{\\frac{2(\\rho u)_{i+\\half,j,1}}{\\rho_{i,j,1}+\\rho_{i+1,j,1}}\\right\\}^2\n + \\left\\{\\frac{(\\rho v)_{i,j-\\half,1} + (\\rho v)_{i,j+\\half,1} + (\\rho v)_{i+1,j-\\half,1} + (\\rho v)_{i+1,j+\\half,1}}{2(\\rho_{i,j,1}+\\rho_{i+1,j,1})}\\right\\}^2 \\nonumber \\\\\n &+ \\left\\{\\frac{(\\rho w)_{i,j,1+\\half} + (\\rho w)_{i+1,j,1+\\half}}{2(\\rho_{i,j,1}+\\rho_{i+1,j,1})}\\right\\}^2, \\\\\n U_{i,j+\\half,1}^2 &=\n \\left\\{\\frac{(\\rho u)_{i-\\half,j,1} + (\\rho u)_{i+\\half,j,1} + (\\rho u)_{i-\\half,j+1,1} + (\\rho u)_{i+\\half,j+1,1}}{2(\\rho_{i,j,1}+\\rho_{i,j+1,1})}\\right\\}^2 \\nonumber\\\\\n &+ \\left\\{\\frac{2(\\rho v)_{i,j+\\half,1}}{\\rho_{i,j,1}+\\rho_{i,j+1,1}}\\right\\}^2\n + \\left\\{\\frac{(\\rho w)_{i,j,1+\\half} + (\\rho w)_{i,j+1,1+\\half}}{2(\\rho_{i,j,1}+\\rho_{i,j+1,1})}\\right\\}^2, \\\\\n U_{i,j,1}^2 &=\n \\left\\{\\frac{(\\rho u)_{i-\\half,j,1} + (\\rho u)_{i+\\half,j,1}}{2\\rho_{i,j,1}}\\right\\}^2\n + \\left\\{\\frac{(\\rho v)_{i,j-\\half,1} + (\\rho v)_{i,j+\\half,1}}{2\\rho_{i,j,1}}\\right\\}^2\n + \\left\\{\\frac{(\\rho w)_{i,j,1+\\half}}{2\\rho_{i,j,1}}\\right\\}^2,\n\\end{align}\nIt is here of note that $(\\rho w)_{i,j,\\half}=0$.\nThe potential temperatures $\\theta$ are:\n\\begin{align}\n \\theta_{i,j,1} &= \\frac{(\\rho \\theta)_{i,j,1}}{\\rho_{i,j,1}}, \\\\\n \\bar{\\theta}_{i+\\half,j,1} &= \\frac{\\theta_{i,j,1}+\\theta_{i+1,j,1}}{2}, \\\\\n \\bar{\\theta}_{i,j+\\half,1} &= \\frac{\\theta_{i,j,1}+\\theta_{i,j+1,1}}{2}.\n\\end{align}\n\nThe roughness lengths, $z_0, z_t$, and $z_q$ are calculated from\neqs. (\\ref{eq: z_0}), (\\ref{eq: z_t}), and (\\ref{eq: z_q}),\nin which the friction velocity $u_*$ is estimated as:\n\\begin{equation}\n u_* = \\sqrt{C_{m0}} U,\n\\end{equation}\nwhere $C_{m0}$ is a constant bulk coefficient,\nand we use $1.0 \\times 10^{-3}$ as its value.\n\nFrom eq. (\\ref{eq: Ri_B0}), the $Ri_{Bt}$, which is the first guess of the $Ri_{B0}$, is:\n\\begin{equation}\n Ri_{Bt} = \\frac{gz_1(\\theta_1-\\theta_{sfc})}{\\bar{\\Theta}U^2},\n\\end{equation}\nwith the assumption that $\\theta_{z_t} = \\theta_{sfc}$.\nThe estimation of $\\hat{\\Psi}_h$ is calculated with $Ri_{Bt}$ from\neqs. (\\ref{eq: Psi}), (\\ref{eq: F_i unstable}), and (\\ref{eq: F_i stable}).\nThe final estimation of $Ri_{B0}$ is obtained from eq. (\\ref{eq: Ri_B0 estimation}),and the final estimation of $\\Psi_h$ is obtained with $Ri_{B0}$.\n\nNow we can calculate the bulk coefficients, $C_m, C_h$, and $C_e$ for moments, heat, and vapor:\n\\begin{align}\n C_m &= \\frac{k^2}{\\ln(z_1/z_0)}F_m(Ri_{B0}), \\\\\n C_h &= \\frac{k^2}{R\\ln(z_1/z_0)}F_h(Ri_{B0})\\left\\{\\frac{R\\ln(z_0/z_t)}{\\Psi_h}+1\\right\\}^{-1}, \\\\\n C_e &= \\frac{k^2}{R\\ln(z_1/z_0)}F_h(Ri_{B0})\\left\\{\\frac{R\\ln(z_0/z_e)}{\\Psi_h}+1\\right\\}^{-1}.\n\\end{align}\nThe fluxes are:\n\\begin{align}\n \\overline{\\rho u'w'} &= - C_m U \\rho u, \\\\\n \\overline{\\rho v'w'} &= - C_m U \\rho v, \\\\\n \\overline{\\rho w'w'} &= - C_m U \\rho w, \\\\\n \\overline{\\rho \\theta'w'} &= - C_h U \\{\\rho \\theta - \\rho \\theta_{sfc}\\}, \\\\\n \\overline{\\rho q'w'} &= -C_e U \\rho ( q - q_{evap} ),\n\\end{align}\nwhere $q_{evap}$ is the saturation value at the surface.\n\n\n", "meta": {"hexsha": "e59b0f8c3441efe5b81372056ccec5f6334b419d", "size": 9211, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/descriptions/surface_flux.tex", "max_stars_repo_name": "slayoo/scale", "max_stars_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2020-06-14T11:12:31.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T05:29:55.000Z", "max_issues_repo_path": "doc/descriptions/surface_flux.tex", "max_issues_repo_name": "slayoo/scale", "max_issues_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-07-29T03:38:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-30T05:08:47.000Z", "max_forks_repo_path": "doc/descriptions/surface_flux.tex", "max_forks_repo_name": "slayoo/scale", "max_forks_repo_head_hexsha": "ca4b476ad55cb728b2009f0427ce3f7161ecfcf7", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-07-10T10:39:45.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-28T22:20:41.000Z", "avg_line_length": 41.490990991, "max_line_length": 210, "alphanum_fraction": 0.6348930626, "num_tokens": 3708, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8774767938900121, "lm_q2_score": 0.702530051167069, "lm_q1q2_score": 0.6164538169094659}} {"text": "\\section{M/D/1 Queues}\n\\label{sec:M-D-1-queues}\n\nA $M/D/1$ is a queue where \n(i) the arrival process is Poissonian with rate $\\lambda$,\n(ii) the service process is deterministic,\n(iii) there is one servers,\n(iv) the buffer has infinite capacity,\n(v) the scheduling policy is FCFS.\n\n\\begin{figure}[tp]\n\\label{fig:M-D-1-queue}\t\n\t\\centering\n\t\\includegraphics{fig/M-D-1-Queue}\n\t\\caption{An M/D/1 queue and its corresponding CTMC.}\n\t\\end{figure}", "meta": {"hexsha": "d888c010e747a2c2b9e2eb8ceac2ac9f03aa6052", "size": 440, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "performance-modeling/sec/m-d-1-queues.tex", "max_stars_repo_name": "gmarciani/research", "max_stars_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-07-27T13:31:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-20T12:54:12.000Z", "max_issues_repo_path": "performance-modeling/sec/m-d-1-queues.tex", "max_issues_repo_name": "gmarciani/research", "max_issues_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "performance-modeling/sec/m-d-1-queues.tex", "max_forks_repo_name": "gmarciani/research", "max_forks_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-02-17T13:30:49.000Z", "max_forks_repo_forks_event_max_datetime": "2018-02-17T13:30:49.000Z", "avg_line_length": 27.5, "max_line_length": 58, "alphanum_fraction": 0.7204545455, "num_tokens": 141, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8577681122619883, "lm_q2_score": 0.7185943985973772, "lm_q1q2_score": 0.616387360766911}} {"text": "\\input{../../../doc/latex/Styles/wpg_notation_define.tex}\n\\chapter{WPG v.04}\n\nThis version of the pattern generator is based on triple integrator with\npiece-wise constant \\acs{CoP} velocity. The control input is position of the\n\\acs{CoP}. Apart from that it is very similar to the \\acs{WPG} proposed in\n\\cite{Herdt2010auro}, for more information refer to \\cite{Sherikov2016phd}.\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Variables}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Footstep positions}\nAny preview horizon contains $1$ fixed and $0$ or more variable footstep\npositions. Let $M$ be the number of variable steps in the preview horizon,\n$\\FRAME{p_j}$ -- frame fixed to $j$-th footstep with $j = 0 \\dots M$ ($0$-th\nfootstep is fixed), $\\fp_j = (p^x_j, p^y_j)$ -- position of the $j$-th footstep\non the ground plane.\n\n\nCoordinate transformation matrix from the frame fixed to the $j$-th footstep to\nthe global frame is defined as\n%\n\\begin{equation}\n \\M[p_j][]{R} =\n \\begin{bmatrix}\n \\cos{\\theta_j} & -\\sin{\\theta_j} \\\\\n \\sin{\\theta_j} & \\cos{\\theta_j} \\\\\n \\end{bmatrix}\n\\end{equation}\n%\nOrientations $\\theta_j$ of the footsteps are predetermined to avoid\nnolinearity. Position of the $j$-th footstep in the global frame can be found\nas\n%\n\\begin{equation}\n \\fp_j = \\fp_0 + \\sum_{i=1}^{j} \\M[p_{i-1}][]{R} \\fp[p_{i-1}]_j,\n\\end{equation}\n%\nwhich leads to\n%\n\\begin{equation}\n \\begin{bmatrix}\n \\fp_0\\\\\n \\fp_1\\\\\n \\vdots\\\\\n \\fp_M\\\\\n \\end{bmatrix}\n =\n \\V{1}_M\n \\kron\n% \\begin{bmatrix}\n% \\M{I}\\\\\n% \\M{I}\\\\\n% \\vdots\\\\\n% \\M{I}\\\\\n% \\end{bmatrix}\n \\fp_0\n +\n \\begin{bmatrix}\n \\M{0} & \\M{0} & \\dots & \\M{0} \\\\\n \\M[p_{0}][]{R} & \\M{0} & \\dots & \\M{0} \\\\\n \\vdots & \\vdots & \\dots & \\vdots \\\\\n \\M[p_{0}][]{R} & \\M[p_{1}][]{R} & \\dots & \\M[p_{M-1}][]{R} \\\\\n \\end{bmatrix}\n \\begin{bmatrix}\n \\fp[p_0]_1\\\\\n \\vdots\\\\\n \\fp[p_{M-1}]_M\\\\\n \\end{bmatrix}.\n\\end{equation}\n%\nIt is preferable to express variable footstep positions in a frame fixed to the\npreceding footstep in order to impose simple bounds on variable footstep\npositions.\n\n\nThe number of sampling intervals in the preview horizon is denoted by $N \\ge\nM$. The footstep position corresponding to $k$-th ($k = 1,\\dots,N$) sampling\ninterval is denoted as $\\fph_k = (\\hat{p}^x_k, \\hat{p}^y_k)$. These positions\ncan be found using a selection matrices as\n%\n\\begin{equation}\n\\begin{split}\n \\underbrace{\n \\begin{bmatrix}\n \\fph_1 \\\\\n \\vdots \\\\\n \\fph_N\n \\end{bmatrix}\n }_{\\FPh}\n & =\n \\underbrace{\n \\begin{bmatrix}\n \\M{I} & \\M{0} & \\dots & \\M{0} \\\\\n \\M{I} & \\M{0} & \\dots & \\M{0} \\\\\n \\M{0} & \\M{I} & \\dots & \\M{0} \\\\\n \\M{0} & \\M{I} & \\dots & \\M{0} \\\\\n \\M{0} & \\M{I} & \\dots & \\M{0} \\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n \\M{0} & \\M{0} & \\dots & \\M{I} \\\\\n \\end{bmatrix}\n }_{\\M{I}_{\\mathit{fps}}}\n \\begin{bmatrix}\n \\fp_0 \\\\\n \\fp_1 \\\\\n \\vdots\\\\\n \\fp_M \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n \\M{I} \\\\\n \\M{I} \\\\\n \\M{0} \\\\\n \\M{0} \\\\\n \\M{0} \\\\\n \\vdots \\\\\n \\M{0} \\\\\n \\end{bmatrix}\n \\fp_0\n +\n \\begin{bmatrix}\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\vdots & \\ddots & \\vdots \\\\\n \\M{0} & \\dots & \\M{I} \\\\\n \\end{bmatrix}\n \\begin{bmatrix}\n \\fp_1 \\\\\n \\vdots\\\\\n \\fp_M\\\\\n \\end{bmatrix}\\\\\n & =\n% \\underbrace{\n% \\begin{bmatrix}\n% \\M{I} \\\\\n% \\M{I} \\\\\n% \\M{I} \\\\\n% \\M{I} \\\\\n% \\M{I} \\\\\n% \\vdots \\\\\n% \\M{I} \\\\\n% \\end{bmatrix}\n% }_{\\M{V}_0}\n \\underbrace{\n \\V{1}_N\n \\kron\n \\fp_0\n }_{\\V{V}_0}\n +\n \\underbrace{\n \\begin{bmatrix}\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\vdots & \\ddots & \\vdots \\\\\n \\M{0} & \\dots & \\M{I} \\\\\n \\end{bmatrix}\n \\begin{bmatrix}\n \\M[p_{0}][]{R} & \\M{0} & \\dots & \\M{0} \\\\\n \\vdots & \\vdots & \\dots & \\vdots \\\\\n \\M[p_{0}][]{R} & \\M[p_{1}][]{R} & \\dots & \\M[p_{M-1}][]{R} \\\\\n \\end{bmatrix}\n }_{\\M{V}}\n \\underbrace{\n \\begin{bmatrix}\n \\fp[p_0]_1\\\\\n \\vdots\\\\\n \\fp[p_{M-1}]_M\\\\\n \\end{bmatrix}\n }_{\\FP},\n\\end{split}\n\\end{equation}\n%\nwhere $\\M{V}$ has the following structure\n%\n\\begin{equation}\n \\M{V}\n =\n \\begin{bmatrix}\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{0} & \\dots & \\M{0} \\\\\n \\M[p_{0}][]{R} & \\dots & \\M{0} \\\\\n \\M[p_{0}][]{R} & \\dots & \\M{0} \\\\\n \\M[p_{0}][]{R} & \\dots & \\M{0} \\\\\n \\vdots & \\ddots & \\vdots \\\\\n \\M[p_{0}][]{R} & \\dots & \\M[p_{M-1}][]{R} \\\\\n \\end{bmatrix}\n =\n \\underbrace{\n \\begin{bmatrix}\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{0} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\M{I} & \\dots & \\M{0} \\\\\n \\vdots & \\ddots & \\vdots \\\\\n \\M{I} & \\dots & \\M{I} \\\\\n \\end{bmatrix}\n }_{\\M{I}_R \\kron \\M{I}}\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\n\\end{equation}\n%\nwhere\n%\n\\begin{equation}\n \\M{I}_R\n =\n \\begin{bmatrix}\n 0 & \\dots & 0 \\\\\n 0 & \\dots & 0 \\\\\n 1 & \\dots & 0 \\\\\n 1 & \\dots & 0 \\\\\n 1 & \\dots & 0 \\\\\n \\vdots & \\ddots & \\vdots \\\\\n 1 & \\dots & 1 \\\\\n \\end{bmatrix}.\n\\end{equation}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Positions of the CoP}\nPosition of the center of pressure in the end of $k$-th ($k = 1,\\dots,N$)\nsampling interval is denoted as $\\cop_k = (z^x_k, z^y_k)$. The current\n\\acs{CoP} position is $\\cop_0$. In order to obtain simple bounds on the\n\\acs{CoP} positions instead of general constraints, their positions are\nexpressed in the local frames fixed to the respective feet, \\emph{i.e.}\n%\n\\begin{equation}\n \\cop_k = \\fph_k + \\M[\\hat{p}_k][]{R} \\cop[\\hat{p}_k]_k.\n\\end{equation}\n%\nOrientation matrices $\\M[\\hat{p}_k][]{R}$ can also be selected with\n$\\M{I}_{\\mathit{fps}}$\n%\n\\begin{equation}\n \\begin{bmatrix}\n \\M[\\hat{p}_1][]{R}\\\\\n \\vdots\\\\\n \\M[\\hat{p}_N][]{R}\\\\\n \\end{bmatrix}\n =\n \\M{I}_{\\mathit{fps}}\n \\begin{bmatrix}\n \\M[p_0][]{R}\\\\\n \\vdots\\\\\n \\M[p_M][]{R}\\\\\n \\end{bmatrix}\n\\end{equation}\n%\nAll positions of the \\acs{CoP} within the preview horizon are\n%\n\\begin{equation}\n \\begin{split}\n \\begin{bmatrix}\n \\cop_1 \\\\\n \\vdots \\\\\n \\cop_N \\\\\n \\end{bmatrix}\n & =\n \\FPh\n +\n \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n \\underbrace{\n \\begin{bmatrix}\n \\cop[\\hat{p}_1]_1 \\\\\n \\vdots \\\\\n \\cop[\\hat{p}_N]_N \\\\\n \\end{bmatrix}\n }_{\\V{Z}}\n =\n \\M{V}_0\n +\n \\M{V} \\FP\n +\n \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n \\CoP\\\\\n & =\n \\V{1}_N\n \\kron\n \\fp_0\n +\n \\left(\n \\M{I}_R \\kron \\M{I}\n \\right)\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\n \\FP\n +\n \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n \\CoP\n \\end{split}\n\\end{equation}\n%\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Model of the system}\n\nThe model has the following form\n%\n\\begin{align}\n \\cstate_{k+1} =& \\M{A} \\cstate_k + \\M{B} \\cop_{k+1}\\\\\n \\dcop_k =& \\M{D} \\cstate_k + \\M{E} \\cop_{k+1}.\n\\end{align}\n%\n\nAfter condensing we obtain\n%\n\\begin{align}\n \\begin{bmatrix}\n \\cstate_1 \\\\\n \\vdots\\\\\n \\cstate_{N}\\\\\n \\end{bmatrix}\n =&\n \\M{U}_x \\cstate_0\n +\n \\M{U}_u\n \\begin{bmatrix}\n \\cop_1 \\\\\n \\vdots \\\\\n \\cop_N \\\\\n \\end{bmatrix} \\\\\n%\n \\underbrace{\n \\begin{bmatrix}\n \\dcop_0 \\\\\n \\vdots\\\\\n \\dcop_{N-1}\\\\\n \\end{bmatrix}\n }_{\\dCoP}\n =&\n \\M{O}_x \\cstate_0\n +\n \\M{O}_u\n \\begin{bmatrix}\n \\cop_1 \\\\\n \\vdots \\\\\n \\cop_N \\\\\n \\end{bmatrix} \\\\\n\\end{align}\n%\nor\n%\n\\begin{align}\n \\cState\n =&\n \\M{U}_x \\cstate_0\n +\n \\M{U}_u\n \\left(\n \\M{V}_0\n +\n \\M{V} \\FP\n +\n \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n \\CoP\n \\right)\\\\\n%\n \\dCoP\n =&\n \\M{O}_x \\cstate_0\n +\n \\M{O}_u\n \\left(\n \\M{V}_0\n +\n \\M{V} \\FP\n +\n \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n \\CoP\n \\right)\\\\\n\\end{align}\n%\n\nThe unknowns $\\CoP$ and $\\FP$ can be grouped together\n%\n\\begin{align}\n \\cState\n =&\n \\underbrace{\n \\begin{bmatrix}\n \\M{U}_u \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}} & \\M{U}_u \\M{V} \\\\\n \\end{bmatrix}\n }_{\\M{S}}\n \\underbrace{\n \\begin{bmatrix}\n \\CoP\\\\\n \\FP\\\\\n \\end{bmatrix}\n }_{\\V{X}}\n +\n \\underbrace{\n \\M{U}_x \\cstate_0\n +\n \\M{U}_u \\M{V}_0\n }_{\\V{s}}\\\\\n%\n \\dCoP\n =&\n \\underbrace{\n \\begin{bmatrix}\n \\M{O}_u \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}} & \\M{O}_u \\M{V} \\\\\n \\end{bmatrix}\n }_{\\M{S}_{\\dot{z}}}\n \\underbrace{\n \\begin{bmatrix}\n \\CoP\\\\\n \\FP\\\\\n \\end{bmatrix}\n }_{\\V{X}}\n +\n \\underbrace{\n \\M{O}_x \\cstate_0\n +\n \\M{O}_u \\M{V}_0\n }_{\\V{s}_{\\dot{z}}}\\\\\n\\end{align}\n%\nor\n%\n\\begin{align}\n \\cState\n =&\n \\begin{bmatrix}\n \\left(\\M{I} \\bkron \\M{U}_u^x \\right) \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n &\n \\left(\\M{I} \\bkron \\M{U}_u^x \\right)\n \\left(\n \\M{I}_R \\kron \\M{I}\n \\right)\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\\\\\n \\end{bmatrix}\n \\V{X}\n \\\\\n &\n +\n \\left(\\M{I} \\bkron \\M{U}_x^x \\right)\n \\cstate_0\n +\n \\left(\\M{I} \\bkron \\M{U}_u^x \\right)\n \\left(\\V{1}_N \\kron \\fp_0 \\right)\n \\\\\n =&\n \\begin{bmatrix}\n \\left(\\M{I} \\bkron \\M{U}_u^x \\right) \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n &\n \\left(\n \\M{I}\n \\bkron\n \\left(\n \\M{U}_u^x\n \\M{I}_R\n \\right)\n \\right)\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\\\\\n \\end{bmatrix}\n \\V{X}\n \\\\\n &\n +\n \\left(\\M{I} \\bkron \\M{U}_x^x \\right)\n \\cstate_0\n +\n \\left(\n \\M{I}\n \\bkron\n \\left(\n \\M{U}_u^x\n \\V{1}_N\n \\right)\n \\right)\n \\fp_0\n \\\\\n\\end{align}\n%\nand\n%\n\\begin{align}\n \\dCoP\n =&\n \\begin{bmatrix}\n \\left(\\M{I} \\bkron \\M{O}_u^x \\right) \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n &\n \\left(\\M{I} \\bkron \\M{O}_u^x \\right)\n \\left(\n \\M{I}_R \\kron \\M{I}\n \\right)\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\\\\\n \\end{bmatrix}\n \\V{X}\n \\\\\n &\n +\n \\left(\\M{I} \\bkron \\M{U}_x^x \\right) \\cstate_0\n +\n \\left(\\M{I} \\bkron \\M{O}_u^x \\right)\n \\left(\\V{1}_N \\kron \\fp_0 \\right)\n \\\\\n = &\n \\begin{bmatrix}\n \\left(\\M{I} \\bkron \\M{O}_u^x \\right) \\diag{k = 1 \\dots N}{\\M[\\hat{p}_k][]{R}}\n &\n \\left(\n \\M{I}\n \\bkron\n \\left(\n \\M{O}_u^x\n \\M{I}_R\n \\right)\n \\right)\n \\diag{k = 0...M-1}{ \\M[p_{k}][]{R} }\\\\\n \\end{bmatrix}\n \\V{X}\n \\\\\n &\n +\n \\left(\\M{I} \\bkron \\M{U}_x^x \\right) \\cstate_0\n +\n \\left(\n \\M{I}\n \\bkron\n \\left(\n \\M{O}_u^x\n \\V{1}_N\n \\right)\n \\right)\n \\fp_0\n \\\\\n\\end{align}\n%\n\nVelocity of the \\acs{CoM} can be expressed as\n%\n\\begin{equation}\n \\cVel =\n \\diag{N}{\\M{I}_{v}} \\cState =\n \\diag{N}{\\M{I}_{v}} \\left( \\M{S}\\V{X} + \\V{s} \\right)=\n \\M{S}_v \\V{X} + \\V{s}_v\n\\end{equation}\n%\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Constraints}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{CoP positions}\nSimple bounds in the case when supports are rectangular.\n\n\\begin{equation}\n \\ubarV{z}_k \\le \\cop[\\hat{p}_k]_k \\le \\barV{z}_k, \\quad k = 1 \\dots N\\\\\n\\end{equation}\n\n\\begin{equation}\n \\ubarV{Z} \\le \\CoP \\le \\barV{Z}\n\\end{equation}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Foot positions}\nSimple bounds when feasible regions are rectangular.\n\n\\begin{equation}\n \\ubarV{p}_j \\le \\fp[p_{j-1}]_j \\le \\barV{p}_j, \\quad j = 1 \\dots M\n\\end{equation}\n\n\\begin{equation}\n \\ubarV{P} \\le \\FP \\le \\barV{P}\n\\end{equation}\n\nOnly initial and final double supports are handled, the respective constraints\ncan also be represented as simple bounds provided that the feet are aligned.\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Objectives}\nThree objectives are minimized: difference between the actual and reference\n\\acs{CoM} velocity\n%\n\\begin{equation}\n \\NORME{\\cVel - \\cVel_{ref}} = \\NORME{\\M{S}_v \\V{X} + \\V{s}_v - \\cVel_{ref}},\n\\end{equation}\n%\nthe \\acs{CoP} velocity\n%\n\\begin{equation}\n \\NORME{\\dCoP} = \\NORME{\\M{S}_{\\dot{z}} \\V{X} + \\V{s}_{\\dot{z}}},\n\\end{equation}\n%\ndistance between the \\acs{CoP} positions and the centers of the feet\n%\n\\begin{equation}\n \\NORME{\\CoP}.\n\\end{equation}\n%\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{QP}\n\\begin{equation}\n\\begin{split}\n \\MINIMIZE{\\CoP, \\FP} & \\frac{\\alpha}{2} \\NORM{\\cVel - \\cVel_{ref}}^2 +\n \\frac{\\beta}{2} \\NORM{\\dCoP}^2 +\n \\frac{\\gamma}{2} \\NORM{\\CoP}^2\\\\\n \\SUBJECTTO & \\ubarV{Z} \\le \\CoP \\le \\barV{Z} \\\\\n & \\ubarV{P} \\le \\FP \\le \\barV{P}\n\\end{split}\n\\end{equation}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Reference velocity}\n\\begin{equation}\n\\begin{split}\n &\\NORM{\\cVel - \\cVel_{ref}}^2\n =\n \\NORM{\\M{S}_v \\V{X} + \\V{s}_v - \\cVel_{ref}}^ 2\n = \\\\\n &\n \\V{X}^T \\M{S}_v^T \\M{S}_v \\V{X}\n +\n \\V{X}^T \\M{S}_v^T \\V{s}_v\n -\n \\V{X}^T \\M{S}_v^T \\cVel_{ref}\\\\\n &\n +\n \\V{s}_v^T \\M{S}_v \\V{X}\n +\n \\cancel{\\V{s}_v^T \\V{s}_v}\n -\n \\cancel{\\V{s}_v^T \\cVel_{ref}} \\\\\n &\n -\n \\cVel_{ref}^T \\M{S}_v \\V{X}\n -\n \\cancel{\\cVel_{ref}^T \\V{s}_v}\n +\n \\cancel{\\cVel_{ref}^T \\cVel_{ref}}\n\\end{split}\n\\end{equation}\n\nOmitting the constant terms we obtain\n\\begin{equation}\n \\frac{\\alpha}{2} \\V{X}^T \\M{S}_v^T \\M{S}_v \\V{X}\n -\n \\alpha \\cVel_{ref}^T \\M{S}_v \\V{X}\n +\n \\alpha \\V{s}^T_v \\M{S}_v \\V{X}\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{CoP velocity}\n\\begin{equation}\n \\NORM{\\dCoP}^2 =\n \\NORM{\\M{S}_{\\dot{z}} \\V{X} + \\V{s}_{\\dot{z}}}^2 =\n \\V{X}^T \\M{S}_{\\dot{z}}^T \\M{S}_{\\dot{z}} \\V{X}\n +\n \\V{X}^T \\M{S}_{\\dot{z}}^T \\V{s}_{\\dot{z}}\n +\n \\V{s}_{\\dot{z}}^T \\M{S}_{\\dot{z}} \\V{X}\n +\n \\cancel{\\V{s}_{\\dot{z}}^T \\V{s}_{\\dot{z}}}\n\\end{equation}\n\nOmitting the constant terms we obtain\n\\begin{equation}\n \\frac{\\beta}{2}\n \\V{X}^T \\M{S}_{\\dot{z}}^T \\M{S}_{\\dot{z}} \\V{X}\n +\n \\beta\n \\V{s}_{\\dot{z}}^T \\M{S}_{\\dot{z}} \\V{X}\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Displacement from the reference CoP}\n\\begin{equation}\n \\frac{\\gamma}{2} \\NORM{\\CoP}^2\n = \\frac{\\gamma}{2} \\CoP^T \\CoP\n\\end{equation}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Objective in the matrix form}\n\\begin{equation}\n\\begin{split}\n \\MINIMIZE{\\V{X}} & \\frac{1}{2} \\V{X}^T\\M{H}\\V{X} + \\V{h}^T \\V{X}\n\\end{split}\n\\end{equation}\n\n\\begin{equation}\n \\M{H} =\n \\alpha \\M{S}_v^T \\M{S}_v\n +\n \\beta \\M{S}_{\\dot{z}}^T \\M{S}_{\\dot{z}}\n +\n \\gamma\n \\begin{bmatrix}\n \\M{0} & \\M{0} & \\M{0}\\\\\n \\M{0} & \\M{I} & \\M{0}\\\\\n \\M{0} & \\M{0} & \\M{0}\\\\\n \\end{bmatrix}\\\\\n\\end{equation}\n\n\\begin{equation}\n \\V{h} =\n -\n \\alpha \\cVel_{ref}^T \\M{S}_v\n +\n \\alpha \\V{s}_v^T \\M{S}_v\n +\n \\beta \\V{s}_{\\dot{z}}^T \\M{S}_{\\dot{z}}\n\\end{equation}\n\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Hierarchical least squares problem}\n\\begin{description}\n \\item[Level 1:]\n \\begin{equation}\n \\begin{split}\n & \\ubarV{Z} \\le \\CoP \\le \\barV{Z} \\\\\n & \\ubarV{P} \\le \\FP \\le \\barV{P}\n \\end{split}\n \\end{equation}\n\n \\item[Level 2:]\n \\begin{equation}\n \\begin{split}\n & \\sqrt{\\frac{\\alpha}{2}} \\M{S}_v \\V{X} = \\sqrt{\\frac{\\alpha}{2}} \\left(\\cVel_{ref} - \\V{s}_v\\right) \\\\\n & \\sqrt{\\frac{\\beta}{2}} \\M{S}_{\\dot{z}} \\V{X} = \\sqrt{\\frac{\\beta}{2}} \\V{s}_{\\dot{z}}\\\\\n & \\sqrt{\\frac{\\gamma}{2}} \\CoP = 0\\\\\n \\end{split}\n \\end{equation}\n\\end{description}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Swing foot trajectory}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Polynomial and boundary conditions}\nTrajectory is generated using cubic polynomial of the form\n\\begin{equation}\n at^3 + bt^2 + ct + d = y_{swing},\n\\end{equation}\nwhere $t$ is time instance; $a,b,c,d$ are coefficients; and $y_{swing}$ is position of the swing\nfoot at time $t$.\n\nDerivatives of the cubic polynomial are\n\\begin{equation}\n\\begin{split}\n & 3at^2 + 2bt + c = \\dot{y}_{swing},\\\\\n & 6at + 2b = \\ddot{y}_{swing},\\\\\n & 6a = \\dddot{y}_{swing}.\\\\\n\\end{split}\n\\end{equation}\n\nThere are four boundary conditions for the polynomial, the first two are defined for\nthe current swing foot state at $t_i = 0$:\n\\begin{equation}\\label{wpg04.eq.swtraj_bc1}\n\\begin{split}\n & d = y_{swing,i}, \\\\\n & c = \\dot{y}_{swing,i}, \\\\\n\\end{split}\n\\end{equation}\nwhere $y_{swing,i}$ and $\\dot{y}_{swing,i}$ are initial position and velocity; the other two\nconditions for landing time instance $t_f$ are:\n\\begin{equation}\\label{wpg04.eq.swtraj_bc2}\n\\begin{split}\n & at_f^3 + bt_f^2 + ct_f + d = y_{swing,f},\\\\\n & 3at^2 + 2bt + c = \\dot{y}_{swing,f},\\\\\n\\end{split}\n\\end{equation}\nwhere $y_{swing,f}$ and $\\dot{y}_{swing,f}$ are final position and velocity.\nTrajectories along $z$ axis and $x,y$ axes are computed separately. Final position\n$y_{swing,f}^z$ for trajectory along $z$ axis is set to the step height during the first\nhalf of the support and to $0$ during the second half:\n\\begin{equation}\n y_{swing,f}^z =\n \\left\\{\n \\begin{array}{ll}\n h_{step} & t \\le \\frac{1}{2}T_{support}; \\\\\n 0 & t > \\frac{1}{2}T_{support}. \\\\\n \\end{array}\n \\right.\n\\end{equation}\nThe final $x,y$ positions are set to the next landing position computed as\n\\begin{equation}\n \\begin{bmatrix}\n y_{swing,f}^x\\\\\n y_{swing,f}^y\\\\\n \\end{bmatrix}\n =\n \\M{V}_{land}\\FP,\n\\end{equation}\nwhere\n\\begin{equation}\n \\M{V}_{land} =\n \\begin{bmatrix}\n \\M[p_0][]{R} & \\M{0}\n \\end{bmatrix}.\n\\end{equation}\nVelocity at the end of trajectory is set to zero.\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Computation of the desired acceleration}\n\nWhole body control requires current acceleration (at time $t_i$) of the swing foot,\nwhich can be found as\n\\begin{equation}\\label{wpg04.eq.swtraj_acc1}\n 2b = \\ddot{y}_{swing,i}.\n\\end{equation}\nHence it is necessary to find coefficient $b$ from equations~\\eqref{wpg04.eq.swtraj_bc1}\nand~\\eqref{wpg04.eq.swtraj_bc2}.\n\nSubstitution of~\\eqref{wpg04.eq.swtraj_bc1} to~\\eqref{wpg04.eq.swtraj_bc2} using $\\dot{y}_{swing,f} = 0$\nyields\n\\begin{equation}\\label{wpg04.eq.swtraj_bc3}\n\\begin{split}\n & at_f^3 + bt_f^2 + \\dot{y}_{swing,i}t_f + y_{swing,i} = y_{swing,f},\\\\\n & 3at_f^2 + 2bt_f + \\dot{y}_{swing,i} = 0.\\\\\n\\end{split}\n\\end{equation}\n\nTrivial algebraic operations on~\\eqref{wpg04.eq.swtraj_bc3} and~\\eqref{wpg04.eq.swtraj_acc1}\nlead to the following equation:\n\\begin{equation}\\label{wpg04.eq.swtraj_acc2}\n \\ddot{y}_{swing,i}\n =\n \\frac{6(y_{swing,f} - y_{swing,i})}{t_f^2} - \\frac{4\\dot{y}_{swing,i}}{t_f}\n =\n \\frac{6}{t_f^2}y_{swing,f} -\n \\underbrace{\n \\frac{6}{t_f^2}y_{swing,i} - \\frac{4}{t_f} \\dot{y}_{swing,i}\n }_{\\mbox{constant}}.\n\\end{equation}\n\nConsequently\n\\begin{equation}\n \\V{\\ddot{y}}_{swing}\n =\n \\begin{bmatrix}\n \\ddot{y}_{swing,i}^x\\\\\n \\ddot{y}_{swing,i}^y\\\\\n \\ddot{y}_{swing,i}^z\\\\\n \\end{bmatrix}\n =\n \\underbrace{\n \\frac{6}{t_f^2}\n \\begin{bmatrix}\n \\M{V}_{land} \\\\\n \\M{0}\n \\end{bmatrix}\n }_{\\M{V}_{sa}}\n \\FP\n +\n \\underbrace{\n \\frac{6}{t_f^2}\n \\begin{bmatrix}\n \\M{0} \\\\\n y_{swing,f}^z\n \\end{bmatrix}\n -\n \\frac{6}{t_f^2}\n \\V{y}_{swing,i}\n -\n \\frac{4}{t_f}\n \\V{\\dot{y}}_{swing,i}\n }_{\\V{b}_{sa}},\n\\end{equation}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Computation of the initial jerk}\nJerk at the beginning of the swing foot trajectory characterizes behaviour of the\npolynomial near this point. High values of the jerk indicate rapid change of\nacceleration, which in turn increases error in trajectory tracking using piece-wise\nconstant acceleration. Due to this reason we penalize the jerk at the starting point\nof trajectory. Obviously, only $x,y$ components are penalized, since $z$ component\nis directly computed.\n\nThe jerk can be computed as:\n\\begin{equation}\\label{wpg04.eq.swtraj_jerk}\n \\dddot{y}_{swing,i}\n =\n - \\frac{12(y_{swing,f} - y_{swing,i})}{t_f^3} + \\frac{6\\dot{y}_{swing,i}}{t_f^2}\n =\n - \\frac{12}{t_f^3}y_{swing,f}\n +\n \\underbrace{\n \\frac{12}{t_f^3}y_{swing,i} + \\frac{6}{t_f} \\dot{y}_{swing,i}\n }_{\\mbox{constant}}.\n\\end{equation}\n\nTherefore $x,y$ components of the jerk are expressed as\n\\begin{equation}\n \\V{\\dddot{y}}_{swing}^{x,y}\n =\n \\begin{bmatrix}\n \\dddot{y}_{swing,i}^x\\\\\n \\dddot{y}_{swing,i}^y\\\\\n \\end{bmatrix}\n =\n \\underbrace{\n -\n \\frac{12}{t_f^3}\n \\M{V}_{land}\n }_{\\M{V}_{sj}}\n \\FP\n +\n \\underbrace{\n \\frac{12}{t_f^3}\n \\V{y}_{swing,i}\n +\n \\frac{6}{t_f^2}\n \\V{\\dot{y}}_{swing,i}\n }_{\\V{b}_{sj}}\n =\n \\underbrace{\n \\begin{bmatrix}\n \\M{0} & \\M{0} & \\M{V}_{sj}\n \\end{bmatrix}\n }_{\\M{A}_{sj}}\n \\V{X}_{mpc}\n +\n \\V{b}_{sj}\n .\n\\end{equation}\n\\input{../../../doc/latex/Styles/wpg_notation_undefine.tex}\n", "meta": {"hexsha": "f6026b5c8bee1155c28b1775e91d57d11d05e81d", "size": 23596, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "extra_modules/wpg04/doc/wpg04.tex", "max_stars_repo_name": "bip-team/humoto-module-vertical-CoM", "max_stars_repo_head_hexsha": "665d7c192eaed69135e7b25f1b56bd531317f4bb", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "extra_modules/wpg04/doc/wpg04.tex", "max_issues_repo_name": "bip-team/humoto-module-vertical-CoM", "max_issues_repo_head_hexsha": "665d7c192eaed69135e7b25f1b56bd531317f4bb", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-11-06T21:22:03.000Z", "max_issues_repo_issues_event_max_datetime": "2020-05-15T22:17:26.000Z", "max_forks_repo_path": "extra_modules/wpg04/doc/wpg04.tex", "max_forks_repo_name": "bip-team/humoto-modules-comanoid", "max_forks_repo_head_hexsha": "665d7c192eaed69135e7b25f1b56bd531317f4bb", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.1288604899, "max_line_length": 115, "alphanum_fraction": 0.4574080353, "num_tokens": 8800, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8577681195338728, "lm_q2_score": 0.718594386544335, "lm_q1q2_score": 0.6163873556537312}} {"text": "\\documentclass[a4paper,12pt]{article}\n\\usepackage{amsmath}\n\\begin{document}\n\n% From http://www.eecs.tufts.edu/~khan/Courses/Spring2013/EE194/Lecs/Lec9and10.pdf\n% and http://www.cs.unc.edu/~wens/WAFR2014-Sun.pdf\n% (which references http://maeresearch.ucsd.edu/skelton/publications/weiwei_ilqg_CDC43.pdf)\n% and http://arl.cs.utah.edu/pubs/ISRR2013.pdf\n\n\\section{Backwards Pass}\n\nLet's start with some definitions for the standard LQR controller.\n$c_t(\\boldsymbol{x}, \\boldsymbol{u})$ is the cost for the iteration $t$ given that we are at $\\boldsymbol{x}$ and are going to apply $\\boldsymbol{u}$ for one cycle.\n$v_t(\\boldsymbol{x})$ is the optimal cost-to-go for the starting point $\\boldsymbol{x}$ at step $t$.\n$v_N(\\boldsymbol{x})$ may be defined to be a different final cost from $v_t(\\boldsymbol{x})$ if need be, where $N$ is the horizon.\n\n$$\\begin{array}{rcl}\nc_t(\\boldsymbol{x}, \\boldsymbol{u}) &=&\n \\frac{1}{2}\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} Q_t & P_t^T \\\\ P_t & R_t \\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix} +\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} \\boldsymbol{q}_t \\\\ \\boldsymbol{r}_t \\end{bmatrix} +\n q_t \\\\\nv_{t + 1}(\\boldsymbol{x}) &=&\n \\frac{1}{2} \\boldsymbol{x}^T S_{t + 1} \\boldsymbol{x} + \\boldsymbol{x}^T \\boldsymbol{s}_{t + 1} + s_{t + 1} \\\\\n\ng_t(\\boldsymbol{x}_t, \\boldsymbol{u}_t) = \\boldsymbol{x}_{t + 1}(\\boldsymbol{u}_t) &=&\n A_{t} \\boldsymbol{x}_t + B_t \\boldsymbol{u}_t + \\boldsymbol{c}_{t} \\\\\n\nv_{t}(\\boldsymbol{x}_t, \\boldsymbol{u}_t) &=&\n v_{t+1}(\\boldsymbol{x}_{t+1}) +\n c_t(\\boldsymbol{x}_t, \\boldsymbol{u}_t) \\\\\\\n &=&\n v_{t+1}(g_t(\\boldsymbol{x}_{t}, \\boldsymbol{u}_t)) +\n c_t(\\boldsymbol{x}_t, \\boldsymbol{u}_t)\n \\\\\n\\end{array}$$\n\nNow, let's calculate $v_t(\\boldsymbol{x})$ given $v_{t+1}(\\boldsymbol{x})$ and the step cost from $c_t(\\boldsymbol{x})$.\nThis tells us the cost of applying $\\boldsymbol{u}$ starting from $\\boldsymbol{x}$.\nWe will then optimize over all $\\boldsymbol{u}$ to find the optimal next state and cost for that state.\nThis lets us then use dynamic programing methods to work out the optimal $\\boldsymbol{x}$ and $\\boldsymbol{u}$ at each timestep.\n\nWe want to get a quadratic solution of the form\n$$v_{t}(\\boldsymbol{x}) =\n \\frac{1}{2} \\boldsymbol{x}^T S_{t} \\boldsymbol{x} + \\boldsymbol{x}^T \\boldsymbol{s}_{t} + s_{t}$$ for each step.\n\n$$\\begin{array}{rcl}\nv_{t}(\\boldsymbol{x}, \\boldsymbol{u}) &=&\n \\frac{1}{2} \\left( A_{t}\\boldsymbol{x} + B_{t}\\boldsymbol{u} + \\boldsymbol{c}_{t}\\right)^T\n S_{t + 1}\n \\left( A_{t}\\boldsymbol{x} + B_{t}\\boldsymbol{u} + \\boldsymbol{c}_{t}\\right) +\n \\left( A_{t}\\boldsymbol{x} + B_{t}\\boldsymbol{u} + \\boldsymbol{c}_{t}\\right)^T \\boldsymbol{s}_{t + 1} +\n s_{t + 1} + \\\\ &&\n \\frac{1}{2}\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} Q_{t} & P_{t}^T \\\\ P_{t} & R_{t} \\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix} +\n \\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} \\boldsymbol{q}_{t} \\\\ \\boldsymbol{r}_{t} \\end{bmatrix} +\n q_{t}\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T A_{t}^T S_{t + 1} A_{t} \\boldsymbol{x} +\n \\boldsymbol{u}^T B_{t}^T S_{t + 1} A_{t} \\boldsymbol{x} +\n \\boldsymbol{c}_t^T S_{t + 1} A_{t} \\boldsymbol{x} + \\right. \\\\ &&\n \\left. \\boldsymbol{x}^T A_{t}^T S_{t + 1} B_{t} \\boldsymbol{u} +\n \\boldsymbol{u}^T B_{t}^T S_{t + 1} B_{t} \\boldsymbol{u} +\n \\boldsymbol{c}_t^T S_{t + 1} B_{t} \\boldsymbol{u} + \\right. \\\\ &&\n \\left. \\boldsymbol{x}^T A_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{u}^T B_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t\n \\right) + \\\\ &&\n \\boldsymbol{x}^T A_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{u}^T B_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} +\n s_{t + 1} + \\\\ &&\n \\frac{1}{2} \\left(\\boldsymbol{x}^T Q_t \\boldsymbol{x} + \\boldsymbol{u}^T P_t \\boldsymbol{x} + \\boldsymbol{x}^T P_t^T \\boldsymbol{u} + \\boldsymbol{u}^T R_t \\boldsymbol{u} \\right) + \\boldsymbol{x}^T \\boldsymbol{q}_t + \\boldsymbol{u}^T \\boldsymbol{r}_t + q_t\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} A_{t} + Q_t \\right) \\boldsymbol{x} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} \\right. + \\\\ &&\n \\left. \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) \\boldsymbol{u} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) \\boldsymbol{u} \\right) + \\\\ &&\n \\boldsymbol{x}^T \\left(A_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{q}_t \\right)+\n\n \\boldsymbol{u}^T \\left( B_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{r}_t \\right) +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} +\n q_t + \\\\ &&\n \\frac{1}{2} \\left( \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{x}^T A_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{u}^T B_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T S_{t + 1} B_{t} \\boldsymbol{u} +\n \\boldsymbol{c}_t^T S_{t + 1} A_{t} \\boldsymbol{x} \\right)\n\\end{array}$$\n\nNow, let's find the optimal $\\boldsymbol{u}$. Do this by evaluating $\\frac{\\partial}{\\partial \\boldsymbol{u}} v_t(\\boldsymbol{x}, \\boldsymbol{u}) = 0$.\n\n$$\\begin{array}{rcl}\n\\frac{\\partial}{\\partial \\boldsymbol{u}} v_{t}(\\boldsymbol{x}, \\boldsymbol{u}) &=& \\frac{1}{2} \\left( 2 \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) + 2 \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) + 2 \\boldsymbol{c}_t^T S_{t + 1} B_{t} \\right) + \\\\&&\\boldsymbol{s}_{t + 1}^T B_{t} + \\boldsymbol{r}_t^T \\\\\n\n\\frac{\\partial}{\\partial \\boldsymbol{u}} v_{t}(\\boldsymbol{x}, \\boldsymbol{u}) &=& \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) + \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) + \\boldsymbol{c}_t^T S_{t + 1} B_{t} + \\boldsymbol{s}_{t + 1}^T B_{t} + \\boldsymbol{r}_t^T \\\\\n0 &=& \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) + \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) + \\boldsymbol{c}_t^T S_{t + 1} B_{t} + \\boldsymbol{s}_{t + 1}^T B_{t} + \\boldsymbol{r}_t^T\n\\end{array}$$\n$$\\begin{array}{rcl}\n\\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) &=& - \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) - \\boldsymbol{c}_t^T S_{t + 1} B_{t} - \\boldsymbol{s}_{t + 1}^T B_{t} - \\boldsymbol{r}_t^T \\\\\n\\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) \\boldsymbol{u} &=& - \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} - B_t^T S_{t + 1} \\boldsymbol{c}_t - B_{t}^T \\boldsymbol{s}_{t + 1}- \\boldsymbol{r}_t \\\\\n\\boldsymbol{u} &=& - \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) ^{-1} \\left( \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} + B_t^T S_{t + 1} \\boldsymbol{c}_t + B_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{r}_t \\right)\n\\end{array}$$\n\nThis gives us the optimal $\\boldsymbol{u}$. There are some substitutions defined here from the 2013 paper which make it easier to read.\nNote: the 2014 paper uses different letters for these same quantities\\dots\n\\\\\n$$\\begin{array}{rcl}\nC_t &=& B^T_t S_{t + 1} A_t + P_t \\\\\nE_t &=& B_{t}^T S_{t + 1} B_{t} + R_t \\\\\nL_t &=& - E_t ^{-1} C_t \\\\\n\n\\boldsymbol{e}_t &=& B_t^T S_{t + 1} \\boldsymbol{c}_t + B_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{r}_t \\\\\n\\boldsymbol{l}_t &=& - E^{-1}_t \\boldsymbol{e}_t \\\\\n\nD_t &=& A_t^T S_{t + 1} A_t + Q_t \\\\\n\\boldsymbol{d}_t &=& A_t^T S_{t + 1} \\boldsymbol{c}_t + A_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{q}_t\n\\\\\n\\boldsymbol{u} &=& L_t \\boldsymbol{x} + \\boldsymbol{l}_t\n\\end{array}$$\n\nWith these, we can simplify $\\boldsymbol{u}$ a bit and make it look like the 2014 paper has it.\n\n$$\\begin{array}{rcl}\n \\boldsymbol{u} &=& -E_t^{-1} \\left( C_t \\boldsymbol{x} + \\boldsymbol{e}_t \\right) \\\\\n \\boldsymbol{u} &=& -E_t^{-1} C_t \\boldsymbol{x} - E_t^{-1} \\boldsymbol{e}_t\n\\end{array}$$\n\nFor reference, here are some equivalences between the symbols used in the 2013\npaper (the ones we use) and the 2014 paper.\nTODO(Brian): Figure out where the ones in the 2014 paper are defined instead of\nguessing by pattern-matching.\n\\\\ \\begin{tabular}{ | r | l | }\n \\hline\n 2013 paper (ours) & 2014 paper \\\\\n \\hline\n $C_t$ & $E_t$ \\\\\n $D_t$ & $C_t$ \\\\\n $\\boldsymbol{d}_t$ & $\\boldsymbol{c}_t$ \\\\\n $E_t$ & $D_t$ \\\\\n $\\boldsymbol{e}_t$ & $\\boldsymbol{d}_t$ \\\\\n \\hline\n\\end{tabular} \\\\\n\nNow, let's solve for the new cost function.\n\n$$\\begin{array}{rcl}\nv_{t}(\\boldsymbol{x}, \\boldsymbol{u}) &=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} A_{t} + Q_t \\right) \\boldsymbol{x} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} \\right. + \\\\ &&\n \\left. \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) \\boldsymbol{u} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) \\boldsymbol{u} \\right) + \\\\ &&\n \\boldsymbol{x}^T \\left(A_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{q}_t \\right)+\n\n \\boldsymbol{u}^T \\left( B_{t}^T \\boldsymbol{s}_{t + 1} +\n \\boldsymbol{r}_t \\right) +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} +\n q_t + \\\\ &&\n \\frac{1}{2} \\left( \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{x}^T A_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{u}^T B_{t}^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T S_{t + 1} B_{t} \\boldsymbol{u} +\n \\boldsymbol{c}_t^T S_{t + 1} A_{t} \\boldsymbol{x} \\right)\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} A_{t} + Q_t \\right) \\boldsymbol{x} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} \\right. + \\\\ &&\n \\left. \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) \\boldsymbol{u} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) \\boldsymbol{u} \\right) + \\\\ &&\n\n \\boldsymbol{x}^T \\left(A_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{q}_t +\n \\frac{1}{2} A_{t}^T S_{t + 1} \\boldsymbol{c}_t \\right) + \\\\ &&\n\n \\boldsymbol{u}^T \\left( B_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{r}_t +\n \\frac{1}{2} B_{t}^T S_{t + 1} \\boldsymbol{c}_t \\right) + \\\\ &&\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\left( A_t \\boldsymbol{x} +\n B_t \\boldsymbol{u} + \\boldsymbol{c}_t \\right) + \\\\ &&\n\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} A_{t} + Q_t \\right) \\boldsymbol{x} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} A_{t} + P_t \\right) \\boldsymbol{x} \\right. + \\\\ &&\n \\left. \\boldsymbol{x}^T \\left( A_{t}^T S_{t + 1} B_{t} + P_t^T \\right) \\boldsymbol{u} +\n \\boldsymbol{u}^T \\left( B_{t}^T S_{t + 1} B_{t} + R_t \\right) \\boldsymbol{u} \\right) + \\\\ &&\n\n \\boldsymbol{x}^T \\left(A_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{q}_t +\n A_{t}^T S_{t + 1} \\boldsymbol{c}_t \\right) + \\\\ &&\n\n \\boldsymbol{u}^T \\left( B_{t}^T \\boldsymbol{s}_{t + 1} + \\boldsymbol{r}_t +\n B_{t}^T S_{t + 1} \\boldsymbol{c}_t \\right) + \\\\ &&\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T D_t \\boldsymbol{x} +\n \\boldsymbol{u}^T C_t \\boldsymbol{x} +\n \\boldsymbol{x}^T C_t^T \\boldsymbol{u} +\n \\boldsymbol{u}^T E_t \\boldsymbol{u} \\right) + \\\\ &&\n\n \\boldsymbol{x}^T \\boldsymbol{d}_t +\n \\boldsymbol{u}^T \\boldsymbol{e}_t + \\\\ &&\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T D_t \\boldsymbol{x} +\n \\left( L_t \\boldsymbol{x} + \\boldsymbol{l}_t \\right)^T C_t \\boldsymbol{x} \\right. + \\\\ &&\n \\left. \\boldsymbol{x}^T C_t^T \\left( L_t \\boldsymbol{x} + \\boldsymbol{l}_t \\right) +\n \\left( L_t \\boldsymbol{x} + \\boldsymbol{l}_t \\right)^T E_t \\left( L_t \\boldsymbol{x} + \\boldsymbol{l}_t \\right) \\right) + \\\\ &&\n\n \\boldsymbol{x}^T \\boldsymbol{d}_t +\n \\left( L_t \\boldsymbol{x} + \\boldsymbol{l}_t \\right)^T \\boldsymbol{e}_t + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t\n\\\\\n&=&\n \\frac{1}{2} \\left(\n \\boldsymbol{x}^T D_t \\boldsymbol{x} + \\right.\n \\boldsymbol{x}^T L_t^T C_t \\boldsymbol{x} + \\boldsymbol{l}_t^T C_t \\boldsymbol{x} + \\\\ &&\n\n \\left. \\boldsymbol{x}^T C_t^T L_t \\boldsymbol{x} + \\boldsymbol{x}^T C_t^T \\boldsymbol{l}_t +\n\n \\boldsymbol{x}^T L_t^T E_t L_t \\boldsymbol{x} +\n \\boldsymbol{l}_t^T E_t L_t \\boldsymbol{x} +\n \\boldsymbol{x}^T L_t^T E_t \\boldsymbol{l}_t +\n \\boldsymbol{l}_t^T E_t \\boldsymbol{l}_t \\right) + \\\\&&\n\n \\boldsymbol{x}^T \\boldsymbol{d}_t +\n \\boldsymbol{x}^T L_t^T \\boldsymbol{e}_t + \\boldsymbol{l}^T_t \\boldsymbol{e}_t + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t\n\\\\\n&=&\n \\frac{1}{2}\n \\boldsymbol{x}^T \\left(\n D_t +\n L_t^T C_t +\n C_t^T L_t +\n L_t^T E_t L_t\n \\right) \\boldsymbol{x} + \\\\ &&\n\n \\boldsymbol{x}^T \\left(\n \\frac{1}{2} \\left( C_t^T \\boldsymbol{l}_t +\n L_t^T E_t \\boldsymbol{l}_t \\right) +\n \\boldsymbol{d}_t +\n L_t^T \\boldsymbol{e}_t \\right) +\n \\frac{1}{2} \\left( \\boldsymbol{l}_t^T E_t L_t + \\boldsymbol{l}_t^T C_t \\right) \\boldsymbol{x} + \\\\ &&\n\n \\frac{1}{2} \\boldsymbol{l}_t^T E_t \\boldsymbol{l}_t +\n \\boldsymbol{l}^T_t \\boldsymbol{e}_t +\n\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t +\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t\n\\\\\n&=&\n \\frac{1}{2}\n \\boldsymbol{x}^T \\left(\n D_t +\n L_t^T C_t +\n C_t^T L_t +\n L_t^T E_t L_t\n \\right) \\boldsymbol{x} + \\\\ &&\n\n \\boldsymbol{x}^T \\left(\n C_t^T \\boldsymbol{l}_t +\n L_t^T E_t \\boldsymbol{l}_t +\n \\boldsymbol{d}_t +\n L_t^T \\boldsymbol{e}_t \\right) + \\\\ &&\n\n \\frac{1}{2} \\boldsymbol{l}_t^T E_t \\boldsymbol{l}_t +\n \\boldsymbol{l}^T_t \\boldsymbol{e}_t +\n\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t +\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t\n\\end{array}$$\n\nOk, let's now pull out the 3 variables of interest and do further simplification.\n\n$$\\begin{array}{rcl}\nv_{t}(\\boldsymbol{x}) &=&\n \\frac{1}{2} \\boldsymbol{x}^T S_{t} \\boldsymbol{x} + \\boldsymbol{x}^T \\boldsymbol{s}_{t} + s_{t} \\\\\nS_t &=&\n D_t +\n L_t^T C_t +\n C_t^T L_t +\n L_t^T E_t L_t \\\\\n &=&\n D_t -\n C_t^T E_t^{-1} C_t -\n C_t^T E_t^{-1} C_t +\n C_t^T E_t^{-1} C_t \\\\\n &=& D_t - C_t^T E_t^{-1} C_t\n\\\\\n\\boldsymbol{s}_t &=&\n \\left( C_t^T + L_t^T E_t \\right) \\boldsymbol{l}_t +\n \\boldsymbol{d}_t +\n L_t^T \\boldsymbol{e}_t \\\\\n\n &=&\n \\left( C_t^T - C_t^T E_t^{-1} E_t \\right) \\boldsymbol{l}_t +\n \\boldsymbol{d}_t -\n C_t^T E_t^{-1} \\boldsymbol{e}_t \\\\\n &=& \\boldsymbol{d}_t - C_t^T E_t^{-1} \\boldsymbol{e}_t\n\\\\\ns_t &=&\n \\frac{1}{2} \\boldsymbol{l}_t^T E_t \\boldsymbol{l}_t +\n \\boldsymbol{l}^T_t \\boldsymbol{e}_t +\n\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t +\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t\n\\\\\n&=&\n\\frac{1}{2} \\boldsymbol{e}_t^T E_t^{-1} E_t E_t^{-1} \\boldsymbol{e}_t -\n \\boldsymbol{e}_t^T E_t^{-1} \\boldsymbol{e}_t +\n\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1} + q_t +\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t\n\\\\\n&=&\nq_t - \\frac{1}{2} \\boldsymbol{e}_t^T E_t^{-1} \\boldsymbol{e}_t +\n\n \\frac{1}{2} \\boldsymbol{c}_t^T S_{t + 1} \\boldsymbol{c}_t +\n \\boldsymbol{c}_t^T \\boldsymbol{s}_{t + 1} + s_{t + 1}\n\\end{array}$$\n\nFor better or worse, everything but the constant term matches both the 2013\npaper (which has no constant term) and the 2014 paper.\nThe constant term does not match the 2014 paper, but we're pretty sure it's\ncorrect (extended\\_lqr\\_derivation.py verifies what we have).\nThe 2014 paper has this for the constant term instead:\n$$\\begin{array}{rcl}\ns_t &=&\n q_t - \\frac{1}{2} \\boldsymbol{e}_t^T E_t^{-1} \\boldsymbol{e}_t\n\\end{array}$$\n\n\\section{Forwards Pass}\n\nLet's define the forwards pass to build the cost-to-come function.\nWe must use the same cycle cost, $c_t(\\boldsymbol{x}, \\boldsymbol{u})$, as the backwards pass.\nThe initial cost $v_0(\\boldsymbol{x})$ needs to evaluate to $0$ at $\\boldsymbol{x}_0$ since\nthere is no cost needed to get from $\\boldsymbol{x}_0$ to $\\boldsymbol{x}_0$.\n\n$$\\begin{array}{rcl}\n\\bar v_{t}(\\boldsymbol{x}) &=&\n \\frac{1}{2} \\boldsymbol{x}^T \\bar S_{t} \\boldsymbol{x} + \\boldsymbol{x}^T \\boldsymbol{\\bar s}_{t} + \\bar s_{t} \\\\\n\n\\bar g_t(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u}) = \\boldsymbol{x}_{t}(\\boldsymbol{u}) &=&\n \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t} \\\\\n\n\\bar c_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u}) &=& c_t(\\bar g_{t}(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u}), \\boldsymbol{u}) \\\\\n\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u}) &=&\n \\bar v_{t}(\\boldsymbol{x}_{t}) +\n \\bar c_t(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u}_t)\n \\\\\n&=&\n \\bar v_{t}(\\bar g_t(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u})) +\n c_t(\\bar g_{t}(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u}), \\boldsymbol{u}) \\\\\n \\\\\n\\end{array}$$\n\nIt is important to note that the optimal $\\boldsymbol{u}$ used to get from $\\boldsymbol{x}_t$ to $\\boldsymbol{x}_{t+1}$\nwhen evaluating $g_t(\\boldsymbol{x}, \\boldsymbol{u})$ is the same optimal $\\boldsymbol{u}$\nused to get from $\\boldsymbol{x}_{t+1}$ to $\\boldsymbol{x}_t$ when evaluating $\\bar g_{t}(\\boldsymbol{x}_{t + 1}, \\boldsymbol{u})$.\nConveniently, this means that the $\\boldsymbol{u}_t$, $L_t$, and $\\boldsymbol{l}_t$ calculated in both the forwards and backwards passes is the same.\n\nLike before, we want to get a quadratic solution of the form\n$$\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}) =\n \\frac{1}{2} \\boldsymbol{x}_{t+1}^T \\bar S_{t + 1} \\boldsymbol{x}_{t+1} + \\boldsymbol{x}_{t+1}^T \\boldsymbol{\\bar s}_{t + 1} + \\bar s_{t + 1}$$ for each step.\n\n$$\\begin{array}{rcl}\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u}) &=&\n \\frac{1}{2} \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T\n \\bar S_{t}\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) + \\\\ &&\n \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) ^T \\boldsymbol{\\bar s}_{t} + \\bar s_{t} + \\\\&&\n \\frac{1}{2}\n \\begin{bmatrix} \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} Q_t & P_t^T \\\\ P_t & R_t \\end{bmatrix}\n \\begin{bmatrix} \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) \\\\ \\boldsymbol{u} \\end{bmatrix} + \\\\&&\n \\begin{bmatrix} \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n \\begin{bmatrix} \\boldsymbol{q}_t \\\\ \\boldsymbol{r}_t \\end{bmatrix} +\n q_t \\\\\n\n &=&\n \\frac{1}{2} \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T\n \\bar S_{t}\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) + \\\\ &&\n \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) ^T \\boldsymbol{\\bar s}_{t} + \\bar s_{t} + \\\\&&\n\n \\frac{1}{2} \\left(\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T Q_t \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) + \\right. \\\\&&\n \\boldsymbol{u}^T P_t \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) + \\\\&&\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T P_t^T \\boldsymbol{u} + \\\\&&\n \\left. \\boldsymbol{u}^T R_t \\boldsymbol{u} \\right) + \\\\&&\n\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T \\boldsymbol{q}_t + \\boldsymbol{u}^T \\boldsymbol{r}_t + q_t \\\\\n\n &=&\n \\frac{1}{2} \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T\n \\left( \\bar S_{t} + Q_t \\right)\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) + \\\\ &&\n \\left(\\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) ^T \\boldsymbol{\\bar s}_{t} + \\bar s_{t} + \\\\&&\n\n \\frac{1}{2} \\left(\n \\boldsymbol{u}^T P_t \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right) \\right. + \\\\&&\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T P_t^T \\boldsymbol{u} + \\\\&&\n \\left. \\boldsymbol{u}^T R_t \\boldsymbol{u} \\right) + \\\\&&\n\n \\left( \\bar A_{t} \\boldsymbol{x}_{t + 1} + \\bar B_t \\boldsymbol{u} + \\boldsymbol{\\bar c}_{t}\\right)^T \\boldsymbol{q}_t + \\boldsymbol{u}^T \\boldsymbol{r}_t + q_t \\\\\n\n &=& \\frac{1}{2} \\left(\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t + \\right. \\\\&&\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t + \\\\&&\n \\left. \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t \\right) + \\\\&&\n\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\boldsymbol{\\bar s}_t +\n \\boldsymbol{u}^T \\bar B_t^T \\boldsymbol{\\bar s}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t +\n\n \\bar s_t + \\\\ &&\n\n \\frac{1}{2} \\left( \\boldsymbol{u}^T P_t \\bar A_{t} \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{u}^T P_t \\bar B_{t} \\boldsymbol{u} +\n \\boldsymbol{u}^T P_t \\boldsymbol{\\bar c}_{t} \\right) + \\\\&&\n \\frac{1}{2} \\left( \\boldsymbol{x}_{t+1}^T \\bar A^T_t P_t^T \\boldsymbol{u} +\n \\boldsymbol{u}^T \\bar B^T_{t} P_t^T \\boldsymbol{u} +\n \\boldsymbol{\\bar c}_t^T P^T_t \\boldsymbol{u} \\right) + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{u}^T R_t \\boldsymbol{u} + \\\\&&\n\n \\boldsymbol{x}^T_{t + 1} \\bar A^T_t \\boldsymbol{q}_t +\n \\boldsymbol{u}^T \\bar B^T_t \\boldsymbol{q}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t +\n\n \\boldsymbol{u}^T \\boldsymbol{r}_t + q_t\n\\end{array}$$\n\nNow, let's find the optimal $\\boldsymbol{u}$.\nDo this by evaluating\n$\\frac{\\partial}{\\partial \\boldsymbol{u}} \\bar v_t(\\boldsymbol{x}, \\boldsymbol{u}) = 0$.\n\n$$\\begin{array}{rcl}\n\\frac{\\partial}{\\partial \\boldsymbol{u}} \\bar v_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u})\n &=& \\frac{1}{2} \\left(\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n\n \\left(\\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1}\\right)^T + \\right. \\\\&&\n\n 2 \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\left( \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t \\right)^T + \\\\&&\n \\left. \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\right) + \\\\&& \n \\boldsymbol{\\bar s}_t ^T \\bar B_t + \\\\&&\n\n \\frac{1}{2} \\left( P_t \\bar A_{t} \\boldsymbol{x}_{t + 1}\\right) ^T +\n \\frac{1}{2} \\boldsymbol{x}_{t+1}^T \\bar A^T_t P_t^T + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{u}^T \\left( P_t \\bar B_{t} + \\bar B_t^T P_t^T \\right) +\n \\frac{1}{2} \\boldsymbol{u}^T \\left( \\bar B^T_{t} P_t^T + P_t \\bar B_t \\right) + \\\\&&\n \\frac{1}{2} \\left( P_t \\boldsymbol{\\bar c}_{t} \\right)^T +\n \\frac{1}{2} \\boldsymbol{\\bar c}^T_{t} P_t^T +\n\n \\boldsymbol{u}^T R_t + \\\\&&\n\n \\left(\\bar B^T_t \\boldsymbol{q}_t\\right)^T +\n\n \\boldsymbol{r}_t^T \\\\\n\n &=&\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t + \\\\&& \n\n \\boldsymbol{\\bar s}_t ^T \\bar B_t + \\\\&&\n\n \\boldsymbol{x}_{t+1}^T \\bar A^T_t P_t^T + \\\\&&\n\n \\boldsymbol{u}^T \\left( \\bar B^T_{t} P_t^T + P_t \\bar B_t \\right) + \\\\&&\n \\boldsymbol{\\bar c}^T_{t} P_t^T +\n\n \\boldsymbol{u}^T R_t + \\\\&&\n\n \\boldsymbol{q}_t^T \\bar B_t +\n \\boldsymbol{r}_t^T \\\\\n0 &=&\n \\boldsymbol{x}_{t + 1}^T \\left( \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t + \\bar A^T_t P_t^T \\right) + \\\\&&\n\n \\boldsymbol{u}^T \\left(\n \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\bar B^T_{t} P_t^T + P_t \\bar B_t + R_t\n \\right) + \\\\&&\n\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\boldsymbol{\\bar s}_t^T \\bar B_t +\n \\boldsymbol{\\bar c}^T_{t} P_t^T +\n \\boldsymbol{q}_t^T \\bar B_t +\n \\boldsymbol{r}_t^T \\\\\n\\end{array}$$\n\n$$\\begin{array}{rcl}\n \\boldsymbol{u}^T \\left(\n \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\bar B^T_{t} P_t^T + P_t \\bar B_t + R_t\n \\right)\n &=&\n - \\boldsymbol{x}_{t + 1}^T \\left( \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t + \\bar A^T_t P_t^T \\right) - \\\\&&\n\n\n \\left( \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\boldsymbol{\\bar s}_t^T \\bar B_t +\n \\boldsymbol{\\bar c}^T_{t} P_t^T +\n \\boldsymbol{q}_t^T \\bar B_t +\n \\boldsymbol{r}_t^T \\right) \\\\\n\n\\left(\n \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\bar B^T_{t} P_t^T + P_t \\bar B_t + R_t\n\\right) \\boldsymbol{u}\n &=&\n - \\left( \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t + P_t \\bar A_t \\right) \\boldsymbol{x}_{t + 1} - \\\\&&\n\n \\left( \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\bar B_t^T \\boldsymbol{\\bar s}_t +\n P_t \\boldsymbol{\\bar c}_{t} +\n \\bar B_t^T \\boldsymbol{q}_t +\n \\boldsymbol{r}_t \\right) \\\\\n\\end{array}$$\n\nSome substitutions to use for simplifying:\n\n$$\\begin{array}{rcl}\n\\bar C_t &=& \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t + P_t \\bar A_t \\\\\n\\bar E_t &=&\n \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t +\n \\bar B^T_{t} P_t^T + P_t \\bar B_t + R_t \\\\\n\\bar L_t &=& - \\bar E_t^{-1} \\bar C_t \\\\\n\n\\boldsymbol{\\bar e}_t &=&\n \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\bar B_t^T \\boldsymbol{\\bar s}_t +\n P_t \\boldsymbol{\\bar c}_{t} +\n \\bar B_t^T \\boldsymbol{q}_t +\n \\boldsymbol{r}_t \\\\\n\\boldsymbol{\\bar l}_t &=& - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\\\\n\n\\bar D_t &=& \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\\\\n\\boldsymbol{\\bar d}_t &=&\n \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\bar A_t^T \\left( \\boldsymbol{\\bar s}_t + \\boldsymbol{q}_t \\right) \\\\\n\n\\boldsymbol{u} &=& - \\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\\\\n\\boldsymbol{u} &=& \\bar L_t \\boldsymbol{x}_{t + 1} + \\boldsymbol{\\bar l}_t \\\\\n\n\\bar E_t \\boldsymbol{u} &=& -\\bar C_t \\boldsymbol{x}_{t + 1} - \\boldsymbol{\\bar e}_t\n\\end{array}$$\n\nNow, let's solve for the new cost function.\n\n$$\\begin{array}{rcl}\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u})\n &=& \\frac{1}{2} \\left(\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t + \\right. \\\\&&\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{u}^T \\bar B_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t + \\\\&&\n \\left. \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar A_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\bar B_t \\boldsymbol{u} +\n \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t \\right) + \\\\&& \n\n \\boldsymbol{x}_{t + 1}^T \\bar A_t^T \\boldsymbol{\\bar s}_t +\n \\boldsymbol{u}^T \\bar B_t^T \\boldsymbol{\\bar s}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t +\n\n \\bar s_t + \\\\ &&\n\n \\frac{1}{2} \\left( \\boldsymbol{u}^T P_t \\bar A_{t} \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{u}^T P_t \\bar B_{t} \\boldsymbol{u} +\n \\boldsymbol{u}^T P_t \\boldsymbol{\\bar c}_{t} \\right) + \\\\&&\n \\frac{1}{2} \\left( \\boldsymbol{x}_{t+1}^T \\bar A^T_t P_t^T \\boldsymbol{u} +\n \\boldsymbol{u}^T \\bar B^T_{t} P_t^T \\boldsymbol{u} +\n \\boldsymbol{\\bar c}_t^T P^T_t \\boldsymbol{u} \\right) + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{u}^T R_t \\boldsymbol{u} + \\\\&&\n\n \\boldsymbol{x}^T_{t + 1} \\bar A^T_t \\boldsymbol{q}_t +\n \\boldsymbol{u}^T \\bar B^T_t \\boldsymbol{q}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t +\n\n \\boldsymbol{u}^T \\boldsymbol{r}_t + q_t \\\\\n\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}, \\boldsymbol{u})\n &=& \\frac{1}{2} \\left(\n \\boldsymbol{x}_{t + 1}^T \\bar D_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{x}_{t + 1}^T \\bar C_t^T \\boldsymbol{u} +\n \\boldsymbol{u}^T \\bar C_t \\boldsymbol{x}_{t + 1} +\n \\boldsymbol{u}^T \\bar E_t \\boldsymbol{u} \\right) + \\\\&&\n\n \\boldsymbol{x}_{t + 1}^T \\boldsymbol{\\bar d}_t +\n\n \\boldsymbol{u}^T \\boldsymbol{\\bar e}_t + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t + \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t + \\bar s_t + q_t \\\\\n\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1})\n &=& \\frac{1}{2} \\left(\n \\boldsymbol{x}_{t + 1}^T \\bar D_t \\boldsymbol{x}_{t + 1} + \\right. \\\\ &&\n \\boldsymbol{x}_{t + 1}^T \\bar C_t^T \\left( -\\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right) + \\\\&&\n \\left( -\\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right)^T \\bar C_t \\boldsymbol{x}_{t + 1} + \\\\&&\n \\left. \\left( -\\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right)^T \\bar E_t \\left( -\\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right) \\right) + \\\\&&\n\n \\boldsymbol{x}_{t + 1}^T \\boldsymbol{\\bar d}_t +\n\n \\left( -\\bar E_t^{-1} \\bar C_t \\boldsymbol{x}_{t + 1} - \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right)^T \\boldsymbol{\\bar e}_t + \\\\&&\n\n \\frac{1}{2} \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t + \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t + \\bar s_t + q_t \\\\\n\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1})\n &=& \\frac{1}{2}\n \\boldsymbol{x}_{t + 1}^T \\left(\\bar D_t - \\bar C_t^T \\bar E_t^{-1} \\bar C_t \\right) \\boldsymbol{x}_{t + 1} + \\\\ &&\n\n \\boldsymbol{x}_{t + 1}^T \\left(\\boldsymbol{\\bar d}_t\n - \\bar C_t^T \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right)\\\\&&\n\n - \\frac{1}{2} \\boldsymbol{\\bar e}_t^T \\bar E_t^{-1} \\boldsymbol{\\bar e}_t +\n \\frac{1}{2} \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t + \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t + \\bar s_t + q_t \\\\\n\\end{array}$$\n\nOk, let's now pull out the 3 variables of interest and do further simplification.\n\n$$\\begin{array}{rcl}\n\\bar v_{t + 1}(\\boldsymbol{x}_{t+1}) &=& \\frac{1}{2} \\boldsymbol{x}_{t + 1}^T \\bar S_{t + 1} \\boldsymbol{x}_{t + 1} + \\boldsymbol{x}_{t + 1}^T \\boldsymbol{\\bar s}_{t + 1} + \\bar s_{t+1} \\\\\n \\bar S_{t + 1} &=& \\bar D_t - \\bar C_t^T \\bar E_t^{-1} \\bar C_t \\\\\n \\boldsymbol{\\bar s}_{t + 1} &=&\n \\left(\\boldsymbol{\\bar d}_t - \\bar C_t^T \\bar E_t^{-1} \\boldsymbol{\\bar e}_t \\right)\\\\\n\n\\bar s_{t+1} &=&\n - \\frac{1}{2} \\boldsymbol{\\bar e}_t^T \\bar E_t^{-1} \\boldsymbol{\\bar e}_t +\n \\frac{1}{2} \\boldsymbol{\\bar c}_t^T \\left( \\bar S_t + Q_t \\right) \\boldsymbol{\\bar c}_t +\n \\boldsymbol{\\bar c}_t^T \\boldsymbol{\\bar s}_t + \\boldsymbol{\\bar c}_t^T \\boldsymbol{q}_t + \\bar s_t + q_t \\\\\n\\end{array}$$\n\n\\newpage\n\\section{Detailed work for a few transformations}\n\n$$\\begin{array}{l}\n\\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n\\begin{bmatrix} Q_{t} & P_{t}^T \\\\ P_{t} & R_{t} \\end{bmatrix}\n\\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix} \\\\\n\n\\begin{bmatrix} \\boldsymbol{x}_1 & \\boldsymbol{x}_2 & \\boldsymbol{u}_1 & \\boldsymbol{u}_2 \\end{bmatrix}\n\\left[\\begin{array}{cc|cc}\n Q_{t11} & Q_{t12} & P_{t11} & P_{t21} \\\\\n Q_{t21} & Q_{t22} & P_{t12} & P_{t22} \\\\\n \\hline\n P_{t11} & P_{t12} & R_{t11} & R_{t12} \\\\\n P_{t21} & P_{t22} & R_{t21} & R_{t22} \\\\\n\\end{array}\\right]\n\\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix}\n\\\\\n\\begin{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t11} \\\\ Q_{t21} \\\\ P_{t11} \\\\ P_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t12} \\\\ Q_{t22} \\\\ P_{t12} \\\\ P_{t22} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t11} \\\\ P_{t12} \\\\ R_{t11} \\\\ R_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t21} \\\\ P_{t22} \\\\ R_{t12} \\\\ R_{t22} \\end{bmatrix} &\n\\end{bmatrix}\n\\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix}\n\\\\\n\\begin{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t11} \\\\ Q_{t21} \\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t12} \\\\ Q_{t22} \\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t11} \\\\ P_{t12} \\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t21} \\\\ P_{t22} \\end{bmatrix}\n\\end{bmatrix}\n\\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} + \\\\ \\quad\n\\begin{bmatrix}\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t11} \\\\ P_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t12} \\\\ P_{t22} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} R_{t11} \\\\ R_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} R_{t12} \\\\ R_{t22} \\end{bmatrix} &\n\\end{bmatrix}\n\\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\\\ \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix}\n\\\\\n\\begin{bmatrix} \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t11} \\\\ Q_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} Q_{t12} \\\\ Q_{t22} \\end{bmatrix}\\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} +\n\\begin{bmatrix} \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t11} \\\\ P_{t12} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t21} \\\\ P_{t22} \\end{bmatrix}\\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} + \\\\ \\quad\n\\begin{bmatrix} \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t11} \\\\ P_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} P_{t12} \\\\ P_{t22} \\end{bmatrix}\\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{x}_1 \\\\ \\boldsymbol{x}_2 \\end{bmatrix} +\n\\begin{bmatrix} \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} R_{t11} \\\\ R_{t21} \\end{bmatrix} &\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix} \\cdot\n \\begin{bmatrix} R_{t12} \\\\ R_{t22} \\end{bmatrix}\\end{bmatrix}\n \\begin{bmatrix} \\boldsymbol{u}_1 \\\\ \\boldsymbol{u}_2 \\end{bmatrix}\n\\\\\n\\boldsymbol{x}^T Q_t \\boldsymbol{x} + \\boldsymbol{u}^T P_t \\boldsymbol{x} + \\boldsymbol{x}^T P_t^T \\boldsymbol{u} + \\boldsymbol{u}^T R_t \\boldsymbol{u}\n\\end{array}$$\n\\\\ \\\\\n$$\\begin{array}{l}\n\\begin{bmatrix} \\boldsymbol{x} \\\\ \\boldsymbol{u} \\end{bmatrix}^T\n\\begin{bmatrix} \\boldsymbol{q}_{t} \\\\ \\boldsymbol{r}_{t} \\end{bmatrix}\n\\\\\n\\begin{bmatrix} \\boldsymbol{x}_1 & \\boldsymbol{x}_2 & \\boldsymbol{u}_1 & \\boldsymbol{u}_2 \\end{bmatrix}\n\\begin{bmatrix} \\boldsymbol{q}_{t1} \\\\ \\boldsymbol{q}_{t2} \\\\ \\boldsymbol{r}_{t1} \\\\ \\boldsymbol{r}_{t1} \\end{bmatrix}\n\\\\\n\\boldsymbol{x}^T \\boldsymbol{q}_t + \\boldsymbol{u}^T \\boldsymbol{r}_t\n\\end{array}$$\n\n\\end{document}\n", "meta": {"hexsha": "52d09496b895c37aec13beeeaff4c50351f1e4ec", "size": 37811, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "y2014/control_loops/python/extended_lqr.tex", "max_stars_repo_name": "Ewpratten/frc_971_mirror", "max_stars_repo_head_hexsha": "3a8a0c4359f284d29547962c2b4c43d290d8065c", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2021-06-18T03:22:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T15:23:43.000Z", "max_issues_repo_path": "y2014/control_loops/python/extended_lqr.tex", "max_issues_repo_name": "Ewpratten/frc_971_mirror", "max_issues_repo_head_hexsha": "3a8a0c4359f284d29547962c2b4c43d290d8065c", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 10, "max_issues_repo_issues_event_min_datetime": "2021-06-18T03:22:19.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-18T22:14:15.000Z", "max_forks_repo_path": "y2014/control_loops/python/extended_lqr.tex", "max_forks_repo_name": "Ewpratten/frc_971_mirror", "max_forks_repo_head_hexsha": "3a8a0c4359f284d29547962c2b4c43d290d8065c", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2021-08-19T19:20:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-08T07:33:18.000Z", "avg_line_length": 49.29726206, "max_line_length": 341, "alphanum_fraction": 0.591362302, "num_tokens": 16185, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382236515259, "lm_q2_score": 0.7154239897159438, "lm_q1q2_score": 0.6163651132575617}} {"text": "\\subsection{Full Decomposition}\n\\SecLabel{full}\n\nThe full decomposition of $\\pi$ is obtained from the decompositions of mini-block ciphers $T$ and $U$. First, let me describe the whitening linear layers. \n\nRecall the whitening linear layer $L^{\\top}$ $L$ was described in the beginning of \\SecRef{multi}. Let $L_{in} \\in \\linbij{8}$ be given by\n$$\nL_{in} \\eqdef l_{\\swaplsb} \\circ l_{\\eta} \\circ l_{\\delta} \\circ L^{\\top},\n$$\nwhere $L$ was described in the beginning of \\SecRef{multi}, $l_{\\eta}, l_{\\delta}, l_{\\swaplsb} \\in \\linbij{8}$ are given by\n\\begin{align*}\n l_{\\delta}(x, k) &\\eqdef (x \\oplus \\delta(k), k), \\\\\n l_{\\eta}(x, k) &\\eqdef (\\eta(x), \\delta(k)), \\\\\n l_{\\swaplsb}(x, k) &\\eqdef (x, \\swaplsb(k)).\n\\end{align*}\nLet $L_{out} \\in \\linbij{8}$ be given by\n$$\nL_{out} \\eqdef \\invtop{L} \\circ l_{\\alpha},\n$$\nwhere $l_{\\alpha} \\in \\linbij{8}$ is given by\n\\begin{align*}\n l_{\\alpha}(x, k) &\\eqdef (x \\oplus \\alpha(k), k), \\\\\n\\end{align*}\n\nThe matrix representations of $L_{in}$ and $L_{out}$ are as follows:\n\\begin{equation}\n \\Label{eq:lin}\nL_{in} = \\matb{\n0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\\n0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\\\\n1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\\\\n1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\\\\n0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\\\n0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\\n},~~\nL_{out} = \\matb{\n1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\\n1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\\\\n1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\\\\n0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n}.\n\\end{equation}\n\nThe final decomposition of $\\pi$ using the linear layers $L_{in}, L_{out}$ and the nonlinear components are given in \\FigRef{final}. An algorithmic representation of the decomposition is shown in Algorithm~\\Ref{alg:pimulti}.\n\n\\FigTex{final.tex}\n\n\\begin{algorithm}\n \\caption{\\Label{alg:pimulti}\n Computing $\\pi$: $v = \\pi(u)$ using finite field multiplications.\n }\n \\begin{algorithmic}[1]\n \\State{$(x, k) \\gets L_{in}(u)$}\n \\If{$r = 0$}\n \\State{$x \\gets \\zeta_0(x)$}\n \\Else\n \\State{$x \\gets \\zeta_+(x \\fmult \\inv(k))$} \\Comment{in $\\fielde{4} \\simeq \\field{}[X]/(X^4 + X^3 + 1)$}\n \\EndIf\n \\State{$(x, k) \\gets (k, x)$}\n \\State{$x \\gets \\beta(x \\fmult \\gamma(k))$} \\Comment{in $\\fielde{4} \\simeq \\field{}[X]/(X^4 + X^3 + 1)$}\n \\State{$v \\gets L_{out}(x,k)$}\n \\State{\\Return{$v$}}\n \\end{algorithmic}\n\\end{algorithm}", "meta": {"hexsha": "6acc3171598c052c29972c0764ec5ef7f3e96299", "size": 2555, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis-source/9strKuz/12full.tex", "max_stars_repo_name": "hellman/thesis", "max_stars_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 19, "max_stars_repo_stars_event_min_datetime": "2019-05-16T19:55:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-31T15:36:12.000Z", "max_issues_repo_path": "thesis-source/9strKuz/12full.tex", "max_issues_repo_name": "hellman/thesis", "max_issues_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-08-09T11:26:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-09T11:26:45.000Z", "max_forks_repo_path": "thesis-source/9strKuz/12full.tex", "max_forks_repo_name": "hellman/thesis", "max_forks_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-05T19:40:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-05T19:40:16.000Z", "avg_line_length": 36.5, "max_line_length": 224, "alphanum_fraction": 0.5307240705, "num_tokens": 1115, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382200964035, "lm_q2_score": 0.7154239897159439, "lm_q1q2_score": 0.6163651107141419}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\chapter{Statistics}\n\\label{chap:stats}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Expectation Value and Variance}\n\\label{stats:expval_and_var}\n\nThe expectation value \\cref{eq:stats:exp_relations} and variance \\cref{eq:stats:var_relations} are introductory, yet essential, statistical measures.\nTheir definitions and interesting properties are reproduced here for reference.\nNote that $s$ is the unbiased sample variance,\nand the sample mean $\\bar{x}$ has $\\mu_{\\bar{x}} = \\mu$, and standard error $\\sigma_{\\bar{x}} = \\sigma / \\sqrt{n}$,\nwhere $\\mu$ and $\\sigma$ are from the parent population.\n\n\\begin{subequations}\\label{eq:stats:exp_relations}\n\\begin{align}\n\\expvalE{X} = \\expval{X} &= \\sum_{j=1}^{m} x_{j} \\, p_{j} = \\int_{-\\infty}^{\\infty} x f\\left(x\\right) \\, \\dd{x} \\label{eq:stats:exp_relations:def} \\\\\n\\bar{x} = \\mu &= \\frac{1}{n} \\sum_{j=1}^{m} x_{j}\\,,\\,\\text{for uniform}~p_{j} \\label{eq:stats:exp_relations:mean} \\\\\n\\expval{X+Y} &= \\expval{X} + \\expval{X} \\label{eq:stats:exp_relations:add} \\\\\n\\expval{a X} &= a \\expval{X} \\label{eq:stats:exp_relations:mult} \\\\\n\\expval{a} &= a \\,\\, \\implies \\, \\expval{\\expval{X}} = \\expval{X} \\label{eq:stats:exp_relations:self} \\\\\n\\expval{X Y}^{2} &\\leq \\expval{X^{2}} \\expval{Y^{2}} \\label{eq:stats:exp_relations:cbs_ineq}\n\\end{align}\n\\end{subequations}\n\n\\begin{subequations}\\label{eq:stats:var_relations}\n\\begin{align}\n\\sigma_{X}^{2} = \\variance{X} &= \\expval{\\left(x-\\expval{x}\\right)^{2}} = \\expval{X^{2}} - \\expval{X}^{2} \\label{eq:stats:var_relations:def} \\\\\ns^{2} &= \\frac{1}{n-1} \\sum_{j=1}^{m} \\left( x_{j} - \\expval{x}\\right)^{2} \\label{eq:stats:var_relations:sample} \\\\\ns_{\\bar{x}} &= \\frac{s}{\\sqrt{n}} \\label{eq:stats:var_relations:standard_error_of_mean} \\\\\n\\variance{X+a} &= \\variance{X} \\label{eq:stats:var_relations:add} \\\\\n\\variance{a X} &= a^{2} \\, \\variance{X} \\label{eq:stats:var_relations:mult} \\\\\n\\variance{a X \\pm b Y} &= a^{2} \\, \\variance{X} + b^{2} \\, \\variance{Y} \\pm 2 \\, ab \\, \\cov{X}{Y} \\label{eq:stats:var_relations:linear} \\\\\n\\variance{X \\mid Y} &= \\expval{\\left(X - \\expval{X \\mid Y}\\right)^{2} \\mid Y} \\label{eq:stats:var_relations:conditional1} \\\\\n\\variance{X} &= \\expval{\\variance{X \\mid Y}} + \\variance{\\expval{X \\mid Y}} \\label{eq:stats:var_relations:conditional2}\n\\end{align}\n\\end{subequations}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Covariance and Correlation}\n\\label{stats:corr_covar}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Covariance}\n\\label{stats:corr_covar:covariance}\n\nThe covariance between two variables $u$ and $v$,\n\n\\begin{equation}\\label{eq:stats:covar}\n\\begin{split}\n\\sigma_{u,v}^{2} = \\cov{u}{v} &= \\frac{1}{m}\\sum_{j=1}^{m}\\left(u_{j}-\\expval{u}\\right)\\left(v_{j}-\\expval{v}\\right) \\\\\n&= \\expval{\\left(u-\\expval{u}\\right)\\left(v-\\expval{v}\\right)} \\\\\n&= \\expval{u v} - \\expval{u}\\expval{v}\\,,\n\\end{split}\n\\end{equation}\n\n\\noindent is a measure of their joint variability,\n\\ie a measure of any linear relationship which may exist between them.\nIt is helpful to remember the following covariance relations:\n\n\\begin{subequations}\\label{eq:stats:covar_relations}\n\\begin{align}\n\\cov{X}{X} &= \\variance{X}, \\label{eq:stats:covar_relations:var} \\\\\n\\cov{X + a}{Y + b} &= \\cov{X}{Y}, \\label{eq:stats:covar_relations:add} \\\\\n\\cov{a\\,X}{b\\,Y} &= ab\\,\\cov{X}{Y}, \\label{eq:stats:covar_relations:mult} \\\\\n\\cov{X}{Y}^{2} &\\leq \\variance{X}\\variance{Y}. \\label{eq:stats:covar_relations:inequality}\n\\end{align}\n\\end{subequations}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Pearson Correlation}\n\\label{stats:corr_covar:pearson}\n\nThe Pearson correlation coefficient,\n\n\\begin{equation}\\label{eq:stats:corr:pearson}\n\\rho_{u,v} = \\corr{u}{v} = \\frac{\\sigma_{u,v}^{2}}{\\sigma_{u}\\sigma_{v}} = \\frac{\\cov{u}{v}}{\\sigma_{u}\\sigma_{v}}\\,,\n\\end{equation}\n\n\\noindent is a convenient dimensionless version, normalized to $-1 \\leq \\rho \\leq 1$.\nExample distributions can be found in \\cref{fig:stats:corr_ex:pearson}.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.95\\textwidth]{figures/stats/corr_ex}\n\\caption{\nExample distributions for\nuncorrelated ($\\rho \\approx 0$),\ncorrelated ($\\rho \\approx 1$),\nand anticorrelated ($\\rho \\approx -1$)\nvariables $u$ and $v$ \\cite{DougNotes}.\n}\n\\label{fig:stats:corr_ex:pearson}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Spearman Correlation}\n\\label{stats:corr_covar:spearman}\n\nThe Spearman rank correlation coefficient $r_{s}$ \\cref{eq:stats:corr:spearman} is a non-parametric measure\nused to quantify how well two variables are monotonically related,\n\\ie measure the degree of rank ordering between two variables across the available data points.\nIn contrast to the Pearson correlation coefficient,\na linear relationship between the variables is not assumed,\nonly that they increase or decrease in a consistent manner.\nThe Spearman correlation is simply the Pearson correlation of the variable's ranks,\nwhere the rank function $R\\left(x_{i}\\right)$ returns the order,\n$1, 2, 3, \\ldots, m$, of a data point $x_{i}$ after being sorted\\footnote{Ties\ncan be handled via dense ranking $1,2,2,3$, fractional ranking $1,2.5,2.5,4$, or arbitrarily broken with ordinal ranking $1,2,3,4$, \\ie row numbering.}.\nA comparison to the Pearson correlation coefficient can be found in \\cref{fig:stats:corr_ex:spearman}.\n\n\\begin{equation}\\label{eq:stats:corr:spearman}\nr_{s} = \\rho_{R\\left(u\\right),R\\left(v\\right)} = \\frac{\\cov{R\\left(u\\right)}{R\\left(v\\right)}}{\\sigma_{R\\left(u\\right)}\\sigma_{R\\left(v\\right)}}\\,.\n\\end{equation}\n\n\\begin{figure}[H]\n \\centering\n \\begin{subfigure}[c]{0.48\\textwidth}\\centering\n \\includegraphics[width=\\textwidth]{figures/stats/spearman_corr_non_para}\n \\caption{Non-Parametric}\n \\label{fig:stats:corr_ex:spearman:non_para}\n \\end{subfigure}\n ~\n \\begin{subfigure}[c]{0.48\\textwidth}\\centering\n \\includegraphics[width=\\textwidth]{figures/stats/spearman_corr_outliers}\n \\caption{Outliers}\n \\label{fig:stats:corr_ex:spearman:outliers}\n \\end{subfigure}\n\\caption{\nComparisons of the Spearman rank correlation and Pearson correlation coefficients,\nadapted from\n\\href{https://en.wikipedia.org/wiki/File:Spearman_fig1.svg}{Skbkekas} and\n\\href{https://en.wikipedia.org/wiki/File:Spearman_fig3.svg}{Skbkekas}.\nNote that non-parametric Spearman rank correlation\nonly measures the monotonic nature of the data, not its linearity,\nand is more resilliant to outliers.\n\\label{fig:stats:corr_ex:spearman}\n}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Covariance Matrix}\n\\label{stats:corr_covar:covar_matrix}\n\nThe covariance matrix,\n\n\\begin{align}\n \\mathbf{M} = \\begin{pmatrix}\n \\sigma_1^2 & \\cov{1}{2} & \\cov{1}{3} & \\ldots \\\\\n \\cov{1}{2} & \\sigma_2^2 & \\cov{2}{3} & \\ldots \\\\\n \\cov{1}{3} & \\cov{2}{3} & \\sigma_3^2 & \\ldots \\\\\n \\vdots & \\vdots & \\vdots & \\ddots\n \\end{pmatrix}\\,,\n\\end{align}\n\n\\noindent with elements $M_{ij} = \\expval{\\left(u_{i} - \\expval{u}_{i}\\right)\\left(u_{j}-\\expval{u}_{j}\\right)}$\nis the higher dimensional extension of the covariance.\nWe can visualize the covariance between variables with\nGaussian error ellipses given by the probability distribution\n\n\\begin{equation}\\label{eq:stats:P_error_ellipse_k}\nP\\left(x_{1},x_{2},\\ldots,x_{k}\\right) = \\frac{1}{(2\\pi)^{k/2}}\\frac{1}{\\abs{\\mathbf{M}}^{1/2}}\\exp\\left[-\\frac{1}{2}\\left(\\vb{x}-\\vb*{\\mu}\\right)^{\\transpose}\\mathbf{M}\\left(\\vb{x}-\\vb*{\\mu}\\right)\\right]\\,,\n\\end{equation}\n\n\\noindent where the ellipse semi-axes are directed along the eigenvectors of $\\mathbf{M}$.\nIn two dimensions it is easier to see the equation of the error ellipse itself:\n\n\\begin{equation}\\label{eq:stats:P_error_ellipse_2}\n\\begin{split}\nP\\left(u,v\\right) &= \\frac{1}{2\\pi\\sigma_{u}\\sigma_{v}}\\frac{1}{\\sqrt{1-\\rho^{2}}}\\exp\\bigg\\{-\\frac{1}{2}\\bigg[ \\\\\n&\\frac{1}{(1-\\rho)^{2}}\\left(\\frac{\\left(u-\\expval{u}\\right)^{2}}{\\sigma_{u}^{2}}+\\frac{\\left(v-\\expval{v}\\right)^{2}}{\\sigma_{v}^{2}}-\\frac{2\\rho \\left(u-\\expval{u}\\right)\\left(v-\\expval{v}\\right)}{\\sigma_{u}\\sigma_{v}}\\right)\\bigg]\\bigg\\}\\,.\n\\end{split}\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Central Limit Theorem (CLT)}\n\\label{stats:CLT}\n\nThe central limit theorem (CLT) states that\nif we take samples of size $n$ from an independent random variable multiple times,\nfrom any distribution\\footnote{Where the mean is defined, \\ie not the Cauchy or other pathological distributions.},\nthe sample means will tend to the normal distribution.\nIn practice, we typically require $\\num{30} \\lesssim n$ points per sample before we say the CLT applies.\nThe CLT is an important result in statistics as it lets us\ntreat many problems in a normally distributed framework,\nin particular finding confidence intervals for the sample mean,\nand hypothesis testing for sample means with a \\ttest or ANOVA.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Bayes' Theorem}\n\\label{stats:Bayes_rule}\n\nBayes' theorem follows from the probability of the intersection of two events $A$ and $B$:\n\n\\begin{equation}\\label{eq:stats:intersection}\nP\\left(A \\cap B\\right) = P\\left(A \\mid B\\right) P\\left(B\\right) = P\\left(B \\mid A\\right) P\\left(A\\right).\n\\end{equation}\n\n\\noindent Dividing by $P\\left(B\\right)$ we have:\n\n\\begin{equation}\\label{eq:stats:Bayes_rule}\n\\begin{split}\nP\\left(A \\mid B\\right) &= \\frac{P\\left(B \\mid A\\right) P\\left(A\\right)}{P\\left(B\\right)}\\,, \\\\\n&= \\frac{P\\left(B \\mid A_{i}\\right) P\\left(A_{i}\\right)}{\\sum_{j} P\\left(B \\mid A_{j}\\right)P\\left(A_{j}\\right)}\\,, \\\\\n\\text{Posterior} &= \\frac{\\text{Likelihood} \\times \\text{Prior}}{\\text{Normalization}}\\,.\n\\end{split}\n\\end{equation}\n\n\\subsubsection{Example: Medical Testing}\n\\label{stats:Bayes_rule:medical_test}\n\nExample: Testing for disease with a \\SI{2}{\\percent} incidence rate in the wider population.\nThe test has a \\SI{99}{\\percent} true positive rate and a \\SI{5}{\\percent} false positive rate.\nWhat is the probability an individual has the disease if their test is positive?\n\n% https://www.wolframalpha.com/input/?i=(0.99*0.02)%2F((0.99*0.02)%2B(0.05*(1%E2%88%920.02)))\n\\begin{equation}\\label{eq:stats:Bayes_rule:medical_test_1}\n\\begin{split}\nP\\left(\\text{Infected} \\mid +\\right) &= \\frac{P\\left(+ \\mid \\text{Infected}\\right) P\\left(\\text{Infected}\\right)}{P\\left(+\\right)}\\,, \\\\\n &= \\frac{P\\left(+ \\mid \\text{Infected}\\right) P\\left(\\text{Infected}\\right)}{\nP\\left(+ \\mid \\text{Infected}\\right)P\\left(\\text{Infected}\\right) + P\\left(+ \\mid \\text{Healthy}\\right)P\\left(\\text{Healthy}\\right)}\\,, \\\\\n&= \\frac{\\num{0.99} \\times \\num{0.02}}{\\num{0.99} \\times \\num{0.02} + \\num{0.05} \\times \\left(1-\\num{0.02}\\right)}\\,, \\\\\n&\\approx \\num{0.288}\\,.\n\\end{split}\n\\end{equation}\n\n\\noindent And if we then run a second, independent, test which also comes back positive?\n\n% https://www.wolframalpha.com/input/?i=(0.99*0.288)%2F((0.99*0.288)%2B(0.05*(1%E2%88%920.288)))\n\\begin{equation}\\label{eq:stats:Bayes_rule:medical_test_2}\n\\begin{split}\nP\\left(\\text{Infected} \\mid ++\\right) &= \\frac{P\\left(+ \\mid \\text{Infected}\\right) P\\left(\\text{Infected} \\mid +\\right)}{\nP\\left(+ \\mid \\text{Infected}\\right)P\\left(\\text{Infected} \\mid +\\right) + P\\left(+ \\mid \\text{Healthy}\\right)P\\left(\\text{Healthy} \\mid +\\right)}\\,, \\\\\n&= \\frac{\\num{0.99} \\times \\num{0.288}}{\\num{0.99} \\times \\num{0.288} + \\num{0.05} \\times \\left(1-\\num{0.288}\\right)}\\,, \\\\\n&\\approx \\num{0.889}\\,.\n\\end{split}\n\\end{equation}\n\n\\noindent Note that if we ran both tests the first time we would still have:\n\n% https://www.wolframalpha.com/input/?i=(0.99%5E2*0.02)%2F((0.99%5E2*0.02)%2B(0.05%5E2*(1%E2%88%920.02)))\n\\begin{equation}\\label{eq:stats:Bayes_rule:medical_test_3}\n\\begin{split}\nP\\left(\\text{Infected} \\mid ++\\right) &= \\frac{P\\left(++ \\mid \\text{Infected}\\right) P\\left(\\text{Infected}\\right)}{P\\left(++\\right)}\\,, \\\\\n&= \\frac{\\num{0.99}^{2} \\times \\num{0.02}}{\\num{0.99}^{2} \\times \\num{0.02} + \\num{0.05}^{2} \\times \\left(1-\\num{0.02}\\right)}\\,, \\\\\n&\\approx \\num{0.889}\\,.\n\\end{split}\n\\end{equation}\n\n\\subsubsection{Example: Biased Coin}\n\\label{stats:Bayes_rule:biased_coin}\n\nConsider the case of a bag of $n$ fair coins and $m$ biased coins.\nLet $P\\left(H \\mid \\stcomp{F}\\right) \\equiv p_{H}$ be the \\apriori probability of heads $H$ for an biased coin $\\stcomp{F}$.\nDrawing one coin from the bag, you flip it multiple times recording $h$ heads and $t$ tails.\nWhat is the probability you have drawn an biased coin, $P\\left(\\stcomp{F} \\mid h,t\\right)$?\n\n\\begin{equation}\\label{eq:stats:Bayes_rule:biased_coin_setup}\n\\begin{gathered}\nP\\left(F\\right) = \\frac{n}{n+m}\\,,\\quad P\\left(\\stcomp{F}\\right) = \\frac{m}{n+m}\\,, \\\\\nP\\left(H \\mid F\\right) = \\frac{1}{2}\\,,\\quad P\\left(h,t \\mid F\\right) = \\left(\\frac{1}{2}\\right)^{h}\\,\\left(\\frac{1}{2}\\right)^{t} = \\frac{1}{2^{h+t}}\\,, \\\\\nP\\left(h,t \\mid \\stcomp{F}\\right) = P\\left(H \\mid \\stcomp{F}\\right)^{h} P\\left(\\stcomp{H} \\mid \\stcomp{F}\\right)^{t} = p_{H}^{h} \\left(1-p_{H}\\right)^{t}.\n\\end{gathered}\n\\end{equation}\n\n\\begin{equation}\\label{eq:stats:Bayes_rule:biased_coin_solution}\n\\begin{split}\nP\\left(\\stcomp{F} \\mid h,t\\right) &= \\frac{\nP\\left(h,t \\mid \\stcomp{F}\\right) P\\left(\\stcomp{F}\\right)}{\nP\\left(h,t \\mid \\stcomp{F}\\right) P\\left(\\stcomp{F}\\right) + P\\left(h,t \\mid F\\right) P\\left(F\\right)} \\\\\n&= \\frac{\nm\\,p_{H}^{h} \\left(1-p_{H}\\right)^{t}}{\nm\\,p_{H}^{h} \\left(1-p_{H}\\right)^{t} + n\\,2^{-h-t}}\\,.\n\\end{split}\n\\end{equation}\n\nSome example values are provided in \\cref{tab:Bayes_rule:biased_coin}.\n\n\\begin{table}[H]\n\\centering\n\\begingroup\n\\renewcommand*{\\arraystretch}{1}\n\\input{tables/coin_examples/bayes_biased_coin/bayes_biased_coin_example.tex}\n\\endgroup\n\\caption{\n$P\\left(\\stcomp{F} \\mid h,t\\right)$ for various values of $h$, $t$, and $p_{H}$ when $m = 50$, $n = 50$.\n}\n\\label{tab:Bayes_rule:biased_coin}\n\\end{table}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Coin Problems}\n\\label{stats:coin_problems}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Making an Biased Coin Fair}\n\\label{stats:coin_problems:biased_to_fair}\n% https://fivethirtyeight.com/features/can-you-make-an-unfair-coin-fair/\n% https://fivethirtyeight.com/features/can-you-snatch-defeat-from-the-jaws-of-victory/\n% https://youtu.be/-SANBbv0-Hw\n% https://math.stackexchange.com/a/146614 for additional references\n\nIf we are given an biased coin with $P\\left(H\\right) = p \\neq \\num{0.5}$\nwe can construct a fair coin\\footnote{The solution is credited to John von Neumann, who really did work on everything!} by\nflipping the biased coin $m$ times and selecting outcomes with an equal probability of occurring, while disregarding all other outcomes.\nFor example, with $m=2$ flips we have the outcomes listed in \\cref{tab:coin_problems:biased_to_fair:m2_outcomes}:\n\n\\begin{table}[H]\n\\centering\n\\begingroup\n\\renewcommand*{\\arraystretch}{1}\n\\input{tables/coin_examples/biased_to_fair/m2_outcomes.tex}\n\\endgroup\n\\caption{\nMaking an biased coin fair in $m=2$ flips.\n}\n\\label{tab:coin_problems:biased_to_fair:m2_outcomes}\n\\end{table}\n\nNote that we are disregarding $p^{2} + \\left(1-p\\right)^{2}$ percent of flips\nand this method will become increasingly inefficient as $p$ moves away from $\\num{0.5}$.\nWe can improve the efficiency by increasing $m$ and reassigning the\noutcomes\\footnote{For $m = \\num{4}$, disregard $P\\left(HHHH \\parallel TTTT\\right) = p^{4} + \\left(1-p\\right)^{4} < p^{2} + \\left(1-p\\right)^{2}$,\\newline\n$HT \\parallel HHHT \\parallel HHTT \\parallel TTHT \\to H$,\\newline\n$TH \\parallel TTTH \\parallel TTHH \\parallel HHTH \\to T = H^{-1}$.} such that\nwe decrease the average number of flips required.\nIf we need to simulate something with more outcomes, like a 6 sided die,\nwe can increase $m$ until we have enough equal outcomes, or combinations of outcomes.\n\nWe can also ask, for what values of $p$ is a fair outcome ensured in $m$ flips?\nTo solve, create $P\\left(\\text{Outcome}\\right)$ for all possible outcomes of $m$ flips.\nThen, sum all possible combinations\\footnote{Excluding the combination with all outcomes where $\\sum P\\left(\\text{Outcome}\\right) = 1$.} of $P\\left(\\text{Outcome}\\right)$\nand try to solve $\\sum P\\left(\\text{Outcome}\\right) = \\num{0.5}$.\nThis partitions the outcomes into two assignments without disregarding any outcomes, \\ie no flips are wasted.\nFor example, with $m=2$ and \\cref{tab:coin_problems:biased_to_fair:m2_outcomes}, we solve\n$p^{2}$, $p\\left(1-p\\right)$, $\\left(1-p\\right)^{2}$,\n$p^{2} + p\\left(1-p\\right)$, $p\\left(1-p\\right) + \\left(1-p\\right)^{2}$, and $p^{2} + \\left(1-p\\right)^{2} = \\num{0.5}$\nseparately for $0 t\\right) = P_{\\text{Poisson}}\\left(k=0;\\,\\lambda_{\\text{Poisson}} \\equiv \\lambda\\,t\\right) = e^{-\\lambda t}$,\n$P\\left(t\\right) = \\dv{t} P\\left(T \\leq t\\right) = \\dv{t} 1 - e^{-\\lambda t} = \\lambda e^{-\\lambda t}$.\nHere we have used the Poisson distribution to find the CDF,\nthen differentiated to find the PDF.\n$\\lambda$ is now a rate, unlike $\\lambda_{\\text{Poisson}}$!} we\narrive at the exponential distribution,\n\n\\begin{equation}\\label{eq:stats:exp:P}\nP\\left(x;\\,\\lambda\\right) = \\begin{cases}\n\\lambda e^{-\\lambda x} & x \\geq 0 \\,, \\\\\n0 & x < 0 \\,,\n\\end{cases}\n\\end{equation}\n\n\\noindent with rate parameter $\\lambda$ events per unit $x$.\nA plot of the distribution can be found in \\cref{fig:dist:exp}.\n\nThe mean and standard deviation of the exponential distribution are identical and equal to $1/\\lambda$:\n\n\\begin{equation}\\label{eq:stats:exp:mean_variance}\n\\expval{x} = \\sigma = \\frac{1}{\\lambda}\\,, \\quad \\sigma^{2} = \\frac{1}{\\lambda^{2}}\\,.\n\\end{equation}\n\nThe exponential distribution is memoryless\\footnote{The exponential (geometric) distribution is the only real number (integer) memoryless probability distribution.} \\cref{eq:stats:exp:memoryless},\n\\ie the time until a future event occurs does not depend on the time elapsed at the present.\n\n\\begin{equation}\\label{eq:stats:exp:memoryless}\nP\\left(T > t + s \\mid T > s\\right) = P\\left(T > t\\right)\\,, \\quad \\forall s,t \\geq 0.\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Example: Cars Driving By}\n\\label{stats:cars}\n\nAfter watching a patch of road,\nyou notice that on average cars are spaced \\SI{5}{\\minute} apart.\nAssume the cars are traveling independently\\footnote{Not a good assumption\nin real life as cars will clump up in traffic, \\ie become correlated.\nBetter examples include\nradioactive decay,\nphoton arrival from a distant star,\nsupport tickets being created\\ldots} of one another.\n\n\\subsubsection{Questions}\n\\label{stats:cars:questions}\n\n\\begin{enumerate}[noitemsep]\n \\item What is $P\\left(\\text{Observing \\num{5} cars in the next \\SI{6}{\\minute}}\\right)$?\\label{item:stats:cars:1}\n \\item If \\SI{20}{\\percent} of the cars are red, what is $P\\left(\\text{Observing \\num{5} red cars in the next \\SI{6}{\\minute}}\\right)$?\\label{item:stats:cars:2}\n \\item What is $P\\left(\\text{Observing \\num{2} or more cars in the next \\SI{10}{\\minute}}\\right)$?\\label{item:stats:cars:3}\n \\item \\num{100} cars have just driven by, how many are expected in the next hour?\\label{item:stats:cars:4}\n \\item If we start watching the road at a random time, how long should we expect to wait until we see our first car?\\label{item:stats:cars:5}\n\\end{enumerate}\n\n\\subsubsection{Solutions}\n\\label{stats:cars:solutions}\n\n\\begin{itemize}[noitemsep]\n % https://www.wolframalpha.com/input/?i=%281.2%5E5+%2F+5%21%29*e%5E-1.2\n \\item[\\cref{item:stats:cars:1}.] $k = 5$, $\\lambda = \\frac{1\\,\\text{car observed}}{\\SI{5}{\\minute}} \\times \\SI{6}{\\minute} = \\num{1.2}$, $P_{\\text{Poisson}} = \\frac{\\num{1.2}^{5}}{5!} e^{-1.2} = \\num{0.006}$.\n % https://www.wolframalpha.com/input/?i=%280.24%5E5+%2F+5%21%29*e%5E-0.24\n \\item[\\cref{item:stats:cars:2}.] $\\lambda = \\frac{\\num{0.2} \\times 1\\,\\text{car observed}}{\\SI{5}{\\minute}} \\times \\SI{6}{\\minute} = \\num{0.24}$, $P_{\\text{Poisson}} = \\num{5.2e-6}$.\n % https://www.wolframalpha.com/input/?i=1+-+3e%5E-2\n \\item[\\cref{item:stats:cars:3}.] $\\lambda = \\frac{1}{\\SI{5}{\\minute}} \\times \\SI{10}{\\minute} = \\num{2}$, $P\\left(2 \\leq k \\right) = 1 - P_{\\text{Poisson}}\\left(k < 2 \\right) = 1 - \\left( \\frac{2^{0}}{0!} e^{-2} + \\frac{2^{1}}{1!} e^{-2}\\right) \\newline = 1 - 3 e^{-2} = \\num{0.6}$.\n \\item[\\cref{item:stats:cars:4}.] The cars are independent, so the prior $k$ cars are irrelevant -- gambler's fallacy. $\\expval{k} = \\lambda = \\frac{1}{\\SI{5}{\\minute}} \\times \\SI{60}{\\minute} = \\num{12}$.\n \\item[\\cref{item:stats:cars:5}.] We can use the exponential distribution with $\\lambda_{\\text{Exp}} = \\frac{1}{\\SI{5}{\\minute}}$, $\\expval{t} = \\frac{1}{\\lambda} = \\SI{5}{\\minute}$,\\newline as one would expect\\ldots\n\\end{itemize}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Gaussian Distribution}\n\\label{stats:gaus}\n\nIn the other direction, for common processes $\\mu \\equiv n\\,p \\gg 1$,\nthe binomial distribution becomes\\footnote{See\none \\href{http://scipp.ucsc.edu/~haber/ph116C/NormalApprox.pdf}{derivation here}.} the\nwell-known Gaussian, or normal, distribution,\n\n\\begin{equation}\\label{eq:stats:gaus:P}\nP\\left(x;\\,\\mu,\\sigma\\right) = \\frac{1}{\\sqrt{2\\pi}\\,\\sigma} \\exp\\left( -\\frac{1}{2} \\left(\\frac{x-\\mu}{\\sigma}\\right)^{2} \\right)\\,,\n\\end{equation}\n\n\\noindent with mean $\\mu$ and variance $\\sigma^{2}$.\nA plot of the distribution can be found in \\cref{fig:dist:gaus}.\n\nIt is helpful to remember that\n\n\\begin{equation}\\label{eq:stats:gaus:sigmas}\n\\begin{split}\n\\mu \\pm 1\\sigma &\\approx \\SI{68}{\\percent}\\,, \\\\\n\\mu \\pm 2\\sigma &\\approx \\SI{95}{\\percent}\\,, \\\\\n\\mu \\pm 3\\sigma &\\approx \\SI{99}{\\percent}\\,,\n\\end{split}\n\\end{equation}\n\n\\noindent while the full width at half maximum (FWHM) $\\Gamma = 2\\sqrt{2 \\ln{2}}\\,\\sigma \\approx \\num{2.355} \\sigma$.\nNote that the standard normal distribution has $\\mu = 0$ and $\\sigma =1$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Student's \\texorpdfstring{$t$}{t}-Distribution}\n\\label{stats:t_dist}\n\nIf we take a sample of size $m$ from a normal distribution\nwe arrive at Student's \\tdist.\nLetting $\\nu = m-1$ be the number of degrees of freedom,\nStudent's \\tdist has the form:\n\n\\begin{equation}\\label{eq:stats:t_dist:P}\nP\\left(t;\\,\\nu\\right) = \\frac{\n\\Gamma\\left(\\frac{\\nu+1}{2}\\right)\n}{\n\\sqrt{\\nu \\pi}\\,\\Gamma\\left(\\frac{\\nu}{2}\\right)\n} \\left(1+\\frac{t^{2}}{\\nu}\\right)^{-\\frac{\\nu+1}{2}}\\,,\n\\end{equation}\n\n\\noindent where $\\Gamma\\left(z\\right)$ is the gamma function\\footnote{$\\Gamma\\left(z\\right) = \\int_{0}^{\\infty} x^{z-1} e^{-x} \\, \\dd{x}$,\nwhich simplifies to $\\Gamma\\left(n\\right)=\\left(n-1\\right)!$ for integer $n$.}.\nThe distribution, \\cref{fig:dist:student_t}, has a mean of \\num{0} for $\\nu > 1$,\nand variance of $\\infty$ for $\\nu =2$ and $\\nu / \\left(\\nu-2\\right)$ for $\\nu > 2$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{\\texorpdfstring{$\\chi^{2}$-Distribution}{Chi-Squared Distribution}}\n\\label{stats:chi2_dist}\n\nThe \\chiSqdist with $\\nu$ degrees of freedom,\n\\cref{eq:stats:chi2_dist:P} and \\cref{fig:dist:chi2},\nis created by summing the squares of $\\nu$ independent standard normal random variables.\nIt therefore has a mean of $\\nu$ and variance of $2\\nu$, and is\nprimary useful for conducting {\\chiSqtest}s.\n\n\\begin{equation}\\label{eq:stats:chi2_dist:P}\nP\\left(x;\\,\\nu\\right) = \\frac{\nx^{\\frac{\\nu}{2} - 1} e^{-\\frac{x}{2}}\n}{\n2^{\\frac{\\nu}{2}} \\Gamma\\left(\\frac{\\nu}{2}\\right)}.\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{\\texorpdfstring{$F$}{F}-Distribution}\n\\label{stats:F_dist}\n\nThe \\Fdist with degrees of freedom $d_{1}$ and $d_{2}$,\n\\cref{eq:stats:F_dist:X,eq:stats:F_dist:P} and \\cref{fig:dist:F},\nis formed by dividing two $\\chi^{2}$-distributed random variables\n$S_{1}$ and $S_{2}$ with their respective degrees of freedom:\n\n\\begin{subequations}\\label{eq:stats:F_dist}\n\\begin{align}\nX_{\\Fdist} &= \\frac{S_{1}/d_{1}}{S_{2}/d_{2}}, \\label{eq:stats:F_dist:X} \\\\\nP\\left(x;\\,d_{1},d_{2}\\right) &= \\frac{1}{x B\\left(d_{1}/2, d_{2}/2\\right)} \\sqrt{\\frac{\\left(d_{1} x\\right)^{d_{1}} d_{2}^{d_{2}}}{\\left(d_{1} x + d_{2}\\right)^{d_{1}+d_{2}}}}, \\label{eq:stats:F_dist:P}\n\\end{align}\n\\end{subequations}\n\n\\noindent where $B\\left(x,y\\right)$ is the beta function\\footnote{$B\\left(x,y\\right) = \\int_{0}^{1} t^{x-1} \\left(1-t\\right)^{y-1} \\, \\dd{t} = \\Gamma\\left(x\\right)\\Gamma\\left(y\\right) / \\Gamma\\left(x+y\\right)$.}.\n\nThe \\Fdist has a complicated\nmean and variance\\footnote{Mean: $d_{2}/\\left(d_{2}-2\\right)$ for $2 < d_{2}$.\nVariance: $2 d_{2}^{2}\\left(d_{1}+d_{2}-2\\right)/\\left(d_{1}\\left(d_{2}-2\\right)^{2}\\left(d_{2}-4\\right)\\right)$ for $4 < d_{2}$.},\nand is primary useful for conducting {\\Ftest}s such as ANOVA.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Geometric Distribution}\n\\label{stats:geo_dist}\n\nThe geometric distribution gives the probability of\nrequiring $k$ independent Boolean trials to observe one success\\footnote{With the success occurring in trial $k$.},\n\\cref{eq:stats:geo:P_trials} and \\cref{fig:dist:geometric_trials},\nor the probability of observing $k$ failed trials before the first success,\n\\cref{eq:stats:geo:P_failures} and \\cref{fig:dist:geometric_failures}.\nAs usual $p$ is the probability of success for any single trial.\nNote that $k$ is indexed differently depending on the framing of the problem.\n\n\\begin{subequations}\\label{eq:stats:geo}\n\\begin{align}\nP_{\\text{Trials}}\\left(k;\\,p\\right) &= \\left(1-p\\right)^{k-1}\\,p,\\quad k \\in \\left\\{1,2,3,\\ldots\\right\\}, \\label{eq:stats:geo:P_trials} \\\\\nP_{\\text{Failures}}\\left(k;\\,p\\right) &= \\left(1-p\\right)^{k}\\,p,\\quad k \\in \\left\\{0, 1,2,3,\\ldots\\right\\}. \\label{eq:stats:geo:P_failures}\n\\end{align}\n\\end{subequations}\n\nThe mean and standard deviation of the geometric distribution are:\n\n\\begin{subequations}\\label{eq:stats:geo_dist:mean_variance}\n\\begin{gather}\n\\expval{k}_{\\text{Trials}} = \\frac{1}{p}\\,, \\quad \\expval{k}_{\\text{Failures}} = \\frac{1-p}{p}\\,, \\label{eq:stats:geo_dist:mean} \\\\\n\\sigma^{2} = \\frac{1-p}{p^{2}}\\,. \\label{eq:stats:geo_dist:variance}\n\\end{gather}\n\\end{subequations}\n\nLike the exponential distribution, its continuous analogue, the geometric distribution is memoryless \\cref{eq:stats:exp:memoryless}.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Hypergeometric Distribution}\n\\label{stats:hypergeo_dist}\n\nThe hypergeometric distribution, \\cref{eq:stats:hypergeometric:P} and \\cref{fig:dist:hypergeometric},\ngives the probability of obtaining $k$ ``successes'' in $n$ draws, {\\em without replacement},\nfrom a population of size $N$ which contains $K$ successes.\nOne example is getting $k$ red balls, \\ie successes,\nout of $n$ draws from a bag of $K$ red and $N-K$ white balls.\n\n\\begin{equation}\\label{eq:stats:hypergeometric:P}\nP\\left(k;\\,N,K,n\\right) = \\frac{\\binom{K}{k}\\binom{N-K}{n-k}}{\\binom{N}{n}}\\,.\n\\end{equation}\n\nThe mean and standard deviation of the hypergeometric distribution are:\n\n\\begin{subequations}\\label{eq:stats:hypergeometric:neg:mean_variance}\n\\begin{gather}\n\\expval{k} = n K / N = n p, \\label{eq:stats:hypergeometric:mean} \\\\\n\\sigma^{2} = n \\frac{K}{N} \\frac{N-K}{N} \\frac{N-n}{N-1} = n p \\left(1-p\\right) \\frac{N-n}{N-1}\\,. \\label{eq:stats:hypergeometric:variance}\n\\end{gather}\n\\end{subequations}\n\nThe hypergeometric distribution should not be confused with the geometric distribution,\nand is far more similar to the binomial distribution where $p = K/N$ and we draw with replacement.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Kurtosis and Skewness}\n\\label{stats:kurtosis_skewness}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Confidence Intervals}\n\\label{stats:CI}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Hoeffding's Inequality}\n\\label{stats:CI:hoeffding}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Outliers}\n\\label{stats:outliers}\n% TODO\n% interquartile range (IQR)\n\n\\begin{equation}\\label{eq:stats:outlier_IQR_def}\n\\expval{x} \\pm \\num{1.5}\\, \\text{IQR}\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Bootstrapping}\n\\label{stats:bootstrapping}\n\nThe bootstrap method can be a great solution when presented with difficult problems\naround measuring uncertainties or even conducting hypothesis tests.\nWith bootstrapping we can estimate the uncertainty on any well defined quantity,\neven when the uncertainty itself does not have an explicit expression,\n\\eg finding the confidence intervals for the median\nor coefficient of determination $R^{2}$ of linear regression.\n\nWe begin with an original sample of size $n$ drawn from the larger population,\nand generate $1 \\ll m$ bootstrap samples by resampling $n$ points from the sample with replacement,\nas illustrated in \\cref{fig:bootstrapping}.\nThe statistic of interest is computed on each of the bootstrapped samples\nbefore being combined into an overall distribution.\nAssuming the original population is independent and identically distributed (\\iid)\\footnote{Note\nthat the \\iid assumption is less stringent than\nthe usual assumption of normality via the central limit theorem (CLT) of \\cref{stats:CLT},\nallowing bootstrapping to be used in cases were $n$ may be to small for other methods.},\nthe resulting bootstrapped distribution will approximate the true sampling distribution\nand we can use it to estimate the uncertainty on the statistic.\nThe \\SI{95}{\\percent} confidence interval of the statistic in question\ncan be found by simply\\footnote{There are\nmore advanced methods for determining the confidence interval not covered here,\nparticularly for asymmetric distributions.} identifying\nthe boundaries of the central \\SI{95}{\\percent} of bootstrapped sample values.\nLikewise, the standard error of the statistic is the standard deviation of the bootstrapped distribution.\n\nIf the bootstrapped distribution is not symmetrical,\nwhich may be the case when working with measures of variance like the standard deviation,\nthe bootstrapped result may be biased.\nIn these cases, one potential correction is to shift the bootstrapped distribution\nsuch that the original sample statistic and mean of the bootstrapped distribution agree.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.7\\textwidth]{figures/stats/bootstrapping.jpeg}\n\\caption{\nIllustration of the bootstrapping method\nby \\href{https://towardsdatascience.com/bootstrapping-statistics-what-it-is-and-why-its-used-e2fa29577307}{Trist'n Joseph}.\nNote that the $m$ resamples are done with replacement.\nThe final bootstrapped distribution is built by measuring the statistic in question on all of the bootstrapped samples,\nand can be used to estimate the statistic's uncertainty.\n}\n\\label{fig:bootstrapping}\n\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Hypothesis Testing}\n\\label{stats:bootstrapping:hypo}\n\nAs we can construct confidence intervals, bootstrapping can also be used for hypothesis testing.\nN{a\\\"i}vely we can test the null hypothesis by checking if the value in question,\ntypically \\num{0}, lands within the confidence interval or not.\nWe can also estimate the \\pvalue of a parameter by shifting the original sample distribution such that\nit satisfies the null hypothesis, \\eg shift all the data points by $\\delta$ such that the sample has $\\expval{x} = \\num{0}$.\nWe then bootstrap the shifted sample to produce the bootstrapped distribution as normal.\nThe \\pvalue is then the proportion of the bootstrapped distribution\nwith values differing from the null hypothesis by the observed difference or more,\n\\eg $P\\left(\\delta \\leq \\abs{x}\\right)$ in the prior example.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Poisson Bootstrapping}\n\\label{stats:bootstrapping:poisson}\n\nFor large sample sizes $n$, instead of drawing $m$ bootstrap samples with replacement,\nwhich can be computationally expensive,\nwe can use the single sample,\nbut give each data point a weight $w$ drawn\nfrom the Poisson distribution with $\\lambda = 1$.\nThe set of weights is then generated $m$ times.\nFormally, the methods are equivalent as for large $n$\nthe binomial probability of including a given data point approaches the Poisson probability with $\\lambda = 1$.\nPoisson Bootstrapping is commonly used in particle physics,\nin particular see the direct balance $\\gamma\\text{+Jet}$ calibration\nin the dissertation \\cite{mepland_dissertation}, Appendix E.1.4.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Bayesian Statistics}\n\\label{stats:Bayes}\n% TODO\n\n% \\cref{stats:Bayes_rule}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Bayesian Conjugate Prior}\n\\label{stats:Bayes:conjugate_prior}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Bayes Estimator}\n\\label{stats:Bayes:estimator}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Approximate Bayesian Computation (ABC)}\n\\label{stats:Bayes:ABC}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Bias of a Predictor}\n\\label{stats:bias}\n% TODO\n\n\\begin{equation}\\label{eq:stats:bias}\n\\bias{\\hat{f}\\left(x\\right)} = \\expval{\\hat{f}\\left(x\\right)} - f\\left(x\\right)\n\\end{equation}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Kullback-Leibler Divergence}\n\\label{stats:kld}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Causal Inference}\n\\label{stats:causal_inference}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Markov Chains}\n\\label{misc:markov_chains}\n% TODO\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Markov Chain Monte Carlo (MCMC)}\n\\label{misc:MCMC}\n% TODO\n\n% Markov Chain Monte Carlo (MCMC) is a Monte Carlo (MC) method\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{\\pymcThree}% would rather have the \\textsc caps than italics\n\\label{misc:MCMC:PyMC3}\n% TODO\n% TODO \\pymcThree\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Sampling from Probability Distributions}\n\\label{misc:sampling_prob_dist}\n\nMost programming languages have built-in functions to generate\npseudo-random numbers from common probability distributions,\nsuch as the Poisson \\cref{eq:stats:poisson:P} or Gaussian \\cref{eq:stats:gaus:P} distributions.\nHowever, in some cases we may wish to sample from an unsupported esoteric function.\nIn these situations we can turn to inverse transform sampling and rejection sampling,\nto give just two examples from many possible computational methods,\nto construct the desired distribution from an existing random number generator.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Inverse Transform Sampling}\n\\label{misc:sampling_prob_dist:inverse}\n% https://www.youtube.com/watch?v=9ixzzPQWuAY\n\nIf we know the explicit form of the target probability density function (PDF), $X = P\\left(x\\right)$,\ncan integrate it to find the cumulative distribution function (CDF), $F_{X}\\left(x\\right) = \\int_{-\\infty}^{x} P\\left(t\\right) \\, \\dd{t}$,\nand furthermore can invert the CDF\\footnote{The inverse CDF\nis known as the percent point function, \\texttt{ppf},\nin \\href{https://docs.scipy.org/doc/scipy/reference/stats.html}{\\scipy}.}, $F^{-1}_{X}\\left(u\\right)$ for $0 \\leq u \\leq 1$,\nwe can explicitly transform the uniform distribution $U\\left(x\\right)$ into $P\\left(x\\right)$ as:\n\n\\begin{equation}\\label{eq:stats:sampling_prob_dist:inverse}\nF^{-1}_{X}\\left(U\\right) = P\\left(x\\right) = X\\,.\n\\end{equation}\n\nTo prove the method, assuming that $F^{-1}_{X}$ exists, we can do:\n\n\\begin{subequations}\\label{eq:stats:sampling_prob_dist:inverse_proof}\n\\begin{align}\nP\\left(F^{-1}_{X}\\left(U\\right) \\leq x\\right) &= P\\left(U \\leq F_{X}\\left(x\\right) \\right) \\label{eq:stats:sampling_prob_dist:inverse_proof:inverse} \\\\\n&= F_{X}\\left(x\\right)\\,,\\label{eq:stats:sampling_prob_dist:inverse_proof:def} \\\\\nP\\left(U \\leq y\\right) &= y\\,, \\label{eq:stats:sampling_prob_dist:inverse_proof:U}\n\\end{align}\n\\end{subequations}\n\n\\noindent where in \\cref{eq:stats:sampling_prob_dist:inverse_proof:inverse} we have applied $F$ to both sides of the inner inequality,\nand in \\cref{eq:stats:sampling_prob_dist:inverse_proof:def} we have used the definition of the uniform distribution \\cref{eq:stats:sampling_prob_dist:inverse_proof:U}.\nAs $P\\left(F^{-1}_{X}\\left(U\\right) \\leq x\\right) = F_{X}\\left(x\\right) = P\\left(X \\leq x\\right)$\nwe can compare terms and see \\cref{eq:stats:sampling_prob_dist:inverse}.\n\nThe inverse sampling method can be seen graphically \\cref{fig:stats:sampling_prob_dist:inverse} being used to generate the normal distribution.\n\n%\\begin{figure}\n%\\centering\n%\\includegraphics[width=0.7\\textwidth]{figures/stats/inverse_transform_sampling_normal_dist}\n%\\caption{\n%Example application of the inverse sampling method to generate the normal distribution, adapted from \\href{https://en.wikipedia.org/wiki/File:Inverse_transform_sampling.png}{Olivier Ricou}.\n%}\n%\\label{fig:stats:sampling_prob_dist:inverse}\n%\\end{figure}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Rejection Sampling}\n\\label{misc:sampling_prob_dist:reject}\n% https://www.youtube.com/watch?v=OXDqjdVVePY\n\nWhile inverse transform sampling is a computationally efficient method\nits assumptions will not be met by all interesting PDFs,\nin particular if we do not have an explicit, invertible CDF.\nIn these cases we can turn to rejection sampling instead,\nwhich can handle more general PDFs at the cost of a lower computational efficiency.\n\nIn rejection sampling, we assume we know the PDF of the random variable to be sampled from,\nup to a normalization constant $A$ \\cref{eq:stats:sampling_prob_dist:reject:X}.\nWe then choose a different PDF, $g\\left(x\\right)$ which is easy for us to sample from.\nAny $g\\left(x\\right)$ covering the range of $x$ will do,\nbut the method is more efficient the closer we can get $g\\left(x\\right)$ to match $f\\left(x\\right)$.\n$g\\left(x\\right)$ is then scaled by a known constant $M$\nsuch that it is always larger than $f\\left(x\\right)$ \\cref{eq:stats:sampling_prob_dist:reject:f_condition}\nas illustrated in \\cref{fig:stats:sampling_prob_dist:reject}.\nFinally, we sample random $x$ values from $g\\left(x\\right)$ and accept them with probability \\cref{eq:stats:sampling_prob_dist:reject:P_accept}.\nThe accepted $x$ values will have the same distribution as $X$.\n\n\\begin{subequations}\\label{eq:stats:sampling_prob_dist:reject}\n\\begin{align}\nX = P\\left(x\\right) &= \\frac{1}{A} f\\left(x\\right), \\label{eq:stats:sampling_prob_dist:reject:X} \\\\\n\\forall x, \\quad f\\left(x\\right) & \\leq M g\\left(x\\right), \\label{eq:stats:sampling_prob_dist:reject:f_condition} \\\\\nP\\left(\\text{Accept} \\mid x\\right) &= \\frac{f\\left(x\\right)}{M g\\left(x\\right)}\\,. \\label{eq:stats:sampling_prob_dist:reject:P_accept}\n\\end{align}\n\\end{subequations}\n\n\\begin{figure}[H]\n \\centering\n \\savebox{\\largestimage}{\n \\includegraphics[width=0.47\\textwidth,trim={3.0cm 0.5cm 3.0cm 0.1cm},clip]{figures/stats/inverse_transform_sampling_normal_dist}% trim={ }\n }% Store largest image in a box\n\n \\begin{subfigure}[b]{0.48\\textwidth}\\centering\n \\usebox{\\largestimage}\n \\vspace{0.01cm}\n \\caption{Inverse Sampling}\n \\label{fig:stats:sampling_prob_dist:inverse}\n \\end{subfigure}\n ~\n \\begin{subfigure}[b]{\\wd\\largestimage}\\centering\n \\raisebox{\\dimexpr.5\\ht\\largestimage-.5\\height}{% Adjust vertical height of smaller image\n \\includegraphics[width=\\textwidth]{figures/stats/rejection_sampling}}\n \\caption{Rejection Sampling}\n \\label{fig:stats:sampling_prob_dist:reject}\n \\end{subfigure}\n\\caption{\nIllustrations of the inverse sampling and rejection sampling methods, adapted from \\href{https://en.wikipedia.org/wiki/File:Inverse_transform_sampling.png}{Olivier Ricou} and \\href{https://www.data-blogger.com/2016/01/24/the-mathematics-behind-rejection-sampling/}{Kevin Jacobs}.\n \\label{fig:stats:sampling_prob_distr}\n}\n\\end{figure}\n\nFor a proof of why the rejection sampling method works see \\cref{eq:stats:sampling_prob_dist:reject:proof} below\nwhich hinges on Bayes' Theorem from \\cref{stats:Bayes_rule} in \\cref{eq:stats:sampling_prob_dist:reject:proof:x_given_Accept}\nalong with other basic definitions from probability theory.\n\n\\begin{subequations}\\label{eq:stats:sampling_prob_dist:reject:proof}\n\\begin{align}\nP\\left(x \\mid \\text{Accept}\\right) &= \\frac{P\\left(\\text{Accept} \\mid x\\right) P\\left(x\\right)}{P\\left(\\text{Accept}\\right)} = \\frac{1}{P\\left(\\text{Accept}\\right)} \\frac{f\\left(x\\right)}{M \\cancel{g\\left(x\\right)}} \\cancel{g\\left(x\\right)}, \\label{eq:stats:sampling_prob_dist:reject:proof:x_given_Accept} \\\\\nP\\left(\\text{Accept}\\right) &= \\int_{-\\infty}^{\\infty} P\\left(\\text{Accept} \\mid x\\right) g\\left(x\\right) \\, \\dd{x} \\label{eq:stats:sampling_prob_dist:reject:proof:P_Accept1} \\\\\n&= \\int_{-\\infty}^{\\infty} \\frac{f\\left(x\\right)}{M g\\left(x\\right)} g\\left(x\\right) \\, \\dd{x} = \\frac{1}{M} \\int_{-\\infty}^{\\infty} f\\left(x\\right) \\, \\dd{x} = \\frac{A}{M}, \\label{eq:stats:sampling_prob_dist:reject:proof:P_Accept2} \\\\\n\\implies P\\left(x \\mid \\text{Accept}\\right) &= \\frac{f\\left(x\\right)/M}{A/M} = \\frac{1}{A} f\\left(x\\right) = P\\left(x\\right) = X. \\label{eq:stats:sampling_prob_dist:reject:proof:conclusion}\n\\end{align}\n\\end{subequations}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Metropolis-Hastings Algorithm}\n\\label{misc:sampling_prob_dist:metropolis_hastings}\n% TODO\n", "meta": {"hexsha": "192560dc569c9f6a5ef61877df3b809b729e9fe6", "size": 49115, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/stats.tex", "max_stars_repo_name": "mepland/data_science_notes", "max_stars_repo_head_hexsha": "f529a86490110fc6a30d1af6d37c0add2517244f", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-05-30T15:15:50.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-23T01:01:08.000Z", "max_issues_repo_path": "sections/stats.tex", "max_issues_repo_name": "mepland/data_science_notes", "max_issues_repo_head_hexsha": "f529a86490110fc6a30d1af6d37c0add2517244f", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, 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YES\n2. YES", "lm_q1_score": 0.7154239836484144, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.6163650978564735}} {"text": "\\section{Hierarchical Information Content}\n\\label{sec:hic}\n\nWe construct a measure for the structural complexity of a system. As motivated\nin section~\\ref{sec:finding_complexity}, we want the complexity to grow at\nthe compositions of ordered and disordered subsystems in a hierarchical\nstructure. We construct the overall complexity from the sum of the complexity at\neach composition.\n\nAssume we are given a hierarchical system with $L$ levels. We construct the\nglobal system complexity from the local complexity computed at each of the\n$L-1$ compositions between consecutive levels. The local complexity measures\nare given by the difference in mutual information of the system at level\n$\\ell$ and that of the subsystem at level $\\ell - 1$. We denote by $X^{\\ell}$ the\nstate variable at the $\\ell$-th level of the hierarchy. Following\n\\citet{simon1991architecture}, we define the span of a level as the\nnumber of elements from the $\\ell - 1$ subsystem used to construct the state\n$X^\\ell$. For now we assume the span to be constant with a value of two for\nevery level. We later generalize to larger spans.\n\nAs an example, the state variable $X^\\ell$ could be a molecule in an artificial\nchemistry constructed from two atoms. Alternatively, $X^{\\ell}$ could be an $n\n\\times n$ state of a cellular automata constructed from two $(\\nicefrac{n}{2})\n\\times n$ sub-states. As an example from nature, $X^{\\ell}$ could be a string\nof $2^\\ell$ base pairs of DNA constructed from two substrings of length $2^{\\ell\n- 1}$.\n\nThe complexity at the $\\ell$-th composition is computed from the squared\ndifference of the mutual information of the span between level $\\ell$ and level\n$\\ell + 1$:\n\\begin{equation}\n C_\\ell = \\left[ I(X^{\\ell + 1}; Y^{\\ell + 1}) - I(X^{\\ell}; Y^{\\ell}) \\right]^2\n\\end{equation}\nThe mutual information can be computed from the Shannon entropy and conditional\nentropy when $X^\\ell$ is discrete or the differential entropy when $X^\\ell$ is\ncontinuous~\\citep{cover1999elements}:\n\\begin{equation}\n\\label{eq:mutual_information}\nI(X^\\ell; Y^\\ell) = H(X^\\ell) - H(X^\\ell \\mid Y^\\ell).\n\\end{equation}\nThe mutual information is large when the $\\ell$-th level is highly ordered and\nsmall when it is highly disordered. Hence, the values $C_\\ell$ will be large\nexactly at the transition points between order and disorder (or vice-versa). We\nuse the level complexities $C_\\ell$ to construct the overall system complexity.\n\n\\begin{definition}[Hierarchical Information Content]\n\\label{def:hic}\n The Hierarchical Information Content (HIC) of a hierarchical system $S$ with\n $L$ levels is given by:\n \\begin{equation}\n \\label{eq:hic}\n \\hic(S) = \\sum_{l=1}^{L-1} C_\\ell = \\sum_{l=1}^{L-1} \\left[ I(X^{\\ell+1}; Y^{\\ell+1}) - I(X^\\ell; Y^\\ell) \\right]^2.\n \\end{equation}\n\\end{definition}\n\n\\paragraph{Generalizing to larger spans:} Using a larger span requires\ngeneralizing the mutual information to multiple variables. The interaction\ninformation is one such generalization~\\citep{mcgill1954multivariate}. The\ninteraction information of a set of variables is defined recursively as:\n\\begin{equation}\n \\label{eq:interaction_information}\n I(X_1^\\ell; \\ldots; X_{n+1}^\\ell) = I(X_1^\\ell; \\ldots; X_n^\\ell) - I(X_1^\\ell; \\ldots; X_n^\\ell \\mid X_{n+1}^\\ell),\n\\end{equation}\nwhere the conditional mutual information is given by:\n\\begin{equation}\n \\label{eq:conditional_mutual_information}\n I(X; Y \\mid Z) = \\sum_{Z} P(Z) I(X; Y).\n\\end{equation}\n\n\\paragraph{Estimating information criteria:} Computing the mutual information\nin equation~\\ref{eq:mutual_information} requires computing the entropy of the\nstate variables $H(X^\\ell)$ and the conditional entropy $H(X^\\ell \\mid\nY^\\ell)$. Estimating these quantities directly requires models for the\ndistributions $P(X^\\ell)$ and $P(X^\\ell \\mid Y^\\ell)$. For state variables with\na small domain these distributions can likely be estimated efficiently from\ncounts. However, for larger state variables sample efficiency becomes an issue.\nIn these cases more sophisticated density estimation techniques should be used.\nSuitable techniques can be found in most modern statistical learning or machine\nlearning texts (see for example \\citet{friedman2001elements}).\n\n\\paragraph{Specifying the hierarchy:} In some cases, the hierarchical structure\nof the system may be self-evident. This might be the case if the system has a\nnatural tree-like structure. However, in the general case, both the boundaries\nbetween levels and the sub-modules which make up the state at the next level up\nmust be specified. This is introduces a degree subjectivity to the HIC which\neffects its ability to faithfully capture the complexity of the system.\n\nAs an example consider measuring the HIC of a sequence of DNA. We might define\nthe state $X^\\ell$ to be a subsequence of $2^\\ell$ nucleotides. On the other\nhand there may be some natural structure in the DNA to take advantage of. For\nexample, codons are sequences of three nucleotides which each code for a\nspecific amino acid. These would make a good definition for a state variable.\n\n\\subsection{Examples}\n\nWe attempt to build intuition for the definition of HIC through some simple\nexamples.\n\n\\begin{example}\n \\label{ex:constant}\n As a first example, consider the state $S = [0, 0, \\ldots, 0]$,\n a constant sequence of all $0$s. At any level for any neighborhood, the mutual\n information $I(X^\\ell; Y^\\ell) = 0$. Hence the terms $C_\\ell = 0$ for all\n $\\ell$ and the overall $\\hic(S) = 0$.\n\\end{example}\n\n\\begin{example}\n \\label{ex:uniform}\n Let $S = [x_1, x_2, \\ldots]$ consist of a sequence of independent draws from a\n multinomial uniform distribution over $K$ categories. Assuming a span of $2$,\n the variable $X^\\ell$ consists of $2^\\ell$ of the $x_i$ primitives and hence\n can take on any of $K^{(2^\\ell)}$ values. Consider the mutual information of level\n $\\ell$:\n \\begin{equation}\n I(X^\\ell; Y^\\ell) = H(X^\\ell) - H(X^\\ell \\mid Y^{\\ell}) = 2^\\ell \\log K - 2^\\ell \\log K = 0\n \\end{equation}\n Hence all of the $C_\\ell = 0$ and $\\hic(S) = 0$.\n\\end{example}\n\nAs examples~\\ref{ex:constant} and \\ref{ex:uniform} show, the HIC behaves as\nexpected when the system exhibits complete order or complete disorder. In both\nof these examples the HIC was zero because the mutual information at each level\nwas zero. In the following example, we see slightly more interesting behavior,\nwhere the mutual information at each level is nonzero, but the resulting HIC\nis.\n\n\\begin{example}\n \\label{ex:repeats}\n Let $S = [0, 1, 0, 1, \\ldots]$ consist of a sequence of alternating zeros and\n ones.\n\n The first level $X^1$ takes on the two values $\\{0, 1\\}$ with equal\n probability hence $H(X^1) = \\log 2$. However given the neighbor $Y^1$ the\n value of $X^1$ is deterministic and hence $H(X^1 \\mid Y^1) = 0$. Thus the\n overall mutual information is $I(X^1; Y^1) = \\log 2$.\n\n The second level $X^2$ takes on two values $\\{[0, 1], [1, 0]\\}$ with\n equal probability, and we have $H(X^2) = \\log 2$. Again $X^2$ is fully\n determined by $Y^2$ thus $H(X^2 \\mid Y^2) = 0$ and $I(X^2, Y^2) = \\log 2$.\n Combining these to produce $C^2$ yields:\n \\begin{equation}\n C_2 = (I(X^2; Y^2) - I(X^1; Y^1))^2 = (\\log 2 - \\log 2)^2 = 0.\n \\end{equation}\n The above argument generalizes to all levels of the hierarchy so $\\hic(S) = 0$.\n\\end{example}\n\nIn example \\ref{ex:repeats} we used overlapping windows to construct $X^2$. If\nwe had chosen disjoint windows the HIC would be small but nonzero. Depending on\nthe problem this may be desirable in that the alternating sequence should be\nconsidered slightly more complex than a constant sequence. This is where\nsubjectiveness comes in. The designer should select from these choices\nbased on the setting at hand.\n\n\\subsection{Comparison to Statistical Complexity}\n\\label{statistical_complexity}\n\nThe statistical complexity of \\citet{crutchfield1989inferring} measures the\ncomplexity of a system by observing the number of distinct future distributions\ngiven the present and past states. Statistical complexity infers a set of\n``causal states'' by collapsing the state variables representing the current\nand past time into equivalence classes which all lead to indistinguishable\nfuture distributions. The entropy of the distribution of these causal states\nthen serves as a measure of the complexity of the process. For futures which\nare ordered or unchanging, the system will have one or a small number of causal\nstates and the entropy will be small. Similarly, for futures which are\nuniformly random, the future distributions will be indistinguishable given the\npast and present, and the states will similarly collapse into a single or small\nnumber of causal states. Thus random processes will also have low statistical\ncomplexity.\n\nLet $X_t$ represent the state variable at the $t$-th time step for a system\nwith $N$ state variables per time step. The past light cone\n$\\mathcal{X}_{\\textrm{P}}$ of $X_t$ with a history of $T_{\\textrm{P}}$ time\nsteps is given by:\n\\begin{equation}\n \\mathcal{X}_{\\textrm{P}} =\n \\{X_{i, j} \\mid i = t - 1, \\ldots, t - T_{\\textrm{P}}\n \\;\\;\\land\\;\\; j = 1, \\ldots, N\n \\;\\;\\land\\;\\; X_{i, j} \\rightarrow X_t \\},\n\\end{equation}\nwhere $A \\rightarrow B$ means the state $A$ can influence $B$ (\\emph{i.e.} $B$\nis a function of $A$). The future light cone $\\mathcal{X}_{\\textrm{F}}$ of\n$X_t$ with a future of $T_{\\textrm{F}}$ time steps is similarly defined as:\n\\begin{equation}\n \\mathcal{X}_{\\textrm{F}} =\n \\{X_{i, j} \\mid i = t + 1, \\ldots, t + T_{\\textrm{F}}\n \\;\\;\\land\\;\\; j = 1, \\ldots, N\n \\;\\;\\land\\;\\; X_t \\rightarrow X_{i, j} \\}.\n\\end{equation}\nIn words, the past light cone of $X_t$ is the set of states which can influence\n$X_t$ and the future light cone is the set of states which can be influenced by\n$X_t$.\n\nEach past light cone and state variable $X_t$ defines a conditional\ndistribution over future light cones $P(\\mathcal{X}_{\\textrm{F}} \\mid\n\\mathcal{X}_{\\textrm{P}}, X_t)$. We can measure the similarity between two such\nconditional distributions using a statistical divergence $D(P_i \\| P_j)$. A\ncausal state $C$ represents a set of states $C = \\{(\\mathcal{X}_{\\textrm{P}},\nX_t)\\}$ where $D(P_i \\| P_j) = 0$ for every pair $(\\mathcal{X}_{\\textrm{P}},\nX_t)_i$ and $(\\mathcal{X}_{\\textrm{P}}, X_t)_j$ in $C$. In other words, the\ndistribution of future light cones of $(\\mathcal{X}_{\\textrm{P}}, X_t)_i$ and\n$(\\mathcal{X}_{\\textrm{P}}, X_t)_j$ are indistinguishable. In practice, we\nrelax $D(P_i \\| P_j) < \\tau$ for some predefined threshold due to finite\nsample sizes. Furthermore, we only require that $D(P_i \\| P_j) < \\tau$ be\nsatisfied for a single $(\\mathcal{X}_{\\textrm{P}}, X_t)_j$ in $C$ instead of\nall such states.\n\nUnlike HIC, statistical complexity is a measure of process complexity. We can\nuse it to observe how the complexity of a system changes over time. We can also\nuse statistical complexity as a measure of the global complexity of the system\nby applying it to every state at every time step; as we do in this work.\n\nPerhaps more importantly, statistical complexity can be computed in many\ncases without much difficulty. The distributions of future light cones given\npast light cones and state variables can be estimated for reasonable values of\n$T_\\textrm{P}$ and $T_\\textrm{F}$. \\citet{shalizi2004quantifying} estimate the\nlocal statistical complexity of circular cellular automata and show\nqualitatively that the measure correlates well to with the perceived complexity\nof the automata.\n", "meta": {"hexsha": "bc54c01948d3e616d159d2d5a0c8404ff4688033", "size": 11541, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/hic.tex", "max_stars_repo_name": "awni/hic", "max_stars_repo_head_hexsha": "74ba29fe55e8592424557558253208f25df640c8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/hic.tex", "max_issues_repo_name": "awni/hic", "max_issues_repo_head_hexsha": "74ba29fe55e8592424557558253208f25df640c8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/hic.tex", "max_forks_repo_name": "awni/hic", "max_forks_repo_head_hexsha": "74ba29fe55e8592424557558253208f25df640c8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 52.4590909091, "max_line_length": 120, "alphanum_fraction": 0.7341651503, "num_tokens": 3273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8856314647623016, "lm_q2_score": 0.6959583187272711, "lm_q1q2_score": 0.6163625852279419}} {"text": "%\n% 111\n%\n\\chapter{The Theory of Residues; Application to the Evaluation of Definite Integrals}\n\n\\Section{6}{1}{Residues}\n\nIf the function $f(z)$ has a pole of order $m$ at $z=a$, then, by the\ndefinition of a pole, an equation of the form\n$$\nf(z)\n=\n\\frac{a_{-m}}{ (z-a)^{m} }\n+ \\frac{ a_{-m+1} }{ (z-a)^{m-1} }\n+ \\cdots\n+ \\frac{a_{-1}}{z-a}\n+ \\phi(z),\n$$\nwhere $\\phi(z)$ is analytic near and at $a$, is true near $a$.\n\nThe coefficient $a_{-1}$ in this expansion is called the \\emph{residue}\\index{Residues!defined} of the\nfunction $f(z)$ relative to the pole $a$.\n\nConsider now the value of the integral $\\int_{\\alpha} f(z) \\dmeasure z$, where the path of\nintegration is a circle\\footnote{The existence of such a circle is\n implied in the definition of a pole as an isolated singularity.}\n$\\alpha$, whose centre is the point $a$ and whose radius $\\rho$ is so small\nthat $\\phi(z)$ is analytic inside and on the circle.\n\nWe have\n$$\n\\int_{\\alpha} f(z) \\dmeasure z\n=\n\\sum_{r=1}^{m}\na_{r}\n\\int_{\\alpha} \\frac{\\dmeasure z}{ (z-a)^{r}}\n+\n\\int_{\\alpha} \\phi(z) \\dmeasure z.\n$$\n\nNow $\\int_{\\alpha} \\phi(z) \\dmeasure z = 0$ by \\hardsectionref{5}{2};\nand (putting $z-a = \\rho e^{i\\theta}$) we have, if $r \\neq 1$,\n$$\n\\int_{\\alpha} \\frac{ \\dmeasure z}{ (z-a)^{r} }\n=\n\\int_{0}^{2\\pi} \\frac{\\rho e^{i\\theta} i \\dmeasure \\theta}{ \\rho^{r}\n e^{ri\\theta}}\n=\n\\rho^{-r+1}\n\\int_{0}^{2\\pi} e^{(1-r)i\\theta} i \\dmeasure \\theta\n=\n\\rho^{-r+1}\n\\thebracket{ \\frac{e^{(1-r) i \\theta}}{1-r} }_{0}^{2\\pi}\n=\n0.\n$$\n\nBut, when $r = 1$, we have\n$$\n\\int_{\\alpha}\n\\frac{\\dmeasure z}{z - a}\n=\n\\int_{0}^{2\\pi} i \\dmeasure \\theta\n=\n2\\pi i.\n$$\n\nHence finally\n$$\n\\int_{\\alpha} f(z) \\dmeasure z = 2 \\pi i a_{-1}.\n$$\n\nNow let $C$ be any contour, containing in the region interior to it a\nnumber of poles $a,b,c,\\ldots$ of a function $f(z)$, with residues\n$a_{-1}, b_{-1}, c_{-1},\\ldots$\nrespectively; and suppose that the function $f(z)$ is\nanalytic throughout $C$ and its interior, except at these poles.\n\nSurround the points $a,b,c,\\ldots$ by circles\n$\\alpha,\\beta,\\gamma,\\ldots$\nso small that their respective centres are the only singularities\ninside or on each circle; then the function $f(z)$ is analytic in the\nclosed region bounded by $C$, $\\alpha,\\beta,\\gamma,\\ldots$.\n%\n% 112\n%\n\nHence, by \\hardsectionref{5}{2} corollary TODO,\n\\begin{align*}\n \\int_{C} f(z) \\dmeasure z\n =&\n \\int_{\\alpha} f(z) \\dmeasure z\n + \\int_{\\beta} f(z) \\dmeasure z\n + \\cdots\n \\\\\n =&\n 2\\pi i a_{-1} + 2\\pi i b_{-1} + \\cdots.\n\\end{align*}\n\nThus we have the \\emph{theorem of residues}, namely that\n\\emph{if $f(z)$ be analytic throughout a contour $C$ and its interior\n except at a number of poles inside the contour, then\n $$\n \\int_{C} f(z) \\dmeasure z = 2\\pi i \\sum R,\n $$\nwhere $\\sum R$ denotes the sum of the residues of the function $f(z)$ at\nthose of its poles which are situated within the contour $C$.}\n\nThis is an extension of the theorem of \\hardsubsectionref{5}{2}{1}.\n\n%\\begin{Remark} TODO:\\alpha or a?\nNote. If $a$ is a simple pole of $f(z)$ the residue of $f(z)$ at that pole\nis $\\lim_{z\\rightarrow a} (z-a) f(z)$.\n%\\end{Remark}\n\n\\Section{6}{2}{The evaluation of definite integrals.}\n\nWe shall now apply the result of \\hardsectionref{6}{1} to evaluating\nvarious classes of definite integrals; the methods to be employed in\nany particular case may usually be seen from the following typical\nexamples.\n\n\\Subsection{6}{2}{1}{The evaluation of the integrals of certain\n periodic functions taken between the limits $0$ and\n $2\\pi$.\\index{Integrals!of periodic functions}\\index{Periodic\n functions!integrals involving}}\nAn integral of the type\n$$\n\\int_{0}^{2\\pi} R(\\cos \\theta, \\sin \\theta) \\dmeasure \\theta,\\index{Trigonometrical integrals}\n$$\nwhere the integrand is a rational function of $\\cos\\theta$ and\n$\\sin\\theta$ finite on the range of integration, can be evaluated by\nwriting $e^{i\\theta}=z$; since\n$$\n\\cos\\theta = \\half(z+z^{-1}),\n\\quad\n\\sin\\theta = \\frac{1}{2i} (z-z^{-1}),\n$$\nthe integral takes the form $\\int_{C} S(z) \\dmeasure z$, where $S(z)$\nis a rational function of $z$ finite on the path of integration $C$, the\ncircle of radius unity whose centre is the origin.\n\nTherefore, by \\hardsectionref{6}{1}, \\emph{the integral is equal to\n $2\\pi i$ times the sum of the residues of $S(z)$ at those of its\n poles which are inside that circle.}\n\n\\begin{wandwexample}\n If $0 < p < 1$,\n $$\n \\int_{0}^{2\\pi}\n \\frac{\\dmeasure \\theta}{ 1 - 2p\\cos\\theta + p^{2}}\n =\n \\int_{C} \\frac{ \\dmeasure z}{ i (1-pz) (z-p) }.\n $$\n The only pole of the integrand inside the circle is a simple pole at\n $p$; and the residue there is\n $$\n \\lim_{z \\rightarrow p} \\frac{z-p}{ i (1-pz)(z-p)}\n =\n \\frac{1}{i (1-p^{2})}.\n $$\n%\n% 113\n%\n Hence\n $$\n \\int_{0}^{2\\pi} \\frac{\\dmeasure\\theta}{1 - 2p\\cos\\theta + p^{2}}\n =\n \\frac{2\\pi}{1-p^{2}}.\n $$\n\\end{wandwexample}\n\n\\begin{wandwexample}\n If $0 < p < 1$,\n \\begin{align*}\n \\int_{0}^{2\\pi} \\frac{ \\cos^{2} 3\\theta}{1-2p\\cos 2\\theta + p^{2}}\n \\dmeasure \\theta\n =&\n \\int_{C} \\frac{\\dmeasure z}{iz}\n \\theparen{ \\half z^{3} + \\half z^{-3}}^{2}\n \\frac{1}{ (1-pz^{2})(1-pz^{-2})}\n \\end{align*}\n where $\\sum R$ denotes the sum of the residues of\n $ \\frac{ (z^{6} + 1)^{2} }{ 4z^{5} (1-pz^{2})(z^{2}-p)} $ at its poles\n inside $C$; these poles are $0, -p^{\\half}, p^{\\half}$;\n and the residues at them are\n $ -\\frac{1+p^{2}+p^{4}}{4p^{3}},\n \\frac{(p^{3}+1)^{2}}{ 8p^{3}(1-p^{2})},\n \\frac{(p^{3}+1)^{2}}{ 8p^{3}(1-p^{2})}$; %TODO:same?\n and hence the integral is equal to\n $$\n \\frac{\\pi (1-p+p^{2})}{1-p}.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n If $n$ be a positive integer,\n $$\n \\int_{0}^{2\\pi} e^{\\cos\\theta} \\cos(n\\theta - \\sin\\theta)\n \\dmeasure \\theta\n =\n \\frac{2\\pi}{n!},\n \\quad\n \\int_{0}^{2\\pi}\n e^{\\cos\\theta} \\sin(n\\theta - \\sin\\theta) \\dmeasure \\theta\n =\n 0.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n If $a>b>0$,\n $$\n \\int_{0}^{2\\pi}\n \\frac{\\dmeasure \\theta}{ (a+b\\cos\\theta)^{2}}\n =\n \\frac{2\\pi a}{ (a^{2}-b^{2})^{3/2} },\n \\quad\n \\int_{0}^{2\\pi}\n \\frac{\\dmeasure\\theta}{ (a+b\\cos^{2}\\theta)^{2}}\n =\n \\frac{\\pi(2a+b)}{ a^{3/2} (a+b)^{3/2}}.\n $$\n\\end{wandwexample}\n\\Subsection{6}{2}{2}{The evaluation of certain types of integrals\n taken between the limits $-\\infty$ and $+\\infty$.}\nWe shall now evaluate $\\int_{-\\infty}^{\\infty} Q(x) \\dmeasure x$,\nwhere $Q(z)$ is a function such that (i) it is analytic when the\nimaginary part of $z$ is positive or zero (except at a finite number\nof poles), (ii) it has no poles on the real axis and (iii) as\n$\\absval{z}\\rightarrow\\infty$, $zQ(z) \\rightarrow 0$ uniformly for all\nvalues of $\\arg z$ such that $0 \\leq \\arg z \\leq \\pi$; provided that\n(iv) when $x$ is real, $x Q(x) \\rightarrow 0$, as\n$x \\rightarrow \\pm\\infty$, in such a way\\footnote{The condition\n $x Q(x)\\rightarrow 0$ is not in itself sufficient to secure the\n convergence of\n $\\int^{\\infty} Q(x) \\dmeasure x$;\n consider $Q(x) = (x \\log x)^{-1}$.}\nthat $\\int_{0}^{\\infty} Q(x) \\dmeasure x$ and\n$\\int_{-\\infty}^{0} Q(x) \\dmeasure x$ both converge.\n\nGiven $\\eps$, we can choose $\\rho_{0}$ (independent of $\\arg z$) such\nthat $\\absval{z Q(z)} < \\eps/\\pi$ whenever\n$\\absval{z} > \\rho_{0}$ and $0 \\leq \\arg z \\leq \\pi$.\n\nConsider $\\int_{C} Q(z) \\dmeasure z$ taken round a contour $C$\nconsisting of the part of the real axis joining the points $\\pm\\rho$\n(where $\\rho > \\rho_{0}$) and a semicircle $\\Gamma$, of radius $\\rho$,\nhaving its centre at the origin, above the real axis.\n\nThen, by \\hardsectionref{6}{1}, $\\int_{C} Q(z) \\dmeasure z = 2 \\pi i\n\\sum R$, where $\\sum R$ denotes the sum of the residues of $Q(z)$ at\nits poles above the real axis.\\footnote{$Q(z)$ has no poles above the\n real axis outside the contour.}\n\n%\n% 114\n%\nTherefore\n$$\n\\absval{\n \\int_{-\\rho}^{\\rho} Q(z) \\dmeasure z\n -\n 2 \\pi i \\sum R\n}\n=\n\\absval{\n \\int_{\\Gamma} Q(z) \\dmeasure z\n}.\n$$\n\nIn the last integral write $z = \\rho e^{i\\theta}$, and then\n\\begin{align*}\n \\absval{ \\int_{\\Gamma} Q(z) \\dmeasure z }\n &=\n \\absval{\n \\int_{0}^{\\pi} Q(\\rho e^{i\\theta}) \\rho e^{i\\theta} i\n \\dmeasure\\theta\n }\n \\\\\n &<\n \\int_{0}^{\\pi} (\\eps / \\pi) \\dmeasure \\theta\n \\\\\n &= \\eps\n\\end{align*}\nby \\hardsubsectionref{4}{6}{2}.\n\nHence\n$$\n\\lim_{\\rho\\rightarrow\\infty}\n\\int_{-\\rho}^{\\rho} Q(z) \\dmeasure z\n=\n2\\pi i \\sum R.\n$$\n\nBut the meaning of $\\int_{-\\infty}^{\\infty} Q(x) \\dmeasure x$;\nis $\\lim_{\\rho,\\sigma\\rightarrow\\infty}\\int_{-\\rho}^{\\sigma} Q(x)\n\\dmeasure x$;\nand since\n$\\lim_{\\sigma\\rightarrow\\infty}\\int_{0}^{\\sigma} Q(x) \\dmeasure x$\nand\n$\\lim_{\\rho\\rightarrow\\infty}\\int_{-\\rho}^{0} Q(x) \\dmeasure x$\nboth exist, this double limit is the same as\n$\\lim_{\\rho\\rightarrow\\infty}\\int_{-\\rho}^{\\rho} Q(x) \\dmeasure x$.\n\nHence we have proved that\n$$\n\\int_{-\\infty}^{\\infty} Q(x) \\dmeasure x = 2\\pi i \\sum R.\n$$\n\nThis theorem is particularly useful in the special case when $Q(x)$ is a\nrational function.\n\n%\\begin{smallfont}\n[Note. Even if condition (iv) is not satisfied, we still have\n$$\n\\int_{0}^{\\infty}\n\\thebrace{\n Q(x) + Q(-x)\n} \\dmeasure x\n=\n\\lim_{\\rho\\rightarrow\\infty} \\int_{-\\rho}^{\\rho} Q(x) \\dmeasure x\n= 2 \\pi i \\sum R.]\n$$\n%\\end{smallfont}\n\n\\begin{wandwexample}\n The only pole of $(z^{2} + 1)^{-3}$ in the upper half plane is a\n pole at $z=i$ with residue there $-\\frac{3}{16} i$. Therefore\n $$\n \\int_{-\\infty}^{\\infty} \\frac{\\dmeasure x}{ (x^{2}+1)^{3} }\n =\n \\frac{3}{8} \\pi.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\nIf $a > 0, b > 0$, shew that\n$$\n\\int_{-\\infty}^{\\infty}\n\\frac{x^{4} \\dmeasure x}{ (a + bx^{2})^{4} }\n=\n\\frac{\\pi}{ 16 a^{3/2} b^{5/2}}.\n$$\n\\end{wandwexample}\n\\begin{wandwexample}\n By integrating $\\int e^{-\\lambda z^{2}} \\dmeasure z$ round a\nparallelogram whose corners are $-R, R, R + ai, -R + ai$ and making $R\n\\rightarrow \\infty$, shew that, if $\\lambda > 0$, then\n $$\n \\int_{-\\infty}^{\\infty} e^{-\\lambda x^{2}} \\cos (2\\lambda a x)\n\\dmeasure x = e^{-\\lambda a^{2}} \\int_{-\\infty}^{\\infty} e^{-\\lambda\nx^{2}} \\dmeasure x = 2 \\lambda^{-\\half} e^{-\\lambda a^{2}}\n\\int_{0}^{\\infty} e^{-x^{2}} \\dmeasure x.\n $$\n\\end{wandwexample}\n\\Subsubsection{6}{2}{2}{1}{Certain infinite integrals involving sines\nand cosines.}\nIf $Q(z)$ satisfies the conditions (i), (ii) and (iii) of\n\\hardsectionref{6}{2}{2}, and $m > 0$,\nthen $Q(z) e^{miz}$ also satisfies those conditions.\n\n%\n% 115\n%\n\nHence\n$\n\\int_{0}^{\\infty}\n\\thebrace{\n Q(x) e^{mix} + Q(-x) e^{-mix}\n}\n\\dmeasure x\n$\nis equal to $2\\pi i \\sum R'$, where\n$\\sum R'$ means the sum of the residues of $Q(z) e^{mix}$ at its poles\nin the upper half plane; and so\n\\begin{itemize}\n\\item % (i)\n If $Q(x)$ is an even function\\index{Even functions}, i.e. if $Q (- x) = Q (x)$,\n $$\n \\int_{0}^{\\infty} Q(x) \\cos(mx) \\dmeasure x\n =\n \\pi i \\sum R'.\n $$\n\\item %(ii)\n If $Q(x)$ is an odd function\\index{Odd functions},\n $$\n \\int_{0}^{\\infty} Q(x) \\sin(mx) \\dmeasure x\n =\n \\pi \\sum R'.\n $$\n\\end{itemize}\n\\Subsubsection[Jordan's lemma.]{6}{2}{2}{2}{Jordan's lemma.\\footnote{TODO:Jordan, Cours d'Aiiahjse, ii. (1894), pp. 285, 286.}\\index{Jordan's lemma}}\n\nThe results of \\hardsubsubsectionref{6}{2}{2}{1} are true if $Q(z)$ be\nsubject to the less stringent condition $Q(z) \\rightarrow 0$ uniformly\nwhen\n$O \\leq \\arg z \\leq \\pi$ as $\\absval{z} \\rightarrow \\infty$ in\nplace of the condition $z Q(z) \\rightarrow 0$ uniformly.\n\nTo prove this we require a theorem known as Jordan's lemma, viz.\n\n\\emph{If $Q(z) \\rightarrow 0$ uniformly with regard to $\\arg z$ as\n $\\absval{z} \\rightarrow \\infty$ when $0 \\leq \\arg z \\leq \\pi$,\n and if $Q(z)$ is analytic when both $\\absval{z} > c$ (a constant)\n and $0 \\leq \\arg z \\leq \\pi$, then\n $$\n \\lim_{\\rho\\rightarrow\\infty}\n \\theparen{ \\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z }\n =\n 0,\n $$\n where $\\Gamma$ is a semicircle of radius $\\rho$ above the real axis with centre at\n the origin.}\n\nGiven $\\eps$, choose $\\rho_{0}$ so that $\\absval{Q(z)} < \\eps/\\pi$\nwhen $\\absval{z} > \\rho_{0}$ and\n$0 \\leq \\arg z \\leq \\pi$;\nthen, if $\\rho > \\rho_{0}$,\n$$\n\\absval{ \\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z }\n=\n\\absval{\n \\int_{0}^{\\pi}\n e^{mi (\\rho \\cos\\theta + i\\rho \\sin\\theta)}\n Q(\\rho e^{i\\theta})\n \\rho e^{i \\theta}\n i \\dmeasure \\theta\n}.\n$$\nBut $\\absval{ e^{mi\\rho\\cos\\theta} } = 1$, and so\n\\begin{align*}\n \\absval{ \\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z }\n <&\n \\int_{0}^{\\pi} (\\eps/\\pi) \\rho e^{-m\\rho\\sin\\theta} \\dmeasure\\theta\n \\\\\n =&\n (2\\eps/\\pi)\n \\int_{0}^{\\half\\pi} \\rho e^{-m\\rho\\sin\\theta} \\dmeasure\\theta.\n\\end{align*}\n\nNow $\\sin\\theta \\geq 2\\theta/\\pi$, when\\footnote{This inequality\n appears obvious when we draw the graphs $y = \\sin x$, $y = 2x/\\pi$;\n it may be proved by shewing that $(\\sin\\theta)/\\theta$ decreases as\n $\\theta$ increases from $0$ to $\\half\\pi$.\n} $0 \\leq \\theta \\leq \\half\\pi$, and so\n\\begin{align*}\n \\absval{ \\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z }\n <&\n (2\\eps/\\pi)\n \\int_{0}^{\\half\\pi}\n \\rho e^{-2m\\rho\\theta/\\pi} \\dmeasure\\theta\n \\\\\n =&\n (2\\eps/\\pi) \\cdot (\\pi/2m)\n \\thebracket{ -e^{-2m\\rho\\theta/\\pi} }_{0}^{\\half\\pi}\n \\\\\n <&\n \\eps/m.\n\\end{align*}\n\n%\n% 116\n%\n\nHence\n$$\n\\lim_{\\rho\\rightarrow\\infty} \\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z = 0.\n$$\n\nThis result is Jordan's lemma.\n\nNow\n$$\n\\int_{0}^{\\rho}\n\\thebrace{\n e^{mix} Q(x) + e^{-mix} Q(-x)\n}\n\\dmeasure x\n=\n2 \\pi i \\sum R'\n-\n\\int_{\\Gamma} e^{miz} Q(z) \\dmeasure z,\n$$\nand, making $\\rho\\rightarrow\\infty$, we see at once that\n$$\n\\int_{0}^{\\infty}\n\\thebrace{\n e^{mix} Q(x) + e^{-mix} Q(-x)\n}\n\\dmeasure x\n=\n2 \\pi i \\sum R',\n$$\nwhich is the result corresponding to the result of \\hardsubsubsectionref{6}{2}{2}{1}.\n\n\\begin{wandwexample}\n Shew that, if $a > 0$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{ \\cos x}{x^{2} + a^{2}} \\dmeasure x\n =\n \\frac{\\pi}{2a} e^{-a}.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $a \\geq 0, b \\geq 0$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{ \\cos 2ax - \\cos 2bx}{x^{2}}\n \\dmeasure x\n =\n \\pi (b-a)\n $$\n (Take a contour consisting of a large semicircle of radius $\\rho$, a\n small semicircle of radius $\\delta$, both having their centres at the\n origin, and the parts of the real axis joining their ends; then\n make $\\rho \\rightarrow \\infty, \\delta \\rightarrow 0$.)\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $b > 0, m \\geq 0$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{3x^{2} - a^{2}}{ (x^{2} + b^{2})^{2} }\n \\cos mx\n \\dmeasure x\n =\n \\frac{\\pi e^{-mb}}{4b^{3}}\n \\thebrace{\n 3b^{2} - a^{2} - mb(3b^{2}+a^{2})\n }.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $k > 0, a > 0$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{ x \\sin ax }{x^{2} + k^{2}}\n \\dmeasure x\n =\n \\half \\pi e^{-ka}.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $m \\geq 0$, $a > 0$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{ \\sin mx }{ x (x^{2} + a^{2})^{2} }\n \\dmeasure x\n =\n \\frac{\\pi}{2a^{4}}\n -\n \\frac{\\pi e^{-ma}}{4a^{3}}\n \\theparen{ m + \\frac{2}{a} }.\n $$\n (Take the contour of example TODO.)\n\\end{wandwexample}\n\\begin{wandwexample}\nShew that, if the real part of $z$ be positive,\n$$\n\\int_{0}^{\\infty} (e^{-t} - e^{-tz}) \\frac{\\dmeasure t}{t} = \\log z.\n$$\n[We have\n\\begin{align*}\n \\int_{0}^{\\infty} (e^{-t} - e^{-tz}) \\frac{\\dmeasure t}{t}\n =&\n \\lim_{\\delta\\rightarrow 0,\\, \\rho\\rightarrow\\infty}\n \\thebrace{\n \\int_{\\delta}^{\\rho} \\frac{e^{-t}}{t} \\dmeasure t\n -\n \\int_{\\delta}^{\\rho} \\frac{e^{-tz}}{t} \\dmeasure t\n }\n \\\\\n =&\n \\lim_{\\delta\\rightarrow 0,\\, \\rho\\rightarrow\\infty}\n \\thebrace{\n \\int_{\\delta}^{\\rho} \\frac{e^{-t}}{t} \\dmeasure t\n -\n \\int_{\\delta z}^{\\rho z} \\frac{e^{-u}}{u} \\dmeasure u\n }\n \\\\\n =&\n \\lim_{\\delta\\rightarrow 0,\\, \\rho\\rightarrow\\infty}\n \\thebrace{\n \\int_{\\delta}^{\\delta z} \\frac{e^{-t}}{t} \\dmeasure t\n -\n \\int_{\\rho}^{\\rho z} \\frac{e^{-t}}{t} \\dmeasure t,\n },\n\\end{align*}\nsince $t^{-1} e^{-t}$ is analytic inside the\nquadrilateral whose corners are $\\delta, \\delta z, \\rho z, \\rho$.\n\n%\n% 117\n%\n\nNow $\\int_{\\rho}^{\\rho z} t^{-1} e^{-t} \\dmeasure t \\rightarrow 0$\nwhen $\\Re(z) > 0$; and\n$$\n\\int_{\\delta}^{\\delta z} t^{-1} e^{-t} \\dmeasure t\n= \\log z\n-\n\\int_{\\delta}^{\\delta z} t^{-1} (1 - e^{-t}) \\dmeasure t\n\\rightarrow\n\\log z,\n$$\nsince $t^{-1} (1 - e^{-t}) \\rightarrow 1$ as $t \\rightarrow 0$.]\n\\end{wandwexample}\n\\Subsection{6}{2}{3}{Principal values of\n integrals.\\index{Integrals!principal values of}\\index{Principal part\n of a function!value of an integral}}\n%\\begin{smallfont}\nIt was assumed in TODO that the function $Q(x)$ had no\npoles on the real axis; if the function has a finite number of \\emph{simple}\npoles on the real axis, we can obtain theorems corresponding to those\nalready obtained, except that the integrals are all principal values\n(\\hardsectionref{4}{5}) and $\\sum R$ has to be replaced by\n$\\sum R + \\half \\sum R_{0}$, where $\\sum R_{0}$ means the\nsum of the residues at the poles on the real axis. To obtain this\nresult we see that, instead of the former contour, we have to take as\ncontour a circle of radius $\\rho$ and the portions of the real axis joining\nthe points\n$$\n-\\rho, \\ a-\\delta_{1};\n\\quad\na+\\delta_{1}, \\ b-\\delta_{2};\n\\quad\nb+\\delta_{2}, \\ c-\\delta_{3},\n\\ \\ldots\n$$\nand small semicircles above the real axis of radii\n$\\delta_{1},\\delta_{2},\\ldots$ with\ncentres $a, b, c, \\ldots$ where $a, b, c, \\ldots$ are the poles of\n$Q(z)$ on the real axis; and then we have to make\n$\\delta_{1}, \\delta_{2}, \\ldots \\rightarrow 0$;\ncall these semicircles $\\gamma_{1},\\gamma_{2},\\ldots$.\nThen instead of the equation\n$$\n\\int_{-\\rho}^{\\rho} Q(z) \\dmeasure z\n+\n\\int_{\\Gamma} Q(z) \\dmeasure z\n=\n2 \\pi i \\sum R,\n$$\nwe get\n$$\nP \\int_{-\\rho}^{\\rho} Q(z) \\dmeasure z\n+\n\\sum_{n}\n\\lim_{\\delta_{n} \\rightarrow 0}\n\\int_{\\gamma_{n}} Q(z) \\dmeasure z\n+\n\\int_{\\Gamma} Q(z) \\dmeasure z\n=\n2 \\pi i \\sum R.\n$$\n\nLet $a'$ be the residue of $Q(z)$ at $a$; then writing\n$z = a + \\delta_{1} e^{i\\theta}$ on $\\gamma_{1}$ we get\n$$\n\\int_{\\gamma_{1}} Q(z) \\dmeasure z\n=\n\\int_{\\pi}^{0}\nQ(a + \\delta_{1} e^{i\\theta})\n\\delta_{1} e^{i\\theta} i \\dmeasure \\theta.\n$$\nBut $Q(a + \\delta_{1} e^{i\\theta}) \\rightarrow a'$\nuniformly as $\\delta_{1} \\rightarrow 0$; and therefore\n$\n\\lim_{\\delta_{1} \\rightarrow 0} \\int_{\\gamma_{1}} Q(z) \\dmeasure z\n=\n- \\pi i a'\n$;\nwe thus get\n$$\nP \\int_{-\\rho}^{\\rho} Q(z) \\dmeasure z\n+\n\\int_{\\Gamma} Q(z) \\dmeasure z\n=\n2 \\pi i \\sum R\n+\n\\pi i \\sum R_{0},\n$$\nand hence, using the arguments of \\hardsubsectionref{6}{2}{2}, we get\n$$\nP \\int_{-\\infty}^{\\infty} Q(x) \\dmeasure x\n=\n2 \\pi i \\theparen{ \\sum R + \\half \\sum R_{0} }.\n$$\n\nThe reader will see at once that the theorems of TODO have\nprecisely similar generalisations.\n\nThe process employed above of inserting arcs of small circles so as to\ndiminish the area of the contour is called \\emph{indenting} the contour.\n%\\end{smallfont}\n\\Subsection{6}{2}{4}{Evaluation of integrals of the form\n $\\int_{0}^{\\infty} x^{a-1} Q(x) \\dmeasure x.$}\nLet $Q(x)$ be a rational function of $x$ such that it has no poles on the\npositive part of the real axis and\n$x^{a} Q(x) \\rightarrow 0$ both when $x \\rightarrow 0$ and when\n$x \\rightarrow \\infty$.\n\n%\n% 118\n%\n\nConsider $\\int (-z)^{a-1} Q(z) \\dmeasure z$ taken round the contour\n$C$ shewn in the figure TODO, consisting of the arcs of circles of\nradii $\\rho,\\delta$ and the straight lines joining their end points;\n$(-z)^{a-1}$ is\nto be interpreted as\n$$\n\\exp \\thebrace{ ( a-1) \\log (- z) }\n$$\nand\n$$\n\\log (-z) = \\log \\absval{z} + \\arg (-z),\n$$\nwhere\n$$\n-\\pi \\leq \\arg (-z) \\leq \\pi; %TODO:is overlap correct?\n$$\nwith these conventions the integrand is one-valued and analytic on and\nwithin the contour save at the poles of $Q(z)$.\n\nHence, if $\\sum r$ denote the sum of the residues of $(-z)^{a-1} Q(z)$\nat all its poles,\n$$\n\\int_{C} (-z)^{a-1} Q(z) \\dmeasure z = 2 \\pi i \\sum r.\n$$\n\nOn the small circle write $-z = \\delta e^{i\\theta}$, and the integral\nalong it becomes\n$- \\int_{\\pi}^{-\\pi} (-z)^{a} Q(z) i \\dmeasure \\theta$,\nwhich tends to zero as $\\delta \\rightarrow 0$.\n\nOn the large semicircle write $-z = \\rho e^{i\\theta}$, and the\nintegral along it becomes $- \\int_{-\\pi}^{\\pi} (-z)^{n} Q(z) i\n\\dmeasure\\theta$, which tends to zero as $\\rho \\rightarrow \\infty$.\n\nOn one of the lines we write $-z = x e^{\\pi i}$ on the other\n$-z = x e^{-\\pi i}$ and $(-z)^{a-1}$ becomes $x^{a-1} e^{\\pm (a-1) \\pi i}$.\n\nHence\n$$\n\\lim_{\\delta\\rightarrow 0,\\ \\rho\\rightarrow\\infty}\n\\int_{\\delta}^{\\rho}\n\\thebrace{\n x^{a-1} e^{-(a-1)\\pi i} Q(x)\n -\n x^{a-1} e^{(a-1)\\pi i} Q(x)\n}\n\\dmeasure x\n=\n2 \\pi i \\sum r;\n$$\nand therefore\n$$\n\\int_{0}^{\\infty} x^{a-1} Q(x) \\dmeasure x\n=\n\\pi \\cosec (a\\pi) \\sum r.\n$$\n\n%TODO: change corollary environment\nCorollary. If $Q(x)$ have a number of simple poles on the positive\npart of the real axis, it may be shewn by indenting the contour that\n$$\nP \\int_{0}^{\\infty} x^{a-1} Q(x) \\dmeasure x\n=\n\\pi \\cosec (a\\pi) \\sum r\n-\n\\pi \\cot (1\\pi) \\sum r',\n$$\nwhere $\\sum r'$ is the sum of the residues of $z^{a-1} Q(z)$ at these poles.\n\\begin{wandwexample}\nIf $0 < a < 1$,\n$$\n\\int_{0}^{\\infty} \\frac{x^{a-1}}{1+x} \\dmeasure x\n=\n\\pi \\cosec a\\pi,\n\\quad\nP \\int_{0}^{\\infty} \\frac{x^{a-1}}{1+x} \\dmeasure x\n=\n\\pi \\cot a\\pi\n$$\n\\end{wandwexample}\n%\n% 119\n%\n\\begin{wandwexample}\n If $0 < z < 1$ and $-\\pi < a < \\pi$,\n $$\n \\int_{0}^{\\infty} \\frac{t^{z-1}}{ t+e^{ia} } \\dmeasure t\n =\n \\frac{ \\pi e^{i(z-1)a}}{\\sin \\pi z}.\n $$\n\\addexamplecitation{Minding.}\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $- 1 < z < 3$, then\n $$\n \\int_{0}^{\\infty} \\frac{ x^{z} }{ (1+x^{2})^{2} } \\dmeasure x\n =\n \\frac{ \\pi (1-z) }{ 4 \\cos \\half\\pi z}.\n $$\n\\end{wandwexample}\n\\begin{wandwexample}\n Shew that, if $-1 < p < 1$ and $-\\pi < \\lambda < \\pi$, then\n $$\n \\int_{0}^{\\infty}\n \\frac{ x^{-p} \\dmeasure x }{ 1 + 2x \\cos\\lambda + x^{2} }\n =\n \\frac{\\pi}{\\sin p\\pi} \\frac{\\sin p\\lambda}{\\sin \\lambda}.\n $$\n \\addexamplecitation{Euler.}\n\\end{wandwexample}\n\\Section{6}{3}{Cauchy's integral.}\nWe shall next discuss a class of contour-integrals which are\nsometimes found useful in analytical investigations.\n\nLet $C$ be a contour in the $z$-plane, and let $f(z)$ be a function\nanalytic inside and on $C$. Let $\\phi(z)$ be another function which is\nanalytic inside and on $G$ except at a finite number of poles; let the\nzeros of $\\phi(z)$ in the interior\\footnote{$\\phi(z)$ must not have\n any zeros or poles on $C$.} of $C$ be $a_{1},a_{2},\\ldots$, and let\ntheir degrees of multiplicity be $r_{1},r_{2},\\ldots$; and let its\npoles in the interior of $C$ be $b_{1},b_{2},\\ldots$, and let their\ndegrees of multiplicity be $s_{1},s_{2},\\ldots$.\n\nThen, by the fundamental theorem of residues,\n$ \\frac{1}{2\\pi i} \\int_{C} f(z) \\frac{\\phi'(z)}{\\phi(z)} \\dmeasure z $\nis equal to the sum of the residues of\n$f(z) \\frac{\\phi'(z)}{\\phi(z)}$\nat its poles inside $C$.\n\nNow\n$f(z) \\frac{\\phi'(z)}{\\phi(z)}$\ncan have singularities only at the poles and zeros of $\\phi(z)$.\nNear one of the zeros, say $a_{1}$, we have\n$$\n\\phi(z)\n=\nA (z-a_{1})^{r_{1}}\n+ B (z-a_{1})^{r_{1}+1}\n+ \\cdots.\n$$\n\nTherefore\n$$\n\\phi'(z)\n=\nA r_{1} (z-a_{1})^{r_{1}-1}\n+ B (r_{1}+1) (z-a_{1})^{r_{1}}\n+ \\cdots,\n$$\nand\n$$\nf(z) = f(a_{1}) + (z-a_{1}) f'(a_{1}) + \\cdots.\n$$\n\nTherefore\n$\n\\thebrace{f(z) \\frac{\\phi'(z)}{\\phi(z)}\n -\n \\frac{ r_{1} f(a_{1}) }{ z - a_{1} }\n}\n$ is analytic at $a_{1}$.\n\nThus the residue of $f(z) \\frac{\\phi'(z)}{\\phi(z)}$, at the point\n$z=a_{1}$, is $r_{1} f(a_{1})$.\n\nSimilarly the residue at $z=b_{1}$ is\n$-s_{1} f(b_{1})$; for near $z=b_{1}$, we have\n$$\n\\phi(z)\n=\nC (z-b_{1})^{-s_{1}}\n+ D (z-b_{1})^{-s_{1} + 1}\n+ \\cdots,\n$$\nand\n$$\nf(z)\n=\nf(b_{1})\n+ (z - b_{1}) f'(b_{1})\n+ \\cdots,\n$$\nso\n$\nf(z) \\frac{ \\phi'(z) }{\\phi(z)}\n+\n\\frac{ s_{1} f(b_{1}) }{ z-b_{1} }\n$\nis analytic at $b_{1}$.\n\nHence\n$$\n\\frac{1}{2\\pi i}\n\\int_{C} f(z) \\frac{ \\phi'(z) }{\\phi(z)} \\dmeasure z\n=\n\\sum r_{1} f(a_{1})\n-\n\\sum s_{1} f(b_{1}),\n$$\nthe summations being extended over all the zeros and poles of\n$\\phi(z)$.\n\n\\Subsection{6}{3}{1}{The number of roots of an equation contained\n within a contour.\\index{Roots of an equation, number of!(inside a contour)}}\nThe result of the preceding paragraph can be at once applied to find\nhow many roots of an equation $\\phi(z) = 0$ lie within a contour $C$.\n\nFor, on putting $f(z) = 1$ in the preceding result, we obtain the\nresult that\n$$\n\\frac{1}{2\\pi i}\n\\int_{C} \\frac{ \\phi'(z) }{\\phi(z)} \\dmeasure z\n$$\nis equal to the excess of the number of zeros over the number of poles\nof $\\phi(z)$ contained in the interior of $C$, each pole and zero\nbeing reckoned according to its degree of multiplicity.\n\n%\n% 120\n%\n\\begin{wandwexample}\n Shew that a polynomial $\\phi(z)$ of degree $m$ has $m$ roots.\n \\index{Equation of degree $m$ has $m$ roots}\\index{Roots of an\n equation!number of}\n\nLet\n$$\n\\phi(z)\n=\na_{0} z^{m} + a_{1} z^{m-1} + \\cdots + a_{m},\n\\quad\n(a_{0} \\neq 0).\n$$\nThen\n$$\n\\frac{ \\phi'(z) }{ \\phi(z) }\n=\n\\frac{ m a_{m}z^{m-1} + \\cdots + a_{m-1} }{ a_{0}z^{m} + \\cdots a_{m}}.\n$$\nConsequently, for large values of $\\absval{z}$,\n$$\n\\frac{ \\phi'(z) }{ \\phi(z) }\n=\n\\frac{m}{z} + O\\theparen{ \\frac{1}{z^{2}} }\n$$\nThus, if $C$ be a circle of radius $\\rho$ whose centre is at the origin, we\nhave\n$$\n\\frac{1}{2 \\pi i} \\int_{C} \\frac{\\phi'(z)}{\\phi(z)} \\dmeasure z\n=\n\\frac{m}{2 \\pi i} \\int_{C} \\frac{ \\dmeasure z }{z}\n+\n\\frac{1}{2 \\pi i} \\int_{C} O\\theparen{ \\frac{1}{z^{2}} } \\dmeasure z\n=\nm + \\frac{1}{2 \\pi i} \\int_{C} O\\theparen{ \\frac{1}{z^{2}} } \\dmeasure z.\n$$\n\nBut, as in \\hardsectionref{6}{2}{2},\n$$\n\\int_{C} O\\theparen{ \\frac{1}{z^{2}} } \\dmeasure z\n\\rightarrow\n0\n$$\nas $\\rho\\rightarrow\\infty$; and hence as $\\phi(z)$ has no poles in the\ninterior of $C$, the total number of zeros of $\\phi(z)$ is\n$$\n\\lim_{\\rho\\rightarrow\\infty}\n\\frac{1}{2\\pi i}\n\\int_{C} \\frac{ \\phi'(z) }{\\phi(z)} \\dmeasure z\n=\nm.\n$$\n\\end{wandwexample}\n\\begin{wandwexample}\nIf at all points of a contour $C$ the inequality\n$$\n\\absval{ a_{k} z^{k} }\n>\n\\absval{\n a_{0}\n + a_{1} z\n + \\cdots\n + a_{k-1} z^{k-1}\n + a_{k+1} z^{k+1}\n + \\cdots\n + a_{m} z^{m}\n}\n$$\nis satisfied, then the contour contains $k$ roots of the equation\n$$\na_{m} z^{m} + a_{m-1} z^{m-1} + \\cdots a_{1} z + a_{0} = 0.\n$$\n\nFor write\n$$\nf(z)\n=\na_{m} z^{m} + a_{m-1} z^{m-1} + \\cdots a_{1} z + a_{0}.\n$$\n\nThen\n\\begin{align*}\n f(z)\n &=\n a_{k} z^{k}\n \\theparen{ 1 +\n \\frac{ a_{m}z^{m}\n + \\cdots\n + a_{k+1}z^{k+1}\n + a_{k-1}z^{k-1}\n + \\cdots\n + a_{0}\n }{ a_{k} z^{k} }\n }\n \\\\\n &= a_{k} z^{k} (1 + U),\n\\end{align*}\nwhere $\\absval{U} \\leq a \\leq 1$ on the contour, $a$ being\nindependent\\footnote{$\\absval{U}$ is a continuous function of $z$ on\n $C$, and so \\emph{attains} its upper bound\n (\\hardsubsectionref{3}{6}{2}). Hence its upper bound $a$ must be\n less than $1$.} of $z$.\n\nTherefore the number of roots of $f(z)$ contained in $C$\n$$\n= \\frac{1}{2 \\pi i} \\int_{C} \\frac{f'(z)}{f(z)} \\dmeasure z\n= \\frac{1}{2 \\pi i}\n\\int_{C} \\theparen{ \\frac{k}{z}\n + \\frac{1}{1+U} \\frac{ \\dd U }{ \\dd z }\n} \\dmeasure z\n$$\n\nBut $\\int_{C} \\frac{\\dd z}{z} = 2 \\pi i$; and, since $\\absval{U} < 1$,\nwe can expand $(1 + U)^{-1}$ in the uniformly convergent series\n$$\n1 - U + U^{2} - U^{3} + \\cdots,\n$$\nso\n$$\n\\int_{C} \\frac{1}{1+U} \\frac{\\dd U}{\\dd z} \\dmeasure z\n=\n\\thebracket{\n [U - \\half U^{2} + \\frac{1}{3} U^{3} - \\cdots]\n}_{C}\n=\n0.\n$$\n\nTherefore the number of roots contained in $C$ is equal to $k$.\n\\end{wandwexample}\n\\begin{wandwexample}\nFind how many roots of the equation\n$$\nz^{6} + 6z + 10 = 0\n$$\nlie in each quadrant of the Argand diagram.\n\\addexamplecitation{Clare, 1900.}\n\\end{wandwexample}\n%\n% 121\n%\n\n\\Section{6}{4}{Connexion between the zeros of a function and the zeros of its\nderivative.}\\index{Zeros of a function!connected with zeros of its derivative}\nMacdonald\\footnote{TODO} has shewn that \\emph{if $f(z)$ be a function of $z$\nanalytic throughout the interior of a single closed contour $C$, defined\nby the equation $\\absval{f(z)} = M$, where $M$ is a constant, then the\nnumber of zeros of $f(z)$ in this region exceeds the number of zeros of\nthe derived function $f'(z)$ in the same region by unity.}\n\nOn $C$ let $f(z) = M e^{i\\theta}$; then at points on $C$\n$$\nf'(z) = M e^{i\\theta} i \\frac{\\dd \\theta}{\\dd z},\n\\quad\nf''(z)\n=\nM e^{i\\theta}\n\\thebrace{\n i \\frac{ \\dd^{2} \\theta }{ \\dd z^{2} }\n -\n \\theparen{ \\frac{\\dd \\theta}{\\dd z} }^{2}\n}.\n$$\nHence, by \\hardsectionref{6}{3}{1}, the excess of the number of zeros\nof $f(z)$ over the number of zeros of $f'(z)$\ninside\\footnote{$f'(z)$ does not vanish on $C$ unless $C$ has a node\n or other singular point; for, if $f = \\phi + i\\psi$,\n where $\\phi$ and $\\psi$ are real, since\n $i \\frac{\\partial f}{\\partial x} = \\frac{\\partial f}{\\partial y}$,\n it follows that if $f'(z) = 0$ at any point, then\n $\\frac{\\partial \\phi}{\\partial x},\n \\frac{\\partial \\phi}{\\partial y},\n \\frac{\\partial \\psi}{\\partial x},\n \\frac{\\partial \\psi}{\\partial y}$\n all vanish; and these are sufficient conditions for a singular point\n on $\\phi^{2} + \\psi^{2} = M^{2}$.\n} $C$ is\n$$\n\\frac{1}{2\\pi i}\n\\int_{C} \\frac{ f'(z) }{f(z)} \\dmeasure z\n-\n\\frac{1}{2\\pi i}\n\\int_{C} \\frac{ f''(z) }{ f'(z) } \\dmeasure z\n=\n-\\frac{1}{2 \\pi i}\n\\int_{C} \\theparen{\n \\left.\n \\frac{ \\dd^{2} \\theta }{ \\dd z^{2} }\n \\right/\n \\frac{\\dd\\theta}{\\dd z}\n } \\dmeasure z.\n$$\nLet $s$ be the arc of $C$ measured from a fixed point and let $\\psi$\nbe the angle the tangent to $C$ makes with $Ox$; then\n\\begin{align*}\n-\\frac{1}{2 \\pi i}\n\\int_{C} \\theparen{\n \\left.\n \\frac{ \\dd^{2} \\theta }{ \\dd z^{2} }\n \\right/\n \\frac{\\dd\\theta}{\\dd z}\n } \\dmeasure z\n &=\n - \\frac{1}{2 \\pi i}\n \\thebracket{ \\log \\frac{\\dd\\theta}{\\dd z} }_{C}\n \\\\\n &=\n - \\frac{1}{2 \\pi i}\n \\thebracket{\n \\log \\frac{\\dd\\theta}{\\dd s}\n -\n \\log \\frac{\\dd z}{\\dd s}\n }_{C}\n .\n\\end{align*}\nNow $\\log \\frac{\\dd \\theta}{\\dd s}$ is purely real and its initial\nvalue is the same as its final value; and\n$\\log \\frac{\\dd z}{\\dd s} = i \\psi$; hence the excess of the number of\nzeros of $f(z)$ over the number of zeros of\n$f'(z)$ is the change in $\\psi/2\\pi$ in describing the curve $C$; and it is\nobvious\\footnote{For a formal proof, see TODO} that if $C$ is any\nordinary curve, $\\psi$ increases by $2\\pi$ as the point of contact of\nthe tangent describes the curve $C$; this gives the required result.\n\\begin{wandwexample}\n Deduce from Macdonald's result the theorem that a\n polynomial of degree $n$ has $n$ zeros.\n\\end{wandwexample}\n\\begin{wandwexample}\n Deduce from Macdonald's result that if a function $f(z)$,\n analytic for real values of $z$, has all its coefficients real, and all\n its zeros real and different, then between two consecutive zeros of\n $f(z)$ there is one zero and one only of $f'(z)$.\n\\end{wandwexample}\nREFERENCES.\n\nM. C. Jordan, Cours dJ Analyse, 11. (Paris, 1894), Ch. vi.\n\nE. Goursat, Cours d' Analyse (Paris, 1911), Ch. xiv.\n\nE. LiNDELOF, Le Calcid des Residus (Paris, 1905), Ch. il.\n\n%\n% 122\n%\n\nMiscellaneous Examples.\n\\begin{enumerate}\n\\item A function $\\phi(z)$ is zero when $z=0$, and is real when $z$\n is real, and is analytic when $\\absval{z} \\leq 1$; if $f(x,y)$ is\n the coefficient of $i$ in $\\phi(x + iy)$, prove that if $-1 < x < 1$,\n $$\n \\int_{0}^{2\\pi}\n \\frac{x \\sin\\theta }{ 1 - 2x\\cos\\theta + x^{2}}\n f(\\cos\\theta, \\sin\\theta)\n \\dmeasure\\theta\n =\n \\pi \\phi(x).\n $$\n \\addexamplecitation{Trinity, 1898.}\n\\item\n By integrating $\\frac{e^{\\pm aiz}}{e^{2\\pi z}-1}$ round a contour formed by the rectangle whose\n corners are $0, R, R+i, i$ (the rectangle being indented at $0$ and\n $i$) and making $R\\rightarrow\\infty$,\n shew that\n $$\n \\int_{0}^{\\infty}\n \\frac{ \\sin ax }{e^{2 \\pi x} - 1}\n \\dmeasure x\n =\n \\frac{1}{4}\n \\frac{ e^{a} + 1 }{ e^{a} - 1 }\n -\n \\frac{1}{2a}.\n $$\n \\addexamplecitation{Legendre.}\n\n\\item\n By integrating $\\log (-z) Q(z)$ round the contour of\n \\hardsubsectionref{6}{2}{4}, where $Q(z)$ is a rational function such\n that $z Q(z) \\rightarrow 0$ as $\\absval{z} \\rightarrow 0$ and as\n $\\absval{z} \\rightarrow \\infty$, shew that if $Q(z)$ has no poles\n on the positive part of the real axis,\n $\\int_{0}^{\\infty} Q(x) \\dmeasure x$ is equal to minus\n the sum of the residues of $\\log(-z) Q(z)$ at the poles of\n $Q(z)$; where the imaginary part of $\\log(-z)$ lies between $\\pm\\pi$.\n\n\\item\n Shew that, if $a > 0, b > 0$,\n $$\n \\int_{0}^{\\infty}\n e^{a \\cos bx} \\sin (a \\sin bx) \\frac{\\dmeasure x}{x}\n =\n \\half \\pi \\theparen{ e^{a} - 1 }.\n $$\n\n\\item\n Shew that\n $$\n \\int_{0}^{\\half\\pi}\n \\frac{a \\sin 2x}{1 - 2a \\cos 2x + a^{2}}\n x \\dmeasure x\n =\n \\begin{cases}\n \\frac{1}{4} \\pi \\log (1+a), & (-1 1).\n \\end{cases}\n $$\n% \\addexamplecitation{Cauchy.}\n\\item Shew that\n $$\n \\int_{0}^{\\infty}\n \\frac{\\sin \\phi_{1}x}{x}\n \\frac{\\sin \\phi_{2}x}{x}\n \\cdots\n \\frac{\\sin \\phi_{n}x}{x}\n \\cos a_{1} x\n \\cdots\n \\cos a_{m} x\n \\frac{ \\sin ax }{x}\n \\dmeasure x\n =\n \\frac{\\pi}{2} \\phi_{1} \\phi_{2} \\cdots \\phi_{n},\n $$\n if $\\phi_{1}, \\phi_{2}, \\ldots \\phi_{n}, a_{1}, a_{2}, \\ldots a_{m}$\n be real and $a$ be positive and\n $$\n a\n >\n \\absval{\\phi_{1}}\n + \\absval{\\phi_{2}}\n + \\cdots\n + \\absval{\\phi_{n}}\n + \\absval{a_{1}}\n + \\cdots\n + \\absval{a_{m}}.\n $$\n \\addexamplecitation{TODO Stormer, Acta Math, xix.}\n %\n\\item\n If a point $z$ describes a circle $C$ of centre $a$, and if\n $f(z)$ be analytic throughout $C$ and its interior except at a\n number of poles inside $C$, then the point $u=f(z)$ will\n describe a closed curve $\\gamma$ in the $u$-plane.\n Shew that if to each element of $\\gamma$ be attributed\n a mass proportional to the corresponding element of $C$, the centre of\n gravity of $\\gamma$ is the\n point $r$, where $r$ is the sum of the residues of\n $\\frac{f(z)}{z-a}$ at its poles in the interior of $C$.\n \\addexamplecitation{TODO Amigues, Noiiv. Ann. de Math. (.3), xii. (1893), pp. 142-148.}\n %\n\\item Shew that\n $$\n \\int_{-\\infty}^{\\infty}\n \\frac{\\dmeasure x}{ (x^{2}+b^{2}) (x^{2}+a^{2})^{2} }\n =\n \\frac{\\pi (2a+b)}{ 2a^{3}b(a+b)^{2}}.\n $$\n\\item\n Shew that\n $$\n \\int_{0}^{\\infty}\n \\frac{\\dmeasure x}{(a+bx^{2})^{n}}\n =\n \\frac{\\pi}{ 2^{n} b^{\\half} }\n \\frac{1 \\cdot 3 \\cdots (2n-3)}{1 \\cdot 2 \\cdots (n-1)}\n \\frac{1}{ a^{n-\\half} }.\n $$\n%\n% 123\n%\n\n\\item\n If $F_{n}(z)= \\prod_{m=1}^{n-1} \\prod_{p=1}^{n-1} (1-z^{mp})$, shew that the series\n $$\n f(z)\n =\n - \\sum_{n=2}^{\\infty}\n \\frac{F_{n}(zn^{-1})}{ (z^{n}n^{-n} -1) n^{n-1} }\n $$\n is an analytic function when $z$ is not a root of any of the equations\n $z^{n} = n^{n}$; and that the sum of the residues of $f(z)$ contained in the\n ring-shaped space included between two circles whose centres are at\n the origin, one having a small radius and the other having a radius\n between $n$ and $n + 1$, is equal to the number of prime numbers less than\n $n + 1$. \\addexamplecitation{TODO Laurent, A' oiiv. Aim. de Math. (3), xviii. (1899), pp.\n 234-241.}\n\n\\item\n If $A$ and $B$ represent on the Argand diagram two given roots (real\n or imaginary) of the equation $f(z) = 0$ of degree $n$, with real or\n imaginary coefficients, shew that there is at least one root of the\n equation $f'(z) = 0$ within a circle whose centre is the middle point\n of $AB$ and whose radius is $\\half AB \\cot \\frac{\\pi}{n}$.\n \\addexamplecitation{TODO Grace, Proc. Camb. Phil. Soc. XI.}\n\\item\n Shew that, if $0<\\nu<1$,\n $$\n \\frac{e^{2\\pi i \\nu x}}{1 - e^{2\\pi i x}}\n =\n \\frac{1}{2\\pi i}\n \\lim_{n\\rightarrow\\infty}\n \\sum_{k=-n}^{n} \\frac{e^{2k\\nu\\pi i}}{k-x}.\n $$\n [Consider\n $\n \\int \\frac{e^{(2\\nu-1) z \\pi i}}{\\sin \\pi z}\n \\frac{\\dmeasure z}{z-x}\n $\n round a circle of radius $n+\\half$; and make $n\\rightarrow\\infty$.]\n \\addexamplecitation{TODO Kronecker, Journal fiir Math, cv.}\n\\item\n Shew that, if $m > 0$, then\n \\begin{align*}\n \\int_{0}^{\\infty}\n \\frac{\\sin^{n} mt}{t^{n}} \\dmeasure t\n \\\\\n =\n \\frac{\\pi m^{n-1}}{2^{n} (n-1)!}\n \\theparen{\n n^{n-1}\n -\n \\frac{n}{1} (n-2)^{n-1}\n +\n \\frac{n(n-1)}{2} (n-4)^{n-1}\n -\n \\frac{n(n-1)(n-2)}{3!} (n-6)^{n-1}\n +\n \\cdots\n }.\n \\end{align*}\n Discuss the discontinuity of the integral at $m = 0$.\n\n\\item\n If $A + B + C + \\cdots = 0$ and $a, b, c, \\ldots$ are positive, shew that\n $$\n \\int_{0}^{\\infty}\n \\frac{A \\cos ax + B \\cos bx + \\cdots + K \\cos kx}{x}\n \\dmeasure x\n =\n -A \\log a\n -B \\log b\n - \\cdots\n -K \\log k.\n $$\n \\addexamplecitation{Wolstenholme.}\n\\item\n By considering\n $\n \\int \\frac{e^{x(k+ti)}}{k+ti} \\dmeasure t\n $\n taken round a rectangle indented at the origin, shew\n that, if $k > 0$,\n $$\n i \\lim_{\\rho\\rightarrow\\infty}\n \\int_{-\\rho}^{\\rho}\n \\frac{e^{x(k+ti)}}{k+ti} \\dmeasure t\n =\n \\pi i\n +\n \\lim_{\\rho\\rightarrow\\infty}\n P \\int_{-\\rho}^{\\rho} \\frac{e^{xti}}{t} \\dmeasure t,\n $$\n and hence deduce, by using the contour of\n \\hardsubsubsectionref{6}{2}{2}{2} example\n TODO:citeEx2, or its reflexion in the real axis\n (according as $x \\geq 0$ or $x < 0$), that\n $$\n \\lim_{\\rho\\rightarrow\\infty}\n \\frac{1}{\\pi}\n \\int_{-\\rho}^{\\rho}\n \\frac{e^{x(k+ti)}}{k+ti} \\dmeasure t\n =\n 2,\\, , 1,\\, \\textrm{or } 0,\n $$\n according as $x>0$, $x = 0$, or $x < 0$.\n\n [This integral is known as Cauchy's discontinuous factor.]\n%\n\\item\n Shew that, if $0 < a < 2, b > 0, r > 0$, then\n $$\n \\int_{0}^{\\infty}\n x^{a-1}\n \\sin (\\half a\\pi - bx )\n \\frac{r \\dmeasure x}{x^{2} + r^{2}}\n =\n \\half \\pi r^{a-1} e^{-br}.\n $$\n%\n% 124\n%\n\n\\item\n Let $t>0$ and let\n $\n \\sum_{n=-\\infty}^{\\infty}\n e^{-n^{2} \\pi t}\n =\n \\psi(t).\n $\n\n By considering\n $\n \\int \\frac{e^{-z^{2} \\pi t}}{e^{2\\pi i z} - 1} \\dmeasure z\n $\n round a rectangle whose corners are\n $\\pm (N+\\half) \\pm i$, where $N$ is an integer,\n and making $N \\rightarrow \\infty$, shew that\n $$\n \\psi(t)\n =\n \\int_{-\\infty - i}^{\\infty - i}\n \\frac{e^{-z^{2} \\pi t}}{e^{2\\pi i z} - 1} \\dmeasure z\n -\n \\int_{-\\infty + i}^{\\infty + i}\n \\frac{e^{-z^{2} \\pi t}}{e^{2\\pi i z} - 1} \\dmeasure z.\n $$\n\n By expanding these integrands in powers of\n $e^{-2\\pi i z}, e^{2\\pi i z}$\n respectively\n and integrating term-by-term, deduce from \\hardsubsectionref{6}{2}{2} example TODO that\n $$\n \\psi(t)\n =\n \\frac{1}{ \\sqrt{\\pi t} }\n \\psi(1/t)\n \\int_{-\\infty}^{\\infty} e^{-x^{2}} \\dmeasure x.\n $$\n\n Hence, by putting $t = 1$ shew that\n $$\n \\psi(t) = t^{-\\half} \\psi(1/t).\n $$\n\n (This result is due to TODOPoisson, Journal de VEcole foly technique, xii.\n (cahier xix), (1823), p. 420; see also Jacobi, Journal fiir Math,\n xxxvi. (1848), p. 109 [Ges. Werke, ii. (1882), p. 188].)\n\n\\item\n Shew that, if $t>0$,\n $$\n \\sum_{-\\infty}^{\\infty}\n e^{-n^{2} \\pi t - 2 n \\pi a t}\n =\n t^{-\\half}\n e^{\\pi a^{2} t}\n \\thebrace{\n 1\n +\n 2\n \\sum_{n=1}^{\\infty}\n e^{-n^{2} \\pi / t} \\cos 2n\\pi a\n }.\n $$\n \\addexamplecitation{TODOPoisson, Mem. de I'Acad. des Sci. vi. (1827), p. 592; Jacobi,\n Journal fur Math. ill. (1828), pp. 403-404 [Ges. Werke, i. (1881), pp.\n 264-265]; and Landsberg, Journal fur Math. CXI. (1893), pp. 234-253;\n see also \\hardsubsectionref{21}{5}{1}.)}\n\n\\end{enumerate}", "meta": {"hexsha": "2e2b7d0d432880ece967d950605e14eb07616441", "size": 38807, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/tex/wandw-ch06.tex", "max_stars_repo_name": "CdLbB/Whittaker-and-Watson", "max_stars_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/tex/wandw-ch06.tex", "max_issues_repo_name": "CdLbB/Whittaker-and-Watson", "max_issues_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/tex/wandw-ch06.tex", "max_forks_repo_name": "CdLbB/Whittaker-and-Watson", "max_forks_repo_head_hexsha": "5fefdd2c36b4e48cc078a88afc77df289e680a73", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 26.0975117687, "max_line_length": 149, "alphanum_fraction": 0.5987321875, "num_tokens": 15619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217432062975979, "lm_q2_score": 0.853912747375134, "lm_q1q2_score": 0.6163057241889199}} {"text": "\n\\subsection{Atomic formulae}\n\nAtomic formulae are those without operators taking more than one input.\n\nLiterals, and negative literals, are types of atomic formula.\n\nA literal is a formula with no operators.\n\n\\(\\theta \\)\n\nThese are also known as positive literals.\n\nNegative literals are the negation of a literal.\n\n\\(\\neg \\theta \\)\n\n\\subsection{Well-formed formulae}\n\nA well-formed formula is one which can be given a truth value.\n\nThe following is not a well-formed formula:\n\n\\(\\theta \\land \\)\n\n", "meta": {"hexsha": "4ccad85ba902894957eee51a276b6f9710f46ff9", "size": 498, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/logic/propositionalLogic/02-05-formulae.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/logic/propositionalLogic/02-05-formulae.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/logic/propositionalLogic/02-05-formulae.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 19.1538461538, "max_line_length": 71, "alphanum_fraction": 0.7550200803, "num_tokens": 111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8539127455162774, "lm_q2_score": 0.721743200312399, "lm_q1q2_score": 0.6163057177364653}} {"text": "In this section it is explained how to impose an external flux or atmospheric exchange of one or a few constituents into the model. This feature can be used to for example incorporate the flux of dissolved oxygen as a result of aeration in a water quality model or to model the influx of contaminant, organic matter and nutrients separate from the inflows. Several influx models are provided in GIFMod as described below: \n\\begin{itemize}\n\\item \\textbf{Constant influx model: } In this model the flux is assumed to be constant: \n\\begin{equation}\n\\label{eq:28}\nJ_{ext} = k_{ext}\n\\end{equation}\nSo the external influx of a constituent will be calculated as: \n\\begin{equation}\n\\label{eq:29}\n\\dot{m} = A_s k_{ext}\n\\end{equation}\n\n\\item \\textbf{Constant rate model: } In this model the flux is assumed to be proportional to the difference between the concentration of a constituent and a saturation concentration $C_s$: \n\\begin{equation}\n\\label{eq:30}\nJ_{ext} = k_{ext}(C_s-C)\n\\end{equation}\nWhich is applied volumetrically. So the external influx of a constituent will be calculated as: \n \\begin{equation}\n\\label{eq:30a}\n\\dot{m} = k_{ext}(C_s-C) \\sout{V}\n\\end{equation}\n\n\\item \\textbf{Free surface model: } The free surface model is used for open-channel reaeration where the flow depth and velocity affect the rate of aeration. The for of the flux equation is consistent with the forms suggested by \\citep{oconnor1956,churchill1964,owens1964some,isaacs1968,langbein1967}:\n\\begin{equation}\n\\label{eq:31}\nJ_{ext} = k_{ext}v^{\\alpha_{ext}}H^{\\beta_{ext}}(C_s-C)\n\\end{equation}\n\n\n\\item \\textbf{Soil model: } Aeration in soils is expected to be limited by the area of air-water interface. In this model it is assumed that the area of the air-water interface is proportional to the moisture content and air content of the soil. The external flux equation for soils is therefore considered as: \n\n\\begin{equation}\n\\label{eq:32}\nJ_{ext} = k_{ext}\\theta (\\theta_s-\\theta) (C_s-C)\n\\end{equation}\nWhich is applied volumetrically. So the external influx of a constituent will be calculated as: \n \\begin{equation}\n\\label{eq:33}\n\\dot{m} = k_{ext}\\theta (\\theta_s-\\theta) (C_s-C) \\sout{V}\n\\end{equation}\n\\end{itemize}\n\\subsubsection{External Flux Properties: }\nAn external flux model must be added by right-clicking \\textbf{Project Explorer}$\\rightarrow$\\textbf{Water Quality}$\\rightarrow$\\textbf{External Fluxes} and clicking on \\textbf{Add External Flux} and then be assigned to individual blocks through the \\textbf{External Flux} property of the block. \n\nBelow properties that can be assigned to a \\textbf{External flux} model are described: \n\n\\begin{itemize}\n\\item \\textbf{Name: } This is the name of the external flux model. The name is used to assign the model to individual blocks. \n\\item \\textbf{Coefficient: } Here the value of $k_{ext}$ in Eqs. (\\ref{eq:28}, \\ref{eq:30}, \\ref{eq:31}, \\ref{eq:32}) can be entered. This is needed for all external flux models. \n\\item \\textbf{Constituent: } This indicate the constituent that will be released as a result of the external flux. \n\\item \\textbf{Depth exponent: } The value of $\\beta_{ext}$ in the free surface model. \n\\item \\textbf{Model: } Here the type of the model to be used for the external flux will be selected. \n\\item \\textbf{Saturation: } The value of saturation concentration $C_s$ used in the \\textbf{constant rate}, \\textbf{free surface}, \\textbf{soil} will be entered here. \n\\textbf{Solid: } Indicates the solid phase (soil or particles) that the external flux will be assigned to. If left empty the influx occur as dissolved. \n\\item \\textbf{Transfer Rate Expression: } This property allows the user to type in an expression to be used for the calculation of the external flux rates. \n\\item \\textbf{Velocity Exponent: } The value of $\\alpha_{ext}$ in free surface model (Eq. \\ref{eq:31}). \n\\end{itemize}\n\n\\input{Aeration_ex.tex}", "meta": {"hexsha": "ab75da4e2510c31426dcc21624b66824297b0300", "size": 3878, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "GIFMod User's Manual/External_Flux.tex", "max_stars_repo_name": "ArashMassoudieh/GIFMod_", "max_stars_repo_head_hexsha": "1fa9eda21fab870fc3baf56462f79eb800d5154f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-11-20T19:32:27.000Z", "max_stars_repo_stars_event_max_datetime": "2018-08-28T06:08:45.000Z", "max_issues_repo_path": "GIFMod User's Manual/External_Flux.tex", "max_issues_repo_name": "ArashMassoudieh/GIFMod_", "max_issues_repo_head_hexsha": "1fa9eda21fab870fc3baf56462f79eb800d5154f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-07-04T05:40:30.000Z", "max_issues_repo_issues_event_max_datetime": "2017-07-04T05:43:37.000Z", "max_forks_repo_path": "GIFMod User's Manual/External_Flux.tex", "max_forks_repo_name": "ArashMassoudieh/GIFMod_", "max_forks_repo_head_hexsha": "1fa9eda21fab870fc3baf56462f79eb800d5154f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-11-09T22:00:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-08-30T10:56:08.000Z", "avg_line_length": 63.5737704918, "max_line_length": 422, "alphanum_fraction": 0.7607013925, "num_tokens": 1069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8539127455162773, "lm_q2_score": 0.721743200312399, "lm_q1q2_score": 0.6163057177364651}} {"text": "\\section{Extensions to the Problem}\nFor the bulk of this thesis, we have looked at many-to-one matching problems with one-sided preferences, as this setting makes the most sense for the student-seminar application. However, there is a wide range of similar problems and extensions that are also worth mentioning. This section will present some of those problems and key results from the literature. \n\n\\subsection{Two-Sided Preferences}\\label{extensions:two-sided}\nIn many practical one-to-many matching schemes, both parties will express preferences over the other. For instance, many countries use such schemes for matching students to university programs, which can be viewed as an instance of a one-to-many matching problem with two-sided preferences. A similar problem exists in the United States, where graduating medical students need to be matched to residency positions in hospitals. \n\nSince we are considering two-sided preferences now, it is not possible to simply transform this problem into a weighted bipartite graph anymore, since edges now have two weights. It could be possible to use the product of those two weights as an edge weight; however, in practice different mechanisms are being used that do not rely on the graph representation of the problem.\n\n\\subsubsection{Stable Matchings}\nIn the context of two-sided preferences, \\emph{Stability} is a commonly used optimality criteria that matchings should fulfill. It can be seen as the two-sided preference extension to Pareto-optimality and is defined as follows. Given a set $H$ of hospitals and a set $R$ of residents, a matching $M$ of an instance $I$ is stable iff there is no pair $(r_i, h_j) \\in M, r_i \\in R, h_j \\in H$ with:\n\\begin{itemize}\n \\item $r_i$ and $h_j$ are on each other's preference lists\n \\item either $r_i$ is assigned in $M$, or $r_i$ prefers $h_j$ to $M(r_i)$\n \\item either $h_j$ is undersubscribed in $M$, or $h_j$ prefers $r_i$ to it is least preferred assigned resident in $M(h_j)$\n\\end{itemize}\nSimply put, in a stable matching $M$, no pair of agents $(r_i, h_j)$ should have an incentive to give up their current match to get matched to each other.\n\n\\subsubsection{Algorithm}\nGale and Shapley \\cite{GaleShapleyOrig} proved that for every instance of the problem, there exists a stable matching that can be computed using a linear-time algorithm. There are two variants of the algorithm, one that computes a resident-optimal, hospital pessimal matching and one that does the opposite. That means that in the first case, each resident is matched to the best hospital among all stable matchings, while each hospital is matched to the worst set of residents among all stable matchings \\cite{Gusfield}. In the context of student seminar matchings, it would make sense to use the resident/student optimal algorithm, which goes as follows \\cite{Gusfield}:\n\\begin{algorithm} % enter the algorithm environment\n \\caption{Resident-oriented Deferred Acceptance Algorithm} \n \\label{alg:resident-oriented-algorithm} \n \\begin{algorithmic} % enter the algorithmic environment\n \\Require set of residents with preferences $R$, set of hospitals $H$\n \\Ensure Stable Matching M\n \\While{Some resident $r$ is free and has a non-empty preference list} \n \\State $h:=$ first hospital on $r$'s preference list\n \\If{$h$ is fully subscribed}\n \\State $r':=$ worst resident provisionally assigned to h\n \\State unassign $r'$ from $h$\n \\EndIf\n \\State provisionally assign $r$ to $h$\n\n \\If{$h$ is fully subscribed}\n $s:=$ worst resident provisionally assigned to $h$\n \\ForEach{successor $s'$ of $s$ on $h$'s list}\n \\State remove $s'$ and $h'$ from each other's list\n \\EndFor\n \\EndIf\n\n \\EndWhile\n \\end{algorithmic}\n\\end{algorithm}\n\nAn advantage of using this algorithm for matching students to seminars is that the order in which students are processed does not matter. Gusfield shows that for any permutation of the input, each hospital is assigned the same number of residents, the same set of residents is unassigned and each undersubscribed hospital is matched to the same set of residents \\cite{Gusfield}.\n\n\\subsection{Many-to-Many Matchings}\nA natural extension of the many-to-one matching setting is the many-to-many matching problem. It could very well be a requirement that students need to be matched to more than one seminar. In that case, we need to differentiate between one- and two-sided preferences again:\n\n\\subsubsection{One-sided Preferences}\nMany-to-many matching scenarios with one-sided preferences are also commonly referred to as the \\emph{Course allocation} problem, which is a combinatorial assignment problem with the goal of assigning students to courses, based on the students' preferences \\cite{CourseAllocation}. Compared to the student and seminar scenario we now allow students to get matched to more than one course. \n\nA common mechanism for solving course allocation problems is using an \\emph{Auction model} or \\emph{Bidding points mechanism} in which students are given a set amount of artificial currency, which they then use to bid on seats in classes. After all bids are submitted, the system assigns the seats to the highest bidders. The bids can be used to infer students' preferences; however, the true preferences may differ significantly, which makes this mechanism less desirable \\cite{Bidding}.\n\nAnother commonly used mechanism is the \\emph{Draft}, in which students are asked to pick a course with remaining seats based on a draft order. Intuitively, this mechanism is an extension of RSD (Section \\ref{algo-rsd}), where the whole process is repeated until no student makes a pick anymore. Such draft mechanisms are commonly used in course allocation settings; one example being Harvard Business School \\cite{CourseAllocation}. However, it has been shown that students have successfully manipulated the draft due to the fact that this mechanism is not strategy-proof.\n\nA mechanism that is more robust towards strategic manipulation is a \\emph{proxy bidding mechanism} presented by Kominers et al. \\cite{CourseAllocation}. The mechanism uses proxies that act on behalf of the students by using their true preferences in the draft.\n\n\\subsubsection{Two-sided Preferences}\nFor the case of two-sided preferences, we have to slightly modify the definition of Stability again to accommodate for the fact that each entity can be assigned more than once: Given the set of hospitals $H$ and residents $R$, let $h \\in H$ and $r \\in R$, so that $h$ and $r$ are acceptable to each other, unmatched and the following holds true: \n\\begin{itemize}\n \\item either $h$ has unfilled places or prefers $r$ to one of his matched partners \\textbf{and}\n \\item either $r$ has unfilled places or prefers $h$ to one of his matched partners\n\\end{itemize}\nUsing this definition of stability, Gusfield \\cite{Gusfield} proposes that an algorithm can be constructed using ideas from the hospital-oriented and resident-oriented (see Algorithm \\ref{alg:resident-oriented-algorithm}) algorithm that finds a stable matching that's optimal for either of the sets. Again, such a stable matching exists for any instance of the problem \\cite{Gusfield}.\n\n\\subsection{Online Variants}\\label{sec:online-variants}\nWhen solving the \\emph{online-variant} of the problem, the whole input is not available from the start. That means that the input needs to be processed piece by piece, or more formally: Given a bipartite weighted graph $(U, V, E)$, where $U$ is known to the algorithm, vertices in $V$ are unknown, but arrive one at a time, while also revealing their incident edges, find a matching that maximizes some objective function. These algorithms could be of interest in the case of a first-come first-serve course allocation system, or in other areas such as DVD-rental or online-advertisement allocation systems \\cite{Mehta:Online}.\n\nIn the case of student-seminar assignments, we would assume that the set of courses and their capacities is known beforehand, and the students arrive later. One of the algorithms we have seen in Section \\ref{chapter:algorithms} can be used for this problem, namely the RSD-algorithm. As a matter of fact, the algorithm will produce the same results for the offline and online case, given that the order in which students are processed is identical.\n\n\\subsubsection{Online Maximum Cardinality Matching}\nThere has been lots of research in particular on finding maximum-cardinality matchings with online inputs. The online-inputs are classified by how much information the algorithm possesses about the input order. For now, we will only consider the \\emph{adversarial order}, where we assume no knowledge of the query sequence, which means that only $U$ is known at the beginning of the algorithm, while we have no knowledge of $V$ and $E$ or the order they appear in \\cite{Mehta:Online}. To measure performance, we will use the \\emph{competitive ratio} of an algorithm which is defined as follows. Given an instance of the problem $I$, the value of the objective function for the online algorithm is given as $ALG(I)$, and the value of the objective function for the best offline algorithm is given as $OPT(I)$. The competitive ratio is now computed as follows: $C.R.=\\frac{ALG(I)}{OPT(I)}$ \\cite{Mehta:Online}.\n\nA simple algorithm for this online problem is a greedy algorithm, which matches arriving vertices to any available neighbor, or a random approach that matches arriving vertices to a random neighbor. These mechanisms achieve a competitive ratio of $\\frac{1}{2}$ \\cite{Mehta:Online}. An optimal, yet simple algorithm was introduced by Karp et al. \\cite{Karp:Online}, which achieves a competitive ratio of $1 - \\frac{1}{e} \\simeq 0.63$. The algorithm, called Ranking, begins by permuting the known vertices of $U$ in a random permutation $\\pi$, i.e. we assign a random priority number to each $u \\in U$. Each incoming vertex $v \\in V$ is then assigned to an available neighbor, with the smallest value of $\\pi(u)$. In detail, the algorithm looks like this:\n\n\\begin{algorithm} % enter the algorithm environment\n \\caption{Ranking} \n \\label{alg:ranking} % and a label for \\ref{} commands later in the document\n \\begin{algorithmic} % enter the algorithmic environment\n \\State \\textbf{Offline:} Pick a random, uniform permutation $\\pi$ of U\n \\ForEach {arriving vertex $v \\in V $}\n \\If{$v$ has no available neighbors}\n \\State continue\n \\EndIf\n \\State Match $v$ to the neighbor $u \\in U$ with the smallest value $\\pi(u)$\n \\EndFor\n \\end{algorithmic}\n\\end{algorithm}", "meta": {"hexsha": "32ba5d2f41405dca1af219e31155b459f91d0c0d", "size": 10724, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/7_extensions.tex", "max_stars_repo_name": "aaronoe/bachelorarbeit", "max_stars_repo_head_hexsha": "b267dc2eb69cc7a1c2421b76277f69517957375d", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/7_extensions.tex", "max_issues_repo_name": "aaronoe/bachelorarbeit", "max_issues_repo_head_hexsha": "b267dc2eb69cc7a1c2421b76277f69517957375d", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/7_extensions.tex", "max_forks_repo_name": "aaronoe/bachelorarbeit", "max_forks_repo_head_hexsha": "b267dc2eb69cc7a1c2421b76277f69517957375d", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 120.4943820225, "max_line_length": 908, "alphanum_fraction": 0.7657590451, "num_tokens": 2445, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833945721304, "lm_q2_score": 0.7956580927949806, "lm_q1q2_score": 0.6163035464359232}} {"text": "\\documentclass{article}\n\\usepackage[margin=2.5cm, includefoot, footskip=30pt]{geometry}\n\\pagestyle{plain}\n\\setlength{\\parindent}{0em}\n\\setlength{\\parskip}{1em}\n\\renewcommand{\\baselinestretch}{1}\n\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\n\\newcommand{\\R}{\\mathbb{R}}\n\n\\title{The effect of homophily in hierarchical systems}\n\n\\renewcommand{\\S}{\\mathcal{S}}\n\n\\begin{document}\n\n\\maketitle\n\n\n\\section{Mathematical formulation of base model}\n\n\\begin{itemize}\n \\item Given a hierarchical system with \\(K\\) levels.\n \\item Level \\(0\\leq k < K\\) has capacity \\(C_k\\)\n \\item The first level (\\(k=0\\)) has the most capacity and capacity is\n monotonically decreasing: \\(C_0 \\geq C_1 \\geq \\dots \\geq C_{k-2} \\geq C_{k-1} = 1\\).\n \\item There are 2 types of agents: \\(j\\in\\{0, 1\\}\\).\n\\end{itemize}\n\nConsider a state space \\(\\S\\):\n\n\\begin{equation}\\label{eqn:state_space}\n \\S = \\left\\{s \\in \\mathbb{Z} ^ {K \\times 2} _ {\\geq 0}\\left|\n \\begin{array}{l}\n s_{i0} + s_{i1} \\leq C_i\\text{ for all }0\\leq i \\leq K - 1\\\\\n s_{K-1} = (1, 0)\\\\\n \\sum_{i=0}^{K - 1}s_{i0} + s_{i1} \\in\\left\\{\\sum_{i=0}^{K - 1}C_i, \\sum_{i=0}^{K - 1}C_i - 1\\right\\}\\\\\n\n \\end{array}\n \\right.\\right\\}\n\\end{equation}\n%TODO comment on top row\nWhere \\(s_{ij}\\) denotes the number of individuals of type \\(j\\) at level \\(i\\).\n\nFor example,\n\n\\begin{itemize}\n \\item Let \\(K = 3\\)\n \\item Let \\(C = (4, 3, 1)\\)\n\\end{itemize}\n\nThen:\n\n\\[\n s = \\begin{pmatrix}\n 3 & 1 \\\\\n 2 & 1 \\\\\n 1 & 0 \\\\\n \\end{pmatrix}\n\\]\n\ncorresponds to a system with 3 agents of first type and 1 of second type at the\nfirst level, 2 of first type and 1 of second type at the second level and 1 of\neach type at the 3rd level.\n\nThe constraints on \\(\\S\\) ensure that either all positions are filled or a\nsingle position is available. Thus at any stage either all spots are full and\nsomeone will retire or there will be a spot available and someone will be\nhired/promoted.\n\nThe size of the state space is then given by:\n\n\\begin{equation}\n |S| = \\prod_{i=0}^{K - 2}\\left(2C_i + 1\\right)\n\\end{equation}\n\nGiven two elements \\(s^{(1)}, s^{(2)}\\in \\S\\) the transition rates are given by:\n\n\\begin{equation}\\label{eqn:transition_rates}\n Q_{s_1, s_2} =\n \\begin{cases}\n \\mu_{ij},& \\text{ if }s^{(2)} - s^{(1)} = -e_{ij}\n \\text{ and }s^{(1)}_{i0} + s^{(1)}_{i1} = C_i\\text{ for all }i\\\\\n \\text{max}(rs_{i + 1,j} + s_{i + 1, \\bar j}, 1),& \\text{ if }s^{(2)} - s^{(1)} = e_{ij} - e_{i-1, j}\n \\text{ and }s^{(1)}_{i0} + s^{(1)}_{i1} < C_{i}\n \\text{ and }i > 0\\\\\n \\lambda_{j},&\\text{ if }s^{(2)} - s^{(1)} = e_{0j}\n \\text{ and }s^{(1)}_{00} + s^{(1)}_{01} = C_{0} - 1\\\\\n \\end{cases}\n\\end{equation}\n\nWhere:\n\n\\begin{itemize}\n \\item \\(\\mu_{ij}\\) is the retirement rate of agents of type \\(j\\) at level\n \\(i\\).\n \\item \\(r > 1\\) is a constant that reflects the homophily effect.\n \\item \\(\\lambda_j\\) is the hiring rate of individuals of type \\(j\\).\n\\end{itemize}\n\n% TODO Add some examples.\n\n\\section{Mathematical formulation of model with competence}\n\nThe base model described in the previous section does not take into account the\ncompetence of the individuals. Let's consider the above example of:\n\n\\begin{itemize}\n \\item \\(K = 3\\) and \\(C = (4, 3, 1)\\)\n\\end{itemize}\n\nA possible state \\(s\\) for the above configuration when competence is taken into\naccount is:\n\n\\[\n s = \\begin{pmatrix}\n (1, 0.1) & (1, 0.2) & (1, 0.3) & (0, 0.5) \\\\\n (1, 0.2) & (1, 0.3) & (0, 0.8) \\\\\n (1, 0.7) \\\\\n \\end{pmatrix}\n\\]\n\nNote that this is a possible state for the configurations of \\(K\\) and \\(C\\) but\nnot the only one. The competence of each individual can be any real number between\n\\([0, 1]\\), thus there are an infinite number of possible states.\n\nThe set of possible states, denoted as \\(S\\), is given by,\n\n\\begin{equation}\\label{eqn:state_space}\n \\S = \\left\\{\n s \\in (\\{0, 1\\}, \\R^{[0,1]}) ^ {K \\times C} \\left|\n \\begin{array}{l}\n s_{i,j} = (l_{i, j}, c_{i, j}) \\quad \\forall i \\in [0, K -1] \\quad \\forall j \\in [C_i - 1, C_i] \\\\\n \\text{ where } l_{i, j} \\in \\{0, 1\\} \\text{ and } c_{i, j} \\in \\R^{[0,1]} \\\\\n |s_{k - 1}| = 1 \\\\\n l_{k -1, 1} = 0\n \\end{array}\n \\right.\\right\\}\n\\end{equation}\n\nwhere \\(s_{i, j}\\) is a tuple of \\((l_{i, j}, c_{i, j})\\) where \\(l_{i, j}\\) is the\ntype of individual \\(j\\) at level \\(i\\) and \\(c_{i, j}\\) is the individual's\ncompetence.\n\nPromotion in the competence model assumes that individuals of the same type give a random\nbonus to the competence of individuals like them.\nThus for a state \\(s\\) where a promotion is happening at level \\(i\\), the promotion\nprobability of an individual \\(j\\) at level \\(i - 1\\) is given by:\n\n\\begin{equation}\n P(i, j) =\n \\frac{\n \\displaystyle \\sum_{t = 1} ^ {C_{i -1} - 1} \\text{max}((1 - |l_{i-i, t} - l_{i, j}|) \\gamma_t, 1) \\ c_{i, j}}\n {\n \\displaystyle \\sum_{\\bar{j}=1}^{C_i} \\sum_{t = 1} ^ {C_{i -1} - 1} \\text{max}((1 - |l_{i-i, t} - l_{i, j}|) \\gamma_t, 1) \\ c_{i, \\bar{j}}\n },\n\\end{equation}\n\nwhere \\(\\gamma_{t} > 1\\) is randomly sampled at each \\(t\\) from \\([0, \\Gamma]\\).\n\nThus, the transition rates between two states are given by:\n\n\\begin{equation}\\label{eqn:transition_rates}\n \\begin{cases}\n \\mu_{ij},& |s_i| = C_i\\text{ for all }i\\\\\n P(i, j), & |s_i| < C_{i} \\text{ and }i > 0\\\\\n \\lambda_{(l, c)},& |s_{0}| = C_0 - 1\\\\\n \\end{cases}\n\\end{equation}\n\nWhere:\n\n\\begin{itemize}\n \\item \\(\\mu_{ij}\\) is the retirement rate of the agent \\(j\\) at level\n \\(i\\).\n \\item \\(\\lambda_{(l, c)}\\) is the hiring rate of an individual \\(l\\) that has\n a competence \\(c\\).\n\\end{itemize}\n\nThe competence of the system is the total competence of the individuals in the\nsystem regarding their type.\n\n\\begin{equation}\\label{eqn:state_space}\n \\sum_{i = 0} ^ {K - 1} \\sum_{j=1} ^ {C_i} c_{i, j}\n\\end{equation}\n\n\n\\end{document}\n\n\n", "meta": {"hexsha": "b8947dda75ebf1ee036bc8c88cc55102b86959e2", "size": 6102, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main.tex", "max_stars_repo_name": "drvinceknight/HierarchicalPromotion", "max_stars_repo_head_hexsha": "8fce38c4dc9b21f50a8ef769482fd6a82cf0e6a3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "main.tex", "max_issues_repo_name": "drvinceknight/HierarchicalPromotion", "max_issues_repo_head_hexsha": "8fce38c4dc9b21f50a8ef769482fd6a82cf0e6a3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2019-10-01T06:47:05.000Z", "max_issues_repo_issues_event_max_datetime": "2020-11-18T13:10:20.000Z", "max_forks_repo_path": "main.tex", "max_forks_repo_name": "drvinceknight/HierarchicalPromotion", "max_forks_repo_head_hexsha": "8fce38c4dc9b21f50a8ef769482fd6a82cf0e6a3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.6165803109, "max_line_length": 141, "alphanum_fraction": 0.5806293019, "num_tokens": 2212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8652240756264638, "lm_q2_score": 0.7122321842389469, "lm_q1q2_score": 0.61624043323956}} {"text": "\\documentclass[12pt]{article}\r\n\\usepackage{amsmath}\r\n\\usepackage{txfonts}\r\n\r\n%- Abbreviations.\r\n\\newcommand{\\psf}{\\mathbf{H}}\r\n\\newcommand{\\G}{\\mathbf{G}}\r\n\\newcommand{\\p}{\\mathbf{p}}\r\n\\newcommand{\\phat}{\\mathbf{\\hat p}}\r\n\\newcommand{\\hhat}{\\mathbf{\\hat h}}\r\n\\newcommand{\\x}{\\mathbf{x}}\r\n\\newcommand{\\xhat}{\\mathbf{\\hat x}}\r\n\\newcommand{\\y}{\\mathbf{y}}\r\n\\newcommand{\\rvect}{\\mathbf{r}}\r\n\\newcommand{\\W}{\\mathbf{W}}\r\n\\newcommand{\\A}{\\mathbf{A}}\r\n\\newcommand{\\avec}{\\mathbf{a}}\r\n\\newcommand{\\F}{\\mathbf{F}}\r\n\\newcommand{\\V}{\\mathbf{V}}\r\n\\newcommand{\\diag}{\\mathrm{diag}}\r\n\r\n\\newcommand{\\ddx}{\\frac{\\partial}{\\partial \\mathbf{x}}}\r\n\\newcommand{\\ddxi}{\\frac{\\partial}{\\partial x_i}}\r\n\r\n\\newcommand{\\Nsn}{\\textbf{XXX}}\r\n\\newcommand{\\Nspec}{\\textbf{YYY}}\r\n\r\n\r\n\\newcommand{\\rem}[1]{\\textbf{[\\textsl{#1}]}}\r\n\\newcommand{\\szf}[1]{\\textbf{\\textsl{SZF: #1}}}\r\n\r\n% -----------------------------------------------------------------------------\r\n% Document\r\n\r\n\\begin{document}\r\n\r\n\\title{CubeFit $\\chi^2$ gradient calculation}\r\n\\author{S.~Bongard, Kyle~Barbary, Clare~Saunders}\r\n\\maketitle\r\n\\begin{abstract}\r\nIn CubeFit, one of the fitting steps is to fit the galaxy model to the\r\ndata while holding other parameters (such as positions) fixed. The\r\ngalaxy model has on the order of one million parameters. The ability to\r\nanalytically calculate the gradient (a vector of length $\\sim$ 1\r\nmillion) on the fit $\\chi^2$ is essential in order to carry out the fit\r\nefficiently. This note explains the derivation of the gradient function.\r\n\\end{abstract}\r\n\r\n\r\n\\section{The Gradient}\r\n\r\nThe $\\chi^2$ in the galaxy fit is given by\r\n\r\n\\begin{equation}\r\n\\chi^2 = (\\y - \\G\\x)^T \\W (\\y - \\G\\x)\r\n\\end{equation}\r\n\r\n\\noindent where $\\y$ is the data, $\\x$ is the model parameters, $\\W$\r\nis the weight matrix (inverse covariance matrix; diagonal in our case)\r\nand $\\G$ is the operator that translates the model into the data\r\nframe. $\\G$ includes the effect of PSF convolution, shift operation\r\nand cropping. Note that we are supposing that we can represent all\r\nthese operations as a matrix $\\G$, but we will not show this until a\r\nlater section. Remember also that this is all in real space. Carrying\r\non,\r\n\r\n\\begin{eqnarray*}\r\n \\label{eq:5}\r\n \\ddx \\chi^{2} & = & \\ddx \\left( (\\y - \\G\\x)^{T} \\W (\\y - \\G\\x) \\right)\\\\\r\n & = & \\ddx \\left( (\\G \\x)^{T} \\W \\G \\x \\right)\r\n - \\ddx \\left( (\\G \\x)^{T} \\W \\y \\right)\r\n - \\ddx \\left( \\y^{T} \\W \\G \\x \\right) \\\\\r\n & = & \\ddx \\left( \\x^{T} \\G^{T} \\W \\G \\x \\right)\r\n - \\ddx \\left( \\x^{T} \\G^{T} \\W \\y \\right)\r\n - \\ddx \\left( (\\G^{T} \\W \\y)^{T} \\x \\right)\r\n\\end{eqnarray*}\r\n\r\n\\noindent where in the second line we have simply separated terms, dropping the\r\nconstant term $\\y^T \\W \\y$, and in the third line we have used the property\r\nthat $(\\mathbf{A}\\mathbf{B})^T = \\mathbf{B}^T\\mathbf{A}^T$ and assumed that\r\n$\\W = \\W^T$ (the weight matrix is symmetric).\r\n\r\nWe note that $(\\G^{T} \\W \\y)^{T}$ and $\\G^{T} \\W \\y$ are vectors ($\\y$ is a\r\nvector). The first two terms of the equation are thus of the form\r\n$\\ddx \\x^T \\avec$ and $\\ddx \\avec^T \\x$. These are both simply equal to\r\n$\\avec$ as $\\x^T \\avec$ and $\\avec^T \\x$ are dot products between the\r\nconstant vector $\\avec$ and $\\x$, so the partial derivative with respect to\r\n$\\x$ is just $\\avec$. Making this substitution, we have\r\n\r\n\\begin{equation}\r\n\\ddx \\chi^{2} = \\ddx \\left( \\x^{T} \\G^{T} \\W \\G \\x \\right) - 2 \\G^T \\W \\y \r\n\\end{equation}\r\n\r\n\\noindent The first term has the form $\\ddx \\x^T\\A\\x$, with\r\n$\\A = \\G^T \\W \\G$. It helps to write this out:\r\n\r\n\\begin{eqnarray*}\r\n \\ddxi \\x^{T}\\A\\x & = & \\ddxi \\left( \\sum_{j} x_{j} \\sum_l A_{j,l} x_{l} \\right) \\\\\r\n & = & \\sum_l A_{i,l} x_{l} + \\sum_j A_{j,i} x_{j} \\\\\r\n & = & \\left( \\A \\x + \\A^{T} \\x␇ \\right)_{i}\r\n\\end{eqnarray*}\r\n\r\n\\noindent or without the $i$ subscripts,\r\n\r\n\\begin{equation}\r\n \\ddx \\x^T\\A\\x = (\\A + \\A^T) \\x\r\n\\end{equation}\r\n\r\n\\noindent Substituting this in (with $\\A = \\G^T \\W \\G$), we finally get:\r\n\r\n\\begin{eqnarray*}\r\n\\ddx \\chi^{2} & = & \\left( \\G^T \\W \\G \\x + (\\G^T \\W \\G)^T \\x \\right)\r\n - 2 \\G^T \\W \\y \\\\\r\n & = & 2 \\G^T \\W \\G \\x - 2 \\G^T \\W \\y \\\\\r\n & = & -2 \\G^T \\W ( \\y - \\G \\x ) \\\\\r\n & = & -2 \\G^T \\W \\rvect \\\\\r\n\\end{eqnarray*}\r\n\r\n\\noindent where $\\rvect$ is shorthand for the residual\r\n$\\y - \\G \\x$. But what is the matrix $\\G$ and what is $\\G^T$?\r\nFor that we need to digress about\r\nFourier transforms as matricies (in the next section).\r\n\r\n\\section{Discrete Fourier Transform: Matrix Form}\r\n\r\nThe discrete Fourier transform of a vector $x_n$ is\r\n\r\n\\begin{equation}\r\n\\hat{x}_k = \\sum_{n=0}^{N-1} x_n e^{-i 2 \\pi k n / N}\r\n\\end{equation}\r\n\r\n\\noindent We can write this in matrix form,\r\n\r\n\\begin{equation}\r\n\\xhat = \\left( \\begin{array}{cccc}\r\n e^{-i 2\\pi 0 \\cdot 0 / N} & e^{-i 2\\pi 0\\cdot 1 / N} & e^{-i 2\\pi 0 \\cdot 2 / N} & \\cdots \\\\\r\n e^{-i 2\\pi 1 \\cdot 0 / N} & e^{-i 2\\pi 1 \\cdot1 / N} & e^{-i 2\\pi 1 \\cdot 2 / N} & \\cdots \\\\\r\n e^{-i 2\\pi 2 \\cdot 0 / N} & e^{-i 2\\pi 2 \\cdot 1 / N} & e^{-i 2\\pi 2 \\cdot 2 / N} & \\cdots \\\\\r\n \\vdots & \\vdots & \\vdots & \\ddots\r\n \\end{array} \\right)\r\n \\left( \\begin{array}{c} x_0 \\\\ x_1 \\\\ x_2 \\\\ \\vdots \\end{array} \\right)\r\n\\end{equation}\r\n\r\n\\noindent For a shorthand, we will call the square matrix $\\mathbf{F}$\r\n\r\n\\begin{equation}\r\n\\xhat = \\F \\x\r\n\\end{equation}\r\n\r\nSimilarly, the inverse transform can be written\r\n\r\n\\begin{equation}\r\nx_n = \\frac{1}{N} \\sum_{k=0}^{N-1} \\hat{x}_k e^{i 2 \\pi k n / N}\r\n\\end{equation}\r\n\r\n\\noindent or\r\n\r\n\\begin{equation}\r\n \\mathbf{x} = \\frac{1}{N} \\left( \\begin{array}{cccc}\r\n e^{i 2\\pi 0 \\cdot 0 / N} & e^{i 2\\pi 1 \\cdot 0 / N} & e^{i 2\\pi 2 \\cdot 0 / N} & \\cdots \\\\\r\n e^{i 2\\pi 0 \\cdot 1 / N} & e^{i 2\\pi 1 \\cdot 1 / N} & e^{i 2\\pi 2 \\cdot 1 / N} & \\cdots \\\\\r\n e^{i 2\\pi 0 \\cdot 2 / N} & e^{i 2\\pi 1 \\cdot 2 / N} & e^{i 2\\pi 2 \\cdot 2 / N} & \\cdots \\\\\r\n \\vdots & \\vdots & \\vdots & \\ddots\r\n \\end{array} \\right)\r\n \\left( \\begin{array}{c} \\hat{x}_0 \\\\ \\hat{x}_1 \\\\ \\hat{x}_2 \\\\ \\vdots \\end{array} \\right)\r\n\\end{equation}\r\n\r\n\\noindent or as shorthand,\r\n\r\n\\begin{equation}\r\n\\x = \\F^{-1} \\xhat\r\n\\end{equation}\r\n\r\n\\noindent Now, note that the matrix $\\F$ has the property\r\n\r\n\\begin{equation}\r\n\\F^{-1} = \\frac{1}{N} \\bar{\\F}\r\n\\end{equation}\r\n\r\n\\noindent where $\\bar{\\F}$ is the element-wise complex conjugate of $\\F$.\r\n\r\n\r\n\\section{The $\\G$ matrix }\r\n\r\nThe $\\G$ matrix represents the operations applied to the model to put\r\nit \"in the data frame\". These are (1) convolve model by PSF (2) shift\r\nmodel to align with data (3) crop model indicies that extend past the\r\ndata (model is spatially larger than data). Convolution in real space\r\nis equivalent to element-wise multiplication (the Hadamard product) in\r\nFourier space (and this is indeed how the convolution is implemented\r\nin CubeFit). While element-wise multiplication can be written\r\n$\\mathbf{a} \\circ \\mathbf{b}$, we will find it useful to write it as\r\n$\\diag(\\mathbf{a}) \\mathbf{b}$, where diag() is an operator\r\nthat translates a vector into a square diagonal matrix with the same\r\nentries.\r\n\r\nConvolution of the galaxy model $\\x$ by a real-space PSF $\\p$ can thus be\r\nwritten as\r\n\\begin{equation}\r\n\\F^{-1} \\diag(\\F \\p) \\F \\x\r\n\\end{equation}\r\n\r\nSimilarly, spatial shifts are implemented as element-wise multiplication in\r\nFourier space. the key point being that we can\r\nsubsume this shift vector (or rather the Fourier transform of it) into\r\n$\\F \\p = \\phat$. Therefore, we can write $\\G$ as\r\n\r\n\\begin{equation}\r\n\\G = \\F^{-1} \\diag(\\phat) \\F\r\n\\end{equation}\r\n\r\nWe are now interested in calculating $\\G^T$ as it appears in the\r\nequation for $\\ddx \\chi^{2}$. Simply taking the transpose, we get\r\n$\\G^{T} = \\F^T \\diag(\\phat)^{T} \\F^{-1T}$, but we would like\r\nto rewrite this in a form that has the Fourier transforms in the same\r\norder as in $\\G$. Since $\\G$ is a real operator, we have $\\G =\r\n\\bar{\\G}$, and thus $\\G^T = \\bar{\\G}^T$. We will use this along with\r\nthe relation between the forward and reverse Fourier transform derived\r\nabove ($\\F^{-1} = \\frac{1}{N} \\bar{F}$), and the fact that the\r\ntranspose of a diagonal matrix is the matrix itself. We find:\r\n\r\n\\begin{eqnarray*}\r\n \\G^{T} & = & \\F^{T} \\diag(\\phat)^{T} \\F^{-1T}\\\\\r\n & = & \\overline{\\F^{T} \\diag(\\phat)^{T} \\F^{-1T}} \\\\\r\n & = & \\bar{\\F}^T \\overline{\\diag(\\phat)^T} \\bar{\\F}^{-1T} \\\\\r\n & = & N\\F^{-1} \\overline{\\diag(\\phat)} \\frac{1}{N}\\F \\\\\r\n & = & \\F^{-1} \\overline{\\diag(\\phat)} \\F\\\\\r\n\\end{eqnarray*}\r\n\r\n\\noindent If we now write the residual $\\rvect = \\y - \\G\\x$, we\r\nfinally find that the derivative of $\\chi^{2}$ is:\r\n\r\n\\begin{eqnarray*}\r\n \\ddx \\chi^2 & = & -2 \\G^T \\W \\rvect \\\\\r\n & = & -2 \\F^{-1} \\overline{\\diag(\\phat)} \\F \\W \\rvect\r\n\\end{eqnarray*}\r\n\r\n\\noindent In CubeFit, the operator $\\G^T = \\F^{-1}\r\n\\overline{\\diag(\\phat)} \\F$ is currently implemented by the method\r\n\\verb|AtmModel.gradient_helper()|.\r\n\r\n\\section{Fitting galaxy and sky simultaneously}\r\n\r\nIn another step in CubeFit, we fit the galaxy model to multiple epochs\r\nof data simultaneously, while adjusting the sky for each epoch on each\r\niteration. On each iteration, the sky is calculated by taking the\r\nweighted average of the residual ($\\y - \\G \\x$) at each wavelength and\r\nsubtracting that to form the new residual. We need to take this into\r\naccount in the $\\chi^2$ gradient: qualitatively, increasing any given\r\nmodel pixel will result in a lower sky estimate, and the sky estimate\r\nfeeds back into the $\\chi^2$. The actual $\\chi^2$ we're using involves\r\nthe residual $\\y - \\G\\x - \\mathrm{avg}(\\y - \\G\\x)$. where avg() is a\r\nweighted average over wavelength. Fortunately, this weighted average\r\ncan be represented by a matrix multiply. For example,\r\n\r\n\\begin{equation}\r\n \\left( \\begin{array}{cccc}\r\n \\frac{w_0}{w_0 + w_1} & \\frac{w_1}{w_0 + w_1} & 0 & 0 \\\\\r\n \\frac{w_0}{w_0 + w_1} & \\frac{w_1}{w_0 + w_1} & 0 & 0 \\\\\r\n 0 & 0 & \\frac{w_2}{w_2 + w_3} & \\frac{w_3}{w_2 + w_3} \\\\\r\n 0 & 0 & \\frac{w_2}{w_2 + w_3} & \\frac{w_3}{w_2 + w_3} \\\\\r\n \\end{array} \\right)\r\n \\left( \\begin{array}{c} x_0 \\\\ x_1 \\\\ x_2 \\\\ x_3 \\end{array} \\right)\r\n\\end{equation}\r\n\r\n\\noindent yields a vector whose first two entries are the weighted\r\naverage across the first two entries of $\\x$ and whose last two\r\nentries are the weighted average across the last two entries of\r\n$\\x$. Call this matrix $\\V$. The weighted average of the residual is then\r\n$\\V (\\y - \\G\\x)$ and the $\\chi^2$ for each epoch is\r\n\r\n\\begin{eqnarray*}\r\n\\chi^2 &=& (\\y - \\G\\x - \\V(\\y - \\G\\x))^T \\W (\\y - \\G\\x - \\V(\\y - \\G\\x)) \\\\\r\n &=& ((I - \\V)(\\y - \\G\\x))^T \\W ((I - \\V)(\\y - \\G\\x)) \\\\\r\n &=& (\\y - \\G\\x)^T (I - \\V)^T \\W (I - \\V) (\\y - \\G\\x)\r\n\\end{eqnarray*}\r\n\r\n\\noindent where in the third line we have used the equality\r\n$(\\mathbf{A}\\mathbf{B})^T = \\mathbf{B}^T \\mathbf{A}^T$. This is the\r\nsame as the original $\\chi^2$ equation, but with $\\W$ replaced by $(I\r\n- \\V)^T \\W (I - \\V)$. This is a constant, so our original result for the\r\n$\\chi^2$ gradient holds under this replacement:\r\n\r\n\\begin{eqnarray*}\r\n\\W \\rvect &\\to& (I - \\V)^T \\W (I - \\V) \\rvect \\\\\r\n &\\to& (I - \\V)^T \\W \\mathbf{r'} \\\\\r\n &\\to& \\W \\mathbf{r'} - \\V^T \\W \\mathbf{r'}\r\n\\end{eqnarray*}\r\n\r\n\\noindent where $\\mathbf{r'} = (I - \\V)(\\y - \\G\\x)$ is the residual after\r\nsubtracting the ``sky.''\r\n\r\nPractically, how do we calculate $\\V^T \\W \\mathbf{r'}$? In other words, if\r\nmultiplication by $\\V$ is equivalent to a weighted average by wavelength,\r\nwhat is $\\V^T$? In our minimal example from above, it would be\r\n\r\n\\begin{equation}\r\n \\V^T \\x\r\n = \\left( \\begin{array}{cccc}\r\n \\frac{w_0}{w_0 + w_1} & \\frac{w_0}{w_0 + w_1} & 0 & 0 \\\\\r\n \\frac{w_1}{w_0 + w_1} & \\frac{w_1}{w_0 + w_1} & 0 & 0 \\\\\r\n 0 & 0 & \\frac{w_2}{w_2 + w_3} & \\frac{w_2}{w_2 + w_3} \\\\\r\n 0 & 0 & \\frac{w_3}{w_2 + w_3} & \\frac{w_3}{w_2 + w_3} \\\\\r\n \\end{array} \\right)\r\n \\left( \\begin{array}{c} x_0 \\\\ x_1 \\\\ x_2 \\\\ x_3 \\end{array} \\right)\r\n = \\left( \\begin{array}{c}\r\n \\frac{w_0}{w_0 + w_1} (x_0 + x_1) \\\\\r\n \\frac{w_1}{w_0 + w_1} (x_0 + x_1) \\\\\r\n \\frac{w_2}{w_2 + w_3} (x_2 + x_3) \\\\\r\n \\frac{w_3}{w_2 + w_3} (x_2 + x_3)\r\n \\end{array} \\right)\r\n\\end{equation}\r\n\r\n\\noindent Looking at the result, we can see this is equivalent to, for\r\neach wavelength, taking the ratio of the sum of the input vector and\r\nthe sum of $\\W$ and multiplying the result by $\\W$. In Python (where\r\nwe are representing $\\W$ and $\\x$ with 3-d arrays), $\\V^T \\W \\mathbf{r'}$\r\nwould be\r\n\r\n\\begin{verbatim}\r\ntmp = np.sum(w * rprime, axis=(1, 2)) / np.sum(w, axis=(1, 2))\r\nvtwrprime = w * tmp[:, None, None]\r\n\\end{verbatim}\r\n\r\n\\noindent where $\\W$ is \\verb|w| and $\\mathbf{r'}$ is \\verb|rprime|, and the\r\nfull replacement for $\\W \\rvect$ would be:\r\n\r\n\\begin{verbatim}\r\nw * rprime - vtwrprime\r\n\\end{verbatim}\r\n\r\n\\section{Fitting SN and Data Positions}\r\n\r\nIn the final step of CubeFit, we fit simultaneously the position of\r\nthe data with respect to the galaxy and the position of the SN with\r\nrespect to the galaxy. Here, we calculate the analytic gradient of the\r\n$\\chi^2$ function used in this fit.\r\n\r\n\\begin{eqnarray*}\r\n\\chi^2 &=& \\sum_{t, \\lambda} \\sum_i w_i (d_i - (scene)_i)^2 \\\\\r\n &=& \\sum_{t, \\lambda} \\sum_i w_i (d_i - (sky + g_i + SN s_i))^2\r\n\\end{eqnarray*}\r\n\r\n\\noindent where the first sum is over epochs and wavelengths and the\r\nsecond sum is over pixels at a give wavelength. $g_i$ and $s_i$ depend\r\non the positions that we are optimizing, as do $sky$ and $SN$, which\r\nare determined based on the galaxy model, $\\vec{g}$ and SN model,\r\n$\\vec{s}$. The derivative of $\\chi^2$ with respect to the position\r\n$x_j$ of the data is\r\n\r\n\\begin{eqnarray*}\r\n\\frac{d\\chi^2}{dx_\\ast} &=& \\sum_{t, \\lambda} \\sum_i -2 w_i (d_i - (scene)_i) \\frac{d(scene)_i}{dx_\\ast} \\\\\r\n &=& \\sum_{t, \\lambda} \\sum_i -2 w_i (d_i - (scene)_i) \\frac{d}{dx_\\ast} (sky + g_i + SN s_i) \\\\\r\n &=& \\sum_{t, \\lambda} \\sum_i -2 w_i (d_i - (scene)_i) \\left(\\frac{d(sky)}{dx_\\ast} + \\frac{dg_i}{dx_\\ast} + \\frac{d(SN)}{dx_\\ast} s_i + SN \\frac{ds_i}{dx_\\ast}\\right)\r\n\\end{eqnarray*}\r\n\r\n\\noindent If $x_\\ast$ is an SN position parameter then $dg_i/dx_\\ast = 0$,\r\nbut otherwise all terms above are nonzero: $sky$ and $SN$\r\ndepend on $x_\\ast$ through their dependence on $\\vec{g}$ and\r\n$\\vec{s}$. The derivatives $dg_i/dx_\\ast$ and $ds_i/dx_\\ast$ are fairly\r\nsimple to compute. $\\vec{g}$ and $\\vec{s}$ are computed via\r\nmultiplication by a phase shift array in Fourier space. The phase\r\nshift array is the only part that depends on position and everything\r\nelse is just multiplication. So all we need to do is to subsitute the\r\nderivative on the phase shift array, then carry out all calculations\r\nas we normally would.\r\n\r\nWe must also determine the gradient in $sky$ and $SN$ with respect to\r\nposition. We start by defining some quantities for\r\nconvenience (in the left column), and taking their derivatives with respect to\r\ndata and SN positions. (Note that $s_i$ is affected by both positions, $g_i$ is only affected by the data position.)\r\n\r\n\\begin{displaymath}\r\n%denom = AC - B^2\r\n\\begin{array}{ccc}\r\nA = \\sum_i w_i s_i^2 & \\frac{dA}{dx_\\ast} = \\sum_i 2 w_i s_i \\frac{ds_i}{dx_\\ast} & (x_\\ast \\in \\{x_{data}, x_{sn}\\}) \\\\\r\nB = \\sum_i w_i s_i & \\frac{dB}{dx_\\ast} = \\sum_i w_i \\frac{ds_i}{dx_\\ast} \\\\\r\nC = \\sum_i w_i & \\frac{dC}{dx_\\ast} = 0 \\\\\r\nD = \\sum_i w_i d_i & \\frac{dD}{dx_\\ast} = 0 \\\\% wd\r\nE = \\sum_i w_i d_i s_i & \\frac {dE}{dx_\\ast} = \\sum_i w_i d_i \\frac{ds_i}{dx_\\ast} \\\\ % wdsn\r\nF = \\sum_i w_i g_i & \\frac{dF}{dx_{data}} = \\sum_i w_i \\frac{dg_i}{dx_{data}} & \\frac{dF}{dx_{sn}} = 0 \\\\ % wgal\r\nG = \\sum_i w_i g_i s_i & \\frac{dG}{dx_{data}} = \\sum_i w_i \\left(g_i \\frac{ds_i}{dx_{data}} + s_i \\frac{dg_i}{dx_{data}} \\right) & \\frac{dG}{dx_{sn}} = \\sum_i w_i g_i \\frac{ds_i}{dx_{sn}} % wgalsn\r\n\\end{array}\r\n\\end{displaymath}\r\n\r\n\\noindent where $i$ ranges over all pixels at the given wavelength,\r\n$s_i$ is the SN model (scaled to 1) $g_i$ is the galaxy model, $d_i$\r\nis the data and $w_i$ is the weight of the data. The SN amplitude, a\r\nscalar here, is given by\r\n\r\n\\begin{equation}\r\nSN = (-DB + EC + FB - GC) / (AC - B^2)\r\n\\end{equation}\r\n\r\n\\noindent (Derivation not shown.) Taking the derivative with\r\nrespect to $x_\\ast$,\r\n\r\n\\begin{eqnarray*}\r\n\\frac{d(SN)}{dx_\\ast} &=& \\frac{-D'B - DB' + E'C + EC' + F'B + FB' - G'C - GC'}{AC - B^2} + \\\\\r\n & & (SN) \\frac{A'C + AC' - 2BB'}{AC-B^2} \\\\\r\n &=& \\frac{-DB' + E'C + F'B + FB' - G'C}{AC - B^2} + (SN) \\frac{A'C - 2BB'}{AC-B^2}\r\n\\end{eqnarray*}\r\n\r\nSimilar to the SN amlitude, the sky is given by\r\n\r\n\\begin{equation}\r\nsky = (DA - EB - FA + GB) / (AC - B^2)\r\n\\end{equation}\r\n\r\n\\noindent The derivative is\r\n\r\n\\begin{eqnarray*}\r\n\\frac{d(sky)}{dx_\\ast} &=& \\frac{D'A + DA' - E'B - EB' - F'A - FA' + G'B + GB'}{AC - B^2} + \\\\\r\n & & + (sky) \\frac{A'C + AC' - 2BB'}{AC-B^2} \\\\\r\n &=& \\frac{DA' - E'B - EB' - F'A - FA' + G'B + GB'}{AC - B^2} + (sky) \\frac{A'C - 2BB'}{AC-B^2}\r\n\\end{eqnarray*}\r\n\r\n\r\n\\end{document}\r\n", "meta": {"hexsha": "88ccce31ee36a565339f6b3b73f3d6a4268bd071", "size": 17187, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/gradient.tex", "max_stars_repo_name": "kbarbary/cubefit", "max_stars_repo_head_hexsha": "d242985d2cf9fafb297da9903ba8ede3ac8a57a3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2015-07-23T22:49:50.000Z", "max_stars_repo_stars_event_max_datetime": "2017-08-04T07:26:31.000Z", "max_issues_repo_path": "docs/gradient.tex", "max_issues_repo_name": "kbarbary/cubefit", "max_issues_repo_head_hexsha": "d242985d2cf9fafb297da9903ba8ede3ac8a57a3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 22, "max_issues_repo_issues_event_min_datetime": "2015-06-02T00:08:11.000Z", "max_issues_repo_issues_event_max_datetime": "2018-02-27T01:28:58.000Z", "max_forks_repo_path": "docs/gradient.tex", "max_forks_repo_name": "kbarbary/cubefit", "max_forks_repo_head_hexsha": "d242985d2cf9fafb297da9903ba8ede3ac8a57a3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2015-09-24T16:41:49.000Z", "max_forks_repo_forks_event_max_datetime": "2019-02-25T19:20:12.000Z", "avg_line_length": 40.9214285714, "max_line_length": 198, "alphanum_fraction": 0.6014429511, "num_tokens": 6277, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711832583695, "lm_q2_score": 0.7248702880639791, "lm_q1q2_score": 0.6161188564545755}} {"text": "\\subsection{Position space representation}\n\nFirst we define the time integral of the Lagrangian of the classical oscillator given in Eq.~(\\ref{eq_5}), over a period $T=2\\pi/\\omega$ as\n\\begin{equation} \\label{eq_b1}\n \\Delta_{\\varepsilon} = \\frac{1}{T} \\int_0^T dt' \\; L(\\zeta,\\dot{\\zeta},t').\n\\end{equation}\nAdditionally, performing this integral, we can obtain a more simplified result\n\\begin{equation} \\label{eq_b2}\n \\Delta_{\\varepsilon} = \\frac{(eE)^2}{4m_e(\\omega_0^2 - \\omega^2)}.\n\\end{equation}\nNext, we define another parameter\n\\begin{equation} \\label{eq_b3}\n \\xi =\n \\int_0^t dt' \\; L(\\zeta,\\dot{\\zeta},t') -\n \\Delta_{\\varepsilon} t,\n\\end{equation}\nand after simplifying, we can identify\n\\begin{equation} \\label{eq_b4}\n \\xi =\n \\frac{(eE)^2\\qty(3\\omega^2 - \\omega_0^2)}{8m_e\\omega(\\omega_0^2 - \\omega^2)^2} \\sin(2\\omega t),\n\\end{equation}\nwhich is a periodic function in time. Using these parameters, we can factorize the wave function given in Eq.~(\\ref{eq_2}) as linearly time dependent part and periodic time dependent part as follows\n\\begin{equation} \\label{eq_b5}\n \\begin{aligned}\n \\psi_{\\alpha}&(x,y,t) =\n \\exp(\\frac{i}{\\hbar}\\qty[-\\epsilon_nt + \\Delta_{\\varepsilon} t ])\n \\frac{1}{\\sqrt{L_x}} \\chi_n\\big(y - y_0 - \\zeta(t)\\big)\n \\\\\n & \\times\n \\text{exp}\\bigg(\n \\frac{i}{\\hbar}\\bigg[\n p_x x +\n \\frac{eEy}{\\omega}\\cos(\\omega t) \\\\\n & \\quad+\n m_e\\dot{\\zeta(t)}\\big[y-\\zeta(t)\\big]\n + \\int_0^{t}dt'L(\\zeta,\\dot{\\zeta},t') - \\Delta_{\\varepsilon} t \\bigg]\n \\bigg).\n \\end{aligned}\n\\end{equation}\nThis leads to separate the linear time dependent phase component as the quasienergies\n\\begin{equation} \\label{eq_b6}\n \\varepsilon_{n} =\n \\hbar \\omega_0\\qty(n + \\frac{1}{2}) - \\Delta_{\\varepsilon},\n\\end{equation}\nwhile rest of the components as time-periodic Floquet modes\n\\begin{equation} \\label{eq_b7}\n \\begin{aligned}\n \\phi_{n,m}(x,y,t) = &\n \\frac{1}{\\sqrt{L_x}} \\chi_{n}\\left(y - y_0 - \\zeta(t)\\right)\n \\text{exp}\\bigg(\n \\frac{i}{\\hbar}\\bigg[\n p_x x \\\\\n & +\n \\frac{eE[y - y_0]}{\\omega}\\cos(\\omega t) \\\\\n & +\n m_e\\dot{\\zeta}(t)\\big[y - y_0 -\\zeta(t)\\big]\n + \\xi \\bigg]\\bigg).\n \\end{aligned}\n\\end{equation}\n\n\\subsection{Momentum space representation}\n\nWe perform continuous Fourier transform over the considering confined space $A=L_xL_y$ on the Floquet modes given in Eq.~(\\ref{eq_7}) to realize the Floquet modes in momentum space\n\\begin{equation} \\label{eq_b8}\n \\begin{aligned}\n \\phi_{n,m}&(k_x,k_y,t) \\\\\n & =\n \\exp(\n \\frac{-i\\gamma(t)}{\\hbar}\n y_0)\n \\exp(\\frac{-i}{\\hbar}\n \\qty[\n m_e \\dot{\\zeta}(t) \\zeta(t) - \\xi\n ])\\\\\n & \\quad\\times\n \\int_{-L_y/2}^{L_y/2} dy\\; \\exp(-i\\qty[k_y - \\gamma(t)]y)\n \\chi_{n}\\qty[y - \\mu(t)] \\\\\n & \\quad\\times\n \\frac{1}{\\sqrt{L_x}}\n \\int_{-L_x/2}^{L_x/2} dx\\;\n \\exp(-ik_x x)\n \\exp( \\frac{i p_x }{\\hbar}x ).\n \\end{aligned}\n\\end{equation}\nHere we used new two parameters\n\\begin{equation} \\label{eq_b9a}\n \\mu(t) = \\frac{eE\\sin(\\omega t)}{m_e(\\omega_0^2 - \\omega^2)} + y_0,\n\\end{equation}\nand\n\\begin{equation} \\label{eq_b9b}\n \\gamma(t) =\n \\frac{eE\\omega_0^2\\cos(\\omega t)}{\\hbar\\omega(\\omega_0^2 - \\omega^2)}.\n\\end{equation}\nSubsequenty, using the Fourier transform identity \\cite{bruus04}\n\\begin{equation} \\label{eq_b10}\n \\int_{L_x} dx\\;\n \\exp( -ik_x x + \\frac{i p_x }{\\hbar}x ) =\n L_x \\delta_{k_x,\\frac{p_x}{\\hbar}},\n\\end{equation}\nwe can derive\n\\begin{equation} \\label{eq_b11}\n \\begin{aligned}\n \\phi_{n,m}&(k_x,k_y,t) =\n \\Phi_{n,m}(k_y,t)\n \\delta_{k_x,\\frac{p_x}{\\hbar}}\\\\\n & \\times\n \\exp(\n \\frac{-i\\gamma(t)}{\\hbar}\n y_0)\n \\exp(\\frac{-i}{\\hbar}\n \\qty[\n m_e \\dot{\\zeta}(t) \\zeta(t) - \\xi\n ]),\n \\end{aligned}\n\\end{equation}\nwhere we can define $\\Phi_{n,m}(k_y,t)$ as\n\\begin{equation} \\label{eq_b12}\n \\begin{aligned}\n \\Phi_{n,m}(k_y,t) =\n \\sqrt{L_x}\n \\int_{-L_y/2}^{L_y/2} dy\\; &\n \\chi_{n}\\qty[y - \\mu(t)] \\\\\n & \\times\n \\exp(\n -i\\qty[k_y - \\gamma(t)]\n y).\n \\end{aligned}\n\\end{equation}\nSubstituting ${k'_y} = k_y -\\gamma(t)$ with $y' = y -\\mu(t)$, and assuming that the size of the considered 2DEG sample in $y$-direction is considerably large ($L_y \\rightarrow \\infty$), we can obtain\n\\begin{equation} \\label{eq_b13}\n \\Phi_{n,m}({k'_y} ,t) =\n {\\sqrt{L_x}} e^{-i {k'_y}\\mu}\n \\int_{-\\infty}^{\\infty} dy'\\;\n \\chi_{n}\\qty(y')\n \\exp(-i{k'_y} y').\n\\end{equation}\nMoreover, we can identify that the above integral represents the Fourier transform of $\\{\\chi_n\\}$ functions. In addition, using the symmetric conditions of the Fourier transform for Gauss-Hermite functions $\\theta_n(x)$ \\cite{celeghini21}\n\\begin{equation} \\label{eq_b14}\n \\mathcal{FT}[\\theta_n(\\kappa x),x,k] = \\frac{i^n}{|\\kappa|}\\theta_n(k/\\kappa),\n\\end{equation}\nwe can simply the Eq.~(\\ref{eq_b13}) as\n\\begin{equation} \\label{eq_b15}\n \\Phi_{n,m}({k'_y} ,t) =\n \\sqrt{L_x}e^{-i {k'_y}\\mu}\n \\tilde{\\chi}_{n}\\qty({k'_y}),\n\\end{equation}\nwith\n\\begin{equation} \\label{eq_b16}\n \\tilde{\\chi}_{n}\\qty(k) =\n \\frac{i^n}{\\sqrt{2^{n} n! \\sqrt{\\pi}}}\n \\qty(\\frac{1}{\\kappa})^{1/2}\n e^{-\\frac{k^2}{2 \\kappa^2}}\n \\mathcal{H}_{\\alpha} \\qty(\\frac{k}{\\kappa}).\n\\end{equation}\nFinally, substitute Eq.~(\\ref{eq_b15}) back into Eq.~(\\ref{eq_b11}) and this leads to\n\\begin{equation} \\label{eq_b17}\n \\begin{aligned}\n \\phi_{n,m}(k_x,k_y,t) = &\n {\\sqrt{L_x}}\n \\tilde{\\chi}_{n}\\qty(k_y - b\\cos(\\omega t)) \\\\\n & \\times\n \\text{exp}\\left(\n i\\xi\n -ik_y \\qty[d\\sin(\\omega t) + \\frac{\\hbar k_x}{eB}]\n \\right),\n \\end{aligned}\n\\end{equation}\nwhere\n\\begin{equation} \\label{eq_b18}\n b \\equiv\n \\frac{eE\\omega_0^2}{\\hbar\\omega(\\omega_0^2 - \\omega^2)},\n\\end{equation}\nand\n\\begin{equation} \\label{eq_b19}\n d \\equiv\n \\frac{eE}{m_e(\\omega_0^2 - \\omega^2)}.\n\\end{equation}\nIt is necessary to notice that $k_x$ is quantized with $k_x = 2\\pi m/L_x ~,~ m \\in \\mathbb{Z}$.\n", "meta": {"hexsha": "7ac572fe09b127150b92303b3c21b9786e010a6f", "size": 5905, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/v1.0/sections/appendix_b.tex", "max_stars_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_stars_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "paper/v1.0/sections/appendix_b.tex", "max_issues_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_issues_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/v1.0/sections/appendix_b.tex", "max_forks_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_forks_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.9888268156, "max_line_length": 239, "alphanum_fraction": 0.617273497, "num_tokens": 2342, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8499711908591638, "lm_q2_score": 0.7248702702332475, "lm_q1q2_score": 0.6161188468085572}} {"text": "\\section{Example Proof for Monoid Laws}\n\\label{sec:example}\nIn this section we use equational reasoning (see section \\ref{sec:equationalreasoning}) to prove the left identity law of the monoid type class (see section \\ref{sec:monoid}) for a new type.\nThe following example is based on a blog post of Gabriel Gonzales \\cite{gonzales14}. We simplified the types of a plugin and extended the example with an additional plugin implementation.\n\nSuppose we want to build a plugin system. A plugin in our example is an \\verb|IO| action that takes a \\verb|Char| value and does some work with it (e.g. log to a file, potentially with side effects). Hence, a plugin is of type \\verb|Char -> IO ()|. Listing \\ref{lst:logtoplugin} shows the definition of the plugin \\verb|logto|. It writes the first argument of type \\verb|Char| to a file named \\verb|log.txt|.\n\n\\begin{lstlisting}[caption={Definition of a plugin that writes a character to a file.}, label={lst:logtoplugin}]\nimport System.IO\nlogto :: Char -> IO ()\nlogto c = do\n handle <- openFile \"log.txt\" WriteMode\n hPutChar handle c\n\\end{lstlisting}\n\nListing \\ref{lst:applyingaSinglePlugin} shows an application of the \\verb|logto| plugin. We read a character \\verb|c| from the standard input and call \\verb|logto c|.\n\n\\begin{lstlisting}[caption={Calling the {\\ttfamily logto} IO action in main}, label={lst:applyingaSinglePlugin}]\nmain = do\n c <- getChar\n logto c\n\\end{lstlisting}\n\nListing \\ref{lst:print2stdout} shows an additional plugin that prints the character to the standard output. \n\n\\begin{lstlisting}[caption={Definition of the plugin {\\ttfamily print2stdout}}, label={lst:print2stdout}]\nprint2stdout :: Char -> IO ()\nprint2stdout c = putChar c\n\\end{lstlisting}\nIn order to extend the program from listing \\ref{lst:applyingaSinglePlugin}, we want to be able to compose several plugins.\nListing \\ref{lst:composition} gives an example of the composition of the plugins \\verb|logto| and \\verb|printstdout|. We apply the \\verb|mappend| function from the \\verb|Monoid| type class to \\verb|logto| and \\verb|print2stdout|. Both arguments are of type \\verb|Char -> IO ()|. The return value is also of type \\verb|Char -> IO ()|. Listing \\ref{lst:composition} only works if \\verb|mappend| is implemented for the type \\verb|Char -> IO ()|. We will give an implementation later in this section.\n\n\\begin{lstlisting}[caption={Composition of two plugins}, label={lst:composition}]\ncomposedPlugin :: Char -> IO ()\ncomposedPlugin = logto `mappend` print2stdout\n\\end{lstlisting}\n\nIn addition, we demand that the order in which we evaluate the plugins must not matter. The plugins have to work independently of each other. For example, the behavior of the plugins \\verb|composed1| and \\verb|composed2| from listing \\ref{lst:evaluateorder} should be the same (\\verb|donothing| is another plugin of type \\verb|Char -> IO ()|).\n\n\\begin{lstlisting}[caption={The order in which we evaluate the plugins does not matter}, label={lst:evaluateorder}]\ncomposed1 = logto `mappend` (print2stdout `mappend` donothing)\ncomposed2 = (logto `mappend` print2stdout) `mappend` donothing\n\\end{lstlisting}\n\nThe monoid laws state that \\verb|mappend| must be associative. Hence, if we can prove that our implementation of \\verb|mappend| satisfies the monoid laws, we can combine them, using the monoid function \\verb|mappend| and we are able to add plugins without concerning about the order of evaluation. In addition, they are easier to use because they will behave as expected.\n\nThe plugins are of type \\verb|Char -> IO ()|. Instead of writing a specialized instance for \\verb|Char -> IO ()|, we use the general implementation of section \\ref{sec:monoid}.\nThe instance implementation is repeated for convenience:\n\n\\begin{lstlisting}[caption={Monoid instance},label={lst:monoidinstance2}]\n{-# LANGUAGE FlexibleInstances #-}\n{-# LANGUAGE OverlappingInstances #-}\n\ninstance (Applicative f, Monoid a) => Monoid (f a) where\n mempty = pure mempty\n mappend = liftA2 mappend\n\\end{lstlisting}\nThe generalization has the advantage of not having that to prove the type class law only once for all types that match the general type declaration. The process of verifying a program is cumbersome and time consuming. Hence, the generalization of proofs is desirable.\n\nWe will verify that the definition \\ref{lst:monoidinstance2} satisfies the left identity law of the \\verb|Monoid| type class in two steps. First, we describe, why the instance implementation of listing \\ref{lst:monoidinstance2} makes the type \\verb|Char -> IO ()| part of the \\verb|Monoid| type class. Then we prove that the left identity law from section \\ref{sec:monoid} holds for the \\verb|Monoid| instance implementation of listing \\ref{lst:monoidinstance2}. We will assume that the instance implementation of listing \\ref{lst:monoidinstance2} is a \\verb|Monoid| in the first step.\n\n\\subsection{Generalization of the plugin type }\n\\label{sec:generalization}\n\nIn order to use the instance implementation of listing \\ref{lst:monoidinstance2}, the type \\verb|(->) Char| (that is the type of a Haskell function) has to be part of the type class \\verb|Applicative| and the type \\verb|IO ()| has to be part of the type class \\verb|Monoid|. \\verb|IO ()| is a \\verb|Monoid| if the type \\verb|IO| is an \\verb|Applicative| and \\verb|()| is a \\verb|Monoid|.\n\nHere is an overview of all the steps of the argumentation:\n\\begin{enumerate}\n\\item Show that \\verb|Char -> IO ()| is a \\verb|Monoid|.\n\\item Show that \\verb|(->) Char| is an \\verb|Applicative| and \\verb|IO ()| is a \\verb|Monoid|.\n\\item Show that \\verb|IO| is an \\verb|Applicative| and \\verb|()| is a \\verb|Monoid|.\n\\end{enumerate}\n\nWe show the required properties in reversed order.\n\n\\begin{etaremune}\n\\item The standard library provides a \\verb|Monoid| instance for \\verb|()| \\cite{monoid} and an \\verb|Applicative| instance for \\verb|IO| \\cite{control.applicative}. All type class instances of the standard library obey the corresponding type class laws \\cite{Lipovaca}. \n\n\\item If the implementation from listing \\ref{lst:monoidinstance2} satisfies the monoid laws then \\verb|IO ()| is a \\verb|Monoid| because \\verb|IO| is part of the \\verb|Applicative| type class \\cite{control.applicative} and \\verb|()| is a \\verb|Monoid|. The type \\verb|(->) r| (that's the type of Haskell functions) is a part of the \\verb|Applicative| type class \\cite{control.applicative}.\n\n\\item \nIf the implementation from listing \\ref{lst:monoidinstance2} satisfies the monoid laws then \\verb|Char -> IO ()| is a \\verb|Monoid| because \\verb|(->) Char| is an \\verb|Applicative| and \\verb|IO ()| is a \\verb|Monoid|. The compiler will use the instance implementation for the \\verb|Monoid| type class from listing \\ref{lst:monoidinstance2} for the type \\verb|Char -> IO ()| because it matches the type\n\\begin{verbatim}\n(Applicative f, Monoid b) => Monoid (f b)\n\\end{verbatim}\n\n\\end{etaremune} \n\nNotice that we rely heavily on the assumption that listing \\ref{lst:monoidinstance2} satisfies the monoid laws. The next section will prove that the implementation is correct.\n\n\\subsection{Proof}\n\\label{sec:exampleproof}\n\nIn this section we will show that the implementation in listing \\ref{lst:monoidinstance2} satisfies the left identity law of the \\verb|Monoid| type class (see section \\ref{sec:monoid}).\nThe left identity law demands that:\n\\begin{verbatim}\nmappend mempty x = x\n\\end{verbatim}\n\nWe use equational reasoning to show that the left-hand side is equal to \\verb|x|. First we use the definitions of \\verb|mappend| and \\verb|mempty| of listing \\ref{lst:monoidinstance2} to substitute the left-hand side. Furthermore we look up the definition of \\verb|liftA2| in the source code \\cite{control.applicative} to evaluate the expression. \nListing \\ref{lst:lifta2} shows the definition of \\verb|liftA2|.\n\n\\begin{lstlisting}[caption={Function definition of {\\ttfamily liftA2}},label={lst:lifta2}]\nliftA2 f x y = (pure f <*> x) <*> y\n\\end{lstlisting}\n\n\\begin{verbatim}\nmappend mempty x -- apply def. mappend\n= liftA2 mappend mempty x -- apply def. mempty\n= liftA2 mappend (pure mempty) x -- apply def. of liftA2\n= (pure mappend <*> pure mempty) <*> x\n\\end{verbatim}\n\nTo resolve this expression further, we use the laws of the \\verb|Applicative| type class described in the appendix \\ref{sec:applicatives}.\nOne law of the \\verb|Applicative| type class states:\n\\begin{verbatim}\npure f <*> pure x = pure (f x)\n\\end{verbatim}\nWe can use this property to substitute the left-hand side. Next we write \\verb|mappend mempty| as lambda function \\verb|\\a -> mappend mempty a| and use the monoid law\n\\begin{verbatim}\nmappend mempty x = x\n\\end{verbatim}\nto simplify the expression. In the last step we use the first applicative law\n\\begin{verbatim}\npure id <*> v = v\n\\end{verbatim}\n to rewrite the expression as \\verb|x|.\n\\begin{verbatim}\n(pure mappend <*> pure mempty) <*> x -- 3. applicative law\n= pure (mappend mempty) <*> x -- transform to lambda\n= pure (\\a -> mappend mempty a) <*> x -- 1. monoid law \n= pure (\\a -> a) <*> x -- a -> a = id\n= pure id <*> x -- 1. applicative law\n= x\n\\end{verbatim}\n\nThat completes the proof.\n\nThe example demonstrated several ideas:\n\\begin{itemize}\n\\item Type classes allow us to generalize definitions. A proof for the generalization is valid for all specializations.\n\\item To prove a type class law we can use equational reasoning.\n\\item Type class laws (or properties) allow to prove further properties.\n\\end{itemize}\n\n", "meta": {"hexsha": "8a1bbe7f6bd375e14af8fbfa12390fc1abcd9280", "size": 9539, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "example.tex", "max_stars_repo_name": "Hofmaier/robertson", "max_stars_repo_head_hexsha": "a9659af0af3c5780230e8fe3cb64350f57fc8226", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "example.tex", "max_issues_repo_name": "Hofmaier/robertson", "max_issues_repo_head_hexsha": "a9659af0af3c5780230e8fe3cb64350f57fc8226", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "example.tex", "max_forks_repo_name": "Hofmaier/robertson", "max_forks_repo_head_hexsha": "a9659af0af3c5780230e8fe3cb64350f57fc8226", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 65.3356164384, "max_line_length": 585, "alphanum_fraction": 0.7456756473, "num_tokens": 2540, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867825403176, "lm_q2_score": 0.7718434978390746, "lm_q1q2_score": 0.6160752781648355}} {"text": "\\section{Conclusion}\nWe are interested in answering a basic question of how to forecast a time series variable of interest when there is uncertainly about parameter instability. Specifically, which model should be used for forecasting: the break model or the stable one? If uncertainty is strong and we decide to combine forecasts from two models, what is the optimal rule in terms of some information criterion about assigning weights? Built upon Hansen's Mallows' model averaging method, we propose using the cross-validation criterion to combine forecasting models. In the literature of model selection, CV is shown to be robust to heteroscedasticity than other information criteria, such as, AIC, BIC and Mallows'. Without assuming conditional heteroscedasticity, we show that CV model averaging is approximately equivalent to Mallows' model averaging. But in many empirical applications related to macroeconomic time series or financial time series, researchers usually can not avoid explicitly dealing with heteroscedasticity for analysis and forecast. This motivates our generalization of the model averaging method by allowing for conditional heteroscedasticity.\n\nResearchers have found that in many applications, equally weighted forecasts perform better than other complex combination methods. This forecast combination puzzle has cast double on the use of complicated model averaging methods. Both CV and Cp weights are easy to compute and do not rely on direct weight estimation as in the Granger-Ramanathan forecast combination. This feature should be appealing to practitioners and professional forecasters because simplicity can help reduce the excess noise introduced by using complex weighting methods. This may help explain why our methods forecast better than equal weight in both controlled simulation and in forecasting U.S. and Taiwan quarterly GDP growth rates out-of-sample.\n\n", "meta": {"hexsha": "eff3d0475380f0b260285e9f116529324c1f3490", "size": 1900, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Tex/conclusion.tex", "max_stars_repo_name": "anwenyin/ooscombo", "max_stars_repo_head_hexsha": "4f747c7ba0c7bde2a4ae13fdc112a24e01f25bf7", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Tex/conclusion.tex", "max_issues_repo_name": "anwenyin/ooscombo", "max_issues_repo_head_hexsha": "4f747c7ba0c7bde2a4ae13fdc112a24e01f25bf7", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Tex/conclusion.tex", "max_forks_repo_name": "anwenyin/ooscombo", "max_forks_repo_head_hexsha": "4f747c7ba0c7bde2a4ae13fdc112a24e01f25bf7", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 316.6666666667, "max_line_length": 1149, "alphanum_fraction": 0.8357894737, "num_tokens": 339, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.7981867801399694, "lm_q1q2_score": 0.6160752763121424}} {"text": "\\section{2D DP}\n\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest common subsequence}\n \\begin{exampleblock}{LCS: longest common subsequence (Problem 7.5)}\n\t\\[\n\t X = X_{1} \\cdots X_{m} \\quad Y = Y_{1} \\cdots Y_{n}\n\t\\]\n\n\t\\begin{enumerate}[(1)]\n \\item Find (the length of) an LCS of $X$ and $Y$\n \\end{enumerate}\n \\end{exampleblock}\n\n \\begin{align*}\n\tX &= \\langle A,\\textcolor{blue}{B},\\textcolor{blue}{C},\\textcolor{blue}{B},D,\\textcolor{blue}{A},B \\rangle \\\\\n\tY &= \\langle \\textcolor{blue}{B},D,\\textcolor{blue}{C},A,\\textcolor{blue}{B},\\textcolor{blue}{A} \\rangle \\\\\n\tZ &= \\langle B,C,B,A \\rangle\n \\end{align*}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest common subsequence}\n \\begin{description}\n\t\\item[Subproblem:] $L[i,j]$: the length of an LCS of $X[1 \\cdots i]$ and $Y[1 \\cdots j]$\n\t\\item[Goal:] $L[m,n]$\n\t \\pause\n\t\\item[Make choice:] Is $X_{i} = Y_{j}$?\n\t\\item[Recurrence:] \\textcolor{red}{(Proof!)}\n\t \\begin{displaymath}\n\t\tL[i,j] = \\left\\{ \\begin{array}{ll}\n\t\t L[i-1, j-1] + 1 & \\textrm{if $X_{i} = Y_{j}$}\\\\\n\t\t \\max \\set{L[i-1,j], L[i,j-1]} & \\textrm{if $X_{i} \\neq Y_{j}$}\n\t\t\\end{array} \\right.\n\t \\end{displaymath}\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tL[0,j] &= 0, \\; 0 \\le j \\le n \\\\\n\t\tL[i,0] &= 0, \\; 0 \\le i \\le m \n\t \\end{align*}\n\t\\item[Time:] $\\Theta(mn)$\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest common subsequence}\n \\begin{exampleblock}{Longest common subsequence (Problem 7.5)}\n\t\\[\n\t X = X_{1} \\cdots X_{m} \\quad Y = Y_{1} \\cdots Y_{n}\n\t\\]\n\n\t\\begin{enumerate}[(1)]\n\t \\setcounter{enumi}{1}\n\t \\item Allowing repetition of $X$\n\t \\item Allowing repetition $\\le k$ of $X$\n\t\\end{enumerate}\n \\end{exampleblock}\n\n \\pause\n \\begin{displaymath}\n\tL[i,j] = \\left\\{ \\begin{array}{ll}\n\t L[\\textcolor{red}{i}, j-1] + 1 & \\textrm{if $X_{i} = Y_{j}$}\\\\\n\t \\max \\set{L[i-1,j], L[i,j-1]} & \\textrm{if $X_{i} \\neq Y_{j}$}\n\t\\end{array} \\right.\n \\end{displaymath}\n\n \\pause\n \\[\n\tX \\implies X^{(k)} \\triangleq X_{1}^{(k)} \\cdots X_{m}^{(k)}\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest common substring}\n \\centerline{What about longest common \\emph{substring}?}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Shortest common supersequence}\n \\begin{exampleblock}{Shortest common supersequence (Problem 7.6)}\n\t\\[\n\t X = X_{1} \\cdots X_{m} \\quad Y = Y_{1} \\cdots Y_{n}\n\t\\]\n\n\t\\begin{itemize}\n\t \\item Find (the length of) a shortest common subsequence of $X$ and $Y$\n\t\\end{itemize}\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Shortest common supersequence}\n \\begin{description}\n\t\\item[Subproblem:] $L[i,j]$: the length of an SCS of $X[1 \\cdots i]$ and $Y[1 \\cdots j]$\n\t\\item[Goal:] $L[m,n]$\n\t \\pause\n\t\\item[Make choice:] Is $X_{i} = Y_{j}$?\n\t\\item[Recurrence:] \n\t \\begin{displaymath}\n\t\tL[i,j] = \\left\\{ \\begin{array}{ll}\n\t\t L[i-1, j-1] + 1 & \\textrm{if $X_{i} = Y_{j}$}\\\\\n\t\t \\min \\set{L[i-1,j] + 1, L[i,j-1] + 1} & \\textrm{if $X_{i} \\neq Y_{j}$}\n\t\t\\end{array} \\right.\n\t \\end{displaymath}\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tL[0,j] &= j, \\; 0 \\le j \\le n \\\\\n\t\tL[i,0] &= i, \\; 0 \\le i \\le m \n\t \\end{align*}\n \\end{description}\n\n \\pause\n \\begin{alertblock}{Remark}\n\t\\[\n\t \\max(m,n) \\le L(m,n) \\le m+n\n\t\\]\n \\end{alertblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Shortest common supersequence}\n \\[\n\tX[1 \\dots m] \\qquad Y[1 \\dots n]\n \\]\n\n \\begin{align*}\n\tr &= |\\text{LCS}(X,Y)| \\\\\n\tt &= |\\text{SCS}(X,Y)| \n \\end{align*}\n\n \\pause\n \\[\n\tX \\cap Y = \\emptyset \\implies r = 0 \\quad t = m + n\n \\]\n\n \\pause\n \\[ \n\tX \\prec Y \\implies r = m \\quad t = n\n \\]\n\n \\pause\n \\[\n\tr + t = m + n\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String shuffling}\n \\begin{exampleblock}{String shuffling (Problem 7.7)}\n\t\\begin{itemize}\n\t \\item Strings $X[1 \\dots m], Y[1 \\dots n], Z[1 \\dots k]$\n\t \\item Is $X \\oplus Y = Z$?\n\t\\end{itemize}\n\n\t\\begin{enumerate}[(1)]\n\t \\item Reduced to LCS:\n\t\t\\[\n\t\t (Z' \\triangleq Z \\setminus \\text{LCS}(X, Z)) = Y\n\t\t\\]\n\t \\item $O(mn)$\n\t \\item Minimum \\# deleted characters to ensure $X \\oplus Y = Z$ in $O(mnk)$\n\t\\end{enumerate}\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String shuffling}\n \\[\n\t(Z' \\triangleq Z \\setminus \\text{LCS}(X, Z)) = Y\n \\]\n\n \\vspace{0.60cm}\n \\begin{columns}\n\t\\column{0.50\\textwidth}\n\t \\[\n\t\t\\text{``YES''} \\implies X \\oplus Y = Z\n\t \\]\n\n\t \\uncover<2->{\n\t\t\\begin{align*}\n\t\t X &= \\textcolor{red}{A}C \\\\\n\t\t Y &= CB \\\\\n\t\t Z &= CB\\textcolor{red}{A} \n\t\t\\end{align*}\n\t }\n\t\\column{0.50\\textwidth}\n\t \\[\n\t\t\\text{``NO''} \\implies X \\oplus Y \\neq Z\n\t \\]\n\t \n\t \\uncover<3->{\n\t\t\\begin{align*}\n\t\t X &= \\textcolor{red}{A} \\\\\n\t\t Y &= AB \\\\\n\t\t Z &= \\textcolor{red}{A}BA \n\t\t\\end{align*}\n\t }\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String shuffling}\n \\begin{description}\n\t\\item[Subproblem:] $S[i,j]$: Is $X[1 \\dots i] \\oplus Y[1 \\dots j] = Z[1 \\dots r \\triangleq i + j]$?\n\t\\item[Goal:] $S[m,n]$\n\t \\pause\n\t\\item[Make choice:] Is $Z[r] = X[i] \\lor Z[r] = Y[j]$?\n\t\\item[Recurrence:] \n\t \\begin{align*}\n\t\tS[i,j] = &(Z[r] = X[i] \\land S[i-1,j]) \\lor \\\\\n\t\t\t&(Z[r] = Y[j] \\land S[i,j-1])\n\t \\end{align*}\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tS[0,j] &= (Z = Y), \\; \\forall 0 \\le j \\le n \\\\\n\t\tS[i,0] &= (Z = X), \\; \\forall 0 \\le i \\le m\n\t \\end{align*}\n\t\\item[Time:] $O(mn)$\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String shuffling}\n \\begin{description}\n\t\\item[Subproblem:] $D[i,j,r]$: minimum \\# deleted characters to ensure $X[1 \\dots i] \\oplus Y[1 \\dots j] = Z[1 \\dots r]$\n\t\\item[Goal:] $D[m,n,k]$\n\t \\pause\n\t\\item[Make choice:] Is $Z[r] = X[i] \\lor Z[r] = Y[j]$?\n\t\\item[Recurrence:] \n\t \\[\n\t\tD[i,j,r] = \\min \\left\\{ \\begin{array}{ll}\n\t\t D[i-1, j, r-1] & \\text{if } Z[r] = X[i] \\\\\n\t\t D[i, j-1, r-1] & \\text{if } Z[r] = Y[j] \\\\\n\t\t 1 + D[i-1, j, r] & \\\\\n\t\t 1 + D[i, j-1, r] & \\\\\n\t\t 1 + D[i, j, r-1] & \\\\\n\t\t\\end{array} \\right.\n\t \\]\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tD[0,j,r] &= j + r - 2|\\text{LCS}(Y[1 \\dots j], Z[1 \\dots r])| \\\\\n\t\tD[i,0,r] &=\ti + r - 2|\\text{LCS}(X[1 \\dots i], Z[1 \\dots r])| \\\\\n\t\tD[i,j,0] &= i + j\n\t \\end{align*}\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest contiguous substring both forward and backward}\n \\begin{exampleblock}{Longest contiguous substring both forward and backward (Problem 7.8)}\n \\begin{itemize}\n \\item String $T[1 \\cdots n]$\n \\item Find a longest contiguous substring (LCS) both forward and backward\n \\end{itemize}\n \\end{exampleblock}\n\n \\begin{center}\n\td\\textcolor{blue}{ynam}icprogramming\\textcolor{blue}{many}times\n \\end{center}\n\n \\begin{itemize}\n\t\\item Subproblem $L[i]$: the length of an LCS in $T[1 \\cdots i]$\n\t\\item Subproblem $L[i,j]$: the length of an LCS in $T[i \\cdots j]$\n \\end{itemize}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest contiguous substring both forward and backward}\n \\begin{description}\n\t\\item[Subproblem:] $L[i,j]$: the length of an LCS \\textcolor{red}{starting with $T_{i}$ and ending with $T_{j}$}\n\t\\item[Goal:] $\\max\\limits_{1 \\le i \\le j \\le n} L[i,j]$\n\t \\pause\n\t\\item[Make choice:] Is $T_{i} = T_{j}$?\n\t\\item[Recurrence:] \n\t \\begin{displaymath}\n\t\tL[i,j] = \\left\\{ \\begin{array}{ll}\n\t\t 0 & \\textrm{if $T_{i} \\neq T_{j}$} \\\\\n\t\t L[i+1,j-1] + 1 & \\textrm{if $T_{i} = T_{j}$}\n\t\t\\end{array} \\right.\n\t \\end{displaymath}\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tL[i,i] &= 0, \\; 0 \\le i \\le n \\\\\n\t\tL[i,i+1] &= \\left\\{ \\begin{array}{ll}\n\t\t 1 & \\text{if } T_{i} = T_{i+1} \\\\\n\t\t 0 & \\text{if } T_{i} \\neq T_{i+1}\n\t\t \\end{array} \\right.\n\t \\end{align*}\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}[fragile]{Longest contiguous substring both forward and backward}\n Code: three ways of filling the table\n \\fignocaption{width = 0.50\\textwidth}{figs/three-ways-filling-table.png}\n\n \\begin{algorithmic}\n\t\\ForAll{$d \\gets 2 \\dots n-1$}\n\t \\ForAll{$i \\gets 1 \\dots n-d$}\n\t \\State $j \\gets i + d$\n\t\t\\State $\\dots$\n\t \\EndFor\n \\EndFor\n\n\t\\State \\Return $\\max\\limits_{1 \\le i \\le j \\le n} L[i,j]$\n \\end{algorithmic}\n\n % \\begin{center}\n % \\begin{verbatim}\n % for d = 2 to n-1\n % for i = 1 to n-d\n % j = i + d\n % ...\n % return max_{1 <= i <= j <= n} L[i,j]\n % \\end{verbatim}\n % \\end{center}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Longest palindrome subsequence}\n \\begin{exampleblock}{Longest palindrome subsequence (Problem 7.10)}\n\t\\begin{enumerate}[(1)]\n \\item Find (the length of) a longest palindrome subsequence of $S[1 \\cdots n]$\n \\end{enumerate}\n \\end{exampleblock}\n\n \\begin{description}\n\t\\item[Subproblem:] $L[i,j]$: the length of an LSP of $S[i \\cdots j]$\n\t\\item[Goal:] $L[1,n]$\n\t \\pause\n\t\\item[Make choice:] Is $S[i] = S[j]$?\n\t\\item[Recurrence:] \n\t \\begin{displaymath}\n\t\tL[i,j] = \\left\\{ \\begin{array}{ll}\n\t\t L[i+1, j-1] + 2 & \\text{if } S[i] = S[j] \\\\\n\t\t \\max\\set{L[i+1,j], L[i,j-1]} & \\text{if } S[i] \\neq S[j]\n\t\t\\end{array} \\right.\n\t \\end{displaymath}\n\t \\pause\n\t\\item[Init:]\n\t \\begin{align*}\n\t\tL[i,i] &= 1, \\; \\forall 1 \\le i \\le n \\\\\n\t\t\\textcolor{red}{L[i,i+1]} &= 2, \\; \\text{if } S[i] = S[i+1], \\; \\forall 1 \\le i \\le n-1\n\t \\end{align*}\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Palindrome splitting}\n \\begin{exampleblock}{Palindrome splitting (Problem 7.10)}\n\t\\begin{enumerate}[(1)]\n\t \\setcounter{enumi}{1}\n\t \\item Split a string $S[1 \\dots n]$ into minimum number of palindromes (\\# cuts)\n \\end{enumerate}\n \\end{exampleblock}\n\n \\begin{description}\n\t\\item[Subproblem:] $C[i,j]$: minimum number of cuts for string $S[i \\dots j]$\n\t\\item[Goal:] $C[1,n] + 1$\n\t \\pause\n\t\\item[Make choice:] Where is the first cut?\n\t\\item[Recurrence:] \n\t \\[\n\t\tC[i,j] = \\left\\{ \\begin{array}{ll}\n\t\t 0\t\\; \\text{ if } S[i \\dots j] \\text{ is a palindrome} & \\\\\n\t\t \\min\\limits_{i+1 \\le k \\le j-1} C[i,k-1] + 1 + C[k,j] & \\text{ o.w.}\n\t\t\\end{array} \\right.\n\t \\]\n\t \\pause\n\t\\item[Init:] $C[i,i] = 0$\n\t\\item[Time:] $O(n^3)$\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Palindrome splitting}\n \\begin{exampleblock}{Palindrome splitting (Problem 7.10)}\n\t\\begin{enumerate}[(1)]\n\t \\setcounter{enumi}{1}\n\t \\item Split a string $S[1 \\dots n]$ into minimum number of palindromes\n \\end{enumerate}\n \\end{exampleblock}\n\n \\begin{description}\n\t\\item[Subproblem:] $P[i]$: minimum number of palindromes for $S[1 \\cdots i]$\n\t\\item[Goal:] $P[n]$\n\t \\pause\n\t\\item[Make choice:] Where does the last palindrome start from?\n\t\\item[Recurrence:] \n\t \\[\n\t\tP[i] = \\min_{\\begin{subarray}{c}\n\t\t\t1 \\le k \\le i \\\\\n\t\t\tS[k \\dots i] \\text{ is a palindrome}\n\t\t \\end{subarray}} P[k-1] + 1\n\t \\]\n\t \\pause\n\t\\item[Init:] $P[0] = 1$\n\t\\item[Time:] $O(n^3)$ \\emph{vs.} $O(n^2)$\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String splitting}\n \\begin{exampleblock}{String splitting (Problem 7.11)}\n \\begin{itemize}\n \\item Split a string $S$ into many pieces\n \\item Cost $|S| = n \\implies n$\n \\item Given locations of $m$ cuts: $\\textcolor{gray}{C_{0}}, C_{1}, \\cdots, C_{m}, \\textcolor{gray}{C_{m+1}}$\n \\item Find the minimum cost of splitting $S$ into $m+1$ pieces $S_{0} \\cdots S_{m}$\n \\end{itemize}\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{String splitting}\n \\begin{description}\n\t\\item[Subproblem:] $C[i,j]$: the minimum cost of splitting substring $S_{i} \\cdots S_{j-1}$ using cuts $C_{i+1} \\cdots C_{j-1}$\n\t\\item[Goal:] $C[0,m+1]$\n\t \\pause\n\t\\item[Make choice:] What is the first cut in $C_{i+1} \\cdots C_{j-1}$?\n\t\\item[Recurrence:] \n\t \\[\n\t\tC[i,j] = \\min_{i < k < j} \\left(C[i,k] + C[k,j] + l(S_{i} \\cdots S_{j-1}) \\right)\n\t \\]\n\t \\pause\n\t\\item[Init:] $C[i, i+1] = 0$\n \\end{description}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "3a007fb44311f84c71b8ac1b3a986a0655316739", "size": 11727, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-dp-20170619/sections/2d-dp.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-dp-20170619/sections/2d-dp.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/alg-tutorial-dp-20170619/sections/2d-dp.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 28.1223021583, "max_line_length": 128, "alphanum_fraction": 0.5595634007, "num_tokens": 4662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.7981867801399695, "lm_q1q2_score": 0.6160752763121424}} {"text": "\\section{Ensemble Methods}\n\n% ===\n\\subsection{Bagging\n\\enskip\\normalfont\\sffamily (\\textbf{B}ootstrap \\textbf{agg}regation)}\n\n\\begin{enumerate}\n \\item Draw $M$ bootstrap sets $Z'_1, \\ldots, Z'_M$\n \\item Train $M$ base models $b^{(1)}, \\ldots, b^{(M)}$\n \\item \\textbf{Aggregate:}\n $\\bbar^{(M)}(\\bm x) =\n \\left\\{\n \\begin{array}{@{}l@{\\enskip}l@{}}\n \\frac1M \\sum_{t\\leq M} b^{(t)}(\\bm x) & \\text{regr.} \\\\\n \\mathrm{sign}\\paren{ \\sum_t b^{(t)}(\\bm x) } & \\text{class.}\n \\end{array}\\right.$\n\\end{enumerate}\n\n\\emph{Why it works:}\nSmall \\textit{variance} (weak learners),\nsmall \\textit{covariance} (almost indep. since $Z'_i \\neq Z'_j$).\n\\\\\\vspace{-2pt}For finite range $y$ and large enough $M$:\n\\\\\\enskip\n$\\E*[\\substack{Y\\mid X\\\\ Z, Z'}]{(y - \\bbar^{(M)}(\\bm x))^2}\n\\leq \\E*[\\substack{Y\\mid X\\\\ Z, Z'}]{(y - b(\\bm x))^2}$\n\n\\emph{Random Forest:}\\enskip\nAt each splitting step, u.a.r. choose $m$ of $p$ features and split only one (best) feature.\n$\\to$ reduce \\textit{correlation} between trees.\n\n\\textbf{Validation:}\\enskip\n\\textit{Out-of-bag error} $\\to$ validate each $\\bm x_i$ with trees that didn't use it for training.\n\n% ===\n\\subsection{Boosting}\n\n\\textit{Sequentially} train weak learners on all data, but \\rotatebox[origin=c]{45}{$\\to$}weight of misclass. samples (\\rotatebox[origin=c]{-45}{$\\!\\to$}bias).\n\n% \\emph{FSAM {\\normalfont\\sffamily (forward stagewise additive modeling)}:}\n\n\\emph{AdaBoost:}\\enskip\nStat. learning (\\textit{forward stagewise additive modeling}) with \\textbf{exp. loss}, trains max-margin ($= y_i \\bbar(\\bm x_i)$), self-avg. and interpolating (\\rotatebox[origin=c]{-45}{$\\!\\to$}overfitting) classifiers.\n\n\\begin{highlightbox}\n \\textit{\\scriptsize [Init]:} $\\bbar^{(0)} \\leftarrow 0$,\\enskip $w_i \\leftarrow 1/n \\:\\forall i\\leq n$ \\\\\n for $t = 1 \\ldots M$:\n \\begin{itemize}[leftmargin=25pt]\n \\item[\\textit{\\scriptsize [Train]:}]\n $b^{(t)} = \\arg\\min_b \\mathcal L^w(b) \\color{gray} = \\sum_i w_i \\mathbb I \\{b(\\bm x_i) \\neq y_i\\}$\n \\item[\\textit{\\scriptsize [Eval]:}]\n $\\mathrm{err}_t = \\mathcal L^w(b^{(t)})$\n \\item[\\textit{\\scriptsize [Aggr]:}]\n $\\bbar^{(t)} = \\bbar^{(t-1)} + \\alpha_t b^{(t)}$;\\enskip\n $\\alpha_t = \\frac12 \\log(\\frac{1}{\\mathrm{err}_t} - 1)$\n \\vspace{-2pt}\n \\item[\\textit{\\scriptsize [Reweight]:}]\n $w_i = w_i \\cdot \\exp(\\alpha_t \\mathbb I\\brace{b^{(t)}(\\bm x_i) \\neq y_i})$\n {\\small normalize!}\n \\end{itemize}\n Return $\\bbar^{(M)}(\\bm x) = \\mathrm{sign}(\\sum_t \\alpha_t b^{(t)}(\\bm x))$\n\\end{highlightbox}\n\n% ===\n", "meta": {"hexsha": "ca2051be6c0e7c7ee69c07ab2c20e4bd39400cc2", "size": 2644, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/AML20/sections/07_ensembles.tex", "max_stars_repo_name": "tstreule/eth-cheat-sheets", "max_stars_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-26T23:11:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T23:11:57.000Z", "max_issues_repo_path": "src/AML20/sections/07_ensembles.tex", "max_issues_repo_name": "tstreule/eth-cheat-sheets", "max_issues_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/AML20/sections/07_ensembles.tex", "max_forks_repo_name": "tstreule/eth-cheat-sheets", "max_forks_repo_head_hexsha": "c61f9fd3b13edf405f790581b4d5eacb50b4f1c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.3125, "max_line_length": 219, "alphanum_fraction": 0.5945537065, "num_tokens": 967, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.6160752744594493}} {"text": "\\documentclass[parskip=half]{scrartcl}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage[citestyle=authoryear-icomp,bibstyle=authoryear]{biblatex}\n\\usepackage{booktabs}\n\\usepackage[style=british]{csquotes}\n\\usepackage{fontspec}\n\\usepackage{mathtools}\n\\usepackage{ntheorem}\n\\usepackage{siunitx}\n\n\\usepackage{tikz}\n\\usetikzlibrary{angles,calc,intersections,quotes}\n\n\\usepackage[p,osf]{cochineal}\n\\usepackage[cochineal,vvarbb]{newtxmath}\n\\usepackage[cal=boondoxupr,calscaled=1.0]{mathalfa}\n\n\\newfontfamily\\dispfamily{Cabin}[Scale=0.95]\n\\addtokomafont{disposition}{\\dispfamily}\n\n\\MakeAutoQuote{«}{»}\n\\DeclareMathOperator{\\Res}{Res}\n\n\\addbibresource{cable.bib}\n\n\\newcommand{\\Int}[2]{\\int_{#1}^{#2}\\!}\n\\newcommand{\\D}{\\mathop{}\\!d}\n\n\\newcommand{\\cZ}{\\mathcal{Z}}\n\\newcommand{\\cC}{\\mathcal{C}}\n\n\\theoremstyle{nonumberplain}\n\\newtheorem{lemma*}{Lemma}\n\\newenvironment{proof}{\\textbf{Proof.}}{\\hfill$\\square$}\n\n\\title{Analytic solutions to the cable equation}\n\n\\author{Sam Yates}\n%\\date{July 29, 2019}\n\n\\sisetup{mode=text}\n\n\\begin{document}\n\n% Work-around for siunitx vs newtxmath bug\n\\ExplSyntaxOn\n \\cs_undefine:N \\c__siunitx_minus_tl\n \\tl_const:Nn \\c__siunitx_minus_tl { - }\n\\ExplSyntaxOff\n\n\\maketitle\n\n\\section{The cable equation}\n\nThe cable equation describes the evolution of the potential\non a long, thin, conducting cable in a conducting medium, separated\nby a leaky dielectric. It assumes that the behaviour\ncan be modelled entirely as a one dimensional problem,\nin terms of the linear conductivity of the cable $\\sigma$,\nand the conductance $g$ and capacitance $c$ per unit length of the\ndielectric. The potential $v(x, t)$ then satisfies\n\\begin{equation}\n (\\sigma v')' = c \\dot v + g v.\n\\end{equation}\nwhere $f'$ denotes the derivative of $f$ with respect to the first\nvariable, and $\\dot{f}$ the derivative with respect to the second.\nThe quantities $\\sigma$, $g$, and $c$ are functions of position,\nand will depend on the electrical and geometrical properties of\nthe system.\n\nFor a cable of radius $r(x)$ with constant bulk resistivity\n$R_L$, areal capacitance $C_M$ and areal surface resistivity\n$R_M$, these parameters are given by\n\\begin{align}\n \\sigma(x) &= \\pi r(x)^2 / R_L, \\\\\n g(x) &= 2 \\pi r(x) \\sqrt{1 + r'(x)^2} / R_M, \\\\\n c(x) &= 2 \\pi r(x) \\sqrt{1 + r'(x)^2} C_M.\n\\end{align}\nLetting $\\theta(x) = \\arctan r'(x)$, the cable equation becomes\n\\begin{equation}\n \\label{eq:constelec}\n \\frac{R_M}{2 R_L}\\left(\n 2\\sin\\theta\\cdot v' + r \\cos\\theta\\cdot v''\n \\right) =\n R_M C_M \\dot v + v.\n\\end{equation}\n\n\\section{Analytic solutions for $v(x, t)$}\n\nTwo important special cases for the cable equation are the cylinder,\nwith $r(x)$ constant and $\\theta(x)=0$, and the conical frustum,\nwith $\\theta(x)$ constant and $r(x)=x\\tan\\theta$.\n\nGiven constant electrical properties, as in \\eqref{eq:constelec},\na zero voltage at $t=0$, and Neumann boundary conditions, what\nis $v(x, t)$? The Rallpack 1 model \\autocite{bhalla1992} is\nan example of such a passive cable model, and is discussed in\nAppendix \\ref{ap:rallpack}.\n\n\\subsection{The cylinder}\n\nSuppose the cable lies on the interval $[0, B]$, with initial\nvoltage zero, zero current\nat $x=0$, and fixed injected current $I$ at $x=B$. The boundary\nconditions are then $v(x,0) = 0$, $v'(0, t) = 0$, and $v'(B, t) = -E$,\nwhere $-E = I\\cdot R_L/\\pi r^2$.\n\nA change of variables gives\n\\begin{equation}\n v(x, t) = -\\lambda E\\cdot u(x/\\lambda, t/\\tau),\n\\end{equation}\nwhere\n\\begin{equation}\n \\lambda = \\sqrt{\\frac{R_M r}{2 R_L}}, \\quad \\tau = R_M C_M, \\quad b = B/\\lambda,\n\\end{equation}\nand $u$ satisfies\n\\begin{equation}\n \\begin{aligned}\n u''(x, t) &= \\dot u(x, t) + u(x, t),\\\\\n u(x, 0) &= 0,\\\\\n u'(0, t) &= 0,\\\\\n u'(b, t) &= 1.\n \\end{aligned}\n \\label{eq:ucyl}\n\\end{equation}\n\nThe corresponding time-invariant problem has solution $\\hat u(x)$ where\n\\begin{equation}\n \\hat u'' = \\hat u,\\quad \\hat u'(0) = 0,\\quad \\hat u'(b) = 1,\n\\end{equation}\ngiving\n\\begin{equation}\n \\hat u(x) = \\frac{\\cosh x}{\\sinh b}.\n\\end{equation}\n\nConsider separable transient solutions $u(x,t)-\\hat u(x)$, with zero derivative\nat $x=0$ and $x=b$, of the form $\\psi(x)\\eta(t)$. From \\eqref{eq:ucyl},\n$\\eta(t)=e^{-\\lambda t}$ for some $\\lambda$ and $\\psi$ satisfies\n\\begin{gather}\n \\label{eq:efncyl}\n \\psi'' + (\\lambda-1)\\psi = 0,\\\\\n \\label{eq:efncylbc}\n \\quad \\psi'(0)=\\psi'(b)=0.\n\\end{gather}\nThis is a regular Sturm-Liouville problem\n\\autocite[see e.g.][Theorem~2.1, p.~146]{reid1980} with discrete non-negative eigenvalues\n$\\lambda_k-1$ and corresponding orthogonal eigenfunctions $\\psi_k$,\n\\begin{equation}\n \\langle \\psi_k, \\psi_{k'} \\rangle =\n \\Int{0}{b} \\psi_k(x)\\psi_{k'}(x) \\D x =0\\quad\\text{if $k\\neq k'$}.\n\\end{equation}\n\nThe least eigenvalue corresponds to $\\lambda_0 = 1$ with eigenfunction\n$\\psi_0(x)=1$.\n\nFor $\\lambda>1$, \\eqref{eq:efncyl} gives\n$\\psi(x) = c_1\\cos\\sqrt{\\lambda-1}x + c_2\\sin\\sqrt{\\lambda-1}x$. From $\\psi'(0)=0$,\nwe have $c_2=0$. From $\\psi'(b)=1$, it follows $c_1=0$ or $\\sqrt{\\lambda-1}b = k\\pi$\nfor some positive integer $k$. Consequently we have eigenvalues and eigenfunctions,\n\\begin{gather}\n \\label{eq:lkcyl}\n \\lambda_k = 1+\\left(\\frac{k\\pi}{b}\\right)^{\\mathrlap{2}},\\\\\n \\psi_k(x) = \\cos \\frac{k\\pi x}{b}.\n\\end{gather}\nfor integers $k\\geq 0$, including the constant $\\lambda=1$ case.\n\nThe time-dependent solution $u(x, t)$ then is\n\\begin{equation}\n u(x,t) = \\hat u(x) - \\sum_{k=0}^{\\infty} a_k e^{-\\lambda_k t}\\psi_k(x),\n\\end{equation}\nwith $a_k$ determined by the initial conditions,\n\\begin{equation}\n 0 = \\langle u(x,0), \\psi_k(x) \\rangle\n = \\langle \\hat u(x), \\psi_k(x) \\rangle\n - a_k\\langle \\psi_k(x), \\psi_k(x)\\rangle,\n\\end{equation}\nwith\n\\begin{align}\n \\langle \\hat u(x), \\psi_k(x) \\rangle\n = \\frac{1}{\\sinh b}\\Int{0}{b}\\cosh x\\cos\\frac{k\\pi}{b}x \\D x\n = (-1)^k\\lambda_k^{-1},\n \\intertext{and}\n \\langle \\psi_k(x), \\psi_k(x) \\rangle\n = \\Int{0}{b}\\cos^2 \\frac{k\\pi}{b}x \\D x\n =\n \\begin{cases}\n b & \\text{if $k=0$,}\\\\\n \\frac{b}{2} & \\text{otherwise.}\\\\\n \\end{cases}\n\\end{align}\nConsequently,\n\\begin{equation}\n \\label{eq:cylseries}\n u(x,t) = \\frac{\\cosh x}{\\sinh b} - \\frac{1}{b}e^{-t} -\n \\frac{2}{b}\\sum_{k=1}^{\\infty} (-1)^k \\lambda_k^{-1} e^{-\\lambda_k t} \\cos\\frac{k\\pi x}{b},\n\\end{equation}\nwith $\\lambda_k$ given in \\eqref{eq:lkcyl}. A discussion on the residuals in this series can\nbe found in Appendix~\\ref{ap:cylcomp}.\n\nNote that the time derivative $\\dot u(x,t)$ can be expressed in terms of Jacobi theta functions:\n\\begin{equation}\n \\begin{aligned}\n \\dot u(x,t) &= \\frac{1}{b}e^{-t}\\left(1+2\\sum_{k=1}^{\\infty} (-1)^ke^{-\\frac{\\pi^2 k^2 t}{b}}\\cos\\frac{k\\pi x}{b}\\right)\\\\\n &= \\frac{1}{b}e^{-t}\\theta_4\\left(\\frac{\\pi x}{2 b}\\middle| \\frac{i\\pi t}{b^2}\\right).\n \\end{aligned}\n\\end{equation}\n\n\\subsection{The tapered cable}\n\nConsider the cable as a conical frustum on the interval $[A, B]$ with radius\n$r(x) = x \\tan\\theta$ and boundary conditions $v'(A, t) = 0$, $v'(B, t) = -E$,\nand $v(x, 0) = 0$ (see Figure \\ref{fig:tapered}). A fixed injecting current $I$\nat $x=B$ corresponds to $-E=I\\cdot R_L/\\pi r(B)^2$. Reparameterizing gives\n\\begin{equation}\n v(x, t) = -\\lambda E\\cdot u(x/\\lambda, t/\\tau),\n\\end{equation}\nwhere\n\\begin{equation}\n \\lambda = \\frac{R_M\\sin\\theta}{2 R_L}, \\quad \\tau = R_M C_M, \\quad a = A/\\lambda, \\quad b = B/\\lambda,\n\\end{equation}\nand $u$ satisfies\n\\begin{equation}\n \\label{eq:conu}\n \\begin{aligned}\n x u''(x, t) + 2 u'(x, t) &= \\dot u(x, t) + u(x, t),\\\\\n u(x, 0) &= 0,\\\\\n u'(a, t) &= 0,\\\\\n u'(b, t) &= 1.\n \\end{aligned}\n\\end{equation}\n\n\\begin{figure}[tbh]\n \\begin{tikzpicture}[scale=0.8]\n \\coordinate (origin) at (0,0);\n \\coordinate (b0) at (15,0);\n \\coordinate (b1) at ($ (b0) + (0,1.8) $);\n \\coordinate (a0) at (5,0);\n\n \\path [name path=tangent] (origin) -- (b1);\n \\path [name path=avert] (a0) -- +(0,2);\n \\path [name intersections={of=tangent and avert, by=a1}];\n\n \\coordinate (a-1) at ($ (a1)!2!(a0) $);\n \\coordinate (b-1) at ($ (b1)!2!(b0) $);\n\n \\draw [name path=axis, dashed] (origin) -- (b0);\n\n \\draw let \\p1 = (a1), \\n2 = {0.2*\\y1} in\n (a1) arc (90:270:\\n2 and \\y1);\n\n \\draw let \\p1 = (a1), \\n2 = {0.2*\\y1} in\n [densely dotted] (a-1) arc (-90:90:\\n2 and \\y1);\n\n \\draw let \\p1 = (b1), \\n2 = {0.2*\\y1} in\n (b0) ellipse (\\n2 and \\y1);\n\n \\draw (a1) -- (b1) (a-1) -- (b-1);\n \\draw [dotted] (origin) -- (a1) (origin) -- (a-1);\n\n \\pic [draw, \"$\\theta$\", angle radius = 15ex, angle eccentricity=1.2] {angle = b0--origin--b1};\n\n \\coordinate (lright) at ($ (b1)!2!(b0) - (0,40pt) $);\n \\coordinate (lleft) at ($ (origin) + (lright) - (b0) $);\n\n \\path (lleft) node (x0label) { $x=0$ };\n \\path (lright) node (xblabel) { $x=B =b\\lambda $ };\n \\path ($ (lleft)!(a0)!(lright) $) node (xalabel) { $x=A =a\\lambda$ };\n\n \\draw [dotted] ($ (a-1)-(0,2ex) $) -- ($ (xalabel)+(0,2ex) $);\n \\draw [dotted] ($ (b-1)-(0,2ex) $) -- ($ (xblabel)+(0,2ex) $);\n \\draw [dotted] ($ (origin)-(0,2ex) $) -- ($ (x0label)+(0,2ex) $);\n \\end{tikzpicture}\n \\caption{Tapered cable coordinates.}\n \\label{fig:tapered}\n\\end{figure}\n\nProceeding similarly to the cylinder problem, the solution can be presented as a sum of\na time-invariant solution $\\hat u$ and a series of transients that are solutions to\na Sturm--Liouville problem.\n\nThe time-invariant problem is\n\\begin{equation}\n \\begin{gathered}\n x\\hat u''(x) + 2\\hat u'(x) - \\hat u(x) = 0,\\\\\n \\hat u'(a) = 0,\\quad \\hat u'(b) = 1.\n \\end{gathered}\n\\end{equation}\n\nMultiplying by $x$ gives a differential equation of the form\n\\begin{equation}\n \\label{eq:conicmu}\n x^2 g''(x) + 2x g'(x) + \\mu x g(x) = 0\n\\end{equation}\nfor some $\\mu$. This has solutions in terms of Bessel functions.\nIf $\\cC_\\nu$ is a solution to the Bessel equation\n\\begin{equation}\n z^2 f''(z) + z f'(z) - (z^2 - \\nu^2) f(z) = 0,\n\\end{equation}\nthen $g(x) = x^{-\\frac{1}{2}}\\cC_1(\\alpha x^\\frac{1}{2})$ satisfies\n\\begin{equation}\n \\label{eq:besg}\n x^2 g''(x) + 2x g'(x) + \\frac{\\alpha^2}{4} x g(x) = 0.\n\\end{equation}\nSimilarly, if $\\cZ_\\nu$ is a solution to the modified Bessel equation\n\\begin{equation}\n z^2 f''(z) + z f'(z) - (z^2 + \\nu^2) f(z) = 0,\n\\end{equation}\nthen $g(x) = x^{-\\frac{1}{2}}\\cZ_1(\\alpha x^\\frac{1}{2})$ satisfies\n\\begin{equation}\n \\label{eq:modbesg}\n x^2 g''(x) + 2x g'(x) - \\frac{\\alpha^2}{4} x g(x) = 0.\n\\end{equation}\n\nThe time-invariant solution $\\hat u(x)$ satisfies \\eqref{eq:modbesg} with $\\alpha=2$, and so\ncan be written as\n\\begin{equation}\n \\hat u(x) = \\frac{1}{\\sqrt{x}} \\left(c_1 I_1(2\\sqrt{x}) - c_2 K_1(2\\sqrt{x})\\right),\n\\end{equation}\nwith derivative\n\\begin{equation}\n \\hat u'(x) = \\frac{1}{x} \\left(c_1 I_2(2\\sqrt{x}) + c_2 K_2(2\\sqrt{x})\\right).\n\\end{equation}\nSolving for the boundary conditions $\\hat u'(a)=0$, $\\hat u'(b)=1$ then determines the coefficients\n$c_1$ and $c_2$, giving\n\\begin{equation}\n \\label{eq:conuhat}\n \\hat u(x) =\n \\frac{b}{\\sqrt{x}}\\cdot\n \\frac{K_2(2\\sqrt{a})I_1(2\\sqrt{x}) + I_2(2\\sqrt{a})K_1(2\\sqrt{x})}\n {K_2(2\\sqrt{a})I_2(2\\sqrt{b})-I_2(2\\sqrt{a})K_2(2\\sqrt{b})}.\n\\end{equation}\n\nSeparable transient solutions $u(x,t)-\\hat u(x)$ will be of the form $e^{-\\lambda t}\\psi(x)$,\nwhere $\\psi(x)$ satisfies\n\\begin{gather}\n \\label{eq:conpsi1}\n x\\psi''(x) + 2\\psi'(x) + (\\lambda-1)\\psi(x) = 0\\\\\n \\label{eq:conpsi2}\n \\psi'(a) = \\psi'(b) = 0.\n\\end{gather}\nMultiplying \\eqref{eq:conpsi1} by $x$ gives\n\\begin{equation}\n \\label{eq:conpsils}\n x^2\\psi''(x) + 2x\\psi'(x) + (\\lambda-1)x \\psi(x) =\n \\frac{d}{dx}(x^2\\psi'(x)) + (\\lambda-1)x \\psi(x) = 0,\n\\end{equation}\nwhich together with \\eqref{eq:conpsi2} forms a regular Sturm--Liouville problem with\nnon-negative eigenvalues $\\lambda_k-1$ and corresponding orthogonal eigenfunctions\n$\\psi_k(x)$ with respect to the weight function $\\rho(x) = x$.\n\nFor $\\lambda_0=1$, the eigenfunction is $\\psi_0(x)=1$. For $\\lambda_k>0$,\nequation \\eqref{eq:conpsils} is of the form \\eqref{eq:besg}, with $\\alpha=2\\sqrt{\\lambda_k-1}$.\nWriting $\\omega_k$ for $\\sqrt{\\lambda_k-1}$, we have\n\\begin{equation}\n \\psi_k(x) = \\frac{1}{\\sqrt{x}} \\left(c_1^{(k)} J_1(2\\omega_k \\sqrt{x})) + c_2^{(k)} Y_1(2\\omega_k \\sqrt{x})\\right),\n\\end{equation}\nwith derivative\n\\begin{equation}\n \\psi_k'(x) = - \\frac{\\omega_k}{x} \\left(c_1^{(k)} J_2(2\\omega_k\\sqrt{x})) + c_2^{(k)} Y_2(2\\omega_k\\sqrt{x})\\right).\n\\end{equation}\n\nThe boundary conditions $\\psi_k'(a) = \\psi_k'(b) = 0$ imply\n\\begin{gather}\n f_2\\left(\\sqrt{\\tfrac{b}{a}}, 2\\omega_k\\sqrt{a}\\right) = 0,\n \\intertext{where $f_\\nu$ is the cross-product Bessel function}\n f_\\nu(q, x) = J_\\nu(qx)Y_\\nu(x)-J_\\nu(x)Y_\\nu(qx).\n\\end{gather}\nThe eigenvalues for $k>0$ are then $\\lambda_k=1+\\omega_k^2$ with\n\\begin{equation}\n \\label{eq:conomegak}\n \\omega_k = \\frac{\\chi_{2, k}\\left(\\sqrt{\\scriptstyle \\frac{b}{a}}\\right)}{2\\sqrt{a}}.\n\\end{equation}\nwhere $\\chi_{\\nu, k}(q)$ is the $k$th positive root of $f_\\nu(q, x)=0$. These solutions\ncan be computed efficiently via a Newton--Raphson scheme \\autocite{sorolla2013}.\n$c_1^{(k)}$ and $c_2^{(k)}$ are then determined up to scale; for example, we can\nset\n\\begin{equation}\n c_1^{(k)} = J_2(2\\omega_k\\sqrt{x})^{-1},\\qquad\n c_2^{(k)} = -Y_2(2\\omega_k\\sqrt{x})^{-1},\n\\end{equation}\ngiving\n\\begin{equation}\n \\label{eq:conpsik}\n \\psi_k(x) =\\frac{1}{\\sqrt{x}}\\left(\n \\frac{J_1(2\\omega_k\\sqrt{x})}{J_2(2\\omega_k\\sqrt{a})}\n -\n \\frac{Y_1(2\\omega_k\\sqrt{x})}{Y_2(2\\omega_k\\sqrt{a})}\n \\right).\n\\end{equation}\n\nThe complete solution then will be\n\\begin{equation}\n \\label{eq:conuxt}\n u(x,t) = \\hat u(x) - \\sum_{k=0}^{\\infty} a_k e^{-\\lambda t} \\psi_k(x),\n\\end{equation}\nwhere the $a_k$ will be determined by the initial conditions.\n\nAs the $\\psi_k$ are\neigenfunctions of the Sturm-Liouville problem (\\ref{eq:conpsils}, \\ref{eq:conpsi2}), they\nare orthogonal with respect to the inner product with weight function $\\rho(x)=x$,\n\\begin{equation}\n \\langle f, g \\rangle = \\Int{a}{b} xf(x)g(x) \\D x.\n\\end{equation}\nTaking the inner product of $\\langle u(x,0), \\psi_k(x)\\rangle$ then gives for each $k$,\n\\begin{equation}\n a_k = \\frac{\\Int{a}{b} x \\hat u(x)\\psi_k(x) \\D x}{\\Int{a}{b} x \\psi_k(x)^2 \\D x}.\n\\end{equation}\n\nFor $k=0$,\n\\begin{equation}\n \\begin{aligned}\n a_0 &= \\frac{1}{b-a} \\cdot \\Int{a}{b} x \\hat u(x) \\D x\\\\\n &= \\frac{1}{b-a} \\cdot \\left. x^2\\hat u'(x)\\right|_a^b\\\\\n &= \\frac{b^2}{b-a}.\n \\end{aligned}\n\\end{equation}\n(See Appendix \\ref{ap:conid} for the derivation of this and the following integrals.)\n\nFor $k>0$,\n\\begin{equation}\n \\begin{aligned}\n \\Int{a}{b} x \\hat u(x)\\psi_k(x) \\D x\n &= \\frac{1}{1+\\omega_k^2} \\cdot \\left. (x^2\\hat u'(x)\\psi_k(x)-x^2\\hat u(x)\\psi'_k(x))\\right|_a^b\\\\\n &= \\lambda_k^{-1} b^2\\psi_k(b),\n \\end{aligned}\n\\end{equation}\nand\n\\begin{equation}\n \\begin{aligned}\n \\Int{a}{b} x \\psi_k(x)^2 \\D x\n &= \\left.\n x^2\\psi_k(x)^2 + \\frac{x^2}{\\omega_k^2}\\psi_k(x)\\psi'_k(x)\n + \\frac{x^3}{\\omega_k^2}\\psi'_k(x)^2\n \\right|_a^b\\\\\n &= b^2\\psi_k(b)^2 - a^2\\psi_k(a)^2,\n \\end{aligned}\n\\end{equation}\ngiving\n\\begin{equation}\n a_k = \\frac{1}{\\lambda_k}\\cdot \\frac{b^2\\psi_k(b)}{b^2\\psi_k(b)^2-a^2\\psi_k(a)^2}.\n\\end{equation}\n\nThis gives the time-dependent solution\n\\begin{equation}\n u(x,t) = \\hat u(x) - \\frac{b^2}{b-a}e^{-t} -\n \\sum_{k=1}^{\\infty} \\lambda_k^{-1} e^{-\\lambda_k t}\n \\frac{b^2\\psi_k(b)\\psi_k(x)}{b^2\\psi_k(b)^2-a^2\\psi_k(a)^2}\n\\end{equation}\nwhere $\\hat u$ is given by \\eqref{eq:conuhat}, $\\psi_k$ is given by \\eqref{eq:conpsik},\nand $\\lambda_k = 1 + \\omega_k^2$ is given by \\eqref{eq:conomegak}.\n\n\n\\section{Approximating the gradient}\n\nIn a finite volume discretization there\nwill be a computation of the approximation to the gradient\n$v'(x)$ as a linear combination of two (or potentially more)\nvalues of the discrete approximation of the voltage for points\nnear $x$.\n\nIf the coefficients in this approximation are constant in\ntime, they cannot account for source terms or values of the\nvoltage outside the points in question. The most faithful\nsuch approximation should then reproduce the exact gradient $v'(x)$\nin the source-free steady state form of the cable equation,\n\\begin{equation}\n (\\sigma v')' = 0.\n\\end{equation}\n\nThe voltage $v$ is then determined by its values at\nany two distinct points $a$ and $b$,\n\\begin{equation}\n \\frac{v(x) - v(a)}{v(b) - v(a)} =\n \\frac{\\displaystyle \\Int{a}{x} \\sigma(z)^{-1} \\D z}\n {\\displaystyle \\Int{a}{b} \\sigma(z)^{-1} \\D z},\n\\end{equation}\nand correspondingly, the gradient is\n\\begin{equation}\n \\sigma(x) v'(x) =\n \\frac{v(b) - v(a)}\n {\\displaystyle\\Int{a}{b} \\sigma(z)^{-1} \\D z}.\n\\end{equation}\nIf $x$ is a point of discontinuity in $\\sigma$,\nthe flux $\\sigma(x)v'(x)$ is nonetheless well defined.\n\nFor a finite volume approximation, however, the available\nestimates may correspond instead to surface area-weighted means\nover control volumes. Consider two adjacent control volumes\non $[a, m]$ and $[m, b]$, with mean voltages $\\bar v_a$ and\n$\\bar v_b$ given in terms of a weight function $w(x)$:\n\\begin{align}\n \\bar v_a &= w_a^{-1} \\Int{a}{m} w(x)v(x) \\D x,\\\\\n \\bar v_b &= w_b^{-1} \\Int{m}{b} w(x)v(x) \\D x,\n\\end{align}\nwith normalizing constants $w_a$ and $w_b$. With $\\sigma v' = \\kappa$\nconstant,\n\\begin{equation}\n \\begin{aligned}\n \\bar v_a\n &= w_a^{-1} \\Int{a}{m} w(x) \\left( v(m) + \\kappa \\Int{m}{x} \\sigma(y)^{-1} \\D y \\right) \\D x \\\\\n &= v(m) + \\kappa \\cdot w_a^{-1} \\Int{a}{m} w(x) \\Int{m}{x} \\sigma(y)^{-1} \\D y \\D x,\n \\end{aligned}\n\\end{equation}\nand similarly, \n\\begin{equation}\n \\begin{aligned}\n \\bar v_b\n &= v(m) + \\kappa \\cdot w_b^{-1} \\Int{m}{b} w(x) \\Int{m}{x} \\sigma(y)^{-1} \\D y \\D x,\n \\end{aligned}\n\\end{equation}\nTaking the difference then gives a solution for $\\kappa$,\n\\begin{equation}\n \\kappa = (\\bar v_b - \\bar v_a) \\cdot \\left[\n w_b^{-1}\\!\\Int{m}{b} w(x)\\! \\Int{m}{x} \\sigma(y)^{-1} \\D y \\D x\\,-\\,\n w_a^{-1}\\!\\Int{a}{m} w(x)\\! \\Int{m}{x} \\sigma(y)^{-1} \\D y \\D x\n \\right]^{\\mathrlap{-1}}.\n\\end{equation}\n\n\\newpage\n\\appendix\n\\section{The Rallpack 1 model}\n\\label{ap:rallpack}\n\nThe Rallpack suite \\autocite{bhalla1992} is a set of three models for the\nvalidation and benchmarking of cable-based neuron simulators. The first of\nthese comprises a cylindrical passive cable of length $L$ and diameter $d$,\nwith a constant current applied at $x=0$. The membrane resistive current\ndensity is given by $J = R_M^{-1}(v-E)$, where $E$ is a fixed reversal potential.\nThe values of the electrial and geometric parameters are given in\nTable~\\ref{tbl:rallpack1}.\n\nThe membrane potential can then be expressed as\n\\begin{equation}\n v(x, t) = E - \\lambda I \\cdot u(\\frac{b-x}{\\lambda}, \\frac{t}{\\tau}),\n\\end{equation}\nwhere\n\\begin{equation}\n \\lambda = \\sqrt{\\frac{R_M d}{4 R_L}},\\quad\n \\tau = R_M C_M,\\quad\n b = L/\\lambda,\n\\end{equation}\nand $u(x,t)$ is given by \\eqref{eq:cylseries}.\n\n\\begin{table}[htb]\n \\centering\n \\begin{tabular}{lll}\n \\toprule\n\tParameter & {} & {Value} \\\\\n \\midrule\n\tCable diameter & $d$ & \\SI{1.0}{\\um} \\\\\n\tCable length & $L$ & \\SI{1.0}{\\mm} \\\\\n\tBulk (axial) resistivity & $R_L$ & \\SI{1.0}{\\ohm\\m} \\\\\n\tMembrane resistivity & $R_M$ & \\SI{4.0}{\\ohm\\m\\squared} \\\\\n\tMembrane specific capacitance & $C_M$ & \\SI{0.01}{\\F\\per\\m\\squared} \\\\\n\tMembrane reversal potential & $E_M$ & \\SI{-65.0}{\\mV} \\\\\n\tInjected current & $I$ & \\SI{0.1}{\\nA} \\\\\n \\bottomrule\n \\end{tabular}\n \\caption{Rallpack 1 parameters}\n \\label{tbl:rallpack1}\n\\end{table}\n\n\n\\section{Series computation for the voltage on the uniform cylinder}\n\\label{ap:cylcomp}\n\nHow many terms in the series \\eqref{eq:cylseries} need to be computed for a given tolerance?\nWe obtain bounds on the remainders in the series as follows.\n\nLet $u_n(x,t)$ be the partial sum up to $k=n-1$ in \\eqref{eq:cylseries}, and $r_n(x,t) = u(x,t) - u_n(x,t)$\nbe the remainder,\n\\begin{equation}\n r_n(x,t) = -\\frac{2}{b}\\sum_{k=n}^{\\infty} (-1)^k\\lambda_k^{-1}e^{-\\lambda_k t}\\cos\\frac{k\\pi x}{b}.\n\\end{equation}\nThen $|r_n(x,t)|\\leq R_n(t)$ for all $x$, with\n\\begin{equation}\n R_n(t) = \\frac{2}{b}\\sum_{k=n}^{\\infty} \\lambda_k^{-1}e^{-\\lambda_k t}\\\\\n = \\frac{2t}{b}\\sum_{k=n}^{\\infty} l(k)^{-1}e^{-l(k)}\\\\\n\\end{equation}\nwhere\n\\[\n l(z) = \\left(1+\\left(\\frac{z\\pi}{b}\\right)^2\\right)\\cdot t.\n\\]\nThe summand is monotonically decreasing in $k$, so\n\\begin{equation}\n \\label{eq:rnigamma}\n R_n(t)\n \\leq \\frac{2t}{b}\\Int{n}{\\infty} l(k)^{-1}e^{-l(k)}\\D k\n = \\frac{\\sqrt{t}}{b} \\Int{\\lambda_n t}{\\infty} e^{-z}z^{-\\frac{3}{2}} \\D z.\n\\end{equation}\n\nFor $a>0$, $s>0$, integrating by parts gives\n$\\displaystyle \\Int{a}{\\infty} e^{-z}z^{-s}\\D z < e^{-a}a^{-s}$,\nand so\n\\begin{equation}\n R_n(t) < \\frac{1}{b t}\\lambda_n^{-3/2}e^{-\\lambda_n t}.\n\\end{equation}\n\n\\section{Identities and integrals for tapered cable eigenfunctions}\n\\label{ap:conid}\n\n\\newcommand{\\Fp}{\\smash{F^{\\mathrlap{\\prime}}}\\mkern-2.0mu}\n\\newcommand{\\Gp}{\\smash{G^{\\mathrlap{\\prime}}}\\mkern-2.0mu}\n\nLet\n\\begin{equation}\n \\begin{aligned}\n F_k(z) &= z^{-\\frac{k}{2}}\\cZ_k(2z^{1/2}),&\n G_k(z) &= z^{-\\frac{k}{2}}\\cC_k(2\\omega z^{1/2}),\n \\end{aligned}\n\\end{equation}\nwhere $C_k$ and $Z_k$ are solutions to the Bessel and modified Bessel equations\nrespectively,\n\\begin{equation}\n \\begin{aligned}\n \\cZ_k(z) &= c_1 I_k(z) + c_2 e^{i\\pi k} K_k(z),&\n \\cC_k(z) &= c_3 J_k(z) + c_4 Y_k(z)\n \\end{aligned}\n\\end{equation}\nfor some constants $c_1$, $c_2$, $c_3$, $c_4$.\n\n$F_k$ and $G_k$ are solutions of the differential equations\n\\begin{gather}\n z f''(z) + (k+1) f'(z) - z = 0\\\\\n \\intertext{and}\n z g''(z) + (k+1) g'(z) + \\omega^2 z = 0\n\\end{gather}\nrespectively, following the identity (2) of \\autocite[p.~512]{lommel1879}.\n\nApplying the derivative and recurrence relations for Bessel functions\n\\autocite[\\S 10.6, \\S10.29]{nistdlmf} then gives the following identities,\n\\begin{align}\n \\label{eq:fgrel}\n \\Fp_k(z) &= F_{k+1}(z), & z \\Fp_k(z) &= -k F_k(z) + F_{k-1}(z),\\\\\n \\Gp_k(z) &= -\\omega G_{k+1}(z), & z \\Gp_k(z) &= -k G_k(z) + \\omega G_{k-1}(z).\n\\end{align}\nWhen $k$ is an integer, these imply\n\\begin{equation}\n \\begin{aligned}\n F_{-k}(z) &= z^k F_k(z),\\\\\n G_{-k}(z) &= (-z)^k G_k(z),\n \\end{aligned}\n \\qquad \\text{for $k\\in\\mathbb{Z}$}.\n\\end{equation}\n\nIn the following, derivatives will be respect to $z$, and the argument $z$ to\n$F_k$ and $G_k$ will be omitted where it is unambiguous.\n\n\\subsection*{The integral $\\Int{}{} z F_1 G_1 \\D z$}\n\nThe relations \\eqref{eq:fgrel} give the identities\n\\begin{equation}\n \\begin{aligned}\n \\label{eq:fgrel2}\n (z^kF_k)' &= z^{k-1}F_{k-1}\\\\\n (z^kG_k)' &= \\omega z^{k-1}G_{k-1}.\n \\end{aligned}\n\\end{equation}\nConsequently\n\\begin{equation}\n \\begin{aligned}\n (z^2 F_2 G_1)' &= z F_1 G_1 + z^2 F_2 \\Gp_1 = z F_1 G_1 - \\omega z^2 F_2 G_2 \\\\\n (z^2 F_1 G_2)' &= z F_1 G_1 + z^2 \\Fp_1 G_2 = z F_1 G_1 + z^2 F_2 G_2.\n \\end{aligned}\n\\end{equation}\nCancelling the $F_2G_2$ term gives\n\\begin{equation}\n (z^2 F_2 G_1 + \\omega z^2 F_1 G_2)' =\n (1+\\omega^2) z F_1 G_1,\n\\end{equation}\nand so\n\\begin{equation}\n (1+\\omega^2) \\Int{}{} z F_1 G_1 dz = z^2 F_2 G_1 + \\omega z^2 F_1 G_2 = z^2 \\Fp_1 G_1 - z^2 F_1 \\Gp_1.\n\\end{equation}\n\n\\subsection*{The integral $\\Int{}{} z F_1 \\D z$}\n\nAs observed in \\eqref{eq:fgrel2}, $(z^2 F_2)' = z F_1$, and so\n\\begin{equation}\n \\Int{}{} z F_1 dz = z^2 F_2 = z^2 \\Fp_1.\n\\end{equation}\n\n\\subsection*{The integral $\\Int{}{} z G_1^2 \\D z$}\n\nFollowing the technique of \\autocite[p.533]{lommel1879},\n\\begin{equation}\n \\begin{aligned}\n (z^a G_k G_{m+1})'\n &= a z^{a-1} G_k G_{m+1} + z^a(\\Gp_k G_{m+1} + G_k \\Gp_{m+1})\\\\\n &= a z^{a-1} G_k G_{m+1} - \\omega z^a G_{k+1}G_{m+1} + z^{a-1} G_k(\\omega G_m - (m+1) G_{m+1})\\\\\n &= (a-m-1) z^{a-1} G_k G_{m+1} + \\omega z^{a-1} (zG_{k+1}G_{m+1} - G_k G_m).\n \\end{aligned}\n\\end{equation}\nThe last term is symmetric in $k$ and $m$, so\n\\begin{equation}\n (z^a G_k G_{m+1} - z^a G_m G_{k+1})' =\n (a-m-1) z^{a-1} G_k G_{m+1} - (a-k-1) z^{a-1} G_m G_{k+1}.\n\\end{equation}\nLetting $a=k+1$ then gives,\n\\begin{equation}\n (z^{k+1} G_k G_{m+1} - z^{k+1} G_{k+1} G_m)' = (k-m) z^k G_k G_{m+1},\n\\end{equation}\nand in particular,\n\\begin{equation}\n \\label{eq:intzg1sq}\n \\Int{}{} z G_1^2 \\D z = z^2 G_1^2 - z^2 G_0 G_2.\n\\end{equation}\n\nThe result \\eqref{eq:intzg1sq} can be expressed in terms of just $G_1$ and $\\Gp_1$ by\napplication of the recurrence relationships:\n\\begin{equation}\n \\label{eq:intzg1sqbis}\n \\Int{}{} z G_1^2 \\D z = z^2 G_1^2 + \\frac{z^2}{\\omega^2} G_1 \\Gp_1 + \\frac{z^3}{\\omega^2} G^{\\prime 2}_1.\n\\end{equation}\n\n\n\\printbibliography\n\\end{document}\n", "meta": {"hexsha": "0c6d34a249bdbb9ec012169292aa395602202fde", "size": 25130, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/math/cable_equation/cable_equation.tex", "max_stars_repo_name": "kanzl/arbor", "max_stars_repo_head_hexsha": "86b1eb065ac252bf0026de7cf7cbc6748a528254", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 53, "max_stars_repo_stars_event_min_datetime": "2018-10-18T12:08:21.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-26T22:03:51.000Z", "max_issues_repo_path": "doc/math/cable_equation/cable_equation.tex", "max_issues_repo_name": "kanzl/arbor", "max_issues_repo_head_hexsha": "86b1eb065ac252bf0026de7cf7cbc6748a528254", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 864, "max_issues_repo_issues_event_min_datetime": "2018-10-01T08:06:00.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-31T08:06:48.000Z", "max_forks_repo_path": "doc/math/cable_equation/cable_equation.tex", "max_forks_repo_name": "kanzl/arbor", "max_forks_repo_head_hexsha": "86b1eb065ac252bf0026de7cf7cbc6748a528254", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 37, "max_forks_repo_forks_event_min_datetime": "2019-03-03T16:18:49.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T10:39:51.000Z", "avg_line_length": 34.854368932, "max_line_length": 130, "alphanum_fraction": 0.6144846797, "num_tokens": 9824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.7981867729389246, "lm_q1q2_score": 0.6160752707540629}} {"text": "\\subsubsection{Repeated Linear Factors}\r\n\\noindent\r\nIf $Q(x)$ has repeated roots, it factors into\r\n\\begin{equation*}\r\n\tQ(x) = R(x)(x-a)^k\\text{, }k \\geq 2\\text{ and }R(a) \\neq 0\r\n\\end{equation*}\r\nWhen making the common denominator for each repeated root of multiplicity $k$, we do\r\n\\begin{equation*}\r\n\t\\frac{P(x)}{R(x)(x-a)^k} = \\left(\\text{Decomposition of }R(x)\\right) + \\frac{A_1}{x-a}+\\ldots+\\frac{A_k}{(x-a)^k}\r\n\\end{equation*}\r\nYou would then multiply each side by the denominator like in the linear factors case and solve for the coefficients. The only additional difficulty is that you might have to use previous results or solve a system of linear equations to get some of the constants.\r\n\r\n\\ifodd\\includeBackgroundReviewExamples\\input{./backgroundReview/algebraPreCalc/repeatedLinearFactors_example.tex}\\fi", "meta": {"hexsha": "829fb0cb403f34eb4ba272a59fd5271bf8b523a1", "size": 816, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/backgroundReview/algebraPreCalc/repeatedLinearFactors.tex", "max_stars_repo_name": "rawsh/Math-Summaries", "max_stars_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "diffEq/backgroundReview/algebraPreCalc/repeatedLinearFactors.tex", "max_issues_repo_name": "rawsh/Math-Summaries", "max_issues_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "diffEq/backgroundReview/algebraPreCalc/repeatedLinearFactors.tex", "max_forks_repo_name": "rawsh/Math-Summaries", "max_forks_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.7692307692, "max_line_length": 263, "alphanum_fraction": 0.7414215686, "num_tokens": 239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.828938825225204, "lm_q2_score": 0.743168019989179, "lm_q1q2_score": 0.616040825434771}} {"text": "\\documentclass[12pt]{article}\n\\usepackage{amssymb,latexsym,amsmath}\n%\\usepackage{epsfig}\n%\\usepackage{graphicx} \n%\\DeclareGraphicsExtensions{.pdf, .jpg, .tif, .eps, .jpg}\n\n\\newif\\ifpdf\n\\ifx\\pdfoutput\\undefined\n\\pdffalse % we are not running PDFLaTeX \n\\else\n\\pdfoutput=1 % we are running PDFLaTeX \n\\pdftrue\n\\fi\n\\ifpdf\n\\usepackage[pdftex]{graphicx}\n\\else\n\\usepackage{graphicx}\n\\fi\n\\ifpdf\n\\DeclareGraphicsExtensions{.pdf, .jpg, .tif}\n\\else\n\\DeclareGraphicsExtensions{.eps, .jpg}\n\\fi\n\n\n\n\\textwidth = 6.5 in\n\\textheight = 9 in\n\\oddsidemargin = 0.0 in\n\\evensidemargin = 0.0 in\n\\topmargin = -0.5 in\n\\headheight = 0.0 in\n\\headsep = 0.0 in\n\\parskip = 0.2 in\n\\parindent = 0.0 in\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\n\\newcommand{\\khat}{\\hat{\\mathbf k}}\n\\newcommand{\\uv}{\\mathbf u}\n%\\newcommand{\\omegav}{\\mathbf \\omega}\n\\newcommand{\\omegav}{\\mathbf w}\n\\newcommand{\\w}{\\mathbf w}\n\\newcommand{\\grad}{\\nabla}\n\\newcommand{\\curl}{\\grad \\times}\n\n\\DeclareMathOperator{\\Span}{span}\n\n\n\\title{The Boussinesq equations and Coordinate scalings in the\n LANL-Sandia DNS code} \\author{Mark Taylor, Beth Wingate, Susan Kurien}\n\n\\begin{document}\n\\maketitle\n\n\\section{Scaling of the Domain}\n\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=4.in]{box}\n\\caption{ }\n\\label{F:box}\n\\end{center}\n\\end{figure}\n\n\n\nWe would like to solve the NS equations in the box pictured in\nFig.~\\ref{F:box}. The equations are\n\\[\n\\frac{ \\partial \\uv }{\\partial t} + (\\curl \\uv + f \\khat) \\times \\uv + \n\\grad \\pi + N \\theta \\khat = \\nu \\Delta \\uv\n\\]\n\\[\n\\grad \\cdot \\uv = 0\n\\]\n\\[\n\\frac{ \\partial \\theta }{\\partial t} + \\uv \\cdot \\grad \\theta - \nN \\uv \\cdot \\khat = \\kappa \\Delta \\theta\n\\]\nThe Schmidt number is $\\nu / \\kappa$. \n\nWe refer to the coordinate system in the box in Fig.~\\ref{F:box} as\nthe primted coordinate system, and we now introduce a \nchange of variables, into the {\\em non-primed} coordinate\nsystem. In the non-primed coordinate system, the domain because a \nunit cube:\n\\[\nx = x', \\qquad y=y', \\qquad z = \\frac{z'}{h},\n\\]\nThe only change to our code is that:\n\\[\n\\frac{\\partial}{\\partial z'} = \\frac{1}{h} \\frac{\\partial}{\\partial z}\n\\]\nAnd thus \n\\[\n\\grad = \n\\begin{pmatrix} \\dfrac{\\partial}{\\partial x} \\\\[5mm]\n \\dfrac{\\partial}{\\partial y} \\\\[5mm]\n \\dfrac{1}{h} \\dfrac{\\partial}{\\partial z} \n\\end{pmatrix}\n \\qquad\n\\Delta = \\left(\\frac{\\partial}{\\partial x}\\right)^2 + \n \\left(\\frac{\\partial}{\\partial y}\\right)^2 +\n \\left(\\frac{1}{h}\\frac{\\partial}{\\partial z}\\right)^2 \n\\]\nand\n\\[\n\\grad \\cdot \\uv = \\frac{\\partial u_1 }{\\partial x} + \n\\frac{\\partial u_2 }{\\partial y} + \n\\frac{1}{h} \\frac{\\partial u_3 }{\\partial z}\n\\]\nand vorticity $\\omegav$ is given by\n\\begin{equation}\n\\label{E:vor}\n\\omegav = \\curl \\uv = \n\\begin{pmatrix}\n\\dfrac{\\partial u_3}{\\partial y} - \\dfrac{1}{h} \\dfrac{\\partial u_2}{\\partial z} \\\\[5mm]\n\\dfrac{1}{h} \\dfrac{\\partial u_1}{\\partial z} - \\dfrac{\\partial u_3}{\\partial x} \\\\[5mm]\n\\dfrac{\\partial u_2}{\\partial x} - \\dfrac{\\partial u_1}{\\partial y}\n\\end{pmatrix} \n\\end{equation}\n\n\nThe divergence free condition on $\\uv$ is solved by the choice\nof $\\pi$. This involves, \nfor some vector field ${\\mathbf v}$, \nsolving an equation of the form \n\\[\n\\Delta \\pi = \\grad \\cdot {\\mathbf v}\n\\] \nThe remaining equations we write in component form:\n\\[\n\\begin{pmatrix} u_1 \\\\\n u_2 \\\\\n u_3 \\\\\n\\end{pmatrix}_t\n + (\\omegav + f \\khat) \\times \n\\begin{pmatrix} u_1 \\\\\n u_2 \\\\\n u_3 \\\\\n\\end{pmatrix}\n+ \\grad \\pi +\n\\begin{pmatrix} 0 \\\\\n 0 \\\\\n N \\theta \\\\\n\\end{pmatrix}\n= \\nu \n\\begin{pmatrix} \\dfrac{\\partial^2 u_1}{\\partial x^2} + \n \\dfrac{\\partial^2 u_1}{\\partial y^2} + \n \\dfrac{1}{h^2}\\dfrac{\\partial^2 u_1}{\\partial z^2} \\\\[5mm]\n \\dfrac{\\partial^2 u_2}{\\partial x^2} + \n \\dfrac{\\partial^2 u_2}{\\partial y^2} + \n \\dfrac{1}{h^2}\\dfrac{\\partial^2 u_2}{\\partial z^2} \\\\[5mm]\n \\dfrac{\\partial^2 u_3}{\\partial x^2} + \n \\dfrac{\\partial^2 u_3}{\\partial y^2} + \n \\dfrac{1}{h^2}\\dfrac{\\partial^2 u_3}{\\partial z^2}\n\\end{pmatrix}\n\\]\n\\[\n\\frac{ \\partial \\theta }{\\partial t} + \\uv \\cdot \\grad \\theta - \nN u_3 = \\kappa \\left(\n \\dfrac{\\partial^2 \\theta}{\\partial x^2} + \n \\dfrac{\\partial^2 \\theta}{\\partial y^2} + \n \\dfrac{1}{h^2}\\dfrac{\\partial^2 \\theta}{\\partial z^2}\n\\right)\n\\]\n\n\n\nThus the only changes needed for our parallel DNS code are is that\nall derivatives in $z$ need to be scaled by $h$.\n\n\n\\section{Hyper Viscosity}\nConsider our scaled hyper viscosity term\n\\[\n\\frac{ \\partial \\uv }{\\partial t} = \n\\sqrt{E(k_\\textrm{max})} \\, k_\\textrm{max}^{-2\\eta + 1.5} \\Delta^\\eta \\uv\n\\]\nWe first verify that the units are correct. The units of\nenergy are $m^2/s^2$, so the units of the energy in a single\nspherical shell $E(k)$ is $m^3/s^2$. Thus we have\n\\[\n\\frac{m}{s^2} = \\frac{m^{1.5}}{s} \\,\\frac{1}{m^{-2\\eta+1.5}} \n \\,\\frac{1}{m^{2\\eta}}\n \\, \\frac{m}{s}\n\\]\nIn the frequency domain, \n\\[\n\\frac{ \\partial u_3 }{\\partial t} = \n\\sqrt{E(k_\\textrm{max})} \\, k_\\textrm{max}^{-2\\eta + 1.5} \n( (2\\pi l)^2 + (2\\pi m)^2 + (2\\pi n/h)^2)^\\eta u_3\n\\]\n\n\\section{Resolution Condition}\n\nThe assumptions made in the hydrostatic equations implicitly assume\n(acording to Leslie) that \n\\[\n \\max{ k_h } << \\min{k_z}\n\\]\nwhich works out to (assuming 2/3 dealiasing)\n\\[\n \\frac{N_x}{3} << \\frac{1}{h}\n\\]\n\n\n\n\\section{Computing Scalars}\nIntegrals per unit volume are approximated by the sum\n\\[\n = \\frac{1}{h} \\iiint f g \\, dx' \\, dy' \\, dz' = \\iiint f g \\, dx \\, dy \\, dz\n=\\frac{1}{N_x N_y N_z} \\sum f g\n\\]\nWe also have that \n\\[\n\\frac{1}{N_x N_y N_z} \\sum f g = \\sum \\hat{f} \\hat{g}\n\\]\nso integrals can be computed in Fourier or grid space. \n\nAll other scalars (kinetic energy, helicity, various dissipation\nterms) are computed in physical coordinates. So for example\n\\[\nk = <\\uv,\\uv> = + + \n\\]\nand the helicity \n\\[\n h = w_1 u_1 + w_2 u_2 + w_3 u_3\n\\]\nAs in example, using integer wave numbers $(l,m,n)$ (defined in the\nnext section), we have\n\\[\n< \\omegav , \\Delta \\uv> = \\sum (2 \\pi)^2 (l^2 + m^2 + (n/h)^2) \n\\, \\uv \\cdot \\omegav\n\\]\n\n\n\n\n\n\\section{Computing 3D Energy Spectra (in a spherical shell)}\nIf we choose to discretize our domain so that the \nmaximum wave number (in the primed coordinate system) is\nthe same in all diretions, then\n\\[\n\\Delta x' = \\Delta y' = \\Delta z'\n\\]\nOr\n\\[\nN_x = N_y = N_z/h\n\\]\nwhere $N_x$ is the number of grid points in the $x$ direction.\nThe highest wave number is given by:\n\\[\nk_x = \\pi N_x, \\qquad k_y = \\pi N_y, \\qquad k_z = \\pi N_z/h \n\\]\n(Note that $\\pi$ appears in the wave numbers since our domain is\nof length 1 in the $x$ and $y$ diretions.) The wave number spacing\nis given by\n\\[\n\\Delta k_x = 2 \\pi, \\qquad \\Delta k_y = 2 \\pi, \\qquad \\Delta k_z = 2 \\pi / h \n\\]\nIn our DNS code, we index our arrays of Fourier coefficients\nwith integers $(l,m,n)$, with $l=0 \\dots N_x/2$, \n$m=0 \\dots N_y/2$ and \n$n=0 \\dots N_z/2$. The wave number spacing is given by\ngiven by\n\\[\nk_x = l \\Delta k_x,, \\qquad k_y = m \\Delta k_y, \\qquad k_z = n \\Delta k_z\n\\]\nWe index the arrays containing power spectra\nsuch as $E(k)$ with integer wave number $k$, represents a shell or anulus\nof some chosen thickness centered around the wave number associated\nwith $k$.\n\nThe total energy is\n\\[\nE = \\sum_{l,m,n} | {\\hat u_1}(l,m,n) |^2 +\n | {\\hat u_2}(l,m,n) |^2 + \n | {\\hat u_3}(l,m,n) |^2\n\\]\n\nThe standard energy spectrum is now, for integers $k$:\n\\[\nE(k) = \\sum_{l,m,n} | {\\hat u_1}(l,m,n) |^2 +\n | {\\hat u_2}(l,m,n) |^2 + \n | {\\hat u_3}(l,m,n) |^2\n\\]\nwhere the sum is taken over all integers $(l,m,n)$ such that\n\\[\n(\\Delta k_z)^2 (k-1/2)^2 \\le (\\Delta k_x l)^2 + (\\Delta k_y m)^2 + (\\Delta k_z n)^2 < (\\Delta k_z)^2 (k+1/2)^2 \n\\]\nwhere we have chosen to use a shell thickness of $\\Delta k_z$.\nScaling out the $2\\pi$, we get\n\\[\n(k-1/2)^2 \\le h^2 \\left( l^2 + m^2 + (n/h)^2 \\right) < (k+1/2)^2\n\\] \n\n\n\n\n\n\\section{Computing 2D Energy Spectra (in an annulus)}\nWe also compute a 2D spectra $E(k_h,k_z)$ by summing squares of Fourier\ncoefficients over annular regions in \n$x$ and $y$ of radius $k_h$. \nwhere \n\\[\nk_h = \\sqrt{(k_x^2 + k_y^2)} = \\sqrt{(2\\pi l)^2 + (2\\pi m)^2}. \n\\]\nFrom this quantity, we can get $E(k_h,0)$, which \nrepresents the energy spectrum of the 2D field\nobtained by averaging the original 3D field in the direction of rotation,\nand then summing the energy within spherical shells (an annulus for 2D fields).\nWe are also interested in \n\\begin{equation}\nE(k_h) = \\sum_{k_z=0}^{\\pi N_z/h} E(k_h,k_z) \n\\label{E:SPECb}\n\\end{equation}\nwhich represents \nthe energy spectrum of the full 3D field, but summed over\nthe annular region between concentric cylinders\n(whose axis are in the $z$ direction). As with the spherical\nshell spectrum, Eq.~\\ref{E:SPECb} has the property that\nit partitions the energy into a set of discrete wave numbers, so\nthat the total energy \n\\[\nE = \\sum_{k_h} E(k_h)\n\\]\n\n\nThe 2D shell (annulus) thickness is now\ndetermined by the horizontal wavenumber spacings. In our code,\nwe use an integer index $k$ defined by $k_h = k \\Delta k_x$ and\ndefine\n\\begin{eqnarray*}\nE(k,j) = \\sum_{l,m} | {\\hat u_1}(l,m,n) |^2 +\n | {\\hat u_2}(l,m,n) |^2 + \n | {\\hat u_3}(l,m,n) |^2\n\\end{eqnarray*}\nwhere the sum is taken over fixed $n$ and all integers $(l,m)$ such that \n\\[\n(\\Delta k_h)^2 (j-1/2)^2 \\le (\\Delta k_x l)^2 + (\\Delta k_y m)^2 < (\\Delta k_h)^2 (j+1/2)^2 \n\\]\nand with $k_x=k \\Delta k_x$ for integers $k$. \nwe have chosen to use a shell thickness of $\\Delta k_h =\\Delta k_x = \\Delta\nk_y = 2\\pi$. Scaling out the $2\\pi$, we get\n\\[\n(j-1/2)^2 \\le l^2 + m^2 < (j+1/2)^2.\n\\] \n\n\\section{Running the Model for the Boussinesq equations}\n\n\\noindent\nThis section discusses important elements of the inputfile and how to\nbuild the executable.\n\nWhen you've chosen scalar type = 4 in the gridsetup script (see\nbelow), you need to add a line to the input file right after the line\nfor the smagorinsky coefficient,\n\\noindent\n\\begin{verbatim}\n.\n.\n.\n107.08 ! Bous. parameter Fr = .21/bous=107.08\n0e0 ! alpha (for NS-alpha model). alpha>1 is in units of delx\n.0 ! smagorinsky\n1 ! max number of schmidt numbers for passive scalars\n1. 4 ! schmidt number = nu/kappa, ratio of viscous coeffs,type of scalar\n0 ! compute_structure_functions\n.\n.\n. \n\\end{verbatim}\n\n\nNote that the type in this file must match the type in the gridsetup\ncommand. Also note that if you want to the code to compute the\ntwo-point correlations of potential vorticity then you need to set the\nvalue of \\texttt{compute structure functions} to 1.\n\n\\noindent\nTo set up the model you need to use the following:\n\n\\begin{description}\n \n\\item[1)] Example to set the FORTRAN arrays sizes before compiling:\n ($32^3$ simulation on 8 processors, prognostic variables: $u,v,w$\n and $\\theta$). Note the 4 on the end, which\n indicates the type of scalar equations for the code to use \\\\\n \n \\begin{verbatim}\n ./gridsetup.py 1 1 8 32 32 32 2 2 2 0 0 0 4\n \\end{verbatim}\n\n \\noindent\n This one sets up the problem for $128^3$ on 32 processors\n like this:\n\n \\begin{verbatim}\n ./gridsetup.py 1 1 32 128 128 128 2 2 2 0 0 0 4\n \\end{verbatim}\n\n\\item[2)] Then one makes the Boussinesq equations by\n typing\\\\\n\n \\begin{verbatim}\n gmake dnsb\n \\end{verbatim}\nthis is identical to the dns code, except that it compiles in\nthe Boussinesq diagnostics (aspect\\_diag.F90) \ninstead of the usual turbulence\ndiagnostics (turb\\_diag.F90). \n\\end{description}\n\n\\noindent\n\n\\section{Comparing with Smith \\& Waleffe (SW99 and SW02)}\n\n\\subsection{Scalings and definitions}\n\n\\subsubsection*{$\\underline{\\text{Wave numbers}}$}\nWave numbers are dimensional and should be computed $k_f = 2 \\pi\nk_f^m$, where the superscript $m$ stands for Mark. \n\n\\subsubsection*{$\\underline{\\text{Rossby Number}}$}\nSW02 computes this Rossby number,\n\\begin{equation}\n Ro = \\frac{(k_f^2 \\epsilon_f)^{1/3}}{\\text{fcor}}, \n\\end{equation}\nwhere each number in the equation is computed as,\n\\begin{description}\n \\item[1)] $k_f = 2 \\pi k_f^m$\n \\item[2)] $\\epsilon_f$ is computed as an average through the run using\n scalars.m. It can also be found for each time in the output file\n if one looks at the $d/dt$ line and hunts for the f parameter (for\n the forcing).\n \\item[3)] fcor is an input to the code. There is some question as to\n whether we should divide by $fcor/2$ or $fcor$, see SW02 Eq 3.2\n\n\\end{description}\n\n\\subsubsection*{$\\underline{\\text{Froude number}}$}\n\nModifications to the dns code to accomodate this is to include the\naddition of a parameter {\\bf{bous}} which multiplies the vertical\nvelocity in the density anomoly equation. The equations we actually\nsolve in the code are,\nWe would like to solve the NS equations in the box pictured in\nFig.~\\ref{F:box}. The equations are\n\\[ \n\\frac{ \\partial \\uv }{\\partial t} + (\\curl \\uv + \\text{fcor} ~\\khat) \\times \\uv\n+ \n\\grad \\pi + \\text{bous}~ \\theta \\khat = \\nu \\Delta \\uv \n\\] \n\\[ \n\\grad \\cdot \\uv = 0 \n\\] \n\\[ \n\\frac{ \\partial \\theta }{\\partial t} + \\uv \\cdot \\grad \\theta - \n\\text{bous} ~\\uv \\cdot \\khat = \\kappa \\Delta \\theta.\n\\] \nTherefore to compute a consistent Froude number,\n\\begin{equation}\nFr = \\frac{(\\epsilon_f k_f^2)^{1/3}}{\\text{bous}}.\n\\end{equation}\nBasically, $\\text{bous}$ in the code is the is the same as $N$ in SW02 eq\n2.1-2.3.\n\n\n\n\\subsubsection*{$\\underline{\\text{Time Scale}}$}\nTo be consistent with SM multiply the time in the output file by the\ntime scale from SM, $t_{sw02} = (\\epsilon_f (2 \\pi k_f)^2)^{1/3}$\n\n\n\\subsection{Test of rotation only, $\\text{bous} = 0$ (no stratification)}\n\nCompare figure (\\ref{fig:SW99Fig3}) and figure (r16ketime).\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{SW99Fig3}\n\\caption{From Smith and Waleffe 1999}\n\\label{fig:SW99Fig3}\n\\end{center}\n\\end{figure}\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{r16ketime}\n\\caption{From sandia/lanl dns code for different kinds of hypeviscosity}\n\\label{fig:r16ketime}\n\\end{center}\n\\end{figure}\n\n\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{SW99Fig4}\n\\caption{From Smith and Waleffe 1999}\n\\label{fig:SW99Fig4}\n\\end{center}\n\\end{figure}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{SW99Fig10}\n\\caption{From Smith and Waleffe 1999}\n\\label{fig:SW99Fig10}\n\\end{center}\n\\end{figure}\n\n\n\\subsection{Test of stratification only, $fcor=0$ (no rotation) }\n\nCompare figure (\\ref{fig:SW02Fig2}) to figure (\\ref{fig:t900})\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{SW02Fig2}\n\\caption{From Smith and Waleffe 2002}\n\\label{fig:SW02Fig2}\n\\end{center}\n\\end{figure}\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=4.in]{t900}\n\\caption{From the sandia/lanl dns code for 3 different types of hyperviscosity.}\n\\label{fig:t900}\n\\end{center}\n\\end{figure}\n\n\n\n\n\nFirst test, $f=0, N\\ne0$. Choose $bous=107.08$ with hyperviscosity. Then\n\\begin{equation}\nFr = \\frac{\\left((.5)(2 \\pi 24)^{2}\\right)^{1/3}}{107.08}= .21\n\\end{equation} \nForcing is: \n\\begin{verbatim}\nsto_high_24.\n\\end{verbatim}\nA plot of the spectrum is in Figure \\ref{fig:norot} and should be\ncompared to SW02's Figure 2.\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{frp21}\n\\caption{Spectrum from $200^3$ simulation of LANL/Sandia DNS code with\n no rotation and $Fr=.21$. This is Smith and Waleffe 2002 Figure 3.}\n\\label{fig:norot}\n\\end{center}\n\\end{figure}\n\n\\subsection{Test of nontrivial rotation and stratification $N\\ne0, f\\ne0$ - }\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=6.in]{SW02Fig4}\n\\caption{From Smith and Waleffe 2002}\n\\label{fig:SW02Fig4}\n\\end{center}\n\\end{figure}\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[angle=-90,width=4.in]{SW02Fig5}\n\\caption{From Smith and Waleffe 2002}\n\\label{fig:SW02Fig5}\n\\end{center}\n\\end{figure}\n\n\n\\begin{eqnarray}\nFr &=& \\frac{\\left((.5)(2 \\pi 24)^{2}\\right)^{1/3}}{107.08}= .21 \\\\ \nRo &=& \\frac{\\left((.5)(2 \\pi 24)^{2}\\right)^{1/3}}{1.07077} = 21\n\\end{eqnarray} \nThis means $\\text{bous}=108.07$ and $\\text{fcor} =1.07077 $ . We try\nto produce the plots in Smith and Waleffe Figure 4 and 5.\n\nForcing is\n\\begin{verbatim}\nsto_high_24.\n\\end{verbatim}\n\n\\section{Test of projection routines in unit aspect ratio. March 4, 2009, SK}\n\nThe projection of the boussinesq data onto the Craya-Herring normal\nmodes is in spectrum.F90 in the compute\\_project\\_ch subroutine. This\nroutine is called from analysis\\_spectral.F90 and decomposes a given\nsnapshot (gridspace) into its two wave and one vortical mode (SW02) in fourier space and computes the power spectra of those modes.\n\nAll our test cases were run to compare with the unit aspect ratio\ncases in SW02. We match the resolution ($128^3$), Ro, Fr and forcing\nwavenumber $k_f = 24$. Our energy input rate $\\epsilon_f = 0.5$.\n\n\\subsection{Case SW02: $128^3$ gridpoints, $N/f \\rightarrow \\infty$, $Fr = 0.21$, $Ro \\rightarrow \\infty$}\nFigure \\ref{Froude_time} shows the time evolution of the kinetic energy of the flow with the same nondimensional time and energy units used in SW02 namely:\n$$T = t(\\varepsilon_f k_f^2)^{1/3}$$\n$$energy = E(\\varepsilon_f/k_f)^{-2/3}$$\nThis case was run upto T = 1088 in these units. This figure should be compared with Fig. 1 in SW02 for the $Fr = 0.21$, $k_f = 24$ case.\n\\begin{figure}[ht]\n\\centering\n \\includegraphics[scale = 0.5]{Comp_SW02_fig1_Nfinf_kf24}\n\\caption{Evolution in time of the kinetic energy. To be compared with $Fr = 0.21$, $k_f = 24$ curve in Figure 1 of SW02.\\label{Froude_time}}\n\\end{figure}\n\nThe figure \\ref{Froude_spec} shows the total and projected spectra for\nthis case at T=1088, to be compared with SW02, Fig. 2, plotted at T = 900.\nThe total and $k_h=0$ agree very well qualitatively.\n\\begin{figure}[ht]\n\\centering\n \\includegraphics[scale = 0.7]{Comp_SW02_fig2}\n \\caption{Total energy and the projected wave, vortical and $k_h=0$\n spectral distributions at T = 690. To be compared with Figure 2 of\n SW02. \\label{Froude_spec}}\n\\end{figure}\n\n\\subsection{Case SW02: $128^3$ gridpoints, $N/f = 100$, $Fr = 0.21$, $Ro = 21$} \nFigure \\ref{NF100_time} shows the time evolution of the kinetic energy of the flow with the same non-dimensional time and energy units used in SW02 namely:\n$$T = t(\\varepsilon_f k_f^2)^{1/3}$$\n$$energy = E(\\varepsilon_f/k_f)^{-2/3}$$\nThis case was run upto T = 1151 in these units. Note that SW02\nruns this case upto T=2200 (not shown in their Fig. 4) and plots their\nspectra at T=2200. Comparing Fig. \\ref{NF100_time} and the $N/f = 100$\ncurve of SW02 Fig. 4 upto time T = 800 (the maximum time-scale in\nSW02) we notice slight differences which can probably be attributed to\nthe slight difference in our peak forcing wavenumbers (ours is\n$k_f=24$, vs. SW02 $k_f=20$) and the fact that our $\\epsilon_f \\simeq\n0.5$ whereas SW02 fixed theirs at $\\epsilon_f = 1$. But the overall\nshape of the curve, the time at which it starts to grow after an\ninitial equilibrium ($\\approx T = 200$) is the same.\n\\begin{figure}[ht]\n\\centering\n \\includegraphics[scale = 0.5]{Comp_SW02_fig4_Nf100}\n\\caption{Evolution in time of the kinetic energy. To be compared with $N/f = 100$ curve in Figure 4 of SW02.\\label{NF100_time}}\n\\end{figure}\n\nThe energy spectra of the total energy as well as the wave, vortical\nand $k_h=0$ VSHF modes is given in Fig. \\ref{NF100_spec}. This is\ncomputed at our latest time of $T=1151$, to be compared with the SW02\nFig. 5 computed at T = 2200. The differences can probably be\nattributed both to the later time of SW02 and the slight differences\nin our forcing wavenumber, energy input rate $\\epsilon_f$. But the\nqualitative agreement is good. The $k_h = 0$ VSHF modes carry most of\nthe energy in the peak low-wavenumbers. An unexplained difference is\nthe 'peaky' nature of the low-mode energy distribution ($5< k < 10$)\nin SW02 Fig. 5. This does not appear in our runs. We show in addition\nthe wave and vortical mode spectra which SW02 does not show. We do not\nshow the potential energy spectrum which SW02 does show.\n\\begin{figure}[ht]\n\\centering\n \\includegraphics[scale = 0.7]{Comp_SW02_fig5}\n\\caption{Total energy and the projected wave, vortical and $k_h=0$ spectral distributions at T = 1151. To be compared with Figure 5 of SW02. \\label{NF100_spec}}\n\\end{figure}\n\n\\subsection{Case SW02: $128^3$ gridpoints, $N/f = 10$, $Fr = 0.21$, $Ro = 2.1$}\nFigure \\ref{NF10_time} shows the time evolution of the kinetic energy\nof the flow with the same non-linear time and energy units used in\nSW02. This case was run upto about T = 690 in those units. Note that SW02\nruns this case upto T=605. Comparing Fig. \\ref{NF10_time} and the $N/f = 10$\ncurve of SW02 Fig. 4 upto time T = 605 (the maximum time-scale in\nSW02) we notice slight differences which can probably be attributed to\nthe fact that our $\\epsilon_f \\simeq\n0.5$ whereas SW02 fixed theirs at $\\epsilon_f = 1$. But the overall\nshape of the curve, the time at which it starts to grow after an\ninitial equilibrium ($\\approx T = 200$) is the same.\n\\begin{figure}[ht]\n\\centering\n \\includegraphics[scale = 0.5]{Comp_SW02_fig4_Nf10}\n\\caption{Evolution in time of the kinetic energy (NOT the total energy as it says in the title field). To be compared with $N/f = 10$ curve in Fig. 4 of SW02. \\label{NF10_time}}\n\\end{figure}\n\nThe energy spectra of the total energy for this run as well as the\nwave, vortical and $k_h=0$ VSHF modes is given in Fig.\n\\ref{NF10_spec}. This is computed at our latest time of $T=690$, to be\ncompared with the SW02 Fig. 9 computed at T = 605. The differences can\nprobably be attributed both to the slightly earlier time of SW02 and\nthe slight differences in our energy input rate $\\epsilon_f$. The\nqualitative agreement is good. An unexplained difference is the\n'peaky' nature of the low-mode energy distribution ($10< k < 15$) in\nSW02 Fig. 9, This does not appear in our runs although we do have a\nsmooth peaking of the energy in the same wavenumber range. The $k_h =\n0$ (VSHF) modes carry most of the energy in the peak low-wavenumbers.\nThe PV (vortical) mode spectra also agree well in that they do not\ncontribute to the most energetic low wave numvers (corresponding to\nthe peak in the $k_h=0$ spectrum but they make the dominant\ncontribution to the low wavenumbers $k < 8$ but the energy in these\nlow modes is an order of magnitude less than the peak in $10< k <15$.\nWe also show in addition the wave modes which SW02 does not show. We\ndo not show the potential energy spectrum which SW02 does show.\n\\begin{figure}[ht]\n\\centering \n \\includegraphics[scale = 0.7]{Comp_SW02_fig9}\n \\caption{Total energy and the projected wave, vortical and $k_h=0$\n spectral distributions computed at $T= 690$. To be compared with\n Fig. 9 of SW02. \\label{NF10_spec}}\n\\end{figure}\n\n\\section{Test of projection routines and non-unit aspect ratio. July 9, 2009. SK}\nWe are trying to recover Figure 1 (dashed line) and Figure 2 in\nRemmel, Sukhatme and Smith, 2009. Since our domain parameters our\ndifferent from those in Remmel et al 2009, we had to define the\nappropriate scales as follows:\n\n\\subsection{Scales}\n\\begin{table}[ht]\n\\begin{tabular}{lll}\n\\hline\\\\\nParameter & RSS09 & Sandia-LANL code\\\\\n\\hline\\\\\naspect ratio $A$ & 0.2 & 0.2\\\\\ndimension ($L_x \\times L_y\\times L_z$) & $10\\pi \\times 10\\pi \\times 2\\pi $ & $1 \\times 1 \\times A$ \\\\ \nresolution ($N_x \\times N_y \\times N_z$) & $80 \\times 400 \\times 400$ & $80 \\times 400 \\times 400$ \\\\\n$\\Delta k_x \\times \\Delta k_y \\times \\Delta k_z = \\frac{2\\pi}{L_x} \\times \\frac{2\\pi}{L_y} \\times \\frac{2\\pi}{L_z}$ & \n$0.2 \\times 0.2 \\times 1$ & $2\\pi \\times 2\\pi \\times 10\\pi$ \\\\\nenergy input rate $\\epsilon_f$ & 1 & 1 \\\\\nsmallest 3D $k$-shell & $\\Delta k = \\Delta k_z = 1$ & $\\Delta k = \\Delta k_z = 10\\pi$\\\\ \nforcing shell $k_f = 4$ & $k_f = 4\\times \\Delta k_z = 4$ & $k_f = 4 \\times \\Delta k_z = 4\\times 10\\pi$\\\\\ndamped modes (fat shells) & $k \\leq 2.5$ & $k \\leq 2.5\\times 10\\pi $ \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\nWith the above definitions, the various scales and $Ro$ and $Fr$ are\ndefined for each of the two simulations:\n\\begin{itemize}\n\\item typical velocity based on energy input rate and forcing scale\n (which by choice is the same as the {\\it vertical scale}) $U =\n ({\\epsilon_f}{H})^{1/3} = \\displaystyle(\\frac{\\epsilon_f}{k_f})^{1/3}$\n\\item typical vertical length scale $H = \\displaystyle\\frac{1}{k_f}$; \n\\item typical horizontal length scale $L = \\displaystyle\\frac{H}{A} = \\displaystyle\\frac{1}{A k_f}$;\n\\item $Ro = \\displaystyle\\frac{U}{Lf} = 0.05 \\implies f = \\displaystyle\\frac{U}{L Ro} = \\displaystyle\\frac{A \\epsilon_f^{1/3} k_f^{2/3} } {Ro}$;\n\\item $Fr = \\displaystyle\\frac{U}{HN} = 0.05 \\implies N = \\displaystyle\\frac{U}{H Fr} = \\displaystyle\\frac{\\epsilon_f^{1/3}k_f^{2/3}}{Fr}$;\n\\item Nonlinear time $\\tau = (\\epsilon_f k_f^2)^{(-1/3)}$. We will use subscript $RSS$ or $S-L$ to denote the RSS09 or Sandia-LANL timescales respectively.\n\\item Energy scale ${\\cal E} = (\\epsilon_f /k_f)^{(2/3)}$. Again the corresponding subscripts will be used.\n\\end{itemize}\n\nRSS09 forces the eigenmodes randomly at each time-step. Sandia-LANL\ncode forces $u,v,w$ and $\\theta$ randomly at each time-step. These\nshould be equivalent.\n\nRemmel et al have a uniform drag added to the rhs of the equations, \n$-0.1 b$ where $b$ is any\nof $u,v,w,\\theta$, for all modes $k < 2.5$. In order to choose the\ncorresponding factor for the Sandia-LANL simulation we note that the\ndrag coefficient $d_r = 0.1$ has units of $T^{-1}$. Assuming the\nnon-dimensional product $\\tau d_r$, should be the same in each of our\nsimulations we can determine $d_r$ for the Sandia-LANL code. \n\nIn the RSS09 simulation, $\\tau d_r \\simeq 25$, and\ntherefore for the Sandia-LANL code the corresponding \n\\begin{itemize}\n\\item $d_r = 25/\\tau = 1$.\n\\end{itemize}\nIn order to get RSS09's dimensional time, we rescale our time by\n$\\tau_{RSS}/\\tau_{S-L}$. In order to get RSS09's dimensional energy,\nwe rescale our energy by ${\\cal E}_{RSS}/{\\cal E}_{S-L}$.\n\n\\subsection{Results}\nUnless otherwise stated, all the scalings are as defined above for the\nSandia-LANL domain convention. The input file is:\n\n\\begin{verbatim}\n\n8 ! input file type\n1 ! file I/O header type 1-5\nns_uvw ! equations\nzero ! initial condition: KH-blob, KH-anal, iso12\n0 ! init condition subtype\nsto_high_t4 ! forcing: none, iso, iso12, iso12w, iso23w\n2.0 1 ! forcing parameters (not used for iso12, n.n=f^2, n width of gaussian)\nvalue ! viscosity type (value, smallest, kediff, Re)\n0e-5 ! viscosity coefficient. (300)\nhyper16_imp ! hyper16 ! hyper16_imp ! hyper vis\n2.500 ! hyper vis value\nhypo0 ! hypo vis\n1 ! hypo vis value\n100 ! fcor rotation\n0.2 ! Z sscaling (aspect ratio)\n500.0 ! Bous. parameter \n0e0 ! alpha (for NS-alpha model). alpha>1 is in units of delx\n.0 ! smagorinsky\n1 ! max number of schmidt numbers for passive scalars\n1. 4 ! schmidt number = nu/kappa, ratio of viscous coeffs, type\n0 ! compute_structure functions (set to 1 to compute strfns)\n0 ! compute pdf functions\nfft-dealias ! derivative method (fft, fft-dealias, fft-sphere, 4th)\nperiodic ! x bc (periodic, no-slip)\nperiodic ! y bc\nperiodic ! z bc\n5 ! time to run\n1.5 ! adv cfl (0 = disabled, use min_dt) .8\n0.25 ! vis cfl (0 = disabled, use min_dt) .01\n0.000010 ! min_dt\n.1 ! max_dt\n0.0 ! restart_dt (0 = restart output disabled)\n.1 ! diag_dt (.scalars)\n.1 ! model_dt (.spec,.scalars-bous, .bisostr, .isostr)\n-50 ! screen_dt (0 = disables, +ve = timesteps, -ve = time)\n.2 ! output_dt (field data, 0 = output disabled, except for custom out times)\n0 ! also output the vorticity\n0 ! n_output_custom number of custom output times\n00.00 ! custom output times, one per line\n\n\n\n\\end{verbatim}\n\n\nFigure \\ref{Fig1_H} is to be compared with RSS09 Fig. 1. \n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale = .7]{RSS09_Fig1_UH}\n\\caption{The time has been rescaled by $\\tau_{RSS}/\\tau_{S-L}$ and the energy has been rescaled by ${\\cal E}_{RSS}/{\\cal E}_{S-L}$. \\label{Fig1_H}}\n\\end{figure}\n\n\nFigure \\ref{Fig2_H} is to be compared with RSS09 Fig. 2.\n\\begin{figure}[ht]\n\\centering\n\\includegraphics[scale = .7]{RSS09_Fig2_UH}\n\\caption{Total, wave and vortical energy spectra at rescaled time = 30.\\label{Fig2_H}}\n\\end{figure}\n\n\n\\section{Bousinesq scalars we're interested in computing}\n\nWe'd like to compute the following (dimensional) quantities integrated\naround in the volume:\n\n\\begin{itemize}\n\\item[1.] Total energy and total energy dissipation\n\n\\begin{equation}\n\\frac{\\partial}{\\partial t} \\frac{1}{2} \\int_V \n \\left[ (u^2 + v^2 + w^2) + \\theta^2\\right] \n= \\frac{1}{2}\\int_V \\left[ \\nu \\nabla^2 (u^2 + v^2 + w^2) \n + \\kappa \\nabla^2 \\theta^2\n \\right] \n\\end{equation}\n\n\\item[2.] Potential energy, $\\frac{1}{2} \\theta^2$ \n\n\\item[3.] Kinetic energy, $\\frac{1}{2} ( u^2 + v^2 + w^2 )$ \n \n\\item[4.] Potential vorticity (q) and its production-dissipation which\n should globally integrate to zero I think.\n \\begin{equation}\n \\frac{\\partial}{\\partial t} \\int_V \n \\left[ q \\right] \n = \\int_V \\left[ \\nu (\\nabla^2 \\mathbf\\omega) \\cdot \\mathbf{\\nabla}\n (\\bar{\\theta})\n + \\kappa (\\nabla^2 \\mathbf{\\nabla} \\theta) \\cdot \n (\\mathbf\\omega \n + \\text{fcor} ~{\\bf{\\hat{z}}} )) \n \\right] \n \\end{equation}\n \n Here $\\mathbf = \\mathbf\\nabla \\times \\mathbf u$ is the vorticity and\n $\\bar\\theta = \\theta_o -\\text{bous}~ z + \\theta(x,y,z)$. Also, $q =\n (\\mathbf\\omega + \\text{fcor}~\\mathbf{\\hat{z}}) \\cdot \\mathbf\\nabla\n \\bar\\theta.$\n\n\\item[5.] Potential enstrophy $Q = q^2/2$ and its\n production-dissipation. This is the important production-dissipation\n rate we need for our two-point correlation.\n\n\\item[6.] Potential vorticity production-dissipation\n\n\\item[7.] Potential enstrophy\n \\begin{equation} \n \\frac{\\partial}{\\partial t} \\int_V \\left[ Q \\right] \n = \\int_V \\left[ \\nu ~q (~\\nabla^2 ~\\mathbf\\omega) \\cdot \\mathbf{\\nabla}\n (\\bar{\\theta}) \n + \\kappa ~q ~\\nabla^2 (\\mathbf{\\nabla} \\theta) \\cdot \n (\\mathbf\\omega \n + \\text{fcor} ~{\\bf{\\hat{z}}} )) \n \\right] \n \\end{equation} \n\n\\item[8.] two-point correlation (coming)\n\n\\end{itemize}\n\n\\section{Comparing SW to KSW}\nIn the two-point correlation paper the equations we use are scaled\ndifferently than than we use in Mark's code (which are the same\nequations used in SW). To make comparisions with our results in the\ntwo-point correlation paper we have to keep in mind the following\nscalings. Note, these are relationships between {\\sl{dimensional}}\nquantities, not nondimensional quantities.\n\\begin{gather}\n\\tilde\\rho = (\\frac{b \\rho_o}{g})^{1/2} ~\\theta\\\\\n(\\frac{g}{\\rho_o b})^{1/2}~\\rho = (\\frac{g \\rho_o}{b}) - \\text{bous}\n~z + \\theta = \\bar\\theta \\\\\nN = \\text{bous} = (\\frac{g b}{\\rho_o})^{1/2}\\\\\npe_{\\text{ksw}} = \\frac{1}{2} \\theta^2 = \\frac{g}{\\rho_o b} \\tilde\\rho^2\\\\\nq_{\\text{ksw}} = (\\frac{g}{b \\rho_o})^{1/2} q_{\\text{SW}}\n\\end{gather}\n\n\\end{document}\n", "meta": {"hexsha": "6071d754a816effd072a310846f61034f6779449", "size": 31651, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "rotation_doc/rotation.tex", "max_stars_repo_name": "fxie2/dns", "max_stars_repo_head_hexsha": "868b5463dc74e0833b4f31674bd5a2f01f492244", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2017-04-07T16:07:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-28T17:17:30.000Z", "max_issues_repo_path": "rotation_doc/rotation.tex", "max_issues_repo_name": "mkha1785/dns", "max_issues_repo_head_hexsha": "868b5463dc74e0833b4f31674bd5a2f01f492244", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-05-10T17:15:47.000Z", "max_issues_repo_issues_event_max_datetime": "2017-05-10T17:15:47.000Z", "max_forks_repo_path": "rotation_doc/rotation.tex", "max_forks_repo_name": "mkha1785/dns", "max_forks_repo_head_hexsha": "868b5463dc74e0833b4f31674bd5a2f01f492244", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 15, "max_forks_repo_forks_event_min_datetime": "2015-07-21T10:54:29.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-09T06:57:04.000Z", "avg_line_length": 34.7813186813, "max_line_length": 177, "alphanum_fraction": 0.6685096837, "num_tokens": 10688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.828938825225204, "lm_q2_score": 0.7431680143008301, "lm_q1q2_score": 0.6160408207194777}} {"text": "\\section{Tactic}\nAlthough proofs can be automatically checked given the system above, it requires user to input the whole\nterm of the specific type, which is tedious, exhaustive, mostly anti-intuitive.\\par\nSo in this part, we introduced some tactics into the system \nto assist users in better writing the proof. With the help of\ntactics, the proof becomes more human-readable and intuitive.\\par\n\n\\subsection{\\it apply}\nThis tactic applies to any goal. The purpose of this tactic is to extract premises of\ncurrent goal.\n\n\\subsection*{Usage}\n\\begin{itemize}\n\\item $\\it apply\\ \\tt t.$ : $\\tt t$ is a term of a type whose consequent is the goal.\n\\item $\\it apply\\ \\tt t\\ \\it in\\ \\tt H.$ : $\\tt t$ is a term of a type whose antecedent is $\\tt H$.\n\\end{itemize}\n\n\\subsection*{Implementation}\nThe tactic {\\it apply} tries to match the current goal against the \nfinal consequent of the type of term. \nIf it succeeds, it returns as many subgoals as \nthe number of non-dependent premises of the type of term.\\par\nHere is an example:\n\\begin{center}\n\\begin{minted}{coq}\nn : nat\nm : nat\no : nat\ne : eq nat n m -> eq nat m o -> eq nat n o \n(* Current Goal *) \neq nat n o \n\n(* $\\Downarrow$ apply n $\\Downarrow$ *) \n\nn : nat n : nat \nm : nat m : nat \no : nat o : nat \ne : eq nat n m -> eq nat m o -> eq nat n o e : eq nat n m -> eq nat m o -> eq nat n o \n(* SubGoal 1 *) (* SubGoal 2 *) \neq nat m o e : eq nat n m\n\\end{minted}\n\\end{center}\nBecause \\mintinline{coq}|e| has a type of \\mintinline{coq}|eq nat n m -> eq nat m o -> eq nat n o| \nand the goal is \\mintinline{coq}|eq nat n o|,\nthe goal is convertible to the \\textbf{consequent} of the type of \\mintinline{coq}|e|.\nWhen applied with \\mintinline{coq}|e|,\nall of the non-dependent premises of the type of \\mintinline{coq}|e| become subgoals.\n\nThe tactic {\\it apply $\\tt tm$ in $\\tt H$} tries to match the hypothesis $\\tt H$ against the \nantecedent of the type of $\\tt tm$. \nIf it succeeds, the hypothesis $\\tt H$ will be replaced by the consequent of $\\tt tm$.\n\\begin{center}\n\\begin{minted}{coq}\nn : nat\nm : nat\no : nat\nH : eq nat n m \ne : eq nat n m -> eq nat m o -> eq nat n o \n(* Current Goal *) \neq nat m o -> eq nat n o \n\n(* $\\Downarrow$ apply e in H $\\Downarrow$ *)\n\nn : nat \nm : nat \no : nat \nH : eq nat m o -> eq nat n o\ne : eq nat n m -> eq nat m o -> eq nat n o\n(* SubGoal 1*) \neq nat m o -> eq nat n o \n\\end{minted}\n\\end{center}\nBecause \\mintinline{coq}|e| has a type of \n\\mintinline{coq}|eq nat n m -> eq nat m o -> eq nat n o| \nand $\\tt H$ is \\mintinline{coq}|eq nat n m|,\n$\\tt H$ is convertible to the \\textbf{antecedent} of the type of the term.\nWhen applied with \\mintinline{coq}|e|,\n$\\tt H$ is replaced by the consequent of $\\tt e$.\n\n\\subsection{\\it destruct}\n\nWhen {\\it destruct} is applied on a term, this term must be inductive type. Then {\\it destruct} will divide current goal\ninto several classified subgoals based on the constructors of the inductive type. During this process, no induction \nhypothesis is generated by {\\it destruct}; the {\\it destruct} uses {\\tt match} to handle classified discussion.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it destruct}\\ \\tt t.$ : $\\tt t$ is a term of inductive type. \n\\end{itemize}\n\n\\subsubsection*{Implementation}\n\nThis tactic not only divides current goal based on the constructors of the inductive\ntype, but requires the system to handle several equivalent relations proposed by the constructors as well.\\par\nHere is an example:\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\nn, m : nat \ne : eq nat n m n : nat \nf : eq nat m m (* destruct e => *) f : eq nat n n \n(* Current Goal *) (* Current Goal *)\neq nat m n eq nat n n \n\\end{minted}\n\\end{minipage}\n\\end{center}\nFrom the constructor \\mintinline{coq}|eq_refl : eq T x x| and the type of \\mintinline{coq}|e : eq nat n m|,\nwe have \\mintinline{coq}|m = n|, so any occurrence of \\mintinline{coq}|m| should be replaced by \\mintinline{coq}|n|.\\par\nIf some term depends on the destructed term, we should rewrite it.\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\nn, m : nat \ne : eq nat n m \nf : eq nat m m \n(* Current Goal *) \neq nat m n \n\n(* $\\Downarrow$ destruct m $\\Downarrow$ *) \n\nn : nat n : nat \ne0 : eq nat n O e0 : eq nat n (S m0) \nf0 : eq nat O O f0 : eq nat (S m0) (S m0) \n(* SubGoal 1 *) (* SubGoal 2 *)\neq nat O n eq nat (S m0) n \n\\end{minted}\n\\end{minipage}\n\\end{center}\nThe two cases above are viewed from user's perspective.\nOn the other hand, in the system we also need to construct the term for back substitution.\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\nn, m : nat \ne : eq nat (S n) (S m) \n(* Current Goal*) \neq nat (S m) (S n) \n\n(* $\\Downarrow$ destruct (S n) $\\Downarrow$ *) \n\n(* The Proof Object *)\n(fun (n:nat)(m:nat)(e:eq nat (S n) (S m)) => \n match (S n) as n0 in nat return (eq nat n0 (S m) -> eq nat (S m) n0) with \n | O => fun e0 : eq nat O (S m) => (* SubGoal 1*) \n | S n0 => fun e0 : eq nat (S n0) (S m) => (* SubGoal 2*) \n end e) \n\\end{minted}\n\\end{minipage}\n\\end{center}\nAfter building the {\\it return type}, we can build the branches by applying {\\it type of branch}.\n\n\\subsection{\\it exact}\n\nThis tactic checks if the input term is actually of the goal's type. If so, the proof of the current goal is finished and the \nconstructed proof object is exactly the input term.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it exact}\\ \\tt t.$ : $\\tt t$ is a term of the goal's type. \n\\end{itemize}\n\n\\subsubsection*{Implementation}\nIt's easy to implement. Here is an example:\n\\begin{center}\n\\begin{minipage}{0.8\\textwidth}\n\\begin{minted}{coq}\nm : nat \n(* Current Goal *) (* exact (eq_refl nat m) => *) (* No More Goals *)\neq nat m m \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it exists}\n\nThis tactic handles the exists qualifier. In our system, $\\tt\\exists (x:T),P\\ x$ is represented by $\\tt ex\\ T\\ P$.\n\nHere is the definition of $\\tt ex$.\n\\begin{center}\n\\begin{minipage}{0.6\\textwidth}\n\\begin{minted}{coq}\nInductive ex (A:Type) (P:A -> Type) : Type :=\n| ex_intro : forall (x:A), P x -> ex A P. \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nThe proof of $\\tt\\exists (x:T), P\\ x$ needs the user to provide a value $\\tt x$\nand prove that it satisfies the predicate $\\tt P$.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it exists}\\ \\tt t.$ : If the goal is $\\tt ex\\ (x:T),P\\ t$, and $\\tt t$ is a term of type $\\tt T$. \n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThe tactic automatically generates the subgoal $\\tt P\\ x$.\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\n(* Current Goal *) \nex nat (fun (n:nat) => eq nat (S (S O)) (plus n n))\n\n(* $\\Downarrow$ exists S O $\\Downarrow$ *)\n\n(* SubGoal *)\neq nat (S (S O)) (plus (S O) (S O))\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nThe proof object built is\n\\begin{center}\n\\begin{minipage}{0.8\\textwidth}\n\\begin{minted}{coq}\nex_intro nat (fun (n:nat) => \n eq nat (S (S O)) (plus n n)) (S O) (* SubGoal *)\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it induction}\n\nThis tactic applies to any goal. \nThe argument term must be of inductive type and the tactic {\\it induction} generates subgoals, \none for each possible form of term, i.e., one for each constructor of the inductive type.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it induction}\\ \\tt t.$ : $\\tt t$ is a term of some inductive type.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\n\nThe tactic generates the predicate by the goal and automatically generates \nthe subgoals for every constructor of the inductive type.\nThe equivalence relations introduced by the constructors are also handled.\n\nIf the term is a hypothesis, {\\it induction} will find the dependencies, \nand try to move them and replace the subterms to solve the dependencies.\nAnd then erase the original hypothesis.\n\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\n \nn : nat \n(* Current Goal *) \neq nat (plus n O) n \n\n(* $\\Downarrow$ induction n $\\Downarrow$ *)\n\n(* SubGoal 1 *) \neq nat (plus O O) O\n\nn0 : nat \ne : eq nat (plus n0 O) n0 \n(* SubGoal 2 *) \neq nat (plus (S n0) O) (S n0)\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nAfter {\\it induction}, all we need to prove is the property holds for \\mintinline{coq}|O| and \nfor any \\mintinline{coq}|n|, if \\mintinline{coq}|P n| holds,\nthen the two subgoals are generated and \\mintinline{coq}|n| is replaced by the constructors.\n\nThe case above is viewed from user's perspective.\nOn the other hand, in the system we also need to construct the term for back substitution.\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\nn : nat \n(* Current Goal*) \neq nat (plus n O) n \n\n(* $\\Downarrow$ induction n $\\Downarrow$ *) \n\n(* The Proof Object *)\n(fun (n:nat) => \n nat_rect \n (fun (n0:nat) => eq nat (plus n0 O) n0)\n (* SubGoal 1 *)\n (fun (n0:nat) (e:eq nat (plus n0 O) n0) => (* SubGoal 2 *))\n n) \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it intro}\n\nIf the current goal is a dependent product $\\tt \\forall x:T,\\ U$, \\textit{intro} puts \n$\\tt x:T$ in the context, and the new subgoal is $\\tt U$.\\par\nIf the current goal is a non-dependent product $\\tt \\forall \\_:T,\\ U$, \\textit{intro} renames\n$\\tt \\_$ into $\\tt s$ not in $\\tt T,U$ and context. Then put $\\tt s:T$ in the context, \nand the new subgoal becomes $\\tt U$.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it intro}\\ .$ : introduce the outermost argument of current goal and rename them automatically.\n\\item ${\\it intro}\\ a_1\\ a_2\\ \\cdots\\ a_n.$ : introduce $n$ outermost arguments of current goal and\ntry to rename them into $a_1,a_2,\\cdots,a_n$.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nSince the context automatically preserves the index, we\ncan safely move the arguments out into the context without worrying about the index.\n\nIf the argument is dependent in either the consequent or some hypotheses of the goal,\nthe argument is replaced by the appropriate constructor form in each of the resulting subgoals\nand induction hypotheses are added to the local context.\n\nHere is an example:\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\n P : Type\n(* Current Goal *) (* intro P => *) (* SubGoal *) \nforall (x:Type) (f:False), x forall (f:False), P\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it intros}\n\nIf the current goal is a dependent (or non-dependent) product with $m$ arguments, namely,\n$\\tt \\forall x_1:T_1,\\ \\forall x_2:T_2,\\ \\cdots\\ \\forall x_m:T_m,\\ U$ where $\\tt U$ is not a product type.\nThen \\textit{intros} will put these $m$ arguments into the context and rename automatically if necessary.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it intros}.$ : introduce all the argument of current goal and rename them automatically.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nKeep calling {\\it intro} until all the goal has no arguments.\\par\nHere is an example:\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\n x : Type\n f : False \n(* Current Goal *) (* intros => *) (* SubGoal *) \nforall (x:Type) (f:False), x x\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it left right}\nThese tactics apply to a goal that is of the disjunction type. Then we can proof one of the two sides to solve the goal.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n \\item ${\\it left}.$\n \\item ${\\it right}.$\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThe disjunction type in our system is defined as follows:\n\\begin{center}\n\\begin{minipage}{0.6\\textwidth}\n\\begin{minted}{coq}\nInductive or (A:Type) (B:Type) : Type :=\n| or_introl : A -> or A B\n| or_intror : B -> or A B. \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nThe tactic ${\\it left}$ builds the proof object with the \\mintinline{coq}|or_introl| constructor.\nThe tactic ${\\it right}$ builds the proof object with the \\mintinline{coq}|or_intror| constructor.\nHere is an example:\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\n(* Current Goal *)\nor (eq nat O O) (eq nat (S O) (S O))\n\n(* $\\Downarrow$ left $\\Downarrow$ *)\n\n(* The Proof Object *)\nor_introl (eq nat O O) (eq nat (S O) (S O)) (* SubGoal 1 *)\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it split}\nThis tactic applies to a goal that is of the conjunction type. Then we need to proof the both sides to solve the goal.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it split}$.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThe conjunction type in our system is defined as follows:\n\\begin{center}\n\\begin{minipage}{0.6\\textwidth}\n\\begin{minted}{coq}\nInductive and (A:Type) (B:Type) : Type :=\n| conj : A -> B -> and A B. \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nThe proof object is built with the \\mintinline{coq}|conj| constructor.\nHere is an example:\n\\begin{center}\n\\begin{minipage}{\\textwidth}\n\\begin{minted}{coq}\n(* Current Goal *)\nand (eq nat O O) (eq nat (S O) (S O))\n\n(* $\\Downarrow$ split $\\Downarrow$ *)\n\n(* The Proof Object *)\nconj (eq nat O O) (eq nat (S O) (S O)) (* SubGoal 1 *) (* SubGoal 2 *)\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\n\\subsection{\\it reflexivity}\nThis tactic applies to a goal that has the form $\\tt eq\\ T\\ t\\ u$, it check that $\\tt t$ and $\\tt u$ \nare convertible and then solves the goal.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it reflexivity}.$\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThis tactic simply checks if the terms are convertible and build the proof object with \\mintinline{coq}|eq_refl| constructor.\n\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\nn : nat\n(* Current Goal *) \neq nat n (plus O n) \n\n(* reflexivity *)\n\n(* The Proof Object *)\nfun (n:nat) => eq_refl nat n\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it rewrite}\nThis tactic applies to any goal.\nThe type of term must have the form $\\tt eq\\ T\\ t\\ u$.\nThen the subterms $\\tt t$ in the goal will be rewritten by $\\tt u$.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it rewrite\\ \\tt H}.$ : $\\tt H$ should have the form $\\tt eq\\ T\\ t\\ u$, \nall the subterms $\\tt t$ in the goal will be replaced by $\\tt u$.\n\\item ${\\it rewrite\\ \\texttt{<-}\\ \\tt H}.$ : $\\tt H$ should have the form $\\tt eq\\ T\\ t\\ u$, \n all the subterms $\\tt u$ in the goal will be replaced by $\\tt t$.\n\\item ${\\it rewrite\\ [\\texttt{<-}/\\texttt{->}]\\ \\tt H\\ \\it in\\ \\tt H1}.$ : \n $\\tt H$ should have the form $\\tt eq\\ T\\ t\\ u$, for the {\\tt ->} case, all the \n subterms $\\tt t$ in $\\tt H1$ will be replaced by $\\tt u$.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThis tactic builds the proof object with \\mintinline{coq}|eq_rect| or \\mintinline{coq}|eq_rect_r|.\n\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\n\\begin{minted}{coq}\na : nat a : nat \nb : nat b : nat \ne : eq nat a b e : eq nat a b\n(* Current Goal *) (* rewrite e => *) (* SubGoal *)\neq nat b a eq nat b b\n\n(* Proof Object *)\nfun (a:nat) (b:nat) (e:eq nat a b) =>\n (fun (e0:eq nat b b) =>\n eq_rect_r nat b (fun (n:nat) (e1:eq nat b n) =>\n eq nat b n) e0 a e) (* SubGoal *)\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\\subsection{\\it simpl}\nThis tactic applies to any goal, it tries to reduce a term to something still readable instead of fully normalizing it. \nIt performs a sort of strong normalization with a key difference:\n\\begin{center}\n\\begin{minipage}{0.7\\textwidth}\nit unfold a constant if and only if it leads to a $\\iota$-reduction, i.e., reducing a {\\tt match} or unfolding a {\\tt Fix}.\n\\end{minipage}\n\\end{center}\n\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it simpl}.$\n\\item ${\\it simpl\\ in}\\ \\tt H.$ : $\\tt H$ is a hypothesis.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\n\nThe tactic first applies $\\beta\\iota$-reduction, then it expands constants and tries to reduce \nfurther using $\\beta\\iota$-reduction, but when no $\\iota$ rule is applied after unfolding,\n$\\iota$-reductions are not applied. For instance, trying to use $\\it simpl$ on \n\\mintinline{coq}|eq nat (plus n O) n| changes nothing.\n\\begin{center}\n\\begin{minipage}{0.8\\textwidth}\n\\begin{minted}{coq}\nn : nat e : nat\n(* Current Goal *) (* simpl => *) (* Current Goal *)\neq nat (plus n O) (plus O n) eq nat (plus n O) n\n\\end{minted}\n\\end{minipage}\n\\end{center}\n\nThe tactic can also be applied to the hypotheses, and the same reduction rules applies.\n\\begin{center}\n\\begin{minted}{coq}\nn : nat n : nat \ne : eq nat (plus n O) (plus O n) e : eq nat (plus n O) n\n(* Current Goal *) (* simpl in e => *) (* Current Goal *) \neq nat (plus n O) n eq nat (plus n O) n \n\\end{minted} \n\\end{center}\n\nThe tactic would never change the current built proof object and won't modify the proof objects built after it.\nIt is transparent in the proof tree.\n\n\\subsection{\\it exact}\n\nThis tactic checks if the input term is actually of the goal's type. If so, the proof of the current goal is finished and the \nconstructed proof object is exactly the input term.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it exact}\\ \\tt t.$ : $\\tt t$ is a term of the goal's type. \n\\end{itemize}\n\n\\subsubsection*{Implementation}\nIt's easy to implement. Here is an example:\n\\begin{center}\n\\begin{minipage}{0.8\\textwidth}\n\\begin{minted}{coq}\nm : nat \n(* Current Goal *) (* exact eq_refl nat m => *) (* No more goals *)\neq nat m m \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\n\\subsection{\\it symmetry}\nThis tactic applies to a goal that has the form $\\tt eq\\ T\\ t\\ u$ \nand changes it into $\\tt eq\\ T\\ u\\ t$.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it symmetry}$.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nThis tactic builds the proof object use the pre-defined theorem \\mintinline{coq}|eq_sym|, here is an example for the proof object building.\n\\begin{center}\n\\begin{minipage}{0.6\\textwidth}\n\\begin{minted}{coq}\nm : nat\nn : nat\ne : eq nat m n \n(* Current Goal*) \neq nat n m \n\n(* $\\Downarrow$ symmetry $\\Downarrow$ *) \n\n(* The proof object *)\nfun (m:nat) (n:nat) (e:eq nat m n) => \n eq_sym nat m n (* SubGoal *) \n\\end{minted}\n\\end{minipage}\n\\end{center}\n\n\n\\subsection{\\it unfold}\nThis tactic applies to any goal, the argument should denote a defined constant (i.e., not a hypothesis or a axiom).\nThe tactic applies the $\\delta$ rule to each occurrence of the constant in the current goal and then replaces it with its $\\beta\\iota$-normal form.\n\n\\subsubsection*{Usage}\n\\begin{itemize}\n\\item ${\\it unfold}\\ \\tt i.$ : $\\tt i$ is an identifier. \n\\item ${\\it unfold}\\ \\tt i\\ \\it in\\ \\tt H.$ : $\\tt i$ is an identifier and $\\tt H$ is a hypothesis.\n\\end{itemize}\n\n\\subsubsection*{Implementation}\nIt finds the occurrences of the constant and applies the $\\delta$ rule, the nameless index is automatically preserved.\n\n\n\\begin{center}\n\\begin{minipage}{0.8\\textwidth}\n\\begin{minted}{coq}\nDefinition addTwo (n : nat) : nat := S (S n).\n\nn : nat\nm : nat\ne : eq nat m n\n(* Current Goal *)\neq nat (addTwo n) (S (S m))\n\n(* $\\Downarrow$ unfold addTwo $\\Downarrow$ *)\n\nn : nat\nm : nat\ne : eq nat m n\n(* SubGoal *)\neq nat (S (S n)) (S (S m))\n\n(* $\\Downarrow$ symmetry $\\Downarrow$ *) \n\n(* The proof object *)\n(fun (n:nat) (m:nat) (e:eq nat m n) => (* SubGoal *)) \n\\end{minted}\n\\end{minipage}\n\\end{center}\n", "meta": {"hexsha": "4f435803c2b74fe966bd53a013ee811fe7372dbe", "size": 21670, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/report/tactic.tex", "max_stars_repo_name": "lsrcz/mini-prover", "max_stars_repo_head_hexsha": "0aa4cdf3b495ddf6707f27dcbee810d519b43177", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 11, "max_stars_repo_stars_event_min_datetime": "2018-05-31T05:55:09.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-30T20:17:52.000Z", "max_issues_repo_path": "tex/report/tactic.tex", "max_issues_repo_name": "lsrcz/mini-prover", "max_issues_repo_head_hexsha": "0aa4cdf3b495ddf6707f27dcbee810d519b43177", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/report/tactic.tex", "max_forks_repo_name": "lsrcz/mini-prover", "max_forks_repo_head_hexsha": "0aa4cdf3b495ddf6707f27dcbee810d519b43177", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.672, "max_line_length": 147, "alphanum_fraction": 0.5976926627, "num_tokens": 6057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.743168019989179, "lm_q2_score": 0.82893881677331, "lm_q1q2_score": 0.6160408191535937}} {"text": "\\documentclass{article}\n\n\\usepackage{enumitem}\n\\usepackage[english]{babel}\n\\usepackage[utf8]{inputenc}\n\\usepackage{parskip}\n\\usepackage{graphicx}\n\\usepackage{mathtools}\n\\usepackage{mathrsfs}\n\\usepackage{amsmath,amsthm,hyperref}\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\\DeclareMathOperator{\\Tr}{tr}\n\\usepackage{bbm}\n\\usepackage{dsfont}\n\\DeclareMathOperator*{\\argmax}{arg\\,max}\n\\DeclareMathOperator*{\\argmin}{arg\\,min}\n\\DeclareMathOperator{\\Tr}{Tr}\n\n% Margins\n\\usepackage[top=2.5cm, left=3cm, right=3cm, bottom=4.0cm]{geometry}\n% Colour table cells\n\\usepackage[table]{xcolor}\n\n% Get larger line spacing in table\n\\newcommand{\\tablespace}{\\\\[1.25mm]}\n\\newcommand\\Tstrut{\\rule{0pt}{2.6ex}} % = `top' strut\n\\newcommand\\tstrut{\\rule{0pt}{2.0ex}} % = `top' strut\n\\newcommand\\Bstrut{\\rule[-0.9ex]{0pt}{0pt}} % = `bottom' /\n\n% my new commands\n\\newcommand\\partialkj{\\frac{\\partial^2}{\\partial\\theta_k\\partial\\theta_j}}\n\\makeatletter\n\\newcommand*\\bigcdot{\\mathpalette\\bigcdot@{.5}}\n\\newcommand*\\bigcdot@[2]{\\mathbin{\\vcenter{\\hbox{\\scalebox{#2}{$\\m@th#1\\bullet$}}}}}\n\\makeatother\n\\newcommand{\\minus}{\\scalebox{0.5}[1.0]{$-$}}\n\\newcommand{\\zero}{\\scalebox{0.6}[0.75]{$^{(0)}$}}\n\\newcommand{\\supx}[1]{\\scalebox{0.6}[0.75]{$^{(#1)}$}}\n\\newcommand{\\supi}{\\scalebox{0.6}[0.75]{$^{(i)}$}}\n\\newcommand{\\sumim}{\\begin{align*}\n \\sum_{i=1}^m\n\\end{align*}\n}\n\\newcommand{\\sumzi}{\\begin{align*}\n \\sum_{z\\supi}\n\\end{align*}\n}\n\n\\newcommand{\\bigDash}{\\scalebox{3.0}[1.0]{$-$}}\n\n\n\n%%%%%%%%%%%%%%%%%\n% Title %\n%%%%%%%%%%%%%%%%%\n\\title{Problem Set #4: Unsupervised Learning and Reinforcement Learning}\n\\author{Eitan Joseph \\and Caroline Wang}\n\\date{\\today}\n\n\\begin{document}\n\\maketitle\n\n%%%%%%%%%%%%%%%%%\n% Problem 1 %\n%%%%%%%%%%%%%%%%%\n\\section*{Problem 1}\n\\textbf{EM for supervised learning}\\\\\nIn class we applied EM to the unsupervised learning setting. In particular, we represented $p(x)$ by marginalizing over a latent random variable \\begin{align*}\n p(x) = \\sum_{z} p(x,z) = \\sum_{z} p(x|z)p(z)\n\\end{align*}\nHowever, EM can also be applied to the supervised learning setting, and in this problem we discuss a “mixture of linear regressors” model; this is an instance of what is often call the Hierarchical Mixture of Experts model. We want to represent $p(y|x)$, $x\\in \\mathbb{R}^n$ and $y\\in\\mathbb{R}$, and we do so by again introducing a discrete latent random variable\n\\begin{align*}\n p(y|x) = \\sum_z p(y,z|x) = \\sum_{z} p(y|x,z) p(z|x) \n\\end{align*}\nFor simplicity we’ll assume that z is binary valued, that p(y|x, z) is a Gaussian density, and that $p(y|x)$ is given by a logistic regression model. More formally\n\\begin{align*}\n p(z|x;\\phi ) =& g(\\phi^Tx)^z(1-g(\\phi^Tx))^{1-z}\\\\\n p(y|x,z = i; \\theta_i) = &\\frac{1}{\\sqrt{2\\pi}\\sigma}\\exp\\left(\\frac{-(y-\\theta_i^Tx)^2}{2\\sigma^2}\\right)\\quad i = 0,1\n\\end{align*}where $\\sigma$ is a known parameter and $\\phi, \\theta_0, \\theta_1 \\in \\mathbb{R}$ are parameters of the model (here we use the subscript on $\\theta $ to denote two different parameter vectors, not to index a particular entry in these vectors).\\\\\nIntuitively, the process behind model can be thought of as follows. Given a data point $x$, we first determine whether the data point belongs to one of two hidden classes $z=0$ or $z = 1$, using a logistic regression model. We then determine $y$ as a linear function of $x$ (different linear functions for different values of z) plus Gaussian noise, as in the standard\nlinear regression model. For example, the following data set could be well-represented by the model, but not by standard linear regression.\n\\begin{enumerate}[label=(\\alph*)]\n \\item Suppose $x$, $y$, and $z$ are all observed, so that we obtain a training set\n${(x\\supx{1}, y\\supx{1}, z\\supx{1}), . . . , (x\\supx{m}, y\\supx{m}, z\\supx{m})}$. Write the log-likelihood of the parameters,\nand derive the maximum likelihood estimates for $\\phi$, $\\theta_0$, and $\\theta_1$. Note that because $p(z|x)$ is a logistic regression model, there will not exist a closed form estimate of $\\phi$. In this case, derive the gradient and the Hessian of the likelihood with respect to $\\phi$; in practice, these quantities can be used to numerically compute the ML estimate.\\\\\\\\\n \\textit{answer: }\\\\\\\\\n The log-likelihood can be written as\n \\begin{align*}\n \\ell(\\phi,\\theta_0, \\theta_1) =& \\sumim \\log p(y\\supi|x\\supi;\\phi,\\theta_0, \\theta_1)\\\\\n = & \\sumim \\log p(y\\supi |x\\supi, z\\supi;\\theta_0, \\theta_1 )p(z\\supi | x\\supi;\\phi)\\\\\n =& \\sumim \\mathds{1}\\{z\\supi=0\\}\\log \\left((1-g(\\phi^Tx\\supi))\\left(\\frac{1}{\\sqrt{2\\pi}\\sigma}\\exp\\left(\\frac{-(y\\supi-\\theta_k^Tx\\supi)^2}{2\\sigma^2}\\right)\\right)\\right)\\\\ &+ \\sumim \\mathds{1}\\{z\\supi=1\\}\\log \\left((g(\\phi^Tx\\supi))\\left(\\frac{1}{\\sqrt{2\\pi}\\sigma}\\exp\\left(\\frac{-(y\\supi-\\theta_1^Tx\\supi)^2}{2\\sigma^2}\\right)\\right)\\right)\n \\end{align*}\n The maximum likelihood estimation for $\\theta_k$ can be derived by taking the gradient with respect to $\\theta_k$ and setting the result to zero.\\begin{align*}\n \\nabla_{\\theta_0}\\ell(\\phi,\\theta_0, \\theta_1) \\overset{\\mathrm{set}}{=}0\n \\end{align*}\n Then by extracting the non-relevant constants to $\\theta_k$: \\begin{align*}\n &{}\\nabla_{\\theta_k}\\sumim-(y-\\theta_k^Tx\\supi)^2 = 0\\\\\n &{}\\sumim-2(y\\supi-\\theta_k^Tx\\supi)\\cdot x\\supi=0\\\\\n \\implies&{}\\sumim x\\supi y\\supi= \\sumim \\theta_k^Tx\\supi^Tx\\supi \\\\\n \\implies&{}X^T\\Vec{y} = \\theta_k^T X^TX\\\\\n \\implies&{}\\theta_k = (X^TX)^{-1}X^T\\Vec{y}\n \\end{align*}Therefore the specific estimations for $\\theta_0$ and $\\theta_1$ are: \\begin{align*}\n \\theta_0 = (X_0^TX_0)^{-1}X_0^T\\Vec{y_0}\\\\\n \\theta_1 = (X_1^TX_1)^{-1}X_1^T\\Vec{y_1}\n \\end{align*}\n Where $X_0$, $y_0$, $X_1$, $y_1$ are the associated matrices and vectors derived in each of the two separate summations that combine to equal all of $z$.\\\\\n \n The next step is to find the gradient vector and the Hessian matrix respect with respect to $\\phi$.\\\\\n We can once again remove the terms not relating to $\\phi$:\\begin{align*}\n \\nabla_{\\phi } \\ell(\\phi,\\theta_0, \\theta_1)=& \\nabla_{\\phi}\\sumim \\mathds{1}\\{z\\supi=0\\}\\log (1-g(\\phi^Tx\\supi))+ \\sumim \\mathds{1}\\{z\\supi=1\\}\\log((g(\\phi^Tx\\supi))\\\\\n =& \\nabla_{\\phi} \\sumim (1-z\\supi) \\log (1-g(\\phi^Tx\\supi))+ z\\supi \\log((g(\\phi^Tx\\supi))\n \\end{align*}\n From previous classes, the derivative of a sigmoid is known to be $\\frac{\\partial}{\\partial z}g(z) = g(z)(1-g(z))$, therefore \\begin{align*}\n \\nabla_{\\phi }\\ell(\\phi,\\theta_0, \\theta_1) = &\\sumim - \\frac{1-z\\supi}{(1-g(\\phi^Tx\\supi))}\\frac{\\partial}{\\partial \\phi}g(\\phi^Tx\\supi)\\cdot x\\supi + \\frac{z\\supi}{g(\\phi^Tx\\supi)}\\frac{\\partial}{\\partial \\phi}g(\\phi^Tx\\supi)\\cdot x\\supi\\\\\n =& \\sumim - \\frac{1-z\\supi}{(1-g(\\phi^Tx\\supi))}g(\\phi^Tx\\supi)(1-g(\\phi^Tx\\supi))\\cdot x\\supi + \\frac{z\\supi}{g(\\phi^Tx\\supi)}g(\\phi^Tx\\supi)(1-g(\\phi^Tx\\supi))\\cdot x\\supi\\\\\n =& \\sumim -(1-z\\supi)g(\\phi^Tx\\supi) \\cdot x\\supi + z\\supi(1-g(\\phi^Tx\\supi))\\cdot x\\supi\\\\\n =& \\sumim x\\supi (z\\supi -g(\\phi^Tx\\supi))\n \\end{align*}\n Finally, this tells us that the gradient is\\begin{align*}\n \\nabla_{\\phi }\\ell(\\phi,\\theta_0, \\theta_1) = X^T(\\Vec{z}-\\Vec{g})\\quad \\text{where} \\quad \\Vec{g}_i = g(\\phi^Tx\\supi)\n \\end{align*}\n Based on the linear algebra review sheet, the Hessian Matrix can be derived by taking the derivative of the gradient, which is equivalent to looking at each $i$th entry of the gradient vector, taking the gradient of that entry, and setting that to be the $i$th column of the Hessian. The $i$th entry of the gradient vector is \\begin{align*}\n x\\supi^T \\cdot (\\Vec{z} -\\Vec{g})\n \\end{align*}and after taking the derivative with respect to that $i$th entry we get \\begin{align*}\n \\nabla_{\\phi} x\\supi^T (\\Vec{z} -\\Vec{g}) = x\\supi^T (\\Vec{g}\\cdot (1-\\Vec{g}))\\cdot x\\supi \n \\end{align*}Which gives the matrix \\begin{align*}\n H = X^TDX\\quad \\text{where} \\quad D_{ii} = g(\\phi^Tx\\supi)(1-g(\\phi^Tx\\supi))\n \\end{align*}\n \\item Now suppose $z$ is a latent (unobserved) random variable. Write the log-likelihood of the parameters, and derive an EM algorithm to maximize the log-likelihood. Clearly specify the E-step and M-step (again, the M-step will require a numerical solution, so find the appropriate gradients and Hessians). \\\\\\\\\n \\textit{answer: }\\\\\\\\\n The log-likelihood can now be written as\n \\begin{align*}\n \\ell(\\phi,\\theta_0, \\theta_1) \n &{}= \\sumim\\log p(y\\supi|x\\supi;\\phi,\\theta_0, \\theta_1)\\\\\n &{}= \\sumim\\log p(y\\supi|x\\supi,z\\supi;\\theta_0, \\theta_1) p(z\\supi|x\\supi;\\phi)\\\\\n &{}= \\sumim\\log \\left(\\left(1-g(\\phi^Tx\\supi)\\right)^{1-z\\supi}\\left(\\frac{1}{\\sqrt{2\\pi}\\sigma}\\exp\\left(\\frac{-(y\\supi-\\theta_k^Tx\\supi)^2}{2\\sigma^2}\\right)\\right)\\right) \\\\&+ \\log \\left(\\left(g(\\phi^Tx\\supi)^{z\\supi}\\right)\\left(\\frac{1}{\\sqrt{2\\pi}\\sigma}\\exp\\left(\\frac{-(y\\supi-\\theta_1^Tx\\supi)^2}{2\\sigma^2}\\right)\\right)\\right)\n \\end{align*}\n Since each $z\\supi$ is unobserved they cannot be separated explicitly into two cases - this is because we have no way of knowing how to separate the summation into two groups for each cluster of $z\\supi$ values when their values are hidden.\\\\\n \n In order to derive the new EM algorithm, instead of explicitly maximizing $\\ell$ we will use the same approach as explained in class which will be to repeatedly construct a lower-bound on $\\ell$ (E-step), and then optimize that lower-bound (M-step). \\\\\n \n In order to accomplish this, we first define $Q_i$ to be the distribution over $z_i$ for all $i$ (implying $\\sum_z Q_i(z) = 1$ and $Q_i(z) \\geq 0$). Next we write the following:\n \\begin{align}\n \\sumim\\log p(y\\supi|x\\supi;\\phi,\\theta_0, \\theta_1)&{}=\n \\sumim \\log\\sum_{z\\supi} Q_i(z\\supi)\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\\\\\n &{}\\geq\\sumim\\sum_{z\\supi} Q_i(z\\supi)\\log\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\n \\end{align}\n Where step (2) utilizes Jensen's Inequality.\\\\\\\\\n We note here that the term\n \\begin{align*}\n \\sum_{z\\supi} Q_i(z\\supi)\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\n \\end{align*}\n is just the expectation over the quantity \n \\begin{align*}\n \\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\n \\end{align*}\n according to the distribution $Q_i$. By Jensen's Inequality we know that\n \\begin{align*}\n f\\left(\\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\\right]\\right) \\geq \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[f\\left(\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{Q_i(z\\supi)}\\right)\\right]\n \\end{align*}\n Since (2) holds true for any set of distributions $Q_i$ we can choose a specific distribution for $Q_i$ which makes the inequality hold with equality at $\\phi$, $\\theta_0$, $\\theta_1$. Since Jensen's Inequality holds with equality over a constant valued random variable it suffices to find $Q_i(z\\supi) = c \\times p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi) $ for some $c \\in \\mathbb{R}, c \\neq 0$. We can solve for $Q_i(z\\supi)$ using many iterations of Bayes' Theorem as follows:\n \\begin{align*}\n Q_i(z\\supi) &{}=\n \\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{\\sum_{z\\supi}p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}\\\\\n &{}=\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi|x\\supi;\\phi)}{p(y\\supi|x\\supi;\\theta_0, \\theta_1)}\\\\\n &{}=\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_0, \\theta_1)p(z\\supi, x\\supi;\\phi)}{p(y\\supi|x\\supi;\\theta_0, \\theta_1)p(x\\supi;\\phi)}\\\\\n &{}=\\frac{p(y\\supi, x\\supi, z\\supi;\\phi, \\theta_0, \\theta_1)}{p(y\\supi, x\\supi;\\phi, \\theta_0, \\theta_1)}\\\\\n &{}= p(z\\supi|x\\supi, y\\supi;\\phi, \\theta_0, \\theta_1)\n \\end{align*}\n With this solved we can now write the following EM algorithm\n \\begin{align*}\n \\text{(E-step) For each i, j set}\\\\\n &{}w_j^{(i)} := p(z\\supi = j | x\\supi, y\\supi;\\phi,\\theta_j)\\\\\n \\text{(M-step) Update the parameters of}\\\\\n &{}\\sumim\\sum_j w_j^{(i)}\\log\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_j)p(z\\supi|x\\supi;\\phi)}{w_j^{(i)}}\n \\end{align*}\n In order to maximize the equation in the M-step with respect to the parameters $\\phi$, $\\theta_0$, $\\theta_1$ we need to take the gradient with respect to each parameter and set the resulting expression to zero. We can first generalize the $\\theta$s as one gradient and solve\n \\begin{align*}\n \\nabla_{\\theta_k}\\sumim\\sum_j w_j^{(i)}\\log\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_j)p(z\\supi|x\\supi;\\phi)}{w_j^{(i)}} \\overset{set}= 0\n \\end{align*}\n After dividing out all the non variable multipliers and taking the gradient we are left with the equation\n \\begin{align*}\n &{}\\sumim-2(y\\supi-\\theta_k^Tx\\supi)x\\supi^T w_k^{(i)} = 0\\\\\n \\implies&{}\\sumim x\\supi^T w_k^{(i)}y\\supi - x\\supi^T w_k^{(i)}\\theta_k^Tx\\supi = 0\\\\\n \\implies&{}\\sumim x\\supi^T w_k^{(i)}\\theta_k^Tx\\supi = \\sumim x\\supi^T w_k^{(i)}y\\supi\\\\\n \\implies&{}\\sumim x\\supi^T w_k^{(i)}\\theta_k^Tx\\supi = \\sumim x\\supi^T w_k^{(i)}y\\supi\\\\\n \\implies&{}\\theta_k^T\\sumim x\\supi^T w_k^{(i)}x\\supi = \\sumim x\\supi^T w_k^{(i)}y\\supi\\\\\n \\implies&{}\\theta_k^T = \\sumim (x\\supi^T w_k^{(i)}x\\supi)^{-1}x\\supi^T w_k^{(i)}y\\supi\\\\\n \\implies&{}\\theta_k = (X^T W X)^{-1}X^TWy \\quad \\text{where} \\quad W_{i,i} = w_k^{(i)}\\\\\n \\end{align*}\n Finally we must derive the gradient vector and Hessian matrix \\begin{align*}\n &\\nabla_{\\phi } \\sumim\\sum_j w_j^{(i)}\\log\\frac{p(y\\supi|x\\supi, z\\supi;\\theta_j)p(z\\supi|x\\supi;\\phi)}{w_j^{(i)}}\\\\\n =&\\nabla_{\\phi }\\sumim w_0^{(i)}\\log (1-g(\\phi^Tx\\supi))+ w_1^{(i)} \\log((g(\\phi^Tx\\supi))\n \\end{align*}\n Since $w_0^{(i)}=1-w_1^{(i)}$ we have\\begin{align*}\n &\\nabla_{\\phi }\\sumim (1-w_1^{(i)})\\log (1-g(\\phi^Tx\\supi))+ w_1^{(i)} \\log((g(\\phi^Tx\\supi))\\\\\n =&\\sum_ix\\supi (w_1^{(i)} -g(\\phi^Tx\\supi))\n \\end{align*}The matrix representation turns out to be \\begin{align*}\n \\nabla_{\\phi } = X^T (\\Vec{w}-\\Vec{g})\\quad \\text{where} \\quad\\Vec{g}_i = g(\\phi^Tx\\supi)\n \\end{align*}\n Finding the Hessian for $\\phi$ in part (b), yields the same result as part (a) \\begin{align*}\n H = X^TDX\\quad \\text{where}\\quad D_{i,i} = g(\\phi^Tx\\supi)(1-g(\\phi^Tx\\supi))\n \\end{align*}\n\\end{enumerate}\n\\section*{Problem 2}\n\\textbf{Factor Analysis and PCA}\\\\\nIn this problem we look at the relationship between two unsupervised learning algorithms we discussed in class: Factor Analysis and Principle Component Analysis.\\\\\nConsider the following joint distribution over (x, z) where $z\\in \\mathbb{R}^k$ is a latent random variable\\begin{align*}\n z\\quad &\\sim \\quad \\mathscr{N}(0,I)\\\\\n x|z\\quad &\\sim \\quad \\mathscr{N}(Uz,\\sigma^2I)\n\\end{align*}\nwhere $U\\in \\mathbb{R}^k$ is a model parameters and $\\sigma^2$ is assumed to be a known constant. This model is often called Probabilistic PCA. Note that this is nearly identical to the factor analysis model except we assume that the variance of $x|z$ is a known scaled identity matrix rather than the diagonal parameter matrix $\\Phi$\b, and we do not add an additional $\\mu$ term to\nthe mean (though this last difference is just for simplicity of presentation). However, as we will see, it turns out that as $\\sigma^2 \\xrightarrow[]{}0$, this model is equivalent to PCA.\\\\\nFor simplicity, you can assume for the remainder of the problem that $k = 1$, i.e., that $U$ is a column vector in $\\mathbb{R}^n$.\n\\begin{enumerate}[label=(\\alph*)]\n\\item Use the rules for manipulating Gaussian distributions to determine the joint distribution over $(x, z)$ and the conditional distribution of $z|x$. [Hint: for later parts of\nthis problem, it will help significantly if you simplify your solution for the conditional distribution using the identity we first mentioned in problem set #1: $(\\lamba I+BA)^{-1}B =\nB(\\lambada I + AB)^{-1}.]$\\\\\\\\\n \\textit{answer:}\\\\\\\\\n According to the definition of $x$, the expectation of $x$ is \\begin{align*}\n E[x] = E[Uz+\\epsilon] = UE[z] + E[\\epsilon] = 0\n \\end{align*} \n where $\\epsilon \\sim \\mathscr{N}(0,\\sigma^2I)$ indicates the added covariance noise.\n We can now obtain the four different components of the matrix $\\Sigma$ using the substitution $x=Uz+\\epsilon$\\begin{align*}\n \\Sigma_{zz} =& E[zz^T] = I\\\\\n \\Sigma_{xz} =& E[(Uz+\\epsilon)z^T] = (UE[zz^T] + E[\\epsilon z^T]) = U\\\\\n \\Sigma_{zx} =& E[z(Uz+\\epsilon)^T] = U^TE[zz^T] + E[z\\epsilon^T] = U^T\\\\\n \\Sigma_{xx} =& E[(Uz+\\epsilon)(Uz+\\epsilon)^T] = UU^TE[zz^T]+UE[z\\epsilon^T] + U^TE[\\epsilon z^T]+E[\\epsilon\\epsilon^T] = UU^T +\\sigma^2I\n \\end{align*}\n Therefore our join distribution is written\n \\begin{align*}\n \\begin{bmatrix}z \\\\ x\\end{bmatrix}\\quad \\sim \\quad \\mathscr{N}\\left(\\begin{bmatrix}0 \\\\ 0\\end{bmatrix}, \\begin{bmatrix}I & U^T \\\\ U & UU^T+\\sigma^2I\\end{bmatrix}\\right)\n \\end{align*} Then we can apply the formula to define the conditional distribution $z|x \\sim \\mathscr{N}(\\mu_{z|x}, \\Sigma_{z|x})$ such that \\begin{align*}\n \\mu_{z|x} =& \\mu_z+\\Sigma_{zx}\\Sigma_{xx}^{-1}x = U^T(UU^T+\\sigma^2I)^{-1}x = \\frac{U^Tx}{UU^T+\\sigma^2}\\\\\n \\Sigma_{z|x} =& \\Sigma_{zz}-\\Sigma_{zx}\\Sigma_{xx}^{-1}\\Sigma_{xz} = I-U^T(UU^T+\\sigma^2I)^{-1}U = 1-\\frac{U^TU}{UU^T+\\sigma^2}\n \\end{align*}\n In conclusion, \\begin{align*}\n z|x \\quad \\sim \\quad \\mathscr{N}\\left(\\frac{U^Tx}{UU^T+\\sigma^2}, 1-\\frac{U^TU}{UU^T+\\sigma^2}\\right)\n \\end{align*}\n\\item Using these distributions, derive an EM algorithm for the model. Clearly state the E-step and the M-step of the algorithm.\\\\\\\\\n \\textit{answer:}\\\\\\\\\n For the E-step, we need to compute $Q_i(z\\supi)$ where \n \\begin{align*}\n Q_i(z\\supi) = p(z\\supi |x\\supi ;U) \n \\end{align*}\n For the M-step we need to maximize the equation \\begin{align*}\n &\\sumim \\int_{z\\supi}Q_i(z\\supi)\\log\\frac{p(x\\supi,z\\supi: U)}{Q_i(z\\supi)}\\\\\n =&\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\log p(x\\supi |z\\supi; U)+\\log p(z\\supi ) -\\log(Q_i(z\\supi))\\right]\n \\end{align*}Then we take the gradient with respect to $U$ and set it to $0$ in order to maximize the equation above.\n \\begin{align*}\n &\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[\\log p(x\\supi |z\\supi; U)]\\\\\n =&\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\log\\frac{1}{\\sqrt{2\\pi^n}|\\sigma^2I|}\\exp \\left(-\\frac{1}{2\\sigma^2}(x\\supi -Uz\\supi)^T(x\\supi-Uz\\supi)\\right)\\right]\\\\\n =&\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\frac{1}{2}\\log|\\sigma^2I|-\\frac{n}{2}\\log(2\\pi)-\\frac{1}{2\\sigma^2}(x\\supi -Uz\\supi)^T(x\\supi-Uz\\supi)\\right]\n \\end{align*}We can drop the terms not dependent on $U$ before taking the gradient. \\begin{align*}\n &\\nabla_U\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\frac{1}{2\\sigma^2}(x\\supi -Uz\\supi)^T(x\\supi-Uz\\supi)\\right]\\\\\n =&-\\frac{1}{2\\sigma^2}\\sumim \\nabla_U\\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[\\Tr z\\supi^TU^TUz\\supi-2\\Tr z\\supi^TUx\\supi\\right]\\\\\n =&-\\frac{1}{2\\sigma^2}\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[2z\\supi z\\supi^TU^T-2z\\supi^Tx\\supi\\right]\\\\\n =&-\\frac{1}{\\sigma^2}\\sumim \\left[U\\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi z\\supi^T]-x\\supi\\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi^T]\\right]\\overset{\\mathrm{set}}{=}0\\\\\n \\implies &U = \\left(\\sumim x\\supi \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi^T]\\right)\\left(\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi z\\supi^T]\\right)^{-1}\n \\end{align*}From the definition of $Q_i$ being Gaussian we know that \\begin{align*}\n \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi^T] =& \\mu_{z\\supi|x\\supi}^T\\\\\n \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}[z\\supi z\\supi^T] =& \\mu_{z\\supi|x\\supi}\\mu_{z\\supi|x\\supi}^T\n \\end{align*}Therefore, the final update of the M-step is \\begin{align*}\n U=\\left(\\sumim \\:x\\supi\\mu_{z\\supi|x\\supi}^T\\right)\\left(\\sumim \\mu_{z\\supi|x\\supi}\\mu_{z\\supi|x\\supi}^T\\right)^{-1}\n \\end{align*}\n\\item As $\\sigma^2 \\xrightarrow[]{}0$, show that if the EM algorithm convergences to a parameter vector $U^*$ (and such convergence is guaranteed by the argument presented in class), then $U^*$ must be an eigenvector of the sample covariance matrix $\\Sigma = \\frac{1}{m}\\sum x\\supi x\\supi^T$\n— i.e., $U^*$ must satisfy\n\\begin{align*}\n \\lambda U^* = \\Sigma U^*\n\\end{align*}\n \\textit{answer:}\\\\\\\\\n The first thing to notice is that as $\\sigma^2 \\xrightarrow[]{}0$ the E-step only needs to compute the means and not the variances, since $\\Sigma_{z|x}\\xrightarrow[]{}0$ as well. If we let $w \\in \\mathbb{R}^m$ be the vector that contains all the means such that $w_i = \\mu_{z\\supi|x\\supi}$ then we can compute the new E-step to be\n \\begin{align}\n w_i = \\mu_{z\\supi|x\\supi} =&{} U^T(\\sigma^2 I - UU^T)^{-1}(Uz\\supi + \\epsilon)\\\\\n =&{} U^T (UU^T)^{-1}x\\supi\\\\\n =&{} \\frac{U^Tx\\supi}{UU^T}\\\\\n =&{} \\frac{x\\supi^TU}{U^TU}\n \\end{align}\n Where step (6) uses the fact that step (5) yields a Real. When this is written in matrix form we see that\n \\begin{align*}\n w = \\frac{XU}{U^TU}\n \\end{align*}\n Next we need to derive the new M-step. We can use the result from (b) in order to obtain the new update rule\n \\begin{align}\n U &{}= \\left(\\sumim \\:x\\supi\\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[z\\supi^T\\right]\\right)\\left(\\sumim \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[z\\supi z\\supi^T\\right]\\right)^{-1}\\\\\n &{}= \\left(\\sumim \\:x\\supi\\mu_{z\\supi|x\\supi}^T\\right)\\left(\\sumim \\mu_{z\\supi|x\\supi}\\mu_{z\\supi|x\\supi}^T\\right)^{-1}\\\\\n &{}= \\left(\\sumim \\:x\\supi w_i\\right)\\left(\\sumim w_i w_i\\right)^{-1}\n \\end{align}\n Where step (8) is derived from the definition of $Q_i$ being Guassian with mean $\\mu_{z\\supi|x\\supi}$ and variance $\\Sigma_{z\\supi|x\\supi}$ yielding\n \\begin{align*}\n \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[z\\supi^T\\right] &{}= \\mu_{z\\supi|x\\supi}^T\\\\\n \\mathop{\\mathbb{E}}_{z\\supi \\mathtt{\\sim} Q_i}\\left[z\\supi z\\supi^T\\right] &{}= \\mu_{z\\supi|x\\supi}\\mu_{z\\supi|x\\supi}^T + \\Sigma_{z\\supi|x\\supi}\n \\end{align*}\n Now to write expression (9) in matrix form we see that\n \\begin{align*}\n \\left(\\sumim w_i w_i\\right)^{-1} &{}= \\frac{1}{w^Tw}\n \\end{align*}\n however to write the numerator in matrix form will be a little more difficult. We want to represent that each column of $X$ from $i\\in{1,...,m}$ gets multiplied by its the $i$th entry of $w$ and then summed together. We cannot simply write this as $X^Tw$ so we will instead have to construct a new diagonal matrix $W$ where $W_{i,i} = w_i$. Then, by multiplying $XW$ we get\n \\begin{align*}\n XW &{}=\n \\begin{bmatrix}\n | & | & &|\\\\\n x\\supx{1} & x\\supx{2} & \\dots & x\\supx{m}\\\\\n | & | & &|\n \\end{bmatrix}_{n\\times m}\n \\begin{bmatrix}\n w_1 & & &\\\\\n & w_2 & &\\\\\n & & \\ddots &\\\\\n & & & w_m\n \\end{bmatrix}_{m\\times m}\\\\\n &{}= \\begin{bmatrix}\n | & | & &|\\\\\n w_1x\\supx{1} & w_2x\\supx{2} & \\dots & w_mx\\supx{m}\\\\\n | & | & &|\n \\end{bmatrix}_{n\\times m}\n \\end{align*}\n Next we just need to sum up each column of $XW$ which we can accomplish by constructing new vector $\\Vec{1}\\in \\mathbb{R}^m$ such that $\\Vec{1}_i = 1$ for all $i\\in{1,...,m}$ and transforming it according to $XW$ as follows\n \\begin{align*}\n XW\\Vec{1} &{}= \\begin{bmatrix}\n | & | & &|\\\\\n w_1x\\supx{1} & w_2x\\supx{2} & \\dots & w_mx\\supx{m}\\\\\n | & | & &|\n \\end{bmatrix}_{n\\times m}\n \\begin{bmatrix}\n 1\\\\\n \\vdots\\\\\n 1\n \\end{bmatrix}_{m\\times1}\\\\\n &{}= \\left(\\sumim \\:x\\supi w_i\\right)\n \\end{align*}\n Finally we can write the expression for U in matrix form\n \\begin{align*}\n U = \\frac{XW\\Vec{1}}{w^T w}\n \\end{align*}\n We know that $U$ has converged when its value has not changed after the E-step, so we can substitute our value for $w$ in the E-step and set it equal to our original U.\n \\begin{align*}\n &{}U = \\frac{XW\\Vec{1}}{\\frac{U^TX^T XU }{(U^TU)^2}}\\\\\n \\implies&{}(U^TX^T XU)U = (U^TU)^2 XW\\Vec{1}\\\\\n \\implies&{}X^TXU = U^TU XW\\Vec{1}\\\\\n \\implies&{}X^TXU = UU^T XW\\Vec{1} = U^T XW\\Vec{1}U\\\\\n \\implies&{}\\Sigma U = \\lambda U\n \\end{align*}\n where we know that $\\Sigma = X^TX$ and we take $\\lambda = U^T XW\\Vec{1} \\in \\mathbb{R}$ and therefore proving the desired result.\n\\end{enumerate}\n\\section*{Problem 4}\n \\textbf{Convergence of Policy Iteration}\\\\\n In this problem we show that the Policy Iteration algorithm, described in the lecture notes, is guaranteed to find the optimal policy for an MDP. First, define $B^\\pi$ to be the Bellman operator for policy $\\pi$, defined as follows: if $V^\\prime = B(V)$, then\\begin{align*}\n V^\\prime (s) = R(s)+\\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V(s^\\prime)\n \\end{align*}\n\\begin{enumerate}[label=(\\alph*)]\n\\item Prove that if $V_1(s) \\leq V_2(s$) for all $s\\in \\mathcal{S}$, then $B(V_1)(s) \\leq B(V_2)(s)$ for all $s\\in \\mathcal{S}$.\\\\\\\\\n\\textit{answer:}\\\\\\\\\nGiven that $V_1(s) \\leq V_2(s$) for all $s\\in \\mathcal{S}$, we can also write the inequality\n\\begin{align*}\n &R(s) + V_1(s^\\prime) \\leq R(s) + V_2(s^\\prime)\\\\\n \\implies& R(s) + \\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V_1(s^\\prime) \\leq R(s) + \\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V_2(s^\\prime)\n\\end{align*}\nHowever, it turns out that\n\\begin{align*}\n &R(s) + \\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V_1(s^\\prime) = B(V_1)(s)\\\\\n &R(s) + \\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V_2(s^\\prime) = B(V_2)(s)\n\\end{align*}\nSo we have shown that $B(V_1)(s) \\leq B(V_2)(s)$ for all $s \\in \\mathcal{S}$\n\\item Prove that for any $V$, \\begin{align*}\n ||B^\\pi(V)-V^\\pi ||_\\infty\\leq \\gamma ||V-V^\\pi ||_\\infty\n\\end{align*}where $||V||_\\infty = \\max_{s\\in \\mathcal{S}}|V(s)|$. Intuitively, this means that applying the Bellman operator $B^\\pi $ to any value function $V$, brings that value function “closer” to the value function for $\\pi, V^\\pi $. This also means that applying $B^\\pi $ repeatedly (an infinite number of times):\\begin{align*}\n B^\\pi(B^\\pi (B^\\pi\\dots B^\\pi(V)\\dots ))) \n\\end{align*}will result in the value function $V^\\pi$.\\\\\\\\\n\\textit{answer:}\\\\\\\\\nWe can first observe that \n\\begin{align*}\n ||B^\\pi(V)-V^\\pi ||_\\infty &= \\max_{s\\in \\mathcal{S}}\\big |R(s)+\\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V(s^\\prime)-R(S)-\\gamma \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )V^\\pi(s^\\prime)\\big |\\\\\n &= \\gamma \\max_{s\\in \\mathcal{S}}\\big |\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )(V(s^\\prime)-V^\\pi (s^\\prime))\\big |\n\\end{align*}\nThen since $P_{s,a}$ is a probability distribution it must be true that \\begin{align}\n P_{s,\\pi(s)}(s^\\prime )\\geq 0\\\\\n \\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime) = 1\n\\end{align} \nwe can apply the fact that for any $a, x\\in \\mathbb{R}^n$, if $\\sum_i a_i = 1$, and $a_i\\geq 0$, then $\\sum_i a_ix_i\\leq \\max_i x_i$ to (10) and (11) to write\n\\begin{align*}\n \\gamma \\max_{s\\in \\mathcal{S}}\\big |\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime )(V(s^\\prime)-V^\\pi (s^\\prime))\\big |\\leq \\gamma\\max_{s^\\prime\\in \\mathcal{S}} |V(s^\\prime)-V^\\pi(s^\\prime)|\n\\end{align*}\nWhich is, by definition, the same inequality as\n\\begin{align*}\n ||B^\\pi(V)-V^\\pi ||_\\infty\\leq \\gamma ||V-V^\\pi ||_\\infty\n\\end{align*}\n\\item Now suppose that we have some policy $\\pi$, and use Policy Iteration to choose a new policy $\\pi\\prime$ according to\n\\begin{align*}\n \\pi^\\prime(s) = \\argmax_{a\\in A}\\sum_{s^\\prime\\in S}P_{s,a}(s^\\prime)V^\\pi (s^\\prime)\n\\end{align*}\nShow that this policy will never perform worse that the previous one — i.e., show that for all $s \\in \\mathcal{S}$, $V^\\pi(s) \\leq V^{\\pi^\\prime}(s)$.\\\\\\\\\n\\textit{answer:}\\\\\\\\\nWe know that\n\\begin{align*}\n V^\\pi(s) &{}= R(s) + \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime)V^\\pi(s^\\prime)\\\\\n &{}\\leq\\\\\n B^{\\pi^\\prime}(V^\\pi)(s) &{}= R(S)+ \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi^\\prime(s)}(s^\\prime)V^\\pi(s^\\prime)\n\\end{align*}\ndue to the fact that\n\\begin{align*}\n \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi(s)}(s^\\prime)V^\\pi(s^\\prime) \\leq \\gamma\\max_{a\\in A}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)V^\\pi(s^\\prime) = \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi^\\prime(s)}(s^\\prime)V^\\pi(s^\\prime)\n\\end{align*}\nWe can further expand this to show\n\\begin{align*}\n B^{\\pi^\\prime}(V^\\pi)(s) &{}= R(S)+ \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi^\\prime(s)}(s^\\prime)V^\\pi(s^\\prime)\\\\\n &{}= R(S)+ \\gamma\\max_{a\\in \\mathcal{A}}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)V^\\pi(s^\\prime)\\\\\n &{}= R(S)+ \\gamma\\max_{a\\in \\mathcal{A}}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)(R(s^\\prime) + \\gamma\\sum_{s^{\\prime\\prime} \\in \\mathcal{S}}P_{s^\\prime,\\pi(s^\\prime)}(s^{\\prime\\prime})V^\\pi(s^{\\prime\\prime}))\\\\\n &{}\\leq R(S)+ \\gamma\\max_{a\\in\\mathcal{ A}}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)(R(s^\\prime) + \\gamma\\max_{a^\\prime \\in \\mathcal{A}}\\sum_{s^{\\prime\\prime} \\in \\mathcal{S}}P_{s^\\prime,a^\\prime}(s^{\\prime\\prime})V^\\pi(s^{\\prime\\prime}))\\\\\n &{}= B^{\\pi^\\prime}(B^{\\pi^\\prime}(V^\\pi))(s)\n\\end{align*}\nUsing part (b) we also know that\\begin{align*}\n ||B^\\pi(V)-V^\\pi ||_\\infty\\leq \\gamma ||V-V^\\pi ||_\\infty\n\\end{align*}\nimplies that applying $B^\\pi$ repeatedly (an infinite number of times) results in the convergent identity \\begin{align*}\n B^{\\pi^\\prime}(B^{\\pi^\\prime} (B^{\\pi^\\prime}\\dots B^{\\pi^\\prime}(V)\\dots ))) = V^{\\pi^\\prime}\n\\end{align*} which finally proves the desired result that $V^\\pi(s) \\leq V^{\\pi^\\prime}(s)$ for all $s \\in \\mathcal{S}$.\n\\item Use the proceeding exercises to show that policy iteration will eventually converge (i.e., produce a policy $\\pi^\\prime = \\pi$). Furthermore, show that it must converge to the optimal policy $\\pi^*$. For the later part, you may use the property that if some value function satisfies\n\\begin{align*}\n V(s) = R(s) + \\gamma\\max_{a\\in A}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)V(s^\\prime)\n\\end{align*}\nthen $V = V^*$.\\\\\\\\\n\\textit{answer:}\\\\\\\\\nGiven that $|S|+|A| < \\infty$ (i.e. that the number of states and the number of \\\\\\m actions are both finite), we let $\\pi_n$ be the resulting policy after the $n$th iteration of the Policy Iteration, and let\n\\begin{align*}\n \\lim_{n\\xrightarrow[]{}\\infty}\\pi_n = \\pi_{\\infty}\n\\end{align*}\nWe know from part (c) that $V^\\pi(s) \\leq V^{\\pi^\\prime}(s)$ for all s, where $\\pi^\\prime$ is the resulting policy after one iteration of Policy Iteration, therefore we have that\n\\begin{align*}\n V^{\\pi_n}(s) \\leq V^{\\pi_\\infty} \\quad\\forall s \\in \\mathcal{S}, \\quad \\forall n < \\infty\n\\end{align*}\nHowever, since there are only $k = |A|^{|S|}$ possibly policies it must be the case that\n\\begin{align*}\n \\lim_{n\\xrightarrow[]{}\\infty}\\pi_n = \\pi_{k}\n\\end{align*}\nTherefore Policy Iteration converges to $\\pi_{k}$ and further,\n\\begin{align*}\n V^{\\pi_{k}}(s) =&{} R(s) + \\gamma\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,\\pi_{|A|^|S|}}(s^\\prime)V^{\\pi_{k}}\\\\\n =&{} R(s) + \\gamma\\max_{a\\in \\mathcal{A}}\\sum_{s^\\prime \\in \\mathcal{S}}P_{s,a}(s^\\prime)V^{\\pi_{k}}\n\\end{align*}\nwhich proves that $V^{\\pi_{k}} = V^*$\n\\end{enumerate}\n\\end{document}\n", "meta": {"hexsha": "571e916f9461bfc264cc1aef9e6bb82d3d112916", "size": 32013, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Unsupervised Learning and Reinforcement Learning/Problem Set 4 - Unsupervised Learning and Reinforcement Learning.tex", "max_stars_repo_name": "EitanJoseph/Standford-Machine-Learning", "max_stars_repo_head_hexsha": "5b1609a3fc1c7f32494a70ebc3f89d2ed8aed941", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-07-01T02:53:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-01T02:53:32.000Z", "max_issues_repo_path": "Unsupervised Learning and Reinforcement Learning/Problem Set 4 - Unsupervised Learning and Reinforcement Learning.tex", "max_issues_repo_name": "EitanJoseph/Standford-Machine-Learning", "max_issues_repo_head_hexsha": "5b1609a3fc1c7f32494a70ebc3f89d2ed8aed941", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Unsupervised Learning and Reinforcement Learning/Problem Set 4 - Unsupervised Learning and Reinforcement Learning.tex", "max_forks_repo_name": "EitanJoseph/Standford-Machine-Learning", "max_forks_repo_head_hexsha": "5b1609a3fc1c7f32494a70ebc3f89d2ed8aed941", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.9393258427, "max_line_length": 507, "alphanum_fraction": 0.630806235, "num_tokens": 12213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680199891789, "lm_q2_score": 0.8289388146603365, "lm_q1q2_score": 0.6160408175832992}} {"text": "\\subsection{Exponents}\n\nEigenmath requires parentheses around negative exponents.\nFor example,\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\n10^(-3)\n\\end{Verbatim}\n\n\\noindent\ninstead of\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\n10^-3\n\\end{Verbatim}\n\n\\noindent\nThe reason for this is that the binding of the negative sign is not always\nobvious.\nFor example, consider\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\nx^-1/2\n\\end{Verbatim}\n\n\\noindent\nIt is not clear whether the exponent should be $-1$ or $-1/2$.\nSo Eigenmath requires\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\nx^(-1/2)\n\\end{Verbatim}\n\n\\noindent\nwhich is unambiguous.\n\n\\bigskip\n\\noindent\nIn general, parentheses are always required when the exponent\nis an expression.\nFor example, \\verb$x^1/2$ is evaluated as $(x^1)/2$ which\nis probably not the desired result.\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\nx^1/2\n\\end{Verbatim}\n\n\\noindent\n$\\displaystyle \\tfrac{1}{2}x$\n\n\\bigskip\n\\noindent\nUsing \\verb$x^(1/2)$ yields the desired result.\n\n\\begin{Verbatim}[formatcom=\\color{blue}]\nx^(1/2)\n\\end{Verbatim}\n\n\\noindent\n$\\displaystyle x^{1/2}$\n", "meta": {"hexsha": "4341faa51c9b0cdde6e128415e24d4f9ab20bc75", "size": 1087, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/exponents.tex", "max_stars_repo_name": "franko/eigenmath", "max_stars_repo_head_hexsha": "3fca24d09686220e430108dac4864a192d138f13", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-01-27T02:41:22.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-02T19:05:49.000Z", "max_issues_repo_path": "doc/exponents.tex", "max_issues_repo_name": "franko/eigenmath", "max_issues_repo_head_hexsha": "3fca24d09686220e430108dac4864a192d138f13", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2020-01-29T14:09:35.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-13T16:17:03.000Z", "max_forks_repo_path": "doc/exponents.tex", "max_forks_repo_name": "franko/eigenmath", "max_forks_repo_head_hexsha": "3fca24d09686220e430108dac4864a192d138f13", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-01-27T02:38:38.000Z", "max_forks_repo_forks_event_max_datetime": "2020-01-27T02:38:38.000Z", "avg_line_length": 17.8196721311, "max_line_length": 74, "alphanum_fraction": 0.739650414, "num_tokens": 346, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.743168019989179, "lm_q2_score": 0.828938806208442, "lm_q1q2_score": 0.6160408113021216}} {"text": "\\lab{Finite Volume Methods}{Finite Volume Methods}\n\\label{lab:finitevolume}\n\nWhen solving a PDE numerically, how do we deal with discontinuous initial data?\nThe Finite Volume method has particular strength in this area.\nIt is commonly used for hyperbolic PDEs whose solutions can spontaneously develop discontinuities as they evolve in time.\nThese solutions are often called shock waves.\n\n\\section*{Conservation Laws}\nConsider the conservation law\n\\begin{align}\n\\label{eqn:1D_continuous}\nu_t + f(u)_x = 0,\n\\end{align}\nwhere $u$ is a (spatially) one-dimensional conserved quantity, and $f(u)$ is the flux of $u$.\nThe continuous integral formulation of \\eqref{eqn:1D_continuous} states that\n\\begin{align*}\n\t\\frac{d}{dt}\\int_a^b u(x,t) dx + \\int_a^b f(u)_x \\,dx = 0.\n\\end{align*}\n$\\frac{d}{dt}\\int_a^b u(x,t) dx$ may be thought of as the time evolution of the total `mass' of $u$ across the domain $[a,b]$, and is dependent only on the flux through the boundaries, since\n\\begin{align*}\n\\frac{d}{dt}\\int_a^b u(x,t) dx = f(u(a))-f(u(b)).\n\\end{align*}\nThis fact is an important idea utilized by finite volume methods, which generally consider the evolution of $u$ not at a given point, but instead in volume-averaged regions.\nFor example, let $\\{x_i\\}$ be a grid of equally spaced points with spacing $\\Delta x$, and let $C_i$ be the $i$-th `volume' (subinterval) defined by $(x_{i-1/2},x_{i+1/2})$.\nWe are interested in the evolution of the volume average of $u$ over this interval,\n\\begin{align*}\nU_i^n = \\frac{1}{\\Delta x}\\int_{C_i} u(x,t^n)dx,\n\\end{align*}\nwhere $\\{t^n\\}$ is the time discretization.\n\n\\begin{figure}\n\\centering\n\\includegraphics[trim= 20mm 195mm 0mm 30mm, clip]{flux_form.pdf}\n\\caption{A schematic of the fluxes for the finite volume method as indicated by \\eqref{eqn:flux_form}.}\n\\end{figure}\n\nThe evolution of these volume-averaged quantities will depend only on the flux through the cell edges, so that\n\\begin{align}\n\\label{eqn:1D_semi_continuous}\n\\frac{d}{dt}\\int_{C_i} u(x,t) dx = f(u(x_{i-1/2},t))-f(u(x_{i+1/2},t)).\n\\end{align}\nWe can then construct a time-stepping method where $\\sum_i U_i^n\\Delta x$ (the total `mass' of the system) is conserved from one time step $n$ to the next.\n\nLet $F^n_{i-1/2} = \\frac{1}{\\triangle t}\\int_{t^n}^{t^{n+1}} f(u(x_{i-1/2},t)) \\, dt$.\nThen\n\\begin{align*}\n\\int_{t^n}^{t^{n+1}} \\left[ \\frac{d}{dt}\\int_{C_i} u(x,t)\\, dx \\right]\n&{}= \\int_{C_i} u(x,t^{n+1}) - u(x,t^{n})\\, dx ,\\\\\n&{ }= \\triangle t \\left(F^n_{i-1/2} - F^n_{i+1/2}\\right).\n\\end{align*}\n\nThus, by integrating \\eqref{eqn:1D_semi_continuous} in time, we may approximate the evolution of the cell (`volume') averages with the method\n\\begin{align}\n\\label{eqn:flux_form}\nU_i^{n+1} = U_i^n - \\frac{\\Delta t}{\\Delta x} \\left(F^n_{i+1/2}-F^n_{i-1/2}\\right).\n\\end{align}\nwhere $U_i^n = \\frac{1}{\\triangle x} \\int_{C_i} u(x,t^n) \\, dt$.\n% where\n% \\begin{gather}\n% F_{i-1/2}^n = \\frac{1}{\\Delta t} \\int_{t_n}^{t_{n+1}}f(u(x_{i-1/2},t))dt.\n% \\end{gather}\nThis formulation guarantees the conservation properties that are so desirable for conservation laws, if the time-averaged fluxes $F_{i-1/2}^n$ can be discretized in a natural way.\n\nThe key contribution of finite volume methods is the computation of $F_{i-1/2}^n$.\nFor a truly nonlinear $f(u)$ this can be rather complicated and messy, and typically will involve solving what is usually referred to as the Riemann problem for the conservation law.\nThe interested student can look at \\cite{Le2002} for a very thorough introduction and discussion on the subject.\nWe will consider the linear problem in one dimension.\nThe analog to higher dimensions is obtained by considering the eigenvector decomposition of any linear system.\nNonlinear equations complicate things further.\n\n\\section*{The linear advection equation and upwinding}\nThe simplest conservation law describes the advection or transport of a quantity.\nThe PDE is given by\n\\begin{align}\nu_t + au_x = 0, \\label{eqn:fv_advection}\n\\end{align}\nand describes the motion of a concentration of some constituent $u$ by a constant velocity one-dimensional `wind' $a>0$.\n%(a slight generalization of this would be to consider $a=a(x)$, but the general concepts are exactly the same as those viewed here, just a more local consideration is necessary).\nIn higher dimensions this is an important problem in many fields, for example the transport of chemicals in the atmosphere and oceans, proper mixing of various properties in metallurgy, and the passing of information along a network.\n\nNote that whenever $u(x,t)$ is a solution of the advection equation, then $u(x-at,t_0)$ (for any fixed $t_0$) is also a solution.\nThus, if $u(x,0) = u_0(x)$ then the solution for all time can be represented by $u(x,t) = u_0(x-at)$.\nThis is an important property of \\eqref{eqn:fv_advection}, and gives a new meaning to the term advection: this equation merely takes the initial conditions and passively transports them with velocity $a$.\n\nFor this equation the computation of the flux appears straightforward: $F_{i-1/2}^n = a\\overline{U}^n_{i-1/2}$ where the $\\overline{U}^n_{i-1/2}$ refers to the time average of $U_{i-1/2}$ over the interval $t_n$ to $t_{n+1}$.\nLet us determine how to approximate this time average.\n% To understand how this might be done, note from Figure \\ref{something} that for the advection equation with $a>0$ the flux that determines $U_i^{n+1}$ will be dependent on the value of $U_{i-1}^n$ so that it makes sense that we approximate the flux here as $F_{i-1/2}^n = aU_{i-1}^n$ (recall that the $U_i^n$ are cell averages of the continuous quantity $u(x)$).\nNote from Figure \\ref{fig:fv_LW_reconstruction} that when $a>0$ the flux that determines $U_i^{n+1}$ will be dependent on the value of $U_{i-1}^n$. Thus, one possibility is to approximate the flux by $F_{i-1/2}^n = aU_{i-1}^n$.\nUsing this approximation of the flux together with the flux differencing formula \\eqref{eqn:flux_form} yields the first order upwind method, given by\n\\begin{align*}\nU_i^{n+1} = U_i^n - \\frac{a\\Delta t}{\\Delta x}\\left(U_i^n-U_{i-1}^n\\right).\n\\end{align*}\n% (recall that the $U_i^n$ are cell averages of the continuous quantity $u(x)$).\n% Similarly, if $a<0$ then $F_{i-1/2}^n = aU_{i}^n$.\n% This is called the first order upwind method\n% and gives a completely discretized version of the problem as \\eqref{eq:flux_form} with\n% \\begin{align*}\n% F_{i-1/2}^n = a^- U_i^n + a^+U_{i-1}^n,\n% \\end{align*}\n% where $a^- = \\min (a,0)$ and $a^+ = \\max (a,0)$.\n% Note that this formulation allows for an easy extension to the case where $a=a(x)$ is a more general function of $x$.\n% From now on we will consider the simpler case where $a>0$ is constant.\n% From now on we will consider $a>0$ only, the generalization to other cases is relatively straightforward.\n% In either case, the flux differencing formula \\eqref{eq:flux_form} yields\n% \\begin{align*}\n% U_i^{n+1} = U_i^n - \\frac{a\\Delta t}{\\Delta x}\\left(U_i^n-U_{i-1}^n\\right)\n% \\end{align*}\n% for the upwind method.\nAnother way to derive the upwind method is to instead suppose that what we want to do is reconstruct $u(x)$ at each time step $n$ inside each cell $(x_{i-1/2},x_{i+1/2})$from the mean values in that cell and its surrounding neighbors. This reconstructed $\\tilde{u}(x)^n$ is then defined piecewise for each cell $i$.\nThe solution at the next time step can be found as $\\tilde{u}^{n+1}(x) = \\tilde{u}^n(x-a\\triangle t)$ which allows us to determine the fluxes $F^n_{i-1/2}$ once we have settled on a method for determining $\\tilde{u}^n(x)$ in each cell.\nThe simplest approach is\n\\[\\tilde{u}^n(x) = U_i^n \\text{ for }x \\in (x_{i-1/2},x_{i+1/2})\\]\nThis leads to fluxes given by\n\\begin{align}\nF_{i-1/2}^n &= \\frac{a}{\\triangle t}\\int_{0}^{t_{n+1}-t_n} \\tilde{u}^n(x_{i-1/2},t)\\, dt , \\\\\n&= \\frac{a}{\\triangle t}\\int_{0}^{\\triangle t} \\tilde{u}^n(x_{i-1/2}-at)\\, dt ,\\\\\n&= aU_{i-1}^n.\\nonumber\n\\end{align}\n\n\\begin{figure}\n\\includegraphics[width=\\textwidth, trim = 20mm 65mm 20mm 65mm, clip]\n{LW_reconstruction.pdf}\n\\caption{The piecewise linear reconstruction for the upwind and Lax-Wendroff methods.\nThe solid lines represent the simplest reconstruction of the cell averages leading to the upwind method, and the dashed lines are those whose slope is obtained via the Lax-Wendroff method.\nNote that the LW method introduces a spurious maximum at $i+3/2$ (the cell edge between $U_{i+1}$ and $U_{i+2}$) and the minimum at $i-3/2$ will be unphysical exaggerated.\nThe upwind method avoids this difficulties, but clearly loses a significant amount of the available information.\nThis provides the motivation for the slope limiters.}\n\\label{fig:fv_LW_reconstruction}\n\\end{figure}\n\nThe following code solves the problem\n\\begin{align}\n\\begin{split}\n\tu_t +a u_x &= 0, \\quad 0 < x < 1, \\\\\n\tu(x,t) &= f(x),\\\\\n\tu(0,t) &= u(1,t),\n\\end{split}\\label{eqn:fv_exercise}\n\\end{align}\nwhere $f$ represents a signal with two parts: one is smooth and the other is discontinuous.\nNotice that this PDE has periodic boundary conditions.\nEssentially we are evolving the signal around the unit circle.\nThis allows us to evolve the signal much further to test our numerical methods, since we only have to discretize the interval $[0,1]$ instead of a much larger domain.\nTo see how to implement the boundary conditions, consider a grid $0 = x_0< x_1< \\ldots< x_{N-1}< x_N = 1$ of evenly spaced points.\nSince $x_N = x_0$, it is sufficient to track $x_0, \\ldots , x_{N-1}$.\n\n\\begin{lstlisting}\nimport numpy as np\nfrom matplotlib import pyplot as plt\nfrom math import floor\n\ndef upwind(u0, a, xmin, xmax, t_final, nt):\n \"\"\" Solve the advection equation with periodic\n boundary conditions on the interval [xmin, xmax]\n using the upwind finite volume scheme.\n Use u0 as the initial conditions.\n a is the constant from the PDE.\n Use the size of u0 as the number of nodes in\n the spatial dimension.\n Let nt be the number of spaces in the time dimension\n (this is the same as the number of steps if you do\n not include the initial state).\n Plot and show the computed solution along\n with the exact solution. \"\"\"\n dt = float(t_final) / nt\n # Since we are doing periodic boundary conditions,\n # we need to divide by u0.size instead of (u0.size - 1).\n dx = float(xmax - xmin) / u0.size\n lambda_ = a * dt / dx\n u = u0.copy()\n for j in xrange(nt):\n # The Upwind method. The np.roll function helps us\n # account for the periodic boundary conditions.\n u -= lambda_ * (u - np.roll(u, 1))\n # Get the x values for the plots.\n x = np.linspace(xmin, xmax, u0.size+1)[:-1]\n # Plot the computed solution.\n plt.plot(x, u, label='Upwind Method')\n # Find the exact solution and plot it.\n distance = a * t_final\n roll = int((distance - floor(distance)) * u0.size)\n plt.plot(x, np.roll(u0, roll), label='Exact solution')\n # Show the plot with the legend.\n plt.legend(loc='best')\n plt.show()\n\n# Define the initial conditions.\n# Leave off the last point since we're using periodic\n# boundary conditions.\nnx = 30\nnt = nx * 3 // 2\nx = np.linspace(0., 1., nx+1)[:-1]\nu0 = np.exp(-(x - .3)**2 / .005)\narr = (.6 < x) & (x < .7 )\nu0[arr] += 1.\n\n# Run the simulation.\nupwind(u0, 1.2, 0, 1, 1.2, nt)\n\\end{lstlisting}\n\nTry running the previous code block with \\li{nx} set to $30$, $60$, $120$, and $240$.\nYou will notice that the numerical solution diffuses with time.\nIt diffuses especially fast at the points of discontinuity.\n\n\\begin{comment}\nAnother way to derive the upwind method is to instead suppose that what we want to do is reconstruct $u(x)$ at each time step $n$ inside each cell $(x_{i-1/2},x_{i+1/2})$ from the mean values in that cell and its surrounding neighbors.\nThis reconstructed $\\tilde{u}(x)^n$ is then defined piecewise for each cell $i$.\nThe solution at the next time step can be found as $\\tilde{u}^{n+1}(x) = \\tilde{u}^n(x-a\\Delta t)$ which allows us to determine the fluxes $F_{i-1/2}^n$ once we have settled on a method for determining $\\tilde{u}^n(x)$ in each cell.\nThe simplest approach is\n\\begin{gather*}\n\\tilde{u}^n(x) = U^n_i \\mbox{ for } x \\in (x_{i-1/2},x_{i+1/2}).\n\\end{gather*}\n% This leads to fluxes given by\n% \\begin{gather}\n% F_{i-1/2}^n = \\frac{a}{\\Delta t} \\int_{t_n}^{t_{n+1}}\\tilde{u}(x_{i-1/2}-a(t-t_n),t)dt = \\frac{a}{\\Delta t} \\int_{t_n}^{t_{n+1}}U_i^n dt = aU_i^n.\n% \\end{gather}\nIn either case, we can see that using the flux differencing formula \\eqref{eq:flux_form} that\n\\begin{gather}\nU_i^{n+1} = U_i^n - \\frac{a\\Delta t}{\\Delta x}\\left(U_i^n-U_{i-1}^n\\right)\n\\end{gather}\nfor the upwind method.\n\\end{comment}\n\n\\section*{Piecewise linear reconstruction and slope limiters}\nThe upwind method is formally only first order, and actually does relatively poorly in terms of actually transporting the initial data with velocity $a$.\nYou can notice from the example code that the upwind method has errors that are `diffusive' meaning that the initial data is diffused as time evolves, losing the peaks and fine details.\nThis is because the error for the upwind method is on the order of the second derivative of $u$ which is of a diffusive nature.\nTo get an improved method, consider a better reconstruction inside each cell, i.e.\n\\begin{gather}\n\\tilde{u}^n(x) = U_i^n + m_i^n(x-x_i) \\mbox{ for } x \\in (x_{i-1/2},x_{i+1/2})\n\\end{gather}\nwhere the slope of this linear reconstruction $m_i^n$ is determined as a function of the neighboring cell averages at time $n$ and $U_i^n$ itself.\nThen the flux is given by\n\\begin{align}\n\\begin{split}\nF_{i-1/2}^n\n&= \\frac{a}{\\triangle t}\\int_{0}^{t_{n+1}-t_n} \\tilde{u}^n(x_{i-1/2}-at)\\, dt ,\\\\\n&= \\frac{a}{\\triangle t}\\int_{0}^{\\triangle t} U_{i-1}^n + m_i^n(x_{i-1/2} - at - x_i), \\\\\n&= a \\left( U_{i-1}^n + \\frac{m_{i-1}^n}{2} (\\triangle x - a \\triangle t)\\right).\n\\end{split}\n\\label{eqn:finite_volume_improved_flux}\n\\end{align}\nOne of the most natural approaches is to just estimate the slope depending on the cell $i$ and a neighboring cell $i+1$ or $i-1$.\nThis leads to two popular methods, the Lax-Wendroff method and the Beam-Warming method (that really is the name).\nThe Lax-Wendroff method has a slope chosen as\n\\begin{gather}\nm_i^n = \\frac{U^n_{i+1}-U_{i}^n}{\\Delta x}.\n\\end{gather}\n% This leads to\n% \\begin{gather}\n% U_i^{n+1} = U_i^n - \\frac{\\Delta t}{2\\Delta x} a \\left(U_{i+1}^n-U_{i-1}^n\\right) + \\frac{1}{2}\\left(\\frac{\\Delta t}{\\Delta x}\\right)^2 a^2 \\left(U_{i-1}^n - 2U_i^n+U_{i+1}^n\\right)\n% \\end{gather}\nwhich it turns out is formally second-order accurate.\nIt turns out though that the errors for this method are dispersive, meaning that near very steep gradients, the method will generate very rapid oscillations (due to the third derivative of $u$ not being approximated accurately).\nAnother way to consider how these errors arise is to notice from Figure \\ref{fig:fv_LW_reconstruction} that if the piecewise linear reconstruction is advocated by some positive wind $a$ then there will be places where the discontinuous nature of the reconstruction will introduce spurious maxima or minima into the solution.\nThese become the spurious waves seen in simulations using the Lax-Wendroff method.\n\nA solution to this dilemma between balancing the diffusive and dispersive errors comes from constructing slopes $m_i^n$ that ensure no such non-monotonic transport takes place.\nThe basic idea is to constrain the slope so that the reconstructed piecewise linear function $\\tilde{u}^n(x)$ will not generate unphysical extremal values when it is advocated by some finite wind $a$.\nThe Minmod limiter chooses the slope as\n\\begin{gather}\nm_i^n = minmod\\left(\\frac{U_i^n-U_{i-1}^n}{\\Delta x},\\frac{U_{i+1}^n-U_{i}^n}{\\Delta x}\\right)\n\\end{gather}\nwhere\n\\begin{gather}\nminmod(a,b) = \\left\\{\\begin{array}{ccc}a \\mbox{ if } |a| < |b| \\mbox{ and } ab>0\\\\\nb \\mbox{ if } |b|<|a| \\mbox{ and } ab>0\\\\ 0 \\mbox{ if } ab < 0 .\\end{array}\\right.\n\\end{gather}\n\n% \\begin{itemize}\n% \\item Minmod limiter\n% \\begin{gather}\n% m_i^n = minmod\\left(\\frac{U_i^n-U_{i-1}^n}{\\Delta x},\\frac{U_{i+1}^n-U_{i^n}}{\\Delta x}\\right)\n% \\end{gather}\n% where\n% \\begin{gather}\n% minmod(a,b) = \\left\\{\\begin{array}{ccc}a \\mbox{ if } |a| < |b| \\mbox{ and } ab>0\\\\\n% b \\mbox{ if } |b|<|a| \\mbox{ and } ab>0\\\\ 0 \\mbox{ if } ab < 0 .\\end{array}\\right.\n% \\end{gather}\n% \\item Superbee limiter\n% \\begin{gather}\n% m_i^n = maxmod(m^{(1)}_i,m^{(2)}_i)\n% \\end{gather}\n% where the maxmod function selects $m^{(k)}_i$ with the larger modulus, and\n% \\begin{gather}\n% m^{(1)}_i = minmod \\left(\\frac{U_{i+1}^n-U_i^n}{\\Delta x},2\\frac{U_i^n-U_{i-1}^n}{\\Delta x}\\right)\\\\\n% m^{(2)}_i = minmod \\left( 2\\frac{U_{i+1}^n-U_i^n}{\\Delta x},\\frac{U_i^n-U_{i-1}^n}{\\Delta x}\\right).\n% \\end{gather}\n% \\item Monotonized central-difference limiter\n% \\begin{gather}\n% m_i^n = minmod\\left(\\frac{U_{i+1}^n-U_{i-1}^n}{2\\Delta x},2\\frac{U_i^n-U_{i-1}^n}{\\Delta x},2\\frac{U_{i+1}^n-U_i^n}{\\Delta x}\\right)\n% \\end{gather}\n% \\end{itemize}\n\n% \\begin{problem}\n% Code up the Superbee and Minmod %MC\n% limiters for an advection problem with constant wind.\n% Use different initial data that is smooth or has steep gradients (such as the square wave in the example code) to see which of the three limiters may be more optimal.\n% \\end{problem}\n\n\\begin{problem}\nImplement the Lax Wendroff method and use it to solve \\eqref{eqn:fv_exercise}.\nCompare it to the upwind method for $N = 30, 60, 120$, and 240.\nYou should be able to tell that the Lax Wendroff method approximates the smooth portion of the signal much better, as it does not struggle with diffusion.\nUnfortunately, it has some difficulty with the discontinuous portion, where unphysical oscillations are seen.\nRecall that we saw something similar in the waves lab when there were discontinuous initial conditions.\n\nHint: Use equations \\ref{eqn:finite_volume_improved_flux} and \\ref{eqn:flux_form}.\n\\end{problem}\n\n\\begin{problem}\nImplement the Minmod method and use it to solve \\eqref{eqn:fv_exercise}.\nCompare it to the upwind and Lax Wendroff methods for $N = 30, 60, 120$, and 240.\nBe sure to vectorize the minmod operation.\n\\end{problem}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=\\textwidth]{FiniteVolume_Comparisons.pdf}\n\\caption{Solutions of \\eqref{eqn:fv_exercise} at time $t = 1.2$ using various methods.\nHere the advection coefficient is $a = 1.2$, and there are $N = 100$ subintervals in space, 150 subintervals in time.}\n\\label{fig:fv_Comparisons}\n\\end{figure}\n\n\\section*{Beyond piecewise linear reconstructions}\nAs you can imagine, using a linear approximation is not the only option.\nThere are a host of high order finite volume methods that consider polynomial reconstructions of $\\tilde{u}^n$ inside each cell.\nThe key is then to use some nonlinear limiting technique that will ensure that when $\\tilde{u}^n(x)$ is advocated that no new extrema are introduced.\nChoosing the correct limiter for the given application then becomes an art unto itself. ", "meta": {"hexsha": "94e27d92186b392a21ccc6a22e9a07ca5bfe61f5", "size": 18870, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Labs/FiniteVolume/finitevolume.tex", "max_stars_repo_name": "jessicaleete/numerical_computing", "max_stars_repo_head_hexsha": "cc71f51f35ca74d00e617af3d1a0223e19fb9a68", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2016-10-18T19:54:25.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-09T20:12:38.000Z", "max_issues_repo_path": "Labs/FiniteVolume/finitevolume.tex", "max_issues_repo_name": "jessicaleete/numerical_computing", "max_issues_repo_head_hexsha": "cc71f51f35ca74d00e617af3d1a0223e19fb9a68", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Labs/FiniteVolume/finitevolume.tex", "max_forks_repo_name": "jessicaleete/numerical_computing", "max_forks_repo_head_hexsha": "cc71f51f35ca74d00e617af3d1a0223e19fb9a68", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-05-14T16:07:59.000Z", "max_forks_repo_forks_event_max_datetime": "2020-06-20T09:05:06.000Z", "avg_line_length": 57.3556231003, "max_line_length": 364, "alphanum_fraction": 0.7138844727, "num_tokens": 5995, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8289388040954683, "lm_q1q2_score": 0.6160408003012408}} {"text": "%!TEX root = ../main.tex\n\n\\section{Vectors problems}\n\\label{sec:vec_problems}\n\n\\vspace{-2mm}\n\nYou learned a bunch of vector formulas and you saw some vector diagrams,\nbut did you really learn how to solve problems with vectors?\nThere is only one way to find out: test yourself by solving problems.\n\nI've said it before and I don't want to repeat myself too much,\nbut it's worth saying again: the more problems you solve, the better you'll understand the material.\nIt's now time for you to try the following vector problems to make sure you're on top of things.\n\n\\medskip\n\n{ \\small\n\n\\begin{problems}{ch3}\n\n\t\\begin{problem}\n\t\tGiven the vectors $\\vec{u}=(1,1,1)$, $\\vec{v} = (2,3,1)$, and $\\vec{w}=(-1,-1,2)$,\n\t\tcompute the following products:\n\t\t\n\t\t\\threecol\n\t\t\t\\textbf{a)}~$\\vec{u} \\cdot \\vec{v}$\n\n\t\t\t\\textbf{b)}~$\\vec{u} \\cdot \\vec{w}$\n\t\t\t\n\t\t\t\\textbf{c)}~$\\vec{v} \\cdot \\vec{w}$\n\t\t\\endthreecol\n\t\t\n\t\t\\threecol\n\t\t\t\\textbf{d)}~$\\vec{u} \\times \\vec{v}$\n\n\t\t\t\\textbf{e)}~$\\vec{u} \\times \\vec{w}$\n\t\t\t\n\t\t\t\\textbf{f)}~$\\vec{v} \\times \\vec{w}$\n\t\t\\endthreecol\n\t\t\n\t\t\\begin{answer}\\textbf{a)}~$6$.\n\t\t\t\t\t\\textbf{b)}~$0$.\n\t\t\t\t\t\\textbf{c)}~$-3$.\n\t\t\t\t\t\\textbf{d)}~$(-2, 1, 1)$.\n\t\t\t\t\t\\textbf{e)}~$(3, -3, 0)$.\n\t\t\t\t\t\\textbf{f)}~$(7, -5, 1)$.\\end{answer}\n\t\\end{problem}\n\n\n\t\\begin{problem}\n\t\tGiven the vectors $\\vec{p} =(1,1,0,3,3)$ and $\\vec{q}=(1,2,3,4,5)$, calculate the following expressions:\n\t\t\\threecol\n\t\t\t\\textbf{a)}~$\\vec{p}+\\vec{q}$\n\t\t\t\n\t\t\t\\textbf{b)}~$\\vec{p}-\\vec{q}$\n\t\t\t\n\t\t\t\\textbf{c)}~$\\vec{p} \\cdot \\vec{q}$\n\t\t\\endthreecol\n\t\t\n\t\t\\begin{answer}\\textbf{a)}~$(2,3,3,7,8)$.\n\t\t\t\t\t\\textbf{b)}~$(0,-1,-3,-1,-2)$.\n\t\t\t\t\t\\textbf{c)}~$30$.\\end{answer}\n\t\\end{problem}\n\n\n\t\\begin{problem}\n\t\tFind a unit vector that is perpendicular to both $\\vec{u}=(1, 0, 1)$ and $\\vec{v} =(1, 2, 0)$.\n\t\t\\begin{hint}\n\t\t\tUse the cross product.\n\t\t\\end{hint}\n\t\t\\begin{answer}$(-\\frac{2}{3}, \\frac{1}{3}, \\frac{2}{3})$ or $(\\frac{2}{3}, -\\frac{1}{3}, -\\frac{2}{3})$.\\end{answer}\n\t\t\\begin{solution}See \\href{http://bit.ly/1cOa8yo}{\\texttt{bit.ly/1cOa8yo}} for calculations.\n\t\t\\end{solution}\n\t\\end{problem}\n\n\t\\begin{problem}\n\t\tFind a vector that is orthogonal to both $\\vec{u}_1 = (1, 0, 1)$ and $\\vec{u}_2 =(1, 3, 0)$, \n\t\t and whose dot product with the vector $\\vec{v}= (1, 1, 0)$ is equal to $8$.\n\t\t%\t\\begin{hint}\n\t\t%\t\\end{hint}\n\t\t\\begin{answer}$(12, -4, -12)$.\\end{answer}\n\t\t\\begin{solution}\n\t\t\tAny multiple of the vector $\\vec{u}_1 \\times \\vec{u}_2 = (-3,1,3)$ \n\t\t\tis perpendicular to both $\\vec{u}_1$ and \t$\\vec{u}_2$.\n\t\t\tWe must find a multiplier $t \\in \\mathbb{R}$ such that \n\t\t\t$t(-3,1,3) \\cdot (1, 1, 0) = 8$.\n\t\t\tComputing the dot product we find $-3t + t = 8$, so $t=-4$.\n\t\t\tThe vector we're looking for is $(12, -4, -12)$.\n\t\t\tSee \\href{http://bit.ly/1nmYH8T}{\\texttt{bit.ly/1nmYH8T}} for calculations.\n\t\t\\end{solution}\n\t\\end{problem}\n\n\n\\end{problems}\n\n} % /small\n\n", "meta": {"hexsha": "cc881beb0147510892b7feff7f4201db39760f7f", "size": 2833, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sources/original/problems/chapter3_problems.tex", "max_stars_repo_name": "minireference/sample-book", "max_stars_repo_head_hexsha": "83e827d7e0c3f5fea1d08815810f0b08bef503e3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 27, "max_stars_repo_stars_event_min_datetime": "2020-10-19T21:21:18.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-27T16:42:13.000Z", "max_issues_repo_path": "sources/original/problems/chapter3_problems.tex", "max_issues_repo_name": "minireference/sample-book", "max_issues_repo_head_hexsha": "83e827d7e0c3f5fea1d08815810f0b08bef503e3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sources/original/problems/chapter3_problems.tex", "max_forks_repo_name": "minireference/sample-book", "max_forks_repo_head_hexsha": "83e827d7e0c3f5fea1d08815810f0b08bef503e3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-09-12T19:03:04.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-12T19:03:04.000Z", "avg_line_length": 28.6161616162, "max_line_length": 118, "alphanum_fraction": 0.5979527003, "num_tokens": 1194, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7090191337850933, "lm_q2_score": 0.8688267830311354, "lm_q1q2_score": 0.6160148131140248}} {"text": "In this section, we first define the problem we address, then the mathematical notation we use, and finally our evaluation metrics.\n\n%In this section, we will define our addressing problem with the notation. Also, the background about this paper will be provided.\n\n\n\\subsection{Problem definition}\n\nThe problem we address in this paper is related to the need of an AUI \nto ease the task of monitoring alerts in\nlarge-scale displayed networks. The new interface is expected\nto filter irrelevant information to provide localized context for\neach event or alert (defined loosely as a relevant related content localized in time, space, and/or keyword usage). Hence, we define the problem we are addressing in this\npaper as a problem of filter optimization.\nWe assume that selection of information elements to display in visual interface is obtained via settings of three filters that can jointly express a global filter. These three sub-filters are the following:\n\n\n%The problem we address in this paper is the proper display of portion in the network for user's easy to investigate large-scale network elements. The new system is expected to filter information to provide localized and relevant element in the interface. This problem is able to be defined as a filter search/recommendation problem, We expect to employ standard IR theory to solve this problem based on our argument about the similarity of AUI and IR system. We assume that display of the elements in the network is obtained via three filters that can jointly express a global filter. These three sub-filters are:\n\\begin{itemize}\n\\item Time: expresses the time window of a retrieved set. \n\\item Keyword: expresses a set of keyword to include or exclude from a retrieved set.\n\\item Location: expresses a bounding box of a retrieved set. The location can be expressed as longitude/latitude values, or pure coordinates obtained using a given display layout. \n\\end{itemize}\n\nThe problem we study is how to optimize one or multiple filter selection settings to maximize retrieval of relevant content. \n%best set of elements that is assessed using one of the metrics described above, i.e., expected precision, expected recall, or expected F1-score\n\n\\subsection{Mathematical Notation}\nThroughout this paper, we present all algorithms for Greedy and Optimal search using the following mathematical notation:\n\n\\begin{itemize}\n\\item An element $j$, for which three types of metadata are associated: (i) a textual content, which is composed of a set of terms of size $n$, (ii) a timestamp $t_e$, which may represent the creation date of $j$, and (iii) a position coordinates $(x_{e},y_{e})$.\n\\item Three variables $I(j)$, $B(j)$ and $S(j)$ are associated to each element $j$. $I(j)$ refers to whether element $j$ is in the retrieved set (E). $B(j)$ is the boolean variable about relevance of element $j$. $S(j)$ is a probabilistic score of element $j$ being relevant. Note that $I(j)$ is correlated with the UI system. $B(j)$ and $S(j)$ are independent of the system. %$\\emph{I(j), B(j)} \\in \\{0, 1\\}$, $\\emph{S(j)} \\in \\left[0, 1\\right]$. \nAlso, $B(j)$ follows a \\emph{Bernoulli} distribution with parameter $S(j)$, and hence, the expectation of $B(j)$ is $S(j)$, i.e., \n $\\mathbb{E_S}[B(j)] = S(i)$.\n\n\n\n\n%which allows to derive the expectation of \\emph{B(j)} as follows:\n%\\begin{equation}\n % \\mathbb{E_S}[B(j)] = 0*(1 - S(i))+ 1*S(i) = S(i)\n%\\end{equation}\n\n\\item A global collection of elements $GC$ with size $m$. %Two subsets of $GC$ are particularly important in this research: retrieved set $E$ and relevant set $RS$. \n\\item A retrieved set $E$ with $E \\subseteq GC$, for which we use $E^*$ to refer to any subset, i.e., $E^*\\subseteq E$. This variable depends on our user-interface filter system. Note that size of $E$ can be represented the sum of $I(j)$ among the global collection. Therefore, we have $|E| = \\sum_{j=1}^m I(j)$.\n\\item A relevant set $RS$ of $|RS|$ elements, which contains all relevant elements. This is independent of the UI-filter system. Note that size of $RS$ is equal to the sum of $B(j)$ among the global collection. Therefore, we have $|RS| = \\sum_{j=1}^m B(j)$. %However, $B(j)$ is not available for our estimation of $RS$ size in practice. We have to use expected $RS$ size $|RS|$, $\\mathbb{E_S}|RS|$, to approximate $|RS|$.\n%\\begin{equation}\n %|RS| \\approx \\mathbb{E_S}|RS| = \\sum_{j=1}^m \\mathbb{E_S}[B(j)] = \\sum_{j=1}^m S(j)\n%\\end{equation}\n\n\\item A keyword query $Q_k=\\{\\neg t_{1}^{*},\\dots \\neg t_{k}^{*}\\}$, which expresses query term exclusion, i.e., find elements that don't contain the terms $t_{1}^{*},\\dots t_{k}^{*}$.\n\\item A time query $Q_t=[t_{start},t_{end}]$, which expresses the search for elements in the time window $[t_{start},t_{end}]$.\n\\item A position query $Q_p=[(x_{min},y_{min}),(x_{max},y_{max})]$, which expresses the search of elements falling in the bounding box represented by the lower and upper bound coordinates -- respectively $(x_{min},y_{min})$ and $(x_{max},y_{max})$.\n\\item A query filter $Q$, which combines the three sub-filters $Q_k$, $Q_t$, and $Q_p$ in a conjunction, i.e., $Q=[Q_k\\wedge Q_t\\wedge Q_p]$. \n\n\\end{itemize}\n\n\n\n\n\n\n\n%The goal is to explore an algorithm to obtain an optimal filter setting to retrieve a set of elements with maximum score of the specific metric.\n\n\n\n%\\subsection{Comparison to standard IR search}\n%The comparison between information retrieval for web search and information retrieval for filtering in AUIs can be summarized in Table \\ref{tbl:Comparaison2IR}. Obviously , there are important differences between web search and filtering for AUIs, especially in the indirect selection of results through filter settings. This unexplored field provides new possibilities and challenges of research in this novel information retrieval setting:\n%\\begin{itemize}\n%\\item Evaluation metrics and human factors: Are there new evaluation metrics specific to this AUI filtering setting? What evaluation metrics correlate with AUI user performance? \n%\\item Optimization and algorithms: How do we optimally select filter settings to maximize evaluation metrics in expectation? How can we interpret simple heuristics like average and cumulative relevance? (Answer: expected precision.) How do MILPs, relaxed LP approximations with guarantees, or greedy approaches compare in terms of time and metric quality? Are there properties of different filters (1D for time, 2D for bounding box, or discrete choices for property selection) that lend themselves to specialized greedy approaches? \n%\\item Robustness: As the signal-to-noise ratio varied in the quality of the relevance scoring, how do various algorithms perform? \n%\\item Explanation: Can we provide explanations for filter settings to allow users to understand the reasons for the suggestions? \n%\\item Personalization and learning: Can we learn from observations of manual adaption of the suggested filter settings to understand how to improve the third-party scoring systems? \n%\\item Collaborative filtering: Can we generalize learning across multiple\n%users in a collaborative filtering approach? \n%\\item Learning from implicit feedback: How can we leverage implicit user\n%feedback such as clicks and dwell time to indirectly measure the relevance\n%of filtered content and improve system performance.\n%\\end{itemize}\n\n%\\subsection{Optimization technique definition}\n\n%The \\emph{greedy algorithm} is an algorithmic paradigm that aims to obtain a global optimum of a problem in terms of making the locally optimal decision at each step \\cite{Black2005}. In a search problem, a greedy algorithm does not in general produce an optimal solution, but it still yields locally optimal solutions that approximate a global optimum in a reasonable time.\n\n%The greedy algorithms described in this paper are coupled with a \\emph{Top-down} search strategy, which basically begins with the whole search space, and then partition into several sub-spaces in a lower level for the local optimization search heuristic.\n\n%An \\emph{optimization-based} search is the problem of finding the best solution from all feasible solutions. Usually, the standard form of an optimization problem is defined as the minimization/maximization of a given objective function subject to a set of constraints. \n\n%The search problem will be transformed into Mixed integer linear programming (MILP), which involves problems in which only some of the variables, $x_{i}$, are constrained to be integers, while other variables are allowed to be non-integers.\n\n\n\n\\subsection{Evaluation metrics}\n\nWe adopt the Boolean relevance framework of IR and thus assume that all information elements $j$ have a ground truth relevance assessment $B(j)$ available at evaluation time. \n%%However, unlike previous document-based methods for estimating relevance at retrieval-time, we do not rely on text associated with information elements and instead rely on a third-party to supply an estimated probability (score) of relevance $S(j)$. \n%Traditionally, Boolean labels indicate the relevance of documents is used to evaluate retrieval sets (e.g., precision). \n%In practice, IR models are based on probabilistic scores to measure relevance of the elements, such as TF-IDF and cosine similarity. \n%In practice, we are able to employ third-party technique to provide probabilistic scores to measure relevance of the document in IR system, such as TF-IDF and cosine similarity. Consequently,\nBecause filters are equivalent to Boolean retrieval (they either select or do not select content as relevant) and we have a probabilistic estimate of relevance $S(j)$, we propose to evaluate\nexpected variants of standard precision, recall, and F1-score.\n%Boolean retrieval evaluation metrics based on the relaxation of Boolean labels to probabilistic scores resulting in expected metrics, i.e., \\emph{expected precision} (EP), \\emph{expected recall} (ER) and \\emph{expected F1-Score} (EF1).\nSpecifically because the filter that selects all information elements would maximize expected recall, the filter that selects the highest probability singleton information element would maximize expected precision, we focus on expected F1-score as a trade-off to balance expected precision and recall.\n\nGiven a retrieved information element set $E$, a relevant element set $RS$ selected by a filter setting, the global information element collection $GC$ with size $m$, the precision of a retrieved set $P(E)$ is defined as follows:\n\\begin{equation}\n P(E) = \\dfrac{\\sum_{j \\in RS} B(j)}{|E|} = \\dfrac{\\sum_{j=1}^m B(j)I(j)}{\\sum_{j=1}^m I(j)} \n\\end{equation}\nThe relaxation of the Boolean label $B(j)$ to probabilistic score $S(j)$ gives the following definition of \\emph{expected precision}: \n\\begin{equation}\nEP(E)=\\mathbb{E_{S}}[\\dfrac{\\sum_{j=1}^{m}B(j)I(j)}{\\sum_{j=1}^{m}I(j)}]=\\dfrac{\\sum_{j=1}^{m}\\mathbb{E_{S}}[B(j)]I(j)}{\\sum_{j=1}^{m}I(j)}=\\dfrac{\\sum_{j=1}^{m}S(j)I(j)}{\\sum_{j=1}^{m}I(j)}\n\\end{equation}\nSimilarly the recall of a retrieved set $R(E)$ is defined as:\n\\begin{equation}\n R(E) = \\dfrac{\\sum_{j \\in RS} B(j)}{|RS|} = \\dfrac{\\sum_{j=1}^m B(j)I(j)}{|RS|} \n\\end{equation}\nTaking a 1st order Taylor expansion, we have the following expectation approximation %$\\mathbb{E}(X/Y)\\approx\\dfrac{\\mathbb{E}(X)}{\\mathbb{E}(Y)}$ \n$\\mathbb{E}(X/Y)\\approx \\mathbb{E}(X)/ \\mathbb{E}(Y)$ for two dependent random variables $X$ and $Y$~\\cite{Kempen2000}. Hence, \nwe can now define an \\emph{approximated expected recall}: \n%Given a retrieved element set $E$ and a relevant element set $RS$ among a global element collection $GC$ with size $m$, we propose the following definition of \\emph{expected precision} (EP), \\emph{expected recall} (ER) and \\emph{expected F1-Score} (EF1):\n\\begin{equation}\n \t\\emph{ER(E)}=\\mathbb{E_{S}}[\\dfrac{\\sum_{j=1}^{m}B(j)I(j)}{|RS|}] \\approx \\dfrac{\\sum_{j=1}^{m}\\mathbb{E_{S}}[B(j)]I(j)}{\\mathbb{E_{S}}[|RS|]} = \\dfrac{\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m S(j)} \n\\end{equation}\nFinally, we define the \\emph{approximated expected F1-Score} using the \\emph{expected precision} and the \\emph{approximated expected recall} as follows: \n\\begin{align}\n \\emph{EF1(E)} \\approx \\dfrac{2\\times EP\\times ER}{EP+ER} = \\dfrac{2\\times\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m I(j) + \\sum_{j=1}^m S(j)}\n\\end{align}\n \n \n%Conventional metrics are based on Boolean relevance label $B(j)$ instead of probabilistic score $S(j)$. To link standard precision with our expected precision, The definition of standard precision $P(RS)$ is as follows:\n\n\n%Now, we derive expectation of precision $P(RS)$ as follows:\n%\\begin{align}\n%\t\\mathbb{E_S}[P(E)] &= \\mathbb{E_S}[\\dfrac{\\sum_{j=1}^m B(j)I(j)}{\\sum_{j=1}^m I(j)}] = \\dfrac{\\sum_{j=1}^m %\\mathbb{E_S}[B(j)]I(j)}{\\sum_{j=1}^m I(j)} \\notag \\\\\n% &= \\dfrac{\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m I(j)} = EP(E)\n \t%\\mathbb{E_S}[R(RS)] &= \\dfrac{\\sum_{j=1}^m \\mathbb{E_S}[B(j)] I(j)}{\\sum_{j=1}^m \\mathbb{E_S}[B(j)]} = \\dfrac{\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m S(j)} = ER(RS) \\\\\n %\\mathbb{E_S}[F1(RS)] &= \\dfrac{2*\\sum_{j=1}^m \\mathbb{E_S}[B(j)]I(j)}{\\sum_{j=1}^m I(j) + \\sum_{j=1}^m \\mathbb{E_S}[B(j)]} \\notag \\\\\n %&= \\dfrac{2*\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m I(j) + \\sum_{j=1}^m S(j)} = EF1(RS)\n%\\end{align}\n\n\n\n\n%\\begin{equation}\n%\\emph{\\ensuremath{EP(RS)}}=\\dfrac{{\\displaystyle \\sum_{j\\in RS}S(j)}}{|RS|} = \\dfrac{\\sum_{j=1}^m S(j)I(j)}{\\sum_{j=1}^m I(j)} \n%\\end{equation}\n\n%Analogously, we define the \\emph{expected recall} (ER) as follows:\n%\\begin{equation}\n%\\emph{\\ensuremath{ER(RS)}}=\\dfrac{{\\displaystyle \\sum_{j\\in RS}S(j)}}{|RE|} = \\dfrac{\\sum_{i=1}^m S(i)I(i)}{\\sum_{i=1}^m S(i)} = \\dfrac{\\sum_{i=1}^m S(i)I(i)}{C} \n%\\end{equation}\n\n%\\noindent where $|RE|$ is the size of the entire relevant element set. Finally, we define the \\emph{expected F1-Score} (EF1) as follows:\n\n%\\begin{equation}\n%\\emph{\\ensuremath{EF1(RS)}}=\\dfrac{2\\times EP\\times ER}{EP+ER} = \\dfrac{2*\\sum_{i=1}^m S(i)I(i)}{\\sum_{i=1}^m I(i) + C}\n%\\end{equation}\n\nIn Figure \\ref{fig:F1_vs_EF1}, we experimentally show that while EF1 is only an approximation of the true expectation, EF1 serves as an excellent surrogate for F1, which is our main concern when considering filter optimization. That is, maximizing EF1 score with respect to a noisy classifier (noise decreases to 0 as $\\lambda \\to 1$) is strongly correlated with maximizing F1 score evaluated on the ground truth. Specifically, while the EF1 and F1 scores are not perfectly calibrated along the diagonal, there is a linear correlation in that as the EF1 score increases for a scenario, the F1-score proportionally increases on average (as shown by the best fit linear regression in the plots).\n\n\n\n\\begin{figure}[H]\n\\begin{centering}\n\\par\\end{centering}\n\\begin{centering}\n\\includegraphics[width=8.5cm]{imgs/Enron_results/scatter_plot_EF1\\lyxdot vs\\lyxdot F1}\n\\par\\end{centering}\n\\caption{Scatter plot showing EF1-Score vs. F1-Score.}\n\\label{fig:F1_vs_EF1}\n\\end{figure}\n\n\n\n\n\n", "meta": {"hexsha": "f06c9ebd3736d4f4e0824908d45eb7dc0237fbe4", "size": 14927, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documents/SIGIR2018/Framework.tex", "max_stars_repo_name": "D3Mlab/visir", "max_stars_repo_head_hexsha": "cd1860984dee8d7aba368857e734ad11c14124c8", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-10T07:40:04.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-10T07:40:04.000Z", "max_issues_repo_path": "Documents/SIGIR2018/Framework.tex", "max_issues_repo_name": "D3Mlab/viz-ir", "max_issues_repo_head_hexsha": "cd1860984dee8d7aba368857e734ad11c14124c8", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Documents/SIGIR2018/Framework.tex", "max_forks_repo_name": "D3Mlab/viz-ir", "max_forks_repo_head_hexsha": "cd1860984dee8d7aba368857e734ad11c14124c8", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 82.9277777778, "max_line_length": 696, "alphanum_fraction": 0.7391974275, "num_tokens": 4150, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8688267694452331, "lm_q2_score": 0.7090191276365463, "lm_q1q2_score": 0.6160147981393379}} {"text": "\\section{Homogeneous Systems}\r\n\\noindent\r\nHomogeneous systems of linear equations have the form\r\n\\begin{equation*}\r\n\t\\vec{x}' = A\\vec{x}\r\n\\end{equation*}\r\nWe'll see how to find solutions to these types of systems using the eigenvalue method. Finding solutions to these systems will allow us to find homogeneous solutions to heterogeneous systems of equations.\r\n\r\n% Eigenvalue Method\r\n\\input{./linearSystems/homogeneousSystems/eigenvalueMethod.tex}", "meta": {"hexsha": "e511b9d8cf422859a19598941a4c8805ff76c529", "size": 447, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/linearSystems/homogeneousSystems/homogeneousSystems.tex", "max_stars_repo_name": "rawsh/Math-Summaries", "max_stars_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "diffEq/linearSystems/homogeneousSystems/homogeneousSystems.tex", "max_issues_repo_name": "rawsh/Math-Summaries", "max_issues_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "diffEq/linearSystems/homogeneousSystems/homogeneousSystems.tex", "max_forks_repo_name": "rawsh/Math-Summaries", "max_forks_repo_head_hexsha": "3ad58ef55c176f7ebaf145144e0a4eb720ebde86", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.7, "max_line_length": 205, "alphanum_fraction": 0.7897091723, "num_tokens": 103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8128673178375734, "lm_q2_score": 0.7577943712746406, "lm_q1q2_score": 0.6159862780504274}} {"text": "\\section{Coordinates, Differentials and Gradients}\r\n\\subsection{Differentials and First-Order Changes}\r\nRecall that if $f=f(u_1,\\ldots,u_n)$, then we write $\\mathrm df=(\\partial f/\\partial u_i)\\,\\mathrm du_i$ (the summation convention is being used).\r\nThose $\\mathrm du_i$ are formal objects called differential forms which are quite abstract geometrical notions that are way beyond the scope of this course.\r\nThese differential forms are taken as linearly independent the same way as vectors are.\r\nSimilarly, if $\\underline{x}=\\underline{x}(u_1,u_2,\\ldots,u_n)$, then $\\mathrm d\\underline{x}=(\\partial \\underline{x}/\\partial u_i)\\,\\mathrm du_i$.\r\n\\begin{example}\r\n If $f(u,v,w)=u^2-v^2+e^w$, then $\\mathrm df=2u\\,\\mathrm du-2v\\,\\mathrm dv+e^w\\,\\mathrm dw$.\r\n If $\\underline{x}=(u^2,v^2,w^2)^\\top$, then $\\mathrm d\\underline{x}=(2u\\,\\mathrm du,2v\\,\\mathrm dv,2w\\,\\mathrm dw)^\\top$.\r\n\\end{example}\r\nDifferential forms give a great tool to describe first-order changes.\r\nIf we perturb a multivariable function $f(u_1,\\ldots,u_n)$, then we can have\r\n$$f(u_1+\\epsilon_1,\\ldots,u_n+\\epsilon_n)=f(u_1,\\ldots,u_n)+\\frac{\\partial f}{\\partial u_i}\\epsilon_i+o(\\|\\underline{\\epsilon}\\|)$$\r\nWe can get the chain rule ``for free'' by using this notion.\r\nSuppose we change our coordinates by $v_i=v_i(u_1,\\ldots,u_n)$ and $F(u_1,\\ldots,u_n)=f(v_1,\\ldots,v_n)$, so\r\n$$\\frac{\\partial F}{\\partial u_i}\\,\\mathrm du_i=dF=df=\\frac{\\partial f}{\\partial v_j}\\,\\mathrm dv_j=\\frac{\\partial f}{\\partial v_j}\\frac{\\partial v_j}{\\partial u_i}\\,\\mathrm du_i$$\r\nTherefore\r\n\\begin{theorem}\r\n $$\\frac{\\partial F}{\\partial u_i}=\\frac{\\partial f}{\\partial v_j}\\frac{\\partial v_j}{\\partial u_i}$$\r\n\\end{theorem}\r\nNote that the summation convention is implicitly used.\r\n\\subsection{Coordinates in Line Elements}\r\nSay $u,v$ are coordinates for $\\mathbb R^2$ by relating them to Cartesians in the form $x=x(u,v),y=y(u,v)$ such that these smooth functions can be inverted smoothly to give $u=u(x,y),v=v(x,y)$.\r\n\\begin{example}\r\n Consider the polar coordinate $(r,\\theta)$ with the relationship $x=r\\cos\\theta,y=r\\sin\\theta$.\r\n We can invert to have $r=\\sqrt{x^2+y^2},\\tan\\theta=x/y$.\r\n \\footnote{This is not quite invertible at $(x,y)=(0,0)$.\r\n Just saying.}\r\n\\end{example}\r\n\\begin{example}\r\n 1. The Cartesian coordinate in $\\mathbb R^2$ is $\\underline{x}=\\underline{x}(x,y)=(x,y)^\\top$.\r\n \\footnote{For future reference, any column vector in this course are implicitly written as per the standard basis unless otherwise specified.}\r\n Note that $\\underline{x}_x,\\underline{x}_y$ give the standard basis, so $\\mathrm d\\underline{x}=(\\mathrm dx,\\mathrm dy)^\\top$.\\\\\r\n 2. The polar coordinate defined above, $\\underline{x}=(r\\cos\\theta,r\\sin\\theta)^\\top$ has\r\n $$\\underline{x}_r=(\\cos\\theta,\\sin\\theta),\\underline{x}_\\theta=(-r\\sin\\theta,r\\cos\\theta)$$\r\n which becomes an orthonormal basis (which depends on $(r,\\theta)$) if we normalize.\r\n So the line element has\r\n $$\\mathrm d\\underline{x}=\\begin{pmatrix}\r\n \\cos\\theta\\\\\\sin\\theta\r\n \\end{pmatrix}\\,\\mathrm dr+r\\begin{pmatrix}\r\n -\\sin\\theta\\\\\\cos\\theta\r\n \\end{pmatrix}\\,\\mathrm d\\theta$$\r\n So small change in the line element can result in a large change in the coordinate $\\mathrm d\\theta$.\r\n The basis vectors above give the rotation basis.\r\n\\end{example}\r\n\\begin{definition}\r\n We say $u,v,w$ are set of orthogonal curvilinear coordinates for $\\mathbb R^3$ if the unit vectors $\\underline{e}_u=\\underline{x}_u/\\|\\underline{x}_u\\|,\\underline{e}_v=\\underline{x}_v/\\|\\underline{x}_v\\|,\\underline{e}_w=\\underline{x}_w/\\|\\underline{x}_w\\|$ always forms a right-handed system of orthonormal vectors.\r\n\\end{definition}\r\n\\begin{definition}\r\n The scale factors are\r\n $$h_{u}=\\left|\\frac{\\partial \\underline{x}}{\\partial u}\\right|,h_{v}=\\left|\\frac{\\partial \\underline{x}}{\\partial v}\\right|,h_{w}=\\left|\\frac{\\partial \\underline{x}}{\\partial w}\\right|$$\r\n\\end{definition}\r\nSo the factors are the scaling factors for a little change in the corresponding coordinates.\r\n\\begin{definition}\r\n The cylindral polar coordinates are $(\\rho,\\phi,z)$ where\r\n $$\\underline{x}=\\underline{x}(\\rho,\\phi,z)=(\\rho\\cos\\phi,\\rho\\sin\\phi,z)^\\top$$\r\n\\end{definition}\r\n\\begin{definition}\r\n The spherical polar coordinates are $(r,\\theta,\\phi)$ where\r\n $$\\underline{x}=\\underline{x}(r,\\theta,\\phi)=(r\\cos\\phi\\sin\\theta, r\\sin\\phi\\sin\\theta,r\\cos\\theta)$$\r\n\\end{definition}\r\n\\subsection{The Gradient Operator}\r\nFor $f:\\mathbb R^3\\to\\mathbb R$, define the gradientof $f$, $\\nabla f$, by\r\n$$f(\\underline{x}+\\underline{h})=f(\\underline{x})+\\nabla f(\\underline{x})\\cdot\\underline{h}+o(\\underline{h})$$\r\n\\begin{definition}\r\n The directional derivative of $f$ in direction $\\underline{v}$ by\r\n $$D_{\\underline{v}}f(\\underline{x})=\\lim_{t\\to 0}\\frac{f(\\underline{x}+t\\underline{v})-f(\\underline{x})}{t}$$\r\n So $f(\\underline{x}+t\\underline{v})=f(\\underline{x})+tD_{\\underline{v}}f(\\underline{x})+o(t)$.\r\n\\end{definition}\r\nSo if we let $\\underline{h}=t\\underline{v}$, we have\r\n$$f(\\underline{x}+t\\underline{v})=f(\\underline{x})+\\nabla f(\\underline{x})\\cdot t\\underline{v}+o(t)$$\r\nSo we have $\\nabla f(\\underline{x})\\cdot \\underline{v}=D_{\\underline{v}}f(\\underline{x})$.\r\nBy Cauchy-Schwartz, to maximize a dot product we will have to make the two vectors parallel, so\r\n\\begin{proposition}\r\n $\\nabla f(\\underline{x})$ is the direction of greatest increase of $f$ at $\\underline{x}$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Cauchy-Schwartz.\r\n\\end{proof}\r\n\\begin{example}\r\n 1. $f(\\underline{x})=|\\underline{x}|^2/2$, so\r\n $$f(\\underline{x}+\\underline{h})=(\\underline{x}+\\underline{h})\\cdot(\\underline{x}+\\underline{h})/2=f(\\underline{x})+\\underline{x}\\cdot\\underline{h}+o(\\underline{h})$$\r\n Hence $\\nabla f(\\underline{x})=\\underline{x}$.\\\\\r\n 2. For generic curve $t\\mapsto \\underline{x}(t)$ and a function $F$, we want to evaluate $(F\\circ \\underline{x})^\\prime$, so\r\n $$F(\\underline{x}(t+\\delta t))=F(\\underline{x}(t)+\\underline{\\delta x})=F(\\underline{x}(t))+\\nabla f(\\underline{x}(t))\\cdot\\underline{\\delta x}+o(\\underline{\\delta x})$$\r\n where $\\underline{\\delta x}=\\underline{x}(t+\\delta t)-\\underline{x}(t)=t\\underline{x}^\\prime(t)+o(t)$.\r\n So plugging it in we have\r\n $$\\frac{\\mathrm dF}{\\mathrm dt}=\\nabla F(\\underline{x}(t))\\cdot \\underline{x}^\\prime(t)$$\r\n 3. Consider a surface in $\\mathbb R^3$ by $S=\\{\\underline{x}:f(\\underline{x})=0\\}$ where $f:\\mathbb R^3\\to\\mathbb R$.\r\n Take curve $t\\mapsto \\underline{x}(t)$ such that $\\underline{x}\\in S$ for all $t$, then $0=\\underline{x}^\\prime(t)\\cdot\\nabla F(\\underline{x}(t))$, so $\\nabla F$ is perpendicular to the tangent to the curve.\r\n Hence necessarily $\\nabla F$ is the normal to $S$.\r\n\\end{example}\r\n\\subsection{Computing the Gradient}\r\nFor general orthogonal curvilinear coordinate, it might be hard to calculate $\\nabla f$ since we do not usually know how to change the coordinates to accomodate the change $\\underline{h}$.\\\\\r\nBut it is easy in Cartesians.\r\nJust evaluating the directional derivatives at the basis vectors reveals:\r\n\\begin{proposition}\r\n $$\\nabla f=\\begin{pmatrix}\r\n \\partial f/\\partial x\\\\\r\n \\partial f/\\partial y\\\\\r\n \\partial f/\\partial z\r\n \\end{pmatrix}$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n We have\r\n $$(\\nabla f(\\underline{x}))_i=\\nabla f(\\underline{x})\\cdot\\underline{e_i}=D_{\\underline{e_i}}f(\\underline{x})$$\r\n As desired.\r\n\\end{proof}\r\n\\begin{example}\r\n Again we take $f(\\underline{x})=|\\underline{x}|^2/2$, then $(\\nabla f(\\underline{x}))_i=x_i$, so $\\nabla f(\\underline{x})=\\underline{x}$ as before.\r\n\\end{example}\r\nIn Cartesians, we know the line elements $\\mathrm d\\underline{x}=\\mathrm dx_i\\underline{e_i}$ which allows us to calculate easily.\r\nBut we have $\\mathrm df=D_{\\underline{e_i}}f\\,\\mathrm dx_i$ in any coordinate.\r\nSo immediately we have\r\n\\begin{proposition}\r\n $\\mathrm df=\\nabla f\\cdot\\mathrm d\\underline{x}$\r\n\\end{proposition}\r\nwhich is coordinate independent.\r\n\\begin{proof}\r\n Just calculate the right hand side.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Let $u,v,w$ be a set of curvilinear coordinates, then we have\r\n $$\\nabla f=\\frac{1}{h_u}\\frac{\\partial f}{\\partial u}\\underline{e_u}+\\frac{1}{h_v}\\frac{\\partial f}{\\partial v}\\underline{e_v}+\\frac{1}{h_w}\\frac{\\partial f}{\\partial w}\\underline{e_w}$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n $\\mathrm df=\\nabla f\\cdot\\mathrm d\\underline{x}$ independent of coordinate.\r\n We also know that $\\mathrm d\\underline{x}=h_u\\underline{e_u}\\,\\mathrm du+h_v\\underline{e_v}\\,\\mathrm dv+h_w\\underline{e_w}\\,\\mathrm dw$.\r\n Write $\\nabla f=(\\nabla f)_u\\underline{e_u}+(\\nabla f)_v\\underline{e_v}+(\\nabla f)_w\\underline{e_w}$, then\r\n $$\\nabla f\\cdot \\mathrm d\\underline{x}=h_u(\\nabla f)_u\\,\\mathrm du+h_v(\\nabla f)_v\\,\\mathrm dv+h_w(\\nabla f)_w\\,\\mathrm dw$$\r\n In addition,\r\n $$\\mathrm df=\\frac{\\partial f}{\\partial u}\\,\\mathrm du+\\frac{\\partial f}{\\partial v}\\,\\mathrm dv+\\frac{\\partial f}{\\partial w}\\,\\mathrm dw$$\r\n But they are equal.\r\n Since $\\mathrm du,\\mathrm dv,\\mathrm dw$ are linearly independent, we have\r\n $$(\\nabla f)_u=\\frac{1}{h_u}\\frac{\\partial f}{\\partial u},(\\nabla f)_v=\\frac{1}{h_v}\\frac{\\partial f}{\\partial v},(\\nabla f)_w=\\frac{1}{h_w}\\frac{\\partial f}{\\partial w}$$\r\n As desired.\r\n\\end{proof}\r\n\\begin{example}\r\n 1. For cylindral coordinates $(\\rho,\\phi,z)$, we have\r\n $$\\nabla f=\\frac{\\partial f}{\\partial\\rho}\\underline{e_\\rho}+\\frac{1}{\\rho}\\frac{\\partial f}{\\partial\\phi}\\underline{e_\\phi}+\\frac{\\partial f}{\\partial z}\\underline{e_z}$$\r\n Then in the previous example where $f(\\underline{x})=|\\underline{x}|^2/2=(\\rho^2+z^2)/2\\implies\\nabla f=\\rho\\underline{e_\\rho}+z\\underline{e_z}=\\underline{x}$\\\\\r\n 2. For spherical coordinates $(r,\\theta,\\phi)$, we can do the same thing,\r\n $$\\nabla f=\\frac{\\partial f}{\\partial r}\\underline{e_r}+\\frac{1}{r}\\frac{\\partial f}{\\partial\\theta}\\underline{e_\\theta}+\\frac{1}{r\\sin\\theta}\\frac{\\partial f}{\\partial \\phi}\\underline{e_\\phi}$$\r\n Using the same example $f(\\underline{x})=r^2/2\\implies \\nabla f=r\\underline{e_r}=\\underline{x}$.\r\n\\end{example}\r\n\\begin{note}\r\n We talked about the functions about position vectors under different coordinates,\r\n $$f(\\underline{x})=f(\\underline{x}(x,y,z))=f(\\underline{x}(r,\\theta,\\phi))$$\r\n So when we are talking about $f$, sometimes we are telling\r\n $$\\tilde{f}(x,y,z)=f(\\underline{x}(x,y,z)),\\tilde{\\tilde{f}}(r,\\theta,\\rho)=f(\\underline{x}(r,\\theta,\\rho))$$\r\n Or other coordinate we might find interesting.\r\n We are actually talking about a pullback here which might be clear in a couple of years' time.\r\n\\end{note}", "meta": {"hexsha": "2ef9594b63660ea163902fdc2812ea86bb16da16", "size": 10679, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/diff.tex", "max_stars_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_stars_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2/diff.tex", "max_issues_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_issues_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2/diff.tex", "max_forks_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_forks_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.7973856209, "max_line_length": 320, "alphanum_fraction": 0.6776851765, "num_tokens": 3444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872243177518, "lm_q2_score": 0.8221891327004132, "lm_q1q2_score": 0.6158913752787721}} {"text": "\\problemname{Limbo: Part 1}\n\n\\noindent Dominick Cobb and Arthur are ``extractors\" who perform corporate espionage. Using experimental military technology that gives them access to shared dream worlds, they infiltrate their targets' subconscious to extract valuable information. Cobb and his entourage are contacted by a mysterious Japanese syndicate and given a seemingly impossible task: instead of extracting information, do the opposite -- plant a new idea in the target's mind.\\\\\n\nCobb notes that when using dream-sharing technology, time slows down by different factors depending on whose dream is being entered. Their target is to convince Cepshun, the heir of an energy conglomerate, to dissolve his father's company. Before Cobb starts, he notes that it would be important to figure out the passage of time to ensure that the operation can be completed smoothly.\\\\\n\nIn Cepshun's dreams, there are multiple different layers of consciousness, each of which controls a different area of Cepshun's dreams and memories. Cobb needs to move between several dreams to reach his destination dream -- some of them belonging to Cepshun, and some of them belonging to his entourage. The factor by which time slows down on each level is best visualized by all of the positive integers arranged in an infinite triangle, where the first row has the single integer $1$, the second row has the next two integers, the third row has the next three, and so forth. The first $4$ levels are depicted below:\n\n\\begin{verbatim}\n 1\n 2 3\n 4 5 6\n 7 8 9 10\n etc...\n\\end{verbatim}\n\nThe initial time factor for reality is $1$. To enter the next dream level, Cobb can either descend left (into a dream by Cepshun) or right (into a dream by his entourage). For example, if he descends right, left, right, then he'll end up at dreams with time factors of $3$, then $5$, and finally $9$. Note the interesting property that for any pair of whole numbers $L$ and $R$, Cobb will always end up on the same time factor if he moves left $L$ times total and right $R$ times total, regardless of the order in which he makes the moves.\\\\\n\nCobb wants to know the slowness of time inside his target dream, which he can reach by descending left $L$ times and right $R$ times.\n\n\\section*{Input}\nThe first line of input consists of a single integer $T$ ($1 \\leq T \\leq 100$), the number of test cases.\\\\\n$T$ lines follow, each of which is a test case consisting of two space-separated integers, $L$ and $R$ ($0 \\leq L, R \\leq 10^{9}$), specifying the number of times Cobb needs to descend left and right, respectively.\n\n\\section*{Output}\nFor each test case, print, on a separate line, the factor by which time slows down in the target's dream at $(L, R)$.\\\\\nNote: the answer can be large, and may not necessarily fit in a 32-bit integer.\\\\\n", "meta": {"hexsha": "780c66ca9130ab47022da50658f8f1b903bfe36b", "size": 2816, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "limbo1/problem_statement/problem.en.tex", "max_stars_repo_name": "csecutsc/utscode2", "max_stars_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2018-09-30T15:06:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-09T06:50:25.000Z", "max_issues_repo_path": "limbo1/problem_statement/problem.en.tex", "max_issues_repo_name": "csecutsc/utscode2", "max_issues_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "limbo1/problem_statement/problem.en.tex", "max_forks_repo_name": "csecutsc/utscode2", "max_forks_repo_head_hexsha": "469367528cc5697e0c5c0ccee28420591335d899", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-06-17T04:10:51.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-17T04:10:51.000Z", "avg_line_length": 100.5714285714, "max_line_length": 618, "alphanum_fraction": 0.7631392045, "num_tokens": 680, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581626286833, "lm_q2_score": 0.8354835452961425, "lm_q1q2_score": 0.6158835151570027}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Written By Michael Brodskiy\n% Class: Analytic Geometry & Calculus III (Math-292)\n% Professor: V. Cherkassky\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass[12pt]{article} \n\\usepackage{alphalph}\n\\usepackage[utf8]{inputenc}\n\\usepackage[russian,english]{babel}\n\\usepackage{titling}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{enumitem}\n\\usepackage{amssymb}\n\\usepackage[super]{nth}\n\\usepackage{everysel}\n\\usepackage{ragged2e}\n\\usepackage{geometry}\n\\usepackage{fancyhdr}\n\\geometry{top=1.0in,bottom=1.0in,left=1.0in,right=1.0in}\n\\newcommand{\\subtitle}[1]{%\n \\posttitle{%\n \\par\\end{center}\n \\begin{center}\\large#1\\end{center}\n \\vskip0.5em}%\n\n}\n\\usepackage{hyperref}\n\\hypersetup{\ncolorlinks=true,\nlinkcolor=blue,\nfilecolor=magenta, \nurlcolor=blue,\ncitecolor=blue,\n}\n\n\\urlstyle{same}\n\n\n\\title{Lecture XI Notes}\n\\date{\\today}\n\\author{Michael Brodskiy\\\\ \\small Professor: V. Cherkassky}\n\n% Mathematical Operations:\n\n% Sum: $$\\sum_{n=a}^{b} f(x) $$\n% Integral: $$\\int_{lower}^{upper} f(x) dx$$\n% Limit: $$\\lim_{x\\to\\infty} f(x)$$\n\n\\begin{document}\n\n\\maketitle\n\n\\subsection{Second Order Partial Derivatives}\n\nAlthough the order of second partial differentiation usually does not matter, sometimes the second order partial derivatives may not be equal. Differentiating $\\frac{\\partial f}{\\partial x}$ by $\\frac{\\partial}{\\partial y}$ results in $\\frac{\\partial^2f}{\\partial y \\partial x}$. If one differentiates in the reversed order, this yields $\\frac{\\partial^2f}{\\partial x \\partial y}$, where the order of partial differentiation is read from right to left.\\\\\n\n\\textit{Example Where $\\frac{\\partial^2f}{\\partial y \\partial x}=\\frac{\\partial^2f}{\\partial x \\partial y}$:}\n\n$$f(x,y)=x^3+x^2y^3-2y^2$$\n$$\\frac{\\partial f}{\\partial x}=3x^2+2xy^3$$\n$$\\frac{\\partial f}{\\partial y}=3x^2y^2-4y$$\n$$\\frac{\\partial^2f}{\\partial y \\partial x}=6xy^2$$\n$$\\frac{\\partial^2f}{\\partial x \\partial y}=6xy^2$$\n\\begin{center}$\\therefore \\frac{\\partial^2f}{\\partial x \\partial y}=\\frac{\\partial^2f}{\\partial y \\partial x}$\\footnote{In most cases, $\\frac{\\partial^2f}{\\partial x \\partial y}=\\frac{\\partial^2f}{\\partial y \\partial x}$}\\end{center}\n\n\\subsection{Clairaut's Theorem}\n\nIf $\\frac{\\partial^2f}{\\partial x \\partial y}$ and $\\frac{\\partial^2f}{\\partial y \\partial x}$ are continuous at point $(a, b)$, then $\\frac{\\partial^2f}{\\partial x \\partial y}=\\frac{\\partial^2f}{\\partial y \\partial x}$ at $(a, b)$\\footnote{This rule applies to any nth order derivative}\n\n\\subsection{Laplace's Differential Equations}\n\nNamed after Pierre Laplace, the partial differential equation looks like this:\n\n$$\\frac{\\partial^2u}{\\partial x^2} + \\frac{\\partial^2u}{\\partial y^2}=0$$\n\nAny function that fits the criteria is called a \\bold{harmonic function}\n\n\\subsection{Wave Equations}\n\nAny wave, whose displacement is defined by the function $u(x,t)$ must satisfy the wave equation:\n\n$$\\frac{\\partial^2u}{\\partial t^2}=a^2\\frac{\\partial^2u}{\\partial x^2}$$\n\n\\section{Tangent Plane Approximation and Total Differentials $-$ 14.4}\n\nThe tangent plane approximation function is given by:\n\n$$z-z_o=f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)$$\n\nThe definition of a total differential is:\n\n$$dz = \\frac{\\partial z}{\\partial x}dx + \\frac{\\partial z}{\\partial y}dy$$\n\n$dz$ represents the change in the plane approximation\n\n\\end{document}\n", "meta": {"hexsha": "e206927213085a0d856e21ebb7893e90be5579af", "size": 3660, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture Notes/Lecture11.tex", "max_stars_repo_name": "MDBrodskiy/Vector_Calculus", "max_stars_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T15:51:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T15:51:52.000Z", "max_issues_repo_path": "Lecture Notes/Lecture11.tex", "max_issues_repo_name": "MDBrodskiy/Vector_Calculus", "max_issues_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture Notes/Lecture11.tex", "max_forks_repo_name": "MDBrodskiy/Vector_Calculus", "max_forks_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.9696969697, "max_line_length": 454, "alphanum_fraction": 0.643442623, "num_tokens": 1074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581626286834, "lm_q2_score": 0.8354835432479661, "lm_q1q2_score": 0.6158835136471729}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\\markright{renyi}\n\\section*{\\hspace*{-1.6cm} renyi}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nMeasure Renyi information.\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\nR = renyi(tfr)\nR = renyi(tfr,t)\nR = renyi(tfr,t,f)\nR = renyi(tfr,t,f,alpha)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty renyi} measures the Renyi information relative to a 2-D\n density function {\\ty tfr} (which can be eventually a\n time-frequency representation). Renyi information of order $\\alpha$\n is defined as\\,:\n\\begin{eqnarray*}\nR_x^{\\alpha} = \\frac{1}{1-\\alpha}\\\nlog_2\\left\\{\\int_{-\\infty}^{+\\infty}\\int_{-\\infty}^{+\\infty}\n\\mbox{tfr}_x^{\\alpha}(t,\\nu)\\ dt\\ d\\nu\\right\\}\n\\end{eqnarray*}\n\nThe result produced by this measure is expressed in {\\it bits} : if one\nelementary signal yields zero bit of information ($2^0$), then two well\nseparated elementary signals will yield one bit of information ($2^1$),\nfour well separated elementary signals will yield two bits of information\n($2^2$), and so on.\\\\\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8.5cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty tfr} & {\\ty (M,N)} 2-D density function (or mass function). Eventually\n {\\ty tfr} can be a time-frequency representation, in which case\n its first row must correspond to the lower frequencies\\\\\n {\\ty t} & abscissa vector parametrizing the {\\ty tfr} matrix. {\\ty t} can be a\n non-uniform sampled vector (eventually a time vector)\n & {\\ty (1:N)}\\\\ \n {\\ty f} & ordinate vector parametrizing the {\\ty tfr} matrix. {\\ty f} can be a\n non-uniform sampled vector (eventually a frequency vector)\n & {\\ty (1:M)}\\\\ \n {\\ty alpha} & rank of the Renyi measure & {\\ty 3}\\\\\n \\hline {\\ty R} & the alpha-rank Renyi measure (in bits if {\\ty tfr} is a time- \n frequency matrix).\\\\\n\\hline\n\\end{tabular*}\n\n\\end{minipage}\n\n\\newpage\n\n{\\bf \\large \\sf Examples}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\n\\begin{verbatim}\n s=atoms(64,[32,.25,16,1]); [tfr,t,f]=tfrsp(s); \n R1=renyi(tfr,t,f,3) \n ans =\n 0.9861\n\n s=atoms(64,[16,.2,16,1;48,.3,16,1]); [tfr,t,f]=tfrsp(s); \n R2=renyi(tfr,t,f,3) \n ans =\n 1.9890\n \n\\end{verbatim}\nWe can see that if {\\ttfamily R} is set to 0 for one elementary atom by\nsubtracting {\\ttfamily R1}, we obtain a result close to 1 bit for two atoms\n({\\ttfamily R2-R1}=1.0029).\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Reference}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n[1] W. Williams, M. Brown, A. Hero III, ``Uncertainty, information and\n time-frequency distributions'', SPIE Advanced Signal Processing\n Algorithms, Architectures and Implementations II, Vol. 1566,\n pp. 144-156, 1991.\n\\end{minipage}\n\n\n", "meta": {"hexsha": "cc288b20d8b0970a4f6745d46f049d251367e291", "size": 3339, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/renyi.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/renyi.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/renyi.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 30.3545454545, "max_line_length": 86, "alphanum_fraction": 0.616951183, "num_tokens": 1110, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835371034368, "lm_q2_score": 0.7371581510799252, "lm_q1q2_score": 0.6158834994688855}} {"text": "\n\\subsection{Chapter 2}\n\n\\begin{p}%{55}\n{Given a manifold $M$, define charts for $TM$ starting from charts $\\varphi_\\alpha:U_\\alpha\\rightarrow \\R^n$ as follows. Let $V_\\alpha$ be the subset of $TM$ given by $V_\\alpha=\\{v\\in TM:\\pi(v)\\in U_\\alpha\\}$. Show that every point in $TM$ lies in some set $V_\\alpha$. Define\nmaps $\\psi_\\alpha:V_\\alpha\\rightarrow \\R^n\\times \\R^n$ by $\\psi_\\alpha(v)=(\\varphi_\\alpha(\\pi(v)),(\\varphi_\\alpha)_*v)$, where we think of $(\\varphi_\\alpha)_*v$, which is really a tangent vector to $\\R^n$, as\na vector in $\\R^n$. Give $TM$ the topology in which open sets are the unions of sets of the form $O\\subset V_\\alpha$ such that $\\psi_\\alpha(O)\\subset \\R^n\\times \\R^n$ is open. Check that $\\psi_\\alpha$ are charts, so that $TM$ is a manifold. Check that $\\pi:TM\\rightarrow M$ is smooth.}\n\\end{p}\n{Every point $v$ in $TM$ is in some $V_\\alpha$ since $v$ is the tangent vector at some point $p$ (as given by $p=\\pi(v)$), and every point $p$ is in some set $U_\\alpha$. To establish that $\\psi_\\alpha$ are charts and $TM$ is a smooth manifold, note that $\\varphi_\\alpha$ is smooth and so is the pushforward, so $\\psi_\\alpha$ is smooth, too. Transition functions\nfrom one chart to another are likewise smooth, since these are compositions of smooth $\\psi_\\alpha$ and $\\psi^{-1}_\\beta$ which overlap on some open set. The projection $\\pi$ is smooth because we can define \nthe $\\R^{2n}\\rightarrow \\R^n$ version of the map by just ignoring the last $n$ coordinates. But this map is just $\\varphi_\\alpha\\circ\\pi\\circ\\psi_\\alpha^{-1}$, so $\\pi$ must be smooth.}\n\n\\begin{p}%{56}\n{Given bundles $\\pi:E\\rightarrow M$ and $\\pi':E'\\rightarrow M'$, show that the maps $\\psi:E\\rightarrow E'$ and $\\phi:M\\rightarrow M'$ are a bundle morphism iff $\\pi'\\circ\\psi=\\phi\\circ\\pi$. Show that\n$\\psi$ uniquely determines $\\phi$.}\n\\end{p}\n{Bundle morphisms respect the bundle structure, i.e.~the split into fiber and base. If $\\pi'\\circ\\psi=\\phi\\circ\\pi$, then the fiber in $E$ over $p$ must be mapped to the fiber in $E'$ over $p'=\\phi(p)$, or else the relationship would not hold. Conversely, if the fibers are mapped to the appropriate base\npoints, then it doesn't matter if we do the tangent space map first or second.}\n\n\\begin{p}%{57}\n{Check that $\\phi_*$ is smooth when we make the tangent bundle into a manifold as in the previous exercise.}\n\\end{p}\n{(typo on the ``previous'' exercise?) A map between two manifolds is smooth if it takes smooth functions on the first manifold to smooth functions on the second. Points in the manifold in \nquestion are vectors $v$, so functions of them are 1-forms $\\om$. Using an exact 1-form $df$ we have\n$df(v)=v(f)$. When $v=\\phi_*(u)$ we get $df(\\phi_*(u))=\\phi_*(u)(f)=u(\\phi^*f))=u(f\\circ\\phi)$, which is \nsmooth.}\n\n\\begin{p}%{58}\n{Show that if $\\phi:M\\rightarrow M'$ is a diffeomorphism, then $\\phi_*:TM\\rightarrow TM'$ is a bundle isomorphism.}\n\\end{p}\n{When $\\phi$ is an isomorphism, the tangent spaces $T_pM$ and $T_\\phi(p)M'$ are \nisomorphic, since the smooth functions on $M$ get mapped isomorphically to smooth functions on $M'$. Since\nthe pushforward is smooth and linear, it's an isomorphism of the tangent spaces.}\n\n\\begin{p}%{59}\n{Show that for any manifold $M$, the tangent bundle $\\pi:TM\\rightarrow M$ is locally trivial.}\n\\end{p}\n{The induced charts $\\psi_\\alpha$ give us a trivialization, with $\\R^n$ as the standard fiber.}\n\n\\begin{p}%{60}\n{Describe a bundle that is not locally trivial.}\n\\end{p}\n{}\n\n\\begin{p}%{61}\n{Check that the tangent bundle of a manifold is a vector bundle.}\n\\end{p}\n{To be a vector bundle, the \nlocal trivialization must be fiberwise linear. The standard fiber for a tangent bundle is $\\R^n$, and the \ntrivialization is linear because it is just the pushforward (exercise 17, part I).}\n\n\\begin{p}%{62}\n{A 1-dimensional bundle is called a (real or complex) line bundle. Check that the M\\\"obius strip is a real line bundle if we regard the standard fiber as being $\\R$.}\n\\end{p}\n{Locally the M\\\"obius strip is just $\\R^2$, so the trivialization is linear.}\n\n\\begin{p}%{63}\n{Show that if a vector bundle morphism is a diffeomorphism, its inverse is a vector bundle morphism.}\n\\end{p}\n{}\n\n\\begin{p}%{64}\n{Show that a (smooth (typo?)) section of the tangent bundle is a vector field.}\n\\end{p}\n{A section of the tangent bundle\nassigns to each point in the base space vector in its tangent space. From this we obtain a \nvector field from the pointwise action of the tangent vectors. The output of the vector field is again a \nsmooth function since the\nsection is smooth; the directional derivative (function value) changes smoothly from point to point since\nthe tangent vectors do.}\n\n\\begin{p}%{65}\n{Show that $\\Gamma(E)$ is a module over $C^\\infty(M)$.}\n\\end{p}\n{The first two conditions are defined \nin the text. The other two are simply: $(fg)s=f(gs)$ since $(fg)s:=((fg)s)(p)=(fg)(p)s(p)=f(p)g(p)s(p)=f(gs)(p)$ for\n$f,g\\in C^\\infty (M)$. }\n\n\\begin{p}%{66}\n{Show that every section of the M\\\"obius strip (viewed as a real line bundle over $S^1$) vanishes somewhere. Conclude that hte M\\\"obius strip hos no basis of sections, hence is not trivial.}\n\\end{p}\n{can't cut the M\\\"obius strip and make a cylinder unless there's a zero somewhere.}\n\n\\begin{p}%{67}\n{Show that the dual vector bundle really is a vector bundle. Also, show that given a basis\nof section $e_i$ of a vector bundle $E$, there is a unique dual basis $e^i$ of sections of $E^*$ such that for each point $p\\in M$, $e^i(p)$ is the basis of $E^*_p$ dual to the basis $e_i(p)$ of $E_p$.}\n\\end{p}\n{The basic idea, I think, is that since $E|_U$ is locally equivalent to \n$U\\times \\mathbb{R}^n$ for any open set $U$, then since $E^*$ just replaces \n$E_p$ with $E^*_p$ everywhere, \n$E^*|_U$ must be locally equivalent to $U\\times (\\mathbb{R}^n)^*=U\\times \\mathbb{R}^n$. \nThis makes $E^*$ a manifold. To see that it is a vector bundle, just use the canonical dual basis when \nmapping $E_p^*$ to $\\mathbb{R}^n$. \n}\n\n\\begin{p}%{68}\n{Show that if $s$ is a section of a vector bundle $E$ over $M$ and $\\lambda$ is a \nsection of $E^*$, there is a smooth function $\\lambda(s)$ on $M$ given by \n\\[\\lambda(s)(p)=\\lambda(p)(s(p))\\] for all $p\\in M$. Show that $\\lambda(s)$ depends $C^\\infty(M)$-linearly on $\\lambda$ and $s$.}\n\\end{p}\n{Locally, the vector bundles are trivial, so locally the section\nlooks like a function from $M$ to $V$ ($V^*$), where $E_p=\\{p\\}\\times V$. We can then \napply $\\lambda$ to $s$ pointwise to obtain the function. It is smooth since the sections are smooth. \n$\\lambda(s)$ is linear over $\\lambda$ and $s$ since the action of the cotangent vector on the tangent vector is linear in each argument. We get the $C^\\infty(M)$ part since the action is pointwise and smooth.}\n\n\\begin{p}%{69}\n{Show that a section of the cotangent bundle is the same as a 1-form.}\n\\end{p}\n{A 1-form is a\nmap taking vector fields to functions on the manifold. Interpreting vector fields as sections of\nthe tangent bundle, means that a 1-form is a map from sections of $TM$ to functions on $M$. By the\nprevious exercise we see that a section of $T^*M$ is just such a map, so a section of the cotangent\nbundle is a 1-form.}\n\n\\setcounter{p}{76}\n\n\\begin{p}%{77}\n{Show that the conditions $g_{\\alpha\\alpha}=1$ and $g_{\\alpha\\beta}g_{\\beta\\gamma}g_{\\gamma\\alpha}=1$ imply $g_{\\beta\\alpha}^{-1}=g_{\\alpha\\beta}$. Show that for any sequences $\\alpha_1,\\dots,\\alpha_n$ and $\\beta_1,\\dots,\\beta_m$ with $\\alpha_1=\\beta_1$ and $\\alpha_n=\\beta_m$, they \nimply $$g_{\\alpha_1\\alpha_2}\\cdots g_{\\alpha_{n-1}\\alpha_n}=g_{\\beta_1\\beta_2}\\cdots g_{\\beta_{m-1}\\beta_m}.$$\n}\n\\end{p}\n{Letting $\\gamma=\\alpha$ in the cocyle condition yields $g_{\\alpha\\beta}g_{\\beta\\alpha}g_{\\alpha\\alpha}=1$. Then using $g_{\\alpha\\alpha}=1$ we obtain\n$g_{\\alpha\\beta}g_{\\beta\\alpha}=1$, so $g_{\\beta\\alpha}^{-1}=g_{\\alpha\\beta}$. Now\nobserve that $g_{\\alpha_1\\alpha_{n-1}}g_{\\alpha_{n-1}\\alpha_n}=g_{\\alpha_1\\alpha_n}$ since\nmultiplying by the inverse of the righthandside gives $g_{\\alpha_1\\alpha_{n-1}}g_{\\alpha_{n-1}\\alpha_n}g_{\\alpha_n\\alpha_1}=1$. By induction we can expand this expression to\n$g_{\\alpha_1\\alpha_2}\\cdots g_{\\alpha_{n-1}\\alpha_n}$. Similarly, we can start with\n$g_{\\beta_1\\beta_{m-1}}g_{\\alpha_{m-1}\\beta_m}=g_{\\alpha_1\\alpha_n}$ and obtain $g_{\\beta_1\\beta_2}\\cdots g_{\\beta_{m-1}\\beta_m}$, so they must be equal.}\n\n\n\\newpage\n\\includepdf[pages=-]{src/gfkg223.pdf}\n\\includepdf[pages=-]{src/gfkg224.pdf}\n\\includepdf[pages=-]{src/gfkg225.pdf}\n\n\n\n\\setcounter{p}{85}\n\\begin{p}%86\n{Using a local trivialization of $E$ over $U_\\alpha\\subseteq M$ write the $G$-connection $D$ as the standard flat connection plus a vector potential: $D=D^0+A$. Show that the vector potential $A'$ for $D'$ is given in local coordinates by $$A'_\\mu=gA_\\mu g^{-1}+g\\partial\\mu g^{-1}.$$ Show that\nsince $A_\\mu$ lives in $\\mathfrak{g}$, so does $A_\\mu'$. (Hint: show that if $A_\\mu$ lives in $\\mathfrak{g}$ and $g\\in G$, then $g A_\\mu g^{-1}$ lives in $\\mathfrak{g}$. Also show that if \n$g\\in\\mathcal{G}$, $g\\partial_\\mu g^{-1}$ lives in $\\mathfrak{g}$.) Conclude that $D'$ is a $G$-connection.}\n\\end{p}\n{$D'_\\mu(s)=gD_\\mu(g^{-1}s)=gD_\\mu^0(g^{-1}s)+gA_\\mu(g^{-1}s)$. The local trivialization \nmeans that we have chosen a basis of sections $\\{e_j\\}$ such that $D_\\mu^0(s^je_j)=(\\partial_\\mu s^j)e_j$. The action of $g$ can be represented by $g=\\rho(g)^j_k e_j\\otimes e^k$ and the corresponding\nflat covariant derivative is just $D_\\mu^0(g)=(\\partial_\\mu \\rho(g)^j_k)e_j\\otimes e^k$. Inserting these expressions into $D'_\\mu(s)$ gives \n\\begin{align}\nD'_\\mu(s)&=g D_\\mu(g^{-1}s)=gD_\\mu^0(g^{-1}s)+gA_\\mu(g^{-1}s)\\\\\n&=g\\left(\\partial_\\mu \\rho(g^{-1})^k_j s^j\\right) e_k+gA_\\mu g^{-1}s\n=D_\\mu^0(s)+g\\left(\\partial_\\mu \\rho(g^{-1})^k_j \\right)s^j e_k+gA_\\mu g^{-1}s\\\\\n&=D_\\mu^0(s)+gD_\\mu^0(g^{-1})s+gA_\\mu g^{-1}s=D_\\mu^0(s)+(g\\partial_\\mu g^{-1})s+gA_\\mu g^{-1}s,\n\\end{align}\nwhere in the last equality we've abuse notation somewhat. Hence the vector potential has changed\naccording to the prescribed form. For $g\\in\\mathcal{G}$, $g\\partial_\\mu g^{-1}$ lives in $\\mathfrak{g}$ for the following reason. Really this expression is $\\rho(g)^j_\\ell \\partial_\\mu \\rho(g^{-1})^{\\ell}_k(x^\\nu)$ and $\\rho(g)(x^\\nu)=\\exp\\left(-c_j(x^\\nu)h^j\\right)$, where\n$h^j$ is a generator set for the Lie algebra $\\mathfrak{g}$. Thus, $g\\partial_\\mu g^{-1}=(\\partial_\\mu c_j(x^\\nu))g h^jg^{-1}$, which is\nan element of $\\mathfrak{g}$. \n }\n\n \\begin{p}%87\n \\end{p}\n\n \\newpage\n\n\\includepdf[pages=-]{src/gfkg226.pdf}\n\n", "meta": {"hexsha": "0a2f505ba8bf56d3f4423fe1efec05ad45973c96", "size": 10479, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/II2.tex", "max_stars_repo_name": "joerenes/Baez-Muniain-solutions", "max_stars_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-04-13T12:10:03.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-19T18:18:34.000Z", "max_issues_repo_path": "src/II2.tex", "max_issues_repo_name": "joerenes/Baez-Muniain-solutions", "max_issues_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-04-13T12:15:30.000Z", "max_issues_repo_issues_event_max_datetime": "2017-04-13T20:19:44.000Z", "max_forks_repo_path": "src/II2.tex", "max_forks_repo_name": "joerenes/Baez-Muniain-solutions", "max_forks_repo_head_hexsha": "e1e38de9acab877bc4200af59c7910d42de748ca", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.5090909091, "max_line_length": 361, "alphanum_fraction": 0.6949136368, "num_tokens": 3494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581510799253, "lm_q2_score": 0.8354835309589073, "lm_q1q2_score": 0.6158834949393956}} {"text": "\\documentclass[a4paper,12pt]{extarticle}\n\n\\usepackage[T1]{fontenc}\n\\usepackage[utf8]{inputenc}\n\\usepackage{graphicx}\n\\usepackage{xcolor}\n\n\\renewcommand\\familydefault{\\sfdefault}\n\\usepackage{tgheros}\n\\usepackage[defaultmono]{droidmono}\n\n\\usepackage{amsmath,amssymb,amsthm,textcomp}\n\\usepackage{enumerate}\n\\usepackage{multicol}\n\\usepackage{tikz}\n\n\\usepackage{geometry}\n\\geometry{left=25mm,right=25mm,%\nbindingoffset=0mm, top=20mm,bottom=20mm}\n\n\n\\linespread{1.3}\n\n\\newcommand{\\linia}{\\rule{\\linewidth}{0.5pt}}\n\n% custom theorems if needed\n\\newtheoremstyle{mytheor}\n {1ex}{1ex}{\\normalfont}{0pt}{\\scshape}{.}{1ex}\n {{\\thmname{#1 }}{\\thmnumber{#2}}{\\thmnote{ (#3)}}}\n\n\\theoremstyle{mytheor}\n\\newtheorem{defi}{Definition}\n\n% my own titles\n\\makeatletter\n\\renewcommand{\\maketitle}{\n\\begin{center}\n\\vspace{2ex}\n{\\huge \\textsc{\\@title}}\n\\vspace{1ex}\n\\\\\n\\linia\\\\\n\\@author \\hfill \\@date\n\\vspace{4ex}\n\\end{center}\n}\n\\makeatother\n%%%\n\n% custom footers and headers\n\\usepackage{fancyhdr}\n\\pagestyle{fancy}\n\\lhead{}\n\\chead{}\n\\rhead{}\n\\lfoot{Intro to Kalman Filter}\n\\cfoot{}\n\\rfoot{Page \\thepage}\n\\renewcommand{\\headrulewidth}{0pt}\n\\renewcommand{\\footrulewidth}{0pt}\n%\n\n% code listing settings\n\\usepackage{listings}\n\\lstset{\n language=Python,\n basicstyle=\\ttfamily\\small,\n aboveskip={1.0\\baselineskip},\n belowskip={1.0\\baselineskip},\n columns=fixed,\n extendedchars=true,\n breaklines=true,\n tabsize=4,\n prebreak=\\raisebox{0ex}[0ex][0ex]{\\ensuremath{\\hookleftarrow}},\n frame=lines,\n showtabs=false,\n showspaces=false,\n showstringspaces=false,\n keywordstyle=\\color[rgb]{0.627,0.126,0.941},\n commentstyle=\\color[rgb]{0.133,0.545,0.133},\n stringstyle=\\color[rgb]{01,0,0},\n numbers=left,\n numberstyle=\\small,\n stepnumber=1,\n numbersep=10pt,\n captionpos=t,\n mathescape=true,\n escapeinside={\\%*}{*)}\n}\n\n%%%----------%%%----------%%%----------%%%----------%%%\n\n\\begin{document}\n\n\\title{ Kalman Filter }\n\n\\author{ Dhruv Patel }\n\n\\maketitle\n\n\\section*{ What is Kalman Filter ? }\n\n\nKalman Filter or simply \\emph{KF}, is one of the many techniques to implement Bayes Filters. It was invented by Swerling (1958) and Kalman (1960) as a technique for filtering and prediction in \\emph{linear Gaussian Systems}. The Kalman Filter implements belief computation for continuous states. It is not applicable to discrete or hybrid state spaces.\n\n% Pseudo Code - Kalman Filter\n\n\\begin{lstlisting}\nAlgorithm Kalman Filter($\\mu_{t-1}, \\Sigma_{t-1}, u_t, z_t$):\n\t# Prediction Step : ref. as Motion Model\n\t$\\overline{\\mu}_t = A_t * \\mu_{t-1} + B_t * u_t$\n\t$\\overline{\\Sigma}_t = A_t * \\Sigma_{t-1} * A_t^T + R_t$\n\t# Update Step : ref. as Measurement Model\n\t$K_t = \\overline{\\Sigma}_t * C_t^T * (C_t * \\Sigma_t * C_t^T + Q_t)^{-1}$\n\t$\\mu_t = \\overline{\\mu}_t + K_t * (z_t - C_t * \\overline{\\mu}_t)$\n\t$\\Sigma_t = ( I - K_t * C_t ) * \\overline{\\Sigma}_t$\n\treturn $\\mu_t , \\Sigma_t$\n\t\n\\end{lstlisting}\n\n\nThe Kalman Filter represents beliefs by the moments parameterization : At time $t$, the belief $bel(x_t)$is represented by the mean $\\mu_t$ and the covariance $\\Sigma_t$. The input of the Kalman filter is the belief at time $t-1$, represented by $\\mu_{t-1}$ and $\\Sigma_{t-1}$. To update these parameters, Kalman filters require the control $u_t$ and the measurement $z_t$. The output is the belief at time $t$, represented by $\\mu_t$ and $\\Sigma_t$. This predicted belief $\\mu_t$ and $\\Sigma_t$ is calculated representing the belief $\\overline{bel}(x_t)$ one time step later, but before incorporating the measurement $z_t$. This belief is obtained by incorporating the control $u_t$. The update of the covariance considers the fact that states depend on previous states through the linear matrix $A_t$. This matrix is multiplied twice into the covariance, since the covariance is a quadratic matrix.\n\n\nThe belief $\\overline{bel}(x_t)$ is subsequently transformed into the desired belief $bel(x_t)$, by incorporating the measurement $z_t$. The variable $K_t$, is called \\emph{Kalman Gain}. It specifies the degree to which the measurement is incorporated into the new state estimate. The key concept here is the \\emph{innovation}, which is the difference between the actual measurement $z_t$ and the expected measurement $C_t$. Finally, the new covariance of the posterior belief is calculated, adjusting for the information gain resulting from the measurement.\n\n\nThe \\emph{Kalman Filter} is computationally quite efficient. In many applications - such as the robot mapping applications - the measurement space is much lower dimensional than the state space, and the update is dominated by the $O(n^2)$ operations.\n\n\n\\section*{ A brief explanation about the variables }\n\nThe state transition probability $ p( x_t | u_t, x_{t-1} ) $ must be a linear function in its arguments with added Gaussian noise. This is expressed by the following equation : \n\\begin{equation} \\label{eq1}\nx_t = A_t * x_{t-1} + B_t * u_t + \\varepsilon_t\n\\end{equation}\nHere, $x_t$ and $x_{t-1}$ are state vectors, and $u_t$ is the control vector at time $t$. These vectors are column vectors. They are of the form\n\\begin{equation}\nx_t = \\left( \\begin{array}{c} x_{1,t} \\\\ x_{2,t} \\\\ . \\\\ . \\\\ x_{n,t} \\end{array} \\right)\n\\mbox{~and~}\nu_t = \\left( \\begin{array}{c} u_{1,t} \\\\ u_{t,2} \\\\ . \\\\ . \\\\ u_{m,t} \\end{array} \\right)\n\\end{equation}\n$A_t$ and $B_t$ are matrices. $A_t$ is a square matrix of size $ n \\times n $, where $n$ is the dimension of the state vector $x_t$. $B_t$ is of size $ n \\times m$, with $m$ being the dimension of the control vector $u_t$. By multiplying the state and control vector with the matrices $A_t$ and $B_t$, respectively, the state transition function becomes \\emph{linear} in its arguments. Thus, Kalman Filters assume linear system dynamics.\n\nThe random variable $\\varepsilon_t$ in (1) is a Gaussian random vector that models the uncertainty introduced by the state transition. It is of the same dimension as the state vector. It has zero mean, and its covariance will be denoted by $R_t$. A state transition probability of the form (1) is called a \\emph{linear Gaussian}, to reflect the fact that it is linear in its arguments with additive Gaussian noise. \n\nThe measurement probability $ p(z_t | x_t ) $ must also be \\emph{linear} in its arguments, with added Gaussian noise :\n\\begin{equation}\nz_t = C_t * x_t + \\delta_t\n\\end{equation}\nHere, $C_t$ is a matrix of size $k \\times n$, where $k$ is the dimension of the measurement vector $z_t$. The vector $\\delta_t$ describes the measurement noise. The distribution of $\\delta_t$ is a multivariate Gaussian with zero mean and covariance $Q_t$.\n\n\n\\vfill\n\n\\section*{ Given Problem }\n\nHere, we have been given a simple target tracking problem in one-dimensional space. The state contains three components : position (one-dimensional), velocity and acceleration and it can be expressed as below.\n\\begin{equation}\nx = \\left( \\begin{array}{c} x_k \\\\ \\dot{x}_k \\\\ \\ddot{x}_k \\end{array} \\right)\n\\end{equation}\nwhere, $x_k$ is the position, $\\dot{x}_k$ is the velocity and $\\ddot{x}_k$ is the acceleration at time '$k$'. $T$ is the size of sample time step.\n\nThe process equation for target motion is given by the following kinematic equation:\n\\begin{equation}\n\\left[ \\begin{array}{c} x_k \\\\ \\dot{x}_k \\\\ \\ddot{x}_k \\end{array} \\right] = \n\\left[ \\begin{array}{ccc} 1 & T & \\frac1{2}T^2 \\\\ 0 & 1 & T \\\\ 0 & 0 & 1 \\end{array} \\right] * \\left[ \\begin{array}{c} x_{k-1} \\\\ \\dot{x}{k-1} \\\\ \\ddot{x}{k-1} \\end{array} \\right] + \\upsilon_{k-1}\n\\end{equation}\nand \n\\begin{equation}\n\\upsilon_{k-1} \\sim N(0, Q)\n\\end{equation}\nwhere,\n$ Q = \\sigma^2 * \\left[ \\begin{array}{ccc} \\frac{T^4}{4} & \\frac{T^3}{2} & \\frac{T^2}{2} \\\\ \\frac{T^3}{2} & 2T^3 & T^2 \\\\ \\frac{T^2}{2} & T^2 & T^2 \\end{array} \\right] $, $\\sigma$ represents intensity of Gaussian noise.\n\nTargets are usually tracked with the help of sensors such as radars or lidars which provide the position of the target. Hence, the measurement equation can be written as:\n\\begin{equation}\ny_k = \n\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\end{array} \\right] * \\left[ \\begin{array}{c} x_k \\\\ \\dot{x}_k \\\\ \\ddot{x}_k \\end{array} \\right] + \\omega\n\\end{equation}\nwhere, $y_k$ is the measurement and $\\omega \\sim N(0, R) $ is the measurement noise. \n\n\\hfill\n\nFor the given problem, obtain the estimate of its state over a period of $ 20 ~sec $. Assume time step $ T = 0.1 ~sec $. \n\n\n\n\n\n\n\n\n\n\n\n\n\\end{document}", "meta": {"hexsha": "016a2c8b37563f7f5f40c6b2ee814006bbfedc65", "size": 8432, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "LaTeX/Kalman_Filter.tex", "max_stars_repo_name": "ruvate/kalman.filter", "max_stars_repo_head_hexsha": "37ee3a45308c8de89e040ad2130fe737cdcaa327", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "LaTeX/Kalman_Filter.tex", "max_issues_repo_name": "ruvate/kalman.filter", "max_issues_repo_head_hexsha": "37ee3a45308c8de89e040ad2130fe737cdcaa327", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "LaTeX/Kalman_Filter.tex", "max_forks_repo_name": "ruvate/kalman.filter", "max_forks_repo_head_hexsha": "37ee3a45308c8de89e040ad2130fe737cdcaa327", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-10-29T02:51:03.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-29T02:51:03.000Z", "avg_line_length": 42.5858585859, "max_line_length": 900, "alphanum_fraction": 0.7056451613, "num_tokens": 2587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920116079209, "lm_q2_score": 0.7690802423634961, "lm_q1q2_score": 0.6157964063459351}} {"text": "\\chapter{Fourier Transform}\n\\begin{enumerate}\n\t\\item Let $F(k)$ is the Fourier exponential transform of $f(x)$ and $G(k)$ be the Fourier Transform of $g(x)=f(x+a)$. Then $G(k)$ is given by\n\t(Use the Fourier integral $F(k)=\\frac{1}{\\sqrt{2 \\pi}} \\int_{-\\infty}^{+\\infty} f(x) e^{i k x} d x$ )\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$e^{i a k} f(k)$\n\t\t\\task[\\textbf{b.}]$e^{-i a k} F(-k)$\n\t\t\\task[\\textbf{c.}] $e^{-i a k} F(k)$\n\t\t\\task[\\textbf{d.}] $e^{-i a k} f(k)$\n\t\\end{tasks}\n\t\\item The value of a function is given by $f(x)=\\left\\{\\begin{array}{cc}1 & 02\\end{array}\\right.$, the Fourier cosine transform of $f(x)$ is\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\sqrt{\\frac{2}{\\pi}}\\left(\\frac{2 \\sin \\omega+\\sin 2 \\omega}{\\omega}\\right)$\n\t\t\\task[\\textbf{b.}]$\\sqrt{\\frac{2}{\\pi}}\\left(\\frac{2 \\sin \\omega-\\sin 2 \\omega}{\\omega}\\right)$\n\t\t\\task[\\textbf{c.}]$\\sqrt{\\frac{2}{\\pi}}\\left(\\frac{\\sin \\omega-\\sin 2 \\omega}{\\omega}\\right)$\n\t\t\\task[\\textbf{d.}] $\\sqrt{\\frac{2}{\\pi}}\\left(\\frac{2 \\sin \\omega}{\\omega}\\right)$\n\t\\end{tasks}\n\t\\item Consider the function $\\delta_{n}(x)=\\frac{n}{\\sqrt{\\pi}} \\exp \\left(-n^{2} x^{2}\\right)$. For $n \\rightarrow \\infty$ The Fourier transform of the function is given by\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\delta(x)=\\int_{-\\infty}^{+\\infty} e^{-i k x} d k$\n\t\t\\task[\\textbf{b.}]$\\delta(x)=\\frac{1}{2 \\pi} \\int_{-\\infty}^{+\\infty} e^{i k x} d k$\n\t\t\\task[\\textbf{c.}]$\\delta(x)=\\frac{1}{2 \\pi} \\int_{-\\infty}^{+\\infty} e^{-i k x} d k$\n\t\t\\task[\\textbf{d.}] 0\n\t\\end{tasks}\n\t\\item For the function $f(t)=\\delta(t-x)$, the Fourier cosine integral is defined as $g_{c}(\\omega)=\\sqrt{\\frac{2}{\\pi}} \\int_{0}^{\\infty} f(t) \\cos \\omega t d t$. Its Fourier transform is:\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}] $\\sqrt{\\frac{2}{\\pi}} \\cos \\omega x$\n\t\t\\task[\\textbf{b.}]$\\sqrt{\\frac{2}{\\pi}}$\n\t\t\\task[\\textbf{c.}]$\\sqrt{\\frac{2}{\\pi}} \\sin \\omega x$\n\t\t\\task[\\textbf{d.}] $i \\sqrt{\\frac{2}{\\pi}}$\n\t\\end{tasks}\n\t\\item Inverse cosine transform of $g_{c}(\\omega)=\\sqrt{\\frac{2}{\\pi}} \\cos \\omega x$ is:\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$\\delta(t+x)$\n\t\t\\task[\\textbf{b.}]$\\delta(t-x)$\n\t\t\\task[\\textbf{c.}]$\\delta(t-x)(t+x)$\n\t\t\\task[\\textbf{d.}] $\\delta^{2}(x-t)(x+t)$\n\t\\end{tasks}\n\t\\item $f(t)=\\frac{\\hbar}{2 \\pi i} \\int_{-\\infty}^{+\\infty} \\frac{e^{-i \\omega t} d \\omega}{\\left(E_{0}-i \\Gamma / 2-\\hbar \\omega\\right)}$ The value of $f(t)$ is given by\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}](a) $f(t)= \\begin{cases}e^{-\\Gamma t / 2 \\hbar} e^{-i E_{0} t / \\hbar} & , t>0 \\\\ 0 & , t<0\\end{cases}$\n\t\t\\task[\\textbf{b.}]$f(t)= \\begin{cases}e^{\\sqrt{1 / 2 \\hbar}} e^{-1 E_{0} t / \\hbar} & , t>0 \\\\ 0 & , t<0\\end{cases}$\n\t\t\\task[\\textbf{c.}]$f(t)= \\begin{cases}e^{-\\Gamma t / 2 \\hbar} e^{-i E_{0} t / \\hbar} & , t>0 \\\\ 0 & , t<0\\end{cases}$\n\t\t\\task[\\textbf{d.}] $f(t)= \\begin{cases}e^{-\\Gamma t / 2 h} e^{-i E_{0} t / h} & , t>0 \\\\ e^{-i E_{0} t / \\hbar} & , t<0\\end{cases}$\n\t\\end{tasks}\n\t\\item The Fourier transform of function $h(t)=t e^{-t^{2}}$ is \n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}] $j \\pi f e^{-\\pi^{2} f^{2}}$\n\t\t\\task[\\textbf{b.}]$-j \\pi f e^{-\\pi^{2} f^{2}}$\n\t\t\\task[\\textbf{c.}]$j \\pi f e^{-\\pi^{2} f^{2} / 4}$\n\t\t\\task[\\textbf{d.}] $-j \\pi f e^{-\\pi^{2} f^{2} / 4}$\n\t\\end{tasks}\n\t\\item The graph of a real periodic function $f(x)$ for the range $[-\\infty, \\infty]$ is shown below\n\t\\begin{figure}[H]\n\t\t\\centering\n\t\t\\includegraphics[height=3cm,width=5cm]{FT-Assignment-05}\n\t\\end{figure}\n\tWhich of the following graphs represents the real part of its Fourier transform?\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]\t\n\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.2cm,width=3cm]{FT-Assignment-01}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{b.}]\n\t\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.2cm,width=3cm]{FT-Assignment-02}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{c.}]\n\t\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.2cm,width=3cm]{FT-Assignment-03}\n\t\t\\end{figure}\n\t\t\\task[\\textbf{d.}] \n\t\t\t\\begin{figure}[H]\n\t\t\t\\centering\n\t\t\t\\includegraphics[height=2.2cm,width=3cm]{FT-Assignment-04}\n\t\t\\end{figure}\n\t\\end{tasks}\n\t\\item The Fourier transform of the function $h(t)=\\left\\{\\begin{array}{cc}\\beta e^{-a t}, & t>0 \\\\ 0, & t<0\\end{array}\\right.$, is given by\n\t \\begin{tasks}(2)\n\t\t\\task[\\textbf{a.}]$H(f)=\\frac{\\alpha}{\\sqrt{\\alpha^{2}+(2 \\pi f)^{2}}} e^{j \\tan ^{-1}\\left[-2 \\pi f^{\\prime \\alpha]}\\right.}$\n\t\t\\task[\\textbf{b.}]$H(f)=\\frac{\\beta}{\\sqrt{\\alpha^{2}+(2 \\pi f)^{2}}} e^{j \\tan ^{-1}[-2 \\pi f / \\beta]}$\n\t\t\\task[\\textbf{c.}]$H(f)=\\frac{\\beta}{\\sqrt{\\alpha^{2}+(2 \\pi f)^{2}}} e^{\\tan ^{-1}[2 \\pi f / \\alpha]}$\n\t\t\\task[\\textbf{d.}] $H(f)=\\frac{\\beta}{\\sqrt{\\alpha^{2}+(2 \\pi f)^{2}}} e^{j \\tan ^{-1}[-2 \\pi f / \\alpha]}$\n\t\\end{tasks}\n\t\\item The Fourier transform of $f(x)=\\left\\{\\begin{array}{cl}-1 & -1 1$ can be written as the product of primes (easy).\n\t\\ii Show that if $p_1 \\dots p_m = q_1 \\dots q_n$ for some primes (not necessarily unique),\n\tthen $p_1 = q_i$ for some $i$, say $q_1$.\n\t\\ii Divide both sides by $p_1$ and use induction.\n\\end{enumerate}\nWhat happens if we try to repeat the proof here?\nWe get step 1 for free, because we're using a better definition of ``prime''.\nWe can also do step 3, since it follows from step 1.\nBut step 2 doesn't work,\nbecause for abstract Dedekind domains\nwe don't really have a notion of size.\nAnd step 4 doesn't work because we don't yet have a\nnotion of what the inverse of a prime ideal is.\n\nWell, it turns out that we \\emph{can} define the inverse $\\ka\\inv$ of an ideal,\nand I'll do so by the end of this chapter.\nYou then need to check that $\\ka \\cdot \\ka\\inv = (1) = \\mathcal A$.\nIn fact, even this isn't easy.\nYou have to check it's true for prime ideals $\\kp$,\n\\emph{then} prove prime factorization,\nand then prove that this is true.\nMoreover, $\\ka\\inv$ is not actually an ideal, so you need to\nwork in the field of fractions $K$ instead of $\\mathcal A$.\n\nSo the main steps in the new situation are as follows:\n\\begin{enumerate}[(1)]\n\t\\ii First, show that every ideal $\\ka$ divides $\\kp_1 \\dots \\kp_g$\n\tfor some finite collection of primes.\n\t(This is an application of Zorn's Lemma.)\n\t\\ii Define $\\kp\\inv$ and show that $\\kp \\kp\\inv = (1)$.\n\t\\ii Show that a factorization exists (again using Zorn's Lemma).\n\t\\ii Show that it's unique, using the new inverse we've defined.\n\\end{enumerate}\n\nFinally, let me comment on how nice this is if $\\mathcal A$ is a PID (like $\\ZZ$).\nThus every element $a \\in \\mathcal A$ is in direct correspondence with an ideal $(a)$.\nNow suppose $(a)$ factors as a product of ideals $\\kp_i = (p_i)$, say,\n\\[ (a) = (p_1)^{e_1} (p_2)^{e_2} \\dots (p_n)^{e_n} . \\]\nThis verbatim reads \\[ a = u p_1^{e_1} p_2^{e_2} \\dots p_n^{e_n} \\]\nwhere $u$ is some unit (recall \\Cref{def:unit}).\nHence, Dedekind domains which are PID's satisfy unique factorization\nfor \\emph{elements}, just like in $\\ZZ$.\n(In fact, the converse of this is true.)\n\n\\section{The factoring algorithm}\nLet's look at some examples from quadratic fields.\nRecall that if $K = \\QQ(\\sqrt{d})$, then\n\\[\n\t\\OO_K =\n\t\\begin{cases}\n\t\t\\ZZ[\\sqrt d] & d \\equiv 2,3 \\pmod 4 \\\\\n\t\t\\ZZ\\left[ \\frac{1+\\sqrt d}{2} \\right] & d \\equiv 1 \\pmod 4.\n\t\\end{cases}\n\\]\nAlso, recall that the norm of $a+b\\sqrt{-d}$ is given by $a^2+db^2$.\n\n%In what follows, we are going to often use the trick that\n%\\[ \\ZZ[\\alpha] \\cong \\ZZ[x] / (f) \\]\n%where $f$ is the minimal polynomial of $\\alpha$.\n\n\\begin{example}[Factoring $6$ in the integers of $\\QQ(\\sqrt{-5})$]\n\tLet $\\OO_K = \\ZZ[\\sqrt{-5}]$ arise from $K = \\QQ(\\sqrt{-5})$.\n\tWe've already seen that\n\t\\[ (6) = (2) \\cdot (3) = \\left( 1+\\sqrt{-5} \\right)\\left( 1-\\sqrt{-5} \\right) \\]\n\tand you can't get any further with these principal ideals.\n\tBut let\n\t\\[ \\kp = \\left( 1+\\sqrt{-5}, 2 \\right) = \\left( 1-\\sqrt{-5}, 2 \\right)\n\t\t\\quad\\text{and}\\quad \\kq_1 = (1+\\sqrt{-5},3),\n\t\t\\; \\kq_2 = (1-\\sqrt{-5},3). \\]\n\tThen it turns out $(6) = \\kp^2\\kq_1\\kq_2$.\n\tMore specifically, $(2) = \\kp^2$, $(3) = \\kq_1\\kq_2$,\n\tand $(1+\\sqrt{-5}) = \\kp\\kq_1$ and $(1-\\sqrt{-5}) = \\kp\\kq_2$.\n\t(Proof in just a moment.)\n\\end{example}\nI want to stress that all our ideals are computed relative to $\\OO_K$.\nSo for example, \\[ (2) = \\left\\{ 2x \\mid x \\in \\OO_K \\right\\}. \\]\n\nHow do we know in this example that $\\kp$ is prime/maximal?\n(Again, these are the same since we're in a Dedekind domain.)\nAnswer: look at $\\OO_K / \\kp$ and see if it's a field.\nThere is a trick to this: we can express\n\\[ \\OO_K = \\ZZ[\\sqrt{-5}] \\cong \\ZZ[x] / (x^2+5). \\]\n%\\begin{ques}\n%\tConvince yourself this is true.\n%\t(More generally, if $\\theta \\in \\ol\\ZZ$ has minimal polynomial $p$,\n%\tthen $\\ZZ[\\theta] \\cong \\ZZ[x] / (p)$).\n%\\end{ques}\nSo when we take \\emph{that} mod $\\kp$, we get that\n\\[ \\OO_K / \\kp = \\ZZ[x] / (x^2+5, 2, 1+x) \\cong \\FF_2[x] / (x^2+5,x+1) \\]\nas rings.\n\\begin{ques}\n\tConclude that $\\OO_K / \\kp \\cong \\mathbb F_2$,\n\tand satisfy yourself that $\\kq_1$ and $\\kq_2$ are also maximal.\n\\end{ques}\nI should give an explicit example of an ideal multiplication: let's compute\n\\begin{align*}\n\t\\kq_1\\kq_2 &= \\left( (1+\\sqrt{-5})(1-\\sqrt{-5}), 3(1+\\sqrt{-5}), 3(1-\\sqrt{-5}), 9 \\right) \\\\\n\t&= \\left( 6, 3+3\\sqrt{-5}, 3-3\\sqrt{-5}, 9 \\right) \\\\\n\t&= \\left( 6, 3+3\\sqrt{-5}, 3-3\\sqrt{-5}, 3 \\right) \\\\\n\t&= (3)\n\\end{align*}\nwhere we first did $9-6=3$ (think Euclidean algorithm!),\nthen noted that all the other generators don't contribute\nanything we don't already have with the $3$\n(again these are ideals computed in $\\OO_K$).\nYou can do the computation for $\\kp^2$, $\\kp\\kq_1$, $\\kp\\kq_2$ in the same way.\n\nFinally, it's worth pointing out that we should quickly\nverify that $\\kp \\neq (x)$ for some $x$;\nin other words, that $\\kp$ is not principal.\nAssume for contradiction that it is.\nThen $x$ divides both $1+\\sqrt{-5}$ and $2$, in the sense\nthat $1+\\sqrt{-5} = \\alpha_1 x$ and $2 = \\alpha_2 x$\nfor some $\\alpha_1, \\alpha_2 \\in \\OO_K$.\n(Principal ideals are exactly the ``multiples'' of $x$, so $(x) = x \\OO_K$.)\nTaking the norms, we find that $\\Norm_{K/\\QQ}(x)$ divides both \n\\[ \\Norm_{K/\\QQ}(1+\\sqrt{-5}) = 6 \\quad\\text{and}\\quad \\Norm_{K/\\QQ}(2) = 4. \\]\nSince $\\kp \\neq (1)$, $x$ cannot be a unit, so its norm must be $2$.\nBut there are no elements of norm $2 = a^2+5b^2$ in $\\OO_K$.\n\n\\begin{example}[Factoring $3$ in the integers of $\\QQ(\\sqrt{-17})$]\n\tLet $\\OO_K = \\ZZ[\\sqrt{-17}]$ arise from $K = \\QQ(\\sqrt{-17})$.\n\tWe know $\\OO_K \\cong \\ZZ[x] / (x^2+17)$.\n\tNow\n\t\\[\n\t\t\\OO_K / 3\\OO_K \\cong \\ZZ[x] / (3,x^2+17)\n\t\t\\cong \\FF_3[x] / (x^2-1).\n\t\\]\n\tThis already shows that $(3)$ cannot be a prime (i.e.\\ maximal) ideal,\n\tsince otherwise our result should be a field.\n\tAnyways, we have a projection\n\t\\[ \\OO_K \\surjto \\FF_3[x] / \\left( (x-1)(x+1) \\right). \\]\n\tLet $\\kq_1$ be the pre-image of $(x-1)$ in the image, that is,\n\t\\[ \\kq_1 = (3, \\sqrt{-17}-1). \\]\n\tSimilarly, \\[ \\kq_2 = (3, \\sqrt{-17}+1). \\]\n\tWe have $\\OO_K / \\kq_1 \\cong \\FF_3$, so $\\kq_1$ is maximal (prime).\n\tSimilarly $\\kq_2$ is prime.\n\tMagically, you can check explicitly that\n\t\\[ \\kq_1 \\kq_2 = (3). \\]\n\tHence this is the factorization of $(3)$ into prime ideals.\n\\end{example}\n\nThe fact that $\\kq_1 \\kq_2 = (3)$ looks magical, but it's really true:\n\\begin{align*}\n\t\\kq_1\\kq_2\n\t&= (3, \\sqrt{-17}-1) (3, \\sqrt{-17}+1) \\\\\n\t&= (9, 3\\sqrt{-17}+3, 3\\sqrt{-17}-3, 18) \\\\\n\t&= (9, 3\\sqrt{-17}+3, 6) \\\\\n\t&= (3, 3\\sqrt{-17}+3, 6) \\\\\n\t&= (3).\n\\end{align*}\nIn fact, it turns out this always works in general:\ngiven a rational prime $p$, there is an algorithm\nto factor $p$ in any $\\OO_K$ of the form $\\ZZ[\\theta]$.\n\n\\begin{theorem}[Factoring algorithm / Dedekind-Kummer theorem]\n\t\\label{thm:factor_alg}\n\tLet $K$ be a number field.\n\tLet $\\theta \\in \\OO_K$ with $[\\OO_K : \\ZZ[\\theta]] = j < \\infty$,\n\tand let $p$ be a prime not dividing $j$.\n\tThen $(p) = p \\OO_K$ is factored as follows:\n\t\\begin{quote}\n\t\tLet $f$ be the minimal polynomial of $\\theta$ and\n\t\tfactor $\\ol f$ mod $p$ as\n\t\t\\[ \\ol f \\equiv \\prod_{i=1}^g (\\ol f_i)^{e_i} \\pmod p. \\]\n\t\tThen $\\kp_i = (f_i(\\theta), p)$ is prime for each $i$\n\t\tand the factorization of $(p)$ is\n\t\t\\[ \\OO_K \\supseteq (p) = \\prod_{i=1}^g \\kp_i^{e_i}. \\]\n\t\\end{quote}\n\tIn particular, if $K$ is monogenic with $\\OO_K = \\ZZ[\\theta]$ then $j=1$\n\tand the theorem applies for all primes $p$.\n\\end{theorem}\nIn almost all our applications in this book, $K$ will be monogenic; i.e.\\ $j=1$.\nHere $\\ol \\psi$ denotes the image in $\\FF_p[x]$ of a polynomial $\\psi \\in \\ZZ[x]$.\n\n\\begin{ques}\n\tThere are many possible pre-images $f_i$ we could have chosen\n\t(for example if $\\ol{f_i} = x^2+1 \\pmod 3$, we could pick $f_i = x^2 + 3x + 7$.)\n\tWhy does this not affect the value of $\\kp_i$?\n\\end{ques}\n\nNote that earlier, we could check the factorization worked\nfor any particular case. \nThe proof that this works is much the same, but we need one extra tool, the ideal norm.\nAfter that we leave the proposition as \\Cref{prob:prove_factoring_algorithm}.\n\nThis algorithm gives us a concrete way to compute prime factorizations of $(p)$\nin any monogenic number field with $\\OO_K = \\ZZ[\\theta]$. To summarize the recipe:\n\\begin{enumerate}\n\t\\ii Find the minimal polynomial of $\\theta$, say $f \\in \\ZZ[x]$.\n\t\\ii Factor $f$ mod $p$ into irreducible polynomials\n\t${\\ol f_1}^{e_1} {\\ol f_2}^{e_2} \\dots {\\ol f_g}^{e_g}$.\n\t\\ii Compute $\\kp_i = (f_i(\\theta), p)$ for each $i$.\n\\end{enumerate}\nThen your $(p) = \\kp_1^{e_1} \\dots \\kp_g^{e_g}$.\n\n\\begin{exercise}\n\tFactor $(29)$ in $\\QQ(i)$ using the above algorithm.\n\\end{exercise}\n\n%\\begin{remark}\n%\tWhat if $K$ isn't monogenic?\n%\tIt turns out that we can still apply the\n%\tfactoring algorithm to ``almost all primes'' as follows.\n%\tSuppose $K$ is a number field and $\\alpha \\in \\OO_K$ such that\n%\t$[\\OO_K : \\ZZ[\\alpha]] = j < \\infty$.\n%\tThen as long as $p \\nmid j$, we can apply the above algorithm\n%\twith $f$ the minimal polynomial of $\\alpha$.\n%\tThe formulation we presented above was the special case where $j=1$.\n%\\end{remark}\n\n\\section{Fractional ideals}\n\\prototype{Analog to $\\QQ$ for $\\ZZ$, allowing us to take inverses of ideals.\nPrime factorization works in the nicest way possible.}\nWe now have a neat theory of factoring ideals of $\\mathcal A$,\njust like factoring the integers.\nNow note that our factorization of $\\ZZ$ naturally gives a way to factor\nelements of $\\QQ$; just factor the numerator and denominator separately.\n\nLet's make the analogy clearer.\nThe analogue of a rational number is as follows.\n\n\\begin{definition}\n\tLet $\\mathcal A$ be a Dedekind domain with field of fractions $K$.\n\tA \\vocab{fractional ideal} $J$ of $K$ is a set of the form \n\t\\[ J = \\frac{1}{x} \\cdot \\ka \\quad \\text{where $x \\in \\mathcal A$, and $\\ka$ is an integral ideal.} \\]\n\tFor emphasis, ideals of $\\mathcal A$ will be sometimes referred to as \\vocab{integral ideals}.\n\\end{definition}\n\nYou might be a little surprised by this definition:\none would expect that a fractional ideal should be of the form $\\frac{\\ka}{\\kb}$\nfor some integral ideals $\\ka$, $\\kb$.\nBut in fact, it suffices to just take $x \\in \\mathcal A$ in the denominator.\nThe analogy is that when we looked at $\\OO_K$, we found that we only needed\ninteger denominators: $\\frac{1}{4-\\sqrt3} = \\frac{1}{13}(4+\\sqrt3)$.\nSimilarly here, it will turn out that we only need to look at $\\frac1x \\cdot \\ka$\nrather than $\\frac{\\ka}{\\kb}$, and so we define it this way from the beginning.\nSee \\Cref{prob:fractional_ideal_alt_def} for a different equivalent definition.\n\n\\begin{example}[$\\frac52\\ZZ$ is a fractional ideal]\n\tThe set \\[ \\frac52 \\ZZ = \\left\\{ \\frac52n \\mid n \\in \\ZZ \\right\\} = \\half (5) \\]\n\tis a fractional ideal of $\\ZZ$.\n\\end{example}\n\nNow, as we prescribed, the fractional ideals form a multiplicative group:\n\\begin{theorem}[Fractional ideals form a group]\n\tLet $\\mathcal A$ be a Dedekind domain and $K$ its field of fractions.\n\tFor any integral ideal $\\ka$, the set\n\t\\[ \\ka\\inv = \\left\\{ x \\in K\n\t\t\t\\mid x\\ka \\subseteq (1) = \\mathcal A \\right\\} \\]\n\tis a fractional ideal with $\\ka \\ka\\inv = (1)$.\n\\end{theorem}\n\\begin{definition}\n\tThus nonzero fractional ideals of $K$ form a group under multiplication\n\twith identity $(1) = \\mathcal A$.\n\tThis \\vocab{ideal group} is denoted $J_K$.\n\\end{definition}\n\n\\begin{example}[$(3)\\inv$ in $\\ZZ$]\n\tPlease check that in $\\ZZ$ we have\n\t\\[ (3)\\inv = \\left\\{ \\frac 13 n \\mid n \\in \\ZZ \\right\\} = \\frac 13 \\ZZ. \\]\n\\end{example}\n\nIt follows that every fractional ideal $J$ can be uniquely written as\n\\[ J = \\prod_i \\kp_i^{n_i} \\cdot \\prod \\kq_i^{-m_i} \\]\nwhere $n_i$ and $m_i$ are positive integers.\nIn fact, $\\ka$ is an integral ideal if and only if all its exponents are nonnegative,\njust like the case with integers.\nSo, a perhaps better way to think about fractional ideals is\nas products of prime ideals, possibly with negative exponents.\n\n\\section{The ideal norm}\nOne last tool is the ideal norm,\nwhich gives us a notion of the ``size'' of an ideal.\n\\begin{definition}\n\tThe \\vocab{ideal norm} (or absolute norm)\n\tof a nonzero ideal $\\ka \\subseteq \\OO_K$ is defined as\n\t$\\left\\lvert \\OO_K / \\ka \\right\\rvert$ and denoted $\\Norm(\\ka)$.\n\\end{definition}\n\\begin{example}[Ideal norm of $(5)$ in the Gaussian integers]\n\tLet $K = \\QQ(i)$, $\\OO_K = \\ZZ[i]$.\n\tConsider the ideal $(5)$ in $\\OO_K$.\n\tWe have that \n\t\\[ \\OO_K / (5) \\cong \\{ a+bi \\mid a,b \\in \\Zc5 \\} \\]\n\tso $(5)$ has ideal norm $25$,\n\tcorresponding to the fact that $\\OO_K/(5)$ has $5^2=25$ elements.\n\\end{example}\n\n\\begin{example}[Ideal norm of $(2+i)$ in the Gaussian integers]\n\tYou'll notice that \\[ \\OO_K / (2+i) \\cong \\FF_5 \\]\n\tsince mod $2+i$ we have both $5 \\equiv 0$ and $i \\equiv -2$.\n\t(Indeed, since $(2+i)$ is prime we had better get a field!)\n\tThus $\\Norm\\left( (2+i) \\right) = 5$; similarly $\\Norm\\left( (2-i) \\right) = 5$.\n\\end{example}\n\nThus the ideal norm measures how ``roomy'' the ideal is:\nthat is, $(5)$ is a lot more spaced out in $\\ZZ[i]$ than it is in $\\ZZ$.\n(This intuition will be important when we will actually view $\\OO_K$ as a lattice.)\n\n\\begin{ques}\n\tWhat are the ideals with ideal norm one?\n\\end{ques}\n% Regrettably, ideals with more elements have smaller norms.\n\nOur example with $(5)$ suggests several properties of the ideal norm\nwhich turn out to be true:\n\\begin{lemma}[Properties of the absolute norm]\n\tLet $\\ka$ be a nonzero ideal of $\\OO_K$.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii $\\Norm(\\ka)$ is finite.\n\t\t\\ii For any other nonzero ideal $\\kb$, $\\Norm(\\ka\\kb) = \\Norm(\\ka)\\Norm(\\kb)$.\n\t\t\\ii If $\\ka = (a)$ is principal, then $\\Norm(\\ka) = \\Norm_{K/\\QQ}(a)$.\n\t\t% \\ii The ideal norm $\\Norm(\\ka)$ equals the GCD of $\\Norm_{K/\\QQ}(\\alpha)$ across\n\t\t% all $\\alpha \\in \\ka$. In particular, if $\\ka = (\\alpha)$ is principal\n\t\t% then $\\Norm(\\ka) = \\Norm_{K/\\QQ}(\\alpha)$.\n\t\\end{enumerate}\n\\end{lemma}\nI unfortunately won't prove these properties, though we already did (a) in our proof that $\\OO_K$\nwas a Dedekind domain.\n% Anyways, this lemma provides another way to show an ideal isn't principal.\n\n%\\begin{example}[Example of an Ideal Norm]\n%\tLet $K = \\QQ(\\sqrt 3)$, so $\\OO_K = \\ZZ[\\sqrt 3]$.\n%\t% \\ii The norm of the principal ideal $(4+\\sqrt 3)$ is the norm of the element, $(4+\\sqrt3)(4-\\sqrt3) = 13$.\n%\t% \\ii\n%\tConsider $\\ka = (2, \\sqrt3)$.\n%\t\tWe need to have \n%\t\t\\[ \\Norm(\\ka) \\mid \\gcd(\\Norm_{K/\\QQ}(2), \\Norm_{K/\\QQ}(3)) = \\gcd(4, 3) = 1. \\]\n%\t\tSo in fact, $\\Norm(\\ka) = 1$. Surprise!\n%\t\tThis actually means $\\ka = (1)$,\n%\t\twhich in hindsight also follows from the fact that\n%\t\t$\\sqrt 3 \\in \\ka$, thus $\\sqrt 3 \\cdot \\sqrt 3 \\in \\ka$ and $3 \\in \\ka$;\n%\t\tbut $2 \\in \\ka$ so $1 \\in \\ka$.\n%\\end{example}\n\nThe fact that $\\Norm$ is completely multiplicative lets us also consider the norm\nof a fractional ideal $J$ by the natural extension\n\\[ J = \\prod_i \\kp_i^{n_i} \\cdot \\prod_i \\kq_i^{-m_i} \n\t\\quad \\implies \\quad\n\t\\Norm(J) \\defeq \\frac{\\prod_i \\Norm(\\kp_i)^{n_i}}{\\prod_i \\Norm(\\kq_i)^{m_i}}. \\]\nThus $\\Norm$ is a natural group homomorphism $J_K \\to \\QQ \\setminus \\{0\\}$.\n\n\n\\section\\problemhead\n\\begin{problem}\n\tShow that there are three different factorizations of $77$ in $\\OO_K$,\n\twhere $K = \\QQ(\\sqrt{-13})$.\n\\end{problem}\n\n\\begin{problem}\n\tLet $K = \\QQ(\\cbrt 2)$;\n\ttake for granted that $\\OO_K = \\ZZ[\\cbrt 2]$.\n\tFind the factorization of $(5)$ in $\\OO_K$.\n\\end{problem}\n\n\\begin{problem}\n\t[Fermat's little theorem]\n\tLet $\\kp$ be a prime ideal in some ring of integers $\\OO_K$.\n\tShow that for $\\alpha \\in \\OO_K$,\n\t\\[ \\alpha^{\\Norm(\\kp)} \\equiv \\alpha \\pmod{\\kp}. \\]\n\t\\begin{hint}\n\t\tCopy the proof of the usual Fermat's little theorem.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tIf $\\alpha \\equiv 0 \\pmod{\\kp}$ it's clear, so assume this isn't the case.\n\t\tThen $\\OO_K/\\kp$ is a finite field with $\\Norm(\\kp)$ elements.\n\t\tLooking at $(\\OO_K/\\kp)^\\ast$, it's a multiplicative group with $\\Norm(\\kp)-1$ elements,\n\t\tso $\\alpha^{\\Norm(\\kp)-1} \\equiv 1 \\pmod{\\kp}$, as desired.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{dproblem}\n\tLet $\\mathcal A$ be a Dedekind domain with field of fractions $K$,\n\tand pick $J \\subseteq K$.\n\tShow that $J$ is a fractional ideal if and only if\n\t\\begin{enumerate}[(i)]\n\t\t\\ii $J$ is closed under addition and multiplication\n\t\tby elements of $\\mathcal A$, and\n\t\t\\ii $J$ is finitely generated as an abelian group.\n\t\\end{enumerate}\n\tMore succinctly: $J$ is a fractional ideal $\\iff$ $J$ is a finitely generated $\\mathcal A$-module.\n\t\\label{prob:fractional_ideal_alt_def}\n\t\\begin{hint}\n\t\tClear denominators!\n\t\\end{hint}\n\t\\begin{sol}\n\t\tSuppose it's generated by some elements in $K$; we can write them as\n\t\t$\\frac{\\beta_i}{\\alpha_i}$ for $\\alpha_i, \\beta_i \\in \\mathcal A$.\n\t\tHence\n\t\t\\[ J = \\left\\{ \\sum_i \\gamma_i \\cdot \\frac{\\beta_i}{\\alpha_i}\n\t\t\\mid \\alpha_i, \\beta_i, \\gamma_i \\in \\OO_K \\right\\}. \\]\n\t\tNow ``clear denominators''. Set $\\alpha = \\alpha_1 \\dots \\alpha_n$,\n\t\tand show that $\\alpha J$ is an integral ideal.\n\t\\end{sol}\n\\end{dproblem}\n\n\\begin{problem}\n\t\\label{prob:prove_factoring_algorithm}\n\tIn the notation of \\Cref{thm:factor_alg}, let $I = \\prod_{i=1}^g \\kp_i^{e_i}$.\n\tAssume for simplicity that $K$ is monogenic, hence $\\OO_K = \\ZZ[\\theta]$.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Prove that each $\\kp_i$ is prime.\n\t\t\\ii Show that $(p)$ divides $I$.\n\t\t\\ii Use the norm to show that $(p) = I$.\n\t\\end{enumerate}\n\t\\begin{hint}\n\t\t(a) is straightforward.\n\t\tFor (b) work mod $p$.\n\t\tFor (c) use norms.\n\t\\end{hint}\n\t\\begin{sol}\n\tFor part (a), note that the $\\kp_i$ are prime\n\tjust because \n\t\\[ \\OO_K / \\kp_i\n\t\t\\cong (\\ZZ[x] / f) / (p, f_i)\n\t\t\\cong \\FF_p[x] / (f_i) \\]\n\tis a field, since the $f_i$ are irreducible.\n\n\tWe check (b).\n\t\tComputing the product modulo $p$ yields\\footnote{%\n\t\t\tFor example, suppose we want to know that $(3, 1+\\sqrt{7})(3, 1-\\sqrt{7})$ is contained in $(3)$.\n\t\t\tWe could do the full computation and get $(9, 3+3\\sqrt{7}, 3-3\\sqrt{7}, 6)$.\n\t\t\tBut if all we care about is that every element is divisible by $3$, we could have just taken ``mod $3$''\n\t\t\tat the beginning and looked at just $(1+\\sqrt{7})(1-\\sqrt{7}) = (6)$;\n\t\t\tall the other products we get will obviously have factors of $3$.\n\t\t}\n\t\t\\[ \\prod_{i=1}^{g} (f_i(\\theta))^{e_i}\n\t\t\t\\equiv (f(\\theta)) \\equiv 0 \\pmod p \\]\n\t\tso we've shown that $I \\subseteq (p)$.\n\n\tFinally, we prove (c) with a size argument.\n\t\tThe idea is that $I$ and $(p)$ really should have the same size;\n\t\tto nail this down we'll use the ideal norm.\n\t\tSince $(p)$ divides $I$, we can write\n\t\t$ (p) = \\prod_{i=1}^g \\kp_i^{e_i'} $\n\t\twhere $e_i' \\le e_i$ for each $i$.\n\t\tRemark $\\OO_K / (p) \\cong \\Zc p[x] / (f)$ has size $p^{\\deg f}$.\n\t\tSimilarly, $\\OO_K / (\\kp_i)$ has degree $p^{\\deg f_i}$ for each $i$.\n\t\tCompute $\\Norm( (p) )$ using the $e_i'$ now and compare the results.\n%\t\tHence\n%\t\t\\[ p^{\\deg f} = \\Norm(p)\n%\t\t= \\prod_{i=1}^r \\Norm(\\kp_i)^{e_i'}\n%\t\t= p^{e_1' \\deg f_1 + \\dots + e_g' \\deg f_g}. \\]\n%\t\tHence\n%\t\t\\[ e_1' \\deg f_1 + \\dots e_g' \\deg f_g = \\deg f. \\]\n%\t\tOn the other hand, it's obvious that \\[ \\deg f = e_1 \\deg f_1 + \\dots + e_g \\deg f_g. \\]\n%\t\tAs $e_i' \\le e_i$ for each $i$, we can only have $e_i = e_i'$.\n\t\\end{sol}\n\\end{problem}\n", "meta": {"hexsha": "a08e58b16a48e5dc5b5d5d5c35266687f20fd8a3", "size": 31188, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/alg-NT/dedekind.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/alg-NT/dedekind.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/alg-NT/dedekind.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.0773480663, "max_line_length": 110, "alphanum_fraction": 0.6686866744, "num_tokens": 10524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920020959544, "lm_q2_score": 0.7690802423634961, "lm_q1q2_score": 0.6157963990304696}} {"text": "\\documentclass{article}\n\n\\begin{document}\n\n\\section*{1.}\n\n\\[ (1 + i_1)^{m_1} = (1 + i_2)^{m_2} \\]\n\n\\[ (1 + 0.18)^{1} = (1 + i_2)^{12} \\]\n\n\\[ (1 + 0.18)^{1/12} = 1 + i_2 \\]\n\n\\[ (1 + 0.18)^{1/12} - 1 = i_2 \\]\n\n\\section*{2.}\n\n\n\n\n\n\\end{document}", "meta": {"hexsha": "269af217ea5f3cef93e5ccdf80b1e58c45a37bd7", "size": 244, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "parcial_1.tex", "max_stars_repo_name": "Youngermaster/Tutorials", "max_stars_repo_head_hexsha": "4fb5a01ba0bf3e8820f61d88ff398f3a09776f29", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "parcial_1.tex", "max_issues_repo_name": "Youngermaster/Tutorials", "max_issues_repo_head_hexsha": "4fb5a01ba0bf3e8820f61d88ff398f3a09776f29", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "parcial_1.tex", "max_forks_repo_name": "Youngermaster/Tutorials", "max_forks_repo_head_hexsha": "4fb5a01ba0bf3e8820f61d88ff398f3a09776f29", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 11.619047619, "max_line_length": 40, "alphanum_fraction": 0.4385245902, "num_tokens": 132, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8006920020959544, "lm_q2_score": 0.7690802423634961, "lm_q1q2_score": 0.6157963990304696}} {"text": "\\documentclass[twoside]{MATH77}\n\\usepackage{multicol}\n\\usepackage[fleqn,reqno,centertags]{amsmath}\n\\begin{document}\n\\begmath 10.2 Trigonometric, Cosine, and Sine Fourier Transforms\n\n\\silentfootnote{$^\\copyright$1997 Calif. Inst. of Technology, \\thisyear \\ Math \\`a la Carte, Inc.}\n\n\\subsection{Purpose}\n\nThis subroutine computes discrete trigonometric, cosine, and sine transforms\nin up to six dimensions using the FFT. In the one dimensional case, the\nvalues $y$ and the Fourier coefficients $\\alpha $ and $\\beta $ are related\nby:\n\\begin{multline}\ny_j=\\frac 12\\alpha _0+\\,\\sum_{k=1}^{(N/2)-1}\\left[ \\alpha _k\\cos \\frac{2\\pi\njk}N+\\beta _k\\sin \\frac{2\\pi jk}N\\right]\\\\\n+\\frac 12\\alpha _{N/2}(-1)^j,\\quad j=0,1,...,N-1\\tag{1S}\n\\end{multline}\\vspace{-10pt}\n\\renewcommand{\\theequation}{\\arabic{equation}A}\n\\begin{equation}\n\\begin{split}\n\\alpha _k&=\\frac 2N\\sum_{j=0}^{N-1}y_j\\cos \\frac{2\\pi jk}N,\\quad\nk=0,1,...,\\frac N2\\\\\n\\beta _k&=\\frac 2N\\sum_{j=0}^{N-1}y_j\\sin \\frac{2\\pi jk}N,\\quad\nk=1,2,...,\\frac N2-1\n\\end{split}\n\\end{equation}\\vspace{-10pt}\n\\begin{multline}\ny_j=\\frac 12\\alpha _0+\\sum_{k=1}^{N-1}\\alpha _k\\cos \\frac{\\pi jk}N+\\frac\n12\\alpha _N(-1)^j,\\\\\nj=0,1,...,N\\tag{2S}\n\\end{multline}\\vspace{-20pt}\n\\begin{multline}\n\\alpha _k=\\frac 2N\\left[ \\frac 12y_0+\\sum_{j=1}^{N-1}y_j\\cos \\frac{\\pi jk}%\nN+\\frac 12y_N(-1)^k\\right] ,\\\\\nk=0,1,...,N\n\\end{multline}\\vspace{-20pt}\n\\begin{equation}\n\\hspace{-5pt}y_j=\\sum_{k=1}^{N-1}\\beta _k\\sin \\frac{\\pi jk}N,\\quad j=1,2,...,N-1%\n\\tag{3S}\n\\end{equation}\\vspace{-10pt}\n\\begin{equation}\n\\hspace{-5pt}\\beta _k=\\frac 2N\\sum_{j=1}^{N-1}y_j\\sin\n\\frac{\\pi jk}N,\\quad k=1,2,...,N-1\n\\end{equation}\nwhere $N =2^{\\text{M}(1)}$, ($N_i=2^{\\text{M}(i)}$, $i=1$, ..., ND in the\nmulti-dimensional case.) In the equation labels above, the numeral 1, 2,\nor~3 denotes the Trigonometric, Cosine, or Sine transform, respectively,\nwhile the letter S or A denotes Synthesis or Analysis.\n\\renewcommand{\\theequation}{\\arabic{equation}}\n\n\\subsection{Usage}\n\n\\subsubsection{Program Prototype, Single Precision, One-Dimensional Transform}\n\n\\begin{description}\n\\item[\\bf INTEGER] \\ {\\bf M}(1){\\bf , ND, MS}\n\\item[\\bf REAL] \\ {\\bf A}$(\\geq \\mu $)\\quad $[\\mu =2^{\\text{M}(1)}+1$\nfor a Cosine transform, and $=2^{\\text{M}(1)}$ for a Trigonometric transform or\nSine transform.]\n\\item[\\bf REAL] \\ {\\bf S}$(\\geq \\nu {-}1$)\\quad $[\\nu =2^{\\text{M}(1)-2}$ for\na Trigonometric transform and $=2^{\\text{M}(1)-1}$ for a Cosine or Sine\ntransform.]\n\\item[\\bf CHARACTER TCS, MODE] \\\n\\end{description}\n\nOn the initial call set MS to~0 to indicate the array S() does not yet\ncontain a sine table. Assign values to A(), TCS, MODE, M(), and ND = 1\nfor a one-dimensional transform.\n$$\n\\fbox{{\\bf CALL STCST(A, TCS, MODE, M, ND, MS, S)}}\n$$\nOn return A() will contain computed values. S() will contain the sine\ntable used in computing the Fourier transform. MS may have been changed.\n\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\\item[A()] [inout] If MODE selects analysis, A() contains $y$ on input and $%\n\\alpha $\\ and/or\\ $\\beta $ on output. If MODE selects synthesis, A()\ncontains $\\alpha $\\ and/or\\ $\\beta $ on input and $y$ on output. Let $N\n=2^{\\text{M}(1)}.$\n\nWhen A() contains $y$'s, the element $y_j$ is stored in A($j+1)$.\nThe range of $j$ is $[0, N-1]$ for the Trigonometric transform, $%\n[0, N]$ for the Cosine transform, and $[1, N-1]$ for\nthe Sine transform.\n\nFor the Trigonometric transform, the $\\alpha $'s and $\\beta $'s\nare stored as A($1)=\\alpha _0$, A($2)=\\alpha _{N/2}$, and ${%\n\\textstyle A}(2k+1)=\\alpha _k$, A($2k+2)=\\beta _k$, $k=1$, 2, ..., ($N/2)-1.$\n\nFor the Cosine transform, the $\\alpha $'s are stored as A($k+1)=\\alpha _k$,\n$k=0$, 1, ..., $N$.\n\nFor the Sine transform, the $\\beta $'s are stored as A($k+1)=\\beta\n_k$, $k=1$, 2, ..., $N-1$. Although the value contained in A(1) is\nirrelevant for the Sine transform, the subroutine does access this location,\nso A(1) must contain a valid floating-point number on entry and its value\nwill generally be changed on return.\n\n\\item[TCS] [in] The character variable TCS selects the type of transform to\nbe done.\n\n$^{\\prime }\\text{T}^{\\prime }$ or $^{\\prime }\\text{t}^{\\prime }$ selects the Trigonometric\ntransform (Formula 1A or~1S).\n\n$^{\\prime }\\text{C}^{\\prime }$ or $^{\\prime }\\text{c}^{\\prime }$ selects the Cosine\ntransform (Formula 2A or~2S).\n\n$^{\\prime }\\text{S}^{\\prime }$ or $^{\\prime }\\text{s}^{\\prime }$ selects the Sine\ntransform (Formula 3A or~3S).\n\n\\item[MODE] [in] The character variable MODE selects Analysis or Synthesis.\n\n$^{\\prime }\\text{A}^{\\prime }$ or $^{\\prime }\\text{a}^{\\prime }$ selects Analysis (Formula\n1A, 2A, or~3A).\n\n$^{\\prime }\\text{S}^{\\prime }$ or $^{\\prime }\\text{s}^{\\prime }$ selects Synthesis\n(Formula 1S, 2S, or~3S).\n\n\\item[M()] [in] Defines $N=2^{\\text{M}(1)}$. The number of real data\npoints is $N$+1, $N$, and $N{-}$1, for the Cosine, Trigonometric, and Sine\ntransforms respectively. Require 0~$\\leq \\text{M(1)} \\leq 31\\ (\\leq 30$\nfor the cosine or sine transform). If M(1) is~0, no action is taken.\n\n\\item[ND] [in] Number of dimensions, =1 for one-dimensional transforms.\n\n\\item[MS] [inout] Gives the state of the sine table in S(). Let\n$\\text{MS}_{in}\\text{ and MS}_{out}$ denote the values of MS on entry\nand return respectively. If the sine table has not previously been\ncomputed, set $\\text{MS}_{in} = 0$ or $-$1 before the call. Otherwise\nthe value of $\\text{MS}_{out}$ from the previous call using the same\nS() array can be used as $\\text{MS}_{in}$ for the current call.\n\nCertain error conditions described in Section E cause the subroutine\nto set $\\text{MS}_{out} = -2$ and return. Otherwise, with M(1) $>$ 0, the\nsubroutine sets $\\text{MS}_{out} = \\max (\\text{M}(1), \\text{MS}_{in})$\nfor the Trigonometric transforms and to $\\text{MS}_{out} = \\max\n(\\text{M}(1)+1, \\text{MS}_{in})$ for the Cosine or Sine transforms.\n\nIf $\\text{MS}_{out} > \\max (2, \\text{MS}_{in}),$ the subroutine sets\nNT = $2^{\\text{MS}_{out}-2}$ and fills S() with NT $-$ 1 sine values.\n\nIf $\\text{MS}_{in}=-1$, the subroutine returns after the above\nactions, not transforming the data in A(). This is intended to allow\nthe use of the sine table for data alteration before a subsequent Fourier\ntransform, as discussed in Section G of Chapter~16.0.\n\n\\item[S()] [inout] When the sine table has been computed, S($j)=\\sin \\pi\nj/(2\\times \\text{NT})$, $j=1$, 2, ..., NT $-$ 1, see MS above.\n\\end{description}\n\n\\subsubsection{Program Prototype, Multi-dimensional Transforms}\n\n{\\bf INTEGER} \\ {\\bf M}($\\geq $ ND){\\bf , ND, MS}\n\n{\\bf REAL} \\ {\\bf A}($\\mu _1$, $\\mu _2$, ..., $\\geq \\mu _{ND})$\\quad $[\\mu\n_k=2^{\\text{M}(k)}+1$ for a Cosine transform and $=2^{\\text{M}(k)}$ for a\nTrigonometric or Sine transform.]\n\n{\\bf REAL} \\ {\\bf S}($\\geq \\max (\\nu _1$, $\\nu _2$, ..., $\\nu _{ND})-1)$%\n\\quad $[\\nu _k=2^{\\text{M}(k)-2}$ for a Trigonometric transform\nand $=2^{\\text{M}(k)-1}$\nfor a Cosine or Sine transform.]\n\n{\\bf CHARACTER TCS*($\\geq $ ND), MODE*($\\geq $ ND)}\n\nOn the initial call set MS to~0 to indicate the array S() does not yet\ncontain a sine table. Assign values to A(), TCS, MODE, M(), and ND.\n$$\n\\fbox{{\\bf CALL STCST(A, TCS, MODE, M, ND, MS, S)}}\n$$\nOn return A() contains the transformed data and S() contains the sine table\nused in computing the Fourier transform. The value of MS may have been\nchanged.\n\n\\subsubsection{Argument Definitions}\n\n\\begin{description}\n\\item[A()] [inout] Array used for input and output data, see Functional\nDescription below for specification of the storage of items in A().\n\n\\item[TCS] [in] The character TCS$(k$:$k)$ selects the type of transform to\nbe done in the $k^{th}$ dimension.\n\n$^{\\prime }\\text{T}^{\\prime }$ or $^{\\prime }\\text{t}^{\\prime }$ selects the Trigonometric\ntransform (Formula 1A or~1S).\n\n$^{\\prime }\\text{C}^{\\prime }$ or $^{\\prime }\\text{c}^{\\prime }$ selects the Cosine\ntransform (Formula 2A or~2S).\n\n$^{\\prime }\\text{S}^{\\prime }$ or $^{\\prime }\\text{s}^{\\prime }$ selects the Sine\ntransform (Formula 3A or~3S).\n\n\\item[MODE] [in] The character MODE$(k$:$k)$ selects Analysis or Synthesis\nin the $k^{th}$ dimension.\n\n$^{\\prime }\\text{A}^{\\prime }$ or $^{\\prime }\\text{a}^{\\prime }$ selects Analysis (Formula\n1A, 2A, or~3A).\n\n$^{\\prime }\\text{S}^{\\prime }$ or $^{\\prime }\\text{s}^{\\prime }$ selects Synthesis\n(Formula 1S, 2S, or~3S).\n\n\\item[M()] [in] In the $k^{th}$ dimension, $k=1$, ..., ND, define $%\nN_k=2^{\\text{M}(k)}$. The index range for the values $y_j$ in the $k^{th}$\ndimension is $[0,N_k-1]$ for the Trigonometric transform, $[0,N_k]$ for the\nCosine transform, and $[1,N_k-1]$ for the Sine transform. Require $0 \\leq\n\\text{M}(k)\\leq 31\\ (\\leq 30$ for the cosine or sine transform). If\nM($k)=0$, no action is taken with respect to dimension $k$.\n\n\\item[ND] [in] Number of dimensions, 1 $\\leq $ ND $\\leq $ 6.\n\n\\item[MS] [inout] As for MS in the one dimensional case above, except,\nif there is no error and M($k) >$ 0 for some $k$, the\nsubroutine sets $\\text{MS}_{out} = \\max\n(\\mu _1$, $\\mu _2$, ..., $\\mu_{ND})$, $\\text{MS}_{in}$), where $\\mu _k=\n\\text{M}(k)$ for the Trigonometric transform,\nand $=\\text{M}(k)+1$ for the Cosine or Sine transform. NT is defined\nin terms of $\\text{MS}_{out}$ as described in Section~B.2 above.\n\n\\item[S()] [inout] When the sine table has been computed, S($j)=\\sin \\pi\nj/(2\\times \\text{NT})$, $j=1$, 2, ..., NT $-$ 1, see MS above.\n\\end{description}\n\n\\subsubsection{Modifications for Double Precision}\n\nChange STCST to DTCST, and the REAL type statements to DOUBLE PRECISION.\n\n\\subsection{Examples and Remarks}\n\nGiven%\n\\begin{equation*}\nf(t)=\\frac{2\\sinh t}{\\sinh \\pi t}\n\\end{equation*}\nobtain an estimate of $\\varphi (\\omega )=\\int_0^Tf(t)\\cos \\omega t\\,dt$ and\ncompare it with $\\varphi (\\omega )=\\sin (1)/(\\cosh \\omega +\\cos (1))$, the true\nsolution. Let $T$ and $\\Omega $ denote the largest values of $t$ and $\\omega $\nto be used in the computation. As was done below Eq.\\,(18) in Chapter~16.0,\nwe introduce the approximations%\n\\begin{multline*}\n\\varphi (\\omega ) = \\int_0^Tf(t)\\cos \\omega t\\,dt\\\\\n\\approx \\frac TN\\!\\left[ \\frac 12f(t_0)+\\!\\sum_{j=1}^{N-1}f(t_j)\\cos\n\\omega t_j+\\frac 12f(t_N)\\cos \\omega t_N\\right]\n\\end{multline*}\nwhere $t_j=j{T}/N$. With $\\omega _k=k\\pi /{T}$, we get%\n\\begin{equation*}\\hspace{-15pt}\n\\varphi (\\omega _k)\\!\\approx \\!\\frac T2\\frac 2N\\!\\left[ \\frac\n12f(t_0)\\!+\\!\\!\\sum_{j=1}^{N-1}\\!f(t_j)\\cos \\frac{\\pi jk}N\\!+\\!\\frac\n12f(t_N)(-1)^k\\!\\right]\n\\end{equation*}\nwhich except for the factor $T/2$ has the form of Eq.\\,(2A) above. Good\nresults are obtained by balancing the error due to a finite $T$ with the\nerror due to aliasing $(i.e$. use of a finite $\\Omega )$. Thus we want to\nselect $T$ and $\\Omega $ so that%\n\\begin{equation}\n\\int_T^\\infty |f(t)|\\,dt\\approx \\frac 2{\\pi -1}e^{-(\\pi -1)T}\\text{,\nand}\n\\end{equation}\\vspace{-10pt}\n\\begin{equation}\n\\int_\\Omega^\\infty |\\varphi (\\omega )|\\,d\\omega \\approx 2\\sin (1)e^{-\\Omega}\n\\end{equation}\nare of the same order of magnitude. Since%\n\\begin{equation*}\n\\int_T^\\infty |f(t)|\\,dt<10^{-4}\\int_0^T|f(t)|\\,dt\\quad \\text{for }T\\geq 10.\n\\end{equation*}\nthere is no point in choosing $T>10$. With $T=10$, $N=64$ gives $\\Omega =(%\nN \\pi /{T})=6.4\\pi $. The right hand sides of Equations (4) and (5) are\nthen $\\approx 4.7\\times 10^{-10}$ and $3.1\\times 10^{-9}$ respectively.\n\nThe program at the end of this chapter carries out the above calculations,\nbut prints results only for every $10^{th}$ value of $k$ to save space.\n\n\\subsection{Functional Description}\n\nThe one-dimensional transforms computed by this subroutine are given by\nequations (1S), (1A), ..., (3S), (3A). The multi-dimensional transform is\naccomplished by applying the appropriate one-dimensional transforms in each\ndimension. To define the relation between input and output contents of the\nA() array in the multi-dimensional case we introduce a 4-argument function,\nT(*,*,*,*), in which the first argument can take the values $%\n^{\\prime }\\text{T}^{\\prime }$,$^{\\prime }\\text{C}^{\\prime }$,\nor $^{\\prime }\\text{S}^{\\prime }$\nto denote Trigonometric, Cosine, or Sine, and the second argument can take\nthe values $^{\\prime }\\text{A}^{\\prime }$ or $^{\\prime }\\text{S}^{\\prime }$\nto denote Analysis or Synthesis. The third and fourth arguments are integers.\nThere is also an implied argument, M, which together with the first\nargument contributes to the definition of $N$ and $\\mu $.\n\\begin{equation*}\n\\hspace{-15pt}N=2^M\n\\end{equation*}\\vspace{-30pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{T}^{\\prime }{,}^{\\prime }\n\\text{S}^{\\prime },j,k)=\\begin{cases}\n1/2 & k=0\\\\\n\\cos (\\pi jk/N)& k=2,4,...,N\\!-\\!2\\\\\n\\sin (\\pi j(k-1)/N)& k=3,5,...,N\\!-\\!1\\\\\n(1/2)(-1)^j& k=1\n\\end{cases}\n\\end{equation*}\\vspace{-10pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{T}^{\\prime }{,}^{\\prime }\n\\text{A}^{\\prime },j,k)\\!=\\!\\frac 2N\n\\begin{cases}\n\\cos (\\pi jk/N)&\\!\\!j\\!=\\!0,2,...,N\\!-\\!2 \\\\\n\\sin (\\pi (j\\!-\\!1)k/N)&\\!\\!j\\!=\\!3,5,...,N\\!-\\!1 \\\\\n(-1)^k,&\\!\\!j\\!=\\!1\n\\end{cases}\n\\end{equation*}\\vspace{-15pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{C}^{\\prime }{,}^{\\prime }\n\\text{S}^{\\prime },j,k)=\n\\begin{cases}\n1/2& k=0 \\\\\n\\cos (\\pi jk/N)& k=1,2,...,N-1 \\\\\n(1/2)(-1)^j& k=N\n\\end{cases}\n\\end{equation*}\\vspace{-15pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{C}^{\\prime }{,}^{\\prime }%\n\\text{A}^{\\prime },j,k)=\\frac 2N\\text{T(}%\n^{\\prime }\\text{C}^{\\prime },^{\\prime }\\text{S}^{\\prime },j,k)\n\\end{equation*}\\vspace{-15pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{S}^{\\prime }{,}^{\\prime }%\n\\text{S}^{\\prime },j,k)=\\sin \\frac{\\pi jk}N\n\\end{equation*}\\vspace{-15pt}\n\\begin{equation*}\n\\hspace{-15pt}\\text{T(}^{\\prime }\\text{S}^{\\prime }{,}^{\\prime }%\n\\text{A}^{\\prime },j,k)=\\frac 2N\\text{T(}^{\\prime }\\text{S}^{\\prime},%\n^{\\prime }\\text{S}^{\\prime },j,k)\n\\end{equation*}\\vspace{-15pt}\n\\begin{equation*}\n\\hspace{-15pt}\\mu _i=\n\\begin{cases}\n2^{\\text{M}(i)}& \\text{for a Trigonometric or Sine transform} \\\\\n2^{\\text{M}(i)}+\\!1& \\text{for a Cosine transform}\n\\end{cases}\n\\end{equation*}\n\nThe array A has its contents replaced according to the formula%\n\\begin{multline*}\n\\hspace{-5pt}A(j_1,j_2,...,j_{ND})=\\sum_{k_1=1}^{\\mu _1}\\cdots\n\\sum_{k_{ND}=1}^{\\mu_{ND}}A(k_1,k_2,...,k_{ND})\\\\\\hspace{-15pt}\n\\times \\text{T(TCS(1:1), MODE(1:1), }j_1-1,k_1-1)\\times \\cdots\\\\\n\\hspace{-15pt}\\times \\text{T(TCS(ND:ND), MODE(ND:ND), }j_{ND}-1,k_{ND}-1),\n\\end{multline*}\nwhere $j_i=1,2,...,\\mu _i$ and storage conventions with respect to each\ndimension are defined as for the one dimensional transforms.\n\nThe computational procedure for the trigonometric transform in the\none-dimensional case is almost identical to the procedure used in SRFT1.\n(See the second paragraph below Eq.\\,10 in Chapter~16.0 to see why this is\nso.)\n\nThe procedure for the cosine transform uses the trigonometric transform and\nthe identity%\n\\begin{equation*}\n\\hspace{-0pt}\\cos \\frac{\\pi j(2k+1)}N=\\frac{\\sin (2\\pi j(k+1)/N)-\\sin\n(2\\pi jk/N)}{2\\sin (\\pi j/N)}.\n\\end{equation*}\nThis is used to transform%\n\\begin{equation*}\ny_j=\\frac 12\\eta _0+\\frac 12\\eta _N(-1)^j+\\sum_{k=1}^{N-1}\\eta _k\\cos \\frac{%\n\\pi jk}N\n\\end{equation*}\nto%\n\\begin{multline*}\ny_j=\\frac 12\\eta _0+\\frac 12\\eta _N(-1)^j+\\sum_{k=1}^{(N/2)-1}\\eta _{2k}\\cos\n\\frac{2\\pi jk}N\\\\\n+\\frac 1{2\\sin (\\pi j/N)}\\sum_{k=1}^{(N/2)-1}(\\eta\n_{2k-1}-\\eta _{2k+1})\\sin \\frac{2\\pi jk}N.\n\\end{multline*}\nLet $Y_j$ be the result from the trigonometric transform with $\\alpha\n_k=\\eta _{2k}$, $k=0$, 1, ..., $N/2$ and $\\beta _k=\\eta _{2k-1}-\\eta _{2k+1}$%\n, $k=1,2,...,(N/2)-1$. It follows that%\n\\begin{eqnarray*}\n y_j+y_{N-j} & = & Y_j+Y_{N-j} \\\\\n y_j-y_{N-j} & = & \\frac{Y_j-Y_{N-j}}{2\\sin (\\pi j/N)}\n\\end{eqnarray*}\nand thus one can compute $y_j$ from $Y_j$. The computational\nprocedure is to compute the $\\alpha $'s and $\\beta $'s\nfrom the $\\eta $'s (the $\\alpha $'s require no\ncomputation), use the trigonometric transform to get the $Y$'s, and\nthe $Y$'s to compute the $y$'s. The inverse transform is\nexactly the same except for a factor of 2/$N$ as is clear from\nEqs.\\,(2S) and (2A).\n\nThe sine transform is obtained in a very similar way to the cosine\ntransform. Details can be found in \\cite{Krogh:1970:SCT}.\n\n\\bibliography{math77}\n\\bibliographystyle{math77}\n\n\\subsection{Error Procedures and Restrictions}\n\nRequire 1 $\\leq $ ND $\\leq $ 6, and $0 \\leq \\text{M}(k)\\leq 31$ for the\nTrigonometric transform, $0 \\leq \\text{M}(k)\\leq 30$ for the Cosine and\nSine transforms. Require that TCS and MODE have only the allowed values.\nOn violation of any of these conditions the subroutine issues an error\nmessage using the error processing procedures of Chapter~19.2 with\nseverity level $=2$ to cause execution to stop. A return will be made\nwith MS $=-2$ instead of stopping if the statement ``CALL\\ ERMSET($-1$)''\nis executed before calling this subroutine.\n\nIf the sine table does not appear to have valid data, an error message is\nprinted, and the sine table and then the transform are computed.\n\n\\subsection{Supporting Information}\n\nThe source language is ANSI Fortran~77.\n\n\\begin{tabular}{@{\\bf}l@{\\hspace{5pt}}l}\n\\bf Entry & \\hspace{.35in} {\\bf Required Files}\\vspace{2pt} \\\\\nDTCST & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\n DFFT, DTCST, ERFIN, ERMSG, IERM1, IERV1\\rule[-5pt]{0pt}{8pt}}\\\\\nSTCST & \\parbox[t]{2.7in}{\\hyphenpenalty10000 \\raggedright\n ERFIN, ERMSG, IERM1, IERV1, SFFT, STCST}\\\\\n\\end{tabular}\n\nSubroutine designed and written by: Fred T. Krogh, JPL, August~1969, revised\nJanuary~1988.\n\n\n\\begcode\n\n\\medskip\\\n\\lstset{language=[77]Fortran,showstringspaces=false}\n\\lstset{xleftmargin=.8in}\n\n\\centerline{\\bf \\large DRSTCST}\\vspace{10pt}\n\\lstinputlisting{\\codeloc{stcst}}\n\n\\vspace{30pt}\\centerline{\\bf \\large ODSTCST}\\vspace{10pt}\n\\lstset{language={}}\n\\lstinputlisting{\\outputloc{stcst}}\n\\end{document}\n", "meta": {"hexsha": "10307ff79c1f84d5b974b8f97c8b62243176462c", "size": 17746, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/doctex/ch10-02.tex", "max_stars_repo_name": "jacobwilliams/math77", "max_stars_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2016-01-04T03:17:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-25T19:17:42.000Z", "max_issues_repo_path": "doc/doctex/ch10-02.tex", "max_issues_repo_name": "jacobwilliams/math77", "max_issues_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2019-01-17T02:48:32.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-21T16:04:58.000Z", "max_forks_repo_path": "doc/doctex/ch10-02.tex", "max_forks_repo_name": "jacobwilliams/math77", "max_forks_repo_head_hexsha": "b562d09e191e99eba8a5bedfec45acf7461203b1", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2016-01-07T09:26:45.000Z", "max_forks_repo_forks_event_max_datetime": "2018-11-25T05:32:54.000Z", "avg_line_length": 40.423690205, "max_line_length": 98, "alphanum_fraction": 0.6622900935, "num_tokens": 6581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920020959543, "lm_q2_score": 0.7690802370707283, "lm_q1q2_score": 0.6157963947925925}} {"text": "\\chapter*{Appendix C: Nonlinear Pyramidal Wavelet Transform (WT-PMT)}\n\\addcontentsline{toc}{chapter}{Appendix C: Pyramidal Wavelet Transform (PMWT)}\n\nOne of the advantages of the pyramidal wavelet transform over the pyramidal\nmedian transform (PMT) is the ability to have robust noise estimation in the \ndifferent scales, while the advantage of the PMT is a better separation\nof the structures in the scales. Using the PMT, \na strong structure (like a bright star, cosmic rays, bad pixels, etc.) will \nnot be spread over all scales as when using a wavelet transform. In fact, when\nthere is no signal in a given region, a wavelet transform would be better, and\nif a strong signal appears, it is the PMT that we would like to use. So the \nidea arises \nto try to merge both transforms, and to adapt the analysis at each position\nand at each scale, depending on the amplitude of the coefficient we measure.\n\nA possible algorithm to perform this on an image $I$ is the following:\n\\begin{enumerate}\n\\item Let $c_j = I$, and $j=0$.\n\\item Filter the the image $c_j$ by the median: we get $m_j$\n\\item Set $c_j^* = c_j$ and $d_j= c_j - m_j$\n\\item For all pixels $k$ do \\\\\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \n if $\\mid d_j(k) \\mid > k \\sigma_j$ then $c_j^*(k) = m_j(k)$\n\\item Smooth $c_j^*$ by a B$_3$-spline: we get $b_j$\n\\item Set $w_j = c_j - b_j$\n\\item $c_{j+1} = dec(b_j)$ (where the decimation operation, dec, entails 1 pixel replacing\neach 2 $\\times$2 subimage).\nLet $j=j+1$ and return to step 2 if $j < N_s$ ($N_s$ being the number of scales).\n\\end{enumerate}\nIn this algorithm, the linear filtering relative to the wavelet transform is\nnot applied to the strong features contained in the image. Indeed, significant\npixel values are detected at step 4, and are replaced by the median. Regions\ncontaining no bright object are treated as if the pyramidal wavelet \ntransform is used. The parameter $k$ used at step 4 must be large enough\nin order to be sure that noise is not filtered by the median ($k=5$ seems\nhigh enough). As this transform merges the wavelet transform and the PMT, \nwe call it the PMWT transform. \nPMWT takes more time than the PMT, but this algorithm has the\nadvantages of the PMT without the drawbacks. The reconstruction is the\nsame as for the PMT.\n\n\n\\chapter*{Appendix D: Half Pyramidal Wavelet Transform}\n\\addcontentsline{toc}{chapter}{Appendix D: Half Pyramidal Wavelet Transform}\n\nAt each iteration of the pyramidal transform, there is a smoothing and \na decimation of the image. But the smoothing is not strong enough to\nreduce the cut-off frequency by a factor two. The decimation then violates\nShannon's theorem. To avoid this problem, which may create artifacts, \nBijaoui \\cite{wave:bijaoui97}\nproposed not to decimate the image at the first iteration of the algorithm.\nThis means that the two first scales have the same size as the original image\n(instead of just one in the standard pyramidal transform). Then the two first\niterations are identical as in the \\`a trous algorithm. For the\nfollowing iterations, there is a decimation and the filtering is done\nas for the second iteration (i.e.\\ by taking pixels interspersed by \none pixel for the filtering).\n\n\\newpage\n$ $\n\n", "meta": {"hexsha": "56f5b5e98288bab0eb57feb8b20058b02f95c478", "size": 3194, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr1/annex_wavemed.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr1/annex_wavemed.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr1/annex_wavemed.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.5161290323, "max_line_length": 91, "alphanum_fraction": 0.7570444584, "num_tokens": 853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920068519376, "lm_q2_score": 0.76908023177796, "lm_q1q2_score": 0.6157963942124481}} {"text": "%{\n\\begin{octave}\n%}\nsheffieldmlToolboxes;\n%{\n\\end{octave}\n\n\\title{SheffieldML - MATLAB Software}\n\n\\begin{document}\n\\section{Examples}\n\n\\section{Gaussian Process Models}\n\n\\subsection{Functions from Gaussians}\n\nThis example shows how points which look like they come from a\nfunction to be sampled from a Gaussian distribution. The sample is 25\ndimensional and is from a Gaussian with a particular covariance.\n\n\\begin{octave}\n%}\ndemGpSample\n%{\n\\end{octave}\n\n\\begin{center}\n\\inputdiagram{gpSample}\\hfill\\inputdiagram{gpCovariance}\n\n\\emph{Left} A single, 25 dimensional, sample from a Gaussian\ndistribution. \\emph{Right} the covariance matrix of the Gaussian\ndistribution.\n\n\\end{center}\n\n\n\\subsection{Joint Distribution over two Variables}\n\nGaussian processes are about conditioning a Gaussian distribution\non the training data to make the test predictions. To illustrate this\nprocess, we can look at the joint distribution over two variables.\n\n\\begin{octave}\n%}\n demGpCov2D([1 2])\n%{\n\\end{octave}\n\nGives the joint distribution for $f_1$ and\n$f_2$. The plots show the joint distributions as well\nas the conditional for $f_2$ given\n$f_1$.\n\n\\begin{center}\n\\inputdiagram{demGpCov2D1_2_3} 0.5\\textwidth \\inputdiagram{demGpCov2D1_5_3} 0.5\\textwidth \n\n\\emph{Left} Blue line is\ncontour of joint distribution over the variables $f_1$\nand $f_2$. Green line indicates an observation of\n$f_1$. Red line is conditional distribution of\n$f_2$ given $f_1$. \\emph{Right} Similar\nfor $f_1$ and $f_5$. \n\\end{center}\n\n\n\n\\subsection{Different Samples from Gaussian Processes}\n\nA script is provided which samples from a Gaussian process with the\nprovided covariance function.\n\n\\begin{octave}\n%}\n gpSample('rbf', 10, [1 1], [-3 3], 1e5)\n%{\n\\end{octave}\n\nwill give 10 samples from an RBF covariance function with a\nparameter vector given by [1 1] (inverse width 1, variance 1) across\nthe range -3 to 3 on the $x$-axis. The random seed will be set to\n1e5.\n\n\\begin{octave}\n%}\n gpSample('rbf', 10, [16 1], [-3 3], 1e5)\n%{\n\\end{octave}\n\nis similar, but the inverse width is now set to 16 (length scale 0.25).\n\n\\begin{center}\\inputdiagram{gpSampleRbfSamples10Seed100000InverseWidth1Variance1} 0.5\\textwidth \\inputdiagram{gpSampleRbfSamples10Seed100000InverseWidth16Variance1} 0.5\\textwidth \\emph{Left} samples from an RBF style covariance function\nwith length scale 1. \\emph{Right} samples from an RBF style covariance\nfunction with length scale 0.25. \n\\end{center}\n\nOther covariance functions can be sampled, an interesting one is\nthe MLP covariance which is non stationary and can produce point\nsymmetric functions,\n\n\n\\begin{octave}\n%}\n gpSample('mlp', 10, [100 100 1], [-1 1], 1e5)\n%{\n\\end{octave}\n\ngives 10 samples from the MLP covariance function where the `bias\nvariance' is 100 (basis functions are centered around the origin\nwith standard deviation of 10) and the `weight variance' is\n100.\n\n\\begin{octave}\n%}\n gpSample('mlp', 10, [100 1e-16 1], [-1 1], 1e5)\n%{\n\\end{octave}\n\ngives 10 samples from the MLP covariance function where the `bias\nvariance' is approximately zero (basis functions are placed on\nthe origin) and the `weight variance' is 100.\n\n\\begin{center}\\inputdiagram{gpSampleMlpSamples10Seed100000WeightVariance100BiasVariance100Variance1} 0.5\\textwidth \\inputdiagram{gpSampleMlpSamples10Seed100000WeightVariance100BiasVariance1e-16Variance1}\n0.5\\textwidth \\emph{Left} samples from an MLP style covariance\nfunction with bias and weight variances set to 100. \\emph{Right}\nsamples from an MLP style covariance function with weight variance 100\nand bias variance approximately zero. \n\\end{center}\n\n\n\\subsection{Posterior Samples}\n\nGaussian processes are non-parametric models. They are specified by their covariance function and a mean function. When combined with data observations a posterior Gaussian process is induced. The demos below show samples from that posterior.\n\n\\begin{octave}\n%}\n gpPosteriorSample('rbf', 5, [1 1], [-3 3], 1e5)\n%{\n\\end{octave}\n\nand \n\n\\begin{octave}\n%}\n gpPosteriorSample('rbf', 5, [16 1], [-3 3], 1e5)\n%{\n\\end{octave}\n\n\\begin{center}\\inputdiagram{gpPosteriorSampleRbfSamples5Seed100000InverseWidth1Variance1bw} 0.5\\textwidth \\inputdiagram{gpPosteriorSampleRbfSamples5Seed100000InverseWidth16Variance1bw} 0.5\\textwidth \n\n\\emph{Left} samples from the posterior induced by an RBF style covariance function\nwith length scale 1 and 5 `training' data points taken from a sine wave. \\emph{Right} Similar but for a length scale of 0.25. \n\\end{center}\n\n\\subsection{Simple Interpolation Demo}\n\nThis simple demonstration plots, consecutively, an increasing\nnumber of data points, followed by an interpolated fit through the\ndata points using a Gaussian process. This is a noiseless system, and\nthe data is sampled from a GP with a known covariance function. The\ncurve is then recovered with minimal uncertainty after only nine data\npoints are included. The code is run with\n\n\\begin{octave}\n%}\n demInterpolationGp\n%{\n\\end{octave}\n\n\\begin{center}\n\n\\inputdiagram{demInterpolation3} 0.5\\textwidth \\inputdiagram{demInterpolation4} 0.5\\textwidth\n\nGaussian process prediction \\emph{left} after two points with a new\ndata point sampled \\emph{right} after the new data point is included\nin the prediction. \n\n\\inputdiagram{demInterpolation7} 0.5\\textwidth \\inputdiagram{demInterpolation8} 0.5\\textwidth\n\nGaussian process prediction \\emph{left} after five points with a four\nnew data point sampled \\emph{right} after all nine data points are\nincluded. \n\\end{center}\n\n\\subsection{Simple Regression Demo}\n\nThe regression demo very much follows the format of the\ninterpolation demo. Here the difference is that the data is sampled\nwith noise. Fitting a model with noise means that the regression will\nnot necessarily pass right through each data point.\n\nThe code is run with\n\n\\begin{octave}\n%}\n demRegressionGp\n%{\n\\end{octave}\n\n\n\\begin{center}\n\\inputdiagram{demRegression3} 0.5\\textwidth \\inputdiagram{demRegression4} 0.5\\textwidth\n\nGaussian process prediction \\emph{left} after two points with a new\ndata point sampled \\emph{right} after the new data point is included\nin the prediction. \n\n\\inputdiagram{demRegression7} 0.5\\textwidth \\inputdiagram{demRegression8} width 0.5\\textwidth\n\nGaussian process prediction \\emph{left} after five points with a four\nnew data point sampled \\emph{right} after all nine data points are\nincluded. \n\\end{center}\n\n\\subsection{Optimizing Hyper-parameters}\n\nOne of the advantages of Gaussian processes over pure kernel\ninterpretations of regression is the ability to select the hyper\nparameters of the kernel automatically. The demo\n\n\\begin{octave}\n%}\n demOptimiseGp\n%{\n\\end{octave}\n\nshows a series of plots of a Gaussian process with different length\nscales fitted to six data points. For each plot there is a\ncorresponding plot of the log likelihood. The log likelihood peaks for\na length scale equal to 1. This was the length scale used to generate\nthe data.\n\n\\begin{center}\\inputdiagram{demOptimiseGp1} 0.33\\textwidth \\inputdiagram{demOptimiseGp3} 0.33\\textwidth \\inputdiagram{demOptimiseGp5}\n0.33\\textwidth \\inputdiagram{demOptimiseGp7} 0.33\\textwidth \\inputdiagram{demOptimiseGp9} 0.33\\textwidth \\inputdiagram{demOptimiseGp11} 0.33\\textwidth \\inputdiagram{demOptimiseGp13} 0.33\\textwidth \\inputdiagram{demOptimiseGp15} 0.33\\textwidth \\inputdiagram{demOptimiseGp17}\n0.33\\textwidth\n\nFrom top left to bottom right, Gaussian process\nregression applied to the data with an increasing length scale. The\nlength scales used were 0.05, 0.1, 0.25, 0.5, 1, 2, 4, 8 and\n16. \n\n\\inputdiagram{demOptimiseGp18} 0.5\\textwidth \n\nLog-log plot of\nthe log likelihood of the data against the length scales. The log\nlikelihood is shown as a solid line. The log likelihood is made up of\na data fit term (the quadratic form) shown by a dashed line and a\ncomplexity term (the log determinant) shown by a dotted line. The data\nfit is larger for short length scales, the complexity is larger for\nlong length scales. The combination leads to a maximum around the true\nlength scale value of 1.\n\n\\end{center}\n\n\\subsection{Regression over Motion Capture Markers}\n\nAs a simple example of regression for real data we consider a motion capture data set. The data is from Ohio State University. In the example script we perform Gaussian process regression with time as the input and the x,y,z position of the marker attached to the left ankle. To demonstrate the behavior of the model when the marker is lost, we remove data from This code can be run with\n\n\\begin{verbatim} \ndemStickGp1 \n\\end{verbatim} \n\n\\begin{octave}\n%}\nload('demStickGp1');\n% Plot results\nfillColor = [0.7 0.7 0.7];\ntTest = linspace(0, 2, 200)';\n[mu, varSigma] = gpPosteriorMeanVar(model, tTest);\n\nfor i = 1:length(outputIndex);\n figure\n fill([tTest; tTest(end:-1:1)], ...\n [mu(:, i); mu(end:-1:1, i)] ...\n + 2*[sqrt(varSigma(:, i)); -sqrt(varSigma(end:-1:1, i))], ...\n fillColor,'EdgeColor',fillColor)\n hold on;\n plot(t(testIndex), y(testIndex, outputIndex(i)), 'ko');\n b=plot(tTrain, yTrain(:, i), 'k.');\n a = plot(tTest, mu(:, i), 'k-');\n set(gca, 'xlim', [0 2])\n set(a, 'linewidth', 2);\n fileName = ['dem' capName 'Gp' num2str(experimentNo) 'Out' num2str(i)];\n printLatexPlot(fileName, '../tex/diagrams', 0.3*textWidth, options);\nend\n%{\n\\end{octave}\n\nThe code will optimize hyper parameters and show plots of the posterior process through the training data and the missing test points.\n\nThe result of the script is given in the plot below. \n\n\\begin{center}\n\\inputdiagram{demStickGp1Out1} 0.3\\textwidth \\inputdiagram{demStickGp1Out2} 0.3\\textwidth\"> <\\inputdiagram{demStickGp1Out3} width\n=\"0.3\\textwidth\"> Gaussian process regression through the x (left), y (middle) and z (right) position of the left ankle. Training data is shown as black spots, test points removed to simulate a lost marker are shown as circles, posterior mean\nprediction is shown as a black line and two standard deviations are\ngiven as grey shading.\\end{center}\n\nNotice how the error bars are tight except in the region where the training data is missing and in the region where the training data disappears.\n\n\\subsection{Sparse Pseudo-input Gaussian Processes}\n\nThe sparse approximation used in this toolbox is based on the\nSparse Pseudo-input Gaussian Process model described by Snelson\nand Ghahramani. Also provided are the extensions suggested by Quiñonero-Candela\nand Rasmussen. They provide a unifying terminology for describing\nthese approximations which we shall use in what follows.\n\nThere are three demos provided for Gaussian process regression in\n1-D. They each use a different form of likelihood approximation. The\nfirst demonstration uses the `projected latent variable'\napproach first described by Csato\nand Opper and later used by Seeger\net al.. In the terminology of Quiñonero-Candela and\nRasmussen (QR-terminology) this is known as the `deterministic\ntraining conditional' (DTC) approximation.\n\nTo use this approximation the following script can be run.\n\n\\begin{octave} \n%}\ndemSpgp1dGp1 \n%{\n\\end{octave} \n\nThe result of the script is\ngiven in the plot below. \n\n\\begin{center}<\\inputdiagram{demSpgp1dGp1} width\n=\"0.5\\textwidth\"> Gaussian process using the DTC approximation with nine\ninducing variables. Data is shown as black spots, posterior mean\nprediction is shown as a black line and two standard deviations are\ngiven as grey shading.\\end{center}\n\nThe improved approximation suggested by Snelson and Ghahramani, in\nQR-terminology this is known as the fully independent training\nconditional (FITC). To try this approximation run the following script\n\n\\begin{octave}\n%} \ndemSpgp1dGp2 \n%{\n\\end{octave}\n\nThe result of the script is given on the left of the plot below.\n\n\\begin{center}\\inputdiagram{demSpgp1dGp2} width=\"0.49\\textwidth\">\n\n\\emph{Left}: Gaussian process using the FITC approximation with nine\ninducing variables. Data is shown as black spots, posterior mean\nprediction is shown as a black line and two standard deviations are\ngiven as grey shading. \\emph{Right}: Similar but for the PITC\napproximation, again with nine inducing variables.\\end{center}\n\nAt the Sheffield\nGaussian Process Round Table Lehel Csato pointed out that the\nBayesian Committee Machine of Schwaighofer\nand Tresp can also be viewed within the same framework. This idea\nis formalised in Quiñonero-Candela\nand Rasmussen's review. This approximation is known as the\n`partially independent training conditional' (PITC) in\nQR-terminology. To try this approximation run the following script\n\n\\begin{octave}\n%}\n demSpgp1dGp3\n%{\n\\end{octave}\n\nThe result of the script is given on the right of the plot above.\n\nFinally we can compare these results to the result from the full\nGaussian process on the data with the correct hyper-parameters. To do\nthis the following script can be run.\n\n\\begin{octave}\n%}\n demSpgp1dGp4\n%{\n\\end{octave}\n\nThe result of the script is given in the plot below.\n\n\\begin{center}\\inputdiagram{demSpgp1dGp4} width=\"0.5\\textwidth\"> Full Gaussian\nprocess on the toy data with the correct hyper-parameters. Data is\nshown as black spots, posterior mean prediction is shown as a black\nline and two standard deviations are given as grey shaded\narea.\\end{center}\n\n\\section{Gaussian Process Latent Variable Model Examples}\n\nThe three approximations outlined above can be used to speed up learning in the GP-LVM. They have the advantage over the IVM approach taken in the original GP-LVM toolbox that the algorithm is fully convergent and the final mapping from latent space to data space takes into account all of the data (not just the points in the active set).\n\nAs well as the new sparse approximation the new toolbox allows the GP-LVM to be run with dynamics as suggested by Wang et al..\n\nFinally, the new toolbox allows the incorporation of `back constraints' in learning. Back constraints force the latent points to be a smooth function of the data points. This means that points that are close in data space are constrained to be close in latent space. For the standard GP-LVM points close in latent space are constrained to be close in data space, but the converse is not true.\n\nVarious combinations of back constraints and different approximations are used in the exmaples below.\n\n\\subsection{Oil Data}\n\nThe `oil data' is commonly used as a bench mark for visualisation algorithms. For more details on the data see this page.\n\nThe C++ implementation of the GP-LVM has details on training the full GP-LVM with this data set. Here we will consider the three different approximations outlined above.\n\n\\subsubsection{FITC Approximation}\n\nIn all the examples we give there will be 100 points in the active set. We first considered the FITC approximation. The script \\texttt{demOilFgplvm1.m} runs the FITC approximation giving the result on the left of the figure shown below.\n\n\\begin{center}\\inputdiagram{demOilFgplvm1} width=\"0.49\\textwidth\">\\inputdiagram{demOilFgplvm2} width=\"49%\">\n\\emph{Left}: GP-LVM on the oil data using the FITC approximation without back constraints. The phases of flow are shown as green circles, red crosses and blue plusses. One hundred inducing variables are used. \\emph{Right}: Similar but for a back-constrained GP-LVM, the back constraint is provided by a multi-layer perceptron with 15 hidden nodes.\\end{center}\n\nBack constraints can be added to each of these approximations. In the example on the right we used a back constraint given by a multi-layer perceptron with 15 hidden nodes. This example can be recreated with \\texttt{demOilFgplvm2.m}.\n\n\\subsubsection{DTC Approximation}\n\nThe other approximations can also be used, in the figures below we give results from the DTC approximation. The can be recreated using \\texttt{demOil3.m} and \\texttt{demOil4.m}.\n\n\\begin{center}\\inputdiagram{demOilFgplvm3} width=\"49%\">\\inputdiagram{demOilFgplvm4} width=\"49%\">\n\\emph{Left}: GP-LVM on the oil data using the DTC approximation without back constraints. The phases of flow are shown as green circles, red crosses and blue plusses. One hundred inducing variables are used. \\emph{Right}: Similar but for a back-constrained GP-LVM, the back constraint is provided by a multi-layer perceptron with 15 hidden nodes.\\end{center}\n\n\\subsubsection{PITC Approximation}\n\nWe also show results using the PITC approximation, these results can be recreated using the scripts \\texttt{demOilFgplvm5.m} and \\texttt{demOilFgplvm6.m}.\n\n\\begin{center}\\inputdiagram{demOilFgplvm5} width=\"49%\">\\inputdiagram{demOilFgplvm6} width=\"49%\">\n\\emph{Left}: GP-LVM on the oil data using the PITC approximation without back constraints. The phases of flow are shown as green circles, red crosses and blue plusses. One hundred inducing variables are used. \\emph{Right}: Similar but for a back-constrained GP-LVM, the back constraint is provided by a multi-layer perceptron with 15 hidden nodes.\\end{center}\n\n\\subsubsection{Variational DTC Approximation}\n\nFinally we also show results using the variational DTC approximation of Titsias, these results can be recreated using the scripts \\texttt{demOilFgplvm7.m} and \\texttt{demOilFgplvm8.m}.\n\n\\begin{center}\\inputdiagram{demOilFgplvm7} width=\"49%\">\\inputdiagram{demOilFgplvm8} width=\"49%\">\n\\emph{Left}: GP-LVM on the oil data using the variational DTC approximation without back constraints. The phases of flow are shown as green circles, red crosses and blue plusses. One hundred inducing variables are used. \\emph{Right}: Similar but for a back-constrained GP-LVM, the back constraint is provided by a multi-layer perceptron with 15 hidden nodes.\\end{center}\n\n\n\\subsection{Back Constraints and Dynamics}\n\nFirst we will demonstrate the dynamics functionality of the toolbox. We raw x-y-z values from a motion capture data set, the \\texttt{Figure Run 1} example available from Ohio State University. To run without dynamics use the script:\n\\texttt{\n demStickFgplvm1\n\\end{octave}\n\nThe results are given on the left of the figure below.\n\n\\begin{center}\\inputdiagram{demStickFgplvm1} width=\"49%\">\nGP-LVM on the motion capture data without dynamics in the latent space. \\end{center}\n\nNotice that the sequence (which is a few strides of a man running) is split into several sub-sequences. These sub-sequences are aligned to the strides of the man. By introducing a dynamics prior, we can force the sequence to link up. Samples from the dynamics prior used are shown in the plot below.\n\n\\begin{center}\\inputdiagram{dynamicsSamp1} width=\"49%\">\\inputdiagram{dynamicsSamp2} width=\"49%\">\n\\inputdiagram{dynamicsSamp3} width=\"49%\">\\inputdiagram{dynamicsSamp4} width=\"49%\">\nSamples from the dynamics prior which is placed over the latent space. This prior has \\emph{Left}: GP-LVM on the motion capture data without dynamics in the latent space. \\emph{Right}: GP-LVM with dynamics. Samples from the dynamics prior used are given in the figure above.\\end{center}\n\nThis prior is used in the model to obtain the results below,\n\n\\begin{octave}\n demStickFgplvm2\n\\end{octave}\n\n\\begin{center}\\inputdiagram{demStickFgplvm2} width=\"49%\">\nGP-LVM with dynamics. Samples from the dynamics prior used are given in the figure above.\\end{center}\n\nNote now the circular form of the latent space. \n\nBack constraints can also be used to achieve a similar effect,\n\n\\begin{octave}\n demStickFgplvm3\n\\end{octave}\n\n\\begin{center}\\inputdiagram{demStickFgplvm3} width=\"49%\">\nGP-LVM with back constraints. A RBF kernel mapping was used to form the back constraints with the inverse width set to 1e-4 (\\emph{i.e.}length scale set to 100).\\end{center}\n\n\\subsection{Loop Closure in Robotics}\n\nIn on-going work with Dieter Fox and Brian Ferris at the University of Washington we are interested in loop closure for robotic navigation, included as an example is a data set of a robot completing a loop while reading wireless access point signal strengths. To produce a neat track and close the loop it turns out it is necessary to use dynamics and back constraints as seen in the images below. These results can be recreated with \\texttt{demRobotWireless1.m} through \\texttt{demRobotWireless4.m}.\n}\n\n\\begin{center}\\inputdiagram{demRobotWireless1} width=\"49%\">\\inputdiagram{demRobotWireless2} width=\"49%\">\n\\inputdiagram{demRobotWireless3} width=\"49%\">\\inputdiagram{demRobotWireless4} width=\"49%\">\nUse of back constraints and dynamics to obtain loop closure in a robot navigation example. \\emph{Top Left}: GP-LVM without back constraints or dynamics, \\emph{Top right}: GP-LVM with back constraints, no dynamics, \\emph{Bottom Left}: GP-LVM with dynamics, no back constraints, \\emph{Bottom right}: GP-LVM with back constraints and dynamics. \\end{center}\n\n\\subsection{Vocal Joystick and Vowel Data}\n\nAnother ongoing piece of work with Jeff Bilmes and Jon Malkin involves embedding vowel sounds in a two dimensional space as part of vocal joystick system. Jon has provided a simple data set of 2,700 examples of different vowels. These are embedded in a two dimensional latent space with and without back constraints.\n\n\\begin{center}\\inputdiagram{demVowels2} width=\"49%\">\\inputdiagram{demVowels3} width=\"49%\">\n\\emph{Left}: embedding of the vowel data without back constraints, \\emph{Right}: embedding of the vowel data with back constraints. $/a/$ - red cross, $/ae/$ - green circle, $/ao/$ - blue plus, $/e/$ - cyan asterix, $/i/$ - magenta square, $/ibar/$ - yellow diamond, $/o/$ - red down triangle, $/schwa/$ - green up triangle, $/u/$ - blue left triangle.\n\\end{center}\n\n\\subsection{Larger Human Motion Data Sets}\n\nFor an AISTATS paper we recreated an experiment from Taylor et al.'s NIPS paper. They created a data set from a motion capture data in the CMU data base of running and walking. The data set can now be recreated using the DATASETS toolbox. We repeated missing data experiments by Taylor et al.. The model learning for these experiments can be recreated with:\n\n\\begin{octave}\n demCmu35gplvm1\n\\end{octave}\n\nfor the four dimensional latent space, \\texttt{demCmu35gplvm2} for the three dimensional latent space and \\texttt{demCmu35gplvm3} for the five dimensional latent space. The test data reconstruction can then be performed for all models with \\texttt{demCmu35gplvmReconstruct}. Taylor et al.'s nearest neighbour results can be recreated using \\texttt{demCmu35TaylorNearestNeighbour}.\n\nData was pre-processed by mapping angles to be between -180 and 180 and scaling the data such that the variance of each dimension was one.\nThe quality of the trained model was evaluated using a missing data problem with a test sequence of data. The model was required to fill in either upper body angles or right leg angles. Results for the GP-LVM and nearest neighbour in both scaled space and original angle space are given in the table below.\n\\begin{center}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LegLegBodyBody
CumulativeRMSCumulativeRMS
ScaledAnglesScaledAngles
GP-LVM ($q$=3)11.43.4016.92.49
GP-LVM ($q$=4)9.73.3820.72.72
GP-LVM ($q$=5)13.44.2523.42.78
Scaled NN13.54.4420.82.62
Nearest Neighbour14.04.1130.93.20
\n\\end{center}\nThe cumulative scaled error is a recreation of the error reported in Taylor et al. which was the average (across angles) cumulative sum (across time) of the squared errors in the down-scaled (\\emph{i.e.} variance one) space of angles. We also present the root mean squared angle error for each joint which we find to be a little easier to interpret.\n\nTaylor et al. used a slightly different representation of\nthe data set which included the absolute $x$ and $z$\nposition of the root node and rotation around the $y$-axis. For\nthis data set, this information does help, principally because the\nsubject seems to start in roughly the same position at the beginning\nof each sequence. However, in general absolute position will not help,\nso we discarded it in favour of a representation of these values in\nterms of differences between frames. Finally Taylor et al.\nconcatenated two frames to form each data point for the model. We\nchose not to do this as we wanted to test the ability of the Gaussian\nprocess dynamics to fully recreate the data set. There results are given in their paper and summarised below.\n\n\\begin{center}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LegBody
CumulativeCumulative
ScaledScaled
Binary Latent Variable Model11.78.8
Scaled NN22.220.5
\n\\end{center}\nFinally we show a plot of reconstructions of two of the angles in the data.\n\n\\begin{center}\\inputdiagram{demCmu35gplvmLegReconstruct1_8} width=\"0.5\\textwidth\">\\inputdiagram{demCmu35gplvmLegReconstruct1_9} width=\"0.5\\textwidth\">\\end{center}\nPrediction for first two angles of the right hip joint (see plots in Taylor et al. for comparison). Dotted line is nearest neighour in scaled space, dashed line is GP-LVM with 4-D latent space.\\end{center}\n\n\n\n\\section{Informative Vector Machines}\n\n\\section{\\texttt{demClassificationOneIvm1}}\n\nThe first example given is \\texttt{demClassificationOneIvm1} which is a simple classification data set, where only one direction of the input is relevant in determining the decision boundary. An ARD MLP kernel is used in combination with a linear kernel. The ARD parameters in the linear and MLP kernel are constrained to be the same by the line:\n\n\\begin{octave}\n\n% Constrain the ARD parameters in the MLP and linear kernels to be the same.\n\nmodel.kern = cmpndTieParameters(model.kern, {[4, 7], [5, 8]});\n\n\\end{octave}\n\nThe resulting classification is shown below.\n\n\\begin{center}\\inputdiagram{demClassificationOneIvm1}\n\nDecision boundary from the \\texttt{demClassificationOneIvm1.m} example. Postive class is red circles, negative class green crosses and active points are yellow dots. Decision boundary shown in red, contours at 0.25 and 0.75 probability shown in blue.\\end{center}\n\n\n\n\n\\section{\\texttt{demClassificationTwoIvm1}}\n\nThe second example attempts to learn a Gaussian process give data that is sampled from a Gaussian process. The code is \\texttt{demClassificationTwoIvm1}. The underlying Gaussian process is based on an RBF kernel with variance inverse width 10. The IVM learns an inverse width of 15 and gives the classification is shown below.\n\n\\begin{center}\\inputdiagram{demClassificationTwoIvm1}\n\nDecision boundary from the \\texttt{demClassificationTwoIvm1.m} example. Postive class is red circles, negative class green crosses and active points are yellow dots. Decision boundary shown in red, contours at 0.25 and 0.75 probability shown in blue.\\end{center}\n\n\n\\section{\\texttt{demOrderedOneIvm1}}\n\nIn this example the ordered categorical noise model is used (ordinal regression). The data is a simple data set for which a linear one dimensional model suffices. The IVM is given a combination of an RBF and linear kernel with ARD.For the ordered categorical case there are several parameters associated with the noise model (in particular the category widths), these are learnt too. The model learns that the system is linear and only one direction is important. The resulting classification is given below.\n\n\n\n\\begin{center}\\inputdiagram{demOrderedOneIvm1}\n\nDecision boundary from the \\texttt{demOrderedOneIvm1.m} example. Class 0 - red cross, Class 1 - green circles, Class 2 - blue crosses, Class 3 - cyan asterisks, Class 4 - pink squares, Class 5 - yellow diamonds. Class 6 - red triangles. Active points are yellow dots, note that because the kernel is linear by now the most informative points tend to be at the extrema. Decision boundaries shown in red, contours at 0.25 and 0.75 probability shown in blue.\\end{center}\n\n\n\n\\section{\\texttt{demOrderedTwoIvm1}}\n\nAnother example with the ordered categorical noise model, here the data is radial, the categories being along the radius of a circle. The IVM is given a combination of an RBF and linear kernel with ARD. Again there are several parameters associated with the noise model, and these are learnt using \\texttt{ivmOptimiseNoise}. The resulting classification is given below.\n\n\n\n\\begin{center}\\inputdiagram{demOrderedTwoIvm1}\n\nDecision boundary from the \\texttt{demOrderedTwoIvm1.m} example. Class 0 - red cross, Class 1 - green circles, Class 2 - blue crosses, Class 3 - cyan asterisks, Class 4 - pink squares, Class 5 - yellow diamonds. Class 6 - red triangles. Active points are yellow dots, note that because the kernel is linear by now the most informative points tend to be at the extrema. Decision boundaries shown in red, contours at 0.25 and 0.75 probability shown in blue.\\end{center}\n\n\n\n\n\n\\section{\\texttt{demRegressionOneIvm1}}\n\nIn this example the Gaussian noise model is used (standard regression). The data is sampled from a Gaussian process, only one input dimension is important. The IVM is given a combination of an RBF and linear kernel with ARD. The resulting regression is given below.\n\n\n\n\\begin{center}\\inputdiagram{demRegressionOneIvm1}\n\nRegression from the example \\texttt{demRegressionOneIvm1.m}. Targets are red dots and active points are yellow dots.\\end{center}\n\n\n\n\\section{\\texttt{demRegressionTwoIvm1}}\n\nA second example with Gaussian noise, sampled from a Gaussian process, but this time with differing length scales.\n\n\n\n\\begin{center}\\inputdiagram{demRegressionTwoIvm1}\n\nRegression from the example \\texttt{demRegressionTwoIvm1.m}. Targets are red dots and active points are yellow dots.\\end{center}\n\n\n\n\\section{Benchmark Data Sets}\n\n\n\nThe function \\texttt{ivmGunnarData} allows you to test the IVM on Gunnar Raetsch's benchmark data sets. Download the data sets, from here and expand the ringnorm data set into '\\$DATASETSDIRECTORY/gunnar/ringnorm'. Then run the following script.\n\n\n\n\n\n\\begin{octave}\n%}\nivmGunnarData('ringnorm', 1, {'rbf', 'bias', 'white'}, 1, 100)\n%{\n\\end{octave}\n\\begin{verbatim}\n\nFinal model:\n\nIVM Model:\n\n Noise Model:\n\n Probit bias on process 1: 0.0439\n\n Probit Sigma2: 0.0000\n\n Kernel:\n\n Compound kernel:\n\n RBF inverse width: 0.0866 (length scale 3.3984)\n\n RBF variance: 1.2350\n\n Bias Variance: 8.2589\n\n White Noise Variance: 0.0000\n\nTest Error 0.0183\n\nModel likelihood -56.7120\n\n\\end{verbatim}\n\n\n\nYou can try any of the data sets by replacing ringnorm with the relevant data set (note that they don't all work with only 100 active points inas in the example above, for example the 'banana' data set needs 200 active points to get a reasonable result,\n\n\n\n\\begin{octave}\n\n ivmGunnarData('banana', 1, {'rbf', 'bias', 'white'}, 1, 200)\n\\end{octave}\n\n\\begin{verbatim}\nFinal model:\n\nIVM Model:\n\n Noise Model:\n\n Probit bias on process 1: 0.1067\n\n Probit Sigma2: 0.0000\n\n Kernel:\n\n Compound kernel:\n\n RBF inverse width: 1.6411 (length scale 0.7806)\n\n RBF variance: 0.2438\n\n Bias Variance: 0.0000\n\n White Noise Variance: 0.0148\n\nTest Error 0.1129\n\nModel likelihood 175.3588\n\n\\end{verbatim}\n\n\n\n\n\n\\begin{center}\n\\inputdiagram{demBanana1}\n\nDecision boundary from the banana example. Postive class is red circles, negative class green crosses and active points are yellow dots. Decision boundary shown in red, contours at 0.25 and 0.75 probability shown in blue.\n\\end{center}\n\n\n\n\\section{Null Category Noise Model}\n\n\n\n\\section{Examples}\n\n\n\nThe toy data example in the papers can be recreated using:\n\n\n\n\\begin{octave}\n\n demUnlabelledOneIvm1\n\n\\end{octave}\n\n\n\nand leads to the decision boundary given below. A standard IVM based classifier can be run on the data using\n\n\n\n\\begin{octave}\n\n demUnlabelledOneIvm2\n\n\\end{octave}\n\n\n\n\\begin{center}\\inputdiagram{demUnlabelledOneIvm1}\\inputdiagram{demUnlabelledOneIvm2}\n\nThe null category noise model run on toy data. \\emph{Top}: using the null category, the true nature of the decision boundary is recovered. \\emph{Bottom}: the standard IVM, does not recover the true decision boundary.\\end{center}\n\n\n\nThe other USPS digit classification example given in the NIPS paper can be re-run with:\n\n\n\n\\begin{octave}\n\n demThreeFiveIvm\n\n\\end{octave}\n\n\n\nBe aware that this code can take some time to run. The results, in the form of averaged area under ROC curve against probability of missing label, can be plotted using\n\n\n\n\n\n\\begin{octave}\n%}\n demThreeFiveResults\n%{\n\\end{octave}\n\n\\begin{center}\n\n\\inputdiagram{demThreeFive} \n\nPlot of average area under ROC curve against probability of label being present. The red line is the standard IVM based classifier, the blue dotted line is the null category noise model based classifier, the green dash-dot line is the a normal SVM and the mauve dashed line is the transductive SVM.\n\n\\end{center}\n\n\n\n\n\\end{document}\n%}", "meta": {"hexsha": "ba05aaa7ed3f6703c6696f39f73a670c6816b080", "size": 35259, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "html/vignette.tex", "max_stars_repo_name": "lawrennd/GPmat", "max_stars_repo_head_hexsha": "c96e814f47ad0ab84dd5c0329eb4dd6eb991e5fc", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-07-30T00:07:33.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-30T00:07:33.000Z", "max_issues_repo_path": "html/vignette.tex", "max_issues_repo_name": "lawrennd/GPmat", "max_issues_repo_head_hexsha": "c96e814f47ad0ab84dd5c0329eb4dd6eb991e5fc", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "html/vignette.tex", "max_forks_repo_name": "lawrennd/GPmat", "max_forks_repo_head_hexsha": "c96e814f47ad0ab84dd5c0329eb4dd6eb991e5fc", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-05-28T09:59:44.000Z", "max_forks_repo_forks_event_max_datetime": "2018-06-06T18:04:53.000Z", "avg_line_length": 42.4807228916, "max_line_length": 659, "alphanum_fraction": 0.7711506282, "num_tokens": 9689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802264851919, "lm_q2_score": 0.8006920092299293, "lm_q1q2_score": 0.6157963918034375}} {"text": "\n\\subsection{Monte-Carlo Tree Search}\n\n\\subsubsection{Introduction}\n\nIn search tree, each node has wins/total.\n\nSo start with just root in 0/0.\n\nAlgo:\n\n\\begin{itemize}\n\\item Start at root\n\\item Take n choices to arrive at node which has not been explored (or until w/l state)\n\\item play randomly from there\n\\item Back prob up (eg if win, then 1/1 for path back to root, or 0/1 if loss)\n\\end{itemize}\n\nWins/simulation count\n\nSo deterministic when choosing paths, up until play randomly.\n\n", "meta": {"hexsha": "a23fcda2129f2f30f6b6655db6fc6baa56a9ae13", "size": 487, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/ai/gameTheoryDeep/03-01-MCTS.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/ai/gameTheoryDeep/03-01-MCTS.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/ai/gameTheoryDeep/03-01-MCTS.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.1739130435, "max_line_length": 87, "alphanum_fraction": 0.7535934292, "num_tokens": 133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8006919925839875, "lm_q2_score": 0.7690802264851919, "lm_q1q2_score": 0.6157963790013727}} {"text": "%!TEX root = ../notes.tex\n\\section{February 8, 2022}\n\n\\subsection{Dirichlet Convolutions}\n\n\\begin{definition}[Dirichlet Convolution]\n Let $f, g$ be arithmetic functions. Then the \\ul{Dirichlet convolution/product} of $g$ and $g$ it\n \\begin{align*}\n (f * g)(n) : & = \\sum_{d_1d_2=n}f(d_1)g(d_2) \\\\\n & = \\sum_{d\\mid n}f(d) g(n/d)\n \\end{align*}\n\\end{definition}\n\nWe do check that this has properties that we want it to have, like associativity:\n\\begin{align*}\n ((f * g) * h)(n) & = (f * (g * h))(n) \\\\\n & = \\sum_{d_1d_2d_3=n} f(d_1)g(d_2)h(d_3)\n\\end{align*}\nIt is also clearly commutative.\n\nWe also have that this product has a multiplicative identity.\n\n\\begin{definition}\n Let $I : \\ZZ_+\\to \\{0, 1\\}$ be given by\n \\[I(n) = \\begin{cases}\n 1 & \\text{if } n = 1 \\\\\n 0 & \\text{otherwise}\n \\end{cases}\\]\n Then $I$ is an identity for $*$, in the sense that $f * I = I * f = f$.\n\\end{definition}\n\n\\begin{lemma}\n If $f$ is an arithmetic function such that $f(1)\\neq 0$, then there exists an arithmetic function $g$ such that $f * g = I$.\n\n It is given recursively by\n \\begin{align*}\n g(1) & = \\frac{1}{f(1)} \\\\\n g(n) & = -\\frac{1}{f(1)}\\cdot \\sum_{d\\mid n, d < n} g(d) f(n/d)\n \\end{align*}\n\\end{lemma}\n\\begin{proof}\n We want to show that given $g$ and $g$ defined as above, we have that $f * g = I$.\n\n $n=1$: \\[g(1)\\cdot f(1) = \\frac{1}{f(1)}f(1) = 1\\]\n\n $n > 1$: \\begin{align*}\n \\sum_{d\\mid n}g(d) f(n/d) & = g(n)\\cdot f(1) + \\sum_{d\\mid n, n 1$, we have $(i * \\mu)(n) = \\displaystyle \\sum_{d\\mid n}\\mu(d) = 0$ from above.\n\\end{proof}\n\nWe see here that summatory functions can be seen as Dirichlet products: the summatory function $F$ of $f$ is $F = f * i$. What we said about summatory functions being multiplicative boils down to Dirichlet convolutions preserving multiplicativity.\n\n\\recall that summatory functions inherit multiplicativity. In fact, this holds for Dirichlet products as well. If $f, g$ are multiplicative, then so is $f * g$.\n\nThe proof is parallel to the proof for summatory functions, for \\cref{lemma:summatory-preserves-multiplicativity}.\n\n\\begin{theorem}[M\\\"obius Inversion]\n Let\n \\begin{align*}\n F(n) & = \\sum_{d\\mid n} f(d) \\\\\n \\intertext{Then we have }\n f(n) & = \\sum_{d\\mid n}\\mu(d)\\cdot F(n/d) = \\mu * F.\n \\end{align*}\n\\end{theorem}\n\\begin{proof}\n $F = f * i$, then \\[F * \\mu = (f * i) * \\mu = f * (i * \\mu) = f * I = f.\\]\n which was simpler than I expected\\dots\n\\end{proof}\n\n\\begin{corollary}\\label{cor:summatory-multiplicativity-original-multiplicative}\n If $F$ is the summatory function of $f$, and $F$ is multiplicative, then $f$ is also multiplicative, as $f = \\mu * F$ and $\\mu$ is multiplicative and convolutions with multiplicative functions are multiplicative.\n\\end{corollary}\n\\begin{corollary}\n \\Cref{cor:summatory-multiplicativity-original-multiplicative} gives another proof that $\\phi$ is multiplicative, as \\[\\sum_{d\\mid n}\\phi(d) = \\phi * i = \\mathrm{id}.\\]\n\\end{corollary}\n\n\\subsection{Applications of M\\\"obius Inversion}\n\\subsubsection{Cyclotomic Polynomials}\n\n\\recall the $n^\\mathrm{th}$ cyclotomic polynomial $\\Phi_n(x)$ is the unique irreducible polynomial in $\\ZZ[x]$ dividing $x^n - 1$ but no $x^k - 1$ for $k < n$.\n\nThus \\[\\Phi_n(x) = \\prod_{\\substack{1\\leq k < n \\\\ (k, n) = 1}} \\left(x - e^{2\\pi i k / n}\\right)\\]\nas the roots of this polynomial are exactly the primitive $n^\\mathrm{th}$ roots of unity. We have that \\[\\prod_{d\\mid n}\\Phi_d(x) = x^n - 1.\\]\nBy M\\\"obius inversion, if\n\\begin{align*}\n G(n) & = \\prod_{d\\mid n}g(d),\n \\intertext{then we have that}\n g(n) & = \\prod_{d\\mid n}G(d)^{\\mu(n/d)}\n\\end{align*}\nIn particular, taking $G(n) = x^n - 1$ (with particular $x\\in\\CC$) as an arithmetic function, we have \\[\\Phi_n(x) = \\prod_{d\\mid n}(x^d - 1)^{\\mu(n/d)}\\qquad (x\\in \\CC)\\]\nApplying this identity for enough $x\\in \\CC$ yields this as an identity of polynomials.\n\n\\subsubsection{Dynatomic Polynomials}\nThe roots of cyclatomic polynomials are roots of unity. Dynatomic polynomials have as roots the periodic points (of certain periods) of a polynomial.\n\n\\begin{definition}\n Let $K$ be a field, and let $f\\in K[x]$ of degree $d\\geq 2$. Let \\[f^n = \\underbrace{f\\circ f\\circ \\cdots \\circ f}_{n\\text{ times}}\\]\n then $P\\in \\overline{K}$\\footnote{Algebraic numbers in field $K$, the field you get by adjoining all roots of polynomials in $K[x]$.} is said to be \\ul{periodic} under $f$ if\n \\[f^n(P) = P\\qquad \\text{for some $n\\geq 1$}\\]\n\\end{definition}\n\\begin{example}\n Let $f(x) = x^2 - 1$. $0$ is a period point under $f$:\n \\[0\\longmapsto -1\\longmapsto 0\\]\n and its \\ul{period} is $2$.\n\n \\begin{remark*}\n If $n$ is the smallest positive integer such that $f^n(p) = p$ ($p$ periodic), then we call $n$ the \\ul{exact period} of $p$ under $f$.\n \\end{remark*}\n\\end{example}\n\\begin{definition}\n The $n^\\mathrm{th}$ dynatomic polynomial of $f$ is \n \\[\\Phi_{f,n}(x) := \\prod_{d\\mid n}\\left(f^d(x) - x\\right)^{\\mu(n/d)}\\]\n\\end{definition}\nWe hope that $\\Phi_{f, n}(x)$ has as its roots the points of exact period $n$\\dots This hope is dashed\\dots \n\\begin{example}\n $f(x) = x^2 - \\frac{3}{4}$. \n \\begin{align*}\n f^2(x) - x &= \\left(x - \\frac{3}{2}\\right)\\left(1 - \\frac{1}{2}\\right)^3 \\\\\n f(x) - x &= \\left(x-\\frac{3}{2}\\right)\\left(x + \\frac{1}{2}\\right)\n \\end{align*}\n Thus \\[\\frac{f^2(x) - x}{f(x) - x} = \\left(x + \\frac{1}{2}\\right)^2\\]\n But $x=-\\frac{1}{2}$ is \\ul{fixed} under $f$. \n\\end{example}", "meta": {"hexsha": "84dd41298cb6e7696c535f1ae18bc9153a81ff61", "size": 8357, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/2022-02-08.tex", "max_stars_repo_name": "jchen/math1560-notes", "max_stars_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_stars_repo_licenses": ["BSL-1.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2022-02-02T15:41:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-03T20:28:48.000Z", "max_issues_repo_path": "lectures/2022-02-08.tex", "max_issues_repo_name": "jchen/math1560-notes", "max_issues_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_issues_repo_licenses": ["BSL-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lectures/2022-02-08.tex", "max_forks_repo_name": "jchen/math1560-notes", "max_forks_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_forks_repo_licenses": ["BSL-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.7539267016, "max_line_length": 247, "alphanum_fraction": 0.5784372382, "num_tokens": 2957, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006919925839875, "lm_q2_score": 0.7690802264851918, "lm_q1q2_score": 0.6157963790013726}} {"text": "\\section{Dynamic Programming}\n\\lst{Knapsack}{$\\mathcal{O}(n\\sum_{i = 1}^n p_i)$}{dynamicProgramming/knapsack.cc}\n\\lst{TSP}{$\\mathcal{O}(n 2^n)$}{dynamicProgramming/tsp.cc}\n\\lst{Subset Sum}{$\\mathcal{O}(n\\sum_{i = 1}^n v_i)$}{dynamicProgramming/subSetSum.cc}\n\\lst{Edit Distance}{$\\mathcal{O}(nm)$}{dynamicProgramming/editDistance.cc}\n\\lst{Longest Increasing Subsequence}{$\\mathcal{O}(n\\log n)$}{dynamicProgramming/lis.cc}\n\\lst{Longest Common Subsequence}{$\\mathcal{O}(nm)$}{dynamicProgramming/lcs.cc}\n", "meta": {"hexsha": "bdd0ffa493d757f6b87ed02d9fb79e70cbc99ef8", "size": 500, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "document/dynamicProgramming.tex", "max_stars_repo_name": "Zeldacrafter/CompProg", "max_stars_repo_head_hexsha": "5367583f45b6fe30c4c84f3ae81accf14f8f7fd3", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2020-02-06T15:44:57.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-21T03:51:21.000Z", "max_issues_repo_path": "document/dynamicProgramming.tex", "max_issues_repo_name": "Zeldacrafter/CompProg", "max_issues_repo_head_hexsha": "5367583f45b6fe30c4c84f3ae81accf14f8f7fd3", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "document/dynamicProgramming.tex", "max_forks_repo_name": "Zeldacrafter/CompProg", "max_forks_repo_head_hexsha": "5367583f45b6fe30c4c84f3ae81accf14f8f7fd3", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 62.5, "max_line_length": 87, "alphanum_fraction": 0.734, "num_tokens": 171, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8933094088947399, "lm_q2_score": 0.6893056231680122, "lm_q1q2_score": 0.6157631987800374}} {"text": "\\hypertarget{k-nearest-neighbor-k-nn}{%\n\\chapter{k-Nearest Neighbor (k-NN)}\\label{k-nearest-neighbor-k-nn}}\n\\section{Introduction}\nK-NN algorithm assumes the similarity between the new case/data and\navailable cases and put the new case into the category that is most\nsimilar to the available categories. K-NN algorithm stores all the\navailable data and classifies a new data point based on the similarity.\nThis means when new data appears then it can be easily classified into a\nwell suite category by using K- NN algorithm. K-NN algorithm can be used\nfor Regression as well as for Classification but mostly it is used for\nthe Classification problems.\n\nK-NN is a non-parametric algorithm, which means it does not make any\nassumption on underlying data. It is also called a lazy learner\nalgorithm because it does not learn from the training set immediately\ninstead it stores the dataset and at the time of classification, it\nperforms an action on the dataset. K-NN algorithm at the training phase\njust stores the dataset and when it gets new data, then it classifies\nthat data into a category that is much similar to the new data.\n\\section{Creating Theoretical Model}\nStep-1: Select the number K of the neighbors. \\(k\\)=5 is preferred.\n\nStep-2: Calculate the Euclidean distance of \\(k\\) number of neighbors.\n\\[\nEuclidean\\ Distance = \\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\n\\]\n\nStep-3: Take the K nearest neighbors as per the calculated Euclidean\ndistance. We may use other means of finding sum of distance calculations\nas well with Minkowski by setting \\(p\\) to \\(1\\) and \\(2\\) for using\nManhattan and Euclidean, respectivly.\n\\[\n||x_1-x_2|| = (\\sum_{i=1}^{n}|x_i-y_i|^p)^\\frac{1}{p}\n\\]\n\nStep-4: Among these k neighbors, count the number of the data points in\neach category.\n\nStep-5: Assign the new data points to that category where the neighbor is maximum.\n\\section{Implementation}\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{1}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{k+kn}{import} \\PY{n+nn}{numpy} \\PY{k}{as} \\PY{n+nn}{np}\n\\PY{k+kn}{import} \\PY{n+nn}{pandas} \\PY{k}{as} \\PY{n+nn}{pd}\n\\PY{k+kn}{from} \\PY{n+nn}{sklearn}\\PY{n+nn}{.}\\PY{n+nn}{model\\PYZus{}selection} \\PY{k+kn}{import} \\PY{n}{train\\PYZus{}test\\PYZus{}split}\n\\PY{k+kn}{from} 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opacityfill=0]\n\\prompt{Out}{outcolor}{10}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\narray([1], dtype=int64)\n\\end{Verbatim}\n\\end{tcolorbox}\n \n \\begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]\n\\prompt{In}{incolor}{11}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\n\\PY{n}{make\\PYZus{}predection}\\PY{p}{(}\\PY{l+m+mi}{5}\\PY{p}{,} \\PY{l+m+mi}{100}\\PY{p}{)}\n\\end{Verbatim}\n\\end{tcolorbox}\n\n \\begin{tcolorbox}[breakable, size=fbox, boxrule=.5pt, pad at break*=1mm, opacityfill=0]\n\\prompt{Out}{outcolor}{11}{\\boxspacing}\n\\begin{Verbatim}[commandchars=\\\\\\{\\}]\narray([0], dtype=int64)\n\\end{Verbatim}\n\\end{tcolorbox}", "meta": {"hexsha": "2ca571cb546b3fec47f9d476916c51bf672eebd8", "size": 13160, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MCA/Machine Learning/Project file/Tex source/k-Nearest Neighbor.tex", "max_stars_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_stars_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2018-03-13T12:34:51.000Z", "max_stars_repo_stars_event_max_datetime": "2018-10-02T18:54:22.000Z", "max_issues_repo_path": "MCA/Machine Learning/Project file/Tex source/k-Nearest Neighbor.tex", "max_issues_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_issues_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "MCA/Machine Learning/Project file/Tex source/k-Nearest Neighbor.tex", "max_forks_repo_name": "muhammadmuzzammil1998/CollegeStuff", "max_forks_repo_head_hexsha": "618cec9ebfbfd29a2d1e5a182b90cfb36b38a906", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.5098039216, "max_line_length": 563, "alphanum_fraction": 0.5886778116, "num_tokens": 5798, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8723473879530491, "lm_q2_score": 0.7057850278370112, "lm_q1q2_score": 0.6156897254899868}} {"text": "\\FloatBarrier\n\\section{Combined Dynamics} \\label{sec:combined}\n\nIt is possible to define a set of dynamics to run a combination of the dynamics. \nThe resulting dynamic is defined as \n\\begin{equation}\n\\dot{ x } = \\sum_{d\\in \\mathcal{D}} \\gamma_d V_d( x ),\n\\end{equation}\nwhere $\\mathcal{D}=\\{ Logit, RD, Smith, BNN \\}$ denotes the set of available dynamics, $V_d()$ is the differential equation of the $d\\th$ dynamic and $\\gamma_d$ is the weight assigned to it.\nThe dynamics should be defined in a cell array, e.g., \n\\begin{lstlisting}\ndynamics = {'bnn', 'rd'};\n\\end{lstlisting}\nThe combination is made making a linear combination between each dynamic listed in the cell array. The weight assigned to each dynamic is defined in the vector \\verb|gamma|. In this case we assign \n\\begin{lstlisting}\ngamma = [.25, .75]; \n\\end{lstlisting}\n\nFig. \\ref{fig:rps_combined} shows an example of the combined dynamics for the rock-paper-scissors game. Note that the evolution of the system is not confined to a limit cycle, as happened with the replicator dynamics in Fig. \\ref{fig:finite1}.\n\n\\begin{figure}[h]\n \\centering\n \\begin{subfigure}[b]{0.4\\textwidth}\n\t \\includegraphics[width=\\textwidth]{./images/test_combined.eps}\n\t \\caption{Simplex.}\n\t \\label{fig:test_combined_simplex}\n \\end{subfigure}\n ~ \n \\begin{subfigure}[b]{0.45\\textwidth}\n\t \\includegraphics[width=\\textwidth]{./images/test_combined_ev.eps}\n\t \\caption{Evolution of the strategies in time.}\n\t \\label{fig:test_combined_ev}\n \\end{subfigure}\n \\caption{Evolution of the combination of replicator dynamics and BNN dynamics.}\n \\label{fig:rps_combined}\n\\end{figure}\n\n\n\n", "meta": {"hexsha": "d0f73d2dc62a1c9a1c1adf5b37d5d122343551ad", "size": 1630, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/combined_dynamics.tex", "max_stars_repo_name": "carlobar/PDToolbox_matlab", "max_stars_repo_head_hexsha": "fea827a80aaa0150932e6e146907f71a83b7829b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 28, "max_stars_repo_stars_event_min_datetime": "2017-08-13T09:50:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-08T09:22:42.000Z", "max_issues_repo_path": "docs/combined_dynamics.tex", "max_issues_repo_name": "sjtudh/PDToolbox_matlab", "max_issues_repo_head_hexsha": "fea827a80aaa0150932e6e146907f71a83b7829b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-07-25T13:04:08.000Z", "max_issues_repo_issues_event_max_datetime": "2020-07-03T21:16:17.000Z", "max_forks_repo_path": "docs/combined_dynamics.tex", "max_forks_repo_name": "sjtudh/PDToolbox_matlab", "max_forks_repo_head_hexsha": "fea827a80aaa0150932e6e146907f71a83b7829b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 17, "max_forks_repo_forks_event_min_datetime": "2015-07-16T00:40:13.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-26T10:20:34.000Z", "avg_line_length": 40.75, "max_line_length": 244, "alphanum_fraction": 0.7404907975, "num_tokens": 459, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8031737963569016, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6154669787282334}} {"text": "\\section{Alternative Estimators}\nIn this section we present a detailed overview of two alternative estimators which in theory exhibit better finite sample properties. \n\\par Both of these estimators, the Jackknife IV estimator and the LIML estimator, fall under the broader class of k-class estimators. These estimators are partially robust ie less sensitive to weak instruments; with weak instruments, they are more reliable than 2SLS estimates. Conventional asymptotic approximations are used for their inference.\n\n\\subsection{Jackknife Instrumental Variables Estimator}\nFor an intuitive sense of the functioning of the jackknife instrumental variables estimator, consider the term `jackknife' as presented in statistical literature. The jackknife estimator of a parameter is found by leaving out each observation from a dataset, calculating the estimate and computing the average of these calculations. Given a sample of size $N$, the jackknife estimate is found by aggregating the estimates of each sub-sample of size $(N-1)$. In a similar vein, the estimator we describe in this section replaces the usual fitted values from the reduced form regression of the two-stage least squares by `omit-one' fitted values.\n\\par The key feature of the JIVE estimator is that it eliminates the correlation between the fitted values and the structural equation errors, as explain in detail below. The fitted value of the standard 2SLS estimator is only asymptotically independent of the structural error ($\\varepsilon_i$), but the JIVE is independent even in finite samples.\n\\par The derivation we present in this section is adopted from \\cite{angrist1999jackknife}. To derive the expression formally, consider again the expressions for $\\hat\\beta_{OPT}$ and $\\hat\\beta_{2SLS}$ from section 2.2, equations (2.4) and (2.5). $\\hat\\beta_{OPT}$ employed the optimal instrument $\\mathbf Z\\pi$ and $\\hat\\beta_{2SLS}$ used an estimate of $\\mathbf Z\\pi$, denoted by $\\mathbf Z\\hat \\pi$. Now we introduce the new estimator for $\\beta$ by using a different estimate of the optimal instrument, $Z_{i}\\tilde{\\pi}$. \n\n\\par From the formulation of approximate bias of the 2SLS estimator we presented in section 2.4.2 (equation 2.11), it is clear that increasing the number of instruments while keeping the explanatory/predictive power of the instruments constant, leads to an increase in the bias of $\\hat\\beta_{2SLS}$. However, we can see that for the optimal estimator, increasing the number of instruments while keeping $Z_i\\pi$ fixed will have no effect on the properties of $\\hat\\beta_{OPT}$. Thus, for finite samples in the presence of many instruments, $\\hat\\beta_{2SLS}$ fares much worse than $\\hat\\beta_{OPT}$. \n\\par Recall the representation of the 2SLS estimator using the projection matrix $P_Z$ from Section 2 (equation 2.8). We can see that the first-stage fitted values $\\mathbf Z\\hat{\\pi}$ can be written as:\n\n\\begin{equation}\n\\mathbf Z\\hat{\\pi} = P_Z\\mathbf X = \\mathbf Z \\pi + P_Z\\eta\n\\end{equation}\n\nThe term $P_Z\\eta$ in the above equation is correlated with the error term of the first-stage $\\eta$ (see equation 1.2) and hence with the structural error term $\\varepsilon$. Put differently, since $\\hat \\pi$ is estimated on the full sample which includes the $i$th observation, it is correlated with $\\eta_i$, which is correlated with $\\varepsilon_i$. This correlation is given by:\n\n\\begin{equation}\nE[\\varepsilon_{i}Z_{i}\\hat{\\pi}]\n=E[E[\\varepsilon_i Z_i\\hat\\pi|\\mathbf Z]]\n=E[Z_i(\\mathbf Z^\\prime \\mathbf Z)^{-1} Z_i^\\prime] \\cdot E[\\varepsilon_i \\eta_i|\\mathbf Z]]\n=E[Z_i(\\mathbf Z^\\prime \\mathbf Z)^{-1} Z_i^\\prime \\cdot \\sigma_{\\varepsilon\\eta}^\\prime]\n=(K/N)\\cdot\\sigma_{\\varepsilon\\eta}'\n\\end{equation}\n\nDue to this correlation between $\\hat \\pi$ and $\\varepsilon_i$, $\\hat\\beta_{2SLS}$ is biased for $\\beta$. While the correlation disappears asymptotically, it holds implications in finite samples. \n\n\\par In devising the new instrument, we attempt to keep the correlation in (4.2) equal to zero. The problem stems from $\\hat \\pi$ being estimated on the full sample which includes the $i$th observation. For the new estimator which has the constructed instrument $Z_{i}\\tilde{\\pi}$, $\\tilde\\pi$ is estimated not on the full sample but on the sample with the $i$th observation removed. Therefore, the new estimated instrument will be independent of $\\varepsilon_i$ even in finite samples.\n\n \nThe $i$th row of the estimated instrument for 2SLS, $\\mathbf Z\\hat{\\pi}$, where $\\hat{\\pi}=(\\mathbf Z'\\mathbf Z)^{-1}(\\mathbf Z'\\mathbf X)$, is given by\n\\begin{equation}\nZ_{i}\\hat{\\pi}= Z_{i}({\\mathbf Z}'\\mathbf Z)^{-1}(\\mathbf Z'\\mathbf X)\n\\end{equation}\n \nRemove the $i$th row from the matrices of regressors and instruments and denote them by $\\mathbf X(i)$ and $\\mathbf Z(i)$. Thus JIVE removes the dependence between $Z_{i}\\hat{\\pi}$ and the regressor $X_{i}$.\n\nThe corresponding estimate of $\\pi$ and the constructed instrument will be:\n\\begin{equation}\n\\tilde{\\pi}(i)=(\\mathbf Z(i)'\\mathbf Z(i))^{-1}(\\mathbf Z(i)'\\mathbf X(i))\n\\end{equation}\n\n\\begin{equation}\nZ_{i}\\tilde{\\pi}(i)=Z_{i}(\\mathbf Z(i)'\\mathbf Z(i))^{-1}(\\mathbf Z(i)'\\mathbf X(i)Y)\n\\end{equation}\n$\\varepsilon_{i}$ and $X_{j}$ are independent when $i\\neq j$, so it follows that\n\\begin{equation}\nE[\\varepsilon_{i}Z_{i}\\tilde{\\pi}(i)]=E[Z_{i}(\\mathbf Z(i)'\\mathbf Z(i))^{-1}(\\mathbf Z(i)')E[\\mathbf X(i)\\varepsilon_{i}|\\mathbf Z]]=0\n\\end{equation}\n\nHence, now we have $E[\\varepsilon_{i}Z_{i}\\tilde{\\pi}(i)]=0$, so we have removed the correlation between the fitted values and the structural error presented in equation (4.2).\n\\par The JIVE estimator is equal to:\n\n\\begin{equation}\n\\hat{\\beta}_{JIVE}=(\\mathbf {\\hat{X}}_{JIVE}'\\mathbf X)^{-1}(\\mathbf {\\hat{X}}_{JIVE}'\\mathbf Y)\n\\end{equation}\nwhere $\\hat{X}_{JIVE}$ is $N\\times L$ dimensional matrix with the $i$ th row $Z_{i}\\tilde{\\pi}(i)$. \\\\\n\n\n\\par \\cite{blomquist1999small} summarize the construction of the JIVE estimator by providing the following algorithm: \n\n\n%\\vspace{0.6cm}\n\\begin{algorithm}\n\\caption{Jackknife Instrumental Variables Estimator}\\label{alg:euclid}\n\\begin{algorithmic}[1]\n\\State Use all observations but the $i$th to estimate parameters of the first-stage equation.\n\\State Combine the estimated first stage parameters with the instruments for the $i$th observation, $Z_i$, to construct a fitted value for the $i$th observation $X_i$.\n\\State Repeat steps 1 and 2 for all N observations.\n\\State Regress Y on the fitted values and the exogenous regressors.\n\\end{algorithmic}\n\\end{algorithm}\n\nThe estimator for $Z_i\\pi$, $Z_{i}\\tilde{\\pi}(i)$ is consistent.\n$\\hat{\\beta}_{JIVE}$ has the same probability limit and first-order asymptotic distribution as $\\hat{\\beta}_{OPT}$ and $\\hat{\\beta}_{2SLS}$, under conventional fixed model asymptotics (\\cite{stock2002survey}).\n\n\\subsection{Limited Information Maximum-Likelihood Estimator}\n\nLimited Information Maximum Likelihood (LIML) is an alternative method to estimate the parameters of the structural equation. It was formalized by \\cite{anderson1949estimation}.\\\\\nWe follow the same model as described in Section 2, and extend it as shown below: \n\\begin{equation}\n \\mathbf{Y} = \\mathbf{X}\\beta + \\varepsilon = \\mathbf{X_0}\\beta_0 + \\mathbf{X_1}\\beta_1 + \\varepsilon\n\\end{equation}\n\\begin{equation}\n \\mathbf{X_1} = \\mathbf{Z}\\pi + \\eta = \\mathbf{Z_0}\\pi_0 + \\mathbf{Z_1}\\pi_1 + \\eta\n\\end{equation}\nwhere $\\mathbf{Z_0} = \\mathbf{X_0}$.\\\\\nHere, $\\mathbf{X_0}$ is the matrix of exogenous variables and $\\mathbf{X_1}$ of endogenous variables. $\\mathbf{Z_0} = \\mathbf{X_0}$ is the matrix of included exogenous variables, $\\mathbf{Z_1}$ is the matrix of excluded exogenous variables.\nBecause the LIML estimator is based on the structural equation for $\\mathbf{Y}$ combined with the first-stage equation for $\\mathbf{X_1}$, it is called `limited information'. LIML estimator is given by:\n\\begin{equation}\n\\hat\\beta_{LIML}=(\\mathbf X'(\\mathbf I-\\hat\\kappa \\mathbf M_Z)\\mathbf X)^{-1}\\mathbf X'(\\mathbf I-\\hat\\kappa \\mathbf M_Z)\\mathbf Y\n\\end{equation}\n\\par The LIML estimator has some excellent properties when the number of excluded instruments (number of columns in matrix $\\mathbf X_1$) in the first-stage equation and the sample size are large. Although the LIML estimator and the 2SLS estimator are asymptotically equivalent in the standard large sample theory, they are quite different in case of many instruments or many weak instruments. It has no finite moments, which implies that its density tends to have very thick tails. \\cite{anderson2010asymptotic} show that the LIML estimator shows asymptotic optimality with many weak instruments.", "meta": {"hexsha": "0c104c89e7ed6e03e2425163b17718c7cdf3de67", "size": 8631, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ch04.tex", "max_stars_repo_name": "mchandra12/research_module_econometrics", "max_stars_repo_head_hexsha": "2e65411b69d6924495c5e321857b64d06f1ce7ff", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ch04.tex", "max_issues_repo_name": "mchandra12/research_module_econometrics", "max_issues_repo_head_hexsha": "2e65411b69d6924495c5e321857b64d06f1ce7ff", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ch04.tex", "max_forks_repo_name": "mchandra12/research_module_econometrics", "max_forks_repo_head_hexsha": "2e65411b69d6924495c5e321857b64d06f1ce7ff", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 91.8191489362, "max_line_length": 645, "alphanum_fraction": 0.7550689376, "num_tokens": 2430, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.8031737940012418, "lm_q1q2_score": 0.6154669726363837}} {"text": "\\documentclass{article}\n\\usepackage{eecstex}\n\\usepackage{pgfplots}\n\\usepackage{physics}\n\n\\DeclareMathOperator{\\rect}{rect}\n\n\\title{Homework 01}\n\\author{Bryan Ngo}\n\\date{2022-01-19}\n\n\\begin{document}\n\n\\maketitle\n\n\\setcounter{section}{1}\n\n\\section{}\n\n\\subsection{}\n\n\\begin{theorem}\n The system \\(L\\) is not time invariant.\n\\end{theorem}\n\\begin{proof}\n We can represent a time-delayed signal as a linear combination of the other 3 signals.\n We can represent the 3 input signals in a matrix\n \\begin{equation}\n \\bm{X} =\n \\begin{bmatrix}\n -2 & -2 & 0 \\\\\n 1 & 1 & 1 \\\\\n -2 & 0 & 1 \n \\end{bmatrix}\n \\end{equation}\n Then, we can find the linear combination necessary to create a time delayed signal (for example, \\(x_2[n - 1]\\)) by solving the system of equations\n \\begin{equation}\n \\bm{Xa} =\n \\begin{bmatrix}\n 0 \\\\\n -2 \\\\\n 1\n \\end{bmatrix}\n \\end{equation}\n where we get \\(\\bm{a} = \\begin{bmatrix}\n -\\frac{3}{2} & \\frac{3}{2} & -2\n \\end{bmatrix}^\\top\\).\n Thus, we have\n \\begin{equation}\n x_2[n - 1] = -\\frac{3}{2} x_1[n] + \\frac{3}{2} x_2[n] - 2x_3[n]\n \\end{equation}\n Calculating \\(-\\frac{3}{2} y_1[n] + \\frac{3}{2} y_2[n] - 2y_3[n]\\),\n we get the following plot:\n \\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(n\\), ylabel={\\(L\\{x_2[n - 1]\\}\\)},\n title={Time invariance counterexample},\n axis lines=middle\n ]\n \\addplot[\n ycomb,\n color=blue,\n mark=*\n ]\n coordinates {\n (-2, -4)\n (-1, 2)\n (0, 3)\n (1, -9)\n (2, -4)\n (3, -3)\n };\n \\end{axis}\n \\end{tikzpicture}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(n\\), ylabel={\\(y_2[n - 1]\\)},\n title={Expected output signal},\n axis lines=middle\n ]\n \\addplot[\n ycomb,\n color=blue,\n mark=*\n ]\n coordinates {\n (0, -1)\n (1, 1)\n (2, -3)\n (3, -0)\n (4, -1)\n };\n \\end{axis}\n \\end{tikzpicture}\n \\end{center}\n So the system is \\emph{not} time-invariant.\n\\end{proof}\n\n\\subsection{}\n\nWe can represent the delta function as\n\\begin{equation}\n \\delta[n] = \\frac{1}{2} (x_1[n] - x_2[n] + 2x_3[n])\n\\end{equation}\nUsing the properties of linear systems, we can find the impulse response\n\\begin{equation}\n h[n] = \\frac{1}{2} (y_1[n] - y_2[n] + 2y_3[n])\n\\end{equation}\nwhich gives us the following plot:\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(n\\), ylabel={\\(L\\{\\delta[n]\\}\\)},\n title={Impulse response},\n axis lines=middle\n ]\n \\addplot[\n ycomb,\n color=blue,\n mark=*\n ]\n coordinates {\n (-2, 2)\n (-1, 1)\n (0, -2)\n (1, 3)\n (2, 2)\n (3, 1)\n };\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\newpage\n\\section{}\n\n\\begin{align}\n y[n] - ay[n - 1] &= x[n] \\\\\n y[0] &= 1\n\\end{align}\n\n\\subsection{}\n\nBy counterexample, consider \\(x[n] = \\delta[n]\\) and \\(n = 1\\),\n\\begin{equation}\n y[1] = a\n\\end{equation}\nThen, consider \\(x[n - 1] = \\delta[n - 1]\\) and \\(n = 1\\),\n\\begin{equation}\n y[1] - a y[0] = \\delta[0] \\implies y[1] = 1 + a\n\\end{equation}\nwhereas \\(y[n - 1] = y[0] = 1\\) by the initial condition.\nSo the system is not time-invariant.\n\n\\subsection{}\n\nGiven the initial condition \\(y[0] = 1\\), scaling the input does not change this fact, so the system \\emph{cannot} be linear.\n\n\\subsection{}\n\nThe system is still not time-invariant due to the same counterexample as above.\nThe system is now linear as we can see because we no longer have an affine transformation at \\(y[0]\\).\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n y[n] = h[n] \\ast (\\cos[\\omega_0 n] x[n]) = \\sum_{k \\geqslant 0} \\frac{1}{1 + k} \\cos[\\omega_0 (n - k)] x[n - k]\n\\end{equation}\n\n\\subsection{}\nProving linearity, suppose we are given the system responses \\(x_1[n] \\iff y_1[n]\\) and \\(x_2[n] \\iff y_2[n]\\),\n\\begin{align}\n T\\{ax_1[n] + bx_2[n]\\} &= h[n] \\ast (\\cosine[\\omega_0 n] (ax_1[n] + bx_2[n])) \\\\\n &= h[n] \\ast (\\cos[\\omega_0 n] ax_1[n] + \\cos[\\omega_0 n] bx_2[n]) \\\\\n &= a (h[n] \\ast \\cos[\\omega_0 n] x_1[n]) + b (h[n] \\ast \\cos[\\omega_0 n] x_2[n]) \\\\\n &= ay_1[n] + by_2[n]\n\\end{align}\nFor time invariance,\n\\begin{align}\n T\\{x[n - n_0]\\} &= h[n] \\ast (\\cosine[\\omega_0 n] x[n - n_0]) \\\\\n y[n - n_0] &= h[n - n_0] \\ast (\\cosine[\\omega_0 (n - n_0)] x[n - n_0])\n\\end{align}\nThe two are clearly not the same, so our system is not time invariant and thus not LTI.\n\n\\subsection{}\n\nWe can bound the cosine function with 1.\nSimilar to Example 2.18 in Oppenheim \\& Schafer, if \\(|x[n]| < B_x\\),\n\\begin{equation}\n |y[n]| = \\sum_{k \\geqslant 0} \\frac{1}{1 + k} B_x = B_x \\sum_{k \\geqslant 1} \\frac{1}{k}\n\\end{equation}\nwhich is a divergent harmonic series.\n\n\\subsection{}\n\nThe system is causal since the \\(y[n]\\) consists of a memoryless multiplication by a cosine function and convolution with a causal LTI system, which makes the system overall causal.\n\n\\newpage\n\\section{}\n\n\\subsection{}\n\nWe can model the frequency response as\n\\begin{equation}\n H(e^{j \\omega}) =\n \\begin{cases}\n 1 & |\\omega| < \\frac{\\pi}{2} \\\\\n 0 & \\text{elsewhere}\n \\end{cases}\n\\end{equation}\nFinding the inverse Fourier transform,\n\\begin{align}\n h[n] &= \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi H(e^{j \\omega}) e^{j \\omega n} \\, d\\omega \\\\\n &= \\frac{1}{2\\pi} \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} e^{j \\omega n} \\, d\\omega = \\frac{1}{2\\pi} \\eval{\\frac{e^{j \\omega n}}{jn}}_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\\\\n &= \\frac{1}{2\\pi j n} \\left(e^{j \\frac{\\pi}{2} n} - e^{-j \\frac{\\pi}{2} n}\\right) = \\frac{\\sin\\left[\\frac{\\pi}{2} n\\right]}{\\pi n}\n\\end{align}\n\n\\subsection{}\n\nThe system is not causal, since \\(h[-1] = \\frac{1}{\\pi} \\neq 0\\).\n\n\\subsection{}\n\nThis is a specific case of Example 2.18 in Oppenheim \\& Schafer.\nEach term of the absolute sum of the sinc function only decreases on the order of \\(\\frac{1}{n}\\), so we can treat the sum as the harmonic series, which diverges.\nThus, the low pass filter is not BIBO stable.\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n y[n] = \\frac{1}{6} \\sum_{k = 0}^5 x[n - k]\n\\end{equation}\n\n\\subsection{}\n\nUsing Equation 2.123 from Oppenheim \\& Schafer with \\(M = 5\\),\n\\begin{align}\n H(e^{j \\omega}) &= \\frac{1}{6} \\frac{\\sin(3\\omega)}{\\sin\\left(\\frac{\\omega}{2}\\right)} e^{-j \\frac{5}{2} \\omega} \\\\\n |H(e^{j \\omega})| &= \\frac{1}{6} \\left|\\frac{\\sin(3\\omega)}{\\sin\\left(\\frac{\\omega}{2}\\right)}\\right| \\\\\n \\angle H(e^{j \\omega}) &= -\\frac{5}{2} \\omega \\\\\n\\end{align}\nThe zero crossings are \\(\\omega = \\frac{k\\pi}{3}\\) for nonzero \\(k \\in \\Z\\), by solving for the equation \\(\\sin(3\\omega) = 0\\).\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(\\omega\\), ylabel={\\(|H(e^{j \\omega})|\\)},\n title={Magnitude of Frequency Response},\n axis lines=middle,\n ymin=0, ymax=1,\n width=0.4\\textwidth\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=n, y=q6a-mag\n ]{q6a-mag.csv};\n \\end{axis}\n \\end{tikzpicture}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(\\omega\\), ylabel={\\(\\angle H(e^{j \\omega})\\)},\n title={Phase of Frequency Response},\n axis lines=middle,\n width=0.4\\textwidth\n ]\n \\addplot[color=blue] table[\n col sep=comma,\n x=n, y=q6a-ang\n ]{q6a-ang.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\n\\subsection{}\n\nLet \\(\\omega_1 = \\frac{\\pi}{3}\\).\nThen, the output \\(y[n]\\) is as shown:\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(n\\), ylabel={\\(y[n]\\)},\n title={Convolution Sum},\n axis lines=middle,\n width=0.4\\textwidth\n ]\n \\addplot[ycomb, mark=*, color=blue] table[\n col sep=comma,\n x=n, y=q6b\n ]{q6b.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\nThe transient at the start of the convolution sum comes from at the beginning of the MAF, where the positive terms from the first 2 inputs of the cosine function.\nThe settling of the convolution sum comes from the fact that the sum of a sinusoid without a vertical shift over 1 period is 0. Since the MAF has a window of size 6, and the cosine has a period of 6 as well, once the window is filled by the cosine, it we get 0.\nFor any value of the convolution sum afterwards, it remains 0 due to the MAF now being saturated with a constant 1 period of the cosine, which has a sum of 0.\n\n\\newpage\n\\section{}\n\n\\begin{equation}\n y[n] = \\mathcal{P}\\{x[n - 2], x[n - 1], x[n], x[n + 1], x[n + 2]\\}\n\\end{equation}\n\n\\subsection{}\n\nWe can transform the problem into a least-squares problem:\n\\begin{equation}\n \\begin{bmatrix}\n x[-2] \\\\\n x[-1] \\\\\n x[0] \\\\\n x[1] \\\\\n x[2]\n \\end{bmatrix}\n = \n \\begin{bmatrix}\n 1 & -2 & 4 \\\\\n 1 & -1 & 1 \\\\\n 1 & 0 & 0 \\\\\n 1 & 1 & 1 \\\\\n 1 & 2 & 4\n \\end{bmatrix}\n \\begin{bmatrix}\n a_0 \\\\\n a_1 \\\\\n a_2\n \\end{bmatrix}\n\\end{equation}\nBy the least squares formula \\(\\bm{\\hat{a}} = (\\bm{P}^\\top \\bm{P})^{-1} \\bm{P}^\\top \\bm{x}\\), we get that\n\\begin{align}\n \\begin{bmatrix}\n a_0 \\\\\n a_1 \\\\\n a_2\n \\end{bmatrix}\n &=\n \\frac{1}{35} \\begin{bmatrix}\n -3 & 12 & 17 & 12 & -3 \\\\\n -7 & -3.5 & 0 & 3.5 & 7 \\\\\n 5 & -2.5 & -5 & -2.5 & 5\n \\end{bmatrix}\n \\begin{bmatrix}\n x[-2] \\\\\n x[-1] \\\\\n x[0] \\\\\n x[1] \\\\\n x[2]\n \\end{bmatrix} \\\\\n \\implies y[0] &= a_0 = \\frac{1}{35} (-3x[-2] + 12x[-1] + 17x[0] + 12x[1] - 3x[2])\n\\end{align}\n\n\\subsection{}\n\n\\begin{equation}\n \\begin{bmatrix}\n x[-1] \\\\\n x[0] \\\\\n x[1] \\\\\n x[2] \\\\\n x[3]\n \\end{bmatrix}\n = \n \\begin{bmatrix}\n 1 & -2 & 4 \\\\\n 1 & -1 & 1 \\\\\n 1 & 0 & 0 \\\\\n 1 & 1 & 1 \\\\\n 1 & 2 & 4\n \\end{bmatrix}\n \\begin{bmatrix}\n a_0 \\\\\n a_1 \\\\\n a_2\n \\end{bmatrix}\n\\end{equation}\nwhich gives us the solution\n\\begin{equation}\n y[1] = a_0 = \\frac{1}{35} (-3x[-1] + 12x[0] + 17x[1] + 12x[2] - 3x[3])\n\\end{equation}\n\n\\subsection{}\n\nThe system is linear since it entirely consists of additions and scalar multiplications.\nThe system is time-invariant since for any shift \\(n_0\\), the corresponding \\(x[n - n_0]\\) is simply shifted and put into the solution, which is what is expected for a time-shifted output.\nThe system is stable since if \\(x[n]\\) is bounded by \\(B_x\\), \\(y[n]\\) is bounded by \\(B_x\\).\nSince the system is LTI, the system has a frequency response.\nWe do not actually need to perform a regression for every \\(n\\), since the coefficient matrix multiplying the set of \\(x[n - k]\\) is constant.\n\n\\subsection{}\n\nWe can see that points further from the operating point have the largest weights, while weights further away have smaller and even negative weights.\nThis means that it better represents a distribution around a local region, and does not flatten out features as drastically as a standard MAF.\n\n\\end{document}\n", "meta": {"hexsha": "2af6020e3f931f831d428ae348b3686712d6f540", "size": 11667, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw01/hw01.tex", "max_stars_repo_name": "bdngo/ee-123", "max_stars_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw01/hw01.tex", "max_issues_repo_name": "bdngo/ee-123", "max_issues_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw01/hw01.tex", "max_forks_repo_name": "bdngo/ee-123", "max_forks_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.3879093199, "max_line_length": 261, "alphanum_fraction": 0.542555927, "num_tokens": 3901, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149868676283, "lm_q2_score": 0.8152324960856175, "lm_q1q2_score": 0.6154312290765378}} {"text": "\\section{Arbitrage theory in continuous markets}\n\nWe now develop an heuristic model for continuous-time markets. We shall omit most technical definitions, details and proofs. It is an amalgam of the presentations found in \\textcite{bjork2004arbitrage}, \\textcite{brigo2007interest}, \\textcite{duffie2010dynamic}, \\textcite{jameswebber2000interest}, \\textcite{musielarutkowski2005martingale} and \\textcite{wu2009interest}. A rigorous treatment for the subject can be found, for example, in \\textcite{musielarutkowski2005martingale}. The fundamentals of this model are the same as the simple discrete model introduced earlier, but due to technicalities, it is not as intuitive.\n\nWe consider a probability space $(\\Omega, \\F, \\Pf)$ and a finite time interval $\\left[ 0, T' \\right]$. Here $\\Pf$ is the physical probability measure on $(\\Omega, \\F)$, $\\Omega$ is the sample path space with a $\\sigma$-algebra $\\F$. The flow of new information is handled with a filtration\\footnote{A filtration $(\\F_t)$ is a collection $\\sigma$-sub-algebras of $\\F$ with $\\F_s \\subseteq \\F_t$ for all $0 \\leq s \\leq t \\leq T$.} $(\\F_t)$. We also assume some technical conditions for the filtrations. Every $\\Pf$-null set\\footnote{$\\Pf$-null set is a set $A \\in \\F$ with $\\Pf (A) = 0$.} must be a member of $\\F_0$ and\n \\begin{align}\n \\F_t = \\bigcap_{t < s} \\F_s\n \\end{align}\nfor all $t \\geq 0$.\n\nWe assume that the price $S_i$ of a market assets are modeled with It\\'{o}-processes\\footnote{See \\ref{sec:itoprocess}}, so\n \\begin{align}\n \\label{generalassetpriceprocess}\n S_i(t,\\omega) = S_i (0) + \\int_0^t \\mu (t,\\omega) \\dx t + \\int_0^t \\sigma(t, \\omega ) \\dx W_i (t, \\omega),\n \\end{align}\nwhere $S_i(0)$ is a deterministic constant and $W_i$ is a Brownian motion. The second integral is an It\\'{o}-integral and the functions $\\mu$ and $\\sigma$ are assumed to be $\\F_t$-adapted and to satisfy technical conditions so that integrability and the existence of solutions are always guaranteed\\footnote{It is usually assumed, for example, that $\\int_0^{T^*} \\left| \\mu(t, \\omega) \\right| \\dx t < \\infty$ and $\\int_0^{T^*} \\sigma(t, \\omega)^2 \\dx t < \\infty$ almost surely}. We usually write the Equation \\ref{generalassetpriceprocess} as\n \\begin{align}\n \\dx S_i(t, \\omega) = \\mu (t,\\omega) \\dx t + \\sigma(t, \\omega ) \\dx W_i (t, \\omega),\n \\end{align}\nbut we note that Brownian motions are $\\Pf$-surely not differentiable. We assume that the assets all these assets can be bought and sold freely, and the trading shall not affect price process.\n\nWe also shall omit $\\omega$ from the argument of functions for readability, when it is not absolutely necessary.\n\nSuppose that there are $n$ tradable assets and let $S = (S_1, S_2, \\ldots, S_n)$ be the price vector for these assets. The trading strategy (or a portfolio) $w$ is a predictable stochastic process $w(t,\\omega) \\in \\R^n$. This means that while the strategy is a random variable, it is not omniscient and uses only information available up to that point. It is left-continuous, so given the history up to that point, it is deterministic. The corresponding portfolio value process is given by\n \\begin{align}\n V^w (t) = (w(t))^{\\top} S(t)\n \\end{align}\nand a self-financing trading strategy satisfies\n\t\\begin{align}\n\t\tV^w(t) - V^w(0) = \\int_0^t w^{\\top}(u) \\ \\dx S(u),\n\t\\end{align}\nor\n \\begin{align}\n \\dx V^w = (w(t))^{\\top} \\dx S(t) .\n \\end{align}\nThis means that for self-financing portfolio, there is no inflow or outflow of money from the portfolio and all the value changes are due to changes of prices. An arbitrage\\footnote{We note that this type of arbitrage is not strong enough to disallow doubling strategies.} is a self-financing portfolio with a value process $V^w (t)$ such that\n \\begin{align}\n V^w(0) &= 0, \\\\\n \\Pf (V^w(t) \\geq 0 ) &= 1 \\textup{ and } \\\\\n \\Pf (V^w(t) > 0 ) &> 0.\n \\end{align}\n for some $0 \\leq t \\leq T'$. An arbitrage is an opportunity to gain without any associated risk. A market without arbitrage opportunities is arbitrage-free. \n \nA contingent claim (or $T$-derivative or $T$-claim) is a $\\F_T$-measurable random variable $X$. We say that the is attainable if there is such self-financing trading strategy $w$ that\n \\begin{align}\n V^w(T) = X(T)\n \\end{align}\nalmost surely. The strategy $w$ is thus replicating the pay-off of $X$. As usually, we say that the market is complete if every contingent claim is attainable.\n \nAs earlier, a deflator is strictly positive It\\^{o}-process and a num\\'{e}raire is an asset with always positive price process. If $D$ is deflator, then the relative (or discounted) market price process is\n \\begin{align}\n S^D(t) = ( S_i(t) / D(t) )\n \\end{align}\nfor all $0 \\leq t \\leq T'$. A measure $\\Pm$ on the space $(\\Omega, \\F)$ is an equivalent martingale measure (EMM) with respect to the deflator $D$ if it is equivalent\\footnote{Meaning that the measures have the same null sets.} to $\\Pf$ and if\n \\begin{align}\n S^D(s) = \\E_{\\Pm} ( \\ S^D(t) \\ | \\ \\F_s \\ )\n \\end{align}\nfor all $0 \\leq s < t$. This means that discounted price processes are martingales under an EMM. A weaker condition is an equivalent local martingale measure (ELMM) which requires only that the measures are equivalent and discounted price process are local martingale under the ELMM.\n\nWe will only consider trading strategies that will satisfy the condition that\n \\begin{align}\n \\int_0^t w(s)^{\\top} \\dx S^D(s)\n \\end{align}\nare martingales under the the measure $\\Pm$ as this guarantees that no doubling strategies are permissable\\footnote{Thus the expected values of portfolios are always bounded.}.\n \nThe first fundamental theorem of asset pricing says that if an equivalent martingale measure exists, then the market is arbitrage-free. As we have seen, in discrete time model, these are equivalent conditions. But in continuous models, this in not true. \\textcite{delbaenschachermayer1994general} showed that under certain conditions, the absence of arbitrage implies that an equivalent local martingale measure exists. We shall not delve in these technicalities further in this thesis.\n\nThe second fundamental theorem of asset pricing ties the uniqueness of this EMM to the ability to hedge every derivative contract. If we assume that the market is arbitrage free and an equivalent martingale measure exists for a deflator $D$, then the market is complete if and only if the equivalent martingale measure for deflator $D$ is unique.\n\nAs in discrete setting, we have three distinct cases: the model has no EMM, it has several EMMs or the EMM is unique. \n\nIdeally the EMM is unique and the model is complete and it allows no arbitrage. Traditionally the discounting in the EMM has been done with respect to the risk-free rate, but \\textcite{gemanelkarouirochet1995changes} showed that the choice of the so-called num\\'{e}raire is actually arbitrary as long as it is strictly positive non-dividend paying asset process. Different num\\'{e}raires produces different EMMs, but if we assume that payoffs are square integrable random variables, then the change of num\\'{e}raire does not change replicating portfolios. Thus the price is unique as long as the derivative can be hedged. If $S_0$ is the num\\'{e}raire, $X$ is a $T$-derivate with replicating self-financing portfolio's value process given by $V^w(t)$, then we have that $X(T) = V^w(T)$ and every discounted value processes of self-financing portfolio is martingale under the EMM $\\Pm_0$ associated with discounting process $S_0$. Thus\n \\begin{align}\n \\frac{V^w(t) }{ S_0(t) } = \\E_{\\Pm_0} \\left( \\frac{V^w(T)}{S_0(T)} | \\F_t \\right) = \\E_{\\Pm_0} \\left( \\frac{X(T)}{S_0(T)} | \\F_t \\right)\n \\end{align}\nmeaning that \n \\begin{align}\n \\label{discountedexpectedvalue}\n V^w(t) = S_0(t) \\E_{\\Pm_0} \\left( \\frac{V^w(T)}{S_0(T)} | \\F_t \\right) = S_0(t) \\E_{\\Pm_0} \\left( \\frac{X(T)}{S_0(T)} | \\F_t \\right) .\n \\end{align}\nSince there are no arbitrage opportunities, $X(t) = V^w(t)$, where $V^w(t)$ is given by the equation (\\ref{discountedexpectedvalue}). This has to hold even if the replicating portfolio is not unique.\n\nIf the model has several EMMs, then arbitrage is not possible but there are derivatives that may not be hedged. In this model, a claim that can be replicated has a unique price. Claims that may not be replicated may have multiple prices corresponding to different EMMs. \n\nThis usually means that calibrating prices are taken carefully selected from the most liquid instruments. We can also use more generalized versions of hedging. We could take the set of all sensible portfolio strategies that promise almost surely a payoff that is equal or greater than the contingent claim. A reasonable price candidate for a derivative is the minimum maintenance price of these portfolio strategies.\n\nThe worst case is the absence of EMM. This means that the model has arbitrage and pricing cannot be done.\n\nThe num\\'{e}raire can be freely chosen. \\cite{gemanelkarouirochet1995changes} showed that if the market is arbitrage-free, $M$ and $N$ are arbitrary num\\'{e}raires, then there exists such EMMs $\\Pm_M$ and $\\Pm_n$ that\n\t\\begin{align}\n\t\t\\frac{X(t)}{M(t)} &= \\E_{\\Pm_M} \\left( \\frac{X(T)}{M(T)} | \\F_t \\right), \\\\\n\t\t\\frac{X(t)}{N(t)} &= \\E_{\\Pm_N} \\left( \\frac{X(T)}{N(T)} | \\F_t \\right)\n\t\\end{align}\nfor any asset $X$ and $0 \\leq t \\leq T$. The Radon-Nikod\\'{y}m derivate\\footnote{See \\ref{sec:radonnikodymtheorem}} is\n\t\\begin{align}\n\t\t\\xi(t) = \\frac{\\dx \\Pm_M}{\\dx \\Pm_N} = \\frac{M(t) N(0)}{M(0) N(t)} .\n\t\\end{align}\nThis implies that\n\t\\begin{align}\n\t\tX(t) &= M(t) \\E_{\\Pm_M} \\left( \\frac{X(T)}{M(T)} | \\F_t \\right) \\\\ \n\t\t&= N(t) \\E_{\\Pm_N} \\left( \\frac{X(T)}{N(T)} | \\F_t \\right)\n\\end{align}\nand the price is unique, if the claim can be replicated. Also\n\t\\begin{align}\n\t\t\\label{expectationbyradonnikodymderivative}\n\t\t\\E_{\\Pm_M} \\left( Y(T) | \\F_t \\right) = \\frac{\\E_{\\Pm_N} \\left( Y(T) \\xi(T) | \\F_t \\right)}{\\xi(t)} \n\t\\end{align}\nholds for any random variable $Y$.\n\n\\subsection{Risk-free measure}\n\nThe bank account $\\Bank(t) > 0$ is a common num\\'{e}raire and the EMM $\\Pm_0$ induced by it is often called as the risk-free measure. If $X$ is a portfolio and the market is arbitrage-free, then\n\t\\begin{align}\n\t\t\\label{pricingunderriskneutralmeasure}\n\t\tX(t) &= \\Bank(t) \\E_{\\Pm_0} \\left( \\frac{X(T)}{B(T)} | \\F_t \\right) \\\\\n\t\t\t&= \\E_{\\Pm_0} \\left( \\frac{B(t)}{B(T) } X(T) | \\F_t \\right) \\\\\n\t\t\t&= \\E_{\\Pm_0} \\left( D(t,T) X(T) | \\F_t \\right),\n\t\\end{align}\nwhere $D(t,T)$ is the stochastic discount factor. If $X(t) = \\Bond(t,T)$, then $\\Bond(T,T)=1$ and we note that\n\t\\begin{align}\n\t\t\\Bond(t,T) = \\E_{\\Pm_0} \\left( D(t,T) \\ | \\ \\F_t \\right) .\n\t\\end{align}\nThis shows that price of a bond is expected value of the corresponding stochastic discount factor under the risk-free measure (or any other EMM). Also if\n\t\\begin{align}\n\t\t\\Bank(t) = \\e^{ \\int_0^t r(s) \\ \\dx s },\n\t\\end{align}\nthen\n\t\\begin{align}\n\t\tD(t,T) = \\frac{\\Bank(t)}{\\Bank(T)} = \\e^{ - \\int_t^T r(s) \\ \\dx s },\n\t\\end{align}\nhence\n\t\\begin{align}\n\t\tX(t) &= \\E_{\\Pm_0} \\left( \\e^{ - \\int_t^T r(s) \\ \\dx s } X(T) | \\F_t \\right) . \n\\end{align}\t\n\nUnder the risk-neutral measure, the discounted process $X(t) / B(t)$ will be a martingale.\n\t\n\\subsection{Black-Scholes--model}\n\nIn the celebrated Black-Scholes--model there are two assets, a stock and a bank account, with given dynamics\n \\begin{align}\n \\label{stockprocessinblackscholes}\n \\dx S(t) &= \\mu S(t) \\dx t + \\sigma S(t) \\dx W(t) \\\\\n \\dx \\Bank (t) &= r \\Bank(t) \\dx t, \n \\end{align}\nunder the physical measure $\\Pf$, where $\\mu, r$ and $\\sigma > 0$ are given constants and the Brownian motion $W(t)$ is the sole source of uncertainty. This means that the stock price follows geometric Brownian motion and $\\Bank (t) = \\Bank (0) \\e^{rt}$. To simplify the notation we assume that $\\Bank (0) = 1$. We denote the discounted stock price as $S^\\Bank(t) = S(t) / \\Bank (t)$ and the discount factor is deterministic $\\DF (t,T) = \\e^{r(T-t)}$. Now $g(t,x) = \\e^{-rt} x$ and a simple application of It\\^{o}'s lemma\\footnote{See \\ref{sec:itoprocess}.} yields that\n \\begin{align}\n \\dx S^\\Bank(t) &= ( \\frac{\\partial g}{\\partial t} + \\mu \\frac{\\partial g}{\\partial x} + \\frac{1}{2} \\sigma^2 \\frac{\\partial^2 g}{\\partial x^2} ) \\dx t + \\sigma \\frac{\\partial g}{\\partial x} \\dx W(t) \\\\\n &= ( -r \\e^{-rt} S(t) + \\e^{-rt} \\mu S(t) + 0 ) \\dx t + \\sigma \\e^{-rt} S(t) \\dx W(t) \\\\\n &= ( \\mu - r ) S^\\Bank(t) \\dx t + \\sigma S^\\Bank(t) \\dx W(t) .\n \\end{align}\nWe can now use Girsanov's theorem\\footnote{See \\ref{sec:girsanov}.} to change the probability measure in order to make the discounted stock price process driftless. We note the bank account discounted by itself is trivially driftless under any measure. Now the market price of risk\n\t\\begin{align}\n\t\t\\lambda = \\frac{r - \\mu}{\\sigma}\n\t\\end{align}\nis the only possible Girsanov kernel that makes the new measure an equivalent martingale measure. So there is an unique EMM $\\Pm$ for the deflator $B(t)$. hence Black-Scholes model is arbitrage-free and complete. Under this measure $\\dx W^{\\Pm}(t) = \\dx W(t) - \\frac{r - \\mu}{\\sigma} \\dx t$ is a Brownian motion. Therefore\n \\begin{align}\n \\dx S^\\Bank(t) &= \\sigma S^\\Bank(t) \\dx W^{\\Pm}(t), \\\\\n \\dx \\Bank (t) &= r \\Bank (t) \\dx t, \\\\\n \\label{BlackScholesProcessesUnderEMM}\n \\dx S(t) &= r S(t) \\dx t + \\sigma S(t) \\dx W^{\\Pm}(t)\n \\end{align}\nunder the EMM $\\Pm$. In other words, the discounted stock price is a martingale and under this measure the drift of the stock price is changed to the risk-free rate $r$. Since the bank account is deterministic, the change of measure does not affect it. The arbitrage-free price of the derivate is at time $0$ is given by\n \\begin{align}\n X(0) &= \\E_{\\Pm} \\left( X(T) / \\Bank (0,T) | \\F_0 \\right) \\\\ &= \\e^{-rT} \\E_{\\Pm} \\left( X(T) | \\F_0 \\right) .\n \\end{align}\nSince the stock price process under the risk-free measure is a geometric Brownian motion, we know that\n \\begin{align}\n \\log (S(T) ) &= \\log (S(0) ) + \\left( r - \\frac{1}{2} \\sigma^2 \\right) T + \\sigma W^{\\Pm}(T) \\\\ &\\sim N \\left( \\log (S(0) ) + \\left( r - \\frac{1}{2} \\sigma^2 \\right), \\sigma^2 T \\right)\n \\end{align}\nunder the measure $\\Pm$. Similarly, we could use the stock price $S(t)$ as the numerator. \n\nIf we pick $X(T) = (S(T)-K)^+$, the price of a call option on the stock at the time $T$, then\n\t\\begin{align}\n\t\t\\E_{\\Pm} \\left( (S(T)-K)^+ | \\F_0 \\right) &= \\E_{\\Pm} \\left( (S(T)-K) \\1_{ \\{ S(T) > K\\} } | \\F_0 \\right) \\\\\n\t\t&= \\E_{\\Pm} \\left( S(T) \\1_{ \\{ S(T) > K\\} } | \\F_0 \\right) - K \\Pm \\left( S(T) > K | \\F_0 \\right)\n\t\\end{align}\nand the standard calculations will yield that\n \\begin{align}\n X(T) &= S(0) N(d_+) - \\e^{-rT} K N(d_-) , \\\\\n d_{\\pm} &= \\frac{\\log S/K + (r \\pm \\sigma^2/2 )T }{\\sigma \\sqrt{T}},\n \\end{align}\nwhich is the celebrated Black-Scholes formula.\n\nWe shall also give a heuristic derivation of the Black-Scholes differential equation. If $X(t,S(t))$ is a smooth value process of a derivative, then by It\\'{o}'s lemma, we have that\n \\begin{align}\n \\label{itolemmatoderivateinblackscholes}\n \\dx X = \\left( \\frac{\\partial X}{\\partial t} + \\mu S \\frac{\\partial X}{\\partial S} + \\frac{1}{2} \\sigma^2 S^2 \\frac{\\partial X^2}{\\partial S^2} \\right) \\dx t + \\sigma S \\frac{\\partial X}{\\partial S} \\dx W .\n \\end{align}\nNow we assume that we can replicate this derivative with a combination of $\\delta_S(t,S)$ stocks and $\\delta_B(t,S)$ bonds. This portfolio has a value \n\t\\begin{align}\n\t\tV(t,S) = \\delta_S S(t) + \\delta_B B(t)\n\t\\end{align}\nand the value follows the process\n\t\\begin{align}\n\t\t\\dx V &= \\delta_S \\dx S + \\delta_B \\dx B \\\\\n\t\t\t&= \\delta_S (\\mu S \\dx t + \\sigma S \\dx W ) + \\delta_B r B \\dx t \\\\\n\t\t\t&= (\\delta_S \\mu S + \\delta_B rB) \\dx t + \\delta_S \\sigma S \\dx W \\\\\n\t\t\t&= (\\delta_S \\mu S + r (V - \\delta_S S)) \\dx t + \\delta_S \\sigma S \\dx W\n\t\\end{align}\nAs $\\dx V = \\dx X$, then the coefficients of $\\dx W$ terms must coincide. Thus\n\t\\begin{align}\n\t\t\\label{replicating_delta}\n\t\t\\delta_S \\sigma S = \\sigma S \\frac{\\partial X}{\\partial S}, \t\t\n\t\\end{align}\nwhich implies that $\\delta_S = \\frac{\\partial X}{\\partial S}$.\n\nNow we consider a portfolio of one derivative and $-\\frac{\\partial X}{\\partial S}$ stocks. The value of this portfolio is\n \\begin{align}\n V = X - \\frac{\\partial X}{\\partial S} S\n \\end{align}\nand if the portfolio is self-financing, then\n \\begin{align}\n \\dx V &= \\dx X - \\frac{\\partial X}{\\partial S} \\dx S \\\\\n &= \\left( \\frac{\\partial X}{\\partial t} + \\frac{1}{2} \\sigma^2 S^2 \\frac{\\partial X^2}{\\partial S^2} \\right) \\dx t\n \\end{align}\nafter substitution of equations \\ref{stockprocessinblackscholes} and \\ref{itolemmatoderivateinblackscholes}. Since there is no diffusion, the portfolio is riskless and the absence of arbitrage implies that \n \\begin{align}\n \\dx V = rV \\dx t = \\left( rX - r S\\frac{\\partial X}{\\partial S} \\right) \\dx t .\n \\end{align}\nBy equating these, we get the Black-Scholes differential equation\n \\begin{align}\n \t\\label{BlackScholesDifferentialEquation}\n \\frac{\\partial X}{\\partial t} + r S\\frac{\\partial X}{\\partial S} + \\frac{1}{2} \\sigma^2 S^2 \\frac{\\partial X^2}{\\partial S^2} - r X = 0\n \\end{align}\nor equivalently\n \t\\begin{align}\n\t\t\\frac{\\partial X}{\\partial t} + \\frac{1}{2} \\sigma^2 S^2 \\frac{\\partial X^2}{\\partial S^2} = r ( X - S\\frac{\\partial X}{\\partial S}) .\n\t\\end{align}\nThe left side is a linear combination of \"theta\", the time decay of the value, and \"gamma\", the second derivative of the value with respect to the price of the underlying. The right side of the equation contains the replicating portfolio.\n\nThe derivative with pay-out $h(S(T))$ satisfies Equation \\ref{BlackScholesDifferentialEquation}. In order to apply Feynman-Kac theorem\\footnote{See \\ref{sec:faynmankac}.}, we need to find a process with drift $rS(t)$ and diffusion $\\sigma S(t)$ under some measure. Under the EMM, the discounted asset price is a martingale, so it grows with the rate risk-free rate $r$. So the price process under the EMM is process we need in order to use Feynman-Kac theorem. We have the EMM $\\Pm$ and, by Equation \\ref{BlackScholesProcessesUnderEMM},\n\t\\begin{align}\n\t\t\\dx S(t) &= r S(t) \\dx t + \\sigma S(t) \\dx W^{\\Pm}(t) ,\t\t\n\t\\end{align}\nwhere $W^{\\Pm}(t)$ is a Brownian motion under $\\Pm$. We may now apply Feynman-Kac theorem and it follows that\n\t\\begin{align}\n\t\th(S(t)) = \\e^{ -r(T-t) } \\E_{\\Pm} ( h(S_T) \\ | \\ \\F_t) .\n\t\\end{align}\n\t\n\\iffalse\n\t\n\\subsubsection{Black-Scholes--model with continuous dividend yield}\n\nIf $X(t)$ is the price of call option on stock $S(t)$ with strike $K$ at the time $T$ that pays continuous dividend yield $q$, then the extension of Black-Scholes--model gives the price\n\t\\begin{align}\n\t\tX(t) &= \\e^{-r(T-t)} \\left( F(t)N(d_+) - KN(d_-) \\right), \\\\\n\t\tF(t) &= S(t) \\e^{(r-q)(T-t)}, \\\\\n\t\td_{\\pm} &= \\frac{\\log \\frac{S(t)}{K} + (r-q \\pm \\frac{1}{2} \\sigma^2)(T-t)}{\\sigma \\sqrt{T-t}}\t\t\n\t\\end{align}\n\t\nSuppose now that assets $S_i(t)$ pay continuous dividend yields $q_i$. Suppose that an option gives right to swap asset $S_1$ with asset $S_2$ at the time $T$. \n\n\\fi\n\n\\subsection{Black-76--model}\n\\label{blackformula}\n\nBlack model (or Black-76--model) is an extension of Black-Scholes--model (\\cite{black1976pricing}) and it is used to price futures. It also assumes that the risk-free interest rate is a constant. The model assumes that futures price of an asset follows log-normal distribution with constant volatility parameter. The price of a call option on a future contract has price at time $t$ is given by the Black's formula\n\\begin{align}\n\\e^{-r(T-t)} \\left( F N(d_+) - K N(d_-) \\right)\n\\end{align}\nwith\n\\begin{align}\nd_{\\pm} = \\frac{\\log \\frac{F}{K} \\pm \\frac{\\sigma^2}{2} (T-t) }{ \\sigma \\sqrt(T-t) } .\n\\end{align}\nHere $K$ is the strike price at the maturity $T$, $r$ is the risk-free rate and $\\sigma$ is the constant volatility of the log-normal distribution. The futures price process is $F(t)$. The Black's formula is used to price interest rate caps, floors and swaptions and the market practice is to quite these instruments in terms of Black's volatilities.\n\t\n\\subsection{$T$-forward measure}\n\nSince $\\Bond(t,T) > 0$, $T$-bond is a num\\'{e}raire. The EMM induced by this as called $T$-forward measure $\\Pm_T$. Since $\\Bond(T,T) = 1$, we have that\n\\begin{align}\nX(t) = \\Bond(t,T) \\E_{\\Pm_T} \\left( X(T) | \\F_t \\right)\n\\end{align}\nfor every $0 \\leq t \\leq T$ and attainable claim $X$. The forward rate was defined as\n\\begin{align}\n\\Rflt(t,T,S) &= \\frac{1}{\\dayc(T,S)} \\left( \\frac{\\Bond(t,T)}{\\Bond(t,S)} - 1 \\right) .\n\\end{align}\nThus\n\\begin{align}\n\\Rflt(t,T,S) \\Bond(t,S) &= \\frac{\\Bond(t,T)-\\Bond(t,S)}{\\dayc(T,S)} ,\n\\end{align}\nwhere the right side is a bond portfolio. As $\\Bond(T,T) = 1$, we know that \n\\begin{align}\n\\frac{\\Bond(t,T)-\\Bond(t,S)}{\\dayc(T,S)} &= \\Bond(t,T) \\E_{\\Pm_T} \\left( \\Rflt(T,T,S) | \\F_t \\right) \n\\end{align}\nNow we have that shown that\n\\begin{align}\n\\Rflt(t,T,S) &= \\E_{\\Pm_T} \\left( \\Rflt(T,S) | \\F_t \\right) \n\\end{align}\nmeaning that the forward rate is expected value of spot rate under the $T$-forward measure.\t\n\nSimilarly\n\\begin{align}\nP(t,T) \\E_{\\Pm_T} \\left( r(T) | \\F_t \\right) &= \\E_{\\Pm_0} \\left( D(t,T) r(T) | \\F_t \\right) \\\\\n&= \\E_{\\Pm_0} \\left( r(T) \\e^{-\\int_t^T r(s) \\dx s} | \\F_t \\right) \\\\\n&= \\E_{\\Pm_0} \\left( \\frac{\\partial}{\\partial T} \\e^{-\\int_t^T r(s)} | \\F_t \\right) \\\\\n&= \\frac{\\partial}{\\partial T} \\E_{\\Pm_0} \\left( \\e^{-\\int_t^T r(s)} | \\F_t \\right) \\\\\n&= \\frac{\\partial}{\\partial T} P(t,T) ,\n\\end{align}\nwhich implies that\n\\begin{align}\nf(t,T) = \\E_{\\Pm_T} \\left( r(T) | \\F_t \\right) \n\\end{align}\nmeaning that the instantaneous forward rate is the expected value of the short-rate under $T$-forward measure. \n\nThe Equation \\ref{expectationbyradonnikodymderivative} implies that \n\\begin{align}\n\\xi (t) = \\frac{\\Bond(t,T)}{\\Bond(0,T) \\Bank(t)}\n\\end{align}\nis the Radon-Nikod\\'{y}m-derivative of $T$-forward measure $\\Pm_T$ with respect to risk-free measure $\\Pm_0$.\n\n\n\\subsection{Change of num\\'{e}raire}\n\nNow we consider an arbitrage free market model with assets $N$ and $M$, which are num\\'{e}raires. If $X$ is a contingent $T$-claim, then we know that arbitrage free price of $X$ at the time $t$ must be\n \\begin{align}\n V_t (X) &= N(t) \\E_{\\Pm_N} \\left( \\frac{X(T)}{N(T)} | \\F_t \\right) \\\\\n &= M(t) \\E_{\\Pm_M} \\left( \\frac{X(T)}{M(T)} | \\F_t \\right)\n \\end{align}\nwhere $\\Pm_N, \\Pm_M$ are the martingale measure corresponding to the num\\'{e}raires $N$ and $M$. Thus\n \\begin{align}\n \\E_{\\Pm_N} \\left( \\frac{X(T)}{N(T)} | \\F_t \\right) &= \\E_{\\Pm_M} \\left( \\frac{M(t)}{N(t)} \\frac{X(T)}{M(T)} | \\F_t \\right) \\\\\n &= \\E_{\\Pm_M} \\left( L_T(t) \\frac{X(T)}{N(T)} | \\F_t \\right),\n \\end{align}\nwhere\n \\begin{align}\n L_T(t) = \\frac{N(T) / N(t)}{ M(T) / M(t) } = \\frac{M(t)}{N(t)} \\frac{N(T)}{M(T)} .\n \\end{align}\nNow $L_T(t)$ is a $\\Pm_N$-martingale \nSince $X$ is an arbitrary $\\F_T$-measurable random variable, we have heuristically shown the following fundamental result. For the proof, see \\cite{gemanelkarouirochet1995changes}.\n \n\\begin{thm}\nLet $\\Pm_N$ be the EMM associated with num\\'{e}raire $N$ and $\\Pm_M$ be the EMM associated with num\\'{e}raire $M$. Under some technical conditions, the Radon-Nikod\\'{y}m derivate of $\\Pm_M$ with respect to $\\Pm_N$ is\n \\begin{align}\n \\frac{\\dx \\Pm_N}{\\dx \\Pm_M} = \\frac{N(T) / N(t)}{ M(T) / M(t) } .\n \\end{align}\n\\end{thm}\n\nSince $Z(t,T) > 0$ for all pairs $(t,T)$ we may use zero-coupon bonds as num\\'{e}raire. We denote $\\Pm_T$ as the equivalent martingale measure given by the $T$-bond. Thus if $X$ is a contingent $T$-claim, then we know that arbitrage free price of $X$ at the time $t$ must be\n \\begin{align}\n V_t (X) &= Z(t,T) \\E_{\\Pm_T} \\left( X(T) \\right) .\n \\end{align}\n\n\\iffalse\n \nTähän jotakin siitä, että Björk s. 357.\n\nIf $X(T) = (S(T) - K)^+$ is a final pay-out of a contingent claim $X$ for some asset $S$ and fixed $K$, then\n \\begin{align}\n V_t (X) &= \\E_{\\Pm} \\left( \\frac{X(T)}{B(T)} | \\F_t \\right) \\\\\n &= \\E_{\\Pm} \\left( \\frac{ (S(T) - K) \\1_{ \\{S_T > K \\} } }{B(T)} | \\F_t \\right) \\\\\n &= \\E_{\\Pm} \\left( \\frac{ S(T) }{ B(T) } \\1_{ \\{S_T > K \\} } | \\F_t \\right) - K \\E_{\\Pm} \\left( \\frac{\\1_{ \\{S_T > K \\} } }{B(T)} | \\F_t \\right) .\n \\end{align}\nIf $S$ is also a num\\'{e}raire we may write\n \\begin{align}\n \\E_{\\Pm} \\left( \\frac{ S(T) }{ B(T) } \\1_{ \\{S_T > K \\} } | \\F_t \\right) &= \\E_{\\Pm_S} \\left( L_S(T) \\frac{ S(T) }{ B(T) } \\1_{ \\{S_T > K \\} } | \\F_t \\right) \\\\\n &= \\E_{\\Pm_S} \\left( \\frac{ S(t) }{ B(t) } \\1_{ \\{S_T > K \\} } | \\F_t \\right) \\\\\n &= \\frac{ S(t) }{ B(t) } \\Pm_S ( S_T > K )\n \\end{align}\nwhere we used likelihood ratio\n \\begin{align}\n L_S (T) = \\frac{ S(t) }{ B(t) } \\frac{ B(T) }{ S(T) }\n \\end{align}\nto change measure from risk-free measure $\\Pm$ to the martingale measure $\\Pm_S$ with num\\'{e}raire $S$. The second term is\n \\begin{align}\n \\E_{\\Pm} \\left( \\frac{\\1_{ \\{S_T > K \\} } }{B(T)} | \\F_t \\right) &= \\E_{\\Pm_T} \\left( L_T(T) \\frac{\\1_{ \\{S_T > K \\} } }{B(T)} | \\F_t \\right) \\\\\n &= \\E_{\\Pm_T} \\left( \\frac{ Z(t,T) }{ B(t) } \\1_{ \\{S_T > K \\} } | \\F_t \\right) \\\\\n &= \\frac{ Z(t,T) }{ B(t) } \\Pm_T ( S_T > K )\n \\end{align}\nwhere we used $Z(T,T) = 1$ and likelihood ratio\n \\begin{align}\n L_T (t) = \\frac{ Z(t,T) }{ B(t) } \\frac{ B(T) }{ Z(T,T) }\n \\end{align}\nto change measure from risk-free measure $\\Pm$ to the $T$-forward measure $\\Pm_T$. Thus\n \\begin{align}\n V_t (X) &= \\frac{ S(t) }{ B(t) } \\Pm_S ( S_T > K ) - K \\frac{ Z(t,T) }{ B(t) } \\Pm_T ( S_T > K )\n \\end{align}\nand especially\n \\begin{align}\n V_0 (X) &= S(0) \\Pm_S ( S_T > K ) - K Z(0,T) \\Pm_T ( S_T > K ) .\n \\end{align}\n \n \\fi\n \n\n \n\n", "meta": {"hexsha": "f4fdb2f6013c1f3d36085278c4454251e34eeaf4", "size": 25686, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "market.tex", "max_stars_repo_name": "mrytty/gradu-public", "max_stars_repo_head_hexsha": "537337ab3dc49be9f1f4283706b0f4dcbc8cb059", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "market.tex", "max_issues_repo_name": "mrytty/gradu-public", "max_issues_repo_head_hexsha": "537337ab3dc49be9f1f4283706b0f4dcbc8cb059", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "market.tex", "max_forks_repo_name": "mrytty/gradu-public", "max_forks_repo_head_hexsha": "537337ab3dc49be9f1f4283706b0f4dcbc8cb059", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 66.030848329, "max_line_length": 934, "alphanum_fraction": 0.6624620416, "num_tokens": 8864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.815232489352, "lm_q2_score": 0.7549149923816046, "lm_q1q2_score": 0.6154312284884017}} {"text": "% !TEX root=../main.tex\n\n\\begin{comment}\nConsider a feature matrix $\\X \\in \\Real^{N \\times D}$,\nwhere $N$ denotes the number of data points and $D$ the number of features for each point.\nFor example, $N$ could be the number of images in some photo collection, and $D$ the number of pixels used to represent each image.\nThe goal of anomaly detection is to determine which rows of $\\X$ are anomalous, in the sense of being dissimilar to all other rows. We will use $\\X_{n :}$ to denote the $n$th row of $\\X$.\n\\end{comment}\n\n%\\subsection{A tour of anomaly detection methods}\n\nAnomaly detection is a well-studied topic in Data Science~\\cite{chandola2007outlier,charubook}. Unsupervised anomaly detection aims at discovering\nrules to separate normal and anomalous data in the absence of labels. One-Class SVM (OC-SVM) is a popular unsupervised approach to detect anomalies, which constructs\na smooth boundary around the majority of probability mass of data~\\cite{Scholkopf:2001}. OC-SVM will be described in detail in Section~\\ref{sec:ocsvm}.\nIn recent times, several approaches of feature selection and feature extraction methods have been proposed for complex, high-dimensional\ndata for use with OC-SVM ~\\cite{cao2003comparison,neumann2005combined}.\nFollowing the unprecedented success of using deep autoencoder networks, as feature extractors, in tasks as diverse as visual, speech anomaly detection~\\cite{chong2017abnormal,marchi2017deep}, several hybrid models that combine feature extraction using deep learning and OC-SVM\nhave appeared~\\cite{sohaib2017hybrid,erfani2016high}. The benefits of leveraging pre-trained transfer learning representations for anomaly detection in hybrid models was made evident by the results obtained, using two publicly available~\\footnote{Pretrained-models:http://www.vlfeat.org/matconvnet/pretrained/.} pre-trained CNN models: ImageNet-MatConvNet-VGG-F (VGG-F) and ImageNet-MatConvNet-VGG-M (VGG-M)~\\cite{andrews2016transfer}. However, these hybrid OC-SVM approaches are decoupled in the sense that the feature learning is task agnostic and not customized for anomaly detecion. Recently a deep model which trains a neural network by minimizing the volume of a hypersphere that encloses the network representations of the data is proposed ~\\cite{pmlrv80ruff18a}, our approach differs from this approach by combining the ability of deep networks to extract progressively rich representation of data alongwith the one-class objective, which obtains the hyperplane to separate all the normal data points from the origin.\n %In this paper, we formulate and evaluate a new architecture for anomaly detection in complex, high-dimensional domains by integrating the one-class objective into neural network arhitecture. The proposed one-class neural network (OC-NN) model is trained using standard back-propogation algorithm to learn, rich differentiable features customized for anomaly detection. To the best of our knowledge, this is the first method proposed which\n%combines the ability of deep networks to extract progressively rich representation of data\n%with the one-class objective of creating a tight envelope around normal data to improve the performance\n%for anomaly detection.\n\n\\begin{comment}\n\\subsection{Deep hybrid OC-SVM Models.}\nA popular hybrid approach followed in unsupervised anomaly detection is to couple feature reduction and extraction methods alongwith SVMs~\\cite{cao2003comparison,neumann2005combined,shen2008feature,widodo2007combination}. Several hybrid models have been proposed to leverage the representational power of deep learning to address scalability issues of SVMs~\\cite{song2017hybrid,you2017hybrid,sohaib2017hybrid}. Deep learning architectures such as sparse stacked autoencoder (SAE)-based deep neural networks (DNNs)~\\cite{sohaib2017hybrid}, deep belief networks (DBNs)~\\cite{erfani2016high} are adopted for extracting robust features. DBNs are generative models consisting of multiple layers of latent variables (\"hidden units\"). The advantages of using DBNs are two-fold: (a) their ability to perform non-linear dimensionality reduction (b) learn high dimensional data representations. A DBN can be trained efficiently in a greedy layer-wise fashion by using a Restricted Boltzmann Machine (RBM)~\\cite{hinton2010practical}. A Restricted Boltzmann machine model can be defined by joint energy function on the hidden variable $h_k$ with $K$ dimension and visible input variable $x_d$ with $D$ dimension as seen in Equation~\\ref{energy}. The model parameters $W^P_{dk}$ encode the pairwise compatibility between $x_d$ and $h_k$ while the parameters $W^B_k$ and $W^C_d$ are biases which, selectively activates either $h_k$ or $x_d$.\n\n\\begin{align}\n\\label{energy}\n E_W(\\mbf{x},\\mbf{h})&=\n-\\sum_{d=1}^D\\sum_{k=1}^K W^P_{dk}x_dh_k - \\sum_{k=1}^KW^B_k h_k - \\sum_{d=1}^D W^C_d x_d\n\\end{align}\n\nIn the context of anomaly detection for images, the $\\mbf{x}$ variables may represent (e.g pixels of the image). The hidden units are binary vector of length $K$ less than original dimension of image $D$. Training an RBM implies finding the values of the parameters such that the energy is minimised. One possible approach is to maximise the log-likelihood of $x$ that is estimated by its gradient with respect to the model parameters using Contrastive Divergence (CD). The DBN consisting of stack of RBMs model encodes rich feature representations of the input within hidden units. While a hybrid model consisting of DBN, trained to extract generic underlying features, as inputs to one-class SVM trained are shown to be effective~\\cite{erfani2016high}. The experiments in the aforementioned work were performed on real-datasets comprising 1D inputs, synthetic data or texture images, which have lower dimensionality and different data distribution compared to colour images or long protein sequences.\n\\end{comment}\n\n%%%%%\n\\subsection{Robust Deep Autoencoders for anomaly detection}\nBesides the hybrid approaches which use OC-SVM with deep learning features another approach for anomaly detection is to use deep autoencoders.\nInspired by RPCA~\\cite{xu2010robust}, unsupervised anomaly detection techniques such as robust deep autoencoders can be used to separate\nnormal from anomalous data~\\cite{zhou2017anomaly,chalapathy2017robust}.\nRobust Deep Autoencoder (RDA) or Robust Deep Convolutional Autoencoder (RCAE) decompose input data $X$ into two parts $X = L_D + S$, where $L_D$ represents the latent representation the hidden layer of the autoencoder. The matrix $S$ captures noise and outliers which are hard to reconstruct as shown in Equation~\\ref{eqn:robust-ae}. The decomposition is carried out by optimizing the objective function shown in Equation~\\ref{eqn:robust-ae}.\n\n\\begin{equation}\n\t\\label{eqn:robust-ae}\n\t\\min_{\\theta, S} + ||L_D - D_{\\theta}(E_{\\theta}(L_D)) ||_{2}+ \\lambda \\cdot \\| S^T \\|_{2,1}\n\\end{equation}\n \\hspace{1.8cm} $s.t. \\hspace{0.2cm} X - L_D -S = 0 $\n\n\n\nThe above optimization problem is solved using a combination of backpropagation and Alternating Direction Method of Multipliers (ADMM) approach~\\cite{boyd2004convex}. In our experiments we have carried out a detailed comparision between OC-NN and approaches based on robust autoencoders.\n\n\\begin{comment}\nNotice that in the constraint of Equation~\\ref{eqn:robust-ae} we split the input data X into two\nparts, $L_D$ and $S$. $L_D$ is the input to an autoencoder $D_{\\theta}(E_{\\theta}(L_D))$ and\nwe train this autoencoder by minimizing the reconstruction error\n $||L_D - D_{\\theta}(E_{\\theta}(L_D)) ||_{2}$ through back-propagation. $S$, on the other\nhand, contains noise and outliers which are learnt applying proximal methods following Alternating Direction Method of Multipliers (ADMM) ~\\cite{boyd2004convex}. While $\\lambda$ tuning parameter is selected in a semi-supervised fashion through grid-search for optimal $F1$ score.\nRDA could be leveraged to detect anomalies similar to OC-SVM setting. The results obtained using our proposed approach of OC-NN are comparable to the results obtained on anomaly detection task on both MNIST and Cifar-10 datasets, as illustrated in Table~\\ref{tbl:mnist-usps-anomaly-results-summary} and Table~\\ref{tbl:cifar-10-pfam-anomaly-results-summary}.\n\\end{comment}\n\n\\subsection{One-Class SVM for anomaly detection}\n\\label{sec:ocsvm}\n\nOne-Class SVM (OC-SVM) is a widely used approach to discover anomalies in an unsupervised fashion~\\cite{scholkopf2002support}. OC-SVMs are a special case of support vector machine, which learns a hyperplane to separate all the data points from the origin in a reproducing kernel Hilbert space (RKHS) and maximises the distance from this hyperplane to the origin. Intuitively in OC-SVM all the data points are considered as positively labeled instances and the origin\nas the only negative labeled instance. More specifically, given a training data $\\X$, a set without any class information, and $\\Phi(\\X)$ a RKHS map function\nfrom the input space to the feature space $F$, a\nhyper-plane or linear decision function $f(\\X_{n :})$ in the feature space $F$ is constructed as\n$f( \\X_{n :} ) = w^T \\Phi(\\X_{n :}) - \\bias$, to separate as many as possible of the mapped vectors\n${ \\Phi(\\X_{n :}), n: 1,2,...,N}$ from the origin. Here $w$ is the norm perpendicular to the hyper-plane and $\\bias$ is the bias of the hyper-plane. In order to\nobtain $w$ and $\\bias$, we need to solve the\nfollowing optimization problem,\n\n\\begin{equation}\n\\label{eqn:ocsvm-objective}\n \\min_{w,\\bias} \\frac{1}{2} \\| w \\|_{2}^2+ \\frac{1}{\\nu} \\cdot \\frac{1}{N} \\sum_{n = 1}^N \\max( 0, \\bias - \\langle w, \\Phi(\\X_{n :}) \\rangle ) - \\bias.\n\\end{equation}\n\nwhere $\\nu \\in (0,1)$, is a parameter that\ncontrols a trade off between maximizing the distance of the\nhyper-plane from the origin and the number of data points\nthat are allowed to cross the hyper-plane (the false positives).\n%When $\\nu$ is small, fewer data fall on the same side of the hyper-plane as the origin in the\n%feature space $F$.\n\n\n\n", "meta": {"hexsha": "95d8dd1b99ba1b2c8fc62a496279d96c208c5028", "size": 9995, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ARXIV_DAD_Survey/src/background.tex", "max_stars_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey", "max_stars_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 107, "max_stars_repo_stars_event_min_datetime": "2019-01-11T12:06:24.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T12:03:57.000Z", "max_issues_repo_path": "ARXIV_DAD_Survey/src/background.tex", "max_issues_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey_Arxiv_WorkingDocument", "max_issues_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ARXIV_DAD_Survey/src/background.tex", "max_forks_repo_name": "raghavchalapathy/Deep-Learning-for-Anomaly-Detection-A-Survey_Arxiv_WorkingDocument", "max_forks_repo_head_hexsha": "aa775990a4b23306885979c4ef8e8cb3ed00441b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 27, "max_forks_repo_forks_event_min_datetime": "2019-01-15T02:42:12.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-06T07:59:29.000Z", "avg_line_length": 112.3033707865, "max_line_length": 1428, "alphanum_fraction": 0.7845922961, "num_tokens": 2532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.815232489352, "lm_q2_score": 0.7549149813536518, "lm_q1q2_score": 0.6154312194980561}} {"text": "\\section{Poisson and Laplace Equations}\r\n\\subsection{The Boundary Value Problem}\r\nMany problems in mathematical physics can be reduced to the form:\r\n$$\\nabla^2\\phi=F$$\r\nwhere $F$ is known.\r\nThis form of equations are called Poisson's Equation.\r\nIn particular, if $F\\equiv 0$, this is called Laplace's Equation.\r\nWe would want to solve them in either all of $\\mathbb R^n$ or on some domain $\\Omega\\subset\\mathbb R^n$.\r\nWe are, sometimes, only interested in the cases $n=2,3$.\r\nNote that we require $\\phi$ to be well-defined and smooth on all of $\\Omega$.\r\nFor example, it is true that $\\nabla^2(1/|\\underline{x}|)=0$ for $\\underline{x}\\neq\\underline{0}$, however it is not true for any domain containing $\\underline{0}$.\r\nWe would want to solve this PDE subject to some boundary conditions, so $\\phi$ will have predetermined behaviour in $\\partial\\Omega$ or as $|\\underline{x}|\\to\\infty$ when working in $\\mathbb R^n$.\\\\\r\nTwo widely studied boundary conditions are the Dirichlet Problem\r\n$$\\begin{cases}\r\n \\nabla^2\\phi=F\\text{, on $\\Omega$}\\\\\r\n f\\text{, on $\\partial\\Omega$}\r\n\\end{cases}$$\r\nand the Neumann Problem\r\n$$\\begin{cases}\r\n \\nabla^2\\phi=F\\text{, on $\\Omega$}\\\\\r\n \\partial\\phi/\\partial\\underline{n}=\\underline{n}\\cdot\\nabla\\phi=g\\text{, on $\\partial\\Omega$}\r\n\\end{cases}$$\r\nBeware that we must interpret boundary data (or boundary conditions) correctly.\r\nWe want $\\phi$ or $\\partial\\phi/\\partial\\underline{n}$ to approach the boundary data continuously as $\\underline{x}$ tends towards the boundary.\r\nSo apart from requiring $\\phi$ to be $C^2$ in $\\Omega$, it must also extend to $\\partial\\Omega$ continuously.\r\n\\begin{example}[Non-example]\r\n If we want to solve the Navier-Stokes Equation\r\n $$\\frac{\\partial\\underline{u}}{\\partial t}+(\\underline{u}\\cdot\\nabla)\\underline{u}-\\nu\\nabla^2\\underline{u}=-\\nabla p,\\nabla\\cdot\\underline{u}=0,\\underline{u}(\\underline{x},0)=\\underline{u_0}(\\underline{x})$$\r\n but ignore the condition on continuous extension we said earlier, then the following solution satisfies the equation\r\n $$\\underline{u}=\\begin{cases}\r\n 0\\text{, if $t>0$}\\\\\r\n \\underline{u_0}\\text{, if $t=0$}\r\n \\end{cases},p\\equiv 0$$\r\n But obviously we are not getting a million for it.\r\n\\end{example}\r\n\\begin{example}\r\n Let $r=|\\underline{x}|$, we consider the Dirichlet Problem\r\n $$\\begin{cases}\r\n \\nabla^2\\phi=r\\text{, for $ra$, we have\r\n $$\\phi^\\prime(r)=-\\frac{1}{4\\pi\\epsilon_0}\\frac{Q}{r^2}\\implies\\underline{E}(r)=\\frac{1}{4\\pi\\epsilon_0}\\frac{Q}{r^2}\\underline{e_r}$$\r\n Let $a\\to 0$ in such a way that $a^3\\rho_0$ remains constant (so we change $\\rho_0$), so $Q$ remains constant.\r\n Therefore the electric field induced by a point charge $Q$ at $\\underline{x}=\\underline{0}$ would have\r\n $$\\underline{E}(\\underline{x})=\\frac{1}{4\\pi\\epsilon_0}\\frac{Q}{r^2}\\underline{e_r}=\\frac{1}{4\\pi\\epsilon_0}\\frac{Q}{|\\underline{x}|^3}\\underline{x},\\phi(\\underline{x})=\\frac{1}{4\\pi\\epsilon_0}\\frac{Q}{|\\underline{x}|}$$\r\n for $r>0$ and $\\phi(\\infty)=0$.\r\n\\end{example}\r\nJust an aside, if we want the existece of solutions to $-\\nabla^2\\phi=\\epsilon_0^{-1}\\rho(\\underline{x})$, Gauss proved that it suffices to consider equations of the form $K\\sigma=f$ which looks like linear algebra, but in a infinite dimensional vector spaces, where we no longer have sequential compactness.\r\nThis induces the study of functional analysis and operator theory.\\\\\r\nSuppose we have $F=F(\\rho)$ where $\\rho=x^2+y^2$.\r\nTo solve $\\nabla^2\\varphi=F(\\rho)$, it is natural to try solutions of the form $\\varphi=\\varphi(\\rho)$.\r\nIn this case, $\\nabla\\varphi=\\varphi^\\prime(\\rho)\\underline{e_\\rho}$, so obviously we want to integrate it over a cylinder $V$ of height $1$, $V=\\{0\\le\\rho\\le R,0\\le\\phi\\le 2\\pi,z_0\\le z\\le z_0+1\\}$, as $\\underline{e_\\rho}\\perp\\underline{e_z}$,\r\n$$Q(R)=\\int_VF\\,\\mathrm dV=\\int_{\\partial V}\\nabla\\varphi\\cdot\\mathrm d\\underline{S}=\\varphi^\\prime(R)2\\pi R\\implies\\varphi^\\prime(R)=\\frac{Q(R)}{2\\pi R}$$\r\nNote that by calculation we have\r\n$$Q(R)=2\\pi\\int_0^RF(\\rho)\\rho\\,\\mathrm d\\rho\\implies\\varphi^\\prime(\\rho)=\\frac{1}{R}\\int_0^RF(\\rho)\\rho\\,\\mathrm d\\rho$$\r\n\\begin{example}\r\n Due to conflicts of notation we write $s^2=x^2+y^2$ instead of $\\rho$ (as we want to do electromagnetism).\r\n Suppose we have charge density\r\n $$\\rho(s)=\\begin{cases}\r\n \\rho_0\\text{, for $s\\le a$}\\\\\r\n 0\\text{, otherwise}\r\n \\end{cases}$$\r\n For electrostatic potential $\\varphi$, we still have $-\\nabla^2\\varphi=\\epsilon_0^{-1}\\rho$, so\r\n $$\\varphi^\\prime(R)=-\\frac{1}{\\epsilon_0R}\\int_0^R\\rho(s)s\\,\\mathrm ds=\\begin{cases}\r\n -\\epsilon_0^{-1}R^{-1}\\rho_0R^2/2\\text{, if $R\\le a$}\\\\\r\n -\\epsilon_0^{-1}R^{-1}\\rho_0a^2/2\\text{, otherwise}\\\\\r\n \\end{cases}$$\r\n Then shrink $a\\to 0$ with $\\rho_0a^2$ fixed, then we have $\\underline{E}\\propto s^{-1}\\underline{e_s}$, which is similar to the superposition a bunch of point charges in a line.\r\n\\end{example}\r\n\\subsection{The Superposition Principle}\r\nFor a linear problem, we can (most of the time) solve them by their defining property of being linear.\r\nSay if we have $L\\psi_n=F_n$, then $L(\\sum_n\\psi_n)=\\sum_nF_n$.\r\nThis allows us to superpose solutions.\r\nIf we write a forcing term $F$ as $\\sum_nF_n$ and solve for $\\psi_n$ individually, then we can obtain the required solution by summing up all of them.\r\n\\begin{example}\r\n 1. Recall the solutions for an electric potential $\\phi$ and electric field $\\underline{E}=-\\nabla\\phi$.\r\n For charge distribution with spherical symmetry, can be found by e.g. Gauss's flux method.\r\n If we shrink the radius to $0$, we obtain a point charge $Q_{\\underline{a}}$ for at $\\underline{a}$ where\r\n $$\\phi(\\underline{x})=\\frac{Q_{\\underline{a}}}{4\\pi\\epsilon_0}\\frac{1}{|\\underline{x}-\\underline{a}|}$$\r\n and $\\rho(\\underline{x})=Q_{\\underline{a}}\\delta(\\underline{x}-\\underline{a})$.\r\n For consistency with Gauss's Law, we shall (and indeed can) obtain\r\n $$\\nabla^2\\left( -\\frac{1}{4\\pi}\\frac{1}{|\\underline{x}-\\underline{a}|} \\right)=\\delta(\\underline{x}-\\underline{a})$$\r\n Now consider the electric potential due to $2$ charges $Q_{\\underline{a}},Q_{\\underline{b}}$ at $\\underline{x}=\\underline{a},\\underline{b}$.\r\n The charge distribution would be $Q_{\\underline{a}}\\delta(\\underline{x}-\\underline{a})+Q_{\\underline{b}}\\delta(\\underline{x}-\\underline{b})$, so we can superpose the solutions correspondingly in Gauss's Law to get the potential\r\n $$\\phi(\\underline{x})=\\frac{Q_{\\underline{a}}}{4\\pi\\epsilon_0}\\frac{1}{|\\underline{x}-\\underline{a}|}+\\frac{Q_{\\underline{b}}}{4\\pi\\epsilon_0}\\frac{1}{|\\underline{x}-\\underline{b}|}$$\r\n 2. We want to find the potential outside the solid sphere $|\\underline{x}|\\le R$ of uniform charge density $\\rho_0$, from which several spheres $|\\underline{x}-\\underline{a_i}|\\le R_i$ with $i=1,2,\\ldots,n$ are removed (given that the spheres do not cross the boundary).\r\n To find the solution, we can superpose the solution for the charge distribution $\\rho_0$ for $|\\underline{x}|\\le R$ and the solutions for the charge distribution $-\\rho_0$ for $|\\underline{x}-\\underline{a_i}|\\le R_i$.\r\n So\r\n $$\\phi(\\underline{x})=\\frac{1}{4\\pi\\epsilon_0}\\left( \\frac{Q}{|\\underline{x}|}-\\sum_{i=1}^n\\frac{Q_i}{|\\underline{x}-\\underline{a_i}|} \\right),Q=\\frac{4}{3}\\pi R^3\\rho_0,Q_i=\\frac{4}{3}\\pi R_i^3\\rho_0$$\r\n for $|\\underline{x}|>R$.\r\n\\end{example}\r\n\\subsection{Integral Solutions}\r\nIn the examples above, we found solutions by superposing (or superimposing) potentials corresponding to charges in different points, which gives solutions of the form\r\n$$\\sum_i\\frac{Q_i}{4\\pi\\epsilon_0}\\frac{1}{|\\underline{x}-\\underline{a_i}|}$$\r\nThis leads to a more general form of superposition of potentials, by thinking of each infinitesimal part as individual point charges, which is just a integral in the following form:\r\n$$\\int_{\\mathbb R^3}\\frac{F(\\underline{y})}{|\\underline{x}-\\underline{y}|}\\,\\mathrm dV(\\underline{y})$$\r\nUp to some factor.\r\n\\begin{proposition}\r\n The unique solution to the Dirichlet problem\r\n $$\\begin{cases}\r\n \\nabla^2\\phi=F\\text{, in $\\mathbb R^3$}\\\\\r\n \\phi(\\underline{x})\\to 0\\text{ as $|\\underline{x}|\\to\\infty$}\r\n \\end{cases}$$\r\n (Assuming $F$ decreases sufficiently rapidly as $|\\underline{x}|\\to\\infty$) is\r\n $$\\phi(\\underline{x})=-\\frac{1}{4\\pi}\\int_{\\mathbb R^3}\\frac{F(\\underline{y})}{|\\underline{x}-\\underline{y}|}\\,\\mathrm dV$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n The solution can be verified by using\r\n $$\\nabla\\left( -\\frac{1}{4\\pi}\\frac{1}{|\\underline{x}-\\underline{a}|} \\right)=\\delta(\\underline{x}-\\underline{a})$$\r\n And differentiating under the integral sign.\r\n\\end{proof}\r\nTo justify (in an applied way, of course) the identity used above, we use the divergence theorem,\r\n\\begin{align*}\r\n \\int_{|\\underline{x}|\\le R}\\nabla^2\\left( \\frac{1}{|\\underline{x}|} \\right)\\,\\mathrm dV&=\\int_{|\\underline{x}|=R}\\nabla\\left( \\frac{1}{r} \\right)\\cdot\\mathrm d\\underline{S}\\\\\r\n &=-\\frac{1}{R^2}\\int_{|\\underline{x}|=R}\\underline{e_r}\\cdot\\underline{e_r}\\,\\mathrm dS\\\\\r\n &=\\frac{1}{R^2}4\\pi R^2\\\\\r\n &=4\\pi\r\n\\end{align*}\r\nwhich is true for any $R>0$, so it is natural (or maybe not) to write the identity.\\\\\r\nThe solution can be regarded as the sum of contributions to the potential to any possible volume elements (that actually contribute).\r\n\\subsection{Harmonic Functions}\r\nIf $\\phi=\\phi(\\underline{x})$ satisfies Laplace's Equation $\\nabla^2\\phi=0$, then $\\phi$ is harmonic.\r\n\\begin{proposition}[Mean-Value Property]\r\n If $\\varphi$ is harmonic on some open $\\Omega\\subset\\mathbb R^3$, then\r\n $$\\varphi(\\underline{a})=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}-\\underline{a}|=r}\\varphi(\\underline{x})\\,\\mathrm dS$$\r\n where $\\underline{a}\\in\\Omega$ and the ball centered at $\\underline{a}$ with radius $r$ is contained in $\\Omega$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Define a function\r\n \\begin{align*}\r\n f(r)&=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}-\\underline{a}|=r}\\varphi(\\underline{x})\\,\\mathrm dS\\\\\r\n &=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}|=r}\\varphi(\\underline{x}+\\underline{a})\\,\\mathrm dS\\\\\r\n &=\\frac{1}{4\\pi r^2}\\int_0^{2\\pi}\\int_0^\\pi \\varphi(\\underline{a}+r\\underline{e_r})r^2\\sin\\theta\\,\\mathrm d\\theta\\,\\mathrm d\\phi\\\\\r\n &=\\frac{1}{4\\pi}\\int_0^{2\\pi}\\int_0^\\pi \\varphi(\\underline{a}+r\\underline{e_r})\\sin\\theta\\,\\mathrm d\\theta\\,\\mathrm d\\phi\r\n \\end{align*}\r\n So differentiating this gives\r\n \\begin{align*}\r\n f^\\prime(r)&=\\frac{1}{4\\pi}\\int_0^{2\\pi}\\int_0^\\pi \\nabla\\varphi(\\underline{a}+r\\underline{e_r})\\cdot\\underline{e_r}\\sin\\theta\\,\\mathrm d\\theta\\,\\mathrm d\\phi\\\\\r\n &=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}|=r}\\nabla\\varphi(\\underline{a}+\\underline{x})\\cdot\\mathrm d\\underline{S}\\\\\r\n &=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}|\\le r}\\nabla^2\\varphi(\\underline{a}+\\underline{x})\\,\\mathrm dV\\\\\r\n &=0\r\n \\end{align*}\r\n So $f$ is constant.\r\n Let $r\\to 0$ gives $f\\equiv \\varphi(\\underline{a})$.\r\n\\end{proof}\r\n\\begin{proposition}\r\n For a smooth function $\\varphi$\r\n $$\\nabla^2\\varphi=\\lim_{r\\to 0}\\frac{6}{r^2}\\left(\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}-\\underline{a}|=r}\\varphi(\\underline{x})\\,\\mathrm dS-\\varphi(\\underline{a})\\right)$$\r\n\\end{proposition}\r\n\\begin{proof}\r\n Consider the function\r\n $$g(r)=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}-\\underline{a}|=r}\\varphi(\\underline{x})\\,\\mathrm dS-\\varphi(\\underline{a})$$\r\n then use the same trick as above, we have\r\n $$g^\\prime(r)=\\frac{1}{4\\pi r^2}\\int_{|\\underline{x}|\\le r}\\nabla^2\\varphi(\\underline{x})\\,\\mathrm dV$$\r\n But\r\n \\begin{align*}\r\n \\int_{|\\underline{x}|\\le r}\\nabla^2\\varphi(\\underline{x})\\,\\mathrm dV&=\\int_{|\\underline{x}|\\le r}\\nabla^2\\varphi(\\underline{a})\\,\\mathrm dV+\\left( \\int_{|\\underline{x}|\\le r}\\nabla^2\\varphi(\\underline{x})-\\nabla^2\\varphi(\\underline{a})\\,\\mathrm dV \\right)\\\\\r\n &=\\frac{4\\pi r^3}{3}\\nabla^2\\varphi(\\underline{a})+o(r^3)\r\n \\end{align*}\r\n as $r\\to 0$.\r\n So\r\n $$g^\\prime(r)=\\frac{r}{3}\\nabla^2\\varphi(\\underline{a})+o(r)$$\r\n But then $g^\\prime(r)=g^\\prime(0)+rg^\\prime(0)+o(r)$ by Taylor's Theorem, so $g^\\prime(0)=0,g^{\\prime\\prime}(0)=3^{-1}\\nabla^2\\varphi(\\underline{a})$.\r\n So\r\n $$g(r)=g(0)+rg^\\prime(0)+r^2\\frac{g^{\\prime\\prime}(0)}{2}+o(r^2)=\\frac{r^2}{6}\\nabla^2\\varphi(\\underline{a})+o(r^2)$$\r\n So taking the stated limit gives the solution.\r\n\\end{proof}\r\nSo the Laplacian measures how much the value of the function at the point differs from the average of the values on the infinitesimal sphere centered at the same point.\r\n\\begin{proposition}[Maximum Principle]\r\n If $\\varphi$ is harmonic on an open, path-connected $\\Omega\\in\\mathbb R^3$, then there is an $\\underline{a}\\in\\Omega$ such that $\\varphi(\\underline{x})\\le\\varphi(\\underline{a})$ throughout $\\Omega$, then $\\varphi$ is constant.\r\n\\end{proposition}\r\n\\begin{proof}\r\n If $\\varphi$ is harmonic, then the mean-value property holds, so\r\n $$\\varphi(\\underline{a})=\\frac{1}{4\\pi\\epsilon^2}\\int_{|\\underline{x}-\\underline{a}|=\\epsilon}\\varphi(\\underline{x})\\,\\mathrm dS$$\r\n for $\\epsilon$ sufficiently small.\r\n If $\\varphi(\\underline{x})\\le\\varphi(\\underline{a})$ throughout $\\Omega$, then\r\n $$0=\\frac{1}{4\\pi\\epsilon^2}\\int_{|\\underline{x}-\\underline{a}|=\\epsilon}\\varphi(\\underline{a})-\\varphi(\\underline{x})\\,\\mathrm dS$$\r\n But the integrand is nonnegative, so we must have $\\varphi(\\underline{x})=\\varphi(\\underline{a})$ for any $\\underline{x}$ in the sphere we are integrating, hence the ball enclosed.\r\n Now take any $\\underline{y}\\in\\Omega$ and consider a path joining $\\underline{a}$ and $\\underline{y}$.\r\n Then by compactness, there is a finite collection of spheres in $\\Omega$ whose interior covers the path.\r\n But then each adjascent two of them will share at least one interior points, and if $f$ achieve $f(\\underline{a})$ at that point, then $f(\\underline{x})\\le f(\\underline{a})$ for any $\\underline{x}$ in each ball, so use the argument inductively gives $f(\\underline{y})=f(\\underline{a})$.\r\n\\end{proof}\r\nOne can also prove the last part without introducing compactness:\r\nAssuming that we have already shown that for any $\\underline{b}\\in\\Omega$ with $\\varphi(\\underline{b})=\\alpha=\\varphi(\\underline{a})$, there is some $\\epsilon>0$ such that $\\varphi(\\underline{x})=\\varphi(\\underline{b})=\\alpha$ for any $|\\underline{x}-\\underline{b}|<\\epsilon$ (by e.g. the first part of our proof above).\r\nConsider the set $\\varphi^{-1}(\\{\\alpha\\})$, which is closed by continuity and open by above, which contradicts the path-connectedness (hence connectedness) of $\\Omega$.", "meta": {"hexsha": "3a04b0cb18b72c93e6a638408f17d1ca9d18875b", "size": 19412, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "7/poislap.tex", "max_stars_repo_name": "david-bai-notes/IA-Vector-Calculus", "max_stars_repo_head_hexsha": "466dbf395800c80f263dbf32161d20a7c1092c5b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, 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YES\n2. YES", "lm_q1_score": 0.7310585786300049, "lm_q2_score": 0.8418256532040707, "lm_q1q2_score": 0.6154238654856433}} {"text": "\\documentclass[12pt]{article}\n\\newcommand{\\m}{\\mathrm{m}}\n\\newcommand{\\cm}{\\mathrm{cm}}\n\\newcommand{\\s}{\\mathrm{s}}\n\\newcommand{\\dd}{\\mathrm{d}}\n\\newcounter{problem}\n\\begin{document}\\thispagestyle{empty}\n\n\\section*{NYU General Physics 1---Problem set 10}\n\n\\paragraph{Problem~\\theproblem:}\\refstepcounter{problem}%\nConsider a function $$x(t) = A\\,\\sin(\\omega\\,t + \\phi) \\quad,$$ where\n$A$, $\\omega$, and $\\phi$ are constants. Now consider the\n``differential equation'' $$m\\,\\frac{\\dd^2 x}{\\dd t^2} = -k\\,x\n\\quad.$$ Take two derivatives of $x(t)$ and plug the answer into the left-hand\nside of the differential equation. Under what conditions will the\ngiven $x(t)$ satisfy the differential equation? That is, what needs\nto be true about $A$, $\\omega$, and $\\phi$? How is the story\ndifferent if you have $\\cos$ instead of $\\sin$?\n\n\\paragraph{Problem~\\theproblem:}\\refstepcounter{problem}%\nConsider a string stretched in the $x$ direction, waving transversely\n(look it up), with the $y$ displacement being a function of position\n$x$ and time $t$ according to\n$$ y(x,t) = A\\,\\cos(\\frac{2\\pi\\,x}{\\lambda} - \\frac{2\\pi\\,t}{T}) $$\nwhere $A$ is an amplitude, $\\lambda$ is the wavelength, and $T$ is\nthe period. For definiteness, set $A=1\\,\\cm$, $\\lambda=0.75\\,\\m$, and\n$T=0.25\\,\\s$.\n\n\\textsl{(a)} Draw a picture of $y(t)$ over the time period $0{\n \\input{\\texfigdir/sigmoid4.pstex_t}\n }{\\alt<3>{\n \\input{\\texfigdir/sigmoid3.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/sigmoid2.pstex_t}\n }{\n \\input{\\texfigdir/sigmoid1.pstex_t}\n }}}\n }\n \\caption{Sigmoid function: $g(ax)=1/(1+e^{-ax})$ for $a=1,2,3,4$}\n \\label{f:sigmoid}\n \\end{figure}\n\\end{frame}\n\n\n\\input{nextTime.tex}\n\n\\subsection{Decision Boundary}\n\n\\begin{frame}\n \\frametitle{Decision Boundary}\n\n The \\structure{decision boundary} $\\delta(\\vec x)=0$ (zero level set) in feature space separates the two classes. \\\\[.3cm]\n Points $\\vec x$ on the decision boundary satisfy:\n \n \\begin{eqnarray*}\n {p(y=0|\\vec{x})}&=& {p(y=1|\\vec x)}\n \\end{eqnarray*}\n \n and thus\n \n \\begin{eqnarray*}\n \\log \\frac{p(y=0|\\vec{x})}{p(y=1|\\vec x)} &=& \\pause \\log 1 = 0 \\quad.\n \\end{eqnarray*}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n \\begin{lemma}\n The decision boundary is given by $F(\\vec x) = 0$.\n \\end{lemma}\n \\pspread\n \n \\structure{Proof:}\n%\n \\begin{eqnarray*}\n \\log \\frac{p(y=0|\\vec{x})}{p(y=1|\\vec x)} &=& F(\\vec x) = 0\\\\[.3cm] \\pause\n \\frac{p(y=0|\\vec{x})}{p(y=1|\\vec x)} &=& e^{F(\\vec x)} \\\\[.3cm] \\pause\n {p(y=0|\\vec{x})}&=& e^{F(\\vec x)} {p(y=1|\\vec x)} \\\\[.3cm]\n % {p(y=0|\\vec{x})}&=& e^{F(\\vec x)}\\left( {1-p(y=0|\\vec x)}\\right)\n \\end{eqnarray*} \n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n Now we use that the posteriors sum up to one:\n \n \\begin{eqnarray*}\n {p(y=0|\\vec{x})}&=& e^{F(\\vec x)}\\left( {1-p(y=0|\\vec x)}\\right) \\\\ \\\\\n \\pause {p(y=0|\\vec{x})}&=& \\frac{e^{F(\\vec x)}}{1+e^{F(\\vec x)}} \\\\ \\\\\n \\pause {p(y=0|\\vec{x})}&=& \\frac{1}{1+e^{-F(\\vec x)}} \n \\end{eqnarray*} \n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n \\begin{figure}\n \\resizebox{1\\linewidth}{!}{\n \\alt<4->{\n \\input{\\texfigdir/logit3.pstex_t}\n }{\\alt<3>{\n \\input{\\texfigdir/logit2.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/logit1.pstex_t} \n }{\n \\input{\\texfigdir/logit0.pstex_t}\n }}}\n }\n \\caption{Two Gaussians and their posteriors: {\\color{bl3} $\\sigma_0$}={\\color{gr3}$ \\sigma_1$}= 0.25, {\\color{bl3} $\\mu_0=-2$},\n {\\color{gr3} $\\mu_1=1$}}\n \\label{f:logit}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n \\begin{ovalblock}{Example}\n \\footnotesize\n Let us assume both classes have normally distributed $d$-dimensional feature vectors:\n \n \\begin{eqnarray*}\n p(\\vec x | y) &=& \\frac{1}{ \\sqrt{ \\det{(2 \\pi \\mat{\\Sigma}_y})} } \n e^{-\\frac{1}{2} (\\vec{x} - \\vec{\\mu}_y)^T \\mat{\\Sigma}_y^{-1} (\\vec{x} - \\vec{\\mu}_y) }\n \\end{eqnarray*}\n \\pause\n \n Then we can write the posterior of $y=0$ in terms of a logistic function:\n \n \\begin{eqnarray*}\n p(y=0|\\vec x) &=& \\frac{1}{1+e^{-F(\\vec x)}} = \\frac{1}{1+e^{ - \\left( \\vec x^T \\mat A \\vec x + \\vec \\alpha^T\\vec x + \\alpha_0 \\right) } } \\\\ \\\\\n \\pause F(\\vec x) &=& \\log \\frac{p(y=0|\\vec x)}{p(y=1|\\vec x)} = \\log \\frac{p(y=0)p(\\vec x | y=0) }{p(y=1)p(\\vec x|y =1)}\n \\end{eqnarray*}\n \\end{ovalblock}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n\n \\begin{ovalblock}{Example cont.}\n \\footnotesize\n \\begin{eqnarray*}\n F(\\vec x) &=&\n \\log\\frac{p(y=0)}{p(y=1)} +\n \\log\n \\frac{\n \\frac{1}{ \\sqrt{ \\det{( 2 \\pi \\mat{\\Sigma}_0)}} }\n e^{ -\\frac{1}{2} (\\vec{x} - \\mat{\\mu}_0)^T \\mat{\\Sigma}_0^{-1} (\\vec{x} - \\vec{\\mu}_0) } \n }{\n \\frac{1}{ \\sqrt{ \\det{(2 \\pi \\mat{\\Sigma}_1)}} }\n e^{ -\\frac{1}{2} (\\vec{x} - \\vec{\\mu}_1)^T \\mat{\\Sigma}_1^{-1} (\\vec{x} - \\vec{\\mu}_1)}\n }\n \\end{eqnarray*}\n \\pause\n \n This function has the constant component:\n% \n \\begin{eqnarray*}\n c &=& \\log \\frac{p(y=0)}{p(y=1)}+\\frac{1}{2}\n \\log \\frac{ { \\det{(2\\pi\\mat{\\Sigma}_1)}}}\n { { \\det{(2\\pi\\mat{\\Sigma}_0)}}}\n \\end{eqnarray*}\n \\pause \n \n We observe:\n \\begin{itemize}\n \\small\n \\item Priors imply a constant offset of the decision boundary. \n \\item If priors and covariance matrices of both classes are identical, this offset is $c=0$.\n \\end{itemize}\n \\end{ovalblock}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n\n \\begin{ovalblock}{Example cont.}\n \\small\n Furthermore we have:\n \\begin{eqnarray*}\n \\log{ \n \\frac{e^{ - \\frac{1}{2} (\\vec{x} - \\vec{\\mu}_0)^T \\mat{\\Sigma}_0^{-1} (\\vec{x} - \\vec{\\mu}_0)} }\n {e^{ - \\frac{1}{2} (\\vec{x} - \\vec{\\mu}_1)^T \\mat{\\Sigma}_1^{-1} (\\vec{x} - \\vec{\\mu}_1)} }\n } & = & \\\\[.3cm]\n & & \\pause \\hspace{-5cm} =\n \\frac{1}{2} \n \\left(\n (\\vec{x} - \\vec{\\mu}_1)^T \\mat{\\Sigma}_1^{-1} (\\vec{x} - \\vec{\\mu}_1) - (\\vec{x} - \\vec{\\mu}_0)^T \\mat{\\Sigma}_0^{-1} (\\vec{x} - \\vec{\\mu}_0)\n \\right) \\\\[.3cm]\n & & \\hspace{-5cm} \\pause =\n \\frac{1}{2} \n \\left(\n \\vec{x}^T (\\mat{\\Sigma}_1^{-1} - \\mat{\\Sigma}_0^{-1}) \\vec{x} \n -2 (\\vec{\\mu}_1^T \\mat{\\Sigma}_1^{-1} - \\vec{\\mu}_0^T \\mat{\\Sigma}_0^{-1} ) \\vec{x} + \\right.\\\\\n & & \\left. \\hspace{-4cm} \n + \\vec{\\mu}_1^T \\mat{\\Sigma}_1^{-1} \\vec{\\mu}_1 - \\vec{\\mu}_0^T \\mat{\\Sigma}_0^{-1} \\vec{\\mu}_0\n \\right)\n \\end{eqnarray*}\n \\end{ovalblock}\n\\end{frame}\n \n \n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n \\begin{ovalblock}{Example cont.}\n \\small\n Now we have:\n \\begin{eqnarray*}\n \\mat{A} &=& \\frac{1}{2} (\\mat{\\Sigma}_1^{-1} - \\mat{\\Sigma}_0^{-1}) \\\\[.5cm]\n \\vec{\\alpha}^T &=& \\vec{\\mu}_0^T \\mat{\\Sigma}_0^{-1} - \\vec{\\mu}_1^T \\mat{\\Sigma}_1^{-1} \\\\[.5cm]\n \\alpha_0 &=& \\log \\frac{p(y=0)}{p(y=1)} + \\frac{1}{2}\n \\left(\n \\log\n \\frac{ \\det{( 2 \\pi \\mat{\\Sigma}_1)}}\n { \\det{( 2 \\pi \\mat{\\Sigma}_0)}} + \n \\vec{\\mu}_1^T \\mat{\\Sigma}_1^{-1} \\vec{\\mu}_1 -\n \\vec{\\mu}_0^T \\mat{\\Sigma}_0^{-1} \\vec{\\mu}_0\n \\right)\n \\end{eqnarray*}\n \\end{ovalblock}\n\\end{frame}\n \n \n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n \n \\begin{figure}\n \n \\resizebox{.6\\linewidth}{!}{\n \\alt<4->{\n \\input{\\texfigdir/plot_decision_boundary.pstex_t}\n }\n{\\alt<3>{\n \\input{\\texfigdir/gaussian2.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/gaussian1.pstex_t}\n }{\n \\input{\\texfigdir/gaussian0.pstex_t}\n }}}\n }\n \\caption{Two Gaussian sample sets and the decision boundary}\n \\end{figure}\n\\end{frame}\n \n \n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n\n \\structure{Quadratic polynomials in the 2 variables $x_1$ and $x_2$}\n\n \\begin{eqnarray*}\n F(\\vec{x}) \n &=& \\vec x^T \\mat A \\vec x + \\vec \\alpha^T\\vec x + \\alpha_0 \\\\\n &=& a x_1^2 + b x_1 x_2 + c x_2^2 + d x_1 + e x_2 + f \\stackrel{!}{=} 0\n \\end{eqnarray*}\n \\vspace{-0.75cm}\n\n \\begin{figure}\n \\copyrightbox[b]{\n \\subfloat[circles and ellipses]{\n \\makebox[.3\\linewidth]{\n \\href{http://en.wikipedia.org/wiki/File:Conic_sections_with_plane.svg}{\n \\includegraphics[height=3cm]{\\pngdir/conic_section_circle_ellipse.\\png}\n }\n }\n }\n \\subfloat[parabolas]{\n \\makebox[.3\\linewidth]{\n \\href{http://en.wikipedia.org/wiki/File:Conic_sections_with_plane.svg}{\n \\includegraphics[height=3cm]{\\pngdir/conic_section_parabola.\\png}\n }\n }\n }\n \\subfloat[hyperbolas]{\n \\makebox[.3\\linewidth]{\n \\href{http://en.wikipedia.org/wiki/File:Conic_sections_with_plane.svg}{\n \\includegraphics[height=3cm]{\\pngdir/conic_section_hyperbola.\\png}\n }\n }\n }\n }{Pbroks13, \\href{https://creativecommons.org/licenses/by/3.0} {CC BY 3.0}, via Wikimedia Commons}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n\n \\structure{Posterior probability}\n\n \\begin{center}\n \\resizebox{.85\\linewidth}{!}{\n \\input{\\texfigdir/plot_posterior.pstex_t}\n }\n \\end{center}\n\\end{frame}\n\n\\input{nextTime.tex}\n \n\\begin{frame}\n \\frametitle{Decision Boundary in Distributions with Equal Dispersion}\n \n \\begin{ovalblock}{Example cont.}\n \\small\n If both classes share the same covariances i.\\,e.\\ $\\mat\\Sigma= \\mat\\Sigma_0=\\mat\\Sigma_1$, then the argument of the sigmoid function is linear in the components of $\\vec{x}$.\n \n \\begin{eqnarray*}\n \\mat{A} &=& \\mat{0} \\\\[.5cm]\n \\vec{\\alpha}^T &=& (\\vec{\\mu}_0 - \\vec{\\mu}_1)^T \\mat{\\Sigma}^{-1} \\\\[.5cm]\n \\alpha_0 &=& \\log \\frac{p(y=0)}{p(y=1)} + \n \\frac{1}{2}(\\vec{\\mu}_1 - \\vec{\\mu}_0)^T \\mat{\\Sigma}^{-1} (\\vec{\\mu}_1 - \\vec{\\mu}_0) \n \\end{eqnarray*}\n \\end{ovalblock}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary in Distributions with Equal Dispersion \\cont}\n \n \\begin{figure}\n \\resizebox{.6\\linewidth}{!}{\n \\alt<2>{\n \\input{\\texfigdir/gaussian3.pstex_t}\n }{\n \\input{\\texfigdir/gaussian5.pstex_t}\n }\n } \n \\caption{Identical covariances lead to linear decision boundary}\n \\end{figure}\n\\end{frame}\n\n\n% \\begin{frame}\n% \\frametitle{Decision Boundary \\cont}\n% \n% \\begin{figure}\n% \\includegraphics[width=0.7\\textwidth]{\\psdir/gauss_both.\\ps}\n% \\caption{Quadratic and linear decision boundary in comparison}\n% \\end{figure}\n% \\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary in Distributions with Equal Dispersion \\cont}\n\n \\structure{Note:}\n \n \\begin{itemize}\n \\item If the class conditionals are Gaussians and share the same covariance, the argument of the exponential function is affine in $\\vec x$.\n \\item This result is even true for a more general family of pdfs and not limited to Gaussians.\n \\end{itemize}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Decision Boundary in Distributions with Equal Dispersion \\cont}\n\n \\begin{citeblock}{Definition}\n \nThe {\\em exponential family} is a class of pdf's that can be written in the following canonical form\n \\begin{eqnarray*}\n p(\\vec x;\\vec \\theta, \\phi) &=& e^{\\frac{\\vec\\theta^T \\cdot \\vec x -b(\\vec \\theta)}{a(\\phi)}+c(\\vec x,\\phi)}\n \\end{eqnarray*}\n where $\\vec \\theta\\in \\real^d$ is the {\\em location parameter vector}, $\\phi$ the {\\em dispersion parameter}.\n \\end{citeblock}\n\\end{frame}\n\n\n% \\begin{ovalblock}{Example}\n% Binomial, Poisson, hypergeometric, exponential probability density functions or Gaussians belong to the exponential family.\n% \\end{ovalblock}\n\n\n\\begin{frame}\n \\frametitle{Exponential Family}\n\n \\structure{Gaussian Probability Density Function \\phantom{\\cont}}\n\n \\begin{displaymath}\n \\mathcal{N}(\\vec{x}; \\vec{\\mu},\\mat{\\Sigma}) = \\frac{1}{ \\sqrt{ \\det{(2 \\pi \\mat{\\Sigma}_y})} } \n e^{-\\frac{1}{2} (\\vec{x} - \\vec{\\mu})^T \\mat{\\Sigma}^{-1} (\\vec{x} - \\vec{\\mu}) } \n \\end{displaymath}\n%\n \\begin{figure}\n \\centering\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss1.pstex_t} \n }\n }\n \\quad\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss2.pstex_t}\n }\n }\n \\quad\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss3.pstex_t}\n }\n }\n\n \\caption{Gaussian probability density functions with $\\vec\\mu = (0, 0)^T$}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Exponential Family}\n\n \\structure{Gaussian Probability Density Function \\cont}\n%\n \\begin{displaymath}\n \\mathcal{N}(\\vec{x}; \\vec{\\mu},\\mat{\\Sigma}) = \\frac{1}{ \\sqrt{ \\det{(2 \\pi \\mat{\\Sigma}_y})} } \n e^{-\\frac{1}{2} (\\vec{x} - \\vec{\\mu})^T \\mat{\\Sigma}^{-1} (\\vec{x} - \\vec{\\mu}) } \n \\end{displaymath}\n%\n \\begin{figure}\n \\centering\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss4.pstex_t}\n }\n }\n \\quad\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss5.pstex_t}\n }\n }\n \\quad\n \\subfloat{\n \\resizebox{.3\\linewidth}{!}{\n \\input{\\texfigdir/gauss6.pstex_t}\n }\n }\n\n \\caption{Gaussian probability density functions with $\\vec\\mu = (0, 0)^T$}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Exponential Family \\cont}\n\n \\structure{Exponential Probability Density Function}\n%\n \\begin{displaymath}\n f_\\lambda(x) = \\left\\{ \n \\begin{array}{ll}\n \\lambda e^{-\\lambda x} & x \\ge 0 \\\\\n 0 & x < 0 \\\\\n \\end{array}\n \\right.\n \\end{displaymath}\n%\n \\begin{figure}\n \\resizebox{.65\\linewidth}{!}{\n \\alt<4->{\n \\input{\\texfigdir/exponential4.pstex_t}\n }{\\alt<3>{\n \\input{\\texfigdir/exponential3.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/exponential2.pstex_t}\n }{\n \\input{\\texfigdir/exponential1.pstex_t}\n }}}\n }\n \\caption{Exponential probability density functions}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Exponential Family \\cont}\n\n \\structure{Binomial Probability Mass Function}\n%\n \\begin{displaymath}\n B(k; p,n) = {n \\choose k} p^k (1-p)^{n-k}\n \\end{displaymath}\n%\n \\begin{figure}\n \\resizebox{.65\\linewidth}{!}{\n \\alt<3->{\n \\input{\\texfigdir/binomial3.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/binomial2.pstex_t}\n }{\n \\input{\\texfigdir/binomial1.pstex_t}\n }}\n }\n \\caption{Binomial probability mass functions for $n=20$}\n \\end{figure}\n\\end{frame}\n\n\n\\begin{frame}\n \\frametitle{Exponential Family \\cont}\n\n \\structure{Poisson Probability Mass Function}\n%\n \\begin{displaymath}\n P_\\lambda(X=k) = \\frac{\\lambda^k}{k!} e^{-\\lambda}\n \\end{displaymath}\n%\n \\begin{figure}\n \\resizebox{.70\\linewidth}{!}{\n \\alt<3->{\n \\input{\\texfigdir/poisson3.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/poisson2.pstex_t}\n }{\n \\input{\\texfigdir/poisson1.pstex_t}\n }}\n }\n \\caption{Poisson probability mass functions}\n \\end{figure}\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Exponential Family \\cont}\n\n \\structure{Hypergeometric Probability Mass Function}\n\n \\begin{displaymath}\n h(k; N,M,n) = \\frac{{M \\choose k}{N-M \\choose n-k}}{{N \\choose n}}\n \\end{displaymath}\n%\n \\begin{figure}\n \\resizebox{.65\\linewidth}{!}{\n \\alt<3->{\n \\input{\\texfigdir/hypergeometric3.pstex_t}\n }{\\alt<2>{\n \\input{\\texfigdir/hypergeometric2.pstex_t}\n }{\n \\input{\\texfigdir/hypergeometric1.pstex_t}\n }}\n }\n \\caption{Hypergeometric probability mass functions}\n \\end{figure}\n\\end{frame}\n\n\\begin{frame}\n \\frametitle{Decision Boundary \\cont}\n\n \\begin{lemma}\n If all class-conditional densities are members of the same exponential family of probability density functions with equal dispersion $\\phi$, the decision boundary $F(\\vec x)=0$ is linear in the components of $\\vec x$.\n \\end{lemma}\n\\end{frame}\n\n\n\\subsection{Lessons Learned}\n\n\\begin{frame}\n \\frametitle{Lessons Learned}\n\n \\begin{itemize}\n \\item Posteriors can be rewritten in terms of a logistic function. \\pause\n \\item Given the decision boundary $F(\\vec x)=0$, we can write down the posterior $p(y|\\vec x)$ right away. \\pause\n \\item Decision boundary for normally distributed feature vectors for each class is a quadratic function. \\pause\n \\item If Gaussians share the same covariances, the decision boundary is a linear function.\n \\end{itemize}\n\\end{frame}\n\n\\input{nextTime.tex}\n\n\\subsection{Further Readings}\n\n\\begin{frame}\n \\frametitle{Further Readings}\n \n \\begin{itemize}\n \\item T. Hastie, R. Tibshirani, and J. Friedman: \\\\\n \\structure{The Elements of Statistical Learning --}\\\\\n \\structure{ Data Mining, Inference, and Prediction},\\\\\n 2nd edition, Springer, New York, 2009. \\\\[.3cm]\n \\item David W. Hosmer, Stanley Lemeshow: \\\\\n \\structure{Applied Logistic Regression}, 2nd Edition, \\\\\n John Wiley \\& Sons, Hoboken, 2000.\n \\end{itemize}\n\\end{frame}\n\n\n\\subsection{Comprehensive Questions}\n\n\\begin{frame}\n \\frametitle{Comprehensive Questions}\n \n \\begin{itemize}\n \\item How can we model the posterior probabilities? \\\\[1cm] \\pause\n \\item Formulate the criterion for the decision boundary! \\\\[1cm] \\pause\n \\item Describe the shape of the decision boundary for a Gaussian with different and same class covariances! \\\\[1cm] \\pause\n \\item What effect does a change of the priors have on the decision boundary?\n \\end{itemize}\n\\end{frame}\n\n\n\n", "meta": {"hexsha": "b6a02d7b778c2df398445de2b51273e0de0d1c57", "size": 19589, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "03_logistic_regression.tex", "max_stars_repo_name": "akmaier/pr-slides", "max_stars_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2021-01-11T07:27:09.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-31T19:21:31.000Z", "max_issues_repo_path": "03_logistic_regression.tex", "max_issues_repo_name": "akmaier/pr-slides", "max_issues_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "03_logistic_regression.tex", "max_forks_repo_name": "akmaier/pr-slides", "max_forks_repo_head_hexsha": "c322ad388993ff7891b959764e4481a05a444dff", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-06-21T06:06:43.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-18T18:47:28.000Z", "avg_line_length": 28.9350073855, "max_line_length": 222, "alphanum_fraction": 0.5777732401, "num_tokens": 6873, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321889812553, "lm_q2_score": 0.7634837689358857, "lm_q1q2_score": 0.6150871000194765}} {"text": "\\section{Multivariate amplitude distributions}\n\\label{sec:epochs}\n\nIn Sect. \\ref{subsec:key_concepts} we introduce the fundamental quantities\nused in the distribution definitions. An example of the non-stationarity in\nfinancial markets is shown in Sect. \\ref{subsec:non_stationarity}. Finally, in\n\\ref{subsec:epochs} we explain the methodology to see how the returns are to a\ngood approximation multivariate Gaussian distributed or multivariate Algebraic\ndistributed.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Key concepts}\\label{subsec:key_concepts}\n\nConsider time series $S_{k} \\left( t \\right)$, $k = 1, 2, \\ldots, K$ of stock\nprices for $K$ companies. The values $S_{k} \\left( t \\right)$ are taken in\nfixed time steps $\\Delta t$. In general, the data contain an exponential\nincrease due to the drift. Thus, to measure the correlations independently of\nthis trend, it is better to use logarithmic differences instead of returns\n\\begin{equation}\n G_{k} \\left( t \\right) = \\ln S_{k} \\left( t + \\Delta t \\right) -\n \\ln S_{k} \\left(t \\right) = \\ln \\frac{S_{k} \\left( t + \\Delta t \\right)}\n {S_{k} \\left(t \\right)}.\n\\end{equation}\nAnyway, logarithmic differences and returns almost coincide if the time steps\n$\\Delta t$ are sufficiently short \\cite{subtle_nature,empirical_facts}\n\\begin{equation}\n G_{k} \\left(t\\right) \\approx r_{k} \\left(t\\right)\n = \\frac{S_{k} \\left( t + \\Delta t \\right) - S_{k} \\left( t \\right)}\n {S_{k} \\left( t \\right)}.\n\\end{equation}\nThe returns are well known to have distributions with heavy tails, the smaller\n$\\Delta t$, the heavier \\cite{non_stationarity_fin_guhr}. Furthermore, the\nsample standard deviations $\\sigma_{k}$, referred to as volatilities, strongly\nfluctuate for different time windows of the same length $T$\n\\cite{non_stationarity_fin_guhr,volatility_change}.\n\nThe mean of the logarithmic differences reads \\cite{exact_distributions_guhr}\n\\begin{equation}\n \\left\\langle G_{k} \\left( t \\right) \\right\\rangle_{T} = \\frac{1}{T}\n \\sum_{t = 1}^{T} G_{k} \\left( t \\right).\n\\end{equation}\nTo compare the different $K$ companies, it is necessary to normalize the time\nseries. The normalized time series are defined by\n\\cite{non_stationarity_fin_guhr,exact_distributions_guhr}\n\\begin{equation}\n M_{k} \\left( t \\right) = \\frac{G_{k} \\left( t \\right) - \\left\\langle\n G_{k} \\left( t \\right) \\right\\rangle} {\\sqrt{\\left\\langle G_{k}^{2}\n \\left( t \\right) \\right\\rangle_{T} - \\left\\langle G_{k} \\left( t \\right)\n \\right\\rangle^2_{T}}},\n\\end{equation}\nwhere\n\\begin{equation}\n \\sigma_{k} = \\sqrt{\\left\\langle G_{k}^{2}\n \\left( t \\right) \\right\\rangle_{T} - \\left\\langle G_{k} \\left( t \\right)\n \\right\\rangle^2_{T}}\n\\end{equation}\nis the volatility of the $k$ company in the time window of length $T$. These\nvalues can be viewed as the elements of a $K \\times T$ rectangular matrix $M$.\nWith these normalizations and rescalings, correlations can be measured in such\na way that all companies and all stocks are treated on equal footing.\n\nThe correlation coefficient for the stocks $k$ and $l$ is defined as\n\\cite{non_stationarity_fin_guhr}\n\\begin{equation}\n C_{kl} = \\left\\langle M_{k} \\left( t \\right) M_{l} \\left( t \\right)\n \\right\\rangle_{T} = \\frac{1}{T} \\sum_{t=1}^{T} M_{k} \\left( t \\right) M_{l}\n \\left( t \\right),\n\\end{equation}\nwhich can be written as\n\\begin{equation}\n C_{kl} = \\frac{\\left\\langle G_{k} \\left( t \\right) G_{l} \\left( t \\right)\n \\right\\rangle_{T} - \\left\\langle G_{k} \\left( t \\right) \\right\\rangle_{T}\n \\left\\langle G_{l} \\left( t \\right) \\right\\rangle_{T}}\n {\\sigma_{k} \\sigma_{l}}.\n\\end{equation}\nThe coefficients $C_{kl}$ are the elements of a $K \\times K$ square matrix $C$,\nthe correlation matrix. The limiting values of these correlation coefficients\n\\begin{equation}\n C_{kl}^{\\text{lim}} =\n \\left\\{\n \\begin{array}{cc}\n +1 & \\text{completely correlated} \\\\\n 0 & \\text{completely uncorrelated}\\\\\n -1 & \\text{completely anticorrelated}\n \\end{array}\n \\right. .\n\\end{equation}\nThe time average of $C_{kl}$ can be viewed as the matrix product of the\nrectangular matrix $M$ ($K \\times T$) with its transpose matrix $M^{\\dagger}$\n($T \\times K$), divided by $T$. Thus, the correlation matrix can be written in\nthe form\n\\begin{equation}\n C = \\frac{1}{T} M M^{\\dagger}.\n\\end{equation}\nThe correlation matrix $C$ is real and symmetric. Using the correlation matrix\n$C$ is possible to define the covariance matrix\n\\cite{exact_distributions_guhr,credit_risk_guhr,portfolio_distributions_guhr,asset_correlations_guhr,stochastic_cov_guhr}\n\\begin{equation}\n \\Sigma = \\sigma C \\sigma ,\n\\end{equation}\nwhere the diagonal matrix $\\sigma$ contains respectively, the volatilities\n$\\sigma_{k}$, $k = 1, \\ldots, K$.\n\nTo ease the comparison between works, we will use covariance matrices as in\n\\cite{non_stationarity_fin_guhr}.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Non-stationarity}\\label{subsec:non_stationarity}\n\nIn general, in non-stationary complex systems crucial parameters or\ndistributions of observables change in an erratic, unpredictable way over time.\nIn financial markets, the change of the $K \\times K$ correlation matrix $C$ as\na whole in time is an example of non-stationary. Fig.\n\\ref{fig:correlation_matrices} shows two large correlation matrices of\nlogarithmic differences of stock prices for companies in the Standard \\&\nPoor's 500 (S\\&P 500) stock market index ordered according to the Global\nIndustry Classification Standard (GICS). The time series were measured in\nsuccessive quarters. The stripes in these correlation matrices indicate the\nstructuring of the market in industrial sectors\n\\cite{state_crisis_7,non_stationarity_fin_guhr}\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.6\\columnwidth]\n {figures/05_correlation_matrix.png}\n \\caption{Correlation matrices of $K = 421$ companies for the fourth quarter\n of 2005 (left) and the first quarter of 2006 (right), the darker,\n the stronger the correlation. The companies are sorted according\n to industrial sectors.}\n \\label{fig:correlation_matrices}\n\\end{figure}\n\nIn Fig. \\ref{fig:correlation_matrices} it can be seen that the matrices look\ndifferent for the successive quarters, because the business relations between\nthe companies and the market expectations of the traders change in time.\nDespite the difference, the coarse structure remains similar, indicating some\nstability of the industrial sectors.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Choice of amplitude in each of the epochs}\\label{subsec:epochs}\n\nWe use the same methodology from T. A. Schmitt et al.\n\\cite{non_stationarity_fin_guhr} to show that to a good approximation, the\nreturns can be multivariate Gaussian distributed or multivariate algebraic\ndistributed, depending on the epochs window length choice. All of this, if the\ncovariance matrix $\\Sigma$ is fixed within the epochs.\n\nWe start from the assumption that the $K$ dimensional vectors\n$r \\left( t \\right) = \\left( r_{1} \\left( t \\right), \\ldots, r_{K} \\left( t \\right) \\right)$\nfor a fixed return interval $\\Delta t$ is given by the multivariate Gaussian\ndistribution\n\\begin{equation}\\label{eq:gaussian_distribution}\n P_{G} \\left( r| \\Sigma_{ep} \\right) =\n \\frac{1}{\\sqrt{\\det 2 \\pi \\Sigma_{ep} }}\n \\exp \\left( - \\frac{1}{2} r^{\\dagger} \\Sigma_{ep}^{-1} r \\right).\n\\end{equation}\nTo test the assumption, we divide the time series in windows of length $T$\nwhere the sampled covariances can be viewed as constant within these windows.\n\nTo carry out the data analysis, whether for the multivariate Gaussian\ndistribution or the multivariate algebraic distribution, we choose all pairs of\nreturns, normalize the epochs (mean $\\mu = 0$ and variance $\\sigma^{2} = 1$)\nand create two-component vectors\n$\\left( r_{k} \\left( t \\right), r_{l} \\left( t \\right) \\right)$. For each pair,\nwe evaluate the $2 \\times 2$ sample covariance matrix\n$\\Sigma^{\\left(k, l \\right)}$. Then we diagonalize the covariance matrix as\n$\\Sigma = U \\Lambda U^{\\dagger}$, such that\n$\\Sigma^{-1/2} = U \\Lambda^{-1/2} U^{\\dagger}$, where $U$ is an orthogonal\n$K \\times K$ matrix and $\\Lambda$ is the diagonal matrix of the eigenvalues\n$\\Lambda_{K}$. We rotate the two-component returns vectors into the eigenbasis\nof $\\Sigma^{\\left(k, l \\right)}$,\n\\begin{equation}\n \\left(\\tilde{r}_{1} \\left(t \\right), \\tilde{r}_{2} \\left(t \\right) \\right)\n = U \\left(r_{k} \\left(t \\right), r_{l} \\left(t \\right) \\right)\n\\end{equation}\nand normalize the axis with the eigenvalues as\n\\begin{equation}\n \\frac{\\tilde{r}_{1} \\left(t \\right)}{\\sqrt{\\Lambda_{1}}} \\text{ and }\n \\frac{\\tilde{r}_{2} \\left(t \\right)}{\\sqrt{\\Lambda_{2}}}.\n\\end{equation}\nFinally, we aggregate all the components into a single univariate distribution,\nas all of them are now comparable.\n\nNow we need to find the one dimensional function related to the multivariate\nGaussian distribution in Eq. \\ref{eq:gaussian_distribution} to compare with the\nempirical single univariate distributions. In the Gaussian case, we know that\n\\begin{align}\n P_{G}^{\\left( ij \\right)} \\left( r^{\\left( ij \\right)} |\n \\Sigma_{ep}^{\\left( ij \\right)} \\right)\n &= \\int d \\left[ r \\right]_{\\ne i,j} P_{G} \\left( r | \\Sigma_{ep} \\right)\\\\\n &= \\frac{1}{\\sqrt{\\det 2 \\pi \\Sigma_{ep}^{\\left( ij \\right)}}}\n \\exp \\left( -\\frac{1}{2} r^{\\left( ij \\right) \\dagger}\n \\Sigma_{ep}^{\\left( ij \\right)^{ - 1}} r^{\\left( ij \\right)} \\right)\n\\end{align}\nis the corresponding bivariate model where\n\\begin{equation}\n r^{\\left(ij\\right)}=\\left[\\begin{array}{c}\n r_{i}\\\\\n r_{j}\n \\end{array}\\right]\\text{ and } \\Sigma_{ep}^{\\left(ij\\right)}\n =\\left[\\begin{array}{cc}\n \\Sigma_{ep,ii} & \\Sigma_{ep,ij}\\\\\n \\Sigma_{ep,ji} & \\Sigma_{ep,jj}\n \\end{array}\\right]\n\\end{equation}\nare the two-component return vector and the $2 \\times 2$ covariance matrix.\n$\\Sigma_{ep,ij}$ is just the $\\left( ij \\right)$ element of $\\Sigma_{ep}$.\nDiagonalizing\n\\begin{equation}\n \\Sigma_{ep}^{\\left(ij\\right)}=U^{\\left(ij\\right)}\n \\Lambda_{ep}^{\\left(ij\\right)}U^{\\left(ij\\right)\\dagger}\n\\end{equation}\nwith the $2 \\times 2$ rotation matrix $U^{\\left( ij \\right)}$ and the\neigenvalue matrix\n\\begin{equation}\n \\Lambda_{ep}^{\\left( ij \\right)} = \\text{diag}\n \\left( \\Lambda_{ep,1}^{\\left( ij \\right)},\n \\Lambda_{ep,2}^{\\left( ij \\right)} \\right).\n\\end{equation}\nWe rotate into the eigenbasis\n\\begin{equation}\n \\bar{r}^{\\left(ij\\right)}=\\left[\\begin{array}{c}\n \\bar{r}_{1}^{\\left(ij\\right)}\\\\\n \\bar{r}_{2}^{\\left(ij\\right)}\n \\end{array}\\right]=U^{\\left(ij\\right)\\dagger}r^{\\left(ij\\right)}\n\\end{equation}\nsuch that\n\\begin{align}\n P_{G}^{\\left(ij\\right)}\\left(r^{\\left(ij\\right)}|\n \\Sigma_{ep}^{\\left(ij\\right)}\\right)\n &=P_{G}^{\\left(ij\\right)}\\left(\\bar{r}^{\\left(ij\\right)}|\n \\Lambda_{ep}^{\\left(ij\\right)}\\right)\\\\\n &=\\frac{1}{\\sqrt{2\\pi\\Lambda_{ep,1}^{\\left(ij\\right)}}}\n \\exp\\left(-\\frac{\\bar{r}_{1}^{\\left(ij\\right)^{2}}}{2\\Lambda_{ep,1}\n ^{\\left(ij\\right)}}\\right)\n \\,\\frac{1}{\\sqrt{2\\pi\\Lambda_{ep,2}^{\\left(ij\\right)}}}\n \\exp\\left(-\\frac{\\bar{r}_{2}^{\\left(ij\\right)^{2}}}{2\\Lambda_{ep,2}\n ^{\\left(ij\\right)}}\\right).\n\\end{align}\nThe univariate rotated distribution is\n\\begin{align}\n \\tilde{P}_{G}^{\\left(ij\\right)}\\left(\\tilde{r}\\right)\n &=\\intop_{-\\infty}^{+\\infty}d\\bar{r}_{2}^{\\left(ij\\right)}\n P_{G}^{\\left(ij\\right)}\\left(\\bar{r}^{\\left(ij\\right)}|\n \\Lambda_{ep}^{\\left(ij\\right)}\\right)\\vert_{\\bar{r}_{1}^{\\left(ij\\right)}/\n \\sqrt{\\Lambda_{ep,1}^{\\left(ij\\right)}}=\\tilde{r}}\\,\n \\frac{d\\bar{r}_{1}^{\\left(ij\\right)}}{d\\tilde{r}}\\\\\n & =\\frac{1}{\\sqrt{2\\pi}}\\exp\\left(-\\frac{1}{2}\\tilde{r}^{2}\\right),\n\\end{align}\nwhich is the standard Gaussian distribution with mean $\\mu = 0$ and variance\n$\\sigma^2 = 1$.\n\nTo check the validity of our assumption, in Fig.\n\\ref{fig:gaussian_agg_returns_epoch} we plot the daily adjusted closing\naggregated returns distribution of 200 companies from the S\\&P 500 dataset,\nwith a $\\Delta t = 1d$ and an epoch window length $T = 25d$. For the\nunivariate distribution case we observe a good agreement with the one\ndimensional Gaussian distribution, as expected.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.6\\columnwidth]\n {figures/05_gaussian_agg_returns_short_epoch.png}\n \\caption{Aggregated distribution of returns ($\\tilde{r}$) for fixed\n covariance of 200 companies selected from the S\\&P 500\n dataset. $\\Delta t = 1d$ and epoch window length $T=25d$.}\n \\label{fig:gaussian_agg_returns_epoch}\n\\end{figure}\n\nTo model heavy tails, we assume that the $K$ dimensional vectors\n$r \\left( t \\right) = \\left( r_{1} \\left( t \\right), \\ldots, r_{K} \\left( t \\right) \\right)$\nfor a fixed return interval $\\Delta t$ is given by the multivariate algebraic\ndistribution\n\\begin{align}\n P_{A} \\left( r | \\Sigma_{ep} \\right)\n &= \\frac{\\alpha_{K1lm}}{\\left( 1 + \\frac{1}{m} r^{\\dagger}\n \\Sigma^{-1}_{ep} r \\right)^{l}}\\label{eq:algebraic_distribution} \\\\\n \\alpha_{K1lm} &= {\\sqrt{\\frac{2}{m}}}^{K}\n \\frac{\\Gamma \\left( l \\right)}{\\Gamma \\left( l - K/2 \\right)}\n \\frac{1}{\\sqrt{\\det 2 \\pi \\Sigma_{ep}}}.\n\\end{align}\nTo test the new assumption and compare the different time step data, we repeat\nthe same method to rotate and scale the returns to then get the aggregate\ndistribution of returns. Once more, we need to find the corresponding one\ndimensional functions related to the multivariate algebraic distribution in Eq.\n\\ref{eq:algebraic_distribution} to compare with the empirical single univariate\ndistributions.\n\nWe know that the algebraic model from Eq. \\ref{eq:algebraic_distribution}\nrelates to the Gaussian model from Eq. \\ref{eq:gaussian_distribution} via\n\\begin{equation}\n P_{A}\\left(r|\\Sigma_{ep}\\right)\n =\\intop_{0}^{\\infty}\\chi_{2\\left(l-K/2\\right)}^{2}\\left(z\\right)\\,\n P_{G}\\left(r|\\frac{m}{2}\\Sigma_{ep}\\right)dz.\n\\end{equation}\nWe immediately find\n\\begin{align}\n P_{A}^{\\left(ij\\right)}\\left(r^{\\left(ij\\right)}|\n \\Sigma_{ep}^{\\left(ij\\right)}\\right)\n & =\\int d\\left[r\\right]_{\\ne i,j}P_{A}\\left(r|\\Sigma_{ep}\\right)\\\\\n & =\\intop_{0}^{\\infty}dz\\,\\chi_{2\\left(l-K/2\\right)}^{^{2}}\\left(z\\right)\n \\frac{1}{\\sqrt{\\det2\\pi\\frac{m}{2}\\Sigma_{ep}^{\\left(ij\\right)}}}\n \\exp\\left(-\\frac{z}{2m}r^{\\left(ij\\right)\\dagger}\n \\Sigma_{ep}^{\\left(ij\\right)^{-1}}r^{\\left(ij\\right)}\\right)\\\\\n & =\\frac{1}{\\sqrt{\\det2\\pi\\Sigma_{ep}^{\\left(ij\\right)}}}\n \\intop_{0}^{\\infty}dz\\,\\chi_{2\\left(l-K/2\\right)}^{^{2}}\\left(z\\right)\n \\frac{z}{m}\\exp\\left(-\\frac{z}{2m}r^{\\left(ij\\right)\\dagger}\n \\Sigma_{ep}^{\\left(ij\\right)^{-1}}r^{\\left(ij\\right)}\\right).\n\\end{align}\nUsing\n\\begin{equation}\n \\chi_{q+2}^{2}\\left(z\\right)=\\frac{1}{q}\\chi_{q}^{2}\\left(z\\right)z\n\\end{equation}\nwe have\n\\begin{align}\n P_{A}^{\\left(ij\\right)}\\left(r^{\\left(ij\\right)}|\n \\Sigma_{ep}^{\\left(ij\\right)}\\right)\n &=\\frac{2\\left(l-K/2\\right)}{m\\sqrt{\\det2\\pi\\Sigma_{ep}^{\\left(ij\\right)}}}\n \\intop_{0}^{\\infty}dz\\,\\chi_{2\\left(l+1-K/2\\right)}^{2}\\left(z\\right)\n \\exp\\left(-\\frac{z}{2m}r^{\\left(ij\\right)\\dagger}\n \\Sigma_{ep}^{\\left(ij\\right)^{-1}}r^{\\left(ij\\right)}\\right)\\\\\n & =\\frac{2\\left(l-K/2\\right)}{m\\sqrt{\\det2\\pi\\Sigma_{ep}\n ^{\\left(ij\\right)}}}\\frac{1}{2^{l+1-K/2}\\Gamma\\left(l+1-K/2\\right)}\n \\intop_{0}^{\\infty}dz\\,z^{l-K/2}\n \\exp\\left[-\\frac{z}{2}\\left(1+\\frac{1}{m}r^{\\left(ij\\right)\\dagger}\n \\Sigma_{ep}^{\\left(ij\\right)^{-1}}r^{\\left(ij\\right)}\\right)\\right]\\\\\n &=\\frac{2l-K}{m\\sqrt{\\det2\\pi\\Sigma_{ep}^{\\left(ij\\right)}}}\n \\frac{1}{\\left(1+\\frac{1}{m}r^{\\left(ij\\right)\\dagger}\n \\Sigma_{ep}^{\\left(ij\\right)^{-1}}r^{\\left(ij\\right)}\\right)^{l+1-K/2}}\n \\label{eq:using_chi}.\n\\end{align}\nRotating into the eigenbasis, we can integrate out one of the variables\n\\begin{equation}\n \\tilde{P}_{A}^{\\left(ij\\right)}\\left(\\bar{r}_{1}^{\\left(ij\\right)}|\n \\Lambda_{ep,1}^{\\left(ij\\right)}\\right)\n =\\intop_{-\\infty}^{+\\infty}d\\bar{r}_{2}^{\\left(ij\\right)}\n P_{A}^{\\left(ij\\right)}\\left(\\bar{r}^{\\left(ij\\right)}|\n \\Lambda_{ep}^{\\left(ij\\right)}\\right).\n\\end{equation}\nUsing Eq. \\ref{eq:using_chi}, we have\n\\begin{align}\n \\tilde{P}_{A}^{\\left(ij\\right)}\\left(\\bar{r}_{1}^{\\left(ij\\right)}|\n \\Lambda_{ep,1}^{\\left(ij\\right)}\\right)\n &=\\frac{2\\left(l-K/2\\right)}{\\sqrt{m}\\sqrt{2\\pi\\Lambda_{ep,1}\n ^{\\left(ij\\right)}}}\\frac{1}{2^{l+1-K/2}\\Gamma\\left(l+1-K/2\\right)}\n \\intop_{0}^{\\infty}dz\\,z^{l-K/2-1/2}\\exp\\left[-\\frac{z}{2}\n \\left(1+\\frac{1}{m}\\frac{\\bar{r}_{1}^{\\left(ij\\right)^{2}}}\n {\\Lambda_{ep,1}^{\\left(ij\\right)}}\\right)\\right]\\\\\n &=\\frac{2\\left(l-K/2\\right)}{\\sqrt{m}\\sqrt{2\\pi\\Lambda_{ep,1}^\n {\\left(ij\\right)}}}\\frac{2^{l+1/2-K/2}\n \\Gamma\\left(l+1/2-K/2\\right)}{2^{l+1-K/2}\n \\Gamma\\left(l+1-K/2\\right)}\\frac{1}{\\left(1+\\frac{1}{m}\n \\frac{\\bar{r}_{1}^{\\left(ij\\right)^{2}}}\n {\\Lambda_{ep,1}^{\\left(ij\\right)}}\\right)^{l-\\left(K-1\\right)/2}}\\\\\n &=\\sqrt{\\frac{2}{m}}\\frac{\\Gamma\\left(l-\\left(K-1\\right)/2\\right)}\n {\\Gamma\\left(l-K/2\\right)}\\frac{1}{\\sqrt{2\\pi\\Lambda_{ep,1}\n ^{\\left(ij\\right)}}}\\frac{1}{\\left(1+\\frac{1}{m}\n \\frac{\\bar{r}_{1}^{\\left(ij\\right)^{2}}}\n {\\Lambda_{ep,1}^{\\left(ij\\right)}}\\right)^{l-\\left(K-1\\right)/2}}.\n\\end{align}\nGoing over to the rescaled variable\n\\begin{equation}\n \\tilde{r} = \\frac{\\bar{r}^{\\left( ij \\right)}}{\\sqrt{\\Lambda_{ep,1}\n ^{\\left( ij \\right)}}}\n\\end{equation}\nwe arrive at\n\\begin{equation}\n \\tilde{P}_{A}^{\\left(ij\\right)}\\left(\\tilde{r}\\right)\n =\\frac{1}{\\sqrt{2\\pi}}\\sqrt{\\frac{2}{m}}\n \\frac{\\Gamma\\left(l-\\left(K-1\\right)/2\\right)}{\\Gamma\\left(l-K/2\\right)}\n \\frac{1}{\\left(1+\\frac{1}{m}\\tilde{r}^{2}\\right)^{l-\\left(K-1\\right)/2}}\n\\end{equation}\nwhich is the algebraic distribution to be fitted to the aggregated returns.\nIn this case we have three parameters: $K$, $l$ and $m$. $K$ is the number of\ncompanies used and is known from the data. $l$ and $m$ are the shape parameters\nof the distribution and have to be determined by fitting. In the model, the\nexpectation value $\\langle r r^{\\dagger} \\rangle$ serves as an estimator for\nthe sample covariances. We find\n\\begin{equation}\n \\langle r r^{\\dagger} \\rangle_{Y} = \\beta_{Y} \\Sigma_{ep}\n\\end{equation}\nwith\n\\begin{equation}\n \\beta_{Y}=\\begin{cases}\n 1, & \\text{if }Y=G\\\\\n \\frac{m}{2l-K-2}, & \\text{if }Y=A\n\\end{cases}.\n\\end{equation}\nDue to its very definition, we have $\\beta_{G} = 1$ for the multivariate\nGaussian, but a different value $\\beta_{A}$ for the algebraic distribution. The\nrelation in the latter case then suggest the useful fixing\n\\begin{equation}\\label{eq:m_relation}\n m = 2l - K - 2\n\\end{equation}\nalso for finite values of $l$ and $m$. Only with this choice, $\\Sigma_{ep}$ can\nbe estimated by the sample covariance matrix, otherwise only up to some factor.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=\\columnwidth]\n {figures/05_algebraic_agg_returns_short_epoch.png}\n \\caption{Semilog plot (left) and loglog plot (right) of the aggregated\n distribution of returns ($\\tilde{r}$) for fixed covariance of 200\n companies selected from the S\\&P 500 dataset. $\\Delta t = 1d$ and\n epoch window length $T=55d$. The empirical data is compared with\n the best fit of one dimensional algebraic distribution with shape\n parameter $l=104$.}\n \\label{fig:algebraic_agg_returns_epoch}\n\\end{figure}\n\nHaving these details in mind and to check the validity of our assumption, in\nFig. \\ref{fig:algebraic_agg_returns_epoch} we plot the daily adjusted closing\naggregated returns distribution of 200 companies from the S\\&P 500 dataset,\nwith a $\\Delta t = 1d$ and an epoch window length $T = 55d$. Using Eq.\n\\ref{eq:m_relation} to compute $m$, we found that $l = 104$ fits well with the\naggregated returns. We plot the one dimensional Gaussian distribution as a\nreference.\n\nWe know that the aggregated returns have internal Gaussian and algebraic\nstructures depending on the epochs window length. The larger the value of $T$,\nthe heavier the tails and the smaller the stationary assumption within the\nepochs. However, during the epochs analysis we found an unusual behavior when\nwe used small epochs window lengths.\n\nUsing the methodology described in the work of Schmitt et al.\n\\cite{non_stationarity_fin_guhr}, we wanted to find which was the best value of\nthe epochs window length to fit the Gaussian and algebraic case. We used epochs\nwindow lengths $T = 10, 25, 40, 55$ days and $\\Delta t = 1d$. In Fig.\n\\ref{fig:window_comparison} can be seen the difference between epochs\nwindow lengths of the aggregated distribution of returns for fixed covariance\nof different amounts of random companies selected from the S\\&P 500 dataset.\nThe first interesting characteristic is that with few companies there is enough\ninformation to obtain the general behavior. This, because we are selecting\npairs of stocks from the dataset, such that the order of the selected stocks\ndoes not matter and without repetition, i.e. a combination. Thus, even with\na small number of companies, the results are statistically relevant. Then, We\nchoose random numbers of companies within our dataset to show this\ncharacteristic. We can see how for $T=25d$ (top right) there is a good agreement\nwith the Gaussian distribution, and how for values larger than $T=25d$, as\n$T=40d$ (bottom left) and $T=55d$ (bottom right), the aggregated returns have\nfat tails. However, for $T=10d$ (top left), the aggregated returns have a\nplatykurtic behavior. This particular behavior is unusual and makes us wonder\nif somehow there is an artifact in the methodology we are using, that is\naffecting the other epochs window lengths as well.\n\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.8\\columnwidth]\n {figures/05_window_comparison.png}\n \\caption{Aggregated distribution of returns ($\\tilde{r}$) for fixed\n covariance of different number of companies selected from the S\\&P\n 500 dataset. $\\Delta t = 1d$ and different epochs window lengths\n $T=10d$ (top left), $T=25d$ (top right), $T=40d$ (bottom left) and\n $T=55d$ (bottom right).}\n \\label{fig:window_comparison}\n\\end{figure}\n\nWe consider two different causes for this situation in the small epochs window\nlengths. First, it could be an ergodicity defect because only the length of the\ntime window of the epochs shows an effect. Second, the platykurtic behavior\nlike in local normalization plots \\cite{local_normalization} could be a\ndetrending method. It is possible that we get this behavior from normalizing\neach epoch to mean $\\mu = 0$ and $\\sigma^{2}=1$.\n\nTo check what is the real cause of this behavior we will use simulations.\nWith these simulations we know which parameters we are using, and we have a\ncontrolled environment. Thus, we can evaluate where the problem is with the\nmethod.", "meta": {"hexsha": "8b06d6ad2570920bb807a1294814d4ff95c9b8da", "size": 23089, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "paper/exact_distributions_financial_paper/sections/05_epochs.tex", "max_stars_repo_name": "juanhenao21/exact_distributions_financial", "max_stars_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-20T18:24:22.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-15T07:25:50.000Z", "max_issues_repo_path": "paper/exact_distributions_financial_paper/sections/05_epochs.tex", "max_issues_repo_name": "juanhenao21/exact_distributions_financial", "max_issues_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "paper/exact_distributions_financial_paper/sections/05_epochs.tex", "max_forks_repo_name": "juanhenao21/exact_distributions_financial", "max_forks_repo_head_hexsha": "02eb058e5f963fbccb9029aae3fb6e15def7a93a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.3354700855, "max_line_length": 121, "alphanum_fraction": 0.6836155745, "num_tokens": 7478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637505099168, "lm_q2_score": 0.7154239897159438, "lm_q1q2_score": 0.6150240702039764}} {"text": "% ------------------------------------------------------------ %\n%\n% CUHK Mathematics\n% MATH3060: Mathematical Analysis III\n%\n% ------------------------------------------------------------ %\n\n\\documentclass[a4paper,12pt]{article}\n\\usepackage{standalone}\n\\input{sty/setup.sty}\n\n\\begin{document}\n\\title{MATH3060: Mathematical Analysis III}\n\\input{sty/cover.sty}\n\n\\remark{}\n\n\\input{sty/header.sty}\n\n\\section{Fourier Series}\n\\subsection{Introduction to Fourier Series}\n\\subsubsection{Trigonometric Series}\n\\begin{dft}\n A \\textbf{trigonometric series} on $[-\\pi,\\pi]$ is a series of functions in the form\n\n $$\\sum_{n=0}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\n where $a_{n},b_{n}\\in\\R$. Furthermore, if $a_{n}=0$ for all $n$, the series is called a \\textbf{sine series}. Similarly, if $b_{n}=0$ for all $n$, the series is called a \\textbf{cosine series}.\n\\end{dft}\\n\n\nNote that it is possible to pull out the zeroth term of the sum, making the series in the form\n\n$$a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\nhence it can be assumed that $b_{0}=0$.\\n\n\n\\begin{pst}\n Let $(a_{n}),(b_{n})$ be infinite series. If\n \n $$\\abs{a_{n}},\\abs{b_{n}}\\leq\\frac{C}{n^{s}}$$\\s\n \n for some $C>0$ and $s>1$, then their corresponding series $\\sum_{n=0}^{\\infty}\\abs{a_{n}}$ and $\\sum_{n=0}^{\\infty}\\abs{b_{n}}$ are convergent.\n\\end{pst}\n\n\\begin{pst}\n If $\\sum_{n=0}^{\\infty}\\abs{a_{n}}$ and $\\sum_{n=0}^{\\infty}\\abs{b_{n}}$ are convergent, then by Weierstrass M-test,\n \n $$\\sum_{n=0}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n \n is uniformly and absolutely convergent.\n\\end{pst}\n\n\\begin{pst}\n Let $\\phi(x)=a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$ be a continuous function on $[-\\pi,\\pi]$. If $\\sum\\abs{a_{n}},\\sum\\abs{b_{n}}<\\infty$, then $\\phi(x)$ is $2\\pi$-periodic.\n\\end{pst}\n\n\\propdisp\n\n\\subsubsection{Fourier Series}\n\\begin{dft}\n Let $f$ be a $2\\pi$-periodic function on $\\R$ which is Riemann integrable on $[-\\pi,\\pi]$, then the \\textbf{Fourier series} (or \\textbf{Fourier expansion}) of $f$ is the trigonometric series\n\n $$a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\n with Fourier coefficients of $f$\n\n $$a_{0}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(y)\\diff y$$\\s\n\n $$a_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}f(y)\\cos(ny)\\diff y$$\\s\n\n $$b_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}f(y)\\sin(ny)\\diff y$$\\s\n\\end{dft}\\n\n\nNote that $a_{0}$ is actually the average of $f$ over $[-\\pi,\\pi]$. Fourier series depends on the global information of $f$ on $[-\\pi,\\pi]$ instead of a point in $f$. Fourier series also depends only on $f\\!\\mid\\!_{(-\\pi,\\pi)}$, which means the end points of the closed interval are independent.\\n\n\n\\begin{pst}\n Let $f_{1},f_{2}$ are Fourier series where $f_{1}\\equiv f_{2}$ almost everywhere on $[-\\pi,\\pi]$, then $f_{1}$ and $f_{2}$ are the same Fourier series.\n\\end{pst}\n\n\\subsubsection{Motivation of Fourier Series}\nRecall the form of a Fourier series as\n\n$$f(x)=a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\nfor all $x\\in\\R$. If $f$ is uniformly convergent,\n\n$$\\begin{aligned}[t]\n &\\int_{-\\pi}^{\\pi}f(x)\\cos(mx)\\diff x\\\\\n =\\;&a_{0}\\int_{-\\pi}^{\\pi}\\cos(mx)\\diff x+\\sum_{n=1}^{\\infty}\\brr{a_{n}\\int_{-\\pi}^{\\pi}\\cos(nx)\\cos(mx)\\diff x+b_{n}\\int_{-\\pi}^{\\pi}\\sin(nx)\\cos(mx)\\diff x}\n\\end{aligned}$$\\s\n\nNote that\n\n$$\\int_{-\\pi}^{\\pi}\\cos(mx)\\diff x=\\begin{cases}\n 2\\pi\\erm{if }m=0\\\\\n 0\\erm{if }m\\neq 0\n\\end{cases}$$\\s\n\n$$\\int_{-\\pi}^{\\pi}\\cos(nx)\\cos(mx)\\diff x=\\begin{cases}\n \\pi\\erm{if }m=n\\\\\n 0\\erm{if }m\\neq n\n\\end{cases}$$\\s\n\n$$\\int_{-\\pi}^{\\pi}\\sin(nx)\\cos(mx)\\diff x=0\\;\\forall m,n\\geq 1$$\\s\n\nwhich will deduce\n\n$$a_{0}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)\\diff x,\\;a_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}f(x)\\cos(mx)\\diff x$$\\s\n\nUsing similar method but instead of $\\cos(mx)$, $\\sin(mx)$ will deduce\n\n$$b_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}f(x)\\sin(mx)\\diff x$$\n\n\\subsection{Complex Fourier Series}\n\\subsubsection{Definition of Complex Fourier Series}\n\\begin{dft}\n Let $f$ be a $2\\pi$-periodic function on $\\C$ which is Riemann integrable on $[-\\pi,\\pi]$, then its \\textbf{complex Fourier series} is a Fourier series of the form\n\n $$\\sum_{-\\infty}^{\\infty}c_{n}e^{inx}$$\n\n where $\\brc{c_{n}}_{-\\infty}^{\\infty}$ is a \\textbf{bisequence} of complex numbers defined by\n\n $$c_{n}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)e^{-inx}\\diff x$$\\s\n \n for all integers $n$. Moreover, $\\sum_{-\\infty}^{\\infty}c_{n}e^{inx}$ is said to be convergent at $x$ if\n\n $$\\lim_{N\\to+\\infty}\\sum_{-N}^{N}c_{n}e^{inx}$$\\s\n\n exists.\n\\end{dft}\\n\n\nNote that for a complex-valued function $f=u+iv$,\n\n$$\\int_{a}^{b}f=\\int_{a}^{b}u+i\\int_{a}^{b}v$$\\s\n\nIn other words, $f$ is said to be integrable if both $u$ and $v$ are integrable.\n\n\\subsubsection{Motivation of Complex Fourier Series}\nRecall the form of a complex Fourier series as\n\n$$f(x)=\\sum_{-\\infty}^{\\infty}c_{n}e^{inx}$$\\s\n\nfor all $x\\in\\C$. If $f$ converges nicely,\n\n$$\\int_{-\\pi}^{\\pi}e^{-imx}\\diff x=\\sum_{-\\infty}^{\\infty}c_{n}\\int_{-\\pi}^{\\pi}e^{i(n-m)x}\\diff x$$\\s\n\nNote that\n\n$$\\int_{-\\pi}^{\\pi}e^{i(n-m)x}\\diff x=\\begin{cases}\n 2\\pi\\erm{if }n=m\\\\\n 0\\erm{if }n\\neq m\n\\end{cases}$$\\s\n\nwhich will deduce\n\n$$c_{n}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)e^{-inx}\\diff x$$\n\n\\subsubsection{Relations between Real and Complex Fourier Series}\nIn this section the relationship between (real) Fourier series and complex Fourier series for a real-valued function $f$ is discussed. Note that\n\n$$\\begin{aligned}[t]\n c_{n}&=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)e^{-inx}\\diff x\\\\\n &=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)(\\cos(nx)-i\\sin(nx))\\diff x\\\\\n &=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)\\cos(nx)\\diff x-\\frac{i}{2\\pi}\\int_{-\\pi}^{\\pi}f(x)\\sin(nx)\\diff x\n\\end{aligned}$$\\s\n\nwhich will deduce\n\n$$c_{n}=\\begin{cases}\n (a_{n}-ib_{n})/2&\\erm{if }n\\geq 1\\\\\n 0&\\erm{if }n=0\\\\\n (a_{-n}+ib_{-n})/2&\\erm{if }n\\leq -1\n\\end{cases}$$\\s\n\n\\begin{pst}\n Let $f$ be a real-valued function, then its complex Fourier coefficient $c_{-n}=\\overline{c_{n}}$.\n\\end{pst}\\n\n\n\\begin{pst}\n Let $f$ be a $2\\pi$-periodic real function which is differentiable on $[-\\pi,\\pi]$ with $f'$ integrable on $[-\\pi,\\pi]$. Denote the Fourier coefficients of $f$ and $f'$ by $\\brc{a_{n}(f), b_{n}(f); c_{n}(f)}$ and $\\brc{a_{n}(f'), b_{n}(f'); c_{n}(f')}$ respectively, then\n \n $$\\begin{cases}\n a_{n}(f')=nb_{n}(f)\\\\\n b_{n}(f')=-na_{n}(f)\n \\end{cases}\\text{ and }c_{n}(f')=inc_{n}(f)$$\n\\end{pst}\\n\n\nNote that one of the advantages of using complex Fourier series is to compute derivatives with more convenience.\n\n\\subsection{Fourier Series and Extensions}\n\\subsubsection{Extensions of Periodic Functions}\nFor any Riemann integrable function $f$ on $[-\\pi,\\pi]$, one can define the Fourier coefficients to form a Fourier series. On the other hand, we can restrict $f$ to $(-\\pi,\\pi]$ and extend periodically to a $2\\pi$-periodic function $\\tilde{f}$ on $\\R$. The function $f$ and its extension $\\tilde{f}$ have the same Fourier series.\\n\n\nHere the $\\sim$ symbol in\n\n$$f(x)\\sim a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\nrepresents $f(x)$ has the Fourier series on the right hand side. The equal sign $=$ is not used since the series may not converge.\\n\n\n\\begin{exm}\n Let $f(x)=x$ be a function in $[-\\pi,\\pi]$. Find the Fourier series of $f$.\\n\n\n \\ans The Fourier coefficients are\n\n $$a_{0}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}x\\diff x=0$$\\s\n\n $$a_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}x\\cos(nx)\\diff x=0$$\\s\n\n $$b_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}x\\sin(nx)\\diff x=(-1)^{n+1}\\frac{2}{n}$$\\s\n\n Therefore\n\n $$f(x)\\sim\\sum_{n=1}^{\\infty}(-1)^{n+1}\\frac{2}{n}\\sin(nx)$$\\s\n\n which is a sine series.\n\\end{exm}\\n\n\nWith the example above, the following proposition can be introduced by observation:\\n\n\n\\begin{pst}\n Let $f$ be a real function, then its Fourier series is a sine series if $f$ is odd, and it is a cosine series if $f$ is even.\n\\end{pst}\\n\n\nNote that the Fourier series may not be the same as the original function, especially when there are discontinuous points like $\\pm\\pi$. Also, the convergence of Fourier series is not clear since some terms like $\\sum(1/n)$ does not converge. \n\n\\subsubsection{Big-O and Little-O Notations}\n\\begin{dft}\n Let $\\brc{x_{n}}$ be a sequence, then the \\textbf{big-O notation}, denoted by $O$, paired with $x_{n}$ is defined as\n\n $$x_{n}=O(n^{s})\\Leftrightarrow\\abs{x_{n}}\\leq Cn^{s}$$\\s\n\n for some constant $C>0$, as $n\\to\\infty$. Similarly, the \\textbf{little-O notation}, denoted by $o$, paired with $x_{n}$ is defined as\n\n $$x_{n}=o(n^{s})\\Leftrightarrow\\frac{\\abs{x_{n}}}{n^{s}}\\to 0$$\\s\n\n as $n\\to\\infty$.\n\\end{dft}\\n\n\n\\begin{exm}\n Find the correlation of\n \n $$x_{n}=\\frac{2(-1)^{n+1}}{n}\\sin(nx)$$\\s\n\n using big-O notation.\\n\n\n \\ans Since $\\abs{x_{n}}\\leq 2/n$, $x_{n}=O(1/n)$.\n\\end{exm}\\n\n\n\\begin{exm}\n Find the correlation of\n\n $$x_{n}=\\log(n)$$\\s\n\n using little-O notation.\\n\n\n \\ans Since $\\abs{\\log(n)}/n\\to 0$ as $n\\to\\infty$, $x_{n}=o(n)$.\n\\end{exm}\n\n\\subsubsection{Fourier Series of Unusual Periodic Functions}\nLet $f$ be a $2T$-periodic function. Note that $g(x)=f(Tx/\\pi)$ is a $2\\pi$-periodic function, then\n\n$$g(x)\\sim a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$$\\s\n\nwith\n\n$$a_{0}=\\frac{1}{2\\pi}\\int_{-\\pi}^{\\pi}g(x)\\diff x$$\\s\n\n$$a_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}g(x)\\cos(nx)\\diff x$$\\s\n\n$$b_{n}=\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}g(x)\\sin(nx)\\diff x$$\\s\n\nalong with the substitution $y=Tx/\\pi$ implies\n\n$$f(y)\\sim a_{0}+\\sum_{n=1}^{\\infty}\\brr{a_{n}\\cos\\brr{\\frac{n\\pi}{T}y}+b_{n}\\sin\\brr{\\frac{n\\pi}{T}y}}$$\\s\n\nwith\n\n$$a_{0}=\\frac{1}{2T}\\int_{-T}^{T}f(y)\\diff y$$\\s\n\n$$a_{n}=\\frac{1}{T}\\int_{-T}^{T}f(y)\\cos\\brr{\\frac{n\\pi}{T}y}\\diff y$$\\s\n\n$$b_{n}=\\frac{1}{T}\\int_{-T}^{T}f(y)\\sin\\brr{\\frac{n\\pi}{T}y}\\diff y$$\\s\n\nSuch Fourier series is called Fourier series of $2T$-periodic function $f$.\n\n\\subsection{Convergence of Fourier Series}\n\\subsubsection{Riemann-Lebesgue Lemma}\nRecall the definition of a step function on $[-\\pi,\\pi]$ as a function of the form\n\n$$s(x)=\\sum_{j=0}^{N-1}s_{j}\\chi_{I_{j}}$$\\s\n\nwhere $-\\pi=a_{0}0$ depending on $s$ such that\n \n $$\\abs{a_{n}(s)},\\abs{b_{n}(s)}\\leq\\frac{C}{n}$$\\s\n\n for all $n\\geq 1$. $a_{n}(s)$ and $b_{n}(s)$ are Fourier coefficients of $s$.\\n\n\n \\prf Let\n \n $$s(x)=\\sum_{j=0}^{N-1}s_{j}\\chi_{I_{j}}$$\\s\n \n be the step function, then for $n\\geq 1$,\n\n $$\\begin{aligned}[t]\n \\pi a_{n}(s)&=\\int_{-\\pi}^{\\pi}s(x)\\cos(nx)\\diff x\\\\\n &=\\sum_{j=0}^{N-1}s_{j}\\int_{a_{j}}^{a_{j+1}}\\cos(nx)\\diff x\\\\\n &=\\sum_{j=0}^{N-1}s_{j}\\frac{\\sin(na_{j+1})-\\sin(na_{j})}{n}\n \\end{aligned}$$\\s\n\n which implies $\\abs{a_{n}(s)}\\leq\\frac{C}{n}$. The proof for $\\abs{b_{n}(s)}$ is similar.\n\\end{pst}\\n\n\n\\begin{pst}\n Let $f$ be integrable on $[-\\pi,\\pi]$, then for all $\\epsilon>0$, there exists a step function $s$ such that $s\\leq f$ on $[-\\pi,\\pi]$ and\n\n $$\\int_{-\\pi}^{\\pi}(f-s)<\\epsilon$$\\s\n\n \\prf Since $f$ is Riemann integrable, the function can be approximated with Darboux lower sum. For all $\\epsilon>0$, there exists a partition $-\\pi=a_{0}0$, there exists a step function $s$ such that $s\\leq f$ and\n\n $$\\int_{-\\pi}^{\\pi}(f-s)<\\frac{\\epsilon}{2}$$\\s\n\n On the other hand, by \\rpst[\\sctd{2}], there exists $n_{0}>0$ such that\n\n $$\\abs{a_{n}(s)}<\\frac{\\epsilon}{2}$$\\s\n\n for all $n\\geq n_{0}$. For instance, $n_{0}=[2C/\\epsilon]+1$ with the constant $C$ in \\rpst[\\sctd{2}]. Note that\n\n $$\\begin{aligned}[t]\n \\abs{a_{n}(f)-a_{n}(s)}&=\\frac{1}{\\pi}\\abs{\\int_{-\\pi}^{\\pi}(f-s)(x)\\cos(nx)\\diff x}\\\\\n &\\leq\\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}(f-s)\\erm{as }f\\geq s\\\\\n &\\leq\\frac{\\epsilon}{2\\pi}\n \\end{aligned}$$\\s\n\n Hence,\n\n $$\\begin{aligned}[t]\n \\abs{a_{n}(f)}&\\leq\\abs{a_{n}(s)}-\\abs{a_{n}(f)-a_{n}(s)}\\\\\n &<\\frac{\\epsilon}{2}+\\frac{\\epsilon}{2\\pi}<\\epsilon\n \\end{aligned}$$\\s\n\n for all $n\\geq n_{0}$, which means $a_{n}(f)\\to 0$ as $n\\to\\infty$. The proof for $b_{n}(f)$ is similar to that above.\n\\end{thm}\n\n\\subsubsection{Lipschitz Continuity at Points}\n\\begin{dft}\n Let $f\\sim a_{0}+\\sum_{n=1}^{\\infty}(a_{n}\\cos(nx)+b_{n}\\sin(nx))$ be a function which has a Fourier series, then the \\textbf{$n$-th partial sum} of Fourier series of $f$, denoted by $(S_{n}f)(x)$, is given by\n\n $$(S_{n}f)(x)=a_{0}+\\sum_{k=1}^{n}(a_{k}\\cos(kx)+b_{k}\\sin(kx))$$\n\\end{dft}\\n\n\n\\begin{dft}\n Let $f$ be a function on $[a,b]$, then $f$ is called \\textbf{Lipschitz continuous} at a point $x_{0}\\in[a,b]$ if there exists $L>0$ and $\\delta>0$ such that\n\n $$\\abs{f(x)-f(x_{0})}\\leq L\\abs{x-x_{0}}$$\\s\n\n for all $\\abs{x-x_{0}}<\\delta$.\n\\end{dft}\\n\n\nNote that both $L$ and $\\delta$ may depend on the point $x_{0}$. Below is a proposition on extending Lipschitz continuity from a point to an interval:\\n\n\n\\begin{pst}\n If $f$ is Lipschitz continuous at $x_{0}\\in[a,b]$ and $f$ is bounded on $[a,b]$, then there exists $L'>0$ which may depends on $x_{0}$ such that\n\n$$\\abs{f(x)-f(x_{0})}\\leq L'\\abs{x-x_{0}}$$\\s\n\nfor all $x\\in[a,b]$.\\n\n\n\\prf By \\rdft[\\sctd{1}], there exists $L,\\delta>0$ such that\n\n$$\\abs{f(x)-f(x_{0})}\\leq L\\abs{x-x_{0}}$$\\s\n\nfor all $\\abs{x-x_{0}}<\\delta$. If $\\abs{x-x_{0}}\\geq\\delta$, then\n\n$$\\begin{aligned}[t]\n \\abs{f(x)-f(x_{0})}&\\leq\\abs{f(x)}+\\abs{f(x_{0})}\\\\\n &\\leq 2M\\leq \\frac{2M\\abs{x-x_{0}}}{\\delta}\n\\end{aligned}$$\\s\n\nwhere $M=\\sup_{[a,b]}\\abs{f}\\geq 0$. Pick $L'=\\max\\brc{L,2M/\\delta}>0$, then\n\n$$\\abs{f(x)-f(x_{0})}\\leq L'\\abs{x-x_{0}}$$\\s\n\nfor all $x\\in[a,b]$.\n\\end{pst}\n\n\\subsubsection{Dirichlet Kernels}\n\\begin{dft}\n The \\textbf{Dirichlet kernel}, denoted by $D_{n}(z)$, is defined by\n\n $$D_{n}(z)=\\begin{cases}\n (\\sin((n+1/2)z))/(2\\pi\\sin(z/2))\\erm{if }z\\neq 0\\\\\n (2n+1)/(2\\pi)\\erm{if }z=0\n \\end{cases}$$\n\\end{dft}\\n\n\n\\begin{pst}\n Below are the properties of Dirichlet kernels:\n\n \\begin{alist}\n \\item Integral of a Dirichlet kernel $\\int_{-\\pi}^{\\pi}D_{n}(z)\\diff z=1$.\n \\item $D_{n}(z)$ is even, continuous, $2\\pi$-periodic on $[-\\pi,\\pi]$ and\n \n $$D_{n}\\brr{\\frac{2k\\pi}{2n+1}}=0$$\\s\n\n for all $k=-n,-n+1,\\cdots,n$.\n \\item The maximum\n \n $$\\max_{[-\\pi,\\pi]}D_{n}(z)=D_{n}(0)=\\frac{2n+1}{2\\pi}$$\n \\item For all $0<\\delta<\\pi/2$,\n \n $$\\int_{0}^{\\delta}\\abs{D_{n}(z)}\\diff z\\to+\\infty$$\\s\n\n as $n\\to+\\infty$\n \\end{alist}\n\n \\prf\\prt[a]{zr} This can be achieved by integrating\n\n $$\\int_{-\\pi}^{\\pi}\\brr{\\frac{1}{2}+\\sum_{k=1}^{n}\\cos(kz)}\\diff z$$\\s\n \n \\prtc[d]{zr} Let $0<\\delta<\\pi/2$, then for all $n\\in\\N$, there exists $N\\in\\N$ such that\n\n $$N<\\frac{n+1/2}{\\pi}\\delta\\leq N+1$$\\s\n\n where $N\\to\\infty$ as $n\\to\\infty$. Note that\n \n $$\\begin{aligned}[t]\n \\int_{0}^{\\delta}\\abs{D_{n}(z)}\\diff z&=\\int_{0}^{\\delta}\\frac{\\abs{\\sin(n+1/2)z}}{2\\pi\\abs{\\sin(z/2)}}\\diff z\\\\\n &=\\int_{0}^{(n+1/2)\\delta}\\frac{\\abs{\\sin(t)}}{2\\pi\\abs{\\sin(t/(2n+1))}}\\brr{\\frac{2\\diff t}{2n+1}}\\erm{where }t=\\brr{n+\\frac{1}{2}}z\\\\\n &=\\frac{1}{\\pi}\\int_{0}^{(n+1/2)\\delta}\\frac{\\sin(t)}{t}\\frac{t/(2n+1)}{\\abs{\\sin(t/(2n+1))}}\\diff t\\\\\n &\\geq\\frac{1}{\\pi}\\int_{0}^{(n+1/2)\\delta}\\frac{\\sin(t)}{t}\\diff t\\erm{since }\\frac{\\sin(x)}{x}<1\\text{ for }00$ and $\\delta>0$ such that\n \n $$\\begin{cases}\n \\abs{f(x)-f(x^{+}_{0})}\\leq L(x-x_{0})\\erm{where }00$ such that\n\n $$\\abs{f(x)-f(y)}\\leq L\\abs{x-y}$$\\s\n\n for all $x,y\\in[a,b]$.\n\\end{dft}\\n\n\nNote that Lipschitz condition is uniform since $L$ is independent of any choice of $x,y$. Also, if $f$ satisfies a Lipschitz condition, $f$ is Lipschitz continuous at every point on $[a,b]$.\\n\n\n\\begin{pst}\n Let $f$ be a $2\\pi$-periodic function satisfying a Lipschitz condition, then its Fourier series converge uniformly to $f$ itself.\n\\end{pst}\n\n\\subsection{Weierstrass Approximation Theorem}\n\\subsubsection{Piecewise Linear Functions}\nRecall that a continuous function is piecewise linear if there exists a partition such that the function is linear within each subinterval.\\n\n\n\\begin{pst}\n Let $f$ be a continuous function on $[a,b]$, then for all $\\epsilon>0$, there exists a continuous and piecewise linear function $g$ with $g(a)=f(a)$, $g(b)=f(b)$ such that\n\n $$\\nrm{f-g}_{\\infty}<\\epsilon$$\\s\n\n where\n\n $$\\nrm{f-g}_{\\infty}=\\sup_{[a,b]}\\abs{f(x)-g(x)}$$\n\\end{pst}\n\n\\subsubsection{Trigonometric Polynomials}\n\\begin{dft}\n A \\textbf{trigonometric polynomial} is of the form $P(\\cos(x),\\sin(x))$ where $P(x,y)$ is a polynomial of $2$ variables.\n\\end{dft}\\n\n\nNote that a trigonometric polynomial is a finite Fourier series, and vice versa.\\n\n\n\\begin{pst}\n Let $f$ be a continuous function on $[0,\\pi]$, then for all $\\epsilon>0$, there exists a trigonometric polynomial $h$ such that $\\nrm{f-h}_{\\infty}<\\epsilon$.\n\\end{pst}\n\n\\subsubsection{General Theorem}\nBelow is the \\textbf{Weierstrass Approximation Theorem}:\\n\n\n\\begin{thm}\n Let $f\\in C[a,b]$, then for all $\\epsilon>0$, there exists a polynomial $q$ such that $\\nrm{f-q}_{\\infty}<\\epsilon$.\n\\end{thm}\n\n\\subsection{Mean Convergence of Fourier Series}\n\\subsubsection{Bracket Products}\n\\begin{dft}\n Let $f,g$ be Riemann integrable functions on $[-\\pi,\\pi]$, then the \\textbf{bracket product} (or \\textbf{$L^{2}$-product}, \\textbf{$L^{2}$ inner product}) of $f$ and $g$ is given by\n\n $$\\bra{f,g}_{2}=\\int_{-\\pi}^{\\pi}f(x)g(x)\\diff x$$\n\\end{dft}\\n\n\nNote that for complex functions, the bracket product is defined by\n\n$$\\bra{f,g}_{2}=\\int_{-\\pi}^{\\pi}f\\overline{g}$$\\s\n\n\\begin{dft}\n Let $f,g$ be Riemann integrable functions on $[-\\pi,\\pi]$, then the \\textbf{$L^{2}$-norm} of $f$ is given by\n\n $$\\nrm{f}_{2}=\\sqrt{\\bra{f,f}_{2}}$$\\s\n\n Also, the \\textbf{$L^{2}$-distance} between $f$ and $g$ is given by $\\nrm{f-g}_{2}$.\n\\end{dft}\n\n\\subsubsection{Mean Convergence}\n\\begin{dft}\n Let $f,f_{n}$ be Riemann integrable functions on $[-\\pi,\\pi]$, then $f_{n}\\to f$ in \\textbf{$L^{2}$-sense} if $\\nrm{f_{n}-f}_{2}\\to 0$ as $n\\to\\infty$.\n\\end{dft}\\n\n\nThis definition brings out why such idea is called mean convergence:\n\n$$\\lim_{n\\to\\infty}\\int_{-\\pi}^{\\pi}(f_{n}-f)^{2}\\to 0$$\\s\n\nis actually a variation of root mean square. Note that $L^{2}$-norm and $L^{2}$-distance are not norm and distance in a strict sense since\n\n$$\\begin{cases}\n \\nrm{f}_{2}=0&\\not\\Rightarrow f=0\\\\\n \\nrm{f-g}_{2}=0&\\not\\Rightarrow f=g\n\\end{cases}$$\\s\n\nin $R[-\\pi,\\pi]$. It is only true for almost everywhere. Also, although it is not hard to show that $f_{n}\\to f$ uniformly implies $\\nrm{f_{n}-f}_{2}\\to 0$, its converse does not hold. Consider the following counterexample:\\n\n\n\\begin{exm}\n Let\n\n $$f_{n}(x)=\\begin{cases}\n 1\\erm{if }x\\in[0,1/n]\\\\\n 0\\erm{otherwise}\n \\end{cases}$$\\s\n\n be a function, then $\\nrm{f_{n}}_{2}^{2}=\\int_{-\\pi}^{\\pi}f_{n}^{2}=1/n$, which tends to $0$ as $n\\to\\infty$. In this way, $f_{n}\\to 0$ in $L^{2}$-sense. However, $f_{n}\\not\\to 0$ uniformly or even pointwisely.\n\\end{exm}\n\n\\subsection{Applications to Fourier Series}\n\\subsubsection{Minimizers}\nConsider the functions on $[-\\pi,\\pi]$\n\n$$\\begin{cases}\n \\varphi_{0}=\\frac{1}{\\sqrt{2\\pi}}\\\\\n \\varphi_{n}=\\frac{1}{\\sqrt{\\pi}}\\cos(nx)\\\\\n \\psi_{n}=\\frac{1}{\\sqrt{\\pi}}\\sin(nx)\n\\end{cases}$$\\s\n\nNote that\n\n$$\\begin{cases}\n \\bra{\\varphi_{m},\\varphi_{n}}_{2}=\\begin{cases}\n 1\\erm{if }m=n\\\\\n 0\\erm{if }m\\neq n\n \\end{cases}\\\\\n \\bra{\\psi_{m},\\psi_{n}}_{2}=\\begin{cases}\n 1\\erm{if }m=n\\\\\n 0\\erm{if }m\\neq n\n \\end{cases}\\\\\n \\bra{\\varphi_{m},\\psi_{n}}_{2}=0\\erm{for all }m,n\n\\end{cases}$$\\s\n\ntherefore\n\n$$\\brc{\\frac{1}{\\sqrt{2\\pi}},\\frac{1}{\\sqrt{\\pi}}\\cos(nx),\\frac{1}{\\sqrt{\\pi}}\\sin(nx)}_{n=1}^{\\infty}$$\\s\n\ncan be regarded as an orthogonal basis in $R[-\\pi,\\pi]$.\\n\n\n\\begin{dft}\n The $(2n+1)$ dimensional vector subspace of $R[-\\pi,\\pi]$ spanned by the first $(2n+1)$ trigonometric functions, denoted by $E_{n}$, is defined by\n\n $$E_{n}=\\mathrm{span}\\brc{\\frac{1}{\\sqrt{2\\pi}},\\frac{1}{\\sqrt{\\pi}}\\cos(kx),\\frac{1}{\\sqrt{\\pi}}\\sin(kx)}_{k=1}^{n}$$\n\\end{dft}\\n\n\nIn general, if there is an orthogonal set (or ortogonal family) $\\brc{\\phi_{n}}_{n=1}^{\\infty}$ in $R[-\\pi,\\pi]$, let\n\n$$S_{n}=\\mathrm{span}\\bra{\\phi_{1},\\phi_{2},\\cdots,\\phi_{n}}$$\\s\n\nbe an $n$-dimensional subspace spanned by the first $n$ functions in the orthogonal set, then for any $f\\in R[-\\pi,\\pi]$, the \\textbf{minimization problem} is\n\n$$\\inf\\brc{\\nrm{f-g}_{2}\\srm g\\in S_{n}}$$\\s\n\n\\begin{pst}\n The unique minimizer of\n\n $$\\inf\\brc{\\nrm{f-g}_{2}\\srm g\\in S_{n}}$$\\s\n\n is attained at the function\n\n $$g=\\sum_{k=1}^{n}\\bra{f,\\phi_{k}}_{2}\\phi_{k}\\in S_{n}$$\\s\n\n \\prf Note that to minimize $\\nrm{f-g}_{2}$ is equivalent to minimize $\\nrm{f-g}_{2}^{2}$. For all $g\\in S_{n}$,\n\n $$g=\\sum_{k=1}^{n}\\beta_{k}\\phi_{k}\\Rightarrow\\nrm{f-g}_{2}^{2}=\\int_{-\\pi}^{\\pi}\\abs{f-\\sum_{k=1}^{n}\\beta_{k}\\phi_{k}}^{2}$$\\s\n\n Let $\\Phi(\\beta)=\\nrm{f-g}_{2}^{2}$, then\n\n $$\\begin{aligned}[t]\n \\Phi(\\beta)&=\\int_{-\\pi}^{\\pi}\\abs{f-\\sum_{k=1}^{n}\\beta_{k}\\phi_{k}}^{2}\\\\\n &=\\brr{\\int_{-\\pi}^{\\pi}f^{2}}-2\\sum_{k=1}^{\\infty}\\brr{\\frac{\\beta_{k}}{\\sqrt{2}}}\\brr{\\sqrt{2}\\bra{f,\\phi_{k}}_{2}^{2}}+\\sum_{k=1}^{n}\\beta_{k}^{2}\\\\\n &\\geq\\brr{\\int_{-\\pi}^{\\pi}f^{2}}-\\sum_{k=1}^{\\infty}\\brr{\\frac{\\beta_{k}^{2}}{2}+2\\bra{f,\\phi_{k}}_{2}^{2}}+\\sum_{k=1}^{n}\\beta_{k}^{2}\\erm{since }2ab\\leq a^{2}+b^{2}\\\\\n &=\\brr{\\int_{-\\pi}^{\\pi}f^{2}}-2\\sum_{k=1}^{n}\\bra{f,\\phi_{k}}_{2}^{2}+\\frac{1}{2}\\sum_{k=1}^{n}\\beta_{k}^{2}\\to\\infty\n \\end{aligned}$$\\s\n\n as\n\n $$\\nrm{\\beta}=\\sqrt{\\sum_{k=1}^{n}\\beta_{k}^{2}}\\to\\infty$$\\s\n\n Hence $\\Phi(\\beta)$ attains its minimum at some finite point $\\beta$. By some calculus, the minimum required is given by $\\beta_{k}=\\bra{f,\\phi_{k}}_{2}$ for all $1\\leq k\\leq n$.\n\\end{pst}\\n\n\nNote that the minimizer $g$ of $\\nrm{f-g}_{2}$ over $S_{n}$ is called the \\textbf{orthogonal projection} of $f$ on $S_{n}$, denoted by $P_{n}(f)$. With the notation of orthogonal projection,\n\n$$\\mathrm{dist}(f,S_{n})=\\nrm{f,P_{n}(f)}_{2}$$\\s\n\n\\begin{crl}\n For a $2\\pi$-periodic function $f$ integrable on $[-\\pi,\\pi]$ and $n\\geq 1$, $\\nrm{f-S_{n}(f)}_{2}\\leq\\nrm{f-g}_{2}$ where $S_{n}(f)$ represents the $n$-th partial sum of the Fourier series of $f$, for all\n\n $$g=\\alpha_{0}+\\sum_{k=1}^{n}(\\alpha_{k}\\cos(kx)+\\beta_{k}\\sin(kx))$$\\s\n\n with real coefficients.\\n\n\n \\prf By the definition of Fourier coefficients $S_{n}(f)=P_{n}(f)$ of $E_{n}$,\n\n $$\\begin{cases}\n a_{0}=\\frac{1}{\\sqrt{2\\pi}}\\bra{f,\\frac{1}{\\sqrt{2\\pi}}}_{2}\\\\\n a_{n}\\cos(nx)=\\frac{1}{\\sqrt{\\pi}}\\bra{f,\\frac{1}{\\sqrt{\\pi}}\\cos(nx)}_{2}\\cos(nx)\\\\\n b_{n}\\sin(nx)=\\frac{1}{\\sqrt{\\pi}}\\bra{f,\\frac{1}{\\sqrt{\\pi}}\\sin(nx)}_{2}\\sin(nx)\n \\end{cases}$$\n\\end{crl}\n\n\\propdisp\n\\subsubsection{Measure Zeroes of Fourier Series}\n\\begin{thm}\n Let $f$ be $2\\pi$-periodic integrable function on $[-\\pi,\\pi]$, then the $n$-th partial sum of the Fourier series of $f$ converges to $f$ in $L^{2}$-sense. In other words,\n\n $$\\lim_{n\\to\\infty}\\nrm{S_{n}(f)-f}_{2}=0$$\\s\n\n \\prf For any $\\epsilon>0$, there exists a $2\\pi$-periodic Lipschitz continuous function $g$ such that $\\nrm{f-g}_{2}<\\epsilon/2$. This can be achieved by finding a step function approximating $f$. By \\rpst[\\sctd{11}], there exists $N>0$ such that\n\n $$\\nrm{g-S_{N}(g)}_{\\infty}<\\frac{\\epsilon}{2\\sqrt{2\\pi}}$$\\s\n\n where $\\nrm{\\cdot}_{\\infty}$ represents the uniform convergence. This induces\n\n $$\\nrm{g-S_{N}(g)}_{2}=\\sqrt{\\int_{-\\pi}^{\\pi}(g-S_{N}(g))^{2}}\\leq\\sqrt{2\\pi\\nrm{g-S_{n}(g)}_{\\infty}^{2}}=\\frac{\\epsilon}{2}$$\\s\n\n By the corollary of \\rpst[\\sctd{1}],\n\n $$\\nrm{f-S_{N}(f)}_{2}\\leq\\nrm{f-S_{N}(g)}_{2}\\leq\\nrm{f-g}_{2}+\\nrm{g-S_{n}(g)}_{2}<\\frac{\\epsilon}{2}+\\frac{\\epsilon}{2}=\\epsilon$$\\s\n\n Finally, since $E_{N}\\subset E_{n}$ for all $n\\geq N$,\n\n $$\\nrm{f-S_{n}(f)}_{2}\\leq\\nrm{f-S_{N}(f)}_{2}<\\epsilon$$\\s\n\n for any $n\\geq N$, thus\n\n $$\\lim_{n\\to\\infty}\\nrm{S_{n}(f)-f}_{2}=0$$\n\\end{thm}\\n\n\n\\begin{crl}\n Let $f_{1}$ and $f_{2}$ be $2\\pi$-periodic integrable functions on $[-\\pi,\\pi]$ with the same Fourier series, then $f_{1}=f_{2}$ almost everywhere, or $f_{1}=f_{2}$ except a set of measure zero. Furthermore, if $f_{1}$ and $f_{2}$ are both continuous on $[-\\pi,\\pi]$, then $f_{1}=f_{2}$.\\n\n\n \\prf Let $f=f_{1}-f_{2}$, then $a_{n}(f)=b_{n}(f)=0$ gives $S_{n}(f)=0$ for any $n\\geq 0$. Therefore\n\n $$\\lim_{n\\to\\infty}\\nrm{S_{n}(f)-f}_{2}=0\\Rightarrow \\nrm{f}_{2}=0$$\\s\n\n and by theory of Riemann integrals, $f=0$ almost everywhere. If $f_{1},f_{2}$ are continuous, $f^{2}\\mathrm{cts}\\geq 0\\Rightarrow f^{2}\\equiv 0$.\n\\end{crl}\\n\n\nNote that a set $E$ is said to be measure zero if for any $\\epsilon>0$, there exists countably many intervals $\\brc{I_{k}}$ such that $E\\subset\\bigcup_{k}I_{k}$ and $\\sum_{k}\\abs{I_{k}}<\\epsilon$.\n\n\\subsubsection{Parserval's Identity}\nBelow is the \\textbf{Parserval's Identity}:\\n\n\n\\begin{pst}\n Let $f$ be a $2\\pi$-periodic function $f$ integrable on $[-\\pi,\\pi]$, then\n\n $$\\nrm{f}_{2}^{2}=2\\pi a_{0}^{2}+\\pi\\sum_{n=1}^{\\infty}(a_{n}^{2}+b_{n}^{2})$$\\s\n\n where $a_{0},a_{n},b_{n}$ are Fourier coefficients of $f$.\\n\n\n \\prf Note that\n\n $$\\begin{cases}\n \\sqrt{2\\pi}a_{0}=\\bra{f,\\frac{1}{\\sqrt{2\\pi}}}_{2}\\\\\n \\sqrt{\\pi}a_{n}=\\bra{f,\\frac{1}{\\sqrt{\\pi}}\\cos(nx)}_{2}\\erm{for all }n\\geq 1\\\\\n \\sqrt{\\pi}b_{n}=\\bra{f,\\frac{1}{\\sqrt{\\pi}}\\sin(nx)}_{2}\\erm{for all }n\\geq 1\n \\end{cases}$$\\s\n\n then by corollary of \\rpst[\\sctd{2}],\n\n $$\\begin{aligned}[t]\n \\bra{f,S_{N}(f)}_{2}&=\\bra{(f-S_{N}(f))+S_{N}(f),S_{N}(f)}_{2}\\\\\n &=\\bra{S_{N}(f),S_{N}(f)}_{2}\\erm{since }(f-S_{N}(f))\\text{ is orthogonal}\\\\\n &=\\int_{-\\pi}^{\\pi}\\brr{a_{0}+\\sum_{k=1}^{\\infty}a_{k}\\cos(kx)+b_{k}\\sin(kx)}^{2}\\diff x\n \\end{aligned}$$\\s\n\n Finally by \\rthm[\\sctd{1}],\n\n $$\\begin{aligned}[t]\n 0&=\\lim_{N\\to\\infty}\\nrm{f-S_{N}(f)}_{2}^{2}\\\\\n &=\\lim_{N\\to\\infty}(\\nrm{f}_{2}^{2}-2\\bra{f,S_{N}(f)}_{2}+\\nrm{S_{N}(f)}_{2}^{2})\\\\\n &=\\lim_{N\\to\\infty}(\\nrm{f}_{2}^{2}-2\\nrm{S_{N}(f)}_{2}^{2}+\\nrm{S_{N}(f)}_{2}^{2})\\\\\n &=\\lim_{N\\to\\infty}(\\nrm{f}_{2}^{2}-\\nrm{S_{N}(f)}_{2}^{2})\n \\end{aligned}$$\\s\n\n therefore\n\n $$\\nrm{f}_{2}^{2}=\\lim_{N\\to\\infty}\\brr{2\\pi a_{0}^{2}+\\pi\\sum_{n=1}^{N}(a_{n}^{2}+b_{n}^{2})}$$\n\\end{pst}\n\n\\pagebreak\n\n\\section{Metric Spaces}\n\\subsection{Introduction of Metric Spaces}\n\\subsubsection{Definition of Metrics Spaces}\n\\begin{dft}\n Let $X$ be a nonempty set, then a \\textbf{metric} on $X$ is a function\n\n $$d:X\\times X\\to[0,+\\infty)$$\\s\n\n such that the following properties is satisfied for all $x,y,z\\in X$:\n\n \\begin{alist}\n \\item Metric is nonnegative, or $d(x,y)\\geq 0$. Equality holds if and only if $x=y$.\n \\item Metric is symmetric, or $d(x,y)=d(y,x)$.\n \\item Metric is subadditive (satisfies triangle inequality), or $d(x,y)\\leq d(x,z)+d(z,y)$.\n \\end{alist}\n\n The pair $(X,d)$ is called a \\textbf{metric space}.\n\\end{dft}\\n\n\n\\begin{dft}\n Let $(X,d)$ be a metric space, then the \\textbf{metric ball} of radius $r$ centered at $x$, denoted by $B_{r}(x)$, is defined as\n\n $$B_{r}(x)=\\brc{y\\in X\\srm d(x,y)0$ such that\n\n $$d(x,y)\\leq C\\rho(x,y)$$\\s\n \n for all $x,y\\in X$. $d$ and $\\rho$ are \\textbf{equivalent} to each other if $d$ is stronger and weaker than $\\rho$ at the same time. In other words, there exists $C_{1},C_{2}>0$ such that\n\n $$d(x,y)\\leq C_{1}\\rho(x,y)\\leq C_{2}d(x,y)$$\\s\n\n for all $x,y\\in X$.\n\\end{dft}\\n\n\nNote that the equivalence of metrics defined above is an equivalent relation.\\n\n\n\\begin{pst}\n Let $d$ and $\\rho$ be different metrics defined on $X$. If $\\rho$ is stronger than $d$ and a sequence $\\brc{x_{n}}$ converges in $(X,\\rho)$, then the sequence also converges in $(X,d)$ with the same limit. If $\\rho$ is equivalent to $d$, then $\\brc{x_{n}}$ converges in $(X,\\rho)$ if and only if it converges in $(X,d)$ also.\n\\end{pst}\\n\n\n\\begin{exm}\n Recall the metrics $d_{1}$, $d_{2}$ and $d_{\\infty}$ on $\\R^{n}$, then\n\n $$\\begin{cases}\n d_{1}(x,y)\\leq nd_{\\infty}(x,y)\\leq nd_{1}(x,y)\\\\\n d_{2}(x,y)\\leq\\sqrt{n}d_{\\infty}(x,y)\\leq\\sqrt{n}d_{2}(x,y)\n \\end{cases}$$\\s\n\n shows that $d_{1}$, $d_{2}$ and $d_{\\infty}$ are equivalent metrics.\n\\end{exm}\\n\n\n\\begin{exm}\n Recall the metrics $d_{1}$ and $d_{\\infty}$ on $C[a,b]$, then\n\n $$d_{1}(f,g)\\leq(b-a)d_{\\infty}(f,g)$$\\s\n\n shows that $d_{\\infty}$ is stronger than $d_{1}$. However, $d_{1}$ is not stronger than $d_{\\infty}$, so $d_{1}$ and $d_{\\infty}$ are not equivalent.\n\\end{exm}\n\n\\subsubsection{Continuity in Metric Spaces}\n\\begin{dft}\n Let $f:(X,d)\\to(Y,\\rho)$ be a mapping between two metric spaces, then $f$ is \\textbf{continuous} at a point $x\\in X$ if $f(x_{n})\\to f(x)$ in $(Y,\\rho)$ whenever $x_{n}\\to x$ in $(X,d)$. $f$ is continuous on a set $E\\in X$ if it is continuous at every point in $E$.\n\\end{dft}\\n\n\n\\begin{pst}\n Let $f:(X,d)\\to(Y,\\rho)$ be a mapping between two metric spaces and $x_{0}\\in X$ be a point, then $f$ is continuous at $x_{0}$ if and only if for any $\\epsilon>0$, there exists $\\delta>0$ such that\n\n $$\\rho(f(x),f(x_{0}))<\\epsilon\\erm{for all }\\brc{x\\in X\\srm d(x,x_{0})<\\delta}$$\n\\end{pst}\\n\n\n\\begin{pst}\n Let $f:(X,d)\\to(Y,\\rho)$ and $g:(Y,\\rho)\\to(Z,m)$ be mappings between metric spaces, then if $f$ is continuous at $x$ and $g$ is continuous at $f(x)$, then $g\\circ f$ is also continuous at $x$. Similarly, if $f$ is continuous at $X$ and $g$ is continuous at $Y$, then $g\\circ f$ is also continuous at $X$.\n\\end{pst}\\n\n\n\\begin{exm}\n Let $(X,d)$ be a metric space and $A\\subset X$ be a nonempty set. Further define $\\rho_{A}:X\\to\\R$ by\n\n $$\\rho_{A}(x)=\\underset{y\\in A}{\\inf}d(y,x)$$\\s\n\n which is the shortest distance from $x$ to the subset $A$. Show that\n\n $$\\abs{\\rho_{A}(x)-\\rho_{A}(y)}\\leq d(x,y)$$\\s\n\n for any $x,y\\in X$.\\n\n\n \\ans For fixed $x,y\\in X$, along with the definition of $\\rho_{A}$, for all $\\epsilon>0$, there exists $z\\in A$ such that $\\rho_{A}(y)+\\epsilon>d(z,y)$. Hence\n\n $$\\rho_{A}(x)\\leq d(z,x)\\leq d(z,y)+d(y,x)0$ such that\n\n $$B_{\\epsilon}(x)=\\brc{y\\srm d(x,y)<\\epsilon}\\subset G$$\n\\end{dft}\\n\n\nNote that $\\epsilon$ may vary depending on the choice of $x$, and the empty set $\\phi$ is considered an open set. Therefore, the proposition applies:\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space and $G_{\\alpha}$ be a collection of open sets, then the following are true:\n\n \\begin{alist}\n \\item $X$ and $\\phi$ are open sets.\n \\item Arbitrary union of open sets $\\bigcup_{\\alpha}G_{\\alpha}$ is an open set.\n \\item Finite intersection of open sets $\\bigcap_{i=1}^{n}G_{i}$ is an open set.\n \\end{alist}\n\\end{pst}\n\n\\subsubsection{Closed Sets}\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a set $F\\in X$ is called an \\textbf{closed set} if $X\\setminus F$ is an open set.\n\\end{dft}\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space and $F_{\\alpha}$ be a collection of closed sets, then the following are true:\n\n \\begin{alist}\n \\item $X$ and $\\phi$ are closed sets.\n \\item Finite union of closed sets $\\bigcup_{j=1}^{n}F_{j}$ is an closed set.\n \\item Arbitrary intersection of closed sets $\\bigcap_{\\alpha}F_{\\alpha}$ is an closed set.\n \\end{alist}\n\\end{pst}\\n\n\n\\begin{crl}\n Let $(X,d)$ be a metric space, then $X$ and $\\phi$ are both open and closed.\n\\end{crl}\n\n\\subsubsection{Applications of Open and Closed Sets}\n\\begin{pst}\n Let $(X,d)$ be a metric space, then a sequence $\\brc{x_{n}}$ converges to $x$ if and only if for all open set $G$ containing $x$, there exists $n_{0}$ such that $x_{n}\\in G$ for all $n\\geq n_{0}$.\n\\end{pst}\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space, then a set $A\\subset X$ is closed if and only if whenever $\\brc{x_{n}}\\subset A$ and $x_{n}\\to x$ as $n\\to\\infty$ implies that $x\\in A$.\n\\end{pst}\\n\n\n\\begin{pst}\n Let $f:(X,d)\\to(Y,\\rho)$ be a mapping between metric spaces, then the following applies:\n\n \\begin{alist}\n \\item $f$ is continuous at $x$ if and only if for all open set $G\\subset Y$ containing $f(x)$, $f^{-1}(G)$ contains $B_{\\epsilon}(x)$ for some $\\epsilon>0$.\n \\item $f$ is continuous at $x$ if and only if for all open set $G\\subset Y$, $f^{-1}(G)$ is open in $X$.\n \\end{alist}\n\\end{pst}\\n\n\nIn this case, $f$ is also continuous at $x$ if and only if for all closed set $F\\subset Y$, $f^{-1}(F)$ is closed in $X$.\n\n\\propdisp\n\\subsection{Points in Metric Space}\n\\subsubsection{Boundary Points and Closures}\n\\begin{dft}\n Let $E$ be a set in a metric space $(X,d)$, then a point $x\\in X$ (which is not necessary in $E$) is called a \\textbf{boundary point} of $E$ if for all open set $G\\subset X$ containing $x$,\n\n $$G\\cap E\\neq\\phi\\text{ and }G\\setminus E\\neq\\phi$$\\s\n\n In other words, this is satisfied when $G\\cap(X\\setminus E)\\neq\\phi$. The \\textbf{boundary} of $E$, denoted by $\\partial E$, is the set of boundary points of $E$. The \\textbf{closure} of $E$, denoted by $\\overline{E}$, is defined as $\\overline{E}=E\\cup\\partial E$.\n\\end{dft}\\n\n\nFor the conditions of a boundary point in the definition above, it suffices to check $G$ of the form $B_{\\epsilon}(x)$ for all small $\\epsilon>0$, or even $B_{1/n}(x)$ for all $n\\geq 1$. Also note that $X$ and $X\\setminus E$ shares the same boundary no matter the choice of $E$, or\n\n$$\\partial E=\\partial X\\setminus E\\erm{for all }E\\subset X$$\\s\n\n\\subsubsection{Properties of Boundaries and Closures}\n\\begin{pst}\n Below are some properties of boundaries and closures:\n\n \\begin{alist}\n \\item The boundary of an empty set is an empty set, or $\\partial\\phi=\\phi$.\n \\item For all $E\\subset X$, $\\partial E$ is a closed set.\n \\item If $E$ is a closed set, $\\overline{E}=E$.\n \\end{alist}\n\\end{pst}\\n\n\n\\begin{pst}\n Let $E$ be a subset of a metric space $(X,d)$, then the following applies:\n\n \\begin{alist}\n \\item $x\\in\\overline{E}$ if and only if $B_{r}(x)\\cap E\\neq\\phi$ for all $r>0$.\n \\item If $A\\subset B$, $\\overline{A}\\subset\\overline{B}$ for all $A,B\\subset(X,d)$.\n \\item $\\overline{E}$ is closed.\n \\item $\\overline{E}$ is the smallest closed set containing $E$, or $\\overline{E}=\\cap\\brc{G\\subset E\\srm G\\text{ is closed}}$.\n \\end{alist}\n\\end{pst}\n\n\\subsubsection{Interior Points}\n\\begin{dft}\n Let $E$ be a subset of a metric space $(X,d)$, then a point $x$ is called an \\textbf{interior point} of $E$ if there exists an open set $G$ such that $x\\in G$ and $G\\subset E$. The \\textbf{interior} of $E$, denoted by $E^{0}$, is the set of interior points of $E$.\n\\end{dft}\\n\n\n\\begin{pst}\n Below are the properties of interiors:\n\n \\begin{alist}\n \\item Interior of $E$, $E^{0}$ is open.\n \\item Interior of $E$ is the set without boundary, or $E^{0}=E\\setminus\\partial E$.\n \\item Interior of $E$, $E^{0}=X\\setminus\\overline{X\\setminus E}$.\n \\item Interior of $E$, $E^{0}=\\cup\\brc{G\\subset E\\srm G\\text{ is open}}$.\n \\end{alist}\n\\end{pst}\n\n\\subsection{Elementary Inequalities for Functions}\n\\subsubsection{Young's Inequality}\n\\begin{thm}\n By \\textbf{Young's Inequality}, for $a,b>0$ and $p>1$,\n\n $$ab\\leq\\frac{a^{p}}{p}+\\frac{b^{q}}{q}\\erm{with }\\frac{1}{p}+\\frac{1}{q}=1$$\\s\n\n Equality holds when $a^{p}=b^{q}$.\n\\end{thm}\\n\n\nNote that $q=\\frac{p}{p-1}>1$ is called the \\textbf{conjugate} of $p$. Also specifically if $p=2$, the inequality reduces to $2ab\\leq a^{2}+b^{2}$.\n\n\\subsubsection{Holder's Inequality}\nFor the following inequality, denote the norm\n\n$$\\nrm{f}_{p}=\\brr{\\int_{a}^{b}\\abs{f(x)}^{p}\\diff x}^{1/p}$$\\s\n\n\\begin{thm}\n Let $f,g\\in R[a,b]$ be Riemann integrable functions and $p>1$, then by \\textbf{Holder's Inequality},\n\n $$\\int_{a}^{b}\\abs{f(x)g(x)}\\diff x\\leq\\brr{\\int_{a}^{b}\\abs{f(x)}^{p}\\diff x}^{1/p}\\brr{\\int_{a}^{b}\\abs{f(x)}^{q}\\diff x}^{1/q}$$\\s\n\n where $q$ is the conjugate of $p$. Equality holds when one of the following conditions is satisfied:\n\n \\begin{alist}\n \\item $f$ or $g$ equals to $0$ almost everywhere.\n \\item There exists a constant $\\lambda>0$ such that $\\abs{g(x)}^{q}=\\lambda\\abs{f(x)}^{p}$ almost everywhere.\n \\end{alist}\n\\end{thm}\\n\n\nNote that Holder's Inequality can be written in norm form $\\nrm{fg}_{1}=\\nrm{f}_{p}\\nrm{g}_{q}$. Holder's Inequality also holds for limiting cases $(p,q)\\to(1,\\infty)$ and $(p,q)\\to(\\infty,1)$.\n\n\\subsubsection{Minkowski's Inequality}\n\\begin{thm}\n By \\textbf{Minkowski's Inequality}, for any $f,g\\in R[a,b]$ and $p>1$,\n \n $$\\nrm{f+g}_{p}\\leq\\nrm{f}_{p}+\\nrm{g}_{p}$$\\s\n\n Equality holds when one of the following conditions is satisfied:\n\n \\begin{alist}\n \\item $f$ or $g$ equals to $0$ almost everywhere.\n \\item $\\nrm{f}_{p},\\nrm{g}_{p}>0$ and there exists a constant $\\lambda>0$ such that $g(x)=\\lambda f(x)$ almost everywhere.\n \\end{alist}\n\\end{thm}\n\n\\pagebreak\n\n\\section{Contraction Mapping Principle}\n\\subsection{Complete Metric Space}\n\\subsubsection{Definition of Complete Metric Space}\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a sequence $\\brc{x_{n}}$ in $(X,d)$ is a \\textbf{Cauchy sequence} if for any $\\epsilon>0$, there exists $n_{0}$ such that $d(x_{n},x_{m})<\\epsilon$ for all $n,m>n_{0}$.\n\\end{dft}\\n\n\n\\begin{dft}\n Let $(X,d)$ be a metric space, then the metric space is \\textbf{complete} if every Cauchy sequence in the metric space converges. A subset $E$ is complete if the induce metric subspace $(E,d)$ with $d=d\\!\\mid_{E\\times E}$ is complete. In other words, every Cauchy sequence in $E$ converges with limit in $E$.\n\\end{dft}\\n\n\nNote that convergent sequence is a Cauchy sequence.\n\n\\begin{pst}\n Let $(X,d)$ be a metric space, then the following applies:\n\n \\begin{alist}\n \\item Every complete set in $X$ is closed.\n \\item If $X$ is complete, then every closed set in $X$ is complete.\n \\end{alist}\n\\end{pst}\\n\n\n\\begin{exm}\n Below are the examples of complete metric space:\n\n \\begin{alist}\n \\item $(\\R,\\text{standard})$ is complete.\n \\item $[a,b]$, $(-\\infty,b]$ and $[a,\\infty)$ are complete.\n \\end{alist}\n\n Below are the counterexamples of complete metric space:\n\n \\begin{alist}\n \\item $[a,b)$ where $b$ is finite, is not complete since $x_{n}=b-1/n\\to b\\not\\in[a,b)$.\n \\item $\\Q$ is not complete.\n \\end{alist}\n\\end{exm}\n\n\\subsubsection{Completion of Metric Spaces}\n\\begin{dft}\n A metric space $(X,d)$ is said to be \\textbf{isometrically embedded} in metric space $(Y,\\rho)$ if there exists a mapping $\\Phi:X\\to Y$ such that $d(x,y)=\\rho(\\Phi(x),\\Phi(y))$. If such mapping exists, $\\Phi$ is called an \\textbf{isometric embedding} (or a \\textbf{metric preserving map}) from $(X,d)$ to $(Y,\\rho)$.\n\\end{dft}\\n\n\nNote that $\\Phi$ must be injective and continuous.\n\n\\begin{dft}\n Let $(X,d)$ and $(Y,\\rho)$ be metric spaces, then $(Y,\\rho)$ is called a completion of $(X,d)$ if the following statements are satisfied:\n\n \\begin{alist}\n \\item $(Y,\\rho)$ is complete.\n \\item There exists an isometric embedding $\\Phi$ such that the closure $\\overline{\\Phi(X)}=Y$.\n \\end{alist}\n\\end{dft}\\n\n\n\\begin{exm}\n Let $(X,d)=(\\Q,\\text{induced metric})$ and $(Y,\\rho)=(\\R,\\text{standard})$. Since $\\Q\\subset\\R$, $(Y,\\rho)$ is complete. Further let $\\Phi:(X,d)\\to(Y,\\rho)$ where $\\Phi(q)=q$, since $\\Q$ is dense in $\\R$, $\\overline{\\Phi(\\Q)}=\\overline{\\Q}=\\R$. Therefore, $(Y,\\rho)$ is a completion of $(X,d)$.\n\\end{exm}\\n\n\n\\begin{thm}\n Every metric space has a completion.\n\\end{thm}\\n\n\nNote that the definition of isometric embedding can be extended to bijection as below:\\n\n\n\\begin{dft}\n Let $(X,d)$ and $(Y,\\rho)$ be metric spaces, then they are called \\textbf{isometric} if there exists a bijective isometric embedding between $(X,d)$ and $(Y,\\rho)$.\n\\end{dft}\\n\n\nNote that the inverse of a bijective isometric embedding is also an isometric embedding. Also, the two metric spaces are regarded as the same if they are isometric.\\n\n\n\\begin{thm}\n If metric spaces $(Y,\\rho)$ and $(Y',\\rho')$ are both completions of a metric space $(X,d)$, then $(Y,\\rho)$ and $(Y',\\rho')$ are isometric. In other words, completion is unique up to isometry.\n\\end{thm}\n\n\\subsection{Introduction to Contraction Mapping Principle}\n\\subsubsection{General Theorem}\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a map $T:(X,d)\\to (X,d)$ is called a \\textbf{contraction} if there exists a constant $\\gamma\\in(0,1)$ such that\n \n $$d(Tx,Ty)\\leq\\gamma d(x,y)$$\\s\n\n for all $x,y\\in X$. A point $x\\in X$ is called a \\textbf{fixed point} of $T$ if $Tx=x$.\n\\end{dft}\\n\n\nNote that $Tx$ is the notation for $T(x)$ but not the multiplication. With the definition of a contraction, below is the \\textbf{Contraction Mapping Principle} (or the \\textbf{Banach Fixed Point Theorem}):\\n\n\n\\begin{thm}\n Every contraction in a complete metric space admit a fixed point.\n\\end{thm}\n\n\\subsubsection{Perturbation of Identity}\n\\begin{dft}\n A normed space $(X,\\nrm{\\cdot})$ is a \\textbf{Banach space} if it is complete as a metric space with respect to the induced metric $d(x,y)=\\nrm{x-y}$ for all $x,y\\in X$.\n\\end{dft}\n\n\\begin{exm}\n Below are some examples of Banach space:\n\n \\begin{alist}\n \\item $(\\R^{n},\\nrm{\\cdot}_{p})$ is a Banach space if $p>1$.\n \\item $(C[a,b],\\nrm{\\cdot}_{\\infty})$ is a Banach space.\n \\end{alist}\n\\end{exm}\n\n\\begin{thm}\n Let $(X,\\nrm{\\cdot})$ be a Banach space, and $\\Phi:\\overline{B_{r}(x_{0})}\\to X$ satisfies $\\Phi(x_{0})=y_{0}$. Suppose that $\\Phi=\\mathrm{Id}_{X}+\\Psi$ such that there exists a constant $\\gamma\\in(0,1)$ such that\n\n $$\\nrm{\\Psi(x_{2})-\\Psi(x_{1})}\\leq\\gamma\\nrm{x_{2}-x_{1}}$$\\s\n\n for all $x_{1},x_{2}\\in\\overline{B_{r}(x_{0})}$, then by \\textbf{Perturbation of Identity}, for all $y\\in\\overline{B_{R}(y_{0})}$ where $R=(1-\\gamma)r$, there exists unique $x\\in\\overline{B_{r}(x_{0})}$ such that $\\Phi(x)=y$.\n\\end{thm}\n\n\\begin{exm}\n Show that $3x^{4}-x^{2}+x=-0.05$ has a real root.\\n\n\n \\prf Notice that $3x^{4}-x^{2}+x=0$ has a root $x=0$. Let $\\Phi(x)=x+\\Psi(x)$ where $\\Psi(x)=3x^{4}-x^{2}$, then $\\Phi(0)=0$. For $x_{1},x_{2}\\in\\overline{B_{r}(0)}$,\n\n $$\\begin{aligned}[t]\n \\abs{\\Psi(x_{1})-\\Psi(x_{2})}&=\\abs{3x_{1}^{4}-x_{1}^{2}-3x_{2}^{4}+x_{2}^{2}}\\\\\n &=\\abs{3(x_{1}^{4}-x_{2}^{4})-(x_{1}^{2}-x_{2}^{2})}\\\\\n &=\\abs{3(x_{1}^{3}+x_{1}^{2}x_{2}+x_{2}^{2}x^{1}+x_{2}^{3})-(x_{1}+x_{2})}\\abs{x_{1}-x_{2}}\\\\\n &=\\abs{12r^{3}+2r}\\abs{x_{1}-x_{2}}\n \\end{aligned}$$\\s\n\n Choose $r>0$ such that $\\gamma=12r^{3}+2^{r}<1$ and $R=(1-\\gamma)r\\geq 0.05$ so that $-0.05\\in\\overline{B_{R}(0)}$. Pick $r=1/4$, then $\\gamma=11/16$ and $R=5/64$. By Perturbation of Identity, for all $y\\in\\overline{B_{R}(0)}$, there exists $x\\in\\overline{B_{r}(0)}$ such that $\\Phi(x)=y$. Therefore, there exists a real root for $3x^{4}-x^{2}+x=-0.05$ since $-0.05\\in\\overline{B_{R}(0)}$.\n\\end{exm}\\n\n\nThe example above can be summarized into the following proposition:\\n\n\n\\begin{pst}\n Let $\\Phi(x)=x+\\Psi(x)$ where $\\Psi(x):U\\to\\R^{n}$ be a $C^{1}$-function on some open set $U\\subset\\R^{n}$ containing $0$, such that\n\n $$\\Psi(0)=0\\text{ and }\\lim_{x\\to 0}\\frac{\\partial\\Psi_{i}}{\\partial x_{j}}(x)=0$$\\s\n\n for all $i,j$, then there exists $r>0$ and $R>0$ such that for all $y\\in B_{R}(0)$, $\\Phi(x)=y$ has a unique solution $x\\in B_{r}(0)$.\n\\end{pst}\n\n\\subsection{Inverse Function Theorem}\n\\subsubsection{Introduction to Inverse Function Theorem}\nRecall the chain rule: let $G:U\\subset\\R^{n}\\to\\R^{m}$ and $F:V\\subset\\R^{m}\\to\\R^{l}$ be differentiable functions where $U,V$ open in $\\R^{n},\\R^{m}$ respectively, and $G(U)\\subset V$. Then $H=F\\circ G:U\\to\\R^{l}$ differentiable and $DH(x)=DF(G(x))DG(x)$ where\n\n$$DG(x)=\\brr{\\frac{\\partial G_{i}}{\\partial x_{j}}(x)}_{i,j}$$\\s\n\nand similarly for $DF$ and $DH$. Besides the proposition is required:\\n\n\n\\begin{pst}\n Let $F:B\\to\\R^{n}$ be $C^{1}$ function, where $B$ is a ball in $\\R^{n}$, then for any $x_{1},x_{2}\\in B$,\n\n $$F(x_{1})-F(x_{2})=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))\\diff t}\\cdot(x_{1}-x_{2})$$\\s\n\n in component form $F=(F_{1},\\cdots,F_{n})$. In other words,\n\n $$F_{i}(x_{1})-F_{i}(x_{2})=\\sum_{j=1}^{n}\\brr{\\int_{0}^{1}\\frac{\\partial F_{i}}{\\partial x_{j}}(x_{2}+t(x_{1}-x_{2}))\\diff t}(x_{1}-x_{2})_{j}$$\n\\end{pst}\\n\n\nFinally, recall that if $F:U\\subset\\R^{n}\\to\\R^{m}$ be differentiable at a point $p$ in an open set $U$ of $\\R^{n}$, then\n\n$$F(p+x)-F(p)=DF(p)x+o(\\abs{x})$$\\s\n\nfor all $x=(x_{1},\\cdots,x_{n})$ sufficiently small (or $\\abs{x}$ small) where $o(\\abs{x})$ is a remaining term such that\n\n$$\\frac{o(\\abs{x})}{\\abs{x}}\\to 0\\text{ as }\\abs{x}\\to 0$$\\s\n\n\\begin{dft}\n The condition in Inverse Function Theorem that $DF(x_{0})$ is invertible is called the \\textbf{nondegeneracy condition}.\n\\end{dft}\\n\n\nNote that nondegeneracy condition is necessary for the differentiability of local inverse.\\n\n\n\\begin{pst}\n Let $F:U\\subset\\R^{n}\\to\\R^{n}$ be a $C^{1}$ function where $U$ is an open set and $x_{0}\\in U$. Suppose there exists open $V$ such that $x_{0}\\in V\\subset U$ and $F\\!\\mid\\!_{V}$ has a differentiable inverse, then $DF(x_{0})$ is nonsingular (or invertible).\n\\end{pst}\\n\n\nBelow is the \\textbf{Inverse Function Theorem}:\\n\n\n\\begin{thm}\n Let $F:U\\to\\R^{n}$ be a $C^{1}$ map from an open set $U\\to\\R^{n}$. Suppose $x_{0}\\in U$ and $DF(x_{0})$ is invertible (as a matrix or linear transformation), then there exists open sets $V,W$ containing $x_{0},F(x_{0})$ respectively such that the restriction of $F$ on $V$ is a bijection onto $W$ with a $C^{1}$ inverse.\\n\n \n Moreover, the inverse is $C^{k}$ when $F$ is $C^{k}$ where $1\\leq k\\leq\\infty$, in $V$.\\n\n\n \\prf\\prt[a]{zr} Consider the special case where $x_{0}=0,y_{0}=F(x_{0})=F(0)=0$, then $DF(0)=I$, which is the identity. Let $\\Psi(x)=-x+F(x)$. As $0\\in U$ and $U$ is open, there exists $r_{0}>0$ such that $\\overline{B_{r_{0}}(0)}\\subset U$. Then\n\n $$\\Psi(x_{1})-\\Psi(x_{2})=-x_{1}+F(x_{1})+x_{2}-F(x_{2})$$\\s\n\n By \\rpst[\\sctd{3}],\n\n $$\\begin{aligned}[t]\n \\Psi(x_{1})-\\Psi(x_{2})&=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))\\diff t}\\cdot(x_{1}-x_{2})-(x_{1}-x_{2})\\\\\n &=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))\\diff t-I}\\cdot(x_{1}-x_{2})\\\\\n &=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))-DF(0)\\diff t}\\cdot(x_{1}-x_{2})\n \\end{aligned}$$\\s\n\n Since $F$ is $C^{1}$, for all $\\epsilon>0$, there exists $00$, there exists $00$ as in Part (a), then for all $y\\in B_{r/2}(0)$, there exists $x\\in B_{r}(0)$ such that $F(x)=y$. This is true because of Perturbation of Identity (\\rthm[\\sctd{6}]) with $\\epsilon=1/2$. The local inverse $G$ of $F$,\n\n $$G:B_{r/2}(0)\\to G(B_{r/2}(0))\\subset B_{r}(0)$$\\s\n\n satisfies\n\n $$\\abs{G(y_{1})-G(y_{2})}\\leq\\frac{1}{1-\\epsilon}\\abs{y_{1}-y_{2}}=2\\abs{y_{1}-y_{2}}$$\\s\n\n for all $y_{1},y_{2}\\in B_{r/2}(0)$ with $G(B_{r/2}(0))$ open in $B_{r}(0)$.\\n\n\n \\prtc[c]{zr} Since $DF(0)=I$, assume that $DF(x)$ is invertible for all $x\\in B_{r}(0)$ for $r>0$ given in Part (a). Further let $W=B_{r/2}(0)=B_{R}(0)$, and $V=G(W)\\ni 0$, then $G:W\\to V$ (and similarly $F:V\\to W$). If $G$ is differentiable, by chain rule $DF(G(y))DG(y)=I$ for all $y\\in W$. Rewriting the equation gives $DG(y)=(DF)^{-1}(G(y))$.\\n\n\n For any $y_{1}\\in W$ such that $y_{1}+y\\in W$,\n\n $$y=(y_{1}+y)-y_{1}=F(G(y_{1}+y))-F(G(y_{1}))$$\\s\n\n let $x_{1}=G(y_{1}+y)$ and $x_{2}=G(y_{1})$, then by \\rpst[\\sctd{3}],\n\n $$\\begin{aligned}[t]\n y&=F(x_{1})-F(x_{2})=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))\\diff t}\\cdot(x_{1}-x_{2})\\\\\n &=\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))-DF(x_{2})\\diff t}\\cdot(x_{1}-x_{2})+DF(x_{2})(x_{1}-x_{2})\n \\end{aligned}$$\\s\n\n Hence\n\n $$\\begin{aligned}[t]\n (DF)^{-1}(x_{2})y&=(DF)^{-1}(x_{2})\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))-DF(x_{2})\\diff t}\\cdot(x_{1}-x_{2})\\\\\n &+(x_{1}-x_{2})\n \\end{aligned}$$\\s\n\n In other words, $G(y_{1}+y)-G(y_{1})=(DF)^{-1}(G(y_{1}))y+R$ where\n\n $$R=(DF)^{-1}(x_{2})\\brr{\\int_{0}^{1}DF(x_{2}+t(x_{1}-x_{2}))-DF(x_{2})\\diff t}\\cdot(x_{1}-x_{2})$$\\s\n\n By Part (b), $\\abs{x_{1}-x_{2}}\\leq 2\\abs{y}$, so $\\abs{x_{1}-x_{2}}\\to 0$ as $\\abs{y}\\to 0$ and\n\n $$\\frac{\\abs{R}}{\\abs{y}}\\leq 2\\nrm{(DF)^{-1}(x_{2})}\\int_{0}^{1}\\nrm{DF(x_{2})-DF(x_{2}+t(x_{1}-x_{2}))}\\diff t$$\\s\n\n With the assumption that $F$ is $C^{1}$,\n\n $$\\lim_{\\abs{y}\\to 0}\\frac{\\abs{R}}{\\abs{y}}=0$$\\s\n\n Therefore $G(y_{1}+y)-G(y)=(DF)^{-1}(G(y_{1}))y+o(\\abs{y})$ which implies $G$ is differentiable at $y_{1}\\in W$ and $DG(y_{1})=(DF)^{-1}(G(y_{1}))$.\\n\n\n Finally, for the special case, it is assumed that $DF$ is continuous and invertible on $B_{r}(0)$, then by linear algebra $(DF)^{-1}$ is also continuous. Then $DG(y)=(DF)^{-1}(G(y))$ is also continuous, and implies $G$ is $C^{1}$. Using induction and differentiating the identity $DG(y)=(DF)^{-1}(G(y))$ will finish the fact that $F$ is $C^{k}$ implies $G$ is $C^{k}$.\n\\end{thm}\\n\n\n\\subsubsection{Diffeomorphisms}\n\\begin{dft}\n Let $F:V\\to W$ be a $C^{k}$ map where $V$ and $W$ are open sets in $\\R^{n}$, then $F$ is called a \\textbf{$C^{k}$-diffeomorphism} if $F^{-1}$ exists and is also $C^{k}$.\n\\end{dft}\\n\n\nWith the definition of diffeomorphisms, the Inverse Function Theorem can be rephrased as follows:\n\n\\begin{thm}\n Let $F:U\\to\\R^{n}$ be a $C^{k}$ map from an open set $U\\to\\R^{n}$. Suppose $x_{0}\\in U$ and $DF(x_{0})$ is invertible (as a matrix or linear transformation), then $F$ is a $C^{k}$-diffeomorphism between some open sets $V,W$ of $x_{0},F(x_{0})$ respectively.\n\\end{thm}\\n\n\nAlso, if $F:V\\to W$ is a $C^{k}$-diffeomorphism, then for all function $\\varphi:W\\to\\R$, there corresponds a function $\\psi=\\varphi\\circ F:V\\to\\R$. Conversely, for all function $\\psi=V\\to\\R$, there corresponds a function $\\varphi=\\psi\\circ F^{-1}:W\\to\\R$. Moreover, $\\varphi$ is $C^{k}$ if and only if $\\psi$ is $C^{k}$. Thus every $C^{k}$-diffeomorphism gives rise to a \\textbf{local $C^{k}$-change of coordinates}.\n\n\\subsubsection{Examples of Inverse Function Theorem}\nBelow are some examples about the Inverse Function Theorem:\\n\n\n\\begin{exm}\n Let $F:(0,\\infty),(-\\infty,\\infty)\\to\\R^{2}$ such that $F(r,\\theta)=(r\\cos(\\theta),r\\sin(\\theta))$, then\n \n $$DF=\\begin{rmatrix}\n \\cos(\\theta) & -r\\sin(\\theta)\\\\\n \\sin(\\theta) & r\\cos(\\theta)\n \\end{rmatrix}$$\\s\n\n is invertible for all $(r,\\theta)$. By the Inverse Function Theorem, $F$ is locally invertible at every point $(r,\\theta)\\in(0,\\infty)\\times(-\\infty,\\infty)$. However, $F$ is not globally invertible as $F(r,\\theta+2\\pi)=F(r,\\theta)$ implies it is not injective.\n\\end{exm}\\n\n\n\\begin{exm}\n Let $U$ be an open interval $(a,b)\\in\\R$, then a $C^{1}$ function $f:(a,b)\\to\\R$ with $f'\\neq 0$ implies $f$ is strictly increasing or decreasing, so global inverse exists. Therefore $1$-dimensional case has stronger result than higher dimensions.\n\\end{exm}\n\n\\subsubsection{Implicit Function Theorem}\nA theorem similar to Inverse Function Theorem is the \\textbf{Implicit Function Theorem}:\\n\n\n\\begin{thm}\n Let $U$ be an open set in $\\R^{n}\\times\\R^{m}$, and $F:U\\to\\R^{m}$ is a $C^{1}$ map. Suppose that $(x_{0},y_{0})\\in U$ satisfies $F(x_{0},y_{0})=0$ and $D_{y}F(x_{0},y_{0})$ is invertible in $\\R^{m}$, then the following applies:\n\n \\begin{alist}\n \\item There exists an open set of the form $V_{1}\\times V_{2}\\in U$ containing $(x_{0},y_{0})$ and a $C^{1}$ map\n \n $$\\varphi:V_{1}\\subset\\R^{n}\\times V_{2}\\subset\\R^{m}$$\\s\n \n with $\\varphi(x_{0})=y_{0}$ such that $F(x,\\varphi(x))=0$ for all $x\\in V_{1}$.\n \\item $\\varphi:V_{1}\\to V_{2}$ is $C^{k}$ when $F$ is $C^{k}$ where $1\\leq k\\leq\\infty$.\n \\item Assume $DF_{y}$ is invertible in $V_{1}\\times V_{2}$, then if $\\psi:V_{1}\\to V_{2}$ is another $C^{1}$ map satisfying $F(x,\\psi(x))=0$, then $\\psi\\equiv\\varphi$.\n \\end{alist}\n\\end{thm}\\n\n\nNote that if\n\n$$F=\\begin{rmatrix}\n F_{1}(x_{1},\\cdots,x_{n},y_{1},\\cdots,y_{m})\\\\\n \\vdots\\\\\n F_{m}(x_{1},\\cdots,x_{n},y_{1},\\cdots,y_{m})\n\\end{rmatrix}$$\\s\n\nthen\n\n$$D_{y}F=\\begin{rmatrix}\n \\partial F_{1}/\\partial y_{1} & \\cdots & \\partial F_{1}/\\partial y_{m}\\\\\n \\vdots & \\ddots & \\vdots\\\\\n \\partial F_{m}/\\partial y_{1} & \\cdots & \\partial F_{m}/\\partial y_{m}\n\\end{rmatrix}$$\\s\n\nis an $m\\times m$ matrix and can be regarded as a linear transformation from $\\R^{m}$ to $\\R^{m}$. In general, for a map $F$ such that $DF(x_{0},y_{0})$ has rank $m$, then one can rearrange the independent variables to make the $m\\times m$ surmatrix corresponding to the last $m$ columns of the Jacobian matrix invertible, which is the situation in the theorem. Hence the condition $DF_{y}(x_{0},y_{0})$ is invertible in the Implicit Function Theorem can be generalized to $\\mathrm{rank}DF(x_{0},y_{0})=m$.\n\n\\subsection{Picard-Lindelof Theorem}\n\\subsubsection{Initial Value Problems}\n\\begin{dft}\n Let $f$ be a function defined on\n\n$$R=[t_{0}-a,t_{0}+a]\\times[x_{0}-b,x_{0}+b]$$\\s\n\nwhere $(t_{0},x_{0})\\in\\R^{2}$ and $a,b>0$. An \\textbf{initial value problem} (or \\textbf{Cauchy problem}) is of the form\n\n$$\\begin{cases}\n \\diff x/\\diff t=f(t,x)\\\\\n x(t_{0})=x_{0}\n\\end{cases}$$\n\\end{dft}\\n\n\nThis means one has to find $x(t)$ defined in an interval\n\n$$x:[t_{0}-a',t_{0}+a']\\to[x_{0}-b,x_{0}+b]$$\\s\n\nfor some $00$, but the solution $x(t)=\\tan(t)$ is defined only on $(-\\pi/2,\\pi/2)$. Therefore, even for a nice function $f$, it is still possible that $a'0$ such that for all $(t,x_{1}),(t,x_{2})\\in R$,\n\n $$\\abs{f(t,x_{1})-f(t,x_{2})}\\leq L\\abs{x_{1}-x_{2}}$$\n\\end{dft}\\n\n\nAlso recall the properties related to Lipschitz condition:\\n\n\n\\begin{pst}\n The following statements are true:\n\n \\begin{alist}\n \\item $f(t,\\cdot)$ is Lipschitz continuous in $x$ for all $t\\in[t_{0}-a,t_{0}+a]$.\n \\item If $L$ is a Lipschitz constant for $f$, then any $L'>L$ is also a Lipschitz constant.\n \\item Continuity does not imply Lipschitz continuity. For example, $f(t,x)=tx^{1/2}$ is continuous but not Lipschitz continuous near $0$.\n \\item If $R=[t_{0}-a,t_{0}+a]\\times[x_{0}-b,x_{0}+b]$ and $f(t,x):R\\to\\R$ is $C^{1}$, then $f(t,x)$ satisfies the Lipschitz condition. In fact, for some $y\\in[x_{0}-b,x_{0}+b]$,\n \n $$\\abs{f(t,x_{1})-f(t,x_{2})}=\\abs{\\frac{\\partial f}{\\partial x}(t,y)(x_{2}-x_{1})}$$\\s\n\n Hence $\\abs{f(t,x_{1})-f(t,x_{2})}\\leq L\\abs{x_{1}-x_{2}}$ for\n \n $$L=\\max\\brc{\\abs{\\frac{\\partial f}{\\partial x}(t,x)}\\srm (t,x)\\in R}$$\n \\end{alist}\n\\end{pst}\\n\n\n\\begin{pst}\n Under assumption of \\rthm[\\sctd{1}], every solution $x$ of the initial value problem from $[t_{0}-a',t_{0}+a']$ to $[x_{0}-b,x_{0}+b]$ satisfies the equation\n\n $$x(t)=x_{0}+\\int_{t_{0}}^{t}f(t,x(t))\\diff t$$\\s\n\n Conversely, every $x(t)\\in C[t_{0}-a',t_{0}+a']$ satisfying the equation above is $C^{1}$ and solves the initial value problem.\\n\n\n \\prf This is a result of Fundamental Theorem of Calculus.\n\\end{pst}\\n\n\nBelow is the \\textbf{Picard-Lindelof Theorem}:\\n\n\n\\begin{thm}\n Let $f$ be a continuous function on $R:[t_{0}-a,t_{0}+a]\\times[x_{0}-b,x_{0}+b]$ where $(t_{0},x_{0})\\in\\R^{2}$ and $a,b>0$ If $f$ satisfies Lipschitz condition on $R$ (uniform in $t$), then there exists $a'\\in(0,a]$ and $x\\in C^{1}[t_{0}-a',t_{0}+a']$ such that\n\n $$x_{0}-b\\leq x(t)\\leq x_{0}+b$$\\s\n\n for all $t\\in[t_{0}-a',t_{0}+a']$ and solving the initial value problem. Furthermore, $x$ is the unique solution in $[t_{0}-a',t_{0}+a']$.\\n\n\n \\prf For $a'>0$ to be chosen later, let\n\n $$X=\\brc{\\varphi\\in C[t_{0}-a',t_{0}+a']\\srm\\varphi(t_{0})=x_{0},\\varphi(t)\\in[x_{0}-b,x_{0}+b]}$$\\s\n\n with uniform metric $d_{\\infty}$ on $X$. Note that $X$ is a closed subset in the complete metric space $(C[t_{0}-a',t_{0}+a'],d_{\\infty})$, so $(X,d_{\\infty})$ is complete.\\n\n\n Define $T$ on $X$ by\n\n $$(T\\varphi)(t)=x_{0}+\\int_{t_{0}}^{t}f(s,\\varphi(s))\\diff s$$\\s\n\n Note that it is well-defined since $\\varphi(s)\\in[x_{0}-b,x_{0}+b]$. To show $T\\varphi\\in X$, one requires $(T\\varphi)(t)\\in[x_{0}-b,x_{0}+b]$. Let $M=\\sup\\brc{\\abs{f(t,x)}\\srm (t,x)\\in R}$, then for all $t\\in[t_{0}-a',t_{0}+a']$,\n\n $$\\begin{aligned}[t]\n \\abs{(T\\varphi)(t)-x_{0}}&=\\abs{\\int_{t_{0}}^{t}f(s,\\varphi(s))\\diff s}\\\\\n &\\leq M\\abs{t-t_{0}}\\leq Ma'\n \\end{aligned}$$\\s\n\n Choose $00$. There exists a unique solution $\\mathbf{x}\\in C^{1}[t_{0}-a',t_{0}+a']$ with\n\n $$\\mathbf{x}(t)\\in[x_{1}-b,x_{1}+b]\\times\\cdots\\times[x_{n}-b,x_{n}+b]$$\\s\n\n for all $t\\in[t_{0}-a',t_{0}+a']$ to the initial value problem, where $a'$ satisfies\n\n $$00$, there exists $\\delta>0$ such that $\\abs{f(x)-f(y)}<\\epsilon$ for all $f\\in C$ and $x,y\\in X$ where $d(x,y)<\\delta$.\n\\end{dft}\\n\n\nIn fact, equicontinuity is based on uniform continuity, but at the same time extends $\\delta$ to fulfill every function $f\\in C$. Therefore, equicontinuity implies every function in $C$ is uniformly continuous. Then, it is simple to see that if $C$ is equicontinuous, any $C'\\subset C$ is also equicontinuous.\\n\n\nThere are other ways to show that a set is equicontinuous. Recall that a function $f$ is Holder continuous if there exists a Holder exponent $\\alpha\\in(0,1)$ such that\n\n$$\\abs{f(x)-f(y)}\\leq L\\abs{x-y}^{\\alpha}$$\\s\n\nfor some constant $L$. $f$ is Lipschitz continuous if the equation holds for $\\alpha=1$. A set $C$ is equicontinuous too if every function $f\\in C$ is Holder continuous or Lipschitz continuous.\n\nAnother method for equicontinuity requires the following definition:\\n\n\n\\begin{dft}\n A set $C$ is said to be \\textbf{convex} (in $\\R^{n}$) if $x+t(y-x)\\in C$ for all $x,y\\in C$ and $t\\in[0,1]$.\n\\end{dft}\\n\n\n\\begin{pst}\n Let $C$ be a subset of $C(\\overline{G})$ where $\\overline{G}$ is a convex in $\\R^{n}$. Suppose that each function in $C$ is differentiable and there is a uniform bound on their partial derivatives:\n\n $$\\nrm{\\frac{\\partial f}{\\partial x_{i}}}_{\\infty}\\leq M$$\\s\n\n for all $f\\in C(\\overline{G})$ and $i$, then $C$ is equicontinuous.\n\\end{pst}\\n\n\n\\begin{pst}\n Let $A=\\brc{z_{j}}$ be a countable set and $f_{n}:A\\to\\R$ where $n=1,2,\\cdots$ be a sequence of functions defined on $A$. Suppose for each $z_{j}\\in A$, $\\brc{f_{n}(z_{j})}$ is a bounded sequence in $\\R$, then there exists a subsequence $\\brc{f_{n_{k}}}$ of $\\brc{f_{n}}$ such that for all $z_{j}\\in A$, $\\brc{f_{n_{k}}(z_{j})}$ is convergent.\n\\end{pst}\n\n\\subsubsection{Ascoli's Theorem}\nBelow is the \\textbf{Ascoli's Theorem}:\\n\n\n\\begin{thm}\n Suppose $G$ is a bounded nonempty open set in $\\R^{m}$, then a set $\\mathcal{E}\\subset C(\\overline{G})=C_{b}(\\overline{G})$ is precompact if $\\mathcal{E}$ is bounded (in supnorm) and equicontinuous.\n\\end{thm}\\n\n\nNote that Ascoli's Theorem remains valid for bounded and equicontinuous subsets of $C(G)$ where $G$ is not necessary to take closure. This is because equicontinuity implies uniform continuity of $G$, which can be further extended to uniform continuity of $\\overline{G}$. However, boundedness of the domain $\\overline{G}$ cannot be removed:\\n\n\n\\begin{exm}\n Let $\\overline{G}=[0,\\infty)\\subset\\R$, then take $\\varphi\\in C^{1}[0,1]$ such that $\\varphi\\not\\equiv 0$ and $\\varphi(x)=0$ when $x\\in[0,1]\\setminus[1/2,3/4]$. Further define\n\n $$f_{n}(x)=\\begin{cases}\n \\varphi(x-n)&\\erm{if }x\\in[n,n+1]\\\\\n 0&\\erm{otherwise}\n \\end{cases}$$\\s\n\n It is easy to check that $f_{n}\\in C(\\overline{G})$ and\n\n $$\\nrm{f_{n}}_{\\infty,\\overline{G}}=\\nrm{\\varphi}_{\\infty,[0,1]}>0$$\\s\n\n Thus $\\mathcal{E}=\\brc{f_{n}}$ is a bounded subset of $C(\\overline{G})$. By chain rule,\n\n $$\\nrm{\\frac{\\diff f_{n}}{\\diff x}}_{\\infty,\\overline{G}}=\\nrm{\\frac{\\diff\\varphi}{\\diff x}}_{\\infty,[0,1]}>0$$\\s\n\n Then \\rpst[\\sctd{3}] states that $\\mathcal{E}$ is also equicontinuous.\\n\n \n Suppose there exists a subsequence $\\brc{f_{n_{j}}}$ of $\\brc{f_{n}}$ converges to the same $f\\in C(\\overline{G})$ in $d_{\\infty}$. In other words, $f_{n_{j}}\\to f$ uniformly on $\\overline{G}$ implies pointwise convergence $f_{n_{j}}(x)\\to f(x)$ for all $x\\in\\overline{G}$. However, for fixed $x$, $f_{n}(x)=0$ for all $n\\geq x$, so it is expected to have\n\n $$\\lim_{j\\to+\\infty}f_{n_{j}}(x)\\to 0$$\\s\n\n which shows that $f(x)=0$ for all $x\\in\\overline{G}$. This is a contraction since\n\n $$0<\\nrm{\\varphi}_{\\infty,[0,1]}=\\nrm{f_{n_{j}}}_{\\infty,\\overline{G}}=\\nrm{f_{n_{j}}-f}_{\\infty,\\overline{G}}\\to 0$$\\s\n\n Therefore $\\mathcal{E}$ is bounded and equicontinuous, but Ascoli's Theorem doesn't hold.\n\\end{exm}\n\n\\subsubsection{Arzela's Theorem}\nBelow is the \\textbf{Arzela's Theorem}, which is the converse of Ascoli's Theorem:\\n\n\n\\begin{thm}\n Suppose $G$ is a bounded nonempty open set in $\\R^{m}$, then every precompact set in $C(\\overline{G})$ must be bounded and equicontinuous.\n\\end{thm}\n\n\\subsection{Applications to Ordinary Differential Equations}\n\\subsubsection{Improvement to Picard-Lindelof Theorem}\nConsider the initial value problem\n\n$$\\begin{cases}\n \\diff x/\\diff t=f(t,x)\\\\\n x(t_{0})=x_{0}\n\\end{cases}$$\\s\n\nwith $f$ being continuous (but not necessary Lipschitz) on $R=[t_{0}-a,t_{0}+a]\\times[x_{0}-b,x_{0}+b]$. Of course this is not expected to give a unique result, but existence can be proved. The idea of proof is as follows:\n\n\\begin{nlist}\n \\item By Weierstrass Approximation Theorem (on $\\R^{2}$), there exists a sequence $\\brc{p_{n}}$ of polynomials such that $d_{\\infty}(p_{n},f)\\to 0$ (in $C(R)$).\n \\item By Picard-Lindelof Theorem, since every $p_{n}$ satisfies Lipschitz condition (uniform in $t$), there exists $a'_{n}>0$ with\n \n $$a'_{n}=\\min\\brc{a,\\frac{b}{M_{n}},\\frac{1}{L_{n}}}$$\\s\n\n where $M_{n}=\\nrm{p_{n}}_{\\infty,R}$ and $L_{n}$ be Lipschitz constant of $p_{n}$ on $R$, such that there exists a unique solution $x_{n}\\in C^{1}[t_{0}-a'_{n},t_{0}+a'_{n}]$ to the approximated initial value problem\n\n $$\\begin{cases}\n \\diff x_{n}/\\diff t=p_{n}(t,x_{n})\\\\\n x_{n}(t_{0})=x_{0}\n \\end{cases}$$\\s\n\n for all $t\\in[t_{0}-a'_{n},t_{0}+a'_{n}]$.\n \\item By Ascoli's Theorem, there exists a convergent subsequence $\\brc{x_{n_{k}}}$ of $\\brc{x_{n}}$ such that $x_{n_{k}}\\to x$ for some function $x(t)$. It is hoped that such $x$ is the required solution.\n\\end{nlist}\n\nHowever, since $f$ is not assumed to satisfy the Lipschitz condition, one cannot expect $\\brc{L_{n}}$ is bounded. In fact, $\\brc{L_{n}}$ is unbounded, otherwise $f$ satisfies Lipschitz condition. Here\n\n$$a'_{n}=\\min\\brc{a,\\frac{b}{M_{n}},\\frac{1}{L_{n}}}\\to 0$$\\s\n\nthen there is no proper interval for existence of the solution. On the other hand, as $p_{n}\\to f$ in $(C(R),d_{\\infty})$, $M_{n}\\leq M$ for some $M>0$. Therefore, in order to implement the plan above, it is required to improve Picard-Lindelof Theorem:\\n\n\n\\begin{pst}\n Under the setting of Picard-Lindelof Theorem, there exists a unique solution $x(t)$ on the interval $[t_{0}-a',t_{0}+a']$ with $x(t)\\in[x_{0}-b,x_{0}+b]$, where $a'$ is any number satisfying\n\n $$00$, $B_{\\epsilon}(x)\\cap E\\neq\\emptyset$.\n\\end{dft}\\n\n\nNote that $X$ is naturally dense in $(X,d)$, and if $E$ is dense in $X$, its closure $\\overline{E}=X$.\\n\n\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a set $E\\subset X$ is said to be \\textbf{nowhere dense} if its closure does not contain any ball. In other words, $\\overline{E}$ has empty interior.\n\\end{dft}\\n\n\n\\begin{exm}\n Set of integers $\\Z$ is nowhere dense in $\\R$. However, although set of rationals $\\Q$ has empty interior, its closure $\\overline{\\Q}=\\R$ has nonempty interior, so $\\Q$ is not nowhere dense.\n\\end{exm}\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space and $E\\subset X$ be a set, then $E$ is nowhere dense if and only if $X\\setminus\\overline{E}$ is dense in $X$.\\n\n\n \\prf If $E$ is nowhere dense, for all $x\\in X$ and any $r>0$, $B_{r}(x)\\not\\subset\\overline{E}$, which implies $B_{r}(x)\\cap(X\\setminus\\overline{E})\\neq\\emptyset$, so $X\\setminus\\overline{E}$ is dense. The converse follows the reverse order.\n\\end{pst}\\n\n\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a point $x\\in X$ is called an \\textbf{isolated point} if $\\brc{x}$ is open in $X$.\n\\end{dft}\\n\n\nNote that $\\brc{x}$ is always closed in a metric space. Therefore, $\\brc{x}$ is both open and closed if and only if $x$ is an isolated point.\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space, then the following applies:\n\n \\begin{alist}\n \\item If $E$ is nowhere dense in $X$, $\\overline{E}$ is nowhere dense in $X$. Also, $E'$ is nowhere dense in $X$ if $E'\\subset E$.\n \\item The union of finitely many nowhere dense sets in $X$ is nowhere dense in $X$.\n \\item If $(X,d)$ has no isolated point, then every finite set is nowhere dense.\n \\end{alist}\n\\end{pst}\\n\n\nNow consider the following example in infinite dimensional normed spaces:\\n\n\n\\begin{exm}\n Let $M[a,b]$ be a space of bounded functions on $[a,b]$, then\n \n $$\\nrm{f}_{\\infty}=\\underset{[a,b]}{\\sup}\\abs{f(x)}$$\\s\n\n is well-defined and is a norm on $M[a,b]$. It is clear that $(C[a,b],d_{\\infty})$ is a metric (and vector) subspace of $(M[a,b],d_{\\infty})$. Show that $C[a,b]$ is nowhere dense in $M[a,b]$.\\n\n\n \\ans Note that $C[a,b]$ is closed because uniform limit of continuous functions is continuous. It is left to show that for all $B_{\\epsilon}^{\\infty}(f)\\subset M[a,b]$,\n\n $$B_{\\epsilon}^{\\infty}(f)\\cap(M[a,b]\\setminus C[a,b])\\neq\\emptyset$$\\s\n\n If $f\\in M[a,b]\\setminus C[a,b]$, the result is already achieved. For any $f\\in C[a,b]$, let\n\n $$g(x)=\\begin{cases}\n f(x)+\\epsilon/2\\erm{if }x\\in[a,b]\\cap\\Q\\\\\n f(x)-\\epsilon/2\\erm{if }x\\in[a,b]\\setminus\\Q\n \\end{cases}$$\\s\n\n such that $\\nrm{g-f}_{\\infty}=\\epsilon/2$ implies $g\\in B_{\\epsilon}^{\\infty}(f)$. Since both $[a,b]\\cap\\Q$ and $[a,b]\\setminus\\Q$ are dense in $[a,b]$,\n\n $$\\underset{x\\to a}{\\limsup}g(x)=f(a)+\\frac{\\epsilon}{2}$$\\s\n\n and\n\n $$\\underset{x\\to a}{\\liminf}g(x)=f(a)-\\frac{\\epsilon}{2}$$\\s\n\n shows that $g\\in M[a,b]\\setminus C[a,b]$. Therefore $B_{\\epsilon}^{\\infty}(f)\\cap(M[a,b]\\setminus C[a,b])\\neq\\emptyset$, and $C[a,b]$ is nowhere dense in $M[a,b]$.\n\\end{exm}\n\n\\subsubsection{First Category and Second Category}\n\\begin{dft}\n Let $(X,d)$ be a metric space, then a set $E\\subset X$ is called \\textbf{first category} (or \\textbf{meager}) if it can be expressed as a countable union of nowhere dense sets. If $E$ is not of first category, then $E$ is called \\textbf{second category}.\\n\n\n $E$ is said to be \\textbf{residual} if its complement is of first category.\n\\end{dft}\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space, then the following applies:\n\n \\begin{alist}\n \\item Every subset of a set of first category is of first category.\n \\item The union of countable many sets of first category is of first category.\n \\item If $(X,d)$ has no isolated point, every countable subset of $X$ is of first category.\n \\end{alist}\n\\end{pst}\\n\n\nWith the proposition above, a similar proposition for residual sets can be made by taking complements:\\n\n\n\\begin{pst}\n Let $(X,d)$ be a metric space, then the following applies:\n\n \\begin{alist}\n \\item Every subset containing a residual set is residual.\n \\item The intersection of countable many residual sets is residual.\n \\item If $(X,d)$ has no isolated point, complement of any countable set is residual.\n \\end{alist}\n\\end{pst}\\n\n\n\\begin{exm}\n Let $(\\R,d_{1})$ be a metric space. Since $\\R$ has no isolated points, $\\brc{q}$ is nowhere dense for any $q\\in\\Q$, so $\\Q$ is of first category since it is a countable union of $\\brc{q}$. On the other hand, the set of irrational numbers $\\I=\\R\\setminus\\Q$ is residual in $\\R$.\n\\end{exm}\n\n\\subsubsection{General Theorem}\nBelow is the \\textbf{Baire Category Theorem}:\\n\n\n\\begin{thm}\n Any set of first category in a complete metric space has empty interior. In other words, any countable intersection of open dense sets in a complete metric space is dense.\n\\end{thm}\\n\n\nWith Baire Category Theorem, there are some corollaries to follow:\\n\n\n\\begin{crl}\n Let $(X,d)$ be a complete metric space. Suppose that $X=\\bigcup_{n=1}^{\\infty}E_{n}$ with $E_{n}$ are closed subsets. Then at least one of there $E_{n}$ has nonempty interior.\n\\end{crl}\\n\n\n\\begin{crl}\n A set of first category in a complete metric space cannot be a residual set, and vice versa.\n\\end{crl}\n\n\\subsubsection{Applications of Baire Category Theorem}\n\\begin{pst}\n Let $f\\in C[a,b]$ be differentiable at $x$, then it is Lipschitz continuous at $x$.\\n\n\n \\prf By assumption, for any $\\epsilon=1>0$, there exists $\\delta_{0}>0$ such that for all $y\\in(x-\\delta_{0},x+\\delta_{0})\\setminus\\brc{x}$ and $y\\in[a,b]$,\n\n $$\\abs{\\frac{f(y)-f(x)}{y-x}-f'(x)}<1$$\\s\n\n implies\n\n $$\\abs{f(y)-f(x)}\\leq(1+\\abs{f'(x)})\\abs{y-x}$$\\s\n\n for all $y\\in(x-\\delta_{0},x+\\delta_{0})\\cap[a,b]$. If $[a,b]\\setminus(x-\\delta_{0},x+\\delta_{0})=\\emptyset$, it is already done. Consider for $y\\in[a,b]\\setminus(x-\\delta_{0},x+\\delta_{0})$ if such set is nonempty, $\\abs{y-x}\\geq\\delta_{0}$, hence\n\n $$\\begin{aligned}[t]\n \\abs{f(y)-f(x)}&\\leq\\abs{f(y)}+\\abs{f(x)}\\\\\n &\\leq 2\\nrm{f}_{\\infty}\\\\\n &\\leq\\frac{2\\nrm{f_{\\infty}}}{\\delta_{0}}\\abs{y-x}=L'\\abs{y-x}\n \\end{aligned}$$\\s\n\n Finally, let $L=\\max\\brc{1+\\abs{f'(x)},L'}$, then $\\abs{f(y)-f(x)}\\leq L\\abs{y-x}$ for all $y\\in[a,b]$.\n\\end{pst}\\n\n\nWith the proposition above, the following theorem can be introduced:\\n\n\n\\begin{thm}\n The set of all continuous, nowhere differentiable functions forms a residual set in $C[a,b]$ and hence dense in $C[a,b]$.\n\\end{thm}\n\n\\input{sty/footer.sty}\n\n\\begin{reflist}\n \\item Elias M. Stein, Rami Shakarchi, \\textit{Fourier Analysis: An Introduction (Princeton Lectures in Analysis)}, Princeton, 2003\n \\item Walter Rudin, \\textit{Principles of Mathematical Analysis}, McGraw-Hill (3rd Edition), 1976\n\\end{reflist}\n\n\\end{document}", "meta": {"hexsha": "355cfb1640fc035198fe90f5b8c2615a1fc362e4", "size": 88017, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/math3060.tex", "max_stars_repo_name": "nablamath/notes", "max_stars_repo_head_hexsha": "e268c0e7f723ee870ca6cacf93fc91dca4368e6b", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-03-05T16:43:38.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-05T16:43:38.000Z", "max_issues_repo_path": "tex/math3060.tex", "max_issues_repo_name": "nablamath/notes", "max_issues_repo_head_hexsha": "e268c0e7f723ee870ca6cacf93fc91dca4368e6b", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/math3060.tex", "max_forks_repo_name": "nablamath/notes", "max_forks_repo_head_hexsha": "e268c0e7f723ee870ca6cacf93fc91dca4368e6b", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.5174528302, "max_line_length": 506, "alphanum_fraction": 0.6266971153, "num_tokens": 34790, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154239836484144, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.6150240662740631}} {"text": "\n\n\\section{Calculation}\n\nWe start on the $E_2$ page. \nOn the zeroth column, we get only the cohomology of $U(2)$, hence we at least see\n\n\\adjustbox{scale=0.9, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 & & & & & & & & \\\\\n3 & a_3 & & & & & & & & \\\\\n2 & 0 & & & & & & & & \\\\\n1 & a_1 & & & & & & & & \\\\\n0 & 1 & & & & & & & & \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\n\\end{tikzcd}\n}\n\nSince we know that $H^*(V_2(\\mathbb{C}^4))$ has no elements in degree $1$, \nwe know that $H^2(G_2(\\mathbb{C}^4))\\cong H^1(V_2(\\mathbb{C}^4))$, \nas the generator $a_1$ has to be killed by a differential. \nCall the generator for the group $H^2(G_2(\\mathbb{C}^4))$ that is the image of $a_1$ for $x$. \nBy multiplications by the other generators and itself we then know we have\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 & & a_1a_3x & & a_1a_3x^2 & & a_1a_3x^3 & & a_1a_3x^4 \\\\\n3 & a_3 & & a_3x & & a_3x^2 & & a_3x^3 & & a_3x^4 \\\\\n2 & & & & & & & & & \\\\\n1 & a_1 \\arrow[rrd, \"\\cong\"] & & a_1x & & a_1x^2 & & a_1x^3 & & a_1x^4 \\\\\n0 & 1 & & x & & x^2 & & x^3 & & x^4 \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nWe know that the differential is a derivation, \nhence we get that $d_2(a_1x)=d_2(a_1)x+(-1)^{|x|}a_1d_2(x) = d_2(a_1)x = x^2$ since $d_2(x)=0$. \nThe same calculation for higher powers of $x$ and for $a_1a_3$ and $a_1a_3x$, \ngives us isomorphisms all the way to the right\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 \\arrow[rrd, \"\\cong\"] & & a_1a_3x \\arrow[rrd, \"\\cong\"] & & a_1a_3x^2 \\arrow[rrd, \"\\cong\"] & & a_1a_3x^3 \\arrow[rrd, \"\\cong\"] & & a_1a_3x^4 \\\\\n3 & a_3 & & a_3x & & a_3x^2 & & a_3x^3 & & a_3x^4 \\\\\n2 & & & & & & & & & \\\\\n1 & a_1 \\arrow[rrd, \"\\cong\"] & & a_1x \\arrow[rrd, \"\\cong\"] & & a_1x^2 \\arrow[rrd, \"\\cong\"] & & a_1x^3 \\arrow[rrd, \"\\cong\"] & & a_1x^4 \\\\\n0 & 1 & & x & & x^2 & & x^3 & & x^4 \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nHere we see a problem. \nThe generator $a_3$ never gets killed by a differential, \nbut the cohomology of $V_2(\\mathbb{C}^4))$ has no elements in degree $3$, \nso it cant survive. \nThe only degree a differential starting from $a_3$ can land in is $4$, \nand the only possible non-zero cohomology in degree 4 comes from $H^4(G_2(\\mathbb{C}^4))$, \nwhich by the likes of it so far has already been killed by a $d_2$ differential. \nHence we got to have a $d_4$ differential hitting another generator in $H^4(G_2(\\mathbb{C}^4))$, \nhence we need higher dimensional cohomology in this degree. \nWe call this new generator which is the image of $a_3$ under the $d_3$ differential for $y$. \nLooking back at the $E_2$ page, we then have new elements\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 \\arrow[rrd, \"\\cong\"] & & a_1a_3x \\arrow[rrd, \"\\cong\"] & & {a_1a_3x^2, a_1a_3y} \\arrow[rrd, \"\\cong\"] & & {a_1a_3x^3, a_1a_3xy} \\arrow[rrd, \"\\cong\"] & & {a_1a_3x^4, a_1a_3y^2} \\\\\n3 & a_3 & & a_3x & & {a_3x^2, a_3y} & & {a_3x^3, a_3xy} & & {a_3x^4, a_3y^2} \\\\\n2 & & & & & & & & & \\\\\n1 & a_1 \\arrow[rrd, \"\\cong\"] & & a_1x \\arrow[rrd, \"\\cong\"] & & {a_1x^2, a_1y} \\arrow[rrd, \"\\cong\"] & & {a_1x^3, a_1xy} \\arrow[rrd, \"\\cong\"] & & {a_1x^4, a_1y^2} \\\\\n0 & 1 & & x & & {x^2, y} & & {x^3, xy} & & {x^4, y^2} \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nNow, what does this new information and new elements give us? \nBy the same calculation as previously, we get that $d_2(a_1y) = xy$, \nbut here have more information. \nWe know that $H^*(V_2(\\mathbb{C}^4))$ has a generator in degree $5$, \nand this new generator $a_1y$ is the only generator left in total degree $5$ in the spectral sequence. \nThus it can't die by a $d_2$ differential. \nAnd since it's image is $xy$, this image has to be zero, which gives us the first of two relations on the cohomology ring of our Grassmannian, namely $xy=0$. \nHence, as far as we know so far, our $E_2$ page looks like this: \n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 \\arrow[rrd, \"\\cong\"] & & a_1a_3x \\arrow[rrd, \"\\cong\"] & & {a_1a_3x^2, a_1a_3y} \\arrow[rrd, \"\\cong\"] & & a_1a_3x^3 \\arrow[rrd, \"\\cong\"] & & {a_1a_3x^4, a_1a_3y^2} \\\\\n3 & a_3 & & a_3x & & {a_3x^2, a_3y} & & a_3x^3 & & {a_3x^4, a_3y^2} \\\\\n2 & & & & & & & & & \\\\\n1 & a_1 \\arrow[rrd, \"\\cong\"] & & a_1x \\arrow[rrd, \"\\cong\"] & & {a_1x^2, a_1y} \\arrow[rrd, \"\\cong\"] & & a_1x^3 \\arrow[rrd, \"\\cong\"] & & {a_1x^4, a_1y^2} \\\\\n0 & 1 & & x & & {x^2, y} & & x^3 & & {x^4, y^2} \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nAs far as I'm aware, we can't squeeze any more information out of the $E_2$ page yet. \nLet us pass to the $E_3$ page and see which generator we are left with after all the isomorphisms kill the generators. \nWe have\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & & & & & a_1a_3y & & & & {a_1a_3y^2} \\\\\n3 & a_3 & & & & a_3y & & & & a_3y^2 \\\\\n2 & & & & & & & & & \\\\\n1 & & & & & a_1y & & & & {a_1y^2} \\\\\n0 & 1 & & & & y & & & & y^2 \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nand we see that all the $d_3$ differentials either start or land in places which are zero, \nhence $E_3 = E_4$. On the $E_4$ page however, \nwe know that we at least have one differential, \nnamely the one we used to justify the generator $y$'s existence. \nBy the derivation property of the differentials, \nwe also get that $d_4(a_1a_3y)=a_1y^2$, \nhence we at least have an $E_4$ page looking like\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & & & & & a_1a_3y \\arrow[rrrrddd, \"\\cong\"] & & & & a_1a_3y^2 \\\\\n3 & a_3 \\arrow[rrrrddd, \"\\cong\"] & & & & a_3y & & & & a_3y^2 \\\\\n2 & & & & & & & & & \\\\\n1 & & & & & a_1y & & & & a_1y^2 \\\\\n0 & 1 & & & & y & & & & y^2 \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nWe are also lucky enough to know that $H^*(V_2(\\mathbb{C}^4))$ has a generator in degree $7$, \nwhich we see is $a_3y$ since it is the only one left in the correct total degree. \nBut what happens to the differential $d_4(a_3y) = y^2$ you ask? Do we have $y^2=0$ then, \nsame as last time? \nActually, no. \nSince we know that $a_1y$ and $a_3y$ both are non-zero generators in $H^*(V_2(\\mathbb{C}^4))$, \ntheir product $a_1a_3y^2$ is also a generator in degree $12$, \nhence it has to be non-zero, \nand then $y^2$ has to be non-zero in $H^8(G_2(\\mathbb{C}^4))$. \nBut, it still has to die in the spectral sequence though, \nand the solution is the second relation on the cohomology ring, \nnamely $x^4=y^2$. \nThis ensures that $y^2$ is already killed on the $E_2$ page, \nand hence that $a_3y$ isn't killed on the $E_4$ page.\n\nTo retrospectively correct some things now that we have all the information, \nwe first note that we were lucky we didn't bother writing more that $8$ columns, \nbecause everything in row $0$ and columns above $8$ either is a product of $x$'s and $y$'s, \nor contain powers of $x$ greater or equal to $5$ or powers of $y$ greater or equal to $3$, \nmaking $x^5 = xx^4=xy^2=(xy)y=0$ and $y^3=yx^4=0$. \nHence our generators $a_1x^4$ and $a_1a_3x^4$ never was killed by any $d_2$ differentials, \nsince their images are $x^5=0$ and $a_3x^5=0$ respectively. \nHence we still have $a_1x^4=a_1y^2$ and $a_1a_3x^4=a_1a_3y^2$ on the $E_4$ page. \nAlso, we get that $a_3x^4=a_3y^2$, \nwhich means that $a_3y^2$ was killed by a $d_2$ differential on the $E_2$ page, \nand hence it does not show up on the $E_4$ page. \nThe final form of the $E_2$ page is then\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & a_1a_3 \\arrow[rrd] & & a_1a_3x \\arrow[rrd] & & {a_1a_3x^2, a_1a_3y} \\arrow[rrd] & & a_1a_3x^3 \\arrow[rrd] & & a_1a_3x^4=a_1a_3y^2 \\\\\n3 & a_3 & & a_3x & & {a_3x^2, a_3y} & & a_3x^3 & & a_3x^4=a_3y^2 \\\\\n2 & & & & & & & & & \\\\\n1 & a_1 \\arrow[rrd] & & a_1x \\arrow[rrd] & & {a_1x^2, a_1y} \\arrow[rrd] & & a_1x^3 \\arrow[rrd] & & a_1x^4=a_1y^2 \\\\\n0 & 1 & & x & & {x^2, y} & & x^3 & & x^4=y^2 \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nand the final form of the $E_4$ page is\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & & & & & a_1a_3y \\arrow[rrrrddd, \"\\cong\"] & & & & a_1a_3y^2 \\\\\n3 & a_3 \\arrow[rrrrddd, \"\\cong\"] & & & & a_3y & & & & \\\\\n2 & & & & & & & & & \\\\\n1 & & & & & a_1y & & & & a_1y^2 \\\\\n0 & 1 & & & & y & & & & \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nSince all higher degree differentials, \ni.e. $d_5$ and above miss any generators due to being too long, \nwe have $E_5=E_{\\infty}$. \nWe can also see this because we have arrived at the correct cohomology ring for $V_2(\\mathbb{C}^4)$, \nnamely the final page\n\n\\adjustbox{scale=0.8, center}{\n\\begin{tikzcd}[column sep=small]\n4 & & & & & & & & & a_1a_3x^4=a_1a_3y^2 \\\\\n3 & & & & & a_3y & & & & \\\\\n2 & & & & & & & & & \\\\\n1 & & & & & a_1y & & & & \\\\\n0 & 1 & & & & & & & & \\\\\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \n\\end{tikzcd}\n}\n\nHence we are done, \nand we have found the correct cohomology ring for our Grassmannian $G_2(\\mathbb{C}^4)$. \nBy all our extensive calculations and work, \nwe finally can say that $H^*(G_2(\\mathbb{C}^4)) = \\mathbb{Z}(x,y)/(xy, x^4-y^2)$ or if we add the degrees into the names of the generators, \n$H^*(G_2(\\mathbb{C}^4)) = \\mathbb{Z}(a_2, a_4)/(a_2a_4, a_2^4-a_4^2)$. \nHence the calculation is finished, and we are done. ", "meta": {"hexsha": "826905d5ad976c403741e9dfda4e207cf731c04d", "size": 12893, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/calculation.tex", "max_stars_repo_name": "torgeiraamboe/cohomology_ring_of_grassmannian", "max_stars_repo_head_hexsha": "72268ac4af69ed167b52d8e7f1e6b69ed4632623", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/calculation.tex", "max_issues_repo_name": "torgeiraamboe/cohomology_ring_of_grassmannian", "max_issues_repo_head_hexsha": "72268ac4af69ed167b52d8e7f1e6b69ed4632623", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/calculation.tex", "max_forks_repo_name": "torgeiraamboe/cohomology_ring_of_grassmannian", "max_forks_repo_head_hexsha": "72268ac4af69ed167b52d8e7f1e6b69ed4632623", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 63.5123152709, "max_line_length": 195, "alphanum_fraction": 0.427906616, "num_tokens": 4615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511579973932, "lm_q2_score": 0.7185943805178139, "lm_q1q2_score": 0.6150098326965904}} {"text": "\\section{The \\sortheap algorithm}\n\\Label{sec:sortheap}\n\nWhereas in the \\cxx Standard Library \\cite[\\S 28.7.7.4]{cxx-17-draft}\n\\sortheap works on a range of random access iterators,\nour version operates on an array of \\valuetype.\nWe therefore use the following signature for \\sortheap\n\n\\begin{lstlisting}[style = acsl-block]\n\n void sort_heap(value_type* a, size_type n);\n\\end{lstlisting}\n\nThe function \\sortheap rearranges the elements of a given heap \\inl{a[0..n-1]}\nin increasing order.\nThus, applying \\sortheap to the heap in Figure~\\ref{fig:heap-array}\nproduces the increasing array in Figure~\\ref{fig:sortheap-array-post}.\n\n\\begin{figure}[hbt]\n\\centering\n\\includegraphics[width=0.65\\linewidth]{Figures/sort_heap_array_post.pdf}\n\\caption{\\Label{fig:sortheap-array-post}Array after the call of \\sortheap}\n\\end{figure}\n\n\\FloatBarrier\n\n\\subsection{Formal specification of \\sortheap}\n\nThe following listing shows our specification of \\sortheap.\nThe formal specification of \\sortheap must ensure that the\nresulting array is increasing.\nFurthermore the multiset contained by the array must be the same\nas in the pre-state of the function.\nThe postconditions \\inl{increasing} and \\inl{reorder} express these properties, respectively.\nThe specification effort is relatively simple because we can reuse\n\n\\input{Listings/sort_heap.h.tex}\n\n\\clearpage\n\n\\subsection{Implementation of \\sortheap}\n\nThe implementation of \\sortheap is relatively simple because it relies on\n\\specref{popheap} performing essential work.\nOur implementation of \\sortheap repeatedly calls \\popheap\nto extract the maximum of the shrinking heap and adding it\nto the part of the array that is already in increasing order.\nThe loop invariants of \\sortheap describe the content of the array\nin two parts.\nThe first \\inl{i} elements form a heap and are described by the \\inl{heap}\ninvariant.\nThe last \\inl{n-i} elements are already arranged in increasing order.\n\nAs already mentioned in the introduction of Chapter~\\ref{cha:binary-search},\nwe use the predicate \\logicref{WeaklyIncreasing} for the loop annotation \\inl{increasing}.\nThus, after leaving the loop we have in fact ``only'' shown that \\inl{WeaklyIncreasing(a, n)}\nholds.\nIn order to derive from this fact the final assertion \\inl{increasing} that uses\nthe predicate \\logicref{Increasing} we rely on lemma \\logicref{WeaklyIncreasingIncreasing}.\n\n\\input{Listings/sort_heap.c.tex}\n\nTo verify the property \\inl{reorder} we rely on the lemmas \\logicref{MultisetReorder}\nthat express that the properties \n\n\\begin{itemize}\n\\item \\inl{MultisetReorder\\{K,L\\}(a, 0, i)} and\n\\item \\inl{Unchanged\\{Old,Here\\}(a, i, n)}\n\\end{itemize}\n\nimply the desired loop invariant \\inl{MultisetReorder\\{K,L\\}(a, 0, n)}.\n\n%\\clearpage\n\n", "meta": {"hexsha": "8125998ca8f4deb18616a28df8dab9b6a7df3b0c", "size": 2731, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Informal/heap/sort_heap.tex", "max_stars_repo_name": "fraunhoferfokus/acsl-by-example", "max_stars_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 90, "max_stars_repo_stars_event_min_datetime": "2017-06-14T04:17:53.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-07T06:07:36.000Z", "max_issues_repo_path": "Informal/heap/sort_heap.tex", "max_issues_repo_name": "fraunhoferfokus/acsl-by-example", "max_issues_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 22, "max_issues_repo_issues_event_min_datetime": "2017-10-18T13:30:41.000Z", "max_issues_repo_issues_event_max_datetime": "2021-06-17T07:10:16.000Z", "max_forks_repo_path": "Informal/heap/sort_heap.tex", "max_forks_repo_name": "fraunhoferfokus/acsl-by-example", "max_forks_repo_head_hexsha": "d8472670150fb3ff4360924af2d0eb14bc80d1e2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 19, "max_forks_repo_forks_event_min_datetime": "2017-06-21T13:49:31.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-31T16:27:06.000Z", "avg_line_length": 36.4133333333, "max_line_length": 93, "alphanum_fraction": 0.7832295862, "num_tokens": 755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7185943805178139, "lm_q2_score": 0.855851143290548, "lm_q1q2_score": 0.6150098221283341}} {"text": "\\documentclass[12pt]{cdblatex}\n\\usepackage{examples}\n\n\\begin{document}\n\n\\section*{Curvature of a 2-sphere}\n\nThis examples uses standard methods to compute the scalar curvature of a 2-sphere.\n\n\\begin{cadabra}\n {\\theta, \\varphi}::Coordinate.\n {\\alpha, \\beta, \\gamma, \\delta, \\rho, \\sigma, \\mu, \\nu, \\lambda}::Indices(values={\\varphi, \\theta}, position=independent).\n\n \\partial{#}::PartialDerivative.\n\n g_{\\alpha\\beta}::Metric.\n g^{\\alpha\\beta}::InverseMetric.\n\n Chr := \\Gamma^{\\alpha}_{\\mu\\nu} -> 1/2 g^{\\alpha\\beta} ( \\partial_{\\nu}{g_{\\beta\\mu}}\n + \\partial_{\\mu}{g_{\\beta\\nu}}\n - \\partial_{\\beta}{g_{\\mu\\nu}} ).\n\n Rabcd := R^{\\rho}_{\\sigma\\mu\\nu} -> \\partial_{\\mu}{\\Gamma^{\\rho}_{\\sigma\\nu}}\n - \\partial_{\\nu}{\\Gamma^{\\rho}_{\\sigma\\mu}}\n + \\Gamma^{\\rho}_{\\beta\\mu} \\Gamma^{\\beta}_{\\sigma\\nu}\n - \\Gamma^{\\rho}_{\\beta\\nu} \\Gamma^{\\beta}_{\\sigma\\mu}.\n\n Rab := R_{\\sigma\\nu} -> R^{\\rho}_{\\sigma\\rho\\nu}.\n\n R := R -> R_{\\sigma\\nu} g^{\\sigma\\nu}.\n\n gab:={ g_{\\theta\\theta} = r**2,\n g_{\\varphi\\varphi} = r**2 \\sin(\\theta)**2 }. # cdb(gab,gab)\n\n complete (gab, $g^{\\alpha\\beta}$) # cdb(iab,gab)\n\n evaluate (Chr, gab, rhsonly=True) # cdb(Chr,Chr)\n\n substitute (Rabcd, Chr)\n evaluate (Rabcd, gab, rhsonly=True) # cdb(Rabcd,Rabcd)\n\n substitute (Rab, Rabcd)\n evaluate (Rab, gab, rhsonly=True) # cdb(Rab,Rab)\n\n substitute (R, Rab)\n evaluate (R, gab, rhsonly=True) # cdb(R,R)\n\\end{cadabra}\n\n\\begin{minipage}[t]{0.65\\textwidth}\n\\begin{align*}\n &\\cdb{iab}\\\\[10pt]\n &\\cdb{Chr}\\\\[10pt]\n &\\cdb{Rabcd}\\\\[10pt]\n &\\cdb{Rab}\\\\[10pt]\n &\\cdb{R}\n\\end{align*}\n\\end{minipage}\n\\hskip 1cm\n\\lower16pt\\hbox{%\n\\begin{minipage}[t]{0.35\\textwidth}\n\\begin{latex}\n \\begin{align*}\n &\\cdb{iab}\\\\[10pt]\n &\\cdb{Chr}\\\\[10pt]\n &\\cdb{Rabcd}\\\\[10pt]\n &\\cdb{Rab}\\\\[10pt]\n &\\cdb{R}\n \\end{align*}\n\\end{latex}\n\\end{minipage}}\n\n\\end{document}\n", "meta": {"hexsha": "58260567aecbd6f60c2a447a1d4e925475f4fbfa", "size": 2215, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "cadabra/examples/example-03.tex", "max_stars_repo_name": "leo-brewin/hybrid-latex", "max_stars_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2018-10-12T06:31:49.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T23:16:08.000Z", "max_issues_repo_path": "cadabra/examples/example-03.tex", "max_issues_repo_name": "leo-brewin/hybrid-latex", "max_issues_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "cadabra/examples/example-03.tex", "max_forks_repo_name": "leo-brewin/hybrid-latex", "max_forks_repo_head_hexsha": "2debaf3f97eb551928d08dc4baded7ef7a4ab29a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-06-27T03:29:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-30T17:17:18.000Z", "avg_line_length": 30.3424657534, "max_line_length": 125, "alphanum_fraction": 0.5060948081, "num_tokens": 763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.863391624034103, "lm_q2_score": 0.7122321842389469, "lm_q1q2_score": 0.6149353022394207}} {"text": "\\chapter{Solomonoff theory}\n\nWhy was Raymond J. Solomonoff \\cite{SolAlpProb2011, GacsVitanyiSolomonoff}\ninterested in predicting sequences of bits?\n\nMarcus Hutter approached intelligence from \\emph{algorithmic} complexity theory (Solomonoff induction)\n\\cite{DefineMachIntel}.\nWarren D. Smith approached intelligence from \\emph{computational} complexity theory\n(NP-completeness)\n\\cite{WdsIntel, WdsIntelSlide}\n\n\\section{Generalization of Solomonoff theory}\n\nLet \\(\\Bit = \\{0,1\\}\\).\nLet \\(\\Bits = \\Bit^*\\) where \\(*\\) is the Kleene star.\nLet \\(f : \\Bits \\to \\Bits\\).\nLet \\(p : \\Bits \\to [0,1]\\) be the input distribution.\nDefine \\(q~y\\) as the probability of finding an input \\(x\\) such that \\(f~x = y\\).\nSolomonoff used a universal Turing machine for \\(f\\),\nbut I'm interested on the consequences of relaxing this constraint.\n\nBijection allows us to use \\(\\Nat\\) or any other countable set instead of \\(\\Bits\\).\nLet \\(f : \\Nat \\to \\Nat\\) be a function.\nWe say that \\emph{\\(x\\) \\(f\\)-explains \\(y\\)} iff \\(f~x = y\\).\nLet \\(p : \\Nat \\to [0,1]\\) be the input distribution.\nDefine \\(q~y\\) as the probability of finding an \\(x\\) such that \\(f~x = y\\).\nWe call an input such as \\(x\\) a \\emph{hypothesis}.\n\n\\newcommand\\preimage{\\fun{preimage}}\n\nLet \\(f\\) be a binary relation.\nWe define \\(\\preimage~f~y = \\{ x ~|~ f~x~y \\}\\).\n\n\\section{Generalizing Kolmogorov complexity}\n\nSchmidh\\\"uber generalized Kolmogorov complexity to super Turing machines \\cite{SchmidhuberKolmogorov}.\n\nThe Kolmogorov complexity of a string\nis the length of the shortest program that generates that string.\nThis complexity depends on the machine.\nFormally, we define \\(\\kolmogorov~f~y\\),\nthe Kolmogorov complexity of \\(y\\) with respect to \\(f\\),\nas the length of the shortest \\(x\\) that satisfies \\(f~x = y\\).\n\nWe can relax the constraint of \\(f\\) so that \\(f\\)\nnow only needs to be a surjective relation.\nWe can relax it further so that it is a relation.\nHowever, if \\(f\\) is not surjective,\nthe Kolmogorov complexity will be partial\n(undefined for some inputs).\nThat seems to be as general as logic allows.\n\nWe can relax the measure \\(m : A \\to \\Real\\) into a function \\(m : A \\to B\\)\nand an ordering \\(c\\).\n\n\\[\n \\kolmogorov~c~m~f = \\minimumBy~c \\circ \\map~m \\circ \\preimage~f\n\\]\n\nNote the notation:\nIf \\(f\\) is a binary relation,\nwe overload the notation \\(f\\)\nto also mean the binary predicate\nthat is the indicator function of the relation \\(f\\).\nWe conflate a predicate and its extension.\nWe write \\(f~x~y\\) to mean \\((x,y) \\in f\\).\nThe type of the expression \\(f~x~y\\) is \\(\\Bool\\).\n\nWe can generalize Kolmogorov complexity.\nInstead of a space of strings and the length of strings,\nwe can use a measure space.\nFurthermore, we can generalize the function into relation.\nLet \\(m : A \\to \\Real\\) be a measure on \\(A\\).\nLet \\(f \\subseteq A \\times A\\) be a binary relation.\nFormally,\n\\[\n \\kolmogorov~m~f~y = \\minimum~(\\map~m~(\\preimage~f~y))\n\\]\nor, more pointfreely,\n\\begin{equation}\n \\kolmogorov~m~f = \\minimum \\circ \\map~m \\circ \\preimage~f\n\\end{equation}\n\nWe define \\(\\kolmogorov~m~f~y\\) as the measure of an \\(x\\)\nthat both minimizes \\(m\\) and satisfies \\(f~x~y\\).\nTherefore \\(\\kolmogorov\\) is a constrained optimization problem.\nThe classical Kolmogorov complexity is a special case of \\(\\kolmogorov\\)\nwhere \\(m\\) computes the length of a bitstring\nand \\(f\\) is a universal Turing machine.\n\nThe Kolmogorov complexity is incomputable due to the halting problem.\nIt does not even have a brute-force algorithm.\nHowever, Levin search allows us to approximate it.\n", "meta": {"hexsha": "2e77efbc9b281c9d6bbf6bf152b75ed47bf21139", "size": 3541, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "research/solomonoff.tex", "max_stars_repo_name": "edom/work", "max_stars_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_stars_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "research/solomonoff.tex", "max_issues_repo_name": "edom/work", "max_issues_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_issues_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-12-02T18:37:37.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-16T00:55:32.000Z", "max_forks_repo_path": "research/solomonoff.tex", "max_forks_repo_name": "edom/work", "max_forks_repo_head_hexsha": "df55868caa436efc631e145a43e833220b8da1d0", "max_forks_repo_licenses": ["Apache-2.0", "CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-10-02T15:20:22.000Z", "max_forks_repo_forks_event_max_datetime": "2018-10-02T15:20:22.000Z", "avg_line_length": 37.670212766, "max_line_length": 102, "alphanum_fraction": 0.7127929963, "num_tokens": 1064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.863391617003942, "lm_q2_score": 0.7122321781307374, "lm_q1q2_score": 0.6149352919585369}} {"text": "\n\\subsection{The Fisher equation}\n\nThe Fisher equation shows the relationship between the real and nominal interest rates.\n\n\\((1+i)=(1+r)(1+\\pi )\\)\n\nFor small values:\n\n\\(i\\approx r+\\pi \\)\n\n", "meta": {"hexsha": "63770a9ade598a3dae18e29b348eb3efde4f13a5", "size": 189, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/economics/neoKeynesian/02-01-fisher.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/economics/neoKeynesian/02-01-fisher.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/economics/neoKeynesian/02-01-fisher.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 15.75, "max_line_length": 87, "alphanum_fraction": 0.6984126984, "num_tokens": 49, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8376199633332891, "lm_q2_score": 0.7341195385342972, "lm_q1q2_score": 0.6149131809493492}} {"text": "\\documentclass[t]{beamer}\n\\usetheme{Copenhagen}\n\\setbeamertemplate{headline}{} % remove toc from headers\n\\beamertemplatenavigationsymbolsempty\n\n\\usepackage{amsmath, tikz, bm, tkz-euclide}\n\\usetkzobj{all}\n\n\\title{Trig Functions of Any Angle}\n\\author{}\n\\date{}\n\n\\AtBeginSection[]\n{\n \\begin{frame}\n \\frametitle{Objectives}\n \\tableofcontents[currentsection]\n \\end{frame}\n}\n\n\\begin{document}\n\n\\begin{frame} \n\\maketitle\n\\end{frame}\n\n\\begin{frame}{From Right Triangle Trig to the Coordinate Plane}\nIn this section, adjacent, opposite and hypotenuse\n\\begin{center}\n\\begin{tikzpicture}\n \\tkzDefPoints{0/0/A, 3/0/C, 3/2/B}\n \\tkzMarkRightAngle[color=red](A,C,B)\n \\tkzDrawPolygon(A,B,C)\n \\tkzLabelSegment[below](A,C){adjacent}\n \\tkzLabelSegment[right](B,C){opposite}\n \\tkzLabelSegment[above left](A,B){hypotenuse}\n\\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{From Right Triangle Trig to the Coordinate Plane}\nbecome\n\\begin{center}\n \\begin{tikzpicture}\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$x$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3) node [right] {$y$};\n \\draw [color=red] (2.5,0) rectangle (2.25,0.25);\n \\draw (0,0) -- (2.5,0) -- (2.5,2) -- cycle;\n \\node at (1.25,0) [below] {$\\bm{x}$};\n \\node at (2.5,1) [right] {$\\bm{y}$};\n \\node at (1.25,1) [above] {$\\bm{r}$};\n \\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{From Right Triangle Trig to the Coordinate Plane}\nwhere $r$ represents the radius of the circle shown below.\n\n\\begin{center}\n \\begin{tikzpicture}\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$x$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3) node [right] {$y$};\n \\draw [color=red] (2.5,0) rectangle (2.25,0.25);\n \\draw (0,0) -- (2.5,0) -- (2.5,2) -- cycle;\n \\node at (1.25,0) [below] {$\\bm{x}$};\n \\node at (2.5,1) [right] {$\\bm{y}$};\n \\node at (1.25,1) [above] {$\\bm{r}$};\n \\draw [dashed] (0,0) circle (3.2cm);\n \\node at (0,0) [below left] {$A$};\n \\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{From Right Triangle Trig to the Coordinate Plane}\nThe biggest difference between this section and the last is that in this section, trig ratios can be positive, negative, zero, or undefined, since $x$ and $y$ can each be positive, negative, or zero. \n\\end{frame}\n\n\\section{Calculate the 6 Trig Ratios for a Point in the Coordinate Plane}\n\n\\begin{frame}{The 6 Trig Ratios}\nIn the coordinate plane, the six trig ratios become \\newline\\\\\n\\begin{center}\n \\begin{tabular}{c|c}\n $\\sin A = \\dfrac{y}{r}$ & \\onslide<2->{$\\csc A = \\dfrac{r}{y}$} \\\\[0.35in]\n $\\cos A = \\dfrac{x}{r}$ & \\onslide<2->{$\\sec A = \\dfrac{r}{x}$} \\\\[0.35in]\n $\\tan A = \\dfrac{y}{x}$ & \\onslide<2->{$\\cot A = \\dfrac{x}{y}$} \\\\\n \\end{tabular}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{Example 1a}\nFor the given point, find the exact values of the 6 trig functions for the angle drawn in standard form. \\newline\\\\\n(a) \\quad $(3,-4)$\n\\begin{align*}\n\\onslide<2->{3^2 + 4^2 &= r^2}\t\\\\[8pt]\n\\onslide<3->{r &= \\sqrt{25} = 5} \\\\\n\\end{align*}\n\\begin{tabular}{p{0.4\\textwidth}p{0.4\\textwidth}}\n\\onslide<4->{$\\sin A = \\frac{-4}{5}$}\t&\t\\onslide<7->{$\\csc A = \\frac{5}{-4}$}\t\\\\[8pt]\n\\onslide<5->{$\\cos A = \\frac{3}{5}$}\t&\t\\onslide<8->{$\\sec A = \\frac{5}{3}$}\t\\\\[8pt]\n\\onslide<6->{$\\tan A = \\frac{-4}{3}$}\t&\t\\onslide<9->{$\\cot A = \\frac{3}{-4}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 1b}\n(b)\t\\quad $(-8,-15)$\n\\begin{align*}\n\\onslide<2->{8^2 + 15^2 &= r^2}\t\\\\[8pt]\n\\onslide<3->{r &= \\sqrt{289} = 17} \\\\\n\\end{align*}\n\\begin{tabular}{p{0.4\\textwidth}p{0.4\\textwidth}}\n\\onslide<4->{$\\sin A = \\frac{-15}{17}$}\t\t\t\t\t&\t\\onslide<7->{$\\csc A = \\frac{17}{-15}$}\t\\\\[8pt]\n\\onslide<5->{$\\cos A = \\frac{-8}{17}$}\t\t\t\t\t&\t\\onslide<8->{$\\sec A = \\frac{17}{-8}$}\t\\\\[8pt]\n\\onslide<6->{$\\tan A = \\frac{-15}{-8} = \\frac{15}{8}$}\t&\t\\onslide<9->{$\\cot A = \\frac{8}{15}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 1c}\n(c)\t\\quad $(-1,5)$\n\\begin{align*}\n\\onslide<2->{1^2 + 5^2 &= r^2}\t\\\\[8pt]\n\\onslide<3->{r &= \\sqrt{26}} \\\\\n\\end{align*}\n\\begin{tabular}{p{0.4\\textwidth}p{0.4\\textwidth}}\n\\onslide<4->{$\\sin A = \\frac{5}{\\sqrt{26}} = \\frac{5\\sqrt{26}}{26}$}\t&\t\\onslide<7->{$\\csc A = \\frac{\\sqrt{26}}{5}$}\t\\\\[8pt]\n\\onslide<5->{$\\cos A = \\frac{-1}{\\sqrt{26}} = \\frac{-\\sqrt{26}}{26}$}\t&\t\\onslide<8->{$\\sec A = \\frac{\\sqrt{26}}{-1} = -\\sqrt{26}$}\t\\\\[8pt]\n\\onslide<6->{$\\tan A = \\frac{5}{-1} = -5$}\t&\t\\onslide<9->{$\\cot A = \\frac{-1}{5}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\n\\section{Find the Exact Values of the Trig Ratios of Special Angles in the Coordinate Plane}\n\n\\begin{frame}\nIf we put a 45-45-90 triangle in the second quadrant of the coordinate plane, it would resemble the following:\n\n\\begin{center}\n \\begin{tikzpicture}[scale=0.8]\n \\tkzDefPoints{0/0/A, -2/0/C, -2/2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-3,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\tkzLabelAngle[pos=0.75](C,A,B){$45^\\circ$}\n \\onslide<2->{\\draw [->, >=stealth, color=blue] (0:1) arc (0:135:1) node [midway, above right, color=blue] {$135^\\circ$};}\n \\onslide<4->{\\draw[->, >=stealth] (-1.75,-1) node [below] {Reference Angle} -- (-0.75,0);}\n \\end{tikzpicture}\n\\end{center}\n\\pause\n\nThe hypotenuse, $r$, would have rotated $180^\\circ - 45^\\circ = 135^\\circ$. \n\\end{frame}\n\n\\begin{frame} \nIn the second quadrant, $x$-coordinates are negative and $y$-coordinates are positive ($r$ is always positive). Thus, the values would be\n\n\\begin{center}\n \\begin{tikzpicture}\n \\tkzDefPoints{0/0/A, -2/0/C, -2/2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelAngle[pos=0.75](C,A,B){$45^\\circ$}\n \\node at (-1,0) [below] {$-1$};\n \\node at (-2,1) [left] {$1$};\n \\node at (135:1.75) [right] {$\\sqrt{2}$};\n \\end{tikzpicture}\n\\end{center}\n\\pause\nSo $\\sin 135^\\circ = \\frac{\\sqrt{2}}{2}$, $\\cos 135^\\circ = -\\frac{\\sqrt{2}}{2}$, and $\\tan 135^\\circ = -1$.\n\\end{frame}\n\n\\begin{frame}{Example 2a}\nFind the exact values of the 6 trig functions for each of the following.\t\\newline\\\\\n(a)\t\\quad $135^\\circ = \\frac{3\\pi}{4}$\t\\newline\\\\\n\\begin{minipage}{0.25\\textwidth}\n\\begin{tikzpicture}[scale=0.8]\n \\tkzDefPoints{0/0/A, -2/0/C, -2/2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\onslide<1->{\\tkzLabelAngle[pos=0.75](C,A,B){$45^\\circ$}}\n \\node at (-1,0) [below] {$-1$};\n \\node at (-2,1) [left] {$1$};\n \\node at (135:1.75) [right] {$\\sqrt{2}$};\n \\end{tikzpicture}\n\\end{minipage}\n\\begin{minipage}{0.65\\textwidth}\n\\begin{tabular}{p{0.43\\textwidth}p{0.53\\textwidth}}\n\\onslide<1->{$\\sin 135^\\circ = \\frac{\\sqrt{2}}{2}$}\t\t&\t\\onslide<2->{$\\csc 135^\\circ = \\frac{\\sqrt{2}}{1} = \\sqrt{2}$}\t\\\\[11pt]\n\\onslide<1->{$\\cos 135^\\circ = -\\frac{\\sqrt{2}}{2}$}\t&\t\\onslide<3->{$\\sec 135^\\circ = \\frac{\\sqrt{2}}{-1} = -\\sqrt{2}$}\t\\\\[11pt]\n\\onslide<1->{$\\tan 135^\\circ = -1$}\t\t\t\t\t\t&\t\\onslide<4->{$\\cot 135^\\circ = -1$}\t\\\\\n\\end{tabular}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 2b}\n(b)\t\\quad $225^\\circ = \\frac{5\\pi}{4}$\t\\newline\\\\\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -2/0/C, -2/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-3,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\onslide<2->{\\tkzLabelAngle[pos=-0.75](B,A,C){$45^\\circ$}}\n \\draw [->, >=stealth, color=blue] (0:1.2) arc (0:225:1.2) node [midway, above right, color=blue] {$225^\\circ$};\n\\end{tikzpicture}\n\\end{frame}\n\n\\begin{frame}{Example 2b}\n\\begin{center}\n\\begin{tikzpicture}\n \\tkzDefPoints{0/0/A, -2/0/C, -2/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelAngle[pos=-0.75](C,A,B){$45^\\circ$}\n \\tkzLabelSegment[midway](A,C){$-1$}\n \\tkzLabelSegment[midway, left](C,B){$-1$}\n \\tkzLabelSegment[below](A,B){$\\sqrt{2}$}\n\\end{tikzpicture} \n\\end{center}\n\\begin{tabular}{p{0.5\\textwidth}p{0.5\\textwidth}}\n\\onslide<2->{$\\sin 225^\\circ = \\frac{-1}{\\sqrt{2}} = -\\frac{\\sqrt{2}}{2}$}\t&\t\\onslide<5->{$\\csc 225^\\circ = \\frac{\\sqrt{2}}{-1} = -\\sqrt{2}$}\t\\\\[11pt]\n\\onslide<3->{$\\cos 225^\\circ = \\frac{-1}{\\sqrt{2}} = -\\frac{\\sqrt{2}}{2}$}\t&\t\\onslide<6->{$\\sec 225^\\circ = \\frac{\\sqrt{2}}{-1} = -\\sqrt{2}$}\t\\\\[11pt]\n\\onslide<4->{$\\tan 225^\\circ = \\frac{-1}{-1} = 1$}\t\t\t\t\t\t\t&\t\\onslide<7->{$\\cot 225^\\circ = \\frac{-1}{-1} = 1$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 2c}\n(c) \\quad $315^\\circ = \\frac{7\\pi}{4}$\t\\newline\\\\\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, 2/0/C, 2/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-3,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\onslide<2->{\\tkzLabelAngle[pos=0.75](B,A,C){$45^\\circ$}}\n \\draw [->, >=stealth, color=blue] (0:1.2) arc (0:315:1.2) node [midway, above left, color=blue] {$315^\\circ$};\n\\end{tikzpicture}\n\\end{frame}\n\n\\begin{frame}{Example 2c}\n\\begin{center}\n\\begin{tikzpicture}\n \\tkzDefPoints{0/0/A, 2/0/C, 2/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelAngle[pos=0.75](C,A,B){$45^\\circ$}\n \\tkzLabelSegment[midway](A,C){$1$}\n \\tkzLabelSegment[midway, right](C,B){$-1$}\n \\tkzLabelSegment[below, left, yshift=-0.2cm](A,B){$\\sqrt{2}$}\n\\end{tikzpicture} \n\\end{center}\n\\begin{tabular}{p{0.5\\textwidth}p{0.5\\textwidth}}\n\\onslide<2->{$\\sin 315^\\circ = \\frac{-1}{\\sqrt{2}} = -\\frac{\\sqrt{2}}{2}$}\t&\t\\onslide<5->{$\\csc 315^\\circ = \\frac{\\sqrt{2}}{-1} = -\\sqrt{2}$}\t\\\\[11pt]\n\\onslide<3->{$\\cos 315^\\circ = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}$}\t&\t\\onslide<6->{$\\sec 315^\\circ = \\frac{\\sqrt{2}}{1} = \\sqrt{2}$}\t\\\\[11pt]\n\\onslide<4->{$\\tan 315^\\circ = \\frac{-1}{1} = -1$}\t\t\t\t\t\t\t&\t\\onslide<7->{$\\cot 315^\\circ = \\frac{1}{-1} = -1$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 2d}\n(d) \\quad $120^\\circ = \\frac{2\\pi}{3}$\t\\newline\\\\\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -2/0/C, -2/4/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-3,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\onslide<2->{\\tkzLabelAngle[pos=0.75](B,A,C){$60^\\circ$}}\n \\draw [->, >=stealth, color=blue] (0:1) arc (0:120:1) node [midway, above right, color=blue] {$120^\\circ$};\n\\end{tikzpicture}\n\\end{frame}\n\n\\begin{frame}{Example 2d}\n\\begin{center}\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -2/0/C, -2/4/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelAngle[pos=0.75](C,A,B){$60^\\circ$}\n \\tkzLabelSegment[midway,below](A,C){$-1$}\n \\tkzLabelSegment[midway, left](C,B){$\\sqrt{3}$}\n \\tkzLabelSegment[right, xshift=0.2cm](A,B){$2$}\n\\end{tikzpicture} \n\\end{center}\n\\begin{tabular}{p{0.5\\textwidth}p{0.5\\textwidth}}\n\\onslide<2->{$\\sin 120^\\circ = \\frac{\\sqrt{3}}{2}$}\t\t\t\t\t&\t\\onslide<5->{$\\csc 120^\\circ = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3}$}\t\\\\[11pt]\n\\onslide<3->{$\\cos 120^\\circ = \\frac{-1}{2}$}\t\t\t\t\t\t&\t\\onslide<6->{$\\sec 120^\\circ = \\frac{2}{-1} = -2$}\t\\\\[11pt]\n\\onslide<4->{$\\tan 120^\\circ = \\frac{\\sqrt{3}}{-1} = -\\sqrt{3}$}\t&\t\\onslide<7->{$\\cot 120^\\circ = \\frac{-1}{\\sqrt{3}} = -\\frac{\\sqrt{3}}{3}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 2e}\n(e) \\quad $150^\\circ = \\frac{5\\pi}{6}$\t\\newline\\\\\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -3.5/0/C, -3.5/2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-4,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\onslide<2->{\\tkzLabelAngle[pos=1](B,A,C){$30^\\circ$}}\n \\draw [->, >=stealth, color=blue] (0:1) arc (0:150:1) node [midway, above right, color=blue] {$150^\\circ$};\n\\end{tikzpicture}\n\\end{frame}\n\n\\begin{frame}{Example 2e}\n\\begin{center}\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -3.5/0/C, -3.5/2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelSegment[below](A,C){$-\\sqrt{3}$}\n \\tkzLabelSegment[left](C,B){1}\n \\tkzLabelSegment[above, xshift = 0.2cm](A,B){2}\n \\tkzLabelAngle[pos=1](B,A,C){$30^\\circ$}\n\\end{tikzpicture}\n\\end{center}\n\\begin{tabular}{p{0.5\\textwidth}p{0.5\\textwidth}}\n\\onslide<2->{$\\sin 150^\\circ = \\frac{1}{2}$}\t\t\t\t\t&\t\\onslide<5->{$\\csc 150^\\circ = \\frac{2}{1} = 2$}\t\\\\[11pt]\n\\onslide<3->{$\\cos 150^\\circ = \\frac{-\\sqrt{3}}{2}$}\t\t\t&\t\\onslide<6->{$\\sec 150^\\circ = \\frac{2}{-\\sqrt{3}} = -\\frac{2\\sqrt{3}}{3}$}\t\\\\[11pt]\n\\onslide<4->{$\\tan 150^\\circ = \\frac{1}{-\\sqrt{3}} = -\\frac{\\sqrt{3}}{3}$}\t&\t\\onslide<7->{$\\cot 150^\\circ = \\frac{-\\sqrt{3}}{1} = -\\sqrt{3}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Example 2f}\n(a) \\quad $210^\\circ = \\frac{7\\pi}{6}$\t\\newline\\\\\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -3.5/0/C, -3.5/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\draw [<->, >=stealth] (-4,0) node [left] {$180^\\circ$} -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) node [below] {$270^\\circ$} -- (0,3) node [above] {$90^\\circ$};\n \\onslide<2->{\\tkzLabelAngle[pos=-1](B,A,C){$30^\\circ$}}\n \\draw [->, >=stealth, color=blue] (0:1.5) arc (0:210:1.5) node [midway, above right, color=blue] {$210^\\circ$};\n\\end{tikzpicture}\n\\end{frame}\n\n\\begin{frame}{Example 2f}\n\\begin{center}\n\\begin{tikzpicture}[scale=0.7]\n \\tkzDefPoints{0/0/A, -3.5/0/C, -3.5/-2/B}\n \\tkzDrawPolygon[fill=yellow!40](A,B,C)\n \\tkzMarkRightAngle[color=red](B,C,A)\n \\tkzLabelSegment[above](A,C){$-\\sqrt{3}$}\n \\tkzLabelSegment[left](C,B){$-1$}\n \\tkzLabelSegment[below, xshift = 0.2cm](A,B){2}\n \\tkzLabelAngle[pos=-1](B,A,C){$30^\\circ$}\n\\end{tikzpicture}\n\\end{center}\n\\begin{tabular}{p{0.5\\textwidth}p{0.5\\textwidth}}\n\\onslide<2->{$\\sin 210^\\circ = \\frac{-1}{2}$}\t\t\t\t\t&\t\\onslide<5->{$\\csc 210^\\circ = \\frac{2}{-1} = -2$}\t\\\\[11pt]\n\\onslide<3->{$\\cos 210^\\circ = \\frac{-\\sqrt{3}}{2}$}\t\t\t&\t\\onslide<6->{$\\sec 210^\\circ = \\frac{2}{-\\sqrt{3}} = -\\frac{2\\sqrt{3}}{3}$}\t\\\\[11pt]\n\\onslide<4->{$\\tan 210^\\circ = \\frac{-1}{-\\sqrt{3}} = \\frac{\\sqrt{3}}{3}$}\t&\t\\onslide<7->{$\\cot 210^\\circ = \\frac{-\\sqrt{3}}{-1} = \\sqrt{3}$}\t\\\\\n\\end{tabular}\n\\end{frame}\n\n\\begin{frame}{Quadrantal Angles}\nA \\alert{quadrantal angle} is an angle whose terminal side lies on an axis. \\newline\\\\ \\pause\n\n\nFinding the values of the trig functions for quadrantal angles can be found by using any point on the axes.\n\\end{frame}\n\n\\begin{frame}{Quadrantal Angles}\nFor simplicity, we will use combinations of 0s and 1s. Below is the point we will use for $0^\\circ$.\n\\begin{center}\n \\begin{tikzpicture}[scale=0.8]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (2,0) circle (2pt) node [below] {$(1,0)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [above] {$r=1$} (2,0);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n \\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{Quadrantal Angles}\nSo, for $0^\\circ$, we have $x=1, \\, y=0, \\, \\text{and } r=1$. Thus \\newline\\\\ \n\n\\begin{center}\n\\begin{tabular}{p{0.3\\textwidth}p{0.5\\textwidth}}\n \\onslide<2->{$\\sin 0 = \\frac{0}{1} = 0$} &\n \\onslide<5->{$\\csc 0 = \\frac{1}{0} = \\text{undefined}$} \\\\[18pt]\n \\onslide<3->{$\\cos 0 = \\frac{1}{1} = 1$} &\n \\onslide<6->{$\\sec 0 = \\frac{1}{1} = 1$} \\\\[18pt]\n \\onslide<4->{$\\tan 0 = \\frac{0}{1} = 0$} &\n \\onslide<7->{$\\cot 0 = \\frac{1}{0} = \\text{undefined}$} \\\\\n\\end{tabular}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}{Example 3a}\nFind the exact values of each of the 6 trig functions for the following angles. \\newline\\\\\n(a) \\quad $90^\\circ = \\frac{\\pi}{2}$ \\newline\\\\ \\pause\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (0,2) circle (2pt) node [right] {$(0,1)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [right] {$r=1$} (0,2);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<3->{\\sin 90^\\circ &= \\frac{1}{1} = 1} \\\\[18pt]\n \\onslide<4->{\\cos 90^\\circ &= \\frac{0}{1} = 0} \\\\[18pt]\n \\onslide<5->{\\tan 90^\\circ &= \\frac{1}{0} = \\text{undefined}} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 3a}\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (0,2) circle (2pt) node [right] {$(0,1)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [right] {$r=1$} (0,2);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<2->{\\csc 90^\\circ &= \\frac{1}{1} = 1} \\\\[18pt]\n \\onslide<3->{\\sec 90^\\circ &= \\frac{1}{0} = \\text{undefined}} \\\\[18pt]\n \\onslide<4->{\\cot 90^\\circ &= \\frac{0}{1} = 0} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 3b}\n (b) \\quad $180^\\circ = \\pi$ \\newline\\\\ \\pause\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (-2,0) circle (2pt) node [below] {$(-1,0)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [above] {$r=1$} (-2,0);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<3->{\\sin 180^\\circ &= \\frac{0}{1} = 0} \\\\[18pt]\n \\onslide<4->{\\cos 180^\\circ &= \\frac{-1}{1} = -1} \\\\[18pt]\n \\onslide<5->{\\tan 180^\\circ &= \\frac{0}{-1} = 0} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 3b}\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (-2,0) circle (2pt) node [below] {$(-1,0)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [above] {$r=1$} (-2,0);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<2->{\\csc 180^\\circ &= \\frac{1}{0} = \\text{undefined}} \\\\[18pt]\n \\onslide<3->{\\sec 180^\\circ &= \\frac{1}{-1} = -1} \\\\[18pt]\n \\onslide<4->{\\cot 180^\\circ &= \\frac{-1}{0} = \\text{undefined}} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 3c}\n(c) \\quad $270^\\circ = \\frac{3\\pi}{2}$ \\newline\\\\ \\pause\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (0,-2) circle (2pt) node [right] {$(0,-1)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [left] {$r=1$} (0,-2);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<3->{\\sin 270^\\circ &= \\frac{-1}{1} = -1} \\\\[18pt]\n \\onslide<4->{\\cos 270^\\circ &= \\frac{0}{1} = 0} \\\\[18pt]\n \\onslide<5->{\\tan 270^\\circ &= \\frac{-1}{0} = \\text{undefined}} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Example 3c}\n\\begin{minipage}{0.5\\textwidth}\n\\begin{tikzpicture}[scale=0.7]\n \\draw [<->, >=stealth] (-3,0) -- (3,0) node [right] {$0$};\n \\draw [<->, >=stealth] (0,-3) -- (0,3);\n \\draw [fill=black] (0,-2) circle (2pt) node [right] {$(0,-1)$};\n \\draw [->, >=latex, line width = 1.5, color=blue] (0,0) -- node [left] {$r=1$} (0,-2);\n \\node at (0,3) [right] {$90^\\circ$};\n \\node at (-3,0) [left] {$180^\\circ$};\n \\node at (0,-3) [right] {$270^\\circ$};\n\\end{tikzpicture}\n\\end{minipage}\n\\hspace{0.35cm}\n\\begin{minipage}{0.3\\textwidth}\n\\begin{align*}\n \\onslide<2->{\\csc 270^\\circ &= \\frac{1}{-1} = -1} \\\\[18pt]\n \\onslide<3->{\\sec 270^\\circ &= \\frac{1}{0} = \\text{undefined}} \\\\[18pt]\n \\onslide<4->{\\cot 270^\\circ &= \\frac{0}{-1} = 0} \\\\\n\\end{align*}\n\\end{minipage}\n\\end{frame}\n\n\\begin{frame}{Angles Not Between 0 and $360^\\circ$ or 0 and $2\\pi$}\n For angles not within 1 standard rotation, use \\alert{coterminal angles} to bring the angle within 1 rotation and then apply the rules of the previous notes. \\newline\\\\ \n \n For instance, $\\cos\\left(\\frac{9\\pi}{4}\\right) = \\cos\\left(\\frac{\\pi}{4}\\right)$\n\\end{frame}\n\n\\end{document}\n", "meta": {"hexsha": "7dff7e866edbe317ae5d591f502fbae788c51d10", "size": 21679, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Trig_Functions_of_Any_Angle(BEAMER).tex", "max_stars_repo_name": "BryanBain/Trig_BEAMER", "max_stars_repo_head_hexsha": "3639d0202fc691738d26f922d9e84f0a0d62b42a", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Trig_Functions_of_Any_Angle(BEAMER).tex", "max_issues_repo_name": "BryanBain/Trig_BEAMER", "max_issues_repo_head_hexsha": "3639d0202fc691738d26f922d9e84f0a0d62b42a", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Trig_Functions_of_Any_Angle(BEAMER).tex", "max_forks_repo_name": "BryanBain/Trig_BEAMER", "max_forks_repo_head_hexsha": "3639d0202fc691738d26f922d9e84f0a0d62b42a", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-08-26T15:49:06.000Z", "max_forks_repo_forks_event_max_datetime": "2020-08-26T15:49:06.000Z", "avg_line_length": 40.5214953271, "max_line_length": 200, "alphanum_fraction": 0.5741039716, "num_tokens": 9397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195269001831, "lm_q2_score": 0.8376199633332891, "lm_q1q2_score": 0.6149131712043829}} {"text": "\n\\subsection{The infintesimal jackknife}\n\n\\subsubsection{The jackknife as a weighting}\n\nIn the jackknife we calculate the statistic leaving one observation out.\n\nThis is the same as weighting observations and giving one a weighting of \\(0\\) and the others \\(1\\).\n\n\\subsubsection{The infintesimal jackknife}\n\nFor the infintesimal jackknife we reduce the weight not to \\(0\\), but by an infintesimal amount.\n\n", "meta": {"hexsha": "a11501f3ca6791a6ecd6ba8579a5ca177e8a86f3", "size": 406, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/jackknife/01-02-jackknifeInf.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/jackknife/01-02-jackknifeInf.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/jackknife/01-02-jackknifeInf.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 29.0, "max_line_length": 100, "alphanum_fraction": 0.7832512315, "num_tokens": 94, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8519528170040852, "lm_q2_score": 0.7217432062975979, "lm_q1q2_score": 0.6148911577587991}} {"text": "\\chapter{Kernel density estimation}\n\\section{Bandwidth and Kernel function}\nIn this chapter we are dealing with marks secured by students in various subject, along with explanatory variables about the students viz., gender, ethnicity etc.. For our analysis for student's performance together with scores in mathematics, reading and writing tests, we consider students attendance for the test preparation course. We drop rest of the variables. Following is the sample for first few entries of the dataset;\n\n\\begin{table}[ht]\n\\centering\n\\begin{tabular}{llrrr}\n \\hline\n & test preparation course & math score & reading score & writing score \\\\ \n \\hline\n1 & none & 72 & 72 & 74 \\\\ \n 2 & completed & 69 & 90 & 88 \\\\ \n 3 & none & 90 & 95 & 93 \\\\ \n 4 & none & 47 & 57 & 44 \\\\ \n 5 & none & 76 & 78 & 75 \\\\ \n 6 & none & 71 & 83 & 78 \\\\ \n 7 & completed & 88 & 95 & 92 \\\\ \n 8 & none & 40 & 43 & 39 \\\\ \n 9 & completed & 64 & 64 & 67 \\\\ \n 10 & none & 38 & 60 & 50 \\\\ \n \\hline\n\\end{tabular}\n\\end{table}\n\nIn this exercise we are suppose to implement kernel density estimation. The data entries i.e. math score are integers variables. Kernel density estimation help us visualize the “shape” of data, even though its discrete. Essentially it is sort of continuous replacement for the discrete histogram. Kernel estimation uses a weighted sum of observations dependent on their distance from the variable. The precise definition is as follows, for the given data $X_1, X_2,..., X_n$, a kernel density estimator is \n$$\\widehat{f}_{n,h}(x)= \\frac{1}{nh}\\sum_{1 = j}^n K\\left(\\frac{x-X_j}{h}\\right), \\: x\\in \\RR .$$ \nwhere $K: \\RR \\rightarrow \\RR$, such that $\\int_{-\\infty}^{\\infty}K(x)dx=1$ is known as kernel and $h>0$ is called the bandwidth. $h$ is smoothing parameter which basically governs how many distinct observations are taken into account around certain location. Thus it has a strong influence on the resulting estimate. Examples of some classical kernels is illustrated below in Figure \\ref{fig:types_kernels8}.\n \n\\begin{figure}[thb]\n\\centering\n\\includegraphics[width=0.75\\textwidth, keepaspectratio]{ex8/types_kernels.png}\n\\caption{Classical examples of kernels often used for density estimation. }\n\\label{fig:types_kernels8}\n\\end{figure}\n\nTo visualise and examine the influence of the choice of bandwidth and kernel we have considered the math score variable from the data. First we fit a kernel density with different bandwidths using the Epanechnikov kernel (See Figure \\ref{fig:diff_bandwidth8}) and then use different kernels with a fixed bandwidth Figure\\ref{fig:diff_ker8} ). We can clearly see the influence of the choice of bandwidth. The estimated curves get smoother with higher and higher $h$. For bandwidth $= 2$ it fluctuates highly. \nFor the values of $h$ between $5$ and $12$ seems reasonable where as for for higher values, like $30$ density curve gets too smooth and it certainly does not represent the shape of the data distribution anymore. Since, visually it appears that the most reasonable bandwidth lies between $5$ and $12$, I have fixed $h= 8$ for the next analysis (See Figure \\ref{fig:diff_ker8}). We will now compare different kernels choices with a fix bandwidth $h= 8$.\n\n\\begin{figure}[tbh]\n\\centering\n\\begin{subfigure}[c]{\\textwidth}\n\\includegraphics[width=0.93\\textwidth, keepaspectratio]{ex8/diff_bandwidths.png}\n\\subcaption{Epanechnikov kernel with different bandwidths.}\n\\label{fig:diff_bandwidth8}\n\\end{subfigure}\n\\begin{subfigure}[c]{\\textwidth}\n\\includegraphics[width=0.98\\textwidth, keepaspectratio]{ex8/diff_ker.png}\n\\subcaption{Different kernels with fixed bandwidth $h=8$.}\n\\label{fig:diff_ker8}\n\\end{subfigure}\n\\caption{Kernel density estimations for the math score (first plot) and kernels (second plot).}\n\\label{6fit}\n\\end{figure}\n\n%Comment on the effect of the bandwidth and of the kernel function on kernel density estimators\nIf we observe closely Figure \\ref{fig:types_kernels8}, we can see that, the different shapes of the kernels shows how observations with certain weights at different distances influence the height of the density estimation at specific points. Therefore, the shape and some basic properties of the resulting density estimations depend highly on the choice of the kernel. The rectangular kernel for example will always result in a piece-wise constant estimation, which is not continuous. The triangular kernel will give us a piece-wise linear density and the Gaussian kernel will result in smooth estimations. In Figure \\ref{fig:diff_ker8} we can see that for the rectangular kernel the result is a step function, where as the rest of the kernels seem smoother. But as mentioned before the curve for the triangular kernel is piece-wise linear. Rectangular, triangular and Epanechnikov look quite similar with respect to the shape of the overall distribution. The Gaussian kernel produces a flatter curve that puts more mass to the tails. This might be because, of the differences in support of kernels. The support of the Gaussian kernel $\\RR$ and its $ [ -1,1] $ for the other kernels. Thus, as we can also see in Figure \\ref{fig:types_kernels8} Gaussian kernel puts more mass to the tails.\n\n\\section{Cross validation for optimal bandwidth} \nNow we want to consider a more formal approach to examine the best bandwidth. Above we just did a graphical examination. To get the best fitting curve a common approach is trying to minimise the mean integrated square error (MISE) for the estimator $\\hat{f}_{n,h}(x)$ of $f(x)$, which is given by $$MISE(\\hat{f}_{n,h})=E\\left[\\int\\left(\\hat{f}_{n,h}(x)-f(x)\\right)^2dx\\right].$$ Because the real $f$ is unknown one minimises an unbiased estimator of the $h$-dependent part of the MISE, called the cross-validation criterion: $$CV(h)=\\int\\left(\\hat{f}_{n,h}(x)\\right)^2dx-\\frac{2}{n(n-1)h}\\sum_{i=1}^{n} \\sum_{j\\neq i} K\\left(\\frac{X_j - X_i}{h}\\right).$$\n\\begin{figure}[!tb]\n\\centering\n\\includegraphics[width=0.9\\textwidth, keepaspectratio]{ex8/CV1.png}\n\\caption{CV values for 40 different bandwidths. Red point shows minimum of the curve. Other points show the results for the \\texttt{optimize} function in R with different limits (see legend).}\n\\label{6CV1}\n\\end{figure} \n\nImplementing a function computing the CV for a given estimation we can use the minimising $h_{CV}=\\argmin _{h>0} CV(h)$.\nAfterwards we compared the results with the functions \\texttt{bw.ucv} and \\texttt{bw.bcv} of the package \\texttt{density}, which compute the optimal bandwidth for an unbiased (\\texttt{ucv}) or biased (\\texttt{bcv}) CV implementation for Gaussian kernels. \nTherefore I used the Epanechnikov kernel as well as the Gaussian kernel in my implementation to compare all results for the maths score. To find the minimizer of my function, I used the R function \\texttt{optimize}. Before comparing the optimisation of my own CV function a little note about the disadvantages of using \\texttt{optimize} here is important. \nTo get a first idea of the shape of CV for different $h$ as a curve I took 40 equidistant points in the interval $[1,25]$ - because it seemed as a (maximal) reasonable region for bandwidths in this case - and computed CV there. Afterwards I took the minimum out of this observations as first reference point ($h=1.62$). After using the R implemented optimisation with limits 0 and 30, I was surprised by the result ($h=11.50$). As \\texttt{optimize} just searches for local minima and starting points for the iteration are chosen as a golden section distance between the given limits, the result also depends highly on those borders. Obviously CV is a strongly fluctuating function as I got different results for every trial, none of them was near the first, say, 'scanned minimum'. Figure \\ref{6CV1} shows the results. All in all, without an analytical examination of CV as a function one cannot be sure to really found the optimal bandwidth or even to be close to it. In the following I won't discuss this issue any more- I will use a combined strategy by first scanning the interval $[1, 20]$ with 40 points to get the area of small values visually and by taking the minimum. Afterwards I will use \\texttt{optimize} limited to this region to get a local minimum. This should in total result in a somehow precise and at least acceptable minimum. The results for all three tests and the different methods are shown together in Table \\ref{6table}. For the math score and the best bandwidth with a Epanechnikov kernel we use the above optimal $h=1.62$. \n\\begin{figure}[p]\n\\centering\n\\includegraphics[width=\\textwidth, keepaspectratio]{ex8/CVplots.png}\n\\caption{CV against $h$ for math score and Gaussian kernel (first row), reading score (second row) and writing score (third row). Last two for Epanechnikov kernel (left) and Gaussian kernel (right)}\n\\label{6CVs}\n\\end{figure}\n\nFirst, observe (see Figure \\ref{6CVs}) that for the Gaussian kernel the problem of fluctuating CV function is not that worse but the usage of the \\texttt{optimize} function still improves the optimal bandwidth a bit. For the Epanechnikov kernel we have the same problem for every test. All in all the CV curves look very similar for math, reading and writing score. Therefore we get quite similar results for optimal bandwidths. As expected the values of my own optimisation for the Gauss kernel are similar to those of the R functions but differ from the optimal bandwidth for the Epanechnikov kernel. For every score the resulting CVs just differ on the presented scale between the kernels where always the Epanechnikov kernel gives a better result. Therefore we use this bandwidth an the Epanechnikov kernel for the last part of the exercise. \n\n\\begin{table}[!t]\n\\centering\n\\begin{tabular}{lrrrrrrrr}\n \\hline\n & \\multicolumn{2}{r}{Epa} & \\multicolumn{2}{r}{Gau}& \\multicolumn{2}{r}{\\texttt{bw.bcv}}& \\multicolumn{2}{r}{\\texttt{bw.ucv}} \\\\ \n\\hline\n & h & CV & h & CV & h & CV & h & CV \\\\ \n \\hline\nmath & 1.62 & -0.0197 & 4.83 & -0.0186 & 4.26 & -0.0186 & 4.65 & -0.0186 \\\\ \n reading & 1.41 & -0.0212 & 4.19 & -0.0191 & 4.40 & -0.0191 & 3.77 & -0.0191 \\\\ \n writing & 1.41 & -0.0203 & 4.52 & -0.0184 & 4.18 & -0.0184 & 4.28 & -0.0184 \\\\ \n \\hline\n\\end{tabular}\n\\caption{Optimal bandwidths $h$ and corresponding CV for math, reading and writing score and different methods. First two columns are optimum of self-written function with Epanechnikov kernel (Epa) and Gaussian kernel (Gau) respectively, last two columns with R function in the package \\texttt{density}.}\n\\label{6table}\n\\end{table}\n\n\n\\section{Comparison of groups}\nNow we use the kernel density estimation for a comparison of those students attending a preparation course and those that did not. Therefore we fit a curve for both groups using the optimal bandwidth we got in the previous examinations and the Epanechnikov kernel. For the bandwidth we use a common value equal to the mean of the three optimal bandwidths (this is just for a simplification of the code and does not change the results crucially). \n\nThe result is shown in Figure \\ref{6groups}. I decided to plot the histograms above each other, so that there is no confusion about the positioning and because one can still distinguish the bars with just two groups. We see that for the math score the whole distribution of scores is not changed very much in the shape but just shifted to the right, i.e. to better grades. For the other two scores we can recognise that the distribution of students with the preparation course is a bit more left skewed. This is just slightly visible for the reading score and a bit more distinct for the writing score. This is probably due to the fact that the student get better in total but the limit of the score is at 100. Psychologists call this effect \\textit{ceiling effect}. This is maybe also the reason why the distribution has a smaller variation, especially for the writing score again. Summing up, we recognize better performance of students doing a preparation course. This is especially visible for the writing score. Apparently, one cannot train maths as easily as writing.\n\\begin{figure}[!t]\n\\centering\n\\includegraphics[width=\\textwidth, keepaspectratio]{ex8/groups.png}\n\\caption{Histogram and kernel density estimate for two groups of students in different colors and the three scores, math (left), reading (middle) and writing (right).}\n\\label{6groups}\n\\end{figure} \n", "meta": {"hexsha": "e6a5933b2f9583594502406d5d864d8de139c014", "size": 12406, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ex8/ex8.tex", "max_stars_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_stars_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ex8/ex8.tex", "max_issues_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_issues_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ex8/ex8.tex", "max_forks_repo_name": "dnanad/Advanced-Statistical-Data-Analysis", "max_forks_repo_head_hexsha": "09114b8840466cc3637bd447ff9adf584c811cde", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 114.8703703704, "max_line_length": 1552, "alphanum_fraction": 0.7628566823, "num_tokens": 3231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217432062975979, "lm_q2_score": 0.8519528057272543, "lm_q1q2_score": 0.614891149619823}} {"text": "\\section{Taylor Series}\\label{sec:taylorseries}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %\nWe have seen that some functions can be represented as series, which\nmay give valuable information about the function. So far, we have seen\nonly those examples that result from manipulation of our one\nfundamental example, the geometric series. We would like to start with\na given function and produce a series to represent it, if possible.\n\nSuppose that $\\ds f(x)=\\sum_{n=0}^\\infty a_nx^n$ on some interval of\nconvergence centered at 0. Then we know that we can compute derivatives of $f$ by\ntaking derivatives of the terms of the series. Let's look at the first\nfew in general:\n\\begin{align*}\n f'(x)&=\\sum_{n=1}^\\infty n a_n x^{n-1}=a_1 + 2a_2x+3a_3x^2+4a_4x^3+\\cdots\t\\\\\n f''(x)&=\\sum_{n=2}^\\infty n(n-1) a_n x^{n-2}=2a_2+3\\cdot2a_3x\n +4\\cdot3a_4x^2+\\cdots\t\\\\\n f'''(x)&=\\sum_{n=3}^\\infty n(n-1)(n-2) a_n x^{n-3}=3\\cdot2a_3\n +4\\cdot3\\cdot2a_4x+\\cdots\t\\\\\n\\end{align*}\n\nBy examining these it's not hard to discern the general pattern. The\n$k$th derivative must be\n\\begin{align*}\n f^{(k)}(x)&=\\sum_{n=k}^\\infty n(n-1)(n-2)\\cdots(n-k+1)a_nx^{n-k}\t\\\\\n &=k(k-1)(k-2)\\cdots(2)(1)a_k+(k+1)(k)\\cdots(2)a_{k+1}x+{}\t\\\\\n &\\qquad {}+(k+2)(k+1)\\cdots(3)a_{k+2}x^2+\\cdots\t\\\\\n\\end{align*}\nWe can express this more clearly by using factorial notation:\n\\[\n f^{(k)}(x)=\\sum_{n=k}^\\infty {n!\\over (n-k)!}a_nx^{n-k}=\n k!a_k+(k+1)!a_{k+1}x+{(k+2)!\\over 2!}a_{k+2}x^2+\\cdots\n\\]\n\nWe can solve for $a_n$ by substituting $x=0$ in the formula for $f^{(k)}(x)$:\n\\[f^{(k)}(0)=k!a_k+\\sum_{n=k+1}^\\infty {n!\\over (n-k)!}a_n0^{n-k}=k!a_k,\\]\n\\[a_k={f^{(k)}(0)\\over k!}.\\]\nNote that the original series for $f$ yields $f(0)=a_0$.\n\nSo if a function $f$ can be represented by a series, we can easily find such a series.\nGiven a function $f$, the series\n\\[\\sum_{n=0}^\\infty {f^{(n)}(0)\\over n!}x^n\\]\nis called the \\dfont{Maclaurin \nseries} for $f$.\n\n\\begin{example}{Maclaurin Series}{MacSeriesOne}\nFind the Maclaurin series for $f(x)=1/(1-x)$.\n\\end{example}\n\\begin{solution}\nWe need to\ncompute the derivatives of $f$ (and hope to spot a pattern).\n\\begin{align*}\n f(x)&=(1-x)^{-1}\t\\\\\n f'(x)&=(1-x)^{-2}\t\\\\\n f''(x)&=2(1-x)^{-3}\t\\\\\n f'''(x)&=6(1-x)^{-4}\t\\\\\n f^{(4)}(x)&=4!(1-x)^{-5}\t\\\\\n &\\vdots\t\\\\\n f^{(n)}(x)&=n!(1-x)^{-n-1}\t\\\\\n\\end{align*}\n\nSo\n\\[a_n={f^{(n)}(0)\\over n!}={n!(1-0)^{-n-1}\\over n!}=1\\]\nand the Maclaurin series is\n\\[\\sum_{n=0}^\\infty 1\\cdot x^n=\\sum_{n=0}^\\infty x^n,\\]\nthe geometric series.\n\\end{solution}\n\nA warning is in order here. Given a function $f$ we may be able to\ncompute the Maclaurin series, but that does not mean we have found a\nseries representation for $f$. We still need to know where the series\nconverges, and if, where it converges, it converges to $f(x)$. While\nfor most commonly encountered functions the Maclaurin series does\nindeed converge to $f$ on some interval, this is not true of all\nfunctions, so care is required.\n\nAs a practical matter, if we are interested in using a series to\napproximate a function, we will need some finite number of terms of\nthe series. Even for functions with messy derivatives we can compute\nthese using computer software like Sage. If we want to describe a series\ncompletely, we would like to be able to write down a formula for a typical\nterm in the series. Fortunately, a few of the most important functions are very\neasy.\n\n\\begin{example}{Maclaurin Series}{MacSeriesTwo}\nFind the Maclaurin series for $\\sin x$.\n\\end{example}\n\\begin{solution}\nComputing the first few derivatives is simple: $f'(x)=\\cos x$, $f''(x)=-\\sin x$,\n$f'''(x)=-\\cos x$, $\\ds f^{(4)}(x)=\\sin x$, and then the pattern\nrepeats. The values of the derivative when $x=0$ are:\n1, 0, $-1$, 0, 1, 0, $-1$, 0,\\dots, and so the Maclaurin series is\n\\[\n x-{x^3\\over 3!}+{x^5\\over 5!}-\\cdots=\n \\sum_{n=0}^\\infty (-1)^n{x^{2n+1}\\over (2n+1)!}.\n\\]\n\nWe should always determine the radius of convergence:\n\\[\n \\lim_{n\\to\\infty} {|x|^{2n+3}\\over (2n+3)!}{(2n+1)!\\over |x|^{2n+1}}\n =\\lim_{n\\to\\infty} {|x|^2\\over (2n+3)(2n+2)}=0,\n\\]\nso the series converges for every $x$. Since it turns out that this\nseries does indeed converge to $\\sin x$ everywhere, we have a series\nrepresentation for $\\sin x$ for every $x$.\n\\end{solution}\n\nSometimes the formula for the $n$th derivative of a function $f$ is\ndifficult to discover, but a combination of a known Maclaurin series\nand some algebraic manipulation leads easily to the Maclaurin series\nfor $f$.\n\n\\begin{example}{Maclaurin Series}{MacSeriesThree}\nFind the Maclaurin series for $x\\sin(-x)$.\n\\end{example}\n\\begin{solution}\nTo get from $\\sin x$ to $x\\sin(-x)$ we substitute $-x$ for $x$ and\nthen multiply by $x$. We can do the same thing to the series for $\\sin\nx$:\n\\[\n x\\sum_{n=0}^\\infty (-1)^n{(-x)^{2n+1}\\over (2n+1)!}\n =x\\sum_{n=0}^\\infty (-1)^{n}(-1)^{2n+1}{x^{2n+1}\\over (2n+1)!}\n =\\sum_{n=0}^\\infty (-1)^{n+1}{x^{2n+2}\\over (2n+1)!}.\n\\]\n\\end{solution}\n\nAs we have seen, a power series can be centered at a point\nother than zero, and the method that produces the Maclaurin series can\nalso produce such series.\n\n\\begin{example}{Taylor Series}{TaylorSeriesOne}\nFind a series centered at $-2$ for $1/(1-x)$.\n\\end{example}\n\\begin{solution}\nIf the series is $\\ds\\sum_{n=0}^\\infty a_n(x+2)^n$ then looking at the\n$k$th derivative:\n$$k!(1-x)^{-k-1}=\\sum_{n=k}^\\infty {n!\\over (n-k)!}a_n(x+2)^{n-k}$$\nand substituting $x=-2$ we get\n$\\ds k!3^{-k-1}=k!a_k$ and $\\ds a_k=3^{-k-1}=1/3^{k+1}$, so the series is\n$$\\sum_{n=0}^\\infty {(x+2)^n\\over 3^{n+1}}.$$\n\\end{solution}\n\nSuch a series is called the \n\\dfont{Taylor series} for the function,\nand the general term has the form\n\\[{f^{(n)}(a)\\over n!}(x-a)^n.\\]\n\nA Maclaurin series is simply a Taylor series with $a=0$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\Opensolutionfile{solutions}[ex]\n\\section*{Exercises for \\ref{sec:taylorseries}}\n\n\\begin{enumialphparenastyle}\n\n\\begin{ex}\nFor each function, find the Maclaurin series or Taylor series centered\nat $a$, and the radius of convergence.\n\n\\begin{enumerate}\n\t\\item $\\cos x$\n\t\\item $\\ds e^x$\n\t\\item $1/x$, $a=5$\n\t\\item $\\ln x$, $a=1$\n\t\\item $\\ln x$, $a=2$\n\t\\item $\\ds 1/x^2$, $a=1$\n\t\\item $\\ds 1/\\sqrt{1-x}$\n\t\\item Find the first four terms of the Maclaurin series for $\\tan\n\tx$ (up to and including the $\\ds x^3$ term).\n\t\\item Use a combination of Maclaurin series and algebraic\n\tmanipulation to find a series centered at zero for\n\t$\\ds x\\cos (x^2)$.\n\t\\item Use a combination of Maclaurin series and algebraic\n\tmanipulation to find a series centered at zero for\n\t$\\ds xe^{-x}$.\n\\end{enumerate}\n\\begin{sol}\n\\begin{enumerate}\n\t\\item $\\ds\\sum_{n=0}^\\infty (-1)^n x^{2n}/(2n)!$, $R=\\infty$\n\t\\item $\\ds\\sum_{n=0}^\\infty x^n/n!$, $R=\\infty$\n\t\\item $\\ds\\sum_{n=0}^\\infty (-1)^n{(x-5)^n\\over 5^{n+1}}$, $R=5$\n\t\\item $\\ds\\sum_{n=1}^\\infty (-1)^{n-1}{(x-1)^n\\over n}$, $R=1$\n\t\\item $\\ds\\ln(2)+\\sum_{n=1}^\\infty (-1)^{n-1}{(x-2)^n\\over n 2^n}$, $R=2$\n\t\\item $\\ds\\sum_{n=0}^\\infty (-1)^n(n+1)(x-1)^n$, $R=1$\n\t\\item $\\ds1+\\sum_{n=1}^\\infty {1\\cdot3\\cdot5\\cdots(2n-1)\\over\n\t\tn!2^n} x^n=1+\\sum_{n=1}^\\infty {(2n-1)!\\over 2^{2n-1}(n-1)!\\,n!}x^n$, $R=1$\n\t\\item $\\ds x+x^3/3$\n\t\\item $\\ds\\sum_{n=0}^\\infty (-1)^n x^{4n+1}/(2n)!$\n\t\\item $\\ds\\sum_{n=0}^\\infty (-1)^n x^{n+1}/n!$\n\\end{enumerate}\n\\end{sol}\n\\end{ex}\n\n\\end{enumialphparenastyle}", "meta": {"hexsha": "714bf519a7e580edaf362a9a005ab9c8de5b3d39", "size": 7353, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "9-sequences-and-series/9-10-taylor-series.old.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "9-sequences-and-series/9-10-taylor-series.old.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "9-sequences-and-series/9-10-taylor-series.old.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.5579399142, "max_line_length": 86, "alphanum_fraction": 0.6368829049, "num_tokens": 2912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217431943271999, "lm_q2_score": 0.8519528076067262, "lm_q1q2_score": 0.6148911407781049}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[english]{babel}\n\\usepackage{graphicx}\n\\usepackage{multicol}\n\\usepackage{amsmath}\n\\usepackage{hyperref}\n\\usepackage{amsthm}\n\\usepackage{geometry}\n\\geometry{a4paper} \n\\usepackage{fancyhdr}\n\\usepackage{xcolor}\n\\usepackage{amssymb}\n\\usepackage{multicol}\n\\usepackage{float}\n\\theoremstyle{definition}\n\\newtheorem{definition}{Defintion}[section]\n\\newtheorem{exmp}{Example}[section]\n\\newtheorem{theorem}{Theorem}\n\n\\begin{document}\n\\author{Fractals}\n\\title{\\textbf{Functions}}\n\\maketitle\n\\tableofcontents\n\\noindent\n\\section{Introduction}\n\\subsection{Defintions}\n\n\\begin{definition}\n A function \\(f\\) from a set \\(X\\) to a set \\(Y\\) is a relation that assigns to each element in\n set \\(X\\) exactly one element in set \\(Y\\).\n\\end{definition}\n\n\\begin{definition}\n The domain is the set of \\(X\\) (a.k.a. the input).\n\\end{definition}\n\n\\begin{definition}\n The range is a subset of Y (a.k.a. the output).\n\\end{definition}\n\n\\subsection{Existence of a Function}\n\\begin{theorem}[\\textbf{Vertical Line Test}]\n if you can draw a Vertical line that passes through more than one point of a\n relation on a grap, it's not a function, if you cannot, it's a function.\n\n\\end{theorem}\n\n\n\\begin{exmp}\n what the domain and range of the function \\(f(x) = \\sqrt{16 - x^2}\\)?\n\\end{exmp}\n\n\\(sloution\\) Note that if \\(a < 0 \\), then \\( \\sqrt{a}\\) is undefined for reals,\nThus, \\(16 - x^2 \\ge 0 \\Rrightarrow \\) \\framebox[\\width]{\n \\(-4 \\le x \\le 4\\)\n} since \\(x^2 \\ge 0\\), we have that \\(0 \\le 16 - x^2 \\le 16\\),\nso the range is \\framebox[\\width]{ \\(0 \\le y \\le 4\\) }\n\n\n\\section{Combinations of Functions}\n\\begin{theorem}[\\textbf{common function Combinations}]\n The following are some common combinations of functions:\n \\begin{itemize}\n \\item \\textbf{Sum} \\((f+g)(x) = f(x) + g(x)\\)\n \\item \\textbf{Differnce} \\((f-g)(x) = f(x) - g(x)\\)\n \\item \\textbf{Product} \\((fg)(x) = f(x)g(x)\\)\n \\item \\textbf{Quotient} \\((\\dfrac{f}{g} )(x) = \\dfrac{f(x)}{g(x)} \\)\n where \\(g(x) \\ne 0\\)\n \\item \\textbf{Compostion} \\((f\\circ g)(x) = f(g(x))\\)\n \\end{itemize}\n\n\\end{theorem}\n\n\\begin{exmp}\n if \\(f(x) = 2x + 3\\) and \\(\\quad g(x) = 2x - 3\\), then what is \\(fg(4)\\)?\n\\end{exmp}\n\\(sloution\\) \\( (fg)(x) = (2x+3)(2x-3) \\) Thus \\((fg)(4) = 55\\).\n\\subsection{Domain and Range of a Composite Function}\nThe domain of a composite function is the intersection of domains of the starting\nand final function.\n\n\\noindent\nThe range of a composite function is the range of the final function restricted by the\nstarting function.\n\n\\begin{exmp}\n let \\(f(x) = \\dfrac{1}{x+2}\\) and \\( \\dfrac{x}{x-3} \\).\n Then \\(g(x)\\) is the starting function and \\( f(g(x)) \\) is the final\n function. Find the domain and range of \\(f(g(x))\\).\n\\end{exmp}\n\n\\noindent\n\\(sloution\\) \\(f(g(x)) = \\dfrac{1}{\\dfrac{x}{x-3} + 2}\\) so \\(x \\ne 3\\),\nimpleing the domain is \\framebox[\\width]{ \\(x \\ne 2,3\\) }.\n\n\\begin{exmp}\n If \\(f(x) = \\sqrt{x}\\) and \\(g(x) = x - 1\\), what is the domain and range\n of \\((g\\circ f)(x)\\)?\n\\end{exmp}\n\\noindent\n\\(sloution\\) \\((g\\circ f)(x) = \\sqrt{x} - 1\\) it's obivious that \\(x \\ge 0\\),\nand all other values work, so the domain is \\framebox[\\width]{ \\(0 \\le x < \\infty \\ \\)}\n. Since \\( \\sqrt{x} \\ge 0\\)we have \\((g\\circ f)(x) \\ge -1\\), with no other restrictions,\nso the range is \\framebox[\\width]{ \\([-1,\\infty] \\) }.\n\n\\section{Types of Functions}\n\\subsection{Piecewise-Defined Function}\nA piecewise function is a function that is defined by two or more equations over a\nspecified domain.\n\n\\begin{exmp}\n let \\(f(x) = |x|\\) Then\n \\[ f(x) =\n \\begin{cases}\n x & \\text{if } x \\ge 0 \\\\\n -x & \\text{if } x < 0\n \\end{cases}\n \\]\n\n\\end{exmp}\n\n\\begin{exmp}\n What are the domain and range of the piecewise function as follows?\n \\[\n f(x) =\n \\begin{cases}\n x^2 + 1 & x < 0 \\\\\n x - 1 & x \\ge 0\n \\end{cases}\n \\]\n\\end{exmp}\n\n\\(sloution\\). The domain includes \\(x < 0 \\) and \\(x \\ge 0\\),\nwhich is all values, so the domain is\n\\framebox[\\width]{ \\((-\\infty, \\infty)\\) }. for \\(x \\ge 0\\), we have \\(f(x) = x - 1\\),\nso the range is \\(y \\ge -1\\). For \\(x < 0\\) we have \\(f(x) = x^2 + 1\\), so the range is \\(y > 1\\).\nThus the range together is \\framebox[\\width]{ \\([-1, \\infty )\\) }\n\n\\section{Properties of Functions}\n\\subsection{Odd and Even Functions}\nA function \\(f\\) is even if \\(f(x) = f(-x)\\)\n\n\n\\begin{figure}[ht]\n \\begin{small}\n \\begin{center}\n \\includegraphics[width=0.75\\textwidth]{../out/even.png}\n \\end{center}\n \\caption{Graph of an even function.}\n \\label{fig: Even Function}\n \\end{small}\n\\end{figure}\n\n\\pagebreak\nA function \\(f\\) is odd if \\(f(x) = - f(-x)\\)\n\n\\begin{figure}[H]\n \\begin{small}\n \\begin{center}\n \\includegraphics[width=0.75 \\textwidth]{../out/odd.png}\n \\end{center}\n \\caption{Graph of an odd function.}\n \\label{fig: Odd Function}\n\n \\end{small}\n\\end{figure}\n\nNote that\n\\begin{itemize}\n \\item Reflecting an even function across the y-axis yields the same function, and\n \\item Rotating an odd function across the x-axis also yields the same function.\n\\end{itemize}\n\n\\begin{theorem}[Parity of Functions Comes From Its Components]\n if \\(h(x) = f(x) + g(x)\\)then\n \\(h(x)\\) is even if \\(f(x)\\) and \\(g(x)\\) are both even,\n and \\(h(x)\\) is odd if \\(f(x)\\) and \\(g(x)\\) are both odd.\n\\end{theorem}\n\n\\begin{exmp}\n Is \\(f(x) = x^3 - 2x\\) odd, even, neither? \\\\\n\n \\(sloution\\) Note thr \\(f(-x) = -x^3 + 2x = -f(x)\\) implying it's odd.\n\\end{exmp}\n\n\\section{Inverse Functions}\nAn inverse function is a function that reverses function \\(f\\).\nIf \\(f\\) is a function mapping \\(x\\) to \\(y\\),\nthen the inverse function of \\(f\\) maps \\(y\\) back to x. The\ninverse function of f is usually denoted by \\(f^{-1}\\)\n\n\\begin{itemize}\n \\item If \\(f^{-1}\\) exists, Then \\(f^{-1}(f(x)) = f(f^{-1}(x)) = x\\).\n \\item The graph of the inverse is the graph of the function reflected across\n \\(y = x\\).2\n\\end{itemize}\n\n\n\\begin{exmp}\n what's the inverse function of \\(f(x) = \\dfrac{3x-5}{2}? \\)\n \\\\\n\n \\(sloution\\) if we switch \\(x\\) and \\(y\\) we get \\(x = \\dfrac{3y-5}{2}\n \\Rightarrow \\) \\framebox[1.1\\width]{\\(f^{-1}(x) = \\dfrac{2x+5}{3}\\)}.\n\\end{exmp}\n\\subsection{Existence of an Inverse Function}\n\\begin{definition}[One to one]\n If a function satisfies the property that each x-value corresponds to one y-value, and\n each y-value corresponds to one x-value, then the function is one-to-one.\n\\end{definition}\n\\begin{theorem}[Inverse Function Criterion]\n If a function \\(f\\) is one-to-one, then its inverse is a function. More specifically, f is\n one-to-one if \\(f\\) is increasing/decreasing on its entire domain\n\\end{theorem}\n\\begin{theorem}[Horizontal Line Test]\n If you can draw a horizontal line passing through more than one point of a function\n on a graph, its inverse is not a function. If you cannot, it is a function.\n\\end{theorem}\nThis makes a lot of sense, since the graph of the inverse is just the function flipped\nacross the \\(y = x\\) line. This also makes it a lot easier to draw graphs.\n\n\\begin{exmp}\n Does the function \\(f(x) = \\sqrt{x-2} + 3\\) have an inverse function?\n\n \\begin{figure}[H]\n \\begin{small}\n \\begin{center}\n \\includegraphics[width=0.75 \\textwidth]{../out/inverse.png}\n \\end{center}\n \\caption{Graph of square root function.}\n \\label{fig: invetable Function}\n\n \\end{small}\n \\end{figure}\n\\end{exmp}\n\\section{problmes}\nTO BE ADDED \n\\end{document}\n", "meta": {"hexsha": "e54f45da8aed93c79d97e7a02eabe80d46a4495b", "size": 7665, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "basic/functions/src/functions.tex", "max_stars_repo_name": "GUC-Fractals/math-curriculum", "max_stars_repo_head_hexsha": "a11336def018106bd31e56e5eb9ac9225d4a072a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-10-06T09:36:30.000Z", "max_stars_repo_stars_event_max_datetime": "2021-10-06T09:36:30.000Z", "max_issues_repo_path": "basic/functions/src/functions.tex", "max_issues_repo_name": "GUC-Fractals/math-curriculum", "max_issues_repo_head_hexsha": "a11336def018106bd31e56e5eb9ac9225d4a072a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "basic/functions/src/functions.tex", "max_forks_repo_name": "GUC-Fractals/math-curriculum", "max_forks_repo_head_hexsha": "a11336def018106bd31e56e5eb9ac9225d4a072a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 31.9375, "max_line_length": 98, "alphanum_fraction": 0.6194390085, "num_tokens": 2522, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217431943271999, "lm_q2_score": 0.8519528076067262, "lm_q1q2_score": 0.6148911407781049}} {"text": "% Created 2018-04-11 mié 19:09\n\\documentclass[a4paper]{scrartcl}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{fixltx2e}\n\\usepackage{graphicx}\n\\usepackage{longtable}\n\\usepackage{float}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{marvosym}\n\\usepackage{wasysym}\n\\usepackage{amssymb}\n\\usepackage{hyperref}\n\\tolerance=1000\n\\usepackage{khpreamble}\n\\usepackage{pgfplots}\n\\author{Kjartan Halvorsen}\n\\date{Due Friday 2017-03-15}\n\\title{Computerized control - homework 4}\n\\hypersetup{\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 24.5.1 (Org mode 8.2.10)}}\n\\begin{document}\n\n\\maketitle\n\n\\section*{An observer for the harmonic oscillator model}\n\\label{sec-1}\nThe harmonic oscillator model, which in continuous-time has poles in \\(\\pm i\\omega\\), can be described in discrete-time as the state-space model\n\\begin{equation}\n\\begin{aligned}\nx(kh+h) &= \\underbrace{\\bbm \\cos\\omega{}h & \\frac{1}{\\omega} \\sin\\omega h\\\\ -\\omega\\sin\\omega h & \\cos\\omega h\\ebm}_{\\Phi(h)} x(kh) + \\underbrace{\\bbm \\frac{1}{\\omega^2}(1-\\cos\\omega h)\\\\\\frac{1}{\\omega}\\sin\\omega h\\ebm}_{\\Gamma(h)} u(kh)\\\\\ny(kh) &= \\underbrace{\\bbm 1 & 0\\ebm}_{C} x(kh)\n\\end{aligned}\n\\end{equation}\n\n\\begin{enumerate}\n\\item Determine the eigenvalues of \\(\\Phi(h)\\) (the discrete-time poles of the system). Verify that the poles are the same as obtained through the relationship \\(p = \\mexp{\\lambda h}\\) where \\(\\lambda\\) is the continuous-time pole and $p$ is the discrete-time pole.\n\\item Form the observability matrix \\[W_c = \\bbm C\\\\ C\\Phi(h) \\ebm\\] and determine for which values of $h$ the system becomes unobservable.\n\\item The observer for the system is given by the state-space system\n\\[ \\hat{x}(kh+h) = (\\Phi - KC) \\hat{x}(kh) + \\Gamma(h) u(kh) + Ky(kh) \\]\nLet \\(\\omega h = \\frac{\\pi}{6}\\) and determine a gain vector \\(K = \\bbm k_1 & k_2 \\ebm\\transp \\) that gives a deadbeat observer, i.e. an observer with characteristic polynomial \\(z^2\\) and poles in the origin.\n\\end{enumerate}\n\n\\section*{Active suspension}\n\\label{sec-2}\nA model of the front suspension of a motorcycle is shown in the figure \\ref{fig:quarter}. This model can also be a so-called \\emph{quarter model} for the suspension of a car. The system consists of two masses. The \\emph{sprung mass} is the larger of the two masses, and is about half the mass of the motorcyle and rider. The other mass is the much smaller \\emph{unsprung mass}, which consists of the mass of the wheel, tyre and part (approximately half) of the suspension system. The two masses are connected by two passive elements: a spring and a damper, as well as an active element: a linear force actuator. The unsprung mass is connected to the ground via a spring representing the tyre (sometimes there is also a damper included in the model of the tyre).\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.65\\linewidth]{../figures/active-suspension-nodamper}\n\\caption{Active suspension model. The heights $h_1(t)$, $h_2(t)$ and $h_g(t)$ are measured with respect to a stationary frame of reference. The vehicle is moving to the right.}\n\\label{fig:quarter}\n\\end{center}\n\\end{figure}\n\nThe reference height of the road surface is $h_g = h_0$. When the vehicle is travelling on a horizontal road of this height and the system is in equilibrium, the height of the two masses will be $h_1=h_{1,0}$ and $h_2 = h_{2,0}$, respectively. Introduce deviations from these reference heights\n\\begin{align*}\nh_1(t) &= h_{1,0} + z_1(t)\\\\\nh_2(t) &= h_{2,0} + z_2(t)\\\\\nh_g(t) &= h_{0} + w(t)\n\\end{align*}\n\nA slightly more complex model is discussed \\href{http://ctms.engin.umich.edu/CTMS/index.php?example=Suspension§ion=ControlStateSpace}{here}. That model also includes a damping in the tyre. You are welcome to study the example. \n\n\\subsection*{Determine the state space model}\n\\label{sec-2-1}\nThere are four forces acting on the sprung mass\n\\begin{center}\n\\begin{tikzpicture}\n\\node (M1) [draw, align=center, minimum width=3.5cm,minimum height=2cm] {Body mass};\n\\draw[->, thick] (M1.center) -- node[left, near end] {$m_1g$} ++(0, -3cm);\n\\draw[<-, thick] (M1.south) ++(-1cm, 0) -- node[left, near end] {$F_s$} ++ (0, -3cm);\n\\draw[<-, thick] (M1.south) ++(1cm, 0) -- node[left, near end] {$F_d$} ++ (0, -3cm);\n\\draw[<-, thick] (M1.south) ++(0.2cm, 0) -- node[right, near end] {$u$} ++ (0, -1.76cm);\n\\end{tikzpicture}\n\\end{center}\nThe equation of motion in the vertical direction becomes\n\\begin{equation}\n \\begin{split}\n m_1 \\ddot{z}_1 &= \\sum_i F_i = -m_1g + \\underbrace{m_1g -k_1(z_1-z_2)}_{F_s} + \\underbrace{\\big(-b_1(\\dot{z}_1 - \\dot{z}_2) \\big) }_{F_d} + u = -k_1(z_1 - z_2) - b_1(\\dot{z}_1 - \\dot{z}_2) + u\\\\\n &= -k_1z_1 -b_1\\dot{z}_1 + k_1z_2 + b_1\\dot{z}_2 + u. \n \\end{split}\n \\label{eq:eom1}\n \\end{equation}\nOn the unsprung mass, there are five forces acting:\n\\begin{center}\n\\begin{tikzpicture}\n\\node (M2) [draw, align=center, minimum width=3.5cm,minimum height=2cm] {Suspension mass};\n\\draw[->, thick] (M2.center) -- node[left, near end] {$m_2g$} ++(0, -3cm);\n\\draw[<-, thick] (M2.north) ++(-1cm, 0) -- node[left, near end] {$F_s$} ++ (0, 3cm);\n\\draw[<-, thick] (M2.north) ++(1cm, 0) -- node[left, near end] {$F_d$} ++ (0, 3cm);\n\\draw[<-, thick] (M2.north) ++(0cm, 0) -- node[right, near end] {$u$} ++ (0, 1.76cm);\n\\draw[<-, thick] (M2.south) ++(0.2cm, 0) -- node[right, near end] {$F_t$} ++ (0, -3.76cm);\n\\end{tikzpicture}\n\\end{center}\nThe equation of motion in the vertical direction becomes\n\\begin{equation}\n\\begin{split}\nm_2\\ddot{z}_2 &= -m2_g - \\underbrace{\\big(m_1g -k_1(z_1-z_2)\\big)}_{F_s}\n - \\underbrace{\\big(- b_1(\\dot{z}_1-\\dot{z}_2)\\big)}_{F_d}\n - \\underbrace{\\big( (m_1 + m_2)g - k_2(z_2 - w) \\big)}_{F_t} - u\\\\\n &= (k_1)z_1 + b_1\\dot{z}_1 - (k_1+k_2)z_2 - b_1\\dot{z}_2 + k_2w - u\n \\end{split}\n \\end{equation}\n\nUse the state-vector \n\\[ x = \\bbm x_1\\\\x_2\\\\x_3\\\\x_4 \\ebm = \\bbm z_1\\\\ \\dot{z}_1 \\\\ z_2\\\\ \\dot{z}_2 \\ebm \\]\nand let the velocity of the sprung mass $y(t) = \\dot{z}_1(t)$ be the output of the system. \n\\textbf{Set up the continous-time state-space model}\n\\begin{equation*}\n\\begin{aligned}\n\\dot{x}(t) &= Ax(t) + B_u u(t) + B_w w(t)\\\\\ny(t) &= C x(t)\n\\end{aligned}\n\\end{equation*}\nNote that there are two input signals: The active suspension force $u(t)$ and the disturbance from the road $w(t)$.\n\n\\subsection*{Simulate step-responses}\n\\label{sec-2-2}\nImplement the model in matlab. Use the following values for the parameters\n\\begin{center}\n\\begin{tabular}{l}\n$m1 = 140$\\\\\n$m2 = 14$\\\\\n$k1 = 4.6\\cdot{}10^4$\\\\\n$k2 = 3.0\\cdot{}10^5$\\\\\n$b1 = 10^3$\\\\\n\\end{tabular}\n\\end{center}\nHere is some code to help you\n\\begin{verbatim}\nm1 = 140;\nm2 = 14;\nk1 = 4.6e4;\nk2 = 3.0e5;\nb1 = 1e3;\n\nA = % Your code here\n\nBu = [0 \n 1/m1\n 0\n -1/m2 ];\nBw = [ 0\n 0\n 0\n k2/m2];\nC = % Your code here\nD=[0];\nsys_uy=ss(A,Bu,C,D); % System with only u(t) as input\nsys_wy=ss(A,Bw,C,D); % System with only w(t) as input\n\n% Step responses\nfigure(1)\nclf\nsubplot(121)\nstepplot(1000*sys_uy);\ntitle('Response from u (step 1kN)')\nxlabel('Time [s]')\nylabel('y [m/s]')\nsubplot(122)\nstepplot(0.1*sys_wy);\ntitle('Response from w (step 0.1m)')\nxlabel('Time [s]')\nylabel('y [m/s]')\n\\end{verbatim}\n\n\n\\subsection*{Sample the system}\n\\label{sec-2-3}\n\\begin{enumerate}\n\\item The system is quite oscillative, as the step-responses show. Plot the poles of the continuos-time system using \\texttt{pzmap} \n\\begin{verbatim}\nfigure(2)\npzmap(sys_uy)\n\\end{verbatim}\nChoose a sampling period $h$, such that $\\omega_0h = 0.8$, where $\\omega_0$ is the natural frequency (in radians per second) of the fastest poles.\n\\item Sample the system numerically using \\texttt{c2d} in matlab:\n\\begin{verbatim}\nh = % Your code here\nsys_uy_d = c2d(sys_uy, h); % Zero-order-hold sampling\nsys_wy_d = c2d(sys_wy, h); % Zero-order-hold sampling\n\\end{verbatim}\n\\end{enumerate}\n\n\\subsection*{Determine a state feedback}\n\\label{sec-2-4}\n\\begin{enumerate}\n\\item First determine the desired poles. The system is fourth order, so we need to choose four poles for the closed-loop system. Assume we want the closed-loop system to have more or less the same speed as the two fastest open-loop poles (same distance from the origin), but more damped. Determine the four desired closed-loop continuous-time poles, and then translate them to discrete time:\n\\begin{verbatim}\np_c = [ ] % Your four desired closed-loop poles in continuous-time here\np_d = exp(p_c*h) % The corresponding four discrete-time poles\n\\end{verbatim}\n\\item Determine the feedback gain vector \\(L\\), and form the closed-loop system. Matlab will happily do the calculations for you:\n\\begin{verbatim}\n[Phi, Gamma_u, C, D] = ssdata(sys_uy_d) % Get the matrices\n[Phi, Gamma_w, C, D] = ssdata(sys_wy_d) \nL = place(Phi, Gamma_u, p_d) \nsys_wy_d_closed = ss(Phi-Gamma_u*L, Gamma_w, C, D, h); % Closed-loop system from w to y\n\\end{verbatim}\n\\end{enumerate}\n\n\\subsection*{Simulate the closed-loop response}\n\\label{sec-2-5}\nCheck the performance of your closed-loop system. Let the disturbance (road profile) look as below\n\\begin{center}\n\\begin{tikzpicture}\n\\begin{axis} [\n width = 12cm, height = 4cm,\n xlabel = {$t$ [s]},\n ylabel = {$w(t) [m]$},\n xmin = -0.5, xmax = 10,\n ymin = -0.5, ymax = 0.5,\n ]\n \\draw[thick] (axis cs: -0.5,0) -- (axis cs: 0,0) -- (axis cs: 0,0.3) -- (axis cs: 2,0.3) -- (axis cs: 2,-0.3) -- (axis cs: 4,-0.3) -- (axis cs: 4, 0) -- (axis cs: 10,0);\n \\end{axis}\n \\end{tikzpicture}\n \\end{center}\n\nIn matlab:\n\\begin{verbatim}\nT = ( 0:ceil(10/h) )*h; % Discrete time-vector from 0 to 10 seconds\nw = zeros(size(T));\nind_before_2 = find(T < 2);\nw(ind_before_2) = 0.3;\nind_between_2_and_4 = intersect( find(T>2), find(T<4) );\nw(ind_between_2_and_4) = -0.3;\n\n[y, Tsim, x] = lsim(sys_wy_d_closed, w, T);\n[yo, Tsim, xo] = lsim(sys_wy_d, w, T); % Simulate system without feedback\n\nfigure(3)\nset(gcf, 'position', [100, 100, 900, 700])\nfor i=1:4\n subplot(5,1,i)\n stairs(T, x(:,i), 'color', [0, 0.447, 0.741])\n hold on\n stairs(T, xo(:,i), 'color', [0.85, 0.325, 0.098])\n ylabel(sprintf('state %d', i))\n xlabel('Time [s]')\nend\nlegend('Closed-loop response', 'Open-loop response')\nsubplot(5,1,5)\nstairs(T, -L*x', 'color', [0, 0.447, 0.741])\nylabel('Control signal [N]')\n\\end{verbatim}\n\n\\textbf{Comment on the figures!} For instance: How do the closed-loop system response compare to the open-loop response? What was the maximum active suspension force (\\(\\max u(t)\\))? \n\n\n\\section*{Solutions}\n\\label{sec-3}\n\n\\subsection*{An observer for the harmonic oscillator model}\n\\label{sec-3-1}\n\\begin{enumerate}\n\\item The eigenvalues are given by the roots of \\(\\det (zI - \\Phi)\\). Writing \\(\\cos\\omega h = c\\) and \\(\\sin\\omega h = s\\) for simplicity, we get\n\\begin{equation*}\n\\begin{aligned}\n\\det (zI - \\Phi) &= \\det \\left( \\bbm z & 0\\\\0 & z\\ebm - \\bbm c & \\frac{s}{\\omega}\\\\-\\omega s & c\\ebm \\right) = \\det \\bbm z-c & -\\frac{s}{\\omega}\\\\ \\omega s & z-c\\ebm \\\\\n&= (z-c)^2 + s^2\n\\end{aligned}\n\\end{equation}\nwith roots on the unit circle:\n\\[ z = c \\pm is = \\cos\\omega h \\pm i \\sin\\omega h.\\]\nUsing the relationship \\(z = \\mexp{sh}\\) between the s-plane and the z-plane we get for the continous-time poles \\(s=\\pm i\\omega\\)\n\\[ z = \\mexp{\\pm i\\omega h} = \\cos\\omega h \\pm i\\sin\\omega h \\]\nusing Euler's formula.\n\\item The observability matrix becomes\n\\[ W_c = \\bbm C\\\\C\\Phi \\ebm = \\bbm 1 & 0\\\\ c & \\frac{s}{\\omega},\\] so\n\\[ \\det W_c = \\frac{1}{\\omega} \\sin\\omega h \\]\nwhich is zero for \\(\\omega h = k\\pi, \\; k=1,2,\\ldots\\). So observability is lost for\n\\[ h = \\frac{k\\pi}{\\omega}, \\; k=1,2\\ldots. \\]\n\\item With \\(\\omega=1\\) and \\(h=\\frac{\\pi}{6}\\) we get \\(\\cos\\omega h = \\sqrt{3}/2\\) and \\(\\sin\\omega h = 1/2\\). This gives\n\\end{enumerate}\n\\[ \\Phi = \\frac{1}{2} \\bbm \\sqrt{3} & 1\\\\-1 & \\sqrt{3} \\ebm\\].\nThe characteristic polynomial of the observer becomes\n\\begin{equation*}\n\\begin{aligned}\n\\det \\big(zI - (\\Phi - KC)\\) &= \\det \\left( \\bbm z & 0\\\\0 & z\\ebm - \\frac{1}{2} \\bbm \\sqrt{3} - 2k_1 & 1\\\\-1-2k_2 & \\sqrt{3}\\ebm\\\\\n&= \\det \\bbm z-\\frac{\\sqrt{3}-2k_1}{2} & -\\frac{1}{2}\\\\\\frac{1+2k_2}{2} & z - \\frac{\\sqrt{3}}{2} \\ebm\\\\\n&= (z-\\frac{\\sqrt{3}-2k_1}{2})(z - \\frac{\\sqrt{3}}{2}) + \\frac{1+2k_2}{4}\\\\\n&= z^2 - (\\sqrt{3}-k_1)z + \\frac{1}{4}\\big( \\sqrt{3}(\\sqrt{3}-2k_1) + 1 + 2k_2\\big)\\\\\n&= z^2 - (\\sqrt{3}-k_1)z + \\frac{1}{4}\\big(-2\\sqrt{3}k_1 + 2k_2 + 4\\big)\n\\end{aligned}\n\\end{equation*}\nSetting coefficient equal to the coefficients of the desired characteristic polynomial \\(z^2\\) gies\n\\begin{align*}\n\\sqrt{3}-k_1 &= 0 \\quad \\Rightarrow \\quad k_1 = \\sqrt{3}\\\\\n-2\\sqrt{3}k_1 + 2k_2 + 4 &= 0 \\quad \\Rightarrow \\quad k_2 = 3-2 = 1\n\\end{align*}\n\n\\subsection*{Active suspension}\n\\label{sec-3-2}\n\\subsubsection*{State space model}\n\\label{sec-3-2-1}\nWith the state vector \n \\[ x = \\bbm x_1\\\\x_2\\\\x_3\\\\x_4 \\ebm = \\bbm z_1\\\\ \\dot{z}_1 \\\\ z_2\\\\ \\dot{z}_2 \\ebm \\]\n we can write the four first-order differential equations of the state space model\n \\begin{align*}\n \\dot{x}_1 &= \\dot{z}_1 = x_2\\\\\n \\dot{x}_2 &= \\ddot{z}_1 = \\frac{1}{m_1}\\Big(-k_1z_1 -b_1\\dot{z}_1 + k_1z_2 + b_1\\dot{z}_2 + u\\Big)\\\\\n &= -\\frac{k_1}{m_1}x_1 - \\frac{b_1}{m_1}x_2 + \\frac{k_1}{m_1}x_3 + \\frac{b_1}{m_2}x_4 + \\frac{1}{m_1} u\\\\\n \\dot{x}_3 &= \\dot{z}_2 = x_4\\\\\n \\dot{x}_4 &= \\ddot{z}_2 = \\frac{1}{m_2} \\Big(k_1z_1 + b_1\\dot{z}_1 - (k_1+k_2)z_2 - b_1\\dot{z}_2 + k_2w - u\\Big)\\\\ \n &= \\frac{k_1}{m_2} x_1 + \\frac{b_1}{m_2} x_2 - \\frac{k_1+k_2}{m_2}x_3 + -\\frac{b_1}{m_2}x_4 + \\frac{k_2}{m_2}w - \\frac{1}{m_2}u.\n\\end{align*}\nIn state-space form, this can be written\n\\begin{align*}\n\\dot{x} &= \\bbm 0 & 1 & 0 & 0\\\\ \n-\\frac{k_1}{m_1} & -\\frac{b_1}{m_1} & \\frac{k_1}{m_1} & \\frac{b_1}{m_1}\\\\\n0 & 0 & 0 & 1\\\\\n\\frac{k_1}{m_2} & \\frac{b_1}{m_2} & -\\frac{k_1+k_2}{m_2} & -\\frac{b_1}{m_2}\\ebm\n\\bbm x_1\\\\x_2\\\\x_3\\\\x_4 \\ebm + \\bbm 0\\\\0\\\\0\\\\\\frac{k_2}{m_2} \\ebm w + \\bbm 0\\\\\\frac{1}{m_1}\\\\0\\\\-\\frac{1}{m_2} \\ebm u \\\\\ny &= \\bbm 0 & 1 & 0 & 0 \\ebm x.\n\\end{align*}\n\n\\subsubsection*{step-responses}\n\\label{sec-3-2-2}\nThe figure below shows the step-responses of both the continuous- and the sampled systems.\n\\begin{center}\n\\includegraphics[width=0.6\\linewidth]{active-susp-plant-response}\n\\end{center}\n\n\\subsubsection*{Sampling the system}\n\\label{sec-3-2-3}\n\\begin{enumerate}\n\\item The pole-zero map below shows that the fast poles have a distance of \\(\\omega = 156\\) rad/s to the origin. The rule-of-thumb (chapter 2.9 in the textbook) tells us that we should have \\(\\omega h \\approx 0.2 -- 0.6\\). For instance,\n\\[ h = \\frac{0.6}{156} \\approx \\unit{0.004}{\\second}. \\]\n\\item Sampling the system is straightforward \n\\begin{verbatim}\nh = 0.004;\nsys_uy_d = c2d(sys_uy, h); % The system from input u to y\nsys_wy_d = c2d(sys_wy, h); % From input w to y\n\\end{verbatim}\n\\end{enumerate}\n\\subsubsection*{State feedback}\n\\label{sec-3-2-4}\n\\begin{enumerate}\n\\item Desired closed-loop poles. The idea is to choose the poles to be equally fast, but more damped than the fast poles of the plant. For instance we can choose the set of four poles \n\\[ p_{1,2} = -\\zeta_1\\omega_n \\pm i \\sqrt{1 - \\zeta_1^2}\\omega_n, \\quad\n p_{3,4} = -\\zeta_2\\omega_n \\pm i \\sqrt{1 - \\zeta_2^2}\\omega_n \\]\nwith natural frequency \\(\\omega_n = 156\\), and damping ratio \\(\\zeta_1 = 0.8\\) and \\(\\zeta_2 = 0.6\\).\n\\item The feedback vector is calculated by Matlab and gives\n$\\backslash$[ L = \\bbm l$_{\\text{1}}$ \\& l$_{\\text{2}}$ \\& l$_{\\text{3}}$ \\& l$_{\\text{4}}$ \\ebm = \\bbm 1.89$\\cdot$ 10$^{\\text{6}}$ \\& 3.85 $\\cdot$ 10$^{\\text{4}}$ \\& -3.23$\\cdot$ 10$^{\\text{5}}$ \\& 6.20 $\\cdot$ 10$^{\\text{2}\\ebm}$.\n\\end{enumerate}\n\n\\subsubsection*{Simulations}\n\\label{sec-3-2-5}\nWith the state feedback proposed, the response is as shown in the figure below.\n\\begin{center}\n\\includegraphics[width=\\linewidth]{../matlab/active-suspension-sim-crop}\n\\end{center}\nWe can see that the closed-loop response of \\(z_1\\) and \\(\\dot{z}_1\\) are both very well damped, and has much smaller amplitude than the open-loop system. The control signal shows quite large peak (30-50 kN) at the onset of the disturbances. \n\n\\subsubsection*{Choosing the closed-loop poles too slow.}\n\\label{sec-3-2-6}\nThere is a potential problem with choosing the closed-loop poles (much) slower than the open-loop poles in this exercise. Slow poles in discrete-time means close to 1. The reason is that state feedback does not change the numerator in the pulse transfer function from input to output. So, if the open-loop pulse transfer function from disturbance to output signal is\n\\[ H_w(z) = C (zI-\\Phi)^{-1} \\Gamma_w = \\frac{\\beta(z)}{\\alpha(z)},\\]\nthen the closed-loop transfer function will be \n\\[ H_c(z) = C \\big(zI -(\\Phi-\\Gamma_u L)\\big)^{-1}\\Gamma_w = \\frac{\\beta(z)}{\\alpha_c(z)}.\\]\nThe denominator is obtained by design (choice of pole placement) to have the roots (desired closed-loop poles) \\(p_{d,1}\\), \\(p_{d,2}\\), \\(p_{d,3}\\), \\(p_{d,4}\\), so \n\\[ H_c(z) = \\frac{\\beta(z)}{(z-p_{d,1})(z-p_{d,2})(z-p_{d,3})(z-p_{d,4})}.\\]\nThe static gain becomes\n\\[ H_c(1) = \\frac{\\beta(1)}{(1-p_{d,1})(1-p_{d,2})(1-p_{d,3})(1-p_{d,4})},\\]\nwhich can be much larger than the open-loop static gain \\(H_w(1)\\), if the closed-loop poles are much closer to 1 than the open-loop poles. \n% Emacs 24.5.1 (Org mode 8.2.10)\n\\end{document}", "meta": {"hexsha": "d45dce490c4cd862dbd1cc494e6b7bb372ebe99f", "size": 17233, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "homework/historical/hw4-ss-active-suspension-spring18.tex", "max_stars_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_stars_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-11-07T05:20:37.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-22T09:46:13.000Z", "max_issues_repo_path": "homework/historical/hw4-ss-active-suspension-spring18.tex", "max_issues_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_issues_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-06-12T20:44:41.000Z", "max_issues_repo_issues_event_max_datetime": "2020-06-12T20:49:00.000Z", "max_forks_repo_path": "homework/historical/hw4-ss-active-suspension-spring18.tex", "max_forks_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_forks_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-14T03:55:27.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-14T03:55:27.000Z", "avg_line_length": 46.2010723861, "max_line_length": 762, "alphanum_fraction": 0.6621598097, "num_tokens": 6477, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217431943271999, "lm_q2_score": 0.8519528076067262, "lm_q1q2_score": 0.6148911407781049}} {"text": "\n\\subsection{The equation of exchange}\n\nAs there’s only so much stuff out there, an increase in the amount of money moving around with cause inflation.\n\nThis can be shown as:\n\n\\(MV=PQ\\)\n\nThe amount of money (M) multiplied by the velocity at which money moves (V) must be equal to the average price (P) multiplied by the quantity of stuff (Q). This isn’t a theory, it’s an accounting identity.\n\n", "meta": {"hexsha": "b90e19d1dd6d94e2d64a42eada5148b60638f454", "size": 394, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/economics/money/02-01-equationExchange.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/economics/money/02-01-equationExchange.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/economics/money/02-01-equationExchange.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.8333333333, "max_line_length": 205, "alphanum_fraction": 0.7512690355, "num_tokens": 93, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8175744850834648, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6148262927679573}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{amsthm}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\n\n\n\n\n% collectivised from: https://www.overleaf.com/learn/latex/Theorems_and_proofs#Theorem_styles\n\\theoremstyle{definition}\n\\newtheorem{definition}{Definition}[section]\n\n\\theoremstyle{remark}\n\\newtheorem*{remark}{Remark}\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{lemma}[theorem]{Lemma}\n%\n\\begin{document}\n%Plan\n% 1. Introduction\n% Original problem: US graphs and edge coloration with \\omega = 2. \n% Known for delta 2. we know delta is at most 5. Concept of direction.\n% 2. On a grid \n% \"what we will do here: ... \". RK: if not all on a grid are US, all US are on a grid. Thus any result on on a grid graph is true for US. \n% 2.1 On a grid - definition. \n% 2.2 Additional concepts: Type 1/2 diagonals, order on the diagonals.\n% 2.3 Algorithm \n% 2.4 Complexity\n% 3. Applications\n% 3.1 US graphs: chromatic index of 4\n% 3.2 angles & directions\n\\section{Introduction}\nThe original problem is to study the chromatic index of triangle-free \\textit{unit square intersection graphs},\nwhich are graphs that are obtained for a family of unit squares. \nEach square is represented by a vertex in a graph, and each edge represent the fact that two squares are intersecting. \\\\\nWe already know that when the maximum degree $\\Delta$ is $2$, there may exists graphs of type 2, for instance any odd cycle. We can also observe that\nwhen two unit squares intersects, there is a corner of each inside the other; from this observation follows that there cannot be \na triangle-free unit square graph with $\\Delta \\ge 5$. \\\\\nIn what follows, we provide a proof that there cannot be a graph of type 2 when $\\Delta = 4$. The proof will be done by introducing a new way to\nrepresent the graphs by using a grid; and we will call them \\textit{on a grid} graphs.\n\n\\section{Graphs on-a-grid}\nWe present in this section a proof that unit square graphs are also \\textit{on a grid} graphs. \n\\subsection{On a grid}\n\\begin{definition}[On a grid]\n A graph is said \\textit{on a grid} if:\n \\begin{itemize}\n \\item All the edges are either parallels or perpendicular.\n \\item All the edges are of integer lengths.\n \\item No edge is overlapping or containing an other edge in more than one point.\n \\end{itemize}\n\\end{definition}\n\nInstead of using the usual directions of the cartesian plane, we will rotate all the edges by $\\tfrac{\\pi}{4}$ as it will be more convenient later.\nThe graph represented in Fig.1. is \\textit{on a grid}.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[scale=0.2]{tex_images/on_a_grid_g1.png}\n \\caption{$G_1$ is on a grid.}\n\\end{figure}\n\n\\begin{remark}\n $G_1$ has exactly $10$ edges.\n\\end{remark}\n\n\\subsection{Additional concepts}\n\n\\begin{definition}[Direction of intersection]\n Let $a, b$ be two unit squares. We say that $a$ intersects $b$ in UR direction (resp. UL, DL, DR) if the up-right (resp up-left, down-left, down-right) corner\n of $a$ is inside $b$. \n\\end{definition}\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[scale = 0.3]{tex_images/same_diagonals.png}\n \\caption{Illustration of direction of intersections.}\n\\end{figure}\n\nFor instance, in Fig. 2, $D$ intersect $F$ in direction DR, $G$ in direction UR and $A$ in direction UL. Conversely, $A$ intersect $D$ in direction DR.\n\n\\begin{lemma}\n If $a$ intersect both $b$ and $c$ in the same direction, then $a, b$ and $c$ forms a triangle.\n\\end{lemma}\n\n\\begin{proof}\n Notice that they would all share a common point in a certain corner of $a$.\n\\end{proof}\n\n\\begin{definition}[Diagonals]\n Let $a$ and $b$ be two squares. They are said to be on the same \\textit{Type 1 diagonal} (resp. Type 2) if you can find a sequence $\\{s_1, ... s_k\\}$ of squares such that:\n \\begin{itemize}\n \\item $s_1 = a$\n \\item $s_k = b$\n \\item $\\forall i \\in \\{1, .. , k-1\\}$, $s_i$ intersects $s_{i+1}$ in UL or DR (resp. UR or DL) direction.\n \\end{itemize} \n A diagonal is a set of points (possibly trivial) that are on the same diagonal. \n\\end{definition}\n\n\\begin{remark}\n We will denote the diagonal as follows : $D:= \\{d_1, ... , d_k\\}$. In this case $d_1$ is the highest square of the diagonal, meaning that $d_1$ intersect $d_2$ in DR direction in the case of a type 1 diagonal, and so on.\n\\end{remark}\n\nIn Fig. 2 we can denote the following diagonals:\\\\\nType 1: $\\{(A, D, F), (E), (I), (G, H)\\}$\\\\\nType 2: $\\{(A), (D,G), (F, E, I, H)\\}$\\\\\n\n\n\\begin{lemma}\n Let $D$, $E$ be two diagonals of the same type. If we can find $d \\in D$; $e \\in E$ such that $d$ intersect $e$ in some direction X, then: \\\\\\\n $\\forall d' \\in D$, $\\forall e' \\in E$, either $e'\\cap d' = \\emptyset$ or $d'$ intersect $e'$ in direction X.\n\\end{lemma}\n\n\\begin{proof}\n Without loss of generality, let $D:=\\{d_1, ... d_k\\}$ and $E:= \\{e_1, ... e_l\\}$ be two distinct diagonals of Type 1. \n Suppose $d_1$ intersect $e_1$ in direction DL. \\\\\n Let $(i,j)$ be the smallest couple different from $(1,1)$ such that $d_i \\cap e_j \\ne \\emptyset$. Observe that as $D$ and $E$ are Type 1, the only possibility is that \n $d_i$ intersect $e_j$ in direction UR or DL.\n\\end{proof}\n\nWe can now use the previous lemma to define a partial order on our diagonals.\n\n\\begin{definition}[Order on diagonals]\n Let $D_1$, $D_2$ two diagonals of Type 1 (resp Type 2). We say that $D_1$ is \\textit{directly higher} than $D_2$, or $D_1 >_d D_2$ if:\n $\\exists d_1 \\in D_1, \\exists d_2 \\in D_2$ such that $d_1$ intersects $d_2$ in DL (resp. UR) direction. \\\\ \\\\\n We say that $D_1$ is higher than $D_2$, or $D_1 > D_2$ if we can find a sequence of diagonals $\\{s_1, ... , s_k\\}$ such that:\n \\begin{itemize}\n \\item $s_1 = D_1$\n \\item $s_k = D_2$\n \\item $\\forall i \\in \\{1, ..., k-1\\}, s_i >_d s_{i+1}$\n \\end{itemize} \n\\end{definition}\n\n%% TODO: explain why we cant have D >_d E and E >_d D at the same time \n\n\\begin{remark}\n Note that $>_d$ is not an order as it is not transitive. \n\\end{remark}\n\n\n\\subsection{From unit squares to the grid}\n\nWe present here an algorithm to obtain a on a grid graph given a family of unit squares.\\\\\nStart by identifying all the diagonals: to do so,\n\n\\section{Some applications}\n\\subsection{Edge coloration}\n\nOne can easily see that any on a grid graph has a chromatic index of at most $4$. As every triangle-free unit square graph is also on a grid, \nit follows that no graph of type 2 can exist for these if $\\Delta = 4$.\n\n\\begin{theorem}\n Let $G$ be a unit square graph, with $\\Delta = 4$ and $\\omega = 2$. Then $\\chi'(G) = 4$.\n\\end{theorem}\n\n\\begin{proof}\n It is sufficient to show it is true for all graphs that are on a grid. \\\\\n Let $G$ be on a grid. Delete for a moment all the edges that correspond to type 1 diagonals. There subsits only paths. \n As all paths can be colored using only two colors, assign color 1 or 2 to all the remaining edges. \\\\\n Do the same thing for the type 2 diagonals with colors 3 and 4. \\\\\n As all the edges of $G$ are either on a type 1 or a type 2 diagonal, all the edges are assigned to a color. If two edges are on the same type of diagonals, either they\n have different colors or they are not adjacent. If two edges are on different type of diagonals, they have different colors. Therefore, this coloration is acceptable.\n\\end{proof}\n\n\\subsection{Directions in cycles}\n\n\n\\end{document}", "meta": {"hexsha": "d00b9f783a1146b3edee02080d8092a28ec1edb4", "size": 7481, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/on_a_grid/on_a_grid.tex", "max_stars_repo_name": "Qiselong/Internship-GSCOP21", "max_stars_repo_head_hexsha": "c33b09fb889181d5d40c785e964f7748a82dc61f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/on_a_grid/on_a_grid.tex", "max_issues_repo_name": "Qiselong/Internship-GSCOP21", "max_issues_repo_head_hexsha": "c33b09fb889181d5d40c785e964f7748a82dc61f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/on_a_grid/on_a_grid.tex", "max_forks_repo_name": "Qiselong/Internship-GSCOP21", "max_forks_repo_head_hexsha": "c33b09fb889181d5d40c785e964f7748a82dc61f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.0058823529, "max_line_length": 224, "alphanum_fraction": 0.6862718888, "num_tokens": 2247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8175744761936437, "lm_q1q2_score": 0.6148262860827001}} {"text": "\\documentclass[12pt]{article}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{geometry}\n\\usepackage{enumerate}\n\\usepackage{natbib}\n\\usepackage{float}%稳定图片位置\n\\usepackage{graphicx}%画图\n\\usepackage[english]{babel}\n\\usepackage{a4wide}\n\\usepackage{indentfirst}%缩进\n\\usepackage{enumerate}%加序号\n\\usepackage{multirow}%合并行\n\n\n\\begin{document}\n\\newpage\n\\section{Problem 1}\n\\subsection{(a)}\n$$I_c=I_s \\exp(\\frac{q V_{BE}}{KT}-1)(1+\\frac{V_{CE}}{V_A})=10^{-16}\\exp(\\frac{0.7}{0.0259}-1)(1+\\frac{V_{CE}}{100})=5.47\\cdot10^{-5}*(1+0.01*V_{OUT})$$\n$$V_{OUT}=V_{CC}-R_c*I_c=3-5000I_c$$\n$$V_{OUT}=2.72(V)$$\n$$I_c=\\frac{3-2.9}{5000}=5.6\\cdot10^{-5}(A)$$\n$$gm=\\frac{q\\cdot I_c}{KT}=\\frac{5.6\\cdot10^{-5}}{0.0259}=2.162\\cdot10^{-3}$$\n$$r_0=\\frac{V_A}{I_c}=\\frac{100}{2.07\\cdot 10^{-5}}=1.79\\cdot10^{6}(\\Omega)$$\n$$A_v=-gm(R_c||r_0)=-2.162\\cdot10^{-3}\\cdot\\frac{5000*1.79\\cdot10^{6}}{5000+1.79\\cdot10^{6}}=-10.78$$\n\\subsection{(b)}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.25]{P1.png}\n\\end{figure}\n$$A_v=\\frac{18}{-1.7}=-10.6$$\nThe slope is about -10.6, which is a little bigger than calculated one in (a)\n\\subsection{(c)}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.25]{P2.png}\n\\end{figure}\n$$|A_v|=|\\frac{(2.8-2.9)}{0.69-0.7}|\\approx10$$\nThe value is very close to part (b) and (a) but the absolute value is a bit smaller.\n\\subsection{(d)}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.25]{P3.png}\n\\end{figure}\n$$|A_v|=|\\frac{(2.96-2.9)}{0.65-0.7}|\\approx47.2$$\nThe value is much more greater than calculated one in previous part, I think it's because the amplitude is so big that the input signal can't be considered as a small signal any more. As a result, the derivative for $V_{BE}$ can't be applied any more gm has changed.\n\\end{document}", "meta": {"hexsha": "406edce887063a620308b275c47fb8886caa7f4e", "size": 1757, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "VE311SemiConductorCircuit/HW/HW4/HW/HW.tex", "max_stars_repo_name": "PANDApcd/SemiConductorCircuit", "max_stars_repo_head_hexsha": "430ffaeea0830dc3105883374dd729fe9f86cc55", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "VE311SemiConductorCircuit/HW/HW4/HW/HW.tex", "max_issues_repo_name": "PANDApcd/SemiConductorCircuit", "max_issues_repo_head_hexsha": "430ffaeea0830dc3105883374dd729fe9f86cc55", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "VE311SemiConductorCircuit/HW/HW4/HW/HW.tex", "max_forks_repo_name": "PANDApcd/SemiConductorCircuit", "max_forks_repo_head_hexsha": "430ffaeea0830dc3105883374dd729fe9f86cc55", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.6041666667, "max_line_length": 266, "alphanum_fraction": 0.6841206602, "num_tokens": 721, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.817574471748733, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6148262827400713}} {"text": "\\documentclass[12pt]{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath}\n\\usepackage{hyperref}\n\n\\title{Piecewise Polynomial Path Planner (P4)}\n\\author{Tucker Haydon}\n\\date{Spring 2019}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Introduction}\nThis document describes the underlying theory for the piecewise polynomial path\nplanner.\n\n\\section{Polynomial Theory}\n\\subsection{Problem Statement}\nGiven a set of waypoints and corresponding arrival times, construct a\npiecewise-continuous polynomial trajectory through these waypoints that minimizes\nthe norm of a specified derivative. Furthermore, the trajectory --- or any\nderivative of the trajectory --- may be subject to linear inequality\nconstraints.\n\\bigbreak\n\\noindent Mathematically, the problem may be stated so: given a set of $m$ waypoints \n%\n\\begin{equation}\n \\mathbf{W} = \n \\begin{bmatrix}\n \\vec{w}_{0} & \\vec{w}_{1} & \\vec{w}_{2} & \\hdots & \\vec{w}_{m}\n \\end{bmatrix}\n \\in \\mathcal{R}^a\n\\end{equation}\n%\nand a set of times to arrive at these waypoints \n\\begin{equation}\n \\vec{T} = \n \\begin{bmatrix}\n t_{0} & t_{1} & t_{2} & \\hdots & t_{m}\n \\end{bmatrix}\n\\end{equation}\n%\ndetermine a set of $n$'th-order piecewise-continuous polynomials that smoothly connect these waypoints\n%\n\\begin{equation}\n \\vec{p}(t) = \n \\begin{cases}\n p_{0}(t) = \\sum_{k=0}^{n} p_{0,k} \\cdot (\\frac{1}{k!} t^{k}) & t_{0} \\leq t < t_{1} \\\\ \n p_{1}(t) = \\sum_{k=0}^{n} p_{1,k} \\cdot (\\frac{1}{k!} t^{k}) & t_{1} \\leq t < t_{2} \\\\ \n \\vdots \\\\\n p_{m}(t) = \\sum_{k=0}^{n} p_{m,k} \\cdot (\\frac{1}{k!} t^{k}) & t_{m-1} \\leq t < t_{m}\n \\end{cases}\n\\end{equation}\n%\nwhile minimizing the squared norm of the $r$'th derivative\n%\n\\begin{equation}\n \\min\n \\int_{t_{0}}^{t_{m}} \\frac{d^{r}}{dt^{r}}|| \\vec{p}(t) ||^{2} dt\n\\end{equation}\n%\nsubject to continuity and path constrains (see sections \\ref{} \\& \\ref{}).\n\\bigbreak\n\\noindent This problem construction and the following solution follow those of Mellinger\nand Kumar \\cite{mellinger2011minimum}.\n\n\\subsection{Polynomial Basis}\nTrajectories for quadcopters may be specified with no more than four independent\ndimensions: x, y, z, and yaw. Any more dimensions (roll, pitch) introduce\nconstraints that may render certain specified trajectories infeasible. For\nexample, a quadcopter cannot be simultaneously stationary and have a non-zero\nroll angle \\cite{mellinger2011minimum}.\n\nImportantly, the x, y, z, and yaw dimensions may be specified\n\\textit{independently}: each dimension of the polynomial basis functions may be\nconsidered alone, permitting scalar analysis. The following section uses a\n3rd-order polynomial, but the analysis generalizes to any order polynomial.\n\nConsider a 3rd-order polynomial representing the first section of trajectory.\n\\begin{equation*}\n p_{0}(t) = \\frac{1}{0!}t^{0}x_{0} + \\frac{1}{1!}t^{1}v_{0}\n + \\frac{1}{2!}t^{2}a_{0} + \\frac{1}{3!}t^{3}j_{0}\n\\end{equation*}\n%\nThis expression may be divided into two components: a vector of coefficients and\na vector of basis functions.\n\\begin{equation*}\n p_{0}(t) = \n \\underbrace{\n \\begin{bmatrix}\n x_{0} & v_{0} & a_{0} & j_{0}\n \\end{bmatrix}\n }_{\\text{coefficients}}\n \\underbrace{\n \\begin{bmatrix}\n \\frac{t^{0}}{0!} \\\\[6pt]\n \\frac{t^{1}}{1!} \\\\[6pt]\n \\frac{t^{2}}{2!} \\\\[6pt]\n \\frac{t^{3}}{3!}\n \\end{bmatrix}\n }_{\\text{basis functions}}\n\\end{equation*}\n%\nOr in vector form\n\\begin{equation*}\n p_{0}(t) = \\vec{x}^{T} \\vec{t}\n\\end{equation*}\n\n\\subsection{Polynomial Optimization}\nFor each section in the trajectory, the $r$'th derivative must be minimized.\n\\begin{align*}\n x_{i} &= \\text{arg} \\min_{\\vec{x}} \\int_{t_{i}}^{t_{i+1}} \\bigg[ \\frac{d^{r}\n p(\\tau)}{d \\tau^{r}} \\bigg]^2 dt \\\\\n &= \\text{arg} \\min_{\\vec{x}} \\int_{t_{i}}^{t_{i+1}} \\bigg[ \\frac{d^{r}\n p(\\tau)}{d \\tau^{r}}\\bigg]^T \\bigg[ \\frac{d^{r} p(\\tau)}{d \\tau^{r}}\\bigg]\n dt \\\\\n &= \\text{arg} \\min_{\\vec{x}} \\int_{t_{i}}^{t_{i+1}} \\vec{x}^T \\frac{d^{r}\n \\vec{t}(\\tau)}{d \\tau^{r}}^{T} \\frac{d^{r} \\vec{t}(\\tau)}{d \\tau^{r}}\n \\vec{x} dt \\\\\n &= \\text{arg} \\min_{\\vec{x}} \\vec{x}^{T} \\int_{t_{i}}^{t_{i+1}} \\frac{d^{r}\n \\vec{t}(\\tau)}{d \\tau^{r}}^{T} \\frac{d^{r} \\vec{t}(\\tau)}{d \\tau^{r}} dt\n \\vec{x}\n\\end{align*}\n%\nBy defining\n\\begin{equation*}\n \\mathbf{Q} = \\int_{t_{i}}^{t_{i+1}} \\frac{d^{r} \\vec{t}(\\tau)}{d\n \\tau^{r}}^{T} \\frac{d^{r} \\vec{t}(\\tau)}{d \\tau^{r}} dt\n\\end{equation*}\n%\nand noting that the time-varying components will be integrated out by the\ndefinite integral, the optimization problem may be written as\n\\begin{equation*}\n x_{i} = \\text{arg} \\min_{\\vec{x}} \\vec{x}^{T} \\mathbf{Q} \\vec{x}\n\\end{equation*}\n%\nThe structure of the integrand of $\\mathbf{Q}$ is revealed by first computing the initial couple\nderivatives of $\\vec{t}(\\tau)$.\n\\begin{align*}\n \\vec{t(\\tau)} &= \n \\begin{bmatrix}\n \\frac{\\tau^{0}}{0!} & \\frac{\\tau^{1}}{1!} & \\frac{\\tau^{2}}{2!} & \\hdots\n \\end{bmatrix} \\\\\n \\frac{d \\vec{t(\\tau)}}{d\\tau} &= \n \\begin{bmatrix}\n 0 & \\frac{\\tau^{0}}{0!} & \\frac{\\tau^{1}}{1!} & \\hdots\n \\end{bmatrix} \\\\\n \\frac{d^{2} \\vec{t(\\tau)}}{d\\tau^{2}} &= \n \\begin{bmatrix}\n 0 & 0 & \\frac{\\tau^{0}}{0!} & \\hdots\n \\end{bmatrix} \\\\\n\\end{align*}\n%\nNotice that successive derivatives pre-pad zeros and shift the vector\nto the right. \n\nThe integrand of $\\mathbf{Q}$ is found by taking the outer product of the\nderivative of the time vector with itself.\n\\begin{align*}\n \\vec{t(\\tau)}^{T} \\vec{t(\\tau)} &= \n \\begin{bmatrix}\n \\frac{t^{0}}{0!} \\frac{t^{0}}{0!} & \\frac{t^{0}}{0!} \\frac{t^{1}}{1!} &\n \\frac{t^{0}}{0!} \\frac{t^{2}}{2!} & \\hdots \\\\[6pt]\n \\frac{t^{1}}{1!} \\frac{t^{0}}{0!} & \\frac{t^{1}}{1!} \\frac{t^{1}}{1!} &\n \\frac{t^{1}}{1!} \\frac{t^{2}}{2!} & \\hdots \\\\[6pt]\n \\frac{t^{2}}{2!} \\frac{t^{0}}{0!} & \\frac{t^{2}}{2!} \\frac{t^{1}}{1!} &\n \\frac{t^{2}}{2!} \\frac{t^{2}}{2!} & \\hdots \\\\[6pt]\n \\vdots & \\vdots & \\vdots & \\ddots\n \\end{bmatrix} \\\\\n \\frac{d\\vec{t(\\tau)}}{d\\tau}^{T} \\frac{d\\vec{t(\\tau)}}{d\\tau} &= \n \\begin{bmatrix}\n 0 & 0 & 0 & \\hdots \\\\[6pt]\n 0 & \\frac{t^{0}}{0!} \\frac{t^{0}}{0!} & \\frac{t^{0}}{0!} \\frac{t^{1}}{1!} & \\hdots \\\\[6pt]\n 0 & \\frac{t^{1}}{1!} \\frac{t^{0}}{0!} & \\frac{t^{1}}{1!} \\frac{t^{1}}{1!} & \\hdots \\\\[6pt]\n \\vdots & \\vdots & \\vdots & \\ddots\n \\end{bmatrix} \\\\\n \\frac{d^{2}\\vec{t(\\tau)}}{d\\tau^{2}}^{T} \\frac{d^{2}\\vec{t(\\tau)}}{d\\tau^{2}} &= \n \\begin{bmatrix}\n 0 & 0 & 0 & \\hdots \\\\[6pt]\n 0 & 0 & 0 & \\hdots \\\\[6pt]\n 0 & 0 & \\frac{t^{0}}{0!} \\frac{t^{0}}{0!} & \\hdots \\\\[6pt]\n \\vdots & \\vdots & \\vdots & \\ddots\n \\end{bmatrix} \\\\\n\\end{align*}\n% \nNotice that the outer product of successive derivatives pre-pads columns and rows\nwith zeros and shift rows down and columns to the right. This admits a\nconvenient way to compute the integrand of $\\mathbf{Q}$: take the outer product of\n$\\vec{t}(\\tau)$ with itself and shift the rows and columns by the power of the\nderivative.\n\\begin{equation*}\n \\frac{d^{r}\\vec{t(\\tau)}}{d\\tau^{r}}^{T} \\frac{d^{r}\\vec{t(\\tau)}}{d\\tau^{r}}\n = \\text{shift}(\\vec{t}(\\tau)^{T} \\vec{t}(\\tau), r) \n\\end{equation*}\n\\\\\nTo compute $\\mathbf{Q}$, the matrix integrand must be integrated. Integrating a\nmatrix is as easy as integrating its components element-wise. The polynomial\nstructure of the components of $\\mathbf{Q}$ admit and easy-to-compute integral\nformula:\n\\begin{equation*}\n \\int \\frac{\\tau^{i}}{i!} \\frac{\\tau^{j}}{j!} d\\tau =\n \\frac{\\tau^{i+j+1}}{i!j!(i+j+1)!}\n\\end{equation*}\n\\\\\nFinally, a careful observer might note that the structure of problem permits one\nto swap the shift and integrate operations. Instead of shifting all elements and\nthen integrating, one could integrate all elements and then shift. This\nsimplifies implementation in a for-loop-based subroutine.\n\\bigbreak\n\\noindent In summary, to compute $\\mathbf{Q}$:\n\\begin{enumerate}\n \\item Compute the outer product of $\\vec{t}(\\tau)$ with itself\n \\item Integrate the outer product\n \\item Shift the rows and columns of the integrated matrix by the order of the\n derivative\n\\end{enumerate}\n\n\\subsection{Relaxing Arrival Time}\nUntil this point, one very important topic has been overlooked: timing. Users\nmay specify arbitrary arrival times for waypoints. When times are too large, too\nsmall, or a combination of the both, numerical accuracy problems occur and many\noptimizers struggle to find sufficiently accurate solutions. To prevent this\nfrom occurring, segment time is normalized to $t \\in [0,1]$. Stated otherwise,\nevery segment in a trajectory is assumed to start at $t=0$ and end at $t=1$.\n\n\nThis new timing scheme is referred to as the \\textit{computational time} while\nthe original timing scheme is referred to as the \\textit{physical time}.\nTransforming the problem from physical time to computational time requires one\nto determine the segment index a particular time is, subtracting the start time,\nand normalized by the total segment time. The inverse transform is defined in a\nsimilar manner.\n\n\\subsection{Time Scaling}\nLet $t \\in [0, t_{f}]$ represent the physical time along a segment and $\\tau \\in\n[0, 1]$ represent the computational time along a segment. The two are related by\na scaling factor.\n%\n\\begin{equation}\n t = \\alpha \\tau\n\\end{equation}\n%\nPath constraints are specified using physical time, but the polynomial functions\nand optimization use computational time. These two timing schemes must be\nreconciled. Let a constraint be defined as:\n%\n\\begin{equation*}\n \\frac{dp(\\tau)}{dt} < \\kappa\n\\end{equation*}\n%\nChange of basis using $\\tau = t / \\alpha$ and $d\\tau = dt / \\alpha$. Substitute\nand rearrange:\n%\n\\begin{equation}\n \\frac{dp(\\tau)}{d\\tau} < \\alpha \\kappa\n\\end{equation}\n%\nConstraints on higher derivatives are derived using the same process.\n\\begin{equation}\n \\frac{dp^{n}(\\tau)}{dt^{n}} < \\kappa \\rightarrow\n \\frac{dp^{n}(\\tau)}{d\\tau^{n}} < \\alpha^{n} \\kappa\n\\end{equation}\n%\nContinuity constraints (discussed in the next section) merit special\nconsideration. These constraints require that the $n$th derivative of the end of\na segment match the $n$th derivative of the following segment. Since the time\nscaling of each segment may be different, the constraint manifests so:\n\\begin{equation}\n \\frac{dp_{k}^{n}(\\tau)}{d\\tau^{n}} \\frac{1}{\\alpha_{k}^{n}} =\n \\frac{dp_{k+1}^{n}(\\tau)}{d\\tau^{n}} \\frac{1}{\\alpha_{k+1}^{n}}\n\\end{equation}\n\n\\subsection{Continuity Constraints}\nMost piecewise-continuous problems require some sort of continuity --- the\nposition, velocity, or acceleration must not instantaneously change at endpoints\nof the piecewise trajectory. Typically this is motivated by real-life\nconstraints: a quadcopter cannot instantaneously teleport or change its\nvelocity from 1 m/s to -3 m/s.\n\nConsider the first two segments of a scaler third-order piecewise polynomial\ntrajectory. Let the optimization vector contain the coefficients for both\nsegments of the trajectory.\n\\begin{equation*}\n \\vec{x} =\n \\begin{bmatrix}\n x_{0} & v_{0} & a_{0} & j_{0} & x_{1} & v_{1} & a_{1} & j_{1}\n \\end{bmatrix}^{T}\n\\end{equation*}\n%\nThe new quadratic matrix is a block-diagonal of the segment quadratic matrices.\n\\begin{equation*}\n \\mathbf{Q} = \n \\begin{bmatrix}\n \\mathbf{Q}_{0} & \\mathbf{0} \\\\\n \\mathbf{0} & \\mathbf{Q}_{1}\n \\end{bmatrix}\n\\end{equation*}\n%\nThe new optimization function is:\n\\begin{equation*}\n J = \\vec{x}^T \\mathbf{Q} \\vec{x}\n\\end{equation*}\n\nFor the function to be continuous, the first segment must be equal to the second\nsegment at the boundary point.\n\\begin{equation*}\n p_{0}(t=1) - p_{1}(t=0) = 0 \\\\\n\\end{equation*}\n%\nWritten in matrix form:\n\\begin{equation*}\n \\underbrace{\n \\begin{bmatrix}\n 1/0! & 1/1! & 1/2! & 1/3! & -1/0! & 0 & 0 & 0 \\\\ \n \\end{bmatrix}\n }_{\\mathbf{A}}\n \\underbrace{\n \\begin{bmatrix}\n x_{0} \\\\ \n v_{0} \\\\\n a_{0} \\\\\n j_{0} \\\\\n x_{1} \\\\\n v_{1} \\\\\n a_{1} \\\\\n j_{1}\n \\end{bmatrix}\n }_{\\vec{x}}\n =\n \\vec{0}\n\\end{equation*}\n%\nOr equivalently\n\\begin{equation*}\n \\vec{l} \\leq \\mathbf{A} \\vec{x} \\leq \\vec{u}\n\\end{equation*}\n%\nwhere both $\\vec{l}$ and $\\vec{u}$ are $\\vec{0}$ in this case. Continuity\nconstraints for arbitrary derivatives may be specified in a similar manner by\nleveraging the fact that the derivative of $\\vec{t}$ is a shifted version of\nitself. The derivatives may be formulated in the same $\\vec{l} \\leq \\mathbf{A}\n\\vec{x} \\leq \\vec{u}$. Without further justification, all continuity and\nwaypoint constraints may be formulated as bounded linear constraints.\n\n\n\\subsection{Path Constraints}\nPath constraints are necessary for any non-trivial path. For example, quadcopters\noften have input constraints and can't fly faster than or accelerate faster than\na certain limit. Alternatively, if obstacles are present, the position of the\npath should be constrained to be outside of obstacles.\n\nIn coordination with the continuity constraints, path constraints can be\nformulated in the form:\n\\begin{equation*}\n \\vec{l} \\leq \\mathbf{A} \\vec{x} \\leq \\vec{u}\n\\end{equation*}\nUnfortunately, these constraints are linear --- they cannot be used to describe\nnon-linear constraints. For example, if a quadcopter is limited to a total\nacceleration of 0.5 m/$s^2$, then the two-norm magnitude of its acceleration\nshould be bounded. The two-norm magnitude is a non-linear constraint and cannot\nbe expressed. However, one could approximate it by specifying an arbitrary number of planes\ntangent-to or inscribed-in the hypersphere that delimits the two-norm. It's not\nperfect, but it enables the quadratic programming solver to find a solution\nfast.\n\nFinally, path constraints introduce one additional complication: violation of\ndimensional independence. Previously, I claimed that the dimensions are\nindependent and thus can be considered standalone as scalar polynomials. In the\npresence of path constraints, this is not necessarily true. Two dimensions can\nbe linked together by a path constraint. For example, consider approximating a\ntwo-norm magnitude by requiring:\n\\begin{align*}\n x + y &< 1 \\\\\n x - y &< 1 \\\\\n -x + y &< 1 \\\\\n -x - y &< 1\n\\end{align*}\nClearly these two dimensions are now linked. Despite this dependency, the scalar\nanalysis still holds. The dependency is managed by the quadratic programming\nsolver.\n\nOne last note: if there were no interdimensional dependence, a multi-dimensional\nproblem could be broken down into multiple scalar dimensional problems and\nsolved simultaneously by multiple solvers distributed over multiple threads.\nHowever, it's usually the case that at least one path constraint introduces\ninterdimensional dependence and this technique cannot be leveraged.\n\n\n\n\n\\subsection{Quadratic Programming}\nThe problem of finding these piecewise polynomials can be cast as a quadratic\nprogramming problem:\n%\n\\begin{equation}\n \\begin{split}\n \\min_{\\vec{x}} \\quad& \\frac{1}{2} \\vec{x}^T \\mathbf{Q} \\vec{x} \\\\\n \\text{s.t.} \\quad& \\vec{l} \\leq \\mathbf{A} \\vec{x} \\leq \\vec{u}\n \\end{split}\n\\end{equation}\n%\n\\href{https://osqp.org}{OSQP} is used to solve the QP problem \\cite{osqp}.\n\n\\bibliography{main}\n\\bibliographystyle{IEEEtran}\n\n\\end{document}\n", "meta": {"hexsha": "cf3d455375ac185f92b328da287ab714b29bd533", "size": 15064, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/tex/main.tex", "max_stars_repo_name": "TuckerHaydon/MinimumSnap", "max_stars_repo_head_hexsha": "474ec8edfec45adb4291f945736772c335dc9cc8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2019-05-29T07:14:47.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-31T03:37:48.000Z", "max_issues_repo_path": "doc/tex/main.tex", "max_issues_repo_name": "TuckerHaydon/MinimumSnap", "max_issues_repo_head_hexsha": "474ec8edfec45adb4291f945736772c335dc9cc8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2019-04-16T23:00:18.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-09T18:37:04.000Z", "max_forks_repo_path": "doc/tex/main.tex", "max_forks_repo_name": "TuckerHaydon/MinimumSnap", "max_forks_repo_head_hexsha": "474ec8edfec45adb4291f945736772c335dc9cc8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2019-04-18T21:44:10.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-05T09:55:09.000Z", "avg_line_length": 36.8312958435, "max_line_length": 102, "alphanum_fraction": 0.6776420605, "num_tokens": 5051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.752012562644147, "lm_q1q2_score": 0.6148262769948281}} {"text": "\\chapter*{Part One: Count Based Language Modeling}\n\\addcontentsline{toc}{chapter}{Part One}\n\n\n\n % \\begin{lstlisting}[style=latexFrameTB, caption={Example of Sleek Template packages usage.}, gobble=8]\n % \\usepackage[english]{babel}\n % \\usepackage[noheader]{packages/sleek}\n % \\usepackage{packages/sleek-title}\n % \\end{lstlisting}\n\n\n\\subsection*{Theoretical Problem A}\nThe assignment is to solve for the maximum likelihood estimate (MLE) of a bigram language model, where the likelihood function is given by\n\n\\begin{equation*}\n \\prod_{t=1}^{T} p_{\\theta_{u,v}}(w_t | w_{t-1})\n\\end{equation*}\n\nFor brevity, I will refer to $p_{\\theta_{u,v}}$ as simply \\theta. \n\nIf we define our corpus to consist of many two word pairs, or bigrams, we can re-write the likelihood as\n\n\\begin{equation*}\n \\prod_{t=1}^{T} \\theta^{c(u,v)}\n\\end{equation*}\n\nwhere $c(u,v)$ represents the count of how many times the bigram $u,v$ appears in the corpus.\n\nIt is convenient to rewrite our function in terms of the log likelihood:\n\\begin{equation*}\n \\sum_{t=1}^{T} c(u,v) * \\ln{\\theta}\n\\end{equation*}\n\nWe want to maximize the log likelihood in terms subject to the constraint that the probabilities $\\theta$ sum to 1:\n\\begin{equation*}\n \\sum_{u,v\\in \\mathcal{V}} \\theta = 1\n\\end{equation*}\n\nWe introduce a Lagrangian multiplier to solve the problem in the form \n\\begin{equation*}\n \\mathcal{L}(\\theta,\\lambda) = L(\\theta) - \\lambda f(\\theta)\n\\end{equation*},\nyielding\n\\begin{equation*}\n \\mathcal{L}(\\theta,\\lambda) = \\sum_{t=1}^{T} c(u,v) * \\ln(\\theta) - \\lambda (\\sum_{u,v\\in \\mathcal{V}}\\theta -1)\n\\end{equation*}\n\nTaking the partial derivatives gives the following system of equations:\n\n\\begin{equation*}\n \\frac{\\partial}{\\partial \\theta} = \\frac{c(u,v)}{\\theta} - \\lambda = 0\n\\end{equation*}\n\n\\begin{equation*}\n \\frac{\\partial}{\\partial \\lambda} = \\sum \\theta - 1 = 0\n\\end{equation*}\n\nSolve the system of linear equations:\n\n\\begin{equation*}\n \\theta = \\frac{c(u,v)}{\\lambda}\n\\end{equation*}\n\n\\begin{equation*}\n \\sum \\frac{c(u,v)}{\\lambda} = 1\n\\end{equation*}\n\n\\begin{equation*}\n \\lambda = \\frac{1}{\\sum c(u,v)}\n\\end{equation*}\n\nThe final solution to the system of equations is the MLE of the bigram model:\n\\begin{equation*}\n {\\hat \\theta_{u,v}} = \\frac{c(u,v)}{\\sum_{u,v\\in \\mathcal{V}}c(u,v)}\n\\end{equation*}\n\n\\subsection*{Theoretical Problem B}\n\nThe assignment is to show that for the case of a unigram language model, given a prior distribution of a Dirichlet distribution, setting all values in parameter vector $\\beta$ results in a posterior distribution with mean equal to the Laplace smoothing estimate:\n\n\\begin{equation*}\n \\hat{\\theta}_u = \\frac{c(u) + \\alpha}{T + \\alpha|\\mathcal{V}'}\n\\end{equation*}\n\nThe posterior is defined as:\n\\begin{equation*}\n p(\\pi|D;\\beta) = \\frac{p(D|\\pi)p(\\pi;\\beta)}{p(D)}\n\\end{equation*}\n\nThe density function for the prior is:\n\\begin{equation*}\n p(\\pi;\\beta) = \\frac{1}{B(\\beta)}\\prod_{u \\in \\mathcal{V}}\\pi_u^{\\beta_u - 1}\n\\end{equation*}\n\nThe likelihood model of $p(D|\\pi)$ is given by:\n\\begin{equation*}\n \\prod \\pi_u^{c(u)}\n\\end{equation*}\nwhere we multiply over probabilities over all words in the corpus and account for duplicates with the count term in the exponent. \n\nPlugging the density function (ignoring the normalizing constant $B(\\beta)$) and likelihood model into the posterior gives\n\\begin{equation*}\n p(\\pi|D;\\beta) = \\frac{\\prod \\pi_u^{c(u)} * \\frac{1}{B(\\beta)}\\prod_{u \\in \\mathcal{V}}\\pi_u^{\\beta_u - 1}}{p(D)}\n\\end{equation*}\n\n\\begin{equation*}\n = \\frac{\\prod \\pi_u^{c(u) + \\alpha - 1}}{p(D)}\n\\end{equation*}\n\nThe numerator of this equation is itself a Dirichlet distribution with parameter vector $\\beta' = c(u) + \\alpha -1$. Because we are changing the observed data by applying a smoothing constant, p(D) becomes trivial. Our result is proportional and the $p(D)$ term does not impact the result. Observe that the expected value of the $\\beta'$ distribution is \n\n\\begin{equation*}\n \\mathbb{E_{p(\\pi|D;\\beta')}} = \\frac{c(u) + \\alpha}{\\sum_u^{\\mathcal{V}} c(u) + \\alpha}\n\\end{equation*}\n\nWe have increased the count of every word in our corpus by $\\alpha$, meaning the size of the corpus has increased by $\\alpha$ times the number of unique words. This yields\n\n\\begin{equation*}\n \\mathbb{E_{p(\\pi|D;\\beta')}} = \\frac{c(u) + \\alpha}{T + \\alpha|\\mathcal{V}|} = \\theta_u\n\\end{equation*}\nwhere $T$ is the size of the original corpus. \n\n\\chapter*{Part Two: RNN Language Modeling}\n\\addcontentsline{toc}{chapter}{Part Two}\n\n\\subsection*{Computational Problem A}\n\nIn order to determine if a substring is part of BPL, I observe that one must know three quantities: the number of \"(\", the number of \")\", and if an illegal \")\" has ever been observed. A \")\" is illegal if there is not already a matching \"(\". \n\nMy hidden state is a column vector with elements \\begin{bmatrix}\n \"(\" count \\\\ \")\" count \\\\ Illegal \\\\\n\\end{bmatrix}. The initial hidden state is given by \n\\begin{equation*}\n h_0 = \\begin{bmatrix}\n -1 \\\\ -1 \\\\ 0 \\\\\n \\end{bmatrix}\n\\end{equation*}\n\n\n$x_t$ is a vector representing either \"(\" or \")\", where I map \\begin{bmatrix}\n 1 && 0 && 0\n\\end{bmatrix} to \"(\" and \\begin{bmatrix}\n 0 && 1 && 0\n\\end{bmatrix} to \")\".\n\nW and U are unnecessary in my BPL model, and are set to 1. The update function is given by the piecewise function \n\n\\begin{equation*}\n f (x) = \n \\begin{cases} \n x & x * \\begin{bmatrix}\n 1 && -1 && 0\n \\end{bmatrix}\\geq 0 \\\\\n x + \\begin{bmatrix}\n 0 \\\\ 0 \\\\ 1 \\\\\n \\end{bmatrix} & x * \\begin{bmatrix}\n 1 && -1 && 0\n \\end{bmatrix}< 0 \\\\\n \\end{cases}\n \n\\end{equation*}\n\nIf $h_t \\lbrack 0 \\rbrack - h_t \\lbrack 1 \\rbrack $ is 0 and $h_t \\lbrack 2 \\rbrack $ is 0 then $x_t$ is in BPL.\n\n\\subsection*{Vanilla Recursive Neural Network Implementation}\n\nNote to the teaching assistant: I apologize for some weak analysis in the implementation sections. I finished all my code several days ago, but forgot to check the notebook into Github and lost all of my code on Monday night when Chrome crashed. \n\nMy vanilla RNN is essentially a wrapper around either an RNNCell or an LSTMCell. The forward function takes a matrix of input ids representing a batch of sentences, and maps them to an embedding. We then initialize a hidden state to zeros, and iterate over the embeddings symbol by symbol for the 50 symbol sequence. At every step the previous hidden state $h_{t-1}$ is fed back into the model, as well as the next embedding column. \n\nThe perplexity of just my basic model is shown below in Figure \\label{firsttry}. \n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.4\\textwidth]{resources/png/no_bias_1e3.png}\n \\noskipcaption{Perplexity of a vanilla RNN over 10 epochs with learning rate 1e-3}\n \\label{firsttry}\n\\end{figure}\n\nThe fact that the perplexity gets worse with more training implies that perhaps my optimizer is descending too far, and passing over the maximum for a local maxima. My perplexity rating is far away from the target somewhere in the 70s, so I decided to use a new architecture. Long Short Term Memory or LSTM incorporate a feedback mechanism in addition to the feed-forward method of an RNN. This allows the network to develop connections over longer sequences, potentially allowing for context a vanilla RNN can't achieve. My LSTM has perplexity ~78 as shown in Figure \\label{secondtry}.\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.4\\textwidth]{resources/png/LSTM_05e3.png}\n \\noskipcaption{Perplexity of a LSTM RNN over 10 epochs with learning rate 5e-4}\n \\label{secondtry}\n\\end{figure}\n\n\\subsection*{Sequence to Sequence Model}\nA sequence to sequence model interprets an input sequence through an input neural network called an encoder, and then feeds the final layer of the encoder through a decoder to produce the output sequence. \n\nMy decoder contains a Gated Recurrent Unit, which functions similar to an LSTM but it also can \"forget\". I chose it because some research indicates GRUs function well on smaller vocabularies (\"Empirical Evaluation of Gated Recurrent Neural Networks on Sequence Modeling\", Chung et al), although I did not have time to compare to other architectures. \n\nAdditionally I feed the output of my GRU through a linear layer. Empirically this improved my BLEU score by about 0.5, although I do not have great intuition for why. \n\nThe first attempt at a decoder with a single layer GRU and no linear layer achieved a perplexity score in the mid 40s (Figure \\label{seq1}).\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.4\\textwidth]{resources/png/seq2seq1layer.png}\n \\noskipcaption{Single layer GRU Decoder perplexity}\n \\label{seq1}\n\\end{figure}\n\nAdding a second GRU layer and the hidden layer resulted in a perplexity in the low 30s, with a BLEU score of 6.83 (Figure \\label{seq2}). \n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.4\\textwidth]{resources/png/seq.png}\n \\noskipcaption{Double layer GRU with linear output Decoder perplexity}\n \\label{seq1}\n\\end{figure}\n\nOne thing I noticed about my decoder is the output produces some seemingly unrelated predictions sometimes; the following example is taken from the training data:\n\n Src : Tôi đoán các bạn đều đã biết về lịch sử của " đàn quay " trong tiếng Pháp là " à " \\hline\n Trg : I see you 're all up on the history of hurdy-gurdy -- " . " \\hline\n Pred: I think you know what most of the most popular TEDTalks " is " . " \\hline\n\nI looked it up, and there is a sentence about TEDTalks right next to a sentence relating to hurdy-gurdies in the training data. I suspect my batch data is getting entangled. This could be fixed by duplicating some of the training data but with sentences rearranged more randomly, or by playing with the batch size parameters. \n\nAnother error is that the prediction seems to sometimes follow the most likely path, regardless of sentence context. \n\nSrc : Tôi muốn cho các bạn biết về sự to lớn của những nỗ lực khoa học đã góp phần làm nên các dòng tít bạn thường thấy trên báo . \\hline\nTrg : I 'd like to talk to you today about the scale of the scientific effort that goes into making the headlines you see in the paper . \\hline\nPred: I want to tell you about the most important thing that you 've learned in the first place of the brain that works in the front of the job . \\hline\n\nIn this example it starts out alright, but I suspect addressing a second person after stating \"I want to tell you\" is more common than \"scientific effort...\". If my model had beam search, it could explore multiple paths instead of just taking the likely followup to a shorter sequence. \n\n\n\\chapter*{Appendix}\n\\subsection*{Vanilla RNN}\n\\begin{python}\n class RNN(nn.Module):\n\n def __init__(self, input_size, hidden_size, src_embed, generator, LSTM=True):\n \"\"\"\n Inputs:\n - `input_size`: a positive integer corresponding to the size of the\n word embeddings\n - `hidden_size`: a positive integer representing the dimensionality of\n the RNN's hidden state vector\n - `src_embed`: an nn.Embedding object representing the lookup table for\n input (source) sentences\n - `generator`: a `Generator` object. Essentially a linear mapping\n followed by a softmax. You should not call it within this class; it\n is called in the SimpleLossCompute class above\n \"\"\"\n super(RNN, self).__init__()\n # `input_size`, `hidden_size`, and `output_size` are all int.\n\n self.hidden_size = hidden_size\n self.src_embed = src_embed\n self.generator = generator\n self.use_lstm = LSTM\n # hint: unless you choose to implement the RNN update equations yourself\n # you will want a `self.rnn` module that does that for you, which\n # is where the RNNCell/LSTMCell/GRUCell modules come in handy\n\n self.rnn = nn.RNNCell(input_size, hidden_size, bias=False, device=device)\n self.lstm = nn.LSTMCell(input_size, hidden_size, device=device)\n\n\n def forward(self, input_ids):\n \"\"\"\n Given a sequence of words (represented as IDs), compute and return the\n hidden state at each timestep (equivalently, for each input word).\n Input:\n - `input_ids`: a 2d-tensor of shape\n (batch_size, MAX_SENT_LENGTH_PLUS_SOS_EOS) representing a batch of\n sentences from the dataset (with IDs instead of words)\n\n Returns:\n - `hiddens`: a 3d-tensor of shape\n (batch_size, MAX_SENT_LENGTH_PLUS_SOS_EOS, hidden_size) representing\n the hidden state of the model at each timestep\n \"\"\"\n # hint: pay close attention to the shapes of your tensors; you may find\n # pytorch's `permute()` method for tensors useful\n\n ### Your code here!\n bsize, fsize = input_ids.size()\n\n hsize = self.hidden_size\n\n embedding = self.src_embed(input_ids)\n embedding = embedding.transpose(0,1).contiguous()\n \n hidden = self.init_hidden(bsize)\n output = []\n if self.use_lstm:\n cell_state = self.init_hidden(bsize)\n for i in range(fsize):\n hidden, cell_state = self.lstm(embedding[i], (hidden, cell_state))\n output.append(hidden)\n else:\n for index in range(fsize):\n hidden = self.rnn(embedding[index], hidden)\n output.append(hidden)\n \n output = torch.stack(output, dim=1)\n return output\n\n def init_hidden(self, batch_size):\n \"\"\"\n Input:\n - `batch_size`: a positive integer\n\n Returns:\n - `hidden`: a 2d-tensor of shape (batch_size, hidden_size) representing\n the initial hidden state of the RNN\n \"\"\"\n # Use to initialize hidden state everytime before running a sentence.\n hidden = torch.zeros(batch_size, self.hidden_size).to(device)\n return hidden\n\\end{python}\n\n\\subsection*{Seq2Seq}\n\\begin{python}\n class Decoder(nn.Module):\n \"\"\"An RNN decoder without attention.\"\"\"\n\n def __init__(self, input_size, hidden_size, dropout=0.):\n \"\"\"\n Inputs:\n - `input_size`, `hidden_size`, and `dropout` the same as in Encoder.\n \"\"\"\n super(Decoder, self).__init__()\n\n # hint 1: while the encoder just needed a single RNN layer, more will be\n # required for the decoder.\n # hint 2: think about what you'll need in init_hidden(self, encoder_finals)\n # hint 3: what will you use to compute the pre_output (see docstring of\n # forward function)?\n\n ### Your code here!\n print(\"Creating Decoder with input size {} and hidden size {}\".format(input_size, hidden_size))\n self.bsize = input_size\n self.hsize = hidden_size\n self.nlayers = 2\n \n self.gru = nn.GRU(input_size=input_size, \n hidden_size=hidden_size, \n num_layers=self.nlayers, \n batch_first=True, dropout=dropout, \n bidirectional=False, device=device)\n \n self.lin = nn.Linear(input_size, hidden_size, device=device)\n\n def forward_step(self, prev_embed, hidden):\n \"\"\"Helper function for forward below:\n Perform a single decoder step (1 word).\n\n Inputs:\n - `prev_embed`: a 3d-tensor of shape (batch_size, 1, embed_size)\n representing the padded embedded word vectors at this step in training\n - `hidden`: a 3d-tensor of shape (1, batch_size, hidden_size) representing\n the current hidden state.\n\n Returns:\n - `hidden`: a 3d-tensor of shape (1, batch_size, hidden_size)\n representing the current decoder hidden state.\n - `pre_output`: a 3d-tensor of shape (batch_size, 1, hidden_size)\n representing the total decoder output for one step\n \"\"\"\n # hint: you'll want to do more here than just run self.rnn (think about\n # what you should do to the output of the self.rnn in order to\n # compute the `pre_output`)\n\n pre_output, hidden = self.gru(prev_embed, hidden)\n x = pre_output.to(device)\n pre_output = self.lin(x)\n return hidden, pre_output\n \n def forward(self, inputs, encoder_finals, hidden=None, max_len=None):\n \"\"\"Unroll the decoder one step at a time.\n\n Inputs:\n - `inputs`: a 3d-tensor of shape (batch_size, max_seq_length, embed_size)\n representing a batch of padded embedded word vectors of target\n sentences (for teacher-forcing during training).\n - `encoder_finals`: a 3d-tensor of shape\n (num_enc_layers, batch_size, hidden_size) representing the final\n encoder hidden states used to initialize the initial decoder hidden\n states.\n - `hidden`: a 3d-tensor of shape (1, batch_size, hidden_size) representing\n the value to be used to initialize the initial decoder hidden states.\n If None, then use `encoder_finals`.\n - `max_len`: an int representing the maximum decoding length.\n\n Returns:\n - `hidden`: a 3d-tensor of shape\n (num_layers, batch_size, hidden_size) representing the final hidden\n state for each element in the batch.\n - `pre_output_vectors`: a 3d-tensor of shape\n (batch_size, max_len, hidden_size) representing the raw decoder\n outputs (before mapping to a `trg_vocab_size`-dim vector).\n \"\"\"\n\n # The maximum number of steps to unroll the RNN.\n if max_len is None:\n max_len = inputs.size(1)\n\n # Initialize decoder hidden state.\n if hidden is None:\n hidden = self.init_hidden(encoder_finals)\n\n # hint: you'll want to keep track of the `pre_output` for each timestep,\n # but you only need the final `hidden` state\n\n ### Your code here!\n bsize,fsize,hsize = inputs.size()\n pre_output_vectors = []\n for i in range(fsize):\n input = torch.unsqueeze(inputs[:,i,:], 1)\n hidden, pre_output = self.forward_step(input, hidden)\n pre_output_vectors.append(pre_output)\n\n pre_output_vectors = torch.stack(pre_output_vectors, dim=1).squeeze(2)\n return hidden, pre_output_vectors\n\n def init_hidden(self, encoder_finals):\n \"\"\"Use encoder final hidden state to initialize decoder's first hidden\n state.\n\n Input: `encoder_finals` is same as in forward()\n\n Returns: \n - `decoder_init_hiddens`: a 3d-tensor of shape \n (num_layers, batch_size, hidden_size) representing the initial\n hidden state of the decoder for each element in the batch \n \"\"\"\n # hint: think about whether or not an activation function is needed here\n\n ### Your code here!\n \n decoder_init_hiddens = torch.ones(self.nlayers,1,1,device=device) * encoder_finals\n\n return decoder_init_hiddens\n\\end{python}\n\n", "meta": {"hexsha": "954d814e7cf39aff8ce4b65b75f2aea742abc6e9", "size": 18872, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Homework_1/report/content.tex", "max_stars_repo_name": "RichardMuri/MIT_6864", "max_stars_repo_head_hexsha": "40a7472f8ad03545a0ecec9b63361d4a3136d815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Homework_1/report/content.tex", "max_issues_repo_name": "RichardMuri/MIT_6864", "max_issues_repo_head_hexsha": "40a7472f8ad03545a0ecec9b63361d4a3136d815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Homework_1/report/content.tex", "max_forks_repo_name": "RichardMuri/MIT_6864", "max_forks_repo_head_hexsha": "40a7472f8ad03545a0ecec9b63361d4a3136d815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.6004514673, "max_line_length": 586, "alphanum_fraction": 0.6884802883, "num_tokens": 5018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.6992544210587585, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.6147472595122757}} {"text": "\\documentclass[11pt]{article}\n\\usepackage[top=1.5cm,bottom=1.5cm]{geometry}\n\\usepackage[T1]{fontenc}\n\\usepackage{url}\n\\usepackage{hyperref}\n\\usepackage{xcolor}\n% \n\\usepackage{amsthm}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{ifthen}\n\\usepackage{mathpartir}\n\\usepackage{mathtools}\n\\input{macros}\n% \n\\begin{document}\n\n\\title{CO663: Language $\\lggeT$ --- double}\n\n\\date{\\vspace{-7ex}} \n\\maketitle\n\n% \\pagenumbering{gobble}\n\n\\newcommand{\\lazyeval}[1]{\\framebox{\\parbox[c][#1]{\\textwidth}{\n \\color{white}{h}% \n }}}\n\n\n\\section{Correct implementation}\n\nA function that doubles its argument can be defined as follows:\n%\n\\[\n\\text{double} \\defi \n\\clam{\\tynat}{x}\n{\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\var{v}}}}{\\var{x}}\n}\n\\]\n\n\nWe can convince ourselves that it works as expected by applying it to\nsome parameter (${\\esucc{\\ez}}$ or 1) and evaluate it.\n\\begin{enumerate}\n\\item Instantiate the function: \\[\n \\semrule\n {\\tsemrulename{lav}}\n {\\,}\n {\\jtrans\n {\\capp{\\text{double}}{\\esucc{\\ez}}}\n {\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\var{v}}}}{{\\esucc{\\ez}}}\n }}\n \\]\n\\item Iterate once (and using $\\alpha$-renaming, for safety):\n \\[\n \\semrule\n {\\tsemrulename{rs}}\n {\\,}\n {\n \\jtrans\n {\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\var{v}}}}{{\\esucc{\\ez}}}\n }\n {\n \\subs\n {\\esucc{\\esucc{\\var{v}}}}\n {\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\var{v'}}}}{{{\\ez}}}\n }\n {v}\n }\n }\n \\]\n\\item Applying the substitution, we obtain\n \\[\n {\\esucc{\\esucc{\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\var{v'}}}}{{{\\ez}}}\n }}}\n \\]\n\\item Reducing one more time:\n \\[\n \\semrule\n {\\tsemrulename{ss}}\n {\n \\semrule\n {\\tsemrulename{ss}}\n {\n \\semrule\n {\\tsemrulename{r0}}\n {\\,}\n { \\jtrans{{{\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\var{v'}}}}{{{\\ez}}}\n }}}\n {{{\\ez}}}}\n }\n { \\jtrans{{\\esucc{\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\var{v'}}}}{{{\\ez}}}\n }}}\n {{\\esucc{\\ez}}}}\n }\n {\n \\jtrans{\\esucc{\\esucc{\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\var{v'}}}}{{{\\ez}}}\n }}}\n {\\esucc{\\esucc{\\ez}}}\n }\n \\]\n\\end{enumerate}\n\n\n\n\n\\section{Wrong implementation}\n\nIf we use $\\var{x}$ instead of $\\var{v}$ in the inductive case, we\nget:\n\\[\n\\text{baddouble} \\defi \n\\clam{\\tynat}{x}\n{\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\var{\\underline{x}}}}}{\\var{{x}}}\n}\n\\]\n\nWe can convince ourselves that it \\emph{does not} work as expected by\napplying it to some parameter (${\\esucc{\\ez}}$ or 1) and evaluate it.\n% \n\\begin{enumerate}\n\\item Instantiate the function: \n % \n \\[\n \\semrule\n {\\tsemrulename{lav}}\n {\\,}\n {\\jtrans\n {\\capp{\\text{baddouble}}{\\esucc{\\ez}}}\n {\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\esucc{\\ez}}}}{{\\esucc{\\ez}}}\n }}\n \\]\n\\item Iterate once (and using $\\alpha$-renaming, for safety):\n \\[\n \\semrule\n {\\tsemrulename{rs}}\n {\\,}\n {\n \\jtrans\n {\n \\citer{\\ez}{v}{\\esucc{\\esucc{\\esucc{\\ez}}}}{{\\esucc{\\ez}}}\n }\n {\n \\subs\n {\\esucc{\\esucc{\\esucc{\\ez}}}}\n {\n \\citer{\\ez}{v'}{\\esucc{\\esucc{\\esucc{\\ez}}}}{\\ez}\n }\n {v}\n }\n }\n \\]\n\\item Applying the substitution, we obtain (note that $\\var{v}$ does\n not appear in the term, so there is nothing to replace).\n \\[\n {\\esucc{\\esucc{\\esucc{\\ez}}}}\n \\]\n\\item ${\\esucc{\\esucc{\\esucc{\\ez}}}}$ is a value, so we stop\n here. Notice that the final result is $3$! \n\\end{enumerate}\n\n\n\n\n\\end{document}\n\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: t\n%%% End:\n", "meta": {"hexsha": "77b0d223067a32678e0840eca6cee6b0b0c1932f", "size": 3477, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "slides/double.tex", "max_stars_repo_name": "julien-lange/CO663-slides", "max_stars_repo_head_hexsha": "40bc888c7389ae9554bfddc90d22b078c6e5d5d6", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "slides/double.tex", "max_issues_repo_name": "julien-lange/CO663-slides", "max_issues_repo_head_hexsha": "40bc888c7389ae9554bfddc90d22b078c6e5d5d6", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "slides/double.tex", "max_forks_repo_name": "julien-lange/CO663-slides", "max_forks_repo_head_hexsha": "40bc888c7389ae9554bfddc90d22b078c6e5d5d6", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 19.1043956044, "max_line_length": 69, "alphanum_fraction": 0.54932413, "num_tokens": 1275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702761768248, "lm_q2_score": 0.8479677526147223, "lm_q1q2_score": 0.6146666190268752}} {"text": "\\documentclass{article}\n\\usepackage{eecstex}\n\\usepackage{pgfplots}\n\n\\title{EE 123 HW 04}\n\\author{Bryan Ngo}\n\\date{2022-02-11}\n\n\\begin{document}\n\n\\maketitle\n\n\\setcounter{section}{2}\n\n\\section{}\n\n\\subsection{}\n\nEach circular convolution in the overlap-save method will result in a length \\(2^v - P + 1\\) signal.\nThen, the cost of the FFT and IFFT is \\(\\frac{2^v}{2} \\log_2(2^v) = v 2^{v - 1}\\).\nThen, there is a \\(2^v\\)-pointwise multiplication.\nThe total number of multiplications is \\(2^v (v + 1)\\).\nThus, the FFT for each sample will require\n\\begin{equation}\n \\frac{2^v (v + 1)}{2^v - P + 1}.\n\\end{equation}\ncomplex multiplications per output sample.\n\n\\subsection{}\n\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(v\\), ylabel={Cost},\n title={Complex Multiplications},\n axis lines=middle,\n width=0.4\\textwidth\n ]\n \\addplot[ycomb, mark=*, color=blue] table[\n col sep=comma,\n x=v, y=cost\n ]{q3.csv};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\nwith a minimum cost of \\(v = 12\\).\nThe direct evaluation would cost \\(500\\) complex multiplications per output sample, since that is the length of a given sample.\n\n\\subsection{}\n\n\\begin{equation}\n \\lim_{v \\to \\infty} \\frac{2^v (v + 1)}{2^v - P + 1} = \\lim_{v \\to \\infty} \\frac{v + 1}{1 - \\left(\\frac{P - 1}{2^v}\\right)} = v\n\\end{equation}\nThus, for \\(P = 500\\), the direct method will be more efficient for \\(v > 500\\).\n\n\\newpage\n\\section{Fun with FFT}\n\n\\begin{enumerate}\n \\item For \\(0 \\leqslant k < N\\), \\(H[k] = X_r[k] + W_{2N}^k X_i[k]\\).\n \\item For \\(0 \\leqslant k < N\\), \\(H[k + N] = X_r[k] - W_{2N}^k X_i[k]\\).\n \\item \\(X[k] = \\frac{1}{2} (H[k] + H[k + N]) + \\frac{j}{2} W_{2N}^{-k} (H[k] - H[k + N])\\), which takes 3 multiplications and 3 additions.\n\\end{enumerate}\n\n\\newpage\n\\section{Hadamard Transform}\n\n\\subsection{}\n\n\\begin{equation}\n H_3 =\n \\begin{bmatrix}\n 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\n 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\\\\n 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\\\\n 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\\\\n 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 \\\\\n 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 \\\\\n 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\\\\n 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1\n \\end{bmatrix}\n\\end{equation}\nThe order that represents increasing frequency content is the sequency ordering.\n\n\\subsection{}\n\n\\begin{center}\n \\includegraphics[width=0.8\\textwidth]{q5b.png}\n\\end{center}\n\n\\newpage\n\\section{}\n\n\\subsection{}\n\n\\begin{align}\n X[3k] &= \\sum_{n = 0}^{N - 1} x[n] W_N^{3kn} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{3kn} + \\sum_{n = \\frac{N}{3}}^{\\frac{2N}{3} - 1} x[n] W_N^{3kn} + \\sum_{n = \\frac{2N}{3}}^{N - 1} x[n] W_N^{3kn} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{3kn} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{N}{3}\\right] W_N^{3kn} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{2N}{3}\\right] W_N^{3kn} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\left(x[n] + x\\left[n + \\frac{N}{3}\\right] + x\\left[n + \\frac{2N}{3}\\right]\\right) W_N^{3kn} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\underbrace{\\left(x[n] + x\\left[n + \\frac{N}{3}\\right] + x\\left[n + \\frac{2N}{3}\\right]\\right)}_{x_1[n]} W_{\\frac{N}{3}}^{kn}\n\\end{align}\n\n\\subsection{}\n\n\\begin{align}\n X[3k + 1] &= \\sum_{n = 0}^{N - 1} x[n] W_N^{n(3k + 1)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{n(3k + 1)} + \\sum_{n = \\frac{N}{3}}^{\\frac{2N}{3} - 1} x[n] W_N^{n(3k + 1)} + \\sum_{n = \\frac{2N}{3}}^{N - 1} x[n] W_N^{n(3k + 1)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{n(3k + 1)} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{N}{3}} W_N^{n(3k + 1)} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{2N}{3}} W_N^{n(3k + 1)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\left(x[n] + x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{N}{3}} + x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{2N}{3}}\\right) W_N^{n(3k + 1)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\underbrace{\\left(x[n] + x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{N}{3}} + x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{2N}{3}}\\right) W_N^n}_{x_2[n]} W_{\\frac{N}{3}}^{kn} \\\\\n X[3k + 2] &= \\sum_{n = 0}^{N - 1} x[n] W_N^{n(3k + 2)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{n(3k + 2)} + \\sum_{n = \\frac{N}{3}}^{\\frac{2N}{3} - 1} x[n] W_N^{n(3k + 2)} + \\sum_{n = \\frac{2N}{3}}^{N - 1} x[n] W_N^{n(3k + 2)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} x[n] W_N^{n(3k + 2)} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{2N}{3}} W_N^{n(3k + 2)} + \\sum_{n = 0}^{\\frac{N}{3} - 1} x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{4N}{3}} W_N^{n(3k + 2)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\left(x[n] + x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{2N}{3}} + x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{4N}{3}}\\right) W_N^{n(3k + 2)} \\\\\n &= \\sum_{n = 0}^{\\frac{N}{3} - 1} \\underbrace{\\left(x[n] + x\\left[n + \\frac{N}{3}\\right] W_N^{\\frac{N}{3}} + x\\left[n + \\frac{2N}{3}\\right] W_N^{\\frac{2N}{3}}\\right) W_N^{2n}}_{x_3[n]} W_{\\frac{N}{3}}^{kn}\n\\end{align}\n\n\\subsection{}\n\n\\begin{center}\n \\includegraphics[width=0.7\\textwidth]{q6c.png}\n\\end{center}\n\n\\subsection{}\n\n\\begin{center}\n \\includegraphics[width=0.8\\textwidth]{q6d.png}\n\\end{center}\n\n\\newpage\n\\section{}\n\n\\begin{enumerate}\n \\item \\(|X[k]| \\leqslant N\\) for \\(k = 0\\).\n \\item We want \\(x[n]\\) to be a constant under the DFT, so we can cancel out the complex exponential terms to obtain \\(x[n] = e^{j \\theta} W_N^{-kn}\\) for all \\(\\theta \\in \\R\\), and \\(k, n \\in \\Z\\).\n\\end{enumerate}\n\n\\newpage\n\\section{}\n\n\\subsection{}\n\n\\begin{equation}\n H_4[k] = \\sum_{k = 0}^3 h[n] W_4^{kn} = 1 - W_4^k = 1 - (-j)^k = \\{0, 1 + j, 2, 1 - j\\}\n\\end{equation}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n xlabel=\\(k\\), ylabel={\\(|H_4[k]|\\)},\n title={Magnitude of 4-point DFT},\n axis lines=middle\n ]\n \\addplot[\n ycomb,\n color=blue,\n mark=*\n ]\n coordinates {\n (0, 0)\n (1, 2^0.5)\n (2, 2)\n (3, 2^0.5)\n };\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\nThe DFT is not even, not odd, is conjugate symmetric.\nThe DFT is a high-pass filter since it lets in \\(\\omega = \\pi\\), which is the highest frequency, and blocks out \\(\\omega = 0\\), the DC frequency.\n\n\\subsection{}\n\nIt is not possible to uniquely identify \\(x[n]\\) since the expression\n\\begin{equation}\n X[k] = \\frac{Y[k]}{H[k]}\n\\end{equation}\ninvolves \\(H[0] = 0\\), so the expression is undefined at \\(k = 0\\), meaning that\n\\begin{equation}\n X[k] =\n \\begin{cases}\n C & k \\equiv 0 \\pmod{4} \\\\\n \\frac{Y[k]}{1 - (-j)^k} & k \\in [1, 3] \\pmod{4}\n \\end{cases}\n\\end{equation}\nfor some \\(C \\in \\C\\).\n\n\\subsection{}\n\nUsing Parseval's theorem for the DFT,\n\\begin{align}\n \\sum_{n = 0}^3 |x[n]|^2 &= \\frac{1}{4} \\sum_{k = 0}^3 |X[k]|^2 = D \\\\\n &= \\frac{1}{4} \\left(|X[0]|^2 + |X[1]|^2 + |X[2]|^2 + |X[3]|^2\\right) \\\\\n \\implies X[k] &=\n \\begin{cases}\n \\pm \\sqrt{4D - \\frac{1}{\\sqrt{2}} |Y[1]|^2 - \\frac{1}{2} |Y[2]|^2 - \\frac{1}{\\sqrt{2}} |Y[3]|^2} & k \\equiv 0 \\pmod{4} \\\\\n \\frac{Y[k]}{1 - (-j)^k} & k \\in [1, 3] \\pmod{4}\n \\end{cases}\n\\end{align}\n\n\\subsection{}\n\nUsing the frequency shift property of the DFT,\n\\begin{equation}\n \\tilde{Y}[k] = X[k] \\tilde{H}[(k + 1)_N]\n\\end{equation}\nAssuming nothing else about \\(x[n]\\),\n\\begin{equation}\n X[k] =\n \\begin{cases}\n \\frac{Y[k]}{1 - (-j)^{k + 1}} & k \\in [0, 2] \\pmod{4} \\\\\n C & k \\equiv 3 \\pmod{4}\n \\end{cases}\n\\end{equation}\nfor \\(C \\in \\C\\).\nAssuming the sum holds from the previous part,\n\\begin{equation}\n X[k] =\n \\begin{cases}\n \\frac{Y[k]}{1 - (-j)^{k + 1}} & k \\in [0, 2] \\pmod{4} \\\\\n \\pm \\sqrt{4D - \\frac{1}{\\sqrt{2}} |Y[0]|^2 - \\frac{1}{2} |Y[1]|^2 - \\frac{1}{\\sqrt{2}} |Y[2]|^2} & k \\equiv 3 \\pmod{4} \\\\\n \\end{cases}\n\\end{equation}\n\n\\end{document}\n", "meta": {"hexsha": "462db2a521e40659de50857e42455bf2cd364689", "size": 8028, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw04/hw04.tex", "max_stars_repo_name": "bdngo/ee-123", "max_stars_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hw04/hw04.tex", "max_issues_repo_name": "bdngo/ee-123", "max_issues_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw04/hw04.tex", "max_forks_repo_name": "bdngo/ee-123", "max_forks_repo_head_hexsha": "d10fe34fdb95f2d7785eeaec9c578991aca9054d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.68, "max_line_length": 258, "alphanum_fraction": 0.5159441953, "num_tokens": 3529, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8080672112416737, "lm_q1q2_score": 0.6146568516380954}} {"text": "\\section{Error Analysis\\label{sec:error}}\nAs with other approximation-based quadrature methods, \\qbkix has two primary sources of error: the quadrature error $e_Q$ incurred as a result of evaluating potential at the check points and the extrapolation error $e_E$ due to evaluating the polynomial approximation of the potential at the target point, assuming $\\Pcoarse$ is admissible. Let\n%This is assumes that the approximation error of geometry and boundary data is less than $\\etrg$.\n\\begin{align}\ne_Q(\\vx) &=\\left|\\sum_{s=0}^p (u(c_s) - \\hat{u}(c_s,\\Pfine) )\\ell_s(t_\\vx)\\right|,\n \\label{eq:err-quad}\\\\\ne_E(\\vx) &= \\left|u(\\vx) - \\sum_{s=0}^p u(c_s)\\ell_s(t_\\vx)\\right|,\n \\label{eq:err-ext} \\\\\ne_\\lbl{hedgehog}(\\vx) &\\leq e_Q(\\vx) + e_E(\\vx),\n \\label{eq:err-hedgehog} \\\\\n\\end{align}\nwhere $u(\\vx)$ and $\\hat{u}(\\vx, \\Pfine)$ are defined in \\cref{eq:double_layer,eq:double_layer_disc} and $\\ell_s(t)$ is the $s$-th Lagrange polynomial defined on the points $\\{0,1,\\hdots, p\\}$.\nWe define $t_\\vx$ such that $\\vx = -\\vn(\\vy)(R + t_\\vx r)$, so $t_\\vx = \\frac{\\|\\vx - \\vy\\| - R}{r}$. \nIn this section, we first prove that we achieve high-order accuracy with our singular/near-singular evaluation scheme in \\cref{sec:singular-eval} with respect to extrapolation order $p$ and quadrature order $q$.\n%We then derive a heuristic inspired by the approaches taken in \\cite{aT2, elliott2011estimates} to estimate the quadrature error at a point $\\vx$ in the domain.\n%Our error heuristic applies an estimate similar to \\cite{aT2} in the principal curvature directions, specialized for our algorithmic framework.\n%Using a low-order approximation of the surface closest to $\\vx$, we can accurately estimate the quadrature accuracy due to local surface curvature without performing Newton iterations, which become costly in \\threed.\nWe then detail the impact of surface approximation on overall solution accuracy.\n\n%In this section, we describe how to ensure that $\\err{Q}$ and $\\err{E}$ are $O(\\err{target})$. \n%In practice, we actually want $ \\err{hedgehog} \\leq \\err{target}$; we achieve this by following the discussions in \\cref{sec:algo,sec:extrap_error,sec:quad_error_heuristic} with $\\err{target}/2$ in place $\\err{target}$. \n%We will drop the factors of two for clarity of exposition.\n\n\\subsection{Quadrature error\\label{sec:quad_error}}\nWe briefly state a tensor-product variation of known Clenshaw-Curtis quadrature error results as applied to smooth functions in \\threed. \nThis estimate is derived based on assumptions detailed in \\cref{app:proof_of_error_quad_high_order} that, in general, are difficult to verify in practice and may not hold for all functions we consider.\nFor this reason, we refer to it as a heuristic.\n\n\\begin{heuristic}\n Let the boundary $\\Gammah$ be discretized by quadrature patches over the domains $[-h,h]$ and the boundary condition $f$ in \\cref{eq:pde} be at least $C^k$.\n Apply the $q$-th order Clenshaw-Curtis quadrature rule to the double-layer potential $u(\\vx)$ given in \\cref{eq:double_layer_patches} and let $\\vx$ be in the interior of $\\Omega$. \n Then for all sufficiently large $q$:\n\\begin{align}\n e_\\lbl{Q}(\\vx) &%= \\sum_{i=1}^N R^2_n\\left[\\frac{\\partial G(\\vx,P_i(u,v))}{\\partial \\vn}\\phi(P_i(u,v)) J_{P_i}(u,v)\\right]\n \\lesssim \\frac{128h^{k+1}}{15\\pi k(2q + 1 - k)^{k}} \\tilde{V}, \\\\ \n \\shortintertext{where}\n \\tilde{V} &= \\max_{i=1,\\hdots,N} \\max_{\\alpha,\\beta\\leq k}\\left\\| \\frac{\\partial^{\\alpha+\\beta}}{\\partial u^\\alpha\\partial v^\\beta}\\left(\\frac{\\partial G(\\vx,\\vP_i(s,t))}{\\partial \\vn}\\phi(\\vP_i(s,t)) g_{\\vP_i}(s,t)\\right)\\right\\|_T,\n \\label{eq:error_quad_gen_target}\n\\end{align}\n $g_\\vP$ is the determinant of the metric tensor of a patch $\\vP$ implicit in \\cref{eq:double_layer_patches}, $\\lesssim$ means \"approximately less than or equal to,\" and $\\|\\zeta\\|_T = \\|\\zeta^\\prime/\\sqrt{1-x^2}\\|_1$.\n %\\begin{equation}\n %e_Q \\leq \\frac{128N}{15\\pi(2q + 1 - k)^{k}} \\max_{i} \\tilde{V}_{k_i}\\left( \\frac{\\partial G(\\vx,P_i)}{\\partial \\vn(\\vy)}\\phi(P_i) J_{P_i}\\right)\n %\\end{equation}\n \\label{heuristic:error_quad_high_order}\n\\end{heuristic}\n\nThis heuristic captures the qualitative behavior of the error. \nWe present the derivation of \\Cref{heuristic:error_quad_high_order} in \\cref{app:proof_of_error_quad_high_order}.\n\\edit{}{The presence of the derivatives of $\\phi$ in \\cref{heuristic:error_quad_high_order} largely captures two troublesome cases for integral equations: (a) when the underlying singularities of $f$ are close to $\\Gamma$; and (b) when pieces of non-local are nearly touching.\nBoth of these cases, if not properly resolved with adaptive refinement, can induce singularities in the harmonic extension of $\\phi$ and cause \\qbkix to accumulate error.\nThese sources of error are handled by \\cref{criteria:2,criteria:3} in \\cref{sec:geom_criteria} and the algorithms to enforce them.\n}\n\nAs $\\vx \\to \\Gammah$, the value of $k$ required in \\Cref{heuristic:error_quad_high_order} grows rapidly due to growing higher order derivatives of the integrand. \nSuch large values of $q$ and $k$ imply that smooth quadrature rules are cost-prohibitive; this is the problem that singular/near-singular quadrature schemes like \\qbkix aim to address.\nThis means that \\cref{heuristic:error_quad_high_order} is insufficient for direct application to \\cref{eq:double_layer_patches}, as we know. \nMoreover, this estimate is too loose to determine whether \\qbkix or smooth quadrature is required to evaluate the potential.\nThe assumption in \\cref{sec:adaptive_upsampling} addresses this problem by providing a cheap, reasonably robust criterion for refinement that is motivated by existing analyses \\cite{aT2,barnett2014evaluation} instead of relying on \\Cref{heuristic:error_quad_high_order}.\n\n\n\\subsection{Extrapolation error\\label{sec:extrap_error}}\n%Before beginning a discussion of the extrapolation error of \\qbkix, it is important to keep in mind the well-known relationship between polynomial interpolation of order $p$ and a local expansion of order $p$.\n%For a fixed $p$ in exact arithmetic, letting the interpolation interval size tend to zero produces a Taylor expansion of order $p$ centered at the interval's origin.\nA reasonable critique of \\qbkix is its reliance on an equispaced polynomial interpolant to extrapolate values of $u$ to the target point.\nDespite using the first-kind barycentric interpolation formula \\cite{webb2012stability}, polynomial interpolation and extrapolation in equispaced points is well-known for an exponentially growing Lebesgue constant and poor stability properties as the number of points $p$ increases \\cite{trefethen1991two,platte2011impossibility}.\nRecently \\cite{DT} demonstrated stable extrapolation in equispaced $p+1$ points\nusing least-sqaures polynomials of degree $\\sqrt{p}$.\nHowever, these results are asymptotic in nature and don't tell the full story for small to moderate values of $p$, as in the \\qbkix context.\n\n\\begin{figure}[!htb]\n \\centering\n \\includegraphics[width=.45\\linewidth]{figs/extrapolation_error_schematic.pdf}\n \\mcaption{fig:extrap-err-setup}{Diagram of extrapolation setup}{ The toy setup used to study the extrapolation error of a singular function. We choose a simple point singularity $\\mu(t) = \\frac{1}{\\|t - q\\|}$ where $q = (\\rho, 0, 0)$ (black star) with $\\rho =-.1$.\n We choose samples at the points $t_i = (R+ir, 0,0)$ for $i=0,\\hdots, p$ (black dots) and extrapolate the values $\\mu(t_0),\\hdots, \\mu(t_p)$ to $t=0$ (green dot).\n }\n\\end{figure}\n\nWe begin our discussion with a simple representative experiment in equispaced extrapolation.\n\\Cref{fig:extrap-err-setup} depicts a minimal extrapolation setup in \\threed of a simple singular function $\\mu(t) = 1/\\|t-q\\|$ along a line, with $q = (\\rho, 0, 0)$ and $\\rho = -.1$.\nWe extrapolate exact values of $\\mu$ from $p$ points, located at $t_i = (R + ir,0,0)$, to the origin.\nThis closely mimics the worse-case extrapolation error in \\oned of a function analytic in a Bernstein ellipse with a real axis intercept of $\\rho+R+ rp/2$.\nWe repeat this for a large range of values of $r$ and $R$ for various values of $p$. \nThe log of the relative error is plotted in \\Cref{fig:extrap-err-p6,fig:extrap-err-p8,fig:extrap-err-p10,fig:extrap-err-p12,fig:extrap-err-p14} as a function of the relative extrapolation interval size $rp/R$ and the scaled extrapolation distance $R/\\rho$.\n\nAs mentioned in \\cite[Section 3.4]{RBZ}, the adaptive refinement of $\\Pcoarse$ resolves the boundary data $f$, and therefore $u$ and $\\phi$, on the length scale $L$ of the patch.\nThis means we can reasonably assume that the distance of the nearest singularity is $O(L)$ from $\\Gammah$, i.e., $\\rho = \\lambda L$ for some $\\lambda$.\nIn the context of \\qbkix, we know that $R=bL(P)$ and $r=aL(P)$.\n\\Cref{fig:extrap-err-p6,fig:extrap-err-p8,fig:extrap-err-p10,fig:extrap-err-p12,fig:extrap-err-p14}\nare a study of extrapolation error as a function of $a/b$, $b/\\lambda$ and $p$.\n\n\n\\begin{figure}[!htb]\n \\centering\n \\hfill\n\n %\\captionsetup[subfigure]{labelformat=empty}\n% \\begin{subfigure}{.4\\textwidth}\n% \\centering\n% \\includegraphics[width=\\linewidth]{extrapolation_error_schematic.pdf}\n% \\caption{\\label{fig:extrap-err-setup}}\n% \\end{subfigure}%\n \\begin{subfigure}{.33\\textwidth}\n \\centering\n \\includegraphics[width=\\linewidth]{figs/extrapolation_error_plot_p6.pdf}\n \\caption{\\label{fig:extrap-err-p6}}\n \\end{subfigure}\n \\begin{subfigure}{.33\\textwidth}\n \\centering\n \\includegraphics[width=\\linewidth]{figs/extrapolation_error_plot_p8.pdf}\n \\caption{\\label{fig:extrap-err-p8}}\n \\end{subfigure}%\n \\begin{subfigure}{.33\\textwidth}\n \\centering\n \\includegraphics[width=\\linewidth]{figs/extrapolation_error_plot_p10.pdf}\n \\caption{\\label{fig:extrap-err-p10}}\n \\end{subfigure}\n \\begin{subfigure}{.33\\textwidth}\n \\centering\n \\includegraphics[width=\\linewidth]{figs/extrapolation_error_plot_p12.pdf}\n \\caption{\\label{fig:extrap-err-p12}}\n \\end{subfigure}%\n \\begin{subfigure}{.33\\textwidth}\n \\centering\n \\includegraphics[width=\\linewidth]{figs/extrapolation_error_plot_p14.pdf}\n \\caption{\\label{fig:extrap-err-p14}}\n \\end{subfigure}%\n \\mcaption{fig:extrap-experiment}{Empirical extrapolation error behavior}{We sweep over a range of $R$ and $r$ values to vary \\Cref{fig:extrap-err-setup} and plot the log of the relative error in\n \\Cref{fig:extrap-err-p6,fig:extrap-err-p8,fig:extrap-err-p10,fig:extrap-err-p12,fig:extrap-err-p14}, for values $p=6,8,10,12,14$, in increasing order, from (a) to (e).\nIn these figures, the $x$-axis is the extrapolation distance $R$ normalized by $\\rho$ and the $y$-axis is the ratio $rp/R$.%: the total size of the approximation interval ($rp$) relative to the extrapolation distance $R$.\nThe top of the $y$-axis corresponds to $r=R$; $rp/R = 1$ corresponds to our choice of the parameter $a$.\nAssuming that $\\rho = O(L)$, $r/R = a/b$ and $R/\\rho = b/\\lambda$ for some constant $\\lambda$.\n}\n\\end{figure}\nThere are several important observations to make from these plots: \n\\begin{itemize}\n \\item Extrapolation error decreases as $R/\\rho$ decreases, as expected.\n \\item For a fixed value of $R/\\rho$, the extrapolation error \\textit{decreases} rapidly as $rp$ decreases, up to a certain value $r^*p$.\nThis is somewhat counterintuitive, since this means placing points closer together and extrapolating a further distance relative to $rp$.\nFor a fixed $p$ in exact arithmetic, letting the interpolation interval size tend to zero produces an order $p$ Taylor expansion of the solution $u$ centered at the interval's origin, which accounts for this phenomenon.\n\n%\\note[MJM]{Another interpretation of this result is from the perspective of Bernstein ellipses.\n%Since $u$ is $C^\\infty$ in a large neighborhood around the interval of interest, we can reason that the Bernstein parameter of the corresponding \\oned approximation problem is growing as we decrease $rp$, thus increasing the convergence rate with respect to $R/\\rho$.}\n\n \\item Beyond $r^*p$, the extrapolation error \\textit{increases}.\n The effects of finite precision eventually pollutes the convergence behavior described above. \n Moreover, the spacing $r^*$ appears to be a function of $p$.\n For $p=6$, $r$ can be reduced to $1/p$ without any numerical issues, but\n by $p=14$, only $r>\\frac{1}{2}$ is a safe choice for extrapolation.\n\\end{itemize}\nWe do not aim to rigorously analyze these phenomena in this work. \nWe highlight them to provide empirical evidence that equispaced extrapolation is a reasonable, but not optimal, choice for our problem of singular/near-singular integration and to provide some intuition for our parameter choices.\n\nThe following simple result describes the behavior of the extrapolation error in \\cref{eq:err-ext}.\n\\begin{theorem}\n Let $u(\\vc(t))$ be the solution to \\cref{eq:pde} given by \\cref{eq:double_layer}, restricted to the line $\\vc(t)$ in \\threed intersecting $\\vx$, let $\\vc(t)$ be given by \n \\begin{equation}\n \\vc(t) = \\vsx - (R + tr)\\vn(\\vsx),\n \\label{eq:check_point_param}\n \\end{equation}\n where $\\vsx$ is the closest point on $\\Gammah$ to $\\vx$, $R=bL_{\\vsx}$, $r=aL_{\\vsx}$, $\\vn(\\vsx)$ is the outward surface normal at $\\vsx$, and let $|u^{(p)}(\\vc(t))|$ be bounded above by $C_p$ on the interval $[-R, R+pr]$.\n Let $\\mathfrak{P}(t)$ be the $p$-th order polynomial interpolant of $u(\\vc(t))$ constructed from the check points $\\vc_0, \\hdots, \\vc_p$, where $\\vc_i = \\vc(i)$. Then the extrapolation error associated with \\qbkix behaves according to:\n\\begin{equation}\n |u(\\vc(t_\\vx)) - \\mathfrak{P}(t_\\vx)| \\leq \\frac{C_p}{(p+1)!}|R + rp|^p = \\frac{C_p}{(p+1)!}|b+ap|^p\\cdot | L|^p,\n \\label{eq:extrap_err_init}\n\\end{equation}\nwhere $t_\\vx = \\frac{\\|\\vx - \\vsx\\| - R}{r}$.\n\\label{thm:extrap_error}\n\\end{theorem}\n\n\\begin{proof}\n\n We know that for a smooth function $f$ and points $x_0, \\hdots x_p$ in a \\oned interval $I_0$, for some $\\xi \\in I_0$, the following relation holds for all $x \\in I_0$:\n\\begin{equation}\n f(x) - \\mathfrak{P}(x) = \\frac{f^{(p)}(\\xi)}{(p+1)!}\\prod_{i=0}^p(x-x_i).\n \\label{eq:exterp_err_init}\n\\end{equation}\nLet $\\mathfrak{P}$ be the $p$th order polynomial interpolating the points $x_0,\\hdots x_p$.\nIn the \\qbkix setup, since $R+rp$ is the distance of the furthest check point to $\\vy$, we know that $x - x_i < R +rp$ for each $i$.\nSince $f(t) = u(\\vc(t))$ is harmonic, and therefore $C^\\infty$, in $\\Omega$, $|f^{(p)}(\\xi)|$ can be uniformly bounded on $I_0$ by some constant $C_p$,\nNoting that $R = bL$ and $r=aL$ yields our result.\n\\end{proof}\n\nFor fixed values of $a$ and $b$, as we let $L\\to 0$, the extrapolation error is bounded by $O(L^p)$.\nIn practice, however, this means that we can choose $a$ and $b$ to minimize the constant factor $|b+ap|^p$ in \\cref{thm:extrap_error}.\nSince $p>1$, $a$ must be chosen to balance out the contribution of $p$, yet our extrapolation study shows that we can't simply set $a=0$. \nWe therefore choose $a \\leq 1/p$ for $p=6$ and 8, motivated by \\cref{fig:extrap-err-p6,fig:extrap-err-p8}.\nMoreover, since $b < 1$, we can choose $a \\leq b/p$, which allows $a$ and $b$ to decay at the same rate.\n%In exact arithmetic, $a$ controls how close the convergence order is to $p$.\n%By holding $b$ fixed and letting $a\\to 0$, we recover a $p$th order Taylor expansion of $u$ centered at $c_0$.\n%Convergence to an approximate Taylor series is critical to the success of \\qbkix and can be seen in \\cref{fig:extrap-experiment}:\n%we can improve extrapolation accuracy for a fixed value of $b$ by decreasing $a$, until numerical errors take over.\nThe advantage of choosing $a \\leq b/p$ is that $b$ is a single parameter that controls the accuracy of \\qbkix.\nSince we have fixed the quadrature order $q=20$ to satisfy the assumption in \\cref{sec:adaptive_upsampling}, a smaller value of $b$ will trigger more upsampling in \\cref{alg:adaptive_upsampling}, keeping quadrature error fixed while reducing extrapolation error.\n\nIt is important to keep in mind that \\cref{thm:extrap_error} only provides insight for moderate values of $p$; our conclusions are largely irrelevant for large $p$.\nWe use $p = 6$ and $a = b/6$, leaving the construction of an optimal extrapolation extrapolation scheme to future work. \n\n\\subsection{Geometry approximation error\\label{sec:error_geom}}\nLet $\\theta$ be a \\edit{}{smooth} scalar function defined on the surface of $\\partial \\Omega$ with $|\\theta| \\leq 1$ and let $\\delta$ be a small real constant.\nSuppose the boundary of the domain $\\Omega$ is perturbed by $\\delta$ along the normal field of $\\partial\\Omega$, scaled by $\\theta$, to produce the perturbed domain $\\Omega_\\delta$ with boundary $\\partial \\Omega_\\delta$.\nMore concretely, for $\\vy \\in \\partial\\Omega$ and $\\vy_\\delta\\in \\partial \\Omega_\\delta$, $\\vy_\\delta= \\vy + \\delta\\theta\\vn(\\vy)$.\nWe can define the \\textit{Eulerian shape derivative} of $u$ with respect to $\\theta$, denoted $u_\\theta$, at a point $\\vx \\in \\Omega_\\delta \\cap \\Omega$ as the rate of change in $u$ at $\\vx$ as $\\delta \\rightarrow 0$.\nThis quantity is of interest to us because the solution to \\cite[Equation 2]{morse2020robust} on $\\Omega_\\delta \\cap \\Omega$ can be written as $u + \\delta u_\\theta$, where $u$ is the solution to \\cite[Equation 2]{morse2020robust} on $\\Omega$.\nMoreover, we can compute the shape derivative by solving a Laplace problem on the unperturbed domain \\cite{pironneau1982optimal}:\n\\begin{equation}\n\\Delta u_\\theta= 0\\;\\mbox{in $\\Omega$, } u_\\theta = - \\theta \\frac{\\partial u}{\\partial n} \\mbox{on $\\partial\\Omega$.}\n\\label{eq:shape-deriv}\n\\end{equation} \nwhere $u$ is the solution of the \\cite[Equation 2]{morse2020robust} on $\\Omega$.\nFor small $\\delta$, this means that the error in the solution introduced by a boundary perturbation along the field $\\theta$ can be estimated by $\\delta \\sup_{\\Omega} \\| u_\\theta \\|$.\nAssuming the boundary is smooth and the gradient of the solution $u$ is bounded, then\n\\begin{equation}\n\\| u_\\theta \\| \\leq C_g \\sup_{\\partial\\Omega} \\left| \\theta \\frac{\\partial u}{\\partial n} \\right| \\leq\nC_g \\sup_{\\partial\\Omega} \\left| \\frac{\\partial u}{\\partial n}\\right|\n\\label{eq:shape-deriv-bound}\n\\end{equation}\nfor some real constant $C_g$.\nThe right-hand side of \\cref{eq:shape-deriv-bound} yields a constant $C_g^\\prime$, such that if $\\err{g} < \\zeta\\etrg/C_g^\\prime$ for some $\\zeta < 1$, the change in the solution is less than $\\etrg$ for a sufficiently small $\\err{g}$.\n%\\note[DZ]{Need to find (again) a reference saying this, as this is not true for general boundaries}\nThe constant depends implicitly on the surface geometry: for example, if an area element of $\\partial\\Omega$ is close to a sharp, concave corner, then $\\frac{\\partial u}{\\partial n}$ can be arbitrarily large. \n\n\n\\subsection{Limitations \\label{sec:limitations}}\nOur error discussion reveals several limitations of our method. \nThe first and most apparent shortcoming is that extrapolation instability fundamentally limits convergence order. \nHowever, for reasonable orders of convergence, up to 14, we have discussed an empirical scheme to choose parameters to maximize the available convergence behavior.\nMoreover, low-order surface geometries used in engineering applications will likely limit the convergence rate before it is limited by the extrapolation order, making this a non-issue in practical scenarios.\n\nAnother downside of the chosen extrapolation approach is lack of direct extension of \\linebreak \\qbkix to oscillatory problems like the Helmholtz equation.\nDue to the limitation on the values of $p$, we can't guarantee the ability to resolve high-frequency oscillations in the solution.\nA new extrapolation procedure is required to do so robustly without compromising efficiency.\n\nIn \\cite{wala20193d}, the authors demonstrate a relationship between the truncation error of a \\qbx expansion and the local curvature of $\\Gammah$. \nOur scheme also is susceptible to this form of error and we do not address nor analyze this in this work.\nThis is a subtle problem that requires a detailed analysis of the surface geometry with respect to the chosen extrapolation scheme.\n%We leave this to a future work that produces an optimal extrapolation approach in the boundary integral context.\nAnother limitation is the lack of an accurate error estimate to serve as an upsampling criteria in place of the criteria in \\cref{sec:adaptive_upsampling}, such as \\cite{klinteberg2019accurate}. \nExtending \\cite{klinteberg2019accurate} to \\threed surfaces is non-trivial and whether the size of $\\Pfine$ would be reduced enough to outweigh the added cost of the additional Newton iterations required by their scheme remains to be seen.\n\nFinally, for certain accuracy targets and geometries, the algorithm above may lead to an impractically high number of patches in $\\Pcoarse$ and $\\Pfine$. \nGeometries with nearly-touching non-local regions, as shown in \\cref{fig:torii}, will see large amounts of refinement.\nIf the nearly-touching embeddings $\\gamma_r$ are close enough, i.e., less than $10^{-10}$ apart, there is little hope of an accurate solution with a fixed computational budget.\nWe allow the user to enforce a minimal patch size $L_\\lbl{min}$, limiting the time and memory consumption at the expense of not reaching the requested target accuracy. \n", "meta": {"hexsha": "6f241d1b66c656e43c206899de85480d81613957", "size": 21367, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hedgehog/error.tex", "max_stars_repo_name": "mmorse1217/nyu-thesis-template", "max_stars_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "hedgehog/error.tex", "max_issues_repo_name": "mmorse1217/nyu-thesis-template", "max_issues_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hedgehog/error.tex", "max_forks_repo_name": "mmorse1217/nyu-thesis-template", "max_forks_repo_head_hexsha": "dbbef3f00a1e91d6f481b4c6cb480d40960b13c8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 87.5696721311, "max_line_length": 344, "alphanum_fraction": 0.7410492816, "num_tokens": 6146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.808067204308405, "lm_q2_score": 0.7606506581031359, "lm_q1q2_score": 0.6146568507487494}} {"text": "\\chapter{Statistics - Basic}\n\n\\textit{In any chapter in this book, especially in this chapter, I will assume (1) you have really basic knowledge on statistics, e.g., what ``Gaussian'', ``probability'', ``standard normal distribution'', ``random variable'' means and/or (2) you know how to use Google so that you can find concepts which you don't know. The reasons we cannot touch much of the mathematical details are because what we have to deal with in this class is rather a practical usages of statistics in astronomy, and because I have not been a good math student.}\n\nLet's start with a question. \n\\begin{ex}\n For the \\textbf{same star}, consider the following three scenarios:\n \\begin{itemize}\n \\item Case A: Researcher 1 says $ m_1 = \\m{14.0} \\pm \\m{0.1} $; Researcher 2 says $ m_2 = \\m{15.0} \\pm \\m{1.0} $.\n \\item Case B: Researcher 1 says $ m_1 = \\m{14.0} \\pm \\m{0.1} $; Researcher 2 says $ m_2 = \\m{15.0} \\pm \\m{0.3} $.\n \\item Case C: Researcher 1 says $ m_1 = \\m{14.0} \\pm \\m{0.1} $; Researcher 2 says $ m_2 = \\m{15.0} \\pm \\m{0.1} $.\n \\end{itemize}\n For each of the three cases: Are the two studies coincide? If so, with how much confidence would you say so? Wait, but what does that ``$ \\pm $'' sign means in rigorous mathematical sense?\n\\end{ex}\n\nSome of the previous course takers answered that they coincide, because the ``3-sigma rule'' says, e.g., for case A, $ m_1 = \\m{14.0} \\pm \\m{0.3} $ \\& $ m_2 = \\m{15.0} \\pm \\m{3.0} $ overlaps with each other. Crudely speaking, this makes sense, but it is of course not the ``publication level'' reasoning.\n\nTo give you the answer: It is an ill-defined question. The answer can change based on many factors; to pick one, the number of observations each reasercher used for the determination of the magnitude. If we assume that number is 5 for both researchers, for example, we can give some meaningful answers: We \\textit{cannot reject} $ m_1 = m_2 $ for case A \\& B with 90 \\% confidence, and we \\textit{can reject} $ m_1 = m_2 $ for case C with 90 \\% confidence. Depending on the confidence level, the answer may also change. Note that the expression that \\textit{can/cannot reject} is very important! You should \\textbf{NOT} say you \\textit{accept $ m_1 = m_2 $}, never ever. Let me explain why throughout this chapter.\n\n\n\\section{The $ n $-$ \\sigma $} \\label{sec: n-sigma notation}\nMaybe you are familiar of the 1-$ \\sigma $, 2-$ \\sigma $, and 3-$ \\sigma $ words (cf. \\cref{fig:nsigmawiki}). For example, mean $ \\pm 1$-$ \\sigma $ contains $ 68.27\\cdots \\% $ of the total area of a normal distribution. Similarly 3-$ \\sigma $ conatains $ 99.73\\cdots \\% $ of it, so it is very unlikely to get a sample outside of the mean $ \\pm 3 $-$ \\sigma $. \n\n\\begin{figure}[ht!]\n \\centering\n \\includegraphics[width=0.5\\linewidth]{figs/nsigma_wiki}\n \\caption{A test sampling from Gaussian (normal) distribution (from Wikimedia).}\n \\label{fig:nsigmawiki}\n\\end{figure}\n\nIn this sense, the ``$ n $-$ \\sigma $'' is defined as the $ x $-axis value which gives the integrated area of certain value. More precisely, we have to call this confidence interval (CI):\n\n\\begin{defn}[$ n $-$ \\sigma $ Confidence Interval; CI]\n $ n $-$ \\sigma $ \\textbf{confidence interval} is defined as \\textit{an} interval which contains $ \\Phi(n) - \\Phi(-n) $ of the total area, usually centered at the mean or median, where $ \\Phi := (2\\pi)^{-1/2} \\int_{-\\infty}^{x} e^{t^2/2} dt $ is the cumulative distrubution function of standard normal distribution. \n\\end{defn}\n\nFollowing this definition, 1-$ \\sigma $ CI is the interval such that the integrated area is $ 0.6827\\cdots $ of the total area. Similarly, 2-$ \\sigma $ CI is that with $ 0.9545\\cdots $, 3-$ \\sigma $ CI is that with $ 0.9973\\cdots $, etc. The term CI is not necessarily limited to Gaussian (normal) distribution: You can apply it to any distribution, and give an interval such as $ 0.11 ^{+0.03}_{-0.01} $ as the 1-$ \\sigma $ CI, for instance.\n\nSometimes it is more convenient to call, e.g., 1-$ \\sigma $ CI as the $ 68.27\\cdots \\% $ CI. We also define the \\textit{significance level} $ \\alpha $ as 100\\% minus this percentage, i.e., the 1-$ \\sigma $ CI is of significance level $ \\alpha = 1 - 0.6827\\cdots = 0.3173\\cdots $:\n\\begin{equation}\n \\text{1-}\\sigma \\text{ CI}\n \\quad \\equiv \\quad \n 68.27\\cdots \\% \\text{ CI}\n \\quad \\equiv \\quad \n \\text{CI of significance level } 0.3173 \\cdots ~.\n\\end{equation}\nIn physical and astronomical sciences we often use the terminology of ``$ n $-$ \\sigma $ CI'', but in mathematics and other branches of sciences and engineering, the ``$ k $\\,\\% CI'' or the ``significance level $ \\alpha $'' are used dominantly. Among them, 90\\,\\%, 95\\,\\%, and 99\\,\\% CIs, i.e., CIs with significance level $ \\alpha = 0.10,\\, 0.05,$ and $ 0.01 $ are found frequently.\n\nBecause when we say ``$ n $-$ \\sigma $'', we are just omitting the term ``CI'' at the end of it, this is not $ n $ times the standard deviation ($ \\sigma $): What we mean by ``$ n $-$ \\sigma $'' is actually ``$ n $-$ \\sigma $ CI'' and is \\textit{not necessarily} $ n \\times \\sigma $. They are the same for, e.g., Gaussian distribution, because the $ n $-$ \\sigma $ terminology is \\textit{defined} using it. For Gaussian (normal) distribution, which is a distribution that appears very often in natural sciences, such as 2-$ \\sigma $ is nothing but 2 times the standard deviation ($ \\sigma $). For almost all of the distributions other than Gaussian, this is not the case.\n\n\n\n\\section{The Meaning of $ \\pm $ Sign}\nSimple answer: The $ \\pm $ sign means the 1-$ \\sigma $ confidence interval (CI). It is also called the ``\\textbf{error-bar}''. \n\nMathematically speaking, the $ \\pm $ sign, i.e., the 1-$ \\sigma $ CI, does \\textit{not} fully describe the uncertainty. The best way to show or describe the uncertainty is to show a graph of \\textbf{probability distribution function} (\\textbf{pdf} or p.d.f.) as in \\cref{fig:figposterior01}. In many publications or books, people do not show this, because (1) the distribution is very similar to Gaussian or a widely known distribution and it is clearly written in the text or assumed as all the readers know that, (2) detailed calculation is not of interest and/or does not affect the final result, (3) the writer is lazy. \n\n\\begin{figure}[ht!]\n\\centering\n\\includegraphics[width=1\\linewidth]{figs/fig_posterior01}\n\\caption{The probability distribution function (pdf) of two examples of distributions, namely, Gaussian and an asymmetric distributions. I used a chi-square distribution for the latter for plotting purpose.}\n\\label{fig:figposterior01}\n\\end{figure}\n\n\nThe uncertainty ranges shown in \\cref{fig:figposterior01} ($ 4.0 \\pm 0.1 $ and $ 3.0^{+3.0}_{-2.0} $) are determined such that the lower/upper bounds include $ 68.27 \\cdots \\% $ of the total area. It is trivial for a Gaussian distribution: mean $ \\pm $ 1-$ \\sigma $. But if the distribution is asymmetric (right panel of the figure), there are few choices to set such bounds. First and the most widely used one is to set the bounds such that the distance between upper/lower bounds are minimized (but include $ 68.27 \\cdots \\% $ of the total area). Second is to make it symmetric while include $ 68.27 \\cdots \\% $ of the total area, so that you can use $ \\pm $ sign for simplicity ($ 3.0 \\pm 2.3 $ for example). Third choice would be something like FWHM: find $ x $-axis values such that the pdf value is $ 0.5 \\times \\mathrm{pdf_{max}} $. The last two bounds are simple to calculate but not as accurate as the first choice.\n\nVery assymetric probability distributions appear, e.g., in cosmological sciences\\footnote{Google ``posterior distribution cosmological constants''.}, and exhaustive statiscal analyses are conducted on such models to \\textit{reject} (I repeat; \\textit{reject, not accept}) cosmological models. The reason is that, in cosmology, we have only one single sample (our universe), and the statistical analyses given the observational data is of utmost importance. Of course similar exhaustive statistical analyses should be conducted on any research field if the data/object/target is of such importance. We don't do that in many observational studies purely because (1) we don't have human power to do all that and (2) the results won't change much (but no one in history thoroughly checked this).\n\n\n\n\\section{Caveat on the ``$ n $-$ \\sigma $ CI'' Notation}\nLet me emphasize again: What we mean by ``$ n $-$ \\sigma $'' is actually ``$ n $-$ \\sigma $ CI'' and is \\textit{not necessarily} $ n \\times \\sigma $. \n\nIt maybe is tempting to convert the 1-$ \\sigma $ CI to the 3-$ \\sigma $ CI. For instance, from \\cref{fig:figposterior01}, $ 4.0 \\pm 0.1 $ and $ 3.0^{+3.0}_{-2.0} $ to $ 4.0 \\pm 0.3 $ and $ 3.0^{+9.0}_{-6.0} $. This is true for the former (Gaussian) but wrong for the latter (non-Gaussian). The lower bound of the second parameter now become negative ($ 3.0 - 6.0 = -3.0 $), which is not even physically correct if this parameter is defined to be positive. What you have to do is, get the 3-$ \\sigma $ CI which should contain $ 0.9973\\cdots $ of the total area from the distribution given in \\cref{fig:figposterior01}. There can be at least three choices to do it, which I mentioned in the previous section.\n\n\n\\section{Central Limit Theorem (CLT)}\nThe Central Limit Theorem (CLT) is at the heart of all the observational or experimental sciences. It is stated as\\footnote{For strict definitions of random sample, population, independence, the statement of CLT in mathematical senses, etc, I recommend mathematical statistics textbooks.}\n\n\\begin{thm}[Central Limit Theorem; CLT] \\label{thm: clt}\n Consider a random sample (observation, measurement, etc) with size $ n $ and the mean value of this sample is $ \\bar{X} $. Then $ (\\bar{X} - \\mu) / (\\sigma / \\sqrt{n}) $ approaches the standard normal distribution ($ \\mathcal{Z} $) as $ n \\rightarrow \\infty $, where $ \\mu $ and $ \\sigma^2 $ are the (finite) mean and variance of the population:\n \\begin{equation}\n \\frac{\\bar{X} - \\mu}{\\sigma / \\sqrt{n}} \\sim \\mathcal{Z} \\quad (\\mathrm{as~} n \\rightarrow \\infty)~.\n \\end{equation}\n\\end{thm}\n\nThe standard deviation (second moment) is sometimes undefined. An example is the Cauchy distribution, and I will show you later with an example when this distribution can appear. There is one more important theorem, which we usually skip to mention:\n\n\\begin{thm}[Sample Variance Distribution] \\label{thm: s and sigma}\n Consider a random sample (observation, measurement, etc) with size $ n $ and the sample variance is $ S^2 $. Then $ (n - 1) S^2 / \\sigma^2 $ follows a chi-squared distribution with degrees of freedom $ (n - 1) $, where $ \\sigma^2 $ is the variance of the population:\n \\begin{equation}\n \\frac{(n - 1) S^2}{\\sigma^2 } \\sim \\chi^2_{(n-1)}\n \\end{equation}\n\\end{thm}\n\nAlso the definition of the Student $ t $-distribution\\footnote{This definition is directly copied from Walpole et al. p.177}:\n\n\\begin{defn}[Student $ t $-distribution]\\label{def: t-distn}\n Let $ Z $ be a standard normal random variable and $ V $ a chi-squared random variable with $ v $ degrees of freedom. If $ Z $ and $ V $ are independent, then the distribution of the random variable $ T $, where\n \\begin{equation}\n T = \\frac{Z}{\\sqrt{V /v}}\n \\end{equation}\n is given by the density function\n \\begin{equation}\n h(t) = \\frac{\\Gamma \\qty[ \\frac{v+1}{2} ]}{\\Gamma \\qty[\\frac{v}{2}] \\sqrt{\\pi v}}\n \\qty( 1 + \\frac{t^2}{v} )^{-\\frac{(v+1)}{2}}\n \\quad ~, \\quad\n -\\infty < t < \\infty ~.\n \\end{equation}\n This is known as the Student $ t $-distribution with $ v $ degrees of freedom.\n\\end{defn}\n\nWe can see that the $ t $-distribution approaches the standard normal distribution as the degrees of freedom $ v \\rightarrow \\infty $, because the power term takes the $ e^{-t^2 / 2} $ form. \n\n\n\\begin{thm}[Practical Usage of the CLT] \\label{thm: practical clt}\n Consider a random sample (observation, measurement, etc) with size $ n $ large enough and the mean value of this sample is $ \\bar{X} $. If $ \\mu $, $ \\sigma^2 $, and $ S^2 $ are the (finite) mean and variance of the population, and the sample variance, respectively:\n \\begin{equation}\n \\frac{\\bar{X} - \\mu}{S / \\sqrt{{n}}} \\sim T_{(n - 1)} ~.\n \\end{equation}\n\\end{thm}\n\nSometimes people empirically say $ n \\ge 30 $ is enough. This really depends on the situation and I will skip this issue here. The sketch of the proof of the theorem is simple. By the definition of the $ t $-distribution and Thm \\ref{thm: s and sigma}, \n\\begin{equation}\n T = \\frac{Z}{\\sqrt{V /v}} = \\frac{ \\frac{\\bar{X} - \\mu}{\\sigma / \\sqrt{n}} }\n {\\sqrt{ \\frac{(n - 1) S^2}{\\sigma^2 } / (n-1)}}\n = \\frac{\\bar{X} - \\mu}{S / \\sqrt{{n}}}\n\\end{equation}\nso the theorem is plausible. For your information, the sample variance is defined as\n\\begin{equation}\n S^2 := \\frac{1}{n - 1} \\sum_{i=1}^{n} (X - \\bar{X})^2 \n\\end{equation}\nand the sample standard deviation $ S $ is the square root of this.\n\nNote here the difference between Thm \\ref{thm: clt} and Thm \\ref{thm: practical clt}: $ \\sigma $ is changed to $ S $, and the standard normal distribution is changed to the $ t $-distribution of $ (n - 1) $ degrees of freedom. Since the true variance $ \\sigma^2 $ is unknown, it is difficult to use the CLT (Thm \\ref{thm: clt}) directly. But thanks to Thm \\ref{thm: s and sigma}, we can use the variance of the sample, $ S^2 $, which is measurable, and utilize the $ t $-distribution, which is slightly bothersome than the standard normal distribution but still useful. \n\nAlso be aware that CLT says the \\textit{expectation value of the mean} is normally distributed, not the \\textit{sample} is so.\n\n\\begin{ex}[Coin Tossing and the CLT]\n Consider a coin-tossing experiment and assign $ \\pm 1 $ to heads and tails. Each experiment will give you $ \\pm 1 $, but never $ 0 $. Meanwhile, we know a fair coin should have $ \\mu = 0 $ (same probability of heads and tails). After $ n = 100 $ experiments, say you obtained $ \\bar{X} = 0.01 $ and $ S = 0.1 $, so\n \\begin{equation*}\n T = \\frac{\\bar{X} - \\mu}{S / \\sqrt{{n}}}\n = \\frac{0.01 - \\mu}{0.1 / \\sqrt{100}} \n = 1 - 100 \\mu ~.\n \\end{equation*}\n Since $ T \\sim t_{99} $, the significance level $ \\alpha = 0.05 $ confidence interval, i.e., the confidence interval containing $ 100 (1 - \\alpha) \\% = 95 \\% $, can be\n \\begin{equation*}\n t_{99, 0.025} < T < t_{99, 0.975}\n \\quad \\rightarrow \\quad\n - 1.9842 < 1 - 100 \\mu < 1.9842\n \\quad \\rightarrow \\quad\n \\mu \\in [-0.0098,\\, 0.0298] ~.\n \\end{equation*}\n The notation $ t_{\\nu, x} $ means the input argument of the $ t $-distribution with the degrees of freedom $ \\nu $ such that $ \\int_{-\\infty}^{t_{99, x}} h(t) dt = x $. Thus, to get the two-tail CI of significance level $ \\alpha $, we need to calculate $ t_{\\nu, \\alpha / 2} $ and $ t_{\\nu, 1 - \\alpha / 2} $. For a symmetric distribution like $ t $ here, $ t_{\\nu, 1 - \\alpha / 2} = - t_{\\nu, \\alpha / 2} $. This is very widely used standard notation.\n \n If you just calculated without pondering about the meaning of the caculation, you may be surprised that your next experiment gives either $ +1 $ or $ -1 $, while the expectation is $ \\mu \\in [-0.0098, 0.0298] $. This is because the result from CLT is, as described, about the \\textit{mean} of the samples, not the \\textit{single sample}. Thus, the error-bar from the CLT is not necessarily predicting the possible range of \\textit{future experiments}, but it just confines the \\textit{position of the mean}.\n\\end{ex}\nThe range which predicts the future experiments is called the prediction interval. You may learn the prediction interval (PI) to clarify this difference.\n\nThe two-tail significance level $ \\alpha $ interval of the $ t $-distribution calculable in python by\n\\begin{python}\nfrom scipy.stats import t\nalpha = 0.95\nnu = 99\n\n# Note that the alpha below is NOT the significance level, \n# but the area under the curve!\nlo, hi = t.interval(alpha=alpha, df=nu)\nprint(f\"Confidence Interval: [{lo:.4f}, {hi:.4f}]\")\n\\end{python}\n\n\n\\subsubsection*{Note: Uncertainty of Median}\nThe 1-$ \\sigma $ error-bar of the median is more difficult to handle than that of the mean (CLT: Thm \\ref{thm: clt}). But we can get a result from simplifying assumptions, which are not necessarily true for real observations:\n\n\\begin{thm}[Uncertainty of Median]\n Consider $ n (\\gg 1) $ samples are independently drawn from $ X \\sim \\mathcal{G}(\\mu, \\sigma^2) $, a general continuous distribution with finite mean and variance $ \\mu $ and $ \\sigma^2 $ with pdf $ p(x) $. The uncertainty of the median estimator is \n \\begin{equation}\\label{eq: err median general}\n \\Delta \\mathrm{med} \\approx \\frac{1}{2 p(\\nu_0) \\sqrt{n - 1}}\n \\end{equation}\n and if $ \\mathcal{G} = \\mathcal{N} $, i.e., a normal distribution,\n \\begin{equation}\\label{eq: err median}\n \\Delta \\mathrm{med} \n \\approx \\sqrt{\\frac{\\pi / 2}{n - 1}} s \n = \\sqrt{\\frac{\\pi}{2} \\frac{n}{n - 1}} \\Delta \\mathrm{mean}\n \\approx 1.25 \\Delta \\mathrm{mean}\n ~,\n \\end{equation}\n where $ s $ is the sample standard deviation and $ \\Delta \\mathrm{mean} = s / \\sqrt{n} $ is the error-bar of the mean from Thm \\ref{thm: practical clt}.\n\\end{thm}\n\\begin{proof}[of general distribution]\nTo prove it, say the true median is $ \\nu_0 $ and samples $ \\{ x_1, \\cdots, x_n \\} $ are sorted as increasing order. For an integer $ m $, set $ n = 2m + 1 $, because $ n \\gg 1 $. The sample median is then $ \\nu = x_{m + 1} $.\nThe probability of having sample median $ \\nu $ is calculable by noting that we have to sample $ m $ of sample with $ x_i \\le \\nu $ and the other $ m $ with $ x_i > \\nu $:\n\\begin{equation*}\n f(\\nu) = \\frac{n!}{m! m!} q^m (1 - q)^{m} ~,\n\\end{equation*}\nwhere $ q = q(\\nu) = \\mathbb{P} \\{ X_i \\le \\nu \\} = \\int_{-\\infty}^{\\nu} p(x) dx $. $ p(x) $ is the pdf of the normal distribution. \n\nThe Taylor expansionof $ q $ at $ \\nu = \\nu_0 $:\n\\begin{equation*}\n\\begin{aligned}\n q(\\nu) &= q(\\nu_0) + \\frac{1}{1!} q'(\\nu_0) (\\nu - \\nu_0) + O\\qty( (\\nu - \\nu_0)^2 )\\\\\n &\\approx \\frac{1}{2} + q'(\\nu_0) (\\nu - \\nu_0) \\\\\n &\\equiv \\frac{1}{2} + p(\\nu_0) (\\nu - \\nu_0) ~.\n\\end{aligned}\n\\end{equation*}\nThen \n\\begin{equation*}\n\\begin{aligned}\n f(\\nu) &\\approx \\frac{n!}{m! m!}\n \\qty[ \\frac{1}{2} + p(\\nu_0) (\\nu - \\nu_0) ]^m\n \\qty[ \\frac{1}{2} - p(\\nu_0) (\\nu - \\nu_0) ]^m \\\\\n &= \n \\frac{n!}{m! m!}\n \\qty[ \\frac{1}{4} \\qty( 1 - 4 p(\\nu_0)^2 (\\nu - \\nu_0 )^2) ]^m \\\\\n &= \n \\frac{n!}{m! m! 4^m}\n \\qty[ 1 - \\frac{4 m p(\\nu_0)^2 (\\nu - \\nu_0 )^2}{m} ]^m \\\\\n &\\approx \n \\frac{n!}{m! m! 4^m}\n e^{ -4 m p(\\nu_0)^2 (\\nu - \\nu_0 )^2 } ~.\n\\end{aligned}\n\\end{equation*}\nThis has the form of Gaussian distribution ($ \\propto e^{-(x - \\mu)^2 / 2\\sigma^2} $) with mean $ \\nu_0 $ and variance $ \\frac{1}{4 (n-1) p(\\nu_0)^2} $ since $ m = (n - 1) / 2 $. Thus, \n\\begin{equation}\n \\nu \\sim \\mathcal{N} \\qty( \\nu_0, \\frac{1}{4 (n-1) p(\\nu_0)^2} )\n \\quad\\rightarrow \\quad\n \\Delta \\mathrm{med} \\approx \\frac{1}{2 p(\\nu_0) \\sqrt{n - 1}} ~.\n\\end{equation}\nQ.E.D.\n\\end{proof}\n\n\\begin{proof}[of Gaussian distribution]\nNow if $ p(x) $ is a Gaussian, median is equal to mean, so the pdf becomes $ p(\\nu_0) = p(\\mu) = \\frac{1}{\\sqrt{2\\pi} \\sigma} $, and\n\\begin{equation*}\n \\nu \\sim \\mathcal{N} \\qty( \\nu_0, \\frac{\\pi / 2}{n - 1} \\sigma^2 ) ~.\n\\end{equation*}\nIf we denote $ A = \\sqrt{(n - 1) / (\\pi / 2)} $, the statistic $ Z = A \\frac{\\nu - \\nu_0}{\\sigma} $ will follow the standard normal distribution. On the other hand, $ V = \\frac{(n - 1) s^2}{\\sigma^2} $ follows the chi-squared distribution of degrees of freedom $ (n - 1) $ by Thm \\ref{thm: s and sigma}. Then \n\\begin{equation*}\n T = \\frac{Z}{\\sqrt{V / (n - 1)}}\n = A \\frac{\\nu - \\nu_0}{s}\n\\end{equation*}\nwill follow a $ t $-distribution with degrees of freedom $ (n - 1) $ by Def \\ref{def: t-distn}. $ T $ is nearly a Gaussian when $ n $ is large (rule-of-thumb: when $ n \\gtrsim 30 $), so \n\\begin{equation}\n \\nu \\mathrel{\\dot{\\sim}} \\mathcal{N} \\qty( \\nu_0, \\frac{\\pi / 2}{n - 1} s^2 )\n \\quad\\rightarrow \\quad\n \\Delta \\mathrm{med} \\approx \\sqrt{\\frac{\\pi / 2}{n - 1}} s ~.\n\\end{equation}\nQ.E.D.\n\\end{proof}\n\n\n\n\\section{Meaning of the Confidence Interval}\nI circunvented the definition of the CI so far. It is defined as\n\n\\begin{defn}[Confidence Interval; CI] \n A confidence interval of significance level $ \\alpha $ of a parameter $ X $ is the interval of $ X $ value such that if we conduct the identical parameter estimation process in many \\textit{parallel universes} (i.e., ensemble) which have the identical true value $ X_\\mathrm{true} $, fraction of $ (1 - \\alpha) $ of such universes could have calculated a CI such that the $ X_\\mathrm{true} $ is included in that CI.\n\\end{defn}\n\nThis is not a practically meaningful definition, but philosophically important. Let me give an example to elaborate the meaning of this.\n\n\\begin{ex}[Meaning of the Confidence Interval]\n Imagine there are many \\textit{parallel universes} and the same observer observes the same star in each universe. Because of random errors (which we will learn later but that includes Poisson photon noise, readout noise, etc), the observer at each universe will obtain slightly different observational results. The observers will observe the star for $ n $ times, and calculate their own mean and error-bar as $ \\bar{X} $ and $ S / \\sqrt{n} $ from their observations. Note again that $ S / \\sqrt{n} $ is the uncertainty of the \\textit{mean} value.\n\n For visualization, the true magnitude of the star is $ \\mu = \\m{14.0} $ and true standard deviation of the observation $ \\sigma = \\m{0.2} $ with $ n = 9 $ are used for the generation of \\cref{fig:figclt01}. The universe ID 0 of both the upper and lower panels show roughly $ m = \\m{14.0} \\pm \\m{0.1} $. That is, when the true magnitude and true standard deviation are given as $ \\m{14.0} $ and $ \\m{0.2} $, an observer at one of the universes (in our case universe ID 0) got $ m = \\m{14.0} \\pm \\m{0.1} $, and that is our universe. \n\\end{ex}\n\n\\begin{figure}[ht!]\n \\centering\n \\includegraphics[width=1\\linewidth]{figs/fig_clt01}\n \\caption{Simulation to show the concept of CI and CLT. Black crosses are the observation in each universe, and the circles are the mean and its uncertainty from the CLT $ S / \\sqrt{n} $. Red circles are the univserses where the confidence interval contains the true mean $ \\mu = \\m{14.0} $ and blue circles are the others. In the title, the fraction of the universes which contains the true mean within their own confidence intervals out of \\texttt{n\\_univ} simulated universes is shown. This fraction approaches $ 0.6827\\cdots $ as the number of universe gets infinity.}\n \\label{fig:figclt01}\n\\end{figure}\n\nI will close this section with an excerpt from Walpole p.234:\n\n\\textit{The interpretation of a CI is often misunderstood. It is tempting to conclude that the parameter falls inside the CI with probability of(, e.g.,) 0.95. (But this is not ture.) A CI merely suggests that if the experiment is conducted and data are observed again and again, about 95\\% of such intervals will contain the true parameter.}\n\n\n\n\\section{Answer to the Question}\nIn statistics, the \\textit{null hypothesis}, denoted $ H_0 $, is the one that is the simplest, and that has value if it is \\textit{rejected} (not \\textit{accepted}). The \\textit{alternative hypothesis}, $ H_1 $, is another possibility if $ H_0 $ is not true. In the simplest case, $ H_1 $ is the complementary of $ H_0 $. For example, $ H_0: \\mu = 0 $ and $ H_1: \\mu \\neq 0 $. \n\nLet's consider the opening question: $ m = \\m{14.0} \\pm \\m{0.1} $. Now you understand that $ \\m{14.0} $ means that the mean value\\footnote{It can actually be the median value, most probable value, or whatever the representative value.} from the observation, and $ \\pm \\m{0.1} $ means the uncertainty or the error-bar of the Gaussian probability distribution of that mean value, $ S / \\sqrt{n} $. If it were not Gaussian, the author should have given more information. Note that the sample standard deviation is $ \\sqrt{n} $ times the error-bar.\n\nTo answer the question, we set the hypotheses\\footnote{There are some ``standard'' ways to set the hypotheses, and it may seem very sudden if you are not familiar with these. Please refer to basic applied statistics textbooks for more examples, e.g., p.246 and p.264 of Walpole et al.}:\n\\begin{equation}\n H_0 : m_1 = m_2 \n \\sep\n H_1 : m_1 \\neq m_2 ~,\n\\end{equation}\nwhere $ m_1 = \\m{14.0} \\pm \\m{0.1} $ and $ m_2 $ is the other (from researcher 2). $ m_1 $ and $ m_2 $ here means the ``true'' magnitude of the star, not the measured value! If it were the measured value, they are just different, since one is 14.0 and the other is 15.0. The two true means may be different because (1) researchers may have mistakenly observed different star, (2) the star is maybe a variable (including binary), (3) explanet occulted some part of the star, or any other scenario is possible.\n\nNow we have to choose which formula to use. When two measurements with $ \\bar{x}_1 \\pm s_1 $ and $ \\bar{x}_2 \\pm s_2 $ from the number of observations $ n_1 $ and $ n_2 $ are given, and $ \\mu_1 $ and $ \\mu_2 $ are the true means of two sampling distributions that will be tested under the hypothesis testing, there are two different formulae:\n\\begin{itemize}\n\\item The true variances are unknown and different (Satterthwhite approximation\\footnote{SatterthwhiteFE (1946, Biometrics Bulletin, 2, 110), ``An Approximate Distribution of Estimates of Variance Components.''}):\n\\begin{equation}\n t = \\frac{(\\bar{x}_1 - \\bar{x}_2) - (\\mu_1 - \\mu_2) }{\\sqrt{s_1^2 / n_1 + s_2^2 / n_2}}\n\\end{equation}\nfollows the $ t $-distribution with degrees of freedom\n\\begin{equation}\n \\nu \\approx \n \\frac{\\qty( s_1^2 / n_1 + s_2^2 / n_2 )^2}\n {\\frac{ \\qty(s_1^2 / n_1)^2 }{n_1 - 1}\n + \\frac{ \\qty(s_2^2 / n_2)^2 }{n_2 - 1}} ~.\n\\end{equation}\n\\item The true variances are unknown but the same:\n\\begin{equation}\n t = \\frac{(\\bar{x}_1 - \\bar{x}_2) - (\\mu_1 - \\mu_2) }{s_p \\sqrt{1 / n_1 + 1 / n_2}}\n\\end{equation}\nfollows the $ t $-distribution with degrees of freedom $ \\nu = n_1 + n_2 - 2 $ and\n\\begin{equation}\n s_p = \\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{\\nu} ~.\n\\end{equation}\n\\end{itemize}\nWhich one will you choose? To answer our original question, it is more reasonable to assume the true variances are not necessarily identical, because they may have used different instruments at different sky conditions. So I will use the first choice.\n\nTo apply the formula, you need the number of observations from each researcher, i.e., the $ n_1 $ and $ n_2 $ values. As I mentioned, it is an ill-defined question since this numbers are not given. So let me assume $ n_1 = n_2 = n = 5 $. The question can be reformulated as $ \\bar{x}_1 = 14.0 $, $ \\bar{x}_2 = 15.0 $, $ s_1 = 0.1 \\sqrt{n} $, and $ s_2 = 1.0\\sqrt{n},\\, 0.3\\sqrt{n}, \\, 0.1\\sqrt{n} $, where I dropped the magnitude sign for brevity. Under the null hypothesis ($ H_0: \\mu_1 = \\mu_2 $), $ \\mu_1 - \\mu_2 = 0 $. For the three $ s_2 $ values, \n\\begin{equation}\n\\begin{aligned}\n t &= \\frac{(14.0 - 15.0) - 0}{\\sqrt{(s_1^2 + s_2^2) / n}}\n \\quad &&\\rightarrow \\quad\n t &&= & -0.995 ,\\, && -3.162 ,\\, && -7.07 \\\\\n \\nu &= (s_1^2 + s_2^2) (n-1)\n \\quad &&\\rightarrow \\quad\n \\nu &&= & 4.08, \\, && 4.32, \\, && 8.0 \\\\\n &&&&&\\approx & 4 ,\\, && 4 ,\\, && 8 \\\\\n\\end{aligned}\n\\end{equation}\nThe values are in the order for $ s_2 = (1.0,\\, 0.3, \\, 0.1)\\sqrt{n} $ cases. The (two-tail) significance level $ \\alpha $ test is listed in \\cref{tab: CI}.\n\n\\begin{table}[ht!]\n\\centering\n\\caption{The confidence interval calculation for the example question.}\n\\label{tab: CI}\n\\begin{tabular}{c|c||c|c}\nCI \\% & significance level $ \\alpha $ and $ \\frac{\\alpha}{2} $& $ \\nu = 4 $ case & $ \\nu = 8 $ case \\\\\n\\hline\n90 \\% & 0.10 \\& 0.05 & $ [-2.1318, 2.1318] $ & $ [-1.8595, 1.8595] $ \\\\\n95 \\% & 0.05 \\& 0.025 & $ [-2.7764, 2.7764] $ & $ [-2.3060, 2.3060] $ \\\\\n99 \\% & 0.01 \\& 0.005 & $ [-4.6041, 4.6041] $ & $ [-3.3554, 3.3554] $ \\\\\n\\end{tabular}\n\\end{table}\n\nThe answers to the questions are, assuming $ n_1 = n_2 = 5 $,\n\\begin{itemize}\n\\item Case A ($ t = -0.995 $ with $ \\nu = 4 $): The $ t $ value is inside of all three CIs. Thus, we cannot reject $ H_0 $ under 90 \\%, 95 \\%, and 99 \\% CI criterion (cannot reject under the significance level $ \\alpha = 0.10,\\, 0.05,\\, 0.01 $).\n\\item Case B ($ t = -3.162 $ with $ \\nu = 4 $): The $ t $ value is inside of 90 \\% and 95 \\% CI but outside of the 99 \\% CI. Thus, we reject $ H_0 $ with $ \\alpha = 0.10 $ and $ 0.05 $, but cannot reject it with $ \\alpha = 0.01 $.\n\\item Case C ($ t = -7.07 $ with $ \\nu = 8 $): We reject $ H_0 $ with all the three $ \\alpha = 0.10,\\, 0.05,\\, 0.01 $.\n\\end{itemize}\n\\vspace{\\parskip}\n\nBy \\textit{rejecting the null hypothesis}, $ H_0: m_1 = m_2 $, we mean that it is unlikely that the true magnitude value of the two studies are identical. By \\textit{failing in rejecting the null hypothesis}, we mean that it is impossible to conclude whether $ m_1 $ and $ m_2 $ are the same under the given significance level. \n\nThe results are as expected: For Case A, the error-bar of $ m_2 $ is too large, which means it is almost impossible to reject the claim that $ m_1 = m_2 $, so the we always fail to reject the null hypothesis. For Case C, the error-bars are too small and thus $ m_1 $ is of course vastly different from $ m_2 $, so the null hypothesis is always rejected.\n\n%\\footnote{You may wonder what if we have $ n = 1 $. The $ \\nu $ is undetermined and all the calculation seem to meaningless, but of course we can have error-bars for each single observation. That is not considered in this example since here we are simply assuming the mean is inferred from $ n $ observations under the CLT, and we need $ n > 1 $ to calculate the sample variance $ S^2 $. I will not consider the error-bars of each observations. Such a detailed statistical analyses should follow some well-developed and specific statistical routines available in your research field.}.\n%In this example we do not care about the error-bar of each observation but the final error-bar of the mean value from researchers. If we want to consider the error-bars of every observations, this kind of classical statistical method gets very complicated and almost impossible to give meaningful results. You then need Bayesian approach and Monte Carlo simulation technique.\n\n\n\n\n\\section{Probability Distributions and Poisson Noise}\nThere are a myriad of probability distributions (PDs) in math, due to hard works of mathematicians lived in the era of frequentist statistics. In this section, I will just give some definitions and properties of few of them, which are frequently visited by astronomical literatures.\n\n\\subsection{Binomial Distribution}\nFirst, let me define Bernoulli process:\n\\begin{defn}[Bernoulli Process]\nA Bernoulli process is a process (experiments) which has the following properties (Walpole p.102):\\newpage\n\\begin{enumerate}\n \\item Repeated trials where each trial (outcome) is classified as either a success (1) or failure (0).\n \\item The success probability ($ p $) should remain constant throughout the process.\n \\item Each trial (outcome) must be independent to each other.\n\\end{enumerate}\n\\end{defn}\n\nA coin-tossing experiment can be a Bernoulli process: The probability of success (heads; H) $ p = 0.5 $ and failure (tails; T) is $ q = 1 - p = 0.5 $. Each coin-tossing is independent of each other. The probability to have the outcome ``HHT'' in this order is $ p \\cdot p \\cdot q = 1/8 $.\n\nA binomial distribution arises when we describe the probability distribution of the number $ X $ which is the number of success in $ n $ Bernoulli trials (Walpole p.102). \n\n\\begin{defn}\nA random variable $ X $ follows a binomial distribution ($ X \\sim \\mathrm{Binom}(n, p) $) has the probability distribution\n \\begin{equation}\\label{def: binom}\n b(x; n, p) = \\binom{n}{x} p^x (1 - p)^{n - x}\n \\end{equation}\nfor the success probability $ p $ of $ n $ Bernoulli trials.\n\\end{defn}\n\nFor the reason why we use a word ``binomial'', see math textbooks such as Walpole p.103. This distribution has mean, variance, and standard deviation of:\n\\begin{thm}[Binomial Distribution Mean and Variance] \\label{thm: binom mean var std}\nA binomial distribution with parameters $ n $ and $ p $, it has\n \\begin{equation}\\label{eq: binom mean var std}\n \\mathrm{mean} = np \\quad;\\quad\n \\mathrm{var} = npq \\quad;\\quad\n \\mathrm{std} = \\sqrt{npq} ~.\n \\end{equation}\n\\end{thm}\n\\begin{proof}[of Binomial Dsitribution Mean and Variance]\nConsider $ O_i $ is the outcome of each Bernoulli trial, i.e., 1 if success and 0 if failure, of the $ i $-th trial. Then $ X = \\sum_{i=1}^{n} O_i $. The expected value of $ X $ is $ E(X) = E(\\sum_{i=1}^{n} O_i) = E(O_1) + \\cdots + E(O_n) $ since they are independent, and this becomes $ E(X) = np $.\n\nFor variance, we use a relationship $ \\mathrm{Var}(O_i) = E(O_i^2) - (E(O_i))^2 $. But since the value $ O_i^2 $ is identical to $ O_i $, $ E(O_i^2) = E(O_i) $, so $ \\mathrm{Var}(O_i) = p(1-p) \\equiv pq $ for $ q := 1-p $. $ X = \\sum_{i=1}^{n} O_i $ gives $ \\mathrm{Var}(X) = \\sum_{i=1}^{n} \\mathrm{Var}(O_i) = npq $.\n\\end{proof}\n\n\n\\subsection{Poisson Distribution}\nThe Poisson process is a fancy naming for some special ``counting''. In astronomy, we count photons (well, actually the CCD counts the photoelectrons) over time. This is why Poisson process is dealt with so much weight in astronomy. In social sciences we can count the death toll over time, in experiments we could count the lattice of a randomly shaped crystal over the distance. A formal definition of Poisson process is\\footnote{See, e.g., \\url{http://dept.stat.lsa.umich.edu/~ionides/620/notes/poisson_processes.pdf} and \\url{https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php}}\n\n\\begin{defn}[Poisson Process 1] \\label{def: Pois1}\nThe Poisson process $ N(t) $ for $ t \\ge 0 $ ($ t $ can be time, distance, or similar things), with rate $ \\lambda $ is defined by\n\\begin{enumerate}\n\\item $ N(0) = 0 $\n\\item $ N(t) $ has independent increment\n\\item $ N(t_2) - N(t_1) $ follows Poisson distribution of rate $ \\lambda (t_2 - t_1) $ for $ t_1 < t_2 $.\n\\end{enumerate}\n\\end{defn}\n\nThe definition of Poisson distribution is given below. It can be proven that it is identical to \\newpage\n\\begin{defn}[Poisson Process 2] \\label{def: Pois2}\nThe Poisson process $ N(t) $ for $ t \\ge 0 $ ($ t $ can be time, distance, or similar things), with rate $ \\lambda $ is defined by\n \\begin{enumerate}\n \\item $ N(0) = 0 $\n \\item $ N(t) $ has independent increment\n \\item $ \\mathbb{P}(N(h) = 1) = \\lambda h + o(h) $ for small $ h $.\n \\item $ \\mathbb{P}(N(h) \\ge 2) = o(h) $ for small $ h $.\n \\end{enumerate}\n\\end{defn}\nHere $ o(h) $ is any function such that $ \\lim_{h \\rightarrow 0} o(h) / h = 0 $. The last two items in this second definition can be re-phrased like this:\n\\begin{itemize}\n\\item [3.] For a very small interval $ h $, the probability of one single ``counting'' (Poisson process) occurs in that interval is proportional to the length of $ h $.\n\\item [4.] For a very small interval $ h $, the probability of two or more ``counting'' (Poisson process) occurs in that interval is nearly 0.\n\\end{itemize}\nThese must be independent of whether the counting happened outside of the interval, because the second item of the definition states the increment is independent.\n\n\n\\begin{defn}[Poisson Distribution] \\label{def: Pois pdf}\nA random variable $ X $ which describes a Poisson process of rate $ \\lambda $ follows the Poisson distribution ($ X \\sim \\mathrm{Pois}(\\lambda t) $) and has the following pdf\n\\begin{equation} \\label{eq: Pois pdf}\n p(x; \\lambda t) = \\frac{(\\lambda t)^{x}}{x!} e^{-\\lambda t}\n\\end{equation}\nfor non-negative integer $ x $.\n\\end{defn}\n\n\\begin{thm}[Poisson Distribution Mean and Variance] \\label{thm: Pois mean std}\nThe mean, variance, and the standard deviation of the Poisson distribution $ p(x; \\lambda t) $ are\n\\begin{equation}\\label{eq: Pois mean std}\n \\mathrm{mean} = \\lambda t\n \\sep\n \\mathrm{var} = \\lambda t\n \\sep\n \\mathrm{std} = \\sqrt{\\lambda t} ~.\n\\end{equation}\n\\end{thm}\nThe reason we use $ \\lambda t $ not just $ \\lambda $ is clear from Thm \\ref{thm: Pois mean std}: the mean changes as $ t $ changes, and we want to express $ p(x; \\mathrm{mean}) $ rather than $ p(x; \\mathrm{mean} / t) $. For instance, if the photon influx is $ \\lambda = \\SI{10}{photons / s} $, the mean of photon count is a function of time as $ \\lambda t $, so for $ t = \\SI{10}{s} $, the pdf is $ p(x; \\SI{100}{photons}) $. If the particles are distributed with mean density along the line of sight $ \\lambda = \\SI{0.1}{particle / m} $, the mean of particles along the line of sight is a function of the reaching distance $ \\lambda d $, so the pdf for $ d = \\SI{5}{m} $ is $ p(x; \\SI{0.5}{particles}) $. We, therefore, \\textbf{frequently denote a Poisson distribution of \\textit{mean} $ m $}, instead of \\textit{rate} $ \\lambda $.\n\n\\subsection{Relationships Between Distributions}\nThere are some relationships I want to emphasize because they will be used to understand statistical processes in astronomy. \n\n\\begin{thm}[Binomial to Poisson]\nFor $ X \\sim \\mathrm{Binom}(n, p) $, if (1) $ n \\rightarrow \\infty $, (2) $ p \\rightarrow 0 $ and (3) $ np \\xrightarrow{n \\rightarrow \\infty} \\mu < \\infty $, \n \\begin{equation}\\label{eq: binom to pois}\n X \\simdot \\mathrm{Pois}(x; \\mu = np)\n \\end{equation}\n\\end{thm}\n\nAlthough I did not describe the Gaussian (normal) distribution, I am assuming you are familiar with it. One thing to note is that, both binomial and Poisson distributions are discrete ones, while normal distribution is continuous one. Therefore, the approximations given below, which approximate binomial and Poisson to normal distribution, should be used with care when the discreteness is important.\n\n\\begin{thm}[Binomial to Normal] \\label{thm: binom to normal}\nFor $ X \\sim \\mathrm{Binom}(n, p) $, it has mean $ np $ and variance $ npq $ (see Thm \\ref{thm: binom mean var std}). Then, if $ n \\rightarrow \\infty $,\n \\begin{equation}\\label{eq: binom to normal}\n X \\simdot \\mathcal{N}(np, npq)\n \\end{equation}\n\\end{thm}\n\n\\begin{thm} [Poisson to Normal] \\label{thm: pois to normal}\nFor $ X \\sim \\mathrm{Pois}(\\lambda t) $, it has mean $ \\lambda t $ and variance $ \\lambda t $ (see Thm \\ref{thm: Pois mean std}). Then, if $ \\lambda t \\rightarrow \\infty $,\n \\begin{equation}\\label{eq: pois to normal}\n X \\simdot \\mathcal{N}(\\lambda t, \\lambda t)\n \\end{equation}\n\\end{thm}\nAlthough I will omit here, the proof uses Stirling's formula $ x! \\approx \\sqrt{2\\pi x} x^x e^{-x} $ and $ \\ln[ (1 + \\varepsilon)^{\\lambda t (1 + \\varepsilon) + 1 / 2 }] \\approx \\lambda t \\varepsilon + \\lambda t \\varepsilon^2 / 2 $, where $ x = \\lambda t ( 1 + \\varepsilon ) $ with $ \\lambda t \\gg 1 $ and $ \\varepsilon \\ll 1 $. Stirling's formula has relative error of less than 1 \\% when $ x = 1 $ and is almost negligible for any $ x $ of interest in many cases. Considering all the approximations in the proof, we can safely assume that Poisson distribution is quite Gaussian (normal) in many astronomical contexts, unless $ x $ is very small. In practical sense, $ \\lambda t > 100 $ is large enough, and in astronomical uses, an error, i.e., error-bar, arise from small mean ($ \\lambda t \\lesssim 50 $), is not very large compared to other approximations we use in date reduction.\n\n\n\\begin{ex}[Babies in Hospital 1]\n\\textit{(Problem provided by prof. Jae-Kwang Kim at the Dept. of Math at KAIST in 2018)}\n\nThere are two hospitals A/B, and new babies are born every day. At A/B, $ N_A = 45 $ and $ N_B = 15 $ babies are born everyday, respectively. They both recorded the number of days when $ \\ge 60\\,\\% $ of the babies have XX chromosome, say $ N'_A $ and $ N'_B $. Which will have higher value? For simplicity, assume a baby can only have either XX or XY chromosome, and the probability of being either of these is the same ($ P_X = P_Y = 0.5 $).\n\nFor a hospital, assuming the events are independent, the probability of having $ \\ge 60\\,\\% $ of XX babies is\n\\begin{equation*}\n p_{A, B} \n = \\sum_{k= \\lceil 0.6N_{A, B} \\rceil}^{N_{A, B}} \\binom{N_{A, B}}{k} P_X^k P_Y^{N_{A, B}-k}\n = 0.5^{N_{A, B}} \\sum_{k= \\lceil 0.6N_{A, B} \\rceil}^{N_{A, B}} \\binom{N_{A, B}}{k} ~,\n\\end{equation*}\nfor the case of hospitals A/B, respectively. A simple calculation gives this probability are $ p_A = 0.0899 $ ($ \\lceil 0.6N_A \\rceil = 27 $ for A) and $ p_B = 0.304 $ ($ \\lceil 0.6N_B \\rceil = 9 $ for B) for hospitals A and B, respectively. For 1 year, 365 days, we can calculate $ N'_A = 32.8 $ and $ N'_B = 111.0 $. Therefore, $ N'_B > N'_A $. \n\\end{ex}\n\nIn the problem above, we found hospital B will have more XX babies than A. But with what confidence? In principle, you can just calculate using binomial distribution, but there's a simpler workaround.\n\nDefine a random variable $ N_i $, such that it is 1 when $ \\ge 60\\,\\% $ of the babies have XX chromosome on the $ i $-th date; otherwise, 0. Then the expectation of $ N_i $ is $ E(N_i) = p_{A, B} $, for the case of hospitals A/B, respectively. Since $ N_i^2 $ must have identical value as $ N_i $, $ E(N_i^2) = p_{A, B} $. Then the variance becomes $ V(N_i) = E(N_i^2) - E^2(N_i) = p_{A, B} - p_{A, B}^2 = p_{A, B}(1 - p_{A, B}) \\equiv p_{A, B} q_{A, B} $.\n\nThe total number of days is $ N_{A, B} = \\sum_{i=1}^{365} N_i $, and if $ N_i $ are all independent, \n\\begin{equation*}\n\\begin{aligned}\n \\mu_{A, B} &= E(N_{A, B}) &&= E\\qty(\\sum_{i} N_i) = np_{A, B} \\\\\n \\sigma_{A, B} &= V(N_{A, B}) &&= V\\qty(\\sum_{i} N_i) = n \\times V(N_i) = np_{A, B}q_{A, B}\n\\end{aligned}\n\\end{equation*}\nwhere $ n = 365 $. Note that $ N $ follows binomial distribution, which is approximated as Gaussian distribution in this case (roughly speaking when $ p $ is not close to 0 or 1 and $ np $ is large enough). Therefore, $ N_{A, B} \\simdot \\mathcal{N} \\qty( \\mu_{A, B}, \\sigma_{A, B}^2 ) $, or\n\\begin{equation*}\n N_A \\simdot \\mathcal{N}(32.8,\\, 29.9) = \\mathcal{N}(32.8,\\, 5.46^2) \\quad;\\quad\n N_B \\simdot \\mathcal{N}(111.0,\\, 77.2) = \\mathcal{N}(111.0,\\, 8.79^2)\n\\end{equation*}\nSince this is a case when $ \\mu $ and $ \\sigma $ are known, we can use standard normal $ z $ statistic for confidence calculation. Take null hypothesis as $ H_0 : d = N_A - N_B = 0 $ and then $ z = \\frac{\\mu_A - \\mu_B}{\\sqrt{\\sigma_A^2 + \\sigma_B^2}} = 12.4 $. That means, we are confident with 12.4-$ \\sigma $, which is much larger than 99.999 \\%. More realistic null hypothesis is $ H_0 : d = N_A - N_B < 0 $, but this will have only small effect to the result.\n\n\n\\subsection{Usage in Astronomy}\nAs mentioned, photon counting is a key process in observational astronomy, and it is tightly bound to Poisson process. I will try to clarify this concept by the examples below.\n\n\\begin{ex}[Photon Counting and Poisson Process]\nThe photon counting process of astronomy resembles the Poisson Process. If $ N(t) $ is the number of photons during the exposure time $ t $, it is trivial that the first two items of Def \\ref{def: Pois2} is satisfied. The number of photons arriving in an infinitesimal time interval $ h = dt $ must be 0 or 1, but not larger than 1. To illustrate, if a total 10,000 photon should come during 10 sec of exposure, we can take $ h = 1\\,\\mathrm{ns} \\,(\\ll \\SI{10}{s} / 10,000 = \\SI{1}{ms}) $ to meet this condition. Thus, photon counting is a Poisson process.\n\\end{ex}\n\nIn reality, however, photon counting is done by electronic devices, such as CCD. Before going deeper into the subtlety, let me give another example dealing with photon noise:\n\n\\begin{ex}[Photon Counting and Poisson Noise] \\label{ex: poisnoise}\nWe saw the photon counting is a Poisson process; let me further assume the photoelectron counting process on CCD is, too (photoelectrons are generated by these photons when they excite bound electrons). Thus, we can use \\cref{eq: Pois pdf,eq: Pois mean std}. \n\nFor example, if we collected $ \\lambda t = 10,000 $ electrons during the exposure time $ t $, the 1-$ \\sigma $ uncertainty (1-$ \\sigma $ CI) or the standard deviation of this counting is $ \\sqrt{10,000} = 100 $, so we denote $ N = 10,000 \\pm 100 $. Since the $ \\lambda t $ is large enough, Thm \\ref{thm: pois to normal} says this is just a Gaussian distribution of mean 10,000 and standard deviation 100 (1 \\% of the mean). Since magnitude is $ (\\mathrm{const}) - 2.5 \\lg (\\mathrm{count}) $, differentiation gives a first-order estimation of the uncertainty in the magnitude $ \\Delta m = \\qty | -\\frac{2.5}{\\ln 10} \\frac{\\Delta (\\mathrm{count})}{\\mathrm{count}} | = \\m{0.011} $ (error-bar), which is small enough for some scientific purposes. \n\nNote that the term $ \\frac{\\Delta (\\mathrm{count})}{\\mathrm{count}} = \\frac{1}{\\sqrt{\\mathrm{count}}} $ is inversely proportional to the square root of count itself. The more we collect the count, i.e., the longer the exposure time, we will get smaller error-bar.\n\\end{ex}\n\nNow a tricky part comes in:\n\\begin{ex}[Photon Counting and Poisson Noise in ADU] \\label{ex: poisnoise adu}\nIn CCD, the number of photon ($ N_\\gamma $) will generate certain number of photoelectrons ($ N_e $), and this is measured by the electric potential (roughly speaking, $ \\mathrm{potential} = V_e \\propto N_\\mathrm{e} e $ where $ e $ is the electron charge). This voltage will be translated into integer numbers by electric circuit, and this integer is said to have a unit ADU (analog-to-digital unit) or DN (data number). In summary, a conversion of this order happens: $ N_\\gamma \\rightarrow N_e \\rightarrow V_e \\rightarrow N_\\mathrm{ADU} $. We define the (electron) gain $ g \\,[\\mathrm{electrons/ADU}] $ such that $ N_e = g N_\\mathrm{ADU} $. \n\nIf $ g = \\SI{2}{e/ADU} $ and $ N_e = 10,000 $ as in the previous example, $ N_\\mathrm{ADU} = 5,000 $. If $ N_\\mathrm{ADU} $, i.e., the ADU counting is a Poisson process, we must hace $ N_\\mathrm{ADU} = 5000 \\pm 71 $, which gives magnitude error $ \\Delta m_\\mathrm{ADU} = \\qty | -\\frac{2.5}{\\ln 10} \\frac{71}{5000} | = \\m{0.015} $. The error-bar has changed! What has just happened?\n\nThe key here is that we only checked the photon counting process is a Poisson process. Rigorously speaking, photoelectron counting and ADU counting are \\textbf{not Poisson processes}.\n\\end{ex}\n\nThen what process it is for photoelectron counting?\n\\begin{ex}[Photoelectron Follows a Binomial Distribution]\\label{ex: photoelectron binom}\nAny device which collect photon does not actually collect photon, but they output \\textit{something} which is proportional to the number of photons incident to it. In CCD, the output is the number of electrons which is measured by electric potential. The key here is the \\emph{conversion of photon into \\emph{something} must be a probabilistic process}. In CCD, the conversion happens with a paramter called the quantum efficiency (QE). Therefore, for a fixed number of photon, $ N_\\gamma $, which follows a Poisson distribution $ N_\\gamma \\sim \\mathrm{Pois}(N_\\gamma) $, the outcome, $ N_e $, will follow a binomial distribution $ N_e \\sim \\mathrm{Binom}(N_\\gamma, QE) $. Verbally put, ``photoelectron will follow a binomial distribution with the paramter $ n $ which follows a Poisson distribution and the parameter $ p $ which is believed to be fixed thanks to engineering process''.\n\n% https://www.facebook.com/groups/astro.r/permalink/2836140943148559/\nAlthough I will not give a proof, $ N_e \\sim \\mathrm{Binom}(n=\\mathrm{Pois}(N_\\gamma), p=QE) $ is approximated as $ N_e \\simdot \\mathrm{Pois}(N_\\gamma QE) $. In observation, since $ N_e \\approx N_\\gamma QE \\gg 1 $, $ N_e \\simdot \\mathcal{N}(N_e, N_e) $ (Thm \\ref{thm: pois to normal}).\n\\end{ex}\n\nHow about the ADU count?\n\n\\begin{ex}[ADU is Approximately Gaussian]\nThe ADU counting is a very strange thing: For electron gain $ g $, ADU increase by 1 for every $ g $ electrons, i.e., a step function which is very difficult to deal with in analytical sense. Regardless of its final distribution, it is clear that this conversion to the original $ N_e $ will destroy the original distribution. Fortunately, the count $ N_e $ or $ N_\\mathrm{ADU} $ is large enough in most observations, so that the discreteness of conversion is ignorable. Hence, we can treat them as if a smooth continuous convertable variables by $ N_e = g N_\\mathrm{ADU} $. Then \n\\begin{equation}\\label{eq: adu distribution}\n N_\\mathrm{ADU} \\simdot \\mathcal{N}(N_e/g, N_e/g^2) \\quad\\mathrm{or}\\quad\n N_\\mathrm{ADU} \\simdot \\mathcal{N}(N_\\mathrm{ADU}, N_\\mathrm{ADU}/g)\n\\end{equation}\nThat means $ N_\\mathrm{ADU} $ follows a Gaussian, but not like usual Poisson distribution (because ADU counting is not a Poisson process). Going back to Ex \\ref{ex: poisnoise adu}, we have to divide the error-bar of the count by the gain, so $ \\Delta m_\\mathrm{ADU} = \\qty | -\\frac{2.5}{\\ln 10} \\frac{\\sqrt{5000/g}}{5000} | = \\m{0.011} $, which is identical result as Ex \\ref{ex: poisnoise}.\n\\end{ex}\nFor a mathematical treatment of the ADU conversion (step function), you may find some research papers useful\\footnote{See, e.g., Merline W. J. \\& Howell S. B. 1995, Exp. Astron. \\textbf{6}, \\textit{163}, ``A REALISTIC MODEL FOR POINT-SOURCES IMAGED ON ARRAY DETECTORS: THE MODEL AND INITIAL RESULTS''.}.\n\n\n\\section{Error Propagation}\nWhen we have to calculate the uncertainty of a parameter derived from other parameters that have their own uncertainties, what we usually use is the \\textit{error propagation}. For example, $ \\mathrm{g'} = 14.0 \\pm 0.1 $ and $ \\mathrm{r'} = 15.0 \\pm 0.1 $, what is the uncertainty of the $ \\mathrm{g' - r'} $ color? Many astronomers will say it is $ -1.0 \\pm 0.14 $, where the error here is $ \\sqrt{\\sigma_\\mathrm{g'}^2 + \\sigma_\\mathrm{r'}^2} $ ($ \\sigma $ is the error-bar). If the temperature is $ T = 100 \\pm 1 \\si{K} $, what is the fractional uncertainty of bolometric luminosity ($ L = \\sigma_\\mathrm{SB} T^4 $)? Now people will say $ \\Delta L / L \\approx \\frac{4 \\sigma_\\mathrm{SB} T^3 \\Delta T }{\\sigma_\\mathrm{SB} T^4} = 4 \\Delta T / T = 4 \\% $. \n\nThe way how we do this is that, for a function $ f = f(x| a, b, c) $ where $ (a, b, c) $ is a set of parameters:\n\\begin{equation}\\label{eq: error propagation}\n (\\Delta f)^2 \\approx \n \\qty(\\frac{\\partial f}{\\partial a} \\Delta a)^2\n + \\qty(\\frac{\\partial f}{\\partial b} \\Delta b)^2\n + \\qty(\\frac{\\partial f}{\\partial c} \\Delta c)^2 ~.\n\\end{equation}\nHere, $ (\\Delta f)^2 $ is the variance of the function $ f $, $ \\Delta X $ is the standard error (such that $ X \\pm \\Delta X $ is the 1-$ \\sigma $ CI for $ X $), and we frequently use $ \\Delta f $, the standard error of $ f $, as \\emph{the error-bar}. \nIf any parameters are dependent, we need a covariance term $ \\sigma_{ab} $ for instance, and this is one of the reasons why the above error-propagation is only an approximation. This formula itself is similar to getting the gradient. In different notation, e.g., sum of pixel values, $ I := \\sum_{i} I_i $, which is reasonable to assume $ I_i $'s are independent, the identical formula becomes\n\\begin{equation}\\label{eq: error propagation 2}\n (\\Delta I)^2 \n \\equiv \\left ( \\Delta \\left ( \\sum I_i\\right ) \\right )^2 \n = \\sum_{i} (\\Delta I_i)^2 ~.\n\\end{equation}\n\nNote that this is only an approximation, and in reality, we need more complicated calculation. Also it is impossible to know how accurate this approximation is, before you conduct detailed calculation. The reason astronomers use it in spite of the pitfall is because most cases we are only interested in the \\emph{rough} estimation of the error-bar. \n\n\n\n\n\\section{The Chi-Square Minimization}\nYou may have heard of it, or simply the \\textit{least square} something. Least square or the chi-square minimization is a process to find the set(s) of model parameters which has the biggest (or bigger than threshold) \\textit{likelihood} in Bayesian statistics. The set of ``best'' model parameters is the one with the largest likelihood, so we call it the maximum likelihood estimator, \\textbf{MLE}.\n\nConsider independent and identically distributed (\\textit{i.i.d.}) random variables\\footnote{Random variables $ X_{i} $ for $ i = 0, \\cdots, N $ are called i.i.d. if they (1) follow identical probability distribution, say $ f(x) $, and (2) $ f(X = x) = f(X_1 = x_1) \\times \\cdots \\times f(X_N = x_N) $, where $ x = (x_1, \\cdots , x_N) $ (same for $ X $). }, e.g., flux as a function of $ \\lambda $. It is independent because the measurement at each wavelength itself is independent from that of any other wavelength. It is identically distributed because and we will assume each error-bar is Gaussian. The second assumption is not necessarily correct, and actually that happens many times (e.g., flux is Gaussian distributed but magnitude is not, because it is log of Gaussian distribution).\n\nWe will study Bayes' theorem later, but let me introduce it first to grap the meaning of chi-square statistic:\n\\begin{align*}\n P(\\mathbf{\\theta}|D, I) &= \\frac{P(D|\\mathbf{\\theta}, I) P(\\mathbf{\\theta}|I)}{P(D, I)} \\\\\n P(\\theta|D, I) &\\propto P(D | \\theta, I) P(\\theta|I)\n\\end{align*}\nWhen we have data points, $ (x_i, y_i) $, which are all independent, and $ y \\sim N(y_i, \\sigma_i^2) $. Then one of the most na\\\"{i}ve goodness-of-fit statistic variables can be the square-sumed error:\n\\begin{equation}\\label{eq: sse}\n \\mathrm{SSE} = \\sum_{i} (y_i - f(x_i| \\theta))^2\n\\end{equation}\nSince $ y_i $ are all Gaussian, the probability of obtaining such data is obtained from the multiplication law:\n\\begin{equation*}\n P(D = \\{y_i\\}) = \\prod_i A e^{-(f(x_i|\\theta)-y_i)^2/2\\sigma_i^2} \n\\end{equation*}\nor taking log:\n\\begin{equation}\n \\ln P(D = \\{y_i\\}) = \\ln A - \\sum_i \\frac{ (f(x_i|\\theta) - y_i)^2}{2\\sigma_i^2}\n\\end{equation}\nNow getting parameters which maximize $ P $ is identical to (1) getting them which maximize $ \\ln P $, and this is also identical to (2) getting them which mininize the summation in the exponent. Therefore, the exponent, which is chi-square statistic:\n\\begin{equation}\\label{eq: chi-square statistic}\n \\chi^2 = \\sum_i \\frac{(y_i - f(x_i | \\theta))^2}{\\sigma_i^2} ~,\n\\end{equation}\nshould be minimized to get the most probable answer to the fitting problem. If the error-bars of all data points are identical, $ \\sigma_i $ is nothing but a constant, so minimizing chi-square is identical to minimizing SSE.\n\n\n\n\\section{Interpolation}\nWhen we fit an analytic function to the data, we usually use a statistic (e.g., the chi-square value or the BIC) that indicates the goodness-of-fit (also called figure of merit). But what if we don't know or not interested in the true analytic function behind the scene? For example, for aperture trace or sense function in spectroscopy, we have absolutely no idea what is the true analytic function. In this situation, \\textbf{interpolation} comes in. \n\n\\begin{defn}[Interpolation]\nAn interpolation function is a function that smoothly connects \\textit{all} the observed data points.\n\\end{defn}\n\nFor instance, a linear interpolation is a function which connects all the data points with linear segments as in \\cref{fig:interpolation}. In cubic spline, you fit a 3rd order polynomial function (total 4 unknowns). For points $ (x_i, y_i) $ where $ i = 0 \\ldots n $, we want to obtain $ n $ spline curves $ S_i(x) $ for $ i = 0 \\ldots n-1 $. All such interpolation functions should be \\textit{smoothly connected} to each other by\n\\begin{itemize}\n \\item $ S_i (x_i ) = y_i \\quad(i = 0 , \\ldots , n-1). $\n \\item $ S_{i-1} (x_i ) = y_i \\quad(i = 1 , \\ldots , n). $\n \\item $ S^\\prime_i (x_i) = S^\\prime_{i-1} (x_i)\\quad(i = 1 , \\ldots , n-1). $\n \\item $ S^{\\prime\\prime}_i (x_i) = S^{\\prime\\prime}_{i-1} (x_i) \\quad(i = 1 , \\ldots , n-1). $\n \\item $ S^{\\prime\\prime}_0 (x_0) = S^{\\prime\\prime}_{n-1} (x_n) =0 $.\n\\end{itemize}\n\n\\begin{figure}\n \\centering\n \\includegraphics[width=0.7\\linewidth]{figs/interpolation}\n \\caption{A display of linear/cubic spline interpolations and spline fittings.}\n \\label{fig:interpolation}\n\\end{figure}\n\nIn python, you can use\n\\begin{python}\nfrom scipy.interpolate import UnivariateSpline\n# linear spline interpolation\ninterp_1 = UnivariateSpline(x, y, s=0, k=1)\n# cubic spline interpolation\ninterp_3 = UnivariateSpline(x, y, s=0, k=3)\n\n# use it as y_interp_1 = interp_1(x_values)\n\\end{python}\nThe parameter \\texttt{s} is the smoothing factor. That is, \n\\begin{equation}\n \\sum_{i=1}^{n} w_i^2 (y_i - S(x_i))^2 \\le s ~.\n\\end{equation}\nIf no error-bar is present, we can just ignore $ w_i $ (weights), and if there exists $ \\sigma_i $, we can use $ w_i = 1/\\sigma_i $, and the above equation becomes nothing but a chi-square. In that case, the $ \\chi^2 $ smaller than the total degrees of freedom (roughly the number of data points) is a good measure, so \\pyth{s = len(x)} is a good guess for spline fitting. If you put $ s = 0 $, it means $ S(x_i) $ must be the same as $ y_i $, i.e., the fitted function must go through all the data points: this is the interpolation. \n\nFrom the figure, cubic spline interpolation is more smooth and the linear one looks too discrete, so you may always want to use cubic version. A caveat is that cubic spline has large fluctuation when there is any large scatter, as seen from $ x = 5 $ and $ x = 6 $. This ``shooting'' effect is severe when you fit a sense function (spectroscopic flux calibration) to a spectroscopic standard star: if the star has an absorption line, the cubic fit will be problematic near that wavelength region. \n\n\n\\section{Change of Variables}\nIn a formal error-propagation calculation, the change of variables is very important, because sometimes the physical paramters which describe the function you are interested in may be different from the observables which you obtain from expeirments or observations. We actually do the change of variables without much mathematical instroduction. An example is when $ m = -2.5 \\lg I $, we differentiate it to get $ \\Delta m = \\frac{2.5}{\\ln 10} \\frac{\\Delta I}{I} \\approx 1.086 \\frac{\\Delta I}{I}$ (as in Ex \\ref{ex: poisnoise}). This way, we changed the error of the variable $ I $ to that of $ m $. Formally, if a conversion happens from parameters (or random variables) $ X_1,\\,\\cdots,\\, X_M $ to $ Y_1,\\,\\cdots,\\, Y_M $, we use Jacobian\n\\begin{equation}\\label{eq: change_of_var}\n\\begin{aligned}\n f(Y_1,\\,\\cdots,\\, Y_M) \\delta Y_1 \\cdots \\delta Y_M \n &= g(X_1,\\,\\cdots,\\, X_M) \\delta X_1 \\cdots \\delta X_M \n \\\\\n f(Y_1,\\,\\cdots,\\, Y_M) \n &= \\qty| \\pdv{(Y_1,\\,\\cdots,\\, Y_M)}{(X_1,\\,\\cdots,\\, X_M)} | g(X_1,\\,\\cdots,\\, X_M)\n\\end{aligned}\n\\end{equation}\nwhere, e.g., \n\\begin{equation}\\label{eq: jacobian}\n \\qty| \\pdv{(Y_1,\\, Y_2)}{(X_1,\\, X_2)} | :=\n \\mqty| \\pdv{Y_1}{X_1} & \\pdv{Y_1}{X_2} \\\\ \\pdv{Y_2}{X_1} & \\pdv{Y_2}{X_2} | ~.\n\\end{equation}\n\n\\begin{ex}[Cauchy distribution: Lighthouse Problem]\\label{ex: lighthouse}\n\\textit{(This is taken from Sivia (\\S 2.4), which is recited from GullSF 1988, which also adopted it from ``Cambridge Part 1A examples sheet''.)}\n\nConsider a lighthouse at $ (x, y) = (\\alpha, \\beta) $. The lighthous emits an infinitely thin lightray to a uniformly random direction $ \\theta $ (angle counterclockwise from $ -y $ direction), as in \\cref{fig: lighthouse}. What is the probability distribution of $ x_k $ for the $ k $-th position on the shore which receives the light?\n\nSicne $ \\theta $ is uniformly distributed, the pdf is $ f(\\theta_k | \\alpha, \\beta) = 1/\\pi $ (uniform over $ \\pm \\pi/2 $; You may use $ 1/2\\pi $ considering uniform over $ \\pm \\pi $, but that will change the result only by up to a multiplication of a constant, 2). Since $ \\tan \\theta_k = (x_k - \\alpha)/\\beta $, the change of variable gives\n\\begin{equation*}\n g(x_k | \\alpha, \\beta) = \\pdv{\\theta_k}{x_k} f(\\theta_k | \\alpha, \\beta)\n = \\frac{1}{\\pi} \\frac{\\beta}{\\beta^2 + (x_k - \\alpha)^2} ~.\n\\end{equation*}\nThis is the Cauchy distribution with the ``$ \\gamma $'' parameter equals $ \\beta $. \n\nMathematically, Cauchy distribution does not have a mean (simple proof can be $ \\lim_{a\\rightarrow \\infty} \\int_{-a}^{+a} g(x) dx = 0$ is not equal to $ \\lim_{a\\rightarrow \\infty} \\int_{-2a}^{+a} g(x) dx = \\infty $). Also the standard deviation cannot be defined since the mean of $ x^2 $ also diverges. If $ \\mu $ and $ \\sigma $ are undefined, the CLT in the form of Thm \\ref{thm: clt} cannot be used.\n\\end{ex}\n\n\n\\begin{figure}\n\\centering\n\n\n\\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt \n\n\\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]\n%uncomment if require: \\path (0,300); %set diagram left start at 0, and has height of 300\n\n%Shape: Axis 2D [id:dp9054512614951153] \n\\draw (29,194.2) -- (420.5,194.2)(68.15,34) -- (68.15,212) (413.5,189.2) -- (420.5,194.2) -- (413.5,199.2) (63.15,41) -- (68.15,34) -- (73.15,41) ;\n%Shape: Star [id:dp4847863660698495] \n\\draw (192.25,81) -- (195.85,88.15) -- (203.9,89.29) -- (198.08,94.85) -- (199.45,102.71) -- (192.25,99) -- (185.05,102.71) -- (186.42,94.85) -- (180.6,89.29) -- (188.65,88.15) -- cycle ;\n%Straight Lines [id:da9248807400548928] \n\\draw [dash pattern={on 0.84pt off 2.51pt}] (192.25,93) -- (192.25,195) ;\n%Straight Lines [id:da9932130816919462] \n\\draw [dash pattern={on 0.84pt off 2.51pt}] (192.25,93) -- (67.5,93) ;\n%Straight Lines [id:da5299276212835915] \n\\draw (192.25,93) -- (289.11,192.57) ;\n\\draw [shift={(290.5,194)}, rotate = 225.79] [color={rgb, 255:red, 0; green, 0; blue, 0 } ][line width=0.75] (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29) ;\n\n% Text Node\n\\draw (417,199.4) node [anchor=north west][inner sep=0.75pt] {$x$};\n% Text Node\n\\draw (51,21.4) node [anchor=north west][inner sep=0.75pt] {$y$};\n% Text Node\n\\draw (187,198.4) node [anchor=north west][inner sep=0.75pt] {$\\alpha $};\n% Text Node\n\\draw (52,82.4) node [anchor=north west][inner sep=0.75pt] {$\\beta $};\n% Text Node\n\\draw (194,111.4) node [anchor=north west][inner sep=0.75pt] {$\\theta _{k}$};\n% Text Node\n\\draw (282,198.4) node [anchor=north west][inner sep=0.75pt] {$x_{k}$};\n% Text Node\n\\draw (356,180) node [anchor=north west][inner sep=0.75pt] [align=left] {sea};\n% Text Node\n\\draw (356,196) node [anchor=north west][inner sep=0.75pt] [align=left] {land};\n\n\n\\end{tikzpicture}\n\n\\label{fig: lighthouse}\n\\caption{A schematic diagram for the lighthouse problem.}\n\\end{figure}\n", "meta": {"hexsha": "0e81a27b909665dc87d8437f0218aff41b133c60", "size": 62501, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Books/chaps/01_stats.tex", "max_stars_repo_name": "ysBach/SNU_AOclass", "max_stars_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-03-23T06:14:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-14T01:49:51.000Z", "max_issues_repo_path": "Books/chaps/01_stats.tex", "max_issues_repo_name": "ysBach/SNU_AOclass", "max_issues_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2020-05-04T17:21:49.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-24T11:41:55.000Z", "max_forks_repo_path": "Books/chaps/01_stats.tex", "max_forks_repo_name": "ysBach/SNU_AOclass", "max_forks_repo_head_hexsha": "e2e364b08c2e6e129c267db9cbd76cfd0ab77527", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 5, "max_forks_repo_forks_event_min_datetime": "2020-05-10T14:19:34.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-14T09:18:08.000Z", "avg_line_length": 84.8046132972, "max_line_length": 924, "alphanum_fraction": 0.6893969696, "num_tokens": 19724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506418255928, "lm_q2_score": 0.8080672135527632, "lm_q1q2_score": 0.6146568446271278}} {"text": "\\section{Solving Equations}\r\n\\noindent\r\nNow that we can take the Laplace transform and its inverse of most of the common functions we've seen when solving differential equations, we can use Laplace transforms as another tool to solve differential equations.\\\\\r\n\r\n\\noindent\r\nThere is a general pattern to how we'll use Laplace transforms to solve differential equations.\r\n\\begin{enumerate}[label=\\arabic*)]\r\n\t\\item If needed, arrange the equation into a convenient form in the $t$ domain.\r\n\t\\item Take the Laplace transform of both sides.\r\n\t\\item Use algebra, especially partial fraction decomposition and completing the square, to arrange items into a convenient form in the $s$ domain.\r\n\t\\item Take the inverse Laplace transform of both sides and solve.\r\n\\end{enumerate}\r\n\r\n% Examples\r\n\\input{./laplaceTransforms/solvingEquations/solvingEquations_examples.tex}", "meta": {"hexsha": "84912f0580c5b4e84419a351b7ab9bca48f9e93f", "size": 861, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/laplaceTransforms/solvingEquations/solvingEquations.tex", "max_stars_repo_name": "aneziac/Math-Summaries", "max_stars_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2020-03-26T06:20:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T17:38:45.000Z", "max_issues_repo_path": "diffEq/laplaceTransforms/solvingEquations/solvingEquations.tex", "max_issues_repo_name": "aneziac/Math-Summaries", "max_issues_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 26, "max_issues_repo_issues_event_min_datetime": "2020-03-28T17:44:18.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T04:47:03.000Z", "max_forks_repo_path": "diffEq/laplaceTransforms/solvingEquations/solvingEquations.tex", "max_forks_repo_name": "aneziac/Math-Summaries", "max_forks_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2020-04-10T05:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-17T15:21:12.000Z", "avg_line_length": 57.4, "max_line_length": 220, "alphanum_fraction": 0.7897793264, "num_tokens": 190, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8080671950640463, "lm_q2_score": 0.760650658103136, "lm_q1q2_score": 0.614656843717022}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\n\\marginpar{Monday\\\\ 2020-11-9, \\\\ compiled \\\\ \\today}\n\nWe have a parameter vector \\(\\vec{x}\\), and in a step in our MCMC we want to move to a new parameter \\(\\vec{y}\\). \nWe start off with a transition proposal distributed according to the distribution \\(Q\\), which can depend on \\(\\vec{x}\\).\nThis \\(Q\\) may, for example, be a simple Gaussian. \n\nThen, we have an acceptance/rejection step: we accept the step with a probability \n%\n\\begin{align}\n\\alpha (y | x) = \\min \\qty(\\frac{p(\\vec{y}) Q(\\vec{x} | \\vec{y})}{p(\\vec{x}) Q(\\vec{y} | \\vec{x})}, 1)\n\\,,\n\\end{align}\n%\nwhere \\(p\\) is the posterior.\nIf we are using a Gaussian \\(Q\\), it is symmetric, therefore we only have \\(\\alpha = \\min (p(\\vec{y}) / p (\\vec{x}), 1)\\). \nIf the posterior is higher in \\(\\vec{y}\\) than in \\(\\vec{x}\\) we always accept the step, else we accept it sometimes. \n\nWe need to check whether the MC is ergodic: the transition probability reads \n%\n\\begin{align}\nT(y | x) &= \\alpha Q = \n\\min \\qty(\\frac{p(\\vec{y}) Q(\\vec{x} | \\vec{y})}{p(\\vec{x}) Q(\\vec{y} | \\vec{x})}, 1)\nQ(\\vec{y} | \\vec{x}) \\\\\n&= \\frac{1}{p(\\vec{x}) Q(\\vec{y} |\\vec{x})} \\min \\qty[p(\\vec{y} ) Q(\\vec{x} | \\vec{y}), p(\\vec{x}) Q(\\vec{y} | \\vec{x})]\nQ(\\vec{y} | \\vec{x}) \\\\\n&= \\frac{1}{p(\\vec{x})} \\min \\qty[p(\\vec{y} ) Q(\\vec{x} | \\vec{y}), p(\\vec{x}) Q(\\vec{y} | \\vec{x})]\n\\,,\n\\end{align}\n%\ntherefore we have \n%\n\\begin{align}\nT(\\vec{y} | \\vec{x}) p(\\vec{x}) = \\min \\qty[p(\\vec{y} ) Q(\\vec{x} | \\vec{y}), p(\\vec{x}) Q(\\vec{y} | \\vec{x})]\n= T(\\vec{x} | \\vec{y}) p(\\vec{y})\n\\,,\n\\end{align}\n%\nwhich is precisely the \\textbf{detailed balance} condition, with the stationary distribution being precisely \\(p\\), the posterior. \n\nSince we only need to compute ratios of the posterior, we never need to compute the evidence. \n\n\\subsection{Gibbs sampling}\n\nWe start from a parameter vector \\(\\vec{x}\\), and we only update the \\(i\\)-th parameter, keeping the others fixed, according to the conditional probability for that parameter given all the others: \\(\\mathbb{P}(\\theta _i | \\theta_1 , \\hat{\\theta}_i, \\theta _n)\\). \nHere we always accept the step. \n\nWe update each parameter separately (using the old values), and then move the full parameter vector together. \nIt is a special case of Metropolis-Hastings with \\(Q\\) being the conditional probability and acceptance probability equal to 1. \nWe might want to use it if we have easy access to the conditional distribution. \n\nThe issue is always convergence: we have convergence theorems for \\(N \\to \\infty \\), but we need \\emph{fast} convergence practically. \n\nWe can compute averages and standard deviations as \n%\n\\begin{align}\n\\expval{\\theta _i} &\\approx \\frac{1}{N} \\sum _{k=1}^{N} \\theta^{(k)}_i \\\\\n\\sigma^2_{\\theta _i} &\\approx \\frac{1}{N} \\sum _{k=1}^{N} \\abs{\\theta^{(k)}_i}^2 - \\expval{\\theta _i}^2\n\\,.\n\\end{align}\n\nThis is not done over \\emph{all} the points, but only over the ones calculated after convergence. \n\nMarginalization is done as such: \n%\n\\begin{align}\np(\\theta_2) = \\int \\dd{\\theta _1} p(\\theta_1 , \\theta _2 ) \n= \\int \\dd{\\theta _1} p(\\theta_1 | \\theta _2) p(\\theta_1 )\n\\,.\n\\end{align}\n\nWe can simply build a marginal histogram by neglecting a coordinate and then binning. \n\n\\subsection{Convergence}\n\nHow do we verify convergence? This is the complicated part.\nWe can do a trace plot: \n%\n\\begin{align}\n\\Tr[ - \\log \\mathcal{P}] = - \\sum _{k=1}^{M} \\frac{1}{k} \\log \\mathcal{P}(\\theta^{(k)}_i | \\vec{d})\n\\,,\n\\end{align}\n%\nwhich should converge to a fixed maximum value after some time. \nHowever, we might not have covered a representative region of the full posterior. \n\nA standard technique is to start \\(M\\) different chains, and for each compute the average separately. \n\nWe can compute the average of each chain, \\(\\overline{\\theta}^{J}\\) (where \\(J = 1 \\dots M\\)); and the total average \\(\\overline{\\theta}\\) over all the chains. \nWe can also estimate the variance of each chain, \\(w^{J}\\): \n%\n\\begin{align}\nw^{J} = \\frac{1}{n-1} \\sum_{i=1}^{n} \\qty(\\theta^{J}_{i} - \\overline{\\theta}^{J})^2\n\\,,\n\\end{align}\n%\nwhere the \\(n-1\\) is the Bessel correction, needed because we use the estimated average instead of the true average. \n\nThrough manipulation, we can show that \n%\n\\begin{align}\nw^{J} &= \\frac{n}{n-1} s_J - \\frac{n}{n-1} \\var{\\overline{\\theta}^{J}}\n\\,,\n\\end{align}\n%\nwhere \\(\\var{\\overline{\\theta}^J} = (\\overline{\\theta}^{J} - \\mu )^2\\) is an estimator for the variance of the \\emph{mean} of the parameter vector, while \\(s_J\\) is the estimator of the true variance of the chain. \nFlipping the expression we can find \n%\n\\begin{align}\ns_J = \\frac{n-1}{n} w^{J} + \\var{\\overline{\\theta}^{J}}\n\\,.\n\\end{align}\n\nWe can estimate the variance of the chain mean as \n%\n\\begin{align}\n\\widehat{\\var{\\overline{\\theta}}} = \\frac{1}{M-1} \\sum _{J=1}^{M} (\\overline{\\theta}^{J}- \\overline{\\theta})^2 = \\frac{B}{n}\n\\,,\n\\end{align}\n%\nand then we can average the single chain variance estimators: \n%\n\\begin{align}\nW = \\frac{1}{M} \\sum _{J=1}^{M} w^{J}\n\\,,\n\\end{align}\n%\nso then we get the estimator \n%\n\\begin{align}\n\\hat{V} = \\frac{n-1}{n} W + \\frac{B}{n}\n\\,,\n\\end{align}\n%\nand then we can define \\(\\hat{R} = \\hat{V} / W\\): we expect \\(B/n\\) to become irrelevant asymptotically. \nIf \\(\\hat{R}\\) is of order 1 then we can say that we have converged. \nTypically, we say that when \\(\\hat{R} \\lesssim \\num{1.02}\\) we have converged. \n\n\\todo[inline]{Add more calculations}\n\nThese tests are not fool-proof: we try to falsify the hypothesis of convergence, and at a certain point we stop trying to do so. \n\n\\end{document}\n", "meta": {"hexsha": "b6e3d7cecfcaa155da1c51ae9f836a925da162f4", "size": 5569, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ap_third_semester/astrostatistics_cosmology/nov09.tex", "max_stars_repo_name": "jacopok/notes", "max_stars_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-10-10T13:10:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T14:52:50.000Z", "max_issues_repo_path": "ap_third_semester/astrostatistics_cosmology/nov09.tex", "max_issues_repo_name": "jacopok/notes", "max_issues_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ap_third_semester/astrostatistics_cosmology/nov09.tex", "max_forks_repo_name": "jacopok/notes", "max_forks_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-03T16:20:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-06T16:11:07.000Z", "avg_line_length": 38.1438356164, "max_line_length": 263, "alphanum_fraction": 0.6581073801, "num_tokens": 1864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.7853085859124002, "lm_q1q2_score": 0.6146315645932251}} {"text": "% $Id$\n%\n% Author: David Fournier\n% Copyright (c) 2008 Regents of the University of California\n%\n\nThe Vector Autoregressive Moving Average Garch process combines\nthe Vector Autoregressive Moving Average (\\textsc{varma})\nprocesses with the Vector Garch (generalized \nautoregressive conditional heteroscedastic) processes.\n\n\n\\section{Formulation of the \\textsc{varma garch} process}\n\nThe \\textsc{varma garch} process of type $(p,q,r,s)$ is given by a \nseries $Y_t$ for $t=-p+1,\\ldots,n$, where for each value of $t$,\n$Y_t$ is an $m$-dimensional vector.\nFor $t>0$, the $Y_t$ are assumed to satisfy a relationship of the form\n\\begin{equation}\n Y_t=\\mu +\\sum_{l=1}^p A_l (Y_{t-l}-\\mu)\n +\\sum_{l=0}^q B_l \\epsilon_{t-l} \n \\label{eq1}\n\\end{equation}\nwhere the $\\mu$ is an $m$-dimensional vector, \n$A_l$ and $B_l$ are $m\\times m$ matrices, $B_0$ is the identity\nmatrix, and the\n$\\epsilon_t$ are multivariate (normal) random vectors,\nwith means~$0$. Covariance matrices $\\Sigma_t$ and\n $\\hbox{\\rm E}(\\epsilon_t\\epsilon_{t^\\prime})=0$ if $t\\ne t^\\prime$.\nLet\n\\begin{equation}\n r_t=Y_t-\\mu -\\sum_{l=1}^p A_l (Y_{t-l}-\\mu)\n \\label{eqc1}\n\\end{equation}\nbe the vector of model residuals or ``shocks.''\\footnote{For the moving\naverage model ($q>0$), one might argue that since the previous values\nof $r_t$ have been observed, the shock part of $r_t$, say,\n$d_t$, is given by $d_t=r_t-\\sum_{l=1}^q B_l r_{t-l}$,\nbut this has not been done at present.}\nThese residuals are\nassumed to contribute to the covariance matrix in the next time period.\nThe $\\Sigma_t$ evolve according to one of several relationships.\n\n\\subsubsection{The \\textsc{dvec} relationship}\n\\begin{equation}\n \\Sigma_t=\\Omega+\n \\sum_{l=1}^r F_l\\bigotimes r_{t-l}r_{t-l}^\\prime \n +\\sum_{l=1}^sG_l\\bigotimes \\Sigma_{t-l}\n \\label{eqc}\n\\end{equation}\nIn equation~(\\ref{eqc}), the matrices \n$F_l$ and $G_l$ are symmetric and the operator\n`$\\bigotimes$' denotes the element-wise product of matrices.\nThis parameterization does not restrict the resulting matrix to\nbe positive definite, so some care is necessary to ensure the\nstability of the resulting~model.\n\n\\subsubsection{The \\textsc{bekk} relationship}\n\\begin{equation}\n\\Sigma_t=\\Omega+\n \\sum_{l=1}^r F_l\\, r_{t-l}\\, r_{t-l}^\\prime\\, F_l^\\prime \n + \\sum_{l=1}^s\\, G_l\\, \\Sigma_{t-l}\\, G_l^\\prime\n \\label{eq2}\n\\end{equation}\n\n\\subsubsection{\\textsc{dveci} and \\textsc{bekkai} parameterizations}\nThe basic \\textsc{dvec} and \\textsc{bekk} parameterizations can be extended by modifying\ncomponents of the $r_t$ to reflect the asymmetric response\nto positive and negative values.\n\\begin{align}\n \\nonumber \\eta_{ij} =&\\epsilon_{ij}/\\alpha_j &\\textrm{if } \\epsilon_{ij} \\ge 0\\\\\n \\eta_{ij} =&\\epsilon_{ij}\\alpha_j &\\textrm{if } \\epsilon_{ij} < 0\n\\end{align}\nThis modified form will be referred to as the \\textsc{dveci} and \\textsc{bekkai} parameterizations.\n\\begin{equation}\n\\Sigma_t=\\Omega+\n \\sum_{l=1}^r F_t\\bigotimes \\eta_{t-l}\\, \\eta_{t-l}^\\prime \n +\\sum_{l=1}^sG_l\\bigotimes \\Sigma_{t-l}\n \\label{eqd}\n\\end{equation}\n\\begin{equation}\n \\Sigma_t=\\Omega+\n \\sum_{l=1}^r F_l\\, \\eta_{t-l}\\, \\eta_{t-l}^\\prime\\, F_l^\\prime \n + \\sum_{l=1}^sG_l\\, \\Sigma_{t-l}\\, G_l^\\prime\n \\label{eqe}\n\\end{equation}\n\n\n\\section{Setting a value for $\\Sigma_1$}\n\nThe value for the parameters in $\\Sigma_1$ are often poorly determined \nand simply letting them be free parameters can lead to \ninstability and initial\ntransient effects in the model. To stabilize the parameterization, we have \ncalculated $\\Sigma_1$ through $\\Sigma_{\\max\\{r,s\\}}$ from the condition\n\\begin{equation}\n \\widehat\\Omega=\\Sigma+\\sum_{l=1}^q B_l\\Sigma B_l^\\prime \n\\end{equation}\nwhere \n\\begin{equation}\n \\widehat\\Omega=\\frac{1}{n}\\sum_{t=1}^n\\hat\\epsilon_t\\, \\hat\\epsilon_t^\\prime \n\\end{equation} \ndenotes the empirical covariance matrix formed from the\nmodels residuals \n\\begin{equation}\n \\hat\\epsilon_t=Y_t-\\mu-\\sum_{l=1}^p A_l (Y_{t-l}-\\mu).\n\\end{equation}\n\n\n\\section{Ensuring that the $\\Sigma_t$ are positive definite}\n\nThe \\textsc{dvec} parameterization can produce matrices that are not\npositive definite, and the \\textsc{bekk} parameterization can produce matrices\nthat are almost not positive definite (much as a positive number can\nget arbitrarily close to zero). At worst, this will lead to a failure\nin the model to converge and at best, it makes the estimation somewhat\nunstable. To improve model performance, the \\textsc{bekk} and \\textsc{dvec}\noperations are followed by a modification of the resulting $\\Sigma_t$ that\n makes them more positive definite. The first problem is to get a notion of what is meant by ``small'' for a particular problem. \nThis is accomplished by first scaling the \n$\\Sigma_t$ to produce a matrix $\\Lambda_t$ where\n\\begin{equation}\n \\Lambda_{ij}=\\frac{\\Sigma_{t_{ij}}}{\\sqrt{\\Sigma_{t_{ij}}\\Sigma_{t_{ij}}}}\n\\end{equation}\nThe terms $\\Lambda_{ii}$ are then bounded above {$1.0\\e{-3}$, i.e., they are replaced in a differentiable fashion with numbers that are $\\ge 1.0\\e{-3}$ using\nthe \\texttt{posfun} function.\nIn addition, the correlation matrix \n$\\Lambda_{ij}/\\sqrt{\\Lambda_{t_{ij}}\\Lambda_{t_{ij}}}$ is\ndecomposed via a Choleski decomposition, the divisors of\n which is forced to be $> 0.3$ in a differentiable fashion using the\n\\texttt{posfun} function.\n\\X{\\fontindexentry{tt}{posfun} function}\nThe\nabove operations leave a matrix that is sufficiently positive definite\nand close enough to $\\Sigma_1$ unchanged.\n\n\n\\section{Missing data}\n\nMissing data points are included into the model as parameters to\nbe estimated. If there are a substantial number of missing data\npoints, this will induce bias into the estimates.\n\n\n\\section{The likelihood function}\n\nThe model was fit by maximum-likelihood or, more correctly, by\nfinding the mode of the Bayesian posterior distribution.\nA robust likelihood function that is a mixture of a normal\ndistribution and a Cauchy distribution is employed. The amount of robustness\ncan be changed by the user.\n\n\n\\section{Model selection}\n\nModel selection consists of fitting the model to the data for various values\nof the parameters $(p,q,r,s)$ and trying to determine the simplest model\nthat adequately fits the data, if any. \n\nThe two criteria which are used for this are the likelihood ratio test\nand investigation of the residuals in the form of the Box-Ljung statistic.\nThe likelihood-ratio test is used for general model selection, while\nthe Box-Ljung statistic is used to investigate whether or not the model\ncan adequately fit the changes in the covariance matrices $\\Sigma_t$\nthat occur over time.\n\n\n\\section{The Box-Ljung statistic}\n\nThe following Box-Ljung\nstatistic was employed \nto test the ability of the model to model the time varying covariance\nstructure of the time series. This statistic is calculated from the\nestimated standardized residuals $z_t$, for $t=1,\\ldots n$,\nwhere for each $t$, $z_t$ is an $m$-dimensional vector.\nThe $z_i$ are obtained in the calculations necessary to calculate the\nlog-likelihood~\\mbox{function}.\n\\begin{align}\n \\hat\\mu_j&=\\frac{1}{ n}\\sum_{i=1}^n z_{ij}\\\\\n z_{ij}^{\\prime}&=z_{ij}-\\hat\\mu_j\\\\\n \\hat\\sigma_{jk}&=\\frac{1}{ n}\\sum_{i=1}^n z_{ij}^\\prime z_{ik}^\\prime\\\\\n \\gamma_{ijk}&=\n \\frac{\n \\frac{1}{ n-l}\\sum_{i=1}^{n-l}\n (z_{ij}^\\prime z_{ik}^\\prime-\\hat\\sigma_{jk})\n (z_{i,j+l}^\\prime z_{i,k+l}^\\prime-\\hat\\sigma_{jk})\n }{ \n \\frac{1}{ n}\\sum_{i=1}^n\n (z_{ij}^\\prime z_{ik}^\\prime-\\hat\\sigma_{jk})^2\n }\n\\end{align}\n\nUnder the null hypothesis that the model is adequate, and if the\n$z_i$ are normally distributed, then the sum\n$$LB(K)_{ij}=n\\sum_{k=1}^k w_k \\gamma_{ijk}^2$$\nis asymptotically distributed as a $\\chi^2$ random variables \nwith~$K$ degrees of freedom. Here, $w_k=(n+2)/(n-k)$.\n\n\n\\section{Analysis of simulated data}\n\nOne method to get an idea how well a statistical model works is to\nuse it with simulated data where the true values of the parameters being estimated\nare known.\nA simple simulator that can generate data sets is included with the\n\\texttt{mgarch} package. The simulator generated a 4-dimensional set of\n1,000 observations. A type~$1,1,1,1$ process was simulated and\nanalyzed.\n\nThe following plots show the actual and predicted values for the\ndiagonal variance and correlation terms for the analysis with\na type $1,1,1,1$~model.\n\n\\includegraphics[height=2.5in, width=\\textwidth]{covarplot11.png}\n\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{covarplot11.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{covarplot22.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{covarplot33.png}\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{covarplot44.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot21.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot31.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot32.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot41.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot42.png}\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{corrplot43.png}\n\n\n\n\\section{Analysis of real data}\n\nThe data consist of daily observations of the German Mark/US Dollar and Japanese Yen/US dollar exchange rates, as well as the SP-500 and Tokyo \n (\\textsc{tokyose}) stock exchange indices. For this data set, $m=4$, and there\nwere $1301$ time periods with $211$ missing values. \n\n\\subsection{Model Parameters log-likelihood directory $p,q,r,s$.}\n\\begin{lstlisting}\nVARMA\n0,0,0,0 221 -1382.56 000\n1,0,0,0 241 -1322.47 100\n1,1,0,0 253 -1308.57 110\n\nVARMA with DVEC\n0,0,1,1 241 -1050.87 \n1,1,2,1 283 -928.518 \n1,2,1,1 289 -924.755 \n\nVARMA with DVECI\n1,1,2,1a 287 -887.764 \n1,2,1,1a 293 -888.531 \n\\end{lstlisting}\n\n\\bigskip\n\\subsection{Ljung-Box statistic ($\\chi^2$ with 10 degrees of freedom).}\n\\begin{lstlisting}\nVARMA\n0,0,0,0 138.953 52.902 26.351 19.171\n 52.902 222.502 154.708 59.681\n 26.351 154.708 60.328 87.031\n 19.171 59.681 87.031 68.889\n\n\n1,1,0,0 154.977 74.511 19.329 11.472\n 74.511 177.016 130.295 59.084\n 19.329 130.295 47.499 72.724\n 11.472 59.084 72.724 49.329\n\nVARMA with DVEC\n0,0,1,1 3.728 13.180 17.682 17.633\n 13.180 11.496 25.846 6.016\n 17.682 25.846 7.610 6.791\n 17.633 6.016 6.791 13.398\n\n\n1,1,1,1 4.239 9.519 10.866 10.885\n 9.519 6.434 21.734 8.047\n 10.866 21.734 4.089 8.318\n 10.885 8.047 8.318 12.814\n\n1,2,1,1 5.660 10.348 11.029 11.557\n 10.348 7.011 21.727 7.039\n 11.029 21.727 4.819 7.908\n 11.557 7.039 7.908 10.684\n\n1,1,2,1 6.920 11.698 11.464 12.593\n 11.698 4.914 24.677 8.179\n 11.464 24.677 2.399 8.855\n 12.593 8.179 8.855 10.985\n\nVARMA with DVECI\n1,2,1,1a 4.352 9.658 9.402 8.542\n 9.658 7.646 17.516 8.416O\n 9.402 17.516 5.795 7.108\n 8.542 8.416 7.108 10.483\n\n1,1,2,1a 4.785 17.077 8.760 10.487\n 17.077 6.718 18.563 9.736\n 8.760 18.563 3.270 8.525\n 10.487 9.736 8.525 10.253\n\\end{lstlisting}\n\nWhile the model $1,1,2,1a$ produced almost as high a\nlog-likelihood value as model $1,2,1,1a$, the superior\nperformance of the latter model with respect to the \nBox-Ljung statistic might prompt us to consider it the model of choice.\n\nThe following plots show the actual and predicted values for the diagonal variance and\ncorrelation terms. %fixed\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/covarplot11.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/covarplot22.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/covarplot33.png}\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/covarplot44.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot21.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot31.png}\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot32.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot41.png}\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot42.png}\n\n\n\\includegraphics[height=2.5in, width=\\textwidth]{example/corrplot43.png}\n\n\n\\bigskip\n\n\n\\section{Input format}\n\nBy default, the stand-alone version of the model attempts to read in the\ndata from a file named \\texttt{mgarch.dat}. This can be changed by a command line option.\nA reasonable command line option would be:\n\\begin{lstlisting}\n mgarch -ind datafile -nox -nohess\n\\end{lstlisting}\nThe command\n\\begin{lstlisting}\n mgarch -?\n\\end{lstlisting}\nwill print out a list of command line options.\n\nPart of a data file is shown below. The first line describes the data\nand specifies the form of the model. The \\texttt{delta} flag determines whether or not\nthe parameters that measure asymmetric response in the \\textsc{arch} component of the\nmodel are estimated. The robustness number controls the amount of\nrobustness in the likelihood function. A value between~0.0 and~0.5 is\nprobably appropriate. A value of~10,000 is used to indicate missing values\nin the~data.\n\\begin{lstlisting}\n# number of dimension p q r s delta robustness\n# observations flag\n 1301 4 0 0 1 1 0 0\n-0.473592 -0.30815 -0.199577 -0.154677\n0.140859 -0.256788 -0.823379 -0.947582\n-0.80579 -0.833361 2.439394 1.208645\n-1.542773 -0.855767 0.469603 -0.433346\n10000 10000 1.445218 0.462298\n6.014853 4.964597 -0.189343 -0.231408\n-1.064758 -1.700819 -3.143447 0.62605\n-0.608465 -2.618972 1.641306 2.127487\n-5.242424 2.902805 2.60543 -0.668981\n2.205787 2.026769 -1.42811 -0.141505\n // ..........................................\n\n10000 10000 -1.072871 0.365977\n-0.208759 1.51892 -0.59516 -0.029556\n-2.413508 -2.578193 2.045828 -0.2362\n0.890598 0.119672 -0.415587 0.029485\n-2.797967 0.943489 -1.377787 0.283592\n1.405865 1.915963 -0.696567 0.450668\n10000 10000 -0.31709 0.255885\n2.971906 -1.704494 0.835319 0.286408\n-3.451449 1.233715 -1.144007 -0.322157\n1.342827 0.376409 0.279213 0.226602\n\\end{lstlisting}\n\n\n\\section{Output files}\n\nSince the model produces a lot of output, it is a good idea to run it in its own directory,\nso files can be easily deleted. The independent variables of the\noptimization are in a file named \\texttt{mgarch.par}. (\\texttt{mgarch.bar} is an equivalent binary file.)\nA more user-friendly report is in the file \\texttt{mgarch.rep}.\nThe estimated covariances and correlations are in files named \\texttt{covar.XX} and \\texttt{correl.XX}.\n\nThe model selection criteria have identified the model~1.1.1a\nas the best of the models considered for these data.\n\nSome parameters of interest are:\n\\begin{lstlisting}\n# alpha:\n 0.394461 0.538978 1.19337 0.969918\n# A1:\n -0.0531276 0.772529 -0.187692 -0.131215\n 0.0898897 -0.399746 -0.142294 -0.106079\n 0.0740065 0.248345 -0.0629680 -0.212928\n 0.0918460 -0.0575289 -0.0700786 0.0537302\n# B1:\n -0.00719450 -0.344793 0.115020 0.00516664\n -0.0943230 0.474330 0.147845 0.0936166\n -0.0860047 -0.243294 0.0187986 0.217280\n -0.0815394 0.0843289 0.00782384 -0.0539808\n# F:\n -0.109526 -0.0292127 -0.122395 0.0548869\n 0.000332929 -0.0964482 0.0153865 -0.0346829\n -0.0148848 -0.0146167 -0.117961 -0.0529385\n -0.00504663 -0.000547068 0.00413882 -0.133956\n# G:\n 0.930026 -0.269741 0.136401 0.427047\n -0.0345549 -0.981606 0.00281003 -0.00411475\n 0.0976783 -0.0132467 -0.976596 0.0306139\n 0.117971 -0.0134465 0.00647385 -0.955173\n\\end{lstlisting}\nFor the four parameters $\\alpha_i$, a value $<1$ indicates\nthat negative values seem to have a larger effect on changes\nin the covariance structure than do positive ones. This effect seems\nto be much larger in the dollar cross rates than in the\nstock exchanges indices.\n\n\n\\section{The code for the \\textsc{bekkgarch} model}\n\nThe code for the \\textsc{bekkgarch} model follows. Additional\ncomments have been added to the~code.\n\\begin{lstlisting}\nDATA_SECTION\n // This section describes the data inputs to the model.\n // By default they are read in from the file bekkgarch.dat.\n init_int na // number of time periods\n init_int m // dimension of the vector time series\n init_int p // degree of autoregression must be >=0\n init_int q // degree of moving average mult be >=0\n init_int ra // degree of arch must be>=0\n init_int sg // degree of arch must be>=0\n init_int delta_switch // turns on asymmetric response to shocks\n init_number robustness // amount of robustness probably something \n // between 0 and .05 is right\n int n\n int msquared \n !! n=na-p; // number of obs for conditional likelihood\n init_matrix cY(-p+1,n,1,m) // the vector time series of observations\n int nmiss\n LOC_CALCS \n msquared=m*m;\n int ii=0;\n int j;\n int i;\n for (i=-p+1;i<=n;i++)\n for (j=1;j<=m;j++)\n if (cY(i,j)==10000) ii++;\n nmiss=ii;\n END_CALCS \n ivector rowmiss(1,nmiss)\n ivector colmiss(1,nmiss)\n LOC_CALCS \n ii=1;\n for (i=1;i<=n;i++)\n for (j=1;j<=m;j++)\n if (cY(i,j)==10000) {\n rowmiss(ii)=i;\n colmiss(ii)=j;\n ii++;\n }\n END_CALCS \n int m1\n int beginSigma\n matrix Idm(1,m,1,m)\n ivector beginF(1,ra)\n ivector beginG(1,sg)\n ivector beginA(1,p)\n ivector beginB(1,q)\n int nstart\n LOC_CALCS\n nstart=0;\n if (nmiss>0) nstart=1;\n m1=m*(m+1)/2; // \"size\" of n x n symmetric matrix\n Idm=identity_matrix(1,m); \n if (p) beginA.fill_seqadd(1+nstart,1);\n if (q) beginB.fill_seqadd(p+1+nstart,1);\n if (ra) beginF.fill_seqadd(p+q+1+nstart,1);\n if (sg) beginG.fill_seqadd(p+q+1+nstart,1);\n beginSigma=p+q+1+nstart;\n END_CALCS \nPARAMETER_SECTION\n LOC_CALCS\n int mm=m;\n int dstart;\n if (delta_switch>0)\n dstart=p+q+2+nstart;\n else\n dstart=-1;\n END_CALCS\n // the initial values for the time series and disturbance as\n // estimated parameters\n init_bounded_vector mudev(1,m,-100,100,-1)\n init_bounded_vector delta(1,m,.1,1.9,dstart)\n vector mu(1,m)\n vector esd(1,m)\n vector muemp(1,m) // empirical covariance matrix\n matrix Y(-p+1,n,1,m) // the vector time series of observations\n init_matrix_vector A1(1,p,1,mm,1,mm,beginA) // pars for the AR coff matrices\n 3darray A(1,p,1,m,1,m) // pars for the AR coff matrices\n init_matrix_vector B1(1,q,1,mm,1,mm,beginB) // pars for the MA coff matrices\n 3darray B(0,q,1,m,1,m) // has the additional B(0)=Id\n 3darray Bt(0,q,1,m,1,m) // has the additional B(0)=Id\n 4darray cov(1,n,0,q,1,m,1,m) // the m x m blocks for the covariance\n 3darray TB_S(0,q,1,m,1,m)\n matrix Omega(1,m,1,m)\n 3darray B_S(0,q,1,m,1,m)\n init_bounded_vector v_Sigma(1,m1,-10,10.1,beginSigma); // pars for Sigma\n matrix Semp(1,m,1,m) \n matrix SSemp(1,m,1,m) \n matrix ch_Sigma(1,m,1,m) \n init_bounded_vector_vector Fcoff(1,ra,1,msquared,-1.000,1.0,beginF) \n 3darray F(1,ra,1,m,1,m) \n 3darray tF(1,ra,1,m,1,m) \n init_bounded_vector_vector Gcoff(1,sg,1,msquared,-.98,.98,beginG) \n LOC_CALCS\n if (ra)\n {\n int mmin,mmax;\n mmin=Fcoff(1).indexmin();\n mmax=Fcoff(1).indexmax();\n for (int ij=mmin;ij<=mmax;ij++) if (value(Fcoff(1)(ij))==0.0) \n Fcoff(1)(ij)=0.02;\n }\n if (sg)\n {\n int mmin,mmax;\n mmin=Gcoff(1).indexmin();\n mmax=Gcoff(1).indexmax();\n for (int ij=mmin;ij<=mmax;ij++) if (value(Gcoff(1)(ij))==0.0) \n Gcoff(1)(ij)=0.001;\n }\n END_CALCS\n 3darray G(1,sg,1,m,1,m) \n 3darray tG(1,sg,1,m,1,m) \n 3darray Sigma(1,n,1,m,1,m) \n init_bounded_vector missvals(1,nmiss,-5.0,5.0);\n vector arpart(1,m)\n matrix r(1,n,1,m)\n 3darray rr(1,n,1,m,1,m)\n vector vecr(1,n*m) // VEC[r]\n vector y(1,n*m)\n number ldet\n objective_function_value f\n matrix Yv(-p+1,n,1,m) // after subtracting off the mean\n !! int q1m=(q+1)*m; \n !!CLASS banded_symmetric_dvar_matrix S(1,n*m,q1m);\nPROCEDURE_SECTION\n int t; int i; int j;\n fill_matrices_with_independent_parameters();\n Y=cY;\n add_missing_values();\n calculate_time_series_mean();\n calculate_the_residuals(); \n calculate_the_empirical_covariance();\n SSemp=get_initial_sigma();\n dvariable fpen=calculate_the_sub_variances_BEKK();\n calculate_the_sub_covariances();\n calculate_the_covariance_matrix();\n int ierr=0;\n // choleski decomposition of a banded symmetric matrix\n // produces a banded lower triangular matrix\n dvariable fpen1=0.0;\n banded_lower_triangular_dvar_matrix blt=choleski_decomp_positive(S,\n 1.e-6,fpen1);\n fpen+=fpen1;\n // solve for y=inv(blt)*vecr\n y=solve(blt,vecr);\n int ss=0;\n dvariable lno=ln_det(Sigma(1),ss);\n f+=norm2(log(delta));\n for (i=1;i<=q;i++) f+=norm2(B1(i));\n f+=norm2(v_Sigma);\n for (i=1;i<=ra;i++) f+=norm2(F(i));\n for (i=1;i<=sg;i++) f+=norm2(G(i));\n dvariable lndet=0.0;\n for (i=1;i<=n*m;i++) lndet+=log(blt(i,i)); \n // robust log-likelihood function -- mixture of normal and \n // tiny bit of cauchy \n dvar_vector y2= square(y);\n if (robustness>1.e-20) \n f+= lndet - sum(log(mfexp(-0.5*y2)+robustness/(1.0+y2)));\n else\n f+= lndet - sum(log(mfexp(-0.5*y2)+.0001/(1.0+y2)));\n\n f+=fpen;\n (*ad_printf)(\"f = %lf\\n\",value(f));\nFUNCTION fill_matrices_with_independent_parameters\n int ii=1; int i=1; int iii;\n double d=sqrt(0.1);\n if (!sg) d=1.0;\n ch_Sigma.initialize();\n // this is the choleski decomp parameterization of the\n // covariance matrix\n for (i=1;i<=m;i++) \n for (int j=1;j<=i;j++) {\n if (i==j)ch_Sigma(i,i)+=d;\n ch_Sigma(i,j)+=v_Sigma(ii++);\n } \n for (iii=1;iii<=ra;iii++) {\n if (iii==1 || active(Fcoff(iii))) {\n ii=1;\n F(iii).initialize();\n tF(iii).initialize();\n for (i=1;i<=m;i++) {\n for (int j=1;j<=m;j++) \n F(iii)(i,j)=Fcoff(iii)(ii++);\n }\n }\n tF(iii)=trans(F(iii));\n } \n\n for (iii=1;iii<=sg;iii++) {\n if (iii==1 || active(Gcoff(iii))) {\n ii=1;\n G(iii).initialize();\n tG(iii).initialize();\n for (i=1;i<=m;i++) {\n for (int j=1;j<=m;j++) {\n G(iii)(i,j)=Gcoff(iii)(ii++);\n } \n if (iii==1) G(iii)(i,i)+=0.90;\n }\n }\n tG(iii)=trans(G(iii));\n }\n B.initialize();\n A.initialize();\n // B(0) is the identity matrix\n B(0)=Idm;\n for (i=1;i<=p;i++) A(i)=A1(i);\n for (i=1;i<=q;i++) B(i)=B1(i);\n for (i=0;i<=q;i++) Bt(i)=trans(B(i));\n \nFUNCTION calculate_the_residuals \n int t; int j;\n mu=muemp+mudev;\n for (t=-p+1;t<=0;t++) Yv(t)=Y(t)-mu;\n for (t=1;t<=n;t++) {\n Yv(t)=Y(t)-mu;\n calculate_autoregressive_part(t);\n r(t)=Yv(t)-arpart;\n }\n int ii=0;\n // this corresponds to the VEC operator\n for (int i=1;i<=n;i++) \n for (j=1;j<=m;j++) vecr(++ii)=r(i,j);\n \nFUNCTION void calculate_the_residuals2(dvar_matrix& e)\n int t; int j;\n mu=muemp+mudev;\n for (t=-p+1;t<=0;t++) Yv(t)=Y(t)-mu;\n for (t=1;t<=n;t++) {\n Yv(t)=Y(t)-mu;\n calculate_autoregressive_part(t);\n e(t)=Yv(t)-arpart;\n for (j=1;j<=q;j++) {\n if (t<=j) break;\n e(t)-=B(j)*e(t-j);\n } \n }\n \nFUNCTION calculate_the_sub_covariances\n int i; int k; int l;\n int qq=0;\n for ( l=1;l<=q;l++) {\n if (!active(B1(l))) break;\n qq=l;\n }\n cov.initialize();\n for (i=1;i<=n;i++) {\n for (int l=0;l<=qq;l++) {\n if (i<=l) break;\n for (int k=0;k<=qq-l;k++) {\n int ilk=i-l-k;\n if (ilk<1) ilk=1;\n cov(i,l)+=B(l+k)*Sigma(ilk)*Bt(k);\n }\t\n }\n } \t\n \nFUNCTION calculate_the_empirical_covariance\n int i;\n Semp.initialize();\n ivector sgn(1,m);\n esd.initialize();\n for (i=1;i<=n;i++) {\n esd+=square(r(i));\n }\n esd/=n;\n esd=sqrt(esd);\n if (active(delta)) {\n dvar_vector mult_neg=elem_div(esd,delta);\n dvar_vector mult_pos=elem_prod(esd,delta);\n for (i=1;i<=n;i++) {\n sgn.initialize();\n dvar_vector sr=sfabs(elem_div(r(i),esd));\n for (int j=1;j<=m;j++)\n if (r(i,j)<0) \n sr(j)=-sr(j)*mult_neg(j);\n else\n sr(j)=sr(j)*mult_pos(j);\n rr(i)=outer_prod(sr,sr);\n }\n Semp=empirical_covariance(r);\n }\n else\n {\n for (i=1;i<=n;i++) {\n rr(i)=outer_prod(r(i),r(i));\n Semp+=rr(i);\n }\n Semp/=n;\n }\n for (int j=1;j<=m;j++)\n esd(j)=sqrt(Semp(j,j));\n \n \nFUNCTION dvariable calculate_the_sub_variances_diagonal_vector_garch(void)\n int i; int k; int ii; int jj;\n dvar_vector norms(1,n);\n dvariable fpen=0.0;\n dvariable fpen1;\n Omega=ch_Sigma*SSemp*trans(ch_Sigma);\n Sigma.initialize();\n // set the first Sigma equal to the empirical covariance\n int rsmax=mymax(ra,sg);\n Sigma(1)=SSemp;\n //cout << Sigma(1) << endl;\n dvar_matrix SS=scale(Sigma(1),esd);\n //cout << SS << endl;\n fpen+=positivize_sigma(SS);\n //cout << SS << endl;\n dvariable ns=norm(SS);\n norms(1)=ns;\n fpen1=0.0;\n dvariable bn=mf_upper_bound(ns,1000.0,fpen1);\n if (fpen1>0.0) {\n SS*=(bn/ns);\n fpen+=fpen1;\n }\n Sigma(1)=unscale(SS,esd);\n for (i=2;i<=rsmax;i++) {\n Sigma(i)=Sigma(1);\n norms(i)=norms(1);\n }\n\n int mmin=Sigma(1).indexmin(); \n int mmax=Sigma(1).indexmax(); \n dvar_vector s(mmin,mmax);\n for (i=rsmax+1;i<=n;i++) {\n Sigma(i)=Omega;\n if (ra) Sigma(i)+=elem_prod(F(1),rr(i-1));\n for (ii=2;ii<=ra;ii++) {\n if (active(Fcoff(ii)))\n Sigma(i)+=elem_prod(F(ii),rr(i-ii));\n }\n \n if (sg) Sigma(i)+=elem_prod(G(1),Sigma(i-1));\n for (ii=2;ii<=sg;ii++) {\n if (active(Gcoff(ii)))\n Sigma(i)+=elem_prod(G(ii),Sigma(i-ii));\n }\n \n // \"positivize\" the\n // correlation matrix\n dvar_matrix SS=scale(Sigma(i),esd);\n fpen+=positivize_sigma(SS);\n dvariable ns=norm(SS);\n norms(i)=ns;\n fpen1=0.0;\n dvariable bn=mf_upper_bound(ns,1000.0,fpen1);\n if (fpen1>0.0) {\n SS*=(bn/ns);\n fpen+=fpen1;\n }\n Sigma(i)=unscale(SS,esd);\n }\n dvector trend(1,n);\n trend.fill_seqadd(-1,2.0/(n-1));\n cout << \"norms*trend/norm(norms)\" << endl; \n dvariable npen=norms*trend/norm(norms); \n cout << norms*trend/norm(norms) << endl; \n fpen+=npen;\n if (fpen>1.0)\n cout << \" fpen = \" << fpen << endl;\n return fpen;\n \nFUNCTION dvariable calculate_the_sub_variances_BEKK(void)\n int i; int k; int ii; int jj;\n dvar_vector norms(1,n);\n dvariable fpen=0.0;\n dvariable fpen1;\n Omega=ch_Sigma*SSemp*trans(ch_Sigma);\n Sigma.initialize();\n // set the first Sigma equal to the empirical covariance\n int rsmax=mymax(ra,sg);\n Sigma(1)=SSemp;\n dvar_matrix SS=scale(Sigma(1),esd);\n fpen+=positivize_sigma(SS);\n dvariable ns=norm(SS);\n norms(1)=ns;\n fpen1=0.0;\n dvariable bn=mf_upper_bound(ns,1000.0,fpen1);\n if (fpen1>0.0) {\n SS*=(bn/ns);\n fpen+=fpen1;\n }\n Sigma(1)=unscale(SS,esd);\n for (i=2;i<=rsmax;i++) {\n Sigma(i)=Sigma(1);\n norms(i)=norms(1);\n }\n\n int mmin=Sigma(1).indexmin(); \n int mmax=Sigma(1).indexmax(); \n dvar_vector s(mmin,mmax);\n for (i=rsmax+1;i<=n;i++) {\n Sigma(i)=Omega;\n if (ra) Sigma(i)+=F(1)*rr(i-1)*tF(1);\n for (ii=2;ii<=ra;ii++) {\n if (active(Fcoff(ii)))\n Sigma(i)+=F(ii)*rr(i-ii)*tF(ii);\n }\n \n if (sg) Sigma(i)+=G(1)*Sigma(i-1)*tG(1);\n for (ii=2;ii<=sg;ii++) {\n if (active(Gcoff(ii)))\n Sigma(i)+=G(ii)*Sigma(i-ii)*tG(ii);\n }\n \n // \"positivize\" the\n // correlation matrix\n dvar_matrix SS=scale(Sigma(i),esd);\n fpen+=positivize_sigma(SS);\n dvariable ns=norm(SS);\n norms(i)=ns;\n fpen1=0.0;\n dvariable bn=mf_upper_bound(ns,1000.0,fpen1);\n if (fpen1>0.0) {\n SS*=(bn/ns);\n fpen+=fpen1;\n }\n Sigma(i)=unscale(SS,esd);\n }\n dvector trend(1,n);\n trend.fill_seqadd(-1,2.0/(n-1));\n cout << \"norms*trend/norm(norms)\" << endl; \n dvariable npen=norms*trend/norm(norms); \n cout << norms*trend/norm(norms) << endl; \n fpen+=npen;\n if (fpen>1.0)\n cout << \" fpen = \" << fpen << endl;\n {\n //ofstream ofs(\"sigma\");\n \n //for (int i=1;i<=n;i++)\n //ofs << Sigma(i) << endl << endl;\n }\n return fpen;\n \nFUNCTION dvar_matrix scale(dvar_matrix& M,dvar_vector& sd)\n int mmin=sd.indexmin();\n int mmax=sd.indexmax();\n dvar_matrix SM(mmin,mmax,mmin,mmax);\n for (int i=mmin;i<=mmax;i++)\n for (int j=mmin;j<=mmax;j++)\n SM(i,j)=M(i,j)/(sd(i)*sd(j));\n return SM;\n \nFUNCTION dvar_matrix unscale(dvar_matrix& M,dvar_vector& sd)\n int mmin=sd.indexmin();\n int mmax=sd.indexmax();\n dvar_matrix SM(mmin,mmax,mmin,mmax);\n for (int i=mmin;i<=mmax;i++)\n for (int j=mmin;j<=mmax;j++)\n SM(i,j)=M(i,j)*(sd(i)*sd(j));\n return SM;\n \nFUNCTION dvar_matrix get_initial_sigma(void)\n int i,j,k,l,ll,m2,r,s;\n m2=m*m;\n dvar_matrix M(1,m2,1,m2);\n \n dvar_vector v=VEC(Semp);\n M=identity_matrix(1,m2);\n for (ll=1;ll<=q;ll++) \n {\n for (i=1;i<=m;i++) \n {\n for (j=1;j<=m;j++) \n {\n int col=(i-1)*m+j;\n for (r=1;r<=m;r++)\n {\n for (s=1;s<=m;s++) \n {\n int row=(r-1)*m+s;\n M(row,col)+=B(ll)(r,i)*B(ll)(s,j);\n }\n }\n }\n }\n } \n v=solve(M,v);\n dvar_matrix tmp= MAT(v,m,m);\n return tmp;\n \nFUNCTION calculate_the_covariance_matrix\n int ioffset; int joffset; int i1; int i; int j1;\n int k; int l;\n int qq=0;\n for ( l=1;l<=q;l++) {\n if (!active(B1(l))) break;\n qq=l;\n }\n S.initialize();\n for (i=1;i<=n;i++) {\n ioffset=(i-1)*m; \n for (int k=0;k<=qq;k++) {\n //if (k>0 && !active(B1)) break;\n joffset=(i-k-1)*m; \n if (joffset<0) break;\n for (i1=1;i1<=m;i1++) { \n int up;\n if (k==0) \n\t up=i1;\n\telse \n\t up=m;\n for (j1=1;j1<=up;j1++) { \n int i2=i1+ioffset;\n\t int j2=j1+joffset;\n S(i1+ioffset,j1+joffset)=cov(i,k,i1,j1);\n } \t\n } \n } \n if (S(1,1) < 0)\n cout << S(1,1) << endl;\n }\n\nFUNCTION void calculate_autoregressive_part(int t)\n // The user can put in any (nonlinear) function desired here\n arpart.initialize();\n for (int j=1;j<=p;j++) arpart+=A(j)*Yv(t-j);\n \nREPORT_SECTION\n int i; int ii; int jj, t;\n for (ii=1;ii<=m;ii++) \n {\n for (jj=1;jj<=ii;jj++) \n {\n ofstream ofs((char*)(\"covar.\" + str(ii) +str(jj)));\n ofstream ofs1((char*)(\"correl.\" + str(ii) +str(jj)));\n dvar_matrix Covariance(1,m,1,m);\n for (i=1;i<=n;i++) {\n Covariance=Sigma(i);\n for(int j=1;j<=q;j++)\n if ( (i-j)>0 ) Covariance+=B(j)*Sigma(i-j)*trans(B(j));\n ofs << Covariance(ii,jj) << endl;\n ofs1 << Covariance(ii,jj)/\n sqrt(Covariance(ii,ii)*Covariance(jj,jj))<< endl;\n }\n }\n }\n \n {\n dvar_matrix ymat=MAT(y,n,m);\n for (int i=1;i<=m;i++) {\n ofstream ofs1((char*)(\"yres.\" + str(i)));\n dvector tmp(1,n);\n for (t=1;t<=n;t++) \n tmp(t)=value(ymat(t,i));\n ivector hist=histogram(-20,20,81,tmp);\t \n ofs1 << column_print(hist) << endl;\n } \n } \n report << \"The means\" << endl;\n report << mu << endl;\n for (i=1;i<=p;i++) {\n report << \"A(\"<< i << \")\" << endl;\n report << setfixed() << setprecision(3) << A(i) << endl;\n } \n report << endl;\n for (i=1;i<=q;i++) {\n report << \"B(\"<< i << \")\" << endl;\n report << setfixed() << setprecision(3) << B(i) << endl;\n } \n report << endl;\n report << \"delta\" << endl;\n report << setfixed() << setprecision(3) << delta << endl;\n report << endl;\n report << \"Omega\" << endl;\n report << setfixed() << setprecision(3) << Omega << endl;\n report << endl;\n for (i=1;i<=ra;i++) {\n report << \"F(\"<< i << \")\" << endl;\n report << setfixed() << setprecision(3) << F(i) << endl;\n } \n report << endl;\n for (i=1;i<=sg;i++) {\n report << \"G(\"<< i << \")\" << endl;\n report << setfixed() << setprecision(3) << G(i) << endl;\n } \n report << endl;\n //report << setfixed() << setw(8) << setprecision(1) << S << endl;\n { // calculate predicted observations for next 20 time periods\n // for graphs results are in t.1 t.2 etc\n int npreds=20;\n dvar_matrix e(1,n,1,m);\n calculate_the_residuals2(e);\n \n dvar_matrix Z(n+1,n+npreds,1,m);\n Z.initialize();\n \n for (int t=n+1;t<=n+npreds;t++) {\n Z(t)+=mu;\n for (int i=1;i<=p;i++) {\n if (t-i>n)\n Z(t)+=A(i)*(Z(t-i)-mu); \n else \n Z(t)+=A(i)*(Y(t-i)-mu); \n }\n for (i=1;i<=q;i++) {\n if (t-i<=n && t-i>0)\n Z(t)+=B(i)*e(t-i); \n }\t \n } \t \n for (i=1;i<=m;i++) {\n ofstream ofs1((char*)(\"pred\" + str(i)));\n for (t=-p+1;t<=n;t++) \n ofs1 << Y(t,i) << endl;\n for (t=n+1;t<=n+npreds;t++) \n ofs1 << Z(t,i) << endl;\n } \n } \n dmatrix T(1,m,1,m);\n dmatrix chi(1,m,1,m);\n T.initialize();\n dmatrix Aut(1,m,1,m);\n Aut.initialize();\n const int K=10;\n d3_array gamma(1,K,1,m,1,m);\n dmatrix cr(1,n,1,m);\n for (i=1;i<=n;i++)\n {\n cr(i)=value(y((i-1)*m+1,i*m).shift(1));\n T+=outer_prod(cr(i),cr(i));\n }\n T=T/n;\n for (i=2;i<=n;i++)\n {\n Aut+=outer_prod(cr(i-1),cr(i));\n }\n Aut=Aut/(n-1);\n for (i=1;i<=m;i++)\n {\n for (int j=1;j<=m;j++)\n {\n Aut(i,j)/=sqrt(T(i,i)*T(j,j));\n }\n }\n gamma.initialize();\n for (int j=1;j<=m;j++)\n {\n for (int k=1;k<=m;k++)\n {\n for (int l=1;l<=10;l++)\n {\n double tmp=0;\n for (i=1;i<=n-l;i++)\n {\n gamma(l,j,k)+=(cr(i,j)*cr(i,k)-T(j,k))*(cr(i+l,j)*cr(i+l,k)-T(j,k));\n tmp+=square(cr(i,j)*cr(i,k)-T(j,k));\n }\n gamma(l,j,k)/=tmp;\n }\n }\n }\n \n chi.initialize();\n for (int l=1;l<=K;l++)\n {\n chi+=n*(n+2)/(n-l)*square(gamma(l));\n }\n \n report << \"Covariance of standardized residuals\" << endl;\n report << T << endl;\n report << \"Lag 1 autocorellation of standardized residuals\" << endl;\n report << Aut << endl;\n report << \"Ljung Box statistic based on chi squared with \" << K \n << \" degrees of freedom\" << endl;\n report << chi << endl;\n \nFUNCTION add_missing_values\n for (int ii=1;ii<=nmiss;ii++) { \n Y(rowmiss(ii),colmiss(ii))=missvals(ii);\n } \t \n ofstream ofs(\"testy\");\n ofs << Y << endl;\n \nFUNCTION calculate_time_series_mean\n muemp.initialize();\n for (int i=-p+1;i<=n;i++) muemp+=Y(i);\n muemp/=(n+p);\n \nFUNCTION dvariable positivize_sigma(dvar_matrix& TS)\n int ii,jj;\n dvariable fpen=0.0;\n int mmin=TS.indexmin();\n int mmax=TS.indexmax();\n dvar_vector s(mmin,mmax);\n for (ii=mmin;ii<=mmax;ii++)\n s(ii)=sqrt(posfun(TS(ii,ii),1.e-3,fpen));\n for (ii=mmin;ii<=mmax;ii++)\n for (jj=mmin;jj<=mmax;jj++) \n TS(ii,jj)/=(s(ii)*s(jj));\n TS=positive_definite_matrix(TS,.3,fpen);\n for (ii=mmin;ii<=mmax;ii++)\n for (jj=mmin;jj<=mmax;jj++) \n TS(ii,jj)*=(s(ii)*s(jj));\n return fpen;\n\nGLOBALS_SECTION\n // some C++ compilers don't supply this!\n int mymax(int x,int y)\n {\n if (x>y) \n return x;\n else\n return y;\n }\n\nTOP_OF_MAIN_SECTION\n \n ofstream ofs(\"Error.log\");\n arrmblsize=5000000; \n gradient_structure::set_GRADSTACK_BUFFER_SIZE(560000);\n gradient_structure::set_CMPDIF_BUFFER_SIZE(15000000);\n gradient_structure::set_MAX_NVAR_OFFSET(1000);\n\\end{lstlisting}\n\n", "meta": {"hexsha": "b8040a5539141e90f71a22e4bad9167e96a0b1fb", "size": 35504, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/manuals/admb/magarch-april2000.tex", "max_stars_repo_name": "johnrsibert/admb", "max_stars_repo_head_hexsha": "063ec863a9f23f6c6afbc7d481af0476b8d63645", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 79, "max_stars_repo_stars_event_min_datetime": "2015-01-16T14:14:22.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-24T06:28:15.000Z", "max_issues_repo_path": "docs/manuals/admb/magarch-april2000.tex", "max_issues_repo_name": "johnrsibert/admb", "max_issues_repo_head_hexsha": "063ec863a9f23f6c6afbc7d481af0476b8d63645", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 172, "max_issues_repo_issues_event_min_datetime": "2015-01-21T01:53:57.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-29T19:57:31.000Z", "max_forks_repo_path": "docs/manuals/admb/magarch-april2000.tex", "max_forks_repo_name": "johnrsibert/admb", "max_forks_repo_head_hexsha": "063ec863a9f23f6c6afbc7d481af0476b8d63645", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 22, "max_forks_repo_forks_event_min_datetime": "2015-01-15T18:11:54.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-11T21:47:51.000Z", "avg_line_length": 30.4755364807, "max_line_length": 158, "alphanum_fraction": 0.6179585399, "num_tokens": 12479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085909370423, "lm_q2_score": 0.7826624738835052, "lm_q1q2_score": 0.6146315645447551}} {"text": "\\chapter{Normalised LMS and Affine Projection Algorithm}\n\\label{ch:NLMS_APA}\n\\section{Review of the Basics}\n\\subsection{Inversion of a 2 by 2 Block Matrix}\nLet $\\vect{A}$ and $\\vect{M}_1=\\vect{D}-\\vect{C}\\vect{A}^{-1}\\vect{B}$ be non-singular matrices. Then\n\\bmath\n \\bbmtx\n \\vect{A} & \\vect{B}\\\\\n \\vect{C} & \\vect{D}\n \\ebmtx^{-1} =\n \\bbmtx\n \\vect{A}^{-1}+\\vect{A}^{-1}\\vect{B}\\vect{M}_1^{-1}\\vect{C}\\vect{A}^{-1} & -\\vect{A}^{-1}\\vect{B}\\vect{M}_1^{-1}\\\\\n -\\vect{M}_1^{-1}\\vect{C}\\vect{A}^{-1} & \\vect{M}_1^{-1}\n \\ebmtx\\ .\n \\label{eq:block_inv1}\n\\emath\nLet instead $\\vect{D}$ and $\\vect{M}_2=\\vect{A}-\\vect{B}\\vect{D}^{-1}\\vect{C}$ be non-singular matrices. Then\n\\bmath\n \\bbmtx\n \\vect{A} & \\vect{B}\\\\\n \\vect{C} & \\vect{D}\n \\ebmtx^{-1} =\n \\bbmtx\n \\vect{M}_2^{-1} & -\\vect{M}_2^{-1}\\vect{B}\\vect{D}^{-1}\\\\\n -\\vect{D}^{-1}\\vect{C}\\vect{M}_2^{-1} & \\vect{D}^{-1}+\\vect{D}^{-1}\\vect{C}\\vect{M}_2^{-1}\\vect{B}\\vect{D}^{-1}\n \\ebmtx\\ .\n \\label{eq:block_inv2}\n\\emath\nThe matrices $\\vect{M}_1$ and $\\vect{M}_2$ are said to be the \\textit{Schur complement} of $\\vect{A}$ and $\\vect{D}$, respectively.\n\n\\subsection{The Method of Lagrange Multipliers}\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/const_opt}\n \\caption{Optimisation problem with a single equality constraint. The points $\\vect{x}_u$ and $\\vect{x}_c$ are the unconstrained and the constrained minimum, respectively.}\n \\label{fig:const_opt}\n\\end{figure}\n\\noindent The method of Lagrange multipliers can be used to solve optimisation problems with equality constraints\\footnote{The method of Lagrange multipliers can also be used to solve the more general problem with inequality constraints, but we do not consider this here.}\n\\begin{gather}\n \\begin{alignedat}{2}\n &\\underset{\\vect{x}}{\\text{optimise}} &\\quad f(\\vect{x}) &\\\\\n &\\text{subject to}&\\quad \\vect{h}(\\vect{x}) &= \\vect{0}\n \\end{alignedat}\n \\label{eq:constr_opt}\n\\end{gather}\nConsider the optimisation problem in Fig.~\\ref{fig:const_opt} with the cost function $f(\\vect{x})$ and the constraints $\\vect{h}(\\vect{x})=\\vect{0}$. At the constrained optimum $\\vect{x}_c$, the gradients of $f(\\vect{x})$ and $\\vect{h}(\\vect{x})$ are parallel. Thus,\n\\begin{alignat*}{2}\n \\vect{\\nabla}f(\\vect{x}) &= \\lambda \\vect{\\nabla}h(\\vect{x}) &\\qquad &\\text{(single constraint)}\\\\\n \\vect{\\nabla}f(\\vect{x}) &= \\sum_{m=1}^M\\lambda_m \\vect{\\nabla}h_m(\\vect{x})= \\vect{\\nabla}\\vect{h}(\\vect{x})\\vect{\\lambda} &\\qquad &\\text{(multiple constraints)}\n\\end{alignat*}\nThe vector $\\vect{\\lambda}$ contains the Lagrange multipliers, and they ensure that the gradients have the same direction and length.\n\nIf we define the Lagrangian function\n\\bmath\n L(\\vect{x},\\vect{\\lambda}) = f(\\vect{x}) - \\vect{\\lambda}^T\\vect{h}(\\vect{x})\\ ,\n\\emath\nthen its critical points satisfy\n\\begin{alignat}{2}\n \\vect{\\nabla}_{\\vect{x}} L(\\vect{x},\\vect{\\lambda}) &= \\vect{0}\\quad\\iff&\\quad\\vect{\\nabla}f(\\vect{x}) &= \\vect{\\nabla}\\vect{h}(\\vect{x})\\vect{\\lambda}\\label{eq:grad_lagr_func}\\\\\n \\vect{\\nabla}_{\\vect{\\lambda}} L(\\vect{x},\\vect{\\lambda}) &= \\vect{0}\\quad\\iff&\\quad\\vect{h}(\\vect{x}) &= \\vect{0}\\ .\\label{eq:grad_lagr_func_const}\n\\end{alignat}\n\\eq{eq:grad_lagr_func} is the same as what we previously derived from Fig.~\\ref{fig:const_opt}, and \\eq{eq:grad_lagr_func_const} is simply the constraints of our optimisation problem. Thus, the solutions to the system of equations given by \\eq{eq:grad_lagr_func} and \\eq{eq:grad_lagr_func_const} are indeed the critical points of our optimisation problems. In order to find the optimum, we need to classify the critical points as either minimums, maximums, or saddle points. The classification of the critical points for any pair of cost function and constraints is beyond the scope of these lecture notes. We refer the interested reader to \\cite[pp.~294--308]{Antoniou2007}. Alternatively, we may formulate the constrained optimisation problem in \\eq{eq:constr_opt} as the sequential unconstrained optimisation problem \\cite[ch.~5]{Boyd2004}\n\\bmath\n \\max_{\\vect{\\lambda}}\\left\\{\\underset{\\vect{x}}{\\text{optimise}}\\ L(\\vect{x},\\vect{\\lambda}) \\right\\}\\ .\n \\label{eq:constr_opt_nested}\n\\emath\nWe could solve this optimisation problem by applying the five step recipe for unconstrained optimisation in Sec.~\\ref{ssec:optimisation} to first the inner optimisation problem and then the outer optimisation problem.\n\nWe are here not concerned with these general methods. Instead, we consider the simpler problem of minimising a quadratic cost function with linear equality constraints.\n\n\\subsubsection{Quadratic Cost Function with Linear Equality Constraints}\nWe consider the following equality constrained minimisation problem\n\\begin{gather}\n \\begin{alignedat}{2}\n &\\min_{\\vect{x}} &\\quad f(\\vect{x}) &= \\frac{1}{2}\\vect{x}^T\\vect{Px}+\\vect{q}^T\\vect{x}+r\\\\\n &\\text{subject to}&\\quad \\vect{h}(\\vect{x}) &= \\vect{Ax}-\\vect{b}=\\vect{0}\n \\end{alignedat}\n \\label{eq:constr_opt_quad_lin}\n\\end{gather}\nin which the cost function is quadratic and the equality constraints are linear. We assume that $\\vect{P}$ is an $N\\times N$ p.d. matrix so that $f(\\vect{x})$ is a convex function. Moreover, we assume that $\\vect{A}$ is a full rank $M\\times N$ matrix with $M\\leq N$. The Langrangian function is \n\\bmath\n L(\\vect{x},\\vect{\\lambda}) = \\frac{1}{2}\\vect{x}^T\\vect{Px}+\\vect{q}^T\\vect{x}+\\vect{r} - \\vect{\\lambda}^T(\\vect{Ax}-\\vect{b})\n\\emath\nand its gradient w.r.t. $\\vect{x}$ is\n\\bmath\n \\vect{\\nabla}_{\\vect{x}} L(\\vect{x},\\vect{\\lambda}) = \\vect{Px}+\\vect{q}-\\vect{A}^T\\vect{\\lambda}\\ .\n\\emath\nThe system of equations given by \\eq{eq:grad_lagr_func} and \\eq{eq:grad_lagr_func_const} are therefore\n\\bmath\n \\bbmtx\n \\vect{P} & -\\vect{A}^T\\\\\n -\\vect{A} & \\vect{0}\n \\ebmtx\n \\bbmtx\n \\vect{x}\\\\\n \\vect{\\lambda}\n \\ebmtx = \n \\bbmtx\n -\\vect{q}\\\\\n -\\vect{b}\n \\ebmtx\\ .\n\\emath\nSince the cost function $f(\\vect{x})$ is assumed to be convex, there is only one point satisfying this system of equations. By using the $2\\times 2$ block matrix inversion rule in \\eq{eq:block_inv1}, we obtain that this point $(\\vect{x}_c,\\vect{\\lambda}_c)$ is\n\\begin{align}\n \\vect{\\lambda}_c &= (\\vect{AP}^{-1}\\vect{A}^T)^{-1}(\\vect{AP}^{-1}\\vect{q}+\\vect{b})\\label{eq:constr_opt_quad_lin_minlam}\\\\\n \\vect{x}_c &= \\vect{P}^{-1}(\\vect{A}^T\\vect{\\lambda}_c-\\vect{q})\\ .\n \\label{eq:constr_opt_quad_lin_minx}\n\\end{align}\nThus, the solution to the constrained minimisation problem in \\eq{eq:constr_opt_quad_lin} is $\\vect{x}_c$ given by \\eq{eq:constr_opt_quad_lin_minx}.\n\n\\section{Overview over Adaptive Filters based on the Mean-Squared Error Cost Function}\nAn overview over the adaptive filters is shown in Fig.~\\ref{fig:overview}. Recall that we have the following important properties for the steepest descent algorithm and the stochastic gradient methods.\n\\subsubsection{Steepest Descent (SD)}\n\\begin{itemize}\n \\item Since the statistics is assumed known, the gradient and filter coefficients are deterministic functions.\n \\item SD is of limited practical usage, but illustrates nicely how adaptive filters function an how they are analysed.\n \\item SD is the \"ideal\" (first-order) adaptive filter.\n\\end{itemize}\n\\subsubsection{Stochastic Gradient Methods (SGMs)}\n\\begin{itemize}\n \\item Since the statistics is unknown, the gradient and filter coefficients are stochastic processes.\n \\item They are generally easy to implement and hard to analyse.\n \\item They are approximations to the steepest descent algorithm.\n\\end{itemize}\n\n\\subsection{Model for the Analysis of SGMs}\n\\label{ssec:analysis_model}\nIn order to make the SGMs easier to analyse, a model is typically assumed for the relationship between the input signal $u(n)$ and the desired signal $d(n)$. Fig.~\\ref{fig:sys_id_block_diagram} shows one of the most popular models used for the analysis of adaptive filters.\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/sys_id_block_diagram}\n \\caption{Block diagram showing the model used to analyse adaptive filters.}\n \\label{fig:sys_id_block_diagram}\n\\end{figure}\nDefine\n\\begin{align}\n \\vect{w}_o &= \\bbmtx w_{o,0} & \\cdots & w_{o,M-1} \\ebmtx^T\\\\\n \\vect{w}(n) &= \\bbmtx w_0(n) & \\cdots & w_{M-1}(n) \\ebmtx^T\\\\\n \\vect{u}(n) &= \\bbmtx u(n) & \\cdots & u(n-M+1) \\ebmtx^T\\\\\n \\vect{\\Delta}\\vect{w}(n) &= \\vect{w}_o-\\vect{w}(n)\n\\end{align}\nwhere $\\vect{\\Delta}\\vect{w}(n)$ is referred to as the weight error.\n\\subsubsection{Assumptions}\nFrom Fig.~\\ref{fig:sys_id_block_diagram}, we have that\n\\bmath\n d(n) = z(n)+v(n) = \\vect{u}^T(n)\\vect{w}_o + v(n)\\ .\n\\emath\nMoreover, we assume that\n\\begin{enumerate}\n \\item $v(n)$ is white noise with variance $\\sigma_v^2$.\n \\item $v(n)$ and $u(n)$ are uncorrelated.\n \\item $\\vect{u}(n)$ is a white process with correlation matrix $\\vect{R}_u$. This assumption implies that $\\vect{u}(n)$ and $\\vect{w}(n)$ are uncorrelated.\n\\end{enumerate}\n\\subsubsection{Equations}\nWe have that\n \\begin{align}\n e(n) &= d(n) - y(n) = v(n) + \\vect{u}^T(n)\\vect{w}_o - \\vect{u}^T(n)\\vect{w}(n)\\\\\n &= \\vect{u}^T(n)\\vect{\\Delta}\\vect{w}(n)+ v(n)\n \\end{align}\nThus, at the optimum where $\\vect{\\Delta}\\vect{w}(n)=\\vect{0}$, we have that\n\\bmath\n J_1(\\vect{w}_o) = J_\\textup{min} = J_1(v^2(n)) = \\sigma_v^2\\ .\n\\emath\nThe cost function can be written as\n\\begin{align}\n J_1(\\vect{w}(n)) &= E[e^2(n)] = E[v^2(n)] + E[\\vect{\\Delta}\\vect{w}^T(n)\\vect{u}(n)\\vect{u}^T(n)\\vect{\\Delta}\\vect{w}(n)]\\notag\\\\\n &= \\sigma_v^2 + E[E[\\vect{\\Delta}\\vect{w}^T(n)\\vect{u}(n)\\vect{u}^T(n)\\vect{\\Delta}\\vect{w}(n)|\\vect{\\Delta}\\vect{w}(n)]]\\notag\\\\\n &= J_\\textup{min} + E[\\vect{\\Delta}\\vect{w}^T(n)\\vect{R}_u\\vect{\\Delta}\\vect{w}(n)]\\ .\n\\end{align}\nWhen viewed as a function of $n$, the cost function is called the learning curve.\n\n\\subsection{How to Analyse Adaptive Filters}\nAdaptive filters can be analysed in several ways.\n\\begin{description}\n \\item[Transient Performance] How does the filter handle an abrupt change in the statistics of $u(n)$, and how fast does it converge to steady-state?\n \\item[Steady-state Performance] How does the filter perform in a WSS environment after all transients have died out?\n \\item[Tracking Performance] How does the filter handle slow variations in the statistics of $u(n)$ and/or $d(n)$?\n \\item[Numerical Precision Effects] What happens when the filter is implemented on a finite precision computer?\n \\item[Computational Complexity] How much processing time does the algorithm require?\n\\end{description}\n\\subsubsection{Important Values and Functions}\nWhen analysing adaptive filters, we typically quantify their performance in terms of the following values and functions.\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/learning_curve_sgm}\n \\caption{The cost function (same as the MSE), the minimum MSE, and the excess MSE.}\n \\label{fig:learning_curve_sgm}\n\\end{figure}\n\\begin{alignat}{2}\n &J_1(\\vect{w}(n)) &\\quad &\\text{(Mean-Square Error (MSE))}\\\\\n &J_1(\\vect{w}_o) = J_\\textup{min} &\\quad &\\text{(Minimum MSE (MMSE))}\\\\\n &J_\\textup{ex} = J_1(\\vect{w}(\\infty))-J_\\textup{min} &\\quad &\\text{(Excess MSE (EMSE))}\\\\\n &\\mathcal{M} = \\frac{ J_\\textup{ex}}{J_\\textup{min}} &\\quad &\\text{(Misadjustment)}\\\\\n &E[\\|\\vect{\\Delta}\\vect{w}(\\infty)\\|^2] &\\quad &\\text{(Mean-Square Deviation (MSD))}\\\\\n &\\lambda_\\textup{max} &\\quad &\\text{(Maximum Eigenvalue of $\\vect{R}_u$)}\\\\\n &\\lambda_\\textup{min} &\\quad &\\text{(Minimum Eigenvalue of $\\vect{R}_u$)}\\\\\n &\\kappa(\\vect{R}_u) = \\frac{\\lambda_\\textup{max}}{\\lambda_\\textup{min}} &\\quad &\\text{(Condition number of $\\vect{R}_u$)}\n\\end{alignat}\nIn adaptive filtering, some people use the term eigenvalue spread $\\chi(\\vect{R}_u)$ instead of the condition number. The MSE, the EMSE, the MSD, and the misadjustment depend on the adaptive filtering algorithm. Conversely, the remaining parameters depend on the desired signal and the input signal and are thus independent of the adaptive filtering algorithm. In Fig.~\\ref{fig:learning_curve_sgm}, some of the quantities are shown.\n\n\\section{LMS Revisited}\nRecall the steepest descent algorithm\n\\bmath\n \\vect{w}(n+1) = \\vect{w}(n)-\\frac{\\mu}{2}\\vect{g}(\\vect{w}(n))\n\\emath\nwhere\n\\bmath\n \\vect{g}(\\vect{w}(n)) = 2\\vect{R}_u\\vect{w}(n)-2\\vect{r}_{ud} = -2E[\\vect{u}(n)e(n)]\n\\emath\nis the gradient of $J_1(\\vect{w}(n))$. Replacing the gradient with the simple estimate \n\\bmath\n \\hat{\\vect{g}}(\\vect{w}(n)) = -2\\vect{u}(n)e(n)\n\\emath\nleads to the LMS algorithm\n\\begin{align}\n \\vect{w}(n+1) &= \\vect{w}(n) + \\mu\\vect{u}(n)e(n)\\\\\n &= (\\vect{I}-\\mu\\vect{u}(n)\\vect{u}^T(n))\\vect{w}(n) + \\mu\\vect{u}(n)d(n)\\ .\n\\end{align}\nIn order to analyse the LMS algorithm, we use the analysis model in Sec.~\\ref{ssec:analysis_model}.\n\n\\subsection{Transient Analysis}\nWe would like that\n\\begin{alignat*}{2}\n \\lim_{n\\to\\infty} E[\\Delta\\vect{w}(n)] &= 0 &\\qquad &\\text{(Convergence in mean)}\\\\\n \\lim_{n\\to\\infty} E[\\|\\Delta\\vect{w}(n)\\|^2] &= c < \\infty &\\qquad &\\text{(Convergence in mean-square)}\n\\end{alignat*}\n\\subsubsection{Convergence in the Mean}\nWe have that\n\\bmath\n \\vect{\\Delta}\\vect{w}(n+1) = (\\vect{I}-\\mu\\vect{u}(n)\\vect{u}^T(n))\\vect{\\Delta}\\vect{w}(n) + \\mu\\vect{u}(n)v(n)\n\\emath\nfrom which we get\n\\bmath\n E[\\vect{\\Delta}\\vect{w}(n+1)] = (\\vect{I}-\\mu\\vect{R}_u)E[\\vect{\\Delta}\\vect{w}(n)]\\ .\n\\emath\nThus, the LMS algorithm is stable in the mean if (same as for SD)\n\\bmath\n 0 < \\mu < \\frac{2}{\\lambda_\\textup{max}}\\ .\n\\emath\nSince $\\lambda_\\textup{max}$ is unknown and hard to estimate, we can bound it by\n\\bmath\n \\lambda_\\textup{max} \\leq \\textup{tr}(\\vect{R}_u) = ME[u^2(n)] \\leq M S_\\textup{max}\\ .\n\\emath\nEstimating the power $E[u^2(n)]$ or the maximum is the power spectral density $S_\\textup{max}$ is fairly easy.\n\n\\subsubsection{Convergence in the Mean-Square}\nWe assume that $\\vect{u}(n)$ has a Gaussian distribution with zero-mean and covariance matrix $\\vect{R}_u$. Then, as we have shown in Appendix~\\ref{app:lms_analysis}, we get that the LMS algorithm is stable in the mean-square if the step-size satisfies\n\\bmath\n f(\\mu) = \\frac{\\mu}{2}\\sum_{m=1}^M \\frac{\\lambda_m}{1-\\mu\\lambda_m} < 1\\ .\n\\emath\nFor small step-sizes $\\mu \\ll 1/\\lambda_\\textup{max}$, we have\n\\bmath\n f(\\mu) \\approx \\frac{\\mu}{2}\\sum_{m=1}^M\\lambda_m = \\frac{\\mu}{2}\\tr{\\vect{R}_u} < 1\\ .\n\\emath\nThus, stability in the mean-square requires that the step-size satisfies\n\\bmath\n 0 < \\mu < \\frac{2}{\\tr{\\vect{R}_u}}\\ .\n\\emath\n\n\\subsection{Steady-State Analysis}\nFrom Appendix~\\ref{app:lms_analysis}, we have that\n\\begin{alignat}{2}\n &\\text{EMSE:}&\\quad J_\\textup{ex} &= J_1(\\vect{w}(\\infty))-J_\\textup{min} = J_\\textup{min}\\frac{f(\\mu)}{1-f(\\mu)}\\\\\n & &\\quad &\\approx \\frac{\\mu}{2}J_\\textup{min}\\tr{\\vect{R}_u}\\\\\n &\\text{Misadjustment:}&\\quad \\mathcal{M} &= \\frac{ J_\\textup{ex}}{J_\\textup{min}} = \\frac{f(\\mu)}{1-f(\\mu)}\\\\\n & &\\quad &\\approx \\frac{\\mu}{2}\\tr{\\vect{R}_u}\\\\\n &\\text{MSD:} &\\quad E[\\|\\vect{\\Delta}\\vect{w}(\\infty)\\|^2] &\\approx \\frac{\\mu}{2} J_\\textup{min} M\n\\end{alignat}\nHere, the approximations are valid for small step-sizes. We see that the EMSE is approximately proportional to the input power. This is a problem and referred to as \\textit{gradient noise amplification}.\n\n\\section{Normalised LMS Adaptive Filters}\nRecall that we can bound the maximum eigenvalue by\n\\bmath\n \\lambda_\\textup{max} \\leq \\textup{tr}(\\vect{R}_u) = ME[u^2(n)]\\ .\n\\emath\nAn estimate of the power could be\n\\bmath\n \\hat{E}[u^2(n)] = \\frac{1}{M}\\vect{u}^T(n)\\vect{u}(n) = \\frac{1}{M}\\|\\vect{u}(n)\\|^2\\ .\n\\emath\nThis leads to the following bound on the step-size\n\\bmath\n 0 < \\mu < \\frac{2}{\\|\\vect{u}(n)\\|^2}\\quad\\iff\\quad 0 < \\mu\\|\\vect{u}(n)\\|^2 < 2\\ .\n\\emath\nWe may define the time-varying step-size\n\\bmath\n \\mu(n) = \\frac{\\beta}{\\|\\vect{u}(n)\\|^2}\\ ,\\qquad 0<\\beta<2\\ .\n\\emath\nThen, the LMS algorithm is\n\\bmath\n \\vect{w}(n+1) = \\vect{w}(n) + \\frac{\\beta}{\\|\\vect{u}(n)\\|^2}\\vect{u}(n)e(n)\n\\emath\nwhich is called the normalised LMS (NLMS) algorithm. The NLMS bypasses the problem of gradient noise amplification by normalising the gradient estimate by the input power. However, this normalisation causes numerical problems when the input power is close to zero. In order to avoid this, we introduce the regularisation parameter $\\epsilon$, which is a small positive constant, into the NLMS algorithm\n\\bmath\n \\boxed{\\vect{w}(n+1) = \\vect{w}(n) + \\frac{\\beta}{\\epsilon+\\|\\vect{u}(n)\\|^2}\\vect{u}(n)e(n)}\\ .\n\\emath\n\\subsubsection{Fast Computation}\nThe power estimate may be computed very efficiently by writing it in a recursive manner\n\\begin{align}\n \\|\\vect{u}(n)\\|^2 &= \\sum_{m=0}^{M-1}u^2(n-m) = u^2(n) - u^2(n-M) + \\sum_{m=1}^{M}u^2(n-m)\\notag\\\\\n &= u^2(n) - u^2(n-M) + \\|\\vect{u}(n-1)\\|^2\\ .\n\\end{align}\nIt should be noted that the use of this recursion can be problematic in practice due to accumulated rounding errors \\cite[p.~227]{Sayed2003}. These rounding errors may potentially cause the norm to be negative.\n\n\\subsection{Transient Analysis}\nIn order to analyse the NLMS algorithm, we use the analysis model in Sec.~\\ref{ssec:analysis_model}. If $\\sigma_v^2=0$, the NLMS algorithm converges in the mean and the mean-square if \\cite[p.~325]{Haykin2001}\n\\bmath\n \\boxed{0<\\beta<2}\\ .\n\\emath\nMoreover, the optimal step-size is\n\\bmath\n \\beta_o = 1\\ .\n\\emath\nIf $\\sigma_v^2>0$, the optimal step-size is more complicated and can be found in \\cite[p.~325]{Haykin2001}.\n\n\\subsection{Steady-State Analysis}\nAssuming the analysis model in Sec.~\\ref{ssec:analysis_model}, it can be shown that \\cite[pp.~300--302,p.~474]{Sayed2003}\n\\begin{alignat}{2}\n &\\text{EMSE:}&\\quad J_\\textup{ex} &= J_1(\\vect{w}(\\infty))-J_\\textup{min}\\notag\\\\\n & &\\quad &\\approx \\frac{\\beta}{2} J_\\textup{min}\\tr{\\vect{R}_u}E\\left[\\frac{1}{\\|\\vect{u}(n)\\|^2}\\right]\\geq\\frac{\\beta}{2} J_\\textup{min}\\\\\n &\\text{Misadjustment:}&\\quad \\mathcal{M} &= \\frac{ J_\\textup{ex}}{J_\\textup{min}} \\approx \\frac{\\beta}{2}\\tr{\\vect{R}_u}E\\left[\\frac{1}{\\|\\vect{u}(n)\\|^2}\\right]\\geq\\frac{\\beta}{2}\\\\\n &\\text{MSD:} &\\quad E[\\|\\vect{\\Delta}\\vect{w}(\\infty)\\|^2] &\\approx \\frac{\\beta}{2} J_\\textup{min} E\\left[\\frac{1}{\\|\\vect{u}(n)\\|^2}\\right] \\geq \\frac{\\beta J_\\textup{min}}{2\\tr{\\vect{R}_u}}\n\\end{alignat}\nwhere the inequality follows from that $E[x^{-1}]\\geq E[x]^{-1}$. The approximations are valid for small values of $\\beta$ and $\\epsilon$. Note that the approximation of the misadjustment no longer depends on the input power. This is a consequence of the normalisation of the LMS gradient with $\\|\\vect{u}(n)\\|^2$. Moreover, it should be noted that there exist several other approximations to the EMSE and the MSD than presented above \\cite[p.~474]{Sayed2003}.\n\n\\subsection{Computational Cost}\nTable~\\ref{tab:comp_cost_nlms} shows the computational cost of the NLMS algorithm in terms of the number of multiplications, additions or subtractions, and divisions. From the table, we see that the total number of flops is $6M+2$. If we use the fast computation of the normalisation constant, the computational cost is reduced to $4M+7$ flops as shown in Table~\\ref{tab:comp_cost_nlms_fast}. Thus, the NLMS algorithm has a linear complexity in the filter length $\\mathcal{O}(M)$ for both versions of the NLMS algorithm.\n\\begin{table}[htbp]\n \\centering\n \\begin{tabular}{l c c c}\n \\toprule\n Term & $\\times$ & $+$ or $-$ & $/$ \\\\\n \\midrule\n $\\|\\vect{u}(n)\\|^2$ & $M$ & $M-1$ & \\\\\n $\\vect{u}^T(n)\\vect{w}(n)$ & $M$ & $M-1$ & \\\\\n $e(n)=d(n)-\\vect{u}^T(n)\\vect{w}(n)$ & & $1$ & \\\\\n $\\beta e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & $1$ & $1$ & $1$\\\\\n $\\beta\\vect{u}(n)e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & $M$ & & \\\\\n $\\vect{w}(n)+\\beta \\vect{u}(n)e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & & $M$ &\\\\\n \\midrule\n Total & $3M+1$ & $3M$ & $1$ \\\\\n \\bottomrule\n \\end{tabular}\n \\caption{Computational cost of the NLMS algorithm.}\n \\label{tab:comp_cost_nlms}\n\\end{table}\n\\begin{table}[htbp]\n \\centering\n \\begin{tabular}{l c c c}\n \\toprule\n Term & $\\times$ & $+$ or $-$ & $/$ \\\\\n \\midrule\n $\\|\\vect{u}(n)\\|^2$ & 2 & 2 & \\\\\n $\\vect{u}^T(n)\\vect{w}(n)$ & $M$ & $M-1$ & \\\\\n $e(n)=d(n)-\\vect{u}^T(n)\\vect{w}(n)$ & & $1$ & \\\\\n $\\beta e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & $1$ & $1$ & $1$\\\\\n $\\beta\\vect{u}(n)e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & $M$ & & \\\\\n $\\vect{w}(n)+\\beta \\vect{u}(n)e(n)/(\\epsilon+\\|\\vect{u}(n)\\|^2)$ & & $M$ & \\\\\n \\midrule\n Total & $2M+3$ & $2M+3$ & $1$ \\\\\n \\bottomrule\n \\end{tabular}\n \\caption{Computational cost of the NLMS algorithm with fast update of the normalisation factor.}\n \\label{tab:comp_cost_nlms_fast}\n\\end{table}\n\n\\subsection{Another Derivation of the NLMS Algorithm}\nWe have two requirements to the adaptive filter.\n\\begin{enumerate}\n \\item If we filter the input sample at time $n$ through the filter at time $n+1$, the error should be zero. That is,\n \\bmath\n d(n) - \\vect{u}^T(n)\\vect{w}(n+1) = 0\\ .\n \\emath\n There are an infinite number of solutions that satisfy this requirement if $M>1$.\n \\item Among all the filter coefficient vectors $\\vect{w}(n+1)$ satisfying the first requirements, we select the vector resulting in the smallest change from iteration $n$ to $n+1$. That is, $\\|\\vect{w}(n+1)-\\vect{w}(n)\\|^2$ should be as small as possible.\n\\end{enumerate}\nThese two requirements lead to the following constrained minimisation problem\n\\begin{gather}\n \\begin{alignedat}{2}\n &\\min_{\\mathclap{\\vect{w}(n+1)}} &\\quad f(\\vect{w}(n+1))&=\\|\\vect{w}(n+1)-\\vect{w}(n)\\|^2\\\\\n &\\text{s.t.}&\\quad h(\\vect{w}(n+1)) &= d(n) - \\vect{u}^T(n)\\vect{w}(n+1) = 0\n \\end{alignedat}\n \\label{eq:nlms_opt}\n\\end{gather}\nThis optimisation problem has a quadratic cost function and a single linear equality constraint, and it is therefore on the same form as \\eq{eq:constr_opt_quad_lin} with\n\\begin{align}\n \\vect{x} &= \\vect{w}(n+1)\\\\\n \\vect{A} &= -\\vect{u}^T(n)\\\\\n \\vect{b} &= -d(n)\\\\\n \\vect{P} &= 2\\vect{I}\\\\\n \\vect{q} &= -2\\vect{w}(n)\\\\\n \\vect{r} &= \\|\\vect{w}(n)\\|^2\\ .\n\\end{align}\nThus, from \\eq{eq:constr_opt_quad_lin_minlam} and \\eq{eq:constr_opt_quad_lin_minx}, we readily obtain the solution to the optimisation problem in \\eq{eq:nlms_opt}. The solution is given by\n\\begin{align}\n \\lambda &= -\\frac{2}{\\|\\vect{u}(n)\\|^2}e(n)\\\\\n \\vect{w}(n+1) &= \\vect{w}(n)+\\frac{1}{\\|\\vect{u}(n)\\|^2}\\vect{u}(n)e(n)\\ .\n\\end{align}\nThe last equation is identical to the NLMS algorithm with $\\beta=1$ and $\\epsilon=0$.\n\n\\section{Affine Projection Adaptive Filters}\nWe may extend the requirements to the adaptive filter in the following way.\n\\begin{enumerate}\n \\item If we filter the input sample at time $n-k$ for $k=0,1,\\ldots,K-11$, The computational complexity of the APA is higher than that of the NLMS algorithm since we have to invert a $K\\times K$ matrix. Although there exist fast ways of doing this \\cite[pp.~339--340]{Haykin2001}, the APA is more expensive than the NLMS algorithm.\n\n\\subsection{Transient Analysis}\nIn order to analyse the APA, we use the analysis model in Sec.~\\ref{ssec:analysis_model}. If $\\sigma_v^2=0$, the APA converges in the mean and the mean-square if \\cite[p.~337]{Haykin2001}\n\\bmath\n \\boxed{0<\\beta<2}\\ .\n\\emath\nMoreover, the optimal step-size is\n\\bmath\n \\beta_o = 1\\ .\n\\emath\nIf $\\sigma_v^2>0$, the optimal step-size is more complicated and can be found in \\cite[p.~337]{Haykin2001}.\n\n\\subsection{Steady-State Analysis}\nAssuming the analysis model in Sec.~\\ref{ssec:analysis_model}, it can be shown that \\cite[p.~327]{Sayed2003}\n\\begin{alignat}{2}\n &\\text{EMSE:}&\\quad J_\\textup{ex} &= J_1(\\vect{w}(\\infty))-J_\\textup{min}\\notag\\\\\n & &\\quad &\\approx \\frac{\\beta}{2} J_\\textup{min}\\tr{\\vect{R}_u}E\\left[\\frac{K}{\\|\\vect{u}(n)\\|^2}\\right]\\geq\\frac{\\beta}{2} J_\\textup{min}K\\\\\n &\\text{Misadjustment:}&\\quad \\mathcal{M} &= \\frac{ J_\\textup{ex}}{J_\\textup{min}} \\approx \\frac{\\beta}{2}\\tr{\\vect{R}_u}E\\left[\\frac{K}{\\|\\vect{u}(n)\\|^2}\\right]\\geq\\frac{\\beta K}{2}\n\\end{alignat}\nwhere the inequality follows from that $E[x^{-1}]\\geq E[x]^{-1}$. The approximations are valid for small values of $\\beta$ and $\\epsilon$. It should be noted that there exist several other approximations to the EMSE than presented above \\cite[p.~325]{Sayed2003}. For more accurate expressions and for an expression for the mean-square deviation, see \\cite[pp.~510--512]{Sayed2003} and \\cite{Shin2004}.\n\n\\subsection{Computational Cost}\nTable~\\ref{tab:comp_cost_apa} shows the computational cost of the APA algorithm in terms of the number of multiplications and additions or subtractions. We have assumed that the cost of inverting a $K\\times K$ matrix is $K^3$ multiplications and additions \\cite[p.~240]{Sayed2003}. From the table, we see that the total number of flops is $2M(K^2+2K)+2K^3+K^2+M$. Thus, the APA has a complexity of $\\mathcal{O}(MK^2)$. Note that there exist faster ways of implementing the APA \\cite[pp.~339-340]{Haykin2001}.\n\\begin{table}[htbp]\n \\centering\n \\begin{tabular}{@{}l c c c@{}}\n \\toprule\n Term & $\\times$ & $+$ or $-$ \\\\\n \\midrule\n $\\vect{U}^T(n)\\vect{w}(n)$ & $MK$ & $(M-1)K$ \\\\\n $\\vect{e}(n)=\\vect{d}(n)-\\vect{U}^T(n)\\vect{w}(n)$ & & $K$ \\\\\n $\\vect{U}^T(n)\\vect{U}(n)$ & $MK^2$ & $(M-1)K^2$ \\\\\n $\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n)$ & & $K$ \\\\\n $(\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n))^{-1}$ & $K^3$ & $K^3$ \\\\\n $(\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n))^{-1}\\vect{e}(n)$ & $K^2$ & $K(K-1)$ \\\\\n $\\vect{U}(n)(\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n))^{-1}\\vect{e}(n)$ & $MK$ & $M(K-1)$ \\\\\n $\\beta\\vect{U}(n)(\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n))^{-1}\\vect{e}(n)$ & $M$ & \\\\\n $\\vect{w}(n)+\\beta\\vect{U}(n)(\\epsilon\\vect{I}+\\vect{U}^T(n)\\vect{U}(n))^{-1}\\vect{e}(n)$ & & $M$\\\\\n \\midrule\n Total & $M(K^2+2K+1)+K^3+K^2$ & $M(K^2+2K)+K^3$\\\\\n \\bottomrule\n \\end{tabular}\n \\caption{Computational cost of the APA.}\n \\label{tab:comp_cost_apa}\n\\end{table}\n", "meta": {"hexsha": "7fcf8ab77e8c2d416759ecf9937dcff56dc3f896", "size": 28736, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture_notes/lectures/NLMS_APA.tex", "max_stars_repo_name": "jkjaer/adaptiveFilteringLectureNotes", "max_stars_repo_head_hexsha": "194706662078f810c163e403548395a532471d0c", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-07-22T19:46:04.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-21T07:10:32.000Z", "max_issues_repo_path": "lecture_notes/lectures/NLMS_APA.tex", "max_issues_repo_name": "jkjaer/adaptiveFilteringLectureNotes", "max_issues_repo_head_hexsha": "194706662078f810c163e403548395a532471d0c", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture_notes/lectures/NLMS_APA.tex", "max_forks_repo_name": "jkjaer/adaptiveFilteringLectureNotes", "max_forks_repo_head_hexsha": "194706662078f810c163e403548395a532471d0c", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.702811245, "max_line_length": 842, "alphanum_fraction": 0.6689518374, "num_tokens": 10218, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529375, "lm_q2_score": 0.7853085758631159, "lm_q1q2_score": 0.6146315567280273}} {"text": "% This is part of the TFTB Reference Manual.\n% Copyright (C) 1996 CNRS (France) and Rice University (US).\n% See the file refguide.tex for copying conditions.\n\n\n\n\\markright{locfreq}\n\\section*{\\hspace*{-1.6cm} locfreq}\n\n\\vspace*{-.4cm}\n\\hspace*{-1.6cm}\\rule[0in]{16.5cm}{.02cm}\n\\vspace*{.2cm}\n\n\n\n{\\bf \\large \\sf Purpose}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\nFrequency localization characteristics.\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Synopsis}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\n[fm,B] = locfreq(x)\n\\end{verbatim}\n\\end{minipage}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf Description}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n {\\ty locfreq} computes the frequency localization characteristics of\n signal {\\ty x}. The definition used for the averaged frequency\n and the frequency spreading are the following\\,:\n\\begin{eqnarray*}\nf_m &=& \\frac{1}{E_x}\\ \\int_{-\\infty}^{+\\infty} \\nu\\ |X(\\nu)|^2\\ d\\nu\\\\\nB &=& 2\\ \\sqrt{\\frac{\\pi}{E_x}\\ \\int_{-\\infty}^{+\\infty} (\\nu-f_m)^2\\\n|X(\\nu)|^2\\ d\\nu }\n\\end{eqnarray*}\nwhere $E_x$ is the energy of the signal and $X(\\nu)$ the Fourier transform\nof $x(t)$. With this definition (and the one used in {\\ty loctime}), the\nHeisenberg-Gabor inequality writes $B\\ T\\geq 1$.\\\\\n\n\\hspace*{-.5cm}\\begin{tabular*}{14cm}{p{1.5cm} p{8.5cm} c}\nName & Description & Default value\\\\\n\\hline\n {\\ty x} & signal\\\\\n\\hline {\\ty fm} & averaged normalized frequency center\\\\\n {\\ty B} & frequency spreading\\\\\n\\hline\n\\end{tabular*}\n\n\\end{minipage}\n\n\\vspace*{1cm}\n\n{\\bf \\large \\sf Example}\n\\begin{verbatim}\n z=amgauss(160,80,50).*fmconst(160,0.2);\n [fm,B]=locfreq(z); [fm,B]\n ans = \n 0.2000 0.0200\n\\end{verbatim}\n\\vspace*{.5cm}\n\n\n{\\bf \\large \\sf See Also}\\\\\n\\hspace*{1.5cm}\n\\begin{minipage}[t]{13.5cm}\n\\begin{verbatim}\nloctime.\n\\end{verbatim}\n\\end{minipage}\n\n\n\n\n", "meta": {"hexsha": "436e2295563339de03059607cf9a31c67cb07a24", "size": 1899, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tftb/refguide/locfreq.tex", "max_stars_repo_name": "sangyoonHan/extern", "max_stars_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 50, "max_stars_repo_stars_event_min_datetime": "2018-03-28T01:50:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-01T07:24:14.000Z", "max_issues_repo_path": "tftb/refguide/locfreq.tex", "max_issues_repo_name": "sangyoonHan/extern", "max_issues_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tftb/refguide/locfreq.tex", "max_forks_repo_name": "sangyoonHan/extern", "max_forks_repo_head_hexsha": "a3c874538a7262b895b60d3c4d493e5b34cf81f8", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": 21, "max_forks_repo_forks_event_min_datetime": "2018-03-28T01:50:04.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T09:09:40.000Z", "avg_line_length": 22.8795180723, "max_line_length": 76, "alphanum_fraction": 0.6345444971, "num_tokens": 699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.7826624738835051, "lm_q1q2_score": 0.6146315566795574}} {"text": "%!TEX root = ../notes.tex\n\\section{May 3, 2022}\n\\subsection{Minkowski, Lagrange, and Waring Walk Into a Bar\\dots}\n\\subsubsection{Four Squares Theorem and Waring's Problem}\n\\begin{theorem}[Lagrange, 1770]\n Every nonnegative integer can be written as a sum of four square integers.\n\\end{theorem}\nIn the same year, Waring asserted, in his book, that\n\\begin{theorem}[Waring, 1770]\n For every integer $k\\geq 2$, there is a $g(k)$ such that every nonnegative integer can be written as a sum of at most $g(k)$ $k$th powers.\n\\end{theorem}\nHe claimed that $g(3) = 9$ and $g(4) = 19$. We note that $23$ and $239$ require $9$ cubes, and $79$ requires $19$ fourth powers. $g(3) = 9$ is a result of Wieferich-Kemper (1909) and $g(4) = 19$ is a result of Balasubramanian, Dress, Deshouillers (1986). Hilbert proved this theorem in $1909$.\n\nThere is also a `capital $G$' version of this question, which is an asymptotic best bound of $g(k)$.\n\n$4 \\leq G(3) \\leq 7$. This is currently unknown.\n\n$G(4) = 16$, which we do know. So, your mileage may very much vary on these.\n\n\\subsubsection{Lattices and Minkowski's Theorem}\n\\begin{definition}[Lattice]\n Let $e_1, e_2, \\dots, e_n$ be a set of basis vectors for $\\RR^n$. Then the additive subgroup of $(\\RR^n, +)$ generated by $e_1, e_2, \\dots, e_n$ is called a \\ul{lattice}.\n\\end{definition}\n\\begin{example}\n The most obvious lattice in $\\RR^n$ is $\\ZZ^n$, where we take $e_i$ to be the standard basis vectors.\n\n We can take $\\alpha\\ZZ^n$ by some $0\\neq \\alpha\\in\\RR$ is also an obvious lattice.\n\n We can also take things like $\\frac{1}{2}\\ZZ\\times \\ZZ\\subseteq\\RR^2$.\n\\end{example}\n\n\\begin{definition}[Fundamental Domain]\n If $L$ is a lattice generated by $e_1, e_2, \\dots, e_n$ in $\\RR^n$, then the \\ul{fundamental domain} of $L$ is the set\n \\[\\left\\{ \\sum a_i e_i \\mid a_i\\in\\RR, 0\\leq a_i < 1 \\right\\}\\]\n\\end{definition}\n\\begin{example}\n If we take $\\ZZ^2\\subseteq\\RR^2$, then the fundamental domain is the square with corners $(0, 0)$ to $(1, 1)$, with some dotted lines.\n\\end{example}\n\\begin{definition}[Convex, Symmetric]\n A set $X\\subseteq\\RR^n$ is \\ul{convex} if forall $x, y\\in X$,\n \\[\\lambda x + (1 - \\lambda)y \\in X\\]\n for all $0\\leq \\lambda \\leq 1$.\n\n A set $X$ is \\ul{symmetric} if $x\\in X$ implies $-x\\in X$.\n\\end{definition}\n\\begin{example}\n A triangle is convex but not symmetric. An ellipse about the origin is convex and symmetric. An annulus is symmetric but not convex.\n \\begin{center}\n \\includegraphics[width=0.9\\textwidth]{images/shapes.png}\n \\end{center}\n\\end{example}\n\\begin{theorem}[Minkowski, p.140 \\cite{stewart2015algebraic}]\n Let $L$ be an $n$-dimensional lattice in $\\RR^n$ with fundamental domain $T$, and let $X$ be a bounded symmetric convex subset of $\\RR^n$. If\n \\[\\Vol(X) > 2^n\\Vol(T)\\]\n then $X$ contains a non-zero point of $L$.\n\\end{theorem}\n\\begin{example}\n If we have the $\\ZZ^2$ example, if we have a `dotted' open square with area $4$, we just about not contain any nonzero points. If we add anything in, we'll have a nontrivial point.\n \\begin{center}\n \\includegraphics[width=0.4\\textwidth]{images/square.png}\n \\end{center}\n\\end{example}\n\n\\begin{proof}[Proof of 4 squares theorem using Minkowski's Theorem]\n We first prove this statement for primes, then extend to all positive integers. We first have\n \\[2 = 1^2 + 1^2 + 0^2 + 0^2.\\]\n So suppose we're trying to prove $p\\in\\ZZ_+$ is an odd prime.\n \\begin{claim*}\n The equation $r^2 + s^2 + 1\\equiv 0\\pmod{p}$ has a solution $(r, s)\\in\\ZZ^2$.\n \\end{claim*}\n \\emph{Why?} Every element of $\\ZZ/p\\ZZ$ is a sum of $2$ squares\\dots\n\n So let us select such an $r, s$. Consider the lattice $\\Lambda\\subseteq\\ZZ^4$ that is given by\n \\[\\Lambda = A\\ZZ^4,\\]\n where\n \\[A = \\begin{pmatrix}\n p & 0 & r & s \\\\\n 0 & p & s & -r \\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1\n \\end{pmatrix}.\\]\n We make an observation of the particular nature of the points in this lattice.\n\n If $\\bvec{t}= (t_1, t_2, t_3, t_4)\\in\\ZZ^4$, and $\\bvec{x} = (x_1, x_2, x_3, x_4)$ where $\\bvec{x} = A\\bvec{t}$, then\n \\begin{align*}\n x_1^2 + x_2^2 + x_3^2 + x_4^2\n & = (pt_1 + rt_3 + st_4)^2 + (pt_2 + st_3 - rt_4)^2 + t_3^2 + t_4^2 \\\\\n & \\equiv (rt_3 + st_4)^2 + (st_3 - rt_4)^2 + t_3^2 + t_4^2\\pmod{p} \\\\\n & = (1 + r^2 + s^2)(t_3^2 + t_4^2)\\pmod{p} \\\\\n & = 0\\pmod{p}\n \\end{align*}\n Geometrically, the idea is if we can get the norm to be small enough, we know the sum is exactly $p$.\n\n Let's find an appropriate radius for our ball. A $4$-dimensional ball of radius $R$ has volume\n \\[\\pi^2R^4/2\\]\n choose $R$ such that\n \\begin{align*}\n 16p^2 & < \\frac{\\pi^2 R^4}{2}\\approx 4.93 R^4 \\\\\n 4p & < \\approx 2.22 R^2 \\\\\n 2p & < \\approx 1.11 R^2\n \\end{align*}\n The next condition is $R^2 < 2p$. So we want $R$ such that\n \\[R^2 < 2p < \\approx 1.11 R^2\\]\n If we take $R^2 = 1.9p$ works.\n\n Let $X$ be the ball centered at the origin with radius $\\sqrt{1.9p}$. Let $T$ be the fundamental domain for $\\Lambda = A\\ZZ^4$. Then we have $\\Vol(T) = \\det(A) = p^2$.\n\n Then $\\Vol(X) > 2^n\\Vol(T) = 16p^2$. Applying Minkowski's theorem, we conclude that $X$ \\emph{must} contain some nonzero lattice point of $\\Lambda$.\n\n If we let $\\bvec{x} = (x_1, x_2, x_3, x_4)$ be this point. Then since $\\bvec{x}\\in X$, so\n \\[x_1^2 + x_2^2 + x_3^2 + x_4^2 < R^2 < 2p.\\]\n so $p\\mid x_1^2 + x_2^2 + x_3^2 + x_4^2$ and $x_1^2 + x_2^2 + x_3^2 + x_4^2 < 2p$, and it's a nonzero point, so this forces\n \\[x_1^2 + x_2^2 + x_3^2 + x_4^2 = p,\\]\n which is as desired for primes.\n\n In the arbitrary case for arbitrary $n$, note that\n \\begin{align*}\n & (a^2 + b^2 + c^2 + d^2)(A^2 + B^2 + c^2 + D^2) \\\\\n & = (aA - bB - cC - dD)^2 + (aB + bA + cD - dC)^2 + \\\\\n & \\phantom{=}\\ \\ (aC - bD + cA + dB)^2 + (aD + bC - cB + dA)^2.\n \\end{align*}\n so composite $n$ is also a sum of 2 squares.\n\\end{proof}\n\n\\begin{center}\n \\emph{\\dots and that's the course! }\n\\end{center}\n", "meta": {"hexsha": "5b22276fbc936b2fcaaeead172c4eb196aa7b8de", "size": 6197, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lectures/2022-05-03.tex", "max_stars_repo_name": "jchen/math1560-notes", "max_stars_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_stars_repo_licenses": ["BSL-1.0"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2022-02-02T15:41:56.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-03T20:28:48.000Z", "max_issues_repo_path": "lectures/2022-05-03.tex", "max_issues_repo_name": "jchen/math1560-notes", "max_issues_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_issues_repo_licenses": ["BSL-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lectures/2022-05-03.tex", "max_forks_repo_name": "jchen/math1560-notes", "max_forks_repo_head_hexsha": "a3605894c69d4e3dd7f90829523ff3ec3c73a6f4", "max_forks_repo_licenses": ["BSL-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.4140625, "max_line_length": 293, "alphanum_fraction": 0.6078747781, "num_tokens": 2293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624688140728, "lm_q2_score": 0.7853085859124002, "lm_q1q2_score": 0.6146315566310875}} {"text": "\\documentclass[aps,superscriptaddress,preprint,amsmath,floatfix,byrevtex]{article}\n\\usepackage[pdftex]{graphicx}\n\\usepackage{textcomp}\n\\usepackage{amsmath}\n\\usepackage{dcolumn}\n\n\\oddsidemargin 0.1in\n\\evensidemargin 0.1in\n\\textwidth 6.7in\n\n\\newcommand{\\lrbar}{\\bar{l}_{\\rho}^{\\mathrm{2D}}}\n\n\\begin{document}\n\\title{Derivation of the mean radial displacement, $\\lrbar$, in 2 dimensions}\n\\author{Markus Dittrich}\n\\date{\\today}\n\n%\\pacs{}\n\\maketitle\n\n\\section{Introduction}\nIn the following we derive an expression for the mean radial displacement, $\\lrbar$,\nfor diffusing 2D molecules in analogy to the well known expression for $\\bar{l}_r$\nin three dimensions. \n\n\n\\section{Derivation}\nStarting point is the solution of the diffusion equation in 2 dimensions\n\\begin{equation}\n\\frac{\\partial c(\\rho,t)}{\\partial t} = D \\nabla^2 c(\\rho,t)\n\\end{equation}\nwhich, for a point source of $M$ molecules released at the origin at time $t=0$ \ncan be shown to be\n\\begin{equation}\nc(\\rho,t) = \\frac{M}{4 \\pi Dt} e^{- \\frac{\\rho^2}{4Dt}}\n\\end{equation}\n(see also Eqs. 3.1, 3.2 in \\cite{KERR2008}). Therefore, the probability\ndensity for a molecule being in a radial shell of thickness $d\\rho$ around the \norigin is given by\n\\begin{equation}\np^{\\mathrm{2D}} (\\rho,t) = \\frac{(2 \\pi \\rho) d\\rho}{4\\pi Dt} e^{- \\frac{\\rho^2}{4Dt}} \n\\end{equation} \nor, in normalized coordinates $\\tilde{\\rho} = \\frac{\\rho}{\\lambda}$ with the \nnormalization constant $\\lambda = \\sqrt{4Dt}$\n\\begin{equation}\np^{\\mathrm{2D}} (\\tilde{\\rho},t) = 2 e^{-\\tilde{\\rho}^2} (\\tilde{\\rho} d\\tilde{\\rho})\n \\;\\;\\;\\; .\n\\end{equation}\nHence, the mean radial displacement of a 2D molecule is given by\n\\begin{equation}\n\\lrbar = 2 \\lambda \\int_0^{\\infty} \\tilde{\\rho}^2 e^{- \\tilde{\\rho}^2} d\\tilde{\\rho}\n\\end{equation}\nSince\n\\begin{eqnarray}\n\\int_0^{\\infty} t^{2n} e^{-at^2} dt &=& \\frac{\\Gamma(n+\\frac{1}{2})}{2a^{n+\\frac{1}{2}}} \n \\nonumber \\\\\n\\Rightarrow \\int_0^{\\infty} \\tilde{\\rho}^2 e^{- \\tilde{\\rho}^2} d\\tilde{\\rho} &=& \n \\frac{\\Gamma(\\frac{3}{2})}{2} = \\frac{\\sqrt{\\pi}}{4} \n\\end{eqnarray}\nand from this\n\n\\begin{equation}\n\\boxed{\n\\lrbar = 2 \\lambda \\frac{\\sqrt{\\pi}}{4} = \\sqrt{\\pi D t}\n}\n\\end{equation}\n\n\\begin{thebibliography}{99}\n\\bibitem{KERR2008} R. Kerr, T.M. Bartol, B. Kaminsky, M. Dittrich, J.C.J. Chang, S. Baden, \nT.J. Sejnowski, and J.R. Stiles, Fast Monte Carlo Simulation Methods for Biological \nReaction-Diffusion Systems in Solution and on Surfaces (2008),\nSIAM J. Sci. Comput., 30:3126-3149. \n\\end{thebibliography}\n\n\\end{document}\n", "meta": {"hexsha": "bd5db12adb973febafadbf82b6e3b418f8cb53ac", "size": 2517, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/lr_bar_2D.tex", "max_stars_repo_name": "mcellteam/mcell", "max_stars_repo_head_hexsha": "3920aec22c55013b78f7d6483b81f70a0d564d22", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2015-03-25T16:36:01.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-17T14:28:43.000Z", "max_issues_repo_path": "docs/lr_bar_2D.tex", "max_issues_repo_name": "mcellteam/mcell", "max_issues_repo_head_hexsha": "3920aec22c55013b78f7d6483b81f70a0d564d22", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 31, "max_issues_repo_issues_event_min_datetime": "2015-02-12T22:15:18.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-30T22:43:24.000Z", "max_forks_repo_path": "docs/lr_bar_2D.tex", "max_forks_repo_name": "mcellteam/mcell", "max_forks_repo_head_hexsha": "3920aec22c55013b78f7d6483b81f70a0d564d22", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 12, "max_forks_repo_forks_event_min_datetime": "2016-01-15T23:20:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-02-10T06:18:00.000Z", "avg_line_length": 33.1184210526, "max_line_length": 91, "alphanum_fraction": 0.687326182, "num_tokens": 923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085708384736, "lm_q2_score": 0.7826624688140726, "lm_q1q2_score": 0.6146315448332909}} {"text": "\\section{Lines}\nIn this section we will discuss how to represent lines in 3D, and some related problems, including finding the line at the intersection of two planes, and finding the intersection point of a plane and a line.\n\n\\subsection{Line representation}\nUnlike 2D lines and planes, 3D lines don't have a nice representation as a single equation on coordinates like $ax+by=c$ or $ax+by+cz=d$. We could represent them as the intersection of two planes, like\n\\begin{align*}\na_1x+b_1y+c_1z &= d_1 \\\\\na_2x+b_2y+c_2z &= d_2\n\\end{align*}\nbut this representation is not very convenient to work with, and there are many pairs planes we could choose.\n\n\\centerFig{line3d-0}\n\nInstead, we will work with a parametric representation: we take a point $O$ on the line and a vector $\\vv{d}$ parallel to the line and say that the points belonging to the line are all points\n\\[P = O + k \\vv{d}\\]\nwhere $k$ is a real parameter.\n\n\\centerFig{line3d-1}\n\nNote that here, $\\vv{d}$ plays the same role as $\\vv{v}$ did for 2D lines (see section~\\ref{ss:line-rep}).\n\nIf we are given two points $P,Q$ on the line, we can set $O=P$ and $\\vv{d}=\\vv{PQ}$, so we implement the structure like this:\n\\begin{lstlisting}\nstruct line3d {\n p3 d, o;\n // From two points P, Q\n line3d(p3 p, p3 q) : d(q-p), o(p) {}\n // From two planes p1, p2 (requires T = double)\n line3d(plane p1, plane p2); // will be implemented later\n \n // Will be defined later:\n // - these work with T = int\n double sqDist(p3 p);\n double dist(p3 p);\n bool cmpProj(p3 p, p3 q);\n // - these require T = double\n p3 proj(p3 p);\n p3 refl(p3 p);\n p3 inter(plane p) {return o - d*p.side(o)/(d|p.n);}\n};\n\\end{lstlisting}\n\n\\subsection{Distance from a line}\nA point $P$ is on a line $l$ described by $O,\\vv{d}$ if and only if $\\vv{OP}$ is parallel to $\\vv{d}$. This is the case when $\\crossv{d}{OP} = \\vv{0}$.\n\n\\centerFig{line3d-2}\n\nMore generally, this product $\\crossv{d}{OP}$ also gives us information about the distance to the line: the distance from $l$ to $P$ is equal to\n\\[\\frac{\\big\\|\\crossv{d}{OP}\\big\\|}{\\big\\|\\vv{d}\\big\\|}\\]\n%Note that this expression also works for 2D lines; indeed, the norm of $\\crossv{d}{OP}$ is equivalent to the \n\\begin{lstlisting}\ndouble sqDist(p3 p) {return sq(d*(p-o))/sq(d);}\ndouble dist(p3 p) {return sqrt(sqDist(p));}\n\\end{lstlisting}\n\n\\subsection{Sorting along a line}\\label{ss:sort-line3d}\nJust like we did with 2D lines in section~\\ref{ss:sort-line}, we can sort points according to their position along a line $l$. To find out if a point $P$ should come before another point $Q$, we simply need to check whether\n\\[\\vv{d}\\cdot P < \\vv{d}\\cdot Q\\]\nso we can use the following comparator:\n\\begin{lstlisting}\nbool cmpProj(p3 p, p3 q) {return (d|p) < (d|q);}\n\\end{lstlisting}\n\n\\subsection{Orthogonal projection and reflection}\nLet's say we want to project $P$ on a line $l$, that is find the closest point to $P$ on $l$.\nOur usual approach to projecting things, which is to move $P$ perpendicularly until it touches $l$, doesn't work as well here: indeed, there are many possible directions that are perpendicular to $l$.\n\n\\centerFig{line3d-3}\n\nInstead we will start from $O$ and move along the line until we reach the projection of $P$. We have seen in the previous section that taking the dot product with $\\vv{d}$ tells us how far some point is along $l$. In fact, if we compute $\\dotv{d}{OP}$, this tells us the (signed) distance from $O$ to the projection of $P$, multiplied by $\\normv{d}$. So we can find the projection this way:\n\\begin{lstlisting}\np3 proj(p3 p) {return o + d*(d|(p-o))/sq(d);}\n\\end{lstlisting}\n\nOnce we've found the projection $P'$, we can find the reflection $P''$ easily, since it is twice as far in the same direction: we have $P'' = P' + \\vv{PP'}$, which becomes $2P'-P$ if we allow vector operations on points.\n\\begin{lstlisting}\np3 refl(p3 p) {return proj(p)*2 - p;}\n\\end{lstlisting}\n\n\\subsection{Plane-line intersection}\\label{ss:plane-line}\nLet's say we have a plane $\\Pi$, represented by vector $\\vv{n}$ and real $d$, and a line $l$, represented by point $O$ and $\\vv{d}$. To find an intersection between them, we need to find a point $O+k\\vv{d}$ that lies on $\\Pi$, that is, such that\n\\[\\vv{n}\\cdot\\left(O+k\\vv{d}\\right) = d\\]\n\n\\centerFig{line3d-4}\n\nSolving for $k$, we find\n\\[k = \\frac{d - \\vv{n}\\cdot O}{\\dotv{n}{d}} = \\frac{-\\side_{\\Pi}(O)}{\\dotv{n}{d}}\\]\nwhich can be implemented directly:\n\\begin{lstlisting}\np3 inter(plane p) {return o - d*p.side(o)/(p.n|d);}\n\\end{lstlisting}\n\nNote that this is undefined when $\\dotv{n}{d}=0$, that is, when $\\Pi$ and $l$ are parallel.\\footnote{We take a closer look at criteria for parallelism and perpendicularity in section~\\ref{s:angles}.}\n \n\\subsection{Plane-plane intersection}\nIf we are given two non-parallel planes $\\Pi_1$ and $\\Pi_2$ (defined by $\\vv{n_1}$, $d_1$ and $\\vv{n_2}$, $d_2$), how can we find their common line?\n\nFirst we need to find its direction $\\vv{d}$. Clearly, the direction needs to be parallel to both planes, so it must be perpendicular to both $\\vv{n_1}$ and $\\vv{n_2}$. Thus we take $\\vv{d} = \\crossv{n_1}{n_2}$.\n\nThen we need to find an arbitrary point $O$ that is on both planes. Here, we will actually compute the closest such point to the origin. It is given by\n\\[O = \\frac{\\big(d_1\\vv{n_2} - d_2\\vv{n_1}\\big)\\times\\vv{d}}{\\big\\|\\vv{d}\\big\\|^2}\\]\n\nLet's analyze this expression. It is the sum of two vectors,\n\\begin{align*}\n\\vv{v_1} &= \\frac{d_1}{\\big\\|\\vv{d}\\big\\|^2}\\left(\\crossv{n_2}{d}\\right) \\\\\n\\vv{v_2} &= -\\frac{d_2}{\\big\\|\\vv{d}\\big\\|^2}\\left(\\crossv{n_1}{d}\\right)\n\\end{align*}\nBecause $\\vv{v_1}$ is perpendicular to $\\vv{n_2}$, it is parallel to $\\Pi_2$, while $\\vv{v_2}$ is perpendicular to $\\vv{n_1}$ and thus parallel to $\\Pi_1$. So we can see $\\vv{v_1}$ as the vector that leads from the origin to $\\Pi_1$ while staying parallel to $\\Pi_2$, and $\\vv{v_2}$ as the vector that leads from the origin to $\\Pi_2$ while staying parallel to $\\Pi_1$.\n\n\\centerFig{line3d-5}\n\nLet's verify that $O$ is on $\\Pi_1$, that is, $\\vv{n_1}\\cdot O = \\vv{n_1}\\cdot(\\vv{v_1} + \\vv{v_2}) = d_1$. Since $\\vv{v_2}$ is perpendicular to $\\vv{n_1}$, clearly $\\dotv{n_1}{v_2} = 0$. What remains is\n\\begin{align*}\n\\dotv{n_1}{v_1}\n&= \\frac{d_1}{\\big\\|\\vv{d}\\big\\|^2}\\left(\\crossv{n_2}{d}\\right)\\cdot\\vv{n_1} \\\\\n&= \\frac{d_1}{\\big\\|\\vv{d}\\big\\|^2}\\Big(\\crossv{n_1}{n_2}\\Big)\\cdot\\vv{d} \\\\\n&= \\frac{d_1}{\\big\\|\\vv{d}\\big\\|^2}\\left(\\dotv{d}{d}\\right) \\\\\n&= d_1\n\\end{align*}\nwhere between the first two lines, we used the fact that the mixed product is conserved if we swap its arguments in a circular way (see exercise~\\ref{ex:mixed-props}).\n\nWe prove similarly that $O$ is on $\\Pi_2$. Finally we note that both $\\vv{v_1}$ and $\\vv{v_2}$ are perpendicular to $\\vv{d}$, so the vector from the origin to $O$ is perpencular to $\\vv{d}$, the direction of the line. Therefore it must necessarily arrive on the line at the closest point to the origin.\n\nWe can implement this as the following constructor:\n\\begin{lstlisting}\nline3d(plane p1, plane p2) {\n d = p1.n*p2.n;\n o = (p2.n*p1.d - p1.n*p2.d)*d/sq(d);\n}\n\\end{lstlisting}\n\n\\subsection{Line-line distance and nearest points}\nConsider line $l_1$ defined by $O_1 + k\\vv{d_1}$ and line $l_2$ defined by $O_2 + k\\vv{d_2}$.\nIf they are parallel, the distance between them is easy to find: just find the distance from $l_1$ to $O_2$.\nOtherwise, they are either intersecting or skew, in which case the question is a bit more complex.\n\n\\centerFig{line3d-6}\n\nLet's call $C_1$ the point of $l_1$ that is closest to $l_2$, and $C_2$ the point on $l_2$ that is closest to $l_1$.\nDirection $\\vv{C_1C_2}$ should be perpendicular to both $l_1$ and $l_2$.\nIndeed, if it were not, it would be possible to get a smaller distance by moving either $C_1$ or $C_2$.\nSo $\\vv{C_1C_2}$ is parallel to $\\vv{n}=\\crossv{d_1}{d_2}$.\n\n\\centerFig{line3d-7}\n\nBecause of this, we could compute the distance as\n\\[|C_1C_2| = \\frac{\\big|\\dotv{C_1C_2}{n}\\big|}{\\normv{n}}\\]\nWe don't know $C_1$ or $C_2$ yet, but since the dot product doesn't change when one of the vectors moves perpendicular to the other, we can move $C_1$ to $O_1$ and $C_2$ to $O_2$, so that we get\n\\[\\frac{\\big|\\dotv{C_1C_2}{n}\\big|}{\\normv{n}} = \\frac{\\big|\\dotv{O_1C_2}{n}\\big|}{\\normv{n}} = \\frac{\\big|\\dotv{O_1O_2}{n}\\big|}{\\normv{n}}\\]\nwhich gives the following implementation:\n\\begin{lstlisting}\ndouble dist(line3d l1, line3d l2) {\n p3 n = l1.d*l2.d;\n if (n == zero) // parallel\n return l1.dist(l2.o);\n return abs((l2.o-l1.o)|n)/abs(n);\n}\n\\end{lstlisting}\n\nNow, how can we find $C_1$ and $C_2$ themselves? Let's call $\\Pi_2$ the plane that contains $l_2$ and is parallel to $n_2$; its normal is $\\vv{n_2} = \\crossv{d_2}{n}$. Then $C_1$ is the intersection between that plane and line $l_1$, so we can use the formula obtained in section~\\ref{ss:plane-line} to find it:\n\\[C_1 = O_1 + \\frac{\\dotv{O_1O_2}{n_2}}{\\dotv{d_1}{n_2}}\\]\n\n\\centerFig{line3d-8}\n\nThis is implemented in a straightforward way. If we want $C_2$ instead we just have to swap $l_1$ and $l_2$.\n\\begin{lstlisting}\np3 closestOnL1(line3d l1, line3d l2) {\n p3 n2 = l2.d*(l1.d*l2.d);\n return l1.o + l1.d*((l2.o-l1.o)|n2)/(l1.d|n2);\n}\n\\end{lstlisting}\n", "meta": {"hexsha": "99deabfcb028cf590b19bd076ccaec2f943f958d", "size": 9228, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "archives/codelibraries/cp-geo-master/3d/line3d.tex", "max_stars_repo_name": "cbarnson/UVa", "max_stars_repo_head_hexsha": "0dd73fae656613e28b5aaf5880c5dad529316270", "max_stars_repo_licenses": ["Unlicense", "MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-09-07T17:00:26.000Z", "max_stars_repo_stars_event_max_datetime": "2020-08-05T02:08:35.000Z", "max_issues_repo_path": "archives/codelibraries/cp-geo-master/3d/line3d.tex", "max_issues_repo_name": "cbarnson/UVa", "max_issues_repo_head_hexsha": "0dd73fae656613e28b5aaf5880c5dad529316270", "max_issues_repo_licenses": ["Unlicense", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "archives/codelibraries/cp-geo-master/3d/line3d.tex", "max_forks_repo_name": "cbarnson/UVa", "max_forks_repo_head_hexsha": "0dd73fae656613e28b5aaf5880c5dad529316270", "max_forks_repo_licenses": ["Unlicense", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.6511627907, "max_line_length": 390, "alphanum_fraction": 0.683029909, "num_tokens": 3162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.7799928951399098, "lm_q1q2_score": 0.6145807344605521}} {"text": "\\documentclass{article}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n\\begin{document}\n\n\\section*{Limits}\n\n\\[\\lim_{x \\to 0^{+}} \\frac{1}{x} = \\infty\\]\n\n\\[\\lim_{x \\to 0^{-}} \\frac{1}{x} = -\\infty\\]\n\n\\[\\lim_{x \\to \\pm\\infty} \\frac{1}{x} = 0\\]\n\n\n\\end{document}", "meta": {"hexsha": "99e150ac1ab1851b2ad455ca4da8ed6cdb6f4178", "size": 266, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "compendium/mathematics/limits.tex", "max_stars_repo_name": "ZenLulz/LatexCompendium", "max_stars_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-07-30T21:43:55.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-23T20:16:19.000Z", "max_issues_repo_path": "compendium/mathematics/limits.tex", "max_issues_repo_name": "ZenLulz/LatexCompendium", "max_issues_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "compendium/mathematics/limits.tex", "max_forks_repo_name": "ZenLulz/LatexCompendium", "max_forks_repo_head_hexsha": "cc623a88ab05ca90430338333003293baea00f8c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 15.6470588235, "max_line_length": 44, "alphanum_fraction": 0.5902255639, "num_tokens": 116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8872045937171068, "lm_q2_score": 0.6926419894793246, "lm_q1q2_score": 0.6145151548674128}} {"text": "\\chapter{Conclusion and Future Works}\n\\label{chap_conc}\nThis thesis presents a study of deep multi-layer kernels on supervised and unsupervised learning algorithms. In the supervised learning settings, multi-layer kernels are studied on structured output prediction problems and discriminant analysis methods. In the unsupervised learning settings, multi-layer kernels are used for feature learning task. In particular, we used MKMs and MKL for building a feature learning framework with kernel machines whose architecture and training is equivalent to existing deep learning algorithms.\n\nIn the unsupervised feature learning model with MKMs we experimented with single kernel (arc-cosine kernel) and mixed kernels. In some cases, the mixed kernel version was found to be superior than their single kernel counterparts. Features learned by the MKMs were visualized with tSNE algorithm to understand the separability of different classes.\n\nInstead of using single kernel in each layer, a convex combination of multiple kernels were tried out using an unsupervised MKL formulation. The training of this model was done in a greedy layer-by-layer fashion. Experimental results indicates that, this ML-MKL model is superior to single-kernel MKMs. Though we had conducted empirical study on object recognition datasets like \\textit{cifar10} and text classification datasets like \\textit{20-newsgroups}, the results were not very impressive. In the case of \\textit{20-newsgroups} classification, multiple layers might be making the similarity information in the kernel matrix more noisy, since the performance was degrading after each layer is added. The empirical studies on supervised learning settings also gave fruitful results. The KFDA algorithm with multi-layer kernels showed competitive performance with state of the art deep learning algorithms. Structured output learning algorithms also works well when multi-layer kernels were used.\n\n\\section{Future Works}\nThe empirical study conducted in this thesis is giving many insights, from which we can list out some potential future directions to explore.\n\nThe complexity of the model in terms of the number of layers and number of kernels in each layer are set by using cross-validation techniques in our experiments. The optimal number of layers and optimal number of kernels in each layer characterizes the structure of ML-MKL. A thoretical study on the optimal structure of ML-MKL model is a future extension for this work.\n\nThe training of ML-MKL framework in the current implementation is performed in a greedy layer-by-layer fashion, by iteratively adding layers. This can be thought of as unsupervised pre-training done in typical deep learning algorithms. Then we can devise a supervised fine-tuning mechanism for this model, by taking class labels into account, which may change either the kernel parameters or weights.\n\nThe deep learning algorithms are found to be particularly effective for learning complex decision functions. Structured output prediction tasks like parse tree prediction, protein segment alignment prediction etc. need very complex decision functions. Thus, the deep multi-layer kernel machines can be studied in such domains.\n\nThe results from KFDA with multi-layer kernels is having some interesting aspects. The algorithm is performing well, either when using a highly non-linear arc-cosine kernel or when very large number of layers are used in the kernel function. The obscurity behind the need of such complex kernels for \ndiscriminant analysis can be explored further.\n", "meta": {"hexsha": "bc6516cc4f6f500055c418aa9782aa1611afded5", "size": 3558, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/chapter6.tex", "max_stars_repo_name": "akhilpm/Masters-Project", "max_stars_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Thesis/chapter6.tex", "max_issues_repo_name": "akhilpm/Masters-Project", "max_issues_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/chapter6.tex", "max_forks_repo_name": "akhilpm/Masters-Project", "max_forks_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 177.9, "max_line_length": 1000, "alphanum_fraction": 0.8268690275, "num_tokens": 667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7905303186696748, "lm_q2_score": 0.7772998663336158, "lm_q1q2_score": 0.614479111034609}} {"text": "\\documentclass[11 pt]{scrartcl}\n\\usepackage[header, margin, koma]{tyler}\n\n\\newcommand{\\hwtitle}{Hard Counting Problems}\n\n\\newif\\ifproblemsol\n\\problemsolfalse\n\n\\pagestyle{fancy}\n\\fancyhf{}\n\\fancyhead[l]{\\hwtitle{}}\n\\fancyhead[r]{Tyler Zhu}\n\\cfoot{\\thepage}\n\n\\begin{document} \n\\title{\\Large \\hwtitle{}}\n\\author{\\large Tyler Zhu}\n\\date{\\large\\today}\n\n\\maketitle \n\nThese are some pretty arbitrary, hard counting problems. They are more so for testing your general ability to see connections in counting well than how well you bookkeep. Safe travels. \n\n\\section{Warm up}\n\\begin{problem}\n How many cubic polynomials $f(x)$ with positive integer coefficients are there such that $f(1) = 9$?\n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n Let $f(x) = ax^3+bx^2+cx+d$. The only condition on $f(x)$ is that $f(1) = 9$, which means $a + b + c + d = 9$. Since $a,b,c,d$ are positive integers, there are $\\binom{8}{3} = 56$ such polynomials. \n\\end{proof}\n\\fi\n\n\\section{Assorted Candies}\n\\begin{problem}\n Prove $\\sum_{k=0}^n \\binom{n}{k} 2^k = 3^n$ with a combinatorial argument. \n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n Suppose we're trying to create a $n$-bead string with 3 colors; red, blue and green. We can either have 3 chocies for each spot for a total of $3^n$, or pick the $n-k$ spots which will be red, and then fill in the remaining $k$ spots with either blue or green for a total of $\\binom{n}{k}2^k$ for each choice of $k$. \n\\end{proof}\n\\fi\n\n\\begin{problem}\n Let $a_1, a_2, \\dots, a_n$ be a sequence of arbitrary natural numbers. Define $b_k$ to be the number of elements $a_i$ for which $a_i \\geq k$. Prove that $a_1 + a_2 + \\dots + a_n = b_1 + b_2 + \\cdots$. \n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n The idea is to double count. Drawing a picture is the best way to see this: for each $a_i$, draw $a_i$ circles vertically. Then the LHS is counting the number of circles going vertically, while the RHS is counting them horizontally. \n \n Formally, one way to count is simply to sum the $a_i$'s. Another way to count uses the fact that they are natural numbers. All the $b_i$ start at 0. Then, for any given $a_i$, $b_1$ through $b_{a_i}$ will have all their values increased by $1$, increasing the total on the RHS by $a_i$. Hence the total contributions of all $a_i$ to the RHS is just $a_1 + \\dots + a_n$. \n\\end{proof}\n\\fi\n\n\\begin{problem}\n How many ways are there to insert $+$’s between the digits of 111111111111111 (fifteen 1’s) so that the result will be a multiple of 30?\n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n No matter how many $+$'s we insert, the result will always be a multiple of $3$ since there are fifteen 1's. For it to be a multiple of 10, we need exactly 10 numbers, which means we're adding 9 $+$'s. There are 14 gaps, so our answer is $\\binom{14}{9}$.\n\\end{proof}\n\\fi\n\n\n\\begin{problem}\n Compute \n \\[ \\sum_{n_{60}=0}^2\\sum_{n_{59}=0}^{n_{60}}\\dots \\sum_{n_2=0}^{n_3} \\sum_{n_1=0}^{n_2}\\sum_{n_0=0}^{n_1} 1.\\]\n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n Another way of phrasing the problem is to find the number of solutions to $0\\leq n_0 \\leq n_1 \\leq n_2 \\leq \\dots \\leq n_{60} \\leq 2$. This corresponds to the number of right-up walks on a $61\\times 2$ grid from the bottom left to the top right, which we all know to be $\\binom{63}{2}$. \n \n Alternatively, notice that every solution is of the form $(0,\\dots, 0, 1, \\dots, 1, 2, \\dots, 2)$, so we only need to specify the number of 0s, 1s, and 2s. This is equivalent to solving the equation $x+y+z = 61$ for nonnegative $x,y,z$, which (by stars and bars) has $\\binom{63}{2}$ solutions.\n\\end{proof}\n\\fi\n\n\\begin{problem}\n There’s a new (virtual) game show featuring $N$ people where a few lucky contestants get to compete for the ultimate prize: a roll of toilet paper. The game works as follows: everyone lines up in front of a jar of ping pong balls numbered 1 through 100 and one-by-one randomly select a ball until everyone has one. Then the winning number is announced, and anyone with the winning number wins the prize. \n \\alphanum\n \\ii Suppose the winning number is 42. What’s the probability that if $N=3$, then the third person wins the game?\n \\ii If $N = 100$, what’s the probability that the third person wins the game now? \n \\ii You and a friend are watching the game (for $N = 100$) and after all of the balls have been drawn, you both decide to bet on the results. Your friend picks contestant 42, thinking they must have the winning number, but before you can pick a contestant, 98 of them groan in realization that they have losing numbers, which leaves you with no choice but to bet on contestant 20. What’s the probability that you win the bet? \n \\enumend\n\\end{problem}\n\\ifproblemsol\n\\begin{proof}[Solution]\n\\alphanum\n \\ii By symmetry, $\\frac{1}{100}$. One can also compute it out to be $\\frac{99}{100}\\cdot \\frac{98}{99}\\frac{1}{98} = \\frac{1}{100}$. \n \\ii As above, by symmetry, $\\frac{1}{100}$. Computation also works. \n \\ii This is just the Monty Hall problem in disguise, where you choose to switch! Once the other 98 doors (contestant's numbers) were revealed, all of the $\\frac{99}{100}$ probability went into contest 20's chances, so you have a $\\frac{99}{100}$ probability of winning the bet. \n\\enumend\n\\end{proof}\n\\fi\n\n\n\\section{Dessert}\nThese aren't actually relevant to course material; please don't do them unless you really want to! \n\n\\begin{exercise}\n Let $(a_1,a_2, \\dots, a_{12})$ be a permutation of $(1,2,\\dots,12)$ for which \n \\[ a_1>a_2>a_3>a_4>a_5>a_6 \\text{ and } a_60$) order bound of function $\\phi$.\n\t\\item Lipschitz continuous of gradient is local \\emph{upper} second order bound of $\\phi$.\n\t\\item $\\mu \\leq L$.\n\\end{enumerate}\n\n\\subsection{Proximal gradient (PG) for convex problem}\nProximal gradient method solves subproblem which the objective function is second order approximation to $\\phi$.\nThe iteration takes the following form:\n\\begin{equation}\n\t\\begin{split}\n\t\tw_{t+1}&= \\mathop{\\arg \\min}_w \\left\\lbrace \\phi(w_t)+\\nabla \\phi(w_t)^T (w-w_t)+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 \\right \\rbrace\\\\\n\t\t&=\\mathop{\\arg \\min}_w \\left\\lbrace \\nabla \\phi(w_t)^T w+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 \\right \\rbrace\\\\\n\t\t&=\\mathop{\\arg \\min}_w \\left\\lbrace \\frac{1}{2\\alpha_t} \\|w-(w_t-\\alpha_t \\nabla \\phi(x_t))\\|^2_2 \\right \\rbrace\\\\\n\t\t&=Prox_{0} (w_t-\\alpha_t \\nabla \\phi(w_t)).\n\t\\end{split}\n\\end{equation}\nHere $\\alpha_t$ is parameter.\nAnd regularization term can be viewed as $0$.\nIt is equivalent to explicit gradient descent\n\\begin{equation}\n\tw_{t+1}=w_t -\\alpha_t \\nabla \\phi(w_{t}).\n\\end{equation}\n\nHere\n$$\\alpha_t=\\frac{1}{\\lambda_{\\max} (\\nabla^2 \\phi(w))}$$\nwas suggested.\n\n\\subsection{Proximal point algorithm (PPA) for convex problem}\nProximal point algorithm\n\\begin{equation}\n\t\\begin{split}\n\t\tw_{t+1}&= \\mathop{\\arg \\min}_w \\left\\lbrace \\phi(w)+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 \\right \\rbrace\\\\\n\t\t&= Prox_{\\alpha \\phi} (w_t). \n\t\\end{split}\n\\end{equation}\nIt is equivalent to implicit gradient descent\n\\begin{equation}\n\tw_{t+1}=w_t -\\alpha_t \\nabla \\phi(w_{t+1}).\n\\end{equation}\n\nAnd these method can combined with stochastic technique for summation problem, then we can easily obtain proximal stochastic gradient and proximal stochastic point algorithm.\n\n\\subsection{Convex composite problem}\nThe composite problem has the following form\n\\begin{equation}\\label{composite problem}\n\t\\min_w \\phi(w)=f(w)+\\Psi(w),\n\\end{equation}\nwhere $f$ is convex (convexity $\\mu$) and smooth (Lipschitz $L$ of gradient) and the regularizer $\\Psi$ is convex and nonsmooth.\nHere $\\Psi$ can be, for example\n\\begin{enumerate}\n\t\\item $\\ell_1$ regularization: $\\lambda \\|w\\|_1$.\n\t$\\lambda$ is parameter determine the sparsity of solution.\n\tWe will discuss some results when $\\Psi(w)=\\lambda \\|w\\|_1$ in next section.\n\t\\item $\\ell_2$ regularization: $\\frac{\\lambda}{2} \\|w\\|_2$. $\\ell_2$ regularizer also called weight decay.\n\tIt helps to avoid overfitting of machine learning models.\n\t\\item mixed regularization: $\\lambda_1 \\|w\\|_1 + \\frac{\\lambda_2}{2} \\|w\\|_2$.\n\t\\item Indicator function.\n\\end{enumerate}\n\nRecall that the first order necessary condition\n\\begin{equation}\n\t0 \\in \\nabla f(w_\\star) + \\partial \\Psi(w_\\star).\n\\end{equation}\n\\subsection{Gradient descent (GD) for convex composite problem}\nExplicit gradient descent applied for (\\ref{composite problem}) has\n\\begin{equation}\n\tw_{t+1} = w_t -\\alpha_t (\\nabla f(x_t) + \\partial \\Psi(w_{t})).\n\\end{equation}\nIt is equal to\n\\begin{equation}\n\t\\frac{1}{\\alpha_t} (w_{t+1}-w_t) +\\nabla \\Psi(w_t) + \\partial \\Psi(w_{t})=0.\n\\end{equation}\nThe subgradient operator works on $\\Psi(w_t)$, so it will not helps to obtain special structure of next point $w_{t+1}$.\n\n\\subsection{Proximal gradient for convex composite problem}\n\\begin{align}\n\tw_{t+1}&= \\mathop{\\arg \\min}_w \\left\\lbrace f(w_t)+\\nabla f(w_t)^T (w-w_t)+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 +\\Psi(w) \\right \\rbrace\\\\\n\t&=\\mathop{\\arg \\min}_w \\left\\lbrace \\nabla f(w_t)^T w+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 +\\Psi(w) \\right \\rbrace\\\\\n\t&=\\mathop{\\arg \\min}_w \\left\\lbrace \\frac{1}{2\\alpha_t} \\|w-(w_t-\\alpha_t \\nabla \\Psi(w_t))\\|^2_2 +\\Psi(w) \\right \\rbrace\\\\\n\t&=Prox_{\\alpha_t \\Psi} (w_t-\\alpha_t \\nabla f(w_t)).\n\\end{align}\nIt is equivalent to solve the equation (KKT condition of subproblem)\n\\begin{equation}\n\t\\nabla f(w_t) + \\frac{1}{\\alpha_t} (w_{t+1}-w_t) +\\partial \\Psi(w_{t+1}) = 0.\n\\end{equation}\nThat is implicit gradient descent.\n\\begin{equation}\n\tw_{t+1} = w_t -\\alpha_t (\\nabla \\Psi(w_t) + \\partial \\Psi(w_{t+1})).\n\\end{equation}\nWe can also write it in the formulation:\n\\begin{equation}\n\t\\frac{1}{\\alpha_t} (w_{t+1}-w_t) +\\nabla f(w_t) + \\partial \\Psi(w_{t+1})=0,\n\\end{equation}\nand it actually using the approximation\n\\begin{equation}\n\t\\frac{1}{\\alpha_t} (w_{t+1}-w_t) +\\nabla f(w_t) \\approx \\nabla f(w_{t+1}).\n\\end{equation}\nFrom the formula we can see, PG can produce better sparsity than GD.\nBecause the function $\\partial R(\\cdot)$ operates on $w_t$ in GD, while operates on $w_{t+1}$ in PG.\n\n\n\\section{Simple dual averaging and Regularized dual averaging algorithms}\n\n\\subsection{Empirical risk minimization}\nThe expected risk minimization arising from machine learning has the following form:\n\\begin{equation}\\label{expected risk minimization}\n\t\\min_{w}~~\\left\\lbrace \\phi(w)= \\mathbf{E}_z f(w,z)\\right\\rbrace,\n\\end{equation}\nwhere $z$ is an input-output pair of data, $f(w,z)$ is loss function.\n\nAnd the regularized expected risk minimization has the following form\n\\begin{equation}\\label{regularized expected risk minimization}\n\t\\min_{w}~~\\left\\lbrace \\phi(w)= \\mathbf{E}_z f(w,z)+\\Psi(w)\\right\\rbrace,\n\\end{equation}\nwhere $\\Psi(w)$ is a regularization term, for example, $\\ell_1$ regularization in obtaining sparse solutions.\n\nThe empirical risk minimization is approximation of (\\ref{expected risk minimization})\n\\begin{equation}\\label{approximated optmization}\n\t\\min_{w}~~\\left\\lbrace \\phi(w)=\\frac{1}{T}\\sum_{t=1}^{T} f(w,z_t)\\right\\rbrace,\n\\end{equation}\n\nAnd the regularized empirical risk minimization\n\\begin{equation}\\label{approximated regularized optimization}\n\t\\min_{w}~~\\left\\lbrace \\phi(w)=\\frac{1}{T} \\sum_{t=1}^{T} f(w,z_t)+\\Psi(w)\\right\\rbrace,\n\\end{equation}\nwhere $f$ is convex (convexity $\\mu$) and smooth (Lipschitz $L$ of gradient) and $\\Psi$ is convex but nonsmooth.\n%Here $\\Psi$ can be, for example\n%\\begin{enumerate}\n%\t\\item $\\lambda \\|x\\|_1$.\n%\t$\\lambda$ is parameter determine the sparsity of solution.\n%\tWe will take $R(x)=\\lambda \\|x\\|_1$ as example.\n%\t\\item Indicator function.\n%\\end{enumerate}\n\n%\\begin{equation}\n%\\min_w f(w)+\\lambda \\|w\\|_1\n%\\end{equation}\n\n\\subsection{Simple dual averaging (SDA)}\nSimple dual averaging method takes the form\n\\begin{equation}\n\tw_{t+1} = \\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{t}\\sum_{\\tau=1}^{t} \\left + \\frac{\\beta_t}{t} h(w) \\right\\}.\n\\end{equation}\n\nSince $g_\\tau(w_\\tau)$ is constant in current iteration, we use $g_\\tau$ instead for simplicity in the following.\nIf we choose $h(w)=\\frac{1}{2} \\|w-w_0\\|^2_2$, where $w_0=\\arg \\min\\limits_w h(w)$ is also an initial point then we can conclude\n\\begin{equation}\n\tw_{t+1}=w_0 - \\frac{1}{\\beta_t} \\sum_{\\tau=1}^{t} g_\\tau=w_0-\\frac{t}{\\beta_t} \\bar g_t.\n\\end{equation}\nWe assume $\\beta_t=t^\\alpha$.\nThen, we can write SDA as recursion\n\\begin{equation}\\label{RDA}\n\t\\begin{split}\n\t\tw_{t+1} & =w_0- \\frac{1}{t^{\\alpha}} \\sum_{\\tau=1}^{t} g_\\tau\\\\\n\t\t& =w_0 -\\frac{1}{t^{\\alpha}} \\left( \\frac{(t-1)^{\\alpha}}{(t-1)^{\\alpha}} \\sum_{\\tau=1}^{t-1} g_\\tau +g_t \\right)\\\\\n\t\t& =\\left( 1-\\frac{(t-1)^{\\alpha}}{t^{\\alpha}} \\right) w_0+ \\frac{(t-1)^{\\alpha}}{t^{\\alpha}}\\left( w_0 - \\frac{1}{(t-1)^\\alpha}\\sum_{\\tau=1}^{t} g_\\tau \\right) - \\frac{1}{t^{\\alpha}} g_t\\\\\n\t\t& =\\left( 1-\\left( 1-\\frac{1}{t} \\right)^{\\alpha} \\right) w_0 + \\left( 1-\\frac{1}{t} \\right)^{\\alpha} w_t -\\frac{1}{t^{\\alpha}} g_t .\n\t\\end{split}\n\\end{equation}\nHere $\\left( 1-\\left( 1-\\frac{1}{t} \\right)^{\\alpha} \\right) \\rightarrow 0$ and $\\left( 1-\\frac{1}{t} \\right)^{\\alpha} \\rightarrow 1$ as $t \\rightarrow \\infty$.\nSo SDA can be viewed as perturbation of SGD.\nWe proved convergence of SDA in stochastic setting in the appendix.\n\n\\subsection{Regularized dual averaging (RDA)}\t\nEach iteration of regularized dual averaging method takes \n\\begin{equation}\\label{RDA subproblem}\n\tw_{t+1}=\\mathop{\\arg\\min}_w \\left\\{\\frac{1}{t} \\sum_{\\tau=1}^{t} \\left+\\Psi(w)+\\frac{\\beta_t}{t}h(w)\\right\\}.\n\\end{equation}\nThe first term $\\sum_{\\tau=1}^{t} \\left$ is linear function obtained by averaging all previous subgradients.\nThe third term $h(w)$ is strongly convex function, and $\\beta_t$ is a nonnegative and nondecreasing sequence which determines the convergence of algorithm.\nThe closed form solution of subproblem (\\ref{RDA subproblem}) can be found such that the computational effort per iteration is only $O(n)$, the same as the SGD method.\n\n\\begin{enumerate}\n\t\\item Proximal gradient method (PG) can be written in forward backward splitting (FOBOS)\n\t\\begin{equation}\n\t\tw_{t+1}=\\mathop{\\arg \\min}_w \\left\\{ \\left + \\Psi(w)+\\frac{1}{2\\alpha_t}\\|w-w_t\\|_2^2 \\right\\}.\n\t\\end{equation}\n\t\\begin{equation*}\n\t\t\\Updownarrow\n\t\\end{equation*}\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\tw_{t+\\frac{1}{2}} &=w_t - \\alpha_t g_t \\\\\n\t\t\tw_{t+1} &=\\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{2}\\|w-w_{t+\\frac{1}{2}}\\|^2_2 +\\alpha_t \\Psi(w) \\right\\}.\n\t\t\\end{split}\n\t\\end{equation}\n\tHere $\\alpha_t=\\frac{1}{\\sqrt{t}}$ to obtain best convergence rate.\n\t\n\t\\item RDA also can be written in the form of forward-backward splitting (when $h(w)=\\frac{1}{2}\\|w-w_0\\|^2_2$):\n\t\\begin{align}\n\t\tw_{t+1}&=\\mathop{\\arg\\min}_w \\left\\{ \\left<\\bar g_t,w\\right>+\\frac{\\beta_t}{2t}\\|w-w_0\\|^2_2+\\Psi(w)\\right\\}\\\\\n\t\t& =\\mathop{\\arg\\min}_w \\left\\{ \\frac{\\beta_t}{2t} \\frac{2t}{\\beta_t}\\bar g_t^T w +\\frac{\\beta_t}{2t}\\|w-w_0\\|^2_2+\\Psi(w)\\right\\}\\\\\n\t\t& =\\mathop{\\arg\\min}_w \\left\\{ \\frac{\\beta_t}{2t}\\|w-(w_0+\\frac{t}{\\beta_t} \\bar g_t)\\|^2_2+\\Psi(w)\\right\\}\\\\\n\t\t&=\\mathop{\\arg\\min}_w \\left\\{ \\frac{1}{2}\\|w-(w_0+\\frac{t}{\\beta_t} \\bar g_t)\\|^2_2+\\frac{t}{\\beta_t}\\Psi(w)\\right\\}\n\t\\end{align}\n\t\\begin{equation*}\n\t\t\\Updownarrow\n\t\\end{equation*}\n\t\\begin{align}\n\t\tw_{t+\\frac{1}{2}} &=w_0-\\frac{t}{\\beta_t} \\bar g_t \\\\\n\t\tw_{t+1} &= \\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{2}\\|w-w_{t+\\frac{1}{2}}\\|^2_2 + \\frac{t}{\\beta_t}\\Psi(w) \\right\\}.\n\t\\end{align}\n\tThe forward operator is actually SDA method.\n\tHere $\\beta_t = \\sqrt{t}$ when $\\Psi$ is convex, and $\\beta_t=\\ln t$ when $\\Psi$ is strongly convex.\n\\end{enumerate}\n\n\\section{$\\ell_1$ regularization and sparsity}\nIn this section, we will consider a kind special regularized problem with $\\ell_1$ regularization.\n\\begin{equation}\n\t\\min_w f(w)+\\lambda \\|w\\|_1\n\\end{equation}\nSince $\\ell_1$ regularization can produce sparse solution.\n\nHere we give out the closed form solution of proximal gradient method and RDA applied for the $\\ell_1$ regularization\n\n\n\\begin{enumerate}\n\t\\item Proximal gradient:\n\t\\begin{equation}\n\t\tw_{t+1}=\\mathop{\\arg \\min}_w \\left\\lbrace g_t^T w+\\frac{1}{2\\alpha_t} \\|w-w_t\\|^2_2 +\\lambda \\|w\\|_1 \\right \\rbrace\n\t\\end{equation}\n\tThe entry-wise closed form solution is\n\t\\begin{equation}\n\t\tw_{t+1}^{(i)}=\n\t\t\\begin{cases}\n\t\t\tw_t^{(i)}-\\alpha_t (g_t^{(i)} +\\lambda), & w_t^{(i)}-\\alpha_t g_t^{(i)} > \\alpha_t \\lambda\\\\\n\t\t\t0, & |w_t^{(i)}-\\alpha_t g_t^{(i)}| \\leq \\alpha_t \\lambda\\\\\n\t\t\tw_t^{(i)}-\\alpha_t (g_t^{(i)} -\\lambda), & w_t^{(i)}-\\alpha_t g_t^{(i)} < -\\alpha_t \\lambda\n\t\t\\end{cases}\n\t\\end{equation}\n\tHere $\\alpha_t=\\frac{1}{\\sqrt{t}}$.\n\t\n\t\\item RDA:\n\t\\begin{equation}\n\t\tw_{t+1}=\\mathop{\\arg \\min}_w \\left\\lbrace \\bar g_t^T w+\\frac{\\beta_t}{2t} \\|w-w_0\\|^2_2 +\\lambda \\|w\\|_1 \\right \\rbrace\n\t\\end{equation}\n\tThe entry-wise closed form solution:\n\t\\begin{equation}\n\t\tw_{t+1}^{(i)}=\n\t\t\\begin{cases}\n\t\t\tw_0^{(i)}-\\frac{t}{\\beta_t} (\\bar g_t^{(i)} +\\lambda), & w_0^{(i)}-\\frac{t}{\\beta_t}\\bar g_t^{(i)} > \\frac{t}{\\beta_t}\\lambda\\\\\n\t\t\t0, & | w_0^{(i)}-\\frac{t}{\\beta_t}\\bar g_t^{(i)}| \\leq \\frac{t}{\\beta_t} \\lambda\\\\\n\t\t\tw_0^{(i)}-\\frac{t}{\\beta_t} (\\bar g_t^{(i)} -\\lambda), & w_0^{(i)}-\\frac{t}{\\beta_t}\\bar g_t^{(i)} < - \\frac{t}{\\beta_t}\\lambda\n\t\t\\end{cases}\n\t\\end{equation}\n\tHere $\\frac{t}{\\beta_t}=\\sqrt{t}$.\n\t\n\t\\item We can conclude the general case from above. \n\tAssume we have the forward-backward splitting\n\t\\begin{align}\n\t\tw_{t+\\frac{1}{2}} & \\\\\n\t\tw_{t+1} &= \\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{2}\\|w-w_{t+\\frac{1}{2}}\\|^2_2 + \\theta_t\\Psi(w) \\right\\}.\n\t\\end{align}\n\tThen, the entry-wise closed form solution has the form:\n\t\\begin{equation}\n\t\tw_{t+1}^{(i)}=\n\t\t\\begin{cases}\n\t\t\tw_{t+1/2}^{(i)}- \\theta_t \\lambda, & w_{t+1/2}^{(i)} > \\theta_t\\lambda\\\\\n\t\t\t0, & | w_{t+1/2}^{(i)}| \\leq \\theta_t \\lambda\\\\\n\t\t\tw_{t+1/2}^{(i)} +\\theta_t\\lambda, & w_{t+1/2}^{(i)} < - \\theta_t\\lambda\n\t\t\\end{cases}\n\t\\end{equation}\n\tThe coefficient $\\theta_t$ of $\\Psi$ also determine the sparsity of the solution.\n\tThere is a trade-off between fast convergence and sparsity of solution.\n\tLarge $\\theta_t$ in the backward operator would produce more sparse solution while it takes $w_{t+1}$ far away from $w_{t+1/2}$.\n\t\\textcolor{red}{Question: what is the best choice of $\\theta_t$ to guarantee both fast convergence and sparsity of solution.}\n\t\n\t\\item \n\tTry\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\tw_{t+\\frac{1}{2}} &=w_t - \\frac{1}{\\sqrt{t}} g_t \\\\\n\t\t\tw_{t+1} &=\\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{2}\\|w-w_{t+\\frac{1}{2}}\\|^2_2 +\\sqrt{t} \\Psi(w) \\right\\}.\n\t\t\\end{split}\n\t\\end{equation}\n\tThe entry-wise closed form solution is\n\t\\begin{equation}\n\t\tw_{t+1}^{(i)}=\n\t\t\\begin{cases}\n\t\t\tw_t^{(i)}-\\alpha_t g_t^{(i)} -\\sqrt{t}\\lambda, & w_t^{(i)}-\\alpha_t g_t^{(i)} > \\sqrt{t} \\lambda\\\\\n\t\t\t0, & |w_t^{(i)}-\\alpha_t g_t^{(i)}| \\leq \\sqrt{t} \\lambda\\\\\n\t\t\tw_t^{(i)}-\\alpha_t g_t^{(i)} +\\sqrt{t}\\lambda, & w_t^{(i)}-\\alpha_t g_t^{(i)} < -\\sqrt{t} \\lambda\n\t\t\\end{cases}\n\t\\end{equation}\n\tIt's equivalent to\n\t\\begin{equation}\n\t\tw_{t+1}=\\mathop{\\arg \\min}_w \\left\\lbrace g_t^T w+\\frac{\\sqrt{t}}{2} \\|w-w_t\\|^2_2 + t\\lambda \\|w\\|_1 \\right \\rbrace\n\t\\end{equation}\n\tIt's approximation of\n\t\\begin{equation}\n\t\tf(w) + t \\lambda \\|w\\|_1.\n\t\\end{equation}\n\tWe can easily conclude that this iteration will not converge.\n\t\n\t\\item Why RDA takes $\\sqrt{t}$ threshold and Proximal gradient takes $\\frac{1}{\\sqrt{t}}$ and they all converges?\n\\end{enumerate}\n\n\nNow comparing the threshold $\\lambda_{PG}=\\alpha_t \\lambda$ of PG and the threshold $\\lambda_{RDA}=\\lambda$ of RDA:\nwith $\\alpha_t=\\frac{1}{\\sqrt{t}}$, we have $\\lambda_{PG} \\rightarrow 0$ as $t \\rightarrow 0$.\nIt is clear that RDA uses a much more aggressive truncation threshold, thus is able to generate significantly more sparse solutions.\n\n\\section{Convergence analysis}\n\\subsection{Convergence of SDA}\nThroughout we assume\n\\begin{enumerate}\n\t\\item SDA method applied for empirical risk minimization, so in this section, we discuss\n\t\\begin{equation*}\n\t\t\\phi(w)=\\frac{1}{T}\\sum_{t=1}^{T} f(w,z_t)\n\t\\end{equation*}\n\t\n\t\\item\n\t$\\phi(w)$ is differentiable and strongly convex. So there exists a constant $\\mu >0$ such that \n\t\\begin{equation}\n\t\t\\phi(y) \\geq \\phi(x) + \\nabla \\phi(x)^T (y-x)+\\frac{\\mu}{2} \\|y-x\\|^2.\n\t\\end{equation}\n\t\n\t\\item\n\t$\\nabla \\phi$ is Lipschitz so that\n\t\\begin{equation}\n\t\t\\|\\nabla \\phi(x) - \\nabla \\phi(y)\\| \\leq L \\|x-y\\|.\n\t\\end{equation}\n\tAnd we can obtain the following relation:\n\t\\begin{equation}\n\t\t\\phi(y) \\leq \\phi(x)+\\nabla \\phi(x)^T (y-x) +\\frac{L}{2} \\|y-x\\|.\n\t\\end{equation}\n\tThus $\\mu \\leq L$.\n\t\n\t\\item\n\t$g_t(w) = \\nabla f(w,z_t)$ and $\\|g_t\\|\\leq M$ for all $w$ and $z_t$.\n\t\n\t\\item\n\t$\\|w_0- w_\\star\\| \\leq D$.\n\t\n\t\\item Let $\\partial g(w)$ denote the subgradient set of $f$ at $w$, namely,\n\t\n\t\n\\end{enumerate}\nOther important remarks:\n\\begin{enumerate}\n\t\\item $w_0=0$ will not improve the convergence rate of SDA, because $\\|w_0-w_\\star\\|$ not changed in this case.\n\t\n\t\\item Another important technique in our proof is\n\t\\begin{equation}\n\t\t2\\langle a, b\\rangle \\leq \\epsilon a^2 +\\frac{1}{\\epsilon} b^2\n\t\\end{equation}\n\twhere $\\epsilon>0$.\n\t\n\t\\item Cauchy-Shwartz inequality:\n\t\\begin{equation}\n\t\t\\langle a, b\\rangle \\leq \\|a\\| \\|b\\|\n\t\\end{equation}\n\t\n\t\\item $\\|a+b\\| \\leq \\|a\\|+\\|b\\|$\n\\end{enumerate}\nFor simplicity, denote SDA as following iteration\n\\begin{equation}\n\tw_{t+1}=(1-\\lambda_t)w_0+\\lambda_t w_t - \\gamma_t g_t,\n\\end{equation}\nwhere $\\lambda_t= \\left( 1-\\frac{1}{t} \\right)^\\alpha <1$ and $\\gamma_t= \\frac{1}{t^\\alpha}$.\n\n\\subsubsection{Convergence of $\\mathbb{E}\\left[ \\phi(\\bar{w}) \\right] -\\phi(w_\\star)$}\nLet's make no assumptions about strongly convexity of $\\phi(w)$ here, i.e. $\\mu=0$.\n\n\\begin{lemma}\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\\\\\n\t\t\t&\\leq\n\t\t\t\\lambda_t\\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\n\t\t\t+\\gamma_t(\\gamma_t+\\lambda_t-1) M^2 + (1-\\lambda_t)(1-\\gamma_t) D^2.\n\t\t\\end{split}\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\tWe have\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[ \\|(1-\\lambda_t)w_0+\\lambda_t w_t - \\gamma_t g_t - w_\\star\\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[ \\| \\lambda_t(w_t-w_\\star) -\\gamma_t g_t +(1-\\lambda_t)(w_0-w_\\star)\\|^2 \\right]\\\\\n\t\t\\notag\n\t\t&=\\mathbb{E}\\left[ \\|\\lambda_t(w_t-w_\\star) - \\gamma_t g_t\\|^2 \\right]\n\t\t+2\\mathbb{E}\\left[ \\left<\\lambda_t(w_t-w_\\star) - \\gamma_t g_t, (1-\\lambda_t)(w_0-w_\\star) \\right> \\right]\\\\ \\label{ineq}\n\t\t&+\\mathbb{E}\\left[ \\|(1-\\lambda_t)(w_0-w_\\star)\\|^2 \\right].\n\t\\end{align}\n\tThe first term in (\\ref{ineq}) has\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[ \\|\\lambda_t(w_t-w_\\star) - \\gamma_t g_t\\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[ \\| \\lambda_t(w_t-w_\\star) \\|^2 \\right]-\n\t\t2\\mathbb{E}\\left[ \\left<\\lambda_t(w_t-w_\\star),\\gamma_t g_t \\right> \\right]+\\mathbb{E}\\left[ \\|\\gamma_t g_t\\|^2 \\right]\\\\\n\t\t&\\leq \\lambda_t^2 \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\left \\right]\n\t\t+\\gamma_t^2 M^2\\\\\n\t\t\\label{ineq1.1}\n\t\t&=\\lambda_t^2 \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\left<\\nabla \\phi(w_t),w_t-w_\\star \\right> \\right]\n\t\t+\\gamma_t^2 M^2\\\\\n\t\t\\label{ineq1.2}\n\t\t&\\leq \\lambda_t^2 \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]+\\gamma_t^2 M^2.\n\t\\end{align}\n\t(\\ref{ineq1.1}) follows because...\n\t(\\ref{ineq1.2}) is a consequence of the inequality\n\t\\begin{equation}\n\t\t\\left<\\nabla\\phi(w_t), w_t-w_\\star \\right> \\geq \\phi(w_t)-\\phi(w_\\star).\n\t\\end{equation}\n\twhich holds because $\\phi$ is convex.\n\t\n\tThe second term in (\\ref{ineq}) has\n\t\\begin{align}\n\t\t&2\\mathbb{E}\\left[ \\left<\\lambda_t(w_t-w_\\star) - \\gamma_t g_t,(1-\\lambda_t)(w_0-w_\\star) \\right> \\right]\\\\\n\t\t&=2\\lambda_t(1-\\lambda_t)\\mathbb{E}\\left[ \\left \\right]\n\t\t-2\\gamma_t(1-\\lambda_t) \\mathbb{E}\\left[ \\left \\right]\\\\\n\t\t&\\leq \\lambda_t(1-\\lambda_t) \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2+\\|w_0-w_\\star\\|^2 \\right]\n\t\t+\\gamma_t(1-\\lambda_t) \\color{red}{\\mathbb{E}\\left[ \\|g_t\\|^2+\\|w_0-w_\\star\\|^2 \\right]}\\\\\n\t\t&= \\lambda_t(1-\\lambda_t) \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2\\right]+ \\lambda_t(1-\\lambda_t)D^2 +\\gamma_t(1-\\lambda_t)(M^2+D^2).\n\t\\end{align}\n\t\n\tThe last term in (\\ref{ineq}) has\n\t\\begin{align}\n\t\t\\mathbb{E}\\left[ \\|(1-\\lambda_t)(w_0-w_\\star)\\|^2 \\right]\\leq (1-\\lambda_t)^2 D^2.\n\t\\end{align}\n\t\n\tSummary above, we have\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\\\\\n\t\t\\notag\n\t\t&\\leq \\lambda_t^2 \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]+\\gamma_t^2 M^2\n\t\t+\\lambda_t(1-\\lambda_t) \\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2\\right]\\\\\n\t\t&+ \\lambda_t(1-\\lambda_t)D^2 + \\gamma_t(1-\\lambda_t)(M^2+D^2)\n\t\t+ (1-\\lambda_t)^2 D^2\\\\\n\t\t&=\\lambda_t\\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\n\t\t+\\gamma_t(\\gamma_t-\\lambda_t+1) M^2 + (1-\\lambda_t)(1+\\gamma_t) D^2.\n\t\\end{align}\n\\end{proof}\n\n\\begin{theorem}\n\t\\begin{equation}\n\t\t\\mathbb{E}\\left[\\phi(\\bar{w}) \\right]-\\phi(w_\\star) \\leq\n\t\t\\frac{D^2+M^2\\sum_{t=0}^{n}\\frac{\\gamma_t^2}{\\lambda_t}\n\t\t\t+(M^2+D^2)\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}\n\t\t\t+D^2\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}}{2\\sum_{t=0}^{n}\\gamma_t}\n\t\\end{equation}\n\twhere $\\bar{w}:=(\\sum_{t=0}^{n}\\gamma_t)^{-1} \\sum_{t=0}^{n}\\gamma_t w_t$.\n\\end{theorem}\n\\begin{proof}\n\tSince $\\lambda_t<1$, we have\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\lambda_t \\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\\\\\n\t\t\t&\\leq \\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\\\\\n\t\t\t&\\leq\n\t\t\t\\lambda_t\\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t\t-2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\n\t\t\t+\\gamma_t(\\gamma_t - \\lambda_t+1) M^2 + (1-\\lambda_t)(1+\\gamma_t) D^2.\n\t\t\\end{split}\n\t\\end{equation}\n\tArranging the bound: by telescoping $t$ from $0$ to $n$,\n\t\\begin{equation}\\label{ineq2}\n\t\t\\begin{split}\n\t\t\t&2\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\\\\\n\t\t\t&\\leq -\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\n\t\t\t+\\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{\n\t\t\t\t\\gamma_t(\\gamma_t-\\lambda_t+1) M^2 + (1-\\lambda_t)(1+\\gamma_t) D^2\n\t\t\t}{\\lambda_t}\\\\\n\t\t\t&=-\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star \\|^2 \\right]\n\t\t\t+\\mathbb{E}\\left[ \\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{M^2\\gamma_t^2+\\gamma_t(1-\\lambda_t)\\left( M^2+D^2 \\right)+(1-\\lambda_t)D^2}{\\lambda_t},\n\t\t\\end{split}\n\t\\end{equation}\n\twe have\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&2\\sum_{t=0}^{n} \\gamma_t \\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\\\\\n\t\t\t&\\leq\n\t\t\t-\\mathbb{E}\\left[ \\|w_{n+1}-w_\\star\\|^2-\\|w_0-w_\\star\\|^2 \\right]\\\\\n\t\t\t&+M^2\\sum_{t=0}^{n}\\frac{\\gamma_t^2}{\\lambda_t}\n\t\t\t+(M^2+D^2)\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}\n\t\t\t+D^2\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}\n\t\t\\end{split}\n\t\\end{equation}\n\tand then dividing by the sum of $\\gamma_t$, we have for any $n$\n\t\\begin{equation}\n\t\t\\frac{1}{\\sum_{t=0}^{n}\\gamma_t} \\sum_{t=0}^{n}\\gamma_t\\mathbb{E}\\left[ \\phi(w_t)-\\phi(w_\\star) \\right]\n\t\t\\leq\n\t\t\\frac{D^2+M^2\\sum_{t=0}^{n}\\frac{\\gamma_t^2}{\\lambda_t}\n\t\t\t+(M^2+D^2)\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}\n\t\t\t+D^2\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}}{2\\sum_{t=0}^{n}\\gamma_t}.\n\t\\end{equation}\n\tThen, by convexity (Jensen's inequality), we have\n\t\\begin{equation}\n\t\t\\mathbb{E}\\left[\\phi(\\bar{w}) \\right]-\\phi(w_\\star) \\leq\n\t\t\\frac{D^2+M^2\\sum_{t=0}^{n}\\frac{\\gamma_t^2}{\\lambda_t}\n\t\t\t+(M^2+D^2)\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}\n\t\t\t+D^2\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}}{2\\sum_{t=0}^{n}\\gamma_t}\n\t\\end{equation}\n\\end{proof}\n\nSince $\\lambda_t=\\left( 1-\\frac{1}{t} \\right)^\\alpha \\sim 1-\\frac{\\alpha}{t}$ and $\\gamma_t=\\frac{1}{t^\\alpha}$, we have ($\\alpha \\neq 1$)\n\\begin{equation}\n\t\\sum_{t=1}^{n} \\gamma_t = \\int_{1}^{n} \\frac{1}{\\tau^\\alpha} d \\tau\n\t=n^{1-\\alpha}.\n\\end{equation}\nSimilarly, we have $\\sum_{t=0}^{n} \\frac{\\gamma_t^2}{\\lambda_t}=n^{1-2\\alpha}$, $\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}= \\alpha n^{-\\alpha}$, \nand $\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}=\\alpha\\ln n$.\nThen\n\\begin{equation}\n\t\\begin{split}\n\t\t&\\frac{D^2+M^2\\sum_{t=0}^{n}\\frac{\\gamma_t^2}{\\lambda_t}\n\t\t\t+(M^2+D^2)\\sum_{t=0}^{n}\\frac{\\gamma_t(1-\\lambda_t)}{\\lambda_t}\n\t\t\t+D^2\\sum_{t=0}^{n}\\frac{1-\\lambda_t}{\\lambda_t}}{2\\sum_{t=0}^{n}\\gamma_t}\\\\\n\t\t&=\\frac{D^2+M^2 n^{1-2\\alpha}+\\alpha(M^2+D^2)n^{-\\alpha} +\\alpha D^2 \\ln n}{2n^{1-\\alpha}}\\\\\n\t\t&=\\frac{D^2}{2n^{1-\\alpha}} + \\frac{M^2}{2} n^{-\\alpha} +\\frac{\\alpha(M^2+D^2)}{2}n^{-1} + \\frac{\\alpha D^2}{2} \\frac{\\ln n}{n^{1-\\alpha}}.\n\t\\end{split}\n\\end{equation}\nSo, the best choice of $\\alpha$ is $\\frac{1}{2}$.\nIn this case, the simple dual averaging iteration reads\n\\begin{equation}\n\tw_{t+1} = \\mathop{\\arg \\min}_w \\left\\{ \\frac{1}{t}\\sum_{\\tau=1}^{t} \\left + \\frac{\\sqrt{t}}{t} h(w) \\right\\}\n\\end{equation}\nThis result is the same with [Nesterov, 2009]'s suggest, the regularization parameter should be $O(\\frac{1}{\\sqrt{t}})$.\nThe convergence rate is $O(\\frac{\\ln t}{\\sqrt{t}})$.\n\n\\subsubsection{Convergenc of $\\mathbb{E}\\left[ \\|w_{t+1}-w_\\star\\| \\right]$}\nIn this section, we ask for the assumption on the strongly convex of $\\phi$.\n\n\\begin{lemma}\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t\t&\\leq \\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star\\|^2 \\right]\n\t\t\t+\\gamma_t^2\\mathbb{E}\\left[\\|\\nabla \\phi(w_t)-g_t\\|^2 \\right]\n\t\t\\end{split}\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t g_t -w_\\star\\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t) +\\gamma_t(\\nabla \\phi(w_t)-g_t) -w_\\star\\|^2 \\right]\\\\\n\t\t\\notag\n\t\t&=\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star\\|^2 \\right]\\\\\n\t\t\\notag\n\t\t&+2\\gamma_t\\mathbb{E}\\left[\\left<\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star , \\nabla \\phi(w_t)-g_t\\right> \\right]\\\\\n\t\t\\label{ineq2.2}\n\t\t&\\gamma_t^2\\mathbb{E}\\left[\\|\\nabla \\phi(w_t)-g_t\\|^2 \\right]\\\\\n\t\t\\label{ineq2.3}\n\t\t&=\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star\\|^2 \\right]\n\t\t+\\gamma_t^2\\mathbb{E}\\left[\\|\\nabla \\phi(w_t)-g_t\\|^2 \\right]\n\t\\end{align}\n\t(\\ref{ineq2.2}) follows because $\\mathbb{E}\\left[g(w,z_t) \\right]=\\nabla\\phi(w)$ for all $w$ and $g(w,z_t)$ is independent of $z_t$.\n\tThus we have\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\left<\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star , \\nabla \\phi(w_t)-g(w_t,z_t)\\right> \\right]\\\\\n\t\t\t&=\\mathbb{E}_{z_0,\\dots,z_{t-1}}\\left[\\mathbb{E}_{z_t}\\left[\\left< \\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star , \\nabla \\phi(w_t)-g(w_t,z_t) \\right>| z_0,\\dots,z_{t-1} \\right] \\right]\\\\\n\t\t\t&=0\n\t\t\\end{split}\n\t\\end{equation}\n\\end{proof}\nNote that the first term in (\\ref{ineq2.3}) is completely independent of $z_t$ while the second term is a variance term concerning the second moments of the subgradient at the current iterate and at the optimum.\nWe can bound each of these terms separately.\n\n\\begin{lemma}\n\tFor the first term\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star\\|^2 \\right]\\\\\n\t\t\t&\\leq\n\t\t\t(2-\\lambda_t)\\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)\\|^2\\right]\n\t\t\t+(1-\\lambda_t)(2-\\lambda_t)\\mathbb{E}\\left[ \\|w_0-w_\\star\\|^2\\right]\n\t\t\\end{split}\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[\\|\\lambda_t w_t +(1-\\lambda_t)w_0 -\\gamma_t \\nabla\\phi(w_t)-w_\\star\\|^2 \\right]\\\\\n\t\t&=\\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)+(1-\\lambda_t)(w_0-w_\\star)\\|^2\\right]\\\\\n\t\t\\notag\n\t\t&=\\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)\\|^2\\right]\\\\\n\t\t\\notag\n\t\t&+2(1-\\lambda_t)\\mathbb{E}\\left[ \\left<\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)+(1-\\lambda_t)(w_0-w_\\star), w_0-w_\\star\\right>\\right]\\\\\n\t\t&+(1-\\lambda_t)^2\\mathbb{E}\\left[ \\|w_0-w_\\star\\|^2\\right]\\\\\n\t\t\\notag\n\t\t&\\leq \\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)\\|^2\\right]\n\t\t+(1-\\lambda_t)\\left( \\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)\\|^2\\right]+\\mathbb{E}\\left[ \\|w_0-w_\\star\\|^2\\right] \\right)\\\\\n\t\t&+(1-\\lambda_t)^2\\mathbb{E}\\left[ \\|w_0-w_\\star\\|^2\\right]\\\\\n\t\t\\label{ineq2.4}\n\t\t&=(2-\\lambda_t)\\mathbb{E}\\left[ \\|\\lambda_t (w_t-w_\\star)-\\gamma_t \\nabla\\phi(w_t)+\\gamma_t\\nabla\\phi(w_\\star)\\|^2\\right]\n\t\t+(2-\\lambda_t)(1-\\lambda_t)\\mathbb{E}\\left[ \\|w_0-w_\\star\\|^2\\right]\n\t\\end{align}\n\t(\\ref{ineq2.4}) follows because $w_\\star$ is the optimal solution, so $\\nabla\\phi(w_\\star)=0$.\n\\end{proof}\n\n\\begin{lemma}\n\tSince $\\phi$ is strongly convex and has a Lipschitz continuous gradient, it follows that\n\t\\begin{equation}\n\t\t\\mathbb{E}\\left[ \\|\\lambda_t(w_t-w_\\star)-\\gamma_t(\\nabla\\phi(w_t)-\\nabla\\phi(w_\\star))\\|^2\\right]\n\t\t\\leq\n\t\t\\max\\{ |\\lambda_t-\\gamma_t L|,|\\lambda_t-\\gamma_t \\mu| \\}^2\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right].\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\t\\begin{align}\n\t\t&\\mathbb{E}\\left[ \\|\\lambda_t(w_t-w_\\star)-\\gamma_t(\\nabla\\phi(w_t)-\\nabla\\phi(w_\\star))\\|^2\\right]\\\\\n\t\t&=\\lambda_t^2\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right] -2\\lambda_t\\gamma_t\\mathbb{E}\\left[ \\left\\right]\n\t\t+\\mathbb{E}\\left[\\|\\nabla\\phi(w_t)-\\nabla\\phi(w_\\star)\\|^2 \\right]\\\\\n\t\t\\label{ineq2.1}\n\t\t&\\leq \\lambda_t^2 \\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t-2\\lambda_t \\gamma_t \\mu\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t+\\gamma_t^2 L^2 \\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\\\\\n\t\t&=(\\lambda_t^2-2\\lambda_t\\gamma_t\\mu+\\gamma_t^2 L^2)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\\\\\n\t\t&\\leq \\max\\{ |\\lambda_t-\\gamma_t L|,|\\lambda_t-\\gamma_t \\mu| \\}^2\n\t\t\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right].\n\t\\end{align}\n\tThe second term in (\\ref{ineq2.1}) due to the strongly convex of $\\phi$ and the last term of (\\ref{ineq2.1}) due to the Lipschitz continuous of $\\nabla\\phi$.\n\\end{proof}\n\nFor the second term, we must make some assumption about the statistics of the random function\n\\begin{equation}\n\t\\psi(w;z_t):=g(w,z_t)-\\nabla \\phi(w).\n\\end{equation}\n\\begin{lemma}\n\tWhen $t$ is large enough, we have\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\n\t\t\t\\leq (1-2\\gamma_t\\ell+\\gamma_t^2L^2)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{\\alpha}{t}D^2\n\t\t\t+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right].\n\t\t\\end{split}\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\tAbove all, we can obtain the following result\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t\t&\\leq(2-\\lambda_t)\\max\\{ |\\lambda_t-\\gamma_t L|,|\\lambda_t-\\gamma_t \\mu| \\}^2\n\t\t\t\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]+(2-\\lambda_t)(1-\\lambda_t)D^2\\\\\n\t\t\t&+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right]\\\\\n\t\t\t&=(2-\\lambda_t)(\\lambda_t^2-2\\lambda_t\\gamma_t \\mu +\\gamma_t^2 L^2)\n\t\t\t\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]+(2-\\lambda_t)(1-\\lambda_t)D^2\\\\\n\t\t\t&+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right]\\\\\n\t\t\\end{split}\n\t\\end{equation}\n\tSince $\\lambda_t=(1-\\frac{1}{t})^\\alpha\\approx 1-\\frac{\\alpha}{t}$, we have\n\t$$\n\t(2-\\lambda_t)(1-\\lambda_t)\n\t=2-3\\lambda_t+\\lambda_t^2\n\t\\approx 2-3(1-\\frac{\\alpha}{t})+(1-\\frac{2\\alpha}{t})\n\t=\\frac{\\alpha}{t}\n\t$$\n\tand\n\t\\begin{align}\n\t\t&(2-\\lambda_t)(\\lambda_t^2-2\\lambda_t\\gamma_t \\mu +\\gamma_t^2 L^2)\\\\\n\t\t&=2\\lambda_t^2-4\\lambda_t\\gamma_t\\mu+2\\gamma_t^2 L^2\n\t\t-\\lambda_t^3+2\\lambda_t^2 \\gamma_t\\mu-\\lambda_t\\gamma_t^2 L^2\\\\\n\t\t&=(2\\lambda_t^2-\\lambda_t^3)+(-4\\lambda_t+2\\lambda_t^2)\\gamma_t\\mu\n\t\t+(2-\\lambda_t)\\gamma_t^2 L^2\\\\\n\t\t&\\approx (2(1-\\frac{2\\alpha}{t})-(1-\\frac{3\\alpha}{t}))\n\t\t+(-4(1-\\frac{\\alpha}{t})+2(1-\\frac{2\\alpha}{t}))\\gamma_t\\mu\n\t\t+(1+\\frac{\\alpha}{t})\\gamma_t^2 L^2\\\\\n\t\t&=(1-\\frac{\\alpha}{t}) - 2\\gamma_t\\mu + (1+\\frac{\\alpha}{t})\\gamma_t^2 L^2\n\t\\end{align}\n\tSo we have\n\t\\begin{align}\n\t\t\\notag\n\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t&=((1-\\frac{\\alpha}{t}) - 2\\gamma_t\\mu + (1+\\frac{\\alpha}{t})\\gamma_t^2 L^2)\n\t\t\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]+\\frac{\\alpha}{t}D^2\n\t\t+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right]\\\\\n\t\t\\notag\n\t\t&=(1-2\\gamma_t\\mu+\\gamma_t^2L^2)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t+\\frac{\\alpha}{t}D^2\n\t\t+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right]\\\\\n\t\t\\label{ineq2.5}\n\t\t&+\\frac{\\alpha}{t}(\\gamma_t^2L^2-1)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\\\\\n\t\t&\\leq (1-2\\gamma_t\\mu+\\gamma_t^2L^2)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t+\\frac{\\alpha}{t}D^2\n\t\t+\\gamma_t^2 \\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right].\n\t\\end{align}\n\tThe last term in (\\ref{ineq2.5}) would be smaller than zero if $t$ is large enough due to the diminishing $\\gamma_t$.\n\\end{proof}\n\nLet's explore some possibilities on $\\psi$.\n\n\\begin{enumerate}\n\t\\item {$\\psi=0$}\\\\\n\tIn the case when there is no randomness at all and we are just following the gradient.\n\tSo we can have the upper bound\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t\\|w_{t+1}-w_\\star\\|^2\n\t\t\t&\\leq(1-2\\gamma_t\\mu+\\gamma_t^2L^2)\\|w_t-w_\\star\\|^2\n\t\t\t+\\frac{\\alpha}{t}D^2\\\\\n\t\t\t&\\leq \\max\\{ |1-\\gamma_t L|,|1-\\gamma_t \\mu| \\}^2 \\|w_t-w_\\star\\|^2\n\t\t\t+\\frac{\\alpha}{t}D^2.\n\t\t\\end{split}\n\t\\end{equation}\n\tSetting $\\alpha=0$, and $\\gamma_t=\\frac{2}{L+\\mu}$ for all $t$, we find that\n\t\\begin{align}\n\t\t\\|w_{t+1}-w_\\star\\| &\\leq \\left( \\frac{L-\\mu}{L+\\mu} \\right)\\|w_t-w_\\star\\|\\\\\n\t\t&=\\left( \\frac{L-\\mu}{L+\\mu} \\right)^t D\n\t\\end{align}\n\tThat is, a constant step-size policy converges at a linear rate.\n\t\n\t\\item {$\\psi$ bounded}\\\\\n\tThe simplest non-trivial assumption is that the deviations are bounded:\n\t\\begin{equation}\n\t\t\\|\\psi(w;z_t)\\| \\leq P\n\t\\end{equation}\n\tfor some universal constant $P$.\n\tIn this case, we have the upper bound\n\t\\begin{align}\n\t\t\\notag\n\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t&\\leq(1-2\\gamma_t\\mu+\\gamma_t^2L^2)\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t+\\frac{\\alpha}{t}D^2\n\t\t+\\gamma_t^2 P^2\n\t\\end{align}\n\t\\begin{theorem}\n\t\tsimple dual averaging method can achieve $O(1/\\sqrt{t})$ rate by choosing $\\alpha=\\frac{1}{2}$ and \n\t\t\\begin{equation}\n\t\t\t\\gamma_t=\\frac{1}{\\sqrt{t}\\ell}.\n\t\t\\end{equation}\n\t\tAnd\n\t\t\\begin{equation}\n\t\t\t\\mathbb{E}\\left[\\|w_{t}-w_\\star\\|^2 \\right]\\leq\n\t\t\t\\frac{1}{\\sqrt{t}}\n\t\t\t\\left(\\frac{L^2+\\mu^2}{\\mu^2} D^2+\\frac{P^2}{\\mu^2}\\right)\n\t\t\\end{equation}\n\t\\end{theorem}\n\t\\begin{proof}\n\t\tIn this case, it follows by induction.\n\t\t\n\t\tTo verify this inequality, note that for $t=0$, the right hand side is greater than $\\|w_0-w_\\star\\|^2=D^2$ due to $\\mu \\leq L$.\n\t\t\n\t\tAssuming that the inequality holds for $t$, observe\n\t\t\\begin{align}\n\t\t\t\\notag\n\t\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t\t&\\leq (1-\\frac{2}{\\sqrt{t}}+\\frac{L^2}{t\\mu^2})\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{\\alpha}{t}D^2+\\frac{1}{t\\ell^2} P^2\\\\\n\t\t\t&=(1-\\frac{2}{\\sqrt{t}})\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{1}{t}\\left(\\frac{L^2}{\\mu^2}\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&\\leq (1-\\frac{2}{\\sqrt{t}})\\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{1}{t}\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&\\leq (1-\\frac{2}{\\sqrt{t}}) \\frac{1}{\\sqrt{t}} \\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\n\t\t\t+\\frac{1}{t}\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&=\\left( \\sqrt{t}-1 \\right) \\frac{1}{t}\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&=\\frac{(\\sqrt{t}-1)\\sqrt{t+1}}{t} \\frac{1}{\\sqrt{t+1}}\n\t\t\t\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&\\leq \\frac{\\sqrt{t-1}\\sqrt{t+1}}{t} \\frac{1}{\\sqrt{t+1}}\n\t\t\t\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\\\\\n\t\t\t&\\leq \\frac{1}{\\sqrt{t+1}}\n\t\t\t\\left(\\frac{L^2}{\\mu^2}D^2+\\alpha D^2+\\frac{P^2}{\\mu^2}\\right)\n\t\t\t=\\frac{1}{\\sqrt{t+1}}\n\t\t\t\\left(\\frac{L^2+\\mu^2}{\\mu^2} D^2+\\frac{P^2}{\\mu^2}\\right).\n\t\t\\end{align}\n\t\\end{proof}\n\t\n\t\\item {$\\psi$ Lipschitz}\\\\\n\t\n\t\n\tIf we add additional assumptions about the behavior of $\\phi$, we can derive considerably faster convergence.\n\tIn particular, suppose $\\psi$ is Lipschitz in expectation:\n\t\\begin{equation}\n\t\t\\mathbb{E} \\left[ \\|\\psi(w)\\|^2 \\right] \\leq \\theta^2 \\|w-w_\\star\\|^2 \\quad \\forall w.\n\t\\end{equation}\n\tIn this case, we have a bound of the form\n\t\\begin{equation}\n\t\t\\begin{split}\n\t\t\t&\\mathbb{E}\\left[\\|w_{t+1}-w_\\star\\|^2 \\right]\\\\\n\t\t\t&\\leq \\left( \\max\\{ |1-\\gamma_t L|,|1-\\gamma_t \\mu| \\}^2+\\gamma_t^2 \\theta^2 \\right) \\mathbb{E}\\left[\\|w_t-w_\\star\\|^2 \\right]\n\t\t\t+\\frac{\\alpha}{t}D^2.\n\t\t\\end{split}\n\t\\end{equation}\n\tNow we can always select a constant $\\gamma_t$ (i.e. $\\alpha=0$) that provides a linear convergence rate.\n\tIndeed, if $\\theta \\le \\sqrt{\\mu L}$, then setting $\\gamma_t=\\frac{2}{\\mu+L}$ gives\n\t\\begin{equation}\n\t\t\\mathbb{E}\\left[\\|w_{t}-w_\\star\\|^2 \\right]\n\t\t\\leq \\left( 1-\\frac{4(L/\\mu -\\theta^2/\\mu^2)}{(1+L/\\mu)^2} \\right)^t D^2.\n\t\\end{equation}\n\tOtherwise, setting $\\gamma_t=\\frac{\\mu}{\\mu^2 + \\theta^2}$, we achieve\n\t\\begin{equation}\n\t\t\\mathbb{E}\\left[\\|w_{t}-w_\\star\\|^2 \\right]\n\t\t\\leq \\left( 1+\\mu^2/\\theta^2 \\right)^{-t} D^2.\n\t\\end{equation}\n\\end{enumerate}\n\n\\subsection{Convergence of RDA}\nFor clarity, we have the following general assumptions:\n\\begin{itemize}\n\t\\item The regularization term $\\Psi(w)$ is a closed convex function with convexity parameter $\\sigma$ and $\\text{dom}\\Psi$ is closed.\n\t\\item For each $t\\geq 1$, $f_t(w)$ is convex and subdifferentiable on dom$\\Psi$.\n\t\\item $h(w)$ is strongly convex on dom$\\Psi$ and subdifferentiable on rint(dom$\\Psi$) and also satisfies\n\t\\begin{equation}\\label{assumpation7}\n\t\tw_0=\\arg\\min_w h(w)\\in \\text{Arg}\\min_w \\Psi(w).\n\t\\end{equation}\n\t\n\tWithout loss of generality, assume $h(w)$ has convexity parameter 1 and $\\min_w h(w)=0$.\n\t\\item There exist a constant $G$ such that \n\t\\begin{equation}\\label{assumpation12}\n\t\t\\|g_t\\|_{\\ast}\\leq G,~\\forall t\\geq 1.\n\t\\end{equation}\n\t\\item Require $\\{\\beta\\}_t$ be a nonnegative and nondecreasing sequence and \n\t\\begin{equation}\\label{assumpation13}\n\t\t\\max\\{\\sigma,\\beta_1\\}>0.\n\t\\end{equation}\n\\end{itemize}\nFor general convex regularizations $\\Psi(w)$, The RDA method ensure the $R_t(w) = O(\\sqrt{t})$ regret bound and $O(\\frac{1}{\\sqrt{t}})$ convergence rate with $\\beta_t=O(\\sqrt{t})$, which are same to the results of online subgradient method.\nFor strongly convex regularizations, setting $\\beta_t=O(\\ln t)$ yields the improved regret bound $O(\\ln t)$ and $O(\\frac{\\ln t}{t})$ convergence rate. The ideas of the detailed proofs are that we first found the regret bound for RDA method and then prove the convergence of cost function $\\phi(\\bar{w_t})$ and primal variables $w_{t+1}$ based on the regret bound. \n\\subsubsection{Regret bounds}\nThe learner's \\emph{regret} of online learning is the difference between his cumulative loss and the cumulative loss of the optimal fixed hypothesis, which is defined by\n\\begin{equation}\\label{regret}\n\tR_t(w)=\\sum_{\\tau=1}^{t}(f_{\\tau}(w_{\\tau})+\\Psi(w_{\\tau})) - \\sum_{\\tau=1}^{t}(f_{\\tau}(w)+\\Psi(w)),\n\\end{equation}\nand bounded by\n\\begin{equation}\n\t\\Delta_t = \\beta_t D^2 +\\frac{G^2}{2}\\sum\\limits_{\\tau=0}^{t-1}\\frac{1}{\\sigma\\tau+\\beta_\\tau}+\n\t\\frac{2(\\beta_0 -\\beta_1)G^2}{(\\beta_1+\\sigma)^2},\n\\end{equation}\nwhere $\\beta_0=\\max\\{\\sigma, \\beta_1\\}$. We could always set $\\beta_1\\geq \\sigma$ so that $\\beta_0=\\beta_1$ and therefore $\\frac{2(\\beta_0 -\\beta_1)G^2}{(\\beta_1+\\sigma)^2}$ vanishes.\n\\begin{lemma}\\label{lemma_regret}\n\tLet the sequence $\\{w_t\\}_{t\\geq 1}$ and $\\{g_t\\}_{t\\geq 1}$ be generated by RDA method, and assume (\\ref{assumpation12}) and (\\ref{assumpation13}) hold. Then for any $t\\geq 1$ and any $w\\in \\mathcal{F}_D=\\{w\\in\\text{dom}\\Psi|h(w)\\leq D^2\\}$, we have\n\t\\begin{itemize}\n\t\t\\item [(a)] The regret defined in (\\ref{regret}) is bounded by $\\Delta_t$\n\t\t\\begin{equation}\n\t\t\tR_t(w)\\leq \\Delta_t\n\t\t\\end{equation}\n\t\t\\item [(b)] The primal variables are bounded as\n\t\t\\begin{equation}\n\t\t\t\\|w_{t+1}-w\\|^2\\leq \\frac{2}{\\sigma t+\\beta_t}(\\Delta_t-R_t(w)).\n\t\t\\end{equation}\n\t\\end{itemize}\n\\end{lemma}\n\\textbf{Proof of Part (a)} First, we define the following $gap$ sequence which measures the quality of the solutions $w_1,..,w_t$:\n\\begin{equation}\\label{52}\n\t\\delta_t = \\max\\limits_{w\\in \\mathcal{F}_D} \\left\\{ \\sum\\limits_{\\tau=1}^{t} \n\t\\big( \\langle g_{\\tau},w_{\\tau}-w \\rangle + \\Psi(w_{\\tau}) \\big) - t \\Psi(w)\\right\\},~\n\tt=1,2,3,....\n\\end{equation}\nand $\\delta_t$ is an upper bound on the regret $R_t(w)$ for all $w\\in \\mathcal{F}_D$, to see this, we use the convexity of $f_t(w)$ in the following:\n\\begin{equation}\\label{53}\n\t\\delta_t \\geq \\sum_{\\tau=1}^{t} \\left(f_{\\tau}(w_{\\tau}) - f_{\\tau}(w) + \\Psi(w_{\\tau})\\right) -t \\Psi(w) = R_t(w).\t\n\\end{equation}\nThen, We are going to derive an upper bound on $\\delta_t$. For this purpose, we subtract $\\sum_{\\tau=1}^{t} \\langle g_{\\tau},w_0 \\rangle$ in (\\ref{52}), which leads to \n\\begin{equation}\\label{54}\n\t\\delta_t = \\sum\\limits_{\\tau=1}^{t} \\left( \\langle g_{\\tau},w_{\\tau}-w_0 \\rangle + \\Psi(w_{\\tau})\\right) +\n\t\\max\\limits_{w\\in \\mathcal{F}_D} \\left\\{ \\langle s_t,w_0 -w \\rangle - t \\Psi(w) \\right\\},\n\\end{equation}\nthe maximization term in (\\ref{54}) is in fact $U_t(-s_t)$, therefore, by applying Lemma \\ref{lemma9}, we have\n\\begin{equation}\\label{55}\n\t\\delta_t\\leq \\sum\\limits_{\\tau=1}^{t} \\big( \\langle g_{\\tau},w_{\\tau}-w_0 \\rangle + \\Psi(w_{\\tau})\\big) +V_t(-s_t)+\\beta_t D^2.\n\\end{equation}\nNext, we show that $\\Delta_t$ is an upper bound for the right-hand side of inequality (\\ref{55}). We consider $\\tau \\geq 2$ and $\\tau =1$ respectively.\\\\\nFor any $\\tau \\geq 2$, we have\n$$\nV_{\\tau}(-s_{\\tau})+\\Psi(w_{\\tau+1})\\leq V_{\\tau-1}(-s_{\\tau-1})+\\langle-g_{\\tau},w_{\\tau}-w_0\\rangle+\\frac{\\|g_{\\tau}\\|^2_{\\ast}}{2(\\sigma(\\tau-1)+\\beta_{\\tau -1})},\n$$\nwhere (\\ref{49}),(\\ref{42}),(\\ref{48}) and (\\ref{47}) are used. Therefore, we have\n$$\n\\langle g_{\\tau},w_{\\tau}-w_0\\rangle +\\Psi(w_{\\tau+1})\\leq \nV_{\\tau-1}(-s_{\\tau-1}) - V_{\\tau}(-s_{\\tau}) + \\frac{\\|g_{\\tau}\\|^2_{\\ast}}{2(\\sigma(\\tau-1)+\\beta_{\\tau -1})},~~\\forall\\tau\\geq 2.\n$$\nFor $\\tau =1$, we have a similar inequality\n$$\n\\langle g_{1},w_1-w_0\\rangle +\\Psi(w_2)\\leq \nV_{0}(-s_{0}) - V_{1}(-s_{1}) + \\frac{\\|g_{1}\\|^2_{\\ast}}{2\\beta_0} +(\\beta_0-\\beta_1)h(w_2),\n$$\nwhere the last term comes from (\\ref{50}). Summing the above inequalities for $\\tau=1,...,t$ and noting that $V_0(-s_0)=V_0=0$, we arrive at\n$$\n\\sum\\limits_{\\tau=1}^{t} \\big( \\langle g_{\\tau},w_{\\tau}-w_0 \\rangle + \\Psi(w_{\\tau+1}) +V_t(-s_t) \\leq\n(\\beta_0-\\beta_1)h(w_2)+\\frac{1}{2} \\sum\\limits_{\\tau=1}^{t} \\frac{\\|g_{\\tau}\\|^2_{\\ast}}{2(\\sigma(\\tau-1)+\\beta_{\\tau -1})}.\n$$ \nUsing $\\Psi(w_{t+1})\\geq \\Psi(w_0)$ and adding the nonpositive quantity $\\Psi(w_1)-\\Psi(w_{t+1})$ to the left hand side of the above inequality yields\n\\begin{equation}\\label{56}\n\t\\sum\\limits_{\\tau=1}^{t} \\big( \\langle g_{\\tau},w_{\\tau}-w_0 \\rangle + \\Psi(w_{\\tau}) +V_t(-s_t) \\leq\n\t(\\beta_0-\\beta_1)h(w_2)+\\frac{1}{2} \\sum\\limits_{\\tau=1}^{t} \\frac{\\|g_{\\tau}\\|^2_{\\ast}}{2(\\sigma(\\tau-1)+\\beta_{\\tau -1})}.\n\\end{equation}\nCombing (\\ref{53}), (\\ref{55}) and (\\ref{56}) and using Lemma \\ref{lemma12}, we conclude\n$$\nR_t(w)\\leq \\delta_t\\leq \\Delta_t.\n$$\n\\textbf{Proof of Part (b)} First, We start with the optimality condition for the minimization problem in (\\ref{RDA subproblem}): there exist subgradients $b_{t+1}\\in \\partial \\Psi(w_{t+1})$ and $d_{t+1}\\in \\partial h(w_{t+1})$ such that\n\\begin{equation}\\label{57}\n\t\\langle s_t+tb_{t+1}+\\beta_t d_{t+1},w-w_{t+1} \\rangle \\geq 0,~~ \\forall w\\in \\text{dom} \\Psi.\n\\end{equation}\nBy the strong convexity of $h$ and $\\Psi$, we have for any $w\\in \\text{dom} \\Psi$\n\\begin{align}\n\t&\\Psi(w) \\geq \\Psi(w_{t+1})+\\langle b_{t+1},w-w_{t+1} \\rangle+\\frac{\\sigma}{2} \\|w_{t+1}-w\\|^2,\\label{58}\\\\\n\t&h(w) \\geq h(w_{t+1})+\\langle d_{t+1},w-w_{t+1} \\rangle+\\frac{1}{2} \\|w_{t+1}-w\\|^2,\\label{59}\n\\end{align}\nMultiplying both sides of (\\ref{58}) by t and both sizes of (\\ref{59}) by $\\beta_t$, then adding them together, we have\n\\begin{equation}\\label{60}\n\t\\begin{split}\n\t\t\\frac{1}{2}(\\sigma t+\\beta_t)\\|w_{t+1}-w\\|^2 \n\t\t&\\leq \\beta_t h(w)-\\beta_t h(w_{t+1})-\\langle tb_{t+1}+\\beta_t d_{t+1},w-w_{t+1} \\rangle +t\\Psi(w)-t\\Psi(w_{t+1})\\\\\n\t\t&\\leq \\beta_t h(w)-\\beta_t h(w_{t+1})+\\langle s_t,w-w_{t+1} \\rangle +t\\Psi(w)-t\\Psi(w_{t+1})\\\\\n\t\t&\\leq \\beta_t h(w)+\\{\\langle -s_t,w_{t+1}-w_0 \\rangle -t\\Psi(w_{t+1})-\\beta_t h(w_{t+1})\\} +t\\Psi(w)+\\langle s_t,w-w_0 \\rangle\\\\\n\t\t&\\leq \\beta_t h(w)+V_t(-s_t) +t\\Psi(w)+\\langle s_t,w-w_0 \\rangle,\\\\\n\t\\end{split}\n\\end{equation}\nthe second and fourth inequalities hold by using (\\ref{57}) and (\\ref{V}).\\\\\nThen, we rewrite the right-hand side of (\\ref{60}) by expanding $\\langle s_t,w-w_0 \\rangle$, adding and subtracting $\\sum\\limits_{\\tau =1}^{t}\\Psi(w_{\\tau})$ to get\n\\begin{equation*}\n\t\\beta_t h(w)+ \\left\\{ V_t(-s_t)+ \\sum\\limits_{\\tau =1}^{t}\\big( \\langle g_{\\tau} ,w_{\\tau}-w_0\\rangle +\\Psi(w_{\\tau}) \\big) \\right\\} \n\t+\\left\\{ \\sum\\limits_{\\tau =1}^{t}\\big( \\langle g_{\\tau} , w - w_{\\tau}\\rangle +t\\Psi(w)-\\sum\\limits_{\\tau =1}^{t}\\Psi(w_{\\tau}) \\right\\},\n\\end{equation*}\nthe terms in the first braces are exactly the left-hand side of (\\ref{56}) and those in the second braces are bounded by $-R_t(w)$.\\\\\nFinally, putting everything together, using Lemma \\ref{lemma12} and assumption \\ref{assumpation12}, we have\n\\begin{equation}\n\t\\begin{split}\n\t\t\\frac{1}{2}(\\sigma t+\\beta_t)\\|w_{t+1}-w\\|^2 &\\leq \\beta_t h(w)+\n\t\t(\\beta_0-\\beta_1)h(w_2)+\\frac{1}{2}\\sum\\limits_{\\tau =1}^{t}\\frac{\\|g_{\\tau}\\|^2_{\\ast}}{\\sigma(\\tau -1)+\\beta_{\\tau -1}}-\n\t\tR_t(w)\\\\\n\t\t&\\leq \\Delta_t-R_t(w).\n\t\\end{split}\n\\end{equation}\n\n\\subsubsection{Convergence rates}\n\\begin{theorem}\\label{2}\n\tAssume there exists an optimal solution $w^{\\star}$ to the problem (1) that satisfies $h(w^{\\star})\\leq D^2$ for some $D>0$, and let $\\phi^{\\star}=\\phi(w^{\\star}).$ Let the sequences $\\{w_t\\}_{t\\geq 1}$ be generated by RDA method, and assume $\\|g_t\\|_{\\ast}\\leq G$ for some constant $G$. Then for any $t\\geq 1$,\n\t\\begin{itemize}\n\t\t\\item[(a)] the expected cost associated with the random variable $\\bar{w}_t$ is bounded as\n\t\t$$\n\t\t\\mathbf{E}\\phi(\\bar{w}_t)-\\phi^{\\star}\n\t\t\\leq \\frac{1}{t}\\Delta_t.\n\t\t$$\n\t\t\\item[(b)] the primal variables are bounded as \n\t\t$$\n\t\t\\mathbf{E}\\|w_{t+1}-w^{\\star}\\|^2\\leq \\frac{2}{\\sigma t+\\beta_t}\\Delta_t.\n\t\t$$\n\t\\end{itemize}\n\\end{theorem}\n\\textbf{Proof} First, from the definition (\\ref{regret}), We have the regret at $w^{\\star}$\n$$\nR_t(w^{\\star})=\\sum_{\\tau=1}^{t}(f(w_{\\tau},z_{\\tau})+\\Psi(w_{\\tau})) - \\sum_{\\tau=1}^{t}(f(w^{\\star},z_{\\tau})+\\Psi(w^{\\star})),\n$$\nLet $\\mathbf{z}[t]$ denote the collection of i.i.d. random variables $(z_,...,z_t)$. We note that the random variable $w_{\\tau}$, where $1\\leq w \\geq t$, is a function of $(z_1,...,z_{\\tau -1})$ and is independent of $(z_{\\tau},...,z_t)$. Therefore\n$$\n\\mathbf{E}_{\\mathbf{z}[t]}\\left( f(w_{\\tau},z_{\\tau})+\\Psi(w_{\\tau})\\right)=\n\\mathbf{E}_{z[\\tau-1]} \\left( \\mathbf{E}_{\\tau} f(w_{\\tau},z_{\\tau})+\\Psi(w_{\\tau})\\right)=\n\\mathbf{E}_{z[\\tau-1]}\\phi(w_{\\tau})=\n\\mathbf{E}_{z[t]}\\phi(w_{\\tau}),\n$$\nand \n$$\n\\mathbf{E}_{\\mathbf{z}[t]}\\left( f(w^{\\star},z_{\\tau})+\\Psi(w^{\\star})\\right)=\n\\mathbf{E}_{\\tau} f(w^{\\star},z_{\\tau})+\\Psi(w^{\\star}) = \\phi(w^{\\star}) = \\phi^{\\star}.\n$$\nSince $\\phi^{\\star}=\\phi(w^{\\star})=\\min\\limits_w\\phi(w)$, we have the expected regret\n\\begin{equation}\\label{exp_regret}\n\t\\mathbf{E}_{\\mathbf{z}[t]} R_t(w^{\\star})=\\sum_{\\tau=1}^{t} \\mathbf{E}_{\\mathbf{z}[t]} \\phi(w_\\tau)-t\\phi^{\\star}\\geq 0.\n\\end{equation}\nThen, by convexity of $\\phi$, we have \n\\begin{equation}\\label{convexinequ}\n\t\\phi(\\bar{w}_t)=\\phi\\left(\\frac{1}{t} \\sum_{\\tau=1}^{t} w_\\tau \\right) \\leq\n\t\\frac{1}{t}\\sum_{\\tau=1}^{t}\\phi\\left( w_\\tau \\right).\n\\end{equation}\nFinally, from (\\ref{convexinequ}) and (\\ref{exp_regret}), we have\n$$\n\\mathbf{E}_{\\mathbf{z}[t]} \\phi(\\bar{w}_t)-\\phi^{\\star} \\leq \n\\frac{1}{t}\\left(\\sum_{\\tau=1}^{t}\t\\mathbf{E}_{\\mathbf{z}[t]}\\phi( w_\\tau)-t\\phi^{\\star}\\right) =\n\\frac{1}{t}\\mathbf{E}_{\\mathbf{z}[t]} R_t(w^{\\star}).\n$$\nThen part (a) follows from that of Lemma \\ref{lemma_regret}. Similarly and obviously, part (b) follows from that of Lemma \\ref{lemma_regret} and (\\ref{exp_regret}).\n\\\\\nSome conclusions from the theorem:\n\\begin{enumerate}\n\t\\item \n\tFrom the part (a) of Theorem \\ref{2}, the expected cost associated with the random variable $\\bar{w}_t$ is bounded as\n\t\\begin{equation}\\label{convergencephi}\n\t\t\\mathbf{E}\\phi(\\bar{w}_t)-\\phi^{\\star}\n\t\t\\leq \\frac{1}{t} \\left(a \\beta_t +b\\sum\\limits_{\\tau=0}^{t-1}\\frac{1}{\\sigma\\tau+\\beta_\\tau}\\right),\n\t\\end{equation}\n\twhere $a$ and $b$ are some positive constants and $\\sigma$ is the convexity parameter of $\\Psi(w)$. Here, we consider $\\ell_1$ regularization function $\\Psi(w)=\\|w\\|_1$ and it is a convex but not strongly convex function, which means $\\sigma=0$.\\\\\n\tNow, we consider how to choose $\\beta_t$ for $t\\geq 1$ and $\\beta_0 = \\beta_1$. First if $\\beta_t = t$, we have $\\frac{1}{t}\\cdot a t = a$, which means the expected cost does not converge. Then assume $\\beta_t=t^{\\alpha}$ and $\\alpha\\neq 1$, the right hand side of the inequality (\\ref{convergencephi}) becomes \n\t$$\n\t\\frac{1}{t} \\left(a t^{\\alpha} +b\\sum\\limits_{\\tau=0}^{t-1}\\frac{1}{\\tau^{\\alpha}}\\right) \\leq \n\t\\frac{1}{t} \\left[ a t^{\\alpha} +b \\big( 2+ \\sum\\limits_{\\tau=2}^{t-1}\\frac{1}{\\tau^{\\alpha}} \\big) \\right] \\leq\n\t\\frac{1}{t} \\left[ a t^{\\alpha} +b \\big( 2+ \\int_{1}^{t-1} \\frac{1}{\\tau^{\\alpha}} \\big) \\right] \t\t \\sim\n\tO(t^{\\alpha-1} + t^{-\\alpha}),\n\t$$\n\tfrom above, we see that if $0<\\alpha<1$, the expected cost converges and the optimal convergence rate $O(t^{-\\frac{1}{2}})$ achieves when $\\alpha=\\frac{1}{2}$.\n\t\\item \n\tFrom the part (b) of Theorem \\ref{2}, the primal variables are bounded as \n\t$$\n\t\\mathbf{E}\\|w_{t+1}-w^{\\star}\\|^2\\leq \\frac{2}{\\sigma t+\\beta_t}\\left(a \\beta_t +b\\sum\\limits_{\\tau=0}^{t-1}\\frac{1}{\\sigma\\tau+\\beta_\\tau}\\right)\n\t= \\frac{2}{\\beta_t}\\left(a \\beta_t +b\\sum\\limits_{\\tau=0}^{t-1}\\frac{1}{\\beta_\\tau}\\right).\n\t$$\n\tThus $\\mathbf{E}\\|w_{t+1}-w^{\\star}\\|^2$ is bounded by a constant.\n\t\\item \n\tNext, we consider how the coefficients $\\gamma$ of $\\beta_t = \\gamma \\sqrt{t}$ and $\\lambda$ of $\\Psi(w)=\\lambda \\| w \\|_1$ affect the performance of convergence and accuracy. Note that in (\\ref{convergencephi}), $a=D^2$ and $b=G^2$, where $h(w)\\leq D^2$ and $\\|g_t\\|_2 \\leq G$. We rewrite (\\ref{convergencephi}) as\n\t$$\n\t\\mathbf{E}\\phi(\\bar{w}_t)-\\phi^{\\star} = \n\t\\mathbf{E}_{z_\\tau}\\big(f(\\bar{w_t},z_{\\tau})-f(w^{\\star},z_{\\tau})\\big)+\\lambda\\big(\\|\\bar{w_t}\\|_1-\\|{w_t}^{\\star}\\|_1\\big)\n\t\\leq\n\t(\\gamma D^2 +\\frac{G^2}{\\gamma})\\frac{1}{\\sqrt{t}}\n\t$$\n\t\\begin{itemize}\n\t\t\\item \n\t\tMinimizing $\\left(\\gamma D^2 +\\frac{G^2}{\\gamma}\\right)$ yields $\\gamma = \\frac{G}{D}$.\n\t\t\\item \n\t\t$\\lambda$ stays at the left hand side of the inequality above.\n\t\\end{itemize}\n\\end{enumerate}\n\n\n\\subsubsection{Some Lemmas}\nFirst of all, we define some functions:\n\\begin{align}\n\t&U_t(s)=\\max\\limits_{w\\in \\mathcal{F}_D} \\{ \\langle s,w-w_0 \\rangle-t\\Psi(w)\\},\\label{U}\\\\\n\t&V_t(s)=\\max\\limits_{w} \\{ \\langle s,w-w_0 \\rangle-t\\Psi(w) - \\beta_t h(w)\\}\\label{V}.\n\\end{align}\nThe maximum in (\\ref{U}) is always achieved because $\\mathcal{F}_D=\\{w\\in\\text{dom}\\Psi|h(w)\\leq D^2\\}$ is a nonempty compact set. Because of (\\ref{assumpation13}), we have $\\sigma t+\\beta_t \\geq \\beta_0 > 0$ for all $t\\geq0$, which means $t\\Psi(w)+\\beta_t h(w)$ are all strongly convex, therefore the maximum in (\\ref{V}) is always achieved and unique. As a result, we have $\\text{dom} U_t= \\text{dom} V_t = E^{\\ast}$ for all $t\\geq 0$. Moreover, by the assumption (\\ref{assumpation7}), both of the functions are nonnegative.\\\\\nLet $s_t$ denote the sum of the subgradients obtained up to time $t$ in RDA method,that is\n\\begin{equation}\\label{42}\n\ts_t=\\sum\\limits_{\\tau =1}^{t} g_{\\tau} = t \\bar{g}_t,\n\\end{equation}\nand $\\pi_t(s)$ denotes the unique maximizer in the definition of $V_t(s)$\n\\begin{equation}\n\t\\begin{split}\n\t\t\\pi_t(s) &= \\arg\\max\\limits_{w} \\{ \\langle s,w-w_0 \\rangle-t\\Psi(w) - \\beta_t h(w)\\}\\\\\n\t\t&= \\arg\\min\\limits_{w} \\{ \\langle -s,w \\rangle + t\\Psi(w) + \\beta_t h(w)\\},\n\t\\end{split}\n\\end{equation}\nwhich then gives \n\\begin{equation}\n\tw_{t+1} = \\pi_t(-s_t).\n\\end{equation}\n\\begin{lemma}\\label{lemma9}\n\tFor any $s\\in E^{\\ast}$ and $t\\geq 0$, we have\n\t\\begin{equation}\n\t\tU_t(s)0$. Let $h(w)=\\frac{1}{\\sigma}\\Psi(w)$ if $\\sigma\\geq 0$. Then\n\t\\begin{equation}\\label{51}\n\t\th(w_2)< \\frac{2\\|g_1\\|^2_{\\ast}}{(\\beta_1+\\sigma)^2}.\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\tFor $t=1$, we have $w_1=w_0$, $\\Psi(w_1)=\\Psi(w_0)=0$, $h(w_1)=h(w_0)=0$, and $\\bar{g}_1=g_1$. Since $w_2$ is the minimizer in (\\ref{RDA}) for $t=1$, we have\n\t$$\n\t\\langle g_1,w_2 \\rangle +\\Psi(w_2) +\\beta_1 h(w_2) \\leq \\langle g_1,w_1 \\rangle + \\Psi(w_1)+\\beta_1 h(w_1) = \\langle g_1,w_1 \\rangle.\n\t$$\n\tTherefore,\n\t$$\n\t\\Psi(w_2) + \\beta_1 h(w_2)\\leq \\langle g_1,w_1-w_2 \\rangle \\leq \\|g_1\\|_{*}\\|w_2-w_1\\|.\n\t$$\n\tOn the other hand, by strongly convexity of $\\Psi(w)$ and $h(w)$, we have\n\t$$\n\t\\Psi(w_2) + \\beta_1 h(w_2)\\geq \\frac{\\sigma+\\beta_1}{d2} \\|w_2-w_1\\|^2.\n\t$$\n\tCombining the last two inequalities together, we have\n\t$$\n\t\\Psi(w_2) + \\beta_1 h(w_2)\\leq \\frac{2\\|g_1\\|_{*}^{2}}{\\sigma+\\beta_1}.\n\t$$\n\tIf $\\sigma=0$, we must have $\\beta_1>0$. In this case, since $\\Psi(w_2)\\geq 0$, we have\n\t$$\n\th(w_2)\\leq \\frac{2\\|g_1\\|^2_{*}}{\\beta_1^2}=\\frac{2\\|g_1\\|^2_{*}}{(\\sigma+\\beta_1)^2}.\n\t$$\n\tIf $\\sigma>0,$ we have $\\Psi(w) = \\sigma h(w)$ by assumption and therefore\n\t$$\n\t\\Psi(w_2) + \\beta_1 h(w_2) = (\\sigma + \\beta_1) h(w_2)\\leq \\frac{2\\|g_1\\|_{*}^{2}}{\\sigma+\\beta_1},\n\t$$\n\tthen we finished the proof.\n\\end{proof}\n\n\\begin{lemma}\n\tmax-min inequality:\n\t\\begin{equation}\n\t\t\\max_x \\min_y f(x,y) \\leq \\min_y \\max_x f(x,y).\n\t\\end{equation}\n\\end{lemma}\n\\begin{proof}\n\tDenote\n\t\\begin{equation}\n\t\tg(x) = \\min_y f(x,y),\n\t\\end{equation}\n\tand\n\t\\begin{equation}\n\t\th(y) =\\max_x f(x,y).\n\t\\end{equation}\n\tLet\n\t\\begin{equation}\n\t\ta= \\mathop{\\arg \\max}_x g(x),\n\t\\end{equation}\n\tand\n\t\\begin{equation}\n\t\tb = \\mathop{\\arg \\min}_y h(y).\n\t\\end{equation}\n\tThen we have\n\t\\begin{equation}\n\t\t\\max_x \\min_y f(x,y) =\\max_x g(x) =g(a),\n\t\\end{equation}\n\tand\n\t\\begin{equation}\n\t\t\\min_y \\max_x f(x.y) = \\min h(y) = h(b).\n\t\\end{equation}\n\tSo,\n\t\\begin{equation}\n\t\tg(a)= \\min_y f(a,y) \\leq f(a,b),\n\t\\end{equation}\n\tand\n\t\\begin{equation}\n\t\th(b) = \\max_x f(x,b) \\geq f(a,b).\n\t\\end{equation}\n\tFinally, we have\n\t\\begin{equation}\n\t\t\\max_x \\min_y f(x,y) =g(a) \\leq f(a,b) \\leq h(b) =\\min_x \\max_y f(x,y).\n\t\\end{equation}\n\tThis is the so-called weak duality.\n\t\n\tThe strong duality holds when there exist $(a,b)$ such that\n\t\\begin{equation}\n\t\tg(a) = h(b),\n\t\\end{equation}\n\\end{proof}\n\n\\iffalse\n\\section{Alternative Convergence Proofs for SDA and RDA}\n\\subsection{SDA Convergence Analysis}\nIn this section, we analyze the convergence of the SDA algorithm\n\\begin{equation}\\label{SDE_iteration}\n x_n = \\arg\\min_x \\left(\\displaystyle\\sum_{i = 1}^{n-1} \\langle g_i, x\\rangle + \\sqrt{n}\\|x\\|_2^2\\right),~x_1 = 0\n\\end{equation}\nwhere $g_i\\in \\partial f(x_i)$.\n\nLet us first analyze the consequences of the definition of $x_n$ as an argmin. We note that if\n$z$ is any other point, then\n\\begin{equation}\n \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, x_n\\rangle + \\sqrt{n}\\|x_n\\|_2^2 \\leq \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, z\\rangle + \\sqrt{n}\\|z\\|_2^2\n\\end{equation}\nso that\n\\begin{equation}\\label{minimum_property}\n \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, x_n - z\\rangle \\leq \\sqrt{n}\\|z\\|_2^2 - \\sqrt{n}\\|x_n\\|_2^2\n\\end{equation}\n\nThis is crucial in the proof of the following theorem\n\\begin{theorem}\n Assume that $f$ is convex and Lipschitz with constant $M$, i.e. $\\|g\\|_2 \\leq M$ for all $g\\in \\partial f(x)$ and let\n $x^*\\in \\arg\\min_x f(x)$. Then the\n iterates of (\\ref{SDE_iteration}) satisfy\n \\begin{equation}\n f(\\bar{x}_n) - f(x^*) \\leq \\frac{2(\\|x^*\\|_2^2 + M^2)}{\\sqrt{n}}\n \\end{equation}\n where $\\bar{x}_n = \\frac{1}{n}\\sum_{i = 1}^nx_i$ is the average of the first $n$ iterates.\n\\end{theorem}\n\\begin{proof}\nWe consider the average of the first $n$ iterates $\\bar{x}_n = \\frac{1}{n}\\sum_{i = 1}^nx_i$. Note that\n\\begin{equation}\n f(\\bar{x}_n) - f(x^*) \\leq \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n f(x_i) - f(x^*) \\leq \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle\n\\end{equation}\nWe rewrite the sum above as follows\n\\begin{equation}\\label{eq45}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle = \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_{n+1} - x^*\\rangle + \\displaystyle\\sum_{i = 1}^n \\displaystyle\\sum_{j = 1}^i\\langle g_j, x_i - x_{i+1}\\rangle\n\\end{equation}\nFrom observation (\\ref{minimum_property}) we obtain\n\\begin{equation}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_{n+1} - x^*\\rangle \\leq \\sqrt{n+1}\\|x^*\\|_2^2 - \\sqrt{n+1}\\|x_{n+1}\\|_2^2\n\\end{equation}\nand\n\\begin{equation}\n \\displaystyle\\sum_{j = 1}^i\\langle g_j, x_i - x_{i+1}\\rangle \\leq \\langle g_i, x_i - x_{i+1}\\rangle + \\sqrt{i}\\|x_{i+1}\\|_2^2 - \\sqrt{i}\\|x_{i}\\|_2^2\n\\end{equation}\nPlugging this into equation (\\ref{eq45}) and noting that $\\sqrt{i} - \\sqrt{i - 1} \\geq \\frac{1}{2\\sqrt{i}}$ for $i > 1$ to get\n\\begin{equation}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle \\leq \\sqrt{n+1}\\|x^*\\|_2^2 - \\|x_1\\|_2^2 - \\displaystyle\\sum_{i = 2}^{n+1}\\frac{\\|x_i\\|_2^2}{2\\sqrt{i}} + \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle\n\\end{equation}\nThrowing away the negative terms above we see that\n\\begin{equation}\\label{eq61}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle \\leq \\sqrt{n+1}\\|x^*\\|_2^2 + \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle\n\\end{equation}\nFinally, we bound the last sum above as\n\\begin{equation}\\label{eq65}\n \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle \\leq \\displaystyle\\sum_{i=1}^n \\|g_i\\|_2\\|x_i - x_{i+1}\\|_2 \\leq M\\displaystyle\\sum_{i=1}^n\\|x_i - x_{i+1}\\|_2\n\\end{equation}\nIn order to bound this last part, we need to write out $x_i - x_{i+1}$ explicitly. We see that\n$$ x_i = \\frac{1}{2\\sqrt{i}}\\displaystyle\\sum_{j = 1}^{i-1} g_j\n$$\nand thus\n$$ x_i - x_{i+1} = \\left(\\frac{1}{2\\sqrt{i}} - \\frac{1}{2\\sqrt{i+1}}\\right)\\displaystyle\\sum_{j = 1}^{i-1} g_j - \\frac{1}{2\\sqrt{i+1}} g_{i+1}\n$$\nutilizing $\\frac{1}{\\sqrt{i}} - \\frac{1}{\\sqrt{i+1}} \\leq \\frac{1}{2i^{3/2}}$ and the triangle inequality, we see that\n\\begin{equation}\n \\|x_i - x_{i+1}\\|_2 \\leq \\frac{3}{4\\sqrt{i}}M \\leq \\frac{M}{\\sqrt{i}}\n\\end{equation}\nWe now plug this into equation (\\ref{eq65}) to get\n\\begin{equation}\n \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle \\leq M^2 \\displaystyle\\sum_{i=1}^n \\frac{1}{\\sqrt{i}} \\leq 2M^2 \\sqrt{n}\n\\end{equation}\nPluggin this into equation (\\ref{eq61}) we get\n\\begin{equation}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle \\leq \\sqrt{n+1}\\|x^*\\|_2^2 + 2M^2 \\sqrt{n} \\leq 2\\sqrt{n}(\\|x^*\\|_2^2 + M^2)\n\\end{equation}\nThus we obtain\n\\begin{equation}\n f(\\bar{x}_n) - f(x^*) \\leq \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle \\leq \\frac{2(\\|x^*\\|_2^2 + M^2)}{\\sqrt{n}}\n\\end{equation}\nas desired.\n\\end{proof}\n\n\\subsection{RDA Convergence Analysis}\nWe consider the optimization problem\n\\begin{equation}\n \\arg\\min_x f(x) + R(x)\n\\end{equation}\nwhere $R(x)$ is a convex regularization term, which in our application is taken to be $\\|\\cdot\\|_1$. We analyze the convergence of the RDA algorithm\n\\begin{equation}\\label{RDA_iteration}\n x_n = \\arg\\min_x \\left(\\displaystyle\\sum_{i = 1}^{n-1} \\langle g_i, x\\rangle + \\sqrt{n}\\|x\\|_2^2 + nR(x)\\right)\n\\end{equation}\nwith $g_i\\in \\partial f(x_i)$.\n\nNote that by completing the square, (\\ref{RDA_iteration}) can be written as\n\\begin{equation}\n x_{n+\\frac{1}{2}} = \\frac{1}{2\\sqrt{n}}\\displaystyle\\sum_{i=1}^n g_i,~x_n = P_{\\sqrt{n}R}(x_{n+\\frac{1}{2}}) = \\arg\\min_x \\|x - x_{n+\\frac{1}{2}}\\|_2^2 + \\sqrt{n}R(x)\n\\end{equation}\n\nFor example, if $R = \\|\\cdot\\|_1$, we see that\n\\begin{equation}\n P_{\\sqrt{n}R}(x_{n+\\frac{1}{2}}) = \n \\begin{cases}\n x_{n+\\frac{1}{2}} + \\frac{1}{2}\\sqrt{n} & \\text{if $x_{n+\\frac{1}{2}} < -\\frac{1}{2}\\sqrt{n}$} \\\\\n 0 & \\text{if $|x_{n+\\frac{1}{2}}| \\leq \\frac{1}{2}\\sqrt{n}$} \\\\\n x_{n+\\frac{1}{2}} - \\frac{1}{2}\\sqrt{n} & \\text{if $x_{n+\\frac{1}{2}} > \\frac{1}{2}\\sqrt{n}$}\n \\end{cases}\n\\end{equation}\n\nBefore we begin our convergence proof, let us note a simple observation. Since $x_n$ is given by the $\\arg\\min$ in (\\ref{RDA_iteration}), if\n$z$ is any point, we have\n\\begin{equation}\n \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, x_n\\rangle + \\sqrt{n}\\|x_n\\|_2^2 + nR(x_n)\\leq \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, z\\rangle + \\sqrt{n}\\|z\\|_2^2 + nR(z)\n\\end{equation}\nso that\n\\begin{equation}\\label{minimum_property}\n \\displaystyle\\sum_{i = 0}^{n-1} \\langle g_i, x_n - z\\rangle \\leq \\sqrt{n}\\|z\\|_2^2 + nR(z) - \\sqrt{n}\\|x_n\\|_2^2 - nR(x_n)\n\\end{equation}\n\n\nAdditionally, we will need the following well-known lemma.\n\\begin{lemma}\\label{non-expansive}\n The proximal map $P_R$ is non-expansive for any convex function $R$, i.e.\n \\begin{equation}\n \\|P_R(x) - P_R(y)\\|_2 \\leq \\|x - y\\|_2\n \\end{equation}\n\\end{lemma}\n\\begin{proof}\n Let $x^\\prime = P_R(x)$ and $y^\\prime = P_R(y)$. Then the definition of the proximal map implies that $\\Delta x = x - x^\\prime\\in \\partial R(x^\\prime)$ and likewise for the $y$s. Thus we have\n \\begin{equation}\n \\langle x - y, x^\\prime - y^\\prime\\rangle = \\|x^\\prime - y^\\prime\\|_2^2 + \\langle\\Delta x - \\Delta y, x^\\prime - y^\\prime\\rangle \\geq \\|x^\\prime - y^\\prime\\|_2^2\n \\end{equation}\n This last inequality follows from the convexity of $R$ and the fact that $\\Delta x \\in \\partial R(x^\\prime)$ and $\\Delta y \\in \\partial R(y^\\prime)$. The Cauchy-Schwartz inequality\n now implies that\n \\begin{equation}\n \\|x^\\prime - y^\\prime\\|_2 \\leq \\|x - y\\|_2\n \\end{equation}\n as desired.\n\\end{proof}\n\n\nLetting $\\phi(x)$ denote $f(x) + R(x)$, we have the following convergence result.\n\\begin{theorem}\n Assume that $f$ is convex and Lipschitz with constant $M$, i.e. $\\|g\\|_2 \\leq M$ for all $g\\in \\partial f(x)$, and\n that $R$ is convex and Lipschitz with constant $N$ and bounded below. Let\n $x^*\\in \\arg\\min_x \\phi(x)$. Then the\n iterates of (\\ref{RDA_iteration}) satisfy\n \\begin{equation}\n\\phi(\\bar{x}_n) - \\phi(x^*) \\leq \\frac{2\\|x^*\\|_2^2 + 3M^2}{\\sqrt{n}} + \\frac{(R(x^*) - \\inf_x R(x)) + MN\\log(n)}{n}\n\\end{equation}\n where $\\bar{x}_n = \\frac{1}{n}\\sum_{i = 1}^nx_i$ is the average of the first $n$ iterates.\n\\end{theorem}\n\\begin{proof}\nConsider the average of the first $n$ iterates $\\bar{x}_n = \\frac{1}{n}\\sum_{i = 1}^nx_i$. Note that\n\\begin{align}\\label{eq157}\n \\phi(\\bar{x}_n) - \\phi(x^*) \\leq \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n (\\phi(x_i) - \\phi(x^*)) &\\leq \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle\\\\ \n & + \\frac{1}{n}\\displaystyle\\sum_{i = 1}^n (R(x_i) - R(x^*))\n\\end{align}\nsince $\\phi(x) = f(x) + R(x)$ and $g_i\\in\\partial f(x_i)$.\nWe rewrite the sum in (\\ref{eq157}) as follows\n\\begin{equation}\\label{eq45}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle = \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_{n+1} - x^*\\rangle + \\displaystyle\\sum_{i = 1}^n \\displaystyle\\sum_{j = 1}^i\\langle g_j, x_i - x_{i+1}\\rangle\n\\end{equation}\nBy setting $z = x^*$ in observation (\\ref{minimum_property}) we obtain\n\\begin{align}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_{n+1} - x^*\\rangle &\\leq \\sqrt{n+1}\\|x^*\\|_2^2 - \\sqrt{n+1}\\|x_{n+1}\\|_2^2 \\\\\n & + (n+1)R(x^*) - (n+1)R(x_{n+1})\n\\end{align}\nand setting $z = x_{i+1}$ we get\n\\begin{align}\n \\displaystyle\\sum_{j = 1}^i\\langle g_j, x_i - x_{i+1}\\rangle &\\leq \\langle g_i, x_i - x_{i+1}\\rangle + \\sqrt{i}\\|x_{i+1}\\|_2^2 - \\sqrt{i}\\|x_{i}\\|_2^2 \\\\\n & + iR(x_{i+1}) - iR(x_i)\n\\end{align}\nPlugging this into equation (\\ref{eq45}) and noting that $\\sqrt{i} - \\sqrt{i - 1} \\geq \\frac{1}{2\\sqrt{i}}$ for $i > 1$, we get\n\\begin{align}\n \\displaystyle\\sum_{i = 1}^n \\langle g_i, x_i - x^*\\rangle &\\leq \\sqrt{n+1}\\|x^*\\|_2^2 + (n+1)R(x^*) - \\displaystyle\\sum_{i = 1}^{n+1} R(x_i) \\\\\n & - \\displaystyle\\sum_{i = 1}^{n+1}\\frac{\\|x_i\\|_2^2}{2\\sqrt{i}} + \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle\n\\end{align}\nPlugging this into equation (\\ref{eq157}) and throwing away the negative terms we see that\n\\begin{equation}\\label{eq61}\n n(\\phi(\\bar{x}_n) - \\phi(x^*)) \\leq \\sqrt{n+1}\\|x^*\\|_2^2 + R(x^*) - R(x_{n+1}) + \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle\n\\end{equation}\nWe bound the last sum above using Cauchy-Schwartz as\n\\begin{equation}\\label{eq65}\n \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle \\leq \\displaystyle\\sum_{i=1}^n \\|g_i\\|_2\\|x_i - x_{i+1}\\|_2 \\leq M\\displaystyle\\sum_{i=1}^n\\|x_i - x_{i+1}\\|_2\n\\end{equation}\nFinally, we must bound $\\|x_i - x_{i+1}\\|_2$. For this, we recall that by (\\ref{RDA_iteration})\n\\begin{equation}\\label{eq201}\n x_i = \\arg\\min_x \\left(\\displaystyle\\sum_{j = 1}^{i-1} \\langle g_j, x\\rangle + \\sqrt{i}\\|x\\|_2^2 + iR(x)\\right)\n\\end{equation}\nand note that by multiplying the objective in (\\ref{RDA_iteration}) by $\\sqrt{\\frac{i}{i+1}}$ we have\n\\begin{equation}\\label{eq205}\n x_{i+1} = \\arg\\min_x \\left(\\sqrt{\\frac{i}{i+1}}\\displaystyle\\sum_{j = 1}^{i} \\langle g_j, x\\rangle + \\sqrt{i}\\|x\\|_2^2 + \\sqrt{i(i+1)}R(x)\\right)\n\\end{equation}\nThe objective in equation (\\ref{eq205}) (which I will denote $O_{i+1}$) is strongly convex with strong convexity parameter $2\\sqrt{i}$. This means \nthat for any $z$ and $g_z\\in \\partial O_{i+1}(z)$, we have\n\\begin{equation}\\label{eq210}\n \\|z - x_{i+1}\\|_2 \\leq \\frac{1}{\\sqrt{i}}\\|g_z\\|_2\n\\end{equation}\nsince $x_{i + 1} = \\arg\\min_x O_{i+1}(x)$. We wish to set $z = x_i$, but we need a $g_z$ first. Recall that since $x_i$ \nis the minimizer in equation (\\ref{eq201})\nwe have\n\\begin{equation}\n 0 = \\displaystyle\\sum_{j = 1}^{i-1} g_j + 2\\sqrt{i}x_i + i\\lambda\n\\end{equation}\nwith $\\lambda\\in \\partial R(x_i)$. Using this and taking the subdifferential of $O_{i+1}$ we see that\n\\begin{equation}\n \\displaystyle\\sum_{j = 1}^{i-1} \\left(\\sqrt{\\frac{i}{i+1}} - 1\\right)g_j + \\sqrt{\\frac{i}{i+1}}g_i + (\\sqrt{i(i+1)} - i)\\lambda \\in \\partial O_{i+1}(x_i)\n\\end{equation}\nWe can bound the norm of this expression using the triangle inequality and our bounds on the elements in the subdifferentials of $f$ and\n$R$ by\n\\begin{equation}\n \\frac{3}{2}M + \\frac{1}{2\\sqrt{i}}N\n\\end{equation}\nPlugging this into equation (\\ref{eq210}) and setting $z = x_i$, we see that\n\\begin{equation}\n \\|x_i - x_{i+1}\\|_2 \\leq \\frac{3}{2\\sqrt{i}}M + \\frac{1}{2i}N\n\\end{equation}\nWe now plug this into equation (\\ref{eq65}) to get\n\\begin{equation}\n \\displaystyle\\sum_{i=1}^n\\langle g_i, x_i - x_{i+1}\\rangle \\leq \\frac{3M^2}{2} \\displaystyle\\sum_{i=1}^n \\frac{1}{\\sqrt{i}} + \\frac{MN}{2}\\displaystyle\\sum_{i=1}^n \\frac{1}{i}\\leq 3M^2 \\sqrt{n} + MN\\log(n)\n\\end{equation}\nPlugging this into equation (\\ref{eq61}) we get\n\\begin{equation}\n n(\\phi(\\bar{x}_n) - \\phi(x^*)) \\leq \\sqrt{n+1}\\|x^*\\|_2^2 + R(x^*) - R(x_{n+1}) + 3M^2 \\sqrt{n}+ MN\\log(n)\n\\end{equation}\nand finally\n\\begin{equation}\n \\phi(\\bar{x}_n) - \\phi(x^*) \\leq \\frac{2\\|x^*\\|_2^2 + 3M^2}{\\sqrt{n}} + \\frac{(R(x^*) - \\inf_x R(x)) + MN\\log(n)}{n}\n\\end{equation}\nas desired.\n\\end{proof}\n\\fi\n", "meta": {"hexsha": "887a3c9f39fbe3383ba67218c6764219d69797ee", "size": 71239, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/ChRDA.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/ChRDA.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "6DL/ChRDA.tex", "max_forks_repo_name": "liuzhengqi1996/math452", "max_forks_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.6605191257, "max_line_length": 529, "alphanum_fraction": 0.6405480144, "num_tokens": 29695, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706734, "lm_q2_score": 0.7431680143008301, "lm_q1q2_score": 0.6143857623985791}} {"text": "\\documentclass[10pt,a4paper]{article}\n\\usepackage[latin1]{inputenc}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{amssymb}\n\\usepackage{graphicx}\n\\usepackage{hyperref}\n\\begin{document}\n\t\\section{Karatsuba Multiplication}\n\t\n\tTwo numbers $l$ and $r$, can be written in the following way:\t\n\t\\begin{align*}\t\n\tl &= a*10^N + b\\\\ \n\tr &= c*10^N + d\\\\\n\t\\end{align*}\n\tIn this case their multiplication can be written\n\t\\begin{align*}\n\tl*r &= (a*10^N + b)*(c*10^N + d)\\\\ \n\t&= (a*c*10^{2N})+(a*d*10^N)+(b*c*10^N)+(b*d)\\\\ \n\t&= (a*c*10^{2N})+([(a*d)+(b*c)]*10^N)+]+(b*d) \n\t\\end{align*}\n\tbut $(a*d)+(b*c)$ is equal to\n\t\\begin{align*}\n\t&= (a*d)+(b*c)\\\\\n\t&= (a+b)*(c+d) - ac - bd\\\\\n\t&= a*c + a*d + b*c + b*d - a*c - b*d\\\\\n\t&= a*d + b*c\n\t\\end{align*}\n\tnow we can change two multiplications $(a*d)$ and $(b*c)$ to just one: $(a+b)*(c+d)$, given that we have already calculated $a*c$ and $b*d$. \n\t\n\tThis gives us the final form:\t\n\t\\begin{align*}\n\tl*r &= (a*c*10^{2N})\\\\&+([(a+b)*(c+d) - a*c - b*d]*10^N)\\\\&+(b*d)\n\t\\end{align*}\n\t\n\tSee:\\\\\n\t\\url{https://academic.microsoft.com/#/detail/204623740}\\\\\\\\\n\t\\url{https://scholar.google.co.uk/scholar?q=Multiplication+of+Many-Digital+Numbers+by+Automatic+Computers&btnG=&hl=en&as_sdt=0%2C5}\\\\\\\\\n\t\\url{http://cstheory.stackexchange.com/questions/21564/why-did-kolmogorov-publish-karatsubas-algorithm}\\\\\n\t\\\\\n\tThe Art of Computer Programming, Volume II, page 294, section 4.3.3\n\\end{document}", "meta": {"hexsha": "4a3640dcb7cb2a9242b2407c04627378ce72cffb", "size": 1420, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "courses/coursera-princeton-algorithms-1/karatsubamultiplication.tex", "max_stars_repo_name": "xunilrj/sandbox", "max_stars_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 7, "max_stars_repo_stars_event_min_datetime": "2017-04-01T17:18:35.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-12T05:23:23.000Z", "max_issues_repo_path": "courses/coursera-princeton-algorithms-1/karatsubamultiplication.tex", "max_issues_repo_name": "xunilrj/sandbox", "max_issues_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2020-05-24T13:36:50.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-15T06:44:20.000Z", "max_forks_repo_path": "courses/coursera-princeton-algorithms-1/karatsubamultiplication.tex", "max_forks_repo_name": "xunilrj/sandbox", "max_forks_repo_head_hexsha": "f92c12f83433cac01a885585e41c02bb5826a01f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2018-09-20T01:07:39.000Z", "max_forks_repo_forks_event_max_datetime": "2019-02-22T14:55:38.000Z", "avg_line_length": 33.8095238095, "max_line_length": 143, "alphanum_fraction": 0.6197183099, "num_tokens": 573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8267117962054048, "lm_q1q2_score": 0.614385759282418}} {"text": "% arara: indent: {overwrite: true, trace: on}\n% A sample chapter file- it contains a lot of\n% environments, including tabulars, align, etc\n%\n% Don't try and compile this file using pdflatex etc, just\n% compare the *format* of it to the format of the\n% sampleAFTER.tex\n%\n% In particular, compare the tabular and align-type\n% environments before and after running the script\n\n\\section{Polynomial functions}\n \\reformatstepslist{P} % the steps list should be P1, P2, \\ldots\n In your previous mathematics classes you have studied \\emph{linear} and\n \\emph{quadratic} functions. The most general forms of these types of\n functions can be represented (respectively) by the functions $f$\n and $g$ that have formulas\n \\begin{equation}\\label{poly:eq:linquad}\n \tf(x)=mx+b, \\qquad g(x)=ax^2+bx+c\n \\end{equation}\n We know that $m$ is the slope of $f$, and that $a$ is the \\emph{leading coefficient}\n of $g$. We also know that the \\emph{signs} of $m$ and $a$ completely\n determine the behavior of the functions $f$ and $g$. For example, if $m>0$\n then $f$ is an \\emph{increasing} function, and if $m<0$ then $f$ is\n a \\emph{decreasing} function. Similarly, if $a>0$ then $g$ is\n \\emph{concave up} and if $a<0$ then $g$ is \\emph{concave down}. Graphical\n representations of these statements are given in \\cref{poly:fig:linquad}.\n\n \\begin{figure}[!htb]\n \t\\setlength{\\figurewidth}{.2\\textwidth}\n \t\\begin{subfigure}{\\figurewidth}\n \t\t\\begin{tikzpicture}\n \t\t\t\\begin{axis}[\n \t\t\t\t\tframed,\n \t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\txtick={-11},\n \t\t\t\t\tytick={-11},\n \t\t\t\t]\n \t\t\t\t\\addplot expression[domain=-10:8]{(x+2)};\n \t\t\t\\end{axis}\n \t\t\\end{tikzpicture}\n \t\t\\caption{$m>0$}\n \t\\end{subfigure}\n \t\\hfill\n \t\\begin{subfigure}{\\figurewidth}\n \t\t\\begin{tikzpicture}\n \t\t\t\\begin{axis}[\n \t\t\t\t\tframed,\n \t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\txtick={-11},\n \t\t\t\t\tytick={-11},\n \t\t\t\t]\n \t\t\t\t\\addplot expression[domain=-10:8]{-(x+2)};\n \t\t\t\\end{axis}\n \t\t\\end{tikzpicture}\n \t\t\\caption{$m<0$}\n \t\\end{subfigure}\n \t\\hfill\n \t\\begin{subfigure}{\\figurewidth}\n \t\t\\begin{tikzpicture}\n \t\t\t\\begin{axis}[\n \t\t\t\t\tframed,\n \t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\txtick={-11},\n \t\t\t\t\tytick={-11},\n \t\t\t\t]\n \t\t\t\t\\addplot expression[domain=-4:4]{(x^2-6)};\n \t\t\t\\end{axis}\n \t\t\\end{tikzpicture}\n \t\t\\caption{$a>0$}\n \t\\end{subfigure}\n \t\\hfill\n \t\\begin{subfigure}{\\figurewidth}\n \t\t\\begin{tikzpicture}\n \t\t\t\\begin{axis}[\n \t\t\t\t\tframed,\n \t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\txtick={-11},\n \t\t\t\t\tytick={-11},\n \t\t\t\t]\n \t\t\t\t\\addplot expression[domain=-4:4]{-(x^2-6)};\n \t\t\t\\end{axis}\n \t\t\\end{tikzpicture}\n \t\t\\caption{$a<0$}\n \t\\end{subfigure}\n \t\\caption{Typical graphs of linear and quadratic functions.}\n \t\\label{poly:fig:linquad}\n \\end{figure}\n\n Let's look a little more closely at the formulas for $f$ and $g$ in\n \\cref{poly:eq:linquad}. Note that the \\emph{degree}\n of $f$ is $1$ since the highest power of $x$ that is present in the\n formula for $f(x)$ is $1$. Similarly, the degree of $g$ is $2$ since\n the highest power of $x$ that is present in the formula for $g(x)$\n is $2$.\n\n In this section we will build upon our knowledge of these elementary\n functions. In particular, we will generalize the functions $f$ and $g$ to a function $p$ that has\n any degree that we wish.\n\n %===================================\n % Author: Hughes\n % Date: March 2012\n %===================================\n \\begin{essentialskills}\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{problem}[Quadratic functions]\n \tEvery quadratic function has the form $y=ax^2+bx+c$; state the value\n \tof $a$ for each of the following functions, and hence decide if the\n \tparabola that represents the function opens upward or downward.\n \t\\begin{multicols}{2}\n \t\t\\begin{subproblem}\n \t\t\t$F(x)=x^2+3$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$a=1$; the parabola opens upward.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$G(t)=4-5t^2$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$a=-5$; the parabola opens downward.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$H(y)=4y^2-96y+8$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$a=4$; the parabola opens upward.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$K(z)=-19z^2$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m=-19$; the parabola opens downward.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\\end{multicols}\n \tNow let's generalize our findings for the most general quadratic function $g$\n \tthat has formula $g(x)=a_2x^2+a_1x+a_0$. Complete the following sentences.\n \t\\begin{subproblem}\n \t\tWhen $a_2>0$, the parabola that represents $y=g(x)$ opens $\\ldots$\n \t\t\\begin{shortsolution}\n \t\t\tWhen $a_2>0$, the parabola that represents the function opens upward.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tWhen $a_2<0$, the parabola that represents $y=g(x)$ opens $\\ldots$\n \t\t\\begin{shortsolution}\n \t\t\tWhen $a_2<0$, the parabola that represents the function opens downward.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n \\end{essentialskills}\n\n \\subsection*{Power functions with positive exponents}\n \tThe study of polynomials will rely upon a good knowledge\n \tof power functions| you may reasonably ask, what is a power function?\n \t\\begin{pccdefinition}[Power functions]\n \t\tPower functions have the form\n \t\t\\[\n \t\t\tf(x) = a_n x^n\n \t\t\\]\n \t\twhere $n$ can be any real number.\n\n \t\tNote that for this section we will only be concerned with the\n \t\tcase when $n$ is a positive integer.\n \t\\end{pccdefinition}\n\n \tYou may find assurance in the fact that you are already very comfortable\n \twith power functions that have $n=1$ (linear) and $n=2$ (quadratic). Let's\n \texplore some power functions that you might not be so familiar with.\n \tAs you read \\cref{poly:ex:oddpow,poly:ex:evenpow}, try and spot\n \tas many patterns and similarities as you can.\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{pccexample}[Power functions with odd positive exponents]\n \t\t\\label{poly:ex:oddpow}\n \t\tGraph each of the following functions, state their domain, and their\n \t\tlong-run behavior as $x\\rightarrow\\pm\\infty$\n \t\t\\[\n \t\t\tf(x)=x^3, \\qquad g(x)=x^5, \\qquad h(x)=x^7\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\tThe functions $f$, $g$, and $h$ are plotted in \\cref{poly:fig:oddpow}.\n \t\t\tThe domain of each of the functions $f$, $g$, and $h$ is $(-\\infty,\\infty)$. Note that\n \t\t\tthe long-run behavior of each of the functions is the same, and in particular\n \t\t\t\\begin{align*}\n \t\t\t\tf(x)\\rightarrow\\infty & \\text{ as } x\\rightarrow\\infty \\\\\n \t\t\t\t\\mathllap{\\text{and }} f(x)\\rightarrow-\\infty & \\text{ as } x\\rightarrow-\\infty\n \t\t\t\\end{align*}\n \t\t\tThe same results hold for $g$ and $h$.\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t\\begin{figure}[!htb]\n \t\t\\begin{minipage}{.45\\textwidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-1.5,xmax=1.5,\n \t\t\t\t\t\tymin=-5,ymax=5,\n \t\t\t\t\t\txtick={-1.0,-0.5,...,1.0},\n \t\t\t\t\t\tminor ytick={-3,-1,...,3},\n \t\t\t\t\t\tgrid=both,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\tlegend pos=north west,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-1.5:1.5]{x^3};\n \t\t\t\t\t\\addplot expression[domain=-1.379:1.379]{x^5};\n \t\t\t\t\t\\addplot expression[domain=-1.258:1.258]{x^7};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-1,-1)} node[axisnode,anchor=north west]{$(-1,-1)$};\n \t\t\t\t\t\\addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=south east]{$(1,1)$};\n \t\t\t\t\t\\legend{$f$,$g$,$h$}\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{Odd power functions}\n \t\t\t\\label{poly:fig:oddpow}\n \t\t\\end{minipage}%\n \t\t\\hfill\n \t\t\\begin{minipage}{.45\\textwidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-2.5,xmax=2.5,\n \t\t\t\t\t\tymin=-5,ymax=5,\n \t\t\t\t\t\txtick={-2.0,-1.5,...,2.0},\n \t\t\t\t\t\tminor ytick={-3,-1,...,3},\n \t\t\t\t\t\tgrid=both,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\tlegend pos=south east,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.236:2.236]{x^2};\n \t\t\t\t\t\\addplot expression[domain=-1.495:1.495]{x^4};\n \t\t\t\t\t\\addplot expression[domain=-1.307:1.307]{x^6};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-1,1)} node[axisnode,anchor=east]{$(-1,1)$};\n \t\t\t\t\t\\addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=west]{$(1,1)$};\n \t\t\t\t\t\\legend{$F$,$G$,$H$}\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{Even power functions}\n \t\t\t\\label{poly:fig:evenpow}\n \t\t\\end{minipage}%\n \t\\end{figure}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{pccexample}[Power functions with even positive exponents]\\label{poly:ex:evenpow}%\n \t\tGraph each of the following functions, state their domain, and their\n \t\tlong-run behavior as $x\\rightarrow\\pm\\infty$\n \t\t\\[\n \t\t\tF(x)=x^2, \\qquad G(x)=x^4, \\qquad H(x)=x^6\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\tThe functions $F$, $G$, and $H$ are plotted in \\cref{poly:fig:evenpow}. The domain\n \t\t\tof each of the functions is $(-\\infty,\\infty)$. Note that the long-run behavior\n \t\t\tof each of the functions is the same, and in particular\n \t\t\t\\begin{align*}\n \t\t\t\tF(x)\\rightarrow\\infty & \\text{ as } x\\rightarrow\\infty \\\\\n \t\t\t\t\\mathllap{\\text{and }} F(x)\\rightarrow\\infty & \\text{ as } x\\rightarrow-\\infty\n \t\t\t\\end{align*}\n \t\t\tThe same result holds for $G$ and $H$.\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t\\begin{doyouunderstand}\n \t\t\\begin{problem}\n \t\tRepeat \\cref{poly:ex:oddpow,poly:ex:evenpow} using (respectively)\n \t\t\\begin{subproblem}\n \t\t\t$f(x)=-x^3, \\qquad g(x)=-x^5, \\qquad h(x)=-x^7$\n \t\t\t\\begin{shortsolution}\n \t\t\t\tThe functions $f$, $g$, and $h$ have domain $(-\\infty,\\infty)$ and\n \t\t\t\tare graphed below.\n\n \t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\tframed,\n \t\t\t\t\t\t\txmin=-1.5,xmax=1.5,\n \t\t\t\t\t\t\tymin=-5,ymax=5,\n \t\t\t\t\t\t\txtick={-1.0,-0.5,...,0.5},\n \t\t\t\t\t\t\tminor ytick={-3,-1,...,3},\n \t\t\t\t\t\t\tgrid=both,\n \t\t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\t\tlegend pos=north east,\n \t\t\t\t\t\t]\n \t\t\t\t\t\t\\addplot expression[domain=-1.5:1.5]{-x^3};\n \t\t\t\t\t\t\\addplot expression[domain=-1.379:1.379]{-x^5};\n \t\t\t\t\t\t\\addplot expression[domain=-1.258:1.258]{-x^7};\n \t\t\t\t\t\t\\legend{$f$,$g$,$h$}\n \t\t\t\t\t\\end{axis}\n \t\t\t\t\\end{tikzpicture}\n\n \t\t\t\tNote that\n \t\t\t\t\\begin{align*}\n \t\t\t\t\tf(x)\\rightarrow-\\infty & \\text{ as } x\\rightarrow\\infty \\\\\n \t\t\t\t\t\\mathllap{\\text{and }} f(x)\\rightarrow\\infty & \\text{ as } x\\rightarrow-\\infty\n \t\t\t\t\\end{align*}\n \t\t\t\tThe same is true for $g$ and $h$.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$F(x)=-x^2, \\qquad G(x)=-x^4, \\qquad H(x)=-x^6$\n \t\t\t\\begin{shortsolution}\n \t\t\t\tThe functions $F$, $G$, and $H$ have domain $(-\\infty,\\infty)$ and\n \t\t\t\tare graphed below.\n\n \t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\tframed,\n \t\t\t\t\t\t\txmin=-2.5,xmax=2.5,\n \t\t\t\t\t\t\tymin=-5,ymax=5,\n \t\t\t\t\t\t\txtick={-1.0,-0.5,...,0.5},\n \t\t\t\t\t\t\tminor ytick={-3,-1,...,3},\n \t\t\t\t\t\t\tgrid=both,\n \t\t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\t\tlegend pos=north east,\n \t\t\t\t\t\t]\n \t\t\t\t\t\t\\addplot expression[domain=-2.236:2.236]{-x^2};\n \t\t\t\t\t\t\\addplot expression[domain=-1.495:1.495]{-x^4};\n \t\t\t\t\t\t\\addplot expression[domain=-1.307:1.307]{-x^6};\n \t\t\t\t\t\t\\legend{$F$,$G$,$H$}\n \t\t\t\t\t\\end{axis}\n \t\t\t\t\\end{tikzpicture}\n\n \t\t\t\tNote that\n \t\t\t\t\\begin{align*}\n \t\t\t\t\tF(x)\\rightarrow-\\infty & \\text{ as } x\\rightarrow\\infty \\\\\n \t\t\t\t\t\\mathllap{\\text{and }} F(x)\\rightarrow-\\infty & \\text{ as } x\\rightarrow-\\infty\n \t\t\t\t\\end{align*}\n \t\t\t\tThe same is true for $G$ and $H$.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\end{problem}\n \t\\end{doyouunderstand}\n\n \\subsection*{Polynomial functions}\n \tNow that we have a little more familiarity with power functions,\n \twe can define polynomial functions. Provided that you were comfortable\n \twith our opening discussion about linear and quadratic functions (see\n \t$f$ and $g$ in \\cref{poly:eq:linquad}) then there is every chance\n \tthat you'll be able to master polynomial functions as well; just remember\n \tthat polynomial functions are a natural generalization of linear\n \tand quadratic functions. Once you've studied the examples and problems\n \tin this section, you'll hopefully agree that polynomial functions\n \tare remarkably predictable.\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{pccdefinition}[Polynomial functions]\n \t\tPolynomial functions have the form\n \t\t\\[\n \t\t\tp(x)=a_nx^n+a_{n-1}x^{n-1}+\\ldots+a_1x+a_0\n \t\t\\]\n \t\twhere $a_n$, $a_{n-1}$, $a_{n-2}$, \\ldots, $a_0$ are real numbers.\n \t\t\\begin{itemize}\n \t\t\t\\item We call $n$ the degree of the polynomial, and require that $n$\n \t\t\t is a non-negative integer;\n \t\t\t\\item $a_n$, $a_{n-1}$, $a_{n-2}$, \\ldots, $a_0$ are called the coefficients;\n \t\t\t\\item We typically write polynomial functions in descending powers of $x$.\n \t\t\\end{itemize}\n \t\tIn particular, we call $a_n$ the \\emph{leading} coefficient, and $a_nx^n$ the\n \t\t\\emph{leading term}.\n\n \t\tNote that if a polynomial is given in factored form, then the degree can be found\n \t\tby counting the number of linear factors.\n \t\\end{pccdefinition}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{pccexample}[Polynomial or not]\n \t\tIdentify the following functions as polynomial or not; if the function\n \t\tis a polynomial, state its degree.\n \t\t\\begin{multicols}{3}\n \t\t\t\\begin{enumerate}\n \t\t\t\t\\item $p(x)=x^2-3$\n \t\t\t\t\\item $q(x)=-4x^{\\nicefrac{1}{2}}+10$\n \t\t\t\t\\item $r(x)=10x^5$\n \t\t\t\t\\item $s(x)=x^{-2}+x^{23}$\n \t\t\t\t\\item $f(x)=-8$\n \t\t\t\t\\item $g(x)=3^x$\n \t\t\t\t\\item $h(x)=\\sqrt[3]{x^7}-x^2+x$\n \t\t\t\t\\item $k(x)=4x(x+2)(x-3)$\n \t\t\t\t\\item $j(x)=x^2(x-4)(5-x)$\n \t\t\t\\end{enumerate}\n \t\t\\end{multicols}\n \t\t\\begin{pccsolution}\n \t\t\t\\begin{enumerate}\n \t\t\t\t\\item $p$ is a polynomial, and its degree is $2$.\n \t\t\t\t\\item $q$ is \\emph{not} a polynomial, because $\\frac{1}{2}$ is not an integer.\n \t\t\t\t\\item $r$ is a polynomial, and its degree is $5$.\n \t\t\t\t\\item $s$ is \\emph{not} a polynomial, because $-2$ is not a positive integer.\n \t\t\t\t\\item $f$ is a polynomial, and its degree is $0$.\n \t\t\t\t\\item $g$ is \\emph{not} a polynomial, because the independent\n \t\t\t\t variable, $x$, is in the exponent.\n \t\t\t\t\\item $h$ is \\emph{not} a polynomial, because $\\frac{7}{3}$ is not an integer.\n \t\t\t\t\\item $k$ is a polynomial, and its degree is $3$.\n \t\t\t\t\\item $j$ is a polynomial, and its degree is $4$.\n \t\t\t\\end{enumerate}\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{pccexample}[Typical graphs]\\label{poly:ex:typical}\n \t\t\\Cref{poly:fig:typical} shows graphs of some polynomial functions;\n \t\tthe ticks have deliberately been left off the axis to allow us to concentrate\n \t\ton the features of each graph. Note in particular that:\n \t\t\\begin{itemize}\n \t\t\t\\item \\cref{poly:fig:typical1} shows a degree-$1$ polynomial (you might also\n \t\t\t classify the function as linear) whose leading coefficient, $a_1$, is positive.\n \t\t\t\\item \\cref{poly:fig:typical2} shows a degree-$2$ polynomial (you might also\n \t\t\t classify the function as quadratic) whose leading coefficient, $a_2$, is positive.\n \t\t\t\\item \\cref{poly:fig:typical3} shows a degree-$3$ polynomial whose leading coefficient, $a_3$,\n \t\t\t is positive| compare its overall\n \t\t\t shape and long-run behavior to the functions described in \\cref{poly:ex:oddpow}.\n \t\t\t\\item \\cref{poly:fig:typical4} shows a degree-$4$ polynomial whose leading coefficient, $a_4$,\n \t\t\t is positive|compare its overall shape and long-run behavior to the functions described in \\cref{poly:ex:evenpow}.\n \t\t\t\\item \\cref{poly:fig:typical5} shows a degree-$5$ polynomial whose leading coefficient, $a_5$,\n \t\t\t is positive| compare its overall\n \t\t\t shape and long-run behavior to the functions described in \\cref{poly:ex:oddpow}.\n \t\t\\end{itemize}\n \t\\end{pccexample}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{figure}[!htb]\n \t\t\\begin{widepage}\n \t\t\\setlength{\\figurewidth}{\\textwidth/6}\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-10:8]{(x+2)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$a_1>0$}\n \t\t\t\\label{poly:fig:typical1}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4:4]{(x^2-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$a_2>0$}\n \t\t\t\\label{poly:fig:typical2}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-7.5:7.5]{0.05*(x+6)*x*(x-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$a_3>0$}\n \t\t\t\\label{poly:fig:typical3}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.35:5.35,samples=100]{0.2*(x-5)*x*(x-3)*(x+2)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$a_4>0$}\n \t\t\t\\label{poly:fig:typical4}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-5.5:6.3,samples=100]{0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$a_5>0$}\n \t\t\t\\label{poly:fig:typical5}\n \t\t\\end{subfigure}\n \t\t\\end{widepage}\n \t\t\\caption{Graphs to illustrate typical curves of polynomial functions.}\n \t\t\\label{poly:fig:typical}\n \t\\end{figure}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{doyouunderstand}\n \t\t\\begin{problem}\n \t\tUse \\cref{poly:ex:typical} and \\cref{poly:fig:typical} to help you sketch\n \t\tthe graphs of polynomial functions that have negative leading coefficients| note\n \t\tthat there are many ways to do this! The intention with this problem\n \t\tis to use your knowledge of transformations- in particular, \\emph{reflections}-\n \t\tto guide you.\n \t\t\\begin{shortsolution}\n \t\t\t$a_1<0$:\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-10:8]{-(x+2)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$a_2<0$\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4:4]{-(x^2-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$a_3<0$\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-7.5:7.5]{-0.05*(x+6)*x*(x-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$a_4<0$\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.35:5.35,samples=100]{-0.2*(x-5)*x*(x-3)*(x+2)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$a_5<0$\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t\txtick={-11},\n \t\t\t\t\t\tytick={-11},\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-5.5:6.3,samples=100]{-0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\\end{shortsolution}\n \t\t\\end{problem}\n \t\\end{doyouunderstand}\n\n \t\\fixthis{poly: Need a more basic example here- it can have a similar\n \t\tformat to the multiple zeros example, but just keep it simple; it should\n \tbe halfway between the 2 examples surrounding it}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{pccexample}[Multiple zeros]\n \t\tConsider the polynomial functions $p$, $q$, and $r$ which are\n \t\tgraphed in \\cref{poly:fig:moremultiple}.\n \t\tThe formulas for $p$, $q$, and $r$ are as follows\n \t\t\\begin{align*}\n \t\t\tp(x) & =(x-3)^2(x+4)^2 \\\\\n \t\t\tq(x) & =x(x+2)^2(x-1)^2(x-3) \\\\\n \t\t\tr(x) & =x(x-3)^3(x+1)^2\n \t\t\\end{align*}\n \t\tFind the degree of $p$, $q$, and $r$, and decide if the functions bounce off or cut\n \t\tthrough the horizontal axis at each of their zeros.\n \t\t\\begin{pccsolution}\n \t\t\tThe degree of $p$ is 4. Referring to \\cref{poly:fig:bouncep},\n \t\t\tthe curve bounces off the horizontal axis at both zeros, $3$ and $4$.\n\n \t\t\tThe degree of $q$ is 6. Referring to \\cref{poly:fig:bounceq},\n \t\t\tthe curve bounces off the horizontal axis at $-2$ and $1$, and cuts\n \t\t\tthrough the horizontal axis at $0$ and $3$.\n\n \t\t\tThe degree of $r$ is 6. Referring to \\cref{poly:fig:bouncer},\n \t\t\tthe curve bounces off the horizontal axis at $-1$, and cuts through\n \t\t\tthe horizontal axis at $0$ and at $3$, although is flattened immediately to the left and right of $3$.\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t\\setlength{\\figurewidth}{0.25\\textwidth}\n \t\\begin{figure}[!htb]\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-6,xmax=5,\n \t\t\t\t\t\tymin=-30,ymax=200,\n \t\t\t\t\t\txtick={-4,-2,...,4},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-5.63733:4.63733,samples=50]{(x-3)^2*(x+4)^2};\n \t\t\t\t\t\\addplot[soldot]coordinates{(3,0)(-4,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=p(x)$}\n \t\t\t\\label{poly:fig:bouncep}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-3,xmax=4,\n \t\t\t\t\t\txtick={-2,...,3},\n \t\t\t\t\t\tymin=-60,ymax=40,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot+[samples=50] expression[domain=-2.49011:3.11054]{x*(x+2)^2*(x-1)^2*(x-3)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-2,0)(0,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=q(x)$}\n \t\t\t\\label{poly:fig:bounceq}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-2,xmax=4,\n \t\t\t\t\t\txtick={-1,...,3},\n \t\t\t\t\t\tymin=-40,ymax=40,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-1.53024:3.77464,samples=50]{x*(x-3)^3*(x+1)^2};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-1,0)(0,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=r(x)$}\n \t\t\t\\label{poly:fig:bouncer}\n \t\t\\end{subfigure}\n \t\t\\caption{}\n \t\t\\label{poly:fig:moremultiple}\n \t\\end{figure}\n\n \t\\begin{pccdefinition}[Multiple zeros]\\label{poly:def:multzero}\n \t\tLet $p$ be a polynomial that has a repeated linear factor $(x-a)^n$. Then we say\n \t\tthat $p$ has a multiple zero at $a$ of multiplicity $n$ and\n \t\t\\begin{itemize}\n \t\t\t\\item if the factor $(x-a)$ is repeated an even number of times, the graph of $y=p(x)$ does not\n \t\t\t cross the $x$ axis at $a$, but `bounces' off the horizontal axis at $a$.\n \t\t\t\\item if the factor $(x-a)$ is repeated an odd number of times, the graph of $y=p(x)$ crosses the\n \t\t\t horizontal axis at $a$, but it looks `flattened' there\n \t\t\\end{itemize}\n \t\tIf $n=1$, then we say that $p$ has a \\emph{simple} zero at $a$.\n \t\\end{pccdefinition}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{pccexample}[Find a formula]\n \t\tFind formulas for the polynomial functions, $p$ and $q$, graphed in \\cref{poly:fig:findformulademoboth}.\n \t\t\\begin{figure}[!htb]\n \t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\\begin{axis}[framed,\n \t\t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\t\txtick={-4,-2,...,4},\n \t\t\t\t\t\t\tminor xtick={-3,-1,...,3},\n \t\t\t\t\t\t\tytick={-8,-6,...,8},\n \t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\tgrid=both]\n \t\t\t\t\t\t\\addplot expression[domain=-3.25842:2.25842,samples=50]{-x*(x-2)*(x+3)*(x+1)};\n \t\t\t\t\t\t\\addplot[soldot]coordinates{(1,8)}node[axisnode,inner sep=.35cm,anchor=west]{$(1,8)$};\n \t\t\t\t\t\t\\addplot[soldot]coordinates{(-3,0)(-1,0)(0,0)(2,0)};\n \t\t\t\t\t\\end{axis}\n \t\t\t\t\\end{tikzpicture}\n \t\t\t\t\\caption{$p$}\n \t\t\t\t\\label{poly:fig:findformulademo}\n \t\t\t\\end{subfigure}\n \t\t\t\\hfill\n \t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\\begin{axis}[framed,\n \t\t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\t\txtick={-4,-2,...,4},\n \t\t\t\t\t\t\tminor xtick={-3,-1,...,3},\n \t\t\t\t\t\t\tytick={-8,-6,...,8},\n \t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\tgrid=both]\n \t\t\t\t\t\t\\addplot expression[domain=-4.33:4.08152]{-.25*(x+2)^2*(x-3)};\n \t\t\t\t\t\t\\addplot[soldot]coordinates{(2,4)}node[axisnode,anchor=south west]{$(2,4)$};\n \t\t\t\t\t\t\\addplot[soldot]coordinates{(-2,0)(3,0)};\n \t\t\t\t\t\\end{axis}\n \t\t\t\t\\end{tikzpicture}\n \t\t\t\t\\caption{$q$}\n \t\t\t\t\\label{poly:fig:findformulademo1}\n \t\t\t\\end{subfigure}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:findformulademoboth}\n \t\t\\end{figure}\n \t\t\\begin{pccsolution}\n \t\t\t\\begin{enumerate}\n \t\t\t\t\\item We begin by noting that the horizontal intercepts of $p$ are $(-3,0)$, $(-1,0)$, $(0,0)$ and $(2,0)$.\n \t\t\t\t We also note that each zero is simple (multiplicity $1$).\n \t\t\t\t If we assume that $p$ has no other zeros, then we can start by writing\n \t\t\t\t \\begin{align*}\n \t\t\t\t \tp(x) & =(x+3)(x+1)(x-0)(x-2) \\\\\n \t\t\t\t \t & =x(x+3)(x+1)(x-2) \\\\\n \t\t\t\t \\end{align*}\n \t\t\t\t According to \\cref{poly:fig:findformulademo}, the point $(1,8)$ lies\n \t\t\t\t on the curve $y=p(x)$.\n \t\t\t\t Let's check if the formula we have written satisfies this requirement\n \t\t\t\t \\begin{align*}\n \t\t\t\t \tp(1) & = (1)(4)(2)(-1) \\\\\n \t\t\t\t \t & = -8\n \t\t\t\t \\end{align*}\n \t\t\t\t which is clearly not correct| it is close though. We can correct this by\n \t\t\t\t multiplying $p$ by a constant $k$; so let's assume that\n \t\t\t\t \\[\n \t\t\t\t \tp(x)=kx(x+3)(x+1)(x-2)\n \t\t\t\t \\]\n \t\t\t\t Then $p(1)=-8k$, and if this is to equal $8$, then $k=-1$. Therefore\n \t\t\t\t the formula for $p(x)$ is\n \t\t\t\t \\[\n \t\t\t\t \tp(x)=-x(x+3)(x+1)(x-2)\n \t\t\t\t \\]\n \t\t\t\t\\item The function $q$ has a zero at $-2$ of multiplicity $2$, and zero of\n \t\t\t\t multiplicity $1$ at $3$ (so $3$ is a simple zero of $q$); we can therefore assume that $q$ has the form\n \t\t\t\t \\[\n \t\t\t\t \tq(x)=k(x+2)^2(x-3)\n \t\t\t\t \\]\n \t\t\t\t where $k$ is some real number. In order to find $k$, we use the given ordered pair, $(2,4)$, and\n \t\t\t\t evaluate $p(2)$\n \t\t\t\t \\begin{align*}\n \t\t\t\t \tp(2) & =k(4)^2(-1) \\\\\n \t\t\t\t \t & =-16k\n \t\t\t\t \\end{align*}\n \t\t\t\t We solve the equation $4=-8k$ and obtain $k=-\\frac{1}{4}$ and conclude that the\n \t\t\t\t formula for $q(x)$ is\n \t\t\t\t \\[\n \t\t\t\t \tq(x)=-\\frac{1}{4}(x+2)^2(x-3)\n \t\t\t\t \\]\n \t\t\t\\end{enumerate}\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n\n \t\\fixthis{Chris: need sketching polynomial problems}\n \t\\begin{pccspecialcomment}[Steps to follow when sketching polynomial functions]\n \t\t\\begin{steps}\n \t\t\t\\item \\label{poly:step:first} Determine the degree of the polynomial,\n \t\t\tits leading term and leading coefficient, and hence determine\n \t\t\tthe long-run behavior of the polynomial| does it behave like $\\pm x^2$ or $\\pm x^3$\n \t\t\tas $x\\rightarrow\\pm\\infty$?\n \t\t\t\\item Determine the zeros and their multiplicity. Mark all zeros\n \t\t\tand the vertical intercept on the graph using solid circles $\\bullet$.\n \t\t\t\\item \\label{poly:step:last} Deduce the overall shape of the curve, and sketch it. If there isn't\n \t\t\tenough information from the previous steps, then construct a table of values.\n \t\t\\end{steps}\n \t\tRemember that until we have the tools of calculus, we won't be able to\n \t\tfind the exact coordinates of local minimums, local maximums, and points\n \t\tof inflection.\n \t\\end{pccspecialcomment}\n \tBefore we demonstrate some examples, it is important to remember the following:\n \t\\begin{itemize}\n \t\t\\item our sketches will give a good representation of the overall\n \t\t shape of the graph, but until we have the tools of calculus (from MTH 251)\n \t\t we can not find local minimums, local maximums, and inflection points algebraically. This\n \t\t means that we will make our best guess as to where these points are.\n \t\t\\item we will not concern ourselves too much with the vertical scale (because of\n \t\t our previous point)| we will, however, mark the vertical intercept (assuming there is one),\n \t\t and any horizontal asymptotes.\n \t\\end{itemize}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2012\n \t%===================================\n \t\\begin{pccexample}\\label{poly:ex:simplecubic}\n \t\tUse \\crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $p$\n \t\tthat has formula\n \t\t\\[\n \t\t\tp(x)=\\frac{1}{2}(x-4)(x-1)(x+3)\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\t\\begin{steps}\n \t\t\t\t\\item $p$ has degree $3$. The leading term of $p$ is $\\frac{1}{2}x^3$, so the leading coefficient of $p$\n \t\t\t\tis $\\frac{1}{2}$. The long-run behavior of $p$ is therefore similar to that of $x^3$.\n \t\t\t\t\\item The zeros of $p$ are $-3$, $1$, and $4$; each zero is simple (i.e, it has multiplicity $1$).\n \t\t\t\tThis means that the curve of $p$ cuts the horizontal axis at each zero. The vertical\n \t\t\t\tintercept of $p$ is $(0,6)$.\n \t\t\t\t\\item We draw the details we have obtained so far on \\cref{poly:fig:simplecubicp1}. Given\n \t\t\t\tthat the curve of $p$ looks like the curve of $x^3$ in the long-run, we are able to complete a sketch of the\n \t\t\t\tgraph of $p$ in \\cref{poly:fig:simplecubicp2}.\n\n \t\t\t\tNote that we can not find the coordinates of the local minimums, local maximums, and inflection\n \t\t\t\tpoints| for the moment we make reasonable guesses as to where these points are (you'll find how\n \t\t\t\tto do this in calculus).\n \t\t\t\\end{steps}\n\n \t\t\t\\begin{figure}[!htbp]\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\t\t\tymin=-10,ymax=15,\n \t\t\t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\t\t\tytick={-5,5},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:simplecubicp1}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\hfill\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\t\t\tymin=-10,ymax=15,\n \t\t\t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\t\t\tytick={-5,5},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};\n \t\t\t\t\t\t\t\\addplot[pccplot] expression[domain=-3.57675:4.95392,samples=100]{.5*(x-4)*(x-1)*(x+3)};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:simplecubicp2}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\caption{$y=\\dfrac{1}{2}(x-4)(x-1)(x+3)$}\n \t\t\t\t\\label{poly:fig:simplecubic}\n \t\t\t\\end{figure}\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2012\n \t%===================================\n \t\\begin{pccexample}\\label{poly:ex:degree5}\n \t\tUse \\crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $q$\n \t\tthat has formula\n \t\t\\[\n \t\t\tq(x)=\\frac{1}{200}(x+7)^2(2-x)(x-6)^2\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\t\\begin{steps}\n \t\t\t\t\\item $q$ has degree $4$. The leading term of $q$ is\n \t\t\t\t\\[\n \t\t\t\t\t-\\frac{1}{200}x^5\n \t\t\t\t\\]\n \t\t\t\tso the leading coefficient of $q$ is $-\\frac{1}{200}$. The long-run behavior of $q$\n \t\t\t\tis therefore similar to that of $-x^5$.\n \t\t\t\t\\item The zeros of $q$ are $-7$ (multiplicity 2), $2$ (simple), and $6$ (multiplicity $2$).\n \t\t\t\tThe curve of $q$ bounces off the horizontal axis at the zeros with multiplicity $2$ and\n \t\t\t\tcuts the horizontal axis at the simple zeros. The vertical intercept of $q$ is $\\left( 0,\\frac{441}{25} \\right)$.\n \t\t\t\t\\item We mark the details we have found so far on \\cref{poly:fig:degree5p1}. Given that\n \t\t\t\tthe curve of $q$ looks like the curve of $-x^5$ in the long-run, we can complete \\cref{poly:fig:degree5p2}.\n \t\t\t\\end{steps}\n\n \t\t\t\\begin{figure}[!htbp]\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\t\t\tymin=-10,ymax=40,\n \t\t\t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\t\t\tytick={-5,0,...,35},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\\left( 0, \\frac{441}{25} \\right)$};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:degree5p1}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\hfill\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\t\t\tymin=-10,ymax=40,\n \t\t\t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\t\t\tytick={-5,0,...,35},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\\left( 0, \\frac{441}{25} \\right)$};\n \t\t\t\t\t\t\t\\addplot[pccplot] expression[domain=-8.83223:7.34784,samples=50]{1/200*(x+7)^2*(2-x)*(x-6)^2};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:degree5p2}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\caption{$y=\\dfrac{1}{200}(x+7)^2(2-x)(x-6)^2$}\n \t\t\t\t\\label{poly:fig:degree5}\n \t\t\t\\end{figure}\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2012\n \t%===================================\n \t\\begin{pccexample}\n \t\tUse \\crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $r$\n \t\tthat has formula\n \t\t\\[\n \t\t\tr(x)=\\frac{1}{100}x^3(x+4)(x-4)(x-6)\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\t\\begin{steps}\n \t\t\t\t\\item $r$ has degree $6$. The leading term of $r$ is\n \t\t\t\t\\[\n \t\t\t\t\t\\frac{1}{100}x^6\n \t\t\t\t\\]\n \t\t\t\tso the leading coefficient of $r$ is $\\frac{1}{100}$. The long-run behavior of $r$\n \t\t\t\tis therefore similar to that of $x^6$.\n \t\t\t\t\\item The zeros of $r$ are $-4$ (simple), $0$ (multiplicity $3$), $4$ (simple),\n \t\t\t\tand $6$ (simple). The vertical intercept of $r$ is $(0,0)$. The curve of $r$\n \t\t\t\tcuts the horizontal axis at the simple zeros, and goes through the axis\n \t\t\t\tat $(0,0)$, but does so in a flattened way.\n \t\t\t\t\\item We mark the zeros and vertical intercept on \\cref{poly:fig:degree6p1}. Given that\n \t\t\t\tthe curve of $r$ looks like the curve of $x^6$ in the long-run, we complete the graph\n \t\t\t\tof $r$ in \\cref{poly:fig:degree6p2}.\n \t\t\t\\end{steps}\n\n \t\t\t\\begin{figure}[!htbp]\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-5,xmax=10,\n \t\t\t\t\t\t\t\tymin=-20,ymax=10,\n \t\t\t\t\t\t\t\txtick={-4,-2,...,8},\n \t\t\t\t\t\t\t\tytick={-15,-10,...,5},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:degree6p1}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\hfill\n \t\t\t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\t\\begin{axis}[\n \t\t\t\t\t\t\t\txmin=-5,xmax=10,\n \t\t\t\t\t\t\t\tymin=-20,ymax=10,\n \t\t\t\t\t\t\t\txtick={-4,-2,...,8},\n \t\t\t\t\t\t\t\tytick={-15,-10,...,5},\n \t\t\t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\t\t]\n \t\t\t\t\t\t\t\\addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};\n \t\t\t\t\t\t\t\\addplot[pccplot] expression[domain=-4.16652:6.18911,samples=100]{1/100*(x+4)*x^3*(x-4)*(x-6)};\n \t\t\t\t\t\t\\end{axis}\n \t\t\t\t\t\\end{tikzpicture}\n \t\t\t\t\t\\caption{}\n \t\t\t\t\t\\label{poly:fig:degree6p2}\n \t\t\t\t\\end{subfigure}%\n \t\t\t\t\\caption{$y=\\dfrac{1}{100}(x+4)x^3(x-4)(x-6)$}\n \t\t\t\\end{figure}\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{pccexample}[An open-topped box]\n \t\tA cardboard company makes open-topped boxes for their clients. The specifications\n \t\tdictate that the box must have a square base, and that it must be open-topped.\n \t\tThe company uses sheets of cardboard that are $\\unit[1200]{cm^2}$. Assuming that\n \t\tthe base of each box has side $x$ (measured in cm), it can be shown that the volume of each box, $V(x)$,\n \t\thas formula\n \t\t\\[\n \t\t\tV(x)=\\frac{x}{4}(1200-x^2)\n \t\t\\]\n \t\tFind the dimensions of the box that maximize the volume.\n \t\t\\begin{pccsolution}\n \t\t\tWe graph $y=V(x)$ in \\cref{poly:fig:opentoppedbox}. Note that because\n \t\t\t$x$ represents the length of a side, and $V(x)$ represents the volume\n \t\t\tof the box, we necessarily require both values to be positive; we illustrate\n \t\t\tthe part of the curve that applies to this problem using a solid line.\n\n \t\t\t\\begin{figure}[!htb]\n \t\t\t\t\\centering\n \t\t\t\t\\begin{tikzpicture}\n \t\t\t\t\t\\begin{axis}[framed,\n \t\t\t\t\t\t\txmin=-50,xmax=50,\n \t\t\t\t\t\t\tymin=-5000,ymax=5000,\n \t\t\t\t\t\t\txtick={-40,-30,...,40},\n \t\t\t\t\t\t\tminor xtick={-45,-35,...,45},\n \t\t\t\t\t\t\tminor ytick={-3000,-1000,1000,3000},\n \t\t\t\t\t\t\twidth=.75\\textwidth,\n \t\t\t\t\t\t\theight=.5\\textwidth,\n \t\t\t\t\t\tgrid=both]\n \t\t\t\t\t\t\\addplot[pccplot,dashed,<-] expression[domain=-40:0,samples=50]{x/4*(1200-x^2)};\n \t\t\t\t\t\t\\addplot[pccplot,-] expression[domain=0:34.64,samples=50]{x/4*(1200-x^2)};\n \t\t\t\t\t\t\\addplot[pccplot,dashed,->] expression[domain=34.64:40,samples=50]{x/4*(1200-x^2)};\n \t\t\t\t\t\t\\addplot[soldot] coordinates{(20,4000)};\n \t\t\t\t\t\\end{axis}\n \t\t\t\t\\end{tikzpicture}\n \t\t\t\t\\caption{$y=V(x)$}\n \t\t\t\t\\label{poly:fig:opentoppedbox}\n \t\t\t\\end{figure}\n\n \t\t\tAccording to \\cref{poly:fig:opentoppedbox}, the maximum volume of such a box is\n \t\t\tapproximately $\\unit[4000]{cm^2}$, and we achieve it using a base of length\n \t\t\tapproximately $\\unit[20]{cm}$. Since the base is square and each sheet of cardboard\n \t\t\tis $\\unit[1200]{cm^2}$, we conclude that the dimensions of each box are $\\unit[20]{cm}\\times\\unit[20]{cm}\\times\\unit[30]{cm}$.\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n\n \\subsection*{Complex zeros}\n \tThere has been a pattern to all of the examples that we have seen so far|\n \tthe degree of the polynomial has dictated the number of \\emph{real} zeros that the\n \tpolynomial has. For example, the function $p$ in \\cref{poly:ex:simplecubic}\n \thas degree $3$, and $p$ has $3$ real zeros; the function $q$ in \\cref{poly:ex:degree5}\n \thas degree $5$ and $q$ has $5$ real zeros.\n\n \tYou may wonder if this result can be generalized| does every polynomial that\n \thas degree $n$ have $n$ real zeros? Before we tackle the general result,\n \tlet's consider an example that may help motivate it.\n \t%===================================\n \t% Author: Hughes\n \t% Date: June 2012\n \t%===================================\n \t\\begin{pccexample}\\label{poly:ex:complx}\n \t\tConsider the polynomial function $c$ that has formula\n \t\t\\[\n \t\t\tc(x)=x(x^2+1)\n \t\t\\]\n \t\tIt is clear that $c$ has degree $3$, and that $c$ has a (simple) zero at $0$. Does\n \t\t$c$ have any other zeros, i.e, can we find any values of $x$ that satisfy the equation\n \t\t\\begin{equation}\\label{poly:eq:complx}\n \t\t\tx^2+1=0\n \t\t\\end{equation}\n \t\tThe solutions to \\cref{poly:eq:complx} are $\\pm i$.\n\n \t\tWe conclude that $c$ has $3$ zeros: $0$ and $\\pm i$; we note that \\emph{not\n \t\tall of them are real}.\n \t\\end{pccexample}\n \t\\Cref{poly:ex:complx} shows that not every degree-$3$ polynomial has $3$\n \t\\emph{real} zeros; however, if we are prepared to venture into the complex numbers,\n \tthen we can state the following theorem.\n \t%===================================\n \t% Author: Hughes\n \t% Date: June 2012\n \t%===================================\n \t\\begin{pccspecialcomment}[The fundamental theorem of algebra]\n \t\tEvery polynomial function of degree $n$ has $n$ roots, some of which may\n \t\tbe complex, and some may be repeated.\n \t\\end{pccspecialcomment}\n \t\\fixthis{Fundamental theorem of algebra: is this wording ok? do we want\n \tit as a theorem?}\n \t%===================================\n \t% Author: Hughes\n \t% Date: June 2012\n \t%===================================\n \t\\begin{pccexample}\n \t\tFind all the zeros of the polynomial function $p$ that has formula\n \t\t\\[\n \t\t\tp(x)=x^4-2x^3+5x^2\n \t\t\\]\n \t\t\\begin{pccsolution}\n \t\t\tWe begin by factoring $p$\n \t\t\t\\begin{align*}\n \t\t\t\tp(x) & =x^4-2x^3+5x^2 \\\\\n \t\t\t\t & =x^2(x^2-2x+5)\n \t\t\t\\end{align*}\n \t\t\tWe note that $0$ is a zero of $p$ with multiplicity $2$. The other zeros of $p$\n \t\t\tcan be found by solving the equation\n \t\t\t\\[\n \t\t\t\tx^2-2x+5=0\n \t\t\t\\]\n \t\t\tThis equation can not be factored, so we use the quadratic formula\n \t\t\t\\begin{align*}\n \t\t\t\tx & =\\frac{2\\pm\\sqrt{(-2)^2}-20}{2(1)} \\\\\n \t\t\t\t & =\\frac{2\\pm\\sqrt{-16}}{2} \\\\\n \t\t\t\t & =1\\pm 2i\n \t\t\t\\end{align*}\n \t\t\tWe conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\\pm 2i$ (simple).\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n \t%===================================\n \t% Author: Hughes\n \t% Date: June 2012\n \t%===================================\n \t\\begin{pccexample}\n \t\tFind a polynomial that has zeros at $2\\pm i\\sqrt{2}$.\n \t\t\\begin{pccsolution}\n \t\t\tWe know that the zeros of a polynomial can be found by analyzing the linear\n \t\t\tfactors. We are given the zeros, and have to work backwards to find the\n \t\t\tlinear factors.\n\n \t\t\tWe begin by assuming that $p$ has the form\n \t\t\t\\begin{align*}\n \t\t\t\tp(x) & =(x-(2-i\\sqrt{2}))(x-(2+i\\sqrt{2})) \\\\\n \t\t\t\t & =x^2-x(2+i\\sqrt{2})-x(2-i\\sqrt{2})+(2-i\\sqrt{2})(2+i\\sqrt{2}) \\\\\n \t\t\t\t & =x^2-4x+(4-2i^2) \\\\\n \t\t\t\t & =x^2-4x+6\n \t\t\t\\end{align*}\n \t\t\tWe conclude that a possible formula for a polynomial function, $p$,\n \t\t\tthat has zeros at $2\\pm i\\sqrt{2}$ is\n \t\t\t\\[\n \t\t\t\tp(x)=x^2-4x+6\n \t\t\t\\]\n \t\t\tNote that we could multiply $p$ by any real number and still ensure\n \t\t\tthat $p$ has the same zeros.\n \t\t\\end{pccsolution}\n \t\\end{pccexample}\n \t\\investigation*{}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Find a formula from a graph]\n \tFor each of the polynomials in \\cref{poly:fig:findformula}\n \t\\begin{enumerate}\n \t\t\\item count the number of times the curve turns round, and cuts/bounces off the $x$ axis;\n \t\t\\item approximate the degree of the polynomial;\n \t\t\\item use your information to find the linear factors of each polynomial, and therefore write a possible formula for each;\n \t\t\\item make sure your polynomial goes through the given ordered pair.\n \t\\end{enumerate}\n \t\\begin{shortsolution}\n \t\t\\Vref{poly:fig:findformdeg2}:\n \t\t\\begin{enumerate}\n \t\t\t\\item the curve turns round once;\n \t\t\t\\item the degree could be 2;\n \t\t\t\\item based on the zeros, the linear factors are $(x+5)$ and $(x-3)$; since the\n \t\t\t graph opens downwards, we will assume the leading coefficient is negative: $p(x)=-k(x+5)(x-3)$;\n \t\t\t\\item $p$ goes through $(2,2)$, so we need to solve $2=-k(7)(-1)$ and therefore $k=\\nicefrac{2}{7}$, so\n \t\t\t \\[\n \t\t\t \tp(x)=-\\frac{2}{7}(x+5)(x-3)\n \t\t\t \\]\n \t\t\\end{enumerate}\n \t\t\\Vref{poly:fig:findformdeg3}:\n \t\t\\begin{enumerate}\n \t\t\t\\item the curve turns around twice;\n \t\t\t\\item the degree could be 3;\n \t\t\t\\item based on the zeros, the linear factors are $(x+2)^2$, and $(x-1)$;\n \t\t\t based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+2)^2(x-1)$;\n \t\t\t\\item $p$ goes through $(0,-2)$, so we need to solve $-2=k(4)(-1)$ and therefore $k=\\nicefrac{1}{2}$, so\n \t\t\t \\[\n \t\t\t \tp(x)=\\frac{1}{2}(x+2)^2(x-1)\n \t\t\t \\]\n \t\t\\end{enumerate}\n \t\t\\Vref{poly:fig:findformdeg5}:\n \t\t\\begin{enumerate}\n \t\t\t\\item the curve turns around 4 times;\n \t\t\t\\item the degree could be 5;\n \t\t\t\\item based on the zeros, the linear factors are $(x+5)^2$, $(x+1)$, $(x-2)$, $(x-3)$;\n \t\t\t based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+5)^2(x+1)(x-2)(x-3)$;\n \t\t\t\\item $p$ goes through $(-3,-50)$, so we need to solve $-50=k(64)(-2)(-5)(-6)$ and therefore $k=\\nicefrac{5}{384}$, so\n \t\t\t \\[\n \t\t\t \tp(x)=\\frac{5}{384}(x+5)^2(x+1)(x-2)(x-3)\n \t\t\t \\]\n \t\t\\end{enumerate}\n \t\\end{shortsolution}\n \t\\end{problem}\n\n\n \t\\begin{figure}[!htb]\n \t\t\\setlength{\\figurewidth}{0.3\\textwidth}\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-2,ymax=5,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4.5:3.75]{-1/3*(x+4)*(x-3)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-4,0)(3,0)(2,2)} node[axisnode,above right]{$(2,2)$};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:findformdeg2}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-3,xmax=2,\n \t\t\t\t\t\tymin=-2,ymax=4,\n \t\t\t\t\t\txtick={-2,...,1},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.95:1.75]{1/3*(x+2)^2*(x-1)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-2,0)(1,0)(0,-1.33)}node[axisnode,anchor=north west]{$(0,-2)$};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:findformdeg3}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-100,ymax=150,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4.5:3.4,samples=50]{(x+4)^2*(x+1)*(x-2)*(x-3)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-4,0)(-1,0)(2,0)(3,0)(-3,-60)}node[axisnode,anchor=north]{$(-3,-50)$};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:findformdeg5}\n \t\t\\end{subfigure}\n \t\t\\caption{}\n \t\t\\label{poly:fig:findformula}\n \t\\end{figure}\n\n\n\n\n \t\\begin{exercises}\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{problem}[Prerequisite classifacation skills]\n \tDecide if each of the following functions are linear or quadratic.\n \t\\begin{multicols}{3}\n \t\t\\begin{subproblem}\n \t\t\t$f(x)=2x+3$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$f$ is linear.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$g(x)=10-7x$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$g$ is linear\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$h(x)=-x^2+3x-9$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$h$ is quadratic.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$k(x)=-17$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$k$ is linear.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$l(x)=-82x^2-4$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$l$ is quadratic\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$m(x)=6^2x-8$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m$ is linear.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\\end{multicols}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{problem}[Prerequisite slope identification]\n \tState the slope of each of the following linear functions, and\n \thence decide if each function is increasing or decreasing.\n \t\\begin{multicols}{4}\n \t\t\\begin{subproblem}\n \t\t\t$\\alpha(x)=4x+1$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m=4$; $\\alpha$ is increasing.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$\\beta(x)=-9x$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m=-9$; $\\beta$ is decreasing.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$\\gamma(t)=18t+100$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m=18$; $\\gamma$ is increasing.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$\\delta(y)=23-y$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$m=-1$; $\\delta$ is decreasing.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\\end{multicols}\n \tNow let's generalize our findings for the most general linear function $f$\n \tthat has formula $f(x)=mx+b$. Complete the following sentences.\n \t\\begin{subproblem}\n \t\tWhen $m>0$, the function $f$ is $\\ldots$\n \t\t\\begin{shortsolution}\n \t\t\tWhen $m>0$, the function $f$ is $\\ldots$ \\emph{increasing}.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tWhen $m<0$, the function $f$ is $\\ldots$\n \t\t\\begin{shortsolution}\n \t\t\tWhen $m<0$, the function $f$ is $\\ldots$ \\emph{decreasing}.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Polynomial or not?]\n \tIdentify whether each of the following functions is a polynomial or not.\n \tIf the function is a polynomial, state its degree.\n \t\\begin{multicols}{3}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=2x+1$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial (you might also describe $p$ as linear). The degree of $p$ is 1.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=7x^2+4x$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial (you might also describe $p$ as quadratic). The degree of $p$ is 2.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=\\sqrt{x}+2x+1$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is not a polynomial; we require the powers of $x$ to be integer values.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=2^x-45$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is not a polynomial; the $2^x$ term is exponential.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=6x^4-5x^3+9$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial, and the degree of $p$ is $6$.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=-5x^{17}+9x+2$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial, and the degree of $p$ is 17.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=4x(x+7)^2(x-3)^3$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial, and the degree of $p$ is $6$.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=4x^{-5}-x^2+x$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is not a polynomial because $-5$ is not a positive integer.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$p(x)=-x^6(x^2+1)(x^3-2)$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$p$ is a polynomial, and the degree of $p$ is $11$.\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\\end{multicols}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Polynomial graphs]\n \tThree polynomial functions $p$, $m$, and $n$ are shown in \\crefrange{poly:fig:functionp}{poly:fig:functionn}.\n \tThe functions have the following formulas\n \t\\begin{align*}\n \t\tp(x) & = (x-1)(x+2)(x-3) \\\\\n \t\tm(x) & = -(x-1)(x+2)(x-3) \\\\\n \t\tn(x) & = (x-1)(x+2)(x-3)(x+1)(x+4)\n \t\\end{align*}\n \tNote that for our present purposes we are not concerned with the vertical scale of the graphs.\n \t\\begin{subproblem}\n \t\tIdentify both on the graph {\\em and} algebraically, the zeros of each polynomial.\n \t\t\\begin{shortsolution}\n \t\t\t$y=p(x)$ is shown below.\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$y=m(x)$ is shown below.\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\t$y=n(x)$ is shown below.\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-90,ymax=70,\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4.15:3.15,samples=50]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n\n \t\t\tThe zeros of $p$ are $-2$, $1$, and $3$; the zeros of $m$ are $-2$, $1$, and $3$; the zeros of $n$ are\n \t\t\t$-4$, $-2$, $-1$, and $3$.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tWrite down the degree, how many times the curve of each function `turns around',\n \t\tand how many zeros it has\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item The degree of $p$ is 3, and the curve $y=p(x)$ turns around twice.\n \t\t\t\t\\item The degree of $q$ is also 3, and the curve $y=q(x)$ turns around twice.\n \t\t\t\t\\item The degree of $n$ is $5$, and the curve $y=n(x)$ turns around 4 times.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n\n \t\\begin{figure}[!htb]\n \t\t\\begin{widepage}\n \t\t\\setlength{\\figurewidth}{0.3\\textwidth}\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\tytick={-5,5},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=p(x)$}\n \t\t\t\\label{poly:fig:functionp}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\tytick={-5,5},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=m(x)$}\n \t\t\t\\label{poly:fig:functionm}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-5,xmax=5,\n \t\t\t\t\t\tymin=-90,ymax=70,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4.15:3.15,samples=100]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};\n \t\t\t\t\t\\addplot[soldot]coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{$y=n(x)$}\n \t\t\t\\label{poly:fig:functionn}\n \t\t\\end{subfigure}\n \t\t\\caption{}\n \t\t\\end{widepage}\n \t\\end{figure}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Horizontal intercepts]\\label{poly:prob:matchpolys}%\n \tState the horizontal intercepts (as ordered pairs) of the following polynomials.\n \t\\begin{multicols}{2}\n \t\t\\begin{subproblem}\\label{poly:prob:degree5}\n \t\t\t$p(x)=(x-1)(x+2)(x-3)(x+1)(x+4)$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$(-4,0)$, $(-2,0)$, $(-1,0)$, $(1,0)$, $(3,0)$\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$q(x)=-(x-1)(x+2)(x-3)$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$(-2,0)$, $(1,0)$, $(3,0)$\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\n \t\t\t$r(x)=(x-1)(x+2)(x-3)$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$(-2,0)$, $(1,0)$, $(3,0)$\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\t\\begin{subproblem}\\label{poly:prob:degree2}\n \t\t\t$s(x)=(x-2)(x+2)$\n \t\t\t\\begin{shortsolution}\n \t\t\t\t$(-2,0)$, $(2,0)$\n \t\t\t\\end{shortsolution}\n \t\t\\end{subproblem}\n \t\\end{multicols}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: March 2012\n \t%===================================\n \t\\begin{problem}[Minimums, maximums, and concavity]\\label{poly:prob:incdec}\n \tFour polynomial functions are graphed in \\cref{poly:fig:incdec}. The formulas\n \tfor these functions are (not respectively)\n \t\\begin{gather*}\n \t\tp(x)=\\frac{x^3}{6}-\\frac{x^2}{4}-3x, \\qquad q(x)=\\frac{x^4}{20}+\\frac{x^3}{15}-\\frac{6}{5}x^2+1\\\\\n \t\tr(x)=-\\frac{x^5}{50}-\\frac{x^4}{40}+\\frac{2x^3}{5}+6, \\qquad s(x)=-\\frac{x^6}{6000}-\\frac{x^5}{2500}+\\frac{67x^4}{4000}+\\frac{17x^3}{750}-\\frac{42x^2}{125}\n \t\\end{gather*}\n \t\\begin{figure}[!htb]\n \t\t\\begin{widepage}\n \t\t\\setlength{\\figurewidth}{.23\\textwidth}\n \t\t\\centering\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-8,-6,...,8},\n \t\t\t\t\t\tgrid=major,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-5.28:4.68,samples=50]{-x^5/50-x^4/40+2*x^3/5+6};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:incdec3}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txmin=-10,xmax=10,ymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-8,-6,...,8},\n \t\t\t\t\t\tgrid=major,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-6.08:4.967,samples=50]{x^4/20+x^3/15-6/5*x^2+1};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:incdec2}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txmin=-6,xmax=8,ymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-4,-2,...,6},\n \t\t\t\t\t\tytick={-8,-4,4,8},\n \t\t\t\t\t\tminor ytick={-6,-2,...,6},\n \t\t\t\t\t\tgrid=both,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-4.818:6.081,samples=50]{x^3/6-x^2/4-3*x};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:incdec1}\n \t\t\\end{subfigure}\n \t\t\\hfill\n \t\t\\begin{subfigure}{\\figurewidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\tframed,\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\txmin=-10,xmax=10,ymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-4,4,8},\n \t\t\t\t\t\tytick={-8,-4,4,8},\n \t\t\t\t\t\tminor xtick={-6,-2,...,6},\n \t\t\t\t\t\tminor ytick={-6,-2,...,6},\n \t\t\t\t\t\tgrid=both,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-9.77:8.866,samples=50]{-x^6/6000-x^5/2500+67*x^4/4000+17/750*x^3-42/125*x^2};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:incdec4}\n \t\t\\end{subfigure}\n \t\t\\caption{Graphs for \\cref{poly:prob:incdec}.}\n \t\t\\label{poly:fig:incdec}\n \t\t\\end{widepage}\n \t\\end{figure}\n \t\\begin{subproblem}\n \t\tMatch each of the formulas with one of the given graphs.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ is graphed in \\vref{poly:fig:incdec1};\n \t\t\t\t\\item $q$ is graphed in \\vref{poly:fig:incdec2};\n \t\t\t\t\\item $r$ is graphed in \\vref{poly:fig:incdec3};\n \t\t\t\t\\item $s$ is graphed in \\vref{poly:fig:incdec4}.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tApproximate the zeros of each function using the appropriate graph.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ has simple zeros at about $-3.8$, $0$, and $5$.\n \t\t\t\t\\item $q$ has simple zeros at about $-5.9$, $-1$, $1$, and $4$.\n \t\t\t\t\\item $r$ has simple zeros at about $-5$, $-2.9$, and $4.1$.\n \t\t\t\t\\item $s$ has simple zeros at about $-9$, $-6$, $4.2$, $8.1$, and a zero of multiplicity $2$ at $0$.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tApproximate the local maximums and minimums of each of the functions.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ has a local maximum of approximately $3.9$ at $-2$, and a local minimum of approximately $-6.5$ at $3$.\n \t\t\t\t\\item $q$ has a local minimum of approximately $-10$ at $-4$, and $-4$ at $3$; $q$ has a local maximum of approximately $1$ at $0$.\n \t\t\t\t\\item $r$ has a local minimum of approximately $-5.5$ at $-4$, and a local maximum of approximately $10$ at $3$.\n \t\t\t\t\\item $s$ has a local maximum of approximately $5$ at $-8$, $0$ at $0$, and $5$ at $7$; $s$ has local minimums\n \t\t\t\t of approximately $-3$ at $-4$, and $-1$ at $3$.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tApproximate the global maximums and minimums of each of the functions.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ does not have a global maximum, nor a global minimum.\n \t\t\t\t\\item $q$ has a global minimum of approximately $-10$; it does not have a global maximum.\n \t\t\t\t\\item $r$ does not have a global maximum, nor a global minimum.\n \t\t\t\t\\item $s$ has a global maximum of approximately $5$; it does not have a global minimum.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tApproximate the intervals on which each function is increasing and decreasing.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ is increasing on $(-\\infty,-2)\\cup (3,\\infty)$, and decreasing on $(-2,3)$.\n \t\t\t\t\\item $q$ is increasing on $(-4,0)\\cup (3,\\infty)$, and decreasing on $(-\\infty,-4)\\cup (0,3)$.\n \t\t\t\t\\item $r$ is increasing on $(-4,3)$, and decreasing on $(-\\infty,-4)\\cup (3,\\infty)$.\n \t\t\t\t\\item $s$ is increasing on $(-\\infty,-8)\\cup (-4,0)\\cup (3,5)$, and decreasing on $(-8,-4)\\cup (0,3)\\cup (5,\\infty)$.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tApproximate the intervals on which each function is concave up and concave down.\n \t\t\\begin{shortsolution}\n \t\t\t\\begin{itemize}\n \t\t\t\t\\item $p$ is concave up on $(1,\\infty)$, and concave down on $(-\\infty,1)$.\n \t\t\t\t\\item $q$ is concave up on $(-\\infty,-1)\\cup (1,\\infty)$, and concave down on $(-1,1)$.\n \t\t\t\t\\item $r$ is concave up on $(-\\infty,-3)\\cup (0,2)$, and concave down on $(-3,0)\\cup (2,\\infty)$.\n \t\t\t\t\\item $s$ is concave up on $(-6,-2)\\cup (2,5)$, and concave down on $(-\\infty,-6)\\cup (-2,2)\\cup (5,\\infty)$.\n \t\t\t\\end{itemize}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tThe degree of $q$ is $5$. Assuming that all of the real zeros of $q$ are\n \t\tshown in its graph, how many complex zeros does $q$ have?\n \t\t\\begin{shortsolution}\n \t\t\t\\Vref{poly:fig:incdec2} shows that $q$ has $3$ real zeros\n \t\t\tsince the curve of $q$ cuts the horizontal axis $3$ times.\n \t\t\tSince $q$ has degree $5$, $q$ must have $2$ complex zeros.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Long-run behaviour of polynomials]\n \tDescribe the long-run behavior of each of polynomial functions in\n \t\\crefrange{poly:prob:degree5}{poly:prob:degree2}.\n \t\\begin{shortsolution}\n \t\t$\\dd\\lim_{x\\rightarrow-\\infty}p(x)=-\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow\\infty}p(x)=\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow-\\infty}q(x)=\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow\\infty}q(x)=-\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow-\\infty}r(x)=-\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow\\infty}r(x)=\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow-\\infty}s(x)=\\infty$,\n \t\t$\\dd\\lim_{x\\rightarrow\\infty}s(x)=\\infty$,\n \t\\end{shortsolution}\n \t\\end{problem}\n\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[True of false?]\n \tLet $p$ be a polynomial function.\n \tLabel each of the following statements as true (T) or false (F); if they are false,\n \tprovide an example that supports your answer.\n \t\\begin{subproblem}\n \t\tIf $p$ has degree $3$, then $p$ has $3$ distinct zeros.\n \t\t\\begin{shortsolution}\n \t\t\tFalse. Consider $p(x)=x^2(x+1)$ which has only 2 distinct zeros.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tIf $p$ has degree $4$, then $\\dd\\lim_{x\\rightarrow-\\infty}p(x)=\\infty$ and $\\dd\\lim_{x\\rightarrow\\infty}p(x)=\\infty$.\n \t\t\\begin{shortsolution}\n \t\t\tFalse. Consider $p(x)=-x^4$.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tIf $p$ has even degree, then it is possible that $p$ can have no real zeros.\n \t\t\\begin{shortsolution}\n \t\t\tTrue.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tIf $p$ has odd degree, then it is possible that $p$ can have no real zeros.\n \t\t\\begin{shortsolution}\n \t\t\tFalse. All odd degree polynomials will cut the horizontal axis at least once.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: May 2011\n \t%===================================\n \t\\begin{problem}[Find a formula from a description]\n \tIn each of the following problems, give a possible formula for a polynomial\n \tfunction that has the specified properties.\n \t\\begin{subproblem}\n \t\tDegree 2 and has zeros at $4$ and $5$.\n \t\t\\begin{shortsolution}\n \t\t\tPossible option: $p(x)=(x-4)(x-5)$. Note we could multiply $p$ by any real number, and still meet the requirements.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tDegree 3 and has zeros at $4$,$5$ and $-3$.\n \t\t\\begin{shortsolution}\n \t\t\tPossible option: $p(x)=(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tDegree 4 and has zeros at $0$, $4$, $5$, $-3$.\n \t\t\\begin{shortsolution}\n \t\t\tPossible option: $p(x)=x(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tDegree 4, with zeros that make the graph cut at $2$, $-5$, and a zero that makes the graph touch at $-2$;\n \t\t\\begin{shortsolution}\n \t\t\tPossible option: $p(x)=(x-2)(x+5)(x+2)^2$. Note we could multiply $p$ by any real number, and still meet the requirements.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tDegree 3, with only one zero at $-1$.\n \t\t\\begin{shortsolution}\n \t\t\tPossible option: $p(x)=(x+1)^3$. Note we could multiply $p$ by any real number, and still meet the requirements.\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n \t%===================================\n \t% Author: Hughes\n \t% Date: June 2012\n \t%===================================\n \t\\begin{problem}[\\Cref{poly:step:last}]\n \t\\pccname{Saheed} is graphing a polynomial function, $p$.\n \tHe is following \\crefrange{poly:step:first}{poly:step:last} and has so far\n \tmarked the zeros of $p$ on \\cref{poly:fig:optionsp1}. Saheed tells you that\n \t$p$ has degree $3$, but does \\emph{not} say if the leading coefficient\n \tof $p$ is positive or negative.\n \t\\begin{figure}[!htbp]\n \t\t\\begin{widepage}\n \t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\theight=.5\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:optionsp1}\n \t\t\\end{subfigure}%\n \t\t\\hfill\n \t\t\\begin{subfigure}{.45\\textwidth}\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\textwidth,\n \t\t\t\t\t\theight=.5\\textwidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\t\\caption{}\n \t\t\t\\label{poly:fig:optionsp2}\n \t\t\\end{subfigure}%\n \t\t\\caption{}\n \t\t\\end{widepage}\n \t\\end{figure}\n \t\\begin{subproblem}\n \t\tUse the information in \\cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient\n \t\tis positive.\n \t\t\\begin{shortsolution}\n \t\t\tAssuming that $a_3>0$:\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-6.78179:8.35598,samples=50]{1/20*(x+5)*(x-2)*(x-6)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tUse the information in \\cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient\n \t\tis negative.\n \t\t\\begin{shortsolution}\n \t\t\tAssuming that $a_3<0$:\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-6.78179:8.35598,samples=50]{-1/20*(x+5)*(x-2)*(x-6)};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \tSaheed now turns his attention to another polynomial function, $q$. He finds\n \tthe zeros of $q$ (there are only $2$) and marks them on \\cref{poly:fig:optionsp2}.\n \tSaheed knows that $q$ has degree $3$, but doesn't know if the leading\n \tcoefficient is positive or negative.\n \t\\begin{subproblem}\n \t\tUse the information in \\cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading\n \t\tcoefficient of $q$ is positive. Hint: only one of the zeros is simple.\n \t\t\\begin{shortsolution}\n \t\t\tAssuming that $a_4>0$ there are $2$ different options:\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-8.68983:7.31809,samples=50]{1/20*(x+5)^2*(x-6)};\n \t\t\t\t\t\\addplot expression[domain=-6.31809:9.68893,samples=50]{1/20*(x+5)*(x-6)^2};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\begin{subproblem}\n \t\tUse the information in \\cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading\n \t\tcoefficient of $q$ is negative.\n \t\t\\begin{shortsolution}\n \t\t\tAssuming that $a_4<0$ there are $2$ different options:\n\n \t\t\t\\begin{tikzpicture}\n \t\t\t\t\\begin{axis}[\n \t\t\t\t\t\txmin=-10,xmax=10,\n \t\t\t\t\t\tymin=-10,ymax=10,\n \t\t\t\t\t\txtick={-8,-6,...,8},\n \t\t\t\t\t\tytick={-15},\n \t\t\t\t\t\twidth=\\solutionfigurewidth,\n \t\t\t\t\t]\n \t\t\t\t\t\\addplot expression[domain=-8.68983:7.31809,samples=50]{-1/20*(x+5)^2*(x-6)};\n \t\t\t\t\t\\addplot expression[domain=-6.31809:9.68893,samples=50]{-1/20*(x+5)*(x-6)^2};\n \t\t\t\t\t\\addplot[soldot] coordinates{(-5,0)(6,0)};\n \t\t\t\t\\end{axis}\n \t\t\t\\end{tikzpicture}\n \t\t\\end{shortsolution}\n \t\\end{subproblem}\n \t\\end{problem}\n", "meta": {"hexsha": "302fb6ab0d9675e9e5e7b04bde3890f90fecc27f", "size": 69952, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "project/server/dependencies/latexindent.pl-master/test-cases/benchmarks/sampleBEFORE-smaller.tex", "max_stars_repo_name": "digorithm/latex-linter", "max_stars_repo_head_hexsha": "d6b63002cdcecf291e2abc7a399e0d7af4bd9038", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "project/server/dependencies/latexindent.pl-master/test-cases/benchmarks/sampleBEFORE-smaller.tex", "max_issues_repo_name": "digorithm/latex-linter", "max_issues_repo_head_hexsha": "d6b63002cdcecf291e2abc7a399e0d7af4bd9038", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "project/server/dependencies/latexindent.pl-master/test-cases/benchmarks/sampleBEFORE-smaller.tex", "max_forks_repo_name": "digorithm/latex-linter", "max_forks_repo_head_hexsha": "d6b63002cdcecf291e2abc7a399e0d7af4bd9038", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.9410589411, "max_line_length": 158, "alphanum_fraction": 0.5903333715, "num_tokens": 24928, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.743167997235783, "lm_q2_score": 0.826711787666479, "lm_q1q2_score": 0.614385743531311}} {"text": "\\appendix\n\n\\chapter{Pangolin v1.0, a~conservative 2-D transport model for large scale\nparallel calculation}\n\\label{app:paper}\nThis paper was accepted by the \\gls{GMD} journal.\n\n\\input pangolin_paper\n\n\\chapter{Supplement}\n\\section{Proof: number of cells at the poles}\n\\label{sec:pole_area}\n\\def\\dphi{\\Delta\\phi}\n\\def\\dlamb{\\Delta\\lambda}\nHere, we demonstrate why the number of cells at the poles for the Pangolin grid\ncan be approximated by $3$. The latitude spacing is supposed to be \"small\"\nenough and the cells are supposed to be square at the Equator.\nWe first consider the surface of the band near the Equator, defined by \n$\\pi/2 - \\dphi \\le \\phi \\le \\pi/2$:\n\\begin{equation*}\n S_{eq} = \\int_{0}^{2\\pi}\\int_{\\pi/2}^{\\pi/2 - \\dphi} r^2 \\sin\\phi \\mathrm{d}\\lambda\n \\mathrm{d}\\phi \n = 2\\pi r^2 \\sin \\dphi\n\\end{equation*}\nThe surface of the band near the North Pole ($0 \\le \\phi \\le \\dphi$) is:\n\\begin{equation*}\n S_{pole} = \\int_{0}^{2\\pi}\\int_{0}^{\\dphi} r^2 \\sin\\phi \\mathrm{d}\\lambda\n \\mathrm{d}\\phi \n = 2\\pi r^2 (1-\\cos \\dphi)\n\\end{equation*}\nThe number of cells at the Equator is $2\\pi/\\dlamb=2\\pi/\\dphi$ so the area of the cells at\nthe Equator is:\n\\begin{equation*}\n \\mathcal{A}_{eq}=\\frac{S_{eq}}{\\frac{2\\pi}{\\dphi}} = r^2 \\dphi \\sin \\dphi\n\\end{equation*}\nThe grid preserves the areas so the number of cells at the pole is:\n\\begin{equation*}\n n_1 = \\Big \\lfloor\\frac{S_{pole}}{\\mathcal{A}_{eq}} \\Big \\rfloor \n =\\Big \\lfloor\\frac{2\\pi(1-\\cos \\dphi)}{\\dphi \\sin\n \\dphi } \\Big \\rfloor \n\\end{equation*}\nUsing small angles approximations, it gives:\n\\begin{equation*}\n n_1 = \\Big \\lfloor \\frac{2\\pi(\\frac{\\dphi^2}{2})}{\\dphi^2} \\Big \\rfloor = \\lfloor \\pi\n \\rfloor = 3\n\\end{equation*}\n\n\\section{MPI send modes}\n\\label{app:mpi_send}\nThe following is taken from\n\\url{http://www.mcs.anl.gov/research/projects/mpi/sendmode.html}:\n\n% Create a format entry-definition, each with a new paragraph\n\\def\\entry#1#2{\\leftskip=0pt \\texttt{#1}\\par\\noindent\n\\leftskip=1.5cm #2\\par\\indent}\n\n\\entry{MPI\\_Send}{will not return until you can use the send buffer. It may or may\nnot block (it is allowed to buffer, either on the sender or receiver side,\nor to wait for the matching receive). \nMay buffer; returns immediately and you can use the send buffer. A late\nadd-on to the MPI specification. Should be used only when absolutely\nnecessary.}\n\\entry{MPI\\_Ssend}{will not return until matching receive posted.}\n\\entry{MPI\\_Rsend}{may be used \\textbf{only} if matching receive already posted. User\nresponsible for writing a correct program.}\n\\entry{MPI\\_Isend}{\n Nonblocking send. But not necessarily asynchronous. You can\n \\textbf{not} reuse the send buffer until either a successful,\n wait/test or you \\textbf{know} that the message has been received\n (see \\texttt{MPI\\_Request\\_free}). Note also that while the I refers to\n immediate, there is no performance requirement on \\texttt{MPI\\_Isend}.\n An immediate send must return to the user without requiring\n a matching receive at the destination. An implementation is\n free to send the data to the destination before returning,\n as long as the send call does not block waiting for a\n matching receive. Different strategies of when to send the\n data offer different performance advantages and\n disadvantages that will depend on the application. \n }\n\\entry{MPI\\_Ibsend}{buffered nonblocking.}\n\\entry{MPI\\_Issend}{\n Synchronous nonblocking. Note that a Wait/Test will\n complete only when the matching receive is posted. \n}\n\\entry{MPI\\_Irsend}{ As with MPI\\_Rsend, but nonblocking.}\n\n\\chapter{Design}\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.9\\linewidth]{pangolin_classes.pdf}\n \\caption{Classes diagram for Pangolin, with the main classes.}\n\\end{figure}\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.4\\linewidth]{flowcharts/pangolin_run.pdf}\n \\caption{Flowchart representing a full run in Pangolin. Input winds are\n supposed to be divergence free: there were corrected either during their\n generation (analytical winds) or during interpolation (real winds). As\n temporal interpolation is linear, interpolated winds do not need to be corrected\n either}\n \\label{fig:pango_run}\n\\end{figure}\n", "meta": {"hexsha": "c4247f3564cf3a2fad3b45b2cf40d2b50c5898be", "size": 4229, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "appendix.tex", "max_stars_repo_name": "alexDarcy/phd", "max_stars_repo_head_hexsha": "f39ef9de23e8c8e08c86f95b7501e1e91321ad2b", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2016-08-28T10:05:20.000Z", "max_stars_repo_stars_event_max_datetime": "2016-08-28T10:05:20.000Z", "max_issues_repo_path": "appendix.tex", "max_issues_repo_name": "alexDarcy/thesis", "max_issues_repo_head_hexsha": "f39ef9de23e8c8e08c86f95b7501e1e91321ad2b", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "appendix.tex", "max_forks_repo_name": "alexDarcy/thesis", "max_forks_repo_head_hexsha": "f39ef9de23e8c8e08c86f95b7501e1e91321ad2b", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.6634615385, "max_line_length": 90, "alphanum_fraction": 0.7297233389, "num_tokens": 1287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833945721304, "lm_q2_score": 0.7931059560743422, "lm_q1q2_score": 0.6143267037114389}} {"text": "\\section{Overview of Optimization Algorithms}\n\\label{sec:51algorithms}\n\n\\paragraph{Problem setting}\n\n\\minitoc{80mm}{7}\n\n\\usenotation{Ëopt}\nGenerally, \\term{unconstrained optimization problems} have the form\n\\begin{equation}\n \\label{eq:unconstrainedOptimization}\n \\xopt = \\vecargmin \\objfun(\\*x),\\quad\n \\*x \\in \\real^d,\n\\end{equation}\nwhere $\\objfun\\colon \\real^d \\to \\real$ is the \\term{objective function.}\n\\term{Constrained optimization problems} are given by\n\\begin{equation}\n \\label{eq:constrainedOptimization}\n \\xopt = \\vecargmin \\objfun(\\*x),\\quad\n \\*x \\in \\real^d\\;\\;\\text{s.t.}\\;\\;\n \\ineqconfun(\\*x) \\le \\*0,\n\\end{equation}\nwhere\n$\\ineqconfun\\colon \\real^d \\to \\real^{m_{\\ineqconfun}}$\n($m_{\\ineqconfun} \\in \\nat$)\nis the \\term{(inequality) constraint function.}\nThis formulation also contains optimization problems\nwith equality constraints $\\eqconfun(\\*x) = \\*0$\nby setting $\\ineqconfun(\\*x) \\ceq (\\eqconfun(\\*x), -\\eqconfun(\\*x))$.\nEquality constraints can also be solved by incorporating them\ninto the unconstrained solver (e.g., see \\cite{Boyd04Convex}\nfor an equality-constrained Newton method).\n\nAs sparse grid surrogates $\\objfun = \\sgintp$ are only defined on the\nunit hyper-cube,\nthe choice of $\\*x$ has to be restricted to $\\clint{\\*0, \\*1}$.\nIn the case of \\eqref{eq:unconstrainedOptimization},\nthis results in a \\term{box-constrained optimization problem.}\nA simple method for applying unconstrained optimization algorithms\nto box-constrained problems is extending $\\sgintp$ to $\\real^d$ by\n$\\sgintp(\\*x) \\ceq +\\infty$ for all $\\*x \\in \\real^d \\setminus \\clint{\\*0, \\*1}$.\nHowever, more sophisticated\n%, algorithm-tailored\napproaches are also available \\cite{More87Optimization}.\n\n\\paragraph{Black-box optimization methods}\n\nProblems of the form \\eqref{eq:unconstrainedOptimization} or\n\\eqref{eq:constrainedOptimization} are \\term{black-box optimization problems,}\nwhere we cannot gain any insight into the structure or algebraic\nproperties of $\\objfun$.\nBlack-box optimization methods perform a series of evaluations\n$\\objfun(\\*x_k)$,\nchoosing the next evaluation point $\\*x_{k+1}$\nbased on the previous function values $\\objfun(\\*x_0), \\dotsc, \\objfun(\\*x_k)$.\nGradient-based methods differ from gradient-free approaches\nin such a way that they also take values\nof the gradient $\\gradient{\\*x}{\\objfun}(\\*x_k)$,\nof the Hessian $\\hessian{\\*x}{\\objfun}(\\*x_k)$, or\nof even higher-order derivatives into account.\n\nA vast range of optimization methods exists in literature.\nSome methods are better suited for specific optimization problems\nthan others.\nHowever, according to the ``no-free-lunch theorem''\nand under some assumptions \\cite{Wolpert97No},\nall methods perform equally well (or equally badly) in the mean of all possible\noptimization problems.\n\n\\paragraph{Local and global optima}\n\nMost optimization methods depend on an initial point $\\*x_0$ and\nonly find local optima,\nwhere \\eqref{eq:unconstrainedOptimization} or\n\\eqref{eq:constrainedOptimization} only holds for $\\*x$\nin a neighborhood of $\\xopt$.\nOne can globalize local methods to increase the probability\nof finding a global optimum with a Monte Carlo multi-start approach:\nThe local method is repeated with different pseudo-random initial points\nand the best local optimum is chosen as the result.\n\nIn the following,\nwe give a brief survey of a small selection of optimization methods\n(see\n\\cref{tbl:optimizationMethod},\n\\cref{fig:optimizationMethodGradientFree}, and\n\\cref{fig:optimizationMethodGradientBased}),\nhighlighting the key ingredients for each method.\n\n\\begin{table}\n \\setnumberoftableheaderrows{1}%\n \\begin{tabular}{%\n >{\\kern\\tabcolsep}=l+l<{\\kern5mm}*{5}{+c}<{\\kern\\tabcolsep}%\n }\n \\toprulec\n \\headerrow\n Method& Type& C& D& S& I\\\\\n \\midrulec\n Nelder--Mead& Simplex heuristic& \\no& 0& \\no& \\yes\\\\\n Differential evolution& Evolutionary& \\no& 0& \\yes& \\yes\\\\\n CMA-ES& Evolutionary& \\no& 0& \\yes& \\yes\\\\\n Simulated annealing& Temperature heuristic& \\no& 0& \\yes& \\no\\\\\n PSO& Swarm heuristic& \\no& 0& \\yes& \\no\\\\\n GP-LCB& Bayesian& \\no& 0& \\yes& \\no\\\\\n \\midrulec\n Gradient descent& Descent& \\no& 1& \\no& \\yes\\\\\n NLCG& Descent& \\no& 1& \\no& \\yes\\\\\n Newton& Newton& \\no& 2& \\no& \\yes\\\\\n BFGS& Quasi-Newton& \\no& 1& \\no& \\yes\\\\\n Rprop& Heuristic& \\no& 1& \\no& \\yes\\\\\n Levenberg--Marquardt& Least sq., trust-region& \\no& 1& \\no& \\yes\\\\\n \\midrulec\n Log-barrier& Interior-point& \\yes& 0+& --& \\yes\\\\\n Squared penalty& Penalty& \\yes& 0+& --& \\yes\\\\\n Augmented Lagrangian& Penalty& \\yes& 0+& --& \\yes\\\\\n SQP& Quadratic subproblems& \\yes& 2& --& \\no\\\\\n \\bottomrulec\n \\end{tabular}\n \\caption[Selection of optimization methods]{%\n Selection of optimization methods.\n The columns show\n if constrained problems are supported (C),\n the order of required derivatives (D),\n if the algorithm is stochastic (S), and\n if the algorithm has been implemented in \\sgpp (I).%\n }%\n \\label{tbl:optimizationMethod}%\n\\end{table}\n\n\n\n\\subsection{Gradient-Free Unconstrained Optimization Methods}\n\\label{sec:511gradientFree}\n\n\\newcommand*{\\optImage}[1]{%\n \\raisebox{-0.5\\height}{\\includegraphics{optimizationMethod_#1}}%\n}\n\n\\begin{figure}\n \\optImage{1}\\quad\\optImage{2}\\quad\\optImage{3}%\n \\\\[3mm]%\n \\optImage{4}\\qquad\\optImage{5}\\qquad\\optImage{6}%\n \\caption[Ideas of various gradient-free optimization methods]{%\n Sketch of the ideas of gradient-free optimization methods.%\n }%\n \\label{fig:optimizationMethodGradientFree}%\n\\end{figure}\n\n\\paragraph{Nelder--Mead}\n\nThe \\term{Nelder--Mead method}\n\\multicite{Nelder65Simplex,Gao12Implementing,Valentin14Hierarchische}\nmaintains a list of $d + 1$ vertices of a $d$-dimensional simplex,\nsorted by ascending function value.\nIn each iteration,\nthe method performs one of the operations\n\\term{reflection,}\n\\term{expansion,}\n\\term{outer contraction,}\n\\term{inner contraction,} and\n\\term{shrinking}\non the vertices.\nTypically, convergence can be detected by\nchecking the size of the simplex,\nas the simplex tends to contract around local minima.\nHowever, there are counterexamples where the method converges to\na non-critical point for an only bivariate objective function\nthat is strictly convex and twice continuously differentiable\n\\cite{McKinnon98Convergence}.\n%For higher dimensionalities, the method quickly suffers from\n%the curse of dimensionality \\cite{Gao12Implementing}.\n\n\\paragraph{Differential evolution}\n\nThe method of \\term{differential evolution}\n\\multicite{%\n Storn97Differential,%\n Zielinski09Optimizing,%\n Valentin14Hierarchische%\n}\nis an evolutionary meta-heuristic algorithm.\nBeing similar to genetic algorithms,\nthe method maintains a \\term{population} of $m$ points\nthat is iteratively updated according to pseudo-random \\term{mutations,}\nwhich are weighted sums of the points of the previous generation.\nThe mutated vector is \\term{crossed over} with the original vector\nentry by entry.\nThe resulting \\term{offspring} are only accepted if they lead to\nan improvement in terms of objective function value.\n%There are various stopping criteria\n%that are based on the function value or on the location of the population\n%\\cite{Zielinski09Optimizing}.\n\n\\paragraph{CMA-ES}\n\n\\term{CMA-ES (covariance matrix adaption, evolution strategy)}\n\\cite{Hansen03Reducing}\nis an evolutionary algorithm that addresses the issue\nthat simple evolution strategies do not prefer a search direction\ndue to the lack of gradients \\cite{Toussaint15Introduction}.\nThe name of the algorithm stems from the fact\nthat it keeps track of the \\term{covariance matrix} of the\nGaussian search distribution.\nAfter $m$ points have been sampled from the current distribution,\nthe mean of the distribution for the next iteration\nis calculated as the weighted mean of the $k$ best samples and\nthe covariance matrix is adapted accordingly.\nAn advantage of the method is that if the population is large enough,\nlocal minima are smoothed out \\cite{Toussaint15Introduction}.\n\n\\paragraph{Simulated annealing}\n\n\\term{Simulated annealing}\n\\multicite{Laarhoven87Simulated,Press07Numerical,Kiranyaz14Multidimensional}\nimitates the cooling of a solid by randomly drawing samples\nfrom a proposal distribution and calculating an \\term{acceptance probability}\nthat depends on the function value improvement as well as\non a \\term{temperature} $T$.\nThis temperature is slowly decreased in the course of the algorithm.\nSimulated annealing is closely connected to the\nMetropolis--Hastings algorithm for drawing random samples of arbitrary\nprobability distributions.\nIf run long enough,\n%(e.g., if the decrease rate of the temperature is small enough),\nsimulated annealing is guaranteed to find the global\noptimum \\cite{Toussaint15Introduction}.\n\n\\paragraph{Particle swarm optimization (PSO)}\n\nThe method of \\term{particle swarm optimization (PSO)}\n\\multicite{Kennedy95Particle,Zielinski09Optimizing,Kiranyaz14Multidimensional}\ncan be seen as another evolutionary algorithm\nthat stems from swarm intelligence.\nFor each \\term{particle} of the population,\nnot only the \\term{position} $\\*x_k$ is stored,\nbut also the current \\term{velocity} $\\*v_k$,\nthe best known position in a neighborhood of $\\*x_k$\n(which may be the whole swarm), and\nthe best known position of the $k$-th particle.\nThe next velocity $\\*v_{k+1}$ is computed as\na pseudo-randomly weighted sum of $\\*v_k$,\nthe vector from $\\*x_k$ to the best neighborhood position, and\nthe vector from $\\*x_k$ to the best own position.\n\n\\paragraph{GP-LCB}\n\n\\term{GP-LCB (Gaussian process, lower confidence bound)}\n\\multicite{Srinivas10Gaussian,Toussaint15Introduction} is an example\nfor a \\term{Bayesian optimization} strategy.\nThe objective function is treated as a stochastic process.\nA \\term{prior distribution} is updated according to the previous function\nevaluations to calculate the \\term{posterior distribution.}\nThe posterior distribution is used to form the \\term{acquisition function,}\nwhich in turn determines the point at which the objective\nfunction is evaluated next.\nThe GP-LCB method is obtained by choosing\n\\term{Gaussian processes} for the family of stochastic processes and\n\\term{lower confidence bounds} (which are the difference of the mean\nand a multiple of the standard deviation) for the acquisition function.\n\n\n\n\\subsection{Gradient-Based Unconstrained Optimization Methods}\n\\label{sec:512gradientBasedUnconstrained}\n\n\\begin{figure}\n \\optImage{7}\\qquad\\optImage{8}\\qquad\\optImage{9}%\n \\\\[3mm]%\n \\optImage{10}\\quad\\optImage{11}\\quad\\optImage{12}%\n \\caption[Ideas of various gradient-based optimization methods]{%\n Sketch of the ideas of gradient-based or constrained\n optimization methods.%\n }%\n \\label{fig:optimizationMethodGradientBased}%\n\\end{figure}\n\nMost gradient-based optimization algorithms determine in\neach iteration $k$ a unit \\term{search direction} $\\*d_k \\in \\real^d$\n($\\norm[2]{\\*d_k} = 1$) to update the current iterate $\\*x_k$:\n\\begin{equation}\n \\*x_k\n \\to \\*x_{k+1}\n \\ceq \\*x_k + \\delta_k \\*d_k,\\qquad\n \\delta_k\n \\ceq \\argmin_{\\delta \\in \\posreal} \\objfun(\\*x_k + \\delta \\*d_k),\n\\end{equation}\nwhere $\\delta_k \\in \\posreal$ is the \\term{step size.}\nThe algorithms essentially differ in the\nchoice of the search direction $\\*d_k$,\nwhich should be oriented like the negative gradient\n($\\innerprod[2]{\\*d_k}{\\gradient{\\*x}{\\objfun}(\\*x_k)} < 0$).\nThe step size $\\delta_k$ can then be determined independently of the\nalgorithm via \\term{line search,}\nfor instance, the \\term{Armijo line search algorithm}\n\\multicite{Nocedal99Numerical,Ulbrich12Nichtlineare,Valentin14Hierarchische},\nwhich uses a heuristic acceptance criterion\nto find $\\delta_k$ with a good enough improvement.\n\n\\paragraph{Gradient descent}\n\n\\term{Gradient descent}\n\\multicite{%\n Ulbrich12Nichtlineare,%\n Valentin14Hierarchische,%\n Toussaint15Introduction%\n}\n%is the simplest gradient-based optimization method,\n%since it\nchooses $\\*d_k\n\\propto -\\gradient{\\*x}{\\!\\objfun}(\\*x_k)$ (i.e., normalized).\nThe method suffers from slow convergence,\nif the Hessian $\\hessian{\\*x}{\\objfun}$ is ill-conditioned:\nOne can show that for strictly convex quadratic functions,\nthe error $\\objfun(\\*x_k) - \\objfun(\\xopt)$\n%to the optimal function value\ncan decrease in each iteration only by the factor of\n$(\\tfrac{\\lambda^{\\max} - \\lambda^{\\min}}{\\lambda^{\\max} + \\lambda^{\\min}})^2$,\n%$(\\lambda^{\\max} - \\lambda^{\\min})^2/(\\lambda^{\\max} + \\lambda^{\\min})^2$,\nwhere $\\lambda^{\\min}$ and $\\lambda^{\\max}$ are the minimum and maximum\neigenvalue of $\\hessian{\\*x}{\\objfun}$, respectively\n\\cite{Ulbrich12Nichtlineare}.\nIf the condition number\n$\\tfrac{\\lambda^{\\max}}{\\lambda^{\\min}}$\n%$\\lambda^{\\max}/\\lambda^{\\min}$\nof $\\hessian{\\*x}{\\objfun}$ is large,\nthen this factor will be very close to one.\n\n\\paragraph{NLCG}\n\nA possible remedy for this issue is the method of\n\\term{non-linear conjugate gradients (NLCG)}\n\\multicite{%\n Nocedal99Numerical,%\n Valentin14Hierarchische,%\n Toussaint15Introduction%\n}.\nIt is equivalent to the CG method for\nsolving symmetric positive definite %\\spd\nlinear systems $\\mat{A} \\*x = \\*b$,\nif we optimize the strictly convex quadratic function\n$\\objfun(\\*x) \\ceq \\frac{1}{2} \\tr{\\*x} \\mat{A} \\*x - \\tr{\\*b} \\*x$\n\\multicite{Reinhardt13Nichtlineare,Valentin14Hierarchische}, i.e.,\nit finds the optimum after only $d$ steps for strictly convex\nquadratic functions.\nThe NLCG method quickly converges even for non-convex objective functions,\nas due to the Taylor theorem,\nthree times continuously differentiable functions\nwith positive definite Hessian are ``similar'' to a\nstrictly convex quadratic function in a neighborhood of $\\xopt$\n\\cite{Valentin14Hierarchische}.\n\n\\paragraph{Newton}\n\nThe \\term{Newton method}\n\\multicite{%\n Ulbrich12Nichtlineare,%\n Valentin14Hierarchische,%\n Toussaint15Introduction%\n}\nreplaces the objective function with the second-order Taylor approximation\ngiven by\n$\\objfun(\\*x_k + \\*d_k)\n\\!\\approx\\! \\objfun(\\*x_k) +\n\\tr{(\\gradient{\\*x}{\\objfun}(\\*x_k))} \\*d_k \\,+\n\\frac{1}{2} \\tr{(\\*d_k)} (\\hessian{\\*x}{\\objfun}(\\*x_k)) \\*d_k$\nand determines the search direction such that $\\*x_k + \\*d_k$ is\nthe minimum of the approximation, i.e.,\n$\\*d_k \\propto\n-(\\hessian{\\*x}{\\objfun}(\\*x_k))^{-1} \\gradient{\\*x}{\\objfun}(\\*x_k)$.\nDespite converging for strictly convex quadratic functions in a single step,\nthe Hessian must not be ill-conditioned for the Newton method as well,\nas we have to solve a linear system with the matrix\n$\\hessian{\\*x}{\\objfun}(\\*x_k)$.\nHence, often a \\term{regularization/damping term} $\\lambda \\eye$\nfor some $\\lambda > 0$ is added to the Hessian.\n\n\\paragraph{BFGS}\n\nThe Newton method has the disadvantage that it needs to evaluate the\nHessian $\\hessian{\\*x}{\\objfun}$,\nwhich may be unavailable or too expensive.\n\\term{Quasi-Newton methods} such as the method of\n\\term{BFGS (Broyden, Fletcher, Goldfarb, Shanno)}\n\\multicite{%\n Nocedal99Numerical,%\n Ulbrich12Nichtlineare,%\n Toussaint15Introduction%\n}\napproximate the Hessian by a solution of the secant equation\n$\\hessian{\\*x}{\\objfun}(\\*x_k) (\\*x_k - \\*x_{k-1}) \\approx\n\\gradient{\\*x}{\\objfun}(\\*x_k) - \\gradient{\\*x}{\\objfun}(\\*x_{k-1})$.\nAs the solution is not unique for $d > 1$,\nQuasi-Newton methods differ in which solution to choose.\nThe BFGS method performs a simple rank-one update.\n\n\\paragraph{Rprop}\n\n\\term{Rprop (resilient propagation)}\n\\multicite{Riedmiller93Direct,Toussaint15Introduction}\nconsiders the gradient entries $(\\gradient{\\*x}{\\objfun}(\\*x_k))_t$\nof each dimension $t = 1, \\dotsc, d$ separately\nand updates the entries $x_{k,t}$ of $\\*x_k$\naccording to the sign of the respective gradient entry,\nwhile adapting the step size dimension-wise.\nAlthough the algorithm is independent of the exact direction\nof $\\gradient{\\*x}{\\objfun}(\\*x_k)$,\nit was found to often work robustly in machine learning scenarios\n\\cite{Toussaint15Introduction}.\n\n\\paragraph{Levenberg--Marquardt}\n\nThe \\term{Levenberg--Marquardt method}\n\\multicite{Nocedal99Numerical,Freund07Stoer,Toussaint15Introduction}\ncan only solve \\term{non-linear least-squares problems,} i.e.,\nthe objective function must be of the form\n$\n \\objfun(\\*x)\n = \\norm[2]{\\*\\phi(\\*x)}^2\n = \\sum_{i=1}^{m_{\\*\\phi}} \\abs{\\phi_i(\\*x)}^2\n$\nfor some function $\\*\\phi\\colon \\real^d \\to \\real^{m_{\\*\\phi}}$.\nIt is an improvement over the \\term{Gauss--Newton method}\n(which is in turn a slight modification of the Newton method)\nand can be obtained by replacing the line search in the\nGauss--Newton method with a \\term{trust-region approach.}\n\n\n\n\\subsection{Constrained Optimization Methods}\n\\label{sec:513gradientBasedConstrained}\n\nMethods for constrained optimization usually\nsolve a series of unconstrained \\term{auxiliary problems} with an arbitrary\nunconstrained optimization method.\nThe auxiliary function to be minimized is\nthe sum of the objective function and \\term{penalty terms,}\nwhich penalize if the current point $\\*x_k$ is near the boundary\nof the feasible domain or even outside.\nThe penalty terms slowly increase to enforce\nthe feasibility of the final result.\nConstrained optimization methods can roughly be divided\ninto \\term{interior-point or barrier methods,}\nwhere $\\*x_k$ always stays in the feasible domain,\nand \\term{penalty methods,}\nwhere intermediate solutions $\\*x_k$ may be infeasible,\nin which case the penalty term is applied.\n%(the penalty vanishes inside the domain).\n\nAt least for the interior-point methods,\na feasible initial solution $\\*x_0$ is required.\nWe can find an initial solution by solving another auxiliary problem\n\\cite{Toussaint15Introduction}, for instance\n\\begin{equation}\n \\min_{(\\*x, s) \\in \\real^{d+1}} s\n \\quad\\text{s.t.}\\quad\n s \\ge 0,\\;\\;\n \\ineqconfun(\\*x) \\le s \\cdot \\*1_{m_{\\ineqconfun}},\n\\end{equation}\nwhere $\\*1_{m_{\\ineqconfun}} \\in \\real^{m_{\\ineqconfun}}$\nis the all-one vector.\nAn initial solution for this problem can be explicitly given\n(for example, $\\*x_0 = \\*0$ and\n$s_0 = \\max(\\max(\\ineqconfun(\\*x_0)), 0)$).\n\n\\paragraph{Log-barrier}\n\nThe \\term{log-barrier method}\n\\multicite{Boyd04Convex,Reinhardt13Nichtlineare,Toussaint15Introduction}\nis an interior-point method that adds\na logarithmic \\term{barrier function term} to the objective function\nnear the boundary.\nThe method solves\n$\\min\\, [\\objfun(\\*x) - \\mu_k \\sum_{i=1}^{m_{\\ineqconfun}} \\log(-g_i(\\*x))]$\nfor some decreasing $\\mu_k \\in \\posreal$.\n\n\\paragraph{Squared penalty}\n\nThe \\term{squared penalty method}\n\\multicite{%\n Polak71Computational,%\n Ulbrich12Nichtlineare,%\n Toussaint15Introduction%\n}\nreplaces the constrained problem with the penalized problem\n$\\min\\, [\\objfun(\\*x) + \\mu_k \\norm[2]{\\nonnegpart{\\ineqconfun(\\*x)}}^2]$,\nwhere $\\mu_k \\in \\posreal$ is a penalty parameter and\n$\\nonnegpart{\\cdot} \\ceq \\vecmax(\\cdot, \\*0)$ denotes the non-negative part.\nWith increasing $\\mu_k$, the constraint violation of the solution of\nthe penalized problem decreases, although it may happen\nthat it never vanishes.\n\n\\paragraph{Augmented Lagrangian}\n\nThe method of the \\term{augmented Lagrangian}\n\\multicite{Reinhardt13Nichtlineare,Toussaint15Introduction}\nconsiders the auxiliary problem\n\\begin{equation}\n \\min_{\\*x \\in \\real^d} \\bracket*{\n \\objfun(\\*x) + \\mu_k \\sum_{i=1}^{m_{\\ineqconfun}} [\\lambda_{k,i} > 0]\n ((\\ineqconfun[i](\\*x))_{+})^2 + \\tr{\\*\\lambda_k} \\ineqconfun(\\*x)\n },\n\\end{equation}\nwhere $[\\lambda_{k,i} > 0] \\in \\{0, 1\\}$ is defined as one\nif and only if $\\lambda_{k,i} > 0$, and\n$\\*\\lambda_k \\in \\nonnegreal^{m_{\\ineqconfun}}$ is an estimate of the\n\\term{Lagrangian multipliers.}\nThey are updated according to the penalty of the previous iteration,\ngenerating a ``virtual gradient'' that drastically decreases\nthe necessary magnitude of the penalty parameter $\\mu_k$\nto achieve feasibility of the solution \\cite{Toussaint15Introduction}.\n\n\\paragraph{Sequential quadratic programming (SQP)}\n\n\\term{Sequential quadratic programming (SQP) methods}\n\\multicite{%\n Ulbrich12Nichtlineare,%\n Reinhardt13Nichtlineare,%\n Toussaint15Introduction%\n}\nare one of the most powerful method classes for constrained optimization.\nThey are motivated by the \\term{Karush--Kuhn--Tucker (KKT) conditions,}\nwhich are necessary to hold in any optimal point\n(similarly to critical points in unconstrained optimization).\nThe Newton method can be employed to solve the KKT conditions when\nwritten as a non-linear system of equations.\nThe linear system of the resulting \\emph{Newton--Lagrange method}\nis equivalent to the KKT conditions of a\n\\term{quadratic programming (QP) problem,}\nfor which objective and constraint functions have the\nform $\\objfun(\\*x) = \\frac{1}{2} \\tr{\\*x} \\mat{Q} \\*x + \\tr{\\*d} \\*x$ and\n$\\ineqconfun(\\*x) = \\mat{A} \\*x - \\*b$, respectively.\n%SQP methods solve one QP problem in each iteration (``sequentially'')\n%to determine the search direction.\n", "meta": {"hexsha": "90fb0a171c9188105ba67ba47d1d43764e1d3169", "size": 21182, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/document/51algorithms.tex", "max_stars_repo_name": "valentjn/thesis", "max_stars_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2022-01-15T19:50:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-15T20:16:10.000Z", "max_issues_repo_path": "tex/document/51algorithms.tex", "max_issues_repo_name": "valentjn/thesis", "max_issues_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/document/51algorithms.tex", "max_forks_repo_name": "valentjn/thesis", "max_forks_repo_head_hexsha": "65a0eb7d5f7488aac93882959e81ac6b115a9ea8", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.6666666667, "max_line_length": 81, "alphanum_fraction": 0.734680389, "num_tokens": 6119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059462938815, "lm_q2_score": 0.774583389368527, "lm_q1q2_score": 0.6143266920086478}} {"text": "\\section{Projects}\n\\textbf{SimpleNavierStokes.jl} (2020) \\\\\n{A \\link{https://github.com/Emadmasroor/SimpleNavierStokes.jl}{Julia package}, \\link{https://emadmasroor.github.io/blog/2020/12/16/CFD-tutorial-in-julia/}{blog post}, and open-source \\link{https://nextjournal.com/emadmasroor/CFD-tutorial-in-Julia}{notebook} to serve as a beginner's tutorial for writing incompressible Navier-Stokes solvers using the $\\omega-\\psi$ formulation.}\n", "meta": {"hexsha": "834384da11843d8fb5a02899c711bfd5a2b502b8", "size": 439, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/projects.tex", "max_stars_repo_name": "Emadmasroor/Simple-CV", "max_stars_repo_head_hexsha": "cd687dd5d96f7d5148cf3968b10ab73f36b3e5a5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sections/projects.tex", "max_issues_repo_name": "Emadmasroor/Simple-CV", "max_issues_repo_head_hexsha": "cd687dd5d96f7d5148cf3968b10ab73f36b3e5a5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/projects.tex", "max_forks_repo_name": "Emadmasroor/Simple-CV", "max_forks_repo_head_hexsha": "cd687dd5d96f7d5148cf3968b10ab73f36b3e5a5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 109.75, "max_line_length": 378, "alphanum_fraction": 0.7813211845, "num_tokens": 139, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833737577158, "lm_q2_score": 0.7931059609645724, "lm_q1q2_score": 0.6143266909912938}} {"text": "\\chapter{Word Factoring}\n\\label{chap:factoring}\n\\index{exact string matching|see{word fac\\-toring}}\n\\index{string matching|see{word factoring}}\n\\index{word factoring|(}\n\nLet us call \\emph{alphabet}\\index{word factoring!alphabet} a\nnon\\hyp{}empty, finite set of symbols, called\n\\emph{letters}\\index{word factoring!letter} and set in a\nsans\\hyp{}serif type, for example \\word{a}, \\word{b} etc. A\n\\emph{word}\\index{word factoring!word} is a finite series of letters,\nlike \\word{word}; in particular, a letter is a word, as in English. We\ndenote repetition of a letter or word with an exponent, for instance,\n\\(\\word{a}^3 = \\word{aaa}\\). Just as letters can be joined to make up\nwords, so can words: the word \\(u \\cdot v\\) is made of the letters of\nword~\\(u\\) followed by the letters of word~\\(v\\), for instance, if\n\\(u=\\word{back}\\) and \\(v=\\word{up}\\), then \\(u \\cdot v =\n\\word{backup}\\). This operation is associative: \\((u \\cdot v) \\cdot w\n= u \\cdot (v \\cdot w)\\). As a shorthand, the operator may be omitted:\n\\((uv)w = u(vw)\\). Catenation on words behave like a\nnon\\hyp{}commutative product, so it has a neutral element\n\\(\\varepsilon\\), called the \\emph{empty word}: \\(u \\cdot \\varepsilon =\n\\varepsilon \\cdot u = u\\).\n\nA word~\\(x\\) is a \\emph{factor}\\index{word factoring!factor} of a\nword~\\(y\\) if there exists two words \\(u\\)~and~\\(v\\) such that \\(y =\nuxv\\). The word~\\(x\\) is a \\emph{prefix}\\index{word factoring!prefix}\nof~\\(y\\), noted \\(x \\prefeq y\\), if~\\(u=\\varepsilon\\), that is, if \\(y\n= xv\\). Moreover, it is a \\emph{proper prefix}, noted \\(x \\pref y\\),\nif \\(v \\neq \\varepsilon\\). Given \\(y = uxv\\), the word~\\(x\\) is a\n\\emph{suffix}\\index{word factoring!suffix} of~\\(y\\) if \\(v =\n\\varepsilon\\). Furthermore, it is a \\emph{proper suffix} if \\(u \\neq\n\\varepsilon\\). Let \\(a\\)~be any letter and \\(x\\), \\(y\\)~any word, then\nthe prefix relation is easy to define by an inference system as\n\\begin{mathpar}\n\\inferrule*{}{\\varepsilon \\prefeq y}\n\\qquad\n\\inferrule{x \\prefeq y}{a \\cdot x \\prefeq a \\cdot y}\n\\end{mathpar}\n\nThe purpose being to write a functional program for factoring, we need\nto translate words and operations on them into terms of the\nlanguage. A letter is translated into a constant data constructor; for\nexample, \\word{a} becomes \\(\\fun{a}()\\). A word of more than one\nletter is mapped to a stack of mapped letters, such as \\word{hi} in\n\\([\\fun{h}(),\\fun{i}()]\\). The catenation of a letter and a word is\ntranslated as a push, like \\(\\word{a} \\cdot \\word{bed}\\) becomes\n\\([\\fun{a}(),\\fun{b}(),\\fun{e}(),\\fun{d}()]\\). The catenation of two\nwords is associated to stack catenation, so \\(\\word{ab} \\cdot\n\\word{cd}\\) leads to\\index{cat@\\fun{cat/2}}\n\\(\\fun{cat}([\\fun{a}(),\\fun{b}()],[\\fun{c}(),\\fun{d}()])\\).\n\n% The transposition of a word is the reversal of the stack implementing\n% the word.\n\nAs usual, the translation of the inference system defining\n\\((\\prefeqName)\\) into a function \\fun{pre/2}\\index{pre@\\fun{pre/2}}\nrequires that the cases corresponding to the axioms evaluate in\n\\(\\fun{true}()\\) and the cases left unspecified \\((\\nprefeqName)\\)\nevaluate in \\(\\fun{false}()\\):\n\\begin{equation*}\n\\fun{pre}(\\el,y) \\rightarrow \\fun{true}();\\quad\n\\fun{pre}(\\cons{a}{x},\\cons{a}{y}) \\rightarrow \\fun{pre}(x,y);\\quad\n\\fun{pre}(x,y) \\rightarrow \\fun{false}().\n\\end{equation*}\nThe inference system is now a formal specification for the program.\n\nA letter in a word can be uniquely characterised by a natural number,\ncalled \\emph{index}\\index{word factoring!index}, assuming that the\nfirst letter has index~\\(0\\) \\citep{Dijkstra_1982}. If\n\\(x=\\word{top}\\), then the letter at index~\\(0\\) is written\n\\(\\ind{x}{0}=\\word{t}\\) and the one at index~\\(2\\) is\n\\(\\ind{x}{2}=\\word{p}\\). A factor~\\(x\\) of~\\(y\\) can be identified by\nthe index of~\\(\\ind{x}{0}\\) in~\\(y\\). The end of the factor can also\nbe given; for example, \\(x=\\word{sit}\\) is a factor of\n\\(y=\\word{curiousity}\\) at index~\\(6\\), written \\(\\ind{y}{6,8} = x\\),\nmeaning \\(\\ind{y}{6} = \\ind{x}{0}\\), \\(\\ind{y}{7} = \\ind{x}{1}\\) and\n\\(\\ind{y}{8} = \\ind{x}{2}\\). Given two words \\(p\\)~and~\\(t\\),\ndetermining whether \\(p\\)~is a factor of~\\(t\\) is called\n\\emph{factoring~\\(p\\) in~\\(t\\)}.\n\nFactor matching is common in text editing, although it is usually\nbetter known as \\emph{exact string matching} in the academic field of\n\\emph{stringology} or \\emph{text\n algorithmics} \\citep{CharrasLecroq_2004,CrochemoreHancartLecroq_2007}\n\\citep[\\S{}32]{CLRS_2009}. Because of the asymmetric nature of\nfactoring, the word~\\(p\\) is called the \\emph{pattern}\\index{word\n factoring!pattern} and the word~\\(t\\) is the \\emph{text}\\index{word\n factoring!text}.\n\n\\section{Na\\\"{\\i}ve factoring}\n\\label{sec:naive_factoring}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$|(}\n\nIn section~\\ref{def:linear_search}, \\vpageref{def:linear_search}, we\nintroduced the linear search\\index{linear search}, that is, the\nstepwise search for the occurrence of an item in a stack. We can\ngeneralise it to search for a series of items occurring consecutively\nin a stack, that is, to solve the factoring problem. This approach is\nqualified as being na\\\"{\\i}ve because it is a simple extension of a\nsimple idea and it is implied that it is not the most efficient.\n\nEverything starts with~\\(\\ind{p}{0}\\) and~\\(\\ind{t}{0}\\) being\ncompared, then, assuming \\(\\ind{p}{0} = \\ind{t}{0}\\), letters\n\\(\\ind{p}{1}\\) and~\\(\\ind{t}{1}\\) are, in turn, compared etc. until\none of the words is exhausted or a mismatch occurs. Assuming that\n\\(p\\)~is shorter than~\\(t\\), the former case means that \\(p\\)~is a\nprefix of~\\(t\\). In the latter case, \\(p\\)~is shifted so\n\\(\\ind{p}{0}\\) is aligned with \\(\\ind{t}{1}\\) and the comparisons are\nresumed from there. If \\(p\\)~cannot be shifted anymore because its end\nwould surpass the end of~\\(t\\), then it is not a factor. The essence\nof this procedure is summed up in \\fig~\\vref{fig:naive},\n\\begin{figure}[t]\n\\centering\n\\includegraphics[bb=75 621 352 715]{naive}\n\\caption{Na\\\"{\\i}vely matching pattern~\\(p\\) against text~\\(t\\)\n(failure in grey)\\label{fig:naive}}\n\\end{figure}\nwhere \\(\\ind{p}{i} \\neq \\ind{t}{j}\\) (the letters\n\\word{a}~and~\\word{b} are not relevant in themselves).\n\n% Wrapping figure better declared before a paragraph\n%\n%\\begin{wrapfigure}[15]{r}[0pt]{0pt}\n% [15] vertical lines\n% {r} right placement\n% [0pt] of margin overhang\n%\\end{wrapfigure}\n\n\\Fig~\\vref{fig:loc0}\n\\begin{figure}[t]\n\\centering \n\\includegraphics[bb=71 548 253 721]{loc0}\n\\caption{Na\\\"{\\i}ve factoring with \\fun{loc\\(_0\\)/2}\n\\label{fig:loc0}}\n\\end{figure}\nshows an abstract program implementing this scheme. The call\n\\(\\fun{loc}_0(p,t)\\)\\index{loc0@\\fun{loc\\(_0\\)/2}} evaluates in\n\\(\\fun{absent}()\\) if the pattern~\\(p\\) is not a factor of the\ntext~\\(t\\), otherwise in \\(\\fun{factor}(k)\\), where \\(k\\)~is the index\nin~\\(t\\) where~\\(p\\) occurs first. Conceptually, this design consists\nin combining a linear search for the first letter of the pattern and a\nprefix check for the rest of the pattern and text. It is important to\nverify whether the invariants implicit in general do not break in the\npresence of limit cases. For instance, in stack processing, set the\ndifferent stacks to be empty and interpret the result of single\nrewrites and entire evaluations. We have \\(\\fun{pre}(\\el,t)\n\\twoheadrightarrow \\fun{true}()\\), because \\(t = \\varepsilon \\cdot\nt\\). Accordingly, \\(\\fun{loc}_0(\\el,t) \\twoheadrightarrow\n\\fun{factor}(0)\\).\n\n\\paragraph{Refinements}\n\nWhile this program composition is intuitive, it is too long. We may\nremark that, after a call to \\fun{pre/2}\\index{pre@\\fun{pre/2}}\nevaluates in \\(\\fun{true}()\\), the interpretation ends with\n\\(\\fun{factor}(j)\\). Dually, a value \\(\\fun{false}()\\) is followed by\nthe call\n\\(\\fun{loc}_0(p,t,j+1)\\)\\index{loc0@\\fun{loc\\(_0\\)/3}}. Therefore,\ninstead of calling \\fun{pre/2} and then inspecting the resulting value\nto decide what to do next, we could have \\fun{pre/2} take the\nlead. This entails that it needs to receive additional arguments to be\nable to end with \\(\\fun{factor}(j)\\) or resume with\n\\(\\fun{loc}_0(p,t,j+1)\\), as expected. The corresponding code is shown\nin \\fig~\\ref{fig:loc1}.\n\\begin{figure}[t]\n\\begin{equation*}\n\\boxed{%\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{loc}_1(p,t) & \\rightarrow & \\fun{loc}_1(p,t,0).\\\\\n\\\\\n\\fun{loc}_1(\\cons{a}{p},\\el,j) & \\rightarrow & \\fun{absent}();\\\\\n\\fun{loc}_1(p,t,j) & \\rightarrow & \\fun{pre}_1(p,t,p,t,j).\\\\\n\\\\\n\\fun{pre}_1(\\el,t,p',t',j) & \\rightarrow & \\fun{factor}(j);\\\\\n\\fun{pre}_1(\\cons{a}{p},\\cons{a}{t},p',t',j)\n & \\rightarrow & \\fun{pre}_1(p,t,p',t',j);\\\\\n\\fun{pre}_1(p,t,p',\\cons{a}{t'},j) & \\rightarrow & \\fun{loc}_1(p',t',j+1).\n\\end{array}}\n\\end{equation*}\n\\caption{Refinement of \\fig~\\vref{fig:loc0}\n\\label{fig:loc1}}\n\\end{figure}\n\nFurther examination reveals that we can merge\n\\fun{loc\\(_1\\)/3}\\index{loc1@\\fun{loc\\(_1\\)/3}} and\n\\fun{pre\\(_1\\)/5}\\index{pre1@\\fun{pre\\(_1\\)/5}} into\n\\fun{pre/5}\\index{pre@\\fun{pre/5}} in\\index{word factoring!na\\\"{\\i}ve\n $\\sim$!program} \\fig~\\vref{fig:loc}. This kind of progressive\ndesign, where a program is transformed into a guided series of\nequivalent programs is called a \\emph{refinement}. Here, each\nrefinement is more efficient than the preceding, but less legible than\nthe original, so each step must be cautiously checked.\n\\begin{figure}[h]\n\\begin{equation*}\n\\boxed{%\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{loc}(p,t) & \\xrightarrow{\\smash{\\pi}} & \\fun{pre}(p,t,p,t,0).\\\\\n\\\\\n\\fun{pre}(\\el,t,p',t',j) & \\xrightarrow{\\smash{\\rho}}\n & \\fun{factor}(j);\\\\\n\\fun{pre}(p,\\el,p',t',j) & \\xrightarrow{\\smash{\\sigma}}\n & \\fun{absent}();\\\\\n\\fun{pre}(\\cons{a}{p},\\cons{a}{t},p',t',j)\n & \\xrightarrow{\\smash{\\tau}}\n & \\fun{pre}(p,t,p',t',j);\\\\\n\\fun{pre}(p,t,p',\\cons{b}{t'},j) & \\xrightarrow{\\smash{\\upsilon}}\n & \\fun{pre}(p',t',p',t',j+1).\n\\end{array}}\n\\end{equation*}\n\\caption{Refinement of \\fig~\\vref{fig:loc1}\n\\label{fig:loc}}\n\\end{figure}\n\n\\paragraph{Termination}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!termination|(}\n\nWe want to show that the index in the text always increases, whether a\ncomparison fails or not, so we choose a lexicographic order on the\ndependency pairs of \\fun{pre/5}\\index{termination!dependency pair}\nmade of the fourth and second arguments (definition~\\eqref{def:lexico}\n\\vpageref{def:lexico}), where \\(s \\succ t\\) if~\\(t\\) is the immediate\nsubstack of~\\(s\\). The third rule satisfies \\((t',\\cons{a}{t}) \\succ\n(t',t)\\). The fourth rule is also ordered, because \\((\\cons{b}{t'},t)\n\\succ (t',t')\\).\\index{word factoring!na\\\"{\\i}ve\n $\\sim$!termination|)}\\hfill\\(\\Box\\)\n\n\\paragraph{Completeness}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!completeness|(}\n\\index{induction!example|(}\n\nNote how, in rule~\\(\\sigma\\), the pattern~\\(p\\) can not be empty\nbecause the rules are ordered and that case would always match\nrule~\\(\\rho\\). The completeness of the definition of \\fun{pre/5}\ndeserves some attention and we need to justify why the call\n\\(\\fun{pre}(\\cons{a}{p},\\cons{b}{t},p',\\el,j)\\), with \\(a \\neq b\\),\ncannot happen. Perhaps surprisingly, a more general statement is\neasier to establish:\n\\begin{center}\n\\(\\fun{loc}(p,t) \\twoheadrightarrow \\fun{pre}(p_0,t_0,p_0',t_0',j)\\)\n\\textsl{implies} \\(t_0' \\succcurlyeq t_0\\),\n\\end{center}\nwhere (\\(\\succcurlyeq\\)) is the reflexive substack relation. Let us\nprove this property by \\emph{induction on the length of the\n derivation}\\index{induction!$\\sim$ on the length of the\n derivation}. More precisely, we want to establish the proposition\n\\begin{equation*}\n\\pred{Comp}{n} \\colon \\fun{loc}(p,t) \\xrightarrow{\\smash{n}}\n\\fun{pre}(p_0,t_0,p_0',t_0',j) \\Rightarrow t_0' \\succcurlyeq t_0.\n\\index{Comp@\\predName{Comp}}\n\\end{equation*}\n\\begin{itemize}\n\n \\item The basis \\(\\pred{Comp}{0}\\) is easy to prove without\n induction by means of rule~\\(\\pi\\): \\(\\fun{loc}(p,t)\n \\xrightarrow{\\smash{\\pi}} \\fun{pre}(p,t,p,t,0)\\) and \\(t\n \\succcurlyeq t\\) trivially holds.\n\n \\item The induction hypothesis is \\(\\pred{Comp}{n}\\) and we want\n to show that, under this assumption, \\(\\pred{Comp}{n+1}\\) holds\n as well. In other words, let us suppose that \\(\\fun{loc}(p,t)\n \\xrightarrow{\\smash{n}} \\fun{pre}(p_0,t_0,p_0',t_0',j)\\) implies\n \\(t_0' \\succcurlyeq t_0\\) and we want to prove that\n \\(\\fun{pre}(p_0,t_0,p_0',t_0',j) \\rightarrow\n \\fun{pre}(p_1,t_1,p_1',t_1',k)\\) implies \\(t_1' \\succcurlyeq\n t_1\\). This rewrite can only be by means of~\\(\\tau\\)\n or~\\(\\upsilon\\). \n \\begin{itemize}\n\n \\item If~\\(\\tau\\), the induction hypothesis on the\n left\\hyp{}hand side entails \\(t' \\succcurlyeq \\cons{a}{t}\\),\n so \\(t' \\succcurlyeq t\\) in the right\\hyp{}hand side;\n\n \\item otherwise, the right\\hyp{}hand side of~\\(\\upsilon\\)\n trivially satisfies \\(t' \\succcurlyeq t'\\).\n\n \\end{itemize}\n\\end{itemize}\nIn summary, \\(\\pred{Comp}{0}\\) is true and \\(\\pred{Comp}{n}\n\\Rightarrow \\pred{Comp}{n+1}\\). Therefore, the induction principle\nyields \\(\\forall n.\\pred{Comp}{n}\\), which, in turn, entails our\nformulation with~(\\(\\twoheadrightarrow\\)). Note how, in this case,\nthis proof technique reduces to mathematical induction\non~\\(n\\).\\hfill\\(\\Box\\)\n\\index{induction!example|)}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!completeness|)}\n\n\\addcontentsline{toc}{subsection}{Cost}\n\\paragraph{Cost}\n\nIn the following cost analysis, let~\\(m\\) be the length of the\npattern~\\(p\\) and \\(n\\)~be the length of the text~\\(t\\). Moreover, as\nit is common with search algorithms, we discriminate on \\(p\\)~being a\nfactor of~\\(t\\) or not.\n\n\\paragraph{Minimum cost}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!minimum cost}\n\nIf \\(m \\leqslant n\\), the best case happens when the pattern is a\nprefix of the text, so the evaluation trace is \\(\\pi\\tau^m\\rho\\) and\n\\(\\B{\\fun{loc}}{m,n} = m + 2\\). If \\(m > n \\), the minimum cost is\n\\(\\B{\\fun{loc}}{m,n} = \\len{\\pi\\tau^n\\sigma} = n + 2\\). We can gather\nthese two cases in one formula:\n\\begin{equation*}\n\\B{\\fun{loc}}{m,n} = \\min\\{m,n\\} + 2.\\index{loc@$\\B{\\fun{loc}}{m,n}$}\n\\end{equation*}\n\n\\paragraph{Maximum cost}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!maximum cost|(}\n\nTo find the maximum cost, let us investigate the cases where the\npattern is a factor of the text and when it is not.\n\\begin{itemize}\n\n \\item \\emph{The text contains the pattern.} The discovery of the\n pattern must be delayed as much as possible, therefore the worst\n case is when \\(w\\)~is a suffix of~\\(t\\) and every mismatch\n involves the last letter of the pattern. An example is \\(p =\n \\word{a}^{m-1}\\word{b}\\) and \\(t = \\word{a}^{n-1}\\word{b}\\). The\n evaluation trace corresponding to this case is \\(\\pi\n (\\tau^{m-1}\\upsilon)^{n-m} \\tau^m\\rho\\), whose length is\n \\(mn-m^2+m+2\\).\n\n \\item \\emph{The text does not contain the pattern.} The pattern is\n not the prefix of any suffix of the text. The most delayed\n comparison failure should occur at the last letter of the pattern,\n like \\(p = \\word{a}^{m-1}\\word{b}\\) and \\(t = \\word{a}^{n}\\). The\n cost is \\(\\len{\\pi(\\tau^{m-1}\\upsilon)^{n-m+1}\\tau^{m-1}\\sigma} =\n mn - m^2 + 2m + 1\\).\n\n\\end{itemize}\nTherefore, the maximum cost is \\(\\W{\\fun{loc}}{m,n} = mn - m^2 + 2m +\n1\\) \\index{loc@$\\W{\\fun{loc}}{m,n}$}, when the pattern is not a factor\nof the text and \\(m \\geqslant 1\\). The previous analysis suggests an\nimprovement for that case, but would make the case when the text\ncontains the pattern the worst: just after rule~\\(\\tau\\), let us add\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!maximum cost|)}\n\\begin{equation*}\n\\belowdisplayskip=0pt\n\\fun{pre}([a],[b],p',t',j) \\rightarrow \\fun{absent}();\\index{pre@\\fun{pre/5}}\n\\end{equation*}\n\n\\paragraph{Average cost}\n\\index{word factoring!na\\\"{\\i}ve $\\sim$!average cost|(}\n\nLet us suppose that \\(0 < m \\leqslant n\\) and that the letters of\n\\(p\\)~and~\\(t\\) are chosen from the same alphabet, whose cardinal\nis~\\(\\breve{a} > 1\\). Na\\\"{\\i}ve factoring consists in matching a\npattern against the prefixes of the suffixes of a text, by decreasing\nlengths. Let~\\(\\OM{\\breve{a}}{m}\\) \\index{loc@$\\OM{\\breve{a}}{n}$} be\nthe average number of letter comparisons for comparing two words of\nlength~\\(m\\) over the alphabet~\\(\\breve{a}\\). The average number\n\\(\\OM{\\fun{loc}}{m,n}\\) of comparisons for na\\\"{\\i}ve factoring is\n\\begin{equation}\n\\OM{\\fun{loc}}{m,n} = (n-m+1)\\OM{\\breve{a}}{m} + \\OM{\\breve{a}}{m-1},\n\\label{eq:loc}\n\\end{equation}\nbecause there are \\(n-m+1\\) suffixes of length at least~\\(m\\) and\n\\(1\\)~suffix of length~\\(m-1\\), against which the pattern is matched.\n\nThe determination of \\(\\OM{\\breve{a}}{m}\\) is achieved by fixing the\npattern~\\(p\\) and letting the text~\\(t\\) vary over all possible\nletters. There are \\(\\breve{a}^{m}\\)~comparisons between\n\\(\\ind{p}{0}\\) and~\\(\\ind{t}{0}\\), as much as there are different\ntexts; if \\(\\ind{p}{0}=\\ind{t}{0}\\), there are\n\\(\\breve{a}^{m-1}\\)~comparisons between \\(\\ind{p}{1}\\)\nand~\\(\\ind{t}{1}\\), as much as there are different \\(\\ind{t}{1,m-1}\\)\netc. In total, there are\n\\begin{equation*}\n\\breve{a}^{m} + \\breve{a}^{m-1} + \\dots +\n\\breve{a} = \\breve{a}(\\breve{a}^{m} - 1)/(\\breve{a}-1)\n\\end{equation*}\ncomparisons. There are \\(\\breve{a}^m\\)~possible texts, hence the\naverage is\n\\begin{equation*}\n\\OM{\\breve{a}}{m}\n = \\frac{\\breve{a}(\\breve{a}^{m}-1)}{\\breve{a}^m(\\breve{a}-1)}\n = \\frac{\\breve{a}}{\\breve{a}-1}\\left(1\n - \\frac{1}{\\breve{a}^{m}}\\right)\n < \\frac{\\breve{a}}{\\breve{a}-1}\n \\leqslant 2.\n\\end{equation*}\nSince \\(\\OM{\\breve{a}}{1} = 1\\), we draw the following bounds from\nequation~\\eqref{eq:loc}:\n\\begin{equation*}\nn - m + 2 \\leqslant \\OM{\\fun{loc}}{m,n} < 2(n-m+2) \\leqslant 2n + 4.\n\\end{equation*}\nNa\\\"{\\i}ve factoring is thus efficient in average, but its hypothesis\nis unlikely to apply to random English texts. Moreover, notice how the\naverage cost gets down as the alphabet grows since \\(\\lim_{\\breve{a}\n \\to \\infty}\\OM{\\breve{a}}{m} = 1\\).\\index{word\n factoring!na\\\"{\\i}ve $\\sim$!average cost|)} \\index{word\n factoring!na\\\"{\\i}ve $\\sim$|)}\n\n\n\\section{Morris-Pratt algorithm}\n\\index{word factoring!Morris-Pratt|(}\n\nIn case of mismatch, the na\\\"{\\i}ve algorithm resumes comparing the\nfirst letters of~\\(p\\) without using the information of the partial\nsuccess, to wit, we know \\(\\ind{p}{0,i-1} = \\ind{t}{j-i,j-1}\\) and\n\\(\\ind{p}{i} \\neq \\ind{t}{j}\\) (see \\fig~\\vref{fig:naive}). The\nattempt at matching~\\(p\\) with \\(\\ind{t}{j-i+1,j-1}\\) could reuse\n\\(\\ind{t}{j-i+1,j-1} = \\ind{p}{1,i-1}\\), in other words,\n\\(\\ind{p}{0,i-2}\\)~is compared to~\\(\\ind{p}{1,i-1}\\), i.e., the\npattern~\\(p\\) is compared to a part of itself. If we know an\nindex~\\(k\\) such that \\(\\ind{p}{0,k-1} = \\ind{p}{i-k,i-1}\\), that is,\n\\(\\ind{p}{0,k-1}\\)~is a \\emph{border} of~\\(\\ind{p}{0,i-1}\\) (also\nknown as a \\emph{side}\\index{side|see{word factoring, border}}), then\nwe can resume by comparing~\\(\\ind{t}{j}\\)\nwith~\\(\\ind{p}{k}\\). Clearly, the greater~\\(k\\), the more comparisons\nare skipped, so we want to find the \\emph{maximum borders} of the\nprefixes of~\\(p\\).\n\n\\paragraph{Border}\n\\index{word factoring!border|(}\n\nThe border of a non\\hyp{}empty word~\\(y\\) is a proper prefix of~\\(y\\)\nwhich is also a suffix. For example, the word \\word{abacaba} has three\nborders: \\(\\varepsilon\\), \\word{a}~and~\\word{aba}. The last one is the\nmaximum border and we write\n\\(\\Border{}{\\underline{\\word{aba}}\\word{c}\\underline{\\word{aba}}} =\n\\word{aba}\\). Another example is \\(\\Border{}{\\word{abac}} =\n\\varepsilon\\), because \\(\\word{abac} =\n\\underline{\\varepsilon}\\word{abac}\\underline{\\varepsilon}\\). Maximum\nborders can overlap; consider, for example,\n\\(\\Border{}{\\underline{\\word{aaa}}\\word{a}} =\n\\Border{}{\\word{a}\\underline{\\word{aaa}}} = \\word{aaa}\\).\n\nThe speed\\hyp{}up brought by Morris and Pratt to the na\\\"{\\i}ve search\nis depicted in \\fig~\\vref{fig:mp}.\n\\begin{figure}[b]\n\\centering\n\\includegraphics[bb=74 621 367 717]{mp}\n\\caption{Morris-Pratt algorithm (failure in grey)\n\\label{fig:mp}}\n\\end{figure}\nNotice that, contrary to na\\\"{\\i}ve factoring, letters in the text are\ncompared in a strictly increasing order (never having to\nbacktrack). Consider the complete run in \\fig~\\vref{fig:mp_ex} where,\nin the end,\n\\begin{figure}[t]\n\\centering\n\\includegraphics{mp_ex}\n\\caption{Morris\\hyp{}Pratt algorithm at work (no match found)\n\\label{fig:mp_ex}}\n\\end{figure}\n\\(p\\)~is not found to be a factor of~\\(t\\). As usual, letters on a\ngrey background correspond to mismatches. It is clear that\n\\(\\Border{}{a} = \\varepsilon\\), for all letters~\\(a\\).\n\nWe now have the choice of finding either \\(\\Border{}{ay}\\) or\n\\(\\Border{}{ya}\\), where \\(y\\)~is a non\\hyp{}empty word. Since we are\ninterested in knowing the maximum borders of all the prefixes of a\ngiven pattern, the latter is more suitable (\\(y \\pref ya\\)). The idea\nis to recursively consider \\(\\Border{}{y} \\cdot a\\): if it is a prefix\nof~\\(y\\), then \\(\\Border{}{ya} = \\Border{}{y} \\cdot a\\); otherwise, we\nseek the maximum border of the maximum border of~\\(y\\), namely,\n\\(\\Border{2}{y} \\cdot a\\) etc. until \\(\\Border{q}{y} \\cdot a\\) is a\nprefix of~\\(y\\) or else \\(\\Border{q}{y} = \\varepsilon\\). For example,\n\\(\\Border{}{y \\cdot \\word{a}} = \\Border{3}{y} \\cdot \\word{a}\\) in\n\\fig~\\vref{fig:max_B}.\n\\begin{figure}[b]\n\\centering\n\\includegraphics{max_B}\n\\caption{\\(\\Border{}{y \\cdot \\word{a}}\n = \\Border{}{\\Border{}{y} \\cdot \\word{a}}\n = \\Border{}{\\Border{2}{y} \\cdot \\word{a}}\n = \\Border{3}{y} \\cdot \\word{a}\\).\n\\label{fig:max_B}}\n\\end{figure}\n\n\\par\\vskip\\baselineskip\n\nFormally, for all words~\\({y \\neq \\varepsilon}\\) and any\nletter~\\(a\\),\n\\begin{equation}\n \\Border{}{a} := \\varepsilon;\\qquad\n \\Border{}{y \\cdot a} := \\left\\{\n \\begin{aligned}\n & \\Border{}{y} \\cdot a,\n && \\text{if \\(\\Border{}{y} \\cdot a \\prefeq y\\)};\\\\\n & \\Border{}{\\Border{}{y} \\cdot a},\n && \\text{otherwise.}\n \\end{aligned}\n \\right.\n\\label{eq:Border}\n\\end{equation}\nConsider the following examples where \\(y\\)~and~\\(\\Border{}{y}\\) are\ngiven:\n\\begin{align*}\n y &= \\word{abaabb},\n& \\Border{}{y} &= \\varepsilon,\n& \\Border{}{y \\cdot \\word{b}}\n &= \\Border{}{\\Border{}{y} \\cdot \\word{b}}\n = \\Border{}{\\word{b}} = \\varepsilon;\\\\\n y &= \\word{baaaba},\n& \\Border{}{y} &= \\word{ba},\n& \\Border{}{y \\cdot \\word{a}}\n &= \\Border{}{y} \\cdot \\word{a}\n = \\word{baa};\\\\\n y &= \\word{abbbab},\n& \\Border{}{y} &= \\word{ab},\n& \\Border{}{y \\cdot \\word{a}}\n &= \\Border{}{\\Border{}{y} \\cdot \\word{a}}\n = \\Border{2}{y} \\cdot \\word{a}\n = \\word{a}.\n\\end{align*}\n\\index{word factoring!border|)}\n\n\\paragraph{Failure function}\n\\index{word factoring!failure function|(}\n\nLet us note~\\(\\wlen{y}\\) the length of a word~\\(y\\). For a given\nword~\\(x\\), let us define a function \\(\\MPfailureName_{x}\\) on all its\nprefixes as\n\\begin{equation}\n \\MPfailure{x}{}{\\wlen{y}}\n:= \\wlen{\\Border{}{y}},\\,\\; \\text{for\n all \\(x\\) and \\(y \\neq \\varepsilon\\) such\n that \\(y \\prefeq x\\)}.\\label{eq:MP_failure}\n\\end{equation}\nFor reasons which will be clear soon, this function is called the\n\\emph{failure function} of~\\(x\\). An equivalent definition is\n\\begin{equation*}\n \\MPfailure{x}{}{i}\n= \\wlen{\\Border{}{\\ind{x}{0,i-1}}}, \\,\\;\n\\text{for all \\(x\\) and \\(i\\) such that\n\\(0 < i \\leqslant \\wlen{x}\\)}.\n\\end{equation*}\nFor example, \\fig~\\vref{fig:mp_fail} shows the table of the maximum\nborders for the prefixes of the word \\word{abacabac}.\n\\begin{figure}\n\\centering\n\\includegraphics{mp_fail}\n\\caption{Failure function of \\word{abacabac}\n\\label{fig:mp_fail}}\n\\end{figure}\nIn \\fig~\\vref{fig:mp}, the length of the maximum border is~\\(k\\), so\n\\(k=\\MPfailure{p}{}{i}\\) and \\(\\smash{\\Ind{p}{\\MPfailure{p}{}{i}}}\\)\nis the first letter to be compared with~\\(\\ind{t}{j}\\) after the\nshift. Also, the figure assumes that \\({i>0}\\), so the border in\nquestion is defined. Equations \\eqref{eq:Border} \\vpageref{eq:Border}\ndefining the maximum border can be unfolded as follows:\n\\begin{align*}\n \\Border{}{ya}\n&= \\Border{}{\\Border{}{y} \\cdot a},\n& \\Border{}{y} \\cdot a &\\nprefeq y;\\\\\n \\Border{}{\\Border{}{y} \\cdot a}\n&= \\Border{}{\\Border{2}{y} \\cdot a},\n& \\Border{2}{y} \\cdot a &\\nprefeq \\Border{}{y};\\\\\n&\\;\\;\\smash{\\vdots} & &\\;\\;\\smash{\\vdots}\\\\\n \\smash{\\Border{}{\\Border{p-1}{y} \\cdot a}}\n&= \\smash{\\Border{}{\\Border{p}{y} \\cdot a}},\n& \\smash{\\Border{p}{y} \\cdot a} &\\nprefeq \\smash{\\Border{p-1}{y}};\n\\end{align*}\nand \\(\\varepsilon \\not\\in \\{y,\\Border{}{y}, \\dots,\n\\Border{p-1}{y}\\}\\). By transitivity, the equations entail\n\\(\\Border{}{ya} = \\Border{}{\\Border{p}{y} \\cdot a}\\). Two cases are\npossible: either \\(\\Border{p}{y} = \\varepsilon\\), so \\(\\Border{}{ya} =\n\\Border{}{a} = \\varepsilon\\), or the unfolding resumes until we find\nthe smallest \\(q > p\\) such that \\(\\Border{}{\\Border{q-1}{y} \\cdot a}\n= \\Border{}{\\Border{q}{y} \\cdot a}\\) with \\(\\Border{q}{y} \\cdot a\n\\prefeq \\Border{q-1}{y}\\). Because a border is a proper prefix, that\nis to say, \\(\\Border{}{y} \\pref y\\), we have \\(\\Border{2}{y} =\n\\Border{}{\\Border{}{y}} \\pref \\Border{}{y}\\), yielding \\(\\Border{q}{y}\n\\cdot a \\prefeq \\Border{q-1}{y} \\pref \\dots \\pref \\Border{}{y} \\pref\ny\\). Therefore \\(\\Border{q}{y} \\cdot a \\prefeq y\\), since \\({q > 0}\\),\nand \\(\\Border{}{ya} = \\Border{q}{y} \\cdot a\\). This reasoning\nestablishes that\n\\begin{equation*}\n\\Border{}{ya} = \n\\left\\{\n \\begin{aligned}\n & \\Border{q}{y} \\cdot a,\n && \\text{if \\(\\Border{q}{y} \\cdot a \\prefeq y\\)};\\\\\n & \\varepsilon, \n && \\text{otherwise;}\n \\end{aligned}\n\\right.\n\\end{equation*}\nwith the additional constraint that \\(q\\)~must be as small as\npossible. This form of the definition of~\\(\\BorderName\\) is simpler\nbecause it does not contain an embedded call like\n\\(\\Border{}{\\Border{}{y} \\cdot a}\\). We can now take the lengths of\neach sides of the equations, leading to\n\\begin{equation*}\n\\wlen{\\Border{}{ya}} = \n\\left\\{\n \\begin{aligned}\n & \\wlen{\\Border{q}{y} \\cdot a} = 1 + \\wlen{\\Border{q}{y}},\n && \\text{if \\(\\Border{q}{y} \\cdot a \\prefeq y\\)};\\\\\n & \\wlen{\\varepsilon} = 0,\n && \\text{otherwise.}\n \\end{aligned}\n\\right.\n\\end{equation*}\nIf~\\({ya \\prefeq x}\\), then \\(\\wlen{\\Border{}{ya}} =\n\\MPfailure{x}{}{\\wlen{ya}} = \\MPfailure{x}{}{\\wlen{y}+1}\\). Let \\(i :=\n\\wlen{y} > 0\\).\n\\begin{equation*}\n\\MPfailure{x}{}{i+1} =\n\\left\\{\n \\begin{aligned}\n & 1 + \\wlen{\\Border{q}{y}},\n && \\text{if \\(\\Border{q}{y} \\cdot a \\prefeq y\\)};\\\\\n & 0,\n && \\text{otherwise.}\n \\end{aligned}\n\\right.\n\\end{equation*}\nWe need to work on~\\(\\wlen{\\Border{q}{y}}\\) now. From the definition\nof~\\(\\MPfailureName\\) by equation~\\eqref{eq:MP_failure}\n\\vpageref{eq:MP_failure}, we deduce\n\\begin{equation}\n\\MPfailure{x}{q}{\\wlen{y}} = \\wlen{\\Border{q}{y}},\n\\,\\; \\text{with \\(y \\prefeq x\\)},\n\\label{eq:MP_failure_p}\n\\end{equation}\nwhich we can prove by complete induction on~\\(q\\). Let us call this\nproperty \\(\\pred{P}{q}\\). Trivially, we have \\(\\pred{P}{0}\\). Let us\nsuppose \\(\\pred{P}{n}\\) for all \\(n \\leqslant q\\): this is the\ninduction hypothesis. Let us suppose \\(y \\prefeq x\\) and prove now\n\\(\\pred{P}{q+1}\\):\n\\begin{equation*}\n \\MPfailure{x}{q+1}{\\wlen{y}}\n= \\MPfailure{x}{q}{\\MPfailure{x}{}{\\wlen{y}}} \n= \\MPfailure{x}{q}{\\wlen{\\Border{}{y}}}\n\\doteq \\wlen{\\Border{q}{\\Border{}{y}}}\n= \\wlen{\\Border{q+1}{y}},\n\\end{equation*}\nwhere \\((\\smash{\\doteq})\\)~is a valid application of the induction\nhypothesis because \\(\\Border{}{y} \\pref y \\prefeq x\\). This proves\n\\(\\pred{P}{q+1}\\) and the induction principle entails that\n\\(\\pred{P}{n}\\) holds for all \\(n \\geqslant 0\\). Therefore, equation\n\\eqref{eq:MP_failure_p} allows us to refine our definition of\n\\(\\MPfailure{x}{}{i+1}\\) as follows, with \\(i>0\\):\n\\begin{equation*}\n\\MPfailure{x}{}{i+1} =\n\\left\\{\n \\begin{aligned}\n & 1 + \\MPfailure{x}{q}{i},\n && \\text{if \\(\\Border{q}{y} \\cdot a \\prefeq y\\)};\\\\\n & 0,\n && \\text{otherwise.}\n \\end{aligned}\n\\right.\n\\end{equation*}\nThere is one part of the definition \\eqref{eq:Border}\n\\vpageref{eq:Border} that we did not use: \\({\\Border{}{a} :=\n \\varepsilon}\\). It implies \\(\\MPfailure{x}{}{1} =\n\\MPfailure{x}{}{\\wlen{a}} = \\wlen{\\Border{}{a}} = \\wlen{\\varepsilon} =\n0\\) and, since the definition of~\\(\\MPfailureName\\) implies\n\\(\\MPfailure{x}{}{1} = 1 + \\MPfailure{x}{}{0}\\), so\n\\(\\MPfailure{x}{}{0} = -1\\). Property `\\(\\Border{q}{y} \\cdot a \\prefeq\ny\\) and \\({ya \\prefeq x}\\) and \\({\\wlen{y} = i}\\)' implies any of the\nequalities \\(\\smash{\\Ind{y}{\\wlen{\\Border{q}{y}}}} = a \\Leftrightarrow\n\\smash{\\Ind{y}{\\MPfailure{x}{q}{i}}} = a \\Leftrightarrow\n\\smash{\\Ind{x}{\\MPfailure{x}{q}{i}}} = \\smash{\\Ind{x}{\\wlen{y}}}\n\\Leftrightarrow \\smash{\\Ind{x}{\\MPfailure{x}{q}{i}}} = \\ind{x}{i}\\).\nWe now know\n\\begin{equation*}\n \\MPfailure{x}{}{0} = -1\\quad\\text{and}\\quad\n \\MPfailure{x}{}{i+1} =\n \\left\\{\n \\begin{aligned}\n & 1 + \\MPfailure{x}{q}{i},\n && \\text{if \\(\\Ind{x}{\\MPfailure{x}{q}{i}} = \\ind{x}{i}\\)};\\\\\n & 0,\n && \\text{otherwise;}\n \\end{aligned}\n \\right.\n\\end{equation*}\nwhere \\(q\\)~is the smallest nonzero natural satisfying the\ncondition. This can be further simplified into\n\\begin{equation*}\n\\MPfailure{x}{}{0} = -1\n\\quad \\text{and} \\quad\n\\MPfailure{x}{}{i+1} = 1 + \\MPfailure{x}{q}{i},\n\\end{equation*}\nwhere~\\(i \\geqslant 0\\) and \\(q>0\\)~is the smallest natural such that\n\\(\\MPfailure{x}{q}{i} = -1\\) or \\(\\smash{\\Ind{x}{\\MPfailure{x}{q}{i}}}\n= \\ind{x}{i}\\).\n\n\\mypar{Preprocessing}\n\\index{word factoring!Morris-Pratt preprocessing!definition|(}\n\nThe function call\n\\(\\fun{fail}_0(x,i)\\)\\index{fail0@\\fun{fail\\(_0\\)/2}}, defined in\n\\fig~\\vref{fig:fail0}, implements \\(\\MPfailure{x}{}{i}\\).\n\\begin{figure}[t]\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\framebox[\\columnwidth]{\\vbox{%\n\\begin{gather*}\n\\fun{fail}_0(x,0) \\rightarrow -1;\\quad\n\\fun{fail}_0(x,i) \\rightarrow\n 1 + \\fun{fp}(x,\\fun{nth}(x,i-1),\\fun{fail}_0(x,i-1)).\\\\\n\\fun{nth}(\\cons{a}{x},0) \\rightarrow a;\\quad\n\\fun{nth}(\\cons{a}{x},i) \\rightarrow \\fun{nth}(x,i-1).\n\\\\\n\\inferrule*{}{\\fun{fp}(x,a,-1) \\rightarrow -1;}\n\\quad\n\\inferrule\n {\\fun{nth}(x,k) \\twoheadrightarrow a}\n {\\fun{fp}(x,a,k) \\twoheadrightarrow k}\n\\,;\\quad\n\\fun{fp}(x,a,k) \\rightarrow \\fun{fp}(x,a,\\fun{fail}_0(x,k)).\n\\end{gather*}\n}}\n\\caption{The failure function \\(\\MPfailureName\\) as \\fun{fail\\(_0\\)/2}\n\\label{fig:fail0}}\n\\end{figure}\nThe function \\fun{fp/3}\\index{fp@\\fun{fp/3}} (\\emph{fixed point})\ncomputes \\(\\MPfailure{x}{q}{i-1}\\), starting with\n\\(\\fun{fail}_0(x,i-1)\\) and\n\\(\\fun{nth}(x,i-1)\\)\\index{nth@\\fun{nth/2}}, which denotes\n\\(\\ind{x}{i-1}\\) and is needed to check the condition\n\\(\\smash{\\Ind{x}{\\MPfailure{x}{q}{i-1}}} = \\ind{x}{i-1}\\). The\nequality test \\(\\MPfailure{x}{q}{i-1} = -1\\) is performed by the first\nrule of \\fun{fp/3}.\\index{word factoring!failure function|)}\n\nThe algorithm of Morris and Pratt requires that \\(\\MPfailure{x}{}{i}\\)\nbe computed for all indexes~\\(i\\) of the pattern~\\(x\\) and, since it\ndepends on the values of some calls \\(\\MPfailure{x}{}{j}\\), with \\(j <\ni\\), it is more efficient to compute \\(\\MPfailure{x}{}{i}\\) for\nincreasing values of~\\(i\\) and store them, so they can be reused\ninstead of being recomputed. This technique is called\n\\emph{memoisation}\\index{memoisation} (not to be confused with\nmemorisation). In this instance, the evaluation of\n\\(\\MPfailure{x}{}{i}\\) relies on the memo\n\\([\\pair{\\ind{x}{i-1}}{\\MPfailure{x}{}{i-1}},\n\\pair{\\ind{x}{i-2}}{\\MPfailure{x}{}{i-2}}, \\dots,\n\\pair{\\ind{x}{0}}{\\MPfailure{x}{}{0}}]\\). The memoising version of\n\\fun{fail\\(_0\\)/2} is named \\fun{fail/2}\\index{fail@\\fun{fail/2}} in\n\\fig~\\ref{fig:fail}.\n\\begin{figure}[b]\n\\centering\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\framebox[0.85\\columnwidth]{\\vbox{%\n\\begin{gather*}\n\\fun{fail}(p,0) \\rightarrow -1;\n\\quad\n\\fun{fail}(\\cons{\\pair{a}{k}}{p},i) \\rightarrow \n\\fun{fp}(p,a,k,i-1).\n\\\\\n\\inferrule*{}{\\fun{fp}(p,a,-1,i) \\rightarrow 0;}\\quad\n\\inferrule\n {\\fun{suf}(p,i-k-1) \\twoheadrightarrow \\cons{\\pair{a}{k'}}{p'}}\n {\\fun{fp}(p,a,k,i) \\twoheadrightarrow k+1}\\,;\\\\\n\\inferrule\n {\\fun{suf}(p,i-k-1) \\twoheadrightarrow \\cons{\\pair{b}{k'}}{p'}}\n {\\fun{fp}(p,a,k,i) \\twoheadrightarrow \\fun{fp}(p',a,k',k)}\\,.\n\\\\\n\\fun{suf}(p,0) \\rightarrow p;\\quad\n\\fun{suf}(\\cons{a}{p},i) \\rightarrow \\fun{suf}(p,i-1).\n\\end{gather*}\n}}\n\\caption{The failure function with memoisation\\label{fig:fail}}\n\\end{figure}\nHere, we work with the memo~\\(p\\), which is a reversed prefix, instead\nof~\\(x\\), so we need to know its length~\\(i\\) in order to know how\nmany letters must be discarded by \\fun{suf/2}\\index{suf@\\fun{suf/2}}\n(\\emph{suffix}): \\(\\fun{suf}(x,i-k-1)\\) instead of\n\\(\\fun{fail}_0(x,k)\\). Thanks to the memo, \\fun{fp/4} does not need to\ncall \\fun{fail/2}, just to look in~\\(p\\) with \\fun{suf/2}. Note that,\nas a small improvement, we also moved the increment: instead of \\(1 +\n\\fun{fp}(\\dots)\\) and \\(\\dots \\twoheadrightarrow k\\), we do now\n\\(\\fun{fp}(\\dots)\\) and \\(\\dots \\twoheadrightarrow k+1\\).\n\nLet us name \\fun{pp/1}\\index{pp@\\fun{pp/1}} (\\emph{preprocessing}) the\nfunction computing the stack \\([\\pair{\\ind{x}{0}}{\\MPfailure{x}{}{0}},\n\\pair{\\ind{x}{1}}{\\MPfailure{x}{}{1}}, \\dots,\n\\pair{\\ind{x}{m-1}}{\\MPfailure{x}{}{m-1}}]\\) for a pattern~\\(x\\) of\nlength~\\(m\\). Its definition is shown in \\fig~\\vref{fig:pp},\n\\begin{figure}[t]\n\\begin{equation*}\n\\boxed{%\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{pp}(x) & \\rightarrow & \\fun{pp}(x,\\el,0).\\\\\n\\\\\n\\fun{pp}(\\el,p,i) & \\rightarrow & \\fun{rev}(p);\\\\\n\\fun{pp}(\\cons{a}{x},p,i)\n & \\rightarrow\n & \\fun{pp}(x,\\cons{\\pair{a}{\\fun{fail}(p,i)}}{p},i+1).\n\\end{array}}\n\\end{equation*}\n\\caption{Preprocessing of a pattern~\\(y\\) by \\fun{pp/1}\n\\label{fig:pp}}\n\\end{figure}\nwhere \\fun{rev/1}\\index{rev@\\fun{rev/1}} is the reversal function\n(definition~\\eqref{def:rev} \\vpageref{def:rev}), and\n\\fun{pp/1}\\index{pp@\\fun{pp/1}} simply calls the failure function\n\\fun{fail/2} for each new index~\\(i\\) on the current memo~\\(p\\) and\ncreates a new memo by pairing the failure index with the current\nletter and pushing on the current memo\n(\\(\\cons{\\pair{a}{\\fun{fail}(p,i)}}{p}\\)). The stack reversal at the\nend is necessary because the memo contains the letters in reversed\norder with respect to the pattern. For example, the example in\n\\fig~\\vref{fig:mp_fail} leads to the evaluation\n\\begin{equation*}\n\\fun{pp}(x) \\twoheadrightarrow\n[\\pair{\\word{a}}{-1}, \\pair{\\word{b}}{0},\n\\pair{\\word{a}}{0}, \\pair{\\word{c}}{1},\n\\pair{\\word{a}}{0}, \\pair{\\word{b}}{1},\n\\pair{\\word{a}}{2}, \\pair{\\word{c}}{3}],\n\\end{equation*}\nwhere \\(x=\\word{abacabac}\\). If \\(x=\\word{ababaca}\\), then\n\\begin{equation*}\n\\belowdisplayskip=0pt\n\\fun{pp}(x) \\twoheadrightarrow\n[\\pair{\\word{a}}{-1}, \\pair{\\word{b}}{0},\n\\pair{\\word{a}}{0}, \\pair{\\word{b}}{1},\n\\pair{\\word{a}}{2}, \\pair{\\word{c}}{3},\n\\pair{\\word{a}}{0}].\n\\end{equation*}\n\\index{word factoring!Morris-Pratt preprocessing!definition|)}\n\n%\\addcontentsline{toc}{subsection}{Cost}\n\\paragraph{Minimum cost} \n\\index{word factoring!Morris-Pratt preprocessing!minimum cost|(}\n\nIt is clear from the definition~\\eqref{eq:Border} \\vpageref{eq:Border}\nthat the determination of the maximum border of a non\\hyp{}empty word\nrequires finding the maximum borders of some or all proper prefixes,\nso, if the word contains \\(n\\)~letters, at least \\(n-1\\) comparisons\nare needed, as the border of the first letter alone needs no\ncomparison. This lower bound is tight, as the following reasoning\nshows. Let us call \\emph{positive comparison}\\index{word\n factoring!comparison!positive $\\sim$} a successful prefix test as\nfound in the definition of~\\(\\BorderName\\), that is, \\(\\Border{}{y}\n\\cdot a \\prefeq y\\). Dually, a \\emph{negative comparison}\\index{word\n factoring!comparison!negative $\\sim$} is a failed prefix test. In\norder to minimise the number of calls to evaluate \\(\\Border{}{ya}\\),\nwe may notice that a positive comparison only entails the evaluation\nof~\\(\\Border{}{y}\\), whilst a negative comparison requires two:\n\\(\\Border{}{\\Border{}{y} \\cdot a}\\). Therefore, the first idea may be\nto assume that only positive comparisons occur:\n\\begin{equation*}\n\\Border{}{x} \\!\\eqn{n-2}\\! \n\\Border{}{\\ind{x}{0,n-2}} \\cdot \\ind{x}{n-1}\n\\!\\eqn{n-1}\\!\\! \\dots\n\\eqn{0}\\! \\Border{}{\\ind{x}{0}} \\cdot \\ind{x}{1,n-1}\n\\!=\\! \\ind{x}{1,n\\!-\\!1},\n\\end{equation*}\nwhere \\((\\smash{\\eqn{i}})\\)~implies \\(\\Border{}{\\ind{x}{0,i}} \\cdot\n\\ind{x}{i+1} \\prefeq \\ind{x}{0,i}\\), for \\(0 \\leqslant i \\leqslant\nn-2\\). Firstly, \\(i=0\\) and the corresponding positive comparison\nyields \\(\\ind{x}{0} = \\ind{x}{1}\\). Unfolding the other comparisons\nyields \\(\\ind{x}{0} = \\ind{x}{1} = \\dots = \\ind{x}{n-1}\\), so a best\ncase is \\(x=a^n\\), for any letter~\\(a\\).\n\nBut there is another case,\nbecause the outermost call to~\\(\\BorderName\\) after a negative\ncomparison does not imply a comparison if its argument is a single\nletter:\n\\begin{align*}\n\\Border{}{x}\n&\\eqn{n-2} \\Border{}{\\Border{}{\\ind{x}{0,n-2}} \\cdot \\ind{x}{n-1}}\\\\\n&\\eqn{n-3} \\Border{}{\\Border{}{\\Border{}{\\ind{x}{0,n-3}} \\cdot\n \\ind{x}{n-2}} \\cdot \\ind{x}{n-1}}\\\\\n&\\;\\;\\;\\smash{\\vdots}\\\\\n&\\;\\mathrel{\\stackrel{\\smash[t]{0}}{=}}\n \\Border{}{\\Border{}{\\ldots \\Border{}{\\Border{}{\\ind{x}{0}} \\cdot\n \\ind{x}{1}} \\dots} \\cdot \\ind{x}{n-1}}\\\\\n&\\,\\,\\doteq\n \\Border{}{\\Border{}{\\ldots \\Border{}{\\Border{}{\\ind{x}{1}} \\cdot\n \\ind{x}{2}} \\dots} \\cdot \\ind{x}{n-1}}\\\\\n&\\;\\;\\;\\smash{\\vdots}\\\\\n&\\,\\,\\doteq \\Border{}{\\ind{x}{n-1}} = \\varepsilon.\n\\end{align*}\nwhere \\((\\smash{\\eqn{i}})\\)~implies \\(\\Border{}{\\ind{x}{0,i}} \\cdot\n\\ind{x}{i+1} \\nprefeq \\ind{x}{0,i}\\), for \\(0 \\leqslant i \\leqslant\nn-2\\) and \\((\\smash{\\doteq})\\) involves no comparisons. Starting with\n\\(i=0\\) yields \\(\\ind{x}{1} \\neq \\ind{x}{0}\\), then \\(i=1\\) leads to\n\\(\\ind{x}{2} \\neq \\ind{x}{0}\\) etc. so the consequences of all these\nnegative comparisons are \\(\\ind{x}{0} \\neq \\ind{x}{i}\\), for \\(1\n\\leqslant i \\leqslant n-2\\). The number of negative comparisons is\n\\(n-1\\), thus is minimal, but the shape of the word is different than\npreviously, as the first letter must differ from all the\nfollowing. Let \\(\\OB{\\fun{pp}}{n}\\)\\index{pp@$\\OB{\\fun{pp}}{n}$} be\nthe minimum number of comparisons involved in the evaluation of\n\\(\\fun{pp}(x)\\)\\index{pp@\\fun{pp/1}}, where the length of the\npattern~\\(x\\) is~\\(n\\). It is the same as the number of comparisons to\nevaluate \\(\\Border{}{\\ind{x}{0,n-2}}\\) when \\(\\ind{x}{0,n-2}\\)~is a\nbest case. Therefore,\n%\\begin{equation*}\n\\(\\OB{\\fun{pp}}{n} = n - 2\\).\n%\\end{equation*}\n\\index{word factoring!Morris-Pratt preprocessing!minimum cost|)}\n\n\\paragraph{Maximum cost}\n\\index{word factoring!Morris-Pratt preprocessing!maximum cost|(}\n\nThe determination of the maximum border of a word implies finding the\nmaximum borders of some or all proper prefixes, so, if we want to\nmaximise the number of comparisons, we may want to compute as many\nborders as possible. In order to do so, evaluating \\(\\Border{}{x}\\)\nwould lead to finding the maximum border of a factor of\nlength~\\(n-1\\), where \\(n\\)~is the length of~\\(x\\). The best case\n\\(x=a^{n}\\) showed that \\(\\Border{}{x} = \\ind{x}{1,n-1}\\), which fits\nour purpose, except we would like \\(\\Border{}{\\ind{x}{1,n-1}}\\). In\nother words, we add the constraint that the first comparison is\nnegative:\n\\begin{align*}\n\\Border{}{x}\n&\\eqn{n-1} \\Border{}{\\Border{}{\\ind{x}{0,n-2}} \\cdot \\ind{x}{n-1}}\\\\\n&\\eqn{n-2} \\Border{}{\\Border{}{\\ind{x}{0,n-3}} \\cdot \\ind{x}{n-2,n-1}}\\\\\n&\\;\\;\\;\\smash{\\vdots}\\\\\n&\\;\\mathrel{\\stackrel{\\smash[t]{1}}{=}}\n \\Border{}{\\Border{}{\\ind{x}{0}} \\cdot \\ind{x}{1,n-1}}\n= \\Border{}{\\ind{x}{1,n-1}},\n\\end{align*}\nwhere \\((\\smash{\\eqn{n-1}})\\) supposes \\(\\Border{}{\\ind{x}{0,n-2}}\n\\cdot \\ind{x}{n-1} \\nprefeq \\ind{x}{0,n-2}\\), and\n\\((\\smash{\\eqn{i}})\\), with \\(1 \\leqslant i \\leqslant n-2\\),\ncorresponds to \\(\\Border{}{\\ind{x}{0,i}} \\cdot \\ind{x}{i+1} \\prefeq\n\\ind{x}{0,i}\\). These constraints imply \\(\\ind{x}{0} = \\ind{x}{1} =\n\\dots = \\ind{x}{n-2} \\neq \\ind{x}{n-1}\\), that is to say,\n\\(x=a^{n-1}b\\), with \\(a \\neq b\\). Up to now, the number of\ncomparisons is~\\(n-1\\), as in the minimal case, but the evaluation\ncontinues as follows:\n\\begin{equation*}\n\\Border{}{a^{i}b}\n\\eqn{i} \\Border{}{\\Border{}{a^{i}} \\cdot b}\\\\\n\\doteq \\Border{}{\\Border{}{a^{i-1}} \\cdot ab}\\\\\n\\doteq \\dots \\doteq \\Border{}{a^{i-1}b},\n\\end{equation*}\nfor \\(1 \\leqslant i \\leqslant n-2\\) and \\((\\smash{\\eqn{i}})\\) entails\nthe negative comparisons \\(\\Border{}{a^{i}} \\cdot b \\nprefeq a^{i}\\)\nand the positive comparisons \\((\\smash{\\doteq})\\), which we do not\ncount because we have in mind to find\n\\(\\OW{\\fun{pp}}{n}\\)\\index{pp@$\\OW{\\fun{pp}}{n}$}, so repeated\nevaluations of the same border do not entail repeated comparisons\nthanks to memoisation\\index{memoisation}. Thus, we have \\(n-2\\)\nnegative comparisons until \\(\\Border{}{b} = \\varepsilon\\), which, with\nthe \\(n-1\\) earlier positive comparisons, sum up \\(2n-3\\). Since\n\\(\\OW{\\fun{pp}}{n}\\) is the number of comparisons to compute\n\\(\\Border{}{\\ind{x}{0,n-2}}\\) without repetitions, we have\n\\begin{equation*}\n\\belowdisplayskip=0pt\n\\OW{\\fun{pp}}{n} = 2(n-1)-3 = 2n - 5.\n\\end{equation*}\n\\index{word factoring!Morris-Pratt preprocessing!maximum cost|)}\n\n\\vspace*{-20pt}\n\n\\mypar{Search}\n\\index{word factoring!Morris-Pratt!search|(}\n\nWe found above that \\(n-2 \\leqslant \\OC{\\fun{pp}}{n} \\leqslant 2n -\n5\\), where the bounds are tight if \\(n \\geqslant 3\\). To make use of\nthe value of \\(\\fun{pp}(p)\\), we could start by modifying the linear\nsearch in section~\\ref{sec:naive_factoring}, in particular the program\nin \\fig~\\ref{fig:loc} on page~\\pageref{fig:loc}, while keeping an eye\non \\fig~\\vref{fig:mp}. The result is displayed in\n\\fig~\\vref{fig:mp_def}. Note how the first argument of\n\\fun{mp/5}\\index{mp@\\fun{mp/5}}, \\(p\\), is the working copy and the\nthird, \\(p'\\), is the original which remains invariant (it is used to\nreset~\\(p\\) after a letter mismatch). Indexes~\\(i\\), \\(j\\) and~\\(k\\)\nare the same as in \\fig~\\vref{fig:mp}. The latter is none other than\nthe value computed by the failure function; variables \\(i\\)~and~\\(j\\)\nare incremented each time a letter in the pattern is successfully\nmatched against a letter in the text (third rule of \\fun{mp/5}) and\n\\(j\\)~is also incremented each time there is a mismatch of the first\nletter of the pattern (fourth rule of \\fun{mp/5}).\\index{word\n factoring!Morris-Pratt!search|)}\n\\begin{figure}[h]\n\\centering\n\\abovedisplayskip=0pt\n\\belowdisplayskip=0pt\n\\framebox[0.8\\columnwidth]{\\vbox{%\n\\begin{gather*}\n\\inferrule\n {\\fun{pp}(p) \\twoheadrightarrow p'}\n {\\fun{mp}(p,t) \\twoheadrightarrow \\fun{mp}(p',t,p',0,0)}.\n\\\\\n\\begin{array}{r@{\\;}l@{\\;}l}\n\\fun{mp}(\\el,t,p',i,j) & \\rightarrow & \\fun{factor}(j-i);\\\\\n\\fun{mp}(p,\\el,p',i,j) & \\rightarrow & \\fun{absent}();\\\\\n\\fun{mp}(\\cons{\\pair{a}{k}}{p},\\cons{a}{t},p',i,j)\n & \\rightarrow & \\fun{mp}(p,t,p',i+1,j+1);\\\\\n\\fun{mp}(\\cons{\\pair{a}{-1}}{p},\\cons{b}{t},p',0,j)\n & \\rightarrow & \\fun{mp}(p',t,p',0,j+1);\\\\\n\\fun{mp}(\\cons{\\pair{a}{k}}{p},t,p',i,j)\n & \\rightarrow\n & \\fun{mp}(\\fun{suf}(p',k),t,p',k,j).\n\\end{array}\n\\end{gather*}}}\n\\caption{Morris-Pratt algorithm (search phase)\\label{fig:mp_def}}\n\\end{figure}\n\n\n\\addcontentsline{toc}{subsection}{Cost}\n\\paragraph{Minimum cost} \n\\index{word factoring!Morris-Pratt!minimum cost|(}\n\nLet\n\\(\\smash[t]{\\OB{\\fun{mp/5}}{m,n}}\\)\\index{mp@$\\OB{\\fun{mp/5}}{m,n}$}\nbe the minimum number of comparisons performed during the evaluation\nof \\fun{mp/5}, where \\(m\\)~is the length of the pattern and \\(n\\)~is\nthe length of the text. Just as with na\\\"{\\i}ve factoring, the best\ncase is when the pattern is a prefix of the text, so\n\\(\\OB{\\fun{mp/5}}{m,n} = m\\). Taking into account the preprocessing\nstage, the minimum number of comparisons\n\\(\\OB{\\fun{mp}}{m,n}\\)\\index{mp@$\\OB{\\fun{mp}}{m,n}$} of\n\\fun{mp/2}\\index{mp@\\fun{mp/2}} is\n\\begin{equation*}\n\\belowdisplayskip=0pt\n\\OB{\\fun{mp}}{m,n} = \\OB{\\fun{pp}}{m} + \\OB{\\fun{mp/5}}{m,n} = (m-2)\n+ m = 2m - 2.\n\\end{equation*}\n\\index{word factoring!Morris-Pratt!minimum cost|)}\n\n\\vspace*{-10pt}\n\n\\paragraph{Maximum cost}\n\\index{word factoring!Morris-Pratt!maximum cost|(}\n\nSince the Morris\\hyp{}Pratt algorithm only reads the text forwards,\nthe worst case must maximise the number of times the letters of the\ntext~\\(t\\) are compared with a letter in the pattern~\\(p\\). Therefore,\nthe first letter of the pattern cannot differ from all the letters of\nthe text, otherwise each letter of the text would be compared exactly\nonce. Let us assume the exact opposite: \\(\\ind{p}{0} = \\ind{t}{i}\\),\nwith \\(i \\geqslant 0\\). But this would also imply one comparison per\nletter in the text. The way to force the pattern to shift as little as\npossible is to further impose \\(\\ind{p}{1} \\neq \\ind{t}{i}\\), for\n\\(i>0\\). In short, this means that \\(ab \\prefeq p\\), with letters\n\\(a\\)~and~\\(b\\) such that \\(a \\neq b\\) and \\(t=a^n\\). A simple\ndrawing is enough to reveal that this configuration leads to the\nmaximum number of comparisons \\(\\OW{\\fun{mp/5}}{m,n} = 2n -\n1\\)\\index{mp@$\\OW{\\fun{mp/5}}{m,n}$}, as each letter in the text is\ncompared twice, except the first, which is compared once. Taking into\naccount the preprocessing stage, the maximum number of comparisons\n\\(\\OW{\\fun{mp}}{m,n}\\) \\index{mp@$\\OW{\\fun{mp}}{m,n}$} of\n\\fun{mp/2}\\index{mp@\\fun{mp/2}} satisfies\n\\begin{equation*}\n\\belowdisplayskip=0pt\n\\OW{\\fun{mp}}{m,n} = \\OW{\\fun{pp}}{m} + \\OB{\\fun{mp/5}}{m,n} = (2m-5)\n+ (2n-1) = 2(n + m - 3).\n\\end{equation*}\n\\index{word factoring!Morris-Pratt!maximum cost|)}\n\n\\vspace*{-10pt}\n\n\\mypar{Metaprogramming}\n\\index{word factoring!Morris-Pratt!metaprogramming|(}\n\nThe previous study leads to programs for the preprocessing and search\nphases that somewhat obscure the main idea supporting the algorithm of\nMorris and Pratt, to wit, the use of the maximum borders of the proper\nprefixes of the pattern and the forward\\hyp{}only reading of the\ntext. The reason for that somewhat unfortunate situation is that, for\nefficiency imperatives, we have to memoise\\index{memoisation} the\nvalues of the failure function and, instead of working with the\noriginal pattern, we proceed with a version of it augmented with these\nvalues. Also, the utilisation of stacks for modelling the pattern\nslows down and obfuscates the reading of the letters and the shifts.\n\nIf the pattern is fixed, a more legible approach is available,\nconsisting in the modification of the preprocessing stage so that a\ndedicated program is output. This kind of taylored method, where a\nprogram is the result of the execution of another, is called\n\\emph{metaprogramming}\\index{metaprogramming}. Of course, it is an\noption only if the time needed to output, compile and execute a\nprogram is amortised in the long run, which implies for the problem at\nhand that the pattern and the text are expected to be significantly\nlong or that the search is likely to be repeated with the same pattern\non other texts (or the remainder of the same text after an occurrence\nof the pattern has been found).\n\n% Wrapping figure better declared before a paragraph\n%\n\\setlength{\\intextsep}{0pt}\n\\begin{wrapfigure}[29]{r}[0pt]{0pt}\n% 29 vertical lines\n% {r} mandatory right placement\n% [0pt] of margin overhang\n\\centering\n\\includegraphics[bb=71 360 238 723]{metaprog}\n\\caption{Factoring \\word{abacabac} %in~\\(t\\) with a metaprogram\n\\label{fig:metaprog}}\n\\end{wrapfigure}\nThere is a graphical way to represent the contents of the table in\n\\fig~\\vref{fig:mp_fail} called \\emph{deterministic finite automaton}\n\\index{word factoring!Morris-Pratt!automaton} \\index{finite\n automaton!deterministic $\\sim$} and shown in\n\\fig~\\vref{fig:abacabac}.\n\\begin{figure}[b]\n\\centering\n\\includegraphics[bb=80 651 390 710]{abacabac}\n\\caption{Morris-Pratt automaton for the pattern \\word{abacabac}\n\\label{fig:abacabac}}\n\\end{figure}\nWe will here only describe informally automata; for a full treatment,\nsee \\cite{VanLeeuwen_1990c, HopcroftMotwaniUllman_2003,\n Sakarovitch_2003}. Consider that the circles, called\n\\emph{states}\\index{states}, contain the values of~\\(i\\) from the\ntable. The edges, called \\emph{transitions}\\index{transition}, between\ntwo states are of two kinds: either solid and carrying a letter,\ncalled \\emph{label}\\index{label}, or dotted and going backwards. The\nsuccession of states throughout solid edges make the word\n\\(x=\\word{abacabac}\\). The rightmost state is distinguished by a\ndouble circling because it marks the end of~\\(x\\). There is a back\nedge between state~\\(i\\) and~\\(j\\) only if \\(\\MPfailure{x}{}{i} = j\\).\nThe leftmost state has an incoming, solid edge without a source and a\ndotted, outgoing edge. The former simply denotes the beginning of the\nword and the latter corresponds to the special value\n\\(\\MPfailure{x}{}{0} = -1\\). What matters for us is that the\nintuitive support brought by an automaton also can be implemented\nintuitively, with each state corresponding to one function and the\noutgoing transitions associated with different rules of the definition\nof the state function. The example \\word{abacabac} is shown in\n\\fig~\\ref{fig:metaprog}. The automaton states, \\(0\\), \\(1\\), through\n\\(7\\), correspond to the functions \\fun{zero/2}, \\fun{one/2}\netc. through \\fun{seven/2}. Notice how\n\\fun{mp\\(_0\\)/1}\\index{mp0@\\fun{mp\\(_0\\)/1}} sets the index to~\\(0\\)\nwhen initialising the first state, that is, calling\n\\fun{zero/2}\\index{zero@\\fun{zero/2}}. The index~\\(j\\) plays the same\nrole as in \\fig~\\ref{fig:mp} on page~\\pageref{fig:mp}. The first rule\nof each function corresponds to a rightwards transition in the\nautomaton in \\fig~\\vref{fig:abacabac} and the second rule is a\nbackwards transition, that is, a failure, except in \\fun{zero/2},\nwhere it means that the pattern is shifted by one letter. The function\n\\fun{zero/2} has a third rule handling the case when the pattern is\nabsent in the text. We could add a similar rule to the other\nfunctions, as an optimisation, but we opt for brevity and let\nsuccessive failure rules bring us back to\n\\fun{zero/2}\\index{zero@\\fun{zero/2}}. The first rule of\n\\fun{seven/2}\\index{seven@\\fun{seven/2}} is special as well, because\nit is used when the pattern has been found. Note the index \\(j-6\\),\nclearly showing that the length of the pattern is part of the program,\nwhich is hence a metaprogram.\\index{word\n factoring!Morris-Pratt!metaprogramming|)} \\index{word\n factoring!Morris-Pratt|)}\n\n\\mypar{Knuth's variant}\n\\index{word factoring!Knuth-Morris-Pratt|(}\n\nIn \\fig~\\vref{fig:mp}, if \\(a = \\word{a}\\), then the sliding would\nimmediately lead to a comparison failure. Hence let us compare\n\\(\\smash{\\Ind{p}{\\MPfailure{p}{}{i}}}\\) to~\\(\\ind{t}{j}\\) only if\n\\(\\smash{\\Ind{p}{\\MPfailure{p}{}{i}}} \\neq \\ind{p}{i}\\). Else, we\nconsider the maximum border of the maximum border etc. until we find\nthe smallest~\\(q\\) such that \\(\\smash{\\Ind{p}{\\MPfailure{p}{q}{i}}}\n\\neq \\ind{p}{i}\\). This is an improvement by\n\\cite{KnuthMorrisPratt_1977}. There is an updated reprint by\n\\cite{Knuth_2010} and a treatment based on automata theory by\n\\cite{CrochemoreHancartLecroq_2007}, in its section~2.6. See also an\ninteresting derivation of the program by algebraic refinements in the\nbook by \\cite{Bird_2010}. In terms of the search automaton, when a\nfailure occurs at state~\\(i\\) on the letter~\\(a\\), we follow the back\nedge to state \\(\\MPfailure{x}{}{i}\\), but, if the normal transition\nis~\\(a\\) again, we follow another back edge etc. until there is a\ntransition different from~\\(a\\) or we have to shift the pattern. The\nimprovement proposed by Knuth consists in replacing \n\\begin{figure}[b]\n\\centering\n\\includegraphics[bb=82 646 388 720]{kmp_abac}\n\\caption{Knuth-Morris-Pratt automaton for the pattern \\word{abacabac}\n\\label{fig:kmp_abac}}\n\\end{figure}\nall these successive failure transitions by only one. For example, the\nautomaton in \\fig~\\ref{fig:kmp_abac} is Knuth's optimisation of the\none in \\fig~\\ref{fig:abacabac}.\n\n\\index{word factoring!Knuth-Morris-Pratt|)}\n\\index{word factoring|)}\n\n\\paragraph{Exercises}\n\n\\begin{enumerate}\n\n \\item Find \\(\\M{\\fun{loc}}{m,n}\\).\n\n \\item Prove that \\fun{pp/1} and \\fun{mp/2} terminate.\n\n \\item Prove \\(\\fun{loc/2} = \\fun{mp/2}\\) (correctness of the\n algorithm of Morris and Pratt).\n\n \\item Find \\(\\B{\\fun{pp}}{m}\\) and \\(\\W{\\fun{pp}}{m}\\). (Mind\n the cost of \\fun{suf/2}.)\n\n \\item Find \\(\\B{\\fun{mp}}{m,n}\\) and \\(\\W{\\fun{mp}}{m,n}\\).\n\n \\item Find a simple modification to avoid calling \\fun{rev/1} in\n \\fig~\\vref{fig:pp}.\\label{factoring:trick}\n\n \\item Modify \\fun{fail/2} so that \\fun{mp/2} implements the\n Knuth\\hyp{}Morris\\hyp{}Pratt algorithm. Study the best and worst\n cases of this variant and show that \\(\\OW{\\fun{pp}}{m} = 2m-6\\),\n for \\(m \\geqslant 3\\).\n\n \\item Write the metaprogram corresponding to the automaton in\n \\fig~\\vref{fig:kmp_abac}.\n\n \\item Write a function \\fun{rlw/2}\\index{rlw@\\fun{rlw/2}}\n (\\emph{remove the last word}) such that \\(\\fun{rlw}(w,t)\\) is\n rewritten into the text~\\(t\\) if the word~\\(w\\) is absent,\n otherwise into~\\(t\\) without the last occurrence of~\\(t\\) in it.\n\n\\end{enumerate}\n", "meta": {"hexsha": "8e1282ac5accb45261b2a19463ecd2f6b7096a68", "size": 52813, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "factoring.tex", "max_stars_repo_name": "rinderknecht/Book", "max_stars_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "factoring.tex", "max_issues_repo_name": "rinderknecht/Book", "max_issues_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "factoring.tex", "max_forks_repo_name": "rinderknecht/Book", "max_forks_repo_head_hexsha": "6f302ab1319c8ae9b3ea690c45fdb3d2b6fbca16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.8282157676, "max_line_length": 77, "alphanum_fraction": 0.6608789503, "num_tokens": 18719, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.7931059462938815, "lm_q1q2_score": 0.6143266878816387}} {"text": "\\documentclass[]{article}\n\\usepackage{kotex}\n\\usepackage{amsmath}\n\n%opening\n\\title{Systematic Unvertainty Note}\n\\author{}\n\n\\begin{document}\n\n\\maketitle\n\n중력가속도 실험에 대한 Systematic Uncertainty Estimation\n\nBased on Monte-Carlo Simulation\n\n\\section{Equation of Datapoints}\n\n\\paragraph{Basic Equation}\n\\begin{equation}\nh = \\frac{1}{2} g t^2\n\\end{equation}\n\\begin{equation}\nt = \\sqrt{2 \\frac{g}{h}}\n\\end{equation}\n\n\\paragraph{Basic Equation with uncertainty}\n\n\\begin{equation}\nh+\\delta h = \\frac{1}{2} g \\left(t+\\delta t\\right)^2\n\\end{equation}\n\n\\begin{equation}\nt = \\sqrt{2 \\frac{g}{h+\\delta h}} + \\delta t\n\\end{equation}\n\nwith following random distribution\n\n\\begin{table}[!h]\n\t\\begin{tabular}{c|cc}\n\t\tRandom Variable & Distribution & Reason\\\\\\hline\n\t\t$\\delta h$ & Gaussian Disribution & Measured by human-eye\\\\\n\t\t$\\delta t$ & Uniform Disribution & Measured by mechanical clock\n\t\\end{tabular}\t\n\\end{table}\n\nWith overlapped uncertainty (depending on measuring methods), the $\\delta h$ has different value for each datapoints.\n\n\n\\section{Case Overlapped Length Dependencies}\nIf measuring length for each data-points has dependencies with each previous data-point. Uncertainty for each data-points is needed to declared especially.\n\n\\subsection{Error Propagation}\n\n\\paragraph{Basic Equation of uncertainty}\n\\begin{equation}\n\\delta f = \\sqrt{\\sum_i\\left(\\frac{\\partial f}{\\partial x_i} \\delta x_i \\right)^2}\n\\end{equation}\n\n\\paragraph{Overlapped Uncertainty, Propagated for Summation}\n\n\\begin{equation}\n\\delta (f+g) = \\sqrt{(\\delta g)^2 + (\\delta f)^2 }\n\\end{equation}\nFor independent 2 formula $f$ and $g$.\n\nOverlapping uncertainties. With measuring length period $l$, number $i$, single measuring uncertainty $\\delta l_i$, Initial offset $b$ and its uncertainty $\\delta b$\n\nFirst term\n\\begin{align}\nh_0 &= b \\\\\nh_1 &= b + l + \\delta l_1 \\\\\nh_2 &= b + 2l + \\delta l_1 +\\delta l_2\n\\end{align} \n\nwith regularity, following correlation can be assumed.\n\n\\begin{equation}\nh_n = b + n \\cdot l + \\sum_i{\\delta l_i}\n\\end{equation}\n\nUncertainty is \n\\begin{equation}\n\\delta h_i = \\sqrt{\\sum_i{\\left(\\delta l\\right)^2}}\n\\end{equation}\n$\\delta l$ is constant for each $n$, \n\n\\begin{equation}\n\\delta h_n = \\sqrt{n} \\delta l\n\\end{equation}\n\n\n\n\t\n\n\\end{document}\n", "meta": {"hexsha": "f7725cb05970781a48cc0b1d346927ebd94d8ce6", "size": 2239, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "assets/code/systematic_uncertainty_note/template_Article.tex", "max_stars_repo_name": "Isaac-Kwon/Isaac-Kwon.github.io", "max_stars_repo_head_hexsha": "9f4145dc3748c9c0364476c90efd78311d4fb62d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "assets/code/systematic_uncertainty_note/template_Article.tex", "max_issues_repo_name": "Isaac-Kwon/Isaac-Kwon.github.io", "max_issues_repo_head_hexsha": "9f4145dc3748c9c0364476c90efd78311d4fb62d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "assets/code/systematic_uncertainty_note/template_Article.tex", "max_forks_repo_name": "Isaac-Kwon/Isaac-Kwon.github.io", "max_forks_repo_head_hexsha": "9f4145dc3748c9c0364476c90efd78311d4fb62d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.0824742268, "max_line_length": 165, "alphanum_fraction": 0.7213041536, "num_tokens": 698, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.793105941403651, "lm_q2_score": 0.7745833789613196, "lm_q1q2_score": 0.6143266799667383}} {"text": "\\input{PreambleCommon}\r\n\\input{../WeekTitles}\r\n\r\n\\newcommand{\\Fe}{ F_{\\mbox{ext}} }\r\n\\newcommand{\\Fg}{ F_{\\mbox{grav}} }\r\n\r\n\\begin{document}\r\n\\setfont\r\n\\pagestyle{fancy}\r\n\\renewcommand{\\Week}{8 }\r\n\\renewcommand{\\WeekTitle}{\\WeekTitleEight }\r\n\r\n\\fancyhead[LE,RO]{Week \\Week} % default, usually only for first page\r\n\\fancyfoot{}\r\n\\sectionbox{Week \\#\\Week: \\WeekTitle}\r\n\r\n\\vspace{5mm}\r\n\\goals\r\n\\begin{itemize}\r\n\\item Express real world situations in terms of second order linear\r\n differential equations.\r\n\\item Describe the difference between homogeneous and nonhomogeneous\r\n second order linear differential equations.\r\n\\item Use MATLAB to solve linear and nonlinear second order\r\n differential equations, both homogeneous and nonhomogeneous.\r\n\\end{itemize}\r\n\\vspace{5mm}\r\n\r\n\\newpage\r\n\r\n\\topic{Generating Numerical Solutions with MATLAB}\r\n\\subsection*{Generating Numerical Solutions with MATLAB}\r\n\r\n\\problem Search for ``ordinary differential equations'' in MATLAB help. \\vsc\r\n\\vsc\r\n\r\nWhat form of differential equation does MATLAB assume we have?\r\n\r\n\\vsc\r\n\\vsc\r\n\r\n\r\n\\newpage\r\n\r\nNote that MATLAB has many different differential equation solvers. If\r\nusing MATLAB after this course, you may have to do some reading to\r\nidentify the properties of the equation you have, and what the most\r\nappropriate solving tool is.\r\n\r\n\\problem Which of the solvers is recommended as a ``first try''\r\nsolver?\r\n\r\n\\vsc\r\n\\vsc\r\n\r\n\\newpage\r\n\r\n\\subsection*{ode45}\r\nThe first solver to reach for in MATLAB is \\verb#ode45#. To run it, we need\r\n\\begin{itemize} \r\n\\item the DE function {\\bf in form $\\displaystyle \\frac{dy}{dt} = f(t, y)$};\r\n\\vfill\r\n\\item the time span for the solution/simulation, $[t_0, t_{\\mbox{end}}]$; and\r\n\\vfill\r\n\\item the initial condition ($y(t_0))$\r\n\\vfill\r\n\\end{itemize}\r\n\r\n\\problem What will \\verb#ode45# compute for us, given that\r\ninformation?\r\n\\vfill\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\topic{Example - Temperature Model}\r\n\\subsection*{Temperature Model - Newton's Law of Heating and Cooling}\r\n\r\nThe temperature of an object, $y$, changes at a rate proportional to\r\nthe temperature difference between object, $y$, and its environment,\r\n$T_{\\mbox{ext}}$.\r\n\r\n\\problem Translate this law into a mathematical statement.\r\n\r\n\\vfill Solution interpretation: a {\\bf solution} to a differential\r\nequation is the same as a {\\bf prediction}. \\vspace{1.5in}\r\n\r\n\\newpage \r\n\r\n{\\bf In MATLAB}\r\n\r\n$$\\frac{dy}{dt} = -k (y - T_{\\mbox{ext}})$$\r\n\r\n\r\n\\begin{verbatim}\r\n% Define temp DE in the form dy/dt = f(t, y)\r\n% and set other constants\r\nk = 0.7; % /min\r\n\r\nT_ext = 20; % external/environment temp\r\n\r\nDE = @(t, y) -k (y - T_ext);\r\n\r\n\\end{verbatim}\r\n\r\n\\newpage\r\n\r\n\\subsection*{Example - Step 2}\r\n\\begin{verbatim}\r\n% Solve based on initial condition\r\ny0 = 100; % initial temperature\r\n\r\ntspan = [0, 30]; % interval for solution\r\n\r\n[t, y] = ode45(DE, tspan_, y0);\r\n\r\nplot(t, y);\r\n\\end{verbatim}\r\n\r\n\\newpage\r\n\r\nTo decipher and work with the output, it is critical that you\r\nunderstand what MATLAB provides. \r\n\\begin{itemize}\r\n\\item The final values of \\verb#t# and \\verb#y# are \r\n\\vfill\r\n\\item \\verb#t# starts at\r\n\\vfill\r\n\\item \\verb#t# ends at\r\n\\vfill\r\n\\item \\verb#y# starts at\r\n\\vfill\r\n\\end{itemize}\r\n\r\n\\newpage\r\n\r\n\\subsection*{Basic Process}\r\nHow does MATLAB do it?\r\n\\vfill\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\topic{Second-Order Linear Equations - Spring System Intro}\r\n\\section*{Second-Order Linear Equations - Spring System}\r\n\r\nSo far we have seen examples of {\\bf first-order DEs}, or equations\r\nwith first derivatives of some unknown function. \r\n\r\nIn the following examples, we will expand our study to differential\r\nequations with {\\bf second or higher derivatives}.\r\n\r\nOne classic source of differential equations of this type comes from\r\nanalyzing the forces on a block at the end of a spring.\r\n\\begin{center}\r\n\\includegraphics[width=0.5\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\\newpage\r\n\\begin{center}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\nWhile the mathematics behind this simple system will be very\r\ninteresting in their own right, we should also note at the outset that\r\nthe simple spring/mass model can be applied to a wide variety of\r\nnot-so-obviously related real-world problems.\r\n\r\n\\begin{center}\r\n\\includegraphics[width=0.30\\linewidth]{graphics/notes_08_SpringSystemCar}\r\n\\includegraphics[width=0.30\\linewidth]{graphics/notes_08_Sound-Attenuation} \\\\\r\n\\includegraphics[width=0.30\\linewidth]{graphics/notes_08_Tuned_Mass_Damper_atop_Taipei_101_-_27_March_2008}\r\n\\end{center}\r\n\r\n\\newpage\r\n\r\n\\topic{Spring System Analysis}\r\n\\subsection*{Spring System Analysis}\r\n\r\n\\begin{center}\r\n\\includegraphics[width=0.5\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\n\\problem In this system, how would you describe $x$ in words?\r\n\r\n\\newpage\r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\\problem Draw a free-body diagram for the mass. Indicate the magnitude of the forces, assuming \r\n\\begin{itemize}\r\n\\item the mass of the block is $m$ kg, and\r\n\\item the spring constant (in $N/m$) is\r\n given by the constant $k$.\r\n\\end{itemize}\r\n\r\n\\newpage\r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\nLet us work with our intuition about this system before beginning the mathematics.\r\n\r\n\\problem If the spring is very stiff, is $k$ large or small? \\vfill\r\n\r\n\\newpage \r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\n{\\bf Definition:}: {\\em Period} is the length of time to\r\ncomplete one full cycle/oscillation. \\vspace{0.2in}\r\n\r\n\\problem If we increase the stiffness of the spring, do you expect the\r\n{\\em period} of the oscillations to increase or decrease? Why?\r\n\r\n\\vfill\r\n\r\nIf we increase the mass, do you expect\r\nthe \\emph{period} of the oscillations to increase or decrease? Why?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\n\\problem If we know $k$ and $m$, and assume that damping is\r\nnegligible, should we be able to determine the exact period of the\r\noscillations? \\vfill\r\n\r\n\r\n\\vfill\r\n\r\nFrom the work so far, can we easily find the formula for the period?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\nThe spring system is an excellent introduction to higher-order differential equations because \r\n\\begin{itemize}\r\n\\item we all have an intuition about how it \\emph{should} work physically,\r\n\\item the mathematics and physics are simple, and\r\n\\item there's no obvious way to predict critical features (e.g. the\r\n period) from the given information.\r\n\\end{itemize}\r\nWe clearly need some new tools!\r\n\r\n\\newpage\r\n\\topic{Spring System as a DE}\r\n\\subsection*{Spring System as a DE}\r\n\\begin{center}\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\n\\problem Use Newton's second law, $F = ma$, to construct an equation\r\ninvolving the position $x(t)$. \\vfill\r\n\r\nWhat order of differential equation does $F=ma$ produce for this\r\nspring/mass system?\r\n\r\n\\vspace{1.3in}\r\n\r\n\\newpage\r\n\r\nTo simplify matters temporarily, let us assume that both\r\n $k = 1 $ N/m and $m = 1$ kg. \r\n\r\n \\problem Rewrite the previous differential equation using those\r\n constants. \\vspace{1.5in}\r\n\r\n This differential equation invites us to find a function $x(t)$\r\n whose second derivative is its own negative. What function(s) would\r\n satisfy that? \\vfill\r\n\r\n\\newpage\r\n\r\n\\problem Having found two (and more) solutions to the differential\r\nequation for the spring/mass system, how does this family of solutions\r\nmap back to the spring system?\r\n\r\n\\begin{center}\r\n\\includegraphics[width=0.4\\linewidth]{graphics/notes_08_block}\r\n\\end{center}\r\n\r\n\\vfill\r\n\r\nIs our solution an exact or a numerical solution?\r\n\\vspace{1.3in}\r\n\r\n\r\n\\newpage\r\n\r\n\r\n\\topic{Mechanical Vibrations - Spring-Mass Wystem}\r\n\\section*{Mechanical Vibrations - Spring-Mass System}\r\n\r\nWe now consider an extension to the spring/mass system.\r\n\r\nConsider a mass $m$ hanging on the end of a vertical spring; the\r\nspring has a naturally stretched length corresponding to $x=0$.\r\n\r\nWe have now added a damper, which exerts a force proportional to the\r\nvelocity of the oscillating mass.\r\n\r\n\r\n\\includegraphics[width=0.4\\linewidth]{graphics/notes_08_hanging_mass}\r\n\r\n\\newpage\r\n\r\n\\problem Using Newton's second law, build a differential equation that\r\ngoverns the system.\r\n\r\n\\includegraphics[width=0.35\\linewidth]{graphics/notes_08_hanging_mass}\r\n\r\n\r\n% \\begin{itemize}\r\n% \\item Gravity acts downward and has a magnitude of $mg$ where $g$ is the\r\n% acceleration due to gravity; $F_g = mg$.\r\n% \\item The spring acts upward and, by Hooke's law, has a magnitude proportional\r\n% elongation. If $y$ represents the displacement of the mass from the\r\n% equilibrium position, then $F_s = -k(\\ell +y)$.\r\n% \\item Assuming that possible damping forces (i.e.\\ a dashpot) are directly\r\n% proportional to the velocity of the mass, we have $F_d = - cy'$.\r\n% \\item There may be an external driving force $F(t)$.\r\n% \\end{itemize}\r\n% Summing the forces, we find that $m y'' = F_g + F_s + F_d + F(t) = mg -k(\\ell\r\n% +y) -cy' + F(t)$. At equilibrium, we have $mg = k\\ell$ so the standard form of\r\n% the equation is\r\n% \\[\r\n% y'' + \\tfrac{c}{m} y' + \\tfrac{k}{m} y = \\tfrac{1}{m} F(t) \\, . \r\n% \\]\r\n\r\n\r\n\\newpage\r\n\\problem Consider a system with a mass of $0.5 \\; \\text{kg}$ with\r\nspring constant $k = 2 \\; \\text{N} \\cdot \\text{m}^{-1}$, with a\r\ndamping coefficient of $c = 0.5$ N/(m/s). Assume the mass is\r\ndisplaced $0.4 \\; \\text{m}$ from equilibrium and released.\r\n\r\n\\problem Write the differential equation dictates the motion of the\r\nmass.\r\n\r\n\\vspace{1in}\r\n\r\nCan we find an easy solution to this differential equation as we did with the undamped\r\ncase?\r\n\\vfill\r\n\r\n\r\nCan MATLAB be used to build a numerical solution for the differential\r\nin its current form?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\r\n\\topic{Converting Higher-Order DEs to 1st Order Systems}\r\n\\subsection*{Converting Higher-Order DEs to 1st Order Systems}\r\nTo solve a higher-order DE using MATLAB, we need to convert it first\r\ninto a larger {\\bf first-order system} of differential equations.\r\n\r\n\r\n\\vspace{0.3in}\r\nNotation: In this section, \r\n\\begin{itemize} \r\n\\item vectors with be written with vector hats, e.g.\r\n $\\vec{w}, \\vec{y}$,\r\n\\item elements of a vector will be noted with subscripts, e.g. $w_1$,\r\n $y_2$, and\r\n\\item other scalars will be in lower-case, e.g. $c$, $\\lambda$.\r\n\\end{itemize}\r\n\r\n\\newpage\r\n\r\n\\begin{minipage}[t]{0.4\\linewidth}\r\n\\vspace{0pt}\r\nDamped Spring DE:\r\n\r\n$$ m x'' = -c x' -kx $$\r\n\\end{minipage}\r\n\\begin{minipage}[t]{0.6\\linewidth}\r\n\\vspace{0pt}\r\n\\begin{center}\r\n\\includegraphics[width=0.5\\linewidth]{graphics/notes_08_hanging_mass}\r\n\\end{center}\r\n\\end{minipage}\r\n\r\n\\problem For the spring/mass DE, define a new vector of 2 variables,\r\n$\\vec{w}$, that will allow the conversion of the second-order system\r\nto a first-order system.\r\n\r\n\\vfill\r\n\r\nAlso see the MATLAB help menu entry on ``Ordinary Differential\r\nEquations'', referring to ``Higher Order ODEs''.\r\n\r\n\\newpage\r\n$$ m x'' = -c x' -kx $$\r\n\\problem Define the derivative, $\\ds \\frac{d \\vec{w}}{dt}$, in terms\r\nof $\\vec{w}$ itself, making use of the DE as necessary.\r\n\r\n\\vfill\r\n\\vfill\r\n\r\nThis is now a {\\bf first-order system} of differential equations. \\\\\r\nThe variable we are most interested in for this example is $w_1 = x$,\r\nthe {\\em position} of the mass.\r\n\r\n\\newpage\r\n\r\n\\topic{MATLAB Function Files}\r\n\\subsection*{MATLAB Function Files}\r\n\r\nOur differential equation is now sufficiently complicated that writing\r\nit out in one line is difficult. We will need another MATLAB tool to\r\nproceed: a MATLAB function file.\r\n\r\n\r\n{\\bf Syntax} A MATLAB function file\r\n\\begin{itemize}\r\n\\item has as its first line the form \\\\\r\n\\verb#function ret = f(....)# \r\n\\item \\verb#ret# is the name of the {\\bf return variable} and must be\r\n defined before the end of the function;\r\n\\item \\verb#f# should be the same as the filename, with the \\verb#.m# extension;\r\n\\item the \\verb#....# can be any list of input variables.\r\n\\end{itemize}\r\n\r\n\\newpage\r\n\\problem Define a new {\\bf MATLAB function \\texttt{.m} file} called\r\n\\verb#hypotenuse# that takes in two lengths, \\verb#a# and \\verb#b#,\r\nand returns the length of the hypotenuse for a right-angle triangle\r\nwith side lengths $a$ and $b$.\r\n\r\n\\newpage\r\n\\problem Define a new {\\bf MATLAB function \\texttt{.m} file} called\r\n\\verb#springDE# which computes the derivative of\r\n$\\ds \\frac{d \\vec{w}}{dt}$.\r\n\r\n\\begin{minipage}[t]{0.4\\linewidth}\r\n\\vspace{0pt}\r\n$$ m x'' = -c x' -kx $$\r\n\\end{minipage}\r\n\\begin{minipage}[t]{0.6\\linewidth}\r\n\\vspace{0pt}\r\n\\begin{align*}\r\n \\frac{dw_1}{dt} & = w_2 \\\\\r\n \\frac{dw_2}{dt} & = \\left(\\frac{1}{m}\\right) (-c w_2 - k w_1)\r\n\\end{align*}\r\n\\end{minipage}\r\n\r\n\r\n\r\n\\newpage \r\n\\problem Write a MATLAB script that graphs the solution\r\n(i.e. predicted motion) for the system described earlier:\r\n\r\nA system with a mass of $0.5 \\; \\text{kg}$ with spring constant\r\n$k = 2 \\; \\text{N} \\cdot \\text{m}^{-1}$, with a damping coefficient of\r\n$c = 0.5$ N/(m/s). Assume the mass is displaced $0.4 \\; \\text{m}$\r\nfrom equilibrium and released.\r\n\r\nInclude the graphical output from MATLAB on the next page.\r\n\r\n\\newpage\r\n\\hfill\r\n\\includegraphics[width=0.3\\linewidth]{graphics/notes_08_hanging_mass}\r\n\r\n\r\n\\newpage\r\n\\topic{Unforced Spring/Mass System - Patterns of Behaviour}\r\n\\subsection*{Unforced Spring/Mass System - Patterns of Behaviour}\r\n\r\nUsing MATLAB, we can now easily study the impact of different masses,\r\nspring constants, or damping strength on the behaviour of a\r\nspring/mass system.\r\n\r\n\\begin{minipage}[t]{0.4\\linewidth}\r\n\\vspace{0pt}\r\nDamped Spring DE:\r\n\r\n$$ m x'' = -c x' -kx $$\r\n\\end{minipage}\r\n\\begin{minipage}[t]{0.6\\linewidth}\r\n\\vspace{0pt}\r\n\\begin{center}\r\n\\includegraphics[width=0.5\\linewidth]{graphics/notes_08_hanging_mass}\r\n\\end{center}\r\n\\end{minipage}\r\n\r\n\\newpage\r\n\r\nA famous result in physics captures important differences in behaviour\r\ndepending on the relationship between $c$, $k$ and $m$.\r\n\\begin{itemize}\r\n\\item $c$ is 0, the system will be {\\bf undamped};\r\n\\item $c$ is {\\bf less} than $\\sqrt{4km}$, the system will be {\\bf under-damped};\r\n\\item $c$ is {\\bf equal} to $\\sqrt{4km}$, the system will be {\\bf\r\n critically damped};\r\n\\item $c$ is {\\bf greater} than $\\sqrt{4km}$, the system will be {\\bf\r\n over-damped};\r\n\\end{itemize}\r\n\r\n\\problem Using a spring constant of $k = 25$ N/m and mass $m = 1$ kg,\r\nfind the critical damping level.\r\n\r\n\\newpage\r\n\\problem Use MATLAB to obtain graphs for all four spring/mass cases.\r\n\r\n \\begin{minipage}[h]{0.475\\linewidth}\r\n \\begin{center}\r\n \r\n \\vspace{0pt} \r\n \\includegraphics[width=0.9\\linewidth]{graphics/notes_08_spring_mass_axes}\\\\\r\nUndamped \r\n\r\n\\hrulefill\r\n\r\n \\includegraphics[width=0.9\\linewidth]{graphics/notes_08_spring_mass_axes}\\\\\r\nCritically Damped\r\n \\end{center}\r\n \\end{minipage}\r\n \\begin{minipage}[h]{0.475\\linewidth}\r\n \\begin{center}\r\n \r\n \\vspace{0pt} \r\n \\includegraphics[width=0.9\\linewidth]{graphics/notes_08_spring_mass_axes}\\\\\r\nUnder-Damped \r\n\r\n\\hrulefill\r\n\r\n \\includegraphics[width=0.9\\linewidth]{graphics/notes_08_spring_mass_axes}\\\\\r\nOver-Damped\r\n \\end{center}\r\n \\end{minipage}\r\n\r\n\\newpage\r\n\\topic{Spring/Mass with Periodic External Forces }\r\n\\subsection*{Spring/Mass with Periodic External Forces }\r\n\r\nWe now add another component to our spring system model: \\\\{\\bf\r\n external forces}.\r\n\r\n \\includegraphics[width=0.25\\linewidth]{graphics/notes_08_hanging_mass}\r\n\r\n \\problem Write out the DE for the position of the mass, given the\r\n addition now of an external force, given by\r\n $F_{\\mbox{ext}} = A \\sin(Bt)$. \\vfill\r\n\r\nWhat are analogous systems to this that you might have encountered\r\nbefore?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\problem Write a MATLAB script and a function that will let you\r\nsimulate the motion of the mass given an external force of the form\r\n$F_{\\mbox{ext}} = A \\sin(Bt)$.\r\n\r\n$$m x'' = -kx -c x' + A \\sin(Bt)$$\r\n\r\n\\vfill\r\n\r\nOnce you have the MATLAB code written, we can explore the solutions\r\nof/predictions for the spring/mass system for different scenarios, by\r\nchanging parameters in our MATLAB script.\r\n\r\n\\newpage\r\n\r\n\\topic{Simulations of the Forced Spring/Mass System}\r\n\\subsection*{Simulations of the Forced Spring/Mass System}\r\n\r\nFor the following systems, use $m = 1$ kg and $k = 25$ N/m, unless\r\notherwise specified.\r\n\r\n{\\bf Unforced Motion}\r\n\r\n\\problem Setting $F_{\\mbox{ext}} = 0$, review the effect of increasing\r\nthe damping coefficient $c$ on the behaviour of the mass.\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n{\\bf Forced, but Undamped Motion}\r\n\r\nNow we turn off the damping (by setting $c=0$) to zero, and\r\nstart to apply the external force. \\\\\r\n\\problem Set $F_{\\mbox{ext}} = 10 \\sin(1.0 t)$ and display a graph of the resulting oscillations. \\\\\r\n\r\n\\vfill\r\n\r\nExperiment with the frequency in $\\Fe$, in the range 0.5 to 3, or well\r\nabove 5. Does the motion of the mass over time make sense?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\topic{Simulations - Resonance, Near-Resonance and Practical Resonance}\r\n\r\n{\\bf Resonance}\r\n\r\nIn unforced, undamped systems, there is a {\\bf natural frequency} for\r\nthe mass, given by $\\omega = \\sqrt{\\frac{k}{m}}$.\r\n\r\n\\problem Set $\\Fe = 1 \\sin(5t)$. What is special about the frequency\r\n5 rad/s?\r\n\r\n\\vfill\r\n\r\nDescribe the graph of the solution.\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\nGive the physical reasons for the graph seen in the\r\nsimulation.\\\\\r\n\r\n\\vfill\r\n\r\nThe condition of having the external force at {\\em exactly} the same\r\nfrequency as the natural vibrations is called {\\bf resonance.} Note:\r\nfor true resonance, there must be {\\em no damping.}\r\n\r\n\\newpage\r\n\r\n{\\bf Near-Resonance}\r\n\r\nAdmittedly, resonance is a unique event requiring {\\bf perfect}\r\nmatching of the stimulating force with the natural frequency.\r\n\r\n\\problem Explore frequencies in $\\Fe$ {\\em close to} 5. Describe the\r\nresulting solutions, both for their amplitude and any other response.\r\n\r\n\\vfill\r\n\r\nGive a physical reason for the shape of the graph for near-resonance\r\nresponse.\r\n\r\n\\vfill\r\n\r\nNote again that there is {\\em no damping} with this response.\r\n\r\n\\newpage\r\n\r\n{\\bf Adding in Friction - Practical Resonance}\r\n\r\nSet $F_{\\mbox{ext}} = 1 \\sin(5t)$ again, but set damping $c$ to 0.1.\r\n\r\n\\problem Compare this solution to the one without damping.\r\n\r\n\\vfill\r\n\r\nGradually increase the amount of damping. Describe how the solution\r\nchanges.\r\n\r\n\\vfill\r\n\r\nHow is this more realistic than the undamped case?\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\topic{Transient and Steady-State Solutions}\r\n\\subsection*{Transient and Steady-State Solutions}\r\n\r\nMore generally, when there is damping in the spring/mass system and an\r\noscillatory external force, we can break the solution into two parts:\r\n{\\em transient} and {\\em steady state}.\r\n\r\n\\problem Set $F_{\\mbox{ext}} = 1 \\sin(t)$, and $c = 1$ for damping,\r\nwhich is fairly high for an oscillating system. Describe the\r\nresulting behaviour.\r\n\r\n\\vfill\r\n\r\nDescribe specifically the {\\bf long run} behaviour.\r\n\r\n\\vfill\r\n\r\n\\newpage\r\n\r\n\\problem Describe an analogous physical scenario where you might have\r\nobserved transient oscillations transitioning into steady state\r\noscillations.\r\n\r\n\r\n\r\n\\newpage\r\n\r\nReminder: we study the simple damped spring/mass system in such depth\r\nbecause\r\n\\begin{itemize}\r\n\\item the system displays all the interesting mathematical solution\r\n forms as we vary just a few parameters, \r\n\\item we hope that the simplicity of the system, and its resemblance\r\n to familiar systems like swings or car shock absorbers, help to you\r\n associate the mathematics with the real-world behaviour.\r\n\\item the DE for the spring/mass system is either identical or similar\r\n to those for a surprising number of other real-world scenarios.\r\n\\end{itemize}\r\n\r\n\r\n\r\n\\newpage\r\n\r\n\\topic{Classifying Higher-Order DEs}\r\n\\subsection*{Classifying Higher-Order DEs}\r\n\r\nWhile working with DEs in MATLAB, we haven't had to distinguish\r\nbetween types of equations: so long as they can be re-written in the\r\nform $\\ds \\frac{dy}{dt} = f(t, y)$, or the vector form\r\n$\\ds \\frac{d\\vec{w}}{dt} = f(t, \\vec{w})$, MATLAB can produce a\r\nnumerical solution or prediction. \\\\[1ex]\r\n\r\nHowever, when looking at references or mathematical derivations, where\r\nDEs are solved exactly using by-hand methods, the method used is\r\nusually chosen based on the form of the DE, so being able to\r\ncategorize equations is a helpful skill.\r\n\r\n\\newpage\r\n\r\n\\problem Distinguish between the following classifications of\r\ndifferential equations, and given an example of each.\r\n\r\n{\\bf Linear vs Non-Linear}\r\n\\vfill\r\n\r\n{\\bf Linear Homogeneous and Linear Non-Homogeneous}\r\n\\vfill\r\n\r\n\r\n \r\n\r\n\\newpage \r\nClassify the following DEs based on the terms {\\em order}, {\\em\r\n homogeneous} and {\\em linear}.\r\n\r\n\\vfill\r\n\r\n$x^2 y'' + x y' + y = 10$ \\vfill\r\n\r\n $ 100 y'' + y = 4x^3$\r\n \\vfill\r\n\r\n $ (y'')^3 + y = 4 e^{x}$\r\n \\vfill\r\n\r\n $ 4 y^{(4)} - 10 y' + y = 0 $ \\vfill\r\n\r\n\r\n\\end{document}\r\n\r\n\r\n", "meta": {"hexsha": "3f91342483378948a2362be468a1969caed666bf", "size": 20979, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/notes08.tex", "max_stars_repo_name": "aableson/MNTCP01", "max_stars_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-11-27T16:10:35.000Z", "max_stars_repo_stars_event_max_datetime": "2015-11-27T16:10:35.000Z", "max_issues_repo_path": "Notes/notes08.tex", "max_issues_repo_name": "aableson/MNTCP01", "max_issues_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/notes08.tex", "max_forks_repo_name": "aableson/MNTCP01", "max_forks_repo_head_hexsha": "1845fe6ac290b008b070f00f1b68856bdbdfc584", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.4594240838, "max_line_length": 108, "alphanum_fraction": 0.7107107107, "num_tokens": 5839, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833737577158, "lm_q2_score": 0.7931059438487663, "lm_q1q2_score": 0.6143266777336749}} {"text": "\\documentclass[11pt,american,usenames,dvipsnames,svgnames,x11names,table]{article}\n\\input{../tex/portrait-1col}\n%\\input{../tex/landscape-2col}\n\\input{../tex/head}\n\\input{../tex/symbols}\n\\begin{document}\n\\input{../tex/clj-listings} \n\n\n\\title{Regression costs for decision trees}\n\n\n\\author{\\textsc{John Alan McDonald }}\n\n\n\\date{\\today}\n\\maketitle\n\nThe purpose of this document is work thru an alternative to $L_2$ cost\nthat is a bit more efficient to compute, and gives the same results when\nchoosing split predicates in decision tree growing.\n\n\\section{\\label{sub:Decision-trees}Greedy decision trees}\n\nA general binary decision tree consists of \n\\begin{itemize}\n\\item internal \\emph{split} nodes, each containing a predicate that determines\nwhether a record goes to the left or right child of that node.\n\\item terminal \\emph{leaf} nodes, each containing a leaf model function\nwhose value is the tree's prediction for any record that ends up in\nthat node. \n\\end{itemize}\n\nGreedy split optimization --- choose the best out of all feasible\nsplits, and repeat on the resulting child nodes until there are no\nfeasible splits --- is the most common way of growing decision trees.\nIt depends on several things:\n\\begin{enumerate}\n \\item A cost function $c$ used to define 'best'. \n \\item An enumeration of splits to consider. Pure greedy splitting\n considers all 'feasible' splits on all attributes.\n \\begin{enumerate} \n \\item For categorical attributes, that, in general, means considering every \n partition of the categories into $2$ subsets. However, for some\n important cost functions (eg Gini, $L_2$), it can be shown that the\n optimal split can be found by sorting the categories by the\n corresponding score function (eg the response mean for $L_2$ cost),\n and then considering only splits by score.\n \\item For numerical attributes, the most general split would come\n from treating the distinct values like the categories of a\n categorical variable. However, no one does that, mostly because\n there are usually too many distinct values. Instead, only splits by\n $\\leq$ vs $>$ one of the distinct values are considered.\n\\end{enumerate}\n\\item A feasibility test that determines whether a given split on a\ngiven attribute is allowed. The most common case here is to require both\nchildren of the split contain some minimum number of training records.\n\\end{enumerate}\n\n\\section{\\label{sec:numerical}Cost functions for $L_2$ numerical\nregression}\n\nLet $\\mathcal{T} = \\{ \\left( y,\\mathbf{x} \\right) \\}$ be the training\ndata in the node to be split.\nIt is a set of pairs of predictor record $\\mathbf{x}$ and ground truth\nresponse $y$, where $y\\in\\mathbb{R}$ for numerical regression.\nWe are considering splits on some particular predictor field $x_k$,\nwhich might be numerical or categorical.\n\nThe cost function for $L_2$ regression is the sum of squared deviations\nfrom the mean: \n\n$L_{2}\\left(\\mathcal{T}\\right)\n= \\sum_{y\\in\\mathcal{T}}\\,\\left(y-\\bar{y}_{\\mathcal{T}}\\right)^{2}$,\nwhere $\\bar{y}_{\\mathcal{T}}\n= \\frac{1}{\\#\\mathcal{T}}\\sum_{y\\in\\mathcal{T}}\\,y$.\n\nNote that computing this \\textit{accurately}, in an online fashion, for\nmoderate $\\#\\mathcal{T}$, the number of records in $\\mathcal{T}$, allowing for\nthe updating/downdating needed for fast split optimization, \nrequires some care. \n\nHowever, a little bit of algebra will let us use a simpler alternative\nto get the same splits. \n\n\nAny split partitions the training y-values\n$\\mathcal{T=}\\left\\{ y\\right\\} $ into left and right subsets: \n$\\mathcal{T}=\\mathcal{L}\\uplus\\mathcal{R}$.\nThe split cost is:\n\n\\begin{align*}\nc\\left(\\mathcal{L},\\mathcal{R}\\right)= & L_{2}\\left(\\mathcal{L}\\right)\n+ L_{2}\\left(\\mathcal{R}\\right)\\\\\n= & \\sum_{y\\in\\mathcal{L}}\\,\\left(y-\\bar{y}_{\\mathcal{L}}\\right)^{2}+\\sum_{y\\in\\mathcal{R}}\\,\\left(y-\\bar{y}_{\\mathcal{R}}\\right)^{2}\\\\\n= & \\sum_{y\\in\\mathcal{L}}\\left[y^{2}-2\\bar{y}_{\\mathcal{L}}y+\\bar{y}_{\\mathcal{L}}^{2}\\right]+\\sum_{y\\in\\mathcal{R}}\\left[y^{2}-2\\bar{y}_{\\mathcal{R}}y+\\bar{y}_{\\mathcal{R}}^{2}\\right]\\\\\n= & \\sum_{\\mathcal{L}\\uplus\\mathcal{R}}y^{2}-\\frac{\\left(\\sum_{\\mathcal{L}}y\\right)^{2}}{\\#\\mathcal{L}}-\\frac{\\left(\\sum_{\\mathcal{R}}y\\right)^{2}}{\\#\\mathcal{R}}\n\\end{align*}\n\nSince $\\sum_{\\mathcal{L}\\uplus\\mathcal{R}}y^{2}$ doesn't depend on\nthe split, minimizing $c\\left(\\mathcal{L},\\mathcal{R}\\right)$ is\nequivalent to minimizing $-\\left[\\frac{\\left(\\sum_{\\mathcal{L}}y\\right)^{2}}{\\#\\mathcal{L}}+\\frac{\\left(\\sum_{\\mathcal{R}}y\\right)^{2}}{\\#\\mathcal{R}}\\right]$,\nso we can use $\\frac{-\\left(\\sum_{\\mathcal{T}}y\\right)^{2}}{\\#\\mathcal{T}}$\nas our cost function in split optimization.\n\n\\section{Cost functions for $L_2$ vector-valued regression}\n\nLet $\\mathcal{T} = \\{ \\left( \\mathbf{y}, \\mathbf{x} \\right) \\}$ be the\ntraining data in the node to be split.\nHere the ground truth response $\\mathbf{y}$ is a vector, \n$\\mathbf{y}\\in\\mathbb{R}^m$, rather than a single number.\n\nThe cost function for $L_2$ vector-valued regression is the sum of\nsquared $L_2$ distances from the mean vector: \n\\begin{align*}\nL_{2}\\left(\\mathcal{T}\\right) \n= &\n\\sum_{y\\in\\mathcal{T}}\\,\\|\\mathbf{y}-\\bar{\\mathbf{y}}_{\\mathcal{T}}\\|_{2}^{2}\\\\\n= &\n\\sum_{y\\in\\mathcal{T}}\\sum_{i=0}^{m-1}\\,\\left(y_i-\\bar{y_i}_{\\mathcal{T}}\\right)^{2}\n\\end{align*}\n\nFollowing the same reasoning as in section~\\ref{sec:numerical}, we get\nfor a simpler cost:\n$$\nc\\left(\\mathcal{L},\\mathcal{R}\\right) = \n-\\left[\n\\frac{\n\\sum_{i=0}^{m-1} \\, \\left(\\sum_{\\mathcal{L}}y_i\\right)^{2}}\n{\\#\\mathcal{L}}\n\\; + \\;\n\\frac{\\sum_{i=0}^{m-1} \\,\n\\left(\\sum_{\\mathcal{R}}y_i\\right)^{2}}{\\#\\mathcal{R}}\\right] \n$$\n\n%\\bibliographystyle{plainurl}\n%\\bibliography{bib,cactus}\n\n\\end{document}\n", "meta": {"hexsha": "fb2dd76c283b346639d7b1a07176147b33bbd179", "size": 5665, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/regression-cost/regression-cost-xelatex.tex", "max_stars_repo_name": "wahpenayo/taiga", "max_stars_repo_head_hexsha": "9d142149240f3f5db5dd8fbfbc0ce0552b81d490", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2017-09-02T01:14:37.000Z", "max_stars_repo_stars_event_max_datetime": "2018-04-30T14:29:40.000Z", "max_issues_repo_path": "doc/regression-cost/regression-cost-xelatex.tex", "max_issues_repo_name": "wahpenayo/taiga", "max_issues_repo_head_hexsha": "9d142149240f3f5db5dd8fbfbc0ce0552b81d490", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2017-10-24T18:35:41.000Z", "max_issues_repo_issues_event_max_datetime": "2017-10-24T18:35:41.000Z", "max_forks_repo_path": "doc/regression-cost/regression-cost-xelatex.tex", "max_forks_repo_name": "wahpenayo/taiga", "max_forks_repo_head_hexsha": "9d142149240f3f5db5dd8fbfbc0ce0552b81d490", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.4642857143, "max_line_length": 187, "alphanum_fraction": 0.7085613416, "num_tokens": 1798, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.833324607730178, "lm_q2_score": 0.7371581684030623, "lm_q1q2_score": 0.6142920415195784}} {"text": "\\chapter{Mappings between endofuncoids and topological spaces}\n\nOreder topologies reversely to set-theoretic inclusion. That is\nfor topologies~$t$ and~$s$ we set\n$t\\sqsubseteq s\\Leftrightarrow t\\supseteq s$. (Intuitively:\nThe less is the topology, the lesser are its open sets.)\n\nLet's study mappings between topological spaces and endofuncoids.\n\n\\begin{defn}\n Let $t$ be a topology.\n \\begin{enumerate}\n \\item $F^{\\fullstar} t = \\bigsqcup_{x \\in \\Ob t} \\left( \\{ x \\} \\times\n \\bigsqcap^{\\mathscr{F}} \\setcond{ E \\in t }{ x \\in\n E } \\right)$;\n \n \\item $(F_{\\fullstar} t) E = \\bigcap \\setcond{ D \\in t }{ E \\subseteq D }$.\n \\end{enumerate}\n\\end{defn}\n\n\\begin{prop}\n Let $t$ be a topology.\n \\begin{enumerate}\n \\item $F^{\\fullstar} t$ is complete, reflexive, transitive funcoid.\n \n \\item $F_{\\fullstar} t$ is co-complete, reflexive, transitive funcoid.\n \n \\item $F^{\\fullstar}$ and $F_{\\fullstar}$ are injections.\n \n \\item $F_{\\fullstar} t = (F^{\\fullstar} t)^{- 1}$.\n \\end{enumerate}\n\\end{prop}\n\n\\begin{proof}\n By theorem~\\bookref{top-two}.\n\\end{proof}\n\n\\begin{defn}\n Let $f$ be an endofuncoid.\n \\[ T f = \\setcond{ E \\in \\subsets \\Ob f }{\n \\forall x \\in E : \\supfun{f} \\{ x \\} \\sqsubseteq E } . \\]\n\\end{defn}\n\n\\begin{prop}\n $T f$ is a topology.\n\\end{prop}\n\n\\begin{proof}\n \n \\begin{description}\n \\item[Union of open sets is open] $S \\subseteq T f \\Rightarrow \\forall E\n \\in S \\forall x \\in E : \\supfun{f} x \\sqsubseteq E \\Rightarrow\n \\forall x \\in \\bigcup S : \\supfun{f} x \\sqsubseteq \\bigcup S$\n \n \\item[Intersection of two open sets is open] Let $X, Y \\in T f$. Then\n $\\forall x \\in X : \\supfun{f} x \\sqsubseteq X$ and $\\forall x \\in Y\n : \\supfun{f} x \\sqsubseteq Y$. So if $x \\in X \\cap Y$ then $\\langle\n f \\rangle x \\sqsubseteq X$ and $\\supfun{f} x \\sqsubseteq Y$, so\n $\\supfun{f} x \\sqsubseteq X \\cap Y$. So $X \\cap Y \\in T f$.\n \n \\item[$\\Ob f$ is an open set] Obvious.\n \\end{description}\n\\end{proof}\n\n\\begin{obvious}\n$T f = \\setcond{ E \\in \\subsets \\Ob f }{ \\langle\n\\Compl f \\rangle E \\sqsubseteq E }$.\n\\end{obvious}\n\nIn some reason when starting this research I assumed that the\nfollowing funcoid (for every endofuncoid~$f$) is a Kuratowski closure:\n\\[\n1 \\sqcup \\CoCompl f \\sqcup (\\CoCompl f)^2 \\sqcup \\ldots.\n\\]\nIt is not true:\n\n\\begin{example}\nThere exists such a co-complete endofuncoid~$f$ that\n$1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots$ is not transitive that\nis\n\\[(1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots)\\circ\n(1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots) \\ne\n1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots\n\\]\n\\end{example}\n\n\\begin{proof}\nTake $f=\\cl\\circ g$ where $g$ is the principal funcoid which maps\nevery real number~$a$ into the closed interval\n$\\left[ \\frac{- 1 - | a |}{2} ; \\frac{1 + | a |}{2} \\right]$.\n\nTake $X = \\left[ - \\frac{1}{2} ; \\frac{1}{2} \\right]$. $\\langle f^n\n\\rangle^{\\ast} X = \\left[ - 1 + \\frac{1}{2^{n + 1}} ; 1 - \\frac{1}{2^{n + 1}}\n\\right]$.\n\nWe have $\\langle 1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots \\rangle^{\\ast} X =] - 1 ;\n1 [$;\n\n$\\langle 1 \\sqcup f \\sqcup f^2 \\sqcup \\ldots \\rangle^{\\ast} \\langle 1 \\sqcup f\n\\sqcup f^2 \\sqcup \\ldots \\rangle^{\\ast} X = [- 1 ; 1]$.\n\nThus follows our inequality.\n\\end{proof}\n\nThat $F^{\\star}$ and $F_{\\star}$ are functors (if we map morphisms to\nthemselves except of changing the objects) follows from\nconjecture~\\bookref{top-cont}.\n\n\\begin{thm}\n$T$ (if we map morphisms to\nthemselves except of changing the objects) is a functor.\n\\end{thm}\n\n\\begin{proof}\n Based on \\url{https://math.stackexchange.com/a/2792239/4876}\n\n Let $f : \\mu \\rightarrow \\nu$ that is $f \\circ \\mu \\sqsubseteq \\nu \\circ f$.\n We need to prove $f : T \\mu \\rightarrow T \\nu$ that is $E \\in T \\nu\n \\Rightarrow \\langle f^{- 1} \\rangle^{\\ast} E \\in T \\mu$.\n\n Suppose $E \\in T \\nu$ that is $\\langle \\nu \\rangle^{\\ast} E \\sqsubseteq E$.\n We will prove $\\langle \\mu \\rangle^{\\ast} \\langle f^{- 1} \\rangle^{\\ast} E\n \\sqsubseteq \\langle f^{- 1} \\rangle^{\\ast} E$.\n\n \\fxnote{Can we use arbitrary filters rather than atoms?}\n\n Really, let atom $y \\sqsubseteq \\langle \\mu \\rangle^{\\ast} \\langle f^{- 1}\n \\rangle^{\\ast} E$. Then there exists atom $x \\sqsubseteq \\langle f^{- 1}\n \\rangle^{\\ast} E$ such that $x \\mathrel{[\\mu]^{\\ast}} y$.\n\n $x \\mathrel{[f \\circ \\mu]^{\\ast}} \\langle f \\rangle y$ and thus $x\n \\mathrel{[\\nu \\circ f]^{\\ast}} \\langle f \\rangle y$, so $\\langle f \\rangle x\n \\mathrel{[\\nu]^{\\ast}} \\langle f \\rangle y$. But $\\langle f \\rangle x\n \\sqsubseteq E$, so $\\langle f \\rangle y \\sqsubseteq \\langle \\nu\n \\rangle^{\\ast} E \\sqsubseteq E$, that is $\\langle \\mu \\rangle^{\\ast} \\langle\n f^{- 1} \\rangle^{\\ast} E \\sqsubseteq E$.\n\\end{proof}\n\n\\begin{prop}\n $f \\in \\mathrm{C} (\\mu, \\nu) \\Rightarrow f \\in \\mathrm{C} (\\mu^n, \\nu^n)$\n for every endofuncoids $\\mu$ and $\\nu$ and positive natural\n number~$n$.\n \\fxnote{Move this proposition to the book.}\n\\end{prop}\n\n\\begin{proof}\n $f \\circ \\mu \\sqsubseteq \\nu \\circ f$; $f \\circ \\mu \\circ \\mu \\sqsubseteq\n \\nu \\circ f \\circ \\mu$; $f \\circ \\mu^2 \\sqsubseteq \\nu^2 \\circ f$; $f \\circ\n \\mu^3 \\sqsubseteq \\nu^3 \\circ f$...\n\\end{proof}\n\n\\begin{prop}\n For every endofuncoid $\\mu$:\n \\begin{enumerate}\n \\item $F_{\\fullstar} T \\mu \\sqsupseteq \\Compl \\mu$;\n \\item $F^{\\fullstar} T \\mu \\sqsupseteq \\Compl \\mu$;\n \\item $F_{\\fullstar} T \\mu \\sqsupseteq \\CoCompl \\mu$;\n \\item $F^{\\fullstar} T \\mu \\sqsupseteq \\CoCompl \\mu$;\n \\end{enumerate}\n\\end{prop}\n\n\\begin{proof}\n We will prove only the first two as the rest are dual.\n \n $\\langle F_{\\fullstar} T \\mu \\rangle^{\\ast} E =\n \\bigcap \\setcond{ D \\in T \\mu\n }{ D \\supseteq E } = \\bigcap \\setcond{ D \\in\n \\subsets \\Ob \\mu }{ \\langle \\Compl\n \\mu \\rangle^{\\ast} D \\sqsubseteq D \\wedge D \\supseteq E } \\sqsupseteq\n \\bigsqcap \\setcond{ \\langle \\Compl \\mu \\rangle^{\\ast} D }{\n D \\in \\subsets \\Ob \\mu, \\langle \\Compl \\mu\n \\rangle^{\\ast} D \\sqsubseteq D \\wedge D \\supseteq E } \\sqsupseteq\n \\langle \\Compl \\mu \\rangle^{\\ast} E$.\n \n $\\langle F^{\\fullstar} T \\mu \\rangle^{\\ast} \\{ x \\} = \\bigsqcap^{\\mathscr{F}}\n \\setcond{ E \\in T \\mu }{ x \\in E } =\n \\bigsqcap^{\\mathscr{F}} \\setcond{ E \\in \\Ob \\mu }{\n x \\in E, \\left\\langle \\Compl \\mu \\right\\rangle^{\\ast} E\n \\sqsubseteq E } \\sqsupseteq \\bigsqcap^{\\mathscr{F}} \\setcond{ \\langle\n \\Compl \\mu \\rangle^{\\ast} E }{ E \\in \\Ob\n \\mu, x \\in E, \\langle \\Compl \\mu \\rangle^{\\ast} E \\sqsubseteq E }\n \\sqsupseteq \\langle \\Compl \\mu \\rangle^{\\ast} \\{ x \\}$.\n\\end{proof}\n\n\\begin{lem}\n For every endofuncoid $\\mu$:\n \\begin{enumerate}\n \\item $F_{\\fullstar} T \\mu \\sqsubseteq 1 \\sqcup \\Compl \\mu \\sqcup\n (\\Compl \\mu)^2 \\sqcup \\ldots$;\n \n \\item $F^{\\fullstar} T \\mu \\sqsubseteq 1 \\sqcup \\CoCompl \\mu \\sqcup\n (\\CoCompl \\mu)^2 \\sqcup \\ldots$\n \\end{enumerate}\n\\end{lem}\n\n\\begin{proof}\n We will prove only the first as the second is dual.\n \n $\\langle 1 \\sqcup \\Compl \\mu \\sqcup (\\Compl \\mu)^2 \\sqcup \\ldots\n \\rangle^{\\ast} E = E \\sqcup \\langle \\Compl \\mu \\rangle^{\\ast} E \\sqcup\n \\langle (\\Compl \\mu)^2 \\rangle^{\\ast} E \\sqcup \\ldots$\n \n Take $D = E \\sqcup \\langle \\Compl \\mu \\rangle^{\\ast} E \\sqcup \\langle\n (\\Compl \\mu)^2 \\rangle^{\\ast} E \\sqcup \\ldots$ We have $\\langle\n \\Compl \\mu \\rangle^{\\ast} D \\sqsubseteq \\langle \\Compl \\mu\n \\rangle^{\\ast} E \\sqcup \\langle (\\Compl \\mu)^2 \\rangle^{\\ast} E \\sqcup\n \\ldots \\sqsubseteq D$. So\n \n $\\bigcap \\setcond{ D \\in \\subsets \\Ob \\mu }{\n \\langle \\Compl \\mu \\rangle^{\\ast} D \\sqsubseteq D \\wedge D \\supseteq E\n } \\subseteq D \\sqsubseteq \\langle 1 \\sqcup \\Compl \\mu \\sqcup\n (\\Compl \\mu)^2 \\sqcup \\ldots \\rangle^{\\ast} E$.\n\\end{proof}\n\n\\begin{thm}\n If we restrict the functor $T$ only to complete endofuncoids (=~complete\n endoreloids), then $T$ is a left adjoint of both $F_{\\fullstar}$ and\n $F^{\\fullstar}$.\n\\end{thm}\n\n\\begin{proof}\n We will prove only for $F_{\\fullstar}$ as the other is dual.\n \n We will disprove $f \\in \\mathrm{C} (T \\mu, s) \\Leftrightarrow f \\in\n \\mathrm{C} (\\mu, F_{\\fullstar} s)$ what is equivalent (because $F_{\\fullstar}$ is\n full and faithful) to\n \\[ f \\in \\mathrm{C} (F_{\\fullstar} T \\mu, F_{\\fullstar} s) \\Leftrightarrow f \\in\n \\mathrm{C} (\\mu, F_{\\fullstar} s) ; \\]\n $F_{\\fullstar} T \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f\n \\Leftrightarrow \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f$.\n \n $F_{\\fullstar} T \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f \\Rightarrow\n \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f$ because $F_{\\fullstar} T \\mu\n \\sqsupseteq \\mu$.\n \n If $\\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f$ then $\\mu^n\n \\sqsubseteq f^{- 1} \\circ (F_{\\fullstar} s)^n \\circ f = f^{- 1} \\circ F_{\\fullstar}\n s \\circ f$. Also obviously $1 \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ\n f$. Thus\n \\[ 1 \\sqcup \\mu \\sqcup \\mu^2 \\sqcup \\ldots \\sqsubseteq f^{- 1} \\circ\n F_{\\fullstar} s \\circ f \\]\n and so $1 \\sqcup \\Compl \\mu \\sqcup (\\Compl \\mu)^2 \\sqcup \\ldots\n \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f$. So $F_{\\fullstar} T \\mu\n \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ F$.\n\\end{proof}\n\n\\fxnote{$F$ and $T$ are also a Galois connection, isn't it?}\n\n\\begin{example}\n $T$ is a not left adjoint of both $F_{\\star}$ and $F^{\\star}$, with bijection\n which preserves the ``function'' part of the morphism.\n\\end{example}\n\n\\begin{proof}\n We will disprove only from $F_{\\fullstar}$ as the other is dual.\n \n We will disprove $f \\in \\mathrm{C} (T \\mu, s) \\Leftrightarrow f \\in \\mathrm{C}\n (\\mu, F_{\\fullstar} s)$ what is equivalent\n (because $F_{\\star}$ is full and faithful) to\n \\[ f \\in \\mathrm{C} (F_{\\fullstar} T \\mu, F_{\\fullstar} s) \\Leftrightarrow f \\in\n \\mathrm{C} (\\mu, F_{\\fullstar} s) ; \\]\n $F_{\\fullstar} T \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f\n \\Leftrightarrow \\mu \\sqsubseteq f^{- 1} \\circ F_{\\fullstar} s \\circ f$.\n \n This equivalence does not hold: Take $s$ the discrete space\n on~$\\mathbb{R}$, $f=\\id_{\\mathbb{R}}$, and\n $\\rsupfun{\\mu}X=X$ for finite sets~$X$ and $\\rsupfun{\\mu}X=\\top$\n for infinite~$X$.\n\\end{proof}\n", "meta": {"hexsha": "f7ef077c610d0f17bf6a7ab1f2058612ba8df540", "size": 10015, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chap-fcd-top.tex", "max_stars_repo_name": "vporton/algebraic-general-topology", "max_stars_repo_head_hexsha": "d1d02a6515a6dabbc5d30b0c00a3e6a9878b36b1", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2018-06-26T00:14:44.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-03T04:56:16.000Z", "max_issues_repo_path": "chap-fcd-top.tex", "max_issues_repo_name": "vporton/algebraic-general-topology", "max_issues_repo_head_hexsha": "d1d02a6515a6dabbc5d30b0c00a3e6a9878b36b1", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2019-12-30T07:16:23.000Z", "max_issues_repo_issues_event_max_datetime": "2020-03-13T02:05:02.000Z", "max_forks_repo_path": "chap-fcd-top.tex", "max_forks_repo_name": "vporton/algebraic-general-topology", "max_forks_repo_head_hexsha": "d1d02a6515a6dabbc5d30b0c00a3e6a9878b36b1", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.0925925926, "max_line_length": 85, "alphanum_fraction": 0.6261607589, "num_tokens": 3908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8333245870332531, "lm_q2_score": 0.7371581741774411, "lm_q1q2_score": 0.614292031074603}} {"text": "\n\\input{\"/Users/brandonwilliams/Documents/LaTeX Includes/notespreamble.tex\"}\n\\input{\"/Users/brandonwilliams/Documents/LaTeX Includes/extrapackages.tex\"}\n\\input{\"/Users/brandonwilliams/Documents/LaTeX Includes/extracommands.tex\"}\n\n\n\n\\begin{document}\n\n\n\\title{Homotopy Theory}\n\\author{Brandon Williams \\\\ \\texttt{mbw@math.sunysb.edu}}\n\\maketitle\n\n\n\n\\tableofcontents\n\n\n\n\n\n\n\n\\newpage\n\\section{Some Category Theory}\n\n\nA \\textbf{category} $\\mathscr C$ consists of a class of \\textbf{objects}, also denoted by $\\mathscr C$, a class of \\textbf{morphisms} $\\hom_{\\mathscr C}(X,Y)$ for any two objects $X$ and $Y$, and a binary operation $\\hom_{\\mathscr C}(X,Y) \\times \\hom_{\\mathscr C}(Y,Z) \\rightarrow \\hom_{\\mathscr C}(X,Z)$ called \\textbf{composition} and denoted by $(f,g) \\mapsto g \\circ f$, such that the following holds:\n\\begin{enumerate}\n\t\\item For morphisms $f \\in \\hom_{\\mathscr C}(X,Y), g \\in \\hom_{\\mathscr C}(Y,Z)$ and $g \\in \\hom_{\\mathscr C}(Z,W)$ we have associativity composition: $h \\circ (g \\circ f) = (h \\circ g) \\circ f$. \n\t\\item For each object $X$ there is a unique morphism $\\id_X \\in \\hom_{\\mathscr C}(X,X)$ such that for any morphism $f \\in \\hom_{\\mathscr C}(X,Y)$ we have $f \\circ \\id_X = f = \\id_Y \\circ f$.\n\\end{enumerate}\nWe will usually write $f : X \\rightarrow Y$ or $X \\stackrel{f}{\\rightarrow} Y$ as a shorthand for saying that $f$ is a morphism from $X$ to $Y$, and when the category $\\mathscr C$ is clear from context we will write $\\hom(X,Y)$ to denote the class of morphisms. A \\textbf{small category} is one in which the objects and morphisms are sets, and a \\textbf{locally small category} is one in which the morphisms are sets. The \\textbf{opposite} of a category $\\mathscr C$ is a category $\\mathscr C^\\op$ with the same objects, but $\\hom_{\\mathscr C^\\op}(X,Y) := \\hom_{\\mathscr C}(Y,X)$. Intuitively we are just ``flipping'' the arrows in $\\mathscr C$.\n\nA morphism $f : X \\rightarrow Y$ is said to be an \\textbf{isomorphism} if there is a morphism $g : Y \\rightarrow X$ such that $g \\circ f = \\id_X$ and $f \\circ g = \\id_Y$. A category where all morphisms are isomorphisms is called a \\textbf{groupoid}. An isomorphism is said to be an \\textbf{automorphism} if the domain and codomain of $f$ are the same object. The class of automorphisms of an object $X$ is denoted by $\\Aut_{\\mathscr C}(X)$, or just $\\Aut(X)$ when the category is clear from context. Many categories have their own notation for the class of automorphisms, such as $\\Homeo(X)$ for the category of topological spaces and $\\Diff(X)$ for the category of smooth manifolds.\n\nAn object $X$ is said to be \\textbf{initial} if there is a unique morphism from $X$ to any other object, and said to be \\textbf{terminal} if there is a unique morphism from any other object to $X$. We say that $X$ is a \\textbf{zero object} if it is both initial and terminal. If there is a family of morphism $0_{XY} : X \\rightarrow Y$ such that for all morphisms $f : X \\rightarrow Y$ and $g : Z \\rightarrow W$ we have $0_{YW} \\circ f = g \\circ 0_{XZ}$, then we call this family a \\textbf{family of zero morphisms}. If the category has a zero object, then these morphisms are unique, and so we call $0_{XY}$ \\emph{the} \\textbf{zero morphism} from $X$ to $Y$. \n\nA \\textbf{functor} between categories $F : \\mathscr C \\rightarrow \\mathscr D$ assigns an object $F(X) \\in \\mathscr D$ to each object $X \\in \\mathscr C$ and a morphism $F(f) : F(X) \\rightarrow F(Y)$ to each morphism $f : X \\rightarrow Y$ such that\n\\begin{enumerate}\n\t\\item $F(\\id_X) = \\id_{F(X)}$.\n\t\\item If $f : X \\rightarrow Y$ and $g : Y \\rightarrow Z$, then $F(g \\circ f) = F(g) \\circ F(f)$.\n\\end{enumerate}\nA \\textbf{cofunctor} between categories $F : \\mathscr C \\rightarrow \\mathscr D$ is a functor in the above sense, except it assigns a morphism $F(f) : F(Y) \\rightarrow F(X)$ to each morphism $f : X \\rightarrow Y$ such that\n\\begin{enumerate}\n\t\\item[$2^\\prime$.] If $f : X \\rightarrow Y$ and $g : Y \\rightarrow Z$, then $F(g \\circ f) = F(f) \\circ F(g)$.\n\\end{enumerate}\nIt is easy to see that a cofunctor $F : X \\rightarrow Y$ is just a functor $F : X^\\op \\rightarrow Y$, or equivalently $F : X \\rightarrow Y^\\op$.\n\nA \\textbf{natural transformation} $\\eta$ between functors $F,G : \\mathscr C \\rightarrow \\mathscr D$, written as $\\eta : F \\rightarrow G$, assigns each object $X \\in \\mathscr C$ a morphism $\\eta(X) : F(X) \\rightarrow G(X)$ such that for any morphism $f : X \\rightarrow Y$ we have the following commutative diagram\n\\[\n\\xymatrix\n@R=3pc\n@C=3pc\n{\n\tF(X) \\ar[r]^{F(f)} \\ar[d]_{\\eta(X)} & F(Y) \\ar[d]^{\\eta(Y)} \\\\\n\tG(X) \\ar[r]_{G(f)} & G(Y)\n}\n\\]\nWe can similarly define natural transformations between cofunctors. The natural transformation $\\eta$ is said to be a \\textbf{natural isomorphism} if $\\eta(X)$ is an isomorphism for all $X$, in which case the functors $F$ and $G$ are said to be \\textbf{naturally isomorphic}.\n\n\\begin{example}\nOne comes across the statement in linear algebra that a vector space $V$ is ``naturally'' isomorphic to its double dual $V^{**}$. In categorical terminology this means that the identity functor $\\id : \\Vect \\rightarrow \\Vect$ on the category of vector spaces is naturally isomorphic to the double dual functor $F : \\Vect \\rightarrow \\Vect$, where $F(V) = V^{**}$ and $F(f) = f^{**}$. \n\\end{example}\n\nMany times we can only deal with objects that are unique only up to isomorphism, and so we usually need some kind of extra naturality when dealing with these objects. For example, it is trivial to see that if $A$ is isomorphic to $B$, then there is a bijective correspondence $\\hom_{\\mathscr C}(A,C) \\leftrightarrow \\hom_{\\mathscr C}(B,C)$ for any object $C$. For if $f : A \\rightarrow B$ is an isomorphism, then the correspondence $h \\mapsto h \\circ f$ is a bijection $\\hom_{\\mathscr C}(A,C) \\rightarrow \\hom_{\\mathscr C}(B,C)$. So these hom-sets only depend on the isomorphism class of the objects. But, if $f : C \\rightarrow C'$ is a morphism, how are all the hom-sets $\\hom_{\\mathscr C}(A,C), \\hom_{\\mathscr C}(B,C), \\hom_{\\mathscr C}(A,C')$ and $\\hom_{\\mathscr C}(B,C')$ related? Well, in general they won't be related in any natural way, but if the bijective we described above is denoted by $\\varphi$, then we have the following commutative diagram\n\\[\n\\xymatrix\n{\n\t\\hom(A,C) \\ar[r]^\\varphi \\ar[d]_{f_*} & \\hom(B,C) \\ar[d]^{f_*} \\\\\n\t\\hom(A,C') \\ar[r]_\\varphi & \\hom(B,C')\n}\n\\]\nwhere $f_*$ is the induced map. So, we say that $\\varphi$ is a natural bijective correspondence.\n\nThere are many useful constructions in category theory, such as products, coproducts, direct limits, inverse limits, etc., but these are all special cases of limits and colimits. Let $\\mathscr D$ be a small category and $\\mathscr C$ any category. A \\textbf{$\\mathscr D$-shaped diagram in $\\mathscr C$} is a functor $F : \\mathscr D \\rightarrow \\mathscr C$. A morphism of $\\mathscr D$-shaped diagrams $F$ and $F'$ is a natural transformation $F \\rightarrow F'$, and so they form a category denoted by $\\mathscr D(\\mathscr C)$. There is a canonical functor $\\Delta : \\mathscr C \\rightarrow \\mathscr D(\\mathscr C)$, called the diagonal functor, that assigns each object $X$ the constant diagram, which maps every object of $\\mathscr D$ to $X$ and every morphism to the identity. Note that a morphism $F : \\Delta(X) \\rightarrow \\Delta(Y)$ of constant diagrams must assign each object of $\\mathscr D$ the same morphism $X \\rightarrow Y$ by the commutativity of the natural transformation square, so we can think of morphisms of constant diagrams as just morphisms of the respective objects.\n\nGiven a $\\mathscr D$-shaped diagram $F$ in $\\mathscr C$, we define its \\textbf{limit} to be an object $\\lim F$ in $\\mathscr C$ and morphism $\\pi : \\Delta(\\lim F) \\rightarrow F$ such that for any object $X$ of $\\mathscr C$ and morphism of diagrams $\\varepsilon : \\Delta(X) \\rightarrow F$ there is a unique morphism $\\tilde\\varepsilon : \\Delta(X) \\rightarrow \\Delta(\\lim F)$ such that the following commutes\n\\[\n\\xymatrix\n{\n\t\\Delta(X) \\ar@{-->}[d]_{\\tilde\\varepsilon} \\ar@/_2.3pc/[dd]_{\\varepsilon} \\\\\n\t\\Delta(\\lim F) \\ar[d]_{\\pi} \\\\\n\tF\n}\n\\]\nThe existence and uniqueness of the morphism $\\tilde\\varepsilon$ for any $\\varepsilon$ ensures that limits are unique up to isomorphism, and so this is called the \\textbf{universal property} of the limit.\n\nWe can dualize this construction to define the \\textbf{colimit} of a $\\mathscr D$-shaped diagram in $\\mathscr C$. In particular, the colimit consists of an object $\\colim F$ in $\\mathscr C$ and morphism $\\iota : F \\rightarrow \\Delta(\\colim F)$ such that for any object $X$ of $\\mathscr C$ and morphism $\\eta : F \\rightarrow \\Delta(X)$ there is a unique morphism $\\tilde\\eta : \\Delta(\\colim F) \\rightarrow \\Delta(X)$ such that the following commutes\n\\[\n\\xymatrix\n{\n\t\\Delta(X) \\\\\n\t\\Delta(\\colim F) \\ar@{-->}[u]^{\\tilde\\eta} \\\\\n\tF \\ar[u]^\\iota \\ar@/^2.5pc/[uu]^{\\eta}\n}\n\\]\nAgain we have that colimits are unique up to isomorphism.\n\n\\begin{example}[Products]\n\\label{categorical product as a limit}\nLet $\\mathscr D$ be a category with two objects and no non-identity morphisms. Then a $\\mathscr D$-shaped diagram $F$ is simply a pair of objects $X,Y$ in $\\mathscr C$, and let us denote the limit by $P$. The morphism $\\pi : \\Delta(P) \\rightarrow F$ is just a pair of morphisms $\\pi_X : P \\rightarrow X$ and $\\pi_Y : P \\rightarrow Y$. Then the universal property of limits says that for any other object $P'$ and morphisms $\\pi_X' : P' \\rightarrow X$ and $\\pi_Y' : P' \\rightarrow Y$ there is a unique morphism $\\epsilon : P' \\rightarrow P$ such that the following diagram commutes\n\\[\n\\xymatrix\n@C=1pc\n{\n\t & P' \\ar@{-->}[d]_{\\varepsilon} \\ar@/_1.5pc/[ddl]_{\\pi_X'} \\ar@/^1.5pc/[ddr]^{\\pi_Y'} \\\\\n\t & P \\ar[dl]_{\\pi_X} \\ar[dr]^{\\pi_Y} & \\\\\n\tX & & Y\n}\n\\]\nIn category theory $P$ is most commonly known as the \\textbf{categorical product} of $X$ and $Y$ in $\\mathscr C$, and $\\pi_X$ and $\\pi_Y$ are called the natural projections. We denote the product, which is unique up to isomorphism, by $X \\times_{\\mathscr C} Y$, or just $X \\times Y$ when the category is understood.\n\nTherefore the product in category theory is just a special case of a limit when $\\mathscr D$ is the discrete category with two objects. More generally, suppose we have a collection of objects $\\lcb X_\\alpha \\rcb$ indexed by a set $\\Lambda$. We can think of $\\Lambda$ as a category whose objects are the elements of $\\Lambda$ and there are no non-identity morphisms. Then the product of the objects $\\lcb X_\\alpha \\rcb$ is just the limit of the $\\Lambda$-shaped diagram that maps each index $\\alpha$ to $X_\\alpha$.\n\\end{example}\n\n\\begin{example}[Coproducts]\nLet $\\mathscr D$ be the discrete category with two objects as in \\cref{categorical product as a limit} so that a $\\mathscr D$-shaped diagram $F$ is just a pair of objects $X,Y$ in $\\mathscr C$. Let $C$ denote the colimit of $F$, and let $\\iota_X : X \\rightarrow C$ and $\\iota_Y : Y \\rightarrow C$ be the morphisms provided by the colimit. Then the universal property of colimits says that for any other object $C'$ and morphisms $\\iota_X' : X \\rightarrow C'$ and $\\iota_Y' : Y \\rightarrow C'$ there is a unique morphism $\\eta : C \\rightarrow C'$ such that the following diagram commutes\n\\[\n\\xymatrix\n@C=1pc\n{\n\t & C' \\\\\n\t & C \\ar@{-->}[u]^{\\eta} & \\\\\n\tX \\ar[ur]^{\\iota_X} \\ar@/^1.5pc/[uur]^{\\iota_X'} & & Y \\ar[ul]_{\\iota_Y} \\ar@/_1.5pc/[uul]_{\\iota_Y'}\n}\n\\]\nIn category theory $C$ is most commonly known as the \\textbf{categorical coproduct} of $X$ and $Y$ in $\\mathscr C$, and $\\iota_X$ and $\\iota_Y$ are called the natural inclusions. We denote the coproduct, which is unique up to isomorphism, by $X \\sqcup_{\\mathscr C} Y$, or just $X \\sqcup Y$ when the category is understood.\n\\end{example}\n\n\\begin{example}[Equalizers]\n\\label{equalizer as a limit}\nLet $\\mathscr D$ be the category with two objects and two non-identity morphisms with same domain and range. We can write this schematically as $\\mathscr D = \\lcb \\xymatrix{ \\cdot \\ar@/_0.5pc/[r] \\ar@/^0.5pc/[r] & \\cdot } \\rcb$. Then a $\\mathscr D$-shaped diagram $F$ is simply a pair of objects $X,Y$ in $\\mathscr C$ and a pair of morphisms $f,g : X \\rightarrow Y$, which we can write as the diagram\n\\[\n\\xymatrix\n{\n\tX \\ar@/_0.5pc/[r]_g \\ar@/^0.5pc/[r]^f & Y\n}\n\\]\nLet $E$ denote the the limit of this diagram, and let $\\pi : F \\rightarrow \\Delta(E)$ be the morphism provided by the limit. This is just a pair of morphisms $\\pi_X : E \\rightarrow X$ and $\\pi_Y : E \\rightarrow Y$, but $\\pi_Y$ is determined from $\\pi_X,f,g$ since the commutativity of the natural transformation square requires that $\\pi_Y = f \\circ \\pi_X = g \\circ \\pi_Y$. The universal property of limits says that for any other object $E'$ and morphism $\\pi_X' : E' \\rightarrow X$ there is a unique morphism $\\varepsilon : E' \\rightarrow E$ such that the following diagram commutes\n\\[\n\\xymatrix\n{\n\tE \\ar[r]^{\\pi_X} & X \\ar@/_0.5pc/[r]_g \\ar@/^0.5pc/[r]^f & Y \\\\\n\tE' \\ar@{-->}[u]^{\\varepsilon} \\ar[ur]_{\\pi_X'}\n}\n\\]\nIn category theory $C$ is most commonly known as the \\textbf{equalizer} of $f,g : X \\rightarrow Y$ in $C$.\n\\end{example}\n\n\\begin{example}[Coequalizers]\nDualizing everything in \\cref{equalizer as a limit} gives us the definition of the \\textbf{coequalizer} of $f,g : X \\rightarrow Y$ as a colimit. Specifically, the coequalizer is an object $C$ and morphism $\\iota_Y : Y \\rightarrow C$ such that for any other object $C'$ and morphism $\\iota_Y' : Y \\rightarrow C$ there is a unique morphism $\\eta : C \\rightarrow C'$ such that the following diagram commutes\n\\[\n\\xymatrix\n{\n\tX \\ar@/_0.5pc/[r]_g \\ar@/^0.5pc/[r]^f & Y \\ar[r]^{\\iota_Y} \\ar[rd]_{\\iota'_Y} & C \\ar@{-->}[d]^{\\eta} \\\\\n\t & & C'\n}\n\\]\n\\end{example}\n\n\\begin{example}[Pullbacks]\n\\label{pullback as a limit}\nConsider the category $\\mathscr D = \\lcb \\cdot \\rightarrow \\cdot \\leftarrow \\cdot \\rcb$, that is, $\\mathscr D$ has three objects and two non-identity morphisms. A $\\mathscr D$-shaped diagram $F$ is just a diagram of the form $Y \\stackrel{g}{\\rightarrow} Z \\stackrel{f}{\\leftarrow} X$. Let $P$ be the pullback of this diagram with induced morphisms $f' : P \\rightarrow Y$ and $g' : P \\rightarrow X$. Technically we should also include another morphism $P \\rightarrow Z$, but it uniquely determined by $f,g,f',g'$ due to the commutativity square of the natural transformation. The universal property of this limit says that for any other object $P'$ and morphisms $f'' : P' \\rightarrow Y$ and $g'' : P' \\rightarrow X$ there is a unique morphism $\\epsilon : P' \\rightarrow P$ such that the following diagram commutes\n\\[\n\\xymatrix\n{\n\tP' \\ar@{-->}[rd]_{\\epsilon} \\ar@/_2pc/[rddd]_{f''} \\ar@/^2pc/[rrrd]^{g''} & & \\\\\n\t & P \\ar[rr]^{g'} \\ar[dd]_{f'} & & X \\ar[dd]^{f} \\\\\n\t & & & \\\\\n\t & Y \\ar[rr]_{g} & & Z\n}\n\\]\nIn category theory we say that $P$ is the \\textbf{pullback} of $X \\stackrel{f}{\\rightarrow} Z \\stackrel{g}{\\leftarrow} Y$, and will be denoted by $X \\times_Z Y$. By the universal property there is a natural isomorphism $X \\times_Z Y \\cong Y \\times_Z X$.\n\\end{example}\n\n\\begin{example}[Pushouts]\nDualizing the construction in \\cref{pullback as a limit} gives us the definition of the \\textbf{pushout} of $Y \\stackrel{g}{\\leftarrow} Z \\stackrel{f}{\\rightarrow} X$. In particular, the pushout of this diagram is an object $P$ with morphisms $X \\stackrel{g'}{\\rightarrow} P$ and $Y \\stackrel{f'}{\\rightarrow} P$ such that for any other object $P'$ and morphisms $X \\stackrel{g''}{\\rightarrow} P'$ and $Y \\stackrel{f''}{\\rightarrow} P$ there is a unique morphism $\\eta : P \\rightarrow P'$ such that the following diagram commutes\n\\[\n\\xymatrix\n{\n\tZ \\ar[rr]^{f} \\ar[dd]_{g} & & X \\ar[dd]^{g'} \\ar@/^2pc/[rddd]^{g''} \\\\\n\t & & \\\\\n\tY \\ar[rr]_{f'} \\ar@/_2pc/[rrrd]_{f''} & & P \\ar@{-->}[rd]^{\\eta} \\\\\n\t & & & P'\n}\n\\]\nThe pushout will be denoted by $X \\cup_Z Y$. By the universal property there is a natural isomorphism $X \\cup_Z Y \\cong Y \\cup_Z X$.\n\\end{example}\n\n\\begin{example}[Inverse Limits]\nLet $\\Lambda$ be a partially ordered set with upper bounds. This means there is a binary relation $\\leq$ on $\\Lambda$ that is reflexive, antisymmetric, transitive and such that for any elements $i,j$ in $\\Lambda$ there is an element $k$ such that $i \\leq k$ and $j \\leq k$. Such a set is sometimes called a directed set. Consider the category $\\underleftarrow\\Lambda$ whose objects are the elements of $\\Lambda$ and such that there is a \\emph{unique} morphism $f_{ij} : j \\rightarrow i$ if $i \\leq j$. A $\\underleftarrow\\Lambda$-shaped diagram $F : \\underleftarrow\\Lambda \\rightarrow \\mathscr C$ in $\\mathscr C$ is called an \\textbf{inverse system} over $\\Lambda$ in $\\mathscr C$ (older references say projective system). The limit of this diagram is called the \\textbf{inverse limit} of the inverse system, and is written $\\invlim F$. \n\nLet us peel away the abstract nonsense for a moment to see what this definition really means. A $\\underleftarrow\\Lambda$-shaped diagram in $\\mathscr C$ is just a collection of objects $A_i$, indexed by $\\Lambda$, and one morphism $f_{ij} : A_j \\rightarrow A_i$ for each $i \\leq j$ such that $f_{ii} = \\id_{A_i}$ and $f_{ik} = f_{ij} \\circ f_{jk}$. The limit of this diagram is an object $A$ with morphisms $\\pi_i : A \\rightarrow A_i$ for each $i \\in \\Lambda$, such that $f_{ij} \\circ \\pi_j = \\pi_i$ (this comes from commutativity of the natural transformation square). The object $A$ has to be universal in the sense that if $A'$ is another object with morphisms $\\pi_i' : A' \\rightarrow A_i$ and $f_{ij} \\circ \\pi_j' = \\pi_i'$, then there is a unique morphism $\\varepsilon : A' \\rightarrow A$ such that $\\pi_i \\circ \\varepsilon = \\pi_i'$ for all $i \\in \\Lambda$. \n\\end{example}\n\n\\begin{example}[Direct Limits]\nDualizing the above gives the definition of direct limits in a category, but we spell out the details to make it more clear. Let $\\Lambda$ be a partially ordered set with upper bounds. Consider the category $\\underrightarrow\\Lambda$ whose objects are the elements of $\\Lambda$ and such that there is a \\emph{unique} morphism $f_{ij} : i \\rightarrow j$ if $i \\leq j$. A $\\underrightarrow\\Lambda$-shaped diagram $F : \\underrightarrow\\Lambda \\rightarrow \\mathscr C$ in $\\mathscr C$ is called a \\textbf{direct system} over $\\Lambda$ in $\\mathscr C$. The colimit of this diagram is called the \\textbf{direct limit} of the direct system, and is written $\\dirlim F$. \n\nLet us see what this means on a lower level. A $\\underrightarrow\\Lambda$-shaped diagram in $\\mathscr C$ is just a collection of objects $A_i$, indexed by $\\Lambda$, and one morphism $f_{ij} : A_i \\rightarrow A_j$ for each $i \\leq j$ such that $f_{ii} = \\id_{A_i}$ and $f_{ik} = f_{jk} \\circ f_{ij}$. The colimit of this diagram is an object $A$ with morphisms $\\iota_i : A_i \\rightarrow A$ for each $i \\in \\Lambda$, such that $\\pi_j \\circ f_{ij} = \\pi_i$ (this comes from commutativity of the natural transformation square). The object $A$ has to be universal in the sense that if $A'$ is another object with morphisms $\\iota_i' : A_i \\rightarrow A'$ and $\\iota_j' \\circ f_{ij} = \\iota_i'$, then there is a unique morphism $\\eta : A \\rightarrow A'$ such that $\\eta \\circ \\iota_i = \\iota_i'$ for all $i \\in \\Lambda$. \n\\end{example}\n\n\nLet us describe how $\\lim$ can be thought of as a functor $\\lim : \\mathscr D(\\mathscr C) \\rightarrow \\mathscr C$. The functor assigns a diagram $F$ in $\\mathscr D(\\mathscr C)$ the object $\\lim F$. For two diagrams $F,G$ let $\\pi_F,\\pi_G$ denote the morphisms from their limits to $F$. If $t : F \\rightarrow G$ is a morphism of these diagrams, then $t \\circ \\pi_F : \\Delta(\\lim F) \\rightarrow G$, so by the universal property of limits there is a unique morphism $\\tilde\\varepsilon : \\Delta(\\lim F) \\rightarrow \\Delta(\\lim G)$ such that the following commutes\n\\[\n\\xymatrix\n{\n\t\\Delta(\\lim F) \\ar@{-->}[d]_{\\tilde\\varepsilon} \\ar@/_2.3pc/[dd]_{t \\circ \\pi_F} \\\\\n\t\\Delta(\\lim G) \\ar[d]_{\\pi_G} \\\\\n\tG\n}\n\\]\nBut we remarked earlier that a morphism between constant diagrams must map every object to the same morphism, so we define $\\lim(t)$ be this unique morphism for $\\tilde\\varepsilon$. Doing something similar for $\\colim$ shows how it can be turned into a functor $\\mathscr D(\\mathscr C) \\rightarrow \\mathscr C$. \n\nGiven a functor $F : \\mathscr C \\rightarrow \\mathscr C'$ between categories and a diagram $G : \\mathscr D \\rightarrow \\mathscr C$ in $\\mathscr C$, we get an induced diagram $F \\circ G$ in $\\mathscr C'$. The functor $F$ is said to be \\textbf{continuous} if $F(\\lim G) = \\lim(F \\circ G)$ for all diagrams $G$, and otherwise is said to be \\textbf{discontinuous}. Similarly, if $F(\\colim G) = \\colim(F \\circ G)$ for all diagrams $G$, then $F$ is said to be \\textbf{cocontinuous}, and otherwise is said to be \\textbf{discocontinuous}. For example, the forgetful functor $\\Ab \\rightarrow \\Set$ on the category of abelian groups is continuous, however the forgetful functor $\\Setp \\rightarrow \\Set$ on the category of pointed sets is discocontinuous (coproducts are not preserved). A category such that limits of all diagrams exists is called \\textbf{complete}, and if all colimits exists then it is called \\textbf{cocomplete}. The following is a surprising theorem.\n\\begin{thm}\n\\label{complete and cocomplete characterization}\nA category is complete (cocomplete) if and only if all products (coproducts) and all equalizers (coequalizers) exist.\n\\end{thm}\n\n\\begin{example}\nWe will show a specific example of \\cref{complete and cocomplete characterization} by describing how a pullback can be formed as a product and equalizer, which is why we chose the notation $X \\times_Z Y$ for pullbacks. Suppose we have two morphisms $f : X \\rightarrow Z$ and $g : Y \\rightarrow Z$. Let $P$ be the product of $X$ and $Y$, with projections $\\pi_X : P \\rightarrow X$ and $\\pi_Y : P \\rightarrow Y$, and let $E$ be the equalizer of $f \\circ \\pi_X$ and $g \\circ \\pi_Y$, with induced natural map $\\pi : E \\rightarrow P$. We claim that $E$ with maps $\\pi_X \\circ \\pi$ and $\\pi_Y \\circ \\pi$ is the pullback of $f$ and $g$. To show this we need to verify the universal property of the limit. So, let $E'$ be any other object with maps $f' : E' \\rightarrow Y$ and $g' : E' \\rightarrow X$. By the universal property of products we have that there is a unique morphism $\\varepsilon : E' \\rightarrow P$. But, now by the universal property of equalizers, we have that there is a unique morphism $\\epsilon' : E' \\rightarrow E$. The diagram looks like the following\n\\[\n\\xymatrix\n{\n\tE' \\ar@/_2.5pc/[rrdddd]_{f'} \\ar@/^2.5pc/[rrrrdd]^{g'} \\ar@{-->}[rd]^{\\varepsilon'} \\ar@{-->}@/_1.5pc/[rrdd]_{\\varepsilon} & & & \\\\\n\t & E \\ar@{-->}[rd]^{\\pi} & & \\\\\n\t & & P \\ar[rr]^{\\pi_Y} \\ar[dd]_{\\pi_X} & & X \\ar[dd]^{f} \\\\\n\t & & & & \\\\\n\t & & Y \\ar[rr]_{g} & & Z\n}\n\\]\nWe need to verify that $\\pi_Y \\circ \\pi \\circ \\varepsilon' = g'$, and this is true since $\\pi \\circ \\varepsilon' = \\varepsilon$ by universality of equalizers, and then $\\pi_Y \\circ \\varepsilon = g'$ by universality of products. We also need to verify that $\\pi_X \\circ \\pi \\circ \\varepsilon' = f'$, but this is similar to what we just showed. Therefore $E$ is the pullback of $f$ and $g$. By dualizing this we can also show that pushouts can be expressed as a coproduct and coequalizer.\n\\end{example}\n\nIn set theory we define the exponential of sets $Y^X$ to be the set of functions $X \\rightarrow Y$. There is an obvious map $e : Y^X \\times X \\rightarrow Y$ defined by $e(f,x) = f(x)$, and for any space $Z$ with function $Z \\times X \\rightarrow Y$ there is a unique $\\lambda : Z \\rightarrow Y^X$ such that $e \\circ (\\lambda \\times \\id) = g$, just take $\\lambda(z)(x) = g(z,x)$. We can easily generalize this to other categories. Let $\\mathscr C$ be a category with all finite products. The \\textbf{exponential object} of $X$ and $Y$ is an object denoted by $Y^X$ and morphism $e : Y^X \\times X \\rightarrow Y$ such that for any other object $Z$ and morphism $g : Z \\times X \\rightarrow Y$, there is a unique morphism $\\lambda : Z \\rightarrow Y^X$ such that the following diagram commutes\n\\[\n\\xymatrix\n{\n\tZ \\times X \\ar[rd]^{g} \\ar@{-->}[d]_{\\lambda \\times \\id_X} \\\\\n\tY^X \\times X \\ar[r]_-{e} & Y\n}\n\\]\nWe say that a category is \\textbf{cartesian closed} if it has all finite products and if all exponential objects exist.\n\n\n\nLet $\\mathscr C$ and $\\mathscr D$ be two locally small categories, and consider two functors $F : \\mathscr C \\rightarrow \\mathscr D$ and $G : \\mathscr D \\rightarrow \\mathscr C$. We say that $F$ is a \\textbf{left adjoint} of $G$ (or equivalently, $G$ is a \\textbf{right adjoint} to $F$) if the bifunctors \n\\[ \\hom_{\\mathscr D}(F(-),-) : \\mathscr C^\\op \\times \\mathscr D \\rightarrow \\Set \\]\n\\[ \\hom_{\\mathscr C}(-,G(-)) : \\mathscr C^\\op \\times \\mathscr D \\rightarrow \\Set \\]\nare naturally isomorphic. This means for every object $X$ in $\\mathscr C$ and object $Y$ in $\\mathscr D$, there is an isomorphism of sets\n\\[ \\hom_{\\mathscr D}(F(X),Y) \\cong \\hom_{\\mathscr C}(X,G(Y)) \\]\nwhich is natural in $X$ and $Y$ (meaning, morphisms on $X$ and $Y$ give us the obvious commuting diagrams). Sometimes the hom-sets of $\\mathscr C$ and $\\mathscr D$ can be given extra structure, in which case we can usually say the above isomorphism of sets is actually an isomorphism of the extra structures too. Adjoint functors are unique up to natural isomorphism. \n\n\\begin{example}\nLet $F : \\Grp \\rightarrow \\Set$ be the forgetful functor on the category of groups, and let $G : \\Set \\rightarrow \\Grp$ be the functor that sends a set to the free group generated by its elements. Then $F$ is a left adjoint of $G$.\n\\end{example}\n\n\\begin{example}\n\\label{forgetful functor on Top}\nThe forgetful functor $F : \\Top \\rightarrow \\Set$ on the category of topological spaces has both a left and right adjoint. Its left adjoint is the functor $D : \\Set \\rightarrow \\Top$ that gives a set the discrete topology, and its right adjoint is the functor $I : \\Set \\rightarrow \\Top$ that gives a set the indiscrete topology. A similar statement holds for the forgetful functor $\\Topp \\rightarrow \\Setp$ on the category of pointed topological spaces. Therefore, by the upcoming \\cref{adjoint functor continuity}, both of these forgetful functors are continuous and cocontinuous.\n\\end{example}\n\n\\begin{example}\n\\label{product-exponential adjoint}\nSuppose all exponential objects $Y^X$ exist for a fixed object $X$. Then the functor $-\\times X : \\mathscr C \\rightarrow \\mathscr C$ is the left adjoint of $-^X : \\mathscr C \\rightarrow \\mathscr C$. \n\\end{example}\n\nAdjoint functors have an amazing property.\n\\begin{thm}\n\\label{adjoint functor continuity}\nIf $F$ has a left (right) adjoint, and hence is a right (left) adjoint, then $F$ is continuous (cocontinuous).\n\\end{thm}\n\n\n\n\n\n\n\\newpage\n\\section{Some General Topology}\n\n\n\\subsection{Categories, Limits and Colimits in Topology}\n\nThe category of topological spaces and continuous maps will be denoted by $\\Top$, and the category of based topological spaces with base-point preserving continuous maps will be denoted by $\\Topp$. When it is understood that we are working in $\\Top$ or $\\Topp$, we will sometimes denote $\\hom_{\\Top}(X,Y)$ by $\\hom(X,Y)$ and $\\hom_{\\Topp}(X,Y)$ by $\\hom_*(X,Y)$. We saw in \\cref{forgetful functor on Top} that the forgetful functors on $\\Top$ and $\\Topp$ have left and right adjoints, hence the forgetful functors are continuous and cocontinuous, and so $\\Top $ and $\\Topp$ are complete and cocomplete. This means that the underlying set of any limit or colimit in $\\Top$ (resp. $\\Topp$) is just the limit or colimit taken in $\\Set$ (resp. $\\Setp$) and topologized by the universal property. For example, the product of topological spaces is just the cartesian product of their underlying sets, and the unversal property gives the product the Tychonoff topology. The product of based spaces $(X,x_0)$ and $(Y,y_0)$ is the cartesian product $X \\times Y$ with base point $(x_0,y_0)$. The coproduct of topological spaces is just the disjoint union of their underlying sets. \n\nHowever, the coproduct of pointed topological spaces $(X,x_0)$ and $(Y,y_0)$ is a little more complicated. If we just take the disjoint union $X \\sqcup Y$, then both $x_0$ and $y_0$ are in the coproduct, and which should we choose for the base point? Neither base point will suffice, so instead we glue together $X$ and $Y$ at their base points. This construction is called the \\textbf{wedge sum}, its the coproduct in the category of based sets, and is denoted by $X \\vee Y = \\lcb (x,y) \\in X \\times Y \\st x=x_0 \\text{ or } y=y_0 \\rcb$, with base point $(x_0,y_0)$. \n\nUnfortunately, the categories $\\Top$ and $\\Topp$ have too many pathological spaces, for example indiscrete spaces. Even the nicer category $\\Haus$ of Hausdorff spaces is still a little too unwieldy. This prevents $\\Top$ and $\\Haus$ from being cartesian closed, which is very important to have for homotopy theory since we like to think of homotopies as 1-parameter families of maps, i.e. we want $\\hom_{\\Top}(X \\times I,Y) \\cong \\hom_{\\Top}(X,Y^I)$. So we would like to find a subcategory of $\\Top$ where exponential objects exists, but also contains all the spaces we usually work with. Steenrod found such a category of spaces (see \\cite{steenrod67}). \n\nWe say that a topological space $X$ is \\textbf{compactly generated} if a subset $C$ is closed if and only if $C \\cap K$ is closed in $K$ for every compact set $K$ in $X$. Most spaces we work with are compactly generated, such as locally compact spaces, metric spaces, manifolds, and CW-complexes. Let $\\CGTop$ denote the full subcategory of $\\Top$ consisting of compactly generated topological spaces, and let $\\CGHaus$ denote the full subcategory of $\\CGTop$ consisting of Hausdorff spaces. Similarly define $\\CGTopp$ and $\\CGHausp$. There is a functor $k : \\Haus \\rightarrow \\CGHaus$ that re-topologizes a space $X$ by declaring a subset $C$ closed if and only if $C \\cap K$ is closed in $K$ for all compact subsets $K$ of $X$. One can show that $k$ is the identity on the subcategory $\\CGHaus$ and $k$ is the right adjoint of the inclusion functor $\\CGHaus \\rightarrow \\Haus$. The product in $\\CGHaus$ is not just the product in $\\Haus$ (the product of compactly generated spaces is not necessarily compactly generated), but we do have $X \\times_{\\CGHaus} Y = k(X \\times_\\Haus Y)$. Further, most importantly, the category $\\CGHaus$ \\emph{is} cartesian closed. The exponent object of $X$ and $Y$ is $k(Y^X)$, where $Y^X = \\hom_{\\Top}(X,Y)$ is given the compact-open topology. Similarly, $\\CGHausp$ is cartesian closed. We now have\n\\begin{prop}\n\\sloppyspace\n\\begin{enumerate}\n\t\\item For a space $X$ in $\\CGHaus$, the functor $-\\times X : \\CGHaus \\rightarrow \\CGHaus$ is the left adjoint of $-^X : \\CGHaus \\rightarrow \\CGHaus$. In particular, the function spaces $Z^{(X \\times Y)}$ and $(Z^Y)^X$ are naturally homeomorphic. \n\t\\item For a based space $X$ in $\\CGHausp$, the functor $-\\times X : \\CGHausp \\rightarrow \\CGHausp$ is the left adjoint of $-^X : \\CGHausp \\rightarrow \\CGHausp$. In particular, the function spaces $(Z,z_0)^{(X,x_0)\\times(Y,y_0)} = ((Z,z_0)^{(Y,y_0)})^{(X,x_0)}$.\n\\end{enumerate}\n\\end{prop}\nWe now see that $\\CGHaus$ and $\\CGHausp$ are very nice categories to do homotopy theory. Any time we consider a topological space, we are going to immediately apply the functor $k$ so that we can assume we always assume we are working in $\\CGHaus$. In fact, we are going to adopt this doctrine to the extreme by \\emph{redefining} $\\Top$ to be the category of compactly generated, Hausdorff spaces, and similarly for $\\Top_*$. \n\nThe category $\\Top^2$ consists of pairs of topological spaces $(X,A)$ such that $A \\subseteq X$, and a morphism $f : (X,A) \\rightarrow (Y,B)$ is a continuous map $f : X \\rightarrow Y$ such that $f(A) \\subseteq f(B)$. Similarly $\\Topp^2$ consists of pairs of topological spaces $(X,A)$ with base point in $A$ (and hence $X$ too), and a morphism is a morphism of pairs that preserves the base point. The product of pairs of spaces $(X,A)$ and $(Y,B)$ is simply $(X \\times Y,A \\times B)$, and the exponential $(Y,B)^{(X,A)}$ is the set of continuous maps $(X,A) \\rightarrow (Y,B)$. We could also define the category of $n$-tuples of spaces $\\Top^n$, but they will not be of use in these notes.\n\n\n\n\n\\subsection{Functors in Topology}\n\nSince $\\Top$ is cartesian closed, for a fixed space $X$ we get a functor $-^X : \\Top \\rightarrow \\Top$. For a map $f : Y \\rightarrow Z$, the induced map $f^X : Y^X \\rightarrow Z^X$ is defined by $f^X(g)(x) = f(g(x))$. If $X$ is based, then we get a functor $-^{(X,x_0)} : \\Topp \\rightarrow \\Topp$, where $(Y,y_0)^{(X,x_0)}$ is based at the constant map $y \\mapsto x_0$. For a based map $f : (Y,y_0) \\rightarrow (Z,z_0)$, the induced map is defined the exact same way as in the unbased case.\n\nWe saw earlier that the regular cartesian product is the categorical product in $\\Topp$ and serves as the left adjoint of $-^X$. There is another operation that is useful in $\\Topp$ and is the left adjoint of $X^-$. Given two pointed spaces $(X,x_0)$ and $(Y,y_0)$, we define their \\textbf{smash product} to be $X \\wedge Y = X \\times Y / X \\vee Y$, which is based at the equivalence class of $(x_0,y_0)$. If $f : X \\rightarrow Z$ and $g : Y \\rightarrow W$ are based, continuous maps, then $f \\wedge g : X \\wedge Y \\rightarrow Z \\wedge W$ is the based, continuous map that sends $[x,y]$ to $[f(x),g(y)]$. It is easy to show that this is well-defined, and so $\\wedge : \\Topp \\times \\Topp \\rightarrow \\Topp$ is a bifunctor. The following proposition makes precise the statement that the smash product in $\\Topp$ is like the tensor product in $\\RMod{R}$.\n\\begin{prop}\n\\label{wedge and exponential adjoint}\nFor a fixed space $Y$ in $\\Topp$, the functor $- \\wedge Y : \\Topp \\rightarrow \\Topp$ is the left adjoint of $Y^- : \\Topp \\rightarrow \\Topp$. In particular, the spaces\n\\[ \\hom_*(X \\wedge Y,Z) \\cong \\hom_*(X,Y^Z) \\]\nare naturally homeomorphic.\n\\end{prop}\n\nThe \\textbf{free path space} functor is just exponentiation by the unit interval, $-^I$, and the \\textbf{based path space} is just the based exponentiation by the unit interval based at 0, $-^{(I,0)}$. The \\textbf{free loop space} functor $\\Omega$ is just exponentiation by the circle, $\\Omega = -^{S^1}$, and the \\textbf{based loop space} functor $\\Omega_*$ is just exponentiation by the based circle, $\\Omega_* = -^{(S^1,1)}$, or equivalently, $\\Omega_* = -^{(I,\\lcb 0,1 \\rcb)}$, which is sometimes easier to user.\n\nThe \\textbf{free suspension} functor $\\Sigma : \\Top \\rightarrow \\Top$ assigns a space $X$ to $X \\times I / X \\times \\lcb 0,1 \\rcb$. Intuitively, we take the cylinder on $X$ and then pinch the ends. Given a continuous function $f : X \\rightarrow Y$, the induced map $\\Sigma f : \\Sigma X \\rightarrow \\Sigma Y$ is defined by $f([x,t]) = [f(x),t]$, which is easily verified to be well-defined. The \\textbf{based suspension} functor $\\Sigma_* : \\Topp \\rightarrow \\Topp$ assigns a space $X$ to $\\Sigma X / \\lcb x_0 \\rcb \\times I$, that is, we identify all the points $\\lcb x_0 \\rcb \\times I$ in the free suspension. We base $\\Sigma_* X$ at the equivalence class of $(x_0,0)$. Induced maps are defined in the exact same way, and are clearly well-defined. It is easy to see that for a based space $X$, the based suspesion can be written as $\\Sigma_* X = X \\wedge S^1$. Note that some people refer to $\\Sigma$ as the unreduced suspension and $\\Sigma_*$ as the reduced suspension.\n\nBy \\cref{wedge and exponential adjoint} we have\n\\begin{cor}\n$\\Sigma_*$ is the left adjoint of $\\Omega_*$. In particular, the spaces\n\\[ \\hom_*(\\Sigma_* X,Y) \\cong \\hom_*(X,\\Omega_* Y) \\]\nare naturally isomorphic.\n\\end{cor}\n\n\n\n\\subsection{The Homotopy Category}\n\n\nWe want to consider homotopies in each of our categories $\\Top,\\Topp,\\Top^2$ and $\\Topp^2$. This can be done in one fell swoop by letting $I$ simultaneously denote 1.) just the unit interval, 2.) the unit interval based at 0, 3.) the pair of spaces $(I,I)$ and finally, 4.) the pair of spaces $(I,I)$ based at 0. Which interpretation of $I$ we use depends on which category we are working in. If $\\mathscr C$ is any of our four categories of spaces, two morphisms $f,g : X \\rightarrow Y$ in $\\mathscr C$ are said to be homotopic in $\\mathscr C$ if there is a map $H : X \\times I \\rightarrow Y$ in $\\mathscr C$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$, and we write $f \\simeq g$. We will usually denote $H(-,t)$ by $H_t$. A homotopy in $\\Top$ is called a \\textbf{free homotopy} and in $\\Topp$ it is called a \\textbf{based homotopy}. A homotopy between maps $(X,A) \\rightarrow (Y,B)$ in $\\Top^2$ that fixes $A$ for all time, i.e. $H_t|A = \\id_A$, is called a \\textbf{homotopy relative $A$} and we write $f \\simeq g \\ (\\rel A)$. Note that a homotopy in $\\Topp$ could just as easily be defined as a map $H : X \\wedge I_+ \\rightarrow Y$ such that $H_0 = f$ and $H_1 = g$. \n\nThe notion of homotopy defines an equivalence relation on $\\hom_{\\mathscr C}(X,Y)$, and we call an equivalence class of maps a \\textbf{homotopy class}. The set of homotopy classes of between $X$ and $Y$ in $\\Top,\\Topp,\\Top^2,\\Topp^2$ is denoted by $[X,Y], [X,Y]_*, [(X,A),(Y,B)]$ and $[(X,A),(Y,B)]_*$ respectively. Note that $[X,Y]_*$ for based spaces is the same thing as $[(X,x_0),(Y,y_0)]$. We do not have a notation for the equivalence classes of maps up to homotopy (rel $A$), but it will be an important ingredient in these notes.\n\nLet $\\hTop$ denote the category whose objects are topological spaces, and $\\hom_{\\hTop}(X,Y) = [X,Y]$, and similarly for $\\Topp,\\Top^2,\\Topp^2$. These categories are called the homotopy category and based homotopy category of topological spaces and pairs of topological spaces. An isomorphism in one of these homotopy categories is called a \\textbf{homotopy equivalence}.\n\nThe functors $\\Omega$ and $\\Sigma$ (and their based counterparts) descend to functors on $\\hTop$ (and $\\hTopp$).\n\\begin{prop}\n\\label{suspension and loop space are adjoint}\n$\\Sigma_*$ is the left adjoint of $\\Omega_*$ in $\\hTopp$. In particular, the sets $[\\Sigma_* X,Y]_*$ and $[X,\\Omega_* Y]_*$ are naturally isomorphic.\n\\end{prop}\n\\begin{proof}\nNote that $[X,Y]_*$ is naturally isomorphic to $[S^0,\\hom_*(X,Y)]_*$, so \n\\[ [\\Sigma_*X,Y]_* \\cong [S^0,\\hom_*(\\Sigma_*X,Y)]_* \\cong [S^0,\\hom_*(X,\\Omega_*Y)]_* \\cong [X,\\Omega_*Y]_* \\]\nand all isomorphisms are natural.\n\\end{proof}\n\n\n\n\n\n\n\n\\subsection{$H$-groups and $H$-cogroups}\n\\label{H-groups and H-cogroups}\n\n\nTopological groups, spaces with a group structure such that the group operations are continuous, are very useful in geometry, but like most things from geometry they are far too restrictive for homotopy theory. We can reformulate the axioms of a group in terms of commutativity of certain digrams, and then define a homotopy theoretic topological group to be a space in which these diagrams only commute up to homotopy. Let $(X,x_0)$ be a based space, and let $x_0$ also denote the constant map $X \\rightarrow X$ that maps everything to $x_0$. A map $\\mu : X \\times X \\rightarrow X$ is said to be a \\emph{multiplication with homotopy identity $x_0$} if the following diagram\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX \\ar[r]^-{\\id \\times x_0} \\ar[rd]_\\id & X \\times X \\ar[d]_\\mu & X \\ar[l]_-{x_0 \\times \\id} \\ar[ld]^\\id \\\\\n\t & X &\n}\n\\]\ncommutes up to homotopy: $\\mu \\circ (\\id \\times x_0) \\simeq \\id \\simeq \\mu \\circ (x_0 \\times \\id)$. Note that $\\id \\times x_0$ is the map induced by $\\id$ and $x_0$ from the universal property of products. We say that $\\mu$ is \\emph{homotopy associative} if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX \\times X \\times X \\ar[r]^-{\\id \\times \\mu} \\ar[d]_-{\\mu \\times \\id} & X \\times X \\ar[d]^\\mu \\\\\n\tX \\times X \\ar[r]_\\mu & X\n}\n\\]\nA map $\\nu : X \\rightarrow X$ is said to be a \\emph{homotopy inverse} for $\\mu$ if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX \\ar[r]^-{\\nu \\times \\id} \\ar[rd]_{x_0} & X \\times X \\ar[d]_\\mu & X \\ar[l]_-{\\id \\times \\nu} \\ar[dl]^{x_0} \\\\\n\t& X &\n}\n\\]\nWe say that $(X,x_0)$ is an \\textbf{$H$-group} if it has a multiplication $\\mu$ with homotopy identity and inverse $\\nu$ such that $\\mu$ is homotopy associative. Further, we say that $X$ is a \\textbf{commutative $H$-group} if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n{\n\tX \\times X \\ar[rr]^-\\tau \\ar[rd]_\\mu & & X \\times X \\ar[dl]^\\mu \\\\\n\t & X\n}\n\\]\nwhere $\\tau$ is the flip map $\\tau(x,y) = (y,x)$. We think of $H$-groups as groups where the axioms only have to hold up to homotopy. An \\textbf{$H$-homomorphism} between $H$-groups $(X,x_0,\\mu_X)$ and $(Y,y_0,\\mu_Y)$ is a based map $f : X \\rightarrow Y$ such that the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=3pc\n{\n\tX \\times X \\ar[r]^-{\\mu_X} \\ar[d]_{f \\times f} & X \\ar[d]^f \\\\\n\tY \\times Y \\ar[r]_{\\mu_Y} & Y\n}\n\\]\n\nBy reversing all arrows and replacing products with coproducts we get the dual notion of $H$-cogroups, but we will also spell out the details. Let $(X,x_0)$ be a pointed space and let $x_0$ also denote the constant map. A map $\\mu' : X \\rightarrow X \\vee X$ is said to be a \\emph{comultiplication with homotopy coidentity $x_0$} if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX & X \\vee X \\ar[l]_-{\\id \\vee x_0} \\ar[r]^-{x_0 \\vee \\id} & X \\\\\n\t & X \\ar[lu]^\\id \\ar[ru]_\\id \\ar[u]^{\\mu'}\n}\n\\]\nwhere $\\id \\vee x_0$ is the map induced by $\\id$ and $x_0$ from the universal property of coproducts. We say that $\\mu$ is \\emph{homotopy coassociative} if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX \\vee X \\vee X & X \\vee X \\ar[l]_-{\\id \\vee \\mu'} \\\\\n\tX \\vee X \\ar[u]^{\\mu' \\vee \\id} & X \\ar[u]_{\\mu'} \\ar[l]^-{\\mu'}\n}\n\\]\nA map $\\nu' : X \\rightarrow X$ is said to be a \\emph{homotopy coinverse} for $\\mu'$ if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=4pc\n{\n\tX & X \\vee X \\ar[l]_{\\nu' \\vee \\id} \\ar[r]^{\\id \\vee \\nu'} & X \\\\\n\t & X \\ar[ul]^{x_0} \\ar[ur]_{x_0} \\ar[u]^{\\mu'}\n}\n\\]\nWe say that $(X,x_0)$ is an \\textbf{$H$-cogroup} if it has a comultiplication $\\mu'$ with homotopy coidentity and coinverse $\\nu'$ such that $\\mu'$ is homotopy coassociative. Further, we say that $X$ is a \\textbf{cocommutative $H$-cogroup} if the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n{\n\tX \\vee X \\ar[rr]^-\\tau & & X \\vee X \\\\\n\t& X \\ar[ul]^{\\mu'} \\ar[ur]_{\\mu'}\n}\n\\]\nwhere $\\tau$ is the flip map $\\tau(x,x_0) = (x_0,x)$ and $\\tau(x_0,x) = (x,x_0)$. An \\textbf{$H$-cohomomorphism} between $H$-cogroups $(X,x_0,\\mu_X')$ and $(Y,y_0,\\mu_Y')$ is a based map $f : X \\rightarrow Y$ such that the following diagram commutes up to homotopy\n\\[\n\\xymatrix\n@R=3pc\n@C=3pc\n{\n\tX \\vee X \\ar[d]_{f \\vee f} & X \\ar[l]_-{\\mu_X'} \\ar[d]^f \\\\\n\tY \\vee Y & Y \\ar[l]^-{\\mu_Y'}\n}\n\\]\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\comment{\n\\note{should all the stuff down here just be in the next section?}\n\nGiven two paths $\\alpha,\\beta : I \\rightarrow X$ such that $\\alpha(1) = \\beta(0)$, we can define their juxtaposition to be the path $\\beta*\\alpha : I \\rightarrow X$ defined by\n\\[ (\\beta * \\alpha)(t) = \\begin{cases} \\alpha(2t) & 0 \\leq t \\leq 1/2 \\\\ \\beta(2t-1) & 1/2 \\leq t \\leq 1 \\end{cases} \\]\nThis just means to traverse $\\alpha$ at double speed for half the time, and then traverse $\\beta$ at double speed for the rest of the time. If $\\alpha : I \\rightarrow X$ is a path from $a$ to $b$, then we define its \\emph{reverse} to be the path $\\overline\\alpha(t) = \\alpha(1-t)$, which goes from $b$ to $a$. Note that the loops $\\overline\\alpha * \\alpha$ and $\\alpha * \\overline\\alpha$ are homotopic to the constant loop relative end points. For a topological space $X$ we define $\\Pi(X)$, called the \\textbf{fundamental groupoid}, to be the category with objects points in $X$ and morphisms homotopy classes of paths relative end points. There is a lot more that can be said about these operations and constructions, but it will be covered in much greater generality in \\cref{H-groups and H-cogroups}. \n\n% This operation is not associative, but we do have $\\gamma * (\\beta * \\alpha) \\simeq (\\gamma * \\beta) * \\alpha (\\rel \\lcb 0,1 \\rcb)$, whenever this is defined. If we let $P_{a,b} X$ denote the space of paths that start at $a$ and end at $b$, then this operation is a map $* : P_{a,b} X \\times P_{b,c} X \\rightarrow P_{a,c} X$. In fact, this operation descends to the space of homotopy classes of paths relative end points. \n}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\n\n\n\\begin{prop}\n\\sloppyspace\n\\begin{enumerate}\n\t\\item The based loop space $\\Omega_* X$ has a canonical $H$-group structure. Also, $\\Omega_* f$ is an $H$-group homomorphism for any based map $f : X \\rightarrow Y$.\n\t\\item The based suspension $\\Sigma_* X$ has a canonical $H$-cogroup structure. Also, $\\Sigma_* f$ is an $H$-cogroup cohomomorphism for any based map $f : X \\rightarrow Y$.\n\\end{enumerate}\n\\end{prop}\n\\begin{proof}\n\\sloppyspace\n\\begin{enumerate}\n\n\t\\item Let $\\alpha,\\beta$ be two paths in $X$ based at $x_0$. We define their product $\\mu(\\alpha,\\beta) = \\beta * \\alpha$ to be the path\n\t\\[ (\\beta * \\alpha)(t) = \\begin{cases} \\alpha(2t) & 0 \\leq t \\leq 1/2 \\\\ \\beta(2t-1) & 1/2 \\leq t \\leq 1 \\end{cases} \\]\n\tThis just means to traverse $\\alpha$ at double speed for half the time, and then traverse $\\beta$ at double speed for the rest of the time. First we show that the constant loop $x_0$ is a homotopy identity for $*$. We can find a homotopy $H$ between $\\alpha$ and $\\alpha * x_0$ by defining\n\t\\[ H(s,t) = \\begin{cases} x_0 & 0 \\leq s \\leq \\frac{t}{2} \\\\ \\alpha\\left( \\frac{2s-t}{2-t} \\right) & \\frac{t}{2} \\leq s \\leq 1 \\end{cases} \\]\n\tA similar homotopy can be defined to show $\\alpha \\simeq x_0 * \\alpha$, so $x_0$ is a homotopy identity for $*$. We define the homotopy inverse by $\\nu(\\alpha)(t) = \\overline{\\alpha}(t) = \\alpha(1-t)$. To show this is indeed the inverse, we use the following homotopy between $x_0$ and $\\alpha * \\overline\\alpha$\n\t\\[ H(s,t) = \\begin{cases} \\alpha(2st) & 0 \\leq s \\leq \\frac{1}{2} \\\\ \\alpha(2t(1-s)) & \\frac{1}{2} \\leq s \\leq 1 \\end{cases} \\]\n\tWe can do something similar to show $x_0 \\simeq \\overline\\alpha * \\alpha$. Finally, to show $*$ is homotopy associative we use the following homotopy between $\\gamma * (\\beta * \\alpha)$ and $(\\gamma * \\beta) * \\alpha$\n\t\\[ H(s,t) = \\begin{cases} \\end{cases} \\]\n\tFor a based map $f : X \\rightarrow Y$ we clearly have $f(\\beta*\\alpha) = f(\\beta) * f(\\alpha)$, so $\\Omega_* f$ is an $H$-group homomorphism.\n\t\n\t\\item Thinking of $\\Sigma_* X$ as $X \\times I / X \\times \\lcb 0,1 \\rcb \\cup \\lcb x_0 \\rcb \\times I$, we define the coproduct of $[x,t]$ to be\n\t\\[ \\mu'([x,t]) = \\begin{cases} ([x,2t],x_0) & 0 \\leq t \\leq \\frac{1}{2} \\\\ (x_0,[x,2t-1]) & \\frac{1}{2} \\leq t \\leq 1 \\end{cases} \\]\n\tThe inverse is defined by $\\nu'([x,t]) = [x,1-t]$. The homotopies that show these operations make $\\Sigma_* X$ into an $H$-cogroup are essentially the same as the homotopies we used for $\\Omega_* X$.\n\t\n\\end{enumerate}\n\\end{proof}\n\n\n\\begin{prop}\nLet $(X,x_0)$ be an $H$-cogroup and $(Y,y_0)$ an $H$-group. Then for any based spaces $(Z,z_0)$ and $(W,w_0)$ and based map $f : Z \\rightarrow W$ we have\n\\begin{enumerate}\n\t\\item The set $[(Z,z_0),(Y,y_0)]$ has a canonical group structure, and the induced map $f_* : [(W,w_0),(Y,y_0)] \\rightarrow [(Z,z_0),(Y,y_0)]$ is a homomorphism of groups.\n\t\\item The set $[(X,x_0),(Z,z_0)]$ has a canonical group structure, and the induced map $f_* : [(X,x_0),(Z,z_0)] \\rightarrow [(X,x_0),(W,w_0)]$ is a homomorphism of groups.\n\t\\item The group structures induced on $[(X,x_0),(Y,y_0)]$ by the above are \\emph{equal} (not just isomorphic), and this group is abelian.\n\\end{enumerate}\n\\end{prop}\n\\begin{proof}\n\\sloppyspace\n\\begin{enumerate}\n\n\t\\item Let $\\mu$ and $\\nu$ denote the $H$-group operations on $Y$. Given two maps $f,g : (Z,z_0) \\rightarrow (Y,y_0)$, we define their product $g * f$ as the composition\n\t\\[ Z \\stackrel{\\Delta}{\\longrightarrow} Z \\times Z \\stackrel{f \\times g}{\\longrightarrow} Y \\times Y \\stackrel{\\mu}{\\longrightarrow} Y \\]\n\twhere $\\Delta$ is the diagonal map $\\Delta(z) = (z,z)$. Note that we have switched the order of $f$ and $g$ in this composition since intuitively we want to think of $g * f$ as first apply $f$, then apply $g$. The inverse of $f$ is defined as $f^{-1} = \\nu \\circ f$, and the identity is the constant map $[c_{x_0}]$ at $x_0$. This operation descends to the homotopy classes of maps $[Z,Y]_*$, for if $F$ is a based homotopy between $f$ and $f'$ and $G$ is a based homotopy between $g$ and $g'$, then\n\t\\[ K(z,t) = \\mu(F(z,t),G(z,t)) \\]\n\tis a homotopy between $g * f$ and $g' * f'$, and\n\t\\[ I(z,t) = \\nu(F(z,t)) \\]\n\tis a homotopy between $\\nu \\circ f$ and $\\nu \\circ f'$. Show that these operations satisfy the group axioms is obvious.\n\t\n\t\\item Let $\\mu'$ and $\\nu'$ denote the $H$-cogroup operations on $X$. Given two maps $f,g : (X,x_0) \\rightarrow (Z,z_0)$, we define their product $g \\cdot f$ as the composition\n\t\\[ X \\stackrel{\\mu'}{\\longrightarrow} X \\vee X \\stackrel{f \\vee g}{\\longrightarrow} Z \\vee Z \\stackrel{\\Delta'}{\\longrightarrow} Z \\]\n\twhere $\\Delta'$ is the folding map $\\Delta'(z,z_0)=z=\\Delta'(z_0,z)$. Note that we have switched the order of $f$ and $g$ in this composition since intuitively we want to think of $g \\cdot f$ as first apply $f$, then apply $g$. The inverse of $f$ is defined as $f^{-1} = f \\circ \\nu'$, and the identity is the constant map $c_{x_0}$ at $x_0$. Showing that this operation descends to $[X,Z]_*$ is essentially the same as before, and the group axioms obviously hold.\n\t\n\t\\item We can see from the above that the homotopy class of the constant map $[x_0]$ is a mutual identity for $*$ and $\\cdot$. A very lengthly calculation shows that the operations distribute over each other, i.e. $([f] \\cdot [g]) * ([h] \\cdot [k]) = ([f] * [h]) \\cdot ([g] \\cdot [k])$. So the result follows from the following general lemma.\n\t\n\\end{enumerate}\n\\end{proof}\n\n\\begin{lem}[Eckmann-Hilton Principle]\n\\label{Eckmann-Hilton Principle}\n\\label{set with two binary operations}\nIf a set $X$ has two binary operations $\\cdot$ and $*$ defined such that the following holds.\n\\begin{enumerate}\n\t\\item There exists a mutual identity $e$, i.e. $x \\cdot e = e \\cdot x = x = x * e = e * x$.\n\t\\item The operations distribute over each other, i.e. $(x \\cdot y) * (w \\cdot z) = (x * w) \\cdot (y * z)$.\n\\end{enumerate}\nThen the binary operations are equal, associative and commutative.\n\\end{lem}\n\\begin{proof}\nThe operations are equal since\n\\[ x \\cdot y = (x * e) \\cdot (y * e) = (x * y) \\cdot (e * e) = x * y \\]\nAssociativity follows from\n\\[ x * (y * z) = (x \\cdot e) * (y \\cdot z) = (x * y) \\cdot (e * z) = (x \\cdot y) \\cdot z \\]\nCommutativity follows from\n\\[ x \\cdot y = (x * e) \\cdot (y * e) = (e * x) \\cdot (y * e) = (e \\cdot y) * (x * e) = y * x = y \\cdot x \\]\n\\end{proof}\n\nThus we have that $[-,(Y,y_0)] : \\hTopp \\rightarrow \\Grp$ is a cofunctor if $Y$ is an $H$-group, and $[(X,x_0),-] : \\hTopp \\rightarrow \\Grp$ is a functor if $X$ is an $H$-cogroup. \n\n\n\n\n\n\n\\newpage\n\\section{Fibrations}\n\n\\subsection{The Universal Test Diagram}\n\nIn geometry one naturally comes across fiber bundles when a collection of spaces is parameterized by another space. We want such a thing in homotopy theory too, but requiring local triviality is far too restrictive for the homotopy category. However, fiber bundles have another property that is homotopy theoretic, called the homotopy lifting property. A continuous map $p : E \\rightarrow B$ is said to have the \\textbf{homotopy lifting property} with respect to a space $X$ if for every continuous map $H : X \\times I \\rightarrow B$ and $f : X \\rightarrow E$ there is a map $\\tilde{H} : X \\times I \\rightarrow E$ such that the following diagram commutes\n\\begin{equation}\n\\label{fibration test diagram}\n\\xymatrix\n@R=2.5pc\n@C=2.5pc\n{\n\tX \\ar[d]_{i_0} \\ar[r]^f & E \\ar[d]^{p} \\\\\n\tX \\times I \\ar[r]_H \\ar@{-->}[ru]^{\\tilde H} & B\n}\n\\end{equation}\nwhere $i_0$ is the inclusion of $X$ into $X \\times \\lcb 0 \\rcb$. We call this a \\emph{test diagram} for the homotopy lifting problem. Commutativity of the lower triangle means that $\\tilde H_t$ is a left of $H_t$ for all time $t$, and commutativity of the upper triangle means that the initial lift $\\tilde H_0$ is precisely the given map $f$. We say that $p : E \\rightarrow B$ is a \\textbf{fibration} if it has the homotopy lifting property with respect to all spaces $X$, and we say it is a \\textbf{Serre fibration} if it has the homotopy lifting property with respect to all discs $D^n$. Clearly such a map must be surjective. The fiber over a point $b \\in B$ in the base is denoted by $E_p = p^{-1}(b)$.\n\n\\begin{example}\n\\label{loop space and path space fibrations}\nThe prototypical example of a fibration is the \\emph{path space fibration}. Let $e_0 : B^I \\rightarrow B$ be evaluation at 0, $e_0(\\alpha)=\\alpha(0)$. Consider the homotopy lifting problem\n\\begin{equation}\n\\label{path space fibration test diagram}\n\\xymatrix\n@R=2.5pc\n@C=2.5pc\n{\n\tX \\ar[d]_{i_0} \\ar[r]^f & B^I \\ar[d]^{e_0} \\\\\n\tX \\times I \\ar[r]_-H \\ar@{-->}[ur]^{\\tilde H} & B\n}\n\\end{equation}\nWe need to construct $\\tilde H$ such that $\\tilde H(x,0)(s) = f(x)(s)$ and $\\tilde H(x,t)(0) = H(x,t)$. The following function accomplishes this\n\\begin{equation}\n\\label{homotopy interpolation trick}\n\\tilde H(x,t)(s) = \\begin{cases} H(x,1-(1+t)s) & 0 \\leq s \\leq \\frac{1}{1+t} \\\\ f(x)((1+t)s) & \\frac{1}{1+t} \\leq s \\leq 1 \\end{cases}\n\\end{equation}\nThis function is continuous since on the overlaps $s=\\frac{1}{1+t}$ we have that the first piece is $H(x,0)$ and the second piece is $f(x)(0)$, which are equal by commutativity of \\eqref{path space fibration test diagram}. Therefore $e_0 : B^I \\rightarrow I$ is a fibration. The fiber $p^{-1}(b)$ of any point in the base is the space of paths in $B$ that start at $b$, i.e. $(B,b)^{(I,0)}$. \n\nThe same argument shows that $e_0 : (B,b_0)^{(I,1)} \\rightarrow (B,b_0)$ is a fibration, called the \\emph{based path space fibration}. The fiber over some point $b \\in B$ is the space of paths starting at $b$ and ending at $b_0$. In particular, the fiber over $b_0$ is just the based loop space $\\Omega_* (B,b_0)$.\n\\end{example}\n\n\\begin{prop}\nA fiber bundle over a paracompact space is a fibration.\n\\end{prop}\n\n\\begin{prop}\nThe pullback of a fibration is a fibration.\n\\end{prop}\n\\begin{proof}\nLet $p : E \\rightarrow B$ be a fibration and let $g : Y \\rightarrow B$ be continuous. Consider the following test diagram for a homotopy lifting problem juxtaposed with the pullback square\n\\[\n\\xymatrix\n@R=3pc\n@C=3pc\n{\n\tX \\ar[d]_{i_0} \\ar[r]^-f & E \\times_B Y \\ar[d]_{p'} \\ar[r]^-{g'} & E \\ar[d]^p \\\\\n\tX \\times I \\ar[r]_-H & Y \\ar[r]_g & B\n}\n\\]\nThen the homotopy lifting problem $g \\circ H$ and $g' \\circ f$ can be solved since $E \\rightarrow B$ is a fibration, so we get a map $G : X \\times I \\rightarrow E$ such that $G \\circ i_0 = g' \\circ f$ and $p \\circ G = g \\circ H$. But, the maps $G$ and $H$ induce a map $\\tilde H : X \\times I \\rightarrow E \\times_B Y$ uniquely determined by the property that $p' \\circ \\tilde H = H$ and $g' \\circ \\tilde H = G$ (by the universal property of pullbacks). See the diagram\n\\[\n\\xymatrix\n@R=1pc\n@C=1pc\n{\n\t& & & & & \\\\\n\t& X \\ar[dd]_{i_0} \\ar[rr]^-f & & E \\times_B Y \\ar[dd]_{p'} \\ar[rr]^-{g'} & & E \\ar[dd]^p \\\\\n\t& \\\\\n\t& X \\times I \\ar[rr]_-H \\ar@{-->}[uurr]^{\\tilde H} \\ar@{-->}`l/2pt[l]`u[luuu]`r[rrrruuu]^{G}[rrrruu] & & Y \\ar[rr]_g & & B\n}\n\\]\nBy the universal property of pullbacks we have that $f$ is the unique map that satisfies $p' \\circ f = H \\circ i_0$. However, $\\tilde H \\circ i_0$ satisfies this equation since $p' \\circ \\tilde H \\circ i_0 = H \\circ i_0$, hence $\\tilde H \\circ i_0 = f$, and we have shown that $p' : E \\times_B Y \\rightarrow Y$ is a fibration.\n\\end{proof}\n\nThere is a universal test diagram such that if it has a solution, then all test diagrams will have a solution. First note that by using the adjoint of $- \\times I$, which is $-^I$, we can rewrite the test diagram \\eqref{fibration test diagram} as\n\\begin{equation}\n\\label{alternative fibration test diagram}\n\\xymatrix\n{\n\tE \\ar[dd]_p & & E^I \\ar[dd]^{p^I} \\ar[ll]_{e_0} \\\\\n\t & X \\ar[ul]^f \\ar[rd]_H \\ar@{-->}[ru]^{\\tilde H} & \\\\\n\tB & & B^I \\ar[ll]^{e_0}\n}\n\\end{equation}\nwhere $e_0$ denotes the evaluation at $t=0$ map. \n\n\\begin{prop}\nLet $p : E \\rightarrow B$ be a map and let $e_0' : E \\times_B B^I \\rightarrow E$ and $p' : E \\times_B B^I \\rightarrow B^I$ denote the induced maps by the pullback of $E \\stackrel{p}{\\rightarrow} B \\stackrel{e_0}{\\leftarrow} B^I$. If \\eqref{alternative fibration test diagram} can be solved for $X = E \\times_B B^I$, $f = e_0'$ and $H = p'$, then any test diagram can be solved.\n\\end{prop}\n\\begin{proof}\nLet $\\tilde H$ be a solution of \\eqref{alternative fibration test diagram} for $X = E \\times_B B^I$, $f=e_0'$ and $H=p'$. Then take any other space $X$ and consider its test diagram. By the universal property of pullbacks there is a unique map $\\varepsilon : X \\rightarrow E \\times_B B^I$ making the following commute\n\\[\n\\xymatrix\n{\n\tE \\ar[ddd]_p & & & E^I \\ar[ddd]^{p^I} \\ar[lll]_{e_0} \\\\\n\t & & E \\times_B B^I \\ar[ull]^{e_0'} \\ar[rdd]_{p'} \\ar@{-->}[ru]^{\\tilde H} & \\\\\n\t & X \\ar@{-->}[ur]^\\varepsilon \\ar[uul]^f \\ar[drr]_H & & \\\\\n\tB & & & B^I \\ar[lll]^{e_0}\n}\n\\]\nClearly $\\tilde H \\circ \\varepsilon$ is now a solution of the diagram for $X$. \n\\end{proof}\n\n\nTherefore if \\ref{alternative fibration test diagram} can be solved for $E \\times_B B^I$, then it can be solved for any space, and so $p : E \\rightarrow B$ would be a fibration. If $f : X \\rightarrow Y$ is any map (not necessarily a fibration), then the pullback $X \\times_Y Y^I$ is called the \\textbf{mapping path space} of $f$, and concretely is defined as\n\\begin{equation}\n\\label{mapping path space}\nP_f = \\lcb (x,\\alpha) \\in X \\times Y^I \\st f(x) = \\alpha(0) \\rcb\n\\end{equation}\n\nFor a map $p : E \\rightarrow B$ (not necessarily a fibration), let $s : E^I \\rightarrow P_p$ be the function $s(\\alpha) = (\\alpha(0), p \\circ \\alpha)$. A function $\\lambda : P_p \\rightarrow E^I$ such that $s \\circ \\lambda = \\id$ is called a \\textbf{path lifting function}. Given the initial data $(e,\\alpha) \\in E \\times B^I$ such that $\\alpha(0)=p(e)$, it gives us a lift to a path $\\tilde\\alpha = \\lambda(e,\\alpha)$ such that $\\tilde\\alpha(0)=e$ and $p \\circ \\tilde\\alpha = \\alpha$. \n\\begin{cor}\nA map $p : E \\rightarrow B$ is a fibration if and only if it has a path lifting function.\n\\end{cor}\n\\begin{proof}\nPath lifting functions provide solutions to the universal test diagram and conversely.\n\\end{proof}\n\n\n\\begin{example}\nLet $p : D^2 \\rightarrow D^1$ be the map $p(x,y) = x$, i.e. the projection of the unit disc onto the unit interval. This is clearly not a fiber bundle since $p^{-1}(1)$ is a point and $p^{-1}(0)$ is an interval. However, it is a fibration as it is easy to construct a path lifting function. \n\\end{example}\n\n\\subsection{The Homotopy Fiber}\n\nLocal triviality in fiber bundles gives us that the fiber above all points in the base space are homeomorphic (assuming the base is path connected). We will not have this with our weakened notion of fibration, but we get something good enough for homotopy theory. Let $\\alpha : I \\rightarrow B$ be a path, and consider the diagram\n\\begin{equation}\n\\label{translation of fibers diagram}\n\\xymatrix\n@R=3pc\n@C=3pc\n{\n\tE_{\\alpha(0)} \\ar@{^{(}->}[rr]^-i \\ar[d]_{i_0} & & E \\ar[d]^p \\\\\n\tE_{\\alpha(0)} \\times I \\ar[r]_-{\\pi_I} \\ar@{-->}[urr]^{\\tilde H} & I \\ar[r]_\\alpha & B\n}\n\\end{equation}\nso that the composition of the bottom row is just the constant homotopy $(e,t) \\mapsto \\gamma(t)$. We can solve this homotopy lifting problem since $p$ is a fibration, so let $\\tilde H$ be this lifted homotopy. Then, since $p \\circ \\tilde H_1 = \\alpha(1)$ we have that $\\tilde H_1$ maps $E_{\\alpha(0)}$ to $E_{\\alpha(1)}$. Let $\\alpha_*$ denote this map; it is called the \\emph{translation of fibers} along $\\alpha$.\n\\begin{prop}\n\\label{fibration fibers are homotopy equivalent}\nThe map $\\alpha_* : E_{\\alpha(0)} \\rightarrow E_{\\alpha(1)}$ depends only on the homotopy class of $\\alpha$ relative end points, and is a homotopy equivalence. Further, if $\\gamma$ is a curve with $\\alpha(1) = \\gamma(0)$, then $(\\gamma * \\alpha)_* \\simeq \\gamma_* \\circ \\alpha_*$.\n\\end{prop}\n\\begin{proof}\nLet $\\alpha,\\beta : I \\rightarrow B$ be curves from $a$ to $b$ that are homotopic relative end points. Let $F : I \\times I \\rightarrow B$ be such a homotopy, i.e. $F_0 = \\alpha$ and $F_1 = \\beta$. These curves determine homotopies $\\tilde H^\\alpha, \\tilde H^\\beta : E_a \\times I \\rightarrow E$ that lift \n$\\alpha \\circ \\pi_I$ and $\\beta \\circ \\pi_I$ respectively. We want to show that $\\tilde H_1^\\alpha$ is homotopic to $\\tilde H_1^\\beta$, and that both of these maps are homotopy equivalences. If we let $J$ be the space\n\\[ J = \\lcb 0,1 \\rcb \\times I \\cup I \\times \\lcb 0 \\rcb \\subset I^2 \\]\nthen there is a homeomorphism of pairs $\\varphi : (I^2,I \\times \\lcb 0 \\rcb) \\rightarrow (I^2,J)$. Note that this homeomorphism of pairs implies $\\iota \\circ \\varphi = i_0$, where $\\iota$ is the inclusion $E_a \\times J \\hookrightarrow E_a \\times I \\times I$ and $i_0$ is the inclusion $E_a \\times I \\times \\lcb 0 \\rcb \\hookrightarrow E_a \\times I \\times I$. \n\nLet $f : E_a \\times J \\rightarrow E$ be the map that is $\\tilde H^\\alpha$ on $E_a \\times \\lcb 0 \\rcb \\times I$, $\\tilde H^\\beta$ on $E_a \\times \\lcb 1 \\rcb \\times I$, and $i \\circ \\pi_I$ on $E_a \\times I \\times \\lcb 0 \\rcb$. Then the following homotopy lifting problem\n\\[\n\\xymatrix\n@C=2.5pc\n@R=3pc\n{\n\tE_a \\times I \\times \\lcb 0 \\rcb \\ar[d]_{i_0} \\ar[r]^-\\varphi & E_a \\times J \\ar[r]^-f \\ar[dl]_{\\iota} & E \\ar[d]^p \\\\\n\tE_a \\times I \\times I \\ar[r]_-{\\pi_{I \\times I}} \\ar@{-->}[rru]^{\\tilde F} & I \\times I \\ar[r]_-F & B\n}\n\\]\ncan be solved to give a map $\\tilde F : E_a \\times I \\times I \\rightarrow E$ such that the whole diagram commutes. But then we see that $\\tilde F = f$ when restricted to $E_a \\times J$, so we have $\\tilde F(e,0,1) = f(e,0,1) = \\tilde H_1^\\alpha$ and $\\tilde F(e,1,1) = f(e,1,1) = \\tilde H_1^\\beta$, hence $\\tilde F(-,-,1) : E_a \\times I \\rightarrow E_b$ is a homotopy between $\\tilde H_1^\\alpha$ and $\\tilde H_1^\\beta$.\n\nIf $c_a$ denotes the constant path at $a$, then $c_{a*} : E_a \\rightarrow E_a$ is just the identity since the constant homotopy $(e,t) \\mapsto e$ is a lift in \\eqref{translation of fibers diagram}. If $\\gamma$ is a curve with $\\alpha(1)=\\gamma(0)$ and $\\tilde H^\\alpha$ and $\\tilde H^\\gamma$ are lifts like in \\eqref{translation of fibers diagram}, then $\\tilde H^\\gamma * \\tilde H^\\alpha$ is a lift of $\\gamma * \\alpha$, where this juxtaposition happens in the second variable. Therefore $(\\gamma * \\alpha)_* \\simeq \\gamma_* \\circ \\alpha_*$. Finally, we have $\\id_{E_a} \\simeq (c_a)_* \\simeq (\\alpha * \\overline\\alpha)_* = \\alpha_* \\circ \\overline\\alpha_*$, and similarly the other way around, so $\\alpha_*$ and $\\overline\\alpha_*$ are homotopy inverses of each other.\n\\end{proof}\n\n\\begin{cor}\n\\label{fiber of a fibration}\nAll fibers of a fibration over a path connected space are homotopy equivalent.\n\\end{cor}\n\n\\begin{cor}\nThe association $\\alpha \\mapsto \\alpha_*$ defines a functor $\\Pi(B) \\rightarrow \\hTop$. In particular, it defines a group homomorphism $\\pi_1(B) \\rightarrow \\Aut_{\\hTop}(E_a)$.\n\\end{cor}\n\nIn view of \\cref{fiber of a fibration}, we call the pre-image of any point in the base the \\textbf{fiber} of the fibration, although this definition only makes sense \\emph{up to homotopy}. A morphism of fibrations $p : E \\rightarrow B$ and $p' : E' \\rightarrow B'$ is a pair of maps $f : E \\rightarrow E'$ and $f' : B \\rightarrow B'$ such that $f' \\circ p = p' \\circ f$. Thus we can define the category of fibrations, denoted by $\\Fib{}$, and it is easy to see that the categorical product in $\\Fib{}$ is just $E \\times E' \\stackrel{p \\times p'}{\\longrightarrow} B \\times B'$. We will usually be concerned with the morphisms of fibrations that are of the form $B=B'$ and $f'=\\id$. We denote the full subcategory of fibrations over $B$ by $\\Fib{B}$. By thinking of the identity map on $I$ as a fibration, we can define the notation of a homotopy between morphisms in $\\Fib{}$, which we will call a \\textbf{fiber homotopy}, as well as fiber homotopy equivalences. The following is a useful result, but the proof is long, so we omit it.\n\n\\begin{prop}\n\\label{homotopy equivalence of fibrations is a fiber homotopy equivalence}\nIf $p : E \\rightarrow B$ and $p' : E' \\rightarrow B$ are fibrations, and $f : E \\rightarrow E'$ a homotopy equivalence such that $p = p' \\circ f$, then $f$ is a fiber homotopy equivalence.\n\\end{prop}\nThis theorem is somewhat surprising. Since $f$ is a homotopy equivalence, we can find a homotopy inverse $g : E' \\rightarrow E$, but there is no guarantee that $g$ preserves the fibers. This proposition ensures that we can homotope $g$ to make it preserve the fibers.\n\nThere is a canonical way of converting any map $f : X \\rightarrow Y$ into a fibration, at least up to homotopy. Consider the maps $h : X \\rightarrow P_f$, $h(x) = (x,c_{f(x)})$, and $p : P_f \\rightarrow Y$, $p(x,\\alpha)$, $p(x,\\alpha)=\\alpha(1)$, where $c_{f(x)}$ denotes the constant path at $f(x)$. Then the composition $X \\stackrel{h}{\\rightarrow} P_f \\stackrel{p}{\\rightarrow} Y$ is equal to $f$. \n\\begin{prop}\n\\label{canonical fibration}\n$h$ is a homotopy equivalence and $p$ is a fibration. If $f$ is already a fibration, then $h$ is a fiber homotopy equivalence between the fibrations $f$ and $p$.\n\\[\n\\xymatrix\n{\n\tX \\ar[rr]^h \\ar[rd]_f & & P_f \\ar[ld]^p \\\\\n\t& Y &\n}\n\\]\n\\end{prop}\n\\begin{proof}\nLet $k : P_f \\rightarrow X$ be defined by $k(x,\\alpha) = x$. Then $k \\circ h = \\id$, and $h \\circ k \\simeq \\id$ via the homotopy $H : P_f \\times I \\rightarrow X$\n\\[ H((x,\\alpha),t) = (x,\\alpha^t) \\]\nwhere $\\alpha^t$ is the path defined by $\\alpha^t(s)=\\alpha(st)$. We can easily see that $H_0 = h \\circ k$ and $H_1 = \\id$, so $h$ is a homotopy equivalence.\n\nTo see that $p$ is a fibration, consider the following test diagram\n\\[\n\\xymatrix\n@R=2.5pc\n@C=2.5pc\n{\n\tX \\ar[r]^g \\ar[d]_{i_0} & P_f \\ar[d]^p \\\\\n\tX \\times I \\ar[r]_-H \\ar@{-->}[ru]^-{\\tilde H} & Y\n}\n\\]\nWe can write $g(x) = (g_1(x),g_2(x))$, where $g_1 : X \\rightarrow X$ and $g_2 : X \\rightarrow Y^I$, and such that $f(g_1(x)) = g_2(x)(0)$. Similarly, the lift $\\tilde H : X \\times I \\rightarrow P_f$ that we need to construct will have two components $\\tilde H(x,t) = (\\tilde H_1(x,t),\\tilde H_2(x,t))$, where $\\tilde H_1 : X \\times I \\rightarrow X$ and $\\tilde H_2 : X \\times I \\rightarrow Y^I$. Commutativity of the upper triangle means we need $\\tilde H_1(-,0)=g_1, \\tilde H_2(-,0)=g_2$, and commutativity of the lower triangle means we need $\\tilde H_2(x,t)(1)=H(x,t)$. \n\nSo, we can just take $\\tilde H_1$ to be the constant homotopy $\\tilde H_1(x,t) = g_1(x)$. To define $\\tilde H_2$ we use the standard trick for interpolating between two functions that we used in \\eqref{homotopy interpolation trick}. In particular, we define\n\\[ \\tilde H_2(x,t)(s) = \\begin{cases} g_2(x)((1+t)s) & 0 \\leq s \\leq \\frac{1}{1+t} \\\\ H(a,(1+t)s-1) & \\frac{1}{1+t} \\leq s \\leq 1 \\end{cases} \\]\nIt is easy to see that this satisfies the conditions we need, and it is continuous on the overlap $s = \\frac{1}{1+t}$ since $g_2(x)(1) = H(a,0)$ by commutativity of the square in the test diagram.\n\\end{proof}\n\nSo, every map can be factored into a homotopy equivalence followed by a fibration, and we call it the \\textbf{canonical fibration} of $f$. We define the \\textbf{homotopy fiber} of any map $f : X \\rightarrow Y$ to be the fiber of its canonical fibration, and we denote it by $F_f$. Note that the homotopy fiber is precisely the pullback of the based path space fibration $e_0 : (Y,y_0)^{(I,1)} \\rightarrow (Y,y_0)$ along the map $f$, which is just\n\\begin{equation}\n\\label{homotopy fiber of a map}\nF_f = \\lcb (x,\\alpha) \\in X \\times Y^I \\st \\alpha(0) = f(x) \\text{ and } \\alpha(1)=y_0 \\rcb\n\\end{equation}\nIf $f$ is already a fibration, then the homotopy fiber is homotopy equivalent to the fiber of the fibration, and clearly fibrations that are fiber homotopy equivalent have homotopy equivalent fibers. \\Cref{canonical fibration} gives another meaning to the universality of the mapping path space $P_f$ as it is used to canonically convert any map into a fibration.\n\n\n\\begin{prop}\n\\label{homotopy fiber of fiber inclusion}\nLet $(E,e_0) \\stackrel{p}{\\rightarrow} (B,b_0)$ be a fibration with fiber $F=p^{-1}(b_0)$ based at $e_0$. Then the homotopy fiber of $F \\hookrightarrow E$ is homotopy equivalent to $\\Omega_* B$.\n\\end{prop}\n\\begin{proof}\nVerifying this directly is difficult because a concrete description of the homotopy fiber of $F \\hookrightarrow E$ in terms of $E$ and $B$ is quite messy. Instead, we find a simpler fibration that is fiber homotopy equivalent to the canonical fibration of $F \\hookrightarrow E$, and its fiber will be easier to compute. Since $p$ is a fibration, we can find a path lifting function $\\lambda : P_p \\rightarrow E^I$ such that $s \\circ \\lambda = \\id$, where $s(\\alpha)=(\\alpha(0),p \\circ \\alpha)$.\n\nConvert the inclusion $i : F \\hookrightarrow E$ into a fibration $q : P_i \\rightarrow E$, where $P_i \\simeq F$, and let $F_p$ be the homotopy fiber of $p : E \\rightarrow B$, with map $\\pi : F_p \\rightarrow E$ defined by $\\pi(e,\\beta)=e$. We already know that $\\pi$ is a fibration (being a pullback of the based path space fibration), and we claim that $\\pi$ is fiber homotopy equivalent to $q$. We define maps $g : P_i \\rightarrow F_p$ and $h : F_p \\rightarrow P_i$ by\n\\[ g(e,\\alpha) = (e,c_{b_0}) \\]\n\\[ h(e,\\beta) = \\left( \\lambda(e,\\beta)(1), c_{\\lambda(e,\\beta)(1)} \\right) \\]\nFirst we check that these maps do in fact map into their respective targets. If $(e,\\alpha) \\in P_i$, then $e \\in F$ and $\\alpha(0)=e$, hence $c_{b_0}(0)=b_0=p(e)$ and $c_{b_0}(1)=b_0$, so $g(e,\\alpha) \\in F_p$. On the other hand, if $(e,\\beta) \\in F_p$, then $\\beta(0)=e$ and $\\beta(1)=b_0$, so $\\lambda(e,\\beta)(1) \\in F$ (since $\\lambda$ liftings $\\beta$ and $\\beta$ ends at $b_0$), and clearly $c_{\\lambda(e,\\beta)(1)}(0) = \\lambda(e,\\beta)(1)$, therefore $h(e,\\beta) \\in P_i$. These maps are actually homotopy inverses of each other. \n\nIf we form the compositions\n\\[ g \\circ h(e,\\beta) = \\left( \\lambda(e,\\beta)(1), c_{b_0} \\right) \\]\n\\[ h \\circ g(e,\\alpha) = \\left( \\lambda(e,c_{b_0})(1), c_{\\lambda(e,c_{b_0})(1)} \\right) \\]\nthen we see that each of these components can easily be homotoped to the identity maps. For example, a homotopy $g \\circ h \\simeq \\id$ is given by\n\\[ H((e,\\beta),t) = (\\lambda(e,\\beta)(t),\\beta^t) \\]\nwhere for any path $\\gamma$ we define $\\gamma^t$ to be the path $\\gamma^t(s)=\\gamma(t+(1-t)s)$ so that $\\gamma^0=\\gamma$ and $\\gamma^1=c_{\\gamma(1)}$. Therefore $P_i$ is homotopy equivalent to $F_p$. Further, $g$ is a fibration map since $\\pi \\circ g = q$, so $g$ is actually a fiber homotopy equivalence by \\cref{homotopy equivalence of fibrations is a fiber homotopy equivalence}, and so the fibers of $\\pi$ and $q$ are homotopy equivalent. But, we can clearly see that\n\\[ \\pi^{-1}(e_0) = \\lcb (e,\\beta) \\in E \\times B^I \\st \\beta(0) = p(e) \\text{ and } \\beta(1) = b_0 \\text{ and } e = e_0 \\rcb \\]\nand so there is an obvious identification of $\\pi^{-1}(e_0)$ with $\\Omega_* B$.\n\\end{proof}\n\nStarting with a fibration $p : E \\rightarrow B$, we can form the sequence $\\Omega_* B \\hookrightarrow F_p \\stackrel{\\pi}{\\rightarrow} E \\stackrel{p}{\\rightarrow} B$. If we continued this and converted $\\Omega_* B \\hookrightarrow F_p$ into a fibration, and so on, we would get a sequence of maps\n\\[ \\cdots \\longrightarrow F_\\theta \\longrightarrow F_\\eta \\stackrel{\\theta}{\\longrightarrow} F_\\pi \\stackrel{\\eta}{\\longrightarrow} F_p \\stackrel{\\pi}{\\longrightarrow} E \\stackrel{p}{\\longrightarrow} B \\]\nBy \\cref{homotopy fiber of fiber inclusion} we have $F_\\pi \\simeq \\Omega_* B, F_\\eta = \\Omega_* E$, and so on. This means that, up to homotopy, we get a sequence of maps\n\\begin{equation}\n\\label{fiber sequence}\n\\cdots \\longrightarrow \\Omega_*^n F \\longrightarrow \\Omega_*^n E \\longrightarrow \\Omega_*^n B \\longrightarrow \\Omega_*^{n-1} F \\longrightarrow \\cdots \\longrightarrow \\Omega_* B \\longrightarrow F \\longrightarrow E \\longrightarrow B\n\\end{equation}\nsuch that any two consecutive maps is a fibration preceded by the inclusion of its fiber.\n\n\n\n\n\n\\subsection{Fiber Sequences}\n\n\nA composition $(X,x_0) \\stackrel{f}{\\rightarrow} (Y,y_0) \\stackrel{g}{\\rightarrow} (Z,z_0)$ is said to be an \\textbf{exact sequence} of pointed sets if $\\image f = g^{-1}(z_0)$. \n\n\\begin{prop}\nLet $p : E \\rightarrow B$ be a fibration with fiber $F = p^{-1}(b_0)$, and let $i : F \\rightarrow E$ be the inclusion of any fiber into $E$. Base the space of homotopy classes of maps $[Y,B]$ at constant at $b_0$. For any space $Y$, the induced sequence of sets\n\\[ [Y,F] \\stackrel{i_*}{\\longrightarrow} [Y,E] \\stackrel{p_*}{\\longrightarrow} [Y,B] \\]\nis exact.\n\\end{prop}\n\\begin{proof}\nFor $[g] \\in [Y,F]$ we have $p_* \\circ i_*([g]) = [p \\circ i \\circ g] = [c_{b_0}]$, hence $\\image i_* \\subseteq g_*^{-1}([c_{b_0}])$. For the other inclusion, let $[g] \\in [Y,E]$ such that $p_*[g] = [c_{b_0}]$. Let $G : Y \\times I \\rightarrow B$ be a homotopy between $p \\circ g$ and $c_{b_0}$, and let $\\tilde G : X \\times I \\rightarrow E$ be a solution of the following homotopy lifting problem\n\\[\n\\xymatrix\n@R=2.5pc\n@C=2.5pc\n{\n\tY \\ar[r]^g \\ar[d]_{i_0} & E \\ar[d]^p \\\\\n\tY \\times I \\ar[r]_-G \\ar@{-->}[ru]^{\\tilde G} & B\n}\n\\]\nThen $p \\circ \\tilde G(-,1) = G(-,1) = b_0$ and $H(-,1) : Y \\rightarrow F$, so $[g] = i_*[H(-,1)]$. \n\\end{proof}\n\\begin{prop}\n\\label{based fiber exact sequence}\nLet $p : E \\rightarrow B$ be a fibration with fiber $F = p^{-1}(b_0)$, and base $E$ and $F$ at the same point $e_0 \\in F$. Let $i : F \\rightarrow E$ be the inclusion of any fiber into $E$. For any space $(Y,y_0)$, the induced sequence of sets\n\\[ [Y,F]_* \\stackrel{i_*}{\\longrightarrow} [Y,E]_* \\stackrel{p_*}{\\longrightarrow} [Y,B]_* \\]\nis exact.\n\\end{prop}\n\nUsing \\eqref{fiber sequence} and \\cref{based fiber exact sequence} we now have\n\\begin{prop}[Puppe]\nLet $(Y,y_0)$ be a based space and $(E,e_0) \\rightarrow (B,b_0)$ a fibration with fiber $F = \\pi^{-1}(b_0)$ based at $e_0$. Then there is a long exact sequence of sets (for $i \\geq 0$), of groups (for $i \\geq 1$) and of abelian groups (for $i \\geq 2$)\n\\[ \\cdots \\rightarrow [Y,\\Omega_*^n F]_* \\rightarrow [Y,\\Omega_*^n E]_* \\rightarrow [Y,\\Omega_*^n B]_* \\rightarrow \\cdots \\rightarrow [Y,\\Omega_* B]_* \\rightarrow [Y,F]_* \\rightarrow [Y,E]_* \\rightarrow [Y,B]_* \\]\n\\end{prop}\n\n\n\n\n\n\n\n\n\\newpage\n\\section{Cofibrations}\n\n\\subsection{The Universal Test Diagram}\n\nIn homology one comes across certain ``nice'' pairs of spaces $(X,A)$ such that the homology of $(X,A)$ is essentially equal to the homology of the quotient $X/A$. These nice pairs of spaces also have the property that a homotopy defined on $A$ with a specified extension to $X$ at time $t=0$ can be extended to $X$ for all time. More generally, a map $i : A \\rightarrow X$ is said to have the \\textbf{homotopy extension property} with respect to a space $Y$, if for every $H : A \\times I \\rightarrow A$ and $f : X \\rightarrow X$ there exists a map $\\tilde H : X \\times I \\rightarrow X$ such that the following diagrams commutes\n\\[\n\\xymatrix\n@R=2.5pc\n@C=2.5pc\n{\n\tA \\ar[rr]^{i_0} \\ar[dd]_i & & A \\times I \\ar[dl]_H \\ar[dd]^{i \\times \\id} \\\\\n\t& Y & \\\\\n\tX \\ar[rr]_{i_0} \\ar[ru]^f & & X \\times I \\ar@{-->}[ul]_{\\tilde H}\n}\n\\] \nWe call this a \\emph{test diagram} for the homotopy extension problem. We say that $i : A \\rightarrow X$ is a \\textbf{cofibration} if it has the homotopy extension property with respect to all spaces $Y$. We call the quotient $X / i(A)$ the \\textbf{cofiber} of the cofibration.\n\n\\begin{prop}\nIf $X$ is a CW-complex and $A$ a subcomplex, then the inclusion $i : A \\hookrightarrow X$ is a cofibration.\n\\end{prop}\n\nRecall that a subspace $A \\subseteq X$ is said to be a retract of $X$ if there is a map $r : X \\rightarrow A$ such that $r|_A = \\id_A$. \n\n\n\n\n\\subsection{The Homotopy Cofiber}\n\n\n\\subsection{Cofiber Sequences}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\newpage\n\\section{Homotopy Groups}\n\n\n\n\n\\newpage\n\\section{Obstruction Theory}\n\n\n\n\\newpage\n\\section{Classifying Spaces}\n\n\n\\subsection{The Brown Representability Theorem}\n\n\nLet $\\mathscr C$ be a locally small category. A cofunctor $F : \\mathscr C \\rightarrow \\Set$ is said to be \\textbf{representable} if there is a natural isomorphism of functors\n\\[ \\eta : F \\rightarrow \\hom_{\\mathscr C}(-,X) \\]\nfor some object $X$. The object $X$ is called a \\textbf{classifying object} for $F$, or if $\\mathscr C$ is a category of topological spaces we will call $X$ a \\textbf{classifying space}. For example, we will see later that the cohomology functor $H^n(-;G)$ is representable with classifying space $K(G,n)$ so that $H^n(X;G)$ is naturally isomorphic to $[X,K(G,n)]_*$. This turns a problem in understand a functor into understand morphisms between objects.\n\nWe are only going to be concerned with representability of functors in the homotopy category of CW-complexes. Consider a cofunctor $F : \\hCWp \\rightarrow \\Setp$. For a collection of based spaces $\\lcb X_\\alpha \\rcb$, let $i_\\beta : X_\\beta \\rightarrow \\bigvee_\\alpha X_\\alpha$ be the natural inclusions induced by the coproduct. The maps $F(i_\\beta) : F(\\bigvee_\\alpha X_\\alpha) \\rightarrow F(X_\\beta)$ then induce a map $f : F(\\bigvee_\\alpha X_\\alpha) \\rightarrow \\prod_\\beta F(X_\\beta)$ by the universal property of products. We say that $F$ satisfies the \\textbf{wedge axiom} if $f$ is a bijection for all collections $\\lcb X_\\alpha \\rcb$. We say that $F$ satisfies the \\textbf{weak wedge axiom} if $f$ is a bijection for all finite collections $\\lcb X_i \\rcb_{i=1}^n$. \n\nFor $(X;A,B)$ a CW-triad, i.e. $A$ and $B$ are subcomplexes of $X$ with $X = A \\cup B$, consider the following inclusions\n\\[\n\\xymatrix\n@R=.7pc\n{\n\t & A \\ar[rd]^{i_A} \\\\\n\tA \\cap B \\ar[ru]^{j_A} \\ar[rd]_{j_B} & & X \\\\\n\t & B \\ar[ru]_{i_B}\n}\n\\]\nWe say that $F$ satisfies the \\textbf{Mayer-Vietoris axiom} if for every CW-triad $(X;A,B)$ and every $x \\in F(A),y \\in F(B)$ such that\n\\[ F(j_A)(x) = F(j_B)(y) \\in F(A \\cap B) \\]\nthere exists an element $z \\in F(X)$ such that\n\\[ F(i_A)(z) = x \\ \\ \\ \\ \\ \\text{and} \\ \\ \\ \\ \\ F(i_B)(z) = y \\]\n\nThe functor $H^n(X-;G)$ satisfies both of these axioms. Note that the Mayer-Vietoris axiom is exactly what gives exactness at the middle group\n\\[ \\cdots \\longleftarrow H^k(A \\cap B) \\longleftarrow H^k(A) \\oplus H^k(B) \\longleftarrow H^k(X) \\longleftarrow \\cdots \\]\nin the Mayer-Vietoris sequence.\n\nWe will sketch the proof of the following amazing theorem through many propositions and lemmas.\n\\begin{thm}[Brown Representability Theorem]\nA cofunctor $F : \\hCWp \\rightarrow \\Setp$ is representable if and only if $F$ satisfies the wedge and Mayer-Vietoris axioms.\n\\end{thm}\n\\begin{proof}(Sufficiency)\nSuppose $F$ is representable with classifying space $X$. We will show that $[-,X]_*$ satisfies the wedge and Mayer-Vietoris axioms, which implies $F$ does too. \\todo{finish later}\n\\end{proof}\n\nFor a based CW-complex $(Y,y_0)$ and element $u \\in F(Y)$ we can define a natural transformation $T_u : [-,Y]_* \\rightarrow F$. For based CW-complex $(X,x_0)$, let $T_u(X) : [X,Y]_* \\rightarrow F(X)$ be the morphisms that maps $[f]$ to $F([f])(u)$. It is easy to see that $T_u$ is a natural transformation. We will inductively build spaces $Y_n$ and elements $u_n \\in F(Y_n)$ such that $T_{u_n}$ gets ``closer'' to being a natural isomorphism of functors. Taking the limit will give us a classifying space for the functor.\n\n\n\\subsection{Applications}\n\n\n\n\n\n\n\n\\newpage\n\\section{Simplicial Objects}\n\n\n\\subsection{The Simplicial Category}\n\n\nWe will model simplicial objects on a monoidal category called the simplicial category. We will describe this category in detail before discussing simplicial sets and objects. Let $\\Simp$ denote the category whose objects are the finite, totally ordered sets $[n] = \\lcb 0,1,\\ldots,n \\rcb$ for each $n \\geq 0$, and whose morphisms are regular set-theoretic functions $f : [n] \\rightarrow [m]$ such that $f(i) \\leq f(j)$ for all $i \\leq j$ in $[n]$. We will also add the object $[-1]=\\emptyset$ to the category, and so there is exactly one morphism from $\\emptyset$ to any other object (the empty function). This category is called the simplicial category, and clearly $\\emptyset$ is an initial object and $[0]$ is a terminal object. For each object $[n]$ in $\\Simp$ and $0 \\leq i \\leq n$ there is a collection of particularly simple morphisms:\n\\[ \\delta_i : [n-1] \\rightarrow [n] \\ \\ \\ \\ \\ \\delta_i(j) = \\begin{cases} j & j < i \\\\ j+1 & j \\geq i \\end{cases} \\]\n\\[ \\sigma_i : [n+1] \\rightarrow [n] \\ \\ \\ \\ \\ \\sigma_i(j) = \\begin{cases} j & j \\leq i \\\\ j-1 & j > i \\end{cases} \\]\nMore easily stated, $\\delta_i$ is the inclusion of $[n-1]$ into the complement of $i$ in $[n]$, and $\\sigma_i$ is the surjection of $[n+1]$ onto $[n]$ such that $\\sigma_i(i) = \\sigma_i(i+1) = i$. The morphisms $\\delta_i$ are called the face maps, and the morphisms $\\sigma_i$ are called the degeneracy maps. It is easy to check that these morphisms satisfy the following relations:\n\\begin{align*}\n\t\\delta_j \\circ \\delta_i &= \\delta_i \\circ \\delta_{j-1} \\ \\ \\ \\ \\ \\ i j+1 \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t \\end{cases}\n\\end{align*}\nThe face and degeneracy maps generate all of the morphisms of $\\Simp$ in the sense that every morphism $f : [n] \\rightarrow [m]$ can be written as the composition of some $\\delta_i$'s and $\\sigma_j$'s. \n\nLet $f : [m] \\rightarrow [m']$ and $g : [n] \\rightarrow [n']$ be two morphisms in $\\Simp$. We define a bifunctor $+ : \\Simp \\times \\Simp \\rightarrow \\Simp$ by\n\\[ [m] + [n] = [m + n + 1] \\]\n\\[ f+g : [m+n+1] \\rightarrow [m'+n'+1] \\]\n\\[ (f+g)(j) = \\begin{cases} f(i) & 0 \\leq i \\leq m \\\\ g(i-m-1) + m' + 1 & m < i \\leq m+n+1 \\end{cases} \\]\nWith this definition we clearly have $\\emptyset + [n] = [n] = [n] + \\emptyset$ and $\\id_{\\emptyset} + f = f = f + \\id_{\\emptyset}$. This makes $\\Simp$ into a strict monoidal category with $\\emptyset$ as the unit. \n\n\n\n\n\\subsection{Simplicial Objects}\n\n\nWith the simplicial category defined we can say what we mean by a simplicial object. A simplicial object $X$ in a category $\\mathscr C$ is just a cofunctor $X : \\Simp \\rightarrow \\mathscr C$. When there is no cause for confusion we will sometimes use the notation $X_n$ to denote the object $X([n])$ of $\\mathscr C$, as well as $d_i = X(\\delta_i)$ and $s_i = X(\\sigma_i)$ to denote the induced face and degeneracy maps. We will let $s\\mathscr C$ denote the category of simplicial objects in $\\mathscr C$ with morphisms given by natural transformations, i.e. $s\\mathscr C$ is the functor category $\\mathscr C^{\\Delta}$. For example, $\\sSet$ is the category of simplicial sets and $\\sGrp$ is the category of simplicial groups. \n\nThis is a complicated definition, heavily encoded in abstract nonsense, so let us examine the definition in the case of a simplicial set. The idea of a simplicial set was around long before categories, so let us define this object in the classical sense and see why it is equivalent to the above definition. A simplicial set is a collection of sets $\\lcb K_n \\rcb_{n \\geq 0}$ with maps $d_i : K_n \\rightarrow K_{n-1}$ and $s_i : K_n \\rightarrow K_{n+1}$, $0 \\leq i \\leq n$, which satisfy the following:\n\\begin{align*}\n\td_i \\circ d_j &= d_{j-1} \\circ d_i \\ \\ \\ \\ \\ \\ i < j \\\\\n\ts_i \\circ s_j &= s_{j+1} \\circ s_i \\ \\ \\ \\ \\ \\ i \\leq j \\\\\n\td_i \\circ s_j &= \\begin{cases} s_{j-1} \\circ d_i & i < j \\\\ \\id & i=j \\text{ or } i=j+1 \\\\ s_j \\circ d_{i-1} & i > j+1 \\end{cases}\n\\end{align*}\nA simplicial map $f : K \\rightarrow L$ between simplicial sets is a collection of maps $f_n : K_n \\rightarrow L_n$ such that\n\\[ f_n \\circ d_i = d_i \\circ f_{n+1} \\]\n\\[ f_n \\circ s_i = s_i \\circ f_{n-1} \\]\nWe will see how a functor $X : \\Delta \\rightarrow \\Set$ gives such an object, and how a natural transformation between two such functors gives a morphism. Immediately we have that $X_n = X([n])$ is a set, and $d_i = X(\\delta_i) : X_n \\rightarrow X_{n-1}$ and $s_i = X(\\sigma_i) : X_{n+1} \\rightarrow X_n$. These maps satisfy the required identities:\n\\begin{align*}\n d_i \\circ d_j = X(\\delta_i) \\circ X(\\delta_j) = X(\\delta_j \\circ \\delta_i) = X(\\delta_i \\circ \\delta_{j-1}) = X(\\delta_{j-1}) \\circ X(\\delta_i) = d_{j-1} \\circ d_i \\ \\ \\ \\ &i \\cdot <\\frac{dx}{dt}, \\frac{dy}{dt}, \\frac{dz}{dt}> \\\\\n &= \\nabla w \\cdot \\frac{d\\vec{r}}{dt} \\\\\n \\end{split} \\\\\n \\nabla w = \\\\\n \\frac{d\\vec{r}}{dt} = <\\frac{dx}{dt}, \\frac{dy}{dt}, \\frac{dz}{dt}> \\\\\n\\end{gather*}\nThe vector $$ is called gradient, denoted as $\\nabla w$. \nGradients on different positions can have different directions and magnitude.\n\n\\subsection{A Property of Gradient}\n\n\\begin{theorem}\n The gradient of a function is always perpendicular to the level surfaces of \n the function, where a level surface means the set of points on the function \n domain whose function values are equal to a constant.\n\\end{theorem}\nNotice that the level surfaces refer to points in the function domain, rather \nthan on the function graph. It doesn't include the dimension of the function \nvalue.\n\n\\begin{example}\n Examine the relation between the gradient and the level surfaces for the \n function $w = a_1x + a_2y + a_3z$.\n\n \\begin{gather*}\n \\frac{\\partial w}{\\partial x} = a_1 \\\\\n \\frac{\\partial w}{\\partial y} = a_2 \\\\\n \\frac{\\partial w}{\\partial z} = a_3 \\\\\n \\nabla w = \\\\\n \\end{gather*}\n\n The level surfaces of the function $w = a_1x + a_2y + a_3z = c$, where $c$ is \n any constant, are a series of planes parallel to each other. For any level \n surfaces of the function, the gradient $\\nabla w$ is a normal vector to it. \n Therefore, the gradient is always perpendicular to the level surfaces for the \n function $w = a_1x + a_2y + a_3z$.\n\\end{example}\n\n\\begin{example}\n Examine the relation between the gradient and the level surfaces for the \n function $w = x^2 + y^2$.\n\n \\begin{gather*}\n \\frac{\\partial w}{\\partial x} = 2x \\\\\n \\frac{\\partial w}{\\partial y} = 2y \\\\\n \\nabla w = <2x, 2y> = \\\\\n \\end{gather*}\n\n The level surfaces of the function $w = x^2 + y^2 = c$, where $c$ is any \n constant, are a series of circles whose centers are the origin. At any point \n $(x_0, y_0)$ in the function domain, the gradient vector is $$, \n which has the same direction as the radius through $(x_0, y_0)$ on the level \n surface $x^2 + y^2 = x_0^2 + y_0^2$. A circle's radius is always perpendicular \n to the circle, hence the gradient vector is perpendicular to the level surface. \n Therefore, the gradient is always perpendicular to the level surfaces for the \n function $w = x^2 + y^2$.\n\\end{example}\n\nProof of the theorem:\n\nSuppose there is a curve $\\vec{r} = \\vec{r}(t)$ that always stays on a level \nsurface of a function $f$. The velocity vector $\\frac{d\\vec{r}}{dt}$ is tangent \nto the curve, and therefore tangent to the level surface on which the curve \nstays.\n\nBy chain rule, on any level surfaces it satisfies the following equation:\n\\begin{equation*}\n \\frac{df}{dt} = \\nabla f \\cdot \\frac{d\\vec{r}}{dt} = 0\n\\end{equation*}\nTherefore, at any points of a curve that stays on a level surface:\n\\begin{equation*}\n \\nabla f \\perp \\frac{d\\vec{r}}{dt}\n\\end{equation*}\nThe same reasoning applies to any curves on a level surface, so at any points on \na level surface, the gradient $\\nabla f$ is perpendicular to velocity vectors of \nevery direction, which are tangent to the level surface. Therefore, at any \npoints on a level surface, the gradient $\\nabla f$ is perpendicular to the \ntangent plane to the level surface.\n\n\\subsection{Application of Gradient}\n\nWe can use gradients to find the equation of the tangent line of a function \ngraph at any points.\n\n\\begin{example}\n Find the equation of the tangent plane to the surface $x^2 + y^2 - z^2 = 4$ at \n the point (2, 1, 1).\n\n Solution:\n\n The surface $x^2 + y^2 - z^2 = 4$ is a level surface of the function \n $w = x^2 + y^2 - z^2$.\n \\begin{gather*}\n \\frac{\\partial w}{\\partial x} = 2x \\\\\n \\frac{\\partial w}{\\partial y} = 2y \\\\\n \\frac{\\partial w}{\\partial z} = -2z \\\\\n \\nabla w = <2x, 2y, -2z> = \\\\\n \\end{gather*}\n Hence the gradient of the function at the point (2, 1, 1) is $<2, 1, -1>$. \n According to the property of gradients, the gradient is perpendicular to the \n tangent plane of the corresponding level surface at the point. Therefore, the \n equation of the tangent plane is \n \\[ 2x + y - z = c, \\textnormal{where c is a constant} \\]\n Then we can substitute the coordinate of the point (2, 1, 1) into the plane \n equation:\n \\[ c = 2 \\times 2 + 1 \\times 1 - 1 \\times 1 = 4 \\]\n Therefore, the equation of the tangent plane to the surface \n $x^2 + y^2 - z^2 = 4$ at the point (2, 1, 1) is\n \\[ 2x + y - z = 4 \\]\n\\end{example}\n\n\\section{Directional Derivatives}\n\n\\subsection{Introduce Directional Derivatives}\n\nHow to calculate the partial derivative towards any direction $\\vec{u}$, rather \nthan only along the $x$ axis and $y$ axis?\n\nSuppose there is a multivariable function $f(x, y)$, and an unit vector \n$\\hat{\\vec{u}} = $. The partial derivative along a direction is the rate \nof change of the function value $f(x, y)$ over the arclength, hence we need the \ndifferential of the arclength along the direction of $\\hat{\\vec{u}}$, which is \ndenoted by $ds$. \n\nBy chain rule, we have\n\\begin{gather*}\n df = \\frac{\\partial f}{\\partial x}dx + \\frac{\\partial f}{\\partial y}dy \\\\\n \\frac{df}{ds} = \\frac{\\partial f}{\\partial x}\\frac{dx}{ds} + \\frac{\\partial f}{\\partial y}\\frac{dy}{ds} \\\\\n \\frac{df}{ds} = <\\frac{\\partial f}{\\partial x}, \\frac{\\partial f}{\\partial y}> \\cdot <\\frac{dx}{ds}, \\frac{dy}{ds}> \\\\\n \\frac{df}{ds} = \\nabla f \\cdot \\frac{d\\vec{r}}{ds} \\\\\n\\end{gather*}\nTherefore, we need to describe the position vector $\\vec{r}$ along the straight \nline trajectory with the direction $\\vec{u}$ using the parameter $s$, the \narclength.\n\\begin{gather*}\n \\begin{cases}\n x(s) = x_0 + as \\\\\n y(s) = y_0 + bs \\\\\n \\end{cases} \\\\\n \\frac{d\\vec{r}}{ds} = \\\\\n \\frac{d\\vec{r}}{ds} = \\vec{u} \\\\\n\\end{gather*}\nTherefore,\n\\begin{equation*}\n \\begin{split}\n \\frac{df}{ds} &= \\nabla f \\cdot \\frac{d\\vec{r}}{ds} \\\\\n &= \\nabla f \\cdot \\vec{u} \\\\\n \\end{split}\n\\end{equation*}\nwhich is the formula for directional derivatives.\n\n\\begin{definition}\n The instantaneous rate of change of a multivariable function $f$ at a given \n direction $\\vec{u}$ is called the directional derivative, whose notation is \n $\\frac{df}{ds}|_{\\vec{u}}$. The formula of directional derivatives is \n \\[ \\frac{df}{ds}|_{\\vec{u}} = \\nabla f \\cdot \\vec{u} \\]\n\\end{definition}\n\n\\subsection{Geometric Interpretation of Directional Derivatives}\n\nFor a multivariable function $f$ with two independent variables, the geometric \ninterpretation of the function toward a direction $\\vec{u}$ is the slope of the \nslice of the function graph by a vertical plane parallel to $\\vec{u}$.\n\n\\begin{example}\n Verify the formula of directional derivatives by finding the partial \n derivative along the $x$ axis.\n\n Suppose there is a multivariable function $f(x, y)$.\n \\begin{equation*}\n \\begin{split}\n \\frac{\\partial f}{\\partial x} &= \\nabla f \\cdot \\vec{i} \\\\\n &= <\\frac{\\partial f}{\\partial x}, \\frac{\\partial f}{\\partial y}> \\cdot \\vec{i} \\\\\n &= \\frac{\\partial f}{\\partial x} \\\\\n \\end{split}\n \\end{equation*}\n\\end{example}\n\n\\subsection{Max/Min of Directional Derivatives}\n\nAccording to the geometric interpretation of dot product of vectors:\n\\begin{equation*}\n \\begin{split}\n \\frac{df}{ds}|_{\\vec{u}} &= \\nabla f \\cdot \\vec{u} \\\\\n &= |\\nabla f| \\times |\\vec{u}| \\times \\cos\\theta \\\\\n &= \\cos\\theta \\times |\\nabla f| \\\\\n \\end{split}\n\\end{equation*}\nWe can see that the directional derivative changes as the angle $\\theta$ between \nits direction and the gradient vector changes. It is obvious that at a given \npoint:\n\\begin{itemize}\n \\item The directional derivative with the maximum value has the same direction \n as the gradient, and the maximum value is the magnitude of the gradient. \n Since the directional derivative along the direction of the gradient is \n positive and has the maximum value, the gradient points to the direction \n where the function value increases fastest.\n \\item The directional derivative with the minimum value has the opposite \n direction to the gradient, and the minimum value is the negative value of \n the magnitude of the gradient.\n\\end{itemize}\n\n\\end{document}", "meta": {"hexsha": "bc3b8db527fe686c4d0ac924e1b4662966ce1015", "size": 9180, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture12.tex", "max_stars_repo_name": "jinxinwangstd/multivariable-calculus-mit", "max_stars_repo_head_hexsha": "d165ef6ff085fafd0fb89b8027fc5f5e0dbfa1e9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lecture12.tex", "max_issues_repo_name": "jinxinwangstd/multivariable-calculus-mit", "max_issues_repo_head_hexsha": "d165ef6ff085fafd0fb89b8027fc5f5e0dbfa1e9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lecture12.tex", "max_forks_repo_name": "jinxinwangstd/multivariable-calculus-mit", "max_forks_repo_head_hexsha": "d165ef6ff085fafd0fb89b8027fc5f5e0dbfa1e9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.8983050847, "max_line_length": 120, "alphanum_fraction": 0.6687363834, "num_tokens": 2840, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8104789155369047, "lm_q1q2_score": 0.6141763602306411}} {"text": "\\subsection{Air Thermoregulation}\n\\label{sec:airthermoregulation}\n\n\\textbf{Purpose}: Maintaining desired leaf-zone air temperature and circulating air.\n\n\\textbf{Function}:\n\\begin{itemize}\n \\item \\textbf{Inputs}: Power, air temperature control signal (\\ref{sec:automation}), air circulation control signal (\\ref{sec:automation});\n \\item \\textbf{Outputs}: $\\pm$Heat to environment, $\\mp$heat to surroundings, internal air circulation, internal air temperature sensor signal (\\ref{sec:automation});\n\\end{itemize}\n\n\\textbf{Method}:\n\\begin{enumerate}\n \\item \\textit{Testing}:\n \\begin{itemize}\n \\item Heat pump direction and magnitude respond to control signal as expected;\n \\item Fans operate as expected;\n \\item Heat pump power exceeds maximum heat loss (temperature extremes)\\footnote{i.e. if X Watts leave the system at MAX$\\degree$C internal, and Y Watts enter the system at MIN$\\degree$C internal, the heat pump must transfer >X, >Y Watts.};\n \\item Heat pump power exceeds that required to reach temperature extremes in under 120 seconds given the system's heat capacity;\n \\end{itemize}\n \\item \\textit{Process}:\n \\begin{enumerate}\n \\item Air is circulated throughout the environment;\n \\item Temperature is measured, sent to control module;\n \\item Control module controls heat pump speed and direction (heating vs. cooling);\n \\end{enumerate}\n\\end{enumerate}\n\n\\textbf{Calculations}:\n\nAssuming an atmospheric pressure $P$ of 101.325kPa, a surroundings temperature range $T_{surr}$ of 22$\\degree$C, a system target temperature range $T_{sys-min}$, $T_{sys-max}$ of 10-35$\\degree$C, a molar mass of dry air\\footnote{Water vapour has a maximum concentration of 30g/kg at 30$\\degree$C, or 3\\%, which is negligible for mass and heat capacity calculations.} $M$ of 28.97 $\\frac g{mol}$, a specific heat capacity of dry air $c_p$ of $1.006 \\frac{J}{g*\\text{K}}$, a 4-unit (2x2 units, 16 faces) expanded configuration, and a face insulation RSI per mm of $0.0328\\text{m}^2~ \\degree \\text{C}~\\text{W}^{-1}~\\text{mm}^{-1}$ (See Section \\ref{sec:housing}):\\\\\n\\vspace{.05cm}\n\\begin{gather*}\n \\label{eqn:heatloss}\n Q_{loss}=\\frac{(T_{surr}-T_{sys-max}) * A}{\\text{RSI per mm} * \\ell}=\\frac{(22\\degree \\text{C}-35\\degree \\text{C}) * (16 \\text{ faces} * 0.5\\text{m} * 0.5\\text{m})}{0.0328 \\text{m}^2~ \\degree \\text{C}~\\text{W}^{-1}~\\text{mm}^{-1} * 25.4 \\text{mm}}=-62.42 W\\\\\\vspace{.05cm}\\\\\n \\label{eqn:heatgain}\n Q_{gain}=\\frac{(T_{surr}-T_{sys-min}) * A}{\\text{RSI per mm} * \\ell}=\\frac{(22\\degree \\text{C}-10\\degree \\text{C}) * (16 \\text{ faces} * 0.5\\text{m} * 0.5\\text{m})}{0.0328 \\text{m}^2~ \\degree \\text{C}~\\text{W}^{-1}~\\text{mm}^{-1} * 25.4 \\text{mm}}=57.61 W\\\\\\vspace{.05cm}\\\\\n \\label{eqn:airmass}\n m_{air}=\\frac{P*V*M}{R*T_{avg}}=\\frac{101325\\text{Pa}*(0.5\\text{m}*0.5\\text{m}*0.5\\text{m}*4\\text{ units})*28.97\\frac g{mol}}{8.314\\frac{J}{\\text{mol}*K}*300\\text{K}}=588.4g\\\\\n \\vspace{1.5cm}\\\\\n \\text{Continued on next page}\n\\end{gather*}\n\\vspace{1cm}\n\n\\begin{gather*}\n \\label{eqn:heating}\n W_{heating}=\\frac{m*c_p*(T_{surr}-T_{sys-max})}{t}=\\frac{588.4g*1.006\\frac{J}{g*\\text{K}}*(22\\degree \\text{C}-35\\degree \\text{C})}{120\\text{ sec}}=-64.13\\text{W}\\\\\n \\label{eqn:cooling}\n W_{cooling}=\\frac{m*c_p*(T_{surr}-T_{sys-min})}{t}=\\frac{588.4g*1.006\\frac{J}{g*\\text{K}}*(22\\degree \\text{C}-10\\degree \\text{C})}{120\\text{ sec}}=59.19\\text{W}\n\\end{gather*}\n\n$\\therefore$ A thermoelectric system able to transfer at least 70W (such as \\cite{peltier}, which transfers up to 85W) will supply enough power to heat/cool the system from ambient to extremes in 120 seconds and maintain temperature.\n\n\\begin{gather*}\n \\label{eqn:thermalresistance-hot}\n R_{\\theta~Peltier-Surr}=R_{\\theta~Peltier-Sink}+R_{\\theta~Sink-Air}\\le\\frac{T_{h~max} - T_{surr}}{Q_{max}}=\\frac{50\\degree C - 22\\degree C}{85W}=0.329\\degree \\text{C W}^{-1}\\\\\n \\label{eqn:thermalresistance-cold}\n R_{\\theta~Peltier-Sys}=R_{\\theta~Peltier-Sink}+R_{\\theta~Sink-Air}\n\\end{gather*}\n\n\\textbf{Features}:\n\\begin{itemize}\n \\item \\textit{Circulation Fans}: Located in growth environment to circulate air even temperature distribution.\n \\item \\textit{Temperature Sensors}: SHT31 \\cite{sht31} sensors on breakout boards located throughout the growth environment to measure air temperature. Informs a \\textbf{PID control loop} (\\ref{sec:automation}).\n \\item \\textit{Heat Pump}: Pumps heat in or out of the growth environment. Is comprised of:\n \\begin{itemize}\n \\item \\textit{Peltier Device}: 85W bidirectional solid-state \\textbf{thermoelectric device} (aka Peltier tile) \\cite{peltier} pumps heat from one face to the other. Better space efficiency, less complexity (no liquids, pressurized fluids, etc.), and more precise than other methods.\n \\item \\textit{Peltier Driver Circuit}: Controls magnitude and direction of Peltier device heat pump via a \\textbf{dimmable voltage source} and \\textbf{MOSFET H-bridge}, respectively. See Figure \\ref{fig:peltierdriver}.\n \\item \\textit{Heat Sinks}: Aluminum blocks with fins hold and exchange heat between air and Peltier devices. One set on each side of the Peltier (inside and outside environment) builds \"heat pump\". Mating face coated with thermal compound for better transfer.\n \\item \\textit{Heat Sink Fans}: Located on both sets of heat sinks for better heat dissipation.\n \\end{itemize}\n\\end{itemize}\n\n\\begin{figure}[h]\n \\centering\n \\includegraphics[width=0.8\\textwidth]{images/thermosim.png}\n \\hfill\n \\caption{Peltier driver circuit simulation (live version: \\cite{thermo-falstad})}\n \\label{fig:peltierdriver}\n\\end{figure}", "meta": {"hexsha": "c61132f32b57cd87c2b64ebe727ef23cb9c69c34", "size": 5665, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/solutionoverview/tex/subsystems/AirThermoregulation.tex", "max_stars_repo_name": "PeaPodTech/PeaPod", "max_stars_repo_head_hexsha": "19f956f4e7612d81c203a616ff505c4dc68b9964", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "docs/solutionoverview/tex/subsystems/AirThermoregulation.tex", "max_issues_repo_name": "PeaPodTech/PeaPod", "max_issues_repo_head_hexsha": "19f956f4e7612d81c203a616ff505c4dc68b9964", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-10-30T05:06:16.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-30T05:06:16.000Z", "max_forks_repo_path": "docs/solutionoverview/tex/subsystems/AirThermoregulation.tex", "max_forks_repo_name": "PeaPodTech/PeaPod", "max_forks_repo_head_hexsha": "19f956f4e7612d81c203a616ff505c4dc68b9964", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-10-24T02:21:18.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-24T02:21:18.000Z", "avg_line_length": 70.8125, "max_line_length": 663, "alphanum_fraction": 0.7043248014, "num_tokens": 1874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8104788995148792, "lm_q2_score": 0.7577943712746406, "lm_q1q2_score": 0.6141763480892405}} {"text": "\\documentclass[]{revtex4-2}\n\\usepackage{amsmath}\n\n\\begin{document}\n \\title{Assignment 5: Physics of Compact Objects}\n \\author{Md Arif Shaikh}\n \\affiliation{International Centre for Theoretical Sciences, Bengaluru}\n \\email{arifshaikh.astro@gmail.com}\n \\date{\\today}\n \\maketitle\n \\section*{Problem 1}\n Show that, in the white dwarf interior, the Coulomb energy per free electron is\n \\begin{equation}\n \\label{eq:coulomb-energy}\n \\epsilon_C = - \\frac{9}{10} \\left(\\frac{4\\pi}{3}\\right)^{1/3}Z^{2/3}e^2 n_e^{1/3},\n \\end{equation}\n where $Z$ is the atomic number, $e$ is the elementary charge and $n_e$ is the number density of free electrons.\n\n\\vspace{0.5cm}\n \n\\noindent {\\bfseries Answer:} As temperature decreases, $T\\to0$, the ions are located in a lattice that maximizes the inter-ion separation. We can consider a spherical shell of the lattice of volume $4\\pi r_0^3/3 = 1/n_N$, where $n_N$ is the number density of the nuclei. The total energy of any one sphere is the sum of potential energies due to electron-electron ($e-e$) interactions and electron-ion ($e-i$) interactions.\n\nTo assemble a uniform sphere of $Z$ electrons requires energy\n\\begin{equation}\n \\label{eq:e-e}\n E_{e-e} = \\int_0^{r_0} \\frac{q dq}{r} = \\int_0^{r_0} - \\frac{Ze (r^3/r_0^3)d ( - Ze (r^3/r_0^3))}{r} = Z^2e^2 \\frac{3}{r_0^6}\\int_0^{r_0} r^4 dr = Z^2e^2 \\frac{3}{r_0^6}\\frac{r_0^5}{5} = \\frac{3}{5}\\frac{Z^2e^2}{r_0}.\n\\end{equation}\nOn the other hand, to assemble the electron sphere about the central nucleus of charge $Ze$ requires energy\n\\begin{equation}\n \\label{eq:e-i}\n E_{e-i} = Ze \\int_0^{r_0} \\frac{dq}{r} = - Ze \\int_0^{r_0} \\frac{3 dr Z e r}{r_0^3} = - \\frac{3}{2}\\frac{Z^2e^2}{r_0}.\n\\end{equation}\nSo, the total Coulomb energy of the shell is\n\\begin{equation}\n \\label{eq:E_tot}\n E_C = E_{e-e} + E_{e-i} = \\frac{3}{5}\\frac{Z^2e^2}{r_0} - \\frac{3}{2}\\frac{Z^2e^2}{r_0} = - \\frac{9}{10}\\frac{Z^2e^2}{r_0}.\n\\end{equation}\nSo the Coulomb energy per electron is\n\\begin{equation}\n \\label{eq:epsilon_c}\n \\epsilon_C = \\frac{E_C}{Z} = -\\frac{9}{10}\\frac{Z e^2}{r_0} = - \\frac{9}{10}\\frac{Z e^2}{(3 Z/4\\pi n_e)^{1/3}} = - \\frac{9}{10} \\left(\\frac{4\\pi}{3}\\right)^{1/3}Z^{2/3}e^2n_e^{1/3},\n\\end{equation}\nwhere we have used\n\\begin{equation}\n \\label{eq:n_e}\n \\frac{4\\pi}{3} r_0^3 n_e = Z. \n\\end{equation}\n\\end{document}", "meta": {"hexsha": "5c59fb69fa1a7b6c6271b4d92768fe2cde1f602b", "size": 2352, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment-5/Assignment-5-prob-1.tex", "max_stars_repo_name": "md-arif-shaikh/Physics-Of-Compact-Objects", "max_stars_repo_head_hexsha": "103617c8d6f02a03727ece5e6dce9dcfb8e9482d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment-5/Assignment-5-prob-1.tex", "max_issues_repo_name": "md-arif-shaikh/Physics-Of-Compact-Objects", "max_issues_repo_head_hexsha": "103617c8d6f02a03727ece5e6dce9dcfb8e9482d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment-5/Assignment-5-prob-1.tex", "max_forks_repo_name": "md-arif-shaikh/Physics-Of-Compact-Objects", "max_forks_repo_head_hexsha": "103617c8d6f02a03727ece5e6dce9dcfb8e9482d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-04-17T02:18:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-04-17T02:18:16.000Z", "avg_line_length": 49.0, "max_line_length": 424, "alphanum_fraction": 0.6611394558, "num_tokens": 925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943712746404, "lm_q2_score": 0.8104788995148792, "lm_q1q2_score": 0.6141763480892404}} {"text": "\\include{config/config}\n\n\\begin{document}\n% ========== Edit your name here\n\\title{MATH 2901 Basic Probability Lecture Notes 2}\n\\author{Instructor: Richard Kleeman}\n\\date{}\n\\maketitle\n\n%\\medskip\n\n% ========== Contents begin here ==============\n\\section{Random variables}\n\\begin{definition}\n\\textbf{A random variable} is a function $X: \\Omega \\to \\R$ with the property that $\\{ \\omega\\in \\Omega \\spacevert X(\\omega) \\leq x\\} \\in \\mathcal{F}$ for each $x \\in \\R$. Such a function is said to be $\\mathbf{\\mathcal{F}}$\\textbf{-measurable}. \n\\end{definition}\n\n\\begin{example}\nTossing two dice. We define \n\\begin{equation*}\n X = \\begin{cases} 1 & \\text{get double ($i=j$)} \\\\ 0 & \\text{not double $(i\\neq j)$} \\end{cases}.\n\\end{equation*}\nThen $X = (i,j) \\in \\R$.\n\\end{example}\n\n\\begin{remark}\nWe are interested in $g(x) = \\Prob\\{ \\omega \\spacevert X(\\omega) = x \\}$. But sometimes it doesn't work very well. Probability triplet is defined as $\\{ \\Omega, \\mathcal{F}, \\Prob \\}$. If $\\Omega$ is countable, then $g(x)$ is okay. But if $\\Omega$ is uncountable like intervals, then $\\Prob(X=x)$ doesn't make much sense because the cardinality is too large.\n\\end{remark}\n\n\\begin{definition}\nThe \\textbf{distribution function} of a random variable $X$ is the function $F: \\R \\to \n[0, 1]$ given by $F(x) = \\Prob(X \\leq x)$. \n\\end{definition}\n\n\\begin{example}\nThe distribution function of preceeding example is \n\\begin{equation*}\n F(x) = \\begin{cases} 0 & x \\leq 0, \\\\ \\frac{30}{36} & 0 < x \\leq 1, \\\\ 1 & x > 1. \\end{cases}\n\\end{equation*}\n\\end{example}\n\nNotice that $\\{ \\omega \\spacevert X(\\omega) \\leq x\\}$ defines an event. It is an element in the corresponding $\\sigma$-field. Denote $A(x) = \\{ \\omega \\spacevert X(\\omega) \\leq x \\}$. Along with $A(x)$, we can define\n\\begin{equation*}\n A^c(x) = \\{ \\omega \\spacevert X(\\omega) > x \\}, \\quad A(x, y) = A^c(x) \\cap A(y) = \\{ \\omega \\spacevert x < X(\\omega) \\leq y \\}.\n\\end{equation*}\nTow points worth noting:\n\\begin{enumerate}[(a)]\n \\item $F$ must be defined for \\textbf{all} $x \\in \\R$.\n \\item $A(x)$ should \\textbf{belongs to} $\\mathcal{F}$. Otherwise, we cannot talk about the probability of $\\Prob(A(x))$. Then the definition of distribution function is meaningless.\n\\end{enumerate}\n\n\\begin{lemma}\nA distribution function $F$ has the following properties:\n\\begin{enumerate}[(a)]\n \\item $\\lim_{x \\to -\\infty}F(x) = 0$, $\\lim_{x\\to\\infty} F(x) = 1$,\n \\item if $x < y$, then $F(x) \\leq F(y)$,\n \\item $F$ is right-continuous, that is $F(x+h) \\to F(x)$ as $h \\downarrow 0$. (left-continuous is not necessary)\n\\end{enumerate}\n\\end{lemma}\n\n\\begin{example}\n\\textbf{Indicator functions.} A particular class of Bernoulli variables is very useful in probability theory. Let $A$ be an event and let $I_A : \\Omega \\to \\R$ be the indicator function of $A$; that is,\n\\begin{equation*}\n I_A(\\omega) = \\begin{cases} 1 & \\text{if $\\omega \\in A$}, \\\\ 0 & \\text{if $\\omega \\in A^c$}. \\end{cases}\n\\end{equation*}\nThen $I_A$ is a Bernoulli random variable taking the values 1 and 0 with probabilities $\\Prob(A)$ and $\\Prob(A^c)$ respectively. Suppose $\\{B_i \\spacevert i \\in I\\}$ is a family of disjoint events with $A \\subseteq \\bigcup_{i\\in I} B_i$. Then \n\\begin{equation*}\n I_A = \\sum_{i} = I_{A\\cap B_i},\n\\end{equation*}\nan identity which is often useful.\n\\end{example}\n\n\\begin{lemma}\nLet $F$ be the distribution function of $X$. Then\n\\begin{enumerate}[(a)]\n \\item $\\Prob(X>x) = 1-F(x)$, \n \\item $\\Prob(x < X \\leq y) = F(y) - F(x)$,\n \\item $\\Prob(X = x) = F(x) - \\lim_{y \\uparrow x} F(y)$.\n\\end{enumerate}\n\\end{lemma}\n\nA random variable $X$ with distribution function $F$ is said to have two ``tails\" given by \n\\begin{equation*}\n T_1 (X) = \\Prob(X > x) = 1 - F(x), \\quad T_2(X) = \\Prob(X \\leq -x) = F(-x), \n\\end{equation*}\nwhere $x$ is large and positive. The rates at which the $T_i$ decay to zero as $x\\to\\infty$ have a substantial effect on the existence or non-existence of certain associated quantities called the ``moments\" of the distribution. \n\n\n\\section{Different random variables}\n\\subsection{Discrete random variables}\n\\begin{definition}\nThe random variable $X$ is called \\textbf{discrete} if it takes values in some \\textbf{countable} subset $\\{ x_1, x_2, \\dots \\}$, only, of $\\R$. The discrete random variable $X$ has \\textbf{(probability) mass function} $f : \\R \\to [0, 1]$ given by $f(x) = \\Prob(X = x)$. \n\\end{definition}\nWe shall see that the distribution function of a discrete variable has jump discontinuities \nat the values $x_1 , x_2, \\dots$ and is constant in between; such a distribution is called \\emph{atomic}.\n\n\\subsection{Continuous random variables}\n\\begin{definition}\nThe random variable $X$ is called \\textbf{continuous} if its distribution function can \nbe expressed as \n\\begin{equation*}\n F(x) = \\int_{-\\infty}^x f(u) du \\qquad x \\in \\R\n\\end{equation*}\nfor some \\textbf{integrable} function $f : \\R \\to [0, \\infty)$ called the \\textbf{(probability) density function} of $X$. \n\\end{definition}\nSome point worth noting:\n\\begin{enumerate}[(a)]\n \\item The density function $f(x)$ is not unique. We can add some separate points to $f$, and it doesn't affect the integration.\n \\item $F$ must be \\textbf{absolutely continuous}. This implies $F$ is continuous. We can also deduce that the probability at certain point must be zero. \\emph{i.e.}, $\\Prob(X=x) = 0$.\n\\end{enumerate}\n\nThere is another sort of random variable, called ``singular\".\n\n\n\\section{Random vectors}\n\\subsection{Definition}\nThe random vector is a function $X : \\Omega \\to \\R^n$. For example, $\\mathbf{X} = (X, Y)$ for $n=2$. We can also define the distribution function for such $X$. But we need to first introduce the ordering in $\\R^n$.\n\\begin{center}\n By definition, $(x_1, y_1) < (x_2, y_2)$ if and only if $x_1 < x_2$ \\textbf{AND} $y_1 < y_2$.\n\\end{center}\n\n\\begin{definition}\nThe \\textbf{joint distribution function} of a random vector $\\mathbf{X} = (X_1, X_2 \\dots, X_n)$ on the probability space $\\{ \\Omega, \\mathcal{F}, \\Prob \\}$ is the function $F_{\\mathbf{X}} : \\R^n \\to [0, 1]$ given by $F_{\\mathbf{X}}(x) = \\Prob(\\mathbf{X} \\leq \\mathbf{x})$ for $\\mathbf{x} \\in \\R^n$. \n\\end{definition}\n\n\\begin{remark}\nThe joint probability $\\Prob(\\mathbf{X} \\leq \\mathbf{x}) = \\Prob(X_1 \\leq x_1, \\dots, X_n \\leq x_n)$. $\\{\\mathbf{X} \\leq \\mathbf{x} \\}$ is an abbreviation for the event $\\{\\omega \\in \\Omega \\spacevert \\mathbf{X}(w) \\leq \\mathbf{x}\\}$.\n\\end{remark}\n\n\\begin{lemma}\nThe joint distribution function $F_{X, Y}$ of the random vector $(X, Y)$ has the following properties: \n\\begin{enumerate}[(a)]\n \\item $\\lim_{x, y\\to -\\infty} F_{X, Y}(x, y) = 0, \\lim_{x,y\\to\\infty} F_{X, Y}(x,y) = 1$,\n \\item if $(x_1, y_1) \\leq (x_2, y_2)$, then $F_{X,Y}(x_1, y_1) \\leq F_{X, Y}(x_2, y_2)$,\n \\item $F_{X,Y}$ is continuous from above, in that \n \\begin{equation*}\n F_{X,Y}(x+u, y+v) \\to F_{X,Y}(x,y) \\quad as \\quad u, v \\downarrow 0.\n \\end{equation*}\n\\end{enumerate}\n\\end{lemma}\n\n\\subsection{Marginalization} \n\\begin{gather*}\n \\lim_{y\\to\\infty} F_{X,Y} = F_X(x) = \\Prob(X \\leq x), \\\\\n \\lim_{x\\to\\infty} F_{X,Y} = F_Y(y) = \\Prob(Y \\leq y).\n\\end{gather*}\nThe functions $F_X$ and $F_Y$ are called the ``marginal\" distribution functions of $F_{X,Y}$. $F_{X,Y}$ can determine two marginals $F_X$ and $F_Y$, but \\textbf{converse is NOT true}.\n\n\\subsection{Discrete and continuous distribution}\n\\begin{definition}\nThe random variables $X$ and $Y$ on the probability space $\\{ \\Omega, \\mathcal{F}, \\Prob \\}$ are called \\textbf{(jointly) discrete} if the vector $(X, Y)$ takes values in some \\textbf{countable} subset of $\\R^2$ only. The jointly discrete random variables $X, Y$ have \\textbf{joint (probability) mass function} $f : \\R^2 \\to [0,1]$ given by $f(x, y) = \\Prob(X = x, Y = y)$. \n\\end{definition}\n\n\\begin{definition}\nThe random variables $X$ and $Y$ on the probability space $\\{ \\Omega, \\mathcal{F}, \\Prob \\}$ are called \\textbf{(jointly) continuous} if their joint distribution function can be expressed as \n\\begin{equation*}\n F_{X,Y}(x,y) = \\int_{u = -\\infty}^x \\int_{v = -\\infty}^y f(u,v)du dv \\qquad x, y\\in\\R,\n\\end{equation*}\nfor some \\textbf{integrable} function $f : \\R^2 \\to [0, \\infty)$ called the \\textbf{joint (probability) density function} of the pair $(X, Y)$. \n\\end{definition}\n\n\\end{document}", "meta": {"hexsha": "9fd4795bd3546e0d02f7a004e824f7ead8949ab9", "size": 8312, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "_drafts/MATH 2901/notes_2.tex", "max_stars_repo_name": "yuhan-zhao/freshman21-v1", "max_stars_repo_head_hexsha": "e4c5f7983a768554399193f47e8426976205330f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "_drafts/MATH 2901/notes_2.tex", "max_issues_repo_name": "yuhan-zhao/freshman21-v1", "max_issues_repo_head_hexsha": "e4c5f7983a768554399193f47e8426976205330f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "_drafts/MATH 2901/notes_2.tex", "max_forks_repo_name": "yuhan-zhao/freshman21-v1", "max_forks_repo_head_hexsha": "e4c5f7983a768554399193f47e8426976205330f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.3086419753, "max_line_length": 374, "alphanum_fraction": 0.6627767084, "num_tokens": 2847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147276, "lm_q2_score": 0.8198933315126792, "lm_q1q2_score": 0.6141716107541824}} {"text": "\\documentclass[gr-notes.tex]{subfiles}\n\n\\begin{document}\n\n\\setcounter{chapter}{7}\n\n\\chapter{The Einstein field equations}\n\n\\setcounter{section}{5}\n\n\\section{Exercises}\n\n\\textbf{3}\n\n(a) Calculate in geometrized units:\n\n(i) the Newtonian potential of the Sun at its surface\n%\n\\begin{displaymath}\n \\phi =\n -G M_\\Sun / R_\\Sun \\approx\n -\\SI{1.476e3}{m} / \\SI{6.960e8}{m} \\approx\n \\num{-2.12e-6}\n\\end{displaymath}\n\n(ii) the Newtonian potential of the Sun at the radius of Earth's orbit\n%\n\\begin{displaymath}\n \\phi =\n -G M_\\Sun / \\SI{1}{AU} \\approx\n -\\SI{1.476e3}{m} / \\SI{1.496e11}{m} \\approx\n \\num{-9.866e-9}\n\\end{displaymath}\n\n(iii) the Newtonian potential of the Earth at its surface\n%\n\\begin{displaymath}\n \\phi =\n -G M_\\Earth / R_\\Earth \\approx\n -\\SI{4.434e-3}{m} / \\SI{6.371e6}{m} \\approx\n \\num{-9.660e-10}\n\\end{displaymath}\n\n(iv) the Earth's orbital velocity\n\nHere I use the result from part (c), and find that\n%\n\\begin{displaymath}\n v =\n \\sqrt{-\\phi} \\approx\n \\num{9.933e-5}\n\\end{displaymath}\n\n(b) If the potential due to the Sun at Earth's orbital radius is greater than the Earth's potential at its surface (as is shown above), then why do we feel the Earth's gravity more than the Sun's?\n\nWe don't feel the potential directly, we feel the gravitational acceleration it produces. Acceleration is obtained from the potential via $\\vb{a} = -\\grad\\phi$, and in the case of a circular orbit in a Newtonian potential:\n%\n\\begin{displaymath}\n a =\n -\\grad\\phi =\n -\\pdv{r} (-G M / r) =\n -G m / r^2 =\n \\phi / r.\n\\end{displaymath}\n%\nSo in the two cases mentioned, we need to divide by the radius once more, to obtain the acceleration.\n%\n\\begin{align*}\n a_\\Sun &=\n \\phi_\\Sun / \\SI{1}{AU} \\approx\n \\SI{-6.595e-20}{\\per\\meter}\n \\\\\n a_\\Earth &=\n \\phi_\\Earth / R_\\Earth \\approx\n \\SI{-1.092e-16}{\\per\\meter}\n\\end{align*}\n%\nAs you can see, the acceleration due to the Earth is greater by a factor of $10^4$.\n\n(c) Show that a circular orbit in a Newtonian potential has an orbital velocity $v^2 = -\\phi$.\n\nWe saw above that $a = \\phi / r$, and we also know that centripetal acceleration is given by $a = -v^2 / r$. Equating the two we get $v^2 = -\\phi$.\n\n\n\\textbf{8}\n\n(a) Show that $\\tensor{R}{^\\alpha_{\\beta\\mu\\nu}} = \\eta^{\\alpha\\sigma} R_{\\alpha\\beta\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2}$.\n\n\\begin{align*}\n \\tensor{R}{^\\alpha_{\\beta\\mu\\nu}} &=\n g^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu} =\n (\\eta^{\\alpha\\sigma} + h^{\\alpha\\sigma}) R_{\\sigma\\beta\\mu\\nu} =\n \\eta^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu} + h^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu}\n \\\\\n h^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu} &=\n \\frac{1}{2} h^{\\alpha\\sigma}\n (h_{\\sigma\\nu,\\beta\\mu} + h_{\\beta\\mu,\\sigma\\nu} -\n h_{\\sigma\\mu,\\beta\\nu} - h_{\\beta\\nu,\\sigma\\mu}) =\n \\order*{[h_{\\alpha\\beta}]^2}\n\\end{align*}\n\n(b) Find $R_{\\alpha\\beta}$ to first order in $h_{\\mu\\nu}$.\n\n\\begin{align*}\n \\tensor{R}{^\\alpha_{\\beta\\mu\\nu}} &\\approx\n \\eta^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu}\n \\\\\n \\tensor{\\delta}{^\\mu_\\alpha} \\tensor{R}{^\\alpha_{\\beta\\mu\\nu}} &\\approx\n R_{\\beta\\nu} \\approx\n \\tensor{\\delta}{^\\mu_\\alpha} \\eta^{\\alpha\\sigma} R_{\\sigma\\beta\\mu\\nu} \\approx\n \\eta^{\\mu\\sigma} R_{\\sigma\\beta\\mu\\nu}\n\\end{align*}\n\n(c) Show that $g_{\\alpha\\beta} R = \\eta_{\\alpha\\beta} \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2}$.\n\n\\begin{align*}\n R &=\n g^{\\mu\\nu} R_{\\mu\\nu} =\n (\\eta^{\\mu\\nu} + h^{\\mu\\nu}) R_{\\mu\\nu} =\n \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\eta^{\\mu\\gamma} \\eta^{\\nu\\lambda} R_{\\mu\\nu} =\n \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2}\n \\\\\n g_{\\alpha\\beta} R &=\n g_{\\alpha\\beta} \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2} =\n (\\eta_{\\alpha\\beta} + h_{\\alpha\\beta})\n \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2} =\n \\eta_{\\alpha\\beta} \\eta^{\\mu\\nu} R_{\\mu\\nu} + \\order*{[h_{\\alpha\\beta}]^2}\n\\end{align*}\n\n(d) Use this to show that $G_{\\alpha\\beta} = R_{\\alpha\\beta} - \\frac{1}{2} \\eta_{\\alpha\\beta} R$.\n\n\\begin{displaymath}\n G_{\\alpha\\beta} =\n R_{\\alpha\\beta} - \\frac{1}{2} g_{\\alpha\\beta} R =\n R_{\\alpha\\beta} - \\frac{1}{2} (\\eta_{\\alpha\\beta} \\eta^{\\mu\\nu} R_{\\mu\\nu}) =\n R_{\\alpha\\beta} - \\frac{1}{2} \\eta_{\\alpha\\beta} R\n\\end{displaymath}\n\n(e) Now use this to simplify the calculation of Equation 8.32.\n\nI got stuck here. I began by expanding the expression in (d) using the results from previous sections, shown in Figure \\ref{fig:ch8-problem8e-i}. Then I expanded Equation 8.32, to get it in a more similar form, in Figures \\ref{fig:ch8-problem8e-ii} and \\ref{fig:ch8-problem8e-iii}. I did this with the hope of matching terms in the two equations, but was only able to match one. I believe something that would help me get further is Equation 8.33, $\\tensor{\\bar{h}}{^{\\mu\\nu}_{,\\nu}} = 0$.\n\n\\begin{figure}[ht]\n \\centering\n \\begin{subfigure}[b]{0.8\\textwidth}\n \\includegraphics[width=\\textwidth]{img/ch8_problem_8e_part3}\n \\caption{}\n \\label{fig:ch8-problem8e-i}\n \\end{subfigure}\n\n \\begin{subfigure}[b]{0.5\\textwidth}\n \\includegraphics[width=\\textwidth]{img/ch8_problem_8e_part1}\n \\caption{}\n \\label{fig:ch8-problem8e-ii}\n \\end{subfigure}\n ~\n \\begin{subfigure}[b]{0.3\\textwidth}\n \\includegraphics[width=\\textwidth]{img/ch8_problem_8e_part2}\n \\caption{}\n \\label{fig:ch8-problem8e-iii}\n \\end{subfigure}\n \\caption{Incomplete solution to Problem 8.8 (e)}\n\\end{figure}\n\n\n\\textbf{9}\n\n(a)\n\n% I start this by making a simplification on Equation 8.32, by using Equation 8.33. In 8.32 we have the term $\\tensor{\\bar{h}}{_{\\mu\\nu}^{,\\mu\\nu}}$. If we can get this into the form of 8.33, we can cancel it out. This is easy to show\n% %\n% \\begin{displaymath}\n% \\tensor{\\bar{h}}{_{\\mu\\nu}^{,\\mu\\nu}} =\n% \\eta^{\\mu\\alpha} \\eta^{\\nu\\beta} \\tensor{\\bar{h}}{_{\\mu\\nu,\\mu\\nu}} =\n% \\tensor{\\bar{h}}{^{\\alpha\\beta}_{,\\alpha\\beta}} = 0.\n% \\end{displaymath}\n% %\n% So Equation 8.32 then becomes\n% %\n% \\begin{displaymath}\n% G_{\\alpha\\beta} =\n% -\\frac{1}{2} [\n% \\tensor{\\bar{h}}{_{\\alpha\\beta,\\mu}^{,\\mu}} -\n% \\tensor{\\bar{h}}{_{\\alpha\\mu,\\beta}^{,\\mu}} -\n% \\tensor{\\bar{h}}{_{\\beta\\mu,\\alpha}^{,\\mu}} +\n% \\order*{[h_{\\alpha\\beta}]^2}\n% ]\n% \\end{displaymath}\n\nI start by making a slight rewrite of Equation 8.32, changing the second $\\bar{h}$ term.\n%\n\\begin{displaymath}\n \\eta_{\\alpha\\beta} \\tensor{\\bar{h}}{_{\\mu\\nu}^{,\\mu\\nu}} =\n \\eta_{\\alpha\\beta} \\eta^{\\mu\\alpha} \\tensor{\\bar{h}}{_{\\mu\\nu,\\alpha}^{,\\nu}} =\n \\tensor{\\bar{h}}{_{\\beta\\nu,\\alpha}^{,\\nu}}\n\\end{displaymath}\n%\nSo now the Einstein tensor can be written as\n%\n\\begin{displaymath}\n G_{\\alpha\\beta} =\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{\\alpha\\beta,\\mu}^{,\\mu}} +\n \\tensor{\\bar{h}}{_{\\beta\\nu,\\alpha}^{,\\nu}} -\n \\tensor{\\bar{h}}{_{\\alpha\\mu,\\beta}^{,\\mu}} -\n \\tensor{\\bar{h}}{_{\\beta\\mu,\\alpha}^{,\\mu}} +\n \\order*{[h_{\\alpha\\beta}]^2}\n ].\n\\end{displaymath}\n%\nFor $G_{00}$ we then have\n%\n\\begin{align*}\n G_{00} &=\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{00,\\mu}^{,\\mu}} +\n \\tensor{\\bar{h}}{_{0\\nu,0}^{,\\nu}} -\n \\tensor{\\bar{h}}{_{0\\mu,0}^{,\\mu}} -\n \\tensor{\\bar{h}}{_{0\\mu,0}^{,\\mu}} +\n \\order*{[h_{00}]^2}\n ]\n \\\\ &=\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{00,\\mu}^{,\\mu}} -\n \\tensor{\\bar{h}}{_{0\\mu,0}^{,\\mu}} +\n \\order*{[h_{00}]^2}\n ]\n \\\\ &=\n -\\frac{1}{2} [\n (\\tensor{\\bar{h}}{_{00,0}^{,0}} + \\tensor{\\bar{h}}{_{00,i}^{,i}}) -\n (\\tensor{\\bar{h}}{_{00,0}^{,0}} + \\tensor{\\bar{h}}{_{0i,0}^{,i}}) +\n \\order*{[h_{00}]^2}\n ]\n \\\\ &=\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{00,i}^{,i}} -\n \\tensor{\\bar{h}}{_{0i,0}^{,i}} +\n \\order*{[h_{00}]^2}\n ],\n\\end{align*}\n%\nwhich contains no second time derivatives. For $G_{0i}$ I encountered a problem:\n%\n\\begin{align*}\n G_{0i} &=\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{0i,\\mu}^{,\\mu}} +\n \\tensor{\\bar{h}}{_{i\\nu,0}^{,\\nu}} -\n \\tensor{\\bar{h}}{_{0\\mu,i}^{,\\mu}} -\n \\tensor{\\bar{h}}{_{i\\mu,0}^{,\\mu}} +\n \\order*{[h_{0i}]^2}\n ]\n \\\\ &=\n -\\frac{1}{2} [\n \\tensor{\\bar{h}}{_{0i,\\mu}^{,\\mu}} -\n \\tensor{\\bar{h}}{_{0\\mu,i}^{,\\mu}} +\n \\order*{[h_{0i}]^2}\n ]\n \\\\ &=\n -\\frac{1}{2} [\n (\\tensor{\\bar{h}}{_{0i,0}^{,0}} + \\tensor{\\bar{h}}{_{0i,j}^{,j}}) -\n \\tensor{\\bar{h}}{_{0\\mu,i}^{,\\mu}} +\n \\order*{[h_{0i}]^2}\n ],\n\\end{align*}\n%\nwhich retains a second time derivative in the $\\tensor{\\bar{h}}{_{0i,0}^{,0}}$ term.\n\n(b)\n\nAccording to Schutz's solution it is not a contradiction, due in part to Equation 8.33. I don't fully understand the reason, though.\n\n\n\\textbf{11}\nWrite the gauge transformation and Lorentz gauge condition in four-tensor notation for Maxwell's equations. Draw an analogy with linearized gravity.\n\nFirst we rewrite $\\phi \\to \\phi - \\pdv*{f}{t}$ as $-A_0 \\to -A_0 - f_{,0}$, and cancelling the negatives we get $A_0 \\to A_0 + f_{,0}$. Combining this with $A_i \\to A_i + f_{,i}$, it is obvious that the gauge transformation generalizes to $A_\\alpha \\to A_\\alpha + f_{,\\alpha}$.\n\nThe Lorentz gauge condition is just slightly less obvious. We start by noting that $A_0 = -\\phi$, and therefore (in Minkowski space) $A^0 = \\eta^{0\\mu} A_\\mu = g^{00} A_0 = (-1) (-\\phi) = \\phi$. Then the Lorentz gauge condition becomes $\\phi_{,0} + \\grad_i A^i = \\tensor{A}{^0_{,0}} + \\tensor{A}{^i_{,i}} = \\tensor{A}{^\\alpha_{,\\alpha}} = 0$. This is analogous to $\\tensor{\\bar{h}}{^{\\mu\\nu}_{,\\nu}}$.\n\n\n\n\\textbf{13}\nGive a physical justification for $\\abs{T^{00}} \\gg \\abs{T^{0i}} \\gg \\abs{T^{ij}}$ in a Newtonian system.\n\nThe first inequality is easy to see. $T^{00} = E / V = p^0 / V$, and $T^{0i} = p^i / V$. In the Newtonian limit, $\\abs{p^0} \\gg \\abs{p^i}$, and so it follows that $\\abs{T^{00}} \\gg \\abs{T^{0i}}$.\n\nThe second inequality is less obvious. In the Newtonian limit, forces must be relatively small, or else objects would be accelerated to relativistic speeds. By this argument, the stresses must also be relatively small, and so $T^{0i} \\gg T^{ij}$.\n\n\\textbf{17}\n\n(a) First I need to convert the orbital period into meters.\n%\n\\begin{displaymath}\n T =\n \\SI{200}{days} \\times\n \\frac{\\SI{24}{hours}}{\\SI{1}{day}} \\times\n \\frac{\\SI{3600}{seconds}}{\\SI{1}{hour}} \\times\n c \\approx\n \\SI{5.18e15}{\\meter}\n\\end{displaymath}\n%\nThen I use the potential to find the speed, which I relate to the circumference and orbital period, and solve for the mass.\n%\n\\begin{gather*}\n \\phi = -G M / r\n \\\\\n v^2 = -\\phi\n\\end{gather*}\n\\begin{align*}\n M &= v^2 r / G = C^3 / (2 \\pi T^2 G) \\approx\n (\\SI{6e11}{\\meter})^3 / (2 \\pi (\\SI{5.18e15}{\\meter})^2 G)\n \\\\ &\\approx\n \\SI{1.281e3}{\\meter} \\times \\frac{1 M_\\Sun}{\\SI{1.476e3}{m}} \\approx\n 0.868 M_\\Sun\n\\end{align*}\n\n(b)\n\nUsing the above formula, I get a distribution of mass estimates, shown in Figure \\ref{fig:ch8-problem17b}. Closer to the black hole, the Newtonian approximation breaks down, and the ``effective mass'' blows up. Far from the black hole, we can see that the effective mass is in agreement for all of the sattelites, and so the Newtonian approximation is working again. Thus, I use the furthest sattelite to find that the black hole's mass is $68 M_\\Sun$.\n\n\\begin{figure}[ht]\n \\centering\n \\includegraphics[width=0.6\\textwidth]{img/ch8_problem_17b}\n \\caption{Black hole mass estimates in Problem 8.17, as a function of sattelite circumference.}\n \\label{fig:ch8-problem17b}\n\\end{figure}\n\n\n\n\\end{document}", "meta": {"hexsha": "c6100bff1ee48c913bd5a8c628d31bfdf7023c72", "size": 11214, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes/textbook/tex/gr-ch8-notes.tex", "max_stars_repo_name": "dwysocki/ASTP-760", "max_stars_repo_head_hexsha": "da116f346bfa40fa3f9620782a2c85e9e9654e37", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "notes/textbook/tex/gr-ch8-notes.tex", "max_issues_repo_name": "dwysocki/ASTP-760", "max_issues_repo_head_hexsha": "da116f346bfa40fa3f9620782a2c85e9e9654e37", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes/textbook/tex/gr-ch8-notes.tex", "max_forks_repo_name": "dwysocki/ASTP-760", "max_forks_repo_head_hexsha": "da116f346bfa40fa3f9620782a2c85e9e9654e37", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.8791540785, "max_line_length": 489, "alphanum_fraction": 0.6182450508, "num_tokens": 4352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390747, "lm_q2_score": 0.7956581073313275, "lm_q1q2_score": 0.6141235366466297}} {"text": "\\chapter{Cryptol prelude functions}\n\n\\commentout{\n\\begin{code}\nprimsPlaceHolder=1;\n\\end{code}\n}\n\n\\paragraph*{Bitwise and logical operations}\n\\begin{Verbatim}\n True, False : Bit\n &&, ||, ^ : {a} (Logic a) => a -> a -> a\n ~ : {a} (Logic a) => a -> a\n ==>, /\\, \\/ : Bit -> Bit -> Bit\n\\end{Verbatim}\n\\paragraph*{Comparisons}\n\\begin{Verbatim}\n ==, != : {a} (Eq a) => a -> a -> Bit\n <, >, <=, >= : {a} (Cmp a) => a -> a -> Bit\n <$, >$, <=$, >=$ : {a} (SignedCmp a) => a -> a -> Bit\n min, max : {a} (Cmp a) => a -> a -> a\n ===, !== : {a, b} (Eq b) => (a -> b) -> (a -> b) -> a -> Bit\n\\end{Verbatim}\n\\paragraph*{Arithmetic}\n\\begin{Verbatim}\n +, -, * : {a} (Ring a) => a -> a -> a\n negate : {a} (Ring a) => a -> a\n fromInteger : {a} (Ring a) => Integer -> a\n ^^ : {a, e} (Ring a, Integral e) => a -> e -> a\n abs : {a} (Cmp a, Ring a) => a -> a\n /, % : {a} (Integral a) => a -> a -> a\n toInteger : {a} (Integral a) => a -> Integer\n lg2 : {n} (fin n) => [n] -> [n]\n /$, %$ : {n} (fin n, n >= 1) => [n] -> [n] -> [n]\n carry : {n} (fin n) => [n] -> [n] -> Bit\n scarry, sborrow : {n} (fin n, n >= 1) => [n] -> [n] -> Bit\n zext : {m, n} (fin m, m >= n) => [n] -> [m]\n sext : {m, n} (fin m, m >= n, n >= 1) => [n] -> [m]\n ratio : Integer -> Integer -> Rational\n /. : {a} (Field a) => a -> a -> a\n recip : {a} (Field a) => a -> a\n floor : {a} (Round a) => a -> Integer\n ceiling : {a} (Round a) => a -> Integer\n trunc : {a} (Round a) => a -> Integer\n roundAway : {a} (Round a) => a -> Integer\n roundToEven : {a} (Round a) => a -> Integer\n\\end{Verbatim}\n\\paragraph*{GF(2) polynomial arithmetic}\n\\begin{Verbatim}\n pdiv : {u, v} (fin u, fin v) => [u] -> [v] -> [u]\n pmod : {u, v} (fin u, fin v) => [u] -> [1 + v] -> [v]\n pmult : {u, v} (fin u, fin v) => [1 + u] -> [1 + v] -> [1 + u + v]\n\\end{Verbatim}\n\\paragraph*{Sequences}\n\\begin{Verbatim}\n take : {front, back, a} (fin front) => [front + back]a -> [front]a\n drop : {front, back, a} (fin front) => [front + back]a -> [back]a\n # : {front, back, a} (fin front) => [front]a -> [back]a -> [front + back]a\n join : {parts, each, a} (fin each) => [parts][each]a -> [parts * each]a\n split : {parts, each, a} (fin each) => [parts * each]a -> [parts][each]a\n groupBy : {each, parts, a} (fin each) => [parts * each]a -> [parts][each]a\n transpose : {rows, cols, a} [rows][cols]a -> [cols][rows]a\n reverse : {n, a} (fin n) => [n]a -> [n]a\n head : {n, a} [1 + n]a -> a\n tail : {n, a} [1 + n]a -> [n]a\n last : {n, a} (fin n) => [1 + n]a -> a\n\\end{Verbatim}\n\\paragraph*{Indexing, updates}\n\\begin{Verbatim}\n @ : {n, a, ix} (Integral ix) => [n]a -> ix -> a\n ! : {n, a, ix} (fin n, Integral ix) => [n]a -> ix -> a\n @@ : {n, k, ix, a} (Integral ix) => [n]a -> [k]ix -> [k]a\n !! : {n, k, ix, a} (fin n, Integral ix) => [n]a -> [k]ix -> [k]a\n update : {n, a, ix} (fin ix) => [n]a -> [ix] -> a -> [n]a\n updateEnd : {n, a, ix} (fin n, Integral ix) => [n]a -> ix -> a -> [n]a\n updates : {n, k, ix, a} (Integral ix, fin k) => [n]a -> [k]ix -> [k]a -> [n]a\n updatesEnd : {n, k, ix, a} (fin n, Integral ix, fin k) => [n]a -> [k]ix -> [k]a -> [n]a\n\\end{Verbatim}\n\\paragraph*{Shifting, rotating}\n\\begin{Verbatim}\n >>>, <<< : {n, ix, a} (fin n, Integral ix) => [n]a -> ix -> [n]a\n >>, << : {n, ix, a} (Integral ix, Zero a) => [n]a -> ix -> [n]a\n >>$ : {n, ix} (fin n, n >= 1, Integral ix) => [n] -> ix -> [n]\n\\end{Verbatim}\n\\paragraph*{Functional programming}\n\\begin{Verbatim}\n iterate : {a} (a -> a) -> a -> [inf]a\n repeat : {n, a} a -> [n]a\n map : {n, a, b} (a -> b) -> [n]a -> [n]b\n zip : {n, a, b} [n]a -> [n]b -> [n](a, b)\n zipWith : {n, a, b, c} (a -> b -> c) -> [n]a -> [n]b -> [n]c\n foldl : {n, a, b} (fin n) => (a -> b -> a) -> a -> [n]b -> a\n foldl' : {n, a, b} (fin n, Eq a) => (a -> b -> a) -> a -> [n]b -> a\n foldr : {n, a, b} (fin n) => (a -> b -> b) -> b -> [n]a -> b\n foldr' : {n, a, b} (fin n, Eq b) => (a -> b -> b) -> b -> [n]a -> b\n scanl : {n, b, a} (b -> a -> b) -> b -> [n]a -> [1 + n]b\n scanr : {n, a, b} (fin n) => (a -> b -> b) -> b -> [n]a -> [1 + n]b\n sum : {n, a} (fin n, Eq a, Ring a) => [n]a -> a\n product : {n, a} (fin n, Eq a, Ring a) => [n]a -> a\n and, or : {n} (fin n) => [n] -> Bit\n all, any : {n, a} (fin n) => (a -> Bit) -> [n]a -> Bit\n curry : {a, b, c} ((a, b) -> c) -> a -> b -> c\n uncurry : {a, b, c} (a -> b -> c) -> (a, b) -> c\n elem : {n, a} (fin n, Eq a) => a -> [n]a -> Bit\n\\end{Verbatim}\n\\paragraph*{Miscellaneous}\n\\begin{Verbatim}\n deepseq : {a, b} Eq a => a -> b -> b\n rnf : {a} Eq a => a -> a\n length : {n, a, b} (fin n, Literal n b) => [n]a -> b\n zero : {a} (Zero a) => a\n\\end{Verbatim}\n\\paragraph*{Representing exceptions}\n\\begin{Verbatim}\n undefined : {a} a\n error : {a, n} (fin n) => String n -> a\n assert : {a, n} (fin n) => Bit -> String n -> a -> a\n trace : {n, a, b} [n][8] -> a -> b -> b\n traceVal : {n, a} [n][8] -> a -> a\n\\end{Verbatim}\n\\todo[inline]{\\texttt{error} and \\texttt{undefined} are not covered in\n the book at the moment.}\n\n\\todo[inline]{What is the state of debugging (\\texttt{trace},\n \\texttt{ASSERT}), randomness (\\texttt{random}), and pretty-printing\n (\\texttt{format}) built-ins?}\n%\\paragraph*{Debugging}\n%\\begin{Verbatim}\n% trace : {a b c} ([a][8],b,c) -> c\n% ASSERT : {a b} (Bit,[a][8],b) -> b\n%\\end{Verbatim}\n%\\paragraph*{Generating random numbers}\n%\\begin{Verbatim}\n% random : {a b} (32 >= a) => [a] -> b\n%\\end{Verbatim}\n%\\paragraph*{Pretty printing}\n%\\begin{Verbatim}\n% format\n%\\end{Verbatim}\n\n% (%$) : {a} (Arith a) => a -> a -> a\n% (/$) : {a} (Arith a) => a -> a -> a\n% number : {val, rep} (Literal val rep) => rep\n% elem : {n, a} (fin n, Cmp a) => a -> [n]a -> Bit\n% fromThenTo :\n% {first, next, last, a, len} (fin first, fin next, fin last,\n% Literal first a, Literal next a, Literal last a, first != next,\n% lengthFromThenTo first next last == len) =>\n% [len]a\n% fromTo :\n% {first, last, a} (fin last, last >= first, Literal last a) =>\n% [1 + (last - first)]a\n% fromZ : {n} (fin n, n >= 1) => Z n -> Integer\n% infFrom : {a} (Arith a) => a -> [inf]a\n% infFromThen : {a} (Arith a) => a -> a -> [inf]a\n% splitAt :\n% {front, back, a} (fin front) =>\n% [front + back]a -> ([front]a, [back]a)\n% toInteger : {bits} (fin bits) => [bits] -> Integer\n% trace : {n, a, b} (fin n) => [n][8] -> a -> b -> b\n% traceVal : {n, a} (fin n) => [n][8] -> a -> a\n% undefined : {a} a\n% zext : {m, n} (fin m, m >= n) => [n] -> [m]\n% sext : {m, n} (fin m, m >= n, n >= 1) => [n] -> [m]\n% carry : {n} (fin n) => [n] -> [n] -> Bit\n% scarry : {n} (fin n, n >= 1) => [n] -> [n] -> Bit\n% sborrow : {n} (fin n, n >= 1) => [n] -> [n] -> Bit\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: \"../main/Cryptol\"\n%%% End: \n", "meta": {"hexsha": "e9616faff62bbab836e1b3d8bfc4f9319f26ee14", "size": 7409, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/ProgrammingCryptol/prims/Primitives.tex", "max_stars_repo_name": "emlisa3162/cryptol", "max_stars_repo_head_hexsha": "5a668a4594e386a084f6f0ceb231c69946971a50", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 773, "max_stars_repo_stars_event_min_datetime": "2015-01-08T15:43:54.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-23T04:26:02.000Z", "max_issues_repo_path": "docs/ProgrammingCryptol/prims/Primitives.tex", "max_issues_repo_name": "emlisa3162/cryptol", "max_issues_repo_head_hexsha": "5a668a4594e386a084f6f0ceb231c69946971a50", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1050, "max_issues_repo_issues_event_min_datetime": "2015-01-02T23:10:55.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-30T17:02:34.000Z", "max_forks_repo_path": "docs/ProgrammingCryptol/prims/Primitives.tex", "max_forks_repo_name": "axon-terminal/cryptol", "max_forks_repo_head_hexsha": "571f0dd249a72f830abd511caca87a971a91d07e", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 117, "max_forks_repo_forks_event_min_datetime": "2015-01-01T18:45:39.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-06T15:40:57.000Z", "avg_line_length": 42.8265895954, "max_line_length": 99, "alphanum_fraction": 0.4235389391, "num_tokens": 3020, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339837155239, "lm_q2_score": 0.7310585786300049, "lm_q1q2_score": 0.6138947325623826}} {"text": "\\chapter{Complete UFC code for Poisson's equation}\n\\index{Poisson's equation}\n\\index{example code}\n\nIn this section, a simple example is given of UFC code generated by\nthe form compilers\nFFC~\\cite{www:ffc,logg:article:07,logg:article:09,logg:article:10,logg:article:11}\nand SyFi~\\cite{www:syfi} for Poisson's equation. The code presented\nbelow is generated for the bilinear form of Poisson's equation for\nstandard continuous piecewise linear Lagrange finite elements on a\ntwo-dimensional domain $\\Omega$,\n\\begin{equation}\n a(v, u) = \\int_{\\Omega} \\nabla v \\cdot \\nabla u \\dx.\n\\end{equation}\n\nAlthough FFC and SyFi are two different form compilers, with very\ndifferent approaches to code generation, both generate code that\nconforms to the UFC specification and may thus be used interchangeably\nwithin any UFC-based system.\n\nIn the generated code, we have omitted the two functions\n\\texttt{evaluate\\_basis} and\n\\texttt{evaluate\\_basis\\_derivatives}\\footnotemark{} to save space.\n\n\\footnotetext{For FFC, this may be done by using the compiler flags\n \\texttt{-fno-evaluate\\_basis} and\n \\texttt{-fno-evaluate\\_basis\\_derivatives}.}\n\n\\section{Code generated by FFC}\n\\index{FFC}\n\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/poisson_ffc.h}\n\\normalsize\n\n\\section{Code generated by SyFi}\n\\index{SyFi}\n\nIn the following we list the complete code for the finite element,\nthe dofmap and the variational form for computing a stiffness matrix\nbased on linear Lagrangian elements in 2D.\n\nThe code below is generated for the assembler in PyCC and it therefore\nincludes some PyCC files, since the option \\texttt{SFC.options.include\\_from = \"pycc\"} was used during the code generation. If PyCC is not present, the option \\texttt{SFC.options.include\\_from = \"syfi\"} can be used, and this will result in some additional files that define the numbering scheme.\n\n\\subsection{Header file for linear Lagrange element in 2D}\n\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/fe_Lagrange_1_2D.h}\n\\normalsize\n\n\\subsection{Source file for linear Lagrange element in 2D}\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/fe_Lagrange_1_2D.cpp}\n\\normalsize\n\n\\subsection{Header file for the dofmap}\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/dof_map_Lagrange_1_2D.h}\n\\normalsize\n\n\\subsection{Source file for the dofmap}\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/dof_map_Lagrange_1_2D.cpp}\n\\normalsize\n\n\\subsection{Header file for the stiffness matrix form}\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/form__stiffness_form__Lagrange_1_2D.h}\n\\normalsize\n\n\n\\subsection{Source file for the stiffness matrix form}\n\\scriptsize\n\\VerbatimInput[frame=single,rulecolor=\\color{blue}]{code/stiffness_syfi/form__stiffness_form__Lagrange_1_2D.cpp}\n\\normalsize\n", "meta": {"hexsha": "35a28923226be8a670a2d83f457b31f061a33585", "size": 2934, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/ufc-2.0.5/doc/manual/chapters/examples.tex", "max_stars_repo_name": "szmurlor/fiver", "max_stars_repo_head_hexsha": "083251420eb934d860c99dcf1eb07ae5b8ba7e8c", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/ufc-2.0.5/doc/manual/chapters/examples.tex", "max_issues_repo_name": "szmurlor/fiver", "max_issues_repo_head_hexsha": "083251420eb934d860c99dcf1eb07ae5b8ba7e8c", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/ufc-2.0.5/doc/manual/chapters/examples.tex", "max_forks_repo_name": "szmurlor/fiver", "max_forks_repo_head_hexsha": "083251420eb934d860c99dcf1eb07ae5b8ba7e8c", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.1038961039, "max_line_length": 295, "alphanum_fraction": 0.80334015, "num_tokens": 845, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8397339676722393, "lm_q2_score": 0.7310585786300048, "lm_q1q2_score": 0.6138947208338016}} {"text": "\\documentclass{article}\n\n\\usepackage{mathrsfs,amssymb,amsmath,amsfonts}\n\n\\newcommand*\\conj[1]{\\overline{#1}}\n\n\\begin{document}\n\n\\section{Let $e_0, e_1, ...$ be an orthonormal basis for $\\mathscr{H}$ and let $\\alpha_0, \\alpha_1, ...$ be complex numbers. Define $\\mathscr{D} = \\{ h \\in \\mathscr{H} : \\sum_{0}^{\\infty} | \\alpha_n \\langle h, e_n \\rangle |^2 < \\infty \\}$ and let $Ah= \\sum_{0}^{\\infty} \\alpha_n \\langle h, e_n \\rangle e_n$ for $h$ in $\\mathscr{D}$. Then $A \\in \\mathscr{C}(\\mathscr{H})$ with dom $A=\\mathscr{D}$. Also, dom $A^*=\\mathscr{D}$ and $A^*h=\\sum_{0}^{\\infty} \\conj{\\alpha_n} \\langle h, e_n \\rangle e_n$ for all $h$ in $\\mathscr{D}$}\n\nExercise X.1.2: Prove claims in X.1.9\n\nSince Hilbert Spaces in the chapter are assumed to be separable, to show that $\\mathscr{D}$ is dense we must show that for every $h \\in \\mathscr{H}$, there exists a sequence $h_{\\bullet}$ in $\\mathscr{D}$ that converges to $h$. Let $h = \\sum_{0}^{\\infty} \\beta_n e_n$ where all but finite number of $\\beta_n$ are 0. ???\n\nNext we have to show that $A$ is closed, that is $\\{ h \\oplus Ah : h \\in \\mathscr{D}\\}$ is closed in $\\mathscr{H} \\oplus \\mathscr{H}$. Again since $\\mathscr{H}$ is separable, this amounts to showing that the limit of an arbitrary sequence $h_{\\bullet}$ of $\\mathscr{D}$ is still in $h \\oplus Ah$. Again not sure where to go from here.\n\nTo show that $A^*h=\\sum_{0}^{\\infty} \\conj{\\alpha_n} \\langle h, e_n \\rangle e_n$, start with $\\langle Ah, k \\rangle = \\langle h, A^*k \\rangle$. \n\n$ \\langle \\sum_{0}^{\\infty} \\alpha_n \\langle h, e_n \\rangle e_n, k \\rangle$\n\n\\end{document}", "meta": {"hexsha": "b01aea4887b80cb58c06b8acda9f1eb601ec2471", "size": 1590, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "analysis/10_Unbounded_Operators/problems.tex", "max_stars_repo_name": "lukemassa/math-exercises", "max_stars_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "analysis/10_Unbounded_Operators/problems.tex", "max_issues_repo_name": "lukemassa/math-exercises", "max_issues_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "analysis/10_Unbounded_Operators/problems.tex", "max_forks_repo_name": "lukemassa/math-exercises", "max_forks_repo_head_hexsha": "765b84eb0a1b5ab59576172e2a814a1862a4f129", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 75.7142857143, "max_line_length": 527, "alphanum_fraction": 0.6610062893, "num_tokens": 590, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396211, "lm_q2_score": 0.7690802476562641, "lm_q1q2_score": 0.6138696846999433}} {"text": "\\documentclass{tufte-handout}\n\n\\usepackage{xcolor}\n\\usepackage{graphicx}\n\n% set hyperlink attributes\n\\hypersetup{colorlinks}\n\n\\usepackage{amsmath}\n\n% ============================================================\n\n% define the title\n\\title{SOC 4930/5050: Week 10 Equations}\n\\author{Christopher Prener, Ph.D.}\n\\date{October 30\\textsuperscript{th}, 2017}\n% ============================================================\n\\begin{document}\n% ============================================================\n\\maketitle % generates the title\n% ============================================================\n\n\\vspace{5mm}\n\\section{Pearson's r}\n\\begin{subequations}\n\\paragraph{Calculating r}\n\\begin{equation}\n\\scalebox{2} {$ r =\\frac { \\sum _{ i=1 }^{ n }{ (x-\\overline { x } )(y-\\overline { y } ) } }{ (n-1){s}_{x}{s}_{y} } $}\n\\end{equation}\n\n\\vspace{3mm}\n\\paragraph{Percent of Variance Explained}\n\\begin{equation}\n\\scalebox{2} {$ {r}^{2} $}\n\\end{equation}\n\n\\vspace{3mm}\n\\paragraph{Calculating Degrees of Freedom (v)}\n\\begin{equation}\n\\scalebox{2} {$ v=n-2 $}\n\\end{equation}\n\n\\vspace{3mm}\n\\paragraph{Calculating t}\n\\begin{equation}\n\\scalebox{2} {$ t=\\frac{r}{\\sqrt{\\frac{1-{r}^{2}}{v}}} $}\n\\end{equation}\n\\end{subequations}\n\n% ============================================================\n\\end{document}", "meta": {"hexsha": "f816f70fa7dab5eb1e5ff4c4b4856c734dc15768", "size": 1285, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week-10-equations.tex", "max_stars_repo_name": "slu-soc5050/Equations", "max_stars_repo_head_hexsha": "6398ba3a3d351d95a37d3245fcc6eabb91fdd032", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "week-10-equations.tex", "max_issues_repo_name": "slu-soc5050/Equations", "max_issues_repo_head_hexsha": "6398ba3a3d351d95a37d3245fcc6eabb91fdd032", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "week-10-equations.tex", "max_forks_repo_name": "slu-soc5050/Equations", "max_forks_repo_head_hexsha": "6398ba3a3d351d95a37d3245fcc6eabb91fdd032", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.1960784314, "max_line_length": 119, "alphanum_fraction": 0.5182879377, "num_tokens": 379, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7690802423634963, "lm_q1q2_score": 0.613869680475326}} {"text": "\\documentclass[12pt]{article}\n\\usepackage{amsmath}\n\n\\title {Jacobian Free Newton's Method for Non-Linear Systems with Composit Residual Functions via MAUD}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Problem Definition}\n\nComposite resids arise when solving a non-linear system whos residual evaluation is a function of two\nor more different components. For example, consider a non linear system with state variable vector, $u$, and\nresidual vector, $r$, computed via two components, $Y$ and $R$. Assume $u$ and $r$ are of size $n$.\n\n\\begin{equation}\n u \\longrightarrow Y \\longrightarrow y \\longrightarrow R \\longrightarrow r\n\\end{equation}\n\n$y$ and $r$ are explicit functions of $u$ and $y$ respectively\n\n\\begin{align}\n y = & Y(u) \\notag \\\\\n r = & R(y)\n\\end{align}\n\nA composite residual function, $G$, is then defined\n\n\\begin{equation}\n G(u) = R(Y(u))\n\\end{equation}\n\nThe problem definition is then given as follows\n\n\\begin{align}\n Find &\\ u \\notag \\\\\n s.t. &\\ G(u) = 0\n\\end{align}\n\n\\section{Traditional Newtons Method}\n\nThe Newton update, $\\Delta u$, is computed via\n\\begin{equation}\n \\frac{\\partial G}{\\partial u} \\Delta u = -G(u)\n\\end{equation}\n\nExpanding the partial derivative of $G$ out to account for the composite residual function\ngives\n\n\\begin{equation}\n \\frac{\\partial R}{\\partial y} \\frac{\\partial Y}{\\partial u} \\Delta u = -R\\left(Y(u)\\right)\n \\label{expanded_newton_iter}\n\\end{equation}\n\nIn order to compute $\\Delta u$, the partial derivatives\n$\\frac{\\partial R}{\\partial y}$ and $\\frac{\\partial Y}{\\partial u}$ must first be computed.\nThen a matrix multiplication is performed ($\\mathcal{O}(n^3)$) to get $\\frac{\\partial G}{\\partial u}$.\nLastly the solution to a size $n$ linear system is found.\n\n\\section{MAUD Formulation}\n\nRepresenting the composite residual as a system of non linear equations using the MAUD yeilds\ntwo separate, but coupled, residual functions with $y$ and $r$ as state variables\nand $u$ as an input.\n\n\\begin{align}\n A(u,y,r) & = y - Y(u) \\label{MAUD_y} \\\\\n B(u,y,r) & = r - R(y) \\label{MAUD_r}\n \\label{maud1}\n\\end{align}\n\nThe original problem formulation seeks to drive $r$ to 0. So we'll just assume\nthat $r$ is 0, then find values of $u$ and $y$ that drive\nthe remaining resids to 0. This gives a new system of equations:\n\n\\begin{align}\n C(u,y) & = y - Y(u) = 0 \\label{y_resid} \\\\\n D(r,y) & = R(y) = 0 \\label{r_resid}\n \\label{maud2}\n\\end{align}\n\nYou can compute a Newton update for the non-linear system via\n\n\\begin{equation}\n \\begin{bmatrix}\n \\frac{\\partial C}{\\partial u} & \\frac{\\partial C}{\\partial y} \\\\\n \\frac{\\partial D}{\\partial u} & \\frac{\\partial D}{\\partial y}\n \\end{bmatrix}\n \\begin{bmatrix}\n \\Delta u \\\\\n \\Delta y\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n -C(u,y) \\\\\n -D(u,y)\n \\end{bmatrix}\n\\end{equation}\n\n\\begin{equation}\n \\begin{bmatrix}\n -\\frac{\\partial Y}{\\partial u} & 1 \\\\\n 0 & \\frac{\\partial R}{\\partial y}\n \\end{bmatrix}\n \\begin{bmatrix}\n \\Delta u \\\\\n \\Delta y \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n Y(u) - y \\\\\n - R(y)\n \\end{bmatrix}\n\\end{equation}\n\nExpanding out the linear system yields\n\n\\begin{align}\n -\\frac{\\partial Y}{\\partial u} \\Delta u + \\Delta y & = Y(u) - y \\label{Delta_u} \\\\\n \\frac{\\partial R}{\\partial y} \\Delta y & = -R(y) \\label{Delta_y}\n\\end{align}\n\n\nMultiplying both sides by of Eqn.~\\ref{Delta_u} by $\\frac{\\partial R}{\\partial y}$ yields\n\n\\begin{equation}\n -\\frac{\\partial R}{\\partial y} \\frac{\\partial Y}{\\partial u} \\Delta u + \\frac{\\partial R}{\\partial y} \\Delta y = Y(u) - y\n\\end{equation}\n\nNow, assuming that you always executed $Y$ first and then $R$, you know that $Y(u) - y$ will always be identically 0.\n\n\\begin{equation}\n -\\frac{\\partial R}{\\partial y} \\frac{\\partial Y}{\\partial u} \\Delta u + \\frac{\\partial R}{\\partial y} \\Delta y = 0\n \\label{newton_update_1}\n\\end{equation}\n\nFinally, substituting Eqn.~\\ref{Delta_y} into Eqn.~\\ref{newton_update_1} yeilds\n\n\\begin{equation}\n \\frac{\\partial R}{\\partial y} \\frac{\\partial Y}{\\partial u} \\Delta u = -R(y)\n \\label{maud_Delta_u_update}\n\\end{equation}\n\nWe now note that Eqn.~\\ref{maud_Delta_u_update} and Eqn.~\\ref{expanded_newton_iter} are equivalent, hence\nthe resulting $\\Delta u$ value computed via the traditional method and the MAUD method are also\nequivalent.\n\nHowever, the MAUD provides a computational savings. The partial derivatives $\\frac{\\partial R}{\\partial y}$\nand $\\frac{\\partial Y}{\\partial u}$ must still be computed. But the newton update, $\\Delta u$ can now be found\ndirectly via a single solution to a linear system of size $n$ without the expensive matrix\nmultiplication step needed with the traditional method.\n\n\\end{document}\n", "meta": {"hexsha": "b11c58ea2c61375a8ca94f718b6ce8b433fdb455", "size": 4658, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "theory/newton_solve_eqivilence.tex", "max_stars_repo_name": "naylor-b/OpenMDAO1", "max_stars_repo_head_hexsha": "49d82f6601b33db9bdcf7d146d030d55e3b62ef4", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2018-01-11T20:13:59.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T03:46:05.000Z", "max_issues_repo_path": "theory/newton_solve_eqivilence.tex", "max_issues_repo_name": "naylor-b/OpenMDAO1", "max_issues_repo_head_hexsha": "49d82f6601b33db9bdcf7d146d030d55e3b62ef4", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2017-10-19T23:14:14.000Z", "max_issues_repo_issues_event_max_datetime": "2020-11-22T17:30:57.000Z", "max_forks_repo_path": "theory/newton_solve_eqivilence.tex", "max_forks_repo_name": "naylor-b/OpenMDAO1", "max_forks_repo_head_hexsha": "49d82f6601b33db9bdcf7d146d030d55e3b62ef4", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2018-04-12T22:13:33.000Z", "max_forks_repo_forks_event_max_datetime": "2020-05-07T10:02:59.000Z", "avg_line_length": 30.2467532468, "max_line_length": 123, "alphanum_fraction": 0.6878488622, "num_tokens": 1473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396211, "lm_q2_score": 0.7690802423634961, "lm_q1q2_score": 0.6138696804753259}} {"text": "%\\ctparttext{\\color{black}\\begin{center}\n%\t\tEsta es una descripción de la parte de informática.\n%\\end{center}}\n\n%\\part{Parte de informática}\n\\chapter{Quantum Mechanics Model}\n\n\nQuantum Mechanics are a mathematical framework in which quantum physics are developed. In this section, we will develop a quantum mechanics model in order to understand quantum computing. The Quantum Postulates will be our guidance. They provide a connection between the physical world and the mathematical formalization. We will provide context and formalization for each postulate, so both the mathematical precision and intuition notions are developed at the same time. This development is based on \\cite{Nielsen2002}, \\cite{Manzano2020} and \\cite{Bayens2019}.\n\n\n\\section{Postulate 1: State Space}\n\n\nThe first postulate sets the environment in which we will operate: The State Space. It will be a Hilbert space associated to a physicial system. Let us rigorously define the necessary concepts using the Bra-ket notation. We will start by revisiting the required linear algebra.\n\n\n\\subsection{Bra-ket notation}\n\n\nLet $V$ be a complex vector space. That is, a vector space over $\\C$. We will restrict our study to finite complex vector spaces. If $z$ is a vector in $V$, we will denote its coordinates either as $z = (z_1, z_2, \\dotsc, z_n)$ or by column notation:\n\n$$ z = \n\\begin{pmatrix}\n\tz_1\\\\\n\tz_2 \\\\\n\t\\vdots \\\\\n\tz_n\n\\end{pmatrix}\n$$\n\nSince $V$ is a vector space we have two basic operations: (vector) addition and scalar multiplication.\n\nIn quantum mechanics, the usual notation is the Dirac's, also known as \\emph{bra-ket} notation. In this context, vectors in a complex vector space are denoted as $|\\varphi\\ra$ and are known as \\emph{kets}. The only exception to this is the zero vector, which will be denoted as $0 = (0, \\dotsc, 0)$ instead of $|0\\ra$ since $|0\\ra$ will be used as something completely different. A \\emph{vector subspace} $W$ of $V$ is a subset of $V$ closed for addition and scalar multiplication.\n\nA \\emph{base} of a vector space is a set of vectors $|v_1\\ra, \\dotsc, |v_n\\ra$ such that they are linearly independent and any given vector $|v\\ra$ can be written as a linear combination of them: $|v\\ra = \\sum_{i=1}^n \\alpha_i|v_i\\ra$. The \\emph{dimension} of a vector space is the number of elements in any of its bases, which is independent from the chosen base.\n\n\\begin{definition}\n\tGiven two complex vector spaces $V$ and $W$, a \\emph{linear operator} is an application $M: V \\rightarrow W $ that is linear in its inputs:\n\t\n\t$$ M \\Big( \\alpha |u\\ra + \\beta |w\\ra \\Big) = \\alpha M (|u\\ra) + \\beta M (|w\\ra) $$\n\\end{definition}\n\nIf $V$ to $W$ have dimensions $n$ and $m$ respectively, there is a bijection between the operators from $V$ to $W$ and the $n$ by $m$ matrices. Given an operator $M$, the obtained matrix $M'$ is called the \\emph{matrix representation} of the linear operator. Furthermore, $M(|u\\ra) = M' \\cdot |u\\ra$, so we usually denote the linear operator and its matrix representation by the same letter, and $M(|u\\ra)$ simply as $M|u\\ra$.\n\nWe will usually refer to linear operators simply as \\emph{operators}.\n\n\n\\subsection{Inner product and Hilbert Spaces}\n\n\nLet us define another operation within the complex vector spaces.\n\n\\begin{definition}\n\tLet $V$ be a complex vector space. An inner product $\\la \\cdot | \\cdot \\ra: V^2 \\rightarrow \\C$ is a function such that:\n\t\n\t\\quad 1) $\\la \\cdot | \\cdot \\ra$ is sesquilinear. That is,\n\t\n\t\\qquad 1.1) $\\la \\cdot | \\cdot \\ra$ is conjugate symmetric: for all $u,v$ in $V$, $\\la u | v \\ra = \\overline{\\la v | u \\ra}$.\n\t\n\t\\qquad 1.2) $\\la \\cdot | \\cdot \\ra$ is linear on the second variable: for all $u,v,w$ in $V$ and $\\alpha, \\beta$ in $\\C$:\n\t\n\t$$ \\la u | \\alpha v + \\beta w \\ra = \\alpha \\la u | v \\ra + \\beta \\la u | w \\ra $$\n\t\n\t\\quad 2) $\\la \\cdot | \\cdot \\ra$ is definite positive. That is, for all $u$ in $V$, $\\la u | u \\ra \\geq 0$ and $\\la u | u \\ra = 0 \\Longleftrightarrow v = 0$.\n\\end{definition}\n\nGiven this properties it can easily be proven that $\\la \\cdot | \\cdot \\ra$ is also conjugate linear on the first variable. That is, for all $u,v,w$ in $V$ and $\\alpha, \\beta$ in $\\C$:\n\n$$ \\la \\alpha u + \\beta v | w \\ra = \\overline{\\alpha} \\la u | w \\ra + \\overline{\\beta} \\la v | w \\ra $$\n\nWe will sometimes denote the inner product $\\la \\cdot | \\cdot \\ra$ as $( \\cdot , \\cdot )$ to simplify notation.\n\nTwo vectors are said to be \\emph{orthonormal} if their inner product is zero. We define the norm of a vector $|v\\ra$ by:\n\n$$\\parallel |v\\ra \\parallel \\ = \\sqrt{ \\la v|v\\ra }$$\n\nA \\emph{unit vector} is a vector $|v\\ra$ such that $\\parallel |v\\ra \\ \\parallel \\ = 1 $. We also say that $|v\\ra$ is \\emph{normalized}, and we can normalize any vector except the zero vector by dividing it by its norm.\n\nA base $|v_1\\ra, \\dotsc, |v_n\\ra$ is said to be \\emph{orthonormal} if every vector is a unit vector and they are pairwise orthogonal. That is, $\\la v_i|v_j\\ra = \\delta_{ij}$ where\n\n\\[\n\\delta_{ij} = \n\\begin{cases}\n\t1 & \\text{if } i = j \\\\\n\t0 & \\text{if } i \\neq j\n\\end{cases}\n\\]\n\n\\begin{definition}\n\tAn \\emph{inner product space} is a vector space with an associated inner product. A \\textbf{Hilbert Space} is an inner product space that is also complete.\n\\end{definition}\n\nHausdorff's Theorem states that every finite normed space is complete, therefore every finite inner product space over $\\C$ is a Hilbert space \\cite{Paya2020}. Again, by Hausdorff's theorem, we know that every $n$ dimensional Hilbert space is isomorphic to $\\C^n$. Thus, $\\C^n$ is the canonical $n$ dimensional Hilbert space. Our study will be focused on these spaces.\n\nLet $\\alpha = a + i \\cdot b\\in \\C$. We define the \\emph{conjugate}, $\\bar \\alpha$, as $\\bar \\alpha = a - i \\cdot b$. The canonical inner product in $\\C^n$ is:\n\n$$ \\la u|v \\ra \\ = \\sum_{i=1}^n \\overline{u_i}v_j $$\n\nwhere $u = (u_1, \\dotsc, u_n)$ and $v = (v_1, \\dotsc, v_n)$, for every $u,v$ in $\\C^n$.\n\n\n\\subsection{Postulate 1 statement}\n\n\nThe reader should be familiar by now with the notation and the necessary linear algebra to formulate the first postulate.\n\n\\begin{postulate}\n\tAssociated to any isolated physical system is a complex vector space with an inner product (that is, a Hilbert space) known as the \\emph{state space} of the system. The system is completely described by its \\emph{state vector}, which is a unit vector in the system’s state space.\n\\end{postulate}\n\nAn important concern with this postulate is that it does not tell us which is the state space of a given system, nor its state vector. Although up to this point we cannot formally assure this, in quantum computing the state space will be fixed: $\\C^{2^n}$ for an n-qubits system. Our evolving state vector will be a vector $2^n$-vector.\n\nLet us start by modulating a simpler system: a single qubit system.\n\n\\begin{definition}\n\tA state vector of the state space $\\C^{2}$ is called a \\textbf{qubit}. Thus, $\\C^{2}$ may be called a single qubit state space.\n\\end{definition}\n\nSuppose $|0\\ra$, $|1\\ra$ form an orthonormal basis of a 2-dimensional Hilbert space. Then, any state vector in this state space may be described as:\n\n$$ |\\varphi\\ra = \\alpha|0\\ra + \\beta|1\\ra $$\n\nwhere $\\alpha,\\beta$ are complex numbers called \\emph{amplitudes}. Thus, the condition that $|\\varphi\\ra$ is a unit vector, $\\la\\varphi|\\varphi\\ra = 1$ is equivalent to $|\\alpha|^2 + |\\beta|^2 = 1$. This is known as the \\emph{normalization condition}.\n\nWe will always think of $|0\\ra$, $|1\\ra$ as a previously fixed orthonormal base. A linear combination of state vectors $\\sum_i a_i |\\varphi_i\\ra$ is called a \\emph{superposition} of the states $|\\varphi_i\\ra$ with amplitudes $a_i$ respectively. For example, the state\n\n$$ \\frac{|0\\ra - |1\\ra}{\\sqrt 2} $$\n\nis a superposition of the states $|0\\ra$ and $|1\\ra$ with amplitudes $1/\\sqrt 2$ and $-1/\\sqrt 2$ respectively.\n\n\n\\subsection{Quantum Computation Perspective: The Quantum Bit}\n\n\nThe bit is the minimum measure of information on classical computation and classical information theory. Everything in these fields is built from scratch based on bits. Likewise, quantum computing and quantum information theory are built upon the \\textbf{qubit}.\n\nWe have reached the qubit definition from quantum physics and pure mathematics, describing the qubit as a mathematical object independent of its physical implementation. By describing them as mathematical entities we will be able to explore their properties mathematically without having to worry about the physics underneath. This allows us to construct the quantum computing and quantum information theories independently of the physical implementation.\n\nSo, intuitively, what is a qubit? Just like the classical bit, a qubit has a state. For the bit, the two only possible states are either $0$ or $1$. A qubit can take the states $|0\\ra$ and $|1\\ra$ -corresponding to the classical states $0$ and $1$- or it can be in a \\emph{linear combination} of them:\n\n$$ |\\varphi\\ra = \\alpha |0\\ra + \\beta |1\\ra $$\n\nWhere $\\alpha$ and $\\beta$ are complex numbers. Thus, we can describe a qubit as a vector in a two-dimensional complex Hilbert space (the canonical $\\C^{2}$), where $|0\\ra$ and $|1\\ra$ form an orthonormal basis called the \\emph{computational basis}. $|0\\ra$ and $|1\\ra$ will be called \\emph{computational basis states}.\n\nAt this point, the reader may ask themselves if a qubit may even physically exist, not just as a mathematical entity. After all, the first postulate states that given a \\emph{physical system}, there is an associated state space and vector states that describe the system. However, we define a qubit from state space ($\\C^{2}$) without considering a physical system. \n\nThe answer is positive: there are numerous physical systems such that their associated state spaces are ($\\C^{2}$). Thus, modeling a qubit. More intuitive examples are provided in section \\ref{qubit-examples} and a precise physical implementation is explained in section \\ref{sec:quantum-annealing-dwave}.\n\n\n\\section{Postulate 2: Measurement}\n\n\nThe second postulate describes how states are 'measured', that is, how an outside observer may look inside the system. In classical physics, consider a simple system of a moving particle. 'Measuring' would be recording, for instance, the particle mass and speed at a given time. That is, someone \\textbf{outside} the system would look \\textbf{into} the system to record some information. Lastly, in the classical computation model measuring a bit is simply retrieving its content.\n\nIn quantum physics, measuring has some unexpected and sometimes counter-intuitive properties. Let us provide some linear algebra context before formulating the second postulate.\n\n\n\\subsection{Outer product}\n\n\n\\begin{definition}\n\tLet $V, W$ be two vector spaces and $|v\\ra \\in V, |w\\ra \\in W$. We define the \\emph{outer product} between $|v\\ra$ and $|w\\ra$, $|w\\ra\\la v|$, as the only linear operator such that for any $|v'\\ra \\in V$, \n\t\n\t$$ (|w\\ra\\la v|) \\ |v'\\ra = |w\\ra\\la v|v'\\ra = \\la v|v'\\ra |w\\ra$$\n\\end{definition}\n\nThese identities provide a dual interpretation: the already known product of a complex value $\\la v|v'\\ra$ with a vector $|w\\ra$, and the application of the new operator, the outer product $|w\\ra\\la v|$ to the vector $|v'\\ra$. The outer product is defined so that this duality occurs.\n\nLet us consider linear combinations of outer products. By definition, $\\sum_i a_i |w_i\\ra\\la v_i|$ is the operator that transforms $|v'\\ra$ into $\\sum_i a_i |w_i\\ra\\la v_i|v'\\ra = \\sum_i a_i \\la v_i|v'\\ra |w_i\\ra$.\n\nThe most important result concerning outer products is the \\emph{completeness relation}:\n\n\\begin{proposition}[Completeness relation]\n\tLet $|i\\ra$ be any orthonormal basis of a finite vector space $V$. Then:\n\t\n\t$$ \\sum_i |i\\ra\\la i| = I $$\n\\end{proposition}\n\n\\begin{proof}\n\tLet $|v\\ra \\in H$. $|v\\ra$ can be expressed as $ \\sum_i v_i |i\\ra$ for some complex numbers $v_i$. Notice that $\\la i|v\\ra = v_i$. Therefor:\n\t\n\t$$|v\\ra = \\sum_i v_i |i\\ra = \\sum_i \\la i|v\\ra |i\\ra = \\sum_i |i\\ra\\la i|v\\ra = \\bigg( \\sum_i |i\\ra\\la i|\\bigg) |v\\ra$$\n\t\n\tSince $|v\\ra$ was arbitrary, this proves that $ \\sum_i |i\\ra\\la i | = I $.\n\\end{proof}\n\n\\begin{corollary}[Cauchy-Schwarz inequality]\n\tFor any two vectors $|v\\ra, |w\\ra$ in a Hilbert space,\n\t\n\t$$|\\la v|w\\ra|^2 \\leq \\la v|v\\ra\\la w|w\\ra$$\n\t\n\twhere the equality occurs if and only if $|v\\ra$ and $|w\\ra$ are linearly dependant\n\\end{corollary}\n\\begin{proof} \n\tWe provide a proof supposed that our Hilbert space is finite.\n\t\n\tLet $|v\\ra, |w\\ra$ be two vectors of a finite Hilbert space $H$. Since $H$ is finite, using the Gram-Schmidt procedure we may obtain a basis $|i\\ra$ where the first vector is $|w\\ra / \\sqrt{\\la w|w\\ra}$. Using the completeness relation $\\sum_i |i\\ra \\la i| = I$:\n\t\n\t$$\\la v|v\\ra \\la w|w\\ra = \\la v|I|v\\ra \\la w|w\\ra = \\la v| \\sum_i (|i\\ra \\la i|) |v\\ra \\la w|w\\ra = \\sum_i \\la v|i\\ra \\la i|v\\ra \\la w|w\\ra $$\n\t\n\tThe sum of a list of positive numbers is obviously greater than its first element, so:\n\t\n\t$$ \\sum_i \\la v|i\\ra \\la i|v\\ra \\la w|w\\ra \\geq \\frac{\\la v|w\\ra \\la w|v\\ra}{\\la w|w\\ra} \\la w|w\\ra = \\la v|w\\ra \\la w|v\\ra = | \\la w|v\\ra |^2 $$\n\t\n\tLastly, the equality occurs if and only if $\\la v|i\\ra = 0$ for every $|i\\ra \\neq |w\\ra / \\sqrt{\\la w|w\\ra}$. But since $|i\\ra$ is a base, this means $|v\\ra$ and $|w\\ra$ are linearly dependent.\t\n\\end{proof}\n\n\n\\subsection{Unitary and Hermitian operators}\n\n\nAnother way of looking at the inner product is the \\emph{adjoint}.\n\n\\begin{definition}\n\tLet $A$ be an operator between $\\C^n$ and $\\C^m$, finite dimensional Hilbert spaces. That is, $A \\in \\mathcal{M}_{n{\\times}m}(\\C)$. Then, its \\emph{adjoint} or \\emph{conjugate transpose} $A^\\dagger$ is defined by:\n\t\n\t$$ (A^\\dagger)_{ij} = \\bar A_{ji} $$\n\\end{definition}\n\nIf $|v\\ra$ is a vector, we can compute its adjoint by seeing it as a matrix. By convention, we will denote $|v\\ra^\\dagger = \\la v|$. Adjoints of vector are usually called \\emph{bras}, making given sense to the \\emph{bra-ket} notation since $\\la v| \\cdot |v\\ra = \\la v|v\\ra$, where $\\cdot$ denotes the dot product.\n\nSome useful algebraic identities associated to adjoints are:\n\n\\begin{itemize}\n\t\\item Given an operator $A \\in \\mathcal{M}_{n}(\\C)$, $A^\\dagger$ is the only operator such that for any two vectors $|u\\ra, |v\\ra \\in \\C^n$: $( |u\\ra, A|v\\ra) = ( A^\\dagger |u\\ra, |v\\ra)$\n\t\\item For any two operators $A,B$, $(AB)^\\dagger = B^\\dagger A^\\dagger$.\n\t\\item As a corollary, for any vector $|v\\ra$ and for any operator $A$, $(A|v\\ra)^\\dagger = \\la v|A^\\dagger$.\n\\end{itemize}\n\n\\begin{definition}\n\tAn operator $A$ is said to be \\emph{normal} if $AA^\\dagger = A^\\dagger A$.\n\\end{definition}\n\nCharacterization of normal operators is provided in Theorem \\ref{spectral-decomposition-theorem}. There are two particular cases of normal operators that will be of special interest to us:\n\n\\begin{definition}\n\tAn operator $A$ is said to be \\emph{Hermitian} if its adjoint is itself: $A^\\dagger = A$.\n\\end{definition}\n\n\\begin{definition}\n\tA matrix $U$ is said to be \\emph{unitary} if $UU^\\dagger = I$. Similarly, an operator is said to be \\emph{unitary} if $UU^\\dagger = I$. A unitary operator $U$ also fulfills that $U^\\dagger U = I$.\n\\end{definition}\n\nClearly, Hermitian and unitary operators are also normal. The importance of unitary matrices and operators in quantum computing lies in the following\n\n\\begin{proposition}\n\tUnitary operators preserve inner product between vectors. Thus, they also preserve the norm of a vector.\n\\end{proposition}\n\n\\begin{proof}\n\tLet $|u\\ra, |v\\ra \\in \\C^n$ and $U \\in \\mathcal{M}_n(\\C)$ be a unitary operator. Then:\n\t\n\t$$ ( U|u\\ra, U|v\\ra) = \\la u|U^\\dagger U|v\\ra = \\la u|I|v\\ra = \\la u|v\\ra = ( |u\\ra, |v\\ra) $$\n\t\n\tWhich proves the proposition.\n\\end{proof}\n\nAn important type of Hermitian operators is the projectors.\n\n\\begin{definition}\n\tLet $W$ be a $k$-dimensional subspace\t of the $d$-dimensional vector space $V$. Let $|1\\ra, \\dots, |d\\ra$ be an orthonormal base of $V$ where $|1\\ra, \\dots, |k\\ra$ is an orthonormal base of $W$. The \\emph{projector} onto the subspace $W$ is defined by:\n\t\n\t$$ P = \\sum_{i=1}^k |i \\ra\\la i| $$\n\\end{definition}\n\nIt can easily be shown that this definition is independent from the chosen base $|1\\ra, \\dots, |k\\ra$. Since $|v \\ra\\la v|$ is Hermitian for any vector $|v\\ra$, $P$ is also Hermitian: $P = P^\\dagger$. We will often refer to the subspace onto which $P$ projects simply as $P$ for comodity. The \\emph{orthonormal completement} of $P$ is $Q \\equiv I - P$. It can be verified the vector subspace $Q$ is spanned by the base $|k+1\\ra, \\dots, |d\\ra$.\n\nAs the reader may already imagine, n-qubits systems will be represented as vector states or \\emph{certain} state spaces. That is, as unitary vectors of certain Hilbert spaces. Thus, our definition of transformations on qubits must preserve their norm. These will be the qubits gates, which will be represented as unitary operators.\n\nOn the other hand, Hermitian operators will be key in order to study how a quantum system evolves with time using the Schrodinger equation in the third Postulate.\n\nFinally, projectors will be used on as particular way of measurement and will be further discussed in section \\ref{projective-measurement}, once systems with multiple qubits have been introduced.\n\n\n\\subsection{Postulate 2 statement}\n\n\nAs previously discussed, the second postulate describes how a quantum system may be measured.\n\n\\begin{postulate}\n\tQuantum measurement are described by a collection $\\{M_m\\}$ of \\emph{measurement operators}. These act on the state space associated to the physical system being measured. The index $m$ refers to the measurement outcomes that may occur. That is, if $|\\varphi\\ra$ is the vector state before measure, then the probability of the result $m$ occurring is:\n\t\n\t$$p(m) = \\la\\varphi|M_m^\\dagger M_m|\\varphi\\ra $$\n\t\n\tand the state of the system after the measurement is:\n\t\n\t$$ \\frac{M_m|\\varphi\\ra}{\\sqrt{p(m)}}$$\n\t\n\tFinally, the measurement operator satisfy the \\emph{completeness equation}: \n\t\n\t$$\\sum_m M_m^\\dagger M_m = I$$\n\\end{postulate}\n\nThe completeness equation is equivalent to the fact that the probabilities of the different possible outcomes add up to one:\n\n$$ \\sum_m p(m) = \\sum_m \\la\\varphi|M_m^\\dagger M_m|\\varphi\\ra = \\la\\varphi|\\sum_m (M_m^\\dagger M_m)|\\varphi\\ra = \\la\\varphi|\\varphi\\ra = 1 $$\n\nwhich holds for every state vector since they are unitary. Reciprocally, this equation occurring for every state vector $|\\varphi\\ra$ implies the completeness equation.\n\nThe reader may have already noticed a huge difference between classic and quantum measurement: The measured state \\textbf{changes} after the measurement. This means that we interfere with the system by merely looking into it! It will ultimately translate into huge differences between classical and quantum computing, such that we will generally not be able to clone a qubit (see theorem \\ref{no-cloning-theorem}).\n\nLet us look at an important example: \\emph{measurement of a qubit on the computational basis}. That is, measuring a qubit with two possible outcomes: $|0\\ra$ and $|1\\ra$. Although this is a particular case of projective measurement (further explained in section \\ref{projective-measurement}), it is worth introducing it now to deepen our understanding of qubits.\n\nTo obtain such results we use the operators $M_0 = |0\\ra \\la 0|$ and $M_1 = |1\\ra \\la 1|$. In order for $\\{M_0, M_1\\}$ to be a correct collection of measurement operators they must satisfy the completeness equation. Observe that each operator is Hemitian: $M_i^\\dagger = (|i\\ra \\la i|)^\\dagger = (\\la i|)^\\dagger (|i\\ra)^\\dagger = |i\\ra \\la i| = M_i$. Furthermore, $M_i^2 = |i\\ra \\la i|i\\ra \\la i| = |i\\ra \\la i| = M_i$.\n\nFinally, the computational basis is an orthonormal basis and therefore the completeness relation tells us that:\n\n$$ \\sum_i |i\\ra\\la i| = 1 $$\n\nWe can see that the completeness equation holds:\n\n$$ \\sum_i M_i^\\dagger M_i = \\sum_i M_i^2 = \\sum_i M_i = \\sum_i |i\\ra\\la i| = 1 $$\n\nLet's measure using this operators, also called \\emph{measure in the computational basis}, to better understand the measurement. Suppose the state being measured is $|\\varphi\\ra = \\alpha|0\\ra + \\beta|1\\ra$. Then, the probability of obtaining the outcome 0 is:\n\n$$ p(0) = \\la\\varphi|M_0^\\dagger M_0|\\varphi\\ra = \\la\\varphi|M_0|\\varphi\\ra = |a|^2 $$\n\nSimilarly, the probability of obtaining the outcome 1 is $p(1) = |b|^2$. Naturally, $ |a|^2 + |b|^2 = 1$. What happens after measuring? The post-measurement state will be, respectively if we measured 0 or 1:\n\n$$ \\frac{M_0|\\varphi\\ra}{|a|} = \\frac{a}{|a|}|0\\ra$$\n\n$$ \\frac{M_1|\\varphi\\ra}{|b|} = \\frac{b}{|b|}|1\\ra$$\n\nFactors like $a/|a|$ are known as \\emph{phases}. An importan result of quantum mechanics assures that multiplying by factors like these does not affect the state vector \\cite{Nielsen2002}, so we virtually obtained $|0\\ra$ and $|1\\ra$. We may appreciate now how dividing by $\\sqrt{p(m)}$ in the post-measurement state is only done so the resulting vector is a unit vector.\n\nSo if we measured a 0, the post-measurement vector will be $|0\\ra$ and vice-versa. Any measurements performed after the first one will yield exactly the same result. This behavior is called \\emph{qubit collapsing}.\n\nWe defined measurement to be independent of any basis so we may measure in the most convenient basis at each point. For example, we may use the following other basis:\n\n$$ |+\\ra = \\frac{1}{\\sqrt{2}}|0\\ra + \\frac{1}{\\sqrt{2}}|1\\ra = \\frac{|0\\ra + |1\\ra}{\\sqrt{2}} $$\n\n$$ |-\\ra = \\frac{1}{\\sqrt{2}}|0\\ra - \\frac{1}{\\sqrt{2}}|1\\ra = \\frac{|0\\ra - |1\\ra}{\\sqrt{2}} $$\n\nIt can easily be proven that $\\{|+\\ra, |-\\ra\\}$ is a basis and therefore it satisfies the completeness equation. In fact, the operators $M_+ = |+\\ra\\la +|$ and $M_- = |-\\ra\\la -|$ satisfy the completeness equation. Using this operators, measuring the qubit $|\\varphi\\ra = \\alpha|0\\ra + \\beta|1\\ra$ will output $+$ with probability:\n\n$$ p(+) = \\la\\varphi|M_+^\\dagger M_+|\\varphi\\ra = \\la\\varphi|M_+|\\varphi\\ra = \\la\\varphi|+\\ra\\la+|\\varphi\\ra = \\la\\varphi|+\\ra^2 = $$\n\n$$ = \\bigg ( \\frac{\\alpha}{\\sqrt{2}}\\la0|0\\ra + \\frac{\\alpha}{\\sqrt{2}}\\la1|0\\ra + \\frac{\\beta}{\\sqrt{2}}\\la0|1\\ra + \\frac{\\beta}{\\sqrt{2}}\\la1|1\\ra \\bigg )^2 = \\frac{(\\alpha + \\beta)^2}{2} $$\n\n$$ p(-) = \\frac{(\\alpha - \\beta)^2}{2} $$\n\nAnd, naturally, $p(+) + p(-) = 1$. So $|\\varphi\\ra$ will collapse to $|+\\ra$ with probability $(\\alpha + \\beta)^2/2$ when measured with this operators. Observe that the post-measurement state will never be $|0\\ra$ nor $|1\\ra$ in this case, we made the qubit collapse to the chosen state basis.\n\nFinally, there is a technical fineness between the first and second postulates worth mentioning. The first postulate was stated for an \\emph{isolated} physical system, but by measuring the system we interfere with it. However, measuring devices are also quantum systems, so together the measured and the measuring system form a larger isolated system (it may be necessary to include more quantum systems, but this can be done).\n\n\n\\subsection{Real life qubit examples}\n\\label{qubit-examples}\n\n\nIn classical computation, we may know the state of a bit by consulting it. That is, what can simply retrieve that information from the bit. The first difficulty we find in quantum computing is that once we \\emph{measure} a qubit it \\emph{collapses} to either $|0\\ra$ with probability $|\\alpha|^2$, or to $|1\\ra$ with probability $|\\beta|^2$. The obtained output reflects the qubit state \\emph{after} it has collapsed to either one of these states, so the outcome may only be either $|0\\ra$ or $|1\\ra$. This means that we may never retrieve directly the values $\\alpha$ and $\\beta$. Thus, being unable to clone a qubit. This is the main idea behind the no-cloning theorem \\ref{no-cloning-theorem}.\n\nWe can, however, initialize qubits in a certain state and apply some operations to them in order to alter their coefficients, thus knowing their exact value. However, once a single measurement is done, the qubit collapses and the $\\alpha$ and $\\beta$ values are 'lost'.\n\nSuperposition and collapsing might be counter-intuitive concepts, so let us look at them with an analogy. We can think of a perfect coin being tossed as the following qubit:\n\n$$ |+\\ra = \\frac{1}{\\sqrt{2}} |0\\ra + \\frac{1}{\\sqrt{2}} |1\\ra $$\n\nThis does \\textbf{not} represent a coin that has landed somehow on its side, but a spinning coin that has not landed yet. Upon measuring it, we 'make the coin land' and see the result: either heads or tails, and neither of the states in between. This example also describes the qubit collapse: once the coin has landed, we will see the same result every time we look at it -obviously-, just like every time a qubit is measured after the first measurement, the outcome will be the same since it has already collapsed. We will return to this state, also known as the Bell state, in section \\ref{postulate-4-section}.\n\nOn the other hand, this was quite an inaccurate example since the system is not really isolated, although it was interesting intuition-wise. One of the first (accurate) qubit models ever proposed was the Schrodinger's Cat \\cite{Schrodinger1935} \\cite{Trimmer1980}. In this hypothetical experiment, a cat would be locked in a room for an hour with a device that during that hour would \\emph{perhaps} trigger, killing the cat. On the other hand, with equal probability, it would not trigger at all. After the whole hour elapses, the cat would be alive and dead with equal probability, ending up in a halfway state. In this case, our computational bases would be the states alive and dead, and we achieve the state $|+\\ra$ after that hour. Once we open the room and check on the cat, our qubit collapses to either state and stays on it until further disturbance.\n\nAlthough a physical implementation is discussed in section \\ref{sec:quantum-annealing-dwave}, we cannot proceed any further without providing a more accurate and reproducible description of a qubit than 'a coin being tossed' and such a hypothetical cat experiment. A possible realization of a qubit is an electron in a single atom's orbit, as seen in Figure \\ref{fig 1.1}. An electron in an orbit may be in the so-called \\emph{ground} and \\emph{excited} states, $|0\\ra$ and $|1\\ra$ respectively, depending on its energy. By shining light to the electron with a certain energy and for a certain amount of time, one can make the electron move from the ground state to the excited state and vice versa. But most interestingly, one can apply the light to the electron during a smaller amount of time, moving the electron somehow 'halfway' between both states.\n\n\\begin{figure}[h]\n\t\\includegraphics[scale=.4]{atom.png}\n\t\\centering\n\t\\caption{Qubit represented by two electron orbits in an atom, \\cite{Nielsen2002}.}\n\t\\label{fig 1.1}\n\\end{figure}\n\n\n\\section{Postulate 3: Evolution}\n\\label{sec:postulate-3}\n\nThe third postulate of quantum mechanics modulates the evolution of a quantum system. That is, the evolution of the state vector that describes the system. In order to properly formalize it, the concepts of eigenvectors, eigenvalues, and Hermitian operators are required.\n\n\n\\subsection{Eigenvalues and eigenvectors}\n\n\n\\begin{definition}\n\tLet $V$ be a vector space and $A$ an operator on $V$. An \\emph{eigenvector} is a non zero vector $|v_\\lambda\\ra$ such that $A|v_\\lambda\\ra = \\lambda|v_\\lambda\\ra$ for a complex value $\\lambda$ called the associated \\emph{eigenvalue}.\n\\end{definition}\n\nEigenvalues and their associated eigenvectors will usually be denoted with the same letter for simplicity: $\\lambda$ and $|\\lambda\\ra$. We assume the reader is familiar with eigenvectors and values basic notions. For instance, that they may be calculated using the \\emph{characteristic equation}: $|I - \\lambda A| = 0$.\n\n\\begin{definition}\n\tA \\emph{diagonal representation} of an operator $A$ is a representation $\\sum_i \\lambda_i |\\lambda_i\\ra\\la\\lambda_i|$ where the $|\\lambda_i\\ra$ form an orthonormal set of A's eigenvectors and $\\lambda_i$ are the respective eigenvalues. An operator is said to be \\emph{diagonalizable} if it allows a diagonal representation. The \\emph{spectrum} of an operator is the set of its eigenvalues.\n\\end{definition}\n\n\\begin{exampleth}\n\tAs an example of this, let us consider the following matrix:\n\t\n\t$$ Z = \n\t\\begin{pmatrix}\n\t\t1 & 0 \\\\\n\t\t0 & -1 \n\t\\end{pmatrix}\n\t$$\n\t\n\tThis matrix is called the \\emph{Z Pauli} matrix. It is relevant for quantum computing and it will be introduced later on along with the rest of the Pauli matrices. For now, Let's compute its diagonalizable representation. Since it is already diagonal we can infer that its eigenvalues are $\\{1, -1\\}$. Computing the diagonal representation we realize that a pair orthonormal eigenvectors are $\\{|0\\ra, |1\\ra\\}$ respectively. Therefore:\n\t\n\t$$ Z = \n\t\\begin{pmatrix}\n\t\t1 & 0 \\\\\n\t\t0 & -1 \n\t\\end{pmatrix} = \n\t|0\\ra\\la0| - |1\\ra\\la1|\n\t$$\n\\end{exampleth}\n\nNormal operators have significant relevancy thanks to the following result:\n\n\\begin{theorem}[Spectral Decomposition Theorem]\n\\label{spectral-decomposition-theorem}\n\tAn operator $A$ is normal if and only if it is diagonalizable.\n\\end{theorem}\n\\begin{proof}\n\tSee Box 2.2 in page 72, \\cite{Nielsen2002}.\n\\end{proof}\n\nSince Hermitian and unitary operators are normal, it follows the next\n\n\\begin{corollary}\n\tAny Hermitian operator is diagonalizable. Any unitary operator is diagonalizable.\n\\end{corollary}\n\n\n\\subsection{Postulate 3 statement}\n\\label{postulate-3-section}\n\n\\begin{postulate}\n\tThe evolution of a \\emph{closed} quantum system is described by a \\emph{unitary transformation}. That is, the state $|\\varphi\\ra$ of the system at time $t_1$ is related to the state $|\\varphi'\\ra$ of the system at time $t_2$ by a unitary operator $U$ which depends only on the times $t_1$ and $t_2$,\n\t\n\t$$ |\\varphi'\\ra = U|\\varphi\\ra $$\n\\end{postulate}\n\nJust like the first postulate does not provide the state space or state vector of the system, the third postulate does not provide the unitary transformation that concretes this evolution. For our quantum computing case, we will be the ones to define the unitary transformation to the system. That is, the quantum circuit that transforms our qubit.\n\n\\begin{exampleth} Let's consider the \\emph{X Pauli matrix}, also known as the \\emph{bit flip} matrix:\n\t\n\t$$ \\sigma_1 = X = \n\t\\begin{pmatrix}\n\t\t0 & 1 \\\\\n\t\t1 & 0 \n\t\\end{pmatrix}\n\t$$\n\t\n\tIt is called the bit flip matrix because it takes $|0\\ra$ to $|1\\ra$ and vice-versa:\n\t\n\t$$ X|0\\ra = \n\t\\begin{pmatrix}\n\t\t0 & 1 \\\\\n\t\t1 & 0 \n\t\\end{pmatrix}\n\t\\begin{pmatrix}\n\t\t1 \\\\\n\t\t0 \n\t\\end{pmatrix}=\n\t\\begin{pmatrix}\n\t\t0 \\\\\n\t\t1 \n\t\\end{pmatrix} =\n\t|1\\ra\n\t$$\n\t\n\tThis product that we just computed is precisely what postulate 3 states: the evolution of our state vector following the unitary operator $X$: $X|0\\ra = |1\\ra$. Even though for an arbitrary system we do not know the specific unitary transformation the system follows, we can create systems that follow certain desired transformations. These are the basics of quantum gates and quantum circuits. \n\\end{exampleth}\n\nThe description of the system evolution provided by Postulate 3 only bears information for those fixed times $t_1$ and $t_2$. A continuous time-description of this evolution is provided by the Schrodinger equation, which provides a redefinition of the second postulate.\n\n\\begin{postulate 3'}\n\\label{postulate-3'}\n\tThe time evolution of the state of a \\emph{closed} quantum system is described by the Schrodinger equation:\n\t\n\t$$ i \\hbar \\frac{d|\\varphi\\ra}{dt} = H|\\varphi\\ra $$\n\t\n\twhere $\\hbar$ is \\emph{Planck’s constant}, $i$ is the imaginary unit and $H$ is a fixed Hermitian operator known as the $Hamiltonian$.\n\\end{postulate 3'}\n\nThere are several notes to make about this postulate. First, the Hamiltonian is fixed for the given system and it is not be confused with the \\emph{Hadamard quantum gate}, also represented by an $H$. Second, $\\hbar$ is a physical constant that can be absorbed into the Hamiltonian for our purposes, simplifying the equation. Finally, this is a differential equation, so by knowing the initial state space of the system and the exact Hamiltonian we may know the exact evolution of the system.\n\nLet us study the Hamiltonian in general. Since it is a Hermitian operator, it allows a spectral decomposition by theorem \\ref{spectral-decomposition-theorem}:\n\n$$ H = \\sum_E E |E\\ra\\la E| $$\n\nwhere $E$ are the eigenvalues and $|E\\ra$ the respective normalized eigenvectors. the states $|E\\ra$ are usually referred to as \\emph{energy eigenstates} or \\emph{stationary states}, and $E$ is the \\emph{energy} of the state $|E\\ra$. Furthermore, the lowest energy is called the \\emph{ground energy state} while the corresponding eigenstate is called the \\emph{ground state}.\n\n\\begin{exampleth} Suppose a single qubit system has a the following Hamiltonian:\n\t\n\t$$ H = \\hbar \\omega X = \\hbar \\omega\n\t\\begin{pmatrix}\n\t\t0 & 1 \\\\\n\t\t1 & 0 \n\t\\end{pmatrix}\n\t$$\n\t\n\tWhere $X$ is the first Pauli matrix, and $\\omega$ is a positive parameter. H eigenenergy states are the same as $X$ eigenstates: $(|0\\ra + |1\\ra)/2$ and $(|0\\ra - |1\\ra)/2$ with respective energies $\\hbar\\omega$ and $-\\hbar\\omega$. Thus, the ground state is $(|0\\ra - |1\\ra)/2$ with ground state energy $-\\hbar\\omega$.\n\\end{exampleth}\n\nLet us deduce the connection between the Hamiltonian perpestive of dinamics, Postulate 3', and the unitary operator perspective, postulate 3. We can solve the Schrodinguer equation, which can be proven to be:\n\n$$ |\\varphi(t_2)\\ra = \\text{exp} \\bigg( \\frac{-iH(t_2-t_1)}{\\hbar} \\bigg) |\\varphi(t_1)\\ra $$\n\nWe define:\n\n$$ U(t_1, t_2) \\equiv \\text{exp} \\bigg( \\frac{-iH(t_2-t_1)}{\\hbar} \\bigg) $$\n\nWhich is a unitary operator. In fact, any unitary operator $U$ may be expressed as $U = \\text{exp(iK)}$ for some Hermitian operator $K$. Thus, we obtained: \n\n$$ |\\varphi(t_2)\\ra = U(t_1, t_2) |\\varphi(t_1)\\ra $$\n\nFollowing this procedure, we have proven that there is a one to one correspondence between the continuous time-varying postulate 3' using the Hamiltonian and the more stationary discrete-time version using the unitary operator. Although the discrete-time vision is usually used in quantum computing, Quantum Annealing -the specific application of quantum mechanics used in this thesis- rests mostly on the Hamiltonian point of view. We will deepen in this architecture on section \\ref{sec:quantum-annealing-dwave}.\n\nIt is worth mentioning that both versions of this postulate assume our physical system to be \\emph{closed}. That is, there is no interaction with the system coming from the exterior. In reality, the only real closed system is the universe as a whole. However, we may recreate sufficiently closed systems so that they can be described with approximations as being closed. \n\nFurthermore, this severely interferes with postulate 2, where an outsider to the system may interfere with it by measuring it. The quantum system will evolve following postulate 3 until measurement is applied. Then they will evolve following the behavior described on postulate 2.\n\nIn practice, obtaining the Hamiltonian for a given quantum system is a really laborious work and usually needs experimental data \\cite{Nielsen2002}. However, for our computational purposes, we will be the ones designing the Hamiltonian such that our system evolves as desired. In particular, chapter \\ref{sec:qubo} describes in detail the construction of Hamiltonians for QUBO problems.\n\n\\subsection{Quantum Computing perspective: Quantum Gates}\n\nAlthough quantum annealing does not make use of quantum gates, they are the basis of quantum computing and they fully rely on postulate 3. Because of their importance, a brief overview of quantum gates is provided o¡in this section. However, a profound understanding of them will not be necessary to understand the rest of the thesis.\n\nQuantum computing is built upon the most simple operation we can compute on single qubits: quantum gates. Since state vectors are unit two-dimensional vectors, the operations we apply to them must preserve the norm. Thus, quantum gates will be represented by 2x2 unitary matrices. We have already introduce some of the most famous gates, the Pauli matrices:\n\n$$ X =\n\\begin{pmatrix}\n\t0 & 1 \\\\\n\t1 & 0 \n\\end{pmatrix}; \\quad\nY =\n\\begin{pmatrix}\n\t0 & -i \\\\\n\ti & 0 \n\\end{pmatrix}; \\quad\nZ =\n\\begin{pmatrix}\n\t1 & 0 \\\\\n\t0 & -1 \n\\end{pmatrix}\n$$\n\nSome of the others most important gates are the Hadamard gate (denoted as H), the phase gate (denoted as S), and the $\\pi/8$ gate (denoted as T):\n\n$$ H = \\frac{1}{\\sqrt 2}\n\\begin{pmatrix}\n\t1 & 1 \\\\\n\t1 & -1 \n\\end{pmatrix}; \\quad\nS =\n\\begin{pmatrix}\n\t1 & 0 \\\\\n\t0 & i \n\\end{pmatrix}; \\quad\nT =\n\\begin{pmatrix}\n\t1 & 0 \\\\\n\t0 & e^{i\\pi/4} \n\\end{pmatrix}\n$$\n\nAlthough we have not described the fundamentals of multiple qubits systems (postulate 4 is required for this), it is worth mentioning a simple two-qubits gate for the purpose of this section. Just like in classical computing, conditional operations are essential for the construction of complex quantum algorithms. The most basic conditional operation is the \\emph{controlled-NOT} gate, also referred to as the CNOT gate. This gate takes two bits $|c\\ra|t\\ra$, a control qubit, and a target qubit. It flips the target qubit if the first qubit is set to $|1\\ra$, producing the operation $|c\\ra|t\\ra \\rightarrow |c\\ra|c \\oplus t\\ra$, where $\\oplus$ denotes the exclusive-or operation. Therefore, if the control qubit is $|c\\ra = \\alpha|0\\ra + \\beta|1\\ra$, it will virtually flip the target qubit with probability $|\\beta|^2$ and leave it how it is with probability $|\\alpha|^2$.\n\nOne of the most important results of classical computation theory is that any boolean function may be constructed by only using AND, OR, and NOT gates, or by simply using NAND gates \\cite{Lau2006}. There is an equivalent result concerning quantum gates \\cite{Bayens2019}.\n\n\\begin{theorem}\n\tAny unitary matrix can be approximated by a combination of the Hadamard, CNOT, and $\\pi/8$ gates.\n\\end{theorem}\n\nIn this case, \\{H, CNOT, T\\} is called a \\emph{universal gate set}. In practice, is it efficient to build such any gate using only this gate set? The answer to the question is positive, as stated by the \\textbf{Solovay-Kitaev theorem} \\cite{Dawson2006}. However, this is out of the scope of this project. We provide a sketch proof of the previous theorem from \\cite{Bayens2019}.\n\n\\begin{proof}\n\tSee \\cite{Bayens2019}, Theorem 4.2\n\\end{proof}\n\nFurther study of quantum circuits falls out of the scope of this thesis since we will use the D-Wave architecture, which relies on quantum annealing instead of quantum circuits.\n\n\n\\section{Postulate 4: Composite systems}\n\n\nIn this section, we introduce the last postulate, which lets us understand the state space associated with a physical system composed of other minor systems. This will allow us to study multiple-qubits systems and state some of the most important results in quantum mechanics and quantum computing: The \\emph{Heisenberg uncertainty principle} and the no-cloning theorem.\n\n\n\\subsection{Tensor product}\n\n\nLet $V$ and $W$ be complex vector spaces with dimensions $m$ and $n$ respectively. Then, $V \\otimes W$, read '$V$ tensor $W$', is a $mn$ complex vector space. Let $|v\\ra$ and $|w\\ra$ be vectors in $V$ and $W$ respectively. Then, $|v\\ra \\otimes |w\\ra$ is in $V \\otimes W$. Furthermore, any element of $V \\otimes W$ may be expressed as a linear combinations of tensor products $|v\\ra \\otimes |w\\ra$ of elements from $V$ and $W$. We may describe elements of $V \\otimes W$ using the following equivalent notations: $|v\\ra \\otimes |w\\ra$, $|v\\ra|w\\ra$, $|v, w\\ra$ and even $|vw\\ra$.\n\nLet $|i\\ra$ and $|j\\ra$ be basis for $V$ and $W$ respectively. Then, $|i\\ra \\otimes |j\\ra = |ij\\ra$ is a basis for $V \\otimes W$. For example, consider the complex vector space $\\C^2$. Then\n\n$$ |0\\ra \\otimes |0\\ra + |1\\ra \\otimes |1\\ra = |00\\ra + |11\\ra $$\n\nis an element in $\\C^2 \\otimes \\C^2 \\cong \\C^4$ .\n\nAlthough the tensor product is generally described as an abstract construct, we may provide a more visual perspective using the \\emph{Kronecker product}. This is a matrix representation for the tensor product of finite vector spaces. Suppose $A = \\{a\\}_{ij}$ is a $n \\times m$ matrix and $B = \\{b\\}_{ij}$ is an $p \\times q$ matrix. Then $A \\otimes B$ is a $np \\times mq$, matrix with the following representation:\n\n$$ A \\otimes B = \n\\begin{pmatrix}\n\ta_{11}B & a_{12}B & \\dotsc & a_{1m}B \\\\\n\ta_{21}B & a_{22}B & \\dotsc & a_{2m}B \\\\\n\t\\vdots & \\vdots & \\vdots & \\vdots \\\\\n\ta_{n1}B & a_{n2}B & \\dotsc & a_{nm}B \\\\\n\\end{pmatrix}\n$$\n\nFor example:\n\n$$ \\begin{pmatrix}\n\t0 & 1 \\\\\n\t2 & 3\n\\end{pmatrix} \\otimes\n\\begin{pmatrix}\n\t10 & 20 \\\\\n\t30 & 40\n\\end{pmatrix} =\n\\begin{pmatrix}\n\t0 \\times \\begin{pmatrix} 10 & 20 \\\\ 30 & 40 \\end{pmatrix} &\n\t1 \\times \\begin{pmatrix} 10 & 20 \\\\ 30 & 40 \\end{pmatrix} \\\\\n\t2 \\times \\begin{pmatrix} 10 & 20 \\\\ 30 & 40 \\end{pmatrix} &\n\t3 \\times \\begin{pmatrix} 10 & 20 \\\\ 30 & 40 \\end{pmatrix}\n\\end{pmatrix} = \n$$\n$$ \n\\begin{pmatrix}\n\t0 \\times 10 & 0 \\times 20 & 1 \\times 10 & 1 \\times 20 \\\\\n\t0 \\times 30 & 0 \\times 40 & 1 \\times 30 & 1 \\times 40 \\\\\n\t2 \\times 10 & 2 \\times 20 & 3 \\times 10 & 3 \\times 20 \\\\\n\t2 \\times 30 & 2 \\times 40 & 3 \\times 30 & 3 \\times 40\n\\end{pmatrix} = \n\\begin{pmatrix}\n\t0 & 0 & 10 & 20 \\\\\n\t0 & 0 & 30 & 40 \\\\\n\t20 & 40 & 30 & 60 \\\\\n\t60 & 80 & 90 & 120\n\\end{pmatrix}\n$$\n\nBy definition, the tensor product satisfies:\n\n\\begin{itemize}\n\t\\item Let $\\alpha$ be a complex number. For any $|v\\ra$ in $V$ and $|w\\ra$ in $W$,\n\t$$ \\alpha(|v\\ra \\otimes |w\\ra) = (\\alpha|v\\ra) \\otimes |w\\ra = |v\\ra \\otimes (\\alpha|w\\ra) $$\n\t\n\t\\item For any $|v_1\\ra$, $|v_2\\ra$ in $V$ and $|w\\ra$ in $W$,\n\t$$ (|v_1\\ra + |v_2\\ra) \\otimes |w\\ra = |v_1\\ra \\otimes |w\\ra + |v_2\\ra \\otimes |w\\ra $$\n\t\n\t\\item For any $|v\\ra$ in $V$ and $|w_1\\ra$, $|w_2\\ra$ in $W$,\n\t$$ |v\\ra \\otimes (|w_1\\ra + |w_2\\ra) = |v\\ra \\otimes |w_1\\ra + |v\\ra \\otimes |w_2\\ra $$\n\\end{itemize}\n\nThe inner products of the spaces $V$ and $W$ may be use to extend and natural inner product the tensor space. Define:\n\n$$ \\bigg ( \\sum_i a_i |v_i\\ra \\otimes |w_i\\ra, \\sum_j b_j |v_j'\\ra \\otimes |w_j'\\ra \\bigg ) \\equiv\n\t\t\\sum_{ij} \\overline{a_i} b_j \\la v_i|v_j' \\ra \\la w_i|w_j' \\ra $$\n\nWith this product and the previous properties in mind, it can easily be proven that the tensor product of Hilbert spaces is a Hilbert space.\n\n\\begin{proposition}\n\tLet $H_1$ be and $H_2$ be two Hilbert spaces with respective orthonormal bases $B_1 = \\{|v_i\\ra\\}_{i=1,\\dots,n}$ and $B_2 = \\{|w_j\\ra\\}_{j=1,\\dots,m}$. Then, $H_1 \\otimes H_2$ is a Hilbert space with inner product given by\n\t\n\t$$ ( |v\\ra \\otimes |w\\ra, |v'\\ra \\otimes |w'\\ra) = \\la vw|v'w'\\ra = \\la v|v'\\ra\\la w|w'\\ra $$\n\t\n\tFor any $|v\\ra, |v'\\ra$ in $V$ and $|w\\ra, |w'\\ra$ in $W$. Furthermore, $B_1 \\otimes B_2 = \\{|v_i w_j\\ra\\}_{i,j}$ is an orthonormal basis of $H_1 \\otimes H_2$.\n\\end{proposition}\n\n\\begin{proof}\n\tBy definition, $H_1 \\otimes H_2$ is a complex vector space with $dim(H_1 \\otimes H_2) = dim(H_1) \\cdot dim(H_2)$. Furthermore, $B_1 \\otimes B_2 = \\{|v_i\\ra \\otimes |w_j\\ra\\}_{i,j} = \\{|v_i w_j\\ra\\}_{i,j}$ is a base of it. Let $|v_i w_j\\ra, |v_k w_l\\ra \\in B_1 \\otimes B_2$:\n\t\n\t$$ \\la v_i w_j|v_k w_l \\ra = \\la v_i|v_k \\ra \\la w_j|w_l \\ra = \\delta_{ik} \\delta_{jl} $$\n\t\n\twhich equals one if and only if $i = k$ and $j = l$, and zero otherwise. Thus, proving that $B_1 \\otimes B_2$ is orthonormal.\n\t\n\tFinally, for $H_1 \\otimes H_2$ to be a Hilbert space the product defined above needs to be, in fact, an inner product. It suffies to prove that it is definite positive. Let $|v\\ra \\in H_1 \\otimes H_2$. Since $B_1 \\otimes B_2$ is an orthonormal basis, there exist $\\alpha_{ij} \\in \\C$ such that:\n\t\n\t$$ |v\\ra = \\sum_{i,j} \\alpha_{ij} |v_i w_j\\ra $$\n\t\n\tBy linearity:\n\t\n\t$$ \\la v|v \\ra = \\sum_{i,j,k,l} \\overline{\\alpha_{ij}} \\alpha_{kl} \\la v_i w_j|v_k w_l \\ra = \\sum_{i,j,k,l} \\overline{\\alpha_{ij}} \\alpha_{kl} \\ \\delta_{ik} \\delta_{jl} = \\sum_{i,j} |\\alpha_{ij}|^2 \\geq 0 $$\n\t\n\twhere the equality holds if and only if $\\alpha_{ij} = 0 \\ \\forall i,j$.\n\\end{proof}\n\nFrom the inner product, the tensor space $H_1 \\otimes H_2$ naturally inherits the notions of adjoint, unitary, normality and Hermicity. We will denote a vector $|\\varphi\\ra$ tensored with itsel n times, $|\\varphi\\ra \\otimes \\overplace{n}{\\dotsc} \\otimes |\\varphi\\ra$, as $|\\varphi\\ra^{\\otimes n}$, and equivalently with Hilbert spaces: $H \\otimes \\overplace{n}{\\dotsc} \\otimes H$, as $H^{\\otimes n}$. For our quantum computing purposes, the following case holds particular relevance.\n\n\\begin{corollary}\n\tThe tensor product of $\\C^2$ with itself n times, $\\C^2 \\otimes \\overplace{n}{\\dotsc} \\otimes \\C^2$, is isomorphic to $\\C^{2^n}$.\n\\end{corollary}\n\n\\begin{proof}\n\tLet $H = \\C^2 \\otimes \\overplace{n}{\\dotsc} \\otimes \\C^2$. Since $dim(H) = dim(\\C^2)^n = 2^n$, $H$ is a Hilbert space with dimension $2^n$. Thus, by Hausdorff's Theorem \\cite{Paya2020}, it is isomorphic to the canonical complex vector space of dimension $2^n$, $\\C^{2^n}$.\n\\end{proof}\n\nIn the same way that the inner product was naturally extended to the tensor space, we may extend operators ensuring linearity. Let $A: V \\longrightarrow V'$ and $B: W \\longrightarrow W'$ two operators between Hilbert spaces. Then, we define $A \\otimes B: V \\otimes W \\longrightarrow V' \\otimes W'$ using the following equation:\n\n$$ (A \\otimes B)(|v\\ra \\otimes |w\\ra) \\equiv A|v\\ra \\otimes B|w\\ra $$\n\nWe extend this definition to every element of $V \\otimes W$ ensuring linearity:\n\n$$ (A \\otimes B) \\bigg(\\sum_i a_i |v_i\\ra \\otimes |w_i\\ra \\bigg) \\equiv \\sum_i a_i A|v_i\\ra \\otimes B|w_i\\ra $$\n\nIt can be shown that $A \\otimes B$ is a well-defined linear operator. In fact, any linear operator mapping $V \\otimes W$ to $V' \\otimes W'$ may be expressed as a linear combination of tensor product of linear operators mapping $V$ to $V'$ and $W$ to $W'$:\n\n$$ C = \\sum_i a_i A_i \\otimes B_i $$\n\nwhere by definition:\n\n$$ \\bigg( \\sum_i a_i A_i \\otimes B_i \\bigg)|v\\ra \\otimes |w\\ra = \\sum_i a_i A_i|v\\ra \\otimes B_i|w\\ra $$\n\n\n\\subsection{Postulate 4 statement}\n\\label{postulate-4-section}\n\nReaching the core of this section, suppose we consider two (or more) distinct physical systems. In order to describe the state of the composite system we need to make use of tensor product.\n\n\\begin{postulate}\n\tThe state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|\\varphi_i\\ra$, then the joint state of the total system is $|\\varphi_1\\ra \\otimes \\dotsc \\otimes |\\varphi_n\\ra$.\n\\end{postulate}\n\nLet us provide an intuitive notion of why is the tensor product used in order to describe composite systems. After all, we would expect that there exists a \\emph{somehow canonical way} os describing the product composition of systems, just like the cartesian product is used with vector spaces. Let us refer again to the so-called \\emph{superposition principle of quantum mechanics}. Let $|\\varphi\\ra$ and $|\\psi\\ra$ be two vector states of a system. Then, any superposition of the states, $\\alpha|\\varphi\\ra + \\beta|\\psi\\ra$, should also be another vector state, where $|\\alpha|^2 + |\\beta|^2 = 1$. Suppose we now have two systems $X$ and $Y$, with two vector states $|\\varphi_x\\ra$ and $|\\varphi_y\\ra$. Then, we could describe the state of the composite system $XY$ as some state $|\\varphi_x\\ra|\\varphi_y\\ra$. Studying the superposition principle on composite systems leads us naturally to the notion of tensor product. It is important to note that this development is not thorough, since we are not taking the superposition principle as a fundamental component of our description of quantum mechanics. It does help up to build an intuitive notion on why this tensor product is used.\n\nWe now introduce the \\emph{Bell State} or \\emph{EPR pair:}\n\n$$ |+\\ra = \\frac{ |00\\ra + |11\\ra }{ \\sqrt 2 } $$\n\nAlthough it may seem harmless at first glance, this state has been responsible for many surprises during the development of quantum physics \\cite{Bellt1964}. Let us have a first look into it, although we will come back to it later on.\n\nThe first thing to notice about this state is that it can be expressed as a single tensor product of state vector: $|+\\ra \\neq |a\\ra|b\\ra$. This means that this state is not describing two independent physical systems put together with their respective vector states independently, it describes a \\textbf{relation} between both systems too. It means that those systems are somehow interfering with each other. So we may affect one by disrupting the other.\n\nThis phenomenon is called \\emph{entaglement}. A state is called \\emph{entangled} when it cannot be expressed as the product of states of its component systems. This concept rests at the heart of the disparity between classic physics and quantum physics and it is key in quantum computing. It was deeply studied first by Einstein, Podolsky, and Rosen (EPR) \\cite{Einstein1935} and second by John Bell \\cite{Bellt1964}.\n\n\n\\subsection{Projective measurement}\n\\label{projective-measurement}\n\n\nIn this section, we explain an important particular case of measurement. In fact, this method of projective measurement is equivalent to Postulate 2, when they are combined with the capacity to perform unitary operations. It has huge relevancy in quantum computing and it is explained here because of its relation with composite systems. Let us state the alternative postulate:\n\n\\begin{postulate 2'}\n\tA projective measurement is described by an \\emph{observable}, $M$, a Hermitian operator on the state space of the system being described. The observable has a spectral decomposition:\n\t\n\t$$ M = \\sum_m m P_m$$\n\t\n\twhere $P_m$ is the projector onto the eigenspace with eigenvalue $m$. The possible outcomes of the measurement correspond to the eigenvalues, $m$, of the observable. Upon measuring the state $|\\varphi\\ra$, the probability of the result $m$ ocurring is:\n\t\t\n\t$$ p(m) = \\la \\varphi | P_m | \\varphi \\ra $$\n\t\n\tand the state of the system after the measurement, givent that outcome $m$ ocurred, is:\n\t\n\t$$ \\frac{P_m |\\varphi\\ra}{\\sqrt{p(m)}} $$\n\\end{postulate 2'}\n\nProjective measurements can be seen as a particular case of Postulate 2. Suppose the measurement operators from such postulate, in addition to satisying the completeness relation $\\sum_m M_m M_m^\\dagger = I$, also fulfill that $M_m$ are \\emph{orthogonal projectors}. That is, $M_m$ is Hermitian and $M_m M_{m'} = \\delta_{mm'} M_m \\ \\forall m,m'$. With these aditional restricions it holds that $M_m M_m^\\dagger = M_m M_m = M_m$. By choosing $P_m = M_m$ we see that projective measurements is a particular case of Postulate 2.\n\nThanks to the way projective measurements are expressed, they have nice and handy properties. Let us compute the average value of a projective measurement:\n\n\\begin{equation}\n\t\\begin{split}\n\t\tE(M)\t& = \\sum_m m p(m) \\\\\n\t\t\t\t& = \\sum_m m \\la \\varphi | P_m | \\varphi \\ra \\\\\n\t\t\t\t& = \\la \\varphi | \\bigg( \\sum_m m P_m \\bigg) | \\varphi \\ra \\\\\n\t\t\t\t& = \\la \\varphi | M | \\varphi \\ra\n\t\\end{split}\n\\end{equation}\n\nThe average value of an operator is usually written as $\\la M \\ra \\equiv \\la \\varphi | M | \\varphi \\ra$. The standar deviation associated to observations of $M$ is then written as:\n\n\\begin{equation}\n\t\\begin{split}\n\t[\\Delta(M)]^2\t& = \\big\\la (M - \\la M \\ra )^2 \\big\\ra\\\\\n\t\t\t\t\t& = \\la M^2 \\ra - \\la M \\ra^2\n\t\\end{split}\n\\end{equation}\n\nThis means that if we prepare our state $|\\varphi\\ra$ multiple times and measure it, the measures will follow a normal distribution with expected value $\\la M \\ra$ and standard deviation $\\Delta(M) = \\sqrt{\\la M^2 \\ra - \\la M \\ra^2}$. This formulation of standard deviation using observables provides us with an elegant proof of perhaps the most famous quantum mechanics result: The \\emph{Heisenberg uncertainty principle}. Such proof can be found in Box 2.4 from \\cite{Nielsen2002}.\n\nThere are two additional notations worth mentioning regarding projective measurement. Instead of providing an observable $M$, we will sometimes provide a complete set of orthogonal projectors. That is, a set $P_m$ such that $\\sum_i P_m = I$ and $P_m P_{m'} = \\delta_{mm'} P_m$. The corresponding implicit observable used in this case is $M = \\sum_m m P_m$. Finally, another widely used notation is to 'measure in a basis $|m\\ra$', where $|m\\ra$ form an orthonormal basis. This simply means that the chosen projectors are $P_m = |m \\ra\\la m|$ and the observable is $M = \\sum_m m |m \\ra\\la m|$. This is the usual notation in quantum computing.\n\n\n\\subsection{Quantum Computing perspective: Multiple qubits}\n\n\nSuppose we have a pair of qubits. In the classical case, two bits can be in four possible states: 00, 01, 10, and 11. Similarly, the two qubits computational basis states are $|00\\ra$, $|01\\ra$, $|10\\ra$ and $|11\\ra$. We arrive at this natural notation by knowing that each qubit has $\\C^2$ as their associated state space and using the tensor product. Just like in the single qubit case, our two qubits system may be in a superposition of these four states:\n\n$$ |\\varphi\\ra = \\alpha_{00} |00\\ra + \\alpha_{01} |01\\ra + \\alpha_{10} |10\\ra + \\alpha_{11} |11\\ra $$\n\nCorrespondingly, the measurement of this system will result in either 00, 01, 10 or 11 since we are using the observable $\\sum_{x \\in \\{0,1\\}^2} x |x\\ra\\la x|$, where $\\{0,1\\}^2$ are the strings of length two where each character is either $0$ or $1$. In fact, it will yield state $x$ with probability $|\\alpha_x|^2$, being $\\alpha_x$ the coefficient associated with the state $|x\\ra$. The condition of the probabilities adding up to one is also called the \\emph{normalization condition} and can be expressed as $\\sum_{x \\in \\{0,1\\}^2} |\\alpha_x|^2 = 1$ for the two qubits case.\n\nThe fundamental differences with the single qubit case start on measurement. Of course, we can measure both qubits at the same time, but we could also measure only one of them. In order to achieve this we can use the projectors $P_0 = |00\\ra\\la00| + |01\\ra\\la01|$ and $P_1 = |10\\ra\\la10| + |11\\ra\\la11|$. Upon measuring, the system will collapse to $P_0|\\varphi\\ra / \\sqrt{p_0}$ with probability:\n\n$$ p_0 = \\la\\varphi|P_0|\\varphi\\ra = |\\alpha_{00}|^2 + |\\alpha_{01}|^2 $$\n\nSince these are the coefficients associated with the first qubit being $0$, this is the probability of first qubit being 0. Furthermore, our system will collapse to:\n\n$$ |\\varphi'\\ra = \\frac{ \\alpha_{00} |00\\ra + \\alpha_{01} |01\\ra }{ \\sqrt{|\\alpha_{00}|^2 + |\\alpha_{01}|^2} } $$\n\nNote the normalization term $\\sqrt{|\\alpha_{00}|^2 + |\\alpha_{01}|^2}$, which appears so the post-measurement state still satisfies the normalization condition. Naturally, after obtaining $0$ in the first qubit we can still obtain either $0$ or $1$ in the second qubit, with probabilities \n\n$$ \\frac{ |\\alpha_{00}|^2 }{ |\\alpha_{00}|^2 + |\\alpha_{01}|^2 } \\ \\text{ and } \\ \n\\frac{ |\\alpha_{01}|^2 }{ |\\alpha_{00}|^2 + |\\alpha_{01}|^2 } $$\n\nrespectively, adding up to $1$. Correspondingly, the first qubit being measured will yield $1$ with probability $p_1 = |\\alpha_{10}|^2 + |\\alpha_{11}|^2$.\n\nAdditionally, the first qubit independently should satisfy the normalization condition. That is, its probabilities of being $0$ and $1$ upon measurement must add up to $1$. But those are $p_0$ and $p_1$, which add up to one because of the normalization condition for $|\\varphi\\ra$, as expected.\n\nLet us come back to the previously introduced \\emph{Bell State}:\n\n$$ |+\\ra = \\frac{ |00\\ra + |11\\ra }{ \\sqrt 2 } $$\n\nWhen measuring using the previously defined projectors $P_0$ and $P_1$ we obtain that the first qubit will collapse to either 0 or 1 with equal probability $p_0 = p_1 = 1/2$. Most importantly, in the first case the qubit will collapse to the state:\n\n$$ \\frac{ |00\\ra / \\sqrt 2 }{ \\sqrt p_0 } = |00\\ra $$\n\nMeaning that the second qubit also collapsed to 0. Note that we only affected the first qubit and the second one was disturbed. These are the consequences of entanglement, key in the appearance of impressive phenomenoms and discussions such as the EPR Paradox \\cite{Einstein1935}, \\cite{Bellt1964}.\n\nLet us finally consider the more general case. In an n-qubits system our computational (orthonormal) basis would consist of the sates $|x_1 x_2 \\dotsc x_n\\ra$, where $x_i \\in \\{0,1\\}$. As we already know, in a single qubit system we have two amplitudes $\\alpha_0$ and $\\alpha_1$. We have four amplitudes for a 2-qubits system, eight for a 3-qubits system... And $2^n$ for an n-qubits system. This means that the number of amplitudes grows exponentially as we add qubits to the system. An immense increment compared to the classical case where the quantity of information that our system holds grows linearly with the numbers of bits. Of course, it is not that simple, since there are huge limitations on how we may access this information in the quantum realm such as how a qubit collapses upon measurement and the no-cloning theorem. However, we can already glimpse the power of quantum computing versus the classical one.\n\n\n\\section{The no-cloning theorem}\n\n\nA first exposure to the basis of quantum computing is fundamentally incomplete without the no-cloning theorem. It lets us understand one of the fundamental inconveniences of quantum computing: the impossibilty of cloning arbitrary states.\n\nWe proceed by formalizing the copying mechanism. Let $H$ be a state space and $|\\varphi\\ra \\in H$ the state to be copied. Given another state $|\\varphi'\\ra \\in H$, we would like to copy $|\\varphi\\ra$ into $|\\varphi'\\ra$. Since we can prepare $|\\varphi'\\ra$ at will before the copying occurs, let $|\\varphi'\\ra = |x_0\\ra$ be a fixed state. Then, our copying operation will take both states and produce the following result:\n\n$$ (|\\varphi\\ra \\otimes |x_0\\ra) \\xrightarrow{U} |\\varphi\\ra \\otimes |\\varphi\\ra $$\n\nOperations between n-qubits (n-qubits gates) must preserve their norm, thus they must be unitary $n \\times n$ matrices. In our case, $U$ is a $2 dim(H) \\times 2 dim(H)$ unitary matrix. Let us now state and prove the theorem.\n\n\\begin{theorem}[No-cloning Theorem]\n\\label{no-cloning-theorem}\n\tThere is no unitary operator $U: H^{\\otimes 2} \\longrightarrow H^{\\otimes 2}$ and state $|x_0\\ra \\in H$ such that for any arbitrary state $|\\varphi\\ra \\in H$ it holds\n\t\n\t$$ U |\\varphi\\ra |x_0\\ra = |\\varphi\\ra |\\varphi\\ra $$\n\\end{theorem}\n\\begin{proof}\n\tSuppose there exists such unitary operator $U$ and such state $|x_0\\ra$. Let $|\\varphi\\ra, |\\psi\\ra \\in H$. We may apply the copying operators to both of them:\n\t\n\t$$ U |\\varphi\\ra |x_0\\ra = |\\varphi\\ra |\\varphi\\ra $$\n\t$$ U |\\psi\\ra |x_0\\ra = |\\psi\\ra |\\psi\\ra $$\n\t\n\tTaking the inner product of both equations and using that $U$ is unitary results in:\n\t\n\t\\begin{equation*}\n\t\t\\begin{gathered}\n\t\t\t\\bigg( U |\\varphi\\ra |x_0\\ra, U |\\psi\\ra |x_0\\ra \\bigg) = \\bigg( |\\varphi\\ra |\\varphi\\ra, |\\psi\\ra |\\psi\\ra \\bigg) \\iff \\\\\n\t\t\t\\bigg( |\\varphi\\ra |x_0\\ra, |\\psi\\ra |x_0\\ra \\bigg) = \\la\\varphi|\\psi\\ra \\la\\varphi|\\psi\\ra \\iff \\\\\n\t\t\t\\la\\varphi|\\psi\\ra \\la x_0|x_0 \\ra = \\la\\varphi|\\psi\\ra^2\n\t\t\\end{gathered}\n\t\\end{equation*}\n\t\n\tSince $|x_0\\ra$ is a normalized vector, $\\la x_0|x_0 \\ra = 1$, and we obtain:\n\t\n\t$$\\la\\varphi|\\psi\\ra = \\la\\varphi|\\psi\\ra^2$$\n\t\n\tWhich only holds if $\\la\\varphi|\\psi\\ra$ equals to either $0$ or $1$, which means $|\\varphi\\ra$ and $|\\psi\\ra$ are either equal or orthonormal. But those vectors were arbitrary, thus such a general cloning operator is impossible. \n\\end{proof}\n", "meta": {"hexsha": "deb105544927e5cbc11860f81bfc9d24911a0913", "size": 59538, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/chapters/chapter1_QuantumMechanicsModel.tex", "max_stars_repo_name": "Ocete/TFG", "max_stars_repo_head_hexsha": "baf0853276aef8f6fd6c5cccbe69ede7ca7987c3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-09-05T22:40:37.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-05T22:40:37.000Z", "max_issues_repo_path": "thesis/chapters/chapter1_QuantumMechanicsModel.tex", "max_issues_repo_name": "Ocete/TFG", "max_issues_repo_head_hexsha": "baf0853276aef8f6fd6c5cccbe69ede7ca7987c3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/chapters/chapter1_QuantumMechanicsModel.tex", "max_forks_repo_name": "Ocete/TFG", "max_forks_repo_head_hexsha": "baf0853276aef8f6fd6c5cccbe69ede7ca7987c3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-07-29T18:23:25.000Z", "max_forks_repo_forks_event_max_datetime": "2021-07-29T18:23:25.000Z", "avg_line_length": 69.0696055684, "max_line_length": 1185, "alphanum_fraction": 0.7211192852, "num_tokens": 17881, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.798186768138228, "lm_q2_score": 0.7690802476562641, "lm_q1q2_score": 0.6138696773157014}} {"text": "\\problemname{Conquest}\n\nNomads, Kingdoms, and Tribes are on the islands of the great seas.\nBridges span between islands allowing travel between them.\nIt is possible to get from every island to every other island\nthrough some sequence of bridges.\nThe islands were at peace until everything changed when the\nSpanning Nation attacked!\n\nInitially the Spanning Nation occupies island 1. From that point\nforward, the Spanning Nation can attack any island that is directly\nconnected to some island already conquered by the\nSpanning Nation. Thankfully wars are resolved without any fighting.\nThe Spanning Nation only attacks an island if the island's army is\nstrictly smaller than the Spanning Nation's army. The smaller island\narmy will simply concede and join the Spanning Nation's army.\n\nAs the tactical advisor of the Spanning Nation, determine the maximum\npossible army size the Spanning Nation can have after making a series\nof attacks.\n\n\\section*{Input}\n\nThe first line contains the integer $N$ ($1 \\leq N \\leq 200\\,000$),\nwhich is the number of islands, and $M$ ($0 \\leq M \\leq 200\\,000$),\nthe number of bridges.\n\nThe next $M$ lines describe the bridges. Each of these lines contains\ntwo distinct integers $u$ and $v$ ($1 \\leq u, v \\leq N$), indicating\nthat there is a bridge between the islands $u$ and $v$. There is at\nmost one bridge between any pair of islands.\n\nThe next $N$ lines describe the islands' army size in order. Each of these\nlines contains a single integer $s$ ($0 \\leq s \\leq 1\\,000$), which is the\narmy size of this island.\n\n\\section*{Output}\n\nDisplay the maximum possible army size of the Spanning Nation.\n\n%%% Local Variables:\n%%% mode: latex\n%%% TeX-master: t\n%%% End:\n", "meta": {"hexsha": "2b4e426d7108c1136d8ee28d8507f51e328a507e", "size": 1695, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problems/conquest/problem_statement/problem.tex", "max_stars_repo_name": "icpc/na-rocky-mountain-2020-public", "max_stars_repo_head_hexsha": "d77cd0dfd9bd707f34497977251c4cc583647fef", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-03-11T21:49:18.000Z", "max_stars_repo_stars_event_max_datetime": "2021-03-19T22:31:57.000Z", "max_issues_repo_path": "problems/conquest/problem_statement/problem.tex", "max_issues_repo_name": "icpc/na-rocky-mountain-2020-public", "max_issues_repo_head_hexsha": "d77cd0dfd9bd707f34497977251c4cc583647fef", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problems/conquest/problem_statement/problem.tex", "max_forks_repo_name": "icpc/na-rocky-mountain-2020-public", "max_forks_repo_head_hexsha": "d77cd0dfd9bd707f34497977251c4cc583647fef", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2021-03-11T18:15:28.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-24T00:15:32.000Z", "avg_line_length": 37.6666666667, "max_line_length": 74, "alphanum_fraction": 0.7646017699, "num_tokens": 423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7690802317779601, "lm_q1q2_score": 0.613869672026091}} {"text": "\\chapter{Bonus: A hint of Pontryagin duality}\n\\label{ch:pontryagin}\n\nIn this short chapter we will give statements\nabout how to generalize our Fourier analysis\n(a bonus chapter \\Cref{ch:fourier})\nto a much wider class of groups $G$.\n\n\\section{LCA groups}\n\\prototype{$\\TT$, $\\RR$.}\nEarlier we played with $\\RR$,\nwhich is nice because in addition to being a topological space,\nit is also an abelian group under addition.\nThese sorts of objects which are both groups and spaces have a name.\n\\begin{definition}\n\tA group $G$ is a \\vocab{topological group}\n\tis a Hausdorff\\footnote{Some authors omit the Hausdorff condition.}\n\ttopological space equipped also with a group operation $(G, \\cdot)$,\n\tsuch that both maps\n\t\\begin{align*}\n\t\tG \\times G &\\to G \\quad\\text{ by }\\quad (x,y) \\mapsto xy \\\\\n\t\tG &\\to G \\quad\\text{ by }\\quad x \\mapsto x\\inv\n\t\\end{align*}\n\tare continuous.\n\\end{definition}\n\nFor our Fourier analysis, we need some additional conditions.\n\\begin{definition}\n\tA \\vocab{locally compact abelian (LCA) group} $G$\n\tis one for which the group operation is abelian,\n\tand moreover the topology is \\emph{locally compact}:\n\tfor every point $p$ of $G$,\n\tthere exists a compact subset $K$ of $G$\n\tsuch that $K \\ni p$, and $K$ contains some open neighborhood of $p$.\n\\end{definition}\n\nOur previous examples all fall into this category:\n\\begin{example}\n\t[Examples of locally compact abelian groups]\n\t\\listhack\n\t\\begin{itemize}\n\t\t\\ii Any finite group $Z$ with the discrete topology is LCA.\n\t\t\\ii The circle group $\\TT$ is LCA and also in fact compact.\n\t\t\\ii The real numbers $\\RR$ are an example of an LCA group\n\t\twhich is \\emph{not} compact.\n\t\\end{itemize}\n\\end{example}\n\nThese conditions turn out to be enough\nfor us to define a measure on the space $G$.\nThe relevant theorem, which we will just quote:\n\\begin{theorem}\n\t[Haar measure]\n\tLet $G$ be a locally compact abelian group.\n\tWe regard it as a measurable space\n\tusing its Borel $\\sigma$-algebra $\\SB(G)$.\n\tThere exists a measure $\\mu \\colon \\SB(G) \\to [0,\\infty]$,\n\tcalled the \\vocab{Haar measure},\n\tsatisfying the following properties:\n\t\\begin{itemize}\n\t\t\\ii $\\mu(gS) = \\mu(S)$ for every $g \\in G$ and measurable $S$.\n\t\tThat means that $\\mu$ is ``translation-invariant''\n\t\tunder translation by $G$.\n\t\t\\ii $\\mu(K)$ is finite for any compact set $K$.\n\t\t\\ii if $S$ is measurable, then $\\mu(S) = \\inf\\left\\{ \\mu(U) \\colon U \\supseteq S \\text{ open} \\right\\}$.\n\t\t\\ii if $U$ is open, then $\\mu(U) = \\sup\\left\\{ \\mu(S) \\colon S \\supseteq U \\text{ measurable} \\right\\}$.\n\t\\end{itemize}\n\tMoreover, it is unique up to scaling by a positive constant.\n\\end{theorem}\n\n\\begin{remark}\n\tNote that if $G$ is compact, then $\\mu(G)$ is finite (and positive).\n\tFor this reason the Haar measure on a LCA group $G$\n\tis usually normalized so $\\mu(G) = 1$.\n\\end{remark}\n\nFor this chapter, we will only use the first two properties at all,\nand the other two are just mentioned for completeness.\nNote that this actually generalizes the chapter where\nwe constructed a measure on $\\SB(\\RR^n)$,\nsince $\\RR^n$ is an LCA group!\n\nSo, in short: if we have an LCA group,\nwe have a measure $\\mu$ on it.\n\n\\section{The Pontryagin dual}\nNow the key definition is:\n\\begin{definition}\n\tLet $G$ be an LCA group.\n\tThen its \\vocab{Pontryagin dual} is the abelian group\n\t\\[ \\wh G \\defeq \\left\\{ \\text{continuous group homomorphisms }\n\t\t\t\\xi : G \\to \\TT \\right\\}. \\]\n\tThe maps $\\xi$ are called \\vocab{characters}.\n\tIt can be itself made into an LCA group.\\footnote{If you must\n\t\tknow the topology, it is the \\vocab{compact-open topology}:\n\t\tfor any compact set $K \\subseteq G$\n\t\tand open set $U \\subseteq \\TT$,\n\t\twe declare the set of all $\\xi$ with $\\xi\\im(K) \\subseteq U$ to be open,\n\t\tand then take the smallest topology\n\t\tcontaining all such sets. We won't use this at all.}\n\\end{definition}\n\\begin{example}\n\t[Examples of Pontryagin duals]\n\t\\listhack\n\t\\begin{itemize}\n\t\t\\ii $\\wh{\\ZZ} \\cong \\TT$,\n\t\tsince group homomorphisms $\\ZZ \\to \\TT$ are determined by the image of $1$.\n\t\t\\ii $\\wh{\\TT} \\cong \\ZZ$.\n\t\tThe characters are given by $\\theta \\mapsto n\\theta$ for $n \\in \\ZZ$.\n\t\t\\ii $\\wh{\\RR} \\cong \\RR$.\n\t\tThis is because a nonzero continuous homomorphism\n\t\t$\\RR \\to S^1$ is determined by the fiber above $1 \\in S^1$.\n\t\t(Algebraic topologists might see covering projections here.)\n\t\t\\ii $\\wh{\\ZZ/n\\ZZ} \\cong \\ZZ/n\\ZZ$,\n\t\tcharacters $\\xi$ being determined by the image $\\xi(1) \\in \\TT$.\n\t\t\\ii $\\wh{G \\times H} \\cong \\wh G \\times \\wh H$.\n\t\\end{itemize}\n\\end{example}\n\\begin{exercise}\n\t[$\\wh Z \\cong Z$, for those who read \\Cref{sec:FTFGAG}]\n\tIf $Z$ is a finite abelian group, show that $\\wh Z \\cong Z$,\n\tusing the results of the previous example.\n\tYou may now recognize that the bilinear form\n\t$\\cdot \\colon Z \\times Z \\to \\TT$\n\tis exactly a choice of isomorphism $Z \\to \\wh Z$.\n\tIt is not ``canonical''.\n\\end{exercise}\n\nTrue to its name as the dual,\nand in analogy with $(V^\\vee)^\\vee \\cong V$ for vector spaces $V$, we have:\n\\begin{theorem}\n\t[Pontryagin duality theorem]\n\tFor any LCA group $G$, there is an isomorphism\n\t\\[ G \\cong \\wh{\\wh G} \\qquad \\text{by} \\qquad\n\t\tx \\mapsto \\left( \\xi \\mapsto \\xi(x) \\right). \\]\n\\end{theorem}\n\nThe compact case is especially nice.\n\\begin{proposition}\n\t[$G$ compact $\\iff$ $\\wh G$ discrete]\n\tLet $G$ be an LCA group.\n\tThen $G$ is compact if and only if $\\wh G$ is discrete.\n\\end{proposition}\n\\begin{proof}\n\t\\Cref{prob:LCA_compact}.\n\\end{proof}\n\n\\section{The orthonormal basis in the compact case}\nLet $G$ be a compact LCA group,\nand work with its Haar measure.\nWe may now let $L^2(G)$ be the space of\nsquare-integrable functions to $\\CC$, i.e.\n\\[ L^2(G) = \\left\\{ f \\colon G \\to \\CC\n\t\\quad\\text{such that}\\quad \\int_G |f|^2 < \\infty \\right\\}. \\]\nThus we can equip it with the inner form\n\\[ \\left< f,g \\right> = \\int_G f \\cdot \\ol{g}. \\]\nIn that case, we get all the results we wanted before:\n\\begin{theorem}\n\t[Characters of $\\wh G$ forms an orthonormal basis]\n\t\\label{thm:god}\n\tAssume $G$ is LCA and compact (so $\\wh G$ is discrete).\n\tThen the characters\n\t\\[ (e_\\xi)_{\\xi \\in \\wh G}\n\t\t\\qquad\\text{by}\\qquad e_\\xi(x) = e(\\xi(x)) = \\exp(2\\pi i \\xi(x)) \\]\n\tform an orthonormal basis of $L^2(G)$.\n\tThus for each $f \\in L^2(G)$ we have\n\t\\[ f = \\sum_{\\xi \\in \\wh G} \\wh f(\\xi) e_\\xi \\]\n\twhere\n\t\\[ \\wh f(\\xi) = \\left< f, e_\\xi \\right>\n\t\t= \\int_G f(x) \\exp(-2\\pi i \\xi(x)) \\; d\\mu. \\]\n\\end{theorem}\nThe sum $\\sum_{\\xi \\in \\wh G}$ makes sense since $\\wh G$ is discrete.\nIn particular,\n\\begin{itemize}\n\t\\ii Letting $G = Z$ for a finite group $G$\n\tgives ``Fourier transform on finite groups''.\n\t\\ii The special case $G = \\ZZ/n\\ZZ$ has its\n\t\\href{https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Definition}%\n\t{own Wikipedia page}: the ``discrete-time Fourier transform''.\n\t\\ii Letting $G = \\TT$ gives the ``Fourier series'' earlier.\n\\end{itemize}\n\n\\section{The Fourier transform of the non-compact case}\nIf $G$ is LCA but not compact, then Theorem~\\ref{thm:god} becomes false.\nOn the other hand, it's still possible to define $\\wh G$.\nWe can then try to write the Fourier coefficients anyways:\nlet \\[ \\wh f(\\xi) = \\int_G f \\cdot \\ol{e_\\xi} \\; d\\mu \\]\nfor $\\xi \\in \\wh G$ and $f \\colon G \\to \\CC$.\nThe results are less fun in this case, but we still have, for example:\n\\begin{theorem}\n\t[Fourier inverison formula in the non-compact case]\n\tLet $\\mu$ be a Haar measure on $G$.\n\tThen there exists a unique Haar measure $\\nu$ on $\\wh G$\n\t(called the \\vocab{dual measure}) such that:\n\twhenever $f \\in L^1(G)$ and $\\wh f \\in L^1(\\wh G)$, we have\n\t\\[ f(x) = \\int_{\\wh G} \\wh f(\\xi) \\xi(x) \\; d\\nu \\]\n\tfor almost all $x \\in G$ (with respect to $\\mu$).\n\tIf $f$ is continuous, this holds for all $x$.\n\\end{theorem}\nSo while we don't have the niceness of a full inner product from before,\nwe can still in some situations at least write $f$ as integral\nin sort of the same way as before.\n\nIn particular, they have special names for a few special $G$:\n\\begin{itemize}\n\t\\ii If $G = \\RR$, then $\\wh G = \\RR$,\n\tyielding the\n\t``\\href{https://en.wikipedia.org/wiki/Fourier_transform}{(continuous) Fourier transform}''.\n\t\\ii If $G = \\ZZ$, then $\\wh G = \\TT$,\n\tyielding the\n\t``\\href{https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform}{discrete time Fourier transform}.\n\\end{itemize}\n\n\\section{Summary}\nWe summarize our various flavors of Fourier analysis\nfrom the previous sections in the following table.\nIn the first part $G$ is compact,\nin the second half $G$ is not.\n\\[\n\t\\begin{array}{llll}\n\t\t\\hline\n\t\t\\text{Name} & \\text{Domain }G & \\text{Dual }\\wh G \n\t\t\t& \\text{Characters} \\\\ \\hline\n\t\t\\text{Binary Fourier analysis} & \\{\\pm1\\}^n\n\t\t\t& S \\subseteq \\left\\{ 1, \\dots, n \\right\\}\n\t\t\t& \\prod_{s \\in S} x_s \\\\\n\t\t\\text{Fourier transform on finite groups} & Z\n\t\t\t& \\xi \\in \\wh Z \\cong Z & e( i \\xi \\cdot x) \\\\\n\t\t\\text{Discrete Fourier transform} & \\ZZ/n\\ZZ & \\xi \\in \\ZZ/n\\ZZ\n\t\t\t& e(\\xi x / n) \\\\\n\t\t\\text{Fourier series} & \\TT \\cong [-\\pi, \\pi] & n \\in \\ZZ\n\t\t\t& \\exp(inx) \\\\ \\hline\n\t\t\\text{Continuous Fourier transform} & \\RR & \\xi \\in \\RR\n\t\t \t& e(\\xi x) \\\\\n\t\t\\text{Discrete time Fourier transform} & \\ZZ & \\xi \\in \\TT \\cong [-\\pi, \\pi]\n\t\t \t& \\exp(i \\xi n) \\\\\n\t\\end{array}\n\\]\nYou might notice that the \\textbf{various names are awful}.\nThis is part of the reason I got confused as a high school student:\nevery type of Fourier series above has its own Wikipedia article.\nIf it were up to me, we would just use the term ``$G$-Fourier transform'',\nand that would make everyone's lives a lot easier.\n\n\n\n\\section{\\problemhead}\n\\begin{problem}\n\tIf $G$ is compact, so $\\wh G$ is discrete,\n\tdescribe the dual measure $\\nu$.\n\t\\begin{hint}\n\t\tYou can read it off \\Cref{thm:god}.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tIt is the counting measure.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t\\label{prob:LCA_compact}\n\tShow that an LCA group $G$ is compact\n\tif and only if $\\wh G$ is discrete.\n\t(You will need the compact-open topology for this.)\n\t\\begin{hint}\n\t\tAfter Pontryagin duality,\n\t\twe need to show $G$ compact implies $\\wh G$ discrete\n\t\tand $G$ discrete implies $\\wh G$ compact.\n\t\tBoth do not need anything fancy:\n\t\tthey are topological facts.\n\t\\end{hint}\n\\end{problem}\n\n\\endinput\n\\section{Peter-Weyl}\nIn fact, if $G$ is a Lie group, even if $G$ is not abelian\nwe can still give an orthonormal basis of $L^2(G)$\n(the square-integrable functions on $G$).\nIt turns out in this case the characters are attached to complex\nirreducible representations of $G$\n(and in what follows all representations are complex).\n\nThe result is given by the Peter-Weyl theorem.\nFirst, we need the following result:\n\\begin{lemma}\n\t[Compact Lie groups have unitary reps]\n\tAny finite-dimensional (complex) representation $V$ of a compact Lie group $G$\n\tis unitary, meaning it can be equipped with a $G$-invariant inner form.\n\tConsequently, $V$ is completely reducible:\n\tit splits into the direct sum of irreducible representations of $G$.\n\\end{lemma}\n\\begin{proof}\n\tSuppose $B : V \\times V \\to \\CC$ is any inner product.\n\tEquip $G$ with a right-invariant Haar measure $dg$.\n\tThen we can equip it with an ``averaged'' inner form\n\t\\[ \\wt B(v,w) = \\int_G B(gv, gw) \\; dg. \\]\n\tThen $\\wt B$ is the desired $G$-invariant inner form.\n\tNow, the fact that $V$ is completely reducible follows from the fact\n\tthat given a subrepresentation of $V$, its orthogonal complement\n\tis also a subrepresentation.\n\\end{proof}\n\nThe Peter-Weyl theorem then asserts that the finite-dimensional irreducible\nunitary representations essentially give an orthonormal basis for $L^2(G)$,\nin the following sense. Let $V = (V, \\rho)$ be such a representation of $G$,\nand fix an orthonormal basis of $e_1$, \\dots, $e_d$ for $V$ (where $d = \\dim V$).\nThe $(i,j)$th \\vocab{matrix coefficient} for $V$ is then given by \n\\[ G \\taking{\\rho} \\GL(V) \\taking{\\pi_{ij}} \\CC \\]\nwhere $\\pi_{ij}$ is the projection onto the $(i,j)$th entry of the matrix.\nWe abbreviate $\\pi_{ij} \\circ \\rho$ to $\\rho_{ij}$.\nThen the theorem is:\n\\begin{theorem}\n\t[Peter-Weyl]\n\tLet $G$ be a compact Lie group.\n\tLet $\\Sigma$ denote the (pairwise non-isomorphic) irreducible finite-dimensional\n\tunitary representations of $G$.\n\tThen\n\t\\[ \\left\\{ \\sqrt{\\dim V} \\rho_{ij}\n\t\t\t\\; \\Big\\vert \\; (V, \\rho) \\in \\Sigma,\n\t\t\t\\text{ and } 1 \\le i,j \\le \\dim V \\right\\} \\]\n\tis an orthonormal basis of $L^2(G)$.\n\\end{theorem}\nStrictly, I should say $\\Sigma$ is a set of representatives of \nthe isomorphism classes of irreducible unitary representations,\none for each isomorphism class.\n\nIn the special case $G$ is abelian,\nall irreducible representations are one-dimensional.\nA one-dimensional representation of $G$ is a map\n$G \\injto \\GL(\\CC) \\cong \\CC^\\times$,\nbut the unitary condition implies it is actually a map $G \\injto S^1 \\cong \\TT$,\ni.e.\\ it is an element of $\\wh G$.\n", "meta": {"hexsha": "039c5ad97c06985b2a2d21dc54c5da84ee39191d", "size": 12690, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "corpus/napkin/tex/measure/pontryagin.tex", "max_stars_repo_name": "aDotInTheVoid/ltxmk", "max_stars_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_stars_repo_licenses": ["Apache-2.0", "MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "corpus/napkin/tex/measure/pontryagin.tex", "max_issues_repo_name": "aDotInTheVoid/ltxmk", "max_issues_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_issues_repo_licenses": ["Apache-2.0", "MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "corpus/napkin/tex/measure/pontryagin.tex", "max_forks_repo_name": "aDotInTheVoid/ltxmk", "max_forks_repo_head_hexsha": "ee461679e51e92a0e4b121f28ae5fe17d5e5319e", "max_forks_repo_licenses": ["Apache-2.0", "MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.2228915663, "max_line_length": 106, "alphanum_fraction": 0.6885736801, "num_tokens": 4109, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802370707281, "lm_q2_score": 0.7981867681382279, "lm_q1q2_score": 0.6138696688664667}} {"text": "\\newpage\n\\subsection{Goal}\n\nExperimental study of the time complexity of different algorithms.\n\n\\subsection{Formulation of the problem}\n\nFor each n from 1 to 2000, measure the average computer execution time of programs implementing the algorithms and functions below for five runs.\nPlot the data obtained showing the average execution time as a function of n. Conduct the theoretical analysis of the time complexity of the algorithms in question and compare the empirical and theoretical time complexities.\n\n\\paragraph{I.}\nGenerate an $n$-dimensional random vector $v=[v_1,v_2,...,v_n]$ with non-negative elements.\nFor $v$, implement the following calculations and algorithms:\n\\begin{itemize}\n \\item $f(v) = const$ (constant function);\n \\item $f(v) = \\sum^{n}_{k = 1}{v_k}$ (the sum of elements);\n \\item $f(v) = \\prod^{n}_{k = 1}{v_k}$ (the product of elements);\n \\item supposing that the elements of $v$ are the coefficients of a polynomial $P$ of degree $n - 1$, calculate the value $P(1.5)$ by a direct calculation of $P(x) = \\sum^{n}_{k = 1}{v_kx^{k - 1}}$ (i.e. evaluating each term one by one) and by Horner's method by representing the polynomial as $P(x) = v_1 + x(v_2 + x(v_3 + ...))$;\n \\item Bubble Sort of the elements of $v$;\n \\item Quick Sort of the elements of $v$;\n \\item Timsort of the elements of $v$.\n\\end{itemize}\n\n\\paragraph{II.}\nGenerate random matrices $A$ and $B$ of size $n \\times n$ with non-negative elements. Find the usual matrix product for $A$ and $B$.\n\n\\paragraph{III.}\nDescribe the data structures and design techniques used within the algorithms.\n\n\\subsection{Brief theoretical part}\n\nTime complexity is characteristic of algorithm, which provides information about the amount of time necessary to process given amount of data.\n\nThe time complexity of the algorithm is estimated by counting the number of elementary operations (addition, multiplication, etc.) performed by the algorithm for a given amount of data. It is assumed that each elementary operation requires a fixed amount of time.\n\nA \\textbf{constant function}, \\textbf{the sum of elements}, \\textbf{the product of elements}, and \\textbf{the calculation of the polynomial} are calculated in a single pass through the loop.\nThey use an iterative strategy.\n\n\\textbf{Bubble sort} is a simple sorting algorithm.\nThis sorting algorithm is comparison-based algorithm in which each pair of adjacent elements is compared and the elements are swapped if they are not in order.\nAn iterative pattern is also used here.\n\n\\textbf{Quick Sort} is a sorting algorithm, which is commonly used in computer science.\nQuick Sort is a divide and conquer algorithm.\nIt creates two empty arrays to hold elements less than the pivot value and elements greater than the pivot value, and then recursively sort the sub arrays.\nThere are two basic operations in the algorithm, swapping items in place and partitioning a section of the array.\n\n\\textbf{TimSort} is a stable, adaptive iterative merge sort that requires a small number of comparisons when working with partially sorted arrays, while providing performance comparable to traditional merge sorting when working with random arrays.\n\n\\subsection{Results}\n\n\\paragraph{Constant function $f(v) = 1$.}\n\nWith the exception of one peak, which is related to the computing platform, the theoretical and empirical estimates are the same - $O(1)$ (Figure \\ref{ris:constant_function}).\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/constant_function.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the function $f(v) = \\sum^{n}_{k = 1}{v_k}$.}\n \\label{ris:constant_function}\n\\end{figure}\n\n\\paragraph{Sum and product of elements.}\n\nDespite the $y$-axis spikes, these functions are approximated by a straight line and coincide with the theoretical estimate - $O(n)$ (Figure \\ref{ris:sum_function}, \\ref{ris:prod_function}).\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/sum_function.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the function $f(v) = \\prod^{n}_{k = 1}{v_k}$.}\n \\label{ris:sum_function}\n\\end{figure}\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/prod_function.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the function $P(x) = \\sum^{n}_{k = 1}{v_kx^{k - 1}}$.}\n \\label{ris:prod_function}\n\\end{figure}\n\n\\paragraph{Calculation of the polynomial.}\n\nThe direct calculation and the Horner's method also correspond to the theoretical estimate - $O(n)$ (Figure \\ref{ris:naive_poly_function}, \\ref{ris:horner_method_function}). It is worth noting that the result, when calculated directly, despite the use of long arithmetic, overflows the buffer when the length of vector is $>1750$.\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/naive_poly_function.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that\nimplements the function $P(x) = \\sum^{n}_{k = 1}{v_kx^{k - 1}}$.}\n \\label{ris:naive_poly_function}\n\\end{figure}\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/horner_method_function.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the function $P(x) = v_1 + x(v_2 + x(v_3 + ...))$.}\n \\label{ris:horner_method_function}\n\\end{figure}\n\n\\paragraph{Bubble Sort.}\n\nBubble sorting has an average time complexity of $O(n^2)$, which is shown on Figure \\ref{ris:bubble_sort}.\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/bubble_sort.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the Bubble Sort.}\n \\label{ris:bubble_sort}\n\\end{figure}\n\n\\paragraph{Quick Sort and TimSort.}\n\nQuick sort and Timsort in the average case have a theoretical complexity of $O(n\\log{n})$ and are approximated by linear functions . This is confirmed by an empirical experiment (Figure \\ref{ris:quick_sort}, \\ref{ris:timsort}).\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/quick_sort.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that\nimplements the Quick Sort.}\n \\label{ris:quick_sort}\n\\end{figure}\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/timsort.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the TimSort.}\n \\label{ris:timsort}\n\\end{figure}\n\n\\paragraph{Matrix product.}\n\nIn the \\textit{numpy} library of the \\textit{Python} programming language, the \\textit{matmul} function implements a matrix product using the \\textit{Strassen} method and has a theoretical complexity of $O(n^{2.81})$ (Figure \\ref{ris:matmul}).\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/matmul.png}\n \\caption{Empirical and theoretical time complexities of an algorithm that implements the matrix product.}\n \\label{ris:matmul}\n\\end{figure}\n\n\\subsection{Conclusion}\n\nAs a result of this task, algorithms with different time complexity are implemented.\nIt is shown that the empirical time for the algorithms described above coincides with the theoretical time.\nStrategies for implementing algorithms are also described.\n\n\\subsection{Appendix}\n\nThe source code is located \\href{https://github.com/vanSultan/anal_dev_algo/tree/lab_01}{here}: \\url{https://github.com/vanSultan/anal_dev_algo/tree/lab_01}.\n", "meta": {"hexsha": "b879b8c4b90b451ac66e7609a0d7cf03fd89462c", "size": 7515, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lab_01/report/lab_01_body.tex", "max_stars_repo_name": "vanSultan/anal_dev_algo", "max_stars_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lab_01/report/lab_01_body.tex", "max_issues_repo_name": "vanSultan/anal_dev_algo", "max_issues_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lab_01/report/lab_01_body.tex", "max_forks_repo_name": "vanSultan/anal_dev_algo", "max_forks_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.1, "max_line_length": 334, "alphanum_fraction": 0.7552894212, "num_tokens": 1880, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154240079185319, "lm_q2_score": 0.8577681104440172, "lm_q1q2_score": 0.6136678994385647}} {"text": "% !TEX root = hott_intro.tex\n\n\\section{Propositions, sets, and the higher truncation levels}\n\\sectionmark{Truncation levels}\\label{chap:hierarchy}\n\n\\index{truncated type|(}\n\\index{truncation level|(}\n%Not all types have interesting higher groupoid structure. For example, we will see below that two natural numbers can only be equal in at most one way. Voevodsky articulated a useful notion to detect the homotopical complexity of types, which allows us to distinguish between contractible types (also called \\emph{$(-2)$-types}), \\emph{propositions} (also called \\emph{$(-1)$-types}), \\emph{sets} (\\emph{$0$-types}), and \\emph{$k$-types} for higher $k$.\n\n%We will see [later] that there are types that are not $k$-types for any $k$.\n\n\\subsection{Propositions and subtypes}\n\n\\index{proposition|(}\n\\begin{defn}\nA type $A$ is said to be a \\define{proposition} if there is a term of type\\index{is-prop(A)@{$\\isprop(A)$}}\n\\begin{equation*}\n\\isprop(A)\\defeq\\prd{x,y:A}\\iscontr(x=y).\n\\end{equation*}\nGiven a universe $\\UU$, we define $\\prop_\\UU$\\index{Prop@{$\\prop$}} to be the type of all small propositions, i.e.,\n\\begin{equation*}\n \\prop_\\UU\\defeq\\sm{X:\\UU}\\isprop(X).\n\\end{equation*}\n\\end{defn}\n\n\\begin{eg}\\label{eg:prop_contr}\n Any contractible type is a proposition by \\cref{ex:prop_contr}\\index{contractible type!is a proposition}\\index{is a proposition!contractible type}. In particular, the unit type is a proposition.\n\n However, propositions do not need to be inhabited: the empty type is also a proposition, since\\index{empty type!is a proposition}\\index{is a proposition!empty type}\n\\begin{equation*}\n\\prd{x,y:\\emptyt}\\iscontr(x=y)\n\\end{equation*}\nfollows from the induction principle of the empty type.\n\\end{eg}\n\nIn the following lemma we prove that in order to show that a type $A$ is a proposition, it suffices to show that any two terms of $A$ are equal. In other words, propositions are types with \\define{proof irrelevance}.\n\n\\begin{thm}\\label{lem:isprop_eq}\n Let $A$ be a type. Then the following are equivalent:\n \\begin{enumerate}\n \\item The type $A$ is a proposition.\n \\item Any two terms of type $A$ can be identified, i.e., there is a dependent function\\index{is-prop'(A)@{$\\isprop'(A)$}}\\index{is-prop(A)@{$\\isprop(A)$}!is-prop(A) iff is-prop'(A)@{$\\isprop(A)\\leftrightarrow\\isprop'(A)$}}\n \\begin{equation*}\n \\isprop'(A)\\defeq\\prd{x,y:A}\\id{x}{y}.\n \\end{equation*}\n \\item The type $A$ is contractible as soon as it is inhabited, i.e., there is a function\\index{is-prop(A)@{$\\isprop(A)$}!is-prop(A) iff A to is-contr(A)@{$\\isprop(A)\\leftrightarrow(A\\to\\iscontr(A))$}}\n \\begin{equation*}\n A \\to \\iscontr(A).\n \\end{equation*}\n \\item The map $\\const_\\ttt : A\\to\\unit$ is an embedding.\\index{is-prop(A)@{$\\isprop(A)$}!is-prop(A) iff is-emb(const star)@{$\\isprop(A)\\leftrightarrow\\isemb(\\const_\\ttt)$}}\n \\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\n To show that (i) implies (ii), let $A$ be a proposition. Then its identity types are contractible, so the center of contraction of $\\id{x}{y}$ is identification $\\id{x}{y}$, for each $x,y:A$. \n\n To show that (ii) implies (iii), suppose that $A$ comes equipped with $p:\\prd{x,y:A}\\id{x}{y}$. Then for any $x:A$ the dependent function $p(x):\\prd{y:A}\\id{x}{y}$ is a contraction of $A$. Thus we obtain the function\n \\begin{equation*}\n \\lam{x}(x,p(x)):A\\to\\iscontr(A).\n \\end{equation*}\n\n To show that (iii) implies (iv), suppose that $A\\to\\iscontr(A)$. We first make the simple observation that\n \\begin{equation*}\n (X\\to \\isemb(f))\\to \\isemb(f)\n \\end{equation*}\n for any map $f:X\\to Y$, so it suffices to show that $A\\to\\isemb(\\const_\\ttt)$. However, assuming we have $x:A$, it follows by assumption that $A$ is contractible. Therefore, it follows by \\cref{ex:contr_equiv} that the map $\\const_\\ttt:A\\to\\unit$ is an equivalence. Since it is an equivalence, it is an embedding by \\cref{cor:emb_equiv}.\n\n To show that (iv) implies (i), note that if $A\\to\\unit$ is an embedding, then the identity types of $A$ are equivalent to contractible types and therefore they must be contractible.\n\\end{proof}\n\nIn the following lemma we show that propositions are closed under equivalences.\n\n\\begin{lem}\\label{lem:prop_equiv}\nLet $A$ and $B$ be types, and let $e:\\eqv{A}{B}$. Then we have\\index{proposition!closed under equivalences}\n\\begin{equation*}\n\\isprop(A)\\leftrightarrow\\isprop(B).\n\\end{equation*}\n\\end{lem}\n\n\\begin{proof}\nWe will show that $\\isprop(B)$ implies $\\isprop(A)$. This suffices, because the converse follows from the fact that $e^{-1}:B\\to A$ is also an equivalence. \n\nSince $e$ is assumed to be an equivalence, it follows by \\cref{cor:emb_equiv} that\n\\begin{equation*}\n\\apfunc{e} : (x=y)\\to (e(x)=e(y))\n\\end{equation*}\nis an equivalence for any $x,y:A$. If $B$ is a proposition, then in particular the type $e(x)=e(y)$ is contractible for any $x,y:A$, so the claim follows from \\cref{thm:contr_equiv}.\n\\end{proof}\n\n In set theory, a set $y$ is said to be a subset\\index{subset} of a set $x$, if any element of $y$ is an element of $x$, i.e., if the condition\n \\begin{equation*}\n \\forall_z (z\\in y)\\to (z\\in x)\n \\end{equation*}\n holds. We have already noted that type theory is different from set theory in that terms in type theory come equipped with a \\emph{unique} type. Moreover, in set theory the proposition $x\\in y$ is well-formed for any two sets $x$ and $y$, whereas in type theory the judgment $a:A$ is only well-formed if it is derived using the postulated inference rules. Because of these differences we must find a different way to talk about subtypes.\n\n Note that in set theory there is a correspondence between the subsets of a set $x$, and the \\emph{predicates} on $x$. A predicate on $x$ is just a proposition $P(z)$ that varies over the elements $z\\in x$. Indeed, if $y$ is a subset of $x$, then the corresponding predicate is the proposition $z\\in y$. Conversely, if $P$ is a predicate on $x$, then we obtain the subset\n \\begin{equation*}\n \\{z\\in x\\mid P(z)\\}\n \\end{equation*}\n of $x$. Now we have the right idea of subtypes in type theory: they are families of propositions.\n\n\\begin{defn}\nA type family $B$ over $A$ is said to be a \\define{subtype}\\index{subtype} of $A$ if for each $x:A$ the type $B(x)$ is a proposition. When $B$ is a subtype of $A$, we also say that $B(x)$ is a \\define{property}\\index{property} of $x:A$.\n\\end{defn}\n\nWe will show in \\cref{thm:subtype} that a type family $B$ over $A$ is a subtype of $A$ if and only if the projection map $\\proj 1:\\big(\\sm{x:A}B(x)\\big)\\to A$ is an embedding.\n\n\\index{proposition|)}\n\n\\subsection{Sets}\n\n\\index{set|(}\n\\begin{defn}\n A type $A$ is said to be a \\define{set} if it comes equipped with a term of type\n \\index{is-set(A)@{$\\isset(A)$}}\\index{is a set}\n\\begin{equation*}\n\\isset(A)\\defeq \\prd{x,y:A}\\isprop(\\id{x}{y}).\n\\end{equation*}\n\\end{defn}\n\n\\begin{lem}\nA type $A$ is a set if and only if it satisfies \\define{axiom K}\\index{axiom K}, i.e., if and only if it comes equipped with a term of type\\index{is-set(A)@{$\\isset(A)$}!is-set(A) iff axiom-K(A)@{$\\isset(A)\\leftrightarrow\\axiomK(A)$}}\n\\begin{equation*}\n\\axiomK(A)\\defeq\\prd{x:A}\\prd{p:\\id{x}{x}}\\id{\\refl{x}}{p}.\n\\end{equation*}\n\\end{lem}\n\n\\begin{proof}\nIf $A$ is a set, then $\\id{x}{x}$ is a proposition, so any two of its elements are equal. \nThis implies axiom $K$. \n\nFor the converse, if $A$ satisfies axiom $K$, then for any $p,q:\\id{x}{y}$ we have $\\id{\\ct{p}{q^{-1}}}{\\refl{x}}$, and hence $\\id{p}{q}$. This shows that $\\id{x}{y}$ is a proposition, and hence that $A$ is a set.\n\\end{proof}\n\n\\begin{thm}\\label{lem:prop_to_id}\nLet $A$ be a type, and let $R:A\\to A\\to\\UU$ be a binary relation on $A$ satisfying\n\\begin{enumerate}\n\\item Each $R(x,y)$ is a proposition,\n\\item $R$ is reflexive, as witnessed by $\\rho:\\prd{x:A}R(x,x)$,\n\\item There is a map\n \\begin{equation*}\n R(x,y)\\to (x=y)\n \\end{equation*}\n for each $x,y:A$.\n\\end{enumerate}\nThen any family of maps\n\\begin{equation*}\n\\prd{x,y:A}(\\id{x}{y})\\to R(x,y)\n\\end{equation*}\nis a family of equivalences. Consequently, the type $A$ is a set.\n\\end{thm}\n\n\\begin{proof}\nLet $f:\\prd{x,y:A}R(x,y)\\to(\\id{x}{y})$. \nSince $R$ is assumed to be reflexive, we also have a family of maps\n\\begin{equation*}\n\\pathind_x(\\rho(x)):\\prd{y:A}(\\id{x}{y})\\to R(x,y).\n\\end{equation*}\nSince each $R(x,y)$ is assumed to be a proposition, it therefore follows that each $R(x,y)$ is a retract of $\\id{x}{y}$. Therefore it follows that $\\sm{y:A}R(x,y)$ is a retract of $\\sm{y:A}x=y$, which is contractible. We conclude that $\\sm{y:A}R(x,y)$ is contractible, and therefore that any family of maps\n\\begin{equation*}\n \\prd{y:A}(x=y)\\to R(x,y)\n\\end{equation*}\nis a family of equivalences.\n\nNow it also follows that $A$ is a set, since its identity types are equivalent to propositions, and therefore they are propositions by \\cref{lem:prop_equiv}. \n\\end{proof}\n\n\\begin{defn}\n A map $f:A\\to B$ is said to be \\define{injective}\\index{injective function} if for any $x,y:A$ there is a map\n \\begin{equation*}\n (f(x)=f(y))\\to (x=y).\n \\end{equation*}\n\\end{defn}\n\n\\begin{cor}\\label{cor:is-emb-is-injective}\n Any injective map into a set is an embedding.\\index{is an embedding!injective map into a set}\n\\end{cor}\n\n\\begin{proof}\n Let $f:A\\to B$ be an injective map between sets. Now consider the relation\n \\begin{equation*}\n R(x,y)\\defeq (f(x)=f(y)).\n \\end{equation*}\n Note that $R$ is reflexive, and that $R(x,y)$ is a proposition for each $x,y:A$. Moreover, by the assumption that $f$ is injective, we have\n \\begin{equation*}\n R(x,y)\\to (x=y)\n \\end{equation*}\n for any $x,y:A$. Therefore we are in the situation of \\cref{lem:prop_to_id}, so it follows that the map $\\apfunc{f} : (x=y)\\to (f(x)=f(y))$ is an equivalence.\n\\end{proof}\n\n\\begin{thm}\\label{thm:eq_nat}\nThe type of natural numbers is a set.\\index{is a set!natural numbers}\\index{natural numbers!is a set}\n\\end{thm}\n\n\\begin{proof}\nWe will apply \\cref{lem:prop_to_id}. Note that the observational equality $\\EqN:\\N\\to(\\N\\to\\UU)$ on $\\N$ (\\cref{defn:obs_nat}) is a reflexive relation by \\cref{ex:obs_nat_eqrel}, and moreover that $\\EqN(n,m)$ is a proposition for every $n,m:\\N$ (proof by double induction).\nTherefore it suffices to show that\n\\begin{equation*}\n\\prd{m,n:\\N}\\EqN(m,n)\\to (\\id{m}{n}).\n\\end{equation*}\nThis follows from the fact that observational equality is the \\emph{least} reflexive relation, which was shown in \\cref{ex:obs_nat_least}.\n\\end{proof}\n\n\\begin{comment}\n\\begin{thm}[Hedberg]\\label{thm:dec_eq}\nAny type with decidable equality is a set.\n\\end{thm}\n\n\\begin{proof}\nLet $A$ be a type, and let $d:\\prd{x,y:A}(\\id{x}{y})+\\neg(\\id{x}{y})$ be the witness that $A$ has decidable equality.\nWe first construct a reflexive binary relation $E:A\\to A\\to\\type$ such that each $E(x,y)$ is a proposition.\nFor every $x,y:A$, we first define a type family $E'(x,y):((\\id{x}{y})+\\neg(\\id{x}{y}))\\to\\type$ by\n\\begin{align*}\nE'(x,y,\\inl(p)) & \\defeq \\unit \\\\\nE'(x,y,\\inr(p)) & \\defeq \\emptyt.\n\\end{align*}\nNote that $E'(x,y,q)$ is a proposition for each $x,y:A$ and $q:(\\id{x}{y})+\\neg(\\id{x}{y})$. \nNow we set $E(x,y)\\defeq E'(x,y,d(x,y))$. Then $E$ is clearly reflexive, and a family of propositions.\nTherefore it remains to show that $E$ implies identity. \n\nSince $E$ is defined as an instance of $E'$, it suffices to construct a term of type\n\\begin{equation*}\n\\prd{x,y:A}\\prd{q:(\\id{x}{y})+\\neg(\\id{x}{y})} E'(q)\\to (\\id{x}{y}). \n\\end{equation*}\nBy induction of disjoint sums, it suffices to construct terms of types\n\\begin{align*}\n& \\prd{x,y:A}\\prd{p:\\id{x}{y}} \\unit\\to (\\id{x}{y}) \\\\\n& \\prd{x,y:A}\\prd{p:\\neg(\\id{x}{y})} \\emptyt\\to (\\id{x}{y}).\n\\end{align*}\nIn the first case, we take $\\lam{x}\\lam{y}\\lam{p}{t}p$, and the second case is by induction on the empty type.\n\\end{proof}\n\\end{comment}\n\\index{set|)}\n\n\\subsection{General truncation levels}\n\n\\begin{defn}\nWe define $\\istrunc{} : \\Z_{\\geq-2}\\to\\UU\\to\\UU$ by induction on $k:\\Z_{\\geq -2}$, taking\\index{is-trunc k(A)@{$\\istrunc{k}(A)$}}\n\\begin{align*}\n\\istrunc{-2}(A) & \\defeq \\iscontr(A) \\\\\n\\istrunc{k+1}(A) & \\defeq \\prd{x,y:A}\\istrunc{k}(\\id{x}{y}).\\qedhere\n\\end{align*}\nFor any type $A$, we say that $A$ is \\define{$k$-truncated}\\index{k-truncated type@{$k$-truncated type}|see {truncated type}}, or a \\define{$k$-type}\\index{k-type@{$k$-type}}, if there is a term of type $\\istrunc{k}(A)$. We say that a map $f:A\\to B$ is $k$-truncated if its fibers are $k$-truncated.\\index{k-truncated map@{$k$-truncated map}|see {truncated map}}\\index{truncated map}\n\\end{defn}\n\n%For the rest of this section, let $k:\\Z_{\\geq-2}$.\n\n\\begin{thm}\\label{thm:istrunc_next}\nIf $A$ is a $k$-type, then $A$ is also a $(k+1)$-type.\\index{is-trunc k(A)@{$\\istrunc{k}(A)$}!is-trunc k(A) to is-trunc k+1(A)@{$\\istrunc{k}(A)\\to\\istrunc{k+1}(A)$}}\n\\end{thm}\n\n\\begin{proof}\nWe have seen in \\cref{eg:prop_contr} that contractible types are propositions. This proves the base case.\nFor the inductive step, note that if any $k$-type is also a $(k+1)$-type, then any $(k+1)$-type is a $(k+2)$-type, since its identity types are $k$-types and therefore $(k+1)$-types.\n\\end{proof}\n\n\\begin{thm}\\label{thm:ktype_eqv}\nIf $e:\\eqv{A}{B}$ is an equivalence, and $B$ is a $k$-type, then so is $A$.\\index{truncated type!closed under equivalences}\n\\end{thm}\n\n\\begin{proof}\nWe have seen in \\cref{ex:contr_equiv} that if $B$ is contractible and $e:\\eqv{A}{B}$ is an equivalence, then $A$ is also contractible. This proves the base case.\n\nFor the inductive step, assume that the $k$-types are stable under equivalences, and consider $e:\\eqv{A}{B}$ where $B$ is a $(k+1)$-type. In \\cref{cor:emb_equiv} we have seen that\n\\begin{equation*}\n\\apfunc{e}:(\\id{x}{y})\\to(\\id{e(x)}{e(y)})\n\\end{equation*}\nis an equivalence for any $x,y$. Note that $\\id{e(x)}{e(y)}$ is a $k$-type, so by the induction hypothesis it follows that $\\id{x}{y}$ is a $k$-type. This proves that $A$ is a $(k+1)$-type.\n\\end{proof}\n\n\\begin{cor}\\label{cor:emb_into_ktype}\nIf $f:A\\to B$ is an embedding, and $B$ is a $(k+1)$-type, then so is $A$.\\index{truncated type!closed under embeddings}\n\\end{cor}\n\n\\begin{proof}\nBy the assumption that $f$ is an embedding, the action on paths\n\\begin{equation*}\n\\apfunc{f}:(\\id{x}{y})\\to (\\id{f(x)}{f(y)})\n\\end{equation*}\nis an equivalence for every $x,y:A$. Since $B$ is assumed to be a $(k+1)$-type, it follows that $f(x)=f(y)$ is a $k$-type for every $x,y:A$. Therefore we conclude by \\cref{thm:ktype_eqv} that $\\id{x}{y}$ is a $k$-type for every $x,y:A$. In other words, $A$ is a $(k+1)$-type.\n\\end{proof}\n\n\\begin{thm}\nLet $B$ be a type family over $A$. Then the following are equivalent:\\index{truncated family of types}\n\\begin{enumerate}\n\\item For each $x:A$ the type $B(x)$ is $k$-truncated. In this case we say that the family $B$ is \\define{$k$-truncated}.\n\\item The projection map\n\\begin{equation*}\n\\proj 1 : \\Big(\\sm{x:A}B(x)\\Big)\\to A\n\\end{equation*}\nis $k$-truncated.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nBy \\cref{ex:proj_fiber} we obtain equivalences\n\\begin{equation*}\n\\eqv{\\fib{\\proj 1}{x}}{B(x)}\n\\end{equation*}\nfor every $x:A$. Therefore the claim follows from \\cref{thm:ktype_eqv}.\n\\end{proof}\n\n\\begin{thm}\\label{thm:trunc_ap}\nLet $f:A\\to B$ be a map. The following are equivalent:\n\\begin{enumerate}\n\\item The map $f$ is $(k+1)$-truncated.\n\\item For each $x,y:A$, the map\n\\begin{equation*}\n\\apfunc{f} : (x=y)\\to (f(x)=f(y))\n\\end{equation*}\nis $k$-truncated. \n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nFirst we show that for any $s,t:\\fib{f}{b}$ there is an equivalence\n\\begin{equation*}\n\\eqv{(s=t)}{\\fib{\\apfunc{f}}{\\ct{\\proj 2(s)}{\\proj 2(t)^{-1}}}}\n\\end{equation*}\nWe do this by $\\Sigma$-induction on $s$ and $t$, and then we calculate\n\\begin{align*}\n(\\pairr{x,p}=\\pairr{y,q}) & \\eqvsym \\Eqfib_f((x,p),(y,q)) \\\\\n & \\jdeq \\sm{\\alpha:x=y} p=\\ct{\\ap{f}{\\alpha}}{q} \\\\\n & \\eqvsym \\sm{\\alpha:x=y} \\ct{\\ap{f}{\\alpha}}{q}=p \\\\\n& \\eqvsym \\sm{\\alpha:x=y} \\ap{f}{\\alpha}=\\ct{p}{q^{-1}} \\\\\n& \\jdeq \\fib{\\apfunc{f}}{\\ct{p}{q^{-1}}}.\n\\end{align*}\nBy these equivalences, it follows that if $\\apfunc{f}$ is $k$-truncated, then for each $s,t:\\fib{f}{b}$ the identity type $s=t$ is equivalent to a $k$-truncated type, and therefore we obtain by \\cref{thm:ktype_eqv} that $f$ is $(k+1)$-truncated.\n\nFor the converse, note that we have equivalences\n\\begin{align*}\n\\fib{\\apfunc{f}}{p} & \\eqvsym ((x,p)=(y,\\refl{f(y)})).\n\\end{align*}\nIt follows that if $f$ is $(k+1)$-truncated, then the identity type $(x,p)=(y,\\refl{f(y)})$ in $\\fib{f}{f(y)}$ is $k$-truncated for any $p:f(x)=f(y)$. We conclude by \\cref{thm:ktype_eqv} that the fiber $\\fib{\\apfunc{f}}{p}$ is $k$-truncated. \n\\end{proof}\n\n\\begin{cor}\\label{cor:prop_emb}\nA map is an embedding if and only if its fibers are propositions.\\index{is an embedding!-1-truncated map@{$(-1)$-truncated map}}\n\\end{cor}\n\n\\begin{cor}\\label{thm:subtype}\nA type family $B$ over $A$ is a subtype if and only if the projection map\n\\begin{equation*}\n\\proj 1 : \\Big(\\sm{x:A}B(x)\\Big)\\to A\n\\end{equation*}\nis an embedding.\n\\end{cor}\n\n\\begin{thm}\nLet $f:\\prd{x:A}B(x)\\to C(x)$ be a family of maps. Then the following are equivalent:\n\\begin{enumerate}\n\\item For each $x:A$ the map $f(x)$ is $k$-truncated.\n\\item The induced map \n\\begin{equation*}\n\\tot{f}:\\Big(\\sm{x:A}B(x)\\Big)\\to\\Big(\\sm{x:A}C(x)\\Big)\n\\end{equation*}\nis $k$-truncated.\n\\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\nThis follows directly from \\cref{lem:fib_total,thm:ktype_eqv}.\n\\end{proof}\n\n\\begin{exercises}\n\\exercise\n \\begin{subexenum}\n \\item Show that $\\succN:\\N\\to\\N$ is an embedding.\\index{succ N@{$\\succN$}!is an embedding}\\index{is an embedding!succ N@{$\\succN$}}\n \\item Show that $n\\mapsto m+n$ is an embedding, for each $m:\\N$.\\index{add N@{$\\addN$}!add N(m) is an embedding@{$\\addN(m)$ is an embedding}}\\index{is an embedding!add N(m)@{$\\addN(m)$}} Moreover, conclude that there is an equivalence\n \\begin{equation*}\n \\fib{\\addN(m)}{n}\\simeq (m\\leq n).\n \\end{equation*}\n \\item Show that $n\\mapsto mn$ is an embedding\\index{mul N@{$\\mulN$}!mul N(m) is an embedding if m>0@{$\\mulN(m)$ is an embedding if $m>0$}}\\index{is an embedding!mul N(m) for m > 0@{$\\mulN(m)$ for $m>0$}}, for each $m>0$ in $\\N$. Conclude that the divisibility relation\\index{d {\"| n}@{$d\\mid n$}!is a proposition if d>0@{is a proposition if $d>0$}}\\index{is a proposition!d {\"| n} for d>0@{$d\\mid n$ for $d>0$}}\n \\begin{equation*}\n d\\mid n\n \\end{equation*}\n is a proposition for each $d,n:\\N$ such that $d>0$. \n \\end{subexenum}\n\\exercise \\label{ex:diagonal}Let $A$ be a type, and let the \\define{diagonal} of $A$ be the map $\\delta_A:A\\to A\\times A$ given by $\\lam{x}(x,x)$. \n\\begin{subexenum}\n\\item Show that\n\\begin{equation*}\n{\\isequiv(\\delta_A)}\\leftrightarrow{\\isprop(A)}.\n\\end{equation*}\n\\item Construct an equivalence $\\eqv{\\fib{\\delta_A}{(x,y)}}{(x=y)}$ for any $x,y:A$.\n\\item Show that $A$ is $(k+1)$-truncated if and only if $\\delta_A:A\\to A\\times A$ is $k$-truncated.\n\\end{subexenum}\n\\exercise \\label{ex:istrunc_sigma}\n\\begin{subexenum}\n\\item Let $B$ be a type family over $A$. Show that if $A$ is a $k$-type, and $B(x)$ is a $k$-type for each $x:A$, then so is $\\sm{x:A}B(x)$. Conclude that for any two $k$-types $A$ and $B$, the type $A\\times B$ is also a $k$-type. Hint: for the base case, use \\cref{ex:contr_in_sigma,ex:contr_equiv}.\n\\item Show that for any $k$-type $A$, the identity types of $A$ are also $k$-types.\n\\item Show that any maps $f:A\\to B$ between $k$-types $A$ and $B$\nis a $k$-truncated map.\n\\item Use \\cref{ex:proj_fiber} to show that for any type family $B:A\\to \\UU$, if $A$ and $\\sm{x:A}B(x)$ are $k$-types, then so is $B(x)$ for each $x:A$. \n\\end{subexenum}\n\\exercise \\label{ex:eq_bool}Show that $\\bool$ is a set by applying \\cref{lem:prop_to_id} with the observational equality on $\\bool$ defined in \\cref{ex:obs_bool}.\n\\exercise \\label{ex:set_coprod}Show that for any two $(k+2)$-types $A$ and $B$, the disjoint sum $A+B$ is again a $(k+2)$-type. Conclude that $\\mathbb{Z}$ is a set.\n\\exercise Use \\cref{ex:contr_retr,ex:retr_id} to show that if $A$ is a retract of a $k$-type $B$, then $A$ is also a $k$-type.\n\\exercise Show that a type $A$ is a $(k+1)$-type if and only if the map $\\const_x:\\unit\\to A$ is $k$-truncated for every $x:A$.\n\\exercise Consider a commuting triangle\n\\begin{equation*}\n\\begin{tikzcd}[column sep=tiny]\nA \\arrow[rr,\"h\"] \\arrow[dr,swap,\"f\"] & & B \\arrow[dl,\"g\"] \\\\\n& X\n\\end{tikzcd}\n\\end{equation*}\nwith $H: f \\htpy g \\circ h$, and suppose that $g$ is $k$-truncated. Show that $f$ is $k$-truncated if and only if $h$ is $k$-truncated.\n% Suppose that $h$ is $k$-truncated and surjective. Show that $f$ is $(k+1)$-truncated if and only if $g$ is $(k+1)$-truncated.\n\\end{exercises}\n\\index{truncated type|)}\n\\index{truncation level|)}\n", "meta": {"hexsha": "b04152f7e04556b6edbaa33bc65ba600638d3e60", "size": 20697, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/hierarchy.tex", "max_stars_repo_name": "hemangandhi/HoTT-Intro", "max_stars_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 333, "max_stars_repo_stars_event_min_datetime": "2018-09-26T08:33:30.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-22T23:50:15.000Z", "max_issues_repo_path": "Book/hierarchy.tex", "max_issues_repo_name": "hemangandhi/HoTT-Intro", "max_issues_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 8, "max_issues_repo_issues_event_min_datetime": "2019-06-18T04:16:04.000Z", "max_issues_repo_issues_event_max_datetime": "2020-10-16T15:27:01.000Z", "max_forks_repo_path": "Book/hierarchy.tex", "max_forks_repo_name": "hemangandhi/HoTT-Intro", "max_forks_repo_head_hexsha": "09c710bf9c31ba88be144cc950bd7bc19c22a934", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 30, "max_forks_repo_forks_event_min_datetime": "2018-09-26T09:08:57.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-16T00:33:50.000Z", "avg_line_length": 50.2354368932, "max_line_length": 454, "alphanum_fraction": 0.6751703145, "num_tokens": 7296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154239957834733, "lm_q2_score": 0.8577681086260461, "lm_q1q2_score": 0.6136678877288783}} {"text": "% !TEX root = index.tex\n\\maketitle\n\\tableofcontents\n\\newpage\n\n\\setcounter{section}{-1}\n\\section{Introduction \\& Motivation}\n\\epigraph{In mathematics you don't understand things. You just get used to them.}{John von Neumann}\n\nCohomology was introduced by Poincare in a series of papers named \\emph{Analysis Situs} and now forms the basis of modern Algebraic Topology.\nTo a topological space $X$ we can associate a sequence of vector spaces denoted\n\\begin{align*}\n \\check H^i(X)\n\\end{align*}\nfor each $i \\in \\Z_{\\ge 0}$ called it's \\textbf{Cech Cohomology} (pronounced \\emph{check cohomology}). In a very loose sense, the dimension of $\\check H^i(X)$ measures the $i^{th}$ dimensional holes in $X$.\\\\\n\nWhy care about the $i^{th}$ dimensional holes? We can use these to rigorously distinguish between spaces. For example, most proofs of the fact that $\\R^m$ is not homeomorphic\\footnote{Homeomorphism is the isomorphism for topological spaces.} to $\\R^n$ if $m \\neq n$ use some cohomology computation. We'll also see that a torus $S^1 \\times S^1$ has two \\emph{1-dimensional holes} which distinguishes it from a sphere $S^2$ which has none.\n\n\\begin{thm}\n Two topological spaces $X$, $Y$ are homeomorphic only if\n \\begin{align*}\n \\check H^i(X) \\cong \\check H^i(Y)\n \\end{align*} for all non-negative integers $i$.\n\\end{thm}\n\\begin{remark}\n The above statement is not an \\emph{if and only if} statement. The other direction is easily shown to be false. You'll be able to come up with examples by yourself in a couple of days.\n\\end{remark}\n\nComputing the cohomology requires multiple steps. The goal of this class is to develop the relevant machinery and actually do some cohomology computations.\n\\begin{align*}\n \\xymatrix@R-2pc{\n X \\ar@{~>}[r] & \\U \\ar@{~>}[r] & \\L^\\bullet(\\U) \\ar@{~>}[r] & \\check H^*(X)\\\\\n \\mbox{Topological Space} & \\mbox{Good cover of } X & \\mbox{Cech Complex of } \\U & \\mbox{Cech Cohomology of } X\n }\n\\end{align*}\n\n\\noindent \\textbf{Note:} The number of stars (*) on the problems indicate their difficulty level. The non-starred marked problems are compulsory, the starred problems are optional.\n\n% We'll gradually define Cech cohomology. The strategy is the following.\n% \\begin{enumerate}\n% \t\\item For a space $ X$, find a good open cover $ \\U$\n% \t\\item Find the Cech complex (to be defined) $ \\L^\\bullet(\\U)$ associated $ \\U$\n% \t\\item Find the cohomology of the complex $H^*(\\L^\\bullet(\\U))$ which then equals the singular cohomology of the space $H^*(X, \\F)$.\n% \\end{enumerate}\n%\n% \\begin{remark}\n% \tWhen $X$ is a nice enough space (which includes most spaces you can think of) the Cech cohomology $\\check H^i(X; \\F)$ equals another purely topological invariant of $X$ called the \\textbf{singular homology}. And thus Cech cohomology provides a purely algebraic way of computing a topological invariant.\n% \\end{remark}\n", "meta": {"hexsha": "f8eae96ed04fe42c491405f0db321b733c74caa1", "size": 2848, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "00.tex", "max_stars_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_stars_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "00.tex", "max_issues_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_issues_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "00.tex", "max_forks_repo_name": "apurvnakade/mc2018-cohomology-via-sheaves", "max_forks_repo_head_hexsha": "46b9ad5b473f98d3f9bb41449fe691b6b14fd7f4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 58.1224489796, "max_line_length": 437, "alphanum_fraction": 0.7243679775, "num_tokens": 857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7549149978955811, "lm_q2_score": 0.8128673133042217, "lm_q1q2_score": 0.6136457261124432}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{geometry}\n\\usepackage[utf8]{inputenc}\n\\usepackage{enumitem}\n\\usepackage{physics}\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{1em}\n\\relpenalty=10000\n\\binoppenalty=10000\n\n\\begin{document}\n\\section*{Geometric inequalities}\n\t\\begin{enumerate}\n\t\t\\item %vorrat 48\n\t\tProve the inequalities for $x \\in \\left(\\frac{-\\pi}{2},\\frac{\\pi}{2}\\right)$\n\t\t$$|\\tan x| \\geq |x| \\geq |\\sin x| \\geq \\frac{2}{\\pi}|x|$$\n\t\t\\item %vorrat 49\n\t\tProve the inequalities for $0 < \\alpha < \\beta < \\frac{\\pi}{2}$\n\t\t$$\\frac{\\sin \\beta}{\\sin \\alpha} < \\frac{\\beta}{\\alpha} < \\frac{\\tan \\alpha}{\\tan \\beta}$$\n\t\t\\item % vorrat 52\n\t\tProve the inequalities where $a,b,c$ are lengths of triangles sides\n\t\t\\begin{enumerate}\n\t\t\t\\item $a^2 + b^2 + c^2 < 2 (ab + bc + ca)$\n\t\t\t\\item $a^4 + b^4 + c^4 < 2 (a^2b^2 + b^2c^2 + c^2a^2)$\n\t\t\\end{enumerate} \n\t\t\\item % vorrat 53\n\t\tProve that $R \\geq 2r$ given that $R$ is the radius of circumcircle and $r$ is the radius of incircle of a triangle.\t\n\t\t\\item \\emph{(Ptolemy's inequality).}\n\t\tProve the inequality for any four points $A,B,C,D$ on a plane\n\t\t$$AC \\cdot BD \\leq AB \\cdot CD + AD \\cdot BC$$\t\n\t\tWhen does the equality hold?\n\t\t\\item \\emph{(Erdős-Mordell inequality).} % vorrat 62\n\t\tLet $P$ be a point in the interior of $\\triangle ABC$ and $X$, $Y$,$Z$ projections of $P$ to $BC$,$AC$,$AB$ respectively. Prove that\n\t\t$$PA+PB+PC \\geq 2(PX+PY+PZ)$$\n\t\tWhen does the equality hold?\n\t\t\\item %vorrat 63\n\t\tPoints $A,B,C,X,Y,Z,P$ are defined as before. Prove that\n\t\t$$AP \\cdot BC + BP \\cdot AC + CP \\cdot AB \\geq 4S_{\\triangle ABC} $$\n\t\t\\item %vorrat 64\n\t\tPoints $A,B,C,X,Y,Z,P$ are defined as before. Prove that\n\t\t$$AP \\cdot XP + BP \\cdot YP + CP \\cdot ZP \\geq 2(XP \\cdot YP + YP \\cdot ZP + ZP \\cdot XP) $$\n\t\t\\item %vorrat 65, imo 96\n\t\tGiven a convex hexagon $ABCDEF$ such that $AB \\parallel DE$, $BC \\parallel EF$, $CD \\parallel FA$ and with perimeter $P$. Let $R_A, R_C, R_E$ be the radii of circles $FAB,BCD,DEF$ respectively. Prove that\n\t\t$$R_A + R_C + R_C \\geq \\frac{P}{2} $$\n\t\t\n\t\\end{enumerate}\n\\end{document}", "meta": {"hexsha": "bd571c80fa87faf4e2298ee681a986a665541111", "size": 2101, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "15_geometric_inequalities.tex", "max_stars_repo_name": "ZhaoWanLong/maths-olympiad", "max_stars_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-21T21:57:43.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-21T21:57:43.000Z", "max_issues_repo_path": "15_geometric_inequalities.tex", "max_issues_repo_name": "ZhaoWanLong/maths-olympiad", "max_issues_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "15_geometric_inequalities.tex", "max_forks_repo_name": "ZhaoWanLong/maths-olympiad", "max_forks_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-08T07:04:43.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-08T07:04:43.000Z", "avg_line_length": 42.8775510204, "max_line_length": 206, "alphanum_fraction": 0.6496906235, "num_tokens": 847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042767, "lm_q2_score": 0.7549149813536518, "lm_q1q2_score": 0.6136457229329347}} {"text": "\\chapter{Geostrophic balance and coastal flow}\n\\label{chap:geostrophic}\n\n\\includegraphics[width=6.5in]{figs/Geostrophic/HaidaEddy}\n\n\\section{Geostrophic balance}\n\nThe second balance that involves the Coriolis force is the \\Wikiref{Geostrophic balance}, and is when the Coriolis force is in ``balance'' with the pressure gradient:\n\n\\begin{equation}\n \\frac{du}{dt} = \\overunderbraces{\\br{2}{geostrophic}}\n {- \\frac{1}{\\rho_0}\\frac{dP}{dx} & +fv & + \\frac{1}{\\rho}\\frac{d\\tau_{xz}}{dz}}\n {&\\br{2}{Ekman}}\n\\end{equation}\n\\begin{equation}\n \\frac{dv}{dt} = \\overunderbraces{\\br{2}{geostrophic}}\n {- \\frac{1}{\\rho_0}\\frac{dP}{dy} & -fu & + \\frac{1}{\\rho}\\frac{d\\tau_{yz}}{dz}}\n {&\\br{2}{Ekman}}\n\\end{equation}\n\\begin{marginfigure}\n \\includegraphics{figs/Geostrophic/BalanceSketch}\n \\caption{Sketch of geostrophic force-balance for a sea-level high and a sea-level low in the northern hemisphere, looking from above, and from the south for the high. The flow will be clockwise around a hight and counter-clockwise.}\n \\label{fig:BalanceSketch} \n\\end{marginfigure}\nin which case the equations of motion simplify to:\n\\begin{eqnarray*}\n -\\frac{1}{\\rho_0}\\frac{dP}{dx} & \\approx & -fv\\\\\n -\\frac{1}{\\rho_0}\\frac{dP}{dy} & \\approx & +fu\\\\ \n\\end{eqnarray*}\nIn the northern hemisphere ($f>0$) this says that water turns to the right as in moves from high pressure to low. This gives a clockwise circulation around an isolated high, or a counter-clockwise around an isolated low (\\fref{fig:BalanceSketch})\n\nCrucially, if we know the pressure gradient, we can directly calculate the geostrophic flow perpendicular to that pressure gradient from the equations above. So, suppose the high in the cover figure is 16 cm, over 100 km. The pressure difference due to this flow gives us an estimate of the velocity (assuming $f = 10^{-4}\\ \\mathrm{rad\\,s^{-1}}$)\n\\begin{equation}\n |u| = \\left|\\frac{1}{f\\rho} \\frac{dP}{dx}\\right| = \\frac{1}{f\\rho}\\frac{\\Delta P}{\\Delta x} = \\frac{g}{f} \\frac{\\eta_C - \\eta_E}{\\Delta x} \\approx 0.16 \\ \\mathrm{m\\,s^{-1}}\n\\end{equation}\n\nThe equation applies everywhere in the fluid, and is only made complex by the presence of water with different densities making the flow have a pressure gradient that changes with depth. An observation of this can be seen in \\fref{fig:InteriorEddy}, where there is a sea-surface high in the middle, but the pressure gradient inside the eddy is the other direction. This causes the pressure gradient force to decrease with depth, and hence the geostrophic flow decays with depth as well. \n\n\\begin{figure}\n\\begin{center}\n \\includegraphics[width=2in]{figs/Geostrophic/InteriorEddy}\n \\caption{Observations of density inside an Eddy. Note how the density contours tend to counteract the high sea-surface height. Blue circles with crosses are into the page, and with dots are out of the page, and the size is sketched to be proportional to the speed of the flow.}\n \\label{fig:InteriorEddy} \n\\end{center}\n\\end{figure}\n\nAs discussed in \\fref{chap:EquationofState}, calculating the pressure gradient is simply a matter of calculating the pressure at two locations, and differencing, where the pressure is given by:\n\\begin{equation}\n P(z) / \\rho_0 = \\frac{1}{\\rho_0}\\int_z^\\eta \\rho g\\ \\mathrm{d}z \\approx g\\eta + \\frac{1}{\\rho_0}\\int_z^0 \\rho g\\ \\mathrm{d}z\n\\end{equation}\nSo imagine one flank of the eddy, and that the eddy is made of two densities $\\rho_1 < \\rho_2$ (\\fref{fig:CounterPressure}). In the upper layer \n\\begin{equation}\n -\\frac{1}{\\rho_0} \\frac{dP}{dx}(z=z_1) = -\\frac{1}{\\Delta x} g (\\eta_B - \\eta_A).\n\\end{equation}\nIn the lower layer, things are a bit more complicated, so we just do it step by step:\n\\begin{equation}\n P_A(z=z_2) = \\rho_1 g (\\eta_A - Z_A) + \\rho_2 g (Z_A - z_2)\n\\end{equation}\n\\begin{equation}\n P_B(z=z_2) = \\rho_1 g (\\eta_B - Z_B) + \\rho_2 g (Z_B - z_2)\n\\end{equation}\nSo the total pressure gradient in the lower layer is approximated as \n\\begin{equation}\n -\\frac{1}{\\rho_0}\\frac{P_B - P_A}{\\Delta x} = \\overbrace{-\\frac{1}{\\Delta x} g (\\eta_B - \\eta_A)}^{Surface} - \\overbrace{\\frac{1}{\\Delta x} g \\frac{\\left(\\rho_2 - \\rho_1\\right)}{\\rho_0} (Z_B - Z_A)}^{Internal}\n \\label{eq:InternalExternal}\n\\end{equation}\n\nAccording to the geostrophic balance, then, the flow in the upper layer is into the page, and the one in the bottom layer is either into the page, but more slow, or even out of the page. \n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics{figs/Geostrophic/CounterPressure}\n \\caption{Two-layer approximation of pressure gradients in an eddy.}\n \\label{fig:CounterPressure} \n \\end{center}\n\\end{figure}\n\n\\subsection{Thermal wind}\n\nThe example above makes it clear that lateral density differences can drive velocity differences in the vertical if the flow is in geostrophic balance. This can easily be extended to the continuous case with some manipulation of the momentum equations. If we take the vertical derivative of the x- and y-momentum equations we arrive at:\n\\begin{eqnarray}\n -\\frac{1}{\\rho_0}\\frac{d^2 P}{dx dz} & = & -f \\frac{dv}{dz} \\\\\n -\\frac{1}{\\rho_0}\\frac{d^2 P}{dy dz} & = & f \\frac{du}{dz} \n\\end{eqnarray}\nbut $P = \\int_z^{\\eta}\\rho g \\mathrm{d}z$, so we get the \\Wikiref{thermal wind} relation.\n\\begin{eqnarray}\n -\\frac{1}{\\rho_0}\\frac{d \\rho}{dx} & = & -f \\frac{dv}{dz} \\\\\n -\\frac{1}{\\rho_0}\\frac{d \\rho}{dy} & = & f \\frac{du}{dz} \n\\end{eqnarray}\nso, if we know the horizontal pressure gradient, we can immediately say how the velocity changes with depth (though we do not know the absolute velocity). \n\n\\subsection{Level of no motion and ``dynamic height''}\n\nBack in the pre-satellite days, all oceanographers would have would be measurements of the density under the ocean, like those in \\fref{fig:InteriorEddy}. Estimating the sea-surface height was impossible; imagine detecting 15 cm over 100 km in an ocean that is otherwise also moving due to tides, surface waves, relative to a poorly-constrained geoid. However, oceanographers noted that most ocean currents, like the eddy, had tilted density surfaces, and that the currents at depth go to approximately zero. \n\nUnder these assumptions, oceanographers would assume that the pressure gradient at a ``level of no motion'' was zero, so that the geostrophic flow was zero there as well. In order to get this level of no motion the pressure gradients inferred from the interior density measurements were assumed to be balanced by a pressure gradient from the surface at that depth. So, for instance, in the simple two-layer case the \\fref{eq:InternalExternal}, the \\emph{Surface} term was equal and opposite in sign to the \\emph{Internal} term, if the deep layer was assumed to be not moving. Note that this allows an estimate of the surface pressure gradient, and using the geostrophic balance, the surface velocities. This surface pressure gradient is of course supplied by the sealevel interface being tilted, and calculating these gradients is like calculating a ``dynamic height'' of the ocean. \n\nAn example in \\fref{fig:DynamicHeight}, calculated in the 1980s looks strikingly similar to the sea surface heights in \\fref{chap:winddriven}. In particular, where there is Ekman convergence in the subtropics there is a high of up to 2 m, and a low in the sub-polar regions where there is Ekman divergence. The highs and lows are displaced to the west sides of the basins. The steepest change in dynamic height correspond to the strongest currents, again largely concentrated on the west side of the basins, and associated with the Gulf Stream, Kurishio, East Australian current etc. \n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics{figs/Geostrophic/DynamicHeight1500}\n \\includegraphics{figs/Geostrophic/DynamicHeight2500}\n \\caption{Dynamic height with a level of no motion at 1500 m (top) and 2500 m (bottom). Note these are quite similar, strongly indicative that the dynamics are largely above these depths. }\n \\label{fig:DynamicHeight} \n \\end{center}\n\\end{figure}\n\nJust to summarize again how this is done:\n\\begin{enumerate}\n \\item An interior pressure gradient is calculated at two ``stations'' as a function of depth using the density data from CTDs or other measurements.\n \\item At a ``depth of no motion'' the pressure gradient is assumed to be zero, so a surface pressure gradient is added to the two points to make that be the case.\n \\item If there are multiple stations in a line the surface pressure gradient is integrated to make a dynamic height with an arbitrary zero at some location. \n\\end{enumerate}\n \nWe will come back to \\emph{why} the dynamic height looks the way it does in the next chapter.\n\n\\section{Coastal upwelling, complete picture}\n\nUsing the Ekman and geostrophic balances we are now in a position to explain the coastal upwelling example in detail (i.e. \\fref{fig:PerlinFig3a}, one snapshot in detail in \\fref{fig:ShelfDetail}). \n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics{figs/Geostrophic/ShelfDetail}\n \\caption{Data from \\fref{fig:PerlinFig3a}.}\n \\label{fig:ShelfDetail} \n \\end{center}\n\\end{figure}\n\nFirst, the wind starts to blow towards the south; this sets up a southwards windstress $\\tau_y^w$ (\\fref{fig:CoastalStep1}, red circle). When this flow is in a steady-state Ekman balance there will be an equal and opposite Coriolis force to the north (\\fref{fig:CoastalStep1}, magenta circle). That implies an offshore Ekman transport to the west given by \n\\begin{equation}\n u_{Ek}D = \\frac{\\tau_y^w}{\\rho_0 f}\n\\end{equation}\n\n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics[width=2.85in]{figs/Geostrophic/CoastalStep1}\n \\caption{On a continental slope, initially at rest, a wind stress blows from north to south (dotted red circle)}\n \\label{fig:CoastalStep1} \n \\end{center}\n\\end{figure}\n\nThe offshore Ekman transport drops the sealevel at the coast (\\fref{fig:CoastalStep2}, surface interface). This creates a surface pressure gradient force that acts to push water back towards the coast (\\fref{fig:CoastalStep2}, red arrow). Again, in \\emph{steady state} geostrophic balance, the pressure gradient force is balanced by a Coriolis force offshore, which must be generated by a geostrophic velocity to the south. Note that this evolves in time with the deep water \\emph{initially} moving onshore, but then being turned south (to the right) by the Coriolis force. \n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics[width=2.85in]{figs/Geostrophic/CoastalStep2}\n \\caption{After the Ekman transport has been active for a while, the sea level drops at the coast (green surface), and an interior geostrophic balanced flow is set up.}\n \\label{fig:CoastalStep2} \n \\end{center}\n\\end{figure}\n\nAt this point the flow cannot be in steady state because we are transporting water offshore in the surface Ekman layer, but not having any onshore transport to replace it. Without this return flow, the sea surface will continue to increase its tilt. The return flow happens in the bottom Ekman layer. The interior geostrophic flow moves south, so there is a corresponding bottom stress to the north $\\tau_y^B$(\\fref{fig:CoastalStep3}, red circle at bottom). In a steady Ekman balance, this implies a Coriolis force to the south, which implies a net Ekman transport onshore (\\fref{fig:CoastalStep3}, blue arrows, bottom), given by \n\\begin{equation}\n u_{Ek}D = \\frac{\\tau_y^B}{\\rho_0 f}\n\\end{equation}\n\nThe offshore Ekman transport drops the sealevel at the coast (\\fref{fig:CoastalStep3})\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics[width=2.85in]{figs/Geostrophic/CoastalStep3}\n \\caption{}\n \\label{fig:CoastalStep3} \n \\end{center}\n\\end{figure}\n\nWhen is the whole flow balanced? When $\\tau^B = \\tau^W$ so that the transport in the bottom Ekman layer is the same as the transport in the surface Ekman layer. So, if we know the wind stress, and have a good estimate of the bottom \\Wikiref{drag co-efficient}, $C_D\\approx10^{-3}$, then we can guess what the mean geostrophic velocity is, and hence the size of the sea surface tilt. For the Oregon shelf, $\\tau_y^W\\approx -0.15\\ \\mathrm{N\\,m^{-2}}$ (minus means to the south). Than means that the geostrophic velocity $v_g \\approx 0.35\\ \\mathrm{m\\,s^{-1}}$ in order for $\\rho_0 C_D |v_g|^2$ to be large enough to provide the bottom stress to match. This is approximately the same as the observations near the bottom. \n\nWe can further calculate the sea-surface height drop across the 20-km width of the shelf from $g \\frac{d\\eta}{dx} = f v_g$ or $\\Delta \\eta = 0.08\\ \\mathrm{m}$ across the shelf. \n\n\\subsection{Internal dynamics on shelf}\n\nThe final thing we notice about the shelf observations is that the flow is slower towards the sea-floor than at the surface. This is also largely explained by the geostrophic balance. Note that the isopycnals tilt up towards the coast, counteracting the surface pressure gradient (also sketched in \\fref{fig:CoastalTilt}). This means that the surface tilt is actually larger than we calculated above if the bottom velocity is to remain $0.38\\ \\mathrm{m\\,s^{-1}}$, leading to faster velocities near the surface. \n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics[width=2.85in]{figs/Geostrophic/CoastalTilt}\n \\caption{Sketch of pressure gradient changes if there are titled isopycnals (cyan line) in the section, as observed in the coastal data. The tilted isopycnal counteracts the surface tilt, slowing the barotropic flow. }\n \\label{fig:CoastalTilt} \n \\end{center}\n\\end{figure}\n\n\\clearpage\n\\section{Exercises}\n\n\\paragraph{Geostrophic balance two stations}\n\nCalculate the geostrophic flow at two stations 50 km apart, assuming $f=10^{-4}\\ \\mathrm{rad\\,s^{-1}}$. \nApproximate the density measured at each location as 50-m blocks of homogeneous density:\n\n\\begin{tabular}{l|cc}\n $z\\ \\mathrm{[m]}$ & $\\rho_A \\mathrm{[kg\\,m^{-3}]}$ & $\\rho_B \\mathrm{[kg\\,m^{-3}]}$ \\\\\n \\hline\n 0--50 & 1021& 1020 \\\\\n 50--100& 1022& 1021\\\\\n 100--150& 1024& 1023\\\\\n 150--200& 1025& 1024\\\\ \n\\end{tabular}\n\n\\begin{itemize}\n \\item If the surface pressure gradient is zero, what is the internal pressure gradient at 50, 100, 150, and 200 m? (do not drop too many decimal places!)\n \\item What is the direction and strength of the \\emph{geostrophic} velocity at the 4 depths?\n \\item If there is a pressure gradient, what must its direction and strength be so that the geostrophic flow is zero at 200 m?\n \\item How large must the surface difference be at the two stations? \n \\item What is the new flow at the other depths in the presence of this surface gradient?\n\\end{itemize}\n\n\\paragraph{Geostrophic flow Gulf Stream}\n\nConsider the temperature section across the Gulf Stream (\\fref{fig:GulfStreamExample}). Assuming a level of no motion at approximately 1500 m, what does \n\\begin{itemize}\n \\item the sea-surface tilt look like?\n \\item the interior velocity look like (just a sketch)?\n \\item where would you expect the Gulf Stream to be the fastest?\n\\end{itemize}\n\n\\begin{figure}[hbt]\n \\begin{center}\n \\includegraphics[width=1.8in]{figs/Geostrophic/GulfStreamLoc}\n \\includegraphics[width=2.7in]{figs/Geostrophic/GulfStreamTemp}\n \\caption{Location of measurements in the Gulf Stream, and plot of temperature across the section \\citep{johnsetal95}.}\n \\label{fig:GulfStreamExample} \n \\end{center}\n\\end{figure}\n\n", "meta": {"hexsha": "3ab25ece79b6f87ca1eff7bf7fceb104d9132b52", "size": 15445, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chap10Geostrophic.tex", "max_stars_repo_name": "jklymak/Eos314Text", "max_stars_repo_head_hexsha": "5f58722b9ec9d44647425ba0461069ef2f218767", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-10-11T23:01:49.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-20T16:31:29.000Z", "max_issues_repo_path": "Chap10Geostrophic.tex", "max_issues_repo_name": "jklymak/Eos314Text", "max_issues_repo_head_hexsha": "5f58722b9ec9d44647425ba0461069ef2f218767", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Chap10Geostrophic.tex", "max_forks_repo_name": "jklymak/Eos314Text", "max_forks_repo_head_hexsha": "5f58722b9ec9d44647425ba0461069ef2f218767", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-12-06T12:30:43.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-20T16:31:30.000Z", "avg_line_length": 68.0396475771, "max_line_length": 890, "alphanum_fraction": 0.7416639689, "num_tokens": 4447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673178375734, "lm_q2_score": 0.7549149758396752, "lm_q1q2_score": 0.6136457116062133}} {"text": "\n\\subsection{Indirect utility functions}\n\nThe normal utility function is:\n\n\\(U=f(\\mathbf x)\\)\n\nWe have our demand:\n\n\\(x_{di}=x_{di}(I, \\mathbf p)\\)\n\n\\subsection{The indirect utility function}\n\nWe can plut this in to get:\n\n\\(U=g(I, \\mathbf p)\\)\n\n", "meta": {"hexsha": "3f52ece8c663bd4ac4ca35db1c94b2f9a32a5d9f", "size": 245, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/economics/consumer/05-01-indirect.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/economics/consumer/05-01-indirect.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/economics/consumer/05-01-indirect.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 13.6111111111, "max_line_length": 42, "alphanum_fraction": 0.6775510204, "num_tokens": 73, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.86153820232079, "lm_q2_score": 0.7122321781307374, "lm_q1q2_score": 0.6136152303817761}} {"text": "\\section{Glossary}\n\\smallskip \\hrule height 2pt \\smallskip\n \n \\begin{itemize}\n \t\\item \\textbf{complementary CDF} (CCDF, or survival function) - can ask how often the random variable is above a particular level.\n\t\tHas applications in statistical hypothesis testing, for example, because the one-sided p-value \n\t\t\tis the probability of observing a test statistic at least as extreme as the one observed.\n \t\\item \\textbf{confidence interval} - An interval that represents the expected range of an estimator if \n\t\tan experiment is repeated many times.\n\t\\item \\textbf{effect size} - any statistic that quantitatively measures the strength of a phenomenon.\n\t\\item \\textbf{jitter} - random noise added to data for purposes of visualization. \n \t\\item \\textbf{oversampling} - The technique of increasing the representation of a sub-population in order to avoid errors due to small sample sizes. \n\t\\item \\textbf{p-value} - the \\underline{\\textbf{p}}robability that, using a given statistical model, the statistical summary \n\t\t(such as the sample mean difference between two compared groups) would be the same as or more extreme \n\t\tthan the actual observed results\n\t\\item \\textbf{percentile rank} - The percentage of values in a distribution that are less than or equal to a given value.\n\t\\item \\textbf{quantizing} - The opposite of smoothing is discretizing, or quantizing.\n\t\\item \\textbf{rank} The index where an element appears in a sorted list.\n\t\\item \\textbf{raw moment} - A statistic based on the sum of data raised to a power.\r\n\t\tE.g. the correlation between two variables, the regression coefficient in a regression, the mean difference, ...\n\t\\item \\textbf{standard error}: - The RMSE of an estimate, which quantifies variability due to sampling error \n\t\t(but not other sources of error).\r\t\\item \\textbf{standard score} A value that has been standardized so that it is expressed in standard deviations from the mean.\n\n \\end{itemize}\n", "meta": {"hexsha": "0995eb53b7b492540392e1841c4d397491972653", "size": 1932, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/glossary.tex", "max_stars_repo_name": "JanetMatsen/Statistics_for_Data_Science", "max_stars_repo_head_hexsha": "b1f626226dfdd444b508203d21c0f66a343c7097", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/glossary.tex", "max_issues_repo_name": "JanetMatsen/Statistics_for_Data_Science", "max_issues_repo_head_hexsha": "b1f626226dfdd444b508203d21c0f66a343c7097", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/glossary.tex", "max_forks_repo_name": "JanetMatsen/Statistics_for_Data_Science", "max_forks_repo_head_hexsha": "b1f626226dfdd444b508203d21c0f66a343c7097", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 77.28, "max_line_length": 163, "alphanum_fraction": 0.7727743271, "num_tokens": 467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006919925839875, "lm_q2_score": 0.766293653760418, "lm_q1q2_score": 0.6135651925338933}} {"text": "% Created 2020-07-14 mar 10:56\n% Intended LaTeX compiler: pdflatex\n\\documentclass[presentation,aspectratio=169]{beamer}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{graphicx}\n\\usepackage{grffile}\n\\usepackage{longtable}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{amssymb}\n\\usepackage{capt-of}\n\\usepackage{hyperref}\n\\usepackage{khpreamble}\n\\usepackage{amssymb}\n\\DeclareMathOperator{\\shift}{q}\n\\DeclareMathOperator{\\diff}{p}\n\\usetheme{default}\n\\author{Kjartan Halvorsen}\n\\date{\\today}\n\\title{Control computarizado - Asignación de polos}\n\\hypersetup{\n pdfauthor={Kjartan Halvorsen},\n pdftitle={Control computarizado - Asignación de polos},\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 26.3 (Org mode 9.3.6)}, \n pdflang={English}}\n\\begin{document}\n\n\\maketitle\n\n\\section{Intro}\n\\label{sec:org70d5d08}\n\n\n\\section{2-dof controller}\n\\label{sec:org70b8192}\n\n\\begin{frame}[label={sec:orga124e5e}]{Two-degree-of-freedom controller}\n\\begin{center}\n\\includegraphics[width=0.8\\linewidth]{../../figures/2dof-block-explicit-no-delay}\n\\end{center}\n\\end{frame}\n\n\\section{Problem 5.3}\n\\label{sec:orga34f588}\n\\begin{frame}[label={sec:org4d6f90f}]{Åström \\& Wittenmark problem 5.3}\nConsider the system given by the pulse-transfer function\n\\[ H(z) = \\frac{z+0.7}{z^2 -1.8z + 0.81} \\]\nUse polynomial design (RST) to determine a controller such that the closed-loop system from command input to output has the characteristic polynomial\n\\[ A_c(z) = z^2 - 1.5z + 0.7. \\]\nLet the observer polynomial have as low order as possible, and place all observer poles in the origin (deadbeat observer). Consider three cases\n\\begin{description}\n\\item[{(a)}] Positional control with cancellation of the process zero\n\\item[{(b)}] Positional control with no cancellation of the zero\n\\item[{(c)}] Incremental controller with no cancellation of the zero\n\\end{description}\n\\end{frame}\n\n\\begin{frame}[label={sec:org368a5ad}]{Why cancel the process zero?}\nBode plots of closed-loop systems (from reference signal to output) with and without cancellation of the process zero:\n\n\\begin{center}\n\\includegraphics[width=0.6\\linewidth]{../../figures/aw5_3_bode}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}[label={sec:org91db4f3}]{Preliminary exercise}\nWhich of the closed-loop responses below corresponds to (I) Positional control with zero cancellation (II), Positional control without zero cancellation, (III) Incremental control without zero cancellation.\n\\begin{center}\n\\includegraphics[width=0.45\\linewidth]{../../figures/aw5_3_refstep}\n\\includegraphics[width=0.45\\linewidth]{../../figures/aw5_3_diststep}\n\\end{center}\n\\end{frame}\n\\end{document}", "meta": {"hexsha": "6a0ffd31b0f4610aae468e33f62bddbd4820cb34", "size": 2706, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "polynomial-design/slides/lecture-problem-solving-RST-sp.tex", "max_stars_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_stars_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-11-07T05:20:37.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-22T09:46:13.000Z", "max_issues_repo_path": "polynomial-design/slides/lecture-problem-solving-RST-sp.tex", "max_issues_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_issues_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-06-12T20:44:41.000Z", "max_issues_repo_issues_event_max_datetime": "2020-06-12T20:49:00.000Z", "max_forks_repo_path": "polynomial-design/slides/lecture-problem-solving-RST-sp.tex", "max_forks_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_forks_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-14T03:55:27.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-14T03:55:27.000Z", "avg_line_length": 34.253164557, "max_line_length": 207, "alphanum_fraction": 0.766075388, "num_tokens": 835, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936324115011, "lm_q2_score": 0.8006920116079209, "lm_q1q2_score": 0.6135651900179054}} {"text": "% !TEX root = main.tex\n%----------------------------------------------------------------------\n\n%----------------------------------------------------------------------\n\\chapter{The Bivariate Normal Distribution}\n\n%The normal distribution is based on the Gaussian integral\n%\\[\n%\\int_{-\\infty}^{\\infty} e^{-x^2} = \\sqrt{\\pi},\n%\\]\n\n%----------------------------------------------------------------------\n\\section{Bivariate transformations}\n\n% definition\n\\begin{definition}\\label{def:jacobian}\nLet $h:\\R^2\\to\\R^2$ and let $(u,v) = h(x,y)$. The \\emph{Jacobian determinant} of the transformation $h$ is the determinant of its $2\\times 2$ matrix of partial derivatives:\n\\[\nJ = \\begin{vmatrix}\n\\displaystyle\\frac{\\partial u}{\\partial x} & \\displaystyle\\frac{\\partial u}{\\partial y} \\\\[2ex]\n\\displaystyle\\frac{\\partial v}{\\partial x} & \\displaystyle\\frac{\\partial v}{\\partial y} \\\\[2ex]\n\\end{vmatrix}\n\\]\n\\end{definition}\n\n% theorem: continuous\n\\begin{theorem}\\label{thm:bivariate_transformation_pdf}\nLet $U$ and $V$ be jointly continuous random variables, let $f_{U,V}$ be their joint PDF, let $g:\\R^2\\to\\R^2$ be an injective transform over the support of $f_{U,V}$ and let $(X,Y) = g(U,V)$. Then the joint PDF of $X$ and $Y$ is given by\n\\[\nf_{X,Y}(x,y) = |J| f_{U,V}\\big[ g^{-1}(x,y) \\big]\n\\] \nwhere $J$ is the Jacobian determinant of the transformation $g^{-1}$,\n\\[\nJ = \n\\begin{vmatrix}\n\\displaystyle\\frac{\\partial u}{\\partial x} & \\displaystyle\\frac{\\partial u}{\\partial y} \\\\[2ex]\n\\displaystyle\\frac{\\partial v}{\\partial x} & \\displaystyle\\frac{\\partial v}{\\partial y} \\\\[2ex]\n\\end{vmatrix}\n\\]\nwhere $(u,v) = g^{-1}(x,y)$.\n\\end{theorem}\n\n% remark\n\\begin{remark}\nThe absolute value $|J|$ is a scale factor, which ensures that the transformed PDF $f_{X,Y}(x,y)$ integrates to one.\n\\end{remark}\n\n% example: jointly continuous\n\\begin{example}\nLet $U$ and $V$ be continuous random variables, and let $X=U+V$ and $Y=U-V$. \n\\ben\n\\it Find the joint PDF of $X$ and $Y$ in terms of the joint PDF of $U$ and $V$.\n\\it If $U,V\\sim\\text{Exponential}(1)$ are independent, find the joint PDF of $X$ and $Y$.\n\\een\n\\end{example}\n\n\\begin{solution}\n\\ben\n\n\\it % <<< (i)\n\\bit\n\\it The transformation $g:\\R^2\\to\\R^2$ is defined by $g(u,v) = (u+v,u-v)$. \n\\it To compute the inverse transformation $g^{-1}:\\R^2\\to\\R^2$, we solve the equations %$x=u+v$ and $y=u-v$ for $u$ and $v$.\n\\[\nx=u+v \\quad\\text{and}\\quad y=u-v.% \\quad \\text{for $u$ and $v$.}\n\\]\n\\it This yields $u = \\frac{1}{2}(x+y)$ and $v = \\frac{1}{2}(x-y)$.\n\\it Thus the inverse transformation is \n\\[\n(u,v) = g^{-1}(x,y) = \\left[\\frac{1}{2}(x+y),\\frac{1}{2}(x-y)\\right].\n\\]\n\\eit\nThe Jacobian determinant is given by\n\\[\nJ = \n\\begin{vmatrix}\n\\displaystyle\\frac{\\partial u}{\\partial x} & \\displaystyle\\frac{\\partial u}{\\partial y} \\\\[2ex]\n\\displaystyle\\frac{\\partial v}{\\partial x} & \\displaystyle\\frac{\\partial v}{\\partial y} \\\\\n\\end{vmatrix}\n=\n\\begin{vmatrix}\n1/2\t& 1/2 \\\\\n1/2 & -1/2 \\\\\n\\end{vmatrix}\n= -\\frac{1}{4} - \\frac{1}{4} = -\\frac{1}{2}.\n\\]\nHence the joint PDF of $X$ and $Y$ is\n\\begin{align*}\nf_{X,Y}(x,y)\n\t& = |J|f_{U,V}(u,v) \\\\\n\t& = \\left|-\\frac{1}{2}\\right| f_{U,V}\\left[\\frac{1}{2}(x+y),\\frac{1}{2}(x-y)\\right] \\\\\n\t& = \\frac{1}{2}f_{U,V}\\left[\\frac{1}{2}(x+y),\\frac{1}{2}(x-y)\\right].\n\\end{align*}\n\n\n\n\\it % <<< (ii)\nLet $U$ and $V$ be independent with $U,V\\sim\\text{Exponential}(1)$.\n\nBy independence, the joint PDF of $U$ and $V$ is \n\\[\nf_{U,V}(u,v) \n\t= \\begin{cases}\n\t\te^{-(u+v)}\t\t& u,v > 0 \\\\\n\t\t0\t\t\t\t& \\text{otherwise.}\n\\end{cases}\n\\]\nTo compute the support of $f_{X,Y}$, since $u>0$ and $v>0$ we have $x>0$, so\n\\bit\n\\it $\\min(y) = \\min(u-v) = -x$ (which occurs when $u=0$ and $v=x$), and\n\\it $\\max(y) = \\max(u-v) = x$ (which occurs when $u=x$ and $v=0$).\n\\eit\nThus, substituting for $u+v = \\frac{1}{2}(x+y) + \\frac{1}{2}(x-y) = x$, we obtain\n\\[\nf_{X,Y}(x,y) = \\begin{cases}\n\t\\frac{1}{2}e^{-x}\t& \\text{for } x > 0 \\text{ and } -x < y < x, \\\\\n\t0\t\t\t\t\t& \\text{otherwise.}\n\\end{cases}\n\\]\n\\een\n\\end{solution}\n\n\n%----------------------------------------------------------------------\n\\section{The bivariate normal distribution}\n%----------------------------------------------------------------------\n\n%%----------------------------------------------------------------------\n%\\section{The sum of two random variables}\n%%----------------------------------------------------------------------\n%% theorem\n%\\begin{theorem}\\label{thm:convolution}\n%Let $X$ and $Y$ be jointly continuous random variables, and let $f_{X,Y}$ be their joint PDF. The random variable $X+Y$ has the following PDF:\n%\\[\n%f_{X+Y}(t) \n%\t= \\int_{-\\infty}^{\\infty} f_{X,Y}(x,t-x)\\,dx\n%\t= \\int_{-\\infty}^{\\infty} f_{X,Y}(t-y,y)\\,dy.\n%\\]\n%\\end{theorem}\n%\n%\\begin{proof}\n%Let $A = \\{(x,y):x+y\\leq z\\}\\subset\\R^2$. Then\n%\\[\n%\\prob(X+Y\\leq z)\n%\t= \\iint_A f(x,y)\\,dxdy\n%\t= \\int_{x=-\\infty}^{\\infty} \\int_{y=-\\infty}^{z-x}f_{X,Y}(x,y)\\,dy\\,dx\n%\\]\n%We change the variable of integration (in the inner integral), making the substitution $y=t-x$:\n%\\begin{align*}\n%F_{X+Y}(z) = \\prob(X+Y\\leq z)\n%\t& = \\int_{x=-\\infty}^{\\infty} \\int_{t=-\\infty}^{z}f_{X,Y}(x,t-x)\\,dt\\,dx \\\\\n%\t& = \\int_{t=-\\infty}^{z} \\int_{x=-\\infty}^{\\infty}f_{X,Y}(x,t-x)\\,dx\\,dt\n%\\end{align*}\n%where the final equality follows by reversing the order of integration. Thus the PDF of $X+Y$ is \n%\\[\n%f_{X+Y}(t) = \\int_{-\\infty}^{\\infty} f_{X,Y}(x,t-x)\\,dx \\qquad\\text{as required.}\n%\\]\n%\\end{proof}\n%\n%% corollary\n%\\begin{corollary}\\label{cor:convolution_independent}\n%If $X$ and $Y$ are independent, the PDF of $X+Y$ is the \\emph{convolution} of the marginal PDFs:\n%\\[\n%f_{X+Y}(t) \n%\t= \\int_{-\\infty}^{\\infty} f_X(x)f_Y(t-x)\\,dx\n%\t= \\int_{-\\infty}^{\\infty} f_X(t-y)f_Y(y)\\,dy.\n%\\]\n%\\end{corollary}\n\n%\\begin{remark}\n%The function $f_{X+Y}$ is called the \\emph{convolution} of $X$ and $Y$, and is often written as $f_{X+Y}=f_X\\ast f_Y$.\n%\\end{remark}\n\n%%----------------------------------------------------------------------\n%\\section{The sum of two standard normal variables}\n%%----------------------------------------------------------------------\n%% lemma: sums of normal random variables}\n%\\begin{lemma}\\label{lem:sum_of_standard_normal_variables}\n%If $U,V\\sim N(0,1)$ are independent, then $U+V\\sim N(0,2)$.\n%\\end{lemma}\n%\n%\\begin{proof}\n%Since $U$ and $V$ are independent, their joint PDF is\n%\\[\n%f(u,v)=f_U(u)f_V(v) = \\frac{1}{2\\pi}\\exp\\left(-\\frac{1}{2}(u^2 + v^2)\\right) \\qquad u,v\\in\\R.\n%\\]\n%\n%Let $W=U+V$. Since $U$ and $V$ are independent, by Corollary~\\ref{cor:convolution_independent} we have\n%\\begin{align*}\n%f_W(w) \n%\t& = \\int_{-\\infty}^{\\infty} f_U(u)f_V(w-u)\\,du \\\\\n%\t& = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} \\exp\\left[-\\frac{1}{2}\\left(u^2 + (w-u)^2\\right)\\right]\\,du \\\\\n%\t& = \\frac{1}{2\\pi} e^{-\\frac{1}{4}w^2} \\int_{-\\infty}^{\\infty} \\exp\\left[-\\left(u-\\frac{w}{2}\\right)^2\\right]\\,du\n%\\end{align*}\n%We change the variable of integration, by making the substitution $t = \\displaystyle\\sqrt{2}\\left(u-\\frac{w}{2}\\right)$:\n%\\[\n%f_W(w) \n%\t= \\frac{1}{2\\sqrt{\\pi}} e^{-\\frac{1}{4}w^2} \\int_{-\\infty}^{\\infty} \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}w^2}\\,dv\n%\t= \\frac{1}{2\\sqrt{\\pi}} e^{-\\frac{w^2}{4}},\n%\\]\n%which is the PDF of the $N(0,2)$ distribution\n%\\end{proof}\n%\n%Lemma~\\ref{lem:sum_of_standard_normal_variables} is a special case of the following theorem.\n%% theorem\n\n%%----------------------------------------------------------------------\n%\\section{The standard bivariate normal distributrion} \n%%----------------------------------------------------------------------\n\n\\begin{theorem}\\label{thm:sum_of_normal_variables}\nif $X\\sim N(\\mu_1,\\sigma_1^2)$ and $Y\\sim N(\\mu_2,\\sigma_2^2)$ are independent, then\n\\[\nX+Y\\sim N(\\mu_1+\\mu_2, \\sigma_1^2+\\sigma_2^2).\n\\]\n\\end{theorem}\n\n%The following corollary will be useful later on:\n% corollary\n\\begin{corollary}\\label{cor:lin_comb_std_normal}\nIf $U,V\\sim N(0,1)$ are independent, then $aU+bV\\sim N(0,a^2+b^2)$ for all $a,b\\in\\R$.\n\\end{corollary}\n%If $U$ and $V$ are independent standard normal variables, their joint PDF of is\n%\n%\\[\n%f(u,v)=f_U(u)f_V(v) = \\frac{1}{2\\pi}\\exp\\left(-\\frac{1}{2}(u^2 + v^2)\\right) \\qquad u,v\\in\\R.\n%\\]\n\n% definition: standard bivariate normal\n\\begin{definition}\\label{def:standard_bivariate_normal}\nA pair of random variables $U$ and $V$ have the \\emph{standard bivariate normal distribution} if their joint PDF $f:\\R^2\\to[0,\\infty)$ can be written as\n\\[\nf_{U,V}(u,v) = \\frac{1}{2\\pi\\sqrt{1-\\rho^2}}\\exp\\left(-\\frac{1}{2(1-\\rho^2)}\\big(u^2 - 2\t\\rho uv + v^2\\big)\\right)\n\\]\nwhere $\\rho$ is a constant satisfying $-1 < \\rho < 1$.\n\\end{definition}\n\n% definition: bivariate normal\n\\begin{definition}\\label{def:bivariate_normal}\nA pair of random variables $X$ and $Y$ are said to have \\emph{bivariate normal distribution} with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$ and correlation $\\rho$, if their joint PDF can be written as\n%\\small\n\\[\nf_{X,Y}(x,y)\n\t= \\frac{1}{2\\pi\\sigma_1\\sigma_2\\sqrt{1-\\rho^2}}\n\t\t\\exp\\left(-\\frac{1}{2(1-\\rho^2)}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \n\t\t\t-2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) \n\t\t\t\t+\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2 \\right]\\right)\n\\]\n\\normalsize\n\\end{definition}\n\n% technical result\nThe following lemma can be used to derive many properties of the bivariate normal distribution.\n\\begin{lemma}\\label{lem:trick}\nLet $U,V\\sim N(0,1)$ be independent, let $\\rho\\in(-1,+1)$. Then the random variables\n\\begin{align*}\nX & = \\mu_1 + \\sigma_1 U, \\\\\nY & = \\mu_2 + \\sigma_2\\big(\\rho U +\\sqrt{1-\\rho^2}V\\big)\n\\end{align*}\nhave bivariate normal distribution with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$, and correlation $\\rho$.\n\\end{lemma}\n\n% proof\n\\begin{proof}\nTo find the joint PDF of $X$ and $Y$, let $g(u,v)$ denote the transformation:\n\\[\ng(u,v) = \\big[\\mu_1 + \\sigma_1 u, \\mu_2 + \\sigma_2(\\rho u +\\sqrt{1-\\rho^2}v)\\big].\n\\]\nThe inverse transformation is\n\\[\ng^{-1}(x,y) \n\t= \\left(\\frac{x-\\mu_1}{\\sigma_1}, \\frac{1}{\\sqrt{1-\\rho^2}}\n \\left[\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) -\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\right]\\right)\n\\]\nThe joint PDF of $X$ and $Y$ is $f_{X,Y}(x,y) = |J|f_{U,V}(u,v)$, where $J$ is the Jacobian determinant of the inverse transformation:\n\\[\nJ=\\begin{vmatrix}\n\\displaystyle\\frac{\\partial u}{\\partial x} & \\displaystyle\\frac{\\partial u}{\\partial y} \\\\[2ex]\n\\displaystyle\\frac{\\partial v}{\\partial x} & \\displaystyle\\frac{\\partial v}{\\partial y} \n\\end{vmatrix}\n=\n\\begin{vmatrix}\n\\displaystyle\\frac{1}{\\sigma_1} \t\t& \\displaystyle0 \\\\[2ex]\n\\displaystyle\\frac{1}{\\rho\\sigma_1}\t& \\displaystyle\\frac{1}{\\sigma_2\\sqrt{1-\\rho^2}} \n\\end{vmatrix}\n=\n\\frac{1}{\\sigma_1\\sigma_2\\sqrt{1-\\rho^2}}\n\\]\n\nBecause $U$ and $V$ are independent,\n\\[\nf_{U,V}(u,v)=f_U(u)f_V(v) = \\frac{1}{2\\pi}\\exp\\left(-\\frac{1}{2}(u^2 + v^2)\\right) \\qquad u,v\\in\\R.\n\\]\nand since\n\\begin{align*}\nu^2 +v^2 \n\t& = \\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \n\t\t\t+ \\frac{1}{1-\\rho^2}\\left[ \n\t\t\t\t\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2 \n\t\t\t\t\t- 2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_{2}}\\right) \n\t\t\t\t\t\t+\\rho^2\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \\right] \\\\\n\t& = \\frac{1}{1-\\rho^2}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \n\t\t\t\t-2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) \n\t\t\t\t\t+\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2\\right]\n\\end{align*}\nit follows that\n%\\small\n\\[\nf_{X,Y}(x,y)\n\t= \\frac{1}{2\\pi\\sigma_1\\sigma_1\\sqrt{1-\\rho^2}}\n\t\t\\exp\\left(\\frac{-1}{2(1-\\rho^2)}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \n\t\t\t-2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) \n\t\t\t\t+\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2 \\right]\\right)\n\\]\n\\normalsize\nas required.\n\\end{proof}\n\nThe following theorem shows that if $X$ and $Y$ have bivariate normal distribution, then any linear combination of $X$ and $Y$ is normally distributed.\n% theorem: linear combination\n\\begin{theorem}\\label{thm:lin_comb_bivar_normal}\nLet $X$ and $Y$ have bivariate normal distribution with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$, and correlation $\\rho$. Then\n%If $X\\sim N(\\mu_1,\\sigma_1^2)$ and $Y\\sim N(\\mu_2,\\sigma_2^2)$ then\n\\[\naX + bY \\sim N\\big(a\\mu_1 + b\\mu_2, a^2\\sigma_1^2 + 2ab\\sigma_1\\sigma_2\\rho + b^2\\sigma_2^2\\big)\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $Z=aX+bY$, let $U$ and $V$ be independent standard normal random variables, and let\n\\begin{align*}\nX' & = \\mu_1 + \\sigma_1 U \\\\\nY' & = \\mu_2 + \\sigma_2\\big(\\rho U +\\sqrt{1-\\rho^2}V\\big)\n\\end{align*}\n\nBy Lemma~\\ref{lem:trick}, $X$ and $Y$ have the same joint distribution as $X'$ and $Y'$, so $Z=aX+bY$ has the same distribution as \n\\[\nZ' = aX'+bY' = (a\\mu_1 + b\\mu_2) + (a\\sigma_1 + b\\sigma_2\\rho)U + b\\sigma_2\\sqrt{1-\\rho^2} V\n\\]\n\nBecause $U,V\\sim N(0,1)$ are independent, it follows by Corollary~\\ref{cor:lin_comb_std_normal} that\n\\[\nZ' \\sim N\\left(a\\mu_1 + b\\mu_2, a^2\\sigma_1^2 + 2ab\\sigma_1\\sigma_2\\rho + b^2\\sigma_2^2\\right),\n\\]\nso $Z=aX+bY$ has normal distribution, as required.\n\\end{proof}\n\n%----------------------------------------------------------------------\n\\section{Properties of the bivariate normal distribution} \n%----------------------------------------------------------------------\n\n% theorem: properties\n\\begin{theorem}\nLet $X$ and $Y$ have bivariate normal distribution with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$, and correlation $\\rho$. Then\n\\ben\n\\it $X\\sim N(\\mu_1,\\sigma_1^2)$ and $Y\\sim N(\\mu_2,\\sigma_2^2)$,\n\\it $\\rho$ is the correlation coefficient of $X$ and $Y$, and\n\\it $X$ and $Y$ are independent if and only if $\\rho=0$.\n\\een\n\\end{theorem}\n\n% proof\n\\begin{proof}\n%By Lemma~\\ref{lem:trick}, let us write\nLet $U,V\\sim N(0,1)$ and define\n\\begin{align*}\nX & = \\mu_1 + \\sigma_1 U \\\\\nY & = \\mu_2 + \\sigma_2\\big(\\rho U +\\sqrt{1-\\rho^2}V\\big)\n\\end{align*}\n\n\n\\ben\n\\it % (i)\nIn the proof of Theorem~\\ref{thm:lin_comb_bivar_normal}:\n\\bit \n\\it taking $a=1$ and $b=0$ yields $X\\sim N(\\mu_1,\\sigma_1^2)$, and\n\\it taking $a=0$ and $b=1$ yields $Y\\sim N(\\mu_2,\\sigma_2^2)$.\n\\eit\n\\it % (ii)\nUsing the fact that $\\cov(aX+b,cY+d)=ac\\cov(X,Y)$ for all $a,b,c,d\\in\\R$,\n\\begin{align*}\n\\cov(X,Y)\n\t& = \\cov\\big[\\mu_1 + \\sigma_1 U, \\mu_2 + \\sigma_2(\\rho U + \\sqrt{1-\\rho^2}V)\\big] \\\\\n\t& = \\sigma_1\\sigma_2\\cov(U, \\rho U + \\sqrt{1-\\rho^2}V) \\\\\n\t& = \\sigma_1\\sigma_2\\big[\\rho\\expe(U^2) + \\sqrt{1-\\rho^2}\\expe(UV)\\big] \\\\\n\t& = \\sigma_1\\sigma_2\\rho.\n\\end{align*}\nThus $\\rho = \\displaystyle\\frac{\\cov(X)}{\\sqrt{\\var(X)}\\sqrt{\\var(Y)}}$ as required.\n\n\n\\it % (iii)\nIf $X$ and $Y$ are independent, they are uncorrelated. If $X$ and $Y$ are uncorrelated then $\\rho=0$, so the joint PDF of $X$ and $Y$ satisfies\n\\begin{align*}\nf_{X,Y}(x,y)\n\t& = \\frac{1}{2\\pi\\sigma_1\\sigma_2}\\exp\\left(-\\frac{1}{2}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 + \\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2 \\right]\\right) \\\\\n\t& = \\frac{1}{\\sqrt{2\\pi}\\sigma_1}\\exp\\left(-\\frac{1}{2}\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2\\right)\n\t\t\t\\times \\frac{1}{\\sqrt{2\\pi}\\sigma_2}\\exp\\left(-\\frac{1}{2}\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2\\right) \\\\\n\t& = f_X(x)f_Y(y).\t\t\n\\end{align*}\nBecause this holds for all $x,y\\in\\R$, it follows that $X$ and $Y$ are independent. \n\\een\n\\end{proof}\n\n%----------------------------------------------------------------------\n\\section{Conditional distributions} \n%----------------------------------------------------------------------\n\n% theorem\n\\begin{theorem}\nLet $X$ and $Y$ have bivariate normal distribution with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$, and correlation $\\rho$.\nThen the conditional distribution of $Y$ given $X=x$ is also normal, with conditional mean and variance given by\n\\begin{align*}\n\\expe(Y|X=x)\t& = \\mu_2 + \\rho\\left(\\frac{\\sigma_2}{\\sigma_1}\\right)(x - \\mu_1), \\\\[2ex]\n\\var(Y|X=x) \t& = \\sigma_2^2(1-\\rho^2),\n\\end{align*}\nand the conditional mean and variance of $Y$ given $X$ is \n\\begin{align*}\n\\expe(Y|X)\t& = \\expe(Y) + \\frac{\\cov(X,Y)}{\\var(X)}\\big[X - \\expe(X)\\big], \\\\[2ex]\n\\var(Y|X) \t& = \\var(Y)(1-\\rho^2).\n\\end{align*}\n\\end{theorem}\n\n% proof\n\\begin{proof}\nLet $U,V\\sim N(0,1)$ be independent, and define the random variables\n\\begin{align*}\nX & = \\mu_1 + \\sigma_1 U, \\\\\nY & = \\mu_2 + \\sigma_2\\big[\\rho U +\\sqrt{1-\\rho^2}V\\big] \\\\\n\t& = \\mu_2 + \\sigma_2\\left[\\rho \\left(\\frac{X-\\mu_1}{\\sigma_1}\\right) +\\sqrt{1-\\rho^2}V\\right].\n\\end{align*}\nIf $X$ is fixed at $x$, then $Y$ is a linear transformation of $V$, so the conditional distribution of $Y$ given that $X=x$ is a normal distribution. \nFurthermore, since $\\expe(V)=0$ and $\\var(V)=1$ we have\n\\begin{align*}\n\\expe(Y|X=x) \t& = \\mu_2 + \\rho\\left(\\frac{\\sigma_2}{\\sigma_1}\\right)(x - \\mu_1) \\\\[2ex]\n\\var(Y|X=x)\t& = \\sigma_2^2(1-\\rho^2)\n\\end{align*}\nas required.\n\\end{proof}\n\n% remarks\n\\begin{remark}\n\\bit\n\\it\nGiven $X=x$, the conditional mean is obtained by adjusting $\\expe(Y)$ by an amount proportional to the difference between $X=x$ and its mean $\\expe(X)$. The size of this adjustment is determined by (1) the size of $Y$ relative to $X$, expressed by the ratio $\\sigma_2/\\sigma_1$, and (2) the correlation coefficient $\\rho$, which quantifies the linear dependence between $X$ and $Y$. Note that $\\rho=0$ implies that $\\expe(Y|X=x)=0$ for all $x$ (which is not surprising, given that uncorrelated normal variables are independent).\n\\it \nThe conditional variance $\\var(Y|X=x)$ quantifies the variability in $Y$ that is \\emph{not} explained by the fact that $X$ takes the value $x$. The squared correlation coefficient $\\rho^2$ thus quantifies the proportion of the overall variance $\\var(Y)$ accounted for by the fact that $X=x$. \n\\it\nThis idea of `explained' and `unexplained' variance is apparent in the law of total variance:\n\\[\n\\var(Y) = \\expe\\big[\\var(Y|X)\\big] + \\var\\big[\\expe(Y|X)\\big] \n\\]\n\t\\bit\n\t\\it $\\var(Y)$ is the total variance,\n\t\\it $\\expe\\big[\\var(Y|X)\\big]$ is the variance explained by $X$, and \n\t\\it $\\var\\big[\\expe(Y|X)\\big]$ is the variance not explained by $X$. \n\t\\eit\n\\eit\n\\end{remark}\n\n%----------------------------------------------------------------------\n\\section{Mulivariate transformations}\n%----------------------------------------------------------------------\n\n% definition\n\\begin{definition}\\label{def:jacobian}\nLet $h:\\R^n\\to\\R^n$ be a transformation of $n$ variables, and let $(x_1,x_2,\\ldots,x_n) = h(y_1,y_2,\\ldots,y_n)$. The \\emph{Jacobian} (or \\emph{Jacobian determinant}) of the transformation $h$ is the determinant of its $n\\times n$ matrix of partial derivatives:\n\\[\nJ = \n\\begin{vmatrix}\n\\displaystyle\\frac{\\partial x_1}{\\partial y_1} & \\displaystyle\\frac{\\partial x_1}{\\partial y_2} & \\cdots & \\displaystyle\\frac{\\partial x_1}{\\partial y_n}\\\\[2ex]\n\\displaystyle\\frac{\\partial x_2}{\\partial y_1} & \\displaystyle\\frac{\\partial x_2}{\\partial y_2} & \\cdots & \\displaystyle\\frac{\\partial x_2}{\\partial y_n}\\\\[2ex]\n\\vdots & \\vdots & \\ddots & \\vdots \\\\[2ex]\n\\displaystyle\\frac{\\partial x_n}{\\partial y_1} & \\displaystyle\\frac{\\partial x_n}{\\partial y_2} & \\cdots & \\displaystyle\\frac{\\partial x_n}{\\partial y_n}\\\\[2ex]\n\\end{vmatrix}\n\\]\n\\end{definition}\n\n% theorem: continuous\n\\begin{theorem}\\label{thm:transf_injective_joint_continuous}\nLet $X=(X_1,X_2,\\ldots,X_n)$ be a vector of continuous random variables, let $f_X(x_1,x_2\\ldots,x_n)$ be their joint PDF, let $g:\\R^n\\to\\R^n$ be an injective transformation, and let $Y=(Y_1,Y_2,\\ldots,Y_n)$ be the continuous random vector defined by $Y=g(X)$, i.e.\\\n\\[\n(Y_1,Y_2,\\ldots,Y_n) = g(X_1,X_2,\\ldots,X_n)\n\\]\nIf the Jacobian of the inverse transformation $g^{-1}$ is continuous and non-zero over the range of the transformation, the joint PDF of $Y$ is\n\\[\nf_Y(y_1,y_2,\\ldots,y_n) = |J| f_X(x_1,x_2,\\ldots,x_n)\n\\]\nwhere \n\\[\n(x_1,x_2,\\ldots,x_n)=g^{-1}(y_1,y_2,\\ldots,y_n)\n\\]\nis the solution of $(y_1,y_2,\\ldots,y_n)=g(x_1,\\ldots,x_n)$.\n\\end{theorem}\n\n% remark\n\\begin{remark}\nThe scale factor $|J|$ ensures that $f_Y(y_1,y_2,\\ldots,y_n)$ integrates to one.\n\\end{remark}\n\n\n%==================================================================================================\n\\section{The multivariate normal distribution} \n%==================================================================================================\nThe bivariate normal distribution can be formulated using matrix notation. Let\n\\[\n\\mathbf{Z}\t\t= \\begin{pmatrix} X \\\\ Y \\end{pmatrix},\\quad\n\\mathbf{\\mu}\t= \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}\\quad\\text{and}\\quad\n\\Sigma\t\t\t= \\begin{pmatrix} \\sigma^2_1 & \\rho\\sigma_1\\sigma_2 \\\\ \\rho\\sigma_1\\sigma_2 & \\sigma^2_2 \\end{pmatrix}\n\\]\n\nThe matrix $\\Sigma$ is called the \\emph{variance-covariance} matrix.\n\\bit\n\\it The determinant of $\\Sigma$ is \n\\[\n|\\Sigma| = (1-\\rho^2)\\sigma^2_1\\sigma^2_2.\n\\]\n\\it The inverse of $\\Sigma$ is\n\\[\n\\Sigma^{-1} = \\frac{1}{(1-\\rho^2)\\sigma^2_1\\sigma^2_2}\n\\begin{pmatrix} \n\t\\sigma^2_2 \t\t\t\t& -\\rho\\sigma_1\\sigma_2 \\\\\n\t-\\rho\\sigma_1\\sigma_2\t& \\sigma^2_1\n\\end{pmatrix}.\n\\]\n\\eit\n\nLet $\\mathbf{z} = (x,y)^T$ and consider the quadratic form \n\\begin{align*}\n\\big(\\mathbf{z}-\\mathbf{\\mu})^T \\Sigma^{-1}(\\mathbf{z}-\\mathbf{\\mu}\\big) \n\t& = \t\\frac{1}{(1-\\rho^2)\\sigma^2_1\\sigma^2_2}\\ \n\t\t\t\\begin{pmatrix} x-\\mu_1 \\\\ y-\\mu_2\t\\end{pmatrix}^T\n\t\t\t\\begin{pmatrix} \\sigma^2_2 & -\\rho\\sigma_1\\sigma_2 \\\\ -\\rho\\sigma_1\\sigma_2 & \\sigma^2_1 \\end{pmatrix}\n\t\t\t\\begin{pmatrix} x-\\mu_1 \\\\ y-\\mu_2\t\\end{pmatrix} \\\\\n\t& = \\frac{1}{(1-\\rho^2)}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2\n\t\t\t -2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) \n\t\t\t \t+\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2\\right]\n\\end{align*}\n\nThe PDF $f(x,y)$ of the bivariate normal distribution can therefore be written as\n\\[\nf(x,y) = \\frac{1}{2\\pi|\\Sigma|^{1/2}}\\exp\\left(-\\frac{1}{2}(\\mathbf{z}-\\mathbf{\\mu})^T \\Sigma^{-1} (\\mathbf{z}-\\mathbf{\\mu})\\right).\n\\]\t\t\n\nUsing this notation, we define the \\emph{multivariate normal distribution} as follows.\n\n\n\n% definition: multivariate normal\n\\begin{definition}\nA random vector $\\mathbf{Z}=(X_1,X_2,\\ldots,X_n)^T$ is said to have \\emph{multivariate normal distribution} with mean vector $\\mathbf{\\mu}= (\\mu_1,\\mu_2,\\ldots,\\mu_n)^T$ and variance-covariance matrix\n\\[\n\\Sigma\t\t\t= \\begin{pmatrix} \n\t\\sigma^2_1\t\t \t\t\t& \\rho_{12}\\sigma_1\\sigma_2 \t& \\ldots & \\rho_{1n}\\sigma_1\\sigma_n \\\\ \n\t\\rho_{21}\\sigma_1\\sigma_2 \t& \\sigma^2_2\t\t\t\t\t& \\ldots & \\rho_{2n}\\sigma_2\\sigma_n \\\\\n\t\\vdots\t\t\t\t\t\t& \\vdots\t\t\t\t\t\t& \\ddots & \\vdots \\\\\n\t\\rho_{n1}\\sigma_1\\sigma_n\t\t \t& \\rho_{n2}\\sigma_n\\sigma_2\t& \\ldots & \\sigma^2_n \\\\\n\t\\end{pmatrix}\n\\]\nif its joint density function can be written as\n\\[\nf(\\mathbf{z}) = \\frac{1}{2\\pi|\\Sigma|^{1/2}}\\exp\\left(-\\frac{1}{2}(\\mathbf{z}-\\mathbf{\\mu})^T \\Sigma^{-1} (\\mathbf{z}-\\mathbf{\\mu})\\right)\n\\]\t\t\n\\end{definition}\n\nwhere $\\mu_i = \\expe(X_i)$, $\\sigma_i^2 = \\var(X_i)$ and $\\rho_{ij} = \\displaystyle\\frac{\\expe(X_iX_j)}{\\sigma_i\\sigma_j}$.\n\n\n\\begin{exercise}\n\\begin{questions}\n%----------------------------------------\n% GS p100\n\\question\nLet $X$ and $Y$ have standard bivariate normal distribution, with joint PDF given by\n%\\[\n%f(x,y)\n%\t= \\frac{1}{2\\pi\\sigma_1\\sigma_2\\sqrt{1-\\rho^2}}\n%\t\t\\exp\\left(-\\frac{1}{2(1-\\rho^2)}\\left[\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)^2 \n%\t\t\t-2\\rho\\left(\\frac{x-\\mu_1}{\\sigma_1}\\right)\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right) \n%\t\t\t\t+\\left(\\frac{y-\\mu_2}{\\sigma_2}\\right)^2 \\right]\\right)\n%\\]\n\\[\nf(x,y) = \\frac{1}{2\\pi\\sqrt{1-\\rho^2}}\\exp\\left(-\\frac{1}{2(1-\\rho^2)}(x^2 - 2\\rho xy + y^2)\\right)\n\\]\nwhere $\\rho$ is a constant satisfying $-1 < \\rho < 1$. \n\\begin{parts}\n\\part\nCheck that $f(x,y)$ is indeed a joint PDF, by verifying that $f(x,y)\\geq 0$ and $\\displaystyle\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty} f(x,y)\\,dxdy = 1$.\n\\begin{answer}\nTODO\n\\end{answer}\n\\part\nCheck that $\\cov(X,Y) = \\displaystyle\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty} xy f(x,y)\\,dxdy = \\rho$.\n\\begin{answer}\nTODO\n\\end{answer}\n\\part\nShow that if $X$ and $Y$ are uncorrelated, then they are independent.\n\\begin{answer}\nTODO\n\\end{answer}\n\\end{parts}\n%----------------------------------------\n\\question\n% GS p110\nLet $X$ and $Y$ have standard bivariate normal distribution. Find the conditional distribution of $Y$ given $X=x$, and hence show that $\\expe(Y|X) = \\rho X$.\n\\begin{answer}\nThe conditional distribution of $Y$ given $X=x$ is $N(\\rho x, 1-\\rho^2)$.\n\\end{answer}\n%----------------------------------------\n% GS 4.7.5\n\\question\nLet $X$ and $Y$ have standard bivariate normal distribution. Show that $X$ and $\\displaystyle Z=\\frac{Y-\\rho X}{\\sqrt{1-\\rho^2}}$ are independent standard normal random variables.\n\\begin{answer}\nTODO\n\\end{answer}\n%----------------------------------------\n% GS 4.7.6\n\\question\nLet $X$ and $Y$ have standard bivariate normal distribution, and let $Z=\\max\\{X,Y\\}$. Show that $\\expe(Z)=\\sqrt{(1-\\rho)/\\pi}$ and $\\expe(Z^2)=1$.\n\\begin{answer}\nTODO\n\\end{answer}\n%----------------------------------------\n\\question\nLet $U,V\\sim N(0,1)$. Show that the random variables $X = U+V$ and $Y = U-V$ are independent.\n\n\\begin{answer}\n\\bit\n\\it The transformation is $g(u,v) = (u+v,u-v)$. \n\\it To compute the inverse transformation, consider $x=u+v$ and $y=u-v$. \n\\it Solving these, we obtain $u = \\frac{1}{2}(x+y)$ and $v = \\frac{1}{2}(x-y)$.\n\\it Thus the inverse transformation is $(u,v) = g^{-1}(x,y) = \\left(\\frac{1}{2}(x+y),\\frac{1}{2}(x-y)\\right)$\n\\eit\n\nThe Jacobian determinant of $g^{-1}(x,y)$ is\n\\[\nJ = \n\\begin{vmatrix}\n\\frac{\\partial u}{\\partial x} & \\frac{\\partial u}{\\partial y} \\\\\n\\frac{\\partial v}{\\partial x} & \\frac{\\partial v}{\\partial y} \n\\end{vmatrix}\n=\n\\begin{vmatrix}[r]\n\\frac{1}{2} & \\frac{1}{2} \\\\\n\\frac{1}{2} & -\\frac{1}{2}\n\\end{vmatrix}\n=\n-\\frac{1}{2}\n\\]\nThe joint PDF of $U$ and $V$ is \n\\[\nf(u,v) = \\frac{1}{2\\pi \\sqrt{1-\\rho ^{2}}} \\exp\\left( -\\frac{u^2 + v^2 - 2\\rho uv}{2(1-\\rho^2)} \\right)\n\\]\n\nNow,\n\\begin{align*}\nu^2 + v^2 - 2\\rho uv\n\t& = \\left(\\frac{1}{2}(x+y)\\right)^2 +\\left(\\frac{1}{2}(x-y)\\right)^2 - 2\\rho\\left(\\frac{1}{2}(x+y)\\right)\\left(\\frac{1}{2}(x-y)\\right) \\\\\n\t& = \\frac{1}{2}x^2(1-\\rho ) + \\frac{1}{2}y^2(1+\\rho )\n\\end{align*}\t\n\nThe joint PDF of $X$ and $Y$ is therefore\n\\begin{align*}\nf(x,y) \n\t& = \\frac{1}{2} \\frac{1}{2\\pi\\sqrt{1-\\rho ^{2}}}\\exp\\left( -\\frac{x^{2} (1-\\rho )}{4(1-\\rho ^{2} )} -\\frac{y^{2} (1+\\rho )}{4(1-\\rho ^{2})}\\right) \\\\\n\t& = \\frac{1}{\\sqrt{4\\pi(1+\\rho)}}\\exp\\left(-\\frac{x^{2}}{4(1+\\rho)}\\right) \\times \\frac{1}{\\sqrt{4\\pi(1-\\rho)}}\\exp\\left(-\\frac{y^{2}}{4(1-\\rho )}\\right) \\\\\n\\end{align*}\nThis is the product of the PDF of a $N\\big(0,2(1+\\rho)\\big)$ variable and the PDF of a $N\\big(0,2(1-\\rho)\\big)$ variable. Thus $X$ and $Y$ are independent. \n\\end{answer}\n\n\\question\nLet $X$ and $Y$ have bivariate normal distribution with means $\\mu_1$ and $\\mu_2$, variances $\\sigma_1^2$ and $\\sigma_2^2$, and correlation $\\rho$. Show that the conditional distribution of $Y$ given $X=x$ is\n\\[\nN\\left(\\mu_2 + \\rho\\left(\\frac{\\sigma_2}{\\sigma_1}\\right)(x-\\mu_1), \\sigma_2^2(1-\\rho^2)\\right).\n\\]\n\\begin{answer}\nTODO\n\\end{answer}\n%----------------------------------------\n\\question\n\\begin{parts}\n\\part % << (i)\nLet $X$ and $Y$ be jointly continuous random variables, and let $f_{X,Y}$ be their joint PDF. Show that the PDF of the random variable $X+Y$ can be written as\n\\[\nf_{X+Y}(t) \n\t= \\int_{-\\infty}^{\\infty} f_{X,Y}(x,t-x)\\,dx\n\t= \\int_{-\\infty}^{\\infty} f_{X,Y}(t-y,y)\\,dy.\n\\]\n\\begin{answer}\nLet $A = \\{(x,y):x+y\\leq z\\}\\subset\\R^2$. Then\n\\[\n\\prob(X+Y\\leq z)\n\t= \\iint_A f(x,y)\\,dxdy\n\t= \\int_{x=-\\infty}^{\\infty} \\int_{y=-\\infty}^{z-x}f_{X,Y}(x,y)\\,dy\\,dx\n\\]\nWe change the variable of integration (in the inner integral), making the substitution $y=t-x$:\n\\begin{align*}\nF_{X+Y}(z) = \\prob(X+Y\\leq z)\n\t& = \\int_{x=-\\infty}^{\\infty} \\int_{t=-\\infty}^{z}f_{X,Y}(x,t-x)\\,dt\\,dx \\\\\n\t& = \\int_{t=-\\infty}^{z} \\int_{x=-\\infty}^{\\infty}f_{X,Y}(x,t-x)\\,dx\\,dt\n\\end{align*}\nwhere the final equality follows by reversing the order of integration. Thus the PDF of $X+Y$ is \n\\[\nf_{X+Y}(t) = \\int_{-\\infty}^{\\infty} f_{X,Y}(x,t-x)\\,dx \\qquad\\text{as required.}\n\\]\n\\end{answer}\n\\part % << (ii)\nHence, or otherwise, show that if $U,V\\sim N(0,1)$ are independent, then $U+V\\sim N(0,2)$. (This is a special case of Theorem~\\ref{thm:sum_of_normal_variables}.)\n\\begin{answer}\nBy part (a), if two random variables $X$ and $Y$ are independent, the PDF of $X+Y$ is the \\emph{convolution} of the marginal PDFs:\n\\[\nf_{X+Y}(t) \n\t= \\int_{-\\infty}^{\\infty} f_X(x)f_Y(t-x)\\,dx\n\t= \\int_{-\\infty}^{\\infty} f_X(t-y)f_Y(y)\\,dy.\n\\]\n$U$ and $V$ are independent, so their joint PDF is\n\\[\nf(u,v)=f_U(u)f_V(v) = \\frac{1}{2\\pi}\\exp\\left(-\\frac{1}{2}(u^2 + v^2)\\right) \\qquad u,v\\in\\R.\n\\]\nLet $W=U+V$. Then because $U$ and $V$ are independent, \n\\begin{align*}\nf_W(w) \n\t& = \\int_{-\\infty}^{\\infty} f_U(u)f_V(w-u)\\,du \\\\\n\t& = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} \\exp\\left[-\\frac{1}{2}\\left(u^2 + (w-u)^2\\right)\\right]\\,du \\\\\n\t& = \\frac{1}{2\\pi} e^{-\\frac{1}{4}w^2} \\int_{-\\infty}^{\\infty} \\exp\\left[-\\left(u-\\frac{w}{2}\\right)^2\\right]\\,du\n\\end{align*}\nWe change the variable of integration, by making the substitution $t = \\displaystyle\\sqrt{2}\\left(u-\\frac{w}{2}\\right)$:\n\\[\nf_W(w) \n\t= \\frac{1}{2\\sqrt{\\pi}} e^{-\\frac{1}{4}w^2} \\int_{-\\infty}^{\\infty} \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}w^2}\\,dv\n\t= \\frac{1}{2\\sqrt{\\pi}} e^{-\\frac{w^2}{4}},\n\\]\nwhich is the PDF of the $N(0,2)$ distribution\n\\end{answer}\n\\end{parts}\n\n\n%----------------------------------------\n\\end{questions}\n\\end{exercise}\n%----------------------------------------------------------------------\n\n", "meta": {"hexsha": "5ff35661102f80f00bfefd9d6f07db3cb547441f", "size": 29267, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "L5/MA2500/14_bivariate_normal.tex", "max_stars_repo_name": "gillardjw/notes", "max_stars_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "L5/MA2500/14_bivariate_normal.tex", "max_issues_repo_name": "gillardjw/notes", "max_issues_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "L5/MA2500/14_bivariate_normal.tex", "max_forks_repo_name": "gillardjw/notes", "max_forks_repo_head_hexsha": "58b3f7e8e2c289a88905bda689c95483bee04490", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-11-04T05:13:05.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-04T05:13:05.000Z", "avg_line_length": 39.4433962264, "max_line_length": 528, "alphanum_fraction": 0.6003690163, "num_tokens": 11167, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7662936430859597, "lm_q2_score": 0.800691997339971, "lm_q1q2_score": 0.6135651876314199}} {"text": "\n\\subsection{Frequent patterns}\n\n\\subsubsection{Frequent patterns}\n\nA frequent itemset is one where the support is above a minimum.\n\nWe know that if an itemset is frequent, then all its subsets are also frequent.\n\nWe look for frequent patterns, \\(F\\), between the items \\(I\\).\n\nAn example of a frequent pattern is \\(\\{a,b\\}\\).\n\n", "meta": {"hexsha": "b689474cdb55a0b79272de10745dd71ef3c7884e", "size": 328, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/statistics/association/01-04-frequent.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/statistics/association/01-04-frequent.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/statistics/association/01-04-frequent.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 23.4285714286, "max_line_length": 79, "alphanum_fraction": 0.737804878, "num_tokens": 80, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8006919925839874, "lm_q2_score": 0.7662936324115011, "lm_q1q2_score": 0.6135651754399863}} {"text": "%gln\n\\section{General linear group}\n\nAs the most prominent example of a functor, we will now discuss the following construction. Let $R$, again, denote a unital commutative ring.\n\\begin{defi}\nLet $\\mathfrak{F}$ be a finite category and $\\mathrm{UCRng}$ the category of unital commutative rings. The bifunctor\n$$\\bao{rrcl}\n\\mathrm{Gl} : &\\mathfrak{F} \\times \\mathrm{UCRng} &\\longrightarrow &\\mathrm{Grp},\\\\\n& (I,R) &\\longmapsto& Gl(I,R) := Gl_{|I|}(R),\\\\\n\\ea$$\nis a pairing of a finite set $I$, a ring $R$ and its group of $R$ module automorphism on the generic module $R^I$.\n\\end{defi}\n\\begin{coro}\nWe immediately get\n$$\\bao{rcl}\nGl_0(R) &=& Gl(\\emptyset,R) = \\{1_R\\} \\ \\mathrm{and}\\\\\nGl_1(R) &=& Gl(\\mathcal{P}(\\emptyset),R)\\\\ &=& R^\\times = U(R)\\\\ &:=& \\{r \\in R : \\exists r' \\in R,\\ r r' = 1\\}.\\\\\n\\ea\\\\$$\n\\end{coro}\nBoth statements are clearly a consequence of our above definition.\n\\begin{theo}\nLet $I \\in \\mathfrak{F}$ be fixed with $|I| = i$ and let $R \\in \\mathrm{UCRng}$ be an ID then\n$$\\mathrm{Gl}_i \\simeq \\mathrm{Hom}_{\\mathrm{UCRng}}(\\mathbb{Z}[X,Y]/\\left<\\det X \\cdot Y - 1\\right>,\\ \\cdot\\ )$$\nas functors.\n\\end{theo}\nBefore we proceed in proofing the theorem, let us dwell in the above formular for a moment. Firstly, $X$ stands for $\\{X_{kl} : 1 \\leq k, l \\leq i\\}$ - a set of $i^2$ variables and\n$$\\mathbb{Z}[X,Y]/\\left<\\det X \\cdot Y - 1\\right> = \\mathbb{Z}[X_{kl},Y : 1 \\leq k, l \\leq i]/\\left<\\det \\left(\\bao{ccc}\nX_{11} & \\ldots & X_{1i}\\\\\n\\vdots&&\\vdots\\\\\nX_{i1}&\\ldots&X_{ii}\\ea\\right) \\cdot Y - 1\\right>$$\nfor the ring of the polynomial ring over $\\mathbb{Z}$ in $i^2 + 1$ indeterminants, factored out the ideal generated by $\\det X \\cdot Y - 1$. Thus, the right hand side claims that the class of all (unital commutative) ring homomorphisms into some ring $R$ are functorially isomorphic to $i$-th general linear group $\\mathrm{Gl}_i(R)$. %In classical algebraic geometry we would get the zero set:\n\n%$$Z(\\det X \\cdot Y - 1) = \\left\\{x \\in \\mathbb{A}^{i^2}_R \\simeq M_i(R) : \\det x \\in \\mathrm{Gl}_1(R)\\right\\}$$\n%and its spectral interpretation:\n%$$\\mathfrak{Z}(\\det X \\cdot Y - 1) = \\left\\{\\prm \\in \\mathrm{Spec}(\\mathbb{Z}[X,Y]/\\left<\\det X Y - 1\\right>) : \\det X Y - 1 \\in \\prm\\right\\}$$\n\nGiven some ring homomorphism $\\varphi : A := \\mathbb{Z}[X,Y]/\\left<\\det X Y - 1\\right> \\longrightarrow R$, we get\n$$\\mathrm{Spec}(\\varphi) = \\left[\\prm \\in \\mathrm{Spec}(R) \\longmapsto \\varphi^{-1}(\\prm) := \\{f \\in A : \\varphi(f) \\in \\prm\\}\\right],$$\nin particular\n$$\\varphi^{-1}(\\prm) \\in \\mathrm{Spec}(A).$$\nClearly, $\\mrm{Spec}(A)$ contains the elements\n$$\\prm_\\mu := \\left\\{\\sum_{kl} \\lambda_{kl} (X_{kl} - \\mu_{kl}) : \\mu = \\left(\\bao{ccc}\\mu_{11}&\\ldots&\\mu_{1i}\\\\\\vdots&&\\vdots\\\\\\mu_{i1}&\\ldots&\\mu_{ii}\\\\\\ea\\right), \\lambda_{kl} \\in A,\\ \\lambda_\\mu (X - \\mu) \\in \\ker eval_{\\mu}\\right\\}$$\nmapping the monomial $X$ to $\\mu$. As $A$ is a UFD and $(X_{kl} - \\mu_{kl})$ cannot have any other divisors we conclude $$\\ker eval_\\mu \\in \\spec{A},$$\nas $\\det \\mu \\cdot Y - 1 \\in R[Y]$ has roots in $R$ (i.e.: for all $\\det \\mu \\in \\mrm{Gl}_1(R)$) and more importantly, $(0) \\in \\spec{R}$.\nHere, we are using the following diagram:\n$$\\xymatrix{\nA \\ar[rr]^{eval_\\mu}\\ar[rrd]_f&&R[Y]\\ar[d]^{eval_{\\det \\mu^{-1}}}\\\\\n&&R.\\\\\n}$$\nThus, the pairing\n$$\\bao{rrcl}\nV :& \\mrm{Gl}_i(R) \\subset \\affr{i^2+1} &\\longrightarrow& \\spec{\\zz[X,Y]},\\\\\n&\\left(\\mu_{k,l},\\frac{1}{\\det \\mu}\\right)_{1\\leq k,l \\leq i} &\\longmapsto& \\left\n\\ea$$\nlets us consider all points in $\\mrm{Gl}_i(R)$ as an algebraic set also as (families of) prime ideals in $\\zz[X,Y]$.", "meta": {"hexsha": "007935780349bf6c477ec49c7fc23e5e636d1b49", "size": 3597, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "grp_scheme/gln.tex", "max_stars_repo_name": "gmuel/texlib", "max_stars_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "grp_scheme/gln.tex", "max_issues_repo_name": "gmuel/texlib", "max_issues_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "grp_scheme/gln.tex", "max_forks_repo_name": "gmuel/texlib", "max_forks_repo_head_hexsha": "1a3fab54f2e03d9ce656f9b8a5b58e26c3c93a02", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 65.4, "max_line_length": 393, "alphanum_fraction": 0.6313594662, "num_tokens": 1436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.828938825225204, "lm_q2_score": 0.7401743735019595, "lm_q1q2_score": 0.6135592756325157}} {"text": "\\section{How fast will Raft elect a leader when split votes are possible?}\n\\label{leaderelection:split:total}\n\nGiven a split vote rate, we can estimate the total election time.\nRaft will elect a leader as soon as an election term\nsuccessfully completes without a split vote. When a split vote occurs,\nit's likely that all servers have reset their timers, since servers do\nthis when they grant a vote (this isn't quite true when logs\ndiffer; see Section~\\ref{leaderelection:logsdiff}). Thus, the\nnext election term has the same probability of success as an entirely\nnew election and will take just as long. In other words, each election\nterm is essentially memoryless, and the number of election terms\nrequired in an election can be modeled as a geometric distribution,\nwhere the probability of success is the probability that a split vote\ndoes not occur. Therefore, Raft elections are expected to complete in\n$\\dfrac{1}{1-\\text{split vote rate}}$ election terms on average.\n\nIf a split vote occurs in a particular election term, the election term\ntakes about $1+M_s$ time units plus a one-way network latency to reset\nthe server's election timers. We do not include the time for the\ncandidate to record its own vote on disk, since this time can be\noverlapped with the RequestVote messages (with this optimization, the\ncandidate may not count its own vote towards leadership until the vote\nis durably recorded). After the vote is split, the cluster must wait\nanother election timeout before the next election term begins. This\nrepeats for each split vote, then the time for an election with no split\nvotes (from Section~\\ref{leaderelection:nosplit}) is additional. Thus, the\ntotal time for an election, $E_s$, is:\n\\begin{align*}\nE_s &= \\Big(\\sum_\\text{split votes} \\text{time for split vote}\\Big)\n +\n \\Big(\\text{time for election with no split vote}\\Big) \\\\\n%\nE_s &= \\Big(\\sum_\\text{split votes} (1 + M_s + L)\\Big)\n +\n \\Big(1 + M_s + 2L + W - U(0,\\dfrac{1}{2})\\Big) & \\\\\n%\n\\Ex[E_s] &= \\Big((\\frac{1}{1-\\text{split vote rate}} - 1) \\times\n (1 + \\frac{1}{s+1} + \\Ex[L])\\Big)\n +\n \\Big(1 + \\frac{1}{s+1} + 2\\Ex[L] + \\Ex[W] - \\dfrac{1}{4}\\Big) \\\\\n%\n\\Ex[E_s] &= \\frac{1}{1-\\text{split vote rate}} \\times\n \\Big(1 + \\frac{1}{s+1} + \\Ex[L]\\Big)\n + \\Ex[L] + \\Ex[W] - \\dfrac{1}{4}\n\\end{align*}\nwhere $L$ is the one-way network latency and $W$ is the latency for a\ndurable disk write.\n\nHoward~\\cite{Howard:2014} suggests an optimization to decrease the time\nfor an election after split votes occur. The optimization separates\nfollowers' timeouts from candidates' timeouts, where candidates select\nsmaller timeouts from a distribution with a smaller range. This results\nin faster iterations once split votes have occurred, though it risks\nadditional split votes. The remainder of this chapter does not use this\noptimization.\n\n\\begin{figure}\n\\centering\n\\includegraphics[height=5.5in]{leaderelection/overall}\n\\hspace{-2em}\n\\vcaption[expected overall election time]{\nThe expected total election times for various clusters,\nas defined by $\\Ex[E_s]$, with a fixed one-way network latency.\nIt excludes the time to write to stable\nstorage (which is usually negligible). The timeout range and\nexpected overall election time are presented as multiples of the one-way\nnetwork latency ($l$), since $l$ is typically fixed in a given\ndeployment.\n}\n\\label{fig:leaderelection:theory:overall}\n\\end{figure}\n\nFigure~\\ref{fig:leaderelection:theory:overall} plots the expected time to\nelect a leader when the network latency is fixed, by combining the\nformula for $\\Ex[E_s]$ with the formula for $Pr(D_{c,s} \\leq l)$.\nFrom the graphs, a Raft cluster with a sufficiently broad timeout range\nwill usually elect a leader within 20 times the one-way network latency,\neven when running with a bare majority of available servers. This\nsuggests that most datacenter Raft deployments should be able to achieve\ntypical leader election times under \\SI{100}{\\milli\\second}. Even worst\ncase global deployments, with one-way latencies of\n\\SI{200}{\\milli\\second},\nshould be able to typically elect leaders within \\SI{4}{seconds}. (Election\ntimes may be larger if some servers are deployed on other planets.)\n\nEach of the curves has a knee. If the timeout range is chosen to be too\nshort, too many servers time out before others are able to collect\nvotes, resulting in poor election times. Once timeout ranges are\nsufficiently large (about 3--8 times the network latency, depending on\nthe cluster), the curves become linear with a slight upward slope:\nelections complete after few or no split votes, but they must wait\nlonger for each timeout to elapse.\n\nThe graphs provide insight into how to configure election timeouts: a\nconservative setting is probably best in practice. The minimum point on\nthe graphs represents the best average election time possible for each\ngiven cluster configuration. However, attaining this minimum time is\nquite risky, since the minimum is close to the knee in the curve. If the\nnetwork latency turns out to be slightly higher than anticipated in\npractice, that might push the system into the left region of the graph\nwhere election times skyrocket. It is better to configure systems\nfarther to the right, trading off a slightly higher average election\ntime in exchange for a more robust system. Thus, we recommend using a\ntimeout range that is ten times the one-way network latency (even if the\ntrue network latency is five times greater than anticipated, most clusters would\nstill be able to elect a leader in a timely manner).\n", "meta": {"hexsha": "1ae608ae8b4be24e40a8c649ac15d90f32ec7bf6", "size": 5548, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "leaderelection/splittotal.tex", "max_stars_repo_name": "ahrtr/dissertation", "max_stars_repo_head_hexsha": "7ad82c250a28c4a4d2406f43756ab2f8837292b3", "max_stars_repo_licenses": ["CC-BY-3.0", "CC-BY-4.0"], "max_stars_count": 788, "max_stars_repo_stars_event_min_datetime": "2015-01-28T14:25:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-28T06:58:22.000Z", "max_issues_repo_path": "leaderelection/splittotal.tex", "max_issues_repo_name": "ahrtr/dissertation", "max_issues_repo_head_hexsha": "7ad82c250a28c4a4d2406f43756ab2f8837292b3", "max_issues_repo_licenses": ["CC-BY-3.0", "CC-BY-4.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2015-09-27T18:10:33.000Z", "max_issues_repo_issues_event_max_datetime": "2016-05-24T23:22:22.000Z", "max_forks_repo_path": "leaderelection/splittotal.tex", "max_forks_repo_name": "ahrtr/dissertation", "max_forks_repo_head_hexsha": "7ad82c250a28c4a4d2406f43756ab2f8837292b3", "max_forks_repo_licenses": ["CC-BY-3.0", "CC-BY-4.0"], "max_forks_count": 125, "max_forks_repo_forks_event_min_datetime": "2015-02-12T13:47:25.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-24T02:46:49.000Z", "avg_line_length": 51.3703703704, "max_line_length": 80, "alphanum_fraction": 0.7664023071, "num_tokens": 1382, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388125473628, "lm_q2_score": 0.7401743735019594, "lm_q1q2_score": 0.6135592662487024}} {"text": "%% -*- coding:utf-8 -*-\n\\chapter{Tensor product. Structure of finite K-algebras}\nThis is a digression on commutative algebra. We introduce and study\nthe notion of tensor product of modules over a ring. We prove a\nstructure theorem for finite algebras over a field (a version of the\nwell-known \"Chinese remainder theorem\").\n\n\n\\section{Definition of tensor product}\n\n\\subsection{Summary for previous lectures}\nWe considered finite \\mynameref{def:fextension1} $L$ i.e\n$\\left[L:K\\right] < \\infty$. We also saw that if\n$L$ is generated by a finite number of \\mynameref{def:degsepelem}s \n$\\alpha_1, \\dots, \\alpha_r$ then the number of\n\\mynameref{def:homomorphism}s over $K$ from $L$ to $\\bar{K}$ denoted by\n$\\left|Hom_K\\left(L, \\bar{K}\\right)\\right|$ is equal to\n$\\left[L:K\\right]$. In general\n\\[\n\\left[L:K\\right]_{sep} = \n\\left|Hom_K\\left(L, \\bar{K}\\right)\\right| \\le\n\\left[L:K\\right].\n\\]\n\nFor $L = K\\left(\\alpha\\right)$ it is clear because the number of\nhomomorphisms is equal to the number of roots of the\n\\mynameref{def:minpolynomial} $P_{min}\\left(\\alpha, K\\right)$. In\ngeneral one can use induction and multiplicativity of the degree\n$\\left[L:K\\right]$ and number of homomorphisms (see theorem\n\\mynameref{thm:lec3_3}). Thus separable extension was exactly an\nextension which had the right number of homomorphisms into the\nalgebraic closure.\n\nOur next goal is to characterize the separability in the terms of\ntensor product.\n\n\\subsection{Tensor product}\n\n\\begin{definition}[Tensor product]\n Let $A$ is a ring, $N$, $M$ are $A$-\\mynameref{def:module}s. The\n tensor product $M \\otimes_A N$\n is another $A$-\\mynameref{def:module} together with an\n $A$-bilinear map $\\phi: M \\times N \\to M \\otimes_A N$ which has\n ``\\mynameref{def:universalproperty}'' defined below\n \\label{def:tensorproduct}\n\\end{definition}\n\n\\begin{definition}[Universal property]\n $A$-bilinear map $\\phi: M \\times N \\to M \\otimes_A N$ has\n ``universal property'' if\n $\\forall P$ - $A$-\\mynameref{def:module} and\n for $A$-bilinear $f: M \\times N \\to P$ (\n i.e.\n \\(\n \\forall m, f_m:\n N \\xrightarrow[n \\to f(m,n)]{} P\n \\)\n and\n \\(\n \\forall n, f_n:\n M \\xrightarrow[m \\to f(m,n)]{} P\n \\)\n are \\mynameref{def:homomorphism}s of $A$-modules\n ), then\n $\\exists! \\tilde{f}$ - homomorphism of $A$-modules such that\n $f = \\tilde{f} \\circ \\phi$\n \\footnote{\n That means that we have a \\mynameref{def:commutativediagram} there\n }\n\n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { M \\times N & & P \\\\\n & M \\otimes_A N & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-1-1) edge node[auto] {$ f $} (m-1-3)\n edge node[auto] {$ \\phi $} (m-2-2)\n (m-2-2) edge node[auto] {$ \\tilde{f} $} (m-1-3);\n \\end{tikzpicture}\n \\label{def:universalproperty}\n\\end{definition}\n\nThe property characterize the pair\n$\\left(\\phi, M \\otimes N\\right)$.\nReally if have another pair\n$\\left(\\overline{\\phi}, \\overline{M \\otimes N}\\right)$\nlike this one then by definition we have mutually inverse\nhomomorphisms of $A$-modules between them\n\n\\begin{lemma}[About uniqueness of object defined by universal\n property]\n \\footnote{\n It is out of the lecture video and can be considered as an\n explanation for the claim about having mutually inverse\n homomorphisms of $A$-modules. The proof was taken from\n \\cite{bib:KeithConradTensorProduct1}.\n }\n If we have two objects $\\left(\\phi, M \\otimes N\\right)$\n and $\\left(\\overline{\\phi},\\overline{M \\otimes N}\\right)$\n which both satisfies \\mynameref{def:universalproperty} than there is\n an unique \\mynameref{def:isomorphism} between them: \n \\[\n \\left(\\phi, M \\otimes N\\right) \\cong \\left(\\overline{\\phi},\n \\overline{M \\otimes N}\\right)\n \\]\n\\begin{proof}\nLet $P = \\overline{M \\otimes N}$ and $f =\n\\overline{\\phi}$. In \nthe case we can consider the following diagram\n\n\\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { & & M \\otimes_A N & \\\\\n M \\times N & & \\overline{M \\otimes_A N} \\\\\n & & M \\otimes_A N & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-2-1) edge node[auto] {$ \\phi $} (m-1-3)\n (m-1-3) edge node[auto] {$ g = \\tilde{\\overline{\\phi}} $} (m-2-3)\n (m-2-1) edge node[auto] {$ \\overline{\\phi} $} (m-2-3)\n (m-2-1) edge node[auto] {$ \\phi $} (m-3-3)\n (m-2-3) edge node[auto] {$ \\bar{g} = \\tilde{\\phi} $} (m-3-3);\n\\end{tikzpicture}\n\nAs soon as we fixed $\\overline{ M \\otimes_A N}$\nwe 2 unique homomorphisms (which are defined by the fixed\n$\\overline{ M \\otimes_A N}$) - \n$g : M \\otimes_A N \\to \\overline{ M \\otimes_A N}$\nand\n$\\bar{g} : \\overline{M \\otimes_A N} \\to M \\otimes_A N$.\nBoth $g$ and $\\bar{g}$ are linear and, as mentioned above, the pair is\nunique (if we fix $g$ we will have only one $\\bar{g}$ that corresponds\nto $g$). The composition $g \\circ \\bar{g}$ maps $M \\otimes_A N$ to\nitself. I.e. $\\forall x \\in M \\times N$ we have\n$\\bar{g}\\left(g\\left(\\phi\\left(x\\right)\\right)\\right) =\n\\phi\\left(x\\right)$. If $y = \\phi\\left(x\\right)$ then\n$\\bar{g}\\left(g\\left(y\\right)\\right) = y$. The last equation holds\n$\\forall y \\in Im\\left(\\phi\\right)$ i.e. $\\forall y \\in M \\otimes_A\nN$. Therefore $\\bar{g} = g^{-1}$. \n\\footnote {\n Another explanation is the following: thus if we fix $g$ and choose\n $\\bar{g} = g^{-1}$ we will get $g \\circ \\bar{g} = id_{M \\otimes_A\n N}$ that satisfied all requirements. The choice is final because\n we don't have a possibility to choose any other $\\bar{g}$ (it should\n be unique).\n}\n\nThus we have an\n\\mynameref{def:isomorphism} and the isomorphism is unique as soon as the\nfunction $g$ is unique due the \\mynameref{def:universalproperty}.\n\nWe just prove an isomorphism existence between\n$M \\otimes N$ and $\\overline{M \\otimes N}$ but the tensor product is\ncharacterized not only by the module $M \\otimes N$\nbut also a bilinear map $\\phi$. Let $P = \\overline{M \\otimes N}$ thus\nwe can get that $\\bar{\\phi} = \\tilde{\\bar{\\phi}} \\circ \\phi$ is\ndetermined by the unique\nrelation $\\phi \\to \\bar{\\phi}$ as soon as\n$\\tilde{\\bar{\\phi}}$ is unique.\nAnalogues one can get the unique relation\n$\\bar{\\phi} \\to \\phi$.\n\\end{proof}\n\\label{lem:universalpropertyuniqueness}\n\\end{lemma}\n\nThe uniqueness does not mean existence and we should proof that such\nobject exists.\n\\begin{lemma}[About tensor product existence]\n Tensor product defined via \\mynameref{def:universalproperty} exists\n \\begin{proof}\n Lets consider $\\mathcal{E}$ the maps (functions) from\n $M \\times N$ to $A$ as sets which are $0$ almost everywhere\n (i.e. outside of a finite set). For example we can consider delta\n functions:\n \\[\n \\delta_{m,n} : M \\times N \\to A\n \\]\n such that\n \\begin{eqnarray}\n \\delta_{m,n}(m,n) = 1,\n \\nonumber \\\\\n \\delta_{m,n}(m',n') = 0 \\mbox{ if } (m,n) \\ne (m',n')\n \\nonumber \n \\end{eqnarray}\n Then $\\mathcal{E}$ is a A-\\mynameref{def:freemodule} with basis\n $\\delta_{m,n}$. Thus we have a map of sets $M \\times N \\to\n \\mathcal{E}$ such that $(m,n) \\to \\delta_{m,n}$ which is not bilinear\n but we can make it bilinear by means of changing $\\mathcal{E}$.\n \n Let $\\mathcal{F} \\subset \\mathcal{E}$ a submodule generated by\n $\\delta_{m+m',n} - \\delta_{m,n} - \\delta_{m',n}$,\n $\\delta_{m,n+n'} - \\delta_{m,n} - \\delta_{m,n'}$,\n $\\delta_{am,n} - a\\delta_{m,n}$,\n $\\delta_{m,an} - a\\delta_{m,n}$.\n \\footnote{\n The basis is chosen to be a bilinear $\\mod \\mathcal{F}$, for instance\n $\\delta_{m+m',n} = \\delta_{m,n} + \\delta_{m',n} \\mod \\mathcal{F}$\n }\n \n It can be shown that $M \\times N \\to \\mathcal{E}/\\mathcal{F}$ is\n bilinear\n \\footnote{\n Follows from the basis choice\n }\n and has the desired \\mynameref{def:universalproperty}.\n\n Really lets we have the following bilinear map:\n $f: M \\times N \\to P$. Then we can consider the following linear\n map (\\mynameref{def:homomorphism})\n $f': \\mathcal{E} \\to P$ that sends $\\delta_{m,n}$ to\n $f(n,m)$. Using the fact that $f$ is bilinear we can get\n \\begin{eqnarray}\n f'(\\delta_{m+m',n}) =\n f(m+m', n) = f(m, n) + f(m', n) =\n \\nonumber \\\\\n =\n f'(\\delta_{m,n}) + f'(\\delta_{m',n}).\n \\nonumber\n \\end{eqnarray}\n With the same approach one can get the following relations\n \\begin{eqnarray}\n f'(\\delta_{m,n+n'}) = f'(\\delta_{m,n}) + f'(\\delta_{m,n'}),\n \\nonumber \\\\\n f'(\\delta_{am,n}) = a f'(\\delta_{m,n}),\n \\nonumber \\\\\n f'(\\delta_{m,an}) = a f'(\\delta_{m,n})\n \\nonumber\n \\end{eqnarray}\n with the $f'$ linearity we have\n \\begin{eqnarray}\n f'(\\delta_{m+m',n}) = f'(\\delta_{m,n} + \\delta_{m',n}),\n \\nonumber \\\\\n f'(\\delta_{m,n+n'}) = f'(\\delta_{m,n} + \\delta_{m,n'}),\n \\nonumber \\\\\n f'(\\delta_{am,n}) = a f'(\\delta_{m,n}),\n \\nonumber \\\\\n f'(\\delta_{m,an}) = a f'(\\delta_{m,n})\n \\nonumber\n \\end{eqnarray}\n\n The kernel $\\ker f' = \\mathcal{F}$ thus if we want to have a\n homomorphism to $P$ we have to replace $\\mathcal{E}$ with\n $\\mathcal{E}/\\mathcal{F}$ that is also denoted by\n $M \\otimes_A N$. In the case we will replace $f'$ with\n $\\tilde{f}\\left(\\delta_{m,n} \\mod \\mathcal{F}\\right) = f(m,n)$. As\n soon as the images for the basis is fixed the mapping is unique.\n \\end{proof}\n \\label{lem:tensorproductexistence}\n\\end{lemma}\n\nWe will denote $\\phi\\left(m,n\\right) = \\delta_{m,n} \\mod \\mathcal{F}$ as\n$m \\otimes n$. I.e our tensor product can be considered as the \n$\\left(\\otimes, M \\otimes_A N\\right)$ pair.\n\n\n\\begin{remark}\n Wrong idea is to define $M \\otimes_A N$\n as a set of $m \\otimes n$. I.e.\n $M \\otimes_A N \\neq \\{m \\otimes n\\}$. The $M \\otimes_A N$ is\n generated by $m \\otimes n$ i.e. \n $\\forall x \\in M \\otimes_A N$ we have\n $x = \\sum_{i = 1}^k m_i \\otimes n_i$ i.e. each element is a finite\n sum of $m \\otimes n$ and I cannot reduce these further\n \\footnote{\n i.e. $\\exists x \\in M \\otimes_A N$ such that\n $\\exists! m \\in M, n \\in N: x = m \\otimes n$ but\n $\\exists m_1, \\dots, m_k \\in M, n_1, \\dots, n_k \\in N :\n x = \\sum_{i = 1}^k m_i \\otimes n_i$\n }. \n\n\\end{remark}\n\n\\section{Tensor product of modules}\n\n\\subsection{Advantages of the universal property}\nNow, you can ask why haven't I just defined the tensor product by\nthis construction? Why am I talking of this universal property? \nAnd the answer is because it is easier to prove things this way. \nSo advantages of the universal property is as follows: the proofs\nbecome easy.\n\n\\subsection{Several examples of universal property usage}\n\n\\begin{example}[Commutativity proof]\n We want to prove that\n \\[\n M \\otimes_A N \\cong N \\otimes_A M\n \\]\n\n We have the following bilinear map:\n $M \\times N \\to N \\otimes_A M$ for which the pair\n $(m,n)$ is mapped to $n \\otimes m$. Thus from\n \\mynameref{def:universalproperty} we have that there is a linear map\n (homomorphism)\n $\\alpha: M \\otimes_A N \\to N \\otimes_A M$:\n \n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { M \\times N & & N \\otimes_A M \\\\\n & M \\otimes_A N & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-1-1) edge node[auto] {$ (m,n) \\to n \\otimes m $} (m-1-3)\n (m-1-1) edge node[description] {$ (m,n) \\to m \\otimes n $} (m-2-2)\n (m-2-2) edge node[description] {$ \\alpha $} (m-1-3);\n \\end{tikzpicture}\n\n \n With the same\n construction we can also get the inverse map $a^{-1}$ that sends\n $N \\otimes_A M$ to $M \\otimes_A N$:\n \n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { M \\times N & & M \\otimes_A N \\\\\n & N \\otimes_A M & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-1-1) edge node[auto] {$ (m,n) \\to m \\otimes n $} (m-1-3)\n (m-1-1) edge node[description] {$ (m,n) \\to n \\otimes m $} (m-2-2)\n (m-2-2) edge node[description] {$ \\alpha^{-1} $} (m-1-3);\n \\end{tikzpicture}\n \n\\end{example}\n\nAlso\n\\begin{corollary}\n\\[\nA \\otimes_A M \\cong M\n\\]\n\\begin{proof}\n For the proof\n \\footnote{\n The proof is missed in the lectures\n }\n lets look at $A$. \n Really $A$ can be considered as $A$-module because all\n requirements from definition \\ref{def:module} are satisfied. The\n following diagrams shows that there exist 2 homomorphisms:\n $\\alpha: A \\otimes_A M \\to M$ and\n $\\alpha^{-1}: M \\to A \\otimes_A M$ as result there is a homomorphism\n $A \\otimes_A M \\cong M$:\n\n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { A \\times M & & M \\\\\n & A \\otimes_A M & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-1-1) edge node[auto] {$ (a,m) \\to a \\cdot m $} (m-1-3)\n (m-1-1) edge node[description] {$ (a,m) \\to a \\otimes_A m $} (m-2-2)\n (m-2-2) edge node[description] {$ \\alpha $} (m-1-3);\n \\end{tikzpicture}\n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { A \\times M & & A \\otimes_A M \\\\\n & M & \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-1-1) edge node[auto] {$ (a,m) \\to a \\otimes_A m $} (m-1-3)\n (m-1-1) edge node[description] {$ (a,m) \\to a \\cdot m $} (m-2-2)\n (m-2-2) edge node[description] {$ \\alpha^{-1} $} (m-1-3);\n \\end{tikzpicture}\n \n In the diagrams $m \\in M$, as usual, and $a \\in A$.\n \n\\end{proof}\n\\end{corollary}\n\nIf we have that $M$ is generated by $e_1, e_2, \\dots$ and\n$N$ is generated by $\\epsilon_1, \\epsilon_2, \\dots$ than\n$M \\otimes_A N$ is generated by pairs $e_i \\otimes \\epsilon_j$. It's\nobvious.\n\nMore complex fact is the following\n\\begin{proposition}\n Let $M$ and $N$ are \\mynameref{def:freemodule}s\n with corresponding basises $e_1, e_2, \\dots, e_n$ and\n $\\epsilon_1, \\epsilon_2, \\dots, \\epsilon_m$ than\n $M \\otimes_A N$ is also free module with basis $e_i \\otimes\n \\epsilon_j$ where $1 \\le i \\le n$ and\n $1 \\le j \\le m$.\n \\begin{proof}\n Lets define\n $f_{i_0,j_0}: M \\times N \\to A$ as a map that sends\n $\\left(\\sum a_i e_i, \\sum b_j \\epsilon_j\\right)$ to\n $a_{i_0} b_{j_0}$. It's bilinear\n \\footnote{\n for example\n $\\left(\\sum (a_i + a_i') e_i, \\sum b_j \\epsilon_j\\right)$ is\n sent to $(a_{j_0} + a_{j_0}') b_{j_0}$.\n }\n so it factors through the tensor product\n $\\tilde{f}_{i_0, j_0} : M \\otimes_A N \\to A$. The map\n $\\tilde{f}_{i_0, j_0}$ sends $e_{i_0} \\otimes \\epsilon_{j_0}$ to 1\n and all others to 0.\n \\footnote{\n Because $f_{i_0, j_0} = \\tilde{f}_{i_0, j_0} \\phi$ i.e.\n \\begin{eqnarray}\n a_{i_0} b_{j_0} =\n f_{i_0,j_0}\\left(\\sum a_i e_i, \\sum b_j\\epsilon_j\\right) =\n \\nonumber \\\\\n = \\tilde{f}_{i_0,j_0}\\left(\n \\phi\\left(\\sum a_i e_i, \\sum b_j\\epsilon_j\\right)\\right) = \n \\nonumber \\\\\n =\n \\tilde{f}_{i_0,j_0}\\left(\\sum a_i e_i \\otimes \\sum b_j\n \\epsilon_j\\right) =\n \\sum_{i,j} a_i b_j \\tilde{f}_{i_0,j_0}(e_i \\otimes \\epsilon_j).\n \\nonumber\n \\end{eqnarray}\n }\n So if\n \\[\n \\sum \\alpha_{ij} e_i \\otimes \\epsilon_j = 0\n \\]\n then applying $\\tilde{f}_{i_0, j_0}$ for all indices one can get\n that $\\forall i,j: \\alpha_{ij} = 0$.\n \\footnote{\n because $\\tilde{f}$ should be linear.\n }\n \\end{proof}\n \\label{prop:lec4_1}\n\\end{proposition}\n\nIn particular for the \\mynameref{def:vectorspace} the tensor product is\ndefined in the same way (as just proved in the proposition\n\\ref{prop:lec4_1}): the tensor product of 2 vector spaces with\nbasises $e_1, e_2, \\dots, e_n$ and\n$\\epsilon_1, \\epsilon_2, \\dots, \\epsilon_m$ is another vector space\nwith the following basis $e_i \\otimes \\epsilon_j$ i.e. the definition\ndoes not take into consideration the \\mynameref{def:universalproperty}.\n\n\\begin{proposition}[Associative]\n \\[\n \\left(M_1 \\otimes_A M_2\\right) \\otimes_A M_3\n \\cong\n M_1 \\otimes_A \\left(M_2 \\otimes_A M_3\\right)\n \\]\n \\begin{proof}\n There is just a scratch of the proof.\n Introduce\n $M_1 \\otimes_A M_2 \\otimes_A M_3$\n as a universal object for 3-linear maps and show that 2 considered\n parts are isomorphic each other.\n \\end{proof}\n \\label{prop:lec4_Associative}\n\\end{proposition}\n\n\\section{Base change}\n\nLet $A$ is a \\mynameref{def:ring} and $B$ is $A$-algebra. Let also $M$\nis an $A$-\\mynameref{def:module} and $N$ is $B$-module.\n\nI can of course make $N$ into $A$-module (just forgetting the\nadditional $A$-algebra structure). But we can also make $B$-module on\n$M$ (that is not a trivial thing) by considering $B \\otimes_A M$\n\\footnote{\n In other words we can make a $B$-module from $A$-module $M$\n}\n.\nWe can introduce $B$-module structure on $B \\otimes_A M$ by\n\\footnote{\n I.e. we introduced $B$-algebra operations for objects from\n $B \\otimes_A M$. See also definition \\ref{def:kalgebra}.\n}\n\\[\nb \\cdot \\left(b' \\otimes m \\right) = \\left( b \\cdot b' \\right) \\otimes m \n\\]\n\n\\begin{example}[The complexification of a real vector space]\n We can ``make'' $\\mathbb{R}^{2n}$ from $\\mathbb{C}^n$ by forgetting\n the complex structure.\n \\footnote{\n In the case we have ring $A = \\mathbb{R}$ and $B = \\mathbb{C}$ -\n $A$ algebra. $A$ - module is the following vector space\n $M = \\mathbb{R}^{n}$ and $B$ - module is $N = \\mathbb{C}^n$.\n }\n The $\\mathbb{C}^n$ has the following basis $e_1, \\dots, e_n$.\n The $\\mathbb{R}^{2n}$ has the following one\n $e_1, \\dots, e_n, i e_1, \\dots, i e_n$. Now we forgot about\n multiplication rules for $i = \\sqrt{-1}$ and denote $i e_i$ as\n $v_i$. In the case the basis for $\\mathbb{R}^{2n}$ is the following\n one: $e_1, \\dots, e_n, v_1, \\dots, v_n$.\n\n But we can also do the following constructions\n \\[\n \\mathbb{R}^n \\rightarrow\n \\mathbb{C}^n = \\mathbb{C} \\otimes \\mathbb{R}^n \\rightarrow\n \\mathbb{R}^{2n}\n \\]\n for the $\\mathbb{C}^n$ basis we have\n $1_{\\mathbb{C}} \\otimes e_1, \\dots, 1_{\\mathbb{C}} \\otimes e_n$ and\n for\n $\\mathbb{R}^{2n}$ -\n $1 \\otimes e_1, \\dots, 1 \\otimes e_n, i \\otimes e_1, \\dots, i\n \\otimes e_n$.\n \\footnote{\n some additional clarification:\n $\\forall x \\in \\mathbb{C} \\otimes \\mathbb{R}^n$ we have\n \\[\n x = \\sum_{i=1}^n c_i \\otimes r_i e_i =\n \\sum_{i=1}^n c_i r_i 1_{\\mathbb{C}} \\otimes e_i =\n \\sum_{i=1}^n c_i' 1_{\\mathbb{C}} \\otimes e_i,\n \\]\n where $c_i,c_i'=c_i r_i \\in \\mathbb{C}, r_i \\in \\mathbb{R}$. Thus\n we just got $\\mathbb{C}^n$. \n From other side we can write $c_i$\n as follows: $c_i = a_i + i b_i$, where $a_i, b_i \\in \\mathbb{R}$.\n Therefore\n \\[\n x = \\sum_{i=1}^n a_i r_i 1 \\otimes e_i + \\sum_{i=1}^n b_i r_i i \\otimes e_i.\n \\]\n I.e. $x \\in \\mathbb{R}^{2n}$ and the basis in $\\mathbb{R}^{2n}$ is\n formed by $1 \\otimes e_1, \\dots, 1 \\otimes e_n, i \\otimes e_1,\n \\dots, i \\otimes e_n$.\n }\n\n\\end{example}\n\n\\begin{proposition}\nIn general we have the following. If $M$ - free $A$ - module with\nbasis $e_1, \\dots, e_n$ then $B \\otimes_A M$ is a free $B$ module with\nbasis $1_B \\otimes e_1, \\dots, 1_B \\otimes e_n$.\n\\label{prop:lec4_Addon}\n\\begin{proof}\n The proof is the same as at proposition \\ref{prop:lec4_1}. Again we\n construct certain bilinear maps and say that those factor over the\n tensor product and this implies that certain families are linearly\n independent.\n\n Really lets define bilinear map $f_{i_0}: B \\times M \\to A$ such that\n \\[\n f_{i_0}\\left(b, \\sum_{i=1}^n m_i e_i\\right) = b m_{i_0} e_{i_0}\n \\]\n so there exists a linear map $\\tilde{f}_{i_0}$ (homomorphism) such\n that $f_{i_0} = \\tilde{f}_{i_0} \\phi$ or\n \\begin{eqnarray}\n f_{i_0}\\left(b, \\sum_{i=1}^n m_i e_i\\right) =\n \\tilde{f}_{i_0} \\left(\n \\phi\\left(b, \\sum_{i=1}^n m_i e_i\\right)\n \\right)\n \\nonumber \\\\\n =\n \\tilde{f}_{i_0}\\left(b \\otimes \\sum_{i=1}^n m_i e_i\\right) =\n b \\tilde{f}_{i_0}\\left(1_B \\otimes \\sum_{i=1}^n m_i e_i \\right) =\n b m_{i_0}\n \\nonumber\n \\end{eqnarray}\n i.e. it sends $1_B \\otimes e_{i_0}$ to 1 and all others $1_B \\otimes\n e_i$ to 0.\n Thus the following sum $\\sum \\alpha_i 1_B \\otimes e_i$ is equal to 0\n if and only if $\\alpha_i = 0$ i.e. $\\alpha_i 1_B \\otimes e_i$ forms a basis.\n\\end{proof}\n\\end{proposition}\n\n\\begin{remark}\nWe have the following maps.\n\\begin{itemize}\n \\label{item:lec4_maps}\n\\item For A - modules:\n $\\alpha: M \\xrightarrow[m \\to 1_B \\otimes_A m]{} B \\otimes_A M$\n which makes a $B$-module from an $A$-module.\n\\item For B - modules:\n $\\mu: B \\otimes_A N \\xrightarrow[b \\otimes n \\to b n]{} N$.\n\\end{itemize}\n\\label{rem:lec4_maps}\n\\end{remark}\n\n\\begin{theorem}[Base-change]\n Let $A$ is a \\mynameref{def:ring} and $B$ is $A$-algebra. Let also $M$\n is an $A$-\\mynameref{def:module} and $N$ is $B$-module.\n \\[\n Hom_A\\left(M, N\\right)\n \\leftrightarrow\n Hom_B\\left(B \\otimes_A M, N\\right)\n \\]\n I.e. the homomorphisms\n \\footnote{\n $A$-homomorphisms between $A$ modules $Hom_A\\left(M, N\\right)$ are the\n same as $B$-homomorphisms between $B$ modules\n $Hom_B\\left(B \\otimes_A M, N\\right)$.\n }\n are the same or in other words the\n corresponding groups of homomorphisms are isomorphic:\n \\[\n Hom_A\\left(M, N\\right)\n \\cong\n Hom_B\\left(B \\otimes_A M, N\\right)\n \\]\n \\label{thm:basechange}\n \n \\begin{proof}\n First of all we have\n \\footnote{\n One homomorphism from $Hom_B\\left(B \\otimes_A M, N\\right)$\n }\n \\mynameref{def:homomorphism} $f: B \\otimes_A M \\to N$. We also have\n the following map (see remark \\ref{rem:lec4_maps}):\n $\\alpha: M \\to B \\otimes_A M$. \n Thus $f \\cdot \\alpha: M \\to N$ i.e.\n we can set the following relation\n \\[\n \\hat{f} : Hom_B\\left(B \\otimes_A M, N\\right)\n \\to Hom_A\\left(M, N\\right)\n \\]\n such that $\\hat{f}\\left(f\\right) = f \\alpha$.\n\n In other direction we have $g: M \\to N$ thus\n $id_B \\otimes g: B \\otimes_A M \\to B \\otimes_A N$ but\n ( see remark \\ref{rem:lec4_maps}) we have\n $\\mu: B \\otimes_A N \\to N$ i.e. we have the following relation\n \\[\n \\hat{g} : Hom_A\\left(M, N\\right) \\to\n Hom_B\\left(B \\otimes_A M, N\\right)\n \\]\n such that\n \\[\n \\hat g(g) = \\mu \\cdot (id_B \\otimes g).\n \\] \n And we can check that those maps ($\\hat{f}$ and $\\hat{g}$) are\n mutually inverse. For the proof\n \\footnote{\n It's missed in the lectures\n }\n the fact consider the following diagram\n \n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { M & & N \\\\\n & B \\otimes_A M & & B \\otimes_A N \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n ([yshift=1ex]m-1-1.east) edge node[description] {$ f \\alpha $}\n ([yshift=1ex]m-1-3.west); \n \\path[->]\n ([yshift=-1ex]m-1-1.east) edge node[description] {$ g $}\n ([yshift=-1ex]m-1-3.west);\n \\path[->]\n (m-1-1) edge node[description] {$ \\alpha $} (m-2-2)\n (m-2-2) edge node[description] {$ f $} (m-1-3)\n (m-2-2) edge node[description] {$ id_B \\otimes g $}\n (m-2-4)\n (m-2-4) edge node[description] {$ \\mu $} (m-1-3)\n ;\n \\end{tikzpicture}\n \n One can conclude, as soon as the diagram commutes\n \n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { M & & N \\\\\n & B \\otimes_A M & & B \\otimes_A N \\\\ };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n ([yshift=-1ex]m-1-1.east) edge node[description] {$ g $}\n ([yshift=-1ex]m-1-3.west);\n \\path[->]\n (m-1-1) edge node[description] {$ \\alpha $} (m-2-2)\n (m-2-2) edge node[description] {$ id_B \\otimes g $}\n (m-2-4)\n (m-2-4) edge node[description] {$ \\mu $} (m-1-3)\n ;\n \\end{tikzpicture}\n\n \n \\[\n \\hat{f}\\left(\\hat{g}\\left(g\\right)\\right) =\n \\mu \\cdot (id_B \\otimes g) \\cdot \\alpha = g.\n \\]\n I. e. $\\hat{f}\\circ\\hat{g} = id$\n \\footnote {\n Operation $\\circ$ is defined as follows $(\\hat{a} \\circ \\hat{b})(x) =\n \\hat{a}(\\hat{b}(x))$ where $\\hat{a}, \\hat{b}$ are 2 maps acting\n on a set $X$ and $x \\in X$.\n }\n or in other words $\\hat{f}$ and\n $\\hat{g}$ are mutually inverse. \n \\end{proof}\n\\end{theorem}\n\n\\section{Examples. Tensor product of algebras}\n\n\\begin{proposition}\n If $I \\subset A$ - is an \\mynameref{def:ideal} so my $B$ - $A$ algebra\n will be $B = A/I$ then\n \\[\n A/I \\otimes_A M \\cong M/IM\n \\]\n where $IM$ is a sub-module of $M$.\n \\label{prop:lec4_prop2}\n \\begin{proof}\n We have map $\\alpha: M \\to B \\otimes_A M = A/I \\otimes_A M$\n (see remark \\ref{rem:lec4_maps}) \n which sends $m$ to $\\bar{1} \\otimes m$.\n \\footnote{\n $\\bar{1} = 1_A + I$\n }\n The map sends $IM$ to $0$\n because\n $\\forall i \\in I, m \\in M : im \\to \\bar{1} \\otimes im = \\bar{i} \\otimes m$\n because the tensor product is over $A$ and everything is $A$\n linear and as result $\\bar{1} \\otimes im = \\bar{i} \\otimes m$,\n but $\\bar{i} \\otimes m = \\bar{0} \\otimes m = 0$.\n \\footnote{\n because\n $\\bar{i} = 0 \\mod I$\n }\n Thus $\\alpha$ sends $IM$ to 0. So $\\alpha$ induces\n $\\bar{\\alpha}: M/IM \\to A/I \\otimes_A M$ such that\n $\\bar{\\alpha}\\left(\\bar{m}\\right) = \\bar{1} \\otimes m$.\n\n For other direction we apply \\mynameref{thm:basechange}. The\n following map (projection) of $A$-modules\n \\[\n M \\xrightarrow{m \\to \\bar{m}} M/IM\n \\]\n gives us the following map of $B$-modules\n \\footnote{\n the \\mynameref{thm:basechange} says that homomorphism of\n $A$-modules $M \\to N$ is isomorphic to homomorphism of $B$\n modules $B\\otimes_A M \\to N$. Using $N = M/IM$ we can get that\n $B\\otimes_A M \\to M/IM$ is isomorphic (i.e. the same) to\n $M \\to M/IM$. We also can get the same result just ignore $B$:\n $B \\otimes_A M \\to M \\to M/IM$.\n }\n \\[\n \\bar{\\beta}: B \\otimes_A M \\to M/IM\n \\]\n i.e.\n \\[\n \\bar{\\beta}: A/I \\otimes_A M \\to M/IM\n \\]\n that sends $\\bar{a} \\otimes m$ to $\\bar{am}$\n Ones check again that this inverse to $\\bar{\\alpha}$.\n \\footnote{\n For example\n \\(\n \\bar{\\beta}\\left(\\bar{\\alpha}\\left(\\bar{m}\\right)\\right) =\n \\bar{\\beta}\\left(\\bar{1} \\otimes m\\right) = \\overline{1 \\cdot m} = \\bar{m}\n \\)\n }\n \\end{proof}\n\\end{proposition}\n\nSeveral examples:\n\\begin{example}\n Let $\\mathbb{Z}/2\\mathbb{Z} \\otimes_\\mathbb{Z}\n \\mathbb{Z}/3\\mathbb{Z}$ what will we obtain?\n \\[\n \\mathbb{Z}/2\\mathbb{Z} \\otimes_\\mathbb{Z} \\cong\n ^{\\mathbb{Z}/3\\mathbb{Z}}/_{(2) \\cdot \\mathbb{Z}/3\\mathbb{Z}}\n \\]\n but 2 is invertible: $2^{-1} = -1 \\mod 3$\n \\footnote{\n I.e. there exist an invertible element $2^{-1} \\in\n \\mathbb{Z}/3\\mathbb{Z} = \\mathbb{F}_3$ therefore\n $1_{\\mathbb{F}_3} \\in 2 \\cdot \\mathbb{F}_3$ or\n $2 \\cdot \\mathbb{F}_3 = \\mathbb{F}_3$ and as result\n $(2) \\cdot \\mathbb{F}_3 = \\mathbb{F}_3$. I.e. $(2)$ is not a\n \\mynameref{def:properideal} in $\\mathbb{F}_3$ because it is equal\n to $\\mathbb{F}_3$.\n }\n thus\n $(2)\\mathbb{Z}/3\\mathbb{Z} =\\mathbb{Z}/3\\mathbb{Z}$ and as result\n \\[\n \\mathbb{Z}/2\\mathbb{Z} \\otimes_\\mathbb{Z} \\cong\n ^{\\mathbb{Z}/3\\mathbb{Z}}/_{ \\mathbb{Z}/3\\mathbb{Z}} = 0\n \\]\n\\end{example}\n\n\\begin{example}\n Another obvious example\n \\footnote {\n $B \\otimes_A A\\left[X\\right]$ has the following $B$-basis:\n $\\{1_B \\otimes X^i\\}$ thus\n $\\forall b \\in B \\otimes_A A\\left[X\\right]$ we can get\n \\[\n b = \\sum b_i \\cdot 1_B \\otimes X^i\n \\]\n and there is an obvious isomorphism\n \\[\n f: B \\otimes_A A\\left[X\\right]\n \\xrightarrow[b \\to \\sum b_i X^i]{}\n B\\left[X\\right]\n \\]\n }\n \\[\n B \\otimes_A A\\left[X\\right] \\cong B\\left[X\\right]\n \\]\n and more interesting one\n \\[\n B \\otimes_A A\\left[X\\right]/(P) \\cong B\\left[X\\right]/(P),\n \\]\n there $(P)$ becomes an ideal generated by $P$ in\n $B\\left[X\\right]$.\n \\label{ex:lec4_1}\n\\end{example}\n\n\\subsection{Tensor product of A-algebras}\n\nLet $B, C$ are $A$-algebras. The following maps form an algebra\nstructure on $A$:\n\\[\n\\alpha: A \\to B\n\\]\n\\[\n\\beta: A \\to C\n\\]\nNew $A$-algebra $B \\otimes_A C$: is a ring with respect to the\nfollowing operation\n\\footnote{\n that makes it $A$-algebra (see \\mynameref{def:kalgebra})\n}\n\\begin{equation}\n\\left(b \\otimes c \\right) \\cdot \\left(b' \\otimes c'\\right) =\n\\left(b \\cdot b'\\right) \\otimes \\left(c \\cdot c'\\right)\n\\label{eq:lec4_algebra_tensor_product}\n\\end{equation}\n\nThe tensor product has the following \n\\begin{definition}[Universal property]\n Let we have the following maps\n \\begin{eqnarray}\n \\alpha: A \\to B,\n \\nonumber \\\\\n \\beta: A \\to C,\n \\nonumber \\\\\n \\phi: B \\xrightarrow[b \\to b \\otimes 1_C ]{} B \\otimes_A C,\n \\nonumber \\\\\n \\psi: C \\xrightarrow[c \\to 1_B \\otimes c ]{} B \\otimes_A C\n \\nonumber\n \\end{eqnarray}\n\n Then for any $A$-algebra $D$ one has\n \\[\n Hom_A\\left(B \\otimes_A C, D\\right)\n \\leftrightarrow\n Hom_A\\left(B, D\\right) \\times\n Hom_A\\left(C, D\\right)\n \\]\n i.e. if I have some \\mynameref{def:homomorphism}\n $h \\in Hom_A\\left(B \\otimes_A C, D\\right)$ this is the same as\n giving 2 homomorphisms\n $f \\in Hom_A\\left(B, D\\right)$ and\n $g \\in Hom_A\\left(C, D\\right)$ such that all maps in the following\n diagram commute (see \\mynameref{def:commutativediagram}).\n \n \\begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]\n \\matrix (m) [matrix of math nodes, row sep=3em,\n column sep=2.5em, text height=1.5ex, text depth=0.25ex]\n { D & & B \\otimes C & \\\\\n & B & & C & \\\\\n & & A & & \\\\\n };\n %\\draw[double,double distance=5pt] (m-1-1) – (m-1-3);\n \\path[->]\n (m-2-2) edge node[description] {$ \\phi $} (m-1-3)\n (m-2-4) edge node[description] {$ \\psi $} (m-1-3)\n (m-3-3) edge node[description] {$ \\alpha $} (m-2-2)\n (m-3-3) edge node[description] {$ \\beta $} (m-2-4)\n (m-1-3) edge [bend right=15] node[description] {$ h $}\n (m-1-1);\n \\path[->, dashed]\n (m-2-2) edge [bend left=15] node[description] {$ f $} (m-1-1)\n (m-2-4) edge [bend left=15] node[description] {$ g $} (m-1-1)\n ;\n \\end{tikzpicture}\n\n Thus if we have $h$ then we can define\n $f = h \\cdot \\phi$ and $g = h \\cdot \\psi$. And conversely if I have\n $f$ and $g$ then I can define $h$ by the following rule:\n \\[\n h\\left(b \\otimes c\\right) = f(b) \\cdot g(c)\n \\]\n\\end{definition}\nThe main point for us is that the tensor product of the $A$-algebras is\nitself an $A$-algebra by this very simple rule, component-wise\nmultiplication (see (\\ref{eq:lec4_algebra_tensor_product})). \n\nLet consider next example.\nWe will start with the following\n\\[\n\\mathbb{C} \\cong {\\mathbb{R}\\left[X\\right]}/{\\left(X^2 + 1\\right)}\n\\]\ntherefore with result from example \\ref{ex:lec4_1} one can get\n\\[\n\\mathbb{C} \\otimes_\\mathbb{R} \\mathbb{C}\n\\cong\n\\mathbb{C} \\otimes_\\mathbb{R}\n{\\mathbb{R}\\left[X\\right]}/{\\left(X^2 + 1\\right)}\n\\cong\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X^2 + 1\\right)}\n\\]\nbut by \\mynameref{thm:chineseremainder} \n\\[\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X^2 + 1\\right)}\n\\cong\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X + i\\right)}\n\\times\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X - i\\right)}\n\\cong\n\\mathbb{C} \\times \\mathbb{C}\n\\]\nAs result we have that\n$\\mathbb{C} \\otimes_\\mathbb{R} \\mathbb{C}$ is not a field\nbecause it\nhas zero divisors. How we can get the zero divisors? \nThe element $\\overline{X + i}$ is a zero divisor\nin\n\\(\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X^2 + 1\\right)}\n\\)\nbecause\n\\[\n\\left(X + i\\right) \\left(X - i\\right) \\equiv 0 \\mod {\\left(X^2 + 1\\right)}\n\\].\n\nAnother proof (not a part of the lecture) of the fact that $\\mathbb{C}\n\\otimes_\\mathbb{R} \\mathbb{C}$ is not a field consider the following\none\n\\[\n\\mathbb{C}\n\\otimes_\\mathbb{R} \\mathbb{C}\n\\cong\n^{\\mathbb{C}\\left[X\\right]}/_{\\left(X^2 + 1\\right)}\n\\]\nbut the polynomial $X^2 + 1 = \\left(X + i\\right)\\left(X - i \\right)$\nis reducible in $\\mathbb{C}\\left[X\\right]$ i.e. is not a\n\\mynameref{def:maxideal} (see theorem \\ref{thm:irreducibleideal}) and\nwith the theorem \\ref{thm:maxideal} the quotient by the polynomial is\nnot a field (see also claim \\ref{claim:lec1_sec14}). \n\n\n\\section{Relatively prime ideals. Chinese remainder theorem}\n\n\\begin{definition}[Relatively prime ideals]\n Let $A$ - \\mynameref{def:ring} and $I,J$ are \\mynameref{def:ideal}s.\n $I$ and $J$ are relatively prime if $I + J = A$.\n \\label{def:relprimeideals}\n\\end{definition}\n\n\\begin{lemma}\n \\begin{enumerate}\n \\item If $I,J$ are relatively prime then $IJ = I \\cap J$\n \\item If $I_1, \\dots, I_k$ relatively prime with $J$ then\n $\\prod_{i =1}^k I_i = I_1 \\dots I_k$ is also\n relatively prime with $J$.\n \\item If $I, J$ relatively prime then $I^k$ and $J^l$ are also\n relatively prime for any $l$ and $k$.\n \\end{enumerate}\n \\begin{proof}\n \\begin{enumerate}\n \\item\n The following one $IJ \\subset I \\cap J$ is clear\n \\footnote {\n Assuming that $I$ and $J$ commute we have if\n $x \\in IJ$ then $x \\in I$ and if\n $x \\in JI$ then $x \\in J$ i.e. $x \\in I \\cap J$.\n }\n If $I$ and $J$ are relatively prime then $1_A = i + j$ for some\n $i \\in I$ and $j \\in J$. Thus $\\forall x \\in I \\cap J$ we have the\n following ones:\n $x i \\in IJ$ and $x j \\in IJ$ and as result\n \\[\n x = x i + x j \\in IJ\n \\]\n i.e. $I \\cap J \\subset IJ$.\n \\item Suppose for simplicity that $k = 2$. In the case we have\n $1_A = i_1 + j_1 = i_2 + j_2$ where $i_1 \\in I_1, i_2 \\in I_2$ and\n $j_1, j_2 \\in J$.\n we also have\n \\[\n 1_A = \\left(i_1 + j_1\\right) \\left(i_2 + j_2\\right) =\n i_1 i_2 + \\left(j_1 i_2 + j_2 i_1 + j_1 j_2\\right)\n \\in I_1 I_2 + J\n \\]\n thus $\\forall x \\in A$ we have\n \\[\n x = 1_A x =\n i_1 i_2 x + \\left(j_1 i_2 + j_2 i_1 + j_1 j_2\\right) x\n \\in I_1 I_2 + J,\n \\]\n i.e. $I_1 I_2 + J = A$ therefore $I_1 I_2$ and $J$ are\n relatively prime. \n \\item is obvious\n \\footnote{\n It follows from the 2 because we can assume $I_i = I$ and will\n get that $\\forall k, I^k$ is relatively prime with $J$. From\n other side we can assume $I_i = J$ and $J = I^k$ and conclude\n that $J^l$ is relatively prime with $I^k$. \n }\n \\end{enumerate}\n \\end{proof}\n \\label{lem:aboutrelprimeideals}\n\\end{lemma}\n\n\\begin{theorem}[Chinese remainder]\n \\label{thm:chineseremainder}\n \n Let $I_1, \\dots, I_n$ - ideals and map\n $\\pi: A \\to A/I_1 \\times \\dots \\times A/I_n$ defined as follows\n \\begin{equation}\n \\pi(a) = \\left(a \\mod I_1, \\dots, a \\mod I_n\\right)\n \\label{eq:lec4_crt_map}\n \\end{equation}\n The kernel $\\ker \\pi = I_1 \\cap \\dots \\cap I_n$.\n\n The $\\pi$ is \\mynameref{def:surjection} if and only if $I_1, \\dots,\n I_n$ are pairwise relatively prime.\n\n In that case\n \\[\n A/\\cap I_k\n \\cong\n A/\\prod I_k\n \\cong\n \\prod \\left(A/I_k\\right)\n \\]\n \\footnote{\n See \\mynameref{thm:firstisomorphism} where \n $G = A$,\n $H = A/I_1 \\times \\dots \\times A/I_n$, $\\phi = \\pi$ and with\n lemma \\ref{lem:aboutrelprimeideals} (as soon as $I_k$ pairwise\n relatively prime)\n \\[\n \\ker \\phi = \\ker \\pi = I_1 \\cap \\dots \\cap I_n = I_1 \\dots I_n.\n \\]\n }\n \\begin{proof}\n Let $\\pi$ is \\mynameref{def:surjection}. In the case\n $\\exists a_i \\in A$ such that\n \\[\n \\pi(a_i) = \\left(0, \\dots, 1 (\\mbox{ in }i\\mbox{-th place }),\n 0, \\dots, 0\\right)\n \\]\n i.e.\n $a_i \\mod I_j = 0$ or $a_i \\in I_j$ for $i \\ne j$. We also have\n $a_i \\mod I_i = 1$ thus $1 - a_i = k I_i$ i.e. $1 - a_i \\in I_i$. \n Thus $\\forall j, \\exists a_i \\in I_j, a_k \\in I_i$ such that\n $1 = a_i + a_k$ thus $A = I_j + I_i$ i.e. $I_i$ relatively prime\n with any $I_j$.\n\n Conversely if $I_i$ is relatively prime with any $I_j$ where $j\n \\ne i$ then it also relatively prime with the product (see lemma\n \\ref{lem:aboutrelprimeideals}) $\\prod_{j \\ne i} I_j$. In the case\n $\\exists x_i \\in I_i, y_i \\in \\prod_{j \\ne i} I_j$ such that\n $1 = x_i + y_i$ in the case\n \\[\n \\pi(y_i) = \\left(0, \\dots, 1 (\\mbox{ in }i\\mbox{-th place }),\n 0, \\dots, 0\\right)\n \\]\n and $\\forall b_i \\in A/I_i$\n \\[\n \\pi \\left(\\sum_{i = 1}^n b_i y_i\\right) =\n \\left(b_1, \\dots, b_n\\right)\n \\]\n i.e. $\\pi$ is surjective.\n \\end{proof}\n\\end{theorem}\n\nLet $K$ is a field and $A$ is a finite (finite dimensional vector\nspace) $K$-algebra.\n\n\\begin{proposition}\n \\begin{enumerate}\n \\item If $A$ is an \\mynameref{def:integraldomain} then $A$ is a\n field.\n \\item (replacing the first one) Any\n \\mynameref{def:primeideal} of $A$ is a \\mynameref{def:maxideal}\n \\end{enumerate}\n \\begin{proof}\n Well, I shall prove only the first part, the second part is just\n a consequence of definitions. In fact, a factor over a prime\n ideal, a quotient over a prime ideal is an integral domain, and a\n quotient over a maximal ideal is a field.\n \\footnote{\n i.e. prime ideal is a maximal ideal \n }\n If you don't know this,\n please look it up in any book.\n\n Lets prove the first part. \\mynameref{def:integraldomain} means\n that there is no zero divisors i.e. $\\forall a \\in A$\n \\footnote{\n $a \\ne 0_A$\n }\n multiplication by $a$ is \\mynameref{def:injection}. $A$ is finite\n dimensional \\mynameref{def:vectorspace} (see above) that implies\n that $\\times a$ is an \n \\mynameref{def:isomorphism},\n \\footnote{\n $\\times a$ sends a vector space into another vector space with\n the same dimension. But with \\mynameref{lem:vsisomorphism} one\n can get that the spaces are isomorphic each others and as\n result the operation $\\times a$ is an \\mynameref{def:isomorphism}.\n }\n in particular \\mynameref{def:surjection} i.e. $\\exists b \\in A$\n such that $b \\times a = 1$ i.e. $a$ is invertible therefore\n $A$ is field. \n \\end{proof}\n \\label{prop:lec4_ideals}\n\\end{proposition}\n\n\\section{Structure of finite algebras over a field. Examples}\n\n\\begin{myremark}\n Let $A$ is a $K$-algebra and $m$ is a maximal ideal of $A$. Then\n $A/m$ is also $K$-algebra.\n \\label{rem:lec5_quotientkalgebra}\n \\begin{proof}\n Lemma \\ref{lem:lec1_homkalgebra} says that there is a ring homomorphism\n between $K$ and $A$: $f_K: K \\to A$. There is also a canonical\n homomorphism $f_A: A \\xrightarrow[a \\to \\bar{a}]{} A/m$. Therefore\n we can define $f = f_A f_K: K \\to A/m$ and therefore with\n lemma about \\mynameref{lem:lec1_homkalgebra} can conclude that $A/m$ is a\n $K$-algebra.\n \\footnote{\n Thanks Zonglin Jiang for the proof.\n }\n\n Note that if $A$ is a field then $f_A$ is injection, $m=\\{0_K\\}$\n and $A = A/m$. \n \\end{proof}\n\\end{myremark}\n\n\\begin{theorem}[Structure of finite $K$-algebra]\n Let $A$ be a finite $K$-algebra i.e.\n $\\dim_K A < \\infty$. Then\n \\begin{enumerate}\n \\item There are only finitely many \\mynameref{def:maxideal}s\n $m_1, \\dots, m_r$ in $A$\n \\item Let $J = m_1 \\cap \\dots \\cap m_r = m_1 \\dots m_r$.\n \\footnote{\n Since the ideals are relatively prime the intersection is the\n same as the product of the ideals\n }\n Then\n $J^n = 0$ for some $n$\n \\item\n $A \\cong A/{m_1^{n_1}} \\times \\dots \\times A/{m_r^{n_r}}$ for\n some $n_1, \\dots, n_r$. \n \\end{enumerate}\n \\label{thm:structurefinitekalgebra}\n \\begin{proof}\n \\begin{enumerate}\n \\item\n Let $m_1, \\dots, m_i$ are maximal ideals. By\n \\mynameref{thm:chineseremainder} we have\n \\footnote{\n Maximal ideals are relatively prime because\n in a commutative ring with unity, every\n \\mynameref{def:maxideal} is a\n \\mynameref{def:primeideal} see also proposition\n \\ref{prop:lec4_ideals}. \n }\n \\[\n A/m_1 \\dots m_i \\cong\n A/m_1 \\times \\dots \\times A/m_i.\n \\]\n We know that $A$ as well as $A/m_1 \\dots m_i$ and $A/m_k$ are\n finite dimensional $K$-\\mynameref{def:vectorspace}\n \\footnote {\n See remark \\ref{rem:lec5_quotientkalgebra} and take into\n consideration \n the theorem statement (see above) that says \n $\\dim_K A < \\infty$. $A/m_1$ is a projection i.e.\n $\\dim_K A/m_1 < \\dim_K A < \\infty$.\n }\n thus we have the\n following relations\n \\[\n \\dim_K A \\ge \\dim_K A/m_1 \\dots m_i =\n \\sum_{j=1}^i dim_K A/m_j \\ge i.\n \\]\n Therefore if $N$ the number of maximal ideals then\n $\\dim_K A \\ge N$ i.e. the number of maximal ideal is limited by\n the vector space dimension.\n \\item\n $J = m_1 \\cap \\dots \\cap m_r = m_1 \\dots m_r$ is finite\n dimensional vector space over $K$\n \\footnote{\n We have $m_i \\subset A$ i.e. $\\dim_K m_i \\le \\dim_K A <\n \\infty$. From other side $J = m_1 \\dots m_r = (m_1 \\cap \\dots \\cap\n m_r) \\subset m_i, \\forall i \\in \\{1, \\dots, r\\}$.\n Thus $\\dim_K J \\le \\dim_K m_i \\le \\dim_K A <\n \\infty\\le $\n }\n as well as its powers\n $J^k$.\n We have the following sequence\n \\footnote {\n Let $j \\in J \\subset A$ and $x \\in JJ$.\n Then $\\exists j \\in J$ such that $x = j j$\n but $j j \\in J$ because $\\forall y \\in J: j y \\in J$.\n As result $J^2 \\subseteq J$.\n }\n \\[\n \\dots \\subseteq J^k \\subseteq \\dots \\subseteq\n J^2 \\subseteq J.\n \\]\n and the sequence should stop somewhere\n \\footnote {\n On each step we should decrease the dimension if we don't\n stop. The dimension is limited and as result the sequence\n should stop.\n }\n i.e.\n $\\exists n$ such that $J^n = J^{n+1}$. We claim that $J^n = 0$\n in the case. Indeed if not we have the following basis of $J^n$:\n $e_1, \\dots, e_s$. And as soon as $J^n = J J^n$ we can write a\n vector $e_i \\in J^n$ as a vector from $J^n$ multiplied on an\n object from $J$ i.e.\n \\[\n e_i = \\sum \\lambda_{ij} e_j,\n \\]\n there $e_j \\in J^n, \\lambda_{ij} \\in J$. Thus if\n $M = id - \\lambda_{ij}$\n \\[\n M \\cdot \\left(\n \\begin{array}{c}\n e_1 \\\\\n \\vdots \\\\\n e_s\n \\end{array}\n \\right) = 0.\n \\]\n It's possible over ring\n to find a matrix $\\tilde{M}$ such that\n \\(\n \\tilde{M} M = \\det M \\cdot id,\n \\)\n \\footnote{\n If $M$ is a matrix over a field we can consider the following\n 2 cases:\n \\begin{enumerate}\n \\item $\\det M = 0$: we can assume that $\\tilde{M} = 0$\n \\item $\\det M \\ne 0$: at the case $\\exists M^{-1}$ and\n therefore $\\tilde{M} = \n \\det M \\cdot M^{-1}$ will work.\n \\end{enumerate}\n For more common case we have to look at Adjugate matrix\n \\cite{wiki:adjugatematrix} \n } \n i.e.\n \\begin{equation}\n \\det M \\cdot \\left(\n \\begin{array}{c}\n e_1 \\\\\n \\vdots \\\\\n e_s\n \\end{array}\n \\right) = 0.\n \\label{eq:lec4_det}\n \\end{equation}\n But $\\det M = 1 + \\lambda$ where $\\lambda \\in J$.\n \\footnote{\n Because the $\\det$ consists of the following \n items $\\prod (1-\\lambda_{ii}) = 1 + (-1)^s\\prod \\lambda_{ii}$\n and $\\prod \\lambda_{ij}$. The sum of the items ($\\det$)\n consists of $1$ and another sum in which all items are from\n $J$. Thus the second sum is an element of $J$ i.e.\n $\\det M = 1 + \\sum \\prod \\lambda_{ij} = 1 + \\lambda$.\n }\n Since $J = m_1 \\cap \\dots \\cap m_r$ then\n $\\forall i: \\lambda \\in m_i$\n so $\\nexists i$ such that $1 + \\lambda \\in m_i$\n \\footnote{\n We have that $m_i$ is a \\mynameref{def:maxideal} and therefore (by\n its definition) it is a \\mynameref{def:properideal} i.e.\n $1 \\notin m_i$. From other side if $1 + \\lambda \\in m_i$ then\n $1 + \\lambda - \\lambda \\in m_i$ as soon as $\\lambda \\in\n m_i$. I.e. we have a contradiction. \n }\n thus $1 + \\lambda$ is invertable\n \\footnote {\n $0 \\in m_i$ thus if $1 + \\lambda \\notin m_i$ then\n $1 + \\lambda \\ne 0$.\n }\n therefore $e_1 = \\dots = e_s = 0$\n \\footnote{\n Because $\\det M = 1 + \\lambda \\ne 0$\n }\n \\item Using part 2\n $\\exists n_1, \\dots, n_r$ such that\n $m_1^{n_1} \\dots m_r^{n_r} = 0$ (for example we can assume\n $n_i = n$). Then by \\mynameref{thm:chineseremainder} \n \\[\n A \\cong\n A/{m_1^{n_1}} \\times \\dots \\times A/{m_r^{n_r}}.\n \\]\n We used the following facts:\n \\begin{itemize}\n \\item $A = A /m_1^{n_1} \\dots m_r^{n_r}$\n \\footnote{\n Because $A = A/\\{0\\}$. For example if $I = \\{0\\}$ and\n $x \\in A$ then $\\bar{x} \\in A/I$ if $\\bar{x} = x + I$. In\n our case $\\bar{x} = x + \\{0\\} = x$ i.e. $\\forall x \\in A$ we\n have $x \\in A/\\{0\\}$.\n (See also \\mynameref{def:quotientring})\n }\n \\item $m_i^{n_i}$ are pairwise relatively prime\n \\footnote{\n As soon as $\\{m_i\\}$ - \\mynameref{def:maxideal}s and as result\n \\mynameref{def:primeideal}s then with lemma\n \\ref{lem:aboutrelprimeideals} one can get that\n $\\forall i \\ne j$ $m_i^{n_i}$ is relatively prime with\n $m_j^{n_j}$.\n }\n \\end{itemize}\n \\end{enumerate}\n \\end{proof}\n\\end{theorem}\n\n\\begin{remark}\n The $n_i$s are not uniquely defined. For example (see also example\n \\ref{ex:lec5_1}) \n \\[\n A = ^{K\\left[X\\right]}/_{\\left(X^2 \\left(X+1\\right)^3\\right)}.\n \\]\n We have 2 ideals there:\n $m_1 = (X)$ and $m_2 = (X+1)$.\n We of course have\n \\[\n A \\cong A/m_1^2 \\times A/m_2^3\n \\]\n but also we have\n \\[\n A \\cong A/m_1^3 \\times A/m_2^3\n \\]\n as soon as $m_1^2 = m_1^3$ in $A$:\n $(X)^2 \\subset (X)^3$ but also\n $(X)^3 \\subset (X)^2$\n \\footnote{\n $(X)^3 \\subset (X)^2$ - this is true for any polynomial\n $P\\left(X\\right)$ because if\n $P\\left(X\\right) \\in (X)^3$ then\n $P\\left(X\\right) = X^3 P'\\left(X\\right) =\n X^2 \\bar{P}\\left(X\\right) \\in (X)^2$ where\n $\\bar{P}\\left(X\\right) = X P'\\left(X\\right)$\n\n $(X)^2 \\subset (X)^3$ is the more complex one and it does not true\n for any polynomial ring but this is true for our $A$.\n Let $P\\left(X\\right) \\in \\left(X\\right)^2 \\subset\n K\\left[X\\right]$ then\n $P\\left(X\\right) = X^2 P'\\left(X\\right)$ but\n $X^2 P'\\left(X\\right) \\equiv 0 \\mod X^2$.\n From other side\n $X^4 P'\\left(X\\right) \\equiv 0 \\mod X^2$\n therefore\n \\[\n P\\left(X\\right) \\equiv X^3 X P'\\left(X\\right) \\mod X^2\n \\]\n i.e. $P\\left(X\\right) \\in (X)^3$.\n }\n \\label{rem:lec5_4}\n\\end{remark}\n\nSeveral examples:\n\\[\n\\mathbb{C} \\otimes_\\mathbb{R} \\mathbb{C} =\n\\mathbb{C} \\times \\mathbb{C}.\n\\]\n\nAnother example\n\\[\n\\mathbb{Q}\\left(\\sqrt{2}\\right)\n\\otimes_{\\mathbb{Q}}\n\\mathbb{Q}\\left(\\sqrt{3}\\right) =\n\\mathbb{Q}\\left(\\sqrt{2}, \\sqrt{3}\\right)\n\\]\nAnd you see that those algebras are Cartesian products of fields.\nSo all $n_i$'s may be taken equal to 1\n\\footnote{\n Because $A/m^n$ (where $m$ is a maximal ideal) is a field if $n=1$\n}.\nIn other words, we don't have\n\\mynameref{def:nilpotent}s in our algebra\n\\footnote{\n Cartesian products of fields has no nilpotent elements except 0\n \\cite{ribenboim2001classical}.\n}. So, it is a reduced\nalgebras. Reduced, by definition, is without nilpotents. \nIt's general phenomena because the presence of nilpotents is due to the\ninseparability of extensions come from inseparable extensions. \n", "meta": {"hexsha": "59ac8ba904bcaf3fe2532ff5a6ab1e6c058d0795", "size": 48575, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lecture4.tex", "max_stars_repo_name": "JiuziLau/courseragalois", "max_stars_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2016-06-21T07:34:33.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-23T00:14:44.000Z", "max_issues_repo_path": "lecture4.tex", "max_issues_repo_name": "JiuziLau/courseragalois", "max_issues_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2016-06-28T21:26:22.000Z", "max_issues_repo_issues_event_max_datetime": "2016-06-30T05:36:37.000Z", "max_forks_repo_path": "lecture4.tex", "max_forks_repo_name": "JiuziLau/courseragalois", "max_forks_repo_head_hexsha": "06423d7609caf8e083fe4c5a442ec01ea27018ef", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-08-08T07:47:20.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-25T06:16:07.000Z", "avg_line_length": 34.8207885305, "max_line_length": 80, "alphanum_fraction": 0.5995470921, "num_tokens": 17747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.828938806208442, "lm_q1q2_score": 0.6135592520547173}} {"text": "% LaTeX file for a 1 page document\n\\documentclass[12pt]{article}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{physymb}\n\\usepackage{braket}\n\\usepackage{array} \n\\title{EP 307 Assignment 4}\n\n\\author{Manish Goregaokar\\\\120260006}\n\n\\begin{document}\n\\maketitle\n\n\\section*{Problem 1}\n$$\\prod_{i\\neq s}\\frac{\\hat A-a_i}{a_s-a_i}$$\n\nIf $\\Ket{a}=\\sum c_i\\Ket{a_i}$, applying this operator we have \\begin{align*}\n\\prod_{i\\neq s}\\frac{\\hat A-a_i}{a_s-a_i}\\left(\\sum c_i\\Ket{a_i}\\right)&=\\sum_j\\left(\\prod_{i\\neq s}\\frac{\\hat A-a_i}{a_s-a_i}\\right)c_j\\Ket{a_j}\\\\\n&=\\left(\\prod_{i\\neq s}\\frac{\\hat A-a_i}{a_s-a_i}\\right)c_s\\Ket{a_s}\\\\&\\qquad\\text{Since the other terms have numerator cancelling out for $i=j$}\\\\\n&=\\left(\\prod_{i\\neq s}\\frac{a_s-a_i}{a_s-a_i}\\right)c_s\\Ket{a_s}\\\\\n&=c_s\\Ket{a_s}\n\\end{align*}\nThus the operator is $\\frac{\\Ket{a_s}\\Bra{a_s}}{\\Braket{a_s|a_s}}=p_s$ (projection operator)\n\\section*{Problem 2}\n\\newcommand{\\Tr}{\\mathrm{Tr}}\nTo prove:\n$$\\Tr(XY)=\\Tr(YX)$$\n\nThe trace of an operator can be denoted by $\\sum_i \\Braket{i|A|i}$, where $\\Bra{i},\\Ket{i}$ are basis vectors corresponding to the $i$th element being $1$ and the rest $0$. \n\nThus,\n\n\\begin{align*}\n\\Tr(AB) \n&= \\sum_i \\Braket{i|AB|i}\\\\\n&= \\sum_i \\Braket{i|AIB|i}\\\\\n&= \\sum_i\\sum_j \\Braket{i|A\\Ket{j}\\Bra{j}B|i}\\\\\n&= \\sum_i\\sum_j \\Braket{j|A\\Ket{i}\\Bra{i}B|j}\\\\\n&= \\sum_j \\Braket{j|BIA|j}\\\\\n&= \\Tr(BA)\n\\end{align*}\n\n\\section*{Problem 7}\n\nApplying $\\hat{A}=\\hat{Q}\\hat{C}+\\hat{C}\\hat{Q}$ to an eigenstate $\\Ket{\\psi_q}$, we get:\n\n\\begin{align*}\n\\hat{A}\\Ket{\\psi_q} &=(\\hat{Q}\\hat{C}+\\hat{C}\\hat{Q})\\Ket{\\psi_q}\\\\\n&= \\hat{Q}\\hat{C}\\Ket{\\psi_q}+\\hat{C}\\hat{Q}\\Ket{\\psi_q} \\\\\n&= \\hat{Q}\\Ket{\\psi_{-q}}+q\\hat{C}\\Ket{\\psi_q}\\\\\n&= -q\\Ket{\\psi_{-q}}+q\\Ket{\\psi_{-q}}\\\\\n&=0\n\\end{align*}\n\nIf $\\hat A\\Ket{\\psi_q}=0$, then $\\hat A\\left(\\sum a_q\\Ket{\\psi_q}\\right)=0$. We can say that the eigenvalue of $\\hat A$ is 0.\n\nFor a state $\\Ket{\\psi_q}$ to be an eigenstate of $\\hat C$, $c\\Ket{\\psi_q}=\\hat C\\Ket{\\psi_q}=\\Ket{\\psi_{-q}}$\n\nSince $\\Ket{\\psi_{-q}},\\Ket{\\psi_q}$ have different eigenvalues (except when $q=0$), they are linearly independent, and thus the only possible value of $c$ is 0, which isn't an eigenstate.\n\nSo the only common eigenstate is $\\Ket{\\psi_0}$, provided that it is not a null vector.\n\\section*{Problem 9}\n\\newcommand{\\kpsi}{\\Ket{\\psi}}\n\\begin{align*}\n[\\hat x, \\exp\\left(\\frac{i\\hat p a}{\\hbar}\\right)]\\kpsi &= [\\hat x, \\exp\\left(\\frac{i(-i)\\hbar\\partial_x a}{\\hbar}\\right)]\\kpsi\\\\\n&=[\\hat x, \\exp(a\\partial_x)]\\kpsi\\\\\n&= x \\sum_i \\frac1{i!}a^i\\partial^i_x\\kpsi - \\sum_i\\frac1{i!} a^i\\partial^i_x (x\\kpsi) \\\\\n&= x \\sum_i \\frac1{i!}a^i\\partial^i_x\\kpsi - \\sum_i\\frac1{i!} a^i(x\\partial^i_x (\\kpsi) + \\partial_x^{i-1}i\\kpsi)\\\\\n&= - \\sum_i\\frac1{i!} a^i \\partial_x^{i-1}i\\kpsi\\\\\n&= - \\sum_i\\frac1{(i-1)!} a^i \\partial_x^{i-1}\\kpsi\\\\\n&= - a\\exp(a\\partial_x)\\Ket{\\psi}\\\\\n&= -a\\exp\\left(ia\\frac{\\hat p}{\\hbar}\\right)\\kpsi\\\\\n\\therefore [\\hat x, \\exp\\left(\\frac{i\\hat p a}{\\hbar}\\right)] &= -a\\exp\\left(ia\\frac{\\hat p}{\\hbar}\\right)\n\\end{align*}\n\n\\section*{Problem 11}\nThe first allowed state (ground state) in the new system will be the first odd wavefunction, i.e. when $n=1$. We need to calculate the probability that the current state ($n=0$) becomes $n=1$ in that region. Note that since the space is halved over a symmetric function, the wavefunctions will be normalized by an extra factor of $\\frac1{\\sqrt2}$\n\nSo, the overlap is \n\\begin{align*}\n\\int_0^\\infty \\psi_0(x)\\frac1{\\sqrt2}\\psi_1(x)\n&= \\int_0^\\infty \\left(\\frac{\\alpha}{\\sqrt{\\pi}}\\right)^{\\frac12}e^{-\\alpha^2x^2/2}H_0(\\alpha x)\\left(\\frac{\\alpha}{\\sqrt{\\pi}2}\\right)^\\frac12 e^{-\\alpha^2x^2/2}H_2(\\alpha x)\\\\\n&= \\frac{\\alpha}{\\sqrt\\pi\\sqrt2}\\cdot \\int_0^\\infty e^{-\\alpha^2x^2}\\cdot 2\\alpha x\\\\\n&=\\frac{2\\alpha^2}{\\sqrt{2\\pi}} \\int_0^\\infty e^{-\\alpha^2x^2}\\cdot x\\\\\n&=\\frac{2\\alpha^2}{\\sqrt{2\\pi}}\\frac{1}{2\\alpha^2}\n&=\\frac{1}{\\sqrt{2\\pi}}\n\\end{align*}\n\nThus the probability is $\\left(\\frac1{\\sqrt{2\\pi}}\\right)^2=\\boxed{\\frac{1}{2\\pi}}$\n\n\\section*{Problem 13}\n$$\\hat{A}=\\begin{pmatrix}a&0&0\\\\0&-a&0\\\\0&0&-a\\end{pmatrix},\\quad\\hat{B}=\\begin{pmatrix}b&0&0\\\\0&0&ib\\\\0&ib&0\\end{pmatrix}$$\n\nTo find eigenvectors of $\\hat B$, we can multiply it with the vector $X=\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}$. We get $\\hat BX=\\begin{pmatrix}bx\\\\ibz\\\\iby\\end{pmatrix}=cX$, so for an eigenvector we either have $c=b,ibz=y,iby=z$ (giving $y=z=0$ for a general $b$, and eigenvector $\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}$. If $c\\neq b$ we have $x=0$, and $y=\\pm z$ with eigenvalues $\\pm ib$\n\nTherefore the eigenvectors are:\n\n\\begin{enumerate}\n\\item $\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}$ or $\\Ket{1}$ with eigenvalue $b$\n\\item $\\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix}$ or $\\Ket{2}+\\Ket{3}$ with eigenvalue $ib$\n\\item $\\begin{pmatrix}0\\\\1\\\\-1\\end{pmatrix}$ or $\\Ket{2}-\\Ket{3}$ with eigenvalue $-ib$\n\\end{enumerate}\n\\hfill Ans. (a)\n\nNow,\n$$\\hat A\\hat B=\\left(\n\\begin{array}{ccc}\n a & 0 & 0 \\\\\n 0 & -a & 0 \\\\\n 0 & 0 & -a \\\\\n\\end{array}\n\\right) \\left(\n\\begin{array}{ccc}\n b & 0 & 0 \\\\\n 0 & 0 & i b \\\\\n 0 & i b & 0 \\\\\n\\end{array}\n\\right)=\\left(\n\\begin{array}{ccc}\n a b & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n\\end{array}\n\\right)$$\n\nand \n\n$$\\hat B\\hat A= \\left(\n\\begin{array}{ccc}\n b & 0 & 0 \\\\\n 0 & 0 & i b \\\\\n 0 & i b & 0 \\\\\n\\end{array}\n\\right)\\left(\n\\begin{array}{ccc}\n a & 0 & 0 \\\\\n 0 & -a & 0 \\\\\n 0 & 0 & -a \\\\\n\\end{array}\n\\right)=\\left(\n\\begin{array}{ccc}\n a b & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n 0 & 0 & 0 \\\\\n\\end{array}\n\\right)$$\n\nAs $\\hat A\\hat B=\\hat B\\hat A$ in matrix form, the operators commute.\n\\hfill Ans. (b)\n\nThe eigenkets of $\\hat A$ can be easily seen to be $\\Ket{1}$ with eigenvalue $a$, and $l\\Ket{2}+m\\Ket{3}$ with eigenvalue $-a$ ($\\forall~l,m$)\n\nWe can see that the eigenkets of $B$ are also eigenkets of $A$. They are also orthogonal, after normalizing we have:\\\\~\\\\\n\n\\begin{tabular}{ >{$}c<{$} >{$}c<{$} >{$}c<{$} }\n\\text{Orthonormal Eigenket} & \\text{Eigenvalue with }\\hat A & \\text{Eigenvalue with }\\hat B \\\\\n\\Ket{1} & b & a\\\\\n\\frac{\\Ket{2}+\\Ket{3}}{\\sqrt{2}} & ib & -a\\\\\n\\frac{\\Ket{2}-\\Ket{3}}{\\sqrt{2}} & -ib & -a\n\\end{tabular}\n\n\nThe eigenkets are not completely determined from the eigenvalues from individual eigenvalues, as the eigenvalue $-a$ has multiplicity 2 and thus has an entire space of eigenkets.\n\nHowever, knowing the eigenvalues from both operators completely specifies the eigenket, barring a constant.\n\\hfill Ans. (c)\n\\section*{Problem 16}\n\n$\\psi(x)=\\frac{1}{\\sqrt{2a}}$ in $[-a,a]$. The momentum space function can be found via the fourier transform, \\begin{align*}\n\\tilde{\\psi}(p)&=\\frac1{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty e^{-ikx}\\psi(x)dx\\\\\n&= \\frac1{\\sqrt{2\\pi}}\\int_{-a}^a\\frac{1}{\\sqrt{2a}} e^{-ikx}dx\\\\\n&=\\frac{i}{2k\\sqrt{a\\pi}} \\left.e^{-ikx}\\right|_{-a}^a\\\\\n&=\\frac{i}{2k\\sqrt{a\\pi}} \\left(e^{-ika}-e^{ika}\\right)\\\\\n&=\\frac{i}{2k\\sqrt{a\\pi}} \\cdot -2i\\sin ka\\\\\n\\therefore \\tilde{\\psi}(p)&=\\frac{\\sin \\left(\\frac{ap}{\\hbar} \\right)}{p\\sqrt{a\\pi}}\\qquad\\text{(Already normalized)}\n\\end{align*}\n\\hfill Ans.\n\nNow, \\begin{align*}\\sigma_x&=\\sqrt{\\Braket{x^2}-\\Braket{x}^2}\\\\\n&=\\sqrt{\\int \\psi^*(x)x^2\\psi(x)dx-0}\\\\\n&= \\sqrt{\\int_{-a}^a \\left(\\frac{1}{\\sqrt{2 a}}\\right)^2 x^2 dx}\\\\\n&= \\frac{a}{\\sqrt{3}}\n\\end{align*}\n\nAnd\n\n\\begin{align*}\\sigma_p&=\\sqrt{\\Braket{p^2}-\\Braket{p}^2}\\\\\n&=\\sqrt{\\int \\tilde{\\psi}^*(p)p^2\\tilde{\\psi}(p)dp-0}\\\\\n&= \\sqrt{\\int_{-\\infty}^\\infty \\left(\\frac{\\sin \\left(\\frac{a p}{\\hbar }\\right)}{\\sqrt{\\pi } \\sqrt{a} p}\\right)^2 p^2 dp - 0}\\qquad\\text{(The wavefunction is odd so the second term vanishes)}\\\\\n&=\\sqrt{\\infty}\n\\end{align*}\n\nThe product of the two uncertantainties is greater than $\\frac\\hbar2$\n\\section*{Problem 17}\n\n$$\\psi(x,t)=\\frac{1}{\\sqrt{2 \\pi \\hbar }}\\int_{-\\infty }^{\\infty } \\phi(p) e^{-\\frac{t \\left(i p^2\\right)}{2 m}+i p x} \\, dp$$\n\nNow, we can rewrite this as $\\frac{1}{\\sqrt{2 \\pi \\hbar }}\\int_{-\\infty }^{\\infty } e^{i p x} \\left(\\phi(p) e^{-\\frac{t \\left(i p^2\\right)}{2 m}}\\right) \\, dp$\n\nwhich is an inverse fourier transform.\n\nThus, $\\tilde{\\psi}(p,t)=\\mathcal F(\\psi(x,t))=\\phi(p) e^{-\\frac{t \\left(i p^2\\right)}{2 m}}$\\\\\nNow,\n\\begin{align*}\n\\Braket{p}&=\\int \\tilde{\\psi}^*(p,t)p\\tilde{\\psi}(p,t)dp\\\\\n&=\\int \\phi^*(p) e^{\\frac{t \\left(i p^2\\right)}{2 m}}p\\phi(p) e^{-\\frac{t \\left(i p^2\\right)}{2 m}}dp\\\\\n&= \\int \\phi^*(p) p \\phi(p) dp\n\\end{align*}\n\nThus, $\\Braket{\\hat p}$ is independant of time\\hfill Ans. (a)\n\nNote that $\\phi(p)=\\tilde{\\psi}(p,t=0)$\nNow, \n\n\\begin{align*}\n\\Braket{x}= \\int \\tilde{\\psi}^*(p,t)i\\hbar\\ud{}{p}\\tilde{\\psi}(p,t)dp\\\\\n&=\\int \\phi^*(p)e^{\\frac{t \\left(i p^2\\right)}{2 m}}i\\hbar\\left(\\pd{\\phi}{p}e^{-\\frac{t \\left(i p^2\\right)}{2 m}}+\\phi(p)\\frac{2pt}{2m}e^{-\\frac{t \\left(i p^2\\right)}{2 m}}\\right)dp\\\\\n&=\\int\\left( \\phi^*(p)i\\hbar\\pd{\\phi}{p} + \\frac{t}{m}\\phi^*(p)p\\phi(p)\\right) dp\\\\\n&= \\Braket{i\\hbar\\pd{}{p}}_{t=t_0} + \\frac{t}{m}\\Braket{p}_{t=t_0}\\\\\n\\therefore \\Braket{x}&=\\Braket{x}_{t=0} + \\frac{t}{m}\\Braket{p}_{t=0}\n\\end{align*}\n\nSince this is linear, we can shift by $t_0$ to get $\\boxed{\\Braket{x}=\\Braket{x}_{t=t_0} + \\frac{t-t_0}{m}\\Braket{p}_{t=t_0}}$\\hfill Ans. (b)\n\\section{Problem 19}\n$$\\hat H = \\frac{\\hat p^2}{2m}+V(x)$$\n\n\\begin{align*}\n[H,\\hat x]&=\\frac{\\hat p^2}{2m}x+V(x)x-x\\frac{\\hat p^2}{2m}+xV(x)\\\\\n&=-x\\frac{\\hat p^2}{2m}\\\\\n[[H,\\hat x],x]&=-x\\frac{\\hat p^2}{2m}x - -x^2\\frac{\\hat p^2}{2m}\\\\\n&=\\frac{x^2\\hat p^2}{2m}\n\\end{align*}\n\\end{document}\n", "meta": {"hexsha": "f0b8e47c2ec6b42540bd58fa08a70643ca2869e2", "size": 9354, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 4/assign4.tex", "max_stars_repo_name": "CourseResources/CourseResources", "max_stars_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-05-28T05:59:31.000Z", "max_stars_repo_stars_event_max_datetime": "2015-05-28T05:59:31.000Z", "max_issues_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 4/assign4.tex", "max_issues_repo_name": "CourseResources/CourseResources", "max_issues_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Course material/EP 307 - Quantum Mechanics/Assignment 4/assign4.tex", "max_forks_repo_name": "CourseResources/CourseResources", "max_forks_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.6355932203, "max_line_length": 384, "alphanum_fraction": 0.6173829378, "num_tokens": 4086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8289388040954684, "lm_q1q2_score": 0.6135592504907484}} {"text": "\\documentclass{article}\n\n\\usepackage{epsfig}\n\\usepackage[export]{adjustbox}% http://ctan.org/pkg/adjustbox\n\\usepackage{caption}\n\\usepackage{subcaption}\n\\usepackage{fullpage}\n\\usepackage{commath}\n\\usepackage{amssymb}\n\\usepackage[space]{grffile}\n\\usepackage{algorithm, algorithmic}\n\n\\def\\E{\\mathbb{E}}\n\\def\\R{\\mathbb{R}}\n\\def\\bin{\\text{bin}}\n\\def\\bet{\\text{beta}}\n\\def\\range{\\text{range}}\n\\def\\vec{\\text{vec}}\n\n\\newcommand\\hide[1]{}\n\n\\title{HMM with Contexts in DNA Methylation Problem}\n\n\\author{Chicheng Zhang}\n\n\\begin{document}\n\\maketitle\n\n\n\\section{The Model}\n\nWe are given sequences $\\cbr{m_t, c_t}$, where at each time $t$ the methylation count is $m_t \\in \\cbr{0,1,\\ldots,n}$, and the context information is $c_t \\in [c]$. We model the data using a hidden Markov model with context information, that is, the underlying dynamics of the hidden states is driven by a Markov chain $\\cbr{y_t}$, where $y_t \\in [m]$.\n\nIt can be described by parameters $(\\pi, T, O)$, where $\\pi$ is the initial probability distribution $\\Pr(y_1)$ and $T$ is the transition matrix of the Markov chain $\\Pr(y_{t+1} | y_t)$, $O$ is the observation matrix, namely the conditional probability of methylation given hidden state and context $\\Pr(m_t | c_t, y_t)$. See Figure~\\ref{fig:contexthmm} for illustration.\n\n\\begin{figure}[h]\n\\centering\n\\scalebox{0.3}\n{\n\\includegraphics{hmm.png}\n}\n\\caption{Modeling DNA methylation using a Hidden Markov model with context information.}\n\\label{fig:contexthmm}\n\\end{figure}\n\nOur objective is to recover parameters $\\pi$, $T$ and $O$ based on iid samples drawn from this model.\n\n\\section{Algorithm based on Method of Moments}\n\nFor each $i$, consider recovering a block of observation matrix $O^{2, i} = \\Pr(m_t | c_t = i, y_t)$ separately. We make the assumption that $O^{2, i}$ is of full column rank. Then, the joint cooccurence tensor of $(m_1, m_2, m_3)$, when the context of observation 2 $c_2$ is $i$, has the following structure:\n\\[ \\E [m_1 \\otimes m_2 \\otimes m_3 | c_2 = i] = \\sum_j \\omega_j O^{1}_j \\otimes O^{2, i}_j \\otimes O^{3}_j \\]\nwhere $\\omega$ is the distribution of $y_2$ given $c_2$, $O^{1} = \\E[m_1 | y_2]$, $O^{2,i} = \\E[m_2 | y_2, c_2 = i]$, $O^{3} = \\E[m_3 | y_2]$.\nAt this point, standard tensor decomposition method can readily be applied to extract the columns of $O^{2, i}$.\n\nAlternatively, we have\n\\[ \\E [m_1 \\otimes m_3] = \\sum_j \\omega_j O^1_j \\otimes O^3_j \\]\nand\n\\[ \\E [m_1 \\otimes m_2 | c_2 = i] = \\sum_j \\omega_j O^{1}_j \\otimes O^{2, i}_j \\]\nThen standard symmetrization and orthogonalization routine can be applied to recover the columns of $O^{2, i}$.\n\n\\section{Binomial Hidden Markov Model}\nIn this section, we consider HMMs whose observation matrix has structured columns, that is, columns which are probability mass functions of binomial distribution. We first take a step back, considering HMMs without context information. (The approach can be easily generalized to the setting with context information.)\n\nIn this model, each element of $O$ can be written as $O_{i,j} = {n \\choose i} p_j^i (1-p_j)^{n-i}$, i.e. the probability that there are $i$ heads out of $n$ trials when flipping a coin with bias $p_j$. Instead of recovering $O$, we now turn to the task of recovering $p_j, j = 1,2,\\ldots,m$. A naive approach would be to first recover the columns of $O$, e.g. $O_j$, then get $p_j$ using the formula:\n\n\\[ p_j = \\sum_{i=0}^n \\frac{i}{n} {n \\choose i} p_j^i (1-p_j)^{n-i} = \\sum_{i=0}^n \\frac{i}{n} O_{i,j} \\]\n\n\\subsection{Exploiting Similarity between Observations}\nSometimes it will be beneficial to consider feature representations of observation other than one-hot encoding. Define feature map $\\phi: \\cbr{0,\\ldots,n} \\to \\R^D$, where $D$ can potentially be very large, even (uncountably) infinite. For example, $\\phi(m) = \\cbr{t^m (1-t)^{n-m}}_{t \\in [0,1]}$, $\\phi(m) = \\cbr{t^m (1-t)^{n-m}}_{t \\in \\cbr{0,0.1, \\ldots, 1}}$, or $\\phi(m) = \\cbr{ \\frac{1}{\\sqrt{2\\pi}} \\exp(-(t-m)^2/2)}_{t \\in \\R}$, etc.\n\nThen, standard method of moments can be applied, resulting in the following representation:\n\\[ \\E [\\phi(m_1) \\otimes \\phi(m_2) \\otimes \\phi(m_3)] = \\sum_j \\omega_j O^{1}_j \\otimes O^{2}_j \\otimes O^{3}_j \\]\nwhere $\\omega$ is the marginal distribution of $y_2$, $O^{1} = \\E[\\phi(m_1) | y_2]$, $O^{2} = \\E[\\phi(m_2) | y_2]$, $O^{3} = \\E[\\phi(m_3) | y_2]$.\n\nNow, to decompose this $D \\times D \\times D$ tensor, dimensionality reduction tricks should be applied. After getting an estimate of $O^{2} = \\E[\\phi(m_2) | y_2]$, we can recover $p_j$'s via algebraic manipulations.\n\n\n\\hide\n{\n\\section{Experiment Results}\n\\subsection{Setting}\nIn this set of experiments, we focus on the methylation data in cell type E1 and chromosome 1. At this point we do not take into account the contexts, and we further assume that the coverage in different segments are non-uniform. Let $l$ be the length of the input sequence, $s$ be the number of base segments merged to a single segment, and $n$ be the number of hidden states assumed.\n\nAlthough the ultimate goal is to recover the transition matrix and observation matrix of the hidden Markov model, at the current stage, we recover $\\E[\\phi(x, y)|h]$ for each value of $h$. Recall that $x = (c, \\mu)$ is the (coverage, methylation) count pair in each segment. The value of $p_h$ for each $h$ can be straightforwardly extracted from $\\E[\\phi(x, y)|h]$. The transition matrix can also be computed subsequently. We leave the recovery of all the parameters for future work.\n\n\n%In subsequent results, we show the entire $\\E[\\phi(x,y)|h]$, as opposed to a single number $p_h \\in [0,1]$. The reason is that we wish to recover more information about the hidden states learned, to get a better understanding of the HMM model.\n\n%-- what features $\\phi$ are you looking at?\n%-- why are you showing the entire $\\phi$ vector and not a single number?\nWe focus on two types of feature mappings $\\phi$, the first one is binning mapping:\n\\[ \\phi_{\\bin}(x,y) = \\begin{cases} {I(\\frac{\\mu}{c} \\in (y,y+h])} & c \\neq 0 \\\\ 0 & c = 0\\end{cases} \\]\nwhere $h$ is the width of the bins.\n\nThe second one is Beta map:\n\\[ \\phi(x,y) = \\frac{1}{B(\\mu+1, c-\\mu+1)}y^\\mu (1-y)^{c-\\mu} \\]\nIn this mapping there is no explicit notion of ``bin width\", but the value of $c$ can be thought of as a parameter controlling it: if the coverage $c$ is large, then the bin width is small, and vice versa.\n\nIn experiments, since it is impossible to write down $\\E[\\phi(x_1, \\cdot) \\otimes \\phi(x_2, \\cdot) \\otimes \\phi(x_3, \\cdot)]$ and perform tensor decomposition, we use discretization: compute an empirical version of $\\E[\\phi(x_1, y_1) \\otimes \\phi(x_2, y_2) \\otimes \\phi(x_3, y_3)]$, where $y_1, y_2, y_3$ is in $\\cbr{0,h,2h,\\ldots,1}$. By decomposing this tensor we are able to recover $\\E[\\phi(x, y) | h]$, where $y$ is in $\\cbr{0,h,2h,\\ldots,1}$. By mild regularity assumption on the distribution of $x$, this can give a fairly accurate estimate of $\\E[\\phi(x, y) | h]$ where $y$ ranges in the $[0,1]$ interval.\n\n\n\\subsection{Effect of Sample Size $l$}\nWe fix $m = 4$, $s = 1$, and vary the sample size (number of triples collected) $l =$ 1000, 10000, 20000, 40000, 80000, 160000, 320000. The samples are drawn without replacement from the original dataset. Figure~\\ref{fig:varyl} shows the result. Note that the value of $m$ is the number of hidden states assumed; for example if $m = 4$, then we get $4$ columns of $\\E[\\phi(x,y)|h]$. We observe that the output of the algorithm stablizes when $l \\geq 80000$.\n\n\\begin{figure}[H]\n\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 10000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 10000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 20000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 20000$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 40000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 40000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 80000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 80000$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 160000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 160000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 320000$}\n \\end{subfigure}\n \\\\\n \\end{tabular}\n\n \\caption{The observation columns recovered $\\E[\\phi(x,y)|h]$ in terms of $h$, for varying sample size $l$. The $x$-axis is the value of $t$, the $y$-axis is the value of $\\E[\\phi(x,y)|h]$.}\n \\label{fig:varyl}\n\\end{figure}\n\n\\subsection{Effect of Specifying the Number of States}\n\nWe fix $l = 320000$, $s = 1$, and vary the number of states $m = 2,3,4,5,6,7,8,9$. Figure~\\ref{fig:varym} shows the columns of $\\E[\\phi(x,y)|h]$ recovered. The spectral algorithm provides reasonable results when $m \\leq 4$; when $m \\geq 5$, the algorithm started to recover observation columns that has a lot of negative entries.\n\n\n\\begin{figure}[H]\n\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 1_n = 30_phi = beta_full.pdf}\n \\caption{$m = 1$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 2_n = 30_phi = beta_full.pdf}\n \\caption{$m = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = beta_full.pdf}\n \\caption{$m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$m = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 5_n = 30_phi = beta_full.pdf}\n \\caption{$m = 5$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns recovered $\\E[\\phi(x,y)|h]$ in terms of $h$, for varying number of states $m$. The $x$-axis is the value of $t$, the $y$-axis is the value of $\\E[\\phi(x,y)|h]$.}\n \\label{fig:varym}\n\\end{figure}\n\n\\subsection{Binning Feature vs. Beta Feature}\n\nWe compare the experimental results using two types of feature maps $\\phi_{\\bin}$ and $\\phi_{\\bet}$. Although the two mappings are substantially different, their respective $\\E[\\phi(x,y)|h]$ have some similarities. In some sense, the beta mapping is performing a ``soft'' binning which takes into account the number count $c$ in $x$: fixing the value of methylation probability $m/c$, if $c$ is larger, then the beta mapping is closer to a binning mapping with smaller bin size. Figure~\\ref{fig:varyphi} shows the columns of the resulting $\\E[\\phi(x,y)|h]$ recovered. Generally, $\\phi_{\\bet}$ produces smoother observation columns.\n\n%-- double check this one experiment -- typo in file naming?\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E2_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = beta_full.pdf}\n \\caption{$\\phi = \\phi_{\\bet}$ is beta mapping, $m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E2_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = binning_igz.pdf}\n \\caption{$\\phi = \\phi_{\\bin}$ is binning mapping, $m = 3$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E2_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$\\phi = \\phi_{\\bet}$ is beta mapping, $m = 4$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E2_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = binning_igz.pdf}\n \\caption{$\\phi = \\phi_{\\bin}$ is binning mapping, $m = 4$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns recovered $\\E[\\phi(x,y)|h]$ in terms of $h$, for $m = 4$ and two types of feature map $\\phi$. The $x$-axis is the value of $t$, the $y$-axis is the value of $\\E[\\phi(x,y)|h]$.}\n \\label{fig:varyphi}\n\\end{figure}\n\n\\subsection{Effect of Number of Segments Combined}\nWe fix $l = 320000$, $m = 4$, and vary the number of merged segments $s = 1,2,3,4,5,6,7,8$. Figure~\\ref{fig:varys} shows the columns of the resulting $\\E[\\phi(x,y)|h]$. It can be seen that the recovered result is fairly insensitive to the choice of $s$.\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 1$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 2_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 3_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 4_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 5_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 5$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns recovered $\\E[\\phi(x,y)|h]$ in terms of $h$, for $m = 4$ and two types of feature map $\\phi$. The $x$-axis is the value of $t$, the $y$-axis is the value of $\\E[\\phi(x,y)|h]$.}\n \\label{fig:varys}\n\\end{figure}\n}\n\n\\section{Possible Variants of Model}\n\n\\begin{figure}\n \\caption{Possible Variants of the Model}\n \\centering\n \\includegraphics[width=0.8\\textwidth]{model_doubt.jpg}\n \\label{fig:modeldoubt}\n\\end{figure}\n\nIn the data given, we are not able to observe the context information on exact positions.\nInstead, the data is grouped every 100 base pairs, where in each group, we observe\n32 numbers $\\cbr{(m_{t,i}, c_{t,i})}_{i=1}^{16}$, two for each context, representing\ntotal coverage and total methylation in context-location specific manner.\n\nWe can model the data in the following two graphical models (see Figure~\\ref{fig:modeldoubt}). The first one models\nthe sequence in each context separately. The hidden states are not only context specific,\nbut also location specific. It is not clear -- at least at this moment -- if we should\nalign the hidden states across different contexts.\n\nThe second model is to put the hidden state as the shared information across different\ncontexts. Then it assumes that given the hidden states, the 16 observations are conditionally\nindependent. We suspect this is too strong, and it might be preferable to assume the\n16 observations to have dependency and directly build a wholistic feature map based\non them altogether.\n\n\\section{Efficient Joint Parameter Estimation of Multiple HMMs}\nIn this section we study the following probem.\nSuppose we are given sequences drawn from two HMMs, and we have the prior knowledge that\nthe parameters of the two HMMs are close to each other.\nFurthermore, suppose we have computed the moments from training data\n(denoted as $P_{ij}, \\tilde P_{ij}$, $P_{ijk}, \\tilde P_{ijk}$ for all $(i,j,k)$ pairs).\nThe goal is to perform method of moments estimating the parameters of the two HMMs.\n\nIt is known that\n\\[ P_{i,j} = \\E[x_i \\otimes x_j] = \\sum_{l=1}^m w_l (O_i)_l \\otimes (O_j)_l \\]\nand\n\\[ P_{i,j,k} = \\E[x_i \\otimes x_j \\otimes x_k] = \\sum_{l=1}^m (O_i)_l \\otimes (O_j)_l \\otimes (O_k)_l \\]\n\n\\subsection{Algorithmic Primitive}\n\\paragraph{Matrix Regularized Least Squares} In subsequent discussions, the following optimization problem is often considered:\n\\[ \\min_{B \\in \\R^{n \\times k}} f(B) = \\| M - A B^T \\|_F^2 + \\lambda \\| B - \\tilde{B} \\|_F^2 \\]\nwhere $M \\in \\R^{m \\times n}$, $A \\in \\R^{m \\times k}$ are given.\n\nThis optimization problem an be solved by noticing it can be decomposed to functions with\nrespect to rows of $B$, that is, $B^{1}, \\ldots, B^{m}$.\nThis gives rise to the following optimization problem:\n\\[ \\min_{B^i \\in \\R^k} \\| M_i - A (B^i)^T \\|^2 + \\lambda \\| B^i - \\tilde B^i \\|^2 \\]\nTaking derivatives, we have\n\\[ 2 A^T A (B^i)^T - 2 A^T M_i + 2 \\lambda ((B^i)^T - (\\tilde B^i)^T) = 0 \\]\nthat is,\n\\[ (B^i)^T = (\\lambda I + A^T A)^{-1} (A^T M_i + \\lambda (\\tilde B^i)^T ) \\]\nhorizonally stacking the solution, we get\n\\[ B^T = (\\lambda I + A^T A)^{-1} (A^T M + \\lambda \\tilde B^T) \\]\nthat is, $B = (M^T A + \\lambda B)(\\lambda I + A^T A)^{-1}$.\n(As a sanity check, when $\\lambda = 0$, $B = (A^+ M)^T$, which is the correct solution.)\n\n\\paragraph{Weighted Extensions} It is sometimes desirable to consider the following\noptimization objective:\n\\[ \\min_{B \\in \\R^{n \\times k}} f(B) = \\| M - A B^T \\|_F^2 + \\lambda \\| C (B - \\tilde{B}) \\|_F^2 \\]\nwhere $C \\in \\R^{p \\times n}$ is a linear transformation (the objective aims to measure the\n$\\ell_2$ similarity of transformed $B$.)\nUnfortunately, the solution does not admit a succinct representation. But we can still\nwrite the solution. First, let $\\vec(B) = ((B^1)^T, (B^2)^T, \\ldots, (B^n)^T)$.\nWe rewrite the objective as:\n\\[ f(B) = \\sum_{i=1}^n \\| M_i - A (B^i)^T \\|_2^2 + \\sum_{j=1}^k \\| C B_j - C B_j \\|_2^2 \\]\nThus, we can formulate the problem as the following least squares problem:\n\\[\n \\begin{bmatrix} M_{11} \\\\ \\ldots \\\\ M_{m1} \\\\ \\ldots \\\\ M_{1n} \\\\ \\ldots \\\\ M_{mn} \\\\ C^1 \\tilde B_1 \\\\ \\ldots \\\\ C^{n'} \\tilde B_1 \\\\ \\ldots \\\\ C^1 \\tilde B_k \\\\ \\ldots \\\\ C^{n'} \\tilde B_k \\end{bmatrix}\n =\n \\begin{bmatrix}\n & A^1 & \\\\ & \\ldots & \\\\ & A^m & \\\\ & & & \\ldots & & & \\\\ & & & & & A^1 \\\\ & & & & & \\ldots \\\\ & & & & & A^m \\\\\n C_{1,1} & & \\ldots & & C_{1,n} \\\\& & \\ldots \\\\ C_{n',1} & & \\ldots & & C_{n',n} \\\\ & & \\ldots \\\\\n & & C_{1,1} & & \\ldots & & C_{1,n} \\\\& & & & \\ldots \\\\ & & C_{n',1} & & \\ldots & & C_{n',n}\n \\end{bmatrix} \\cdot \\vec(B)\n\\]\nLetting the vector in the left hand side as $Y$ and the matrix on the right hand side as $H$, we get $\\vec(B) = H^+ Y$.\n\n\\paragraph{Basic Weighted Least Squares}\nSuppose we would like to solve the following least squares problem:\n\\[ \\min f(\\tau) = \\| T - \\sum_{i=1}^k \\tau_i T_i \\|_F^2 \\]\nThis is equivalent to solving the following equation:\n\\[\n \\begin{bmatrix} T \\cdot T_1 \\\\ \\ldots \\\\ T \\cdot T_k \\end{bmatrix}\n =\n \\begin{bmatrix} T_1 \\cdot T_1 & \\ldots & T_1 \\cdot T_k \\\\ \\ldots & \\ldots & \\ldots \\\\ T_k \\cdot T_1 & \\ldots & T_k \\cdot T_k \\end{bmatrix} \\cdot \\tau\n\\]\nSpecifically, if $T_i = a_i \\cdot b_i \\cdot c_i$, then the equation becomes\n\\[\n \\begin{bmatrix} T(a_1, b_1, c_1) \\\\ \\ldots \\\\ T(a_k, b_k, c_k) \\end{bmatrix}\n =\n \\begin{bmatrix} (a_1 \\cdot a_1) (b_1 \\cdot b_1) (c_1 \\cdot c_1) & \\ldots & (a_1 \\cdot a_k) (b_1 \\cdot b_k) (c_1 \\cdot c_k) \\\\ \\ldots & \\ldots & \\ldots \\\\ (a_k \\cdot a_1) (b_k \\cdot b_1) (c_k \\cdot c_1) & \\ldots & (a_k \\cdot a_k) (b_k \\cdot b_k) (c_k \\cdot c_k) \\end{bmatrix} \\cdot \\tau\n\\]\nthat is,\n\\[ \\tau = ((A^T A) \\circ (B^T B) \\circ (C^T C))^{-1} \\begin{bmatrix} T(a_1, b_1, c_1) \\\\ \\ldots \\\\ T(a_k, b_k, c_k) \\end{bmatrix} \\]\n\n\n\\subsection{Basic Algorithm}\nDefine tensor $T = P_{123}(U_1,U_2,U_3) = \\sum_i w_i (U_1^TO_1)_i \\otimes (U_2^TO_2)_i \\otimes (U_3^TO_3)_i$\n\nGiven rank parameter $m$, we define objective function $F$ as follows:\n\\[ F(\\tau, A, B, C, \\tilde\\tau, \\tilde A, \\tilde B, \\tilde C) = \\| \\sum_{i=1}^m \\tau_i A_i \\otimes B_i \\otimes C_i - T \\|_F^2 + \\| \\sum_{i=1}^m \\tilde\\tau_i \\tilde A_i \\otimes \\tilde B_i \\otimes \\tilde C_i - \\tilde T \\|_F^2 + \\lambda \\| B - \\tilde B \\|_F^2 \\]\n\nWe have the following alternating minimization algorithm for the objective function.\n\n\\begin{algorithm}\n \\caption{Efficient Joint Parameter Estimation; Basic Algorithm}\n\\begin{algorithmic}[H]\n\\STATE Randomly initialize $A, B, C, \\tau, \\tilde A, \\tilde B, \\tilde C, \\tilde \\tau$.\n\\FOR{$t=1,2,\\ldots$:}\n\\STATE Update $A$: $A \\gets \\del{\\del{\\lambda I + (\\tau \\tau^T) \\circ (B^T B) \\circ (C^T C)}^{-1} \\del{\\begin{bmatrix} \\tau_1 T(I, B_1, C_1) \\\\ \\ldots \\\\ \\tau_k T(I, B_k, C_k) \\end{bmatrix} + \\lambda \\tilde A^T}}^T$\n\\STATE Update $B$: $B \\gets \\del{\\del{\\lambda I + (\\tau \\tau^T) \\circ (A^T A) \\circ (C^T C)}^{-1} \\del{\\begin{bmatrix} \\tau_1 T(A_1, I, C_1) \\\\ \\ldots \\\\ \\tau_k T(A_k, I, C_k) \\end{bmatrix} + \\lambda \\tilde B^T}}^T$\n\\STATE Update $C$: $C \\gets \\del{\\del{\\lambda I + (\\tau \\tau^T) \\circ (A^T A) \\circ (B^T B)}^{-1} \\del{\\begin{bmatrix} \\tau_1 T(A_1, B_1, I) \\\\ \\ldots \\\\ \\tau_k T(A_k, B_k, I) \\end{bmatrix} + \\lambda \\tilde C^T}}^T$\n\\STATE Update $\\tau$: $\\tau \\gets ((A^T A) \\circ (B^T B) \\circ (C^T C))^{-1} \\begin{bmatrix} T(A_1, B_1, C_1) \\\\ \\ldots \\\\ T(A_k, B_k, C_k) \\end{bmatrix}$\n\\STATE Update $\\tilde A$: $\\tilde A \\gets \\del{\\del{\\lambda I + (\\tilde \\tau \\tilde \\tau^T) \\circ (\\tilde B^T \\tilde B) \\circ (\\tilde C^T \\tilde C)}^{-1} \\del{\\begin{bmatrix} \\tilde\\tau_1 \\tilde T(I, \\tilde B_1, \\tilde C_1) \\\\ \\ldots \\\\ \\tilde\\tau_k \\tilde T(I, \\tilde B_k, \\tilde C_k) \\end{bmatrix} + \\lambda {\\tilde A}^T}}^T$\n\\STATE Update $\\tilde B$: $\\tilde B \\gets \\del{\\del{\\lambda I + (\\tilde \\tau \\tilde \\tau^T) \\circ (\\tilde A^T \\tilde A) \\circ (\\tilde C^T \\tilde C)}^{-1} \\del{\\begin{bmatrix} \\tilde\\tau_1 \\tilde T(\\tilde A_1, I, \\tilde C_1) \\\\ \\ldots \\\\ \\tilde\\tau_k \\tilde T(\\tilde A_k, I, \\tilde C_k) \\end{bmatrix} + \\lambda {\\tilde B}^T}}^T$\n\\STATE Update $\\tilde C$: $\\tilde C \\gets \\del{\\del{\\lambda I + (\\tilde \\tau \\tilde \\tau^T) \\circ (\\tilde A^T \\tilde A) \\circ (\\tilde B^T \\tilde B)}^{-1} \\del{\\begin{bmatrix} \\tilde\\tau_1 \\tilde T(\\tilde A_1, \\tilde B_1, I) \\\\ \\ldots \\\\ \\tilde\\tau_k \\tilde T(\\tilde A_k, \\tilde B_k, I) \\end{bmatrix} + \\lambda {\\tilde C}^T}}^T$\n\\STATE Update $\\tilde \\tau$: $\\tilde \\tau \\gets ((\\tilde A^T \\tilde A) \\circ (\\tilde B^T \\tilde B) \\circ (\\tilde C^T \\tilde C))^{-1} \\begin{bmatrix} \\tilde T(\\tilde A_1, \\tilde B_1, \\tilde C_1) \\\\ \\ldots \\\\ \\tilde T(\\tilde A_k, \\tilde B_k, \\tilde C_k) \\end{bmatrix}$\n\\ENDFOR\n\\end{algorithmic}\n\\end{algorithm}\n\n\\subsection{Algorithm Exploiting Three Views}\n\n\\subsection{Symmetrization }\n\nLet $U$ be the left singular vectors of $P_{i,1}$. It it known that $\\range(U) = \\range(O)$\nThis serves as the projection matrix for dimensionality reduction.\nWe operate on the following tensor:\n\\[ T = P_{123}(U,U,U)(S_1, I, S_3) \\]\nwhere $S_1 = (U^T P_{23} U)(U^T P_{13} U)^{-1}$ and $S_3 = (U^T P_{21} U)(U^T P_{31} U)^{-1}$.\nIt is known that\n\\[ T = \\sum_i w_i \\]\n\n\n\\end{document}\n", "meta": {"hexsha": "9fbab6861a37bfd88587acf0a091b57f905e041c", "size": 23477, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "kernelbin/model.tex", "max_stars_repo_name": "anapophenic/knb", "max_stars_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-04-21T21:10:22.000Z", "max_stars_repo_stars_event_max_datetime": "2016-08-27T19:15:23.000Z", "max_issues_repo_path": "kernelbin/model.tex", "max_issues_repo_name": "anapophenic/knb", "max_issues_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "kernelbin/model.tex", "max_forks_repo_name": "anapophenic/knb", "max_forks_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.8212435233, "max_line_length": 631, "alphanum_fraction": 0.6587298207, "num_tokens": 8315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289387956435734, "lm_q2_score": 0.740174367770488, "lm_q1q2_score": 0.6135592489859116}} {"text": "\\section{Exponential Functions}\\label{sec:ExpFunctions}\r\nAn \\dfont{exponential function} is a function of the form $f(x)=a^x$, where $a$ is a constant.\r\nExamples are $2^x$, $10^x$ and $(1/2)^x$.\r\nTo more formally define the exponential function we look at various kinds of input values.\r\n\r\nIt is obvious\r\nthat $\\ds a^5=a\\cdot a\\cdot a\\cdot a\\cdot a$ and $\\ds a^3=a\\cdot a\\cdot a$, \r\nbut when we consider an exponential function $\\ds a^x$ we can't\r\nbe limited to substituting integers for $x$. What does $\\ds a^{2.5}$ or\r\n$a^{-1.3}$ or $\\ds a^\\pi$ mean? And is it really true that\r\n$a^{2.5}a^{-1.3}=a^{2.5-1.3}$? The answer to the first question is\r\nactually quite difficult, so we will evade it; the answer to the\r\nsecond question is ``yes.''\r\n\r\nWe'll evade the full answer to the hard question, but we have to know\r\nsomething about exponential functions. You need first to understand\r\nthat since it's not ``obvious'' what $\\ds 2^x$ should mean, we are really\r\nfree to make it mean whatever we want, so long as we keep the behavior\r\nthat {\\em is} obvious, namely, when $x$ is a positive integer.\r\nWhat else do we want to be true about $\\ds 2^x$? We\r\nwant the properties of the previous two paragraphs to be true for all\r\nexponents: $\\ds 2^x2^y=2^{x+y}$ and $\\ds (2^x)^y=2^{xy}$.\r\n\r\nAfter the positive integers, the next easiest\r\nnumber to understand is 0: $\\ds 2^0=1$. You have presumably learned this\r\nfact in the past; why is it true? It\r\nis true precisely because we want $\\ds 2^a2^b=2^{a+b}$ to be true about\r\nthe function $\\ds 2^x$. We need it to be true that $\\ds 2^02^x=2^{0+x}=2^x$,\r\nand this only works if $\\ds 2^0=1$. The same argument implies that $\\ds a^0=1$\r\nfor any $a$.\r\n\r\nThe next easiest set of numbers to\r\nunderstand is the negative integers: for example, $\\ds 2^{-3}=1/2^3$. \r\nWe know that whatever $\\ds 2^{-3}$ means it must be\r\nthat $\\ds 2^{-3}2^{3}=2^{-3+3}=2^0=1$, which means that $\\ds 2^{-3}$ must be\r\n$1/2^3$. In fact, by the same argument, once we know what $\\ds 2^x$ means\r\nfor some value of $x$, $\\ds 2^{-x}$ must be $\\ds 1/2^{x}$ and more generally\r\n$a^{-x}=1/a^x$.\r\n\r\nNext, consider an exponent $1/q$, where $q$ is a positive integer. We\r\nwant it to be true that $\\ds (2^x)^y=2^{xy}$, so $\\ds\r\n(2^{1/q})^q=2$. This means that $\\ds 2^{1/q}$ is a $q$-th root of 2,\r\n$\\ds 2^{1/q}=\\root q\\of{2\\ }$. This is all we need to understand that\r\n$2^{p/q}=(2^{1/q})^p=(\\root q\\of{2\\ })^p$ and\r\n$a^{p/q}=(a^{1/q})^p=(\\root q\\of{a\\ })^p$.\r\n\r\nWhat's left is the hard part: what does $\\ds 2^x$ mean when $x$ cannot be\r\nwritten as a fraction, like $\\ds x=\\sqrt{2\\ }$ or $\\ds x=\\pi$? What we know so\r\nfar is how to assign meaning to $\\ds 2^x$ whenever $x=p/q$. If we were to\r\ngraph $a^x$ (for some $a>1$) at points $x=p/q$ then we'd see something like this:\r\n$$\\includegraphics[width=2.5in]{images/exp1}$$\r\n\r\nThis is a poor picture, but it illustrates a series \r\nof individual points above the rational numbers\r\non the $x$-axis. There are really a lot of ``holes'' in the curve,\r\nabove $x=\\pi$, for example. But (this is the hard part) it is possible\r\nto prove that the holes can be ``filled in'', and that the resulting\r\nfunction, called $\\ds a^x$, really does have the properties we want,\r\nnamely that $\\ds a^xa^y=a^{x+y}$ and $\\ds (a^x)^y=a^{xy}$.\r\nSuch a graph would then look like this:\r\n$$\\includegraphics[width=2.5in]{images/exp2}$$\r\n\r\n\\begin{formulabox}[Three Types of Exponential Functions]\r\nThere are \\ifont{three kinds} of exponential functions $f(x)=a^x$ depending on whether $a>1$, $a=1$ or $00$\r\n\\item \\ifont{Always positive:} $a^x>0$, for all $x$\r\n\\item \\ifont{Exponent rules: }\r\n\\begin{multicols}{2}\r\n\\begin{enumerate}\r\n\t\\item $\\ds{a^xa^y=a^{x+y}}$\r\n\t\\item $\\ds{\\frac{a^x}{a^y}=a^{x-y}}$\r\n\t\\item $\\ds{\\left(a^x\\right)^y=a^{xy}=a^{yx}=\\left(a^y\\right)^x}$\r\n\t\\item $\\ds{a^xb^x=(ab)^x}$\r\n\\end{enumerate}\r\n\\end{multicols}\r\n\\item \\ifont{Long term behaviour:} If $a>1$, then $a^x\\to\\infty$ as $x\\to\\infty$ and $a^x\\to 0$ as $x\\to-\\infty$.\r\n\\end{itemize}\r\n\\end{formulabox}\r\n\r\nThe last property can be observed from the graph. If $a>1$, then as $x$ gets larger and larger, so does $a^x$.\r\nOn the other hand, as $x$ gets large and negative, the function approaches the $x$-axis, that is, $a^x$ approaches $0$.\r\n\r\n\\begin{example}{Reflection of Exponential}{ReflectionExponential}\r\nDetermine an equation of the function after reflecting $y=2^x$ about the line $x=-2$.\r\n\\end{example}\r\n\r\n\\begin{solution} \r\nFirst reflect about the $y$-axis to get $y=2^{-x}$.\r\nNow shift by $2\\times 2=4$ units to the \\ifont{left} to get $y=2^{-(x+4)}$.\r\nSide note: Can you see why this sequence of transformations is the same as reflection in the line $x=-2$?\r\nCan you come up with a general rule for these types of reflections?\r\n\\end{solution}\r\n\r\n\\begin{example}{Determine the Exponential Function}{DetermineExponentialFunction}\r\nDetermine the exponential function $f(x)=ka^x$ that passes through the points $(1,6)$ and $(2,18)$.\r\n\\end{example}\r\n\r\n\\begin{solution} \r\nWe substitute our two points into the equation to get:\r\n$$x=1,y=6\\to6=ka^1$$\r\n$$x=2,y=18\\to18=ka^2$$\r\nThis gives us $6=ka$ and $18=ka^2$.\r\nThe first equation is $k=6/a$ and subbing this into the second gives: $18=(6/a)a^2$.\r\nThus, $18=6a$ and $a=3$.\r\nNow we can see from $6=ka$ that $k=2$.\r\nTherefore, the exponential function is $$f(x)=2\\cdot 3^x.$$\r\n%\\vspace{-1cm}\r\n\\end{solution}\r\n\r\nThere is one base that is so important and convenient that we give it a special symbol.\r\nThis number is denoted by $e=2.71828\\ldots$ (and is an irrational number). \r\nIts \\ifont{importance} stems from the fact that it simplifies many formulas \r\nof Calculus and also shows up in other fields of mathematics.\r\n\r\n\\begin{example}{Domain of Function with Exponential}{DomainFunctionExponential}\r\nFind the domain of $\\ds f(x)=\\frac{1}{\\sqrt{e^x+1}}$.\r\n\\end{example}\r\n\r\n\\begin{solution} \r\nFor domain, we cannot divide by zero or take the square root of negative numbers.\r\nNote that one of the properties of exponentials is that they are always positive!\r\nThus, $e^x+1>0$ (in fact, as $e^x>0$ we actually have that $e^x+1$ is at least one).\r\nTherefore, $e^x+1$ is never zero nor negative, and gives no restrictions on $x$.\r\nThus, the domain is $\\R$.\r\n\\end{solution}\r\n\r\n\r\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r\n\\Opensolutionfile{solutions}[ex]\r\n\\section*{Exercises for \\ref{sec:ExpFunctions}}\r\n\r\n\\begin{enumialphparenastyle}\r\n\r\n%%%%%%%%%%\r\n\\begin{ex}\r\nDetermine an equation of the function $y=a^x$ passing through the point $(3,8)$.\r\n\\begin{sol}\r\n$y=2^x$\r\n\\end{sol}\r\n\\end{ex}\r\n\r\n%%%%%%%%%%\r\n\\begin{ex}\r\nFind the $y$-intercept of $f(x)=4^x+6$.\r\n\\begin{sol}\r\n$y=7$\r\n\\end{sol}\r\n\\end{ex}\r\n\r\n%%%%%%%%%%\r\n\\begin{ex}\r\nFind the $y$-intercept of $f(x)=2\\left(\\frac{1}{2}\\right)^x$.\r\n\\begin{sol}\r\n$y=2$\r\n\\end{sol}\r\n\\end{ex}\r\n\r\n%%%%%%%%%%\r\n\\begin{ex}\r\nFind the domain of $\\ds{y=e^{-x}+e^{\\frac{1}{x}}}$.\r\n\\begin{sol}\r\n$x\\neq 0$\r\n\\end{sol}\r\n\\end{ex}\r\n\r\n\\end{enumialphparenastyle}", "meta": {"hexsha": "27a4ab23959e2b541578fa3dd1b20de3c2823aa3", "size": 7441, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2-functions/2-3-exp-functions.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2-functions/2-3-exp-functions.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2-functions/2-3-exp-functions.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.2784090909, "max_line_length": 120, "alphanum_fraction": 0.6637548717, "num_tokens": 2485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7217431943271999, "lm_q2_score": 0.849971175657575, "lm_q1q2_score": 0.6134609114051437}} {"text": "\n\\subsection{Unitary time}\n\nWhy? What's the interpretation here? Is this an assumption, or just a modelling choice?\n\n\\(\\Psi (t_b-t_a)^* \\Psi (t_b-t_a)=e^{(t_b-t_a)X^*} e^{(t_b-t_a)X}\\)\n\n\\(\\Psi (t_b-t_a)^* \\Psi (t_b-t_a)=e^{(t_b-t_a)(X^*+X)}\\)\n\n\\(X=iH\\)\n\n\\(\\Psi (t_b-t_a)^* \\Psi (t_b-t_a)=e^{(t_b-t_a)(-iH+iH)}=I\\)\n\n\\(\\Psi (t_b-t_a)=e^{(t_b-t_a)iH}\\)\n\n", "meta": {"hexsha": "ce10b4bfdbf6456b8217566d513e988d247dd44d", "size": 351, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/pug/theory/physics/QM/03-06-unitary.tex", "max_stars_repo_name": "adamdboult/nodeHomePage", "max_stars_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/pug/theory/physics/QM/03-06-unitary.tex", "max_issues_repo_name": "adamdboult/nodeHomePage", "max_issues_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 6, "max_issues_repo_issues_event_min_datetime": "2021-03-03T12:36:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-01T22:16:09.000Z", "max_forks_repo_path": "src/pug/theory/physics/QM/03-06-unitary.tex", "max_forks_repo_name": "adamdboult/nodeHomePage", "max_forks_repo_head_hexsha": "266bfc6865bb8f6b1530499dde3aa6206bb09b93", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 21.9375, "max_line_length": 87, "alphanum_fraction": 0.5754985755, "num_tokens": 164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9219218348550491, "lm_q2_score": 0.6654105521116443, "lm_q1q2_score": 0.6134565171346784}} {"text": "\\documentclass{amsart}\n\n\\usepackage{amsthm}\n\\usepackage{amssymb}\n\\usepackage{listings}\n\\usepackage{color}\n\\usepackage{courier}\n\\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue\n\\definecolor{mylilas}{RGB}{170,55,241}\n\\lstset{language=Matlab,%\n basicstyle=\\footnotesize\\ttfamily,\n breaklines=true,%\n morekeywords={matlab2tikz},\n keywordstyle=\\color{blue},%\n morekeywords=[2]{1}, keywordstyle=[2]{\\color{black}},\n identifierstyle=\\color{black},%\n stringstyle=\\color{mylilas},\n commentstyle=\\color{mygreen},%\n showstringspaces=false,%without this there will be a symbol in the places where there is a space\n numbers=none,%\n numberstyle={\\tiny \\color{black}},% size of the numbers\n numbersep=9pt, % this defines how far the numbers are from the text\n emph=[1]{for,end,break},emphstyle=[1]\\color{red}, %some words to emphasise\n %emph=[2]{word1,word2}, emphstyle=[2]{style}, \n}\n\n\n\\theoremstyle{definition}\n\\newtheorem{remark}{Remark}\n\\newtheorem{exercise}{\\color{red}Exercise}\n\\newtheorem{example}{Example}\n\n\\title[Programming finite elements in Matlab]{Programming finite elements in high-level languages with a focus on Matlab}\n\\author{Tom Gustafsson}\n\n\\begin{document}\n\n\\maketitle\n\n\\tableofcontents\n\n\\section{Variational problems}\n\nFinite element method for linear problems concerns the computer\nsolution of the following \\emph{variational problem}: find $u\n\\in U$ such that\n\\begin{equation}\n a(u,v) = l(v),\n\\end{equation}\nfor every $v \\in V$. Here $U$ and $V$ are Hilbert spaces with the\nnorms $\\|\\cdot\\|_U$ and $\\|\\cdot\\|_V$, $a \\in \\mathcal{L}(U \\times V;\n\\mathbb{R})$ is a continuous bilinear form and $l \\in\n\\mathcal{L}(V;\\mathbb{R})$ is a continuous linear form.\n\nRecall that if $U=V$ the existence and uniqueness of the solution\nfollows from the Lax--Milgram theorem if the bilinear form $a$ is\n\\emph{coercive} in $V$. If $a$ is not coercive or if $U\\neq V$, a more\ngeneral theory is required, see Babu\\v{s}ka--Aziz~\\cite{BA72}.\n\n\\begin{remark}\n The uniqueness aspect has no effect on the practical programming\nin the sense that using finite elements one can easily \\emph{try} to\ndiscretize and solve continuous problems that are not\nwell-posed. Usually such approaches lead to singular,\nnon-invertible, matrix systems.\n\\end{remark}\n\n\\section{Galerkin methods}\n\nSuppose that there exist finite dimensional subspaces $U_N \\subset U$\nand $V_M \\subset V$ with the dimensions $N$ and $M$, respectively. The\n\\emph{discrete variational problem} reads: find $u_N \\in U_N$ such\nthat\n\\begin{equation}\n \\label{eq:dweakf}\n a(u_N,v) = l(v),\n\\end{equation}\nfor every $v \\in V_M$.\n\n\\begin{remark}\n If $U_N = V_M$, the resulting method is sometimes referred to as the\n \\emph{Ritz--Galerkin method}. Otherwise, the term\n \\emph{Petrov--Galerkin method} is used.\n\\end{remark}\n\nSuppose that $U_N = \\mathrm{span}\\,\\{\\varphi_i\\}_{i=1}^N$ and $V_N =\n\\mathrm{span}\\,\\{\\xi_i\\}_{i=1}^M$ for some basis functions $\\varphi_i\n\\in U_N$, $i\\in\\{1,\\cdots,N\\}$ and $\\xi_i \\in V_M$,\n$i\\in\\{1,\\cdots,M\\}$.\nThen we may expand the solution as\n\\begin{equation}\n u_N = \\sum_{j=1}^N u_j \\varphi_j,\n\\end{equation}\nwhere $u_j \\in \\mathbb{R}$, $j\\in\\{1,\\cdots,N\\}$, are the degrees of\nfreedom. Substituting this expanded solution to \\eqref{eq:dweakf}\nleads to the matrix system\n\\begin{equation}\n \\label{eq:matrixsystem}\n A x = b,\n\\end{equation}\nwhere $A \\in \\mathbb{R}^{M \\times N}$, $A_{ij} = a(\\varphi_j, \\xi_i)$, $b \\in \\mathbb{R}^M$, $b_i = l(\\xi_i)$ and $x \\in \\mathbb{R}^N$, $x_i = u_i$.\n\n\\begin{exercise}\n Derive \\eqref{eq:matrixsystem} from \\eqref{eq:dweakf}.\n\\end{exercise}\n\nThe programming aspect of the Galerkin method is focused on\nunderstanding and implementing the \\emph{assembly procedure} that\nconstructs the pair $(A,b)$ given a bilinear form $a$, linear form $l$\nand a set of basis functions. It is now \\emph{crucial} to understand\nthat given a basis and a bilinear form, we \\emph{always} have a corresponding\nmatrix that approximates the continuous operator.\n\n\\begin{remark}\n In mathematical literature the finite dimensional spaces $U_N$ and\n$V_M$ are sometimes defined by limiting the set of arbitrary functions\nto, e.g., piecewise polynomial functions. In such cases the basis is\nimplicit in the sense that we must do additional work to find the\nbasis functions or evaluate them pointwise. Thus, from a programming\npoint-of-view an explicit basis is usually more preferable.\n\\end{remark}\n\n\\subsection{On the choice of basis functions}\n\nLet us briefly discuss through examples how the choice of the basis\naffects to the properties of the matrix $A$. In the following examples\n$U=V$ and $U_N = V_M$.\n\n\\begin{example}\n \\label{ex:diagonal}\n Consider the bilinear and linear forms\n \\begin{equation}\n a(w,v) = \\int_0^1 w^\\prime v^\\prime\\,\\mathrm{d}x, \\quad l(v) = \\int_0^1 v\\,\\mathrm{d}x,\n \\end{equation}\n and the basis $\\{\\sin (\\pi (2n-1) x)\\}_{n=1}^N$. In terms of Matlab\n code, we have\n \\begin{lstlisting}\n syms x\n N=5;\n dphi=diff(sin(pi*(1:2:(2*N))*x),x);\n A=int(dphi'*dphi,x,0,1);\n \\end{lstlisting}\n which outputs the matrix\n \\begin{equation}\n\\left(\\begin{array}{ccccc} \\frac{{\\pi}^2}{2} & 0 & 0 & 0 & 0\\\\ 0 & \\frac{9\\, {\\pi}^2}{2} & 0 & 0 & 0\\\\ 0 & 0 & \\frac{25\\, {\\pi}^2}{2} & 0 & 0\\\\ 0 & 0 & 0 & \\frac{49\\, {\\pi}^2}{2} & 0\\\\ 0 & 0 & 0 & 0 & \\frac{81\\, {\\pi}^2}{2} \\end{array}\\right).\n \\end{equation}\n Notice that the matrix is diagonal. This means that the solution\nwill be quick. The corresponding linear form can be assembled and\nthe system solved with\n \\begin{lstlisting}\n b=int(sin(pi*(1:2:(2*N))*x),x,0,1)';\n x=A\\b;\n \\end{lstlisting}\n\\end{example}\n\n\\begin{exercise}\n Draw the solution given by the Galerkin method in the previous example.\n\\end{exercise}\n\n\\begin{remark}\n The usage of symbolic integration is not a good idea in general due\nto a high computational cost in comparison to numerical integration.\n\\end{remark}\n\n\\begin{example}\n Consider the same bilinear form as in Example~\\ref{ex:diagonal}\n and the polynomial basis $\\{x^n(1-x)^n\\}_{n=1}^N$. Now the matrix $A$ is\n \\begin{equation}\n \\left(\\begin{array}{ccccc} \\frac{1}{3} & \\frac{1}{15} & \\frac{1}{70} & \\frac{1}{315} & \\frac{1}{1386}\\\\ \\frac{1}{15} & \\frac{2}{105} & \\frac{1}{210} & \\frac{4}{3465} & \\frac{5}{18018}\\\\ \\frac{1}{70} & \\frac{1}{210} & \\frac{1}{770} & \\frac{1}{3003} & \\frac{1}{12012}\\\\ \\frac{1}{315} & \\frac{4}{3465} & \\frac{1}{3003} & \\frac{4}{45045} & \\frac{1}{43758}\\\\ \\frac{1}{1386} & \\frac{5}{18018} & \\frac{1}{12012} & \\frac{1}{43758} & \\frac{5}{831402} \\end{array}\\right).\n \\end{equation}\n Notice that this time the matrix is full and hence the solution of\nthe linear system requires more effort.\n\\end{example}\n\nBased on the examples, we can deduce that the choice of the basis (or,\nequivalently, the subspace $V_N$) has an effect on\n\\begin{itemize}\n \\item the \\emph{sparsity\n pattern} of the resulting matrix, and\n \\item the \\emph{condition\nnumber}\\footnote{As given by the Matlab function \\emph{cond}.} which for the\nformer matrix is 81 whereas the condition number of the latter matrix\nis approximately $2.1 \\cdot 10^8$.\n\\end{itemize}\nMoreover, let $P_N : V \\rightarrow V_N$ be an orthogonal\nprojection. Using the continuity and coercivity of $a$, we get\n\\begin{align*}\n \\|u-u_N\\|_V^2 &\\leq C^\\prime a(u-u_N,u-u_N) \\\\\n &= C^\\prime a(u-u_N,u-P_N u) \\\\\n &\\leq C \\|u-u_N\\|_V \\|u-P_N u\\|_V,\n\\end{align*}\nfor some $C, C^\\prime >0$. This leads to\n\\begin{equation}\n \\|u-u_N\\|_V \\leq C \\|u-P_N u\\|_V,\n\\end{equation}\nwhich essentially means that the error (in the sense of $\\|u-u_N\\|_V$)\ndepends on \\emph{how well the space $V_N$ can represent functions of\nthe space $V$}. In an ``ideal case'' the exact solution belongs to the\ndiscrete space $V_N$ leading to zero error. In practice this is rarely\nthe case.\n\nEverything we have discussed so far applies to all Galerkin methods.\nNext we focus on a particular Galerkin method called the \\emph{finite\nelement method}.\n\n\\section{Finite element method}\n\nFinite element methods are special Galerkin methods where the basis\nfunctions are defined using a \\emph{mesh}. In the following we\nconsider two dimensional problems but many parts can be\nstraightforwardly generalized to three dimensional problems.\n\n\\subsection{Mesh}\n\nLet $\\Omega \\subset \\mathbb{R}^2$ a polygonal domain. We split the\ndomain $\\Omega$ into non-overlapping triangles or \\emph{elements} $T\n\\in \\mathcal{T}$ that satisfy\n\\begin{equation}\n \\overline{\\Omega} = \\bigcup_{T \\in \\mathcal{T}} \\overline{T}.\n\\end{equation}\nThe collection of elements $\\mathcal{T}$ is called a mesh. One\nfrequently encounters the notion of a \\emph{mesh parameter $h_T$ of a\ntriangle $T$} that describes the maximum edge length of $T$, i.e.,\n\\begin{equation}\n h_T = \\max_{x,y \\in T} \\|x-y\\|,\n\\end{equation}\nwhere $\\|\\cdot\\|$ is the Euclidean norm. The mesh parameter of a\nmesh $\\mathcal{T}$ is then\n\\begin{equation}\n h = \\max_{T \\in \\mathcal{T}} h_T.\n\\end{equation}\n\nLet $N$ be the number of vertices in the mesh and let $M$ be the\nnumber of elements. In high-level array-based languages such as\nMatlab, a convenient way to represent triangular (and other) meshes is\nto use two matrices $p \\in \\mathbb{R}^{2 \\times N}$ and $t \\in\n\\mathbb{R}^{3 \\times M}$. In the former, each column corresponds to\n$x$ and $y$ coordinates of a vertex and in the latter each column\ncorresponds to the three indices of the vertices of an element.\n\n\\begin{example}\n Using Matlab PDE Toolbox we can generate a triangular mesh for the\n unit square with $h\\leq 0.1$ using the command\n\\begin{lstlisting}\n [p,~,t]=initmesh(decsg([3 4 0 1 1 0 0 0 1 1]'),'Hmax',0.1)\n\\end{lstlisting}\n The matrix $t$ given by \\emph{initmesh} has one extra row which\n in case of multiple domains tells us the subdomain in which the\n triangle lies. We do not need this information and therefore we\n strip the last row as follows:\n\\begin{lstlisting}\n t=t(1:3,:);\n\\end{lstlisting}\n The mesh can now be visualized with\n\\begin{lstlisting}\n trimesh(t',p(1,:),p(2,:));\n\\end{lstlisting}\n\\end{example}\n\n\\begin{exercise}\n Visualize two meshes with different mesh parameters.\n\\end{exercise}\n\n\\subsection{Definition of the piecewise linear basis functions}\n\nIn this text we focus on the \\emph{piecewise linear basis}. Each\nbasis function is defined so that it has value 1 in one vertex of the\nmesh and value 0 in all other vertices. Furthermore, the basis\nfunction is linear inside each element.\n\n\\begin{example}\n Let us generate a mesh and visualize all the basis functions.\n \\begin{lstlisting}\n [p,~,t]=initmesh(decsg([3 4 0 1 1 0 0 0 1 1]'),'Hmax',0.5);\n t=t(1:3,:);\n figure;\n % there is one basis function per node, loop over nodes\n for itr = 1:size(p,2)\n subplot(ceil(sqrt(size(p,2))),ceil(sqrt(size(p,2))),itr);\n % the basis function is zero everywhere else ...\n u=zeros(size(p,2),1);\n % ... but one node.\n u(itr)=1;\n % draw the resulting surface\n trisurf(t',p(1,:),p(2,:),u);\n colormap parula;\n shading interp;\n view(2);\n end\n \\end{lstlisting}\n From the resulting figure you can probably see why these\n kind of basis functions are sometimes referred to as\n the \\emph{hat functions}.\n\\end{example}\n\nThese kind of basis functions have some nice properties.\nFor example:\n\\begin{enumerate}\n\\item They are based on a mesh and hence any domain\n that can be meshed is amenable to discretization.\n\\item The basis functions are nonzero on a very small area.\n Therefore, most entries of the resulting system matrix are zero\n and the matrix will be sparse.\n\\item We can easily project an arbitrary function to the\n basis by evaluating it at the vertices of the mesh.\n\\end{enumerate}\n\n\\subsection{Assembly of finite elements}\n\nLet us denote by $\\varphi_i$, $i \\in \\{1, \\cdots, N\\}$ the set of\npiecewise linear basis functions defined by the mesh $\\mathcal{T}$\nof the domain $\\Omega$.\nConsider the bilinear form\n\\begin{equation}\n a(w,v) = \\int_\\Omega \\nabla w \\cdot \\nabla v\\,\\mathrm{d}x.\n\\end{equation}\nOur goal is to construct the matrix $A \\in \\mathbb{R}^{N \\times N}$\nwith the entries $A_{ij} = a(\\varphi_j, \\varphi_i)$. Since $N$ can be\nvery large and a large portion of the values $A_{ij}$ are zero, it is\nnot a good idea to evaluate each $a(\\varphi_j, \\varphi_i)$\nseparately. Instead, we evaluate \\emph{all nonzero terms} of the form\n\\begin{equation}\n \\label{eq:integral}\n \\int_T \\nabla \\varphi_j \\cdot \\nabla \\varphi_i\\,\\mathrm{d}x, \\quad T \\in \\mathcal{T}, \\quad i,j \\in \\{1,\\cdots,N\\},\n\\end{equation}\nand add them together appropriately to \\emph{assemble} $A$.\n\n\\begin{exercise}\n Download the commented version of the Matlab class template with the command\n \\begin{lstlisting}\n unzip('http://super.org/file.zip')\n \\end{lstlisting}\n Read through \\emph{LinearAssembler.m} and try to run the tests with\n \\begin{lstlisting}\n run(LinearAssemblerTests)\n \\end{lstlisting}\n\\end{exercise}\n\nNext we start introducing necessary concepts to implement the methods\nof the class \\verb|LinearAssembly|. We first implement the static\nmethods, secondly we write the constructor which does some\nprecomputations and finally implement the methods that assemble\nbilinear and linear forms.\n\n\\subsection{Reference element}\n\nEvaluation of the integral \\eqref{eq:integral} can be done\nsystematically by mapping it to the \\emph{reference triangle}\n$\\widehat{T}$ defined as\n\\begin{equation}\n \\widehat{T} = \\{ (\\widehat{x}, \\widehat{y}) : \\widehat{x} \\geq 0,~ \\widehat{y} \\geq 0,~ \\widehat{x}+\\widehat{y}\\leq 1 \\}.\n\\end{equation}\nWe end up evaluating integrals of polynomials on the reference\ntriangle and for this we require \\emph{quadrature rules}. The idea of\nquadrature rules is to replace the integral with a sum, i.e., for\na given polynomial $p \\in P_n(\\widehat{T})$\nand triplets $(\\widehat{x}_k,\\widehat{y}_k,w_k)$, $k \\in \\{1, \\cdots, M\\}$,\nwe write\n\\begin{equation}\n \\int_{\\widehat{T}} p(\\widehat{x},\\widehat{y}) \\,\\mathrm{d}\\widehat{x}\\,\\mathrm{d}\\widehat{y} \\approx \\sum_{k = 1}^{M} p(\\widehat{x}_k,\\widehat{y}_k) w_k.\n\\end{equation}\n\\begin{remark}\n For any given polynomial order $n$ the quadrature rule can be exact\n if $M$ and the triplets are chosen right. The construction of exact\n rules is, however, non-trivial. See, e.g., Dunavant~\\cite{D85} where\n an exact rule for second order polynomials is given by the triplets\n \\begin{equation}\n (\\tfrac{1}{3},\\tfrac{1}{3},\\tfrac{1}{6}), \\quad \n (\\tfrac{2}{3},\\tfrac{1}{3},\\tfrac{1}{6}), \\quad \n (\\tfrac{1}{3},\\tfrac{2}{3},\\tfrac{1}{6}).\n \\end{equation}\n\\end{remark}\n\\begin{exercise}\n Implement the method \\verb|localQuadrature|.\n\\end{exercise}\nLet $F_T : \\widehat{T} \\rightarrow T$ be an affine mapping of the form\n\\begin{equation}\n \\begin{bmatrix} x \\\\ y \\end{bmatrix} = F_T(\\widehat{x},\\widehat{y}) = B_T\\begin{bmatrix} \\widehat{x} \\\\ \\widehat{y}\\end{bmatrix}+c_T, \n\\end{equation}\nwhere $B_T \\in \\mathbb{R}^{2 \\times 2}$ and $c_T \\in \\mathbb{R}^{2 \\times 1}$\ndepend on the location and the shape of the triangle $T$.\nNotice that we denote by $(x,y)$ points in the global\ncoordinate system and by $(\\widehat{x},\\widehat{y})$ points in the\ncoordinate system of the reference element.\n\\begin{exercise}\n Find explicit expressions for $B_T$ and $c_T$ in terms of the\nvertices of the element $T$ and implement the method \\verb|affineMappings|.\n\\end{exercise}\n\\begin{exercise}\n Implement the method \\verb|F| which maps points from the local coordinate\n system to the global coordinate system using $B_T$ and $c_T$.\n\\end{exercise}\n\\subsection{Transforming the local basis}\nLet the local basis functions be defined as\n\\begin{equation}\n \\widehat{\\varphi}_1(\\widehat{x},\\widehat{y}) = 1 - \\widehat{x} - \\widehat{y}, \\quad \\widehat{\\varphi}_2(\\widehat{x},\\widehat{y}) = \\widehat{x}, \\quad \\widehat{\\varphi}_3(\\widehat{x},\\widehat{y}) = \\widehat{y}.\n\\end{equation}\nOur goal is to describe the global basis functions $\\varphi_i$,\n$i\\in\\{1,\\cdots,N\\}$, using these local basis functions.\n\\begin{exercise}\n Implement the method \\verb|localBasis|.\n\\end{exercise}\nLet $m$ be the column corresponding to the triangle $T$ in the element\nmatrix $t$. The global basis functions that are nonzero in element $T$\ncan be now defined as\n\\begin{equation}\n \\label{eq:phi}\n \\varphi_{e_i^{\\mathrm{T}}te_m}(F_T(\\widehat{x},\\widehat{y}))= \\widehat{\\varphi}_i(\\widehat{x},\\widehat{y}),\n\\end{equation}\nfor each $i \\in \\{1,2,3\\}$. Using the chain rule, we deduce that the\nderivatives transform as follows:\n\\begin{equation}\n \\label{eq:dphi}\n \\nabla \\varphi_{e_i^{\\mathrm{T}}te_m}(F_T(\\widehat{x},\\widehat{y})) = B_T^{-\\mathrm{T}} \\widehat{\\nabla} \\widehat{\\varphi}_i(\\widehat{x},\\widehat{y}),\n\\end{equation}\n\\begin{exercise}\n Implement \\verb|globalBasis| which maps the local basis to global\nbasis using \\eqref{eq:phi} and \\eqref{eq:dphi}.\n\\end{exercise}\nFinally, the integral \\eqref{eq:integral} is transformed as\n\\begin{equation}\n \\label{eq:integraltransform}\n \\int_T \\nabla \\varphi_{e_j^{\\mathrm{T}}te_m} \\cdot \\nabla \\varphi_{e_i^{\\mathrm{T}}te_m}\\,\\mathrm{d}x = \\int_{\\widehat{T}} B_K^{-\\mathrm{T}}\\widehat{\\nabla} \\widehat{\\varphi}_j \\cdot B_K^{-\\mathrm{T}}\\widehat{\\nabla}\\widehat{\\varphi}_i|\\det B_K|\\,\\mathrm{d}\\widehat{x}.\n\\end{equation}\n\\begin{exercise}\n Implement \\verb|determinant| for computing the determinant in \\eqref{eq:integraltransform}.\n\\end{exercise}\n\n\\begin{thebibliography}{1}\n\\bibitem{BA72} I.~Babu\\v{s}ka, A.~K.~Aziz. The Mathematical Foundations of the Finite Element Method with Applications to Partial Differential Equations. Academic Press (1972).\n\\bibitem{D85} D.~A.~Dunavant. High degree efficient symmetrical Gaussian quadrature rules for the triangle (1985).\n\\end{thebibliography}\n\n\\end{document}", "meta": {"hexsha": "e260736e2e116ce34bfa3a7607c75ffa5cfc8216", "size": 17755, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notes.tex", "max_stars_repo_name": "kinnala/matlab-fem-course", "max_stars_repo_head_hexsha": "27b9f5648cc060ea118b8d356129cf3a7f90ca7d", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-06-05T05:59:37.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-05T05:59:37.000Z", "max_issues_repo_path": "notes.tex", "max_issues_repo_name": "kinnala/matlab-fem-course", "max_issues_repo_head_hexsha": "27b9f5648cc060ea118b8d356129cf3a7f90ca7d", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "notes.tex", "max_forks_repo_name": "kinnala/matlab-fem-course", "max_forks_repo_head_hexsha": "27b9f5648cc060ea118b8d356129cf3a7f90ca7d", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.9101382488, "max_line_length": 465, "alphanum_fraction": 0.707969586, "num_tokens": 5717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195152660688, "lm_q2_score": 0.8354835411997897, "lm_q1q2_score": 0.6133447722783683}} {"text": "\\documentclass{article}\n\\usepackage{ws_template}\n\n\\usepackage{amsmath}\n\\usepackage{listings}\n\n\\title{homework sheet 01}\n\\author{\n\t\\name{Denys Sobchyshak}\\\\\n\t\\imat{03636581}\\\\\n\t\\email{denys.sobchyshak@gmail.com}\n\t\\And\n\t\\name{Sergey Zakharov}\\\\\n\t\\imat{03636642}\\\\\n\t\\email{ga39pad@mytum.de}\n}\n\n\\begin{document}\n\t\\maketitle\n\t\n\t\\section{Problem}\n\tTo find eigenvalues of the matrix A we have to solve $ det(A-\\lambda I)=0 $ which cane be derived from an eigenvelue equation. \\\\\n\t\\[\n\tdet(A-\\lambda I)= det\n\t\\begin{pmatrix} \n\t\t2-\\lambda & -1 & 0 \\\\ \n\t\t-1 & 2-\\lambda & -1 \\\\\n\t\t0 & -1 & 2-\\lambda\n\t\\end{pmatrix}\n\t=(2-\\lambda)^3-2(2-\\lambda)=(2-\\lambda)((2-\\lambda)^2-2)=0\n\t\\]\\\\\n\tThus we can see that the spectrum of the matrix is given by $\\lambda(A)=\\left\\{2, 2-\\sqrt{2},2+\\sqrt{2}\\right\\}$. To find the corresponding eigenvectors we substitute our eigenvalues into the eigenvalue equation.\\\\\n\t\\begin{enumerate}\n\t\t\\item For $\\lambda=2$:\\\\\n\t\t$\n\t\tA-2I=0 \\Rightarrow\n\t\t\\left(\\begin{array}{ccc|c}\n\t\t2-2 & -1 & 0 & 0 \\\\ \n\t\t-1 & 2-2 & -1 & 0\\\\ \n\t\t0 & -1 & 2-2 & 0\n\t\t\\end{array}\\right)\n\t\t\\Rightarrow \\left\\{\n\t\t\\begin{array}{cc}\n\t\tx=-z\\\\\n\t\ty=0\\\\\n\t\t\\end{array}\\right.\n\t\t$\\\\\n\t\tThus using $z=-1$ we get the next eigenvector $v_1=\\begin{pmatrix}1 \\\\0\\\\ -1\\end{pmatrix}$.\n\n\t\t\\item For $\\lambda=2-\\sqrt{2}$:\\\\\n\t\t$\n\t\tA-2I=0 \\Rightarrow\n\t\t\\left(\\begin{array}{ccc|c}\n\t\t2-2+\\sqrt{2} & -1 & 0 & 0 \\\\\n\t\t0 & -1 & 2-2+\\sqrt{2} & 0 \\\\\n\t\t-1 & 2-2+\\sqrt{2} & -1 & 0 \n\t\t\\end{array}\\right)\n\t\t\\Rightarrow \\left\\{\n\t\t\\begin{array}{cc}\n\t\tx=\\frac{y}{\\sqrt{2}}\\\\\n\t\ty=\\sqrt{2}z\\\\\n\t\tz=\\frac{y}{\\sqrt{2}}\\\\\n\t\t\\end{array}\\right.\n\t\t$\\\\\n\t\tThus using $z=1$ we get the next eigenvector $v_2=\\begin{pmatrix}1 \\\\\\sqrt{2}\\\\ 1\\end{pmatrix}$.\n\t\t\n\t\t\\item For $\\lambda=2+\\sqrt{2}$:\\\\\n\t\t$\n\t\tA-2I=0 \\Rightarrow\n\t\t\\left(\\begin{array}{ccc|c}\n\t\t2-2-\\sqrt{2} & -1 & 0 & 0 \\\\ \n\t\t-1 & 2-2-\\sqrt{2} & -1 & 0\\\\ \n\t\t0 & -1 & 2-2-\\sqrt{2} & 0\n\t\t\\end{array}\\right)\n\t\t\\Rightarrow \\left\\{\n\t\t\\begin{array}{cc}\n\t\tx=-\\frac{y}{\\sqrt{2}}\\\\\n\t\ty=-\\sqrt{2}z\\\\\n\t\tz=-\\frac{y}{\\sqrt{2}}\\\\\n\t\t\\end{array}\\right.\n\t\t$\\\\\n\t\tThus using $z=1$ we get the next eigenvector $v_3=\\begin{pmatrix}1 \\\\-\\sqrt{2} \\\\ 1\\end{pmatrix}$.\n\t\\end{enumerate}\n\tThe corresponding python/numpy code looks as follows:\\\\\n\t\\lstinputlisting[language=Python,firstline=21]{scripts/ws1-linear-algebra.py}\n\t\n\t\\section{Problem}\n\tIf $ x_1, x_2 ... x_n$ are linearly independent eigenvectors of $ B \\in R^{n \\times n} $ and $ \\lambda_i $ are its eigenvalues then the following equation is satisfied: $ Bx_i = \\lambda_i x_i $.\n\tThis equation can be rewritten as $ BU = UD $ from here we get matrix $ B = UDU^{-1} $ as follows:\\\\\\\\\n\t$\n\tBU=B\\begin{bmatrix} X_{1} & X_{2} & \\cdots & X_n \\end{bmatrix}\n\t=\\begin{bmatrix} BX_{1} & BX_{2} & \\cdots & BX_n \\end{bmatrix}\n\t=\\begin{bmatrix} \\lambda_1X_{1} & \\lambda_2X_{2} & \\cdots & \\lambda_nX_n \\end{bmatrix}\\\\\n\t=\\begin{bmatrix} \n\t\\lambda_1x_{1,1} & \\lambda_2x_{2,1} & \\cdots & \\lambda_nx_{n,1} \\\\\n\t\\lambda_1x_{1,2} & \\lambda_2x_{2,2} & \\cdots & \\lambda_nx_{n,2} \\\\\n\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\t\\lambda_1x_{1,n} & \\lambda_2x_{2,n} & \\cdots & \\lambda_nx_{n,n}\n\t\\end{bmatrix}\n\t=\\begin{bmatrix} \n\tx_{1,1} & x_{2,1} & \\cdots & x_{n,1} \\\\\n\tx_{1,2} & x_{2,2} & \\cdots & x_{n,2} \\\\\n\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\tx_{1,n} & x_{2,n} & \\cdots & x_{n,n}\n\t\\end{bmatrix}\n\t\\begin{bmatrix} \n\t\\lambda_1 & 0 & \\cdots & 0 \\\\\n\t0 & \\lambda_2 & \\cdots & 0 \\\\\n\t\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\t0 & 0 & \\cdots & \\lambda_n\n\t\\end{bmatrix}\\\\\n\t=UD \\Leftrightarrow BUU^{-1}=B=UDU^{-1}\n\t$\n\t\\section{Problem}\n\t\\begin{itemize}\n\t\t\\item $B \\bar{x} = \\lambda \\bar{x} $. Suppose that $\\lambda$ and $\\bar{x}$ are possibly complex and $x \\not=0$ .\n\t\tWe take conjugate transpose to get $\\bar{\\lambda} x^* = (\\lambda x)^* = (Ax)^* =x^*A^* = x^*A$ (because A is real and symmetric). Then $\\bar{\\lambda}^*x = (x^*A)x = x^*(Ax) = \\lambda x^*x$. $x^*x$ is nonzero and real. So we get $\\bar{\\lambda} = \\lambda$, which means that $\\lambda$ is real.\n\t\t\\item Assume $Av =\\lambda v$ and $ Aw =\\mu w$. Then \\[ \\lambda(v, w) = (\\lambda v, w) = (Av, w) = (v, A^T w) = (v, Aw) = (v, \\mu w) = \\mu (v, w) \\] is possible only if $(v, w) = 0$.\n\t\\end{itemize}\n\t\n\t\\section{Problem}\n\t\\begin{itemize}\n\t\t\\item $ |B| = |UDU^{-1}| = |DU^{-1}U| = |D| = \\prod_i{\\lambda_i} $\n\t\t\\item $ tr(B) = tr(UDU^{-1}) = tr(DU^{-1}U) = tr(D) = \\sum_i{\\lambda_i} $\n\t\\end{itemize}\n\t\n\t\\section{Problem}\n\t\\begin{itemize}\n\t\t\\item $ h(x_0) = w_0 + w^T x_0 = 0 \\Rightarrow w^T x_0 = -w_0 $\n\t\t\\item $ h(x_1) = w_0 + w^T x_1 = 0 \\\\\n\t\th(x_2) = w_0 + w^T x_2 = 0 \\\\\n\t\th(x_1) - h(x_2) = w^T x_1 - w^T x_2 = w^T (x_1 - x_2) = 0 $\n\t\t\\item Normal vector is defined as $\\nabla h(x) = w$. $ \\frac{\\nabla h(x)}{|| \\nabla h(x)||} = \\frac{w}{||w||}$ is a unit normal vector. So $ \\hat{w} = \\frac{w}{||w||}$ is a unit normal vector.\n\t\t\\item Distance can be found by projecting $(x - x_0)$ onto normal vector $w$. \\[ D = |proj_w(x-x_0)| =\\frac{w^T (x - x_0)}{||w||} = \\hat{w}(x-x_0) = \\frac{w^T x - w^T x_0}{||w||} \\overset{w^T x_0 = -w_0}{= \\joinrel = \\joinrel = \\joinrel = \\joinrel = \\joinrel = \\joinrel =} \\frac{w^T x + w_0}{||w||} \\]\n\t\\end{itemize}\n\t\n\\end{document}", "meta": {"hexsha": "6879d5f6c46977080c5cf6cbd8c563a17f44722b", "size": 5120, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tum/machine-learning/ws01-linear-algebra.tex", "max_stars_repo_name": "nyxcalamity/classwork", "max_stars_repo_head_hexsha": "dfe47a40fe57ec5e0ccfd672a8dcaf246386de99", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2017-02-10T19:12:21.000Z", "max_stars_repo_stars_event_max_datetime": "2020-09-03T00:08:40.000Z", "max_issues_repo_path": "tum/machine-learning/ws01-linear-algebra.tex", "max_issues_repo_name": "nyxcalamity/classwork", "max_issues_repo_head_hexsha": "dfe47a40fe57ec5e0ccfd672a8dcaf246386de99", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tum/machine-learning/ws01-linear-algebra.tex", "max_forks_repo_name": "nyxcalamity/classwork", "max_forks_repo_head_hexsha": "dfe47a40fe57ec5e0ccfd672a8dcaf246386de99", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 15, "max_forks_repo_forks_event_min_datetime": "2016-05-02T11:00:28.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-20T03:55:35.000Z", "avg_line_length": 37.3722627737, "max_line_length": 304, "alphanum_fraction": 0.580078125, "num_tokens": 2285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7341195152660688, "lm_q2_score": 0.8354835309589074, "lm_q1q2_score": 0.6133447647603366}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage[top=1in, bottom=1in, left=1in, right=1in]{geometry} \n\\usepackage{graphicx}\n\\usepackage{setspace}\n\\usepackage{bm}\n\\usepackage{amsmath}\n\\usepackage{amssymb,amsmath}\n\\usepackage{listings}\n\\usepackage{color}\n\\usepackage{enumitem}\n\\usepackage{fancyvrb}\n\\usepackage{hyperref}\n\\usepackage{diagbox}\n\\usepackage{float}\n\\geometry{letterpaper}\n\\linespread{1.1}% \\geometry{landscape} % rotated page geometry\n\n\\definecolor{codegreen}{rgb}{0,0.6,0}\n\\definecolor{codegray}{rgb}{0.5,0.5,0.5}\n\\definecolor{codepurple}{rgb}{0.58,0,0.82}\n\\definecolor{backcolour}{rgb}{0.95,0.95,0.92}\n\\definecolor{outcolor}{rgb}{0.545, 0.0, 0.0}\n\n\\lstdefinestyle{mystyle}{\n\tbackgroundcolor=\\color{backcolour}, \n\tcommentstyle=\\color{codegreen},\n\tkeywordstyle=\\color{magenta},\n\tnumberstyle=\\tiny\\color{codegray},\n\tstringstyle=\\color{codepurple},\n\tbasicstyle=\\footnotesize,\n\tbreakatwhitespace=false, \n\tbreaklines=true, \n\tcaptionpos=b, \n\tkeepspaces=true, \n\tnumbers=left, \n\tnumbersep=5pt, \n\tshowspaces=false, \n\tshowstringspaces=false,\n\tshowtabs=false, \n\ttabsize=2\n}\n\n\\lstset{style=mystyle}\n\n\\title{HW 2: State Estimation in Oil \\& Gas Well Drilling}\n\\date{16 Feb. 2018} \n\\author{Franklin Zhao \\\\ SID: 3033030808}\n\n\\begin{document}\n\t\n\t\\maketitle\n\t\\newcommand{\\tabitem}{~~\\llap{\\textbullet}~~}\n\t\\renewcommand\\theequation{\\arabic{equation}}\n\t\\renewcommand{\\figurename}{Fig.}\n\t\\renewcommand\\thesection{Problem \\arabic{section}:}\n\t\\renewcommand\\thesubsection{(\\alph{subsection})}\n\t\\onehalfspacing\n\t\n\\section{Dynamic System Modeling}\n\\subsection{}\n\\textbf{Modeling objective:} The objective is to formulate a mathmetical model that estimates the drill bit velocity given the table torque $T(t)$.\\\\\n\\textbf{Controllable input:} Table torque $\\bf{T(t)}$\\\\\n\\textbf{Uncontrollable input:} Frictional torque $\\bf{T_f(t)}$\\\\\n\\textbf{Measured output:} Table velocity $\\bf{\\omega_T(t)}$\\\\\n\\textbf{Performance output:} Bit velocity $\\bf{\\omega_B(t)}$\n\\subsection{}\nFor the top/table:\n\\begin{equation}\nJ_T\\ddot{\\theta}_T(t)=-k(\\theta_T(t)-\\theta_B(t))-b\\omega_T(t)+T(t)\n\\end{equation}\nFor the bottom/bit:\n\\begin{equation}\nJ_B\\ddot{\\theta}_B(t)=k(\\theta_T(t)-\\theta_B(t))-b\\omega_B(t)-T_f(t)\n\\end{equation}\n\\subsection{}\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\left[\n\\begin{array}{c}\n\\theta_T\\\\\n\\dot\\theta_T\\\\\n\\theta_B\\\\\n\\dot\\theta_B\n\\end{array}\\right]=\\underbrace{\\left[\n\\begin{array}{cccc}\n0&1&0&0\\\\\n-\\frac{k}{J_T}&-\\frac{b}{J_T}&\\frac{k}{J_T}&0\\\\\n0&0&0&1\\\\\n\\frac{k}{J_B}&0&-\\frac{k}{J_B}&-\\frac{b}{J_B}\n\\end{array}\\right]}_\\text{\\textbf{A}}\\left[\n\\begin{array}{c}\n\\theta_T\\\\\n\\dot\\theta_T\\\\\n\\theta_B\\\\\n\\dot\\theta_B\n\\end{array}\\right]+\\underbrace{\\left[\n\\begin{array}{c}\n0\\\\\n\\frac{1}{J_T}\\\\\n0\\\\\n0\n\\end{array}\n\\right]}_\\text{\\textbf{B}}T+\\left[\n\\begin{array}{c}\n0\\\\\n0\\\\\n0\\\\\n-\\frac{1}{J_B}\n\\end{array}\n\\right]T_f\n\\end{equation}\n\\begin{equation}\n\\dot\\theta_T=\\underbrace{\\left[\n\\begin{array}{cccc}\n0&1&0&0\n\\end{array}\n\\right]}_\\text{\\textbf{C}}\\left[\n\\begin{array}{c}\n\\theta_T\\\\\n\\dot\\theta_T\\\\\n\\theta_B\\\\\n\\dot\\theta_B\n\\end{array}\n\\right]\n\\end{equation}\n\\section{Observability Analysis}\n\\subsection{}\n\\lstinputlisting[language=Python]{2a.py}\n\\begin{Verbatim}\nRank of Observability Matrix for four-state system: 3\n\\end{Verbatim}\nSince $rank(\\mathcal{O})=3<4$, $(A,C)$ is not observable.\n\\subsection{}\n\\begin{equation}\n\\frac{\\partial}{\\partial t}\\left[\n\\begin{array}{c}\n\\theta\\\\\n\\dot\\theta_T\\\\\n\\dot\\theta_B\n\\end{array}\\right]=\\underbrace{\\left[\n\t\\begin{array}{ccc}\n\t0&1&-1\\\\\n\t-\\frac{k}{J_t}&-\\frac{b}{J_t}&0\\\\\n\t\\frac{k}{J_B}&0&-\\frac{b}{J_B}\n\t\\end{array}\\right]}_\\text{\\textbf{A}}\\left[\n\\begin{array}{c}\n\\theta\\\\\n\\dot\\theta_T\\\\\n\\dot\\theta_B\n\\end{array}\\right]+\\underbrace{\\left[\n\t\\begin{array}{c}\n\t0\\\\\n\t\\frac{1}{J_T}\\\\\n\t0\n\t\\end{array}\n\t\\right]}_\\text{\\textbf{B}}T+\\left[\n\\begin{array}{c}\n0\\\\\n0\\\\\n-\\frac{1}{J_B}\n\\end{array}\n\\right]T_f\n\\end{equation}\n\\begin{equation}\n\\dot\\theta_T=\\underbrace{\\left[\n\t\\begin{array}{cccc}\n\t0&1&0\n\t\\end{array}\n\t\\right]}_\\text{\\textbf{C}}\\left[\n\\begin{array}{c}\n\\theta\\\\\n\\dot\\theta_T\\\\\n\\dot\\theta_B\n\\end{array}\n\\right]\n\\end{equation}\n\\subsection{}\n\\lstinputlisting[language=Python]{2c.py}\n\\begin{Verbatim}\nRank of Observability Matrix for three-state system: 3\n\\end{Verbatim}\nSince $rank(\\mathcal{O})=3$, $(A,C)$ is observable.\n\\section{Measurement Data}\n\\lstinputlisting[language=Python]{3.py}\nThe plots of table torque and measured table velocity versus time are shown in Figure~\\ref{fig:3}.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{3.png} \n\t\\caption{Table torque and measured table velocity versus time}\n\t\\label{fig:3}\n\\end{figure}\n\\section{Luenberger Observer}\n\\subsection{}\nDenote $\\left[\n\\begin{array}{c}\n\\hat\\theta\\\\\n\\hat{\\dot\\theta}_T\\\\\n\\hat{\\dot\\theta}_B\n\\end{array}\n\\right]$ by $\\hat x(t)$, $\\hat{\\dot\\theta}_T$ by $\\hat y(t)$, $T(t)$ by $u(t)$, and the Luenberger observer equations will be:\n\\begin{equation}\\label{eq:luen1}\n\\dot{\\hat x}(t)=A\\hat x(t)+Bu(t)+L[y(t)-\\hat y(t)],\\ \\ \\ \\hat x(0)=\\hat{x}_0\n\\end{equation}\n\\begin{equation}\\label{eq:luen2}\n\\hat y(t)=C\\hat x(t)\n\\end{equation}\n\\lstinputlisting[language=Python]{4a.py}\n\\begin{Verbatim}\nEigenvalues of open-loop system: \n[-0.08338525+0.29860789j -0.08338525-0.29860789j -0.08322949+0.j ]\n\\end{Verbatim}\n\\subsection{}\nThe eigenvalues I chose were $2.5\\lambda(A)$. The choice was justified in the RMSE error to be computed and converge speed to be shown in part (d). Basically I chose \\{5, 1, 2, 3, 2.5\\} and did the validation. It turned out that $2.5\\lambda(A)$ was the best choice.\n\\lstinputlisting[language=Python]{4b.py}\n\\begin{Verbatim}\nL:\n[[-25.125]\n[ 0.375]\n[ -0.675]]\n\\end{Verbatim}\n\\subsection{}\nWrite Equation~(\\ref{eq:luen1})(\\ref{eq:luen2}) in LTI form, and we obtain $A_{lobs}$, $B_{lobs}$, $C_{lobs}$:\n\\begin{equation}\n\\dot{\\hat x}(t)=\\underbrace{(A-LC)}_{A_{lobs}}\\hat x(t)+\\underbrace{\\left[\n\\begin{array}{cc}\nB&L\n\\end{array}\n\\right]}_{B_{lobs}}\\left[\n\\begin{array}{c}\nu(t)\\\\\ny(t)\n\\end{array}\n\\right]\n\\end{equation}\n\\begin{equation}\n\\hat y(t)=C_{lobs}\\hat x(t)\n\\end{equation}\n\\subsection{}\nThe eigenvalues I chose were $2.5\\lambda(A)$:\n\\begin{Verbatim}\narray([-0.20846314+0.74651974j, -0.20846314-0.74651974j, -0.20807373+0.j])\n\\end{Verbatim}\n\\lstinputlisting[language=Python]{4d.py}\n\\begin{Verbatim}\nLuenberger Observer RMSE: 0.0991557694635\n\\end{Verbatim}\nThe result of Luenberger observer is shown in Figure~\\ref{fig:4}\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{4.png} \n\t\\caption{Luenberger observer result}\n\t\\label{fig:4}\n\\end{figure}\n\\newpage\n\\section{Kalman Filter (KF) Design}\n\\subsection{}\n\\begin{equation}\n\\dot{\\hat x}(t)=A\\hat x(t)+Bu(t)+L(t)(y(t)-\\hat y(t))\n\\end{equation}\n\\begin{equation}\n\\hat y(t)=C\\hat x(t)\n\\end{equation}\n\\begin{equation}\nL(t)=\\Sigma(t)C^TN^{-1}\n\\end{equation}\nwhere $\\Sigma(t)$ is the solution of the following differential equation:\n\\begin{equation}\n\\dot\\Sigma(t)=\\Sigma(t)A^T+A\\Sigma(t)+W-\\Sigma(t)C^TN^{-1}C\\Sigma(t),\\ \\ \\ \\Sigma(0)=\\Sigma_0\n\\end{equation}\nwhere $W$ and $N$ are covariance matrices of $w(t)$ and $n(t)$.\n\\subsection{}\n\\lstinputlisting[language=Python]{5b.py}\n\\begin{Verbatim}\nKalman Filter RMSE: 0.0560275348234 rad/s\n\\end{Verbatim}\nThe value of $W$ I tuned is listed as follow. Basically I determined the magnitude first. Then based on the plots in part (c), I tuned the diagonal elements, and finally tuned other elements. It turned out that the RMSE did not change too much, but $\\Sigma_{33}$ changed dramatically. \n\\begin{Verbatim}\nW = np.matrix([[0.0006, 0.005, 0.001], \n\t [0.02, 0.003, 0.004], \n\t [0.004, 0.001, 0.003]])\n\\end{Verbatim}\n\\subsection{}\n\\lstinputlisting[language=Python]{5c.py}\nThe result of Kalman Filter is shown in Figure~\\ref{fig:5}.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{5.png} \n\t\\caption{Kalman filter result}\n\t\\label{fig:5}\n\\end{figure}\n\\subsection{}\n\\lstinputlisting[language=Python]{5d.py}\n\\begin{Verbatim}\narray([-0.08707987+0.2849352j, -0.08707987-0.2849352j, -0.34475403+0.j])\n\\end{Verbatim}\nThe 1st and 2nd poles are very close to the eigenvalues I chose in the Luenberger observer, but the 3rd one is a bit more different.\n\\section{Extended Kalman Filter (EKF) Design}\nReplace Hooke's law with nonlinear spring torque relationship, and the EKF can be obtained:\n\\begin{equation}\n\\underbrace{\\frac{\\partial}{\\partial t}\\left[\n\\begin{array}{c}\n\\theta\\\\\n\\dot\\theta_T\\\\\n\\dot\\theta_B\n\\end{array}\n\\right]}_{\\hat x(t)}=\\underbrace{\\left[\n\\begin{array}{c}\n\\hat\\theta_T - \\hat\\theta_B\\\\\n\\frac{-k_1\\theta-k_2\\theta^3-b\\dot\\theta_T+T(t)}{J_T}\\\\\n\\frac{k_1\\theta+k_2\\theta^3-b\\dot\\theta_B}{J_B}\n\\end{array}\t\n\\right]}_{f(x(t),u(t))}+w(t),\\ \\ \\ x(0)=x_0\n\\end{equation} \n\\begin{equation}\n\\dot\\theta_T=\\underbrace{C\\left[\n\\begin{array}{c}\n\\theta\\\\\n\\dot\\theta_T\\\\\n\\dot\\theta_B\n\\end{array}\n\\right]}_{h(x(t),u(t))}+n(t)\n\\end{equation}\nHence, $F(t)$ and $H(t)$ can be derived:\n\\begin{equation}\nF(t)=\\frac{\\partial f}{\\partial x}(\\hat x(t),u(t))=\\left[\n\\begin{array}{ccc}\n0&1&-1\\\\\n\\frac{-k_1-3k_2\\hat\\theta^2}{J_T}&\\frac{-b}{J_T}&0\\\\\n\\frac{k_1+3k_2\\hat\\theta^2}{J_B}&0&\\frac{-b}{J_B}\n\\end{array}\n\\right]\n\\end{equation}\n\\begin{equation}\nH(t)=\\frac{\\partial h}{\\partial x}(\\hat x(t),u(t))=C=\\left[\n\\begin{array}{ccc}\n0&1&0\n\\end{array}\\right]\n\\end{equation}\n\\lstinputlisting[language=Python]{6.py}\n\\begin{Verbatim}\nExtended Kalman Filter RMSE: 0.0458628285791 rad/s\n\\end{Verbatim}\nThe result of EKF is shown in Figure~\\ref{fig:6}.\n\\begin{figure}[H]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{6.png} \n\t\\caption{EKF result}\n\t\\label{fig:6}\n\\end{figure}\n\\end{document}", "meta": {"hexsha": "c51fa3bd60d000254219f7a322fd5d06931f54f2", "size": 9674, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW2 - State Estimation/ZHAO_FRANKLIN_HW2.tex", "max_stars_repo_name": "QinganZhao/Energy-Systems-and-Control", "max_stars_repo_head_hexsha": "ef55c598c2475f4fb3aa9cc9cdae89457e7bae16", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-01-08T02:54:02.000Z", "max_stars_repo_stars_event_max_datetime": "2019-01-08T02:54:02.000Z", "max_issues_repo_path": "HW2 - State Estimation/ZHAO_FRANKLIN_HW2.tex", "max_issues_repo_name": "QinganZhao/Energy-Systems-and-Control", "max_issues_repo_head_hexsha": "ef55c598c2475f4fb3aa9cc9cdae89457e7bae16", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW2 - State Estimation/ZHAO_FRANKLIN_HW2.tex", "max_forks_repo_name": "QinganZhao/Energy-Systems-and-Control", "max_forks_repo_head_hexsha": "ef55c598c2475f4fb3aa9cc9cdae89457e7bae16", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-03-07T11:59:05.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-07T11:59:05.000Z", "avg_line_length": 27.64, "max_line_length": 285, "alphanum_fraction": 0.6903039074, "num_tokens": 3574, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837743174788, "lm_q2_score": 0.8031737987125612, "lm_q1q2_score": 0.6132101632739733}} {"text": "\\subsection{A Gluing Construction}\r\nIn previous parts, we saw that we can make $\\log$ is a complete analytic function.\r\nBut we are not entirely satisfied, as it is just a bunch of function elements related together, instead of a genuin function.\r\nWe shall construct a space as a ``bigger domain'' $R$ at which we can realise $\\log$ as a geniune function.\\\\\r\nThe idea is to consider the function elements we defined earlier and glue them together.\r\nWe define\r\n$$R=\\left( \\coprod_{n\\in\\mathbb Z}U_{I(n)} \\middle)\\right/\\sim$$\r\nwhere $z_1\\in U_{I(m)}$ and $z_2\\in U_{I(n)}$ have $z_1\\sim z_2$ iff $z_1=z_2$ as elements of $\\mathbb C$ and $f_{I(m)}(z_1)=f_{I(n)}(z_2)$.\r\nWe give $R$ the quotient topology.\r\nOne can imagine $R$ as an ``infinite multi-storey carpark'' that spirals up and down.\r\n\\begin{remark}\r\n Since $F_{I(m)}\\approx F_{I(n)}$ for all $m,n\\in\\mathbb Z$, it follows that $R$ is path-connected by following the sequence of direct analytic continuations.\r\n\\end{remark}\r\nThen, with this construction, we can extend all the $f_{I(n)}$ on the $U_{I(n)}$ to a global function $f:R\\to\\mathbb C$ by $f([z])=f_{I(n)}(z)$ for $z\\in U_{I(n)}$.\r\n\\begin{proposition}\r\n $f$ is well-defined.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Follows directly from our definition of the equivalence relation $\\sim$.\r\n\\end{proof}\r\nSimilarly, the natural inclusions $U_{I(n)}\\hookrightarrow\\mathbb C_\\star$ can be extended to a global function $\\pi:R\\to\\mathbb C_\\star$ with $\\pi([z])=z$.\r\nOne can also easily verify that $\\pi$ is well-defined.\r\nThere is a very nice relationship between $f,\\pi$ and the usual exponential map.\r\nIndeed, $\\exp\\circ f=\\pi$, which basically tells us $f$ has basically everything we want from $\\log$.\\\\\r\nWe can use these global functions together.\r\nDefine $\\Phi([z])=(\\pi([z]),f([z]))$, then $\\Phi$ is injective by definition of $\\sim$.\r\nTherefore $R$ is Hausdorff as $\\mathbb C^2$ is.\r\n\\begin{remark}\r\n $\\Phi$ indeed identifies $R$ with the graph $\\{(w,z)\\in\\mathbb C^2:w=\\exp(z)\\}$.\r\n So we can view $R$ alteratively as ``flipping'' the graph of $\\exp$.\r\n\\end{remark}", "meta": {"hexsha": "0a83a01ebeade11e3531d13ce0d68d731c143d9c", "size": 2093, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/glue.tex", "max_stars_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_stars_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "2/glue.tex", "max_issues_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_issues_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "2/glue.tex", "max_forks_repo_name": "david-bai-notes/II-Riemann-Surfaces", "max_forks_repo_head_hexsha": "cbda76f7189c679c4aaccf030b70d310823ead3f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.5161290323, "max_line_length": 165, "alphanum_fraction": 0.6927854754, "num_tokens": 647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837635542924, "lm_q2_score": 0.8031737963569016, "lm_q1q2_score": 0.613210152830756}} {"text": "\\section{Empirical relevance}\n\\begin{frame}{Empirical relevance: definition}\n The \\alert{empirical relevance} $r_i(t)$ of node $i$ at time $t$ is defined as:\n\n \\[\n r_i(t) = \\frac{n_i(t)}{n_i^{PA}(t)}\n \\]\n\n \\begin{itemize}\n \\item $n_i(t) = \\frac{\\Delta k_i^{in}(t, \\Delta t)}{L(t, \\Delta t)}$: ratio between:\n\n \\begin{itemize}\n \\begin{footnotesize}\n \\item $\\Delta k_i^{in}(t, \\Delta t)$: \\# of incoming links received by node $i$ in the time window $[t, t+\\Delta t]$\n \\item $L(t, \\Delta t)$: total \\# of links created within the same time window\n \\end{footnotesize}\n \\end{itemize}\n\n \\item $n_i^{PA}(t) = \\frac{k_i^{in}(t)}{\\sum_j k_j^{in}(t)}$: expected value of $n_i(t)$ according to preferential attachment alone\n \\end{itemize}\n\n $r_i(t) > 1$ \\alert{$(< 1)$}: node $i$ at time $t$ outperforms \\alert{(underperforms)} in the competition for incoming links with respect to its preferencial attachment weight.\n\\end{frame}\n\n\\section{The Extended Fitness Model}\n\n\\begin{frame}{The Extended Fitness Model}\n \\begin{itemize}\n \\item \\textbf{PageRank's under-performance in time-dependent networks is a general feature.}\n \\item We can validate this using a model more compatible with the idea that \\emph{a node is important if it's pointed by other important nodes.}\n \\end{itemize}\n \\begin{center}\n \\alert{Extended Fitness Model} (EFM)\n \\end{center}\n \\begin{itemize}\n \\item \\textbf{High-fitness nodes are more sensitive to fitness} than low-fitness nodes, when choosing their outgoing links.\n \\item High-fitness nodes are then more likely to be pointed by other high-fitness nodes than low-fitness nodes.\n \\item EFM is \\alert{more favorable} to PageRank than RM.\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}{EFM:\\@ sensitivity to fitness}\n Probability $\\Pi_{i;j}^{in}(t)$ that a link created by node $j$ at time $t$ ends in node $i$:\n \\[\n \\Pi_{i;j}^{in}(t) \\sim (k_i^{in}(t)+1)^{1-\\eta_j} \\, \\eta_i^{\\eta_j} \\, f_R(t-\\tau_i)\n \\]\n \\begin{itemize}\n \\item Fitness $\\eta \\in [0, 1]$ to prevent negative exponents\n \\item $\\Pi^{in}$ depends on the fitness of the target \\emph{and of the source} nodes (difference with RM).\n \\item $k_i^{in}(t)$: indegree of node $i$ at time $t$.\n \\end{itemize}\n\\end{frame}\n\n\\begin{frame}{PageRank vs.\\ indegree: correlation with fitness in EFM}\n \\begin{figure}\n \\begin{columns}\n\n \\column{0.6\\textwidth}\n \\includegraphics[width=1.0\\textwidth]{figures/PageRankEFM_heatmap}\n\n \\column{0.4\\textwidth}\n \\begin{footnotesize}\n \\begin{itemize}\n \\item $\\rho(A) = 2A^{-3}, \\; A \\in [1, \\infty]$\n \\item $H$ nodes: high fitness; $\\eta \\in [10^{-5}, 1]$\n \\item $(N-H)$ nodes: low fitness; $\\eta \\in [0, 10^{-5}]$\n \\item $N=10000$, $H=250$\n \\end{itemize}\n \\end{footnotesize}\n \\end{columns}\n \\caption{Comparison of performance of PageRank and indegree (EFM data).}\n \\end{figure}\n\\end{frame}\n", "meta": {"hexsha": "41de768fa81b0a2b0dc2628659856282db7a6b66", "size": 3142, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/appendix.tex", "max_stars_repo_name": "pietrodn/csr_pagerank", "max_stars_repo_head_hexsha": "5c037e0e085b697c9c79950c9c4fe5cbcc494c6d", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2017-11-09T14:36:59.000Z", "max_stars_repo_stars_event_max_datetime": "2017-11-09T14:36:59.000Z", "max_issues_repo_path": "sections/appendix.tex", "max_issues_repo_name": "pietrodn/csr_pagerank", "max_issues_repo_head_hexsha": "5c037e0e085b697c9c79950c9c4fe5cbcc494c6d", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/appendix.tex", "max_forks_repo_name": "pietrodn/csr_pagerank", "max_forks_repo_head_hexsha": "5c037e0e085b697c9c79950c9c4fe5cbcc494c6d", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 42.4594594595, "max_line_length": 180, "alphanum_fraction": 0.6107574793, "num_tokens": 953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.868826769445233, "lm_q2_score": 0.7057850278370111, "lm_q1q2_score": 0.6132049256584442}} {"text": "\\documentclass[../../main.tex]{subfiles}\n\\begin{document}\n\\subsubsection{Append}\n\nThe operation $append$ will return a Collection with a Value added at a specified numeric Index.\n\\begin{schema}{Append[Collection, V, \\nat]}\n coll?, coll! : Collection \\\\\n v? : V \\\\\n idx? : \\nat \\\\\n append~\\_ : Collection \\cross V \\cross \\nat \\bij Collection\n \\where\n \\#~idx? = 1 \\\\\n coll! = append(coll?, v?, idx?) @ \\\\\n \\t1 let ~ ~ coll' == front(\\{~ i : \\nat ~| ~ i \\in 0~..~idx?\\} \\extract coll?) \\cat v? \\\\\n \\t1 \\ \\ \\ ~~~ coll'' == \\{~ j : \\nat ~| ~ j \\in idx?~..~\\#~coll?\\} \\extract coll? \\\\\n \\t1 = coll' \\cat coll'' \\implies \\\\\n \\t2 (front(coll') \\cat v? \\cat coll'') ~ \\land \\\\\n \\t2 (~ v? \\mapsto idx? \\in coll!) ~\\land \\\\\n \\t2 (~\\#~ coll! = \\#~ coll? + 1)\n\\end{schema}\n$append$ results in the composition of $coll'$ and $coll''$ such that\n$$coll! = coll' \\cat coll'' ~ \\land ~idx? \\mapsto v? \\in coll!$$\n\\begin{itemize}\n\\item $coll'$ is the items in $coll?$ up to and including $idx?$ but the value at $idx?$ is replaced with $v?$ such that\n $idx? \\mapsto coll?_{idx?} \\not \\in coll'$\n\\item $coll''$ is the items in $coll?$ from $idx?$ to $\\#~coll? \\implies coll?_{idx?} \\in coll''$\n\\end{itemize}\n The following example illustrates these properties.\n\\begin{argue}\n X = \\langle x_{0}, x_{1}, x_{2} \\rangle \\\\\n \\t1 x_{0} = 0 \\\\\n \\t1 x_{1} = foo \\\\\n \\t1 x_{2} = \\langle a, b, c \\rangle \\\\\n \\t1 v? = bar \\\\\n append(X, v?, 0) = \\langle bar, 0, foo, \\langle a, b, c \\rangle \\rangle \\\\\n append(X, v?, 1) = \\langle 0, bar, foo, \\langle a, b, c \\rangle \\rangle \\\\\n append(X, v?, 2) = \\langle 0, foo, bar, \\langle a, b, c \\rangle \\rangle \\\\\n append(X, v?, 3) = \\langle 0, foo, \\langle a, b, c \\rangle, bar \\rangle \\\\\n append(X, v?, 4) = append(X, v?, 3) \\iff 3 \\not \\in \\dom X\n\\end{argue}\n\\end{document}\n", "meta": {"hexsha": "985033ec70e75a5804188b0ce9bd6ee2b64f642c", "size": 1811, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/operations/collections/append.tex", "max_stars_repo_name": "yetanalytics/dave", "max_stars_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2018-08-17T00:38:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T02:32:37.000Z", "max_issues_repo_path": "docs/operations/collections/append.tex", "max_issues_repo_name": "adlnet/dave", "max_issues_repo_head_hexsha": "9339713fac747118e462e4fc7e1ecd54e5d916e3", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 95, "max_issues_repo_issues_event_min_datetime": "2018-08-31T18:57:51.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-15T16:40:01.000Z", "max_forks_repo_path": "docs/operations/collections/append.tex", "max_forks_repo_name": "yetanalytics/dave", "max_forks_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2019-09-28T06:48:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-31T16:20:47.000Z", "avg_line_length": 43.119047619, "max_line_length": 120, "alphanum_fraction": 0.5726118167, "num_tokens": 684, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8688267762381844, "lm_q2_score": 0.7057850216484837, "lm_q1q2_score": 0.6132049250760493}} {"text": "\\section{Defining data for Oriented TQFT}\nThis is the defining data for $(d + 1)$-dimensional oriented TQFT.\nIt Closed $d$-dimensional $\\Sigma^d$\nhas vector space $V(\\Sigma^d)$.\nWhen $M^{d+1}$ is closed, meaning it has no boundary,\n$Z(M^{d+1})\\in\\mathbb{C}$ is just a number.\nIf $M^{d+1}$ has a boundary,\nthen $Z(M^{d+1})\\in V(\\partial M^{d+1})$.\nThis is a state.\n$Z$ is the path integral on the spacetime manifold.\nYou can think of it as a wave function over boundary conditions.\n\nThese are the axioms.\nAll manifolds have orientation.\n\n\\begin{enumerate}\n \\item $Z$ and $V$ are functorial with respect to orientation-preserving\n diffeomorphisms of $M^{d+1}$ and $\\Sigma^d$.\n \\item $V$ should be ``involutory'' in the sense that\n $V(\\bar{\\Sigma^d}) = V(\\Sigma^d)^*$.\n Here the $*$ means dual vector space.\n Over bar means hermitian conjugate.\n \\item $Z$ and $V$ should be multiplicative.\n \\item Non-triviality.\n \\item $Z$ is hermitian.\n\\end{enumerate}\n\nLet's explain the axioms.\n\\begin{axiom}[Axiom 1]\n Let $f:\\Sigma_1^d \\to \\Sigma_2^d$ be a diffeomorphism.\n This defines an isomorphism\n $\\mathcal{U}_f: V(\\Sigma_1^d)\\to V(\\Sigma_2^d)$.\n If we also have $g: \\Sigma_2^d \\to \\Sigma_3^d$,\n then\n $\\mathcal{U}_g \\mathcal{U}_f = \\mathcal{U}_{gf}$.\n\n If $f$ extends to\n $\\tilde{f}: M_1^{d+1}\\to M_2^{d+1}$\n where $\\partial M_1 = \\Sigma_1$\n and $\\partial M_2 = \\Sigma_2$,\n then\n $\\mathcal{U}_f\\left( Z\\left( M_1^{d+1} \\right) \\right)\n = Z\\left( M_2^{d=1} \\right)$.\n\\end{axiom}\nSuppose we have a diffeomorphism from one circle $\\Sigma_1 = S^1$ to another\ncircle.\nThen there will be an induced map $\\mathcal{U}_f$ from vector spaces between\nthen too.\n\n\\begin{question}\n Isn't $Z$ a path integral that is a number?\n Why is it a vector space?\n\\end{question}\nYou can think of $Z\\left( M_1^{d+1} \\right)$\nas a state,\nbecause like in path integrals you have boundary conditions,\nwhich are variable.\n\n\n\\begin{question}\n What is a diffeomorphism?\n\\end{question}\nA diffeomorphism $f:X_1 \\to X_2$ is a smooth and continuous map.\n\n\\begin{question}\n What is a manifold?\n\\end{question}\nManifolds can be broken up into charts $\\mathcal{U}_\\alpha$\nwith coordinate charts $\\phi: \\mathcal{U}_\\alpha\\to \\mathbb{R}^d$.\n\n\\begin{question}\nWhat is the relation between $\\mathcal{U}_{f}$ and\n$\\mathcal{U}_{\\tilde{f}}$?\n\\end{question}\nThey are actually the same.\n\n\\begin{question}\n The extension from $f$ to $\\tilde{f}$ is not unique?\n\\end{question}\nIt's not unique.\n\n\\begin{question}\n $\\mathcal{U}_f$ acting on the partition function\n gives the other partition function.\n This is the sum of all paths.\n\\end{question}\nDon't think of it like that.\nThis thing is only acting on the partition function.\nNow you're saying $Z$ is the sum of something\nand how does $U_{f}$ act on terms in the sum.\nI'm hesitant to say anything general about that.\nThere's no general clean answer.\nJust think of it as an abstract object $Z$\nwhich gives a state if there is a boundary\nand a number otherwise.\n\nLet's talk about axiom 3.\nImagine if I build a manifold by gluing simpler manifolds.\nThen I can build my vector spaces by some tensor products and multiplications.\nRecall that $\\sqcup$ is the disjoint union.\nThen we can state the axiom.\n\\begin{axiom}[Axiom 3]\n Let\n \\begin{align}\n V\\left( \\Sigma_1^d \\sqcup \\Sigma_2^d \\right)\n =\n V\\left( \\Sigma_1^d \\right) \\otimes\n V\\left( \\Sigma_2^d \\right)\n \\end{align}\n Suppose the boundary is\n \\begin{align}\n \\partial M^{d+1} =\n \\Sigma_1 \\sqcup \\bar{\\Sigma}_2\n \\end{align}.\n Then\n \\begin{align}\n Z\\left( M_1^{d=1} \\right) &\\in\n V\\left( \\Sigma_1 \\right) \\otimes V\\left( \\bar{\\Sigma}_2 \\right)\\\\\n &\\simeq V\\left( \\Sigma_1 \\right) \\otimes \\overbar{V(\\Sigma_2)}\n \\end{align}\n so we can think of $Z\\left( M^{d+1}_1 \\right)$\n as a state in\n $V\\left( \\Sigma_1 \\right)\\otimes V\\left( \\overbar{\\Sigma_2} \\right)$\n or as a linear map from\n $V\\left( \\Sigma_2 \\right)\\to V\\left( \\Sigma_1 \\right)$.\n\\end{axiom}\nI can think of this as a state in a tensor product of two spaces.\n\\begin{align}\n Z\\left( M_1^{d+1} \\right)\n = \\sum_{n,m} c_{n,m}\n \\ket{n}_{V(\\Sigma_1)}\n \\otimes \\ket{m}_{V(\\overbar{\\Sigma_2})}\n\\end{align}\nor you can think of it as\n\\begin{align}\n Z\\left( M_1^{d+1} \\right)\n = \\sum_{n,m} c_{n,m}\n \\ket{n}_{V(\\Sigma_1)}\n \\bra{m}_{V(\\overbar{\\Sigma_2})}\n\\end{align}\n\n\\begin{question}\nThis only applies when the boundary is a disjoint union of two manifolds?\n\\end{question}\nYou can always think of the boundary as $\\partial M^{d+1}=\\Sigma_1 \\sqcup\n\\emptyset$\n\n\\begin{question}\n Why do we need the complex conjugate?\n\\end{question}\nIt's a state on $V(\\Sigma_1)\\otimes V(\\ovebar{\\Sigma_2})$.\nActually maybe it shouldn't be there,\nput a question mark there.\nActually, there should be no complex conjugate.\n\nWe can think of the path integral as a state on the vector space\n$V(\\Sigma_1)\\otimes\\overbar{V(\\Sigma_2)}$\nor as a map from\n$V(\\Sigma_2)\\to V(\\Sigma_1)$.\n\nI can think of a state as a map from the vector space to the empty manifold.\nI'm just moving pieces around no each side.\n\nIt's really nothing more than thinking of it as bras and kets.\nJust think of this as a definition.\nI have a state that lives on a tensor product of two vector spaces.\nI can think of it as an operator from one vector space to another.\nIt's really just that simple.\nThere's nothing more to that.\nForget that we called it the path integral.\nYou can think of any tensor product state as an operator from one space to\nanother.\n\n\\begin{question}\n When you have 3 disjoint components of $\\Sigma$,\n then how do you define orientation?\n How do you decided which orientation to use?\n\\end{question}\nI don't think you need extra information about what the boundary orientation\nlooks like.\nIf this has + orientation, then are you saying that?\n\nActually I think there should be a star.\n\\begin{align}\n Z\\left( M_1^{d+1} \\right)\n = \\sum_{n,m} c_{n,m}^*\n \\ket{n}_{V(\\Sigma_1)}\n \\bra{m}_{V(\\overbar{\\Sigma_2})}\n\\end{align}\nWe'll come back to this.\n\nThe path integral is a state on the boundary.\n\nWe've done the multiplicative axiom for $V$.\nWe haven't gotten to the multiplicative axiom for $Z$ yet.\nWe want to glue together $M$s with common boundaries.\n\nSuppose that we have some manifold $M_1$\nand another manifold $M_2$.\nThe boundary $\\partial M_1 = \\Sigma_1 \\sqcup \\overbar{\\Sigma_2}$\nand the boundary\n$\\partial M_2 = \\Sigma_2 \\sqcup \\overbar{\\Sigma_3}$.\nThen $M=M_1 \\cup_{\\Sigma_2} M_2$.\nIf you think of these $Z$ as linear maps,\nthen we just compose the maps.\n\\begin{align}\n Z(M) = Z(M_2)\\circ Z(M_1): V(\\Sigma_1) \\to V(\\Sigma_3)\n\\end{align}\nor equivalently,\nwe can think of $Z(M)$ as the inner product of these two states,\nwhere the inner product\n\\begin{align}\n Z(M) =\n {\\langle Z(M_1), Z(M_1) \\rangle}_{V(\\Sigma_2)}\n \\in V(\\Sigma_1) \\otimes V(\\overbar{\\Sigma_3})\n\\end{align}\n\n\\begin{question}\n The bar just means going counterclockwise right?\n\\end{question}\nYes.\n\n\nActually there are many ways of glueing manifolds.\nWhen I identify points together,\nI can actually identify with any points I want.\nSo in general we need a gluing map\n$\\phi: \\Sigma_2 \\to \\Sigma_2$\nwith\n$M = M_1 \\cup_\\phi M_2$.\n\nI'm going to use Axiom 1\nand this $\\phi$ map is going to define a corresponding map\n$\\mathcal{U}_\\phi: V(\\Sigma_2)\\to V(\\Sigma_2)$\nso\n\\begin{align}\n Z\\left( M_1 \\cup_\\phi M_2 \\right)\n =\n Z(M_2) \\circ \\mathcal{U}_\\phi \\circ Z(M_1).\n\\end{align}\nOr equivalently\n$\\langle Z(M_2), \\mathcal{U}_\\phi Z(M_1)\\rangle_{V(\\Sigma_2)}$.\n$\\mathcal{U}_\\phi$ is an isomorphism between these two vector spaces.\n\nClosed $M^{d=1}$\ncan be obtained by gluing together\n$M_1$, $M_2$ along the their common boundary.\nFor example,\nif you wanted to calculate the path integral on a sphere,\nyou blue toegether the two on a common boundary.\n\nIf you want a torus,\nou glue\n\\begin{align}\n T^2 = (S_1\\otimes I) \\union (S^1\\times I).\n\\end{align}\n\nThat was the only non-trivial axiom,\nthe rest are pretty simple.\n\nFunnily enough,.\nThis is about what happens when we have a non-empty manifold.\nthe next axiom is called non-triviality\n\n\\begin{axiom}[Axiom 4: non-triviality]\n Consider a boundary $\\Sigma^d = \\emptyset$.\n Then\n V(\\emptyset) = V(\\emptyset) \\otimes V(\\emptyset)\n and\n \\begin{align}\n V(\\emptyset) =\n \\begin{cases}\n 0 & \\text{ordinal}\\\\\n \\mathbb{C} & 1-dimensional\n \\end{cases}\n hence we have\n \\begin{align}\n V(\\phi) = \\mathbb{C}\n \\end{align}\n \\end{align}\n If $M^{d+1}$ is closed,\n $\\partial M^{d+1} = \\emptyset$,\n then\n $Z(M^{d+1}) \\in V(\\emptyset) \\simeq \\mathbb{C}$.\n Consider the case\n \\begin{align}\n M^{d+1} = \\emptyset.\n \\end{align}\n Then\n \\begin{align}\n Z(\\emptyset) = Z\\emptyset) Z(\\emptyset)\n \\end{align}\n which implies\n \\begin{align}\n Z(\\emptyset) =\n \\begin{cases}\n 0 &\\\\\n 1\n \\end{cases}.\n \\end{align}\n Take $Z(\\emptyset)=1$.\n\\end{axiom}\n\nconsider $M^{d+1}= \\sigma^d \\times I$.\nThen\n\\begin{align}\n Z\\left( \\Sigma^d \\times I \\right)\n =\n \\begin{cases}\n 0\\\\\n II_{V(\\Sigma^d)}\n \\end{cases}\n\\end{align}\nTake\n$Z(\\Sigma^d\\times I) = II_{V(\\Sigma^d)}\n= \\sum_{n=1}^{\\dim(V(\\Sigma^d))}\n\\ket{n}\\bra{n}$\nwhere $\\{\\ket{n}\\}$ is orthonormal basis in $V(\\Sigma^d)$.\nOr,\n\\begin{align}\n Z(\\Sigma^d\\times I) =\n \\sum_n\n \\ket{n}_{V(\\Sigma^d)}\n \\otimes\n \\ket{n}_{V(\\Sigma)}^d.\n\\end{align}\n\nThe last axiom is simple.\n\\begin{axiom}[Hermitian]\n $Z(\\overbar{M^{d+1}}) = Z(M^{d+1})^\\dagger$.\n\\end{axiom}\n\n\\section{Non-trivial consequences}\n\\subsection{Homotopy invariance of maps}\nThis has some non-trivial consequences.\nConsider $M_1^{d+1}=\\Sigma^d\\times I$\nand consider $M_2^{d+1}=\\Sigma^2\\times I$\nwith homotopy\n$F: M_1\\to M_2$.\nThen\n$\\mathcal{F}(\\Sigma^d, t)\n= f_t(\\Sigma^d)$\nfor $f_t: \\Sigma_d\\to \\Sigma^d$.\nThis defines a continuous family\n$\\mathcal{U}_{f_t}: \\Sigma^d\\to \\Sigma^d$ satisfying\n$\\mathcal{U}_{f_t} = \\mathcal{U}_{f_{t'}}$.\n\n[diagram here]\n\nIf I have a homotopy group with a homotopy of maps.\n\n\\subsection{Mapping Class Group}\nThat leads us to the important group for a manifold.\nFor every manifold, there is a mapping class group $\\Sigma^d$.\nColloquially, it's the group of ``large'' diffeomorphisms.\nYou would take all diffeomorphisms and mod out all those that are continuously\n\\begin{align}\n \\mathrm{MCG}(\\Sigma^d) = \\frac{\\mathrm{Diff}(\\Sigma^d)}{\\mathrm{diff}_0(\\Sigma^d)}\n\\end{align}\nTwo diffeomorphisms are equivalent if they can be connected to each other.\n\nThis is the group taking all possible diffeomorphisms\nand modding out all the ones equivalent to the identity.\n\nThe mapping class group is related to the homotopy group,\nbut not the same thing.\nYou may not understand this,\nbut\n\\begin{align}\n \\mathrm{MCG}(\\Sigma^d)\n = \\mathrm{out}(\\pi_1(\\Sigma^d))\n\\end{align}\nIt's isomorphic to the group of all outer maps between loops.\nAn example of such a map that maps a non-trivial loop on an torus\nto the other type of non-trivial loop around the torus.\n\nIn TQFT,\nall my maps are smooth.\nIf my maps aren't smooth,\nI don't have something that takes me everywhere from the left to everywhere on\nthe right.\nIf I have a smooth map,\nthat's what allows me to take vector spaces on the left to vector space on the\nright.\nThe only reason I can go from left to right is because $\\mathcal{F}$ is a\ndiffeomorphism.\n\nThe punchline is this.\nFor each $g\\in \\mathrm{MCG}(\\Sigma^d)$,\nwe get some $\\mathcal{U}_g: V(\\Sigma^d) \\to V(\\Sigma^d)$.\nIf you figure out the $\\mathcal{U}_g$,\nyou can completely figure out the TQFT.\nThese $\\mathcal{U}_g$ are very important,\nand can basically characterise the TQFT.\n\nIs this the $S$ and $T$ matrices?\nYes.\nFor every element in the MCG of a torus,\nthere is a linear map that acts on the ground state subspace on the system,\nhence the $S$ and $T$ matrices on the subspace.\nDon't worry if you've never heard of it.\n\nDo you have a concise reference to recommend for the last two lectures?\nThere's an article from Atiyah 1988.\nIt's basically called TQFT.\nMy whole representation is based on this article.\n\n\\begin{question}\n Is it readable?\n\\end{question}\nIt's as readable as my lectures are understandable.\nIf you don't understand my lecture,\nthen I'm not sure.\nIt's not too bad,\nit doesn't assume you know that much.\n\nWe're basically done,\nbut there's one more consequence to talk about.\nThese TQFTs are very baby versions of TQFTs,\nand in real applications, we need some extra structure.\n", "meta": {"hexsha": "78168a09fc0dd8cc34081ddc96c456b7a5a33e9d", "size": 12659, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "phys733/lecture4.tex", "max_stars_repo_name": "ehua7365/umdphysnotes", "max_stars_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-12-11T12:53:46.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-11T12:53:46.000Z", "max_issues_repo_path": "phys733/lecture4.tex", "max_issues_repo_name": "ehua7365/umdphysnotes", "max_issues_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "phys733/lecture4.tex", "max_forks_repo_name": "ehua7365/umdphysnotes", "max_forks_repo_head_hexsha": "00e4e2b6aba3d03baaec5caa36903e5135b014de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.0688836105, "max_line_length": 86, "alphanum_fraction": 0.6817284146, "num_tokens": 4085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424450764199, "lm_q2_score": 0.724870282120402, "lm_q1q2_score": 0.6131985388201672}} {"text": "\n\\chapter{\\projmr3 \\ Multi-Channel Data Set}\n\\index{multi-channel}\n\\section{Introduction}\n\nModern image processing applications often involve multispectral data.\nThis can be color images in R, G and B coordinates, observations of the same \narea but at different times, etc. Much effort in the compression application\ndomain\nhas been expended in recent years in order to compress efficiently such\ndata sets. The challenge is to have a data representation which takes\ninto account at the same time both the spatial and the spectral \n(or temporal) correlation. A three-dimensional transform-based coding\ntechnique has been proposed in \\cite{mc:saghri95}, consisting of\na one-dimensional spectral Karhunen-Lo\\`eve transform \n\\cite{ima:karhunen47} (KLT)\nand a two-dimensional spatial discrete cosine transform (DCT).\nThe KLT is used to decorrelate the spectral domain and the \nDCT is used to decorrelate the spatial domain.\nAll images are first decomposed into blocks, and each block uses its own\nKarhunen-Lo\\`eve transform instead of one single transform \nfor the whole image.\nLee \\cite{mc:lee99} has improved on this approach by introducing a varying\nblock size. The block size is defined using a quadtree, followed by \nbit allocation\nfor each block. The DCT transform can also be replaced by a wavelet \ntransform \\cite{mc:epstein92,mc:tretter95} (WT). \n\nWe present in this chapter how a Wavelet-KLT transform can be used for\nnoise removal. Decorrelating first the data in the spatial domain using \nthe WT and following that in the \nspectral domain, by the KLT, has the advantage of\nproviding us with robust noise modeling in the WT-KLT space, and hence to\nbe able \nto filter the transformed data in an efficient and effective \nway. We show also that\nthe correlation matrix can be computed by different methods based on \nthe noise modeling. These methods are evaluated with a set of different \nimages. \n \nWe first present the Wavelet-KLT transform and how to model the \nnoise in this space. We then describe three different methods to \nfilter the coefficients, and finally illustrate these approaches with\n a set of experiments.\n\n% We are interested in having an entropy measurement \n% of such data sets, knowing that the correlation between two images \n% can be relatively high. A 3-dimensional wavelet transform is not \n% appropriate because the third dimension is generally completely different\n% from the two first. The standard case is the data set where we have \n% 2-dimensional spatial information versus a frequency band.\n \n\n\\section{The Wavelet-Karhunen-Lo\\`eve transform}\n\\subsection{Definition}\nThe Karhunen-Lo\\`eve transform, also often referred to as eigenvector,\nHotelling transform, or Principal Component Analysis (PCA)\n \\cite {ima:karhunen47,ima:loeve48,ima:hotelling33}\nallows us to transform discrete signals into a sequence of uncorrelated \ncoefficients.\nConsidering a vector $D={d_1,..., d_L}$ \nof $L$ signals or images of dimension N \n(i.e., $N$ pixels per image), \nwe denote $M = \\left\\{m_1, ..., m_L\\right\\}$ the \nmean vector of the population ($m_i$ is the mean of the ith\nsignal $d_i$). The covariance matrix $C$ of $D$ is defined by \n$C = (D-M)(D-M)^t$, and is of order $L \\times L$. \nEach element $c_{i,i}$ of $C$ is\nthe variance of $d_i$, and each element $c_{i,j}$ is the \ncovariance between $d_i$ and\n$d_j$. The KLT method consists of \napplying the following transform to all\nvectors $x_i = \\left\\{d_1(i),...,d_L(i)\\right\\}$ ($i=1..N$):\n\\begin{eqnarray}\ny_i = \\Lambda^{-\\frac{1}{2}}A(x_i-M)\n\\label{eqn_trans}\n\\end{eqnarray}\nwhere $\\Lambda$ is the diagonal matrix of eigenvalues of the covariance \nmatrix $C$, and $A$ is a matrix whose rows are formed \nfrom the eigenvectors\nof $C$ \\cite{ima:gonzalez93}, ordered following \ndecreasing order of eigenvalues. \n\nBecause the rows of $A$ are orthonormal vectors, $A^{-1} = A^t$, and\nany vector $x_i$ can be recovered from its corresponding $y_i$ by:\n\\begin{eqnarray}\nx_i = \\Lambda^{\\frac{1}{2}}A^t y_i + M\n\\end{eqnarray}\n\nThe $\\Lambda$ matrix multiplication can be seen as a normalization.\nBuilding $A$ from the correlation matrix instead of the covariance matrix\nleads to another kind of normalization, \nand the $\\Lambda$ matrix can be suppressed \n($y_i = A(x_i-M)$ and $x_i = A^t y_i + M$). Then the norm of $y$ will \nbe equal to the norm of $x$. \n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{ \n\\psfig{figure=fig_wtpca.ps,bbllx=1cm,bblly=11cm,bburx=19cm,bbury=27cm,width=12cm,height=10cm,clip=}\n}\n}}\n\\caption{WT-KLT transform flowchart. Each frame of the input data set is first wavelet\ntransformed, and a principal component analysis is applied at each resolution\nlevel.}\n\\label{fig_wtpca}\n\\end{figure}\n\nWe suppose now that we have $L$ observations of the same view, e.g.\\\nat different wavelengths (or at different epochs, etc.), and denote \nas $d_l$ one \nobservation, $W^{(l)}$ its wavelet transform, and $w_{l,j,k}$ one wavelet\ncoefficient at scale $j$ and at position $k$. \nThe standard approach would be to use an orthogonal wavelet transform, and\nto calculate the correlation matrix $C$ from the wavelet coefficients instead\nof the pixel values:\n\\begin{eqnarray}\nC_{m,n} = \n \\frac{\\sum_{j=1}^{J} \\sum_{k=1}^{N_j} w_{m,j,k} w_{n,j,k}}\n {\\sqrt{\\sum_{j=1}^{J} \\sum_{k=1}^{N_j} w^2_{m,j,k}} \\sqrt{\\sum_{j=1}^{J} \\sum_{k=1}^{N_j} w^2_{n,j,k}}}\n\\label{Correlation_matrix1}\n\\end{eqnarray}\nwhere $J$ is the number of bands, and $N_j$ is the number of coefficients\nin the band $j$.\nIn \\cite{mc:lee99}, a more complex approach has been\nproposed, which is to decompose the images into $N_b$ blocks and \napply a KLT o each block separately. \nWe investigate here different approaches for data restoration.\n\n\n\\subsection{Correlation matrix and noise modeling}\nWe introduce a noise model into our calculation of the correlation matrix.\nIndeed, if the input sequence $D$ contains noise, then the \nwavelet coefficient \nare noisy too. Eigenvalues at the high scales are computed with \nnoisy WT coefficients and \nwe may lose the true underlying correlation that exists \nbetween the input images $d_l$.\nThe expression of the correlation matrix has to be modified in order \nto allow us to take into account the noise.\nWe add a weighting term to each wavelet coefficient which \ndepends on the signal to noise ratio.\nThe correlation matrix is calculated by:\n\\begin{eqnarray}\nC_{m,n} =\n {\\frac{ \\sum_{j=1}^{J} \\sum_{k=1}^{N_j} p_j(w_{m,j,k}) w_{m,j,k} p_j(w_{n,j,k}) w_{n,j,k}}\n {\\sqrt{ \\sum_{j=1}^{J} \\sum_{k=1}^{N_j} p^2_j(w_{m,j,k}) w^2_{m,j,k}} \\sqrt{\\sum_{j=1}^{J} \\sum_{k=1}^{N_j} p^2_j(w_{n,j,k}) w^2_{n,j,k}}}}\n\\label{Correlation_matrix2}\n\\end{eqnarray}\nwhere $p_j$ is a weighting function. The standard approach corresponds to the \nspecific case where $p_j(w_m) = 1$ (no weighting). By considering that only\nwavelet coefficients with high signal to noise ratio should be used for \nthe correlation matrix calculation, $p_j$ can be defined by:\n\\begin{eqnarray}\np_j(w) = \\left\\{\n \\begin{array}{ll}\n \\mbox{ 1 } & \\mbox{ if } w \\mbox{ is significant} \\\\\n \\mbox{ 0 } & \\mbox{ if } w \\mbox{ is not significant}\n \\end{array}\n \\right.\n\\end{eqnarray}\nand a wavelet coefficient $w$ is said to be ``significant'' \nif its probability of being \ndue to noise is smaller than a given $\\epsilon$ value. In the case of\nGaussian noise, it suffices to compare the wavelet coefficients $w$ to\na threshold level $t_j$. $t_j$ is generally taken as $\\lambda \\sigma_j$, where\n$\\sigma_j$ is the noise standard deviation at scale $j$, and $\\lambda$ is \nchosen\nbetween 3 and 5. The value of $\\lambda=3$ corresponds to a probability of \nfalse\ndetection of $0.27$\\%, for Gaussian statistics.\n\nThis hard weighting scheme may lead to problems if only a few coefficient are \nsignificant, and can be replaced by a soft weighting one, by\ndefining $p_j(w)$ by:\n\\begin{eqnarray}\np_j(w) = 1 - \\mathrm{Prob}(W > \\mid w \\mid) \n\\end{eqnarray}\nwhere $\\mathrm{Prob}(W > \\mid w \\mid)$ is the probability that a wavelet\ncoefficient is larger than $w$ due to the noise. For Gaussian noise, we have:\n\\begin{eqnarray}\n p_j(w) & = & 1 - \\frac{2}{\\sqrt{2 \\pi} \n\\sigma_j} \\int_{\\mid w \\mid}^{+\\infty} \\exp(-W^2/2\\sigma^2_j) dW \\nonumber \\\\ \n & = & \\mbox{erf}(\\frac{\\mid w \\mid }{\\sqrt{2}\\sigma_j})\n\\end{eqnarray}\n\n\\subsection{Scale and Karhunen-Lo\\`eve transform}\n\nWe can also analyze separately each band of the \nwavelet transform, and then apply one KLT per resolution level. This \nimplies calculating a correlation matrix $C^{(j)}$ for each band $j$.\n\n\\begin{eqnarray}\nC^{(j)}_{m,n} = \n {\\frac{\\sum_{k=1}^{N_j} p_j(w_{m,j,k}) w_{m,j,k} p_j(w_{n,j,k}) w_{n,j,k}}\n {\\sqrt{\\sum_{k=1}^{N_j} p^2_j(w_{m,j,k}) w^2_{m,j,k}} \\sqrt{\\sum_{k=1}^{N_j} p^2_j(w_{n,j,k}) w^2_{n,j,k}}}}\n\\label{Correlation_matrix3}\n\\end{eqnarray}\n\nThis has the advantage to take into account more complex behavior of the\nsignal. Indeed, structures of different sizes may have a different \nspectral behavior (for example, stars and galaxies in astronomical images),\nand a band-by-band independent analysis allows us to better represent\nthis kind of data.\n\n\n\\subsection{The WT-KLT transform}\nThe final WT-KLT algorithm has these steps:\n\\begin{enumerate}\n\\item Estimate the noise standard deviation $\\sigma^{(l)}$ of each input\ndata set $d_l$.\n\\item Calculate the wavelet transform $W^{(l)}$ of its each input\ndata set $d_l$.\n\\item For each band $j$ of the wavelet transform, calculate\nthe correlation matrix $C^{(j)}$ relative to the vector \n$x_j = \\left\\{ W^{(1)}_j, W^{(2)}_j, ..., W^{(L)}_j \\right\\}$, where \n$W^{(l)}_j$ represents the band $j$ of the wavelet transform $W^{(l)}$ of $d_l$.\n\\item For each band $j$, we diagonalize the matrix $C^{(j)}$ and build the\ntransform matrix $A_j$ from the eigenvectors of $C^{(j)}$.\n\\item For each band $j$ and each position $k$, we apply the matrix $A_j$\nto the vector\n$x_{j,k} = \\left\\{ w_{1,j,k}, w_{2,j,k}, ..., w_{L,j,k} \\right\\}$:\n\\begin{eqnarray}\n y_{j,k} = A_j x_{j,k}\n\\end{eqnarray}\n\\item The WT-KLT coefficients $c_{l,j,k}$ are derived from $y_{j,k}$ by\n$c_{l,j,k} = y_{j,k}(l)$. The $l$ index in the transformed coefficients\nno longer represent the observation number, but instead the \neigenvector number.\n$l=1$ indicates the main eigenvector while $l=L$ indicates the last one.\n\\end{enumerate}\nThe mean vector $M$ disappears in this algorithm because the wavelet \ncoefficients are zero mean.\n\nFigre~\\ref{fig_wtpca} shows the flowchart of the WT-KLT transform.\n\n\n\\subsection{The WT-KLT reconstruction algorithms}\n \nThe reconstruction algorithm has these steps:\n\\begin{enumerate}\n\\item For each band $j$ and each position $k$, we apply the matrix $A_j^t$\nto the vector \n$y_{j,k} = \\left\\{ c_{1,j,k}, c_{2,j,k}, ..., c_{L,j,k} \\right\\}$\n\\begin{eqnarray}\n x_{j,k} = A_j^t y_{j,k}\n\\end{eqnarray}\n\\item The wavelet coefficients $w_{l,j,k}$ are derived from $x_{j,k}$ by\n $ w_{l,j,k} = x_{j,k}(l)$.\n\\item An inverse wavelet transform of $W^{(l)}$ furnishes $d_l$. \n\\end{enumerate}\n\n\n\n\n\\subsection{WT-KLT Transform of 1D Multichannel Data: wk1d\\_trans}\n\\index{wk1d\\_trans}\n\\label{sect_mc1d_trans}\nThe program \n{\\em wk1d\\_trans} computes the WT-KLT transform of \na multichannel 1D signal (2D data set). \nThe wavelet transform (\\`a trous algorithm)\nis applied on each individual 1D signal, and the KL transform is applied for\neach resolution level, exept the last (smooth) scale. \nSo we have a set of eigenvectors for \neach resolution level $j$ $\\{e_{j,1}, \\dots, e_{j,N}\\}$.\nThe output is a set \n$N$ multiresolution transform files in FITS format (suffix ``.fits''). \nEigenvectors\nare stored by increasing importance. The ith multiresolution file \n($i = 1 \\dots N$)\ncontains the eigenvectors $\\{e_{1,i}, \\dots, e_{J-1,i}\\}$, where $J$ is the \nmaximum resolution level (last smooth array).\n{\\bf\n\\begin{center}\n USAGE: wk1d\\_trans option data\\_in prefix\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n \\item {\\bf [-n number\\_of\\_scales]} \\\\\n Number of scales used in the multiresolution transform.\n Default is 4.\n\\item {\\bf [-C]} \\\\\nWrite the correlation matrix to disk. \nThe file name is ``Correl\\_Matrix''. Default is not to do this.\n\\item {\\bf [-x CorrelMat\\_Method]}\n{\\small\n\\begin{itemize} \n\\item{0:} One correlation matrix per band. \n\\item{1:} One correlation matrix for all bands. \n\\item{2:} One correlation matrix for all band, except the last scale. \n\\item{3:} Import correlation matrix.\n\\end{itemize}\nDefault is one correlation matrix per band.\n} \n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nInput correlation matrix file name.\nOnly used when option ``-x 3'' is set. Default is not to avail of this.\n\\end{itemize}\n\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\item wk1d\\_trans input.fits wk \\\\\nCalculate the WT-KLT of the input data and store the eigenvectors\nin the files ``wk\\_ev\\_1.fits, ..., wk\\_ev\\_N.fits''. Each file number $i$ \ncontains a 2D data set. A given row $j$ represents eigenvector \nnumber $i$ at resolution level $j$. \n\\item wk1d\\_trans -C input.fits wk \\\\\nDitto, but write the calculated correlation matrix to disk.\n\\end{itemize}\n\n\\subsection{WT-KLT Reconstruction of 1D Multichannel Data: wk1d\\_trec}\n\\index{wk1d\\_trec}\nThe program \n{\\em wk1d\\_trec} computes the WT-KLT transform of \na multichannel 1D signal (2D data set), and reconstructs the data from\na subset of eigenvectors. By default, only the first eigenvector is\nused at each resolution level.\n{\\bf\n\\begin{center}\n USAGE: wk1d\\_trec option data\\_in data\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item {\\bf [-n number\\_of\\_scales]} \\\\\n Number of scales used in the multiresolution transform.\n Default is 4.\n\\item {\\bf [-C]} \\\\\nWrite the correlation matrix to disk. \nThe file name is ``Correl\\_Matrix''. Default is not to do this.\n\\item {\\bf [-x CorrelMat\\_Method]}\n{\\small\n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm \n\\item{0:} One correlation matrix per band. \n\\item{1:} One correlation matrix for all bands. \n\\item{2:} One correlation matrix for all band, except the last scale. \n\\item{3:} Import correlation matrix.\n\\end{itemize}\nDefault is one correlation matrix per band.\n} \n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nInput correlation matrix file name.\nOnly used when option ``-x 3'' is set. Default is not to use this. \n\\item {\\bf [-F NbrEigenVect]} \\\\\nNumber of eigenvectors used for the reconstruction. \nDefault is set to the number of images.\n\\item {\\bf [-K EigenVect\\_Number]} \\\\\nEigenvector number which will not be used for the reconstruction. \n\\end{itemize}\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\item wk1d\\_trec input.fits output.fits \\\\\nComputes the WT-KLT transform and reconstructs the data from\nthe first eigenvector of each resolution level.\n\\item wk1d\\_trec -F 3 input.fits output.fits \\\\\nUse the three first eigenvectors\n\\item wk1d\\_trec -F 7 -K 2 -K 3 -K 4 trans.fits output.fits \\\\\nUse four eigenvectors (i.e., eigenvectors 1,5,6,7).\n\\end{itemize}\n\n\n\\subsection{WT-KLT Transform of 2D Multichannel Data: wk\\_trans}\n\\index{wk\\_trans}\n\\label{sect_mc_trans}\nThe program \n{\\em wk\\_trans} computes the WT-KLT transform of \na multichannel image (3D data set) containing $N$ frames \n($N < 500$). The wavelet transform\nis applied to each frame, and the KL transform is applied to each\nresolution level, exept the last (smooth) scale. \nSo we have a set of eigenvectors for \neach resolution level $j$ $\\{e_{j,1}, \\dots, e_{j,N}\\}$.\nThe output is a set \n$N$ multiresolution transform files (suffice ``.mr''). Eigenvectors\nare stored by increasing importance. The ith multiresolution file (i=1..N)\ncontains the eigenvectors $\\{e_{1,i}, \\dots, e_{J-1,i}\\}$, where $J$ is the \nmaximum resolution level (last smooth array).\nIf the output file name\ngiven by the user does not contain the ``.mr'' suffix, it is automatically\nadded. The ``.mr'' file is a FITS format file, and can be manipulated by\nany package dealing with FITS format.\n{\\bf\n\\begin{center}\n USAGE: wk\\_trans option image\\_in wtklt\\_transform\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\item {\\bf [-t type\\_of\\_multiresolution\\_transform]}\n\\itemsep=0.1truecm\n{\\small \n\\begin{enumerate}\n\\baselineskip=0.4truecm\n\\item linear wavelet transform: \\`a trous algorithm \n\\item B-spline wavelet transform: \\`a trous algorithm \n\\item wavelet transform in Fourier space \n\\item morphological median transform \n\\item morphological minmax transform \n\\item pyramidal linear wavelet transform \n\\item pyramidal B-spline wavelet transform \n\\item pyramidal wavelet transform in Fourier space: \n wavelet = between two resolutions \n\\item pyramidal wavelet transform in Fourier space: \n wavelet = difference between the square of two resolutions\n\\item pyramidal median transform \n\\item pyramidal Laplacian \n\\item morphological pyramidal minmax transform \n\\item decomposition on scaling function \n\\item (bi-) orthogonal wavelet transform. \\\\ \nAntonini 7/9 filters ~\\cite{wave:antonini92} are used by default, with an \n$L_1$ normalization. The filters can be changed using the ``-T'' option, and\nan $L_2$ normalization is obtained by ``-L'' option.\n\\item Feauveau wavelet transform \n\\item Feauveau wavelet transform without undersampling \n\\item G transform (non-redundant morphological min-max algorithm)\n\\item Haar wavelet transform (L2 normalization).\n\\item Half-pyramidal wavelet transform (HPWT)\n\\item Mixed HPWT and Median method\n\\item dyadic wavelet transform \n\\item Mixed WT and PMT method (WT-PMT) \n\\item Undecimated Haar transform: \\`a trous algorithm\n\\item Undecimated (bi-) orthogonal wavelet transform. \\\\\nAntonini 7/9 filters ~\\cite{wave:antonini92} are used by default, with an \n$L_1$ normalization. The filters can be changed using the ``-T'' option, and\nan $L_2$ normalization is obtained by ``-L'' option.\n\\end{enumerate}}\nDefault is 2.\n\\item {\\bf [-T type\\_of\\_filters]} \n{\\small\n\\begin{enumerate}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item Antonini 7/9 filters. \n\\item Daubechies filter 4. \n\\item Biorthogonal 2/6 Haar filters.\n\\item Biorthogonal 2/10 Haar filters.\n\\item Odegard 7/9 filters.\n\\item User's filters.\n\\end{enumerate}}\nDefault is Antonini 7/9 filters. \\\\\n This option is only available if the chosen transform method is\n the (bi-) orthogonal transform (-t 14 or -t 24).\n\\item {\\bf [-L]} \\\\\nUse an $L_2$ normalization. Default is $L_1$.\n% \\item {\\bf [-u]} \\\\\n% Number of undecimated scales used in the Undecimated Wavelet Transform\n% Default is all scales.\n \\item {\\bf [-n number\\_of\\_scales]} \\\\\n Number of scales used in the multiresolution transform.\n Default is 4.\n\\item {\\bf [-C]} \\\\\nWrite the correlation matrix to disk. \nThe file name is ``Correl\\_Matrix''. Default is not to do this.\n% \\item {\\bf [-p]} \\\\\n% Normalize the correlation matrix. Values on the diagonal are 1. \n% Default is no.\n\\item {\\bf [-x CorrelMat\\_Method]}\n{\\small\n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm \n\\item{0:} One correlation matrix per band. \n\\item{1:} One correlation matrix for all bands. \n\\item{2:} One correlation matrix for all band, except the last scale. \n\\item{3:} Import correlation matrix.\n\\end{itemize}\nDefault is one correlation matrix per band.\n} \n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nInput correlation matrix file name.\nOnly used when option ``-x 3'' is set. Default is not to do this.\n\\end{itemize}\n\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\item wk\\_trans cube.fits wk \\\\\nCalculate the WK-KLT of a 3D data set. If the cube contains $N$ frames,\n$N$ files will be created (wk\\_ev\\_1.mr, ..., wk\\_ev\\_N.mr).\n\\item mr\\_extract -b 2 wk\\_ev\\_1.mr b2\\_1 \\\\\nExtract the second band from the multiresolution file ``wk\\_ev\\_1.mr''.\nThe file ``b2\\_1.fits'' contains an image which\ncorresponds to the first eigenvector at the second resolution level.\n\\end{itemize}\n\n\n\\subsection{WT-KLT Reconstruction of 2D Multichannel Data: wk\\_trec}\n\\index{wk\\_trec}\nThe program \n{\\em wk\\_trec} computes the WT-KLT transform of \na multichannel image (3D data set), and reconstructs the data from\na subset of eigenvectors.\nThe output file contains the reconstruction. Most of the options are the\nsame as those described in section~\\ref{sect_mc_trans}.\n{\\bf\n\\begin{center}\n USAGE: wk\\_trans option image\\_in wtklt\\_transform\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\n\\item {\\bf [-t type\\_of\\_multiresolution\\_transform]} \\\\\nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-T type\\_of\\_filters]} \nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-L]} \nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-n number\\_of\\_scales]}\n\\item {\\bf [-w]} \\\\\n Write to disk the eigenvectors multiresolution files.\nFile names are: ``wk\\_ev\\_x.mr'', where x is the eigenvector number.\nDefault is not to do this.\n\\item {\\bf [-C]}\nWrite the correlation matrix to disk. \nThe file name is ``Correl\\_Matrix''. Default is not to do this.\n\\item {\\bf [-x]} \n{\\small\n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm \n\\item{0:} One correlation matrix per band. \n\\item{1:} One correlation matrix for all bands. \n\\item{2:} One correlation matrix for all band, except the last scale. \n\\item{3:} Import correlation matrix.\n\\end{itemize}\nDefault is one correlation matrix per band.\n} \n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nInput correlation matrix file name.\nOnly used when option ``-x 3'' is set. Default is not to do this.\n\\item {\\bf [-F NbrEigenVect]} \\\\\nNumber of eigenvectors used for the reconstruction. \nDefault is set to the number of images.\n\\item {\\bf [-K EigenVect\\_Number]} \\\\\nEigenvector number which will not be used for the reconstruction. \n\\end{itemize}\n\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item wk\\_trec in\\_cube.fits out\\_cube \\\\\nCalculate the WT-KLT transform, and reconstruct only from the \nfirst eigenvector of each scale.\n\\item wk\\_trec -F 5 in\\_cube.fits out\\_cube \\\\\nCalculate the WT-KLT transform, and reconstruct from the \nfirst five eigenvectors of each scale.\n\\item wk\\_trec -F 5 -K 2 in\\_cube.fits out\\_cube \\\\\nDitto, but do not use eigenvector number 2. Only 4 eigenvectors are\nused.\n\\item wk\\_trec -F 5 -K 2 -t24 trans.fits output.fits \\\\\nDitto, but use the undecimated wavelet transform instead of the \n\\`a trous algorithm.\n\\item wk\\_trec -F 5 -K 1 -K 3 -t24 trans.fits output.fits \\\\\nOnly three eigenvectors (i.e., eigenvectors numbered 2,4,5) are used. \n\\end{itemize}\n\n\n\\section{Noise Modeling in the WT-KLT Space}\nSince a WT-KLT coefficient $c$ is obtained by two successive linear \ntransforms, robust noise modeling can be derived in order to know \n the noise standard deviation associated with the $c$ value.\n\n\\subsection{Non-Gaussian noise}\nIf the noise in the data $D$ is Poisson, the Anscombe transformation \n\\cite{rest:anscombe48}\n\\begin{eqnarray}\nt(D) = 2\\sqrt{D + \\frac{3}{8}}\n\\end{eqnarray}\nacts as if the data arose from a\nGaussian white noise model, with $\\sigma = 1$, under the\nassumption that the mean value of $I$ is sufficiently large.\nThe arrival of photons, and their expression by electron counts, on CCD\ndetectors may be modeled by a Poisson distribution. In addition, there is \nadditive Gaussian read-out noise. The Anscombe \ntransformation has been extended to take this combined noise into \naccount. The generalization of the variance stabilizing\nAnscombe formula is derived as \\cite{starck:book98}:\n\\begin{eqnarray}\nt(D) = \\frac{2}{g} \\sqrt{g D + \\frac{3}{8} g^2 + \\sigma^2 - g m}\n\\end{eqnarray}\nwhere $g$ is the electronic gain of the detector, $\\sigma$ and $m$ the standard deviation \nand the mean of the read-out noise. \n \nThis implies that for the filtering of an image with Poisson noise or\na mixture of Poisson and Gaussian noise, we will first pre-transform \nthe data $D$ into another one $t(D)$ with Gaussian noise. Then $t(D)$ \nwill be filtered, and the filtered data will be inverse-transformed.\n\nFor other kinds of noise, modeling must be performed in order to \ndefine the noise probability distribution of the wavelet coefficients \n\\cite{starck:book98}. \nIn the following, we will consider only stationary Gaussian noise.\n\n\\subsection{Noise level on WT-KLT coefficients}\nAssuming a Gaussian noise standard deviation $\\sigma_l$ for each signal \nor image $d_l$, the noise in the wavelet space follows a Gaussian distribution\n$\\sigma_{l,j}$, $j$ being the scale index. For a bi-orthogonal wavelet \ntransform with a $L^2$ normalization, $\\sigma_{l,j} = \\sigma_{l}$ for all $j$.\nSince the WT-KLT coefficients are obtained from a linear transform, we can\neasily derive the noise standard deviation relative to a WT-KLT coefficient\nfrom the noise standard deviation relative to the wavelet coefficients.\nConsidering the noise standard deviation vector \n$s = \\left\\{\\sigma_{1}, ..., \\sigma_{L} \\right\\}$, we apply the\nfollowing transformation:\n\\begin{eqnarray}\ny_j = A_j^2 s^2\n\\end{eqnarray}\nand the noise standard deviation relative to a WT-KLT coefficient \n$ c_{l}(j,k)$ is $\\sqrt{y_j(l)}$. \n\n\n\n\\section{Multichannel Data Filtering}\n\\index{filtering}\n\\subsection{Introduction}\nKLT based filtering methods have been proposed \nin the past \\cite{mc:andrews76,mc:lee91,mc:constant97} for single images.\nThe proposed idea was to decompose the image $I$ of $M \\times N$ pixels\ninto non-overlapping blocks $B_s$ of size $N_b \\times N_b$. Typically, $N_b$ takes\nvalues from 4 to 16. Let $\\lambda_1,\\lambda_2,...\\lambda_n$ be the \nsingular values of the matrix $I$ in decreasing order, and\nassuming that the matrix $I$ is noisy, the rank $r$ of $I$ \nhas been defined as \\cite{mc:constant88}\n\\begin{eqnarray}\n\\lambda_r \\ge \\epsilon_1 > \\lambda_{r+1}\n\\end{eqnarray}\nwhere $\\epsilon_1$ is the norm of the noise. In case of \na Gaussian distribution of zero mean, an upper bound is $\\sqrt{MN}\\sigma$\n\\cite{mc:constant97}.\nA filtered version of $I$ can be obtained by reconstructing each block\nonly from its $r$ first eigenvalues.\nAn original approach has been developed in \\cite{mc:natarajan95,mc:constant97}\nin order to find the optimal\n$\\epsilon$ value, based on the $\\epsilon$-compression ratio curve, using \na lossless compression method like JPEG. It has been found that the maximum \nof the second derivative of the curve furnishes the \noptimal $\\epsilon$ value \\cite{mc:natarajan95}.\n\nIn the case of multichannel data filtering, several different approaches \nmay be considered based on noise modeling. They are \npresented in this section, and evaluated in the next one.\n\n\\subsection{Reconstruction from a subset of eigenvectors}\nThe WT-KLT transform of a data set $D=\\left\\{d_1, ..., d_L\\right\\}$\nconsists of applying a KLT on the wavelet scales. Hence the vector\n$W_j = \\left\\{ W_j^{(1)}, ..., W_j^{(L)}\\right\\}$ of the scales \n$j$ of the wavelet\ntransforms $W^{l}$ can be decomposed uniquely as:\n\\begin{eqnarray}\n W_j = U_j \\Lambda_j^{1 \\over 2} V_j^{-1} = \\sum_{i=1}^L \n \\sqrt{\\lambda_{j,i}} {u}_{j,i} \n {v}_{j,i}^t\n\\end{eqnarray}\nwhere $\\Lambda_j$ is the diagonal matrix of eigenvalues of the correlation\nmatrix $C_j$, $U_j$ and $V_j$ are orthogonal matrices with column vectors\n$u_{j,i}$ and $v_{j,i}$ which are respectively \n the eigenvectors of $W_j W_j^t$ and $W_j^t W_j$.\n\n% \\overrightarrow{u}_{j,i}\nThe filtered wavelet coefficients of the band $j$ can be obtained by:\n\\begin{eqnarray}\n\\tilde W_j = \\sum_{i=1}^r \\sqrt{\\lambda_{j,i}} {u}_{j,i} {v}_{j,i}^t\n\\end{eqnarray}\nwhere $r$ is the rank of the matrix.\n\n\n\\subsection{WT-KLT Coefficient Thresholding}\nHard thresholding can be applied to the WT-KLT coefficients\nin a fashion analogous to the thresholding of wavelet coefficients.\n\n\\subsubsection*{Example: astronomical source detection}\n\nFigure~\\ref{fig_exp_trans1} shows a simulation. We created\na dataset of 18 frames, each of them containing a source (i.e., a point\nsource, or idealized star) at the \nsame position, \nbut at different intensity levels. The source is a small Gaussian, and \nthe source SNR is defined as the ratio between the maximum of the source and \nthe noise standard deviation. Figure~\\ref{fig_exp_trans1} (top) shows the\nevolution of the SNR in the 18 frames. Frames two and ten \nare shown in Figure~\\ref{fig_exp_trans1}, middle \nleft and right. The source \nSNR ratio is respectively three and one. Figure~\\ref{fig_exp_trans1}, bottom\nleft and right, shows respectively frame ten filtered by the \n wavelet transform and the WT-KLT. The WT detects only noise, while the \nWT-KLT clearly identifies the source.\n\n\\begin{figure}[htb]\n\\centerline{\n\\vbox{\n\\hbox{ \n\\psfig{figure=transp1.ps,bbllx=0.1cm,bblly=18cm,bburx=11cm,bbury=25cm,width=8cm,height=5.2cm,clip=}\n}\n\\hbox{ \n\\psfig{figure=transp1.ps,bbllx=0.5cm,bblly=1cm,bburx=22cm,bbury=18cm,width=13.6cm,height=11.2cm,clip=}\n}\n}}\n\\caption{Simulation: the dataset is composed of 18 frames. \nEach of them contains a source (small Gaussian) at the same position, \nbut at different intensity levels. Top, plot of the source maximum value versus\nthe frame number. Middle, frames 2 and 10, and bottom, filtered version\nof the frame 10 by the wavelet transform and wavelet Karhunen-Lo\\`eve transform.}\n\\label{fig_exp_trans1}\n\\end{figure}\n\n\n\\subsection{Multiscale Entropy}\nThe multiscale entropy relative to a set of observations $D(1..M)$ can \nbe written as:\n\\begin{eqnarray}\nH(D) = \\sum_{l=1}^{L} \\sum_{j=1}^{J} \\sum_{k=1}^{N_j} h(c_{l,j,k})\n\\end{eqnarray}\nwhere $J$ is the number of scales used in the wavelet transform\ndecomposition, $L$ the number of observations, $k$ a pixel position,\n $c$ WT-PCA coefficients, and $l$ denotes the eigenvector number.\n\nThe last scale of the wavelet transform is not used, as previously, so\nthis entropy measurement is background independent, which is really\nimportant because the background can vary from one wavelength to another.\n\nAs for a wavelet coefficients in the case of mono-channel data,\nwe know the noise standard deviation relative to a coefficient,\nand coefficients are of zero mean. Therefore, we can apply the same\nfiltering method. The filtered WT-PCA coefficients are found by minimizing\nfor each $c_{l,j,k}$:\n\\begin{eqnarray}\nj(\\tilde c_{l,j,k}) = h_s(c_{l,j,k} -\\tilde c_{l,j,k}) \n + \\alpha h_n(\\tilde c_{l,j,k})\n\\label{eqn_func_mc}\n\\end{eqnarray}\n\n\\subsubsection*{Example}\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{ \n\\psfig{figure=fig_simu20_noi12.ps,bbllx=3cm,bblly=2.8cm,bburx=18cm,bbury=17.4cm,width=7.5cm,height=8cm,clip=}\n}}\n\\caption{Simulation: Root Mean Square Error versus the noise standard deviation.\nSee text.}\n\\label{fig_simu20_noi12}\n\\end{figure}\n\nFigure~\\ref{fig_simu20_noi12} relates to a simulation using the same dataset \nas in the previous example (i.e., \na dataset of 18 frames).\nAdditive noise was used, and the data were filtered. \nWe calculated the Root Mean Square Error\n(RMSE) on each individual frame on a $5 \\times \n5$ square centered on the source.\nHence, the RMSE reflects well the photometric errors, and the addition\nof the 18 RMSE values, which we call IRMSE (Integrated RMSE), \nfurnishes us a reliable measurement of the filtering quality.\nThe simulation was repeated with 12 noise levels, and four\ndifferent filtering methods were compared. Figure~\\ref{fig_simu20_noi12}\nshows the IRMSE versus the noise standard deviation plot. The four\nmethods are (i) multiscale entropy aaplied to \nthe WT-KLT coefficients (diamond),\n(ii) reconstruction from a subset of eigenvectors of the KLT (triangle),\n(iii) multiscale entropy applied to each frame independently (square),\nand (iv) thresholding applied to the wavelet transform of each frame (star).\nThis simulation shows clearly that the approach proposed here, multiscale\nentropy applied to the WT-KLT coefficients, outperforms all other methods.\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{ \n\\psfig{figure=fig_planck_simu8_noi12.ps,bbllx=2.5cm,bblly=2.8cm,bburx=18cm,bbury=17.4cm,width=7.5cm,height=8cm,clip=}\n}}\n\\caption{Planck Simulation: Root Mean Square Error versus the noise standard deviation.\nSee text.}\n\\label{fig_planck_simu8_noi12}\n\\end{figure}\n\nThe same experiments were performed using a simulated Planck data set.\nPlanck is an upcoming European Space Agency mission to carry out measurements\nof the cosmic microwave background.\nThe data set contains ten images, each one being a linear combination of\n6 sky component images (CMB, SZ, free-free, etc.). As in the previous \nsimulation, noise was added, and the data were filtered\nby the four methods. The only difference is that the RMSE is calculated\non the full frames. Figure~\\ref{fig_planck_simu8_noi12} shows \nIRMSE versus the noise standard deviation plot. Diamonds, triangles, \nsquares and stars represent the same methods as before.\nAgain, the multiscale entropy applied to the WT-KLT coefficients outperforms \nthe other methods.\n\n\\subsection{WT-KLT Filtering of 1D Multichannel Data: wk1d\\_filter}\n\\label{sect_mc1d_filter}\n\\index{wk1d\\_filter}\nThe program \n{\\em wk1d\\_filter} filters \na multichannel signal (2D data set).\n{\\bf\n\\begin{center}\n USAGE: wk1d\\_filter option data\\_in data\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n \\itemsep=0.1truecm\n\n\\item {\\bf [-n number\\_of\\_scales]}\\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-g sigma]}\\\\\n The image contains Gaussian noise, and the standard deviation is\ngiven by {\\em sigma}. This option should be set only if the user\nknows the standard deviation of the noise. \n\\item {\\bf [-x]} \\\\\nsee section~\\ref{sect_mc_trans}. \n% \\item {\\bf [-p]} \\\\\n% Normalize the correlation matrix. Values on the diagonal are 1. \n% Default is no.\n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-y NoiseCorrelationType]}\n{\\small\n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item{0:} Compute the correlation matrix with all wavelet coefficients. \n\\item{1:} Compute the correlation matrix only with significant wavelet coefficients. \n\\item{2:} Compute the correlation matrix with weighted wavelet coefficients. \n\\end{itemize}\n} \nDefault is 1. \n\\item {\\bf [-s NSigma]} \\\\\nWavelet coefficients are significant when their absolute values are larger\nthan nsigma * SigmaNoise. This option has an effect when \nthe ``-y'' option is set to 1 or 2.\nDefault {\\em NSigma} value is 3. \n\\item {\\bf [-S NSigma]} \\\\\nThe WT-KLT coefficients are at {\\em NSigma}*SigmaNoise.\nDefault is 3.\n\\item {\\bf [-P]}\\\\\nPositivity constraint. Default is not to enforce this.\n\\item {\\bf [-F NbrEigenVect]} \\\\\nNumber of eigenvectors used for the reconstruction. \nDefault is set to the number of images.\n\\item {\\bf [-K EigenVect\\_Number]} \\\\\nEigenvector number which will not be used for the reconstruction. \n\\item {\\bf [-C]}\\\\\nsee section~\\ref{sect_mc_trans}. \n\\end{itemize}\n\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item wk1d\\_filter input.fits output.fits \\\\\nFilter the 1D multi-channel data set using the WT-KLT transform.\n\\item wk1d\\_filter -S 5 input.fits output.fits \\\\\nDitto, but smooth the WT-KLT at 5 sigma instead of 3. \n\\end{itemize}\n\n\\subsection{WT-KLT Filtering of 2D Multichannel Data: wk\\_filter}\n\\index{wk\\_filter}\n\\label{sect_mc_filter}\nThe program \n{\\em wk\\_filter} filters \na multichannel image (3D data set).\n{\\bf\n\\begin{center}\n USAGE: wk\\_filter option data\\_in data\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item {\\bf [-t type\\_of\\_multiresolution\\_transform]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-T type\\_of\\_filters]} \\\\ \nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-L]} \\\\\nsee section~\\ref{sect_mc_trans}.\n% \\item {\\bf [-u]} \\\\\n% see section~\\ref{sect_mc_trans}. \n\\item {\\bf [-n number\\_of\\_scales]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-g sigma]}\\\\\nNoise standard deviation. Default is automatically calculated.\n\\item {\\bf [-x]} \\\\\nsee section~\\ref{sect_mc_trans}. \n% \\item {\\bf [-p]} \\\\\n% Normalize the correlation matrix. Values on the diagonal are 1. \n% Default is no.\n\\item {\\bf [-O Input\\_CorrelMat\\_FileName]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-y NoiseCorrelationType]}\n{\\small\n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item{0:} Compute the correlation matrix with all wavelet coefficients. \n\\item{1:} Compute the correlation matrix only with significant wavelet coefficients. \n\\item{2:} Compute the correlation matrix with weighted wavelet coefficients. \n\\end{itemize}\n} \nDefault is 1. \n\\item {\\bf [-s NSigma]} \\\\\nWavelet coefficients are significant when their absolute values are larger\nthan nsigma * SigmaNoise. This option has an effect when \nthe ``-y'' option is set to 1 or 2.\nDefault {\\em NSigma} value is 3. \n\\item {\\bf [-S NSigma]} \\\\\nThe WT-KLT coefficients are at {\\em NSigma}*SigmaNoise.\nDefault is 3.\n\\item {\\bf [-P]}\\\\\nPositivity constraint. Default is not to enforce this.\n\\item {\\bf [-b]}\\\\\nMaximum image constraint at 255. Default is not to use this.\n\\item {\\bf [-w]}\\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-F NbrEigenVect]} \\\\\nNumber of eigenvectors used for the reconstruction. \nDefault is set to the number of images.\n\\item {\\bf [-K EigenVect\\_Number]} \\\\\nEigenvector number which will not be used for the reconstruction. \n\\item {\\bf [-C]}\\\\\nsee section~\\ref{sect_mc_trans}. \n\\end{itemize}\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item wk\\_filter in\\_cube.fits out\\_cube.fits \\\\\nFilter the cube using all default options.\n\\item wk\\_filter -S 5 in\\_cube.fits out\\_cube.fits\\\\\nThreshold the WT-KLT coefficients at 5 sigma, instead of 3 sigma.\n\\item wk\\_filter -S 5 -t24 in\\_cube.fits out\\_cube.fits\\\\\nDitto, but use an undecimated wavelet transform.\n\\end{itemize}\n\n\\subsection{WT-KLT Filtering of 2D Multichannel Data by the Multiscale\nEntropy Method: wk\\_memfilter}\n\\index{wk\\_memfilter}\nThe program \n{\\em wk\\_memfilter} filters \na multichannel image (3D data set) by the multiscale entropy method.\n{\\bf\n\\begin{center}\n USAGE: wk\\_memfilter option cube\\_in cube\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item {\\bf [-t type\\_of\\_multiresolution\\_transform]} \\\\\nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-T type\\_of\\_filters]} \\\\ \nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-L]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-n number\\_of\\_scales]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-g sigma]}\\\\\nsee section~\\ref{sect_mc1d_filter}.\n\\item {\\bf [-x]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-O]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-y NoiseCorrelationType]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-s nsigma]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-P]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-b]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-w]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-F NbrEigenVect]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-K EigenVect\\_Number]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-C CvgParam]} \\\\\nConvergence parameter. Default is $1e-2$.\n\\item {\\bf [-G RegulParam]} \n Regularization parameter. Default is 1. \n\\item {\\bf [-U Type\\_of\\_Regularization]} \n{\\small\n\\begin{enumerate}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item Use a fixed user Alpha value.\n\\item Estimate the optimal Alpha.\n\\item Estimate one Alpha value per band.\n\\end{enumerate}}\nDefault is 1.\n\\item {\\bf [-D]} \\\\\nAlpha is modified using the data SNR. Default is no.\n\\item {\\bf [-i MaxIter]} \\\\\n Maximum number of iterations. Default is 20.\n\\end{itemize}\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item wk\\_memfilter -U 3 in\\_cube.fits out\\_cube.fits \\\\\nFiltering using one regularization parameter per resolution level.\n\\item wk\\_memfilter -U 3 -D in\\_cube.fits out\\_cube.fits \\\\\nDitto, but protect wavelet coefficients with high SNR from the regularization.\n\\item wk\\_memfilter -t24 -U 3 -D in\\_cube.fits out\\_cube.fits \\\\\nDitto, but use the undecimated wavelet transform.\n\\end{itemize}\n\n\n\n\\section{Filtering using the Haar-Multichannel Transform}\n\\subsection{Definition}\nWe have seen in \\proj \\ that the Haar transform \npresents some advantages, especially when the data contains Poisson noise.\n\nIn order to decorrelate the information both spatially and in wavelength (or in\ntime), a 2D-wavelet transform\nmust first be performed on each frame of the date cube $D(x,y,z)$. \nWe denote the result $w_{j,k_x,k_y,z}$, \nwhere $j$ is the scale index ($j \\in [1,J]$, and\nwhere $J$ is the number of scales), $k_x,k_y$ the \nspatial position in the scale ($k_x \\in [0,N_x-1]$, and $k_y \\in [0,N_y-1]$),\n and $z$ is the frame number. \nThis set must again be transformed in order to \ndecorrelate the information in the third dimension. We apply a 1D-transform\nto each vector $w_{j,k_x,k_y,z}$, \nand we get a new set of data $u_{j,j^{'},k_x,k_y,k_z}$,\nwhere $j^{'}$ and $k_z$ are respectively the scale index and position\nin the third dimension.\n\nUsing the unnormalized Haar transform, \na coefficient $u_{j+1,j^{'}+1,k,l,t}$ can be written\nas:\n\\begin{eqnarray}\nu_{j+1,j^{'}+1,k_x,k_y,k_z} \n = \\sum_{i=2^{j^{'}+1}k_z}^{2^{j^{'}}k_z+2^{j^{'}}-1} w_{j+1,k_x,k_y,i} -\n \\sum_{i=2^{j^{'}+1}k_z+2^{j^{'}}}^{2^{j^{'}+1}(k_z+1)-1} w_{j+1,k_x,k_y,i}\n\\end{eqnarray}\nWe assume now a constant background rate $\\lambda_i$ for each frame $i$ \nof the data cube. \nEach Haar wavelet coefficient $w_{j+1,k_x,k_y,i}$ is the difference between two\nrandom variables $X_i$ and $Y_i$, which follows a Poisson distribution of\nparameter $\\lambda_{j,i}$, where $\\lambda_{j,i}$ represents the rate per\npixel over $2^{j+1}$ pixels of the $i$th frame, and is equal to $2^{2j}\\lambda-i$.\nThen $u_{j+1,j^{'}+1,k_x,k_y,k_z}$ is the difference of \n two variables, $X=\\sum_i X_i$ and $Y=\\sum_i Y_i$, and both follow a Poisson\n distribution of parameter $\\sum_i \\lambda_{j,i}$.\nThe thresholding method described in \\proj \ncan therefore be used, using the correct $\\lambda$ value.\n\n\\subsubsection*{Filtering}\n\nThe multichannel filtering algorithm is:\n\\begin{enumerate}\n\\item For each frame $D(*,*,z)$, apply the 2D-Haar transform: we obtain \n$w_{j,k,l}(z)$.\n\\item For each scale $j$, and at each position $k_x,k_y$, \nextract the 1D vector\n $w_{j,k_x,k_y,*}$ and compute its wavelet transform. We get a new data set.\n\\item Threshold the coefficients using the correct $\\lambda value$.\n\\item Inverse 1D transform.\n\\item Inverse 2D transform.\n\\end{enumerate}\n\nFor non-constant background, a coarse to fine approach, as explained in\n\\proj, can be used.\n\n\\subsection{WT-KLT Filtering of 2D Multichannel Data by the Multiscale\nHaar Transform: ww\\_filter}\n\\index{ww\\_filter}\nThe program \n{\\em ww\\_filter} filters \na multichannel image (3D data set) by applying a 2D WT following\nby a 1D WT in the third direction.\n{\\bf\n\\begin{center}\n USAGE: ww\\_filter option data\\_in data\\_out\n\\end{center}}\nwhere options are: \n\\begin{itemize} \n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item {\\bf [-t type\\_of\\_multiresolution\\_transform]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-T type\\_of\\_filters]} \\\\ \nsee section~\\ref{sect_mc_trans}.\n\\item {\\bf [-L]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-n number\\_of\\_scales]} \\\\\nsee section~\\ref{sect_mc_trans}. \n\\item {\\bf [-g sigma]}\\\\\nsee section~\\ref{sect_mc1d_filter}.\n\\item {\\bf [-s nsigma]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-S nsigma]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\item {\\bf [-P]} \\\\\nsee section~\\ref{sect_mc_filter}. \n\\end{itemize}\n\\subsubsection*{Examples}\n\\begin{itemize}\n\\baselineskip=0.4truecm\n\\itemsep=0.1truecm\n\\item ww\\_filter -t 18 -T 3 input.fits output.fits \\\\\nFiltering using the Haar wavelet transform.\n\\item ww\\_filter -t 24 -T 3 input.fits output.fits \\\\\nDitto, but use an undecimated wavelet transform.\n\\item ww\\_filter -t 24 -T 3 -S 5 input.fits output.fits \\\\\nIncrease the thresholding level.\n\\end{itemize}\n\n\\newpage\n\n\\section{Independent Component Analysis}\n\\index{independent component analysis}\n\\index{blind source separation}\n\\index{ICA}\n\nThe idea of independent component analysis (ICA) originates from\nthe problem of {\\em blind source separation} (BSS), and consists of\nrecovering unobserved signals or ``sources'' from several observed mixtures\n\\cite{mc:cardoso98}. Assuming that $n$ statistically independent \nsignals $s_1(t), ..., s_n(t)$ are mixed by an unknown $n \\times n$ mixing \nmatrix $A=[a_{ij}]$, we have:\n\\begin{eqnarray}\nX(t) = A S(t)\n\\end{eqnarray}\nwhere $S(t) = [ s_1(t), \\dots, s_n(t) ]^t$, and\n $X(t) = [ x_1(t), \\dots, x_n(t) ]^t$ represents the observed signals, with \n $x_i(t) = \\sum_{j=1}^n a_{ij} s_j(t)$. \nThe challenge is to find how to achieve separation using as our\nonly assumption that\nthe source signals are statistically independent. The solution \nconsists of finding \nan $n \\times n$ separating matrix $B$, $Y(t) = B S(t)$, such that \n$Y$ is an estimate of S.\nThis is achieved by minimizing contrast functions $\\Phi$, which are defined \nin terms of the Kullback-Leibler divergence $K$:\n\\index{Kullback-Leibler}\n\\begin{eqnarray}\n\\Phi(Y) = \\int p_Y(u) \\log{ {p_Y(u) \\over \\prod p_{Y_i}(u_i) }} du\n\\end{eqnarray}\n\nThe mutual information, expressed by the Kullback-Leibler divergence, \nvanishes if and only if the variables\n$Y_i$ are mutually independent, and is strictly positive otherwise.\n\nICA has been used in astronomy to analyze multispectral \nimages \\cite{mc:bijaoui00} of the galaxy 3C 120 and to separate \n\\index{data!\\texttt{3}C120}\nthe Cosmic Microwave Background\n\\index{Cosmic Microwave Background}\nfrom other sky components \\cite{mc:bacci00}.\nAs for PCA, it has been shown \\cite{mc:zibu99,mc:zibu01} that applying \nICA to wavelet transformed signals leads to better quality results,\nespecially in the presence of noise.\n\n\\subsection*{Example}\nFig.~\\ref{fig_ica1} shows three simulated signals. Six observed data sets\nwere derived from the three sources, by a linear combination\n(see Fig.~\\ref{fig_ica2}). Without any knowledge of the mixing matrix,\nthe JADE-ICA \\cite{mc:cardoso98} method was applied and the \nthree signals were\nreconstructed from the six observed data sets (see Fig.~\\ref{fig_ica3}).\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{ \n\\psfig{figure=fig_ica_source.ps,bbllx=2cm,bblly=12.5cm,bburx=20cm,bbury=25.5cm,width=9cm,height=7.5cm,clip=}\n}}\n\\caption{Example of three simulated sources.}\n\\label{fig_ica1}\n\\end{figure}\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{ \n\\psfig{figure=fig_ica_obs_signal.ps,bbllx=2cm,bblly=12.5cm,bburx=20cm,bbury=25.5cm,width=9cm,height=7.5cm,clip=}\n}}\n\\caption{Mixed sources. Each of these six signals is a linear combination\nof the three simulated sources.}\n\\label{fig_ica2}\n\\end{figure}\n\n\\begin{figure}[htb]\n\\centerline{\n\\hbox{ \n\\psfig{figure=fig_ica_demixing.ps,bbllx=2cm,bblly=12.5cm,bburx=20cm,bbury=25.5cm,width=9cm,height=7.5cm,clip=}\n}}\n\\caption{Reconstructed sources from the mixed signals.}\n\\label{fig_ica3}\n\\end{figure}\n\n\\subsection{JADE-ICA IDL Programs}\n \n\\subsubsection{JADE}\n\\label{jade}\n\\index{JADE}\n Apply the Independant Componant Method to a set of vectors using\nthe JADE method.\n{\\bf\n\\begin{center}\n USAGE: jade, ObservSig, NbSource, DeMixingMat, Process, Verbose=Verbose\n\\end{center}}\nwhere\n\\begin{itemize}\n\\item {\\em ObservSig:} Input Data (ObservSig = $A$ \\# original signal).\nObservSig$(i,*)$ = $i$th vector.\n\\item {\\em NbSource:} Input number of sources in the input signal.\n\\item {\\em DeMixingMat:} Output demixing matrix.\n\\item {\\em Process:} Output Reconstructed process = DeMixingMat \\# ObservSig.\nProcess$(i, *)$ = $i$th Process, with $i = 0.. {\\mathrm NbSource} - 1$.\n\\end{itemize}\n\n\n\\subsubsection{JADE1D}\nApply the Independant Componant Method to a set of vectors using\nthe JADE method. The routine is identical to jade routine, except\nthat the input-output vectors are ordered differently ($V(*,i)$ for\nthe $i$th vector, instead of $V(i,*)$).\n{\\bf\n\\begin{center}\n USAGE: jade1d, ObservSig, NbSource, DeMixingMat, Process, Verbose=Verbose\n\\end{center}}\nwhere\n\\begin{itemize}\n\\item {\\em ObservSig:} Input Data (ObservSig = $A$ \\# original signal).\n; ObservSig$(*,i)$ = $i$th vector.\n\\item {\\em NbSource:} Input number of sources in the input signal.\n\\item {\\em DeMixingMat:} Output demixing matrix.\n\\item {\\em Process:} Output Reconstructed process = DeMixingMat \\# ObservSig.\nProcess$(*, i)$ = $i$th Process, with $i = 0.. {\\mathrm NbSource} - 1$.\n\\end{itemize}\n\n\\subsubsection{JADE2D}\nApply the Independant Componant Method to a set of images using\nthe JADE method. The routine is an extension to 2d of jade1d routine.\nThe input set of images must be given with the following syntax:\nObservSig$(*, *, i)$ = $i$th image. If the ``/wave'' keyword is set,\na 2D wavelet transform is first perform on each image, JADE is applied\non the wavelet transformed images, and an inverse wavelet transform\nis applied on the reconstructed process. The wavelet transform is\nperformed by calling the program {\\em mr\\_transform}. \n{\\bf\n\\begin{center}\n USAGE: jade2d, ObservSig, NbSource, DeMixingMat, Process, wave=wave, optw=optw, Verbose=Verbose\n\\end{center}}\nwhere\n\\begin{itemize}\n\\item {\\em ObservSig:} Input Data (ObservSig = $A$ \\# original signal).\nObservSig$(*,*,i)$ = $i$th vector.\n\\item {\\em NbSource:} Input Number of sources in the input signal.\n\\item {\\em DeMixingMat:} Output demixing matrix.\n\\item {\\em Process:} Output Reconstructed process = DeMixingMat \\# ObservSig.\nProcess$(*, *, i)$ = $i$th Process, with $i = 0.. {\\mathrm NbSource} - 1$.\n\\item {\\em wave::} if set, an bi-orthogonal wavelet transform is applied\non each image, and the component separation is performed\non the wavelet transformed images.\nThe inverse wavelet transform is then applied on the\nreconstructed process.\n\\item {optw:} string = options for the wavelet transformation\n\\end{itemize}\n\n\\subsubsection*{Examples:}\n\\begin{itemize}\n\\item jade2d, ObservIma, NbSource, DeMixingMat, RecIma \\\\\nApply the JADE method to {\\em ObservIma}. \n\\item jade2d, ObservIma, NbSource, DeMixingMat, RecIma, /wave, optw='-n5'\\\\\nDitto, but applied a wavelet transform with five scales on each image.\n\\end{itemize}\n", "meta": {"hexsha": "abddb3b19a0f99fd502d1397cb426745f65f0ceb", "size": 50301, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/doc/doc_mra/doc_mr3/mc_3d.tex", "max_stars_repo_name": "sfarrens/cosmostat", "max_stars_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/doc/doc_mra/doc_mr3/mc_3d.tex", "max_issues_repo_name": "sfarrens/cosmostat", "max_issues_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/doc/doc_mra/doc_mr3/mc_3d.tex", "max_forks_repo_name": "sfarrens/cosmostat", "max_forks_repo_head_hexsha": "a475315cda06dca346095a1e83cb6ad23979acae", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.3899765074, "max_line_length": 163, "alphanum_fraction": 0.7329675354, "num_tokens": 15064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8459424411924673, "lm_q2_score": 0.724870282120402, "lm_q1q2_score": 0.6131985360048055}} {"text": "\\chapter{Introduction to Spreadsheets}\n\nFor many real-world problems, spreadsheets are the perfect\ntool. In this chapter, you will be introduced to how to use a\nspreadsheet. There are numerous spreadsheet programs: Google Sheets,\nMicrosoft Excel, Apple Numbers, OpenOffice Calc, etc. All of them are\nvery similar. I will be using Google Sheets, but if you are using one\nof the others, you should be able to follow along.\n\nThe first spreadsheet program (VisiCalc) was introduced in 1979 as a\ntool for finance people to play ``what if'' games. For example, a\ncompany might make a spreadsheet that told them how much more profit\nthey would make if they changed from using an expensive metal to using\na cheaper alloy.\\index{Spreadsheet}\n\nIn honor of its history, let's start by studying a business question:\nI have a friend who dreams of quitting her job to become a cooper. (A\ncooper makes barrels that are used for aging wine and whiskey.) She\nsays:\n\\begin{itemize}\n\\item It costs \\$45 dollars in materials to build one barrel.\n\\item A barrel sells for \\$100 dollars.\n\\item The workshop/warehouse she wants to rent costs \\$2000 per month.\n\\item Taxes would take about 20\\% of her profits\n\\item She needs to make \\$4000 after taxes to live \n\\end{itemize}\n\nShe has asked you, ``How many barrels do I need to make each month to survive?''\n\n\\section{Solving It Symbolically}\n\nMany problems can be solved two ways: symbolically or\nnumerically.\\index{symbolic vs. numeric solutions} To solve this\nproblem symbolically, you would write out the facts as equations or\ninequalities and then do symbol manipulations until you ended up with\nan answer. In this case, you would let $b$ be the number of barrells\nand create the following inequality:\n\n$$(1.0 - 0.2)\\left(b(100 - 45) - 2000\\right) \\geq 4000$$\n\nYou would simplify it:\n\n$$(0.8)\\left(55 b - 2000\\right) \\geq 4000$$\n\nAnd simplify it more:\n\n$$44b - 1600 \\geq 4000$$\n\nIf that is true, then:\n\n$$44b \\geq 5600$$\n\nAnd if that is true, then:\n\n$$b \\geq \\frac{1400}{11}$$\n\n$1400/11$ is about 127.27, so she needs to make and sell 128 barrels\neach month.\n\nThat is a perfect answer, and we didn't need a spreadsheet at all. Two things:\n\\begin{itemize}\n\\item As problems get larger and more realistic, it gets much more difficult to solve them symbolically.\n\\item As soon as you say ``Yes, you need to make and sell 128 barrels\n each month.'' Your friend will ask ``What if I make and sell 200\n barrels? How much money will I make then?''\n\\end{itemize}\n\nSo we will use a spreadsheet to solve the problem numerically.\n\n\\section{Solving It Numerically (with a spreadsheet)}\n\n\nLet's get back to our example. Put labels in the A column:\n\\begin{itemize}\n\\item{Barrels produced (per month)}\n\\item{Materials cost (per barrel)}\n\\item{Sale price (per barrel)}\n\\item{Pre-tax earnings (per month)}\n\\item{Taxes (per month)}\n\\item{Take home pay (per month)}\n\\end{itemize}\n\nFormat them any way you like. It should look something like this:\n\n\\includegraphics[width=0.4\\textwidth]{BarrelLabels.png}\n\nIn the B column, the first four cells are values (not formulas):\n\\begin{itemize}\n\\item{115 formatted as a number with no decimal point}\n\\item{45 formatted as currency}\n\\item{100 formatted as currency}\n\\item{2000 formatted as currency}\n\\end{itemize}\n\nIt should look something like this:\n\n\\includegraphics[width=0.5\\textwidth]{BarrelValues.png}\n\nThe next three cells in the B column will have formulas:\n\\begin{itemize}\n\\item{B1 * (B3 - B2) - B4}\n\\item{0.2 * B5}\n\\item{B5 - B6}\n\\end{itemize}\n\nIt should look something like this:\n\n\\includegraphics[width=0.5\\textwidth]{BarrelFormulas.png}\n\nNow you can share this spreadsheet with your friend and she can put\ndifferent values into the cells for what-if games. Like ``If I can\nget my materials cost down to \\$42 per barrel, what happens to my take\nhome pay?''\n\nSometimes it is nice to show a range of values for a variable or two.\nIn this case, it might be nice to show your friend what the numbers\nlook like if she produces 115, 120, 125, 130, 135, or 140 barrels per\nmonth.\n\nWe have one column, and now we need six. How do we duplicate cells?\n\\begin{enumerate}\n\\item Click B1 to select it and then shift click on B7 to select all seven cells.\n\\item Copy them. (There is probably a menu item for this.)\n\\item Click C1 to select it\n\\item Paste them.\n\\end{enumerate}\n\n\\includegraphics[width=0.5\\textwidth]{BarrelCopyPaste.png}\n\nWe want the first cell in the new column to be 120. You could just\ntype in 120, but lets do something more clever. Put a formula into that\ncell: = B1 + 5. Now the cell should show 120.\n\nWhy did we put in a formula? When we duplicate this column, this cell\nwill always have 5 more barrels than the cell to its left.\n\nNow lets duplicate the second column a few times. The easy way to do\nthis is to select the cells as you did before and drag the lower-right\ncorner to the right until column G is in the selection. When you end\nthe drag, the copies will appear:\n\n\\includegraphics[width=0.8\\textwidth]{BarrelDragPaste.png}\n\nNice, right? Now your friend can easily see how many barrels\ncorresponds to how much take-home pay. You know what she really wants? A graph.\n\n\\section{Graphing}\n\nGraphing is a little different on every different platform. Here is what you want the graph to look like.\n\n\\includegraphics[width=0.8\\textwidth]{BarrelGraph.png}\n\nOn Google Sheets:\\index{spreadsheet!graphs}\n\n\\begin{enumerate}\n\\item Select cells B7 through G7. \n\\item Choose the menu item Insert -> Chart.\n\\item Choose the chart type (Line)\n\\item Add the X-axis to be B1 through G1.\n\\item Under the Customize tab, Set the label for the X-axis to be ``Barrels Made and Sold''.\n\\item Delete the chart title (which is the same as the Y-axis label).\n\\end{enumerate}\n\n\\section{Other Things You Should Know About Spreadsheets}\n\nYour spreadsheet document can have several ``Sheets''. Each has its\nown grid of cells. The sheet has a name; usually you call it\nsomething like ``Salaries''. When you need to use a value from the\n``Salaries'' sheet in another sheet, you can specify ``Salaries!A2''\n-- that is, cell A2 on sheet ``Salaries''. To flip between the sheets\nthere is usually a tab for each at the bottom of the document.\n\n\\includegraphics[width=0.5\\textwidth]{Sheets.png}\n\nBy default, the cell references are relative. That is, when you write\na formula in cell H5 that references the value in cell G4, the cell\nremembers ``The cell that is one up and one to the left of me.''\nThus, if you copy that formula into B9, now that formula reads the\nvalue from A8.\n\nIf you want an absolute reference, you use \\$. If H5 references\n\\$G\\$4, G4 will be used no matter where on the sheet the formula is\ncopied to.\n\nYou can use the \\$ on the row or colum. In \\$A4, the column is\nabsolute and the row is relative. In A\\$4, the row is absolute and\nthe column is relative.\n\n\\section{Challenge: Make a spreadsheet}\n\nYou have a company that bids on painting jobs. Make a\nspreadsheet to help you do bids. Here are the parameters:\n\\begin{itemize}\n\\item The client will tell you how many square meters of wall need to be painted.\n\\item Paint costs \\$0.02 per square meter of wall to cover\n\\item On average, a square meter of wall takes 0.02 hours to paint.\n\\item You can hire painters at \\$15 per hour.\n\\item You add 20\\% to your estimated costs for a margin of error and profit.\n\\end{itemize}\n\nMake a spreadsheet such that when you type in the square meters to be\npainted, the spreadsheet tells you how much you will spend on paint\nand labor. It also tells you what your bid should be.\n", "meta": {"hexsha": "89d6fc7fd0b031ea52f8098ecfe96b3396e3ea2e", "size": 7552, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Modules/Spreadsheets/intro-en_US.tex", "max_stars_repo_name": "rajivjhoomuck/sequence", "max_stars_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2020-06-13T17:19:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-05T00:43:44.000Z", "max_issues_repo_path": "Modules/Spreadsheets/intro-en_US.tex", "max_issues_repo_name": "rajivjhoomuck/sequence", "max_issues_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Modules/Spreadsheets/intro-en_US.tex", "max_forks_repo_name": "rajivjhoomuck/sequence", "max_forks_repo_head_hexsha": "5b39f09b6350922867c3f88beaf3683425715676", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-05T00:43:58.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-05T00:43:58.000Z", "avg_line_length": 37.0196078431, "max_line_length": 106, "alphanum_fraction": 0.7548993644, "num_tokens": 2009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597972, "lm_q2_score": 0.8152324848629214, "lm_q1q2_score": 0.6130650791543605}} {"text": "\\chapter{Conclusion}\nIn conclusion, different MIP exact model have been developed to resolve the TSP problem and \\texttt{subtour\\_callback\\_general} has definitely obtain the best results in term of times. However no exact algorithm can resolve a problem with more than 1000 nodes in less than 600s. This must be considered in particular when the execution time is limited. In this case heuristics can be successfully applied to obtain a good solution (not the best) about 10 times faster. This result are confirmed with the performance profile in fig \\ref{fig:pp_Lbest}.\\\\\nFor the heuristics, \\texttt{best\\_two\\_opt} obtain really good results in term of execution time and cost optimization, indeed the best heuristic are the constructive ones combined with the iteration of \\texttt{best\\_two\\_opt}.\nThe constructive \\texttt{n\\_greedy} achieves the best performances as a heuristic alone but the use of the \\texttt{n\\_grasp} as a warm start brings a more variability.\n\\begin{figure}[!h]\n\t\\centering\n\t\\begin{subfigure}{.49\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lbest_time.png}\n\t\t\\caption{Performance profile in time domain.}\n\t\t\\label{fig:Lbest_time}\n\t\\end{subfigure}\n\\hfill\n\t\\begin{subfigure}{.49\\textwidth}\n\t\t\\centering\n\t\t\\includegraphics[width=\\columnwidth]{../res/Lbest_lb.png}\n\t\t\\caption{Performance profile in solution cost domain.}\n\t\t\\label{fig:Lbest_lb}\n\t\\end{subfigure}\n\t\\caption{Performance profile of the best models executed in the union of \\textit{data light} and \\textit{data average}.}\n\t\\label{fig:pp_Lbest}\n\\end{figure}", "meta": {"hexsha": "1b166e395823805741f9c231e0789892ad403df4", "size": 1567, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Latex/conclusion.tex", "max_stars_repo_name": "Fisher4537/OR2", "max_stars_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Latex/conclusion.tex", "max_issues_repo_name": "Fisher4537/OR2", "max_issues_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Latex/conclusion.tex", "max_forks_repo_name": "Fisher4537/OR2", "max_forks_repo_head_hexsha": "aeefe436c9be70071cfd92bd59d4b67b03e4ed27", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 71.2272727273, "max_line_length": 552, "alphanum_fraction": 0.7849393746, "num_tokens": 415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324803738429, "lm_q2_score": 0.7520125737597974, "lm_q1q2_score": 0.6130650757785171}} {"text": "\\documentclass[letterpaper,12pt]{article}\n% \\usepackage[utf8]{inputenc}\n\\usepackage{graphicx}\n\\usepackage{ifpdf}\n\n% \\usepackage[spanish,mexico]{babel}\n\\usepackage{multicol}\n\\usepackage{tikz}\n\n\\usepackage{amssymb}\n\\usepackage{amsmath}\n\n\n%opening\n\\title{Lecture 2}\n\\author{Isaac Ayala Lozano}\n\n\\begin{document}\n\n% \\maketitle\n\n\n\n\\section{Lecture 2 - Exercise 8}\n\nCalculate the velocity, speed and acceleration for each \nposition vector.\n\n\n% --------------------------\n\n\\subsection{$\\vec{r}(t)= (\\cos (\\omega t), \\mathrm{e}^{\\omega t})$}\n\nThe first derivative of the vector gives us\n\n\\begin{equation*}\n \\vec{v}(t) = (- \\omega \\sin (\\omega t), \\omega \\mathrm{e}^{\\omega t})\n \\end{equation*}\n\nThe norm of the velocity vector gives us the speed.\n \n \\begin{equation*}\n \\left \\| \\vec{v}(t) \\right \\| = \\sqrt{ \\omega^2 \\sin^2 (\\omega t) + \\omega ^2 \n\\mathrm{e}^{2 \\omega t}}\n \\end{equation*}\n\n The second derivative results in \n \n \\begin{equation*}\n \\vec{a}(t) = (-\\omega ^2 \\cos (\\omega t) , \\omega ^2 \n\\mathrm{e}^{\\omega t})\n \\end{equation*}\n\nFigure \\ref{fig: vector 1} shows the plots for all three vectors with \nconditions $\\omega = 1$, and $t \\in \\left [ 0, 2 \\pi \\right ] $.\n \n\\begin{figure}[h]\n \\centering\n\n\\includegraphics[scale=0.65,keepaspectratio=true]{./img/graph01.png}\n \\caption{Position, Velocity and Acceleration plots for \n$\\vec{r}(t)= (\\cos (\\omega t), \\mathrm{e}^{\\omega t})$\n}\n \\label{fig: vector 1}\n\\end{figure}\n\n\n% -----------------------\n\n\\subsection{$\\vec{r}(t) = ( \\cos (\\omega t - \\phi), \\sin (\\omega t - \\phi))$}\n\nUsing the following trigonometric identities\n\n\\begin{equation*}\n \\begin{split}\n \\sin (\\alpha \\pm \\beta) & = \\sin \\alpha \\cos \\beta \\pm \\cos \\alpha \\sin \\beta \n\\\\\n\\cos (\\alpha \\pm \\beta & = \\cos \\alpha \\cos \\beta \\mp \\sin \\alpha \\sin \\beta\n \\end{split}\n\\end{equation*}\n\nthe position vector can be rewritten as follows\n\n\\begin{equation*}\n \\vec{r} = (\n \\cos \\omega t \\cos \\phi + \\sin \\omega t \\sin \\phi,\n \\sin \\omega t \\cos \\phi - \\cos \\omega t \\sin \\phi\n )\n\\end{equation*}\n\nThe velocity, speed and acceleration are as follows\n\n \\begin{equation*}\n \\begin{split}\n \\vec{v}(t) & = ( - \\omega \\cos \\phi \\sin \\omega t + \\omega \\sin \\phi \\cos \n\\omega t, \\omega \\cos \\phi \\cos \\omega t + \\omega \\sin \\phi \\sin \\omega t) \\\\\n& = (- \\omega \\sin (\\omega t + \\phi), \\omega \\cos (\\omega t + \\phi))\n \\end{split}\n \\end{equation*}\n\n\n\\begin{equation*}\n \\left \\| \\vec{v}(t) \\right \\| = \\sqrt{ \\omega ^2 \\sin ^2 (\\omega t + \\phi) + \n\\omega ^ 2 \\cos ^2 (\\omega t + \\phi)}\n\\end{equation*}\n\n \\begin{equation*}\n \\begin{split}\n \\vec{a}(t) & = ( - \\omega ^2 \\cos \\phi \\cos \\omega t - \\omega ^2 \\sin \\phi \n\\sin \\omega t, - \\omega ^2 \\cos \\phi \\sin \\omega t + \\omega ^2 \\sin \\phi \\cos \n\\omega t)\\\\\n& = ( - \\omega ^2 \\cos (\\omega t - \\phi), -\\omega ^ 2 \\sin (\\omega t - \\phi))\n \\end{split}\n \\end{equation*}\n \n Figure \\ref{fig: vector 2} shows the plotted movement equations for the \nsystem. Notice that all three curves overlap when evaluated with $\\omega = 1$, \n $t \\in \\left [ 0, 2 \\pi \\right ] $ and $\\phi = \\frac{\\pi}{4}$.\n \n \\begin{figure}[h]\n \\centering\n\\includegraphics[scale=0.4,keepaspectratio=true]{./img/graph02.png}\n \\caption{Position, Velocity and Acceleration plots for \n$\\vec{r}(t) = ( \\cos (\\omega t - \\phi), \\sin (\\omega t - \\phi))$\n}\n \\label{fig: vector 2}\n\\end{figure}\n\n\n% ----------------------\n\n\n\\subsection{$\\vec{r}(t) = ( c \\cos ^3 (t), c \\sin ^3 (t))$}\n\nApply the chain rule to obtain the derivatives of $\\vec{r}(t)$.\n\n \\begin{equation*}\n \\vec{v}(t) = (- 3 c \\cos ^2 (t) \\sin (t), 3 c \\sin ^ 2 (t) \\cos (t))\n \\end{equation*}\n\n\n\\begin{equation*}\n \\left \\| \\vec{v}(t) \\right \\| = \\sqrt{ 9 c^2 \\cos ^4 (t) \\sin^2 (t) + 9 c^2 \n\\sin ^4 (t) \\cos ^2 (t)}\n\\end{equation*}\n\n \\begin{equation*}\n \\vec{a}(t) = (- 3 c \\cos ^3 (t) - 6 c \\sin ^2 (t) \\cos (t), -3 c \\sin ^3 (t) \n+ 6 c \\cos ^2 (t) \\sin (t))\n \\end{equation*}\n \n All three vectors are plotted in figure \\ref{fig: vector 3} with $t \\in \n\\left [ 0, 2 \\pi \\right ] $, and $c = 1$.\n \n \\begin{figure}[h]\n \\centering\n\n\\includegraphics[scale=0.4,keepaspectratio=true]{./img/graph03.png}\n \\caption{Position, Velocity and Acceleration plots for \n$\\vec{r}(t) = ( c \\cos ^3 (t), and c \\sin ^3 (t))$.\n}\n \\label{fig: vector 3}\n\\end{figure}\n\n\n% ---------------------------\n\n\n\\subsection{$\\vec{r}(t) = ( c (t - \\sin t), c (1 - \\cos t))$}\n \n \\begin{equation*}\n \\vec{v}(t) = (c (1 - \\cos (t) ), c \\sin (t))\n \\end{equation*}\n\n\n\\begin{equation*}\n \\left \\| \\vec{v}(t) \\right \\| = \\sqrt{ c^2 - c^2 \\cos ^2 (t) + c^2 \\sin ^2 (t)}\n\\end{equation*}\n\n \\begin{equation*}\n \\vec{a}(t) = (c \\sin (t), c \\cos (t))\n \\end{equation*}\n \n Figure \\ref{fig: vector 4} shows the plots for all three equations evaluated \nwith $c = 1$ and $t \\in \n\\left [ 0, 2 \\pi \\right ] $.\n \n \\begin{figure}[h]\n \\centering\n\n\\includegraphics[scale=0.4,keepaspectratio=true]{./img/graph04.png}\n \\caption{Position, Velocity and Acceleration plots for \n$\\vec{r}(t) = ( c (t - \\sin t), c (1 - \\cos t))$\n}\n \\label{fig: vector 4}\n\\end{figure}\n \n\\end{document}\n", "meta": {"hexsha": "419fec1971f3040c66e9ca397e1fb3e677f54f5b", "size": 4980, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture2/lecture2.tex", "max_stars_repo_name": "der-coder/CINVESTAV-System-Modeling-2019", "max_stars_repo_head_hexsha": "9345444113885181560940b3b4467c9555b8a85a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Lecture2/lecture2.tex", "max_issues_repo_name": "der-coder/CINVESTAV-System-Modeling-2019", "max_issues_repo_head_hexsha": "9345444113885181560940b3b4467c9555b8a85a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture2/lecture2.tex", "max_forks_repo_name": "der-coder/CINVESTAV-System-Modeling-2019", "max_forks_repo_head_hexsha": "9345444113885181560940b3b4467c9555b8a85a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 25.0251256281, "max_line_length": 80, "alphanum_fraction": 0.5957831325, "num_tokens": 1845, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7025300449389326, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.612850235672055}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n\n\n \n \n\n\\subsection*{maha.m} \n\n\\begin{par}\n\\textbf{Summary:} Point-wise squared Mahalanobis distance (a-b)*Q*(a-b)'. Vectors are row-vectors\n\\end{par} \\vspace{1em}\n\n\\begin{verbatim} function K = maha(a, b, Q)\\end{verbatim}\n \\begin{par}\n\\textbf{Input arguments:}\n\\end{par} \\vspace{1em}\n\\begin{verbatim}a matrix containing n row vectors [n x D]\nb matrix containing n row vectors [n x D]\nQ weight matrix. Default: eye(D) [D x D]\\end{verbatim}\n\\begin{par}\n\\textbf{Output arguments:} K point-wise squared distances [n x n]\n\\end{par} \\vspace{1em}\n\\begin{par}\nCopyright (C) 2008-2013 by Marc Deisenroth, Andrew McHutchon, Joe Hall, and Carl Edward Rasmussen.\n\\end{par} \\vspace{1em}\n\\begin{par}\nLast modified: 2013-03-21\n\\end{par} \\vspace{1em}\n\n\\begin{lstlisting}\nfunction K = maha(a, b, Q)\n\\end{lstlisting}\n\n\n\\subsection*{Code} \n\n\n\\begin{lstlisting}\nif nargin == 2 % assume unit Q\n K = bsxfun(@plus,sum(a.*a,2),sum(b.*b,2)')-2*a*b';\nelse\n aQ = a*Q; K = bsxfun(@plus,sum(aQ.*a,2),sum(b*Q.*b,2)')-2*aQ*b';\nend\n\\end{lstlisting}\n", "meta": {"hexsha": "98fafe506536b35887dcaa98bef7827116309aeb", "size": 1346, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/tex/maha.tex", "max_stars_repo_name": "SJTUGuofei/pilco-matlab", "max_stars_repo_head_hexsha": "a0b48b7831911837d060617903c76c22e4180d0b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 53, "max_stars_repo_stars_event_min_datetime": "2016-12-17T15:15:48.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-09T16:59:27.000Z", "max_issues_repo_path": "doc/tex/maha.tex", "max_issues_repo_name": "sahandrez/quad_pilco", "max_issues_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2020-04-24T11:02:23.000Z", "max_issues_repo_issues_event_max_datetime": "2020-04-24T11:09:45.000Z", "max_forks_repo_path": "doc/tex/maha.tex", "max_forks_repo_name": "sahandrez/quad_pilco", "max_forks_repo_head_hexsha": "2c99152e3a910d147cd0a52822da306063e6a834", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 36, "max_forks_repo_forks_event_min_datetime": "2017-04-19T06:55:25.000Z", "max_forks_repo_forks_event_max_datetime": "2021-05-19T10:19:12.000Z", "avg_line_length": 28.0416666667, "max_line_length": 104, "alphanum_fraction": 0.587667162, "num_tokens": 427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8418256631249076, "lm_q2_score": 0.7279754548076477, "lm_q1q2_score": 0.6128284199821042}} {"text": "\\chapter{Homomorphisms and quotient groups}\n\\section{Generators and group presentations}\n\\prototype{$D_{2n} = \\left< r,s \\mid r^n=s^2=1\\right>$}\n\nLet $G$ be a group.\nRecall that for some element $x \\in G$,\nwe could consider the subgroup\n\\[ \\left\\{ \\dots, x^{-2}, x^{-1}, 1, x, x^2, \\dots \\right\\} \\]\nof $G$.\nHere's a more pictorial version of what we did:\n\\textbf{put $x$ in a box, seal it tightly, and shake vigorously}.\nUsing just the element $x$,\nwe get a pretty explosion that produces the subgroup above.\n\nWhat happens if we put two elements $x$, $y$ in the box?\nAmong the elements that get produced are things like\n\\[ xyxyx, \\quad x^2y^9x^{-5}y^3, \\quad y^{-2015}, \\quad \\dots\\]\nEssentially, I can create any finite product of $x$, $y$, $x\\inv$, $y\\inv$.\nThis leads us to define:\n\\begin{definition}\n\tLet $S$ be a subset of $G$.\n\tThe subgroup \\vocab{generated} by $S$,\n\tdenoted $\\left$,\n\tis the set of elements which can be written as a\n\tfinite product of elements in $S$ (and their inverses).\n\tIf $\\left = G$ then we say $S$ is a set of\n\t\\vocab{generators} for $G$,\n\tas the elements of $S$ together create all of $G$.\n\\end{definition}\n\\begin{exercise}\n\tWhy is the condition ``and their inverses''\n\tnot necessary if $G$ is a finite group?\n\t(As usual, assume Lagrange's theorem.)\n\\end{exercise}\n\n\\begin{example}[$\\ZZ$ is the infinite cyclic group]\n\tConsider $1$ as an element of $\\ZZ = (\\ZZ, +)$.\n\tWe see $\\left<1\\right> = \\ZZ$, meaning $\\{1\\}$ generates $\\ZZ$.\n\tIt's important that $-1$, the inverse of $1$ is also allowed:\n\twe need it to write all integers as the sum of $1$ and $-1$.\n\\end{example}\n\nThis gives us an idea for a way to try and express groups compactly.\nWhy not just write down a list of generators for the groups?\nFor example, we could write\n\\[ \\ZZ \\cong \\left< a \\right> \\]\nmeaning that $\\ZZ$ is just the group generated by one element.\n\nThere's one issue: the generators usually satisfy certain properties.\nFor example, consider $\\Zc{100}$.\nIt's also generated by a single element $x$,\nbut this $x$ has the additional property that $x^{100} = 1$.\nThis motivates us to write\n\\[ \\Zc{100} = \\left< x \\mid x^{100} = 1 \\right>. \\]\nI'm sure you can see where this is going.\nAll we have to do is specify a set of generators and\n\\vocab{relations} between the generators,\nand say that two elements are equal if and only if\nyou can get from one to the other using relations.\nSuch an expression is appropriately called a \\vocab{group presentation}.\n\n\\begin{example}[Dihedral group]\n\tThe dihedral group of order $2n$ has a presentation\n\t\\[ D_{2n} = \\left< r, s\n\t\t\\mid r^n = s^2 = 1, rs = sr\\inv \\right>. \\]\n\tThus each element of $D_{2n}$ can be written uniquely in the form $r^\\alpha$\n\tor $sr^\\alpha$, where $\\alpha = 0, 1, \\dots, n-1$.\n\\end{example}\n\n\\begin{example}[Klein four group]\n\tThe \\vocab{Klein four group},\n\tisomorphic to $\\Zc2 \\times \\Zc2$, is given by the presentation\n\t\\[ \\left< a,b \\mid a^2=b^2=1, ab=ba \\right>. \\]\n\\end{example}\n\n\\begin{example}\n\t[Free group]\n\tThe \\vocab{free group on $n$ elements} is the group\n\twhose presentation has $n$ generators and no relations at all.\n\tIt is denoted $F_n$, so\n\t\\[\n\t\tF_n = \\left< x_1, x_2, \\dots, x_n \\right>.\n\t\\]\n\tIn other words, $F_2 = \\left$ is the set of strings\n\tformed by appending finitely many copies of $a$, $b$, $a\\inv$, $b\\inv$ together.\n\\end{example}\n\\begin{ques}\n\tNotice that $F_1 \\cong \\ZZ$.\n\\end{ques}\n\\begin{abuse}\n\tOne might unfortunately notice that ``subgroup generated by $a$ and $b$''\n\thas exactly the same notation as the free group $\\left$.\n\tWe'll try to be clear based on context which one we mean.\n\\end{abuse}\n\nPresentations are nice because they provide a compact way to write down groups.\nThey do have some shortcomings, though.\\footnote{%\nActually, determining whether two elements of a presentation are equal is undecidable.\nIn fact, it is undecidable to even determine if a group is finite from its presentation.}\n\n\\begin{example}\n\t[Presentations can look very different]\n\tThe same group can have very different presentations.\n\tFor instance consider\n\t\\[ D_{2n} = \\left< x,y \\mid x^2=y^2=1, (xy)^n=1. \\right>. \\]\n\t(To see why this is equivalent, set $x=s$, $y=rs$.)\n\\end{example}\n\n\\section{Homomorphisms}\n\\prototype{The ``mod out by $100$'' map, $\\ZZ \\to \\Zc{100}$.}\n\nHow can groups talk to each other?\n\nTwo groups are ``the same'' if we can write an isomorphism between them.\nAnd as we saw, two metric spaces are ``the same''\nif we can write a homeomorphism between them.\nBut what's the group analogy of a continuous map?\nWe simply drop the ``bijection'' condition.\n\n\\begin{definition}\n\tLet $G = (G, \\star)$ and $H = (H, \\ast)$ be groups.\n\tA \\vocab{group homomorphism} is a map $\\phi : G \\to H$\n\tsuch that for any $g_1, g_2 \\in G$ we have\n\t\\[ \\phi(g_1 \\star g_2) = \\phi(g_1) \\ast \\phi(g_2). \\]\n\\end{definition}\n\n\\begin{example}\n\t[Examples of homomorphisms]\n\tLet $G$ and $H$ be groups.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Any isomorphism $G \\to H$ is a homomorphism.\n\t\tIn particular, the identity map $G \\to G$ is a homomorphism.\n\t\t\\ii The \\vocab{trivial homomorphism} $G \\to H$ sends\n\t\teverything to $1_H$.\n\t\t\\ii There is a homomorphism from $\\ZZ$ to $\\Zc{100}$ by\n\t\tsending each integer to its residue modulo $100$.\n\t\t\\ii There is a homomorphism from $\\ZZ$ to itself by $x \\mapsto 10x$\n\t\twhich is injective but not surjective.\n\t\t\\ii There is a homomorphism from $S_n$ to $S_{n+1}$ by ``embedding'':\n\t\tevery permutation on $\\{1,\\dots,n\\}$ can be thought of as a permutation\n\t\ton $\\{1,\\dots,n+1\\}$ if we simply let $n+1$ be a fixed point.\n\t\t\\ii A homomorphism $\\phi: D_{12} \\to D_6$\n\t\tis given by $s_{12} \\mapsto s_6$\n\t\tand $r_{12} \\mapsto r_6$.\n\t\t\\ii Specifying a homomorphism $\\ZZ \\to G$ is the same as\n\t\tspecifying just the image of the element $1 \\in \\ZZ$. Why?\n\t\\end{enumerate}\n\\end{example}\nThe last two examples illustrates something: suppose we have a presentation of $G$.\nTo specify a homomorphism $G \\to H$, we only have to specify where each generator of $G$ goes, in such a way that the relations are all satisfied.\n\nImportant remark:\nthe right way to think about an isomorphism is as a ``bijective homomorphism''.\nTo be explicit,\n\\begin{exercise}\n\tShow that $G \\cong H$ if and only if there exist\n\thomomorphisms $\\phi \\colon G \\to H$ and $\\psi \\colon H \\to G$\n\tsuch that $\\phi \\circ \\psi = \\id_H$ and $\\psi \\circ \\phi = \\id_G$.\n\\end{exercise}\nSo the definitions of homeomorphism of metric spaces\nand isomorphism of groups are not too different.\n\nSome obvious properties of homomorphisms follow.\n\\begin{fact}\n\tLet $\\phi \\colon G \\to H$ be a homomorphism.\n\tThen $\\phi(1_G) = 1_H$ and $\\phi(g\\inv) = \\phi(g)\\inv$.\n\\end{fact}\n\\begin{proof}\n\tBoring, and I'm sure you could do it yourself if you wanted to.\n\\end{proof}\n\nNow let me define a very important property of a homomorphism.\n\\begin{definition}\n\tThe \\vocab{kernel} of a homomorphism $\\phi \\colon G \\to H$ is defined by\n\t\\[ \\ker \\phi \\defeq\n\t\t\\left\\{ g \\in G : \\phi(g) = 1_H \\right\\}.\n\t\t\\]\n\tIt is a \\emph{subgroup} of $G$\n\t(in particular, $1_G \\in \\ker \\phi$ for obvious reasons).\n\\end{definition}\n\\begin{ques}\n\tVerify that $\\ker\\phi$ is in fact a subgroup of $G$.\n\\end{ques}\nWe also have the following important fact, which we also encourage the reader to verify.\n\\begin{proposition}\n\t[Kernel determines injectivity]\n\tThe map $\\phi$ is injective if and only if $\\ker\\phi = \\{1_G\\}$.\n\\end{proposition}\n\nTo make this concrete, let's compute the kernel of each of our examples.\n\\begin{example}\n\t[Examples of kernels]\n\t\\listhack\n\t\\begin{enumerate}[(a)]\n\t\t\\ii The kernel of any isomorphism $G \\to H$ is trivial,\n\t\tsince an isomorphism is injective.\n\t\tIn particular, the kernel of the identity map $G \\to G$ is $\\{1_G\\}$.\n\t\t\\ii The kernel of the trivial homomorphism $G \\to H$\n\t\t(by $g \\mapsto 1_H$) is all of $G$.\n\t\t\\ii The kernel of the homomorphism $\\ZZ \\to \\Zc{100}$ by $n \\mapsto \\ol n$\n\t\tis precisely \\[ 100\\ZZ = \\{\\dots, -200, -100, 0, 100, 200, \\dots \\}. \\]\n\t\t\\ii The kernel of the map $\\ZZ \\to \\ZZ$ by $x \\mapsto 10x$ is trivial: $\\{0\\}$.\n\t\t\\ii There is a homomorphism from $S_n$ to $S_{n+1}$ by ``embedding'',\n\t\tbut it also has trivial kernel because it is injective.\n\t\t\\ii A homomorphism $\\phi\\colon D_{12} \\to D_6$\n\t\tis given by $s_{12} \\mapsto s_6$ and $r_{12} \\mapsto r_6$.\n\t\tYou can check that\n\t\t\\[ \\ker \\phi = \\left\\{ 1, r_{12}^{3} \\right\\} \\cong \\Zc2. \\]\n\t\t\\ii Exercise below.\n\t\\end{enumerate}\n\\end{example}\n\\begin{exercise}\n\tFix any $g \\in G$.\n\tSuppose we have a homomorphism $\\ZZ \\to G$ by $n \\mapsto g^n$.\n\tWhat is the kernel?\n\\end{exercise}\n\n\\begin{ques}\n\tShow that for any homomorphism $\\phi: G \\to H$,\n\tthe image $\\phi\\im(G)$ is a subgroup of $H$.\n\tHence, we'll be especially interested in the case where $\\phi$ is surjective.\n\\end{ques}\n\n\\section{Cosets and modding out}\n\\prototype{Modding out by $n$: $\\ZZ / (n \\cdot \\ZZ) \\cong \\Zc n$.}\n\\emph{The next few sections are a bit dense.\nIf this exposition doesn't work for you, try \\cite{ref:gowers}.}\n\nLet $G$ and $Q$ be groups, and suppose there exists\na \\emph{surjective} homomorphism \\[ \\phi : G \\surjto Q. \\]\nIn other words, if $\\phi$ is injective then $\\phi : G \\to Q$ is a bijection,\nand hence an isomorphism.\nBut suppose we're not so lucky and $\\ker\\phi$ is bigger than just $\\{1_G\\}$.\nWhat is the correct interpretation of a more general homomorphism?\n\nLet's look at the special case where $\\phi : \\ZZ \\to \\Zc{100}$ is ``modding out by $100$''.\nWe already saw that the kernel of this map is\n\\[ \n\t\\ker \\phi = 100\\ZZ = \\left\\{ \\dots, -200, -100, 0, 100, 200, \\dots \\right\\}.\n\\]\nRecall now that $\\ker \\phi$ is a subgroup of $G$.\nWhat this means is that \\textbf{$\\phi$ is indifferent to the subgroup $100\\ZZ$ of $\\ZZ$}:\n\\[ \\phi(15) = \\phi(2000 + 15) = \\phi(-300 + 15) = \\phi(700 + 15) = \\dots. \\]\nSo $\\Zc{100}$ is what we get when we ``mod out by $100$''. Cool.\n\nIn other words, let $G$ be a group and $\\phi : G \\surjto Q$\nbe a surjective homomorphism with kernel $N \\subseteq G$.\n\\begin{moral}\n\tWe claim that $Q$ should be thought of as the quotient of $G$ by $N$.\n\\end{moral}\nTo formalize this, we will define a so-called\n\\vocab{quotient group} $G/N$\nin terms of $G$ and $N$ only (without referencing $Q$)\nwhich will be naturally isomorphic to $Q$.\n\nFor motivation, let's give a concrete description of $Q$ using just $\\phi$ and $G$.\nContinuing our previous example, let $N = 100\\ZZ$ be our subgroup of $G$.\nConsider the sets\n\\begin{align*}\n\tN &= \\left\\{ \\dots, -200, -100, 0, 100, 200, \\dots \\right\\} \\\\\n\t1+N &= \\left\\{ \\dots, -199, -99, 1, 101, 201, \\dots \\right\\} \\\\\n\t2+N &= \\left\\{ \\dots, -198, -98, 2, 102, 202, \\dots \\right\\} \\\\\n\t&\\vdots \\\\\n\t99+N &= \\left\\{ \\dots, -101, -1, 99, 199, 299, \\dots \\right\\}.\n\\end{align*}\nThe elements of each set all have the same image when we apply $\\phi$,\nand moreover any two elements in different sets have different images.\nThen the main idea is to notice that\n\\begin{moral}\n\tWe can think of $Q$ as the group\n\twhose \\emph{elements} are the \\emph{sets} above.\n\\end{moral}\n\nThus, given $\\phi$ we define an equivalence relation $\\sim_N$\non $G$ by saying $x \\sim_N y$ for $\\phi(x) = \\phi(y)$.\nThis $\\sim_N$ divides $G$ into several equivalence classes in $G$\nwhich are in obvious bijection with $Q$, as above.\nNow we claim that we can write these equivalence classes very explicitly.\n\n% Also, for each $g \\in G$ define $\\ol g$ to be the equivalence class of $g$ under $\\sim_\\phi$.\n\\begin{exercise}\n\tShow that $x \\sim_N y$ if and only if $x = yn$ for some $n \\in N$\n\t(in the mod $100$ example, this means they ``differ by some multiple of $100$'').\n\tThus for any $g \\in G$, the equivalence class of $\\sim_N$ which contains $g$\n\tis given explicitly by \\[ gN \\defeq \\left\\{ gn \\mid n \\in N \\right\\}. \\]\n\\end{exercise}\n%Note that\n%\\[ \\phi(x) = \\phi(y)\n%\t\\implies 1_Q = \\phi(x)\\phi(y)\\inv = \\phi(xy\\inv)\n%\t\\implies xy\\inv \\in \\ker\\phi. \\]\n%Hence another way to describe $\\ol x$ is\n%\\[ \\ol g = gN = \\left\\{ gn \\mid n \\in N \\right\\}. \\]\n\nHere's the word that describes the types of sets we're running into now.\n\\begin{definition}\n\tLet $H$ be any subgroup of $G$ (not necessarily the kernel of some homomorphism).\n\tA set of the form $gH$ is called a \\vocab{left coset} of $H$.\n\\end{definition}\n\\begin{remark}\n\tAlthough the notation might not suggest it,\n\tkeep in mind that $g_1N$ is often equal to $g_2N$ even if $g_1 \\neq g_2$.\n\tIn the ``mod $100$'' example, $3+N = 103+N$.\n\tIn other words, these cosets are \\emph{sets}.\n\n\tThis means that if I write ``let $gH$ be a coset'' without telling you what $g$ is,\n\tyou can't figure out which $g$ I chose from just the coset itself.\n\tIf you don't believe me, here's an example of what I mean:\n\t\\[ x+100\\ZZ = \\left\\{ \\dots, -97, 3, 103, 203, \\dots \\right\\} \\implies x = {?}. \\]\n\tThere's no reason to think I picked $x=3$. (I actually picked $x=-13597$.)\n\t\\label{remark:coset_warning}\n\\end{remark}\n\\begin{remark}\n\tGiven cosets $g_1H$ and $g_2H$,\n\tyou can check that the map $x \\mapsto g_2g_1\\inv x$ is a bijection between them.\n\tSo actually, all cosets have the same cardinality.\n\\end{remark}\n\nSo, long story short,\n\\begin{moral}\n\tElements of the group $Q$ are naturally identified with left cosets of $N$.\n\\end{moral}\nIn practice, people often still prefer to picture elements of $Q$ as single points\n(for example it's easier to think of $\\Zc2$ as $\\{0,1\\}$\nrather than $\\big\\{ \\{\\dots,-2,0,2,\\dots\\}, \\{\\dots,-1,1,3,\\dots\\} \\big\\}$).\nIf you like this picture,\nthen you might then draw $G$ as a bunch of equally tall fibers (the cosets),\nwhich are then ``collapsed'' onto $Q$.\n\n\\begin{center}\n\t\\begin{asy}\n\t\tsize(9cm);\n\t\tfor (int i=1; i<=6; ++i) {\n\t\t\tdraw( (i,0)--(i,3.4) );\n\t\t\tdot( (i,0.6) );\n\t\t\tdot( (i,1) );\n\t\t\tdot( (i,1.7) );\n\t\t\tdot( (i,2.4) );\n\t\t\tdot( (i,2.8) );\n\t\t\tdot( (i,3) );\n\t\t\tlabel(rotate(90)*scale(1.4)*\"$\\Longleftarrow$\", (i+0.02,-0.9), dir(90));\n\t\t\tdot( (i,-1) );\n\t\t}\n\t\tlabel(\"$G$\", (0.5, 3));\n\t\tlabel(\"$Q$\", (0.5,-1));\n\t\\end{asy}\n\\end{center}\n\nNow that we've done this, we can give an \\emph{intrinsic}\ndefinition for the quotient group we alluded to earlier.\n\\begin{definition}\n\tA subgroup $N$ of $G$ is called \\vocab{normal} if it is the\n\tkernel of some homomorphism.\n\tWe write this as $N \\normalin G$.\n\\end{definition}\n\\begin{definition}\n\tLet $N \\normalin G$.\n\tThen the \\vocab{quotient group}, denoted $G/N$\n\t(and read ``$G$ mod $N$''),\n\tis the group defined as follows.\n\t\\begin{itemize}\n\t\t\\ii The elements of $G/N$ will be the left cosets of $N$.\n\t\t\\ii We want to define the product of two cosets $C_1$ and $C_2$ in $G/N$.\n\t\tRecall that the cosets are in bijection with elements of $Q$.\n\t\tSo let $q_1$ be the value associated to the coset $C_1$,\n\t\tand $q_2$ the one for $C_2$.\n\t\tThen we can take the product to be the coset corresponding to $q_1q_2$.\n\n\t\tQuite importantly,\n\t\t\\textbf{we can also do this in terms of representatives of the cosets}.\n\t\tLet $g_1 \\in C_1$ and $g_2 \\in C_2$,\n\t\tso $C_1 = g_1N$ and $C_2 = g_2N$.\n\t\tThen $C_1 \\cdot C_2$ should be the coset which contains $g_1g_2$.\n\t\tThis is the same as the above definition since\n\t\t$\\phi(g_1g_2) = \\phi(g_1)\\phi(g_2) = q_1q_2$;\n\t\tall we've done is define the product in terms of elements of $G$,\n\t\trather than values in $H$.\n\n\t\tUsing the $gN$ notation,\n\t\tand with \\Cref{remark:coset_warning} in mind,\n\t\twe can write this even more succinctly:\n\t\t\\[ (g_1N) \\cdot (g_2N) \\defeq (g_1g_2)N. \\]\n\t\\end{itemize}\n\\end{definition}\nAnd now you know why the integers modulo $n$ are often written $\\ZZ/n\\ZZ$!\n\\begin{ques}\n\tTake a moment to digest the above definition.\n\\end{ques}\nBy the way we've built it, the resulting group $G/N$ is isomorphic to $Q$.\nIn a sense we think of $G/N$ as ``$G$ modulo the condition that $n=1$\nfor all $n \\in N$''.\n\n\\section{(Optional) Proof of Lagrange's theorem}\nAs an aside, with the language of cosets\nwe can now show Lagrange's theorem in the general case.\n\\begin{theorem}\n\t[Lagrange's theorem]\n\t\\label{thm:lagrange_grp}\n\tLet $G$ be a finite group, and let $H$ be any subgroup.\n\tThen $\\left\\lvert H \\right\\rvert$ divides $\\left\\lvert G \\right\\rvert$.\n\\end{theorem}\n\nThe proof is very simple: note that the cosets of $H$\nall have the same size and form a partition of $G$\n(even when $H$ is not necessarily normal).\nHence if $n$ is the number of cosets,\nthen $n \\cdot \\left\\lvert H \\right\\rvert = \\left\\lvert G \\right\\rvert$.\n\n\\begin{ques}\n\tConclude that $x^{\\left\\lvert G \\right\\rvert}=1$\n\tby taking $H = \\left \\subseteq G$.\n\\end{ques}\n\n\\begin{remark}\n\tIt should be mentioned at this point that\n\tin general, if $G$ is a finite group and $N$ is normal,\n\tthen $|G/N| = |G| / |N|$.\n\\end{remark}\n\n\\section{Eliminating the homomorphism}\n\\prototype{Again $\\ZZ/n\\ZZ \\cong \\Zc n$.}\nLet's look at the last definition of $G/N$ we provided.\nThe short version is:\n\\begin{itemize}\n\t\\ii The elements of $G/N$ are cosets $gN$, which you can think\n\tof as equivalence classes of a relation $\\sim_N$\n\t(where $g_1 \\sim_N g_2$ if $g_1 = g_2n$ for some $n \\in N$).\n\t\\ii Given cosets $g_1N$ and $g_2N$ the group operation is \n\t\\[ g_1N \\cdot g_2N \\defeq (g_1g_2)N. \\]\n\\end{itemize}\nQuestion: where do we actually use the fact that $N$ is normal?\nWe don't talk about $\\phi$ or $Q$ anywhere in this definition.\n\nThe answer is in \\Cref{remark:coset_warning}.\nThe group operation takes in two cosets,\nso it doesn't know what $g_1$ and $g_2$ are.\nBut behind the scenes,\n\\textbf{the normal condition guarantees that the group operation can pick\nany $g_1$ and $g_2$ it wants and still end up with the same coset.}\nIf we didn't have this property, then it would be hard to define the\nproduct of two cosets $C_1$ and $C_2$ because it might make a difference\nwhich $g_1 \\in C_1$ and $g_2 \\in C_2$ we picked.\nThe fact that $N$ came from a homomorphism meant we could pick any representatives\n$g_1$ and $g_2$ of the cosets we wanted, because they all had the same $\\phi$-value.\n\nWe want some conditions which force this to be true without referencing $\\phi$ at all.\nSuppose $\\phi \\colon G \\to K$ is a homomorphism of groups with $H = \\ker\\phi$.\nAside from the fact $H$ is a group, we can get an ``obvious'' property:\n\\begin{ques}\n\tShow that if $h \\in H$, $g \\in G$,\n\tthen $ghg\\inv \\in H$.\n\t(Check $\\phi(ghg\\inv) = 1_K$.)\n\\end{ques}\n\\begin{example}[Example of a non-normal subgroup]\n\t\\label{ex:dihedral_normal_subgroup}\n\tLet $D_{12} = \\left$.\n\tConsider the subgroup of order two $H = \\{1,s\\}$\n\tand notice that \\[ rsr\\inv = r(sr\\inv)= r(rs) = r^2s \\notin H. \\]\n\tHence $H$ is not normal, and cannot be the kernel of any homomorphism.\n\\end{example}\nWell, duh -- so what?\nAmazingly it turns out that that this is the \\emph{sufficient} condition we want.\nSpecifically, it makes the nice ``coset multiplication'' we wanted work out.\n\\begin{remark}\n\t[For math contest enthusiasts]\n\tThis coincidence is really a lot like functional equations at the IMO.\n\tWe all know that normal subgroups $H$ satisfy $ghg\\inv \\in H$;\n\tthe surprise is that from the latter seemingly weaker condition,\n\twe can deduce $H$ is normal.\n\\end{remark}\n\nThus we have a new criterion for ``normal'' subgroups which does not\nmake any external references to $\\phi$.\n\\begin{theorem}[Algebraic condition for normal subgroups]\n\tLet $H$ be a subgroup of $G$.\n\tThen the following are equivalent:\n\t\\begin{itemize}\n\t\t\\ii $H \\normalin G$.\n\t\t\\ii For every $g \\in G$ and $h \\in H$, $ghg\\inv \\in H$.\n\t\\end{itemize}\n\\end{theorem}\n\\begin{proof}\n\tWe already showed one direction.\n\n\tFor the other direction, we need to build a homomorphism with kernel $H$.\n\tSo we simply \\emph{define} the group $G/H$ as the cosets.\n\tTo put a group operation, we need to verify:\n\t\\begin{claim}\n\t\tIf $g_1' \\sim_H g_1$ and $g_2' \\sim_H g_2$ then $g_1'g_2' \\sim_H g_1g_2$.\n\t\\end{claim}\n\t\\begin{subproof}\n\t\tBoring algebraic manipulation (again functional equation style).\n\t\tLet $g_1' = g_1h_1$ and $g_2' = g_2h_2$, so we want to show that\n\t\t$g_1h_1g_2h_2 \\sim_H g_1g_2$.\n\t\tSince $H$ has the property, $g_2\\inv h_1g_2$ is some element of $H$, say $h_3$.\n\t\tThus $h_1 g_2 = g_2 h_3$, and the left-hand side becomes $g_1g_2(h_3h_2)$,\n\t\twhich is fine since $h_3h_2 \\in H$.\n\t\\end{subproof}\n\tWith that settled we can just \\emph{define} the\n\tproduct of two cosets (of normal subgroups) by \\[ (g_1H) \\cdot (g_2H) = (g_1g_2)H. \\]\n\n\tThus the claim above shows that this multiplication is well-defined\n\t(this verification is the ``content'' of the theorem).\n\tSo $G/H$ is indeed a group!\n\tMoreover there is an obvious ``projection'' homomorphism\n\t$G \\to G/H$ (with kernel $H$), by $g \\mapsto gH$.\n\\end{proof}\n%Another way to write the condition is\n%\\[ H = gHg\\inv \\defeq \\left\\{ ghg\\inv \\mid h \\in H \\right\\}. \\]\n%You should take a moment to check that these definitions are equivalent.\n\n\\begin{example}[Modding out in the product group]\n\tConsider again the product group $G \\times H$.\n\tEarlier we identified a subgroup\n\t\\[ G' = \\left\\{ (g, 1_H) \\mid g \\in G \\right\\} \\cong G. \\]\n\tYou can easily see that $G' \\normalin G \\times H$.\n\t(Easy calculation.)\n\n\tMoreover, just as the notation would imply, you can check that\n\t\\[ (G \\times H) / (G') \\cong H. \\]\n\tIndeed, we have $(g, h) \\sim_{G'} (1_G, h)$ for all $g \\in G$ and $h \\in H$.\n\\end{example}\n\\begin{example}[Another explicit computation]\n\tLet $\\phi: D_8 \\to \\Zc 4$ be defined by \\[ r \\mapsto \\ol 2, \\quad s \\mapsto \\ol 2. \\] \n\tThe kernel of this map is $N = \\{1,r^2,sr,sr^3\\}$.\n\n\tWe can do a quick computation of all the elements of $D_8$ to get\n\t\\[ \\phi(1) = \\phi(r^2) = \\phi(sr) = \\phi(sr^3) = \\ol 0 \n\t\\text{ and }\n\t \\phi(r) = \\phi(r^3) = \\phi(s) = \\phi(sr^2) = \\ol 2. \\]\n\tThe two relevant fibers are \\[ \\phi\\pre(\\ol 0) = 1N = r^2N = srN = sr^3N = \\{1,r^2,sr,sr^3\\} \\] and \n\t\\[ \\phi\\pre(\\ol 2) = rN = r^3N = sN = sr^2N = \\{r,r^3,s,sr^2\\}. \\]\n\tSo we see that $|D_8/N| = 2$ is a group of order two, or $\\Zc 2$.\n\tIndeed, the image of $\\phi$ is \\[ \\left\\{ \\ol 0, \\ol 2 \\right\\} \\cong \\Zc 2. \\]\n\\end{example}\n\n\\begin{ques}\n\tSuppose $G$ is abelian.\n\tWhy does it follow that any subgroup of $G$ is normal?\n\\end{ques}\n\n\nFinally here's some food for thought:\nsuppose one has a group presentation for a group $G$\nthat uses $n$ generators.\nCan you write it as a quotient of the form $F_n / N$,\nwhere $N$ is a normal subgroup of $F_n$?\n\n\\section{(Digression) The first isomorphism theorem}\nOne quick word about what other sources usually say.\n\nMost textbooks actually \\emph{define} normal using the $ghg\\inv \\in H$ property.\nThen they define $G/H$ for normal $H$ in the way I did above,\nusing the coset definition\n\\[ (g_1H) \\cdot (g_2H) = g_1g_2H. \\]\nUsing purely algebraic manipulations (like I did) this is well-defined,\nand so now you have this group $G/H$ or something.\nThe underlying homomorphism isn't mentioned at all,\nor is just mentioned in passing.\n\nI think this is incredibly dumb.\nThe normal condition looks like it gets pulled out of thin air\nand no one has any clue what's going on,\nbecause no one has any clue what a normal subgroup actually should look like.\n\nOther sources like to also write the so-called first isomorphism theorem.\\footnote{\n\tThere is a second and third isomorphism theorem.\n\tBut four years after learning about them,\n\tI \\emph{still} don't know what they are.\n\tSo I'm guessing they weren't very important.}\nIt goes like this.\n\\begin{theorem}\n\t[First isomorphism theorem]\n\tLet $\\phi : G \\to H$ be a homomorphism.\n\tThen $G / \\ker \\phi$ is isomorphic to $\\phi\\im(G)$.\n\\end{theorem}\nTo me, this is just a clumsier way of stating the same idea.\n\nAbout the only merit this claim has is that if $\\phi$ is injective,\nthen the image $\\phi\\im(G)$ is an \\emph{isomorphic copy}\nof $G$ inside the group $H$.\n(Try to see this directly!)\nThis is a pattern we'll often see in other branches of mathematics:\nwhenever we have an \\emph{injective structure-preserving map},\noften the image of this map will be some ``copy'' of $G$.\n(Here ``structure'' refers to the group multiplication,\nbut we'll see some more other examples of ``types of objects'' later!)\n\nIn that sense an injective homomorphism $\\phi : G \\injto H$\nis an \\emph{embedding} of $G$ into $H$.\n\n\\section\\problemhead\n\\begin{problem}\n\t[18.701 at MIT]\n\tDetermine all groups $G$ for which the map $\\phi : G \\to G$ defined by \n\t\\[ \\phi(g) = g^2 \\] is a homomorphism.\n\t\\begin{hint}\n\t\tWrite it out: $\\phi(ab) = \\phi(a)\\phi(b)$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tAbelian groups: $abab =a^2b^2 \\iff ab= ba$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\tConsider the dihedral group $G = D_{10}$.\n\t\\begin{enumerate}[(a)]\n\t\t\\ii Is $H = \\left< r \\right>$ a normal subgroup of $G$?\n\t\tIf so, compute $G/H$ up to isomorphism.\n\t\t\\ii Is $H = \\left< s \\right>$ a normal subgroup of $G$?\n\t\tIf so, compute $G/H$ up to isomorphism.\n\t\\end{enumerate}\n\t\\begin{hint}\n\t\tYes, no.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tYes to (a): you can check this directly\n\t\tfrom the $ghg\\inv$ definition.\n\t\tFor example, for (a)\n\t\tit is enough to compute $(r^a s) r^n (r^a s)\\inv = r^{-n} \\in H$.\n\t\tThe quotient group is $\\Zc2$.\n\n\t\tThe answer is no for (b) by following\n\t\t\\Cref{ex:dihedral_normal_subgroup}.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\tDoes $S_4$ have a normal subgroup of order $3$?\n\t\\begin{hint}\n\t\tNo.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tA subgroup of order $3$ must be generated by\n\t\tan element of order $3$, since $3$ is prime.\n\t\tSo we may assume WLOG that $H = \\left< (1\\; 2 \\; 3) \\right>$\n\t\t(by renaming elements appropriately).\n\t\tBut then let $g = (3 \\; 4)$; one can check $gHg\\inv \\ne H$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\tLet $G$ and $H$ be finite groups, where $\\left\\lvert G \\right\\rvert = 1000$\n\tand $\\left\\lvert H \\right\\rvert = 999$.\n\tShow that a homomorphism $G \\to H$ must be trivial.\n\t\\begin{hint}\n\t\t$\\gcd(1000,999)=1$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\t$G/\\ker G$ is isomorphic to a subgroup of $H$.\n\t\tThe order of the former divides $1000$;\n\t\tthe order of the latter divides $999$.\n\t\tThis can only occur if $G / \\ker G = \\{1\\}$\n\t\tso $\\ker G = G$.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\tLet $\\CC^\\times$ denote the nonzero complex numbers under multiplication.\n\tShow that there are five homomorphisms $\\Zc5 \\to \\CC^\\times$\n\tbut only two homomorphisms $D_{10} \\to \\CC^\\times$,\n\teven though $\\Zc5$ is a subgroup of $D_{10}$.\n\\end{problem}\n\n\\begin{problem}\n\t\\gim\n\tFind a non-abelian group $G$\n\tsuch that every subgroup of $G$ is normal.\n\t(These groups are called \\vocab{Hamiltonian}.)\n\t\\begin{hint}\n\t\tFind an example of order $8$.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tQuaternion group.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t[PRIMES entrance exam, 2017]\n\t\\gim\n\tLet $G$ be a group with presentation given by\n\t\\[ G = \\left< a,b,c \\mid\n\t\tab = c^2a^4, \\; bc = ca^6, \\; ac = ca^8, \\;\n\t\tc^{2018} = b^{2019}\n\t\t\\right>. \\]\n\tDetermine the order of $G$.\n\t\\begin{hint}\n\t\tTry to show $G$ is the dihedral group of order $18$.\n\t\tThere is not much group theory content here --- just manipulation.\n\t\\end{hint}\n\t\\begin{sol}\n\t\tThe answer is $|G| = 18$.\n\n\t\tFirst, observe that by induction we have\n\t\t\\[ a^n c = ca^{8n} \\]\n\t\tfor all $n \\ge 1$.\n\t\tWe then note that\n\t\t\\begin{align*}\n\t\t\ta(bc) &= (ab)c \\\\\n\t\t\ta \\cdot ca^6 &= c^2 a^4 \\cdot c \\\\\n\t\t\tc a^8 \\cdot a^6 &= c^2 a^4 \\cdot c \\\\\n\t\t\ta^{14} &= c(a^4c) = c^2 a^{32}.\n\t\t\\end{align*}\n\t\tHence we conclude $c^2 = a^{-18}$.\n\t\tThen $ab = c^2a^4 \\implies b = a^{-15}$.\n\n\t\tIn that case, if $c^{2018} = b^{2019}$,\n\t\twe conclude $1 = a^{2018 \\cdot 18 - 2019 \\cdot 15} = a^{6039}$.\n\t\tFinally, \n\t\t\\begin{align*}\n\t\t\tbc &= ca^6 \\\\\n\t\t\ta^{-15} c &= ca^6 \\\\\n\t\t\ta^{-15} c^2 &= c(a^6c) = c^2 a^{48} \\\\\n\t\t\ta^{-33} &= a^{30} \\\\\n\t\t\t\\implies a^{63} &= 1.\n\t\t\\end{align*}\n\t\tSince $\\gcd(6039, 63) = 9$, we find $a^9 = 1$, hence finally $c^2 = 1$.\n\t\tSo the presentation above simplifies to\n\t\t\\[ G = \\left< a,c \\mid a^9=c^2=1, \\; ac = ca^{-1} \\right> \\]\n\t\twhich is the presentation of the dihedral group of order $18$.\n\t\t% a = r, r^9 = 1\n\t\t% b = r^3\n\t\t% c = s, s^2 = 2\n\t\tThis completes the proof.\n\t\\end{sol}\n\\end{problem}\n\n\\begin{problem}\n\t[Homophony group]\n\t\\gim\n The homophony group (of English) is the group\n\twith $26$ generators $a$, $b$, \\dots, $z$\n\tand one relation for every pair of English words\n\twhich sound the same.\n\tFor example $knight = night$ (and hence $k=1$).\n\tProve that the group is trivial.\n\t\\begin{hint}\n\t\tGet yourself a list of English homophones, I guess.\n\t\tDon't try too hard.\n\t\tLetter $v$ is the worst; maybe $felt = veldt$?\n\t\\end{hint}\n\t\\begin{sol}\n\t\tYou can find many solutions by searching ``homophone group'';\n\t\tone is \\url{https://math.stackexchange.com/q/843966/229197}.\n\t\\end{sol}\n\\end{problem}\n", "meta": {"hexsha": "d267945bab0c56b1253348578d924fd8f50c1d73", "size": 28378, 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null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.0402144772, "max_line_length": 146, "alphanum_fraction": 0.6746775671, "num_tokens": 9771, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754371026368, "lm_q2_score": 0.841825655188238, "lm_q1q2_score": 0.6128283992998712}} {"text": "\\chapter{Multi-layer Multiple Kernel Learning}\n\\label{chap_mlmkl}\nIn this chapter we discuss the Multi-layer Multiple Kernel Learning (ML-MKL\\nomenclature{ML-MKL}{Multi-layer Multiple Kernel Learning}) framework developed using an unsupervised MKL algorithm. The organization of this chapter is as follows; section \\ref{sec_mkl} gives the discussion of traditional MKL algorithm in a supervised learning settings, in section \\ref{sec_umkl} the unsupervised MKL formulation is introduced, in section \\ref{sec_rw} the related works in ML-MKL domain is discussed, in section \\ref{sec_mlmkl} the proposed ML-MKL algorithm is discussed along with the experimental results and section \\ref{sec_conc} concludes this chapter.\n\n\n\\section{Multiple Kernel Learning}\n\\label{sec_mkl}\nMultiple Kernel Learning(MKL\\nomenclature{MKL}{Multiple Kernel Learning}) aims at learning a convex combination of a set of predefined base kernels for choosing an optimum kernel(\\cite{mkl} et al.). The primary aim of MKL algorithm is to automate the process of choosing the optimum kernel for the learning task.\n\n\nTypically multiple kernel learning is formulated in a supervised learning settings. Suppose we are given $n$ datapoints $\\mathcal{D} = \\{(x_1, y_1), (x_2, y_2), \\ldots, (x_n, y_n)\\}$, where $x_i \\in \\mathbb{R}^d$ is the $i^{th}$ input data vector and $y_i$ is the label of the class to which $x_i$ belongs to. Then the MKL learning task is formulated as the following optimization problem.\n\\begin{equation}\n\\min_{k \\in \\mathcal{K}} \\min_{f \\in \\mathcal{H}_k} \\lambda \\norm{f}^2_{\\mathcal{H}_k} + \\sum_{i=1}^n l(y_i, f(x_i))\n\\end{equation}\nwhere $l(\\cdot)$ denotes the loss function(commonly used loss function is the hinge loss, defined as $l(t) = \\max(0, 1-t)$), $\\mathcal{H}_k$ is the RKHS corresponding to the kernel $k$, $\\mathcal{K}$ denotes the optimization domain of the candidate kernels, and $\\lambda > 0$ is the regularization parameter.\n\nThe optimizaion domain $\\mathcal{K}$ is the convex combination of a set of predefined base kernels, defined as follows\n\\[ \\mathcal{K} = \\Bigg\\{ k( \\cdot \\textrm{ , } \\cdot ) = \\sum_{t=1}^m \\mu_t k_t(\\cdot \\textrm{ , } \\cdot) \\textrm{ : } \\sum_{t=1}^m \\mu_t = 1, \\mu_t \\geq 0 \\Bigg\\} \\]\nwhere $k_t$ is the $t^{th}$ base kernel and $\\mu_t$ is the weight associated to the $t^{th}$ base kernel. The decision function $f(x)$ can be computed as a linear combination of kernel evaluations on all training samples;\n\\[ f(x) = \\sum_{i=1}^n \\alpha_i k(x_i, x) \\]\nwhere $\\alpha_i$'s are the coefficients. As per the definition of kernel $k$ in MKL, the decision function of the conventional MKL is expressed as\n\\begin{equation*}\n\\begin{aligned}\nf(x) &= \\sum_{i=1}^n \\alpha_i \\sum_{t=1}^m \\mu_t k_t(x_i, x) \\\\\n&= \\sum_{i=1}^n \\sum_{t=1}^m \\alpha_i \\mu_t k_t(x_i, x)\n\\end{aligned}\n\\end{equation*}\nIn this basic formulation, the optimization task has to identify both the optimal kernel $k$ from the domain $\\mathcal{K}$ and the optimal decision function $f$ from the RKHS $\\mathcal{H}_k$ simultaneously. In order to eliminate this problem \\cite{corinna} et al. proposed a two stage kernel learning algorithm which separates the kernel learning task from the decision function learning.\n\n\n\\section{Unsupervised MKL}\n\\label{sec_umkl}\nIn the KPCA based feature extraction stages, we were using only one kernel for the task. With the intuition that by using multiple kernels at each layer we would get more similarity information, we computed a convex combination of multiple kernels following the work in \\cite{zhuang} et al. Since we are using kernels for unsupervised feature extraction, traditional MKLs following supervised paradigm cannot be used here.\n\nThe goal of an unsupervised multiple kernel learning task is to find an optimal linear combination of the $m$ kernel functions as, i.e, $k^*(\\cdot \\textrm{ , } \\cdot) \\in \\mathcal{K}$. %where $K$ is defined as follows:\n%\\[ K = \\Bigg\\{ k( \\cdot \\textrm{ , } \\cdot ) = \\sum_{t=1}^m \\mu_t k_t(\\cdot \\textrm{ , } \\cdot) \\textrm{ : } \\sum_{t=1}^m \\mu_t = 1, \\mu_t \\geq 0 \\Bigg\\} \\] \n%here $k_t$'s are the base kernels. \nIn order to determine the optimality of a linear combination of kernels, we used the following quality criteria\\cite{zhuang} et al.:\n\n\\begin{itemize}\n\\item A good kernel should enable each training instances to be well reconstructed from the localized bases weighted by the kernel values. Formulating this requirment mathematically, for each $x_i$ we expect the optimal kernel should minimize the approximation error $\\norm{x_i-\\sum_j k_{ij}x_j}^2$, where $k_{ij} = k(x_i, x_j)$.\n\\item A good kernel should induce kernel values that are coincided with the local geometry of the training data. This is equivalent to finding the optimal kernel that minimizes the distortion over all trainig data, computed as $\\sum_{i,j}k_{ij} \\norm{x_i-x_j}^2 $.\n\\end{itemize}\n\nIn addition to this, the locality preserving principle can be exploited by using a set of local bases for each $x_i \\in X$ denoted as $B_i$. By fixing the size of the local bases to some constant $N_B$, the optimization problem of unsupervised MKL can be formulated as follows.\n\\[ \\min_{k \\in \\mathcal{K}} \\frac{1}{2}\\sum_{i=1}^n \\norm{x_i - \\sum_{x_j \\in B_i} k_{ij}x_j}^2 + \\gamma* \\sum_{i=1}^n \\sum_{x_j \\in B_i} k_{ij} \\norm{x_i-x_j}^2 \\]\n\nwhere $\\gamma$ is a tuning parameter, which controls the tradeoff between the coding error and the locality distortion. Converting to matrix notations the above problem becomes\n\\begin{equation}\n \\min_{\\mu \\in \\Delta,D} \\frac{1}{2} \\norm{X(I-K \\circ D)}_F^2 + \\gamma* \\textrm{ tr }K\\circ D\\circ M(11^T)\n \\label{obj_mat} \n\\end{equation}\n%\\begin{equation*}\n % \\begin{aligned}\n\\[ \\textrm{subject to } D \\in \\{0,1\\}^{n \\times n} \\]\n\\[ \\norm{d_i}_1 = N_B, i=1,2,\\ldots,n \\]\n\\[ \\Delta = \\Big\\{\\mu : \\mu^T\\textbf{1} = 1, \\mu \\geq 0 \\Big\\} \\textrm{ and } \\]\n\\[ [K]_{i,j} = \\sum_{t=1}^m \\mu_t k^t(x_i, x_j), 1\\leq i,j \\leq n \\]\n %\\end{aligned}\n%\\end{equation*}\n\nThe matrix $D \\in \\{0,1\\}^{n \\times n}$ contains information about local bases of each $x_i$ as a column vector. In particular, each column vector $d_i \\in \\{0, 1\\}^n $ in $D$ has a 1 at those points $j$ where, $x_j \\in B_i$ and zero elsewhere(or $B_i = \\{ x_j : d_j \\neq 0 \\} $). The matrix M is defined as\n\\[ [M]_{ij} = x_i^Tx_i + x_j^Tx_j - 2x_i^Tx_j \\]\n\nIn equation \\ref{obj_mat} the notation `$\\circ$' denotes elementwise multiplication of two matrices, $\\norm{\\cdot}_F^2$ denotes the Frobenius norm of a matrix and `tr' denotes the trace of a matrix.\n\nIn their implementation(\\cite{zhuang}) Zhuang et.al solved the optimization problem in two stages in an alternating fashion, first by solving for $\\mu$ with a fixed $D$ (using convex optimization) and then solving for $D$ by fixing $\\mu$ (using a greedy mixed integer programming formulation). Since we are using a many layer architecture, the alternating optimization strategy is too costly; so we chose to do the optimization across $\\mu$ only by choosing $D$ beforehand. Specifically, the matrix $D$ is computed beforehand by taking $k$ nearest neighbours of $x_i$ from the training set and putting a one in those positions for $d_i$. Rest of the positions are filled with zeros. The resulting optimization problem will be \n\\begin{equation}\n \\begin{aligned}\n \\min_{\\mu \\in \\Delta} \\frac{1}{2} \\norm{X(I-K \\circ D)}_F^2 + \\gamma* \\textrm{ tr }K\\circ D\\circ M(11^T)\n \\end{aligned}\n\\end{equation}\n\\[ \\textrm{subject to } D \\in \\{0,1\\}^{n \\times n}, \\norm{d_i}_1 = N_B, i=1,2,\\ldots,n \\]\n\\[ \\Delta = \\Big\\{\\mu : \\mu^T\\textbf{1} = 1, \\mu \\geq 0 \\Big\\} \\]\n\nThe objective function can be formulated as a convex quadratic programming problem w.r.t to kernel weights $\\mu$ as shown below(derivation of the objective function $J(\\mu)$ is shown in Appendix \\ref{derivation1}).\n\\begin{equation}\nJ(\\mu) = \\mu^T \\Bigg( \\sum_{t=1}^m \\sum_{i=1}^n k_{t,i}k_{t,i}^T \\circ d_i d_i^T \\circ P \\Bigg)^T \\mu + z^T \\mu \n\\end{equation}\nwhere $[z]_t = \\sum_{i=1}^n (2 \\gamma v_i \\circ d_i - 2 p_i \\circ d_i)^T \\mathit{k}_{t,i} $, $P = X^TX$, and $\\mathit{k}_{t,i} = \\Big[ k^t(x_i, x_1), \\ldots, k^t(x_i, x_n) \\Big]^T $ is the $i^{th}$ column of the $t^{th}$ kernel matrix. $p$ and $v$ are columns of $P$ and $M$ corresponding to $x_i$ respectively.\n\n\\section{Related Works}\n\\label{sec_rw}\n\n\\cite{2l_mkl} et al. explored the idea of ML-MKL, in which at each layer they took a non-linear combination of kernels in the previous layer, forming a layered structure. They called such a model as deep multiple kernel learning framework. The architecture of their model is shown in figure \\ref{fig_2l_mkl}. The kernel function obtained at layer $l$ is having the following domain.\n\\[ \\mathcal{K}^{(l)} = \\bigg\\{ k^{(l)}(\\cdot, \\cdot) = g^{(l)}\\Big( \\big[k_1^{(l-1)},\\ldots, k_m^{(l)}(\\cdot, \\cdot) \\big] \\Big) \\bigg\\} \\]\n\n\\begin{figure*}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[width=1.1\\textwidth,height=8.5cm]{figures/2lmkl}\n \\caption{Architecture of ML-MKL model proposed by \\cite{2l_mkl} et al.}\n \\label{fig_2l_mkl}\n\\end{figure*}\n\nwhere $g^{(l)}$ is a function to combine multiple $(l-1)$ level kernels, which must ensure the resulting combination is a valid kernel. Though this combination can be applied for any number of layers, optimization problem is difficult to solve beyond two layers. Hence \\cite{2l_mkl} et al. considered only 2-layer MKL in their empirical study. \n\nIn the 2-layer MKL, \\cite{2l_mkl} et al. defined the kernel domain by using an RBF kernel for the combination function. \n\\[ \\mathcal{K}^{(2)} = \\bigg\\{ k^{(2)}(x_i, x_j;\\bm{\\mu}) = exp\\Big( \\sum_{t=1}^m \\mu_t k_t^{(1)}(x_i, x_j) \\Big) \\bigg\\} \\colon \\bm{\\mu} \\in \\mathbb{R}_+^m \\]\n$\\mu_t$ is the weight of the $t^{(th)}$ antecedent layer kernel $k_t^{(1)}$(superscript (1) is used here because the antecedent layer is layer 1 in a 2-layer architecture). In order to prevent the kernel weight being too large, the kernel weights are also introduced into the optimization objective as a regularization term. Then the optimization objective becomes\n\\begin{equation}\n\\min_{k \\in \\mathcal{K}} \\min_{f \\in \\mathcal{H}_k} \\norm{f}^2_{\\mathcal{H}_k} + \\mathcal{C} \\sum_{i=1}^n l(y_i, f(x_i)) + \\sum_{t=1}^m \\mu_t\n\\end{equation}\nsolving the Lagrangian, we will get the dual objective function as\n\\[ \\min_{\\bm{\\mu}} \\max_{\\bm{\\alpha}} \\sum_{i=1}^n \\alpha_i - \\frac{1}{2}\\sum_{i,j=1}^n \\alpha_i \\alpha_j y_i y_j k^{(2)}(x_i, x_j;\\bm{\\mu}) + \\sum_{t=1}^m \\mu_t \\]\n\\[ \\textrm{s.t } 0\\leq \\alpha_i\\leq \\mathcal{C}, \\sum_{i=1}^n \\alpha_i y_i = 0, \\mu_t \\geq 0, t=1, \\ldots, m \\]\nwhere $\\alpha = [\\alpha_1, \\alpha_2, \\ldots, \\alpha_n]^T$ is a vector of dual variables and $\\mu = [\\mu_1, \\mu_2, \\ldots, \\mu_m]^T$. The final decision function of the 2-layer MKL is given as\n\\[f(x;\\bm{\\alpha}, \\bm{\\mu}) = \\sum_{i=1}^n \\alpha_i y_i k^{(2)}(x_i, x;\\bm{\\mu}) + b \\]\nwhere $b$ is the bias term. Rewriting the optimization objective in terms of $\\alpha$ and $\\mu$, we have\n\\[ \\mathcal{J}(\\alpha, \\mu) = \\frac{1}{2} \\sum_{i,j=1}^n \\alpha_i \\alpha_j y_i y_j k^{(2)}(x_i, x_j;\\bm{\\mu}) - \\sum_{i=1}^n \\alpha_i - \\sum_{t=1}^m \\mu_t \\]\nThe optimization problem is solved in two stages\n\\begin{itemize}\n\\item by fixing $\\mu$ and solve for $\\alpha$.\n\\item by fixing $\\alpha$ and solve for $\\mu$.\n\\end{itemize}\nSince $k^{(2)}$ is positive definite $\\mathcal{J}(\\alpha, \\mu)$ is convex over $\\alpha$, thus an SVM solver can be used to solve the optimization over $\\alpha$. The optimization over $\\mu$ is solved using gradient ascent. In order to address the challenge of choosing the optimal set of base kernels, they proposed to choose base kernels iteratively inside $k^{(2)}$.\n\n\\cite{deep_mkl} et al. used backpropagation algorithm for designing the ML-MKL framework. The architecture of the model studied by them is shown in figure \\ref{fig_deep_mkl}. They defined the kernel domain at level $l$ as\n\n\\begin{figure*}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[width=1\\textwidth,height=6cm]{figures/back_prop_mlmkl}\n \\caption{Architecture of ML-MKL model proposed by \\cite{deep_mkl} et al. The kernels in each layer are linearly combined and the resulting Gram matrix is passed to the next layer as input.}\n \\label{fig_deep_mkl}\n\\end{figure*}\n\n\\[ \\mathcal{K}^{(l)} = \\big\\{ K^{(l)}(K^{(l-1)}; \\mu^{(l)}) = \\sum_{t=1}^m \\mu_t^{(l)} k_t^{(l)}(K^{(l-1)}) \\big\\} \\]\n\\[ \\textrm{s.t } \\mu_t^{(l)} \\geq 0, l = 1, \\ldots, L \\]\nwhere $k_t^{(l)}$ is the $k^{th}$ base kernel at layer $l$ and $\\mu_t^{(l)}$ denotes the weight of the $k^{th}$ base kernel at layer $l$. $L$ is the total number of layers. The feature extracted in the antecedent layer are combined linearly and passed to the next layer as input. The kernel weights are obtained by minimizing the mean squre error(denotes as $\\mathcal{E}$) of the predicted outputs.\n\\[ \\mathcal{E} = \\frac{1}{2n} \\sum_{i=1}^n \\norm{f(x_i) - y_i}^2 \\]\nwhere $f(x_i)$ is the value of the decision function for the input $x_i$, which is computed as\n\\[ f(x) = \\sum_{i=1}^n \\alpha_i y_i K^{(l)}(K^{(l-1)}; \\mu^{(l)}) + b \\]\nwhere $b$ is the bias term. The weights $\\mu^{(l)}$ in each layer are obtained using gradient descent algoritm.\n\\[ \\mu^{(l)} := \\mu^{(l)} + \\eta \\nabla \\mathcal{E} \\textrm{ } l = 1, \\ldots, L \\]\n\\[ \\nabla \\mathcal{E} = \\Bigg\\{ \\frac{\\partial \\mathcal{E}}{\\partial \\mu_1^{(1)}}, \\cdots, \\frac{\\partial \\mathcal{E}}{\\partial \\mu_m^{(1)}}; \\cdots; \\frac{\\partial \\mathcal{E}}{\\partial \\mu_1^{(L)}}, \\cdots, \\frac{\\partial \\mathcal{E}}{\\partial \\mu_m^{(L)}} \\Bigg\\} \\]\nHere $\\eta$ is the learning rate parameter. The error obtained in the final layer are propagated back to all layers using backpropagation algorithm to update the kernel weight parameters $\\mu^{(l)}$ in each layer.\n\n\\section{Multi-layer Multiple Kernel Learning}\n\\label{sec_mlmkl}\nThe architectue of the proposed ML-MKL framework is shown in figure \\ref{fig_mlmkl}. It consists of many layers and in each layer the kernel PCA based feature extraction is performed using the combination of a set of predefined kernels. The dimensionality of the features thus obtained are reduced by using supervised feature selection techniques. The final output can be given to any classifier. Algorithm \\ref{algo1} summarizes the proposed ML-MKL algorithm.\n\n\\begin{algorithm}\n\\caption{ML-MKL Algorithm}\n\\textbf{Input}: data X, true labels y, no. of layers L, base kernels for each layer $K_{base}^{(l)} = \\{k_1^{(l)}, k_2^{(l)}, k_m^{(l)}\\}$, $N_B$, $\\gamma$\\;\n\\textbf{Output}: kernel weights $\\mu^l$ for each layer, predicted labels\\;\n1.Initialize $[M]_{ij} = x_i^Tx_i + x_j^Tx_j - 2x_i^Tx_j$, $\\bm{D} = d_1, d_2, \\ldots, d_n$ as row vectors, where $d_i = \\{1 \\textrm{ if } x_j \\in B_i \\textrm{ else } 0 \\forall x_j \\in X \\}$, $\\mu = \\frac{1}{m}$, $\\bm{P} = X^TX$\\;\n2.\\For{each layer l}{\n a. $\\bm{W} = \\sum_{t=1}^m \\sum_{i=1}^n k_{t,i}^{(l)}k_{t,i}^{(l)^{T}} \\circ d_i d_i^T \\circ P$\\\\\n b. $\\bm{[z]}_t^l = \\sum_{i=1}^n (2 \\gamma v_i \\circ d_i - 2 p_i \\circ d_i)^T \\mathit{k}_{t,i}^{(l)}$\\\\\n c. $\\bm{\\mu}^{*^{l}} = \\mu^{l^{T}}W\\mu^l + z^{l^{T}}\\mu^l$\\\\\n d. $\\bm{K}_{new} = \\sum_{t=1}^m \\mu_t^l * K_t^{(l)}$\\\\\n e. extract principal components with $\\bm{K}_{new}$\\\\\n f. select most informative features for layer $l$($X_{new}$)\\\\\n g. $\\bm{P} = X_{new} ^T X_{new}$\n}\n3.Give the final set of features to any classifier\\;\n\\label{algo1}\n\\end{algorithm}\n\n\\begin{figure*}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[width=1.0\\textwidth,height=5.5cm]{figures/mlmkm}\n \\caption{An ML-MKL with L layers of transformations. Each layer consists of many kernels for feature extraction using kernel PCA and a supervised feature selection module.}\n \\label{fig_mlmkl}\n\\end{figure*}\n\nThe experimental setup used here was the same as described in \\autoref{chap_mkm}. Table \\ref{tab_mlmkl} lists the results obtained with the proposed ML-MKL framework. The classifier used was SVM with arc-cosine kernel. For \\textit{mnist-back-rand} dataset, the best result was obtained with a model consists of 4 layers and in each layer 7 kernels were used. In particular, each layer consisted of a mixture of one arc-cosine kernel and 6 gaussian kernels. For \\textit{mnist-back-image} dataset, the best ML-MKL model obtained had 2 layers and 5 kernels in each layer. In each layer a mixture of one arc-cosine kernel and 4 polynomial kernels were used. In the case of \\textit{mnist-rot-back-image} dataset the best result was fetched by a model having only one layer with 4 arc-cosine kernels in it. For \\textit{rectangles-image} dataset the best result was shown by a one layer ML-MKL model with one arc-cosine kernel and 6 Gaussian kernels.\n\n\n\\renewcommand{\\arraystretch}{2.1}\n\\begin{table*}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Dataset}} & \\multicolumn{7}{ |c| }{\\textbf{Loss in Percentage}} \\\\\n \\cline{2-8}\n &$\\textrm{SVM}_{\\textrm{RBF}}$ & $\\textrm{SVM}_{\\textrm{Poly}}$ & NNet & DBN-3 & SAA-3 & DBN-1 & \\textbf{ML-MKL}\\\\\n \\hline \n \\textit{back-rand} & 14.58 & 16.62 & 20.04 & \\textbf{6.73} & 11.28 & 9.80 & 8.43$\\pm 0.088$\\\\\n \\hline\n \\textit{back-image} & 22.61 & 24.01 & 27.41 & 16.31 & 23.00 & \\textbf{16.15} & 20.92$\\pm 0.092$\\\\\n \\hline\n \\textit{rot-back-image} & 55.18 & 56.41 & 62.16 & \\textbf{47.39} & 51.93 & 52.21 & 51.21$\\pm 0.811$\\\\\n \\hline\n \\textit{rect-image} & 24.04 & 24.05 & 33.20 & 23.69 & 24.05 & \\textbf{22.50} & 22.88$\\pm 0.124$\\\\\n \\hline\n\\end{tabular}\n\\caption{Experimental Results of ML-MKL.}\n\\label{tab_mlmkl}\n\\end{table*}\n\\renewcommand{\\arraystretch}{1}\n\nFigures \\ref{fig_mbr_layers} and \\ref{fig_mbi_layers} illustrates the variation in classifier performance on \\textit{mnist-back-rand} and \\textit{mnist-back-image} datasets respectively when layers were added iteratively to the ML-MKL model. The value shown for each layer was the best error rate obtained after tuning the kernel parameters. The parameters were chosen greedily for each layer (with the expectation that subsequent layers would learn more valuable features from the current one) and no fine-tuning was performed with respect to the entire architecture.\n\n\\begin{figure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.8]{figures/mlmkl_rand}\n \\caption{Change in classifier performance on \\textit{mnist-back-rand} dataset when adding layers iteratively.}\n \\label{fig_mbr_layers}\n\\end{figure}\n\n\\begin{figure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.8]{figures/mlmkl_image}\n \\caption{Change in classifier performance on \\textit{mnist-back-image} dataset when adding layers iteratively.}\n \\label{fig_mbi_layers}\n\\end{figure}\n\nFigure \\ref{tsne_rand_mlmkl} shows the tSNE embedding of the features learned by ML-MKL algorithm. The visualization indicates that classes are well separated.\n\n\\begin{sidewaysfigure}\n \\centering\n \\captionsetup{justification=centering,margin=0.1cm}\n \\includegraphics[scale=0.45]{figures/tsne_rand_mlmklbest}\n \\caption{tSNE embedding of features obtained by ML-MKL algorithm for the \\textit{mnist-back-rand} dataset.}\n \\label{tsne_rand_mlmkl}\n\\end{sidewaysfigure}\n\nTables \\ref{back_rand_kw} and \\ref{back_image_kw} shows the kernel weights of each kernel in the mixture at every layer for \\textit{mnist-back-rand} and \\textit{mnist-back-image} datasets respectively. In both cases $k_1$ is an acr-cosine kernel, and the remaining are Gaussian kernels for \\textit{mnist-back-rand} dataset and polynomial kernel for textit{mnist-back-image} dataset. The results in the table indicates that, the contribution of individual kernels was highly varying in each layer (no single kernel had complete dominance over all layers in the feature learning process). \n\n\n\\renewcommand{\\arraystretch}{2.3}\n\\begin{table*}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Layers}} & \\multicolumn{7}{ |c| }{\\textbf{Kernel Weights}} \\\\\n \\cline{2-8}\n & $k_1$ & $k_2$ & $k_3$ & $k_4$ & $k_5$ & $k_6$ & $k_7$ \\\\\n \\hline \n \\textit{Layer 1} & 0.2007 & 0.1331 & 0.1331 & 0.1332 & 0.1332 & 0.1333 & 0.1333\\\\\n \\hline\n \\textit{Layer 2} & 0.2711 & 0.1160 & 0.1181 & 0.1203 & 0.1225 & 0.1248 & 0.1271\\\\\n \\hline\n \\textit{Layer 3} & 0.1764 & 0.0999 & 0.1125 & 0.1266 & 0.1426 & 0.1607 & 0.1811\\\\\n \\hline\n \\textit{Layer 4} & 0.0598 & 0.0524 & 0.0747 & 0.1071 & 0.1547 & 0.2245 & 0.3269\\\\\n \\hline\n \\textit{Layer 5} & 0.0514 & 0.0493 & 0.0726 & 0.1069 & 0.1569 & 0.2295 & 0.3334\\\\\n \\hline \n\\end{tabular}\n\\caption{Kernel weights in each layer for the \\textit{mnist-back-rand} dataset}\n\\label{back_rand_kw}\n\\end{table*}\n\\renewcommand{\\arraystretch}{1}\n\n\\renewcommand{\\arraystretch}{2.3}\n\\begin{table}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Layers}} & \\multicolumn{5}{ |c| }{\\textbf{Kernel Weights}} \\\\\n \\cline{2-6}\n & $k_1$ & $k_2$ & $k_3$ & $k_4$ & $k_5$\\\\\n \\hline \n \\textit{Layer 1} & 0.2445 & 0.1887 & 0.1888 & 0.1889 & 0.1891\\\\\n \\hline\n \\textit{Layer 2} & 0.3421 & 0.1603 & 0.1630 & 0.1659 & 0.1688\\\\\n \\hline\n \\textit{Layer 3} & 0.2035 & 0.1615 & 0.1843 & 0.2104 & 0.2403\\\\\n \\hline\n \\textit{Layer 4} & 0.0843 & 0.1409 & 0.1877 & 0.2508 & 0.3362\\\\\n \\hline\n\\end{tabular}\n\\caption{Kernel weights in each layer for the \\textit{mnist-back-image} dataset}\n\\label{back_image_kw}\n\\end{table}\n\\renewcommand{\\arraystretch}{1}\n\nIn order to evaluate the contribution of each kernel, exploratory analysis was carried out to monitor the performance of individual kernels in each layer. The individual kernels performance were compared with the combined kernel's performance. Tables \\ref{back_rand_exp} and \\ref{back_image_exp} summarizes the results of this exploratory analysis on \\textit{mnist-back-rand} and \\textit{mnist-back-image} datasets respectively (here $K_{conv}$ is the result of combined kernel). The results clearly indicates that the combination was not always improving the performance in all layers. However, the best result in both datasets were obtained by a combined kernel.\n\n\\renewcommand{\\arraystretch}{2.1}\n\\begin{table*}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Layers}} & \\multicolumn{8}{ |c| }{\\textbf{Loss in Percentage}} \\\\\n \\cline{2-9}\n & $k_1$ & $k_2$ & $k_3$ & $k_4$ & $k_5$ & $k_6$ & $k_7$ & $K_{conv}$\\\\\n \\hline \n \\textit{Layer 1} & 11.46 & 9.92 & 10.06 & 10.04 & 10.1 & 9.87 & 10.06 & 11.62\\\\\n \\hline\n \\textit{Layer 2} & 10.36 & 9.87 & 9.90 & 9.89 & 9.92 & 9.82 & 9.81 & 9.52\\\\\n \\hline\n \\textit{Layer 3} & 9.46 & 11.07 & 10.33 & 9.85 & 9.70 & 9.56 & 9.01 & 9.18\\\\\n \\hline\n \\textit{Layer 4} & 9.26 & 9.06 & 8.95 & 8.96 & 8.93 & 8.72 & 9.00 & \\textbf{8.72}\\\\\n \\hline\n \\textit{Layer 5} & 9.18 & 9.07 & 9.02 & 9.20 & 9.22 & 9.36 & 9.47 & 9.15\\\\\n \\hline \n\\end{tabular}\n\\caption{Individual kernels performance evaluation for \\textit{mnist-back-rand} dataset}\n\\label{back_rand_exp}\n\\end{table*}\n\\renewcommand{\\arraystretch}{1}\n\n\\renewcommand{\\arraystretch}{2.1}\n\\begin{table}\n\\centering\n\\begin{tabular}{|c|c|c|c|c|c|c|}\n \\hline\n \\multirow{2}{*}{\\textbf{Layers}} & \\multicolumn{6}{ |c| }{\\textbf{Loss in Percentage}} \\\\\n \\cline{2-7}\n & $k_1$ & $k_2$ & $k_3$ & $k_4$ & $k_5$ & $K_{conv}$\\\\\n \\hline \n \\textit{Layer 1} & 20.91 & 21.82 & 21.82 & 21.83 & 21.77 & 21.15\\\\\n \\hline\n \\textit{Layer 2} & 21.27 & 21.00 & 21.09 & 21.04 & 21.04 & \\textbf{20.83}\\\\\n \\hline\n \\textit{Layer 3} & 20.95 & 21.18 & 21.02 & 20.99 & 21.05 & 21.03\\\\\n \\hline\n \\textit{Layer 4} & 21.06 & 21.36 & 21.42 & 21.64 & 21.66 & 21.51\\\\\n \\hline\n\\end{tabular}\n\\caption{Individual kernels performance evaluation for \\textit{mnist-back-image} dataset}\n\\label{back_image_exp}\n\\end{table}\n\\renewcommand{\\arraystretch}{1}\n\n\n\n\\section{Conclusion}\n\\label{sec_conc}\nIn this chapter we explored the concept of multiple kernel learning in MKMs. A linear combination of multiple kernels formulated purely from unlabelled data is used in each layer of MKMs. The learning process of the proposed ML-MKL algorithm employs a greedy layerwise training for each layer. Empirical results indicates that using (unsupervised) MKL in MKMs improves the classifier performance. In our experimental analysis the classification accuracy of ML-MKL models was better than learning machines with shallow architectures and was comparable with existing deep architectures.\n\nExploratory analysis performed on the features learned by the ML-MKL model reveals much more interesting facts about the model. The contribution of kernels in each layer is measured in terms of individual kernel performance and the weights assigned to that kernel. However this information is insufficient to determine the optimal structural complexity required for modelling a problem.\n", "meta": {"hexsha": "b43ce505faa6a33b295b3b10274ec330ffe391bb", "size": 24812, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/chapter3.tex", "max_stars_repo_name": "akhilpm/Masters-Project", "max_stars_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Thesis/chapter3.tex", "max_issues_repo_name": "akhilpm/Masters-Project", "max_issues_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/chapter3.tex", "max_forks_repo_name": "akhilpm/Masters-Project", "max_forks_repo_head_hexsha": "cc10673e695cbc0531f6268d729760705890a116", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 74.9607250755, "max_line_length": 944, "alphanum_fraction": 0.7035305497, "num_tokens": 8500, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321983146848, "lm_q2_score": 0.7606506526772884, "lm_q1q2_score": 0.6128046574659036}} {"text": "\\documentclass[../main.tex]{subfiles}\n\n\\begin{document}\n\n\\setcounter{chapter}{3}\n\n\\chapter{Schemes and algebraic spaces I}\n\nIn the previous lecture we saw the notion of a geometric context, and that this notion gives rise to manifolds and geometric spaces. We will now construct a context specific to algebraic geometry in which the manifolds are schemes and geometric spaces are algebraic spaces.\n\n\nIn order to understand the conrtuction of algebraic contexts we commence by recalling some elementary facts about (affine) algebraic varieties, .... Naively an algebraic variety is a set of solutions to a system of plynomial equations. ... An algebraic variety is determined a finite familyof polynomials $P_1, \\cdot P_r \\in \\mathbb C[x_1, \\cdots, x_n]$. X is then a solution a dset of solutions of the system $\\{P_i(x) = 0\\}_{1 \\le i \\le r}$, in ... terms we have\n\n\\[\nX = \\{ x \\in \\mathbb C^n ~|~ P_i(x) = 0 ~ \\forall i\\}\n\\]\nWe also then have the question of defining the ``algebraic functions'' on $X$. ... ring of algebraic functions on $X$ is witten as\n\n\\[\n\\mathcal O(X) = \\mathcal C[x_1, \\cdots x_n] / (P_1, \\cdots, P_r)\n\\]\n\nAn algebraic variety $X$ corresponds to the ring $C[x_1, \\cdots x_n] / (P_1, \\cdots, P_r)$. This ring is a commutative $\\mathbb C$-algebra of finite type (i.e. with a finite number of generators). Conversely, if $A$ is an commutative $C$-algebra of finite type, we can write\n\n\\[\nA \\simeq \\mathbb C[y_1, \\cdots, y_m]/(Q_1, \\cdots, Q_s)\n\\]\n\nand construct an algebraic variety in $\\mathbb C^m$ with the equation $Q_1(y) = \\cdots = Q_s(y) = 0$.\n\n\nthere's a correspondence between affines algebraic varies in $\\mathbb C$ (i.e. subvarieties of $\\mathbb C^n$ for some $n$), and commutative $\\mathbb C$-algebras of finite type. The starting point for the theory of schemes and considering commutative rings A (not necessarily commutative $\\mathbb C$-algebras of finite type) is the considering of the correspondenc between a geometric object $X$ and $A$ a ring of algebraic functions on $X$. The geometric objects associated with the ring are affine schemes, and a scheme is by definition a collection of affine schemes (much like a topological property is a collection of opens in $\\mathbb R^n$). We have the following two principles\n\n\\begin{prcp*}\n Affine schemes are in correspondence with commutative rings\n\\end{prcp*}\n\\begin{prcp*}\n Schemes are obtained from collections of affines schemes\n\\end{prcp*}\n\n\nFrom the combination two principles we obtain the following principles\n\n\\begin{prcp*}\n Schemes are obtained from collections of commutative rings\n\\end{prcp*}\n\n\n\n\\section{Geometric contexts}\n\n\n\n\n\\end{document}", "meta": {"hexsha": "826e01f635ff65dce1231997e0e3eb42aa07e0e0", "size": 2648, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapters/lecture3.tex", "max_stars_repo_name": "jakebian/OTIM-toen-mastercourse", "max_stars_repo_head_hexsha": "61d89ade9f4c08966d671277c13ace3614636ac2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapters/lecture3.tex", "max_issues_repo_name": "jakebian/OTIM-toen-mastercourse", "max_issues_repo_head_hexsha": "61d89ade9f4c08966d671277c13ace3614636ac2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapters/lecture3.tex", "max_forks_repo_name": "jakebian/OTIM-toen-mastercourse", "max_forks_repo_head_hexsha": "61d89ade9f4c08966d671277c13ace3614636ac2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.1454545455, "max_line_length": 683, "alphanum_fraction": 0.7401812689, "num_tokens": 753, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7606506526772884, "lm_q1q2_score": 0.612804643266945}} {"text": "\\chapter{Explicit One-Step Methods and Convergence}\n\n\\section{Introduction}\n\\begin{example}[Euler's method]\n We begin this section with the method which serves as prototype for\n a whole class of schemes which solves an IVP or rather the Volterra\n integral equation numerically. Here, as always for problems with\n infinite dimensional solution spaces, numerical solution refers to\n finding an approximation by applying a discretization method, and\n studying the error of this method.\n \n Consider the following problem: given an IVP of the form of \\slideref{Definition}{IVP},\n calculate the value $u(T)$ at a later point in time $T$.\n \n To this end, we note first of all that for an IVP at the initial\n point $0$, not only the function value $u(0) = u_0$ is known,\n but also the derivative $u'(0) = f(0, u_0)$. Thus we are capable\n to replace the solution $u(t)$ in blue by a straight line $y(t)$ in\n red, which we can see on the left of Figure~\\ref{fig:explicit:Euler}.\n \\begin{figure}[tp]\n \\centering\n \\includegraphics[width=.48\\textwidth]{fig/euler1}\n \\includegraphics[width=.48\\textwidth]{fig/euler2}\n \\caption{Derivation of the Euler method. Left: replacement of the\n\t\t\tsolution of the IVP by a line with slope and initial point given \n\t\t\tby the IVP. Right: Euler method with three subintervals.}\n \\label{fig:explicit:Euler}\n \\end{figure}\n The figure suggests that in general the accuracy of this method may\n not be very good. The first improvement is that we do not draw the\n line through the whole interval from $0$ to $T$. Instead, we\n insert intermediate points and apply the method to each subinterval,\n where we use the result of a previous interval as the initial point\n for the next subinterval. As a result one obtains a\n chain of straight lines and the so-called \\define{Euler method}.\n\\end{example}\n\n\n\\input{definitions/partitioning}\n\n\\begin{definition}\n In the following chapters we will regularly compare the solution of\n an IVP with the results of discretization methods. Therefore, we\n introduce the following convention for notations and symbols.\n \n \\defindex{solution!continuous}\\defindex{solution!exact} The solution\n of the IVP is called the \\textbf{exact} or \\define{continuous\n solution}\\defindex{exact solution}. The term ``continuous''\n indicates here the solution of the non-discretized problem. Its\n symbol is in general $u$ and we set as abbreviation\n \\begin{gather*}\n u_k = u(t_k).\n \\end{gather*}\n If $u$ is vector-valued we also use the alternative superscript\n $u^{(k)}$ and $u_i^{(k)}$ for a entry of the vector $u(t_k)$.\n\n \\defindex{solution!discrete} In general we write the\n \\define{discrete solution} with the\n symbol $y$. We write $y_k$ or $y^{(k)}$ for the value of the\n discrete solution at the point in time $t_k$. In contrast to the\n continuous solution, $y$ only defined at discrete time steps, unless\n for special methods discussed later.\n\\end{definition}\n\n\\input{definitions/one-step}\n\n\\begin{remark}\n \\label{remark:expl:first-step}\n For one-step methods every step is \\emph{per definitionem} similar.\n Therefore, it is sufficient to consider the first step only. Hence,\n we will define and analyze methods by stating the dependence of\n $y_1$ on $y_0$ which then can be transferred to the general step\n from $y_{n-1}$ to $y_{n}$. The general one-step method above then\n reduces to\n \\begin{gather*}\n y_1 = y_0 + h \\verfahren(t_0, y_0).\n \\end{gather*}\n This implies that the values $y_k$ with $k\\ge 2$ are computed\n through formula~\\eqref{eq:explicit:8} with the respective $h_k$ and\n the same increment function.\n\\end{remark}\n\n\\input{theorems/euler-linear-1}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Error analysis}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{remark}\n In Figure~\\ref{fig:explicit:Euler}, we observe that the error\n consists of two parts at a given time $t_{k+1}$. First, an error on\n the interval $I_k$ due to replacing the differential equation by the\n discrete method. Second, we have to add the error which results from\n the fact that our initial value $y_k$ is already not exact due to\n previous errors. This situation is displayed in\n Figure~\\ref{fig:forward-errors}. After one time step, a local error\n has appeared. In the second time step, we already start with an\n erroneous initial value. Therefore, we split the error into the\n local error and an accumulated error. The local error compares\n continuous and discrete solutions on a single interval with the same\n initial value. In the analysis, we will have the options of using\n the exact (right figure) or the approximated initial value (left figure).\n\n \\begin{figure}[tp]\n \\centering\n \\includegraphics[width=.49\\textwidth]{fig/forward-error.tikz}\n \\includegraphics[width=.49\\textwidth]{fig/backward-error.tikz}\n \\caption{Local and accumulated errors. Exact solution in black,\n the Euler method in red. On the left, in blue the exact solution\n of an IVP on the second interval with initial value $y_1$. On\n the right, in purple the second step of the Euler method, but with\n exact initial value $u_1$.}\n \\label{fig:forward-errors}\n \\end{figure}\n\\end{remark}\n\n\\input{definitions/globalerror}\n\n\\input{definitions/truncation-error}\n\n\\begin{example}[Euler method]\n To find out the order of consistency of the Euler method, we \n consider the Taylor expansion of the solution at the point $t_{n-1}$:\n \\begin{equation*}\n u(t_n) = u(t_{n-1}) + h_n u'(t_{n-1}) + \\frac12 h_n^2 u''(\\zeta)\n \\end{equation*}\n \n As a result the truncation error reduces to:\n \n \\begin{align*}\n \\tau_n & = \\frac{u_n - u_{n-1}}{h_n} - F(h;t_{n-1},u(t_{n-1})) \\\\\n & = \\frac{u_{n-1} + h_n f(t_{n-1},u_{n-1})\n + \\frac12 h_n^2 u''(\\zeta) - u_{n-1}}{h_n} - f(t_{n-1}; u_{n-1}) \\\\\n & = \\frac12 h_n u''(\\zeta)\n \\end{align*}\n\n% For $u''(\\zeta)$ however there holds:\n% \\begin{equation*}\n% u''(\\zeta) = \\diffq[\\zeta] f(\\zeta,u(\\zeta)) = \\pdiffq[\\zeta] f(\\zeta,u(\\zeta)) + \\pdiffq[u] f(\\zeta,u(\\zeta)) \\underbrace{u'(\\zeta)}_{= f(\\zeta,u(\\zeta))}\n% \\end{equation*}\n\nUnder the assumption that $f \\in C^1$ on a compact set around the graph\nof $u$, this term is bounded, yielding.\n\\begin{gather*}\n | \\tau_n | \\le \\frac{h_n}{2} \\max_{\\zeta\\in I_n} \\abs{u''(\\zeta)}\n = \\frac{h_n}{2} \\max_{\\zeta\\in I_n} \\bigl|\\partial_x f(\\zeta,\n u(\\zeta)) + \\partial_u f(\\zeta, u(\\zeta)) f(\\zeta, u(\\zeta))\\bigr|\n\\end{gather*}\nHere, we enter the assumption that $f$ is sufficiently smooth to\nconclude that the Euler method is consistent of order 1.\n\\end{example}\n\n\\input{theorems/gronwall-discrete}\n\n\\begin{proof}\n Define the functions $w(t)$, $a(t)$, and $b(t)$ such that for $k\\ge\n 1$ and $t\\in [k-1,k)$ there holds\n \\begin{gather*}\n w(t) = w(t_{k-1}),\\quad\n a(t) = b(t_{k-1}),\\quad\n b(t) = b(t_{k-1}).\n \\end{gather*}\n These functions are bounded and piecewise continuous on any finite\n interval. Thus, they are integrable on $[0,n]$. Therefore, the\n continuous Grönwall inequality of Lemma~\\ref{Lemma:gronwall} applies\n and proves the result.\n\\end{proof}\n\n\\input{theorems/stability-discrete}\n\n\\begin{proof}\n Subtracting the equations\n \\begin{align*}\n \\eta_k(y) &= y_k-y_{k-1} - \\verfahren_{h_k}(t_{k-1},y_{k-1}),\n \\\\\n \\eta_k(y) &= z_k-z_{k-1} - \\verfahren_{h_k}(t_{k-1},z_{k-1}),\n \\end{align*}\n we obtain\n \\begin{multline*}\n y_k-z_k = y_{k-1} - z_{k-1} + \\eta_k(y) - \\eta_k(z)\n \\\\\n + h_k \\bigl(\n F_{h_k}(t_{k-1},y_{k-1})-F_{h_k}(t_{k-1},z_{k-1})\n \\bigr)\n .\n \\end{multline*}\n Recursive application yields\n \\begin{gather*}\n \\abs{y_n-z_n} \\le \\abs{y_0-z_0}\n + \\sum_{k=1}^n \\abs{\\eta_k(y)-\\eta_k(z)}\n + \\sum_{k=1}^n L_h h_{k} \\abs{y_k-z_k}.\n \\end{gather*}\n The estimate now follows from the discrete Grönwall inequality in\n Lemma~\\ref{Lemma:gronwall-discrete}.\n\\end{proof}\n\n\\begin{Corollary*}{finite-precision}{One-step methods with finite precision}\n Let the one-step method $\\verfahren$ be run on a computer, yielding\n a sequence $(z_n)$, such that each time step is executed in finite\n precision arithmetic. Let $(y_n)$ be the mathematically correct\n solution of the one-step method. Then, the difference\n equation~\\eqref{eq:explicit:8} is fulfilled only up to machine\n accuracy $\\epsilon_m$:\n \\begin{align*}\n y_0-z_0 &\\approx \\epsilon_m \\\\\n \\abs{\\eta_k(y) - \\eta_k(z)} &= \\abs{\\eta_k(z)} \\approx \\epsilon_m \\abs{z_k}.\n \\end{align*}\n Then, the error between the true solution of the one-step method\n $(y_n)$ and the computed solution is bounded by\n \\begin{gather*}\n \\abs{y_n-z_n} \\le e^{L_h(t_n-t_0)} n \\epsilon_m \\max_k\\abs{z_k}.\n \\end{gather*}\n\\end{Corollary*}\n\n\\begin{Theorem*}{convergence-one-step}{Convergence of one-step methods}\n Let the one-step method $F(.,.)$ be consistent of order $p$ and\n discretely stable, that is, $F(.,.)$ is Lipschitz continuous in its\n second argument. Let $f(t,u)\\in C^{p}$. Furthermore, let\n $y_0 = u_0$. Let $h = \\max h_n$ and let there be a positive number\n $\\gamma$ such that $\\min h_n = \\gamma h$. Then, the method converges\n with order $p$ and there holds\n \\begin{equation}\n \\label{eq:convergence-onestep:1}\n \\abs{u_n - y_n} \\le c e^{L_h(t_n-t_0)} h^p,\n \\end{equation}\n where the constant $c$ is independent of $h$.\n\\end{Theorem*}\n\n\\begin{proof}\n Again we use the discrete stability theorem: with the definition of\n the order of the method, we obtain\n \\begin{gather}\n \\label{eq:convergence-onestep:2}\n \\abs{\\eta_k(u)-\\eta_k(y)} = \\abs{\\eta_k(u)} \\le c h^{p+1},\n \\end{gather}\n where $c$ depends on the derivatives of $u$ (and thus of $f$), but\n not on $u_k-y_k$. On the other hand, we have\n \\begin{gather*}\n n \\le \\frac{t_n-t_0}{\\min h_n} \\le \\frac{t_n-t_0}{\\gamma h}.\n \\end{gather*}\n Thus, we obtain by summing up~\\eqref{eq:convergence-onestep:2} over all $n$\n \\begin{gather*}\n \\abs{u_n-y_n} \\le e^{L_h(t_n-t_0)}\\sum_{k=1}^n h_k^{p+1}\n \\le c e^{L_h(t_n-t_0)} h^p.\n \\end{gather*}\n\\end{proof}\n\n\\begin{corollary}\n The Euler method converges of first order.\n\\end{corollary}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Runge-Kutta methods}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{intro}\n We are searching for methods which approximate the solution to an\n IVP numerically. In fact we are not solving the IVP, but the Volterra\n integral equation~\\eqref{eq:volterra}. Hence we can consider\n solving differential equations as a quadrature problem; with the\n difficulty that the function, which we integrate, is not known. This\n consideration leads to a class of methods for IVP, the\n Runge-Kutta methods.\n\\end{intro}\n\n\\input{definitions/erk}\n\n\\begin{remark}\n Pursuant to remark~\\ref{remark:expl:first-step} we present the\n formula for the calculation of $y_1$ from $y_0$ on the interval from\n $t_0=0$ to $t_1 = h$. The formula for a later time step $k$ is\n obtained by replacing $y_0$ and $t_0=0$ by $y_k$ and $t_k$,\n respectively to obtain $y_{k+1}$.\n\\end{remark}\n\n\\begin{remark}\n The intermediate values $\\rkg_i$ will not be saved separately in\n typical implementations, because it is possible to execute the\n method with the values $k_i$ alone. Nevertheless, the values\n $\\rkg_i$ are useful for highlighting the structure of the method.\n\\end{remark}\n\n\\input{definitions/butcher-tableau-erk}\n\n\\begin{remark}\n The first row of the tableau is to read in such a manner, that\n $g_1 = y_0$ and $k_1$ is computed directly by $f(t_0, y_0)$. The\n coefficients $a_{1j}$ and $c_0$ do not appear in formulas~\\eqref{eq:explicit:1a}\n and~\\eqref{eq:explicit:1b} (or are considered zero).\n \n The further rows indicate the rules for the computation of the\n further values $k_i$ in each case according to the\n formulas~\\eqref{eq:explicit:1a} and~\\eqref{eq:explicit:1b}. The\n the method is explicit since the computation of $k_i$ only involves\n coefficients with index less than $i$.\n \n The last row below the line is then the short form of\n formula~\\eqref{eq:explicit:1c} and lists quadrature weights.\n\n We see, that the coefficients $a_{ij}$ form the strict lower\n triangle of a square $s\\times s$-matrix $A$. Therefore, in order to\n simplify the summation bounds, we implicitly complete this matrix\n with values $a_{ij} = 0$ for $j\\ge i$.This way, the sum\n in~\\eqref{eq:explicit:1a} can be taken from $1$ to $s$, independent\n of $i$. We will also associate to an $s$-stage method the vector $b\n = (b_1,\\dots,b_s)^T$.\n\\end{remark}\n\n\\begin{example}\n The Euler method \\putindex{Euler method} has the \\putindex{Butcher\n tableau}:\n \\begin{gather*}\n \\begin{array}{c|c}\n 0 & \\\\\n \\hline\n & 1 \\\\\n \\end{array}\n \\end{gather*}\n That leads to the already known formula:\n \\begin{equation*}\n y_1 = y_0 + h f(t_0, y_0)\n \\end{equation*}\n The values $b_1=1$ and $c_1=0$ indicate that this is a quadrature\n rule with a single point at the left end of the interval. Such a\n rule is exact for constant polynomials and thus of order 1.\n\\end{example}\n\n\\input{definitions/rk2}\n\n\\begin{remark}\n The modified Euler method uses an approximation to the value of\n $f(h/2, u(h/2))$ in its quadrature, corresponding to the midpoint\n quadrature rule. The Heun method is constructed analogous to the\n trapezoidal rule. Both quadrature rules are of second order, and so\n are these one-step methods. Both methods were discussed by Runge in\n his article of 1895~\\cite{Runge95}.\n\\end{remark}\n\n\\begin{Lemma}{Heun-consistency}\n The Heun method and the modified Euler method are consistent of second\n order\\footnote{Here and in the following proofs of consistency\n order, we will always assume that all necessary derivatives of $f$\n exist and are bounded. We say ``$f$ is sufficiently smooth''.}.\n\\end{Lemma}\n\n\\begin{proof}\n The proof uses Taylor expansion of the continuous solution $u$ and\n the discrete solution $y$ around $t_0$ with respect to $h$. First,\n abbreviating $f_t = \\partial_t f(t_0,u_0)$ and\n $f_u = \\partial_u f(t_0,u_0)$ and so forth\\footnote{Note that $f_u$,\n $f_{u u}$ and so on are tensors of increasing rank.} and replacing\n $u'(t_0)=f(t_0,u_0) = f$:\n \\begin{multline}\n \\label{eq:explicit:4}\n u_1 = u(t_0+h) = u_0 + h f(t_0, u_0)\n + \\frac{h^2}2\\bigl(f_t + f_u f\\bigr)\n \\\\\n + \\frac{h^3}6\\bigl(f_{t t}+2f_{t u}f+f_{u u}f^2+f_u f_t+f_u^2f\\bigr)\n + \\dots.\n \\end{multline}\n For the discrete solution of the modified Euler step on the other\n hand, there holds\n \\begin{align*}\n y_1 &= u_0 + h f\\left(t_0+\\frac h2, u_0+\\frac h2 f(t_0,u_0)\\right)\n \\\\\n &= u_0 + h f(t_0, u_0) + \\frac{h^2}2 \\bigl(f_t + f_u f\\bigr)\n \\\\&\\hphantom{...}\n + \\frac{h^3}8\\bigl(f_{t t}+2f_{t u}f+f_{u u}f^2+f_u f_t+f_u^2f\\bigr)\n +\\dots.\n \\end{align*}\n Thus, $\\abs{u_1-y_1} = \\mathcal O(h^3)$ and the method is of second\n order. The proof for the Heun method is left as an exercise.\n\\end{proof}\n\n\\begin{Example}{rk3}\n \\defindex{Runge-Kutta method!three-stage}\n The three stage Runge-Kutta method is\n\n \\begin{gather*}\n \\begin{aligned}\n k_1 & = f(t_0, y_0) \\\\\n k_2 & = f(t_0 + \\frac12 h, y_0 + \\frac12 h k_1) \\\\\n k_3 & = f(t_0 + h, y_0 - h k_1 + 2 h k_2) \\\\\n y_{n+1} & = y_0 + h ( \\frac16 k_1 + \\frac46 k_2 + \\frac16 k_3 ) \\\\\n \\end{aligned}\n \\qquad\n \\begin{array}{c|ccc}\n 0 & \\\\\n \\frac12 & \\frac12 \\\\\n 1 & -1 & 2 \\\\\n \\hline\n & \\frac16 & \\frac46 & \\frac16 \\\\\n \\end{array}\n \\end{gather*}\n This method is obviously based on the Simpson rule.\n\\end{Example}\n\n\\begin{remark}\n Computations become tedious very fast, in part due to the sum of\n partial derivatives of $f(t,u)$. This can be simplified by\n considering Runge-Kutta methods for the autonomized ODE (see\n Definition~\\ref{Definition:autonomization})\n \\begin{gather*}\n \\begin{pmatrix} u' \\\\ t' \\end{pmatrix} = \n \\begin{pmatrix} f(t,u) \\\\ 1 \\end{pmatrix}.\n \\end{gather*}\n Then, the Runge-Kutta method~\\eqref{eq:explicit:1} simplifies to\n \\begin{gather}\n \\label{eq:explicit:7a}\n \\begin{split}\n \\rkg_i &= y_0 +\n \\sum_{j=1}^{i-1} \\rka_{ij} h f(\\rkg_j),\\quad i=1,\\dots,\\rks\n \\\\\n y_1 &= y_0 + \\sum_{j=1}^s b_j h f(g_j).\n \\end{split}\n \\end{gather}\n\\end{remark}\n\n\\begin{Lemma}{erk-autonomization}\n An ERK is invariant under autonomization (or in short,\n \\putindex{autonomizable}), if and only if\n \\begin{gather}\n \\label{eq:explicit:9}\n c_i = \\sum_{j=1}^{i-1} a_{ij}, \\quad i=1,\\dots,s.\n \\end{gather}\n\\end{Lemma}\n\n\\begin{proof}\n Observing the last component of the vector $u$ in the previous\n remark and the method applied to it yields the condition.\n\\end{proof}\n\n%HNW p. 145 and pp. 135/137\n\\begin{Lemma}{rk-order-3-4}\n An autonomizable ERK with $s$ stages is consistent of third order, if\n and only if the following conditions are met:\n \\begin{subequations}\n \\label{eq:explicit:11}\n \\begin{align}\n \\label{eq:explicit:12}\n b_1 + \\dots + b_s &= 1, \\\\\n \\label{eq:explicit:13}\n b_1c_1 + \\dots + b_s c_s &= 1/2, \\\\\n \\label{eq:explicit:14}\n b_1c_1^2 + \\dots + b_s c_s^2 &= 1/3, \\\\\n \\label{eq:explicit:15}\n \\sum\\nolimits_{i,j} b_i a_{ij}c_j &= 1/6.\n \\end{align}\n It is consistent of fourth order, if and only if additionally\n \\begin{align}\n \\label{eq:explicit:16}\n b_1c_1^3 + \\dots + b_s c_s^3 &= 1/4, \\\\\n \\label{eq:explicit:18}\n \\sum\\nolimits_{i,j} b_i a_{ij} c_j^2 &= 1/12, \\\\\n \\label{eq:explicit:19}\n \\sum\\nolimits_{i,j,k} b_i a_{ij} a_{jk} c_k &= 1/24,\\\\\n \\label{eq:explicit:17}\n \\sum\\nolimits_{i,j} b_i c_i a_{ij} c_j &= 1/8.\n \\end{align}\n \\end{subequations}\n\\end{Lemma}\n\n\\begin{remark}\n We can rephrase these conditions, such that an ERK is of order $k$ if\n the quadrature with\n support points $c_i$ and corresponding weights $b_i$ is exact for\n polynomials of degree $k-1$:\n \\begin{gather*}\n \\sum_{i=1}^{s} b_i p(c_i) = \\int_0^1 p(t)\\dt,\n \\qquad \\forall p\\in \\P_{k-1}.\n \\end{gather*}\n Furthermore, for $k \\ge 3$\n \\begin{gather*}\n \\sum_{ij} b_i a_{ij} p(c_j) = \\int_0^1\\int_0^t p(s)\\ds\\dt,\n \\qquad \\forall p\\in \\P_{k-2}.\n \\end{gather*}\n Additionally, for $k\\ge 4$\n \\begin{xalignat*}2\n \\sum_{ijk} b_i a_{ij} a_{jk} p(c_k) &= \\int_0^1\\int_0^t\\int_0^s p(r)\\dr \\ds \\dt,\n &&\\forall p\\in \\P_{k-3},\\\\\n \\sum_{ij} b_i p(c_i) a_{ij} q(c_j)\n &= \\int_0^1 p(t) \\int_0^t q(s) \\ds\\dt\n &&\\forall p\\in \\P_{k_1}, q\\in\\P_{k_2}, k_1+k_2 = k-2.\n \\end{xalignat*}\n\\end{remark}\n\n\\input{theorems/taylor-u}\n\\begin{proof}\n Taking derivatives of $u$ and replacing every occurrence of $u'$ by\n $f(u)$. For scalar valued functions, we clarify this at the example\n \\begin{align*}\n u'(t) &= f(u(t)) \\\\\n u''(t) &= \\bigl(u'(t)\\bigr)'\n = f(u(t))' = f'(u(t))u'(t)\n \\\\& = f'(u(t)) f(u(t))\n \\\\\n u^{(3)} &= \\bigl(u''(t)\\bigr)' = \\bigl(f'(u(t)) f(u(t))\\bigr)'\n \\\\&= f''(u(t)) u'(t) f(u(t)) + f'(u(t)) f'(u(t)) u'(t)\n \\\\& = f''(u(t)) f(u(t))^2 + f'(u(t))^2 f(u(t)).\n \\end{align*}\n After the concept is clear, we have to keep track of the vector\n indices and compute with brute force.\n It may be worth noting, that in the 4th order term, we used the fact\n that we can swap summation indices and get\n \\begin{gather*}\n \\sum_{\\lambda,\\mu,\\nu} \\partial_{\\lambda\\mu} f_n \\partial_\\nu f_\\lambda f_\\mu f_\\nu\n = \\sum_{\\lambda,\\mu,\\nu} \\partial_{\\lambda\\mu} f_n \\partial_\\nu f_\\mu f_\\lambda f_\\nu\n = \\sum_{\\lambda,\\mu,\\nu} \\partial_{\\lambda\\nu} f_n \\partial_\\mu f_\\lambda f_\\mu f_\\nu\n .\n \\end{gather*}\n\\end{proof}\n\n\\input{theorems/taylor-y}\n\n\\begin{proof}\n We begin with the observation that for an arbitrary function $\\phi$ holds\n \\begin{gather*}\n \\frac{d^{q}}{d h^{q}}\\bigl(h\\phi(h)\\bigr)\\bigg|_{h=0}\n = \\left[ h \\frac{d^{q}}{d h^{q}} \\phi(h) + q h'\n \\frac{d^{q-1}}{d h^{q-1}} \\phi + \\binom{q}{2}\n h''\\dots\\right]_{h=0}\n = q \\frac{d^{q-1}}{d h^{q-1}} \\phi.\n \\end{gather*}\n Next, we use~\\eqref{eq:explicit:1c} to obtain\n \\begin{align*}\n y(h) &= u_0, \\\\\n y^{(q)}(h)\\big|_{h=0} &= q \\sum_{j=1}^s b_j \\frac{d^{q-1}}{d h^{q-1}} f(g_j)\\bigg|_{h=0}.\n \\end{align*}\n We observe $g_i(0) = u_0$. Further, from~\\eqref{eq:explicit:1a}, we\n obtain\n \\begin{align*}\n g_i(h)\\big|_{h=0} &= u_0, \\\\\n g_{i;n}^{(q)}(h)\\big|_{h=0} &= q \\sum_{j=1}^{i-1} a_{ij}\n \\frac{d^{q-1}}{d h^{q-1}} f_n(g_{j})\\bigg|_{h=0}.\n \\end{align*}\n Here, $g_{i;n}$ refers to the component $k$ of vector\n $g_i$. Finally, we need\n \\begin{align*}\n \\frac{d}{d h} f_n(g_i(h))\\big|_{h=0}\n & = \\sum_{\\lambda} \\partial_{\\lambda} f_n g'_{i;\\lambda}\\\\\n \\frac{d^2}{d h^2} f_n(g_i(h))\\big|_{h=0}\n & = \\sum_{\\lambda,\\mu} \\partial_{\\lambda\\mu} f_n g'_{i;\\lambda} g'_{i;\\mu}\n + \\sum_k \\partial_{\\lambda} f_n g''_{i;\\lambda}.\n \\end{align*}\n Summarizing, we obtain\n \\begin{align*}\n y'_n &= \\sum_{j=1}^s b_j f_n(g_j)\n \\\\\n y''_n &= 2 \\sum_{j=1}^s b_j\\frac{d}{d h} f_n(g_j)\n = \\sum_{j=1}^s \\sum_{k=1}^{j-1} b_j a_{j k}\n \\sum_{\\lambda} \\partial_{\\lambda} f_n f_\\lambda\n \\\\\n y'''_n &= 3 \\sum_{j=1}^s b_j\\frac{d^2}{d h^2} f_n(g_j)\n \\\\&= 3 \\sum_{j=1}^s b_j \\left[\n \\sum_{\\lambda,\\mu} \\partial_{\\lambda\\mu} f_n g'_{j;\\lambda} g'_{j;\\mu}\n + \\sum_\\lambda \\partial_{\\lambda} f_n g''_{j;\\lambda}\\right]\n \\\\&= 3 \\sum_{j=1}^s b_j \\left[\n \\sum_{\\lambda,\\mu} \\partial_{\\lambda\\mu} f_n \\sum_{k=1}^{j-1} a_{j k}\n f_\\lambda \\sum_{k=1}^{j-1} a_{j k}\n f_\\mu + \\sum_{\\lambda} \\partial_{\\lambda} f_n 2 \\sum_{k=1}^{j-1}\n a_{j k} \\sum_{l=1}^{k-1} a_{kl} \\sum_\\mu \\partial_\\mu f_\\lambda f_\\mu\n \\right]\n \\end{align*}\n\\end{proof}\n\n\\begin{proof}[Proof of Lemma~\\ref{Lemma:rk-order-3-4}]\n The proof utilizes Taylor expansion of $u_1$ and $y_1$ provided in\n Lemmas~\\ref{Lemma:taylor-u} and~\\ref{Lemma:taylor-y},\n respectively. Once we have computed these expansions, we compare\n coefficients in front of equal derivatives in order to get the result.\n\\end{proof}\n\n\\begin{remark}\n Butcher introduced a graph theoretical method for order conditions\n based on trees. While this simplifies the process of deriving these\n conditions for higher order methods considerably, it is beyond the\n scope of this course.\n\\end{remark}\n\n\\input{definitions/rk4}\n\n\\begin{remark}[Order conditions and quadrature]\n The order conditions derived by excessive Taylor expansion have a\n very natural interpretation through the analysis of quadrature\n formulas for the Volterra integral equation, where $(h c_i)$ are the\n quadrature points and the other values are quadrature weights.\n First, we observe that\n \\begin{gather*}\n \\sum_i b_i f(g_i) \\quad\\text{approximates}\\quad\n \\frac1h\\int_0^1 f(u(h s)\\ds.\n \\end{gather*}\n In this view,\n conditions~\\eqref{eq:explicit:12}--\\eqref{eq:explicit:14}\n and~\\eqref{eq:explicit:16} state that the formula $\\sum_i b_i\n p(c_i)$ is an exact integral for polynomials of degree up to 3. In a\n previous semester, we have made use of this property to prove that\n the formula is of 4th order.\n\n Equally, we deduce from formula~\\eqref{eq:explicit:1a} for $g_i$ that\n \\begin{gather*}\n \\sum_j a_{ij} f(g_j) \\quad\\text{approximates}\\quad\n \\frac1h\\int_0^{c_i} f(u(h s)\\ds.\n \\end{gather*}\n The condition~\\eqref{eq:explicit:9} that the method be autonomizable\n states nothing but that this be exact for constant functions. For\n higher order, the accuracy of the value of $g_i$ only implicitly\n enters the accuracy of the Runge-Kutta method by integrating this\n value again. Thus, we actually look at approximations of integrals\n of the form\n \\begin{gather*}\n \\int_0^1 \\phi(s) \\int_0^s \\psi(r) \\dr\\ds.\n \\end{gather*}\n Condition~\\eqref{eq:explicit:15} for 3rd order states, that this\n condition must be true for linear polynomials $\\psi(r)$ and constant\n $\\phi(s)$, thus, after the interior integration again a polynomial\n of second order. Equally, conditions~\\eqref{eq:explicit:17}\n and~\\eqref{eq:explicit:18} state this for linear polynomials\n $\\psi(r)$ with linear $\\phi(s)$ and for quadratic polynomials\n $\\psi(r)$ with constant $\\phi(s)$, respectively. Finally,\n condition~\\eqref{eq:explicit:19} states that the quadrature has to\n be exact for any linear polynomial $\\phi(\\tau)$ in\n \\begin{gather*}\n \\int_0^1 \\int_0^s \\int_0^r \\phi(\\tau) \\,\\diffd \\tau\\dr\\ds.\n \\end{gather*}\n\\end{remark}\n\n\\begin{remark}[Butcher barriers]\n The maximal order of an explicit Runge-Kutta method is limited\n through the number of stages, or vice versa, a minimum number of\n stages is required for a certain order. The \\define{Butcher barriers} state\n that in order to achieve order $p$ one requires $\\rks$ stages, where\n $p$ and $\\rks$ relate as follows:\n \\begin{center}\n \\input{tables/butcher-barriers}\n \\end{center}\n \n These order bounds refer to systems of differential equations. For\n a simple equation they may be better. For instance, there exists a\n five-stage method which solves the one dimensional IVP with order 5.\n\n For $p = 10$ there is only known a method with $s = 17$ until now.\n It is possible that there exists a method that needs less stages,\n because currently no proof for a minimal number of stages is\n available.\n\\end{remark}\n\n\\begin{Lemma}{erk-Lipschitz}\n Let $f(t,u)$ admit the uniform \\putindex{Lipschitz condition}. Then,\n every autonomizable ERK which is consistent of order one admits a\n uniform Lipschitz condition.\n\\end{Lemma}\n\n\\begin{proof}\n We observe that the \\putindex{increment function} is\n \\begin{gather}\n \\label{eq:explicit:7}\n \\verfahren(0, y) = \\sum_{j=1}^s \\rkb_j f(h\\rkc_i, \\rkg_i(y)),\n \\end{gather}\n with $g_i(y)$ defined recursively by\n \\begin{gather*}\n g_i(y) = y + h \\sum_{j=1}^{i-1} \\rka_{ij} f(h\\rkc_j, \\rkg_j(y)).\n \\end{gather*}\n Let $L$ be the Lipschitz constant of $f$. Let\n $d_i = \\abs{g_i(x)-g_i(y)}/\\abs{x-y}$. We have\n \\begin{align*}\n d_1 &= 1\n \\\\\n d_2 &= \\abs{x-y + h \\rka_{21}\n \\Bigl(f\\bigl(h\\rkc_2,g_1(x)\\bigr)\n -f\\bigl(\\rkc_2,g_1(x)\\bigr)\n \\Bigr)}/\\abs{x-y}\n \\\\ &\\le (1+ha_{21}L)\n = (1+h\\rkc_{1}L)\n \\\\\n d_3 & \\le \\Bigl(1+\n h L\\bigl(\\rka_{31}+\\rka_{32}(1+ha_{21}L)\\bigr)\\Bigr)\n \\\\ &\\le \\bigl(1+h L c_3(1+h c_2L)\\bigr)\n \\\\\n d_4 & \\le \\Bigl(1+\n h L c_4\\bigl(1+h L c_3(1+h L c_2)\\bigr)\\Bigr)\n \\\\\n d_s & \\le\n \\Bigl(1+ h L c_s\\bigl(\n 1+\\dots(1+h L c_1)\\dots\\bigl)\\Bigl).\n \\end{align*}\n Since $c_i \\le 1$, the factor is bounded by\n $d_s\\le(1+h L)^{s-1}$. Moreover, if $h L \\le 1$, we realize that\n \\begin{gather*}\n d_s = \\Bigl(1+ h L \\bigl(1+\\dots(1+h L)\\dots\\bigl)\\Bigl) \\le s.\n \\end{gather*}\n Finally, we enter this result into~\\eqref{eq:explicit:7} to obtain\n \\begin{align*}\n \\abs{\\verfahren(0,x)-\\verfahren(0,y)}\n & \\le \\sum_{j=1}^s b_j L d_j \\abs{x-y}\n \\\\\n & \\le d_s L \\abs{x-y}.\n \\end{align*}\n Thus, the increment function $\\verfahren$ admits a Lipschitz\n condition with constant $L_h = L(1+h L)^{s-1}$ for general step size\n $h$ and $L_h = s L$ for $h\\le 1/L$.\n\\end{proof}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Estimates of the local error\n and time step control}\n \\label{section:step_size_control}\n\n\\begin{intro}\n In the preceding paragraphs, we have used a crude a priori estimate\n of the local error based on high order derivatives of the right hand\n side $f(t,u)$. In the case of a complex nonlinear system, such an\n estimate is bound to be inefficient, since it involves global bounds\n on the derivatives. Obviously, the local error cannot be computed\n exactly either, because that would require or imply the knowledge of\n the exact solution.\n\n In this section, we discuss two methods which allow an estimate of\n the truncation error from computed solutions. These estimates are\n local in nature and therefore usually much sharper. Thus, they can\n be used to control the step size, which in turn gives good control\n over the balance of accuracy and effort. Nevertheless, it should be\n pointed out that in these estimates there is an implicit assumption\n that the true solution $u$ is sufficiently regular and the step size\n is sufficiently small, such that the local error already follows\n the theoretically predicted order.\n\n Given an estimate for the local error, we can devise an algorithm\n step size control, which controls the local error and thus in a\n certain way the global error.\n\\end{intro}\n\n\\begin{algorithm}[Adaptive step size control]\n Let there be an estimate for the local error based\n on $\\abs{y_1 - \\hat y_1}$ . Then, the following algorithm can be used\n to guarantee that the local error of a one-step method remains below\n a threshold $\\epsilon$ in every time step:\n\n \\begin{enumerate}\n \\item Given $y_{k-1}$, compute $y_k$ and $\\hat y_k$ with time step $h_k$.\n \\item Compute\n \\begin{gather}\n \\label{eq:explicit:30}\n h_{\\text{opt}} = h_k \\left(\\frac{\\epsilon}{y_k - \\hat\n y_k}\\right)^{\\frac1{p+1}}.\n \\end{gather}\n \\item If $h_{\\text{opt}} < h_k$ the time step is rejected: let\n $h_k = h_{\\text{opt}}$ and recompute $y_k$ and $\\hat y_k$.\n \\item If the time step was accepted, let $h_{k+1} = h_{\\text{opt}}$.\n \\begin{enumerate}\n \\item If $t_k+h_{k+1} > t_n$, let $h_{k+1} = t_n-t_k$.\n \\end{enumerate}\n Increase $k$ by one and proceed with the first step.\n \\end{enumerate}\n\\end{algorithm}\n\n\\begin{remark}\n It might happen, that the value $t_k$ is just below $t_n$ with a\n difference close to machine accuracy. As a result, the next time step\n with $h_{k+1} \\approx \\epsilon_m$ would suffer from round-off\n errors. Therefore, it is advisable to avoid this situation by\n expanding the last time step, if $t_n-t_k \\le c h_{k+1}$ where $c$\n is a moderate constant of size around $1.1$.\n\\end{remark}\n\n\\begin{remark}\n This algorithm controls and equilibrates the local\n error. Nevertheless, the global estimate still retains the\n exponential term. The error estimation techniques in this section\n are thus not optimal controlling the global error, which involves\n considerably more effort and will be discussed in a later course.\n\\end{remark}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Extrapolation methods}\n\n\\begin{intro}\n Here, we estimate the \\putindex{local error} by a method called\n \\putindex{Richardson extrapolation}. It is based on computing two\n approximations with the same method, but different step size, say\n an approximation $y_2$ with two steps of size $h$ and an\n approximation $\\hat y_2$ with one step of size $2h$.\n\\end{intro}\n\n\\input{theorems/richardson}\n\n\\begin{proof}\n For the proof we need a refined version of the local error estimates\n as well as the global error estimate in\n Theorem~\\eqref{Theorem:convergence-one-step} which can be obtained by\n adding one more step of Taylor expansion. Then, we get for the local\n error of a method of order $p$ estimates of the form\n \\begin{gather}\n \\label{eq:explicit:21}\n e_1 = u_1 - y_1 = C h^{p+1} + \\mathcal O(h^{p+2}),\n \\end{gather}\n with a constant (vector) $C$ with not necessarily positive\n entries. In the same way, we refine the estimate for error\n propagation from basic Lipschitz continuity to\n \\begin{gather}\n \\label{eq:explicit:22}\n e_{2;\\text{acc}} = \\left( \\identity + h \\frac{\\partial f}{\\partial y} +\n \\mathcal O(h^2)\\right) \\bigl(u_1-y_1).\n \\end{gather}\n The local error on the second interval is of the same structure\n as~\\eqref{eq:explicit:21}, but on the interval starting at $t_1$\n with initial value $y_1 = y_0 + \\mathcal O(h)$.\n\n Thus, we obtain for the error after two steps of size $h$:\n \\begin{multline}\n \\label{eq:explicit:25}\n u_2-y_2\n = \\underbrace{\\bigl(\\identity + \\mathcal O(h)\\bigr) C\n h^{p+1}}_{\\text{local 2}}\n + \\bigl(C+\\mathcal O(h)\\bigr) h^{p+1} + \\mathcal\n O(h^{p+2})\n \\\\ = 2C h^{p+2} + \\mathcal O(h^{p+2}).\n \\end{multline}\n We compare this to a single step for $\\hat y$ with\n \\begin{gather}\n \\label{eq:explicit:23}\n u_2 - \\hat y_2 = C (2h)^{p+1} + \\mathcal O(h^{p+2}).\n \\end{gather}\n Subtracting equations~\\eqref{eq:explicit:25}\n and~\\eqref{eq:explicit:23}, we obtain\n \\begin{gather*}\n y_2 - \\hat y_2 = \\bigl(2-2^{p+1}\\bigr)C h^{p+1} + \\mathcal O(h^{p+2}),\n \\end{gather*}\n such that\n \\begin{gather}\n \\label{eq:explicit:26}\n C h^{p+1} = \\frac{y_2 - \\hat y_2}{2^{p+1}-2} + \\mathcal O(h^{p+2}).\n \\end{gather}\n We enter this result into~\\eqref{eq:explicit:25} to conclude\n \\begin{gather}\n \\label{eq:explicit:27}\n u_2 - y_2 = \\frac{y_2 - \\hat y_2}{2^{p}-1} + \\mathcal O(h^{p+2}).\n \\end{gather}\n Adding $y_2$ on both sides, we see that\n \\begin{gather}\n \\label{eq:explicit:28}\n \\tilde y_2 = y_2 + \\frac{y_2 - \\hat y_2}{2^{p}-1}\n \\end{gather}\n approximates $u_2$ of order $\\mathcal O(h^{p+2})$ and thus one order\n better than $y_2$.\n\\end{proof}\n\n\\begin{remark}\n Formula~\\eqref{eq:explicit:10} can be evaluated after computation of\n $y_2$ and $\\hat y_2$ in order to obtain an estimate for the local\n error of $y_2$. This estimate can be used to control the step size\n control according to the algorithm above. We do not have an\n estimate for the error of $\\tilde y$. Nevertheless, we expect its\n values to be more accurate, such that we should use $\\tilde y$ as\n approximation and initial value for the next time step.\n\\end{remark}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection{Embedded Runge-Kutta methods}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nInstead of estimating the \\putindex{local error} by doubling the step size,\nembedded Runge-Kutta methods use two methods of different order to\nachieve the same effect. The key to efficiency is here, that the\ncomputed stages $g_i$ are the same for both methods, and only the\nquadrature weights $b_i$ differ.\n\n\\begin{definition}[Embedded Runge-Kutta methods]\n \\defindex{Runge-Kutta method!embedded} An embedded $\\rks$-stage\n Runge-Kutta method with orders of consistence $p$ and $\\hat p$\n computes two solutions $y$ and $\\hat y$ with the same function\n evaluations. For this purpose we will first compute contributions\n $g_i$ and $k_i$ for $i=1,\\dots,\\rks$ as in the normal Runge-Kutta\n method of stage $\\rks$. The function values at the end of the time\n step result as follows\n \\begin{gather}\n \\begin{split}\n y_1 &= y_0 + h \\sum \\rkb_i k_i \\\\\n \\hat y_1 &= y_0 + h \\sum \\hat \\rkb_i k_i.\n \\end{split}\n \\end{gather}\n The methods for $y$ and $\\hat y$ are consistent of order $p$ and\n $\\hat p$, respectively. We let $\\hat p < p$, for example\n $\\hat p = p-1$. \n \n This can be achieved by e.g. using the same mathod twice and omitting\n $\\hat{b}_j$ for one $j \\in \\{1, \\dots, s\\}$.\n\\end{definition}\n\n\\input{definitions/embedded-butcher}\n\n\\begin{remark}\n For higher order methods or functions $f(t,u)$ with complicated\n evaluation, most of the work lies in computation of the\n stages. Thus, the additional quadrature for the computation of $\\hat\n y$ is almost for free. Nevertheless, due to the different orders of\n approximation, $y$ is much more accurate and we obtain\n \\begin{gather}\n \\label{eq:explicit:29}\n u_1 - \\hat y_1 = y_1 - \\hat y_1 + \\mathcal O(h^p).\n \\end{gather}\n Thus, $y_1 - \\hat y_1$ is a good estimate for the \\putindex{local\n error} of $\\hat y_1$. This is the error which is used in step size\n control below. Similar to Richardson extrapolation above, we use the\n more accurate value $y_1$ for further computation, even if we do not\n have a computable estimate for its local error.\n\\end{remark}\n\n\\input{definitions/dormand-prince-45.tex}\n\n\\begin{remark}\n The Dormand-Prince method of orders 4 and 5 has become a standard\n tool for the integration of IVP. It is the backbone of\n \\texttt{ode45} in Matlab.\n\\end{remark}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Continuous Runge-Kutta methods}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\begin{intro}\n The Runge-Kutta methods discussed so far compute highly accurate\n approximations to the solution $u(t)$ in the discrete points\n $(t_k)_{k=1,\\dots,n}$. Such approximations are useful, if only the\n value at the interval end $t_n=T$ is needed, or if the time steps\n are sufficiently small in order to generate a plot of the solution\n history. There are some problems though, where the values in\n discrete points are not sufficient:\n \\begin{enumerate}\n \\item The step size control managed to use very large steps from\n which for instance a plot cannot read easily. Thus, we require the\n solution in the continuum between two time steps.\n \\item Accurate approximations inside an interval are required. This\n may be due to the fact that we want to measure the length of a\n period of a periodic solution, or that the equation contains\n switches which change discretely when the solution attains certain\n values.\n \\end{enumerate}\n In all of these cases, we need an interpolation formula for these\n intermediate values. Unfortunately, as the example of the classical\n Runge-Kutta method shows, the values $g_i$ have questionable value\n in this business. Therefore, in order to be better than linear\n interpolation between $y_{k-1}$ and $y_k$, we have to consider\n formulas, which provide the information for accurate interpolation\n with low additional cost.\n\\end{intro}\n\n\\begin{definition}\n \\defindex{Runge-Kutta method!continuous}\n A continuous Runge-Kutta method is a method of the same type as in \n definition~\\ref{Definition:erk}, for which the coefficients $\\rkb_i$ are replaced \n\tby continuous functions $\\rkb_i(\\theta)$ on the interval $[0,1]$.\n For this reason the equation~\\eqref{eq:explicit:1c} is augmented by\n \\begin{equation}\n \\label{eq:explicit:24}\n y(t_0+ \\theta h) = y_0 + \\sum\\limits_{i=1}^{s^*} \\rkb_i (\\theta) k_i.\n \\end{equation}\n Here the stage number $\\rks^*$ may be higher than $\\rks$. \n Then additional intermediate values $k_i$ have to be generated.\n\\end{definition}\n\n\\begin{remark}\n The local error $u(t_0+\\theta h) - y(t_0+\\theta h)$ is of order\n $p^*$ if the derivative $\\partial_\\theta^k y$ approximates\n $\\partial_\\theta^k u$ with an error of order $h^{p^*-k+1}$. Note that\n the first derivatives involve the derivatives of $g_i$ with respect\n to $h$.\n \n For a later time, we observe that the error of the initial value of\n an interval is only $\\mathcal O(h^p)$. Thus, an optimal continuous\n formula balancing the global error of the original method with the\n local error of the continuous method should be of order\n $p^*=p-1$.\n\\end{remark}\n\n\\begin{remark}\n If $\\rks^* > \\rks$ choose $k_{\\rks+1} = k_1 = f(t_n,y_1)$ of the next time step.\n\\end{remark}\n\n\\input{definitions/rk4-continuous}\n\\input{definitions/dormand-prince-45-continuous}\n\n\\begin{remark}\n Collocation methods will provide a natural way to obtain continuous\n method in the next chapter.\n\\end{remark}\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: \"notes\"\n%%% End: \n", "meta": {"hexsha": "5db4d1711d42ac8fab5dbf4a0fa5c69625fc95f1", "size": 39889, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ode/explicit.tex", "max_stars_repo_name": "ahumanita/notes", "max_stars_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ode/explicit.tex", "max_issues_repo_name": "ahumanita/notes", "max_issues_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-05-24T07:31:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-08-31T12:58:14.000Z", "max_forks_repo_path": "ode/explicit.tex", "max_forks_repo_name": "ahumanita/notes", "max_forks_repo_head_hexsha": "73f23770e2b02f1b4a67987744ceffbd9ce797d7", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2018-05-15T19:28:53.000Z", "max_forks_repo_forks_event_max_datetime": "2020-11-05T19:07:29.000Z", "avg_line_length": 39.9289289289, "max_line_length": 159, "alphanum_fraction": 0.6530371782, "num_tokens": 12914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8056321796478255, "lm_q1q2_score": 0.612804643266945}} {"text": "\\documentclass[a4paper]{scrartcl}\n\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n\\usepackage{pgfplots}\n\\usepackage{amsmath, enumerate, amssymb, multirow, fancyhdr, color, graphicx, lastpage, listings, tikz, pdflscape, subfigure, float, polynom, hyperref, tabularx, forloop, geometry, listings, fancybox, tikz, forest, tabstackengine, cancel, bbm}\n\\input kvmacros\n\\geometry{a4paper,left=3cm, right=3cm, top=3cm, bottom=3cm}\n\\pagestyle {fancy}\n\n\n\\begin{document}\n\n\nWe assume $\\sigma$ is the weight matrix of a weighted undirected graph, so $\\sigma_{ij} = \\sigma_{ji}$ and $\\sigma_{ij} = 1$ if there is an edge between $i$ to $j$ and $\\sigma_{ij} = 0$ if there is no edge between $i$ and $j$.\nFor simplicity we assume that nodes are not self connected, that is $\\sigma_{ii} = 0$.\n\n\\section*{(a)}\n\n{\n\\color{green}\nTODO: Note the $\\leq$ instead of the $=$ in the following. This propagates through all formulas below!\n}\n\nFor $M$ to be a matching, it needs to hold by definition, that $$\\forall i: \\sum_j \\sigma_{ij} \\leq 1$$ \nThis is equivalent to saying $$ \\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]} = 1 $$\nwhere $\\mathbbm{1}$ is the indicator function.\n\nSo we find the probability distribution\n\n\\begin{align}\n P(M) = \\frac{1}{Z}\\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]}\n\\end{align}\n\nWhere $Z$ is the normalization wich is the count of all matchings \n\\begin{align}\n Z = \\sum_{M \\in \\mathcal{G}} \\mathbbm{1}_{\\left[ \\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]} = 1 \\right]} \n\\end{align}\nHere $\\mathcal{G}$ is the set of all subsets of the graph.\n\n\\section*{(b)}\n\nThe size of a matching $\\lvert M \\rvert$ is the count of its edges. With the adjecency matrix $\\sigma$ that is the sum of all its elements divided by two: $$ \\lvert M \\rvert = \\frac{\\sum_{i} \\sum_{j} \\sigma_{ij}}{2}$$\n\nWe simply introduce the size of a matching $\\lvert M \\rvert$ as a scaling factor in the probability distribution, to give larger weight to larger matchings:\n\n\\begin{align}\n P(M) = \\frac{1}{Z} \\lvert M \\rvert \\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]}\n\\end{align}\n\nwhich has new normalization factor\n\n\\begin{align}\n Z = \\sum_{M \\in \\mathcal{G}} \\lvert M \\rvert \\cdot \\mathbbm{1}_{\\left[ \\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]} = 1 \\right]} \n\\end{align}\n\n\n\n% We have the size $S = M=\\sum_{ij} \\sigma^{ij}$ and the normalized version $S^*=\\frac{M}{\\lvert E \\rvert}$.\n\n{\\color{red}\nTODO I feel like there would be a way to integrate the scalar $\\lvert M \\rvert$ into the factor (in the product) but i can't find it. (Nevermind i don't think so anymore, with clarification 2 we don't need it anyways)\n}\n\n\\section*{(c)}\n\nThe Graph is given as:\n\n\\begin{tikzpicture}\n [scale=1,auto=left,every node/.style={circle,draw}]\n \\node (n1) at (1,3) {1};\n \\node (n2) at (3,3) {2};\n \\node (n3) at (4,2) {3};\n \\node (n4) at (3,1) {4};\n \\node (n5) at (1,1) {5};\n \\node (n6) at (0,2) {6};\n \n \\foreach \\from/\\to in {n1/n2,n1/n3,n2/n3,n1/n4,n2/n5,n3/n6}\n \\draw (\\from) -- (\\to);\n \n\\end{tikzpicture}\n\nAn ordered drawing of this graph is:\n\n\n\\begin{tikzpicture}\n [scale=1,auto=left,every node/.style={circle,draw}]\n \\node (n1) at (1,4) {1};\n \\node (n2) at (3,3) {2};\n \\node (n3) at (5,4) {3};\n \\node (n4) at (1,1) {4};\n \\node (n5) at (3,1) {5};\n \\node (n6) at (5,1) {6};\n \n \\foreach \\from/\\to in {n1/n2,n1/n3,n2/n3,n1/n4,n2/n5,n3/n6}\n \\draw (\\from) -- (\\to);\n \n\\end{tikzpicture}\n \n\n\n\n\n\nIn general the probability distribution found above is given as\n\\begin{align}\n P(M) = \\frac{1}{Z}\\prod_i \\mathbbm{1}_{\\left[\\sum_j \\sigma_{ij} \\leq 1\\right]}\n\\end{align}\nWhich we can write as \n\\begin{align}\n P(M) = \\frac{1}{Z} \\prod_i \\psi_i(\\partial M)\n\\end{align}\nwith factors\n\\begin{align}\n \\psi_i(\\partial M) = \\mathbbm{1}_{\\left[ \\sum_j \\sigma_{ij} \\leq 1 \\right]}\n\\end{align}\n\nBy $\\partial M$ we denote, that each factor $\\psi_i$ is only connected to all nodes $\\sigma_{ij}$ for which there is a connection from $i$ to $j$, so for all nodes for which $\\sigma_{ij} = 1$ in the initial graph $G$.\n\n\nIn general to draw a factor graph we would start with a graph with all $\\sigma_{ij}$ connected to all factors $\\psi_i$. Then we can omit all $\\sigma_{ij}$ where $\\sigma_{ij} = 0$ in the initial graph $G$. Furthermore we can omit all connections from factors $\\psi_i$ to $\\sigma_{ij}$ if $i$ and $j$ are not connected. \n\nFor the graph given here this yields the factor graph:\n\n\\begin{tikzpicture}\n [scale=1,auto=left,every node/.style={circle,draw}]\n \\node (s12) at (4,6) {$\\sigma_{12}$};\n \\node (s13) at (4,5) {$\\sigma_{13}$};\n \\node (s23) at (4,4) {$\\sigma_{23}$};\n \\node (s14) at (4,3) {$\\sigma_{14}$};\n \\node (s25) at (4,2) {$\\sigma_{25}$};\n \\node (s36) at (4,1) {$\\sigma_{36}$};\n \n \\node[rectangle] (f1) at (1,6) {$\\psi_{1}$};\n \\node[rectangle] (f2) at (1,5) {$\\psi_{2}$};\n \\node[rectangle] (f3) at (1,4) {$\\psi_{3}$};\n \\node[rectangle] (f4) at (1,3) {$\\psi_{4}$};\n \\node[rectangle] (f5) at (1,2) {$\\psi_{5}$};\n \\node[rectangle] (f6) at (1,1) {$\\psi_{6}$};\n \n\n\n \\foreach \\from/\\to in {f1/s12,f1/s13,f1/s14,f2/s12,f2/s23,f2/s25,f3/s13,f3/s23,f3/s36,f4/s14,f5/s25,f6/s36}\n \\draw (\\from) -- (\\to);\n \n\\end{tikzpicture}\n\n\n\n{\\color{red}\nTODO Is this also so simple for the probability distribution scaled by matching size $\\lvert M \\rvert$? (Nevermind, is saw Clarification 2)\n}\n\n\n\n\n\n\n\n\n\n\n\n\\section*{(d)}\n\nThe general definitions are given as:\n\\begin{align}\n \\nu_{i \\rightarrow a}^{(t + 1)}(x_i) &\\cong \\prod_{b \\in \\partial i \\backslash a} \\hat{\\nu}_{b \\rightarrow i}^{(t)}(x_i) \\\\\n \\hat{\\nu}_{a \\rightarrow i}^{(t)}(x_i) &\\cong \\sum_{\\textbf{x}_{\\partial a} \\backslash i} \\psi_a (\\textbf{x}_{\\partial a}) \\prod_{j \\in \\partial a \\backslash i} \\nu_{j \\rightarrow a}^{(t)}(x_j) \\\\\n P(x_i) &= \\nu_i^{(t)}(x_i) \\cong \\prod_{a \\in \\partial i} \\hat{\\nu}_{a \\rightarrow i}^{(t-1)}(x_i)\n\\end{align}\n\n\nIn our case we have messages from variable to factor:\n\\begin{align}\n \\nu_{\\sigma_{ij}}^{(ij) \\rightarrow i} &\\cong \\prod_{b \\in \\partial (ij) \\backslash i} \\hat{\\nu}_{\\sigma_{ij}}^{b \\rightarrow (ij)} \\\\\n &= \\hat{\\nu}_{\\sigma_{ij}}^{j \\rightarrow (ij)}\n\\end{align}\nThis simplification comes, since variable $(ij)$ only has one other neighbouring factor except $i$, which is $j$. \\\\\n\nAnd messages from factor to variable:\n\\begin{align}\n \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} \\cong \\sum_{\\boldsymbol{\\sigma}_{\\partial i} \\backslash (ij)} \n \\psi_i(\\boldsymbol{\\sigma}_{\\partial i}) \\prod_{(ik) \\in \\partial i \\backslash (ij)} \\nu_{\\sigma_{ik}}^{(ik) \\rightarrow i}\n\\end{align}\n\nFor our problem this leads to messages from factor to variable:\n\\begin{align}\n \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} &\\cong \\sum_{\\boldsymbol{\\sigma}_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{il} \\leq 1 \\right]} \\prod_{(ik) \\in \\partial i \\backslash (ij)} \\nu_{\\sigma_{ik}}^{(ik) \\rightarrow i}\\\\\n\\end{align}\n\n\n{\n\\color{green}\nNote here that the sum is over all the values that the $\\sigma$ which are neighbours of $i$ could assume.\n}\n\n\nThe marginal is:\n\\begin{align}\n P(\\sigma_{ij}) &\\cong \\prod_{b \\in \\partial (ij)} \\hat{\\nu}_{\\sigma_{ij}}^{b \\rightarrow (ij)} \\\\\n &= \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} \\cdot \\hat{\\nu}_{\\sigma_{ij}}^{j \\rightarrow (ij)}\n\\end{align}\n\n\n\n{\n\\color{green}\nTODO: I don't know if we have to write something for the second hint, but this would be the following?\n}\n\nIn the following we find some derivations for the adjecent edges if $\\sigma_{ij} = 1$ and $\\sigma_{ij} = 0$:\n\nFor $\\sigma_{ij} = 1$ this leads to \n\\begin{align}\n P(\\sigma_{ij} = 1) &\\cong \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} \\cdot \\hat{\\nu}_{\\sigma_{ij}}^{j \\rightarrow (ij)} \\\\\n &= \\left[ \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{il} \\leq 1 \\right]} \\prod_{(ik) \\in \\partial i \\backslash (ij)} \\nu_{\\sigma_{ik}}^{(ik) \\rightarrow i} \\right]\n \\cdot \\left[ \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{jl} \\leq 1 \\right]} \\prod_{(jk) \\in \\partial j \\backslash (ij)} \\nu_{\\sigma_{jk}}^{(jk) \\rightarrow j} \\right]\n\\end{align}\nWhich can only be $\\geq 0$ if \n\\begin{align}\n \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{il} \\leq 1 \\right]} &= 1 \\\\\n \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_{l \\backslash (l = j)} \\sigma_{il} + 1 \\leq 1 \\right]} &= 1 \\\\\n\\end{align}\nand \n\\begin{align}\n \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{jl} \\leq 1 \\right]} &= 1 \\\\\n \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_{l \\backslash (l = i)} \\sigma_{jl} + 1 \\leq 1 \\right]} &= 1\n\\end{align}\n\nThis is saying that all edges adjecent to $i$ and $j$ except edge $(ij)$ have to be zero, meaning non existant. \n\n\nFor $\\sigma_{ij} = 0$ we find with similar reasoning:\n\n\\begin{align}\n P(\\sigma_{ij} = 0) &\\cong \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} \\cdot \\hat{\\nu}_{\\sigma_{ij}}^{j \\rightarrow (ij)} \\\\\n &= \\left[ \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{il} \\leq 1 \\right]} \\prod_{(ik) \\in \\partial i \\backslash (ij)} \\nu_{\\sigma_{ik}}^{(ik) \\rightarrow i} \\right]\n \\cdot \\left[ \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{jl} \\leq 1 \\right]} \\prod_{(jk) \\in \\partial j \\backslash (ij)} \\nu_{\\sigma_{jk}}^{(jk) \\rightarrow j} \\right]\n\\end{align}\nWhich can only be $\\geq 0$ if \n\\begin{align}\n \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{il} \\leq 1 \\right]} &= 1 \\\\\n \\sum_{\\sigma_{\\partial i} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_{l \\backslash (l = j)} \\sigma_{il} + 0 \\leq 1 \\right]} &= 1 \\\\\n\\end{align}\nand \n\\begin{align}\n \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_l \\sigma_{jl} \\leq 1 \\right]} &= 1 \\\\\n \\sum_{\\sigma_{\\partial j} \\backslash (ij)} \\mathbbm{1}_{\\left[\\sum_{l \\backslash (l = i)} \\sigma_{jl} + 0 \\leq 1 \\right]} &= 1\n\\end{align}\n\nWhich is saying that any one, but no more than one of the edges adjecent to $i$ except $(ij)$ may be 1 and that any one, but no more than one of the edges adjecent to $j$ except $(ij)$ may be 1. \nIf we use $c_i$ and $c_j$ as the count of edges adjecent to $i$ and $j$ except for edge $(ij)$ respectively, we can say that this allows $c_i+1$ configurations for the edges adjecent to $i$ and $c_j + 1$ configurations for the edges adjecent to $j$ (Any one may be 1 or all are 0).\n\nThese derivations are intuitively consistent with the definitions of the matching.\n\n\n\n\n\n{\n\\color{green}\nTODO i don't quite get the following, from the reasoning it makes sense but i don't see how it follows from the formulas\n\n\nFor $\\sigma_{ij}=1$, this leads to:\n\\begin{align}\n P(\\sigma_{ij}=1) = \\prod_{\\sigma_{ik} \\in \\partial a \\backslash i} v_{\\sigma_{ik} \\rightarrow i}(0)\n\\end{align}\nwich intuitively makes sense, since all the others have to be 0 in order for $\\sigma_{ij}$ to be 1.\\\\\nFor $\\sigma_{ij}=0$, this leads to:\n\\begin{align}\n P(\\sigma_{ij}=1) = \\sum_{\\sigma_{ik} \\in \\partial i \\backslash \\sigma_{ij}} (\\mathbbm{1}^{(\\sum_{l \\backslash i} \\sigma_{il} = 1)} \\prod_{\\sigma_{ik} \\in \\partial a \\backslash i} v_{\\sigma_{ik} \\rightarrow i}(\\sigma_{ik}))\\\\\n\\end{align}\nhere the main difference is that the factor $\\Psi_{i}$ doesn't sum over $\\sigma_{ij}$ anymore, since it is set to 0.\n}\n\n\n\n\n\n{\n\\color{red}\nWe have the messages from variable to factor:\n\\begin{align}\n \\hat{v}_{i \\rightarrow \\alpha}(x_i) = \\prod_{\\beta \\in \\partial \\alpha \\backslash i} \\hat{v}_{b \\rightarrow i}(x_i)\n\\end{align}\nand factor to variable:\n\\begin{align}\n \\hat{v}_{\\alpha \\rightarrow i}(x_i) = \\sum_{x \\in \\partial \\alpha \\backslash i} (\\Psi_a(X_{\\partial a}) \\prod_{j \\in \\partial a \\backslash i} v_{j \\rightarrow a}(x_j))\n\\end{align}\n\nThe general BP marginal is:\n\\begin{align}\n P(x_i) = v_i(x_i) = \\prod_{a \\in \\partial i} \\hat{v}_{\\alpha \\rightarrow i}(x_i)\n\\end{align}\n\nFor our problem, this leads to to message from factor to variable:\n\\begin{align}\n \\hat{v}_{i \\rightarrow \\sigma_{ij}}(\\sigma_{ij}) &= \\sum_{\\sigma_{ik} \\in \\partial i \\backslash \\sigma{ij}} (\\mathbbm{1}^{(\\sum_l \\sigma^{il} = 1)} \\prod_{\\sigma_{ik} \\in \\partial a \\backslash i} v_{\\sigma_{ik} \\rightarrow i}(\\sigma_{ik}))\\\\\n\\end{align}\n\nThis means the marginal is:\n\\begin{align}\n P(\\sigma_{ij}) = \\hat{v}_{i \\rightarrow \\sigma_{ij}}(\\sigma_{ij}) \\cdot \\hat{v}_{j \\rightarrow \\sigma_{ij}}(\\sigma_{ij})\n\\end{align}\nFor $\\sigma_{ij}=1$, this leads to:\n\\begin{align}\n P(\\sigma_{ij}=1) = \\prod_{\\sigma_{ik} \\in \\partial a \\backslash i} v_{\\sigma_{ik} \\rightarrow i}(0)\n\\end{align}\nwich intuitively makes sense, since all the others have to be 0 in order for $\\sigma_{ij}$ to be 1.\\\\\nFor $\\sigma_{ij}=0$, this leads to:\n\\begin{align}\n P(\\sigma_{ij}=1) = \\sum_{\\sigma_{ik} \\in \\partial i \\backslash \\sigma_{ij}} (\\mathbbm{1}^{(\\sum_{l \\backslash i} \\sigma_{il} = 1)} \\prod_{\\sigma_{ik} \\in \\partial a \\backslash i} v_{\\sigma_{ik} \\rightarrow i}(\\sigma_{ik}))\\\\\n\\end{align}\nhere the main difference is that the factor $\\Psi_{i}$ doesn't sum over $\\sigma_{ij}$ anymore, since it is set to 0.\n\n\n}\n\n\n\n\n\n\n\\section*{(d)}\n{\n\\color{green}\nTODO: To get the full marginal we would need to do the full message passing. For the equations the following suffices:\n\n}\n\nFor one point marginal, see (c):\n\\begin{align}\n P(\\sigma_{ij}) &\\cong \\hat{\\nu}_{\\sigma_{ij}}^{i \\rightarrow (ij)} \\cdot \\hat{\\nu}_{\\sigma_{ij}}^{j \\rightarrow (ij)}\n\\end{align}\n\n\nFor two point marginal, we have:\n\\begin{align}\n P(\\sigma_{ij}, \\sigma_{kl}) = P(\\sigma_{kl} \\vert \\sigma_{ij}) P(\\sigma_{ij})\n\\end{align}\n\nIn addition to the one point marginal we need to find $P(\\sigma_{kl} \\vert \\sigma_{ij})$ which is defined by:\n\\begin{align}\n P(\\sigma_{kl} \\vert \\sigma_{ij} = x) &\\cong \\prod_{a} \\psi_a(\\boldsymbol{\\sigma}_{\\partial a}) \\mathbbm{1}_{\\left[ \\sigma_{ij} = x \\right]} \\\\\n &= \\prod_{a} \\mathbbm{1}_{\\left[ \\sum_{l} \\sigma_{al} \\leq 1 \\right]} \\mathbbm{1}_{\\left[ \\sigma_{ij} = x \\right]}\n\\end{align}\n\nThis is, we add to the original factor graph a node of degree 1 connected to the variable node $\\sigma_{ij}$ which fixes $\\sigma_{ij} = x$ and run BP on this modified factor graph.\n\n\n\n\n\\end{document}\n\n", "meta": {"hexsha": "4f67579514acdb98f3a2dc820670e44d47a433f9", "size": 14358, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment/assignment.tex", "max_stars_repo_name": "philippvb/APMLA-WS_21-22_material", "max_stars_repo_head_hexsha": "e2e3ccfe393f47b1e78ea481e59d98cb784b1f1e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment/assignment.tex", "max_issues_repo_name": "philippvb/APMLA-WS_21-22_material", "max_issues_repo_head_hexsha": "e2e3ccfe393f47b1e78ea481e59d98cb784b1f1e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment/assignment.tex", "max_forks_repo_name": "philippvb/APMLA-WS_21-22_material", "max_forks_repo_head_hexsha": "e2e3ccfe393f47b1e78ea481e59d98cb784b1f1e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.3314606742, "max_line_length": 318, "alphanum_fraction": 0.6450759159, "num_tokens": 5176, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506418255928, "lm_q2_score": 0.8056321866478979, "lm_q1q2_score": 0.6128046398490794}} {"text": "\\section{Gale Games}\n\\label{gale-games-sect}\n\nWe now apply Theorem \\ref{origgale-thm} to the study of bimatrix games.\nBy Proposition \\ref{nash-to-acl}, solving 2-{\\sc Nash} can be reduced to\n{\\sc Another Completely Labeled Facet}; if the polytope $P^\\Delta$ in\nTheorem \\ref{unit-vector-dual-thm} is cyclic, we can exploit\nTheorem~\\ref{origgale-thm} to translate this special case of 2-{\\sc Nash}\nto a problem on Gale strings.\n\n\nFirst of all, we have to define a labeling on Gale strings such that a\ncompletely labeled Gale string corresponds to a completely labeled facet\nof~$P^\\Delta$.\nWe say that $s\\in G(d,n)$ is a {\\em completely labeled Gale string}\nfor some labeling function $l_s:[n]\\to[d]$ if\n$\\{ l_s(i)\\,\\mid\\,s(i)=\\1\\text{ for }i\\in [n] \\}=[d]$.\nSince a string in $G(d,n)$ has exacty $d$ bits equal to \\1, a Gale string\nis completely labeled if and only if for each $j\\in [d]$ there is exactly\none $i\\in [n]$ such that $s(i)=\\1$ and $l_s(i)=j$.\n\nNotice that, given a labeling $l_s:[n]\\to[d]$, it may not always be\npossible to find a Gale string $s\\in G(d,n)$ that is completely labeled\nby $l_s$.\n\n\\begin{example}\n\\label{no-clgs}\nFor $l_s = 121314$, there are no completely labeled Gale strings.\n\nThe labels $l_s(i)=2,3,4$ appear only once in $l_s$, so we must have\n$s(2)=s(4)=s(6)=1$. We also must have $l_s(i)=1$ for exactly one $i=1,3,5$.\nThe candidate strings are then $s_1=\\1\\10\\10\\1$, $s_2=0\\1\\1\\10\\1$,\n$s_3=0\\10\\1\\1\\1$, but none of these satisfies the Gale Evenness Condition.\n\\end{example}\n\nA {\\em Gale game} is a unit vector game $(U,B)$ where\n$U=[e_{l(1)}\\cdots e_{l(d)}]$ for some labeling $l:[n]\\to [d]$ and\nfor which the dual of the best response polytope is a cyclic polytope\n$P^\\Delta=\\conv\\{ e_1,\\ldots,e_d,c_1,\\ldots,c_n \\}$.\nWe define the problem {\\sc Gale Nash} as in Table~\\ref{gale-nash-pbl}\n\n\\begin{problem}\n{Gale Nash}\n{A Gale game.}\n{A Nash equilibrium of the game.}\n\\label{gale-nash-pbl}\n\\end{problem}\n\nTheorem\n\\ref{unit-vector-dual-thm} gives the labeling (\\ref{vert-labeling-unitv})\nfor the $d+n$ vertices of $P^\\Delta$ as\n\\[\n\\arraycolsep.2em\n\\begin{array}{rcll}\nl_v(-e_i)&=&i\\quad & \\text{ for }i\\in [d],\\\\\nl_v(c_j)&=&l(j)\\quad & \\text{ for }j\\in [n].\n\\end{array}\n\\]\nWe define the labeling $l_s:[d+n]\\to [d]$ of $G(d,n)$ as\n\\begin{equation}\n\\label{gs-labeling-unitv}\n\\arraycolsep.2em\n\\begin{array}{rcll}\nl_s(i)&=&i\\quad & \\text{ for }i\\in [d],\\\\\nl_s(d+j)&=&l(j)\\quad & \\text{ for }j\\in [n].\n\\end{array}\n\\end{equation}\nThen the Gale strings $s\\in G(d,d+n)$ that are completely labeled by $l_s$\ncorrespond exactly to facets of $P^\\Delta$ that are completely labeled by $l_v$,\nwith the facet $F_0$ corresponding to the ``trivial'' completely labeled\nstring $\\1^d 0^n$.\n\n\\begin{example}\n\\label{c46-123432-ex}\nGiven the string of labels $l_s=123432$, there are four associated completely\nlabeled Gale strings in $G(4,6)$:\n$s_A=\\1\\1\\1\\100$, $s_B=\\1\\10\\1\\10$,\n$s_C=\\100\\1\\1\\1$ and\n$s_D=\\10\\1\\10\\1$. These correspond to the completely labeled facets\nfor the labeling shown in Figure \\ref{c46-123432-fig} on the left.\n\\begin{figure}[hbt]\n\\strut\\hfill\n\\includegraphics[width=50ex]{chapter-2/fig-gale-games/123432-facets.pdf}%\n\\hfill\n\\small\n\\begin{tabular}{c | c @{ } c @{ } c @{ } c @{ } c @{ } c @{ } c }\nfacet & {\\bf 1} & {\\bf 2} & {\\bf 3} & {\\bf 4} & {\\bf 3} & {\\bf 2}\\\\\n\\hline\n{\\bf\\it{A}} & \\1 & \\1 & \\1 & \\1 & 0 & 0 \\\\\n{\\bf\\it{B}} & \\1 & \\1 & 0 & \\1 & \\1 & 0 \\\\\n{\\bf\\it{C}} & \\1 & 0 & 0 & \\1 & \\1 & \\1 \\\\\n{\\bf\\it{D}} & \\1 & 0 & \\1 & \\1 & 0 & \\1\n\\end{tabular}\n\\hfill\\strut\n\\caption[A labeling of $C_4(6)$ and its completely labeled facets]{%\nThe cyclic polytope $C_4(6)$, where the labeling of the vertices\ncorresponds to the labeling of $G(4,6)$ given by $l_s=123432$.\n\nThe\ncompletely labeled\nfacets $A$, $B$, $C$ and $D$ correspond respectively to the completely Gale\nstrings $s_A=\\1\\1\\1\\100$, $s_B=\\1\\10\\1\\10$, $s_C=\\100\\1\\1\\1$ and\n$s_D=\\10\\1\\10\\1$.\n}\n\\label{c46-123432-fig}\n\\end{figure}\n\\end{example}\n\nFrom this point forward, we will assume that the labeling\n$l_s:[d+n]\\to [d]$ satisfies\n$l_s(i)\\neq l_s(i+1)$. This can be done without loss of generality,\ngiven the following consideration.\nSuppose that $l_s(i)=l_s(i + 1)$ for some index $i$, and let $s$ be a\ncompletely labeled Gale string for $l_s$. Then only one of\n$s(i)$ and $s(i+1)$ can be equal to \\1 (it is possible that both\nare equal to~0), so $s(i)s(i+1)$ will never be part of a run of even length\nthat ``interferes'' with the Gale Evenness Condition. Therefore, we can\nidentify the indices $i$ and $i + 1$.\n\nWe can now define the problem \\anothergale\\ as in Table\n\\ref{another-gale}. For brevity, we change the notation from $d+n$ to $n$.\n\n\\begin{problem}\n{\\anothergale}\n{A labeling $l:[n]\\to[d]$, where $d= 0\\end{math}. Moreover if \\begin{math}{}q^*q =0\\end{math} then \\begin{math}{}q =0\\end{math}. If\n\\begin{math}{}q* = q\\end{math} then \\begin{math}q\\end{math} is real, and conversely.\n\nThe algorithm assumes only that the field is some subfield of the quaternions, which is closed under this conjugation operation, which is the same thing as saying that if \\begin{math}{}\\alpha\\end{math} is in the field, so is \\begin{math}{}\\alpha^*\\alpha\\end{math}, or equivalently \\begin{math}{}\\alpha + \\alpha^*\\end{math}. This is of course true of the rationals, the reals, any subfield of the reals, the complexes, and the quaternions.\n\nRecall that the adjoint of a matrix \\begin{math}A\\end{math}, denoted \\begin{math}A^*\\end{math}, is the transpose of the matrix obtained from \\begin{math}A\\end{math} by replacing each number \\begin{math}{}a_{ij}\\end{math} by its complex/quaternion conjugate \\begin{math}{}a_{ij}^*\\end{math}.\n\n\\section{Linear Equations}\n\nWe seek all \\begin{math}{}x\\end{math} for which \\begin{math}{}Ax=b\\end{math}. We start by noticing that if\n\\begin{math}{}Ax=b\\end{math}, then also \\begin{math}{}A^*Ax=A^*b\\end{math}. The converse need not be true,\nbut \\begin{math}{}\\{x|Ax=b\\} \\subset \\{x|A^*Ax = A^*b\\}\\end{math}, and in fact it's the simplest sort of subset:\nit's either empty, or it's everything. To see this, note that if \\begin{math}{}A^*Ax_1 = A^*Ax_2 \\end{math},\nthen \\begin{math}{}A^*A(x_1-x_2)=0 \\end{math}, so \\begin{math}{}0=(x_1-x_2)^*A^*A(x_1-x_2)=(A(x_1-x_2))^*(A(x_1-x_2)) \\end{math}. But then \\begin{math}{}A(x_1-x_2)=0 \\end{math}, or \\begin{math}{}Ax_1 = Ax_2 \\end{math}.\n\nIn summary if \\begin{math}{}A^*Ax_1 = A^*Ax_2 = A^*b \\end{math}, then either \\begin{math}{}Ax_1 = Ax_2 = b \\end{math}, or else \\begin{math}{}Ax_1 = Ax_2 \\ne b \\end{math}, in which case there is no solution. \n\nSo having found all solutions to the auxiliary equation, either they are precisely the solutions to the original equation, or the original equation has no solutions, as determined by picking any one of the candidate solutions and computing \\begin{math}{}Ax\\end{math}.\n\nNotice that \\begin{math}{}(A^*A)^* = A^*A^{**} = A^*A\\end{math}, and that \\begin{math}{}x^*(A^*A)x=(Ax)^*(Ax) >= 0\\end{math}. Thus the general problem reduces to the special case considered in the next section.\n\n\\section{Linear Positive Equations}\n\nWe consider then the problem \\begin{math}{}Ax=b\\end{math} in the special case where\n\\begin{math}{}A^*=A\\end{math} and \\begin{math}{}x^*Ax >= 0\\end{math} for all \\begin{math}{}x\\end{math}. Such a matrix is called positive semi-definite.\n\nWe write \\begin{math}{}A\\end{math} as \\begin{math}{}\\begin{bmatrix}\\alpha & b^* \\\\ a & A^{'}\\end{bmatrix}\\end{math}. It is possible\nthat \\begin{math}{}A\\end{math} is 1x1 in which case all expressions involving\n\\begin{math}{}A\\end{math},\n\\begin{math}{}a\\end{math}, or\n\\begin{math}{}b\\end{math} are empty sums, and therefor 0. If \n\\begin{math}{}A\\end{math} is \n\\begin{math}{}n\\end{math} by\n\\begin{math}{}n\\end{math}, then \n\\begin{math}{}A^{'}\\end{math} is \n\\begin{math}{}n-1\\end{math} by\n\\begin{math}{}n-1\\end{math}, and\n\\begin{math}{}a\\end{math} and\n\\begin{math}{}b\\end{math} are \n\\begin{math}{}n-1\\end{math} column vectors, while\n\\begin{math}{}\\alpha\\end{math} is a scalar.\n\nSince \\begin{math}{}A=A^*\\end{math}, \n \\begin{math}{}\n\\begin{bmatrix}\\alpha & b^* \\\\ a & A^{'}\\end{bmatrix} = \n\\begin{bmatrix}\\alpha & b^* \\\\ a & A^{'}\\end{bmatrix}^* = \n\\begin{bmatrix}\\alpha^* & a^* \\\\ b & A^{'*}\\end{bmatrix} \\end{math}, so \n\\begin{math}{}\\alpha^*=\\alpha\\end{math}, \\begin{math}{}a=b\\end{math}, and\n\\begin{math}{}A^{'*} = A^{'}\\end{math}.\n\nMoreover, \n\\begin{math}{}\n0 <= \\begin{bmatrix}1\\\\0\\end{bmatrix}^*\\begin{bmatrix}\\alpha & a^* \\\\ a & A^{'}\\end{bmatrix}\\begin{bmatrix}1\\\\0\\end{bmatrix}\n= \\begin{bmatrix}1&0\\end{bmatrix}\\begin{bmatrix}\\alpha \\\\ a\\end{bmatrix}\n= \\alpha\n\\end{math}\n\n\\subsection{Case 1}\n\\begin{math}{}\\alpha=0\\end{math}\n\nThen in fact it's also true that \\begin{math}{}a=0\\end{math}. For \n\\begin{math}0 <= \\begin{bmatrix}\\chi\\\\a\\end{bmatrix}^*\\begin{bmatrix}0 & a^* \\\\ a & A^{'}\\end{bmatrix}\\begin{bmatrix}\\chi\\\\a\\end{bmatrix} = \\begin{bmatrix}\\chi^* & a^*\\end{bmatrix} \\begin{bmatrix}a^*a\\\\a\\chi+A^{'}a\\end{bmatrix}\n= \\chi^*a^*a + a^*a\\chi + a^*A^{'}a\n\\end{math}.\n\nNotice that it is here that we make use of the assumption that we started with\na positive semi-definite matrix.\n\nIf we restrict attention to rational numbers, then we get the equation of a line. The only lines that have no negative values are horizontal, so the slope \\begin{math}{}2a^*a\\end{math} must be zero, which means \\begin{math}{}a\\end{math} must be zero.\n\nOur equation simplifies to \n\\begin{math}{}\n\\begin{bmatrix}0 & 0 \\\\ 0 & A^{'}\\end{bmatrix}\\begin{bmatrix}\\chi\\\\x^{'}\\end{bmatrix} =\\begin{bmatrix}\\beta\\\\b^{'}\\end{bmatrix}\n\\end{math}, which reduces to the pair of equations \\begin{math}0=\\beta{}\\end{math}, and\n\\begin{math}{}A^{'}x^{'}=b^{'} \\end{math}. The first equation tells us that there can be no solution unless\n\\begin{math}{}\\beta=0\\end{math}. If, however, it is, then any solution \\begin{math}{}x^{'}\\end{math} of the\nsecond equation, which we note, satisfies the same properties we required of \n\\begin{math}{}A\\end{math}, can be combined with any number \\begin{math}{}z_n\\end{math} (where we use the dimension n to tag the variable, to prevent recursive name collisions)\ninto a solution, \\begin{math}{}A\\begin{bmatrix}z_n\\\\x^{'}\\end{bmatrix} = b\\end{math}\n\nIn particular, we can always take \\begin{math}{}z_n\\end{math} to be 0.\n\n\\subsection{Case 2}\n\\begin{math}{}\\alpha>0\\end{math}\n\nThen \n\\begin{math}{}\n\\begin{bmatrix}\\alpha & a^* \\\\ a & A^{'}\\end{bmatrix}\\begin{bmatrix}\\chi\\\\x^{'}\\end{bmatrix}=\\begin{bmatrix}\\beta\\\\b^{'}\\end{bmatrix}\n\\end{math}\nif and only if\n\\begin{math}{}\\alpha\\chi + a^*x^{'} = \\beta\\end{math} and\n\\begin{math}{}a\\chi + A^{'}x^{'} = b^{'}\\end{math}\nif and only if\n\\begin{math}{}\\alpha\\chi= \\beta - a^*x^{'} \\end{math} and\n\\begin{math}{}a\\chi + A^{'}x^{'} = b^{'}\\end{math}\nif and only if\n\\begin{math}{}\\chi= \\alpha^{-1}(\\beta - a^*x^{'})\\end{math} and\n\\begin{math}{}a\\alpha^{-1}(\\beta - a^*x^{'}) + A^{'}x^{'} = b^{'}\\end{math}\nif and only if\n\\begin{math}{}\\chi= \\alpha^{-1}(\\beta - a^*x^{'})\\end{math} and\n\\begin{math}{}a\\alpha^{-1}(- a^*x^{'}) + A^{'}x^{'} = b^{'}-a\\alpha^{-1}\\beta\\end{math}\nif and only if\n\\begin{math}{}\\chi= \\alpha^{-1}(\\beta - a^*x^{'})\\end{math} and\n\\begin{math}{}(A^{'} - a\\alpha^{-1}a^*)x^{'} = b^{'}-a\\alpha^{-1}\\beta\\end{math}\n\nSo if we can solve the second equation, the first equation shows us how to convert it into a solution\nof the original equation. Notice that if \n\\begin{math}{}b =0\\end{math}, then so is the derived right hand side; a homogeneous problem generates\na homogeneous subproblem.\n\nBut this second equation is again of the same form, because\n\\begin{gather}{}(A^{'} - a\\alpha^{-1}a^*)^* = \nA^{'*} - a^{**}\\alpha^{-1}a^* = \nA^{'} - a\\alpha^{-1}a^* \n\\end{gather}\nand moreover\n\\begin{gather}{}\n0 <= \\begin{bmatrix}-\\alpha^{-1}a^*x\\\\x\\end{bmatrix}^*\\begin{bmatrix}\\alpha & a^* \\\\ a & A^{'}\\end{bmatrix}\\begin{bmatrix}-\\alpha^{-1}a^*x\\\\x\\end{bmatrix}\n= \nx^*(A^{'} - a\\alpha^{-1}a^*)x\n\\end{gather}. So the derived submatrix is nonnegative, and positive if the original matrix is.\n\nAs in either case we have reduced to either the base case of dimension one or to a similar problem in one dimension smaller, the solution, whether it exists or\nnot, and whether or not multiple solutions exist, is complete.\n\n\\section{After-thought}\nThis section can be safely skipped by anyone whose interest is confined to\nthe real numbers, the complex numbers, the rational numbers, or even the quaternions.\n\nThe method works provided the scalars are taken from any subfield of the\nquaternions which is closed under the * operation. More precisely, we really\nneed \\begin{math}\\alpha^* = \\alpha\\end{math} if and only if \\begin{math}\\alpha\\end{math} is real, that \\begin{math}\\alpha^*\\alpha >= 0\\end{math}, and that \n\\begin{math}\\alpha^*\\alpha = 0\\end{math} only if \\begin{math}\\alpha=0\\end{math}.\n\nBut the method works, at least in some cases, even if the field is not *-closed.We can take the *-closure of any subfield of the quaternions, by intersecting all *-closed subfields containing it. Since the quaternions are *-closed, that *-closure is a *-closed subfield of the quaternions. We can then find all solutions composed of vectors having entries in the closure; it remains to determine\nwhich of those solutions, if any, lie in the original field. In the case where\n\\begin{math}A\\end{math} is positive, the solution is unique, and a solution whose entries lie in the original field does exist, so must coincide with the solution obtained by this method, which then must necessarily also have entries lying in the original field.\n\n\\section{Implementation Notes}\nAt each level of recursion, we only need to retain\n\\begin{math}{}a\\end{math},\n\\begin{math}{}\\alpha\\end{math}\nand\n\\begin{math}{}\\beta\\end{math}, allowing us to overwrite\n\\begin{math}{}A^{'}\\end{math}\nwith \\begin{math}{}(A^{'} - a\\alpha^{-1}a^*)\\end{math}\nand\n\\begin{math}{}b^{'}\\end{math} with\n\\begin{math}{}b^{'}-a\\alpha^{-1}\\beta\\end{math}\n.\n\nWe conclude with a C-language implementation of a solver for the positive semi-definite case, where\n\\begin{math}{}x^*Ax >=0 \\end{math} for all \\begin{math}{}x\\end{math}.\n\n\\begin{verbatim}\n$ cat positive_solver.h \nextern\ndouble * positive_solver(double ** A, double * b, int n);\n\n$ cat positive_solver.c\n/* \n * Andrew Winkler\n\nThis code solves the equation Ax=b, where A is a positive\nsemi-definite matrix, so that x^t A x >= 0 for all x,\nprovided a solution exists. If A is positive definite,\nit will be an isomorphism, but in general conditions\nmust be imposed on b, for it to lie in the range of A.\n\nIt has the virtue of dramatic simplicity - there's no need\nto explicitly construct the Cholesky decomposition, no need\nto do the explicit back-substitutions. Yet it's essentially\nequivalent to that more labored approach, so its\nperformance/stability/memory, etc. should be at least as good.\n\nThere are two kinds of checks, both of which are disabled.\nThis means that a solution will be generated; it is up to\nthe caller to verify that the equation Ax=b is satisfied to\nthe desired precision. If it does not, then no solution exists.\n\nThe first kind of check is that Avec is 0 whenever A[0][0]\nis 0, which will always be true if A is positive semi-definite,\nbut which could become false because of any small perturbation\ncaused by roundoff error upstream in the computation of A.\nThe second kind of check is that b[0] is 0 whenever A[0][0] is 0,\nwhich is necessary for b to lie in the range of A. The checks are\ndisabled because the code does not deal with real numbers,\nbut rather floating point numbers; it's up to the user to decide\nwhat is \"close enough\" to zero.\n\n*/\n\n#include \n#include \n#include \"positive_solver.h\"\n\nvoid _consistency_checker(double * v, int n) {\n for (int i=0; i\ndouble * positive_solver(\n double ** A, \n double * b, \n int n\n ) \n{\n double * x = (double *) malloc(n * sizeof(double));\n _positive_solver(A, b, x, n);\n return x;\n}\n\n\\end{verbatim}\n\\end{document}\n", "meta": {"hexsha": "c003f6415ffc219e3d689c2095e02d474bcc09c1", "size": 13854, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "OPERATORS/AX_EQUALS_B/doc/linear_solver_orig.tex", "max_stars_repo_name": "subramon/qlu", "max_stars_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "OPERATORS/AX_EQUALS_B/doc/linear_solver_orig.tex", "max_issues_repo_name": "subramon/qlu", "max_issues_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 7, "max_issues_repo_issues_event_min_datetime": "2020-07-29T16:48:25.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-26T23:47:22.000Z", "max_forks_repo_path": "OPERATORS/AX_EQUALS_B/doc/linear_solver_orig.tex", "max_forks_repo_name": "subramon/qlu", "max_forks_repo_head_hexsha": "2fb8a2b3636dd11e2dfeae2a6477bd130316da47", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2015-05-14T22:34:13.000Z", "max_forks_repo_forks_event_max_datetime": "2015-05-14T22:34:13.000Z", "avg_line_length": 47.1224489796, "max_line_length": 438, "alphanum_fraction": 0.6800202108, "num_tokens": 4590, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8056321796478255, "lm_q2_score": 0.7606506418255928, "lm_q1q2_score": 0.6128046345244698}} {"text": "\\documentclass{amsart}\n\n\\usepackage{psfrag}\n\\usepackage{graphicx}\n\\usepackage{url}\n\\usepackage{amsmath,amsfonts}\n%\\usepackage{fullpage}\n\n\\newcommand\\br{{\\mathbf r}}\n\\newcommand\\w{{\\mathbf w}}\n\\newcommand\\bL{{\\mathbf L}}\n\\newcommand\\boldf{{\\mathbf f}}\n\\newcommand\\bz{{\\mathbf z}}\n\\newcommand\\bnabla{{\\boldsymbol \\nabla}}\n\\newcommand\\e{{\\mathbf e}}\n\\newcommand\\q{{\\mathbf q}}\n\n\\begin{document}\n\n\\section{Differential equations on the unit sphere}\n\nThe theoretical underpinnings of this package are described in \\cite{montgomerysmith:10}.\n\nThe objective of the software is to provide a systematic spherical harmonic method to numerically solve partial differential equations on the two\ndimensional sphere, $S = \\{\\br = (x,y,z):|\\br| = x^2+y^2+z^2 = 1\\}$, or in spherical coordinates, $\\br = (x,y,z) = (\\sin\\theta\n\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta)$, where $0\\le\\theta\\le\\pi$, $0\\le\\phi\\le2\\pi$. The partial differential equations considered here are of the\nform\n\\begin{equation}\n\\label{pde}\n\\frac\\partial{\\partial t}\\psi = F(\\br,\\bnabla) \\psi,\n\\end{equation}\nwhere $\\psi$ is a function (for example, the probability distribution function) defined on the sphere. Here $\\bnabla$ is the gradient operator\nrestricted to the sphere that is defined as\n\\begin{equation}\n\\begin{split}\n\\bnabla &= (\\nabla_x,\\nabla_y,\\nabla_z) \\\\\n&= \\left(\\cos\\theta\\cos\\phi\\frac{\\partial}{\\partial\\theta} - \\csc\\theta\\sin\\phi \\frac{\\partial}{\\partial\\phi},\\cos\\theta\\sin\\phi\\frac{\\partial}{\\partial\\theta} + \\csc\\theta\\cos\\phi \\frac{\\partial}{\\partial\\phi},-\\sin\\theta\\frac{\\partial}{\\partial\\theta} \\right) \\\\\n&= \\left((1-x^2)\\frac\\partial{\\partial x}-xy\\frac\\partial{\\partial y}-xz\\frac\\partial{\\partial z},\n -xy\\frac\\partial{\\partial x}+(1-y^2)\\frac\\partial{\\partial y}-yz\\frac\\partial{\\partial z},\\right.\\\\\n&\\qquad\\qquad\\left.\n -xz\\frac\\partial{\\partial x}-yz\\frac\\partial{\\partial y}+(1-z^2)\\frac\\partial{\\partial\n z}\\right) .\n\\end{split}\n\\end{equation}\nIn the above, $F$ is a polynomial in six variables, with the proviso that it matters in which order the terms of each monomial part are written. We\nbelieve that this includes many of the published, if not all, of the partial differential equations that describe the evolution of\nthe orientation distribution function for short fiber suspensions.\n\nThe spherical harmonic approach converts equation~\\eqref{pde} to a\nsystem of ordinary differential equations written as\n\\begin{equation}\n\\label{spherical-pde}\n\\frac\\partial{\\partial t} \\hat\\psi_l^m = \\sum_{l'=0}^\\infty \\sum_{m'=-l'}^{l'} c_{l,l'}^{m,m'} \\hat\\psi_{l'}^{m'} ,\n\\end{equation}\nwhere $\\hat\\psi_l^m$ ($l \\ge 0$, $|m| \\le l$) are the spherical harmonic coefficients of $\\psi$, defined in\nequation~\\eqref{spherical-harmonic-coefficient} below, and $c_{l,l'}^{m,m'}$ are the coefficients defined through equation~\\eqref{decompose F}.\n\nThe software provides a systematic algorithm for calculating the coefficients $c_{l,l'}^{m,m'}$ appearing in equation~\\eqref{spherical-pde} (and also in equation~\\eqref{decompose F} below). This algorithm makes it possible to readily solve all differential\nequations on the unit sphere of the form shown in equation~\\eqref{pde}.\n\n\\section{Description of the model equations}\n\nTo illustrate the software, we consider variations of Jeffery's equation \\cite{jeffery:23} which describes the motion of fibers in a\nmoving fluid with vorticity $\\w$ and rate of deformation tensor $\\Gamma$. Jeffery's equation is\noften written in terms of the fiber orientation distribution function $\\psi$ and the fiber aspect ratio parameter $-1 \\le \\lambda\\le1$ as\n\\begin{equation}\n\\label{jeffery's}\n\\frac\\partial{\\partial t}\\psi = J \\psi := -\n\\tfrac12\\bnabla\\cdot(\\w \\times \\br \\psi + \\lambda(\\Gamma\\cdot\\br -\n\\Gamma:\\br\\br\\br) \\psi) ,\n\\end{equation}\nwhere we note that the right hand side is in the form of equation~\\eqref{pde}. A variation of equation~\\eqref{jeffery's} is Jeffery's equation with\nrotary diffusion as expressed by Bird \\cite{bird:87b} as\n\\begin{equation}\n\\label{jeffery's-folgar-tucker}\n\\frac\\partial{\\partial t}\\psi = J \\psi + \\bnabla\\cdot\\bnabla(D_r\n\\psi) ,\n\\end{equation}\nwhere $D_r$ captures the effect of fiber interaction and depends upon the flow kinetics. Folgar and Tucker \\cite{folgar:84} selected $D_r = C_I \\dot{\\gamma}$ where $\\dot{\\gamma} =\n\\left(\\frac12\\Gamma:\\Gamma\\right)^{1/2}$ and $C_I$ is a constant that depends upon the volume fraction and aspect ratio of the\nfibers.\n\nAnother example is anisotropic diffusion such as that proposed by Koch \\cite{koch:95} in the\ndifferential equation\n\\begin{equation}\n\\label{koch} \\frac\\partial{\\partial t}\\psi = J \\psi +\n\\bnabla\\cdot(I-\\br\\br)\\cdot D_r\\cdot\\bnabla\\psi = J \\psi +\n(\\bnabla-2\\br)\\cdot D_r\\cdot\\bnabla\\psi\n\\end{equation}\nwhere the anisotropic diffusion matrix $D_r$ is defined in terms of the model parameters $C_1$ and $C_2$ (see Koch \\cite{koch:95} for more detail) as\n\\begin{equation}\n\\label{kochDr}\nD_r = C_1 {\\dot\\gamma}^{-1} (\\Gamma:\\mathbb A:\\Gamma)I + C_2\n{\\dot\\gamma}^{-1} \\Gamma:\\mathcal A:\\Gamma .\n\\end{equation}\nIn the above, the $(I-\\br\\br)$ term serves to project vectors onto the surface of the sphere. We note that this term is not explicitly included in\nKoch's formula in \\cite{koch:95}, however it's existence is implied in the paragraph following the original introduction.\n\nThe anisotropic diffusion matrix $D_r$ defined in equation~\\eqref{kochDr} is written in terms of the 2nd, 4th and 6th order moment tensors of $\\psi$\nwhich are, respectively, defined as\n\\begin{equation}\n\\label{momenttensors}\n A := \\int_S \\psi \\br\\br \\, d\\br ,\\qquad\n\\mathbb A := \\int_S \\psi \\br\\br\\br\\br \\, d\\br ,\\qquad \\text{and}\n\\qquad \\mathcal A := \\int_S \\psi \\br\\br\\br\\br\\br\\br \\, d\\br .\n\\end{equation}\nIn these integrals, we adopt the common definition for the integral of a function $F(\\br)$ over the surface of the unit sphere as\n\\begin{equation}\n\\int_S F(\\br) \\, dr := \\int_{\\theta=0}^\\pi \\int_{\\phi=0}^{2\\pi}\nF(\\br) \\sin\\theta\\,d\\phi\\,d\\theta ,\n\\end{equation}\nand we note here that the moment tensors $A$, $\\mathbb A$, and $\\mathcal{A}$ can be expressed in terms of spherical harmonics as shown below in\nequation~\\eqref{moment tensor}.\n\nCare must be exercised in not confusing the differential geometry on the surface of the sphere $S$ with the differential geometry\nin the three dimensional space $\\mathbb R^3$ it is embedded in. Thus, for example, integration by parts\nis the slightly unexpected form\n\\begin{equation}\n\\label{by-parts}\n\\int_S \\boldf \\cdot \\bnabla g\\, d\\br = - \\int_S (\\bnabla\\cdot (I-\\br\\br)\\cdot\\boldf) g\\, d\\br = \\int_S ((2\\br - \\bnabla)\\cdot\\boldf)g \\, d\\br ,\n\\end{equation}\nwhich reduces to the usual integration by parts when $\\boldf$ is\ntangential to the surface of the sphere. We also recall the so\ncalled angular momentum operator\n\\begin{equation}\n\\label{ang-momentum-ops}\n\\bL = -i\\br\\times\\bnabla = (L_x,L_y,L_z) =\n-i\\left(y\\frac\\partial{\\partial z} - z\\frac\\partial{\\partial y} ,\nz\\frac\\partial{\\partial x} - x\\frac\\partial{\\partial z} ,\nx\\frac\\partial{\\partial y} - y\\frac\\partial{\\partial x} \\right) ,\n\\end{equation}\nwhere, as usual, $i$ denotes the complex number satisfying $i^2=-1$. Integration by parts for the angular momentum operator is more straightforward, that is,\n\\begin{equation}\n\\label{by-parts-l}\n\\int_S \\boldf \\cdot \\bL g\\, d\\br = -\\int_S (\\bL\\cdot\\boldf) g\\, d\\br .\n\\end{equation}\nNote that $\\bL\\times\\bL = \\bnabla\\times\\bnabla = i\\bL$, and in particular $\\nabla_x$, $\\nabla_y$ and $\\nabla_z$\ndo not commute with each other. We also use the formulae $\\br\\cdot\\bnabla f=\\br\\cdot\\bL f=\\bL\\cdot(\\br f) = 0$ and $\\bnabla\\cdot(\\br f) = 2f$.\n\n\\section{Spherical harmonics solutions}\n\nAny square integrable function $\\psi$ defined on the unit sphere may be written as a Fourier series like representation in terms of the spherical\nharmonics $Y_l^m(\\theta,\\phi)$ for $l \\ge 0$ and $|m| \\le l$ as\n\\begin{equation}\n\\label{spherical-series}\n\\psi = \\sum_{l=0}^\\infty \\sum_{m=-l}^l\n\\hat\\psi_l^m Y_l^m ,\n\\end{equation}\nwhere $Y_l^m$ are the spherical harmonic functions as defined in \\cite{weisstein:06a}.\nThe coefficients $\\hat\\psi_l^m$ in equation~\\eqref{spherical-series}\nare evaluated from\n\\begin{equation}\n\\label{spherical-harmonic-coefficient}\n\\hat\\psi_l^m = \\int_S \\psi\n\\bar Y_l^m d\\br .\n\\end{equation}\n\nTo obtain the system of ordinary differential equations~\\eqref{spherical-pde}, we integrate equation~\\eqref{pde} against $\\bar Y_l^m$ over the unit\nsphere to obtain its adjoint or weak form as\n\\begin{equation}\n\\label{weakform}\n\\frac\\partial{\\partial t} \\int_S \\psi \\bar Y_l^m\n\\,d\\br = \\int_S \\psi F^*(\\br,\\bnabla) \\bar Y_l^m \\,d\\br ,\n\\end{equation}\nwhere the right hand side follows from integration by parts and the term $F^*$ denotes any polynomial $F$ in which each monomial term is written in\nreverse order with the substitution of $2\\br - \\bnabla$ for $\\bnabla$. Therefore, applying equation~\\eqref{weakform}, Jeffery's\nequation~\\eqref{jeffery's} is defined by\n\\begin{equation}\nJ^* = \\tfrac12(\\w\\cdot(\\br\\times\\bnabla) + \\lambda\\br\\cdot\\Gamma\\cdot\\bnabla),\n\\end{equation}\nSimilarly, the two extensions of Jeffery's equation that include diffusion appearing in equations~\\eqref{jeffery's-folgar-tucker} and~\\eqref{koch}\n(expressed as $\\frac\\partial{\\partial t}\\psi = F \\psi$), respectively become\n\\begin{equation}\n\\label{FStarFT}\nF^* = J^* + D_r \\bnabla\\cdot\\bnabla ,\n\\end{equation}\nand\n\\begin{equation}\nF^* = J^* + (\\bnabla-2\\br)\\cdot D_r\\cdot\\bnabla .\n\\end{equation}\nIn all cases, we can decompose\n\\begin{equation}\n\\label{decompose F}\nF^*(\\br,\\bnabla) \\bar Y_l^m = \\sum_{l'=0}^\\infty \\sum_{m'=-l'}^{l'} c_{l,l'}^{m,m'} \\bar Y_{l'}^{m'} ,\n\\end{equation}\nand therefore obtain the spherical harmonic representation shown in equation~\\eqref{spherical-pde} above.\n\n\\section{The ``spherical'' program algorithm}\n\nAn automated algorithm is implemented to compute the coefficients $c_{l,l'}^{m,m'}$ in equation~\\eqref{decompose F} for\nany differential equation satisfying the criteria given above (and, therefore, the $\\mathcal C_{l,m}$ in equation~\\eqref{moment tensor}). The\nalgorithm is to recursively apply the replacement rules described in equation~\\eqref{rules} until no further substitutions can be made. Here ``op''\ndenotes any of $z$, $L_z$, $L_+$, or $L_-$ operations, and $\\mathcal Y$ denotes any linear combination of the $\\bar Y_l^m$'s.\n\\begin{equation}\n\\label{rules}\n\\begin{split}\n\\text{op}(\\mathcal Y \\pm c \\bar Y_l^m) &\\to \\text{op}(\\mathcal Y) \\pm c \\, \\text{op}(\\bar Y_l^m), \\\\\nx \\mathcal Y &\\to z i L_y(\\mathcal Y) - i L_y(z\\mathcal Y), \\\\\ny \\mathcal Y &\\to i L_x(z\\mathcal Y) - z i L_x(\\mathcal Y), \\\\\nz\\bar Y_l^m &\\to \\sqrt{\\frac{(l+m)(l-m)}{(2l-1)(2l+1)}} \\bar Y_{l-1}^m + \\sqrt{\\frac{(l+m+1)(l-m+1)}{(2l+1)(2l+3)}} \\bar Y_{l+1}^m ,\\\\\n\\nabla_x \\mathcal Y &\\to z i L_y(\\mathcal Y) - y i L_z(\\mathcal Y), \\\\\n\\nabla_y \\mathcal Y &\\to x i L_z(\\mathcal Y) - z i L_x(\\mathcal Y), \\\\\n\\nabla_z \\mathcal Y &\\to y i L_x(\\mathcal Y) - x i L_y(\\mathcal Y), \\\\\nL_x (\\mathcal Y) &\\to \\tfrac 12 \\left(L_+ (\\mathcal Y) + L_- (\\mathcal Y)\\right), \\\\\nL_y (\\mathcal Y) &\\to -\\tfrac i2 \\left(L_+ (\\mathcal Y) - L_- (\\mathcal Y)\\right), \\\\\nL_z \\bar Y_l^m &\\to -m\\bar Y_l^m ,\\\\\nL_+ \\bar Y_l^m &\\to - \\sqrt{(l+m)(l-m+1)}\\bar Y_l^{m-1}, \\\\\nL_- \\bar Y_l^m &\\to - \\sqrt{(l-m)(l+m+1)}\\bar Y_l^{m+1}\n\\end{split}\n\\end{equation}\nIt is important to note that $x \\bar Y_l^m$, $y \\bar Y_l^m$, $z \\bar Y_l^m$, $\\nabla_x\\bar Y_l^m$, $\\nabla_y \\bar Y_l^m$, and $\\nabla_z \\bar Y_l^m$\ninvolve $\\bar Y_{l'}^{m'}$ for $l'$ and $m'$ that differ from $l$ and $m$, respectively, by at most one. The script described in the appendix uses\nthe recursive algorithm in equation~\\eqref{rules} to create threaded functions in the programming language\nC which in turn are used in an iterative procedure for computing the solution to the differential equation in equation~\\eqref{pde}.\n\nMoment tensors may be computed using this algorithm as well. For example, to compute the 6th order moment tensor $\\mathcal A$ in\nequation~\\eqref{momenttensors} we simply expand\n\\begin{equation}\n\\label{expand Y_0^0} \\br\\br\\br\\br\\br\\br \\bar Y_0^0 = \\sum_{l=0}^6\n\\sum_{m=-l}^l \\mathcal C_{l,m} \\bar Y_l^m,\n\\end{equation}\nwhere $\\mathcal C_{l,m}$ is the tensor of rank six composed of coefficients calculated by applying $x$, $y$ and $z$ to $\\bar Y_0^0$ six times using the algorithm. It follows that since $\\bar Y_0^0 = 1/\\sqrt{4\\pi}$, the 6th order orientation tensor\nbecomes\n\\begin{equation}\n\\label{moment tensor} \\mathcal A = \\sqrt{4\\pi} \\int_S \\psi\n\\br\\br\\br\\br\\br\\br \\bar Y_0^0 \\, d\\br = \\sqrt{4\\pi} \\sum_{l=0}^6\n\\sum_{m=-l}^l \\mathcal C_{l,m} \\hat\\psi_l^m ,\n\\end{equation}\nwhere the orthogonality of the spherical harmonics are used to simplify the final result.\n\n\\section{The ``spherical'' software package}\n\nThe program developed to implement our spherical harmonics\nsimulation procedure described above which computes the coefficients\nin equation~\\eqref{spherical-pde}, and then convert these terms into\na C program, may be found at\n\\url{https://github.com/stephenmontgomerysmith/spherical}. The\nprogram is written in \\emph{perl}, and makes use of either the\ncommercial computer algebra system \\emph{Mathematica} by Wolfram\nResearch, or the open-source computer algebra system \\emph{Maxima}.\nThe scripts are designed to work in a Unix like environment.\n\nThe program that evaluates spherical harmonics coefficients is a\n\\emph{perl} script which applies the rules in~\\eqref{rules}. It is\ninvoked with the command sequence ``{\\tt perl expand-method.pl}''.\nThe user may type in an expression involving {\\tt x}, {\\tt y}, {\\tt\nz}, {\\tt dx}, {\\tt dy}, {\\tt dz}, {\\tt lx}, {\\tt ly}, or {\\tt lz}\nrepresenting ``multiply by $x$,'' ``multiply by $y$,'' ``multiply by\n$z$,'' $\\nabla_x$, $\\nabla_y$, $\\nabla_z$, $L_x$, $L_y$, or $L_z$\nrespectively, or any number, including {\\tt I} for $i$, and any\nappropriate combination of parentheses, addition and subtraction,\nand ``{\\tt *}'' which represents composition of the operators. It\noutputs a sequence of lines, which when added together, represents\nthe effect of the expression upon $\\bar Y_l^m$. Each line has three\nentries separated by semicolons, $l'$;$m'$;$c$, and represents\n$c\\bar Y_{l+l'}^{m+m'}$. The third entry is written in the\nprogramming language C, using complex numbers as described in\n\\cite{c:99}. So, for example, the input\n\\begin{verbatim}\n%perl expand-method.pl\n> -2*I*(dy-y)\n\\end{verbatim}\nyields the output\n\\begin{verbatim}\n-1;-1;((l*sqrt(-1+l+m)*sqrt(l+m))/(sqrt(-1+2*l)*sqrt(1+2*l)))+(0)*I\n-1;1;((l*sqrt(-1+l-m)*sqrt(l-m))/(sqrt(-1+2*l)*sqrt(1+2*l)))+(0)*I\n1;-1;(((1+l)*sqrt(1+l-m)*sqrt(2+l-m))/(sqrt(1+2*l)*sqrt(3+2*l)))+(0)*I\n1;1;(((1+l)*sqrt(1+l+m)*sqrt(2+l+m))/(sqrt(1+2*l)*sqrt(3+2*l)))+(0)*I\n\\end{verbatim}\nThis result reflects the equation\n\\begin{equation*}\n\\begin{split}\n-2i(\\nabla_y - y)\\bar Y_l^m = &\n\\tfrac{l\\sqrt{(l+m-1)(l+m)}}{\\sqrt{(2l-1)(2l+1)}} \\bar Y_{l-1}^{m-1}\n+ \\tfrac{l\\sqrt{(l-m-1)(l-m)}}{\\sqrt{(2l-1)(2l+1)}} \\bar Y_{l-1}^{m+1} \\\\\n& + \\tfrac{(l+1)\\sqrt{(l-m+1)(l-m+2)}}{\\sqrt{(2l+1)(2l+3)}} \\bar Y_{l+1}^{m-1}\n+ \\tfrac{(l+1)\\sqrt{(l+m+1)(l+m+2)}}{\\sqrt{(2l+1)(2l+3)}} \\bar Y_{l+1}^{m+1} .\n\\end{split}\n\\end{equation*}\n\nThe program ``{\\tt perl expand-method.pl}'' is called by another\nscript invoked as ``{\\tt perl expand-iterate.pl }'' which\nconverts a configuration file into a C program to numerically solve\nthe spherical harmonics differential equation. For example,\nJeffery's equation (written here in weak form) with Koch diffusion\ncan be written as below\n\\begin{verbatim}\nvoid compute_psidot(complex* psidot, complex* psi,\n double lambda, double w[3], double g[3][3]) {\n double a4[3][3][3][3], a6[3][3][3][3][3][3], Dr[3][3];\n\n tensor4(psi,a4); tensor6(psi,a6);\n ... compute Dr from g, a4 and a6 ...\n\n @spherical_iterate {\n psidot[@index] =\n /* Jeffery's equation */\n w[0]*@method(psi,0.5*y*dz-0.5*z*dy)\n + w[1]*@method(psi,0.5*z*dx-0.5*x*dz)\n + w[2]*@method(psi,0.5*x*dy-0.5*y*dx)\n + lambda*g[0][0]*@method(psi,0.5*x*dx)\n + lambda*g[0][1]*@method(psi,0.5*x*dy+0.5*y*dx)\n + lambda*g[0][2]*@method(psi,0.5*x*dz+0.5*z*dx)\n + lambda*g[1][1]*@method(psi,0.5*y*dy)\n + lambda*g[1][2]*@method(psi,0.5*y*dz+0.5*z*dy)\n + lambda*g[2][2]*@method(psi,0.5*z*dz)\n /* Koch diffusion */\n + Dr[0][0]*@method(psi,(dx-2*x)*dx)\n + Dr[0][1]*@method(psi,(dx-2*x)*dy+(dy-2*y)*dx)\n + Dr[0][2]*@method(psi,(dx-2*x)*dz+(dz-2*z)*dx)\n + Dr[1][1]*@method(psi,(dy-2*y)*dy)\n + Dr[1][2]*@method(psi,(dy-2*y)*dz+(dz-2*z)*dy)\n + Dr[2][2]*@method(psi,(dz-2*z)*dz);\n }\n}\n\\end{verbatim}\n\nTo calculate the coefficients for the moment tensors, as in\nequation~\\eqref{expand Y_0^0}, if the environmental variable {\\tt\nTENSOR} is set to a non-zero value, then {\\tt expand-method.pl}\napplies the input expression to $\\bar Y_0^0$, and outputs the\nanswers in floating point format. So, for example the input\n\\begin{verbatim}\n%env TENSOR=1 perl expand-method.pl\n> x*y*z*z\n\\end{verbatim}\nyields\n\\begin{verbatim}\n2;-2;(0)+(-0.026082026547865053022)*I\n2;2;(0)+(0.026082026547865053022)*I\n4;-2;(0)+(-0.030116930096841707924)*I\n4;2;(0)+(0.030116930096841707924)*I\n\\end{verbatim}\nThis results reflects the 1233 component of 4th order moment tensor\n\\begin{equation*}\n\\mathbb A_{1233} =\ni\\sqrt{4\\pi}\\left(\\tfrac{\\sqrt{30}}{210}(\\hat\\psi_2^2-\\hat\\psi_2^{-2})\n+\\tfrac{\\sqrt{10}}{105}(\\hat\\psi_4^2-\\hat\\psi_4^{-2})\\right).\n\\end{equation*}\nThis is used by ``{\\tt perl make-tensor.pl $n$}'' and ``{\\tt perl\nmake-reverse-tensor.pl $n$}'', which create C programs to convert,\nrespectively, spherical harmonic coefficients to moment tensors of\nrank $n$, and vice-versa.\n\n\\bibliographystyle{plain}\n\\bibliography{info_08}\n\n\\end{document}\n", "meta": {"hexsha": "73cac62665fa7b9f328d1915529d174c2fb2aa4c", "size": 17725, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/pde-sphere-software.tex", "max_stars_repo_name": "stephenmontgomerysmith/spherical", "max_stars_repo_head_hexsha": "d7ac6ed2b216e317d10051d8909c9978ca0270a1", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doc/pde-sphere-software.tex", "max_issues_repo_name": "stephenmontgomerysmith/spherical", "max_issues_repo_head_hexsha": "d7ac6ed2b216e317d10051d8909c9978ca0270a1", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doc/pde-sphere-software.tex", "max_forks_repo_name": "stephenmontgomerysmith/spherical", "max_forks_repo_head_hexsha": "d7ac6ed2b216e317d10051d8909c9978ca0270a1", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.787965616, "max_line_length": 264, "alphanum_fraction": 0.6990126939, "num_tokens": 6072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8976953030553434, "lm_q2_score": 0.6825737344123242, "lm_q1q2_score": 0.6127432353708888}} {"text": "We wish to predict useful dependencies for a proof goal, or conjecture.\nFor this we define rankings, which are generated by predictors:\n\n\\begin{definition}\\defgls{rankings}\n A ranking $r \\in \\rankings$ is a partial function which yields the likelyhood a\n dependency $\\phi \\in \\depset$ is useful in the current context.\n\n $$\n r : \\depset \\rightharpoonup \\mathbb{R}\n $$\n\\end{definition}\n\n\\begin{definition}\\defgls{predictors}\n A predictor $P \\in \\predictors$ is a function which given an unproven conjecture $c \\in \\types$\n computes the likelyhood a dependency $\\phi \\in \\depset$ is useful in proof of conjecture $c$.\n In other words, for a conjecture $c \\in \\types$ it yields a ranking $r \\in \\rankings$.\n $$\n P : \\types \\rightarrow \\rankings\n $$\n\\end{definition}\n\nA weak total ordered subset of definitions useful for conjecture $c$ is formed by sorting on the resulting value of type $\\mathbb{R}$.\nIn the next few chapters we define a few of these predictors.\n\n\\subsubsection{\\knn}\n\\input{document-approach-predictors-knn}\n\n\\subsubsection{\\knnadaptive}\n\\input{document-approach-predictors-knnadaptive}\n\n\\subsubsection{\\nb}\n\\input{document-approach-predictors-nb}\n\n\\subsubsection{Adarank}\n\\input{document-approach-predictors-adarank}\n\n\\subsubsection{Ensembles}\n\\input{document-approach-predictors-ensemble}\n", "meta": {"hexsha": "800997be9b865a1069ae8ec920fde87a8bc315b0", "size": 1322, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documents/thesis/document-approach-predictors.tex", "max_stars_repo_name": "Wassasin/premiseselection", "max_stars_repo_head_hexsha": "c07c7d2d52605fd3d960ec4b5d952eb0aae4bb5b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-02-11T14:59:32.000Z", "max_stars_repo_stars_event_max_datetime": "2019-02-11T14:59:32.000Z", "max_issues_repo_path": "documents/thesis/document-approach-predictors.tex", "max_issues_repo_name": "Wassasin/premiseselection", "max_issues_repo_head_hexsha": "c07c7d2d52605fd3d960ec4b5d952eb0aae4bb5b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "documents/thesis/document-approach-predictors.tex", "max_forks_repo_name": "Wassasin/premiseselection", "max_forks_repo_head_hexsha": "c07c7d2d52605fd3d960ec4b5d952eb0aae4bb5b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.8974358974, "max_line_length": 134, "alphanum_fraction": 0.7556732224, "num_tokens": 372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.897695283896349, "lm_q2_score": 0.682573734412324, "lm_q1q2_score": 0.6127432222934623}} {"text": "\\section{Methodology for Library Evaluation}\n\nThe three dimensions of library evaluation are (i)efficiency, (ii)scalability, and (iii)reliability. The overall evaluation of the library is the geometric mean of numerical measurements of each dimension. Standard numerical measurements for evaluating efficiency has been found in literature. A similar approach may apply to the evaluation of scalability. A numerical measure of reliability, unfortunately, requires further study. The first section of this part will review methodology of hardware efficiency evaluation, which I believe applies to software efficiency evaluation as well. In the rest three sections of this part, approaches to evaluate each dimension will be examined respectively.\n\n\\subsection{Measuring Hardware Efficiency}\nIn practice, hardware efficiency is measured by its performance regarding to a benchmark suite, which contains a set of test programs. Performance is defined as the reciprocal of mean execution time of all tests\\cite{HePa06}. This project will use geometric mean instead of weighted arithmetic mean for three reasons.\n\nFirstly and most importantly, as proved by Fleming and Wallace \\cite{Fleming}, geometric mean is the only correct average of normalised measurements. As users often compare the performance of one system to another, performance is usually normalised to a reference system. \n\n\\begin{comment}\n Precisely, let the benchmark suite has $n$ tests indexed from $1$ to $n$, performance of system A is calculated as:\n\n $performance_A$ \n= $\\frac{time_{ref}}{time_{A}}$\n= $\\frac{\\sqrt[n]{\\prod^{n}_{i=1} time_{ref_i}}}{\\sqrt[n]{\\prod^{n}_{i=1} time_{A_i}}}$\n= $\\sqrt[n]{\\prod^{n}_{i=1}{\\frac{time_{ref_i}}{time_{A_i}}}}$\n\n\\hspace{0.8 cm}where $time_{S_i}$ is the execution time of program $i$ on system $S$, \n\n\\hspace{1.0 cm} and $time_{S}$ is the time measurement of system $S$.\n\\end{comment}\n\nSecondly, the order of calculating means and normalisation is flexible in practice. Therefore, the suite developer will only implement a generic tool for measuring unnormalised execution time. The work of comparing performance of two interested platforms will be left to users. Besides, users are free to extend the benchmark suite. When new tests are added to the benchmark suite, performance measure for old examples becomes a valid partial measure.\n\nFinally, statistic tools could be employed to verify the soundness of chosen suite. For example, the geometric standard deviations is an indicator to assess the variability of the chosen suite.\n\n\n\\subsection{Measuring Library Efficiency}\nThe Methodology of measuring hardware efficiency also applies to measuring library efficiency, where both the library implementation and the running platform may differ. \n\nTo test the efficiency of different distributed programming libraries, a set of benchmarking examples need to be identified. To avoid unintentional optimisation in the TAkka implementation, benchmarks are scrupulously selected from existing applications written in other OTP-like libraries and modifications are made at the minimal level. \n\nTheoretically, any program that involves a time measurement could be a candidate efficiency test. Such examples could be easily found in both the correctness test suite and the scalability test suite. However, we would like to exclude examples whose results significantly determine the overall measure. To this end, statistic methods such as principal component analysis will be employed.\n\n\\subsection{Measuring Scalability}\n\nTo assess the scalability of different OTP-like libraries, suitable examples from the BenchErl suite\\cite{RELEASE} have been reimplemented using Akka and TAkka. A summary of selected examples will be given at \\S\\ref{scalability}.\n\nSimilar to efficiency evaluation, geometric mean of scalability measures of all examples could be the indicator of the library scalability. Unfortunately, numerical evaluation for scalability is not found in the literature. This project will use linear regression techniques to analysis the scalability of an OTP-like library in two aspects: (i) the degrees to which the performance speedup deviate from the ideal linear relationship regarding to the number of active CPU cores. (ii) the slope of the linear equation, assuming that the relationship between performance speedup and the number of active CPU cores is linear.\n\nFinally, scalability of OTP-like libraries will be tested on a variety of platforms. The three candidate platforms are Beowulf cluster, Google App Engine, and Amazon EC2.\n\n\n\\subsection{Measuring Reliability}\n\nIdentify types of failures and their impact is a crucial part of reliability evaluation. The next two months will be devoted to the study of possible failures that could be handled at the language level. Methodology for numerically evaluating reliability requires more studies and careful verifications.\n\n\n", "meta": {"hexsha": "143929925600692d1fabd7b534f19fdabb2a264d", "size": 4921, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "s1024484/ReviewReport/Oct2012/evaluation.tex", "max_stars_repo_name": "Jiansen/TAkka", "max_stars_repo_head_hexsha": "d2410190552aeea65c1da5f0ae05f08ba1f4d102", "max_stars_repo_licenses": ["BSD-Source-Code"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-09-11T14:35:53.000Z", "max_stars_repo_stars_event_max_datetime": "2019-06-27T06:36:09.000Z", "max_issues_repo_path": "s1024484/ReviewReport/Oct2012/evaluation.tex", "max_issues_repo_name": "Jiansen/TAkka", "max_issues_repo_head_hexsha": "d2410190552aeea65c1da5f0ae05f08ba1f4d102", "max_issues_repo_licenses": ["BSD-Source-Code"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "s1024484/ReviewReport/Oct2012/evaluation.tex", "max_forks_repo_name": "Jiansen/TAkka", "max_forks_repo_head_hexsha": "d2410190552aeea65c1da5f0ae05f08ba1f4d102", "max_forks_repo_licenses": ["BSD-Source-Code"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 100.4285714286, "max_line_length": 707, "alphanum_fraction": 0.8047144889, "num_tokens": 1009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8244619350028204, "lm_q2_score": 0.7431680086124812, "lm_q1q2_score": 0.6127137344128389}} {"text": "\\section{Asymptotic Analysis for Closed Systems}\n\\label{sec:Asymptotic-Analysis-Closed-Systems}\n\n\\begin{theorem}[Asymptotic Bounds for Closed Systems]\n\t\\label{thm:Asymptotic-Bounds-Closed-Systems}\n\tFor any closed system with load $N$, we have that\n\t\n\t\\begin{equation}\n\t\\label{eqn:Asymptotic-Bounds-Closed-Systems}\n\t\\begin{split}\n\tX & \\leq \\min \\Big\\{ \\frac{N}{D + \\expected{Z}} , \\frac{1}{D_{max}} \\Big\\} \\\\\n\t\\expected{R} & \\geq \\max\\{D , N \\cdot D_{max} - \\expected{Z} \\}\n\t\\end{split}\n\t\\end{equation}\n\t\n\twhere the first term in each clause is an asymptote for small $N$, and the second one for large $N$.\n\t\n\t\\begin{proof}\n\t\tSee \\cite{harchol2013performance} for a formal demonstration.\n\t\\end{proof}\n\\end{theorem}\n\nIt is possible to get even tighter bounds by using the \\textit{Balanced Bounds technique}, described in \\cite{lazowska1984quantitative}.\n\nThe lower and the upper bound intercept in \n\\begin{equation}\n\\label{eqn:Asymptotic-Bounds-Closed-Systems-Interception}\nN^{*} = \\frac{D + \\expected{Z}}{D_{max}}\n\\end{equation}\nThat is the point beyond which there must be some queueing in the system, because $\\expected{R} > D$.\n\nIf $N > N^{*}$, performance can be improved ($X$ increased and $\\expected{R}$ decreased) only decreasing $D_{max}$.\n\nIf $N >> N^{*}$, decreasing some $D_{i} < D_{max}$ have no effect.\nIf $N << N^{*}$, decreasing some $D_{i} < D_{max}$ have little effect.\n\nIn a batch system ($Z=0$), $N^{*}$ decreases, meaning that the domination of $D_{max}$ occurs with smaller load.\n\nThe bottleneck server is the server with demand $D_{max}$. It is the key limiting factor to performance improvement. Thus, the first step in improving performance is to identify the bottleneck server.\n\nThe asymptotic bounds hold for open systems only if $\\lambda . X \\rightarrow \\frac{1}{D_{max}}$.", "meta": {"hexsha": "1494fc9de213a1099646042400a004d3b6a04d99", "size": 1799, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "performance-modeling/sec/asymptotic-analysis.tex", "max_stars_repo_name": "gmarciani/research", "max_stars_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-07-27T13:31:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-20T12:54:12.000Z", "max_issues_repo_path": "performance-modeling/sec/asymptotic-analysis.tex", "max_issues_repo_name": "gmarciani/research", "max_issues_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "performance-modeling/sec/asymptotic-analysis.tex", "max_forks_repo_name": "gmarciani/research", "max_forks_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-02-17T13:30:49.000Z", "max_forks_repo_forks_event_max_datetime": "2018-02-17T13:30:49.000Z", "avg_line_length": 43.8780487805, "max_line_length": 200, "alphanum_fraction": 0.7165091718, "num_tokens": 551, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430645886584, "lm_q2_score": 0.7371581684030623, "lm_q1q2_score": 0.6126838991730835}} {"text": "%\\section{A result on expectation}\n%\\begin{lemma} \\label{MC} \n%Let $\\lambda\\ge 0$ be a probability density function such that \n%\\begin{equation}\n% \\label{density}\n%\\int_{\\mathbb R^d}\\lambda(\\omega)dx =1.\n%\\end{equation}\n%\\begin{equation}\n% \\label{E}\n%\\mathbb{E}g:=\\int_{\\mathbb{R}^d}\n% g(\\omega)\\lambda(\\omega)d\\omega\n%\\end{equation}\n%and\n%\\begin{equation}\n% \\label{En}\n%\\mathbb{\\bar E}h:=\\int_{\\mathbb{R}^d\\times \\mathbb{R}^d\\times\\ldots\\times \\mathbb{R}^d}\n%h(\\omega_1,\\omega_2,\\ldots,\\omega_n) \\lambda(\\omega_1) \\lambda(\\omega_n)\\ldots \\lambda(\\omega_n)\n%d\\omega_1d\\omega_2\\ldots d\\omega_n.\n%\\end{equation}\n%\n%Then for any $g\\in L^\\infty(\\mathbb{R}^d)$, we have\n% \\begin{equation}\n% \\mathbb{\\bar E}(\\mathbb{E}g-\\frac1n\\sum_{i=1}^n\n% g(\\omega_i))^2=\\frac{1}{n}(\\mathbb{E}(g-\\mathbb{E}(g))^2) \n%\\le \\frac{1}{n}\\mathbb{E}(g^2)\\le \\frac{1}{n}\\|g\\|^2_{L^\\infty},\n% \\end{equation}\n% where \n%\\end{lemma}\n%\\begin{proof}\n% \\begin{equation}\n% \\label{eqn}\n% \\begin{aligned}\n% &\\quad \\mathbb{\\bar E}\\left(\\mathbb{E} g-\\frac1n\\sum_{i=1}^ng(\\omega_i)\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\left(n\\mathbb{E} g-\\sum_{i=1}^ng(\\omega_i)\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\left(\\sum_{i=1}^n(\\mathbb{E} g-g(\\omega_i))\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\sum_{i,j=1}^n(\\mathbb{E} g-g(\\omega_i))(\\mathbb{E} g-g(\\omega_j))\\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\sum_{i=1}^n(\\mathbb{E} g-g(\\omega_i))^2\\\\\n%&+\\frac{1}{n^2} \\mathbb{\\bar E}\\sum_{i\\neq j}^n((\\mathbb{E}g)^2-\\mathbb{E}(g)(g(\\omega_i)+\n%g(\\omega_j))+g(\\omega_i)g(\\omega_j)))\n% \\end{aligned}\n% \\end{equation}\n%We note that\n%$$\n%\\mathbb{\\bar E} g(\\omega_i)=\\mathbb{\\bar E} g(\\omega_j) =\\mathbb E(g) \n%$$\n%and, for $i\\neq j$, \n%$$\n%\\mathbb{\\bar E} ( g(\\omega_i)g(\\omega_j)) =\\mathbb{\\bar E} (\n%g(\\omega_i))\\mathbb{\\bar E}(g(\\omega_j))\n%=[\\mathbb E(g)]^2.\n%$$\n%Thus\n% \\begin{equation}\n%\\mathbb{\\bar E}\\sum_{i\\neq j}^n((\\mathbb{E}g)^2-\\mathbb\n% E(g)(g(\\omega_i)+ g(\\omega_j))+(g(\\omega_i)g(\\omega_j)))=0.\n% \\end{equation}\n%Furthermore\n%$$\n%\\mathbb{\\bar E}(\\mathbb{E} g-g(\\omega_i))^2\n%=\\mathbb{\\bar E}(\\mathbb{E} g-g(\\omega))^2 \n%=[\\mathbb{E} g]^2-2[\\mathbb E g(\\omega)]^2 +\\mathbb E(g^2) \n%=\\mathbb E(g^2) -[\\mathbb E(g)]^2 \n%\\le\\mathbb E(g^2) \n%$$\n%Consequently\n% \\begin{equation}\n% \\label{eqn}\n% \\begin{aligned}\n% &\\quad \\mathbb{\\bar E}\\left(\\mathbb{E} g-\n% \\frac1n\\sum_{i=1}^ng(\\omega_i)\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\left(n\\mathbb{E} g-\n% \\sum_{i=1}^ng(\\omega_i)\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\mathbb{\\bar E}\\left(\\sum_{i=1}^n(\\mathbb{E}\n% g-g(\\omega_i))\\right)^2 \\\\\n% &=\\frac{1}{n^2} \\sum_{i=1}^n(\\mathbb{\\bar E}(\\mathbb{E} g-g(\\omega_i))^2) \\\\\n% &=\\frac{1}{n}(\\mathbb{E}(g-\\mathbb{E}(g))^2) \\\\\n% &=\\frac{1}{n}(\\mathbb{E}(g^2)-(\\mathbb{E}(g))^2) \\\\\n% &\\le\\frac1n \\mathbb E(g^2)\\\\\n% &\\le\\frac1n\\|g\\|^2_{L^\\infty}\n% \\end{aligned}\n% \\end{equation}\n%\\end{proof}\n%\n%The above result be extended to the following general case easily.\n%\\begin{lemma} \\label{MC1} Given $1\\le i\\le n$, let $\\Omega $ be a\n% probability space with a probability measure $\\mu$ with density\n% function $\\lambda\\ge 0$ ($1\\le i\\le n$), namely \n%$$\n%\\mu(G)=\\int_G \\lambda(\\omega)d\\mu, \\quad \n%\\mu(\\Omega)=1, \\quad G\\subset \\Omega.\n%$$\n%Define\n%\\begin{equation}\n%\\label{Ei}\n%\\mathbb{E}g:=\\int_{\\Omega} g(\\omega)\\lambda(\\omega)d\\mu\n%\\end{equation}\n%and\n%\\begin{equation}\n% \\label{En}\n%\\mathbb{\\bar E}h:=\\int_{\\Omega\\times\\Omega\\times\\ldots\\times\\Omega}\n%h(\\omega_1,\\omega_2,\\ldots,\\omega_n) \\lambda(\\omega_1) \\lambda(\\omega_n)\\cdots \\lambda(\\omega_n)\n%d\\mu_1d\\mu_2\\ldots d\\mu_n.\n%\\end{equation}\n%Then for any $g\\in L^\\infty(\\Omega)$, we have\n% \\begin{equation}\n% \\mathbb{\\bar E}(\\mathbb{E}g-\\frac1n\\sum_{i=1}^n\n% g(\\omega_i))^2=\\frac{1}{n}(\\mathbb{E}(g-\\mathbb{E}(g))^2) \n%\\le \\frac{1}{n}\\|g\\|^2_{L^\\infty}.\n% \\end{equation}\n%\\end{lemma} \n%\\section{Fourier transformation and Barron spaces}\n%Given a bounded domain $B\\subset\\mathbb R^d$, we consider the function\n%$$\n%f: B\\mapsto \\mathbb R\n%$$\n%Let \n%$$\n%f^e: \\mathbb R^d\\mapsto \\mathbb R\n%$$ \n%be any extension of $f$ so that\n%$$\n%f^e|_B=f(x), \\quad x\\in B. \n%$$\n%Most time, we will drop the superscript $``e\"$ to still use $f$ to\n%denote an extension of $f$. \n%\n%Consider the Fourier transform:\n%\\begin{equation}\n% \\label{Fourier}\n% \\hat f(\\omega)=\\frac{1}{(2\\pi)^d}\\int_{\\mathbb{R}^d}e^{-i\\omega\\cdot x}f(x)dx\n% \\quad \\forall \\omega \\in \\mathbb R^d,\n%\\end{equation}\n%Using the Fourier inversion formula, we can have a Fourier\n%representation of $f(x)$ as follows\n%\\begin{equation}\n% \\label{eqn1}\n% f(x)=\\int_{\\mathbb{R}^d}e^{i\\omega\\cdot x}\\hat{f}(\\omega)d\\omega\n% \\quad \\forall x \\in B,\n%\\end{equation}\n%From \\eqref{eqn1}, we have\n%\\begin{equation}\n% \\label{f0}\n% f(x_B)=\\int_{\\mathbb{R}^d}e^{i\\omega\\cdot x_B}\\hat{f}(\\omega)d\\omega.\n%\\end{equation}\n%Let us write\n%\\begin{equation}\n% \\label{theta-omega}\n% \\hat{f}(\\omega)=e^{i\\theta(\\omega)}|\\hat{f}(\\omega)|. \n%\\end{equation}\n%Since $f(x)$ is real-valued, it follows from \\eqref{eqn1} and\n%\\eqref{f0} that, for $x, x_B\\in B$\n% \\begin{equation}\n% \\label{eqn4.3}\n% \\begin{aligned}\n%f(x)-f(x_B)\n%&={\\rm Re}\\int_{\\mathbb{R}^d}\n%(e^{i\\omega\\cdot x}-e^{i\\omega\\cdot x_B}) \n%\\hat{f}(\\omega)d\\omega \\\\\n%&={\\rm Re}\\int_{\\mathbb{R}^d}\n%(e^{i\\omega\\cdot x}-e^{i\\omega\\cdot x_B}) \n% e^{i\\theta\n% (\\omega)}|\\hat{f}(\\omega)|d\\omega \\\\\n% &=\\int_{\\mathbb{R}^d}(\\cos(\\omega\\cdot\n% x+\\theta(\\omega))-\\cos(\\omega\\cdot x_B+\\theta(\\omega)))|\\hat{f}(\\omega)|d\\omega \\\\\n%&=\\int_{\\mathbb{R}^d}(\\cos(\\omega\\cdot(x-x_B)+\\theta_B(\\omega))-\\cos(\\theta_B(\\omega)))|\\hat{f}(\\omega)|d\\omega \\\\\n%&=\\int_{\\mathbb{R}^d}g(x,\\omega)|\\hat{f}(\\omega)|d\\omega \\\\\n%&=\\|f\\|_{B^m}\\int_{\\mathbb{R}^d}|\\omega|_B^{-m}g(x,\\omega)\\lambda^{m}(\\omega)d\\omega \n%%\\\\&=\\int_{\\mathbb{R}^d}\\|f\\|_{B^m}\n%%(\\cos(\\omega\\cdot x+\\theta(\\omega))-\\cos(\\omega\\cdot x_B+\\theta(\\omega)))\\lambda(\\omega)d\\omega \\\\\n%% &=\\int_{\\mathbb{R}^d} g(x,\\omega)\\lambda(\\omega)d\\omega,\n%\\end{aligned}\n% \\end{equation}\n%where\n%$$\n%\\theta_B(\\omega)=\\omega\\cdot x_B+\\theta(\\omega)\n%$$\n%and $g: B\\times \\mathbb{R}^d\\rightarrow \\mathbb{R}$\n%is given by\n%\\begin{equation}\\label{gz}\n% g(x,\\omega):=\n%\\cos(\\omega\\cdot (x-x_B)+\\theta_B(\\omega)) -\\cos(\\theta_B(\\omega)),\n%\\end{equation}\n%\n%\\begin{equation}\n%\\label{lambda}\n%\\lambda^m(\\omega)=\\frac{|\\omega|_B^m|\\hat{f}(\\omega)|}{\\|f\\|_B^m}\n%\\end{equation}\n%is a probability distribution density function. \n%\n%We define \n%\\begin{equation}\n%\\|f\\|_{B^m}:=\\int_{\\mathbb R^d}|\\omega|_B^m|\\hat{f}(\\omega)|d\\omega,\n%\\end{equation}\n%\\begin{equation}\n%\\|f\\|_{B^m}:=\\inf_{f^e|_B=f}\\int_{\\mathbb R^d}|\\omega|_B^m|\\hat{f^e}(\\omega)|d\\omega,\n%\\end{equation}\n%where \n%$$\n%|\\omega|_B:=\\sup\\limits_{x\\in B}|\\omega\\cdot(x-x_B)|\n%$$\n%We note that, if, for example, $B=B_r=\\{x:|x|\\le r\\}$, we then have $|\\omega|_{B_r}\n%=r\\cdot |\\omega|$ if $x_B=0$. In general\n%$$\n%|\\omega|_B\\le {\\rm diam}(B)|\\omega|. \n%$$\n%\n%We then obtain the following result. \n%\\begin{lemma}\n%\\label{lem2.2}\n%\\begin{equation}\n%\\frac{f(x)-f(x_B)}{\\|f\\|_{B^m}}\n%=\\int_{\\mathbb R^d} \\frac{g(x,\\omega)}{|\\omega|^m}\\lambda(\\omega)d\\omega =\\mathbb E_\\omega(\\frac{g(x, \\omega)}{|\\omega|_B^m})\n%\\end{equation}\n%and \n%$g$ is given by \\eqref{gz} and satisfies\n%\\begin{equation}\\label{g-bound1}\n%|g(x,\\omega)|\\le \\min(2,|\\omega|_B)\\quad \\forall x\\in B, \\omega\\in \\mathbb\n%R^d. \n%\\end{equation}\n%\\end{lemma}\n%\n%\\begin{theorem}\n%\\label{thm:cos} For a probability measure $\\mu$ on $B$ and every\n% function $f$ with $\\|\\hat f\\|_{L^1}< \\infty$, there exist\n% $\\omega_1,\\ldots,\\omega_n\\in \\mathbb R^d$ such that\n%\\begin{equation}\n% \\int_{B}\\left|f(x)-f(x_B)-\\frac{\\|\\hat f\\|_{L^1}}{n}\n% \\sum_{i=1}^ng(x,\\omega_i)\\right|^2\\/d\\mu(x) \\le \\frac{4\\|\\hat f\\|^2_{L^1}}{n}.\n%\\end{equation}\n%\\end{theorem}\n%\\begin{theorem}\n% \\label{thm2.1} For a probability measure $\\mu$ on $B$ and every\n% function $f$ with $\\|f\\|_{B}< \\infty$, there exist\n% $\\omega_1,\\ldots,\\omega_n\\in \\mathbb R^d$ such that\n%\\begin{equation}\n%\\int_{B}\\left|f(x)-f(x_B)-\\frac{\\|f\\|_{B}}{n} \\sum_{i=1}^n\\frac{g(x,\\omega_i)}{|\\omega_i|}\\right|^2\\/d\\mu(x) \\le \\frac{\\|f\\|^2_B}{n}.\n%\\end{equation}\n%\\end{theorem}\n%\\begin{proof}\n%Denote\n%$$\n%\\tilde f(x)=\\frac{f(x)-f(x_B)}{\\|f\\|_{B^m}}\n%$$\n%By Lemma \\ref{MC} and Lemma \\ref{lem2.2}, we have for $m=0,1$,\n%\\begin{equation}\n% \\label{eqn3.10}\n% \\begin{aligned}\n% &\\quad \\mathbb{\\bar E}\\left(\\tilde f(x)-\\frac1n\\sum_{i=1}^n\\frac{g(x,\\omega_i)}{|\\omega|_B^m}\\right)^2 \\\\\n% &=\\mathbb{\\bar E}(\\mathbb{E}_\\omega(\\frac{g(x,\\omega)}{|\\omega|_B^m})-\n% \\frac1n\\sum_{i=1}^n \\frac{g(x,\\omega_i)}{|\\omega|_B^m})^2 \\\\\n% &\\le\\frac{1}{n}\\max_{x\\in B, \\omega\\in \\mathbb R^d}\\frac{|g(x,\\omega)|^2}{|\\omega|_B^{2m}}\\\\\n% &\\le \\frac{4^{1-m}}{n}\n% \\end{aligned}\n%\\end{equation}\n%\n%By the Fubini's theorem,\n% \\begin{equation}\n% \\label{eqn3.11}\n% \\begin{aligned}\n% &\\quad \\mathbb{\\bar E}\\int_B(\\tilde f(x)-\\frac1n\\sum_{i=1}^n\\frac{g(x,\\omega_i)}{|\\omega_i|_B^m})^2 \\mu(dx)\\\\\n% &=\\int_B \\mathbb{\\bar E} (\\tilde f(x)-\\frac1n\\sum_{i=1}^n\\frac{g(x,\\omega_i)}{|\\omega_i|_B^m})^2 \\mu(dx)\\\\\n% &\\le \\frac{4^{1-m}}{n}.\n% \\end{aligned}\n% \\end{equation}\n%Therefore, there exists a sequence of $\\{\\omega_i\\}_{i=1}^n$ where\n%$\\omega_i\\in \\mathbb{R}^d$, such that \n%$$\n%\\int_B (\\tilde\n%f(x)-\\frac1n\\sum_{i=1}^n\\frac{g(x,\\omega_i)}{|\\omega_i|_B^m})^2\n%\\mu(dx)\n%\\le \\frac{4^{1-m}}{n}.\n%$$\n%This complete the proof. \n%\\end{proof}\n%Recalling \\eqref{gz}, we have\n%\\begin{eqnarray*}\n%f_n\n%&\\equiv& f(0)+\\frac{\\| f\\|_B}{n} \\sum_{i=1}^n\\frac{g(\\cdot,\\omega_i)}{|\\omega_i|_B}\\\\\n%&=& f(0)+\\frac1n \\sum_{i=1}^n \\frac{\\|f\\|_B}{|\\omega_i|_B}(\\cos(\\omega_i\\cdot x+\\theta(\\omega_i)) -\\cos(\\theta(\\omega_i)))\\\\\n%&=& \n%\\frac1n \\sum_{i=1}^n \\frac{\\|f\\|_B}{|\\omega_i|_B}(\\cos(\\omega_i\\cdot\n%x+\\theta(\\omega_i)) +\\left(\n%f(0)-\\frac1n \\sum_{i=1}^n \\frac{\\|f\\|_B}{|\\omega_i|_B}\\cos(\\theta(\\omega_i)) \\right)\\\\\n%&=& \n%\\sum_{i=1}^n a_i\\cos(\\omega_i\\cdot x+b_i) +c\n%\\end{eqnarray*}\n%where\n%$$\n%a_i=\\frac1n \\frac{\\|f\\|_B}{|\\omega_i|_B}, b_i=\\theta(\\omega_i), c=f(0)-\\frac1n \\sum_{i=1}^n \\frac{\\|f\\|_B}{|\\omega_i|_B}\\cos(\\theta(\\omega_i)).\n%$$\n%\n%\\begin{theorem}\n% \\begin{equation}\n%\\min_{a_i,b_i,c\\in \\mathbb R^1, \\omega_i\\in \\mathbb R^d}\n%\\int_{B}|f(x)-\\sum_{i=1}^n a_i\\cos(\\omega_i\\cdot x+b_i)\n%-c|^2\\mu(dx)\\le \\frac{4\\|\\hat f\\|^2_{L^1(\\mathbb R^d)}}{n}.\n% \\end{equation}\n% \\end{theorem}\n\n\\section{An improved analysis}\n\\subsection{Heaviside Function}\nDefine $g_i: [-1,1]\\mapsto \\mathbb R$ as\nfollows:\n\\begin{equation}\n \\label{psi}\ng_i(t)=\\frac{1}{|\\omega_i|_B}[\\cos(|\\omega_i|_Bt+\\theta_B(\\omega_i)) -\\cos(\\theta_B(\\omega_i))],\n\\end{equation}\nIn view of \\eqref{gz}, we have\n\\begin{equation}\n \\label{gpsi}\ng_i(s_i)=\\frac{g(x,\\omega_i)}{|\\omega_i|_B}, \\quad s_i=\\omega_i^B\\cdot(x-x_B),\\quad \\omega^B=\\frac{\\omega}{|\\omega|_B}\n\\end{equation}\nNow, we take an integer\n\\begin{equation}\n \\label{k}\nk\\ge \\sqrt{n} \n\\end{equation}\nand consider a partition of $[-1,1]$ with the following grid points\n$$\nt_j=jh_k, j=-k:k\n$$\nwith \n$$\n h_k=\\frac{1}{k}\\le \\frac{1}{\\sqrt{n}}.\n$$\nWe first take a piecewise constant interpolation for $g_i$ on $[0,1]$ to get\n$$\ng_{i,k}(t)=(\\Pi_kg_i)(t)=\\sum_{j=0}^{k-1}g_i(t_j) M_j(t), \n$$\nwhere\n$$\nM_j(t)=M_0(\\frac{t-t_j}{h_k})\n$$\nand\n\\begin{equation}\n \\label{cardinal}\nM_0(x)=\n\\left\\{\n \\begin{array}{ll}\n0 & x\\le0 \\\\\n1 & 0< x\\le1 \\\\\n0 & x > 1 \n \\end{array}\n\\right.\n\\end{equation}\nWe note that\n$$\nM_0(x)=H(x)-H(x-1)\n$$\nwhere $H$ is the Heaviside function\nThus\n$$\nM_0(\\frac{t-t_{j}}{h_k})\n=H(\\frac{t-t_{j}}{h_k})-H(\\frac{t-t_{j}}{h_k}-1)=H(\\frac{t-t_{j}}{h_k})-H(\\frac{t-t_{j+1}}{h_k})\n\\equiv H_{j}(t)-H_{j+1}(t).\n$$\nThus, since $g_i(t_0)=0, H_k=0$, we have\n\\begin{equation} \\label{gi0}\ng_{i,k}(t)=\\sum_{j=0}^{k-1}g_i(t_j) M_j(t)\n=\\sum_{j=1}^{k-1}(g_i(t_j) - g_i(t_{j-1})) H_{j}(t), \\quad t\\in [0,1]\n\\end{equation}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\nNow we consider\n\\begin{equation}\nh_i(t) = g_i(-t), \\quad t\\in [0,1].\n\\end{equation}\nSimilar to \\eqref{gi0}, we have\n$$\n(\\Pi_kh_i)(t)=\\sum_{j=1}^{k-1}(h_i(t_j) - h_i(t_{j-1}))H_j(t)=\n=\\sum_{j=1}^{k-1}(g_i(-t_j) - g_i(-t_{j-1}))H_j(t)\n$$\nNamely\n$$\n(\\Pi_k g_i)(-t)=\\sum_{j=1}^{k-1}(g_i(-t_j) - g_i(-t_{j-1}))H_j(t), \\quad t\\in [0,1]\n$$\nor\n\\begin{equation}\\label{gi1}\n(\\Pi_k g_i)(t)=\\sum_{j=1}^{k-1}(g_i(-t_j) - g_i(-t_{j-1}))H_j(-t), \\quad t\\in [-1,0]\n\\end{equation}\nBy combining \\eqref{gi0} and \\eqref{gi1}, we get a piecewise constant\ninterpolation of $g_i$ on $[-1,1]$ as follows:\n\\begin{eqnarray}\ng_{i,k}(t)&=&\n\\sum_{j=1}^{k-1}(g_i(-t_j) - g_i(-t_{j-1}))H_j(-t)+\\sum_{j=1}^{k-1}(g_i(t_j) - g_i(t_{j-1})) H_{j}(t)\\nonumber \\\\ \n&=&\\sum_{j=1}^{k-1}[a_{ij}^-H_j(-t)+a_{ij}^+H_{j}(t)] \\label{gih}\n\\quad t\\in [-1,1]\n\\end{eqnarray}\nwhere \n$$\na_{ij}^{\\pm}=g_i(\\pm t_j) - g_i(\\pm t_{j-1})\n$$\nIt is easy to see that\n\\begin{equation}\n|g_i(t)-g_{i,k}(t)|\\le h_k, \\quad t\\in [-1,1].\n\\end{equation} \n\n\\begin{equation}\n\\|\\frac1n \\sum_{i=1}^n \\frac{g(\\cdot,\\omega_i)}{|\\omega_i|_B}-f^*\\|_{L^2(\\mu,B)}\\le h_k\n\\end{equation}\nwhere\n\\begin{equation}\n \\label{fstar}\nf^*(x)=\n\\frac1n\\sum_{i=1}^ng_{i,k}(\\omega_i^B\\cdot (x-x_B)).\n\\end{equation}\nBy the approximation in last section, we have \n\\begin{equation}\n \\|\\tilde f-f^*\\|_{L^2(\\mu,B)}\\le \\frac{2}{\\sqrt{n}}\n\\end{equation}\nLet us rewrite\n$$\nf^*(x)\n=\\sum_{i=1}^n\\sum_{j=1}^{k-1}[\\gamma_{ij}^- f_{ij}^-+ \\gamma_{ij}^+f_{ij}^+]\n$$\nwhere\n$$\n\\gamma_{ij}^{\\pm}=\\frac{|a^\\pm_{ij}|}{nd_i}, \nf^\\pm_{i,j}=d_i\\/{\\rm sign}(a^\\pm_{ij})H_j(\\pm \\omega_i^B\\cdot (x-x_B))\n$$\nand \n$$\nd_i=\\sum_{j=1}^{k-1}(|a^-_{ij}|+|a^+_{ij}|)\\le 2\n$$\nBy definition\n\\begin{equation}\n \\label{gammaij}\n\\sum_{i=1}^n\\sum_{j=1}^{k-1}[\\gamma_{i,j}^-+\\gamma_{i,j}^+]=1.\n\\end{equation}\nWith re-numeration as\n$$\np_\\ell=\\gamma_{ij}^{\\pm}, f_\\ell = f_{ij}^{\\pm}, 1\\le \\ell \\le N=2n(k-1)\n$$\nWe have\n$$\nf^*(x)=\\sum_{\\ell=1}^N p_\\ell f_\\ell\n$$\nConsider \n$$\n\\mathcal N=\\{1,2,\\ldots, N\\}\n$$\nand \n$$\n\\bar f: \\mathcal N\\mapsto \\mathbb R^1\n$$\nsuch that\n$$\n\\bar f(\\ell)=f_{\\ell}, \\ell\\in \\mathcal N\n$$\nWith the probability measure\n$$\n\\mu(\\mathcal M)=\\sum_{m\\in \\mathcal M}p_m \\quad \\mathcal M\\subset\\mathcal N.\n$$\nBy definition. \n$$\n\\mathbb E(\\bar f) = f^*(x).\n$$\nBy the basic result on expectation in Lemma \\ref{MC1}, we have\n$$\n\\sum_{\\ell_1,\\ldots \\ell_n=1}^Np_{\\ell_1}\\cdots p_{\\ell_n}\\left(f^*(x)-\n{1\\over n}\\sum_{i=1}^n f_{\\ell_i}\\right)^2\n=\\mathbb E_{\\mathcal N^n} \\left(\\mathbb E(\\bar f)-{1\\over\n n}\\sum_{i=1}^n \\bar f(\\ell_i)\\right)^2\\\\\n\\le\\frac1n \\|\\bar f\\|_{\\infty}^2 \\le\\frac4n.\n$$\nBy taking the $L^2(\\mu,B)$ on the above inequality, we get\n$$\n\\sum_{\\ell_1,\\ldots \\ell_n=1}^Np_{\\ell_1}\\cdots p_{\\ell_n}\\|f^*-{1\\over n}\\sum_{i=1}^n f_{\\ell_i}\\|_{L^2(\\mu,B)}^2\n\\le\\frac4n.\n$$\nThus, there exisit $\\ell_1^*, \\ldots, \\ell_n^*\\in \\mathcal N$ such that\n$$\n\\|f^*-{1\\over n}\\sum_{i=1}^n f_{\\ell_i^*}\\|_{L^2(\\mu,B)}^2\n\\le\\frac4n.\n$$\nwhere\n$$\nf_n(x)={1\\over n}\\sum_{i=1}^n f_{\\ell_i^*}(x).\n$$\nThen we have \n\\begin{equation}\n\\|\\tilde f-f_n\\|^2_{L^2(\\mu,B)}\\le \\frac{9}{n}.\n\\end{equation}\nConsequently\n\\begin{equation}\n\\left\\|f(x)-f(x_B)-\\|f\\|_Bf_n\\right\\|^2_{L^2(\\mu,B)}\\le \\frac{9\\|f\\|^2_B}{n}.\n\\end{equation}\n\n\\subsection{Piecewise linear function}\nThe proof here is almost the same as the proof for Heaviside function in the last part. \n\nNow we take a piecewise linear interpolation for $g_i$ on $[0,1]$, since $g_i(t_0)=0$ , we get\n$$\ng_{i,k}(t)=(\\Pi_kg_i)(t)=\\sum_{j=1}^{k}[g_i(t_{j})-g_i(t_{j-1})]\\sigma_{j-1}(t), \\quad t\\in [0,1]\n$$\nwhere\n$$\n\\sigma_j(t)=M_0(\\frac{t-t_j}{h_k})\n$$\nand\n\\begin{equation}\n\\label{cardinal}\nM_0(x)=\n\\left\\{\n\\begin{array}{ll}\n0 & x\\le0 \\\\\nx & 0< x\\le1 \\\\\n1 & x > 1 \n\\end{array}\n\\right.\n\\end{equation}\n%We note that\n%$$\n%M_0(x)=ReLU(x)-ReLU(x-1)\n%$$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%\nConsider\n\\begin{equation}\nh_i(t) = g_i(-t), \\quad t\\in [0,1].\n\\end{equation}\nSimilarly, we have\n$$\n(\\Pi_kh_i)(t)=\\sum_{j=1}^{k}(h_i(t_j) - h_i(t_{j-1}))\\sigma_{j-1}(t)=\n=\\sum_{j=1}^{k}(g_i(-t_j) - g_i(-t_{j-1}))\\sigma_{j-1}(t)\n$$\nNamely\n$$\n(\\Pi_k g_i)(-t)=\\sum_{j=1}^{k}(g_i(-t_j) - g_i(-t_{j-1}))\\sigma_{j-1}(t), \\quad t\\in [0,1]\n$$\nor\n\\begin{equation}\\label{gi1}\n(\\Pi_k g_i)(t)=\\sum_{j=1}^{k}(g_i(-t_j) - g_i(-t_{j-1}))\\sigma_{j-1}(-t), \\quad t\\in [-1,0]\n\\end{equation}\nCombine together we get a piecewise linear\ninterpolation of $g_i$ on $[-1,1]$ as follows:\n\\begin{eqnarray}\ng_{i,k}(t)&=&\n\\sum_{j=1}^{k}(g_i(-t_j) - g_i(-t_{j-1}))\\sigma_{j-1}(-t)+\\sum_{j=1}^{k}(g_i(t_j) - g_i(t_{j-1}))\\sigma_{j-1}(t)\\nonumber \\\\ \n&=&\\sum_{j=1}^{k}[a_{ij}^-\\sigma_{j-1}(-t)+a_{ij}^+\\sigma_{j-1}(t)] \\label{gih}\n\\quad t\\in [-1,1]\n\\end{eqnarray}\nwhere \n$$\na_{ij}^{\\pm}=g_i(\\pm t_j) - g_i(\\pm t_{j-1})\n$$\n%It is easy to see that\n%\\begin{equation}\n%|g_i(t)-g_{i,k}(t)|\\le h_k, \\quad t\\in [-1,1].\n%\\end{equation} \n%\\begin{equation}\n%\\|\\frac1n \\sum_{i=1}^n \\frac{g(\\cdot,\\omega_i)}{|\\omega_i|_B}-f^*\\|_{L^2(\\mu,B)}\\le h_k\n%\\end{equation}\n%where\n%\\begin{equation}\n%\\label{fstar}\n%f^*(x)=\n%\\frac1n\\sum_{i=1}^ng_{i,k}(\\omega_i^B\\cdot (x-x_B)).\n%\\end{equation}\n%\n%By Theorem \\ref{jones}, we have \n%\\begin{equation}\n%%\\|\\tilde f-f^*\\|_{1,B}\\le \\frac{C_1(d,B)}{\\sqrt{n}}\n% \\|\\tilde f-f^*\\|_{L^2(\\mu,B)}\\le \\frac{2}{\\sqrt{n}}\n%\\end{equation}\n%here $C_1(d,B)=\\sqrt{\\mu(B)}[\\sqrt{(d+{\\rm diam}(B))}+{\\rm diam}(B)(1+d)]$.\n\n\n%Let us rewrite\n%$$\n%f^*(x)\n%=\\sum_{i=1}^n\\sum_{j=1}^{k}[\\gamma_{ij}^- f_{ij}^-+ \\gamma_{ij}^+f_{ij}^+]\n%$$\n%where\n%$$\n%\\gamma_{ij}^{\\pm}=\\frac{|a^\\pm_{ij}|}{nd_i}, \n%f^\\pm_{i,j}=d_i\\/{\\rm sign}(a^\\pm_{ij})\\sigma_{j-1}(\\pm \\omega_i^B\\cdot (x-x_B))\n%$$\n%and \n%$$\n%d_i=\\sum_{j=1}^{k}(|a^-_{ij}|+|a^+_{ij}|)\\le 2\n%$$\n%By definition\n%\\begin{equation}\n%\\label{gammaij}\n%\\sum_{i=1}^n\\sum_{j=1}^{k}[\\gamma_{i,j}^-+\\gamma_{i,j}^+]=1.\n%\\end{equation}\n%With re-numeration as\n%$$\n%p_\\ell=\\gamma_{ij}^{\\pm}, f_\\ell = f_{ij}^{\\pm}, 1\\le \\ell \\le N=2nk\n%$$\n%We have\n%$$\n%f^*(x)=\\sum_{\\ell=1}^N p_\\ell f_\\ell\n%$$\n%Consider \n%$$\n%\\mathcal N=\\{1,2,\\ldots, N\\}\n%$$\n%and \n%$$\n%\\bar f: \\mathcal N\\mapsto \\mathbb R^1\n%$$\n%such that\n%$$\n%\\bar f(\\ell)=f_{\\ell}, \\ell\\in \\mathcal N\n%$$\n%With the probability measure\n%$$\n%\\lambda(\\mathcal M)=\\sum_{m\\in \\mathcal M}p_m \\quad \\mathcal M\\subset\\mathcal N.\n%$$\n%By definition. \n%$$\n%\\mathbb E(\\bar f) = f^*(x).\n%$$\n%\n%By the basic result on expectation:\n%\\begin{equation}\n%\\begin{aligned}\n%&\\sum_{\\ell_1,\\ldots \\ell_n=1}^Np_{\\ell_1}\\cdots p_{\\ell_n}\\left((f^*(x)-\n%{1\\over n}\\sum_{i=1}^n f_{\\ell_i})^2\\right)\\\\\n%=&\\mathbb E_{\\mathcal N^n} \\left(\\mathbb E(\\bar f)-{1\\over\n%\tn}\\sum_{i=1}^n \\bar f(\\ell_i)\\right)^2\n%\\le\\frac1n \\|\\bar f\\|_{\\infty}^2 \\le\\frac{4}{n}.\n%\\end{aligned}\n%\\end{equation}\n%\n%By taking the $L^2(\\mu,B)$ on the above inequality, we get\n%$$\n%\\sum_{\\ell_1,\\ldots \\ell_n=1}^Np_{\\ell_1}\\cdots p_{\\ell_n}\\|f^*-{1\\over n}\\sum_{i=1}^n f_{\\ell_i}\\|_{L^2(\\mu,B)}^2\n%\\le\\frac4n.\n%$$\n%Thus, there exisit $\\ell_1^*, \\ldots, \\ell_n^*\\in \\mathcal N$ such that\n%$$\n%\\|f^*-f_n(x)\\|_{L^2(\\mu,B)}^2\n%\\le\\frac4n.\n%$$\n%where\n%$$\n%f_n(x)={1\\over n}\\sum_{i=1}^n f_{\\ell_i^*}(x).\n%$$\n%Then we have \n%\\begin{equation}\n%\\|\\tilde f-f_n\\|^2_{L^2(\\mu,B)}\\le \\frac{9}{n}.\n%\\end{equation}\n\n%Consequently\n%\\begin{equation}\n%\\left\\|f(x)-f(x_B)-\\|f\\|_Bf_n\\right\\|^2_{L^2(\\mu,B)}\\le \\frac{9\\|f\\|^2_B}{n}.\n%\\end{equation}\n\nFollow the procedure in last section, and notice that $\\sigma(x)=ReLU(x)-ReLU(x-1)$, we obtain the following theorem.\n\n\n\\begin{theorem}\n\tFor a probability measure $\\mu$ on B and every function with $\\|f\\|_B<\\infty$, there exists $\\omega_1,\\dots,\\omega_n\\in\\mathbb{R}^d$ such that \n$$\n\\left\\|f(x)-f_n(x)\\right\\|^2_{L^2(\\mu,B)}\\le \\frac{C\\|f\\|^2_B}{n}.\n$$\nwhere $f_n(x)=\\sum_{i=1}^{n}a_iReLU(\\omega_i x+b_i)+c$.\n\\end{theorem}\n", "meta": {"hexsha": "09664d6f63e9e7756d01165db569a75533c88479", "size": 19538, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "6DL/X-Approx.tex", "max_stars_repo_name": "liuzhengqi1996/math452", "max_stars_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "6DL/X-Approx.tex", "max_issues_repo_name": "liuzhengqi1996/math452", "max_issues_repo_head_hexsha": "635b6ce53cb792e316abf4f47396f2e4f0686815", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, 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YES\n2. YES", "lm_q1_score": 0.8311430562234878, "lm_q2_score": 0.7371581684030623, "lm_q1q2_score": 0.6126838930066297}} {"text": "% !TEX root = main.tex\n\n\\section{Mathematics}\n\n\\begin{itemize}\n\\item\nAn inline equation with TeX delimiters (dollars): $\\sum_{n=1}^{\\infty} a_n = 1$.\n\\item\nAn inline equation with LaTeX delimiters: \\(\\sum_{n=1}^{\\infty} a_n = 1\\).\n\\end{itemize}\n\n\\bigskip\nAn \\texttt{equation} environment:\n\\begin{equation}\n\\sum_{n=1}^{\\infty} a_n = 1.\n\\end{equation}\n\nAn \\texttt{equation*} environment:\n\\begin{equation*}\n\\sum_{n=1}^{\\infty}\\frac{1}{n^2} = \\frac{\\pi^2}{6}.\n\\end{equation*}\n\nA \\texttt{display} environment with LaTeX delimiters:\n\\[\nx = \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}.\n\\]\n\nA \\texttt{display} environment with TeX delimiters (double-dollars):\n$$\nx = \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}.\n$$\n\nA \\texttt{display} environment containing a \\texttt{text} command, which itself contains (inline) mathmode:\n\\[\nE=mc^2 \\text{ where $c$ is the speed of light}.\n\\]\n\n\\subsection{Labelled equations}\n\nAn equation with a label:\n\\begin{equation}\\label{eq:euler}\ne^{i\\pi}+1=0.\n\\end{equation}\n\nEquation -\\eqref{eq:euler}- is due to Euler.\n\nAn \\texttt{align} environment:\n\\begin{align}\\label{eq:polar}\nx & = r\\sin\\theta \\\\\ny & = r\\cos\\theta\n\\end{align}\n\n%\\bigskip\n%A labelled \\texttt{equation} environment:\n%\\begin{equation}\\label{eq:einstein}\n%E = mc^2\n%\\end{equation}\n%Equation -\\eqref{eq:einstein}- is due to AE. \n\n\\bigskip\nAn {\\tt equation} containing three {\\tt pmatrix} environments:\n\\begin{equation}\n\\begin{pmatrix}0 & 1 \\\\ 1 & 0\\end{pmatrix}\n\\qquad\n\\begin{pmatrix}0 & -i \\\\ i & 0\\end{pmatrix}\n\\qquad\n\\begin{pmatrix}1 & 0 \\\\ 0 & -1\\end{pmatrix}\n\\end{equation}\n\n", "meta": {"hexsha": "f3cafca718b02253832e4b078206faa17c7b6b21", "size": 1540, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/test_article/_maths.tex", "max_stars_repo_name": "imagingbook/latextree", "max_stars_repo_head_hexsha": "272ee1594b3bdea39a043fb2ac2b86ac9a1728e8", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2020-02-16T22:41:21.000Z", "max_stars_repo_stars_event_max_datetime": "2020-02-16T22:41:21.000Z", "max_issues_repo_path": "tex/test_article/_maths.tex", "max_issues_repo_name": "imagingbook/latextree", "max_issues_repo_head_hexsha": "272ee1594b3bdea39a043fb2ac2b86ac9a1728e8", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/test_article/_maths.tex", "max_forks_repo_name": "imagingbook/latextree", "max_forks_repo_head_hexsha": "272ee1594b3bdea39a043fb2ac2b86ac9a1728e8", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-09-11T09:38:25.000Z", "max_forks_repo_forks_event_max_datetime": "2019-09-11T15:30:21.000Z", "avg_line_length": 22.0, "max_line_length": 107, "alphanum_fraction": 0.6811688312, "num_tokens": 583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430478583168, "lm_q2_score": 0.7371581626286834, "lm_q1q2_score": 0.6126838820408407}} {"text": "% !TeX root = ../main.tex\n\\documentclass[../main.tex]{subfiles}\n\\begin{document}\n\\section{Geometric transformation}\n\\begin{lemma}\\label{M:transform:distance}\nFor any point \\(p,q\\in N\\)\nand element \\(X\\in X\\left(n,\\lambda\\right)\\),\n\\[\n\\norm{\\left(Xp\\right)\\left(Xq\\right)}=\\norm{pq}\\text{.}\n\\]\n\\end{lemma}\n\\begin{proof}[\\proofof{M:transform:distance}]\nIf \\(\\lambda>0\\),\n\\begin{align*}\nX^{-1}&=X^T\\\\\n&=X^TI\\\\\nI&=X^TIX\\\\\ng&=X^TgX\\text{.}\n\\end{align*}\nIf \\(\\lambda\\to 0\\),\n\\begin{align*}\nX&\\to I\\\\\nX^T\n&\\to I^T\\\\\n&=I\\\\\ng&=g\\\\\ng&=IgI\\\\\ng&\\to X^TgX\\text{.}\n\\end{align*}\nIf \\(\\lambda<0\\),\n\\begin{align*}\nX^{-1}&=gX^Tg\\\\\nI&=gX^TgX\\\\\ng^{-1}&=X^TgX\\\\\ng&=X^TgX\\text{.}\n\\end{align*}\n\\begin{align*}\n\\norm{pq}\n&=\\sqrt{g\\left(q-p,q-p\\right)}\\\\\n&=\\sqrt{\\left(q-p\\right)^Tg\\left(q-p\\right)}\\\\\n\\norm{\\left(Xp\\right)\\left(Xq\\right)}\n&=\\sqrt{g\\left(Xq-Xp,Xq-Xp\\right)}\\\\\n&=\\sqrt{g\\left(X\\left(q-p\\right),X\\left(q-p\\right)\\right)}\\\\\n&=\\sqrt{\\left(X\\left(q-p\\right)\\right)^Tg\\left(X\\left(q-p\\right)\\right)}\\\\\n&=\\sqrt{\\left(q-p\\right)^TX^TgX\\left(q-p\\right)}\\\\\n&=\\sqrt{\\left(q-p\\right)^Tg\\left(q-p\\right)}\\\\\n&=\\norm{pq}\\qedhere\n\\end{align*}\n\\end{proof}\n\\end{document}", "meta": {"hexsha": "59b0a87e628bf0ac95aa3867912c0303eb1a6c57", "size": 1144, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sections/transformation.tex", "max_stars_repo_name": "30MA19-02/proof", "max_stars_repo_head_hexsha": "24d95d419e5632a2cc08a9c9bc94e89e262afc04", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sections/transformation.tex", "max_issues_repo_name": "30MA19-02/proof", "max_issues_repo_head_hexsha": "24d95d419e5632a2cc08a9c9bc94e89e262afc04", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sections/transformation.tex", "max_forks_repo_name": "30MA19-02/proof", "max_forks_repo_head_hexsha": "24d95d419e5632a2cc08a9c9bc94e89e262afc04", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 22.88, "max_line_length": 74, "alphanum_fraction": 0.6092657343, "num_tokens": 534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430478583168, "lm_q2_score": 0.7371581568543043, "lm_q1q2_score": 0.6126838772415056}} {"text": "\\chapter{Mathematical derivations of MOFA+} \\label{appendix:mofa}\n\n\\section{Deriving the variational inference algorithm}\n\nThe theoretical foundations for the variational inference scheme are described in \\Cref{section:variational_inference}. Just to brief, we need to define a variational distribution of a factorised form and subsequently look for the member of this family that most closely resembles the true posterior using the KL divergence as a \\textit{distance} metric. Following the mean-field principle, in MOFA+ we factorised the variational distribution as follows:\n\n\\begin{equation} \\begin{aligned}\n\tq(\\bfX) &= q \\left(\\{\\widehat{\\bfZ^\\bfg}, \\bfS^\\bfg, \\bf{\\alpha^g}, \\bf{\\theta^g}\\}, \\{\\widehat{\\bfW^m}, \\bfS^m, \\bf\\alpha^m, \\bf\\theta^m\\}, \\{ \\bf\\tau^{gm}\\} \\right)\\\\\n\t&= \\prod_{g=1}^{G} \\prod_{n=1}^{N_g} \\prod_{k=1}^{K} q(\\hat{z}_{nk}^g, s_{nk}^g) \\prod_{g=1}^{G} \\prod_{k=1}^{K} q(\\alpha_k^g) \\prod_{g=1}^{G} \\prod_{k=1}^{K} q(\\theta_k^g) \\\\\n\t&\\times \\prod_{m=1}^{M} \\prod_{d=1}^{D_m} \\prod_{k=1}^{K} q(\\hat{w}_{kd}^m,s_{kd}^m) \\prod_{m=1}^{M} \\prod_{k=1}^{K} q(\\alpha_k^m) \\prod_{m=1}^{M} \\prod_{k=1}^{K} q(\\theta_k^m) \\\\\n\t&\\times \\prod_{g=1}^{G} \\prod_{m=1}^{M} \\prod_{d=1}^{D_m} q(\\tau_d^{gm})\n\\end{aligned} \\end{equation}\n\nHowever, inspired by \\cite{Titsias2011}, we did not adopt a fully factorised distribution as $\\hat{w}_k^m$ and $s_k^m$ can hardly be assumed to be independent.\n\nTo derive the variational updates we can proceed in two ways, as described in \\Cref{section:variational_inference}. One option is to use exploit the mean-field assumption and use calculus of variations to find the optimal distribution $q(\\bfX)$ that maximises the lower bound $\\Lagr(\\bfX)$\\cite{Bishop2006,Murphy}. The alternative and possibly easier approach is to define a parametric form for the distribution $q(\\bfX)$ with some parameters $\\bTheta$ to be of the same form as the corresponding prior distribution $p(\\bfX)$. Then, one can find the gradients with respect to the parameters to obtain the coordinate ascent optimisation scheme. In our derivations we followed the first approach, but because we used conjugate priors the second one should converge to the same result.\\\\\nBelow we give the explicit update equations for every hidden variable in the MOFA+ model which are applied at each iteration of the variational inference algorithm.\n\n\\section{Variational update equations}\n\n\\paragraph*{Factors}\n\nFor every group $g$, sample $n$ and factor $k$: \\\\\n\nPrior distribution $p(\\hat{z}_{nk}^g,s_{nk}^g)$:\n\\begin{align}\n\tp(\\hat{z}_{nk}^g,s_{nk}^g) &= \\mathcal{N} (\\hat{z}_{nk}^g \\,|\\, 0, ~ 1/\\alpha_k^g)\\, \\text{Ber}(s_{nk}^g \\,|\\,\\theta_k^g)\n\\end{align}\n\nVariational distribution $q(\\hat{z}_{nk}^g,s_{nk}^g)$:\\\\\n\nUpdate for $q(s_{nk}^g)$:\n\\begin{equation}\n\tq(s^g_{nk}) = \\mathrm{Ber}(s^g_{nk}|\\gamma^g_{nk})\n\\end{equation}\nwith\n\\begin{equation} \\begin{aligned}\n\t&\\gamma^g_{nk} = \\frac{1}{1+\\exp(-\\lambda_{nk}^g)}\\\\\n\t& \\lambda_{nk}^g = \\la \\ln\\frac{\\theta}{1-\\theta} \\ra + 0.5\\ln\\frac{\\la\\alpha_k^g\\ra}{\\la\\tau_d^{gm}\\ra} - 0.5\\ln\\left( \\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la (w_{kd}^m)^2 \\ra + \\frac{\\la\\alpha_k^g\\ra}{\\la\\tau_d^{gm}\\ra} \\right) \\\\\n\t&+ \\frac{\\la\\tau_d^{gm}\\ra}{2} \\frac{ \\left( \\sum_{m=1}^M\\sum_{d=1}^{D_m} y_{nd}^{gm} \\la w^m_{kd} \\ra - \\sum_{j \\neq k} \\la s_{nj}^g\\hat{z}_{nj}^g\\ra \\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la w^m_{kd} \\ra \\la w^m_{jd} \\ra \\right)^2} {\\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la (w_{kd}^m)^2 \\ra + \\frac{\\la\\alpha_k^g\\ra}{\\la\\tau_d^{gm}\\ra} }\n\\end{aligned} \\end{equation}\n\nUpdate for $q(\\hat{z}_{nk}^g)$:\n\\begin{equation} \\begin{aligned}\n q(\\hat{z}_{nk}^g|s_{nk}^g=0) &= \\mathcal{N} \\left(\\hat{z}_{nk}^g \\middle| 0, 1/\\alpha_k^g \\right) \\\\\n q(\\hat{z}_{nk}^g|s_{nk}^g=1) &= \\mathcal{N} \\left( \\hat{z}_{nk}^g \\middle| \\mu_{z_{nk}^g}, \\sigma_{z_{nk}^g}^2\\right)\n \\end{aligned} \\end{equation}\nwith\n\\begin{equation} \\begin{aligned}\n \t\\mu_{z_{nk}^g} &= \\frac{ \\sum_{m=1}^M\\sum_{d=1}^{D_m} y_{nd}^{m,g} \\la w^m_{kd} \\ra - \\sum_{j \\neq k} \\la s_{nj}^g\\hat{z}_{nj}^g \\ra \\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la w^m_{kd} \\ra \\la w^m_{jd} \\ra } { \\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la (w_{kd}^m)^2 \\ra + \\frac{\\la\\alpha_k^g\\ra}{\\la\\tau_d^{gm}\\ra} }\\\\\n \t\\sigma_{z_{nk}^g}^2 &= \\frac{ \\la\\tau_d^{gm}\\ra^{-1} } { \\sum_{m=1}^M\\sum_{d=1}^{D_m} \\la (w_{kd}^m)^2 \\ra + \\frac{\\la\\alpha_k^g\\ra}{\\la\\tau_d^{gm}\\ra} }\n\\end{aligned} \\end{equation}\n\n\n\\paragraph*{ARD prior on the factors}\n\nFor every group $g$ and factor $k$: \\\\\n\nPrior distribution:\n\\begin{align}\n\tp(\\alpha_k^g) &= \\Gdist{\\alpha_k^g}{a_0^\\alpha, b_0^\\alpha}\n\\end{align}\n\nVariational distribution $q(\\alpha_k^g)$:\n\\begin{equation}\n q(\\alpha^g_{k}) = \\Gdist{\\alpha_k^g}{\\hat{a}_{gk}^{\\alpha}, \\hat{b}_{gk}^{\\alpha}}\n\\end{equation}\nwhere:\n\\begin{equation} \\begin{aligned}\n\t\\hat{a}_{gk}^\\alpha &= a_0^\\alpha + \\frac{N_g}{2}\\\\\n\t\\hat{b}_{gk}^\\alpha &= b_0^\\alpha +\\frac{ \\sum_{n=1}^{N_g} \\la (\\hat{z}_{nk}^g)^2 \\ra }{2}\n\\end{aligned} \\end{equation}\n\n\\paragraph*{Sparsity parameter of the Factors}\n\nFor every group $g$ and factor $k$: \\\\\n\nPrior distribution:\n\\begin{align}\n\tp(\\theta_k^g) &= \\Bdist{\\theta_k^g}{a_0^\\theta,b_0^\\theta}\n\\end{align}\n\nVariational distribution:\n\\begin{equation}\n\tq(\\theta_k^g) = \\Bdist{\\theta_k^g}{\\hat{a}_{gk}^{\\theta}, \\hat{b}_{gk}^{\\theta}}\n\\end{equation}\nwhere\n\\begin{equation}\n \\begin{aligned}\n \t\\hat{a}_{gk}^\\theta &= \\sum_{n=1}^{N_g} \\la s^g_{nk}\\ra + a_0^\\theta\\\\\n \t\\hat{b}_{gk}^\\theta &= b_0^\\theta - \\sum_{n=1}^{N_g} \\la s^g_{nk}\\ra + N_g\n \\end{aligned}\n\\end{equation}\n\n\\paragraph*{Feature weights}\nFor every view $m$, feature $d$ and factor $k$: \\\\\n\nPrior distribution $p(\\hat{w}_{kd}^m,s_{kd}^m)$:\n\\begin{align}\n\tp(\\hat{w}_{kd}^m,s_{kd}^m) &= \\mathcal{N} (\\hat{w}_{kd}^m \\,|\\, 0, ~ 1/\\alpha_k^m)\\, \\text{Ber}(s_{kd}^m \\,|\\,\\theta_k^m)\n\\end{align}\n\nVariational distribution $q(\\hat{w}_{kd}^m,s_{kd}^m)$:\\\\\n\nUpdate for $q(s_{kd}^m)$:\n\\begin{equation}\n\tq(s^m_{kd}) = \\mathrm{Ber}(s^m_{kd}|\\gamma^m_{kd})\n\\end{equation}\nwith\n\\begin{equation} \\begin{aligned}\n\t&\\gamma^m_{kd} = \\frac{1}{1+\\exp(-\\lambda_{kd}^m)}\\\\\n\t& \\lambda_{kd}^m = \\la \\ln\\frac{\\theta}{1-\\theta} \\ra + 0.5\\ln\\frac{\\la\\alpha_k^m\\ra}{\\la\\tau_d^{gm}\\ra} - 0.5\\ln\\left( \\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la (z_{nk}^g)^2 \\ra + \\frac{\\la\\alpha_k^m\\ra}{\\la\\tau_d^{gm}\\ra} \\right) \\\\\n\t&+ \\frac{\\la\\tau_d^{gm}\\ra}{2} \\frac{ \\left( \\sum_{g=1}^G\\sum_{n=1}^{N_g} y_{nd}^{gm} \\la z^g_{nk} \\ra - \\sum_{j \\neq k} \\la s_{jd}^m\\hat{w}_{jd}^m\\ra \\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la z^g_{nk} \\ra \\la z^g_{nj} \\ra \\right)^2} {\\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la (z_{nk}^g)^2 \\ra + \\frac{\\la\\alpha_k^m\\ra}{\\la\\tau_d^{gm}\\ra} }\n\\end{aligned} \\end{equation}\n\nUpdate for $q(\\hat{w}_{kd}^m)$:\n\\begin{equation} \\begin{aligned}\n q(\\hat{w}_{kd}^m|s_{kd}^m=0) &= \\mathcal{N} \\left(\\hat{w}_{kd}^m \\middle| 0, 1/\\alpha_k^m \\right) \\\\\n q(\\hat{w}_{kd}^m|s_{kd}^m=1) &= \\mathcal{N} \\left( \\hat{w}_{kd}^m \\middle| \\mu_{w_{kd}^m}, \\sigma_{w_{kd}^m}^2\\right)\n \\end{aligned} \\end{equation}\nwith\n\\begin{equation} \\begin{aligned}\n \t\\mu_{w_{kd}^m} &= \\frac{ \\sum_{g=1}^G\\sum_{n=1}^{N_g} y_{nd}^{gm} \\la z^g_{nk} \\ra - \\sum_{j \\neq k} \\la s_{jd}^m\\hat{w}_{jd}^m \\ra \\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la z^g_{nk} \\ra \\la z^g_{nj} \\ra } { \\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la (z_{nk}^g)^2 \\ra + \\frac{\\la\\alpha_k^m\\ra}{\\la\\tau_d^{gm}\\ra} }\\\\\n \t\\sigma_{w_{kd}^m}^2 &= \\frac{ \\la\\tau_d^{gm}\\ra^{-1} } { \\sum_{g=1}^G\\sum_{n=1}^{N_g} \\la (z_{nk}^g)^2 \\ra + \\frac{\\la\\alpha_k^m\\ra}{\\la\\tau_d^{gm}\\ra} }\n\\end{aligned} \\end{equation}\n\n \t% Taken together this means that we can update $q(\\hat{w}_{kd}^m,s_{kd}^m)$ using:\n \t% \\begin{equation*}\n \t% q(\\hat{w}_{kd}^m|s_{kd}^m) q(s_{kd}^m) = \\Ndist{ \\hat{w}_{kd}^m } { s_{kd}^m \\mu_{w_{kd}^m}, s_{kd}^m\\sigma_{w_{kd}^m}^2 + (1-s_{kd}^m)/\\alpha_k^m} (\\new{\\gamma_{kd}^m})^{s_{kd}^m} (1-\\new{\\gamma_{kd}^m})^{1-s_{kd}}\n \t% \\end{equation*}\n\n\n\\paragraph*{ARD prior on the weights}\n\nFor every view $m$ and factor $k$: \\\\\n\nPrior distribution $p(\\alpha_k^m)$:\n\\[\n\tp(\\alpha_k^m) = \\Gdist{\\alpha_k^m}{a_0^\\alpha, b_0^\\alpha}\n\\]\nVariational distribution $q(\\alpha_k^m)$:\n\\begin{equation}\n q(\\alpha^m_{k}) = \\Gdist{\\alpha_k^m}{\\hat{a}_{mk}^{\\alpha}, \\hat{b}_{mk}^{\\alpha}}\n\\end{equation}\nwhere:\n\\begin{equation} \\begin{aligned}\n\t\\hat{a}_{mk}^\\alpha &= a_0^\\alpha + \\frac{D_m}{2}\\\\\n\t\\hat{b}_{mk}^\\alpha &= b_0^\\alpha +\\frac{ \\sum_{d=1}^{D_m} \\la (\\hat{w}_{kd}^m)^2 \\ra }{2}\n\\end{aligned} \\end{equation}\n\n\n\\paragraph*{Sparsity parameter of the weights}\n\nFor every view $m$ and factor $k$: \\\\\n\nPrior distribution:\n\\[\n\tp(\\theta_k^m) = \\Bdist{\\theta_k^m}{a_0^\\theta,b_0^\\theta}\n\\]\n\nVariational distribution:\n\\begin{equation}\n\tq(\\theta_k^m) = \\Bdist{\\theta_k^m}{\\hat{a}_{mk}^{\\theta}, \\hat{b}_{mk}^{\\theta}}\n\\end{equation}\nwhere\n\\begin{equation}\n \\begin{aligned}\n \t\\hat{a}_{mk}^\\theta &= \\sum_{d=1}^{D_m} \\la s^m_{kd}\\ra + a_0^\\theta\\\\\n \t\\hat{b}_{mk}^\\theta &= b_0^\\theta - \\sum_{d=1}^{D_m} \\la s^m_{kd}\\ra + D_m\n \\end{aligned}\n\\end{equation}\n\n\n\\paragraph*{Noise (Gaussian)}\n\nFor every view $m$, group $g$ and feature $d$:\\\\\n\nPrior distribution $p(\\tau_d^{gm})$:\n\\[\n\tp(\\tau_d^{gm}) = \\Gdist{\\tau_{dg}^m}{a_0^\\tau,b_0^\\tau},\n\\]\n\nVariational distribution $q(\\tau_d^{gm})$:\n\\begin{equation}\n\tq(\\tau_d^{gm}) = \\Gdist{\\tau_d^{gm}}{\\hat{a}_{d}^{gm} , \\hat{b}_{d}^{gm}}\n\\end{equation}\nwhere:\n\\begin{equation} \\begin{aligned}\n\t\\hat{a}_{d}^{gm} &= a_0^{\\tau} + \\frac{N_g}{2}\\\\\n\t\\hat{b}_{d}^{gm} &= b_0^{\\tau} + \\frac{1}{2} \\sum_{n=1}^{N_g} \\la\\left(y_{nd}^{gm} - \\sum_k^{K} w_{kd}^m z_{nk}^{g}\\right)^2 \\ra\n\\end{aligned} \\end{equation}\n\n\n\\section{Evidence Lower Bound}\n\nAlthough computing the ELBO is not necessary in order to estimate the posterior distribution of the parameters, it is used to monitor the convergence of the algorithm. As shown in \\Cref{eq_elbo2}, the ELBO can be decomposed into a sum of two terms: (1) the expected log likelihood under the current estimate of the posterior distribution of the parameters and (2) the KL divergence between the prior and the variational distributions of the parameters:\\\\\n\n\\begin{equation} \\begin{aligned}\n \\Lagr = \\E_{q(X)} \\ln p(Y|X) - \\KL\\left(q(X) \\middle|\\middle| p(X) \\right)\n\\end{aligned} \\end{equation}\n\n\n\\paragraph*{Log likelihood term}\n\nAssuming a Gaussian likelihood:\n\\begin{equation} \\begin{aligned}\n\t\\E_{q(X)} \\ln p(Y|X) = & -\\sum_{m=1}^M \\frac{ND_m}{2} \\ln(2\\pi) + \\sum_{g=1}^G \\frac{N_g}{2} \\sum_{m=1}^M \\sum_{d=1}^{D_m} \\la \\ln(\\tau_{d}^{gm}) \\ra \\\\\n\t&-\\sum_{g=1}^G \\sum_{m=1}^M \\sum_{d=1}^{D_m} \\frac{\\la \\tau_{d}^{gm} \\ra}{2} \\sum_{n=1}^{N_g} \\big( y_{nd}^{m,g} - \\sum_{k=1}^{K}\\la s_{kd}^m \\hat{w}_{kd}^m \\ra \\la z_{nk}^g \\ra \\big)^2\n\\end{aligned} \\end{equation}\n\n\n\\paragraph*{KL divergence terms}\n\nNote that $\\KL\\left(q(X) \\middle|\\middle| p(X) \\right) = \\E_q(q(X)) - \\E_q(p(X))$. \\\\\nBelow, we will write the analytical form for these two expectations.\n\n\\paragraph*{Weights}\n\n\\begin{equation} \\begin{aligned}\n \\E_q[\\ln p(\\hat{W},S)] =& -\\sum_{m=1}^{M}\\frac{KD_m}{2}\\ln(2\\pi) + \\sum_{m=1}^{M}\\frac{D_m}{2}\\sum_{k=1}^{K} \\ln(\\alpha_k^m) - \\sum_{m=1}^{M} \\frac{\\alpha_k^m}{2} \\sum_{d=1}^{D_m} \\sum_{k=1}^{K} \\la (\\hat{w}_{kd}^m)^2 \\ra \\\\\n & + \\la \\ln(\\theta) \\ra \\sum_{m=1}^{M} \\sum_{d=1}^{D_m} \\sum_{k=1}^{K} \\la s_{kd}^m \\ra + \\la \\ln(1-\\theta) \\ra \\sum_{m=1}^{M} \\sum_{d=1}^{D_m}\\sum_{k=1}^{K} (1- \\la s_{kd}^m \\ra)\n\\end{aligned} \\end{equation}\n\n\\begin{equation} \\begin{aligned}\n\t\\E_q[\\ln q(\\hat{W}, S)] =&-\\sum_{m=1}^{M}\\frac{KD_m}{2}\\ln(2\\pi) + \\frac{1}{2}\\sum_{m=1}^{M}\\sum_{d=1}^{D_m}\\sum_{k=1}^{K}\\ln(\\la s_{kd}^m \\ra \\sigma_{w_{kd}^m}^2 + (1-\\la s_{kd}^m \\ra)/\\alpha_k^m) \\\\\n\t&+ \\sum_{m=1}^{M} \\sum_{d=1}^{D_m} \\sum_{k=1}^{K} (1-\\la s_{kd}^m \\ra) \\ln(1 - \\la s_{kd}^m \\ra) - \\la s_{kd}^m \\ra \\ln \\la s_{kd}^m \\ra\n\\end{aligned} \\end{equation}\n\n\\paragraph*{Factors}\n\n\\begin{equation} \\begin{aligned}\n \\E_q[\\ln p(\\hat{Z},S)] =& -\\sum_{g=1}^{G}\\frac{N_g K}{2}\\ln(2\\pi) + \\sum_{g=1}^{G}\\frac{N_g}{2}\\sum_{k=1}^{K} \\ln(\\alpha_k^g) - \\sum_{g=1}^{G} \\frac{\\alpha_k^g}{2} \\sum_{n=1}^{N_g} \\sum_{k=1}^{K} \\la (\\hat{z}_{nk}^g)^2 \\ra \\\\\n & + \\la \\ln(\\theta) \\ra \\sum_{g=1}^{G} \\sum_{n=1}^{N_g} \\sum_{k=1}^{K} \\la s_{nk}^g \\ra + \\la \\ln(1-\\theta) \\ra \\sum_{g=1}^{G} \\sum_{n=1}^{N_g}\\sum_{k=1}^{K} (1- \\la s_{nk}^g \\ra)\n\\end{aligned} \\end{equation}\n\n\\begin{equation} \\begin{aligned}\n\t\\E_q[\\ln q(\\hat{Z}, S)] =&-\\sum_{g=1}^{G}\\frac{N_g K}{2}\\ln(2\\pi) + \\frac{1}{2}\\sum_{g=1}^{G}\\sum_{n=1}^{N_g}\\sum_{k=1}^{K}\\ln(\\la s_{nk}^g \\ra \\sigma_{z_{nk}^g}^2 + (1-\\la s_{nk}^g \\ra)/\\alpha_k^g) \\\\\n\t&+ \\sum_{g=1}^{G} \\sum_{n=1}^{N_g} \\sum_{k=1}^{K} (1-\\la s_{nk}^g \\ra) \\ln(1 - \\la s_{nk}^g \\ra) - \\la s_{nk}^g \\ra \\ln \\la s_{nk}^g \\ra\n\\end{aligned} \\end{equation}\n\n\\paragraph*{ARD prior on the weights}\n\n\\begin{equation} \\begin{aligned}\n\t\\E_q [\\ln p(\\balpha)] &= \\sum_{m=1}^{M}\\sum_{k=1}^{K}\\Big(a_0^\\alpha\\ln b_0^\\alpha + (a_0^\\alpha - 1) \\la \\ln \\alpha_k \\ra - b_0^\\alpha \\la \\alpha_k \\ra - \\ln \\Gamma(a_0^\\alpha) \\Big) \\\\\n\t\\E_q [\\ln q(\\balpha)] &= \\sum_{m=1}^{M}\\sum_{k=1}^{K} \\Big( \\hat{a}_{k}^\\alpha \\ln \\hat{b}_{k}^\\alpha + (\\hat{a}_{k}^\\alpha - 1) \\la \\ln \\alpha_k \\ra - \\hat{b}_{k}^\\alpha \\la \\alpha_k \\ra - \\ln \\Gamma(\\hat{a}_{k}^\\alpha) \\Big)\n\\end{aligned} \\end{equation}\n\n\\paragraph*{Sparsity parameter of the weights}\n\n\\begin{equation} \\begin{aligned}\n\t &\\E_q\\left[ \\ln p(\\btheta) \\right] = \\sum_{m=1}^M \\sum_{k=1}^K\\sum_{d=1}^{D_m}\\left( (a_0 - 1) \\times \\la \\ln(\\pi^m_{d, k}) \\ra + (b_0 -1) \\la \\ln(1 - \\pi^m_{d, k}) \\ra - \\ln (\\mathrm{B} (a_0, b_0))\\right) \\\\\n\t &\\E_q\\left[ \\ln q(\\btheta) \\right] = \\sum_{m=1}^M \\sum_{k=1}^K\\sum_{d=1}^{D_m}\\left( (a^m_{k,d} - 1) \\times \\la \\ln(\\pi^m_{d, k}) \\ra + (b^m_{k,d} -1) \\la \\ln(1 - \\pi^m_{d, k}) \\ra - \\ln (\\mathrm{B} (a^m_{k,d}, b^m_{k,d})) \\right) \\\\\n\\end{aligned} \\end{equation}\n\n\\paragraph*{ARD prior on the Factors}\n\n\\begin{equation} \\begin{aligned}\n\t\\E_q [\\ln p(\\balpha)] &= \\sum_{g=1}^{G}\\sum_{k=1}^{K}\\Big(a_0^\\alpha\\ln b_0^\\alpha + (a_0^\\alpha - 1) \\la \\ln \\alpha_k \\ra - b_0^\\alpha \\la \\alpha_k \\ra - \\ln \\Gamma(a_0^\\alpha) \\Big) \\\\\n\t\\E_q [\\ln q(\\balpha)] &= \\sum_{g=1}^{G}\\sum_{k=1}^{K} \\Big( \\hat{a}_{k}^\\alpha \\ln \\hat{b}_{k}^\\alpha + (\\hat{a}_{k}^\\alpha - 1) \\la \\ln \\alpha_k \\ra - \\hat{b}_{k}^\\alpha \\la \\alpha_k \\ra - \\ln \\Gamma(\\hat{a}_{k}^\\alpha) \\Big)\n\\end{aligned} \\end{equation}\n\n\\paragraph*{Sparsity parameter of the Factors}\n\\begin{equation} \\begin{aligned}\n\t &\\E_q\\left[ \\ln p(\\btheta) \\right] = \\sum_{g=1}^G \\sum_{k=1}^K\\sum_{n=1}^{N_g}\\left( (a_0 - 1) \\times \\la \\ln(\\pi^g_{n, k}) \\ra + (b_0 -1) \\la \\ln(1 - \\pi^g_{n, k}) \\ra - \\ln (\\mathrm{B} (a_0, b_0))\\right) \\\\\n\t &\\E_q\\left[ \\ln q(\\btheta) \\right] = \\sum_{g=1}^G \\sum_{k=1}^K\\sum_{n=1}^{N_g}\\left( (a^g_{k,n} - 1) \\times \\la \\ln(\\pi^g_{n, k}) \\ra + (b^g_{k,n} -1) \\la \\ln(1 - \\pi^g_{n, k}) \\ra - \\ln (\\mathrm{B} (a^g_{k,n}, b^g_{k,n})) \\right) \\\\\n\\end{aligned} \\end{equation}\n\n\\paragraph*{Noise}\n\n\\begin{equation} \\begin{aligned}\n\t\\E_q [\\ln p(\\btau)] &= \\sum_{m=1}^{M} D_m a_0^\\tau \\ln b_0^\\tau + \\sum_{g=1}^{G}\\sum_{m=1}^{M}\\sum_{d=1}^{Dm} (a_0^\\tau - 1) \\la \\ln \\tau_d^{gm} \\ra - \\sum_{g=1}^{G}\\sum_{m=1}^{M}\\sum_{d=1}^{Dm} b_0^\\tau \\la \\tau_d^{gm} \\ra - \\sum_{m=1}^{M} D_m \\ln \\Gamma(a_0^\\tau)\\\\\n\t\\E_q [\\ln q(\\btau)] &= \\sum_{g=1}^{G} \\sum_{m=1}^{M} \\sum_{d=1}^{D_m} \\left( \\hat{a}_{dgm}^\\tau \\ln \\hat{b}_{dgm}^\\tau + (\\hat{a}_{dgm}^\\tau - 1) \\la \\ln \\tau_d^{gm} \\ra - \\hat{b}_{dgm}^\\tau \\la \\tau_d^{gm} \\ra - \\ln \\Gamma(\\hat{a}_{dgm}^\\tau) \\right)\n\\end{aligned} \\end{equation}\n", "meta": {"hexsha": "107fe5edeee8b7210251c5a977cf6f25485793c0", "size": 15372, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Appendix/appendix1.tex", "max_stars_repo_name": "rargelaguet/thesis", "max_stars_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 15, "max_stars_repo_stars_event_min_datetime": "2021-01-08T13:01:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T07:24:40.000Z", "max_issues_repo_path": "Appendix/appendix1.tex", "max_issues_repo_name": "rargelaguet/thesis", "max_issues_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Appendix/appendix1.tex", "max_forks_repo_name": "rargelaguet/thesis", "max_forks_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-01-09T04:47:49.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-04T08:25:50.000Z", "avg_line_length": 56.102189781, "max_line_length": 785, "alphanum_fraction": 0.6037600833, "num_tokens": 6988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8311430311279739, "lm_q2_score": 0.7371581684030623, "lm_q1q2_score": 0.6126838745072666}} {"text": "\\subsection{Quivers}\\label{subsec:quivers}\n\n\\begin{definition}\\label{def:quiver}\n A \\term{quiver} or \\term{directed multigraph} is an extension of \\hyperref[def:undirected_multigraph]{undirected multigraphs}. A quiver consists of the following:\n \\begin{thmenum}\n \\thmitem{def:quiver/vertices} A set \\( V \\) of \\term{vertices}.\n \\thmitem{def:quiver/arcs} A disjoint from \\( V \\) set \\( A \\) of \\term{arcs}. It is conventional to call them arcs or even directed edges rather than simply edges.\n \\thmitem{def:quiver/head} A \\term{head} function \\( h: A \\to V \\).\n \\thmitem{def:quiver/tail} A \\term{tail} function \\( t: A \\to V \\).\n \\thmitem{def:quiver/adjacency} We say that the arc \\( a \\) is a \\term{successor} of \\( b \\) and that \\( b \\) is a \\term{predecessor} of \\( a \\) if \\( t(a) = h(b) \\). We also say that \\( a \\) and \\( b \\) are \\term{consecutive}.\n\n For vertices, we say that \\( w \\) is a successor of \\( v \\) and that \\( w \\) is consecutive to \\( w \\) if there exists an arc from \\( u \\) to \\( v \\).\n\n \\thmitem{def:quiver/forgetful} For each quiver \\( Q = (V, A, h, t) \\), we have an \\enquote{underlying} undirected multigraph\n \\begin{equation}\\label{eq:def:quiver/forgetful}\n \\begin{aligned}\n &G: A \\to V, \\\\\n &G(a) \\coloneqq \\set{ h(a), t(a) }.\n \\end{aligned}\n \\end{equation}\n\n All properties of \\( G \\) are inherited by \\( Q \\), although some of them do not really fit or need adaptation.\n\n We denote this graph by \\( U(Q) \\coloneqq G \\) and will regard it as a forgetful functor --- see \\fullref{def:multigraph_orientation}.\n\n \\thmitem{def:quiver/opposing} One adaptation is that the arcs \\( a \\) and \\( b \\) are called \\term{opposing} they are successors of each other and the term \\enquote{parallel} is reserved for arcs which are parallel in the sense of \\fullref{def:hypergraph/parallel_hyperedges} but not opposing.\n\n \\thmitem{def:quiver/simple} If there are no parallel and opposing arcs or loops, we call the quiver \\term{simple}. The term \\term{simple directed graph} is more commonly used.\n\n For a simple directed graph, we often write \\( G = (V, E) \\), where \\( E \\) is an \\hyperref[def:binary_relation/irreflexive]{irreflexive} binary relation.\n\n We say that the simple directed graph is \\term{symmetric} if this relation is symmetric. For many purposes, symmetric directed graphs can be regarded as undirected --- see \\fullref{rem:symmetric_directed_graphs}.\n\n \\thmitem{def:quiver/submodel} The \\term{subquivers} of \\( Q \\) are the precisely \\hyperref[def:undirected_multigraph/submodel]{subgraphs} of \\( G \\). In particular, the \\term{full subquivers} are the \\hyperref[def:undirected_multigraph/submodel]{full subgraphs}, that are maximal among the subgraphs with the same vertices.\n \\end{thmenum}\n\n \\begin{figure}\n \\begin{equation}\\label{eq:fig:def:quiver}\n \\begin{aligned}\n \\includegraphics[page=1]{output/def__quiver.pdf}\n \\end{aligned}\n \\end{equation}\n \\caption{A quiver with a pair of parallel arcs, a pair of opposing arcs and a loop. Removing the dashed arcs makes it a simple directed graph.}\\label{fig:def:quiver}\n \\end{figure}\n\\end{definition}\n\n\\begin{example}\\label{ex:infinite_integer_graphs}\n A very simple example of an infinite quiver is the \\hyperref[def:relation_closures/transitive]{transitive reduction} of the positive integers, i.e. the simple directed graph\n \\begin{equation}\\label{eq:ex:infinite_integer_graphs/positive}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__infinite_integer_graphs.pdf}\n \\end{aligned}\n \\end{equation}\n\n Since the graph is simple, we have\n \\begin{equation*}\n \\deg(n) = \\begin{cases}\n 1 = 0 + 1, &n = 0, \\\\\n 2 = 1 + 1, &n > 0 \\\\\n \\end{cases}\n \\end{equation*}\n and thus \\( \\deg(G) = 2 \\).\n\n The \\hyperref[def:categorical_diagram]{categorical diagrams} corresponding to this quiver is used to define direct limits in \\fullref{def:direct_and_inverse_limits/direct}.\n\n Another related graph is based on the negative integers:\n \\begin{equation}\\label{eq:ex:infinite_integer_graphs/negative}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__infinite_integer_graphs.pdf}\n \\end{aligned}\n \\end{equation}\n\n The \\hyperref[def:categorical_diagram]{categorical diagrams} corresponding to this quiver is used to define inverse limits in \\fullref{def:direct_and_inverse_limits/inverse}.\n\n Finally, the union of the two with zero added gives us the following quiver:\n \\begin{equation}\\label{eq:ex:infinite_integer_graphs/two_sided}\n \\begin{aligned}\n \\includegraphics[page=3]{output/ex__infinite_integer_graphs.pdf}\n \\end{aligned}\n \\end{equation}\n\n All three graphs are infinite but \\hyperref[def:hypergraph/degree]{locally finite}.\n\\end{example}\n\n\\begin{definition}\\label{def:category_of_small_quivers}\n Suppose that we are given a \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\), which is safe to assume to be the smallest suitable one as explained in \\fullref{def:large_and_small_sets}.\n\n We denote the \\hyperref[def:category]{category} of \\( \\mscrU \\)-small \\hyperref[def:hypergraph]{quivers} by \\( \\ucat{Quiv} \\) or, if the universe is clear from the context, simply by \\( \\cat{Quiv} \\). See \\fullref{def:category_size} for a further discussion of universes and categories.\n\n A more \\enquote{categorical} definition, in which this category arises naturally, is given in \\fullref{ex:quivers_as_functors}.\n\n For simple directed graphs, we denote the category using \\( \\cat{SimpQuiv} \\).\n\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(\\cat{Quiv}) \\) is the set of all \\( \\mscrU \\)-small quivers, i.e. the quivers \\( Q = (V, A, h, t) \\) such that \\( V \\) and \\( A \\) are both members of \\( \\mscrU \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{Quiv}(Q, R) \\) from \\( Q \\) to \\( R \\) is the set quiver homomorphisms. Given two quivers \\( Q = (V_Q, A_Q, h_Q, t_Q) \\) and \\( R = (V_R, A_R, h_R, t_R) \\), a \\term{quiver homomorphism} is a pair of functions\n \\begin{equation}\\label{eq:def:category_of_small_quivers/homomorphism}\n \\begin{cases}\n f_V: V_Q \\to V_R \\\\\n f_A: A_Q \\to A_R\n \\end{cases}\n \\end{equation}\n such that\n \\begin{align}\n h_R \\bincirc f_A &= f_V \\bincirc h_Q \\label{eq:def:category_of_small_quivers/homomorphism/head} \\\\\n t_R \\bincirc f_A &= f_V \\bincirc t_Q \\label{eq:def:category_of_small_quivers/homomorphism/tail}\n \\end{align}\n\n See \\fullref{ex:quivers_as_functors} for a categorical justification of this definition, including reducing the above conditions to the diagram \\eqref{eq:ex:quivers_as_functors/index/diagram}.\n\n Clearly every quiver homomorphism is a \\hyperref[eq:def:category_of_small_hypergraphs/homomorphism]{hypergraph homomorphism}.\n\n Note that \\enquote{quiver embedding} commonly refers to an embedding of its \\hyperref[def:quiver_geometric_realization/undirected]{geometric realization}, hence we will avoid the term when referring to injective quiver homomorphisms. Furthermore, it should be clarified whether we mean \\enquote{injective on vertices} or \\enquote{injective on arcs}, which is an important distinction in category theory --- see \\fullref{def:functor_invertibility}.\n\n \\item The \\hyperref[def:category/composition]{composition of quiver morphisms} is pointwise composition --- the same as composition of hypergraph homomorphisms.\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the quiver \\( Q = (V, A, h, t) \\) is the pair of \\hyperref[def:multi_valued_function/identity]{identity functions} \\( (\\id_V, \\id_A) \\).\n \\end{itemize}\n\\end{definition}\n\n\\begin{definition}\\label{def:multigraph_orientation}\\mcite{nLab:oriented_graph}\n Given an \\hyperref[def:undirected_multigraph]{undirected (multi)graph} \\( G = (V, E, \\mscrE) \\), constructing a quiver requires a choice of head and tail for every edge. We can view undirected graphs as equivalence classes of quivers. A head function is merely a \\hyperref[def:choice_function]{choice function} on the set \\( \\set{ \\mscrE(e) \\given e \\in E } \\), and a choice function can be provided by the \\hyperref[def:zfc/choice]{axiom of choice} even for infinite graphs.\n\n For a given choice function \\( c \\), we denote by \\( O_c(G) = (V, E, h, t) \\) the quiver obtained by identifying the head \\( h(e) \\) with \\( c(\\mscrE(e)) \\) and the tail \\( t(e) \\) with the other endpoint of \\( e \\), if one exists, and with \\( h(e) \\) for loops. We call \\( O_c(G) \\) the \\term{orientation} induced by the choice function \\( c \\).\n\n Note that \\( O_c \\) is not left adjoint to \\( U \\) --- see \\fullref{ex:def:category_adjunction/multgph_quiv}.\n\\end{definition}\n\n\\begin{remark}\\label{rem:graphs_linear_algebra_and_topology}\n As we shall see, the \\hyperref[def:hypergraph/adjacency]{adjacency} and \\hyperref[def:hypergraph/incidence]{incidence} of a quiver can be studied using linear algebra via \\hyperref[def:graph_adjacency_matrix]{adjacency} and \\hyperref[def:hypergraph_incidence_matrix]{incidence matrices}, while the \\hyperref[def:quiver_connectedness]{connectedness} can be studied using \\hyperref[def:quiver_connectedness]{topology} via \\hyperref[def:quiver_geometric_realization/embedding]{graph embeddings}.\n\\end{remark}\n\n\\begin{definition}\\label{def:quiver_incidence_matrix}\\mcite[ch. 1, sec. 2.1]{GondranMinoux1984Graphs}\n Let \\( Q = (A, E, h, t) \\) be a finite quiver. Its \\term{incidence matrix}\n \\begin{equation*}\n M = \\seq{ M_{va} }_{v \\in V, a \\in A}\n \\end{equation*}\n has elements\n \\begin{equation*}\n M_{va} \\coloneqq \\begin{cases}\n 1, &v = h(e) \\\\\n -1, &v = h(e) \\\\\n 0, &\\T{otherwise.}\n \\end{cases}\n \\end{equation*}\n\n Compare this definition to \\fullref{def:hypergraph_incidence_matrix}.\n\\end{definition}\n\n\\begin{example}\\label{ex:quiver_matrices}\n The \\hyperref[def:quiver_incidence_matrix]{incidence matrix} of the simple directed graph \\eqref{eq:fig:def:quiver} is\n \\begin{equation}\\label{ex:quiver_matrices/incidence}\n \\begin{blockarray}{cccccccc}\n & e_1 & e_2 & e_3 & e_4 & e_5 & e_6 & e_7 \\\\\n \\begin{block}{c(ccccccc)}\n a & 1 & 1 & & & & & \\\\\n b & \\fbox{-1} & & 1 & & & & \\\\\n c & & \\fbox{-1} & & 1 & 1 & & \\\\\n d & & & \\fbox{-1} & \\fbox{-1} & & 1 & \\\\\n e & & & & & \\fbox{-1} & & 1 \\\\\n f & & & & & & \\fbox{-1} & \\fbox{-1} \\\\\n \\end{block}\n \\end{blockarray}\n \\end{equation}\n\n It can be read column-by-column. Every column contains exactly two nonzero elements whose rows correspond to the head (positive) and tail (negative).\n\n To obtain the incidence matrix for the underlying undirected graph \\eqref{eq:fig:def:undirected_multigraph}, we need to simply flip the sign of the boxed elements above.\n\n The \\hyperref[def:graph_adjacency_matrix]{adjacency matrix} is,\n \\begin{equation}\\label{ex:quiver_matrices/adjacency}\n \\begin{blockarray}{cccccccc}\n & a & b & c & d & e & f \\\\\n \\begin{block}{c(ccccccc)}\n a & & 1 & 1 & & & \\\\\n b & \\fbox{1} & & & 1 & & \\\\\n c & \\fbox{1} & & & 1 & 1 & \\\\\n d & & \\fbox{1} & \\fbox{1} & & & 1 \\\\\n e & & & \\fbox{1} & & & 1 \\\\\n f & & & & \\fbox{1} & \\fbox{1} & \\\\\n \\end{block}\n \\end{blockarray}\n \\end{equation}\n where the boxed elements are nonzero only in the adjacency matrix for \\eqref{eq:fig:def:undirected_multigraph} and not \\eqref{eq:fig:def:quiver}.\n\n The matrix can be read either column-by-column or row-by-row.\n \\begin{itemize}\n \\item The \\( v \\)-th column lists the vertices \\( u \\) such that there is an arc from \\( u \\) to \\( v \\).\n \\item The \\( u \\)-th row lists the vertices \\( v \\) such that there is an arc from \\( v \\) to \\( u \\).\n \\end{itemize}\n\\end{example}\n\n\\begin{proposition}\\label{thm:graph_undirected_iff_adjacency_matrix_is_symmetric}\n Every square finite matrix over the two-element field \\hyperref[thm:ring_of_integers_modulo_zero_divisors]{\\( \\BbbF_2 \\)} corresponds to a \\hyperref[def:quiver]{quiver} without parallel arcs.\n\n If the matrix is symmetric, we obtain its underlying \\hyperref[def:undirected_multigraph]{undirected graph}.\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{remark}\\label{rem:symmetric_directed_graphs}\n It is sometimes convenient to conflate symmetric \\hyperref[def:quiver/simple]{simple directed graphs} with \\hyperref[def:undirected_multigraph]{simple undirected graphs}.\n\n This is partially justified by \\fullref{thm:graph_undirected_iff_adjacency_matrix_is_symmetric}.\n\\end{remark}\n\n\\begin{definition}\\label{def:quiver_path}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver}.\n\n \\begin{thmenum}\n \\thmitem{def:quiver_path/directed} A \\term{directed path} or simply \\term{path} in \\( Q \\) a finite or infinite sequence of \\hyperref[def:quiver/adjacency]{consecutive arcs} with a specified head vertex. Formally, it is a sequence \\( p = (v, e_1, e_2, \\ldots) \\), where \\( h(e_k) = t(e_{k-1}) \\) for every index \\( k \\).\n\n This is a simplification of \\fullref{def:undirected_multigraph_path} since every arc has a head and tail. It is also too restrictive for some purposes.\n\n The definitions of \\hyperref[def:undirected_multigraph_path/endpoints]{path endpoints}, \\hyperref[def:undirected_multigraph_path/empty]{empty path}, \\hyperref[def:undirected_multigraph_path/length]{length}, \\hyperref[def:undirected_multigraph_path/domain]{domain}, \\hyperref[def:undirected_multigraph_path/concatenation]{concatenation} and \\hyperref[def:undirected_multigraph_path/characteristic_vector]{characteristic vector} of a directed path are inherited from \\fullref{def:undirected_multigraph_path}.\n\n \\thmitem{def:quiver_path/cycle} A \\term{directed cycle} is a directed path whose endpoints coincide. The term \\term{circuit} is also used, for example in \\cite[ch. 1, sec. 3.2]{GondranMinoux1984Graphs}. A quiver without cycles is called \\term{acyclic} and a simple acyclic directed graph is commonly abbreviated as \\term{DAG}.\n\n \\thmitem{def:quiver_path/simple} Similarly to \\fullref{def:undirected_multigraph_path/simple}, we say that a directed path is \\term{simple} if it contains to directed cycles.\n\n \\thmitem{def:quiver_path/undirected} If \\( p = (v, e_1, e_2, \\ldots) \\) is a path in the \\hyperref[def:undirected_multigraph]{undirected multigraph} \\( U(Q) \\),\n \\begin{equation*}\n D(p) \\coloneqq (v, e_1, e_2, \\ldots)\n \\end{equation*}\n is a sequence of adjacent arcs in \\( Q \\) (with a specified head vertex).\n\n The head and tail of \\( D(p) \\) are defined as the head and tail of \\( p \\). This definition obviously differs from \\fullref{def:quiver_path}.\n\n We say that the arc \\( e_k \\) of \\( D(p) \\) is \\term{positively oriented} if the head \\( h_k \\) of the edge \\( e_k \\) in \\( Q \\) equals the head \\( h(e_k) \\) of the arc \\( e_k \\) in \\( D(p) \\). If \\( e_k \\) is not positively oriented, we say that it is \\term{negatively oriented}.\n\n We conflate \\( p \\) and \\( D(p) \\) where this does not cause confusion.\n\n Undirected paths are also called \\term{chains} in \\cite[ch. 1, sec. 3.2]{GondranMinoux1984Graphs}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:def:quiver_path}\n An example of an infinite \\hyperref[def:quiver_path/directed]{directed path} is the reduced positive integer graph \\eqref{eq:ex:infinite_integer_graphs/positive} regarded as a path of the reduced integer graph \\eqref{eq:ex:infinite_integer_graphs/two_sided}. It is also simple since the degree of \\eqref{eq:ex:infinite_integer_graphs/positive} is \\( 2 \\).\n\n Now consider again the quiver \\eqref{eq:fig:def:quiver}. The solid lines in \\eqref{fig:ex:def:quiver_path} describe a path.\n \\begin{equation}\\label{fig:ex:def:quiver_path}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__def__quiver_path.pdf}\n \\end{aligned}\n \\end{equation}\n\n \\begin{itemize}\n \\item This path corresponds to the sequence\n \\begin{equation*}\n p = \\parens{ a, \\underbrace{a \\to c}_{e_2}, \\underbrace{c \\to d}_{e_4}, \\underbrace{d \\to f}_{e_6} }.\n \\end{equation*}\n\n \\item Its \\hyperref[def:undirected_multigraph_path/characteristic_vector]{characteristic vector} is\n \\begin{equation*}\n \\vect{p}\n =\n \\begin{blockarray}{ccccccr}\n e_1 & e_2 & e_3 & e_4 & e_5 & e_6 & e_7 \\\\\n \\begin{block}{(ccccccr)}\n 0 & 1 & 0 & 1 & 0 & 1 & 0 \\\\\n \\end{block}\n \\end{blockarray}\n {}^T.\n \\end{equation*}\n\n Furthermore, \\( p \\) can be identified from the characteristic vector.\n\n \\item The \\hyperref[def:quiver_path]{endpoints} are \\( h(p) = h(e_1) = a \\) and \\( t(p) = t(e_7) = f \\).\n\n \\item It is a \\hyperref[def:undirected_multigraph_path/simple]{simple path} because \\( \\deg(a) = \\deg(f) = 1 \\) and \\( \\deg(c) = \\deg(d) = 2 \\) in the induced subgraph.\n\n \\item The \\hyperref[def:undirected_multigraph_path/converse]{converse undirected path}\n \\begin{equation*}\n p^{-1} = (f, d \\to f, c \\to d, a \\to c),\n \\end{equation*}\n is not a directed path because every arc is negatively oriented.\n \\end{itemize}\n\\end{example}\n\n\\begin{definition}\\label{def:quiver_condensation}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver}. We say that the vertex \\( w \\) is \\term{reachable} from \\( v \\) if there exists a \\hyperref[def:quiver_path/directed]{directed path} from \\( v \\) to \\( w \\).\n\n Define the equivalence relation\n \\begin{equation*}\n v \\sim w \\iff v \\T{is reachable from} w \\T{and} w \\T{is reachable from} u.\n \\end{equation*}\n\n Now define the binary relation \\( \\widetilde{A} \\) on the \\hyperref[def:equivalence_relation/quotient]{quotient set} \\( \\widetilde{V} \\coloneqq V / {\\sim} \\) as\n \\begin{equation*}\n ([u], [v]) \\in \\widetilde{A} \\iff v \\T{is reachable from} w \\T{but not vice versa}.\n \\end{equation*}\n\n It is well-defined because if \\( ([v_1], [w_1]) \\in \\widetilde{A} \\), \\( v_2 \\in [v_1] \\) and \\( w_2 \\in [w_1] \\), then by the transitivity of reachability we have that \\( w_2 \\) is reachable from \\( v_2 \\) and thus \\( ([v_2], [w_2]) \\in \\widetilde{A} \\).\n\n As discussed in \\fullref{def:quiver/simple}, irreflexive relations over sets can be regarded as simple directed graphs. Therefore, the pair \\( \\widetilde{G} \\coloneqq (\\widetilde{V}, \\widetilde{A}) \\) is a directed graph. It is called the \\term{condensation} of the quiver \\( Q \\).\n\n Compare this definition to \\fullref{def:undirected_multigraph_connectedness/condensation}.\n\\end{definition}\n\n\\begin{example}\\label{ex:def:quiver_condensation}\n The \\hyperref[def:quiver_condensation]{condensation} of the quiver in \\eqref{eq:fig:def:quiver} is (isomorphic to) the graph itself. If we add the arc \\( f \\to a \\) to \\eqref{eq:fig:def:quiver}, the condensation would be an edgeless quiver with a single vertex.\n\n We can add new arcs \\( e_8 \\) and \\( e_9 \\) to the quiver in \\eqref{eq:fig:def:quiver} to make the example more interesting:\n \\begin{equation}\\label{eq:ex:def:quiver_condensation/uncondensed}\n \\begin{aligned}\n \\includegraphics[page=1]{output/ex__def__graph_condensation.pdf}\n \\end{aligned}\n \\end{equation}\n\n Its condensation is\n \\begin{equation}\\label{eq:ex:def:quiver_condensation/condensed}\n \\begin{aligned}\n \\includegraphics[page=2]{output/ex__def__graph_condensation.pdf}\n \\end{aligned}\n \\end{equation}\n\n These are the \\hyperref[def:quiver_connectedness/strong]{strongly connected components} of \\eqref{eq:ex:def:quiver_condensation/uncondensed}.\n\\end{example}\n\n\\begin{proposition}\\label{thm:graph_condensation_is_acyclic_dag}\n The \\hyperref[def:quiver_condensation]{condensation} of a quiver is a \\hyperref[def:undirected_multigraph_path]{directed acyclic graph}.\n\\end{proposition}\n\\begin{proof}\n Let \\( \\widetilde{G} = (\\widetilde{V}, \\widetilde{E}) \\) be the condensation of the quiver \\( Q = (V, A, h, t) \\). We have already discussed in \\fullref{def:quiver_condensation} that it is a directed graph. It remains to show that it is acyclic.\n\n Aiming at a contradiction, suppose that there exist cosets \\( [u] \\) and \\( [v] \\) and a path\n \\begin{equation*}\n \\parens[\\Big]{ [u] \\to [w_1], \\cdots, [w_n] \\to [v] }\n \\end{equation*}\n in \\( \\widetilde{G} \\) that connects them. We can easily prove by induction that \\( v \\) is reachable from \\( u \\), thus contradicting the definition of \\( \\widetilde{A} \\).\n\n Therefore, \\( \\widetilde{G} \\) is acyclic.\n\\end{proof}\n\n\\begin{definition}\\label{def:quiver_connectedness}\n Let \\( Q = (V, E, h, t) \\) be a quiver and let \\( \\widetilde{G} = (\\widetilde{V}, \\widetilde{E}) \\) be its condensed directed graph.\n\n \\begin{thmenum}\n \\thmitem{def:quiver_connectedness/strong}\\mcite[ch. 1, sec. 3.5]{GondranMinoux1984Graphs} For every coset \\( [v] \\) in \\( \\widetilde{V} \\), the \\hyperref[def:hypergraph/submodel]{subquiver} of \\( Q \\) induced by the vertices \\( [v] \\) is called a \\term{strongly connected component} of \\( Q \\).\n\n The \\term{strong connectivity number} of \\( Q \\) is the cardinality \\( \\card(\\widetilde{V}) \\).\n\n If \\( Q \\) has only one strongly connected component, we say that it itself is \\term{strongly connected}.\n\n \\thmitem{def:quiver_connectedness/weak}\\mcite[ch. 1, sec. 3.3]{GondranMinoux1984Graphs} Similarly, the subquiver \\( Q' \\) of \\( Q \\) is called a \\term{weakly connected component} if is a connected component, in the sense of \\fullref{def:undirected_multigraph_connectedness/components}, of the underlying undirected multigraph \\( U(Q) \\).\n\n The \\term{weak connectivity number} of \\( Q \\) is the connectivity number of \\( U(Q) \\).\n\n If \\( Q \\) has only one weakly connected component, we say that it itself is \\term{weakly connected}.\n \\end{thmenum}\n\n This definition generalizes symmetric and transitive closures of binary relations not commuting --- see \\fullref{ex:thm:def:relation_closures/symmetric_and_transitive}.\n\n Compare this definition to \\fullref{def:undirected_multigraph_connectedness}.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:quiver_connectedness_via_chains}\n Let \\( Q \\) be a \\hyperref[def:quiver]{quiver}.\n\n \\begin{thmenum}\n \\thmitem{thm:quiver_connectedness_via_chains/strong} \\( Q \\) is \\hyperref[def:quiver_connectedness/strong]{strongly connected} if and only if there exists a \\hyperref[def:quiver_path/directed]{directed path} connecting every pair of vertices.\n\n \\thmitem{thm:quiver_connectedness_via_chains/weak} \\( Q \\) is \\hyperref[def:quiver_connectedness/weak]{weakly connected} if and only if there exists an \\hyperref[def:quiver_path/undirected]{undirected path} connecting every pair of vertices.\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n Trivial.\n\\end{proof}\n\n\\begin{remark}\\label{rem:well_founded_graphs}\n We can regard any set \\( A \\) in the sense of \\hyperref[def:zfc]{\\logic{ZFC}} as the simple directed graph \\( (A, \\in) \\).\n\n The \\hyperref[def:zfc/foundation]{axiom of foundation} (via \\fullref{thm:set_membership_is_well_founded}) implies that the relation \\( \\in \\) is well-founded.\n\n In the terminology if graph theory, this well-foundedness means that, for every vertex \\( v \\), there exists no infinite \\hyperref[def:quiver_path]{path} that ends with \\( v \\).\n\n This implies that the graph is \\hyperref[def:undirected_multigraph_path]{acyclic}. For finite graphs the converse holds, but an infinite graph this is not so. For example, the chain \\eqref{eq:ex:infinite_integer_graphs/positive} is acyclic but not well-founded.\n\n Well-founded graphs are important for \\fullref{thm:well_founded_induction}.\n\\end{remark}\n\n\\begin{definition}\\label{def:quiver_free_category}\n Let \\( Q = (V, A, h, t) \\) be a \\hyperref[def:quiver]{quiver}. We define the \\term{free category} \\( F(Q) \\) generated by \\( Q \\) as follows:\n \\begin{itemize}\n \\item The \\hyperref[def:category/objects]{set of objects} \\( \\obj(F(Q)) \\) is the set of vertices \\( V \\).\n\n \\item The \\hyperref[def:category/morphisms]{set of morphisms} \\( \\cat{F(Q)}(v, w) \\) is the set of all paths from \\( v \\) to \\( w \\).\n\n \\item The \\hyperref[def:category/composition]{composition of the morphisms} \\( p \\) and \\( q \\) with \\( h(q) = t(p) \\) is the \\hyperref[def:undirected_multigraph_path/concatenation]{concatenation}\n \\begin{equation*}\n q \\bincirc p = p \\cdot q.\n \\end{equation*}\n\n \\item The \\hyperref[def:category/identity]{identity morphism} on the vertex \\( v \\) is the \\hyperref[def:undirected_multigraph_path/empty]{empty path} at \\( v \\). This is the primary motivation for having a distinct empty path at every vertex.\n \\end{itemize}\n\n Since \\( F(Q) \\) is already defined for every quiver \\( Q \\), if we also define how it acts on \\hyperref[eq:def:category_of_small_quivers/homomorphism]{quiver homomorphisms}, this will make \\( F \\) a \\hyperref[def:functor]{functor} from the category \\hyperref[def:category_of_small_quivers]{\\( \\ucat{Quiv} \\)} of \\( \\mscrU \\)-small quivers to the category \\hyperref[def:category_of_small_quivers]{\\( \\ucat{Cat} \\)} of \\( \\mscrU \\)-small categories, for every \\hyperref[def:grothendieck_universe]{Grothendieck universe} \\( \\mscrU \\) containing \\( Q \\).\n\n For every \\( \\mscrU \\)-small category \\( \\cat{C} \\) and every quiver homomorphism \\( (g_V, g_A): Q \\to U(\\cat{C}) \\), we define the following functor:\n \\begin{equation}\\label{eq:def:quiver_free_category/functor_from_homomorphism}\n \\begin{aligned}\n &G: F(Q) \\to \\cat{C}, \\\\\n &G(v) \\coloneqq g_V(v) \\\\\n &G(v, a_1, \\ldots, a_n) \\coloneqq \\begin{cases}\n \\id_v, &n = 0, \\\\\n g_A(a_n) \\bincirc G(v, a_1, \\ldots, a_{n-1}), &n > 0,\n \\end{cases}\n \\end{aligned}\n \\end{equation}\n\n The functor \\( G \\) \\enquote{evaluates} paths in \\( Q \\) inside \\( \\cat{C} \\). Put \\( F(f_V, f_A) \\coloneqq G \\). Parameterized on \\( (f_V, f_A) \\), this defines \\( F \\) is a functor from \\( \\ucat{Quiv} \\) to \\( \\ucat{Cat} \\).\n\n We will see in \\fullref{ex:def:category_adjunction/quiv_cat} that \\( F \\) is actually left adjoint to the forgetful functor \\( U \\). The new identity loops are an important part of this adjunction.\n\\end{definition}\n", "meta": {"hexsha": "14233c3fa779d47845d22b8d067bb82edcc9e5d1", "size": 26701, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/quivers.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/quivers.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/quivers.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 64.03117506, "max_line_length": 553, "alphanum_fraction": 0.6837571627, "num_tokens": 8240, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624789529376, "lm_q2_score": 0.7826624738835051, "lm_q1q2_score": 0.6125605519931029}} {"text": "\\chapter{Rosetta Stone}\nBayesian statistics has a lot of terminology floating around, and sometimes\ndifferent communities use different terms for the same concept. This appendix\nlists various common terms that are basically synonymous (they mean the same thing).\nI may even throw in the different terms from time to time, intentionally or\nunintentionally!\n\n\\subsection*{Event, Hypothesis, Proposition, Statement}\nThese are all basically the same thing.\nA proposition is something that can be either true or false, such as ``my age is\ngreater than 35'' or ``the number of Aardvarks in my room is either zero or\none''. These are the things that go in our\nprobability statements: If we write $P(A|B)$, $A$ and $B$ are both propositions,\nthat is, statements that may be true or false. Event is the preferred term in\nclassical statistics.\n\n\n\\subsection*{Sampling Distribution, Probability Model for the Data,\nGenerative Model, Likelihood Function, Likelihood}\nThis is the thing that we\nwrite as $p(x|\\theta)$. Sometimes it is called the sampling distribution or\na generative model because\nyou can sometimes think of the data as having been ``drawn from''\n$p(x|\\theta)$ but using the true value of $\\theta$, which you don't actually\nknow. There is a subtlety here, and that is that the word likelihood can be\nused to mean either $p(x|\\theta)$ before $x$ is known (in which case it is the\nthing you would use to predict possible data) or after $x$ is known. In the latter\ncase $p(x|\\theta)$ is only a function of $\\theta$ because $x$ is fixed at the\nobserved value. The term likelihood function is often used at this point.\n\n\\subsection*{Probability Distribution}\nThe term probability distribution is used to refer to either a probability density\nfunction (in the continuous case) or a probability mass function (in the discrete\ncase). The latter gives the probability of each particular value, whereas the\nformer only gives a probability if you integrate it within some region.\n\n\\subsection*{Marginal Likelihood, Evidence, Prior Predictive Probability, Normalising\nConstant}\nThis is the $p(x)$ or $P(x)$ term in the denominator of Bayes' rule,\nand it is also the total of the {\\tt prior} $\\times$ {\\tt likelihood} column\nof a Bayes' Box. This is the probability of getting the data that you actually got,\nbefore you observed it: hence the terminology ``prior predictive probability''.\nIt is also the thing you use to normalise the posterior distribution (make it sum\nor integrate to 1), hence the term normalising constant. Marginal likelihood makes\nsense because it is a probability of data (like the regular likelihood) but\n``marginalised'' (i.e. not caring about) the value of the parameter(s). It is\nalso called ``evidence'' because it can be used to compare different models, or\nto ``patch together'' two Bayes' Boxes after the fact (see the hypothesis testing\nchapter).\n\n", "meta": {"hexsha": "67d6f6708ea500d9157e57cd4a1acc749cacf849", "size": 2871, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "rosetta_stone.tex", "max_stars_repo_name": "xulinpan/stat331", "max_stars_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 55, "max_stars_repo_stars_event_min_datetime": "2015-03-09T18:03:23.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-25T03:36:54.000Z", "max_issues_repo_path": "rosetta_stone.tex", "max_issues_repo_name": "xulinpan/stat331", "max_issues_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2015-07-07T05:00:32.000Z", "max_issues_repo_issues_event_max_datetime": "2015-07-10T08:48:27.000Z", "max_forks_repo_path": "rosetta_stone.tex", "max_forks_repo_name": "xulinpan/stat331", "max_forks_repo_head_hexsha": "7ca22bf3ce2c43d1a3b5fd8ff22842bdeb87ecf5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 11, "max_forks_repo_forks_event_min_datetime": "2015-07-29T14:34:51.000Z", "max_forks_repo_forks_event_max_datetime": "2020-06-04T20:04:47.000Z", "avg_line_length": 56.2941176471, "max_line_length": 85, "alphanum_fraction": 0.7784743992, "num_tokens": 663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7826624688140726, "lm_q2_score": 0.7826624840223698, "lm_q1q2_score": 0.6125605519931027}} {"text": "% Document type\n\\documentclass[]{amsbook}\n%----------------------------------------------------------\n% Packages\n\\usepackage{amsmath}\n\\usepackage{amsthm}\n\\usepackage{amssymb}\n\\usepackage{mathrsfs}\n\\usepackage{mathtools}\n\\usepackage[pdftex, bookmarks, colorlinks=true]{hyperref}\n\\usepackage{tikz-cd}\n\\usepackage{bbm}\n%----------------------------------------------------------\n% User-defined environments\n\\theoremstyle{plain}\n\\newtheorem{acknowledgement}{Acknowledgement}\n\\newtheorem{algorithm}{Algorithm}[section]\n\\newtheorem{axiom}{Axiom}[section]\n\\newtheorem{case}{Case}[section]\n\\newtheorem{claim}{Claim}[section]\n\\newtheorem{conclusion}{Conclusion}[section]\n\\newtheorem{condition}{Condition}[section]\n\\newtheorem{conjecture}{Conjecture}[section]\n\\newtheorem{corollary}{Corollary}[section]\n\\newtheorem{criterion}{Criterion}\n\\newtheorem{definition}{Definition}[section]\n\\newtheorem{example}{Example}[section]\n\\newtheorem{exercise}{Exercise}[section]\n\\newtheorem{lemma}{Lemma}[section]\n\\newtheorem{notation}{Notation}[section]\n\\newtheorem{problem}{Problem}[section]\n\\newtheorem{proposition}{Proposition}\n\\newtheorem{remark}{Remark}[section]\n\\newtheorem{solution}{Solution}[section]\n\\newtheorem{summary}{Summary}[section]\n\\newtheorem{theorem}{Theorem}[section]\n\\numberwithin{equation}{section}\n\n\\theoremstyle{definition}\n\\newtheorem{xca}{Exercise}[section]\n%-----------------------------------------------------------\n% User-defined commands\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\Q}{\\mathbb{Q}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\C}{\\mathbb{C}}\n\n\\newcommand{\\dps}{\\displaystyle}\n\\newcommand{\\mb}[1]{\\mathbf{#1}}\n\\newcommand{\\mbb}[1]{\\mathbb{#1}}\n\\newcommand{\\mf}[1]{\\mathfrak{#1}}\n\\newcommand{\\mc}[1]{\\mathcal{#1}}\n\\newcommand{\\opn}[1]{\\operatorname{#1}}\n\\newcommand{\\tb}[1]{\\textbf{#1}}\n\n\\newcommand{\\catname}[1]{\\mathbf{#1}}\n\\newcommand{\\0}{\\mathbf{0}}\n\\newcommand{\\1}{\\mathbf{1}}\n\n\\newcommand{\\la}{\\langle}\n\\newcommand{\\ra}{\\rangle}\n\n\\newcommand{\\q}{\\quad}\n\\newcommand{\\qq}{\\quad \\quad}\n\\newcommand{\\qqq}{\\quad \\quad \\quad}\n\n\\newcommand{\\oldemptyset}{\\emptyset}\n\\renewcommand{\\emptyset}{\\varnothing}\n%-----------------------------------------------------------\n% Math operators\n\\DeclareMathOperator{\\End}{End}\n\\DeclareMathOperator{\\Hom}{Hom}\n\\DeclareMathOperator{\\lcm}{lcm}\n%-----------------------------------------------------------\n\\begin{document}\n\\frontmatter\n\\title[`Algebra: Chapter 0']{Solutions to Exercises from `Algebra: Chapter 0'}\n\n\\begin{abstract}\nSolutions to exercises from the book `Algebra: Chapter 0' by Paolo Aluffi.\n\\end{abstract}\n\n\\maketitle\n\\tableofcontents\n\n\\chapter*{Preface}\n\n\\markboth{PREFACE}{PREFACE} This document contains my attempt at writing\n(hopefully correct!) solutions to exercises from Aluffi's book, while engaging\nin some self-study of modern abstract algebra with the ultimate aim of teaching\nmyself some modern algebraic geometry.\n\\mainmatter\n\n\\include{chapters/set-theory-and-categories}\n\\include{chapters/groups-first-encounter}\n\n\\end{document}\n", "meta": {"hexsha": "ea8692fb0f3c052be3983fa91ccde7c724f3655d", "size": 3029, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "main.tex", "max_stars_repo_name": "vishallama/algebra-chapter0", "max_stars_repo_head_hexsha": "cef5cc84416953797e93b8da3ccf13d436b192be", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2015-02-05T20:56:28.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-02T09:21:00.000Z", "max_issues_repo_path": "main.tex", "max_issues_repo_name": "vishallama/algebra-chapter0", "max_issues_repo_head_hexsha": "cef5cc84416953797e93b8da3ccf13d436b192be", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "main.tex", "max_forks_repo_name": "vishallama/algebra-chapter0", "max_forks_repo_head_hexsha": "cef5cc84416953797e93b8da3ccf13d436b192be", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.595959596, "max_line_length": 79, "alphanum_fraction": 0.6876857049, "num_tokens": 855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7853085909370422, "lm_q2_score": 0.7799929104825007, "lm_q1q2_score": 0.6125351334718951}} {"text": "\\lab{Algorithms}{ARMA Models}{ARMA Models}\n\\label{lab:arma}\n\\objective{Fit and forecast ARMA models.}\n\nAn $\\text{ARMA}(p,q)$ model is a covariance-stationary discrete stochastic\nprocess $\\{z_t\\}$ that satisfies\n\\begin{align}\n \\label{eq:arma:def}\n z_t - \\mu = \\left(\\sum_{i=1}^p \\phi_{i}(z_{t - i} - \\mu)\\right) + a_t +\n \\left(\\sum_{j=1}^{q} \\theta_{j}a_{t-j} \\right)\n\\end{align}\nwhere $\\mu = E[z_t]$ and $a_t$ are identically Gaussian distributed with\nvariance $\\sigma_a^2$. We note that the assumption that $\\{z_t\\}$ is\ncovariance-stationary is eqivalent to the condition that the roots of the\npolynomial in $B$ \n\\begin{align}\n \\label{eq:arma:characteristic}\n \\phi(B) = 1 - \\sum_{i=1}^p\\phi_iB^i \n\\end{align} \nlie without the unit circle.\n\nThe first sum on the right hand side of \\ref{eq:arma:def} is interpreted as an\n``autoregression'' since it is a linear combination of previously observed\nvalues of $z_t$. The second sum is interpreted as a ``moving average'' of the\ncurrent and previous error terms; though formally similar to an average, note\nthat the $\\theta_j$ need not be positive nor sum to one. We say that an\n$\\text{ARMA}(p,q)$ model is an ``autoregressive moving-average model of order\np, q''.\n\n\\section*{Likelihood via Kalman Filter}\n\nIn a general $\\text{ARMA}(p,q)$ model, the likelihood is a function of the\nunobserved error terms $a_t$ and is not trivial to compute. Simple\napproximations can be made, but these may be inaccurate under certain\ncircumstances. Explicit derivations of the likelihood are possible, but\ntedious. However, when the $\\text{ARMA}$ model is placed in state-space, the\nKalman filter affords a straightforward, recursive way to compute the\nlikelihood.\n\nWe demonstrate a state-space representation of an $\\text{ARMA}(p,q)$ model. If\n$r = \\max(p, q+1)$, we write\n\\begin{align}\n F &= \\begin{bmatrix}\n \\phi_1 & \\phi_2 & \\cdots & \\phi_{r-1} & \\phi_r\\\\\n 1 & 0 & \\cdots & 0 & 0\\\\\n 0 & 1 & \\cdots & 0 & 0\\\\\n \\vdots & \\vdots & \\cdots & \\vdots & \\vdots\\\\\n 0 & 0 & \\cdots & 1 & 0\n \\end{bmatrix}\\\\\n H &= \\begin{bmatrix}\n 1 & \\theta_1 & \\theta_2 & \\cdots & \\theta_{r-1}\n \\end{bmatrix}\\\\\n Q &= \\begin{bmatrix}\n \\sigma_a^2 & 0 & \\cdots & 0\\\\\n 0 & 0 & \\cdots & 0\\\\\n \\vdots & \\vdots & \\cdots & \\vdots\\\\\n 0 & 0 & \\cdots & 0\n \\end{bmatrix}\\\\\n w_t &\\sim \\text{MVN}(0, Q)\n\\end{align}\nwhere $\\phi_i = \\theta_j = 0$ for $i > p, j > q$. Then the linear stochastic\ndynamical system \n\\begin{align}\n x_{t+1} &= Fx_t + w_t\\\\\n z_t &= Hx_t + \\mu\n\\end{align}\ndescribes the same process as the original $\\text{ARMA}$ model.\n\nLet $\\Theta = \\{\\phi_i, \\theta_j, \\mu, \\sigma_a^2\\}$ be the set of parameters\nfor an $\\text{ARMA}(p,q)$ model. Using the chain rule, we can factorize the\nlikelihood as\n\\begin{align}\n \\label{eq:arma:factorized}\n p(\\{z_t\\} | \\Theta) = \\prod_{t=1}^{n} p(z_t | z_{t-1}, \\ldots, z_{1},\n \\Theta)\n\\end{align}\nSince we have assumed that the error terms are Gaussian, each conditional\ndistribution in \\ref{eq:arma:factorized} is also Gaussian, and is completely\ncharacterized by its mean and variance. But these two quantities are easily\nfound via the Kalman filter, namely\n\\begin{align}\n \\text{mean} & \\quad H\\hat{x}_{t|t-1} + \\mu \\\\\n \\text{variance} & \\quad HP_{t|t-1}H^T\n\\end{align}\nwhere $\\hat{x}_{t|t-1}, P_{t|t-1}$ are found during the Predict step. Our\nlikelihood becomes\n\\begin{align}\n \\label{eq:arma:likelihood}\n p(\\{z_t\\} | \\Theta) = \\prod_{t=1}^{n} N(z_t;\\; H\\hat{x}_{t|t-1} + \\mu,\\;\n HP_{t|t-1}H^T)\n\\end{align}\nWe begin the recursion by letting\n\\begin{align}\n \\hat{x}_{1|0} &= \\mathbb{E}(x_1) = 0 \\\\\n \\text{vec}(P_{1|0}) &= \\mathbb{E}\\left[(x_1 - \\mathbb{E}x_1)(x_1 -\n \\mathbb{E}x_1)^T\\right] = \\left[I_{r^2} - (F \\otimes F)\\right]^{-1} \\cdot\n \\text{vec}(Q)\n\\end{align}\nwhere $\\text{vec}$ flattens a matrix and $\\otimes$ is the Kronecker product\n({\\tt numpy.kron}).\n\n\\begin{problem}\n\\label{prob:arma:likelihood}\nWrite a function that computes the log-likelihood of an $\\text{ARMA}(p,q)$\nmodel, given a time series $z_t$. You may reuse your Kalman filter code from\nLab \\ref{lab:kalman}.\n\n\\begin{lstlisting}\ndef arma_likelihood(time_series, phis=array([]), thetas=array([]), mu=0.,\n sigma=1.):\n \"\"\"\n Return the log-likelihood of the ARMA model parameters, given the time\n series.\n\n Parameters\n ----------\n time_series : ndarray of shape (n,1)\n The time series in question\n phis : ndarray of shape (p,)\n The phi parameters\n thetas : ndarray of shape (q,)\n The theta parameters\n mu : float\n The parameter mu\n sigma : float\n The parameter sigma\n\n Returns\n -------\n log_likelihood : float\n The log-likelihood of the model\n \"\"\"\n pass\n\\end{lstlisting}\n\n\\vspace{3mm} \\noindent\nFor example,\n\\begin{lstlisting}\n>>> arma_likelihood(time_series_a, phis=array([0.9]), mu=17., sigma=0.4)\n-77.6035\n\\end{lstlisting}\n\\end{problem}\n\n\\section*{Identification and Fitting}\n\nWhen modeling a data set with an $\\text{ARMA}(p,q)$ model, the order of the\nmodel must be determined as well as the other parameters. The process of\nchoosing $p$ and $q$ is called model identification. Different methods have\nbeen used; for example, Box and Jenkins propose a methodology that involves\nexamining the estimated autocorrelation and partial-autocorrelation functions\nof the data. We will choose $p$ and $q$ that minimize the Akaike information\ncriterion with a correction (AICc), given by\n\\begin{align}\n 2k\\left(1 + \\frac{k+1}{n-k}\\right) - 2 \\ell(\\Theta)\n\\end{align}\nwhere $n$ is the sample size, $k = p + q + 2$ is the number of parameters in\nthe model, and $\\ell(\\Theta)$ is the maximum likelihood for the model class.\n\nTo compute the maximum likelihood for a model class, we need to optimize\n\\ref{eq:arma:likelihood} over the space of parameters $\\Theta$. We can do so\nusing our function from Problem \\ref{prob:arma:likelihood} along with some\noptimization routine, such as {\\tt scipy.optimize.fmin}.\n\n\\begin{problem}\n\\label{prob:arma:mle}\nWrite a function that accepts a time series $\\{z_t\\}$ and returns the\nparameters of the model that minimizes the AICc, given the constraint that $p\n\\leq 3$, $q \\leq 3$.\n\n\\begin{lstlisting}\ndef arma_fit(time_series):\n \"\"\"\n Return the ARMA model that minimizes AICc for the given time series,\n subject to p,q <= 3. \n\n Parameters\n ----------\n time_series : ndarray of shape (n,1)\n The time series in question\n\n Returns\n -------\n phis : ndarray of shape (p,)\n The phi parameters\n thetas : ndarray of shape (q,)\n The theta parameters\n mu : float\n The parameter mu\n sigma : float\n The parameter sigma\n \"\"\"\n pass\n\\end{lstlisting}\n\n\\vspace{3mm} \\noindent\nFor example,\n\\begin{lstlisting}\n>>> arma_fit(time_series_a)\n(array([ 0.9087]), array([-0.5759]), 17.0652..., 0.3125...)\n\\end{lstlisting}\n\\end{problem}\n\n\\begin{problem}\n\\label{prob:arma:data}\nUse your solution from Problem \\ref{prob:arma:mle} to fit models to the data\nfound in {\\tt time\\_series\\_a.txt}, {\\tt time\\_series\\_b.txt}, {\\tt\ntime\\_series\\_c.txt}. Report the fitted parameters $p, q, \\Theta$.\n\\end{problem}\n\n\\section*{Forecasting}\nThe Kalman filter provides a straightforward way to predict future states, by\ngiving the mean and variance of the conditional distribution of future\nobservations.\n\\begin{align}\n z_{t + k} | z_{1}, \\cdots, z_{t} \\sim N(z_k;\\; H\\hat{x}_{t+k|t} + \\mu,\\;\n HP_{t+k|t}H^T)\n\\end{align}\nRecall the relations\n\\begin{align}\n \\hat{x}_{t+k|t} &= F\\hat{x}_{t+k-1|t}\\\\\n P_{t+k|t} &= FP_{t+k-1|t}F^T + Q\n\\end{align}\n\n\\begin{problem}\n\\label{prob:arma:forecast}\nForecast each data set ahead 20 intervals using the parameters discovered from\nProblem \\ref{prob:arma:data}, and plot their expected values along with the\noriginal data set. Also plot the expected values plus and minus $\\sigma_{t+k}$,\nand plus and minus $2\\sigma_{t+k}$ to demonstrate credible intervals. \n\n\\begin{lstlisting}\ndef arma_forecast(time_series, phis=array([]), thetas=array([]), mu=0.,\n sigma=1., future_periods=20):\n \"\"\"\n Return forecasts for a time series modeled with the given ARMA model.\n \n Parameters\n ----------\n time_series : ndarray of shape (n,1)\n The time series in question\n phis : ndarray of shape (p,)\n The phi parameters\n thetas : ndarray of shape (q,)\n The theta parameters\n mu : float\n The parameter mu\n sigma : float\n The parameter sigma\n future_periods : int\n The number of future periods to return\n\n Returns\n -------\n evls : ndarray of shape (future_periods,)\n The expected values of z for times n + 1, ..., n + future_periods\n sigs : ndarray of shape (future_periods,)\n The deviations of z for times n + 1, ..., n + future_periods\n \"\"\"\n pass\n\\end{lstlisting}\n\n\\vspace{3mm} \\noindent\nFor example,\n\\begin{lstlisting}\n>>> arma_forecast(time_series_a, phis, thetas, mu, sigma, 4)\n(array([ 17.3762, 17.3478, 17.322 , 17.2986]),\n array([ 0.3125, 0.3294, 0.3427, 0.3533]))\n\\end{lstlisting}\n\\end{problem}\n\n\n\\begin{comment}\n\nThe Box-Jenkins methodology may be too esoteric, in light of straightforward\nalternatives such as the AICc. I have decided to leave this material out.\n\n\\section*{Identification}\n\nIf we have a time series $\\{z_t\\}$ and wish to fit an $ARMA(p,q)$ to it, we\nmust first identify the order of the model, that is, what $p$ and $q$ should\nbe. We discuss a methodology presented by Box and Jenkins that depends on the\nautocorrelations and partial-autocorrelations of the data. Alternative methods\nexist, such as using the Akaike information criterion with a correction (AICc).\n\n\\subsection*{Autocorrelation}\n\nThe autocorrelation between between $z_{t}, z_{t+k}$ is defined as the\nnormalized covariance between the two random variables, verbosely given as\n\\begin{align}\n \\label{eq:arma:autocorrelation}\n \\text{cor}[z_{t}, z_{t+k}] = \\frac{E[(z_{t} - E[z_{t}])(z_{t+k} -\n E[z_{t+k}])]}{\\sqrt{E[(z_{t} - E[z_{t}])^2(z_{t+k} - E[z_{t+k}])^2]}}\n\\end{align}\nSince the $\\{z_t\\}$ is assumed to be covariance-stationary,\n\\ref{eq:arma:autocorrelation} depends only on the lag $k$, not on $t$. This\njustifies the notation\n\\begin{align}\n \\rho_k = \\text{cor}[z_{t}, z_{t+k}] \\quad \\forall t\n\\end{align}\nWhen $\\rho_k$ is considered a function of $k$, it is called the autocorrelation\nfunction. Since $\\rho_0 = 1, \\rho_{-k} = \\rho_k$, we will be interested in the\ncases when $k > 0$. We can estimate $\\rho_k$ from sample data as\n\\begin{align}\n \\label{eq:arma:autocorrelation_estimate}\n \\hat{\\rho}_k = \\frac{\\sum_{t=1}^{N-k}(z_t - \\hat{\\mu})(z_{t+k} -\n \\hat{\\mu})}{\\sum_{t=1}^{N}(z_t - \\hat{\\mu})^2}\n\\end{align}\nwhere $\\hat{\\mu}$ is the sample mean. This is one of a number of estimates of\nautocorrelation.\n\n\\begin{problem}\n\\label{prob:arma:autocorrelation}\nWrite a function that computes the autocorrelation estimate described in\n\\ref{eq:arma:autocorrelation_estimate}, given a time series and a lag $k$.\nPlot $\\hat{\\rho}_k$ for $1 \\leq k \\leq 20$ for each of the three data sets.\n\\end{problem}\n\n\\subsection*{Partial-Autocorrelation}\n\n\\end{comment}\n", "meta": {"hexsha": "8e4747a1f11631e92c6c0caa63f30aa2907070db", "size": 11230, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Algorithms/TimeSeries/TimeSeries.tex", "max_stars_repo_name": "m4webb/numerical_computing", "max_stars_repo_head_hexsha": "d26e5ace9dbb91cd87440d84f0bd05d4a46e6781", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Algorithms/TimeSeries/TimeSeries.tex", "max_issues_repo_name": "m4webb/numerical_computing", "max_issues_repo_head_hexsha": "d26e5ace9dbb91cd87440d84f0bd05d4a46e6781", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Algorithms/TimeSeries/TimeSeries.tex", "max_forks_repo_name": "m4webb/numerical_computing", "max_forks_repo_head_hexsha": "d26e5ace9dbb91cd87440d84f0bd05d4a46e6781", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-12-08T01:19:23.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-08T01:19:23.000Z", "avg_line_length": 34.984423676, "max_line_length": 79, "alphanum_fraction": 0.6727515583, "num_tokens": 3545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929104825006, "lm_q2_score": 0.785308580887758, "lm_q1q2_score": 0.6125351256335246}} {"text": "\\documentclass[../PHYS306Notes.tex]{subfiles}\n\n\\begin{document}\n\\section{Lecture 3}\n\\subsection{Lecture Notes - Lagrangian Mechanics Part 3}\n\\subsubsection{Review of Variations}\nRecall \\textbf{Hamilton's Principle}, where we consider a trajectory between $t_1$, $t_2$, and variations from that trajectory $\\eta(t)$. The true trajectory taken by the physical system is given by $\\bar{q}(t)$. We parameterize the variations with an $\\alpha$ term, where $q(t) = \\bar{q}(t) + \\alpha\\eta(t)$. $\\bar{q}(t)$ minimizes the action functional $S[q]$ if $S[\\bar{q}] < S[q]$ for any other trajectories $q$. Our functional $S[q]$ is given by:\n\\[S[q] = \\int_{t_1}^{t_2}\nL(q, \\dot{q}, t)dt\\]\nWe found a method to find $\\bar{q}$ by taking the derivative of the action (i.e. a stationary point):\n\\[\\left. \\dod{S[\\bar{q} + \\alpha \\eta]}{\\alpha}\\right|_{\\alpha = 0} = 0\\]\nor alternatively, the statement that the (first order) variation must vanish:\n\\[\\delta S[\\bar{q}] = \\lim_{\\alpha \\rightarrow 0} \\frac{1}{\\alpha}\\left(S[\\bar{q} + \\alpha\\eta] - S[\\bar{q}]\\right) = 0\\]\nWe will explore this more on Monday when we look at constraint forces. \n\n\\subsubsection{Invariance, The Functional Derivative, and Multiple Variables}\n\\begin{enumerate}\n \\item Form invariance of Euler-Lagrange equations. Going from generalized coordinates $q \\rightarrow \\tilde{q}$, then we have that:\n \\[\\LL(\\tilde{q}, \\dot{\\tilde{q}}, t) = \\LL(q, \\dot{q}, t)\\]\n by the definition of the Lagrangian. Alternatively, see that the action is invariant of the choice of general coordinates:\n \\[\\tilde{S}[\\tilde{q}] = \\int_{t_{1}}^{t_2} \\tilde{\\LL}\\left(\\tilde{q}, \\dot{\\tilde{q}}, t\\right) d t=\\int_{t_{1}}^{t_2} \\mathcal{L}(q, \\dot{q}, t) d t=S[q]\\]\n \\item One can take a functional derivative and set this to zero. The variation can be written using a functional derivative (indicated by lowercase $\\delta$):\n \\[\\delta S[q] = \\int \\frac{\\delta S}{\\delta q} \\eta(t) dt\\]\n What is the definition of the functional derivative? It's very similar to the conventional derivative:\n \\[\\frac{\\delta S}{\\delta q(t)} = \\lim_{\\alpha = 0} \\frac{S[q + \\alpha \\delta(t)] - S[q]}{\\alpha} = \\dpd{\\LL}{q} - \\dod{}{t}\\dpd{\\LL}{\\dot{q}}\\]\n Therefore, $\\frac{\\delta S}{\\delta q} = 0$ implies the Euler Lagrange equations. The way you can think about it is like the partial derivative. We are familiar with the total differential of a function (the sum of the partial differentials, see HW1 as an example). Now, imagine we have a function that depends continuously on a function $q$; its like a partial derivative with an index that is continuous (rather than discrete). \n \\item If we have multiple variables, the generalization is straightforwards; for $n$ variables, the action functional becomes:\n \\[S = \\int_{t_1}^{t_2}\\LL(q_1, q_2, \\cdots, q_n, \\dot{q}_1, \\dot{q}_2, \\cdots, \\dot{q}_n, t)dt\\]\n So we would get $n$ Euler-Lagrange equations:\n \\[\\dpd{\\LL}{q_i} = \\dod{}{t}\\dpd{\\LL}{\\dot{q}_i}\\]\n Where $\\dpd{\\LL}{q_i}$ is the generalized force, and $\\dpd{\\LL}{\\dot{q}_i}$ is the generalized momentum. \n\\end{enumerate}\n\n\\subsubsection{Torque, Angular Momentum, and Generalized Momentum Conservation}\nSee derivation in Worksheet three for the relation:\n\\[\\dpd{\\LL}{\\phi} = \\dod{}{t}\\dpd{\\LL}{\\dot{\\phi}}\\]\nWhich corresponds to:\n\\[\\Gamma = \\dod{L}{t}\\]\nSo if there is zero torque, we have conservation of angular momentum. Generally, if $\\LL$ is independent of $q_i$, then the generalized momentum $\\pd{\\LL}{\\dot{q}_i}$ is conserved. We have a conservation law that comes from a certain property in the Lagrangian. We call variables that do not appear in the Lagrangian as a \"cyclic\" or \"ignorable\" variable.\n\\end{document}", "meta": {"hexsha": "53d6182c93246345c7feff7a0df6c54f3cb18ac4", "size": 3694, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture-3/Lecture-Notes-3.tex", "max_stars_repo_name": "RioWeil/PHYS306-notes", "max_stars_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Lecture-3/Lecture-Notes-3.tex", "max_issues_repo_name": "RioWeil/PHYS306-notes", "max_issues_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture-3/Lecture-Notes-3.tex", "max_forks_repo_name": "RioWeil/PHYS306-notes", "max_forks_repo_head_hexsha": "9394a8cd986722b6fdcb57c8846c6b0d52c23188", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 92.35, "max_line_length": 451, "alphanum_fraction": 0.6935571197, "num_tokens": 1156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7799929053683038, "lm_q2_score": 0.785308580887758, "lm_q1q2_score": 0.612535121617302}} {"text": "\\section{Searching}\n\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{$\\max/\\min$ differences (Problem 4.5)}\n \\begin{enumerate}[(a)]\n\t\\item unsorted; $\\max |x - y|$; $O(n)$\n\t\\item sorted; $\\max |x - y|$; $O(1)$\n\t\\item unsorted; $\\min |x - y|$; $O(n \\log n)$\n\t\\item sorted; $\\min |x - y|$; $O(n)$\n \\end{enumerate}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Searching in matrix (Problem 4.6)}\n \\begin{itemize}\n\t\\item $M$: matrix $m \\times n$\n\t\\item row: increasing from left to right\n\t\\item col: increasing from top to down\n\t\\item Is $x \\in M$?\n \\end{itemize}\n\n \\pause\n \\centerline{Divide and conquer.}\n\n \\pause\n \\[\n\tT(m,n) = 3T(\\frac{m}{2}, \\frac{n}{2}) + 1 \n \\]\n\n % \\pause\n % \\[\n % m = n \\implies T(n) = 3T(\\frac{n}{2}) + 1 \\pause \\implies T(n) = \\Theta(n^{\\log_{2}3})\n % \\]\n\n \\pause\n \\centerline{Always checking the lower left corner.}\n\n \\pause\n \\[\n\tT(m, n) = m + n - 1\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Searching in matrix (Problem 4.6)}\n \\[\n\t\\text{Assume } M: n \\times n \n \\]\n\n \\[\n\tW(n) \\le 2n - 1\n \\]\n\n \\pause\n \\vspace{0.30cm}\n \\centerline{$W(n) \\ge 2n - 1$ by adversary argument!}\n\n \\pause\n \\begin{align*}\n\ti + j \\le n - 1 &\\implies x > M_{ij} \\\\\n\ti + j > n - 1 &\\implies x < M_{ij} \\\\\n \\end{align*}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Local minimum (Problem 4.11)}\n \\begin{itemize}\n\t\\item Array $A[0 \\ldots n-1]$\n\t\\item Boundary conditions:\n\t \\begin{align*}\n\t\tA[0] &\\ge A[1] \\\\\n\t\tA[n-2] &\\le A[n-1]\n\t \\end{align*}\n\t\\item Local minimum $A[i]$:\n\t \\[\n\t\tA[i-1] \\ge A[i] \\le A[i+1]\n\t \\]\n\t\\item Goal: Find \\emph{any} local minimum.\n \\end{itemize}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Local minimum (Problem 4.11)}\n \\begin{enumerate}\n\t\\item Checking each element:\n\t \\[\n\t\tT(n) = O(n)\n\t \\]\n\t\\item $\\min A$:\n\t \\[\n\t\tT(n) = O(n)\n\t \\]\n\t\\pause\n\t\\item Required:\n\t \\[\n\t\tT(n) = O(\\log n)\n\t \\]\n\n\t \\pause\n\t \\[\n\t\tT(n) = T(\\frac{n}{2}) + 1\n\t \\]\n \\end{enumerate}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Local minimum (Problem 4.11)}\n 2D local minimum:\n \\begin{itemize}\n\t\\item Matrix $M: n \\times n$\n\t\\item Boundary conditions:\n\t \\[\n\t\t\\infty\n\t \\]\n\t\\item Local minimum $A[i,j]$:\n\t \\begin{align*}\n\t\tA[i,j-1] &\\ge A[i,j] \\le A[i,j+1] \\\\\n\t\tA[i-1,j] &\\ge A[i,j] \\le A[i+1,j]\n\t \\end{align*}\n\t\\item Goal: Find \\emph{any} local minimum.\n \\end{itemize}\n\n \\pause\n \\[\n\tO(n^2) \\pause \\implies O(n \\log n) \\pause \\implies O(n) \\implies \\textcolor{red}{O(\\log n)}\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{$a_i = i$ (Problem 4.2)}\n \\begin{itemize}\n\t\\item Sorted integer sequence $\\set{a_1, a_2, \\ldots, a_n}$:\n\t \\[\n\t\t\\forall i \\neq j: a_i \\neq a_j\n\t \\]\n\t\\item Goal: \n\t \\[\n\t\t\\exists? i: a_i = i\n\t \\]\n \\end{itemize}\n\n \\pause\n \\[\n\tT(n) = O(n)\n \\]\n\n \\pause\n \\[\n\tT(n) = T(\\frac{n}{2}) + 1 = O(\\log n)\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Smallest missing positive integer (Problem 4.3)}\n \\begin{itemize}\n\t\\item Sorted array $A[1 \\ldots n]$:\n\t \\begin{gather*}\n\t\ta_i \\in \\mathbb{Z}^{+} \\\\\n\t\t\\forall i \\neq j: a_i \\neq a_j\n\t \\end{gather*}\n\t\\item Goal: Find the smallest missing positive integer.\n \\end{itemize}\n\n \\pause\n \\[\n\tT(n) = O(n)\n \\]\n\n \\pause\n \\[\n\tT(n) = T(\\frac{n}{2}) + 1 = O(\\log n)\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Computing $\\lceil \\sqrt{N} \\rceil$ (Problem 4.12)}\n \\begin{itemize}\n\t\\item Given an $n$-bit natural number $N$ ($0 \\le N < 2^{n} - 1$)\n\t\\item Goal: Compute $\\lceil \\sqrt{N} \\rceil$ using $O(n)$ additions and shifts.\n \\end{itemize}\n \n \\vspace{0.30cm}\n \\begin{columns}\n \\column{0.55\\textwidth}\n\t \\pause\n\t \\begin{block}{Elementrary operations:}\n\t \\begin{itemize}\n\t \\item $n$-bit + $n$-bit: $O(1)$\n\t \\item $n$-bit shifted by $1$-bit: $O(1)$\n\t \\item $\\Rightarrow x^2: O(n)$\n\t \\end{itemize}\n\t \\end{block}\n\t\\column{0.45\\textwidth}\n\t \\pause\n\t \\begin{enumerate}[<+->]\n\t \\item Na{\\\"i}ve search: $O(2^{n} \\cdot n)$\n\t \\item Binary search: $O(n \\cdot n)$\n\t \\item Binary search in range:\n\t\t \\[\n\t\t\t2^{\\lfloor \\frac{n-1}{2} \\rfloor} \\le \\lceil \\sqrt{N} \\rceil \\le 2^{\\lceil \\frac{n}{2} \\rceil}\n\t\t \\]\n\t\t \\[\n\t\t\t\\lg{(2^{\\lceil \\frac{n}{2} \\rceil} - 2^{\\lfloor \\frac{n-1}{2} \\rfloor})} = O(n)\n\t\t \\]\n\t\t \\[ O(n \\cdot n) \\]\n\t \\end{enumerate}\n \\end{columns}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Computing $\\lceil \\sqrt{N} \\rceil$ (Problem 4.12)}\n \\begin{block}{A Little History:}\n \\begin{description}\n\t \\item[$2007$:] Mid-term problem \\\\\n\t\t\\centerline{$O(n)$ required; NO $O(n)$ solutions, however}\n\t \\item[$\\sim 2013$:] $O(n^2)$\n\t \\pause\n\t \\item[$2014$:] $O(n)$\n\t\t\\fignocaption{width = 0.80\\textwidth}{figs/square-root-cs-se.png}\n \\end{description}\n \\end{block}\n\n \\pause\n \\[\n\tx^2: O(n) \\to O(1)\n \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Computing $\\lceil \\sqrt{N} \\rceil$ (Problem 4.12)}\n Given\n \\[\n\tM = \\lfloor N/4 \\rfloor\n \\]\n\n \\[ \n\tx = \\lceil \\sqrt{M} \\rceil \\textrm{ and } (x, x^2),\n \\]\n\n what is\n \\[ \n y = \\lceil \\sqrt{N} \\rceil \\textrm{ and } (y, y^2)? \n \\]\n\n \\pause\n \\begin{exampleblock}{An Example:}\n \\[\n\t\\begin{array}{lll}\n\t N = 280 & y = \\lceil \\sqrt{280} \\rceil = 17 & y^2 = 289 \\\\ \\pause\n\t M = \\lfloor 280/4 \\rfloor = 70 & x = \\lceil \\sqrt{70} \\rceil = 9 & x^2 = 81 \\\\ \\pause\n\t M = \\lfloor 70/4 \\rfloor = 17 & x = \\lceil \\sqrt{17} \\rceil = 5 & x^2 = 25 \\\\\n\t M = \\lfloor 17/4 \\rfloor = 4 & x = \\lceil \\sqrt{4} \\rceil = 2 & x^2 = 4 \\\\\n\t M = \\lfloor 4/4 \\rfloor = 1 & x = \\lceil \\sqrt{1} \\rceil = 1 & x^2 = 1\n\t\\end{array}\n \\]\n \\end{exampleblock}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Computing $\\lceil \\sqrt{N} \\rceil$ (Problem 4.12)}\n \\begin{algorithm}[H]\n \\caption{Computing $\\lceil \\sqrt{N} \\rceil$.}\n \\begin{algorithmic}[]\n \\Procedure{Sqrt-Root}{$N$}\n\t \\If{$N < 3$}\n\t \\State \\Return $1 \\Rightarrow (1,1); 2 \\Rightarrow (2,4); 3 \\Rightarrow (2,4)$\n\t \\EndIf\n\t \\State $M \\gets \\lfloor N/4 \\rfloor$\n\t \\State $(x, x^2) \\gets \\textsc{Sqrt-Root}{(M)}$\n\t\t\\State \\Return the $(y, y^2)$ with $y^2 \\sim N$:\n \\[\n\t\t\t(y, y^2) = \\left\\{ \\begin{array}{ll}\n\t\t\ty = 2x & y^2 = 4x^2 \\\\\n\t\t\ty = 2x + 1 & y^2 = 4x^2 + 4x + 1 \\\\\n\t\t\ty = 2x - 1 & y^2 = 4x^2 - 4x + 1\n\t\t\t\\end{array} \\right.\n\t \\]\n\t \\EndProcedure\n \\end{algorithmic}\n \\end{algorithm}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Computing $\\lceil \\sqrt{N} \\rceil$}\n \\[ \\xout{T(n) = T(n/4) + O(1) = \\Theta(\\lg n)} \\]\n\n \\[ T(n) = T(n - 2) + O(1) = \\Theta(n) \\]\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}{Space for hashing (Problem 4.4)}\n \\begin{description}\n\t\\item[Key:] $x$\n\t\\item[Node:] $y$\n\t\\item[Load factor:] $\\alpha$\n \\end{description}\n\n \\pause\n \\begin{itemize}\n\t\\item Closed-address hashing\n\t \\[\n\t\th_c + \\alpha y h_c\n\t \\]\n\t\\item Open-address hashing\n\t \\[\n\t\t\\frac{\\alpha h_c}{\\frac{h_c + \\alpha y h_c }{x}} = \\frac{\\alpha x}{1 + \\alpha y}\n\t \\]\n \\end{itemize}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n\\begin{frame}[noframenumbering]\n \\fignocaption{width = 0.50\\textwidth}{figs/thankyou.jpg}\n\\end{frame}\n%%%%%%%%%%%%%%%%%%%%\n", "meta": {"hexsha": "9f924b8e7bee76ffa964b9ed8e551de170ece26c", "size": 6972, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-search-selection-20170410/sections/search.tex", "max_stars_repo_name": "hengxin/algorithm-ta-tutorial", "max_stars_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 45, "max_stars_repo_stars_event_min_datetime": "2017-03-29T08:22:32.000Z", "max_stars_repo_stars_event_max_datetime": "2020-04-02T15:12:15.000Z", "max_issues_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-search-selection-20170410/sections/search.tex", "max_issues_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_issues_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "alg-ta-by-years/alg-ta-2017/algorithm-tutorial-search-selection-20170410/sections/search.tex", "max_forks_repo_name": "courses-at-nju-by-hfwei/algorithm-ta-tutorial", "max_forks_repo_head_hexsha": "0bb0376d96f388671597903fc833f68d7946020e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-10T08:47:12.000Z", "max_forks_repo_forks_event_max_datetime": "2019-06-18T07:58:43.000Z", "avg_line_length": 23.0860927152, "max_line_length": 97, "alphanum_fraction": 0.5243832473, "num_tokens": 2887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.7799929053683038, "lm_q1q2_score": 0.612535121617302}} {"text": "\\chapter{GPS}\n\nSuppose that the Earth is a spheroid\n\\begin{equation}\n \\label{eq:spheroid}\n \\frac{x^2}{a^2} + \\frac{y^2}{a^2} + \\frac{z^2}{b^2} = 1, \\quad a\n \\geq b > 0,\n\\end{equation}\nwith its centre of symmetry in the origin of the Cartesian coordinate\nsystem and axes overlapping axes of coordinate system.\n\nLet $P$ be a point on the ellipsoid~\\eqref{eq:spheroid}. Define\ngeographical coordinates: latitude and longitude as follows. Latitude\nof $P$ is the angle between the normal to the ellipsoid at the point\n$P$ and the plane $z = 0$ (see figure \\ref{fig:spheroid_coordinates}).\nLongitude of $P$ is the angle between the plane $y = 0$ and the plane\nperpendicular to the plane $z = 0$ passing through the point $P$ and\nthe origin of the coordinate system.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\resizebox{6cm}{!}{\\includegraphics{1.pdf}}\n\\end{center}\n\\caption{Cartesian and geographic coordinates on the spheroid.}\n\\label{fig:spheroid_coordinates}\n\\end{figure}\n\nDenote latitude by $-\\frac{\\pi}{2} \\leq \\varphi \\leq \\frac{\\pi}{2}$,\nlongitude by $-\\pi \\leq \\lambda \\leq \\pi$. The $X$ axis passes through\nthe point $(a, 0, 0)$ with latitude $0^\\circ$ and longitude $0^\\circ$\n($\\varphi = 0, \\lambda = 0$), the $Y$ axis passes through the point\n$(0, a, 0)$ with latitude $0^\\circ$ and longitude $90^\\circ$ east\n($\\varphi = 0, \\lambda = \\frac{\\pi}{2}$) and the $Z$ axis passes\nthrough the north pole $(0, 0, b)$ with latitude $90^\\circ$ north\n($\\varphi = \\frac{\\pi}{2}, -\\pi \\leq \\lambda \\leq \\pi$).\n\n\\section{Conversion of coordinates}\n\nLet $P$ be a point on the ellipsoid~\\eqref{eq:spheroid} with latitude\n$0 < |\\varphi| < \\frac{\\pi}{2}$ and longitude $-\\pi \\leq \\lambda \\leq\n\\pi$. We are looking for its coordinates $(x, y, z)$. As the\n$Z$-coordinate depends only on $\\varphi$, take a point $P'$ with\nlatitude $\\varphi$ and longitude $0$, lying on the plane $y = 0$ (see\nfigure \\ref{fig:plane_y_0}). The line passing through $P'$ and\ninclined to the $x$-axis at the angle $\\varphi$ is a normal to the\nellipse\n\\begin{equation} \\label{eq:ellipseXZ}\n \\frac{x^2}{a^2} + \\frac{z^2}{b^2} = 1\n\\end{equation}\nat a point $(x_0, z_0)$ and intersects the $x$-axis at the point\n$(x_1, 0)$.\n\n\\begin{figure}[ht]\n\\begin{center}\n\\resizebox{6cm}{!}{\\includegraphics{2.pdf}}\n\\end{center}\n\\caption{Position of the point $P'$ on the plane $y = 0$.}\n\\label{fig:plane_y_0}\n\\end{figure}\n\n\\noindent Hence it follows that\n\\begin{equation} \\label{eq:first_eq_for_z}\n \\tg \\varphi = \\frac{z_0}{x_0 - x_1}.\n\\end{equation}\nThis normal has the equation\n\\[\nz_0x - (x_0 - x_1)z - z_0x_1 = 0.\n\\]\nOn the other hand, a normal to the ellipse \\eqref{eq:ellipseXZ} at its\npoint $(x_0, z_0)$ has the equation \\cite [pp.~215,~244]\n{bronsztejn-siemiendiajew-musiol-muhlig-2004}\n\\[\na^2z_0x - b^2x_0z - (a^2 - b^2)x_0z_0 = 0.\n\\]\nThese two lines must overlap, so there must be\n\\[\n\\frac{z_0}{a^2z_0} = \\frac{x_0 - x_1}{b^2x_0} = \\frac{z_0x_1}{(a^2 -\n b^2)x_0z_0}\n\\]\nand\n\\begin{equation} \\label{eq:x1}\n x_0 - x_1 = \\frac{b^2}{a^2}x_0, \\quad x_1 = \\frac{a^2 -\n b^2}{a^2}x_0.\n\\end{equation}\nFrom \\eqref{eq:ellipseXZ} we have\n\\begin{equation} \\label{eq:x0_from_XZ}\n x_0 = \\frac{a}{b}\\sqrt{b^2 - z_0^2}.\n\\end{equation}\nUsing \\eqref{eq:x1} and \\eqref{eq:x0_from_XZ} in\n\\eqref{eq:first_eq_for_z} we obtain\n\\[\nz_0 = p \\frac{\\sin \\varphi}{\\sqrt{1 - e^2\\sin^2 \\varphi}},\n\\]\nwhere $p = b^2 / a$ \\cite [p.~214]\n{bronsztejn-siemiendiajew-musiol-muhlig-2004} and\n\\[\ne = \\frac{\\sqrt{a^2 - b^2}}{a}\n\\]\nis the eccentricity of the ellipse~\\eqref{eq:ellipseXZ} \\cite [p.~214]\n{bronsztejn-siemiendiajew-musiol-muhlig-2004}.\nFrom~\\eqref{eq:x0_from_XZ} we obtain\n\\[\nx_0 = a \\frac{\\cos \\varphi}{\\sqrt{1 - e^2\\sin^2 \\varphi}}.\n\\]\n$X$ and $Y$ coordinates can be calculated from the parametric equation\nof the circle $x^2 + y^2 = x_0^2$ for the angle $\\lambda$: $x = x_0\n\\cos \\lambda$, $y = x_0 \\sin \\lambda$.\n\nThe complete coordinates are\n\\begin{align*}\n x &= a \\frac{\\cos \\varphi}{\\sqrt{1 - e^2\\sin^2 \\varphi}} \\cos \\lambda \\\\\n y &= a \\frac{\\cos \\varphi}{\\sqrt{1 - e^2\\sin^2 \\varphi}} \\sin \\lambda \\\\\n z &= p \\frac{\\sin \\varphi}{\\sqrt{1 - e^2\\sin^2 \\varphi}}\n\\end{align*}\nThese formulas are valid also for $\\varphi = 0$ and $\\varphi = \\pm\n\\frac{\\pi}{2}$.\n\nIf a point lies on the normal with distance $h$ from the surface of\nthe ellipsoid, then the corresponding values of $z_0$ and $x_0$ should\nbe increased by $h \\sin \\varphi$ and $h \\cos \\varphi$, respectively.\nTherefore, if a point is specified by coordinates $(\\varphi, \\lambda,\nh)$, corresponding Cartesian coordinates are\n\\begin{align}\n \\label{eq:x_coord}\n x &= \\left( \\frac{a}{\\sqrt{1 - e^2\\sin^2 \\varphi}} + h \\right) \\cos\n \\varphi \\cos \\lambda \\\\\n \\label{eq:y_coord}\n y &= \\left( \\frac{a}{\\sqrt{1 - e^2\\sin^2 \\varphi}} + h \\right) \\cos\n \\varphi \\sin \\lambda \\\\\n \\label{eq:z_coord}\n z &= \\left( \\frac{p}{\\sqrt{1 - e^2\\sin^2 \\varphi}} + h \\right) \\sin\n \\varphi\n\\end{align}\n\n\\section{Parameters of the spheroid}\n\nWe specify the spheroid of the Earth by two parameters\\footnote{See\n \\url{https://en.wikipedia.org/wiki/World_Geodetic_System}.}:\nsemi-major axis $a = 6378137.0$ metres and inverse flattening $1 / f =\n298.257223563$, where $f = (a - b) / a$.\n", "meta": {"hexsha": "570062dc336898c03411daa54edf9ecf999bcf55", "size": 5164, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/gps.tex", "max_stars_repo_name": "shgalus/shg", "max_stars_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 9, "max_stars_repo_stars_event_min_datetime": "2015-05-21T04:14:50.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-31T17:15:15.000Z", "max_issues_repo_path": "doc/gps.tex", "max_issues_repo_name": "shgalus/shg", "max_issues_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2015-05-21T05:31:04.000Z", "max_issues_repo_issues_event_max_datetime": "2015-05-21T05:31:04.000Z", "max_forks_repo_path": "doc/gps.tex", "max_forks_repo_name": "shgalus/shg", "max_forks_repo_head_hexsha": "0318d0126cf12c3236183447d130969c468a02fa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2015-05-21T04:14:44.000Z", "max_forks_repo_forks_event_max_datetime": "2020-08-18T12:35:22.000Z", "avg_line_length": 36.6241134752, "max_line_length": 74, "alphanum_fraction": 0.6719597211, "num_tokens": 1991, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.785308580887758, "lm_q2_score": 0.7799928900257126, "lm_q1q2_score": 0.6125351095686334}} {"text": "\\chapter*{\\uppercase{Abbreviations and Notation} \\label{appendix-notation}}\n\n\\section{Abbreviations}\n\n\\begin{description}[leftmargin=!,labelwidth=0.5in,font=\\normalfont]\n\t\\item\n\\end{description}\n\n\n\\section{Mathematical Notation}\n\\begin{description}[leftmargin=!,labelwidth=0.5in,font=\\normalfont]\n\t\\item\n\\end{description}\n\n\n\\subsection{General Notation}\n\n% EDIT BELOW AND REMOVE WHAT YOU DO NOT NEED\n\\begin{description}[leftmargin=!, labelwidth=0.7in]\n \\item[$x$] italicized, Roman or Greek letter, denotes scalar values\n \\item[$\\bs{x}$] italicized, bold, lowercase Roman or Greek letter, denotes a column vector\n \\item[$\\bs{X}$] italicized, bold, uppercase Roman or Greek letter, denotes a matrix\n\t\\item[$\\bs{X}_{ij}$] \t denotes the component of matrix $\\bs{X}$ occupying row $i$ and column $j$\n\n \\item[$x \\in (a, b)$] the value $x$ is within the interval such that $a < x < b$\n \\item[{$x \\in (a, b]$}] the value $x$ is within the interval such that $a < x \\leq b$\n \\item[$x \\in [a, b)$] the value $x$ is within the interval such that $a \\leq x < b$\n \\item[{$x \\in [a, b]$}] the value $x$ is within the interval such that $a \\leq x \\leq b$\n \\item[$1_{A}\\left(x\\right)$] the indicator function,\\[1_{A}\\left(x \\right) = \\begin{cases} 1 & x \\in A \\\\ 0 & x \\notin A \\end{cases}.\\]\n \\item[$\\bs{1}_n$] an column vector of $n$ $1$s\n \\item[$\\bs{I}$] the identity matrix\n \\item[$\\bs{I}_n$] the $n \\times n$ identity matrix\n \\item[$\\bs{X}^{-1}$] the inverse matrix, i.e., the operator satisfying $\\bs{X}^{-1} \\bs{X} = \\bs{X} \\bs{X}^{-1} = \\bs{I}.$\n \\item[$\\bs{X}^{\\top{T}}$] transpose, i.e. the operator satisfying $X_{i,j} = X_{j, i} \\; \\forall \\; i,j \\in \\mathbb{N}$\n \\item[$\\otimes$] Kronecker product\n \\item[$\\odot$] element-wise multiplication\n\\end{description}\n\n\\subsection{Sets}\n\n\\begin{description}[leftmargin=!, labelwidth=0.8in]\n \\item[$\\set{x, y, z, \\ldots}$] the set comprising the elements of $x,$ $y,$ $z,$ $\\ldots$\n \\item[$\\set{x, y, z} \\backslash x$] the set comprising the elements of $y$ and $z$\\footnote{In other words, the backslash removes an element (or a subset) from the original set.}\n \\item[$\\sett{x_i}{i = 1}{n}$] the set comprising the elements of $x_1, x_2, x_3, \\ldots, x_n$\n \\item[$\\RR$] set of real numbers\n\t\\item[$\\NN$] set of natural numbers\n \\item[$\\bs{x} \\in \\mathbb{R}^n$] $\\bs{x}$ is a vector with $n$ elements, all of which are real numbers\n\\end{description}\n\n\\subsection{Statistical Distributions}\n\\begin{description}[leftmargin=!, labelwidth=0.8in]\n \\item[$\\mathcal{N} \\left( \\mu, \\sigma^2 \\right)$] the uni-variable Gaussian distribution with mean $\\mu$ and variance $\\sigma^2$\n \\item[$\\mathcal{N} \\left( \\bs{\\mu}, \\bs{\\Sigma} \\right)$] the multi-variable Gaussian distribution with mean vector $\\bs{\\mu}$ and variance-covariance matrix $\\bs{\\Sigma}$\n \\item[$\\phi(x)$] the standard Gaussian density function\n\\end{description}\n\n\\subsection{Specialized Notation}\nThis notation refers to values defined explicitly in the context of this thesis.\n\n\\begin{itemize}\n\\itembox{$\\M$} a model that takes input parameters to an (observable) state space\n\\itembox{$u$} an observable state space from which data is to be collected\n\\itembox{$\\param$} a (model) parameter into model $\\M$\n\\itembox{$\\obs$} a Parameter-to-Observables (PtO) map, also denoted $\\obs( u\\lam )$ or $\\obs\\lam$, each component of which is a functional $\\obs_i: u\\lam \\to \\RR$ on the observable state\\footnote{This map represents performing an individual experiment, which may consist of one or more observations (either in space or time). Arranging observational data into a vector defines the map.}\n\\itembox{$\\qoi$} a Quantity of Interest (QoI) map, also denoted $\\qlam$, which acts on observable data to transform the data into a scalar or vector quantity\\footnote{For example, it may be the average of measurements encompassed in $\\obs$. Technically, this map encompasses the following set of compositions: $Q(\\obs(u\\lam))$.}\n\\itembox{$\\data$} data representing an individual output of a (possibly vector-valued) QoI map for a particular parameter $\\param$, i.e. $\\qlam = \\data$\n\\end{itemize}\n\n[TK - add Measure-Theory notation into this section, based on what you have in your newcommands.tex file]\n", "meta": {"hexsha": "13606a15c0063c5e9fdcb2b3216ddf74a223d6e0", "size": 4251, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "notation.tex", "max_stars_repo_name": "mathematicalmichael/thesis", "max_stars_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-04-24T08:05:49.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-28T20:34:29.000Z", "max_issues_repo_path": "notation.tex", "max_issues_repo_name": "mathematicalmichael/thesis", "max_issues_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 59, "max_issues_repo_issues_event_min_datetime": "2019-12-27T23:15:05.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-24T17:52:57.000Z", "max_forks_repo_path": "notation.tex", "max_forks_repo_name": "mathematicalmichael/thesis", "max_forks_repo_head_hexsha": "2906b10f94960c3e75bdb48e5b8b583f59b9441e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.7285714286, "max_line_length": 387, "alphanum_fraction": 0.6937191249, "num_tokens": 1347, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746912, "lm_q2_score": 0.7879311956428947, "lm_q1q2_score": 0.6124588089884305}} {"text": "\\section{Queueing Theory}\n\\label{sec:Queueing-Theory}\n\nThe queueing theory is built on the stochastic modeling and analysis, which represents service demands and interarrivals as random variables.\nIt is the theory behind what makes queues appear and how to make them go away.\nIt is the study of queueing behavior in systems and networks.\n\nQueueing theory applies anywhere that queues come up. It is also at the heart of any computer system.\n\nQueueing theory has its origins in research by Agner Krarup Erlang when he created models to describe the Copenhagen telephone exchange.\n\nQueues results from variability in service time and/or interarrival time distribution.\n\nA major reason why computer scientists are so slow to adopt queueing theory is that standard assumptions often do not fit. In some cases they are very far from reality: for example, in the case of high variability and correlations. However there are often ways to work around these assumptions.\n\nThe goals of a queueing theorist is \\textit{predicting} the system performance and \\textit{designing} it to improve performance.\n\nThere are many questions in computer systems design that lend themselves to a queueing-theoretic solution (e.g., power management, priority scheduling, and so on).\nHowever, there are lots of problems that we can at best only analyze approximately (e.g., mean response time for M/G/k queue). Typically, approximations are very poor when jobs size variability gets high \\cite{gupta2010inapproximability}.\n\n\n\n\n\\subsection{Queue Metrics}\n\\label{sec:Queue-Metrics}\n\nA queue node is described by the following parameters and performance metrics:\n\n\\begin{description}\n\t\\item [Service Order] The scheduling discipline.\n\t\n\t\\item [Average Arrival Rate] ($\\lambda$) The rate at which jobs arrive to the node.\n\t\n\t\\item [Interarrival Time] ($\\tau$) The time between successive job arrivals.\n\t\n\t\\item [Mean Interarrival Time] ($\\expected{\\tau}=\\frac{1}{\\lambda}$) The average time between successive job arrivals.\n\t\n\t\\item [Service Requirement, Size] ($S$) The time taken by a job to run on server.\n\t\n\t\\item [Average Service Rate] ($\\mu$) The rate at which jobs are served by the server.\n\t\n\t\\item [Mean Service Time] ($\\expected{S}=\\frac{1}{\\mu}$) The average time required to serve a job.\n\t\n\t\\item [Waiting Time, Delay] ($T_{Q}$) The time that the job spends in the queue.\n\t\n\t\\item [Response Time, Turnaround Time] ($T$) The time during which the job is in the node. Thus $T=T_{departure}-T_{arrival}$. Note that $\\expected{T}=\\expected{T_{Q}}+\\expected{S}$.\n\t\n\t\\item [Number of System Jobs] ($N$) The number of jobs in the queue plus the ones being served.\n\t\n\t\\item [Number of Queued Jobs] ($N_{Q}$) The number of jobs in the queue.\n\t\n\t\\item [Utilization] ($\\varrho$) The fraction of time the server is busy, the probability of the server to be busy, the mean number of jobs in that node.\n\t\n\t\\item [Throughput] ($X$) The rate of completions at server.\n\\end{description}\n\n\n\n\n\\subsection{Kendall notation}\nA queue node il classified using the Kendall notation \\cite{kendall1953stochastic}.\n\n\\begin{equation}\nA/S/m/c/N/D\n\\end{equation} \n\nwhere \n\n\\begin{description}\n\t\\item [A] the arrival process, the value indicates the probability distribution of inter-arrival times.\n\t\n\t\\item [S] the service process, the value indicates the probability distribution of service times.\n\t\n\t\\item [A] the number of servers.\n\t\n\t\\item [c] the capacity of the system (that is, the number of servers plus the buffer size).\n\t\n\t\\item [N] the size of the population, that is the total number of jobs.\n\t\n\t\\item [D] the scheduling discipline.\n\\end{description}\n\nThe arrival process \\textit{A} and service process \\textit{S} could be equal to \n\n\\begin{description}\n\t\\item [$M$] Exponential distribution.\n\t\n\t\\item [$D$] Deterministic distribution.\n\t\n\t\\item [$E_{n}$] Erlang distribution of order $n$.\n\t\n\t\\item [$H_{n}$] Hyper-exponential distribution of order $n$.\n\t\n\t\\item [$G$] Generic distribution of independent variables.\n\\end{description}\n\nThe scheduling discipline \\textit{D} could be equal to \n\n\\begin{description}\n\t\\item [$FCFS (FIFO)$] first come, first served.\n\t\n\t\\item [$LCFS (LIFO)$] last come, first served.\n\t\n\t\\item [$RO$] random order.\n\t\n\t\\item [$RR$] round robin.\n\t\n\t\\item [$PS$] processor sharing.\n\t\n\t\\item [$P$] custom priority.\n\\end{description}\n\nWhen the final three parameters are not specified (e.g. M/M/1 queue), it is assumed $K=\\infty$, $N=\\infty$ and $D=FIFO$.\n\n\n\n\n\\subsection{Classification of Queueing Networks}\n\\label{sec:Classification-Queueing-Networks}\n\nQueueing networks can be classified into two categories:\n\n\\begin{description}\n\t\\item [Open Networks] They have external arrivals and departures. In such systems, $X=\\lambda$\n\t\n\t\\item [Closed Networks] They have no external arrivals or departures. In such systems, $X=\\mu$. The number of jobs in the system is fixed to $N$, and it is called \\textit{load} or \\textit{multiprogramming level (MPL)}. $\\expected{T}=\\expected{T_{R}}+\\expected{Z}$. Closed networks can be classified in two categories:\n\t\n\t\\begin{description}\n\t\t\\item [Interactive Networks] There are $N$ \\textit{terminals} which submit jobs to a network of queues, called \\textit{central subsystem}. After submission, a terminal waits for job completion, then sleeps for a think-time $Z$, then can submit again. Thus, a terminal oscillates between a submission-state and a think-state. The goal is to limit the rise of $\\expected{T_{R}}$ with $N$.\n\t\t\n\t\t\\item [Batch Networks] They are interactive networks with no think-time ($Z=0$). The goal is to obtain high throughput.\n\t\\end{description}\t\t\n\t\n\\end{description}\n\n%\\begin{figure}[tp]\n%\\label{fig:Open-Network-Sample}\n%\t\\centering\n%\t\\includegraphics{fig/Open-Network-Sample}\n%\t\\caption{An Open network.}\n%\\end{figure}\n\n%\\begin{figure}[tp]\n%\\label{fig:Interactive-Network-Sample}\n%\t\\centering\n%\t\\includegraphics{fig/Interactive-Network-Sample}\n%\t\\caption{An Interactive network.}\n%\\end{figure}\n\n%\\begin{figure}[tp]\n%\\label{fig:Batch-Network-Sample}\n%\t\\centering\n%\t\\includegraphics{fig/Batch-Network-Sample}\n%\t\t\\caption{A Batch network.}\n%\\end{figure}\n\n\\cite{schroeder2006open} proposed the notion of \\textit{partly-open networks}. Here terminals arrive from outside as in open networks, but make request as in closed ones.\n\n\n\n\n\\subsection{Utilization}\n\\label{sec:Utilization}\n\nThe utilization of a queue node is the fraction of time the node is busy, the probability that it is busy, and the mean number of jobs in that node.\n\n\\begin{theorem}[Utilization Law]\n\\label{thm:Utilization-Law}\n\n\tFor every stable queue node, we have that:\n\t\n\t\\begin{equation}\n\t\\label{eqn:Utilization-Law}\n\t\\varrho = \\frac{\\lambda}{\\mu} = X \\cdot \\mathbf{E}[S]\n\t\\end{equation}\n\t\n\t\\begin{proof}\n\t\tLet us consider the definition of throughput ($X=\\frac{C}{\\tau}$) and utilization ($\\varrho=\\frac{B}{\\tau}$). We have that\n\t\t\n\t\t\\begin{equation*}\n\t\t\tX = \\frac{C}{\\tau} = \\frac{C}{B} \\cdot \\frac{B}{\\tau} = \\frac{C}{B} \\cdot \\varrho\n\t\t\\end{equation*}\n\t\t\n\t\tSince $\\frac{B}{C}=\\expected{S}$ and $\\frac{C}{B}=\\mu$, we have that\n\t\t\n\t\t\\begin{equation*}\n\t\t\tX = \\mu \\cdot \\varrho \\equiv \\frac{\\varrho}{\\expected{S}} \\Rightarrow \\varrho = X \\cdot \\expected{S}\n\t\t\\end{equation*}\n\t\\end{proof}\n\\end{theorem}\n\n\n\n\n\\subsection{Stability}\n\\label{sec:Stability}\n\nThe stability condition guarantees that the queue cannot grow unbounded.\n\n\\begin{theorem}\n\\label{thm:Stability}\n\tA queue node is stable if and only if \\footnote{The condition is necessary, but there are more complex networks (not examined here) where it is not sufficient \\cite{bramson2008stability}.} \n\t\n\t\\begin{equation}\n\t\\label{eqn:Stability}\n\t\\lambda < \\mu\n\t\\end{equation}\n\t\n\t\\begin{proof}\n\t\tLet's prove that if $\\lambda > \\mu$ then the queue would grow to infinity over time.\t\t\n\t\tGiven a time $t$, the number of system jobs $N(t)$, the number of arriving jobs $A(t)$ and the number of departing jobs $D(t)$, we have that\n\t\t\n\t\t\\begin{equation*}\n\t\t\\expected{N(t)}=\\expected{A(t)}-\\expected{D(t)} \\geq t(\\lambda - \\mu)\n\t\t\\end{equation*}\n\t\t\n\t\twhere the inequality holds because the server is not always busy, then the departures at time $t$ are less than $\\mu t$. \n\t\tAt the end, we have that\n\t\t\n\t\t\\begin{equation*}\n\t\t\\lambda > \\mu \\Rightarrow \\lim_{t \\to \\infty} \\expected{N(t)} = \\infty\n\t\t\\end{equation*}\n\t\\end{proof}\n\\end{theorem}", "meta": {"hexsha": "944c3fdcf194d0b57dfe55a9502ff01fcbf4a887", "size": 8299, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "performance-modeling/sec/queueing-theory.tex", "max_stars_repo_name": "gmarciani/research", "max_stars_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2017-07-27T13:31:43.000Z", "max_stars_repo_stars_event_max_datetime": "2018-07-20T12:54:12.000Z", "max_issues_repo_path": "performance-modeling/sec/queueing-theory.tex", "max_issues_repo_name": "gmarciani/research", "max_issues_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "performance-modeling/sec/queueing-theory.tex", "max_forks_repo_name": "gmarciani/research", "max_forks_repo_head_hexsha": "7cc526fe7cd9916ceaf8285c4e4bc4dce4028537", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-02-17T13:30:49.000Z", "max_forks_repo_forks_event_max_datetime": "2018-02-17T13:30:49.000Z", "avg_line_length": 37.0491071429, "max_line_length": 388, "alphanum_fraction": 0.7318954091, "num_tokens": 2310, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7772998611746911, "lm_q2_score": 0.7879311906630568, "lm_q1q2_score": 0.6124588051176031}} {"text": "\\section{Numerical Solution for Burgers' Equation}\n\tLet $\\mathcal{H} = L^2 (0, 1)$. Let's consider an initial value problem for (\\ref{burgers_stochastic}) as follows\n\t\\begin{align}\n\t\\label{burgers_stochastic2}\n\td X(\\xi, t) = \\left[\\alpha \\partial_{\\xi}^2 X(\\xi, t) + \\frac{1}{2} \\partial_\\xi \\left(X^2 (\\xi, t)\\right) \\right] dt + dW_t (\\xi, t), \\hspace{0.2cm} \\xi \\in [0, 1] \n\t\\end{align}\n\tThe boundary condition and its initial condition are respectively \n\t\\begin{align*}\n\t%\\label{IC_burgers_stochastic2}\n\tX(0, t) &= X(1, t) = 0 , \\hspace{0.2cm} t > 0 \\\\\n\tX(\\xi, 0) &= x(\\xi), \\hspace{0.2cm} x \\in \\mathcal{H},\n\t\\end{align*}\n\twhere $W$ is a cylindrical Wiener process on $\\mathcal{H}$ as was given by \\ref{cylindrical}, associated to a stochastic basis $(\\Omega, \\mathcal{F}, \\mathbb{P}, \\{\\mathcal{F}_t\\}_{t \\geq 0})$, and as usually $\\alpha > 0$ is the viscosity coefficient. \\\\\n\t\n\tUsing the above problem, we will proceed to develop the implementation of the method described in the previous section.\n\t\n\t\\input{burgers_equation/stochastic/numerical_experiments/Implementation}\n\t\\input{burgers_equation/stochastic/numerical_experiments/Initial_Condition}\n\t\\input{burgers_equation/stochastic/numerical_experiments/Numerical_Experiments}\t\n", "meta": {"hexsha": "9c6ab1b7d246730265447a1af811f83f817c04c3", "size": 1242, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/burgers_equation/stochastic/numerical_experiments/Numerical_Solution.tex", "max_stars_repo_name": "alanmatzumiya/Maestria", "max_stars_repo_head_hexsha": "c5e2a019312fb8f9bc193b04b07b7815e6ed4032", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2020-12-29T10:44:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-12T11:18:45.000Z", "max_issues_repo_path": "docs/burgers_equation/stochastic/numerical_experiments/Numerical_Solution.tex", "max_issues_repo_name": "alanmatzumiya/spectral-methods", "max_issues_repo_head_hexsha": "c5e2a019312fb8f9bc193b04b07b7815e6ed4032", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/burgers_equation/stochastic/numerical_experiments/Numerical_Solution.tex", "max_forks_repo_name": "alanmatzumiya/spectral-methods", "max_forks_repo_head_hexsha": "c5e2a019312fb8f9bc193b04b07b7815e6ed4032", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-03-04T13:29:56.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-04T13:29:56.000Z", "avg_line_length": 62.1, "max_line_length": 255, "alphanum_fraction": 0.7133655395, "num_tokens": 420, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7879311956428946, "lm_q2_score": 0.7772998508568416, "lm_q1q2_score": 0.6124588008586749}} {"text": "\\section{Approach}\nThe user-based collaborative filtering \\cite{Schafer2007} approach was selected to make movie recommendation for a set of users. This meant that recommendations for an individual user would be based on what similar users have liked that the current user have not yet rated/seen. This is based on the theory that users with similar taste are likely to provide good recommendations to each other. The key here lies in how is similarity between users determined, in this instance the Pearson correlation coefficient measurement \\cite{Wikipedia2017} was selected as the primary method to evaluate similarity, it measures similarity of users by making a linear correlation between what user x and y have rated in terms the score they applied to a set that contains the intersection of movies rated by x and y. The inspiration for this approach was drawn from \\cite{Lead} where this was applied in conjunction with a weighted similarity score that makes sure that users you agree most with are rated higher in terms of movie recommendations.\n\nThis technology was selected based on multiple factors. This included the suggestions made by \\cite{Burke2011} about movie recommendation systems and their domain features including the low risk value of the recommendation, the homogeneous nature of the system and the stability in terms of preference that users have in terms of how much can you rely on previous received information e.g. user's rating history to make future predictions. In terms of technology variation, collaborative filtering provides multiple approaches, e.g. user-based or item based filtering \\cite{Yao}. Both of these have merits, the user-based approach increases in computational cost as the number of users increases. The item based however grows whenever new items are added. In this instance the number of users/items is stable since it is not a live system a choice was made to use the user based approach since the dataset contains 671 users and 9125 movies.\n\n\\section{Required Knowledge}\n\nThe knowledge required for this approach to work include user histories of movie ratings from which user similarity can be based upon. To achieve this a number of public available datasets containing user movie ratings were evaluated. This included the Netflix Prize dataset \\cite{bennett0netflix} , the IMDb academic dataset \\cite{IMDb2017} and the MovieLens \\cite{Harper2015} dataset. From these the MovieLens 100K dataset was chosen as being the most appropriate. The reason for this was that the original Netflix dataset only contained data about when a user rated a movie and not how it was rated. The reason for this was that it was the task for the algorithm to predict the number of reviews and not the actual rating a movie would receive in 2006. Similar to this the IMDb dataset contained rating of movie's but they weren't tied to any users, meaning that user-based collaborative filtering wasn't feasible. The MovieLens dataset on the other hand doesn't have any of these missing features, meaning that it contained users which had committed at least 20 ratings with a value between 1.0 and 5.0 to different movies. Because of this the MovieLens dataset was selected as the most appropriate to develop this movie recommendation system.\n\nHowever it was discovered later in the process that there has been developed an altered version of the Netflix dataset which provides actual ratings of movies \\cite{Netflix2009}, this was discovered after the MovieLens dataset had been applied.\n\n\\section{Algorithm}\n\tThe pseudo code for the similarity and recommendation functions can be seen in Algorithm \\ref{alg:simpearsons} and \\ref{alg:recommendations}. The similarity function uses Pearson correlation coefficient to determine user similarity on a shared set of movies. The similarity is calculated by basically drawing a line through the rating values of the shared movies. If two users agree on the rating of their shared set of movies the result would be a positive correlation between them, meaning that they are expected to have similar taste. After the similarity has been calculated the recommendation function is run \\ref{alg:recommendations}. This function finds movie recommendations for each user by ranking the unseen movies that similar users have seen. The proposed rating for the unseen movies is proportional to how similar the user who has seen is to the current user, i.e. \\textit{weighted\\_sum[rating.movie\\_id] + = [rating.value * similarity]}. After this has been completed for every user the proposed movies are ranked based on their estimated rating.\n\t\n\tUpon completion the correlation is stored in the user/user \\textit{df\\_similarities} matrix and later persisted in the underlying database for later retrieval when a user requests movie recommendations. See \\cite{Munk2017} for the full implementation.\n\n\t\\begin{algorithm}[H]\n\t\t\\caption{Recommender.sim\\_pearsons()}\n\t\t\\label{alg:simpearsons}\n\t\t\\begin{algorithmic}\n\t\t\t\\STATE $df\\_movies\\_rated[user] \\leftarrow$ Series containing movies rated by user\n\t\t\t\\STATE $df\\_similarities \\leftarrow$ DataFrame consisting of user Id's on the x and y axis\n\t\t\t\n\t\t\t\\FORALL{user\\_x in users}\n\t\t\t\t\\FORALL{user\\_y in users}\n\t\t\t\t\t\\IF{df\\_similarities[user\\_x][user\\_y] is null \\AND user\\_x != user\\_y}\n\t\t\t\t\t\t\\STATE $intersection \\leftarrow$ shared movies between user\\_x and user\\_y\n\t\t\t\t\t\t\\IF{$intersection \\geq 5$}\n\t\t\t\t\t\t\t\\STATE $x\\_ratings \\leftarrow$ list of user\\_x rating values\n\t\t\t\t\t\t\t\\STATE $y\\_ratings \\leftarrow$ list of user\\_y rating values\n\t\t\t\t\t\t\t\\STATE $pearson \\leftarrow personr(x\\_ratings, y\\_ratings)$\n\t\t\t\t\t\t\t\\STATE $df\\_similarities[user\\_x][user\\_y] \\leftarrow pearson$\n\t\t\t\t\t\t\t\\STATE $df\\_similarities[user\\_y][user\\_x] \\leftarrow pearson$\n\t\t\t\t\t\t\\ENDIF\n\t\t\t\t\t\\ENDIF\n\t\t\t\t\\ENDFOR\n\t\t\t\\ENDFOR\n\t\t\t\n\t\t\t\\STATE $Recommender.users\\_ratings\\_similarities \\leftarrow df\\_similarities$\n\t\t\\end{algorithmic}\n\t\\end{algorithm}\n\n\t\\begin{algorithm}[H]\n\t\t\\caption{Recommender.recommendations() Page 16 \\cite{Lead}}\n\t\t\\label{alg:recommendations}\n\t\t\\begin{algorithmic}\n\t\t\t\\STATE $weighted\\_sum \\leftarrow \\{\\}$\n\t\t\t\\STATE $similarity\\_sum \\leftarrow \\{\\}$\n\t\t\t\\STATE $num\\_rows \\leftarrow length(users)$\n\t\t\t\n\t\t\t\\FORALL{user in users}\n\t\t\t\t\\STATE $df\\_similar\\_users \\leftarrow Recommender.users\\_ratings\\_similarities[user.id]$\n\t\t\t\t\\STATE $df\\_similar\\_users \\leftarrow $ Remove users that aren't similar.\n\t\t\t\t\\STATE $user\\_movies \\leftarrow $ user rated movie ids\n\t\t\t\t\\FORALL{target\\_user\\_id, similarity in df\\_similar\\_users}\n\t\t\t\t\t\\IF{$similarity \\textgreater 0$}\n\t\t\t\t\t\t\\FORALL{rating in users[target\\_user\\_id].ratings}\n\t\t\t\t\t\t\t\\IF{rating.movie\\_id not in user\\_movies}\n\t\t\t\t\t\t\t\t\\STATE $weighted\\_sum.setdefault(rating.movie, 0) $\n\t\t\t\t\t\t\t\t\\STATE $weighted\\_sum[rating.movie\\_id]\\ + =\\ (rating.value * similarity)$\n\t\t\t\t\t\t\t\t\n\t\t\t\t\t\t\t\t\\STATE $similarity\\_sum.setdefault(rating.movie\\_id, 0)$\n\t\t\t\t\t\t\t\t\\STATE $similarity\\_sum[rating.movie_id]\\ + =\\ similarity$\n\t\t\t\t\t\t\t\\ENDIF\n\t\t\t\t\t\t\\ENDFOR\n\t\t\t\t\t\\ENDIF\n\t\t\t\t\\ENDFOR\n\t\t\t\n\t\t\t\t\\STATE Rank the ratings, users who agree alot with the current user will have their ratings affect the final score the most\n\t\t\t\t\\STATE $rankings \\leftarrow [(w\\_score / similarity\\_sum[movie\\_id], movie\\_id)$\n\t\t\t\t\\bindent\n\t\t\t\t \\STATE $for\\ movie\\_id,\\ w\\_score\\ in\\ weighted\\_sum.items()]$ \n\t\t\t\t\\eindent\n\t\t\t\t\\STATE $rankings.sort()$\n\t\t\t\t\\STATE $rankings.reverse()$\n\t\t\t\t\\STATE $Recommender.recoms[user] \\leftarrow rankings$\n\t\t\t\\ENDFOR\n\t\t\\end{algorithmic}\n\t\\end{algorithm}\n\n\n\\section{Evaluation \\& Reflection}\n\nTo evaluate whether the recommendations made by the correlation filter was actually usable to the users a series of offline tests were conducted. This included randomly splitting the set of user ratings into a training and testing set in of either 60/40 or 80/20 distribution. This was then used to apply the training dataset to the recommendation engine to evaluate whether the engine would be accurately be able to predict the rating contained in the test data Beyond this the amount of minimum shared rated movies were also varied between either above 0 or equal to and above 5. The hypothesis was that by sharing more movie ratings the algorithm would be able to better predict the missing ratings.\n\nTo evaluate the precision of the recommendation predictions the root mean squared error(RMSE) was calculated to determine the accuracy of the predicted ratings. The reason for this was that as described in \\cite{Schafer2007} the RMSE emphasizes large errors and penalizes the performance value by how far each individual prediction is from the true value/rating.\n\nThe results of the testing can be seen in \\ref{table:results}. The results are from running the recommendation test 20 times for each configuration on the MovieLens 100K dataset and extracting the top 100 recommendations. The reason for multiple runs is that because the testing data is a randomly selected sample that by chance could extract a very good set of users. i.e they have rated more movies than the average.\n\nAs Table \\ref{table:results} shows the mean RMSE improves when the number of required shared movies is increased. T-tests at the 1\\% level also shows that there is a significant difference in the results gathered i.e. (p-value = 0,00004) for the 80/20 configurations and (p-value = 4.43e-13) for the 60/40's. However, when looking the percentage between the training/test distributions it indicates no significant difference in the achieved predictions with (p-value = 0.58) between the 60/40 1 or more shared movies vs the 80/20's configuration or (p-value = 0.13) with 60/40 5 or more shared movies vs the 80/20's.\n\n\\begin{table}[H]\n\t\\caption{\\label{table} Testing Results}\n\t\\centering\n\t\\label{table:results}\n\t\\begin{tabular}{c c c c}\n\t\t\\hline\n\t\tTrain \\% & Test \\% & Shared Movies & Mean RMSE \\\\\n\t\t\\hline\n\t\t60 & 40 & 1 or more & 1,58840 \\\\\n\t\t60 & 40 & 5 or more & 1,23029 \\\\\n\t\t80 & 20 & 1 or more & 1,54424 \\\\\n\t\t80 & 20 & 5 or more & 1,29246 \\\\\n\t\t\\hline\n\t\\end{tabular}\n\\end{table} \n\nReflecting on these results it shows that in terms of the configuration parameters for the user-based collaborative filtering approach using the Pearson correlation coefficient similarity function the results supports the proposed hypothesis. I.e that an increase in shared movies has a significant impact on the RMSE with a lower prediction average which means that the prediction ability of the system improves. Whether the recommended movies are actually engaging and interesting to the users is another matter and can't be claimed based on these results.\n\nIn terms of overall drawbacks of the user-based approach, whenever a user adds a new rating that users recommendations has to be recalculated. This could pose a significant query time whenever a user attempts to retrieve recommendations. To mitigate this the current recommendations are pre-calculated when the system is launched, however this does mean that if a user rates a new item it won't be reflected in the recommendations before the system is relaunched. To counter this in general the item-based approach does seem preferable in that the similarity of each item can be precomputed, hence whenever a user commits a new rating the recommendations can be updated subsequently.", "meta": {"hexsha": "11c6e9e9559e88f3707664985b5d86f12da45f1b", "size": 11239, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report/datawrangling_body_conf.tex", "max_stars_repo_name": "rasmunk/set11521cw2", "max_stars_repo_head_hexsha": "602332583921c700ffeee28cef421d55268ca218", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "report/datawrangling_body_conf.tex", "max_issues_repo_name": "rasmunk/set11521cw2", "max_issues_repo_head_hexsha": "602332583921c700ffeee28cef421d55268ca218", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report/datawrangling_body_conf.tex", "max_forks_repo_name": "rasmunk/set11521cw2", "max_forks_repo_head_hexsha": "602332583921c700ffeee28cef421d55268ca218", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 102.1727272727, "max_line_length": 1248, "alphanum_fraction": 0.7762256429, "num_tokens": 2621, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.787931185683219, "lm_q2_score": 0.7772998560157663, "lm_q1q2_score": 0.6124587971818981}} {"text": "\\chapter{Constant functions}\n\\section{Defined on $\\mdr$}\n\\begin{lemma}\nLet $f:\\mdr \\rightarrow \\mdr$, $f(x) := c$ with $c \\in \\mdr$ be a constant function.\n\nThen $(x_P, f(x_P))$ is the only point on the graph of $f$ with\nminimal distance to $P$.\n\\end{lemma}\n\nThe situation can be seen in Figure~\\ref{fig:constant-min-distance}.\n\\begin{figure}[htp]\n \\centering\n \\begin{tikzpicture}\n \\begin{axis}[\n legend pos=north west,\n legend cell align=left,\n axis x line=middle,\n axis y line=middle,\n grid = major,\n width=0.8\\linewidth,\n height=8cm,\n grid style={dashed, gray!30},\n xmin=-5, % start the diagram at this x-coordinate\n xmax= 5, % end the diagram at this x-coordinate\n ymin= 0, % start the diagram at this y-coordinate\n ymax= 3, % end the diagram at this y-coordinate\n axis background/.style={fill=white},\n xlabel=$x$,\n ylabel=$y$,\n tick align=outside,\n minor tick num=-3,\n enlargelimits=true,\n tension=0.08]\n \\addplot[domain=-5:5, thick,samples=50, red] {1};\n \\addplot[domain=-5:5, thick,samples=50, green] {2};\n \\addplot[domain=-5:5, thick,samples=50, blue, densely dotted] {3};\n \\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};\n \\addplot[blue, mark = *, nodes near coords=$P_{h,\\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 3)};\n \\addplot[green, mark = x, nodes near coords=$P_{g,\\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 2)};\n \\addplot[red, mark = *, nodes near coords=$P_{f,\\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 1)};\n \\draw[thick, dashed] (axis cs:2,0) -- (axis cs:2,3);\n \\addlegendentry{$f(x)=1$}\n \\addlegendentry{$g(x)=2$}\n \\addlegendentry{$h(x)=3$}\n \\end{axis}\n \\end{tikzpicture}\n \\caption{Three constant functions and their points with minimal distance}\n \\label{fig:constant-min-distance}\n\\end{figure}\n\n\\begin{proof}\nThe point $(x, f(x))$ with minimal distance can be calculated directly:\n\\begin{align}\n d_{P,f}(x) &= \\sqrt{(x - x_P)^2 + (f(x) - y_P)^2}\\\\\n &= \\sqrt{(x^2 - 2x_P x + x_P^2) + (c^2 - 2 c y_P + y_P^2)} \\\\\n &= \\sqrt{x^2 - 2 x_P x + (x_P^2 + c^2 - 2 c y_P + y_P^2)}\\label{eq:constant-function-distance}\\\\\n \\xRightarrow{\\text{Theorem}~\\ref{thm:fermats-theorem}} 0 &\\stackrel{!}{=} (d_{P,f}(x)^2)'\\\\\n &= 2x - 2x_P\\\\\n \\Leftrightarrow x &\\stackrel{!}{=} x_P\n\\end{align}\n\nSo $(x_P,f(x_P))$ is the only point with minimal distance to $P$. $\\qed$\n\\end{proof}\n\nThis result means:\n\n\\[S_0(f, P) = \\Set{x_P} \\text{ with } P = (x_P, y_P)\\]\n\\clearpage\n\n\\section{Defined on a closed interval $[a,b] \\subseteq \\mdr$}\n\\begin{theorem}[Solution formula for constant functions]\nLet $f:[a,b] \\rightarrow \\mdr$, $f(x) := c$ with $a,b,c \\in \\mdr$ and\n$a \\leq b$ be a constant function.\n\nThen the point $(x, f(x))$ of $f$ with minimal distance to $P$ is\ngiven by:\n\\[\\underset{x\\in [a,b]}{\\arg \\min d_{P,f}(x)} = \\begin{cases}\n S_0(f,P) &\\text{if } S_0(f,P) \\cap [a,b] \\neq \\emptyset \\\\\n \\Set{a} &\\text{if } S_0(f,P) \\ni x_P < a\\\\\n \\Set{b} &\\text{if } S_0(f,P) \\ni x_P > b\n \\end{cases}\\]\n\\end{theorem}\n\n\\begin{figure}[htp]\n \\centering\n \\begin{tikzpicture}\n \\begin{axis}[\n legend pos=north west,\n legend cell align=left,\n axis x line=middle,\n axis y line=middle,\n grid = major,\n width=0.8\\linewidth,\n height=8cm,\n grid style={dashed, gray!30},\n xmin=-5, % start the diagram at this x-coordinate\n xmax= 5, % end the diagram at this x-coordinate\n ymin= 0, % start the diagram at this y-coordinate\n ymax= 3, % end the diagram at this y-coordinate\n axis background/.style={fill=white},\n xlabel=$x$,\n ylabel=$y$,\n tick align=outside,\n minor tick num=-3,\n enlargelimits=true,\n tension=0.08]\n \\addplot[domain=-5:-2, thick,samples=50, red] {1};\n \\addplot[domain=-1:3, thick,samples=50, green] {1.5};\n \\addplot[domain=3:5, thick,samples=50, blue, densely dotted] {3};\n \\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};\n\n \\addplot[blue, mark = *, nodes near coords=$P_{h,\\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(3, 3)};\n \\addplot[green, mark = x, nodes near coords=$P_{g,\\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 1.5)};\n \\addplot[red, mark = *, nodes near coords=$P_{f,\\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(-2, 1)};\n\n \\draw[thick, dashed] (axis cs:2,1.5) -- (axis cs:2,2);\n \\draw[thick, dashed] (axis cs:2,2) -- (axis cs:-2,1);\n \\draw[thick, dashed] (axis cs:2,2) -- (axis cs:3,3);\n \\addlegendentry{$f(x)=1, D = [-5,-2]$}\n \\addlegendentry{$g(x)=1.5, D = [-1,3]$}\n \\addlegendentry{$h(x)=3, D = [3,5]$}\n \\end{axis}\n \\end{tikzpicture}\n \\caption{Three constant functions and their points with minimal distance}\n \\label{fig:constant-min-distance-closed-intervall}\n\\end{figure}\n\n\\begin{proof}\n\\begin{align}\n \\underset{x\\in[a,b]}{\\arg \\min d_{P,f}(x)} &= \\underset{x\\in[a,b]}{\\arg \\min d_{P,f}(x)^2}\\\\\n &=\\underset{x\\in[a,b]}{\\arg \\min} \\big ((x-x_P)^2 + \\overbrace{(y_P^2 - 2 y_P c + c^2)}^{\\text{constant}} \\big )\\\\\n &=\\underset{x\\in[a,b]}{\\arg \\min} (x-x_P)^2\n\\end{align}\n\nwhich is optimal for $x = x_P$, but if $x_P \\notin [a,b]$, you want\nto make this term as small as possible. It gets as small as possible when\n$x$ is as similar to $x_p$ as possible. This yields directly to the\nsolution formula.$\\qed$\n\\end{proof}\n", "meta": {"hexsha": "4e0f690d1ddd6705636b60da6e360e4897190d8f", "size": 6070, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "documents/math-minimal-distance-to-cubic-function/constant-functions.tex", "max_stars_repo_name": "RalfGuder/LaTeX-examples", "max_stars_repo_head_hexsha": "a1bf9fe422969be1ca4674394ebd2170c07f7693", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1231, "max_stars_repo_stars_event_min_datetime": "2015-01-07T04:04:25.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-31T17:43:29.000Z", "max_issues_repo_path": "documents/math-minimal-distance-to-cubic-function/constant-functions.tex", "max_issues_repo_name": "DoubleL61/LaTeX-examples", "max_issues_repo_head_hexsha": "cd0d97f85fadb59b7c6e9062b37a8bf7d725ba0c", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 5, "max_issues_repo_issues_event_min_datetime": "2015-05-10T13:10:47.000Z", "max_issues_repo_issues_event_max_datetime": "2021-05-02T21:28:49.000Z", "max_forks_repo_path": "documents/math-minimal-distance-to-cubic-function/constant-functions.tex", "max_forks_repo_name": "DoubleL61/LaTeX-examples", "max_forks_repo_head_hexsha": "cd0d97f85fadb59b7c6e9062b37a8bf7d725ba0c", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 400, "max_forks_repo_forks_event_min_datetime": "2015-01-05T06:22:18.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-19T04:07:59.000Z", "avg_line_length": 43.3571428571, "max_line_length": 139, "alphanum_fraction": 0.5741350906, "num_tokens": 1972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744806385542, "lm_q2_score": 0.7490872243177518, "lm_q1q2_score": 0.612434598374562}} {"text": "\\documentclass{homework}\n\\course{Math 5522H}\n\\author{Jim Fowler}\n\\input{preamble}\n\n\\begin{document}\n\\maketitle\n\n\\begin{inspiration}\n Luck is the residue of design\n \\byline{Branch Rickey}\n \\end{inspiration}\n\n \\section{Terminology}\n\n \\begin{problem}\n Define the \\textbf{residue} of $f$ at the point $z$, which we write $\\Res(f,z)$.\n \\end{problem}\n \\begin{solution}\n The residue of $f:U\\mapsto \\R$ is \n \\[\n \\lim_{\\epsilon\\to 0^+} \\frac{1}{2\\pi i}\\int_{\\gamma_\\epsilon} \\frac{f(w)}{w-z} dz\n \\]\n Where $\\gamma_\\epsilon:[0, 2\\pi ] \\to \\C$ is defined as $\\gamma(t) = \\epsilon e^{it}$.\n\n Equivalently, it is the coefficient of the $\\frac{1}{w-z}$ term in the Laurent expansion of $f$ at $z$.\n \\end{solution}\n \\section{Numericals}\n\n \\begin{problem}\n Evaluate $\\Res(f,z)$ for the function $f(z) = \\displaystyle\\frac{e^z}{z^2-1}$.\n \\end{problem}\n \\begin{solution}\n If $z_0\\neq \\pm 1$, then the function is analytic divided by analytic nonzero, so $\\Res(f,z_0)=0$. We can directly compute the other two cases by multiplying by $z-z_0$ and evaluating at $z_0$:\n \\begin{gather*}\n \\Res(f,1) = f(z)(z-1)|_{z=1} = \\frac{e^z}{z+1}|_{z=1} = \\frac{e}{2}\\\\\n \\Res(f,-1) = f(z)(z+1)|_{z=-1} = \\frac{e^z}{z-1}|_{z=-1} = -\\frac{1}{2e}\n \\end{gather*}\n \\end{solution}\n \\begin{problem}\\label{residues-all-one}\n Evaluate $\\Res(f,z)$ for the\n function $f(z) = \\pi \\cot (\\pi z)$.\n \\end{problem}\n \\begin{solution}\n The only points where the residue might be nonzero are where the denominator is zero, that is, where\n \\[\n 0=\\sin \\pi z = \\frac{e^{i\\pi z}+e^{-i\\pi z}}{2},\n \\]\n so points where $z = k$ for $k\\in \\N$.\n\n Now we can compute the residue at the point $k$ using L'Hopitals rule:\n \\[\n \\lim_{z\\to k} \\pi\\frac{(z-k\\pi)\\cos \\pi z}{\\sin \\pi z}\\big|_{z=k} = \\lim_{z\\to k}\\pi \\frac{\\cos (\\pi z) - (z-k\\pi)\\pi \\sin \\pi z}{\\pi\\cos \\pi z}\\big|_{z=k} = \\frac{\\pi(-1)^k}{\\pi (-1)^k} = 1\n \\]\n \\end{solution}\n \\begin{problem}\\label{residue-coth}Compute $\\Res(f,\\pm bi)$ for the\n function\n \\[\n f(z) = \\frac{\\pi \\cot(\\pi z)}{z^2 + b^2}.\n \\]\n \\end{problem}\n \\begin{solution}\n For any value of $b$ except for when $bi\\in \\Z$, a simple formula gives the answer:\n \\[\n Res(f, \\pm bi) = \\lim_{z\\to \\pm bi}\\frac{\\pi \\cot(\\pi z)}{z \\pm bi} =\\frac{\\pi \\cot(\\pm \\pi bi)}{\\pm2bi}\n \\]\n Otherwise, let $k\\in \\Z = bi$. \n we will write out each term as a power series at $k$ and nab the $\\frac{1}{z\\mp bi}$ term.\n\n Using a few steps of polynomial long division, we first compute the power series for $\\frac{1}{\\sin z}$ at $\\pm k$ as \n \\[\\frac{1}{(z\\mp k)} + \\frac{(z\\mp k)}{6} + O((z\\mp k)^3).\\]\n (This formula can be verified by multiplication with the power series for $\\sin (z\\mp k) = (z\\mp k) - \\frac{(z\\mp n)^3}{6} + O(z^5)$, I don't want to latex out polynomial long division.)\n\n Next we can compute the power series of $\\frac{1}{z\\pm k}$ at $z\\pm k$. Let $w=z\\mp k$\n \\[\n \\frac{1}{z\\pm k} = \\frac{1}{w\\pm 2k} = \\frac{\\mp\\frac{1}{2k}}{1-\\frac{w}{\\pm 2k}}= \\mp \\frac{1}{2k}\\sum_{n=0}^\\infty \\frac{w^n}{(\\pm 2k)^n} = \\mp \\frac{1}{2k}\\sum_{n=0}^\\infty \\frac{(z\\mp k)^n}{(\\pm 2k)^n}\n \\]\n\n Unless $k=0$, in which case the power series of $\\frac{1}{z}$ is $\\frac{1}{z}$.\n\n Knowing the series for $\\cos z$, we can write $f$ as a product of power series,\n\n \\begin{align*}\n \\frac{\\pi\\cot(\\pi (z\\mp k))}{z^2 + b^2} &= \\pi\\cos(\\pi (z\\mp k))\\frac{1}{\\sin(\\pi (z\\mp k))}\\frac{1}{z\\pm k}\\frac{1}{z\\mp k} \\\\\n &= \\pi\\left(1 + O((z\\mp k)^2\\right)\\left(\\frac{1}{\\pi(z\\mp k)} - \\frac{\\pi(z\\mp k)}{6} + O((z\\mp k)^3)\\right)\\\\&\\times \\left(\\mp\\frac{1}{2k}\\sum_{n=0}^\\infty \\frac{(z\\mp k)^{n-1}}{(\\pm 2k)^{n}}\\right) \\\\\n &= \\pi(1)\\left(\\frac{1}{\\pi(z\\mp k)}\\right)\\left(\\frac{\\mp 1}{2k(z\\mp k)} - \\frac{1}{4k^2}\\right) + O(1)\\\\\n &= \\frac{1}{\\pm 2k (z\\mp k)^2} - \\frac{1}{4k^2(z\\mp k)}+ O(1)\n \\end{align*}\n The residue is the coefficient of $\\frac{1}{z\\mp k}$. Since $bi=k$, it is $-\\frac{1}{4(bi)^2}=\\frac{1}{4b^2}$.\n\n If $k=0$, then we can replace the last term in the big product above with $\\frac{1}{z}$ to get a laurant series of $\\frac{1}{z^2} + O(1)$, and so the residue is 0.\n \\end{solution}\n\n\n \\begin{problem}\n Evaluate the integrals\n \\[\n \\int_{-\\infty}^\\infty \\frac{\\sin x}{1+x^2} \\, dx \\mbox{ and }\n \\int_{-\\infty}^\\infty \\frac{\\cos x}{1+x^2} \\, dx.\n \\]\n \\end{problem}\n \\begin{solution}\n Since $e^{ix} = \\cos x + i \\sin x$, we can just integrate\n \\[\n \\int_\\infty^\\infty \\frac{e^{iz}}{1+z^2} dz\n \\]\n and the real part will be the cos integral, the imaginary the sin integral. We can integrate over the following curve:\n\n \\begin{tikzpicture}\n \\draw [<->] (-4, 0) -- (4, 0);\n \\draw [<->] (0, -4) -- (0, 4);\n \\node [below] at (3, 0) {$R$};\n \\node [below] at (1, 0) {$\\gamma_1$};\n \\node [right] at (2, 2.5) {$\\gamma_2$};\n \\draw (0,1) circle[radius=2pt];\n \\node [right] at (0, 1) {$\\Res(\\frac{e^{iz}}{1+z^2}, i)$};\n \\begin{scope}[very thick,decoration={\n markings,\n mark=at position 0.5 with {\\arrow{>}}}\n ] \n \\draw[postaction={decorate}] (-3,0) -- (3,0);\n \\draw[postaction={decorate}] (3, 0) arc [radius=3, start angle=0, end angle=180];\n \\end{scope}\n \\end{tikzpicture}\n\n We can compute the integral over the curve $\\gamma_2$.\n \\begin{align*}\n \\int_{\\gamma_2} \\frac{e^{iz}}{1+z^2}dz = \\int_0^\\pi \\frac{e^{iRcos(\\theta) - Rsin(\\theta)d\\theta}iRe^{i\\theta}}{1+R^2e^{2i\\theta}}d\\theta\\\\\n \\end{align*}\n Note that the $e^{iRcos(\\theta) - Rsin(\\theta)}$ term is never more than $1$ in magnitude, so\n \\begin{align*}\n \\abs{\\lim_{R\\to\\infty} \\int_{\\gamma_2} \\frac{e^{iz}}{1+z^2}dz} \\leq \\lim_{R\\to\\infty} \\int_0^\\pi \\frac{R}{R^2+1}d\\theta \\leq \\lim_{R\\to\\infty} \\frac{\\pi}{R} = 0\n \\end{align*}\n\n By the residue thoerem, the value of the integral around the contour is equal to the value of the residue at $i$, so the value of the integral on the real line is equal to the residue at $i$ times $2\\pi i$.\n\n \\begin{align*}\n \\int_\\infty^\\infty \\frac{e^{iz}}{1+z^2} dz &= 2\\pi i \\Res(\\frac{e^{iz}}{1+z^2}, i)\\\\\n &= 2\\pi i\\lim_{z\\to i} \\frac{e^{iz}(z-i)}{1+z^2} \\\\\n &= 2\\pi i\\lim_{z\\to i} \\frac{e^{iz}}{z+i} = \\frac{\\pi}{e}\n \\end{align*}\n\n Therefore, \n \\[\n \\int_{-\\infty}^\\infty \\frac{\\sin x}{1+x^2} \\, dx = 0 \\mbox{ and }\n \\int_{-\\infty}^\\infty \\frac{\\cos x}{1+x^2} \\, dx = \\frac{\\pi}{e}\n \\] \n \\end{solution}\n \\begin{problem}\n For a real number $\\lambda \\in (0,1)$, evaluate\n \\[\n \\int_{-\\infty}^\\infty \\frac{e^{\\lambda x}}{1 + e^x} \\, dx.\n \\]\n \\end{problem}\n \\begin{solution}\n We shall evaluate the integral on the following contour:\\\\\n \\scalebox{1.5}{\n \\begin{tikzpicture}\n \\draw [<->] (-4, 0) -- (4, 0);\n \\draw [<->] (0, -1) -- (0, 4);\n \\node [below] at (-4, 0) {$-R$};\n \\node [below] at (4, 0) {$R$};\n \\node [below] at (1, 0) {$\\gamma_1$};\n \\node [above] at (-1, 3) {$\\gamma_2$};\n \\node [above] at (0, 3) {$2\\pi i$};\n \\node [right] at (4, 1.5) {$\\gamma_3$};\n \\node [right] at (-4, 1.5) {$\\gamma_4$};\n \\draw (0,1.5) circle[radius=2pt];\n \\node [right] at (0, 1.5) {$\\Res( \\frac{e^{\\lambda x}}{1 + e^x}, \\pi)$};\n \\begin{scope}[very thick,decoration={\n markings,\n mark=at position 0.5 with {\\arrow{>}}}\n ] \n \\draw[postaction={decorate}] (-4,0) -- (4,0);\n \\draw[postaction={decorate}] (4,0) -- (4,3);\n \\draw[postaction={decorate}] (4,3) -- (-4,3);\n \\draw[postaction={decorate}] (-4,3) -- (-4,0);\n \\end{scope}\n \\end{tikzpicture}\n }\n \\begin{align*}\n \\int_{\\gamma_1}dz +\\int_{\\gamma_2}dz &= \\int_{\\gamma_1}dz -\\int_{-\\gamma_2}dz\\\\\n &= \\int_{-R}^R \\frac{e^{\\lambda x}}{1+e^x} - \\frac{e^{\\lambda (x+2\\pi i)}}{1+e^{x+2\\pi i}} dx\\\\\n &= \\int_{-R}^R \\frac{e^{\\lambda x} - e^{\\lambda x +\\lambda2\\pi i}}{1+e^x}dx\\\\\n &= (1 -e^{\\lambda2\\pi i})\\int_{-R}^R \\frac{e^{\\lambda x}}{1+e^x}dx\\\\\n &= (1 -e^{\\lambda2\\pi i})\\int_{\\gamma_1}dz \n \\end{align*}\n \\begin{align*}\n \\abs{\\int_{\\gamma_3}dz} &= \\abs{\\int_0^{2\\pi i} \\frac{ie^{\\lambda R + \\lambda ti}dt}{1 + e^{R+ti}}}\\\\\n &= \\abs{\\int_0^{2\\pi} \\frac{e^{\\lambda R + \\lambda ti} dt}{1 + e^{R+ti}}}\\\\\n &\\leq \\abs{\\int_0^{2\\pi} \\frac{e^{\\lambda R} dt}{e^{R}}} = 2\\pi e^{R(\\lambda - 1)} \\underset{R\\to\\infty}{\\to} 0\n \\end{align*}\n \\begin{align*}\n \\abs{\\int_{\\gamma_4}dz} &= \\abs{\\int_0^{2\\pi} \\frac{-ie^{-\\lambda R + \\lambda ti}dt}{1 + e^{-R+ti}}}\\\\\n &\\leq 2\\pi \\sup \\abs{\\frac{e^{\\lambda R+\\lambda ti}}{1 + e^{R+ti}}}\\\\\n &\\leq 2\\pi \\abs{\\frac{e^{\\lambda R}}{1 + e^{R}}} \\underset{R\\to -\\infty}{\\to} \\frac{0}{1} = 0\n \\end{align*}\n Now by the residue theorem, \n \\[\n (1-e^{\\lambda 2\\pi i})\\int_{\\gamma_1}dz= \\int_{\\gamma_1+\\gamma_2+\\gamma_3+\\gamma_4} dz = 2\\pi i \\Res(\\frac{e^{\\lambda x}}{1+e^x}, i\\pi)\n \\]\n\n We can compute the residue at the simple pole \n \\[\n \\Res(\\frac{e^{\\lambda x}}{1+e^x}, i\\pi) = \\lim_{x\\to i\\pi} \\frac{e^{\\lambda x}(x-i\\pi)}{1-e^x}\\big|_{x=i\\pi} = \\frac{\\lambda e^{\\lambda x}(x-i\\pi) + e^{\\lambda x}}{-e^x}|_{x=i\\pi} = e^{i\\pi\\lambda}\n \\]\n\n And now we can use the last two formulas to solve for the desired integral:\n \\[\n \\int_{\\gamma_1}dz = \\frac{2\\pi i e^{i\\pi \\lambda}}{1 - e^{2\\pi i \\lambda}}\n \\]\n \\end{solution}\n \\begin{problem}\n Evaluate the integral $\\displaystyle\\int_{0}^{2\\pi} \\frac{1}{3 + \\sin^2 x} \\, dx$.\n \\end{problem}\n \\begin{solution}\n Rewrite $\\sin x$ as follows:\n \\[\n \\sin x = \\frac{e^{ix} - \\frac{1}{e^{ix}}}{2i}\n \\]\n\n Now let $\\gamma=e^{ix}$ be the positively oriented unit circle.\n \\begin{align*}\n \\int_{0}^{2\\pi} \\frac{dx}{3 + \\sin^2 x} \\, dx &= \n \\int_{0}^{2\\pi} \\frac{dx}{3 + \\left(\\frac{e^{ix} - \\frac{1}{e^{ix}}}{2i}\\right)^2}\\\\\n &= \\int_{0}^{2\\pi} \\frac{ie^{ix}dx}{ie^{ix}\\left(3 - \\left(\\frac{e^{ix} - \\frac{1}{e^{ix}}}{2}\\right)^2\\right)}\\\\\n &= \\int_\\gamma \\frac{dz}{iz\\left(3 - \\left(\\frac{z - \\frac{1}{z}}{2}\\right)^2\\right)}\n \\end{align*}\n\n We can now try to apply the residue theorem. We need to find poles of the function \n $f(z) = \\frac{-i}{z(3 - (\\frac{z - \\frac{1}{z}}{2})^2)}$ in the unit circle.\n $z=0$ is a removable discontinuity since the second factor grows as fast as $\\frac{1}{z^2}$ there, so the only poles are roots of the second factor. We can solve it with the quadratic formula and expand\n \\begin{align*}\n 3 - (\\frac{z - \\frac{1}{z}}{2})^2 &= \n \\frac{4}{z^2}\\left(3z^2 - (\\frac{z^2 - 1}{2})^2\\right)\\\\\n &= \\frac{4}{z^2}\\left( (z-(2-\\sqrt{3}))(z-(2+\\sqrt{3}))(z-(-2-\\sqrt{3}))(z-(-2+\\sqrt{3}))\\right)\n \\end{align*}\n\n The roots in the unit circle can be found with the quadratic formula as $\\pm (2-\\sqrt{3})$.\n \\begin{align*}\n \\Res(f, 2-\\sqrt{3}) &= \n \\frac{-4iz}{(z-(-2+\\sqrt{3}))(z-(2+\\sqrt{3}))(z-(-2-\\sqrt{3}))}\\big|_{z=2-\\sqrt{3}}\\\\\n &= \\frac{-4iz}{2z\\cdot 4\\cdot 2\\sqrt{3}} = \\frac{-i}{4\\sqrt{3}}\n \\end{align*}\n \\begin{align*}\n \\Res(f, 2+\\sqrt{3}) &= \n \\frac{-4iz}{(z-(2-\\sqrt{3}))(z-(2+\\sqrt{3}))(z-(-2-\\sqrt{3}))}\\big|_{z=-2+\\sqrt{3}}\\\\\n &= \\frac{-4iz}{2z\\cdot 4\\cdot 2\\sqrt{3}} = \\frac{-i}{4\\sqrt{3}}\n \\end{align*}\n Now we can finally apply the residue formula to obtain the value of the desired integral:\n \\begin{align*}\n 2\\pi i \n \\left(\\frac{-i}{4\\sqrt{3}}+ \\frac{-i}{4\\sqrt{3}}\\right) = \\frac{\\pi}{\\sqrt{3}}\n \\end{align*}\n\n \\end{solution}\n \\begin{problem}\n Evaluate the integral $\\displaystyle \\int_0^\\pi \\log \\sin x \\, dx$.\n \\end{problem}\n \\begin{solution}\n \\begin{align*}\n \\int_0^\\pi \\log \\sin x dx &= \\int_0^\\pi \\log \\abs{\\frac{e^{ix} - e^{-ix}}{2i}} dx\\\\\n &= \\int_0^\\pi \\log \\abs{\\frac{(e^{ix})(e^{ix} - e^{-ix})}{i}} dx - \\int_0^\\pi \\log 2 dx \\\\\n &= \\int_0^\\pi \\log \\abs{1 - e^{2ix}} dx - \\pi \\log 2\n \\int_0^\\pi \\log \\sin x dx &= \\frac{1}{2}\\lim_{r^- \\to 1}\\int_0^{2\\pi} \\log \\abs{1 - re^{ix}} dx - \\pi \\log 2\\\\\n \\end{align*}\n Now we use the fact that for functions $f$ holomorphic on a disk of radius $r$ about the point $z_0$, \n \\[\n f(z_0) = \\frac{1}{2\\pi}\\int_0^{2\\pi} f(z_0+re^{i\\theta})d\\theta\n \\]\n Applying this to $z_0=1, f=\\log$:\n \\begin{align*}\n \\lim_{r^- \\to 1}\\int_0^{2\\pi} \\log (1 - e^{ix}) dx &= \\lim_{r^-\\to 1} \\log 1 = 0\n \\end{align*}\n And we conclude that\n \\[\n \\int_0^\\pi \\log \\sin x \\, dx = -\\pi\\log 2\n \\]\n \\end{solution}\n \\begin{problem}\n Evaluate the integral $\\displaystyle\\int_{-\\infty}^{\\infty} \\frac{1-x}{1-x^7} \\, dx$.\n \\end{problem}\n \\begin{solution}\n Let $f = \\frac{1-x}{1-x^7} = \\frac{1}{1+x+x^2+x^3+x^4+x^5+x^6}$ and $\\zeta$ be the 7th root of unity with the smallest counterclockwise nonzero angle to the positive x axis.\n We will compute the integral on the half circle contour as $R\\to\\infty$.\n\n \\begin{tikzpicture}\n \\draw [<->] (-4, 0) -- (4, 0);\n \\draw [<->] (0, -.2) -- (0, 4);\n \\node [below] at (3, 0) {$R$};\n \\node [below] at (1, 0) {$\\gamma_1$};\n \\node [right] at (2, 2.5) {$\\gamma_2$};\n \\draw (1.247, 1.564) circle[radius=2pt];\n \\draw (-.445, 1.95) circle[radius=2pt];\n \\draw (-1.802, 0.868) circle[radius=2pt];\n \\node [below] at (1.247, 1.564) {$\\Res(f, \\zeta)$};\n \\node [right] at (-.445, 1.95) {$\\Res(f, \\zeta^2)$};\n \\node [above] at (-1.802, 0.868) {$\\Res(f, \\zeta^3)$};\n \\begin{scope}[very thick,decoration={\n markings,\n mark=at position 0.5 with {\\arrow{>}}}\n ] \n \\draw[postaction={decorate}] (-3,0) -- (3,0);\n \\draw[postaction={decorate}] (3, 0) arc [radius=3, start angle=0, end angle=180];\n \\end{scope}\n \\end{tikzpicture}\n\n The value of the integral on the outer curve $\\gamma_2$ is 0:\n \\[\n \\int_{\\gamma_2} \\frac{1-z}{1-z^7} \\leq \\pi R \\sup_{|z|=R} \\frac{\\abs{1-z}}{\\abs{1-z^7}} \\in O(R^{-6}) \n \\]\n Using the residue theorem, we can get the value of the integral by multiplying the residues by $2\\pi i$. Unfortunately the residues are nasty to compute, so we will have a nasty expression.\n\n \\begin{align*}\n \\Res(f, \\zeta^p) = \\lim_{z\\to \\zeta^p} \\frac{z-\\zeta^p}{\\prod_{k=1}^6 (z-\\zeta^k)} = \\frac{1}{\\prod_{1\\leq k \\leq 6, k\\neq p}^5 (\\zeta^p-\\zeta^k)}\n \\end{align*}\n It's easy enough to numerically compute that none of these terms are 0, and consequentally that the computation is indeed valid.\n \\begin{code}\n import cmath # using python\n zeta = complex(0.6234898018587336, 0.7818314824680298)\n res = [1/reduce(operator.mul, [pow(zeta, p) - pow(zeta,k) for k in range(1, 7) if k != p], 1) for p in range(1, 4)]\n >> [(-0.12085867654500679+0.02758520424482756j), (-0.09692113342087182-0.2012588073284831j), (0.21777980996587906-0.173673603083655j)]\n \\end{code}\n And then we can get the answer by summing the residues and multiplying.\n \\[\n \\lim_{R\\to \\infty} \\int_{\\gamma_1} fdz = \\int_{-\\infty}^{\\infty} f dz = 2\\pi i \\sum_{p=1}^3{\\Res(f, \\zeta^p)}\n \\]\n \\begin{code}\n sum(res)*2*cmath.pi * complex(0, 1)\n >> (2.182446862280324+2.964688223307337e-15j)\n \\end{code}\n It's about 2.18.\n \\end{solution}\n \\begin{problem}\n Show that $f(z) = z^5 + 15z - 1$ has five zeros in $B_2(0)$, and one zero in $B_{1/15}(0)$.\n \\end{problem}\n \\begin{solution}\n First we will show that on $B_2(0)$, $\\abs{z^5} > \\abs{15z-1}$. By the triangle inequality, $\\abs{15z-1} \\leq \\abs{15z} + \\abs{1} = 31$, and $\\abs{z^5} = 2^5 = 32$.\n\n Now we can apply Rouch\\'e's theorem to say that $z^5$ and $z^5+15z-1$ have the same number of zeros in the ball of radius 2 counted with multiplicity, so $f(z)$ must have $5$ zeros in the ball.\n\n Next consider the ball of radius $\\frac{1}{15}$. The function $15z-(1-10^{-9})$ has exactly one zero in this ball; we hope to apply Rouch\\'e's theorem again by showing that \n \\[\n \\abs{15z-1+10^{-9}} \\geq \\abs{z^5 - 10^{-9}}\n \\]\n We can just apply the triangle inequality again:\n \\[\n \\abs{15z-1+10^{-9}} \\geq 14 + 10^{-9} \\geq \\frac{1}{15}^5 - 10^{-9} \\geq \\abs{z^5 - 10^{-9}}\n \\]\n Then we can confirm that there is one zero in the ball of radius $\\frac{1}{15}$.\n\n \\end{solution}\n \\begin{problem}\\label{form-at-infinity}\n Set $w = 1/z$ and compute $f(w) \\, dw$ in terms of $f(z) \\, dz$.\n \\end{problem}\n \\begin{solution}\n If $w=\\frac{1}{z}$, differentiating both sides we get that $dw = \\frac{-dz}{z^2}$. Thus $f(w) dw = -f(\\frac{1}{z})\\frac{dz}{z^2}$\n\n \\end{solution}\n \\section{Exploration}\n\n \\begin{problem}\n Fix $w \\in \\R$ with $w>0$. By the intermediate value theorem, the\n polynomial $f(z) = z^4 + 4w^3 z - 1$ has a real root in the interval\n $(-\\infty,0)$ and another in $(0,1)$, along with two complex roots\n $a \\pm bi$. Use Gauss-Lucas (\\ref{gauss-lucas}) and Rouch\\'e's\n (\\ref{rouches-theorem}) theorem to describe a subset of $\\C$\n containing $a\\pm bi$.\n \\end{problem}\n \\begin{solution}\n\n We can use Rouch\\'e's theorem to show that the roots lie in the ball of radius $\\max(2w, \\sqrt{2})$. Under these conditions we can prove the inequality $|z^4|>|4w^3z - 1|$, and since $z^4$ has 4 zeros in the ball, so must $f$.\n\n To prove the inequality: Suppose that $w>\\sqrt[4]{1/8}$. Then \n \\begin{align*}\n &8w^4 > 1 \\\\\n \\implies &16w^4 > 8w^3 + 1\\\\\n \\implies &\\abs{z^4} > \\abs{4zw^3 - 1} \\quad \\text{ if } \\abs{z} = 2w\n \\end{align*}\n Otherwise, we must have that \n \\[w<\\sqrt[4]{1/8}< \\sqrt[3]{\\frac{3}{4\\sqrt{2}}}\\]\n which implies that\n \\begin{align*}\n 3 > 4\\sqrt{2}w^3\\\\\n &\\implies 4 > 4\\sqrt{2}w^3 - 1\\\\\n &\\implies \\abs{z^4} > \\abs{4zw^3 - 1} \\quad \\text{ if } \\abs{z} = \\sqrt{2}\n \\end{align*}\n In either case, the inequality is satisfied.\n\n By Gauss-Lucas, the roots of $f$ contain the convex hull of the roots of $f'(z) = 4z^3+4w^3$, which are cubic roots of $w$. WLOG let $b\\geq 0$. This implies that $b > \\frac{\\sqrt{3}w}{2}$. Combining this with the fact that $a+bi$ is contained in the ball of radius $\\max(2w, \\sqrt{2})$, we get a somewhat restricted set of locations for the roots.\n\n We can restrict it a bit more if needed by considering the convex hull further: Let $\\zeta_1 = w(\\frac{\\sqrt{3}}{2}+\\frac{i}{2})$. Since the real roots are all on the line from $-\\max(2w, \\sqrt{2})$ to $1$, $a+bi$ is above the line from 1 to $\\zeta_1$ and the line from $-\\max(2w, \\sqrt{2})$ to $\\zeta_1$. And if $w>1$, then the real part of the roots is greater than $w.$ \n \\end{solution}\n \\begin{problem}\n What is the correct definition of $\\Res(f,\\infty)$? Generally, we\n would shift the viewport by replacing $f$ with the function\n $g(z) = f(1/z)$ and then we study $g$ near $z = 0$ in order to\n investigate ``$f$ near $\\infty$.'' Does \\ref{form-at-infinity} help\n here?\n \\end{problem}\n \\begin{solution}\n Le $\\gamma_R$ be the circle of radius $R$ centered at 0.\n The residue of $f$ at $\\infty$ is kinda given by the formula\n \\[\n \\lim_{R \\to \\infty k} \\int_{\\gamma_R} \\frac{f(w)dw}{w}\n \\]\n Except that the part of the circle that is inside is backwards.\n\n We fix this by inverting everything using the subsitution $w=\\frac{1}{z}$ investigated in $\\ref{form-at-infinity}$ to get the formula\n \\[\n \\lim_{R^+\\to 0} \\int_{\\gamma_R} -zg(z)\\frac{dz}{z^2} = \\lim_{R^+\\to 0} \\int_{\\gamma_R} -g(z)\\frac{dz}{z} \n \\]\n Except that since we inverted everything, we are now travelling on $\\gamma$ in the clockwise direction. To fix this we get rid of the negative sign, and see that we are looking for the coefficent of $\\frac{1}{z}$ in the taylor series of $g(z)$, which is nothing but the coefficient of $z$ in the taylor series of $f(z)$.\n \\end{solution}\n \\begin{problem}\\label{rouches-theorem}Suppose $U$ is an open set containing the closed disk $D_r(z_0)$ and\n $f, g : U \\to \\C$ are holomorphic functions satisfying\n \\[\n \\abs{f(z)} > \\abs{g(z)}\n \\]\n for $z \\in \\partial D_r(z_0)$. Prove \\textbf{Rouch\\'e's theorem}\n that $f$ and $f+g$ have the same number of zeros, counted with\n multiplicity, in $B_r(z_0)$. (This is sometimes called the dog\n leash theorem---can you see why?)\n \\end{problem}\n \\begin{solution}\n The argument principal tells us that for any holomorphic function $f$, the number of roots of $f$ inside of a positively oriented jordan curve $\\gamma$ is given by\n $\\frac{1}{2\\pi i}\\int_\\gamma \\frac{f'}{f} dz.$\n\n In particular, the number of zeros of $f+g = f(1 + \\frac{g}{f})$ is\n \\begin{align*}\n \\frac{1}{2\\pi i}\\int_\\gamma \\frac{(f(1 + \\frac{g}{f}))'}{f(1 + \\frac{g}{f})}dx &=\n \\frac{1}{2\\pi i}\\int_\\gamma \\frac{f'}{f} + \\frac{(1+\\frac{g}{f})'}{1 + \\frac{g}{f}}dx \n \\end{align*}\n Thus it suffices to show that the term \n \\(\n \\int_\\gamma \\frac{(1+\\frac{g}{f})'}{1 + \\frac{g}{f}}dx\n \\)\n vanishes. Since $1+\\frac{g}{f}>0$ on the boundary of $D_r(z_0)$, the function is analytic on the boundary, and since $\\abs{f}>\\abs{g}$, the function $1+\\frac{g}{f}$ is analytic and always positive, so it has no zeros. By the argument principal, the term must evaluate to 0.\n \\end{solution}\n \\begin{problem}\n Use \\ref{rouches-theorem} to give another proof of the Fundamental Theorem of Algebra.\n \\end{problem}\n \\begin{solution}\n Let $p(x) = \\sum_{k=1}^d a_kx^k$ be a degree $d$ polynomial with $a_d\\neq 0$. Then we can choose a circle $B_r(0)$ of sufficently big radius so that the leading term dominates on $\\partial B_r(0)$:\n \\[\n \\abs{a_dx^d} \\geq \\abs{\\sum_{k=1}^d a_kx^k}\n \\]\n We should also ensure that $r > \\sqrt[d]{a_d}$, to observe that $a_dx^d$ has exactly $d$ roots in the circle corresponding to the scaled $d$th roots of unity. By Rouch\\'e's theorem, this means that $p(x)$ also has exactly $d$ roots in this big circle. And, in fact, in any circle of even larger size, so we can even say that $p(x)$ has exactly $d$ roots in $\\C$.\n \\end{solution}\n \\begin{problem}\n It is important to recognize patterns in how residues equip us to evaluate integrals, so that, when presented with a fresh integration problem, we have some ideas about the best tool for the task. Develop a tool for evaluating\n \\[\n \\int_0^{2\\pi} f(\\cos \\theta,\\sin \\theta) \\, d\\theta\n \\]\n via residues. Here $f$ is a rational function, explain how to make\n the substitution $z = e^{i\\theta}$ to rewrite\n $f(\\cos \\theta,\\sin \\theta)$ in terms of $z$, and\n similarly rewrite $d\\theta$ in terms of\n $dz$ to reduce the given integral to a certain contour integral.\n \\end{problem}\n \\begin{solution}\n Let $z=e^{i\\theta}$ so that $dz = ie^{i\\theta} d\\theta$. Then $d\\theta = \\frac{dz}{iz}$ and \n \\begin{gather*}\n \\cos\\theta = \\frac{e^{i\\theta} + e^{-i\\theta}}{2} = \\frac{z+z^{-1}}{2}\\\\\n \\sin\\theta = \\frac{e^{i\\theta} - e^{-i\\theta}}{2i} = \\frac{z-z^{-1}}{2i}\n \\end{gather*}\n So the integral becomes\n \\[\n \\int_\\gamma f(\\frac{z+z^{-1}}{2},\\frac{z-z^{-1}}{2i})\\frac{dz}{iz}\n \\]\n Where $\\gamma$ is the image of $e^{i\\theta}$ for $0\\leq \\theta \\leq 2\\pi$, that is, the positively oriented unit circle \n \\[\\gamma:[0, 2\\pi] \\to \\C\\qquad \\gamma(\\theta)= e^{i\\theta}.\\]\n\n Since $f$ is a rational function, we have a rational function in $z$ so it is meromorphic with finitely many poles $\\{z_k\\}_{k=1}^n$ in the unit circle and the residue theorem can be used to evaluate the integrals:\n \\[\n \\int_0^{2\\pi} f(\\cos \\theta, \\sin\\theta) d\\theta = 2\\pi i \\sum_{k=1}^n \\Res(f(\\frac{z+z^{-1}}{2},\\frac{z-z^{-1}}{2i}\\frac{1}{iz} ), z_k)\n \\]\n \\end{solution}\n \\begin{problem}\\label{summation-theorem}Sometimes the residue calculus\n permits us to sum series.\n\n Suppose $f$ is meromorphic with poles $z_1,\\ldots,z_N$ with $z_j \\not\\in \\Z$. Our goal is a formula\n \\[\n \\sum_{n=-\\infty}^{\\infty} f(n) = - \\sum_{j=1}^N \\Res(g, z_j)\n \\]\n where $g(z) = \\pi \\cot(\\pi z) f(z)$. \\textit{Hint:} Consider a square contour $R$ centered at the origin with side-length $2N+1$. You may assume that $\\pi \\cot(\\pi z)$ is bounded on $R$ independent of $N$. You will then need to impose some bound on $f$ on the contour $R$.\n \\end{problem}\n \\begin{solution}\n Let's compute the contour integral $I=\\int_R g(z)dz$ in two different ways. First, using the residue theorem, we have that \n \\begin{equation}\\label{Residue_eq_q}\n \\frac{I}{2\\pi i} = \\sum_{j=1}^N \\Res(g,z_i) + \\sum_{k=-\\infty} ^\\infty \\Res(g, k).\n \\end{equation}\n We can compute the residue of $g$ at $k$\n \\[\n \\Res(\\pi\\cot(\\pi z)f(z), k) = \\lim_{z\\to k} \\pi\\cos(\\pi z)f(z)\\frac{(z-k)}{\\sin(\\pi z)} = \\lim_{z\\to k} \\pi\\cos(\\pi z)f(z)\\frac{(z-k)}{\\sin(\\pi z)} \n \\]\n\n Thus if we can show that $I=0$, we get the desired equality from \\eqref{Residue_eq_q}.\n \\[\n 0 = \\sum_{j=1}^N \\Res(g,z_i) + \\sum_{k=-\\infty} ^\\infty f(k).\n \\]\n\n Second, let's compute $\\int_R g(z)dz$ directly, aiming for it to be 0.\n \\[\n \\lim_{N\\to\\infty} \\int_R g(z) dz \\leq (4(2N+1))\\sup_{z\\in R}(g(z))\n \\]\n Since $\\pi\\cot(\\pi z)$ is bounded, it suffices for the growth rate of $f(z)$ to be asymptotically slowed than that of $8N+4$: $f(|z|) \\in \\omega(|z|)$.\n\n If so, we can conclude that $I=0$, proving the claim.\n \\end{solution}\n \\begin{problem}\n Apply \\ref{residue-coth} and \\ref{summation-theorem} to evaluate\n \\[\n \\sum_{n=-\\infty}^\\infty \\frac{1}{n^2 + b^2}.\n \\]\n \\end{problem}\n \\begin{solution}\n Let $f(n) = \\frac{1}{n^2+b^2}$. We assume $bi \\not \\in \\Z$ so that the sum is finite. Then in the previous problem, $g(n) = \\frac{\\pi\\cot(\\pi z)}{z^2 + b^2}$. By \\ref{residue-coth}, the residue at $\\pm bi$ is $\\frac{\\pi\\cot(\\pm \\pi bi)}{\\pm 2bi}$, so the sum of the residues is $-\\frac{\\pi \\coth{\\pi b}}{b}$ and thus the sum is $\\frac{\\pi \\coth{\\pi b}}{b}$.\n \\end{solution}\n \\section{Prove or Disprove and Salvage if Possible}\n\n \\begin{problem}\\label{zero-residue-not-enough}Suppose $f : D_r(z_0) \\to \\C$ is holomorphic. The residue\n $\\Res(f,z_0)$ vanishes if and only if the singularity $z_0$ is\n removable.\n \\end{problem}\n \\begin{solution}\n This is true for stupid reasons: if $f$ is holomorphic on a disk surrounding $z_0$, then $f$ has no singularity at $z_0$, so if a singularity exists, it is surely removable. If it is merely meromorphic, a counterexample like $\\frac{1}{z^2}$ suffices, since it has residue $0$ but no removable singularity at $0$.\n\n However, if the singularity is removable, the residue will vanish, since after removing the singularity, $f$ is holomorphic on a small ball containing $z_0$, so the residue must vanish.\n\n\n \\end{solution}\n \\begin{problem}\\label{open-mapping-theorem}If $U \\subset \\C$ is open\n and $f : U \\to \\C$ is holomorphic, then $f(U)$ is\n open. % missing nonconstant, can be proved from rouche\n \\end{problem}\n \\begin{solution}\n No, consider $f(z)=0$. If we don't allow $f$ to be a constant, then it is true.\n\n Choose some arbitrary point $z_0\\in U$ with $f(z_0)=w_0$. Now consider a series of open balls $\\{B_{1/n}(z_0)\\}_{n\\in\\N}$. We can surely find an $n$ where the $n$th ball is contained in $U$ since $U$ is open. Furthermore, by the identity theorem, there is an $n$ where the $n$th ball does not contain any point $z_1$ with $f(z_1)=w_0$ - otherwise, we would have an infinite series of points $\\{z_i\\}$ containing their limit point where $f(z_i)=w$, implying by the identity theorem that $\\forall z\\in U, f(z) = w$. Call the first $n$ satsifying both conditions $N$.\n\n Let $\\delta = (\\inf_{z\\in B_{1/N}(z_0)}\\abs{f(z)-w_0})/2$, so that for all $z$ in this ball,\n \\[\n \\abs{f(z) - w_0 } \\geq 2\\delta.\n \\]\n Choose some $w_1\\in \\partial D_\\delta (f(z_0))$, so that \n \\[\\delta \\geq \\abs{w_0-w_1}.\n \\]\n\n To show that $f(U)$ is open, it suffices to show that any such $w_1$ is in the image of $f$, or equivalently that $f(z) - w_1$ has a zero. We know from the last paragraph that $f(z)-w_0$ has exactly one zero in $B_{1/N}(z_0)$, and that \n \\[\n \\abs{f(z) - w_0} \\geq \\abs{w_0 - w_1}\n \\]\n so by Rouche's theorem, $f(z)-w_0 +w_0-w_1 = f(z) - w_1$ has a zero in $B_{1/N}$. Thus $f(U)$ is open.\n \\end{solution}\n\n\n \\end{document}\n", "meta": {"hexsha": "4e2022ea50db0dd69047177ef08f7d6a22996dfd", "size": 76314, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "problem-solutions/sol9.tex", "max_stars_repo_name": "Alex7Li/math5522h", "max_stars_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "problem-solutions/sol9.tex", "max_issues_repo_name": "Alex7Li/math5522h", "max_issues_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "problem-solutions/sol9.tex", "max_forks_repo_name": "Alex7Li/math5522h", "max_forks_repo_head_hexsha": "9f1fa070997f40e11e981c49e7d6fb9556e128d6", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 136.518783542, "max_line_length": 772, "alphanum_fraction": 0.217142333, "num_tokens": 11232, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.817574471748733, "lm_q2_score": 0.7490872187162397, "lm_q1q2_score": 0.6124345871356573}} {"text": "\n\n\\section{CP Violation and Mixing}\n\\label{sect:cpviolation}\n\\index{CP violation}\n\nThis section discusses how CP violation and mixing work\nin the generator. These two topics are closely related, since\na large fraction of CP violating decays occurs via mixing. Exactly what is \nthe best way of introducing these two effects into EvtGen is still not clear; \nthere are different ways of generating CP violating decays depending on what \none is interested in. For example, one might want to generate a sample of events with \n$B\\rightarrow a_1^+ \\pi^-$, and the $B$'s which tag the flavor of this $B$ \n(e.g., via semileptonic decays) are all $B^0$'s. On the\nother hand, one might want to generate a sample that has the \ncorrect mixture of $B^0$ and $\\bar B^0$ tags, as determined by the decay \ndynamics. From the point of view of implementation these two situations are \nfairly different, and at the moment we are still working on choosing \na consistent way of dealing with them.\n\nThis section starts out describing how mixing works, and\nthen proceeds to describe how CP violation is introduced. Examples\nwill be provided. CP violating decays are\namong the most complicated things this generator does, and \nvarious ways in which one might get the generation of CP\nviolating decays wrong will be pointed out.\n\n\\subsection{Mixing in the $B^0\\bar B^0$ System.}\n\nAs is well know, when $\\Upsilon(4S)$ decays into the $B^0\\bar B^0$ system, \nthe $B$'s are produced in a coherent state. This implies that \nwhen one of the $B$'s decays, and its flavor is determined, the other\n$B$ is known to be in a pure state of the opposite flavor.\nThis state then evolves (oscillates) independently, with the oscillation frequency\ngiven by $\\Delta m /2$, ($\\Delta m$ being $m_{2} - m_{1}$, the mass difference between the two $B$ mass \neigenstates). Assuming $|B_{0}^{\\rm phys}(t = 0)\\rangle = |B^{0}\\rangle$, its time evolution is given by: \n\\begin{equation}\n|B_{\\rm phys}^{0}(t)\\rangle = e^{-\\Gamma t/2}e^{i\\overline{m} t}\\,\\left[ |B^0\\rangle\\,\\cos\\left(\\frac{\\Delta m t}{2}\\right) + i\\frac{q}{p}|\\bar B^0\\rangle\\,\\sin\\left(\\frac{\\Delta m t}{2}\\right)\\right],\n\\end{equation}\nwhere $\\overline{m} \\equiv (m_{1} + m_{2})/2$, $q = e^{i\\phi_{M}}/\\sqrt{2}$, and $p = e^{-i\\phi_{M}}/\\sqrt{2}$ ($\\phi_{M}$ is the mixing angle). Time $t$ in this formula is the time between the two $B$ decays.\n\nBy projecting $|B_{0}^{\\rm phys}(t)\\rangle$ into a pure $B^0$\nor $\\bar B^0$ state, one can obtain the amplitudes for the $B$ to decay with\na given flavor. In turn, squaring them gives the probabilities:\n\\begin{eqnarray}\nP(B_{\\rm phys}^{0}\\rightarrow B^0) & = & \\frac{1}{2} e^{-\\Gamma t} (1+\\cos(\\Delta m t))\\\\\nP(B_{\\rm phys}^{0}\\rightarrow \\bar B^0) & = & \\frac{1}{2} e^{-\\Gamma t} (1-\\cos(\\Delta m t))\\\\\n\\end{eqnarray}\n\nIntegrating these over all times, one obtains the total rates:\n\\begin{eqnarray}\nP(B_{\\rm phys}^{0}\\rightarrow B^0) & = & \\frac{x_{d}^2}{2\\,(1 + x_{d}^{2})}\\\\\nP(B_{\\rm phys}^{0}\\rightarrow \\bar B^0) & = & \\frac{2 + x_{d}^2}{2\\,(1 + x_{d}^{2})}\\\\\n\\end{eqnarray}\nwhere $x_d \\equiv \\frac{\\Delta m}{\\Gamma}$.\n\nConsidering in a similar way the time evolution of the coherently produced state ($\\Psi(t = 0) = \\frac{1}{\\sqrt{2}}\\,\\left [ |B^{0} \\overline{B}^{0} \\rangle - |\\overline{B}^{0} B^{0} \\rangle \\right ]$ ), one can obtain the ratio:\n\\begin{equation}\n{ N(\\Upsilon(4S)\\rightarrow B^0B^0)+N(\\Upsilon(4S)\\rightarrow\\bar B^0\\bar B^0)\n \\over N(\\Upsilon(4S)\\rightarrow B^0\\bar B^0)}= \\frac{x_{d}^{2}}{2 + \\,x_{d}^{2}}.\n\\end{equation}\n\nCurrently, mixing is implemented in two decay models: \n$\\tt VSS\\_MIX$ and $\\tt VSS\\_BMIX$. These models both take the mass\ndifference to be a parameter and use the width assigned in the particle\nproperties table, i.e., from the evt.pdl file in the BaBar environment.\nThese models differ in the number of parameters that\nthe user must supply. The $\\tt VSS\\_MIX$ model requires that the branching\nfractions into mixed and unmixed final states be set by hand, and does not\nrequire that they be consistent with the mass difference and lifetime.\nThe $\\tt VSS_BMIX$ model calculates the branching fractions automatically\nand is therefore safer and easier to use. One can use this model as follows:\n\\begin{verbatim}\nDecay Upsilon(4S)\n0.5 B+ B- VSS;\n0.5 B0 anti-B0 VSS_BMIX;\nEnddecay\n\\end{verbatim}\nBoth models give identical results when the $\\tt VSS\\_MIX$ branching\nfractions are set correctly.\n\n\n\\subsection{Mixing and Non-zero Lifetime Differences.}\n\n{\\it This section should probably be merged with the previous but \nis currently written separately, mostly due to historical reasons, as\nthis is only needed for $B_s$ mixing, which was added later.}\n\nIn this section $B^0$ and $\\bar B^0$ will generically denote any\nneutral meson flavor eigenstate. These states are not eigenstates\nof the full hamiltonian and hence do not have a definite mass or\nlifetime. The states that are eigenstates of the hamiltonian are\n\\begin{eqnarray}\n|B_L\\rangle & = & p|B^0\\rangle +q|\\bar B^0\\rangle, \\\\\n|B_H\\rangle & = & p|B^0\\rangle -q|\\bar B^0\\rangle.\n\\end{eqnarray}\nThe states $B_L$ and $B_H$ have masses, $m_L$ and $m_H$, and widths, $\\Gamma_L$\nand $\\Gamma_H$, respectively such that \n\\begin{equation}\n|B_{L,H}(t)\\rangle=e^{-(\\Gamma_{L,T}/2+im_{L,T})t}|B_{L,H}\\rangle\n\\end{equation}\nwhere $B_{L,H}(t)$ is the state after evolving the state $B_{L,H}$ for\na time $t$.\n\nThis allows us to evaluate several matrix elements that are useful for\ncalculating the mixing rates\n\\begin{eqnarray}\n\\langle B^0|B^0(t)\\rangle & = & {1\\over 2p}(\\langle B^0|B_L(t)\\rangle\n +\\langle B^0|B_H(t)\\rangle)\\nonumber\\\\\n & = & {1\\over 2}(e^{-(\\Gamma_L/2+im_L)t}\n +e^{-(\\Gamma_H/2+im_H)t}),\\\\\n\\langle \\bar B^0|B^0(t)\\rangle & = & {1\\over 2p}(\\langle \\bar B^0|B_L(t)\\rangle\n +\\langle \\bar B^0|B_H(t)\\rangle)\\nonumber\\\\\n & = & {q\\over 2p}(e^{-(\\Gamma_L/2+im_L)t}\n -e^{-(\\Gamma_H/2+im_H)t}),\\\\\n\\langle B^0|\\bar B^0(t)\\rangle & = & {1\\over 2q}(\\langle B^0|B_L(t)\\rangle\n -\\langle B^0|B_H(t)\\rangle)\\nonumber\\\\\n & = & {p\\over 2q}(e^{-(\\Gamma_L/2+im_L)t}\n -e^{-(\\Gamma_H/2+im_H)t}),\\\\\n\\langle \\bar B^0|\\bar B^0(t)\\rangle & = & {1\\over 2q}(\\langle \\bar B^0|B_L(t)\\rangle\n +\\langle \\bar B^0|B_H(t)\\rangle)\\nonumber\\\\\n & = & {1\\over 2}(e^{-(\\Gamma_L/2+im_L)t}\n +e^{-(\\Gamma_H/2+im_H)t}).\n\\end{eqnarray}\n\nNow it is straight forward to calculate the mixing rate. Let $\\chi$ \ndefine the probability that a meson produced, at $t=0$, as a $B^0$ will\ndecay as a $\\bar B^0$. It is now easy to see that \n\\begin{eqnarray}\n\\chi & = &{\\int_0^\\infty |\\langle \\bar B^0|B^0(t)\\rangle|^2dt \\over\n\\int_0^\\infty |\\langle \\bar B^0|B^0(t)\\rangle|^2dt +\n\\int_0^\\infty |\\langle B^0| B^0(t)\\rangle|^2dt }\\nonumber\\\\\n & = & { x^2+y^2 \\over x^2+y^2+\\vert {p\\over q} \\vert^2(2+x^2-y^2)} \n\\end{eqnarray}\nwhere \n\\begin{equation}\nx\\equiv {\\Delta m \\over \\Gamma},\\qquad y\\equiv {\\Delta\\Gamma\\over \\Gamma}\n\\end{equation}\nand\n\\begin{equation}\n\\Gamma\\equiv {\\Gamma_L+\\Gamma_H \\over 2},\\qquad\n\\Delta\\Gamma \\equiv \\Gamma_H-\\Gamma_L,\\qquad\n\\Delta m \\equiv m_H-m_L.\n\\end{equation}\nSimilarly the probability that that a meson produced, at $t=0$ \nas a $\\bar B^0$ will decay as a $B^0$ is given by\n\\begin{eqnarray}\n\\bar \\chi & = &{\\int_0^\\infty |\\langle B^0|\\bar B^0(t)\\rangle|^2dt \\over\n\\int_0^\\infty |\\langle B^0|\\bar B^0(t)\\rangle|^2dt +\n\\int_0^\\infty |\\langle \\bar B^0| \\bar B^0(t)\\rangle|^2dt }\\nonumber\\\\\n & = & { x^2+y^2 \\over x^2+y^2+\\vert {q\\over p} \\vert^2(2+x^2-y^2)}.\n\\end{eqnarray}\nIn the limit of $|p/q|=1$ and $y=0$ we have \n\\begin{equation}\nP_{\\pm}(t)=\\chi =\\bar\\chi={x^2\\over 2(1+x^2)}\n\\end{equation}\nas expected.\nThe time distribution, arbitrary normalization, is given by\n\\begin{equation}\ne^{-\\Gamma_L t}(1+e^{\\Delta\\Gamma t}\\pm 2e^{\\Delta\\Gamma t/2}\\cos\\Delta m t)\n\\end{equation}\nThe minus sign corresponds to the case of mixing, and the plus sign\nto the case of no mixing.\n\n\n\\subsection{CP-violation.}\n\nThe ``first generation'' of models which were implemented in EvtGen\nto deal with CP violation did not solve all the problems that are \ninvolved in simulating the physics of the $B\\bar B$ system. To illustrate how \nthese models work, let us consider the following example (the decay \n$B\\rightarrow \\pi^+\\pi^-$):\n\\begin{verbatim}\nAlias MYB B0\n#\nDecay Upsilon(4S)\n0.70 MYB B0 VSS;\n0.30 MYB B0B VSS;\nEnddecay\n#\nDecay MYB\n1.0 pi+ pi- SSS_CP alpha dm CP |A| arg(A) |Abar| arg(Abar);\nEnddecay\nEnd\n\\end{verbatim}\n\nThe model used in this example is {\\tt SSS\\_CP}. Its first argument is {\\tt alpha}, is the\nrelevant CKM angle (in this case it is $\\alpha$); the second argument {\\tt dm} is\nthe mass difference of the two $B$ mass eigenstates; the third\nargument, {\\tt CP}, specifies the CP of the final state, and is either $+1$ or $-1$.\nThe latter argument could in principle be determined from the particle properties, but at \nthe moment EvtGen does not include the C and P quantum numbers in its particle properties\nlist (this should be fixed). The last 4 arguments are the (complex) amplitudes for \n$B^0$ and $\\bar B^0$ to produce the final state.\n\nIn this example the $\\Upsilon(4S)$ decays with unequal\nprobabilities to ``MYB'' and $B^0$ and to ``MYB'' and $\\bar B^0$. The \n$B^0$ and $\\bar B^0$ simply decay as a $B^0$ and a $\\bar B^0$,\nrespectively, according to the generic decay table. The interesting\nthing is how ``MYB'' decays. In the {\\tt SSS\\_CP} model, ``MYB'' will produce a\n$\\pi^+ \\pi^-$ final state, while the other $B$ will provide\na tag of the flavor that is listed in the decay of the $\\Upsilon(4S)$ ($B^0$ in 70\\% \nof the cases, and $\\bar B^0$ in 30\\% of the cases). Therefore, in this example there will\nbe some number of $B^0$ tags and some number of $\\bar B^0$ tags. \n \nIf one would like to get a sample of events with only $B^0$ tags, \none could do it in the following way:\n\\begin{verbatim}\nAlias MYB B0\n#\nDecay Upsilon(4S)\n1.00 MYB B0 VSS;\nEnddecay\n#\nDecay MYB\n1.0 pi+ pi- SSS_CP dm alpha CP |A| arg(A) |Abar| arg(Abar);\nEnddecay\nEnd\n\\end{verbatim}\n\nHere, the only available tag (as specified by the $\\Upsilon(4S)$ branching fraction) is $B^{0}$.\n\nWhen the $\\Upsilon(4S)$ is decayed and the two $B$'s are produced,\nthey are assigned lifetimes which are sampled from a pure exponential\ndistribution, with a mean of the average lifetime specified\nin the particle properties list. (If the decay model is one that\nincludes the effects of mixing, the time distribution is no longer\na pure exponential; see the previous section. However, as will be\nevident below, it doesn't matter what time distribution is used in\nthe decay of the $\\Upsilon(4S)$.) Since the lifetime distributions\nof both the $B$ that decays to the CP channel and the tag $B$ are\nno longer exponential, the {\\tt SSS\\_CP} model has to regenerate \nlifetimes of {\\it both} $B$'s. However, the model does not change what\ntag one of the $B$ provides. This is important, because it means that this model\ncannot enforce the right fraction of $B^0$ and $\\bar B^0$ tags. {\\bf The only way to control this is through the decay table}.\n\nThe {\\tt SSS\\_CP} model handles decays to a pair of scalar particles.\nOf these, the most important one is probably $B \\rightarrow \\pi\\pi$, but the model also \nworks for any other decay with two scalars in the final state, e.g., $K_s\\eta'$.\nOther models which are implemented in a similar way are {\\tt SVS\\_CP} and {\\tt STS\\_CP},\nwhich handle the decays to a scalar and a vector and to a scalar and a \ntensor, respectively. There is also a model for two vectors\nin the final state: it is called {\\tt SVV\\_CP}. One complication with this\nmodel is that there are three partial waves that contribute to the \nrate with different CP. \n\nAll of these models have one feature in common: they generate the $B$ tags with the relative\nfractions specified in the decay table. It seems logical that the next step in expanding \nthe functionality of the generator should be implementing some way to generate the right \nfraction of $B^0$ and $\\bar B^0$ tags according to the amplitudes that are specified for a \nparticular decay channel. Indeed, this would be a very natural thing to do, because given \nthe amplitudes, the relative fraction of $B^0$ and $\\bar B^0$ tags can be determined. A new class of models which have been recently put into EvtGen does have this feature. These new models are particularly important for decays into final states that are not\nCP eigenstates, for example $B\\rightarrow a_1^+\\pi^-$, or decays with direct CP violation (e.g., due to the presence of penguins).\n\nLet us consider in more detail what the problem is. The time-dependent CP-asymmetry for\nthe decay $B \\rightarrow f$, where $f$ is a CP eigenstate, is defined as:\n\\begin{equation}\nA_f(t) = \\frac{\\Gamma(B^{0}(t) \\rightarrow f) - \\Gamma(\\overline{B}^{0}(t) \\rightarrow f)}{\\Gamma(B^{0}(t) \\rightarrow f) + \\Gamma(\\overline{B}^{0}(t) \\rightarrow f)}\n\\end{equation}\nIf $|A(B^{0} \\rightarrow f)|$ = $|A(\\overline{B}^{0} \\rightarrow f)|$, then the integrated \nasymmetry $\\int_{-\\infty}^{+\\infty}A_f(t)dt = 0$, and the relative fraction of $B^{0}$ ($\\bar B^0$) tags is known (1/2). However, if $|A(B^{0} \\rightarrow f)| \\neq |A(\\overline{B}^{0} \\rightarrow f)|$ (which can happen, e.g., because of penguin pollution), $\\int_{-\\infty}^{+\\infty}A_f(t)dt \\neq 0$, so the fraction of tags must be somehow determined.\n\n\nOne way to get the correct fraction of tags is to use integrated rates. For example, for decays into a CP-eigenstate it can be defined as:\n\\begin{equation}\nfr \\equiv \\frac{\\int_{-\\infty}^{\\infty}\\Gamma(\\overline{B}(t)^{0} \\rightarrow f)dt}{\\int_{-\\infty}^{\\infty}\\left[\\Gamma(B^{0}(t) \\rightarrow f) + \\Gamma(\\overline{B}^{0}(t) \\rightarrow f)\\right]dt} \n\\end{equation}\nso that\n\\begin{equation}\nfr = \\frac{|\\overline{A}_{f}|^2 \\left(1 + |\\overline{r}_{f}|^2 + \\frac{(1 - |\\overline{r}_{f}|^2)}{1+x_{d}^2} \\right)}{|\\overline{A}_{f}|^2 \\left(1 + |\\overline{r}_{f}|^2 + \\frac{(1 - |\\overline{r}_{f}|^2)}{1+x_{d}^2} \\right) + |{A}_{f}|^2 \\left(1 + |r_{f}|^2 + \\frac{(1 - |r_{f}|^2)}{1+x_{d}^2} \\right)},\n\\end{equation}\nwhere $x_{d} \\equiv \\frac{\\Delta m}{\\Gamma}$, $r_{f} = e^{2i\\,\\phi_{M}}\\,\\frac{\\overline{A}}{A}$, and $\\overline{r}_{f} = \\frac{1}{r_{f}}$.\n\nTwo functions in the new version of EvtOtherB return this fraction (fractB0CP for decays into a CP eigenstate and fractB0nonCP for decays into non-CP eigenstates). However, analytical calculations of the integrals can be done only in the simplest cases. For example, for the $B \\rightarrow \\rho \\pi$ decay one would have to integrate over the Dalitz plot as well as over time to get the total rates, which is very hard to do! A better solution is to use the so-called acceptance-rejection method, which is effectively equivalent to numerical integration. This method has been used in the ``second generation'' models which introduce direct CP-violation into EvtGen.\n\n The new models that are available are: {\\tt SSS\\_CP\\_PNG} for the $B \\rightarrow \\pi^{+} \\pi^{-}$ decay with penguin contributions; {\\tt BTO2PI\\_CP\\_ISO} for the three isospin-related $B\\rightarrow \\pi \\pi$ modes; {\\tt BTOKPI\\_CP\\_ISO} for the four isospin-related $B \\rightarrow K \\pi$ modes; and {\\tt SVS\\_CP\\_ISO} for the five isospin-related generic $scalar \\rightarrow (pseudo)scalar + vector$ modes. There is another decay model for the latter case, which also returns the correct fraction of $B^0$ ($\\bar B^0$) tags. It is called {\\tt SVS\\_NONCPEIGEN}. Here is an example of how this model can be used for the $B \\rightarrow a_1 \\pi$ decay:\n\\begin{verbatim}\nAlias MYB B0\nDecay Upsilon(4S)\n1.000 MYB anti-B0 VSS;\nEnddecay\nDecay MYB\n1.000 a_1- pi+ SVS_NONCPEIGEN 1.22 0.51e12 0.0 \n 1.0 0.0 3.0 0.1 \n 3.0 0.1 1.0 0.0;\nEnddecay\nEnd\n\\end{verbatim}\n\nThe first parameter of this model is the corresponding CKM angle (in this case, it is $\\alpha$); the second parameter is $\\Delta m$; the third parameter is the ``flip'' which determines the fraction of ``MYB'' $\\rightarrow f$ to ``MYB'' $\\rightarrow \\overline{f}$ decays, where the state specified in the decay table is considered the ``$f$'' state. If the flip is set to 0, ``MYB'' always decays into the $f$ state. The next four parameters are the absolute values and phases for the amplitudes $B^{0} \\rightarrow f$ and $\\bar B^{0} \\rightarrow f$, respectively, and the last four parameters are those for the amplitudes $B^{0} \\rightarrow \\overline{f}$ and $\\overline{B}^{0} \\rightarrow \\overline{f}$.\n\nIt is important not to confuse the ``flip'' parameter with the number of tags. For example, one would like to generate 10,000 $B\\overline{B}$ events with the flip parameter set to 0.5. This means that one would get 5,000 ``MYB'' $\\rightarrow f$ decays and 5,000 ``MYB'' $\\rightarrow \\overline{f}$ decays, but within each 5,000 event sample the number of $B^{0}$ and $\\bar B^{0}$ tags is determined by the amplitudes specified in the decay file! \n\nAs before, ``MYB'' in this example decays into a CP-mode, while the other B provides a tag. However, the model itself generates the correct fraction of $B^{0}$ ($\\bar B^{0}$) tags, which is determined from the last eight parameters (the amplitudes), and is independent of the $\\Upsilon(4S)$ branching fractions.\n\n\n\n\\subsection{Special CP models}\n\n The models described above are meant to be generic models that can\nbe applied to different decays with appropriate arguments. All of the\nabove models are two body decays. There is, however, many other decays\nthat are not two body decays and these can not be treated quite as\nsimply since there will in general be interference between various\nresonances. As an example consider the decay $B\\rightarrow \\pi^+\\pi^-\\pi^0$\nwhich is dominated by the $\\rho(770)$ resonances.\n\n To handle the $B\\rightarrow 3\\pi$ and the $B\\rightarrow 4\\pi$\ndecays two special models have been written. These models are \ncalled {\\tt BTO3PI\\_CP} and {\\tt BTO4PI\\_CP}. The second\nof model {\\tt BTO4PI\\_CP} needs more work interms of understanding the\nconventions of the phases between the resonances that contributes.\n\n\\newpage\n\n\\subsection{Unified implementation of CP-violating and mixing}\n\nThis section describes a {\\it proposal} for how to unify and \nsimplify the simulation of CP-violating decays and mixing in EvtGen, this \nshould allow writing models that works independently of\nhow the neutral $B$ meson is produced, i.e. in a coherent pair in an\n$\\Upsilon(4S)$ decay or incoherently e.g. in a $p\\bar p$ collision.\n\n\\subsubsection{CP violation}\n\nThe decays that we are considering here are \ncommon final states to a neutral $B$ meson and its antiparticle,\ne.g. $B\\to J/\\psi K_S$\nor $B\\to J/\\psi K^*$. At the $\\Upsilon(4S)$ in the absence of\ndirect CP violation these decays can be simplified as we can make the \nassumption that you have an equal number of $B^0$ and $\\bar B^0$\ntags, or equivalent that there is no time integrated asymmetry.\nHowever, we want to write the code such that no assumptions are\nmade about the numbers of $B^0$ and $\\bar B^0$ tags.\nFor $B$ mesons that are produced incoherently we always have to\nconsider the asymmetries in the rates for an initial $B^0$ or\n$\\bar B^0$. \n\nTo allow generating these asymmetries correctly these models\nwill be allowed to modify the flavor of the $B$ meson that is decayed.\nFor the $\\Upsilon(4S)$ this means changing the the flavor of 'the\nother' $B$ meson in the $\\Upsilon(4S)$ decay, the tag $B$. \nFor incoherent produced $B$\nmesons this means that if a specific flavor is given to \nEvtGen it can return a decay of the opposite flavor. If\ne.g. EvtGen is given a $B^{**}+$ to decay and it decays it to\n$B^0\\pi^+$ the decay of the $B^0$ might decide to flip the \nflavor of the decay chain to allow generating the right \ntime integrated asymmetry. This means that after EvtGen is\ndone the code that called EvtGen will get back a $B^{**}-$\nand needs to be prepared to handle this, e.g. by applying CP\nto the decay chain that produced the $B^{**}+$. Also note the\nmeaning of branching fractions in the decay table, for modes\nthat are common to a $B$ and its anti-particle, the $\\bar B$\nthe above implies that the branching fraction listed for the\n$B$ and the $\\bar B$ are the average branching fractions.\n\nPractically, this is implemented in the {\\tt EvtCPUtil}\nclass, which will handle the cases of coherent and incoherently\nproduced $B$ mesons.\n\nHow is this different from the current implementation, V00-09-38?\nNot very much, there are already some models, e.g. the \n{\\tt SVS\\_NONCPEIGEN}\nthat has the full functionality that allows for direct CP violation \nand will generate the right mixture of $B$ and $\\bar B$ tags.\nAll CP models should convert to this more general model. For \napplying this to $B$ mesons that are produced incoherent, i.e.\nwith a definite flavor, we need to modify the {\\tt EvtCPUtil::otherB()}\nmethod to allow the change of the flavor.\n\n\\subsubsection{Mixing}\n\nIn the $\\Upsilon(4S)$ system mixing is handled by the decay of the\n$\\Upsilon(4S)$ meson via models such as {\\tt VSS\\_BMIX}. This \nimplementation seems sufficient at this point. Note that we\nare not considering the case of the $\\Upsilon(5S)$ at this point.\n\nFor incoherently produced neutral $B$ mesons, $B^0$ or $B_s$, we assume that\nthe flavor is tagged and mixing will be generated by adding a \ndecay like $B^0\\to \\bar B^0$ where the mixed particle have \nzero lifetime. Models that simulate CP violation, or in general\nmodels that handle decays common to the $B$ and the $\\bar B$ \ndecay will remove any mixing, this is done in {\\tt EvtCPUtil}. \n(Models that do not make any assumption about\nthe flavor of the $B$ should probably be using a 'neutral $B$' particle\ninstead of a specific flavor, but at BABAR we have not implemented \nthis as it would be very confusing at this point to a lot of\nanalysis code that uses Monte Carlo truth.)\n\nThis means that models that simulates CP-violating decays should not\nneed to turn of mixing, it will practically be ignored. \n\nThe mixing of incoherently produced $B$ mesons is not done in a \nmodel, this causes some problems as it is hard to control the\nmixing, e.g., turning it of or modifying the parameters. We need to\ninvent a way to control this.\n\n\n", "meta": {"hexsha": "30cafacbd1a8009e695e5ca0f6251e3a85d9f21a", "size": 22692, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "EvtGen1_06_00/doc/evt_cpviolation.tex", "max_stars_repo_name": "klendathu2k/StarGenerator", "max_stars_repo_head_hexsha": "7dd407c41d4eea059ca96ded80d30bda0bc014a4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-12-24T19:37:00.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-28T06:57:20.000Z", "max_issues_repo_path": "EvtGen1_06_00/doc/evt_cpviolation.tex", "max_issues_repo_name": "klendathu2k/StarGenerator", "max_issues_repo_head_hexsha": "7dd407c41d4eea059ca96ded80d30bda0bc014a4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "EvtGen1_06_00/doc/evt_cpviolation.tex", "max_forks_repo_name": "klendathu2k/StarGenerator", "max_forks_repo_head_hexsha": "7dd407c41d4eea059ca96ded80d30bda0bc014a4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 56.73, "max_line_length": 705, "alphanum_fraction": 0.6983959105, "num_tokens": 6917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.7490872131147276, "lm_q1q2_score": 0.6124345858856297}} {"text": "\\documentclass{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage[margin=2.5cm]{geometry}\n\n\n\n\\begin{document}\n\t\\section{Geometry}\n\t\\begin{itemize}\n\t\\item\n\tThe sum of opposite angles in cyclic quadrilateral is $180^\\circ$.\n\t\\item\n\tInscribed angles which subtend to same arc are equal to half of the central angle. \n\t\\item\n\tLimiting case of the above is alternate segment theorem. The angle between tangent and chord is equal to inscribed angle subtending to the same chord.\n\t\\end{itemize}\n\t\n\t\n\t\\begin{enumerate}\n\t\t\\item\n\t\tFind mistake in the \"proof that all triangles are isosceles\".\n\t\t\n\t\t\\item %% LPV 2000 11\n\t\t$PQRS$ is a cyclic quadrilateral in which $\\angle PSR = 90^\\circ$. Points $H$ and $K$ are projections of point $Q$ to lines $PR$ and $PS$ respectively. Prove that line $HK$ intersects segment $SQ$ at its midpoint.\n\t\t\n\t\t\\item %% LPV 2002\n\t\tPoint $P$ is chosen on the circumcircle of triangle $ABC$. Points $K$, $L$ and $M$ are projections of $P$ to lines $AB$, $BC$ and $AC$ respectively. Prove that $K$, $L$ and $M$ lie on the same line.\n \t\t\n\t\t\\item %% problem solving strategies\n\t\tOne of the cross sections in a rectangular box is a regular hexagon. Prove that the rectangular box is a cube.\n\t\t\n\n\t\\end{enumerate}\n\t\\section{Trigonometry}\n\t\t\\begin{itemize}\n\t\t\t\\item\n\t\t\tSine law:\n\t\t\t$$\\frac{a}{\\sin \\alpha}=\\frac{b}{\\sin \\beta}=\\frac{c}{\\sin \\gamma}=2R $$\n\t\t\t\\item\n\t\t\tCosine law:\n\t\t\t$$ c^2=a^2+b^2-2ab\\cos\\gamma $$\n\t\t\t\\item\n\t\t\tTrigonometric functions of sums:\n\t\t\t$$ \\sin(\\alpha+\\beta)=\\sin\\alpha\\cos\\beta+\\cos\\alpha\\sin\\beta $$\n\t\t\t$$ \\cos(\\alpha+\\beta)=\\cos\\alpha\\cos\\beta-\\sin\\alpha\\sin\\beta $$\n\t\t\\end{itemize}\n\t\t\\begin{enumerate}\n\t\t\t\\item %% Prasolov 1.19\n\t\t\tA straight line passing through vertex $A$ of square $ABCD$ intersects side $CD$ at $E$ and line $BC$ at $F$. Prove that $\\frac{1}{|AE|^2}+\\frac{1}{|AF|^2} =\n\t\t\t\\frac{1}{|AB|^2}$.\n\t\t\t\\item\n\t\t\tAn equilateral triangle is inscribed in a circle. An arbitrary point $P$ is chosen on arc $BC$. Prove that $|MA|=|MB|+|MC|$\n\t\t\t\\item %% Prasolov 1.25\n\t\t\tAngles $\\alpha$ and $\\beta$ of $\\triangle ABC$ are related as $3\\alpha+2\\beta=180^\\circ$. Prove that $a^2+bc=c^2$.\n\t\t\t\\item %% Prasolov 12.49\n\t\t\tProve that for non right angle triangle with circumcircle radius $R$ and area $S$ the following holds\n\t\t\t$$ \\tan\\alpha+\\tan\\beta+\\tan\\gamma=\\tan\\alpha\\tan\\beta\\tan\\gamma=\\frac{4S}{a^2+b^2+c^2-8R^2}$$\n\t\t\\end{enumerate}\n\t\n\t\\newpage\n\t\n\t\\section{Functional equations}\n\t\t\\begin{enumerate} \n\t\t\t\\item %% LVS 2016\t\t\n\t\t\tFind all functions defined on all real numbers which satisfy for all $(x,y)$\n\t\t\t$$f(x+y)f(xy)=f(x^2-y^2-1) $$\n\t\t\t\\item %% LVS 2015\n\t\t\tFind all functions defined on all real numbers which satisfy for all $(x,y)$\n\t\t\t$$f(f(x)+f(y))=f(x)+y $$\n\t\t\t\\item %% LVT 2014\n\t\t\tFind all functions defined on all real numbers which satisfy for all $(x,y)$\n\t\t\t$$f(x^2)+f(xy)=f(f(x+y)) $$\n\t\t\t\\item %% problemsolvingstrategies p277\n\t\t\tFind all polynomial functions that satisfy $P(x+1)=P(x)+2x+1$\n\t\t\t\\item%% problemsolvingstrategies p277\n\t\t\tFind all solutions to functional equation $f(x+y) + f(x-y) = 2f(x)\\cos y$\n\t\t\t\n\t\t\t\n\t\t\\end{enumerate}\n\t\n\t\\section{Algebra}\n\t\tMean inequalities:\n\t\t$$\\min (a,b,c) \\leq \\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}} \\leq \\sqrt[3]{abc} \\leq \\frac{a+b+c}{3} \\leq \\sqrt{\\frac{a^2+b^2+c^2}{3}} \\leq \\max (a,b,c)$$\n\t\t\\\\\n\t\tCauchy-Schwarz inequality\n\t\t$$(a_1b_1+...+a_nb_n)^2 \\leq (a_1^2+...+a_n^2)(b_1^2+...+b_n^2) $$\n\t\tGeometric series\n\t\t$$ 1+a+a^2+...a^{n-1}=\\frac{a^n-1}{a-1} $$\n\t\t\n\t\t\n\t\t\\begin{enumerate}\n\t\t\\item\n\t\tProve that\n\t\t$$\\frac{x^2+2}{\\sqrt{x^2+1}} \\geq 2 $$\n\t\t\\item\n\t\tProve that\n\t\t$$ a^2+b^2+c^2 \\geq ab+bc+ac $$\n\t\t\\item\n\t\tFind sum\n\t\t$$ \\frac{1}{1 \\times 2} + \\frac{1}{2 \\times 3} +...+ \\frac{1}{(n-1)\\times n} $$\n\t\t\\item\n\t\tProve that\n\t\t$$ 1+2^2+3^2+...+n^2 = \\frac{n(n+1)(2n+1)}{6} $$\n\t\t\\item\n\t\tLet $n \\geq 3$ be an integer, and let $a_2$, $a_3$, ..., $a_n$ be positive real numbers such that $a_2a_3...a_n = 1$. Prove that\n\t\t$$(1 + a_2)^2(1 + a_3)^3...(1 + a_n)^n > n^n$$\t\t\t\n\t\t\\end{enumerate}\n\t\n\t\\newpage\n\t\n\t\\section{Number Theory}\n\tFermat's Little theorem for prime number $p$ and integer $a$\n\t$$ a^p \\equiv a \\pmod p$$\n\tEuclidian algorithm\n\t$$ \\gcd(a,b)=\\gcd(a-b,b)$$\n\t\\begin{enumerate}\n\t\t\\item\n\t\tProve that $a^4+4b^4$ is not a prime given $a,b$ are positive integers.\n\t\t\\item\n\t\tFind integer solutions $(x,y)$ in terms of prime $p$ for equation\n\t\t$$ \\frac{1}{x}+\\frac{1}{y}=\\frac{1}{p} $$\n\t\t\\item\n\t\tFind all solutions $(x,y,z)$ in integers for\n\t\t$$ x^2 + y^2 + z^2 = 2xyz $$ \n\t\t\\item\n\t\tFind all prime numbers that are the factors of any number in form of\n\t\t$\\underbrace{111...111}_{n}$ for positive integers $n$.\n\t\t\\item\n\t\tPositive integers $a,b$ satisfy\n\t\t$$ a-b = 5b^2-4a^2 > 0$$\n\t\tShow that $a-b$ is a square of an integer.\n\t\\end{enumerate}\n\t\n\t\\section{Combinatorics}\n\t\\begin{enumerate}\n\t\t\\item\n\t\tThere are 6 people in the room. Each pair of people either knows one another or they do not. Prove that there either 3 people among whom nobody knows another or 3 people among whom all people know one another.\n\t\t\\item\n\t\tIs it possible to cover 2018x2018 board with L-shaped pieces (length 3, width 2) without overlapping?\n\t\t\\item\n\t\tIs it possible to find 5 prime numbers, so that sum of each three of those is also a prime?\n\t\t\n\t\t\\item\n\t\tOn the table there are 13 blue chips, 15 red chips and 17 green chips. In each move two chips of different colour are taken from the table and replaced with two chips of the third colour. Is it possible that after some moves there are equal amount of chips of each colour?\n\t\t\n\t\t\\item\n\t\tEach face of a cube is coloured with a different colour. How many distinct colourings are there?\n\t\t \n\t\\end{enumerate}\n\t\n\\end{document}", "meta": {"hexsha": "de0d8d767adeb294cf8cf9090849355068c2117f", "size": 5679, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "01_intro.tex", "max_stars_repo_name": "ZhaoWanLong/maths-olympiad", "max_stars_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-08-21T21:57:43.000Z", "max_stars_repo_stars_event_max_datetime": "2019-08-21T21:57:43.000Z", "max_issues_repo_path": "01_intro.tex", "max_issues_repo_name": "ZhaoWanLong/maths-olympiad", "max_issues_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "01_intro.tex", "max_forks_repo_name": "ZhaoWanLong/maths-olympiad", "max_forks_repo_head_hexsha": "0dcacba8a6d1769bbccfedda89d08fa22c3f55a1", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2022-01-08T07:04:43.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-08T07:04:43.000Z", "avg_line_length": 37.3618421053, "max_line_length": 274, "alphanum_fraction": 0.6536362036, "num_tokens": 2110, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872131147276, "lm_q2_score": 0.8175744761936437, "lm_q1q2_score": 0.6124345858856297}} {"text": "\\section{Trust Flow}\n We can now define indirect trust from $A$ to $B$.\n \\subimport{common/definitions/}{indirecttrust.tex}\n \\noindent Note that $Tr_{A \\rightarrow B} \\geq DTr_{A \\rightarrow B}$. The next result shows $Tr_{A \\rightarrow B}$ is\n finite.\n \\subimport{fc17/theorems/}{convergencetheorem.tex}\n \\subimport{common/proofsketches/}{convergenceproofsketch.tex}\n Proofs of all theorems can be found in Appendix A\\ifdefined\\proceedings \\ of the full version \\cite{trustisrisk}\\fi.\n\n In the setting of \\texttt{TransitiveGame(}$\\mathcal{G}$\\texttt{,}$A$\\texttt{,}$B$\\texttt{)} and $j$ being a turn in which\n the game has converged, we use the notation $Loss_A = Loss_{A, j}$. $Loss_A$ is not the same for repeated executions of this\n kind of game, since the order in which players are chosen may differ between executions and conservative players can choose\n which incoming direct trusts they will steal and how much from each.\n\n Let $G$ be a weighted directed graph. We investigate the maximum flow on it. For an introduction to maximum\n flows see Introduction to Algorithms, p. 708 \\cite{clrs}. Considering each edge's capacity as its weight, a flow\n assignment $X = [x_{vw}]_{\\mathcal{V} \\times \\mathcal{V}}$ with source $A$ and sink $B$ is valid when:\n \\begin{equation}\n \\label{flow1}\n \\forall (v, w) \\in \\mathcal{E}, x_{vw} \\leq c_{vw} \\mbox{ and}\n \\end{equation}\n \\begin{equation}\n \\label{flow2}\n \\forall v \\in \\mathcal{V} \\setminus \\{A,B\\}, \\sum\\limits_{w \\in N^{+}(v)}x_{wv} = \\sum\\limits_{w \\in N^{-}(v)}x_{vw}\n \\enspace.\n \\end{equation}\n The flow value is $\\sum\\limits_{v \\in N^{+}\\left(A\\right)}x_{Av} = \\sum\\limits_{v \\in N^{-}\\left(B\\right)}x_{vB}$. We do not\n suppose skew symmetry in $X$. There exists an algorithm $MaxFlow\\left(A, B\\right)$ that returns the maximum possible flow\n from $A$ to $B$. This algorithm needs full knowledge of the graph and runs in $O\\left(|\\mathcal{V}||\\mathcal{E}|\\right)$\n time \\cite{maxflownm}. We refer to the flow value of $MaxFlow\\left(A, B\\right)$ as $maxFlow\\left(A, B\\right)$.\n\n We will now introduce two lemmas that will be used to prove one of the central results of this work, the Trust Flow\n theorem.\n \\subimport{fc17/lemmas/}{flowgamelemma.tex}\n \\subimport{common/proofsketches/}{flowgameproofsketch.tex}\n \\subimport{fc17/lemmas/}{gameflowlemma.tex}\n \\subimport{common/proofsketches/}{gameflowproofsketch.tex}\n \\subimport{common/theorems/}{trustflowtheorem.tex}\n \\subimport{common/proofs/}{trustflowproof.tex}\n\n \\noindent Note that the maxFlow is the same in the following two cases: If a player chooses the evil strategy and if that\n player chooses a variation of the evil strategy where she does not nullify her outgoing direct trust.\n\n Further justification of trust transitivity through the use of $MaxFlow$ can be found in the sociological work by Karlan et\n al. \\cite{kmrs} where a direct correspondence of maximum flows and empirical trust is experimentally validated.\n\n Here we see another important theorem that gives the basis for risk-invariant transactions between different, possibly\n unknown, parties.\n \\subimport{common/theorems/}{riskinvtheorem.tex}\n \\subimport{common/proofs/}{riskinvproof.tex}\n \\noindent $A$ can reduce her outgoing direct trust in a manner that achieves (\\ref{primetrust}), since\n $maxFlow\\left(A, B\\right)$ is continuous with respect to $A$'s outgoing direct trusts.\n", "meta": {"hexsha": "6f0132bb3d58ad777f84c0fd5c965fabdf05b449", "size": 3417, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "fc17/flow.tex", "max_stars_repo_name": "dionyziz/DecentralizedTrust", "max_stars_repo_head_hexsha": "60f65bff00041e7e940491913bd4ca3f11bf22d9", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2017-03-15T14:33:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-01T14:07:45.000Z", "max_issues_repo_path": "fc17/flow.tex", "max_issues_repo_name": "dionyziz/DecentralizedTrust", "max_issues_repo_head_hexsha": "60f65bff00041e7e940491913bd4ca3f11bf22d9", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2017-03-07T12:25:26.000Z", "max_issues_repo_issues_event_max_datetime": "2017-07-31T14:42:20.000Z", "max_forks_repo_path": "fc17/flow.tex", "max_forks_repo_name": "dionyziz/DecentralizedTrust", "max_forks_repo_head_hexsha": "60f65bff00041e7e940491913bd4ca3f11bf22d9", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-03-07T10:49:58.000Z", "max_forks_repo_forks_event_max_datetime": "2017-08-28T06:32:33.000Z", "avg_line_length": 64.4716981132, "max_line_length": 126, "alphanum_fraction": 0.736318408, "num_tokens": 1023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8175744761936437, "lm_q2_score": 0.7490872075132152, "lm_q1q2_score": 0.6124345813059762}} {"text": "\\section{Excision, and the Eilenberg-Steenrod axioms}\nWe have homotopy invariance and the lexseq of a pair. We claimed that $ H_\\ast(X,A)$ ``depends only on $X-A$''. You have to be careful about this. \n\\begin{definition}\nA triple $(X,A,U)$ where $U\\subseteq A\\subseteq X$ is \\emph{excisive} if $\\overline{U}\\subseteq\\mathrm{Int}(A)$. This is a point-set definition. From an excisive triple you can get a pair $(X-U,A-U)\\subseteq (X,A)$, and this is called an excision.\n\\end{definition}\n\\begin{theorem}\nAn excision induces an isomorphism in homology, i.e., $ H_\\ast(X-U,A-U)\\cong H_\\ast(X,A)$. We might prove this on Wednesday.\n\\end{theorem}\nWhat are some consequences? We'll compute $ H_\\ast(S^n)$ and $ H^\\ast(D^n,S^{n-1})$. Here's the result. We'll use the homeomorphisms $D^n\\simeq \\Delta^n$ and $S^{n-1}\\simeq\\partial\\Delta^n$. Let's write $S^0=\\{0,1\\}$, and $\\iota_n:\\Delta^n\\to\\Delta^n$.\n\\begin{theorem}\n\t\\begin{enumerate}\n\t\\item \\begin{equation*}\n\t H_q(S^n)=\\begin{cases}\\Z = \\langle[c^0_\\ast]\\rangle & q=0,n>0\\\\ \\Z\\oplus\\Z = \\langle[c^0_1],[\\partial\\iota_1]\\rangle & q=n=0 \\\\ \\Z = \\langle[\\partial\\iota_{n+1}]\\rangle & q=n>0 \\\\ 0 & \\text{else} \\end{cases}\n\t\\end{equation*}\n\n\t\\item \\begin{equation*}\n\t H_q(D^n,S^{n-1}) = \\begin{cases}\n\t\\Z=\\langle [\\iota_n]\\rangle & q=n\\\\\n\t0 & \\text{else}\n\t\\end{cases}\n\t\\end{equation*}\n\t\\end{enumerate}\n\\end{theorem}\nIf $n=0$, then we say that $S^{-1}=\\emptyset$. What are the generators of these groups?\n\\begin{proof}\nWe'll use the lexseq, homotopy invariance, and excision. We have the lexseq:\n\\begin{equation*}\n\\xymatrix{ & & \\ar[dll]^\\partial\\\\\n H_q(S^{n-1})\\ar[r] & H_q(D^n)\\ar[r] & H_q(D^n,S^{n-1})\\ar[dll]^\\partial\\\\\n H_{1-1}(S^{n-1})\\ar[r] & H_{q-1}(D^n)\\ar[r] & H_{q-1}(D^n,S^{n-1})\\ar[dll]^\\partial\\\\\n & & &}\n\\end{equation*}\nBut we know that $D^n$ is contractible, so $ H_q(D^n)=\\begin{cases}\\Z & q=n\\\\ 0 & \\text{else}\\end{cases}$. This means that $\\partial: H_q(D^n,S^{n-1})\\cong H_{q-1}(S^{n-1})$ for $q>1$, but when $q=1$, we get $0\\to H_1(D^n,S^{n-1})\\xrightarrow{\\partial} H_0(S^{n-1})\\to H_0(D^n) \\to H_0(D^n,S^{n-1})\\to 0$.\n\nLet's think about the case $n>1$. Then $ H_0(S^{n-1})=\\mathbf{Z}=\\langle[c^0_\\ast]\\rangle$, and $ H_0(D^n)=\\Z=\\langle[c^0_\\ast]\\rangle$, so you have an isomorphism, which means that $ H_0(D^n,S^{n-1})=0$ and $ H_1(D^n,S^{n-1})=0$. Now, let's go to the case $n=1$. Then $ H_0(S^0)=\\Z\\oplus\\Z=\\langle [c^0_\\ast],[\\partial\\iota_1]\\rangle$ and $ H_0(D^0)=\\Z$. But $[\\partial\\iota_1]$ goes to zero, and $[c^0_\\ast]$ goes to the generator of $\\Z= H_0(D^0)$. This means that $ H_1(D^1,S^0)$ is generated by $\\langle[\\iota_1]\\rangle\\cong\\Z$ because the map $ H_1(D^n,S^{n-1})\\to H_0(S^{n-1})$ is the boundary map, so $[\\iota_1]\\mapsto [\\partial\\iota_1]$.\n\nExcision will come into play through the following statement.\n\\begin{prop}\nIf $n>1,q>1$, then $ H_q(D^n,S^{n-1})\\to H_q(D^n/S^{n-1},\\ast)\\cong H_q(S^n,\\ast)\\cong H_q(S^n)$, because $D^n/S^{n-1}\\simeq S^n$. The claim is that this collapse map is an isomorphism.\n\\end{prop}\nThis provides the inductive step. Let's assume we've proved the proposition. Then $ H_q(D^n,S^{n-1})\\cong H_{q-1}(S^{n-1})$. The proposition says that $ H_q(S^n)\\cong H_{q-1}(S^{n-1})$. But we also have the boundary map $\\partial: H_{q+1}(D^n,S^{n-1})\\cong H_q(S^n)$, i.e., $ H_{q+1}(D^n,S^{n-1})\\cong H_q(D^n,S^{n-1})$.\n\nNow I want to prove the proposition. \n\\begin{proof}[Proof of proposition]\nWe want to compare $ H_q(D^n,S^{n-1})$ and $ H_q(D^n/S^{n-1},\\ast)$. We'll use excision to do this. We have $D^n=\\{ x\\in\\mathbf{R}^{n+1} | |x|\\leq 1 \\}$. Let $A=\\{ x|1/3\\leq |x|\\leq 1 \\}$ and $U=\\{ x|2/3<|x|\\leq 1 \\}$, and $\\{ x||x|=1 \\} =S^{n-1}\\subseteq U$. We need a preliminary step. $ H_q(D^n,S^{n-1})\\to H_q(D^n,A)$. We claim that this an isomorphism, but this is true because of the lexseq and the 5-lemma. By excision, $ H_q(D^n,A)\\cong H_q(D^n-U,A-U)$. We can collapse the $(n-1)$-sphere, and $(D^n/S^{n-1}-U/S^{n-1},A/S^{n-1}-U/S^{n-1})=(D^n-U,A-U)$ because you're collapsing something from something that's already been collapsed! Now, we claim that $ H_q(D^n/S^{n-1}-U/S^{n-1},A/S^{n-1}-U/S^{n-1})\\cong H_q(D^n/S^{n-1},A/S^{n-1})$, which is true by excision. THE FOLLOWING PART IS MOST DEFINITELY NOT RIGHT\\footnote{My own comment: this proof can be finished by noticing that $A\\simeq S^{n-1}$ via $\\mathbf{v}\\mapsto\\frac{\\mathbf{v}}{||\\mathbf{v}||}$, and that $D^n/S^{n-1}\\simeq S^n$.}. But also, $ H_q(D^n/S^{n-1},\\text{disk})= H_q(D^n/S^{n-1},A/S^{n-1})$. Because the disk is contractible, using the lexseq and the 5-lemma completes the proof of the proposition.\n\\end{proof}\n\\end{proof}\n``This really turns me on, because I love homology.'' Why should you care about homology?\n\\begin{corollary}\nIf $m\\neq n$, then $S^m\\not\\simeq S^n$ because they have different homology groups. \n\\end{corollary}\n\\begin{corollary}\nIf $m\\neq n$, then $\\mathbf{R}^m\\not\\cong \\mathbf{R}^n$, because they're not homeomorphic.\n\\end{corollary}\n\\begin{proof}\nLet $m,n>0$. Assume we have a homeomorphism $f:\\mathbf{R}^m\\to \\mathbf{R}^n$. This restricts to $\\mathbf{R}^m-\\{0\\}\\to \\mathbf{R}^n-\\{0\\}$, but each of these are homotopy equivalent to spheres, but we can't get a homotopy equivalence between two spheres of different dimension by the above corollary.\n\\end{proof}\n\\subsection{Eilenberg-Steenrod axioms}\n\\begin{definition}\nA homology theory (on $\\mathbf{Top}$) is:\n\\begin{itemize}\n\\item a functor $h_n:\\mathbf{Top}_2\\to\\mathbf{Ab}$ for all $n$. We'll write $h_n(X)=h_n(X,\\emptyset)$\n\\item natural transformations $\\partial:h_n(X,A)\\to h_{n-1}(A)$.\n\\end{itemize}\nsuch that:\n\\begin{itemize}\n\\item if $f_0,f_1:(X,A)\\to (Y,B)$ are homotopic, then $f{0,\\ast}\\simeq f_{1,\\ast}:h_n(X,A)\\to h_n(Y,B)$.\n\\item an excision induces isomorphisms.\n\\item a lexseq:\n\\begin{equation*}\n\\cdots\\to h_{q+1}(X,A)\\xrightarrow{\\partial}h_q(A)\\to h_q(X)\\to h_q(X,A)\\xrightarrow{\\partial}\\cdots\n\\end{equation*}\n\\item (the dimension axiom): $h_n(\\ast)$ is nonzero only in dimension zero. This is like the parallel postulate.\n\\end{itemize}\n\\end{definition}\n\\begin{example}\nOrdinary singular homology satisfies these. \n\\end{example}\n\\begin{theorem}[Brouwer fixed-point theorem]\nIf $f:D^n\\to D^n$, then there is some point $x\\in D^n$ such that $f(x)=x$.\n\\end{theorem}\n\\begin{proof}\nSuppose not. Then you can draw a ray from $x$ to $f(x)$ to the boundary $S^{n-1}$, intersecting at a point $g(x)$. Left to you to check that $g$ is continuous. If $x$ was on the boundary, then $x=g(x)$. This is inconsistent by our computation because otherwise the identity on $ H_{n-1}(S^{n-1})$ would be zero, contradiction!\n\\end{proof}\n", "meta": {"hexsha": "1e84b21bbfc5aaf02b751cec94877d691587f5e9", "size": 6587, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old-905/lec-10-eilenberg-steenrod.tex", "max_stars_repo_name": "ichung/algtop-notes", "max_stars_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-04-26T15:00:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T22:47:06.000Z", "max_issues_repo_path": "old-905/lec-10-eilenberg-steenrod.tex", "max_issues_repo_name": "ichung/algtop-notes", "max_issues_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-03-13T17:54:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-13T17:59:46.000Z", "max_forks_repo_path": "old-905/lec-10-eilenberg-steenrod.tex", "max_forks_repo_name": "ichung/algtop-notes", "max_forks_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-21T18:15:11.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-13T17:38:04.000Z", "avg_line_length": 77.4941176471, "max_line_length": 1177, "alphanum_fraction": 0.6609989373, "num_tokens": 2549, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7310585786300049, "lm_q2_score": 0.8376199714402813, "lm_q1q2_score": 0.6123492657532373}} {"text": "\\documentclass[a4paper]{article}\n\n\\input{temp}\n\n\\setcounter{section}{-1}\n\n\\begin{document}\n\n\\title{Introduction to Discrete Analysis}\n\n\\maketitle\n\n\\newpage\n\n\\tableofcontents\n\n\\newpage\n\n\\section{Introduction}\nasdasd\n\n\\newpage\n\n\\section{The discrete Fourier transform}\n\nLet $N$ be a fixed positive integer. Write $\\omega$ for $e^{2\\pi i/N}$, and $\\Z_N$ for $\\Z/n\\Z$. Let $f:\\Z_N \\to \\C$. Given $r \\in \\Z_N$, define $\\hat{f}(r)$ to be\n\\begin{equation*}\n\\begin{aligned}\n\\frac{1}{N} \\sum_{x \\in \\Z_N} f(x) \\omega^{-rx}\n\\end{aligned}\n\\end{equation*}\nFrom now on we use the notation $\\E_{x \\in \\Z_N}$ for $\\frac{1}{N}\\sum_{x \\in \\Z_N}$, so $\\hat{f}(r) = \\E_x f(x) e^{-\\frac{2\\pi irx}{N}}$.\n\nIf we write $\\omega_r$ for the function $x \\to \\omega^{rx}$, and $\\bra f,g\\ket$ for $\\E_x f(x) \\overline{g(x)}$, then $\\hat{f}(r) = \\bra f,\\omega_r \\ket$. So the discrete fourier transforn is basically expanding the function $f$ in the set of orthonormal basis $\\omega_r$.\n\nLet us write $||f||_p$ for $\\E_x |f(x)|^p)^{1/p}$ (the $L_p$-norm), and call the resulting space $L_p(\\Z_n)$.\n\nImportant convention: we use \\emph{averages} for the 'original functions' in 'physical spaces', and \\emph{sums} for their Fourier transforms in 'frequency space' (referring to $\\E$: $\\bra,\\ket$ is average in the original space but just $\\sum$ in frequency space, i.e. for $\\hat{f},\\hat{g}$ etc.)\n\n\\begin{lemma} (1, Parseval's identity)\\\\\n If $f,g: \\Z_n \\to \\C$, then $\\bra \\hat{f},\\hat{g}\\ket = \\bra f,g\\ket$.\n \\begin{proof}\n \\begin{equation*}\n \\begin{aligned}\n \\bra\\hat{f},\\hat{g}\\ket &= \\sum_r \\hat{f}(r) \\overline{\\hat{g}(r)}\\\\\n &= \\sum_r (\\E_x f(x) \\omega^{-rx}) (\\overline{\\E_y g(y) \\omega^{-ry}})\\\\\n &= \\E_x \\E_y f(x) \\overline{g(y)} \\sum_r \\omega^{-r(x-y)}\\\\\n &= \\E_x \\E_y f(x) \\overline{g(y)} n\\delta_{xy}\\\\\n &= \\bra f,g\\ket\n \\end{aligned}\n \\end{equation*}\n \\end{proof}\n\\end{lemma}\n\n\\begin{lemma} (2, Convolution identity)\\\\\n \\begin{equation*}\n \\begin{aligned}\n \\widehat{f*g}(r) = \\hat{f}(r)\\hat{g}(r)\n \\end{aligned}\n \\end{equation*}\n where\n \\begin{equation*}\n \\begin{aligned}\n (f*g)(x) = \\E_{y+z=x}f(y)g(z) = \\E_y f(y)g(x-y)\n \\end{aligned}\n \\end{equation*}\n \\begin{proof}\n \\begin{equation*}\n \\begin{aligned}\n \\widehat{f*g}(r) &= \\E_x f*g(x)\\omega^{-rx}\\\\\n &=\\E_x \\E_{y+z=x}f(y)g(z)\\omega^{-rx}\\\\\n &=\\E_x\\E_{y+z=x} f(y)g(z) \\omega^{-ry}\\omega^{-rz}\\\\\n &=\\E_y\\E_z f(y) \\omega^{-ry} g(z) \\omega^{-rz}\\\\\n &=\\hat{f}(r) \\hat{g}(r)\n \\end{aligned}\n \\end{equation*}\n \\end{proof}\n\n\\end{lemma}\n\n\\begin{lemma} (3, Inversion formula)\\\\\n \\begin{equation*}\n \\begin{aligned}\n f(x) = \\sum_r \\hat{f}(r) \\omega^{rx}\n \\end{aligned}\n \\end{equation*}\n (note the sign of $\\omega^{rx}$).\n \\begin{proof}\n \\begin{equation*}\n \\begin{aligned}\n \\sum_r \\hat{f}(r) \\omega^{rx} &= \\sum_r \\E_y f(y) \\omega^{r(x-y)}\\\\\n &= \\E_y f(y) \\sum_r \\omega^{r(x-y)}\\\\\n &= \\E_y f(y) n\\delta_{xy}\\\\\n &=f(x)\n \\end{aligned}\n \\end{equation*}\n This is really just the statement that we get the original vector back when we sum up its components.\n \\end{proof}\n\\end{lemma}\n\nFurther observations: If $f$ is real-valued, then $\\hat{f}(-r) = \\E_x f(x) \\omega^{rx} = \\overline{\\E_x f(x) \\omega^{-rx}} = \\overline{\\hat{f}(r)}$.\n\nIf $A \\subset \\Z_n$, write $A$ (instead of $1_A,\\chi_A$) for the characteristic function of $A$. Then $\\hat{A}(0) = \\E_x A(x) = \\frac{|A|}{N}$, the \\emph{density} of $A$.\n\nAlso, $||\\hat{A}||^2_2 = \\bra\\hat{A},\\hat{A}\\ket = \\bra A,A\\ket = \\E_x A(x)^2 = \\E_x A(x) = \\frac{|A|}{N}$, again the density.\n\nLet $f:\\Z_n \\to \\C$. Given $\\mu \\in \\Z_n$, define $f_\\mu(x)$ to be $f(\\mu^{-1}x)$ (so we need $(\\mu,N) = 1$). Then\n\\begin{equation*}\n \\begin{aligned}\n \\hat{f}_\\mu(r) &= \\E_x f_\\mu (x) \\omega^{-rx}\\\\\n &= \\E_x f(x/\\mu) \\omega^{-rx}\\\\\n &=\\E_x f(x) \\omega^{-r\\mu x}\\\\\n ^= \\hat{f}(\\mu r)\n \\end{aligned}\n\\end{equation*}\n\n\\subsection{Roth's theorem}\n\\begin{thm} (4)\n For every $\\delta > 0$, $\\exists N$ s.t. if $A \\subset \\{1,...,N\\}$ is a set of size at least $\\delta N$, then $A$ must contain an arithmetic progression of length 3.\\\\\n This is also true for 4,5,..., but the proof is much harder -- Szemeredi's theorem.\\\\\n Basic strategy of proof: show that if $A$ has density $\\delta$ and no AP of length 3 (3AP), then there's a long AP in $P \\subset \\{1,2,...,n\\}$ s.t.\n \\begin{equation*}\n \\begin{aligned}\n |A \\cap P| \\geq (\\delta+c(\\delta)) |p|\n \\end{aligned}\n \\end{equation*}\n where $c(\\delta)$ is some positive number. But then we can continue this argument to expand $A\\cap P$ to infinity (note that $|A\\cap P|$ is an integer, so each time increase by 1 at least).\n\n The best known relationship between $\\delta$ and the $N$ required is around $\\delta \\sim \\frac{c}{\\log\\log N}$ for some constant $c$.\n\\end{thm}\n\n---Lecture 2---\n\n\\begin{lemma} (5)\\\\\n Let $N$ be odd, $A,B,C \\subset \\Z_N$ have densties $\\alpha,\\beta,\\gamma$.\\\\\n If $\\max_{r \\neq 0} |\\hat{A}(r)| \\leq \\frac{\\alpha(\\beta\\gamma)^{1/2}}{2}$ and $\\frac{\\alpha\\beta\\gamma}{2} > \\frac{1}{N}$, then there exists $x,d \\in \\Z_N$ with $d \\neq 0$ s.t. $(x,x+d,x+2d) \\in A \\times B \\times C$.\n \\begin{proof}\n \\begin{equation*}\n \\begin{aligned}\n \\E_{x,d} A(x) B(x+d) C(x+2d) &= \\E_{x+z=2y} A(x)B(y)C(z)\\\\\n &=\\E_u (\\E_{x+z=u}A(x)C(z)) \\E_{2y=u} B(y)\\\\\n &=\\E_u A * C(u) B_2(u)\\\\\n &=\\bra A*C,B_2\\ket\\\\\n &=\\bra\\widehat{A*C},\\hat{B}_2\\ket\\\\\n &=\\bra \\hat{A}\\hat{C},\\hat{B}_2\\ket\\\\\n &=\\sum_r \\hat{A}(r)\\hat{C}(r)\\hat{B}(-2r)\\\\\n &=\\alpha\\beta\\gamma + \\sum_{r \\neq 0} \\hat{A}(r)\\hat{C}(r)\\hat{B}(-2r)\n \\end{aligned}\n \\end{equation*}\n Recall here the notation is $B_2(u) = B(u/2)$. now\n \\begin{equation*}\n \\begin{aligned}\n |\\sum_{r\\neq 0} \\hat{A}(r)\\hat{B}(-2r)\\hat{C}(r)| &\\leq \\frac{\\alpha(\\beta\\gamma)^{1/2}}{2}\\sum_{r \\neq 0} |\\hat{B}(-2r)||\\hat{C}(r)|\\\\\n &\\leq \\frac{\\alpha(\\beta\\gamma)^{1/2}}{2} \\left(\\sum_r |\\hat{B}(-2r)^2\\right)^{1/2}\\left(\\sum_r |\\hat{C}(r)|^2\\right)^{1/2} \\ \\text{By Cauchy-Schwarz}\\\\\n &=\\frac{\\alpha(\\beta\\gamma)^{1/2}}{2} ||\\hat{B}||_2 ||\\hat{C}||_2\\\\\n &= \\frac{\\alpha(\\beta\\gamma)^{1/2}}{2} ||B||_2||C||_2\\\\\n &= \\frac{\\alpha\\beta\\gamma}{2}\n \\end{aligned}\n \\end{equation*}\n The contribution to $\\E_{x,d} A(x)B(x+d)C(x+2d)$ from $d=0$ is at most $\\frac{1}{N}$, so if $\\frac{\\alpha\\beta\\gamma}{2} > \\frac{1}{N}$, we are done.\n \\end{proof}\n\\end{lemma}\n\nNow let $A$ be a subset of $\\{1,...,N\\}$ with density $\\geq \\delta$ and let $B=C=A \\cap [\\frac{N}{3},\\frac{2N}{3})$. If $B$ has density $<\\frac{\\delta}{5}$ (??), then either $A \\cap [1,\\frac{N}{3}]$ or $A \\cap [\\frac{2N}{3},N]$ has density at least $\\frac{2\\delta}{5}$. In that case we find an AP $P$ of length about $N/3$ such that $|A \\cap P| / |P| \\geq \\frac{6\\delta}{5}$.\\\\\nOtherwise, we find that if $\\max_{r \\neq 0}|\\hat{A}(r)| \\leq \\frac{\\delta}{10}$ and $\\frac{\\delta^3}{50} > \\frac{1}{N}$, then $A \\times B \\times C$ contains a 3AP, so $A$ contains a 3AP.\\\\\nSo if $A$ does not contain a 3AP, then either we find $P$ of length about $N/3$ with $|A\\cap P| / |P| \\geq \\frac{6\\delta}{5}$, or ther exists $r \\neq 0$ s.t. $|\\hat{A}(r)| \\geq \\frac{\\delta}{10}$.\n\n\\begin{defi}\n If $X$ is a finite set and $f:X \\to \\C$, $Y \\subset X$, write $osc(f|_Y)$ to mean $\\max_{y_1,y_2 \\in Y} |f(y_1)-f(y_2)|$ (I think \\emph{amplitude} is a better word for this).\n\\end{defi}\n\n\\begin{lemma} (6)\\\\\n Let $r \\in \\Z_n$ and let $\\varepsilon > 0$. Then there is a partition of $\\{1,2,...,N\\}$ into arithmetic progressions $P_i$ of length at least $c(\\varepsilon)\\sqrt{N}$ such that\n \\begin{equation*}\n \\begin{aligned}\n osc(\\omega_r|_{P_i}) \\leq \\varepsilon\n \\end{aligned}\n \\end{equation*}\n for each $i$.\n \\begin{proof}\n Let $t = \\lfloor \\sqrt{N}\\rfloor$. Of the numbers $1,\\omega^r,...,\\omega^tr$, there must be two that differ by at most $\\frac{2\\pi}{t}$.\\\\\n If $|\\omega^{ar}-\\omega^{br}| \\leq \\frac{2\\pi}{t}$ with $a < b$, then $|1-\\omega^{dr}| \\leq \\frac{2\\pi}{t}$ where $d=b-a$. Then $|\\omega^{urd}-\\omega^{vrd}| \\leq |\\omega^{urd}-\\omega^{(u+1)rd}| + ... + |\\omega^{(v-1)rd}-\\omega^{vrd}| \\leq \\frac{2\\pi}{t}(v-u)$.\\\\\n So if $P$ is a progression with common differnece $d$ and length $l$, then $osc(\\omega_r|_P) \\leq \\frac{2\\pi l}{t}$. So divide up $\\{1,...,N\\}$ into residue classes mod $d$, and partition each residue class into parts of length between $\\frac{\\varepsilon t}{4\\pi}$ and $\\frac{\\varepsilon t}{2\\pi}$ (possible, since $d \\leq t \\leq \\sqrt{N}$).\\\\\n We are done, with $c(\\varepsilon) = \\frac{\\varepsilon}{16}$ (a casual choice).\n \\end{proof}\n\\end{lemma}\nNow let us use the information that $r \\neq 0$ and $|\\hat{A}(r)| \\geq \\frac{\\delta^2}{10}$.\\\\\nDefine the \\emph{balanced function} $f$ of $A$ by $f(x) = A(x)=\\frac{|A|}{N}$ for each $x$.\\\\\nNote that $\\hat{f}(0) = 0$ and $\\hat{f}(r) = \\hat{A}(r)$ for all $r \\neq 0$.\n\nNow let $P_1,...,P_m$ be given by Lemma 6 with $\\varepsilon = \\delta^2/20$. Then\n\\begin{equation*}\n \\begin{aligned}\n \\frac{\\delta^2}{10} &\\leq |\\hat{f}(r)|\\\\\n &= \\frac{1}{N} |\\sum_x f(x) \\omega^{-rx}|\\\\\n &\\leq \\frac{1}{N} \\sum_{i=1}^m |\\sum_{x \\in P_i} f(x) \\omega^{-rx}|\\\\\n &\\leq \\frac{1}{N} \\sum_{i=1}^N \\left[\\left|\\sum_{x \\in P_i} f(x) \\omega^{-rx_i}\\right| + \\left|\\sum_{x \\in P_i} f(x) (\\omega^{-rx} - \\omega^{-rx_i})\\right|\\right] \\ x_i \\in P_i \\text{ arbitrary}\\\\\n &\\leq \\frac{1}{N} \\sum_{i=1}^m |\\sum_{x \\in P_i} f(x)| + \\frac{\\delta^2}{20}\n \\end{aligned}\n\\end{equation*}\nTherefore $\\sum_{i=1}^m \\left|\\sum_{x \\in P_i} f(x)\\right| \\geq \\frac{\\delta^2 N}{20}$.\n\nWe also have $\\sum_{i=1}^m \\sum_{x\\in P_i} f(x) = 0$, so\n\\begin{equation*}\n \\begin{aligned}\n \\sum_{i=1}^m \\left(\\left|\\sum_{x \\in P_i} f(x) \\right| + \\sum_{x \\in P_i}f(x) \\right) \\geq \\frac{\\delta^2}{20} \\sum_{i=1}^m |P_i|\n \\end{aligned}\n\\end{equation*}\nTherefore, \n\\begin{equation*}\n \\begin{aligned}\n &|\\sum_{x \\in P_i} f(x)| + \\sum_{x \\in P_i} f(x) \\geq \\frac{\\delta^2}{20} |P_i|\\\\\n &\\implies \\sum_{x \\in P_i} f(x) \\geq \\frac{\\delta^2}{40}|P_i|\\\\\n &\\implies |A \\cap P_i| \\geq \\left(\\delta+\\frac{\\delta^2}{40}\\right)|P_i|\n \\end{aligned}\n\\end{equation*}\n\n---Lecture 3---\n\nNow let $A \\subset \\Z_N$, $|A| \\geq \\delta N$. Then:\\\\\n$\\bullet$ either $A$ contains a 3AP,\\\\\n$\\bullet$ or $N$ is even,\\\\\n$\\bullet$ or $\\exists P \\subset \\{1,...,N\\}$, $|P| \\geq N/3$ s.t. $|A \\cap P| \\geq \\frac{6\\delta}{5} |P|$,\\\\\n$\\bullet$ or $\\exists P \\subset \\{1,...,N\\}$, $|P|\\geq \\frac{\\delta^2}{640}\\sqrt{N}$ (casual) s.t. $|A\\cap P| \\geq (\\delta+\\frac{\\delta^2}{40})|P|$.\n\nNote that the third case is strictly worse than the fourth.\\\\\nWell if the first is true then we're done. Suppose now the seconds hols. Write $N=N_1+N_2$ with $N_1,N_2$ odd, $N_1,N_2 \\approx\\frac{N}{2}$. Then $A$ has density at least $\\delta$ in one of $\\{1,...,N_1\\}$ or $\\{N_1+1,...,N_1+N_2\\}$.\\\\\nIf (4) holds (note (3) $\\implies$ (4)), then we pass to $P$ and start to again. After $\\frac{40}{\\delta}$ iterations, the density at least doubles. Therefore the toatl number of iterations we can have is at most $\\frac{40}{\\delta} + \\frac{40}{2\\delta}+... \\leq \\frac{80}{\\delta}$.\\\\\nIf $\\frac{\\delta^2}{640} \\sqrt{N} \\geq N^{1/3}$ (to account for the above, and also for the possible use of (2)) at each iteration, and $\\frac{\\delta^3}{25} \\geq N^{-1}$ (which follows from the first condition), then after $\\frac{80}{\\delta}$ iterations we have $N \\geq N^{(1/3)^{80/\\delta}}$, so the argument works provided\n\\begin{equation*}\n \\begin{aligned}\n N^{(1/3)^{80\\delta}} \\geq \\left(\\frac{640}{\\delta^2}\\right)^6\n \\end{aligned}\n\\end{equation*}\ntaking logs and simplify a bit, we need\n\\begin{equation*}\n \\begin{aligned}\n &-\\frac{80}{\\delta} \\log 3 + \\log \\log N \\geq \\log 6 + \\log (\\log 640 + 2 \\log \\frac{1}{\\delta}))\\\\\n &\\Leftarrow \\log\\log N \\geq \\frac{160}{\\delta}\\\\\n &\\Leftarrow \\delta \\geq \\frac{160}{\\log\\log N}\n \\end{aligned}\n\\end{equation*}\n\n\\subsection{Bogolyubov's method}\nLet $K \\subset \\hat{\\Z}_N$ and let $\\delta > 0$. The \\emph{Bohr set} $B(K,\\delta)$ has two definitions (not exactly equivalent, but quite equivalent):\\\\\n$\\bullet$ (1) $B(K,\\delta) = \\{x \\in \\Z_N: rx \\in [-\\delta N,\\delta N] \\forall r \\in K\\}$ (arc-length definition);\\\\\n$\\bullet$ (2) $B(K,\\delta) = \\{x \\in \\Z_N: |1-\\omega^{rx}| \\leq \\delta \\forall r \\in K\\}$ (chord-length definition).\n\n\\begin{defi}\n Let $G$ be an abelian group and let $A,B$ be subsets of $G$. Then write $A + B = \\{a+b: a \\in A, b \\in B\\}$ and the obvious definition for $A-B$. We also write $rA = \\{a_1+...+a_r:a_1,...,a_r\\in A\\}$ (note this might be different than what you think this notation should mean).\n\\end{defi}\n\n\\begin{lemma} (7)\\\\\n Let $A \\subset \\Z_N$ be a set of density $\\alpha$. Then $2A-2A$ contains a Bohr set $B(K,1/4)$ (arc) with $|K| \\leq \\alpha^{-2}$.\n \\begin{proof}\n Observe that $x \\in 2A-2A$ iff $A*A*(-A)*(-A)(x) \\neq 0$ (this makes more sense if we write it as $\\E_{a+b-c-d=x} A(a)A(b)A(c)A(d) \\neq 0$, i.e. we are basically just counting the number of ways $x$ can be written as $a+b-c-d$ where $a,b,c,d \\in A$.)\\\\\n But\n \\begin{equation*}\n \\begin{aligned}\n A*A*(-A)*(-A)(x) &= \\sum_r \\widehat{A*A*(-A)*(-A)}(r) \\omega^{rx} \\text{ inversion formula}\\\\\n &= \\sum |\\hat{A}(r)|^4 \\omega^{rx}\n \\end{aligned}\n \\end{equation*}\n Let $K = \\{r: |\\hat{A}(r)| \\geq \\alpha^{3/2}\\}$. Then $\\alpha = ||\\hat{A}||_2^2 = \\sum_r |\\hat{A}(r)|^2 \\geq \\alpha^3 |K|$.\\\\\n So $|K| \\leq \\alpha^{-2}$.\\\\\n Now suppose that $x \\in B(K,1/4)$. Then\n \\begin{equation*}\n \\begin{aligned}\n \\sum_r |\\hat{A}(r)|^4 \\omega^{rx} &= \\alpha^4 + \\sum_{r \\in K, r \\neq 0} |\\hat{A}(r)|^4 \\omega^{rx} + \\sum_{r \\not\\in K} |\\hat{A}(r)|^4 \\omega^{rx}\n \\end{aligned}\n \\end{equation*}\n The real part of the second term is non-negative since $rx \\in [-N/4,N/4]$ when $r \\in K$.\\\\\n Also \n \\begin{equation*}\n \\begin{aligned}\n \\left|\\sum_{r \\not\\in K} |\\hat{A}(r)|^4 \\omega^{rx}\\right| &\\leq \\sum_{r \\not\\in K} |\\hat{A}(r)|^4\\\\\n &< \\alpha^3 \\sum_{r \\not\\in K} |\\hat{A}(r)|^2\\\\\n &\\leq \\alpha^4\n \\end{aligned}\n \\end{equation*}\n So it follows that the real part of $\\sum_r |\\hat{A}(r)|^4 \\omega^{rx} > 0$, i.e. it is non-zero. So $x \\in 2A - 2A$.\n \\end{proof}\n\\end{lemma}\n\n\\begin{lemma} (8)\\\\\n Let $K \\subset \\Z_N$ and let $\\delta > 0$. Then:\\\\\n (i) $B(K,\\delta)$ (arc) has density at least $\\delta^{|K|}$;\\\\\n (ii) $B(K,\\delta)$ contains a mod-$N$ artihmetic progression of length at least $\\delta N^{1/|K|}$.\n \\begin{proof}\n (i) Let $K = \\{r_1,...,r_k\\}$. Consider the $N$ $k$-tuples $(r_1x,...,r_kx) \\in \\Z_N^k$. If we intersect this set of $k$-tuples with a random 'box' $[t_1,t_1+\\delta N] \\times ... \\times [t_k,t_k+\\delta N]$ (here we are thinking $t_i$ as real numbers), then the expected number of the $k$-tuples in the box is $\\delta^k N$ (since each one has a probability $\\delta^k$).\\\\\n But if $(r_1x,...,r_kx)$ and $(r_1y,...,r_ky)$ belong to this box, then $x-y \\in B(K,\\delta)$.\\\\\n (ii) If we take $\\eta > N^{-1/k}$, then by (i) we get that $|B(K,\\eta)| > 1$, therefore at least 2. So $\\exists x \\in B(K,\\eta)$ s.t. $x \\neq 0$. But then $dx\\in B(K,d\\eta)$ for every $d$.\\\\\n So if $d\\eta \\leq \\delta$ then $dx \\in B(K,\\delta)$. That gives us an AP of length at least $\\frac{\\delta}{\\eta}$. So we get one of length at least $\\delta N^{1/k}$.\n \\end{proof}\n\\end{lemma}\n\n\\begin{defi}\n Let $A,b$ be subsets of Abelian groups and let $\\phi:A \\to B$. Then $\\phi$ is a \\emph{Freiman homomorphism of order $k$} if \n \\begin{equation*}\n \\begin{aligned}\n a_1+...+a_k=a_{k+1}+...+a_{2k} \\implies \\phi(a_1)+...+\\phi(a_k) = \\phi(a_{k+1})+...+\\phi(a_{2k})\n \\end{aligned}\n \\end{equation*}\n If $k=2$, we call this just a \\emph{Freiman homomorphism}. In that case, the condition is equivalent to $a-b=c-d \\implies \\phi(a)-\\phi(b)=\\phi(c)-\\phi(d)$.\\\\\n If $\\phi$ has an inverse which is also a F-homomorphism of order $k$, then $\\phi$ is a F-\\emph{isomorphism} of order $k$.\n\\end{defi}\n\n\\begin{lemma} (9)\\\\\n Assume $0 \\not \\in K$, and $N$ is prime. If $\\delta < 1/4$, then $B(K,\\delta)$ (arc) is Freiman isomorphic to the intersection in $\\R^{|K|}$ of $[-\\delta N, \\delta N]^{|K|}$ with some lattice $\\Lambda$.\n \\begin{proof}\n Let $K = \\{r_1,...,r_k\\}$, and let $\\Lambda = N\\Z^k + \\{(r_1 x,...,r_kx):x \\in \\Z\\}$. Write $\\mathbf{r}$ for $(r_1,...,r_k)$. Claim that $B(K,\\delta) \\cong \\Lambda \\cap [-\\delta N, \\delta N]^k$.\\\\\n Define a map $\\phi:B(K,\\delta) \\to \\Lambda \\cap [-\\delta N, \\delta N]^k$ by sending $x$ to $(\\bra r_1 x\\ket, ...,\\bra r_k x\\ket)$ where $\\bra u\\ket$ means the least-modulus residue $u$ mod $N$.\\\\\n If $x+y = z+w$, then $\\mathbf{r} x +\\mathbf{r} y = \\mathbf{r}z+\\mathbf{r}w$ in $\\Z_N^k$. But for each $i$, $\\bra r_ix\\ket+\\bra r_iy\\ket-\\bra r_iz\\ket-\\bra r_iw\\ket \\in [-4\\delta N, 4\\delta N]$. Since $\\delta< 1/4$, that implies that $\\bra r_ix\\ket +\\bra r_iy\\ket-\\bra r_iz\\ket-\\bra r_iw\\ket =0$. So $\\bra\\mathbf{r}x\\ket+\\bra\\mathbf{r}y\\ket=\\bra\\mathbf{r}z\\ket+\\bra\\mathbf{r}w\\ket$.\\\\\n That already implies that $\\phi$ is an injection.\\\\\n If $\\mathbf{r}x + \\mathbf{a} N \\in [-\\delta N,\\delta N]^k$, then $r_i x \\in [-\\delta N,\\delta N]$ mod $N$ for each $i$. So $x \\in B(K,\\delta)$ and $\\phi(x) = \\mathbf{r}x + \\mathbf{a} N$. So $\\phi$ is a surjection as well.\\\\\n If $\\mathbf{r}x+\\mathbf{a}N + \\mathbf{r}y+\\mathbf{b}N = \\mathbf{r}z+\\mathbf{c}N+\\mathbf{r}w+\\mathbf{d}N$, then $r_1(x+y) = r_1(z_w)$ mod $N$, so $x+y=z+w$ mod $N$. So the inverse of $\\phi$ is also a Freiman homomorphism.\n \\end{proof}\n\\end{lemma}\n\n\\begin{lemma} (10)\\\\\n Let $\\Lambda$ be a lattice and $C$ be a symmetric convex body, both in $\\R^k$. Then $|\\Lambda \\cap C| \\leq 5^k |\\Lambda \\cap \\frac{C}{2}$|.\n \\begin{proof}\n let $x_1,...,x_m$ be a maximal subset of $\\Lambda \\cap C$ such that for all $i \\neq j$, $x_j \\not\\in x_i + \\frac{C}{2}$. Then by maximality, the sets $x_i+\\frac{C}{2}$ over(are?) all of $\\Lambda \\cap C$. Also, the sets $x_i+\\frac{C}{4}$ are disjoint subsets of $\\R^k$, and they are all contained in $C+\\frac{C}{4} = \\frac{5}{4} C$. So $m \\leq \\frac{vol(\\frac{5}{4}C)}{vol(\\frac{1}{4}C)} = 5^k$.\n \\end{proof}\n\\end{lemma}\n\n\\begin{coro} (11)\\\\\n If $N$ is prime, $0 \\not \\in K$, $|K| = k$, $\\delta < 1/4$, then $|B(K,\\delta)| \\leq 5^k |B(K,\\frac{\\delta}{2})|$.\n\\end{coro}\n\n\\newpage\n\n\\section{Sumsets and their structure}\nIt's to be shown that $|A+A| \\leq K|A|$ $\\implies$ $|rA-sA| \\leq K^{r+s}|A|$ (Ruzsa).\n\n\\begin{lemma} (1, Petridis)\\\\\n Let $A_0,B$ be finite subsets of an Abelian group such that $|A_0+B| \\leq K_0|A_0|$. Then there exist a non-empty subset $A \\subset A_0$ and $K \\leq K_0$ s.t. $|A+B+C| \\leq K|A+C|$ for every finite subset $C$ of the group.\n \\begin{proof}\n Let $A$ minimize the ratio $\\frac{|A+B|}{|A|}$, and let the minimal ratio be $K$.\\\\\n Claim: this works. We prove this by induction on $C$.\\\\\n If $C =\\phi$, then the result holds. Now assume it for $C$ and let $x \\not\\in C$. Then $A+(C \\cup \\{x\\}) = (A+C) \\cup [(A+x) \\setminus (A'+x)]$ where $A' = \\{a \\in A:a+x \\in A+C\\}$.\\\\\n This is a disjoint union, so $|A+(C\\cup \\{x\\})| = |A+C|+|A|-|A'|$, $A+B+(C\\cup\\{x\\}) = (A+B+C) \\cup ((A+B+x)\\setminus (A'+B+x))$, since if $a+x \\in A+C$ then $a+B+x \\subset A+B+C$. So\n \\begin{equation*}\n \\begin{aligned}\n |A+B+(C\\cup\\{x\\})| &\\leq |A+B+C|+|A+B|-|A'+B|\\\\\n &\\leq K|A+C|+K|A|-K|A'|\n \\end{aligned}\n \\end{equation*}\n by induction and minimality property of $A$.\n \\end{proof}\n\\end{lemma}\n\n---Lecture 5---\n\n\\begin{coro} (2)\\\\\n If $A,B$ are finite subsets of an abelian group, and $|A+B| \\leq K^r|A|$, then there exists $A'\\subset A$, $A' \\neq \\phi$ such that $|A'+rB| \\leq K|A'|$ for every positive integer $r$.\n \\begin{proof}\n Choose $A'$ as we choose $A$ in lemma 1. Then $|A'+rB| = |A'+B+(r-1)B| \\leq K|A'+(r-1)B|$.\\\\\n And $|A'+B| \\leq K|A'|$, so we are done by induction.\n \\end{proof}\n\\end{coro}\n\n\\begin{coro} (3)\\\\\n If $|A+A| \\leq K|A|$ or $|A-A| \\leq K|A|$, then $|rA| \\leq K^r|A|$.\n \\begin{proof}\n Set $B=A$ or $-A$ in corollary 2. (think about this)\n \\end{proof}\n\\end{coro}\n\n\\begin{lemma} (4, Ruzsa triangle inequality)\\\\\n Let $A,B,C$ be finite subsets of an abliean group. Then $|A||B-C| \\leq |A-B||A-C|$.\n \\begin{proof}\n Define a map $\\phi: A\\times(B-C) \\to (A-B) \\times (A-C)$. Given $(a,x)$ with $a\\in A, x \\in B-C$, choose, \\emph{somehow}, $b(x) \\in B$ and $c(x) \\in C$ s.t. $b(x) - c(x) = x$, and set $\\phi(a,x) = (a-b(x),a-c(x))$. Note that $(a-c(x)) - (a-b(x)) = b(x)-c(x) = x$ (!). And then, having worked out $x$, we know $b(x)$, and $a=a-b(x)+b(x)$, so $a$ is determined too. So $\\phi$ is an injection.\n \\end{proof}\n\\end{lemma}\n\nThe proof was easy, but why is this called the triangle inequality? We can rewrite it as \n\\begin{equation*}\n \\begin{aligned}\n \\frac{|B-C|}{|B|^{1/2}|C|^{1/2}} \\leq \\frac{|A-B|}{|A|^{1/2}|B|^{1/2}} \\cdot \\frac{|A-C|}{|A|^{1/2}|C|^{1/2}}\n \\end{aligned}\n\\end{equation*}\nSo if we define the \\emph{Ruzsa distance} $d(A,B) = \\frac{|A-B|}{|A|^{1/2}|B|^{1/2}}$, then the inequality says $d(B,C) \\leq d(A,B) d(A,C)$.\n\n\\begin{coro} (5)\\\\\n If $|A+B| \\leq K|A|$, then $|rB-sB| \\leq K^{r+s}|A|$ for all $r,s$.\n \\begin{proof}\n Pick $a'$ as before. Then by corollary 2 with $B$ replaced by $-B$, $|A'-rB| \\leq K^r|A'|$ and $|A'-sB| \\leq K^s|A'|$.\\\\\n Therefore, by Rusza triangle inequality (lemma 4), $|A'||rB-sB| \\leq K^{r+s}|A'|^2$, so $|rB-sB| \\leq K^{r+s} |A|$.\n \\end{proof}\n\\end{coro}\n\nOne finally corollary:\n\\begin{coro} (6, Plunnecke's theorem)\\\\\n If $|A+A| \\leq K|A|$ or $|A-A| \\leq K|A$, then $|rA-sA| \\leq K^{r+s} |A|$.\n \\begin{proof}\n Just apply corollary 5 with $B=-A$ or $B=A$.\n \\end{proof}\n\\end{coro}\n\n\\begin{lemma} (7, Ruzsa's embedding lemma)\\\\\n Let $A \\subset \\Z$ be finite and suppose that $|kA-kA| \\leq C|A|$. Then there exists a prime $p \\leq 4C|A|$ and a subset $A'\\subset A$ of size at least $|A|/k$ such that $A'$ is Freiman isomorphic of order $k$ to a subset of $\\Z_p$.\n \\begin{proof}\n Consider the following composition of maps $\\Z \\xrightarrow{\\text{ reduce mod }q}\\Z_q \\xrightarrow{\\times \\text{ by random non-zero } r} \\Z_q \\xrightarrow{\\text{least non-negative residue}} \\Z \\xrightarrow{\\text{reduce mod }p} \\Z_p$, where $q$ is a prime bigger than $diam A$ and $p$ is a prime $\\in (2C|A|,4C|A|]$ by Bertrand's postulate.\\\\\n Let $|phi$ be the composition. THe first, second and fourth parts are group homomorphisms, and thus Freiman homomorphisms of all order. Also, the third map is a Freiman homomorphism of order $k$ if you restrict to a subinterval of $[0,q-1]$ of lemgth $\\leq q/k$. To see this, write $\\bra u \\ket$ for least non-negative residue. Then if $I$ has length $\\leq q/k$ (and therefore $ 2C|A|$, there exists $r$ s.t. we get an Freiman isomorphism of order $k$.\n \\end{proof}\n\\end{lemma}\n\n\\end{document}\n", "meta": {"hexsha": "1330d40b45405f60f69153316d9bd0e077302af4", "size": 25321, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Notes/IDA.tex", "max_stars_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_stars_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_stars_repo_licenses": ["CC-BY-3.0"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-25T17:34:25.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-25T17:34:25.000Z", "max_issues_repo_path": "Notes/IDA.tex", "max_issues_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_issues_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_issues_repo_licenses": ["CC-BY-3.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Notes/IDA.tex", "max_forks_repo_name": "raoxiaojia/raoxiaojia.github.io", "max_forks_repo_head_hexsha": "d20c23a64794b500f2e0356fd01017ee31830fa2", "max_forks_repo_licenses": ["CC-BY-3.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 60.721822542, "max_line_length": 627, "alphanum_fraction": 0.559259113, "num_tokens": 9982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8080672135527632, "lm_q1q2_score": 0.6123487948820894}} {"text": "\\chapter{Summary of Saleena and Paleri}\n\\label{chap:chapter4}\n\nSaleena and Paleri gave an algorithm for \n\\textbf{global value numbering (GVN)}. GVN works by assigning a value \nnumber to variables and expressions. The same value number is \nassigned to those variables and expressions which are provably \nequivalent. A notable difference between Herbrand equivalence and GVN \nis that in Herbrand equivalence we talk about equivalences at a \nparticular program point but in GVN are concerned with equivalence \nbetween expressions at two different program points.\n\nThe data structure used in the algorithm is called \\textbf{value \nexpression} - an expression with value numbers as operands. Two \nexpressions are equivalent if they have same value expression. \nSo, a value expression can be used to represent a set of equivalent \nprogram expressions.\n\n\\section{Notation}\n\\label{sec:Notation}\nInput is a flow graph atmost one assignment statement in each node\nwhich has one of the following forms\n$$x\\; ::=\\; e$$ \n$$e\\; ::=\\; x\\: |\\: c\\: |\\: x\\,op\\,x$$\nThe flow graph also has two additional empty $ENTRY$ and $EXIT$ nodes.\nFor a node $n$, $IN_n$ and $OUT_n$ denotes the input and output \nprogram points of the node.\n\nExpression pool at a program point, is a partition of expressions at \nthat point, in which equivalent expression belongs to the same \npartition. Each class will have a value number which we will consider \nas its first element. For a node $n$, $EIN_n$ and $EOUT_n$ denotes \nthe expression pools at input and output program points of the node.\n\n\\section{Value Expression}\n\\label{sec:ValueExpression}\nThe value expression corresponding to an expression is obtained by \nreplacing actual operands with their corresponding value numbers.\nExample - For the expression-pool $\\{\\{v_1, a, x\\}$, $\\{v_2, b, y \\}\\}$ \nand statement $z ::= x + y$ , the value-expression for $x + y$ will be\n$v_1 + v_2$. Instead of $x + y$, its value-expression is included in \nthe expression-pool, with a new value number ie. the new \nexpression-pool would be \n$\\{\\{v_1, a, x\\},\\{v_2, b, y\\}, \\{v_3, v_1 + v_2, z\\}\\}$.\n\nThe value expression $v_1 + v_2$ represents not just $x + y$ but the\nset of equivalent expressions $\\{a + b, x + b, a + y, x + y\\}$. Its \npresence indicates that an expression from this set is already \ncomputed and this information is enough for detection of redundant \ncomputations. Also, a single binary value expression can represent \nequivalence among any numbre of expressions of any length. Example - \n$v_1 + v_3$ represents, $a + z$, $x + z$, $a + (a + b)$, \n$a + (x + b)$ and so on.\n\n\\section{Algorithm}\n\\label{sec:Algorithm}\nSimilar to Gulwani's algorithm, the algorithm consists of two main \nfunctions - a transfer function for changes in expression pool across \nassignment statements and a confluence function to find the \nexpression pool at points were two branches meet. The algorithm \nstarts with $EOUT_{ENTRY} = \\phi$, and uses transfer and confluence \nfunctions to calculate expression pools at other points. This process \nis repeated till there is any change in the equivalence information.\nFor detailed implementation refer to \\cite{Saleena}.\n", "meta": {"hexsha": "5d4307d82287df3e4f98b5bdc805ba1455d4a933", "size": 3162, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "reports/Rep_Mid_8/chapter4.tex", "max_stars_repo_name": "himanshu520/HerbrandEquivalence", "max_stars_repo_head_hexsha": "bfe056d9d370d9e5fe2782381b872bf102a960ba", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "reports/Rep_Mid_8/chapter4.tex", "max_issues_repo_name": "himanshu520/HerbrandEquivalence", "max_issues_repo_head_hexsha": "bfe056d9d370d9e5fe2782381b872bf102a960ba", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "reports/Rep_Mid_8/chapter4.tex", "max_forks_repo_name": "himanshu520/HerbrandEquivalence", "max_forks_repo_head_hexsha": "bfe056d9d370d9e5fe2782381b872bf102a960ba", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 48.6461538462, "max_line_length": 72, "alphanum_fraction": 0.7504743833, "num_tokens": 800, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7577943822145998, "lm_q2_score": 0.8080672112416737, "lm_q1q2_score": 0.6123487931307586}} {"text": "\\chapter{Sampling with sufficient statistics}\nThis chapter will go in depth of generating samples from the NHPP given its sufficient statistics. The main focus will be a Gibbs sampler with a Metropolis-Hastings step, and discuss possible modifications.\n\\section{Gibbs sampler}\nThe Gibbs sampler is Markov chain Monte Carlo (MCMC) sampler. A MCMC sampler is a way to sample from a distribution by contructing a Markov chain with spesific equilibrium. From \\cite{casella1992explaining} we have that the Gibbs sampler is used when direct sampling is difficult or cannot be done. This sampler samples from a conditional distribution. This can be done with sufficient statistics.\n\\section{Metropolis-Hastings sampler}\nThis is also a MCMC sampler where direct sampling is difficult. However this is used when the conditional posterior is unknown. Hence this uses a proportional density instead, and has an acceptance rejection step to determine if the new sample is from the desired distribution.\n\\section{Gibbs with Hastings step}\n\\label{sec:gibbsmet}\nA Gibbs sampler with a Metropolis-Hastings step is what we used in this project \\cite{gilks1995adaptive}. Since the values of the sufficient statistics cannot differ from the original data set, the sampler must generate new samples with the same values for the sufficient statistics. Given a sample with $N$ points and number of sufficient statistics, one can draw three arbitrary points and draw a new value for one of the points given the two others. By setting one of the points one can calculate the two others. This is where the Metropolis-Hastings step is used to draw a new value. How to find a new value is shown in section \\ref{sec:draw}. Further when this new value is drawn the two other values must be adjusted to fulfill the sufficient statistics. This should be done serveral times for each sample. The reason for this is to make the samples independent.\n\\subsection{Drawing a new value}\n\\label{sec:draw}\nAssume we have chosen three points ($X_1, X_2, X_3$) from our sample and that we have the same sufficient statistics as in section \\ref{sec:sufstat}. It is these three points that are going to be updated. The idea is to draw a new value for one of them as stated in section \\ref{sec:gibbsmet}. Assume $\\tau = 1$. Hence all values in the samples must be between 0 and 1. This makes it easier for the uniform distribution. To draw a new value we need to use tranformation of variables from section \\ref{sec:transvar}. The tranformation is\n\\begin{equation*}\n(Z_1, Z_2, Z_3) = (\\sum_{i=1}^{3} X_i, \\prod_{i=1}^{3} X_i, X_3).\n\\end{equation*}\nThis will yield the probability density $f_{Z_1, Z_2, Z_3}(z_1, z_2, z_3)$. However we are interested in $f_{Z_3 | Z_1, Z_2}(z_3 | z_1, z_2)$, where $Z_3=X_3$ is the variable drawn from this distribution. Fortunately from Bayes rule \\cite{casella2002statistical} the distribution becomes\n\\begin{equation*}\nf_{Z_3 | Z_1, Z_2}(z_3 | z_1, z_2) \\propto f_{Z_1, Z_2, Z_3}(z_1, z_2, z_3).\n\\end{equation*} \nFurther we continue with tranformation of variables. The determinant of the Jacobian matrix then becomes\n\\begin{equation*}\n|J| = X_3 (X_2 - X_1).\n\\end{equation*}\nWe also have the following relations from the tranformation, where we have put $a = Z_1$ and $b = Z_2$\n\\begin{align*}\nX_1 + X_2 + X_3 = a \\\\\nX_1 + X_2 = a - X_3 \\\\\nX_1 \\cdot X_2 \\cdot X_3 = b \\\\\nX_1 \\cdot X_2 = \\frac{b}{X_3}.\n\\end{align*}\nThen $X_1$ and $X_3$ are the roots of\n\\begin{equation*}\nX^2 - (a - X_3) X^2 + \\frac{b}{X_3} = 0,\n\\end{equation*}\nwhich are\n\\begin{equation*}\nX_{(1,2)} = \\frac{(a - X_3) ± \\sqrt{(a - X_3)^2 - \\frac{4b}{X_3}}}{2}.\n\\end{equation*}\nFrom this the jacobian becomes\n\\begin{equation*}\n|J| = X_3\\sqrt{(a - X_3)^2 - \\frac{4b}{X_3}}.\n\\end{equation*}\nFrom paper \\cite{lockhart2007use} the determinant of the inverse jacobian is\n\\begin{equation*}\n|J|^{-1} = \\frac{1}{X_3\\sqrt{(a - X_3)^2 - \\frac{4b}{X_3}}}.\n\\end{equation*}\nSince we have that\n\\begin{equation*}\nf_{X_1, X_2, X_3}(x_1, x_2, x_3) = 1, \\quad 0 \\leq x_i \\leq 1 \\text{ for all }i,\n\\end{equation*}\nthe density to draw new values for $X_3$ becomes\n\\begin{equation*}\n\\Pi \\propto \\frac{1}{X_3\\sqrt{(a - X_3)^2 - \\frac{4b}{X_3}}}.\n\\end{equation*}\nThis is where the Metropolis-Hastings step is used. The requirements for this density are listed below.\n\\begin{align}\n0 \\leq X_3 \\leq \\min(1, a) \\label{eq:req1} \\\\\n(a - X_3)^2 - \\frac{4b}{X_3} \\geq 0 \\label{eq:req2}\\\\ \n(a - X_3) + \\sqrt{(a - X_3)^2 - \\frac{4b}{X_3}} \\leq 1. \\label{eq:req3}\n\\end{align}\nSo when we draw from a proposal density like a uniform distribution we need to make sure that the new $X_3$ fulfills these requirements. If the requirements are not fulfilled then the probability of accepting this new value is zero. If the requirements are fulfilled then the acceptance probability is defined as,\n\\begin{equation}\n\\alpha = \\min\\left(1, \\frac{\\Pi(X_{prop})}{\\Pi(X_{curr})}\\right)\n\\label{eq:alpha}\n\\end{equation}\nAn overview of the algorithm is shown below.\n\\begin{algorithm}\n\\caption{Generate new samples with sufficient statistics}\n\\label{alg:simdata}\n\\begin{algorithmic}\n\\STATE Draw 3 arbitrary indices from $\\{1,2,...,n\\}$\n\\STATE Caclulate sum $a$ and product $b$ of the corresponding $X$'s.\n\\STATE Draw a proposal $X_3$ from $U[0,\\min(a,1)]$.\n\\STATE Check requirements in equations \\ref{eq:req1}, \\ref{eq:req2} and \\ref{eq:req3}.\n\\STATE Calculate $\\alpha$ from equation \\ref{eq:alpha}.\n\\STATE Accept with probability $\\alpha$\n\\STATE Repeat from top desired times to make sample approximately independent.\n\\end{algorithmic}\n\\end{algorithm}\n\\subsection{Possible modifications}\nIn our current version if the sampler generates invalid values for the three new indices in a sample, the algorithm tries over again with three new indices. One possible modification is to not try over again with three new indices, but keep the three indices and draw new random value. However the area for valid values might be small. Hence this modification might not be recommended.\n\\\\\n\\\\\nAnother modification is to use the indices of the entire sample when generating a new sample. This has been done before for related distribution in \\cite{lockhart2007use}. The pros of this method is that you have chance of getting new values for more of the indices. However the cons is that the values for the sub sums and sub products are sensitive to large and small values when a larger number of indices are used.\n", "meta": {"hexsha": "1d94fa9d68ed2fa27916f908e3862a392e1a31fe", "size": 6406, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Thesis/chapters/samplingsufficient.tex", "max_stars_repo_name": "mariufa/ProsjektOppgave", "max_stars_repo_head_hexsha": "3ef2fda314c55322de20f19ca861e4268a5e2d08", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Thesis/chapters/samplingsufficient.tex", "max_issues_repo_name": "mariufa/ProsjektOppgave", "max_issues_repo_head_hexsha": "3ef2fda314c55322de20f19ca861e4268a5e2d08", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Thesis/chapters/samplingsufficient.tex", "max_forks_repo_name": "mariufa/ProsjektOppgave", "max_forks_repo_head_hexsha": "3ef2fda314c55322de20f19ca861e4268a5e2d08", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 75.3647058824, "max_line_length": 868, "alphanum_fraction": 0.7508585701, "num_tokens": 1900, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8080672135527631, "lm_q2_score": 0.7577943658046608, "lm_q1q2_score": 0.6123487816217555}} {"text": "%%!TEX TS-program = latex\n\\documentclass[11pt]{article} %DIF > \n\\usepackage{etex}\n\\usepackage[utf8]{inputenc}\n\\input ../AuxFiles/PreambleNotes.tex\n\n\n\\begin{document}\n\\onehalfspace\n\n\\vspace*{\\fill}\n\\begingroup\n\\centering\n\n\\Large {\\scshape Introduction to Statistics}\\\\\n\n(Lectures 11-12: Confidence Sets)\n\n\\endgroup\n\\vspace*{\\fill}\n\n\\newpage\n\n\n\\section{Confidence Sets}\n\n{\\scshape Overview:} \\noindent Lectures 11-12 will focus on the construction of \\emph{confidence sets}. The actions are subsets of the parameter space. Each action is interpreted as a region that (presumably) contains the parameter selected by nature. The loss function we work with is a convex combination of ``volume’’ and ``coverage’’, in a sense we will make precise.\n\nWe first show that Bayes rules for this problem are ``highest posterior density” sets: the collection of parameters with a sufficiently high posterior probability. We then argue that in large samples, highest posterior density sets can be approximated by ``inverting’’ a Wald test. When the parameter of the model is scalar, inverting the Wald test is tantamount to reporting the interval formed by the Maximum Likelihood estimator $\\pm$ $c$ times its estimated “standard error”.\n\nWe then introduce and analyze two popular algorithms for summarizing the uncertainty in parameter estimation: (parametric) bootstrap confidence intervals and Bayesian ``credible’’ intervals. We use these algorithms to construct confidence/credible regions for one of the possibly many coefficients of the Linear Regression model. \n\n\\subsection{Description of the problem} \n\nLet $X$ be a random variable and let $\\{f(x | \\theta)\\}_{\\theta \\in \\Theta}$ be a statistical model. The problem of constructing a confidence set for $\\theta$ starts as follows. In the first stage nature picks an element of the parameter space, which in this section we model explicitly as a subset of $\\mathbb{R}^{k}$. The econometrician cannot observe the parameter selected by nature but observes data. Based on the data, the econometrician has to suggest a ``region’’ of the parameter space that is likely to contain the true parameter selected by nature. \n\nIn principle, the action space for the econometrician based on the description above is the power set of $\\Theta$. A minor technical restriction is that the econometrician can only report regions that are ``measurable’’. This happens because the ``volume’’ of whatever set is reported will be part of the loss function. Let $\\mathcal{C}$ denote the class of regions that the econometrician can report and let $c$ denote an element of $\\mathcal{C}$. \n\nA \\emph{decision rule/algorithm/strategy} for this problem is a mapping\n\\begin{equation*}\nC:X \\rightarrow \\mathcal{C}\n\\end{equation*}\nThese decision rules will be called ``confidence’’ regions. \n\n\n\\subsection{Loss function}\n\nThe payoff/loss for the econometrician depends on the action taken ($c \\in \\mathcal{C}$) and the true parameter ($\\theta$). The following loss is sometimes used to evaluate confidence sets:\n\\[ \\mathcal{L}_{\\delta}(c, \\theta) = \\delta \\mathbf{1}\\{ \\theta \\notin c \\} + (1-\\delta)\\int_{c} d\\theta, \\]\n\n\\noindent where $\\delta \\in (0,1)$. The first component in the loss function above penalizes the econometrician whenever $\\theta$ does not belong to $c$. The econometrician could avoid the penalty by reporting a ``large’’ region (for example, reporting $\\Theta$). The second component makes large regions unattractive by taking into account their volume. \n\nThe expected loss (risk) of a confidence region $C$ is given by\n\n\\begin{equation} \\label{equation:TypeI}\n\\mathbb{E}_{f(x|\\theta)} [ \\mathcal{L}_{\\delta}(C(x),\\theta) ] = \\delta (1-P_{f(x|\\theta)} ( \\theta \\in C(x) )) + (1-\\delta) \\mathbb{E}_{f(x|\\theta)} [ \\textrm{Vol}(C(x))].\n\\end{equation}\n\n\n\\noindent The term \n\\[ P_{f(x|\\theta)} ( \\theta \\in C(x) ) \\]\nis typically referred to as the \\emph{coverage probability} of $C(\\cdot)$ at $\\theta$. This captures how often the confidence region contains the parameter $\\theta$ when the data is indeed generated by it. The worst coverage probability over the parameter space:\n\n\\[ \\inf_{\\theta \\in \\Theta} P_{f(x|\\theta)} ( \\theta \\in C(x) ) \\] \n\n\\noindent is called the \\emph{confidence level} of the confidence set $C(x)$. \n\n\\subsection{Minimizing Posterior-Loss}\n\nWe now present a rule that minimizes posterior-loss. Any such rule, by construction, cannot be uniformly improved: there is no other region that has better coverage everywhere and also smaller expected volume. Let $\\pi(\\cdot)$ denote a prior over $\\Theta$ and let $\\pi(\\theta | x)$ denote the posterior. \n\nThe posterior loss of an action $c$ is given by:\n\n\\[ \\delta(1- P_{\\pi(\\theta | x)}( \\theta \\in c )) + (1-\\delta) \\int_{c} d \\theta. \\]\n\n\\noindent The term \n\\[ P_{\\pi(\\theta | x)}( \\theta \\in c ) = \\int_{c} \\pi(\\theta | x) d \\theta \\] \n\\noindent is the posterior probability that $\\theta$ falls in a set $c$. Such probability is usually referred to as the \\emph{credibility} of $c$. Algebra shows that the posterior loss can be written as\n\n\\[ \\delta - \\int_{c} \\left( \\delta \\pi(\\theta | x) - (1-\\delta) \\right) d\\theta. \\]\n\n\\noindent Thus, for each $x$, the posterior loss is minimized by choosing\n\\begin{equation} \\label{equation:HPD}\nC^{\\textrm{Bayes}}_{\\delta}(x) = \\{ \\theta \\in \\Theta \\: | \\: \\pi(\\theta | x) \\geq (1-\\delta)/\\delta \\}.\n\\end{equation}\nThe region above is called a ``highest posterior density’’ region. As it is name suggests, the region collects the points in the parameter space for which the posterior density is above certain threshold. \n\nIn general, it is hard to give closed form expressions for the highest posterior density region. However, in large samples it is possible to provide (heuristic) approximations for these regions; exploiting the fact that---regardless of the prior---the posterior distribution is approximately distributed as\n\\[ \\mathcal{N}_{k} \\left( \\widehat{\\theta}_{\\textrm{ML}} \\quad, \\quad \\widehat{I}^{-1} \\right), \\]\n\\noindent where $\\widehat{I}$ is the observed information. This means that highest posterior density regions will be approximately of the form \n\\begin{eqnarray*}\n\\left \\{ \\theta \\in \\Theta \\: \\Bigg | \\: \\frac{1}{ (2 \\textrm{det} \\widehat{I}))^{k/2} } \\exp \\left( -\\frac{1}{2} (\\theta - \\widehat{\\theta}_{\\textrm{ML}}) ’\\widehat{I} (\\theta - \\widehat{\\theta}_{\\textrm{ML}})\\right) \\geq (1-\\delta)/\\delta \\right \\}, \\\\\n\\end{eqnarray*}\nwhich can be written as \n\\[ \\left \\{ \\theta \\in \\Theta \\: \\Bigg | \\: (\\widehat{\\theta}_{\\textrm{ML}} - \\theta) ’\\widehat{I} (\\widehat{\\theta}_{\\textrm{ML}}-\\theta) \\leq \\widehat{c}_{\\delta} \\right \\}, \n\\]\n\\noindent where $\\widehat{c}_{\\delta}$ is some function that depends on $\\delta$ and $\\widehat{I}$ but not on $\\theta$. Recognize the quadratic form as the Wald statistic for the problem\n\\[ \\mathbf{H}_0: \\theta = \\theta_0, \\textrm{ vs. } \\mathbf{H}_1: \\theta \\neq \\theta_0. \\]\nThus, in large samples, highest-posterior density regions are approximately given by those points in the parameter space in which the Wald statistic is below certain threshold that might depend on the data. \n\nWhen $\\theta$ is scalar, the approximation takes the form\n\\[ \\Big[ \\widehat{\\theta}_{\\textrm{ML}} - \\sqrt{\\frac{\\widehat{c}_{\\delta}}{\\widehat{I}}} \\:, \\: \\widehat{\\theta}_{\\textrm{ML}} + \\sqrt{\\frac{\\widehat{c}_{\\delta}}{\\widehat{I}}} \\Big] \\]\n\n\n\\noindent {\\scshape Example:} We derive the highest posterior density region for a linear regression model with known variance. We have shown that if the prior density is that of a $\\mathcal{N}_{k}(0, \\sigma^2 (\\mathbb{I}_n/\\lambda))$, the posterior of $\\beta$ is given by\n\\[ \\beta | Y, X \\sim \\mathcal{N}_{k} ( m , V ), \\]\nwhere \n\\begin{eqnarray*}\nm &\\equiv& (X’X + \\lambda \\mathbb{I}_n)^{-1} X’Y, \\\\\nV & \\equiv & \\sigma^2 (X’X + \\lambda \\mathbb{I}_n)^{-1}.\n\\end{eqnarray*}\n\\noindent The highest-posterior density region is thus \n\\begin{eqnarray*}\nC_{\\delta}(x) &= & \\left \\{ \\beta \\in \\mathbb{R}^{k} \\: \\Big | \\: \\frac{1}{ (2 \\textrm{det} V))^{k/2} } \\exp \\left( -\\frac{1}{2 } (\\beta - m) ’V^{-1} (\\beta - m) \\right) \\geq (1-\\delta)/\\delta \\right \\}, \\\\\n&=& \\left \\{ \\beta \\in \\mathbb{R}^{k} \\: \\Big | \\: (\\beta - m) ’V^{-1} (\\beta - m) \\leq \\widehat{c}_{\\delta,X,\\sigma^2} \\right \\}.\n\\end{eqnarray*}\n\n\\noindent This is an ellipse centered at the posterior mean. In large samples, \n$$ m \\approx \\widehat{\\beta}_{\\textrm{OLS}}, \\quad V \\approx \\sigma^2 (X’X)^{-1}.$$\nHence, the highest-posterior density region is approximately given by \n\\[ \\left \\{ \\beta \\in \\mathbb{R}^{k} \\: \\Big | \\: \\frac{1}{\\sigma^2}(\\widehat{\\beta} - \\beta) ’(X’X) (\\widehat{\\beta} - \\beta) \\leq \\widehat{c}_{\\delta,X,\\sigma^2} \\right \\}. \\]\n\n\\noindent This is set of all parameter values for which the Wald test for the null $\\beta = \\beta_0$ is low enough. \n\n\\subsection{Confidence Regions and Hypothesis Testing}\nThere is a connection between confidence regions and tests of hypothesis that goes beyond the approximation argument above. Any family of tests of size $\\alpha$ can be ``inverted’’ to construct a confidence region with confidence level of at least $1-\\alpha$. Likewise, any confidence region with confidence level $1-\\alpha$ can be transformed into a statistical test with rate of Type 1 error of at most $\\alpha$. \n\n{\\scshape Inverting a family of tests:} Let $\\{\\phi_{\\theta}\\}_{\\theta \\in \\Theta}$ denote a collection of non-randomized tests. Suppose each element in the collection is a test for the problem\n\\[ \\mathbf{H}_0: \\theta =\\theta_0, \\quad \\mathbf{H}_1: \\theta \\neq \\theta_0. \\]\nSuppose further that each test has size of at most $\\alpha$ in the sense that\n\\[ \\mathbb{P}_{f(x|\\theta)} [\\phi_{\\theta}(x)=1] \\leq \\alpha. \\]\n\n\\noindent The confidence region obtained by ``test inversion’’ is given by\n\n\\[ C (x) \\equiv \\{ \\theta \\in \\Theta \\: | \\: \\phi_{\\theta}(x)=0\\}. \\]\n\n\\noindent The coverage of such confidence region at $\\theta$ is given by\n\\begin{eqnarray*}\nP_{f(x|\\theta)} [\\theta \\in C(x)] &=& 1-P_{f(x|\\theta)} [\\theta \\notin C(x)] \\\\\n&=& 1- P_{f(x|\\theta)} [ \\phi_{\\theta}(x) = 1] \\\\\n& \\geq & 1-\\alpha. \n\\end{eqnarray*}\n\\noindent Therefore, \n\\[\\inf_{\\theta \\in \\Theta} P_{f(x|\\theta)} [\\theta \\in C(x)] \\geq 1-\\alpha, \\]\n\\noindent which implies that the confidence region formed by test inversion has confidence level, provided each of the tests $\\phi_{\\theta}$ have size of at $\\alpha$. \\\\\n\n{\\scshape Using a confidence region to test a hypothesis:} A confidence region with $C(x)$ confidence level of at least $1-\\alpha$ can be readily used to test a hypothesis of the form\n\\[ \\mathbf{H}_0: \\theta =\\theta_0 \\quad \\textrm{vs.} \\quad \\mathbf{H}_1: \\theta \\neq \\theta_0. \\]\n\nConsider the test $\\phi$ that rejects the null hypothesis above whenever $\\theta \\notin C(x)$. The rate of Type I error of the test is \n\\begin{eqnarray*}\nP_{f(x| \\theta)} [ \\phi(x)=1 ] &=& P_{f(x| \\theta)} [\\theta \\notin C(x)] \\\\\n&=& 1- P_{f(x| \\theta)} [\\theta \\in C(x)] \\\\\n&\\leq& 1- \\inf_{\\theta \\in \\Theta} P_{f(x| \\theta)} [\\theta \\in C(x)] \\\\\n&\\leq& 1-(1-\\alpha) =\\alpha.\n\\end{eqnarray*}\n\n\n\\subsection{Parametric Bootstrap and Bayesian Credible sets}\n\n\\subsubsection{The Parametric Bootstrap}\n\nSuppose that we have a statistical model $\\{ f(x | \\theta)\\}_{\\theta \\in \\Theta}$ for data $X$. The estimator $\\widehat{\\theta}_{\\textrm{ML}}$ is a random variable, as it depends on the data. If $\\theta$ were known, we could approximate the distribution of $\\widehat{\\theta}_{\\textrm{ML}}$ by Monte-Carlo methods: generate $I$ independent draws from $x \\sim f(x;\\theta)$, and evaluate $\\widehat{\\theta}_{\\textrm{ML}}$ over the $I$ new data sets that we have generated. \\\\\n\n\nThe true parameter that generated the data is unknown; hence it is not possible to generate draws from the model $f(x;\\theta)$. However, since there is an estimator for $\\theta$, one could perform the Monte-Carlo approximation exercise described above by using draws from the model:\n\n\\[ f(x | \\widehat{\\theta}_{\\textrm{ML}}).\\]\n\nIf we do this, we will effectively end with $I$ estimators (one for each new data set). Suppose that the sample size is large enough to guarantee that $\\widehat{\\theta}_{\\textrm{ML}}$ is close to the true $\\theta$ (whatever the true value might be) with high probability. If the parametric model $f(x | \\theta)$ varies smoothly with $\\theta$, the c.d.f. based on the $I$ estimators can be shown to be a reasonable approximate for the distribution of $\\widehat{\\theta}_{\\textrm{ML}}$. If the parameter of interest is some function $g(\\theta)$, the procedure above also provides an approximation for $g(\\widehat{\\theta}_{\\textrm{ML}})$, provided $g$ is smooth. \\\\\n\nThe exercise described above is known as the \\underline{parametric bootstrap} and it is a general statistical technique used to estimate the distribution of $\\widehat{\\theta}_{\\textrm{ML}}$ or any function smooth function $g(\\widehat{\\theta}_{\\textrm{ML}})$. \\\\\n\n\\noindent {\\scshape Example:} Consider the linear regression model with unknown variance\n\\[ Y \\sim \\mathcal{N}_n (X\\beta, \\sigma^2 \\mathbb{I}_n). \\]\nThe parameters of the model are $\\beta = (\\beta_1, \\ldots, \\beta_{k})’$ and $\\sigma^2$. Suppose we want to construct a confidence region for the $\\beta_1$. The ML estimators for $\\beta$ and $\\sigma^2$ are\n\\[ \\widehat{\\beta}_{\\textrm{ML}} = (X’X)^{-1}X’Y, \\quad \\widehat{\\sigma}^2_{\\textrm{ML}} = (Y-X \\widehat{\\beta}_{\\textrm{ML}})’ (Y-X \\widehat{\\beta}_{\\textrm{ML}})/n. \\]\n\n\\noindent The parametric bootstrap algorithm starts by generating $I$ new data sets, independently according to the distribution\n\\[ Y(i) \\sim \\mathcal{N}_{n}( X \\widehat{\\beta}_{\\textrm{ML}}\\:, \\: \\widehat{\\sigma}^2_{\\textrm{ML}} \\mathbb{I}_n). \\]\n\\noindent For each of these data sets, we can compute \n\n\\[ \\widehat{\\beta}_{\\textrm{ML}}(i) = (X’X)^{-1}X’Y(i), \\quad \\widehat{\\sigma}^2_{\\textrm{ML}} = (Y(i)-X \\widehat{\\beta}_{\\textrm{ML}}(i))’ (Y(i)-X \\widehat{\\beta}_{\\textrm{ML}}(i))/n, \\]\n\n\\noindent and collect the first component of $ \\widehat{\\beta}_{\\textrm{ML}}(i)$:\n\n\\[ \\widehat{\\beta}_{\\textrm{ML}}(i)_1 = e_1’ \\widehat{\\beta}_{\\textrm{ML}}(i), \\]\n\n\\noindent where $e_1$ is the first column of the identity matrix of dimension $k$. Let $\\widehat{q}_{\\eta}$ denote the $\\eta$ quantile of $\\{ \\widehat{\\beta}_{\\textrm{ML}}(i)_1\\}_{i=1}^{I}$. The parametric bootstrap confidence interval for $\\beta_1$ (of ``nominal’’ level $1-\\alpha$) for $\\beta_1$ is thus given by\n\\[ C_{\\textrm{Bootstrap}}(Y,X) \\equiv [ \\widehat{q}_{\\alpha/2} \\:, \\: \\widehat{q}_{(1-\\alpha)/2} ]. \\] \n\n\\subsection{Credible Sets based on the quantiles of the posterior}\nThe quantiles of the posterior provide an off-the-shelf algorithm to construct ``credible’’ regions for any real-valued function $g(\\theta)$. Start with the parametric model $\\{f(x|\\theta)\\}$ and let $\\pi(\\theta)$ be a prior. Suppose that we can generate $I$ draws from the posterior distribution $\\pi(\\theta | x)$. Denote these draws $\\{ \\theta(i) \\})_{i=1}^{I}$. For each of these draws we can construct\n\\[ \\Big\\{ g(\\theta(i)) \\Big\\}_{i=1}^{I}. \\]\nLet $\\widehat{p}_{\\eta}$ denote the $\\eta$ quantile of $\\{ g(\\theta(i))\\}_{i=1}^{I}$. The interval with credibility of approximately $1-\\alpha$ is equal to\n\\[ C_{\\textrm{Bayes}} \\equiv [ \\widehat{b}_{\\alpha/2} \\:, \\: \\widehat{b}_{(1-\\alpha)/2} ]. \\] \n\n\\noindent {\\scshape Example:} Let’s go back to the linear regression model with unknown variance. The parameter of interest is still $\\beta_1$. The prior is $\\mathcal{N}_{k}(0, \\sigma^2 (\\mathbb{I}_n/\\lambda))$. The posterior of $\\beta$ (conditional on $\\sigma^2$) is given by\n\\[ \\beta | Y, X, \\sigma^2 \\sim \\mathcal{N}_{k} ( m , V ), \\]\nwhere \n\\begin{eqnarray*}\nm &\\equiv& (X’X + \\lambda \\mathbb{I}_k)^{-1} X’Y, \\\\\nV & \\equiv & \\sigma^2 (X’X + \\lambda \\mathbb{I}_k)^{-1}.\n\\end{eqnarray*}\nThe posterior distribution for distribution for $\\sigma^2$ is an Inverse-Gamma($a_n$,$b_n$) where\n\\[ a_n = a_0 + n/2, \\quad b_n = b_0 + \\frac{1}{2}( (Y-Xm)’(Y-Xm) + \\lambda m’m ). \\]\n\nWe can generate $I$ draws for $\\sigma^2$ from the Inverse-Gamma posterior. For each draw $\\sigma^2(i)$ we can also take a draw from $\\beta | Y,X, \\sigma^2$, and collect the first element $\\beta(i)_1$. The $\\alpha/2$ and $(1-\\alpha)/2$ quantiles, provide an approximation to the credible set for $\\beta_1$. \n\n\\newpage\n\n\n\\newpage\n\n\n\\bibliographystyle{../AuxFiles/ecta}\n\\bibliography{../AuxFiles/BibMaster}\n\n\n\n\\end{document}\n", "meta": {"hexsha": "5bf77c2307b2370b1c6ed1980947d43b966dd59d", "size": 16404, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/Lectures/Lectures11-12.tex", "max_stars_repo_name": "yu45020/Courses-IntroEconometrics-Ph.D", "max_stars_repo_head_hexsha": "84162de869a4c8a7ad93a7e7f3f33e741d4caeb3", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 17, "max_stars_repo_stars_event_min_datetime": "2018-09-10T01:31:49.000Z", "max_stars_repo_stars_event_max_datetime": "2021-09-13T12:01:36.000Z", "max_issues_repo_path": "docs/Lectures/Lectures11-12.tex", "max_issues_repo_name": "yu45020/Courses-IntroEconometrics-Ph.D", "max_issues_repo_head_hexsha": "84162de869a4c8a7ad93a7e7f3f33e741d4caeb3", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-09-07T02:05:29.000Z", "max_issues_repo_issues_event_max_datetime": "2019-10-15T12:46:32.000Z", "max_forks_repo_path": "docs/Lectures/Lectures11-12.tex", "max_forks_repo_name": "yu45020/Courses-IntroEconometrics-Ph.D", "max_forks_repo_head_hexsha": "84162de869a4c8a7ad93a7e7f3f33e741d4caeb3", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 16, "max_forks_repo_forks_event_min_datetime": "2018-09-05T03:12:41.000Z", "max_forks_repo_forks_event_max_datetime": "2021-09-13T12:01:38.000Z", "avg_line_length": 73.8918918919, "max_line_length": 661, "alphanum_fraction": 0.6949524506, "num_tokens": 5049, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7577943712746406, "lm_q2_score": 0.8080672066194946, "lm_q1q2_score": 0.612348780787875}} {"text": "\\subsection{Homogeneity}\r\n\\begin{definition}\r\n\tA differential equation is homogeneous if all terms involve the dependent variable or its derivatives.\r\n\\end{definition}\r\n\r\n\\noindent\r\nBelow is a table of homogeneity and equation numbers.\r\n\\begin{table}[H]\r\n\t\\centering\r\n\t\\begin{tabular}{c|c}\r\n\t\tHomogeneity & Equation Number \\\\\r\n\t\t\\hline\r\n\t\tHomogeneous & 3, 7, 8 \\\\\r\n\t\tHeterogeneous & 1, 2, 4, 5, 6, 9, 10 \\\\\r\n\t\\end{tabular}\r\n\\end{table}\r\n\\noindent\r\nHomogeneous equations have some nice properties, like always having at least a trivial solution of 0.", "meta": {"hexsha": "8973d42511854a59003f6ad69174da282372c091", "size": 550, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "diffEq/basics/classification/homogeneity.tex", "max_stars_repo_name": "aneziac/Math-Summaries", "max_stars_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 39, "max_stars_repo_stars_event_min_datetime": "2020-03-26T06:20:36.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-17T17:38:45.000Z", "max_issues_repo_path": "diffEq/basics/classification/homogeneity.tex", "max_issues_repo_name": "aneziac/Math-Summaries", "max_issues_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": 26, "max_issues_repo_issues_event_min_datetime": "2020-03-28T17:44:18.000Z", "max_issues_repo_issues_event_max_datetime": "2021-10-07T04:47:03.000Z", "max_forks_repo_path": "diffEq/basics/classification/homogeneity.tex", "max_forks_repo_name": "aneziac/Math-Summaries", "max_forks_repo_head_hexsha": "20a0efd79057a1f54e093b5021fbc616aab78c3f", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": 10, "max_forks_repo_forks_event_min_datetime": "2020-04-10T05:41:17.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-17T15:21:12.000Z", "avg_line_length": 30.5555555556, "max_line_length": 104, "alphanum_fraction": 0.7218181818, "num_tokens": 164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8080671950640465, "lm_q2_score": 0.7577943767446202, "lm_q1q2_score": 0.6123487764513326}} {"text": "\\section{Introduction}\\label{sec:intro}\n\n\\mypar{Motivation} The longest common subsequence (LCS) problem is the problem of finding the longest (not necessarily contiguous) subsequence that is shared by two input strings. This problem occurs very widely in practice. For example, the two sequences $ABCD$ and $ACBAD$ have the longest common subsequences $ACD$ and $ABD$. The length of the LCS is unique, although multiple subsequences with that length might exist.\n\nLCS algorithms are at the core of comparison tools like ``diff'', which are ubiquitous when working with source code. These comparisons are also used extensively in version control systems. Outside of the software domain, the longest common subsequence problem is extremely important in biology - LCS is the exact solution to DNA nucleotide \\emph{sequence alignment}. However, it is important to note that heuristic methods (BLAST \\cite{altschul_basic_1990}) are often used today, since they allow fast search of very large databases of genetic sequences.\n\nThere exist various specialized LCS algorithms which achieve faster runtime or lower memory usage than general algorithms. One class of specialized algorithms provides improved runtime when both input sequences have significant similarities (edit distance not too large). This is generally the case when comparing two versions of a text document. For this reason algorithms of this class are nearly universally used by tools similar to ``diff''. However such algorithms are generally sequential.\n\nWe adapt the sequential LCS algorithm of Myers \\cite{myers_anond_1986}, which is by far the most commonly used LCS algorithm for similar sequences, for parallel execution on a distributed memory system.\n\n\\mypar{Related work} Bergroth et al. prepared a survey \\cite{bergroth_survey_2000} of various sequential LCS algorithms. They group the algorithms into three categories: ``row-by-row methods'', ``contour methods'' and ``diagonal methods''. The classic dynamic programming algorithm for LCS \\cite{wagner_string--string_1974} is classified as a row-by-row method. Myers' algorithm \\cite{myers_anond_1986} belongs to the class of diagonal methods. Diagonal methods are designed to be efficient when both input sequences are similar. \n\nMost parallel algorithms for the LCS problem are based on row-by-row sequential algorithms. Lu and Lin \\cite{mi_lu_parallel_1994} presented a parallel algorithm of this class, achieving optimal complexity with a shared-memory machine model. Yang et al. \\cite{jiaoyun_efficient_2010} designed a parallel LCS algorithm for GPUs (graphics processing units) by restructuring the classic DP (dynamic programming) table to reduce data dependencies.\\linebreak These algorithms are fundamentally based on row-by-row methods, which are less efficient than diagonal methods for similar input sequences.\n\nAllison and Dix \\cite{allison_bit-string_1986} present a ``bit-parallel'' LCS algorithm (row-by-row method). Assuming a sufficiently small alphabet, their algorithm allows calculating multiple elements of the DP table in parallel, for example by using vector instructions. Their bit-parallel technique can be applied to other algorithms. We do not incorporate it into our algorithm, but do use vectorization within each DP cell.\n\nSince Myers' algorithm \\cite{myers_anond_1986} is an extremely popular LCS algorithm, we attempted to parallelize it. As far as we are aware, there are no published parallel versions of an LCS algorithm from the ``diagonal methods'' class, which Myers' algorithm belongs to. ", "meta": {"hexsha": "7e0ab4851d369825bf468afd7d4da2c560a3dd58", "size": 3554, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "report_src/sections/01-introduction.tex", "max_stars_repo_name": "tehwalris/mpi-myers-diff", "max_stars_repo_head_hexsha": "9d6dafc9dc16dcf97b4c712dbb8c6dace25eeee5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-11-09T11:30:02.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T17:47:49.000Z", "max_issues_repo_path": "report_src/sections/01-introduction.tex", "max_issues_repo_name": "tehwalris/mpi-myers-diff", "max_issues_repo_head_hexsha": "9d6dafc9dc16dcf97b4c712dbb8c6dace25eeee5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "report_src/sections/01-introduction.tex", "max_forks_repo_name": "tehwalris/mpi-myers-diff", "max_forks_repo_head_hexsha": "9d6dafc9dc16dcf97b4c712dbb8c6dace25eeee5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 209.0588235294, "max_line_length": 592, "alphanum_fraction": 0.8137310073, "num_tokens": 728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.7905303211371899, "lm_q1q2_score": 0.6123316514314282}} {"text": "\\section{Numerical Experiment}\n\\label{sec:4}\nIn this section, we apply various methods to solve the LP form of optimal transport \\ref{lp} and the entropic regularization of optimal transport \\ref{equ:kl}. All of the numerical experiments are conducted on an Intel Core i7-6500U Processor with 2 cores and 4 threads.\n\n\\subsection{Datasets of Optimal Transport}\nWe perform the numerical experiments on various datasets, including randomly generated data, DOTmark\\cite{DOTmark}, ellipse example \\cite{multiscale}, Caffarelli's example \\cite{multiscale} and Gaussian mixture model. \n\n\\subsubsection{Randomly generated data}\nIn order to make the feasible set not empty, we first generate a random matrix $R\\in \\mbR^{m\\times n}$, where $R_{i,j}\\sim \\mcN(0,1)\\, i.i.d.$. and then generate $\\mu_i, \\nu_j$ by\n$$\n\\mu_i=\\frac{\\sum_{j=1}^n|R_{i,j}|}{\\sum_{i=1}^m\\sum_{j=1}^n|R_{i,j}|}, \\, \\forall i=1, 2, \\dots m\\\\\n$$\n$$\n\\nu_j=\\frac{\\sum_{i=1}^m|R_{i,j}|}{\\sum_{i=1}^m\\sum_{j=1}^n|R_{i,j}|}, \\, \\forall j=1, 2, \\dots n\\\\\n$$\nIt is easy to verify that $\\sum_{i=1}^m\\mu_i=\\sum_{j=1}^n\\nu_j=1$. We then generate $C_{i,j} \\sim \\mcN(0,1)\\, i.i.d.$. We denote $C_{\\mathrm{min}}=\\min_{i,j}C_{i,j}$ and let $C_{i,j}=C_{i,j}-C_{\\mathrm{min}}$ to ensure $C_{i,j}\\geq 0$. \n\n\\subsubsection{DOTmark}\nDOTmark \\cite{DOTmark} provides $10$ classes of $10$ different images, each of which is available at the different resolutions from $32\\times32$ to $512\\times512$. Suppose we select two images $x$ and $y$ with same resolution $l\\times l$ from one class. The corresponding parameter for the standard form of LP is given in the following way: $m=n=l^2$, $\\mu_i=x_i$, $\\nu_j=y_j$ and $C_{i,j}=\\|x_i-y_j\\|_p$, where $(x_i)_{1\\leq i\\leq m}$ and $(y_j)_{1\\leq j\\leq n}$ form a regular square grid in $\\mbR^2$. In practice, we take $p=2$. In solving this LP, we actually calculate the Wasserstein distance between $x$ and $y$.\n\n\\subsubsection{Ellipse Example}\nThe ellipse example consists of two uniform samples, source and target data set, of size $n$ from the unit circle with normal distributed noise added with zero mean and deviation 0.1. Slightly different from that in the paper \\cite{multiscale}, we construct the source data example by scaling in the x-Axis by 2.0 and in the y-Axis by 0.5. The target data sample is then scaled in the x-Axis by 0.5 and y-Axis by 2.0. The corresponding parameter for the standard form of LP is as follows. $m = n$, $\\mu_i=1, \\nu_i=1(i=1, ..., n)$ and $C_{i, j}=\\|x_i-y_j\\|_2$, which is the Euclidean distance between two points $x_i$ and $y_j$. Figure \\ref{ellipse} gives an example. \n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=12cm]{final/ellipse1.png}\n\\caption{An Ellipse Example with sample size $n=50$. The darker the line, the greater the $\\pi_{i, j}$.}\n\\label{ellipse}\n\\end{figure}\n\n\\subsubsection{Caffarelli's Example}\nThis example, mentioned in \\cite{multiscale}, consists of two uniform samples on $[-1, 1]^2$ of size $n$. Then any point outside the unit circle are then discarded. The source and target data sample is split along the x-Axis at 0 and shifted by -2 and +2, respectively. Note that the true number of points may be less than $n$. When $n$ is large, about $\\frac{\\pi}{4}n$ points are preserved. For ease of notation, we still assume there are $n$ points. The corresponding parameter for the standard form of LP is as follows. $m = n$, $\\mu_i=1, \\nu_i=1(i=1, ..., n)$ and $C_{i, j}=\\|x_i-y_j\\|_2$, which is the Euclidean distance between two points $x_i$ and $y_j$. Figure \\ref{caff} gives an example.\n\\begin{figure}[h]\n\\centering\n\\includegraphics[width=12cm]{final/caff1.png}\n\\caption{An Caffarelli's Example with sample size $n=50$. The darker the line, the greater the $\\pi_{i, j}$.}\n\\label{caff}\n\\end{figure}\n\n\\subsubsection{Gaussian Mixture Model}\nSuppose $p_\\mu$ and $p_\\nu$ are the probability density function of two different Gaussian mixture models and $N$ is the number of discrete points. In this example, $\\mu$ and $\\nu$ are the normalized discrete distribution of Gaussian mixture models, which satisfy\n$$\n\\mu_i=\\frac{p_\\mu(\\frac{i-1}{N-1})}{\\sum_{i=1}^Np_\\mu(\\frac{i-1}{N-1})}, \\quad \\nu_j=\\frac{p_\\nu(\\frac{j-1}{N-1})}{\\sum_{j=1}^Np_\\nu(\\frac{j-1}{N-1})}\n$$\nAnd $C_{i,j}=|i-j|$. \n\n\\subsection{Different methods on Mosek and Gurobi}\n`prim', `dual', `int' represent primal simplex, dual simplex and interior point method respectively. `(M)' means Mosek, `(G)' means Gurobi. For each method, we record the time and the number of iteration it takes and evaluate their performance. We examine the accuracy of the solution $\\pi$ by the value of the objective function ('objval') \n$$\n\\sum_{i=1}^m\\sum_{j=1}^nC_{i,j}\\pi_{i,j}\n$$\nand the violation of the constraints ('vltcst') \n$$\n\\sum_{i=1}^m|\\mu_i - \\sum_{j=1}^{n}\\pi_{i, j}| + \\sum_{j=1}^n|\\nu_j-\\sum_{i=1}^m\\pi_{i, j}|$$\nIn the table, we only give the exact value of `objval' for prim(M). In `objval' of other methods, we give its relative difference to prim(M). Namely, if `objval' for prim(M) is $f_1$ and `objval' for prim(G) is $f_2$, in `objval' of prim(G), we give $\\frac{f_1-f_2}{f_1}$. \n\n\\subsubsection{Randomly generated data}\nFor simplicity, we only consider the case when $m=n$. We take $m=128,256,512,1024$. \n\n\\input{final/RGD_mg}\n\nFrom Table \\ref{RGD_mg}, we can see on Mosek, dual simplex method and interior point method take much more time than primal simplex does, especially when $m, n$ are large. But on Gurobi, dual simplex method is the fastest method and the time primal simplex method takes is close to the time interior point takes. Interior point method takes much smaller number of iterations than We can observe that compared to other methods, primal simplex method on Mosek gets the largest value of objective function and the largest value of the violation of constraints. On Mosek, dual simplex methods achieves the lowest value of `vltcst', which partly explains why it takes the longest time. On Gurobi, the solution from all methods satisfies constraints perfectly.\n\n\\subsubsection{DOTmark}\nWe then test the performance of Mosek and Gurobi on DOTmark \\cite{DOTmark}. We number the classes of DOTmark with the order of alphabet in the following way.\n\\begin{table}[!htp]\n\\centering\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline\nCauchyDensity&ClassicImages&GRFmoderate&GRFrough&GRFsmooth\\\\\\hline\n1&2&3&4&5\\\\\\hline\nLogGRF&LogitGRF&MicroscopyImages&Shapes&WhiteNoise\\\\\\hline\n6&7&8&9&10\\\\\\hline\n\\end{tabular}\n\\end{table}\n\nWe test on pictures with resolution $32\\times32$. \n\\input{final/DOTmark_mg}\n\nFrom Table \\ref{DOTmark_mg}, we observe that dual simplex method on Mosek takes much longer time than other methods. The cpu-time of primal simplex method on Mosek and primal simplex method and dual simplex method on Gurobi are close. On Mosek and Gurobi, the iteration number vary with different classes of DOTmark. Compared to other methods, primal simplex method on Mosek gets the largest `vltcst' and the largest `vltcst'. On Mosek, dual simplex methods achieves the smallest `vltcst'. On Gurobi , the solution $\\pi$ from all methods satisfies the constraints perfectly.\n\n\\subsubsection{Ellipse Example \\& Caffarelli's Example}\nSimilar to randomly generated data, we only consider the case when $m=n$. We take $m=128,256,512,1024$. \n\\input{final/ellipse_mg}\n\\input{final/caff_mg}\n\nDifferent from the results on randomly generated data, Table \\ref{ellipse_mg} shows that on Ellipse Example, dual simplex method takes much longer time and more iterations than primal simplex method, no matter on Mosek or on Gurobi. Similar to Ellipse Example, Table \\ref{caff_mg} shows that dual simplex method takes much longer time than the primal simplex method, no matter on Mosek or on Gurobi. The iteration number of primal simplex method and dual simplex method are close. Similarly, compared to other methods, primal simplex method on Mosek gets the largest `vltcst' and the largest `vltcst'. On Mosek, dual simplex methods achieves the smallest `vltcst'. On Gurobi , the solution $\\pi$ from all methods satisfies the constraints perfectly.\n\n\\subsection{First order methods}\n`prim(M)' means primal method of Mosek. `ADMM-p' means ADMM primal method \\ref{alg:ADMM}. `ADMM-d' means ADMM dual method \\ref{alg:ADMM-dual}. `ADMM-s' means another ADMM splitting method for primal problem \\ref{alg:ADMM-split}. BADMM means Bregman ADMM \\ref{alg:BADMM}.\n\\subsubsection{Art of Tuning Parameters}\nThe success of ADMM always lies in an appropriate choice of the coefficient $t$ in the quadratic term of the augmented Lagrangian function. Thanks to the normalization of constraints, i.e., letting $\\sum_{i=1}^m\\mu_i = \\sum_{j=1}^n\\nu_j = 1$, the tuning of parameters is relatively easier. Intuitively, the value of $t$ should be dependent on the problem size, $m$ and $n$, and the objective coefficients $C$. In our implementation, for primal problem, we let $t$ to be proportional to $(m+n)\\bar C$, where $\\bar C$ is the mean of the coefficient matrix $C$ ($\\frac{1}{mn}\\sum_{i, j}C_{i, j}$). For dual problem, we set $t$ to be inversely proportional to $(m+n)\\bar C$. For the Bregman ADMM, we set $t$ to be proportional to $\\bar C$. The detailed proportion will be clear in the following. Empirically, we found that such strategy performs well in our experiments.\n\n\\subsubsection{Randomly Generated Data}\n\\input{final/RGD_fo}\nIn Table \\ref{RGD_fo}, we present the numerical results on random generated data. \nDetailed tuning parameters is listed in Table \\ref{param_rgd_fo}.\n\\begin{table}[h]\n\\centering\n\\caption{Tuned Parameters: Random Generated Data}\n\\label{param_rgd_fo}\n\\begin{tabular}{|c|c|c|}\n\\hline \nSolver & $t$ & Stopping Rule \\\\\n\\hline\nADMM-p & $5(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 5e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-d & $1/(8(m+n)\\bar C)$ & $\\text{`vltcst'}\\leqslant$ 2e-04 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-s & $2(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 1e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nBADMM & $10\\bar C$ & $\\|\\pi-\\tilde\\pi\\|_1\\leqslant$ 1e-06 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\nIn this setting, both `ADMM-p' and `ADMM-s' achieves relatively better results. They both satisfy the constraints quite well and achieve lower objective value. `ADMM-p' spends moderate time in all problems because all computation has closed form. `ADMM-s' performs quite well in `objval' and `vltcst', especially when the problem size is larger. However, such accuracy is at the cost of much longer time. In fact, from our observation, much of the time by `ADMM-s' is spent on the projection on the simplex(constraints). There exists more efficient algorithms for projection, but since time is limited, we do not implement them. The projection is quite essential in `ADMM-s' because it imposes $\\pi$ to satisfy constraints, and this may explain why the constraints are easy to satisfy within tolerance.\n\nIt's worth noting that `ADMM-d' has weakness in satisfying constraints. This is because in Section 2.1.2 we regard the ``dual of dual'' as an approximation of initial variables, which is numerically unstable. In our experiments, we found that `ADMM-d' converges faster at the beginning to the optimal value, but then continuously vibrates around optimal value for a long time. `BADMM' appears to perform worse than other three algorithms. The selection of $t$ is inconsistent with other ADMM due to the property of KL divergence. If $t$ is selected to be too big, then the exponential term in \\ref{badmm_pipit_upd} may be so close to 1 that leads to bad precision. Furthermore, in our experiments, we found that when we increase $t$, `BADMM' indeed has slightly better performance in `vltcst' but deteriorates rapidly in `objval'. Our parameter is a balance between the two targets.\n\n\\subsubsection{DOTmark}\nSome detailed tuning parameters is listed in Table \\ref{param_DOTmark_fo}.\n\\begin{table}[htbp]\n\\centering\n\\caption{Tuned Parameters: DOTmark}\n\\label{param_DOTmark_fo}\n\\begin{tabular}{|c|c|c|}\n\\hline \nSolver & $t$ & Stopping Rule \\\\\n\\hline\nADMM-p & $5(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 5e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-d & $1/(8(m+n)\\bar C)$ & $\\text{`vltcst'}\\leqslant$ 2e-04 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-s & $2(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 4e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nBADMM & $10\\bar C$ & $\\|\\pi-\\tilde\\pi\\|_1\\leqslant$ 4e-06 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\n\\input{final/DOTmark_fo}\nTable \\ref{DOTmark_fo} presents the numerical results on DOTmark.\n\nAs to constraints, all methods perform similar to that in the setting of random generated data, since our stopping rule is built on `vltcst' or the norm of difference between $\\pi$ and $\\tilde\\pi$. \n\nAs to objective value, the three traditional ADMM methods are not stable on `objval' and tend to achieve higher value than Mosek. This may be because of the different structure between random generated data and DOTmark. `BADMM' tends to achieve lower objective value. However, the results are not stable and the constraints are not satisfied well.\n\nAs to iteration numbers and time spent, surprisingly, `ADMM-p' and `ADMM-s' need much less iterations than others. This phenomenon implies that we could tighten our stopping rule by decreasing the thresholds. Due to the time limit, further experiments could be conducted in the future.\n\n\\subsubsection{Ellipse Example \\& Caffarelli's Example}\nThe detailed parameters is listed in Table \\ref{param_ellcaff_fo}. \n\\begin{table}[htbp]\n\\centering\n\\caption{Tuning Parameters: Ellipse \\& Caffarelli}\n\\label{param_ellcaff_fo}\n\\begin{tabular}{|c|c|c|}\n\\hline \nSolver & $t$ & Stopping Rule \\\\\n\\hline\nADMM-p & $5(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 5e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-d & $1/(8(m+n)\\bar C)$ & $\\text{`vltcst'}\\leqslant$ 2e-04 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nADMM-s & $2(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 2e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nBADMM & $10\\bar C$ & $\\|\\pi-\\tilde\\pi\\|_1\\leqslant$ 4e-06 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\n\\input{final/ellipse_fo}\n\\input{final/caff_fo}\n\nTable \\ref{ellipse_fo} and \\ref{caff_fo} presents numerical results for Ellipse Example and Caffarelli's Example, respectively.\n\nThe constraints are generally satisfied before we stop iterations. The solution from `BADMM' does not satisfy the constraints very well. However, the optimal value is not within our expectation. The vibration of objective values around optimal values are observed in the three traditional ADMM methods. Among these algorithms, `ADMM-s' performs relatively better on `iter', `objval', and `vltcst', though it takes more time in each iteration. Nevertheless, `objval' of `BADMM' is relatively low.\n\n\\subsection{Algorithms for entropic regularized OT}\n`sinkhorn' represents the Sinkhorn's algorithm with numerical stability and continuation strategy \\ref{alg:sansc}. `ADMM' means the algorithm described in \\ref{alg:ADMM-ER}. Addition to `time', `iter', `objval' and `vltcst', we use `entval' to denote the objective value with extropy regularization term. The values in `objval' are, same as before, compared to `prim(M)'.\n\nThe parameters and stopping criterions of our algorithms are given in Table \\ref{param_er}. In 'sinkhorn', we apply continuation strategy. If the coefficient of the regularized term is $\\epsilon$, we select $10^i\\epsilon(i=4, 3, 2, 1, 0)$ as the sequence and run 4000 iterations in optimizing every subproblem.\n\n\\begin{table}[htbp]\n\\centering\n\\caption{Tuned Parameters: algorithms for entropic regularized OT}\n\\label{param_er}\n\\begin{tabular}{|c|c|c|}\n\\hline \nmethod & $t$ & Stopping Rule \\\\\n\\hline\nADMM & $5(m+n)\\bar C$ & $\\text{`vltcst'}\\leqslant$ 1e-07 or $\\text{`iter'}\\geqslant$ 2e+04 \\\\\\hline\nsinkhorn & / & $\\text{`iter'}\\geqslant$ 2e+04 \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\n\\subsubsection{Gaussian Mixture Model: The effect of continuation strategy}\n\nWe consider to test Sinkhorn's algorithm with numerical stability on Gaussian mixture model and test the effect of continuation strategy. We first take $N=128$ as a first exmaple. Figure \\ref{gmm3} gives the probability density function of two Gaussian mixture models.\n\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=12cm]{final/gmm3.jpg}\n\\caption{The probability distribution function of two Gaussian mixture models $G_1$ and $G_2$. $G_1$ (blue one) has mean $[0.2,0.5]$, variance $[0.05^2, 0.03^2]$ and component proportion $[0.6, 0.4]$; $G_2$ (red one) has mean $[0.6,0.7]$, variance $[0.03^2, 0.05^2]$ and component proportion $[0.3, 0.7]$. }\n\\label{gmm3}\n\\end{figure}\n\nWe then take $\\epsilon=10^{-2},10^{-3},10^{-4},10^{-5}$. If we use the vanilla Sinkhorn's algorithm \\ref{alg:sa} for $\\epsilon=10^{-4},10^{-5}$, $K^{\\epsilon}$ would be a zero matrix. For Sinkhorn's algorithm with numerical stability \\ref{alg:sans}, we take the maximum iteration number to be $10,100,1000,4000$ respectively. For Sinkhorn's algorithm with numerical stability and continuation strategy \\ref{alg:sansc}, we take $\\alpha=0.1$, $\\epsilon_0=\\epsilon\\times 10^5$ and the maximum iteration number to be $2,20,200,800$ respectively. \n\n\\begin{figure}[!htb]\n\\centering\n\\includegraphics[width=12cm]{final/gmm4.jpg}\n\t\\caption{Impact of $\\epsilon$ on the coupling between two gaussian mixture models \\ref{gmm1}. From left to right, $\\epsilon=10^{-2},10^{-3},10^{-4},10^{-5}$. The upper row is without continuation strategy and the lower row is with continuatation strategy. The purple one is the solution to the entropic regularized LP. The green one is the solution to the original LP.}\n\\label{gmm4}\n\\end{figure}\n\nFrom Figure \\ref{gmm4}, we can find that with same iteration number, Sinkhorn's algorithm with numerical stability and continuation strategy converges faster than the one only with numerical stability. This shows that with proper choice of the parameter of continuation, we can achieve acceleration in Sinkhorn's algorithm with continuation strategy.\n\nWe then perform a comprehensive numerical experiment on Gaussian mixture model. Detailed parameters and the stopping criterions are listed in Table \\ref{param_er}. \n\n\\input{final/gmm_er}\nTable \\ref{gmm_er} gives the comprehensive numerical results for Gaussian Mixture Model. The setting is same as Figure \\ref{gmm1}, except that we test examples for different problem sizes, i.e., we let $N=128, 256, 512, 1024$, respectively. We observe that, in most cases, although `sinkhorn' takes much longer time and more iterations than `ADMM', `sinkhorn' outperforms `ADMM' in both `objval' and `vltcst'. \n\n\\subsubsection{Randomly generated data}\nThe parameters and the stopping criterions are listed in Table \\ref{param_er}. Table \\ref{RGD_er} presents numerical results on random generated data. \n\n\\input{final/RGD_er}\n\nWe find that the `objval' and `vltcst' of `sinkhorn' is getting worse with the decrease of $\\epsilon$. This can be explained by the computation complexity of Sinkhorn's algorithm. As shown in Section 3.2, the $\\tau$-approximate solution of the unregularized OT problem takes $O(n^2\\log(n)\\tau^{-3})$ operations where $\\epsilon=\\tau/\\log(n)$. This means that if we decrease the $\\epsilon$ by $0.1$, theoretically we would need $10^3$ times iteration to get a precise solution. Although our improvements of Sinkhorn's algorithm makes it adapt to small $\\epsilon$, but the computation is costly. If we want to use Sinkhorn's algorithm to get an approximation solution of the orignial problem \\ref{lp}, we should pay attention to choose an appropriate $\\epsilon$. \n\nOn the contrary, `ADMM' appears robustness in satisfying constraints within acceptable iteration number. `vltcst' of all test samples are below 1e-06, though it's not comparable to `sinkhorn'. `objval' of ADMM appears to be gradually lower than `prim(M)' when $\\epsilon$ decreases. It's worth noting that in `ADMM' we do not apply continuation strategy. In fact, we observe that with the carefully selected $t$, `ADMM' converges quite fast to the optimal value at the beginning, whatever $\\epsilon$ is, although vibration is unavoidable in the latter steps. Of course, for $\\epsilon$ not large and small or moderate problem sizes, `sinkhorn' performs better than `ADMM'.\n\n\n\\subsubsection{DOTmark}\nDetailed parameters are listed in \\ref{param_er}. (The stopping rule for `ADMM' changes to $\\text{`vltcst'}\\leqslant$ 1e-07 or $\\text{`iter'}\\geqslant$ 2e+04.) Table \\ref{DOTmark_er1} and \\ref{DOTmark_er2} present numerical results on DOTmark.\n\\input{final/DOTmark_er1}\n\\input{final/DOTmark_er2}\n\nSame as before, `objval' and `vltcst' of `sinkhorn' is getting worse with the decrease of $\\epsilon$. All of the results are within 1e-09 in `vltcst', but `objval' deteriorate rapidly when $\\epsilon$ decreases. This can be again explained by the computation complexity of Sinkhorn's algorithm. \n\n'ADMM' can achieve relatively lower objective value on average, and need moderate iterations before `vltcst' arrives at a threshold. The performance is similar to random generated data setting with larger problem sizes. We should note that due to the different problem structures, it seems difficult sometimes for `ADMM' to outperform `prim(M)'.\n\n\\subsubsection{Ellipse Example \\& Caffarelli's Example}\n\\input{final/ellipse_er}\n\\input{final/caff_er}\nTable \\ref{param_er} lists the detailed parameters.Table \\ref{ellipse_er} and Table \\ref{caff_er} present numerical results for Ellipse Example and Caffarelli's Example, respectively. Through our previous experiments, we observe that with the decrease of $\\epsilon$, `ADMM' does not require so much increase of computational complexity as `sinkhorn'. This can be explained by the limit when $\\epsilon\\to0$. When $\\epsilon\\to0$, `ADMM' reduce to ADMM for the primal problem \\ref{lp}. But with $\\epsilon\\to0$, we have shown in Section 3.3.1 that `sinkhorn' will diverge due to the loss of strong convexity. And with the decrease of $\\epsilon$, the strong convexity of problem \\ref{equ:kl} decreases. This also explains why `sinkhorn' requires such enormous iterations to find the optimal solution. In summary, we shall use `sinkhorn' to solve the entropic regularization of OT \\ref{equ:kl} with moderate $\\epsilon$. With large $\\epsilon$, we shall use the original Sinkhorn's algorithm \\ref{alg:sa} because it is faster in computation. `ADMM' seems to be suitable to arbitrary small $\\epsilon$. \n", "meta": {"hexsha": "58a2a137c888b1d40df7de772dc220b1edd6db5f", "size": 22471, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Latex/final/numerical_experiment.tex", "max_stars_repo_name": "YiifeiWang/Optimal-Transport", "max_stars_repo_head_hexsha": "2b39683ec19c0c5c81c2150cd0c3f68a9e915214", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 12, "max_stars_repo_stars_event_min_datetime": "2019-01-17T19:17:11.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-03T13:04:15.000Z", "max_issues_repo_path": "Latex/final/numerical_experiment.tex", "max_issues_repo_name": "YiifeiWang/Optimal-Transport", "max_issues_repo_head_hexsha": "2b39683ec19c0c5c81c2150cd0c3f68a9e915214", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Latex/final/numerical_experiment.tex", "max_forks_repo_name": "YiifeiWang/Optimal-Transport", "max_forks_repo_head_hexsha": "2b39683ec19c0c5c81c2150cd0c3f68a9e915214", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 4, "max_forks_repo_forks_event_min_datetime": "2019-01-21T02:55:52.000Z", "max_forks_repo_forks_event_max_datetime": "2021-06-04T02:16:55.000Z", "avg_line_length": 94.8143459916, "max_line_length": 1093, "alphanum_fraction": 0.7490098349, "num_tokens": 6598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833841649233, "lm_q2_score": 0.7905303186696747, "lm_q1q2_score": 0.612331649520132}} {"text": "\n% This LaTeX was auto-generated from an M-file by MATLAB.\n% To make changes, update the M-file and republish this document.\n\n%%% \\documentclass{article}\n%%% \\usepackage{graphicx}\n%%% \\usepackage{color}\n\n%%% \\sloppy\n%%% \\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n%%% \\begin{document}\n\n \n \n\\subsection*{Four parameter sine wave fitting}\n\n\\begin{par}\nExample for algorithm ThreePSF.\n\\end{par} \\vspace{1em}\n\\begin{par}\nThreePSF is an algorithm for estimating the amplitude, phase and offset of the sine waveform according standard IEEE Std 1241-2000';\n\\end{par} \\vspace{1em}\n\n\\subsubsection*{Contents}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item Generate sample data\n \\item Call algorithm\n \\item Display results\n\\end{itemize}\n\n\n\\subsubsection*{Generate sample data}\n\n\\begin{par}\nTwo quantities are prepared: \\lstinline{t} and \\lstinline{y}, representing 1 second of sinus waveform of nominal frequency 1 kHz, nominal amplitude 1 V, nominal phase 1 rad and offset 1 V sampled at sampling frequency 10 kHz.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nDI = [];\nAnom = 2; fnom = 100; phnom = 1; Onom = 0.2;\nt = [0:1/1e4:1-1/1e4];\nDI.y.v = Anom*sin(2*pi*fnom*t + phnom) + Onom;\nDI.Ts.v = 1e-4;\nDI.f.v = fnom;\n\\end{lstlisting}\n\n\n\\subsubsection*{Call algorithm}\n\n\\begin{par}\nUse QWTB to apply algorithm \\lstinline{ThreePSF} to data \\lstinline{DI}.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nCS.verbose = 1;\nDO = qwtb('ThreePSF', DI, CS);\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\nQWTB: no uncertainty calculation\n\\end{lstlisting} \\color{black}\n \n\n\\subsubsection*{Display results}\n\n\\begin{par}\nResults is the amplitude, phase and offset of sampled waveform.\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nA = DO.A.v\nph = DO.ph.v\nO = DO.O.v\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\n\nA =\n\n 2.0000\n\n\nph =\n\n 1.0000\n\n\nO =\n\n 0.2000\n\n\\end{lstlisting} \\color{black}\n \\begin{par}\nErrors of estimation in parts per milion:\n\\end{par} \\vspace{1em}\n\\begin{lstlisting}[style=mcode]\nAerrppm = (DO.A.v - Anom)/Anom .* 1e6\npherrppm = (DO.ph.v - phnom)/phnom .* 1e6\nOerrppm = (DO.O.v - Onom)/Onom .* 1e6\n\\end{lstlisting}\n\n \\begin{lstlisting}[style=output]\n\nAerrppm =\n\n -8.4377e-09\n\n\npherrppm =\n\n -9.7700e-09\n\n\nOerrppm =\n\n 8.3267e-10\n\n\\end{lstlisting} \\color{black}\n \n\n\n%%% \\end{document}\n \n", "meta": {"hexsha": "76aca5d11fe22cf4a9c5288eb77af0f481202641", "size": 2399, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doc/algs_examples_published/doc_ThreePSF.tex", "max_stars_repo_name": "qwtb/qwtb", "max_stars_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-12-09T13:18:54.000Z", "max_stars_repo_stars_event_max_datetime": "2015-12-09T13:18:54.000Z", "max_issues_repo_path": "doc/algs_examples_published/doc_ThreePSF.tex", "max_issues_repo_name": "qwtb/qwtb", "max_issues_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 18, "max_issues_repo_issues_event_min_datetime": "2015-12-09T13:08:38.000Z", "max_issues_repo_issues_event_max_datetime": "2021-07-13T11:33:41.000Z", "max_forks_repo_path": "doc/algs_examples_published/doc_ThreePSF.tex", "max_forks_repo_name": "qwtb/qwtb", "max_forks_repo_head_hexsha": "f6c79c7dca4065fd85d6f1c05257c1af34e85e34", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2016-11-11T02:12:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-09-17T12:59:18.000Z", "avg_line_length": 19.192, "max_line_length": 225, "alphanum_fraction": 0.6877865777, "num_tokens": 816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833737577158, "lm_q2_score": 0.7905303260722198, "lm_q1q2_score": 0.6123316470268072}} {"text": "% !Mode:: \"TeX:UTF-8\"\n% !TEX program = xelatex\n\\title{Project 5: Solving the FDMs for Two-asset Options}\n\\author{Iydon Liang}\n\\date{\\today}\n\n\\maketitle\\tableofcontents\\clearpage\n\n\\begin{abstract}\n It's clear that how to solve the FDM for one-asset option, either European, American or Path-dependent types. In this report, we discussed the two-asset options case. And gave three methods to determine the price for a two-asset option.\n\\end{abstract}\n\n\\section{Two-asset Option}\n\\[\n \\dd S_i(t) = S_i(t)\\alpha_i\\dd t + S_i(t)\\sigma_i\\dd W_i(t)\n\\]\nwhere $\\dd W_i(t)$ $(i=1,\\ldots,n)$ are standard Brownian motion which satisfy\n\\[\n \\begin{aligned}\n \\EE(\\dd W_i) &= 0 \\\\\n \\var(\\dd W_i) &= \\dd t \\\\\n \\cov(\\dd W_i, \\dd W_j) &= \\delta_{ij} p_{ij} \\dd t.\n \\end{aligned}\n\\]\n\n\\subsection{Pricing by Monte Carlo Method}\nWe know that price of the two stocks follows a bi-variate log-normal distribution. And we can generate it from a bi-variate normal distribution with $\\mu=0$ and\n\\[\n \\Sigma = \\begin{pmatrix}\n 1 & \\rho \\\\\n \\rho & 1\n \\end{pmatrix}.\n\\]\nThen we can sample $(S_1(T), S_2(T))$ from the distribution, and pricing by Monte Carlo method is possible by so.\n\n\n\n\\section{FDMs for Two-asset Option}\n\\[\n \\frac{\\partial V}{\\partial t} + rS_1\\frac{\\partial V}{\\partial S_1} + \\frac{1}{2}\\sigma_1^2S_1^2\\frac{\\partial^2 V}{\\partial S_1^2} + rS_2\\frac{\\partial V}{\\partial S_2} + \\frac{1}{2}\\sigma_2^2S_2^2\\frac{\\partial^2 V}{\\partial S_2^2} - rV = 0\n\\]\n\n\\subsection{Approximate the Differential Quotients}\nSuppose $f(i, j, k) = V(S_1=i\\Delta S, S_2=j\\Delta S, t=k\\Delta k)$,\n\\[\n \\begin{aligned}\n f_t &\\approx \\frac{f(i, j, k+1) - f(i, j, k)}{\\Delta t} \\\\\n f_{S_1} &\\approx \\frac{f(i+1, j, k) - f(i-1, j, k)}{2\\Delta t} \\\\\n f_{S_1S_2} &\\approx \\frac{f(i+1, j, k) - 2f(i, j, k) + f(i-1, j, k)}{\\Delta^2S} \\\\\n f_{S_2} &\\approx \\frac{f(i, j+1, k) - f(i, j-1, k)}{2\\Delta t} \\\\\n f_{S_1S_2} &\\approx \\frac{f(i, j+1, k) - 2f(i, j, k) + f(i, j-1, k)}{\\Delta^2S}\n \\end{aligned}\n\\]\n\n\n\\subsection{Recursion Form}\nSubstitute the approximations in to the BS equation, and put $f(\\cdot, \\cdot, k+1)$ to the left side, put $f(\\cdot, \\cdot, k)$ to the right hand side. Then we get\n\\[\n \\begin{aligned}\n f(i, j, k+1) =& \\quad\\alpha_i f(i-1, j, k) \\\\\n &+ \\beta_i f(i+1, j, k) \\\\\n &+ \\gamma_j f(i, j-1, k) \\\\\n &+ \\theta_j f(i, j+1, k) \\\\\n &+ \\lambda_{1i} f(i, j, k) \\\\\n &+ \\lambda_{2j} f(i, j, k)\n \\end{aligned}\n\\]\n\nHere we divide the coefficient of $f(i, j, k)$ into two parts for the convenience of following steps.\n\\[\n \\begin{aligned}\n \\alpha_i &= \\frac{1}{2}\\Delta t(ri-\\sigma_1^2i^2) \\\\\n \\beta_i &= -\\frac{1}{2}\\Delta t(ri+\\sigma_1^2i^2) \\\\\n \\gamma_j &= \\frac{1}{2}\\Delta t(rj-\\sigma_2^2j^2) \\\\\n \\theta_j &= -\\frac{1}{2}\\Delta t(ri-\\sigma_2^2j^2) \\\\\n \\lambda_{1i} &= \\frac{1}{2} + \\Delta t(\\frac{1}{2}r+\\sigma_1^2i^2) \\\\\n \\lambda_{2j} &= \\frac{1}{2} + \\Delta t(\\frac{1}{2}r+\\sigma_2^2j^2) \\\\\n \\end{aligned}\n\\]\n\n\n\\subsection{Matrix Form}\nThe relationship can be written as a matrix equation\n\\[\n AX + XB = C\n\\]\nwhere matrix $C(i, j) = f(i, j, k+1)$, $X(i, j) = f(i, j, k)$\n\n\\begin{figure}[H]\n \\centering\n \\includegraphics[width=.7\\textwidth]{figures/2019-12-25-matrix-form.png}\n\\end{figure}\n\n\n\\subsection{Solve the Matrix Equation}\nMatrix equation of the form $AX + XB = C$ is called \\textbf{Sylvester equation}. The equation has a unique solution when the eigenvalues of $A$ and $−B$ are distinct. In \\texttt{MATLAB}, there is a function \\textbf{sylvester(A,B,C)}, which can solve the Sylvester equation.\n", "meta": {"hexsha": "75f3b413e9ef164df77d34709c2b2e4303d09383", "size": 3680, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MA216/sections/project-5/main.tex", "max_stars_repo_name": "iydon/homework", "max_stars_repo_head_hexsha": "253d4746528ef62d33eba1de0b90dcb17ec587ed", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2019-10-20T08:18:54.000Z", "max_stars_repo_stars_event_max_datetime": "2021-07-11T12:14:56.000Z", "max_issues_repo_path": "MA216/sections/project-5/main.tex", "max_issues_repo_name": "iydon/homework", "max_issues_repo_head_hexsha": "253d4746528ef62d33eba1de0b90dcb17ec587ed", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2022-01-13T03:04:10.000Z", "max_issues_repo_issues_event_max_datetime": "2022-03-12T00:49:10.000Z", "max_forks_repo_path": "MA216/sections/project-5/main.tex", "max_forks_repo_name": "iydon/homework", "max_forks_repo_head_hexsha": "253d4746528ef62d33eba1de0b90dcb17ec587ed", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2019-11-02T05:46:01.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-12T23:11:28.000Z", "avg_line_length": 37.9381443299, "max_line_length": 273, "alphanum_fraction": 0.6065217391, "num_tokens": 1381, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7745833737577158, "lm_q2_score": 0.7905303162021596, "lm_q1q2_score": 0.6123316393816227}} {"text": "\\documentclass[10pt,preprint]{../aastex}\n\n\\usepackage{amsfonts}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{booktabs}\n\\usepackage{mathrsfs}\n\\usepackage{cite}\n\\usepackage{times}\n\\usepackage{url}\n\\usepackage{hyperref}\n\\usepackage{lineno}\n\\usepackage{yhmath}\n\\usepackage{natbib}\n\\usepackage{../../definitions}\n\\hypersetup{\n bookmarksnumbered = true,\n bookmarksopen=false,\n pdfborder=0 0 0, % make all links invisible, so the pdf looks good when printed\n pdffitwindow=true, % window fit to page when opened\n pdfnewwindow=true, % links in new window\n colorlinks=true, % false: boxed links; true: colored links\n linkcolor=blue, % color of internal links\n citecolor=magenta, % color of links to bibliography\n filecolor=magenta, % color of file links\n urlcolor=cyan % color of external links\n}\n\n\\usepackage{float}\n\\usepackage{graphicx}\n\\newtheorem{remark}{Remark}\n\\graphicspath{{Figures/}}\n\\newcommand{\\figref}[1]{Figure \\ref{#1}}\n\n\\begin{document}\nThroughout this document, lowercase latin indices ($i,j,$ etc.) run from 1 to 3, lowercase Greek indices ($\\mu,\\nu,$ etc.) run from 0 to 3, and uppercase latin indices ($J$) run from 1 to 5.\n\n\\section{Equations}\nWe are solving the general relativistic hydrodynamics (GRH) equations \\citep{RezzollaRelHyd}:\n\\begin{equation}\\label{Eq:GRH}\n\\pd{\\left(\\sqrt{\\gamma}\\,\\bU\\right)}{t}+\\pd{\\left(\\alpha\\,\\sqrt{\\gamma}\\,\\bF^{i}\\right)}{i}=\\sqrt{\\gamma}\\,\\bS,\n\\end{equation}\nwhere $\\bU$ is the vector of conserved variables, defined as\n\\begin{equation}\n\\bU\\longrightarrow U^{J}=\\begin{pmatrix}D\\\\ S_{j}\\\\ E\\end{pmatrix}=\\begin{pmatrix}\\rho\\,W\\\\ \\rho\\,h\\,W^{2}\\,v_{j}\\\\\\rho\\,h\\,W^{2}-p\\end{pmatrix},\n\\end{equation}\n$\\bF^{i}$ is the vector of fluxes of the conserved variables in the $x^{i}$-direction, defined as\n\\begin{equation}\n\\bF^{i}\\longrightarrow \\left(F^{i}\\right)^{J}=\\begin{pmatrix}D\\left(v^{i}-\\eta^{i}\\right)\\\\ S_{j}\\left(v^{i}-\\eta^{i}\\right)+p\\,\\delta^{i}_{~j}\\\\S^{i}-\\eta^{i}\\,E\\end{pmatrix},\n\\end{equation}\n$\\bS$ is the vector of sources, defined as\n\\begin{equation}\n\\bS\\longrightarrow S^{J}=\\begin{pmatrix}0\\\\\\frac{1}{2}\\,\\alpha\\,P^{ik}\\,\\pd{\\gamma_{ik}}{j}+S_{i}\\,\\pd{\\beta^{i}}{j}-E\\,\\pd{\\alpha}{j}\\\\\\alpha\\,P^{ij}\\,K_{ij}-S^{j}\\,\\pd{\\alpha}{j}\\end{pmatrix},\n\\end{equation}\nwhere $\\rho$ is the mass-density of the fluid, $v^{i}$ are the contravariant components of the fluid three-velocity, and $p$ is the pressure of the fluid. The quantity $h$ is the specific enthalpy of the fluid, defined as\n\\begin{equation}\nh\\equiv1+\\frac{e+p}{\\rho},\n\\end{equation}\nwhere $e$ is the internal energy-density of the fluid. The quantity $W$ is the Lorentz factor of the fluid, defined as\n\\begin{equation}\nW\\equiv\\left(1-\\vv{v}\\cdot\\vv{v}\\right)^{-1/2}.\n\\end{equation}\nThe quantity $\\alpha$ is the lapse function, $\\eta^{i}$ are the ratio of the components shift-vector, $\\beta^{i}$, to the lapse function:\n\\begin{equation}\n\\vv{\\eta}\\longrightarrow\\eta^{i}\\equiv\\alpha^{-1}\\,\\beta^{i},\n\\end{equation}\nand $\\gamma_{ij}$ are the covariant components of the spatial three-metric tensor, i.e. $\\gamma^{ij}\\,\\gamma_{jk}=\\delta^{i}_{~k}$, where $\\delta^{i}_{~k}$ are the usual components of the Kronecker delta tensor. The $P^{ij}$ are the components of the stress-tensor, defined as\n\\begin{equation}\n\\bP\\longrightarrow P^{ij}\\equiv\\rho\\,h\\,W^{2}\\,v^{i}\\,v^{j}+p\\,\\gamma^{ij},\n\\end{equation}\nand the $K_{ij}$ are the components of the extrinsic curvature, defined as\n\\begin{equation}\n\\text{FILL IN HERE}.\n\\end{equation}\n\n\\section{Derivation of Middle Wave-Speed Estimate for General Relativistic HLLC Numerical Flux}\nFor a given Riemann problem we assume that a discontinuity breaks up into four distinct regions (see \\figref{Fig:HLLC_RiemannFan}).\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.5]{HLLC_RiemannFan_MB2005}\n\\caption{HLLC Riemann fan from \\citet{Mignone2005}.}\\label{Fig:HLLC_RiemannFan}\n\\end{figure}\n\n\\subsection{Derivation of Rankine-Hugoniot Jump Conditions}\nThis derivation closely follows \\citet{RezzollaRelHyd}.\n\nWe start the derivation by integrating the one-dimensional version of \\eqref{Eq:GRH} in space from a point $x_{L}$ to a point $x_{R}$, that contains a shock, which we define to be at a time-dependent point $x_{L}<\\lambda\\left(t\\right) 0} \\tlnot I(s_i),$$\nand where equality is defined pointwise over the state vectors.\nWe can now strengthen the left hand side of the induction step\nentailment:\n$$\\bigwedge^{k-1}_{i=0} P(s_i) \\tland \\bigwedge ^{k-1}_{i=0}\nT(s_i,s_{i+1}) \\tland C_k(s_0, \\ldots, s_k) \\models P(s_k). $$\nNow, we can make the algorithm complete for bounded state spaces by\nchecking the entailment \n$$I(s_0) \\tland T(s_0,s_1) \\tland \\cdots \\tland T(s_{k-1}, s_{k})\n \\models \\tlnot C_k(s_0,\\ldots, s_{k-1})$$\nafter the $k$-th bounded model checking iteration. If this entailment\nholds, and $P$ is $k$-invariant, then $P$ is invariant. \n\n\\paragraph{Structural Abstraction}\nA limiting factor when it comes to applying $k-$induction in practice\nis the computing resources required by the SMT solver. Abstraction is\noften applied to the model the system being analyzed in order to\nreduce the problem space to a more manageable problem. {\\em\n Structural abstraction}~\\cite{ bh07structural} is a technique that\nconstructs a abstract model that simulates the concrete one and\naccepts at least the behaviors accepted by the concrete model. The\nabstraction can reduce the size of the problem\nspace and render $k-$induction tractable. This\napproach is \nrelated to counterexample guided refinement\n(CEGAR)~\\cite{Clarke2003CAR}, where if a counterexample is produced in\nthe analysis of the abstract model that is inconsistent with the\nconcrete model, the abstraction is refined to eliminate the\ncounterexample. The idea is to replace $T$\nby a weaker approximation $T'$ by removing some clauses. Then, \n\\begin{itemize}\n\\item If $P$ is invariant for $T',$ then $P$ is also invariant for $T.$\n\\item Otherwise, we use the counterexample produced by the SMT solver\n to refine $T'$ by restoring some well-chosen clauses.\n\\end{itemize} \nIndeed, suppose a counterexample $(v_0,\\ldots, v_k)$ is found for $T'$. Then we can\ncheck whether or not\n%$$\\bigwedge_{i=0}^k \\bigwedge_{j=1}^n v_{ij} = s_{ij} \\tland I(s_0)\n%\\tland \\bigwedge_{i=0}^{k-1} T(s_i,s_{i+1})$$ \n\n$$ I(v_0)\n\\tland \\bigwedge_{i=0}^{k-1} T(v_i,v_{i+1})$$ \n\n\\jonathan{I simplified the old formula, which I originally made too complex for subtle reasons (optimal use of incremental SMT solvers)}\n\nis satisfiable. If it is, then the counterexample is valid for $T$.\nOtherwise, the counterexample is spurious. In that case, it is possible to query\nthe SMT solver to obtain an unsatisfiable core of state variables that\ncandidates for refinement.\n\n\n\n\n\n\n\n%\\paragraph{IC3 Algorithm} \n\n%\\jonathan{Concerning this paragraph, I wonder if it's a good thing to give such a brief but formal description. I'm not sure anyone who don't already know the algorithm could understand more than the very basic idea just with this. In my case, it took me several much longer articles to understand the very subtle IC3 algorithm. Maybe we should give up trying to give a formal description, just give the main idea and a reference to a more detailed document in a shorter paragraph. What do you think ? I'm not sure I'm right, I'm really biased on this question.}\n\n%Bounded model checking approaches such as $k$-induction suffer from\n%the drawback of searching for a single strengthening of the invariant\n%to be proved. This monolithic approach often overwhelms the SAT\n%solver due to its many iterations. \\jonathan{Moreover, we should add that k-induction technique is limited to k-inductive properties.} \n\n\n\n\n%Recall that a property $P$ is inductive if \n%\\jonathan{I don't agree with your notations in this paragraph. Instead of :}\n%$$ P(s_{k-1}) \\tland T(s_{k-1},s_k) \\models P(s_k). $$\n%\\jonathan{I would have written $$ P(s) \\tland T(s, s') \\models P(s'). $$ Indeed, you notation suggests the last equation is about a path and not about a single transition. The use of indexes is misleading I think. What is your opinion about this ? }\n% Otherwise there\n%exists states $s_{k-1}$ and $s_k$ such that \n$$P(s_{k-1}) \\tland T(s_{k-1},s_k) \\tland \\tlnot P(s_k),$$\n%where $s_{k-1}$ is said to be the {\\em counterexample to\n% inductiveness} (CTI). The property $P$ is said to be inductive\n%relative to $A$ if and only if\n%$$A(s_{k-1}) \\tland P(s_{k-1}) \\tland T(s_{k-1},s_k) \\models P(s_k). $$\n%Note that if $P$ is inductive relative to $Q$ and $Q$ is inductive\n%relative to $P,$ then $P \\tland Q$ is inductive. The idea of using\n%relative inductiveness to prove inductiveness lead to the development\n%of incremental approaches such as that used in the FSIS model\n%checker~\\cite{Bradley06, Bradley2011}. The basic incremental\n%algorithm used in FSIS is as follows:\n%\\begin{itemize}\n%\\item Try to prove $P$ is inductive.\n%\\item If fails, if $s$ is CTI, then find an inductive invariant\n % $\\phi_0$ such that $\\phi_0(s)$ does not hold. \n%\\item Continue finding such lemmas $\\phi_0, \\phi_1, \\ldots$ until\n% there are no more CTIs. \n%\\item As $P$ is inductive to $\\bigwedge_{i} \\phi_i$ it s proven\n% invariant if it holds for the initial states.\n%\\end{itemize} \n%Note that each $\\phi_i$ has only to be proven inductive relative to\n%the previously discovered CTIs $\\phi_0, \\phi_1, \\ldots, \\phi_{i-1}$\n%or even inductive to $\\{ \\phi_j \\}_{j 0. \n%\\end{eqnarray*}\n %If the sequence satisfies these four constraints and admits a fixed\n %point, then $P$ is invariant. In constructing the sequence, $R_0$\n %is initialized to $\\{ I \\}$ and $R_1 = P.$ We continue in similar\n %fashion, at each step, $k,$ is initialized to $P$ then propagate as\n% many lemmas from the previous frame as possible while ensuring that\n %the an error is not reached within one transition step. \n\n%Just as $R$ represents the reachable good states, the IC3 algorithm\n%maintains a set of bad states $B$ such that each state in $B$ leads to\n%an error in one step. In order to show $B$ is unreachable in $R_k$\n%the predicate $\\tlnot B$ is conjoined to $R_k$ as a lemma. Although\n%$B$ may not be inductive, it is sufficient to show that it is\n%inductive relative to the previous frame and consequently $B$ is not\n%$k-$reachable. This can be accomplished by checking the following two\n%entailments:\n%\\begin{eqnarray*}\n%I(s_{i-1}) & \\models & \\tlnot B(s_{i-1}) \\\\ \n%R_{k-1} (s_{i-1}) \\tland \\tlnot B(s_{i-1}) \\tland T(s_{i-1}, s_i) \n%& \\models & \\tlnot B(s_i).\n%\\end{eqnarray*}\n%If the first entailment fails, then $P$ is not an invariant. If the\n%second fails, then $B$ is reachable from some state of the previous\n%frame. We generalize this state to a set of states $B',$ consisting of\n%states belonging to the previous frame that lead to $B$ in a single\n%transition. Next, the algorithm recursively refines $R_{k-1}$ to\n%remove the states in $B'.$ This backward reachability is performed\n%until one of the following conditions are met:\n%\\begin{itemize}\n%\\item The algorithm reaches $R_0 = I$ whence $P$ is not invariant. \n%\\item All the bad states that are recursively generated have been\n % shown to be $k$-unreachable. \n%\\end{itemize} \n\n%Once the strengthening process completes, meaning that no error state\n%is reachable from the last frame, the IC3 algorithm propagates the new\n%lemmas we discovered from each frame to its successor. This is\n%accomplished by verifying the entailment\n\n%$$ R_j(s_{i-1}) \\tland T(s_{i-1}, s_i) \\models C(s_i)$$\n%for all $C \\in R_j - R_{j+1},$ where $j$ increases from $0$ to $k.$\n%If all the lemmas of $R_{k-1}$ are successfully propagated to $R_k$,\n%the sequence $(R_j)$ reaches a fixed point and $P$ is proven to be\n%invariant. Otherwise, a new frame $R_{k+1}$ is added that is initialized to $P$\n%and the algorithm continues. Numerous improvements to the basic\n%algorithm such as {\\em lemma tightening} can be found in the\n%literature that can produce an invariant that is more tractable to\n%modern SMT solvers. \n\n\n\n", "meta": {"hexsha": "36f9e49fc55d60222ed0b9c4ef8e0b2a7f0fbd81", "size": 15048, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "co-kind-report/mc-kind.tex", "max_stars_repo_name": "Copilot-Language/copilot-discussion", "max_stars_repo_head_hexsha": "caccad918b23dae991095344a845827ddccd6047", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2015-06-10T00:44:21.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-17T13:20:09.000Z", "max_issues_repo_path": "co-kind-report/mc-kind.tex", "max_issues_repo_name": "Copilot-Language/copilot-discussion", "max_issues_repo_head_hexsha": "caccad918b23dae991095344a845827ddccd6047", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 30, "max_issues_repo_issues_event_min_datetime": "2019-04-01T20:24:19.000Z", "max_issues_repo_issues_event_max_datetime": "2021-09-07T22:34:17.000Z", "max_forks_repo_path": "co-kind-report/mc-kind.tex", "max_forks_repo_name": "Copilot-Language/copilot-discussion", "max_forks_repo_head_hexsha": "caccad918b23dae991095344a845827ddccd6047", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 51.8896551724, "max_line_length": 563, "alphanum_fraction": 0.7395002658, "num_tokens": 4402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8558511359371249, "lm_q2_score": 0.7154239897159438, "lm_q1q2_score": 0.6122964342750604}} {"text": "\\subsection{Decomposition of T}\n\\SecLabel{t}\n\nThe decomposition of $T$ follows similar path as that of $U$, but with a couple of differences. all permutations $T_i$ are linearly-equivalent, excluding $T_{\\hex{0}}$. The latter function stands out and does not follow any patterns. Furthermore, all $T_i,i \\ne 0$ are related only by linear layer in the input. Let $N_{\\hex{1}},\\ldots,N_{\\hex{f}}\\colon \\field{4} \\to \\field{4}$ be given by\n$$\nN_k(x) \\eqdef T_k^{-1}(x) \\circ T_{\\hex{1}}(x).\n$$\nFor any $k \\ne 0$, $T_k = T_{\\hex{1}} \\circ N_k$ and each of $N_i$ is affine. The look-up table of $N$ is given in \\TabRef{n}.\n\nIn order to obtain linear mapping from affine, we detach the constant xor \\emph{before} the linear map. It could also be detached \\emph{after}, but detaching \\emph{before} allows to merge it with the outer linear layer $L$. Let $\\delta\\colon \\field{4} \\to \\field{4}$ be given by\n$$\n\\delta(k) \\eqdef \\begin{cases}\n 0, &~\\text{if}~k = 0,\\\\\n N_k^{-1}(0), &~\\text{otherwise}.\n\\end{cases}\n$$\nIt turns out that $\\delta$ is a linear map:\n$$\n\\delta(k_1, k_2, k_3, k_4) = (0, k_1 \\oplus k_3, 0, k_1 \\oplus k_2 \\oplus k_3).\n$$\n\nLet $N'_{\\hex{1}},\\ldots,N'_{\\hex{f}}\\colon \\field{4} \\to \\field{4}$ be given by\n$$\nN'_k(x) = N_k(x \\oplus \\delta(k)).\n$$\nThen all $N'_k$ are linear functions, i.e. $N'_k(0) = 0$ for all $k \\in \\field{4},k\\ne 0$. The codebook of $N'$ is given in \\TabRef{np}.\n\n\\FigTex{n-np.tex}\n\nConsider $N'_{\\hex{2}}$ (other choices are possible, but this one leads to simplest linear layers in the decomposition). It is linear-similar to the same field multiplication chosen in the decomposition of $U$: there exists $\\eta \\in \\linbij{4}$ such that\n$$\nN'_{\\hex{2}} = \\eta \\circ (\\cdot~\\odot X) \\circ \\eta^{-1}.\n$$\nSuch $\\eta$ is given by:\n$$\n\\eta(x_1, x_2, x_3, x_4) \\eqdef (x_1, x_2 \\oplus x_4, x_3 \\oplus x_2, x_4).\n$$\nFurther, all $N'_k$ turn out to be multiplications by a $k$-dependent constant in the finite field. Let $\\varepsilon\\colon \\field{4} \\to \\field{4}$ be given by\n$$\n\\varepsilon(k) \\eqdef \\begin{cases}\n0, & ~\\text{if}~ k = 0,\\\\\n\\eta^{-1} \\circ N'_{k} \\circ \\eta (1), & ~\\text{otherwise}.\n\\end{cases}\n$$\nThen $\\varepsilon$ turns out to be a bijection, and the following holds for $k \\ne 0$:\n$$\n\\eta^{-1} \\circ N'_{k} \\circ \\eta (x) = \\varepsilon(k) \\odot x,\n$$\n$\\varepsilon$ seems to be a complicated permutation without any pattern. Note that when the inverse of $T$ is computed, the field multiplication becomes the field division and thus, the output of $\\varepsilon$ is inverted. Denote the composition of the inversion in the finite field with $\\varepsilon$ by $1/\\varepsilon$ (defining $1/0 = 0$). Surprisingly, it is a linear function. Furthermore, when composed with $\\swaplsb$ which appears here from the decomposition of $U$, it becomes a simple multiplication by a constant in the finite field:\n$$\n1/\\varepsilon \\circ \\swaplsb(k) = k \\odot (X^3 + X^2)\n$$\nfor all $k \\in \\field{4}$.\nIt follows that\n$$\n\\varepsilon(k) = 1/(\\swaplsb(x) \\odot (X^3 + X^2)) = X \\odot (1 / \\swaplsb(k)),\n$$\nwhere $X \\odot (X^3 + X^2) = 1$ in the chosen finite field. The multiplication by constant can be transferred through the main multiplication in $N'$ and merged with $T_{\\hex{1}}$.\n\nWe obtain that, when $k \\ne 0$, $T_k$ can be computed as:\n$$\nT_k(x) = T_{\\hex{1}} \\circ \\eta (\\cdot~\\odot X) \\pround{\n(1/\\swaplsb(k)) \\odot \\eta^{-1} (x \\oplus \\delta(k))\n}.\n$$\nThe addition of $\\delta(k)$, $\\eta^{-1}$ and $\\swaplsb$ can be merged with the outer linear layer $L$. $\\eta$ can be merged with $T_{\\hex{1}}$, and $\\swaplsb$ will cancel out when $T$ is merged with $U$.\n\nLet\n\\begin{align*}\n\\zeta_0 &\\eqdef T_{\\hex{0}} \\circ \\eta,\\\\\n\\zeta_+ &\\eqdef T_{\\hex{1}} \\circ \\eta \\circ (\\odot X).\n\\end{align*}\n\nThen $$\nT_k(x) = \\begin{cases}\n\\zeta_+ \\pround{\n(1/\\swaplsb(k)) \\odot \\eta^{-1} (x \\oplus \\delta(k))\n},&~\\text{if}~k\\ne 0,\\\\\n\\zeta_0\\pround {\n\\eta^{-1} (x \\oplus \\delta(k))\n},&~\\text{if}~k = 0.\n\\end{cases}\n$$\n\nThe final decomposition of $T^{-1}$ and the codebooks of $\\delta, \\eta, \\swaplsb, \\zeta_0, \\zeta_+$ and the field inverse $1/x$ are given are shown in \\FigRef{final-t}. It uses a multiplexer, which chooses its left input branch if the control branch (i.e., $k$) is equal to zero, and its right input branch otherwise.\nThe only differences between $T$ and $T^{-1}$ are using the inverses of $\\zeta_0, \\zeta+$, removing the field inversion, and changing the position of the multiplexer. Note that $\\eta$ is an involution, and the addition of $\\delta(k)$ is involution too. \n\n\\FigTex{final-t.tex}", "meta": {"hexsha": "9dcfeeacf6996eae66a5cf51d9e03d93e6651c91", "size": 4539, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis-source/9strKuz/11t.tex", "max_stars_repo_name": "hellman/thesis", "max_stars_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 19, "max_stars_repo_stars_event_min_datetime": "2019-05-16T19:55:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-31T15:36:12.000Z", "max_issues_repo_path": "thesis-source/9strKuz/11t.tex", "max_issues_repo_name": "hellman/thesis", "max_issues_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-08-09T11:26:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-09T11:26:45.000Z", "max_forks_repo_path": "thesis-source/9strKuz/11t.tex", "max_forks_repo_name": "hellman/thesis", "max_forks_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-05T19:40:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-05T19:40:16.000Z", "avg_line_length": 51.5795454545, "max_line_length": 544, "alphanum_fraction": 0.6664463538, "num_tokens": 1627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8596637433190939, "lm_q2_score": 0.7122321964553656, "lm_q1q2_score": 0.6122801961171999}} {"text": "\\documentclass[12pt]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage{amsthm}\n\\usepackage{hyperref}\n\\usepackage[pdftex]{graphicx}\n%\\usepackage{physymb}\n%\\usepackage{wrapfig}\n\\usepackage{braket}\n\\usepackage{subcaption}\n\\title{PH 304: Assignment 1}\n\n\\author{Manish Goregaokar (120260006)}\n\\date{January 25, 2015}\n\\begin{document}\n\\maketitle\n\\section*{Problem 1}\n\nThe probability of there not being a single 1 in four rolls is $\\frac56\\times\\frac56\\times\\frac56\\times\\frac56 = \\frac{625}{1296}$ since we restrict each die roll to the space of 2-6 (5 possibilities) out of its actual space of 6.\n\nThe inverse of this, i.e. the probability of there being at least one 1 in four rolls, is $\\boxed{1 - \\frac{625}{1296} = \\frac{671}{1296} = 0.517747}\\hfill \\text{Ans (i)}$\n\nWhen a pair of dice are rolled 24 times, the probability of never getting a double 1 is $\\left(\\frac{35}{36}\\right)^{24}$ since the (equiprobable) sample space for a single roll has 36 candidates, of which one is a double-1. Thus the probability of having at least one double 1 is $\\boxed{1-\\left(\\frac{35}{36}\\right)^{24} = 0.491404}\\hfill\\text{Ans (ii)}$\n\\section*{Problem 2}\nLet the probability for a single ameoba to die or have its progeny eventually die out be $d$.\n\n\\begin{align*}\nd & = P(\\text{amoeba dies}) + P(\\text{amoeba lives but descendants die})\\\\\n&= 1-p + p(P(\\text{amoeba or descendants die out})^2)\\\\\n&= 1-p + pd^2\\\\\n\\therefore d &= 1-p + pd^2\\\\\n\\therefore 0 &=pd^2 -d +1-p\\\\\n\\therefore d &= \\frac{1\\pm \\sqrt{1-4p(1-p)}}{2p}\\\\\n &= \\frac{1\\pm \\sqrt{4p^2 - 4p + 1}}{2p}\\\\\n &= \\frac{1\\pm \\sqrt{(2p - 1)^2}}{2p}\\\\\n &= \\frac{1\\pm (2p -1)}{2p}\n\\end{align*}\n\nFor $p<\\frac12$, the solution with the $+$ is the only possible one, and thus $d=1$. For $p>\\frac12$, it is the solution with the $-$, and we get the probability of death as $\\frac{1-p}{p}$.\n\nThus, the minimum value of $p$ for there to be a non-zero probability of survival ($d<1$) is $\\boxed{\\frac12}$.\n\nFor $p=3/4$, $d=\\frac13$, so the probability of survival is $1-d = \\boxed{\\frac23}$\n\\section*{Problem 3}\nGiven: $P(\\text{twin}) = 0.02, P(\\text{identical twin}) = 0.002$\n\nWe want $P(\\text{identical}|\\text{twin})$, which by Bayes' theorem is $$\\frac{P(\\text{twin}|\\text{identical})P(\\text{identical})}{P(\\text{twin})}=\\frac{P(\\text{identical})}{P(\\text{twin})}=0.1$$\n\nThus the probability that he was an identical twin is $\\boxed{0.1}\\hfill \\text{Ans.}$\n\n\\section*{Problem 4}\n\nThe distribution is $P(n|N) = {n\\choose N}p^n(1-p)^{N-n}$, and thus the characteristic function is \n\\begin{align*}\\tilde{P}(k) &= \\Braket{e^{-ikn}}\\\\\n&= \\sum_{n=0}^N e^{-ikn}{n\\choose N}p^n(1-p)^{N-n}\\\\\n&= \\sum_{n=0}^N {n\\choose N}\\left(e^{-ik}\\cdot p\\right)^n(1-p)^{N-n}\\\\\n&= (1 - p(e^{-ik}-1))^N\n\\end{align*}\n\n$\\therefore \\boxed{\\tilde{P}(k) = (1 - p(e^{-ik}-1))^N}\\hfill \\text{Ans.}$\n\nCumulant generating function\n\n\\begin{align*}\nQ(k) & = \\ln\\tilde{P}(k)\\\\\n&= \\ln (1 - p(e^{-ik}-1))^N\\\\\n&= N \\ln (1 - p(e^{-ik}-1))\n\\end{align*}\n\n$\\therefore \\boxed{\\ln\\tilde{P}(k) = N\\ln (1 - p(e^{-ik}-1))}\\hfill \\text{Ans.}$\n~\\\\\nIf $-ik = y$, then $Q(k) = \\ln\\tilde{P}(k) = \\sum_n \\frac{y^n}{n!}\\Braket{x^n}_C$. This means that $\\left.Q^{(j)}(k)\\right|_{k=0} = \\Braket{x^j}_C$. Thus the first moment is $Q(0) = 0$, and the second moment is calculated as:\n\n\\begin{align*}\nQ'(0) &= \\left.\\frac{d}{dk}N\\ln (1 - p(e^{-ik}-1))\\right|_{k=0}\\\\\n&= \\left.\\frac{N p e^y}{1-p \\left(e^y-1\\right)}\\right|_{k=0}\\\\\n&= Np\n\\end{align*}\n\nThus the first two cumulants are $\\boxed{0}$ and $\\boxed{Np}\\hfill$ Ans.\n\\section*{Problem 5}\n$$p(x) = \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{(x-\\lambda)^2}{2\\sigma^2}\\right]$$\n\n\\begin{align*}\n\\therefore \\tilde{p}(k) &= \\int_{-\\infty}^{\\infty} e^{-ikx} \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{(x-\\lambda)^2}{2\\sigma^2}\\right] dx\\\\\n&=\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{(x-\\lambda)^2 + 2ikx\\sigma^2}{2\\sigma^2}\\right] dx\\\\\n&=\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{x^2 + \\lambda^2 - 2x\\lambda + 2ikx\\sigma^2}{2\\sigma^2}\\right] dx\\\\\n&=\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{x^2 - 2x(\\lambda - ik\\sigma^2) + \\lambda^2 }{2\\sigma^2}\\right] dx\\\\\n&=\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{x^2 - 2x(\\lambda - ik\\sigma^2) + (\\lambda - ik\\sigma^2)^2 -2\\lambda ik\\sigma^2 +i^2k^2\\sigma^4 }{2\\sigma^2}\\right] dx\\\\\n&=\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{(x -(\\lambda - ik\\sigma^2))^2 -2\\lambda ik\\sigma^2 +i^2k^2\\sigma^4 }{2\\sigma^2}\\right] dx\\\\\n&=e^{-\\lambda ik + \\frac12 i^2k^2\\sigma^2}\\int \\frac1{\\sqrt{2\\pi\\sigma^2}}\\exp\\left[-\\frac{(x -(\\lambda - ik\\sigma^2))^2}{2\\sigma^2}\\right] dx\\\\\n&= e^{-\\lambda ik +\\frac12 i^2k^2\\sigma^2} \\cdot 1\\\\\n&= e^{-\\lambda ik +\\frac12 i^2k^2\\sigma^2}\n\\end{align*}\n\nThus cumulant generating function is just $Q(k) = \\ln \\tilde{p}(k) = -\\lambda ik + \\frac12 i^2k^2\\sigma^2$. If $-ik = y$, $Q(y) = y + \\frac12 y^2\\sigma^2$\n\nThus the cumulants are $\\Braket{x^n}_C = (y + \\frac12 y^2\\sigma^2)^{(n)}$, giving us :\n\n$\\boxed{\\Braket{x}_C = 1}$, $\\boxed{\\Braket{x^2}_C = \\frac12\\sigma^2}$, and for $n>2$, $\\boxed{\\Braket{x^n}_C = 0}$\n\n\\section*{Problem 6}\n\nIf we break T into $n$ intervals of $dt$ each ($dt = T/n$), probability of exactly $M$ of these having an event is (where $\\lambda$ is the proportionality constant such that individual probability is $\\lambda dt$):\n\n$$P(M)_n = {n\\choose M}(\\lambda dt)^M (1-\\lambda dt)^{n-M}$$\n\nAssuming no simultaneous events, we can limit $n$ to infinity:\n\\begin{align*}\n\\lim_{n\\to\\infty} P(M)_n &= \\lim_{n\\to\\infty} {n\\choose M}(\\lambda dt)^M (1-\\lambda dt)^{n-M}\\\\\n&= \\lim_{n\\to\\infty} \\frac{n(n-1)....(n-m+2)(n-m+1)}{M!}\\left(\\lambda \\frac{T}{n}\\right)^M \\left(1-\\lambda \\frac{T}{n}\\right)^{n-M}\\\\\n&= \\lim_{n\\to\\infty} \\frac{n(n-1)....(n-m+2)(n-m+1)}{n^M}\\frac{\\left(\\lambda T\\right)^M}{M!} \\left(1-\\lambda \\frac{T}{n}\\right)^n\\left(1-\\lambda \\frac{T}{n}\\right)^{-M}\\\\\n&=1\\cdot \\frac{\\left(\\lambda T\\right)^M}{M!} \\cdot e^{-\\lambda T} \\cdot 1\\\\\n&= \\frac{k^M e^{-k}}{M!}\\qquad\\qquad (k = T\\lambda)\n\\end{align*}\n\n\nThus, the probability distribution, in terms of the proportionality constant $\\lambda$ is:\n\n$\\boxed{P(M) = \\frac{(T\\lambda)^M e^{-T\\lambda}}{M!}}$\n\nThe characteristic function can be calculated as (rewriting $T\\lambda$ as $\\lambda$)\n\\begin{align*}\n\\tilde{P}(k) &= \\sum_{M=0}^\\infty \\frac{\\lambda^M e^{-\\lambda}}{M!}e^{-ikM}\\\\\n&= e^{-\\lambda}\\sum_{M=0}^\\infty \\frac{e^{(ln\\lambda-ik) M}}{M!}\\\\\n&= \\exp{e^{ln\\lambda-ik}-\\lambda}\n\\end{align*}\n\nThus the cumulant generating function is $e^{ln\\lambda-ik}-\\lambda = \\lambda (e^{y} - 1)$, giving us $\\boxed{\\Braket{x^n}_C = \\lambda}$.\n\\end{document}", "meta": {"hexsha": "90335493e21ca6a16ab64078d301d2ac3c30363b", "size": 6574, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Course material/PH 304 - Statistical Physics/Assignment 1/assign1.tex", "max_stars_repo_name": "CourseResources/CourseResources", "max_stars_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2015-05-28T05:59:31.000Z", "max_stars_repo_stars_event_max_datetime": "2015-05-28T05:59:31.000Z", "max_issues_repo_path": "Course material/PH 304 - Statistical Physics/Assignment 1/assign1.tex", "max_issues_repo_name": "CourseResources/CourseResources", "max_issues_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Course material/PH 304 - Statistical Physics/Assignment 1/assign1.tex", "max_forks_repo_name": "CourseResources/CourseResources", "max_forks_repo_head_hexsha": "4040bfe499609389d1978823e4e2896bf4ce41e3", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 50.1832061069, "max_line_length": 357, "alphanum_fraction": 0.6282324308, "num_tokens": 2703, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7122321720225278, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.6122801815150942}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath, amssymb}\n\\usepackage[nodayofweek]{datetime}\n\n% Set size of text area with total parameter\n\\usepackage[a4paper, total={155mm, 255mm}]{geometry}\n\n\\title{Polynomial Root Manipulation}\n\\author{Dyson}\n\\date{\\today}\n\n\\newcommand{\\ab}{\\alpha\\beta}\n\\newcommand{\\bg}{\\beta\\gamma}\n\\newcommand{\\ga}{\\gamma\\alpha}\n\\newcommand{\\ag}{\\alpha\\gamma}\n\n\\newcommand{\\sone}{\\alpha + \\beta + \\gamma}\n\\newcommand{\\stwo}{\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha}\n\\newcommand{\\abg}{\\alpha\\beta\\gamma}\n\n\\newcommand{\\Sone}{\\Sigma_1}\n\\newcommand{\\Stwo}{\\Sigma_2}\n\n\\newtheorem{lemma}{Lemma}\n\\newtheorem{sublemma}{Lemma}[lemma]\n\\newtheorem{conjecture}{Conjecture}\n\n\\begin{document}\n\n\\maketitle\n\n% Set paragraph spacing here to avoid messing with title\n\\setlength{\\parindent}{0em}\n\\setlength{\\parskip}{1em}\n\n\\section{The Question}\n\nChallenge question 2 in the mixed exercise at the end of chapter 4 of the Further Maths Core book says:\n\nThe cubic equation $x^3 + 2x^2 -3x - 5 = 0$ has roots $\\alpha$, $\\beta$, and $\\gamma$. Without solving the equation, find an equation that has roots, $\\alpha + \\beta$, $\\beta + \\gamma$, and $\\gamma + \\alpha$.\n\nThis question is completely different to all the prior questions in the chapter. Most of these questions ask for a linear transformation of roots that can be done with simple substitution, but this question requires a full expansion of triple brackets, lots of algebra, and some very interesting problem solving at the end.\n\n\\section{The Answer}\n\n\\subsection{Establishing Lemmas}\n\nFirstly, we want to establish some lemmas. All of these can be shown by expanding $(x - \\alpha)(x - \\beta)(x - \\gamma)$, but we'll just take them as identities here.\n\nFor a cubic equation $ax^3 + bx^2 + cx + d = 0$, for $x \\in \\mathbb{C}$ and $a, b, c, d \\in \\mathbb{R}$ with roots $\\alpha$, $\\beta$, $\\gamma$:\n\n\\begin{lemma}\n$\\sone \\equiv -\\dfrac{b}{a}$\n\\end{lemma}\n\n\\begin{lemma}\n$\\stwo \\equiv \\dfrac{c}{a}$\n\\end{lemma}\n\n\\begin{lemma}\n$\\abg \\equiv -\\dfrac{d}{a}$\n\\end{lemma}\n\nWe can express these with a more compact notation, $\\sone = \\Sone$, $\\stwo = \\Stwo$, representing the sum of single terms and the sum of pairs respectively.\n\nWe can evaluate these for our original equation and get $\\Sone = -2$, $\\Stwo = -3$, and $\\abg = 5$.\n\n\\subsection{Expanding The New Roots}\n\nWe want roots of $\\alpha + \\beta$, $\\beta + \\gamma$, and $\\gamma + \\alpha$, so we have the expression $(x - \\alpha - \\beta)(x - \\beta - \\gamma)(x - \\gamma - \\alpha)$. This can be expanded as such:\n\\begin{gather*}\n(x - \\alpha - \\beta)(x - \\beta - \\gamma)(x - \\gamma - \\alpha)\\\\[0.5em]\n= (x - \\alpha - \\beta)(x^2 - \\gamma x - \\alpha x - \\beta x + \\bg + \\ab - \\gamma x + \\gamma^2 + \\ag)\\\\[0.5em]\n= (x - \\alpha - \\beta)(x^2 - x\\Sone + \\Stwo - \\gamma x + \\gamma^2)\\\\[0.5em]\n= x^3 - x^2\\Sone + x\\Stwo - \\gamma x^2 + \\gamma^2 x - \\alpha x^2 + \\alpha x\\Sone - \\alpha \\Stwo + \\ag x - \\ag^2 - \\beta x^2 + \\beta x\\Sone - \\beta\\Stwo + \\bg x - \\bg^2\n\\end{gather*}\n\nAfter expanding the expression, we need to reorganise and simplify it.\n\\begin{gather*}\nx^3 - x^2\\Sone + \\alpha x\\Sone + \\beta x\\Sone+ x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\gamma x^2 - \\alpha x^2 - \\beta x^2 + \\gamma^2 x + \\ag x + \\bg x - \\ag^2 - \\bg^2\\\\[0.5em]\n= x^3 - x^2\\Sone + \\alpha x\\Sone + \\beta x\\Sone+ x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - x^2(\\sone) + \\gamma x(\\sone) - \\ag^2 - \\bg^2\\\\[0.5em]\n= x^3 - x^2\\Sone + \\alpha x\\Sone + \\beta x\\Sone + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - x^2\\Sone + \\gamma x\\Sone - \\ag^2 - \\bg^2\\\\[0.5em]\n= x^3 - 2x^2\\Sone + (\\sone)x\\Sone + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\ag^2 - \\bg^2\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\ag^2 - \\bg^2\n\\end{gather*}\n\nIt was at this point that I got stuck for about 15 minutes. If we had a $-\\abg$ term at the end, we could factor it out and do\n\\begin{gather*}\nx^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\ag^2 - \\bg^2 - \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\ag^2 + \\bg^2 + \\abg)\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\gamma(\\ag + \\bg + \\ab))\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\gamma\\Stwo)\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -\\Stwo(\\alpha + \\beta + \\gamma)\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\Sone\\Stwo\n\\end{gather*}\n\nWe can then evaluate this and get a definitive equation to answer the original question. But, how do we get that $- \\abg$ term at the end?\n\nI spent ages just staring at my book, trying to work out what I'd done wrong and where I'd dropped this term. After thoroughly checking and making sure that I'd expanded everything correctly, I decided to just add the term myself.\n\n\\subsection{Adding New Terms}\n\nWe know that adding and subtracting a constant from an expression doesn't change its value. So, for any expression $E$, and any constant $k$, $E + k - k = E$.\n\nSo, we can add the $-\\abg$ term and as long as we also add an extra $+\\abg$ term, the value of the expression won't actually change. And we know the value of $\\abg$ for our original equation, so we can evaluate it after the simplification.\n\nThus,\n\\begin{gather*}\nx^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\ag^2 - \\bg^2\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\alpha\\Stwo - \\beta\\Stwo - \\ag^2 - \\bg^2 - \\abg + \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\ag^2 + \\bg^2 + \\abg) + \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\gamma(\\ag + \\bg + \\ab)) + \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -1(\\alpha\\Stwo + \\beta\\Stwo + \\gamma\\Stwo) + \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo -\\Stwo(\\alpha + \\beta + \\gamma) + \\abg\\\\[0.5em]\n= x^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\Sone\\Stwo + \\abg\n\\end{gather*}\n\nWe can then evaluate this expression with the values for $\\Sone$, $\\Stwo$, and $\\abg$ that we obtained from the coefficients of our original equation.\n\\begin{gather*}\nx^3 - 2x^2\\Sone + x\\Sone^2 + x\\Stwo - \\Sone\\Stwo + \\abg\\\\[0.5em]\n= x^3 + 4x^2 + 4x - 3x - 6 + 5\\\\[0.5em]\n= x^3 + 4x^2 + x - 1\n\\end{gather*}\n\nTherefore, our final equation is $x^3 + 4x^2 + x - 1 = 0$. This is correct, as per the answers in the book.\n\n\\end{document}\n", "meta": {"hexsha": "16d2de47a68e5cf4c57c6c76bceab20d69b06b4d", "size": 6330, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Questions/Polynomial_Root_Manipulation.tex", "max_stars_repo_name": "DoctorDalek1963/LaTeX", "max_stars_repo_head_hexsha": "e91a79837bff80f9d361b921acb870a9fcfc3e0b", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Questions/Polynomial_Root_Manipulation.tex", "max_issues_repo_name": "DoctorDalek1963/LaTeX", "max_issues_repo_head_hexsha": "e91a79837bff80f9d361b921acb870a9fcfc3e0b", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Questions/Polynomial_Root_Manipulation.tex", "max_forks_repo_name": "DoctorDalek1963/LaTeX", "max_forks_repo_head_hexsha": "e91a79837bff80f9d361b921acb870a9fcfc3e0b", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 49.0697674419, "max_line_length": 323, "alphanum_fraction": 0.644549763, "num_tokens": 2435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7185943925708562, "lm_q2_score": 0.8519528000888387, "lm_q1q2_score": 0.6122085048788791}} {"text": "\\section{Sturm-Liouville Theory}\r\n\\subsection{Review of Second-Order Linear ODEs}\r\nFor a general inhomogeneous ODE $\\mathcal Ly=f(x)$ where\r\n$$\\mathcal Ly=\\alpha(x)\\frac{\\mathrm d^2y}{\\mathrm dx^2}+\\beta(x)\\frac{\\mathrm dy}{\\mathrm dx}+\\gamma(x)y$$\r\nIn general, the homogeneous equation $\\mathcal Ly=0$ has two linearly independent solutions $y_1,y_2$.\r\nThe complementary function $y_c(x)=Ay_1+By_2$ for constants $A,B$ is then the general solution to $\\mathcal Ly=0$ by linearity.\\\\\r\nIf we can find a particular solution (aka particular integral) $y_p$ to $\\mathcal Ly=f$, then $y_p+y_c=y_p+Ay_1+By_2$ for $A,B$ constants is the general solution to $\\mathcal Ly=f$ again by linearity.\r\nTwo pieces of boundary data is then needed to determine the constants $A,B$.\\\\\r\nThere are several types of boundary conditions.\r\nWe sometimes get the Dirichlet condition of specifying the function's value at the endpoints, or the Neumann consitions of specifying the derivative's values at the endpoints.\r\nSometimes these two types of conditions are mixed.\\\\\r\nThe sort of conditions we often consider are homogeneous conditions, i.e. the function vanishes at the endpoints.\r\nThe reason of it is that it allows the superposition of solutions in a linear DE.\r\nWhat if we come across a nonhomogeneous condition?\r\nWe can use the complementary solution to cancel stuff out.\\\\\r\nSometimes, we specify initial data of the function and its derivative as boundary conditions.\\\\\r\nAnother matter of interest is the general eigenvalue problem.\r\nTo solve $\\mathcal Ly=f$ using eigenvalue decompositions like we did previously, we must first solve (subject to boundary conditions) the related eigenvalue problem\r\n$$\\mathcal Ly=\\alpha(x)\\frac{\\mathrm d^2y}{\\mathrm dx^2}+\\beta(x)\\frac{\\mathrm dy}{\\mathrm dx}+\\gamma(x)y=-\\lambda\\rho(x)y$$\r\nwhere $\\rho$ is nonegative.\r\nThis form often occurs after seperation of variables in a PDE.\r\n\\subsection{Self-Adjoint Operators}\r\n\\begin{definition}\r\n For two functions $f,g:[a,b]\\to\\mathbb C$ we define their inner product to be\r\n $$\\langle f,g\\rangle=\\int_a^bf^*(x)g(x)\\,\\mathrm dx$$\r\n\\end{definition}\r\nWe can guarantee to rewrite the original eigenvalue problem into the Sturm-Liouville form, i.e. $\\mathcal Ly=\\lambda wy$ where we are able to rewrite $\\mathcal Ly=-(py^\\prime)^\\prime+qy$ and $w$ is a nonnegative wavefunction.\r\n\\footnote{The reason why there is a wavefunction there is just for convenience.}\r\nHow to convert a second order linear ODE to this form?\r\nSimply multiply the diffential equation by an integrating factor $F$ that will be specified later and we can write\r\n$$\\frac{\\mathrm d}{\\mathrm dx}(F\\alpha y^\\prime)-F^\\prime\\alpha y^\\prime-F\\alpha^\\prime y^\\prime+F\\beta y^\\prime+F\\gamma y=-\\lambda F\\rho y$$\r\nSo to eiminate the $y^\\prime$ terms, we set\r\n$$F(x)=\\exp\\left(\\int\\frac{\\beta-\\alpha^\\prime}{\\alpha}\\,\\mathrm dx\\right)$$\r\nwhich reduced the equation to\r\n$$(F\\alpha y^\\prime)^\\prime+F\\gamma y=-\\lambda F\\rho y$$\r\nSetting $p=F\\alpha,q=F\\gamma$ and $w=F\\rho\\ge 0$.\r\n\\begin{example}\r\n Consider the Hermite equation that appears in quantum mechanics\r\n $$y^{\\prime\\prime}-2xy^\\prime+2ny=0$$\r\n Then $\\alpha=1,\\beta=-2x,\\gamma=0,\\lambda\\rho=2n$, so the above procedure translates this to the Sturm-Liouville form\r\n $$\\mathcal L=(-e^{-x^2}y^\\prime)^\\prime=2ne^{-x^2}y$$\r\n\\end{example}\r\n\\begin{definition}\r\n Let $\\mathcal L:C\\to C$ be an operator, where $C$ on a class of functions $[a,b]\\to\\mathbb C$ equipped with the inner product we defined previously.\r\n This operator $\\mathcal L$ is self-adjoint if $\\langle y_1,\\mathcal Ly_2\\rangle=\\langle\\mathcal Ly_1,y_2\\rangle$ for any $y_1,y_2\\in C$.\r\n\\end{definition}\r\nIf we let $\\mathbb L$ be the operator in the Strum-Liouville form, then\r\n\\begin{align*}\r\n \\langle y_1,\\mathcal Ly_2\\rangle-\\langle\\mathcal Ly_1,y_2\\rangle&=\\int_a^b[-y_1(py_2^\\prime)^\\prime+y_1qy_2+y_2(py_1^\\prime)^\\prime-y_2qy_1]\\,\\mathrm dx\\\\\r\n &=\\int_a^b[-y_1(py_2^\\prime)^\\prime+y_2(py_1^\\prime)^\\prime]\\,\\mathrm dx\\\\\r\n &=\\int_a^b[-(y_1(py_2^\\prime)^\\prime+y_1^\\prime py_2^\\prime)+(y_2(py_1^\\prime)^\\prime+y_2^\\prime py_1^\\prime)]\\,\\mathrm dx\\\\\r\n &=\\int_a^b[-(py_1y_2^\\prime)^\\prime+(py_1^\\prime y_2)^\\prime]\\,\\mathrm dx\\\\\r\n &=[-py_1y_2^\\prime+py_1^\\prime y_2]_a^b\r\n\\end{align*}\r\nSo for this operator to be self-adjoint, we need some good enough boundary conditions so that enough stuff vanishes.\r\nThis includes homogeneous boundary condition $y(a)=y(b)=0$ or $y^\\prime(a)=y^\\prime(b)=0$ or mixed $y+ky^\\prime=0$ etc..\r\nWe say a Sturm-Liouville problem is regular if the boundary conditions are homogeneous.\r\nPeriodic boundary conditions also work, where we can take $y(a)=y(b)$ and the derivatives are specified (or periodic) at the boundary.\r\nThere can also be singular points of this ODE, where $p(a)=p(b)=0$.\r\nWe can have combinations of above too.\r\n\\subsection{Properties of Self-Adjoint Operators}\r\n\\begin{definition}\r\n The inner product of $y_1,y_2:[a,b]\\to\\mathbb C$ with respect to weight $w:[a,b]\\to\\mathbb R_{\\ge 0}$ is\r\n $$\\langle f,g\\rangle_w=\\int_a^bwf^*g\\,\\mathrm dx=\\langle wf,g\\rangle=\\langle f,wg\\rangle$$\r\n\\end{definition}\r\nAnalogous to the finite dimensional case, we have\r\n\\begin{theorem}\\label{self-adjoint}\r\n For a sufficiently nice self-adjoint operator $\\mathcal L$ on a sufficiently nice space of functions:\\\\\r\n (a) Eigenvalues of $\\mathcal L$ are real.\\\\\r\n (b) Eigenfunctions of it with different eigenvalues are orthogonal with respect to the weight $w$.\\\\\r\n (c) We can take the eigenfunctions as a set of basis for the function space, just like Fourier series.\r\n\\end{theorem}\r\n\\begin{proof}[Proof of (a)]\r\n 1. If $\\mathcal Ly=\\lambda wy$, taking complex conjugate gives $\\mathcal Ly^*=\\lambda^*wy^*$.\r\n Hence as $\\mathcal L$ is self-adjoint,\r\n $$0=\\int_a^b(y^*\\mathcal Ly-y\\mathcal Ly^*)\\,\\mathrm dx=(\\lambda-\\lambda^*)\\int_a^bw|y|^2\\,\\mathrm dx$$\r\n which means $\\lambda=\\lambda^*$, so $\\lambda$ is real.\r\n\\end{proof}\r\nIf $\\lambda$ is non-degenerate (simple), i.e. it has a one-dimensional eigenspace, then $y$ is guaranteed to be real.\r\nEven if it has dimension $2$ (not more because the ODE is second order), we can still find two real functions as basis of the eigenspace.\r\nAlso, by considering $u\\mathcal Lv-v\\mathcal Lu=(-p(uv^\\prime-u^\\prime v))^\\prime$, one can show that a regular Sturm-Liouville problem always has all eigenvalues simple.\r\n\\begin{proof}[Proof of (b)]\r\n Suppose $\\mathcal Ly_m=\\lambda_mwy_m$ and $\\mathcal Ly_n=\\lambda_nwy_n$, then\r\n $$0=\\int_a^by_n\\mathcal Ly_m-y_m\\mathcal Ly_n\\,\\mathrm dx=(\\lambda_m-\\lambda_n)\\int_a^bwy_ny_m\\,\\mathrm dx$$\r\n But $\\lambda_m$ and $\\lambda_n$ are distinct.\r\n The claim follows.\r\n\\end{proof}\r\nAs an aside, we do not really need the weight function in order to formulate Sturm-Liouville theory, since we can do the transformation $\\tilde{y}=\\sqrt{w}y$ and replace $\\mathcal Ly$ by $(1/\\sqrt{w})\\mathcal L(\\tilde{y}/\\sqrt{w})$.\r\nYet the analytic property is generally simpler if we keep $w$.\\\\\r\nWhat?\r\nHow about (c), you say?\r\nBold of you to assume we'll prove it.\r\nWe are just gonna take it (and several other properties we want it to have) as truth and do stuff with this idea.\r\n\\subsection{Eigenfunction Expansions}\r\nSo basically we just want to find an expansion\r\n$$f=\\sum_{n=1}^\\infty a_ny_n$$\r\nwhere $y_n$ is a set of eigenfunctions of some self-adjoint operator.\r\nTheorem \\ref{self-adjoint}(c) shows that we can do it.\r\nTo find the coefficients $a_n$, we can they use the orthogonality to get\r\n$$\\int_a^bwy_mf\\,\\mathrm dx=\\sum_{n=1}^\\infty a_n\\int_a^bwy_ny_m\\,\\mathrm dx=a_m\\int_a^bwy_n^2\\,\\mathrm dx$$\r\nSo\r\n$$a_n=\\left(\\int_a^bwy_nf\\,\\mathrm dx\\middle)\\right/\\left(\\int_a^bwy_n^2\\,\\mathrm dx\\right)$$\r\nIt's a common practice not to normalise the eigenfunctions as it is not really always clean.\r\nOf course, if we want, we can always write down\r\n$$Y_n=y_n\\left/\\sqrt{\\int_a^bwy_n^2\\,\\mathrm dx}\\right.$$\r\nSo we can get rid of the denominator in $a_n$ and the coefficients will have the expression\r\n$$A_n=\\int_a^bwy_nf\\,\\mathrm dx=a_n\\int_a^bwy_n^2\\,\\mathrm dx$$\r\nbut it isn't that useful and can cause some messiness.\r\n\\begin{example}\r\n Recall the particular operator already in Sturm-Liouville form $\\mathcal Ly=y^{\\prime\\prime}$, then (with appropriate boundary conditions) we can easily get the eigenvalues $\\lambda_n=(n\\pi/L)^2$ and eigenfunctions being the trigonometrics.\r\n This just reproduces the Fourier series.\r\n\\end{example}\r\n\\subsection{Completeness and Parseval's Identity}\r\nWe expand\r\n\\begin{align*}\r\n 0&=\\int_a^bw\\left( f(x)-\\sum_{n=1}^\\infty a_ny_n \\right)^2\\,\\mathrm dx\\\\\r\n &=\\int_a^bw\\left( f^2-2f\\sum_{n=1}^\\infty a_ny_n+\\sum_{n=1}^\\infty a_n^2y_n^2 \\right)\\,\\mathrm dx\\\\\r\n &=\\int_a^bwf^2\\,\\mathrm dx-\\sum_{n=1}^\\infty a_n^2\\int_a^bwy_n^2\\,\\mathrm dx\r\n\\end{align*}\r\nHence we have\r\n$$\\int_a^bwf^2\\,\\mathrm dx=\\sum_{n=1}^\\infty a_n^2\\int_a^bwy_n^2\\,\\mathrm dx=\\sum_{n=1}^\\infty A_n^2$$\r\nwhich Parseval's identity in this general case.\r\nEasily our previous Parseval's Theorem on Fourier series is a special case.\\\\\r\nIf some of the eigenfunctions are missing from the series, then this gives\r\n$$\\int_a^bwf^2\\,\\mathrm dx\\ge\\sum_{n=1}^\\infty A_n^2$$\r\nThis is known as Bessel's Inequality.\\\\\r\nConsider the partial sums $\\sum_{n\\le N}a_ny_n$, then we shall have $S_N\\to f$ as $N\\to\\infty$ where we would like the style of convergence to be\r\n$$\\epsilon_N=\\int_a^bw[f(x)-S_N(x)]^2\\,\\mathrm dx\\to 0,N\\to\\infty$$\r\nAn interesting question is that, while we know (maybe) the series converges as we want, if we truncate the sequence in some $N$, would the coefficients $\\{a_n\\}_{n\\le N}$ provide the best approxmation (with respect to the error defined in this way) of that particular partial sum, or a different set of partial coefficient will yield a better result?\r\nTo answer this, we evaluate\r\n$$\\frac{\\partial\\epsilon_N}{\\partial a_n}=-2\\int_a^bwy_n\\left( f-\\sum_{k=1}^Na_ky_k \\right)\\,\\mathrm dx=-2\\int_a^bwfy_n-a_nwy_n^2\\,\\mathrm dx$$\r\nwhich is zero when $a_n$ is of the expression we got earlier.\r\nWe can see it is indeed a minimum by observing that\r\n$$\\frac{\\partial^2\\epsilon_N}{\\partial a_n^2}=2\\int_a^bwy_n^2\\,\\mathrm dx\\ge 0$$\r\nThis answers our question.\r\n\\subsection{Legendre's Equation}\r\nTake the usual spherical polar coordinate\r\n$$\\begin{cases}\r\n x=r\\sin\\theta\\cos\\phi\\\\\r\n y=r\\sin\\theta\\sin\\phi\\\\\r\n z=r\\cos\\theta\r\n\\end{cases}$$\r\nwhere Laplace's equation $\\nabla^2 u=0$ translates to\r\n$$\\frac{1}{r^2}\\frac{\\partial}{\\partial r}\\left( r^2\\frac{\\partial u}{\\partial r} \\right)+\\frac{1}{r^2\\sin\\theta}\\frac{\\partial}{\\partial\\theta}\\left( \\sin\\theta\\frac{\\partial u}{\\partial\\theta} \\right)+\\frac{1}{r^2\\sin^2\\theta}\\frac{\\partial^2u}{\\partial\\phi^2}=0$$\r\nSeperation of variables $u=R(r)\\Theta(\\theta)\\Phi(\\phi)$ then gives\r\n$$\\frac{1}{\\sin\\theta}(\\Theta^\\prime\\sin\\theta)^\\prime+\\left( K-\\frac{m^2}{\\sin^2\\theta} \\right)\\Theta=0$$\r\nwhere $K,m$ are constants which essentially makes it an eigenvalue problem.\r\nNow the transformation $x=\\cos\\theta\\in[-1,1]$ and renaming $\\Theta$ as $y$ then gives Legendre's Equation\r\n$$(1-x^2)y^{\\prime\\prime}-2xy^\\prime+\\lambda y=0$$\r\nwhere $\\lambda$ is a constant which is again intepreted as an eigenvalue.\r\nThis is already in Strum-Liouville form by taking $p=1-x^2,q=0,w=1$.\r\nNow $p=1-x^2$ vanishes at the boundary $\\pm 1$, so this equation has to be self-adjoint.\r\nWe assume that $y$ is bounded near the boundary.\\\\\r\nWe now seek a power series solution to the problem.\r\nIf we set\r\n$$y=\\sum_{n=0}^\\infty c_nx^n$$\r\nThen substitution gives\r\n$$(n+2)(n+1)c_{n+2}-n(n-1)c_n-2nc_n+\\lambda c_n=0\\implies c_{n+2}=\\frac{n(n+1)-\\lambda}{(n+1)(n+2)}c_n$$\r\nThe iteration steps by $2$, so we get two linearly independent solutions\r\n\\begin{align*}\r\n y_{\\text{even}}&=c_0\\left( 1+\\frac{-\\lambda}{2!}x^2+\\frac{(6-\\lambda)(-\\lambda)}{4!}x^4+\\cdots \\right)\\\\\r\n y_{\\text{odd}}&=c_1\\left( x+\\frac{2-\\lambda}{3!}x^3+\\cdots \\right)\r\n\\end{align*}\r\nNote that $c_{n+2}/c_n\\to 1$, so the both series has radius of convergence $1$ but they diverges at $x=\\pm 1$.\r\nHowever, this is not the end of the world!\r\nThese series may not be infinite.\r\nIf $\\lambda=l(l+1)$ for some $l\\in\\mathbb N$, then one of these two series will terminate and give a polynomial solution.\r\nThese polynomials are called Legendre polynomials $P_l(x)$ which are eigenfunctions of the Legendre equation.\r\nConventionally we normalise $P_l$ by requiring $P_l(1)=1$.\r\nOne can check that this restricts $P_l([-1,1])\\subset [-1,1]$ and $|P_l(-1)|=1$.\r\nBy calculation we have\r\n$$P_0(x)=1,P_1(x)=x,P_2(x)=\\frac{3x^2-1}{2},P_3(x)=\\frac{5x^3-3x}{2},\\ldots$$\r\nWe easily observe that $P_l$ has $l$ roots in $[-1,1]$, also $P_l$ is odd if $l$ is odd, and even when $l$ is even.\r\nBy orthogonality and some calculation,\r\n$$\\forall n\\neq m,\\int_{-1}^1P_nP_m\\,\\mathrm dx=0,\\int_{-1}^1P_n^2\\,\\mathrm dx=\\frac{2}{2n+1}$$\r\nThere are several other ways to characterise Legendre polynomails.\r\nOne can prove that we have\r\n$$P_n(x)=\\frac{1}{2^nn!}\\frac{\\mathrm d^n}{\\mathrm dx^n}(x^2-1)^n,\\sum_{n=0}^\\infty P_n(x)t^n=\\frac{1}{\\sqrt{1-2xt+t^2}}$$\r\nWe also have the recursions\r\n$$l(l+1)P_{l+1}(x)=(2l+1)xP_l(x)-lP_{l-1}(x),(2l+1)P_l(x)=\\frac{\\mathrm d}{\\mathrm dx}(P_{l+1}(x)-P_{l-1}(x))$$\r\nIf we take these $P_l$ as a set of eigenfunctions, then any well-behaved $f$ on $[-1,1]$ can be expressed as\r\n$$f(x)=\\sum_{l=0}^\\infty a_lP_l(x),a_l=\\frac{2l+1}{2}\\int_{-1}^1f(x)P_l(x)\\,\\mathrm dx$$\r\n\\begin{example}\r\n We can verify that $f(x)=(15x^2-3)/2=P_0(x)+5P_2(x)$.\r\n\\end{example}\r\n\\begin{example}\r\n The odd equare wave with $f([0,1))=\\{1\\}$ has the expansion\r\n $$\\sum_{m=1}^\\infty (P_{2m}(0)-P_{2m+2}(0))P_{2m+1}(x)$$\r\n\\end{example}\r\n\\subsection{Inhomogeneous ODEs}\r\nConsider the ODE $\\mathcal Ly=f=wF$ (with homogeneous boundary conditions so that $\\mathcal L$ is self-adjoint) where $w\\ge 0$ is our wavefunction.\r\nGiven eigenfunctions $\\{y_n\\}$ satisfying $\\mathcal Ly_n=\\lambda_nwy_n$ for eigenvalues $\\{\\lambda_n\\}$.\r\nWe now try to find a solution in the form $y=\\sum_nc_ny_n$.\r\nTo do this, we expand $F=\\sum_na_ny_n$ where\r\n$$a_n=\\left( \\int_a^bwFy_n\\,\\mathrm dx\\middle) \\right/\\left( \\int_a^bwy_n^2\\,\\mathrm dx \\right)$$\r\nthen\r\n$$w\\sum_na_ny_n=wF=\\mathcal Ly=\\mathcal L\\sum_nc_ny_n=\\sum_nc_n\\mathcal Ly=w\\sum_nc_n\\lambda_ny_n$$\r\nSo take\r\n$$y=\\sum_n\\frac{a_n}{\\lambda_n}y_n$$\r\ngives a particular solution, assuming everything is well-defined and converges nicely enough.\\\\\r\nAn aside:\r\nThe driving force $F$ sometimes induces a linear response term $\\tilde\\lambda wy$, so the solution is $\\mathcal Ly-\\tilde{\\lambda}wy=f$.\r\nThen our particular solution can be\r\n$$y=\\sum_{\\lambda_n\\neq\\tilde{\\lambda}}\\frac{a_n}{\\lambda_n-\\tilde\\lambda}y_n$$\r\nNow back to theme.\r\nIf we expand the expression of $a_n$, we can get\r\n\\begin{align*}\r\n y(x)&=\\sum_{n=1}^\\infty\\frac{a_n}{\\lambda_n}y_n(x)\\\\\r\n &=\\sum_{n=1}^\\infty\\frac{y_n(x)}{\\lambda_nN_n}\\int_a^bw(\\xi)F(\\xi)y_n(\\xi)\\,\\mathrm d\\xi,N_n=\\int_a^bwy_n^2\\,\\mathrm dx\\\\\r\n &=\\int_a^b\\left( \\sum_{n=1}^\\infty\\frac{y_n(x)y_n(\\xi)}{\\lambda_nN_n} \\right)w(\\xi)F(\\xi)\\,\\mathrm d\\xi\\\\\r\n &=\\int_a^bG(x,\\xi)f(\\xi)\\,\\mathrm d\\xi,G(x,\\xi)=\\sum_{n=1}^\\infty\\frac{y_n(x)y_n(\\xi)}{\\lambda_nN_n}\r\n\\end{align*}\r\nThis $G(x,\\xi)$ is called the Green's function of this particular eigenvalue problem of that self-adjint operator.\r\nWorth noting that $G$ does not depend on the forcing term $f$.\r\nThe Green's function also induces\r\n$$\\mathcal L^-(\\phi)=\\int_a^bG(x,\\xi)\\phi(\\xi)\\,\\mathrm d\\xi$$\r\nwhich can be taken as kind of an inverse operator to $\\mathcal L$ since $\\mathcal L(\\mathcal L^-(f))=f$.", "meta": {"hexsha": "fd3d938fd70198fb810216ea4d646656f1a4ec79", "size": 15565, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "2/sturm.tex", "max_stars_repo_name": "david-bai-notes/IB-Methods", "max_stars_repo_head_hexsha": "b60135106d09d1e24d2f7b9c7e3eee1ca69f6907", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, 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YES\n2. YES", "lm_q1_score": 0.7185943805178139, "lm_q2_score": 0.8519528057272543, "lm_q1q2_score": 0.6122084986619898}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section{Comparing Inverse Problems and Solutions}\\label{sec:compare}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% The Bayesian framework poses a different question and seeks a different answer than the Data-Consistent one.\nThe objective of the Bayesian approach is to determine a single ``true'' parameter that explains all of the observed data \\citep{Smith, Concrete, Complete, Stuart10, Tarantola_book}.\nThe philosophical underpinnings of Bayesian inference is akin to asking the following:\n\n\\begin{center}\n \\emph{How does one incorporate collected data to shift prior beliefs about specific parameter values?}\n\\end{center}\n\n% The motivations of Bayesian inference are distinct from the Data-Consistent framework, where we seek a pull-back measure: a description of the uncertainty set that explains the variation in the observations under a given description of error.\nHowever, the Data-Consistent approach reflects the goal of constructing a distribution (or probability measure), which is akin to asking:\n\n\\begin{center}\n \\emph{How does one update initial beliefs in such a way that the modified predictions match the description of uncertainty in observed data?}\n\\end{center}\n\nWe now develop a typical Bayesian inverse problem following the framework described in \\cite{Stuart10, Smith}, in order to make a formal comparison to the Data-Consistent SIP.\n\n\\subsubsection{The Bayesian Inverse Problem}\nLet $d$ denote the ``noisy'' data obtained on $Q(\\paramref)$, which is often represented as\n\\begin{equation*}\n\td = Q(\\paramref) + \\xi,\n\\end{equation*}\nwhere $\\xi$ is a random variable used to model the measurement error that is often assumed to follow a Gaussian distribution.\nThen, the data-likelihood function, often written as a conditional density, $L_\\dspace(\\q \\, |\\, \\param)$, is formed.\nThis describes the differences in relative likelihoods that the data could have been generated from a particular $\\param$.\nIdeally, the largest values of $L_\\dspace(\\q \\, | \\, \\param)$ occur whenever $\\param$ is a close approximation of the true parameter $\\paramref$.\nThe data-likelihood function is distinct from the observed density used in the data-consistent framework.\n\nThe next step in constructing a Bayesian inverse problem is the specification of a prior density denoted by $\\pi_\\text{prior}(\\param)$.\nThe prior describes the different relative likelihoods assumed for the true parameter before data are collected.\nThis is also distinct from the role of the {\\em initial} density used in the data-consistent framework.\n% We choose them to represent the set of feasible parameters, and rely on Monte Carlo sampling for both approaches\\footnote{Priors in Bayesian inference are sometimes chosen for reasons related to Markov-Chain Monte Carlo algorithms in order to ensure their balancing of investigation and exploration, or convergence [TK - cite someone]}.\n\nThe posterior density (i.e., the formal solution to the Bayesian inverse problem) is given by a conditional density, denoted by $\\pi_\\text{post}(\\param\\, | \\, d)$, proportional to the product of the prior and data-likelihood function \\citep{Walpole, Berger, Complete, Smith}.\nIn other words,\n\\begin{equation*}\n\t\\posterior(\\param\\, | \\, \\q) \\propto \\prior(\\param)L_\\dspace(\\q\\, | \\, \\param)\n\\end{equation*}\nThis form of the density follows from Bayes' rule (not from the Disintegration Theorem as with the updated density).\nThe posterior can be interrogated to assess the difference in relative likelihoods of a fixed parameter given the observed data.\nSubsequently, the posterior is often used to produce a ``best'' estimate of the true parameter.\nFor example, the maximum a posteriori (MAP) point is the parameter that maximizes the posterior density.\n\nUnlike the updated density, the posterior is often interrogated indirectly without knowledge of the normalizing constant $C$ required to represent it as a density.\nIn other words, the posterior is a density of the form\n\n\\begin{equation}\\label{eq:sb_post}\n \\posterior\\lam := \\prior\\lam \\frac{L_\\dspace (\\q | \\param)}{ C },\n\\end{equation}\n\n\\noindent where we emphasize the use of $\\posterior$ to distinguish the \\emph{posterior} from the updated density $\\updated$ in \\eqref{eq:updated-pdf}.\nHere, the denominator $C$ (known as the \\emph{evidence} \\cite{Smith}), ensures the posterior density integrates to one, and is given by\n\\[\nC = \\int_\\pspace \\prior\\lam L_\\dspace(\\q | \\param) \\, d\\param.\n\\]\n\nNote that there are no constraints or requirements that likelihood function be a density.\nIn fact, $L_\\dspace$ need not even be in $L^1(\\pspace)$ since it is actually only the product $\\prior(\\param) L_\\dspace (\\q | \\param)$ that is required to be in $L^1(\\pspace)$ to form a posterior.\nIn other words, $L_\\dspace (\\q | \\param)$ and $\\observed(\\qlam)$ can model completely different things with respect to uncertainty in the data.\nAs stated above, $L_\\dspace (\\q | \\param)$ is used to assess the relative likelihood that a single parameter $\\param\\in\\pspace$ explains all of the observed data, whereas $\\observed(\\qlam)$ describes the relative likelihood of a predicted datum associated with $\\param\\in\\pspace$.\nIn the Bayesian framework, there is a different notion of consistency of solutions, referring to certain asymptotic properties of $\\posterior$ in the limit of infinite data \\citep{Barron, Silverman, starktenorio}.\n\n\n\n%%%%%%%%%\n\n\\subsection{Comparison of Data-Consistent and Bayesian Inversion}\\label{sec:bayesian}\n\nWe summarize the posterior and updated densities side-by-side in Table~\\ref{tab:dens_comparisons} and comment on a few notable aspects not mentioned above.\nObserve for the posterior density that the data-likelihood function appears in both the numerator and denominator.\nIn particular, the data-likelihood function informs the {normalizing constant}, commonly referred to as the evidence term, in the denominator.\nThis is in contrast to the denominator of the updated density, which is given by the predicted density, which is in general not a constant, and can be constructed independently of $\\observed$.\n\n\\begin{table}[htbp]\n\\centering\n\\begin{tabular}{|c|c|}\n\\hline\n & \\\\\n$\\displaystyle \\updated(\\param) = \\initial(\\param) \\frac{\\observed(\\q)}{\\predicted(\\q)}\n$\n&\n$\n\t\\displaystyle \\pi_{\\text{post}}(\\param\\,|\\,\\q) = \\frac{\\prior(\\param)L_\\dspace(\\q \\,|\\,\\param)}{\\int_{\\Lambda} L_\\dspace(\\q \\, |\\, \\param) \\prior(\\param) \\, d\\pmeas}\n$\n \\\\ & \\\\ \\hline\n\\end{tabular}\n\\caption{Updated density solving the SIP (left) and posterior density solving the Bayesian inverse problem (right).}\n\t\t\\label{tab:dens_comparisons}\n\\end{table}\n\nA practical implication of this difference is that the updated density only alters the structure of the initial density in what we refer to as the ``data-informed'' parameter directions.\nSpecifically, for a fixed $\\q\\in\\dspace$, let $C_\\q := \\set{\\param\\in\\pspace\\, : \\, \\qlam=\\q}$, i.e., $C_\\q$ is a ``contour'' in parameter space.\nThen, for any $\\param\\in C_\\q$, we immediately have $\\updated(\\param)=r(\\q)\\initial(\\param)$ where $r(\\q)$ is a fixed constant of proportionality for all $\\param\\in C_\\q$.\n%Subsequently, using the Disintegration theorem on both the initial and updated densities produces exactly the same family of conditional densities on the contours in parameter space.\nBy contrast, while the posterior does not have to agree with the prior in any direction in parameter space, the prior does impact the structure of the posterior in all directions.\n\nThe previous paragraph is not\\---and should not be interpreted as\\---a criticism of the Bayesian inverse framework.\nIt is simply meant to highlight that the data-consistent and Bayesian frameworks formulate and solve inverse UQ problems from different perspectives and with different (although at times seemingly compatible) assumptions.\nConsequently, the solutions for an inverse problem formulated under either framework may differ significantly.\nAs the example (adopted from \\cite{BJW18a, BJW18b}) below demonstrates, this is true even if we arbitrarily force the inverse problems to appear as similar as possible.\n\n\\FloatBarrier\n", "meta": {"hexsha": "cdc5c052f4b7e6d4537035bccfad5d3ef1fa0ec2", "size": 8319, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "intro/comparison.tex", "max_stars_repo_name": "mathematicalmichael/thesis", 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"avg_line_length": 76.3211009174, "max_line_length": 338, "alphanum_fraction": 0.7468445727, "num_tokens": 1908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8633916134888613, "lm_q2_score": 0.7090191399336402, "lm_q1q2_score": 0.6121611792217904}} {"text": "\\newpage\n\\chapter[Cauchy's Integral Formula]{Cauchy's Integral Formula and its Applications}\n\\section{Cauchy's Integral Formula}\nWe have seen so far that if $f:\\mathcal{R} \\to \\C$ is holomorphic on a starlit region $\\mathcal{R}$ then\n\\[\n\\int_{\\mathcal{C}} f = 0\n\\]\nfor any closed contour $\\mathcal{C}$ in $\\mathcal{R}$. What happens when $\\mathcal{R}$ is not starlit? \n\nIt may happen that while $\\mathcal{R}$ itself is not starlit, the contour $\\mathcal{C}$ is contained in a starlit subregion $\\mathcal{S}$ of $\\mathcal{R}$.\n\n\\begin{figure}[H]\n\\begin{tabular}{ccc}\n\\includegraphics[scale=1]{ch5_notsc3} & \\quad \\includegraphics[scale=1]{ch5_notsc1} \\quad & \\includegraphics[scale=1]{ch5_notsc2}\n\\end{tabular}\n\\caption{The region $\\mathcal{R}$ consists of the complex plane with a disc centred at the origin removed. This region is not starlit, but there are starlit subregions of $\\mathcal{R}$ containing the contour $\\mathcal{C}$.}\n\\end{figure}\n\nThus if $f$ is holomorphic on $\\mathcal{R}$, then $f$ is also holomorphic on $\\mathcal{S}$ and so Cauchy's Theorem for Starlit Regions (Theorem~\\ref{t:cauchyst}) gives\n\\[\n\\int_{\\mathcal{C}} f=0.\n\\]\n\nIf $\\mathcal{R}$ is not simply connected, then it may not be possible to do this; in particular, when $\\mathcal{C}$ encloses a point $z_0$ at which $f$ is not holomorphic. For example, suppose $f:\\C \\backslash \\set{0} \\to \\C$ is defined by $f(z) = \\dfrac{1}{z}$, and $\\mathcal{C}$ is the contour shown in Figure~\\ref{f:notsc}. \n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.75]{ch5_notsc}\n\\caption{A contour $\\mathcal{C}$ contained in the non-simply connected region $\\C\\backslash \\set{0}$.}\n\\label{f:notsc}\n\\end{figure}\n\nHere $\\C \\backslash \\set{0}$ is not starlit, and there is no starlit subregion of $\\C \\backslash \\set{0}$ containing $\\mathcal{C}$. Nonetheless, in certain circumstances, it is at least possible to replace $\\mathcal{C}$ something easier in order to calculate the integral.\n\n\\begin{definition}\nA contour $\\mathcal{C}$, that is not closed, is said to be \\emph{simple} if $\\mathcal{C}$ does not intersect itself. A \\emph{simple closed contour $\\mathcal{C}$} is one that does not intersect itself except that its start point is the same as its end point.\n\\end{definition}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.3]{ch5_simple_full}\n\\caption{Simple contours.}\n\\end{figure}\n\\begin{figure}[H]\n\\centering\n\\includegraphics[scale=0.3]{ch5_nonsimple_full}\n\\caption{Non-simple contours. Note that the closed, non-simple contour on the right has no obvious orientation - it is neither clockwise nor anticlockwise.}\n\\end{figure}\n\n\nFor a simple closed contour $\\mathcal{C}$, when we speak of $\\mathcal{C}$ being clockwise or anticlockwise, we mean relative to the points enclosed by $\\mathcal{C}$. In other words, a contour $\\mathcal{C}$ is anticlockwise if, as we travel along $\\mathcal{C}$, the region enclosed by $\\mathcal{C}$ always lies to our left. Again, we shall avoid giving a precise definition, and treat this notion informally.\n\n\n\n\n\nThe following Theorem shows us how in some cases, a potentially complicated contour integral may be reduced to an easier one.\n\\begin{theorem}[Shrinking Contour Theorem/Deformation Theorem]\n\\label{t:sc}\nLet $\\mathcal{R}$ be a simply connected region, $\\mathcal{C}$ an anticlockwise, simple closed contour in $\\mathcal{R}$, $z_0$ a point enclosed by $\\mathcal{C}$ and $g$ a function which is holomorphic on $\\mathcal{R} \\backslash \\set{z_0}$. Then\n\\[\n\\int_{\\mathcal{C}} g = \\int_{\\mathcal{C}_r} g\n\\]\nwhere $\\mathcal{C}_r$ is an anticlockwise circular contour with centre $z_0$ and radius $r$, which is small enough so that $\\mathcal{C}_r$ is contained in $\\mathcal{R}$.\n\\end{theorem}\n\\begin{center}\n\\includegraphics[scale=0.75]{ch5_scnotstarlit4}\n\\end{center}\n{\\bf Sketch of Proof}\n\nLet us first consider the case where $\\mathcal{R} = \\C$ as shown:\n\\begin{center}\n\\altgraphics[scale=1]{ch5_shrinking_cont_full}{ch5_shrinking_cont}\n\\end{center}\n\\begin{blankbox}\nDraw a straight line through $z_0$ to get two starlit subregions $\\mathcal{R}_1$ and $\\mathcal{R}_2$, with $g$ holomorphic on both $\\mathcal{R}_1$ and $\\mathcal{R}_2$. Join $\\mathcal{C}$ and $\\mathcal{C}_r$ along this straight line to get two new (simple, closed, anticlockwise) contours $\\mathcal{C}_1$ and $\\mathcal{C}_2$ contained in $\\mathcal{R}_1$ and $\\mathcal{R}_2$ respectively.\n\nBy Cauchy's Theorem for Starlit Regions,\n\\[\n\\int_{\\mathcal{C}_1} g = \\int_{\\mathcal{C}_2} g = 0.\n\\]\nMoreover,\n\\[\n0= \\int_{\\mathcal{C}_1} f + \\int_{\\mathcal{C}_2} f = \\int_{\\mathcal{C}} f - \\int_{\\mathcal{C}_r} f,\n\\]\nas the integrals along the straight line segments cancel, leaving us with\n\\begin{itemize}\n\\item an anticlockwise copy of $\\mathcal{C}$, and\n\\item a clockwise copy of $\\mathcal{C}_r$, i.e., the reverse of $\\mathcal{C}_r$.\n\\end{itemize}\nThus\n\\[\n\\int_{\\mathcal{C}} f = \\int_{\\mathcal{C}_r} f.\n\\]\n\\end{blankbox}\n\nFor more general simply connected regions $\\mathcal{R}$, we may need to partition $\\mathcal{R}$ into starlit subregions $\\mathcal{R}_1,\\mathcal{R}_2,\\ldots, \\mathcal{R}_n$, numbered so that $z_0 \\in \\mathcal{R}_n$. By joining points on $\\mathcal{C}$, we get anticlockwise closed contours $\\mathcal{C}^{(1)},\\mathcal{C}^{(2)},\\ldots,\\mathcal{C}^{(n)}$, with $\\mathcal{C}^{(j)}$ contained in $\\mathcal{R}_j$ for each $j$. Label the regions $\\mathcal{R}_j$ so that $z_0 \\in \\mathcal{R}_n$. \n\n\\begin{center}\n\\altgraphics[scale=0.3]{ch5_deformation1_full}{ch5_deformation2}\n\\end{center}\n\n\\begin{blankbox}\n%\\vspace*{12cm}\nAs before,\n\\[\n\\int_{\\mathcal{C}} g = \\int_{\\mathcal{C}^{(1)}} g + \\int_{\\mathcal{C}^{(2)}} g + \\ldots + \\int_{\\mathcal{C}^{(n)}} g,\n\\]\nas the integrals along the connecting edges cancel in pairs.\n\\begin{comment}\n\\leftimage{\n\\includegraphics[scale=0.3]{ch5_deformation2_full}\n}\n{\nThe edge where $\\mathcal{C}^{(j)}$ and $\\mathcal{C}^{(j+1)}$ intersect is traversed in one direction along $\\mathcal{C}^{(j)}$ and in the opposite direction along $\\mathcal{C}^{(j+1)}$. Thus when we compute the sum of the integral of $g$ along $\\mathcal{C}^{(j)}$ and along $\\mathcal{C}^{(j+1)}$, the contributions made by this connecting edge cancel.\n}\n\\end{comment}\n\nFor each $j \\neq n$, $g$ is holomorphic on $\\mathcal{R}_j$, and $\\mathcal{C}^{(j)}$ is a closed contour contained in the starlit region $\\mathcal{R}_j$. Thus by Cauchy's Theorem for Starlit regions, we have\n\\[\n\\int_{\\mathcal{C}^{(j)}} g = 0\\quad \\text{ for } j=1,2,\\ldots,n-1.\n\\]\n\nCombining these two observations it follows that\n\\[\n\\int_{\\mathcal{C}} g = \\int_{\\mathcal{C}^{(n)}} g,\n\\]\nand thus we need to show that\n\\[\n\\int_{\\mathcal{C}^{(n)}}g = \\int_{\\mathcal{C}_r} g.\n\\]\nThe proof of this is almost identical to the case $\\mathcal{R}=\\C$.\n\\end{blankbox}\n\\begin{comment}\n\nWhile $\\mathcal{R}_n$ is starlit, $\\mathcal{R}_n \\backslash \\set{z_0}$ is not. Thus we split $\\mathcal{R}_n \\backslash \\set{z_0}$ into two starlit subregions, and construct contours $S_1$ and $S_2$ by connecting $\\mathcal{C}^{(n)}$ and $\\mathcal{C}_r$ as shown.\n\\begin{center}\n\\includegraphics[scale=0.25]{ch5_shrinking}\n\\end{center}\n\n\nBy Cauchy's Theorem for Starlit Regions, we again have\n\\[\n\\int_{S_1} g = \\int_{S_2} g = 0.\n\\]\n\nNotice that if we join $S_1$ and $S_2$ we get a contour consisting of\n\\begin{itemize}\n\\item An anticlockwise copy of the circular contour $\\mathcal{C}_r$, \n\\item A clockwise copy of $\\mathcal{C}^{(n)}$, i.e. the reverse of $\\mathcal{C}^{(n)}$, and\n\\item two copies of each of the new edges connecting $\\mathcal{C}^{(n)}$ and $\\mathcal{C}_r$, one in each direction.\n\\end{itemize}\nThus we get\n\\[\n\\int_{S_1} g + \\int_{S_2} g = \\int_{\\mathcal{C}_r} g - \\int_{\\mathcal{C}^{(n)}} g\n\\]\nas the integrals along the new edges connecting $\\mathcal{C}^{(n)}$ and $\\mathcal{C}_r$ cancel, and we are left with $\\mathcal{C}_r$ and the reverse of $\\mathcal{C}^{(n)}$.\n \nHence\n\\[\n0 = \\int_{S_1} g + \\int_{S_2} g = \\int_{\\mathcal{C}_r} g - \\int_{\\mathcal{C}^{(n)}} g\n\\]\nand so\n\\[\n\\int_{\\mathcal{C}_r} g = \\int_{\\mathcal{C}^{(n)}} g = \\int_{\\mathcal{C}} g.\n\\]\n\\blanksoff\n%\\vspace*{9cm}\n\\end{comment}\n\n\\qed\n\nA very similar argument is used to prove Cauchy's Theorem for Simply Connected Regions.\n\n\\begin{theorem}[Cauchy's Theorem for Simply Connected Regions]\nLet $f$ be a function that is holomorphic in a simply connected region $\\mathcal{R}$, and let $\\mathcal{C}$ be a closed contour in $\\mathcal{R}$. Then\n\\[\n\\int_{\\mathcal{C}} f = 0.\n\\]\n\\end{theorem}\n{\\bf Proof}\nThis is identical to the Proof of Theorem~\\ref{t:sc}, except that now, $f$ is holomorphic on the starlit region $\\mathcal{R}_n$, so that\n\\[\n\\int_{\\mathcal{C}} f = \\int_{\\mathcal{C}_r} f =0.\n\\]\n\\qed\n\n\n%\\vspace*{7cm}\n\n\n\n\n\n\\begin{theorem}[Cauchy's Integral Formula]\n\\label{t:cauchyformula}\nLet $\\mathcal{R}$ be a simply connected region, $\\mathcal{C}$ an anticlockwise simple closed contour in $\\mathcal{R}$, $z_0$ a point enclosed by $\\mathcal{C}$ and $f$ a function that is holomorphic on $\\mathcal{R}$. Then\n\\[\n\\int_{\\mathcal{C}} \\frac{f(z)}{z-z_0} \\ dz = 2 \\pi i f( z_0).\n\\]\n\\end{theorem}\n\\begin{proof}\nDefine a new function $g$ by the formula\n\\[\ng(z) = \\frac{f(z)}{z-z_0}\n\\]\nso that $g$ is holomorphic on $\\mathcal{R} \\backslash \\set{z_0}$, i.e., where $f$ is holomorphic and $z-z_0 \\neq 0$. Thus we are trying to show that\n\\[\n\\int_{\\mathcal{C}} g = 2\\pi i f(z_0).\n\\]\nUsing the Shrinking Contour Theorem~\\ref{t:sc}, we may replace $\\mathcal{C}$ with an anticlockwise circular contour $\\mathcal{C}_r$ with centre $z_0$ and radius $r$ to get\n\\[\n\\int_{\\mathcal{C}} g = \\int_{\\mathcal{C}_r} g\n\\]\nwhenever $r>0$ is small enough so that $\\mathcal{C}_r$ is contained in $\\mathcal{R}$. \n\nIf we define $I$ by\n\\[\nI = \\left(\\int_{\\mathcal{C}_r} \\frac{f(z)}{z-z_0}\\ dz \\right) - 2 \\pi i f(z_0),\n\\]\nthen we want to show that $I=0$. To do this, we will write $I$ as a single integral along $\\mathcal{C}_r$ and use the Estimation Lemma (together with the fact that our definition of $I$ does not depend on our choice of sufficiently small $r$).\n\nParametrising $\\mathcal{C}_r$ with $\\gamma:[0,2\\pi] \\to \\C$, $\\gamma(t) = z_0 + r \\exp (it)$, we have\n\\[\n\\int_{\\mathcal{C}_r} \\frac{1}{z-z_0}\\ dz = \\int_0^{2\\pi} \\frac{1}{z_0+r\\exp(it)-z_0} \\cdot i r\\exp (it) \\ dt = \\int_0^{2\\pi} i \\ dt = 2 \\pi i\n\\]\n(which we saw in Exercise Sheet 2). Therefore, since $f(z_0)$ is a constant, we get\n\\[\n2\\pi i f(z_0) = \\int_{\\mathcal{C}_r} \\frac{f(z_0)}{z-z_0}\\ dz.\n\\]\nThus\n\\begin{align*}\nI &= \\int_{\\mathcal{C}_r} \\frac{f(z)}{z-z_0}\\ dz - \\int_{\\mathcal{C}_r} \\frac{f(z_0)}{z-z_0}\\ dz \\\\\n& = \\int_{\\mathcal{C}_r} \\frac{f(z)-f(z_0)}{z-z_0}\\ dz.\n\\end{align*}\nTo apply the Estimation Lemma, we need to find an upper bound for\n\\[\n\\abs{ \\frac{f(z)-f(z_0)}{z-z_0} } \\quad \\text{ where } z \\in \\mathcal{C}_r.\n\\]\nSince $\\mathcal{C}_r$ is a circle with centre $z_0$ and radius $r$, it is clear that $\\abs{z-z_0} = r$ for all $z \\in \\mathcal{C}_r$, and thus we look at $\\abs{f(z)-f(z_0)}$.\n\nSince $f$ is holomorphic on $\\mathcal{R}$, it is continuous on $\\mathcal{R}$ and in particular, continuous at $z_0$. Thus given any $\\epsilon>0$ there exists $\\delta>0$ such that\n\\[\n0 < \\abs{z-z_0} < \\delta \\Rightarrow \\abs{ f(z)-f(z_0) } < \\epsilon.\n\\]\nIn other words, given any $\\epsilon >0$, if we choose $r$ so that $00$. Hence $\\abs{I}=0$ which implies $I=0$, or in other words,\n\\[\n\\int_{\\mathcal{C}} \\frac{f(z)}{z-z_0}\\ dz = 2 \\pi i f(z_0).\n\\]\n\\end{proof}\n\n\\begin{note}\nCauchy's Integral Formula, and its proof, can be easily remembered using the following approximation: if $r$ is small and $z \\in \\mathcal{C}_r$ then $f(z) \\approx f(z_0)$, hence\n\\[\n\\int_{\\mathcal{C}_r} \\frac{f(z)}{z-z_0}\\ dz \\approx \\int_{\\mathcal{C}_r} \\frac{f(z_0)}{z-z_0}\\ dz = f(z_0) \\int_{\\mathcal{C}_r} \\frac{1}{z-z_0}\\ dz = f(z_0) (2\\pi i).\n\\]\n\\end{note}\nSuppose we wish to evaluate the integral $\\int_{\\mathcal{C}} g$ of a continuous function $g$ along some closed contour $\\mathcal{C}$. If we can find some point $z_0$ enclosed by $\\mathcal{C}$ and a function $f$ holomorphic on $\\mathcal{R}$ with \n\\[\ng(z) = \\frac{f(z)}{z-z_0} \\quad \\text{ for all }z \\in \\mathcal{C}\n\\]\nthen our integral may be evaluated using\n\\[\n\\int_{\\mathcal{C}} g = \\int_{\\mathcal{C}} \\frac{f(z)}{z-z_0}\\ dz = 2\\pi i f(z_0).\n\\]\nIn other words, it is enough to know the value of $f(z_0)$ in order to calculate $\\int_{\\mathcal{C}}g$.\n\\begin{example}\nEvaluate the integral\n\\[\n\\int_{\\mathcal{C}} \\frac{\\Log (z)}{z^2+9}\n\\]\nwhere $\\Log$ is the Principal Logarithm function and $\\mathcal{C}$ is the anticlockwise circle with centre $4i$ and radius $3$.\n\\end{example}\n%\\vspace*{5cm}\n\\begin{solution}\nWe know that $\\Log$ is holomorphic on $\\C_{\\pi}$, and moreover, we have\n\\[\nz^2+9 = (z+3i)(z-3i).\n\\]\nIf we write\n\\[\n\\frac{\\Log (z)}{z^2+9} = \\frac{\\Log (z)}{(z+3i)(z-3i)} = \\frac{f(z)}{z-z_0}\n\\]\nwhere $z_0=3i$ and\n\\[\nf(z) = \\frac{\\Log (z)}{z+3i}\n\\]\nthen $z_0$ is enclosed by $\\mathcal{C}$ and $f$ is holomorphic on $\\C_{\\pi} \\backslash \\set{-3i}$, which is a region containing $\\mathcal{C}$. However, it is clear that the region $\\C_{\\pi} \\backslash \\set{-3i}$ is not simply connected, so the hypotheses of Cauchy's Integral Formula are not fully satisfied. We deal with this by replacing $\\C_{\\pi} \\backslash \\set{-3i}$ with a subregion that is simply connected.\n\n\\leftimage{\n\\altgraphics[scale=1]{ch5_cpi5}{ch5_cpi4}\n}{\nIndeed, if we consider the region $\\mathcal{R} = \\set{ z \\in \\C: \\Im (z)>0}$, then we see that\n\\begin{itemize}\n\\item $\\mathcal{R}$ is simply connected,\n\\item $f$ is holomorphic on $\\mathcal{R}$, and\n\\item $\\mathcal{C}$ is a simple, closed anticlockwise contour that is contained in $\\mathcal{R}$ and encloses the point $z_0=3i$.\n\\end{itemize}\n}\n\n\n\n\nTherefore, we can apply Cauchy's Integral Formula, which gives\n\\begin{align*}\n\\int_{\\mathcal{C}} \\frac{\\Log(z)}{z^2+9}\\ dz &= \\int_{\\mathcal{C}} \\frac{f(z)}{z-z_0}\\ dz \\\\\n& = 2 \\pi i f(z_0) \\\\\n& = 2 \\pi i \\left( \\frac{\\Log(3i)}{3i+3i} \\right) \\\\\n& = \\frac{\\pi}{6} \\left( \\log (3) + i \\frac{\\pi}{2} \\right).\n\\end{align*}\n\\end{solution}\n%\\vspace*{5cm}\n\n\n\n%\\vspace*{8cm}\n\n\\begin{theorem}[Cauchy's Integral Formula for Derivatives; proof non examinable]\n\\label{t:cauchyd}\nLet $\\mathcal{R}$ be a simply connected region, $\\mathcal{C}$ a simple, closed anticlockwise contour contained in $\\mathcal{R}$ and $f$ a function that is holomorphic on $\\mathcal{R}$. Then\n\\begin{enumerate}\n\\item $f$ is infinitely differentiable on $\\mathcal{R}$, and\n\\item At every point $z$ in the region enclosed by $\\mathcal{C}$, the $k^{th}$ derivative $f^{(k)}$ of $f$ satisfies\n\\[\nf^{(k)}(z) = \\frac{k!}{2\\pi i} \\int_{\\mathcal{C}} \\frac{f(\\zeta)}{(\\zeta-z)^{k+1}}\\ d \\zeta.\n\\]\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{theorem}[Liouville's Theorem]\nLet $f:\\C \\to \\C$ be holomorphic (i.e. $f$ is holomorphic on the entire complex plane) and suppose that $f$ is bounded, i.e., there exists $M>0$ with $\\abs{f(z)}\\leq M$ for all $z \\in \\C$. Then $f$ is constant.\n\\end{theorem}\n{\\bf Proof: } Exercise Sheet 4.\n\n\\section{Evaluating Real Integrals Using Cauchy's Integral Formula}\nAn important application of Cauchy's Integral Formula (Theorem~\\ref{t:cauchyformula}) is that it allows us to evaluate certain \\emph{real} integrals that would be difficult, if not impossible, to evaluate otherwise. The following example illustrates the general method for doing so, which we shall develop in the next few sections.\n\\begin{example}\n\\label{e:realint1}\nLet $R>1$, and define a contour $\\cont_R$ by joining the line segment $[-R,R]$ and the upper semicircle $\\mathcal{S}_R$ from $R$ to $-R$ via $iR$. Use Cauchy's Integral Formula to evaluate\n\\[\n\\int_{\\cont_R} \\frac{1}{1+z^2}\\ dz.\n\\]\n\\begin{center}\n\\includegraphics[scale=1.5]{ch5_cr1}\n\\end{center}\n\\end{example}\n\\begin{solution}\nThe function $f$ is holomorphic on $\\C \\backslash \\set{i,-i}$. Write\n\\[\n\\frac{1}{1+z^2} = \\frac{1}{(z+i)(z-i)} = \\frac{g(z)}{z-z_0}\n\\]\nwhere\n\\[\ng(z) = \\frac{1}{z+i}\\quad \\text{ and } z_0=i.\n\\]\nThen $g$ is holomorphic on $\\C \\backslash \\set{-i}$ and in particular, on the simply connected region $\\mathcal{R} = \\set{ z \\in \\C: \\Im (z) > - \\frac{1}{2}}$. Moreover, $\\mathcal{C}_R$ is a closed, simple anticlockwise contour in $\\mathcal{R}$ and $z_0=i$ is a point enclosed by $\\mathcal{C}_R$.\n\\begin{comment}\n\\begin{center}\n\\includegraphics[scale=1]{ch5_cr1}\n\\end{center}\n\\end{comment}\nHence by Cauchy's Integral Formula,\n\\begin{align*}\n\\int_{\\mathcal{C}_R} \\frac{1}{1+z^2}\\ dz &= \\int_{\\mathcal{C}_R} \\frac{g(z)}{z-i}\\ dz \\\\\n&= 2 \\pi i g(i) \\\\\n& = 2\\pi i \\frac{1}{i+i} = \\pi.\n\\end{align*}\n(Note that this integral does not depend on the value of $R$ once $R>1$).\n\n\\end{solution}\n\n\\begin{example}\nNow, let's use Example~\\ref{e:realint1} to evaluate the real integral\n\\[\n\\int_{-\\infty}^{+\\infty} \\frac{1}{1+x^2}\\ dx.\n\\]\n\\end{example}\n\\begin{solution}\nHaving evaluated the integral using Cauchy's Integral Formula, we now look at the integrals along the two paths $L_R$ and $S_R$;\n\\[\n\\int_{\\mathcal{C}_R} \\frac{1}{z^2+1}\\ dz = \\int_{L_R} \\frac{1}{z^2+1}\\ dz + \\int_{S_R} \\frac{1}{z^2+1}\\ dz.\n\\]\nparametrise $L_R$ with $\\gamma_L:[-R,R] \\to \\C$, $\\gamma(t)=t$, so that $\\gamma'(t)=1$ and\n\\[\n\\int_{L_R} \\frac{1}{1+z^2}\\ dz = \\int_{-R}^R \\frac{1}{1+t^2}\\ dt.\n\\]\nWe have seen already (Example~\\ref{e:estimation}) that\n\\[\n\\lim_{R \\to \\infty} \\int_{S_R} \\frac{1}{1+z^2}\\ dz = 0.\n\\]\nHence\n\\begin{align*}\n\\pi & = \\lim_{R \\to \\infty} \\int_{\\mathcal{C}_R} \\frac{1}{1+z^2}\\ dz \\\\\n& = \\left( \\lim_{R \\to \\infty} \\int_{L_R} \\frac{1}{1+z^2}\\ dz \\right) + \\left( \\lim_{R \\to \\infty} \\int_{S_R} \\frac{1}{1+z^2}\\ dz \\right) \\\\\n& = \\lim_{R \\to \\infty} \\int_{-R}^R \\frac{1}{1+t^2}\\ dt + 0 \\\\\n& = \\int_{-\\infty}^{\\infty} \\frac{1}{1+t^2}\\ dt.\n\\end{align*}\nSo we have used contour integration to show that\n\\[\n\\int_{-\\infty}^{\\infty} \\frac{1}{1+x^2}\\ dx = \\pi.\n\\]\n\\end{solution}\n\n\n\\begin{example}\n\\label{e:trig1}\nWe shall evaluate the integral\n\\[\n\\int_{\\Gamma} \\frac{2i}{3z^2-10z+3}\\ dz,\n\\]\nwhere $\\Gamma$ is the anticlockwise unit circle $\\set{z: \\abs{z}=1}$, and use it to evaluate the real integral\n\\[\n\\int_0^{2\\pi} \\frac{1}{5-3\\cos(t)}\\ dt.\n\\]\n\\end{example}\n\\begin{solution}\nIf we factorise the denominator, we can write\n\\[\n\\frac{2i}{3z^2-10z+3} = \\frac{2i}{(3z-1)(z-\\frac{1}{3}}.\n\\]\n\nThe point $z_0=\\frac{1}{3}$ is the only point enclosed by $\\Gamma$ at which this function is not holomorphic. If we let\n\\[\ng(z) = \\frac{2i}{3(z-3)}\n\\]\nwe have\n\\[\n\\frac{g(z)}{z-\\frac{1}{3}} = \\frac{2i}{(3z-1)(z-3)}\n\\]\nand $g$ is holomorphic on the simply connected region $\\set{z \\in \\C: \\Re (z) < 2}$, which contains $\\mathcal{C}$. Thus\n\\begin{align*}\n\\int_{\\Gamma} \\frac{2i}{(3z-1)(z-3)}\\ dz & = \\int_{\\Gamma} \\frac{g(z)}{z-\\frac{1}{3}}\\ dz \\\\\n& = 2\\pi i g(\\tfrac{1}{3}) \\\\\n& = 2\\pi i \\left( - \\frac{i}{4} \\right) = \\frac{\\pi}{2}.\n\\end{align*}\n\nFor the second part, we parametrise $\\Gamma$ using $\\gamma:[0,2\\pi]\\to \\C$, $\\gamma(t)=e^{it}$ and use\n\\begin{align*}\n\\int_{\\Gamma} f &= \\int_0^{2\\pi} f ( \\gamma(t) ) \\gamma'(t)\\ dt \\\\\n& = \\int_0^{2\\pi} \\frac{2i}{3e^{2it}-10e^{it}+3}\\cdot i e^{it}\\ dt \\\\\n& = \\int_0^{2\\pi} \\frac{-2}{-10+3(e^{it}-e^{-it})}\\ dt.\n\\end{align*}\nSince $\\cos(t) = \\frac{1}{2}\\brac{e^{it}-e^{-it}}$, it follows that\n\\[\n\\frac{-2}{-10+3(e^{it}+e^{-it})} = \\frac{-2}{-10+6\\cos(t)} = \\frac{1}{5-3\\cos(t)}.\n\\]\nHence\n\\[\n\\int_0^{2\\pi} \\frac{1}{5-3\\cos(t)}\\ dt = \\int_{\\Gamma} \\frac{2i}{3z^2-10z+3} = \\frac{\\pi}{2}.\n\\]\n\\end{solution}\n\n\nWe can generalise the method of Example~\\ref{e:trig1} to evaluate other real trigonometric integrals.\n\\begin{definition}\nA \\emph{Rational Function} $R:U \\to \\R$, where $U \\subseteq \\R^2$, is a function of the form\n\\[\nR(x,y) = \\frac{f(x,y)}{g(x,y)},\n\\]\nwhere $f,g: \\mathbb{R}^2 \\to \\R$ are two polynomials in $x$ and $y$ with real coefficients.\n\\end{definition}\nWe will now describe how to use contour integration to evaluate integrals of the form\n\\[\n\\int_0^{2\\pi} R( \\cos (t), \\sin (t) )\\ dt,\n\\]\nwhere $R$ is a rational function of two real variables. For example, if $R$ is defined by\n$\n\\displaystyle R(x,y) = \\frac{1}{16x^2+25y^2},\n$\nthen we are looking at the integral\n\\[\n\\int_0^{2\\pi} R(\\cos (t) , \\sin (t) )\\ dt = \\int_0^{2\\pi} \\frac{1}{16 \\left( \\cos (t) \\right)^2+25 \\left( \\sin (t) \\right)^2}\\ dt.\n\\]\nNote that $\\cos$ and $\\sin$ are periodic with period $2\\pi$, and that the above integral has limits $0$ and $2\\pi$.\n\n\\begin{example}\n\\label{e:trig2}\nFor any rational function $R$ of two real variables, with $R(\\cos(t),\\sin(t))$ defined for every $t \\in [0,2\\pi]$, we can find a suitable closed contour $\\mathcal{C}$ and complex function $f$ so that\n\\[\n\\int_0^{2\\pi} R ( \\cos(t), \\sin (t) ) = \\int_{\\mathcal{C}} f.\n\\]\n\\begin{blankbox}\nLet $\\mathcal{C}$ be the anticlockwise circle $\\set{z \\in \\C: \\abs{z}=1}$, parametrised using $\\gamma:[0,2\\pi] \\to \\C,\\ \\gamma(t) = \\exp(it)$. Let $f$ denote the complex function we are looking for, so we want\n\\[\n\\int_0^{2\\pi} R ( \\cos(t),\\sin(t) ) \\ dt = \\int_{\\mathcal{C}} f = \\int_0^{2\\pi} f \\left( \\gamma (t) \\right) \\gamma' (t)\\ dt,\n\\]\nor in other words,\n\\begin{equation}\n\\tag{R1}\\label{e:r1}\nf( \\gamma(t) ) = \\frac{1}{\\gamma'(t)} \\cdot R( \\cos(t), \\sin (t) ).\n\\end{equation}\nFor each $z \\in \\mathcal{C}$, $z= \\exp(it)$ for some $t \\in [0,2\\pi]$ and so\n\\begin{align*}\n\\cos(t) & = \\frac{\\exp(it)+\\exp(-it)}{2} = \\frac{z+z^{-1}}{2} \\\\\n\\sin(t) & = \\frac{\\exp(it)-\\exp(-it)}{2i} = \\frac{z-z^{-1}}{2i} \\\\\n\\end{align*}\nHence for $z=\\exp (it) \\in \\mathcal{C}$, we have\n\\begin{equation}\n\\tag{R2}\\label{e:r2}\n R ( \\cos (t), \\sin (t) ) = R \\left( \\frac{z+z^{-1}}{2}, \\frac{z-z^{-1}}{2i} \\right), \n \\end{equation}\n(this is not true for all $z$, only for $z$ lying on the contour $\\mathcal{C}$). Moreover, $\\gamma'(t) = i \\exp (it) = iz$; so combining~\\eqref{e:r1} and~\\eqref{e:r2}, we define\n\\[\nf(z) = R \\left( \\frac{z+z^{-1}}{2}, \\frac{z-z^{-1}}{2i} \\right) \\cdot \\frac{1}{iz}.\n\\]\nThen\n\\begin{align*}\n\\int_{\\mathcal{C}} f & = \\int_0^{2\\pi} f( \\gamma(t)) \\gamma'(t) \\\\\n& = \\int_0^{2\\pi} R \\left( \\tfrac{1}{2}(\\exp(it)+\\exp(-it)),\\tfrac{1}{2i}(\\exp(it)-\\exp(-it)) \\right) \\cdot \\frac{1}{i\\exp(it)} \\cdot i\\exp(it)\\ dt \\\\\n&= \\int_0^{2\\pi} R( \\cos (t) , \\sin (t))\\ dt.\n\\end{align*}\n\\end{blankbox}\n\\end{example}\n\\begin{example}\nUse a suitable contour integral to evaluate\n\\[\n\\int_0^{2\\pi} \\frac{1}{2+\\sin \\theta}\\ d \\theta.\n\\]\n\\begin{solution}\nWith $\\mathcal{C}$ the anticlockwise unit circle, we have\n\\begin{align*}\n\\int_0^{2\\pi} \\frac{1}{2+\\sin \\theta}\\ d \\theta & = \\int_{\\mathcal{C}} \\frac{1}{2+\\frac{z-z^{-1}}{2i}} \\cdot \\frac{1}{iz}\\ dz \\\\\n& = \\int_{\\mathcal{C}} \\frac{2}{z^2+4iz-1}\\ dz \\\\\n& = \\int_{\\mathcal{C}} \\frac{2}{(z-(-2+\\sqrt{3})i)(z-(-2-\\sqrt{3})i)}\\ dz.\n\\end{align*}\nBy Cauchy's Integral Formula, this integral has the value\n\\[\n2\\pi i \\cdot \\frac{2}{(-2+\\sqrt{3})i-(-2-\\sqrt{3})i} = \\frac{2\\pi}{\\sqrt{3}}.\n\\]\n\\end{solution}\n\\end{example}\n\n\\begin{example}\nSet up a contour integral that could be used to evaluate\n\\[\n\\int_0^{2\\pi} \\frac{1}{16 \\cos^2 (t) + 25 \\sin^2 (t)}\\ dt.\n\\]\n\\begin{solution}\nHere $R(x,y) = \\dfrac{1}{16x^2+25y^2}$, so the required integral (with $\\mathcal{C}$ the anticlockwise circle $\\set{z \\in \\C : \\abs{z}=1 }$) is\n\\begin{align*}\n\\int_{\\mathcal{C}} R \\left( \\frac{z+z^{-1}}{2}, \\frac{z-z^{-1}}{2i} \\right) \\cdot \\frac{1}{iz}\\ dz & = \\int_{\\mathcal{C}} \\frac{1}{16\\left(\\frac{1}{2}(z+z^{-1})\\right)^2+25\\left(\\frac{1}{2i}(z-z^{-1}) \\right)^2} \\cdot \\frac{1}{iz}\\ dz \\\\\n& = \\int_{\\mathcal{C}} \\frac{1}{4(z+2+z^{-2})-\\frac{25}{4}(z^2-2+z^{-2})} \\cdot \\frac{1}{iz}\\ dz \\\\\n& = \\int_{\\mathcal{C}} \\frac{4i}{9z^4-82z^2+9}\\ dz.\n\\end{align*}\nWe do not yet know how to evaluate this integral, we shall do this in Chapter 6.\n\\end{solution}\n\\end{example}\n\n\n\\section{Series Representations of Holomorphic Functions}\n\\begin{definition}\nA sequence $z_n$ of complex numbers is said to converge to a complex number $z$ if, for all $\\epsilon>0$ there is a natural number $N$ such that $n \\geq N$ implies that $\\abs{z_n-z}<\\epsilon$.\n\nAn infinite series $\\sum_{k=1}^{\\infty} a_k$ is said to converge to a complex number $s$ if the sequence of partial sums $s_n:= \\sum_{k=1}^n a_k$ converges to $s$. When this occurs we write $\\sum_{k=1}^{\\infty} a_k = s$.\n\\end{definition}\n\nFor a fixed $z_0 \\in \\C$, a \\emph{power series} (centred at $z_0$) is a series of the form $\\displaystyle \\sum_{n=0}^{\\infty} a_n (z-z_0)^n$, where $a_n$ are complex numbers. As with power series in $\\mathbb{R}$, a complex power series may converge for some values of $z$ and diverge for others. Note that if $\\mathcal{S}$ denotes the subset\n\\[\n\\mathcal{S} = \\set{ z \\in \\C: \\sum_{n=0}^{\\infty} a_n(z-z_0)^n \\text{ converges } },\n\\]\nthen we get a function $f:\\mathcal{S} \\to \\C$ defined by\n\\[\nf(z) = \\sum_{n=0}^{\\infty} a_n (z-z_0)^n.\n\\]\n\n\n\\begin{theorem}[Taylor's Theorem]\n\\label{t:taylor}\nLet $\\mathcal{R}$ be a region, $f:\\mathcal{R} \\to \\C$ holomorphic and $z_0$ a point in $\\mathcal{R}$. Then $f$ can be represented by the power series\n\\[\nf(z) = \\sum_{n=0}^{\\infty} \\frac{f^{(n)}(z_0)}{n!} (z-z_0)^n\n\\]\nfor all $z \\in D(z_0;R)$, whenever $R>0$ is small enough so that $D(z_0;R) \\subseteq \\mathcal{R}$.\n\\end{theorem}\n\\begin{center}\n\\includegraphics[scale=1]{ch5_taylordisc_full}\n\\end{center}\nBefore proving Taylor's Theorem, we need the following lemma:\n\\begin{lemma}\n\\label{l:taylor}\nLet $f:\\mathcal{R} \\to \\C$ be a function, $z_0 \\in \\mathcal{R}$ and $R>0$ be such that $D(z_0,R) \\subseteq \\mathcal{R}$. Then for any $00$ with $0 < r < R$ and let $\\mathcal{C}_r$ be the anticlockwise circle with centre $z_0$ and radius $r$. Then by Cauchy's Integral Formula,\n\\[\nf(z) = \\frac{1}{2\\pi i} \\int_{\\mathcal{C}_r} \\frac{f(\\zeta)}{\\zeta-z}\\ d\\zeta\n\\]\nfor all $z$ with $\\abs{z-z_0} \\cdot Q =\n\\begin{pmatrix} \nva_1 & 0 \\\\\n0 & va_2\n\\end{pmatrix}\n\\cdot Q \n= \\frac{1}{\\Delta}\n\\begin{pmatrix} \nva_1 \\cdot (1-d) & va_1 \\cdot b \\\\\nva_2 \\cdot c & va_2 \\cdot (1-a)\n\\end{pmatrix}\n$$\n\n$$ \\cdot Q \n= \\frac{1}{\\Delta}\n\\begin{pmatrix} \n(1- a-c)\\cdot (1-d) & (1-a-c) \\cdot b \\\\\n(1-b-d) \\cdot c & (1-b-d) \\cdot (1-a)\n\\end{pmatrix}\n$$\n\nwhich simplifies by noting that the sum of the column are equal to 1 :\n$$ \\cdot Q \n= \n\\begin{pmatrix} \n1-\\theta_1 & \\theta_2 \\\\\n\\theta_1 & 1-\\theta_2\n\\end{pmatrix}\n$$\n\nwith $\\theta_1 = \\frac{(1-b-d) \\cdot c}{(1-a)(1-d)-bc}$ and $\\theta_2 = \\frac{(1-a-c) \\cdot b}{(1-a)(1-d)-bc}$. \nThis formula reveals that the Leontief matrix allocates demand between the various sector to generate a value added. For example, an increase of $\\delta$ in the demand adressed to sector S1 would generated a value added in sector S1 equal to $(1-\\theta_1) \\cdot \\delta$, and a value added in sector S2 equal to $\\theta_1 \\cdot \\delta$.\n\nWe can now estimate the number of job per unit of final demand. To that end, we define the vector $e$ of job per unit of value added by :\n$(e_i) = (E_i/VA_i)$, where $E_i$ is the number of jobs in sector i and $VA_i$ the value added in sector i.\nThe number of job per unit of domestic demand - which we call domestic employment content $\\pmb{ce}^d$ - is equal to:\n$$(\\pmb{ce^d})^t =\n\\pmb{e}^t \\cdot \\cdot Q \n= \n\\begin{pmatrix} \ne_1 (1-\\theta_1) + e_2 \\theta_1 ;&\ne_1 \\theta_2 + e_2 (1-\\theta_2)\n\\end{pmatrix}\n$$\n\nThis domestic employment content must now be linked to final demand. \nA shift of $\\delta$ in final demand from S1 to S2 leads to a change in domestic demand equals to:\n$$\\pmb{d^d} =\n\\pmb{<1-\\tau_m>} \\cdot \n\\begin{pmatrix} \n-\\delta \\\\\n\\delta\n\\end{pmatrix} \n=\n\\begin{pmatrix} \n- (1 - \\tau_1) \\\\\n1 - \\tau_2\n\\end{pmatrix} \\cdot \\delta\n$$\n\n\\subsubsection{Impacts of shifting investment}\nAn increase of one million euros in final demand addressed to S2 leads to an increase in jobs equal to:\n$$(1-\\tau_2) \\cdot \\left( (1-\\theta_2) e_2+ \\theta_2 e_1 \\right)$$.\n\nAn decrease of one million euros in final demand addressed to S1 leads to an increase in jobs equal to:\n$$(1-\\tau_1) \\cdot \\left( (1-\\theta_1) e_1+ \\theta_1 e_2 \\right)$$.\n\nA shift in final demand of one million euros from S1 to S2 leads to an increase in employment if and only if:\n$$(1-\\tau_2) \\cdot \\left( (1-\\theta_2) e_2+ \\theta_2 e_1 \\right) > (1-\\tau_1) \\cdot \\left( (1-\\theta_1) e_1+ \\theta_1 e_2 \\right)$$\n\nIt then useful to note that the right parenthesis in each side is the domestic employment content, a weighted average of direct employment intensity. The above equation can be re-written as:\n$$(1-\\tau_2) \\cdot ce^d_2> (1-\\tau_1) \\cdot ce^d_1 $$\nso a shift in final demand generates jobs if an only if the product of the employment content and the import rate is higher.\nBut the computation of the employment content shows that the domestic employment content is high if demand generates value added in sector with a high number of jobs per value added.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\clearpage\n\n\\subsection{Closed economy}\n\\label{app:closed_economy}\n\n\\subsubsection{Overview of the closed model}\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=7cm]{figures/overview_closed.pdf}\n\t\\caption{Overview of the closed economy model for the S2 sector}\n\t\\label{fig:overview_closed}\n\\end{figure}\n\n\\subsubsection{Index of sets}\n\\begin{itemize}\n\t\\item $u$: S1, S2, CAP, LAB, HOH, INV\n\t\\item $i(u)$: S1, S2. Alias: $j(u)$.\n\t\\item $h(u)$: CAP, LAB.\n\\end{itemize}\n\n\\subsubsection{Index of variables}\n\\begin{itemize}\n\t\n\t\\item $Z_j$: output of the j-th good\n\t\\item $F_{h,j}$: the h-th factor input by the j-th firm\n\t\\item $FF_h$: the h-th factor supply\n\t\\item $Inc$: Household income\n\t\\item $C_i$: household consumption of the i-th good\n\t\\item $p^x_i$: demand price of the i-th good\n\t\\item $p^z_i$ supply price of the i-th good\n\t\\item $p^f_h$ the h-th factor price\n\t\\item $pc$ consumption price index\n\t\\item $S^p$: private savings\t\n\t\\item $U$: unemployment rate\n\t\\item $UU$: utility (Cobb-Douglas)\n\\end{itemize}\n\n\\subsubsection{Index of parameters}\n\\begin{itemize}\n\t\\item $\\alpha_i$: share parameter in utility function\n\t\\item $\\sigma^z$: elasticity in CES production function\n\t\\item $\\rho^z = \\frac{\\sigma^z - 1}{\\sigma^z}$: param. in CES production function\n\t\\item $\\delta_{h,j}$ share parameter in CES production function\n\t\\item $scZ_j$: scale parameter in CES production function \n\t\\item $\\beta_{h,j}$: share parameter in Cobb-Douglas production function\n\t\\item $b_j$: scale parameter in Cobb-Douglas production function\n\t\\item $a_{h,j}$: Leontief coefficient in production function\n\t\\item $\\gamma$: wage curve elasticity\n\t\\item $U^0$: initial unemployment rate\n\\end{itemize}\n\n\n\n\\subsubsection{Tests}\n\n\\paragraph{Walras' law}\nWalras' law allows to take off one equation. \nIn our numerical application, we remove the market clearing of good for the first sector.\nThen, we check after the run that Walras' law is satisfied.\n\n\\paragraph{Numéraire}\nIn our numerical application, we use the price index of consumption goods as the numéraire. \nWe check that the model is not sensitive to this assumption. We increase the numéraire by 10\\% and rerun the model.\nWith this assumption, we check that all prices have increased by 10\\%, and volumes remain the same.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\clearpage\n\n\\subsection{Simple CGE with two labor skills}\n\\label{app:two_labour_model}\n\n\\subsubsection{Overview of the simple CGE with two labor skills}\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=8cm]{figures/overview_twoLabors.pdf}\n\t\\caption{Overview of the open economy model for the S2 sector}\n\t\\label{fig:overview_twoLabors}\n\\end{figure}\n\n\\subsubsection{Index of sets}\n\\begin{itemize}\n\t\\item $u$: S1, S2, CAP, LAB, HOH, INV\n\t\\item $i(u)$: S1, S2. Alias: $j(u)$\n\t\\item $h(u)$: CAP, LAB\n\t\\item $s$: LABL, LABH \n\\end{itemize}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\n\n\\clearpage\n\n\\subsection{Open economy}\n\\label{app:open_economy_model}\n\n\\subsubsection{Overview of the open economy}\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=8cm]{figures/overview_open.pdf}\n\t\\caption{Overview of the open economy model for the S2 sector}\n\t\\label{fig:overview_open}\n\\end{figure}\n\n\\subsubsection{Index of sets}\n\\begin{itemize}\n\t\\item $u$: S1, S2, CAP, LAB, HOH, INV, EXT\n\t\\item $i(u)$: S1, S2\n\t\\item $h(u)$: CAP, LAB\n\\end{itemize}\n\n\\subsubsection{Index of variables}\n\\begin{itemize}\n\t\\item $Z_j$: output of the j-th good\n\t\\item $FF_h$: factor supply\n\t\\item $F_{h,j}$: the h-th factor input by the j-th firm\n\t\\item $Inc$: Household income\n\t\\item $C_i$: household consumption of the i-th good\n\t\\item $E_i$: exports\n\t\\item $M_i$: imports\n\t\\item $Q_i$: Armingtons composite good\n\t\\item $D_i$: domestic good\n\n\t\\item $pf_h$: the h-th factor price\n\t\\item $p^x_i$: consumption price\n\t\\item $p^z_j$: supply price of the i-th good\n\t\\item $p^q_i$: Armingtons composite good price\n\t\\item $p^e_i$: export price in local currency\n\t\\item $p^m_i$: import price in local currency\n\t\\item $p^d_i$: the i-th domestic good price\n\t\\item $\\epsilon$: exchange rate\n\t\\item $p^c$: price index\n\n\t\\item $S^p$: private savings\n\t\\item $S^f$: foreign savings\n\t\\item $U$: unemployment rate\n\t\\item $UU$: utility (Cobb-Douglas)\n\\end{itemize}\n\n\\subsubsection{Index of parameters}\n\\begin{itemize}\n\t\\item $\\alpha^u_i$: share parameter in utility func.\n\t\\item $\\sigma^z$: elasticity in CES production function\n\t\\item $\\rho^z = \\frac{\\sigma^z - 1}{\\sigma^z}$: param. in CES production function\n\t\\item $\\delta_{h,j}$ share parameter in CES production function\n\t\\item $scZ_j$: scale parameter in CES production function \n\t\\item $\\sigma^q$: elasticity of Armington substitution\n\t\\item $\\rho^q = \\frac{\\sigma^q-1}{\\sigma^q}$: substitution elasticity parameter for Armington\n\t\\item $\\alpha^m_i$: share par. in Armington func.\n\t\\item $\\alpha^d_i$: share par. in Armington func.\n\t\\item $scq_i$: scale par. in Armington func.\n\t\\item $\\gamma$: wage curve elasticity\n\t\\item $U^0$: initial unemployment rate\n\\end{itemize}\n\n\n\\subsubsection{Results for IO and CGE}\n\\label{subsec:open_economy_results}\n\n\\begin{table}[!ht]\n\t\\centering\n\t\\caption{SAM results in an open economy with our IO model}\n\t\\label{tab:SAM_IO_openEconomy}\n\t\\begin{tabular}{llllllll}\n\t\t\\toprule\n\t\t& S1 & S2 & HOH & INV & EXT \\\\\n\t\t\\midrule\n\t\tS1 & & & 8 & 5 & 3 \\\\\n\t\tS2 & & & 12 & 3 & 3 \\\\\n\t\tLAB & 9.60 & 8.53 & & & \\\\\n\t\tCAP & 5.33 & 4.74 & & & \\\\\n\t\tEXT & 1.06 & 4.74 & & & \\\\\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\n\n\\begin{table}[!ht]\n\t\\centering\n\t\\caption{Open economy with fixed exchange rate $\\epsilon$ and endogenous imports}\n\t\\label{tab:SAM_IO_OpenEconomy_noBudget}\n\t\\begin{tabular}{llllll}\n\t\t\\toprule\n\t\t& S1 & S2 & HOH & INV & EXT \\\\\n\t\t\\midrule\n\t\tS1 & & & 10.96 & 5.00 & 0.00 \\\\\n\t\tS2 & & & 16.44 & 3.00 & 0.00 \\\\\n\t\tLAB & 9.41 & 8.59 & & & \\\\\n\t\tCAP & 5.23 & 4.77 & & & \\\\\n\t\tEXT & 1.33 & 6.08 & & & \\\\\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\clearpage\n\n\\subsection{Full model}\n\\label{sec:full_model}\n\n\\subsubsection{Data}\n\\label{app:full_model_data}\n\nAt the 64-sector level of disaggregation, we encounter some issues to calibrate the CGE model:\n\\begin{itemize}\n\t\\item Postal services (CPA\\_H53) have negative capital revenues in 2013. Standard production functions with constant elasticity of substitution cannot account for a negative capital revenue. \n\tSince this sector is not crucial for our analysis of green jobs, we aggregate it with the other transportation service (CPA\\_H50, CPA\\_H51 and CPA\\_H52).\n\t\\item The \"Manufacture of textiles, wearing apparel and leather products\" sector (CPA\\_C13-15) and \"Computer, electronic and optical products\" sector (CPA\\_C26) export more than they produce. To avoid issues with negative values in the calibration of trade within the constant elasticity of transformation (CET) function, we aggregate the textile and leather products with \"Manufacture of wood and of products of wood\" (CPA\\_C16) and computers with electrical equipment (CPA\\_C27)\n\t\\item The \"imputed rent\" sector (CPA\\_L68A) indicates a low but positive remuneration for employees, but no employee. This would lead to an infinite salary. We aggregate this sector with the other real estate services (CPA\\_L68B)\n\\end{itemize}\n\n\n\n\\subsubsection{Overview of the full CGE model}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[width=14cm]{figures/overview_full.pdf}\n\t\\caption{Overview of the fully-fledged model for the S2 sector}\n\t\\label{fig:overview_full}\n\\end{figure}\n\n\\subsubsection{Index of sets}\n\\begin{itemize}\n\t\\item u: SAM entry /S1*S58, CAP, LAB, IDT, TRF, HOH, GOV,\n\t\\item i(u): goods /S1*S58/\n\t\\item h(u): factor /CAP, LAB/\n\t\\item s: labor skill / highSkill, lowSkill /\n\\end{itemize}\n\n\\subsubsection{Index of variables}\n\\begin{itemize} \n\t\\item $Y_j$: composite factor\n\t\\item $Inc$: household income\n\t\\item $N(s,j)$: worker input by skill\n\t\\item $K_j$: capital factor input by the j-th firm\n\t\\item $L_j$: labor input \n\t\\item $KK$: capital supply\n\t\\item $NN(s)$: worker supply\n\t\\item $X(i,j)$: intermediate input\n\t\\item $Z_j$: output of the j-th good\n\t\\item $C_i$: household consumption of the i-th good\n\t\\item $G_i$: government consumption\n\t\\item $I_i$: investment\n\t\\item $E_i$: exports\n\t\\item $M_i$: imports\n\t\\item $Q_i$: Armington composite good\n\t\\item $D_i$: domestic good\n\n\t\\item $p^k$: capital factor price\n\t\\item $p^l_j$: labor price\n\t\\item $p^n_s$: workers price\n\t\\item $p^y_j$: composite factor price\n\t\\item $p^z_j$: supply price of the i-th good\n\t\\item $p^q_i$: Armingtons composite good price\n\t\\item $p^e_i$: export price in local currency\n\t\\item $p^m_i$: import price in local currency\n\t\\item $p^d_i$: the i-th domestic good price\n\t\\item $\\epsilon$: exchange rate\n\n\t\\item $S^p$: private saving\n\t\\item $S^g$: government saving\n\t\\item $S^f$: foreign saving\n\t\\item $T^d$: direct tax\n\t\\item $T^z_j$: production tax\n\t\\item $T^p$: private consumption tax\n\t\\item $T^v$: investment tax\n\n\t\\item $KK$: capital supply\n\t\\item $NN_s$: labor supply\n\t\\item $pc$: consumption price\n\n\t\\item $UU$: utility (Cobb-Douglas)\n\\end{itemize}\n\n\\subsubsection{Index of parameters}\n\\begin{itemize}\n\t\\item $\\sigma^Y$: elasticity of substitution in composite factor production\n\t\\item $\\rho^Y = \\frac{\\sigma^Y - 1}{\\sigma^Y}$: substitution elasticity parameter for capital-labor composite\n\t\\item $\\sigma_i$: elasticity of Armington substitution\n\t\\item $\\eta_i = \\frac{\\sigma^z - 1}{\\sigma^z}$: substitution elasticity parameter for Armington\n\t\\item $\\psi_i$: elasticity of transformation\n\t\\item $\\phi_i= \\frac{\\psi^z - 1}{\\psi^z}$: transformation elasticity parameter\n\t\\item $\\alpha_i$: share parameter in utility func.\n\t\\item $\\delta^l_j$: labor share in CES prod. func.\n\t\\item $\\delta^k_j$: capital share in CES prod. func.\n\t\\item $scY$: scale param. in CES prod. func.\n\t\\item $\\beta\\_N(s,j)$: share parameter in labor function\n\t\\item $scL_j$: scale parameter in labor function\n\t\\item $ax(i,j)$: intermediate input requirement coeff.\n\t\\item $ay_j$: composite fact. input req. coeff.\n\t\\item $\\mu_i$: government consumption share\n\t\\item $\\lambda_i$: investment demand share\n\t\\item $\\delta^m_i$: share par. in Armington func.\n\t\\item $\\delta^d_i$: share par. in Armington func.\n\t\\item $scQ_i$: scale par. in Armington func.\n\t\\item $\\xi^d_i$: share par. in transformation func.\n\t\\item $\\xi^e_i$: share par. in transformation func.\n\t\\item $\\theta_i$: scale par. in transformation func.\n\t\\item $ssg$: average propensity for gov. saving\n\t\\item $\\tau^d$: direct tax rate\n\t\\item $Pop$: population (labor force)\n\t\\item $U^0_s$: initial unemployment\n\t\\item $p^{We}_i$: export price in US dollars\n\t\\item $p^{Wm}_i$: import price in US dollars\n\t\\item $\\tau^z_i$: production tax rate\n\t\\item $\\tau^p$: private consumption tax rate\n\t\\item $\\tau^v$: investment tax\n\t\\item $Pop$: population (labor force)\n\t\\item $\\gamma$: wage curve elasticity\n\\end{itemize}\n\n\n\n\\begin{table}[!ht]\n\t\\centering\n\t\\caption{Comparative equations of IO and CGE models. Identical equations are not repeated to facilitate reading. The main differences with the closed version shown in table \\ref{tab:closedModel} have their equations labeled in bold.}\n\t\\label{tab:fullModel}\n\t\\begin{tabular}{llll}\n\t\t\\toprule\n\t\tNum & Description & Standard CGE \\\\\n\t\t\\midrule\n\t\teq1 & Production function & $Y_j = scY_j \\cdot ( \\delta^k_j \\cdot K_j^{\\rho_Y} + \\delta^l_j \\cdot L_j^{\\rho_Y} )^{1/\\rho_Y}$ \\\\\n\t\teq2 & Capital demand & $ K_j = (scY_j^{\\rho_Y} \\cdot \\delta^k_j \\cdot p^y_j / p^k )^{\\sigma_Y} \\cdot Y_j $ \\\\\n\t\teq3 & Labor demand & $L_j = (scY_j^{\\rho_Y} \\cdot \\delta^l_j \\cdot p^y_j / p^l_j )^{\\sigma_Y} \\cdot Y_j $ \\\\\t\n\t\teq4 & Demand of intermediate goods & $X_{i,j} = ax_{i,j} \\cdot Z_j$ \\\\\n\t\teq5 & Demand of composite factor & $Y_j = ay_j \\cdot Z_j$ \\\\\n\t\teq6 & Condition of zero profit & $p^z_j = ay_j \\cdot p^y_j + \\sum_i ax_{i,j} \\cdot p^q_i$ \\\\\n\t\teq7 & Production of labor composite & $L_j = scL_j \\cdot \\prod_s N_{s,j}^{\\beta^N_{s,j}}$ \\\\\n\t\teq8 & Demand of labor by skill & $N_{s,j} = \\beta^N_{s,j} \\cdot p^l_j \\cdot L_j / p^n_s$ \\\\\n\t\teq9 & Unemployment & $ U_s = 1 - NN_s/Pop_s $\\\\\n\t\teq10 & Labor supply & $\\log( p^n_s / p^c ) = - \\gamma \\cdot log(\\frac{U_s}{U^0_s}) $ \\\\\n\t\teq11 & Capital supply & $KK = \\overline{KK} $ \\\\\n\t\t\\midrule\n\t\teq12 & Direct tax on household & $T^d = \\tau^d \\cdot Inc $\\\\\n\t\teq13 & Tax on production & $T^z_j = \\tau^z_j \\cdot p^z_j \\cdot Z_j$ \\\\\n\t\teq14 & Tax on household consumption & $T^p = \\tau^p \\cdot \\sum_j p^q_j \\cdot C_j $ \\\\\n\t\teq15 & Tax on investment & $T^v = \\tau^v \\cdot \\sum_j p^q_j\\cdot I_j $ \\\\\n\t\teq16 & Government consumption & $G_i = \\mu_i ~ ( T^d + \\sum_j T^z_j + T^p + T^v - S^g ) / p^q_i$ \\\\\n\t\t\\midrule\n\t\teq17 & Private savings & $S^p = \\sum_i (1+\\tau^v) \\cdot p^q_i \\cdot I_i - S^g - \\epsilon \\cdot S^f$ \\\\\n\t\teq18 & Government savings & $S^g = ssg \\cdot (T^d + \\sum_j T^z_j + T^p + T^v)$ \\\\\n\t\t\\midrule\n\t\teq19 & Household income & $Inc = p^k \\cdot KK + \\sum_s p^n_s \\cdot NN_s $ \\\\\n\t\teq20 & Houshold consumption & $C_i = \\alpha_i \\cdot (Inc -S^p - T^d) / ((1+\\tau^p) ~ p^q_i) $ \\\\\n\t\t\\midrule\n\t\teq21 & Export demand & $p^e_i = \\epsilon \\cdot pWe_i$ \\\\\n\t\teq22 & Import supply & $p^m_i =\\epsilon \\cdot p^{Wm}_i$ \\\\\n\t\teq23 & Balance of trade & $\\sum_i p^{We}_i \\cdot E_i +S^f = \\sum_i p^{Wm}_i \\cdot M_i$ \\\\\n\t\teq24 & Trade closure & $S^f = \\overline{S^f}$ \\quad OR \\quad $\\epsilon = \\overline{\\epsilon}$ \\\\\n\t\t\\midrule\n\t\teq25 & Armington function & $Q_i = scQ_i \\cdot (\\alpha^m_i M_i^{\\rho_Q} + \\alpha^d_i D_i^{\\rho_Q} )^{1/\\rho_Q} $ \\\\\n\t\teq26 & Imports demand & $M_i = \\left( scQ_i^{\\rho_Q} \\cdot \\alpha_i^m \\cdot \\frac{p_i^q}{p_i^m} \\right)^{\\sigma_Q} Q_i$ \\\\\n\t\teq27& Domestic demand & $D_i = \\left( scQ_i^{\\rho_Q} \\cdot \\alpha_i^d \\cdot \\frac{p_i^q}{p_i^d} \\right)^{\\sigma_Q} Q_i$ \\\\\n\t\teq28 & CET function & $Z_i = \\theta_i \\cdot (\\xi^e_i \\cdot E_i^{\\phi_i} + \\xi^d_i \\cdot D_i^{\\phi_i} )^{(1/\\phi_i)} $ \\\\\n\t\teq29 & Demand of E & $ E_i = (\\theta_i^{\\phi_i} \\cdot \\xi^e_i\\cdot (1+\\tau^z_i) ~ p^z_i/p^e_i)^{(1/(1-\\phi_i))} ~Z_i $ \\\\\n\t\teq30 & Demand of D & $ D_i = (\\theta_i^{\\phi_i} \\cdot \\xi^d_i \\cdot (1+\\tau^z_i) ~ p^z_i/p^d_i)^{(1/(1-\\phi_i))} ~ Z_i $ \\\\\n\t\t\\midrule\n\t\teq31 & Balance of domestic good & $Q_i = C_i + I_i$ & \\\\\n\t\teq32 & Balance of labor & $NN_s = \\sum_j N_{s,j}$ & \\\\\n\t\teq33 & Balance of capital & $KK = \\sum_j K_j $ & \\\\\n\t\teq34 & Price equality & $p^c_i = p^z_i$ & \\\\ \n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\n\\clearpage\n\n\\subsubsection{Cobb-Douglas elasticity vs Empirical estimates}\n\n\\begin{table}[!ht]\n\t\\centering\n\t\\caption{Comparing job creation using CES vs Cobb-Douglas \\\\\\hspace{\\textwidth} (for a wage curve elasticity $\\gamma$=0.1, $\\sigma_{Armington}$ =2 and $\\psi_{CET}$=2)}\n\t\\label{tab:CobbDouglasError}\n\t\\begin{tabular}{lcccc}\n\t\t\\toprule\n\t\tTechno & Elasticité & Value & Value & Error (\\%) \\\\\n\t & $\\sigma_{KL}$ & with CES & with Cobb-Douglas& Cobb-Douglas vs CES \\\\\n\t\t\\midrule \n\t\tSolar & 0.2 & 5858 & 3542 & -40 \\\\\nSolar & 0.3 &5394 & 3542 & -34 \\\\\nSolar & 0.4 & 5003 & 3542 & -29 \\\\\nSolar & 0.5 & 4670 & 3542 & -24 \\\\\nSolar & 0.6 & 4382 & 3542 & -19 \\\\\nWeatherization & 0.2 & 6589 & 4135 & -37 \\\\\nWeatherization & 0.3 & 6097 & 4135 & -32 \\\\\nWeatherization & 0.4 & 5683 & 4135 & -27 \\\\\nWeatherization & 0.5 & 5331 & 4135 & -22 \\\\\nWeatherization & 0.6 & 5026 & 4135 & -18 \\\\\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\n\\clearpage\n\n\\subsubsection{Trade}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=4cm]{figures/Armington.pdf}\n\t\\caption{Impact of the Armington elasticity (with CET elasticity equal to two)}\n\t\\label{fig:armington}\n\\end{figure}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=4cm]{figures/CET.pdf}\n\t\\caption{Impact of the CET elasticity (with Armington elasticity equal to two)}\n\t\\label{fig:cet}\n\\end{figure}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[height=4cm]{figures/closure.pdf}\n\t\\caption{Impact of the trade closure rule}\n\t\\label{fig:closure}\n\\end{figure}\n\n\\subsubsection{Full sensitivity analysis}\n\\label{sec:full_sensitivity}\n\n\\begin{table}\n\t\\small\n\t\\centering\n\t\\caption{Full sensitivity analysis for solar }\n\t\\begin{tabular}{cccccccc}\n\t\t\\toprule\n\t\tTechno & $\\gamma$ & $\\sigma_{KL}$ & $\\sigma_{\\text{Armington}}$ & $\\psi_{\\text{CET}}$ & Jobs in CGE &Jobs in IO & Ratio \\\\\n\t\t\\midrule\n\t\tSolar & 0.1 & 0.2 & 1.5 & 1.5 & 6035 & 5010 & 0.83 \\\\\n\t\tSolar & 0.1 & 0.2 & 1.5 & 2 & 5945 & 5010 & 0.84 \\\\\n\t\tSolar & 0.1 & 0.2 & 1.5 & 4 & 5612 & 5010 & 0.89 \\\\\n\t\tSolar & 0.1 & 0.2 & 1.5 & 6 & 5314 & 5010 & 0.94 \\\\\n\t\tSolar & 0.1 & 0.2 & 2 & 1.5 & 5945 & 5010 & 0.84 \\\\\n\t\tSolar & 0.1 & 0.2 & 2 & 2 & 5858 & 5010 & 0.86 \\\\\n\t\tSolar & 0.1 & 0.2 & 2 & 4 & 5534 & 5010 & 0.91 \\\\\n\t\tSolar & 0.1 & 0.2 & 2 & 6 & 5244 & 5010 & 0.96 \\\\\n\t\tSolar & 0.1 & 0.2 & 4 & 1.5 & 5613 & 5010 & 0.89 \\\\\n\t\tSolar & 0.1 & 0.2 & 4 & 2 & 5535 & 5010 & 0.91 \\\\\n\t\tSolar & 0.1 & 0.2 & 4 & 4 & 5245 & 5010 & 0.96 \\\\\n\t\tSolar & 0.1 & 0.2 & 4 & 6 & 4984 & 5010 & 1.01 \\\\\n\t\tSolar & 0.1 & 0.2 & 6 & 1.5 & 5316 & 5010 & 0.94 \\\\\n\t\tSolar & 0.1 & 0.2 & 6 & 2 & 5247 & 5010 & 0.95 \\\\\n\t\tSolar & 0.1 & 0.2 & 6 & 4 & 4985 & 5010 & 1.00 \\\\\n\t\tSolar & 0.1 & 0.2 & 6 & 6 & 4748 & 5010 & 1.06 \\\\\n\t\tSolar & 0.1 & 0.4 & 1.5 & 1.5 & 5130 & 5010 & 0.98 \\\\\n\t\tSolar & 0.1 & 0.4 & 1.5 & 2 & 5066 & 5010 & 0.99 \\\\\n\t\tSolar & 0.1 & 0.4 & 1.5 & 4 & 4823 & 5010 & 1.04 \\\\\n\t\tSolar & 0.1 & 0.4 & 1.5 & 6 & 4603 & 5010 & 1.09 \\\\\n\t\tSolar & 0.1 & 0.4 & 2 & 1.5 & 5066 & 5010 & 0.99 \\\\\n\t\tSolar & 0.1 & 0.4 & 2 & 2 & 5003 & 5010 & 1.00 \\\\\n\t\tSolar & 0.1 & 0.4 & 2 & 4 & 4766 & 5010 & 1.05 \\\\\n\t\tSolar & 0.1 & 0.4 & 2 & 6 & 4551 & 5010 & 1.10 \\\\\n\t\tSolar & 0.1 & 0.4 & 4 & 1.5 & 4824 & 5010 & 1.04 \\\\\n\t\tSolar & 0.1 & 0.4 & 4 & 2 & 4767 & 5010 & 1.05 \\\\\n\t\tSolar & 0.1 & 0.4 & 4 & 4 & 4552 & 5010 & 1.10 \\\\\n\t\tSolar & 0.1 & 0.4 & 4 & 6 & 4355 & 5010 & 1.15 \\\\\n\t\tSolar & 0.1 & 0.4 & 6 & 1.5 & 4605 & 5010 & 1.09 \\\\\n\t\tSolar & 0.1 & 0.4 & 6 & 2 & 4553 & 5010 & 1.10 \\\\\n\t\tSolar & 0.1 & 0.4 & 6 & 4 & 4356 & 5010 & 1.15 \\\\\n\t\tSolar & 0.1 & 0.4 & 6 & 6 & 4175 & 5010 & 1.20 \\\\\n\t\tSolar & 0.1 & 0.6 & 1.5 & 1.5 & 4480 & 5010 & 1.12 \\\\\n\t\tSolar & 0.1 & 0.6 & 1.5 & 2 & 4430 & 5010 & 1.13 \\\\\n\t\tSolar & 0.1 & 0.6 & 1.5 & 4 & 4244 & 5010 & 1.18 \\\\\n\t\tSolar & 0.1 & 0.6 & 1.5 & 6 & 4073 & 5010 & 1.23 \\\\\n\t\tSolar & 0.1 & 0.6 & 2 & 1.5 & 4431 & 5010 & 1.13 \\\\\n\t\tSolar & 0.1 & 0.6 & 2 & 2 & 4382 & 5010 & 1.14 \\\\\n\t\tSolar & 0.1 & 0.6 & 2 & 4 & 4200 & 5010 & 1.19 \\\\\n\t\tSolar & 0.1 & 0.6 & 2 & 6 & 4033 & 5010 & 1.24 \\\\\n\t\tSolar & 0.1 & 0.6 & 4 & 1.5 & 4245 & 5010 & 1.18 \\\\\n\t\tSolar & 0.1 & 0.6 & 4 & 2 & 4201 & 5010 & 1.19 \\\\\n\t\tSolar & 0.1 & 0.6 & 4 & 4 & 4034 & 5010 & 1.24 \\\\\n\t\tSolar & 0.1 & 0.6 & 4 & 6 & 3879 & 5010 & 1.29 \\\\\n\t\tSolar & 0.1 & 0.6 & 6 & 1.5 & 4075 & 5010 & 1.23 \\\\\n\t\tSolar & 0.1 & 0.6 & 6 & 2 & 4035 & 5010 & 1.24 \\\\\n\t\tSolar & 0.1 & 0.6 & 6 & 4 & 3880 & 5010 & 1.29 \\\\\n\t\tSolar & 0.1 & 0.6 & 6 & 6 & 3737 & 5010 & 1.34 \\\\\n\t\t\\bottomrule\n\t\t\\end{tabular}\n\\end{table}\t\n\n\\begin{table}\n\t\\small\n\t\t\\centering\n\t\t\\caption{Full sensitivity analysis for weatherization}\n\t\t\\begin{tabular}{cccccccc}\n\t\t\\toprule\n\t\t\tTechno & $\\gamma$ & $\\sigma_{KL}$ & $\\sigma_{\\text{Armington}}$ & $\\psi_{\\text{CET}}$ & Jobs in CGE &Jobs in IO & Ratio \\\\\n\t\t\\midrule\n\t\tWeatherization & 0.1 & 0.2 & 1.5 & 1.5 & 6782 & 8050 & 1.19 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 1.5 & 2 & 6684 & 8050 & 1.20 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 1.5 & 4 & 6321 & 8050 & 1.27 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 1.5 & 6 & 5997 & 8050 & 1.34 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 2 & 1.5 & 6683 & 8050 & 1.20 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 2 & 2 & 6589 & 8050 & 1.22 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 2 & 4 & 6237 & 8050 & 1.29 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 2 & 6 & 5921 & 8050 & 1.36 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 4 & 1.5 & 6321 & 8050 & 1.27 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 4 & 2 & 6236 & 8050 & 1.29 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 4 & 4 & 5921 & 8050 & 1.36 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 4 & 6 & 5637 & 8050 & 1.43 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 6 & 1.5 & 5998 & 8050 & 1.34 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 6 & 2 & 5922 & 8050 & 1.36 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 6 & 4 & 5638 & 8050 & 1.43 \\\\\n\t\tWeatherization & 0.1 & 0.2 & 6 & 6 & 5380 & 8050 & 1.50 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 1.5 & 1.5 & 5824 & 8050 & 1.38 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 1.5 & 2 & 5753 & 8050 & 1.40 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 1.5 & 4 & 5486 & 8050 & 1.47 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 1.5 & 6 & 5244 & 8050 & 1.54 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 2 & 1.5 & 5753 & 8050 & 1.40 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 2 & 2 & 5683 & 8050 & 1.42 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 2 & 4 & 5423 & 8050 & 1.48 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 2 & 6 & 5187 & 8050 & 1.55 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 4 & 1.5 & 5486 & 8050 & 1.47 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 4 & 2 & 5423 & 8050 & 1.48 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 4 & 4 & 5187 & 8050 & 1.55 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 4 & 6 & 4971 & 8050 & 1.62 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 6 & 1.5 & 5245 & 8050 & 1.53 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 6 & 2 & 5188 & 8050 & 1.55 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 6 & 4 & 4972 & 8050 & 1.62 \\\\\n\t\tWeatherization & 0.1 & 0.4 & 6 & 6 & 4774 & 8050 & 1.69 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 1.5 & 1.5 & 5135 & 8050 & 1.57 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 1.5 & 2 & 5080 & 8050 & 1.58 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 1.5 & 4 & 4873 & 8050 & 1.65 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 1.5 & 6 & 4684 & 8050 & 1.72 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 2 & 1.5 & 5080 & 8050 & 1.58 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 2 & 2 & 5026 & 8050 & 1.60 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 2 & 4 & 4824 & 8050 & 1.67 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 2 & 6 & 4639 & 8050 & 1.74 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 4 & 1.5 & 4873 & 8050 & 1.65 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 4 & 2 & 4824 & 8050 & 1.67 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 4 & 4 & 4639 & 8050 & 1.74 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 4 & 6 & 4468 & 8050 & 1.80 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 6 & 1.5 & 4684 & 8050 & 1.72 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 6 & 2 & 4640 & 8050 & 1.74 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 6 & 4 & 4469 & 8050 & 1.80 \\\\\n\t\tWeatherization & 0.1 & 0.6 & 6 & 6 & 4310 & 8050 & 1.87 \\\\\n\t\t\\bottomrule\n\t\\end{tabular}\n\\end{table}\n\n", "meta": {"hexsha": "cabba4ea7cf84fe1b629becc27cfba261ab261da", "size": 26151, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "chapter4/appendix.tex", "max_stars_repo_name": "QPerrier/Dissertation", "max_stars_repo_head_hexsha": "bf7382137cee951292aa7fb0404b9075d7d2a2eb", "max_stars_repo_licenses": ["MTLL"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "chapter4/appendix.tex", "max_issues_repo_name": "QPerrier/Dissertation", "max_issues_repo_head_hexsha": "bf7382137cee951292aa7fb0404b9075d7d2a2eb", "max_issues_repo_licenses": ["MTLL"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "chapter4/appendix.tex", "max_forks_repo_name": "QPerrier/Dissertation", "max_forks_repo_head_hexsha": "bf7382137cee951292aa7fb0404b9075d7d2a2eb", "max_forks_repo_licenses": ["MTLL"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 39.6828528073, "max_line_length": 481, "alphanum_fraction": 0.6260181255, "num_tokens": 10425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355186, "lm_q2_score": 0.7931059536292271, "lm_q1q2_score": 0.6121536817309697}} {"text": "%%%%%%%%%%%%%%%%%%%%%%\n%%%%%%%%%%%%%%%%%%%%%%\n\\section{Discussion}\n\\label{sec:Discussion}\n%%%%%%%%%%%%%%%%%%%%%%\n\nWe have addressed the problem of characterizing the loss surface of neural networks from the perspective\nof gradient descent algorithms. We explored two angles -- topological and geometrical aspects -- that build on top of each other. \n\n\nOn the one hand, we have presented new theoretical results that quantify \nthe amount of uphill climbing that is required in order to progress to lower energy configurations in \nsingle hidden-layer ReLU networks, and proved that this amount converges to zero with overparametrization under mild conditions. On the other hand, we have introduced a dynamic programming algorithm that efficiently approximates geodesics within each level set, providing a tool that not only verifies the connectedness of level sets, but also estimates the geometric regularity of these sets. Thanks to this information, we can quantify how `non-convex' an optimization problem is, and verify that the optimization of quintessential deep learning tasks -- CIFAR-10 and MNIST classification using CNNs, and next word prediction using LSTMs -- behaves in a nearly convex fashion up until they reach high accuracy levels.\n\nThat said, there are some limitations to our framework. In particular, we do not address saddle-point issues that can greatly affect the actual convergence of gradient descent methods. There are also a number of open questions; amongst those, in the near future we shall concentrate on:\n\\begin{itemize}\n\\item \\emph{Extending Theorem \\ref{maintheo} to the multilayer case}. We believe this is within reach, since the main analytic tool we use is that small changes in the parameters result in small changes in the covariance structure of the features. That remains the case in the multilayer case. \n\\item \\emph{Empirical versus Oracle Risk}. A big limitation of our theory is that right now it does not inform us on the differences between optimizing the empirical risk versus the oracle risk. Understanding the impact of generalization error and stochastic gradient in the ability to do small uphill climbs is an open line of research.\n\\item \\emph{Influence of symmetry groups}. Under appropriate conditions, the presence of discrete symmetry groups does not prevent the loss from being connected, but at the expense of increasing the capacity. An important open question is whether one can improve the asymptotic properties by relaxing connectedness to being connected up to discrete symmetry. \n\\item \\emph{Improving numerics with Hyperplane method}. Our current numerical experiments employ a greedy (albeit faster) algorithm to discover connected components and estimate geodesics. We plan to perform experiments using the less greedy algorithm described in Appendix \\ref{sec:ConstrainedAlg}. \n\\end{itemize}\n \n\n%\\begin{itemize}\n%\\item Future: Generalization Error Question. \n%\n%\\end{itemize}\n", "meta": {"hexsha": "cea5ceb75e514d2f54240c324840c2524c47a353", "size": 2946, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Writeup/iclr/conclusions.tex", "max_stars_repo_name": "danielfreeman11/convex-nets", "max_stars_repo_head_hexsha": "252a8230845fb2076221113ac8cabfade5152bfb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-08-09T00:48:46.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-03T09:04:59.000Z", "max_issues_repo_path": "Writeup/iclr/conclusions.tex", "max_issues_repo_name": "danielfreeman11/convex-nets", "max_issues_repo_head_hexsha": "252a8230845fb2076221113ac8cabfade5152bfb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Writeup/iclr/conclusions.tex", "max_forks_repo_name": "danielfreeman11/convex-nets", "max_forks_repo_head_hexsha": "252a8230845fb2076221113ac8cabfade5152bfb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 105.2142857143, "max_line_length": 719, "alphanum_fraction": 0.7942973523, "num_tokens": 591, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718435083355187, "lm_q2_score": 0.7931059487389968, "lm_q1q2_score": 0.6121536779564773}} {"text": "\n\\section{Flow over a smooth downward step}\n\nThis is a standard textbook energy conservation type problem. Water flows over a flat bedded channel, then through a smooth downward step, and continues over a flat bottomed channel with lower bed elevation.\n\nThe bed is represented with a one dimensional domain over $[0,25]$ m, with topography\n\\begin{equation}\nz(x)= \\left\\{ \\begin{array}{ll}\n 0.2 & ~\\textrm{if}\\quad x \\leq 9 ,,\\\\\n 0.2-0.1(x-9) & ~\\textrm{if}\\quad 9 \\leq x \\leq 11\\,,\\\\\n 0 & ~\\textrm{otherwise}\\,,\\\\\n\\end{array} \\right.\n\\end{equation}\nThe stage is imposed at the downstream boundary, and the discharge is imposed at the upstream boundary.\n\nThe analytical height is found by solving the Bernoulli equation. The simplified Bernoulli equation is the following cubic equation\n\\begin{equation}\nh^3 + \\left(z - \\frac{q^2}{2 g H^2} - H \\right) h^2 + \\frac{q^2}{2 g} = 0\\,,\n\\end{equation}\nwhere $H$ is the upstream height and $q=uh$ is the discharge or $x$-momentum. When the height $h$ has been found, the velocity is computed as $u=q/h$\\,.\n\n\\subsection{Results}\nFor our test we impose a discharge-per-unit-width of 1m$^2$/s, and a downstream stage boundary condition of 1.0m. \n\nRepresentatives of the simulation results are given in the following three figures. The boundary regions are excluded from the plot because the inflow boundary takes some time to adapt to the analytical solution (which is to be expected as we don't enforce the depth or velocity directly there). There should be excellent agreement between the numerical and analytical solutions. \n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.9\\textwidth]{stage_plot.png}\n\\end{center}\n\\caption{Stage results}\n\\end{figure}\n\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.9\\textwidth]{xmom_plot.png}\n\\end{center}\n\\caption{Xmomentum results}\n\\end{figure}\n\n\n\\begin{figure}\n\\begin{center}\n\\includegraphics[width=0.9\\textwidth]{xvel_plot.png}\n\\end{center}\n\\caption{Xvelocity results}\n\\end{figure}\n\n\n\\endinput\n", "meta": {"hexsha": "94d76ac614bcd92175e91e910b612f9fad7a2fb6", "size": 1992, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "validation_tests/analytical_exact/subcritical_depth_expansion/results.tex", "max_stars_repo_name": "samcom12/anuga_core", "max_stars_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_stars_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_stars_count": 136, "max_stars_repo_stars_event_min_datetime": "2015-05-07T05:47:43.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-16T03:07:40.000Z", "max_issues_repo_path": "validation_tests/analytical_exact/subcritical_depth_expansion/results.tex", "max_issues_repo_name": "samcom12/anuga_core", "max_issues_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_issues_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_issues_count": 184, "max_issues_repo_issues_event_min_datetime": "2015-05-03T09:27:54.000Z", "max_issues_repo_issues_event_max_datetime": "2021-12-20T04:22:48.000Z", "max_forks_repo_path": "validation_tests/analytical_exact/subcritical_depth_expansion/results.tex", "max_forks_repo_name": "samcom12/anuga_core", "max_forks_repo_head_hexsha": "f4378114dbf02d666fe6423de45798add5c42806", "max_forks_repo_licenses": ["Python-2.0", "OLDAP-2.7"], "max_forks_count": 70, "max_forks_repo_forks_event_min_datetime": "2015-03-18T07:35:22.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-01T07:07:29.000Z", "avg_line_length": 38.3076923077, "max_line_length": 380, "alphanum_fraction": 0.75, "num_tokens": 572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7718434978390746, "lm_q2_score": 0.7931059438487663, "lm_q1q2_score": 0.6121536658571924}} {"text": "% Created 2021-07-14 Wed 11:54\n% Intended LaTeX compiler: pdflatex\n\\documentclass[presentation,aspectratio=169]{beamer}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\\usepackage{graphicx}\n\\usepackage{grffile}\n\\usepackage{longtable}\n\\usepackage{wrapfig}\n\\usepackage{rotating}\n\\usepackage[normalem]{ulem}\n\\usepackage{amsmath}\n\\usepackage{textcomp}\n\\usepackage{amssymb}\n\\usepackage{capt-of}\n\\usepackage{hyperref}\n\\usepackage{khpreamble}\n\\usepackage{amssymb}\n\\usepackage{pgfplotstable}\n\\DeclareMathOperator{\\shift}{q}\n\\DeclareMathOperator{\\diff}{p}\n\\usetheme{default}\n\\author{Kjartan Halvorsen}\n\\date{\\today}\n\\title{System identification}\n\\hypersetup{\n pdfauthor={Kjartan Halvorsen},\n pdftitle={System identification},\n pdfkeywords={},\n pdfsubject={},\n pdfcreator={Emacs 26.3 (Org mode 9.4.6)}, \n pdflang={English}}\n\\begin{document}\n\n\\maketitle\n\n\\section{Intro}\n\\label{sec:org2f040f4}\n\\begin{frame}[label={sec:org0fdcf79}]{A complicated process}\n\\begin{columns}\n\\begin{column}{0.6\\columnwidth}\nFrom Wikipedia \"Cyclonic separation\"\n\\end{column}\n\\begin{column}{0.4\\columnwidth}\n\\begin{center}\n\\includegraphics[height=1.0\\textheight]{../../figures/Vertical-cyclone.jpg}\n\\end{center}\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n\\begin{frame}[label={sec:org5066307}]{System identification}\n\\begin{center}\n \\begin{tikzpicture}[node distance=22mm, block/.style={rectangle, draw, minimum width=15mm, inner sep=10pt}, sumnode/.style={circle, draw, inner sep=2pt},]\n\n \\node[coordinate] (input) {};\n \\node[coordinate, right of=input] (copy) {};\n \\node[coordinate, right of=copy] (midp) {};\n \\node[block, above of=midp, node distance=10mm] (sys) {System};\n \\node[block, below of=midp, node distance=10mm] (mod) {Model};\n \\node[sumnode, right of=midp, node distance=26mm] (sum) {\\tiny $\\Sigma$};\n \\node[coordinate, right of=sum, node distance=22mm] (output) {};\n\n \\draw[-] (input) -- node[above, pos=0.2] {Measured input} (copy);\n \\draw[->] (copy) |- node[above] {} (sys);\n \\draw[->] (copy) |- node[above] {} (mod);\n \\draw[->] (sys) -| node[left, pos=0.9] {$+$} (sum);\n \\draw[->] (mod) -| node[left, pos=0.9] {$-$} (sum);\n \\draw[->] (sum) -- node[above, near end] {Error} (output);\n\n \\draw[thick, red!70!black, ->] (2.7,-2) -- (3.3,-2) -- (5.3, 0);\n \\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\n\\begin{frame}[label={sec:orgbf66964}]{The Auto-Regressive with eXogenous input (ARX) model}\n\\begin{center}\n\\includegraphics[width=0.7\\linewidth]{../../figures/block-arx}\n\\end{center}\n\\[ A(\\shift) y(k) = B(\\shift)u(k) + e(k+n) \\]\n\nThe error signal \\(e(k)\\) is a zero-mean white noise sequence representing perturbations and modeling errors.\n\\end{frame}\n\n\\begin{frame}[label={sec:org43fbae7}]{First-order ARX model with one delay}\n\\begin{center}\n\\includegraphics[width=0.3\\linewidth]{../../figures/block-arx}\n\\end{center}\n\\[ (\\shift + a_1) y(k) = (b_0 \\shift + b_1) \\shift^{-1}u(k) + e(k+1) \\]\n\\[ y(k+1) + a_1y(k) = b_0u(k) + b_1u(k-1) + e(k+1) \\]\n\n\\pause\n\nUsing the model to predict the output one step ahead:\n\\begin{align*}\n \\hat{y}(k+1) &= -a_1y(k) + b_0u(k) + b_1u(k-1) = \\underbrace{\\begin{bmatrix} -y(k) & u(k) & u(k-1) \\end{bmatrix}}_{\\varphi_{k+1}^T} \\underbrace{\\begin{bmatrix} a_1\\\\b_0\\\\b_1\\end{bmatrix}}_{\\theta}\\\\\n &= \\varphi_{k+1}^T\\theta\n \\end{align*}\n\\end{frame}\n\n\n\n\\begin{frame}[label={sec:org82bc73b}]{Parameter estimation - Least squares}\n\\alert{Objective} Given observations \\[\\mathcal{D} = \\{ (u_1,y_1), (u_2, y_2), \\ldots, (u_N, y_N)\\}\\] and model \\(\\mathcal{M}: \\; y(k+1) = -ay(k) + b_0u(k) + b_1u(k-1) + e(k+1)\\), obtain the parameters \\((a_1,\\,b_0,\\,b_1)\\) which gives the best fit of the model to the data.\n\\end{frame}\n\n\n\n\\begin{frame}[label={sec:orgee9dfe7}]{Parameter estimation - Least squares}\nGiven observations \\[\\mathcal{D} = \\{ (u_1,y_1), (u_2, y_2), \\ldots, (u_N, y_N)\\}\\] and model \\(\\mathcal{M}: \\; y(k+1) = -ay(k) + b_0u(k) + b_1u(k-1) + e(k+1)\\).\n\n\\begin{enumerate}\n\\item Form the one-step ahead prediction\n\\[ \\hat{y}_{k+1} = -a_1y_k + b_0u_k + b_1u_{k-1} = \\underbrace{\\begin{bmatrix} -y_k & u_k & u_{k-1} \\end{bmatrix}}_{\\varphi_{k+1}^T} \\underbrace{\\begin{bmatrix} a_1\\\\b_0\\\\b_1\\end{bmatrix}}_{\\theta}\\] and the prediction error\n \\[ \\epsilon_{k+1} = y_{k+1} - \\hat{y}_{k+1} = y_{k+1} - \\varphi_{k+1}^T\\theta.\\]\n\\end{enumerate}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:orgd1a0844}]{Parameter estimation - Least squares}\n\\begin{enumerate}\n\\setcounter{enumi}{1}\n\\item Combine all the observations \\(y_k\\) and predictions \\(\\hat{y}_k\\) on vector form\n\\begin{align*}\n\\epsilon &= \\begin{bmatrix} \\epsilon_3\\\\\\epsilon_4\\\\\\vdots\\\\\\epsilon_N\\end{bmatrix} = \\begin{bmatrix} y_3\\\\ y_4\\\\\\vdots\\\\y_N \\end{bmatrix} - \\begin{bmatrix} \\hat{y}_3\\\\ \\hat{y}_4\\\\\\vdots\\\\\\hat{y}_N \\end{bmatrix}\n = \\begin{bmatrix} y_2\\\\ y_3\\\\\\vdots\\\\y_N \\end{bmatrix} - \\begin{bmatrix} \\varphi_2^T\\theta\\\\ \\varphi_3^T\\theta\\\\\\vdots\\\\\\varphi_N^T\\theta \\end{bmatrix}\\\\\n&= y - \\underbrace{\\begin{bmatrix}\\varphi_1^T\\\\\\varphi_2^T\\\\\\vdots\\\\\\varphi_N^T\\end{bmatrix}}_{\\Phi}\\theta = y - \\Phi\\theta \n\\end{align*}\n\\item Solve \\(\\arg\\min \\; J(\\theta) = \\frac{1}{2}\\epsilon^T\\epsilon = \\frac{1}{2}\\sum_{i=3}^N \\epsilon_i(\\theta)^2\\)\n\\end{enumerate}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:org6ce2d0d}]{The problem with least squares}\n\\begin{columns}\n\\begin{column}{0.4\\columnwidth}\n\\begin{align*}\n \\text{minimize} \\; &\\sum_k g(\\epsilon_k)\\\\\n \\text{where} \\; g(u) &= u^2\n\\end{align*}\n\\end{column}\n\n\\begin{column}{0.6\\columnwidth}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n width=8cm,\n height=6cm,\n ylabel=loss,\n xlabel=penalty,\n ]\n \\addplot[red, thick, no marks, domain=-4:4, samples=201] {x^2};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n\\begin{frame}[label={sec:org49222fc}]{More robust: The Huber loss function}\n\\begin{columns}\n\\begin{column}{0.4\\columnwidth}\nAlso known as \\alert{robust regression}\n\\begin{align*}\n \\text{minimize} \\; &\\sum_k g_{hub}(\\epsilon_k)\\\\\n \\text{where}\\; g_{hub}(u) &= \\begin{cases} u^2 & |u| \\le M\\\\ M(2|u|-M) & |u| > M \\end{cases}\n\\end{align*}\n\\end{column}\n\n\\begin{column}{0.6\\columnwidth}\n\\begin{center}\n \\begin{tikzpicture}\n \\begin{axis}[\n width=8cm,\n height=6cm,\n ylabel=penalty,\n xlabel=residual,\n ]\n \\addplot[red, thick, no marks, domain=-4:4, samples=201] {x^2};\n \\addplot[orange!90!black, ultra thick, no marks, domain=-4:-1, samples=201] {2*abs(x)-1};\n \\addplot[orange!90!black, thin, no marks, domain=-1:1, samples=201] {x^2};\n \\addplot[orange!90!black, ultra thick, no marks, domain=1:4, samples=201] {2*abs(x)-1};\n \\end{axis}\n \\end{tikzpicture}\n\\end{center}\n\\end{column}\n\\end{columns}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:org10b0bf3}]{First-order ARX model without delay}\n\\begin{center}\n\\includegraphics[width=0.4\\linewidth]{../../figures/block-arx}\n\\end{center}\n\\[ (\\shift + a_1) y(k) = (b_0 \\shift + b_1) u(k) + e(k+1) \\]\n\n\\alert{Activity}\n\n\\begin{enumerate}\n\\item Determine the one-step ahead predictor \\(\\hat{y}_{k+1}\\) and the prediction error \\(\\epsilon_{k+1}\\).\n\\item Form the system of equations \\(\\Phi\\theta = y\\)\n\\end{enumerate}\n\\end{frame}\n\n\n\\begin{frame}[label={sec:orgc2d06b4}]{The ARX model}\n\\[ A(\\shift) y(k) = B(\\shift)u(k) + e(k+n)\\]\n\n\\alert{Activity} Fill the empty blocks.\n\n\\begin{center}\n \\begin{tikzpicture}[node distance=22mm, block/.style={rectangle, draw, minimum width=15mm, minimum height=12mm}, sumnode/.style={circle, draw, inner sep=2pt}]\n \n \\node[coordinate] (input) {};\n \\node[block, right of=input, node distance=20mm] (plant) {};\n \\node[sumnode, right of=plant, node distance=24mm] (sum) {\\tiny $\\Sigma$};\n \\node[block, above of=sum, node distance=20mm] (dist) {};\n\n \\node[coordinate, above of=dist, node distance=12mm] (disturbance) {};\n \\node[coordinate, right of=sum, node distance=20mm] (output) {};\n\n \\draw[->] (input) -- node[above, pos=0.3] {$u(k)$} (plant);\n \\draw[->] (plant) -- node[above] {} (sum);\n \\draw[->] (sum) -- node[above, near end] {$y(k)$} (output);\n \\draw[->] (disturbance) -- node[right, pos=0.2] {$e(k)$} (dist);\n \\draw[->] (dist) -- node[above] {} (sum);\n\n \\end{tikzpicture}\n\\end{center}\n\\end{frame}\n\\end{document}", "meta": {"hexsha": "e3bdc88a0c53167e62042bad877e4ec7cd20af2f", "size": 8218, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "system-identification/slides/lecture-sysid.tex", "max_stars_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_stars_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-11-07T05:20:37.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-22T09:46:13.000Z", "max_issues_repo_path": "system-identification/slides/lecture-sysid.tex", "max_issues_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_issues_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 4, "max_issues_repo_issues_event_min_datetime": "2020-06-12T20:44:41.000Z", "max_issues_repo_issues_event_max_datetime": "2020-06-12T20:49:00.000Z", "max_forks_repo_path": "system-identification/slides/lecture-sysid.tex", "max_forks_repo_name": "kjartan-at-tec/mr2007-computerized-control", "max_forks_repo_head_hexsha": "16e35f5007f53870eaf344eea1165507505ab4aa", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-03-14T03:55:27.000Z", "max_forks_repo_forks_event_max_datetime": "2021-03-14T03:55:27.000Z", "avg_line_length": 35.5757575758, "max_line_length": 275, "alphanum_fraction": 0.654295449, "num_tokens": 3099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.793105941403651, "lm_q2_score": 0.7718434978390747, "lm_q1q2_score": 0.6121536639699462}} {"text": "\\subsection{Formal languages}\\label{subsec:formal_languages}\n\nLanguages are used to define formulas for expressing the \\hyperref[def:zfc]{axioms of set theory}. Here, sets are used to formally define languages. A simple way out of this vicious cycle is via the theory-metatheory relationship discussed in \\fullref{rem:metalogic} and \\fullref{rem:set_definition_recursion}. In short, we define languages within the metatheory using the already available concept of set, and we later define formulas, again in the metatheory, which allows us to subsequently formally define sets via axioms within the object logic.\n\n\\begin{definition}\\label{def:formal_language}\n Fix a nonempty set \\( \\mscrA \\).\n\n \\begin{thmenum}\n \\thmitem{def:formal_language/alphabet} We call \\( \\mscrA \\) an \\term{alphabet}.\n\n \\thmitem{def:formal_language/symbol} We call each element of \\( \\mscrA \\) a \\term{symbol}.\n\n \\thmitem{def:formal_language/word} A \\term{string} over \\( \\mscrA \\) is a \\hyperref[def:sequence]{finite sequence} of symbols. If \\( (a, b, c) \\) is a word, for convenience we write it as the string \\( abc \\). This is the reason words are also referred to as \\term{strings}. This notation only makes sense if each symbol of the language is actually represented by one typographic symbol.\n\n The term \\enquote{string} is common in programming practice. In the context of formal languages, strings are often called \\term{words}. We will avoid the later term since it does not correspond to the everyday use of the term \\enquote{word}.\n\n \\thmitem{def:formal_language/empty_word} We denote the empty string by \\( \\varepsilon \\).\n\n \\thmitem{def:formal_language/word_length} The \\term{length} \\( \\len(w) \\) of a word \\( w \\) is the number of elements of the tuple \\( w \\).\n\n \\thmitem{def:formal_language/concatenation} The \\term{concatenation} of the words \\( v = (v_1, \\ldots, v_n) \\) and \\( w = (w_1, \\ldots, w_m) \\) is the word\n \\begin{equation*}\n vw \\coloneqq (v_1, \\ldots, v_n, w_1, \\ldots, w_m).\n \\end{equation*}\n\n We abbreviate \\( \\overbrace{w w \\ldots w}^{k \\T{times}} \\) as \\( w^k \\). This is only a notation. We do not distinguish, formally, between the words \\( aaabbaa \\) and \\( a^3 b^2 a^2 \\), nor between \\( a \\varepsilon b \\) and \\( ab \\).\n\n \\thmitem{def:formal_language/reverse} The \\term{reverse word} of \\( w = (w_1, \\ldots, w_n) \\) is\n \\begin{equation*}\n \\op{rev}(w) \\coloneqq (w_n, \\ldots, w_1).\n \\end{equation*}\n\n \\thmitem{def:formal_language/prefix} The word \\( p = (p_1, \\ldots, p_m) \\) is a \\term{prefix} of \\( w = (w_1, \\ldots, w_n) \\) if\n \\begin{equation*}\n w = (\\underbrace{p_1, \\ldots, p_m}_p, w_{m+1}, \\ldots, w_n).\n \\end{equation*}\n\n \\thmitem{def:formal_language/suffix} The word \\( s \\) is a \\term{suffix} of \\( w \\) if \\( \\op{rev}(s) \\) is a prefix of \\( \\op{rev}(w) \\).\n\n \\thmitem{def:formal_language/subword} The word \\( v \\) is a \\term{subword} of \\( w \\) if there exists a prefix \\( p \\) and a suffix \\( s \\) of \\( v \\) such that\n \\begin{equation*}\n w = pvs.\n \\end{equation*}\n\n \\thmitem{def:formal_language/kleene_star} The \\term{Kleene star} \\( \\mscrA^* \\) of \\( \\mscrA \\) is the set of all words over \\( \\mscrA \\). If we wish to exclude the empty word, like we often do, we instead write \\( \\mscrA^+ \\) for the set of all non-empty words over \\( \\mscrA \\).\n\n \\thmitem{def:formal_language/language} A \\term{language} over \\( \\mscrA \\) is any subset of \\( \\mscrA^* \\). Note that in some contexts like \\hyperref[subsec:propositional_logic]{propositional logic} or \\hyperref[subsec:first_order_logic]{first-order logic} the term \\enquote{language} may refer to the alphabet itself (see \\fullref{rem:propositional_language_is_alphabet}).\n \\end{thmenum}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:kleene_star_is_monoid}\n For any alphabet \\( \\mscrA \\), the Kleene star \\( \\mscrA^* \\) is a \\hyperref[def:unital_magma/monoid]{monoid} under concatenation.\n\\end{proposition}\n\\begin{proof}\n Concatenation is clearly associative and the empty word \\( \\varepsilon \\) is a \\hyperref[def:magma_identity]{two-sided identity} under concatenation.\n\\end{proof}\n\n\\begin{definition}\\label{def:formal_grammar}\\mcite[def. 2.2]{Sipser2013}\n Let \\( V \\) and \\( \\Sigma \\) be disjoint nonempty subsets of some \\hyperref[def:formal_language/alphabet]{alphabet}.\n\n \\begin{thmenum}\n \\thmitem{def:formal_grammar/terminals} We call elements of \\( \\Sigma \\) \\term{terminals}. We denote terminals in abstract grammars using lowercase Greek letters, and we denote words using lowercase Latin letters.\n\n \\thmitem{def:formal_grammar/non_terminals} We call elements of \\( V \\) \\term{non-terminals}. By convention, variables are denoted using capital letters.\n\n \\thmitem{def:formal_grammar/start} We assume that a special \\term{start symbol} \\( S \\in V \\) is fixed.\n\n \\thmitem{def:formal_grammar/production_rules} We define a binary \\hyperref[def:relation]{relation} \\( \\to \\) of \\term{production rules} over \\( (V \\cup \\Sigma)^* \\).\n\n We impose the restriction that no rules of the form \\( \\varepsilon \\to v \\) exist for any word \\( v \\). We do allow, however, production rules of the form \\( v \\to \\varepsilon \\). Such rules are called \\term{\\( \\varepsilon \\)-rules}.\n\n Rules describe transformations that define how a language is \\enquote{generated} starting from \\( S \\). See \\fullref{def:grammar_derivation} and \\fullref{ex:natural_number_arithmetic_grammar/derivation}.\n\n \\thmitem{def:formal_grammar/grammar} The quadruple \\( G \\coloneqq (V, S, \\Sigma, \\to) \\) is called a \\term{formal grammar} or simply a \\term{grammar}.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{definition}\\label{def:chomsky_hierarchy}\n We will define the \\term{Chomsky hierarchy} of \\hyperref[def:formal_grammar]{formal grammars}. We can classify a grammar \\( G = (V, S, \\Sigma, \\to) \\) as follows, based on their rules:\n \\begin{thmenum}\n \\thmitem{def:chomsky_hierarchy/unrestricted} In general, every production rule has the form \\( v \\to w \\), where both \\( v \\) and \\( w \\) are words consisting of terminal and non-terminals, and \\( v \\) is nonempty.\n\n When no additional restrictions are imposed on the rules of the grammar, we call it \\term{unrestricted grammar}. The other levels of the hierarchy are subsets of the unrestricted grammars.\n\n \\thmitem{def:chomsky_hierarchy/non_contracting} The grammar is \\term{non-contracting} if \\( \\len(v) \\leq \\len(w) \\) for every production rule \\( v \\to w \\).\n\n \\thmitem{def:chomsky_hierarchy/context_sensitive} The grammar is \\term{context-sensitive} if every rule has the form \\( aAb \\to w \\) for some non-terminal \\( A \\), arbitrary words \\( a \\) and \\( b \\), and a nonempty word \\( w \\).\n\n The requirement for \\( w \\) to be nonempty is set up so that context-sensitive grammars are non-contracting.\n\n \\thmitem{def:chomsky_hierarchy/context_free} The grammar is \\term{context-free} every rule has the form \\( A \\to w \\) for some non-terminal \\( A \\) and a nonempty word \\( w \\).\n\n Unlike for context-sensitive languages, \\( w \\) is allowed to be empty. Thus, a context-free grammar is non-contracting, however it may not be context-sensitive if it has \\( \\varepsilon \\)-rules.\n\n \\thmitem{def:chomsky_hierarchy/regular} Finally, the grammar is \\term{regular} if every rule has one of the forms\n \\begin{align*}\n &A \\to \\varepsilon, \\\\\n &A \\to B \\tau, \\\\\n &A \\to \\tau B, \\\\\n &A \\to \\tau,\n \\end{align*}\n where \\( A \\) and \\( B \\) are non-terminals and \\( \\tau \\) is a terminal.\n\n Regular grammars are obviously context-free.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:natural_number_arithmetic_grammar/backus_naur_form}\n We define a grammar for primary school notation of multiplication and division of \\hyperref[def:set_of_natural_numbers]{natural numbers}. Note that we consider the numbers in \\( \\BbbN \\) only as symbols, without any regard to semantics.\n\n Let \\( V \\coloneqq \\set{ N, O, M, E } \\) and \\( \\Sigma \\coloneqq \\BbbN \\cup \\set{ \\times, \\div, (, ) } \\). Define the grammar\n \\begin{equation}\\label{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/simple}\n \\begin{aligned}\n N &\\to 0 \\\\\n N &\\to 1 \\\\\n \\phantom{N} &\\vdots \\\\\n N &\\to n \\\\\n \\phantom{N} &\\vdots \\\\\n O &\\to \\times \\\\\n O &\\to \\div \\\\\n E &\\to N \\\\\n E &\\to (E O E)\n \\end{aligned}\n \\end{equation}\n\n It is convenient to use the following shorthands:\n \\begin{equation}\\label{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/shorthand}\n \\begin{aligned}\n N &\\to 0 \\mid 1 \\mid 2 \\mid \\ldots \\\\\n O &\\to \\times \\mid \\div \\\\\n E &\\to N \\mid (E O E)\n \\end{aligned}\n \\end{equation}\n\n We can choose different non-terminals as the starting symbol. The symbol \\( N \\) corresponds to numbers, \\( O \\) corresponds to operations, and \\( E \\) can be either a number or an expression. We say that \\eqref{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/shorthand} specifies a \\term{grammar schema}. With any starting symbol, the grammar is clearly \\hyperref[def:chomsky_hierarchy]{context-free}.\n\\end{example}\n\n\\begin{remark}\\label{rem:backus_naur_form}\n The infinitude of possible rules in \\fullref{ex:natural_number_arithmetic_grammar/backus_naur_form} may not bother us formally, but when dealing with software implementations, we must have a finite number of rules. An example of a nontrivial grammar in the wild is the Python grammar that can be found in \\cite{Python39Grammar}. There are also other advantages of introducing a more convenient metasyntax (a syntax for describing language syntax).\n\n For \\hyperref[def:chomsky_hierarchy/context_free]{context-free grammars}, is often convenient to use the \\term{Backus-Naur form (BNF)}. For \\fullref{ex:natural_number_arithmetic_grammar/backus_naur_form}, the BNF is\n \\begin{bnf*}\n \\bnfprod{nonzero digit} {\\bnfts{1} \\bnfor \\bnfts{2} \\bnfor \\bnfts{3} \\bnfor \\bnfts{4} \\bnfor \\bnfts{5} \\bnfor \\bnfts{6} \\bnfor \\bnfts{7} \\bnfor \\bnfts{8} \\bnfor \\bnfts{9}} \\\\\n \\bnfprod{digit} {\\bnfts{0} \\bnfor \\bnfpn{nonzero digit}} \\\\\n \\bnfprod{number} {\\bnfpn{nonzero digit} \\bnfor \\bnfpn{number} \\bnfsp \\bnfpn{digit}} \\\\\n \\bnfprod{operation} {\\bnfts{\\( \\times \\)} \\bnfor \\bnfts{\\( \\div \\)}} \\\\\n \\bnfprod{expression} {\\bnfpn{number} \\bnfor \\bnfts{(} \\bnfsp \\bnfpn{number} \\bnfsp \\bnfpn{operation} \\bnfsp \\bnfpn{number} \\bnfsp \\bnfts{)}}.\n \\end{bnf*}\n\n The obvious difference is that we explicitly define numbers via their decimal representation, which means that we get a finite amount of rules. Compared to \\eqref{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/simple} some other differences are:\n \\begin{itemize}\n \\item Variables are denoted by \\( \\langle \\)words enclosed in angle brackets\\( \\rangle \\), so that we can name variables more descriptively using more than one symbol.\n \\item Terminals are, by convention, put in \\enquote{quotes}. In human-readable rich text documents like this one, it is sometimes possible to use different fonts, and so instead of using \\enquote{quotes} we specify terminals using an \\texttt{upright typewriter font}.\n \\item Free-text rules can be specified using a normal font. This is also only used in human-readable rich text documents, however this usage is justified because such rules are only beneficial for human understanding and not for machine parsing.\n \\item By convention, the symbol \\( \\Coloneqq \\) is used instead of \\( \\to \\) for specifying transition rules.\n \\item Different rules with the same, source are concatenated as in \\eqref{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/shorthand}.\n \\item In order to fully describe a context-free grammar, we must only specify its Backus-Naur form and its starting variable.\n \\end{itemize}\n\\end{remark}\n\n\\begin{definition}\\label{def:grammar_derivation}\n Fix a \\hyperref[def:formal_grammar]{formal grammar} \\( G = (V, S, \\Sigma, \\to) \\).\n\n \\begin{thmenum}\n \\thmitem{def:grammar_derivation/derivation} We define the binary relation \\( \\Rightarrow \\) on the Kleene star \\( (V \\cup \\Sigma)^* \\) by declaring that, for every two \\hyperref[def:formal_language/word]{words} \\( p \\) and \\( s \\) over \\( V \\cup \\Sigma \\) and every production rule \\( v \\to w \\), we have \\( pvw \\Rightarrow pws \\). We also define the relation \\( \\Rightarrow_L \\) as a restriction of \\( \\Rightarrow \\) to the cases where \\( p \\) contains only terminal symbols and \\( \\Rightarrow_R \\) --- if \\( s \\) contains only terminal symbols.\n\n A \\term{derivation} of the word \\( w_n \\) from \\( w_1 \\) is a \\hyperref[def:quiver_path/directed]{directed path} in the quiver induced by the relation \\( \\Rightarrow \\), i.e.\n \\begin{equation}\\label{eq:def:grammar_derivation/derivation}\n w_1\n \\reloset {u_1 \\to v_1} \\implies\n w_2\n \\reloset {u_2 \\to v_2} \\implies\n \\cdots\n \\reloset {u_{n-2} \\to v_{n-2}} \\implies\n w_{n-1}\n \\reloset {u_{n-1} \\to v_{n-1}} \\implies\n w_n.\n \\end{equation}\n\n A \\term{leftmost derivation} is a derivation performed using \\( \\Rightarrow_L \\) rather than \\( \\Rightarrow \\). \\term{Rightmost derivations} are defined analogously.\n\n We say that \\( w_n \\) is \\term{derivable} from \\( w_1 \\) if there exists a derivation from \\( w_1 \\) to \\( w_n \\).\n\n We denote the \\hyperref[def:relation_closures/transitive]{transitive closure} of \\( \\Rightarrow \\) by \\( \\reloset + \\Rightarrow \\) and the \\hyperref[def:relation_closures/reflexive]{reflexive} closure of \\( \\reloset + \\Rightarrow \\) by \\( \\reloset {*} \\Rightarrow \\). Clearly \\( w_1 \\) is derivable from \\( w_n \\) if and only if \\( w_1 \\reloset {*} \\Rightarrow w_n \\).\n\n The leftmost and rightmost derivations generate the same derivability relation --- the only potential difference is in the order of rule applications in the derivation itself.\n\n \\thmitem{def:grammar_derivation/unambiguous}\\mcite[def. 2.7]{Sipser2013} We say that the word \\( w \\) can be derived \\term{unambiguously} if it has a unique leftmost derivation.\n\n Define the set\n \\begin{equation*}\n D \\coloneqq \\set{ w \\in (V \\cup \\Sigma)^* \\colon S \\reloset + \\Rightarrow w }\n \\end{equation*}\n of all words derivable from the starting symbol \\( S \\).\n\n If every word in \\( D \\) can be derived unambiguously, we say that the grammar itself is \\term{unambiguous}.\n\n In an unambiguous grammar, \\fullref{thm:structural_induction_on_unambiguous_grammars} can be used on the \\hyperref[def:quiver/simple]{simple directed graph} \\( (D, \\Rightarrow_L) \\). Indeed, every word in \\( D \\) is derivable from \\( S \\) and every leftmost derivation is unique.\n\n \\thmitem{def:grammar_derivation/grammar_language} The \\term{language} of the grammar is the set\n \\begin{equation*}\n \\mscrL(G) \\coloneqq \\set{ w \\in \\Sigma^* \\colon S \\reloset + \\Rightarrow w }\n \\end{equation*}\n of all terminal-only words derivable from the starting symbol \\( S \\).\n\n We also say that strings in \\( \\mscrL(G) \\) are \\term{generated} by the grammar \\( G \\).\n\n If a language can be generated by a \\hyperref[def:chomsky_hierarchy/regular]{regular} grammar, we say that the language itself is regular, and similarly for \\hyperref[def:chomsky_hierarchy/context_free]{context-free} and \\hyperref[def:chomsky_hierarchy/context_sensitive]{context-sensitive} grammars.\n\n For other grammars, for example \\hyperref[def:chomsky_hierarchy/unrestricted]{unrestricted} or \\hyperref[def:chomsky_hierarchy/non_contracting]{non-contracting}, such a terminology is not established.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:natural_number_arithmetic_grammar/derivation}\n We continue \\fullref{ex:natural_number_arithmetic_grammar/backus_naur_form}. Depending on our choice of starting symbol, we can derive different sets of words.\n\n For the sake of simplifying our exposition and proof, however, we will assume the simpler grammar described in \\eqref{eq:ex:natural_number_arithmetic_grammar/backus_naur_form/simple}.\n\n Choose the starting symbol to be \\( E \\). We will show that this grammar is unambiguous.\n\n \\Cref{fig:ex:natural_number_arithmetic_grammar/derivation/ambiguous} demonstrates that removing the parentheses makes even this simple grammar ambiguous.\n\n \\begin{figure}\n \\hfill\n \\includegraphics[page=1]{output/ex__natural_number_arithmetic_grammar__derivation.pdf}\n \\hfill\\hfill\n \\caption{The unique way to produce the parenthesized arithmetic expression \\( ((6 \\div 2) \\times 3) \\)}\n \\label{fig:ex:natural_number_arithmetic_grammar/derivation/unambiguous}\n \\end{figure}\n\n \\begin{figure}\n \\hfill\n \\includegraphics[page=2]{output/ex__natural_number_arithmetic_grammar__derivation.pdf}\n \\hfill\\hfill\n \\caption{Different leftmost derivations of the parenthesis-less arithmetic expression \\( 6 \\div 3 \\times 2 \\)}\n \\label{fig:ex:natural_number_arithmetic_grammar/derivation/ambiguous}\n \\end{figure}\n\n We will show that \\( G \\) is unambiguous. Let \\( w \\) be a word in \\( \\mscrL(G) \\). We explicitly build the leftmost derivation of \\( w \\) using recursion on \\( \\len(w) \\):\n \\begin{itemize}\n \\item If \\( \\len(w) = 1 \\), then \\( w = n \\in \\BbbN \\), and the word has been derived as \\( E \\to N \\to n \\).\n\n \\item Assume that \\( w \\) is unambiguously derived for \\( \\len(w) < m + 2 \\) and let \\( \\len(w) = m + 2 \\), then \\( w \\) is necessarily enclosed in parentheses. Let \\( w = ( \\sigma_1 \\ldots \\sigma_m ) \\) be the symbols of \\( w \\). Because of the parentheses, the only possibility for \\( \\sigma_1 \\ldots \\sigma_m \\) is that it consists of two words in \\( \\mscrL(G) \\) with either a multiplication symbol \\( \\times \\) or a division symbol \\( \\div \\) between them. Let \\( k \\) be the index of the operator symbol, that is, the index such that \\( \\sigma_1 \\ldots \\sigma_{k-1} \\) and \\( \\sigma_{k+1} \\ldots \\sigma_m \\) both belong to \\( \\mscrL(G) \\).\n\n By the inductive hypothesis, both \\( \\sigma_1 \\ldots \\sigma_{k-1} \\) and \\( \\sigma_{k+1} \\ldots \\sigma_m \\) are unambiguously derived from \\( E \\). Then \\( w \\) is generated by the rule \\( E \\to (E O E) \\), where the operator symbol \\( \\sigma_k \\) determines the terminal of \\( O \\). Therefore, the derivation of \\( w \\) is also unambiguous.\n \\end{itemize}\n\\end{example}\n\n\\begin{definition}\\label{def:ordered_arborescence}\n An \\term{ordered arborescence} is an \\hyperref[def:arborescence]{arborescence} \\( T = (G, A) \\) with a \\hyperref[def:partially_ordered_set]{partial order} \\( \\leq \\) such that every set of \\hyperref[def:arborescence/ancestry]{siblings} is a \\hyperref[def:partially_ordered_set_chain_and_antichain]{chain}.\n\\end{definition}\n\n\\begin{definition}\\label{def:grammar_syntax_tree}\\mimprovised\n Fix a \\hyperref[def:chomsky_hierarchy/context_free]{context-free} \\hyperref[def:formal_grammar]{formal grammar} \\( G = (V, S, \\Sigma, \\to) \\).\n\n For every word \\( w \\) in \\( \\mscrL(G) \\), we will build an \\hyperref[def:ordered_arborescence]{ordered} \\hyperref[def:arborescence/undirected]{rooted tree} whose \\hyperref[def:arborescence/ancestry]{leaves} are the symbols of \\( w \\) and whose root is \\( S \\). We will call this a \\term{syntax tree} for \\( w \\).\n\n \\begin{figure}\n \\hfill\n \\includegraphics[page=1]{output/alg__grammar_syntax_tree.pdf}\n \\hfill\\hfill\n \\caption{A grammar tree corresponding to the expression \\( (6 \\div (3 \\times 2)) \\) from \\cref{fig:ex:natural_number_arithmetic_grammar/derivation/unambiguous}}\n \\label{fig:def:grammar_syntax_tree}\n \\end{figure}\n\n Fix a derivation\n \\begin{equation}\\label{eq:def:grammar_syntax_tree/derivation}\n S \\Rightarrow w_1 \\Rightarrow \\cdots \\Rightarrow w_{n-1} \\Rightarrow w_n.\n \\end{equation}\n\n We use \\hyperref[rem:natural_number_recursion]{natural number recursion} on \\( n \\) to build the tree. Note that, for the purposes of recursion, we allow \\( w_n \\) to contain non-terminals.\n\n \\begin{itemize}\n \\item In the trivial case where \\( n = 0 \\), and there is no actual derivation, we build a single-vertex tree with root \\( S \\).\n\n \\item Suppose that we can build a tree for all derivations of length \\( m - 1 \\) and fix a derivation \\eqref{eq:def:grammar_syntax_tree/derivation} of length \\( n \\).\n\n First, build a tree \\( T \\) from the derivation\n \\begin{equation*}\n S \\Rightarrow w_1 \\Rightarrow \\cdots \\Rightarrow w_{n-2} \\Rightarrow w_{n-1}.\n \\end{equation*}\n\n There must exist words \\( p \\), \\( s \\) and \\( v \\) and a non-terminal \\( A \\) such that\n \\begin{equation*}\n w_{n-1} = pAs \\Rightarrow pvs = w_n.\n \\end{equation*}\n\n There already exists a leaf for \\( A \\) in \\( T \\). For every symbol in \\( v \\), add a new node as a child of this node.\n \\end{itemize}\n\\end{definition}\n\n\\begin{proposition}\\label{thm:unambiguous_grammar_syntax_trees}\n A \\hyperref[def:chomsky_hierarchy/context_free]{context-free} grammar is \\hyperref[def:chomsky_hierarchy/context_free]{unambiguous} if and only if every word has a unique \\hyperref[def:grammar_syntax_tree]{syntax tree}.\n\\end{proposition}\n\n\\begin{theorem}[Structural induction on unambiguous grammars]\\label{thm:structural_induction_on_unambiguous_grammars}\\mimprovised\n Unlike for the other induction principles in \\fullref{rem:induction}, we will not formulate this one via logical formulas. This will complicate us unnecessarily. We will instead describe how the principle is used in practice.\n\n Let \\( G = (V, S, \\Sigma, \\to) \\) be an \\hyperref[def:chomsky_hierarchy/context_free]{unambiguous} \\hyperref[def:chomsky_hierarchy/context_free]{context-free} \\hyperref[def:formal_grammar]{formal grammar}.\n\n Suppose that we want to prove a statement for every word in \\( \\mscrL(G) \\). It is sufficient to perform the following for every rule \\( A \\to w \\):\n \\begin{displayquote}\n Let \\( A_1, \\ldots, A_n \\) be all non-terminals of \\( w \\) and let \\( u_0, \\ldots, u_n \\) be subwords of \\( w \\) such that\n \\begin{equation*}\n w = u_0 A_1 u_1 A_2 \\ldots A_n u_n.\n \\end{equation*}\n\n Let \\( v_1, \\ldots, v_n \\) be arbitrary words in \\( \\mscrL(G) \\) derivable from \\( A_1, \\ldots, A_n \\), respectively, so that we have the syntax tree\n \\begin{equation*}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__structural_induction_on_unambiguous_grammars.pdf}\n \\end{aligned}\n \\end{equation*}\n\n Then we must prove the statement for the word\n \\begin{equation*}\n u_0 v_1 u_1 \\ldots v_n u_n.\n \\end{equation*}\n \\end{displayquote}\n\n Compare this principle to the more general \\fullref{thm:well_founded_induction}.\n\\end{theorem}\n\\begin{proof}\n Clearly every word in \\( \\mscrL(G) \\) can be obtained in this way.\n The role of non-ambiguity is discussed in the proof of \\fullref{thm:well_founded_induction}.\n\\end{proof}\n", "meta": {"hexsha": "97107fdf6ef5454ee61c6670aecba05bf459bd85", "size": 22793, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/formal_languages.tex", "max_stars_repo_name": "v--/notebook", "max_stars_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/formal_languages.tex", "max_issues_repo_name": "v--/notebook", "max_issues_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/formal_languages.tex", "max_forks_repo_name": "v--/notebook", "max_forks_repo_head_hexsha": "d9bdfbab9f35095db2721f991a3418f58f997a56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 69.7033639144, "max_line_length": 649, "alphanum_fraction": 0.7089896021, "num_tokens": 6811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7931059511841119, "lm_q2_score": 0.7718434873426303, "lm_q1q2_score": 0.6121536631941388}} {"text": "\\section{Discrete Armijo Gradient}\n\\lab{sec:DAG}\nThe Discrete Armijo Gradient algorithm can be used to solve\nproblem $\\mathbf P_c$ defined in~\\eqref{sub:Proc} where\n$f(\\cdot)$ is continuously differentiable.\\\\\n\nThe Discrete Armijo Gradient algorithm approximates gradients by finite differences.\nIt can be used for problems where the cost function is evaluated by\ncomputer code that defines a continuously differentiable function but for\nwhich obtaining analytical expressions for the gradients is impractical\nor impossible.\\\\\n\nSince the Discrete Armijo Gradient algorithm is sensitive to discontinuities in\nthe cost function, we recommend not to use this algorithm if the simulation program\ncontains adaptive solvers with loose precision settings, such as\nEnergyPlus~\\cite{Crawley2001:1}.\nOn such functions, the algorithm is likely to fail. \nIn Section~\\ref{sec:AlgSel}, we recommend algorithms that\nare better suited for such situations.\\\\\n\nWe will now present the Discrete Armijo Gradient algorithm and the\nArmijo step-size subprocedure.\n\n\\noindent\n\\begin{minipage}[b]{\\textwidth}\n\\begin{algorithm}\n[Discrete Armijo Gradient Algorithm]\n~\\\\\n{\\em\n\\begin{tabularx}{\\headwidth}{m{2cm}l}\n\\multicolumn{2}{l}{\\hspace{\\textwidth}~} \\\\ \\\\[-8ex]\\\\\n\\hline \\\\[-2ex]\n \\textbf{Data}:\n & Initial iterate $x_0 \\in \\mathbf X$.\\\\ \n & $\\alpha, \\beta \\in (0, 1)$, $\\gamma \\in (0, \\infty)$, $k^*, k_0 \\in \\mathbb Z$,\\\\\n & $l_{max}, \\kappa \\in \\Na$ (for reseting the step-size calculation).\\\\\n & Termination criteria $\\epsilon_m, \\epsilon_x \\in \\Re_+$, \n $i_{max} \\in \\Na$.\\\\\n \\textbf{Step 0}: \n & Initialize $i=0$ and $m = 0$.\\\\\n \\textbf{Step 1}:\n & \\underline{Compute the {\\it search direction} $h_i$.}\\\\\n & If $\\beta^m < \\epsilon_m$, stop.\\\\\n & Else, set $\\epsilon = \\beta^{k_0 + m }$ and compute, for $j \\in \\{1, \\ldots, n\\}$,\\\\\n & $h_i^j = - \\left({f(x_i + \\epsilon \\, e_j ) - f(x_i)}\\right)/{\\epsilon}$.\\\\\n\\textbf{Step 2} : \n & \\underline{Check descent.} \\\\\n & Compute $\\Delta(x_i; h_i) = \n \\left( f(x_i + \\epsilon \\, h_i ) - f(x_i) \\right) / \\epsilon$.\\\\\n & If $\\Delta(x_i; h_i) < 0$, go to Step 3.\\\\\n & Else, replace $m$ by $m+1$ and go to Step 1.\\\\\n\\textbf{Step 3} :\n & \\underline{Line search.}\\\\\n & Use Algorithm~\\ref{al:ArmijoSubPro} (which requires $k^*, l_{max}$ and $\\kappa$) \n to compute $k_i$.\\\\\n & Set \n\\end{tabularx}\n\\vspace{-1ex}\n \\begin{equation}\n \\hspace{3cm} \\lambda_i = \\argmin_{\\lambda \\in \\{\\beta^{k_i}, \\beta^{k_i-1} \\} } \n f(x_i + \\lambda \\, h_i).\n\\end{equation}\n\\vspace{-1ex}\n\\begin{tabularx}{\\headwidth}{m{2cm}l}\n\\textbf{Step 4} :\n & If $f(x_i + \\lambda_i \\, h_i) - f(x_i) > - \\gamma \\, \\epsilon$, \n replace $m$ by $m+1$ and go to Step 1.\\\\\n\\textbf{Step 5} :\n & Set $x_{i+1} = x_i + \\lambda_i \\, h_i$.\\\\\n & If $\\| \\lambda_i \\, h_i \\| < \\epsilon_x$, stop. Else, replace $i$ by $i+1$ and go to Step 1.\\\\ \n\\hline\n\\end{tabularx}\n}\n~\\\\ \\lab{al:DAG}\n\\end{algorithm}\n\\end{minipage}\n\n\\noindent\n\\begin{minipage}[b]{\\textwidth}\n\\begin{subequations}\n\\begin{algorithm}\n[Armijo Step-Size Subprocedure]\n~\\\\\n{\\em\n\\begin{tabularx}{\\headwidth}{m{2cm}l}\n\\multicolumn{2}{l}{\\hspace{\\textwidth}~} \\\\ \\\\[-8ex]\\\\\n\\hline \\\\[-2ex]\n \\textbf{Data}:\n & Iteration number $i \\in \\Na$, iterate $x_i \\in \\Re^n$, \n search direction $h_i \\in \\Re^n$,\\\\ \n & $k^*, k_{i-1} \\in \\mathbb Z$, $\\alpha, \\beta \\in (0, \\ 1)$, and \n $\\Delta(x_i;h_i) \\in \\Re$ with $\\Delta(x_i;h_i) < 0$,\\\\\n & parameter for restart $l_{max}, \\kappa \\in \\Na$.\\\\\n\\textbf{Step 0}:\n & Initialize $l=0$.\\\\\n & If $i = 0$, set $k' = k^*$, else set $k' = k_{i-1}$.\\\\\n\\textbf{Step 1}:\n & Replace $l$ by $l+1$, and test the conditions\n\\end{tabularx}\n\\vspace{-1ex}\n\\begin{eqnarray}\n\\hspace{2cm} f(x_i + \\beta^{k'} \\, h_i ) - f(x_i) & \\le &\n \\beta^{k'} \\, \\alpha \\, \\Delta(x_i; h_i), \\label{eq:ArmSteSizPro1} \\\\\n\\hspace{2cm} f(x_i + \\beta^{k' - 1} \\, h_i ) - f(x_i) & > &\n \\beta^{k' - 1} \\, \\alpha \\, \\Delta(x_i; h_i).\n\\label{eq:ArmSteSizPro2}\n\\end{eqnarray}\n\\vspace{-1ex}\n\\begin{tabularx}{\\headwidth}{m{2cm}l}\n\\textbf{Step 2}:\n& If $k'$ satisfies \\eqref{eq:ArmSteSizPro1} and \n \\eqref{eq:ArmSteSizPro2}, return $k'$.\\\\\n\\textbf{Step 3}:\n & If $k'$ satisfies \\eqref{eq:ArmSteSizPro2} \n but not \\eqref{eq:ArmSteSizPro1},\\\\\n & \\hspace{1cm} replace $k'$ by $k'+1$.\\\\\n & else,\\\\\n & \\hspace{1cm} replace $k'$ by $k'-1$.\\\\\n & If $l < l_{max}$ or $k_{i-1} \\le k^* + \\kappa$, go to Step 1. Else, go to Step 4.\\\\\n\\textbf{Step 4}:\n & Set $\\mathbf K \\triangleq \\{ k \\in \\mathbb Z \\ | \\ k \\ge k^* \\}$,\n and compute\\\\\n & $k'\n \\triangleq \\min_{k \\in \\mathbf K} \\{ k \\ | \\\n f(x_i + \\beta^k \\, h_i ) - f(x_i) \\le \\beta^k \\, \\alpha \\, \\Delta(x_i; h_i) \\}$.\\\\\n & Return $k'$.\\\\\n\\hline\n\\end{tabularx}\n}\n\\lab{al:ArmijoSubPro}\n\\end{algorithm}\n\\end{subequations}\n\\end{minipage}\n~\\\\\nNote that in Algorithm~\\ref{al:ArmijoSubPro}, as $\\beta \\to 1$, \nthe number of tries to compute the Armijo step-size is likely to go to infinity.\nUnder appropriate assumptions one can show that $\\alpha = 1/2$ yields \nfastest convergence~\\cite{Pol97:1}.\\\\\n\nThe step-size Algorithm~\\ref{al:ArmijoSubPro} requires often only a small number\nof function evaluations.\nHowever, occasionally, once a very small step-size has occurred,\nAlgorithm~\\ref{al:ArmijoSubPro} can trap the Discrete Armijo Gradient algorithm\ninto using a very small step-size for all subsequent iterations.\nHence, if $k_{i-1} > k^* + \\kappa$, we reset the step-size by computing Step 4.\\\\\n\nAlgorithm~\\ref{al:DAG} together with the step-size Algorithm~\\ref{al:ArmijoSubPro}\nhave the following convergence properties~\\cite{Pol97:1}.\n\n\\begin{theorem}\nLet $f \\colon \\Re^n \\to \\Re$ be continuously differentiable and bounded below.\n\\begin{enumerate}\n\\item \nIf Algorithm~\\ref{al:DAG} jams at $x_i$, cycling indefinitely in the loop\ndefined by Steps 1-2 or in the loop defined by Steps 1-4,\nthen $\\nabla f(x_i) = 0$.\n\\item\nIf $\\{ x_i \\}_{i=0}^\\infty$ is an infinite sequence constructed by\nAlgorithm~\\ref{al:DAG} and Algorithm~\\ref{al:ArmijoSubPro}\nin solving~\\eqref{sub:Proc},\nthen every accumulation point $\\widehat x$ of $\\{ x_i \\}_{i=0}^\\infty$\nsatisfies $\\nabla f(\\widehat x) = 0$.\n\\end{enumerate}\n\\rbox\n\\end{theorem}\n\n\nNote that $\\epsilon \\, h_i$ has the same units as the cost function,\nand the algorithm evaluates $x_i + \\lambda \\, h_i$ for some $\\lambda \\in \\Re_+$.\nThus, the algorithm is sensitive to the scaling of the problem variables,\na rather undesirable effect.\nTherefore, in the implementation of \nAlgorithm~\\ref{al:DAG} and Algorithm~\\ref{al:ArmijoSubPro},\nwe normalize the cost function values by replacing, for all $x \\in \\Re^n$,\n$f(x)$ by $f(x)/f(x_0)$, where $x_0$ is the initial iterate.\nFurthermore, we set $x_0 = 0$ and evaluate the cost function for the values\n$\\chi^j + x^j \\, s^j$, $j \\in \\{1, \\ldots, n\\}$, where\n$x^j \\in \\Re$ is the $j$-th component of the design parameter computed in\nAlgorithm~\\ref{al:DAG} or Algorithm~\\ref{al:ArmijoSubPro} and\n$\\chi^j \\in \\Re$ and $s^j \\in \\Re$ are the setting of the parameters\n\\texttt{Ini} and \\texttt{Step}, respectively, for\nthe $j$-th design parameter in the optimization command file \n(see page~\\pageref{par:comFil}).\n\nIn view of the sensitivity of the Discrete Armijo Gradient algorithm\nto the scaling of the problem variables and the cost function values,\nthe implementation of penalty and barrier functions may cause\nnumerical problems if the penalty is large compared to the unpenalized\ncost function value.\n\nIf box-constraints for the independent parameters are specified,\nthen the transformations~\\eqref{sub:traBoxCon} are used.\n\n%-------------------------\n\\subsection{Keywords}\nFor the Discrete Armijo Gradient algorithm, the command file (see page~\\pageref{par:comFil}) must only contain continuous parameters.\\\\\n\nTo invoke the algorithm, \nthe \\texttt{Algorithm} section of the GenOpt command file must have the following form:\n\\begin{lstlisting}\nAlgorithm{\n Main = DiscreteArmijoGradient;\n Alpha = Double; // 0 < Alpha < 1\n Beta = Double; // 0 < Beta < 1\n Gamma = Double; // 0 < Gamma\n K0 = Integer;\n KStar = Integer;\n LMax = Integer; // 0 <= LMax\n Kappa = Integer; // 0 <= LMax\n EpsilonM = Double; // 0 < EpsilonM\n EpsilonX = Double; // 0 < EpsilonX\n}\n\\end{lstlisting}\nThe entries are defined as follows:\n\\begin{codedescription}\n\\item [Main]\nThe name of the main algorithm.\n\\item [Alpha]\nThe variable $\\alpha$ used in Step 1 and in Step 4 of Algorithm~\\ref{al:ArmijoSubPro}.\nA typical value is $\\alpha = 1/2$.\n\\item [Beta]\nThe variable $\\beta$ used in approximating\nthe gradient and doing the line search.\nA typical value is $\\beta = 0.8$.\n\\item [Gamma]\nThe variable $\\gamma$ used in Step 4 of Algorithm~\\ref{al:DAG} \nto determine\nwhether the accuracy of the gradient approximation will be increased.\n\\item [K0]\nThe variable $k_0$ that determines the initial accuracy\nof the gradient approximation.\n\\item [KStar]\nThe variable $k^*$ used to initialize the line search.\n\\item [LMax]\nThe variable $l_{max}$ used in Step 3 of Algorithm~\\ref{al:ArmijoSubPro}\nto determine whether the line search needs to be reinitialized.\n\\item [Kappa]\nThe variable $\\kappa$ used in Step 3 of Algorithm~\\ref{al:ArmijoSubPro}\nto determine whether the line search needs to be reinitialized.\n\\item [EpsilonM]\nThe variable $\\epsilon_m$ used in the determination criteria\n$\\beta^m < \\epsilon_m$ in Step 1 of Algorithm~\\ref{al:DAG}.\n\\item [EpsilonX]\nThe variable $\\epsilon_x$ used in the determination criteria\n$\\| \\lambda_i \\, h_i \\| < \\epsilon_x$ in Step 5 of Algorithm~\\ref{al:DAG}.\n\\end{codedescription}\n\n\n% ===============================================\n\n", "meta": {"hexsha": "1ae2dfe03827297fbc7d259c06068e0b9964417a", "size": 9707, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/manual/algDisArmGra.tex", "max_stars_repo_name": "bergsee/GenOpt", "max_stars_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_stars_repo_licenses": ["BSD-3-Clause-LBNL"], "max_stars_count": 16, "max_stars_repo_stars_event_min_datetime": "2015-08-30T09:47:56.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-06T15:16:18.000Z", "max_issues_repo_path": "src/manual/algDisArmGra.tex", "max_issues_repo_name": "bergsee/GenOpt", "max_issues_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_issues_repo_licenses": ["BSD-3-Clause-LBNL"], "max_issues_count": 10, "max_issues_repo_issues_event_min_datetime": "2016-01-14T00:01:46.000Z", "max_issues_repo_issues_event_max_datetime": "2020-01-21T15:28:52.000Z", "max_forks_repo_path": "src/manual/algDisArmGra.tex", "max_forks_repo_name": "lbl-srg/GenOpt", "max_forks_repo_head_hexsha": "3925277af881cea6e12e3d1bf0285bd657bbcced", "max_forks_repo_licenses": ["BSD-3-Clause-LBNL"], "max_forks_count": 13, "max_forks_repo_forks_event_min_datetime": "2015-08-30T09:47:57.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-01T18:07:07.000Z", "avg_line_length": 38.6733067729, "max_line_length": 135, "alphanum_fraction": 0.6657051612, "num_tokens": 3282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8479677622198946, "lm_q2_score": 0.7217432122827968, "lm_q1q2_score": 0.6120149766168416}} {"text": "\\documentclass[11pt]{article}\n\n\\input{preamble}\n\n%\\usepackage{tikz}\n%\\usepackage{tikz-qtree}\n\\usepackage{qtree}\n\\usepackage{xfrac}\n\n\\title{Week 6: Decision Trees and Variational Autoencoders}\n\\author{\\url{http://mlvu.github.io}}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Decision trees}\n\nImagine you do a newspaper round to help you get through these lean times. On your round, you encounter a number of dogs that either bark or (try to) bite. The dogs are described by the following binary features: \\emph{Heavy}, \\emph{Smelly}, \\emph{Big} and \\emph{Growling}. Consider the following set of examples:\n\n\\begin{center}\n\\begin{tabular}{c c c c | c }\nHeavy & Smelly & Big & Growling & Bites \\\\\n\\hline\n\\rc{No} & \\rc{No} & \\rc{No} & \\rc{No} & \\oc{No} \\\\\n\\rc{No} & \\rc{No} & \\gc{Yes} & \\rc{No} & \\oc{No} \\\\\n\\gc{Yes} & \\gc{Yes}& \\rc{No} & \\gc{Yes} & \\oc{No} \\\\\n\\gc{Yes} & \\rc{No} & \\rc{No} & \\gc{Yes} & \\bc{Yes} \\\\\n\\rc{No} &\\gc{Yes} & \\gc{Yes} & \\rc{No} & \\bc{Yes} \\\\\n\\rc{No} & \\rc{No} & \\gc{Yes} & \\gc{Yes} & \\bc{Yes} \\\\\n\\rc{No} & \\rc{No} & \\rc{No} & \\gc{Yes} & \\bc{Yes} \\\\\n\\gc{Yes} &\\gc{Yes} & \\rc{No} & \\rc{No} & \\bc{Yes} \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\n\\qu What is the entropy of the target value \\emph{Bites} in the data?\n\\ans{\n\\[H(\\text{Bites})= - \\sfrac{5}{8}\\log_2\\sfrac{5}{8} - \\sfrac{3}{8}\\log_2\\sfrac{3}{8} \\approx 0.9544\\]\n}{}\n\n\\qu Which attribute would the ID3 algorithm choose to use for the root of the tree (without pruning)?\n\\ans{Growling}{}\n\n\\qu What is the information gain of the attribute you chose in the previous question?\n\\ans{approximately 0.0487}{}\n\n\\qu Draw the full decision tree that would be learned for this data using ID3 without pruning.\n\\ans{\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.7\\linewidth]{Tree}\n\\caption{Decision tree}\n\\end{figure}\n}{}\n\n\\qu Suppose three new dogs appear in your round as listed in the table below. Classify them using the tree from the previous question.\n\n\\begin{center}\n\\begin{tabular}{c c c c c | c }\nDog & Heavy & Smelly & Big & Growling & Bites \\\\\n\\hline\nBuster & \\gc{Yes} & \\gc{Yes} & \\gc{Yes} & \\gc{Yes} & \\ans{\\oc{No}}{?} \\\\\nPluto & \\rc{No} & \\gc{Yes} & \\rc{No} & \\gc{Yes} & \\ans{\\oc{No}}{?} \\\\\nZeus & \\gc{Yes} & \\gc{Yes} & \\rc{No} & \\rc{No} & \\ans{\\bc{Yes}}{?} \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\n\\qu Someone proposes a new scheme to prevent overfitting: she suggests to set a pre-defined maximum depth for the decision trees. When the standard algorithm reaches this depth, it terminates. Could this help to prevent overfitting? Why (not)?\n\\ans{\nIt ensures smaller, simpler trees (a smaller model space) and therefore can reduce the risk of overfitting. Overfitting is essentially memorizing too much of your data. Smaller trees can memorize less.\n}{}\n\n\\qu In the maximum depth scheme introduced above, how would you determine a good value for the maximum depth for a given data set?\n\n\\ans{\nThe maximum depth is a \\emph{hyperparameter}. We can choose good values for our hyperparameters by splitting our training data into a validation set and trying different values (either by cross validation or just single runs).}{}\n\n\\qu Why can't we apply L1-regularization to this decision tree learning problem?\n\n\\ans{L1 regularization assumes that your model is described by a real valued vector of parameters (i.e. your model space is continuous.) Decision trees have a \\emph{discrete} model space.}{}\n\n\\section{Variational autoencoders}\n\nThe maximum likelihood principle tells us to optimize the quantity $p(x\\mid \\theta)$ as a function of $\\theta$ (the model parameters).For complex models, this does not usually lead to a closed form solution. Instead, we will rewrite the maximum likelihood objective using the following decomposition.\n\n\\[\n\\ln p(\\x\\mid \\theta) = L(\\rc{q}, \\theta) + KL(\\rc{q}, p) \\\\\n\\]\nwith\n\\begin{align*}\n\\rc{q}(\\z\\mid \\x)\\;&\\text{any distribution on $\\z$} \\\\\nKL(\\rc{q}, p)\\;&\\text{the Kullback-Leibler divergence\\footnotemark}\\\\\n&\\text{between $\\rc{q}(\\z \\mid \\x)$ and $p(\\z \\mid \\x, \\theta)$}\\\\\nL(\\rc{q}, \\theta) &= \\E_\\rc{q} \\ln \\frac{p(\\x, \\z \\mid \\theta) }{\\rc{q}(\\z\\mid \\x)}\n\\end{align*}\n\n\\footnotetext{The KL divergence can be defined using $\\log_2$ or using $\\ln$. The first leads to a value in \\emph{bits}, the second to a value in the slightly more abstract unit of \\emph{nats}. That is, the information distance between the two distributions is the same, but expressed in different units. In the context of the VAE, the natural logarithm $\\ln$ is preferred, mainly because it cancels out neatly against the exponent in the pdf of the normal distribution.\n}\n\nWe will first prove that this equality holds. We start with the right hand side, fill in the components, and derive the left-hand side.\n\n\\qu Fill in the blanks. We have written everything in terms of \\emph{expectations} $\\E$ to simplify the notation. The expectation is over the random variable $\\z$, while $\\x$ has some definite value. Note that $\\E f(\\z) + \\E g(\\z) = \\E\\left [f(\\z) + g(\\z)\\right]$.\n\n\\begin{align*}\nL(\\rc{q}, \\theta) + KL(\\rc{q}, p) &= \\mathbb{E}_\\rc{q} \\ln \\frac{p(\\x, \\z \\mid \\theta) }{\\rc{q}(\\z\\mid \\x)} - \\mathbb{E}_\\rc{q} \\ln \\frac{p(\\z \\mid \\x, \\theta)}{\\rc{q}(\\z\\mid \\x)}\\\\\n&= \\ans{\\E_\\rc{q} \\ln p(\\x,\\z\\mid \\theta) - \\E_\\rc{q} \\ln \\rc{q}(\\z\\mid \\x) - \\E_\\rc{q} \\ln p(\\z\\mid \\x, \\theta) + \\E_\\rc{q} \\ln \\rc{q}(\\z \\mid \\x)}{\\ldots}\\\\\n&= \\mathbb{E}_\\rc{q} \\ln p(\\x, \\z \\mid \\theta) - \\mathbb{E}_\\rc{q} \\ln p(\\z \\mid \\x, \\theta) \\\\\n&= \\mathbb{E}_\\rc{q} \\ln \\frac{p(\\x, \\z \\mid \\theta) }{ p(\\z \\mid \\x, \\theta) } = \\ans{\\mathbb{E}_\\rc{q} \\ln \\frac{p(\\z\\mid \\x, \\theta)p(\\x\\mid \\theta) }{ p(\\z \\mid \\x, \\theta)}}{\\ldots}\\\\\n&= \\mathbb{E}_\\rc{q} \\ln p(\\x \\mid \\theta) = \\ln p(\\x \\mid \\theta) \n\\end{align*}\n\nIn EM, we search by alternately (1) optimizing $L(\\rc{q}\\mid \\theta)$ with respect to $\\theta$ and (2) setting $\\rc{q}$ equal to $\\gc{p}$ (so that the KL term becomes zero).\n\n\\qu For the variational autoencoder, we cannot (easily) perform this last step. Why not?\n\\ans{In the variational autoencoder, our model is a neural network that transforms $\\z$ into a distribution on $\\x$. To set the KL divergence term equal to zero, we would have to compute $p(\\z \\mid \\x, \\theta)$: i.e. a probability distribution on $\\z$ that indicates for which $\\z$ our observed $\\x$ is most likely.\n\nWhile sampling techniques exist to approximate this kind of distribution, they are costly and can be very inaccurate.\n}{}\n\n~\\linebreak\nInstead, we \\emph{approximate} $p(\\z \\mid\\x, \\theta)$ with a neural network $\\rc{q_\\bv}(\\z \\mid \\x)$ that produces a distribution on $\\z$ given some $\\x$. We call the neural network computing $p(\\x \\mid \\x, \\theta)$ $\\gc{p_\\bw}(\\x\\mid \\z)$, to make the notation a little more friendly. Here, $\\gc{\\bw}$, stands for all parameters of the \\gc{p} network, and $\\rc{\\bv}$ stands for all parameters of the \\rc{q} network.\\footnotemark \n\n\\footnotetext{We've turned $\\theta$ into $\\gc{\\bw}$ and added parameters $\\rc{\\bv}$ for our approximation $\\rc{q}$ on the conditional distribution on $\\z$. We've also taken the parameters out of the conditional, because we will always talk about the function ``given the parameter''; we will never talk about the probability on the parameters themselves.}\n\nThis gives us an auto-encoder-like structure. An input is mapped to a distribution $\\rc{q_\\bv}(\\z\\mid \\x)$ by the \\rc{encoder}. We sample a single $\\z$ from this distribution and pass it through the \\gc{decoder} $\\gc{p_\\bw}(\\x \\mid \\z)$ to produce a distribution on $\\x$ (see the \\href{https://mlvu.github.io/lectures/51.Deep\\%20Learning2.annotated.pdf}{slides} for diagrams).\n\nTo find a way to train such an architecture, we turn again to our decomposition of the likelihood. In our new notation:\n\\[\n\\ln \\gc{p_\\bw}(\\x) = L(\\rc{\\bv}, \\gc{\\bw}) + KL(\\rc{q}, \\gc{p}) \\p \\\\\n\\]\n\nThe KL divergence term is difficult to compute: it's an expectation, and it contains the function $\\gc{p_\\bw}(\\z \\mid \\x)$ which requires us to invert the \\gc{decoder} neural network (that is, to reason about the inputs given the outputs).\n\nHowever, because the KL divergence is always positive, we know that \n\n\\[\n\\ln \\gc{p_\\bw}(\\x) \\geq L(\\rc{\\bv}, \\gc{\\bw}) \\\\\n\\]\nfor \\emph{any} $\\rc{q_\\bv}$ we choose. This is why $L$ is called the variational \\emph{lower bound}.\\footnotemark~If we choose our parameters $\\gc{\\bw}, \\rc{\\bv}$ to maximize $L$, we are also, indirectly, maximizing $\\ln \\gc{p_\\bw}(\\x)$.\\footnotemark\n\n\\footnotetext{The word \\emph{variational} comes from the fact that one of its arguments, \\rc{q}, is a function (the calculus of functions is called \\emph{variational} calculus). For our purposes, this distinction doesn't matter much, since the function $\\rc{q}$ is defined by a set of parameters $\\rc{\\bv}$, so ultimately we will take the derivative over those parameters, as we are used to.}\n\n\\footnotetext{How close the lower bound $L$ comes to the true value $\\gc{p_\\bw}(\\x)$ depends on how well our encoder network $\\rc{q_\\bv}$ approximates the true conditional distribution on $\\z$: $\\gc{p_\\bw}(\\z \\mid\\x)$. I.e. how small the KL term in the original decomposition is.}\n\nTo do so, we rewrite $L(\\rc{\\bv}, \\gc{\\bw})$ into two separate terms: a KL divergence and an expectation:\n\n\\[\nL(\\rc{\\bv}, \\gc{\\bw}) = - KL(\\rc{q_\\bv}(\\z\\mid\\x), \\gc{p_\\bv}(\\z)) + \\E_\\rc{q_\\bv} \\ln \\gc{p_\\bw}(\\x\\mid \\z)\n\\]\n\n\\qu Show that this equation holds. That is, rewrite the left part into the right. We will assume that all expectations are over $\\rc{q}$.\n\\begin{align*}\n\tL(\\rc{\\bv}, \\gc{\\bw}) &= \\E_\\rc{q} \\ln \\frac{\\gc{p_\\bw}(\\x, \\z)}{\\rc{q_\\bv}(\\z\\mid \\x)} \\\\\n\t&= \\ans{\\E \\ln \\gc{p_\\bw}(\\x, \\z) - \\E \\rc{q_\\bv}(\\z \\mid \\x)}{\\ldots}\\\\\n\t&= \\ans{\\E \\ln \\left[\\gc{p_\\bw}(\\x\\mid \\z)\\gc{p_\\bw}(\\z)\\right] - \\E \\ln \\rc{q_\\bv}(\\z \\mid \\x)}{\\ldots} \\\\\n\t&= \\ans{\\E \\ln \\gc{p_\\bw}(\\x \\mid \\z) + \\E \\ln \\gc{p_\\bw}(\\z) - \\E \\ln \\rc{q_\\bv}(\\z \\mid \\x)}{\\ldots} \\\\\n\t&= \\E \\ln \\gc{p_\\bw}(\\x \\mid \\z) - \\left [ \\E \\ln \\rc{q_\\bv}(\\z \\mid \\x) - \\E \\ln \\gc{p_\\bw}(\\z) \\right ]\\\\\n\t&= \\E \\ln \\gc{p_\\bw} (\\x\\mid \\z) - KL(\\rc{q_\\bv}(\\z\\mid\\x), \\gc{p_\\bw}(\\z)) \n\\end{align*}\n\n\nThus, to optimize our variational autoencoder, we should maximize $L$. In other words, $-L$ is our loss function. The only problem left to solve is that the second term is an expectation (which we cannot compute explicitly). \n\n\\qu How is this solved in practice?\n\n\\ans{We take a single sample from $\\rc{q_\\bv}(\\z\\mid \\x)$ and use $\\ln \\gc{p_\\bw}(\\x\\mid \\z)$ as a (very crude) estimate of the expectation term. \n\nTo let the gradient propagate through the sampling, we add a sample from the standard MVN to the input and transform it to a sample from $\\rc{q_\\bv(\\z\\mid \\x)}$ by multiplying by a matrix $A$ (with $\\Sigma = AA^T$) and adding the mean.}{}\n\n\\end{document}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "b7620b8c08cfe664f4b1068ac048d67a07a37701", "size": 10750, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "week6.tex", "max_stars_repo_name": "mlvu/homework", "max_stars_repo_head_hexsha": "2183b91c2a355279fbe958b1bbc8bd13ea956615", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 4, "max_stars_repo_stars_event_min_datetime": "2019-02-27T13:25:15.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-07T07:04:26.000Z", "max_issues_repo_path": "week6.tex", "max_issues_repo_name": "mlvu/homework", "max_issues_repo_head_hexsha": "2183b91c2a355279fbe958b1bbc8bd13ea956615", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "week6.tex", "max_forks_repo_name": "mlvu/homework", "max_forks_repo_head_hexsha": "2183b91c2a355279fbe958b1bbc8bd13ea956615", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 53.75, "max_line_length": 470, "alphanum_fraction": 0.6705116279, "num_tokens": 3591, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.705785040214066, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.6119408613242983}} {"text": "\\chapter{Simulate simple 2D Brownian motion of $\\textbf{\\textit{E.coli}}$} % Main chapter title\n\n\\label{Part1_chapter} % For referencing the chapter elsewhere, use \\ref{Chapter1} \n\n\\section{Symbols}\n\n\\begin{table}[H]\n\\caption{The symbols used in the model of simulating simple 2D Brownian motion of $E.coli$ }\n\\label{tab:part1_symbols}\n\\centering\n\\begin{tabular}{l l}\n\\toprule\n\n\\tabhead{Symbol} & \\tabhead{Definition} \\\\\n\\midrule\n$X(t)$ & Stochastic processes \\\\\n$X_i$ & The direction of movement $X_i = 1,\\ -1$\\\\\n$B(t)$ & Standard Brownian Motion \\\\\n$N(\\mu,\\sigma)$ & Normal distribution with mean $\\mu$ and standard deviation $\\sigma$ \\\\\n% ${\\rm R^{2}_{i}}$ 改成正常体\n$\\Delta x$ & Space interval \\\\\n$\\Delta t$ & Time interval \\\\\n$\\qquad \\sigma $ & $\\qquad \\sigma = \\Delta x$/$\\sqrt{ \\Delta t}$ \\\\\n$(x,y)$ & Coordinates of $E.coli$ \\\\\n$\\qquad x_t $ & \\qquad Coordinate in x axis at time t \\\\\n$\\qquad y_t $ & \\qquad Coordinate in y axis at time t \\\\\n$v$\t\t\t & The speed of the movement of $E.coli$ \\\\\n\\bottomrule\\\\\n\\end{tabular}\n\\end{table}\n\n\n\\section{Biological background}\n\n$Escherichia \\ coli$ is a Gram-negative, optional anaerobic, rod-shaped, coliform bacterium of the genus Escherichia that is commonly found in the lower intestine of warm-blooded organisms. $E.coli$ are widely used in biological research, cells are typically rod-shaped, and are about 2.0 $\\mu m$ long and 0.25-1.0 $\\mu m$ in diameter, with a cell volume of 0.6-0.7 $\\mu m^3$ . Strains that possess flagella are motile. The flagella have a peritrichous arrangement.It also attaches and effaces to the microvilli of the intestines via an adhesion molecule known as intimin. The thin straight filaments of bacteria called pili, that enable it to attach to specify substrate, and thicker longer helical filaments, called flagella, that enable it to swim.\n\nBrownian motion is the random motion of microscopic particles suspended in a fluid resulting from their collision with the quick atoms or molecules in the liquid or gas. This phenomenon is named after British botanist Robert Brown. In 1827, while looking through a microscope at particles trapped in cavities inside pollen grains in water, Brown noted that the particles moved through the water randomly but failed to explain the mechanisms that caused this movement. He supposed that active molecules were inside those particles thus there was no relationship with the surrounded liquid. \n\nThe Brownian motion is a Gaussian process with time $t$, we can find that, for stochastic processes $\\{X(t),t\\geq0\\}$ :\n\n\\begin{equation*} \n\\begin{aligned} \n\\centering\nX(0) &= 0 \\\\\nX(t) &\\backsim N(0,\\sigma^2t) \\\\ \nX(t) &= \\Delta x(X_1 + ... + X_{[t/\\Delta t]}) \\\\\n\\sigma^2 &= \\frac{(\\Delta x)^2}{\\Delta t} \\\\\n\\end{aligned} \n\\end{equation*}\n\n\\newpage\nNormally, we set $\\sigma=1$ and defines this kind of stochastic processes $\\{X(t),t\\geq0\\}$ as \\textbf{Standard Brownian Motion} and they could be denote as $\\{B(t),t\\geq0\\}$, where :\n\n\\begin{equation*} \n\\begin{aligned} \n\\centering\nB(0) &= 0 \\\\\nB(t) &\\backsim N(0,t) \\\\ \n\\end{aligned} \n\\end{equation*}\n\n\\section{Hypothesis for simulation}\n\nAssuming the $E.coli$ here have no mitosis, so the number of cells maintain constant.\nI use $(x,y)$ to define the location of each cell, assume all the cells start from origin$(0,0)$. They have a random judging for each step in both x and y directions. Set the speed as $50 \\mu m/s$, so for each step,\n\n\n\\begin{equation*} \n\\begin{aligned} \n\\centering\n\\phi_x &\\backsim N(0,1) \\\\\n\\phi_y &\\backsim N(0,1) \\\\\nx_{t+1} &= x_{t} + \\phi_x v \\\\ \ny_{t+1} &= y_{t} + \\phi_y v \\\\ \n\\end{aligned} \n\\end{equation*}\nThe move on both x and y directions are random and it follows the Brownian motion.\\\\\nSo in theory, for a $E.coli$ population large enough, we have: \\\\\n\\begin{equation*} \n\\begin{aligned} \n\\centering\n\\frac{1}{n} \\sum_{i=1}^{n}x_i(t) &\\approx 0 \\\\\n\\frac{1}{n} \\sum_{i=1}^{n}y_i(t) &\\approx 0 \\\\\n\\frac{1}{n} \\sum_{i=1}^{n}x_i(t)^2 &\\approx v^2t = 250t \\\\\n\\frac{1}{n} \\sum_{i=1}^{n}y_i(t)^2 &\\approx v^2t = 250t \\\\\n\\end{aligned} \n\\end{equation*}\n%----------------------------------------------------------------------------------------\n%\\newpage\n\\section{Results of simulation}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=1\\linewidth]{Figures/P1_fig1.png}\n\\caption{The simulation of 10000 $E.coli$ cells for simple 2D Brownian motion form (0,0) after 100s}\n\\label{P1_fig1}\n\\end{figure}\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=1\\linewidth]{Figures/P1_fig2.png}\n\\caption{The ”mean displacement” and ”mean displacement-square” overtime for simple 2D brownic motion}\n\\label{P1_fig2}\n\\end{figure}\n\n\n\\begin{figure}[H]\n\\centering\n\\includegraphics[width=1\\linewidth]{Figures/P1_fig3.png}\n\\caption{Single Bacteria Random Walk Trajectory}\n\\label{P1_fig2}\n\\end{figure}\n\n\n%----------------------------------------------------------------------------------------\n\n\n\n\n\n\n\n%----------------------------------------------------------------------------------------\n\n\n\n%----------------------------------------------------------------------------------------\n\n\n\n\n%The \\code{biblatex} package is used to format the bibliography and inserts references such as this one \\parencite{Reference1}. The options used in the \\file{main.tex} file mean that the in-text citations of references are formatted with the author(s) listed with the date of the publication. Multiple references are separated by semicolons (e.g. \\parencite{Reference2, Reference1}) and references with more than three authors only show the first author with \\emph{et al.} indicating there are more authors (e.g. \\parencite{Reference3}). This is done automatically for you. To see how you use references, have a look at the \\file{Chapter1.tex} source file. Many reference managers allow you to simply drag the reference into the document as you type.\n\n", "meta": {"hexsha": "5f997a7c9a1dac4f09b25cb2c3c6d83e28394da2", "size": 5837, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "HW3/Chapters/Chapter1.tex", "max_stars_repo_name": "c235gsy/Sustech_Systems-Biology", "max_stars_repo_head_hexsha": "bd72b7e7d1238e22901b410b3254a4622d249964", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "HW3/Chapters/Chapter1.tex", "max_issues_repo_name": "c235gsy/Sustech_Systems-Biology", "max_issues_repo_head_hexsha": "bd72b7e7d1238e22901b410b3254a4622d249964", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "HW3/Chapters/Chapter1.tex", "max_forks_repo_name": "c235gsy/Sustech_Systems-Biology", "max_forks_repo_head_hexsha": "bd72b7e7d1238e22901b410b3254a4622d249964", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 43.5597014925, "max_line_length": 751, "alphanum_fraction": 0.6772314545, "num_tokens": 1678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7690802476562643, "lm_q2_score": 0.7956581073313275, "lm_q1q2_score": 0.6119249342360918}} {"text": "\\section{Cubical diagrams}\n\nIn this lecture we study commuting diagrams of the form\n\\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n A' \\arrow[d] & C \\arrow[dl] \\arrow[dr] & B' \\arrow[dl,crossing over] \\arrow[d] \\\\\n A \\arrow[dr] & X' \\arrow[from=ul,crossing over] \\arrow[d] & B \\arrow[dl] \\\\\n & X.\n \\end{tikzcd}\n\\end{equation*}\nIn these diagrams there are six homotopies witnessing that the faces of the cube commute, as well as a homotopy of homotopies witnessing that the cube as a whole commutes.\n\nOnce the basic definitions of cubes are established, we focus on pullbacks and pushouts that appear in different configurations in these cubical diagrams. For example, if all the vertical maps in a commuting cube are equivalences, then the top square is a pullback square if and only if the bottom square is a pullback square. In \\cref{chap:descent} we will use cubical diagrams in our formulation of the universality and descent theorems for pushouts.\n\nIn our first main theorem of this lecture we show that given a commuting cube in which the bottom square is a pullback square, the top square is a pullback square if and only if the induced square of fibers of the vertical maps is a pullback square. This theorem should be compared to \\cref{cor:pb_fibequiv}, where we showed that a square is a pullback square if and only if it induces equivalences on the fibers of the vertical maps.\n\nIn our second main theorem we use the previous result to derive the 3-by-3 properties for pullbacks and pushouts.\n\n\\subsection{Commuting cubes}\n\\begin{defn}\\label{defn:cube}\nA \\define{commuting cube}\\index{commuting cube|textbf}\n\\begin{equation*}\n\\begin{tikzcd}[column sep=large,row sep=large]\n& C' \\arrow[dl,swap,\"{p'}\"] \\arrow[dr,\"{q'}\"] \\arrow[d,swap,\"{h_C}\" near end] \\\\\nA' \\arrow[d,swap,\"{h_A}\"] & C \\arrow[dl,swap,\"{p}\" very near start] \\arrow[dr,\"{q}\" very near start] & B' \\arrow[dl,crossing over,\"{g'}\" near end] \\arrow[d,\"{h_B}\"] \\\\\nA \\arrow[dr,swap,\"f\"] & X' \\arrow[d,swap,\"{h_X}\" near start] \\arrow[from=ul,crossing over,swap,\"{f'}\" near end] & B \\arrow[dl,\"{g}\"] \\\\\n& X,\n\\end{tikzcd}\n\\end{equation*}\nconsists of types and maps as indicated in the diagram, equipped with\n\\begin{enumerate}\n\\item homotopies\n \\begin{align*}\n \\mathsf{top} & : f' \\circ p' \\htpy g' \\circ q' \\\\\n \\mathsf{back\\usc{}left} & : p \\circ h_C \\htpy h_A \\circ p' \\\\\n \\mathsf{back\\usc{}right} & : q \\circ h_C \\htpy h_B \\circ q' \\\\\n \\mathsf{front\\usc{}left} & : f \\circ h_A \\htpy h_X \\circ f' \\\\\n \\mathsf{front\\usc{}right} & : g \\circ h_B \\htpy h_X \\circ g' \\\\\n \\mathsf{bottom} & : f \\circ p \\htpy g \\circ q\n \\end{align*}\n witnessing that the 6 faces of the cube commute,\n\\item and a homotopy \n \\begin{align*}\n % ((((h ·l back-left) ∙h (front-left ·r f')) ∙h (hD ·l top))) ~\n % ((bottom ·r hA) ∙h ((k ·l back-right) ∙h (front-right ·r g')))\n\\mathsf{coh\\usc{}cube} & : \\ct{(\\ct{(f \\cdot \\mathsf{back\\usc{}left})}{(\\mathsf{front\\usc{}left}\\cdot p')})}{(h_X \\cdot \\mathsf{top})} \\\\\n& \\qquad \\htpy \\ct{(\\mathsf{bottom}\\cdot h_C)}{(\\ct{(g \\cdot \\mathsf{back\\usc{}right})}{(\\mathsf{front\\usc{}right}\\cdot q')})}\n\\end{align*}\nfilling the cube.\n\\end{enumerate}\n\\end{defn}\n\nIn the following lemma we show that if a cube commutes, then so do its rotations and mirror symmetries (that preserve the directions of the arrows).\\footnote{The group acting on commuting cubes of maps is the \\emph{dihedral group} $D_3$ which has order $6$.} This fact is obviously true, but there is some `path algebra' involved that we wish to demonstrate at least once.\n\n\\begin{lem}\n Consider a commuting cube\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n A' \\arrow[d] & C \\arrow[dl] \\arrow[dr] & B' \\arrow[dl,crossing over] \\arrow[d] \\\\\n A \\arrow[dr] & X' \\arrow[from=ul,crossing over] \\arrow[d] & B \\arrow[dl] \\\\\n & X.\n \\end{tikzcd}\n \\end{equation*}\n Then the cubes\n\n \\begin{center}\n \\begin{minipage}{.3\\textwidth}\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n C \\arrow[d] & B' \\arrow[dl] \\arrow[dr] & A' \\arrow[dl,crossing over] \\arrow[d] \\\\\n B \\arrow[dr] & A \\arrow[from=ul,crossing over] \\arrow[d] & X' \\arrow[dl] \\\\\n & X\n \\end{tikzcd}\n \\end{equation*}\n \\end{minipage}\n \\begin{minipage}{.3\\textwidth}\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n B' \\arrow[d] & A' \\arrow[dl] \\arrow[dr] & C \\arrow[dl,crossing over] \\arrow[d] \\\\\n X' \\arrow[dr] & B \\arrow[from=ul,crossing over] \\arrow[d] & A \\arrow[dl] \\\\\n & X\n \\end{tikzcd}\n \\end{equation*}\n \\end{minipage}\n\n \\begin{minipage}{.3\\textwidth}\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n C \\arrow[d] & A' \\arrow[dl] \\arrow[dr] & B' \\arrow[dl,crossing over] \\arrow[d] \\\\\n A \\arrow[dr] & B \\arrow[from=ul,crossing over] \\arrow[d] & X' \\arrow[dl] \\\\\n & X.\n \\end{tikzcd}\n \\end{equation*}\n \\end{minipage}\n \\begin{minipage}{.3\\textwidth}\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n A' \\arrow[d] & B' \\arrow[dl] \\arrow[dr] & C \\arrow[dl,crossing over] \\arrow[d] \\\\\n X' \\arrow[dr] & A \\arrow[from=ul,crossing over] \\arrow[d] & B \\arrow[dl] \\\\\n & X.\n \\end{tikzcd}\n \\end{equation*}\n \\end{minipage}\n \\begin{minipage}{.3\\textwidth}\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[d] \\arrow[dr] \\\\\n B' \\arrow[d] & C \\arrow[dl] \\arrow[dr] & A' \\arrow[dl,crossing over] \\arrow[d] \\\\\n B \\arrow[dr] & X' \\arrow[from=ul,crossing over] \\arrow[d] & A \\arrow[dl] \\\\\n & X.\n \\end{tikzcd}\n \\end{equation*}\n \\end{minipage}\n \\end{center}\n also commute.\n\\end{lem}\n\n\\begin{proof}\n We only show that the first cube commutes, which is obtained by a counter-clockwise rotation of the original cube around the axis through $C'$ and $X$. The other cases are similar, and they are formalized in the Agda library for this course.\n\n First we list the homotopies witnessing that the faces of the cube commute:\n \\begin{align*}\n \\mathsf{top}' & \\defeq \\mathsf{back\\usc{}left} \\\\\n \\mathsf{back\\usc{}left}' & \\defeq \\mathsf{back\\usc{}right}^{-1} \\\\\n \\mathsf{back\\usc{}right}' & \\defeq \\mathsf{top}^{-1} \\\\\n \\mathsf{front\\usc{}left}' & \\defeq \\mathsf{bottom}^{-1} \\\\\n \\mathsf{front\\usc{}right}' & \\defeq \\mathsf{front\\usc{}left}^{-1} \\\\\n \\mathsf{bottom}' & \\defeq \\mathsf{front\\usc{}right}. \n \\end{align*}\n Thus, to show that the cube commutes, we have to show that there is a homotopy of type\n \\begin{align*}\n & \\ct{(\\ct{(g \\cdot \\mathsf{back\\usc{}right}^{-1})}{(\\mathsf{bottom}^{-1}\\cdot h_C)})}{(f \\cdot \\mathsf{back\\usc{}left})} \\\\\n & \\qquad\\qquad \\htpy \\ct{(\\mathsf{front\\usc{}right}\\cdot q')}{(\\ct{(h_X \\cdot \\mathsf{top}^{-1})}{(\\mathsf{front\\usc{}left}^{-1}\\cdot p')})}.\n \\end{align*}\n In other words, we need to construct a path\n \\begin{align*}\n & \\ct{(\\ct{\\ap{g}{\\mathsf{back\\usc{}right}(c')^{-1}}}{\\mathsf{bottom}(h_C(c'))^{-1}})}{\\ap{f}{\\mathsf{back\\usc{}left}(c')}} \\\\\n & \\qquad\\qquad \\htpy \\ct{\\mathsf{front\\usc{}right}(q'(c'))}{(\\ct{\\ap{h_X}{\\mathsf{top}(c')^{-1}}}{\\mathsf{front\\usc{}left}(p'(c'))^{-1}})}\n \\end{align*}\n for every $c':C'$. Recall that $\\ap{g}{\\mathsf{back\\usc{}right}(c')^{-1}}=\\ap{g}{\\mathsf{back\\usc{}right}(c')}^{-1}$. Using the action on paths of the concatenation function (twice), we see that\n \\begin{align*}\n & \\ct{(\\ct{\\ap{g}{\\mathsf{back\\usc{}right}(c')^{-1}}}{\\mathsf{bottom}(h_C(c'))^{-1}})}{\\ap{f}{\\mathsf{back\\usc{}left}(c')}} \\\\\n & \\qquad\\qquad \\ct{(\\ct{\\ap{g}{\\mathsf{back\\usc{}right}(c')}^{-1}}{\\mathsf{bottom}(h_C(c'))^{-1}})}{\\ap{f}{\\mathsf{back\\usc{}left}(c')}}.\n \\end{align*}\n Similarly, we have an identification\n \\begin{align*}\n & \\ct{\\mathsf{front\\usc{}right}(q'(c'))}{(\\ct{\\ap{h_X}{\\mathsf{top}(c')}^{-1}}{\\mathsf{front\\usc{}left}(p'(c'))^{-1}})} \\\\\n & \\qquad\\qquad \\htpy \\ct{\\mathsf{front\\usc{}right}(q'(c'))}{(\\ct{\\ap{h_X}{\\mathsf{top}(c')^{-1}}}{\\mathsf{front\\usc{}left}(p'(c'))^{-1}})}\n \\end{align*}\n Therefore it suffices to construct an identification\n \\begin{align*}\n & \\ct{(\\ct{\\ap{g}{\\mathsf{back\\usc{}right}(c')}^{-1}}{\\mathsf{bottom}(h_C(c'))^{-1}})}{\\ap{f}{\\mathsf{back\\usc{}left}(c')}} \\\\\n & \\qquad\\qquad \\htpy \\ct{\\mathsf{front\\usc{}right}(q'(c'))}{(\\ct{\\ap{h_X}{\\mathsf{top}(c')}^{-1}}{\\mathsf{front\\usc{}left}(p'(c'))^{-1}})}\n \\end{align*}\n Now we note that our goal is of the form\n \\begin{equation*}\n \\ct{(\\ct{\\varepsilon^{-1}}{\\delta^{-1}})}{\\beta}=\\ct{\\zeta}{(\\ct{\\gamma^{-1}}{\\beta^{-1}})}, %%% check greek alphabet\n \\end{equation*}\n whereas the assumption that the original cube commutes yields an identification of the form\n \\begin{equation*}\n \\ct{(\\ct{\\alpha}{\\beta})}{\\gamma}=\\ct{\\delta}{(\\ct{\\varepsilon}{\\zeta})}\n \\end{equation*}\n Indeed, in the case that $\\alpha$, $\\beta$, $\\gamma$, $\\delta$, $\\varepsilon$, and $\\zeta$ are general identifications, we can conclude our goal using path induction on all of them.\n\\end{proof}\n\n\\begin{lem}\nGiven a commuting cube as in \\cref{defn:cube} we obtain a commuting square\n\\begin{equation*}\n\\begin{tikzcd}\n\\fib{f_{1\\check{1}1}}{x} \\arrow[r] \\arrow[d] & \\fib{f_{0\\check{1}1}}{f_{\\check{1}01}(x)} \\arrow[d] \\\\\n\\fib{f_{1\\check{1}0}}{f_{10\\check{1}}(x)} \\arrow[r] & \\fib{f_{0\\check{1}0}}{f_{00\\check{1}}(x)}\n\\end{tikzcd}\n\\end{equation*}\nfor any $x:A_{101}$. \n\\end{lem}\n\n\\begin{lem}\nConsider a commuting cube\n\\begin{equation*}\n\\begin{tikzcd}\n& A_{111} \\arrow[dl] \\arrow[dr] \\arrow[d] \\\\\nA_{110} \\arrow[d] & A_{101} \\arrow[dl] \\arrow[dr] & A_{011} \\arrow[dl,crossing over] \\arrow[d] \\\\\nA_{100} \\arrow[dr] & A_{010} \\arrow[d] \\arrow[from=ul,crossing over] & A_{001} \\arrow[dl] \\\\\n& A_{000}.\n\\end{tikzcd}\n\\end{equation*}\nIf the bottom and front right squares are pullback squares, then the back left square is a pullback if and only if the top square is.\n\\end{lem}\n\n\\begin{rmk}\nBy rotating the cube we also obtain:\n\\begin{enumerate}\n\\item If the bottom and front left squares are pullback squares, then the back right square is a pullback if and only if the top square is.\n\\item If the front left and front right squares are pullback, then the back left square is a pullback if and only if the back right square is.\n\\end{enumerate}\nBy combining these statements it also follows that if the front left, front right, and bottom squares are pullback squares, then if any of the remaining three squares are pullback squares, all of them are. Cubes that consist entirely of pullback squares are sometimes called \\define{strongly cartesian}\\index{strongly cartesian cube}.\n\\end{rmk}\n\n\\subsection{Families of pullbacks}\n\n\\begin{lem}\\label{lem:fiberwise-pullback}\nConsider a pullback square\\index{pullback!Sigma-type of pullbacks@{$\\Sigma$-type of pullbacks}}\n \\begin{equation*}\n \\begin{tikzcd}\n C \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\n A \\arrow[r,swap,\"f\"] & X\n \\end{tikzcd}\n \\end{equation*}\n with $H : f \\circ p \\htpy g \\circ h$. Furthermore, consider type families $P_X$, $P_A$, $P_B$, and $P_C$ over $X$, $A$, $B$, and $C$ respectively, equipped with families of maps\n \\begin{align*}\n f' & : \\prd{a:A} P_A(a) \\to P_X(f(a)) \\\\\n g' & : \\prd{b:B} P_B(b) \\to P_X(g(b)) \\\\\n p' & : \\prd{c:C} P_C(c) \\to P_A(p(c)) \\\\\n q' & : \\prd{c:C} P_C(c) \\to P_B(q(c)),\n \\end{align*}\n and for each $c:C$ a homotopy $H'_c$ witnessing that the square\n \\begin{equation}\\label{eq:family-squares-pullback}\n \\begin{tikzcd}\n P_C(c) \\arrow[rr,\"{q'_c}\"] \\arrow[d,swap,\"{p'_c}\"] & &[3em] P_B(q(c)) \\arrow[d,\"{g'_{q(c)}}\"] \\\\\n P_A(p(c)) \\arrow[r,swap,\"{f'_{p(c)}}\"] & P_X(f(p(c))) \\arrow[r,swap,\"{\\tr_{P_X}(H(c))}\"] & P_X(g(q(c)))\n \\end{tikzcd}\n \\end{equation}\n commutes. Then the following are equivalent:\n \\begin{enumerate}\n \\item For each $c:C$ the square in \\cref{eq:family-squares-pullback} is a pullback square.\n \\item The square\n \\begin{equation}\\label{eq:total-square-pullback}\n \\begin{tikzcd}[column sep=huge]\n \\sm{c:C}P_C(c)\n \\arrow[r,\"{\\total[q]{q'}}\"] \\arrow[d,swap,\"{\\total[p]{p'}}\"] &\n \\sm{b:B}P_B(b) \\arrow[d,\"{\\total[g]{g'}}\"] \\\\\n \\sm{a:A}P_A(a) \\arrow[r,swap,\"{\\total[f]{f'}}\"] & \\sm{x:X}P_X(x)\n \\end{tikzcd}\n \\end{equation}\n is a pullback square.\n \\end{enumerate}\n\\end{lem}\n\n\n\\begin{cor}\nConsider a pullback square\n\\begin{equation*}\n\\begin{tikzcd}\nC \\arrow[r,\"q\"] \\arrow[d,swap,\"p\"] & B \\arrow[d,\"g\"] \\\\\nA \\arrow[r,swap,\"f\"] & X,\n\\end{tikzcd}\n\\end{equation*}\nwith $H:f\\circ p\\htpy g\\circ q$, and let $c_1,c_2:C$. Then the square\n\\begin{equation*}\n\\begin{tikzcd}[column sep=8em]\n(c_1=c_2) \\arrow[r,\"\\apfunc{q}\"] \\arrow[d,swap,\"\\apfunc{p}\"] & (q(c_1)=q(c_2)) \\arrow[d,\"\\lam{\\beta}\\ct{H(c_1)}{\\ap{g}{\\beta}}\"] \\\\\n(p(c_1)=p(c_2)) \\arrow[r,swap,\"\\lam{\\alpha}\\ct{\\ap{f}{\\alpha}}{H(c_2)}\"] & f(p(c_1))=g(q(c_2)),\n\\end{tikzcd}\n\\end{equation*}\ncommutes and is a pullback square.\n\\end{cor}\n\n\n\\begin{thm}\n Consider a commuting cube\n \\begin{equation*}\n \\begin{tikzcd}\n & C' \\arrow[dl] \\arrow[dr] \\arrow[d] \\\\\n A' \\arrow[d] & C \\arrow[dl] \\arrow[dr] & B' \\arrow[crossing over,dl] \\arrow[d] \\\\\n A \\arrow[dr] & X' \\arrow[d] \\arrow[from=ul,crossing over] & B \\arrow[dl] \\\\\n & X\n \\end{tikzcd}\n \\end{equation*}\n in which the bottom square is a pullback square. Then the following are equivalent:\n \\begin{enumerate}\n \\item The top square is a pullback square.\n \\item The square\n \\begin{equation*}\n \\begin{tikzcd}\n \\fib{\\gamma}{c} \\arrow[d] \\arrow[r] & \\fib{\\beta}{q(c)} \\arrow[d] \\\\\n \\fib{\\alpha}{p(c)} \\arrow[r] & \\fib{\\varphi}{f(p(c))}\n \\end{tikzcd}\n \\end{equation*}\n is a pullback square for each $c:C$.\n \\end{enumerate}\n\\end{thm}\n\n\n\\subsection{The 3-by-3-properties for pullbacks and pushouts}\n\n\\begin{thm}\n Consider a commuting diagram of the form\n \\begin{equation*}\n \\begin{tikzcd}[column sep=large,row sep=large]\n AA \\arrow[r,\"Af\"] \\arrow[d,swap,\"fA\"] \\arrow[dr,phantom,\"\\Rightarrow\" description] & AX \\arrow[d,swap,\"fX\"] & AB \\arrow[l,swap,\"Ag\"] \\arrow[d,\"gB\"] \\arrow[dl,phantom,\"\\Leftarrow\" description] \\\\\n XA \\arrow[r,\"Xf\"] & XX & XB \\arrow[l,swap,\"Xg\"] \\\\\n BA \\arrow[u,\"gA\"] \\arrow[r,swap,\"Bf\"] & BX \\arrow[u,\"gX\"] & BB \\arrow[u,swap,\"gB\"] \\arrow[l,\"Bg\"]\n \\end{tikzcd}\n \\end{equation*}\n with homotopies\n \\begin{align*}\n ff & : Xf \\circ fA \\htpy Af \\circ fX \\\\\n fg & : Xg \\circ gB \\htpy Ag \\circ fX \\\\\n gf & : \n \\end{align*}\n filling the (small) squares. Furthermore, consider\n pullback squares\n \\begin{equation*}\n \\begin{tikzcd}\n AC \\arrow[r] \\arrow[d] & AB \\arrow[d] & XC \\arrow[r] \\arrow[d] & XB \\arrow[d] & BC \\arrow[r] \\arrow[d] & BB \\arrow[d] \\\\\n AA \\arrow[r] & AX & XA \\arrow[r] & XX & BA \\arrow[r] & BX\n \\end{tikzcd}\n \\end{equation*}\n \\begin{equation*}\n \\begin{tikzcd}\n CA \\arrow[r] \\arrow[d] & BA \\arrow[d] & CX \\arrow[r] \\arrow[d] & BX \\arrow[d] & CB \\arrow[r] \\arrow[d] & BB \\arrow[d] \\\\\n AA \\arrow[r] & XA & AX \\arrow[r] & XX & AB \\arrow[r] & XB.\n \\end{tikzcd}\n \\end{equation*}\n Finally, consider a commuting square\n \\begin{equation*}\n \\begin{tikzcd}\n D_3 \\arrow[r] \\arrow[d] & D_2 \\arrow[d] \\\\\n D_0 \\arrow[r] & D_1.\n \\end{tikzcd}\n \\end{equation*}\n Then the following are equivalent:\n \\begin{enumerate}\n \\item This square is a pullback square.\n \\item The induced square\n \\begin{equation*}\n \\begin{tikzcd}\n D_3 \\arrow[r] \\arrow[d] & C_3 \\arrow[d] \\\\\n A_3 \\arrow[r] & B_3\n \\end{tikzcd}\n \\end{equation*}\n is a pullback square.\n \\end{enumerate}\n\\end{thm}\n\n\\begin{proof}\n First we construct an equivalence\n \\begin{equation*}\n (A_0\\times_{B_0}C_0)\\times_{(A_1\\times_{B_1}C_1)}(A_2\\times_{B_2} C_2) \\eqvsym (A_0\\times_{A_1}A_2)\\times_{(B_0\\times_{B_1} B_2)} (C_0\\times_{C_1}C_2).\n \\end{equation*}\n Now it follows that we have an equivalence\n \\begin{equation*}\n \\mathsf{cone}(f_0,g_0)\n \\end{equation*}\n\\end{proof}\n\n\n\\begin{exercises}\n\\item Some exercises.\n\\end{exercises}\n", "meta": {"hexsha": "c447d9e0d2466081c10135ee22b4933f0b94834e", "size": 16047, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Book/cubical.tex", "max_stars_repo_name": "tadejpetric/HoTT-Intro", "max_stars_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Book/cubical.tex", "max_issues_repo_name": "tadejpetric/HoTT-Intro", "max_issues_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Book/cubical.tex", "max_forks_repo_name": "tadejpetric/HoTT-Intro", "max_forks_repo_head_hexsha": "f4228d6ecfc6cdb119c6e8b0e711fea05b98b2d5", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.9495798319, "max_line_length": 452, "alphanum_fraction": 0.6267838225, "num_tokens": 5952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7956581000631542, "lm_q2_score": 0.76908023177796, "lm_q1q2_score": 0.6119249160125819}} {"text": "\\section{More refined sparse grid and improved estimates}\nThis section presents a modified estimate \\cite{bungartz2004sparse} of the approximation property of sparse grids. Recall \n$$\n\\mathcal V_{\\mathbf{k}} = \\mathcal V_{k_1} \\bigotimes \\mathcal V_{k_2}\\bigotimes\\cdots \\bigotimes \\mathcal V_{k_d},\n$$\nwhere $\\mathbf{k}=(k_1,k_2,\\cdots,k_d)$ is a multi-index. Let \n$$\n\\mathbf{I_l}=\\{\\mathbf{i}\\in \\mathbb{N}^d: \\mathbf{1}\\le \\mathbf{i}\\le 2^\\mathbf{l}-1, i_j \\text{ is odd for all }1\\le j\\le d\\},\n$$\nand the hierarchical basis \n$$\n\\{\\psi_\\mathbf{k, i}^d: \\mathbf{i}\\in \\mathbf{I_k}, |\\mathbf{k}|_\\infty\\le J\\}.\n$$ \n\n\n\n\nRecall $\\mathcal V_{\\mathbf{k}}$ in \\eqref{sparseVk}. Let\n$$\nR_J=\\left\\{\\mathbf{k}: |\\mathbf{k}|_1 - \\frac25 \\log_2\\|2^{\\mathbf{k}}\\|_{l^2}\\le J+d-1- \\frac15 \\log_2 (4^J +4d-4)\\right\\},\n$$\n\\begin{equation}\n \\label{sparseVJ}\nS_J =\\bigoplus_{\\mathbf{k}\\in R_J} \\mathcal V_{\\mathbf{k}}. \n\\end{equation}\nThe following the result is analyzed in \\cite{bungartz2004sparse}.\n \\begin{theorem}\\Label{lm:modifySparse1}\nSuppose that $u\\in W^{2\\mathbf{e}, \\infty}(D_d)$,\n$$\n\\inf_{\\chi\\in S_J} |v-\\chi|_{1, D_d}\\le C h_J \\|u\\|_{2\\mathbf{e},\\infty}.\n$$\nand the dimension of $S_J$ is $\\mathcal{O}(de^dh_J^{-1})$.\n\\end{theorem}\n\\begin{proof}\nFirst, we prove that $S_J$ is a subspace of $S_{J,J+d-1}^d$.\nIf $|\\mathbf{k}|_{1}=J+d-1+i, i \\in \\mathbb{N}$, we can prove \n$$\n\\|2^{\\mathbf{k}}\\|_{l^2}^2=\\sum_{s=1}^{d} 4^{k_{s}}\\le 4^{J+i}+4 d-4\n$$\n by induction with respect to $d$. \n% It is trivial when $d=1$. For $\\mathbf{k}\\in \\mathbb{R}^d$ and $|\\mathbf{k}|_{1}=J+d-1+i$, there exists $1\\le j\\le d$ such that $k_j\\ge 1$, thus\n%\\begin{equation}\n%\\begin{split}\n%\\sum_{s=1}^{d} 4^{k_{s}} &= \\sum_{s=1, s\\neq i}^{d} 4^{k_{s}} + 4^{k_{j}}\\le 4^{J+i+1-k_j}+4 (d-1)-4 + 4^{k_j}\\le 4^{J+i} + 4d-4.\n%\\end{split}\n%\\end{equation}\nFor subspaces $\\mathcal V_{\\mathbf{k}}$ with $|\\mathbf{k}|_{1}=J+d-1+i, i \\in \\mathbb{N},$ we have\n\\begin{equation}\n\\begin{split}\n|\\mathbf{k}|_{1}-\\frac{1}{5} \\cdot \\log _{2}\\left(\\sum_{s=1}^{d} 4^{k_{s}}\\right) & \\geq J+d-1+i-\\frac{1}{5} \\cdot \\log _{2}\\left(4^{J+i}+4 d-4\\right) \\\\\n& \\geq J+d-1+i-\\frac{1}{5} \\cdot \\log _{2}\\left(4^{i}\\left(4^{J}+4 d-4\\right)\\right) \\\\\n&>J+d-1-\\frac{1}{5} \\cdot \\log _{2}\\left(4^{J}+4 d-4\\right)\n\\end{split}\n\\end{equation}\nTherefore, no $\\mathcal V_{\\mathbf{k}}$ with $|\\mathbf{k}|_{1}>J+d-1$ can belong to $V_J .$ Consequently, \n\\begin{equation}\nS_J\\subset S_{J,J+d-1}^d,\\qquad \\left|S_J\\right| \\leq\\left|S_{J,J+d-1}^d\\right|.\n\\end{equation}\n\nThen, we prove that the dimension of $S_J$ is $\\mathcal{O}(h_J^{-1})$. \nNote that $\\left|\\mathcal V_{\\mathbf{k}}\\right| = 2^{|\\mathbf{k}|_{1}-d}$. For any $\\mathbf{k}\\in R_J$ with $|\\mathbf{k}|_{1}=J+d-1-i$, $ \\sum_{j=1}^d 4^{k_j}\\ge {4^J+4d-4\\over 32^i}$. By \\eqref{sparseVJ},\n\\begin{equation}\n\\begin{split}\n\\mbox{dim}S_J&=\\sum_{i=0}^{J-1} \\sum_{|\\mathbf{k}|_{1}=J+d-1-i, \\sum_{j=1}^d 4^{k_j}\\ge {4^J+4d-4\\over 32^i}} \\mbox{dim}\\mathcal V_{\\mathbf{k}}\n\\\\\n&=2^{J-1} \\cdot \\sum_{i=0}^{J-1} 2^{-i} \\sum_{|\\mathbf{k}|_{1}=J+d-1-i, \\sum_{j=1}^d 4^{k_j}\\ge {4^J+4d-4\\over 32^i}}1\n\\\\\n& \\leq 2^{J-1} \\cdot \\lim _{J \\rightarrow \\infty} \\sum_{i=0}^{J-1} 2^{-i} \\sum_{|\\mathbf{k}|_{1}=J+d-1-i, \\sum_{j=1}^d 4^{k_j}\\ge {4^J+4d-4\\over 32^i}}1\n\\\\\n&=2^{J-1} d \\lim _{J \\rightarrow \\infty} \\sum_{i=0}^{J-1} 2^{-i} \\left(\\begin{array}{c}\nd-1+\\lfloor 1.5 i\\rfloor \\\\\nd-1\n\\end{array}\\right).\n\\end{split}\n\\end{equation} \nLet $j=\\lfloor 1.5 i\\rfloor $. Since $\\displaystyle\\sum_{i=0}^{\\infty} x^{i} \\cdot\\left(\\begin{array}{c}k+i \\\\ k\\end{array}\\right)=(1-x)^{-k-1}$ for $k \\in \\mathbb{N}_{0}$ and $0 0, i = 1, \\ldots, r\n\\end{align*}\nThere are $\\binom{n+r-1}{r-1}$ distince nonnegative integer-valued vectors $(x_1, x_2, \\ldots, x_r)$ satisfying the equation\n\\[\n x_1 + x_2 + \\cdots + x_r = n\n\\]\n\n\\section{Conditional Probability}\n\\[\n P(E|F) = \\frac{P(EF)}{P(F)} \\Leftrightarrow P(EF) = P(E|F)P(F)\n\\]\n\\subsection{Generalized Chain Rule}\n\\[\n P(E_1 E_2 \\ldots E_n) = P(E_1)P(E_2|E_1)P(E_3|E_1 E_2)\\ldots P(E_n|E_1 E_2 \\ldots E_{n-1})\n\\]\n\\subsection{Bayes' Theorem}\nThe many shapes and forms of Bayes' Theorem...\n\\[\n P(E) = P(E|F)P(F) + P(E|F^c)P(F^c)\n\\]\n\\[\n P(F|E) = \\frac{P(EF)}{P(E)} = \\frac{P(E|F)P(F)}{P(E)}\n\\]\n\\[\n P(F|E) = \\frac{P(E|F)P(F)}{P(E|F)P(F) + P(E|F^c)P(F^c)}\n\\]\n\\emph{Fully General Form}:\\\\\nIf $F_1, F_2, \\ldots, F_n$ comprise a set of mutually exclusive and exhaustive events, then\n\\[\n P(F_j|E) = \\frac{P(E|F_j)P(F_j)}{\\sum_{i=1}^n P(E|F_i)P(F_i)}\n\\]\n\\emph{That's odd.} \\\\\nThe odds of $H$ given observed evidence $E$:\n\\[\n \\frac{P(H|E)}{P(H^c|E)} = \\frac{P(H)P(E|H)}{P(H^c)P(E|H^c)}\n\\]\n\n\\section{Independence}\n\\subsection{Definition}\nTwo events are independent if $P(EF) = P(E)P(F)$. Otherwise they are dependent. \\\\\nMore generally, events $E_1, E_2, \\ldots, E_n$ are independent if for every subset $E_{1'}, E_{2'}, \\ldots, E_r$ where $r \\leq n$ it holds that\n\\[\n P(E_{1'} E_{2'} \\ldots E_r) = P(E_{1'})P(E_{2'})\\cdots P(E_r)\n\\]\n\\subsection{Conditional Independence}\nTwo events $E$ and $F$ are conditional independent given $G$ if \n\\[\n P(EF|G) = P(E|G)P(F|G)\n\\]\nDependent events can become independent, and vice-versa, by conditioning on additional information.\n\n\\section{Random Distributions}\n\\subsection{Definitions and Properties}\n\\emph{Probability Mass Function}:\n\\[\n p(a) = P(X=a)\n\\]\n\\emph{Probability Density Function}:\n\\begin{align*}\n &P(a \\leq X \\leq b) = \\int_{a}^{b} f(x) dx &P(-\\infty < X < \\infty) = \\int_{-\\infty}^{\\infty} f(x) dx = 1\n\\end{align*}\n\\emph{Cumulative Distribution Function}:\n\\[\n F(a) = F(X \\leq a) \\text{ where } -\\infty < a \\infty\n\\]\n\\begin{align*}\n &F(a) = \\sum_{\\text{all } x\\leq a} p(x) &F(a) = \\int_{-\\infty}^a f(x) dx\n\\end{align*}\nDensity $f$ is the derivative of CDF $F$: $f(a) = \\frac{d}{da} F(a)$\n\n\\subsection{Joint distributions}\n\\emph{Joint Probability Mass Function}:\n\\[\n p_{X,Y}(a,b) = P(X=a,Y=b)\n\\]\nMarginal distributions:\n\\begin{align*}\n &p_X(a) = P(X=a)= \\sum_y p_{X,Y} (a, y) &p_Y(b) = P(Y=b)= \\sum_x p_{X,Y} (x, b)\n\\end{align*}\n\\emph{Joint Cumulative Probability Distribution (CDF)}:\n\\[\n F_{X,Y}(a,b) = F(a,b) = P(X \\leq a, Y \\leq b) \\text{ where } -\\infty < a,b < \\infty\n\\]\nMarginal distributions:\n\\begin{align*}\n F_X(a) = P(X \\leq a) = P(X \\leq a, Y < \\infty) = F_{X,Y}(a,\\infty) \\\\\n F_Y(b) = P(Y \\leq b) = P(X < \\infty, Y \\leq b) = F_{X,Y}(\\infty, b)\n\\end{align*}\n\\emph{Joint Probability Density Function}:\n\\[\n P(a_1 < X \\leq a_2, b_1 < Y \\leq b_2) = \\int_{a_1}^{a_2} \\int_{b_1}^{b_2} f_{X,Y}(x,y) dy dx\n\\]\n\\begin{align*}\n &F_{X,Y}(a,b) = \\int_{-\\infty}^a \\int_{-\\infty}^b f_{X,Y}(x,y) dy dx &f_{X,Y}(a,b) = \\frac{\\partial^2}{\\partial a \\partial b} F_{X,Y} (a,b)\n\\end{align*}\nMarginal density functions:\n\\begin{align*}\n &f_x(a) = \\int_{-\\infty}^{\\infty} f_{X,Y}(a,y)dy \n &f_y(b) = \\int_{-\\infty}^{\\infty} f_{X,Y}(x,b)dx\n\\end{align*}\n\\subsection{Independent Random Variables}\n$n$ random variables $X_1, X_2, \\ldots, X_n$ are called independent if\n\\[\n P(X_1 = x_1, X_2 = x_2, \\ldots, X_n = x_n) = \\prod_{i=1}^n P(X_i = x_i) \\text{ for all } x_1, x_2, \\ldots, x_n\n\\]\nor analogously for continuous random variables if\n\\[\n P(X_1 \\leq a_1, X_2 \\leq a_2, \\ldots, X_n \\leq a_n) = \\prod_{i=1}^n P(X_i \\leq a_i) \\text{ for all } a_1, a_2, \\ldots, a_n\n\\]\n\\subsection{Convolution}\nLet $X$ and $Y$ be independent random variables. The convolution of $F_X$ and $F_Y$ is $F_{X+Y}$:\n\\[\n F_{X+Y}(a) = P(X+Y \\leq a) = \\int_{y=-\\infty}^{\\infty} F_X (a-y)f_Y(y)dy\n\\]\n\\[\n f_{X+Y}(a) = \\int_{y=-\\infty}^{\\infty} f_X (a-y)f_Y(y)dy\n\\]\nIn discrete case, replace $\\int_{y=-\\infty}^{\\infty}$ with $\\sum_{y}$, and $f(y)$ with $p(y)$.\n\\subsection{Conditional Distributions}\n\\emph{Conditional PMF of $X$ given $Y$}:\n\\[\n p_{X|Y}(x|y) = P(X=x|Y=y) = \\frac{p_{X,Y}(x,y)}{p_Y(y)}\n\\]\n\\emph{Conditional PDF of $X$ given $Y$}:\n\\[\n f_{X|Y}(x|y) = \\frac{f_{X,Y}(x,y)}{f_Y(y)}\n\\]\n\\emph{Conditional CDF of $X$ given $Y$}:\n\\begin{align*}\n F_{X|Y}(a|y) = P(X \\leq a, Y=y) &= \\sum_{x\\leq a} p_{X|Y}(x|y) \\\\\n &=\\int_{-\\infty}^{a} f_{X|Y} (x|y) dx\n\\end{align*}\n$n$ random variables $X_1, X_2, \\ldots, X_n$ are conditionally independent given $Y$ if\n\\[\n P(X_1 = x_1, X_2 = x_2, \\ldots, X_n = x_n|Y=y) = \\prod_{i=1}^n P(X_i = x_i|Y=y) \\text{ for all } x_1, x_2, \\ldots, x_n, y\n\\]\nor analogously for continuous random variables if\n\\[\n P(X_1 \\leq a_1, X_2 \\leq a_2, \\ldots, X_n \\leq a_n|Y=y) = \\prod_{i=1}^n P(X_i \\leq a_i|Y=y) \\text{ for all } a_1, a_2, \\ldots, a_n, y\n\\]\nIt is possible to mix continuous and discrete random variables in conditional distributions. For example let $X$ be a continuous random variable and $N$ be a discrete random variable. Then the conditional PDF of $X$ given $N$ and the conditional PMF of $N$ given $X$ are\n\\[\n f_{X|N}(x|n) = \\frac{p_{N|X}(n|x)f_X(x)}{p_N(n)}\n\\]\n\\[\n P_{N|X}(n|x) = \\frac{f_{X|N}(x|n)p_N(n)}{f_X(x)}\n\\]\n\n\\section{Expectation}\n\\subsection{Definitions}\nThe expected value for a discrete random variable $X$ is defined as\n\\[\n E[X] = \\sum_{x: p(x) > 0} x p(x)\n\\]\nFor a continuous random variable $X$, the expected value is\n\\[\n E[X] = \\int_{-\\infty}^{\\infty} x f(x) dx\n\\]\n\\subsection{Properties}\nIf $I$ is an indicator variable for the event $A$, then\n\\[\n E[I] = P(A)\n\\]\nLet $g(X)$ be a real-valued function of $X$.\n\\begin{align*}\n &E[g(X)] = \\sum_{i} g(x_i)p(x_i) &E[g(X)] = \\int_{-\\infty}^{\\infty} g(x) f(x) dx\n\\end{align*}\nLet $g(X,Y)$ be a real-valued function of two random variables.\n\\begin{align*}\n &E[g(X,Y)] = \\sum_y \\sum_x g(x,y) p_{X,Y}(x,y) &E[g(X,Y)] = \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} g(x,y) f_{X,Y}(x,y) dx dy\n\\end{align*}\nLinearity:\n\\[\n E[aX + b] = aE[X] + b\n\\]\n$N$-th Moment of $X$:\n\\[\n E[X^n] = \\sum_{x: p(x) > 0} x^n p(x)\n\\]\nExpected Values of Sums:\n\\[\n E\\left[ \\sum_{i=1}^n X_i \\right] = \\sum_{i=1}^n E[X_i]\n\\]\nBounding Expectation:\\\\\nIf random variable $X \\geq a$ then $E[X] \\geq a$. \\\\\nIf $P(a \\leq X < \\infty) = 1$ then $a \\leq E[X] < \\infty$. \\\\\nIf random variables $X \\geq Y$ then $E[X] \\geq E[Y]$.\n\n\\subsection{Conditional Expectation}\nConditional Expectation of $X$ given $Y=y$:\n\\begin{align*}\n &E[X|Y=y] = \\sum_x x p_{X|Y}(x|y) &E[X|Y=y] = \\int_{-\\infty}^{+\\infty} x f_{X|Y} (x|y) dx\n\\end{align*}\nExpectation of conditional sum:\n\\[\n E\\left[ \\sum_{i=1}^n X_i | Y=y \\right] = \\sum_{i=1}^n E[X_i | Y=y]\n\\]\nExpectation of conditional expectations:\n\\[\n E[E[X|Y]] = E[X]\n\\]\n\n\n\\section{Variance}\n\\subsection{Definition}\nIf $X$ is a random variable with mean $\\mu$ then the variance of $X$, denoted $\\Var(X)$, is:\n\\[\n \\Var(X) = E[(X-\\mu)^2] = E[X^2] - (E[X])^2\n\\]\n\\subsection{Properties}\n\\[\n \\Var(aX + b) = a^2\\Var(X)\n\\]\nIf $X_1, X_2, \\ldots, X_n$ are independent random variables, then\n\\[\n \\Var\\left( \\sum_{i=1}^n X_i \\right) = \\sum_{i=1}^n \\Var(X_i)\n\\]\n\\subsection{Covariance}\n\\[\n \\Cov(X,Y) = E[(X-E[X])(Y-E[Y])] = E[XY] - E[X]E[Y]\n\\]\nIf $X$ and $Y$ are independent, $\\Cov(X,Y) = 0$\n\\emph{Properties}:\n\\begin{align*}\n \\Cov(X,Y) &= \\Cov(Y,X) \\\\\n \\Cov(X,X) &= \\Var(X) \\\\\n \\Cov(aX+b, Y) = a\\Cov(X,Y)\n\\end{align*}\nIf $X_1, X_2, \\ldots, X_n$ and $Y_1, Y_2, \\ldots, Y_m$ are random variables, then\n\\[\n \\Cov\\left( \\sum_{i=1}^n X_i, \\sum_{j=1}^m Y_j \\right) = \\sum_{i=1}^n \\sum_{j=1}^m \\Cov(X_i, X_j)\n\\]\n\\subsection{Correlation}\n\\[\n \\rho(X,Y) = \\frac{\\Cov(X,Y)}{\\sqrt{\\Var(X)\\Var(Y)}}\n\\]\nNote: $-1 \\leq \\rho(X,Y) \\leq 1$. \\\\\nCorrelation measures linearity between $X$ and $Y$.\\\\\nIf $\\rho(X,Y) = 0$, $X$ and $Y$ are uncorrelated.\n\n\\section{Moment Generating Functions}\n\\subsection{Definition}\nMoment Generating Function (MGF) of a random variable $X$, where $-\\infty < t < \\infty$, is\n\\[\n M(t) = E[e^{tX}]\n\\]\n\\begin{align*}\n &\\text{When $X$ is discrete: } &= \\sum_x e^{tx} p(x) \\\\\n &\\text{When $X$ is continuous: } &= \\int_{-\\infty}^{\\infty} e^{tx} f(x) dx\n\\end{align*}\nFor any $n$ random variables $X_1, X_2, \\ldots, X_n$\n\\[\n M(t_1, t_2, \\ldots, t_n) = E[e^{t_1X_1+t_2X_2+\\cdots+t_nX_n}]\n\\]\nThe individual moment generating function is obtained:\n\\[\n M_{X_i}(t) = E[e^{tX}] = M(0, \\ldots, 0, t, 0, \\ldots, 0) \\text{ where $t$ at $i$th place}\n\\]\n\\subsection{Properties}\n\\[\n M^n(t) = \\left( \\frac{d^n}{dt^n} \\right) M(t) = E[X^n e^{nX}]\n\\]\n\\[\n M^n(0) = E[X^n]\n\\]\n\\[\n M_X(t) = M_Y(t) \\text{ iff } X \\sim Y\n\\]\n$X_1, X_2, \\ldots, X_n$ independent if and only if:\n\\[\n M(t_1, t_2, \\ldots, t_n) = M_{X_1}(t_1)M_{X_2}(t_2)\\ldots M_{X_n}(t_n)\n\\]\n\n\\section{Inequalities}\n\\subsection{Boole's Inequality}\nLet $E_1, E_2, \\ldots, E_n$ be events with indicator random variables $X_i$.\n\\[\n \\sum_{i=1}^n P(E_i) \\geq P\\left( \\bigcup_{i=1}^n E_i \\right)\n\\]\n\\subsection{Markov's Inequality}\n$X$ is a nonnegative random variable.\n\\[\n P(X \\geq a) \\leq \\frac{E[X]}{a} \\text{ for all } a > 0\n\\]\n\\subsection{Chebyshev's Inequality}\n$X$ is a random variable with $E[X] = \\mu$ and $\\Var(X) = \\sigma^2$.\n\\[\n P(|X-\\mu| \\geq k) \\leq \\frac{\\sigma^2}{k^2} \\text{ for all } k > 0\n\\]\n\\emph{One-sided inequality}:\n\\[\n P(X \\geq E[X] + a) \\leq \\frac{\\sigma^2}{\\sigma^2 + a^2} \\text{ for any } a > 0\n\\]\n\\[\n P(X \\leq E[X] - a) \\leq \\frac{\\sigma^2}{\\sigma^2 + a^2} \\text{ for any } a > 0\n\\]\n\\subsection{Chernoff Bound}\n$X$ is a random variable with MGF $M(t)$.\n\\[\n P(X \\geq a) \\leq e^{-ta} M(t) \\text{ for all } t > 0 \n\\]\n\\[\n P(X \\leq a) \\leq e^{-ta} M(t) \\text{ for all } t < 0 \n\\]\nIn practice, use the $t$ that minimizes $e^{-ta} M(t)$. \n \n\\subsection{Jensen's Inequality}\nIf $f(x)$ is a convex function ($f''(x) \\geq 0$ for all $x$) then\n\\[\n E[f(x)] \\geq f(E[X])\n\\]\n\n\\end{document}\n\n", "meta": {"hexsha": "1dc8c6dacfa6dfe228981a2fefa25dfc3a9e513d", "size": 11606, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "studysheet.tex", "max_stars_repo_name": "kashizui/Stanford-CS109-Notes", "max_stars_repo_head_hexsha": "e0e0ab8fbc1afbe0711d23b8be4b0c3d4be09cb2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2015-02-24T23:32:55.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-21T14:54:10.000Z", "max_issues_repo_path": "studysheet.tex", "max_issues_repo_name": "kashizui/Stanford-CS109-Notes", "max_issues_repo_head_hexsha": "e0e0ab8fbc1afbe0711d23b8be4b0c3d4be09cb2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "studysheet.tex", "max_forks_repo_name": "kashizui/Stanford-CS109-Notes", "max_forks_repo_head_hexsha": "e0e0ab8fbc1afbe0711d23b8be4b0c3d4be09cb2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 7, "max_forks_repo_forks_event_min_datetime": "2015-06-06T06:07:22.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-25T19:57:35.000Z", "avg_line_length": 31.7103825137, "max_line_length": 270, "alphanum_fraction": 0.5950370498, "num_tokens": 5017, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743735019595, "lm_q2_score": 0.8267118004748677, "lm_q1q2_score": 0.6119108889831621}} {"text": "\\documentclass[../main.tex]{subfiles}\n\n\\begin{document}\n \\section{Geometric Features of Curves}\n We have previously examined curvature, but curves have other geometric properties which may be of interest, so we wish to look at those and define them mathematically. One of these is the principal normal, which is always perpendicual to the tangent vector. The principal normal describes the rate of change of the tangent vector. Another is osculating planes and circles, where we use planes and circles to approximate curves, rather than linear line segments.\n \\begin{definition}{Principal Normal Vector}{}\n Consider a curve parameterized by \\(\\vec{r}(t)\\) on some interval. There exists a vector \\(\\vec{N}\\) of unit length called the \\emph{principal normal vector} which is always perpendicular to \\(\\vec{T}\\), and describes the direction in which \\(\\vec{T}\\) is moving.\n\n In the arc length parameterization, we have\n \\[\n \\vec{T}=\\frac{d\\vec{r}}{ds}\n \\]\n then if \\(\\kappa\\neq0\\), we define\n \\[\n \\vec{N}(s)=\\frac{\\vec{T}'(s)}{|\\vec{T}'(s)|}\n \\]\n which is of unit length and points in the direction in which \\(\\vec{T}\\) is curving. There is also an expression for \\(\\vec{N}\\) in terms of an arbitrary parameterization, using that\n \\[\n \\frac{d\\vec{T}}{ds}=\\frac{d\\vec{T}}{dt}\\frac{dt}{ds}\n \\]\n we have\n \\[\n \\vec{N}(t)=\\frac{\\vec{T}'(t)}{|\\vec{T}'(t)|}\n \\]\n so the formula is independent of parameterization.\n \\end{definition}\n \\begin{theorem}{}{}\n Prove that \\(\\vec{N}\\perp\\vec{T}\\).\n \\tcblower\n Since \\(|\\vec{T}|=1\\), we have \\(\\vec{T}\\cdot\\vec{T}'=0\\), but since \\(\\vec{N}\\) is a scalar multiple of \\(\\vec{T}'\\), it must be that \\(\\vec{N}\\cdot\\vec{T}=0\\) and \\(\\vec{N}\\perp\\vec{T}\\).\n \\end{theorem}\n \\begin{definition}{The Osculating Plane}{}\n Consider a curve described by \\(\\vec{r}(t)\\) on some interval. We define the \\emph{osculating plane} at a point \\(P\\) on the curve as the plane through \\(P\\) which is spanned by \\(\\vec{N}\\) and \\(\\vec{T}\\).\n \\end{definition}\n \\begin{definition}{The Osculating Circle}{}\n Consider a curve described by \\(\\vec{r}(t)\\) on some interval. We define the \\emph{osculating circle} at a point \\(P\\) on the curve to be a circle which lies in the osculating plane, having a radius of \\(\\rho=\\frac{1}{\\kappa}\\), and a centre at \\(P+\\rho\\vec{N}\\).\n \\end{definition}\n The osculating circle represents a best circular approximation of the curve at a given point. Note that a circle of radius \\(a\\) has curvature \\(\\kappa=\\frac{1}{a}\\), so the osculating circle has the same curvature as the curve it approximates.\n \\begin{example}{}{}\n Consider the circle parameterized by \\(\\vec{r}(t)=a\\cos{t}\\i+a\\sin{t}\\j\\) in \\(\\mathbb{R}^3\\). Find an expression for the osculating circle.\n \\tcblower\n It has previously been found that\n \\[\n \\vec{T}(t)=-\\sin{t}\\vec{i}+\\cos{t}\\j\n \\]\n and therefore\n \\[\n \\vec{T}'(t)=-\\cos{t}\\vec{i}-\\sin{t}\\j\n \\]\n but since \\(|\\vec{T}'|=1\\), we have\n \\[\n \\vec{N}(t)=\\vec{T}'(t)\n \\]\n\n The curvature of this circle is \\(\\kappa=\\frac{1}{a}\\), and therefore the radius of the osculating circle is \\(a\\), identical to that o f the original circle. The centre of the osculating circle is at \\(\\vec{r}-a\\vec{N}=\\vec{0}\\), so the osculating circle is precisely equal to the original circle.\n \\end{example}\n \\begin{example}{}{}\n Consider the helix parameterized by \\(\\vec{r}=a\\cos{t}\\i+a\\sin{t}\\j+bt\\k\\). Find an expression for the osculating circle.\n \\tcblower\n We have previously found that\n \\[\n \\vec{T}'(t)=\\frac{1}{\\sqrt{a^2+b^2}}\\left(-a\\cos{t}\\i-a\\sin{t}\\j\\right)\n \\]\n and\n \\[\n |\\vec{T}'(t)|=\\frac{a}{\\sqrt{a^2+b^2}}\n \\]\n but this yields\n \\[\n \\vec{N}=-\\cos{t}\\i-\\sin{t}\\j\n \\]\n which gives the same osculating circle as that of the circle in the previous example, appropriately offset in the z-axis.\n \\end{example}\n \\begin{example}{}{}\n Find an expression for the osculating plane of the helix from the previous example. \n \\tcblower\n We require our osculating plane be spanned by \\(\\vec{N}\\) and \\(\\vec{T}\\), so the plane's normal vector must be\n \\[\n \\vec{B}=\\vec{T}\\times\\vec{N}\n \\]\n where \\(\\vec{B}\\) is known as the \\emph{binormal vector}.\n An expression describing the plane may then be obtained using the usual method.\n \\end{example}\n\\end{document}\n", "meta": {"hexsha": "56803073363b8a9a15877817f08f7b7b35fe8dd0", "size": 5233, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/lec_5.tex", "max_stars_repo_name": "CrashAndSideburns/MATH227-Notes", "max_stars_repo_head_hexsha": "ec5356ff816c2f51828e4f8b64b5ae0a8b0c8cd3", "max_stars_repo_licenses": ["Unlicense"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2022-01-25T04:16:23.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-25T04:16:23.000Z", "max_issues_repo_path": "src/lec_5.tex", "max_issues_repo_name": "CrashAndSideburns/MATH227-Notes", "max_issues_repo_head_hexsha": "ec5356ff816c2f51828e4f8b64b5ae0a8b0c8cd3", "max_issues_repo_licenses": ["Unlicense"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/lec_5.tex", "max_forks_repo_name": "CrashAndSideburns/MATH227-Notes", "max_forks_repo_head_hexsha": "ec5356ff816c2f51828e4f8b64b5ae0a8b0c8cd3", "max_forks_repo_licenses": ["Unlicense"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 61.5647058824, "max_line_length": 470, "alphanum_fraction": 0.547869291, "num_tokens": 1378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8267118026095991, "lm_q1q2_score": 0.6119108810866852}} {"text": "\\subsection{Boolean algebras}\\label{subsec:boolean_algebras}\n\n\\begin{definition}\\label{def:heyting_algebra}\\mcite[10]{BezhanishviliHolliday2019}\n A \\term{Heyting algebra} is a \\hyperref[def:semilattice/bounded]{bounded} \\hyperref[def:semilattice/distributive_lattice]{distributive lattice} \\( X \\) with a binary operation \\( \\rightarrow \\) defined as\n \\begin{equation}\\label{eq:def:heyting_algebra/conditional}\n (x \\rightarrow y) \\coloneqq \\bigvee\\set{ a \\in X \\given a \\wedge x \\leq y }.\n \\end{equation}\n\n In order for the operation to be well-defined, we require that the corresponding join exists for all \\( y \\) and \\( z \\). We call this operation the \\term{conditional} in analogy with the \\hyperref[def:propositional_language/connectives/conditional]{propositional connective}, although this operation is often called \\enquote{implication} because of \\hyperref[def:material_implication]{material implication}.\n\n Heyting algebras are useful for defining truth values for intuitionistic logic -- see \\fullref{def:propositional_heyting_algebra_semantics} --- and also appear as the \\hyperref[def:lindenbaum_tarski_algebra]{Lindenbaum-Tarski algebra} for propositional intuitionistic theories --- see \\fullref{thm:intuitionistic_lindenbaum_tarski_algebra}.\n\n \\begin{thmenum}[series=def:heyting_algebra]\n \\thmitem{def:heyting_algebra/pseudocomplement} For any element \\( x \\), we define its \\term{pseudocomplement} as\n \\begin{equation}\\label{eq:def:heyting_algebra/pseudocomplement}\n \\widetilde x\n \\coloneqq\n (x \\rightarrow \\bot)\n =\n \\bigvee\\set{ a \\in X \\given a \\wedge x = \\bot }.\n \\end{equation}\n \\end{thmenum}\n\n Heyting algebras have the following metamathematical properties:\n \\begin{thmenum}[resume=def:heyting_algebra]\n \\thmitem{def:heyting_algebra/theory} We extend the language of the \\hyperref[def:semilattice/theory]{theory of lattices} with the binary infix functional symbol \\( \\rightarrow \\) and the unary functional symbol \\( \\widetilde{\\anon} \\). By adding the axiom \\eqref{eq:def:heyting_algebra/conditional} to the theory of bounded distributive lattices, we obtain the theory of Heyting algebras.\n\n \\thmitem{def:heyting_algebra/submodel} The Heyting subalgebras are the \\hyperref[def:semilattice/submodel]{bounded sublattices} for which the conditional is well-defined.\n\n \\thmitem{def:heyting_algebra/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} Heyting algebra is the \\hyperref[def:semilattice/trivial]{one-element bounded lattice}.\n\n \\thmitem{def:heyting_algebra/homomorphism} \\hyperref[def:first_order_homomorphism]{Homomorphisms} between Heyting algebras are lattice homomorphisms with the additional requirement that homomorphisms preserve conditionals.\n\n \\thmitem{def:heyting_algebra/category} The \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} for Heyting algebras is denoted by \\( \\cat{Heyt} \\). It is a full subcategory of the \\hyperref[def:semilattice/category]{category \\( \\cat{Lat} \\) of lattices}.\n\n \\thmitem{def:heyting_algebra/opposite} The \\hyperref[def:semilattice/duality]{principle of duality for lattices} does not hold for Heyting algebras.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:topological_space_is_heyting_algebra}\n Somewhat similar to how the power set of a nonempty set is a Boolean algebra, as shown in \\fullref{thm:boolean_algebra_of_subsets}, the topology \\( \\mscrT \\) of a \\hyperref[def:topological_space]{topological space} \\( (X, \\mscrT) \\) is a Heyting algebra. This is actually used in topological semantics --- see \\fullref{def:propositional_topological_semantics}.\n\n Indeed,\n \\begin{itemize}\n \\item \\hyperref[def:semilattice/join]{Arbitrary joins} are given by \\hyperref[def:basic_set_operations/union]{unions \\( \\bigcap \\)}.\n \\item \\hyperref[def:semilattice/meet]{Finite meets} are given by \\hyperref[def:basic_set_operations/intersection]{intersections \\( \\bigcup \\)}.\n \\item The \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{top} is the entire domain \\( X \\).\n \\item The \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{bottom} is the empty set.\n \\item The \\hyperref[eq:def:heyting_algebra/conditional]{conditional \\( U \\leadsto V \\)} is then\n \\begin{equation*}\n \\bigcup\\set[\\Big]{ A \\in T \\given \\underbrace{A \\cap U}_{A \\setminus (X \\setminus U)} \\subseteq V }\n =\n \\bigcup\\set[\\Big]{ A \\in T \\given A \\subseteq V \\cup (X \\setminus U) }\n =\n \\Int((X \\setminus U) \\cup V),\n \\end{equation*}\n which is actually similar to \\fullref{thm:boolean_equivalences/conditional_as_disjunction} despite the fact that arbitrary topologies are not Boolean algebras.\n\n \\item As a result, the \\hyperref[def:heyting_algebra/pseudocomplement]{pseudocomplement} is\n \\begin{equation*}\n \\widetilde U = \\Int(X \\setminus U).\n \\end{equation*}\n \\end{itemize}\n\\end{example}\n\n\\begin{definition}\\label{def:bounded_lattice_complement}\n Let \\( X \\) be a \\hyperref[def:semilattice/bounded]{bounded} \\hyperref[def:semilattice/lattice]{lattice} and fix an element \\( x \\in X \\). A \\term{complement} of \\( x \\) is an element \\( y \\) such that\n \\begin{align}\n x \\vee y = \\top \\label{def:bounded_lattice_complement/join}, \\\\\n x \\wedge y = \\bot \\label{def:bounded_lattice_complement/meet}.\n \\end{align}\n\n Due to the commutativity of both \\( \\vee \\) and \\( \\wedge \\), \\( y \\) is a complement of \\( x \\) if and only if \\( x \\) is a complement of \\( y \\).\n\\end{definition}\n\n\\begin{proposition}\\label{thm:distributive_bounded_lattice_unique_complement}\n In a \\hyperref[def:semilattice/bounded]{bounded} \\hyperref[def:semilattice/distributive_lattice]{distributive lattice} \\( X \\), each \\( x \\in X \\) has at most one complement.\n\n Thus, complementation can be regarded as a \\hyperref[def:partial_function]{partial operation}.\n\\end{proposition}\n\\begin{proof}\n If \\( y \\) and \\( z \\) are both complements of \\( x \\), then\n \\begin{balign*}\n y\n &\\reloset {\\eqref{eq:thm:binary_lattice_operations/identity/meet}} =\n y \\wedge \\top\n = \\\\ &\\reloset {\\eqref{def:bounded_lattice_complement/join}} =\n y \\wedge (z \\vee x)\n = \\\\ &\\reloset {\\eqref{eq:def:semilattice/distributive_lattice/finite/meet_over_join}} =\n (y \\wedge z) \\vee (y \\wedge x)\n = \\\\ &\\reloset {\\eqref{def:bounded_lattice_complement/meet}} =\n y \\wedge z\n = \\\\ &\\reloset {\\eqref{def:bounded_lattice_complement/meet}} =\n (x \\wedge z) \\vee (y \\wedge z)\n = \\\\ &\\reloset {\\eqref{eq:def:semilattice/distributive_lattice/finite/meet_over_join}} =\n (x \\vee y) \\wedge z\n = \\\\ &\\reloset {\\eqref{eq:thm:binary_lattice_operations/identity/meet}} =\n z.\n \\end{balign*}\n\\end{proof}\n\n\\begin{definition}\\label{def:boolean_algebra}\\mcite[48]{Gratzer1978}\n A \\term{Boolean algebra} is a \\hyperref[def:semilattice/bounded]{bounded} \\hyperref[def:semilattice/distributive_lattice]{distributive lattice} in which every element has a \\hyperref[def:bounded_lattice_complement]{complement}. The complement of each element is unique due to \\fullref{thm:distributive_bounded_lattice_unique_complement}. We define a unary function that gives to every element \\( x \\) its complement \\( \\overline \\anon \\). By definition, this function is an \\hyperref[def:set_with_involution]{involution}.\n\n \\begin{thmenum}[series=def:boolean_algebra]\n \\thmitem{def:boolean_algebra/conditional} We also define the binary operation \\term{conditional} (\\( \\rightarrow \\)) via\n \\begin{equation}\\label{eq:def:boolean_algebra/conditional}\n (x \\rightarrow y) \\coloneqq (\\overline x \\vee y)\n \\end{equation}\n in analogy with \\fullref{thm:boolean_equivalences/conditional_as_disjunction}. This operation highlights that Boolean algebras are a special case of \\hyperref[thm:boolean_algebras_are_heyting_algebras]{Heyting algebras}.\n\n \\thmitem{def:boolean_algebra/biconditional} It remains to define a binary operation corresponding to the \\hyperref[def:propositional_language/connectives/biconditional]{propositional biconditional}. Inspired by \\fullref{thm:boolean_equivalences/biconditional_via_conditionals}, define\n \\begin{equation}\\label{eq:def:boolean_algebra/biconditional}\n (x \\leftrightarrow y) \\coloneqq (x \\rightarrow y) \\wedge (y \\rightarrow x).\n \\end{equation}\n \\end{thmenum}\n\n Boolean algebras have the following metamathematical properties:\n \\begin{thmenum}[resume=def:boolean_algebra]\n \\thmitem{def:boolean_algebra/theory} To obtain the theory of Boolean algebras, we replace the unary functional symbol \\( \\widetilde{\\anon} \\) with \\( \\overline \\anon \\) in the language of the \\hyperref[def:heyting_algebra/theory]{theory of Heyting algebras} and then add the axioms \\eqref{def:bounded_lattice_complement/join} and \\eqref{def:bounded_lattice_complement/meet} to the theory. We may also replace \\eqref{eq:def:heyting_algebra/conditional} defining \\( \\rightarrow \\) with the simpler axiom \\eqref{eq:def:boolean_algebra/conditional}.\n\n \\thmitem{def:boolean_algebra/submodel} The Boolean subalgebras are the \\hyperref[def:semilattice/submodel]{bounded sublattices} which are closed under compementation.\n\n \\thmitem{def:boolean_algebra/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} Boolean algebra is the \\hyperref[def:semilattice/trivial]{one-element bounded lattice}.\n\n \\thmitem{def:boolean_algebra/homomorphism} \\hyperref[def:first_order_homomorphism]{Homomorphisms} between Boolean algebras are simply lattice homomorphisms.\n\n Complements are automatically preserved because for any lattice homomorphism \\( \\varphi \\) between the Boolean algebras \\( X \\) and \\( Y \\),\n \\begin{equation*}\n \\varphi(x) \\vee_Y \\varphi(\\overline x) = \\varphi(x \\vee_X \\overline x) = \\varphi(\\top_X) = \\top_Y,\n \\end{equation*}\n and similarly for \\( \\wedge \\), hence, due to \\fullref{thm:distributive_bounded_lattice_unique_complement},\n \\begin{equation*}\n \\varphi(\\overline x) = \\overline {\\varphi(x)}.\n \\end{equation*}\n\n Implications are also automatically preserved because of \\eqref{eq:def:boolean_algebra/conditional}.\n\n \\thmitem{def:boolean_algebra/category} The \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} for Boolean algebras \\( \\cat{Bool} \\) is a full subcategory the \\hyperref[def:heyting_algebra/category]{category \\( \\cat{Heyt} \\) of Heyting algebras}.\n\n \\thmitem{def:boolean_algebra/opposite} The \\hyperref[def:semilattice/duality]{principle of duality for lattices} holds for Boolean algebras without interchanging complements.\n \\end{thmenum}\n\\end{definition}\n\n\\begin{example}\\label{ex:boolean_algebras}\n Examples of \\hyperref[def:boolean_algebra]{Boolean algebras} include:\n\n \\begin{itemize}\n \\thmitem{ex:boolean_algebras/lindenbaum_tarski} The \\hyperref[def:lindenbaum_tarski_algebra]{Lindenbaum-Tarski algebras} for classical logic. We prove in \\fullref{thm:intuitionistic_lindenbaum_tarski_algebra} that it is a Boolean algebra.\n \\thmitem{ex:boolean_algebras/f2} The \\hyperref[thm:finite_fields]{prime field} \\( \\BbbF_2 \\) with suitably defined operations discussed in \\fullref{thm:f2_is_boolean_algebra}.\n \\thmitem{ex:boolean_algebras/power_set} The power set of any set, usually taken to be a space with additional structure (see \\fullref{thm:boolean_algebra_of_subsets}).\n \\end{itemize}\n\\end{example}\n\n\\begin{proposition}\\label{thm:binary_boolean_algebra}\n All two-element \\hyperref[def:boolean_algebra]{Boolean algebras} are isomorphic.\n\\end{proposition}\n\\begin{proof}\n Follows from the requirement that \\hyperref[def:semilattice/homomorphism]{lattice homomorphisms} preserve constants.\n\\end{proof}\n\n\\begin{corollary}\\label{thm:f2_is_boolean_algebra}\n For certain purposes, for example \\hyperref[def:zhegalkin_polynomial]{Zhegalkin polynomials}, we may regard the \\hyperref[thm:finite_fields]{prime field} \\( \\BbbF_2 \\) as a Boolean algebra with \\( 1 \\) as the \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{top} and \\( 0 \\) as the \\hyperref[def:partially_ordered_set_extremal_points/top_and_bottom]{bottom element}.\n\\end{corollary}\n\\begin{proof}\n Follows from \\fullref{thm:binary_boolean_algebra}.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:boolean_algebras_are_heyting_algebras}\n Every \\hyperref[def:boolean_algebra]{Boolean algebra} is a \\hyperref[def:heyting_algebra]{Heyting algebra} with an identification given by \\eqref{eq:def:boolean_algebra/conditional}.\n\\end{proposition}\n\\begin{proof}\n Fix any \\( x, y \\in X \\) in a Boolean algebra \\( X \\). We will show that \\( x \\rightarrow y \\) as defined in \\eqref{eq:def:boolean_algebra/conditional} satisfies \\eqref{eq:def:heyting_algebra/conditional}.\n\n Let\n \\begin{equation*}\n A \\coloneqq \\set{ a \\in X \\given a \\wedge x \\leq y }\n \\end{equation*}\n be the set from \\eqref{eq:def:heyting_algebra/conditional}.\n\n We will show that \\( \\overline x \\vee y \\) is an \\hyperref[def:partially_ordered_set_extremal_points/upper_and_lower_bounds]{upper bound} of \\( A \\).\n\n Fix some \\( a_0 \\in A \\). By definition of \\( A \\), we have\n \\begin{equation*}\n a_0 \\wedge x \\leq y.\n \\end{equation*}\n\n But this means that\n \\begin{equation*}\n \\underbrace{(a_0 \\wedge x) \\vee \\overline x}_{a_0 \\vee \\overline x} \\leq y \\vee \\overline x.\n \\end{equation*}\n\n Since \\( a_0 \\leq a_0 \\vee b \\) for any \\( b \\in X \\), it follows that \\( a_0 \\leq y \\vee \\overline x \\). Therefore, \\( \\overline x \\vee y \\) is indeed an upper bound of \\( A \\).\n\n Also note that\n \\begin{equation*}\n (\\overline x \\vee y) \\wedge x = \\underbrace{(\\overline x \\vee x)}_{\\top} \\wedge (y \\wedge x) = y \\wedge x \\leq y,\n \\end{equation*}\n hence \\( \\overline x \\vee y \\in A \\).\n\n Thus, \\( \\overline x \\vee y \\) is both an upper bound of \\( A \\) and an element of \\( A \\), i.e. it is the least upper bound of \\( A \\). Therefore,\n \\begin{equation*}\n \\overline x \\vee y = \\bigvee A.\n \\end{equation*}\n\\end{proof}\n\n\\begin{theorem}[De Morgan's laws]\\label{thm:de_morgans_laws}\n If \\( X \\) is a \\hyperref[def:boolean_algebra]{Boolean algebra}, the following hold for any finite \\hyperref[def:cartesian_product/indexed_family]{family} \\( \\{ x_k \\}_{k \\in \\mscrK} \\subseteq X \\):\n \\begin{align}\n \\overline{\\bigvee_{k \\in \\mscrK} x_k} = \\bigwedge_{k \\in \\mscrK} \\overline{x_k} \\label{eq:thm:de_morgans_laws/complement_of_join} \\\\\n \\overline{\\bigwedge_{k \\in \\mscrK} x_k} = \\bigvee_{k \\in \\mscrK} \\overline{x_k} \\label{eq:thm:de_morgans_laws/complement_of_meet}\n \\end{align}\n\n If \\( X \\) is \\hyperref[def:semilattice/complete]{complete}, \\( \\mscrK \\) may be taken to be any family, not necessarily finite.\n\\end{theorem}\n\\begin{proof}\n We will only show \\eqref{eq:thm:de_morgans_laws/complement_of_join} since \\eqref{eq:thm:de_morgans_laws/complement_of_meet} is dual.\n\n In order for \\( \\wedge_{k \\in \\mscrK} \\overline{x_k} \\) to be the complement of \\( \\vee_{k \\in \\mscrK} x_k \\), the conditions \\eqref{def:bounded_lattice_complement/join} and \\eqref{def:bounded_lattice_complement/meet} need to be satisfied.\n\n From distributivity, we have\n \\begin{equation*}\n \\parens*{ \\bigwedge_{k \\in \\mscrK} \\overline{x_k} } \\vee \\parens*{ \\bigvee_{m \\in \\mscrK} x_m }\n \\reloset {\\eqref{eq:def:semilattice/distributive_lattice/arbitrary/join_over_meet}} =\n \\bigwedge_{k \\in \\mscrK} \\parens*{ \\overline{x_k} \\vee \\bigvee_{m \\in \\mscrK} x_m }\n =\n \\bigwedge_{k \\in \\mscrK} \\parens*{ \\underbrace{\\overline{x_k} \\vee x_k}_{\\top} \\vee \\bigvee_{\\mathclap{m \\in K \\setminus \\set{k}}} x_m }\n =\n \\bigwedge_{k \\in \\mscrK} \\top\n =\n \\top,\n \\end{equation*}\n which proves \\eqref{def:bounded_lattice_complement/join}. The proof of \\eqref{def:bounded_lattice_complement/meet} is analogous.\n\\end{proof}\n", "meta": {"hexsha": "3c9c5b62474e1aeb7e6600bfd5f4fc12118de3ae", "size": 15794, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/boolean_algebras.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/boolean_algebras.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/boolean_algebras.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 67.2085106383, "max_line_length": 549, "alphanum_fraction": 0.7434468786, "num_tokens": 4954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267118026095991, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6119108810866852}} {"text": "\\documentclass{report}\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\begin{document}\n\\section{Perspective of Probability}\n\\subsection{Abstract}\nIn this issue, we will observe multivariate Gaussian distribution from the perspective of probability.\n\\subsection{Prior Knowledge}\n$$\n\\begin{aligned}\nx &\\backsim N(\\mu, \\sigma^2)\\\\\n\\mu &\\in \\mathcal{R}^p, \\sigma \\in \\mathcal{R}^p\\\\\nx_i &\\backsim N(\\mu_i, \\sigma_i)\\\\\np(x_i) &= \\frac{1}{\\sqrt{2\\pi}\\sigma_i} \\exp(-\\frac{(x_i - \\mu_i)^2}{2\\sigma_i^2})\n\\end{aligned}\n$$\n\\subsection{Derivation}\nFirst, let's assume that each $x_i$ is $iid (independent\\ identically\\ distribution)$ as below:\n$$\n\\begin{aligned}\np(x)\n&=\\prod_{i=1}^p p(x_i)\\\\\n&=\\frac{1}{(2\\pi)^{\\frac{p}{2}}\\prod_{i=1}^p \\sigma_i} \\exp(-\\frac{1}{2}\\sum_{i=1}^p (\\frac{(x_i-\\mu_i)^2}{\\sigma_i^2}))\\\\\n&=\\frac{1}{(2\\pi)^{\\frac{p}{2}}|\\Sigma|^{\\frac{1}{2}}} \\exp(-\\frac{1}{2}\n\\left (\n\\begin{matrix}\nx_1-\\mu_1 & x_2-\\mu_2 & ... & x_p - \\mu_p\n\\end{matrix}\n\\right )\n\\left (\n\\begin{matrix}\n\\frac{1}{\\sigma_1^2} & 0 & ... & 0 \\\\\n... & ... & ... & ... \\\\\n0 & ... & 0 & \\frac{1}{\\sigma_p^2}\n\\end{matrix}\n\\right )\n\\left (\n\\begin{matrix}\nx_1-\\mu_1\\\\\n...&\\\\\n...&\\\\\nx_p-\\mu_p\n\\end{matrix}\n\\right ))\\\\\n&=\\frac{1}{(2\\pi)^{\\frac{p}{2}}|\\Sigma|^{\\frac{1}{2}}} \\exp(-\\frac{1}{2}(x-\\mu)^T \\Sigma^{-1}(x-\\mu))\n\\end{aligned}\n\\label{Gau.pdf}\n$$\nThe above is the probability density function of multivariate Gaussian distribution.\\\\\\\\\nWe know that $\\sigma$ is a positive semidefinite matrix, so we can perform singular value decomposition. So we have:\n$$\n\\begin{aligned}\n\\Sigma\n&=UVU^T\\\\\n&=\n\\left ( \\begin{matrix}\nu_1 & ... & u_p\n\\end{matrix} \\right )\n\\left ( \\begin{matrix}\n\\lambda_1 & 0 & ... & 0\\\\\n... & 0 & ... & ...\\\\\n0 & ... & ... & \\lambda_p\n\\end{matrix} \\right )\n\\left ( \\begin{matrix}\nu_1^T \\\\\n.\\\\\n.\\\\\nu_p^T\n\\end{matrix} \\right )\\\\\n&=\n\\left ( \\begin{matrix}\nu_1\\lambda_1 & ... & u_p \\lambda_p\\\\\n\\end{matrix} \\right )\n\\left ( \\begin{matrix}\nu_1^T\\\\\n.\\\\\n.\\\\\nu_p^T\n\\end{matrix} \\right )\\\\\n&=\\sum_{i=1}^p u_i\\lambda_i u_i^T\n\\end{aligned}\n$$\nthen\n$$\n\\begin{aligned}\n\\Sigma^{-1}\n&=(UVU^T)^{-1}\\\\\n&=(U^T)^{-1}V^{-1}U^{-1}\\\\\n&=UV^{-1}U^T\\\\\n&=\\sum_{i=1}^p u_i \\frac{1}{\\lambda_i} u_i^T\n\\end{aligned}\n$$\nLet's set $\\Delta = (x-\\mu)^T \\Sigma^{-1} (x-\\mu)$\\\\\\\\\nSubstitute the results derived above into:\n$$\n\\begin{aligned}\n\\Delta\n&=(x-\\mu)^T \\Sigma^{-1} (x-\\mu)\\\\\n&=(x-\\mu)^T \\sum_{i=1}^p u_i \\frac{1}{\\lambda_i}u_i^T (x-\\mu)\\\\\n&=\\sum_{i=1}^p(x-\\mu)^T u_i \\frac{1}{\\lambda_i}u_i^T (x-\\mu)\n\\end{aligned}\n$$\nLet's set $y_i=(x-\\mu)^T u_i$\\\\\\\\\nHere, $y_i $ represents the coordinate value of $x$ projected onto the new orthogonal basis $u_i$ after centralization.\\\\\nso:\n$$\n\\Delta=\\sum_{i=1}^p \\frac{y_i^2}{\\lambda_i}\n$$\nNext, let's look at the probability density function of multivariate Gaussian distribution:\n$$\np(x)=\\frac{1}{(2\\pi)^{\\frac{p}{2}}|\\Sigma|^{\\frac{1}{2}}} \\exp(-\\frac{1}{2}(x-\\mu)^T \\Sigma^{-1}(x-\\mu))\n$$\nYou can see that only the exponential part of the formula is related to the variable $x $. The previous factor is to make the probability sum $1 $.\\\\\nTherefore, the probability of Gaussian distribution is directly related to the value of $\\Delta $.\\\\\\\\\nWe assume $p=2$, then:\n$$\n\\frac{y_1^2}{\\lambda_1}+\\frac{y_2^2}{\\lambda_2}=\\Delta\n$$\nWe were surprised to find that this is very similar to the elliptic equation. The value of $\\Delta$ is not fixed, so for different $x$, these sample points form concentric ellipses in the plane. This is one of the properties of Gaussian distribution.\n\\end{document}", "meta": {"hexsha": "d65140ccd0e31c247cb76ffae4d23c8d38b0e83b", "size": 3495, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "EN-TeX_files/Intro_Math/02_fundamentals-of-math_gaussian-distribution_perspective-of-probability.tex", "max_stars_repo_name": "btobab/Machine-Learning-notes", "max_stars_repo_head_hexsha": "bc064bd2fe3817444bb8850340ac1177f5fd71c4", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2021-08-28T18:47:12.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-04T07:36:27.000Z", "max_issues_repo_path": "EN-TeX_files/Intro_Math/02_fundamentals-of-math_gaussian-distribution_perspective-of-probability.tex", "max_issues_repo_name": "btobab/Machine-Learning-notes", "max_issues_repo_head_hexsha": "bc064bd2fe3817444bb8850340ac1177f5fd71c4", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "EN-TeX_files/Intro_Math/02_fundamentals-of-math_gaussian-distribution_perspective-of-probability.tex", "max_forks_repo_name": "btobab/Machine-Learning-notes", "max_forks_repo_head_hexsha": "bc064bd2fe3817444bb8850340ac1177f5fd71c4", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-28T18:47:22.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-28T18:47:22.000Z", "avg_line_length": 29.125, "max_line_length": 250, "alphanum_fraction": 0.6268955651, "num_tokens": 1358, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8267118004748677, "lm_q1q2_score": 0.6119108795066117}} {"text": "\\documentclass{article}\n\\usepackage{tocloft}\n\\include{common_symbols_and_format}\n\\renewcommand{\\cfttoctitlefont}{\\Large\\bfseries}\n\\begin{document}\n\\logo\n\\rulename{True Strength Index} %Argument is name of rule\n\\tblofcontents\n\n\\ruledescription{The True Strength Index (TSI) is a momentum indicator used to identify overbought and oversold signals. It can also be used to indicate trend reversals using bullish and bearish crossover of the signal line and TSI line.}\n\n\\howtotrade\n{The strategy is to identify the oversold and overbought markets and a momentum indicator. \\\\\nBullish Momentum: when the TSI line cross above signal line. \\\\\nBearish Momentum: when signal line cross above TSI line\n}\n\n\n\\ruleparameters\n{Window size for Exponential moving average}{25}{This is the number of time steps over which exponential contributions are sourced.}{$\\lookbacklength$}\n{Window size for Exponential moving average}{13}{This is the number of time steps over which exponential contributions are sourced for smoothed momentum.}{$\\lookbacklength_{s}$}\n{Smoothing Factor}{2}{Smoothing factor represents the weighting applied to the most recent period’s value.}{$S$}\n\\stoptable\n\n\\newpage\n\\section{Equation}\n\\begin{equation}\n \\Delta p_{\\currenttime} = p_{\\currenttime} - p_{\\currenttime - 1}\n\\end{equation}\n\n\\begin{equation}\n EMA(\\Delta p_{\\currenttime}, S, \\lookbacklength) = \\Big(\\Delta p_{\\currenttime} * \\Big(\\frac{S}{1 + \\lookbacklength}\\Big)\\Big) + EMA_{\\currenttime-1} * \\Big(1-\\frac{S}{1 + \\lookbacklength}\\Big)\n\\end{equation}\n\n\\begin{equation}\n EMA(|\\Delta p_{\\currenttime}|, S, \\lookbacklength) = \\Big(|\\Delta p_{\\currenttime}| * \\Big(\\frac{S}{1 + \\lookbacklength}\\Big)\\Big) + EMA_{\\currenttime-1} * \\Big(1-\\frac{S}{1 + \\lookbacklength}\\Big)\n\\end{equation}\n\n\\begin{equation}\n TSI = 100 \\times \\frac{EMA(EMA(\\Delta p_{\\currenttime}, S, \\lookbacklength), S, \\lookbacklength_{s})}{EMA(EMA(|\\Delta p_{\\currenttime}|, S, \\lookbacklength), S,\\lookbacklength_{s})}\n\\end{equation}\n\\\\\n\nwhere: \\\\\n\n$TSI$ is True Strength Index. \\\\\n\n$EMA$ is exponentially weighted moving average. \\\\\n\n$p_{\\currenttime}$ is current stock value. \\\\\n\n$\\Delta p_{\\currenttime}$ is change in stock value. \\\\\n\n$|\\Delta p_{\\currenttime}|$ is absolute change in stock value. \\\\\n\n$\\lookbacklength$ \\ is look back length. \\\\\n\n$\\lookbacklength_{s}$ is look back length for smoothed momentum. \\\\\n\n$S$ is the smoothing factor.\n\n\\hspace{200mm}\n\\hspace{200mm}\n\\keyterms\n\\furtherlinks\n\\end{document}", "meta": {"hexsha": "192266849768cbbc080a96deec1890dd1421cef5", "size": 2465, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/strategies/tex/TrueStrengthIndex.tex", "max_stars_repo_name": "parthgajjar4/infertrade", "max_stars_repo_head_hexsha": "2eebf2286f5cc669759de632970e4f8f8a40f232", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 34, "max_stars_repo_stars_event_min_datetime": "2021-03-25T13:32:54.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-06T23:03:01.000Z", "max_issues_repo_path": "docs/strategies/tex/TrueStrengthIndex.tex", "max_issues_repo_name": "parthgajjar4/infertrade", "max_issues_repo_head_hexsha": "2eebf2286f5cc669759de632970e4f8f8a40f232", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 137, "max_issues_repo_issues_event_min_datetime": "2021-03-25T10:59:46.000Z", "max_issues_repo_issues_event_max_datetime": "2022-01-28T19:36:30.000Z", "max_forks_repo_path": "docs/strategies/tex/TrueStrengthIndex.tex", "max_forks_repo_name": "parthgajjar4/infertrade", "max_forks_repo_head_hexsha": "2eebf2286f5cc669759de632970e4f8f8a40f232", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 28, "max_forks_repo_forks_event_min_datetime": "2021-03-26T14:26:04.000Z", "max_forks_repo_forks_event_max_datetime": "2021-11-10T18:21:14.000Z", "avg_line_length": 37.3484848485, "max_line_length": 238, "alphanum_fraction": 0.738336714, "num_tokens": 730, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7401743620390163, "lm_q2_score": 0.8267117940706734, "lm_q1q2_score": 0.6119108747663913}} {"text": "\\section{Outlook}\n\n\\begin{frame}{Motivation}\n \\begin{block}{Verlinde Formula}\n \\begin{itemize}\n \\item Let $\\mc{N}$ be the moduli space of stable $\\SL_{2}$-bundles over $\\Sigma_{2}$; it is isomorphic to $\\CC\\PP^{3}$ \\cite{NR1969}.\n \\item Geometric quantisation $\\mc{Q}(\\mc{N}) := H^{0}(\\mc{N}; \\mc{L}^{\\otimes k})$; its dimension equals the \\textbf{Verlinde formula} \\cite{verlinde1988, JW1992},\n \\begin{equation*}\n \\begin{split}\n \\dim \\mc{Q}(\\mc{N}) = \\Ver(k) &= \\frac{k^{3}}{6} + k^{2} + \\frac{11k}{6} + 1 \\\\\n &= \\frac{(k+1)(k+2)(k+3)}{3!}.\n \\end{split}\n \\end{equation*}\n \\item Named after Dutch physicist Erik Verlinde, who was working on conformal field theories.\n \\end{itemize}\n \\end{block}\n\\end{frame}\n\n\\begin{frame}{Integration}\n \\begin{block}{Riemann-Roch-Hirzebruch Theorem}\n For toric $X$, lattice point count of $\\Delta_{X}$ equals to Euler characteristic,\n \\vspace*{-12pt}\n $$ \\chi(X) = \\int_{X} e^{c_{1}( \\mc{O}(k) )}\\cdot \\Td{(TX)}. $$\n \\vspace*{-12pt}\n \\end{block}\n \\begin{block}{Example}\n For $X = \\CC\\PP^{3}$, have $e^{c_{1}(\\mc{O}(k))} = 1 + kH + \\tfrac{k^{2}}{2}H^{2} + \\tfrac{k^{3}}{6}H^{3}$, and $\\Td{(T\\CC\\PP^{3})} = 1 + 2H + \\tfrac{11}{6}H^{2} + H^{3}$:\n $$ \\chi(\\CC\\PP^{3}) = \\int_{\\CC\\PP^{3}} \\big( \\tfrac{k^{3}}{6} + k^{2} + \\tfrac{11k}{6} + 1 \\big) \\cdot H^{3} + \\ldots = \\Ver(k). $$\n \\end{block}\n\\end{frame}\n\n\\begin{frame}{Our Direction}\n \\begin{block}{Lattice Points}\n \\begin{itemize}\n \\item So $\\Ver(k) = \\chi(\\CC\\PP^{3}) = \\#( k\\cdot \\Delta_{3} \\cap \\ZZ^{3} )$.\n \\item Anything similar for hypertoric manifolds?\n \\end{itemize}\n \\end{block}\n \\begin{block}{Equivariant Verlinde Formula}\n \\begin{itemize}\n \\item Recently, equivariant Verlinde formula for moduli spaces Higgs bundle, $\\mc{M}$, popped up \\cite{GP2017}.\n \\item $\\dim \\mc{Q} (\\mc{M}) = \\infty$, but $\\mc{M}$ has a $\\CC^{\\ast}$-action.\n \\item Decompose into $\\CC^{\\ast}$-weight spaces:\n \\vspace*{-6pt}\n $$ \\dim \\mc{Q} (\\mc{M}) = \\sum_{n} t^{n} \\cdot \\dim \\mc{Q}_{n}(\\mc{M}), $$\n \\vspace*{-6pt}\n but now $\\dim \\mc{Q}_{n}(\\mc{M}) < \\infty$.\n \\end{itemize}\n \\end{block}\n\\end{frame}\n\n\\begin{frame}{Localisation}\n \\begin{block}{Fixed-Point Formula}\n \\vspace*{-12pt}\n $$ \\sum_{q \\in \\Delta} e^{\\langle q, \\phi \\rangle} = \\sum_{p \\in M^{T}} \\frac{ e^{\\langle p, \\phi \\rangle} }{ \\prod_{k = 1}^{n} (1- e^{\\langle \\alpha_{k}^{p}, \\phi \\rangle}) }, $$\n with edge vectors $\\alpha_{k}^{p}$, and $\\langle \\alpha_{k}^{p}, \\phi \\rangle \\neq 0$, \\cite{barvinok1993}.\n \\begin{itemize}\n \\item Letting $\\phi \\ra 0$ gets the lattice point count (for Delzant $\\Delta$).\n \\end{itemize}\n \\end{block}\n \\begin{block}{Example for $T^{\\ast}\\CC\\PP^{3}$}\n For $(T^{\\ast}\\CC\\PP^{3})_{\\epsilon-\\text{cut}}$, get\n $$ \\frac{(\\epsilon + 1)(\\epsilon + 2)(\\epsilon + 3)}{3!} \\cdot \\frac{(k + \\epsilon +1)(k + \\epsilon + 2)(k + \\epsilon + 3)}{3!}. $$\n Observe that for $\\epsilon = 0$, it becomes $\\Ver(k)$.\n \\end{block} \n\\end{frame}\n\n\\begin{frame}{Other Hypertoric Manifolds}\n \\begin{block}{Non-Convex Core}\n Want to see what happens for hypertoric manifolds with non-convex cores.\n \\end{block}\n \\begin{block}{Example}\n \\vspace{136pt}\n \\end{block} \n\\end{frame}", "meta": {"hexsha": "92c8cba9754b084437ac00e15bb0f5b4963d4d63", "size": 3552, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/summary.tex", "max_stars_repo_name": "bencwbrown/WomenInSTEM-Semester2-2021", "max_stars_repo_head_hexsha": "8222f9ca0696d1f32feb3306f99b9294dd16c75f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "content/summary.tex", "max_issues_repo_name": "bencwbrown/WomenInSTEM-Semester2-2021", "max_issues_repo_head_hexsha": "8222f9ca0696d1f32feb3306f99b9294dd16c75f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/summary.tex", "max_forks_repo_name": "bencwbrown/WomenInSTEM-Semester2-2021", "max_forks_repo_head_hexsha": "8222f9ca0696d1f32feb3306f99b9294dd16c75f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.36, "max_line_length": 187, "alphanum_fraction": 0.5335022523, "num_tokens": 1298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117898012104, "lm_q2_score": 0.7401743620390163, "lm_q1q2_score": 0.6119108716062442}} {"text": "% Written by Anders Sjoqvist and Ulf Lundstrom, 2009\n% The main sources are: tinyKACTL, Beta and Wikipedia\n% Transcribed to plain TeX by Richard Hladík in 2018.\n{\n% Quick fixes\n\\let\\section\\header\n\\def\\subsection#1{\\header{\\subsectionfont#1}}\n\\def\\subsubsection#1{\\header{\\subsubsectionfont#1}}\n\\chapter{Mathematics (text)}\n\n\\section{Equations}\n\n$$ax^2+bx+c=0 \\Rightarrow x = \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$$\n\nThe extremum is given by $x = -b/2a$.\n%\n$$\\eqalign{ax+by&=e\\cr cx+dy&=f\\cr}\n\\,\\Rightarrow\\,\n\\eqalign{x&=\\frac{ed-bf}{ad-bc}\\cr y&=\\frac{af-ec}{ad-bc}\\cr}$$\n%\nIn general, given an equation $Ax = b$, the solution to a variable $x_i$ is given by\n$$x_i = \\frac{\\det A_i'}{\\det A} $$\nwhere $A_i'$ is $A$ with the $i$'th column replaced by $b$.\n\n\\section{Recurrences}\n\nIf $a_n = c_1 a_{n-1} + \\dots + c_k a_{n-k}$, and $r_1, \\dots, r_k$ are distinct roots of $x^k + c_1 x^{k-1} + \\dots + c_k$, there are $d_1, \\dots, d_k$ s.t.\n$$a_n = d_1r_1^n + \\dots + d_kr_k^n. $$\nNon-distinct roots $r$ become polynomial factors, e.g. $a_n = (d_1n + d_2)r^n$.\n\n\\section{Trigonometry}\n\n$$\\eqalign{\n\\sin(v+w)&=\\sin v\\cos w+\\cos v\\sin w\\cr\n\\cos(v+w)&=\\cos v\\cos w-\\sin v\\sin w\\cr\n\\tan(v+w)&=\\frac{\\tan v+\\tan w}{1-\\tan v\\tan w}\\cr\n\\sin v+\\sin w&=2\\sin\\frac{v+w}{2}\\cos\\frac{v-w}{2}\\cr\n\\cos v+\\cos w&=2\\cos\\frac{v+w}{2}\\cos\\frac{v-w}{2}\\cr\n(V+W)\\tan(v-w)/2&=(V-W)\\tan(v+w)/2\\cr}$$\nwhere $V, W$ are lengths of sides opposite angles $v, w$.\n$$\\eqalign{a\\cos x+b\\sin x&=r\\cos(x-\\phi)\\cr\na\\sin x+b\\cos x&=r\\sin(x+\\phi)\\cr}$$\nwhere $r=\\sqrt{a^2+b^2}$, $\\phi=\\mathop{\\rm atan2}(b,a)$.\n\n\\section{Geometry}\n\n\\subsection{Triangles}\n\nSide lengths: $a,b,c$\n\nSemiperimeter: $p=\\frac{a+b+c}{2}$\n\nArea: $A=\\sqrt{p(p-a)(p-b)(p-c)}$\n\nCircumradius: $R=\\frac{abc}{4A}$\n\nInradius: $r=\\frac{A}{p}$\n\nLength of median (divides triangle into two equal-area triangles):$$m_a=\\frac{1}{2}\\sqrt{2b^2+2c^2-a^2}$$\n\nLength of bisector (divides angles in two): $s_a=\\sqrt{bc\\left[1-\\left(\\frac{a}{b+c}\\right)^2\\right]}$\n\nLaw of sines: $\\frac{\\sin\\alpha}{a}=\\frac{\\sin\\beta}{b}=\\frac{\\sin\\gamma}{c}=\\frac{1}{2R}$\n\nLaw of cosines: $a^2=b^2+c^2-2bc\\cos\\alpha$\n\nLaw of tangents: $$\\frac{a+b}{a-b}=\\frac{\\tan\\frac{\\alpha+\\beta}{2}}{\\tan\\frac{\\alpha-\\beta}{2}}$$\n\n\\subsection{Quadrilaterals}\n\nWith side lengths $a,b,c,d$, diagonals $e, f$, diagonals angle $\\theta$, area $A$ and\nmagic flux $F=b^2+d^2-a^2-c^2$:\n\n$$ 4A = 2ef \\cdot \\sin\\theta = F\\tan\\theta = \\sqrt{4e^2f^2-F^2} $$\n\n For cyclic quadrilaterals the sum of opposite angles is $180^\\circ$,\n$ef = ac + bd$, and $A = \\sqrt{(p-a)(p-b)(p-c)(p-d)}$.\n\n\\subsection{Spherical coordinates}\n\\figure{kactl/math/sphericalCoordinates.pdf}\n\n$$\\eqalign{x &= r\\sin\\theta\\cos\\phi\\cr\ny &= r\\sin\\theta\\sin\\phi\\cr\nz &= r\\cos\\theta\\cr}\\;\\eqalign{\n\tr &= \\sqrt{x^2+y^2+z^2}\\cr\n\\theta &= \\mathop{\\rm acos}\\left(z/\\sqrt{x^2+y^2+z^2}\\right)\\cr\n\\phi &= \\mathop{\\rm atan2}\\left(y,x\\right)\\cr}$$\n\n\\section{Derivatives/Integrals}\n\n$$\\eqalign{\n\\frac{d}{dx}\\arcsin x &= \\frac{1}{\\sqrt{1-x^2}}\\cr\n\\frac{d}{dx}\\arccos x &= -\\frac{1}{\\sqrt{1-x^2}}\\cr\n\\frac{d}{dx}\\tan x &= 1+\\tan^2 x\\cr\n\\frac{d}{dx}\\arctan x &= \\frac{1}{1+x^2}\\cr}\\;\\;\\;\\eqalign{\n\\int\\tan ax &= -\\frac{\\ln|\\cos ax|}{a}\\cr\n\\int x\\sin ax &= \\frac{\\sin ax-ax \\cos ax}{a^2} \\cr\n\\int e^{-x^2} &= \\frac{\\sqrt \\pi}{2} \\text{erf}(x)\\cr\n\\int xe^{ax}dx &= \\frac{e^{ax}}{a^2}(ax-1)\\cr}$$\n\nIntegration by parts:\n$$\\int_a^bf(x)g(x)\\d x = [F(x)g(x)]_a^b-\\int_a^bF(x)g'(x)\\d x$$\n\n\\section{Sums}\n\n$$c^a + c^{a+1} + \\dots + c^{b} = \\frac{c^{b+1} - c^a}{c-1},\\; c \\neq 1$$\n%\n$$\\eqalign{\n\t1 + 2 + 3 + \\dots + n &= \\frac{n(n+1)}{2} \\cr\n\t1^2 + 2^2 + 3^2 + \\dots + n^2 &= \\frac{n(2n+1)(n+1)}{6} \\cr\n\t1^3 + 2^3 + 3^3 + \\dots + n^3 &= \\frac{n^2(n+1)^2}{4} \\cr\n\t1^4 + 2^4 + 3^4 + \\dots + n^4 &= \\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \\cr\n}$$\n\n\\section{Series} \n\n$$e^x = 1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\dots,\\quad(-\\infty0$.\n$$f(x) = \\cases{\\lambda e^{-\\lambda x}&$x\\ge 0$,\\cr 0&otherwise.\\cr}$$\n$$\\mu=\\frac{1}{\\lambda},\\quad\\sigma^2=\\frac{1}{\\lambda^2}$$\n\n\\subsubsection{Normal distribution}\nMost real random values with mean $\\mu$ and variance $\\sigma^2$ are well described by ${\\cal N}(\\mu,\\sigma^2)$, where $\\sigma>0$.\n$$ f(x) = \\frac{1}{\\sqrt{2\\pi\\sigma^2}}e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}} $$\nIf $X_1 \\sim {\\cal N}(\\mu_1,\\sigma_1^2)$ and $X_2 \\sim {\\cal N}(\\mu_2,\\sigma_2^2)$ then\n$$ aX_1 + bX_2 + c \\sim {\\cal N}(\\mu_1+\\mu_2+c,a^2\\sigma_1^2+b^2\\sigma_2^2) $$\n\n\\section{Markov chains}\nA \\emph{Markov chain} is a discrete random process with the property that the next state depends only on the current state.\nLet $X_1,X_2,\\ldots$ be a sequence of random variables generated by the Markov process.\nThen there is a transition matrix $\\mathbf{P} = (p_{ij})$, with $p_{ij} = \\Pr(X_n = i | X_{n-1} = j)$,\nand $\\mathbf{p}^{(n)} = \\mathbf P^n \\mathbf p^{(0)}$ is the probability distribution for $X_n$ (i.e., $p^{(n)}_i = \\Pr(X_n = i)$),\nwhere $\\mathbf{p}^{(0)}$ is the initial distribution.\n\n% \\subsubsection{Stationary distribution}\n$\\mathbf{\\pi}$ is a stationary distribution if $\\mathbf{\\pi} = \\mathbf{\\pi P}$.\nIf the Markov chain is \\emph{irreducible} (it is possible to get to any state from any state),\n% TODO: make \\scriptfont\\mathbb work\nthen $\\pi_i = \\frac{1}{{\\mathbb E}(T_i)}$ where ${\\mathbb E}(T_i)$ is the expected time between two visits in state $i$.\n$\\pi_j/\\pi_i$ is the expected number of visits in state $j$ between two visits in state $i$.\n\nFor a connected, undirected and non-bipartite graph, where the transition probability is uniform among all neighbors, $\\pi_i$ is proportional to node $i$'s degree.\n\n% \\subsubsection{Ergodicity}\nA Markov chain is \\emph{ergodic} if the asymptotic distribution is independent of the initial distribution.\nA finite Markov chain is ergodic iff it is irreducible and \\emph{aperiodic} (i.e., the gcd of cycle lengths is 1).\n$\\lim_{k\\rightarrow\\infty}\\mathbf{P}^k = \\mathbf{1}\\pi$.\n\n% \\subsubsection{Absorption}\nA Markov chain is an A-chain if the states can be partitioned into two sets $\\mathbf{A}$ and $\\mathbf{G}$, such that all states in $\\mathbf{A}$ are absorbing ($p_{ii}=1$), and all states in $\\mathbf{G}$ leads to an absorbing state in $\\mathbf{A}$.\nThe probability for absorption in state $i\\in\\mathbf{A}$, when the initial state is $j$, is $a_{ij} = p_{ij}+\\sum_{k\\in\\mathbf{G}} a_{ik}p_{kj}$.\nThe expected time until absorption, when the initial state is $i$, is $t_i = 1+\\sum_{k\\in\\mathbf{G}}p_{ki}t_k$.\n\n\n\\section{Number-theoretical}\n\n\\subsection{Pythagorean Triples}\n The Pythagorean triples are uniquely generated by\n $$a=k\\cdot (m^{2}-n^{2}),\\ \\,b=k\\cdot (2mn),\\ \\,c=k\\cdot (m^{2}+n^{2}),$$\n with $m > n > 0$, $k > 0$, $m \\bot n$, and either $m$ or $n$ even.\n\n\\subsection{Primes}\n\t$p=962592769$ is such that $2^{21} \\mid p-1$, which may be useful. For hashing\n\tuse 970592641 (31-bit number), 31443539979727 (45-bit), 3006703054056749\n\t(52-bit). There are 78498 primes less than $1\\,000\\,000$.\n\n\tPrimitive roots exist modulo any prime power $p^a$, except for $p = 2, a > 2$, and there are $\\phi(\\phi(p^a))$ many.\n\tFor $p = 2, a > 2$, the group ${\\mathbb Z}_{2^a}^\\times$ is instead isomorphic to ${\\mathbb Z}_2 \\times {\\mathbb Z}_{2^{a-2}}$.\n\n\\subsection{Estimates}\n\t$\\sum_{d|n} d = O(n \\log \\log n)$.\n\n\tThe number of divisors of $n$ is at most around 100 for $n < 5e4$, 500 for $n < 1e7$, 2000 for $n < 1e10$, $200\\,000$ for $n < 1e19$.\n\n\\chapter{Combinatorial (text)}\n\n%\\section{The Twelvefold Way}\n%\t\\import{twelvefoldway.tex}\n\n\\section{Permutations}\n%\\subsection{Factorial}\n%\t\\import{factorial.tex}\n%\t\\kactlimport{intperm.h}\n\n\\subsection{Cycles}\n\tLet the number of $n$-permutations whose cycle lengths all belong to the set $S$ be denoted by $g_S(n)$. Then\n\t$$\\sum_{n=0} ^\\infty g_S(n) \\frac{x^n}{n!} = \\exp\\left(\\sum_{n\\in S} \\frac{x^n} {n} \\right)$$\n\n\\subsection{Involutions}\n\tAn involution is a permutation with maximum cycle length 2, and it is its own inverse.\n\t$$a(n) = a(n-1) + (n-1)a(n-2)$$\n\t$$a(0) = a(1) = 1$$\n\t1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496, 35696, 140152\n\n\\subsection{Stirling numbers of the first kind}\n\t$$s(n,k) = (-1)^{n-k}c(n,k)$$\n\t$c(n,k)$ is the unsigned Stirling numbers of the first kind, and they count the number of permutations on $n$ items with $k$ cycles.\n\t$$s(n,k) = s(n-1,k-1) - (n-1) s(n-1,k)$$\n\t$$s(0,0) = 1, s(n,0) = s(0,n) = 0$$\n\t$$c(n,k) = c(n-1,k-1) + (n-1) c(n-1,k)$$\n\t$$c(0,0) = 1, c(n,0)=c(0,n)=0$$\n\n\\subsection{Eulerian numbers}\n\tNumber of permutations $\\pi \\in S_n$ in which exactly $k$ elements are greater than the previous element. $k$ $j$:s s.t. $\\pi(j)>\\pi(j+1)$, $k+1$ $j$:s s.t. $\\pi(j)\\geq j$, $k$ $j$:s s.t. $\\pi(j)>j$.\n\t$$E(n,k) = (n-k)E(n-1,k-1) + (k+1)E(n-1,k)$$\n\t$$E(n,0) = E(n,n-1) = 1$$\n\t$$E(n,k) = \\sum_{j=0}^k(-1)^j\\binom{n+1}{j}(k+1-j)^n$$\n\n\\subsection{Burnside's lemma}\n\tGiven a group $G$ of symmetries and a set $X$, the number of elements of $X$ \\emph{up to symmetry} equals\n\t $$ {\\frac {1}{|G|}}\\sum _{{g\\in G}}|X^{g}|, $$\n\t where $X^{g}$ are the elements fixed by $g$ ($g.x = x$).\n\n\t If $f(n)$ counts \"configurations\" (of some sort) of length $n$, we can ignore rotational symmetry using $G = \\mathbb Z_n$ to get\n\t $$ g(n) = \\frac 1 n \\sum_{k=0}^{n-1}{f(\\text{gcd}(n, k))} = \\frac 1 n \\sum_{k|n}{f(k)\\phi(n/k)}. $$\n\n\\section{Partitions and subsets}\n\\subsection{Partition function}\n\tPartitions of $n$ with exactly $k$ parts, $p(n,k)$, i.e., writing $n$ as a sum of $k$ positive integers, disregarding the order of the summands.\n\t$$p(n,k) = p(n-1,k-1)+p(n-k,k)$$\n\t$$p(0,0)=p(1,n)=p(n,n)=p(n,n-1)=1$$\n\n\tFor partitions with any number of parts, $p(n)$ obeys\n\t$$ p(0) = 1,\\ p(n) = \\sum_{k \\in \\mathbb Z \\setminus \\{0\\}}{(-1)^{k+1} p(n - k(3k-1) / 2)} $$\n\t$$ p(n) \\sim 0.145 / n \\cdot \\exp(2.56 \\sqrt{n}) $$\n\n%\t\\begin{center}\n%\t\\begin{tabular}{c|c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c@{\\ }c}\n%\t\t$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 20 & 50 & 100 \\\\ \\hline\n%\t\t$p(n)$ & 1 & 1 & 2 & 3 & 5 & 7 & 11 & 15 & 22 & 30 & 627 & $\\mathtt{\\sim}$2e5 & $\\mathtt{\\sim}$2e8 \\\\\n%\t\\end{tabular}\n%\t\\end{center}\n%\n%\n\\subsection{Stirling numbers of the second kind}\n\tPartitions of $n$ distinct elements into exactly $k$ groups.\n\t$$S(n,k) = S(n-1,k-1) + k S(n-1,k)$$\n\t$$S(n,1) = S(n,n) = 1$$\n\t$$S(n,k) = \\frac{1}{k!}\\sum_{j=0}^k (-1)^{k-j}\\binom{k}{j}j^n$$\n\n\\subsection{Bell numbers}\n\tTotal number of partitions of $n$ distinct elements.\n\t$$B(n) = \\sum_{k=1}^n \\binom{n-1}{k-1}B(n-k) = \\sum_{k=1}^n S(n,k)$$\n\t$$B(0) = B(1) = 1$$\n\tThe first are 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597.\n\tFor a prime $p$\n\t$$B(p^m+n)\\equiv mB(n)+B(n+1) \\pmod{p}$$\n\n\\subsection{Triangles}\n\tGiven rods of length $1,\\ldots,n$,\n\t$$T(n) = \\frac{1}{24} \\cases{n(n-2)(2n-5) & $n$ even\\cr(n-1)(n-3)(2n-1) & $n$ odd}.$$\n\tis the number of distinct triangles (positive are) that can be constructed, i.e., the \\# of 3-subsets of $[n]$ s.t. $x\\leq y\\leq z$ and $z\\neq x+y$.\n\n\\section{General purpose numbers}\n\\subsection{Catalan numbers}\n\t$$C_n=\\frac{1}{n+1}\\binom{2n}{n}= \\binom{2n}{n}-\\binom{2n}{n+1} = \\frac{(2n)!}{(n+1)!n!}$$\n\t$$C_{n+1} = \\frac{2(2n+1)}{n+2}C_n$$\n\t$$C_0=1, C_{n+1}=\\sum C_iC_{n-i}$$\n\tFirst few are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900.\n\t\t\\itemitem{$\\bullet$}\\# of monotonic lattice paths of a $n\\times n$-grid which do not pass above the diagonal.\n\t\t\\itemitem{$\\bullet$}\\# of expressions containing $n$ pairs of parenthesis which are correctly matched.\n\t\t\\itemitem{$\\bullet$}\\# of full binary trees with with $n+1$ leaves (0 or 2 children).\n\t\t\\itemitem{$\\bullet$}\\# of non-isomorphic ordered trees with $n+1$ vertices.\n\t\t\\itemitem{$\\bullet$}\\# of ways a convex polygon with $n+2$ sides can be cut into triangles by connecting vertices with straight lines.\n\t\t\\itemitem{$\\bullet$}\\# of permutations of $[n]$ with no three-term increasing subsequence.\n\n\\subsection{Super Catalan numbers}\n\tThe number of monotonic lattice paths of a $n\\times n$-grid that do not touch the diagonal.\n\t$$S(n) = \\frac{3(2n-3)S(n-1)-(n-3)S(n-2)}{n}$$\n\t$$S(1)=S(2)=1$$\n\t1, 1, 3, 11, 45, 197, 903, 4279, 20793, 103049, 518859\n\n\\subsection{Motzkin numbers}\n\tNumber of ways of drawing any number of nonintersecting chords among $n$ points on a circle. Number of lattice paths from $(0,0)$ to $(n,0)$ never going below the $x$-axis, using only steps NE, E, SE.\n\t$$M(n) = \\frac{3(n-1)M(n-2)+(2n+1)M(n-1)}{n+2}$$\n\t$$M(0) = M(1) = 1$$\n\t1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634\n\n\\subsection{Narayana numbers}\n\tNumber of lattice paths from $(0,0)$ to $(2n,0)$ never going below the $x$-axis, using only steps NE and SE, and with $k$ peaks.\n\t$$N(n,k) = \\frac{1}{n}\\binom{n}{k}\\binom{n}{k-1}$$\n\t$$N(n,1) = N(n,n) = 1$$\n\t$$\\sum_{k=1}^n N(n,k) = C_n$$\n\t1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50\n\n\\subsection{Schr\\\" oder numbers}\n\tNumber of lattice paths from $(0,0)$ to $(n,n)$ using only steps N,NE,E, never going above the diagonal. Number of lattice paths from $(0,0)$ to $(2n,0)$ using only steps NE, SE and double east EE, never going below the $x$-axis. Twice the Super Catalan number, except for the first term.\n\t1, 2, 6, 22, 90, 394, 1806, 8558, 41586, 206098\n\n}\n", "meta": {"hexsha": "673f66ef8ad0bf10347175c6e08437947130a7f3", "size": 15599, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "export/chapters/math.tex", "max_stars_repo_name": "trinerdi/trinerdi-icpc", "max_stars_repo_head_hexsha": "c305484f2c4b338d0c59ce2f1e0fc47c5ee9d252", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2018-10-20T07:36:27.000Z", "max_stars_repo_stars_event_max_datetime": "2019-10-03T14:32:36.000Z", "max_issues_repo_path": "export/chapters/math.tex", "max_issues_repo_name": "trinerdi/trinerdi-icpc", "max_issues_repo_head_hexsha": "c305484f2c4b338d0c59ce2f1e0fc47c5ee9d252", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 38, "max_issues_repo_issues_event_min_datetime": "2017-10-22T14:11:37.000Z", "max_issues_repo_issues_event_max_datetime": "2020-02-25T13:07:33.000Z", "max_forks_repo_path": "export/chapters/math.tex", "max_forks_repo_name": "trinerdi/trinerdi-icpc", "max_forks_repo_head_hexsha": "c305484f2c4b338d0c59ce2f1e0fc47c5ee9d252", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2019-01-16T06:11:58.000Z", "max_forks_repo_forks_event_max_datetime": "2019-01-16T06:11:58.000Z", "avg_line_length": 46.2878338279, "max_line_length": 289, "alphanum_fraction": 0.6253606, "num_tokens": 6468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117940706735, "lm_q2_score": 0.7401743563075447, "lm_q1q2_score": 0.6119108700281162}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\n\\section*{Fri Oct 18 2019}\n\n\\subsection{Covariant differentiation}\n\nWe want to do derivatives, as in: \n%\n\\begin{equation}\n f(x + \\dd{x} ) = f(x) + \\pdv{f}{x} \\dd{x} \n\\,,\n\\end{equation}\n%\nwhich allow us to ``move around''. In more dimensions, the rule will be \n%\n\\begin{equation}\n f(x + \\dd{x} ) - f(x) = \\dd{x^{\\mu }} \\partial_\\mu f\n\\,.\n\\end{equation}\n%\n\nNow, we want to prove that, in Minkowski spacetime, \\(\\partial_\\mu f\\) is a rank (0,1) tensor if \\(f\\) is a function.\n\nUnder a change of variables \\(x^{\\mu } \\rightarrow x^{\\prime \\mu }\\) we have \n%\n\\begin{equation}\n \\pdv{f(x)}{x^{\\mu }} \\rightarrow \\pdv{f (x')}{x^{\\prime \\mu }} = \\pdv{x^{\\alpha }}{x^{\\prime \\mu }} \\pdv{}{x^{\\alpha }} f(x) \\marginnote{Since \\(x'\\) and \\(x\\) represent the same point, \\(f(x') = f(x)\\).}\n\\,,\n\\end{equation}\n%\nwhich is the transformation law of a covariant vector, or (0,1) tensor.\n\nHowever, for a vector \\(\\dd{x^{\\nu }} \\partial_\\nu A^{\\mu }\\) is \\emph{not} a tensor!\n\nUnder a change of coordinates, \n%\n\\begin{subequations}\n\\begin{align}\n \\pdv{A^{\\mu}(x)}{x^{\\nu }} &= \\pdv{}{x^{\\prime \\nu }} A^{\\mu }(x') \\\\\n &= \\pdv{x^{\\alpha }}{x^{\\prime \\nu }} \\pdv{}{x^{\\alpha }} \\qty(\\pdv{x^{\\prime \\mu }}{x^{\\beta }} A^{\\beta }(x)) \\\\\n &= \\pdv{x^{\\alpha }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\alpha }}{x^{\\beta }} +\n \\pdv{x^{\\alpha }}{x^{\\prime \\nu }} \\pdv{x^{\\prime \\mu }}{x^{\\beta }} \\pdv{A^{\\beta }}{x^{\\alpha }} \n \\,,\n\\end{align}\n\\end{subequations}\n%\nand we can see that the second term is the transformation we want, but the first term spoils the transformation. \n\nThis is not an issue in SR: there, the second derivative vanishes:\n%\n\\begin{equation}\n \\pdv[2]{x^{\\prime \\mu }}{x^{\\alpha }}{x^{\\beta }}\n = \\pdv{}{x^{\\alpha }} \\Lambda^{\\mu }_{\\beta } = 0 \n\\,,\n\\end{equation}\n%\nsince Lorentz matrices are constant.\n\nSo, we construct a \\emph{Covariant Derivative} which transforms as a tensor under diffeomorphisms.\n\nWe denote it as \\(\\nabla_\\nu A^{\\mu }\\).\nFor any tensor \\(T\\) of arbitrary rank \\((p, q)\\) we request \\(\\nabla_\\nu T\\) to be a tensor of rank \\((p, q+1)\\).\nAlso, we request \\(\\nabla_\\mu \\rightarrow \\partial_\\mu\\) for flat spacetime.\n\nWe define the Christoffel symbols: \n%\n\\begin{equation}\n \\Gamma_{\\mu \\nu }^{\\alpha } =\n \\frac{1}{2} g^{\\alpha \\lambda } \\qty(g_{\\lambda \\mu , \\nu }+ g_{\\lambda \\nu , \\mu } - g_{\\mu \\nu , \\lambda })\n\\,,\n\\end{equation}\n%\nwhere we introduced comma notation for partial non-covariant differentiation: the full notation is \n%\n\\begin{align}\n\\partial_{\\mu } x_{A} \\overset{\\text{def}}{=} x_{A, \\mu } \n\\qquad \\text{and} \\qquad\n\\nabla_{\\mu } x_{A} \\overset{\\text{def}}{=} x_{A;\\mu }\n\\,.\n\\end{align}\n%\n\nThey are symmetric in the two lower indices and they are not tensors. Their transformation law is: \n%\n\\begin{equation}\n \\Gamma_{\\nu \\kappa }^{\\mu }\\rightarrow\n \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv{x^{\\beta }}{x^{\\prime \\nu }} \\pdv{x^{\\gamma }}{x^{\\prime \\mu }} \\Gamma^{\\alpha }_{\\beta \\gamma } + \n \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} \n\\,,\n\\end{equation}\n%\nwhich we note is \\emph{not tensorial}!\n\nWe define \n%\n\\begin{equation}\n \\nabla_{\\nu }V_{\\mu } = \\partial_{\\nu } V_\\mu - \\Gamma_{\\nu \\mu }^{\\alpha } V_{\\alpha }\n\\,,\n\\end{equation}\n%\nand \n%\n\\begin{equation}\n \\nabla_{\\nu }V^{\\mu} = \\partial_{\\nu } V_\\mu + \\Gamma_{\\nu \\alpha }^{\\mu } V^{\\alpha }\n\\,.\n\\end{equation}\n%\n\nHow does it transform? For the covariant derivative, we have:\n%\n\\begin{subequations}\n\\begin{align}\n \\nabla_{\\nu }' V_{\\kappa }' &= \\pdv{}{x^{\\prime \\nu }} V'_\\kappa - \\Gamma^{\\prime \\mu }_{\\nu \\kappa } V_{\\mu }' \\\\\n &= \\pdv{x^{\\mu }}{x^{\\prime \\nu }} \\partial_{\\mu } \\qty(\\pdv{x^{\\lambda }}{x^{\\prime \\kappa}}V_{\\lambda }) - \\qty(\\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv{x^{\\beta }}{x^{\\prime \\nu }} \\pdv{x^{\\gamma }}{x^{\\prime \\mu }} \\Gamma^{\\alpha }_{\\beta \\gamma } + \n \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} ) \n \\qty(\\pdv{x^{\\lambda }}{x^{\\prime \\mu }} V_{\\lambda }) \\\\\n &= \\pdv[2]{x^{\\lambda }}{x^{\\prime \\nu }}{x^{\\prime \\mu }} V_{\\lambda }\n + \\pdv{x^{\\lambda }}{x^{\\prime \\mu}} \\pdv{V_{\\lambda }}{x^{\\prime \\nu }} - \n \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\qty(\\pdv{x^{\\beta }}{x^{\\prime \\nu }}\\pdv{x^{\\gamma }}{x^{\\prime \\kappa }} \\Gamma^{\\alpha }_{\\beta \\gamma } + \\pdv[2]{x^{\\alpha }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} ) \\pdv{x^{\\lambda }}{x^{\\prime \\mu }} V_{\\lambda } \\\\\n &=\\pdv[2]{x^{\\lambda }}{x^{\\prime \\nu }}{x^{\\prime \\mu }} V_{\\lambda }\n + \\pdv{x^{\\lambda }}{x^{\\prime \\kappa }} \\pdv{x^{\\alpha }}{x^{\\prime \\nu }} \\pdv{V_{\\lambda } }{x^{\\alpha }}\n - \\qty(\\pdv{x^{\\beta }}{x^{\\prime \\nu }}\\pdv{x^{\\gamma }}{x^{\\prime \\kappa }} \\Gamma^{\\alpha }_{\\beta \\gamma } + \\pdv[2]{x^{\\alpha }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} ) \\delta^{\\lambda }_{\\alpha }V_{\\lambda } \\\\\n &= \\pdv[2]{x^{\\lambda }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} V_{\\lambda } \n + \\pdv{x^{\\lambda }}{x^{\\prime \\kappa }} \\pdv{x^{\\alpha }}{x^{\\prime \\nu }} \\pdv{V_{\\lambda } }{x^{\\alpha }}\n - \\pdv{x^{\\beta }}{x^{\\prime \\nu }} \\pdv{x^{\\gamma }}{x^{\\prime \\kappa }} \\Gamma^{\\alpha }_{\\beta \\gamma }V_{\\alpha }\n -\\pdv[2]{x^{\\lambda }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }} V_{\\lambda } \\\\\n &= \\pdv{x^{\\lambda }}{x^{\\prime \\kappa }} \\pdv{x^{\\alpha }}{x^{\\prime \\nu }}\n \\qty(\\pdv{V_{\\lambda }}{x^{\\alpha } } - \\Gamma^{\\sigma }_{\\alpha \\lambda }V_{\\sigma }) \\\\\n &= \\pdv{x^{\\lambda }}{x^{\\prime \\kappa }} \\pdv{x^{\\alpha }}{x^{\\prime \\nu }}\n \\nabla_{\\alpha }V_{\\lambda }\n \\,,\n\\end{align}\n\\end{subequations}\n%\nwhere we used: relabeling of indices, contraction of the Jacobian matrix with its inverse, the chain rule, the product rule, the transformation law of the Christoffel symbols (some steps are wrong... oh well, there's more interesting stuff to do).\n\nThe derivative of a contravariant tensor is a tensor: this can be proven by noticing that \\(\\nabla_\\mu (A^{\\alpha }B_{\\alpha }) = \\partial_\\mu (A^{\\alpha }B_{\\alpha }) = B_{\\alpha } \\nabla_{\\mu }A^{\\alpha } + A^{\\alpha }\\nabla_{\\mu }B_{\\alpha }\\). Otherwise, we can compute away:\n%\n\\begin{subequations}\n\\begin{align}\n \\nabla_{\\nu }' V^{\\prime \\mu } &= \\pdv{V^{\\prime \\mu }}{x^{\\prime \\nu }} + \\Gamma^{\\prime \\mu }_{\\nu \\kappa }V^{\\prime \\kappa } \\\\\n &= \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{}{x^{\\lambda }} \\qty(\\pdv{x^{\\prime \\mu }}{x^{\\alpha }} V^{\\alpha }) \n + \\qty(\\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv{x^{\\beta }}{x^{\\prime \\nu }} \\pdv{x^{\\gamma }}{x^{\\prime \\mu }} \\Gamma^{\\alpha }_{\\beta \\gamma } \n + \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\prime \\nu }}{x^{\\prime \\kappa }}) \\qty(\\pdv{x^{\\prime \\kappa }}{x^{\\sigma }} V^{\\sigma }) \\\\\n &= \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\lambda }}{x^{\\alpha }} V^{\\alpha }\n + \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv{V^{\\alpha }}{x^{\\lambda }}\n + \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv{x^{\\beta }}{x^{\\prime \\nu }} \\delta^{\\gamma}_{\\sigma } \\Gamma^{\\alpha }_{\\beta \\gamma } V^{\\gamma }+ \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{\\prime \\alpha }}{x^{\\nu }}{x^{\\mu }} \\pdv{x^{\\prime \\kappa }}{x^{\\sigma}} V^{\\sigma } \\\\\n &= \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\lambda }}{x^{\\alpha }} V^{\\alpha }\n + \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\qty(\\pdv{V^{\\alpha }}{x^{\\lambda }} + \\Gamma^{\\alpha }_{\\lambda \\gamma } V^{\\gamma }) \\\\\n & = \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\nabla_{\\lambda }V^{\\alpha }\n + \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\lambda }}{x^{\\alpha }}V^{\\alpha }\n + \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{ \\alpha }}{x^{\\prime \\sigma }}{x^{\\prime \\kappa }} \\pdv{x^{\\prime \\kappa }}{x^{ \\sigma }} V^{\\sigma } \n \\,,\n\\end{align}\n\\end{subequations}\n%\nand we would like to see that the two last terms cancel: is \n%\n\\begin{equation}\n \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\lambda }}{x^{\\alpha }}V^{\\alpha }\n + \\pdv{x^{\\prime \\mu }}{x^{\\alpha }} \\pdv[2]{x^{ \\alpha }}{x^{\\prime \\sigma }}{x^{\\prime \\kappa }} \\pdv{x^{\\prime \\kappa }}{x^{ \\sigma }} V^{\\sigma } \\overset{?}{=} 0 \n\\,,\n\\end{equation}\n%\nfor all \\(V^{\\mu }\\)? Let us factor the vector, changing the indices: \n%\n\\begin{equation}\n \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv[2]{x^{\\prime \\mu }}{x^{\\lambda }}{x^{\\alpha }}\n + \\pdv{x^{\\prime \\mu }}{x^{\\sigma }} \\pdv[2]{x^{ \\sigma }}{x^{\\prime \\alpha }}{x^{\\prime \\kappa }} \\pdv{x^{\\prime \\kappa }}{x^{ \\alpha }} \\overset{?}{=} 0\n\\,,\n\\end{equation}\n%\nwe can rewrite it as \n%\n\\begin{equation}\n \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{}{x^{\\alpha }} \\pdv{x^{\\prime \\mu }}{x^{\\lambda }}\n + \\pdv{x^{\\prime \\mu }}{x^{\\lambda }} \n \\pdv{}{x^{\\prime \\kappa }} \n \\pdv{x^{ \\lambda }}{x^{\\prime \\nu }}\n \\pdv{x^{\\prime \\mu }}{x^{ \\alpha }} \\overset{?}{=} 0\n\\,,\n\\end{equation}\n%\nwhich can be recombined into: \n%\n\\begin{equation}\n \\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{}{x^{\\alpha }} \\pdv{x^{\\prime \\mu }}{x^{\\lambda }}\n + \\pdv{x^{\\prime \\mu }}{x^{\\lambda }} \n \\pdv{}{x^{\\alpha }} \n \\pdv{x^{ \\lambda }}{x^{\\prime \\nu }}\n \\overset{?}{=} 0\n\\,,\n\\end{equation}\n%\nand becomes \n%\n\\begin{equation}\n \\pdv{}{x^{\\alpha }} \\qty(\\pdv{x^{\\lambda }}{x^{\\prime \\nu }} \\pdv{x^{\\prime \\mu }}{x^{\\lambda }} ) \n = \\pdv{}{x^{\\alpha }} \\delta^{\\mu}_{\\nu }\n = 0\n\\,.\n\\end{equation}\n%\n\nFor any order tensor, we add a Christoffel symbol for every index, such as in: \n%\n\\begin{equation}\n \\nabla_{\\mu }V_{\\alpha \\beta } \n = \\partial_\\mu V_{\\alpha \\beta } - \\Gamma^{\\lambda}_{\\mu \\alpha }V_{\\lambda \\beta }- \\Gamma^{\\lambda }_{\\mu \\beta }V_{\\alpha \\lambda }\n\\,,\n\\end{equation}\n%\nor \n%\n\\begin{equation}\n \\nabla_{\\mu }V_{\\alpha}^{\\beta } \n = \\partial_\\mu V_{\\alpha}^{\\beta }\n - \\Gamma^{\\lambda}_{\\mu \\alpha }V_{\\lambda}^{\\beta }\n + \\Gamma^{\\beta }_{\\mu \\lambda }V_{\\alpha}^{\\lambda }\n\\,.\n\\end{equation}\n\n\\begin{bluebox}\n \n\nThe general formula reads: \n%\n\\begin{align}\n\\nabla_{\\pi } T^{\\mu_1 \\dots \\mu_{n}}_{\\nu_1 \\dots \\nu_{p}} =\n\\partial_{\\pi } T^{\\mu_1 \\dots \\mu_{n}}_{\\nu_1 \\dots \\nu_{p}}\n+ \\sum_{i=1}^{n} \\Gamma^{\\mu_{i}}_{\\pi \\sigma }\nT^{\\mu_1 \\dots \\mu_{i-1} \\sigma \\mu_{i+1} \\dots \\mu_{n}}_{\\nu_1 \\dots \\nu_{p}}\n- \\sum _{j=1}^{p} \\Gamma^{\\sigma }_{\\pi \\nu_j}\nT^{\\mu_1 \\dots \\mu_{n}}_{\\nu_1 \\dots \\nu_{i-1} \\sigma \\nu_{i+1} \\dots \\nu_{p}}\n\\,,\n\\end{align}\n%\nwhich can be generalized to tensor \\emph{densities} beyond regular tensors: any tensor tensity \\(\\mathfrak{T}\\) (with any amount of indices, omitted for simplicity) of weight \\(w\\) (that is, which transforms almost like a tensor, with an extra factor of the determinant of the metric raised to the \\(w\\)th power) can be written as \\(\\mathfrak{T} = \\sqrt{-g}^{w} T\\), where \\(g\\) is the determinant of the metric while \\(T\\) is a regular tensor, since as we saw \\(\\sqrt{-g}\\) is a density of weight +1.\n\nThen, we can use this to calculate the covariant derivative of a tensor density: we can express \\(T = \\sqrt{-g}^{-w} \\mathfrak{T}\\), so the rule will be \n%\n\\begin{align}\n\\nabla_{\\mu } \\mathfrak{T} = \\nabla_{\\mu } \\qty(\\sqrt{-g}^{w} \\sqrt{-g}^{-w} \\mathfrak{T})\n= \\sqrt{-g}^{w} \\nabla_{\\mu } \\qty(\\sqrt{-g}^{-w} \\mathfrak{T})\n\\,,\n\\end{align}\n%\nwhere we used the fact that \n%\n\\begin{align}\n\\nabla_{\\mu} \\sqrt{-g} = 0\n\\,,\n\\end{align}\n%\nwhich follows from the fact that the metric is covariantly constant: if we compute it in a LIF, then we get \n%\n\\begin{align}\n\\nabla_{ \\mu} \\qty(\\sqrt{-g})^{w} = \\partial_{\\mu } \\sqrt{-g}^{w}\n- w \\Gamma^{\\alpha }_{\\alpha \\mu} \\sqrt{-g} \n\\,,\n\\end{align}\n%\nwhich is zero:\\footnote{For a derivation of this expression, see \\url{https://yarikraak.nl/tensor_densities.pdf}} there, the coordinate derivatives of the metric are zero, and therefore the Christoffel symbols are also zero. \n\\end{bluebox}\n\n\n\\subsection{Properties of the covariant derivative}\n\n\\begin{itemize}\n \\item The covariant derivative of a tensor is a tensor;\n \\item the covariant derivative obeys the Leibniz rule: \\(\\nabla_{\\mu} (AB) = B \\nabla_{\\mu }A + A \\nabla_{\\mu }B\\);\n \\item the metric is covariantly constant: \\(\\nabla_{\\mu }g_{\\alpha \\beta }=0\\). \n\\end{itemize}\n\nThe fact that the metric is covariantly constant could actually be assumed instead of the explicit expression of the Christoffel symbols in terms of derivatives of the metric. These two are equivalent.\n\nNotice that covariant derivatives do \\emph{not} commute!\n\nWe can check that \\(\\partial_{\\mu} (A^{\\alpha }B_{\\alpha }) = \\nabla_{\\mu } (A^{\\alpha }B_{\\alpha })\\). It is \n%\n\\begin{equation}\n \\qty(\\partial_{\\mu }A_{\\alpha } - \\Gamma^{\\lambda }_{\\mu \\alpha }A_{\\lambda })B^{\\alpha } \n + A_{\\alpha }\\qty(\\partial_{\\mu }B^{\\alpha } + \\Gamma_{\\mu \\lambda }^{ \\alpha }B^{\\lambda })\n\\,,\n\\end{equation}\n%\nexpanding and relabeling indices we get the desired cancellation. Now, for parallel transport:\n\n\\subsection{Parallel transport}\n\nTake a curve \\(x^{\\alpha }(\\lambda )\\) and a vector \\(V^{\\mu }\\) defined at a certain point along the curve.\nFor infinitesimal displacement we will have \\(\\dd{V} = \\dd{x} \\cdot \\nabla V\\): in components \\(\\dd{V^{\\mu }} = \\dd{x^{\\alpha }} \\nabla_{\\alpha }V^{\\mu }\\).\n\nParallel transport means that the vector does not change when it is transported: \\(\\nabla_t V^{\\mu }=0\\) where the index \\(t \\) indicates derivation along the curve's tangent vector \\(t^{\\alpha }= \\dv*{x^{\\alpha }}{\\lambda } \\): more explicitly, \\(V^{\\mu }\\) is parallel-transported along the curve parametrized as \\(x^{\\alpha }(\\lambda )\\) if \n%\n\\begin{align}\n\\dv{}{\\lambda } \\qty(V^{\\mu }(x^{\\alpha }(\\lambda )))\n=\n\\dv{x^{\\alpha }}{\\lambda } \\nabla_{\\alpha } V^{\\mu } \n= 0\n\\,.\n\\end{align}\n\n\\end{document}", "meta": {"hexsha": "4d4cf87f8e49976922e9d84c0bfea3c1dd628059", "size": 13776, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ap_first_semester/general_relativity/18oct.tex", "max_stars_repo_name": "jacopok/notes", "max_stars_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-10-10T13:10:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T14:52:50.000Z", "max_issues_repo_path": "ap_first_semester/general_relativity/18oct.tex", "max_issues_repo_name": "jacopok/notes", "max_issues_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ap_first_semester/general_relativity/18oct.tex", "max_forks_repo_name": "jacopok/notes", "max_forks_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-03T16:20:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-06T16:11:07.000Z", "avg_line_length": 45.1672131148, "max_line_length": 501, "alphanum_fraction": 0.580937863, "num_tokens": 5403, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8267117769928211, "lm_q2_score": 0.7401743563075446, "lm_q1q2_score": 0.6119108573875277}} {"text": "\\subsection{Analytical solution}\nWill not work because of how \\eqref{line} is defined. It would have to be the line between the two points that is on the spherical surface.\n\\begin{equation}\n g(s)=K(f(s))\n\\end{equation}\n\\begin{equation}\n g'(s) = \\underbrace{K'(f(s))}_{\\text{3$\\times$2 Matrix}}\\cdot\\underbrace{f'(s)}_{\\text{2$\\times$1 Vector}}=\\text{3$\\times$1 Vector}\n\\end{equation}\n\\begin{equation}\n g'(a,b,s) = K'(f(a,b,s))\\cdot f'(a,b,s)\n\\end{equation}\n\\begin{equation}\n g'(\\theta_1,\\varphi_1,\\theta_2,\\varphi_2,s)\n\\end{equation}\n\\begin{equation}\n K'(f(s))=K'\n \\begin{pmatrix}\n (1-s)\\cdot\\theta_1 + s\\cdot\\theta_2 \\\\\n (1-s)\\cdot\\varphi_1 + s\\cdot\\varphi_2\n \\end{pmatrix}\n\\end{equation}\ngiven:\n\\begin{equation*}\n A\\dbinom{\\theta_1}{\\varphi_1}, B\\dbinom{\\theta_2}{\\varphi_2}, C\\dbinom{\\theta_3}{\\varphi_3}\n\\end{equation*}\n\\begin{equation}\n \\cos(\\alpha)=\\frac{\\langle g'(A,B,0), g'(A,C,0)\\rangle}{\\|g'(A,B,0))\\|\\cdot \\|g'(A,C,0)\\|}\n\\end{equation}\n\\begin{equation}\n \\cos(\\beta)=\\frac{\\langle g'(B,A,0), g'(B,C,0)\\rangle}{\\|g'(B,A,0))\\|\\cdot \\|g'(B,C,0)\\|}\n\\end{equation}\n\\begin{equation}\n \\cos(\\gamma)=\\frac{\\langle g'(C,A,0), g'(C,B,0)\\rangle}{\\|g'(C,A,0))\\|\\cdot \\|g'(C,B,0)\\|}\n\\end{equation}\n\\begin{lstlisting}[escapechar=@]\n def f(a, b, s) {\n res[0] = (1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0]\n res[1] = (1-s)@$\\cdot$@a[1] + s@$\\cdot$@b[1]\n return res\n }\n\\end{lstlisting}\nParameters tp representing $\\theta\\varphi$\n\\begin{lstlisting}[escapechar=@]\n def k'(tp) {\n res[0, 0] = cos(tp[0])@$\\cdot$@cos(tp[1])\n res[1, 0] = cos(tp[0])@$\\cdot$@sin(tp[1])\n res[2, 0] = -sin(tp[0])\n res[0, 1] = -sin(tp[0])@$\\cdot$@sin(tp[1])\n res[1, 1] = sin(tp[0])@$\\cdot$@cos(tp[1])\n return res\n }\n\\end{lstlisting}\n\\begin{lstlisting}[escapechar=@]\n def f'(a, b) {\n res[0] = b[0] - a[0]\n res[1] = b[1] - a[1]\n retrun res\n }\n\\end{lstlisting}\nUse $f(a, b, s)$ as tp in $k'(tp)$ and multiply resulting matrix with vector $f'(a, b)$. Resulting in an 3$\\times$1 Vector.\n\\begin{equation*}\n \\begin{pmatrix}\n a && b \\\\\n c && d \\\\\n e && f\n \\end{pmatrix}\n \\cdot\n \\begin{pmatrix}\n u \\\\\n v\n \\end{pmatrix}\n =\n \\begin{pmatrix}\n a\\cdot u + b\\cdot v \\\\\n c\\cdot u + d\\cdot v \\\\\n e\\cdot u + f\\cdot v\n \\end{pmatrix}\n\\end{equation*}\n\\begin{lstlisting}[escapechar=@]\n def g'(a, b, s) {\n res[0] = (cos((1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0])@$\\cdot$@\n cos((1-s)@$\\cdot$@a[1] + s@$\\cdot$@b[1]))@$\\cdot$@(b[0] - a[0]) +\n (-sin((1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0])@$\\cdot$@\n sin((1-s)@$\\cdot$@a[1] + s@$\\cdot$@b[1]))@$\\cdot$@(b[1] - a[1])\n res[1] = (cos((1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0])@$\\cdot$@\n sin((1-s)@$\\cdot$@a[1] + s@$\\cdot$@b[1]))@$\\cdot$@(b[0] - a[0]) +\n (sin((1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0]))@$\\cdot$@\n cos((1-s)@$\\cdot$@a[1] + s@$\\cdot$@b[1]))@$\\cdot$@(b[1] - a[1])\n res[2] = (-sin((1-s)@$\\cdot$@a[0] + s@$\\cdot$@b[0]))@$\\cdot$@\n (b[0] - a[0])\n return res\n }\n\\end{lstlisting}\nFor $s$ being $0$:\n\\begin{lstlisting}[escapechar=@]\n def g'(a, b) {\n res[0] = cos(a[0])@$\\cdot$@cos(a[1])@$\\cdot$@(b[0] - a[0]) -\n sin(a[0])@$\\cdot$@sin(a[1])@$\\cdot$@(b[1] - a[1])\n res[1] = cos(a[0])@$\\cdot$@sin(a[1])@$\\cdot$@(b[0] - a[0]) +\n sin(a[0])@$\\cdot$@cos(a[1])@$\\cdot$@(b[1] - a[1])\n res[2] = -sin(a[0])@$\\cdot$@(b[0] - a[0])\n return res\n}\n\\end{lstlisting}", "meta": {"hexsha": "d7927052b89c6ba273239554f0a617d409902756", "size": 3601, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/analytical.tex", "max_stars_repo_name": "paddyez/ingress", "max_stars_repo_head_hexsha": "44b4fde664ee562d516de3dd7e49aa2f1e539d79", "max_stars_repo_licenses": ["WTFPL"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-04-11T11:07:14.000Z", "max_stars_repo_stars_event_max_datetime": "2021-04-11T11:07:14.000Z", "max_issues_repo_path": "tex/analytical.tex", "max_issues_repo_name": "paddyez/ingress", "max_issues_repo_head_hexsha": "44b4fde664ee562d516de3dd7e49aa2f1e539d79", "max_issues_repo_licenses": ["WTFPL"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/analytical.tex", "max_forks_repo_name": "paddyez/ingress", "max_forks_repo_head_hexsha": "44b4fde664ee562d516de3dd7e49aa2f1e539d79", "max_forks_repo_licenses": ["WTFPL"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 34.625, "max_line_length": 139, "alphanum_fraction": 0.4981949458, "num_tokens": 1554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.879146761176671, "lm_q2_score": 0.6959583124210896, "lm_q1q2_score": 0.6118494962789827}} {"text": "\\section{The Euler-Lagrange Equation}\r\n\\subsection{Derivation of the Equation}\r\nWe now move on to the most important theorem of the course, which gives a necessary condition to extremise a functional in the form\r\n$$F[y]=\\int_\\alpha^\\beta f(x,y,y^\\prime)\\,\\mathrm dx$$\r\nwhere $f$ is given.\r\nIn contexts of geometry and physics, $y$ is most likely to carry the meaning of the trajectory of some point.\\\\\r\nWe first assume that a extremum $y$ exists, then when we apply a small perturbation $y\\mapsto y+\\epsilon\\eta$ with $\\eta(\\alpha)=\\eta(\\beta)=0$ to keep the endpoint fixed.\r\nNow we want to compute $F[y+\\epsilon\\eta]$, but first of all we will need a lemma.\r\n\\begin{lemma}\\label{fund_lemma}\r\n If $g:[\\alpha,\\beta]\\to\\mathbb R$ is continuous on $[\\alpha,\\beta]$ and\r\n $$\\int_\\alpha^\\beta g(x)\\eta(x)\\,\\mathrm dx=0$$\r\n for all $\\eta\\in C([\\alpha,\\beta])$ with $\\eta(\\alpha)=\\eta(\\beta)=0$, then $\\forall x\\in [\\alpha,\\beta],g(x)=0$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Assume for sake of contradiction that there exists some $\\bar{x}$ on $(\\alpha,\\beta)$ such that $g(\\bar{x})\\neq 0$.\r\n WLOG $g(\\bar{x})>0$, then by continuity there is an interval $[x_1,x_2]\\subset[\\alpha,\\beta]$ such that $\\exists c>0,\\forall x\\in [x_1,x_2],g(x)>c$.\r\n Set\r\n $$\\eta(x)=\\begin{cases}\r\n (x-x_1)(x_2-x)\\text{, if $x\\in [x_1,x_2]$}\\\\\r\n 0\\text{, otherwise}\r\n \\end{cases}$$\r\n Then\r\n \\begin{align*}\r\n \\int_\\alpha^\\beta g(x)\\eta(x)\\,\\mathrm dx\r\n &=\\int_{x_1}^{x_2}g(x)(x-x_1)(x_2-x)\\,\\mathrm dx\\\\\r\n &\\ge\\int_{x_1}^{x_2}c(x-x_1)(x_2-x)\\,\\mathrm dx\\\\\r\n &>0\r\n \\end{align*}\r\n Contradiction.\r\n So $g$ is zero on $(\\alpha,\\beta)$, and it is also zero at $\\alpha,\\beta$ by continuity, hence $g$ is zero on $[\\alpha,\\beta]$.\r\n\\end{proof}\r\nThe $\\eta$ we have used above is called a bump function, which is $C^2$ as one can verify.\r\nOne can also make a $C^k$ bump function by considering\r\n\\footnote{It is also easy to construct a $C^{\\infty}$ bump function.}\r\n$$\\eta(x)=\\begin{cases}\r\n ((x-x_1)(x_2-x))^{k+1}\\text{, if $x\\in [x_1,x_2]$}\\\\\r\n 0\\text{, otherwise}\r\n\\end{cases}$$\r\nNow back at $F[y+\\epsilon\\eta]$, then\r\n\\begin{align*}\r\n F[y+\\epsilon\\eta]&=\\int_\\alpha^\\beta f(x,y+\\epsilon\\eta,y^\\prime+\\epsilon\\eta^\\prime)\\,\\mathrm dz\\\\\r\n &=F[y]+\\epsilon\\int_\\alpha^\\beta\\left( \\frac{\\partial f}{\\partial y}\\eta+\\frac{\\partial f}{\\partial y^\\prime}\\eta^\\prime \\right)\\,\\mathrm dx+O(\\epsilon^2)\r\n\\end{align*}\r\nWe will analysis the $O(\\epsilon^2)$ remainder later.\r\nFor now, we just observe that for $y$ to be an extremum, the first-order term shall vanish, so we want something like $\\partial F[y+\\epsilon\\eta]/\\partial\\epsilon=0$.\\\\\r\nIntegrate the vanishing first-order coefficient by part,\r\n\\begin{align*}\r\n 0&=\\left.\\frac{\\partial f}{\\partial y^\\prime}\\eta\\right|_\\alpha^\\beta+\\int_\\alpha^\\beta\\left( \\frac{\\partial f}{\\partial y}\\eta-\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime}\\eta \\right)\\,\\mathrm dx\\\\\r\n &=\\int_\\alpha^\\beta\\left( \\frac{\\partial f}{\\partial y}-\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime} \\right)\\eta\\,\\mathrm dx\r\n\\end{align*}\r\nBy the preceding lemma, we must have\r\n$$\\frac{\\partial f}{\\partial y}-\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime}=0$$\r\nThis is known as the Euler-Lagrange equation, which is the necessary condition for an extremum.\r\nThis equation is first developed in 1745 in a letter from Lagrange to Euler.\r\n\\begin{remark}\r\n 1. The Euler-Lagrange equation is a second-order ODE with initial conditions $y(\\alpha)=y_1,y(\\beta)=y_2$.\\\\\r\n 2. Sometimes the LHS is denoted $\\delta F[y]/\\delta y(x)$ and is called the functional derivative.\r\n Some author also write $\\epsilon\\eta=\\delta y$, allowing one to write $F[y+\\delta y]=F[y]+\\delta F[y]$ where\r\n $$\\delta F[y]=\\int_\\alpha^\\beta\\frac{\\delta F[y]}{\\delta y(x)}\\delta y(x)\\,\\mathrm dx$$\r\n 3. Other kinds of boundary conditions (e.g. those on $y^\\prime$) are possible.\\\\\r\n 4. Be careful with derivatives as the notation can be a bit confusing.\r\n The $x,y,y^\\prime$ are independent variables when we are talking about partial derivatives of $f(x,y,y^\\prime)$.\\\\\r\n 5. For any $h(x,y,y^\\prime)$, a somewhat useful formula is\r\n $$\\frac{\\mathrm dh(x,y(x),y^\\prime(x))}{\\,\\mathrm dx}=\\frac{\\partial h}{\\partial x}+\\frac{\\partial h}{\\partial y}y^\\prime+\\frac{\\partial h}{\\partial y^\\prime}y^{\\prime\\prime}$$\r\n For example, for $f(x,y,y^\\prime)=x((y^\\prime)^2-y^2)$, we have $\\mathrm df/\\mathrm dx=(y^\\prime)^2-y^2-2xyy^\\prime+2y^{\\prime\\prime}y^\\prime x$.\r\n\\end{remark}\r\n\\subsection{First Integrals of the Euler-Lagrange Equation}\r\nNow we move on to solve the Euler-Lagrange equations, which is a second order ODE.\r\nIn some special cases, this is a quite easy thing to do.\r\nIn particular, if $f$ does not explicitly depend on some of its variables, then we can simplify the equation to something that is easier to solve.\r\nThese simplifications are often in the form of something being constant.\r\nExpressions like these are called the first integrals of the equations.\r\nNot only are they tools we can use to solve the equation, we can also view them as a conserved quantity of something that is described by a variational problem.\r\nWe will discuss the former in this section.\r\nThe latter will be mentioned later, when we discuss Noether's Theorem.\\\\\r\nAssume that $f$ does not explicitly depend on $y$, then $\\partial f/\\partial y=0$, so the Euler-Lagrange equation can be rewritten as\r\n$$\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime}=0\\implies \\frac{\\partial f}{\\partial y^\\prime}=\\text{const.}$$\r\nwhich is a first order ODE.\r\n\\begin{example}[Geodesics on the Plane]\r\n Consider the geodesics on the Euclidean plane.\r\n We know that we want to extremise the functional\r\n $$F[y]=\\int_\\alpha^\\beta\\sqrt{1+(y^\\prime)^2}\\,\\mathrm dx$$\r\n In this case, the apparent $f$ does not depend on $y$, hence we can obtain the solution by just solving the first integral\r\n $$\\frac{y^\\prime}{\\sqrt{1+(y^\\prime)^2}}=\\text{const.}$$\r\n One can solve this to get $y=mx+c$ for some constants $m,c$, which is our familiar formulation of a straight line, the geodesics of the plane.\r\n\\end{example}\r\n\\begin{example}[Geodesics on a Sphere]\r\n Consider the two-dimensional unit sphere $S^2$ in $\\mathbb R^3$.\r\n Use the spherical polar coordinates\r\n $$x=\\sin\\theta\\sin\\phi,y=\\sin\\theta\\cos\\phi,z=\\cos\\theta,\\theta\\in [0,2\\pi),\\theta\\in[0,\\pi)$$\r\n The line element on $S^2$ inherited from $\\mathbb R^3$ then gives\r\n $$\\mathrm ds^2=\\mathrm dx^2+\\mathrm dy^2+\\mathrm dz^2=\\mathrm d\\theta^2+\\sin^2\\theta\\,\\mathrm d\\phi^2$$\r\n A path restricted on the sphere can then be parameterized in terms of $\\theta,\\phi$.\r\n Suppose we parameterize the curve by $\\phi(\\theta)$, then the length functional is\r\n $$F[\\phi]=\\int_{\\theta_1}^{\\theta_2}\\sqrt{1+(\\phi^\\prime)^2\\sin^2\\theta}\\,\\mathrm d\\theta$$\r\n We observe that the integral inside does not depend on $\\phi$, so we can rewrite the equation to\r\n $$\\frac{\\phi^\\prime\\sin^2\\theta}{\\sqrt{1+(\\phi^\\prime)^2\\sin^2\\theta}}=\\frac{\\partial f}{\\partial \\phi^\\prime}=\\text{const.}=K$$\r\n Seperating the variables yields\r\n $$(\\phi^\\prime)^2=\\frac{K^2}{\\sin^2\\theta(\\sin^2\\theta-K^2)}\\implies\\phi=\\pm\\int\\frac{K\\,\\mathrm d\\theta}{\\sin\\theta\\sqrt{\\sin^2\\theta-K^2}}$$\r\n which shall produce two solutions, each going one way round.\r\n To evaluate the integral, we do the substitution $u=\\cot\\theta$ which produces\r\n $$\\pm\\frac{\\sqrt{1-K^2}}{K}\\cos(\\phi-\\phi_0)=\\cot\\theta$$\r\n for a constant $\\phi_0$.\r\n By considering the geometrical meaning of this equation, it then follows that $\\phi(\\theta)$ describes a great circle, i.e. a circle in $\\mathbb R^3$ which exists as the intersection of $S^2$ and a plane that goes through the origin.\r\n\\end{example}\r\nNow, consider for general case of $f(x,y,y^\\prime)$, we have\r\n\\begin{align*}\r\n \\frac{\\mathrm d}{\\mathrm dx}\\left( f-y^\\prime\\frac{\\partial f}{\\partial y^\\prime} \\right)&=\\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}y^\\prime+\\frac{\\partial f}{\\partial y^\\prime}y^{\\prime\\prime}-y^\\prime\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime}-y^{\\prime\\prime}\\frac{\\partial f}{\\partial y^\\prime}\\\\\r\n &=y^\\prime\\left( \\frac{\\partial f}{\\partial y}-\\frac{\\mathrm d}{\\mathrm dx}\\frac{\\partial f}{\\partial y^\\prime} \\right)+\\frac{\\partial f}{\\partial x}\r\n &=\\frac{\\partial f}{\\partial x}\r\n\\end{align*}\r\nIf $y$ satisfies the Euler-Lagrange Equation.\r\nSo if $f$ does not depend explicitly on $x$, then the above indicates that\r\n$$f-y^\\prime\\frac{\\partial f}{\\partial y^\\prime}=\\text{const.}$$\r\n\\begin{example}[The Brachistochrone Problem]\r\n Consider the functional in the Brachistochrone Problem we defined before, with the initial point assumed to be the origin:\r\n $$F[y]=\\frac{1}{\\sqrt{2g}}\\int_0^\\beta\\frac{\\sqrt{1+(y^\\prime)^2}}{\\sqrt{-y}}\\,\\mathrm dx$$\r\n As $f$ in this case does not depend explicitly on $x$, we know that\r\n $$\\frac{\\sqrt{1+(y^\\prime)^2}}{\\sqrt{-y}}-y^\\prime\\frac{y^\\prime}{\\sqrt{1+(y^\\prime)^2}\\sqrt{-y}}=\\text{const.}=K$$\r\n So\r\n $$y^\\prime=\\pm\\frac{\\sqrt{1+K^2y}}{K\\sqrt{-y}}\\implies x=\\pm K\\int\\frac{\\sqrt{-y}}{\\sqrt{1+K^2y}}\\,\\mathrm dy$$\r\n Set $y=-K^{-2}\\sin^2(\\theta/2)$, so $\\mathrm dy=-K^{-2}\\sin(\\theta/2)\\cos(\\theta/2)$, so\r\n \\begin{align*}\r\n x&=\\pm K\\int(-1)\\frac{1}{K^3}\\frac{\\sin^2(\\theta/2)\\cos(\\theta/2)}{\\sqrt{1-\\sin^2(\\theta/2)}}\\,\\mathrm d\\theta\\\\\r\n &=\\mp\\frac{1}{2K^2}\\int(1-\\cos\\theta)\\,\\mathrm d\\theta\\\\\r\n &=\\mp\\frac{1}{2K^2}(\\theta-\\sin\\theta)+C\r\n \\end{align*}\r\n where $C$ is a constant.\r\n By our initial condition, $\\theta(0)=0$, so $C=0$.\r\n Substitute $\\theta$ for both $x,y$ and we get the parameterised equations\r\n $$\\begin{cases}\r\n x=(\\theta-\\sin\\theta)/(2K^2)\\\\\r\n y=-K^{-2}\\sin^2(\\theta/2)\r\n \\end{cases}$$\r\n which is the equation of a cycloid, i.e. the path traversed by a fixed point on a wheel which rolls on the $x$-axis.\r\n Hence, the Brachistochrone is a cycloid.\r\n\\end{example}\r\n\\subsection{Fermat's Principle}\r\nFermat's Principle postulates that light (or sound) travels along paths between two points that are stationary points of the time variation.\r\n\\footnote{In its original form, however, it said that light travels the path that requires the least time, which is not necessarily true in all cases.}\r\nSuppose the light way is described by $y=y(x)$, then the time functional is\r\n$$F[y]=\\int\\frac{\\mathrm dl}{c}=\\int_\\alpha^\\beta\\frac{\\sqrt{1+(y^\\prime)^2}}{c(x,y)}\\,\\mathrm dx$$\r\nwhere $c$ is the speed of light in the medium.\\\\\r\nFirst assume that $c=c(x)$ does not depend on $y$, then $f$ does not explicitly depend on $y$, in which case we have the first integral\r\n$$\\frac{y^\\prime}{\\sqrt{1+(y^\\prime)^2}c(x)}=\\frac{\\partial f}{\\partial y^\\prime}=\\text{const.}$$\r\nSuppose the light ray has an initial incident angle of $\\theta_1$ upwards, then $\\tan\\theta_1=y^\\prime(\\alpha)$.\r\nLet $\\tan\\theta=y^\\prime$ in general, then the above first integral indicates that $\\sin\\theta/c(x)$ is constant.\r\nThis is Snell's Law.\r\nIf $c$ is increasing, then the path is concave and if $c$ is decreasing it is convex.\r\nAlso, if the light goes through a barrier, to the left of which $c$ is constant at $c_F$ and to the right of which $c_S$ with $c_S 0) s_{\\it f} + \\sum_{c \\in a_{\\it f}} {\\it dist}({\\it f},c) \\right) \\\\\n\\mbox{subject to:} & \\\\\n & \\displaystyle \\sum_{c \\in a_{\\it f}} d_c \\leq cap_{\\it f} \\;\\;\\; ({\\it f} \\in N)\\\\\n & \\displaystyle \\sum_{{\\it f} \\in N} (c \\in a_{\\it f}) = 1 \\;\\;\\; (c \\in M) \n\\end{array}\n$$\n\n\\clearpage\n\\section{Data Format Specification}\n\nThe input consists of $|N| + |M| + 1$ lines. The first line contains two numbers, $|N|$ followed by $|M|$.\nThe first line is followed by $|N|$ lines, where each line encodes the facility's setup cost $s_{\\it f}$, capacity $cap_{\\it f}$, and the location $x_{\\it f}, y_{\\it f}$. The remaining $|M|$ lines capture the customer information, where each line encodes the customer's demand, $d_c$, and location $x_c, y_c$.\n\n\\vspace{0.2cm}\n\\noindent\nInput Format\n\\vspace{-0.2cm}\n\\begin{Verbatim}[frame=single]\n|N| |M|\ns_0 cap_0 x_0 y_0\ns_1 cap_1 x_1 y_1\n...\ns_|N|-1 cap_|N|-1 x_|N|-1 y_|N|-1\nd_|N| x_|N| y_|N|\nd_|N|+1 x_|N|+1 y_|N|+1\n...\nd_|N|+|M|-1 x_|N|+|M|-1 y_|N|+|M|-1\n\\end{Verbatim}\n%\nThe output has two lines. The first line contains two values: $obj$ and $opt$. $obj$ is the cost of the customer to facility assignment (i.e. the objective value) as a real number. $opt$ should be $1$ if your algorithm proved optimality and $0$ otherwise. The next line is a list of $|M|$ values in $N$ -- this is the mapping of customers to facilities.\n\n\\vspace{0.2cm}\n\\noindent\nOutput Format\n\\vspace{-0.2cm}\n\\begin{Verbatim}[frame=single]\nobj opt\nc_0 c_1 c_2 ... c_|M|-1\n\\end{Verbatim}\n%\n%It is essential that the value order in the solution output matches the value order of the input. Otherwise the grader will misinterpret the output.\n\n\\clearpage\n\\paragraph{Examples}\n%\\vspace{0.1cm}\n%\\noindent\n%(based on Figure \\ref{fig:graph})\n\n\\vspace{0.2cm}\n\\noindent\nInput Example\n\\vspace{-0.2cm}\n\\begin{Verbatim}[frame=single]\n 3 4 \n 100 100 1065.0 1065.0\n 100 100 1062.0 1062.0 \n 100 500 0.0 0.0 \n 50 1397.0 1397.0\n 50 1398.0 1398.0 \n 75 1399.0 1399.0 \n 75 586.0 586.0\n\\end{Verbatim}\n\n\\vspace{0.2cm}\n\\noindent\nOutput Example\n\\vspace{-0.2cm}\n\\begin{Verbatim}[frame=single]\n2550.013 0\n1 1 0 2\n\\end{Verbatim}\n%\nThis output represents the assignment of customers to facilities, $a_0 = \\{2\\}, a_1 = \\{0,1\\}, a_2 = \\{3\\}$. That is, customers 0 and 1 are assigned to facility 1, customer 2 is assigned to facility 0, and customers 3 is assigned to facility 2.\n\n\\section{Instructions}\n\n\\input{instructions.tex}\n\n%We use \\texttt{stdout} for output.\n%Output to other stream will be ignored (you may want to send runtime information to \\texttt{stderr}). Your submission will be tested on a department linux machine. If your algorithm is a standalone program, please name it \\texttt{nr},\n%otherwise, please specify the compilation procedure,\n%it is appreciated if you also provide a script that follows the above format to run the program.\n\n\\paragraph{Resources}\nYou will find several facility location problem instances in the \\texttt{data} directory provided with the handout.\n\n%An example output file, \\texttt{blabla.out}, is also provided.\n\n%\\section{Remarks}\n\n\\input{handin.tex}\n\n\\input{grading.tex}\n\n\\input{collaboration.tex}\n\n%\\paragraph{Questions} Please contact the class GTA Carleton (cjc@cs.brown.edu).\n\n\\input{warnings.tex}\n\n%\\paragraph{Hint} \n%The optimal value for \\texttt{data/gc\\_1000\\_5} is near $85$.\n \n\\input{techReqs.tex}\n\n\\end{document}\n\n\n\n\n", "meta": {"hexsha": "bfb979b2cd9ea340c61d60068ecb65f5371746de", "size": 6390, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "handouts/facility.tex", "max_stars_repo_name": "mike715/assignment", "max_stars_repo_head_hexsha": "f69378420ce2bb845abaef0f448eab303aa7a7e7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 101, "max_stars_repo_stars_event_min_datetime": "2016-08-08T05:41:27.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-09T15:04:42.000Z", "max_issues_repo_path": "handouts/facility.tex", "max_issues_repo_name": "sthagen/assignment", "max_issues_repo_head_hexsha": "57d18b188177269c8fe07f3d9bef416720c7b465", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 44, "max_issues_repo_issues_event_min_datetime": "2016-08-07T20:57:56.000Z", "max_issues_repo_issues_event_max_datetime": "2022-02-01T08:07:50.000Z", "max_forks_repo_path": "handouts/facility.tex", "max_forks_repo_name": "sthagen/assignment", "max_forks_repo_head_hexsha": "57d18b188177269c8fe07f3d9bef416720c7b465", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 88, "max_forks_repo_forks_event_min_datetime": "2016-10-05T23:38:45.000Z", "max_forks_repo_forks_event_max_datetime": "2022-02-28T09:28:46.000Z", "avg_line_length": 37.5882352941, "max_line_length": 948, "alphanum_fraction": 0.6968701095, "num_tokens": 2029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389817407017, "lm_q2_score": 0.8198933425148214, "lm_q1q2_score": 0.6117543837199892}} {"text": "\\section{Case Study: Rössler System}\n\nThe well known Rössler system is given by the equations\n\\[\n\t\\begin{array}{rcl}\n\t\t\\dot x &=& -y - z \\\\\n\t\t\\dot y &=& x + a \\cdot y \\\\\n\t\t\\dot z &=& b + z \\cdot (x - c) \\\\\n\t\\end{array}\n\\]\nwith parameters $a$, $b$, and $c$. For this example, $a$ and $b$\nare fixed to $0.1$, while we are varying $c$.\n\nThe system has several periodic solutions for each value of $c$ with different\nperiodicities, though only one is stable at a time. We are interested in the origin and\nbranching of those solutions and, thus, drawing a bifurcation diagram using the map\n\\[\n\t\\mathbf y \\mapsto \\{ \\|\\mathbf y(t)\\| \\ | \\ t \\in [0, 2\\pi), \\ \\mathbf y_1(t) = 0, \\ \\mathbf y'_1(t) > 0 \\} \\ .\n\\]\nHere, $f$ denotes a single periodic solution. Note, that we use only the approximation\ngiven by Galerkin's method.\n\nThe exploration of the bifurcations with the given tools follows roughly these steps:\n\\begin{enumerate}\n\t\\item As a starting point, search for a value of $c$ such that the system has a\n\t\tstable, periodic solution and use the method described in\n\t\t\\autoref{sec:initial} to find it. 4000 iterations in the transient part with step\n\t\tsize of $0.01$, 120 generated intersections with the $(x=0)$-plane and tests for\n\t\tat most $30$ periods were sufficient for all initial solutions of the Rössler\n\t\tsystem. We started at $c=4$ with $64$ samples.\n\t\\item Trace out the branch just by following the newly found solution in both directions\n\t\tusing the predictor-corrector continuation method (\\autoref{sec:cont}). The\n\t\tparameters $\\kappa=0.4$, $\\delta=3.0$, and $\\alpha=10.0$ are a decent choice for\n\t\tthe whole diagram as a good tradeoff between performance and accuracy.\n\t\\item It is possible to detect a pitchfork bifurcation by doubling up the periodicity\n\t\tartificially. Using the described predictor-corrector method, the value\n\t\tof $c$ will eventually converge to the bifurcation point, i.e., a point with a\n\t\tsingular tangent. Since the method is now stuck, reaching the doubly-periodic\n\t\tsolution can be done by solving a slightly perturbed problem for a short time.\n\t\tA short perturbation of strength $0.2$ was enough to reach the doubly-periodic\n\t\tsolution. Also, it might be necessary to increase the sample size due to increased\n\t\tcomplexity in the descending branch.\n\t\\item Remove the perturbation and continue tracing out the bifurcation, i.e. start\n\t\tover in 1 or 2.\n\\end{enumerate}\n\n\\begin{figure}[!ht]\n\t\\centering\n\t\\subfloat[][]{\\includegraphics[width=.4\\textwidth]{img/roessler1a.pdf}}\\quad\n\t\\subfloat[][]{\\includegraphics[width=.4\\textwidth]{img/roessler1b.pdf}}\\\\\n\t\\subfloat[][]{\\includegraphics[width=.4\\textwidth]{img/roessler1c.pdf}}\\quad\n\t\\subfloat[][]{\\includegraphics[width=.4\\textwidth]{img/roessler1d.pdf}}\n\t\\caption{Tracing out the branch of solutions beginning with the solution for\n\t$c=4$ (a). The bifurcation point in $c \\approx 5.376$ is avoided by using a permutation (b),\n\tand the two-periodic solutions can be traced out after removing the inaccurate values\n\tfrom before (c). Eventually, we reach (d), a widely traced out bifurcation diagram.\n\tNote the solutions with periodicity three and descendants, that do not originate in\n\tthe periodicity-one branch.}\n\t\\label{fig:reosslerseq}\n\\end{figure}\n\nSee figure \\ref{fig:reosslerseq} for a visualization of this process, and figures \\ref{fig:roesslerfull}, \\ref{fig:roesslercut} for a depiction of the results.\n\n%\\begin{figure}[!ht]\n\t%\\centering\n\t%\\includegraphics[width=1\\textwidth]{img/roessler2a.pdf}\n\t%\\caption{The bifurcation diagram of the Rössler system for varying $c$. The\n\t\t%even-periodic solutions are colored red, the odd-periodic ones green.}\n%\\end{figure}\n%\n%\n%\\begin{figure}[!ht]\n\t%\\centering\n\t%\\includegraphics[width=1\\textwidth]{img/roessler2b.pdf}\n\t%\\caption{All found solutions for $c=\\dots$ plotted into one diagram.\n\t%Observe how the odd-periodic and even-periodic solutions dodge since they do not\n\t%originate from each other.}\n%\\end{figure}\n\n\\newgeometry{top=0cm}\n\n\\begin{figure}\n\\centering\\makebox[0pt]{\\rotatebox{90}{\\includegraphics{img/roessleroverview.png}}}\n\\caption{An overview of the Rössler bifurcation diagram.}\n\\label{fig:roesslerfull}\n\\end{figure}\n\n\\begin{figure}\n\\centering\\makebox[0pt]{\\rotatebox{0}{\\includegraphics{img/roesslercut20.png}}}\n\\caption{The periodic orbits for $c=20$, together with an underlying trajectory obained through forward intergration (grey) in phase-space. The colors of the periodic solutions match the bifurcation diagram.}\n\\label{fig:roesslercut}\n\\end{figure}\n\n\\restoregeometry\n", "meta": {"hexsha": "97f9dc5dc5f4df978c701e911fcfd84fbfe27af3", "size": 4539, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "doctheory/roessler.tex", "max_stars_repo_name": "285714/ncm", "max_stars_repo_head_hexsha": "fcf289c7ef5f8500ebcb238e36c6a7ee9e054147", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "doctheory/roessler.tex", "max_issues_repo_name": "285714/ncm", "max_issues_repo_head_hexsha": "fcf289c7ef5f8500ebcb238e36c6a7ee9e054147", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "doctheory/roessler.tex", "max_forks_repo_name": "285714/ncm", "max_forks_repo_head_hexsha": "fcf289c7ef5f8500ebcb238e36c6a7ee9e054147", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 47.7789473684, "max_line_length": 208, "alphanum_fraction": 0.7473011677, "num_tokens": 1314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7461389817407017, "lm_q2_score": 0.8198933271118221, "lm_q1q2_score": 0.611754372227211}} {"text": "% !TeX root = ./corona_contact_tracing.tex\n% chktex-file 46\n% !TeX spellcheck = en-GB\n% !TeX encoding = utf8\n\n\\subsection{Probabilistic Health Prediction}\n\nWe model partially infected populations as a graph, where each individual (interchangeably called agent) is a node. Edges of this graph model contacts of two agents. We use graph convolution, as explained below, to update our assumptions on the current health state of all individuals.\n\nThe definition of a graph convolution~\\cite{Kipf2017SemiSupervisedCW} for here is\n\n\\begin{equation}\n\t\\label{eq:graph_convolution}\n\th_{v_i}^{(l+1)} = \\sum_{j\\in A(i)} h_{v_j}^{(l)}\n\\end{equation}\n\nwith feature vector $h$ of node $i$ from iteration $(l)$ to $(l+1)$. This formulation is equivalent to the matrix formulation $h^{(l+1)} = A h^{(l)}$ with $A$ as adjacency matrix as shown in \\cref{sec:consistency}. We consider here adjacency matrices without diagonal elements.\n\n\\begin{itemize}\n \\item Each agent $i$ is modeled by $D$ features $h_{v_i} \\in \\mathbb{R}^D$. Therefore, the feature matrix, $h^{(l)}$, consists of all agents' features at time $(l)$ and is thereby of dimension $N\\times D$ where there are $N$ agents in the population and each agent is described by $D$ features. A three dimensional feature space is used, $D=3$, modeling the three possible health states. The unit vectors of this space are interpreted as following:\n\t\\begin{itemize}\n\t\t\\item $\\vec{e}_0$: susceptible state\n\t\t\\item $\\vec{e}_1$: infected state\n\t\t\\item $\\vec{e}_2$: recovered state\n\t\\end{itemize}\n\tA uniform distribution over these possible states expresses complete uncertainty of the health state of an agent.\n\\end{itemize}\n\nTo accurately model the disease, there is graph contribution and an individual health contribution.\n\n\\subsubsection{Explanation of the graph contribution term}\nThe graph contribution models how infected agents spread the disease through contacts with susceptible agents.\n\n\\begin{equation}\n h_{v_i, m}^{(l+1)}\n\t=\n\t\\underbrace{\n\t\t\\sum_k \\textcolor{red}{\\frac{\\hat{A}_{v_i, k}^{(l)}}{\\sum_j \\hat{A}_{v_i, j}^{(l)}}} h_{k, m}^{(l)} \\textcolor{blue}{\\delta_{m, e_I}}\n\t}_{\\text{Graph}}\n\\end{equation}\n\n\\begin{itemize}\n \\item A sum over all agents' features $h_{k,m}^{(l)}$ is weighted by the normalised infection-adjusted graph connections as shown in red.\n \\item The Kronecker delta, as shown in blue, ensures that only the I feature is added as this is the only one that matters during social contacts between agents.\n \\item The infection-adjusted adjacency matrix $\\hat{A}$ is constructed from $A$ and $I$ which are the regular continuous adjacency matrix and the infection matrix, respectively. These three quantities are explained in the following:\n \\begin{itemize}\n \\item The adjacency matrix $A$ is time dependent, $A^{(l)}$, and inferred from data. In our use case, $A_{ij} = \\frac{1}{dist(v_i, v_j)+\\epsilon}$, hence $A_{ij}$ is large when persons $i$ and $j$ have been in contact. $\\epsilon$ serves as regularization for small distances.\n \\item The infection matrix is constructed as\n \\begin{equation}\n I =\n \\begin{pmatrix}\n 0 & 0 & 0 \\\\\n \\beta & 0 & \\alpha \\\\\n 0 & 0 & 0\n \\end{pmatrix}\n =\n (I_{ij})_{i,j}\n \\end{equation}\n with $i$ as the index of the host state and $j$ is the index of the contact person state. The states that we consider here are ordered as follows: susceptible, infected, recovered. $\\beta$ denotes the probability of infection after contact (also known as attack rate). $\\alpha$ models the probability of being reinfected, which we assume to be zero ($\\alpha=0$) based upon current medical \\todo{cite!}research.\n \\item $\\hat{A}$, with $\\hat{A}_{ij}\\in [0, 1]$, is the adjacency matrix that takes the infection interactions into account and is computed as follows\n \\begin{equation}\n \\hat{A}_{ij} = A_{ij}\\cdot \\frac{ h_{v_1}^T I h_{v_2} + h_{v_2}^T I h_{v_1} }{\\beta}.\n \\end{equation}\n The weighted scalar product of the health states of agents $i$ and $j$ is used to evaluate whether the edge is relevant for the infection dynamics. Only when an infected person and a susceptible have contact, the edge $A_{ij}$ should be considered, otherwise it should be dropped.\tThe sum in the denominator comes from the fact that both, agent $i$ and $j$, can act as host during a contact. The division by $\\beta$ normalises the factor to one to ensure $\\hat{A}_{ij} \\in [0, 1]$. Since $I$ is not symmetric, $p_a$ is a proper normalization because the sum is in $\\{0, p_a\\}$. Note that the fraction has the desired properties for pure $S$-, $I$- and $R$-persons.\n \\end{itemize}\n\\end{itemize}\n\n\n\\subsubsection{Temporal}\n% \\item Explanation of the temporal term:\nThe transition of a persons' health state $h_{v_i}^{(l)}$ is determined by the following assumptions:\n\\begin{itemize}\n\t\\item A susceptible person always stays susceptible\n\t\\item An infected person has a probability $\\gamma$, called recovery rate, to recover. The remaining probability $1-\\gamma$ denotes that the person stays sick.\n\t\\item A recovered person could have a probability to be re-infected, but we assume this to be zero. Thus a recovered person always stays recovered.\n\\end{itemize}\nThus the temporal transition matrix $T$ is:\n\\begin{equation}\n\tT = \n\t\\begin{pmatrix}\n\t\t1 & 0 & 0 \\\\\n\t\t0 & 1-\\gamma & \\gamma \\\\\n\t\t0 & 0 & 1 \\\\\n\t\\end{pmatrix}\n\\end{equation}\nThe temporal update rule based on the health status thus becomes:\n\\begin{equation}\n\tH^{(l+1)} = H^{(l)} T\n\\end{equation}\n\n\\subsubsection{Joint Model}\n\nOur main propagation rule is based on the definition of a graph convolution shown in Eq.~\\eqref{eq:graph_convolution} and reads as follows in component notation\n\\begin{equation}\n\th_{v_i, m}^{(l+1)}\n\t=\n\t\\underbrace{\n\t\t\\sum_k \\textcolor{red}{\\frac{\\hat{A}_{v_i, k}^{(l)}}{\\sum_j \\hat{A}_{v_i, j}^{(l)}}} h_{k, m}^{(l)} \\textcolor{blue}{\\delta_{m, e_I}}\n\t}_{\\text{Graph}}\n\t+\n\t\\underbrace{\n\t\t{(h_{v_i}^{(l)}\\cdot T)}_m\n\t}_{\\text{Temporal}}\n\\end{equation}\nwith $m$ being the index of the health state.\n\nThis propagation rule is based on the following aspects:\n\n\\begin{itemize}\n\t\\item The propagation consists of two parts, first the graph contribution and second the temporal contribution. While the former captures the dynamics of infections based on the social contacts between agents, the former ensures that an infected agent heals over time and becomes resistant against the Corona virus.\n\\end{itemize}", "meta": {"hexsha": "0da4d6a82d98bc85bcaef8c486652d9f09cc9282", "size": 6585, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/probabilistic_health_prediction.tex", "max_stars_repo_name": "PellelNitram/corona_contact_tracing", "max_stars_repo_head_hexsha": "df5a6ba18b84397b721893fb5eb89889dc82ab2f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2020-03-21T20:44:54.000Z", "max_stars_repo_stars_event_max_datetime": "2020-05-14T05:32:49.000Z", "max_issues_repo_path": "docs/probabilistic_health_prediction.tex", "max_issues_repo_name": "PellelNitram/corona_contact_tracing", "max_issues_repo_head_hexsha": "df5a6ba18b84397b721893fb5eb89889dc82ab2f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "docs/probabilistic_health_prediction.tex", "max_forks_repo_name": "PellelNitram/corona_contact_tracing", "max_forks_repo_head_hexsha": "df5a6ba18b84397b721893fb5eb89889dc82ab2f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2020-03-22T15:37:41.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-31T10:11:24.000Z", "avg_line_length": 59.3243243243, "max_line_length": 672, "alphanum_fraction": 0.7053910402, "num_tokens": 1829, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8918110569397306, "lm_q2_score": 0.6859494421679929, "lm_q1q2_score": 0.6117372970270564}} {"text": "\\section{Differentiation}\r\nUntil further notice, we will just study differentiation of a function $f:I\\to\\mathbb R$ where $I$ is an interval.\r\n\\subsection{The Derivative}\r\n\\begin{definition}\r\n If $a\\in I$ and we have\r\n $$\\lim_{x\\to a}\\frac{f(x)-f(a)}{x-a}$$\r\n exists and equals $c\\in\\mathbb R$, we say $f$ is differentiable at $a$ and the derivative is $f^\\prime(a)=c$.\r\n Moreover, if $f$ is differentiable at all $a\\in I$, we say $f$ is differentiable.\r\n\\end{definition}\r\nGeometrically $f^\\prime(a)$ is the slope of the tangent line of $f$ at $a$.\\\\\r\nAnother (formal) way to put this is to observe that the definition is essentially $\\forall \\epsilon>0,\\exists\\delta>0$,\r\n$$|x-a|<\\delta\\implies\\left|\\frac{f(x)-f(a)}{x-a}-c\\right|<\\epsilon\\implies |f(x)-f(a)-c(x-a)|<\\epsilon|x-a|$$\r\nAnother way we can write is\r\n$$f^\\prime(a)=\\lim_{h\\to 0}\\frac{f(a+h)-f(a)}{h}$$\r\nwhere $h\\in I-a$.\r\n\\begin{example}\r\n 1. The constant function is differentiable everywhere and has zero derivative everywhere.\\\\\r\n 2. The identity function is differentiable everywhere and has derivative $1$.\\\\\r\n 3. $f:(0,\\infty)\\to\\mathbb R$ by $x\\mapsto 1/x$ is differentiable and has derivative $f^\\prime(a)=-a^{-2}$.\\\\\r\n 4. (non-example) The function $x\\mapsto |x|$ is not differentiable at $0$.\r\n\\end{example}\r\n\\begin{lemma}\r\n Suppose $f$ is differentiable at $a$ with derivative $f^\\prime(a)$ iff\r\n $$f(a+h)=f(a)+f^\\prime(a)h+\\alpha(h)h$$\r\n where $\\alpha(0)=0$ and $\\alpha$ is continuous at $0$.\r\n\\end{lemma}\r\nNote that given this formula, then $\\alpha(h)$ is completely deterined by $f$ and $f^\\prime(a)$.\r\n\\begin{proof}\r\n Obvious.\r\n\\end{proof}\r\n\\begin{corollary}\r\n $f$ is differentiable at $a$ implies $f$ is continuous at $a$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Follows directly.\r\n\\end{proof}\r\n\\begin{proposition}\r\n If $f,g:I\\to\\mathbb R$ are differentiable at some $a\\in I$, so are $f+g,fg$ with derivatives $f^\\prime(a)+g^\\prime(a),f^\\prime(a)g(a)+f(a)g^\\prime(a)$ respectively.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\n\\begin{theorem}[Chain Rule]\r\n Suppose $f:I_1\\to I_2$ is differentiable at $a\\in I_1$ and $g:I_2\\to\\mathbb R$ is differentiable at $f(a)$, then $(g\\circ f)^\\prime(a)=g^\\prime(f(a))f^\\prime(a)$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Let $b=f(a)$, then by our conditions,\r\n $$f(a+h)=f(a)+f^\\prime(a)h+\\alpha(h)h,g(b+k)=g(b)+g^\\prime(b)k+\\beta(k)k$$\r\n And $\\alpha(0)=\\beta(0)=0$ and they are both continuous at $0$.\r\n Take $k=f(a+h)-f(a)=f^\\prime(a)h+\\alpha(h)h$, so\r\n \\begin{align*}\r\n (g\\circ f)(a)&=g(b+k)\\\\\r\n &=g(b)+g^\\prime(b)k+\\beta(k)k\\\\\r\n &=g(b)+g^\\prime(b)(f^\\prime(a)h+\\alpha(h)h)+\\beta(f^\\prime(a)h+\\alpha(h)h)(f^\\prime(a)h+\\alpha(h)h)\\\\\r\n &=(g\\circ f)(a)+g^\\prime(f(a))f^\\prime(a)h+\\phi(h)h\r\n \\end{align*}\r\n Where $\\phi(h)=g^\\prime(b)\\alpha(h)+\\beta(f^\\prime(a)h+\\alpha(h)h)(f^\\prime(a)+\\alpha(h))$ which is $0$ at $0$ and is continuous at $0$.\r\n The chain rule follows.\r\n\\end{proof}\r\n\\begin{corollary}\r\n Suppose $f,g:I\\to\\mathbb R$ are differentiable at $a$ and $g(a)\\neq 0$, then $f/g$ is differentiable at $a$ and $(f/g)^\\prime=(f^\\prime g-fg^\\prime)/g^2$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate from the chain rule.\r\n\\end{proof}\r\n\\begin{corollary}\r\n The polynomial $p(x)=a_0+a_1x+\\cdots+a_nx^n$ is differentiable at all $x\\in\\mathbb R$, and if $p,q$ are polynomials, $p/q$ is differentiable at $a$ when $q(a)\\neq 0$.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Trivial.\r\n\\end{proof}\r\n\\subsection{Mean Value Theorem}\r\nSuppose we have some $f:I\\to\\mathbb R$.\r\n\\begin{definition}\r\n We say some $c\\in I$ is a global maximum of $f$ if $\\forall x\\in I,f(x)\\le f(c)$ and a global minimum if $\\forall x\\in I,f(x)\\ge f(c)$.\r\n\\end{definition}\r\nWe know that if $I$ is a closed bounded interval, then $f$ has global maxima and minima.\r\n\\begin{definition}\r\n Some $c\\in I$ is called a local maximum of $f$ if $\\exists\\epsilon>0,\\forall x\\in (c-\\epsilon,c+\\epsilon)\\cap I,f(x)\\le f(c)$.\r\n Similarly it is a local minimum if $\\exists\\epsilon>0,\\forall x\\in (c-\\epsilon,c+\\epsilon)\\cap I,f(x)\\ge f(c)$.\r\n\\end{definition}\r\n\\begin{definition}\r\n $c\\in I$ is called an interior point of $I$ if $\\exists \\epsilon>0,(c-\\epsilon,c+\\epsilon)\\subset I$.\r\n\\end{definition}\r\n\\begin{proposition}\r\n Suppose $f:I\\to\\mathbb R$ and $c$ is an interior point of $I$ where $f$ attains local maximum or local minimum at $c$ and $f$ is differentiable at $c$, then $f^\\prime(c)=0$.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Suffice to prove the local maximum case.\r\n Since $c$ is an interior point of $I$, we can choose $\\epsilon'>0,(c-\\epsilon',c+\\epsilon')\\subset I$.\r\n As $c$ is a local maximum, then there is some $\\epsilon$ such that $0<\\epsilon\\le\\epsilon'$ and that $\\forall x\\in (c-\\epsilon,c+\\epsilon),f(x)\\le f(c)$.\r\n We can easily find sequences $(x_n^-)\\in (c-\\epsilon,c),(x_n^+)\\in (c,c+\\epsilon)$ both converging to $c$.\r\n Let $g:(c-\\epsilon,c+\\epsilon)\\setminus\\{c\\}\\to\\mathbb R$ defined by $g(x)=(f(x)-f(c))/(x-c)$, then\r\n \\begin{align*}\r\n 0&\\le\\lim_{n\\to\\infty}\\frac{f(x_n^-)-f(c)}{x_n^--c}\\\\\r\n &=\\lim_{x\\to c^-}g(x)=\\lim_{x\\to c}g(x)=\\lim_{x\\to c^+}g(x)\\\\\r\n &=\\lim_{n\\to\\infty}\\frac{f(x_n^+)-f(c)}{x_n^+-c} \\le 0\r\n \\end{align*}\r\n Hence $f^\\prime(c)=\\lim_{x\\to c}g(x)=0$.\r\n\\end{proof}\r\n\\begin{theorem}[Rolle's Theorem]\r\n Suppose $f:[a,b]\\to\\mathbb R$ is continuous and is differentiable in $(a,b)$ and $f(a)=f(b)=0$, then there is some $c\\in (a,b)$, $f^\\prime(c)=0$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n It is obvious if $f\\equiv 0$, otherwise WLOG $f$ attains some positive value.\r\n Then let $c$ be a global maximum of $f$, then $c$ is a local maximum and an interior point of $f$ (since $f$ is zero on endpoints but we have assumed that $f$ attains some positive value), then $c\\in (a,b)$ and we have $f^\\prime(c)=0$ by the preceding proposition.\r\n\\end{proof}\r\n\\begin{theorem}[Mean Value Theorem]\r\n If $f:[a,b]\\to\\mathbb R$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is some $c\\in (a,b)$ with $f^\\prime(c)=(f(b)-f(a))/(b-a)$.\r\n\\end{theorem}\r\n\\begin{proof}\r\n Let\r\n $$p(x)=\\frac{1}{b-a}((x-a)f(b)-(x-b)f(a))$$\r\n which although looks dreadful is just a linear polynomial.\r\n Note that $p(a)=f(a),p(b)=f(b)$, so it is just a line joining the endpoints.\r\n Then $g(x)=f(x)-p(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, so by Rolle's Theorem, there is some $c\\in (a,b)$ sucb that\r\n $$0=g^\\prime(c)=f^\\prime(c)-\\frac{f(b)-f(a)}{b-a}\\implies f^\\prime(c)=\\frac{f(b)-f(a)}{b-a}$$\r\n As desired.\r\n\\end{proof}\r\n\\begin{proposition}\r\n Suppose $f:[a,b]\\to\\mathbb R$ is continuous and differentiable in $(a,b)$.\\\\\r\n 1. $f^\\prime\\equiv 0\\iff f$ is constant.\\\\\r\n 2. $f^\\prime\\ge 0\\iff f$ is increasing.\\\\\r\n 3. $f^\\prime>0\\implies f$ is strictly increasing.\r\n\\end{proposition}\r\n\\begin{proof}\r\n Immediate from Mean Value Theorem.\r\n\\end{proof}\r\nNote that the converse of the third statement is false in general.\r\nE.g, $f(x)=x^3$.\\\\\r\nNote that all these depends on the least upper bound property since we used the Mean Value Theorem which depends on Rolle's Theorem which depends on maximum value theorem which depends on Bolzano-Weierstrass Theorem which depends on the Monotone Sequence Theorem which depends on the least-upper-bound property.\r\n\\subsubsection{Inverse Function Theorem}\r\n\\begin{lemma}\r\n If $f:[a,b]\\to\\mathbb R$ satisfies $f^\\prime(x)>0$ for all $x\\in [a,b]$, then $f:[a,b]\\to[f(a),f(b)]$ is a bijection.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Injectivity is obvious by Mean Value Theorem.\r\n For surjectivity we can just exploit Intermediate Value Theorem.\r\n\\end{proof}\r\n\\begin{lemma}\r\n Let $I$ be an interval.\r\n Suppose $f:I\\to f(I)$ is a continuous bijection, then $f^{-1}:f(I)\\to I$ is also continuous.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Just check definition.\r\n\\end{proof}\r\n\\begin{theorem}[Inverse Function Theorem]\r\n Suppose $I\\subset R$ is an interval and $f^\\prime(x)>0$ for any $x\\in I$, then the continuous inverse (exists by the preceding lemmas) $f^{-1}:f(I)\\to I$ is differentiable and $(f^{-1})^\\prime(y)=1/f^\\prime(f^{-1}(y))$.\r\n\\end{theorem}\r\n\\begin{example}\r\n 1. $f:x\\mapsto x^n$ on $(0,\\infty)$ satisfies the conditions, so its inverse $f^{-1}:y\\mapsto y^{1/n}$ has $(f^{-1})^\\prime(y)=(1/n)y^{1/n-1}$.\\\\\r\n 2. Consider $\\tan:(-\\pi/2,\\pi/2)\\to\\mathbb R$, it shall be invertible with differentiable inverse $\\tan^{-1}$, then we can calculate, by the formula, that $(\\tan^{-1})(y)=1/(y^2+1)$\r\n\\end{example}\r\n\\begin{proof}[Proof assuming differentiability]\r\n Assuming $f^{-1}$ is differentiable, then we know that $f(f^{-1}(y))=y$, then by the chain rule, we get $f^\\prime(f^{-1}(y))(f^{-1})^\\prime(y)=1$, rearrange to give the formula.\r\n\\end{proof}\r\n\\begin{proof}[Actual proof]\r\n Fix $b\\in f(I)$ and let $a=f^{-1}(b)$.\r\n Suppose $(y_n)$ is a sequence in $f(I)\\setminus \\{b\\}$ with $y_n\\to b$, and let $x_n=f^{-1}(y_n)$.\r\n We know that $f$ is differentiable hence continuous, so $f^{-1}$ is continuous by the preceding lemma.\r\n Therefore $x_n=f^{-1}(y_n)\\to f^{-1}(b)=a$.\r\n Since we know that\r\n $$\\lim_{x\\to a}\\frac{f(x)-f(a)}{x-a}=f^\\prime(a)$$\r\n We have\r\n $$\\frac{f(x_n)-f(a)}{x_n-a}\\to f^\\prime(a)$$\r\n By hypothesis, $f^\\prime(a)\\neq 0$, so\r\n $$\\frac{f^{-1}(y_n)-f^{-1}(b)}{y_n-b}=\\frac{x_n-a}{f(x_n)-f(a)}\\to\\frac{1}{f^\\prime(a)}=\\frac{1}{f^\\prime(f^{-1}(b))}$$\r\n which is true for every sequence $y_n\\to b$, so the theorem is proved.\r\n\\end{proof}\r\n\\subsubsection{L'H\\^opital's Rule}\r\n\\begin{proposition}[L'H\\^opital's Rule]\r\n Suppose $f,g:I\\to\\mathbb R$ are differentiable with $f(a)=g(a)=0$, and $\\exists r>0, g(x)\\neq 0\\neq g^\\prime(x)$ for $0<|x-a|a$.\r\n Consider $h(t)=f(t)g(x)-f(x)g(t),t\\in [a,x]$.\r\n Since $f,g$ are differentiable and we have $h(a)=h(a)=0$, so by Rolle's Theorem, there is some $c\\in (a,x)$ with $f^\\prime(c)g(x)-f(x)g^\\prime(c)=h^\\prime(c)=0$.\r\n Therefore\r\n $$\\frac{f^\\prime(c)}{g^\\prime(c)}=\\frac{f(x)}{g(x)}$$\r\n Same for $x0$, choose $\\delta>0$ such that\r\n $$0<|c-a|<\\delta\\implies\\left|\\frac{f^\\prime(c)}{g^\\prime(c)}-k\\right|<\\epsilon$$\r\n So when $0<|x-a|<\\delta$, one can find $c$ such that $0<|c-a|<|x-a|<\\delta$ and $f^\\prime(c)/g^\\prime(c)=f(x)/g(x)$, which means that\r\n $$\\left|\\frac{f(x)}{g(x)}-k\\right|=\\left|\\frac{f^\\prime(c)}{g^\\prime(c)}-k\\right|<\\epsilon$$\r\n which is as we wanted.\r\n\\end{proof}\r\n\\subsection{Higher Derivatives and Taylor Series}\r\n\\begin{definition}\r\n We say (inductively) that $f:I\\to\\mathbb R$ is $k$ times differentiable with $k^{th}$ derivative $f^{(k)}$ if $f$ is $(k-1)$ times differentiable with differentiable $(k-1)^{th}$ derivative $f^{(k-1)}$ and $f^{(k-1)})^\\prime=f^{(k)}$.\r\n\\end{definition}\r\n\\begin{example}\r\n A polynomial is $k$ times differentiable for any $k\\in\\mathbb N$.\r\n\\end{example}\r\n\\begin{definition}\r\n We say $f$ is $C^k$ if $f$ is $k$ times differentiable and $f^{(k)}$ is continuous.\r\n\\end{definition}\r\n\\begin{example}[Non-example]\r\n Consider\r\n $$f(x)=\\begin{cases}\r\n x^2\\sin(1/x)\\text{, for $x\\neq 0$}\\\\\r\n 0\\text{, for $x=0$}\r\n \\end{cases}$$\r\n So one can verify that $f$ is differentiable but the derivative, namely,\r\n $$f^\\prime(x)=\\begin{cases}\r\n 2x\\sin(1/x)-\\cos(1/x)\\text{, for $x\\neq 0$}\\\\\r\n 0\\text{, for $x=0$}\r\n \\end{cases}$$\r\n is not continuous at $0$.\r\n\\end{example}\r\nNow, given $f:I\\to\\mathbb R$, we want a degree $k$ polynomial that best approximates $f$.\r\nThe answer depends on the meaning of ``best''.\r\nWe can either calculate the total squared difference (we have not defined it and will not care about it), but on the other hand, the theory of differentiation hints that\r\n$$f(x)=f(a)+f^\\prime(a)(x-a)=P_1(x)$$\r\nis the best linear approximation of $f$ around $a$.\r\nWhy ``around $a$''?\r\nsay for example we want to approximate $\\sin x$ over a large interval, then the best approximation intuitively might be a line with slope very close to $0$, which is obviously not the tangent at, e.g., $0$.\r\nAnd we are only interested in the best approximation close to $a$.\\\\\r\nNow back to our best linear approximation.\r\nThe reason we say it is the ``best'' in the sense that $f(x)-P_1(h)=\\alpha(h)h$ with $\\alpha(0)=0$ and $\\alpha$ is continuous at $0$.\r\nWe can also easily know that such a $P_1$ is unique so as for such an $\\alpha$ to exist by working out their coefficients by substitution and differentiating (with appropriate comments on differentiability, of course).\r\nMore generally, for a sufficiently smooth $f$, we can try to approximate $f(a+h)=P_k(a+h)+h^k\\alpha(h)$ where $\\alpha(0)=0$ and $\\alpha$ is continuous at $0$ and $P_k$ is a polynomial.\r\nUsing the same trick, such an $P_k$ is unique.\r\nIn particular, we obtain $P_k^{(r)}=f^{(r)}$ for any $r\\le k$.\r\n\\begin{lemma}\r\n If $a,c_0,c_1,\\ldots,c_k\\in\\mathbb R$, then there is a unique degree $k$ polynomial $P$ with $P^{(i)}(a)=c_i$.\r\n\\end{lemma}\r\n\\begin{proof}\r\n Write out $P(x)=a_0+a_1(x-a)+\\cdots+a_k(x-a)^k$ and differentiating gives $i!a_i=c_i=P^{(i)}(a)$, so the polynomial is uniquely determined.\r\n\\end{proof}\r\n\\begin{definition}\r\n Given $f:I\\to\\mathbb R$ which is $k$ times differentiable and $a$ is an interior point of $I$, then the $k^{th}$ Taylor polynomial of $f$ centered at $a$ is\r\n $$P_k(x)=\\sum_{i=0}^k\\frac{f^{(i)}(a)}{i!}(x-a)^i$$\r\n which is chosen such that $P_k^{(i)}(a)=f^{(i)}(a)$.\r\n\\end{definition}\r\n\\begin{remark}\r\n $P_k(x)$ is the unique polynomial with degree at most $k$ having the property $P_k^{(i)}(a)=f^{(i)}(a)$ by the preceding lemma.\r\n\\end{remark}\r\n\\begin{theorem}[Taylor's Theorem with Remainder]\r\n Suppose $f:I\\to\\mathbb R$ is $(k+1)$ times differentiable, and $[a,x]\\subset I$, then\r\n $$f(x)=\\left(\\sum_{i=0}^k\\frac{f^{(i)}(a)}{i!}(x-a)^i\\right)+\\frac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1}$$\r\n for some $c\\in (a,x)$.\r\n\\end{theorem}\r\nFor $k=0$, this reduces to the mean value theorem.\r\n\\begin{proof}\r\n Fix $x>a$ and define\r\n $$g(t)=f(t)+\\sum_{i=1}^k\\frac{f^{(i)}(t)}{i!}(x-t)^i+\\alpha\\frac{(x-t)^{k+1}}{(k+1)!}$$\r\n where $\\alpha$ is chosen such that $g(a)=f(x)$ (which exists as a solution to a nondegenerate linear equation).\r\n Our goal is to find $c\\in (a,x)$ such that $f^{(k+1)}(c)=\\alpha$.\r\n We have, by definition,\r\n $$f(x)=g(a)=f(a)+\\sum_{i=1}^k\\frac{f^{(i)}(a)}{i!}(x-a)^i+\\alpha\\frac{(x-a)^{k+1}}{(k+1)!}$$\r\n Now $f$ is $k+1$ times differentiable so $g$ is differentiable and $g(a)=f(x)=g(x)$, so by Rolle's Theorem, there is some $c\\in (a,x)$ such that $g^\\prime(c)=0$.\r\n But we have\r\n $$g^\\prime(t)=f^\\prime(t)+\\sum_{i=1}^k\\left(\\frac{f^{(i+1)}(t)}{i!}(x-t)^i-\\frac{f^{(i)}(t)}{(i-1)!}(x-t)^{i-1}\\right)-\\alpha\\frac{(x-t)^k}{k!}$$\r\n The first part of this expression is telescoping, so we can simplify it to\r\n $$g^\\prime(t)=(f^{(k+1)}(t)-\\alpha)\\frac{(x-t)^k}{k!}$$\r\n Now plugging in $t=c$ shows $f^{(k+1)}(c)=\\alpha$, which completes the proof.\r\n\\end{proof}\r\nIf $f$ is infinitely differentiable, one can then form the Taylor series\r\n$$\\sum_{i=0}^\\infty\\frac{f^{(i)}(a)}{i!}(x-a)^i$$\r\nas the formal limit of the Taylor polynomilas $P_k(x)$, which sometimes does converge and equals $f$.\r\nIf we understand the derivatives of $f$ well-enough and that those derivatives do behave nicely, then it is sometimes possible to use Taylor's Theorem to prove that\r\n$$p(x)=\\lim_{n\\to\\infty}P_n(x)=f(x)$$\r\n\\begin{example}\r\n Take $f(x)=\\cos x$ (which, albeit hasn't been properly defined, we shall assume standard facts about), then the Taylor series about $0$ is\r\n $$1-\\frac{x^2}{2!}+\\frac{x^4}{4!}-\\frac{x^6}{6!}+\\cdots=\\sum_{n=0}^\\infty(-1)^n\\frac{x^{2n}}{(2n)!}$$\r\n The series always converges by the ratio test.\r\n Must the series converges to $\\cos x$ then?\r\n The $(2k)^{th}$ partial sum of the series is exactly $P_{2k}(x)$ of $\\cos$ about $0$, then there is $c\\in (0,x)$ with\r\n $$|f(x)-P_{2k}(x)|=\\left|\\frac{f^{2k+1}(c)}{(2k+1)!}x^{2k+1}\\right|\\le\\frac{|x|^{2k+1}}{(2k+1)!}\\to 0$$\r\n as $k\\to\\infty$, so the series does converge to $\\cos x$.\r\n\\end{example}\r\n\\begin{definition}\r\n We say $f:I\\to\\mathbb R$ is $C^\\infty$ (or smooth) if $f$ is $C^k$ for all $k\\in\\mathbb N$.\r\n\\end{definition}\r\nSo for a smooth $f:I\\to\\mathbb R$, one can form the Taylor series (formally)\r\n$$\\sum_{n=0}^\\infty\\frac{f^{(n)}(a)}{n!}(x-a)^n$$\r\nBut even if the series converges for all $x\\in I$, it may not converge to $f(x)$ for $x\\neq a$.\r\nPhilosophically, one cannot fully expect that the behaviour of $f$ near $a$ control its global behaviour.\r\n\\footnote{For complex-valued functions, however, the local behaviour of a nice enough function $f$ can be controlled by its local properties.}\r\nIndeed, for $f_1,f_2:\\mathbb R\\to\\mathbb R$ smooth and $ab}=f_2|_{>b}$.\r\nThis will be proven in example sheet.\r\n\\subsection{Complex Differentiation}\r\n\\begin{definition}\r\n If $c\\in\\mathbb C,r\\in[0,\\infty)$, we define the open ball of radius $r$ centered at $c$ by $B_r(c)=\\{z\\in\\mathbb C:|z-c|0,B_\\epsilon(c)\\subset\\Omega$.\r\n\\end{definition}\r\n\\begin{example}\r\n 1. $\\mathbb C,\\varnothing$ are open.\\\\\r\n 2. Open balls are open.\\\\\r\n 3. (non-example) $\\{z\\in\\mathbb C:|z-a|\\le r\\}$ is not open, but its complement is.\r\n\\end{example}\r\n\\begin{definition}\r\n Suppose $\\Omega\\in\\mathbb C$ is open and $c\\in\\Omega$.\r\n For a function $f:\\Omega\\to\\mathbb C$, we say $f$ is complex differentiable at $c$ if\r\n $$\\lim_{z\\to c}\\frac{f(z)-f(c)}{z-c}$$\r\n exists.\r\n In case it does exist, we write it as $f^\\prime(c)$.\r\n\\end{definition}\r\nAll the rules in real differentiation still applies, but we don't really have mean value theorem.\r\nWe can write $f(x+iy)=u(x,y)+iv(x,y)$ where $u,v:\\tilde{\\Omega}\\to\\mathbb R$ where $\\tilde{\\Omega}=\\{(x,y)\\in\\mathbb R^2:x+iy\\in\\Omega\\}$.\r\n\\begin{definition}\r\n Say $\\tilde{\\Omega}$ is open if $\\Omega$ is.\r\n Then the partial derivatives of $F:\\tilde{\\Omega}\\to\\mathbb R$ at $(a,b)\\in\\tilde{\\Omega}$ are\r\n \\begin{align*}\r\n \\left.\\frac{\\partial F}{\\partial x}\\right|_{(a,b)}&=\\lim_{h\\to 0}\\frac{F(a+h,b)-F(a,b)}{h}\\\\\r\n \\left.\\frac{\\partial F}{\\partial y}\\right|_{(a,b)}&=\\lim_{h\\to 0}\\frac{F(a,b+h)-F(a,b)}{h}\r\n \\end{align*}\r\n Provided that they exist.\r\n\\end{definition}\r\nAlternatively we can define $g_b=F(x,b)$ and so\r\n$$\\left.\\frac{\\partial F}{\\partial x}\\right|_{(a,b)}=g^\\prime_b(a)$$\r\nSimilar for the other component.\r\n\\begin{proposition}\r\n If $f:\\Omega\\to\\mathbb C$ is complex differentiable at $c=a+ib$ with derivative $f^\\prime(c)=\\alpha+i\\beta$, then the partial derivatives of $u,v$ at $(a,b)$ exist and satisfy\r\n \\begin{align*}\r\n \\left.\\frac{\\partial u}{\\partial x}\\right|_{(a,b}&= \\left.\\frac{\\partial v}{\\partial y}\\right|_{(a,b)}=\\alpha\\\\\r\n -\\left.\\frac{\\partial u}{\\partial y}\\right|_{(a,b}&=\\left.\\frac{\\partial v}{\\partial x}\\right|_{(a,b)}=\\beta\r\n \\end{align*}\r\n\\end{proposition}\r\n\\begin{proof}\r\n For any sequence $h_n\\to 0$ in $\\mathbb R\\setminus\\{0\\}$, then it is also a sequence in $\\mathbb C\\setminus\\{0\\}$, so since $f$ is differentiable at $c$,\r\n \\begin{align*}\r\n \\alpha+i\\beta&=f^\\prime(c)\\\\\r\n &=\\lim_{h\\to 0}\\frac{f(c+h)-f(c)}{h}\\\\\r\n &=\\lim_{n\\to\\infty}\\frac{u(a+h_n,b)-u(a,b)}{h_n}+i\\lim_{n\\to\\infty}\\frac{v(a+h_n,b)-v(a,b)}{h_n}\r\n \\end{align*}\r\n So $\\partial u/\\partial x$ and $\\partial v/\\partial x$ exist and are $\\alpha,\\beta$ respectively.\r\n We can get the other half of the equations by replacing $h_n$ by $ih_n$ and casting exactly the same argument.\r\n\\end{proof}\r\n\\begin{remark}\r\n If $f:\\Omega\\to\\mathbb C$ is complex differentiable, then (assuming $C^1$ partial derivatives and symmetry of partial derivatives)\r\n $$\\frac{\\partial^2u}{\\partial x^2}+\\frac{\\partial^2u}{\\partial y^2}=0$$\r\n which is the Laplace Equation.\r\n\\end{remark}\r\n\\begin{proposition}\r\n Suppose $F:B_r((a,b))\\to\\mathbb R$ has partial derivatives with\r\n $$\\frac{\\partial F}{\\partial x}=\\frac{\\partial F}{\\partial y}=0$$\r\n then $F$ is constant.\r\n\\end{proposition}\r\n\\begin{proof}\r\n For $(a',b')\\in B_r((a,b))$, we consider the functions $f_b(x)=F(x,b)$ and $f_a(x)=F(a',x)$ on the reals.\r\n Then they are differentiable and using mean value theorem on $f_b$ wrt $a,a'$ shows $f(a,b)=f(a',b)$, and using it again on $f_a$ wrt $b,b'$ shows $f(a',b)=f(a',b')$, so $f(a,b)=f(a',b')$.\r\n\\end{proof}\r\n\\begin{corollary}\r\n If $f:B_r(c)\\to\\mathbb C$ has derivative zero, then $f$ is constant.\r\n\\end{corollary}\r\n\\begin{proof}\r\n Immediate.\r\n\\end{proof}\r\n\\begin{remark}\r\n Complex differentiable functions $f:\\mathbb C\\to\\mathbb C$ behave very nicely.\r\n Firstly, if $f$ is complex differentiable, so is $f^\\prime$, consequently $f$ is $C^\\infty$.\r\n Even better, its Taylor series always converges to itself.\r\n Secondly, if $f$ is bounded then $f$ is constant.\r\n\\end{remark}", "meta": {"hexsha": "c8c1fb8c7d13f8ea747d860d2f1c22d439eaa377", "size": 21326, "ext": "tex", "lang": "TeX", 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"avg_line_length": 59.4038997214, "max_line_length": 312, "alphanum_fraction": 0.6305448748, "num_tokens": 7541, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702880639791, "lm_q2_score": 0.8438951084436077, "lm_q1q2_score": 0.6117144903533008}} {"text": "\\section{Notation}\n\nA Markov decision process (MDP) $\\mdp$ consists of a tuple $(\\mathcal{S}, \\mathcal{A}, P, R, \\gamma, s_0)$, where $\\mathcal{S}$ is the state space, $\\mathcal{A}$ is the action space, $P$ is the transition function mapping $\\mathcal{S} \\times \\mathcal{A}$ to distributions on $\\mathcal{S}$, $R$ is the scalar reward function on $\\mathcal{S} \\times \\mathcal{A}$, $\\gamma$ is the discount factor, and $s_0$ is the starting state.\nNote that a single start state can be converted to a start distribution by letting $P(s_0, \\cdot)$ be independent of the action taken.\n% This work focuses on the episodic setting, where after some number of steps $T$\nAn \\emph{agent} interacts with an MDP by producing a policy $\\pi$ at each timestep and observing the transitions it visits.\n\nThe \\emph{value} of a state $s$ for a policy $\\pi$ is the (discounted) sum of future rewards obtained by running that policy starting from the state $s$:\n\\begin{align}\n V^{\\pi}(s) = \\E_{\\substack{a_t \\sim \\pi(\\cdot \\mid s_t) \\\\ s_{t+1} \\sim P(s_t, a_t)}} \\sum_{t=0}^\\infty \\gamma^t R(s_t, a_t) ~\\big|~ s_0=s\n\\end{align}\nSimilarly the value of a state-action pair $(s, a)$ is written as\n\\begin{align}\n Q^{\\pi}(s, a) = \\E_{\\substack{a_t \\sim \\pi(\\cdot \\mid s_t) \\\\ s_{t+1} \\sim P(s_t, a_t)}} \\sum_{t=0}^\\infty \\gamma^t R(s_t, a_t) ~\\big|~ s_0=s, a_0 = a\n\\end{align}\nAny MDP admits a deterministic optimal policy $\\pi^*$ with corresponding value functions $V^*$ and $Q^*$ such that $V^*(s) \\ge V^\\pi(s)$ for all $s$ and $\\pi$ \\citep{sutton2018reinforcement}.\n\n\n\\section{Discounting and Resets}\n\nWhen $\\gamma < 1$ we say that the setting is \\emph{discounted}, and for $\\gamma = 1$ we say that it is \\emph{undiscounted}.\nAn environment may have a time limit $T$ such that after every $T$ steps, the state is reset to $s_0$.\nIn this case we call it \\emph{episodic}.\nTo satisfy the Markov property, episodic environments should include the current timestep $t$ in the state.\n\nFor episodic environments, it is natural to define a policy's quality in terms of the total reward earned (on average) in a single episode, with discounted environments preferring rewards obtained earlier in the episode.\nFor non-episodic environments comparisons between policies are less clear-cut; see Section 2 of \\citet{strehl2008analysis} for a discussion.\n\nCommon practice in deep reinforcement learning is to train agents with discounting and resets, but not include the timestep in the state observations and simply ignore the transition from $s_T$ to $s_0$.\nPolicies are typically evaluated by the total undiscounted reward in an episode, despite the conflict with the training setup.\nIn most cases the rewards are truncated to reflect the limited episode duration \\citep{schulman2017proximal,Fujimoto2018AddressingFA,haarnoja2018soft}.\nSince the agent is unable to tell when an episode will end, this effectively introduces noise into value prediction targets, and this noise varies by state depending on how often the agent has ended an episode on that state.\nIn other cases value targets may be bootstrapped from the state $s_T$ as if the environment were not episodic.\nThis has two issues: (1) it introduces bias by treating an estimate of $V(s_T)$ as the true value, when in some states and environments this estimate may never be updated; and (2) by pretending the environment has no resets, it introduces a mismatch between the training and test objectives.\\endnote{This mismatch is decreased if the divergence between the distributions $P(s_T, \\pi(\\cdot \\mid s_T))$ and $s_0$ is smaller. Like most problems in RL, it also becomes smaller if smaller discount factors are used. However in general there are actions which would provide more short-term reward, and thus perform better toward the end of an episode, than the infinite-horizon optimal actions.}\nHowever, it does not introduce noise.\n\nMore fundamentally, practically all discounted policy gradient algorithms drop the discounting term from the state distribution.\n\\citet{Nota2020IsTP} show that this results in following a direction which is not the gradient of any function, and which is not guaranteed to converge to a good solution with respect to the discounted or undiscounted objectives.\nWhile this is deeply worrying, these methods frequently work well in practice.\nThis may be due to their usage with overparameterized neural network models, which are largely invariant to a reweighting of the data \\citep{byrd2018effect,brandfonbrener2021offlinecontextual}.\n\n% This thesis focuses on the episodic setting.\n% Following common practice in deep reinforcement learning, agents are trained with algorithms which use $\\gamma < 1$, but evaluated according to the total reward earned in an episode.\n\n\n\\section{Defining Sample Complexity}\n\nWe say that a policy $\\pi$ is $\\varepsilon$-optimal if $V^*(s_0) \\le V^\\pi(s_0) + \\varepsilon$.\nDefine $\\bm{\\pi} = A(\\mdp, i)$ to be the policy obtained by running the agent (i.e. algorithm) $A$ in the environment $\\mdp$ for $i$ timesteps, being careful to note that the policy $\\bm{\\pi}$ is itself a random variable due to the randomness in the experience collected in those $i$ steps.\nLet the \\emph{sample complexity} of learning a $\\varepsilon$-optimal policy on $\\mdp$ with $A$ be the expected number of steps (indexed as $i$) such that\n\\begin{align}\n V^{\\pi_i}(s_0) < V^*(s_0) - \\varepsilon, && \\text{where } \\pi_i = A(\\mdp, i).\n\\end{align}\nThis definition is related to those proposed by \\citet{Fiechter1994EfficientRL} and \\citet{strehl2008analysis} for PAC learning.\nSometimes it is also useful to consider the \"anytime\" performance of an algorithm $A$ on an MDP $\\mdp$.\nAn algorithm $A_1$ would dominate $A_2$ if $\\forall i$, $V^{A_1(\\mdp, i)}(s_0) \\ge V^{A_2(\\mdp, i)}(s_0)$.\n\n\n\n\\section{Online and Deployment Settings} \\label{sec:regret-deployment}\n\nWhile the overall MDP framework is largely shared in the community, several different objectives for a learning agent are commonly studied.\nThe online and offline settings are perhaps the most studied in the theory community, but the \"learn-and-deploy\" setting has the most relevance for present applications of RL.\n\n\n\\paragraph{The online setting.}\n% \\side{talk about sample complexity}\nHere an RL agent learns by continually interacting with the environment with the goal of maximizing the total reward earned across all time.\nThis gives rise to the explore-exploit tradeoff when acting: at each moment, the agent may choose to take an action which is uninformative but leads to greater short-term reward, or one which will yield more information at the cost of lower reward.\nThe objective for this setting is to minimize the rate of accumulation of \\emph{regret}, which measures the difference between the total reward obtained by an optimal policy and the agent:\n\\begin{align}\n L(A, \\mdp, T) = \\E \\left[ \\sum_{i=1}^T R(s^*_i, a^*_i) - R(s_i, a_i) \\right].\n\\end{align}\nThis setting is appropriate when an agent is being trained \"on the job,\" where mistakes early in learning have just as much deleterious effect as those made later.\nIn this context, sample efficiency controls the total amount of regret the agent will accumulate.\n\n\\paragraph{The learn-and-deploy setting.}\n% \\side{talk about sample complexity}\nThis setting consists of distinct learning and deployment phases.\nIn the learning phase, the agent is not required to perform well and may collect whatever data is most informative.\nIn the deployment phase, the policy is fixed and should be as close to optimal as possible.\nNote that the policy produced at the end of the training procedure need bear no resemblance to those used to collect data.\nFor episodic environments, with a training period consisting of $N$ steps we can write the this objective as\n\\begin{align}\n L(A, \\mdp, N) = V^*(s_0) - V^{\\pi_N}(s_0), && \\text{where } \\pi_N = A(\\mdp, N)\n\\end{align}\nHistorically this setting has not been much discussed, though it is analagous to the task of best-arm identification in bandits \\citep{Russo2016SimpleBA,Kaufmann2016OnTC}.\nIt is also related to the iterative technique of fitted Q iteration \\citep{Ernst2005TreeBasedBM,Riedmiller2005NeuralFQ}.\n\nCrucially, this setting is the one used in practice in nearly every application of reinforcement learning.\nRL algorithms are not yet safe and reliable enough to allow them to update a policy on the fly, especially with a physical system which may be damaged or cause injury.\nFurthermore, most works studying RL implicitly provide results in this setting by evaluating according to a different policy than the one used for training, for example showing learning curves with a deterministic policy \\citep{Mnih2015HumanlevelCT,Lillicrap2016ContinuousCW,Fujimoto2018AddressingFA,haarnoja2018soft}.\nComparisons of final, large-data performance similarly reflect this setting \\citep{Silver2016MasteringTG,Vinyals2019GrandmasterLI,openai2019dota,OpenAI2019SolvingRC}.\n\nIn this setting, sample efficiency reflects the amount of data the agent needs to collect during training time before it can perform near-optimally at deployment time.\n\n\n\\paragraph{The offline setting.}\n% \\side{talk about sample complexity}\nAlso known as the \\emph{batch} setting, this consists only of pure policy optimization given a fixed dataset of environment interactions collected by an extrinsic behavior policy.\nAfter learning from this data in whatever way it sees fit, an algorithm produces a fixed policy with the objective of earning as much reward as possible.\nThough described as a reinforcement learning setting, it does not include any actual reinforcement as the agent never learns from its own interactions with the environment.\nHowever, this makes the offline RL setting uniquely valuable for isolating how much can be learned from particular data.\nThis setting is also appealing as it would in principle allow an agent to be trained in a risk-free way by using data collected from a safe policy, and for free (in terms of samples) if data from one experiment can be repurposed as training data for another.\\endnote{While there are doubtless some settings where this cross-task data reuse is possible, it has quite clear limits. For a policy to be trained to solve task B using data collected from task A, the policy used for task A would have had to actually also solve task B. It could have been done piecewise rather than in a single good trajectory, but unless the tasks are extremely similar it is vanishingly unlikely. Perhaps a more practical application would be to start with a safe but poor policy for solving a task, then incrementally collect new data, refine the policy using offline RL, validate that the new policy is also safe, and then collect data once more.}\nFor this setting, sample efficiency must be redefined to include the behavior policy under which the data was collected.\nIt then reflects the minimum amount of data collected under that behavior policy required for a particular algorithm to recover the optimal policy.\\endnote{Note that this quantity of data should be expected to be extraordinarily large for most behavior policies.}\n\n% \\side{break into \\textbackslash paragraphs}\n% \\side{write math definitions}\n% \\side{write sample complexity definitions}\n% \\begin{itemize}\n% \\item The \\emph{online} or \\emph{regret} setting.\n\n% \\item The \\emph{learn-and-deploy} setting.\n% \\side{Need a better name for learn-and-deploy}\n% This setting consists of distinct learning and deployment phases. In the learning phase, the agent is not required to perform well and may collect whatever data is most informative. In the deployment phase, the policy is fixed and should be as close to optimal as possible. This setting schematically breaks down into pure data collection followed by pure policy optimization, though in practice policy optimization is an essential tool during exploration.\n% \\item The \\emph{offline} or \\emph{batch} setting.\n% This consists only of pure policy optimization given a fixed dataset of environment interactions collected by an extrinsic behavior policy. After learning from this data in whatever way it sees fit, an algorithm produces a fixed policy with the objective of earning as much reward as possible. Though described as a reinforcement learning setting, it does not include any actual reinforcement as the agent never learns from its own interactions with the environment. However, this makes the offline RL setting uniquely valuable for isolating how much can be learned from particular data.\n% \\end{itemize}\n\n\n% \\side{Do I want a section on simulated versus physical envs?}\n% \\section{Simulated and Physical Environments}\n\n\n\\printendnotes", "meta": {"hexsha": "9f746d297fc8d754de12f66a6daa1d5fe5ab1249", "size": 12676, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "content/rl-setting.tex", "max_stars_repo_name": "willwhitney/dissertation", "max_stars_repo_head_hexsha": "a9842f84e53ca47ec849488b6cb9acb8a11336ef", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-06-20T20:31:06.000Z", "max_stars_repo_stars_event_max_datetime": "2021-06-20T20:31:06.000Z", "max_issues_repo_path": "content/rl-setting.tex", "max_issues_repo_name": "willwhitney/doctoral-thesis", "max_issues_repo_head_hexsha": "a9842f84e53ca47ec849488b6cb9acb8a11336ef", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "content/rl-setting.tex", "max_forks_repo_name": "willwhitney/doctoral-thesis", "max_forks_repo_head_hexsha": "a9842f84e53ca47ec849488b6cb9acb8a11336ef", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-25T13:01:43.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-25T13:01:43.000Z", "avg_line_length": 100.6031746032, "max_line_length": 928, "alphanum_fraction": 0.7808456926, "num_tokens": 3050, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951104066293, "lm_q2_score": 0.724870282120402, "lm_q1q2_score": 0.6117144867604812}} {"text": "\\documentclass{article}\n\n\\usepackage{blindtext}\n\\usepackage{multicol}\n\\usepackage{caption}\n\\usepackage{bm}\n\\usepackage{tikz}\n\\usepackage{pgfplots}\n\\usepackage{hyperref}\n\\hypersetup{\n colorlinks=true,\n linkcolor=blue,\n filecolor=magenta,\n urlcolor=cyan,\n}\n\\usepackage{geometry}\n\\geometry{\n\ta4paper,\n\tnoheadfoot=true,\n\tleft=1.0in,\n\tright=1.0in,\n\ttop=1.0in,\n\tbottom=1.0in\n}\n\n% url package\n\\usepackage{hyperref}\n\\usepackage{subcaption}\n\n% Titling and Author\n\\title{Latex Tikz Examples, Annotate on Graph}\n\\author{\\href{https://fanwangecon.github.io/}{Fan Wang}\\thanks{https://fanwangecon.github.io, repository: \\href{https://fanwangecon.github.io/Tex4Econ/}{Tex4Econ}}}\n\\date{\\today}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Annotate Text in Figure}\n\n\\subsection{Slope and Intercept Annotate}\n\nDraw axis below, and a line, middle of page, and annotate the slope of the intercept with arrows. Annotate with box without color.\n\n\\begin{verbatim}\n\\begin{center}\n\\begin{tikzpicture}\n\\draw (0,2) -- (0.25*\\textwidth,2);\n\\draw (0,1) -- (0.5*\\textwidth,1);\n\\end{tikzpicture}\n\\end{center}\n\\end{verbatim}\n\\bigskip\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[->] (-1,0) -- (6,0) node[right] {$N_m$};\n\\draw[->] (0,-1) -- (0,6) node[above] {$N_{m^{\\prime}}$};\n\n% A. Clip area so that line below is only drawn inside finite box\n\\clip (-3,-1) rectangle (6,6);\n\n% B. Draw a line with intercept and slope\n\\draw[line width=0.50mm,domain=-1:6,smooth,variable=\\x, blue] plot ({\\x},{1+\\x*1.25});\n\n% C. Draw Line pointing to Intercept of line\n\\draw[->, line width=0.75mm] (-1.7,2.2) -- (-0.1,1.1);\n\n% D. Draw a transparent text box that wraps text\n\\node[text width=3cm] at (-1.4,3.25) {\\large{$\\bm{\\alpha_i}$ (this is the position of the \\textbf{x-intercept}), wrap text.}};\n\n\\end{tikzpicture}\n\\end{center}\n\n\\subsection{Anote Slope, X intercept and Y intercept}\n\nDraw a line, then point to its slope, x and y intercepts.\n\n\\begin{enumerate}\n \\item Define the slope and y-intercept: $a_i, b_i$\n \\item Define and calculate the x-intercept: $c_i = -\\frac{a_i}{b_i}$\n \\item Define the text that should appear for each element: $a_{desca}$\n \\item Drawing lines pointing to $a$, $b$ or $c$ points:\n \\begin{itemize}\n \\item direction: $d_i \\in \\left\\{NE, SE, SW, NW\\right\\}$\n \\item rotation: $0 <= e_i <= 90$, which direction points to text\n \\item dist one: $f_i > 0 $, straight distance from origin\n \\item dist two: $g_i > 0 $, distance of line\n \\item point start: $h^o_i, h^o_i$, starting point based on the four pieces of information\n \\item point end: $h^d_i, h^d_i$, ending point based on the four pieces of information\n \\end{itemize}\n\\end{enumerate}\n\nSo end up drawing using $a$, $b$ for line, $h^o$ for one point $h*d$ line small segment.\n\n\\begin{verbatim}\n\\end{verbatim}\n\\bigskip\n\\begin{center}\n\\begin{tikzpicture}\n\\draw[->] (-1,0) -- (6,0) node[right] {$N_m$};\n\\draw[->] (0,-1) -- (0,6) node[above] {$N_{m^{\\prime}}$};\n\n% A. Clip area so that line below is only drawn inside finite box\n\\clip (-3,-1) rectangle (6,6);\n\n% B. Draw a line with intercept and slope\n\\draw[line width=0.50mm,domain=-1:6,smooth,variable=\\x, blue] plot ({\\x},{1+\\x*1.25});\n\n% C. Draw Line pointing to Intercept of line\n\\draw[->, line width=0.75mm] (-1.7,2.2) -- (-0.1,1.1);\n\n% D. Draw a transparent text box that wraps text\n\\node[text width=3cm] at (-1.4,3.25) {\\large{$\\bm{\\alpha_i}$ (this is the position of the \\textbf{x-intercept}), wrap text.}};\n\n\\end{tikzpicture}\n\\end{center}\n\n\n\\end{document}\n", "meta": {"hexsha": "163280efad5e45a01e3dace87976dda041a09566", "size": 3479, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "_other/tikz/annotate/fs_tikz_annotate.tex", "max_stars_repo_name": "guohui-jiang/Tex4Econ", "max_stars_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "_other/tikz/annotate/fs_tikz_annotate.tex", "max_issues_repo_name": "guohui-jiang/Tex4Econ", "max_issues_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "_other/tikz/annotate/fs_tikz_annotate.tex", "max_forks_repo_name": "guohui-jiang/Tex4Econ", "max_forks_repo_head_hexsha": "7bdbfb29e956d31239bd592b6392574e4aec5c15", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.9916666667, "max_line_length": 164, "alphanum_fraction": 0.6812302386, "num_tokens": 1207, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702761768248, "lm_q2_score": 0.8438951104066293, "lm_q1q2_score": 0.6117144817447254}} {"text": "\\section{Current Operator in Landau Levels}\n\nNow consider about the current density operator for $N$th Landau level. Since we have already found the extact solution for our time depenedent Hamiltonian and we have identify them as Floquet states with quesi energies. From these solutions we can identify the \\textit{Floquet modes} as given in Eq. \\eqref{3.15} and using quantum numbers we can represent those states as follows\n\\begin{equation} \\label{6.1}\n \\ket{\\phi_{\\alpha}} = \\ket{\\phi_{n,m}} \\equiv \\ket{n,k_x} \\quad\n \\text{where} \\quad\n k_x = m \\frac{2\\pi}{L_x}\n\\end{equation}\nUsing above complete set of eigenstates of Floquet Hamiltonian we can represent the single particle current operator's matrix element as\n\\begin{equation} \\label{6.2}\n \\qty(\\mb{j})_{nm,n'm'} = \\mel{n,k_x}{\\;\\hat{\\mb{j}}\\;}{n',k'_x}\n\\end{equation}\nwhere particle current operator for this system will be\n\\begin{equation} \\label{6.3}\n \\hat{\\mb{j}} = \\frac{1}{m} \\qty(\\hat{\\mb{P}} - e\\qty[\\mb{A}_s + \\mb{A}_d(t)]).\n\\end{equation}\nLet's consider the transverse conductivity in $x$ direction and we can identify that $x$ directional current operator as\n\\begin{equation} \\label{6.4}\n \\hat{j}_x = \\frac{1}{m} \\qty(-i\\hbar\\pdv{x} + eBy).\n\\end{equation}\nNow we can calculate the matrix elements of $x$ directional current operator's matrix in Floquet mode basis as\n\\begin{equation} \\label{6.5}\n \\qty({j}_x)_{nm,n'm'} = \\mel{n,k_x}{\\;\\hat{j}_x\\;}{n',k'_x} =\n \\mel{n,k_x}{\\;\\frac{1}{m} \\qty(-i\\hbar\\pdv{x} + eBy)\\;}{n',k'_x}\n\\end{equation}\nand we can evaluate these using Floquet modes derived in Eq.\\eqref{3.15} as follows\n\\begin{equation} \\label{6.6}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\int dx \\int dy \\;\n \\frac{1}{\\sqrt{L_x}} \\chi_{n}\\big(y - y_0 - \\zeta(t)\\big)\n \\exp(-ik_x x) \\\\\n & \\times\n \\frac{1}{m} \\qty(-i\\hbar\\pdv{x} + eBy)\n \\frac{1}{\\sqrt{L_x}} \\chi_{n'}\\big(y - y_0 - \\zeta(t)\\big)\n \\exp(i k'_x x)\n \\end{aligned}\n\\end{equation}\nand this can be simplified as\n\\begin{equation} \\label{6.7}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\frac{1}{{mL_x}}\n \\int dx \\exp(-i(k_x-k'_x) x)\n \\int dy \\;\n \\chi_{n}\\big(y - y_0 - \\zeta(t)\\big) \\\\\n & \\times\n \\qty(\\hbar k'_x + eBy)\n \\chi_{n'}\\big(y - y_0 - \\zeta(t)\\big)\n \\end{aligned}\n\\end{equation}\nand\n\\begin{equation} \\label{6.8}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n \\int dy \\;\n \\qty(\\hbar k'_x + eBy)\n \\chi_{n}\\big(y - y_0 - \\zeta(t)\\big)\n \\chi_{n'}\\big(y - y_0 - \\zeta(t)\\big).\n \\end{aligned}\n\\end{equation}\nNow let $y - y_0 - \\zeta(t) = \\bar{y}$ and we will get\n\\begin{equation} \\label{6.9}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n \\int d\\bar{y} \\;\n \\qty(\\hbar k'_x + eB\\bar{y} + eBy_0 + eB\\zeta(t))\n \\chi_{n}(\\bar{y})\n \\chi_{n'}(\\bar{y}).\n \\end{aligned}\n\\end{equation}\nusing definition of $y_0$ given in Eq. \\eqref{1.11} this will be modiofied to\n\\begin{equation} \\label{6.10}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n \\int d\\bar{y} \\;\n \\qty(\\hbar k'_x + eB\\bar{y} -\\hbar k'_x + eB\\zeta(t))\n \\chi_{n}(\\bar{y})\n \\chi_{n'}(\\bar{y})\n \\end{aligned}\n\\end{equation}\nand using integral indentities of Gauss-Hermite functions\n\\begin{equation} \\label{6.11}\n \\int d{y} \\;\n \\chi_{n}({y})\n \\chi_{n'}({y}) =\n \\delta_{n',n}\n\\end{equation}\n\\begin{equation} \\label{6.12}\n \\int dy \\;\n y\n \\chi_{n}({y})\n \\chi_{n'}({y}) =\n \\qty(\\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1} + \\sqrt{\\frac{n}{2}}\n \\delta_{n',n-1})\n\\end{equation}\nthis becomes\n\\begin{equation} \\label{6.13}\n \\begin{aligned}\n \\qty({j}_x)_{nm,n'm'} & =\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n eB\n \\qty[\n \\qty(\\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1} + \\sqrt{\\frac{n}{2}}\n \\delta_{n',n-1})\n + \\zeta(t) \\delta_{n',n}\n ]\n \\end{aligned}\n\\end{equation}\n\\textcolor{red}{Due to complexity we can only consider the constant contribution and we allows only the one-cycle averaged current flow} and then we can derive the $s=0$ components of the Fourier series as\n\\begin{equation} \\label{6.14}\n \\begin{aligned}\n \\qty({j}^x_{s=0})_{nm,n'm'} & =\n \\frac{1}{T} \\int_0^T dt \\;\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n eB\n \\qty[\n \\qty(\\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1} + \\sqrt{\\frac{n}{2}}\n \\delta_{n',n-1})\n + \\frac{eE}{m(\\omega_0^2 - \\omega^2)}\\sin(\\omega t) \\delta_{n',n}\n ]\n \\end{aligned}\n\\end{equation}\nand this can be evaluate and get\n\\begin{equation} \\label{6.15}\n \\begin{aligned}\n \\qty({j}^x_{s=0})_{nm,n'm'} & =\n \\frac{eB}{{m}}\n \\delta_{k_x,k'_x}\n \\qty(\\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1} + \\sqrt{\\frac{n}{2}}\n \\delta_{n',n-1})\n \\end{aligned}\n\\end{equation}\n\n\\noindent\nFor electric current operator we can introduce the electron's charge and effective mass\n\\begin{equation} \\label{6.16}\n \\begin{aligned}\n \\qty({j}^x_{s=0})_{nm,n'm'} & =\n \\frac{e^2B}{{m_e}}\n \\delta_{k_x,k'_x}\n \\qty(\\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1} + \\sqrt{\\frac{n}{2}}\n \\delta_{n',n-1})\n \\end{aligned}\n\\end{equation}\n\\hfill$\\blacksquare$\n\n\\noindent\nNext we can consider the transverse conductivity in $y$ direction and we can identify that $y$ directional current operator as\n\\begin{equation} \\label{6.17}\n \\hat{j}_y = \\frac{1}{m} \\qty(-i\\hbar\\pdv{y} - \\frac{eE}{\\omega}\\cos(\\omega t)).\n\\end{equation}\nNow we can calculate the matrix elements of $y$ directional current operator's matrix in Floquet mode basis as\n\\begin{equation} \\label{6.18}\n \\qty({j}_y)_{nm,n'm'} = \\mel{n,k_x}{\\;\\hat{j}_y\\;}{n',k'_x} =\n \\mel{n,k_x}{\\;\\frac{-1}{m} \\qty(i\\hbar\\pdv{y} + \\frac{eE}{\\omega}\\cos(\\omega t))\\;}{n',k'_x}\n\\end{equation}\nand we can evaluate these using Floquet modes derived in Eq.\\eqref{3.15} as follows\n\\begin{equation} \\label{6.19}\n \\begin{aligned}\n \\qty({j}_y)_{nm,n'm'} & =\n \\int dx \\int dy \\;\n \\frac{1}{\\sqrt{L_x}} \\chi_{n}\\big(y - y_0 - \\zeta(t)\\big)\n \\exp(-ik_x x) \\\\\n & \\times\n \\frac{-1}{m} \\qty(i\\hbar\\pdv{y} + \\frac{eE}{\\omega}\\cos(\\omega t))\n \\frac{1}{\\sqrt{L_x}} \\chi_{n'}\\big(y - y_0 - \\zeta(t)\\big)\n \\exp(i k'_x x)\n \\end{aligned}\n\\end{equation}\nthen introducing new variable\n\\begin{equation} \\label{6.20}\n y - y_0 - \\zeta(t) = \\bar{y} \\quad \\longrightarrow \\quad dy = d{\\bar{y}}\n\\end{equation}\nand follwing identity [*Ref: Appendix A Hermite fucntions]\n\\begin{equation} \\label{6.21}\n \\pdv{\\chi_n(y)}{y} = -\\sqrt{\\frac{n+1}{2}} \\chi_{n+1}(y) +\n \\sqrt{\\frac{n}{2}} \\chi_{n-1}(y)\n\\end{equation}\nabove expression can be simplified as previous case (let $y=\\bar{y}$)\n\\begin{equation} \\label{6.22}\n \\begin{aligned}\n \\qty({j}_y)_{nm,n'm'} & =\n \\frac{-1}{{m}}\n \\delta_{k_x,k'_x}\n \\int dy \\;\n \\chi_{n}(y)\n \\qty[-i\\hbar \\qty(\n \\sqrt{\\frac{n'+1}{2}} \\chi_{n'+1}(y) - \\sqrt{\\frac{n'}{2}} \\chi_{n'-1}(y)\n ) + \\frac{eE}{\\omega}\\cos(\\omega t)\n \\chi_{n'}(y)].\n \\end{aligned}\n\\end{equation}\nThen considering Gauss-Hermite integral indentities we can derive that\n\\begin{equation} \\label{6.23}\n \\begin{aligned}\n \\qty({j}_y)_{nm,n'm'} & =\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n \\qty[i\\hbar \\qty(\n \\sqrt{\\frac{n'+1}{2}} \\delta_{n,n'+1}\n - \\sqrt{\\frac{n'}{2}} \\delta_{n,n'-1}\n ) - \\frac{eE}{\\omega}\\cos(\\omega t) \\delta_{n,n'}]\n \\end{aligned}\n\\end{equation}\n\\textcolor{red}{Due to complexity we can only consider the constant contribution and we allows only the one-cycle averaged current flow} and then we can derive the $s=0$ components of the Fourier series as\n\\begin{equation} \\label{6.24}\n \\begin{aligned}\n \\qty({j}^y_{s=0})_{nm,n'm'} & =\n \\frac{1}{T} \\int_0^T dt \\;\n \\frac{1}{{m}}\n \\delta_{k_x,k'_x}\n \\qty[i\\hbar \\qty(\n \\sqrt{\\frac{n'+1}{2}} \\delta_{n,n'+1}\n - \\sqrt{\\frac{n'}{2}} \\delta_{n,n'-1}\n ) - \\frac{eE}{\\omega}\\cos(\\omega t) \\delta_{n,n'}]\n \\end{aligned}\n\\end{equation}\nand this can be evaluate and get\n\\begin{equation} \\label{6.25}\n \\begin{aligned}\n \\qty({j}^y_{s=0})_{nm,n'm'} & =\n \\frac{i\\hbar}{{m}}\n \\delta_{k_x,k'_x}\n \\qty[\n \\sqrt{\\frac{n'+1}{2}} \\delta_{n,n'+1}\n - \\sqrt{\\frac{n'}{2}} \\delta_{n,n'-1}\n ]\n \\end{aligned}\n\\end{equation}\nand this can be re-write as\n\\begin{equation} \\label{6.26}\n \\begin{aligned}\n \\qty({j}^y_{s=0})_{nm,n'm'} & =\n \\frac{i\\hbar}{{m}}\n \\delta_{k_x,k'_x}\n \\qty[\n \\sqrt{\\frac{n}{2}} \\delta_{n',n-1}\n - \\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1}\n ]\n \\end{aligned}\n\\end{equation}\n\n\\noindent\nFor electric current operator we can introduce the electron's charge and effective mass\n\\begin{equation} \\label{6.27}\n \\begin{aligned}\n \\qty({j}^y_{s=0})_{nm,n'm'} & =\n \\frac{ie\\hbar}{{m}}\n \\delta_{k_x,k'_x}\n \\qty[\n \\sqrt{\\frac{n}{2}} \\delta_{n',n-1}\n - \\sqrt{\\frac{n+1}{2}} \\delta_{n',n+1}\n ]\n \\end{aligned}\n\\end{equation}\n\\hfill$\\blacksquare$\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nxx\n", "meta": {"hexsha": "0724eabc38c02d4f2fbd293b181a3f2a6555998f", "size": 8863, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "theory/sec_06.tex", "max_stars_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_stars_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "theory/sec_06.tex", "max_issues_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_issues_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "theory/sec_06.tex", "max_forks_repo_name": "KosalaHerath/magnetic-2DEG-conductivity", "max_forks_repo_head_hexsha": "91c5df1b018579b4b9c91d84f2d60ee482a001de", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 30.0440677966, "max_line_length": 380, "alphanum_fraction": 0.5953965926, "num_tokens": 3543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8438951025545426, "lm_q2_score": 0.7248702821204019, "lm_q1q2_score": 0.6117144810687368}} {"text": "\\documentclass[10pt]{article}\n\\usepackage{amsfonts}\n\\usepackage{amssymb, amsmath}\n\\usepackage{eucal}\n\\usepackage{amscd}\n\\usepackage{url}\n\\usepackage{listings}\n\\urlstyle{sf}\n\\pagestyle{plain}\n\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Q}{\\mathbb{Q}}\n\\newcommand{\\I}{\\mathbb{I}}\n\\newcommand{\\C}{\\mathbb{C}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\Pee}{\\mathbb{P}}\n\\newcommand{\\EuO}{\\mathcal{O}}\n\\newcommand{\\Qbar}{\\overline{\\mathbb{Q}}}\n\\newcommand{\\code}{\\lstinline}\n\n\\newcommand{\\ljk}[2]{\\left(\\frac{#1}{#2}\\right)}\n\\newcommand{\\modulo}[1]{\\;\\left(\\mbox{mod}\\;#1\\right)}\n\\newcommand{\\fr}{\\mathfrak}\n\n\\def\\notdivides{\\mathrel{\\kern-3pt\\not\\!\\kern4.5pt\\bigm|}}\n\\def\\nmid{\\notdivides}\n\\def\\nsubseteq{\\mathrel{\\kern-3pt\\not\\!\\kern2.5pt\\subseteq}}\n\n\\newtheorem{theorem}{Theorem}[section]\n\\newtheorem{lemma}[theorem]{Lemma}\n\\newtheorem{proposition}[theorem]{Proposition}\n\\newtheorem{corollary}[theorem]{Corollary}\n\\newtheorem{property}[theorem]{Property}\n\n\\newenvironment{proof}[1][Proof]{\\begin{trivlist}\n\\item[\\hskip \\labelsep {\\itshape #1}]}{\\end{trivlist}}\n\\newenvironment{definition}[1][Definition]{\\begin{trivlist}\n\\item[\\hskip \\labelsep {\\bfseries #1}]}{\\end{trivlist}}\n\\newenvironment{example}[1][Example]{\\begin{trivlist}\n\\item[\\hskip \\labelsep {\\bfseries #1}]}{\\end{trivlist}}\n\\newenvironment{remark}[1][Remark]{\\begin{trivlist}\n\\item[\\hskip \\labelsep {\\bfseries #1}]}{\\end{trivlist}}\n\\newenvironment{exercise}[1][Exercise]{\\begin{trivlist}\n\\item[\\hskip \\labelsep {\\bfseries #1}]}{\\end{trivlist}}\n\n\\parindent=0pt\n\\parskip 4pt plus 2pt minus 2pt \n\n\\title{Group Theory}\n\n\\author{William B. Hart}\n\n\\begin{document}\n\n\\maketitle\n\n\\tableofcontents\n\n\\section{Introduction to Groups and Subgroups}\n\n\\subsection{Groups and their Basic Properties}\n\n\\subsubsection{Groups}\n\n\\begin{definition}\nA set $G$ with a closed binary operation $\\star : G \\times G \\to G$ is a \\emph{group} if the following rules hold:\n\\begin{enumerate}\n\\item (Associativity) For all $a, b, c \\in G$ we have $(a \\star b) \\star c = a \\star (b \\star c)$.\n\\item (Identity) There exists $e \\in G$ such that for all $g \\in G$ we have $g\\star e = g = e \\star g$.\n\\item (Inverses) For every $g \\in G$ there exists $g' \\in G$ such that $g \\star g' = e = g' \\star g$.\n\\end{enumerate}\nWe sometimes write $(G, \\star)$ to denote the group, in order to emphasise the group operation. In other cases where the operation is clear from the context, we just write $G$.\n\\end{definition}\n\nWe call the element $e$ in the definition above the \\emph{identity} (element) of the group and $g'$ the \\emph{inverse} of $g$.\n\nIn the case that the group operation is addition, we usually write $0$ for the identity element and $-g$ for the inverse of $g$.\n\nWhen the group operation is multiplication, we usually write $1$ for the identity element and $g^{-1}$ for the inverse of $g$.\n\nImplicit in the definition of a group is the closure property:\n\n\\begin{enumerate}\n\\item (Closure) For all $a, b \\in G$ we have $a\\star b \\in G$.\n\\end{enumerate}\n\n\\subsubsection{Basic Results}\n\nTo save on notation we will write $g\\star h$ as $gh$ where this will not cause any confusion.\n\\begin{theorem}\nThe identity element $e$ of a group $G$ is unique.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe inverse of an element $g$ of a group $G$ is unique.\n\\end{theorem}\n\n\\begin{proof}\nClearly $gg'' = e$. The result follows after left multiplication by $g'$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $ab = e$ in a group $G$ then $b = a^{-1}$.\n\\end{theorem}\n\n\\begin{proof}\nThe result follows from $ab = e$ after left multiplication by $a^{-1}$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\\label{idempotent_is_identity}\nSuppose $G$ is a group with identity $e$. If $a \\in G$ is idempotent i.e. $aa = a$, then $a = e$.\n\\end{theorem}\n\n\\begin{proof}\nThe result follows after right multiplication of $a^2 = a$ by $a^{-1}$. $\\square$\n\\end{proof}\n\n\\begin{theorem} \\label{product_inverse}\nSuppose $G$ is a group and $a, b \\in G$. Then $(ab)^{-1} = b^{-1}a^{-1}$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Right Identity and Inverses Suffice for a Group}\n\nIn the definition of a group, it isn't necessary to assume that $e$ is both a left and right identity or that $g'$ is both a left and right inverse of an element $g$. The following theorem shows that a right identity and right inverses suffice in order to have a group.\n\n\\begin{theorem}\nLet $G$ be a set closed under a binary, associative operation. Suppose that:\n\\begin{enumerate}\n\\item (Right identity) There exists $e \\in G$ such that $g\\star e = g$ for all $g \\in G$.\n\\item (Right inverse) For every $g \\in G$ there exists $g' \\in G$ such that $g\\star g' = e$.\n\\end{enumerate}\nThen $e\\star g = g$ and $g'\\star g = e$ and $(G, \\star)$ is a group.\n\\end{theorem}\n\n\\begin{proof}\nAssume $e$ is a right identity and every element $g$ has a right inverse $g'$.\n\nTheorem \\ref{idempotent_is_identity} only relies on right inverses and a right identity, so it holds in the context of the theorem.\n\nIt is easy to check that $g'g$ is idempotent. Thus $g'g = e$, i.e. $g'$ is a left inverse.\n\nThat $eg = g$ follows from $e = gg^{-1}$. $\\square$\n\\end{proof}\n\n\\subsubsection{Cancellation in a Group}\n\n\\begin{theorem}\nLet $G$ be a group and $a, b, c \\in G$. If $ab = ac$ then $b = c$. Similarly if $ac = bc$ then $a = b$.\n\\end{theorem}\n\n\\begin{proof}\nThese follow by left (resp. right) multiplication by $a^{-1}$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Laws of Exponents in a Group}\n\nJust as we do in arithmetic, it is convenient to define integer powers of elements of a group $G$.\n\n\\begin{definition}\nLet $G$ be a group, $g\\in G$ and $n$ a positive integer. We define:\n\n\\begin{align*}\n g^n &= \\underbrace{g g \\cdots g}_{n \\:\\mathrm{factors}}\\\\\ng^{-n} &= \\underbrace{g^{-1} g^{-1} \\cdots g^{-1}}_{n \\:\\mathrm{factors}}\\\\\n g^0 &= e\n\\end{align*}\n\\end{definition}\n\nAll of the usual laws of exponents hold.\n\n\\begin{theorem}\nIf $G$ is a group and $g \\in G$ then for all $m, n \\in \\mathbb{Z}$ we have\n$$g^m g^n = g^{m + n}.$$\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $g \\in G$. Then for any $m, n \\in \\mathbb{Z}$ we have\n$$(g^{m})^n = g^{mn}.$$\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\nWhen the group operation is addition, we write $ng$ for the sum of $n$ copies of $g$. Similarly $-ng$ corresponds to the sum of $n$ copies of $-g$. Similar results to the above hold, i.e. $mg + ng = (m + n)g$ and $m(ng) = (mn)g$.\n\n\\subsubsection{The Order of a Group}\n\n\\begin{definition}\nThe cardinality of a group is called its \\emph{order}. We denote the order of a group $G$ by $|G|$.\n\\end{definition}\n\n\\begin{definition}\nIf a group has finite order we say that it is a \\emph{finite group}. Otherwise we say that it is an \\emph{infinite group}.\n\\end{definition}\n\nThe smallest number of elements a group can have is $1$, since a group must contain an identity element.\n\n\\begin{theorem}\nA set $G$ containing just the identity element is a group under any closed binary operation on $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsection{The Order of Elements of a Group}\n\n\\subsubsection{The Order of Elements of a Group}\n\n\\begin{definition}\nLet $G$ be a group and $g \\in G$. If there is a positive integer $n$ such that $g^n = e$, the least such $n$ is called the \\emph{order} of $g$, denoted ord$(g)$. If no such $n$ exists, the element $g$ is said to have \\emph{infinite order}.\n\\end{definition}\n\nIf the group is written additively the order of an element $g$ is the least positive integer $n$ such that $ng = 0$.\n\n\\begin{theorem}\nThe order of an element $g$ in a group $G$ is $1$ iff it is the identity.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Basic Results about the Order Elements in a Group}\n\n\\begin{theorem}\nLet $G$ be a group and $g \\in G$. Then ord$(g) =$ ord$(g^{-1})$.\n\\end{theorem}\n\n\\begin{proof}\nThe result follows easily from $(g^{-1})^n = (g^n)^{-1}$. $\\square$.\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $a, b \\in G$. Then ord$(ab) =$ ord$(ba)$. \n\\end{theorem}\n\n\\begin{proof}\nThe result follows easily from $(ab)^n = a(ba)^{n-1}b$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $g \\in G$ has finite order $n$, then $g^0$, $g^1$, $\\ldots$, $g^{n - 1}$ are all of the distinct powers of $g$.\n\\end{theorem}\n\n\\begin{proof}\nFirst show that all powers $g^a$ are equivalent to one of the given powers. By the division algorithm\n$$a = nq + r$$\nfor some $0 \\leq r < n$. It follows easily that $g^a = g^r.$\n\nTo show they are distinct, suppose to the contrary that $g^i = g^j$ for $0\\leq i < j < n$. Multiplication by $g^{-i}$ contradicts the minimality of $n$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIn a finite group $G$, all elements have finite order.\n\\end{theorem}\n\n\\begin{proof}\nIf $g \\in G$ is of infinite order then it is easy to see that all powers of $g$ are distinct. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups with every Element of Finite Order can be Infinite}\n\n\\begin{theorem}\nThere are infinite groups in which all elements have finite order.\n\\end{theorem}\n\n\\begin{proof}\nConsider an alphabet $L$ of infinitely many distinct letters $\\{g_1, g_2, \\ldots\\}$. Now consider the set $W$ of all finite words (including the empty word) made from these letters in which\n\\begin{enumerate}\n\\item Each letter occurs only once.\n\\item The letters are in alphabetical order, i.e. $g_i$ appears before $g_j$ if $i < j$.\n\\end{enumerate}\n\nDefine a binary operation on $W$ as follows. For $w_1, w_2 \\in W$, form the word that is obtained by taking all the letters that occur in either $w_1$ or $w_2$, but not both, and putting them in alphabetical order.\n\nIt is easy to show that $W$ is a group under this operation and $w^2 = e$ for all $w \\in W$. But $g_1, g_2, \\ldots \\in W$ are infinitely many distinct elements of $W$. $\\square$\n\\end{proof}\n\n\\subsection{Subgroups}\n\n\\subsubsection{Subgroups}\n\n\\begin{definition}\nA subset $H$ of a group $G$ which is a group under the same operation as $G$ when restricted to the elements of $H$ is called a \\emph{subgroup} of $G$. We write $H \\leq G$.\n\\end{definition}\n\n\\begin{theorem}\nIf $e$ is the identity of a group $G$ and $H \\leq G$ then $e$ is also the identity of $H$.\n\\end{theorem}\n\n\\begin{proof}\nThe identity of $H$ is idempotent in $H$ and therefore also in $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $H \\leq G$ and $g^{-1}$ is the inverse of $g$ in $H$ then it is also the inverse of $g$ in $G$.\n\\end{theorem}\n\n\\begin{proof}\nThis follows immediately from the uniqueness of inverses and the previous theorem. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA non-empty subset $H$ of a group $G$ is a subgroup iff it is closed under the group operation of $G$ and $h^{-1} \\in H$ for all $h \\in H$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{corollary} \\label{subgroup}\nA non-empty subset $H$ of a group $G$ is a subgroup iff $ab^{-1} \\in H$ for every pair of elements $a, b \\in H$.\n\\end{corollary}\n\n\\begin{proof}\nLet $H$ satisfy the given conditions. As $H$ is non-empty it contains an element $a$. Therefore by assumption $aa^{-1} = e \\in H$ and $ea^{-1} = a^{-1} \\in H$.\n\nClosure follows by taking $a = g$ and $b = h^{-1}$ for arbitrary $g, h \\in H$. Thus $H$ is a subgroup by the previous theorem.\n\nThe converse is clear. $\\square$\n\\end{proof}\n\n\\subsubsection{The Intersection and Union of two Subgroups}\n\n\\begin{theorem}\nIf $G$ is a group and $H, K \\leq G$ then $H\\cap K$ is a subgroup of $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $H, K \\leq G$ then $H\\cup K$ is a subgroup of $G$ iff $H \\leq K$ or $K \\leq H$.\n\\end{theorem}\n\n\\begin{proof}\nLet $H \\cup K$ be a subgroup of $G$. Suppose $H \\not\\subseteq K$ and $K \\not\\subseteq H$.\n\nThis implies that there exists $h \\in H$ with $h \\notin K$ and $k \\in K$ with $k \\notin H$. We have that $h, k \\in H\\cup K$ and it is easy to see that $hk \\in H$ or $hk \\in K$.\n\nIn the first case, $k = h^{-1}(hk) \\in H$ which is a contradiction. The second case also gives a contradiction. Thus our assumption is false and either $H \\leq K$ or $K \\leq H$.\n\nThe converse is clear. $\\square$\n\\end{proof}\n\n\\subsubsection{The Product of Two Subgroups}\n\n\\begin{definition}\nIf $G$ is a group and $H, K \\leq G$ we define:\n$$HK = \\{hk \\;|\\; h \\in H, K \\in K\\}.$$\n\\end{definition}\n\nNote that $HK$ is just a set at this point. In fact we will show the following.\n\n\\begin{theorem}\\label{HKKH}\nIf $G$ is a group and $H, K \\leq G$ then $HK \\leq G$ iff $HK = KH$.\n\\end{theorem}\n\n\\begin{proof}\nFirst suppose that $HK = KH$. Clearly $HK$ is non-empty as $1 \\in HK$.\n\nSuppose $h_1, h_2 \\in H$ and $k_1, k_2 \\in K$. Then as $HK = KH$ we have that $k_1h_2 = h'k'$ for some $h' \\in H$ and $k' \\in K$. It follows that\n$(h_1k_1)(h_2k_2) = (h_1h')(k'k_2) \\in HK$. Thus $HK$ is closed under the group operation.\n\nThat inverses exist in $HK$ follows from\n$$(hk)^{-1} = k^{-1}h^{-1} \\in KH = HK.$$\nThus $HK$ is a subgroup of $G$ by Theorem \\ref{subgroup}.\n\nConversely, suppose that $HK \\leq G$ and let $hk$ be an arbitrary element of $HK$ for some $h \\in H$ and $k \\in K$.\n\nIt is easy to see that $(hk)^{-1} \\in KH$. But $x \\to x^{-1}$ is a bijection on the elements of $HK$, thus $HK \\subseteq KH$.\n\nThe reverse inclusion is easy to show since $k, h \\in HK$ for all $k \\in K$ and $h \\in H$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Order of the Product of Two Subgroups}\n\n\\begin{theorem}\nIf $G$ is a group and $H, K \\leq G$ both of finite order then:\n$$|HK| = \\frac{|H||K|}{|H\\cap K|}.$$\n\\end{theorem}\n\n\\begin{proof}\nThere are $|H||K|$ not necessarily distinct products $hk$ with $h \\in H$ and $k \\in K$. We claim there are precisely $|H\\cap K|$ ways of writing a given element $hk$ as a product of an element of $H$ by an element of $K$.\n\nFor each $z \\in H\\cap K,$\n$$hk = (hz)(z^{-1}k) \\in HK.$$\n\nThus there are at least $|H\\cap K|$ distinct such ways of writing $hk$.\n\nConversely, note that if $hk = h'k'$ for $h' \\in H$ and $k' \\in K$ then\n$$h^{-1}h' = k(k')^{-1} \\in H\\cap K.$$\n\nWriting $h^{-1}h' = z = k(k')^{-1}$ we have that $h'k' = (hz)(z^{-1}k)$.\n\nTherefore every product $h'k'$ is of this form which settles our claim.\n\nAs $H\\cap K$ is non-empty, $|H\\cap K| \\neq 0$ and the result follows. $\\square$\n\\end{proof}\n\n\\subsubsection{The Subgroup Generated by a Subset}\n\n\\begin{definition}\nLet $G$ be a group and $S \\subseteq G$. The \\emph{subgroup generated} by $S$ is defined to be the intersection of all subgroups of $G$ containing $S$. We denote it $\\langle S \\rangle$. If $S = \\{g_1, g_2, \\ldots, g_n\\}$ we usually write $\\langle g_1, g_2, \\ldots, g_n \\rangle$.\n\\end{definition}\n\nThe subgroup generated by $S$ is the smallest subgroup of $G$ containing $S$ in the sense that it is contained in any subgroup of $G$ containing $S$.\n\nIt is clearly also the smallest subset of $G$ containing $S$ which is closed under taking inverses and closed under the group operation. In fact, we have the following theorem.\n\n\\begin{theorem}\nLet $G$ be a group and $S$ a subset of $G$. Let $S' = \\{s^{-1} \\;|\\; s \\in S\\}$. Then $\\langle S \\rangle$ is the set of all finite products of elements of $S \\cup S'$.\n\\end{theorem}\n\n\\begin{proof}\nLet the given set be denoted $T$. Then $T$ is clearly closed under the group operation. But if $a \\in T$, it has the form $a = s_1s_2\\ldots s_n$ for some $s_i \\in S\\cup S'$. The inverse of $a$ is also clearly in $T$. Thus $T$ is a subgroup of $G$.\n\nBut all of the elements of $G$ are contained in any subgroup of $G$ containing $S$. Thus $T = \\langle S \\rangle$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Commutator Subgroup of a Group}\n\n\\begin{definition}\nLet $G$ be a group and $g, h \\in G$. We define the \\emph{commutator} of $g$ and $h$ to be $[g, h] = g^{-1}h^{-1}gh$. The subgroup of $G$ generated by the commutators of $G$ is called the \\emph{commutator subgroup} of $G$ and is denoted $[G, G]$.\n\\end{definition}\n\nNote that the commutator subgroup may contain elements that are not commutators.\n\n\\begin{theorem}\nIf $G$ is a group and $g, h \\in G$ then $[g, h] = 1$ iff $g$ and $h$ commute.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $g, h \\in G$ then $[g, h] = hg[h, g]$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ is a group then every $h \\in [G, G]$ can be written as a finite product of commutators.\n\\end{corollary}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\begin{theorem}\nA group $G$ is abelian iff it has only one distinct commutator.\n\\end{theorem}\n\n\\begin{proof}\nWe always have $1 = [1, 1]$. Thus if a group contains only one distinct commutator, $[g, h] = 1$ for all $g, h \\in G$ which implies it is abelian. The converse is also clear. $square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $g, h \\in G$ then $[g, h]^{-1} = [h, g]$.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $g, h, s \\in G$ then $[g, h]^s = [g^s, h^s]$ where $g^s = s^{-1}gs$.\n\\end{theorem}\n\n\\begin{proof}\nWe have\n$$[g, h]^s = s^{-1}g^{-1}h^{-1}ghs = s^{-1}g^{-1}ss^{-1}h^{-1}ss^{-1}gss^{-1}hs = [g^s, h^s].$$\n$\\square$\n\\end{proof}\n\n\\subsection{Abelian Groups}\n\n\\subsubsection{Abelian Groups}\n\n\\begin{definition}\nA group $G$ with operation $*$ is said to be \\emph{abelian} if the following rule holds\n\\begin{itemize}\n\\item (Commutativity) $ab = ba$ for all $a, b \\in G$.\n\\end{itemize}\n\\end{definition}\n\nWe often write the group operation in an abelian group as $+$ as the number systems $\\Z$, $\\Q$, $\\R$ and $\\C$ all form abelian groups under $+$.\n\nA familiar example of a group which is not abelian is the group of invertible $n\\times n$ matrices under multiplication. \n\n\\subsubsection{Basic Theorems about Abelian Groups}\n\nWe give some criteria for a group to be abelian.\n\n\\begin{theorem}\nLet $G$ be a group. If $(ab)^2 = a^2b^2$ for all $a, b \\in G$ then $G$ is abelian.\n\\end{theorem}\n\n\\begin{proof}\nIf $(ab)^2 = a^2b^2$, then $a(ba)b = a(ab)b$ from which the result follows by cancellation. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA group $G$ is abelian iff $(ab)^{-1} = a^{-1}b^{-1}$ for all $a, b \\in G$.\n\\end{theorem}\n\n\\begin{proof}\nWe have $(ab)^{-1} = b^{-1}a^{-1}$ by Theorem \\ref{product_inverse}. If $G$ is abelian this implies $(ab)^{-1} = a^{-1}b^{-1}$.\n\nAssuming the latter, the converse follows from $b^{-1}a^{-1} = (ab)^{-1} = a^{-1}b^{-1}$ and the bijection $z \\leftrightarrow z^{-1}$ between elements of $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group in which $a = a^{-1}$ for all $a \\in G$ then $G$ is abelian.\n\\end{theorem}\n\n\\begin{proof}\nSuppose $a = a^{-1}$ for all $a \\in G$. Then $ab = a^{-1}b^{-1} = (ba)^{-1} = ba$ for all $b \\in G$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Order of a Nonabelian Group is at least 6}\n\n\\begin{theorem}\nIf $G$ is a non-abelian group then $|G| \\geq 6$.\n\\end{theorem}\n\n\\begin{proof}\nLet $G$ be a non-abelian group and suppose that $ab \\neq ba$ for distinct $a, b \\in G$.\n\nClearly $a = 1$ and $b = 1$ lead to a contradiction. Thus $a, b, 1$ are distinct.\n\nBy application of cancellation we have that $ab \\neq a$, $ab \\neq b$. If $ab = 1$ then $a$ and $b$ are inverse which contradicts $ab \\neq ba$. \n\nSimilar arguments apply for $ba$ and we have that $1$, $a$, $b$, $ab$, $ba$ are distinct. Suppose that these are the only elements in $G$.\n\nFirstly, note $a^2$ cannot equal $ab$, $ba$ or $a$, by cancellation. We can't have $a^2 = b$ otherwise $ab = ba$. Therefore $a^2 = 1$.\n\nThe element $aba$ cannot equal $a$, $ab$ or $ba$ by cancellation. We also can't have $aba = 1$ since multiplication by $a$ on the left leads to a contradiction. Similarly we can't have $aba = b$.\n\nThus $aba$ is distinct from all five elements in the group and the group must have order at least $6$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Quaternion Group $Q_8$}\n\nWe give an example of a finite non-abelian group.\n\n\\begin{definition}\nThe quaternion group $Q_8$ is the multipicative group containing eight elements $\\{1, -1, i, -i, j, -j, k, -k\\}$ which satisfy the relations $i^2 = j^2 = k^2 = -1$, $(-1)^2 = 1$ and $ijk = -1$.\n\\end{definition}\n\nThe remaining entries in the multiplication table for $Q_8$ can be worked out from the given relations:\n$$jk = -i^2jk = -i(ijk) = i.$$\nand\n$$ij = -ijk^2 = -(ijk)k = k,$$\nand\n$$ki = -kij^2 = -k(ij)j = -k^2j = j.$$\n\nWe also have:\n$$kj = k(ki) = k^2i = -i,$$\nand\n$$ji = j(jk) = j^2k = -k,$$\nand\n$$ik = i(ij) = i^2j = -j.$$\n\nNote that $ij \\neq ji$, $jk \\neq kj$ and $ik \\neq ki$, so $Q_8$ is nonabelian.\n\n\\subsection{Cyclic Groups}\n\n\\subsubsection{Cyclic Groups}\n\n\\begin{definition}\nA group $G$ is said to be \\emph{cyclic} if there exists an element $g \\in G$ such that every element of the group is of the form $g^n$ for some $n \\in \\Z$. We call $g$ a \\emph{generator} of $G$ and write $G = $.\n\\end{definition}\n\nThe identity in a cyclic group is $g^0$ and the inverse of $g^n$ is $g^{-n}$.\n\n\\subsubsection{Basic Results Regarding Cyclic Groups}\n\n\\begin{theorem}\nCyclic groups are abelian.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe order of a finite cyclic group $G = $ is the smallest positive integer $n$ for which $g^n = 1$.\n\\end{theorem}\n\n\\begin{proof}\nLet $n$ be as in the theorem. Clearly $g^a = g^b$ for all $a \\equiv b \\pmod{n}$. Therefore all elements of $G$ lie in $\\{g^i \\;|\\; 0 \\leq i < n\\}$.\n\nIt is easy to see that no two such elements are equal. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe generator of an infinite cyclic group is unique.\n\\end{theorem}\n\n\\begin{proof}\nLet $G = $ be infinite cyclic and suppose that $g^a$ is also a generator for some $1 \\neq a \\in \\Z$.\n\nThen we must have $g = (g^a)^n = g^{an}$ for some $n \\in Z$.\n\nMultiplying by $g^{-1}$ we get $g^{an - 1} = 1$ which contradicts the infinite order of $g$. $\\square$\n\\end{proof}\n\nA generator of a finite cyclic group is not generally unique.\n\n\\begin{theorem}\nLet $G = $ be a cyclic group of order $n$. Then $g^a$ is a generator of $G$ iff $a$ coprime to $n$.\n\\end{theorem}\n\n\\begin{proof}\nIf gcd$(a, n) = d$ then clearly all powers of $g^a$ lie in $\\{1, g^d, g^{2d}, \\ldots, g^{n - d}\\}$. Thus if $d \\neq 1$ then $g^a$ doesn't generate $G$.\n\nConversely, if gcd$(a, n) = 1$ there exist $s, t \\in Z$ such that $as + nt = 1$ by the extended Euclidean algorithm. As $g^n = 1$, this implies that $(g^a)^s = g$. Thus every power of $g$ is also a power of $g^a$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA subgroup $H$ of a cyclic group $G = $ is cyclic and is generated by the smallest positive power of $g$ in $H$.\n\\end{theorem}\n\n\\begin{proof}\nSuppose that $g^a$ is the smallest positive power of $g$ in $H$. Clearly all powers of $g^a$ are in $H$.\n\nNo other power of $g$ is in $H$ by the division algorithm and the minimality of $a$. $\\square$\n\\end{proof}\n\n\\subsection{The Centre and Centralizers}\n\n\\subsubsection{The Centre of a Group}\n\n\\begin{definition}\nThe \\emph{centre} of a group $G$ is defined to be the subset\n$$Z_G = \\{g \\in G \\;|\\; ga = ag \\;\\;\\mbox{for all}\\;\\; a \\in G\\}.$$\n\\end{definition}\n\nIn other words, the centre is the set of all elements that commute with every element of $G$.\n\n\\begin{theorem}\nThe centre of a group $G$ is an abelian subgroup of $G$.\n\\end{theorem}\n\n\\begin{proof}\nClearly the identity of $G$ is in the centre of $G$. Therefore $Z_G$ is not empty.\n\nIf $g \\in Z_G$ and $a \\in G$ then $ga = ag$ for all $a \\in G$. Multiplying by $g^{-1}$ on the left and right we see that $g^{-1} \\in Z_G$.\n\nIt is easy to see that if $g, h \\in Z_G$ then $gh \\in Z_G$ and so $Z_G \\leq G$. It is clear that $Z_G$ is abelian. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA group $G$ is abelian iff $Z_G = G$.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\subsubsection{The Centralizer of a Subset of a Group}\n\n\\begin{definition}\nLet $G$ be a group and $S$ a nonempty subset of $G$. The \\emph{centralizer} of $S$ in $G$ is the set\n$$C_G(S) = \\{g \\in G \\;|\\; gs = sg \\;\\;\\mbox{for all}\\;\\; s \\in S\\}.$$\nWhen $S = \\{g\\}$ we write $C_G(S) = C_G(g)$.\n\\end{definition}\n\nIn other words, the centralizer of $S$ is the set of all elements that commute with $S$. The centralizer of $g \\in G$ is the set of all elements that commute with $g$.\n\n\\begin{theorem}\nLet $S$ be a subset of a group $G$. Then $C_G(S) \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\nNote that $C_G(S)$ need not be abelian. For example, if $G$ is nonabelian, $C_G(1) = G$ which is not abelian.\n\n\\subsubsection{Basic Results Regarding Centralizers}\n\n\\begin{theorem}\nLet $G$ be a group. Then $Z_G = \\bigcap_{g \\in G} C_G(g)$.\n\\end{theorem}\n\n\\begin{proof}\nClearly $Z_G \\subseteq \\bigcap_{g \\in G} C_G(g)$.\n\nFor the reverse inclusion, if $x \\in \\bigcap_{g \\in G} C_G(g)$ then $x$ commutes with every $g \\in G$. Thus $g \\in Z_G$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ is a group and $S \\subseteq G$ then $Z_G \\subseteq C_G(S)$.\n\\end{corollary}\n\n\\begin{proof}\nClearly $C_G(S) = \\bigcap_{s \\in S} C_G(s)$. The result therefore follows from the theorem. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group. Then $C_G(Z_G) = G$.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $S \\subseteq T \\subseteq G$, then $C_G(T) \\leq C_G(S)$.\n\\end{theorem}\n\n\\begin{proof}\nIt suffices to show that $C_G(T) \\subseteq C_G(S)$. But this is clear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. Then $H \\subseteq C_G(H)$ iff $H$ is abelian.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\subsection{Conjugacy}\n\n\\subsubsection{Conjugate elements}\n\n\\begin{definition}\nLet $G$ be a group. Two elements $a, b \\in G$ are said to be \\emph{conjugate} if $a = g^{-1}bg$ for some $g \\in G$.\n\\end{definition}\n\n\\begin{theorem}\nIf $G$ is an abelian group then $a, b \\in G$ are conjugate iff $a = b$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group. If $a, b \\in G$ are conjugate then they have the same order.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group. Conjugacy of elements is an equivalence relation on $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $G$ be a group and $g \\in G$. The class of all elements of $G$ that are conjugate to $g$ is called the \\emph{conjugacy class} of $g$.\n\\end{definition}\n\nClearly conjugacy classes partition $G$.\n\n\\subsubsection{Conjugacy and Centralizers}\n\n\\begin{theorem}\\label{conjugacy}\nLet $G$ be a group. The elements of the conjugacy class of $g$ are in bijection with the left cosets of $C_G(g)$.\n\\end{theorem}\n\n\\begin{proof}\nSuppose that $h$ and $h'$ are in the conjugacy class of $g$, i.e. $h = aga^{-1}$ and $h' = bgb^{-1}$ for some $a, b \\in G$.\n\nWe have that $h = h'$ iff $(b^{-1}a)g(b^{-1}a)^{-1} = g$. But this holds iff $b^{-1}a \\in C_G(g)$.\n\nBut $b^{-1}a \\in C_G(a)$ iff $aC_G(g) = bC_G(g)$.\n\nWriting $[g]$ for the conjugacy class of $g$ and $\\mathcal{C}$ for the set of left cosets of $C_G(g)$, we can define a map $\\phi : [g] \\to \\mathcal{C}$ by $h \\mapsto aC_G(g)$ where $h = aga^{-1}$.\n\nBy what we have just shown, this is a well-defined map and it is clearly a bijection. $\\square$\n\\end{proof}\n\n\\subsection{Cosets and Lagrange's Theorem}\n\n\\subsubsection{Cosets}\n\n\\begin{definition}\nLet $G$ be a group and $H \\leq G$. A \\emph{left coset} is a set of the form\n$$aH = \\{ah \\;|\\; h \\in H\\},$$\nfor some $a \\in G$. Similarly a \\emph{right coset} is a set of the form\n$$Ha = \\{ha \\;|\\; h \\in H\\}.$$\n\\end{definition}\n\nClearly if $G$ is abelian, the left coset $aH$ coincides with the right coset $Ha$ for all $a \\in G$.\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. Two left (right) cosets of $H$ either coincide or are disjoint.\n\\end{theorem}\n\n\\begin{proof}\nIf $g \\in aH$ then clearly $gH \\in aH$. Similarly if $gH = aH$ then $g.1 \\in aH$. A similar argument applies for right cosets. $\\square$\n\\end{proof}\n\n\\begin{theorem}\\label{cosets}\nLet $G$ be a group and $H \\leq G$. If $g \\in aH$ then $aH = gH$. Similarly if $g \\in Ha$ then $Ha = Hg$.\n\\end{theorem}\n\n\\begin{proof}\nClear, as $1 \\in H$. $\\square$\n\\end{proof}\n\nNote that these theorems imply that the cosets of $H$ in $G$ partition $G$.\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. There is a bijection between the left and right cosets of $H$ in $G$.\n\\end{theorem}\n\n\\begin{proof}\nDefine a map $\\phi : gH \\to Hg^{-1}$ from left to right cosets.\n\nThe map $\\phi$ is well-defined, since $gH = g'H$ iff $g \\in g'H$ iff $g = g'h$ for some $h \\in H$ iff $g'^{-1} = hg^{-1}$ iff $Hg^{-1} = Hg'^{-1}$.\n\nThe map is clearly bijective. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe number of cosets of $H$ in $G$, if it is finite, is called the \\emph{index} of $H$ in $G$, written $[G:H]$. Otherwise we say that $H$ has infinite index in $G$ and write $[G:H] = \\infty$.\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. There is a bijection between any two left (right) cosets of $H$ in $G$.\n\\end{theorem}\n\n\\begin{proof}\nDefine the map $\\psi : gH \\to hH$ which sends $a \\mapsto hg^{-1}a$.\n\nAs the map is invertible, it is bijective. A similar argument holds for right cosets. $\\square$\n\\end{proof}\n\n\\begin{corollary} \\label{cosetsize}\nIf $G$ is a finite group and $H \\leq G$ then $|gH| = |hH|$ for all $g, h \\in G$.\n\\end{corollary}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\subsubsection{Lagrange's Theorem}\n\n\\begin{theorem} (Lagrange)\nIf $G$ is a group and $H \\leq G$ then $|H| \\;|\\; |G|$.\n\\end{theorem}\n\n\\begin{proof}\nFollows from Corollary \\ref{cosetsize} and the fact that cosets partition a group. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ is a finite group and $H \\leq G$ then $[G:H] = |G|/|H|$.\n\\end{corollary}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ is a finite group, the order of $g$ must divide $G$.\n\\end{corollary}\n\n\\begin{proof}\nThe subgroup generated by $g$, $\\langle g \\rangle = \\{g^i \\;|\\; i \\in \\Z\\}$ is a subgroup of $G$. We have already shown that if the order of $g$ is $n$, then there are $n$ distinct powers of $g$, which it is easy to see are the elements of $\\langle g \\rangle$.\n\nThe result then follows from Lagrange's theorem. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf a group $G$ has prime order $p$, then it is cyclic.\n\\end{theorem}\n\n\\begin{proof}\nLet $g$ be any element of $G$ other than the identity. The result follows from the fact that the order of $g$ must divide $p$. Thus $\\langle g \\rangle = G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a finite group and $K \\leq H \\leq G$, then\n$$[G:K] = [G:H][H:I].$$\n\\end{theorem}\n\n\\begin{proof}\n$[G:K] = |G|/|K| = (|G|/|H|)(|H|/|K|) = [G:H][G:K]$. $\\square$\n\\end{proof}\n\n\\begin{theorem} \\label{gtoorder}\nIf $G$ is a finite group of order $n$ and $g \\in G$ then $g^n = e$.\n\\end{theorem}\n\n\\begin{proof}\nWe have that the order of $g$ divides $n$. $\\square$\n\\end{proof}\n\n\\begin{theorem} (Euler)\nIf gcd$(a, n) = 1$ then $a^{\\varphi(n)} = 1$.\n\\end{theorem}\n\n\\begin{proof}\nThe function $\\varphi(n)$ counts the number of integers $i$ in $0 \\leq i < n$ such that $gcd(i, n) = 1$.\n\nIt is easy to see that this set of integers is a group under multiplication modulo $n$. Euler's theorem is an application of Theorem \\ref{gtoorder}. $\\square$\n\\end{proof}\n\n\\begin{corollary} (Fermat's Little Theorem)\nIf $p$ is a prime then $a^{p-1} \\equiv 1 \\pmod{p}$ for all $1 \\leq a < p$.\n\\end{corollary}\n\n\\begin{proof}\nThis is Euler's theorem where $n = p$, as $\\varphi(p) = p - 1$. $\\square$\n\\end{proof}\n\n\\subsection{Normal Subgroups and Normalizers}\n\n\\subsubsection{Normal Subgroups}\n\n\\begin{definition}\nLet $G$ be a group. A subgroup $N$ is said to be \\emph{normal} if $gN = Ng$ for all $g \\in G$. We write $N \\mathrel{\\unlhd} G$.\n\\end{definition}\n\n\\begin{theorem}\\label{equivnormal}\nLet $G$ be a group and $N \\leq G$. The following are equivalent:\n\\begin{enumerate}\n\\item $N \\mathrel{\\unlhd} G$\n\\item $g^{-1}ng \\in N$ for all $n \\in N$ and $g \\in G$\n\\item For all $g, h \\in G$, $gh \\in N$ iff $hg \\in N$\n\\item For all $g \\in G$, $g^{-1}Ng \\subseteq N$\n\\item For all $g \\in G$, $g^{-1}Ng = N$\n\\item All left cosets are right cosets and conversely\n\\item For all $x, y, g, h \\in G$, $x \\in gN$ and $y \\in hN$ implies $xy \\in (gh)N$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\n(1)$\\iff$ (2) We have that $gN = Ng$ iff for all $n \\in N$, $ng \\in gN$. Multiplication on the left by $g^{-1}$ gives the result.\n\n(1)$\\iff$ (3) We have that $gh \\in N$ iff $h \\in g^{-1}N$. This is equal to $Ng^{-1}$ for all $g \\in G$ iff (1) holds. But $h \\in Ng^{-1}$ iff $hg \\in N$.\n\n(1)$\\iff$ (5) Clear.\n\n(5)$\\implies$ (4) Clear.\n\n(4)$\\implies$ (1) Clearly (4) implies $Ng \\subseteq gN$. But replacing $g$ with $g^{-1}$ in (4) also yields $gN \\subseteq Ng$. Thus we have (1).\n\n(1)$\\iff$ (6) Clear.\n\n(1)$\\implies$ (7) If $x \\in gN$ and $y \\in hN$, then (1) implies $xy \\in gNhN = (gh)NN = (gh)N$.\n\n(7)$\\implies$ (1) Another way of writing (7) is that for $g, h \\in G$ and $n, n' \\in N$, $(gn)(hn') = (gh)n''$ for some $n'' \\in N$. Multiplication on the left by $g^{-1}$ and on the right by the inverse of $n'$ yields a statement equivalent to (1). $\\square$\n\\end{proof}\n\n\\begin{theorem}\nAll subgroups of an abelian group are normal.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group then the trivial subgroup $\\{e\\}$ and $G$ itself are normal subgroups of $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group then $Z_G \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nClearly $g^{-1}zg = z$ for all $z \\in Z_G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $H \\mathrel{\\unlhd} G$ and $H \\leq K \\leq G$ then $H \\mathrel{\\unlhd} K$.\n\\end{theorem}\n\n\\begin{proof}\nWe have that $kH = Hk$ for all $k \\in K \\subseteq G$. $\\square$\n\\end{proof}\n\n\\subsubsection{A Subgroup of Order $2$ is Normal}\n\n\\begin{theorem}\nIf $G$ is a group and $H \\leq G$ with $[G:H] = 2$ then $H \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $H$ and $aH$ be the two distinct left cosets. This implies that $a \\notin H$, which in turn implies that $Ha \\neq H$.\n\nAs the number of left and right cosets is equal, the right cosets must be $H$ and $Ha$. But then it is easy to see every left coset of is a right coset and vice versa and so $H$ is normal in $G$. $\\square$\n\\end{proof}\n\n\\subsubsection{Intersection and Products of Normal Subgroups}\n\n\\begin{theorem}\nIf $G$ is a group and $M, N \\mathrel{\\unlhd} G$ then $M \\cap N \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nWe already know $M \\cap N \\leq G$. \n\nIf $a \\in M \\cap N$ then $g{-1}ag \\in M, N$ for all $g \\in G$. Thus $M \\cap N$ is normal in $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $M \\leq G$ and $N \\mathrel{\\unlhd} G$ then $MN, NM \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\nIf $N \\mathrel{\\unlhd} G$ then $g^{-1}Ng \\in N$ for all $g \\in G$. In particular, this holds for all $g \\in M$ and $Ng = gN$ for all $g \\in M$, i.e. $NM = MN$. Thus by Theorem \\ref{HKKH} $MN, NM \\leq G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $M, N \\mathrel{\\unlhd} G$ then $MN, NM \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nBy the previous theorem $MN, NM \\leq G$.\n\nIf $a = nm \\in NM$ for some $n \\in N$ and $m \\in M$ and $g \\in G$ then $g^{-1}ag = (g^{-1}ng)(g^{-1}mg) \\in NM$. Thus $NM \\mathrel{\\unlhd} G$. A similar argument shows that $MN \\mathrel{\\unlhd} G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $H \\leq G$ and $N \\mathrel{\\unlhd} G$ then $H\\cap N \\mathrel{\\unlhd} H$.\n\\end{theorem}\n\n\\begin{proof}\nLet $x \\in H\\cap N$ and $h \\in H$. Then $h^{-1}xh \\in H$.\n\nWe also have $h^{-1}xh \\in N$ as $x \\in N \\mathrel{\\unlhd} G$.\n\nThus $h^{-1}xh \\in H\\cap N$ for all $h \\in H$ and $x \\in H\\cap N$. Thus $H\\cap N \\mathrel{\\unlhd} H$. $\\square$\n\\end{proof}\n\n\\subsubsection{Congugate subgroups}\n\n\\begin{definition}\nLet $G$ be a group and $H \\leq G$. If $g \\in G$ then $g^{-1}Hg$ is called a \\emph{conjugate} of $H$.\n\\end{definition}\n\n\\begin{theorem}\nIf $G$ is a group with $H \\leq G$ and $g \\in G$ then $g^{-1}Hg \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $K = g^{-1}Hg$. It is easy to show that $1 \\in K$, $g^{-1} \\in K$ for any $g \\in K$ and $ab \\in K$ for any $a, b \\in K$. Thus $K$ is a subgroup of $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $H \\leq G$ then $H$ is in bijection with any conjugate subgroup $g^{-1}Hg$.\n\\end{theorem}\n\n\\begin{proof}\nThe map $h \\mapsto g^{-1}hg$ is invertible and hence a bijection. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $G$ be a group and $H \\leq G$. The intersection, $\\cap_{g \\in G} g^{-1}Hg$, of the conjugates of $H$ in $G$ is called the \\emph{normal core} of $H$ in $G$.\n\\end{definition}\n\n\\begin{theorem}\nThe normal core of a subgroup $H$ of a group $G$ is a normal subgroup of $G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $K = \\cap_{g \\in G} g^{-1}Hg$. Clearly $K$ is a subgroup of $G$ as it is an intersection of subgroups of $G$.\n\nAs $M \\to g^{-1}Mg$ is a bijection of conjugates of $H$ we have that $k \\in K$ implies $g^{-1}kg \\in K$. Thus $K$ is normal in $G$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Normalizer of a Subset}\n\n\\begin{definition}\nLet $G$ be a group and $S \\subseteq G$, then the \\emph{normalizer} of $S$ in $G$ is the set\n$$N_G(S) = \\{g \\in G \\;|\\; g^{-1}Sg = S\\}.$$\nIf $S = \\{s\\}$ we usually write $N_G(s) = N_G(S)$.\n\\end{definition}\n\nNote that $N_G(S)$ need not be a normal subgroup of $G$.\n\n\\begin{theorem}\nIf $G$ is a group and $\\emptyset \\neq S \\subseteq G$ then $N_G(S) \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $S \\subseteq G$ then $Z_G \\in N_G(S)$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Basic Theorems Regarding the Normalizer of a Subset}\n\n\\begin{theorem}\nLet $G$ be a group and $g \\in G$. Then $N_G(g) = C_G(g)$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$, then\n$$C_G(H) \\leq N_G(H) \\leq G.$$\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group, then $N_G(Z_G) = G$.\n\\end{theorem}\n\n\\begin{proof}\nWe have that $G = C_G(Z_G) \\leq N_G(Z_G) = G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. Then $H \\leq N_G(H)$.\n\\end{theorem}\n\n\\begin{proof}\nIt is clear that $H \\subseteq N_G(H)$. As both are groups, the result follows. $\\square$\n\\end{proof}\n\n\\subsubsection{The Commutator Subgroup of a Group is Normal}\n\n\\begin{theorem}\nLet $G$ be a group. Then $[G, G] \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nLetting $g^s = s^{-1}gs$. Recall that any element of $[G, G]$ is a finite product of commutators.\n\nThe result follows from\n\\begin{align*}\n([g_1, h_1][g_2, h_2]\\ldots [g_n, h_n])^s &= [g_1, h_1]^s[g_2, h_2]^s\\ldots [g_n, h_n]^s\\\\\n &= [g_1^s, h_1^s][g_2^s, h_2^s]\\ldots [g_n^s, h_n^s].\n\\end{align*} $\\square$\n\\end{proof}\n\n\\subsubsection{Unique Subgroups of a Given Order}\n\n\\begin{theorem}\nLet $G$ be a group and suppose that $H$ is the only subgroup of $G$ of (finite) order $n$. Then $H$ is a normal subgroup of $G$.\n\\end{theorem}\n\n\\begin{proof}\nAs $H$ is in bijection with all its conjugates $g^{-1}Hg$, all its conjugates also have order $n$ and are thus equal to $H$. Thus $H$ is a normal subgroup of $G$. $\\square$\n\\end{proof}\n\n\\subsection{Quotient Groups}\n\n\\subsubsection{Quotient Groups}\n\n\\begin{theorem}\nLet $G$ be a group and $N \\mathrel{\\unlhd} G$. If $aH = a'N$ and $bN = b'N$ for $a, a', b, b' \\in G$ then $abN = a'b'N$.\n\\end{theorem}\n\n\\begin{proof}\nThis follows immediately from Theorem \\ref{equivnormal} (7) and Theorem \\ref{cosets}. $\\square$\n\\end{proof}\n\nThis allows us to define a well-defined operation on cosets of a normal subgroup.\n\n\\begin{theorem}\nLet $G$ be a group an $N \\mathrel{\\unlhd} G$. The cosets of $N$ form a group under the operation\n$$(aN).(bN) = abN.$$\n\\end{theorem}\n\n\\begin{proof}\nThe identity is the coset $N$. The inverse of $gN$ is $g^{-1}N$ and the associative law clearly holds. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe group of cosets of a normal subgroup $N$ of a group $G$ is called the \\emph{quotient group} of $G$ by $N$, which we denote $G/N$.\n\\end{definition}\n\nNote that we have used left cosets in this definition. As $N$ is normal in $G$ we have $gN = Ng$ for all $g \\in G$. Therefore we can equally well give the definition in terms of right cosets. The quotient group in that case is usually denoted $N\\backslash G$.\n\n\\begin{theorem}\nIf $N \\mathrel{\\unlhd} G$ then\n$$|G/N| = [G:N].$$\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and $N \\mathrel{\\unlhd} G$. Then:\n\\begin{enumerate}\n\\item If $G$ is abelian then so is $G/N$\n\\item If $G$ is finite then so is $G/N$\n\\item If $G$ is cyclic then so is $G/N$\n\\item If $G$ is finitely generated then so is $G/N$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\n(1) $(aN)(bN) = abN = baN = (bN)(aN)$.\n\n(2) Clear.\n\n(3) Let $G = \\langle g \\rangle$. The elements of $G/N$ are precisely the elements $g^iN = (gN)^i$ for $i \\in \\mathbb{N}$.\n\n(4) If $G = \\langle g_1, g_2, \\ldots, g_n \\rangle$ then $G/N = \\langle g_1N, g_2N, \\ldots, g_nN \\rangle$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $G/Z_G$ is cyclic then $G$ is abelian.\n\\end{theorem}\n\n\\begin{proof}\nAs $G/Z_G$ is cyclic $G/Z_G = \\langle aZ_G \\rangle$ for some $a \\in G$. Then $g = a^rk_1$ and $h = a^sk_2$ for some $r, s \\in \\Z$ and $k_1, k_2 \\in Z_G$.\n\nThus\n$$gh = a^rk_1a^sk_2 = a^ra^sk_1k_2 = a^sa^rk_1k_2 = a^sk_2a^rk_1 = hg.$$ $\\square$\n\\end{proof}\n\n\\subsubsection{The Abelianisation of a Group}\n\n\\begin{theorem}\nIf $G$ is a group then $G/[G, G]$ is abelian. It is the largest quotient of $G$ which is abelian.\n\\end{theorem}\n\n\\begin{proof}\nThe first part follows from $ab[G, G] = ab[b, a][G, G] = ba[G, G]$.\n\nSuppose that $N$ is a normal subgroup of $G$, but $[g, h] \\notin N$ for some $g, h \\in G$.\n\nClearly $[g, h] \\neq 1$, as $1 \\in N$. Thus $gh \\neq hg$.\n\nBut we also have that $ghN \\neq hgN$, as $g^{-1}h^{-1}gh \\notin N$. Thus $(gN)(hN) \\neq (hN)(gN)$ and $G/N$ is not abelian.\n\nThus if $G/N$ is abelian, $[G, G] \\subseteq N$ which shows the result. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe group $G/[G, G]$ is called the \\emph{abelianisation} of $G$.\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\mathrel{\\unlhd} G$. Then $G/H$ is abelian iff $[G, G] \\leq H$.\n\\end{theorem}\n\n\\begin{proof}\nSuppose $G/H$ is abelian. If $g_1, g_2 \\in G$ then we have $(g_1H)(g_2H) = (g_2H)(g_1H)$. Thus $g_1^{-1}g_2^{-1}g_1g_2 \\in H$. This shows $[G, G] \\in H$. \n\nThe argument can be reverse to show the converse. $\\square$\n\\end{proof}\n\n\\subsubsection{The Correspondence Theorem}\n\n\\begin{theorem}\nIf $G$ is a group and $N \\mathrel{\\unlhd} G$ then there is an order preserving bijection\n$$\\{\\mbox{subgroups of}\\;\\;G\\;\\mbox{containing}\\;N \\} \\leftrightarrow \\{\\mbox{subgroups of}\\;\\;G/N\\}.$$ \n\\end{theorem}\n\n\\begin{proof}\nIf $N \\leq A \\leq G$ then $N \\mathrel{\\unlhd} A$. Thus we can define the map $\\phi : A \\mapsto A/N$. The group Clearly $A/N \\subseteq G/N$. It is easy to show that $A/N \\leq G/N$.\n\nConversely, suppose $H' \\leq G/N$. Let $H = \\{h \\in G \\;|\\; hN \\in H'\\}$. Clearly $N \\subseteq H$. It is also easy to show that $H \\leq G$. We see that $H' = \\phi(H)$ and so $\\phi$ is a surjection.\n\nIf $A/N = B/N$ for $A, B \\leq G$ containing $N$, then $A$ and $B$ consist of precisely the same cosets of $N$ in $G$. Thus $A = B$ and $\\phi$ is injective.\n\nIt is easy to see that $\\phi$ is order preserving. $\\square$\n\\end{proof}\n\n\\begin{definition}\nIf $G$ is a group and $A, B \\leq G$ we write $\\langle A, B \\rangle = \\langle A\\cup B \\rangle$.\n\\end{definition}\n\n\\begin{theorem} \\label{normalcontaining}\nIf $G$ is a group, $A, B \\leq G$ and $N \\mathrel{\\unlhd} G$ then\n\\begin{enumerate}\n\\item If $A \\subseteq B$ for a finite group $A$ then $|B:A| = |B/N:A/N|$\n\\item $\\langle A, B \\rangle/N = \\langle A/N, B/N \\rangle$\n\\item $(A\\cap B)/N = (A/N)\\cap (B/N)$\n\\item $A \\mathrel{\\unlhd} G$ iff $A/N \\mathrel{\\unlhd} G/N$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\n(1) Follows from $|B:A| = |B|/|A|$ and $|B/N| = |B|/|N|$.\n\n(2) $\\langle A, B \\rangle/N$ consists of the cosets $gN$ for $g \\in \\langle A\\cup B \\rangle$. But as $(aN)(bN) = abN$ and $(aN)^{-1} = a^{-1}N$ a set of generators of this group is given by $(A/N)\\cup (B/N)$, which gives the result.\n\n(3) Note that $N \\mathrel{\\unlhd} A\\cap B$ and so the result makes sense. It is clear that both groups consist of the cosets $gN$ for $g \\in A\\cap B$.\n\n(4) We have that $A \\mathrel{\\unlhd} G$ iff $g^{-1}ag \\in A$ for all $a \\in A$ and $g \\in G$. But this holds iff $g^{-1}agN \\in A/N$ which is true iff $(g^{-1}N)(aN)(gN) \\in A/N$ for all $gN \\in G/N$, which is precisely the statement of the result. $\\square$\n\\end{proof}\n\n\\subsection{Homomorphisms}\n\n\\subsubsection{Homomorphisms}\n\n\\begin{definition}\nIf $G$ and $H$ are groups a \\emph{homomorphism} from $G$ to $H$ is a map $\\phi : G \\to H$ such that $\\phi(g_1g_2) = \\phi(g_1)\\phi(g_2)$ for all $g_1, g_2 \\in G$.\n\\end{definition}\n\nOf course this is the multiplicative version of the definition. There is a similar definition with addition instead of multiplication in the additive case.\n\n\\begin{theorem}\nIf $\\phi : G \\to H$ is a homomorphism then $\\phi(g^{-1}) = \\phi(g)^{-1}$ and $\\phi(1) = 1$.\n\\end{theorem}\n\n\\begin{proof}\nFrom $\\phi(1) = \\phi(1.1) = \\phi(1)\\phi(1)$ we see that $\\phi(1) = 1$.\n\nFrom $1 = \\phi(1) = \\phi(gg^{-1}) = \\phi(g)\\phi(g^{-1})$ we see that $\\phi(g^{-1}) = \\phi(g)^{-1}$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $H \\leq G$ then the map $\\iota : H \\to G$ which sends $h \\mapsto h$ for all $h \\in H$ is a homomorphism, called the canonical injection.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is a group and $N \\mathrel{\\unlhd} G$ then the map $\\rho : G \\to G/N$ which maps $g \\mapsto gN$ for all $g \\in G$ is a homomorphism, called the canonical projection.\n\\end{theorem}\n\n\\begin{proof}\nWe have that $\\rho(gh) = ghN = (gN)(hN) = \\rho(g)\\rho(h)$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Kernel and Image of a Homomorphism}\n\n\\begin{definition}\nLet $\\phi : G \\to H$ be a homomorphism. The \\emph{kernel} of $\\phi$ is the set\n$$\\mbox{ker}(\\phi) = \\{g \\in G \\;|\\; \\phi(g) = 1\\}.$$\n\\end{definition}\n\n\\begin{theorem}\nIf $\\phi : G \\to H$ is a homomorphism then ker$(\\phi) \\mathrel{\\unlhd} G$. Moreover, if $N \\mathrel{\\unlhd} G$ is any normal subgroup, it is the kernel of some homomorphism from $G$.\n\\end{theorem}\n\n\\begin{proof}\nIt is easy to see that ker$(\\phi)$ is a subgroup of $G$.\n\nTo see that it is normal, let $g \\in G$ and $k \\in \\mbox{ker}(\\phi)$, then $\\phi(g^{-1}kg) = 1$. Thus $g^{-1}kg \\in \\mbox{ker}(\\phi)$.\n\nTo show the second part of the theorem it is enough to note that the kernel of the canonical projection $\\rho : G \\to G/N$ is $N$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nIf $\\phi : G \\to H$ is a homomorphism, the \\emph{image} of $\\phi$ is the set\n$$\\mbox{im}(\\phi) = \\{h \\in H \\;|\\; \\mbox{there exists}\\;\\; g \\in G \\;\\;\\mbox{such that}\\;\\; \\phi(g) = h\\}.$$\n\\end{definition}\n\n\\begin{theorem}\nIf $\\phi : G \\to H$ is a homomorphism then im$(\\phi) \\leq H$.\n\\end{theorem}\n\n\\begin{proof}\nClear.\n\\end{proof}\n\n\\begin{theorem}\nLet $\\phi : G \\to H$ be a surjective homomorphism and let $N$ be a normal subgroup of $G$. Then $\\phi(N) \\mathrel{\\unlhd} H$.\n\\end{theorem}\n\n\\begin{proof}\nLet $h \\in H$ and $n' \\in \\phi(N)$. As $\\phi$ is surjective, we can find $g \\in G$ such that $h = \\phi(g)$. We also have $n' = \\phi(n)$ for some $n \\in N$ by definition.\n\nAs $N \\mathrel{\\unlhd} G$ we have that $g^{-1}ng \\in N$. Thus $h^{-1}n'h = \\phi(g)^{-1}\\phi(n)\\phi(g) \\in \\phi(N)$. As $h$ and $n'$ were arbitrary, this shows that $\\phi(N)$ is normal in $H$. $\\square$\n\\end{proof}\n\n\\subsubsection{Endomorphisms of a Group}\n\n\\begin{definition}\nLet $G$ be a group. An \\emph{endomorphism} of $G$ is a homomorphism $\\phi : G \\to G$. The set of all endomorphisms of $G$ is denoted End$(G)$.\n\\end{definition}\n\n\\begin{theorem}\nIf $G$ is an abelian group under addition, then End$(G)$ has the structure of an additive abelian group.\n\\end{theorem}\n\n\\begin{proof}\nIf $\\phi_1$ and $\\phi_2$ are elements of End$(G)$ we define $(\\phi_1 + \\phi_2)(g) = \\phi_1(g) + \\phi_2(g)$ for all $g \\in G$. If it easy to check that this gives the structure of an abelian group on End$(G)$, with the identity map serving as the identity of End$(G)$. $\\square$\n\\end{proof}\n\nIn fact, the group End$(G)$ can even be given the structure of a ring, where the multiplication in the ring is given by composition of homomorphisms. However, we will not prove this here.\n\n\\subsubsection{Isomorphisms}\n\n\\begin{definition}\nA homomorphism of groups $\\phi : G \\to H$ is said to be an \\emph{isomorphism} if there exists a homomorphism $\\phi^{-1} : H \\to G$ such that $\\phi^{-1}\\circ \\phi =$ id$_G$. We say that $G$ and $H$ are isomorphic, which we denote by $G \\cong H$.\n\\end{definition}\n\n\\begin{theorem}\nA homomorphism $\\phi : G \\to H$ is an isomorphism iff it is a bijection.\n\\end{theorem}\n\n\\begin{proof}\nLet $\\phi^{-1} : H \\to G$ be the inverse map. Clearly it is a homomorphism, which also composes with $\\phi$ to give the identity. $\\square$\n\\end{proof}\n\n\\subsubsection{Automorphisms}\n\n\\begin{definition}\nIf $G$ is a group, an \\emph{automorphism} of $G$ is an isomorphism $\\phi : G \\to G$. We denote the set of automorphisms of $G$ by Aut$(G)$.\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group and $a \\in G$. The map $\\sigma_a : g \\mapsto a^{-1}ga$, called \\emph{conjugation} by $a$ is an automorphism.\n\\end{theorem}\n\n\\begin{proof}\nThe map is clearly a bijection. We have that\n$$\\sigma_a(g_1g_2) = a^{-1}g_1g_2a = (a^{-1}g_1a)(a^{-1}g_2a) = \\sigma_a(g_1)\\sigma_a(g_2).$$\nThus $\\sigma_a$ is a homomorphism. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $G$ be a group. An automorphism of $G$ that is conjugation $\\sigma_a$ for some $a \\in G$ is said to be an \\emph{inner autormorphism} of $G$. We write Inn$(G)$ for the set of inner autormorphisms of $G$.\n\\end{definition}\n\n\\begin{theorem}\nFor a group $G$, the set Aut$(G)$ is a group under composition.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nFor a group $G$, Inn$(G) \\mathrel{\\unlhd}$ Aut$(G)$.\n\\end{theorem}\n\n\\begin{proof}\nConjugation by $1 \\in G$ is an inner automorphism. It is the identity of Aut$(G)$.\n\nIt is clear that $\\sigma_{g^{-1}}$ is the inverse of $\\sigma_g$ in Aut$(G)$ and is an inner automorphism. It is easy to see that $\\sigma_{ab}$ is the composition of $\\sigma_a$ and $\\sigma_b$ and so Inn$(G)$ is a subgroup of Aut$(G)$.\n\nTo show that it is normal, let $\\sigma_g \\in$ Inn$(G)$ and $\\phi \\in$ Aut$(G)$. Then for $h \\in G$:\n\\begin{align*}\n\\phi^{-1}\\cdot\\sigma_g\\cdot\\phi(h) &= \\phi\\circ \\sigma_g \\circ \\phi^{-1}(h)\\\\\n &= \\phi(g^{-1}\\phi^{-1}(h)g)\\\\\n &= \\phi(g^{-1})h\\phi(g)\\\\\n &= \\sigma_{\\phi(g)}(h) \\in \\mbox{Inn}(G)\n\\end{align*} $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group and let $\\phi : G \\to$ Aut$(G)$ be the map $g \\mapsto \\sigma_g$. Then $\\phi$ is a homomorphism. We refer to this as the natural homomorphism from $G$ to Aut$(G)$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $\\phi : G \\to$ Aut$(G)$ be the natural homomorphism. Then ker$(\\phi) = Z_G$.\n\\end{theorem}\n\n\\begin{proof}\n$\\phi(g) =$ id$_G$ iff $g^{-1}ag = a$ for all $a \\in G$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Automorphism Group of a Cyclic Group}\n\n\\begin{theorem}\nThe automorphism group of $\\mathbb{Z}$ is isomorphic to $C_2$.\n\\end{theorem}\n\n\\begin{proof}\nAn automorphism of $\\mathbb{Z}$ must send $1$ to a generator of $\\mathbb{Z}$. As the only generators are $\\pm 1$, the result follows. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $C_n = \\langle g \\rangle$ be a cyclic group of order $n$. The automorphisms of $C_n$ consist of the maps $a \\mapsto a^k$ for $1 \\leq k < n$ with gcd$(k, n) = 1$.\n\\end{theorem}\n\n\\begin{proof}\nLet $\\sigma \\in$ Aut$(C_n)$. Then $\\sigma(g) = g^k$ for some $k$. In order to be an automorphism $g^k$ must generate $C_n$. Thus gcd$(k, n) = 1$.\n\nConsider any element $a = g^r \\in C_n$. Then $\\sigma(a) = \\sigma(g)^r = (g^k)^r = a^k$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nWe have $|\\mbox{Aut}(C_n)| = \\varphi(n)$.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\nNote that Aut$(C_n)$ is an abelian group since $(a^r)^s = (a^s)^r$.\n\n\\subsection{Characteristic Subgroups}\n\n\\subsubsection{Characteristic Subgroups}\n\n\\begin{definition}\nLet $G$ be a group. A subgroup $H \\leq G$ is said to be a \\emph{characteristic subgroup} if $\\phi(H) = H$ for all $\\phi \\in$ Aut$(G)$.\n\\end{definition}\n\n\\begin{theorem}\nA characteristic subgroup $H$ of a group $G$ is normal in $G$.\n\\end{theorem}\n\n\\begin{proof}\n$H$ is preserved under all inner automorphisms, i.e. under conjugation by elements of $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $H$ is a characteristic subgroup of a group $G$ then every $\\phi \\in$ Aut$(G)$ induces an automorphism of $G/H$ which gives a map Aut$(G) \\to$ Aut$(G/H)$.\n\\end{theorem}\n\n\\begin{proof}\nAs $\\phi$ preserves $H$, the induced map $gH \\mapsto \\phi(g)H$ is well defined and an injection. It is clearly a surjection because $\\phi$ is invertible. It is clearly a homomorphism. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $H$ be a subgroup of $G$ of index $n$. If it is the only subgroup of $G$ of index $n$ then it is a characteristic subgroup.\n\\end{theorem}\n\n\\begin{proof}\nAny automorphism $\\phi$ of $G$ must send $H$ to a subgroup $H'$ of $G$. The induced map $H \\to H'$ must be an isomorphism since $\\phi$ is injective.\n\nLet $g_1H, g_2H, \\ldots, g_nH$ be the set of left cosets of $H$ in $G$, for some fixed set of representatives $g_i$.\n\nThe map $\\phi$ sends each coset $g_iH$ to a coset $\\phi(g_i)H'$ of $H'$ in $G$. As every element of $G$ has a unique representation as $g_ih$ for some $i$ and $h \\in H$ and $\\phi$ is an automorphism, the cosets $\\phi(g_i)H'$ must partition $G$.\n\nThus the index of $H'$ in $G$ is $[G:H]$. But $H$ is the only subgroup with this index, thus $H = H'$. $\\square$\n\\end{proof}\n\nLet us write $H$ char $G$ to denote that $H$ is a characteristic subgroup of $G$.\n\n\\begin{theorem}\nIf $K$ char $H$ and $H$ char $G$ then $K$ char $G$.\n\\end{theorem}\n\n\\begin{proof}\nClearly any automorphism of $G$ induces an automorphism of $H$ which in turn induces an automorphism of $K$. $\\square$\n\\end{proof}\n\n\\subsubsection{A Characteristic Subgroup of a Normal Subgroup is Normal}\n\n\\begin{theorem}\nIf $K$ char $H \\mathrel{\\unlhd} G$ then $K \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nAny inner automorphism of $G$ preserves $H$ because it is normal. Thus it induces an automorphism of $H$, which preserves $K$ because it is a characteristic subgroup of $H$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Centre of a Group is a Characteristic Subgroup}\n\n\\begin{theorem}\nLet $G$ be a group. Then $Z_G$ char $G$.\n\\end{theorem}\n\n\\begin{proof}\nAn automorphism $\\phi$ of $G$ maps $Z_G$ isomorphically to a subgroup $Z$ of $G$. But if $z \\in Z_G$ and $g \\in G$ we have $\\phi(z)\\phi(g) = \\phi(g)\\phi(z)$.\n\nBecause $\\phi$ is an automorphism of $G$, this shows that every element of $Z$ commutes with every element of $G$. Thus $Z \\leq Z_G$. By inverting $\\phi$ we see that $Z_G \\leq Z$. $\\square$\n\\end{proof}\n\n\\subsubsection{Subgroups of Cyclic Groups are Characteristic}\n\n\\begin{theorem}\nA subgroup $H$ of a cyclic group $G$ is characteristic.\n\\end{theorem}\n\n\\begin{proof}\nLet $G = \\langle g \\rangle$. Any automorphism $\\phi$ of $G$ sends $g$ to $g^n$ for some integer $n$ such that $\\langle g \\rangle = \\langle g^n \\rangle$.\n\nIf $H \\leq G$ then $H = \\langle g^k \\rangle$ for some integer $k$.\n\nNow $phi(H) = \\langle \\phi(g^k) \\rangle = \\langle g^{nk} \\rangle$.\n\nClearly $\\langle g^{nk} \\rangle \\subseteq \\langle g^k \\rangle$. Therefore it remains to show that $\\langle g^k \\rangle \\subseteq \\langle g^{nk}$.\n\nAs $\\langle g^n \\rangle = G$ there exists an integer $s$ such that $g = g^{ns}$.\n\nThen $g^k = (g^{ns})^k = (g^{nk})^s \\in \\langle g^{nk} \\rangle$.\n\nThus $\\phi(H) = H$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Commutator Subgroup is Characteristic}\n\n\\begin{theorem}\nIf $G$ is a group then $[G, G]$ char $G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $S$ be the set of commutators of $G$. If $[x, y] \\in S$ and $\\phi \\in$ Aut$(G)$ then $\\phi([x, y] = [\\phi(x), \\phi(y)]$.\n\nThis is clearly a bijective map on $S$. As $[G, G]$ is the smallest subgroup of $G$ containing $S$, we have that $\\phi([G, G]) = [G, G]$. $\\square$\n\\end{proof}\n\n\\subsubsection{Subgroups with Elements of Order Dividing $p$}\n\n\\begin{theorem}\nLet $G$ be a group and $N = \\{x \\in G \\;|\\; x^p = 1\\}$ for $p$ a prime. Then $N$ char $G$.\n\\end{theorem}\n\n\\begin{proof}\nAny automorphism of $G$ sends $N$ to an isomorphic subgroup, in which every element must have order $p$ or $1$. In other words, $N$ is sent to an isomorphic subgroup of $N$. $\\square$\n\\end{proof}\n\n\\subsection{The Group Isomorphism Theorems}\n\n\\subsubsection{The First Group Isomorphism Theorem}\n\n\\begin{theorem}\nLet $\\phi : G \\to H$ be a group homomorphism. Then $G/\\mbox{ker}(\\phi) \\cong \\mbox{im}(\\phi)$.\n\\end{theorem}\n\n\\begin{proof}\nLet $K = \\mbox{ker}(\\phi)$. Define $\\psi : G/K \\to \\phi(G)$ by $\\psi(aK) = \\phi(a)$.\n\nThe map $\\psi$ is well defined, for if $aK = bK$ then $a = bk$ for some $k \\in K$ and $\\phi(a) = \\phi(b)\\phi(k) = \\phi(b)$.\n\nIt is easy to see that $\\psi$ is a homomorphism on $G/K$.\n\nWe can check that $\\psi$ is injective, since if $\\phi(a) = \\phi(b)$ then $ab^{-1} \\in K$ and so $aK = bK$.\n\nClearly $\\psi$ is surjective and thus an isomorphism. $\\square$\n\\end{proof}\n\n\\subsubsection{Quotients of quotient groups}\n\n\\begin{theorem}\nLet $H$ and $K$ be normal subgroups of a group $G$ and let $K \\leq H$. Then $(G/K)/(H/K) \\cong G/H$.\n\\end{theorem}\n\n\\begin{proof}\nAs $K \\mathrel{\\unlhd} G$ we have that $K \\mathrel{\\unlhd} H$. As $H \\mathrel{\\unlhd} G$ and both groups contain $K$, we have that $H/K \\mathrel{\\unlhd} G/K$. Thus the theorem makes sense.\n\nWe define the map $\\phi : G/K \\to G/H$ by $\\phi(gK) = gH$. It is easy to check that $\\phi$ is well-defined and a homomorphism.\n\nWe have that ker$(\\phi) = \\{gK \\in G/K \\;|\\; g \\in H\\} = H/K$. Then by the first isomorphism theorem $(G/K)/(H/K) \\cong G/H$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Second Isomorphism Theorem}\n\n\\begin{theorem}\nLet $H, N$ be subgroups of a group $G$ and let $N \\mathrel{\\unlhd} G$. Then\n$$H/(H\\cap N) \\cong HN/N.$$\n\\end{theorem}\n\n\\begin{proof}\nWe have previously shown that $H\\cap N \\mathrel{\\unlhd} H$ and it is easy to see that $N \\mathrel{\\unlhd} HN$, so the theorem makes sens.\n\nDefine a map $\\phi : H \\to HN/N$ by $\\phi(h) = hN$. It is easy to check that this map is a homomorphism.\n\nWe have that ker$(\\phi) = H\\cap N$ and $\\phi$ is surjective. Thus by the first isomorphism theorem the result follows. $\\square$\n\\end{proof}\n\n\\subsubsection{The Third Isomorphism Theorem}\n\n\\begin{theorem}\nLet $G$ be a group and $N$ a normal subgroup. If $K$ is a normal subgroup of $G$ containing $N$ then\n$$(G/N)/(K/N) \\cong G/K.$$\n\\end{theorem}\n\n\\begin{proof}\nWe have already shown in Theorem \\ref{normalcontaining} that $K/N$ is a normal subgroup of $G/N$. Therefore the quotient $(G/N)(K/N)$ makes sense.\n\nDefine a map $\\Phi : G/N \\to G/K$ by $\\Phi : gN \\mapsto gK$ for all $g \\in G$. It is easy to see that this is well-defined as $N \\subseteq K$. It is also easy to see that it is a surjective homomorphism.\n\nThe kernel is precisely the set of cosets $gN$ with $g \\in K$, i.e. $K/N$. \n\nTherefore, the result follows by the first isomorphism theorem. $\\square$\n\\end{proof}\n\n\\subsection{Direct Products of Groups}\n\n\\subsubsection{The Direct Product}\n\n\\begin{theorem}\nLet $G$ and $H$ be groups. Define a binary operation on $G\\times H$ by\n$$(g_1, h_1)\\cdot (g_2, h_2) = (g_1g_2, h_1h_2),$$\nfor $g_1, g_2 in G$ and $h_1, h_2 \\in H$.\nThen $G\\times H$ with this operation is a group.\n\\end{theorem}\n\n\\begin{proof}\nThe operation is clearly closed.\n\nAssociativity follows from associativity of the groups $G$ and $H$. The element $(1_G, 1_H)$ is an identity, and $(g^{-1}, h^{-1})$ is an inverse of $(g, h)$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe group $G\\times H$ under the operation given in the theorem is called the \\emph{(external) direct product} of $G$ and $H$ and is also denoted $G\\times H$.\n\\end{definition}\n\n\\subsubsection{The Order of a Direct Product}\n\n\\begin{theorem}\nLet $G$ and $H$ be groups and let $g \\in G$ have order $m$ and $h \\in H$ have order $n$. Then $(g, h) \\in G\\times H$ has order $lcm(m, n)$.\n\\end{theorem}\n\n\\begin{proof}\nThe order of $(g, h)$ will be the smallest positive integer which is a multiple of $m$ and $n$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ and $H$ be finite groups. Then $|G\\times H| = |G|\\times |H|$.\n\\end{corollary}\n\n\\begin{proof}\nFollows from the order of a cartesian product of finite sets. $\\square$\n\\end{proof}\n\n\\subsubsection{The Direct Product of a Family of Groups}\n\n\\begin{definition}\nLet $\\{G_i : i \\in I\\}$ be a family of groups indexed by a set $I$. The (external) \\emph{direct product} of the family is\n$$\\prod_{i \\in I} G_i = \\{f : I \\to \\cup_{i \\in I} G_i \\;|\\; f(i) \\in G_i \\forall i \\in I\\}.$$\n\\end{definition}\n\nIf $I$ is a two element set, the definition isn't exactly the same as the definition we already gave, but note that $(g, h) = (f(1), f(2))$ for some $f$ from the second definition.\n\nAlso note that if $I$ is a countably infinite set then we can think of the direct product as sequences of elements, one from each group $G_i$, i.e. the elements are of the form $(g_i)_{i \\in \\mathbb{N}}$ where $g_i \\in G_i$.\n\n\\begin{theorem}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The direct product $\\prod_{i \\in I} G_i$ is a group under the operation $(f_1\\cdot f_2)(i) = f_1(i)f_2(i)$.\n\\end{theorem}\n\n\\begin{proof}\nThe definition makes sense because $f_1(i)f_2(i) \\in G_i$ since $G_i$ is a group. It is clear that the direct product is closed under its operation.\n\nIt is easy to check associativity and that $e(i) = 1_{G_i}$ is an identity for the group. The inverse of $f$ is given by $f^{-1}(i) = f(i)^{-1}$ for all $i \\in I$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Canonical Projections of a Direct Product}\n\n\\begin{definition}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The \\emph{canonical projections} of $\\prod_{i \\in I} G_i$ are the maps $\\pi_j : \\prod_{i \\in I} G_i \\to G_j$ given by $\\pi_j : f \\mapsto f(i)$ for all $f \\in \\prod_{i \\in I} G_i$.\n\\end{definition}\n\n\\begin{theorem}\nThe canonical projections $\\pi_j : \\prod_{i \\in I} G_i \\to G_j$ are surjective homomorphisms.\n\\end{theorem}\n\n\\begin{proof}\nIt is easy to see that the $\\pi_j$ are homomorphisms.\n\nIf $g \\in G_j$ and $f(i) = e_{G_i}$ for all $i \\neq j$ and $f(j) = g$ then we see that $\\pi_j(f) = g$, which shows that $\\pi_j$ is surjective. $\\square$\n\\end{proof}\n\n\\subsubsection{The Restricted Direct Product}\n\n\\begin{definition}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The \\emph{restricted direct product} of the $G_i$ is\n$$\\prod'_{i \\in I} G_i = \\{f \\in \\prod_{i \\in I} G_i \\;|\\; f(i) = e_{G_i} \\;\\;\\mbox{for all but finitely many} i\\}.$$\n\\end{definition}\n\n\\begin{theorem}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The restricted direct product $H = \\prod'_{i \\in I} G_i$ is a normal subgroup of the direct product $G = \\prod_{i \\in I} G_i$.\n\\end{theorem}\n\n\\begin{proof}\nIt is easy to check that the restricted direct product is a subgroup of the direct product.\n\nNote that if $g \\in G$ and $h \\in H$ then $(g^{-1}hg)(i) = g^{-1}(i)h(i)g(i)$ for all $i \\in I$.\n\nBut $h(i) = e_{G_i}$ for all but finitely many $i \\in I$. Thus $(g^{-1}hg)(i) = g^{-1}(i)g(i) = e_{G_i}$ for all but finitely many $i$. This means that $g^{-1}hg \\in H$ and $H \\mathrel{\\unlhd} G$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Canonical Injections of the Restricted Direct Product}\n\n\\begin{definition}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The \\emph{canonical injections} of the restricted product are the maps $\\iota_j : G_j \\to \\prod'_{i \\in I} G_i$ given by\n$$(\\iota_j(g))(i) = \\begin{cases}g, & i = j\\\\e_{G_i}, & i \\neq j\\end{cases}.$$\n\\end{definition}\n\n\\begin{theorem}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The canonical injections $\\iota_j : G_j \\to \\prod'_{i\\in I} G_i$ are injective homomorphisms.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $\\{G_i : i \\in I\\}$ be a family of groups. The images $\\iota_j(G_j)$ of the canonical injections of the restricted product are normal subgroups of the product $\\prod_{i\\in G} G_i$.\n\\end{theorem}\n\n\\begin{proof}\nAs $\\iota_j$ is a homomorphism, its image is a subgroup of the product.\n\nLet $G = \\prod_{i \\in I} G_i$ and $H = \\iota_j(G_j)$ be the image of $\\iota_j$. If $h \\in H$ then there exists an $a \\in G_j$ such that $\\iota_j(a) = h$. But this means that $h(j) = a$ and $h(i) = e_{G_i}$ for all $i \\neq j$.\n\nLet $g \\in G$. Then for $i = j$ we have $(g^{-1}hg)(j) = g^{-1}(j)ag(j) \\in G_j$.\n\nIf $i \\neq j$ then $(g^{-1}hg)(i) = g^{-1}(i)g(i) = e_{G_i}$.\n\nThus we see that $g^{-1}hg = \\iota_j(g^{-1}(j)ag(j)) \\in \\iota_j(G_j) = H$. Thus $H \\mathrel{\\unlhd} G$. $\\square$\n\\end{proof}\n\n\\subsubsection{Internal Direct Products}\n\n\\begin{definition}\nLet $G$ be a group and $H, K \\leq G$. The group $G$ is said to be the \\emph{internal direct product} of $H$ and $K$ if\n\\begin{enumerate}\n\\item $G = \\{hk \\;|\\; h \\in H, k \\in K\\}$\n\\item $G\\cap K = \\{1_G\\}$\n\\item $hk = kh$ for all $h \\in H, k \\in K$\n\\end{enumerate}\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group and $H, K \\leq G$. Then $G$ is the internal direct product of $H$ and $K$ iff the following hold:\n\\begin{enumerate}\n\\item $G = \\{hk \\;|\\; h \\in H, k \\in K\\}$\n\\item $G\\cap K = \\{1_G\\}$\n\\item $H, K \\mathrel{\\unlhd} G$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\nSuppose $G$ is the direct product of $H$ and $K$. Let $g \\in G$ and $k \\in K$. Then by the first condition above, $g = h'k'$ for some $h' \\in H$, $k' \\in K$.\n\nWe have that $g^{-1}kg = k'^{-1}h'^{-1}kh'k'$ which by the third condition above is equal to $k'^{-1}kk' \\in K$. Thus $K$ is a normal subgroup of $G$. A similar argument holds for $H$.\n\nConversely, suppose that the given conditions hold. Let $h \\in H$ and $k \\in K$. As $K$ is normal in $G$ we have $h^{-1}kh = k'$ for some $k' \\in K$. Similarly $k'^{-1}hk' = h'$ for some $h' \\in H$.\n\nRewriting these, we have $h^{-1}k = k'h^{-1}$ and $k'^{-1}h = h'k'^{-1}$. Multiplying these relations we have that $k'^{-1}k = h'h^{-1}$. But the left side is an element of $K$ and the right side an element of $H$. Thus both equal $1$ by the second condition above.\n\nIn other words, $k' = k$ and $h' = h$, which implies that $hk = kh$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $\\{N_i : i \\in I\\}$ be a family of subgroups of a group $G$. The group $G$ is said to be the internal direct product of the family if:\n\\begin{enumerate}\n\\item $N_i \\mathrel{\\unlhd} G$ for all $i$\n\\item $G = \\langle \\cup_{i\\in I} N_i \\rangle$\n\\item $N_i \\cap \\langle \\cup_{j \\neq i} N_i \\rangle = 1_G$\n\\end{enumerate}\n\\end{definition}\n\n\\begin{theorem}\nLet $\\{N_i : i \\in I\\}$ be a family of subgroups of a group $G$. Then $G$ is the internal direct product of the family if:\n\\begin{enumerate}\n\\item $N_i \\mathrel{\\unlhd} G$ for all $i \\in I$\n\\item $g_ig_j = g_jg_i$ for all $g_i \\in N_i$ and $g_j \\in N_j$ for $i \\neq j$\n\\item Every $g \\in G$ has a unique expression as a product of elements from distinct $N_i$'s\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\nThe proof of the second condition is essentially the same as for the theorem above.\n\nAssume $G$ is the direct product of the $N_i$'s. Let $g \\in G$. As the $N_i$'s generate $G$, we can write $g$ as a product $g_{\\lambda_1}g_{\\lambda_2}\\ldots g_{\\lambda_k}$ where $1 \\neq g_{\\lambda_i} \\in G_{\\lambda_i}$, and the $\\lambda_i$ are distinct. \n\nLet $g = h_{\\mu_1}h_{\\mu_2}\\ldots h_{\\mu_m}$ be another such expression and suppose that $\\mu_1 \\neq \\lambda_i$ for all $i$. Then $h_{\\mu_1} \\in N_{\\mu_1} \\cap \\langle \\cup_{i \\neq \\mu_1} N_i \\rangle$, which is trivial.\n\nThis shows that the third condition holds.\n\nConversely, if the conditions in the theorem hold then clearly the third condition of the definition holds, due to uniqueness of expression as a product. The other conditions follow just as for the proof of the previous theorem. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group that is the internal direct product of subgroups $H_1, H_2, \\ldots, H_n$. Suppose $N \\mathrel{\\unlhd} H_1$, then $N \\mathrel{\\unlhd} G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $g = h_1h_2\\ldots h_n \\in G$ for $h_i \\in H_i$ then for $h \\in H_1$ we have $g^{-1}hg = h_1^{-1}hh_1 \\in H_1$. $\\square$ \n\\end{proof}\n\n\\subsubsection{Equivalence of Internal and External Direct Products}\n\n\\begin{theorem}\nLet $G$ be a group and $H, K \\leq G$. If $G$ is the internal direct product of $H$ and $K$ then $G \\cong H\\times K$.\n\\end{theorem}\n\n\\begin{proof}\nDefine a map $f : H\\times K \\to G$ by $(h, k) = hk$.\n\nBy the commutativity in the internal direct product, the map $f$ is a homomorphism.\n\nTo show it is injective, let $(h_1, k_1), (h_2, k_2) \\in H\\times K$. Suppose that $f(h_1, k_1) = f(h_2, k_2)$, i.e. $h_1k_1 = h_2k_2$. We have that $h_2^{-1}h_1 = k_2k_1^{-1}$. As the intersection of $H$ and $K$ is trivial, we have that $h_1 = h_2$ and $k_1 = k_2$.\n\nSurjectivity of $f$ follows from the fact that $G = \\{hk \\;|\\; h \\in H, k \\in K\\}$. $\\square$\n\\end{proof}\n\nA similar proof shows that internal and external direct products of families of subgroups are isomorphic.\n\n\\subsubsection{The Automorphism Group of a Direct Product}\n\n\\begin{theorem}\nIf $A$ and $B$ are finite groups with coprime orders then Aut$(A\\times B) \\cong$ Aut$(A)\\times$Aut$(B)$.\n\\end{theorem}\n\n\\begin{proof}\nLet $\\phi : A\\times B \\to A\\times B$ be an automorphism. Composing with the projections onto $A$ and $B$ we get homomorphisms $\\phi_A : A\\times B \\to A$ and $\\phi_B : A\\times B \\to B$.\n\nBy the first isomorphism theorem, the order of ker$(\\phi_A)$ must be $|B|$. As the order of any nontrivial element $a \\in A$ is coprime to $|B|$ we see that no element of $A\\times \\{1\\}$ can be in the kernel of $\\phi_A$.\n\nThis means that $\\phi_A$ induces an automorphism on $A$, and similarly $\\phi_B$ induces an automorphism on $B$.\n\nIt is now clear that there is a homomorphism from Aut$(A\\times B) \\to$ Aut$(A)\\times$ Aut$(B)$. But any pair of automorphisms $\\rho : A \\to A$, $\\sigma : B \\to B$ induces an automorphism of $A\\times B$, and there is a homomorphism Aut$(A)\\times$ Aut$(B) \\to$ Aut$(A\\times B)$, which is thus an isomorphism. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe automorphism group of $C_p\\times C_p$ for a prime $p$ is isomorphic to GL$_2(\\mathbb{Z}/p\\mathbb{Z})$.\n\\end{theorem}\n\n\\begin{proof}\nAn endomorhism of $C_p\\times C_p$ is entirely characterised by where it sends the generators $x$ and $y$ of the summands. Suppose it maps $(x, 1) \\to (x^a, y^b)$ and $(1, y) \\to (x^c, y^d)$, where $a, b, c, d$ can be taken modulo $p$.\n\nThis is an automorphism iff it is invertible, which happens iff $ad - bc = \\pm 1$. Thus an automorphism corresponds to a matrix in GL$_2(\\mathbb{Z}/p\\mathbb{Z})$ and vice versa.\n\nAn easy computation shows that composition of two automorphisms corresponds to matrix multiplication of matrices in GL$_2(\\mathbb{Z}/p\\mathbb{Z})$. $\\square$\n\\end{proof}\n\n\\subsection{The Semidirect Product}\n\n\\subsubsection{The Semidirect Product}\n\n\\begin{definition}\nLet $H$ and $Q$ be groups and let $\\theta : Q \\to$ Aut$(H)$ be a group homomorphism. The semidirect product $H\\rtimes_{\\theta} Q$ is the set $\\{(h, q)\\;|\\; h \\in H, q \\in Q\\}$ with operation $(h_1, q_1)(h_2, q_2) = (h_1\\theta(q_1)h_2, q_1q_2)$.\n\\end{definition}\n\n\\begin{theorem}\nThe semidirect product $H\\rtimes_{\\theta} Q$ is a group.\n\\end{theorem}\n\n\\begin{proof}\nIt is clear that $(1, 1)$ is the identity.\n\nWe have\n$$(h_1, q_1)[(h_2, q_2)(h_3, q_3)] = (h_1\\theta(q_1)[h_2\\theta(q_2)h_3], q_1q_2q_3)] = (h_1\\theta(q_1)h_2\\theta(q_1q_2)h_3).$$\nSimilarly\n$$[(h_1, q_1)(h_2, q_2)](h_3, q_3) = (h_1\\theta(q_1)h_2\\theta(q_1q_2)h_3, q_1q_2q_3).$$\nThus the associative law holds.\n\nFinally, $(\\theta(q^{-1})h^-1, q^{-1})$ is the inverse of $(h, q)$.\n$\\square$\n\\end{proof}\n\n\\begin{theorem}\nGiven groups $H$ and $Q$ and a group homomorphism $\\theta : Q \\to$ Aut$(H)$ there exist canonical injections $H \\to H\\rtimes_{\\theta} Q$ and $Q \\to H\\rtimes_{\\theta} Q$ given by $h \\mapsto (h, 1)$ and $q \\mapsto (1, q)$. These are group homomorphisms.\n\\end{theorem}\n\n\\begin{proof}\nIt is clear that the first is a homomorphism.\n\nFor the second, it follows from $\\theta(q)1 = 1$. $\\square$\n\\end{proof}\n\nWe will identify $H$ with its image under the first injection, so that it is a subgroup of $H\\rtimes_{\\theta} Q$ and similarly we will identify $Q$ with its image under the second injection, so that it is also a subgroup of $H\\rtimes_{\\theta} Q$.\n\n\\begin{theorem}\nLet $H$ and $Q$ be groups and $\\theta : Q \\to$ Aut$(H)$ be a homomorphism. Then\n\\begin{enumerate}\n\\item $H \\mathrel{\\unlhd} H\\rtimes_{\\theta} Q$\n\\item $HQ = H\\rtimes_{\\theta} Q$\n\\item $H\\cap Q = (1, 1) \\in H\\rtimes_{\\theta} Q$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\nNote that $(h, q) = (h, 1)(1, q)$. Therefore $HQ = H\\rtimes_{\\theta} Q$.\n\nLet $\\rho : H\\rtimes_{\\theta} Q \\to Q$ be defined by $\\rho(h, q) = q$. This is clearly a homomorphism with kernel $H$. Thus $H$ is a normal subgroup of $H\\rtimes_{\\theta} Q$.\n\nClearly $(1, 1)$ is the only element in (the images of) $H$ and $Q$ (in $H\\rtimes_{\\theta} Q$). $\\square$\n\\end{proof}\n\n\\subsubsection{Internal Semidirect Products}\n\n\\begin{definition}\nIf $G$ is a group with subgroups $H$ and $Q$, we say that $G$ is the \\emph{internal direct product} of $H$ and $Q$ if\n\\begin{enumerate}\n\\item $H \\mathrel{\\unlhd} G$\n\\item $HQ = G$\n\\item $H\\cap Q = 1 \\in G$\n\\end{enumerate}\n\\end{definition}\n\n\\begin{lemma}\nLet $G$ be a group with subgroups $H$ and $Q$. Suppose $G = HQ$ and $H\\cap Q = \\{1\\}$. Then every element $g \\in G$ can be written uniquely in the form $hq$ with $h \\in H$ and $q \\in Q$.\n\\end{lemma}\n\n\\begin{proof}\nSuppose $hq = h'q'$. Then $h'^{-1}h = q'q^{-1} \\in H\\cap Q$. Thus $h = h'$ and $q = q'$. $\\square$\n\\end{proof}\n\nIt is clear that any external semidirect product is an internal semidirect product. But we also have the converse.\n\n\\begin{theorem}\nLet $G$ be a group which is the internal semidirect product of subgroups $H$ and $Q$. Then $G \\cong H\\rtimes_{\\theta} Q$ where $\\theta : Q \\to$ Aut$(H)$ is given by\n$$\\theta(q)(h) = qhq^{-1}.$$\n\\end{theorem}\n\n\\begin{proof}\nBy the lemma, every $g \\in G$ can be written uniquely as $hq$ for $h \\in H$ and $q \\in Q$. Thus the map $\\phi : (h, q) \\mapsto hq$ is a bijection from $H\\rtimes_{\\theta} Q$ to $G$.\n\nIt suffices to show $\\phi$ is a homomorphism:\n\\begin{multline*}\n\\phi((h_1, q_1)(h_2, q_2)) = \\phi((h_1\\theta(q_1)h_2, q_1q_2)) =\\\\\n\\phi(h_1q_1h_2q^{-1}, q_1q_2) = h_1q_1h_2q_2 = \\phi((h_1, q_1))\\phi((h_2, q_2)).\n\\end{multline*} $\\square$\n\\end{proof}\n\nNote that if $G$ is the internal direct product of subgroups $H$ and $Q$ with $Q$ also normal in $G$ then the semidirect product is the direct product.\n\nAlso note that whilst any two direct products of groups $H$ and $Q$ are isomorphic, this is not true of the semidirect product as the semidirect product still depends on the action of $Q$ on $H$.\n\nIn the case of the internal semidirect product, one should not confuse the action of $Q$ on $H$ by conjugation in $G$ with inner automorphisms of $H$ (which would correspond to conjugation by elements of $H$). As $H$ is normal in $G$, the action of $Q$ on $H$ must map $H$ to itself, but this map can be any automorphism of $H$, not just an inner automorphism.\n\n\\subsubsection{Dihedral Groups}\n\n\\begin{definition}\nLet $D_n$ be the group generated by $r$ and $s$ with $r^n = 1$, $s^2 = 1$ and $(sr)^2 = 1$. We call $D_n$ the $n$-th \\emph{dihedral group}.\n\\end{definition}\n\n\\begin{theorem}\nThe dihedral group $D_n$ has $2n$ elements, namely $1, r, \\ldots, r^{n-1}$ and $s, sr, \\ldots sr^{n-1}$.\n\\end{theorem}\n\n\\begin{proof}\nFrom $(sr)^2 = 1$ we have $rs = s^{-1}r^{-1} = sr^{n-1}$. Repeated application of this identity allows us to replace any product of $s$ and $r$ with a product having at most one $s$ on the left. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe dihedral group $D_n$ is isomorphic to the semidirect product $C_n\\times C_2$ where the the non-identity element of $C_2$ acts on $C_n$ by inverting elements, i.e. if $C_n = \\langle r \\rangle$ and $C_2 = \\langle s \\rangle$ then $srs^{-1} = r^{-1}$.\n\\end{theorem}\n\n\\begin{proof}\nNote that we have identified $C_n$ and $C_2$ with subgroups of $D_n$ generated by $r$ and $s$ respectively.\n\nWe have $C_n \\mathrel{\\unlhd} D_n$ as it has index 2.\n\nClearly $C_n\\cap C_2 = 1$ and $C_nC_2 = D_n$.\n\nThe given identity follows from $(sr)^2 = 1$ and $s^2 = 1$. $\\square$\n\\end{proof}\n\n\\subsection{Exact Sequences}\n\n\\subsubsection{Exact Sequences}\n\n\\begin{definition}\nAn exact sequence of groups is a sequence of homomorphisms $f_i$ between groups $G_i$:\n$$G_0 \\overset{f_1}{\\longrightarrow} G_1 \\overset{f_2}{\\longrightarrow} G_{2} \\overset{f_{3}}{\\longrightarrow} \\cdots \\overset{f_n}{\\to} G_n$$\nsuch that im$(f_k) =$ ker$(f_{k+1})$ for every $n$.\n\\end{definition}\n\n\\begin{theorem}\nThe sequence $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B$ is exact at $A$ iff $f$ is an injective homomorphism.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe sequence $A \\overset{f}{\\longrightarrow} B \\longrightarrow 0$ is exact at $B$ iff $f$ is an surjective homomorphism.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nThe sequence $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\longrightarrow 0$ is exact at $B$ iff $f$ is an isomorphism.\n\\end{corollary}\n\n\\subsubsection{Short Exact Sequences}\n\n\\begin{definition}\nA \\emph{short exact sequence} is an exact sequence of the form\n$$0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0.$$\n\\end{definition}\n\n\\begin{theorem}\nIf $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ is a short exact sequence of groups then $C \\cong B/\\mbox{im}(f)$.\n\\end{theorem}\n\n\\begin{proof}\nFollows from the first isomorphism theorem and the definition of exactness at $B$. $\\square$\n\\end{proof}\n\n\\subsubsection{Split Short Exact Sequences}\n\n\\begin{definition}\nA short exact sequence $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ is said to be \\emph{split} if there exists a homomorphism $h : C \\to B$ such that the composition $g\\circ h$ on $C$ is the identity map.\n\\end{definition}\n\n\\begin{theorem}\nLet $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ be a short exact sequence of groups. If there exists a homomorphism $t : B \\to A$ such that $t\\circ f$ is the identity map on $A$ then $B$ is a direct sum of $A$ and $C$.\n\\end{theorem}\n\n\\begin{proof}\nDefine $\\theta : B \\to A\\times C$ by\n$$\\theta(b) = (t(b), g(b)).$$\n\nThis is a homomorphism since $t$ and $g$ are homomorphisms.\n\nWe will show that the kernel of $\\theta$ is trivial, so that it is injective. If $b \\in$ ker$(\\theta)$ then $t(b) = 1$ and $g(b) = 1$. From exactness at $B$ we see that $g(b) = 1$ implies $b = f(a)$ for some $a \\in A$. If $t(b) = 1$ then $1 = t(f(a)) = a$. But this also means $b = f(1) = 1$.\n\nTo show that $\\theta$ is surjective, let $(a, c) \\in A\\times C$. As $g$ is surjective, $c = g(b)$ for some $b \\in B$.\n\nLet $x = t(b)^{-1}a \\in A$. We have $t(bf(x)) = t(b)t(f(x)) = t(b)x = a$. By exactness at $B$ we have that $g(bf(x)) = g(b)g(f(x)) = g(b) = c$. Thus $\\theta(bf(x)) = (a, c)$ and $\\theta$ is surjective. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nLet $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ be a short exact sequence of groups. If there exists a homomorphism $t : B \\to A$ such that $t\\circ f$ is the identity map on $A$ then the sequence is split.\n\\end{corollary}\n\n\\begin{proof}\nDefine $h : C \\to B$ as the composition of the canonical injection of $C$ into $A\\times C$ and the inverse of the map $\\theta$ from the proof of the theorem.\n\nLet $c \\in C$ and $b = h(c) = \\theta^{-1}((1, c))$. Then $g(b) = \\rho_2(\\theta(b))$, by definition of $\\theta$, where $\\rho_2$ is the projection onto the second component. But this is clearly $c$. Thus $h$ splits the sequence. $\\square$\n\\end{proof}\n\nThe converse is not true, but we have the following.\n\n\\begin{theorem}\nLet $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ be a short exact sequence of groups. If the sequence is split then $B \\cong A\\rtimes_{\\theta} C$ for some homomorphism $\\theta : C \\to$ Aut$(A)$.\n\\end{theorem}\n\n\\begin{proof}\nLet $h : C \\to B$ split the sequence. It suffices to show that $B$ is the internal direct product im$(f)\\rtimes_{\\theta}$ im$(h)$.\n\nAs im$(f) =$ ker$(g)$, it is a normal subgroup of $B$.\n\nFirstly, we show that im$(f)$im$(h) = B$. Let $b \\in B$. Then $b = yz$ where $z = h(g(b))$ and $y = bz^{-1}$. We have $g(z) = g(h(g(b))) = g(b)$. Thus $g(y) = g(b)g(z)^{-1} = 1$. Thus $y \\in$ ker$(g) =$ im$(f)$. But $z \\in$ im$(h)$ by construction, and so $b \\in$ im$(f)$im$(h)$.\n\nNow we show that if $x \\in$ im$(f)\\cap$ im$(h)$ then $x = 1$. For if $x = f(a) = h(c)$ for $a \\in A$ and $c \\in C$ then $c = g(h(c)) = g(f(a)) = 1$ due to exactness of the sequence. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nLet $0 \\longrightarrow A \\overset{f}{\\longrightarrow} B \\overset{g}{\\longrightarrow} C \\longrightarrow 0$ be a short exact sequence of groups. If $A$, $B$ and $C$ are abelian and the sequence is split, then $C \\cong A\\times B$.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\section{Free Groups and Presentations}\n\n\\subsection{Free Groups}\n\n\\subsubsection{Free Groups}\n\n\\begin{definition}\nLet $S$ be a set of symbols and let $S^{-1}$ be the set of symbols $s^{-1}$ for each symbol in $S$. Let $T = S\\cup S^{-1}$ and let $W$ be the set of words constructed from symbols in $T$, including the empty word. A word can be \\emph{reduced} by removing any pair of adjacent symbols of the form $ss^{-1}$ or $s^{-1}s$ from the word, for any $s \\in S$. A word is said to be \\emph{reduced} if no such adjacent pairs exist in the word. The \\emph{free group} on $S$ is the set $F(S)$ of reduced words with concatenation followed by reduction as the operation.\n\\end{definition}\n\n\\begin{theorem}\nThe free group $F(S)$ on a set $S$ is a group.\n\\end{theorem}\n\n\\begin{proof}\nThe empty word is the identity, and the operation is clearly associative. The inverse of a word is defined to be the reverse of the word with each symbol $s$ replaced with $s^{-1}$ and vice versa. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nEvery group $G$ is a quotient of a free group.\n\\end{theorem}\n\n\\begin{proof}\nLet $S$ be the underlying set of $G$.\n\nDefine a map $F(S) \\to G$ by sending a reduced word in $F(S)$ to the corresponding product of elements and inverses in $G$.\n\nThis map is clearly well-defined and a surjective homomorphism. The result follows from the first isomorphism theorem. $\\square$\n\\end{proof}\n\n\\subsection{Group Presentations}\n\n\\subsubsection{Group Presentations}\n\n\\begin{definition}\nLet $S$ be a set and $R$ be a set of (reduced) words in $F(S)$. Let $N(R)$ be the smallest normal subgroup of $F(S)$ containing $R$ (the \\emph{normal closure} of $R$), then the group $G = F(S)/N(R)$ is said to have \\emph{presentation} $\\langle S \\;|\\; R \\rangle$.\n\\end{definition}\n\n\\begin{theorem}\nLet $R$ be the kernel of the map $F(S) \\to G$ in the previous theorem. Then $\\langle S \\;|\\; R \\rangle$ is a presentation of $G$.\n\\end{theorem}\n\n\\begin{proof}\nThe kernel is normal and so $N(R) = R$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nA presentation $\\langle S \\;|\\; K \\rangle$ of a group $G$ is said to be finitely generated if $S$ is finite. It is said to be \\emph{finitely related} if $K$ is finite, and a \\emph{finite presentation} if both $S$ and $K$ are finite. The group $G$ is also said to be finitely generated or finitely related or finitely presented, respectively, if it has a presentation with the given properties.\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group. There is a unique homomorphism $\\phi : F(G) \\to G$ which is the identity on elements of $G$.\n\\end{theorem}\n\n\\begin{proof}\nThe definition of $\\phi$ on a word in $F(G)$ is fixed in view of $\\phi$ being a homomorphism and from the fact that $gg^{-1}$ and $g^{-1}g$ are equal to the empty word after reduction, for any $g \\in G$, so that $\\phi(g)\\phi(g^{-1}) = \\phi(g^{-1})\\phi(g) = 1$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nEvery group has a presentation and every finite group has a finite presentation.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Fundamental Theorem of Presentations of a Group}\n\n\\begin{theorem}\nLet $G = \\langle S \\;|\\; R \\rangle$ be a presentation of a group. Let $\\theta : S\\cup S^{-1} \\to H$ be a map to any group $H$ such that $\\theta(s^{-1}) = \\theta(s)^{-1}$ for any $s \\in S$. Suppose that for all relations $r = t_1t_2\\ldots t_k \\in R$ with $t_i \\in S\\cup S^{-1}$ we have $\\theta(t_1)\\theta(t_2)\\ldots \\theta(t_k) = 1$. Then $\\theta$ extends uniquely to a homomorphism $\\theta' : G \\to H$.\n\\end{theorem}\n\n\\begin{proof}\nIf such a homomorphism exists, it is clearly unique.\n\nDefine $\\psi : F(X) \\to H$ by $\\psi(x) = \\phi(x)$ for $x \\in X$ and $x \\in X^{-1}$ and by $\\psi(x_1x_2) = \\phi(x_1)\\phi(x_2)$ for $x_1, x_2 \\in X\\cup X^{-1}$. This is clearly a well-defined homomorphism.\n\nNow $\\psi(r) = 1$ for all $r \\in R$ and so $R \\subseteq$ ker$(\\psi)$.\n\nLet $N =$ ker$(\\psi)$ and $R'$ be the normal closure of $R$ in $F(X)$. Then $N \\mathrel{\\unlhd} F(X)$ and $R' \\leq N$. Thus there is a well-defined homomorphism $\\theta' : F(X)/R' \\to H$. $\\square$\n\\end{proof}\n\n\\subsubsection{Presentation of a Direct Product}\n\n\\begin{theorem}\nLet $G_1 = \\langle X_1 \\;|\\; R_1 \\rangle$ and $G_2 = \\langle X_2 \\;|\\; R_2 \\rangle$ be groups with presentations such that $X_1$ and $X_2$ are disjoint. Then $G_1\\times G_2 = \\langle X_1\\cup X_2 \\;|\\; R_1\\cup R_2\\cup [X_1, X_2] \\rangle$.\n\\end{theorem}\n\n\\begin{proof}\nDefine $\\theta : X_1\\cup X_2 \\to G_1\\times G_2$ by $\\theta(x_1) = (x_1, 1)$ and $\\theta(x_2) = (1, x_2)$ for $x_i \\in X_i$ and extend $\\theta$ to $F(X_1\\cup X_2)$ by multiplicativity.\n\nLet $H = \\langle X_1\\cup X_2 \\;|\\; R_1\\cup R_2\\cup [X_1, X_2] \\rangle$. Let $r = t_1t_2\\ldots t_k$ be a relation in $H$ with $t_i \\in X_1\\cup X_2\\cup X_1^{-1}\\cup X_2^{-1}$.\n\nIf $r \\in R_1$ then $\\theta(r) = (r, 1) = (1, 1)$. Similarly if $r \\in R_2$ then $\\theta(r) = (1, 1)$. Finally, if $r = [x_1, x_2]$ for $x_i \\in X_i$ then\n$$\\theta(r) = \\theta(x_1)^{-1}\\theta(x_2)^{-1}\\theta(x_1)\\theta(x_2) = (x_1, 1)^{-1}(x_2, 1)^{-1}(x_1, 1)(x_2, 1) = (1, 1).$$\n\nTherefore by the Fundamental Theorem of Presentations, $\\theta$ extends to a map $H \\to G_1\\times G_2$. It is clearly surjective.\n\nLet $V = \\langle X_1\\cup X_2 \\;|\\; H \\rangle$.\n\nWe will show that ker$(\\theta) = 1$. Because elements of $X_1$ and $X_2$ commute in $V$, any element of $V$ can be written $v = s_1s_2\\ldots s_mt_1t_2\\ldots t_n$ with $s_i \\in X_1\\cup X_1^{-1}$ and $t_i \\in X_2\\cup X_2^{-1}$. Then $\\theta(v) = (s_1s_2\\ldots s_m, t_1t_2\\ldots t_n)$.\n\nThus if $v \\in$ ker$(\\theta)$ then $s_1s_2\\ldots s_m = 1$ and $t_1t_2\\ldots t_n = 1$. Thus $\\alpha = s_1s_2\\ldots s_m$ is in the normal closure of $R_1$ and $\\beta = t_1t_2\\ldots t_n$ in the normal closure of $R_2$.\n\nThus both $\\alpha$ and $\\beta$ are in the normal closure of $H$. Thus both are $1$ in $V$. Thus $v = 1$. $\\square$\n\\end{proof}\n\n\\section{Permutations}\n\n\\subsection{Permutations}\n\n\\subsubsection{Permutations}\n\n\\begin{definition}\nA \\emph{permutation} on a set $S$ of $n$ elements is a bijection $\\phi : S \\to S$. Such a permutation can be defined by its action on the $i$-th element of $S$, which we can denote\n$$\\left(\\begin{array}{cccc}1 & 2 & \\cdots & n\\\\ \\phi(1) & \\phi(2) & \\cdots & \\phi(n)\\end{array}\\right).$$\n\\end{definition}\n\n\\subsubsection{Cycles}\n\n\\begin{definition}\nA \\emph{cycle} $(a_1, a_2, \\cdots, a_k)$ is a permutation $\\phi$ that maps $a_i \\mapsto a_{i+1}$ for $1 \\leq i < k$ and $a_k \\mapsto a_1$.\n\\end{definition}\n\n\\begin{theorem}\nA permution $\\phi$ on the set $S$ of $n$ elements can be written as a product of disjoint cycles.\n\\end{theorem}\n\n\\begin{proof}\nBegin with $1 \\in S$. Then apply $\\phi$ repeatedly until we reach $1$ again (as $\\phi$ is a bijection and $S$ is finite, this must eventually happen): $1 \\mapsto a_2 \\mapsto \\cdots \\mapsto 1$. This is a cycle.\n\nIf any elements of $S$ did not appear in this cycle, let $b_1$ be the first such element. Apply the same trick to get a cycle beginning with $b_1$. As $\\phi$ is a bijection, the elements of this cycle must be distinct from the elements of the first cycle.\n\nApply the same trick until all elements of $S$ appear in a cycle. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $\\alpha = (a_1, a_2, \\ldots, a_s)$ be a cycle. Let $\\sigma$ be a permutation. Then\n$$\\sigma\\alpha\\sigma^{-1} = (\\sigma(a_1), \\sigma(a_2), \\ldots, \\sigma(a_s)).$$\n\\end{theorem}\n\n\\begin{proof}\nLet $b_i = \\sigma(a_i)$ for $i \\in \\{1, 2, \\ldots, s\\}$.\n\nFor $i \\in \\{1, 2, \\ldots, s-1\\}$ we have\n$$(\\sigma\\alpha\\sigma^{-1})(b_i) = (\\sigma\\alpha)(a_i) = \\sigma(a_{i+1}) = b_{i+1}.$$\n\nWe also have\n$$(\\sigma\\alpha\\sigma^{-1})(b_s) = \\sigma\\alpha(a_s) = \\sigma(a_1) = b_1.$$\n\nAs $\\alpha$ leaves any $x \\not\\in \\{a_1, a_2, \\ldots, a_s\\}$ fixed, we have that $(\\sigma\\alpha\\sigma^{-1})(y) = y$ for all $y \\not\\in \\{b_1, b_2, \\ldots, b_s\\}$. $\\square$\n\\end{proof}\n\n\\subsubsection{Transpositions}\n\n\\begin{definition}\nA \\emph{transposition} is a cycle of length $2$.\n\\end{definition}\n\n\\begin{theorem}\nA cycle is a product of transpositions.\n\\end{theorem}\n\n\\begin{proof}\n$(a_1, a_2, \\ldots, a_s) = (a_s, a_{s-1})(a_s, a_{s - 2})\\cdots(a_s, a_1)$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Symmetric Group}\n\n\\begin{definition}\nThe \\emph{symmetric group} on the set $S$ of $n$ elements is the set of permutations of $S$. We denote it $S_n$.\n\\end{definition}\n\n\\begin{theorem}\nThe set $S_n$ is a group under composition of permutations.\n\\end{theorem}\n\n\\begin{proof}\nThe trivial permutation is the identity. As a permutation is a bijection, it is invertible. Thus inverses exist. Composition is clearly associative. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe symmetric group $S_n$ is generated by transpositions of the form $t_i = (i, i + 1)$.\n\\end{theorem}\n\n\\begin{proof}\nWe have already seen that any permutation can be written as a product of disjoint cycles, and each cycle can be written as a product of transpositions.\n\nBut $(a_i, a_j)$ for $j > i$ can be expressed as $(a_{j-1}, a_j)\\ldots (a_{i+1}, a_{i+2})(a_i, a_{i+1})$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\n$|S_n| = n!$.\n\\end{theorem}\n\n\\begin{proof}\nFollows immediately by counting the number of possible permutation. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nTwo elements of $S_n$ are conjugate iff they have disjoint cycle representations with cycles of the same length.\n\\end{theorem}\n\n\\begin{proof}\nLet $\\alpha = c_1c_2\\ldots c_k$ be a product of disjoint cycles in $S_n$. Let $\\sigma$ be any permutation. Then\n$$\\sigma c_1c_2\\ldots c_k\\sigma^{-1} = (\\sigma c_1\\sigma^{-1})(\\sigma c_2 \\sigma^{-1})\\ldots (\\sigma c_k\\sigma^{-1}).$$\n\nBut by the theorem on conjugation of cycles, this will have the same cycle structure as $\\alpha$.\n\nConversely, let $\\alpha = c_1c_2\\ldots c_k$ and $\\beta = d_1d_2\\ldots d_k$ be permutations with the same cycle structure.\n\nLet $a_1, a_2, \\ldots a_s$ be the elements of the cycles of $\\alpha$ in order (including $1$-cycles), and let $b_1, b_2, \\ldots, b_s$ be the elements of the cycles of $\\beta$ in order (including $1$-cycles). Then the permutation that sends $a_i$ to $b_i$ for all $1 \\leq i \\leq s$ will conjugate $\\alpha$ to $\\beta$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Alternating Group}\n\n\\begin{definition}\nA permutation is even if it is a product of an even number of transpositions.\n\\end{definition}\n\n\\begin{theorem}\nLet $\\epsilon$ be the identity permutation. Then $\\epsilon$ is an even permutation.\n\\end{theorem}\n\n\\begin{proof}\nWe will show that if $\\epsilon$ is a product of $k \\geq 2$ transpositions, then it is the product of $k - 2$ transpositions, which will prove the result, as the identity is not a transposition.\n\nLet $\\epsilon = s_1s_2\\ldots s_k$ and suppose that $s_j = (x, a)$ but $x$ does not appear in $s_{j+1}, s_{j+2}, \\ldots, s_k$.\n\nLet $s_{j-1} = (y, b)$. There are four cases:\n\n(i) If $s_{j-1} = (x, a)$ we are done.\n\n(ii) If $s_{j-1} = (x, b)$ with $b \\neq a$ and $b \\neq x$ then $s_{j-1}s_j = (x, a)(a, b)$.\n\n(iii) If $s_{j-1} = (y, a)$ with $y \\neq x$, $y \\neq a$ then $s_{j-1}s_j = (x, y)(y, a)$.\n\n(iv) If $s_{j-1} = (y, b)$ with $y \\neq x$, $y \\neq a$, $b \\neq x$ and $b\\neq a$ then $s_{j-1}s_j = (x, a)(y, b)$.\n\nIn all cases, we may move the first appearance of $x$ back or cancel a pair of transpositions.\n\nAs $x$ cannot appear just once in the identity, we must always eventually end up in case (i). $\\square$\n\\end{proof}\n\n\\begin{corollary}\nEvery permutation is either odd or even, but not both.\n\\end{corollary}\n\n\\begin{proof}\nIf a permutation were both, the identity could be written as a product of an even permutation and the inverse of an odd permutation, which would be odd. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe \\emph{alternating group} on $n$ elements is the subset of $S_n$ consisting of even permutations. We denote it $A_n$.\n\\end{definition}\n\n\\begin{theorem}\nThe set $A_n$ is a group under composition of permutations.\n\\end{theorem}\n\n\\begin{proof}\nIt is clearly a subgroup of $S_n$ because it is closed under composition and taking of inverses. $\\square$\n\\end{proof}\n\n\\begin{theorem}\n$|A_n| = n!/2$.\n\\end{theorem}\n\n\\begin{proof}\nWe show that half of the elements of $S_n$ are in $A_n$.\n\nLet $\\tau$ be a fixed transposition. Define $f : S_n \\to S_n$ by $f(\\sigma) = \\tau\\sigma$. This is a bijection between the sets of odd and even permutations. Note $f(\\tau\\sigma) = \\tau\\tau\\sigma = \\sigma$, which implies that $f$ is surjective, and it is clearly injective. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe group $A_n$ is generated by its $3$-cycles.\n\\end{theorem}\n\n\\begin{proof}\nWe note:\n\n(i) $(a, b)(c, d) = (a, c, b)(a, c, d)$.\n\n(ii) $(a, b)(b, c) = (a, b, c)$.\n\n(iii) $(a, b)(a, b) = (a, b, c)^2$. $\\square$\n\\end{proof}\n\n\\subsubsection{A Presentation of $S_n$}\n\n\\begin{theorem}\nThe symmetric group $S_n$ has the presentation\n$$\\langle T \\;|\\; t_i^2 = 1, (t_it_{i+1})^3 = 1, (t_it_j)^2 = 1 \\;\\;\\mbox{for}\\;\\; |i - j| > 1 \\rangle,$$\nwhere $T = \\{t_1, t_2, \\ldots, t_{n-1}\\}$.\n\\end{theorem}\n\n\\begin{proof}\nLet the group in the theorem be denoted $G_n$.\n\nWe already know that $S_n$ is generated by transpositions $t_i = (i, i+1)$. It is also easy to check that the given relations hold in $S_n$.\n\nThus there is a surjective group homomorphism $\\phi : G_n \\to S_n$ which maps the symbol $t_i$ to the transposition $t_i$.\n\nAs $S_n$ is finite, it only remains to show that $|G_n| \\leq n!$. We will prove this by induction, the basecase $n = 2$ being clear.\n\nSuppose the result is true for $G_n$ and let $H$ be the subgroup of $G_{n + 1}$ generated by $\\{t_1, t_2, \\ldots, t_n\\}$. By induction $|H| \\leq n!$. We will investigate the cosets of $H$ in $G_{n+1}$.\n\nLet $H_n = H$, $H_{n-1} = t_nH$, $H_{n-2} = t_{n-1}t_nH$, $\\ldots$, $H_0 = t_1t_2\\ldots t_nH$. We will show that these are precisely the cosest of $H$ in $G_{n+1}$ which will prove the result.\n\nTo prove this, it suffices to show that $\\{H_0, H_1, \\ldots, H_n\\}$ is permuted by left multiplication by elements of $G_{n+1}$, since all such cosets include the cosets of $H = H_0$.\n\nAs $G_n$ is generated by $t_1, t_2, \\ldots, t_n$ it is enough to show this for left multiplication by these elements. We will compute $t_iH_j$ for all $1 \\leq i < n$ and $0 \\leq j \\leq n$.\n\nIt's clear that $t_iH_i = H_{i-1}$ and $t_iH_{i-1} = H_i$. Therefore, suppose that $j \\neq i, i - 1$. We will show that $t_iH_j = H_j$ in these cases.\n\nLet us denote $t_kt_{k_1}\\ldots t_{l}$ by $P(k, l)$.\n\nFirstly, if $j \\geq i + 1$ we have\n$$t_iH_j = t_iP(j+1, n)H = P(j+1, n)t_iH = P(j+1, n)H = H_j.$$\n\nIn the case $j \\leq i - 2$ we have\n\\begin{multline*}\ny_iH_j = y_iP(j+1, n)H = P(j+1, i-2)t_it_{i-1}t_iP(i+1, n)H =\\\\\n P(j+1, i-2)t_{i-1}t_it_{i-1}P(i+1, n)H = P(j+1, n)t_{i-1}H = P(j+1, n)H = H_j.\n\\end{multline*}\n$\\square$\n\\end{proof}\n\n\\section{Groups Acting on a Set}\n\n\\subsection{Groups Actions}\n\n\\subsubsection{Group Actions}\n\n\\begin{definition}\nLet $A$ be a set and $G$ a group. A \\emph{left group action} of $G$ on $A$ is a map $G\\times A \\to A$ denoted by $(g, a) \\mapsto ga$ for all $g \\in G$ and $a \\in A$ such that the following hold.\n\\begin{enumerate}\n\\item $g_1(g_2a) = (g_1g_2)a$ for all $g_1, g_2 \\in G$ and $a \\in A$\n\\item $1a = a$ for all $a \\in A$\n\\end{enumerate}\nSimilarly, a \\emph{right group action} is a map $A\\times G \\to A$ such that\n\\begin{enumerate}\n\\item $(ag_1)g_2 = a(g_1g_2)$ for all $g_1, g_2 \\in G$ and $a \\in A$\n\\item $a1 = a$ for all $a \\in A$\n\\end{enumerate}\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group acting on a set $A$ on the left. For each $g \\in G$ we let $\\sigma_g : A \\to A$ be defined by $\\sigma_g(a) = ga$ for all $a \\in A$. Then the following hold.\n\\begin{enumerate}\n\\item Each $\\sigma_g$ is a permutation of $A$, i.e. $\\sigma_g \\in S_A$\n\\item The map $\\varphi : G \\to S_A$ defined by $\\varphi(g) = \\sigma_g$ is a group homomorphism\n\\item If $\\psi : G \\to S_A$ is any homomorphism then the map $G\\times A \\to A$ defined by $(g, a) \\mapsto \\psi(g)a$ is a left group action of $G$ on $A$\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\n1. It is easy to see that $\\sigma_g\\circ \\sigma_{g^{-1}}$ is the identity on $A$ and similarly for $\\sigma_{g^{-1}}\\circ \\sigma_g$. Thus $\\sigma_g$ is a bijection and hence a permutation of $A$.\n\n2. It is easy to show that $\\sigma_{g_1g_2} = \\sigma_{g_1}\\circ\\sigma_{g_2}$. Therefore $\\phi(g_1g_2) = \\varphi(g_1)\\circ\\varphi_{g_2}$.\n\n3. Let $g_1, g_2 \\in G$ and $a \\in A$. Then it is easy to check that $g_1(g_2(a)) = \\psi(g_1)(\\psi(g_2)(a))$. Now on account of $\\psi$ being a homomorphism, this is equal to $\\psi(g_1g_2) = (g_1g_2)(a)$. But this is precisely the first requirement for a group representation.\n\nFor $a \\in A$ we have $1a = \\psi(1)(a) = a$. Thus we have a group action of $G$ on $A$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe map $\\varphi : G \\to S_A$ or more generally any homomorphism $\\psi$ as in the theorem is called a \\emph{permutation representation} of $G$ on the set $A$.\n\\end{definition}\n\n\\subsubsection{The Trivial Group Action}\n\n\\begin{theorem}\nLet $G$ be a group and $A$ a set. The map $G\\times A \\to A$ defined by $(g, a) = a$ for all $g \\in G$ and $a \\in A$ is a group action.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe group action of the theorem is called the trivial group action of $G$ on $A$.\n\\end{definition}\n\n\\begin{example}\nLet $\\varphi : G \\to S_A$ be the permutation representation given by $\\varphi(g) = \\;\\;\\mbox{id}_A$. Clearly $\\varphi : G \\to S_A$ is the trivial homomorphism of the group $G$ to the trivial subgroup $\\{\\mbox{id}_A\\}$ of $S_A$.\n\\end{example}\n\n\\subsubsection{Regular Group Actions}\n\nGroups can act on themselves by a left or right regular action.\n\n\\begin{definition}\nLet $G$ be a group. The \\emph{left regular action} of $G$ on itself is the group action defined by $(g, a) \\mapsto ga$ for all $g, a \\in G$, where $ga$ on the right hand side here stands for the group operation. The right regular action of $G$ on itself is defined similarly.\n\\end{definition}\n\n\\begin{theorem}\nThe left and right regular actions of $G$ on itself are left and right group actions, respectively.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{The Action of the Symmetric Group}\n\n\\begin{theorem}\nLet $X$ be a nonempty set and $S_X$ be the symmetric group on $X$. Then the action defined by $(\\sigma, x) = \\sigma(x)$ for $\\sigma \\in S_X$ and $x \\in X$ is a group action.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\nLet $\\varphi : S_X \\to S_X$ be the permutation representation defined by $\\varphi(\\delta) = \\sigma_{\\delta}$ for all $\\delta \\in S_X$ where $\\sigma_\\delta$ is defined by $\\sigma_\\delta(x) = \\delta_x$.\n\nClearly $\\varphi(\\delta) = \\delta$ for all $\\delta \\in S_X$ and so $\\varphi$ is simply the identity homomorphism on $S_X$.\n\n\\subsubsection{The Group Action of Conjugation by a Subgroup}\n\n\\begin{theorem}\nLet $G$ be a group and $H \\leq G$. The map $(h, g) \\mapsto hgh^{-1}$ is a group action of $H$ on $G$. We say that $H$ acts by conjugation on $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Faithful Group Actions}\n\n\\begin{definition}\nLet $A$ be a set and $G$ a group acting on $A$. We say that the action is \\emph{faithful} if the associated representation homomorphism $\\varphi : G \\to S_A$ defined by $\\varphi(g) = \\sigma_g$ where $\\sigma_g(x) = gx$, is injective.\n\\end{definition}\n\n\\begin{theorem}\nLet $A$ be a set and $G$ a group acting on $A$. Then $G$ acts faithfully on $A$ iff for every pair of distinct elements $g_1, g_2 \\in G$ there exists an element $a \\in A$ such that $g_1a \\neq g_2a$.\n\\end{theorem}\n\n\\begin{proof}\nThis just says $\\sigma_{g_1} \\neq \\sigma_{g_2}$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $A$ be a set and $G$ a group acting on $A$. Then $G$ acts faithfully on $A$ iff for every $g \\neq e$ in $G$ there exists $a \\in A$ such that $ga \\neq a$.\n\\end{theorem}\n\n\\begin{proof}\nThis just says that the kernel of $\\varphi$ is trivial. $\\square$\n\\end{proof}\n\n\\subsubsection{Transitive Group Actions}\n\n\\begin{definition}\nLet $A$ be a set and $G$ a group acting on $A$. Then $G$ is said to act \\emph{transitively} on $A$ if for every $a, b \\in A$ there exists $g \\in G$ such that $ga = b$.\n\\end{definition}\n\n\\subsubsection{Regular Group Actions}\n\n\\begin{definition}\nThe action of a group $G$ on a set $A$ is said to be \\emph{regular} if it is faithful and transitive. We say that $A$ is a \\emph{principal homogeneous space} or $G$-torsor.\n\\end{definition}\n\n\\begin{theorem}\nThe action of a group $G$ on itself by left multiplication is regular.\n\\end{theorem}\n\n\\begin{proof}\nIt is clearly transitive and free. $\\square$\n\\end{proof}\n\n\\begin{corollary} (Cayley)\nEvery group $G$ can be embedded in $S_G$.\n\\end{corollary}\n\n\\begin{proof}\nThe associated permutation representation $\\varphi : G \\to S_G$ is injective as the action of a group on itself by left multiplication is faithful. $\\square$\n\\end{proof}\n\n\\subsection{Orbits and Stabilisers}\n\n\\subsubsection{The Orbit and Stabiliser of a Point}\n\n\\begin{definition}\nLet $G$ be a group acting on a nonempty set $A$ and let $a \\in A$. The \\emph{orbit} of $a$ under the action of $G$ is the set\n$$Ga = \\{b \\in A \\;|\\; b = ga \\;\\;\\mbox{for some}\\;\\; g \\in G\\}.$$\n\\end{definition}\n\n\\begin{definition}\nLet $G$ be a group acting on a nonempty set $A$ and let $a \\in A$. The \\emph{stabiliser} of $a$ in $G$ is the set\n$$G_a = \\{g \\in G \\;|\\; ga = a\\}.$$\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group acting on a set $A$. Then $G_a \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\n$G_a$ is clearly closed under the group operation of $G$.\n\nClearly $e \\in G_a$.\n\nIf $g \\in G_a$ then\n$$g^{-1}a = g^{-1}(ga) = (g^{-1}g)a = ea = a.$$\nThus $g^{-1} \\in G_a$ and $G_a$ is a subgroup of $G$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Class Equation for Group Actions}\n\n\\begin{theorem}\nLet $G$ be a group acting on a nonempty set $A$ and let $a \\in A$. Then the orbit $Ga$ is in bijection with the cosets of $G_a$, i.e. $|Ga| = [G:G_a]$.\n\\end{theorem}\n\n\\begin{proof}\nLet $f : G/G_a \\to Ga$ be a map on sets ($G_a$ need not be normal in $G$) defined by $f(gG_a) = ga$.\n\nWe will show for $g_1, g_2 \\in G$ that $g_1G_a = g_2G_a$ iff $g_1a = g_2a$ so that $f$ is well-defined and injective.\n\nClearly we have $g_1a = g_2a$ iff $(g_2^{-1}g_1)a = a$ iff $g_2^{-1}g_1 \\in G_a$ iff $g_1G_a = g_2G_a$.\n\nThe map $f$ is clearly surjective. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group acting on a nonempty set $A$. The set of orbits of the action of $G$ on $A$ partition $A$.\n\\end{theorem}\n\n\\begin{proof}\nIt is clear that two elements $a, b \\in A$ are in the same orbit iff there exists $g \\in G$ such that $a = gb$.\n\nBut this is clearly an equivalence relation. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $G$ be a group acting on a nonempty set $A$. The subset of $A$ \\emph{fixed} by $G$ is the set\n$$A^G = \\{a \\in A \\;|\\; ga = a \\;\\;\\mbox{for all}\\;\\; g \\in G\\}.$$\n\\end{definition}\n\n\\begin{theorem} (Class Equation)\nLet $G$ be a group acting on a finite, nonempty set $A$. Then\n$$|A| = |A^G| + \\sum [G:G_a],$$\nwhere the sum runs over a set of representatives $a$ of the orbits $Ga$ with $|Ga| > 1$.\n\\end{theorem}\n\n\\begin{proof}\nClear from the preceding two theorems. $\\square$\n\\end{proof}\n\nIn particular, note that the size of any orbit divides the order of the group $G$.\n\n\\subsubsection{The Class Equation for Conjugation}\n\n\\begin{theorem}\nLet $G$ be a finite group. Then\n$$|G| = |Z_G| + \\sum [G:C_G(g)],$$\nwhere the sum is over all nontrivial (more than one element) conjugacy classes in $G$.\n\\end{theorem}\n\n\\begin{proof}\nEach element of the centre has trivial conjugacy class and the converse is also true.\n\nThe rest of the sum follows immediately from Theorem \\ref{conjugacy}. $\\square$\n\\end{proof}\n\n\\section{The Sylow Theorems}\n\n\\subsection{$p$-groups}\n\n\\subsubsection{$p$-groups}\n\n\\begin{definition}\nA $p$-group for a prime $p$ is a group in which every element has order a power of $p$.\n\\end{definition}\n\n\\begin{theorem} (Cauchy)\nIf $G$ is a finite group whose order is divisible by a prime $p$ then it has an element of order $p$.\n\\end{theorem}\n\n\\begin{proof}\nWe prove the result first for abelian groups, using induction. If $|G| = p$ then any nontrivial element will do.\n\nLet $a$ be any nontrivial element and let $H = \\langle a \\rangle$. If $p \\;|\\; |H|$ then $a^{|H|/p}$ is of order $p$. If not, then $p$ divides $[G:H]$ and by induction $G/H$ has an element of order $p$. Let that element be $xH$ and let the order of $x$ in $G$ be $m$. As $(xH)^m = H$ we have that $p \\;|\\; m$ and $x^{m/p}$ has order $p$.\n\nWhen $G$ is not abelian, if $p$ divides $|Z_G|$ then by the first part, $Z_G$ contains an element of order $p$ and we are done. If not, as $p \\;|\\; |G|$ we have by the class equation there is some $a \\in G$ such that $[G:C_G(a)]$ is not divisible by $p$. This means that $|C_G(a)|$ is divisible by $p$. As $C_G(a)$ is a proper subgroup it has an element of order $p$ by induction. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $p$ be a prime number. A finite group is a $p$-group iff it has order a power of $p$.\n\\end{theorem}\n\n\\begin{proof}\nIf a group $G$ has order $p^k$, every element of the group must have order dividing $p^k$ and so \n$G$ must be a $p$-group.\n\nIf $G$ does not have order $p^k$ for some $k \\in \\N$ then its order is divisible by some prime $q$ or it is the trivial group. In either case it is not a $p$-group. $\\square$\n\\end{proof}\n\n\\subsubsection{Burnside's theorem for $p$-groups}\n\n\\begin{theorem} (Burnside)\nThe centre $Z_G$ of a nontrivial $p$-group $G$ is nontrivial.\n\\end{theorem}\n\n\\begin{proof}\nIf the centre is trivial, from the class equation, we must have that $[G:C_G(a)]$ is not a power of $p$ for some non-central $a \\in G$. But this means that either $|G|$ is divisible by a prime $q \\neq p$ in which case $G$ is not a $p$-group, or $[G:C_G(a)] = 1$, which means $|C_G| = |G|$. But the latter means that $a$ is central, which is a contradiction. $\\square$\n\\end{proof}\n\n\\begin{theorem} \\label{pgroupsubs}\nIf $G$ is a $p$-group of order $p^k$ it has a normal subgroup of order $p^m$ for all $1 \\leq m \\leq n$.\n\\end{theorem}\n\n\\begin{proof}\nWe proceed by induction. The result is trivial for $k = 0, 1$.\n\nBy the previous theorem, $|Z_G| = p^m$ for some $1 \\leq m \\leq k$. Thus $Z_G$ has an element of order $p$ by Cauchy's theorem. Let $H$ be the subgroup it generates. Note that $H$ is a normal subgroup since it is central. By the induction hypothesis, the result is true for $G/H$.\n\nTherefore by the correspondence theorem $G$ has normal subgroups of order $p^m$ for $2 \\leq m \\leq k$. As $H$ has order $p^1$ the result is proved. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a finite $p$-group and $H$ a proper subgroup of $G$. Then $H$ is a proper subgroup of $N_G(H)$.\n\\end{theorem}\n\n\\begin{proof}\nRecall that $H \\leq N_G(H)$. Suppose to the contrary of the theorem that $H = N_G(H)$ and that $G$ is the smallest group with this property. As $Z_G \\leq N_G(H)$ we have that $Z_G \\leq H$. But $Z_G$ is normal in $H$.\n\nBut $Z_G$ is nontrivial by the theorem above and so $G/Z_G$ furnishes a smaller group where the normalizer of the subgroup $H/Z_G$ is $N/Z_G$ and $N/Z_G = N/Z_G$, which is a contradiction of the minimality of $G$. $\\square$\n\\end{proof}\n\n\\subsubsection{Basic results regarding $p$-groups}\n\n\\begin{theorem}\nLet $G$ be a finite $p$-group. Every nontrivial normal subgroup $N$ of $G$ intersects $Z_G$ nontrivially.\n\\end{theorem}\n\n\\begin{proof}\nAs $N$ is a normal subgroup, it is a union of conjugacy classes. Each conjugacy class has $[G:C_G(a)]$ elements for $a$ in the conjugacy class. But this is a power of $p$ divisible by $p$ unless $a$ is in the centre of $G$.\n\nAs $N$ has order a power of $p$, the number of elements of $N$ in the centre of $G$ must also be divisible by $p$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA group $G$ of order $p^2$ is abelian.\n\\end{theorem}\n\n\\begin{proof}\nIf $G$ is not abelian then $Z_G$ is a proper subgroup of $G$. But it is not trivial by Burnside's theorem and so we must have $|Z_G| = p$.\n\nLet $g \\in G$ but $g \\notin Z_G$. Note that $g \\in C_G(g)$ and $Z_G \\leq C_G(g)$. But this implies $|C_G(g)| = p^2$, which implies $C_G(g) = G$ and thus $C_G(g) = G$ which is a contradiction as this would imply $g \\in Z_G$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe centre of a non-abelian group $G$ of order $p^3$ has order $p$.\n\\end{theorem}\n\n\\begin{proof}\nWe have that $|Z_G| \\in \\{1, p, p^2, p^3\\}$. As $G$ is non-abelian, $|Z_G| \\neq p^3$. As $G$ is a $p$-group $|Z_G| \\neq 1$.\n\nAssume $|Z_G| = p^2$. As $Z_G \\mathrel{\\unlhd} G$ the group $G/Z_G$ has order $p$ and is therefore cyclic. Thus $G$ is abelian.\n\nThe only remaining case is $|Z_G| = p$. $\\square$\n\\end{proof}\n\n\\subsection{Sylow $p$-subgroups and the Sylow Theorems}\n\n\\subsubsection{Sylow $p$-subgroups}\n\n\\begin{definition}\nLet $G$ be a finite group of order $n$ and let $p$ be a prime dividing $n$. A \\emph{Sylow $p$-subgroup of $G$} is a subgroup of order $p^k$ where $p^i \\;|\\; n$ for $1 \\leq i \\leq k$ but not for $i = k + 1$.\n\\end{definition}\n\n\\begin{definition}\nThe set of Sylow $p$-subgroups of a finite group $G$ for which $p \\;|\\; n$ is denoted Syl$_p(G)$. We will denote the number of such groups by $n_p$.\n\\end{definition}\n\n\\subsubsection{Intersection of Sylow Subgroups}\n\n\\begin{theorem}\nLet $G$ be a finite group of order $n$ and let $p$ and $q$ be distinct primes that divide $n$. A Sylow $p$-subgroup $H$ of $G$ and a Sylow $q$-subgroup $K$ of $G$ intersect trivially.\n\\end{theorem}\n\n\\begin{proof}\nThe order of $H$ is a power of $p$ and the order of $K$ is a power of $q$.\n\nBut $H \\cap K$ is a subgroup of both $H$ and $K$, thus it has order dividing a power of $p$ and a power of $q$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Sylow Theorems}\n\n\\begin{theorem} (First Sylow)\nLet $G$ be a finite group of order $n$ and let $p$ be a prime. If $p^k \\;|\\; n$ then $G$ has a subgroup of order $p^k$.\n\\end{theorem}\n\n\\begin{proof}\nFirst we consider the case where $p$ divides the order of $Z_G$. By Cauchy's Theorem, $Z_G$ has an element $a$ of order $p$. As $a \\in Z_G$ the subgroup $\\langle a \\rangle$ is normal in $G$.\n\nBy induction the quotient group $G/\\langle a \\rangle$ has a subgroup $M$ of order $p^{k-1}$. It is easy to see that the preimage of $M$ under the quotient homomorphism is a subgroup of $G$ of order $p^k$.\n\nNow consider the case where $p$ does not divide the order of $Z_G$. From the class equation we see that there is an element $g \\in G$ for which $p$ does not divide $[G:C_G(g)]$. We see that $p^k$ must divide the order of $C_G(g)$ and by induction this must have a subgroup of order $p^k$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $G$ is a finite group of order $n$ and $p$ is a prime for which $p^k \\;|\\; n$ then $G$ has a subgroup of order $p^k$.\n\\end{corollary}\n\n\\begin{proof}\nFollows immediately from the First Sylow Theorem and Theorem \\ref{pgroupsubs}. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nA group $G$ is a $p$-group iff the order of every element of $G$ is a power of $p$.\n\\end{corollary}\n\n\\begin{proof}\nIf $G$ has an element of order divisible by $q$ for some prime $q \\neq p$ then $q$ divides the order of $G$ and it is not a $p$-group.\n\nConversely, if $G$ is not a $p$-group it has order divisible by a prime $q \\neq p$ which means that it has an element of order $q$. $\\square$\n\\end{proof}\n\n\\begin{theorem} (Second Sylow)\nLet $G$ be a finite group, then the number $n_p$ of Sylow $p$-subgroups of $G$ satisfies $n_p \\equiv 1 \\pmod{p}$.\n\\end{theorem}\n\n\\begin{proof}\nLet $P$ be a Sylow $p$-subgroup. Let $g \\in G$ have order a power of $p$ and $gPg^{-1} = P$. We claim $g \\in P$. \n\nConsider the subgroup $R$ of $G$ generated by $P$ and $g$. We have $g \\in N_G(P)$ and so $R \\leq N_G(P)$. Thus $P \\mathrel{\\unlhd} R$.\n\nWe have that $|R| = |R/P||P|$. But $R/P$ is generated by $gP$ which has order a power of $p$. Thus $|R|$ is a power of $p$.\n\nWe note that $G$ acts on Syl$_p(G)$ by conjugation. Let $Q$ be a Sylow $p$-subgroup of $G$ distinct from $P$. Then $Q$ cannot be fixed under conjugation by elements of $P$ because of the claim above.\n\nLet $O$ be the orbit of $Q$ under conjugation by elements of $P$. By the orbit-stabilizer theorem $|O| = [P:P_Q]$ where $P_Q$ is the stabiliser of $Q$ under the action of $P$ by conjugation.\n\nAs $|P|$ is a power of $p$, $|O|$ must be a power of $p$. We cannot have $|O| = 1$ since $Q$ is not fixed by the whole of $P$.\n\nThe set of all Sylow $p$-subgroups is a union of $P$-orbits. The only orbit of order $1$ is $\\{P\\}$. Thus $n_p \\equiv 1 \\pmod{p}$. $\\square$\n\\end{proof}\n\n\\begin{theorem} (Third Sylow)\nAny two Sylow $p$-subgroups of a finite group $G$ are conjugate.\n\\end{theorem}\n\n\\begin{proof}\nLet $P$ be a Sylow $p$-subgroup of $G$. Let $S$ be the set of all $G$-conjugates of $P$. We see that $S$ is invariant under the action of $P$ by conjugation and that $P \\in S$.\n\nAs per the proof of the second Sylow theorem, $|S| \\equiv 1 \\pmod{p}$.\n\nIf $S$ does not cover the whole of $G$, choose $Q$ a Sylow $p$-subgroup of $G$ not in $S$. Let $T$ be the set of $G$-conjugates of $Q$. We also have $|T| \\equiv 1 \\pmod{p}$ by a similar argument to that for $S$ and $P$.\n\nBut $T$ is invariant under the action of $P$ by conjugation and $P \\notin T$. Thus $|T| \\equiv 0 \\pmod{p}$, which is a contradiction. Thus $S$ must cover the whole of $G$. $\\square$\n\\end{proof}\n\n\\begin{theorem} (Fourth Sylow)\nIf $G$ is a finite group of order $n$ and $n = p^km$ for a prime $p$ such that $p$ does not divide $m$ then $n_p$ divides $m$.\n\\end{theorem}\n\n\\begin{proof}\nLet $P$ be a Sylow $p$-subgroup of $G$. The number of conjugates of $P$ is given by $[G:N_G(P)]$ by the orbit-stabiliser theorem.\n\nAs $P \\leq N_G(P)$ we have that $|N_G(P)|$ is divisibly by $p^k$. Thus the number of conjugates of $P$ must divide $m$.\n\nBut there are $n_p$ conjugates of $P$, since all the Sylow $p$-subgroups are conjugate. $\\square$\n\\end{proof}\n\n\\subsubsection{Theorems Regarding Sylow Subgroups}\n\n\\begin{theorem}\nLet $H$ be a $p$-group which is a subgroup of a finite group $G$. Then $H$ is a contained in a Sylow $p$-subgroup of $G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $H$ act on Syl$_p(G)$ by conjugation. The size of the orbit of $P \\in$ Syl$_p(G)$ under this action is $[H:H_P]$ where $H_P$ is the stabiliser of $P$ under the action of $H$. But this is a power of $p$.\n\nAs the size of Syl$_p(G)$ is not a multiple of $p$, there must be an orbit with a single element, $P$ say. Consider the subgroup $HP$ of $G$ generated by $P$ and $H$. We have that $P$ is a normal subgroup of $HP$.\n\nBy the second isomorphism theorem $HP/P \\cong H/(H\\cap P)$. The order of the group on the left is coprime to $p$ because $P$ is a Sylow $p$-subgroup of $G$. The order of the group on the right is a power of $p$ as it divides the order of $H$. \n\nThus both groups are trivial and $H \\subseteq P$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nAn abelian group $G$ is the inner direct product of its Sylow subgroups.\n\\end{theorem}\n\n\\begin{proof}\nThere is a unique Sylow $p$-subgroup for each prime $p$ dividing the order of $G$, as all the Sylow $p$-subgroups are conjugate and $G$ is abelian.\n\nWe can proceed by induction on the number $n$ of primes dividing the order of $G$. Clearly if $n = 1$ the result holds.\n\nLet $P_1, P_2, \\ldots, P_n$ be the Sylow $p$-subgroups of $G$ of order $p_1^{a_1}, \\ldots, p_n^{a_n}$ respectively. We have by induction that $H = P_1P_2\\ldots P_{n-1}$ is the direct product of $P_1, P_2, \\ldots, P_{n-1}$. By induction it has order $m = p_1^{a_1}\\ldots p_{n-1}^{a_{n-1}}$.\n\nWe see that the subgroup of $G$ generated by $H$ and $P_n$ has order divisible by $mp_n^{a_n}$ which is the order of $G$. Thus $HP_n = G$.\n\nWe must have that $H\\cap P_n = \\emptyset$ since every element of $H$ has order dividing $m$ and every element of $P_n$ has order dividing $p_n$.\n\nA similar argument will apply if we replace $P_n$ in the argument by any of the $P_i$. As $G$ is abelian, all the $P_i$ are normal in $G$. Thus $G$ is the direct product of the $P_i$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Frattini Argument}\n\n\\begin{theorem}\nLet $G$ be a finite group, $N \\mathrel{\\unlhd} G$ and $P$ be a Sylow $p$-subgroup of $N$ for some prime $p$. Then\n$$G = NN_G(P).$$ \n\\end{theorem}\n\n\\begin{theorem}\nFor $g \\in G$ we have $gPg^{-1} \\leq gNg^{-1} = N$ because $N \\mathrel{\\unlhd} G$. Thus $gPg^{-1}$ is a Sylow $p$-subgroup of $N$. Thus there exists $n \\in N$ such that $nPn^{-1} = gPg^{-1}$ because Sylow $p$-subgroups are conjugate.\n\nThus $P = (n^{-1}g)P(n^{-1}g)^{-1}$ so that $n^{-1}g \\in N_G(P)$. Note that $g = n(n^{-1}g)$ which establishes the theorem as $g$ was arbitrary.\n\\end{theorem}\n\n\\subsection{Groups of Prescribed Order}\n\n\\subsubsection{Groups of Order $pq$}\n\n\\begin{theorem}\nLet $G$ be a group of order $pq$ where $p < q$ are primes. Then there are two possibilities:\n\\begin{enumerate}\n\\item $G$ is cyclic of order $pq$\n\\item $G$ is non-abelian with generators $a, b$ such that $a^p = b^q = 1$ and $a^{-1}ba = b^r$ for some $r \\not\\equiv 1 \\pmod{q}$, $r^p \\equiv 1 \\pmod{p}$ and $p \\;|\\; q - 1$.\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\nLet $n_p, n_q$ be the number of Sylow $p$-subgroups and $q$-subgroups respectively, of $G$.\n\nBy the Sylow theorems, $n_q = 1 + kq$ for some $k \\in \\Z$ and $n_q$ divides $p$. Thus $n_q = 1$. Thus there is one Sylow $q$-subgroup and it is therefore normal in $G$. Denote it by $Q$.\n\nLet $P$ be any $p$-Sylow subgroup. Then $Q\\cap P = 1$ and $QP = G$, so $G$ is a semidirect product of $Q$ and $P$.\n\nWe will determine the action of $P$ on $Q$ by conjugation.\n\nIf $p$ does not divide $q - 1$ then $n_p = 1 + mp$ does not equal $q$ but divides $q$, so it must equal $1$. Then $G$ is a direct product of $Q$ and $P$ and therefore cyclic of order $pq$.\n\nIf $p$ does divide $q - 1$ then Aut$(Q) \\cong C_{q-1}$, which has a unique subgroup $P'$ of order $p$. The map $\\varphi : P \\to$ Aut$(Q)$ must be either trivial, which is the case above, or it must map to the unique subgroup of order $p$.\n\nLet $a$ and $b$ be generators of $P$ and $Q$ respectively. Let the action of $a$ on $Q$ by conjugation be $x \\mapsto x^r$ where $r \\neq 1 \\pmod{q}$. By the above argument we must have $r^p \\equiv 1 \\pmod{p}$.\n\nChoosing a different such $r$ is equivalent to choosing a different generator for $P$ and gives an isomorphic group $G$. Thus there is a unique group $G$ up to isomorphism in this case, and $G$ is clearly non-abelian. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups of Order $p^2q$}\n\n\\begin{theorem}\nLet $G$ be a group of order 12. Then $G$ has one Sylow $3$-subgroup or it is isomorphic to $A_4$ which has one Sylow $2$-subgroup.\n\\end{theorem}\n\n\\begin{proof}\nBy the Sylow theorems, $n_3 \\;|\\; 4$ and $n_3 \\equiv 1 \\pmod{3}$. Thus $n_3 = 1$, in which case we are done, or $n_3 = 4$.\n\nWe will first show that $A_4$ has four Sylow $3$-subgroups.\n\nWe have $|A_4| = 12$. There are eight elements that are $3$-cycles, three elements that are a pair of disjoint transpositions and the identity element.\n\nIt is easy to see there are four Sylow $3$-subgroups, each consisting of the identity and two $3$-cycles.\n\nSimilarly, there is only one Sylow $2$-subgroup consisting of the identity and three pairs of disjoint $2$-cycles.\n\nReturning to the general case, suppose that $G$ has four Sylow $3$-subgroups, i.e. Syl$_3(G) = \\{H_1, H_2, H_3, H_4\\}$. We have $H_i\\cap H_j = 1$ for $i \\neq j$. $G$ acts on Syl$_3(G)$ by conjugation.\n\nWe claim $N_G(H_i) = H_i$. This follows because all Sylow $3$-subgroups are conjugate. Thus there is only one orbit of the action of $G$, from which the theorem follows by the orbit-stabiliser theorem and the fact that $H_i \\leq N_G(H_i)$.\n\nLet $\\varphi : G \\to S_4$ be a permutation representation associated to the action of $G$ on Syl$_3(G)$. We will show the kernel of the map is trivial, i.e. the representation is faithful.\n\nBut if $g \\in$ ker$\\varphi$ it must be in the normaliser of each of the $H_i$ and thus in each of the $H_i$. But we showed this was impossible as the $H_i$ intersect trivially. Thus the permutation representation is injective.\n\nAs $G$ contains eight distinct elements of order $3$ each including the identity and the two other elements of one of the $H_i$, its image in $S_3$ must also have eight such elements. Thus its intersection with $A_4$ must contain these elements. But $A_4$ can't have a subgroup of more than $6$ elements, so the image must in fact be $A_4$.\n\nThus $G$ is isomorphic to $A_4$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group of order $p^2q$ for distinct primes $p$ and $q$. Then $G$ contains a normal Sylow subgroup.\n\\end{theorem}\n\n\\begin{proof}\nIf $p > q$ then $n_p \\equiv 1 \\pmod{p}$ and $n_p \\;|\\; q$. Thus $n_p = 1$ and the Sylow $p$-subgroup is normal in $G$.\n\nIf $q > p$ then $n_q \\equiv 1 \\pmod{q}$ and $n_q \\;|\\; p^2$. If $n_q = 1$ we are done. Thus the only remaining case is $n_q = p^2$ and $n_q \\equiv 1 \\pmod{q}$.\n\nBut then $p^2 = 1 + kq$ for some $k \\in \\mathbb{Z}$. Thus $q \\;|\\; p + 1$. But $q > p$ and so $q = p + 1$. This implies $p = 2$ and $q = 3$ as both $p$ and $q$ are prime.\n\nThus $|G| = 12$. The result follows by the previous theorem. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups of Order $p^2$}\n\n\\begin{theorem}\nLet $G$ be a group of order $p^2$ for some prime $p$. Then either $G$ is cyclic of order $p^2$ or $G \\cong C_p\\oplus C_p$.\n\\end{theorem}\n\n\\begin{proof}\nRecall that a group of order $p^2$ is abelian.\n\nLet $g \\in G$ be an element other than the identity. If the order of $g$ is $p^2$ then $G$ is cyclic.\n\nOtherwise, $g$ has order $p$. Now let $h \\in G$ be a element that is not in $\\langle g \\rangle$. Clearly $\\langle g \\rangle$ and $\\langle g \\rangle$ intersect trivially.\n\nIt's also clear that $g$ and $h$ generate the whole of $G$. And as the group $G$ is abelian, this means that $G = \\langle g \\rangle \\oplus \\langle h \\rangle$. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups of Order $30$}\n\n\\begin{theorem}\nThere are four isomorphism classes of groups $G$ of order $30$, of which one is abelian.\n\\end{theorem}\n\n\\begin{proof}\nWe note that $n_3$ is either $1$ or $10$ and $n_5$ is either $1$ or $6$. However we cannot have both $n_3 = 10$ and $n_5 = 6$. Thus either $n_3 = 1$ or $n_5 = 1$.\n\nTherefore $G$ has a subgroup $P_3$ or $P_5$ of order $3$ or $5$ respectively which is normal in $G$. Thus $H = P_3P_5$ is a subgroup of $G$ of order $15$.\n\nBy the classification of groups of order $pq$ and the fact that $5 \\not\\equiv 1 \\pmod{3}$ we have that $H$ is cyclic.\n\nAs $H$ has index $2$ in $G$ it must be normal. Thus $G \\cong H\\rtimes_{\\theta} Q$ for some homomorphism $\\theta : Q \\to$ Aut$(H)$.\n\nWe have that Aut$(H) \\cong C_4\\times C_2$. Let $p$ be a generator of $C_4$ and $q$ a generator of $C_2$. Then there are only four possible images of the non-identity element of $Q$ under $\\theta$, namely $(1, 1)$, $(p^2, 1)$, $(1, q)$ and $(p^2, q)$. The only one of these that results in an abelian group is the trivial homomorphism. $\\square$\n\\end{proof}\n\n\\section{Simple Groups}\n\n\\subsection{Simple Groups}\n\n\\subsubsection{Simple Groups}\n\n\\begin{definition}\nA group $G$ is said to be \\emph{simple} if $G$ has no proper, nontrivial, normal subgroups.\n\\end{definition}\n\n\\subsubsection{Cyclic Groups of Prime Order are Simple}\n\n\\begin{theorem}\nThe cyclic group $C_p$ with $p$ prime is simple.\n\\end{theorem}\n\n\\begin{proof}\nThere are no subgroups other than the trivial and improper subgroup. $\\square$\n\\end{proof}\n\n\\subsubsection{Finite Abelian Simple Groups}\n\n\\begin{theorem}\nThe only simple abelian groups are the cyclic groups of prime order.\n\\end{theorem}\n\n\\begin{proof}\nA subgroup of an abelian group is always normal. Therefore if $G$ is a simple abelian group it has no nontrivial, proper subgroups.\n\nIf $g \\in G$ then $\\langle g \\rangle$ is a normal subgroup of $G$. Thus $G$ must be cyclic.\n\nBut any infinite cyclic group or finite cyclic group of non-prime order has a proper, normal subgroup other than the trivial group. $\\square$\n\\end{proof}\n\n\\subsection{Simple Alternating Groups}\n\n\\subsubsection{The Group $A_5$ is Simple}\n\n\\begin{theorem}\nIf $G$ is a group of order $60$ and $n_5 > 1$ then $G$ is simple.\n\\end{theorem}\n\n\\begin{proof}\nIf $n_5 > 1$ then by the Sylow theorems, $n_5 = 6$. This accounts for $25$ of the elements of $G$. Let Syl$_5(G) = \\{P_1, P_2, \\ldots, P_6\\}$. We have $P_i\\cap P_j = \\{1\\}$ for all $i \\neq j$.\n\nSuppose that $G$ has a nontrivial, normal proper subgroup. We claim that $|H|$ is not divisible by $5$. If it were, $H$ would have a subgroup of order $5$. It will be one of the Sylow $5$-subgroups of $G$, $P_1$ say. \n\nBut $H \\mathrel{\\unlhd} G$ and so all the conjugates of $P_1$ must be in $H$. Thus $|H| \\geq 25$.\n\nAs $|H| \\;|\\; |G|$ we must have $|H| = 30$. But we have shown above that all groups of order $30$ have a single Sylow $5$-subgroup.\n\nThus we must in fact have $|H| \\;|\\; 12$.\n\nBut any group $H$ of order $12$ has a normal subgroup $K$ of order $3$ or order $4$. As $H$ is normal in $G$, any conjugate of $K$ in $G$ must be a subgroup of $H$ and since $K$ is the only subgroup of that order, $K$ must be normal in $G$.\n\nSimilarly, if $H$ has order $6$, it contains a normal subgroup of order $3$, and by the same argument as above, this is also normal in $G$.\n\nThus we have shown that if $G$ has a nontrivial, proper normal subgroup, it has a normal subgroup $K$ of order $4$, $3$ or $2$.\n\nThus $G/K$ has order $15$, $20$ or $30$.\n\nIn a group of order $15$, $20$ or $30$ there is a normal Sylow subgroup of order $5$. But the preimage of this group in $G$ is a normal subgroup of order $20$, $15$ or $10$.\n\nBut each of these orders is divisible by $5$, which contradicts what we have proved above. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe group $A_5$ is simple.\n\\end{theorem}\n\n\\begin{proof}\nNote that $|A_5| = 60$.\n\n$A_5$ has one identity element, twenty $3$-cycles, fifteen double $2$-cycles and twenty four $5$-cycles.\n\nThe result then follows from the previous theorem by noting that $n_5 = 6$. $\\square$\n\\end{proof}\n\n\\subsubsection{The Group $A_n$ is Simple for $n \\geq 5$}\n\n\\begin{theorem}\nThe group $A_n$ is simple for $n \\geq 5$.\n\\end{theorem}\n\n\\begin{proof}\nWe will prove the result by induction on $n$, beginning with the fact that $A_5$ is simple by the previous theorem.\n\nSuppose $A_n$ is simple for $5 \\leq n < k$, but also suppose that it is not simple for $n = k$, i.e. there exists $H \\mathrel{\\unlhd} A_k$.\n\nLet $G_i$ be the stabiliser of the $i$-th element of the set $A_k$ is acting on. Each of the $G_i$ are isomorphic to $A_{k-1}$ and therefore simple.\n\nRecall that if $\\sigma = (a_1 a_2 \\ldots a_s)$ is a cycle then $\\alpha\\sigma\\alpha^{-1} = (\\alpha(a_1) \\alpha(a_2) \\ldots \\alpha(a_s))$. Thus it is clear that $\\alpha G_i \\alpha^{-1} = G_{\\alpha_i}$.\n\nWe will show that if $\\sigma \\in H$ and $\\sigma(i) = i$ for some $1 \\leq i \\leq k$ then $\\sigma = 1$.\n\nIf a $\\sigma \\neq 1$ exists in $H$ then $\\sigma \\in H\\cap G_i$. But $H\\cap G_i \\mathrel{\\unlhd} G_i$ since $H$ is a normal subgroup of $A_k$ so any conjugate of it is in $H$, and conjugation by an element of $G_i$ must yield another element of $G_i$.\n\nBut $G_i$ is simple and so $H\\cap G_i$ must be $G_i$ since it is not trivial as $\\sigma \\neq 1$ is an element. Thus $G_i \\leq H$.\n\nBy conjugation we have that $G_{\\alpha(i)} \\leq H$ for each $\\alpha \\in A_k$. Thus $G_i \\leq H$ for all $1 \\leq i \\leq k$.\n\nBut every double transposition fixes some $i$ and is an element of one of the $G_i$. Thus the whole of $A_k$ must be in $H$, i.e. $H = A_k$. But this is a contradiction as we assumed $H$ to be a proper subgroup of $A_k$. Thus $\\sigma = 1$ as claimed.\n\nIn other words, no non-identity elements of $H$ fix an $i$.\n\nBut this implies that there do not exist $\\sigma_1, \\sigma_2 \\in H$ such that $\\sigma_1(i) = \\sigma_2(i)$ for some $1 \\leq i \\leq k$.\n\nWe will now show that no element of $H$ has a cycle decomposition containing a cycle of length $3$ or greater.\n\nSuppose that such a $\\sigma = (a_1 a_2 a_3 \\ldots)$ exists. Let $\\alpha \\in A_k$ fix $a_1$ and $a_2$ but not fix $a_3$.\n\nNow $\\alpha\\sigma\\alpha^{-1} = (a_1 a_2 \\alpha(a_3) \\ldots) \\in H$ as $H$ is normal. But now have that $\\sigma(a_1) = \\alpha\\sigma\\alpha^{-1}(a_1)$. These are distinct permutations because $\\sigma(a_2) \\neq \\alpha\\sigma\\alpha^{-1}(a_2)$. But this contradicts what we have proved above.\n\nThus $H$ consists entirely of elements composed of pairs of pairwise disjoint transpositions.\n\nSuppose that $H$ contains $\\sigma = (a_1 a_2)(a_3 a_4)(a_5 a_6)\\ldots$. Let $\\alpha = (a_1 a_2)(a_3 a_5)$. Then $\\alpha\\sigma\\alpha^{-1} = (a_2 a_1)(a_5 a_4)(a_3 a_6)$. Again we have a contradiction to what we proved above as we have distinct permutations in $H$ that send $a_1$ to the same place.\n\nBut now the only element that can be in $H$ is the identity and $A_k$ must be simple. $\\square$\n\\end{proof}\n\n\\subsubsection{$A_3$ is Simple, $A_4$ is not Simple}\n\n\\begin{theorem}\n$A_3$ is simple.\n\\end{theorem}\n\n\\begin{proof}\n$|A_3| = 3$ and is cyclic group of order $3$ and is therefore simple. $\\square$\n\\end{proof}\n\n\\begin{theorem}\n$A_4$ is not simple.\n\\end{theorem}\n\n\\begin{proof}\n$|A_4| = 12$ and we have already shown that either $n_2 = 1$. $\\square$\n\\end{proof}\n\n\\section{Subnormal Series}\n\n\\subsection{Composition Series}\n\n\\subsubsection{Composition Series}\n\n\\begin{definition}\nLet $G$ be a group. A \\emph{composition series} for $G$ of length $n$ is a finite chain of subgroups\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G$$\nsatisfying the following:\n\\begin{enumerate}\n\\item $G_i \\mathrel{\\unlhd} G_{i + 1}$\n\\item $G_{i+1}/G_i$ is simple\n\\end{enumerate}\nThe quotients $G_{i+1}/G_i$ are called the \\emph{composition factors} of the series.\n\\end{definition}\n\n\\begin{theorem}\nAny finite group $G$ has a composition series.\n\\end{theorem}\n\n\\begin{proof}\nStart with the group $G_n = G$ and at each step take a maximal, normal, proper subgroup, $G_i$ of $G_{i+1}$, which exists because $G$ is finite.\n\nThe quotient $G_{i+1}/G_i$ is simple by the correspondence theorem and the maximality of $G_i$ in $G_{i+1}$.\n\nThe process must terminate as $G$ is finite. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nFor a finite group $G$, the order of $G$ is the product of the orders of the composition factors of $G$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{The Zassenhaus Butterfly Lemma}\n\n\\begin{theorem}\nLet $G$ be a group and let $H$, $H'$, $K$, $K' \\leq G$ be such that $H' \\mathrel{\\unlhd} H$ and $K' \\mathrel{\\unlhd} K$. Then\n$$H'(H\\cap K') \\mathrel{\\unlhd} H'(H\\cap K) \\;\\;\\mbox{and}\\;\\; K'(K\\cap H') \\mathrel{\\unlhd} K'(H\\cap K).$$\nFurthermore,\n$$\\bigl(H'(H\\cap K)\\bigr)/\\bigl(H'(H\\cap K')\\bigr) \\cong \\bigl(K'(H\\cap K) \\bigr)/\\bigl(K'(K\\cap H')\\bigr).$$\n\\end{theorem}\n\n\\begin{proof}\nFirst note that $H \\cap K \\leq H$.\n\nLet $\\phi : H \\to H/H'$ be the natural projection. Then $(\\phi^{-1}\\circ \\phi)(H\\cap K) = H'(H \\cap K)$, is a group.\n\nNote that $H \\cap K' \\mathrel{\\unlhd} H \\cap K$. Hence\n$$(\\phi^{-1}\\circ \\phi)(H\\cap K') = H'(H\\cap K')$$\nis a normal subgroup of\n$$(\\phi^{-1}\\circ \\phi)(H\\cap K) = H'(H\\cap K).$$\n\nNow, let $\\rho : H'(H\\cap K) \\to \\bigl(H'(H\\cap K)\\bigr)/ \\bigl(H'(H\\cap K')\\bigr)$ be the natural projection.\n\nNote that\n$$(H\\cap K)\\cap \\bigl(H'(H\\cap K')\\bigr) = (H'\\cap K)(H \\cap K').$$\n\nThus by the homomorphism theorems,\n$$\\bigl(H'(H\\cap K)\\bigr)/\\bigl(H'(H\\cap K')\\bigr) \\cong (H\\cap K)/\\bigl((H'\\cap K)(H \\cap K')\\bigr).$$\n\nThe other half of the lemma follows by symmetry. $\\square$\n\\end{proof}\n\n\\subsubsection{The Schreier Refinement Theorem}\n\n\\begin{definition}\nLet $G$ be a group. Any series of groups\n$$1 \\mathrel{\\unlhd} G_1 \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} G_n = G$$\nis called a \\emph{subnormal series}.\nA second subnormal series\n$$1 \\mathrel{\\unlhd} H_1 \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} H_m = G$$\nis said to be a \\emph{refinement} of the first if every $G_i$ appears as a $H_j$ in the same order.\n\\end{definition}\n\n\\begin{theorem} (Schreier)\nAny two subnormal series of a group $G$ have refinements that have isomorphic factor groups, up to reordering.\n\\end{theorem}\n\n\\begin{proof}\nLet\n$$1 = H_0 \\leq H_1 \\leq \\cdots \\leq H_n = G$$\nand\n$$1 = K_0 \\leq K_1 \\leq \\cdots \\leq K_m = G$$\nbe two subnormal series for $G$.\n\nWe refine the series of $H_k$ as follows. For $0 \\leq i \\leq n - 1$ and $0 \\leq j \\leq m$ define\n$$H'_{im + j} = H_i(H_{i+1}\\cap K_j).$$\n\nNote that\n$$H'_{im} = H_i(H_{i+1}\\cap K_0) = H_i.$$\nThus every group that appears in the original series of $H_i$ appears in the new series.\n\nNote that when $m \\;|\\; k$ we have $k = im + m = (i + 1)m$, meaning that there is an ambiguity in this case.\n\nFor the series to be well-defined, we'd require\n$$H_i(H_{i+1}\\cap K_m) = H_{i+1}(H_{i+2}\\cap H_0).$$\n\nBut the the left and right hand sides are both $H_{i+1}$, thus the series is well-defined.\n\nWe will show that this is a subnormal series (except for some possible duplication). Using the representation $k = im + j$ in which $j < m$ we have that\n$$H'_{k+1} = H_i(H_{i+1}\\cap K_{j + 1}).$$\n\nAs $K_j \\mathrel{\\unlhd} K_{j+1}$ we have that\n$$H'_k = H_i(H_{i+1}\\cap K_j) \\mathrel{\\unlhd} H_i(H_{i+1}\\cap H_{j+1}) = H'_{k+1}.$$\n\nNote that it is still possible that $H'_k = H'_{k+1}$ for some $k$. \n\nWe apply a similar construction to create a refinement of the series of $K_k$, by defining for $k = in + j$\n$$K'_k = K_i(K_{i+1}\\cap H_j).$$\n\nNow let $k = um + v$ and $l = vn + u$. We have\n\\begin{align*}\n\\frac{H'_{k+1}}{H'_k} &= \\frac{H_u(H_{u+1}\\cap K_{v+1})}{H_u(H_{u+1}\\cap H_v)}\\\\\n&\\cong \\frac{K_v(K_{v+1}\\cap H_{u+1})}{K_v(K_{v+1}\\cap H_U)}\\\\\n&= \\frac{K'_{l+1}}{K'_l},\n\\end{align*}\nwhere the isomorphism follows from Zassenhaus' lemma.\n\nThus the series have isomorphic factors.\n\nNote that it may still be that some of the factors are redundant because of equal terms in the sequences, as noted above. But we can simply remove these redundant elements. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nLet $G$ be a group with a composition series. Any subnormal series of $G$ can be refined to a composition series.\n\\end{corollary}\n\n\\begin{proof}\nSimply take one of the series to be the composition series, in the theorem and the other to be the given subnormal series. As there are no nontrivial refinements of a composition series, the result follows. $\\square$\n\\end{proof}\n\n\\subsubsection{The Jordan-H\\\"{o}lder Theorem}\n\n\\begin{theorem}\nLet $G$ be a group. The composition factors of any two composition series for $G$ are isomorphic in pairs, though the order of the factors may be different.\n\\end{theorem}\n\n\\begin{proof}\nWe proceed by induction on $|G|$. The theorem is true for $|G| = 2$.\n\nAssume the result is true for all groups of order less than $n$ and suppose $G$ is a group of order $n$.\n\nIf $G$ is simple, the theorem is trivially true. Otherwise, suppose that\n$$1 \\mathrel{\\unlhd} A_r \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} A_1 \\mathrel{\\unlhd} A \\mathrel{\\unlhd} G$$\nand\n$$1 \\mathrel{\\unlhd} B_s \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} B_1 \\mathrel{\\unlhd} B \\mathrel{\\unlhd} G$$\nare two composition series for $G$.\n\nIf $A = B$ then $G/A = G/B$ and by the inductive hypothesis all the other quotients are the same in pairs, up to order, and we are done.\n\nIf $A \\neq B$, consider the group $AB$ of $G$. Now $A$ and $B$ are distinct and maximal in $G$, and thus $AB = G$.\n\nNow let $D = A\\cap B$. By the second isomorphism theorem \n\\begin{equation}\\label{gabd}\nG/A \\cong B/D\\;\\;\\mbox{and}\\;\\;G/B \\cong A/D.\n\\end{equation}\nBut $G/A$ and $G/B$ are simple, thus so are $B/D$ and $A/D$. In other words, $D$ is a maximal normal subgroup of $A$ and $B$.\n\nLet $1 \\mathrel{\\unlhd} D_t \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} D_1 \\mathrel{\\unlhd} D$ be a composition series for $D$.\n\nConsider the quotient groups\n\\begin{equation}\\label{comp1}G/A, A/D, D/D_1, \\ldots, D_t, 1,\\end{equation}\nand\n$$G/A, A/A_1, \\ldots, A_r, 1.$$\nBy the induction hypothesis, the theorem is true for $A$ and so the sequences above are isomorphic in pairs. In particular, $t = r$.\n\nSimilarly, the quotients\n\\begin{equation}\\label{comp2}G/B, B/D, D/D_1, \\ldots, D_t, 1\\end{equation}\nis isomorphic in pairs with\n$$G/B, B/B_1, \\ldots, B_s, 1,$$\nand $s = r$.\n\nBut the sequences \\eqref{comp1} and \\eqref{comp2} are clearly isomorphic in pairs by \\eqref{gabd} and so we are done. $\\square$\n\\end{proof}\n\n\\subsubsection{Dedekind's Modular Law}\n\n\\begin{theorem}\nIf $A, B, C \\leq G$ with $A \\subseteq B$ then\n$$A(B\\cap C) = B\\cap AC.$$\n\\end{theorem}\n\n\\begin{proof}\nFirst note that $A \\subseteq B$ and $B\\cap C \\subseteq B$, thus $A(B\\cap C) \\subseteq B$. Also note that $B\\cap C \\subseteq C$, thus $A(B\\cap C) \\subseteq AC$. Thus we have shown that $A(B\\cap C) \\subseteq B\\cap AC$.\n\nFor the other direction, let $b \\in B\\cap AC$. Then $b = ac$ for some $a \\in A$ and $c \\in C$. But then $c = ba^{-1}$ and so $c \\in B$. Thus $c \\in B\\cap C$. Thus $b = ac \\in A(B\\cap C)$. $\\square$\n\\end{proof}\n\n\\subsubsection{Normal Subgroups of a Group with Composition Series}\n\n\\begin{theorem}\nLet $G$ be a group with composition series and $N \\mathrel{\\unlhd} G$, then $N$ has a composition series.\n\\end{theorem}\n\n\\begin{proof}\nSuppose $G$ has a composition series\n$$1 = G_0 \\mathrel{\\unlhd} G_1 \\mathrel{\\unlhd} \\cdots \\mathrel{\\unlhd} G_n = G.$$\nDefine $N_i = N\\cap G_i$. Then\n$$1 = N_0 \\leq N_1 \\leq \\cdots \\leq N_n = N.$$\n\nLet $x \\in N_i$ and $g \\in N_{i+1}$, then $g^{-1}xg \\in G_i\\cap N = N_i$ by the normality of $G_i$ in $G_{i+1}$. Thus we see that the $N_i$ form a subnormal series.\n\nBut\n\\begin{align*}\nN_i/N_{i+1} &= \\frac{N\\cap G_{i+1}}{N\\cap G_i}\\\\\n&= \\frac{N\\cap G_{i+1}}{(N\\cap G_{i+1})\\cap G_i}\\\\\n&\\cong \\frac{(N\\cap G_{i+1})G_i}{G_i}\n\\end{align*}\nby the second isomorphism theorem.\n\nAs $N \\mathrel{\\unlhd} G$ we have that $N\\cap G_{i+1} \\mathrel{\\unlhd} G_{i+1}$. Thus, as $G_i \\mathrel{\\unlhd} G_{i+1}$ we have that $(N\\cap G_{i+1})G_i \\mathrel{\\unlhd} G_{i+1}$.\n\nTherefore, by the correspondence theorem we have that\n$$\\frac{(N\\cap G_{i+1})G_i}{G_i} \\mathrel{\\unlhd} \\frac{G_{i+1}}{G_i}.$$\n\nBut the latter is simple and so either\n$$\\frac{(N\\cap G_{i+1})G_i}{G_i} = 1 \\;\\;\\mbox{or}\\;\\; \\frac{G_{i+1}}{G_i}.$$\n\nApplying this to the isomorphism we proved above, we have that either $N_i = N_{i+1}$ or $N_{i+1}/N_i$ is simple.\n\nTherefore if we delete any repetitions in the subnormal series of groups $N_i$, we obtain a composition series for $N$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group with composition series and $N \\mathrel{\\unlhd} G$, then $G/N$ has a composition series.\n\\end{theorem}\n\n\\begin{proof}\nAs $N$ is normal in $G$ we have\n$$N = G_0N \\leq G_1N \\leq \\cdots \\leq G_nN = G.$$\n\nThus from the correspondence theorem we have\n$$1 = G_0N/N \\leq G_1N/N \\leq \\cdots \\leq G_nN/N = G/N.$$\n\nFor simplicity, write $Q_i = G_iN/N$. If $x \\in G_i$ and $g \\in G_{i+1}$ we have $g^{-1}xg \\in G_i$, thus\n$$(Ng)^{-1}(Nx)(Ng) \\in G_iN/N = Q_i.$$\nThus $Q_i \\mathrel{\\unlhd} Q_{i+1}$.\n\nFrom $G_i \\leq G_{i+1}$ we have by Dedekind's modular law that\n$$G_iN\\cap G_{i+1} = (G_{i+1}\\cap N)G_i.$$\n\nWe then have\n\\begin{align*}\nQ_{i+1}/Q_i &= \\frac{G_{i+1}}{G_iN/N}\\\\\n&= \\frac{G_{i+1}N}{G_iN}\\\\\n&= \\frac{(G_{i+1}G_i)N}{G_iN}\\\\\n&= \\frac{G_{i+1}}{G_{i+1}\\cap G_iN}\\\\\n&= \\frac{G_{i+1}}{(G_{i+1}\\cap N)G_i}\\\\\n&= \\frac{G_{i+1}/G_i}{(G_{i+1}\\cap N)G_i/G_{i+1}},\n\\end{align*}\nwhere we have used in turn, the third isomorphism theorem, $G_i \\leq G_{i+1}$, the second isomorphism theorem, Dedekind's modular law and the third isomorphism theorem.\n\nAs $G_{i+1}/G_i$ is simple, the final quotient above is either trivial or isomorphic to $G_{i+1}/G_i$. The same then applies to $Q_{i+1}/Q_i$.\n\nUpon deleting repetitions, we obtain a composition series for $G/N$. $\\square$\n\\end{proof}\n\n\\subsubsection{Abelian Groups with Composition Series}\n\n\\begin{theorem}\nLet $G$ be an abelian group. If $G$ has a composition series then $G$ is finite.\n\\end{theorem}\n\n\\begin{proof}\nThe composition factors of $G$ must be simple and abelian, thus cyclic of prime order. As the order of $G$ is the product of the orders of the composition factors, the result follows. $\\square$\n\\end{proof}\n\n\\subsection{Solvable Groups}\n\n\\subsubsection{Solvable Groups}\n\n\\begin{definition}\nA group $G$ is said to be \\emph{solvable} if it has a subnormal series\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G,$$\nsuch that $G_{i+1}/G_i$ is abelian.\n\\end{definition}\n\n\\begin{theorem}\nEvery abelian group is solvable.\n\\end{theorem}\n\n\\begin{proof}\n$1 = G_0 \\leq G_1 = G$ is a composition series such that $G_1/G_0$ is abelian. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nThe group $S_n$ is not solvable for $n \\geq 5$.\n\\end{theorem}\n\n\\begin{proof}\nWe know that $A_n$ is simple for $n \\geq 5$ and so the normal series\n$$1 \\leq A_n \\leq S_n$$\nis a composition series. Its composition factors are $C_2$ and $A_n$, the latter of which is not abelian. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA finite group $G$ is solvable iff it has a composition series with composition factors that are cyclic of prime order.\n\\end{theorem}\n\n\\begin{proof}\nSuppose $G$ is solvable so that it has a subnormal series\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G$$\nwith abelian factors $G_{i+1}/G_i$.\n\nAs it is a finite group it has a composition series. Therefore by the Schreier refinement theorem, the subnormal series can be refined to a composition series. Thus between $G_i$ and $G_{i+1}$ we have a refinement\n$$G_i = H_{i,0} \\leq H_{i,1} \\leq \\cdots \\leq H_{i, m} = G_{i+1},$$\nwith $H_{i, j+1}/H_{i, j}$ simple for all $j$.\n\nBut as $G_i \\mathrel{\\unlhd} G_{i+1}$ we have that $G_i \\mathrel{\\unlhd} H_{i, j+1}$ for all $0 \\leq j < m$. Then $H_{i, j+1}/G_i$ is a subgroup of $G_{i+1}/G_i$ and therefore abelian.\n\nBut now by the second isomorphism theorem, $H_{i, j+1}/H_{i, j}$ is a quotient of $H_{i, j+1}/G_i$ and therefore also abelian. As it is also simple, it must be cyclic of prime order.\n\nConversely, suppose $G$ has a composition series with each composition factor cyclic of prime order. Then this composition series is a subnormal series with each factor abelian and $G$ is solvable. $\\square$\n\\end{proof}\n\n\\subsubsection{Derived Subgroups}\n\n\\begin{definition}\nLet $G$ be a group. We call $[G, G]$ the \\emph{first derived subgroup} of $G$, denoted $G'$ or $G^{(1)}$. The \\emph{$n$-th derived subgroup} of $G$ is the group $[G^{(n-1)}, G^{(n-1)}]$. We denote it $G^{(n)}$.\n\\end{definition}\n\n\\begin{theorem}\nLet $G$ be a group. We have that\n$$G^{(n)} \\leq G^{(n-1)} \\leq \\cdots \\leq G^{(1)} \\leq G.$$\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA group $G$ is solvable if there is a finite integer $n$ such that the $n$-th derived subgroup of $G$ is trivial.\n\\end{theorem}\n\n\\begin{proof}\nLet\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G$$\nbe a subnormal series for $G$ with abelian factor groups.\n\nSince $G_n/G_{n-1}$ is abelian we have that the derived subgroup $G^{(1)} \\leq G_{n-1}$.\n\nBy a similar argument, $G_{n-1}^{(1)} \\leq G_{n-2}$. But we also have $G^{(2)} \\leq G_{n-1}^{(1)}$ because of $G^{(1)} \\leq G_{n-1}$. Thus $G^{(2)} \\leq G_{n-2}$.\n\nContinuing in the same way, we eventually find that $G^{(n)} \\leq 1$.\n\nConversely, if $G^{(1)} = 1$ then we have a chain of subgroups\n$$1 = G^{(n)} \\leq G^{(n-1)} \\leq \\cdots \\leq G^{(1)} \\leq G.$$\n\nBut this is a subnormal series. But $G^{(i-1)}/G^{(i)}$ is abelian iff $[G^{(i-1)}, G^{(i-1)}] \\leq G^{(i)}$. But this is trivially true. $\\square$ \n\\end{proof}\n\n\\subsubsection{Every $p$-group is solvable}\n\n\\begin{theorem}\nIf $G$ is a $p$-group then it is solvable.\n\\end{theorem}\n\n\\begin{proof}\nBegin with the group $G_0 = 1$. As $G$ is a $p$-group, $Z(G)$ is non-trivial by Burnside's theorem. Let $G_1 = Z(G)$. Note that $G_1/G_0$ is abelian since it is isomorphic to $Z(G)$.\n\nNow $Z(G)$ is normal in $G$ and the quotient $G/Z(G)$ is a $p$-group.\n\nAgain by Burnside's theorem, the centre $Z(G/G_1)$ is non-trivial. By the correspondence theorem, there is a subgroup $G_2$ of $G$ containing $G_1$ such that $G_2/G_1$ is isomorphic to $Z(G/G_1)$. Note that this quotient is abelian.\n\nWe continue the same construction until $G/G_{n-1}$ is abelian because the centre of $G/G_{n-1}$ is the whole group. We then set $G_n = G$ and we have constructed a subnormal series for $G$ with abelian cofactors. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups of Order $pq$ are Solvable}\n\n\\begin{theorem}\nIf $G$ is a group of order $pq$ with $p$ and $q$ prime, then $G$ is solvable.\n\\end{theorem}\n\n\\begin{proof}\nIf $p = q$ then $G$ is abelian and we are done. Therefore, assume that $p$ and $q$ are distinct primes with $p > q$.\n\nAs $n_p \\;|\\; q$ we either have $n_p = q$ or $n_p = 1$. But as $n_p \\equiv 1 \\pmod{p}$ the first case is impossible since $p > q$. Thus $n_p = 1$ and the unique Sylow $p$-subgroup $G_1$ is normal in $G$.\n\nWe see that\n$$1 = G_0 \\leq G_1 \\leq G_2 = G$$\nis a subnormal series of $G$ with abelian quotients (cyclic groups of order $p$ and $q$). Thus $G$ is solvable. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups of Order $p^2q$ are Solvable}\n\n\\begin{theorem}\nIf $G$ is a group of order $p^2q$ for primes $p$ and $q$ then $G$ is solvable.\n\\end{theorem}\n\n\\begin{proof}\nIf $p = q$ then $G$ is a $p$-group, which is solvable. So assume $p \\neq q$.\n\nIf $p > q$, then by the same argument as for groups of order $pq$, there is a unique Sylow $p$-subgroup $G_1$ of $G$ which is normal in $G$.\n\nThus we have that\n$$1 = G_0 \\leq G_1 \\leq G_2 = G$$\nis a subnormal series. Note that $|G_1/G_0| = p^2$, and any group of order $p^2$ is abelian. Thus $G$ is solvable.\n\nThe remaining case is when $p < q$. As usual, we have that $n_q$ is $1$, $p$ or $p^2$ by the Sylow theorems. However, $n_q = p$ is not possible as $p < q$.\n\nIf $n_q = 1$ then there is a unique Sylow $q$-subgroup $G_1$ of $G$. Then we have that\n$$1 = G_0 \\leq G_1 \\leq G_2 = G$$\nis a subnormal series, and $|G_2/G_1| = p^2$. Thus all factors are abelian and $G$ is solvable.\n\nIf $n_q = p^2$ then $G$ has $p^2(q - 1) = n - p^2$ elements of order $q$. The remaining $p^2$ elements must lie in a Sylow $p$-subgroup $G_1$ of order $p^2$.\n\nAs $G_1$ is the unique subgroup of order $p^2$, it is normal in $G$. By a similar argument to the above, $G$ is solvable. $\\square$\n\\end{proof}\n\n\\subsubsection{Homomorphic Images of Solvable Groups are Solvable}\n\n\\begin{theorem}\nLet $\\phi : G \\to H$ be a group homomorphism and $G$ a solvable group. Then $\\phi(G)$ is solvable.\n\\end{theorem}\n\n\\begin{proof}\nLet $G$ be solvable and\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G$$\nbe a subnormal series for $G$ in which each factor is abelian.\n\nlet $H_k = \\phi(G_k)$ for all $k$. Then\n$$1 = H_0 \\leq H_1 \\leq \\cdots \\leq H_n = \\phi(G)$$\nis a subnormal series with normal quotients.\n\nLet $f : G_{i+1}/G_i \\to H_{i+1}/H_i$ be defined by $f(gG_i) = \\phi(g)H_i$ for all $g \\in G_{i+1}$. It is easy to see that it is a surjective homomorphism.\n\nAs $G_{i+1}/G_i$ is abelian, so is $f(G_{i+1}/G_i) = H_{i+1}/H_i$, for all $0 \\leq i \\leq n - 1$. Thus $\\phi(G)$ is solvable. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nEvery quotient of a solvable group is solvable.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Every Subgroup of a Solvable Group is Solvable}\n\n\\begin{theorem}\nLet $G$ be a solvable group. Every subgroup $H$ of $G$ is solvable.\n\\end{theorem}\n\n\\begin{proof}\nLet\n$$1 = G_0 \\leq G_1 \\leq \\cdots \\leq G_n = G$$\nbe a subnormal series for $G$ with abelian factors. Let $H_i = H\\cap G_i$. We claim that\n$$1 = H_0 \\leq H_1 \\leq \\cdots \\leq H_n = (G\\cap H) = H$$\nis a normal series of $H$.\n\nFirstly, because $G_i \\mathrel{\\unlhd} G_{i+1}$ we have that $H_i \\mathrel{\\unlhd} H_{i+1}$. For if $x \\in H_i$ and $y \\in H_{i+1}$ then $yxy^{-1} \\in H$ and $yxy^{-1} \\in G_i$ due to $G_i$ being normal in $G_{i+1}$. Thus $yxy^{-1} \\in H\\cap G_i = H_i$.\n\nNow we will show that $H_{i+1}/H_i$ is abelian for all $i$.\n\nNote that\n$$H_i = H\\cap G_i = (H\\cap G_{i+1})\\cap G_i = H_{i+1}\\cap G_i.$$\n\nBy the second isomorhism theorem we therefore have\n$$\\frac{H_{i+1}}{H_i} = \\frac{H_{i+1}}{H_{i+1}\\cap G_i} \\cong \\frac{H_{i+1}G_i}{G_i} \\leq \\frac{G_{i+1}}{G_i}.$$\nThe last relation follows because $H_{i+1}G_i \\leq G_{i+1}$ on account of $G_i$ being normal in $G_{i+1}$.\n\nBut $G_{i+1}/G_i$ is abelian, thus $N_{i+1}/N_i$ is isomorphic to a subgroup of an abelian group. $\\square$\n\\end{proof}\n\n\\subsubsection{Direct Products of Solvable Groups}\n\n\\begin{theorem}\nLet $G$ be a group with $H \\mathrel{\\unlhd} G$. If $H$ and $G/H$ are solvable then so is $G$.\n\\end{theorem}\n\n\\begin{proof}\nLet\n$$1 = K^*_0 \\leq K^*_1 \\leq \\cdots \\leq K^*_n = G/H$$\nbe a subnormal series for $G/H$ with abelian factors.\n\nBy the correspondence theorem, we can construct a partial solvable series of $G$ starting at $H$\n$$H = K_0 \\leq K_1 \\leq \\cdots \\leq K_n = G$$\nwith $K_i \\mathrel{K_{i+1}}$ and $K_{i+1}/K_i$ abelian.\n\nBut $H$ is solvable, and so we can complete the series to a subnormal series for $G$ with abelian factors. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $H$ and $K$ are solvable groups then so is $G = H\\times K$.\n\\end{corollary}\n\n\\begin{proof}\nWe have $H \\mathrel{\\unlhd} G$ and $G/H \\cong K$. $\\square$\n\\end{proof}\n\n\\subsubsection{Groups Without Characteristically Simple Subgroups}\n\n\\begin{definition}\nLet $G$ be a group. A subgroup $N$ is said to be a \\emph{minimal normal subgroup} of $G$ if $N \\neq 1$ and if $K \\mathrel{\\unlhd} G$ with $K \\leq N$ then $K = 1$.\n\\end{definition}\n\n\\begin{definition}\nA group $G$ is said to be \\emph{elementary abelian} if it is an abelian group with every element of order $p$ for a fixed prime $p$.\n\\end{definition}\n\n\\begin{theorem}\nEvery minimal normal subgroup of a finite solvable group $G$ is elementary abelian.\n\\end{theorem}\n\n\\begin{proof}\nLet $N$ be a minimal normal subgroup of $G$ and suppose $H$ is a characteristic subgroup of $V$. Then $H \\mathrel{\\unlhd} G$, which violates the minimality of $N$ unless $H = 1$ or $H = N$.\n\nNow $N' = [N, N]$ is characteristic in $N$ and hence $N' = 1$ or $N' = N$. But $N$ is solvable and so $N' \\neq N$ otherwise the derived series would not terminate in $1$.\n\nThus $N' = 1$ and $N$ is abelian. But any Sylow $p$-subgroup of an abelian group is characteristic. Again by the minimality of $N$, this implies that $N$ is an abelian $p$-group.\n\nBut $\\{x \\in N \\;|\\; x^p = 1\\}$ is a characteristic subgroup of $N$. Thus $N$ is elementary abelian. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nA finite group $G$ whose only characteristic subgroups are $1$ and $G$ is either simple or a direct product of isomorphic simple groups.\n\\end{theorem}\n\n\\begin{proof}\nLet $N_1$ be a minimal normal subgroup of $G$ with minimal order amongst all such groups. \n\nConsider all subgroups of $G$ of the form $N_1\\times N_2\\times \\cdots \\times N_n$ for $N_i \\mathrel{\\unlhd} G$ and $N_i \\cong N_j$ for all $i, j$.\n\nLet $M$ be a subgroup of this form with largest possible order. We will show that $M$ char $G$ so that $M = G$. It is enough to show for any automorphism $\\phi \\in$ Aut$(G)$ that $\\phi(N_i) \\leq M$.\n\nWe firstly show that $\\phi(N_i) \\mathrel{\\unlhd} G$. Let $a \\in G$ and $b = \\phi^{-1}(a)$. Then $a\\phi(N_i)a^{-1} = \\phi(bN_ib^{-1}) \\leq \\phi(N_i)$ due to the normality of $N_i$.\n\nSuppose $\\phi(N_i) \\not\\leq M$. Then $|\\phi(N_i)\\cap M| < |\\phi(N_i)| = |N|$. But $\\phi(N_i)\\cap M \\mathrel{\\unlhd} G$ as we showed above that $\\phi(N_i) \\mathrel{\\unlhd} G$. Thus, by the minimality of $|H|$ we have that $\\phi(H_i)\\cap M = 1$.\n\nBut then $M\\times \\phi(H_i)$ violates the maximality of $M$.\n\nThus $\\phi(N_i) \\leq M$ and so $M$ char $G$ and $M = G$.\n\nNow $N_1$ must be simple. If it has a nontrivial normal subgroup $H$ then $H$ is a normal subgroup of $M = G$. But this contradicts the minimal choice of $N$. $\\square$\n\\end{proof}\n\n\\subsubsection{Hall's Theorem on Solvable Groups of Order $mn$}\n\n\\begin{theorem}\nIf $G$ is a solvable group of order $ab$ with gcd$(a, b) = 1$ then $G$ has a subgroup of order $a$. Any two such subgroups are conjugate.\n\\end{theorem}\n\n\\begin{proof}\nThe proof is by induction on $|G|$. For $|G| = 1$ there is nothing to prove.\n\nWe break into two cases. In the first case, assume $G$ contains a normal subgroup $H$ of order $m'n'$ with $m' \\;|\\; m$ and $n' \\;|\\; n$ and $n' < n$.\n\nIn this case, $G/H$ is solvable and has order $(mn)/(m'n') < mn$ as $n' < n$. Thus by the induction hypothesis, $G/H$ has a subgroup $A/H$ of order $m/m'$. By the correspondence theorem $A$ has order $mn'$. But $A$ is now a solvable group of order $mn' < mn$ and so by the inductive hypothesis, $A$ has a subgroup of order $m$ as required.\n\nWe now show that any two subgroups $A$ and $A'$ of order $m$ are conjugate in this case.\n\nAs $H \\mathrel{\\unlhd} G$ we have that $AH \\leq G$. Moreover, $A, H \\leq AH$ and so $|AH| = m\\nu$ where $n' \\;|\\; \\nu \\;|\\; n$.\n\nBut if we apply the second isomorphism theorem to $A$ and $H$ we see that $|AH| \\;|\\; mm'n'$. Thus $\\nu \\;|\\; n'$. Thus $|AH| = mn'$.\n\nA similar computation shows that $|A'H| = mn'$. Thus both $AH/H$ and $A'H/H$ are subgroups of $G/H$ of order $m/m'$.\n\nAs $|G/H| = (mn)/(m'n') < mn$ we have by induction that $AH/H$ and $A'H/H$ are conjugate.\n\nSuppose for example that $(xH)(AH/H)(xH)^{-1} = (A'H/H)$. Then $xAHx^{-1} = A'H$ since $H$ is normal in $G$ and so left and right cosets of $H$ are in bijection, so that both sides of the equation are unions of cosets of $H$. But this shows that $xAx^{-1} \\leq A'H$.\n\nBut we already have that $A' \\leq A'H$, therefore as both $xAx^{-1}$ and $A'$ have order $a$ and $|A'H| < ab$ we have by induction that $xAx^{-1}$ and $A'$ are conjugate, which completes the induction for this case assuming the second case also holds.\n\nThe first case reduces the to a smaller case whenever $G$ has a proper normal subgroup which is not divisible by $n$. Therefore we may assume without loss of generality that every proper normal subgroup $H$ has order divisible by $n$.\n\nIn the case that $H$ is a minimal normal subgroup we have that $H$ is abelian and every element of $H$ is of order $p$ for some prime $p$. In other words, we can assume without loss of generality that $H$ is a Sylow $p$-subgroup of $G$ for some prime $p$ with order $n = p^k$.\n\nAs $H$ is normal, it is the unique $p$-Sylow subgroup. We then have $|G| = mp^k$ with $p \\not |\\; m$. This constitutes the second case, which we shall now prove.\n\nThe group $G/H$ is a solvable group of order $m$. Thus if $K/H$ is a minimal normal subgroup of $G/H$ then $|K/H| = q^n$ for some prime $q \\neq p$, i.e. $|K| = p^kq^l$. Then if $Q$ is a Sylow $q$-subgroup of $K$ then $K = HQ$.\n\nWe let $N^*$ be the normalizer of $Q$ in $G$ and $N = N^*\\cap K$ which is clearly the normalizer of $Q$ in $K$. We claim that $|N^*| = m$.\n\nBy the Frattini argument $G = KN^*$. But as\n$$G/K = KN^*/K \\cong N^*/(N^*\\cap K) = N^*/N$$\nwe have that $|N^*| = |G||N|/|K|$.\n\nAs $K = HQ$ and $Q \\leq N \\leq K$ we have $K = HN$. Thus $|K| = |H||N|/|H\\cap N|$. Thus\n$$|N^*| = |G||N|/|K| = (|G|/|H|)|H\\cap N| = m|H\\cap N|.$$\n\nIf $H\\cap N = 1$ we have $|N^*| = m$, which would complete the argument. Thus we will prove that $H\\cap N = 1$.\n\nFirst we claim that $H\\cap N \\leq Z_K$.\n\nLet $x \\in H\\cap N$. Let $k = hs \\in K = HQ$ be arbitrary, with $h \\in H, s \\in Q$. We see that $x$ commutes with $h$ because $H$ is abelian.\n\nWe have that $(xsx^{-1})s^{-1} \\in Q$ because $x$ is in the normalizer of $Q$. Also $x(sx^{-1}s^{-1}) \\in H$ as $H$ is normal. Thus $xsx^{-1}x^{-1} \\in Q\\cap H$ which is trivial because the orders of $Q$ and $H$ are coprime. This proves the claim.\n\nNow $Z_K$ is characteristic in $K$ and $K$ is normal in $G$ by the correspondence theorem. Thus $Z_K \\mathrel{\\unlhd} G$.\n\nIf $Z_K$ is not trivial then it contains a minimal subgroup which must also be normal in $G$. But $H$ is the unique minimal normal subgroup of $G$ by definition, so $H \\leq Z_K$.\n\nBut $K = HQ$ and so $Q$ char $K$ since it is a $q$-Sylow subgroup of $K$, all of which are conjugate, and $Q$ is preserved under conjugation by any element of $K$ since $H \\leq Z_K$.\n\nNow $Q$ char $K \\mathrel{\\unlhd} G$, thus $Q \\mathrel{\\unlhd} G$. But as $H$ is the unique minimal normal subgroup of $G$, this implies $H \\leq Q$, which contradicts $H\\cap Q = 1$.\n\nThus we must in fact have $Z_K = 1$ and $N^*$ has the required order.\n\nTo show that all groups of order $m$ are conjugate in this case, let $A$ be any other subgroup of order $m$.\n\nAs $K \\mathrel{\\unlhd} G$ we have that $|AK|$ is divisible by $m$ and by $|K| = p^kq^l = nq^l$. Thus $|AK| = mn = |G|$ and so $AK = G$. Then\n$$G/K = AK/K \\cong A/(A\\cap K),$$\nwhich implies $|A\\cap K| = q^l$, i.e. that $A\\cap K$ is a Sylow $q$-subgroup. It is therefore conjugate to $Q$.\n\nIt is easy to show that conjugate subgroups have conjugate normalizers, and so $N^*$ is conjugate to $N_G(A\\cap K)$. Thus $|N_G(A\\cap K)| = m$.\n\nAs $A\\cap K \\mathrel{\\unlhd} A$ we have that $A \\leq N_G(A\\cap K)$ and so $A = N_G(A\\cap K)$. But this implies that $A$ is conjugate to $N^*$. $\\square$\n\\end{proof}\n\n\\subsection{Central Series and Nilpotent Groups}\n\n\\subsubsection{Lower Central Series}\n\n\\begin{definition}\nLet $G$ be a group. The \\emph{lower central series} of $G$ is the series\n$$G = G_1 \\geq G_2 \\geq \\cdots$$\nwhere $G_{i+1} = [G_i, G]$.\n\\end{definition}\n\nClearly $G_{i+1} \\leq G_i$ for all $i$.\n\n\\begin{theorem}\nLet $G$ be a group with lower central series\n$$G = G_1 \\geq G_2 \\geq \\cdots$$\nthen $G_i$ char $G$ for all $i$.\n\\end{theorem}\n\n\\begin{proof}\nNote that $G_1 = G$ and $G_2 = [G, G]$, both of which are characteristic.\n\nThe proof is completed by induction. If $[x, y] \\in [G_i, G]$ and $\\phi \\in$ Aut$(G)$ then $\\phi([x, y]) = [\\phi(x), \\phi(y)]$. But $\\phi(y) \\in G$ and $\\phi(x) \\in G_i$ by the inductive hypothesis. Thus $\\phi([x, y]) \\in [G_i, G]$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $K \\mathrel{\\unlhd} G$ and $K \\leq H \\leq G$. Then $[H, G]$, the group generated by commutators of the form $h^{-1}g^{-1}hg$ for $h \\in H$ and $g \\in G$, is a subgroup of $K$ iff $H/K \\leq Z(G/K)$.\n\\end{theorem}\n\n\\begin{proof}\nAs $K \\mathrel{\\unlhd} G$ we certainly have $K \\mathrel{\\unlhd} H$. We also have that $(hK)(gK) = (gK)(hK)$ iff $h^{-1}g^{-1}hg \\in K$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a group with lower central series\n$$G = G_1 \\geq G_2 \\geq \\cdots$$\nthen $G_i/G_{i+1} \\leq Z(G/G_{i+1})$.\n\\end{theorem}\n\n\\begin{proof}\nAs $[G_i, G] = G_{i+1}$ and $G_{i+1} \\mathrel{\\unlhd} G$ we can apply the previous theorem which gives that $G_i/G_{i+1} \\leq Z(G/G_{i+1})$. $\\square$ \n\\end{proof}\n\n\\subsubsection{Upper Central Series}\n\n\\begin{definition}\nLet $G$ be a group. The \\emph{higher centers} of $G$ are the groups defined by\n$$Z^0(G) = 1, \\;\\; Z^{i+1}(G)/Z^i(G) = Z(G/Z^i(G)),$$\ni.e. $Z^{i+1}(G)$ is the inverse image of the center in the canonical projection $v_i : G \\to G/Z^i(G)$.\n\\end{definition}\n\nNote that $Z^1(G) = Z(G)$ and $Z_{i+1} \\mathrel{\\unlhd} G$ by the correspondence theorem.\n\n\\begin{definition}\nThe \\emph{upper central series} of the group $G$ is\n$$1 = Z^0(G) \\leq Z^1(G) \\leq Z^2(G) \\leq \\cdots .$$\n\\end{definition}\n\n\\begin{definition}\nLet $G$ be a group. A \\emph{central series} for $G$ is a series\n$$1 = H_n \\leq H_{n-1} \\leq \\cdots \\leq H_0 = G,$$\nwhere $H_i$ is a normal subgroup of $G$ and $H_{i-1}/H_i \\leq Z(G/H_i)$ for all $i$.\n\\end{definition}\n\nNote that a central series terminates and a group need not have a central series.\n\n\\begin{theorem}\nSuppose\n$$1 = H_n \\leq H_{n-1} \\leq \\cdots \\leq H_0 = G,$$\nis a central series for a group $G$ and groups $G_i$ form a lower central series and $Z_i$ an upper central series for $G$. Then\n$$G_{i+1} \\leq H_i \\;\\;\\mbox{and}\\;\\; H_{n-i} \\leq Z_{i+1}.$$\n\\end{theorem}\n\n\\begin{proof}\nWe prove the first inclusion by induction.\n\nNote that $G_1 = H_0 = G$. Now suppose that $G_i \\leq H_{i-1}$ for some $i$. If $x \\in G_i$ and $y \\in G$ then\n$$H_ix \\in G_{i-1}/G_i \\leq Z(G/H_i)$$\nby the assumption and the fact that the $H_i$ form a central series. Thus $H_ix$ and $H_iy$ commute.\n\nThus\n$$H_i[x, y] = (H_ix)^{-1}(H_iy)^{-1}(H_ix)(H_iy) = H_i,$$\nand so $[x, y] \\in H_i$.\n\nThus $$G_{i+1} = [G_i, G] \\leq H_i$$ completing the induction.\n\nWe will also prove the second inclusion by induction.\n\nFirst note that $Z_0 = H_n = 1$. Now suppose that $H_{n-i} \\leq Z_i$ for some $i$. As the $H_i$ form a central series we have\n$$H_{n-i-1}/H_{n-i} \\leq Z(G/H_{n-i}).$$\n\nThus if $x \\in H_{n-i-1}$ and $y \\in G$ we have that $H_{n-i}x$ and $H_{n-i}y$ commute. Thus $[x, y] \\in H_{n-i} \\leq Z_i$. Thus $Z_ix$ commutes with $Z_iy$. Thus $Z_ix \\in Z(G/Z_i) = Z_{i+1}/Z_i$.\n\nBut $x$ was an arbitrary element of $H_{n-i-1}$, thus $H_{n-i-1} \\leq Z_{i+1}$ which completes the induction. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nLet $G$ be a group and $G_i$, $Z_i$ form a lower and upper central series as per the theorem. Then the following are equivalent\n\\begin{enumerate}\n\\item $G_{c+1} = 1$ for some $c$\n\\item $Z_c = G$ for some $c$\n\\item $G$ has a central series.\n\\end{enumerate}\nMoreover if the conditions hold for a group $G$, the value of $c$ is the same in the first two conditions.\n\\end{corollary}\n\n\\begin{proof}\nIf $G$ has a central series of length $n$ of groups $H_i$ as per the theorem then $G_{n+1} \\leq H_n = 1$ and $G = H_0 \\leq Z_n$. Thus $(3)$ implies $(1)$ and $(2)$.\n\nIf $Z_c = G$ then\n$$1 = Z_0 \\leq Z_1 \\leq \\cdots \\leq Z_{c-1} \\leq Z_c = G$$\nis a central series for $G$. Thus $(2)$ implies $(3)$.\n\nIf $G_{c+1} = 1$ then\n$$1 = G_{c+1} \\leq G_c \\leq \\cdots \\leq G_1 = G$$\nis a central series for $G$ by a similar argument to the cases in the theorem. \n\nIt is clear that if the lower central series has $G_{c+1} = 1$ then as it is a central series, the upper central series must have $Z_c = G$. Similarly the other way around, thus the final statement of the theorem holds. $\\square$\n\\end{proof}\n\n\\subsubsection{Nilpotent Groups}\n\n\\begin{definition}\nAny group $G$ which satisfies the equivalent conditions of the corollary in the previous section is said to be a \\emph{nilpotent} group. The number $c$ is called the \\emph{nilpotence} class of $G$.\n\\end{definition}\n\n\\begin{theorem}\nA group $G$ is abelian iff its nilpotence class is $1$.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{Nonabelian Groups of Order $p^3$}\n\n\\begin{theorem}\nLet $G$ be a nonabelian group of order $p^3$ for a prime $p$. Then $|Z(G)| = p$, $G/Z(G) \\cong C_p\\times C_p$ and $Z(G) = [G, G]$.\n\\end{theorem}\n\n\\begin{proof}\nWe have previously shown that $|Z(G)| = p$. Now either $G/Z(G)$ is cyclic of order $p^2$ or isomorphic to $C_p\\times C_p$. As $G$ is abelian if $G/Z(G)$ is cyclic we must have the latter.\n\nAs $G/Z(G)$ is abelian, we must have $[G, G] \\leq Z(G)$. As $G$ is nonabelian, $[G, G]$ is nontrivial and so the result follows. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nA nonabelian group of order $p^3$ for a prime $p$ is nilpotent of class $2$.\n\\end{corollary}\n\n\\begin{proof}\nConsider the sequence $1 \\leq Z(G) \\leq G$. This is the upper central series for $G$ since $Z(G/Z(G)) = G/Z(G)$ by the theorem. Thus the nilpotence class is $2$. $\\square$\n\\end{proof}\n\n\\subsubsection{Every Finite $p$-group is Nilpotent}\n\n\\begin{theorem}\nEvery finite $p$-group is nilpotent.\n\\end{theorem}\n\n\\begin{proof}\nThe centre of a finite $p$-group is nontrivial. But for each higher centre $Z_i$ we have that $G/Z_i$ is a $p$-group and hence its centre is nontrivial. Thus the upper central series for $G$ terminates at $G$. $\\square$\n\\end{proof}\n\n\\subsubsection{Every Nilpotent Group is Solvable}\n\n\\begin{theorem}\nEvery nilpotent group $G$ is solvable.\n\\end{theorem}\n\n\\begin{proof}\nWe previously showed that if $G_i$ are the derived subgroups and $Z_i$ the higher centres of $G$ then $G_i \\leq Z_{c-i}$ where $c$ is the nilpotence class. Thus each of the $G_i$ is abelian and $G$ is solvable. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is nontrivial and nilpotent then $Z(G)$ is nontrivial.\n\\end{theorem}\n\n\\begin{proof}\nIf $Z(G)$ is trivial then there is no upper central series for $G$. $\\square$\n\\end{proof}\n\n\\begin{corollary}\n$S_3$ is solvable but not nilpotent.\n\\end{corollary}\n\n\\begin{proof}\n$S_3$ has trivial centre, thus by the theorem is not nilpotent. However, it is easy to see that it is solvable since $A_3$ is a normal subgroup of index $2$ and order $3$. $\\square$\n\\end{proof}\n\n\\subsubsection{Subgroups, Quotients and Products of Nilpotent Groups are Nilpotent}\n\n\\begin{theorem}\nEvery subgroup $H$ of a nilpotent group $G$ of class $c$ is nilpotent with class at most $c$.\n\\end{theorem}\n\n\\begin{proof}\nBy induction $Z_i(H) \\leq Z_i(G)$ for all $i$. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $G$ is nilpotent of class $c$ and $H \\mathrel{\\unlhd} G$ then $G/H$ is nilpotent of class at most $c$.\n\\end{theorem}\n\n\\begin{proof}\nSuppose that $\\phi: G \\to M$ is a group homomorphism and $H, K \\leq G$. Clearly $\\phi([h, k]) = [\\phi(h), \\phi(k)]$ for all $h \\in H$, $k \\in K$. Thus $\\phi([H, K]) = [\\phi(H), \\phi(K)]$.\n\nSuppose now that $\\phi$ is surjective. As $G_{i+1} = [G, G_i]$ and $M_{i+1} = [M, M_i]$ we have inductively that $M_i \\leq \\phi(G_i)$. Thus if $G$ is nilpotent of class $c$, $M$ has lower central series bounded in length by that of $G$ and the result follows. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $H$ and $K$ are nilpotent, so is $H\\times K$.\n\\end{theorem}\n\n\\begin{proof}\nIt is clear that $[H\\times K, H\\times K] \\leq [H, H]\\times [K, K]$. By induction $(H\\times K)_i \\leq H_i\\times K_i$ for all $i$. Thus if $H$ has nilpotence class $c_1$ and $K$ has nilpotence class $c_2$ then $H\\times K$ is nilpotent with class at most max$\\{c_1, c_2\\}$. $\\square$\n\\end{proof}\n\n\\subsubsection{Nilpotent Groups are the Product of their Sylow Subgroups}\n\n\\begin{theorem} \\label{normalizercond}\nIf $G$ is a nilpotent group and $H < G$ then $H < N_G(H)$.\n\\end{theorem}\n\n\\begin{proof}\nThere must exist an $i$ such that $G_{i+1} \\leq H$ but $G_i \\not\\leq H$.\n\nBut $[G_i, H] \\leq [G_i, G] = G_{i+1} \\leq H$. Thus $G_i \\leq N_G(H)$. \n\nRecall that $H \\leq N_G(H)$ for all $H \\leq G$. But if $H = N_G(H)$ then $G_i \\leq H$, a contradiction. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $P$ be a $p$-Sylow subgroup of a group $G$ and suppose that $N_G(P) \\leq H \\leq G$. Then $N_G(H) = H$.\n\\end{theorem}\n\n\\begin{proof}\nClearly $H \\leq N_G(H)$ so it suffices to show the reverse inclusion.\n\nLet $a \\in N_G(H)$. As $P \\leq N_G(P) \\leq H$ we have\n$$aPa^{-1} \\subseteq aHa^{-1} = H,$$\nsince $a \\in N_G(H)$.\n\nWe therefore have that both $P$ and $aPa^{-1}$ are Sylow subgroups of $H$.\n\nAs all Sylow subgroups are conjugate, there exists $b \\in H$ such that\n$$(b^{-1}a)P(b^{-1}a)^{-1} = P.$$\n\nIn other words, $b^{1}a \\in N_G(P) \\leq H$.\n\nAs $b \\in H$ we must have $a \\in H$ and so $H \\leq N_G(H)$ as required. $\\square$ \n\\end{proof}\n\n\\begin{theorem}\nA group $G$ is nilpotent iff it is the direct product of its Sylow subgroups.\n\\end{theorem}\n\n\\begin{proof}\nAs Sylow subgroups are $p$-groups, they are nilpotent. Thus their direct product is nilpotent.\n\nFor the converse, suppose $G$ is nilpotent and that $H$ is a $p$-Sylow subgroup of $G$.\n\nBy the previous theorem $N_G(N_G(P)) = N_G(P)$.\n\nNow suppose that $N_G(P)$ is a proper subgroup of $G$. Then by Theorem \\ref{normalizercond} we have that $N_G(P)$ is a proper subgroup of $N_G(N_G(P))$. As this would be a contradiction, we must have that $G = N_G(P)$, i.e. $P \\mathrel{\\unlhd} G$.\n\nAs Sylow subgroups intersect trivially and clearly generate $G$, the result follows. $\\square$\n\\end{proof}\n\n\\subsubsection{Maximal Subgroups of Nilpotent Groups are of Prime Index}\n\n\\begin{theorem}\nEvery maximal subgroup $H$ of a nilpotent group $G$ is normal and of prime index in $G$.\n\\end{theorem}\n\n\\begin{proof}\nLet $H$ be maximal in $G$. Then as $H < N_G(H)$ we must have $N_G(H) = G$, i.e. $H \\mathrel{\\unlhd} G$. If the quotient $G/H$ were not of prime order it would have a nontrivial proper subgroup which by the correspondence theorem would contradict the maximality of $H$ in $G$. $\\square$\n\\end{proof}\n\n\\section{Fundamental Theorem of Finitely Generated Abelian Groups}\n\n\\subsection{Finite Abelian $p$-groups}\n\n\\subsubsection{Finite Abelian $p$-groups are a Product of Cyclic Groups}\n\n\\begin{theorem}\nLet $G$ be an abelian group of order $p^n$ for a prime $p$ and let $a \\in G$ have maximal order. Then $G$ can be written $\\langle a \\rangle \\times K$ for some $K \\leq G$.\n\\end{theorem}\n\n\\begin{proof}\nWe proceed by induction on $n$. The result is obvious for $n = 1$.\n\nSuppose $a$ has order $p^m$. For any $x \\in G$ we have $|x| = p^i$ for some $i \\leq m$. Thus for all $x \\in G$ we have $x^{p^m} = 1$.\n\nAssume $G \\neq \\langle a \\rangle$, otherwise the result is trivial.\n\nLet $b \\in G$, $b \\notin \\langle a \\rangle$ be chosen such that $|c| < |b|$ implies $c \\in \\langle a \\rangle$. We will show that $\\langle a \\rangle \\cap \\langle b \\rangle = \\{1\\}$.\n\nAs $|b^p| < |b|$ we have that $b^p \\in \\langle a \\rangle$, i.e. $b^p = a^i$ for some $i$.\n\nClearly $|a^i| \\leq p^{m-1}$. Thus $a^i$ is not a generator of $\\langle a \\rangle$ and so gcd$(p^m, i) \\neq 1$. Thus $i = pj$ for some $j$, i.e. $b^p = a^{pj}$.\n\nConsider $c = a^{-j}b$. Clearly $c \\notin \\langle a \\rangle$ otherwise we would have $b \\in \\langle a \\rangle$. But $c^p = 1$ and so $|c| = p$. Therefore by definition $|b| = p$.\n\nAs $b \\notin \\langle a \\rangle$ we must have $\\langle a \\rangle \\cap \\langle b \\rangle = 1$ as claimed.\n\nReturning to the induction, consider the quotient group $\\bar{G} = G/\\langle b \\rangle$.\n\nLet $\\bar{x}$ denote the coset $x\\langle b \\rangle$.\n\nSuppose that $|\\bar{a}| < |a| = p^m$. This means that $\\bar{a}^{p^{m-1}} = \\bar{1}$, i.e. $a^{p^{m-1}}\\langle b \\rangle = \\langle b \\rangle$. But then $a^{p^{m-1}} \\in \\langle a \\rangle \\cap \\langle b \\rangle = \\{1\\}$, which contradicts $|a| = p^m$.\n\nThus we must have that $|\\bar{a}| = |a| = p^m$ and $bar{a}$ is an element of maximal order in $\\bar{G}$.\n\nAs $|\\bar{G}| = p^{n-1}$ we can apply the induction hypothesis, i.e. $\\bar{G}$ can be written $\\langle \\bar{a} \\rangle\\times \\bar{K}$ for some $\\bar{K} \\leq \\bar{G}$.\n\nLet $K = \\{x \\in G \\;|\\; \\bar{x} \\in \\bar{K}\\}$. Then $K \\leq G$ and $|K| = p|\\bar{K}|$.\n\nLet $x \\in \\langle a \\rangle \\cap K$. Then $\\bar{x} \\in \\langle \\bar{a} \\rangle \\cap \\bar{K} = \\bar{1} = \\langle \\bar{b} \\rangle$. Thus $x \\in \\langle a \\rangle \\cap \\langle b \\rangle = \\{1\\}$.\n\nIt is easy to check that $|\\langle a \\rangle K| = |G|$ and thus $G = \\langle a \\rangle K$.\n\nAs $G$ is abelian, the result follows. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nLet $G$ be a finite abelian $p$-group. Then $G$ is a direct product of cyclic groups of prime power order.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsection{Finitely Generated Abelian Groups}\n\n\\subsubsection{The Fundamental Theorem of Finite Abelian Groups}\n\n\\begin{theorem}\nEvery finite abelian group is a direct product of cyclic groups of prime power order. The product is unique up to reordering.\n\\end{theorem}\n\n\\begin{proof}\nA finite abelian group is a direct product of its Sylow subgroups, and this product is clearly unique.\n\nBy the corollary of the previous section, each of the Sylow subgroups is a product of cyclic groups of prime power order. It remains to prove that each such decomposition of a Sylow $p$-subgroup is unique up to reordering.\n\nTo this end, let $G$ be a group with order $p^k$. We will prove the result by induction on $k$, the base case $k = 1$ being trivial.\n\nSuppose that\n$$G = H_1\\times H_2\\times \\cdots \\times H_m = K_1\\times K_2 \\times \\cdots \\times K_n$$\nis a decomposition of $G$ in two different ways into nontrivial cyclic groups of $p$-power order, with\n$$|H_1| \\geq |H_2| \\geq \\cdots \\geq |H_m|$$\n$$|K_1| \\geq |K_2| \\geq \\cdots \\geq |K_m|$$.\n\nThe set $G^p = \\{x^p \\;|\\; x \\in G\\}$ is a proper subgroup of $G$.\n\nWe have that\n$$G^p = H_1^p\\times \\cdots \\times H_{m'}^p = K_1^p \\times \\cdots \\times K_{n'}^p$$\nwhere $m'$ is the largest integer $i$ such that $|H_i| > p$ and $n'$ is the largest integer $j$ such that $|K_i| > p$.\n\nBy Cauchy's theorem $|G^p| < |G|$. Thus the induction hypothesis applies.\n\nWe have that $|H_i| = p|H_i^p|$ and $|K_i| = p|K_i^p|$. Thus $|H_i| = |K_i|$ for $i \\leq m' = n'$.\n\nAs all the remaining $H_i$ and $K_j$ have order $p$, a simple comparison of orders shows that $m = n$ and we are done. $\\square$\n\\end{proof}\n\n\\subsection{Fundamental Theorem of Finitely Generated Abelian Groups}\n\n\\subsubsection{Finitely Generated Abelian Groups}\n\n\\begin{definition}\nAn abelian group $G$ is finitely generated if there exists a set of \\emph{generators} $g_1, g_2, \\ldots, g_n \\in G$ such that every element of $G$ is a product of powers of the $g_i$ and their inverses.\n\\end{definition}\n\n\\begin{theorem}\nEvery finite group is finitely generated.\n\\end{theorem}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\subsubsection{The Fundamental Theorem}\n\n\\begin{theorem}\nEvery subgroup of $\\mathbb{Z}^n$ is finitely generated (by at most $n$ generators).\n\\end{theorem}\n\n\\begin{proof}\nWe proceed by induction on $n$. For $n = 1$ we know that every subgroup is of the form $k\\mathbb{Z}$ for some $k \\geq 0$. The proof is classical and makes use of the fact that if $a$ and $b$ are in the subgroup then $d = $gcd$(a, b)$ is in the subgroup. As $a$ and $b$ are both in the subgroup generated by $d$, we can replace the generators $a$ and $b$ by the single generator $d$.\n\nIn general, let $H$ be a subgroup of $\\mathbb{Z}^n$ and let $F$ be the set of all $f$ such that $(f, a_2, a_3, \\ldots, a_n) \\in H$. Clearly $F$ is a subgroup of $\\mathbb{Z}$. Let $F$ be $k\\mathbb{Z}$ for some $k \\geq 0$. Now choose any element $(k, a_2, a_3, \\ldots, a_n)$ of $H$.\n\nFor any $(h_1, h_2, \\ldots, h_n) \\in K$ we have\n$$(h_1, h_2, \\ldots, h_n) = s(k, a_2, a_3, \\ldots, a_n) + (0, h_2 - sa_2, h_3 - sa_3, \\ldots, h_n - sa_n).$$\n\nIt is easy to check that the set of all elements $(h_2 - sa_2, h_3 - sa_3, \\ldots, h_n - sa_n)$ is a subgroup of $\\mathbb{Z}^{n-1}$, which is finitely generated with at most $n - 1$ generators, by the inductive hypothesis.\n\nIt is then easy to see that $H$ itself must be finitely generated with at most $n$ generators. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nIf $H$ is a subgroup of $\\mathbb{Z}^n$ then\n$$H \\cong d_1\\mathbb{Z} \\oplus d_2\\mathbb{Z} \\oplus \\cdots \\oplus d_r\\mathbb{Z}$$ \nand\n$$\\mathbb{Z}^n/H \\cong \\mathbb{Z}/d_1\\mathbb{Z} \\oplus \\mathbb{Z}/d_2\\mathbb{Z} \\oplus \\cdots \\oplus \\mathbb{Z}/d_r\\mathbb{Z}\\oplus \\mathbb{Z} \\oplus \\mathbb{Z} \\oplus \\cdots \\oplus \\mathbb{Z},$$\nwhere $1 < d_1 \\;|\\; d_2 \\;|\\; \\ldots \\;|\\; d_r$.\n\\end{theorem}\n\n\\begin{proof}\nThe subgroup $H$ is finitely generated by the previous theorem. Let $h_1, h_2, \\ldots, h_m$ be generators of $H$. Write out the matrix whose rows are given by these generators\n$$A = \\left(\\begin{array}{ccc}h_{1,1} & \\cdots & h_{1,n}\\\\ \\vdots & \\vdots & \\vdots\\\\ h_{m,1} & \\cdots & h_{m, n}\\end{array}\\right).$$\n\nWe reduce $A$ to Smith Normal Form by row and column operations (swap rows or columns, multiply rows or columns by $-1$ or add two rows or columns).\n\nRow operations change the generators of $H$ and column operations change the generators of $\\mathbb{Z}^n$.\n\nThe result is a matrix with zeroes everywhere except on the diagonal, where we have values $d_1, d_2, \\ldots d_r$, followed by zeroes.\n\nIt's easy to see that we can arrange things so that $d_i \\;|\\; d_{i+1}$ via row and column operations and using the extended GCD algorithm.\n\nThe result is that we have found bases for both the groups such that one has the required form in terms of the other. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $G$ be a finitely generated abelian group. Then\n$$G \\cong C_{d_1}\\times C_{d_2}\\times \\cdots \\times C_{d_k}\\times \\mathbb{Z}^r$$\nwhere $1 < d_1 \\;|\\; d_2 \\;|\\; \\ldots \\;|\\; d_k$. The $d_i$ are uniquely determined, as is $r$.\n\\end{theorem}\n\n\\begin{proof}\nLet $g_1, g_2, \\ldots, g_n$ be generators of $G$. Consider the homomorphism $\\mathbb{Z}^n \\to G$ sending $(a_1, a_2, \\ldots, a_n)$ to $a_1g_1 + a_2g_2 + \\cdots + a_ng_n$. Let $K$ be the kernel of the homomorphism. By the first isomorphism theorem $G \\cong \\mathbb{Z}/K$.\n\nNow the result follows by applying the previous theorem. Uniqueness follows as per the theorem for finite abelian groups. $\\square$\n\\end{proof}\n\n\\begin{definition}\nThe value $r$ in the previous theorem is called the \\emph{rank} of the finitely generated group $G$.\n\\end{definition}\n\n\\section{Group Transfer}\n\n\\subsection{The Transfer Map}\n\n\\subsubsection{Complements}\n\n\\begin{definition}\nLet $G$ be a group and $K$ a (not necessarily normal) subgroup of $G$. We say that $Q \\leq G$ is a \\emph{complement} of $K$ in $G$ if $K\\cap Q = 1$ and $KQ = G$.\n\\end{definition}\n\nEven a normal subgroup $K$ of a group $G$ need not have a complement. If it does, it need not be unique. However, we have the following result.\n\n\\begin{theorem}\nLet $K$ be a normal subgroup of a group $G$. Any complements of $K$ in $G$, if they exist, are isomorphic.\n\\end{theorem}\n\n\\begin{proof}\nWe have\n$$G/K = KQ/K \\cong Q/(K\\cap Q) = Q/1 \\cong Q.$$ $\\square$\n\\end{proof}\n\nRecall that $G$ is a direct product of normal subgroups $K$ and $Q$ if $Q$ is a complement of $K$ in $G$.\n\nWe also have that $G = K\\rtimes Q$ if $K \\mathrel{\\unlhd} G$ and $K$ has a complement isomorphic to $Q$.\n\n\\begin{definition}\nLet $K$ be a subgroup of a group $G$, then a (right) transversal of $K$ in $G$ is a subset $T$ of $G$ consisting of one element from each right coset of $K$ in $G$. A left transversal is defined similarly.\n\\end{definition}\n\nIf $G$ is a semidirect product of a normal subgroup $K$ with complement $Q$ then $Q$ is a transversal of $K$ in $G$.\n\n\\subsubsection{The Transfer Map}\n\n\\begin{theorem}\nLet $Q$ be a subgroup of finite index $n$ in a group $G$. Suppose that $\\{l_1, l_2, \\ldots, l_n\\}$ and $\\{h_1, h_2, \\ldots, h_n\\}$ are two left traversals of $Q$ in $G$. For fixed $g \\in G$ and each $i$ there is a unique $\\sigma(i)$ and a unique $x_i \\in Q$ with\n$$gh_i = l_{\\sigma(i)}x_i.$$\nMoreover, $\\sigma$ is a permutation of $\\{1, 2, \\ldots, n\\}$.\n\\end{theorem}\n\n\\begin{proof}\nThe left cosets of $Q$ partition $G$, and so there is a unique left coset $l_jQ$ containing $gh_i$. The first part of the theorem follows if we set $\\sigma(i) = j$.\n\nWe will now show that $\\sigma$ is injective (on a finite set) and therefore a permutation. Assume that $\\sigma(i) = \\sigma(k) = j$. Then $gh_i = l_jx_i$ and $gh_k = l_jx_k$. Then $gh_ix_i^{-1} gh_kx_k^{-1}$ and so $h_i^{-1}h_k = x_i^{-1}x_k$ and $h_iQ = h_kQ$, i.e. $i = k$. $\\square$\n\\end{proof}\n\n\\begin{definition}\nLet $Q$ be a subgroup of finite index $n$ in a group $G$. The \\emph{transfer} is the map $V : G \\to H/H'$ defined by\n$$V(g) = \\prod_{i=1}^n x_iQ',$$\nwhere $\\{l_1, l_2, \\ldots, l_n\\}$ is some transversal of $Q$ in $G$ and $gl_i = l_jx_i$.\n\\end{definition}\n\n\\begin{theorem}\nLet $Q$ be a subgroup of finite index in a group $G$, then the transfer is a homomorphism. Moreover it does not depend on the choice of transversal in the definition.\n\\end{theorem}\n\n\\begin{proof}\nLet $\\{l_1, l_2, \\ldots, l_n\\}$ and $\\{h_1, h_2, \\ldots, h_n\\}$ be two transversals of $Q$ in $G$. By the previous theorem we have $gl_i = l_{\\sigma(i)}x_i$, $gh_i = h_{\\tau(i)}y_i$ and $h_i = l_{\\alpha(i)}z_i$ for some permutations $\\sigma$, $\\tau$ and $\\alpha$ and some $x_i, y_i, z_i \\in Q$.\n\nCombining these, we have\n$$gh_i = gk_{\\alpha(i)}z_i = l_{\\sigma(\\alpha(i))}x_{\\alpha(i)}z_i.$$\n\nDefine $j$ by $\\alpha(j) = \\sigma(\\alpha(i))$ we have $h_j = l_{\\sigma(\\alpha(i))}z_j$, from which we get\n$$gh_i = h_jz_j^{-1}x_{\\alpha(i)}z_i.$$\n\nBy uniqueness from the previous theorem, and the definition of $j$ we have $j = \\tau(i)$ and\n$y_i = z_j^{-1}x_{\\alpha(i)}z_i = z_{\\alpha^{-1}(\\sigma(\\alpha(i)))}^{-1}x_{\\alpha(i)}z_i.$\n\nWe can ignore order of factors in the abelian group $Q/Q'$. Therefore\n$$\\prod y_iQ' = \\prod z_{\\alpha^{-1}(\\sigma(\\alpha(i)))}^{-1}x_{\\alpha(i)}z_iQ' = \\prod x_{\\alpha(i)}Q',$$\nsince $\\alpha^{-1}\\sigma\\alpha$ is a permutation and so the inverse of each $z_i$ occurs and cancels it.\n\nBecause $\\alpha$ is a permutation we have $\\prod x_{\\alpha(i)}Q' = \\prod x_iQ'$. But this now precisely establishes the independence of the choice of transversal.\n\nNow we will show $V$ is a homomorphism. Let $g, g' \\in G$ and $\\{l_1, l_2, \\ldots, l_n\\}$ be a left transversal of $Q$ in $G$. Then $gl_i = l_{\\sigma(i)}x_i$ and $g'l_i = l_{\\tau(i)}y_i$ for permutations $\\sigma$ and $\\tau$ and some $x_i, y_i \\in Q$. Then\n$$gg'l_i = gl_{\\tau(i)}y_i = l_{\\sigma(\\tau(i))}x_{\\tau(i)}y_i.$$\n\nThus\n\\begin{align*}\nV(gg') = \\prod x_{\\tau(i)}y_iQ' &= \\left(\\prod x_{\\tau(i)}Q'\\right)\\left(\\prod y_{i}Q'\\right)\\\\\n &= \\left(\\prod x_{i}Q'\\right)\\left(\\prod y_{i}Q'\\right) = V(g)V(g').\n\\end{align*} $\\square$\n\\end{proof}\n\n\\subsubsection{Normal Complements}\n\n\\begin{theorem}\nLet $Q$ be a subgroup of finite index in a group $G$ and let $\\{l_1, l_2, \\ldots l_n\\}$ be a left transversal of $Q$ in $G$. For each $g \\in G$ there exist $h_1, h_2, \\ldots, h_m \\in G$ and $n_1, n_2, \\ldots, n_m > 0$ depending on $g$ such that \n\\begin{itemize}\n\\item $h_i \\in \\{l_1, l_2, \\ldots, l_n\\}$\n\\item $h_i^{-1}g^{n_i}h_i \\in Q$\n\\item $\\sum n_i = n = [G:Q]$\n\\item $V(g) = \\prod (h_i^{-1}g^{n_i}h_i)Q'$.\n\\end{itemize}\nIn other words, the transfer is a product of conjugages of powers of $g$.\n\\end{theorem}\n\n\\begin{proof}\nWe have that $gl_i = l_{\\sigma(i)}x_i$ for some permutation $\\sigma$ and $x_i \\in Q$. Write $\\sigma$ as a product of disjoint cycles\n$$\\sigma = \\alpha_1\\alpha_2\\ldots \\alpha_m,$$\nwith 1-cycles included (so that the sum of the lengths of the cycles is $n$).\n\nSuppose $\\alpha_i = (j_1, j_2, \\ldots, j_r)$. Then\n$$gl_{j_1} = l_{\\sigma(j_1)}x_{j_1} = l_{j_2}x_{j_1}, \\ldots, gl_{j_r} = l_{j_1}x_{j_r}.$$\n\nNow the following product is contained in $Q$\n$$x_{j_r}\\cdots x_{j_1} = (l_{j_1}^{-1}gl_{j_r})\\cdots (l_{j_2}^{-1}gl_{j_1}) = l_{j_1}^{-1}g^rl_{j_1}.$$\n\nAll of the conclusions follow trivially. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $Q$ be an abelian subgroup of finite index in a group $G$. If $Q \\leq Z(G)$ then $V(g) = g^n$ for all $g\\in G$.\n\\end{theorem}\n\n\\begin{proof}\nAs $Q$ is an abelian subgroup, the map $V$ can be regarded as a homomorphism to $Q$. Furthermore, $Q \\leq Z(G)$ implies that $Q \\mathrel{\\unlhd} G$.\n\nIf $g \\in G$ and $h^{-1}g^rh \\in Q$ then by normality of $Q$ we have\n$$g^r = h(h^{-1}g^rh)h^{-1} \\in Q.$$\n\nBut as $Q \\leq Z(G)$ we now have that $h^{-1}g^rh = g^r$. The result now follows from the previous theorem. $\\square$\n\\end{proof}\n\n\\begin{corollary}\nIf $Q$ is a subgroup of finite index in a group $G$, with $Q \\leq Z(G)$, the map $g \\to g^n$ is a homomorphism.\n\\end{corollary}\n\n\\begin{proof}\nClear. $\\square$\n\\end{proof}\n\n\\begin{theorem}\nLet $Q$ be a $p$-Sylow subgroup of a finite group $G$. If $g, h \\in C_G(Q)$ are conjugate in $G$, they are conjugate in $N_G(Q)$.\n\\end{theorem}\n\n\\begin{proof}\nLet $h = \\alpha^{-1}g\\alpha$ for some $\\alpha \\in G$. Thus $h \\in \\alpha^{-1}C_G(Q)\\alpha = C_G(\\alpha^{-1}Q\\alpha)$.\n\nAs $Q$ and $\\alpha^{-1}Q\\alpha$ are contained in $C_G(h)$, these are both Sylow subgroups of $C_G(h)$.\n\nBy the Sylow theorems, $Q = c^{-1}\\alpha^{-1}Q\\alpha c$ for some $c \\in C_G(h)$. Thus $\\alpha c \\in N_G(Q)$ and $c^{-1}\\alpha^{-1}g\\alpha c = c^{-1}hc = h$, showing that $g$ and $h$ are conjugate in $N_G(Q)$. $\\square$\n\\end{proof}\n\n\\begin{theorem} (Burnside Normal Complement)\nLet $G$ be a finite group and let $Q$ be an abelian Sylow subgroup with $Q \\leq Z(N_G(Q))$. Then $Q$ has a complement $K$ which is normal in $G$.\n\\end{theorem}\n\n\\begin{proof}\nAs $Q$ is abelian, we may identify $Q/Q'$ with $Q$ and regard $V$ as a homomorphism $V: G \\to Q$.\n\nWe have that $V(g) = \\prod h_i^{-1}g^{n_i}h_i$ for some powers $n_i$ and $h_i$ in some transversal of $Q$.\n\nIf $g \\in Q$ then $g^{n_i}$ and $h_i^{-1}g^{n_i}h_i$ are also in $Q$, by the theorems above.\n\nAs $Q$ is abelian, $Q \\leq C_G(Q)$ and so the elements $g^{n_i}$ and $h_i^{-1}g^{n_i}h_i$ are conjugate in $C_G(Q)$. Thus by the previous theorem there is a $c \\in N_G(Q)$ such that $h_i^{-1}g^{n_i}h_i = c^{-1}g^{n_i}c$.\n\nBut if $Q \\leq Z(N_G(C))$ then $c^{-1}g^{n_i}c = g^{n_i}$ and so if $n = [G:Q]$, we have that $V(g) = g^n$ for all $g \\in Q$.\n\nAs $Q$ is a Sylow subgroup, gcd$(n, q) = 1$, where $q = |Q|$. Thus by the extended Euclidean algorithm, there exist $\\alpha, \\beta \\in \\Z$ with $\\alpha n + \\beta q = 1$.\n\nThus for $g \\in Q$ we have $g = g^{\\alpha n}g^{\\beta q} = (g^\\alpha)^n$. Thus $V : G \\to Q$ is surjective.\n\nBy the first isomorphism theorem $G/K \\cong Q$ where $K =$ ker$(V)$. Thus $G = KQ$, $K\\cap Q = 1$ because $|K| = n$ and thus gcd$(|K|, |Q|) = 1$.\n\nThus $K$ is a normal complement of $Q$ as required. $\\square$\n\\end{proof}\n\n\\end{document}\n", "meta": {"hexsha": "6d1bd122a06c43430e04a4c73fc79e116fd6d724", "size": 186365, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "GroupTheory.tex", "max_stars_repo_name": "wbhart/ShortMathNotes", "max_stars_repo_head_hexsha": "bb10ca85044cc4767dcdbd5bd41ce530edad3667", "max_stars_repo_licenses": ["BSD-2-Clause"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2019-07-23T15:01:27.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-10T06:53:30.000Z", "max_issues_repo_path": "GroupTheory.tex", "max_issues_repo_name": "wbhart/ShortMathNotes", "max_issues_repo_head_hexsha": "bb10ca85044cc4767dcdbd5bd41ce530edad3667", "max_issues_repo_licenses": ["BSD-2-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "GroupTheory.tex", "max_forks_repo_name": "wbhart/ShortMathNotes", "max_forks_repo_head_hexsha": "bb10ca85044cc4767dcdbd5bd41ce530edad3667", "max_forks_repo_licenses": ["BSD-2-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.8515607456, "max_line_length": 556, "alphanum_fraction": 0.6645990395, "num_tokens": 66430, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702761768248, "lm_q2_score": 0.843895106480586, "lm_q1q2_score": 0.6117144788988533}} {"text": "\\documentclass[12pt]{article}\n\\usepackage{geometry} \n\\geometry{margin=1in}\n\\geometry{a4paper} \n\n\n\\usepackage{textcomp}\n\\usepackage{booktabs}\n\\usepackage{array}\n\\usepackage{paralist}\n\\usepackage{verbatim} \n\\usepackage{subfigure}\n\\usepackage{graphicx,caption}\n\\usepackage{placeins}\n\\usepackage{lipsum}\n\\usepackage{xcolor}\n\\usepackage{dcolumn}\n\\usepackage{sectsty}\n\\allsectionsfont{\\sffamily\\mdseries\\upshape}\n\\usepackage{gensymb,amsmath,mathtools,amssymb}\n\\usepackage{flafter}\n%\\usepackage{parskip}\n\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n\\usepackage{tocbibind}\n\\usepackage[toc,page]{appendix}\n\\captionsetup{width=\\linewidth}\n\\usepackage{bm}\n\\usepackage{url}\n\\usepackage{empheq}\n\n\\newcommand{\\half}{\\frac{1}{2}}\n\n\n\\graphicspath{{./figs/}}\n\n\n\\title{Passive Flight Solution}\n\\author{Devansh Agrawal}\n%\\date{} \n\n\n\\begin{document}\n\n\\maketitle\n\n\n\\section{Problem statement}\n\nThe dynamics of a rocket flying along the vertical axis (without thrust) can be written as\n\n\\begin{align}\n\\dot{h} &= v\\\\\n\\dot{v} &= -\\frac{D}{m} - g = -\\beta v^2 - g\n\\end{align}\nwhere I have defined a modified ballistic coefficient\n\\begin{equation}\n\\beta = \\frac{1}{2} \\rho \\frac{C_D A}{m}\n\\end{equation}\n\nSince there is no thrusting, the mass is constant, and therefore $\\beta$ is also constant. \n\nFurther we assign the following boundary conditions, \n\n\\begin{align}\nh(0) &= h_0\\\\\nv(0) &= v_0\\\\\nv(t_f) &= 0\n\\end{align}\n\nWe wish to solve these equations, and determine the coasting distance. \n\n\n\\section{Coasting Distance}\n\nUsing Mathematica's DSolve function, we can solve this non-linear coupled ode. Mathematica returns two solutions (which have been simplified here)\\footnote{$\\log$ here refers to the natural logarithm}:\n\nSolution 1 (invalid):\n\\begin{align}\nh(t) &=h_0 + \\frac{\\log \\left(\\cos \\left(\\sqrt{\\beta g}(t_f-t)\\right)\\right)}{\\beta }-\\frac{\\log \\left(-\\sqrt{\\frac{g}{g+\\beta v_0^2}}\\right)}{\\beta }\\\\\nv(t) &=\\sqrt{\\frac{g}{\\beta }} \\tan \\left(\\sqrt{\\beta g}(t_f-t)\\right)\n\\end{align}\n\nand solution 2 (valid):\n\\begin{align}\nh(t) &=h_0 + \\frac{\\log \\left(\\sqrt{1+\\frac{\\beta v_0^2}{g}} \\cos \\left(\\sqrt{\\beta g}(t_f-t)\\right)\\right)}{\\beta } \\label{eqn:h}\\\\\nv(t) &= \\sqrt{\\frac{g}{\\beta }} \\tan \\left(\\sqrt{\\beta g}(t_f-t)\\right) \\label{eqn:v}\n\\end{align}\n\nNotice, that due to the second logarithm in solution 1, it must be invalid\\footnote{or if you think about expanding the log(negative square root()) into negative half log(), the expression turns out to be same as solution 2}, and that the velocity in both solutions are identical. Therefore, solution 2 is the accurate flight path. \n\nImmediately we can see from eqn~\\ref{eqn:h} that the remaining height to go for the rocket is \n\\begin{equation}\n\\boxed{\\Delta h = h(t_f) - h_0 = \\frac{1}{\\beta} \\log \\left(\\sqrt{1+\\frac{\\beta v_0^2}{g}} \\right)}\n\\end{equation}\n\nIn the limit that drag is insignificant $\\beta \\rightarrow 0$, and the equation correctly approaches the kinematic result, \n\n\\begin{equation}\n\\lim_{\\beta\\rightarrow0} \\Delta h = \\frac{1}{2} \\frac{v_0^2}{ g}\n\\end{equation}\n\nThe figure shows the results plotted. \n\n\n%\\FloatBarrier\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.8\\linewidth]{coast_regions}\n \\caption{This figure shows the combination of height to target and current flight speeds that will allow the rocket to coast to the target height. We see little dependence on the dry mass in the range of masses we are interested in. It is extremely important not to overshoot this line. The red lines shows the limiting case where the drag force can be neglected, and is a nice proxy for the case that drag cannot be accurately modelled.}\n \\label{fig:}\n\\end{figure}\n%FloatBarrier\n\n%\\FloatBarrier\n\\begin{figure}[htbp]\n \\centering\n \\includegraphics[width=0.8\\linewidth]{impulsiveCorrection.eps}\n \\caption{This figure shows the amount of propellant needed if an impulsive manoeuvre is to be used to get the current (h,v) point onto the coasting line (ie moving sideways on the graph). It is computed for a specific case of 20~kg dry mass, and shows that rather small amounts of propellant are needed (note, the propellant mass needed scales basically linearly with dry mass though). Less propellant should be needed for a non-impulsive manoeuvre. Therefore we could design the open loop trajectory aiming for 40~m/s at 300 m remaining, which gives us on the order 4 seconds to recompute and implement the final thrusting.}\n \\label{fig:}\n\\end{figure}\n\\FloatBarrier\n\n\n\\section{Full Flight Path}\n\nIf we assume $v$ behaves well, we can solve eqn~\\ref{eqn:v} for the final time using $t=0$, and determine the flight profile:\n\n\\begin{empheq}[box = \\fbox]{align}\nt_f &= \\frac{1}{\\sqrt{\\beta g}} \\arctan{\\left(v_0 \\sqrt{\\frac{\\beta}{g}}\\right)}\\\\\nh(t) &= h_0 + \\frac{1}{\\beta} \\log \\left[\\cos \\left(t \\sqrt{\\beta g}\\right)+\\sqrt{\\frac{\\beta v_0^2 }{ g}} \\sin \\left(t \\sqrt{\\beta g}\\right)\\right]\\\\\nv(t) &= \\sqrt{\\frac{g}{\\beta }} \\tan \\left[\\arctan\\left(v_0\\sqrt{\\frac{\\beta }{g}}\\right) - t \\sqrt{\\beta g}\\right]\n\\end{empheq}\n\n\nWhile these expressions look complicated, they are in closed form, and therefore are extremely useful!\n\nNote, these derivations ignore the change in drag coefficient with mach number, and the change in air density with altitude. \n\n\n\n\n%\\bibliographystyle{unsrt}\n%\\bibliography{biblio}\n\n\\end{document}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n", "meta": {"hexsha": "c03d37ac29f54c28e3ba35ad05867dd667170f57", "size": 5353, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "PassiveFlight/passive_flight_solution.tex", "max_stars_repo_name": "icl-rocketry/optimalAscent", "max_stars_repo_head_hexsha": "a841745e4097c55fa4996dddc8a7dc45b2059023", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-08-19T01:06:38.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-06T05:32:08.000Z", "max_issues_repo_path": "PassiveFlight/passive_flight_solution.tex", "max_issues_repo_name": "icl-rocketry/optimalAscent", "max_issues_repo_head_hexsha": "a841745e4097c55fa4996dddc8a7dc45b2059023", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "PassiveFlight/passive_flight_solution.tex", "max_forks_repo_name": "icl-rocketry/optimalAscent", "max_forks_repo_head_hexsha": "a841745e4097c55fa4996dddc8a7dc45b2059023", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-12-06T05:32:10.000Z", "max_forks_repo_forks_event_max_datetime": "2020-12-06T05:32:10.000Z", "avg_line_length": 31.4882352941, "max_line_length": 628, "alphanum_fraction": 0.7257612554, "num_tokens": 1624, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7248702642896702, "lm_q2_score": 0.8438951005915208, "lm_q1q2_score": 0.6117144645985335}} {"text": "\\documentclass[main.tex]{subfiles}\n\\begin{document}\n\n\\marginpar{Monday\\\\ 2020-10-12, \\\\ compiled \\\\ \\today}\n\nWe come back to our noisy dataset \\(x_i = \\mu + n_i\\). \nIf our noise comes from a Gaussian with a \\emph{known}, \\emph{constant} \\(\\sigma \\), then we have \n%\n\\begin{align}\n\\mathscr{L}(x_i | \\mu , \\sigma )\n= \\frac{1}{\\sigma \\sqrt{2 \\pi }} \n\\exp(- \\frac{1}{2} \\frac{(x_i - \\mu )^2}{\\sigma^2})\n\\,.\n\\end{align}\n\nWe also assume that the realizations of the noise are independent. \nSo, the combined likelihood for all the data is given by \n%\n\\begin{align}\n\\mathscr{L}(\\vec{x} | \\mu , \\sigma ) = \\prod_{i=1}^{N} \\mathscr{L}(x_i, \\mu , \\sigma )\n\\,.\n\\end{align}\n\nNow, choosing a prior as we mentioned is difficult in general. Until now we have assumed that our noninformative prior would be constant. \nFor the type of parameter we have now --- \\(\\mu \\) is a \\emph{location} parameter --- works; however we have an issue: a uniform distribution must have a certain range in which it is nonzero, elsewhere it is zero. \nThis range must be finite.\n \nZero cannot be updated: this is a problem.\nWe can, however take a prior that is so wide that it is nonzero in any place the likelihood is measurably nonzero. Practically speaking we only work up to finite precision, so this is not a problem. \n\nWe only case about proportionality: the posterior is proportional to the likelihood, so\n%\n\\begin{align}\nP \\propto \\prod_{i=1}^{N} \\exp(- \\frac{1}{2} \\frac{(x_i - \\mu )^2}{\\sigma^2})\n\\,,\n\\end{align}\n%\ntherefore the log-posterior is \n%\n\\begin{align}\nL = \\log P = - \\frac{1}{2 \\sigma^2} \\sum _{i=1}^{N} (x_i - \\mu ) + \\const\n\\,,\n\\end{align}\n%\nso if we want to find the maximum (log)-posterior: \n%\n\\begin{align}\n\\dv{L}{\\mu } = \\sum_{i=1}^{N} \\frac{x_i- \\mu }{\\sigma^2} =0 \n\\,,\n\\end{align}\n%\nmeaning that \n%\n\\begin{align}\nN \\mu = \\sum_{i=1}^{N} x_i \\implies \\hat{\\mu} = \\frac{1}{N} \\sum_{i=1}^{N} x_i\n\\,.\n\\end{align}\n\nWe can also compute the estimate of the error on the estimate: \n%\n\\begin{align}\n\\dv[2]{L}{\\mu } =- \\sum_{i=1}^{N} \\frac{1}{\\sigma^2} = - \\frac{N}{\\sigma^2}\n\\,,\n\\end{align}\n%\ntherefore \n%\n\\begin{align}\n\\sigma_\\mu = \\qty(- \\dv[2]{L}{\\mu })^{-1/2} = \\frac{\\sigma}{\\sqrt{N}}\n\\,.\n\\end{align}\n\nThe great simplification we made here was that we are dealing with only one parameter.\nIn any realistic astrophysics scenario we have at least a dozen. \n\n\\section{Multiparameter estimation}\n\n\\subsection{A two-parameter example}\n\nWe consider a photon counting experiment. \nWe have a diffraction experiment: we can measure light in \\(M\\) frequency channels, labelled by \\(k\\), and at different angles. \n\nOur model (we need one for parameter estimation) is that for each frequency channel \\(k\\) we expect a Gaussian spatial profile centered around \\(x_0 \\) --- for simplicity we assume \\(x_0 \\) is known, and that the standard deviation \\(w\\) is known as well.\nWe are interested in the amplitude \\(A\\) of this Gaussian peak.\n\nAlso, we will have spatially constant noise with amplitude \\(B\\).\nWe need to estimate \\(A\\) and \\(B\\) jointly; \\(B\\) is a \\emph{nuisance parameter} we would ideally integrate over later. \n\nWe expect that the measured data at a frequency channel will look like \n%\n\\begin{align}\nD_k = n_0 \\qty[ A\\exp(- \\frac{(x_k - x_0)^2}{2 w^2} + B)]\n\\,.\n\\end{align}\n\nThe parameter \\(n_0 \\) accounts for the fact that if we measure for longer we see more photons. \n\nPhotons, both noise and signal, are expected to obey Poissonian statistics: \n%\n\\begin{align}\n\\mathbb{P}(N_k | D_k) = \\frac{D_k^{N_k} e^{- D_k}}{N_k!}\n\\,,\n\\end{align}\n%\nwhich is the likelihood: the probability of the data, given our model. \nOur actual likelihood will be given by \\(\\mathscr{L} = \\prod_k \\mathbb{P}(N_k | D_k)\\).\n\nWhat is our prior \\(\\mathbb{P}(A,B | I)\\)? We know that these parameters cannot be negative, so we just take a uniform bivariate prior on \\(A\\geq 0\\), \\(B \\geq 0\\).\nAs before, this is just meant to say ``uniform over all representable values''. \n\nWe will discuss later that by the nature of \\(A\\) as a scale parameter it would be more reasonable to take \\(\\log A\\) to be uniform.\n\nThe posterior will then be \n%\n\\begin{align}\n\\mathbb{P}(A, B | N_k, I) \\propto\n \\prod_k \\frac{D_k^{N_k} e^{-D_k}}{N_k!}\n\\,.\n\\end{align}\n\nThis then depends on two parameters, and we can maximize it numerically. It is a difficult problem computationally, but not conceptually. \n\nHow can we model this in order to give an error? We can approximate it as a bivariate Gaussian. \n\nIn general, a multivariate Gaussian is given as \n%\n\\begin{align}\n\\mathbb{P}(\\vec{x}) = \\frac{1}{(2 \\pi )^{n/2} \\abs{\\Sigma }^{1/2}}\n\\exp(- \\frac{1}{2} (\\vec{x} - \\vec{\\mu})^{\\top} \\Sigma^{-1} (\\vec{x} - \\vec{\\mu}))\n\\,,\n\\end{align}\n%\nwhere \\(\\Sigma \\) is the \\textbf{covariance matrix} of the two variables. \nOn the diagonal we have the variance of each variable: \\(\\Sigma_{ii} = \\sigma_i^2\\), so \\((\\Sigma^{-1})_{ii} = 1 / \\sigma _i^2\\) as long as the covariance (meaning, the nondiagonal elements) is zero.\n\n\\end{document}\n", "meta": {"hexsha": "e84b4ee8dbf3adda91bfc2e541b97f657ff986fa", "size": 4974, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ap_third_semester/astrostatistics_cosmology/oct12.tex", "max_stars_repo_name": "jacopok/notes", "max_stars_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-10-10T13:10:57.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-13T14:52:50.000Z", "max_issues_repo_path": "ap_third_semester/astrostatistics_cosmology/oct12.tex", "max_issues_repo_name": "jacopok/notes", "max_issues_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "ap_third_semester/astrostatistics_cosmology/oct12.tex", "max_forks_repo_name": "jacopok/notes", "max_forks_repo_head_hexsha": "805ebe1be49bbd14c6b46b24055f9fc7d1cd2586", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2019-10-03T16:20:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-06T16:11:07.000Z", "avg_line_length": 36.3065693431, "max_line_length": 255, "alphanum_fraction": 0.6801367109, "num_tokens": 1589, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7662936484231889, "lm_q1q2_score": 0.6116454580372434}} {"text": "In this report we have introduced Markov Chain Monte Carlo algorithms, with a particular focus on Langevin Monte Carlo methods. We then tested the performance of the algorithms based on a number of metrics and a broad range of parameters. Our main contribution is extending the work of \\cite{Brosse18tULA} and producing a \\textsc{Python} package ready for use by other researchers to implement their own methods to be tested against the algorithms we have presented here. As well as implementing their own methods, it is also written in a way that allows easy extension to different potentials and distributions. As well as this, we have extended the visualisation library of Rogozhnikov \\cite{rogozhnikov2016hmc} to include LMC methods. This allows students and those newly introduced to the field to get an intuitive look at how these algorithms work, and the differences between them. \n\nWe have shown that taming is a viable method of preventing divergence of Langevin-based algorithms and gives insight into a distribution when Metropolised algorithms are unable to, particularly in ill-conditioned problems. It is interesting to note the efficiency of the \\texttt{LM} algorithm in particular, despite its relative simplicity. It does however suffer from divergence problems as \\texttt{ULA}.\n\nThere is no simple answer as to which algorithm is `best'. Depending on the application, and the prior information given on the distribution, one could make the case for any of the algorithms presented here. If the general shape of the distribution is known and computational power is not a restriction, a simple Random Walk Metropolis algorithm may well be the best option, even in the superlinear case.\n\nIf very little is known about maxima of the distribution, the taming method is the best choice. It is able to quickly find the modes of the distribution without divergence, however once at the mode it often over estimates the width of the potential well. This has provoked some investigation into `switching' methods, where the chain is initially started using \\texttt{tULA} for a fixed time to find the mode, before switching to the \\texttt{RWM} algorithm to better explore the well. This is the rationale behind the \\texttt{HPH} algorithm in the package, however it has been omitted from the report due to a lack of theoretical justification. It may well turn out to be similar to existing adaptive time stepping methods, or tempering methods.\n\nHigher order methods seem to come at too great a computational cost, especially in very high dimension, although the theory supporting them suggests that more work is to be done on the numerical implementation of such methods.\n\nIf this report has highlighted anything, it is that Metropolisation is not the final word in Langevin Monte Carlo, and other methods of approximating Langevin dynamics should be exploited and explored. Non-asymptotic bounds are of great importance in this area, as it is known that \\texttt{MALA} converges in the limit but this is of little practical use. \n\n\\subsection{Future Work}\\label{subsec:future}\nThere are many possible ways in which the work presented here could be extended, both from a research and personal perspective. First, there is great scope for the improvement of our program. It is possible to add new analytic distributions to sample from, including those with non-smooth potentials as in \\cite{durmus2018efficient}. Another obvious avenue would be to apply all the methods and analysis here to real (large) datasets, when the gradient is not known analytically. This would give the end user a much clearer impression of which algorithm to use in their given case. It would also slow down all gradient based methods as an unbiased estimator for the gradient would have to be calculated at every iteration. It may also be possible to speed up the higher order methods (\\texttt{HOLA,tHOLA,tHOLAc}) by implementing a parallelised version, breaking up the complex iteration into smaller easier to manage sections. This in general is highly non-trivial due to the inherent dependence on the previous step in a Markov chain. Another important consideration is that of burn-in time. For the majority of our tests, we started from a minimiser of the potential well however our code is certainly not limited to this case. Doing so effectively removes any burn-in time as the chains are started in approximate stationarity. When started far away from the mode of a distribution, initial tests suggest taming is a highly preferable method to Metropolised algorithms. They take many more steps towards the mode, while Metropolised algorithms waste lots of time rejecting moves. In practice, this would greatly reduce the burn-in time -- a key feature of an effective MCMC method. \n\nIt is also important that we test the accuracy of our measures. This is difficult, especially in high dimensions as no methods exist for numerically calculating the 2-Wasserstein metric that many of the theoretical bounds use. Furthermore, using kernel density estimation or histograms with few bins introduces an error that is difficult to quantify and reduce. Here we have only qualitative comparisons between metrics. This is a well known problem in numerical optimal transport and inherent in dealing with high dimensional datasets and problems.\n\nMany other methods exist in MCMC that remain to be tested against the tamed algorithms, or incorporated in to our program. These include Hamiltonian Monte Carlo (HMC), manifold MALA (mMALA), underdamped Langevin Monte Carlo and stochastic gradient Langevin dynamics (SGLD) \\cite{betancourt2017conceptual, Girolami2011,cheng2018,pitfalls}. For dealing with stiff problems, specific methods also exist that could be used as benchmarks for taming algorithms \\cite{abdulle2013weak}. In the next section SGLD will be expanded on due to its popularity in machine learning and high dimensional problems. \n\nFor the wider field, it is important that either `user-friendly' bounds on nonasymptotic error are developed in terms of a numerically implementable metric, or that methods are developed to accurately calculate the 2-Wasserstein metric. Even this will not solve the problem of approximating a high dimensional distribution using samples from a distribution; it is simply infeasible to be able to generate enough samples to get a good representation. \n\n", "meta": {"hexsha": "56b6995b7c97750cb1a09afd1e1db2afd9834d9b", "size": 6372, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "WriteUp/conclusion.tex", "max_stars_repo_name": "Tom271/LangevinMC", "max_stars_repo_head_hexsha": "ed36a17ce9b7d1e39097aeaf5b92f0fa286d5489", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 10, "max_stars_repo_stars_event_min_datetime": "2019-02-07T12:51:19.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-04T13:35:13.000Z", "max_issues_repo_path": "WriteUp/conclusion.tex", "max_issues_repo_name": "swyoon/LangevinMC", "max_issues_repo_head_hexsha": "ed36a17ce9b7d1e39097aeaf5b92f0fa286d5489", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "WriteUp/conclusion.tex", "max_forks_repo_name": "swyoon/LangevinMC", "max_forks_repo_head_hexsha": "ed36a17ce9b7d1e39097aeaf5b92f0fa286d5489", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-01-19T17:44:19.000Z", "max_forks_repo_forks_event_max_datetime": "2021-01-19T17:44:19.000Z", "avg_line_length": 289.6363636364, "max_line_length": 1685, "alphanum_fraction": 0.816698054, "num_tokens": 1294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.7981867777396211, "lm_q2_score": 0.7662936430859597, "lm_q1q2_score": 0.6116454537771374}} {"text": "\\subsection{Semimodules}\\label{subsec:semimodules}\n\nSemimodules are generalizations of monoid actions. Notation and terminology-wise, semimodules are somewhat special in that they are very much influenced by linear algebra and analysis, where vector spaces are crucial.\n\n\\begin{definition}\\label{def:endomorphism_semiring}\n Let \\( X \\) be a monoid or, more generally, an object in a \\hyperref[def:category]{category} that is \\hyperref[def:concrete_category]{concrete} over \\hyperref[def:monoid/category]{\\( \\cat{Mon} \\)}.\n\n Let \\( \\End(X) \\) be the \\hyperref[def:endomorphism_monoid]{endomorphism monoid} over \\( X \\). These are necessarily monoid endomorphisms, however they may carry additional structure like being \\hyperref[def:group/homomorphism]{group homomorphisms}, \\hyperref[def:semimodule/homomorphism]{semimodule homomorphisms}, \\hyperref[def:semilattice/homomorphism]{(semi)lattice homomorphisms} or their \\hyperref[rem:topological_first_order_structures]{continuous counterparts}.\n\n Define addition in \\( \\End(X) \\) pointwise as \\( [f + g](x) \\coloneqq f(x) + g(x) \\). Then \\( \\End(X) \\) with pointwise addition and composition is a \\hyperref[def:semiring]{semiring}, which we call the \\term{endomorphism semiring} over \\( X \\).\n\\end{definition}\n\n\\begin{definition}\\label{def:semimodule}\n Fix a \\hyperref[def:semiring]{semiring} \\( R \\), whose elements we will call \\term{scalars}, and an \\hyperref[rem:additive_magma]{additive} \\hyperref[def:magma/commutative]{commutative} \\hyperref[def:monoid]{monoid} \\( M \\), whose elements we will call \\term{vectors}. See \\fullref{rem:vector_etymology} for a discussion of the term \\enquote{vector}.\n\n We say that \\( M \\) is a \\term{semimodule} over \\( R \\) if they are compatible in any of the equivalent ways listed below. Analogously to \\hyperref[def:monoid_action]{monoid actions}, if \\( R \\) is not commutative, we distinguish between left and right semimodules. Rather than \\enquote{\\( M \\) is a semimodule over \\( R \\)}, it is often more convenient to say \\enquote{\\( M \\) is an \\( R \\)-semimodule}.\n\n \\begin{thmenum}[series=def:semimodule]\n \\thmitem{def:semimodule/action} A left semimodule is a \\hyperref[def:semiring/homomorphism]{homomorphism} from \\( R \\) to the \\hyperref[def:endomorphism_semiring]{endomorphism semiring} \\( \\End(M) \\). A right semimodule is a homomorphism from the \\hyperref[def:semiring/opposite]{dual semiring} \\( R^{-1} \\) to \\( \\End(M) \\).\n\n This definition is concise and natural, but unfortunately not very useful.\n\n \\thmitem{def:semimodule/operation} The usual way to define a left semimodule is via a binary operation \\( \\cdot: R \\times M \\to M \\) called \\term{scalar multiplication} that satisfies the following conditions:\n \\begin{thmenum}\n \\thmitem{def:semimodule/operation/scalar_multiplication_action} Scalar multiplication is a \\hyperref[def:monoid_action]{monoid action} of the multiplicative monoid \\( (R, \\cdot_R) \\) on \\( M \\). The following conditions correspond to \\eqref{eq:def:monoid_action/family/identity} and \\eqref{eq:def:monoid_action/family/compatibility}:\n \\begin{align}\n &1_R \\cdot x = x, \\label{eq:def:semimodule/operation/scalar_multiplication_action/identity} \\\\\n &(r \\cdot_R s) \\cdot x = r \\cdot (s \\cdot x). \\label{eq:def:semimodule/operation/scalar_multiplication_action/compatibility}\n \\end{align}\n\n The second condition can be regarded as a form of associativity.\n\n \\thmitem{def:semimodule/operation/scalar_addition_distributivity} Scalar addition distributes over scalar multiplication:\n \\begin{equation}\\label{eq:def:semimodule/operation/scalar_addition_distributivity}\n (r +_R s) \\cdot x = r \\cdot x + s \\cdot x.\n \\end{equation}\n\n \\thmitem{def:semimodule/operation/vector_addition_distributivity} Vector addition distributes over scalar multiplication:\n \\begin{equation}\\label{eq:def:semimodule/operation/vector_addition_distributivity}\n r \\cdot (x + y) = r \\cdot x + r \\cdot y.\n \\end{equation}\n\n \\thmitem{def:semimodule/operation/absorption} The scalar and vector zeros are compatible:\n \\begin{equation}\\label{eq:def:semimodule/operation/absorption}\n 0_R \\cdot x = 0_M = r \\cdot 0_M.\n \\end{equation}\n \\end{thmenum}\n\n In practice, we use the same symbol for both scalar and vector addition, and we denote both scalar and vector multiplication via juxtaposition.\n \\end{thmenum}\n\n Semimodules have the following metamathematical properties:\n \\begin{thmenum}[resume=def:semimodule]\n \\thmitem{def:semimodule/theory} In order to fit the heterogeneous operation \\( \\cdot \\) into the framework of \\hyperref[def:first_order_semantics/satisfiability]{first-order logic models}, we can extend the \\hyperref[def:monoid/theory]{theory of monoids} by adding, for every semiring element \\( r \\), a unary \\hyperref[def:first_order_language/func]{functional symbol} \\( m_r \\). All conditions can then be reformulated via this operation. For example, \\eqref{eq:def:semimodule/operation/scalar_multiplication_action/compatibility} corresponds to the axiom schema\n \\begin{equation*}\n m_{rs}(\\xi) = m_r(m_s(\\xi)).\n \\end{equation*}\n\n \\thmitem{def:semimodule/homomorphism} A \\hyperref[def:first_order_homomorphism]{first-order homomorphism} between two \\( R \\)-semimodules \\( M \\) and \\( N \\) is a function \\( \\varphi: M \\to N \\) that is a \\hyperref[def:monoid/homomorphism]{monoid homomorphism} and satisfies \\( \\varphi \\bincirc m_r^M = m_r^N \\bincirc \\varphi \\).\n\n This can be expressed more clearly via the following two conditions, which we call \\term{additivity} and \\term{homogeneity}:\n \\begin{align}\n \\varphi(x + y) &= \\varphi(x) + \\varphi(y) \\label{def:semimodule/homomorphism/additive} \\\\\n \\varphi(rx) &= r \\varphi(x) \\label{def:semimodule/homomorphism/homogeneity}\n \\end{align}\n\n Functions satisfying additivity and homogeneity are commonly called \\term{linear}. These are a central object of study in \\hyperref[sec:linear_algebra]{linear algebra} and, to a lesser extent, (linear) \\hyperref[sec:functional_analysis]{functional analysis}.\n\n \\thmitem{def:semimodule/submodel} The set \\( A \\subseteq M \\) is a \\hyperref[thm:substructure_is_model]{submodel} of \\( M \\) if it is a \\hyperref[def:monoid/submodel]{submonoid} of \\( M \\) that is closed under scalar multiplication, i.e. \\( rM = m_r[M] \\subseteq M \\) for every \\( r \\in R \\). We say that \\( A \\) is an \\( R \\)-\\term{sub-semimodule} of \\( M \\). If \\( M \\) is a module over some semiring extension \\( T \\) of \\( R \\), \\( A \\) may not be a \\( T \\)-sub-semimodule. For this reason, we should only use the term \\enquote{sub-semimodule} of the underlying ring is clear from the context.\n\n As a consequence of \\fullref{thm:positive_formulas_preserved_under_homomorphism}, the \\hyperref[def:multi_valued_function/image]{image} of an \\( R \\)-semimodule homomorphism \\( \\varphi: M \\to N \\) is an \\( R \\)-sub-semimodule of \\( M \\).\n\n For an arbitrary set \\( A \\), we denote the \\hyperref[def:first_order_generated_substructure]{generated submodel} by \\( \\linspan{ A } \\) and call it the \\term{linear span} of \\( A \\).\n\n \\Fullref{rem:span_over_different_semirings} shows how it is important to be unambiguous about over which semiring we take the span of \\( A \\). In case of possible ambiguity, we will use subscripts like \\( \\linspan_R A \\).\n\n The linear span can be characterized via \\hyperref[rem:linear_combinations]{linear combinations} --- see \\fullref{ex:def:first_order_substructure/vector_space}.\n\n \\thmitem{def:semimodule/trivial} The \\hyperref[thm:substructures_form_complete_lattice/bottom]{trivial} semimodule is the \\hyperref[def:pointed_set/trivial]{trivial pointed set} \\( \\set{ 0 } \\).\n\n \\thmitem{def:semimodule/bisemimodule} A \\term{bisemimodule} is a triple \\( (R, A, B) \\), where \\( A \\) is a left \\( R \\)-semimodule, \\( B \\) is a right \\( R \\)-semimodule, and the following associativity condition holds for \\( a \\in A \\), \\( r \\in R \\) and \\( b \\in B \\):\n \\begin{equation}\\label{eq:def:semimodule/bimodule/associativity}\n (a \\cdot_A r) \\cdot_B b = a \\cdot_A (r \\cdot_B b).\n \\end{equation}\n\n \\thmitem{def:semimodule/category} For a fixed semiring \\( R \\), the \\hyperref[def:category_of_small_first_order_models]{category of \\( \\mscrU \\)-small models} \\( \\ucat{SMod}_R \\) of left semimodules is \\hyperref[def:concrete_category]{concrete} over \\hyperref[def:monoid]{\\( \\ucat{Mon} \\)}.\n\n Other notations are in use, for example \\( R-\\cat{Mod} \\) in \\cite[158]{Aluffi2009}, that better highlight whether we are considering left or right semimodules. We will prefer \\( \\cat{Mod}_R^{\\opcat} \\) for the category of right modules.\n \\end{thmenum}\n\\end{definition}\n\\begin{proof}\n \\ImplicationSubProof{def:semimodule/action}{def:semimodule/operation} Fix a semiring homomorphism \\( \\varphi: R \\to \\End(M) \\) and define the operation \\( r \\cdot x \\coloneqq \\varphi(r)(x) \\).\n\n We will verify that all conditions from \\fullref{def:semimodule/operation} hold for this operation.\n\n \\begin{itemize}\n \\item By definition, \\( \\varphi \\) is a monoid action of \\( (R, \\cdot) \\) on \\( (M, \\bincirc) \\).\n\n \\item Distributivity of scalar addition holds because \\( \\varphi \\) is a \\hyperref[def:magma/homomorphism]{magma homomorphism} from \\( (R, +) \\) to \\( (M, +) \\).\n\n \\item Distributivity of vector addition holds because, for each \\( r \\), \\( \\varphi(r) \\) is a magma endomorphism of \\( (M, +) \\).\n\n \\item Since \\( \\varphi \\) is a monoid homomorphism from \\( (R, +) \\) to \\( (R, \\cdot) \\), it preserves identities and hence\n \\begin{equation*}\n 0_R \\cdot x = \\varphi(0_R)(x) = [y \\mapsto 0_M](x) = 0_M.\n \\end{equation*}\n\n This proves half of \\fullref{def:semimodule/operation/absorption}.\n\n \\item Since, for each \\( r \\), \\( \\varphi(r) \\) is a monoid endomorphism of \\( (M, +) \\), we have\n \\begin{equation*}\n r \\cdot 0_M = \\varphi(r)(0_M) = 0_M.\n \\end{equation*}\n\n This proves the other half of \\fullref{def:semimodule/operation/absorption}.\n \\end{itemize}\n\n \\ImplicationSubProof{def:semimodule/operation}{def:semimodule/action} Let \\( \\cdot: R \\times M \\to M \\) be an operation satisfying all conditions from \\fullref{def:semimodule/operation}. Define the function \\( \\varphi(r) \\coloneqq (x \\mapsto r \\cdot x) \\). We will show that this is a semiring homomorphism.\n\n It preserves both identities because\n \\begin{equation*}\n \\varphi(0_R) = (x \\mapsto 0) = 0_{\\End(M)}\n \\end{equation*}\n and\n \\begin{equation*}\n \\varphi(1_R) = (x \\mapsto x) = \\id_M.\n \\end{equation*}\n\n We must also show that it preserves both binary operations. Clearly\n \\begin{equation*}\n \\varphi(r + s)\n =\n (x \\mapsto (r + s) x)\n \\reloset {\\eqref{eq:def:semiring/right_distributivity}} =\n (x \\mapsto r x + s x)\n =\n (x \\mapsto r x) + (x \\mapsto s x)\n =\n \\varphi(r) + \\varphi(s).\n \\end{equation*}\n\n For multiplication, we have\n \\begin{equation*}\n \\varphi(rs)\n =\n (x \\mapsto (rs)x)\n \\reloset {\\eqref{eq:def:magma/associative}} =\n (x \\mapsto r(sx))\n =\n \\parens[\\Big]{ x \\mapsto \\varphi(r)\\parens[\\Big]{ \\varphi(s)(x) } }\n =\n \\varphi(r) \\bincirc \\varphi(s).\n \\end{equation*}\n\\end{proof}\n\n\\begin{proposition}\\label{thm:def:semimodule}\n \\hyperref[def:semimodule]{Semimodules} have the following basic properties:\n \\begin{thmenum}\n \\thmitem{thm:def:semimodule/union} The union of a \\hyperref[eq:def:partially_ordered_set/homomorphism/sequence]{monotone sequence}\n \\begin{equation*}\n N_1 \\subseteq N_2 \\subseteq \\cdots\n \\end{equation*}\n of \\( R \\)-\\hyperref[def:semimodule/submodel]{sub-semimodules} of \\( M \\) is also an \\( R \\)-sub-semimodule of \\( M \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n \\SubProofOf{thm:def:semimodule/union} Trivial.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:semiring_is_semimodule}\n Every semiring is a \\hyperref[def:semimodule/bisemimodule]{bisemimodule} over itself with scalar multiplication given by the semiring multiplication.\n\n This result specializes to \\fullref{thm:semiring_is_algebra}.\n\\end{proposition}\n\\begin{proof}\n Fix a semiring \\( R \\). We will show that \\( \\cdot \\) satisfied the conditions in \\fullref{def:semimodule/operation}.\n \\begin{itemize}\n \\item The identity law \\eqref{eq:def:semimodule/operation/scalar_multiplication_action/identity} holds because \\( 1 \\) is a multiplicative identity of \\( M \\).\n \\item The associativity-like law \\eqref{eq:def:semimodule/operation/scalar_multiplication_action/compatibility} follows from associativity of multiplication.\n \\item The two distributivity laws \\eqref{eq:def:semimodule/operation/scalar_addition_distributivity} and \\eqref{eq:def:semimodule/operation/vector_addition_distributivity} follow from left and right distributivity on \\( R \\).\n \\item The absorption law \\eqref{eq:def:semimodule/operation/absorption} follows from absorption on semirings.\n \\end{itemize}\n\n All the above also hold for right semimodules rather than left.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:commutative_monoid_is_semimodule}\n The categories \\( \\hyperref[def:monoid/category]{\\cat{CMon}} \\) of commutative monoids and \\( \\hyperref[def:semimodule/category]{\\cat{SMod}_\\BbbN} \\) of natural number semimodules are \\hyperref[rem:category_similarity/isomorphism]{isomorphic}.\n\n More concretely, every commutative monoid \\( M \\) is a left semimodule over \\( \\BbbN \\) with scalar multiplication given by \\hyperref[rem:additive_magma/multiplication]{recursively defined multiplication}\n \\begin{equation}\\label{eq:thm:commutative_monoid_is_semimodule/operation}\n \\begin{aligned}\n &\\cdot: \\BbbN \\times M \\to M \\\\\n &n \\cdot x \\coloneqq \\begin{cases}\n 0_M, &n = 0, \\\\\n n \\cdot x + x, &n > 1.\n \\end{cases}\n \\end{aligned}\n \\end{equation}\n\n Conversely, in every semimodule over \\( \\BbbN \\), scalar multiplication matches the recursively defined multiplication.\n\n This result specializes to \\fullref{thm:semiring_is_natural_number_algebra} and \\fullref{thm:abelian_group_is_module}.\n\\end{proposition}\n\\begin{proof}\n \\SufficiencySubProof Let \\( M \\) be a commutative monoid. The operation \\( \\cdot: \\BbbN \\times M \\to M \\) defined in \\fullref{thm:semiring_characteristic_homomorphism} satisfies the conditions in \\fullref{def:semimodule/operation} as either a direct consequence of the definition or as a consequence of \\fullref{thm:monoid_distributivity}.\n\n The homomorphisms are thus also compatible.\n\n \\NecessitySubProof Let \\( M \\) be a semimodule over \\( \\BbbN \\). We will use induction to show that \\eqref{eq:thm:semiring_is_semimodule/operation} holds.\n \\begin{itemize}\n \\item For \\( n = 0 \\), this follows from the absorption law \\eqref{eq:def:semimodule/operation/absorption}.\n \\item If \\( n \\cdot x = n \\cdot x + x \\), then by scalar distributivity, \\( (n + 1) \\cdot x = n \\cdot x + 1 \\cdot x \\). The multiplicative identity law \\eqref{eq:def:semimodule/operation/scalar_multiplication_action/identity} then shows that \\( 1 \\cdot x = x \\), which concludes our proof.\n \\end{itemize}\n\n The homomorphisms are thus also compatible.\n\\end{proof}\n\n\\begin{proposition}\\label{thm:functions_over_semimodule}\n For a set \\( A \\) and an \\( R \\)-\\hyperref[def:semimodule]{semimodule} \\( N \\), the set \\( \\fun(A, N) \\) of all functions from \\( A \\) to \\( N \\) is itself an \\( R \\)-semimodule with the following operations:\n \\begin{thmenum}\n \\thmitem{thm:functions_over_semimodule/addition} Pointwise addition\n \\begin{equation*}\n [f + g](x) \\coloneqq f(x) + g(x)\n \\end{equation*}\n\n \\thmitem{thm:functions_over_semimodule/scalar_multiplication} Pointwise scalar multiplication\n \\begin{equation*}\n [t \\cdot f](x) \\coloneqq t \\cdot f(x)\n \\end{equation*}\n \\end{thmenum}\n\n If \\( A \\) is also an \\( R \\)-algebra, we denote the semimodule of all \\( R \\)-\\hyperref[def:semimodule/homomorphism]{linear maps} by \\( \\hom(A, N) \\).\n\n This extends to \\fullref{thm:functions_over_algebra}.\n\\end{proposition}\n\\begin{proof}\n By \\fullref{thm:functions_over_model_form_model}, \\( N \\) is an \\( R \\)-semimodule.\n\\end{proof}\n\n\\begin{definition}\\label{def:function_support}\n The \\term{support} of a function \\( f: S \\to R \\) from any set \\( S \\) to a semiring \\( R \\) is the set\n \\begin{equation*}\n \\supp(f) \\coloneqq \\set{ x \\in S \\given f(x) \\neq 0_R }.\n \\end{equation*}\n\\end{definition}\n\n\\begin{definition}\\label{def:semimodule_direct_product}\n The \\term{direct product} of a family of \\( R \\)-\\hyperref[def:semimodule]{semimodules} \\( \\seq{ M_k }_{k \\in \\mscrK} \\) is their \\hyperref[def:monoid_direct_product]{monoid direct product} \\( \\prod_{k \\in \\mscrK} M_k \\) with the additional componentwise scalar product\n \\begin{equation*}\n r \\cdot \\seq{ x_k }_{k \\in \\mscrK}\n \\coloneqq\n \\seq{ r \\cdot x_k }_{k \\in \\mscrK}.\n \\end{equation*}\n\n As in the case of general monoids, the \\term{direct sum} \\( \\bigoplus_{k \\in \\mscrK} M_k \\) is the submonoid of the direct product consisting only of \\hyperref[def:cartesian_product/tuple]{tuples} with \\hyperref[def:set_finiteness]{finite} \\hyperref[def:function_support]{support}. That is, tuples with only finitely many nonzero components.\n\\end{definition}\n\n\\begin{proposition}\\label{thm:semimodule_categorical_limits}\n We present a refinement to \\fullref{thm:monoid_categorical_limits}.\n\n \\begin{thmenum}\n \\thmitem{thm:semimodule_categorical_limits/product} The \\hyperref[def:discrete_category_limits]{categorical product} of the family \\( \\seq{ M_k }_{k \\in \\mscrK} \\) in the category \\hyperref[def:semimodule/category]{\\( \\cat{SMod} \\)} of semimodules is their \\hyperref[def:semimodule_direct_product]{direct product} \\( \\prod_{k \\in \\mscrK} M_k \\).\n\n \\thmitem{thm:semimodule_categorical_limits/coproduct} The \\hyperref[def:discrete_category_limits]{categorical coproduct} of the family \\( \\seq{ M_k }_{k \\in \\mscrK} \\) in the category \\hyperref[def:semimodule/category]{\\( \\cat{CSMon} \\)} of \\hi{commutative} semimodules is their \\hyperref[def:semimodule_direct_product]{direct sum} \\( \\bigoplus_{k \\in \\mscrK} M_k \\).\n \\end{thmenum}\n\\end{proposition}\n\\begin{proof}\n Simple refinement of \\fullref{thm:monoid_categorical_limits}.\n\\end{proof}\n\n\\begin{definition}\\label{def:free_semimodule}\\mimprovised\n Fix a \\hyperref[def:semiring]{semiring} \\( R \\). We associate with every set \\( A \\) its \\term{free \\( R \\)-semimodule} \\( R^{\\oplus A} \\) over \\( R \\) defined as the set\n \\begin{equation*}\n R^{\\oplus A} \\coloneqq \\bigoplus_{x \\in A} R = \\set{ t: A \\to R \\given t \\T{has finite \\hyperref[def:function_support]{support}} }.\n \\end{equation*}\n\n In the case when \\( R \\) is the semiring \\( \\BbbN \\) of natural numbers, \\( \\BbbN^{\\oplus A} \\) is the set of finite \\hyperref[def:weighted_set/multiset]{multisets} over \\( S \\).\n\n We call \\( A \\) the \\term{basis} of \\( R \\). As with general direct sums, we regard the function \\( t \\) as the indexed family \\( \\seq{ t_x }_{x \\in A} \\), and we call the indexed family a \\term{linear combination} over \\( R \\).\n\n By \\fullref{thm:functions_over_model_form_model}, \\( R^{\\oplus A} \\) inherits addition and multiplication from \\( R \\) and is actually a semiring. Scalar multiplication can be defined as\n \\begin{equation*}\n \\begin{aligned}\n &\\cdot: R \\times R^{\\oplus A} \\to R^{\\oplus A}, \\\\\n &r \\cdot \\seq{ t_x }_{x \\in A} \\coloneqq \\seq{ r \\cdot t_x }_{x \\in A}.\n \\end{aligned}\n \\end{equation*}\n\n Finally, define the canonical inclusion.\n \\begin{equation*}\n \\begin{aligned}\n &\\iota_A: A \\to R^{\\oplus A}, \\\\\n &\\iota_A(x) \\coloneqq \\parens[\\Bigg]\n {\n y \\mapsto \\begin{rcases}\n \\begin{cases}\n 1_R, &y = x \\\\\n 0_R, &y \\neq x\n \\end{cases}\n \\end{rcases}\n }\n \\end{aligned}\n \\end{equation*}\n\n Free right semimodules require trivial adjustments.\n\\end{definition}\n\n\\begin{theorem}[Free semimodule universal property]\\label{thm:free_semimodule_universal_property}\n Fix a semiring \\( R \\) and a set \\( A \\). The \\hyperref[def:free_semimodule]{free \\( R \\)-semimodule} \\( R^{\\oplus A} \\) over \\( R \\) is the unique up to a unique isomorphism semimodule that satisfies the following \\hyperref[rem:universal_mapping_property]{universal mapping property}:\n \\begin{displayquote}\n For every semimodule \\( M \\) over \\( R \\) and every function \\( e: A \\to M \\), there exists a unique \\( R \\)-semimodule homomorphism \\( \\Phi_e: R^{\\oplus A} \\to M \\) such that the following diagram commutes:\n \\begin{equation}\\label{eq:thm:free_semimodule_universal_property/diagram}\n \\begin{aligned}\n \\includegraphics[page=1]{output/thm__free_semimodule_universal_property.pdf}\n \\end{aligned}\n \\end{equation}\n \\end{displayquote}\n\n Via \\fullref{rem:universal_mapping_property}, \\( A \\mapsto R^{\\oplus A} \\) becomes \\hyperref[def:category_adjunction]{left adjoint} to the \\hyperref[def:concrete_category]{forgetful functor}\n \\begin{equation*}\n U: \\cat{SMod}_R \\to \\cat{Set}.\n \\end{equation*}\n\n The function \\( e \\) assigns a value from \\( M \\) for each component of a linear combination, while \\( \\Phi_e \\) evaluates the obtained sum. We call \\( \\Phi_e \\) the \\term{linear extension} or \\term{linear combination evaluation map} corresponding to \\( e \\).\n\\end{theorem}\n\\begin{proof}\n For every function \\( e: A \\to M \\), we want\n \\begin{equation*}\n \\Phi_e(\\iota(x)) = e(x).\n \\end{equation*}\n\n This suggests the definition\n \\begin{equation*}\n \\begin{aligned}\n &\\Phi_e: R^{\\oplus A} \\to M, \\\\\n &\\Phi_e(\\seq{ t_x }_{x \\in A}) \\coloneqq \\sum_{x \\in A}^n t_x \\cdot e(x).\n \\end{aligned}\n \\end{equation*}\n\n We discuss well-definedness of infinitary operations in direct sums in \\fullref{rem:binary_operation_syntax_trees/infinite/direct_sum}.\n\\end{proof}\n\n\\begin{remark}\\label{rem:linear_combinations}\n The \\hyperref[def:free_semimodule]{linear combination} \\( \\sum_{x \\in A} t_x x \\) can instead be written as \\( \\sum_{k=1}^n t_k x_k \\), where \\( x_1, \\ldots, x_n \\) are the values in \\( A \\) for which the scalars \\( t_1, \\ldots, t_n \\) are nonzero (we denote \\( t_{x_k} \\) by \\( t_k \\) for brevity). This is actually the dominating convention, although we sometimes use the former notation and. In vector spaces, we utilize the projection functionals from \\fullref{def:basis_decomposition}.\n\n This issue is discussed in more generality in \\fullref{rem:binary_operation_syntax_trees/infinite/direct_sum}.\n\\end{remark}\n\n\\begin{proposition}\\label{thm:span_via_linear_combinations}\n For a set \\( A \\) in an \\( R \\)-\\hyperref[def:semimodule]{semimodule} \\( M \\), the \\hyperref[def:semimodule/submodel]{linear span} of \\( A \\), defined as the \\( R \\)-submodule generated by \\( A \\) in the sense of \\fullref{def:first_order_generated_substructure}, equals the set of all \\hyperref[rem:linear_combinations]{linear combinations} over \\( A \\).\n\n We say that \\( A \\) is a \\term{generating set} of \\( M \\).\n\n Compare this result to \\fullref{thm:generators_via_polynomials} for algebras.\n\\end{proposition}\n\\begin{proof}\n \\Cref{fig:thm:span_via_linear_combinations} shows a \\hyperref[def:concrete_syntax_tree]{syntax tree} for a given linear combination, which can be traversed and evaluated to obtain a vector in \\( M \\). Furthermore, it must be a member of \\( \\linspan S \\) since the latter is closed under vector addition and scalar multiplication with members of \\( S \\). Hence, the set \\( L \\) of all linear combinations over \\( S \\) is a subset of \\( \\linspan S \\).\n\n Generalizing the syntax tree construction from \\cref{fig:thm:span_via_linear_combinations}, we see that \\( L \\) satisfies \\fullref{def:first_order_substructure/inductive}, and is thus a submodule of \\( M \\). Since \\( \\linspan S \\) is the smallest module containing \\( S \\), we have \\( \\linspan S = L \\).\n\n \\begin{figure}[h]\n \\hfill\n \\includegraphics[page=1]{output/thm__span_via_linear_combinations.pdf}\n \\hfill\\hfill\n \\caption{A linear combination is simply a \\hyperref[def:multi_valued_function/superposition]{superposition} of scalar multiplication and binary addition.}\n \\label{fig:thm:span_via_linear_combinations}\n \\end{figure}\n\\end{proof}\n\n\\begin{remark}\\label{rem:span_over_different_semirings}\n If \\( M \\) is both an \\( R \\)-semimodule and a \\( T \\)-semimodule, \\fullref{thm:span_via_linear_combinations} highlights a fundamental difference between the generated \\( R \\)-sub-semimodule and the generated \\( T \\)-sub-semimodule.\n\n For example, the \\( \\BbbN \\)-sub-semimodule generated by \\( 2 \\) is the semiring \\( 2\\BbbN \\) of even natural numbers, while the \\( \\BbbR_{\\geq 0} \\)-sub-semimodule generated by \\( 2 \\) is \\( \\BbbR_{\\geq 0} \\) itself.\n\\end{remark}\n", "meta": {"hexsha": "61ad6adb29427e4d50a0cfc4c0547242bc50e061", "size": 24749, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/semimodules.tex", "max_stars_repo_name": "v--/anthology", "max_stars_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "src/semimodules.tex", "max_issues_repo_name": "v--/anthology", "max_issues_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/semimodules.tex", "max_forks_repo_name": "v--/anthology", "max_forks_repo_head_hexsha": "89a91b5182f187bc1aa37a2054762dd0078a7b56", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 65.8218085106, "max_line_length": 601, "alphanum_fraction": 0.708796315, "num_tokens": 7682, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7981867777396212, "lm_q2_score": 0.7662936377487305, "lm_q1q2_score": 0.6116454495170317}} {"text": "\\documentclass[a4paper]{article}\n\n\n\\input{../preambles/preamble}\n\\setdefaultlanguage{english}\n%\\setotherlanguages{german,greek}\n\n\\input{../preambles/mathematics}\n\\input{../preambles/physics}\n\n%\\newcommand{\\RomaN}[1]{%\n% \\textup{\\uppercase\\expandafter{\\romannumeral#1}}%\n%}\n\n\n%\\addbibresource{main.bib}\n\\title{Complex Harmonic Oscillators and How to Squeeze Them}\n\\author{YiFan Wang}\n\n\\newcommand{\\Poibr}[1]{\\sbr{#1}_\\text{P}}\n\\newcommand{\\Combr}[1]{\\sbr{#1}_\\text{-}}\n\n\\newcommand{\\tR}{\\Re}\n\\newcommand{\\tI}{\\Im}\n\\newcommand{\\tS}{\\text{S}}\n\\newcommand{\\tD}{\\text{D}}\n\n\n\\begin{document}\n\\maketitle\n\n%\\begin{abstract}\n%Your abstract.\n%\\end{abstract}\n\n\\tableofcontents\n\n\\section{Single complex oscillator}\n\n\\begin{nameddef}{Classical Hamiltonian in canonical coordinates}\nComplex phase space\n\\begin{equation}\nH = \\frac{1}{2}\\pi^+\\pi^- + \\frac{\\Omega^2}{2}\\phi^+\\phi^-,\n\\end{equation}\nThe corresponding Poisson brackets read\n\\begin{equation}\n\\Poibr{\\rfun{f}{\\eta^C}, \\rfun{g}{\\eta^C}} = \\sum_C \\rbr{\n\\frpa{f}{\\phi^C}\\frpa{g}{\\pi^C} - \\frpa{f}{\\pi^C}\\frpa{g}{\\phi^C}},\n\\end{equation}\nwhere $C \\in \\cbr{+, -}$, $\\eta^+ = \\rbr{\\eta^-}^*$ and $\\eta \\in \\cbr{\\pi, \n\\phi}$, so that\n\\begin{equation}\n\\Poibr{\\phi^{C_1}, \\pi^{C_2}} = \\delta^{C_1 C_2},\\qquad\n\\Poibr{\\phi^{C_1}, \\phi^{C_2}} = \\Poibr{\\pi^{C_1}, \\pi^{C_2}} = 0.\n\\end{equation}\n\nReal phase space\n\\begin{equation}\n\\phi^C = \\frac{1}{\\sqrt{2}}\\rbr{ \\phi_\\tR - C \\ii \\phi_\\tI},\\qquad\n\\pi^C = \\frac{1}{\\sqrt{2}}\\rbr{ \\pi_\\tR + C \\ii \\pi_\\tI}.\n%\\phi^* &= \\frac{1}{\\sqrt{2}}\\rbr{\\phi^\\tR - \\ii \\phi^\\tI},\\qquad\n%\\pi^* &&= \\frac{1}{\\sqrt{2}}\\rbr{\\pi^\\tR + \\ii \\pi^\\tI}.\n\\end{equation}\nInverse transformation\n\\begin{equation}\n\\begin{aligned}\n\\phi_\\tR &= \\frac{1}{\\sqrt{2}}\\rbr{\\phi^+ + \\phi^-},\n&\\phi_\\tI &= \\frac{\\ii}{\\sqrt{2}}\\rbr{\\phi^+ - \\phi^-}, \\\\\n\\pi_\\tR &= \\frac{1}{\\sqrt{2}}\\rbr{\\pi^- + \\pi^+},\n&\\pi_\\tI &= \\frac{\\ii}{\\sqrt{2}}\\rbr{\\pi^- - \\pi^+}. \\\\\n\\end{aligned}\n\\end{equation}\n\nOne can verify\n\\begin{equation}\nH = \\sum_{F} \\frac{1}{2} \\pi_F^2 + \\frac{\\Omega^2}{2} \\phi_F^2,\n\\end{equation}\nwhere $F \\in \\cbr{\\tR, \\tI}$, and\n\\begin{equation}\n\\Poibr{\\rfun{f}{\\eta_F}, \\rfun{g}{\\eta_F}} = \\sum_F\n\\frpa{f}{\\phi_F}\\frpa{g}{\\pi_F} - \\frpa{f}{\\pi_F}\\frpa{g}{\\phi_F},\n\\end{equation}\nso that\n\\begin{equation}\n\\Poibr{\\phi_{F_1}, \\pi_{F_2}} = \\delta_{F_1 F_2},\\qquad\n\\Poibr{\\phi_{F_1}, \\phi_{F_2}} = \\Poibr{\\pi_{F_1}, \\pi_{F_2}} = 0.\n\\end{equation}\nhold as well.\n\\end{nameddef} % Classical Hamiltonian in canonical coordinates\n\n\\begin{nameddef}{Ladder coordinates}\nLadder coordinates (ladder `numbers', later to be quantised) in complex phase\nspace\n\\begin{align}\na_\\phi^C &= \\frac{1}{\\sqrt{2}} \\rbr{\\Omega^{+\\frac{1}{2}}\\phi^{C} - C \\ii\n\\Omega^{-\\frac{1}{2}} \\pi^{-C}}, \\\\\na_\\pi^C &= \\frac{1}{\\sqrt{2}} \\rbr{\\Omega^{+\\frac{1}{2}}\\phi^{-C} - C \\ii\n\\Omega^{-\\frac{1}{2}} \\pi^{C}},\n\\end{align}\nwhere $-- = +$, $-+ = -$.\nPoisson brackets? Inverse transformation\n\\begin{equation}\n\\phi^- = \\frac{\\Omega^{-\\frac{1}{2}}}{\\sqrt{2}}\\rbr{a_\\pi^+ + a_\\phi^-},\\qquad\n\\pi^- = \\frac{\\ii\\Omega^{+\\frac{1}{2}}}{\\sqrt{2}}\\rbr{a_\\phi^+ - a_\\pi^-}.\n\\end{equation}\n\n\nLadder coordinates in real phase space\n\\begin{equation}\na_F^C = \\frac{1}{\\sqrt{2}} \\rbr{\\Omega^{+\\frac{1}{2}}\\phi_F - C \\ii\n\\Omega^{-\\frac{1}{2}} \\pi_F}.\n\\end{equation}\nPoisson brackets? Inverse transformation\n\\begin{equation}\n\\phi_F = \\frac{\\Omega^{-\\frac{1}{2}}}{\\sqrt{2}}\\rbr{a_F^+ + a_F^-},\\qquad\n\\pi_F = \\frac{\\ii\\Omega^{+\\frac{1}{2}}}{\\sqrt{2}}\\rbr{a_F^+ - a_F^-}.\n\\end{equation}\n\n\nOne can check that\n\\begin{equation}\na_\\phi^C = \\frac{1}{\\sqrt{2}}\\rbr{a_\\tR^C - C \\ii a_\\tI^C},\\qquad\na_\\pi^C = \\frac{1}{\\sqrt{2}}\\rbr{a_\\tR^C + C \\ii a_\\tI^C}.\n\\end{equation}\n%Inverse transformation\n%\\begin{equation}\n%a_\\phi^C = \\frac{1}{\\sqrt{2}}\\rbr{a_\\tR^C - C \\ii a_\\tI^C},\\qquad\n%a_\\pi^C = \\frac{1}{\\sqrt{2}}\\rbr{a_\\tR^C + C \\ii a_\\tI^C}.\n%\\end{equation}\n\n\n\\end{nameddef} % Ladder coordinates\n\n\n\n\\begin{nameddef}{Quantisation}\nQuantisation in complex canonical coordinates\n\\begin{equation}\nf \\mapsto \\what{f};\\qquad\n\\Poibr{f,g} \\mapsto \\Combr{\\what{f},\\what{g}} = \\ii \\what{\\Poibr{f,g}}.\n\\end{equation}\nAll classical equations listed above can be immediately quantised, since no \nproduct of non-commuting operators appears.\n\n%Quantum Hamiltonian in canonical coordinates\n%\\begin{align}\n%\\what{H} &= \\frac{1}{2} \\what{\\pi}^ + \\what{\\pi}^- + \n%\\frac{\\Omega^2}{2} \\what{\\phi}^+ \\what{\\phi}^-\n%\\\\\n%&= \\sum_{F} \\frac{1}{2} \\what{\\pi}_F^2 + \\frac{\\Omega^2}{2} \\what{\\phi}_F^2.\n%\\end{align}\n\nCommutators of the ladder operators\n\\begin{align}\n\\Combr{\\what{a}_{\\eta_1}^{-C_1}, \\what{a}_{\\eta_2}^{C_2}} &= \n\\delta_{\\eta_1 \\eta_2} \\delta^{C_1 C_2} \\what{1}; \\\\\n\\Combr{\\what{a}_{F_1}^{-C_1}, \\what{a}_{F_2}^{C_2}}\n&= \\delta_{F_1 F_2} \\delta^{C_1 C_2} \\what{1}. \\\\\n\\end{align}\n\nNumber operators\n\\begin{equation}\n\\what{n}_\\eta \\coloneqq \\what{a}_\\eta^+ \\what{a}_\\eta^-,\\qquad\n\\what{n}_F \\coloneqq \\what{a}_F^+ \\what{a}_F^-.\n\\end{equation}\n\n\nAngular momentum operator\n\\begin{align}\n\\what{L} &\\coloneqq\n\\what{\\phi}_\\tR\\what{\\pi}_\\tI - \\what{\\phi}_\\tI\\what{\\pi}_\\tR \\nonumber \\\\\n&= \\ii\\rbr{\\what{a}_\\tI^+\\what{a}_\\tR^- - \\what{a}_\\tR^+\\what{a}_\\tI^-}\n\\nonumber \\\\\n&= \\ii\\rbr{\\what{\\phi}^-\\what{\\pi}^- - \\what{\\phi}^+\\what{\\pi}^+}\n= \\ii\\rbr{\\what{\\pi}^-\\what{\\phi}^- - \\what{\\pi}^+\\what{\\phi}^+}\n\\nonumber \\\\\n&= \\what{n}_\\pi - \\what{n}_\\phi.\n\\label{eq:angular-momentum}\n\\end{align}\n$\\what{L} = \\what{L}^\\dagger$.\n\n\\begin{align}\n\\what{n}_\\phi &= \\frac{1}{2}\n\\rbr{\\Omega^{+\\frac{1}{2}}\\what{\\phi}^+ - \\ii\\Omega^{-\\frac{1}{2}}\\what{\\pi}^-}\n\\rbr{\\Omega^{+\\frac{1}{2}}\\what{\\phi}^- + \\ii\\Omega^{-\\frac{1}{2}}\\what{\\pi}^+}\n\\nonumber \\\\\n&= \\frac{1}{2} \\rbr{\\Omega^{+1} \\what{\\phi}^+\\what{\\phi}^- + \n\\ii\\rbr{\\what{\\phi}^+\\what{\\pi}^+ - \\what{\\pi}^-\\what{\\phi}^-}\n+ \\Omega^{-1} \\what{\\pi}^-\\what{\\pi}^+} \\nonumber \\\\\n&= \\Omega^{-1}\\what{H} - \\frac{1}{2}\\rbr{1+\\what{L}}.\n\\end{align}\nSubstituting \\eqref{eq:angular-momentum} yields the quantum Hamiltonian\n\\begin{empheq}[box=\\fbox]{equation}\n\\what{H} = \\frac{\\Omega}{2}\\rbr{\\what{n}_\\phi + \\what{n}_\\pi + 1}.\n\\end{empheq} % equation\n%Luckily,\n%\\begin{equation}\n%\\Combr{\\what{L},\\what{H}} = 0\n%\\end{equation}\n%so that\n%\\begin{equation}\n%\\Omega \\Combr{\\what{n},\\what{L}} = \\Combr{\\what{n},\\what{H}} = 0\n%\\end{equation}\n%as well.\n\n\n%\\begin{equation}\n%H = \\frac{1}{2}\\what{\\pi}^+\\what{\\pi}^- + \\frac{\\Omega^2}{2}\\phi^+\\phi^-,\n%\\end{equation}\n\\end{nameddef}\n\n\\begin{nameddef}{Wave function}\nOne may choose the \n\\end{nameddef}\n\n\n\\section{Rotating}\n\n\\section{Cohering}\n\n\\section{Single--mode squeezing}\n\n\\section{Double--mode squeezing}\n\n\\\n\n\n\n\n% Let's print the overall heading of the bibliography first:\n%\\printbibheading\n%\\printbibliography\n\n\\end{document}", "meta": {"hexsha": "1de17142e64c2b659ca15ce2f34ab0a055d36ff4", "size": 6654, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old/complex-ho/main.tex", "max_stars_repo_name": "cmp0xff/Notes", "max_stars_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "old/complex-ho/main.tex", "max_issues_repo_name": "cmp0xff/Notes", "max_issues_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "old/complex-ho/main.tex", "max_forks_repo_name": "cmp0xff/Notes", "max_forks_repo_head_hexsha": "afd712c1e42275bf781a030d6c5f1b7f4c6ec57b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.4358974359, "max_line_length": 79, "alphanum_fraction": 0.6211301473, "num_tokens": 2870, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972717658209, "lm_q2_score": 0.7025300636233416, "lm_q1q2_score": 0.6116207567239498}} {"text": "\n\n\\section{Monte Carlo Methods}\n\n\\subsection{Requirement}\n\n\\begin{itemize}\n\t\\item need episode, so need to terminate\n\t\\item start only when episode ends\n\t\\item do not need model\n\\end{itemize}\n\nNote: Monte Carlo Method is not a online method because it is not step-by-step method.\n\n\\subsection{Prediction}\n\n\\subsubsection{First Visit MC}\n\n\\cindex{Monte Carlo} prediction has \\cindex{first visit} and \\cindex{every visit} methods.\n\n\nThe \\cindex{first visit} Algorithm (\\ref{algo:fvmc}) is an unbiased estimate. \n\n\n\n\\begin{algorithm}\n\t\\caption{First visit MC, estimate $v_\\pi$}\\label{algo:fvmc}\n\t\n\t\\begin{algorithmic}[1]\n\t\t\\State $\\forall s \\in \\mathcal{S}, V(s) \\gets random$\n\t\t\\State $\\text{Returns}(s) \\gets []$\n\t\t\n\t\t\\Statex\n\t\t\n\t\t\\Loop\n\t\t\t\\State generate an episode $\\pi: S_0, A_0, R_1, S_1, A_1, R_2, \\dots, S_{T-1}, A_{T-1}, R_T$\n\t\t\t\\State $G \\gets 0$\n\t\t\t\\For{$t \\gets T-1, T-2, \\dots, 0$}\n\t\t\t\t\\State $G \\gets \\gamma G + R_{t+1}$\n\t\t\t\t\n\t\t\t\t\\Comment{loop backward to find first appearance}\n\t\t\t\t\\If{$S_t \\notin \\{S_0, S_1, \\dots, S_{t-1}\\}$}\n\t\t\t\t\t\\State $\\text{Returns}(S_t) \\gets \\text{Returns}(S_t) + G$\n\t\t\t\t\t\\State $V(S_t) \\gets \\text{average}(\\text{Returns}(S_t))$\n\t\t\t\t\\EndIf\n\t\t\t\\EndFor\n\t\t\\EndLoop\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\\subsection{Explore All State-Action Pair}\n\nIn Monte Carlo control, the policy at $S_t$ need to explore all $q(S_t,A_t)$, which means the MC algorithm needs to cover all $\\langle S_t, A_t \\rangle$ pairs. There are two ways to achieve this:\n\n\\begin{itemize}\n\t\\item exploring start. It might not be possible to enumerate all start.\n\t\\item $\\varepsilon$-soft policy.\n\\end{itemize}\n\n\\subsubsection{Exploring Starts}\n\nIt tries all $\\langle S_t, A_t \\rangle$ pairs as the first action. See Algorithm (\\ref{algo:estart}) for detail.\n\nIf model is not available, $q_\\pi$ is preferred than $V_\\pi$ because it does not need transition probability when calculating optimal policy. \n\n\n\\begin{algorithm}\n\t\\caption{first visit MCES (Exploring Starts), estimate $\\pi_*$}\\label{algo:estart}\t\n\t\n\t\\begin{algorithmic}[1]\n\t\t\\State $\\pi(s) \\in \\mathcal{A}(s)$\n\t\t\\State $q(s,a) \\in \\mathbb{R}$\n\t\t\\State $\\text{Returns}(s,a) \\gets []$\n\t\t\n\t\t\\Statex\n\t\t\n\t\t\\Loop\n\t\t\t\\State choose $S_0$ and $A_0$ so all pairs will appear \\Comment{exploring starts}\n\t\t\t\\State generate episode from $\\langle S_0,A_0 \\rangle$: $\\pi: S_0, A_0, R_1,\\dots, S_{T-1}, A_{T-1}, R_T$\n\t\t\t\\State $G \\gets 0$\n\t\t\t\n\t\t\t\\For{$t \\gets T-1, T-2, \\dots, 0$}\n\t\t\t\t\\State $G \\gets \\gamma G + R_{t+1}$\n\t\t\t\t\n\t\t\t\t\\If{$\\langle S_t, A_t \\rangle \\notin \\{\\langle S_0,A_0 \\rangle,\\langle S_1,A_1 \\rangle, \\dots,\\langle S_{t-1}, A_{t-1} \\rangle \\}$}\n\t\t\t\t\t\\State $\\text{Returns}(S_t, A_t) \\gets \\text{Returns}(S_t, A_t) + G$\n\t\t\t\t\t\\State $q(S_t, A_t) \\gets \\text{average}(\\text{Returns}(S_t, A_t))$\n\t\t\t\t\t\\State $\\pi (S_t) \\gets \\underset{a \\in \\mathcal{A}(S_t)}{\\text{argmax}}\\ q(S_t, a) $\n\t\t\t\t\\EndIf\n\t\t\t\\EndFor\n\t\t\\EndLoop\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\\subsubsection{$\\varepsilon$-soft Policy}\n\nIn $\\varepsilon$-soft policy, all $\\langle S_t, A_t \\rangle$ pairs are tried in the middle with non-zero probability:\n\n\\begin{itemize}\n\t\\item \\cindex{$\\varepsilon$-soft} : $\\pi(a|s) >\\displaystyle \\frac{\\varepsilon}{\\mathcal{A}(s)}$.\n\t\\item \\cindex{$\\varepsilon$-greedy} : see Algorithm (\\ref{algo:fvmcsoft}) for detail.\n\\begin{equation*}\n\t\\pi(a|s) = \t\n\t\t\\begin{cases}\n\t\t\t\t\\displaystyle \\frac{\\varepsilon}{\\mathcal{A}(s)} & \\text{, if } a \\text{ is not the greedy choice} \\\\\n\t\t\t\t1 - \\varepsilon + \\displaystyle \\frac{\\varepsilon}{\\mathcal{A}(s)} & \\text{, if } a \\text{ is the greedy choice}\n\t\t\\end{cases}\n\\end{equation*}\n\\end{itemize}\n\n\n\n\\begin{algorithm}\n\t\\caption{first visit MC control ($\\varepsilon$-greedy), estimate $\\pi_*$}\\label{algo:fvmcsoft}\t\n\t\n\t\\begin{algorithmic}[1]\n\t\t\\State $\\pi \\gets $ random $\\varepsilon$-greedy policy\n\t\t\\State $q(s,a) \\in \\mathbb{R}$\n\t\t\\State $\\text{Returns}(s,a) \\gets []$\n\t\t\n\t\t\\Statex\n\t\t\n\t\t\\Loop\n\t\t\t\\State choose $S_0$ and $A_0$ \\Comment{non-exploring start}\n\t\t\t\\State generate episode from $\\langle S_0,A_0 \\rangle$: $\\pi: S_0, A_0, R_1, \\dots, S_{T-1}, A_{T-1}, R_T$\n\t\t\t\\State $G \\gets 0$\n\t\t\t\n\t\t\t\\For{$t \\gets T-1, T-2, \\dots, 0$}\n\t\t\t\t\\State $G \\gets \\gamma G + R_{t+1}$\n\t\t\t\t\n\t\t\t\t\\If{$\\langle S_t, A_t \\rangle \\notin \\{\\langle S_0,A_0 \\rangle,\\langle S_1,A_1 \\rangle, \\dots,\\langle S_{t-1}, A_{t-1} \\rangle \\}$}\n\t\t\t\t\t\\State $\\text{Returns}(S_t, A_t) \\gets \\text{Returns}(S_t, A_t) + G$\n\t\t\t\t\t\\State $q(S_t, A_t) \\gets \\text{average}(\\text{Returns}(S_t, A_t))$\n\t\t\t\t\t\\State $A^* \\gets \\underset{a \\in \\mathcal{A}(S_t)}{\\text{argmax }} q(S_t, a) $\n\t\t\t\t\t\n\t\t\t\t\t\\For{$a \\in \\mathcal{A}(S_t)$} \\Comment{$\\varepsilon$-greedy}\n\t\t\t\t\t\t\\State \\begin{equation*}\n\t\t\t\t\t\t\t\\pi(a|S_t) \\gets \\begin{cases}\n\t\t\t\t\t\t\t\t\\displaystyle 1 - \\varepsilon + \\frac{\\varepsilon}{\\mathcal{A}(s)} & \\text{,if } a = A^* \\\\\n\t\t\t\t\t\t\t\t\\displaystyle \\frac{\\varepsilon}{\\mathcal{A}(s)} & \\text{,if } a \\neq A^*\n\t\t\t\t\t\t\t\\end{cases} \n\t\t\t\t\t\t\\end{equation*}\n\t\t\t\t\t\\EndFor\n\t\t\t\t\\EndIf\n\t\t\t\\EndFor\n\t\t\\EndLoop\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\\subsection{Off-policy prediction via Importance Sampling}\n\n\\subsubsection{Target Policy}\n\n\\begin{description}\n\t\\item[behavior policy] the policy $b$ used to generate behavior\n\t\\item[target policy] the policy $\\pi$ being learned\n\\end{description}\n\nIt has the assumption of \\cindex{coverage}: \n\\begin{equation}\n\t\\pi(a|s) > 0 \\Rightarrow b(a|s) >0\n\\end{equation}\n\n\n\\subsubsection{Importance Sampling}\n\nThe probability of $\\{ A_t, S_{t+1},A_{t+1},\\dots,S_T \\}$ under policy $\\pi$ is (using Monte Carlo property):\n\\begin{equation}\n\t\\mathbb{P} \\{ A_t, S_{t+1},A_{t+1},\\dots,S_T | S_t, A_{t:T-1} \\sim \\pi \\} = \\prod_{k=t}^{T-1} \\pi(A_k|S_k)p(S_{k+1}|S_k,A_k)\n\\end{equation}\n\nThe \\cindex{importance-sampling ratio} is:\n\\begin{equation}\\label{importancsamplingratio}\n\t\\rho_{t:T-1}^{\\pi / b} = \\frac{\\prod\\limits_{k=t}^{T-1} \\pi(A_k|S_k)p(S_{k+1}|S_k,A_k)}{\\prod\\limits_{k=t}^{T-1} b(A_k|S_k)p(S_{k+1}|S_k,A_k)} = \\prod_{k=t}^{T-1} \\frac{\\pi(A_k|S_k)}{b(A_k|S_k)}\n\\end{equation}\n\nLet $\\mathcal{T}$ denotes the set of all timestamps that $s$ is visited, $T(t)$ is the timestamp that the episode terminate following timestamp $t$, $G_t$ is the return between timestamp $t$ and $T(t)$. There are two different importance sampling:\n\\begin{description}\n\t\\item [ordinary importance sampling] \\begin{equation}\n\t\tV(s) = \\frac{\\sum\\limits_{t \\in \\mathcal{T}} \\rho_{t:T-1}^{\\pi/b} G_t}{|\\mathcal{T}|}\n\t\\end{equation}\n\t\\item [weighted importance sampling] \\begin{equation}\n\t\tV(s) = \\frac{\\sum\\limits_{t \\in \\mathcal{T}} \\rho_{t:T-1}^{\\pi/b} G_t}{\\sum\\limits_{t \\in \\mathcal{T}} \\rho_{t:T-1}^{\\pi/b}}\n\t\\end{equation}\n\\end{description}\n\nFor \\cindex{first visit} method, \\cindex{Ordinary importance sampling} is unbiased, with unlimited variance. So \\cindex{weighted importance sampling} is preferred in practice.\n\nFor \\cindex{every visit} method, both sampling is biased which reduces to near zero when the number of sampling increases.\n\n\nIn practice, \\cindex{every visit} is preferred because it does not need to keep trace of which states have been visited.\n\n\\subsection{Incremental Policy Evaluation}\n\nIncremental implementation need the following background. If we want to estimate \n\\begin{equation}\n\tV_n = \\frac{\\sum\\limits_{k=1}^{n-1} W_k G_k}{\\sum\\limits_{k=1}^{n-1} W_k}, n \\geq 2\n\\end{equation}\n\n$V_n$ could be incrementally updated by:\n\\begin{equation}\n\tV_{n+1} = V_n + \\frac{W_n}{C_n} \\Big( G_n - V_n \\Big), n \\geq 1\n\\end{equation}\n\nwhere \n\\begin{equation*}\n\tC_{n+1} = C_n + W_{n+1}\n\\end{equation*}\n\nAlgorithm (\\ref{algo:opmcp}) implements incremental solution of weighted importance sampling:\n\n\\begin{algorithm}\n\t\\caption{off-policy MC policy evaluation, estimate $q_\\pi$}\\label{algo:opmcp}\t\n\t\n\t\\begin{algorithmic}[1]\n\t\t\\State $Q(s,a) \\in \\mathbb{R}$\n\t\t\\State $C(s,a) \\in 0$\n\t\t\n\t\t\\Statex\n\t\t\n\t\t\\Loop\n\t\t\t\\State $b \\gets $ any policy with coverage of $\\pi$\n\t\t\t\\State generate episode following $b$: $S_0, A_0,R_1, \\dots, S_{T-1},A_{T-1},R_T$\n\t\t\t\\State $G \\gets 0$\n\t\t\t\\State $W \\gets 1$\n\t\t\t\\For{$t \\gets T-1, T-2, \\dots, 0$}\n\t\t\t\t\\State $G \\gets \\gamma G + R_{t+1}$\n\t\t\t\t\\State $C(S_t,A_t) \\gets C(S_t,A_t) + W$ \n\t\t\t\t\\State $\\displaystyle Q(S_t,A_t) \\gets Q(S_t,A_t) + \\frac{W}{C(S_t,A_t)} \\Big ( G - Q(S_t,A_t) \\Big)$\n\t\t\t\t\\State $\\displaystyle W \\gets W \\frac{\\pi(A_t|S_t)}{b(A_t|S_t)}$\n\t\t\t\t\n\t\t\t\t\\If{$W = 0$} \\Comment{$\\pi(A_t|S_t) = 0$}, $b$ does not cover $\\pi$\n\t\t\t\t\t\\State exit For loop\n\t\t\t\t\\EndIf\n\t\t\t\\EndFor\n\t\t\\EndLoop\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\n\\subsection{Incremental Policy Control}\n\nThe behavior policy $b$ need to be $\\varepsilon$-soft.\n\nAlgorithm (\\ref{algo:opmcpc}) implements incremental control of weighted importance sampling:\n\n\\begin{algorithm}\n\t\\caption{off-policy MC policy control, estimate $q_*$}\\label{algo:opmcpc}\t\n\t\n\t\\begin{algorithmic}[1]\n\t\t\\State $Q(s,a) \\in \\mathbb{R}$\n\t\t\\State $C(s,a) \\in 0$\n\t\t\\State $\\pi(s) \\gets \\underset{a}{\\text{argmax}} Q(s,a)$\n\t\t\n\t\t\\Statex\n\t\t\n\t\t\\Loop\n\t\t\t\\State $b \\gets $ any $\\varepsilon$-soft policy\n\t\t\t\\State generate episode following $b$: $S_0, A_0,R_1, \\dots, S_{T-1},A_{T-1},R_T$\n\t\t\t\\State $G \\gets 0$\n\t\t\t\\State $W \\gets 1$\n\t\t\t\\For{$t \\gets T-1, T-2, \\dots, 0$}\n\t\t\t\t\\State $G \\gets \\gamma G + R_{t+1}$\n\t\t\t\t\\State $C(S_t,A_t) \\gets C(S_t,A_t) + W$ \n\t\t\t\t\\State $\\displaystyle Q(S_t,A_t) \\gets Q(S_t,A_t) + \\frac{W}{C(S_t,A_t)} \\Big( G - Q(S_t,A_t) \\Big)$\n\t\t\t\t\\State $\\pi(S_t) \\gets \\underset{a}{\\text{argmax}}\\ Q(S_t,a)$\n\t\t\t\t\n\t\t\t\t\\If{$A_t \\neq \\pi(S_t)$}\n\t\t\t\t\t\\State exit For loop\n\t\t\t\t\\EndIf\n\t\t\t\t\n\t\t\t\t\\State $\\displaystyle W \\gets W \\frac{1}{b(A_t|S_t)}$ \\Comment{$\\pi(A_t|S_t) = 1$ because it is greedy}\n\t\t\t\\EndFor\n\t\t\\EndLoop\n\t\\end{algorithmic}\n\\end{algorithm}\n\n\n", "meta": {"hexsha": "f40499a13b399aae9226db7ec38a8367b57bab45", "size": 9615, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "src/reinforcement_learning/rl.4.mc.tex", "max_stars_repo_name": "elvisren/machine-learning-notes", "max_stars_repo_head_hexsha": "d12ac08d30be4341776714ad895116a243ec026f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2020-05-07T03:05:08.000Z", "max_stars_repo_stars_event_max_datetime": "2020-12-04T17:28:22.000Z", "max_issues_repo_path": "src/reinforcement_learning/rl.4.mc.tex", "max_issues_repo_name": "elvisren/machine-learning-notes", "max_issues_repo_head_hexsha": "d12ac08d30be4341776714ad895116a243ec026f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "src/reinforcement_learning/rl.4.mc.tex", "max_forks_repo_name": "elvisren/machine-learning-notes", "max_forks_repo_head_hexsha": "d12ac08d30be4341776714ad895116a243ec026f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2020-10-01T23:34:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-10-01T23:34:47.000Z", "avg_line_length": 34.3392857143, "max_line_length": 247, "alphanum_fraction": 0.6498179927, "num_tokens": 3639, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8705972717658209, "lm_q2_score": 0.7025300636233416, "lm_q1q2_score": 0.6116207567239498}} {"text": "\\section{Parametric Equations}\\label{sec:Parametric Equations}\nWhen we computed the derivative $dy/dx$ using polar coordinates, we\nused the expressions $x=f(\\theta)\\cos\\theta$ and\n$y=f(\\theta)\\sin\\theta$. These two equations completely specify the\ncurve, though the form $r=f(\\theta)$ is simpler. The expanded form has\nthe virtue that it can easily be generalized to describe a wider range\nof curves than can be specified in rectangular or polar coordinates. \n\nSuppose $f(t)$ and $g(t)$ are functions. Then the equations\n$x=f(t)$ and $y=g(t)$ describe a curve in the plane. In the case of\nthe polar coordinates equations, the variable $t$ is replaced by\n$\\theta$ which has a natural geometric interpretation. In general, $t$\nis simply an arbitrary variable, often called in this case a\n\\dfont{parameter}\\index{parameter}, and this method of specifying a curve is known as \n\\dfont{parametric equations}\\index{parametric equations}\\index{polar coordinates!parametric equations}. One\nimportant interpretation of $t$ is {\\it time}. In this interpretation,\nthe equations $x=f(t)$ and $y=g(t)$ give the position of an object at\ntime $t$.\n\n\\begin{example}{Position of a Path}{positionofapath}\n Describe the path of an object that moves so that its\nposition at time $t$ is given by $x=\\cos t$, $\\ds y=\\cos^2 t$.\n\\end{example}\n\n\\begin{solution}\nWe see immediately that $\\ds y=x^2$, so the path lies on this parabola. The path\nis not the entire parabola, however, since $x=\\cos t$ is always\nbetween $-1$ and $1$. It is now easy to see that the object oscillates\nback and forth on the parabola between the endpoints $(1,1)$ and\n$(-1,1)$, and is at point $(1,1)$ at time $t=0$.\n\\end{solution}\n\nIt is sometimes quite easy to describe a complicated path in\nparametric equations when rectangular and polar coordinate expressions\nare difficult or impossible to devise.\n\n\\begin{example}{Wheel}{wheelexample}\n A wheel of radius 1 rolls along a straight line, say the\n$x$-axis. A point on the rim of the wheel will trace out a curve, called a\ncycloid. Assume the point starts at the origin; find\nparametric equations for the curve.\n\\end{example}\n\n\\begin{solution}\nFigure~\\ref{fig:cycloid} illustrates the generation of the curve. The wheel is shown at its\nstarting point, and again after it has rolled through about 490\ndegrees. We take as our parameter $t$ the angle through which the\nwheel has turned, measured as shown clockwise from the line connecting\nthe center of the wheel to the ground. Since the radius is 1, the\ncenter of the wheel has coordinates $(t,1)$. We seek to write the\ncoordinates of the point on the rim as $(t+\\Delta x,1+\\Delta y)$,\nwhere $\\Delta x$ and $\\Delta y$ are as shown in figure~\\ref{fig:blow\n up of wheel}. These values are nearly the sine and cosine of the\nangle $t$, from the unit circle definition of sine and\ncosine. However, some care is required because we are measuring $t$\nfrom a nonstandard starting line and in a clockwise direction, as\nopposed to the usual counterclockwise direction. A bit of thought\nreveals that $\\Delta x=-\\sin t$ and $\\Delta y=-\\cos t$. Thus the\nparametric equations for the cycloid are $x=t-\\sin t$, $y=1-\\cos t$.\n\\end{solution}\n\n%\n%\\figure[!ht]\n%\\centerline{\\vbox{\\beginpicture\n%\\normalgraphs\n%%\\sevenpoint\n%\\setcoordinatesystem units <7truemm,7truemm>\n%\\setplotarea x from 0 to 20, y from 0 to 2.2\n%\\axis left shiftedto x=0 /\n%\\axis bottom shiftedto y=0 /\n%\\setquadratic\n%\\textRed\n%\\plot 0.000 0.000 0.001 0.018 0.009 0.070 0.030 0.156 0.069 0.271\n%0.133 0.412 0.226 0.574 0.351 0.751 0.510 0.937 0.704 1.125\n%0.934 1.309 1.197 1.482 1.491 1.637 1.813 1.771 2.157 1.876\n%2.518 1.951 2.891 1.992 3.267 1.998 3.642 1.969 4.007 1.905\n%4.358 1.809 4.687 1.685 4.991 1.536 5.265 1.368 5.506 1.187\n%5.712 1.000 5.883 0.813 6.019 0.632 6.122 0.464 6.195 0.315\n%6.243 0.191 6.269 0.095 6.281 0.031 6.283 0.002 6.284 0.008\n%6.288 0.049 6.304 0.124 6.337 0.229 6.392 0.363 6.475 0.518\n%6.589 0.691 6.736 0.875 6.919 1.063 7.137 1.249 7.389 1.426\n%7.673 1.588 7.986 1.729 8.323 1.844 8.680 1.930 9.049 1.982\n%9.425 2.000 9.801 1.982 10.170 1.930 10.526 1.844 10.863 1.729\n%11.176 1.588 11.461 1.426 11.713 1.249 11.931 1.063 12.113 0.875\n%12.261 0.691 12.375 0.518 12.457 0.363 12.513 0.229 12.545 0.124\n%12.561 0.049 12.566 0.008 12.566 0.002 12.569 0.031 12.580 0.095\n%12.607 0.191 12.654 0.315 12.727 0.464 12.830 0.632 12.966 0.813\n%13.137 1.000 13.343 1.187 13.584 1.368 13.858 1.536 14.162 1.685\n%14.492 1.809 14.842 1.905 15.208 1.969 15.582 1.998 15.959 1.992\n%16.331 1.951 16.692 1.876 17.037 1.771 17.358 1.637 17.652 1.482\n%17.916 1.309 18.145 1.125 18.340 0.937 18.499 0.751 18.623 0.574\n%18.716 0.412 18.780 0.271 18.820 0.156 18.841 0.070 18.848 0.018\n%18.850 0.000 /\n%\\textBlack\n%\\plot 8.500 0.800 8.442 0.799 8.383 0.816 8.329 0.850 8.286 0.900\n%8.258 0.962 8.248 1.033 8.259 1.105 8.291 1.174 8.343 1.233\n%8.412 1.277 8.493 1.299 8.579 1.298 8.664 1.272 8.740 1.221\n%8.800 1.150 8.839 1.061 8.851 0.963 8.835 0.863 8.791 0.769\n%8.720 0.690 8.627 0.632 8.520 0.602 8.406 0.604 8.294 0.638\n%8.195 0.704 8.116 0.797 8.066 0.910 8.049 1.036 8.068 1.164\n%8.124 1.283 /\n%\\multiput {$\\bullet$} at 0 0 7.70 1.602 /\n%\\put {$t$} [bl] <2pt,2pt> at 8.73 1.23\n%\\setlinear\n%\\plot 7.70 1.602 8.5 1 8.5 0 /\n%\\circulararc 360 degrees from 0 0 center at 0 1 \n%\\circulararc 360 degrees from 8.5 0 center at 8.5 1\n%\\endpicture}}\n%\\caption{{A cycloid.} \\label{fig:cycloid}}\n%\\endfigure\n\n\\figure[H]\n\\[\n\\includegraphics[width=4.5in]{images/7-cycloid} \n\\] \n\\caption{{A cycloid.} \\label{fig:cycloid}}\n\\endfigure\n\n\\figure[H]\n\\centerline{\\vbox{\\beginpicture\n\\normalgraphs\n%\\sevenpoint\n\\setcoordinatesystem units <15truemm,15truemm>\n\\setplotarea x from -1 to 1, y from 0 to 2\n\\multiput {$\\bullet$} at -0.8 1.602 /\n%\\plot 0.000 0.800 -0.058 0.799 -0.117 0.816 -0.171 0.850 -0.214 0.900\n%-0.242 0.962 -0.252 1.033 -0.241 1.105 -0.209 1.174 -0.157 1.233\n%-0.088 1.277 -0.007 1.299 0.079 1.298 0.164 1.272 0.240 1.221\n%0.300 1.150 0.339 1.061 0.351 0.963 0.335 0.863 0.291 0.769\n%0.220 0.690 0.127 0.632 0.020 0.602 -0.094 0.604 -0.206 0.638\n%-0.305 0.704 -0.384 0.797 -0.434 0.910 -0.451 1.036 -0.432 1.164\n%-0.376 1.283 /\n\\putrule from -0.8 1.602 to -0.8 1\n\\putrule from -0.8 1 to 0 1\n\\put {$\\Delta x$} [t] <0pt,-2pt> at -0.4 1\n\\put {$\\Delta y$} [l] <2pt,0pt> at -0.8 1.2\n\\setlinear\n\\plot -0.8 1.602 0 1 0 0 /\n\\circulararc 360 degrees from 0 0 center at 0 1\n\\endpicture}}\n\\caption{{The wheel.} \\label{fig:blow up of wheel}}\n\\endfigure\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\Opensolutionfile{solutions}[ex]\n\\section*{Exercises for \\ref{sec:Parametric Equations}}\n\n\\begin{enumialphparenastyle}\n\n%%%%%%%%%%\n\\begin{ex}\n What curve is described by $\\ds x=t^2$, $\\ds y=t^4$? If $t$ is\ninterpreted as time, describe how the object moves on the curve.\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n What curve is described by $x=3\\cos t$, $y=3\\sin t$? If $t$ is\ninterpreted as time, describe how the object moves on the curve.\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n What curve is described by $x=3\\cos t$, $y=2\\sin t$? If $t$ is\ninterpreted as time, describe how the object moves on the curve.\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n What curve is described by $x=3\\sin t$, $y=3\\cos t$? If $t$ is\ninterpreted as time, describe how the object moves on the curve.\n\\end{ex}\n\n%%%%%%%%%%\n\\begin{ex}\n Sketch the curve described by $\\ds x=t^3-t$, $\\ds y=t^2$. If $t$ is\ninterpreted as time, describe how the object moves on the curve.\n\\end{ex}\n\n%%%%%%%%%%%\n\\begin{ex}\\label{exer:pseudo cycloid}\n A wheel of radius 1 rolls along a straight line, say the\n$x$-axis. A point $P$ is located halfway between the center of the\nwheel and the rim; assume $P$ starts at the point $(0,1/2)$. As the\nwheel rolls, $P$ traces a curve; find parametric equations for the\ncurve.\n\\begin{sol}\n $\\ds x=t-{\\sin(t)\\over2}$, $\\ds t=1-{\\cos(t)\\over2}$\n\\end{sol}\n\\end{ex}\n\n%%%%%%%%%%%\n%\\\\begin{ex}\n% A wheel of radius 1 rolls around the outside of a circle of\n%radius 3. A point $P$ on the rim of the wheel traces out a curve\n%called a \\dfont{hypercycloid}, as indicated in\n%figure~\\ref{fig:hypercycloid and hypocycloid}. Assuming $P$\n%starts at the point $(3,0)$, find parametric equations for the curve.\n%\\begin{sol}\n% $x=4\\cos t-\\cos(4t)$,\\hfill\\break $y=4\\sin t-\\sin(4t)$\n%\\end{sol}\n%\\end{ex}\n%\\exrdef{exer:hypercycloid}\n%\n%\\figure\n%\\vbox{\\beginpicture\n%\\normalgraphs\n%\\sevenpoint\n%\\setcoordinatesystem units <7truemm,7truemm>\n%\\setplotarea x from -5 to 5, y from -5 to 5\n%\\axis left shiftedto x=0 /\n%\\axis bottom shiftedto y=0 /\n%\\multiput {$\\bullet$} at 2.81 4.12 3 0 /\n%\\circulararc 360 degrees from 2.81 4.12 center at 2.16 3.366\n%\\circulararc 360 degrees from 0 3 center at 0 0\n%\\circulararc 360 degrees from 5 0 center at 4 0\n%\\textRed\n%\\setquadratic\n%\\plot 3.000 0.000 3.024 0.002 3.092 0.020 3.200 0.065 3.339 0.150\n%3.495 0.285 3.656 0.474 3.807 0.721 3.931 1.022 4.015 1.373\n%4.045 1.763 4.012 2.182 3.908 2.613 3.730 3.041 3.479 3.450\n%3.160 3.824 2.781 4.148 2.353 4.410 1.890 4.602 1.410 4.717\n%0.927 4.755 0.459 4.719 0.021 4.614 -0.375 4.450 -0.717 4.241\n%-1.000 4.000 -1.220 3.743 -1.378 3.487 -1.478 3.245 -1.531 3.030\n%-1.545 2.853 -1.535 2.721 -1.516 2.637 -1.501 2.600 -1.506 2.607\n%-1.542 2.648 -1.620 2.714 -1.746 2.791 -1.924 2.864 -2.152 2.918\n%-2.427 2.939 -2.740 2.914 -3.079 2.832 -3.432 2.685 -3.782 2.471\n%-4.113 2.187 -4.410 1.839 -4.658 1.434 -4.845 0.983 -4.961 0.500\n%-5.000 0.000 -4.961 -0.500 -4.845 -0.983 -4.658 -1.434 -4.410 -1.839\n%-4.113 -2.187 -3.782 -2.471 -3.432 -2.685 -3.079 -2.832 -2.740 -2.914\n%-2.427 -2.939 -2.152 -2.918 -1.924 -2.864 -1.746 -2.791 -1.620 -2.714\n%-1.542 -2.648 -1.506 -2.607 -1.501 -2.600 -1.516 -2.637 -1.535 -2.721\n%-1.545 -2.853 -1.531 -3.030 -1.478 -3.245 -1.378 -3.487 -1.220 -3.743\n%-1.000 -4.000 -0.717 -4.241 -0.375 -4.450 0.021 -4.614 0.459 -4.719\n%0.927 -4.755 1.410 -4.717 1.890 -4.602 2.353 -4.410 2.781 -4.148\n%3.160 -3.824 3.479 -3.450 3.730 -3.041 3.908 -2.613 4.012 -2.182\n%4.045 -1.763 4.015 -1.373 3.931 -1.022 3.807 -0.721 3.656 -0.474\n%3.495 -0.285 3.339 -0.150 3.200 -0.065 3.092 -0.020 3.024 -0.002\n%3.000 0.000 /\n%\\textBlack\n%\\setcoordinatesystem units <7truemm,7truemm> point at -12 0\n%\\setplotarea x from -3.5 to 3.5, y from -3.5 to 3.5\n%\\axis left shiftedto x=0 /\n%\\axis bottom shiftedto y=0 /\n%\\multiput {$\\bullet$} at 3 0 /\n%%\\circulararc 360 degrees from 2.81 4.12 center at 2.16 3.366\n%\\circulararc 360 degrees from 0 3 center at 0 0\n%\\circulararc 360 degrees from 3 0 center at 2 0\n%\\textRed\n%\\setquadratic\n%\\plot 3.000 0.000 2.985 0.000 2.942 0.003 2.870 0.009 2.771 0.021\n%2.645 0.041 2.496 0.070 2.325 0.110 2.134 0.161 1.927 0.225\n%1.706 0.301 1.474 0.390 1.234 0.492 0.989 0.606 0.744 0.731\n%0.500 0.866 0.261 1.010 0.030 1.160 -0.191 1.314 -0.399 1.471\n%-0.592 1.628 -0.769 1.781 -0.928 1.929 -1.067 2.069 -1.187 2.197\n%-1.287 2.312 -1.367 2.410 -1.427 2.490 -1.469 2.549 -1.492 2.586\n%-1.500 2.598 -1.493 2.585 -1.473 2.546 -1.443 2.481 -1.404 2.389\n%-1.358 2.270 -1.309 2.127 -1.258 1.959 -1.207 1.768 -1.158 1.557\n%-1.113 1.327 -1.074 1.081 -1.043 0.823 -1.019 0.554 -1.005 0.279\n%-1.000 0.000 -1.005 -0.279 -1.019 -0.554 -1.043 -0.823 -1.074 -1.081\n%-1.113 -1.327 -1.158 -1.557 -1.207 -1.768 -1.258 -1.959 -1.309 -2.127\n%-1.358 -2.270 -1.404 -2.389 -1.443 -2.481 -1.473 -2.546 -1.493 -2.585\n%-1.500 -2.598 -1.492 -2.586 -1.469 -2.549 -1.427 -2.490 -1.367 -2.410\n%-1.287 -2.312 -1.187 -2.197 -1.067 -2.069 -0.928 -1.929 -0.769 -1.781\n%-0.592 -1.628 -0.399 -1.471 -0.191 -1.314 0.030 -1.160 0.261 -1.010\n%0.500 -0.866 0.744 -0.731 0.989 -0.606 1.234 -0.492 1.474 -0.390\n%1.706 -0.301 1.927 -0.225 2.134 -0.161 2.325 -0.110 2.496 -0.070\n%2.645 -0.041 2.771 -0.021 2.870 -0.009 2.942 -0.003 2.985 -0.000\n%3.000 0.000 /\n%\\textBlack\n%\\endpicture}\n%\\label{fig:hypercycloid and hypocycloid}\n%\\endfigure{A hypercycloid and a hypocycloid.}\n%\n%\n%%%%%%%%%%%\n%\\\\begin{ex}\n% A wheel of radius 1 rolls around the inside of \n%a circle of radius 3. A point $P$ on the rim of the wheel traces out a\n%curve called a \\dfont{hypocycloid}, as indicated in\n%figure~\\ref{fig:hypercycloid and hypocycloid}. Assuming $P$\n%starts at the point $(3,0)$, find parametric equations for the curve.\n%\\begin{sol}\n% $x=2\\cos t+\\cos(2t)$,\\hfill\\break $y=2\\sin t-\\sin(2t)$\n%\\end{sol}\n%\\end{ex}\n%\\exrdef{exer:hypocycloid}\n%\n%%%%%%%%%%%\n%\\\\begin{ex}\n% An \\dfont{involute\\index{involute}} of a circle is formed\n%as follows: Imagine that a long (that is, infinite) string is wound tightly\n%around a circle, and that you grasp the end of the string and begin to\n%unwind it, keeping the string taut. The end of the string traces out\n%the involute. Find parametric equations for this curve, using a circle\n%of radius 1, and assuming that the string unwinds counter-clockwise\n%and the end of the string is initially at $(1,0)$. \n%Figure~\\ref{fig:involute} shows part of the curve; the\n%dotted lines represent the string at a few different times.\n%\\begin{sol}\n% $x=\\cos t+t\\sin t$,\\hfill\\break $y=\\sin t-t\\cos t$\n%\\end{sol}\n%\\end{ex}\n%\\exrdef{exer:involute of a circle}\n%\n%\\figure\n%\\vbox{\\beginpicture\n%\\normalgraphs\n%\\sevenpoint\n%\\setcoordinatesystem units <7truemm,7truemm>\n%\\setplotarea x from -5 to 2, y from -4 to 3.5\n%\\axis left shiftedto x=0 /\n%\\axis bottom shiftedto y=0 /\n%\\circulararc 360 degrees from 0 1 center at 0 0\n%\\setquadratic\n%\\textRed\n%\\plot 1.000 0.000 1.002 0.000 1.008 0.001 1.018 0.002 1.031 0.005\n%1.048 0.010 1.069 0.018 1.092 0.028 1.118 0.041 1.147 0.058\n%1.178 0.079 1.211 0.105 1.245 0.135 1.280 0.170 1.315 0.210\n%1.350 0.255 1.385 0.306 1.418 0.362 1.449 0.423 1.478 0.490\n%1.504 0.563 1.527 0.640 1.545 0.723 1.559 0.811 1.568 0.903\n%1.571 1.000 1.568 1.101 1.558 1.205 1.541 1.312 1.516 1.422\n%1.484 1.534 1.443 1.647 1.393 1.761 1.335 1.875 1.268 1.989\n%1.191 2.102 1.105 2.212 1.010 2.320 0.905 2.425 0.791 2.526\n%0.668 2.621 0.536 2.711 0.395 2.794 0.246 2.870 0.088 2.939\n%-0.077 2.998 -0.250 3.048 -0.429 3.088 -0.614 3.117 -0.805 3.135\n%-1.000 3.142 -1.199 3.135 -1.402 3.116 -1.606 3.084 -1.812 3.038\n%-2.019 2.978 -2.225 2.903 -2.430 2.815 -2.632 2.712 -2.831 2.594\n%-3.025 2.462 -3.214 2.316 -3.396 2.155 -3.570 1.981 -3.736 1.793\n%-3.892 1.592 -4.037 1.378 -4.171 1.152 -4.292 0.914 -4.399 0.666\n%-4.492 0.408 -4.570 0.141 -4.631 -0.135 -4.676 -0.417 -4.703 -0.706\n%-4.712 -1.000 -4.703 -1.298 -4.675 -1.598 -4.627 -1.901 -4.559 -2.203\n%-4.472 -2.504 -4.364 -2.803 -4.236 -3.099 -4.088 -3.389 -3.920 -3.672 /\n%%-3.733 -3.948 \n%%-3.526 -4.215 -3.300 -4.471 -3.056 -4.715 -2.794 -4.946\n%%-2.515 -5.163 -2.219 -5.363 -1.908 -5.547 -1.583 -5.713 -1.244 -5.860\n%%-0.893 -5.986 -0.531 -6.091 -0.160 -6.174 0.220 -6.234 0.607 -6.271\n%%1.000 -6.283 /\n%\\textBlack\n%\\setlinear\\setdashes <2pt>\n%\\plot 0.54 0.84 1.38 0.3 /\n%\\plot -0.65 -0.757 -3.68 1.86 /\n%\\plot -0.42 0.91 1.4 1.74 /\n%\\endpicture}\n%\\label{fig:involute}\n%\\endfigure{An involute of a circle.}\n%\n%\\end%%%%%%%%%%\n%\\\\begin{ex}\n%s\n\n\\end{enumialphparenastyle}\n", "meta": {"hexsha": "e2812ccbebab412f43253f9097db3d9a0f0e244f", "size": 14893, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "11-polar-coord-parametric-eq/11-4-parametric-equations.tex", "max_stars_repo_name": "TimAlderson/OpenCalc", "max_stars_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "11-polar-coord-parametric-eq/11-4-parametric-equations.tex", "max_issues_repo_name": "TimAlderson/OpenCalc", "max_issues_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "11-polar-coord-parametric-eq/11-4-parametric-equations.tex", "max_forks_repo_name": "TimAlderson/OpenCalc", "max_forks_repo_head_hexsha": "7d0110b6bc4ba42a6b911729420e1406296d6964", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 41.9521126761, "max_line_length": 107, "alphanum_fraction": 0.6738736319, "num_tokens": 6694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.689305616785446, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.6115551158543979}} {"text": "\\documentclass[a4paper, 12pt]{article}\n\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath}\n\\usepackage[]{amsfonts}\n\\usepackage[]{graphicx}\n\n\\title{CS231A Course Notes 1: Camera Models}\n\\author{Kenji Hata and Silvio Savarese}\n\\date{}\n\n\\renewcommand\\emph{\\textbf}\n\n\\begin{document}\n\n\\maketitle\n\n\\section{Introduction}\nThe camera is one of the most essential tools in computer vision. It is the mechanism by which we can record the world around us and use its output - photographs - for various applications. Therefore, one question we must ask in introductory computer vision is: how do we model a camera?\n\n\\section{Pinhole cameras}\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-1.pdf}\n\\caption{A simple working camera model: the pinhole camera model.}\n\\label{fig:simpleCamera}\n\\end{figure}\nLet's design a simple camera system -- a system that can record an image of an object or scene in the 3D world. This camera system can be designed by placing a barrier with a small aperture between the 3D object and a photographic film or sensor. As Figure~\\ref{fig:simpleCamera} shows, each point on the 3D object emits multiple rays of light outwards. Without a barrier in place, every point on the film will be influenced by light rays emitted from every point on the 3D object. Due to the barrier, only one (or a few) of these rays of light passes through the aperture and hits the film. Therefore, we can establish a one-to-one mapping between points on the 3D object and the film. The result is that the film gets exposed by an ``image\" of the 3D object by means of this mapping. This simple model is known as the \\emph{pinhole camera model}.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-2.pdf}\n\\caption{A formal construction of the pinhole camera model.}\n\\label{fig:pinholeCamera}\n\\end{figure}\nA more formal construction of the pinhole camera is shown in Figure~\\ref{fig:pinholeCamera}. In this construction, the film is commonly called the \\emph{image or retinal plane}. The aperture is referred to as the \\emph{pinhole} $O$ or \\emph{center of the camera}. The distance between the image plane and the pinhole $O$ is the \\emph{focal length} $f$. Sometimes, the retinal plane is placed between $O$ and the 3D object at a distance $f$ from $O$. In this case, it is called the \\emph{virtual image} or \\emph{virtual retinal plane}. Note that the projection of the object in the image plane and the image of the object in the virtual image plane are identical up to a scale (similarity) transformation.\n\nNow, how do we use pinhole cameras? Let $P = \\begin{bmatrix}x & y & z\\end{bmatrix}^T$ be a point on some 3D object visible to the pinhole camera. $P$ will be mapped or \\textbf{projected} onto the image plane $\\Pi'$, resulting in point\\footnote{Throughout the course notes, let the prime superscript (e.g. $P'$) indicate that this point is a projected or complementary point to the non-superscript version. For example, $P'$ is the projected version of $P$.} $P' = \\begin{bmatrix}x' & y'\\end{bmatrix}^T$. Similarly, the pinhole itself can be projected onto the image plane, giving a new point $C'$. \n\nHere, we can define a coordinate system $\\begin{bmatrix}i & j & k\\end{bmatrix}$ centered at the pinhole $O$ such that the axis $k$ is perpendicular to the image plane and points toward it. This coordinate system is often known as the \\emph{camera reference system} or \\emph{camera coordinate system}. The line defined by $C'$ and $O$ is called the \\emph{optical axis} of the camera system.\n\nRecall that point $P'$ is derived from the projection of 3D point $P$ on the image plane $\\Pi'$. Therefore, if we derive the relationship between 3D point $P$ and image plane point $P'$, we can understand how the 3D world imprints itself upon the image taken by a pinhole camera. Notice that triangle $P'C'O$ is similar to the triangle formed by $P$, $O$ and $(0,0,z)$. Therefore, using the law of similar triangles we find that:\n\n\\begin{equation}\n P' = \\begin{bmatrix}x'&y'\\end{bmatrix}^T = \\begin{bmatrix}f\\frac{x}{z} & f\\frac{y}{z}\\end{bmatrix}^T\n\\end{equation}\n\nNotice that one large assumption we make in this pinhole model is that the aperture is a single point. In most real world scenarios, however, we cannot assume the aperture can be infinitely small. Thus, what is the effect of varying aperture size? \n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-3.pdf}\n\\caption{The effects of aperture size on the image. As the aperture size decreases, the image gets sharper, but darker.}\n\\label{fig:apertureSize}\n\\end{figure}\n\nAs the aperture size increases, the number of light rays that passes through the barrier consequently increases. With more light rays passing through, then each point on the film may be affected by light rays from multiple points in 3D space, blurring the image. Although we may be inclined to try to make the aperture as small as possible, recall that a smaller aperture size causes less light rays to pass through, resulting in crisper but darker images. Therefore, we arrive at the fundamental problem presented by the pinhole formulation: can we develop cameras that take crisp and bright images?\n\n\\section{Cameras and lenses}\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-4.pdf}\n\\caption{A setup of a simple lens model. Notice how the rays of the top point on the tree converge nicely on the film. However, a point at a different distance away from the lens results in rays not converging perfectly on the film.}\n\\label{fig:lens}\n\\end{figure}\n\nIn modern cameras, the above conflict between crispness and brightness is mitigated by using \\emph{lenses}, devices that can focus or disperse light. If we replace the pinhole with a lens that is both properly placed and sized, then it satisfies the following property: all rays of light that are emitted by some point $P$ are refracted by the lens such that they converge to a single point $P'$ in the image plane. Therefore, the problem of the majority of the light rays blocked due to a small aperture is removed (Figure~\\ref{fig:lens}). However, please note that this property does not hold for all 3D points, but only for some specific point $P$. Take another point $Q$ which is closer or further from the image plane than $P$. The corresponding projection into the image will be blurred or out of focus. Thus, lenses have a specific distance for which objects are ``in focus\". This property is also related to a photography and computer graphics concept known as depth of field, which is the effective range at which cameras can take clear photos.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-5.pdf}\n\\caption{Lenses focus light rays parallel to the optical axis into the focal point. Furthermore, this setup illustrates the paraxial refraction model, which helps us find the relationship between points in the image plane and the 3D world in cameras with lenses.}\n\\label{fig:paraxial}\n\\end{figure}\n\nCamera lenses have another interesting property: they focus all light rays traveling parallel to the optical axis to one point known as the \\emph{focal point} (Figure~\\ref{fig:paraxial}). The distance between the focal point and the center of the lens is commonly referred to as the \\emph{focal length} $f$. Furthermore, light rays passing through the center of the lens are not deviated. We thus can arrive at a similar construction to the pinhole model that relates a point $P$ in 3D space with its corresponding point $P'$ in the image plane. \n\n\\begin{equation}\n P' = \\begin{bmatrix}x'\\\\ y'\\end{bmatrix} = \\begin{bmatrix}z'\\frac{x}{z} \\\\ z'\\frac{y}{z}\\end{bmatrix}\n\\end{equation}\n\nThe derivation for this model is outside the scope of the class. However, please notice that in the pinhole model $z' = f$, while in this lens-based model, $z' = f+z_0$. Additionally, since this derivation takes advantage of the paraxial or ``thin lens\" assumption\\footnote{For the angle $\\theta$ that incoming light rays make with the optical axis of the lens, the paraxial assumption substitutes $\\theta$ for any place $\\sin(\\theta)$ is used. This approximation of $\\theta$ for $\\sin\\theta$ holds as $\\theta$ approaches 0.}, it is called the \\textbf{paraxial refraction model}.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-6.pdf}\n\\caption{Demonstrating how pincushion and barrel distortions affect images.}\n\\label{fig:distortion}\n\\end{figure}\n\nBecause the paraxial refraction model approximates using the thin lens assumption, a number of aberrations can occur. The most common one is referred to as \\emph{radial distortion}, which causes the image magnification to decrease or increase as a function of the distance to the optical axis. We classify the radial distortion as \\emph{pincushion distortion} when the magnification increases and \\emph{barrel distortion}\\footnote{Barrel distortion typically occurs when one uses fish-eye lenses.} when the magnification decreases. Radial distortion is caused by the fact that different portions of the lens have differing focal lengths.\n\n\\section{Going to digital image space}\nIn this section, we will discuss the details of the parameters we must account for when modeling the projection from 3D space to the digital images we know. All the results derived will use the pinhole model, but they also hold for the paraxial refraction model.\n\nAs discussed earlier, a point $P$ in 3D space can be mapped (or projected) into a 2D point $P'$ in the image plane $\\Pi'$. This $\\mathbb{R}^3 \\rightarrow \\mathbb{R}^2$ mapping is referred to as a \\emph{projective transformation}. This projection of 3D points into the image plane does not directly correspond to what we see in actual digital images for several reasons. First, points in the digital images are, in general, in a different reference system than those in the image plane. Second, digital images are divided into discrete pixels, whereas points in the image plane are continuous. Finally, the physical sensors can introduce non-linearity such as distortion to the mapping. To account for these differences, we will introduce a number of additional transformations that allow us to map any point from the 3D world to pixel coordinates.\n\n\\subsection{The Camera Matrix Model and Homogeneous Coordinates}\n\n\\subsubsection{Introduction to the Camera Matrix Model}\nThe camera matrix model describes a set of important parameters that affect how a world point $P$ is mapped to image coordinates $P'$. As the name suggests, these parameters will be represented in matrix form. First, let's introduce some of those parameters.\n\nThe first parameters, $c_x$ and $c_y$, describe how image plane and digital image coordinates can differ by a translation. Image plane coordinates have their origin $C'$ at the image center where the $k$ axis intersects the image plane. On the other hand, digital image coordinates typically have their origin at the lower-left corner of the image. Thus, 2D points in the image plane and 2D points in the image are offset by a translation vector $\\begin{bmatrix}c_x, c_y\\end{bmatrix}^T$. To accommodate this change of coordinate systems, the mapping now becomes:\n\n\\begin{equation}\n P' = \\begin{bmatrix}x'\\\\y'\\end{bmatrix} = \\begin{bmatrix}f\\frac{x}{z}+c_x \\\\ f\\frac{y}{z}+c_y\\end{bmatrix}\n\\end{equation}\n\nThe second effect we must account for that the points in digital images are expressed in pixels, while points in image plane are represented in physical measurements (e.g. centimeters). In order to accommodate this change of units, we must introduce two new parameters $k$ and $l$. These parameters, whose units would be something like $\\frac{\\mathrm{pixels}}{\\mathrm{cm}}$, correspond to the change of units in the two axes of the image plane. Note that $k$ and $l$ may be different because the aspect ratio of a pixel is not guaranteed to be one. If $k=l$, we often say that the camera has \\emph{square pixels}. We adjust our previous mapping to be\n\n\\begin{equation}\n P' = \\begin{bmatrix}x'\\\\y'\\end{bmatrix} = \\begin{bmatrix}fk\\frac{x}{z}+c_x \\\\ fl\\frac{y}{z}+c_y\\end{bmatrix} = \\begin{bmatrix}\\alpha\\frac{x}{z}+c_x \\\\ \\beta\\frac{y}{z}+c_y\\end{bmatrix}\n \\label{eq:finalImage}\n\\end{equation}\n\nIs there a better way to represent this projection from $P\\rightarrow P'$? If this projection is a linear transformation, then it can be represented as a product of a matrix and the input vector (in this case, it would be $P$. However, from Equation~\\ref{eq:finalImage}, we see that this projection $P\\rightarrow P'$ is not linear, as the operation divides one of the input parameters (namely $z$). Still, representing this projection as a matrix-vector product would be useful for future derivations. Therefore, can we represent our transformation as a matrix-vector product despite its nonlinearity? Homogeneous coordinates are the solution.\n\n\\subsubsection{Homogeneous Coordinates}\n\nOne way to solve this problem is to change the coordinate systems. For example, we introduce a new coordinate, such that any point $P' =(x',y')$ becomes $(x',y',1)$. Similarly, any point $P =(x,y,z)$ becomes $(x,y,z,1)$. This augmented space is referred to as the \\emph{homogeneous coordinate system}. As demonstrated previously, to convert a Euclidean vector $(v_1,...,v_n)$ to homogeneous coordinates, we simply append a 1 in a new dimension to get $(v_1,...,v_n,1)$. Note that the equality between a vector and its homogeneous coordinates only occurs when the final coordinate equals one. Therefore, when converting back from arbitrary homogeneous coordinates $(v_1, ... , v_n , w)$, we get Euclidean coordinates $(\\frac{v_1}{w},...,\\frac{v_n}{w})$. Using homogeneous coordinates, we can formulate\n\\begin{equation}\n P_h' = \\begin{bmatrix}\\alpha x + c_xz\\\\\\beta y + c_yz \\\\ z\\end{bmatrix} = \n \\begin{bmatrix}\n \\alpha & 0 & c_x & 0\\\\\n 0 & \\beta & c_y & 0 \\\\ \n 0 & 0 & 1 & 0\n \\end{bmatrix}\n \\begin{bmatrix}x\\\\y\\\\z\\\\1\\end{bmatrix} = \\begin{bmatrix}\n \\alpha & 0 & c_x & 0\\\\\n 0 & \\beta & c_y & 0 \\\\ \n 0 & 0 & 1 & 0\n \\end{bmatrix} P_h\n \\label{eq:homogeneous}\n\\end{equation}\n\nFrom this point on, assume that we will work in homogeneous coordinates, unless stated otherwise. We will drop the $h$ index, so any point $P$ or $P'$ can be assumed to be in homogeneous coordinates. As seen from Equation~\\ref{eq:homogeneous}, we can represent the relationship between a point in 3D space and its image coordinates by a matrix vector relationship:\n\\begin{equation}\n P' = \\begin{bmatrix}x'\\\\y'\\\\ z\\end{bmatrix}=\\begin{bmatrix}\n \\alpha & 0 & c_x & 0\\\\\n 0 & \\beta & c_y & 0 \\\\ \n 0 & 0 & 1 & 0\n \\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\\\1\\end{bmatrix}=\n \\begin{bmatrix}\n \\alpha & 0 & c_x & 0\\\\\n 0 & \\beta & c_y & 0 \\\\ \n 0 & 0 & 1 & 0\n \\end{bmatrix}P = MP \n \\label{eq:canonical}\n % = K\\begin{bmatrix} I&0\\end{bmatrix}P\n\\end{equation}\nWe can decompose this transformation a bit further into \n\\begin{equation}\nP' = MP = \\begin{bmatrix}\n \\alpha & 0 & c_x \\\\\n 0 & \\beta & c_y \\\\ \n 0 & 0 & 1 \n \\end{bmatrix}\\begin{bmatrix}I & 0\\end{bmatrix}P = K\\begin{bmatrix}I & 0\\end{bmatrix}P\n \\label{eq:decomposedHomogenousTransform}\n\\end{equation}\n\nThe matrix $K$ is often referred to as the \\emph{camera matrix}. \n\n\\subsubsection{The Complete Camera Matrix Model}\n\nThe camera matrix $K$ contains some of the critical parameters that describes a camera's characteristics and its model, including the $c_x, c_y, k,$ and $l$ parameters as discussed above. Two parameters are currently missing this formulation: \\emph{skewness} and \\emph{distortion}. We often say that an image is skewed when the camera coordinate system is skewed, meaning that the angle between the two axes is slightly larger or smaller than 90 degrees. Most cameras have zero-skew, but some degree of skewness may occur because of sensor manufacturing errors. Deriving the new camera matrix accounting for skewness is outside the scope of this class and we give it to you below:\n\\begin{equation}\nK = \\begin{bmatrix}x'\\\\y'\\\\ z\\end{bmatrix}=\\begin{bmatrix}\n \\alpha & -\\alpha\\cot\\theta & c_x \\\\\n 0 & \\frac{\\beta}{\\sin\\theta} & c_y \\\\ \n 0 & 0 & 1 \n \\end{bmatrix}\n\\end{equation}\nMost methods that we introduce in this class ignore distortion effects, therefore our class camera matrix $K$ has 5 degrees of freedom: 2 for focal length, 2 for offset, and 1 for skewness. These parameters are collectively known as the \\emph{intrinsic parameters}, as they are unique and inherent to a given camera and relate to essential properties of the camera, such as its manufacturing.\n\n\\subsection{Extrinsic Parameters}\n\nSo far, we have described a mapping between a point $P$ in the 3D camera reference system to a point $P'$ in the 2D image plane using the intrinsic parameters of a camera described in matrix form. But what if the information about the 3D world is available in a different coordinate system? Then, we need to include an additional transformation that relates points from the world reference system to the camera reference system. This transformation is captured by a rotation matrix $R$ and translation vector $T$. Therefore, given a point in a world reference system $P_w$, we can compute its camera coordinates as follows:\n\\begin{equation}\nP = \\begin{bmatrix}R&T\\\\0&1\\end{bmatrix} P_w\n\\end{equation}\nSubstituting this in equation (\\ref{eq:decomposedHomogenousTransform}) and simplifying gives\n\\begin{equation}\nP' = K\\begin{bmatrix}R&T\\end{bmatrix}P_w = MP_w\n\\label{eq:cameramatrix}\n\\end{equation}\n\nThese parameters $R$ and $T$ are known as the \\emph{extrinsic parameters} because they are external to and do not depend on the camera.\n\n\\vspace{12pt}\n\nThis completes the mapping from a 3D point $P$ in an arbitrary world reference system to the image plane. To reiterate, we see that the full projection matrix $M$ consists of the two types of parameters introduced above: \\emph{intrinsic} and \\emph{extrinsic} parameters. All parameters contained in the camera matrix $K$ are the intrinsic parameters, which change as the type of camera changes. The extrinsic paramters include the rotation and translation, which do not depend on the camera's build. Overall, we find that the $3\\times4$ projection matrix $M$ has 11 degrees of freedom: 5 from the intrinsic camera matrix, 3 from extrinsic rotation, and 3 from extrinsic translation.\n\n\\section{Camera Calibration}\nTo precisely know the transformation from the real, 3D world into digital images requires prior knowledge of many of the camera's intrinsic parameters. If given an arbitrary camera, we may or may not have access to these parameters. We do, however, have access to the images the camera takes. Therefore, can we find a way to deduce them from images? This problem of estimating the extrinsic and intrinsic camera parameters is known as \\emph{camera calibration}.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/1-7.pdf}\n\\caption{The setup of an example calibration rig.}\n\\label{fig:calibration}\n\\end{figure}\n\nSpecifically, we do this by solving for the intrinsic camera matrix $K$ and the extrinsic parameters $R,T$ from Equation~\\ref{eq:cameramatrix}. We can describe this problem in the context of a calibration rig, such as the one show in Figure~\\ref{fig:calibration}. The rig usually consists of a simple pattern (i.e. checkerboard) with known dimensions. Furthermore, the rig defines our world reference frame with origin $O_w$ and axes $i_w, j_w, k_w$. From the rig's known pattern, we have known points in the world reference frame $P_1,...,P_n$. Finding these points in the image we take from the camera gives corresponding points in the image $p_1,...,p_n$. \n\nWe set up a linear system of equations from $n$ correspondences such that for each correspondence $P_i, p_i$ and camera matrix $M$ whose rows are $m_1, m_2, m_3$:\n\\begin{equation}\np_i = \\begin{bmatrix}u_i\\\\v_i\\end{bmatrix} = MP_i = \\begin{bmatrix}\\frac{m_1 P_i}{m_3P_i}\\\\\\frac{m_2 P_i}{m_3P_i}\\end{bmatrix}\n\\end{equation}\n\nAs we see from the above equation, each correspondence gives us two equations and, consequently, two constraints for solving the unknown parameters contained in $m$. From before, we know that the camera matrix has $11$ unknown parameters. This means that we need at least $6$ correspondences to solve this. However, in the real world, we often use more, as our measurements are often noisy. To explicitly see this, we can derive a pair of equations that relate $u_i$ and $v_i$ with $P_i$. \n\\begin{align*}\nu_i(m_3P_i) - m_1P_i = 0\\\\\nv_i(m_3P_i) - m_2P_i = 0\n\\end{align*}\n\nGiven $n$ of these corresponding points, the entire linear system of equations becomes\n\\begin{align*}\nu_1(m_3P_1) -& m_1P_1 = 0\\\\\nv_1(m_3P_1) -& m_2P_1 = 0\\\\\n&\\vdots\\\\\nu_n(m_3P_n) -& m_1P_n = 0\\\\\nv_n(m_3P_n) -& m_2P_n = 0\\\\\n\\end{align*}\n\nThis can be formatted as a matrix-vector product shown below:\n\\begin{equation}\n\\begin{bmatrix} \nP_1^T & 0^T & -u_1P_1^T \\\\\n0^T & P_1^T & -v_1P_1^T \\\\\n& \\vdots &\\\\\nP_n^T & 0^T & -u_nP_n^T \\\\\n0^T & P_n^T & -v_nP_n^T \n\\end{bmatrix}\n\\begin{bmatrix}\nm_1^T \\\\ m_2^T \\\\m_3^T\n\\end{bmatrix} = \\mathbf{P}m = 0\n\\label{eq:linearsystem}\n\\end{equation}\n\nWhen $2n > 11$, our homogeneous linear system is overdetermined. For such a system $m=0$ is always a trivial solution. Furthemore, even if there were some other $m$ that were a nonzero solution, then $\\forall k\\in \\mathbb{R},km$ is also a solution. Therefore, to constrain our solution, we complete the following minimization:\n\\begin{equation}\n\\begin{aligned}\n & \\underset{m}{\\text{minimize}}\n & & \\|\\mathbf{P}m\\|^2 \\\\\n & \\text{subject to}\n & & \\|m\\|^2 = 1\n\\end{aligned}\n\\end{equation}\nTo solve this minimization problem, we simply use singular value decomposition. If we let $P = UDV^T$, then the solution to the above minimization is to set $m$ equal to the last column of $V$. The derivation for this solution is outside the scope of this class and you may refer to Section 5.3 of Hartley \\& Zisserman on pages 592-593 for more details.\n\nAfter reformatting the vector $m$ into the matrix $M$, we now want to explicitly solve for the extrinsic and intrinsic parameters. We know our SVD-solved $M$ is known up to scale, which means that the true values of the camera matrix are some scalar multiple of $M$:\n\\begin{equation}\n\\rho M = \\begin{bmatrix}\n\\alpha r_1^T - \\alpha\\cot \\theta r_2^T + c_xr_3^T & \\alpha t_x - \\alpha \\cot \\theta t_y + c_x t_z \\\\\n\\frac{\\beta}{\\sin\\theta}r_2^T + c_yr_3^T & \\frac{\\beta}{\\sin\\theta}t_y + c_yt_z \\\\ r_3^T & t_z\n\\end{bmatrix}\n\\end{equation}\n\nHere, $r_1^T$, $r_2^T$, and $r_3^T$ are the three rows of $R$. Dividing by the scaling parameter gives\n\\[M = \\frac{1}{\\rho}\\begin{bmatrix}\n\\alpha r_1^T - \\alpha\\cot \\theta r_2^T + c_xr_3^T & \\alpha t_x - \\alpha \\cot \\theta t_y + c_x t_z \\\\\n\\frac{\\beta}{\\sin\\theta}r_2^T + c_yr_3^T & \\frac{\\beta}{\\sin\\theta}t_y + c_yt_z \\\\ r_3^T & t_z\n\\end{bmatrix} = \\begin{bmatrix}A & b\\end{bmatrix} =\\begin{bmatrix}\na_1^T \\\\ a_2^T \\\\ a_3^T\n\\end{bmatrix} \\begin{bmatrix}\nb_1\\\\b_2\\\\b_3\n\\end{bmatrix}\n\\]\n\nSolving for the intrinsics gives \n\\begin{equation}\\begin{aligned}\n \\rho &= \\pm \\frac{1}{\\|a_3\\|}\\\\\n c_x &= \\rho^2(a_1\\cdot a_3)\\\\\n c_y &=\\rho^2(a_2\\cdot a_3)\\\\\n \\theta &= \\cos ^{-1} \\left(-\\frac{(a_1 \\times a_3)\\cdot(a_2\\times a_3)}{\\|a_1\\times a_3\\|\\cdot\\|a_2\\times a_3\\|}\\right)\\\\\n \\alpha &= \\rho^2 \\|a_1 \\times a_3\\| \\sin \\theta\\\\\n \\beta &= \\rho^2 \\|a_2 \\times a_3\\| \\sin \\theta\n\\end{aligned}\\end{equation}\n\nThe extrinsics are \n\\begin{equation}\\begin{aligned}\n r_1 &= \\frac{a_2\\times a_3}{\\|a_2\\times a_3\\|}\\\\\n r_2 &= r_3\\times r_1\\\\\n r_3 &= \\rho a_3\\\\\n T &= \\rho K^{-1} b\n\\end{aligned}\\end{equation}\nWe leave the derivations as a class exercise or you can refer to Section 5.3.1 of the Forsyth \\& Ponce textbook.\n\nWith the calibration procedure complete, we warn against degenerate cases. Not all sets of $n$ correspondences will work. For example, if the points $P_i$ lie on the same plane, then the system will not be able to be solved. These unsolvable configurations of points are known as \\emph{degenerate configurations}. More generally, degenerate configurations have points that lie on the intersection curve of two quadric surfaces. Although this outside the scope of the class, you can find more information in Section 1.3 of the Forsyth \\& Ponce textbook. \n\n\\section{Handling Distortion in Camera Calibration}\nSo far, we have been working with ideal lenses which are free from any distortion. However, as seen before, real lenses can deviate from rectilinear projection, which require more advanced methods. This section provides just a brief introduction to handling distortions. \n\nOften, distortions are radially symmetric because of the physical symmetry of the lens. We model the radial distortion with an isotropic transformation:\n\\begin{equation}\n QP_i = \\begin{bmatrix}\n \\frac{1}{\\lambda}&0&0\\\\0 & \\frac{1}{\\lambda} & 0 \\\\ 0 & 0 &1\\end{bmatrix} M P_i = \\begin{bmatrix}\n u_i\\\\v_i\\end{bmatrix}\n = p_i\n\\end{equation}\nIf we try to rewrite this into a system of equations as before, we get \n\n\\begin{align*}\n u_iq_3P_i = q_1P_i\\\\\n v_iq_3P_i = q_2P_i\n\\end{align*}\n\nThis system, however, is no longer linear, and we require the use of nonlinear optimization techniques, which are covered in Section 22.2 of Forsyth \\& Ponce. We can simplify the nonlinear optimization of the calibration problem if we make certain assumptions. In radial distortion, we note that the ratio between two coordinates $u_i$ and $v_i$ is not affected. We can compute this ratio as\n\\begin{equation}\n \\frac{u_i}{v_i} = \\frac{\\frac{m_1P_i}{m_3P_i}}{\\frac{m_2P_i}{m_3P_i}} = \\frac{m_1P_i}{m_2P_i}\n\\end{equation}\n\nAssuming that $n$ correspondences are available, we can set up the system of linear equations:\n\\begin{align*}\n v_1(m_1P_1)-&u_1(m_2P_1) = 0\\\\\n &\\vdots\\\\\n v_n(m_1P_n)-&u_n(m_2P_n) = 0\n\\end{align*}\nSimilar to before, this gives a matrix-vector product that we can solve via SVD:\n\\begin{equation}\n Ln = \\begin{bmatrix}\n v_1P_1^T & -u_1P_1^T\\\\\n \\vdots & \\vdots\\\\\n v_nP_n^T & -u_nP_n^T\\\\ \n \\end{bmatrix}\n \\begin{bmatrix}\n m_1^T\\\\m_2^T\n \\end{bmatrix}\n\\end{equation}\n\nOnce $m_1$ and $m_2$ are estimated, $m_3$ can be expressed as a nonlinear function of $m_1,m_2,$ and $\\lambda$. This requires to solve a nonlinear optimization problem whose complexity is much simpler than the original one.\n\n\\section{Appendix A: Rigid Transformations}\nThe basic rigid transformations are rotation, translation, and scaling. This appendix will cover them for the 3D case, as they are common type in this class. \n\nRotating a point in 3D space can be represented by rotating around each of the three coordinate axes respectively. When rotating around the coordinate axes, common convention is to rotate in a counter-clockwise direction. One intuitive way to think of rotations is how much we rotate around each degree of freedom, which is often referred to as \\emph{Euler angles}. However, this methodology can result in what is known as \\emph{singularities}, or \\emph{gimbal lock}, in which certain configurations result in a loss of a degree of freedom for the rotation. \n\nOne way to prevent this is to use rotation matrices, which are a more general form of representing rotations. Rotation matrices are square, orthogonal matrices with determinant one. Given a rotation matrix $R$ and a vector $v$, we can compute the resulting vector $v'$ as\n\\[v' = Rv\\]\n\nSince rotation matrices are a very general representation of matrices, we can represent a rotation $\\alpha, \\beta, \\gamma$ around each of the respective axes as follows:\n\\[R_x(\\alpha) = \\begin{bmatrix}1 & 0 & 0\\\\ 0 &\\cos\\alpha &-\\sin\\alpha \\\\ 0 &\\sin\\alpha &\\cos\\alpha \\end{bmatrix}\\]\n\\[R_y(\\beta) = \\begin{bmatrix}\\cos\\beta & 0 & \\sin\\beta \\\\ 0 & 1 & 0 \\\\ -\\sin\\beta & 0 &\\cos\\beta \\end{bmatrix}\\]\n\\[R_z(\\gamma) = \\begin{bmatrix}\\cos \\gamma & -\\sin\\gamma &0 \\\\ \\sin\\gamma &\\cos\\gamma & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}\\]\n\nDue to the convention of matrix multiplication, the rotation achieved by first rotating around the z-axis, then y-axis, then x-axis is given by the matrix product $R_xR_yR_z$.\n\nTranslations, or displacements, are used to describe the movement in a certain direction. In 3D space, we define a translation vector $t$ with 3 values: the displacements in each of the 3 axes, often denoted as $t_x, t_y, t_z$. Thus, given some point $P$ which is translated to some other point $P'$ by $t$, we can write it as:\n\\[ P' = P + t = \\begin{bmatrix}P_x\\\\P_y\\\\P_z\\end{bmatrix} + \\begin{bmatrix}t_x\\\\t_y\\\\t_z\\end{bmatrix}\\]\n\nIn matrix form, translations can be written using homogeneous coordinates. If we construct a translation matrix as\n\\[ T = \\begin{bmatrix}\n1 & 0 & 0 & t_x \\\\0 & 1 & 0 & t_y \\\\0 & 0 & 1 & t_z \\\\ 0 & 0 & 0 & 1\n\\end{bmatrix}\\]\nthen we see that $P'=TP$ is equivalent to $P' = P+t$.\n\n\nIf we want to combine translation with our rotation matrix multiplication, we can again use homogeneous coordinates to our advantage. If we want to rotate a vector $v$ by $R$ and then translate it by $t$, we can write the resulting vector $v'$ as:\n\\[ \\begin{bmatrix}v'\\\\1\\end{bmatrix} = \\begin{bmatrix}\nR & t \\\\ 0 & 1\n\\end{bmatrix}\\begin{bmatrix}\nv \\\\ 1\n\\end{bmatrix}\\]\n\nFinally, if we want to scale the vector in certain directions by some amount $S_x, S_y, S_z$, we can construct a scaling matrix \n\\[S = \\begin{bmatrix}\nS_x & 0 & 0 \\\\ 0 & S_y & 0 \\\\ 0 & 0 & S_z\n\\end{bmatrix}\\]\n\nTherefore, if we want to scale a vector, then rotate, then translate, our final transformation matrix would be:\n\\[T = \\begin{bmatrix}\nRS & t \\\\ 0 & 1\n\\end{bmatrix}\\]\n\nNote that all of these types of transformations would be examples of affine transformations. Recall that projective transformations occur when the final row of $T$ is not $\\begin{bmatrix}\n0 & 0 &0 &1\n\\end{bmatrix}$.\n\n\\section{Appendix B: Different Camera Models}\nWe will now describe a simple model known as the \\emph{weak perspective model}. In the weak perspective model, points are first projected to the reference plane using orthogonal projection and then projected to the image plane using a projective transformation.\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/weak_perspective.png}\n\\caption{The weak perspective model: orthogonal projection onto reference plane}\n\\label{fig:weak_perspective}\n\\end{figure}\n\nAs Figure~\\ref{fig:weak_perspective} shows, given a reference plane $\\Pi$ at a distance $z_o$ from the center of the camera, the points $P,Q,R$ are first projected to the plane $\\Pi$ using an orthogonal projection, generating points $P\\_, Q\\_, R\\_$. This is a reasonable approximation when deviations in depth from the plane are small compared to the distance of the camera. \n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/weak_perspective2.png}\n\\caption{The weak perspective model: projection onto the image plane}\n\\label{fig:weak_perspective2}\n\\end{figure}\n\nFigure~\\ref{fig:weak_perspective2} illustrates how points $P\\_, Q\\_, R\\_$ are then projected to the image plane using a regular projective transformation to produce the points $p', q', r'$. Notice, however, that because we have approximated the depth of each point to $z_o$ the projection has been reduced to a simple, constant magnification. The magnification is equal to the focal length $f'$ divided by $z_o$, leading to\n\\begin{align*}\nx'=\\frac{f'}{z_0}x\\ \\ \\ \\ \\ \ny'=\\frac{f'}{z_0}y\n\\end{align*}\nThis model also simplifies the projection matrix\n\\[M = \\begin{bmatrix}\nA & b \\\\ 0 & 1\n\\end{bmatrix}\\]\nAs we see, the last row of $M$ is $\\begin{bmatrix}\n0 & 0& 0 &1\n\\end{bmatrix}$ in the weak perspective model, compared to $\\begin{bmatrix}\nv&1\n\\end{bmatrix}$ in the normal camera model. We do not prove this result and leave it to you as an exercise. The simplification is clearly demonstrated when mapping the 3D points to the image plane. \n\\begin{equation}\n P' = MP = \\begin{bmatrix}\n m_1 \\\\ m_2 \\\\ m_3\n \\end{bmatrix}P = \\begin{bmatrix}\n m_1P \\\\ m_2P \\\\ 1\n \\end{bmatrix}\n\\end{equation}\nThus, we see that the image plane point ultimately becomes a magnification of the original 3D point, irrespective of depth. The nonlinearity of the projective transformation disappears, making the weak perspective transformation a mere magnifier.\n\n\\begin{figure}[h!]\n\\centering\n\\includegraphics[width=0.8\\textwidth]{figures/orthographic.png}\n\\caption{The orthographic projection model}\n\\label{fig:orthographic}\n\\end{figure}\n\n\nFurther simplification leads to the \\emph{orthographic (or affine) projection model}. In this case, the optical center is located at infinity. The projection rays are now perpendicular to the retinal plane. As a result, this model ignores depth altogether. Therefore,\n\\begin{align*}\n x' = x\\\\\n y' = y\n\\end{align*}\nOrthographic projection models are often used for architecture and industrial design.\n\nOverall, weak perspective models result in much simpler math, at the cost of being somewhat imprecise. However, it often yields results that are very accurate when the object is small and distant from the camera. \n\\end{document}\n", "meta": {"hexsha": "7e8e6b433bbe6e07c30c3662e2b8ec306c48900c", "size": 33203, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "01-camera-models/01-camera-models.tex", "max_stars_repo_name": "zishanqin/cs231a-notes", "max_stars_repo_head_hexsha": "b864cfb7c472573ec1fb7da780748348fc320dab", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 287, "max_stars_repo_stars_event_min_datetime": "2017-04-03T00:30:48.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-27T03:52:04.000Z", "max_issues_repo_path": "01-camera-models/01-camera-models.tex", "max_issues_repo_name": "zishanqin/cs231a-notes", "max_issues_repo_head_hexsha": "b864cfb7c472573ec1fb7da780748348fc320dab", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2019-06-26T11:23:10.000Z", "max_issues_repo_issues_event_max_datetime": "2020-01-16T09:00:43.000Z", "max_forks_repo_path": "01-camera-models/01-camera-models.tex", "max_forks_repo_name": "zishanqin/cs231a-notes", "max_forks_repo_head_hexsha": "b864cfb7c472573ec1fb7da780748348fc320dab", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 112, "max_forks_repo_forks_event_min_datetime": "2017-04-09T10:44:14.000Z", "max_forks_repo_forks_event_max_datetime": "2022-03-02T09:19:59.000Z", "avg_line_length": 75.9794050343, "max_line_length": 1053, "alphanum_fraction": 0.746287986, "num_tokens": 9376, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920116079209, "lm_q2_score": 0.7634837581726991, "lm_q1q2_score": 0.6113153461612739}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage{amsmath,amssymb,siunitx,cancel}\n\n\\def\\mhe{m_{\\mathrm{He}}}\n\n\\title{Interstellar Propulsion}\n\\author{Matthias Raives}\n\\date{\\today}\n\\begin{document}\n \n \\maketitle\n \n \\section{Bussard Ramjet}\n This can be derived multiple ways both relativistic and non-relativistic. The terminal velocity is large enough that the relativistic derivation is preferred, but the non-relativistic answers are correct to an order-of-magnitude level (within a factor of 3). Three derivations that I have found are presented here.\n \n \\subsection{Relativistic Monentum Conservation}\n The spacecraft generates thrust from the proton-proton chain:\n \\begin{equation}\n 6\\mathrm{p} \\rightarrow \\mathrm{He}^{4} + 2\\mathrm{p} + \\nu_{e} + \\Delta E\n \\end{equation}\n Since our reactor is perfectly efficient, we'll assume that all the $\\Delta E$ goes into launching the neutrino (which we'll assume is negligible) and helium nucleus out of the spaceship for propulsion, and that we capture the two extra protons produced (so that, in the steady state, we only need to capture 4 protons per reaction).\n \n At terminal velocity, the momentum of the ejected helium nuclei is equal to the total momentum of the protons we're collecting. The momentum of the protons is:\n \\begin{equation}\n p_{1} = \\sum p_{p} = 4\\gamma m_{p}v_{T} = \\frac{4m_{p}v_{T}}{\\sqrt{1-\\frac{v_{T}^{2}}{c^{2}}}},\n \\end{equation}\n and the momentum of the exhaust can be determined by:\n \\begin{align}\n E = \\mhe c^{2} + \\Delta E &= (\\mhe^{2}c^{4} + p_{2}^{2}c^{2})^{1/2}\\\\\n (\\mhe+\\Delta m)^{2}c^{4} &= \\mhe^{2}c^{4} + p_{2}^{2}c^{2}\\\\\n \\cancel{\\Delta m^{2}} + \\mhe\\Delta m &= p_{2}^{2}c^{-2}\\\\\n p_{2} &= c\\sqrt{\\mhe\\Delta m}.\n \\end{align}\n The helium nucleus mass is about $\\mhe\\sim 4 m_{p}$, and the energy produced by the proton-proton chain converts to a mass:\n \\begin{equation}\n \\Delta m = \\frac{\\Delta E}{c^{2}} \\sim 10^{-2}m_{p}\n \\end{equation}\n Setting these momenta equal:\n \\begin{align}\n p_{1} &= p_{2}\\\\\n \\frac{4m_{p}v_{T}}{\\sqrt{1-\\frac{v_{T}^{2}}{c^{2}}}} &= c\\sqrt{\\mhe\\Delta m}\\\\\n \\frac{\\frac{v_{T}}{c}}{\\sqrt{1-\\frac{v_{T}^{2}}{c^{2}}}} &= \\frac{\\sqrt{\\mhe\\Delta m}}{4m_{p}}\\\\\n \\frac{v_{T}}{c} &= \\left(\\frac{1}{\\frac{16m_{p}^{2}}{\\mhe\\Delta m}+1}\\right)^{1/2}\\\\\n \\frac{v_{T}}{c} &= \\left(\\frac{1}{400+1}\\right)^{1/2} \\sim \\frac{1}{20} = 0.05\n \\end{align}\n \n \\subsection{Non-relativistic Momentum Conservation}\n This method is similar to the first, but it uses the non-relativistic formulas. We have the total momentum of the protons:\n \\begin{equation}\n p_{1} = \\sum p_{p} = 4m_{p}v_{T},\n \\end{equation}\n and the exhaust momentum:\n \\begin{equation}\n p_{2} = \\sqrt{2\\mhe\\Delta E} = c\\sqrt{2\\mhe\\Delta m}.\n \\end{equation}\n Setting these equal:\n \\begin{align}\n p_{1} &= p_{2}\\\\\n 4m_{p}v_{T} &= c\\sqrt{2\\mhe\\Delta m}\\\\\n \\frac{v_{T}}{c} &= \\left(\\frac{\\mhe\\Delta m}{8m_{p}^{2}}\\right)^{1/2}\\\\\n \\frac{v_{T}}{c} &= \\frac{\\sqrt{2}}{20} \\sim 0.07\n \\end{align}\n \n \\subsection{Non-relativistic Drag}\n The generalized drag formula is:\n \\begin{equation}\n \\frac{1}{2}\\rho v_{T}^{2}A = F\n \\end{equation}\n where $A$ is the effective collecting area of our craft. In this case, $F$ is the thrust generated by our reactor. We can approximate $F$ by:\n \\begin{equation}\n F = \\frac{\\mathrm{d}p}{\\mathrm{d}t} \\sim \\frac{p}{4\\tau} = \\frac{pv_{T}}{4\\lambda}\n \\end{equation}\n where $\\tau = \\frac{\\lambda}{v_{T}}$ is the mean time between interactions with ISM protons, and $\\lambda$ is the mean free path of our spaceship in the ISM, given by\n \\begin{equation}\n \\lambda = \\frac{m_{p}}{\\rho A}.\n \\end{equation}\n Thus, we have:\n \\begin{equation}\n \\frac{1}{2}\\rho v_{T}^{2}A = \\frac{p\\rho Av_{T}}{4m_{p}}.\n \\end{equation}\n The exhaust momentum $p$ is the same as in the earlier non-relativistic derivation. Solving for $v_{T}$:\n \\begin{equation}\n \\frac{v_{T}}{c} = \\frac{1}{2c}\\frac{p}{m_{p}} = \\left(\\frac{\\mhe\\Delta m}{2m_{p}^{2}}\\right)^{1/2} = \\frac{\\sqrt{2}}{20} \\sim 0.14\n \\end{equation}\n \n \\section{Project Orion}\n \\subsection{Rocket Equation}\n Integral to this question is the Rocket Equation:\n \\begin{equation}\n \\Delta v = v_{e}\\ln\\left(\\frac{m_{0}}{m_{f}}\\right)\n \\end{equation}\n The derivation is as follows. At time $t=0$, the rocket has mass $m+\\Delta m$ and velocity $v$. At time $t=\\Delta t$, the rocket has velocity $v+\\Delta v$, mass $m$, and the spent fuel has mass $\\Delta m$ and velocity $v-v_{e}$, where $v_{e}$ is the rest frame ejecta velocity. By conservation of momentum, then:\n \\begin{align}\n (m+\\Delta m)v &= m(v+\\Delta v) + \\Delta m (v-v_{e})\\\\\n mv + \\Delta m v &= mv + m\\Delta v + \\Delta m v - \\Delta mv_{e}\\\\\n \\Delta m v_{e} = m\\Delta v\\\\\n \\Delta v &= v_{e}\\frac{\\Delta m}{m}\n \\end{align}\n Integrating this yields:\n \\begin{equation}\n \\Delta v = v_{e}\\ln\\left(\\frac{m_{0}}{m_{f}}\\right)\n \\end{equation}\n where $m_0$ is the initial mass of the spacecraft (fuel + payload) and $m_f$ is the final mass (just payload).\n \n \\subsection{Exhaust Velocity}\n In the case of pulse propulsion, the ``ejecta velocity'' is actually the velocity of the explosion, when it pushes against your ship. A nuclear bomb of mass $M$ has yield $E = \\epsilon Mc^{2}$; the ejecta then has velocity:\n \\begin{equation}\n v_{e} = \\sqrt{\\frac{2E}{M}} = \\sqrt{2\\epsilon c^{2}}\n \\end{equation}\n However, only a fraction $f\\leq \\frac{1}{2}$ of the mass is intercepted, so the effective velocity is\n \\begin{equation}\n v_{e} = f\\sqrt{2\\epsilon} c.\n \\end{equation}\n \n \\subsection{Mass Ratio}\n We can calculate $\\epsilon$ as:\n \\begin{equation}\n \\epsilon = \\frac{\\Delta E}{mc^2} \\approx 10^{-3}\n \\end{equation}\n Adopting a fiducial $f=\\frac{1}{2}$, we see:\n \\begin{equation}\n v_e \\approx 0.02c\n \\end{equation}\n and thus\n \\begin{equation}\n \\frac{m_{0}}{m_{f}} = \\exp\\left(\\frac{\\Delta v}{v_{e}}\\right) = e^{2.5} \\approx 10.\n \\end{equation}\n That is, about 90\\% of the mass of the spacecraft must be fuel (i.e., nukes). The answer here will vary considerably because of the exponential, however. Also, modern thermonuclear weapons would increase the efficiency of this design.\n \n\\end{document}\n", "meta": {"hexsha": "c7af88f9ae6821b5bf245341ebebbd7511a3e400", "size": 6702, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Bussard Ramjet/BussardRamjet_Answer.tex", "max_stars_repo_name": "osugoom/questions", "max_stars_repo_head_hexsha": "5ad4fa6de9c9a8c60a3043adacfad41aef24ed4a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Bussard Ramjet/BussardRamjet_Answer.tex", "max_issues_repo_name": "osugoom/questions", "max_issues_repo_head_hexsha": "5ad4fa6de9c9a8c60a3043adacfad41aef24ed4a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Bussard Ramjet/BussardRamjet_Answer.tex", "max_forks_repo_name": "osugoom/questions", "max_forks_repo_head_hexsha": "5ad4fa6de9c9a8c60a3043adacfad41aef24ed4a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2018-01-10T21:05:11.000Z", "max_forks_repo_forks_event_max_datetime": "2018-01-10T21:05:11.000Z", "avg_line_length": 50.7727272727, "max_line_length": 339, "alphanum_fraction": 0.609967174, "num_tokens": 2235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006919925839875, "lm_q2_score": 0.7634837743174788, "lm_q1q2_score": 0.6113153445638055}} {"text": "\\documentclass{article}\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\usepackage{graphicx}\n\\usepackage{float}\n\\usepackage{subcaption}\n\\usepackage{bm}\n\\graphicspath{{./}}\n\\usepackage[left=1cm,right=1cm,top=1cm,bottom=1cm]{geometry}\n\\begin{document}\n\\title{Rotation Invariance using Second-order Convolutional Neural Networks}\n\n\\section{Introduction}\nWe propose a non-trivial filter filter for a Convolutional Neural Network (CNN) that is invariant to rotations of the input image around the center of the filter. \nIn particular we show how to construct such non-trivial rotationally-invariant filter by using two convolutional neural network layers with a nonlinear ReLu after each convolution.\nOur layer has the following two important properties that cannot both be attained by multiplication by a single filter.\nOne, it is rotationally invariant, and two, it is non-trivial. \nWhat we mean by non-trivial is that our single filter is able to distinguigsh a dumbbell from a donut, even if the donut looks exactly like the surface generated by a rotated dumbbell.\nNo actiong by a single linear filter can have both those properties since linearity implies the second property cannot hold given the first.\nFurthermore, the filter does not need to be trained on rotated views of its input image.\nIn \\cite{chengzhouhan}, Cheng et al. build a neural network that is essentially rotation invariant.\nIn \\cite{jaderberg}, Jaderberg builds a neural network that is invariant to rotations (more generally, to affine transformations of the input).\nHowever, both of those papers require that the network be trained on several transformed input representations in order for the network to learn to be rotationally invariant.\nRelatedly, neither of those networks is guaranteed mathematically to be invariant to rotations of the input image (even for square images rotated a quarter turn, for example).\nWe construct a non-trivial rotationally invariant CNN architecture that does not need to be trained on multiple rotated views on an object.\nOur filter is exactly rotationally invariant for rotations by a quarter turn, and the only reason it is only approximately invariant for other rotations is because of the rectangular nature of image pixels, which we compensate for by using gaussian blurring.\nFor sufficiently high-resolution pixels (or for pixels arranged finely enough along grid lines in polar coordinates) our method is guaranteed to be rotationally invariant to arbitrary precision.\nNa\\\"ively, one might expect that rotationally invariant filters satisfy our goal.\nHowever, as pointed out above, since a single filter is linear in its inputs, any rotationally invariant filter will return the same value for an image and for any weighted average of various rotations of that image.\nWe find this problematic because we would like our rotationally invariant filter to be able to distinguish objects like a donut from a dumbbell, even though if you rotate a dumbbell you get an object that looks like a donut.\nWe show that two layers of rotationally invariant second-order filters have the desired properties.\n\n\\section{Related Work}\nIn \\cite{marcos}, they generate exact rotation invariance over a finite number of rotations by taking the maxpool over a set of rotated filters.\n\n\\section{Defining a rotationally invariant filter}\nNo standard first-order filter is able to accomplish that task because first-order filters are linear in their inputs, so if they are invariant to rotation then they are also necessarily equally triggered by an image of that object and by a weighted average of images of rotations of that object.\n\nWe construct an (approximately) rotationally-symmetric filter as follows. \nWe assume a square filter of side length $2h + 1$. \nThe pixels are defined by $\\mathbf x = (x,y),\\text{ for } x,y\\in [-h,h]$ we define the radius $r$ of a pixel as $r_{\\mathbf{x}} = ||x||_2$. \nWe can therefore construct an approximately rotationally-invariant basis filter by setting the filter value $f_n(\\mathbf{x})$ to be equal to\n$f(\\mathbf{x}) = P(n,\\sigma,r_x)$ \nwhere $n$ is the radius parameter for the desired filter, $\\sigma$ is the standard deviation of the blurring gaussian for the filter, and $P(n, \\sigma, r_x)$ is the gaussian probability density value of the point $r_x$ for the gaussian centered at $n$ with standard deviation $\\sigma$.\n\n\\begin{figure}[h]\n\\centering\n\\begin{subfigure}[t]{.3\\textwidth}\n \\centering\n \\includegraphics[width=1in]{basis}\n \\caption{Our rotationally symmetric basis filter of radius 5 pixels}\n\\end{subfigure}\n\\begin{subfigure}[t]{.3\\textwidth}\n \\centering\n \\includegraphics[width=1in]{rotbasis45}\n \\caption{That basis filter rotated by 45 degrees counterclockwise (using bilinear interpolation to fit rotated pixels)}\n\\end{subfigure}\n\\begin{subfigure}[t]{.3\\textwidth}\n \\centering\n \\includegraphics[width=1in]{diffbasisrot45}\n \\caption{The difference in pixel intensity between the basis filter and the rotated basis filter (hardly discernable)}\n\\end{subfigure}\n\\end{figure}\n\n\\bibliography{traversr}{}\n\\bibliographystyle{plain}\n\\end{document}\n", "meta": {"hexsha": "9fcbd5a7e5ca7a49e19c1ea4805a297c0744cf65", "size": 5086, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "writeup/traversr.tex", "max_stars_repo_name": "travers-rhodes/ricnn", "max_stars_repo_head_hexsha": "c5a7a15bff78b8f1495223b2bd70c4a1326e469e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "writeup/traversr.tex", "max_issues_repo_name": "travers-rhodes/ricnn", "max_issues_repo_head_hexsha": "c5a7a15bff78b8f1495223b2bd70c4a1326e469e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "writeup/traversr.tex", "max_forks_repo_name": "travers-rhodes/ricnn", "max_forks_repo_head_hexsha": "c5a7a15bff78b8f1495223b2bd70c4a1326e469e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 74.7941176471, "max_line_length": 296, "alphanum_fraction": 0.799252851, "num_tokens": 1150, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8006920020959543, "lm_q2_score": 0.7634837635542924, "lm_q1q2_score": 0.6113153432080405}} {"text": "\\input{permve-ntnu-latex-assignment.tex}\n\n\\title{\n\\normalfont \\normalsize\n\\textsc{Norwegian University of Science and Technology\\\\IT3708 -- Subsymbolic Methods in AI}\n}\n\n\\author{Per Magnus Veierland\\\\permve@stud.ntnu.no}\n\n\\date{\\normalsize\\today}\n\n\\begin{document}\n\n\\fancyfoot[C]{}\n\\maketitle\n\n\\section*{Intelligence Emerging: EA Rank Errata Suggestion}\n\nIn the description of the rank selection in evolutionary algorithms on page 434 in section A.5.2 I've found a small error concerning the derivation in Equation A.6 and A.7.\n\nEquation A.6 shows that Equation~\\ref{eq:a6a} is equal to Equation~\\ref{eq:a6b}.\n\n\\begin{equation}\n\\label{eq:a6a}\n\\sum_{i=1}^{N}\\Bigg[ \\text{min} + (\\text{max} - \\text{min})\\frac{\\textit{rank}(i, g) - 1}{N - 1}\\Bigg]\n\\end{equation}\n\n\\begin{equation}\n\\label{eq:a6b}\nN \\cdot \\text{min} + (\\text{max} - \\text{min}) \\sum_{i=1}^{N}\\frac{i}{N - 1}\n\\end{equation}\n\nHowever these are not equivalent, and Equation~\\ref{eq:a6b} should instead be:\n\n\\begin{equation}\nN \\cdot \\text{min} + (\\text{max} - \\text{min}) \\sum_{i=1}^{N}\\frac{i - 1}{N - 1}\n\\end{equation}\n\nThis results in the following derivation, which differs from the first two steps of Equation A.7:\n\n\\begin{equation}\n\\sum_{i=1}^{N}\\frac{i - 1}{N - 1} =\n\\frac{1}{N - 1}\\sum_{i=1}^{N}(i - 1) =\n\\frac{1}{N - 1}\\sum_{i=1}^{N - 1}i =\n\\frac{1}{N - 1} \\cdot \\frac{(N - 1)N}{2} =\n\\frac{N}{2}\n\\end{equation}\n\n\\end{document}\n\n", "meta": {"hexsha": "a3fae63193c782320e06d2f46c35c94e89bac57c", "size": 1396, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "errata/permve-intelligence-emerging-errata-ea-rank/permve-intelligence-emerging-errata-ea-rank.tex", "max_stars_repo_name": "pveierland/permve-ntnu-it3708", "max_stars_repo_head_hexsha": "1066d5c1af5c953dbaf129d7e05ce32f2d4292aa", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "errata/permve-intelligence-emerging-errata-ea-rank/permve-intelligence-emerging-errata-ea-rank.tex", "max_issues_repo_name": "pveierland/permve-ntnu-it3708", "max_issues_repo_head_hexsha": "1066d5c1af5c953dbaf129d7e05ce32f2d4292aa", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "errata/permve-intelligence-emerging-errata-ea-rank/permve-intelligence-emerging-errata-ea-rank.tex", "max_forks_repo_name": "pveierland/permve-ntnu-it3708", "max_forks_repo_head_hexsha": "1066d5c1af5c953dbaf129d7e05ce32f2d4292aa", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 27.3725490196, "max_line_length": 172, "alphanum_fraction": 0.6783667622, "num_tokens": 523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7634837527911057, "lm_q2_score": 0.8006919997179627, "lm_q1q2_score": 0.6113153327744851}} {"text": "\n% This LaTeX was auto-generated from MATLAB code.\n% To make changes, update the MATLAB code and republish this document.\n\n\\documentclass{article}\n\\usepackage{graphicx}\n\\usepackage{color}\n\n\\sloppy\n\\definecolor{lightgray}{gray}{0.5}\n\\setlength{\\parindent}{0pt}\n\n\\begin{document}\n\n \n \n\\section*{Author: DAVID LI}\n\n\n\\begin{verbatim}FinalPrep.m --- ELEC 403 Final exam, this matlab file is used to do the\nvarious problems in the ELEC 403 textbook that are related to the final\nexam, of course the focus is on chapter 5 and chapter 7.\\end{verbatim}\n \n\\subsection*{Contents}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item Prob 5.2 with jacobian f(x) = x\\^{}2+2*y\\^{}2+4*x+4*y\n \\item Prob 5.2 with Newton\n \\item Prob 5.4 -- Find a good starting point besides x0 = [1 1]\n \\item 5.4 Using Newton\n \\item 5.4 Using Gauss Newton\n \\item Prob 5.5 -- use a closer point so that it can be done manually\n \\item Chapter 7, using Prob 5.2 with DFP\n \\item Chapter 7, using Prob 5.2 with BFGS\n \\item Examples from the Internet --- Using DFP\n \\item USING BFGS\n\\end{itemize}\n\n\n\\subsection*{Prob 5.2 with jacobian f(x) = x\\^{}2+2*y\\^{}2+4*x+4*y}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item REMEMBER THAT CHAPTER 5 only has SDM, Newton and Gauss-Newton*\n\\end{itemize}\n\\begin{verbatim}\nsyms x_1 x_2\nf = x_1^2+2*x_2^2+4*x_1+4*x_2;\nlatex(f);\ng=gradient(f);\nlatex(g);\nf1 = x_1+2;\nf2 = sqrt(2)*(x_2+1);\nj=jacobian([f1,f2],[x_1,x_2]);\nh = hessian(f);\n%steep_desc3('func52','grad52',[-1; 1],1e-3);\ngauss_newton('func52','grad52','jacob52',[-1; 1],1e-3);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram gauss_newton.m\n\nF_k =\n\n 3\n\n\ngk =\n\n 2\n 8\n\n\nJk =\n\n 1.0000 0\n 0 1.4142\n\n\nHk =\n\n 2.0000 0\n 0 4.0000\n\n\ndk =\n\n -1.0000\n -2.0000\n\n\nak =\n\n 1.0000\n\n\nxk =\n\n -2.0000\n -1.0000\n\n\nF_k1 =\n\n -6\n\n\ngk =\n\n 1.0e-12 *\n\n 0.4450\n -0.2185\n\n\nJk =\n\n 1.0000 0\n 0 1.4142\n\n\ndk =\n\n 1.0e-12 *\n\n -0.2225\n 0.0546\n\n\nak =\n\n 1\n\n\nadk =\n\n 1.0e-12 *\n\n -0.2225\n 0.0546\n\n\ner =\n\n 9\n\n\nxk =\n\n -2\n -1\n\n\nF_k1 =\n\n -6\n\n\ngk =\n\n 0\n 0\n\n\nJk =\n\n 1.0000 0\n 0 1.4142\n\n\ndk =\n\n 0\n 0\n\n\nak =\n\n 1\n\n\nadk =\n\n 0\n 0\n\n\ner =\n\n 0\n\nSolution point:\n\nxs =\n\n -2\n -1\n\nObjective function at the solution point:\n\nfs =\n\n -6\n\nNumber of iterations performed:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Prob 5.2 with Newton}\n\n\\begin{verbatim}\nnewton('func52','grad52','hess52',[0; 0],0.1,1e-3);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram newton.m\n\ngk =\n\n 4\n 4\n\n\nHk =\n\n 2 0\n 0 4\n\n\ndk =\n\n -2\n -1\n\nAlpha: 1.000000000 \n\ner =\n\n 2.2361\n\n\ngk =\n\n 0\n 0\n\n\nHk =\n\n 2 0\n 0 4\n\n\nak =\n\n 1\n\nAlpha: 1.000000000 \nSolution point:\n\nxs =\n\n -2\n -1\n\nObjective function at the solution point:\n\nfs =\n\n -6\n\nNumber of iterations performed:\n\nk =\n\n 2\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Prob 5.4 -- Find a good starting point besides x0 = [1 1]}\n\n\\begin{verbatim}\nsyms x_1 x_2\nf = 5*x_1^2-9*x_1*x_2+4.075*x_2^2+x_1;\nlatex(f);\ng = gradient(f);\nlatex(g);\nsteep_desc3('func54','grad54',[-16; -17.9],1e-3);\nh = hessian(f);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram steep_desc3.m\nAlpha: 0.055180698 \n\t xk+1: -16.115879466 \n\t xk+1: -17.795984384 \nf(xk) = -8.148958e+00 \n\nAlpha: 36.073499423 \n\t xk+1: -16.298584133 \n\t xk+1: -17.999528044 \nf(xk) = -8.149995e+00 \nNorm Value: 2.735160e-01\n\nAlpha: 0.055180697 \n\t xk+1: -16.299131033 \n\t xk+1: -17.999037137 \nf(xk) = -8.150000e+00 \nNorm Value: 7.349077e-04\n\nSolution point:\n\nxs =\n\n -16.299131032840879\n -17.999037136909219\n\nObjective function at the solution point:\n\nfs =\n\n -8.149999976794053\n\nNumber of iterations performed:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{5.4 Using Newton}\n\n\\begin{verbatim}\nh = hessian(f);\nlatex(h)\nnewton('func54','grad54','hess54',[-10; -5],0.1,1e-6);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\nans =\n\n\\left(\\begin{array}{cc} 10 & -9\\\\ -9 & \\frac{163}{20} \\end{array}\\right)\n\n \nProgram newton.m\n\ngk =\n\n -54.0000\n 49.2500\n\n\nHk =\n\n 10.0000 -9.0000\n -9.0000 8.1500\n\n\ndk =\n\n -6.3000\n -13.0000\n\nAlpha: 0.945675721 \n\ner =\n\n 13.6613\n\n\ngk =\n\n -2.9335\n 2.6755\n\n\nHk =\n\n 10.0000 -9.0000\n -9.0000 8.1500\n\n\nak =\n\n 1\n\nAlpha: 1.000000000 \n\ngk =\n\n 1.0e-13 *\n\n 0.5684\n -0.5684\n\n\nHk =\n\n 10.0000 -9.0000\n -9.0000 8.1500\n\n\nak =\n\n 1\n\nAlpha: 1.000000000 \nSolution point:\n\nxs =\n\n -16.300000000000054\n -18.000000000000060\n\nObjective function at the solution point:\n\nfs =\n\n -8.149999999999963\n\nNumber of iterations performed:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{5.4 Using Gauss Newton}\n\n\\begin{verbatim}\nj = jacobian(f,[x_1,x_2]);\ngauss_newton('func54','grad54','jacob54',[-16; -18],1e-3);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram gauss_newton.m\n\nF_k =\n\n -7.7000\n\n\ngk =\n\n 3.0000\n -2.7000\n\n\nJk =\n\n 3.0000 -2.7000\n\n\nHk =\n\n 18.0000 -16.2000\n -16.2000 14.5800\n\n\ndk =\n\n -0.0920\n 0.0828\n\n\nak =\n\n 1.7993\n\n\nxk =\n\n -16.1656\n -17.8511\n\n\nF_k1 =\n\n -8.1494\n\n\ngk =\n\n 0.0037\n 0.0041\n\n\nJk =\n\n 0.0037 0.0041\n\n\ndk =\n\n -60.4078\n -67.1792\n\n\nak =\n\n 0.0022\n\n\nadk =\n\n -0.1338\n -0.1488\n\n\ner =\n\n 0.4494\n\n\nxk =\n\n -16.2994\n -17.9999\n\n\nF_k1 =\n\n -8.1500\n\n\ngk =\n\n 0.0049\n -0.0044\n\n\nJk =\n\n 0.0049 -0.0044\n\n\ndk =\n\n -56.5957\n 50.8910\n\n\nak =\n\n 1.0000e-05\n\n\nadk =\n\n 1.0e-03 *\n\n -0.5660\n 0.5089\n\n\ner =\n\n 5.5388e-04\n\nSolution point:\n\nxs =\n\n -16.300007501945657\n -17.999413346528229\n\nObjective function at the solution point:\n\nfs =\n\n -8.149998557647706\n\nNumber of iterations performed:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Prob 5.5 -- use a closer point so that it can be done manually}\n\n\\begin{par}\nProduce the latex needed for final exam review document\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\nsyms x_1 x_2 x_3\nf = (x_1+5)^2+(x_2+8)^2+(x_3+7)^2+2*x_1^2*x_2^2+4*x_1^2*x_3^2;\nlatex(f)\ng = gradient(f);\nlatex(g)\n\nsteep_desc3('func55','grad55',[0; -8; -7],1e-3)\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim}\nans =\n\n2\\, {x_{1}}^2\\, {x_{2}}^2 + 4\\, {x_{1}}^2\\, {x_{3}}^2 + {\\left(x_{1} + 5\\right)}^2 + {\\left(x_{2} + 8\\right)}^2 + {\\left(x_{3} + 7\\right)}^2\n\n\nans =\n\n\\left(\\begin{array}{c} 4\\, x_{1}\\, {x_{2}}^2 + 8\\, x_{1}\\, {x_{3}}^2 + 2\\, x_{1} + 10\\\\ 4\\, x_{2}\\, {x_{1}}^2 + 2\\, x_{2} + 16\\\\ 8\\, x_{3}\\, {x_{1}}^2 + 2\\, x_{3} + 14 \\end{array}\\right)\n\n \nProgram steep_desc3.m\nAlpha: 0.001538462 \n\t xk+1: -0.015384615 \n\t xk+1: -8.000000000 \n\t xk+1: -7.000000000 \nf(xk) = 2.492308e+01 \n\nAlpha: 0.499585233 \n\t xk+1: -0.015384615 \n\t xk+1: -7.996216159 \n\t xk+1: -6.993378279 \nf(xk) = 2.492302e+01 \nNorm Value: 7.626575e-03\n\nAlpha: 0.001540793 \n\t xk+1: -0.015407926 \n\t xk+1: -7.996216155 \n\t xk+1: -6.993378281 \nf(xk) = 2.492302e+01 \nNorm Value: 2.331078e-05\n\nSolution point:\n\nxs =\n\n -0.015407926169005\n -7.996216155022361\n -6.993378280948492\n\nObjective function at the solution point:\n\nfs =\n\n 24.923018533805848\n\nNumber of iterations performed:\n\nk =\n\n 3\n\n\nans =\n\n -0.0154\n -7.9962\n -6.9934\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Chapter 7, using Prob 5.2 with DFP}\n\n\\begin{itemize}\n\\setlength{\\itemsep}{-1ex}\n \\item REMEMBER THAT CHAPTER 7 only has DFP and BFGS*\n \\item DFP\n \\item BFGS\n\\end{itemize}\n\\begin{verbatim}\ndfp('func52','grad52',[-1; 1],1e-6);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram dfp.m\n------- ITERATION 1 ------\nSTEP 1\n\nxk =\n\n -1\n 1\n\n\nSk =\n\n 1 0\n 0 1\n\n\ngk =\n\n 2\n 8\n\nSTEP 2\n\ndk =\n\n -2\n -8\n\n\nak =\n\n 0.2576\n\n\ndtk =\n\n -0.5152\n -2.0606\n\n\nxk_new =\n\n -1.5152\n -1.0606\n\nSTEP 3\nNorm of delta is 8.76 and epsi is 1.000000e-06\nSTEP 4\n\ngk_new =\n\n 0.9697\n -0.2424\n\n\ngmk =\n\n -1.0303\n -8.2424\n\n\nsg =\n\n -1.0303\n -8.2424\n\n\nsw1 =\n\n 0.2654 1.0615\n 1.0615 4.2461\n\n\nsw2 =\n\n 1.0615 8.4922\n 8.4922 67.9376\n\n\nsw3 =\n\n 68.9991\n\n\nSk =\n\n 0.9998 -0.0625\n -0.0625 0.2578\n\n\ngk =\n\n 0.9697\n -0.2424\n\n------- ITERATION 2 ------\nSTEP 2\n\ndk =\n\n -0.9846\n 0.1231\n\n\nak =\n\n 0.4924\n\n\ndtk =\n\n -0.4848\n 0.0606\n\nSTEP 3\nNorm of delta is 0.49 and epsi is 1.000000e-06 \nSTEP 4\n\ngk_new =\n\n 0\n 0\n\n\ngmk =\n\n -0.9697\n 0.2424\n\n\nsg =\n\n -0.9846\n 0.1231\n\n\nsw1 =\n\n 0.2351 -0.0294\n -0.0294 0.0037\n\n\nsw2 =\n\n 0.9695 -0.1212\n -0.1212 0.0151\n\n\nsw3 =\n\n 0.9846\n\n\nSk =\n\n 0.5000 -0.0000\n -0.0000 0.2500\n\n\ngk =\n\n 0\n 0\n\n------- ITERATION 3 ------\nSTEP 2\n\ndk =\n\n 0\n 0\n\n\nak =\n\n 1\n\n\ndtk =\n\n 0\n 0\n\nSTEP 3\nNorm of delta is 0.00 and epsi is 1.000000e-06 \nsolution point:\n\nxs =\n\n -2\n -1\n\nobjective function at the solution point:\n\nfs =\n\n -6\n\nnumber of iterations at convergence:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Chapter 7, using Prob 5.2 with BFGS}\n\n\\begin{verbatim}\nbfgs('func52','grad52',[-1; 1],1e-6);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram bfgs.m\n------- ITERATION 1 ------\nSTEP 1\n\nxk =\n\n -1\n 1\n\n\nSk =\n\n 1 0\n 0 1\n\n\ngk =\n\n 2\n 8\n\nSTEP 2\n\ndk =\n\n -2\n -8\n\n\nak =\n\n 0.2576\n\n\ndtk =\n\n -0.5152\n -2.0606\n\n\nxk_new =\n\n -1.5152\n -1.0606\n\nSTEP 3\nNorm of delta is 8.76 and epsi is 1.000000e-06\nSTEP 4\n\ngk_new =\n\n 0.9697\n -0.2424\n\n\ngmk =\n\n -1.0303\n -8.2424\n\n\nD =\n\n 17.5152\n\n\nsg =\n\n -1.0303\n -8.2424\n\n\nsw1 =\n\n 0.2654 1.0615\n 1.0615 4.2461\n\n\nsw2 =\n\n 0.5308 2.1230\n 4.2461 16.9844\n\n\nSk =\n\n 1.0142 -0.0643\n -0.0643 0.2580\n\n\nfk =\n\n -5.7576\n\n\ngk =\n\n 0.9697\n -0.2424\n\n------- Iteration 2 ------\nSTEP 2\n\ndk =\n\n -0.9991\n 0.1249\n\n\nak =\n\n 0.4853\n\n\ndtk =\n\n -0.4848\n 0.0606\n\nSTEP 3\nNorm of delta is 0.49 and epsi is 1.000000e-06 \nSTEP 4\n\ngk_new =\n\n 0\n 0\n\n\ngmk =\n\n -0.9697\n 0.2424\n\n\nD =\n\n 0.4848\n\n\nsg =\n\n -0.9991\n 0.1249\n\n\nsw1 =\n\n 0.2351 -0.0294\n -0.0294 0.0037\n\n\nsw2 =\n\n 0.4844 -0.0606\n -0.0606 0.0076\n\n\nSk =\n\n 0.5000 0.0000\n 0.0000 0.2500\n\n\nfk =\n\n -6\n\n\ngk =\n\n 0\n 0\n\n------- Iteration 3 ------\nSTEP 2\n\ndk =\n\n 0\n 0\n\n\nak =\n\n 1\n\n\ndtk =\n\n 0\n 0\n\nSTEP 3\nNorm of delta is 0.00 and epsi is 1.000000e-06 \nsolution point:\n\nxs =\n\n -2\n -1\n\nobjective function at the solution point:\n\nfs =\n\n -6\n\nnumber of iterations at convergence:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{Examples from the Internet --- Using DFP}\n\n\\begin{par}\n$f(x)=-2*x_1^2-10*x_2^2$\n\\end{par} \\vspace{1em}\n\\begin{verbatim}\ndfp('funcNet1','gradNet1',[1; -1],1e-3);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram dfp.m\n------- ITERATION 1 ------\nSTEP 1\n\nxk =\n\n 1\n -1\n\n\nSk =\n\n 1 0\n 0 1\n\n\ngk =\n\n 4\n -20\n\nSTEP 2\n\ndk =\n\n -4\n 20\n\n\nak =\n\n 0.0577\n\n\ndtk =\n\n -0.2308\n 1.1538\n\n\nxk_new =\n\n 0.7692\n 0.1538\n\nSTEP 3\nNorm of delta is 10.58 and epsi is 1.000000e-03\nSTEP 4\n\ngk_new =\n\n 3.0769\n 3.0769\n\n\ngmk =\n\n -0.9231\n 23.0769\n\n\nsg =\n\n -0.9231\n 23.0769\n\n\nsw1 =\n\n 0.0533 -0.2663\n -0.2663 1.3314\n\n\nsw2 =\n\n 0.8521 -21.3018\n -21.3018 532.5444\n\n\nsw3 =\n\n 533.3964\n\n\nSk =\n\n 1.0004 0.0300\n 0.0300 0.0512\n\n\ngk =\n\n 3.0769\n 3.0769\n\n------- ITERATION 2 ------\nSTEP 2\n\ndk =\n\n -3.1705\n -0.2499\n\n\nak =\n\n 0.2699\n\n\ndtk =\n\n -0.8556\n -0.0674\n\nSTEP 3\nNorm of delta is 1.33 and epsi is 1.000000e-03 \nSTEP 4\n\ngk_new =\n\n -0.3456\n 1.7281\n\n\ngmk =\n\n -3.4225\n -1.3488\n\n\nsg =\n\n -3.4644\n -0.1718\n\n\nsw1 =\n\n 0.7321 0.0577\n 0.0577 0.0045\n\n\nsw2 =\n\n 12.0017 0.5951\n 0.5951 0.0295\n\n\nsw3 =\n\n 12.0886\n\n\nSk =\n\n 0.2500 -0.0001\n -0.0001 0.0503\n\n\ngk =\n\n -0.3456\n 1.7281\n\n------- ITERATION 3 ------\nSTEP 2\n\ndk =\n\n 0.0866\n -0.0869\n\n\nak =\n\n 0.9949\n\n\ndtk =\n\n 0.0862\n -0.0865\n\nSTEP 3\nNorm of delta is 0.12 and epsi is 1.000000e-03 \nSTEP 4\n\ngk_new =\n\n 1.0e-03 *\n\n -0.9953\n -0.9953\n\n\ngmk =\n\n 0.3446\n -1.7291\n\n\nsg =\n\n 0.0864\n -0.0870\n\n\nsw1 =\n\n 0.0074 -0.0074\n -0.0074 0.0075\n\n\nsw2 =\n\n 0.0075 -0.0075\n -0.0075 0.0076\n\n\nsw3 =\n\n 0.1801\n\n\nSk =\n\n 0.2501 0.0000\n 0.0000 0.0500\n\n\ngk =\n\n 1.0e-03 *\n\n -0.9953\n -0.9953\n\n------- ITERATION 4 ------\nSTEP 2\n\ndk =\n\n 1.0e-03 *\n\n 0.2489\n 0.0498\n\n\nak =\n\n 0.9997\n\n\ndtk =\n\n 1.0e-03 *\n\n 0.2488\n 0.0498\n\nSTEP 3\nNorm of delta is 0.00 and epsi is 1.000000e-03 \nsolution point:\n\nxs =\n\n 1.0e-10 *\n\n 0.458617078233185\n -0.458617078910812\n\nobjective function at the solution point:\n\nfs =\n\n 2.523955499581143e-20\n\nnumber of iterations at convergence:\n\nk =\n\n 4\n\n\\end{verbatim} \\color{black}\n \n\n\\subsection*{USING BFGS}\n\n\\begin{verbatim}\nbfgs('func52','grad52',[-1; 1],1e-6);\n\\end{verbatim}\n\n \\color{lightgray} \\begin{verbatim} \nProgram bfgs.m\n------- ITERATION 1 ------\nSTEP 1\n\nxk =\n\n -1\n 1\n\n\nSk =\n\n 1 0\n 0 1\n\n\ngk =\n\n 2\n 8\n\nSTEP 2\n\ndk =\n\n -2\n -8\n\n\nak =\n\n 0.2576\n\n\ndtk =\n\n -0.5152\n -2.0606\n\n\nxk_new =\n\n -1.5152\n -1.0606\n\nSTEP 3\nNorm of delta is 8.76 and epsi is 1.000000e-06\nSTEP 4\n\ngk_new =\n\n 0.9697\n -0.2424\n\n\ngmk =\n\n -1.0303\n -8.2424\n\n\nD =\n\n 17.5152\n\n\nsg =\n\n -1.0303\n -8.2424\n\n\nsw1 =\n\n 0.2654 1.0615\n 1.0615 4.2461\n\n\nsw2 =\n\n 0.5308 2.1230\n 4.2461 16.9844\n\n\nSk =\n\n 1.0142 -0.0643\n -0.0643 0.2580\n\n\nfk =\n\n -5.7576\n\n\ngk =\n\n 0.9697\n -0.2424\n\n------- Iteration 2 ------\nSTEP 2\n\ndk =\n\n -0.9991\n 0.1249\n\n\nak =\n\n 0.4853\n\n\ndtk =\n\n -0.4848\n 0.0606\n\nSTEP 3\nNorm of delta is 0.49 and epsi is 1.000000e-06 \nSTEP 4\n\ngk_new =\n\n 0\n 0\n\n\ngmk =\n\n -0.9697\n 0.2424\n\n\nD =\n\n 0.4848\n\n\nsg =\n\n -0.9991\n 0.1249\n\n\nsw1 =\n\n 0.2351 -0.0294\n -0.0294 0.0037\n\n\nsw2 =\n\n 0.4844 -0.0606\n -0.0606 0.0076\n\n\nSk =\n\n 0.5000 0.0000\n 0.0000 0.2500\n\n\nfk =\n\n -6\n\n\ngk =\n\n 0\n 0\n\n------- Iteration 3 ------\nSTEP 2\n\ndk =\n\n 0\n 0\n\n\nak =\n\n 1\n\n\ndtk =\n\n 0\n 0\n\nSTEP 3\nNorm of delta is 0.00 and epsi is 1.000000e-06 \nsolution point:\n\nxs =\n\n -2\n -1\n\nobjective function at the solution point:\n\nfs =\n\n -6\n\nnumber of iterations at convergence:\n\nk =\n\n 3\n\n\\end{verbatim} \\color{black}\n \n\n\n\\end{document}\n \n", "meta": {"hexsha": "4a93c2b8e4a83f79a81f6806349984498c85046b", "size": 14373, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "ELEC403/ELEC403Final/html/FinalPrep.tex", "max_stars_repo_name": "FriendlyUser/Matlab", "max_stars_repo_head_hexsha": "4615540072e198549370bc8b1f2ce502e94a867e", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "ELEC403/ELEC403Final/html/FinalPrep.tex", "max_issues_repo_name": "FriendlyUser/Matlab", "max_issues_repo_head_hexsha": "4615540072e198549370bc8b1f2ce502e94a867e", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2019-11-10T06:00:21.000Z", "max_issues_repo_issues_event_max_datetime": "2019-11-10T06:00:21.000Z", "max_forks_repo_path": "ELEC403/ELEC403Final/html/FinalPrep.tex", "max_forks_repo_name": "FriendlyUser/Matlab", "max_forks_repo_head_hexsha": "4615540072e198549370bc8b1f2ce502e94a867e", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 8.7800855223, "max_line_length": 186, "alphanum_fraction": 0.5373269324, "num_tokens": 6393, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754489059774, "lm_q2_score": 0.8397339696776499, "lm_q1q2_score": 0.6113057135376856}} {"text": "\\section{The Real and Complex Number Systems}\n\n\\subsection{Exercise 1}\nIf $rx = q$ or $r+x = q$ for some rational $q$, then substracting $r$ from $q$ or dividing $q$ by $r$ \nyields $x$ rational, which is a contradiction.\n\n\\subsection{Exercise 2}\nWe can first show that $\\sqrt{3}$ is irrational by seeing that $\\frac{a^2}{b^2} = 3 \\implies 3 | a, 3 | b$.\nThen, since $12 = 3 * 2^2$, we have that $\\sqrt{12}$ is irrational as well.\n\n\\subsection{Exercise 4}\nIf $\\alpha > \\beta$ then $\\alpha$ would be an upper bound as well.\n\n\\subsection{Exercise 5}\n$\\forall x \\in A, \\: -x \\leq \\sup{-A}$ and $\\forall \\epsilon \\in \\mathbb{R}, \\: \\exists x \\in A \\: | \\sup{-A} + \\epsilon < -x \\leq \\sup{-A}$. Negating the last inequality gives $\\inf{A} = -\\sup{-A}$.\n\n\\subsection{Exercise 6}\n(a) Follows from $m = \\frac{np}{q}$.\n\n(b) Put $r = \\frac{m}{n}, \\: s = \\frac{p}{q}$. Then $b^r b^s = b^{\\frac{mq}{nq}} b^{\\frac{np}{nq}}$.\nPulling out $\\frac{1}{nq}$ gives the desired result.\n\n(c) $b^r$ is an upper bound since $b > 1$, and if it were not the supremum we could choose \n$t < r$ such that $b^t > b^r$. This is not possible since again, $b > 1$.\n\n(d) Every element in $B(x + y)$ can be expressed as $b^{s + t} = b^s b^t \\: s \\leq x, \\: t \\leq y$.\nIf $\\sup{B(x+y)} = \\alpha < \\sup{B(x)}\\sup{B(y)}$, then $b^s b^t \\leq \\alpha \\implies B(x) \\leq \\alpha b^{-t}\n\\implies B(y) \\leq \\frac{\\alpha}{B(x)} \\implies B(x) B(y) \\leq \\alpha$.\n\n\\subsection{Exercise 7}\n(a) $b^n - 1 = (b - 1) (b^{n-1} + b^{n-2} + ... + 1) \\geq n (b-1)$ since $b > 1$.\n\n(b) Plug $b^{\\frac{1}{n}}$ into (a).\n\n(c) Plug $n > \\frac{b_-1}{t-1}$ into (b).\n\n(d) Using (c) gives that we can choose $n$ such that\n$b^{\\frac{1}{n}} < y \\dot b^{-w} \\implies b^{w + \\frac{1}{n}} < y$.\n\n(e) We can take the reciprocal of (c) and do the same as in (d).\n\n(f) If $b^x > y$ we can apply (e) for a contradiction, if $b^x < y$ we can apply (d)\nfor a contradiction. \n\n(g) Supremum is unique.\n\n\\subsection{Exercise 8}\nSuppose $(0, 1) < (0, 0)$. Then $(0, -1) < (0, 0)$ after multiplying by $(0, 1)$ twice yields a\ncontradiction. Similarly, assuming the opposite yields $(-1, 0) > (0, 0)$.\n\n\\subsection{Exercise 9}\nDoes exhibit least upper-bound property since you can take $(\\sup{a_i}, \\sup{b_i})$.\n\n\\subsection{Exercise 10}\nException is 0.\n\n\\subsection{Exercise 11}\nTake $w = \\frac{1}{\\abs{z}}z$ and $r = \\abs{z}$ when $\\abs{z} \\neq 0$. \n$w$ and $r$ are not uniquely determined; take $z = 0$ for example.\n\n\\subsection{Exercise 12}\nBy strong induction:\n\\begin{align*}\n \\abs{z_1 + ... + z_{n+1}} &\\leq \\abs{z_1 + ... + z_n} + \\abs{z_{n+1}} \\\\\n &\\leq \\abs{z_1} + ... + \\abs{z_{n+1}}\n\\end{align*}\n\n\\subsection{Exercise 13}\n\\begin{align*}\n \\abs{x - y}^2 &= x\\bar{x} - 2\\abs{x}\\abs{y} + y \\bar{y} \\\\\n &\\geq (\\abs{x} - \\abs{y})^2\n\\end{align*}\n", "meta": {"hexsha": "03800ab2d6d393d65992235b1bd73c2c748a1e9a", "size": 2836, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_1.tex", "max_stars_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_stars_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2019-09-19T07:33:25.000Z", "max_stars_repo_stars_event_max_datetime": "2019-09-19T07:33:25.000Z", "max_issues_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_1.tex", "max_issues_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_issues_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Principles_of_Mathematical_Analysis_Rudin/chapter_1.tex", "max_forks_repo_name": "2014mchidamb/Math-Exercise-Guides", "max_forks_repo_head_hexsha": "5ea4efe2267053695123a2c2d2c5171a672f61d4", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 38.8493150685, "max_line_length": 199, "alphanum_fraction": 0.5782792666, "num_tokens": 1112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7279754489059774, "lm_q2_score": 0.839733963661418, "lm_q1q2_score": 0.6113057091580165}} {"text": "\\section{More about $\\Tor$}\nWhere is everybody? Looks like nobody wants to hear about $\\Tor$. You're the select few.\n\nOn Monday I gave ``axioms'' for $\\Tor$, basically by saying that $\\Tor^R_n(M,-):\\mathbf{Mod}_R\\to\\mathbf{Mod}_R$ is like a homology theory. Today I'm going to show a construction of $\\Tor$, and verify the axioms. Or at least the lexseq business. I also tried to show that it's a reasonable idea to study free resolutions, namely:\n\\begin{equation*}\n\\xymatrix{\\cdots\\ar@{-->}[rr] & & F_2\\ar[dr]\\ar@{-->}[rr]^d & & F_1\\ar[dr]\\ar@{-->}[rr]^d & & F_0\\ar[dr]\\\\\n& K_2\\ar[ur]\\ar[dr] & & K_1\\ar[ur]\\ar[dr] & & K_0\\ar[ur]\\ar[dr] & & N\\ar[dr]\\\\\n0\\ar[ur] & & 0\\ar[ur] & & 0\\ar[ur] & & 0\\ar[ur] & & 0}\n\\end{equation*}\nWhere $F_{i+1}$ surjects onto $K_i$ and the $F_i$ are free $R$-modules. Splicing these exact sequences gives you a exact sequence in the top row, which is a free resolution of $N$. Of course there are a lot of choices involved, so free resolutions aren't unique. The resolution $F_\\bullet$ does \\emph{not} include $N$, and in the following diagram, the top row isn't exact but $\\cdots\\to F_0\\to N\\to 0$ is exact.\n\\begin{equation*}\n\\xymatrix{\\cdots\\ar[r] & F_2\\ar[r] & F_1\\ar[r] & F_0\\ar[r]\\ar[d]^\\epsilon & 0\\\\\n & & & N & }\n\\end{equation*}\nThen, note that:\n\\begin{equation*}\n H_q(F_\\bullet)=\\begin{cases}\nN & q=0\\\\\n0 & q>0\n\\end{cases}\n\\end{equation*}\n\\begin{construction}\nWe construct $\\Tor^R_n(M,N)$ via $ H_n(M\\otimes_R F_\\bullet)$ where $F_\\bullet$ is a free resolution of $N$.\n\\end{construction}\nI have to check that this is well-defined, that it's functorial, and that it satisfies the lexseq. Maybe I should also check what $ H_0(M\\otimes_R F_\\bullet)$ is. I do get $M\\otimes_R F_\\bullet$, because tensoring with $M$ is right exact, i.e., you have an exact sequence $M\\otimes_R F_1\\xrightarrow{p} M\\otimes_R F_0\\to M\\otimes_R N\\to 0$, and the zeroth homology is the cokernel of $p$, which is $M\\otimes_R N$.\n\nThe check that it's well-defined goes like this. I call this the fundamental theorem of homological algebra.\n\\begin{theorem}[Fundamental theorem of homological algebra]\nLet $f:M\\to N$ be an $R$-module homomorphism. Let $\\cdots\\to E_1\\to E_0\\to M\\to 0$ be such that each $E_n$ is free, and $\\cdots\\to F_1\\to F_0\\to N\\to 0$ is exact. The fundamental theorem says that I can lift the map $f:M\\to N$ to a chain map $E_\\bullet\\to F_\\bullet$ (i.e. they commute with the differentials and the augmentations $\\epsilon_N:F_0\\to N$ and $\\epsilon_M:E_0\\to M$), that is unique up to chain homotopy. I.e., they sit in the following commutative diagram:\n\\begin{equation*}\n\\xymatrix{\\cdots\\ar[r] & E_2\\ar[r]\\ar@{-->}[d]^{f_2} & E_1\\ar[r]\\ar@{-->}[d]^{f_1} & E_0\\ar[r]^{\\epsilon_M}\\ar@{-->}[d]^{f_0} & M\\ar[d]^f\\ar[r] & 0\\\\\n\\cdots\\ar[r] & F_2\\ar[r] & F_1\\ar[r] & F_0\\ar[r]^{\\epsilon_N} & N\\ar[r] & 0}\n\\end{equation*}\n\\end{theorem}\nI'm making a big deal about homological algebra in this course on algebraic topology, because this really is a homotopy-theoretic statement. It's part of the homotopy theory of chain complexes. I just want to mention something.\n\\begin{definition}\nA projective $R$-module $P$ is something such that there's a lift:\n\\begin{equation*}\n\\xymatrix{ & M\\ar@{->>}[d]\\\\\nP\\ar@{-->}[ur]\\ar[r] & N}\n\\end{equation*}\n\\end{definition}\nEvery free module is projective, clearly. Anything that's a direct summand in a projective is also projective. Any projective module is a direct summand of a free module.\n\\begin{example}\nLet $k$ be a field. Then $k\\times k$ acts on $k$ via $(a,b)c=ac$. This is an example of a projective that isn't free.\n\\end{example}\n\\begin{remark}\nThis proof uses only that $E_n$ is projective. But if you have a PID, there's no difference between projective and free.\n\\end{remark}\n\\begin{proof}[Proof of the fundamental theorem of homological algebra]\nLet's try to construct $f_0$. Consider:\n\\begin{equation*}\n\\xymatrix{0\\ar[r] & K_0\\ar[r]\\ar@{-->}[d]^{g_0} & E_0\\ar[r]^{\\epsilon_M}\\ar@{-->}[d]^{f_0} & M\\ar[d]^f &\\\\\n0\\ar[r] & L_0=\\ker(\\epsilon_N)\\ar[r] & F_0\\ar@{->>}[r]^{\\epsilon_N} & N\\ar[r] & 0}\n\\end{equation*}\nWe know that $E_0=R\\langle S\\rangle$. What we do is push forward the generators of $E$ via $\\epsilon_M$, push it forward to $f$, and pull it back via $\\epsilon_N$ which makes sense because it's surjective. This gives us $f_0$. You can restrict it to get $g_0$. Now I'm in exactly the same situation.\n\\begin{equation*}\n\\xymatrix{0\\ar[r] & K_1\\ar[r]\\ar@{-->}[d]^{g_1} & E_1\\ar[r]^{\\epsilon_M}\\ar@{-->}[d]^{f_1} & K_0\\ar[d]^{g_0} &\\\\\n0\\ar[r] & L_1\\ar[r] & F_1\\ar[r] & L_0\\ar[r] & 0}\n\\end{equation*}\nAnd $g_1$ exists. Now we need to prove the chain homotopy claim. Suppose I have $f_\\bullet:E_\\bullet\\to F_\\bullet$ and $f^{\\prime}_\\bullet:E_\\bullet\\to F_\\bullet$. Then $f^\\prime_n-f_n$ (which we'll rename $\\ell_n$) is a chain map lifting $0:M\\to N$. Let's rename things, so I have:\n\\begin{equation*}\n\\xymatrix{\\cdots\\ar[r] & E_2\\ar[r]\\ar[d]^{\\ell_2} & E_1\\ar[r]\\ar[d]^{\\ell_1} & E_0\\ar[r]^{\\epsilon_M}\\ar[d]^{\\ell_0} & M\\ar[d]^0\\ar[r] & 0\\\\\n\\cdots\\ar[r] & F_2\\ar[r] & F_1\\ar[r] & F_0\\ar[r]^{\\epsilon_N} & N\\ar[r] & 0}\n\\end{equation*}\nWe want that $\\ell_\\bullet\\simeq 0$. That is, we want $h:E_n\\to F_{n+1}$ such that $dh+hd=\\ell$. To begin with, we consider:\n\\begin{equation*}\n\\xymatrix{ & E_0\\ar[d]^{\\ell_0}\\ar@{-->}[dl]^h & \\\\\nF_1\\ar[r]^d & F_0}\n\\end{equation*}\nAt the beginning, we want $dh=0$. Well, we consider:\n\\begin{equation*}\n\\xymatrix{ & & E_0\\ar[d]^{\\ell_0}\\ar[dl]\\ar@{-->}[dll] & \\\\\nF_1\\ar@{->>}[r] & L_0\\ar[r] & F_0}\n\\end{equation*}\nBecause $F_1\\to L_0$ is a surjection, the lift exists, and we have $dh=\\ell_0$. For the next step, we have:\n\\begin{equation*}\n\\xymatrix{ & & E_1\\ar[r]\\ar[d]^{\\ell_1}\\ar@{-->}[dll] & E_0\\ar[d]^{\\ell_0}\\ar[dl]^h & \\\\\nF_2\\ar@{->>}[r] & L_1\\ar[r] & F_1\\ar[r]^d & F_0}\n\\end{equation*}\nSo what do I want to do here? Ultimately what I want is that $dh=\\ell_1-hd$. Well, $d(\\ell_1-hd)=d\\ell_1-dhd=d\\ell_1-\\ell_0d=0$ where the last equality comes because $\\ell$ is a chain map. So now we can use exactness of $E_\\bullet$ to define $h$. Exactly the same process continues.\n\\end{proof}\nThat's it. We're going to use this several times. I'm glad to have mentioned the notion of projectivity, because we'll use it later. Now, apply $M\\otimes_R -$ to that resolution (???). Suppose I have $f:N\\to N^\\prime$, and get a map $f_\\bullet:F_0\\to F^\\prime_\\bullet$. Apply $M\\otimes_R -$ to this, to get a chain map $M\\otimes_R F_\\bullet\\to M\\otimes_R F^\\prime_\\bullet$ to get a map in homology $ H_\\ast(M\\otimes_R F_\\bullet)\\to H_\\ast(M\\otimes_R F^\\prime_\\bullet)$. How independent is this of the lifting that I chose? Suppose I have two chain maps $1\\otimes f_0,f\\otimes f_0^\\prime:M\\otimes_R F_\\bullet\\to M\\otimes_R F^\\prime_\\bullet$. I can certainly form $1\\otimes h:M\\otimes_R F_n\\to M\\otimes_R F^\\prime_{n+1}$. I know that $dh+hd=f-f^\\prime$. When I tensor, I get $1\\otimes(hd+dh)=1\\otimes(f-f^\\prime)$. But I want that $(1\\otimes h)(1\\otimes d)+(1\\otimes d)(1\\otimes h)=1\\otimes f-1\\otimes f^\\prime$. We use a further property of the tensor product:\n\\begin{enumerate}\n\\setcounter{enumi}{7}\n\\item If $f,f^\\prime:N\\to N^\\prime$, then $1\\otimes(f+f^\\prime)=1\\otimes f+1\\otimes f^\\prime:M\\otimes_R N\\to M\\otimes_R N^\\prime$, and $1\\otimes(rf)=r(1\\otimes f)$. And $M\\otimes_R -$ is an $R$-linear functor. \n\\end{enumerate}\nThere's things called derived functors. In more general cases, you can't use chain complexes, but rather you use simplicial resolutions. There's non-additive homological algebra. Anyway, you check that $1\\otimes f$ and $1\\otimes f^\\prime$ are indeed chain homotopic, and so you're done.\n\nI think I've verified that it's well-defined and functorial. What about the lexseq? Now, I start with a sexseq and want to get an lexseq. Suppose I have a sexseq $0\\to A\\to B\\to C\\to 0$. I should first come up with an sexseq of resolutions. Consider:\n\\begin{equation*}\n\\xymatrix{0\\ar[r] & A\\ar[r] & B\\ar[r] & C\\ar[r] & 0\\\\\n & F^\\prime_0\\ar[u] & & F^{\\prime\\prime}_0\\ar[u] & \\\\\n & F^\\prime_1\\ar[u] & & F^{\\prime\\prime}_1\\ar[u] & \\\\\n & \\vdots\\ar[u] & & \\vdots\\ar[u] & }\n\\end{equation*}\nI want to get a free resolution in the middle. The only thing that I can think of doing is constructing:\n\\begin{equation*}\n\\xymatrix{0\\ar[r] & A\\ar[r]^i & B\\ar[r] & C\\ar[r] & 0\\\\\n0\\ar[r] & F^\\prime_0\\ar[u]^{\\epsilon_A}\\ar[r] & F^\\prime_0\\oplus F^{\\prime\\prime}_0\\ar@{-->}[u]^{\\epsilon_B}\\ar[r] & F^{\\prime\\prime}_0\\ar[u]^{\\epsilon_B}\\ar[r] & 0 \\\\\n & F^\\prime_1\\ar[u] & & F^{\\prime\\prime}_1\\ar[u] & \\\\\n & \\vdots\\ar[u] & & \\vdots\\ar[u] & }\n\\end{equation*}\nIn fact, it's the only choice because you have a free module and the sexseq splits. If I'm going to make this work, this is the only thing I can do. I need the augmentation, though. We can think of $\\epsilon_B$ as a row vector. The first entry obviously has to be $i\\epsilon$. And, well, there's a lift\\footnote{There's some ambiguity here. I have to make a choice. It might seem that the boundary map is made up of the choices, but it \\emph{isn't}! I haven't proved that. It still needs to be proved.} because $F^{\\prime\\prime}_0$ is a free resolution:\n\\begin{equation*}\n\\xymatrix{0\\ar[r] & A\\ar[r]^i & B\\ar[r] & C\\ar[r] & 0\\\\\n0\\ar[r] & F^\\prime_0\\ar[u]^{\\epsilon_A}\\ar[r] & F^\\prime_0\\oplus F^{\\prime\\prime}_0\\ar@{-->}[u]^{\\epsilon_B}\\ar[r] & F^{\\prime\\prime}_0\\ar[u]^{\\epsilon_B}\\ar@{-->}[ul]^{\\overline{\\epsilon}}\\ar[r] & 0 \\\\\n & F^\\prime_1\\ar[u] & & F^{\\prime\\prime}_1\\ar[u] & \\\\\n & \\vdots\\ar[u] & & \\vdots\\ar[u] & }\n\\end{equation*}\nSo that $\\epsilon_B=[i\\epsilon,\\overline{\\epsilon}]$. This is surjective by the Snake lemma. Consider:\n\\begin{equation*}\n\\xymatrix{& 0 & 0 & 0 & \\\\\n0\\ar[r] & A\\ar[r]^i\\ar[u] & B\\ar[r]\\ar[u] & C\\ar[r]\\ar[u] & 0\\\\\n0\\ar[r] & F^\\prime_0\\ar[u]^{\\epsilon_A}\\ar[r] & F^\\prime_0\\oplus F^{\\prime\\prime}_0\\ar@{-->}[u]^{\\epsilon_B}\\ar[r] & F^{\\prime\\prime}_0\\ar[u]^{\\epsilon_B}\\ar@{-->}[ul]^{\\overline{\\epsilon}}\\ar[r] & 0\\\\\n 0\\ar[r] & K^\\prime_0=\\ker\\epsilon_A\\ar[u]\\ar[r] & K_0\\ar[r]\\ar[u] & K^{\\prime\\prime}_0=\\ker\\epsilon_B\\ar[u]\\ar[r] & 0\\\\\n & 0\\ar[u] & 0\\ar[u] & 0\\ar[u] & }\n\\end{equation*}\nThe bottom row is exact by the $3\\times 3$-lemma. It's why I gave it to you for homework! Anyway, I now have a sexseq of free resolutions $0\\to F^\\prime_\\bullet\\to F_\\bullet\\to F^{\\prime\\prime}_\\bullet\\to 0$. Now I want a sexseq $0\\to M\\otimes_R F^\\prime_\\bullet\\to M\\otimes_R F_\\bullet\\to M\\otimes_R F^{\\prime\\prime}_\\bullet\\to 0$, but I don't know this because $M\\otimes_R -$ isn't exact. It's why we got into the business in the whole place. But $0\\to F^\\prime_\\bullet\\to F_\\bullet\\to F^{\\prime\\prime}_\\bullet\\to 0$ is split. And applying any functor gives a splitting of the sexseq, i.e., $M\\otimes_R -$ sends split sexseqs to (split) sexseq. This means that $0\\to M\\otimes_R F^\\prime_\\bullet\\to M\\otimes_R F_\\bullet\\to M\\otimes_R F^{\\prime\\prime}_\\bullet\\to 0$ also splits. Thus we're done proving the lexseq in $\\Tor$.\n\nThe key idea in homological algebra is that free modules are good.\n", "meta": {"hexsha": "fa495d39862ebe0274457090df61ae1ed58179e9", "size": 10964, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "old-905/lec-22-more-on-tor.tex", "max_stars_repo_name": "ichung/algtop-notes", "max_stars_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 5, "max_stars_repo_stars_event_min_datetime": "2017-04-26T15:00:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-05-27T22:47:06.000Z", "max_issues_repo_path": "old-905/lec-22-more-on-tor.tex", "max_issues_repo_name": "ichung/algtop-notes", "max_issues_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 3, "max_issues_repo_issues_event_min_datetime": "2018-03-13T17:54:37.000Z", "max_issues_repo_issues_event_max_datetime": "2018-03-13T17:59:46.000Z", "max_forks_repo_path": "old-905/lec-22-more-on-tor.tex", "max_forks_repo_name": "ichung/algtop-notes", "max_forks_repo_head_hexsha": "3f5d3189e2082716a69fccc1711d02ed848552d2", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 3, "max_forks_repo_forks_event_min_datetime": "2017-10-21T18:15:11.000Z", "max_forks_repo_forks_event_max_datetime": "2019-08-13T17:38:04.000Z", "avg_line_length": 89.868852459, "max_line_length": 959, "alphanum_fraction": 0.674206494, "num_tokens": 4113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125848754472, "lm_q2_score": 0.8128673178375734, "lm_q1q2_score": 0.6112864528478053}} {"text": "\\graphicspath{{Chapter2/Figs/}}\n\n\\section{Natural Gradient descent}\n\n%REPRODUCED WITH SOME MODIFCIATIONS FROM:\n%\thttps://wiseodd.github.io/techblog/2018/03/14/natural-gradient/\n%\tHoffman2013\n%\tAmari1998\n\nThe application of standard gradient descent optimisation to functions that depend on probability distributions (as the KL divergence in in variational inference) has an important limitation, as we will show below. \\\\\n\nConsider the gradient descent optimisation of an ELBO $\\Lagr(\\bfX|\\theta)$, where $\\bfX$ are the hidden variables and $\\theta$ the corresponding parameters. From the definition of a gradient:\n\\[\n\\nabla \\Lagr(\\bfX|\\theta) = \\lim_{h\\to0} \\frac{\\Lagr(\\theta + h) - \\Lagr(\\theta)}{\\|h\\|}\n\\]\nwhere $h$ represents an infinitesimally small positive step in the space of $\\theta$.\\\\\n\nTo find the steepest gradient, one would need to search over all possible directions $d$ in an infinitely small distance $h$, and select the $\\hat{d}$ with the largest gradient:\n\\[\n\\nabla \\Lagr(\\bfX|\\theta) = \\lim_{h\\to0} \\frac{1}{\\|h\\|}\\argmax_{d \\, s.t. \\|d\\|=h} \\Lagr(\\theta+d) - \\Lagr(\\theta)\n\\]\nNotice that the neighborhood of $\\theta$ is measured in terms of its Euclidean norm, and the direction of steepest descent is hence dependent on the Euclidean geometry of the $\\theta$ space. This becomes problematic in variational inference. A small step from $\\theta_t$ to $\\theta^{(t+1)}$ does not guarantee a similar small change from $\\Lagr(\\bfX|\\theta^t)$ to $\\Lagr(\\bfX|\\theta^{(t+1)})$. As an example, consider four random variables:\n\\begin{equation}\n\t\\begin{split}\n\t\t\\psi_1 &\\sim \\Ndist{0}{5} \\\\\n\t\t\\psi_2 &\\sim \\Ndist{10}{5}\n\t\\end{split}\n\t\\qquad\n\t\\begin{split}\n\t\t\\psi_3 &\\sim \\Ndist{0}{1} \\\\\n\t\t\\psi_4 &\\sim \\Ndist{10}{1}\n\t\\end{split}\n\\end{equation}\n\nUsing the euclidean metric, the distance between $\\psi_1$ and $\\psi_2$ is the same as the distance between $\\psi_3$ and $\\psi_4$. However, the distance in distribution space (measured for example by the KL divergence) is clearly much larger between $\\psi_1$ and $\\psi_2$ than between $\\psi_3$ and $\\psi_4$ (Figure X).\n\n\\begin{figure}[]\n\t\\begin{center}\n\t\t\\includegraphics[width=0.75\\textwidth]{rnorm.pdf}\n\t\t\\caption{}\n\t\t\\label{fig:XX}\n\t\\end{center}\n\\end{figure}\n\nHence, rather than using the euclidean distance, it is more appropriate to use a KL divergence as a distance metric:\n\\[\n\t\\nabla \\Lagr(\\bfX|\\theta)_{KL} = \\lim_{h\\to0} \\frac{1}{\\|h\\|}\\argmax_{d \\, s.t. KL[p_\\theta||p_{\\theta+d}]=\\|h\\|} \\Lagr(\\theta+d) - \\Lagr(\\theta)\n\\]\nThe direction of steepest ascent measured by the KL divergence is called the natural gradient.\n\nTo find the optimal $\\hat{d}_{KL}$, one needs to solve the following optimisation problem:\n\\[\n\t\\hat{d}_{KL} = \\argmin_{d} \\Lagr(\\theta+d)\n\\]\nwith the constrain $KL[p_\\theta||p_{\\theta+d}]=c$, where $c$ is an arbitrary constant. Introducing Lagrange multipliers $\\lambda$: \\cite{??}:\n\\[\n\t\\Lagr(\\theta+d) + \\lambda(KL[p_\\theta||p_{\\theta+d}] - c)\n\\]\nBefore taking its derivative, we can simplify the equation by approximating the term $ \\Lagr(\\theta+d)$ with a First-order Taylor series and the term $KL[p_\\theta||p_{\\theta+d}]$ with a second-order Taylor series around $\\theta+d$:\nTO FINMISH\n\\[\nKL[p_\\theta||p_{\\theta+d}] \\approx \nKL[p_\\theta||p_{\\theta}] + \n%d \\d nabla_{\\theta+d} KL[(\\bfX|\\theta)||(\\bfX|\\theta+d)] | _{\\theta} +\n%\\frac{1}{2} d^T \\d nabla^2_{\\theta+d} KL[(\\bfX|\\theta)||(\\bfX|\\theta+d)] | _{\\theta}d\n\\]\nThe first two terms vanish. The third term corresponds to the negative of the Hessian matrix with respect to $\\theta+d$ (the Fisher Information Matrix $F$), evaluated at $\\theta$. Hence, the term $KL[p_\\theta||p_{\\theta+d}]$ simplifies to $\\frac{1}{2} d^T F d$\n\nThe equation to optimise becomes:\n\\[\n\\Lagr(\\theta) + \\nabla_{\\theta} \\Lagr(\\theta)^T d + \\frac{1}{2} \\lambda d^T F d - \\lambda c\n\\]\n\nFinally, by taking the derivative of Equation X with respect to $d$, setting to zero and solving, we obtain the direction of the steepest natural gradient. It corresponds to the standard (euclidean) gradient pre-multiplied by the inverse of the Fisher Information Matrix:\n\\[\n%\t\\frac{\\partial \\Lagr(\\theta+d)}{\\partial d} + \\lambda\\frac{\\partial(KL[p_\\theta||p_{\\theta+d}]}{\\partial d} = 0\n\\hat{d} \\propto F^{-1} \\nabla_{\\theta} \\Lagr(\\theta)\n\\]\n\n\nFINISH HERE. CHECK IF THE BOTTOM IS MEANINGFUL:::\n\nPlugging this expression into [[EQUATION XX]], we find that the optimal $d$ is:\n\\[\n\\hat{d} = -\\frac{1}{\\lambda} F(\\theta)^{-1} \\nabla_{\\theta} \\Lagr(\\theta)\n\\]\nThe constatn factor can be absorbed into the learning rate. The natural gradient is then defined\n\\[\n\\nabla \\Lagr(\\bfX|\\theta)_{KL} = F(\\theta)^{-1} \\nabla_{\\theta} \\Lagr(\\theta)\n\\]\n\n\n% \\cite{[[GITHUBPAGE]]} shows how, in a simple logistic regression problem, the use of natural gradients lead to improved convergence over standard gradient descent.", "meta": {"hexsha": "06f3d5da5ea29bda538e02cc8b7b6e7b6097ac86", "size": 4808, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chapter2/old/natural_gradient_descent.tex", "max_stars_repo_name": "rargelaguet/thesis", "max_stars_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 15, "max_stars_repo_stars_event_min_datetime": "2021-01-08T13:01:16.000Z", "max_stars_repo_stars_event_max_datetime": "2022-01-26T07:24:40.000Z", "max_issues_repo_path": "Chapter2/old/natural_gradient_descent.tex", "max_issues_repo_name": "rargelaguet/thesis", "max_issues_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Chapter2/old/natural_gradient_descent.tex", "max_forks_repo_name": "rargelaguet/thesis", "max_forks_repo_head_hexsha": "ff3f7b996710c06d6924b7e780a4a9531651a3a0", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2021-01-09T04:47:49.000Z", "max_forks_repo_forks_event_max_datetime": "2021-10-04T08:25:50.000Z", "avg_line_length": 51.6989247312, "max_line_length": 440, "alphanum_fraction": 0.7036189684, "num_tokens": 1528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7520125737597971, "lm_q2_score": 0.8128673223709251, "lm_q1q2_score": 0.6112864472213941}} {"text": "\\documentclass[../../main.tex]{subfiles}\n\\begin{document}\n\\subsubsection{At Index}\nThe operation $atIndex$ will return the Value at a\nspecified numeric index within a Collection or\nan empty Collection if there is no value at the specified index.\n\\begin{schema}{AtIndex[Collection, \\nat]}\n idx? : \\nat\\\\\n coll? : Collection \\\\\n atIndex~\\_ : Collection \\cross \\nat \\surj V\n \\where\n \\# ~ idx? = 1 \\\\\n coll! = atIndex(coll?, idx?) = (head ~(~idx? \\extract coll?)) \\iff idx? \\in coll? \\\\\n coll! = atIndex(coll?, idx?) = \\langle \\rangle \\iff idx? \\not \\in coll?\n\\end{schema}\nGiven the definition of the $Collection$ and $V$ free types\n\\begin{zed}\n Collection :== emptyColl ~| ~append \\ldata Collection \\cross Scalar ~\\lor Collection ~\\lor KV \\cross \\nat \\rdata \\\\\n V ::= Scalar ~| ~Collection ~| ~KV\n\\end{zed}\nThe collection member $coll?_{idx?} : V$ is implied from $append$ accepting the argument of type $Scalar ~\\lor Collection ~\\lor KV \\equiv V$ which means each Collection member is of type $V$. Given that extraction ($~\\_~\\extract~\\_$) returns a Collection,\n\\begin{axdef}\n \\seq X : Collection\n \\where\n \\_~\\extract~\\_ : \\power \\nat_1 \\cross \\seq X \\fun \\seq X\n\\end{axdef}\nin order for $atIndex$ to return the collection member without altering its type,\nthe first member of $atIdx'$ must be returned, not $atIdx'$ itself.\n\\begin{axdef}\n atIdx' : Collection \\\\\n coll!, coll?_{idx?} : V \\\\\n \\where\n atIdx' = (~idx? \\extract coll?) \\implies \\langle coll?_{idx?} \\rangle \\\\\n coll! = head(atIdx') = coll?_{idx?}\n\\end{axdef}\nThe $head$ call is made possible by restricting $idx?$ to be a single numeric value.\n\\begin{argue}\n idx?, idx' : \\nat \\\\\n \\t1 \\# ~ idx? = 1 @ (~idx? \\extract coll?) = \\langle coll?_{idx?} \\rangle @ \\\\\n \\t2 (head (~idx? \\extract coll?)) = coll?_{idx?} & expected return given $idx?$ \\\\\n \\t1 \\# ~ idx' \\geq 2 @ (~idx' \\extract coll?) = \\langle coll?_{idx'_{i}}~..~~coll?_{idx'_{j}} \\rangle @ \\\\\n \\t2 (head (~idx' \\extract coll?)) = coll?_{idx'_{i}} & unexpected return given $idx'$\n\\end{argue}\nAdditionally, if the provided $idx? \\not \\in coll?$ then an empty Collection will be returned\ngiven that $head$ must be passed a non-empty Collection.\n\\begin{axdef}\n head : \\seq_1 X \\fun X\n \\where\n idx? \\not \\in coll? \\implies (~idx? \\extract coll?) = \\langle \\rangle ~ \\lnot ~\\seq_1\n\\end{axdef}\nThe properties of $atIndex$ are illustrated in the following examples.\n\\begin{argue}\n X = \\langle x_{0}, x_{1}, x_{2} \\rangle \\\\\n \\t1 x_{0} = 0 \\\\\n \\t1 x_{1} = foo \\\\\n \\t1 x_{2} = \\langle a, b, c \\rangle \\\\\n atIndex(X, 0) = 0 & $head~(\\langle~x_{0}~\\rangle)$\\\\\n atIndex(X, 1) = foo & $head~(\\langle~x_{1}~\\rangle)$\\\\\n atIndex(X, 2) = \\langle a, b, c \\rangle & $head~(\\langle~x_{2}~\\rangle)$\\\\\n atIndex(X, 3) = \\langle \\rangle & $3 \\not \\in X \\implies x_{3} \\not \\in X$\n\\end{argue}\n\\end{document}\n", "meta": {"hexsha": "453e94f08015339aa892a275c4f860f11a19d919", "size": 2840, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "docs/operations/collections/atIndex.tex", "max_stars_repo_name": "yetanalytics/dave", "max_stars_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": 25, "max_stars_repo_stars_event_min_datetime": "2018-08-17T00:38:20.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-09T02:32:37.000Z", "max_issues_repo_path": "docs/operations/collections/atIndex.tex", "max_issues_repo_name": "adlnet/dave", "max_issues_repo_head_hexsha": "9339713fac747118e462e4fc7e1ecd54e5d916e3", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": 95, "max_issues_repo_issues_event_min_datetime": "2018-08-31T18:57:51.000Z", "max_issues_repo_issues_event_max_datetime": "2020-09-15T16:40:01.000Z", "max_forks_repo_path": "docs/operations/collections/atIndex.tex", "max_forks_repo_name": "yetanalytics/dave", "max_forks_repo_head_hexsha": "7a71c2017889862b2fb567edc8196b4382d01beb", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 6, "max_forks_repo_forks_event_min_datetime": "2019-09-28T06:48:47.000Z", "max_forks_repo_forks_event_max_datetime": "2020-03-31T16:20:47.000Z", "avg_line_length": 45.0793650794, "max_line_length": 255, "alphanum_fraction": 0.6433098592, "num_tokens": 963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673269042768, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6112864461127572}} {"text": "\\SubSecDef{verify}{Verifying Algebraic Security}\n\n\\PropRef{composability} shows that we can compose algebraically secure circuits. Large circuits can be constructed from a set of \\emph{gadgets} - small algebraically secure circuits with some useful semantics. In order to design new gadgets we need to be able to check their algebraic security. The simplest way to get a bound on the absolute correlation is based on the algebraic degree of computed functions: the minimum weight of a \\emph{nonzero} function of $n$ bits of degree $d$ is equal to $2^{n-d}$ (see e.g. \\cite{Carlet}). Therefore, we can think about the following algorithm for checking a circuit $C(x,r_C)$: for any fixed input $x$ compute the ANFs of the functions computed in $C(x,\\cdot)$ (functions of $r_C$) and return the maximum observed degree. The degree does not grow when functions are combined linearly. Therefore, the absolute correlation bound can not grow as well, except when the resulting function is constant in which case the absolute correlation is maximal and the gadget may be insecure. As a result, our method for verifying algebraic security splits into two parts:\n\n\\begin{enumerate}\n \\item verify that there is no absolute correlation equal to 1 among restrictions of functions from $\\FUNCSD{1}(C)$ except the constant functions and affine functions of $x$;\n \\item compute the maximum degree among all restrictions of the intermediate functions and compute the corresponding correlation bound.\n\\end{enumerate}\n\nThe second step is straight-forward. We describe an algorithm that solves the first step.\n\nConsider a circuit $C(x,r): \\field{N}\\times\\field{R}\\to\\field{M}$.\nFor all $c \\in \\field{N}$ let $L_c$ be the linear map that returns the restriction $x=c$ of a function $f$ from $\\FUNCSD{1}(C)$ (e.g. if functions are represented as truth table vectors then $L_c$ returns the truth table entries corresponding to the case $x = c$). Note that the domain of $L_c$ is defined to be the subspace $\\FUNCSD{1}(C)$.\n\nWe now give an equivalent condition for the first part of the verification. It serves as a basis for the verification algorithm given in \\AlgRef{verify}.\n\n\\begin{proposition}\nThe circuit $C$ is $\\maxcor$-1-AS for some $\\maxcor < 1$ if and only if for all $c$ the following holds:\n\\begin{equation}\n \\dim{\\ker{L_c}} = N.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor any $c \\in \\field{N}$ let $F_c$ be the subspace of $\\FUNCSD{1}(C)$ containing functions that are constant when $x$ is fixed to $c$. Also let $F = \\bigcup_c F_c$.\n$\\maxcor < 1$ requires that any $f\\in \\FUNCSD{1}(C)$ either belongs to $\\XFUNCSD{1}(C)$ or is non-constant for any fixed $x$. It is equivalent to require that $F$ is equal to $\\XFUNCSD{1}(C)$. Note that each $F_c$ includes $\\XFUNCSD{1}(C)$ as a subset. Therefore, $F = \\bigcup_c F_c$ is equal to $\\XFUNCSD{1}(C)$ if and only if for all $c$ $F_c = \\XFUNCSD{1}(C)$. Since these are linear subspaces then we can compare their dimensions.\n\n$\\XFUNCSD{1}(C)$ is spanned by all $x_i$ and the constant-1 function:\n\\eql{prop-algo-dimX}{\n \\dim{\\XFUNCSD{1}(C)} = N + 1;\n}\n\nThe constant-1 function always belongs to $\\FUNCSD{1}(C)$ and to any of the $F_c$. The subspace of functions that are constant on the restriction can be obtained by adding the constant-1 function to the subspace of functions that are equal to zero on the restriction:\n\\eql{xxx}{\n F_c &= \\ker{L_c} \\oplus \\pset{\\fzero,\\fone}, \\\\\n \\EqLabel{prop-algo-dimFc}\n \\dim{F_c} &= \\dim{\\ker{L_c}} + 1.\n}\n\nBy comparing the dimensions obtained in \\EqRef{prop-algo-dimX},\\EqRef{prop-algo-dimFc} we prove the proposition.\n\\end{proof}\n\n\n\\newcommand\\tbas{\\bas_c}\n\nThe algorithm operates on functions using their truth tables. The truth tables are obtained by evaluating the circuit on all possible inputs and recording the values computed in each node. The set of computed truth tables corresponds to $\\FUNCS(C)$. By removing redundant vectors we can compute a basis $\\bas$ of $\\FUNCSD{1}(C)$ (and also ensure presence of the constant-1 vector). Then, for each $c$ we take the part of each basis vector that corresponds to the fixed $x=c$ (and $r$ taking all possible values). These parts form the subspace $\\Ima{L_c}$. We compute a basis $\\tbas$ of these parts. Finally, we verify that\n\\begin{equation}\n \\dim{\\ker{L_c}} = \\dim{\\FUNCSD{1}(C)} - \\dim{\\Ima{L_c}} = |\\bas| - |\\tbas| = N.\n\\end{equation}\n\n\\FigTex{algo-verify.tex}\n\nThe algorithm is implemented in SageMath~\\cite{sage} and is publicly available in~\\cite{OurWhiteboxCode}.\n\n\\paragraph{Complexity analysis.}\nThe truth tables have size $2^{N+R}$ bits. Computing the basis of $\\FUNCSD{1}(C)$ takes time $\\OO(min(2^{N+R}, |C|)^\\matexp)$. The same holds for $\\Ima{L_c}$ except that the vectors have size $2^{R}$ and for small $R$ this can be done more efficiently. The total complexity is $\\OO(min(2^{N+R}, |C|)^\\matexp + 2^N min(2^{R}, |C|)^\\matexp)$. Recall that by Proposition~\\ref{wba.prop:nonlinear-gates} we should consider only the nonlinear nodes of the circuit. \n\n\n", "meta": {"hexsha": "b5f3f359b65a61a107eab8021f34143f42e3bbcb", "size": 5029, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis-source/9wbCounter/34verify.tex", "max_stars_repo_name": "hellman/thesis", "max_stars_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 19, "max_stars_repo_stars_event_min_datetime": "2019-05-16T19:55:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-12-31T15:36:12.000Z", "max_issues_repo_path": "thesis-source/9wbCounter/34verify.tex", "max_issues_repo_name": "hellman/thesis", "max_issues_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 1, "max_issues_repo_issues_event_min_datetime": "2021-08-09T11:26:45.000Z", "max_issues_repo_issues_event_max_datetime": "2021-08-09T11:26:45.000Z", "max_forks_repo_path": "thesis-source/9wbCounter/34verify.tex", "max_forks_repo_name": "hellman/thesis", "max_forks_repo_head_hexsha": "6ba1c2b241e63c07cf76108481c1b67f21a50f12", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-08-05T19:40:16.000Z", "max_forks_repo_forks_event_max_datetime": "2021-08-05T19:40:16.000Z", "avg_line_length": 86.7068965517, "max_line_length": 1101, "alphanum_fraction": 0.7341419765, "num_tokens": 1407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673178375734, "lm_q2_score": 0.7520125737597972, "lm_q1q2_score": 0.6112864438122567}} {"text": "\\section{Network Level Performance Without Retransmission}\n\\label{sec:op_over_infinite_plane}\nThe transmission success probability $p_s(r)$ paves the way to analyze the network packet loss rate $P_{f}$, which is the base to calculate:\\begin{inparaenum}[1)]\n\t\\item the maximum normalized load $L_{\\text{max}}$ under packet loss rate constraint $P_{f}^{\\text{max}}$;\n\t\\item the normalized spatial throughput (per BS) $S = (1-P_{f}) p\\lambda_{m}/ \\lambda_{b}$\n\\end{inparaenum}\nWe first consider two standard ALOHA transmission approaches in unicast mode (best and nearest BS attach) and then analyze the macro diversity case. For macro diversity case, we also consider two cases: selective ratio and maximum ratio combing.\n\\subsection{Nearest BS attach method}\n\\label{sec:nearest_BS_attach_method}\n%A device can attach to the best BS which the received power of packet transmission is the strongest if the channel state info is available during the communication.\nIn this subsection, we assume that the device attaches to the geographically nearest BS then transmits packets. \n%This method is rational when downlink is seldom used for the sake of cost. %TODO: Add this in cover letter.\nTherefore, network packet loss rate $P_{f,n}$ (index $n$ means nearest) is the expectation of $p_s(r)$ with respect to the distance to the nearest base station $r$ . The PDF of $r$ for a PPP, proved in~\\cite{andrews2011tractable}, is as follows:\n\\begin{align}\n\\label{eq:pdf_nearest_distance}\nf\\left( r\\right) = 2 \\pi \\lambda_b r \\exp(-\\lambda_b \\pi r^2), r \\in \\left[ 0, +\\infty\\right]. \n\\end{align}\nThus:\n\\begin{align}\n\\label{eq:mean_bx+1_step1}\n&P_{f,n}= \\mathbb{E}_{r}\\left[ 1-p_{s}\\left(r\\right) \\right] \\nonumber\\\\\n&= 1 -\\int_{0}^{+\\infty} \\mathbb{E}_{\\chi} \\left[ \\exp(-p \\lambda_{m} \\pi A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} r^2 e^{-\\frac{2}{\\gamma}\\chi}) \\right] 2 \\pi \\lambda_b r e^{-\\lambda_b \\pi r^2} dr \\nonumber\\\\\n&= 1-\\pi \\lambda_b \\mathbb{E}_{\\chi}\\left[ \\int_{0}^{+\\infty} \\exp(-p \\lambda_{m} \\pi A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} r^2 e^{-\\frac{2}{\\gamma}\\chi}-\\lambda_b \\pi r^2)\\right] \\!\\!dr^2 \\nonumber\\\\\n&= 1 -\\mathbb{E}_{\\chi}\\left[\\left( A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} L e^{-\\frac{2}{\\gamma}\\chi}+1 \\right)^{-1} \\right] \\nonumber\\\\\n%&= 1- \\int_{-\\infty}^{+\\infty} \\frac{1}{ \\frac{p\\lambda_{m}AK}{\\pi \\lambda_{b}}e^t+1}\\frac{\\exp\\left\\lbrace -\\frac{t^2}{2 \\left( \\frac{2}{\\gamma}\\sigma\\right) ^2}\\right\\rbrace }{\\sqrt{2\\pi} \\frac{2}{\\gamma}\\sigma} dt \\nonumber\\\\\n&= 1 - \\int_{-\\infty}^{+\\infty} \\!\\!\\! \\frac{\\gamma}{ 2\\sqrt{2\\pi}\\left( e^t+1\\right)\\sigma} \\exp \\left\\lbrace -\\frac{\\gamma^2}{8 \\sigma^2} \\left( t-\\ln(A L \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\!\\! \\frac{2\\sigma^2}{\\gamma^2}} ) \\right) ^2 \\right\\rbrace dt. \n\\end{align}\n%which is actually a logistic-normal integral. According to~\\cite{crooks2009logistic}, this kind of \nIntegral in $(\\ref{eq:mean_bx+1_step1})$ can be accurately approximated by a logistic function~\\cite{crooks2009logistic}. Thus,\n\\begin{align}\n\\label{eq:bs_nst_att_analytical}\nP_{f,n} \n&\\approx 1 - \\frac{1}{1 + \\exp\\left\\lbrace \\left( 1 +\\pi \\sigma^2 / (2\\gamma^2) \\right)^{-1/2} \\ln( A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} L) \\right\\rbrace} \\nonumber\\\\\n&\\approx 1-\\frac{ 1 }{ 1 + \\left( A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} L\\right) ^{C} }\n\\end{align}\nwhere $A$ is defined in $(\\ref{eq:def_ps_2})$, $L=p\\lambda_{m}/\\lambda_{b}$, $C= \\left( 1 +\\pi \\sigma^2 / (2\\gamma^2) \\right)^{-1/2}$. We conducted a Monte-Carlo simulation and found that the maximum difference between $(\\ref{eq:mean_bx+1_step1})$ and $(\\ref{eq:bs_nst_att_analytical})$ is $2.46\\%$ in normalized load interval $\\left[ 0.021, 0.3\\right] $, which proves the accuracy of the proposed approximation formula $(\\ref{eq:bs_nst_att_analytical})$\\qsong{Remember to put the figure into Annexe part. Here I wan to show that, the difference between approximation formula and exact integral is smaller when normalized load increase }.\n\\begin{figure}[!ht]\n\t\\centering\n\t\\includegraphics[width=\\linewidth]{/Users/qsong/Documents/slotted_aloha_related_project/test/comparison_monte_carlo_approximation.eps}\n\t\\caption{Comparison between approximation formula $(\\ref{eq:bs_nst_att_analytical})$ and Monte-Carlo simulation result.}\n\t\\label{fig:comparison_monte_carlo}\n\\end{figure}\n\nFrom $(\\ref{eq:bs_nst_att_analytical})$, the maximum supported normalized load $L_{\\text{max},n}$ and normalized spatial throughput (index $n$ refers to the nearest BS attach method) is obtained:\n\\begin{align}\n\\label{eq:bs_nst_att_max_load}\nL_{\\text{max},n} &= \\frac{1}{A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} } \n\\left( \\frac{P_{f}^{\\text{max}}}{1 - P_{f}^{\\textbf{max}}} \\right) ^{\\frac{1}{C}}, \\\\\n\\label{eq:bs_nst_att_spatial_throughput}\nS_{n} &= \\frac{L}{1 + \\left( A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} L\\right) ^{C} }.\n\\end{align}\n%We should note that $(\\ref{eq:bs_nst_att_spatial_throughput})$ is obtained by $S = (1-P_{f}) p\\lambda_{m}/ \\lambda_{b}$, where $P_{f}$ is a ... This comment is applied to all spatial throughput in BS attach case.\n\n\\subsection{Best BS attach method}\nIn this subsection, we assume that the device attaches to the BS for which the received power averaged over all fading realizations is the strongest (i.e. the BS that maximizes $r^{-\\gamma}\\exp(\\chi)$). Let $P_{f,b} $ be packet loss rate for this case. As shown in~\\cite[lemma 1]{dhillon2014downlink}, attaching to the best BS in a PPP of intensity $\\lambda_{b}$ with shadowing is equivalent to attaching the nearest one in a transformed PPP of intensity $\\lambda_{b} \\mathbb{E}\\left[ e^{-\\frac{2}{\\gamma}\\chi}\\right] = \\lambda_{b} e^{\\frac{2\\sigma^2}{\\gamma^2}}$ without shadowing. \nTherefore, the link level transmission success probability $p_{s}(r)$ is as follows:\n\\begin{align}\n\t\\label{eq:succ_proba_with_modified_r}\n\tp_{s}(r) = \\exp(-p \\lambda_{m} \\pi A e^{\\frac{2\\sigma^2}{\\gamma^2}} \\theta_{T}^{\\frac{2}{\\gamma}} r^2 )\n\\end{align}\nThe distribution of modified $r$ is thus:\n\\begin{align}\n\t\\label{eq:pdf_modified_r}\n\tf(r) = 2 \\pi \\lambda_b e^{\\frac{2\\sigma^2}{\\gamma^2}} \\exp( -\\lambda_b e^{\\frac{2\\sigma^2}{\\gamma^2}} \\pi r^2 ) r \n\\end{align}\nSimilar to the analysis in Sec.~\\ref{sec:nearest_BS_attach_method}, we have:\n\\begin{align}\n\\label{eq:bs_best_att_analytical}\n&P_{f,b}= \\mathbb{E}_{r}\\left[ 1-p_{s}\\left(r\\right) \\right] \\nonumber\\\\\n&= 1 -\\int_{0}^{+\\infty} \\exp(-p \\lambda_{m} \\pi A \\theta_{T}^{\\frac{2}{\\gamma}} e^{\\frac{2\\sigma^2}{\\gamma^2}} r^2 ) 2 \\pi \\lambda_b e^{\\frac{2\\sigma^2}{\\gamma^2}} \\exp( -\\lambda_b e^{\\frac{2\\sigma^2}{\\gamma^2}} \\pi r^2 ) r dr \\nonumber\\\\\n&= 1-\\frac{1}{ 1 + A \\theta_{T}^{\\frac{2}{\\gamma}} L }.\n\\end{align}\nBy inverting $(\\ref{eq:bs_best_att_analytical})$, the maximum supported normalized load $L_{\\text{max}, b}$ (subscript $b$ refers to the best BS attach method) is as follows:\n\\begin{align}\n\tL_{\\text{max},b} &=\\frac{1}{A \\theta_{T}^{\\frac{2}{\\gamma}} } \n\t\\frac{P_{f}^{\\text{max}}}{1 - P_{f}^{\\textbf{max}}}. \n\\end{align}\nThe normalized spatial throughput $S_{b}$ is as follows:\n\\begin{align}\n\tS_{b} = \\frac{L}{1 + A \\theta_{T}^{\\frac{2}{\\gamma}} L }.\n\\end{align}\n\n\\subsection{Selective Combining like Macro Diversity}\n\\label{sec:sc_macro_diversity}\nSince a transmitted packet is received by all surrounding BS, the transmission of this packet fails if and only if the received SINR at each BS is less than the capture ratio. In other words, if the maximum SINR is still less than the capture ratio, the packet transmission is failed. This is why such a scheme is called selective combining (SC) like macro reception diversity. In this section, we evaluate SC-like macro diversity in one-shot random access case.\n\nStrictly speaking, the cumulative interferences at two different BS are correlated: when a device is transmitting, it generates some non-negligible interference on base stations that are not very far. However, it is still rational to assume that the interferences received by different BS are mutually independent.\n%when received power at each BS exhibits obvious diversity such as in urban area suffered by strong shadowing effect. \nThe reasons are in twofold:\\begin{inparaenum}[1)]\n\t\\item the interference correlation coefficient is shown to be close to $0$ if locations of two BS are different with path-loss model $r^{-\\gamma}$~\\cite[lemma 3.5]{haenggi2009interference}; \n\t\\item Each contribution to the cumulative interference is affected by fading and shadowing, which are i.i.d. random variable for different base stations.\n\\end{inparaenum}\nThus, the packet loss rate $P_{f,m}$ (index $m$ refers to macro reception diversity) is the expectation of the product of failure probabilities to each BS:\n\\begin{align}\n\\label{eq:definition_pfm}\nP_{f,m} &= \\mathbb{E}\\left[ \\prod_{r_i \\in \\Phi_{b}} (1-p_{s}(r_i)) \\right], \\text{with } r_i \\in \\left[0, +\\infty\\right],\n\\end{align} \nwhere $r_i$ is the distance between the device and BS with label $i$. Using Campbell theorem for $(\\ref{eq:definition_pfm})$:\n\\begin{align}\n\\label{eq:bs_rx_divers_analytical_before_last}\nP_{f,m} &= \\exp\\left\\lbrace -2\\pi \\lambda_{b} \\int_{0}^{+\\infty} p_{s}(r)rdr \\right\\rbrace.\n\\end{align} \nCombining with $(\\ref{eq:def_ps_2})$ and changing integration order, $(\\ref{eq:bs_rx_divers_analytical_before_last})$ can be further simplified:\n\\begin{align}\n\\label{eq:bs_rx_divers_analytical}\nP_{f,m} \n%&= \\exp\\left\\lbrace -\\left[A \\theta_{T}^{\\frac{2}{\\gamma}} L \\right] ^{-1}\\right\\rbrace.\n&= \\exp\\left\\lbrace -\\frac{1}{A \\theta_{T}^{\\frac{2}{\\gamma}} L }\\right\\rbrace.\n\\end{align}\nFrom $(\\ref{eq:bs_rx_divers_analytical})$, we can easily find the maximum load and the throughput:\n\\begin{align}\n\t\\label{eq:bs_rx_divers_max_load}\n\tL_{\\text{max}, m} &= \\frac{1}{A \\theta_{T}^{\\frac{2}{\\gamma}}} \\frac{1}{\\ln(1/P_{f}^{\\text{max}})}, \\\\\n\t\\label{eq:bs_rx_divers_spatial_throughput}\n\tS_{m} &= L \\left(1 -\\exp\\left\\lbrace -\\frac{1}{A \\theta_{T}^{\\frac{2}{\\gamma}} L } \\right\\rbrace \\right). \n\\end{align}\nThe macro diversity gain against the best BS attach mode $G_{\\text{diversity},b}$ is obtained by inverting $(\\ref{eq:bs_best_att_analytical})$ and $(\\ref{eq:bs_rx_divers_analytical})$:\n\\begin{align}\n\t\\label{eq:macro-diversity-gain}\n\tG_{\\text{diversity},b} &= \\left(1-P_{f}^{\\text{max}}\\right)/ \\left( P_{f}^{\\text{max}} \\ln(1/P_{f}^{\\text{max}}) \\right) .\n\\end{align}\nWe deduce that, whatever the ALOHA type, the macro diversity gain against best BS attach\nonly depends on the packet loss rate target.\n\n\\subsection{Maximum Ratio Combining based Macro diversity}\nDue to the fact that a transmitted packet theoretically can be simultaneously received by all BS (ignorance of background noise), linear combining of signals at each BS can be leveraged so that the output SINR is maximized. Such a scheme is called Maximum Ratio Combining (MRC) based macro diversity\nIn this section, the performance of MRC-based macro diversity in case of one-shot random access is evaluated.\n\nConsider a typical device $x_0$ at origin, it has been proved that if the weigh factors involved in MRC context is well designed (see Fig.~\\ref{fig:mrc_macro_diversity_recpetion_illustration}), the output $\\text{SINR}$ $\\Theta$ of best combiner is expressed as \\qs{We need a reference for this...}:\n\\begin{align}\n\\Theta &= \\sum_{y_j \\in \\Phi_{b}}^{} \\theta_{y_j} \\nonumber\\\\\n&= \\sum_{y_j \\in \\Phi_{b}}^{} \\frac{H_{y_j} \\exp(\\chi_{y_j}) r_{y_j}^{-\\gamma}}{I_{y_j}} \\nonumber\\\\\n&= \\sum_{y_j \\in \\Phi_{b}}^{} \\epsilon_{y_j} r_{y_j}^{-\\gamma}\n\\end{align}\nwhere $\\theta_{y_j}$ is the received SINR at BS $y_j$, $H_{y_{j}}$ and $\\exp(\\chi_{y_j})$ are respectively the Rayleigh fading and shadowing component for link between device $x_0$ and BS $y_j$, $I_{y_j}$ refers to cumulative interference suffered at BS $y_j$. For the sake of notation simplicity, let $\\epsilon_{y_j} = H_{y_j}\\exp(\\chi_{y_j}) / I_{y_j}$. Using similar justification explained in Sec.~\\ref{sec:sc_macro_diversity}, $\\epsilon_{y_j}$ for $j=0,1,2,...$ constitute a series of RV whose element is assumed to be identically independently distributed. \n\nLet $P_{f, \\text{MRC}}$ be the network level packet loss rate in this case. According to capture effect:\n\\begin{align}\n\tP_{f, \\text{MRC}} &= \\mathbb{P} \\left\\lbrace \\Theta < \\theta_{T} \\right\\rbrace \\nonumber \\\\\n\t&= F_{\\Theta} (\\theta_{T}),\n\\end{align}\nwhere $\\theta_{T}$ is the capture ratio, $F_{\\Theta} (\\theta)$ is the CDF of $\\Theta$. It is can be numerically computed from Laplace Transform or Characteristic function of $\\Theta$.\n\nThe Laplace Transform (LT) of $\\Theta$ is by definition as follows\\qsong{To be more restrict, we should explain why we can remove the subscript $y_i$ in the following formula}:\n\\begin{align}\n\\label{eq:lt-sinr-mrc-1}\n\\mathcal{L}_{\\Theta}\\left( s \\right) &= \\mathbb{E}\\left[ e^{-s\\Theta}\\right] \\nonumber\\\\\n&=\\mathbb{E}\\left[ \\exp( -s \\sum_{y_j \\in \\Phi_{b}}^{} \\theta_{y_j} )\\right] \\nonumber\\\\\n&= \\mathbb{E}\\left[ \\prod_{y_j \\in \\Phi_b}^{} \\mathbb{E}_{\\epsilon} \\left[ \\exp( -s \\epsilon r_{y_{j}}^{-\\gamma}) \\right] \\right] \n\\end{align}\nActually, the mathematical operations for interference analysis used in~\\cite{haenggi2009interference} can be reused for $\\eqref{eq:lt-sinr-mrc-1}$. For the sake of clarity, we detail the operations as follows:\n \nApplying Campbell theorem to $\\eqref{eq:lt-sinr-mrc-1}$ and changing the order of integration and expectation operator, we have:\n\\begin{align}\n\\label{eq:lt-sinr-mrc-2}\n\\mathcal{L}_{\\Theta}\\left( s \\right) &= \\exp \\left\\lbrace -\\int_{0}^{+\\infty} \\mathbb{E}_{\\epsilon}\\left[ 1-\\exp(-s\\epsilon r^{-\\gamma} ) \\right] 2 \\pi \\lambda_{b} r dr \\right\\rbrace \\nonumber\\\\\n&= \\exp \\left\\lbrace -\\mathbb{E}_{\\epsilon}\\left[ \\int_{0}^{+\\infty} \\left(1-\\exp(-s\\epsilon r^{-\\gamma} ) \\right) 2 \\pi \\lambda_{b} r dr \\right] \\right\\rbrace \n\\end{align}\n\nLet us focus on the integral $D = \\int_{0}^{+\\infty} \\left(1-\\exp(-s\\epsilon r^{-\\gamma} ) \\right) 2 \\pi \\lambda_{b} r dr $:\n\\begin{align}\n\\label{itg:D}\nD &= \\int_{0}^{+\\infty} \\left(1-\\exp(-s\\epsilon r^{-\\gamma/2} ) \\right) \\pi \\lambda_{b} dr \\nonumber\\\\\n&\\overset{\\mathclap{\\strut\\text{(a)}}} = \\pi \\lambda_{b} \\int_{0}^{+\\infty} \\left(1-\\exp(-x ) \\right) d\\left( -x^{-\\frac{2}{\\gamma}} (s\\epsilon)^{\\frac{2}{\\gamma}}\\right) \\nonumber\\\\\n&\\overset{\\mathclap{\\strut\\text{(b)}}}= \\pi \\lambda_{b} \\int_{0}^{+\\infty} \\exp(-x ) x^{-\\frac{2}{\\gamma}} (s\\epsilon)^{\\frac{2}{\\gamma}} dx \\nonumber\\\\\n&= \\pi \\lambda_{b} (s\\epsilon)^{\\frac{2}{\\gamma}} \\Gamma (1 - \\frac{2}{\\gamma}),\n\\end{align}\nwhere step $(a)$ is obtained with a change of variable $x = s \\epsilon r^{-\\gamma/2}$, step $(b)$ is achieved via integration by parts, $\\Gamma(\\cdot)$ is gamma function.\n\nCombining $\\eqref{eq:lt-sinr-mrc-2}$ and $\\eqref{itg:D}$ , the Laplace Transform of $\\Theta$ is finally simplified as:\n\\begin{align}\n\\label{eq:lt_Theta}\n\\mathcal{L}_{\\Theta}\\left( s \\right) &= \\exp(-\\lambda_{b} \\pi \\mathbb{E}\\left[ \\epsilon ^{\\frac{2}{\\gamma}} \\right] \\Gamma(1-\\frac{2}{\\gamma}) s^{\\frac{2}{\\gamma}}),\n\\end{align}\nwhere $\\mathbb{E}\\left[ \\epsilon ^{\\frac{2}{\\gamma}} \\right] $ is:\n\\begin{align}\n\\label{eq:epsilon_fractional_moment}\n\\mathbb{E}\\left[ \\epsilon ^{\\frac{2}{\\gamma}} \\right] &= \\mathbb{E}\\left[ \\left( H \\exp(\\chi)\\right) ^{\\frac{2}{\\gamma}} \\right] \\mathbb{E}\\left[ I ^{-\\frac{2}{\\gamma}}\\right] \\nonumber\\\\\n&=\\Gamma(1+\\frac{2}{\\gamma}) \\exp( \\frac{2\\sigma^2}{\\gamma^2}) \\mathbb{E}\\left[ I ^{-\\frac{2}{\\gamma}}\\right] \n\\end{align}\n\nSince the Laplace transform of cumulative interference $I$ can be easily obtained and PDF of $I$ is not always exist except $\\gamma=4$, to compute $\\mathbb{E}\\left[ I ^{-\\frac{2}{\\gamma}}\\right]$ (i.e. negative fractional moment calculation problem), one can consider departure from LT. This is a research subject in applied mathematical domain and planned as our future work. The most straightforward way is to rely on Monte-Carlo method.\n\nWith substitution $s = -i \\omega$ in $\\eqref{eq:lt_Theta}$ , the characteristic function (CF) of $\\Theta$ is as follows:\n\\begin{align}\n\\phi_{\\Theta}\\left( \\omega \\right) &= \\exp(-\\lambda_{b} \\pi \\mathbb{E}\\left[ \\epsilon ^{\\frac{2}{\\gamma}} \\right] \\Gamma(1-\\frac{2}{\\gamma}) \\exp(-i\\pi/\\gamma) \\omega^{\\frac{2}{\\gamma}}), \n\\end{align}\nwhere $\\omega \\geq 0$, $i$ is imaginary unit.\n\n\\qsong{This paragraph should be simplified. Since in another Chapter, the same numerical technique should have been presented...}As a continuous random variable, the cumulative distribution function $F_{\\Theta}\\left( \\theta \\right)$ of total SINR $\\Theta$ can be directly derived from its characteristic function $\\phi_{\\Theta}\\left(\\omega\\right)$. Applying mathematical techniques used in finance domain~\\cite{hirsa2012computational}, we seek to calculate the Fourier transform of $e^{-\\eta \\theta} F_{\\Theta}\\left( x \\right)$ where term $e^{-\\eta \\theta}$ is a damping function with $\\eta > 0$. \n\\begin{align}\n\\label{eq:intermediate_formula_1}\n\\int_{-\\infty}^{+\\infty} e^{i\\omega \\theta} e^{-\\eta \\theta} F_{\\Theta}\\left( \\theta \\right) dx = \\frac{1}{\\eta - i\\omega} \\phi_{\\Theta}\\left( \\omega +i\\eta \\right) \n\\end{align}\nApplying Fourier inversion for $\\eqref{eq:intermediate_formula_1}$, we obtain the expression for $F_{\\Theta}\\left( \\theta \\right)$ as follows:\n\\begin{align}\n\\label{eq:pr_c_m_case2}\nF_{\\theta}\\left( \\theta \\right) &= \\frac{e^{\\eta \\theta}}{2\\pi} \\int_{-\\infty}^{+\\infty} e^{-i \\omega \\theta} \\frac{1}{\\eta - i\\omega} \\phi_{Theta}\\left( \\omega +i\\eta\\right) d\\omega \\nonumber\\\\\n&= \\frac{e^{\\eta \\theta}}{\\pi} \\Re\\left\\lbrace \\int_{0}^{+\\infty} e^{-i \\omega \\theta} \\frac{1}{\\eta - i\\omega} \\phi_{\\Theta}\\left( \\omega +i\\eta\\right) d\\omega\\right\\rbrace, \n\\end{align}\nThe packet loss rate $P_{f,\\text{MRC}}$ in case of capture ratio $\\theta_{T}$ can be obtained by substituting $\\theta = \\theta_{T}$ in $\\eqref{eq:pr_c_m_case2}$. The latter can be computed using a single numerical integration.\n\n\n%We need to calculate CDF of $\\Theta$ which is the packet loss rate. However, unlike in Sec.XXX, in general case, no closed-form expression. We rely on numerical method to compute packet loss rate wit low complexity. blabla...\n\n\\subsection{A special case, $\\gamma=4$}\nIn this section, we study a special case where path loss component $\\gamma=4$. Proved in ~\\cite[Eq. 3.17]{haenggi2009interference}, the probability density function of cumulative interference $I$, suffered by device at the origin, in a system without shadowing, is as follows:\n\\begin{align}\nf_{I}(x) = \\frac{p\\lambda_{m}}{4} (\\frac{\\pi}{x})^{\\frac{3}{2}} \\exp(-\\frac{\\pi^4 p^2\\lambda_{m}^2}{16x}), x \\geq 0\n\\end{align}\nAlong side with log-normal shadowing effect, the probability density function of cumulative interference $I$ is obtained just by scaling $\\lambda_{m}$ as $ \\lambda_{m} \\exp(\\frac{\\sigma^2}{8})$:\n\\begin{align}\nf_{I}(x) = \\frac{ p\\lambda_{m} \\exp(\\frac{\\sigma^2}{8}) }{4} (\\frac{\\pi}{x})^{\\frac{3}{2}} \\exp( -\\frac{ \\pi^4 p^2\\lambda_{m}^2 \\exp^2(\\frac{\\sigma^2}{8}) }{ 16x } ), x \\geq 0\n\\end{align} \nHence, the negative fractional moment of $I$ can be calculated:\n\\begin{align}\n\\label{eq:gamma=4_fractional_I}\n\\mathbb{E}\\left[ I ^{-\\frac{1}{2}}\\right] &= \\int_{0}^{+\\infty} x^{-\\frac{1}{2}} f_{I}(x) dx \\nonumber\\\\\n&= \\int_{0}^{+\\infty} \\frac{ p\\lambda_{m} \\exp(\\frac{\\sigma^2}{8}) }{4} \\pi^{\\frac{3}{2}} x^{-2} \\exp( -\\frac{ \\pi^4 p^2\\lambda_{m}^2 \\exp^2(\\frac{\\sigma^2}{8}) }{ 16x } ) dx \\nonumber\\\\\n&= \\frac{4}{ \\pi^{\\frac{5}{2}} p\\lambda_{m} \\exp(\\frac{\\sigma^2}{8}) }\n\\end{align}\nWith substitution of $\\eqref{eq:gamma=4_fractional_I}$ into $\\eqref{eq:epsilon_fractional_moment}$,\n\\begin{align}\n\\label{eq:negative_fractonal_epsilon_4}\n\\mathbb{E}\\left[ \\epsilon ^{\\frac{1}{2}} \\right] & = 4\\Gamma(\\frac{3}{2})\\pi^{-\\frac{5}{2}} p^{-1}\\lambda_{m}^{-1}\\nonumber\\\\\n&=2\\pi^{-2} p^{-1}\\lambda_{m}^{-1}\n\\end{align}\nFrom $\\eqref{eq:negative_fractonal_epsilon_4}$ and $\\eqref{eq:lt_Theta}$, we have:\n\\begin{align}\n\\mathcal{L}_{\\Theta}\\left( s \\right) &= \\exp(-\\lambda_{b} \\pi 2\\pi^{-2}\\lambda_{m}^{-1} \\Gamma(\\frac{1}{2}) s^{\\frac{1}{2}}) \\nonumber \\\\\n&= \\exp(-2L^{-1}\\pi^{-\\frac{1}{2}} s^{\\frac{1}{2}}), \n\\end{align}\nwhich is the Laplace Transform of probability density function of $\\Theta$. According to derivative property of Laplace Transform, the cumulative distribution function $F_{\\Theta} (\\theta) $ of ${\\Theta}$ can be obtained by directly inversing $\\frac{1}{s} \\mathcal{L}_{\\Theta}\\left( s \\right)$.\nWhen $\\gamma = 4$, instead of using numerical integration for $\\eqref{eq:pr_c_m_case2}$, we have closed-form CDF of total SINR $\\Theta$:\n\\begin{align}\nP_{f, \\text{MRC}} = F_{\\Theta} (\\theta_{T}) \n&= \\mathcal{L}^{-1} \\left[ \\frac{1}{s} \\mathcal{L}_{\\Theta}\\left( s \\right) \\right] \\nonumber\\\\\n&=1 -\\erf \\left( \n\\frac{1}{ L \\sqrt{\\pi \\theta_{T}}}\n\\right),\n\\end{align}\nwhere $\\mathcal{L} \\left[ \\cdot \\right] $ is inverse Laplace Transform operator.\n%We verify that the packet loss rate obtained by $(?)$ and $(?)$ are the same.\n\\qsong{Don't forget to talk about throughput, macro diversity gain as like what we have done in previous section.}\n\n%============================================================================================\n%\\input{Chapter5/bs_rx_divers_mrc.tex}\n%Macro diversity gain against the nearest BS attach method $G_{\\text{diversity}, n}$ is obtained by multiplying $\\exp \\left( \\frac{\\sqrt{2}\\sigma}{\\gamma}\\right)^2$ to $(\\ref{eq:macro-diversity-gain})$.\n% depends on path-loss exponent $\\gamma $, shadowing standard deviation $\\sigma$ and packet loss rate constraint $P_f^{\\text{max}}$, but has nothing to do with capture ratio.\n\n\n%Old expression for P_f1\n%\\begin{align}\n%\\label{eq:bs_nst_att_analytical}\n%P_{f1} &\\approx 1 -\\frac{1}{1 + \\exp\\left\\lbrace \\left( 1 +\\frac{\\pi \\sigma_X^2}{8} \\right)^{-\\frac{1}{2}} \\ln(B) \\right\\rbrace} \\nonumber\\\\\n%&=1-\\frac{1}{1 + \\left(B\\frac{p\\lambda_m}{\\lambda_b} \\right) ^{\\left( 1 +\\frac{\\pi \\sigma^2}{2\\gamma^2} \\right)^{-\\frac{1}{2}}}}, \\\\\n%L_{\\text{max}} &= \\frac{\\left( \\frac{P_{f1}^{\\text{max}}}{1-P_{f1}^{\\text{max}}}\\right)^{\\left( 1 +\\frac{\\pi \\sigma^2}{2\\gamma^2} \\right)^{\\frac{1}{2}}} }{B},\n%\\end{align}\n\n%&= \\int_{-\\infty}^{+\\infty} \\frac{1}{ e^t+1}\\frac{1}{\\sqrt{2\\pi} \\frac{2}{\\gamma}\\sigma} \\exp\\left\\lbrace -\\frac{1}{2} \\frac{(t-\\ln(\\frac{A}{\\pi \\lambda_{b}}))^2}{\\left( \\frac{2}{\\gamma}\\sigma\\right)^2}\\right\\rbrace dt.\n\n%\\begin{align}\n%\\label{eq:mean_bx+1_step1}\n%&P_{f}= \\mathbb{E}_{r}\\left[ 1-p_{s}\\left(r\\right) \\right] \\nonumber\\\\\n%&= 1-\\int_{0}^{+\\infty} \\mathbb{E}_{G}\\left[ \\exp(-p\\lambda_{m}AK r^2 e^{-\\frac{2}{\\gamma}G})\\right] 2 \\pi \\lambda_b r e^{-\\lambda_b \\pi r^2} dr \\nonumber\\\\\n%&= 1-\\pi \\lambda_b \\mathbb{E}_{G}\\left[ \\int_{0}^{+\\infty} \\exp(-p\\lambda_{m} A K r^2 e^{-\\frac{2}{\\gamma}G}-\\lambda_b \\pi r^2)\\right] dr^2 \\nonumber\\\\\n%&= 1 -\\mathbb{E}_{G}\\left[ \\frac{1}{\\frac{AKp\\lambda_{m}}{\\pi \\lambda_{b}}e^{-\\frac{2}{\\gamma}G}+1} \\right] \\nonumber\\\\\n%&\\mathbb{E}_{G}\\left[ \\frac{1}{\\frac{p\\lambda_{m}AK}{\\pi \\lambda_{b}}e^{-\\frac{2}{\\gamma}G}+1} \\right] = \\int_{-\\infty}^{+\\infty} \\frac{1}{ \\frac{p\\lambda_{m}AK}{\\pi \\lambda_{b}}e^t+1}\\frac{\\exp\\left\\lbrace -\\frac{t^2}{2 \\left( \\frac{2}{\\gamma}\\sigma\\right) ^2}\\right\\rbrace }{\\sqrt{2\\pi} \\frac{2}{\\gamma}\\sigma} dt \\nonumber\\\\\n%&= \\int_{-\\infty}^{+\\infty} \\frac{\\gamma}{ 2\\sqrt{2\\pi}\\left( e^t+1\\right)\\sigma}e^{\\left\\lbrace -\\frac{\\gamma^2}{8 \\sigma^2} (t-\\ln\\frac{AKp\\lambda_{m}}{\\pi \\lambda_{b}})^2\\right\\rbrace} dt.\n%\\end{align}\n\n%$\\lim\\limits_{\\frac{p\\lambda_{m}}{\\lambda_{b} } +\\infty} \\frac{p\\lambda_{m}}{\\lambda_{b}}(1-P_{f2}) =\\frac{1}{\\frac{K}{\\pi}\\Gamma(1+\\frac{2}{\\gamma}) \\theta_{T}^{\\frac{2}{\\gamma}}}$\n\n%The normalized spatial throughput $S$ converges to a limit value $ \\left[ K\\Gamma(1+\\frac{2}{\\gamma})\\theta_{T}^{\\frac{2}{\\gamma}} \\right] ^{-1}\\pi$, which depends on SINR threshold and path loss exponent.\n\n%The analysis in Sec.~\\ref{secsec:nst_bs_att} related to normalized spatial throughput still holds for this case.\n\n%With $\\ref{eq:bs_nst_att_max_load}$, we note that normalized spatial throughput $S$ converges to $(AK)^{-1}\\pi$. It is worth indicating that normalized spatial throughput diverges to infinity when normalized load increases. However, this divergence is caused by the approximation in $(\\ref{eq:bs_nst_att_analytical})$.\n\n%Surely, the best BS attach method always outperforms than the nearest BS attach method. However, we still first consider the nearest BS attach method due to three reasons:\\begin{inparaenum}[1)]\n%\\item nearest BS attach is a widely studied method in stochastic geometry field;\n%\\item nearest BS attach is the only choice when the downlink is not available (or seldom used for cost purpose) for some cheap systems;\n%\\end{inparaenum} \n\n\n\n%As discussed before, $(\\ref{eq:bs_nst_att_analytical})$ is applicable for best BS attach method. To obtain a general result for both methods, we redefine $A$ is redefined as follows:\n%\\[ A=\n%\\begin{cases} \n%\\Gamma(1+\\frac{2}{\\gamma}) \\exp \\left( \\frac{\\sqrt{2}\\sigma}{\\gamma}\\right) ^2 \\theta_{T}^{\\frac{2}{\\gamma}} , & \\text{for Nearest BS attach }\\\\\n%\\Gamma(1+\\frac{2}{\\gamma})\\theta_{T}^{\\frac{2}{\\gamma}} , & \\parbox[t]{.4\\columnwidth}{for Best BS attach}\n%\\end{cases}\n%\\]\n\n\n\n%\\begin{align}\n%x & \\\\\n%&\\overset{\\mathclap{\\strut\\text{a}}}= foo\\\\\n%& = bar\n%\\end{align}", "meta": {"hexsha": "60217d3d6a48a34b406d2838f6b51d4ab474e791", "size": 25397, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Chapter5/bs_rx_divers_op_plane_oneshot.tex", "max_stars_repo_name": "hansomesong/PhD-Thesis", "max_stars_repo_head_hexsha": "4bdd0a41012030398a23ae16a66d5a02631f76f5", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Chapter5/bs_rx_divers_op_plane_oneshot.tex", "max_issues_repo_name": "hansomesong/PhD-Thesis", "max_issues_repo_head_hexsha": "4bdd0a41012030398a23ae16a66d5a02631f76f5", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Chapter5/bs_rx_divers_op_plane_oneshot.tex", "max_forks_repo_name": "hansomesong/PhD-Thesis", "max_forks_repo_head_hexsha": "4bdd0a41012030398a23ae16a66d5a02631f76f5", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 84.3754152824, "max_line_length": 638, "alphanum_fraction": 0.6824428082, "num_tokens": 8859, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673087708699, "lm_q2_score": 0.7520125793176222, "lm_q1q2_score": 0.6112864415117559}} {"text": "\\chapter{\\label{ch:math}Mathematical preliminaries}\n\nMany methods and applications of natural language processing require\ntopics from mathematics.\nBeing an interdisciplinary field,\nit is often difficult to assume\nthat all students of computational linguistics possess a (fresh) knowledge of\nsome of the mathematical topics and notation.\nThis chapter provides a highly coarse overview\nof some topics in calculus and linear algebra.\nThe discussion here is incomplete and informal.\nThe interested reader should follow\nthe references provided for in-depth treatments of these subjects.\n\nSection~\\ref{sec:linear-intro} introduces some topics from linear algebra.\nWe will mainly introduce vectors, matrices and operations on vectors and matrices.\nAgain, these topics and notation will be important in discussing\nmany of the machine learning methods.\n\n\nIn Section~\\ref{sec:calculus-intro},\nwe will very briefly revisit derivatives and integrals.\nParticularly, derivatives are used for finding maxima or minima of functions,\nwhich is the basis for many of the machine learning methods.\nThe integrals will also come back in our discussion of some of the machine learning methods,\nparticularly in discussion of probabilistic learning and inference.\n\n\\section{\\label{sec:linear-intro}Linear algebra}\n\nIn many NLP methods,\nwe make heavy use of \\term{vectors} and \\emph{matrices},\nwhich are objects studied in \\emph{linear algebra}.\nVectors are used for representing \\term{features} in many machine learning methods.\nOperations on vectors and matrices also have important applications.\nIn this section we will review some of the properties of vectors and matrices,\nand the operations defined on them.\nIf you had a linear algebra course,\nof if you know, for example, matrix multiplication,\nor dot product of vectors, you can safely skip this section.\n\n\\subsection{Vectors}\n\n\\begin{marginfigure}[5ex]\n \\tikzsetnextfilename{vector-magnitude-direction}\n \\begin{tikzpicture}\n \\draw[->,thick] (0,0) -- (1, 1);\n \\draw[blue,dotted] (-1,-1) -- (2, 2)\n node[near end, above,font=\\scriptsize,sloped] {direction};\n \\node[coordinate] (x) at ($(0.5,0)!(0,0)!(1.5,1)$) {};\n \\node[coordinate] (y) at ($(0.5,0)!(1,1)!(1.5,1)$) {};\n \\draw[dotted,gray] (x) -- (0,0);\n \\draw[dotted,gray] (y) -- (1,1);\n \\draw[<->,dotted,blue] (x) -- (y)\n node[midway,below,font=\\scriptsize,sloped] {magnitude};\n \\end{tikzpicture}\n \\caption{\\label{fig:single-vector-example}A graphical representation of a vector.}\n\\end{marginfigure}\nA vector is a mathematical object with a magnitude and a direction.\nGraphically, we can represent or visualize a (two-dimensional) vector\nas in Figure~\\ref{fig:single-vector-example}.\nMore commonly,\nwe represent vectors by an ordered list of number,\nsuch as $(1, 0, 1)$.\nA vector defined with $n$ real numbers is said to be in the vector space $\\mathbb{R}^{n}$.\nWe often write a vector of $n$ real numbers (vectors in $\\mathbb{R}^{n}$) as\n$\\vect{v} = (v_{1}, v_{2}, \\ldots, v_{n})$.\nNote that the $\\vect{v}$ that stands for the vector is typeset in boldface font.\nIt can alternatively be marked with a arrow over it, like $\\vec{v}$.\nOther notations for vectors of $n$ numbers include\n$\\vect{v} = \\langle v_{1}, v_{2}, \\ldots, v_{n}\\rangle$ or\n$\\vect{v} = \\begin{bmatrix} v_{1}\\\\ \\vdots\\\\ v_{n}\\\\ \\end{bmatrix}$.\n\n\\begin{marginfigure}[5ex]\n \\tikzsetnextfilename{example-vectors}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[<->,thick] (0, -4) -- (0, 4);\n \\node[anchor=west] at (0, 4) {y};\n \\draw[<->,thick] (-4, 0) -- (4, 0);\n \\node[anchor=south] at (4, 0) {x};\n \\draw[step=1cm,grid] (-4,-4) grid (4,4);\n \\draw[thick, ->,red] (0,0) -- (1,1);\n \\node[anchor=west,red] at (1,1) {$(1,1)$};\n \\draw[thick, ->,blue] (0,0) -- (1,3);\n \\node[anchor=west,blue] at (1,3) {$(1,3)$};\n \\draw[thick, ->,green] (0,0) -- (-1,-3);\n \\node[anchor=east,green] at (-1,-3) {$(-1,-3)$};\n \\end{tikzpicture}\n \\caption{\\label{fig:vector-examples}Example vectors in 2-dimensional Euclidean space.}\n\\end{marginfigure}\nGeometrically, we represent vectors as arrows as in Figure~\\ref{fig:vector-examples}.\nThe individual numbers on the notation represent their projection to the respective axis.\nIn the example on the right,\nfor example,\nthe green and blue vectors have the same magnitude,\nbut their direction is opposite of each other.\nIf we take the projections of the vector to the $x$ and $y$ axes,\nthey correspond to real first and the second number in our notation respectively.\n\nMany operations on (real) numbers have analogues forms for vectors,\nand they are used in many branches of science,\nas well as machine learning and natural language processing.\n\n\\newthought{Vector norms} are\na generalization of the magnitude of a vector.\nA \\term{norm} assigns a non-negative \\emph{length} or \\emph{size} to a vector.\nNorms are used often in many machine learning methods,\nand they are related to distance metrics which by themselves are useful in comparing objects represented as (feature) vectors.%\n\\footnote{The norm of a vector is the distance from its tail to its tip.}\nThe most familiar norm is the Euclidean norm,\nwhich is also known as L2 (or L$^{\\text{2}}$) norm.\nL2 norm of a vector $\\vect{v} = (v_{1}, \\ldots, v_{n})$ is\n\n\\[\n \\norm{\\vect{v}}_{2} = \\sqrt{v_{1}^{2} + \\ldots + v_{n}^{2}}\\;.\n\\]\n\nThe subscript 2\\ in $\\norm{v}_{2}$ communicates that the norm is L2 norm.\nThe L2 norm is often taken to be the default.\nIf the subscript is omitted then we mean the L2 norm.\nAnother interesting norm for our purposes is the L1 norm,\nwhich is related to the so-called taxi-cab, city-block or Manhattan distance.\nIt is defined as\n\n\\[\n \\norm{\\vect{v}}_{1} = \\lvert{}v_{1}\\rvert + \\ldots + \\lvert{}v_{n}\\rvert\\;.\n\\]\n\n\\begin{marginfigure}[5ex]\n \\tikzsetnextfilename{regularization-l1-l2}\n \\begin{tikzpicture}\n \\draw[step=1cm,grid] (-1,-1) grid (4,4);\n \\draw[->,thick] (0, -1) -- (0, 4);\n \\node[anchor=west] at (0, 4) {y};\n \\draw[->,thick] (-1, 0) -- (4, 0);\n \\node[anchor=south] at (4, 0) {x};\n \\node[anchor=west,red] at (3,3) {$(3,3)$};\n \\draw[very thick,blue] (0,0) -- (3,3);\n \\draw[very thick,red,dotted] (0.05,0) -- (3,0) -- (3,0.05) -- (3,3);\n \\draw[very thick,orange,dotted] (0,0) -- (0,2) -- (1,2) -- (1,3) -- (3,3);\n \\draw[very thick,green,dotted]\n (0,0) -- (1,0) -- (1,1) -- (2,1) -- (2,2) -- (3, 2) -- (3,3);\n \\end{tikzpicture}\n \\caption{\\label{fig:l1-l2}%\n Visualizations of L2(solid \\textcolor{blue}{blue})\n and example L1\n (dotted \\textcolor{green}{green},\n \\textcolor{orange}{orange}\n and \\textcolor{red}{red})\n norms vector $(3,3)$.}\n\\end{marginfigure}\n\nFigure~\\ref{fig:l1-l2} visualizes the L1 and L2 norms\nin two-dimensional Euclidean space.\nNote that both apply to higher dimensional spaces as well.\nSo for our example,\n\\[ \\norm{(3,3)}_{2} = \\sqrt{3^{2} + 3^{2}} = \\sqrt{18} \\approx 4.24\\]\n\\[ \\norm{(3,3)}_{1} = \\lvert{}3\\rvert + \\lvert{}3\\rvert = 6 \\]\\;.\nLike any other vector operation or property,\nvector norms can be generalized to vectors of any dimension.\n\nThe concept of vector norm\ncan also be generalized to any positive integer $p$\nthe L$_{\\text{p}}$ norm for an n-dimensional vector is defined as\n\n\\[\n \\norm{\\vect{v}}_{p} = \\left(\n \\sum_{i=1}^{n} \\lvert{}v_{i}\\rvert^{p}\n \\right)^\\frac{1}{p}\n\\]\n\nIn this course, we will only work with L1 and L2 norms defined above.\n\n\\newthought{Scalar multiplication} is the operation of\nmultiplying a vector with a scalar\n(for our purposes a scalar is a real number).\nGiven a vector $\\vect{v} = (v_{1},\\ldots, v_{n})$,\nits multiplication with scalar $a$ is defined as\n\n\\[\n a \\vect{v} = (a v_{1}, \\ldots, a v_{2})\n\\]\n\nMultiplying a vector with a positive scalar,\nchanges its magnitude, `scales' it,\nbut does not change its direction.\nMultiplying a vector with a negative scalar\nreverses the direction of the original vector.\n\n\\begin{marginfigure}[-5ex]\n \\tikzsetnextfilename{scalar-multiplication}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[<->,thick] (0, -4) -- (0, 4);\n \\draw[<->,thick] (-4, 0) -- (4, 0);\n \\draw[step=1cm,grid] (-4,-4) grid (4,4);\n% \\draw[->,line width=1mm, blue!40] (0,0) -- (2,4);\n \\draw[->,very thick, blue!40] (0,0) -- (2,4);\n \\node[anchor=north west, blue!40] at (2,4) {$2 \\vect{v}$};\n \\draw[->,thick,red] (0,0) -- (1,2);\n \\node[anchor=west, red] at (1,2) {$\\vect{v} = (1,2)$};\n \\draw[->,thick,green] (0,0) -- (-0.5,-1);\n \\node[anchor=east, green] at (-0.5, -1) {$-0.5 \\vect{v}$};\n \\end{tikzpicture}\n \\caption{\\label{fig:scalar-multipl}%\n Scalar multiplication.\n }\n\\end{marginfigure}\n\n\\newthought{Vector addition and subtraction}\nare defined on two vectors with the same dimensionality.\nFor $n$-dimensional vectors $\\vect{v} = (v_{1}, \\ldots, v_{n})$\nand $\\vect{w} = (w_{1}, \\ldots, v_{w})$,\n\\[\n \\vect{v} + \\vect{w} = (v_{1} + w_{1} + \\ldots + v_{n} + w_{n})\n\\]\nThe subtraction is simply addition with a negative vector,\n\\[\n \\vect{v} - \\vect{w} = \\vect{v} + (-\\vect{w}) = (v_{1} - w_{1} + \\ldots + v_{n} - w_{n})\n\\]\n\n\\begin{marginfigure}\n \\tikzsetnextfilename{vector-sum}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[<->,thick] (0, -4) -- (0, 4);\n \\draw[<->,thick] (-4, 0) -- (4, 0);\n \\draw[step=1cm,grid] (-4,-4) grid (4,4);\n \\draw[->,thick,red] (0,0) -- (1,2);\n \\node[anchor=east, red] at (1,2) {$\\vect{v}$};\n \\draw[->,thick,blue] (0,0) -- (2,1);\n% \\draw[->, dotted,blue] (1,2) -- (3,3);\n \\draw[->, dotted,red] (2,1) -- (3,3);\n \\node[anchor=west, blue] at (2,1) {$\\vect{w}$};\n \\draw[->, thick] (0,0) -- (3,3);\n \\node[anchor=south] at (3,3) {$\\vect{v + w}$};\n \\draw[->,thick,purple] (0,0) -- (-2,-1);\n \\draw[->,purple,dotted] (1,2) -- (-1,1);\n \\node[anchor=north, purple] at (-2, -1) {$-\\vect{w}$};\n \\draw[->,thick,black] (0,0) -- (-1,1);\n \\node[anchor=south,xshift=-0.5em] at (-1, 1) {$\\vect{v}-\\vect{w}$};\n \\end{tikzpicture}\n \\caption{\\label{fig:vector-sum}%\n Vector addition and subtraction.\n }\n\\end{marginfigure}\n\n\\newthought{Dot product} is a very important quantity that will come up regularly in this course.\nDot product of two vectors,\n$\\vect{v} = (v_{1}, \\ldots, v_{n})$\nand $\\vect{w} = (w_{1}, \\ldots, v_{w})$,\nis a scalar defined as:\n\n\\[\n \\vect{v}\\cdot\\vect{w} = v_{1} \\times w_{1} + \\ldots + v_{n} \\times w_{n}\n\\]\n\nIt should be emphasized that dot product yields a scalar (real number),\nnot a vector.\nThere are other vector product operations:\n\\emph{outer product} that we will discuss below,\nand \\emph{cross product} defined for vectors in $\\mathbb{R}^{3}$.\nHence, without `dot' the notation $\\vect{v}\\vect{w}$ is ambiguous.\nHowever, it is common to treat $k$-dimensional vectors\nas $k\\times{}1$ matrices for which multiplication is not ambiguous\n(we discuss this notation on page~\\pageref{pageref:vector-matrix-notation}.\n\n\n\\begin{marginfigure}\n \\tikzsetnextfilename{dot-product}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[step=1cm,grid] (-4,-4) grid (4,4);\n \\draw[<->,thick] (0, -4) -- (0, 4);\n \\draw[<->,thick] (-4, 0) -- (4, 0);\n \\coordinate (origin) at (0,0);\n \\draw[->,thick,red] (0,0) -- (2,2);\n \\node[anchor=east, red] (v) at (2,2) {$\\vect{v}$};\n \\draw[->,thick,blue] (0,0) -- (3,1);\n \\node[anchor=west, blue] (w) at (3,1) {$\\vect{w}$};\n \\draw[dashed] (2,2) -- (2.4, 0.8);\n \\draw [decorate,decoration={brace,amplitude=7pt,mirror,raise=4pt}]\n (0,0) -- (2.4, 0.8);\n \\pic [draw, \"{\\small $\\alpha$}\", angle eccentricity=1.5]\n {angle = w--origin--v};\n \\node[anchor=north,rotate=20] at (1.5,-0.1) {$\\norm{v}\\cos\\alpha$};\n \\end{tikzpicture}\n \\caption{\\label{fig:dot-product}%\n Dot product of two vectors.\n }\n\\end{marginfigure}\nThere is an alternative way to define the dot product,\nwhich also leads to a nice geometric interpretation.\nWe can calculate the dot product as\n\n\\[\n \\vect{v}\\cdot\\vect{w} = \\norm{\\vect{v}}\\norm{\\vect{w}} \\cos \\alpha\n\\]\nwhere $\\alpha$ is the angle between the two vectors\n(see Figure~\\ref{fig:dot-product}).\nThis also allows us to interpret the dot product geometrically.\nThe dot product of two vectors is proportional to each vector's magnitude,\nand also to the cosine of the angle between them.\nSince the cosine of the angle will be larger for smaller angles,\nthe dot product will be larger for vectors that point to similar directions\n(keeping the magnitudes constant).\nThe dot product of two orthogonal vectors\n(vectors with a \\num{90}\\textdegree{} angle between them)\nis \\num{0}.\nIf the angle is larger tan \\num{90}\\textdegree{},\nthe dot product is negative.\n\n\n\n\\newthought{Cosine similarity} is a similarity measure related to dot product,\nwhich we will often use for measuring similarities between objects of interest,\ne.g., documents.\nFrom above, we can simply write the cosine of the angle between two vectors as,\n\n\\[ \\cos\\alpha = \\frac{\\vect{v}\\vect{w}}{\\norm{\\vect{v}}\\norm{\\vect{w}}} \\]\n\nNote that by dividing the vectors to their Euclidean (L2) norms,\nwe are scaling them to unit vectors pointing in the same direction.\nAs a result, cosine similarity ignores the magnitudes of the vectors.\nThe cosine similarity for vectors that point to the same direction is $1$\n(regardless of their magnitude) and the vectors\nthat point exact opposite directions have a cosine similarity of $-1$.\n\n\\todo[inline]{Cross product (?)}\n\n\\subsection{Matrices}\n\nMatrices are the second type of mathematical objects we often encounter in various NLP methods.\nA matrix is simply a two-dimensional array of numbers,\nwhich is noted as a rectangular placement of scalars.\nA matrix of $n$ rows and $m$ columns is an $n \\times m$ matrix.\nA real-valued $n \\times m$ matrix is said to be in $\\mathbb{R}^{n\\times m}$.\nWe can think about a matrix as a collection of column or row vectors.\nWe denote matrices with boldface capital letters, like \\vect{A}.\nWhile referring to a matrix' elements,\nwe subscript the element first with its row and then its column.\n \\[ \\vect{A} =\n \\begin{bmatrix}\n a_{1,1} & a_{1,2} & a_{1,3} & \\ldots & a_{1,m} \\\\\n a_{2,1} & a_{2,2} & a_{2,3} & \\ldots & a_{2,m} \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n a_{n,1} & a_{n,2} & a_{n,3} & \\ldots & a_{n,m} \\\\\n \\end{bmatrix}\n \\]\n\nWe will briefly revisit some of the operations on matrices in this section.\n\n\\newthought{Transpose of a matrix} simply replaces its rows by columns.\nTranspose of a matrix $\\vect{A}$ is denoted with $\\vect{A}^{T}$.\n\n\\begin{center}\n If $\\vect{A} = \\begin{bmatrix} a & b \\\\ c & d \\\\ e & f \\\\ \\end{bmatrix}$,\n $\\vect{A}^{T} = \\begin{bmatrix} a & c & e\\\\ b & d & f \\\\ \\end{bmatrix}$.\n\\end{center}\n\n\\newthought{Multiplication by a scalar} is also defined for matrices.\nTo multiply a matrix with a scalar,\neach element of the matrix is multiplied by the scalar.\nFor example,\n\\[\n 2 \\begin{bmatrix} 2 & 1 \\\\ 1 & 4 \\\\ \\end{bmatrix} =\n \\begin{bmatrix}\n 2\\times{}2 & 2\\times{}1 \\\\\n 2\\times{}1 & 2\\times{}4 \\\\\n \\end{bmatrix} =\n \\begin{bmatrix}\n 4 & 2 \\\\\n 2 & 8 \\\\\n \\end{bmatrix}\n\\]\n\n\\newthought{Matrix addition and subtraction} require two matrices of same dimensions.\nTo obtain sum (difference) of two matrices,\neach element of the second matrix is added to (subtracted from)\nthe corresponding element of the first matrix.\nFor example:\n\n \\[\n \\begin{bmatrix}\n 2 & 1 \\\\\n 1 & 4 \\\\\n \\end{bmatrix} +\n \\begin{bmatrix}\n 0 & 1 \\\\\n 1 & 0 \\\\\n \\end{bmatrix} =\n \\begin{bmatrix}\n 2+0 & 1+1 \\\\\n 1+1 & 4+0 \\\\\n \\end{bmatrix} =\n \\begin{bmatrix}\n 2 & 2 \\\\\n 2 & 4 \\\\\n \\end{bmatrix}\n \\]\n\n\\newthought{Matrix multiplication} is a slightly complicated operation.\nThe matrix multiplication $\\vect{A}\\times\\vect{B}$ is defined only\nif \\vect{A} has the same number of columns as the number of rows in \\vect{B}.\nMultiplying a $n\\times{}k$ matrix with a $k\\times{}m$ matrix\nresults in a $n\\times{}m$ matrix.\nNote that both $A\\times{}B$ and $B\\times{}A$ is defined only for square matrices (of the same dimensions).\n\nFor an $n\\times{}k$ matrix \\vect{A}\nand a $k\\times{}m$ \\vect{B},\nif $\\vect{A} \\times \\vect{B} = \\vect{C}$,\n$c_{i,j}$, the element of the resulting matrix \\vect{C} on row $i$ and column $j$,\nis calculated as:\n\n\\[\n c_{ij} = \\sum_{\\ell=0}^{k} a_{i\\ell} b_{\\ell j}\n\\]\n\nFigure~\\ref{fig:matrix-mult} demonstrates the matrix multiplication.\n\n\\begin{figure*}\n \\tikzsetnextfilename{matrix-multiplication}\n \\begin{tikzpicture}[ampersand replacement=\\&]\n \\matrix (A) [matrix of math nodes,\n left delimiter = {[},\n right delimiter = {]}] at (0,0) {%\n \\textcolor{blue}{a_{11}} \\&\n \\textcolor{blue}{a_{12}} \\&\n \\textcolor{blue}{\\ldots} \\&\n \\textcolor{blue}{a_{1k}} \\\\\n a_{21} \\&\n a_{22} \\&\n \\ldots \\&\n a_{2k} \\\\\n \\vdots \\& \\vdots \\& \\ddots \\& \\vdots \\\\\n a_{n1} \\&\n a_{n2} \\&\n \\ldots \\&\n a_{\\textcolor{red}{nk}} \\\\\n };\n \\node[right of=A,anchor=west,xshift=1.5cm] {\\Huge$\\times$};\n \\matrix (B) [matrix of math nodes,\n left delimiter = {[},\n right delimiter = {]}] at (6,0) {%\n b_{11} \\&\n \\textcolor{blue}{b_{12}} \\&\n \\ldots \\&\n b_{1m} \\\\\n b_{21} \\&\n \\textcolor{blue}{b_{22}} \\&\n \\ldots \\&\n b_{2m} \\\\\n \\vdots \\&\n \\textcolor{blue}{\\vdots} \\&\n \\ddots \\&\n \\vdots \\\\\n b_{k1} \\&\n \\textcolor{blue}{b_{k2}} \\&\n \\ldots \\&\n b_{\\textcolor{red}{km}} \\\\\n };\n \\node[right of=B,anchor=west,xshift=1.5cm] {\\Huge$=$};\n \\matrix (C) [matrix of math nodes,\n left delimiter = {[},\n right delimiter = {]}] at (12,0) {%\n c_{11} \\&\n \\textcolor{blue}{c_{12}} \\&\n \\ldots \\&\n c_{1m} \\\\\n c_{21} \\&\n c_{22} \\&\n \\ldots \\&\n c_{2m} \\\\\n \\vdots \\&\n \\vdots \\&\n \\ddots \\&\n \\vdots \\\\\n c_{n1} \\&\n c_{n2} \\&\n \\ldots \\&\n c_{\\textcolor{red}{nm}} \\\\\n };\n \\draw[blue] (A-1-1.south west) rectangle (A-1-4.north east);\n \\draw[blue] (B-1-2.north west) rectangle (B-4-2.south east);\n \\node[below=of C, yshift=5ex,blue]\n {$c_{12} = a_{11} b_{12} + a_{12} b_{22} + \\ldots a_{1k} b_{k2}$};\n \\end{tikzpicture}\n \\caption{Matrix multiplication.\n The calculation of the resulting matrix \\textcolor{blue}{$c_{12}$} is highlighted.\n }\\label{fig:matrix-mult}\n\\end{figure*}\n\nNote that the element $c_{ij}$ is the dot product of\n$i^{th}$ row vector of \\vect{A} and $j^{th}$ column vector of \\vect{B}.\nHence, we can view dot-product as matrix multiplication\nof a row vector (on the left) and column vector (on the right).\nDot product of two vectors \\vect{v} and \\vect{w} is often noted as\n$\\vect{v}\\vect{w}^{T}$.%\n\\footnote{It is a common convention to assume that vectors are column vectors\n unless stated otherwise.}\nTechnically,\nresult of a matrix multiplication of a $1\\times{}k$ vector with a $k\\times{}1$\nvector is a $1\\times{}1$ matrix,\nnot a scalar.\nHowever, this notation is prevalent in machine learning and NLP literature,\nwe will often use this notation during this course.\n\n\\label{pageref:vector-matrix-notation}\nFor example, $\\vect{w} = (2, 2)$ and $\\vect{v} = (2, -2)$,\n\n\\[\n \\vect{v}\\vect{w}^{T} = \\begin{bmatrix}\n 2 \\\\ -2 \\\\\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n 2 & 2 \\\\\n \\end{bmatrix}\n = 2\\times{}2 + 2\\times{}-2 = 4 - 4 = 0\n\\]\n\n\\marginnote{What does the result of dot product ($0$) say about the vectors?}\n\n\\newthought{Outer product} of two vectors with the same dimensionality,\ncan also be defined as matrix multiplication.\nThis time we put the column vector to the left and the row vector to the right.\nSo, in the notation used above, outer product of two matrices \\vect{v} and \\vect{w} is\n$\\vect{v}^{T}\\vect{w}$.\nNote that result of outer product of two $k$-dimensional vectors is\na $k\\times{}k$ matrix, not a scalar.\nThe following is an example of outer product of two $3$-dimensional vectors.\n\n\\[\n \\begin{bmatrix}\n 1 & 2 & 3\\\\\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n 6 \\\\ 5 \\\\ 4 \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n 6 & 5 & 4 \\\\\n 12 & 10 & 8 \\\\\n 18 & 15 & 12 \\\\\n \\end{bmatrix}\n\\]\n\n\\todo[inline]{Relation to SVD / variance ?}\n\n\\newthought{An identity matrix} is a square matrix\nin which all the elements of the main diagonal are ones,\nand all other elements are zeros.\nThe $n\\times{}n$ identity matrix is denoted by $\\vect{I}_{n}$.\nWhen there is no ambiguity, we omit the subscript, and simply write \\vect{I}.\nMultiplying a matrix with a compatible identity matrix does not change the original matrix.\nFor $n \\times m$ matrix \\vect{A},\n\\[\n \\vect{I}_{n} \\vect{A} = \\vect{A} \\quad \\vect{A} \\vect{I}_{m} = \\vect{A}\n\\]\n\n\\begin{marginfigure}\n \\[\n \\vect{I}_{4} =\n \\begin{bmatrix}\n 1 & 0 & 0 & 0\\\\\n 0 & 1 & 0 & 0\\\\\n 0 & 0 & 1 & 0 \\\\\n 0 & 0 & 0 & 1 \\\\\n \\end{bmatrix}\n \\]\n \\caption{\\label{fig:identity-matrix}$4\\times{}4$ identity matrix.}\n\\end{marginfigure}\n\n\\newthought{Multiplying a vector with a matrix} (linearly) transforms it to another\n(possibly a different dimensional) vector.\nThese linear transformations have many applications,\nand they will also be useful for understanding some of the machine learning concepts.\nHere are a few interesting transformations in 2-dimensional space:\n\n\\begin{itemize}\n \\item Identity transformation has no effect on the vector to be transformed\n \\[\n \\begin{bmatrix}\n 1 & 0 \\\\\n 0 & 1 \\\\\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n 1 \\\\\n 2 \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n 1 \\\\\n 2 \\\\\n \\end{bmatrix}\n \\]\n \\item Stretch along x-axis\n \\[\n \\begin{bmatrix}\n 3 & 0 \\\\\n 0 & 1 \\\\\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n 1 \\\\\n 2 \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n 3 \\\\\n 2 \\\\\n \\end{bmatrix}\n \\]\n \\begin{marginfigure}\n \\tikzsetnextfilename{linear-op-stretch}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[step=1cm,grid] (-4,-1) grid (4,4);\n \\draw[->] (0,-1) -- (0, 4);\n \\draw[<->] (-4,0) -- (4,0);\n \\coordinate (v1) at (1,2);\n \\coordinate (v2) at (3,2);\n \\draw[blue,thick,->] (0,0) -- (v1);\n \\node[anchor=south east] at (v1) {$(1,2)$};\n \\draw[orange,thick,->] (0,0) -- (v2);\n \\node[anchor=south west] at (v2) {$(3,2)$};\n \\draw[dotted,->,thick] (v1) to[bend left] (v2);\n \\end{tikzpicture}\n \\caption{Stretch (three times) along x.}\n \\end{marginfigure}\n \\item Multiplying a vector with\n \\[\n \\begin{bmatrix}\n \\cos\\theta & -\\sin\\theta \\\\\n \\sin\\theta & \\cos\\theta \\\\\n \\end{bmatrix}\n \\]\n rotates it with $\\theta$ degrees (counter-clock wise).\n For example, for 90-dgrees rotation,\n \\[\n \\begin{bmatrix}\n 0 & -1 \\\\\n 1 & 0 \\\\\n \\end{bmatrix}\n \\times\n \\begin{bmatrix}\n 1 \\\\\n 2 \\\\\n \\end{bmatrix}\n =\n \\begin{bmatrix}\n -2 \\\\\n 1 \\\\\n \\end{bmatrix}\n \\]\n \\begin{marginfigure}\n \\tikzsetnextfilename{linear-op-rotate}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[step=1cm,grid] (-4,-1) grid (4,4);\n \\draw[->] (0,-1) -- (0, 4);\n \\draw[<->] (-4,0) -- (4,0);\n \\coordinate (v1) at (1,2);\n \\coordinate (v2) at (-2,1);\n \\draw[blue,thick,->] (0,0) -- (v1);\n \\node[anchor=south west] at (v1) {$(1,2)$};\n \\draw[orange,thick,->] (0,0) -- (v2);\n \\node[anchor=south east] at (v2) {$(-2,1)$};\n \\draw[dotted,->,thick] (v1) to[bend right] (v2);\n \\end{tikzpicture}\n \\caption{Rotate 90 degrees.}\n \\end{marginfigure}\n% \\item Shear: $\\begin{bmatrix}1 & k \\\\ 0 & 1 \\\\\\end{bmatrix}$\n\\end{itemize}\n\nThese linear operations can be combined (composed)\nfor more complex transformations.\n\n\\newthought{Solving a set of linear linear equations} has been one of\nthe main applications of linear algebra.\nWe will not discuss how to solve a linear equations here\n(since we rarely do this by hand),\nbut we will demonstrate how a set of linear equations are represented using matrices\nand vectors.\nWe will encounter this in various forms during the course.\n\nThe set of equations,\n \\[\n \\begin{array}{rcrcr}\n 2x_{1} &+& x_{2} &=& 6 \\\\\n x_{1} &+& 4x_{2} &=& 17 \\\\\n \\end{array}\n \\]\n can be written as:\n \\[\n \\underbrace{\\begin{bmatrix} \n 2 & 1 \\\\ \n 1 & 4 \\\\ \n \\end{bmatrix}}_{\\vect{W}}\n \\underbrace{\\begin{bmatrix} \n x_{1} \\\\ \n x_{2} \\\\ \n \\end{bmatrix}}_{\\vect{x}} \n = \n \\underbrace{\\begin{bmatrix} \n 6 \\\\ \n 17 \\\\ \n \\end{bmatrix}}_{\\vect{b}}\n \\]\nwhich allows finding a solution (if one exists)\nusing \\emph{Gaussian elimination}.\n\nFor our purposes,\nimportant part is to realize that\nthis amounts to the matrix/vector operations operations\nwe have been reviewing so far.\n\n\\newthought{Inverse of a matrix} is defined for square matrices. \nInverse of matrix \\vect{W} is denoted by $\\vect{W}^{-1}$.\nMultiplying a matrix with its inverse yields the identity matrix.\n\\[\n \\vect{W}\\vect{W}^{-1} = \\vect{W}^{-1} \\vect{W} = \\vect{I}\n\\]\n\nNow that we have defined the inverse of a matrix,\nwe can solve the set of linear equations $\\vect{W}\\vect{x} = \\vect{b}$,\nusing the inverse matrix.\n\\[ \n \\begin{aligned}\n \\vect{W} \\vect{x} &= \\vect{b} \\\\\n \\vect{W}^{-1} \\vect{W} \\vect{x} &= \\vect{W}^{-1} \\vect{b} \\\\\n \\vect{I} \\vect{x} &= \\vect{W}^{-1} \\vect{b} \\\\\n \\vect{x} &= \\vect{W}^{-1} \\vect{b} \\\\\n \\end{aligned}\n\\]\n\n\\todo[inline]{note `pseudo inverse' ?}\nCalculating inverse of a matrix involves using a set of operations,\ncalled \\emph{elementary row operations},\non the augmented matrix that contains the original matrix and the identity matrix side by side.\nWe will not cover this here, as we rarely do this by hand.\nInterested readers should check any of the linear algebra sources listed at the end of the chapter.\n\n\\newthought{The determinant of a matrix} is a scalar value\nwith some interesting properties and applications, including\n\\begin{itemize}\n \\item A matrix is invertible if it has a non-zero determinant\n \\item A system of linear equations has a unique solution if the coefficient matrix has a non-zero determinant\n\\end{itemize}\n\nWe denote the determinant of a matrix with vertical bars around it,\ndeterminant of \\vect{A} is denoted by $\\lvert\\vect{A}\\rvert$.\nGeometric interpretation of determinant is the (signed) changed in the volume of a unit (hyper)cube caused by the transformation defined by the matrix.\n\nThe determinant of a $2\\times{}2$ matrix can be calculated by the formula:\n\\[\n \\begin{vmatrix}\n a & b \\\\\n c & d \\\\\n \\end{vmatrix} = a d - b c\n\\]\nThe above formula generalizes to higher dimensional matrices through a recursive definition.\n\\todo{definition, geometric interpretation}\n\n\\newthought{Eigenvalues and eigenvectors} of a matrix also has important applications.\n\nAn \\emph{eigenvector}, \\vect{v}\nand corresponding \\emph{eigenvalue}, $\\lambda$,\nof a matrix $\\vect{A}$ is defined such that\n\\[ \\vect{A} \\vect{v} = \\lambda \\vect{v}\\;. \\] \n\nIn (other) words, multiplying a matrix with its eigen vector\nonly changes its magnitude of the vector\nand does not change its direction.\n\nEigenvalues an eigenvectors have many applications from communication theory to quantum mechanics\nA better known example (and close to home) is Google's PageRank algorithm\nWe will return to them while discussing PCA\nand SVD (and maybe other topics/concepts)\n\n\\todo[inline]{Tensors}\n\n\\section{\\label{sec:calculus-intro}Derivatives and integrals}\n\nDifferentiation and integration are two fundamental concepts in calculus.\nThe reason we review some of the basic calculus here has to do\nwith the fact that these operations are often used\nin probability theory and machine learning.\nIn many machine learning problems,\nlearning is achieved through minimizing the error\nor maximizing an objective (e.g., likelihood).\nA particularly important use of derivatives in machine learning is\nto find maxima or minima of error or objective function.\nThis section will give a refresher on these topics,\nand define some notation that we will use throughout the course. \nYou can safely skip this section,\nif you know how to differentiate polynomial functions\nor what a \\emph{gradient} is.\n\nDerivative of a function indicates the rate of change.\nThe familiar example from physics is that\nthe derivative of the velocity of a moving object is it acceleration.\nFor example,\nthe velocity of a car changes proportional to its acceleration of deceleration.\n\n\\marginnote{\n \\begin{tcolorbox}[colback=white,colframe=gray,halign=left,width=\\linewidth]\n A quick refresher on polynomial functions: if $f(x) = x^{n}$,\n \\[\n f'(x) = \\frac{df}{dx} = n x^{n-1}\\;.\n \\]\n For example, for\\\\\n $f(x) = x^{3} + 2 x^{2}$,\n \\[\n f'(x) = \\frac{df}{dx} = 3 x^{2} + 4 x\\;.\n \\]\n \\end{tcolorbox}\n}\nOne of the common ways of denoting a function's derivative is,\nusing the `prime notation'.\nFor example derivative of the function function $f(x)$ written as $f'(x)$.\nAnother common notation is $\\frac{df}{dx}(x)$ or $\\frac{df}{dx}(x)$.\\todo{double check}\nFor multi-variate functions,\nthis notation makes it clear that the derivative is taken with respect to the variable $x$.\n\nIf defined, derivative of a function is another function.\nA well known example is the polynomials,\nwhose derivatives are lower degree polynomials.\nFor example, if $f(x) = x^{2} - 2 x $ then $f'(x) = 2 x - 2$,\nwhich means that the rate of change of a quadratic function doubles\nas $x$ is increased one unit.\nNote that if a polynomial of degree $n$ is differentiated $n$ times,\nit becomes a constant.\nDerivative of a linear function is a constant value,\nsince a linear function changes with the same rate everywhere.\nOn the other hand, derivative of a constant (function) is $0$,\nsince there is no change.\n\n\\begin{marginfigure}\n \\tikzsetnextfilename{func-derivative}\n \\begin{tikzpicture}[scale=0.7]\n \\draw[step=1cm,grid] (-4,-2) grid (4,4);\n \\draw[->] (0,-2) -- (0, 4);\n \\draw[<->] (-4,0) -- (4,0);\n \\begin{scope}\n \\clip(-4,-2) rectangle (4,4);\n \\draw[domain=-2:4,smooth,variable=\\x,blue]\n plot ({\\x},{\\x*\\x - 2*\\x});\n\n \\node[blue,xshift=0.2ex] at (1,2) {$f(x) = x^{2} - 2x$};\n\n \\node[orange,inner sep=0pt,draw,circle,minimum size=2pt]\n at (1,-1) {};\n \\node[orange,anchor=north] at (1,-1) {$f'(1) = 0$};\n \\draw[orange] (-4,-1) -- (4, -1);\n\n \\begin{scope}\n \\clip(-4,-0.5) rectangle (4,4);\n \\node[purple,inner sep=0pt,draw,circle,minimum size=2pt]\n at (3,3) {};\n \\node[purple,anchor=south east] at (3,3) {$f'(3) = 4$};\n \\draw[purple] (4, 7) -- (3,3) -- (2, -1);\n\n \\node[red,inner sep=0pt,draw,circle,minimum size=2pt]\n at (-0.5,1.25) {};\n \\draw[red] (-1.5, 4.25) -- (-0.5,1.25) -- (0.5, -1.75);\n \\node[red,anchor=north east] at (-0.5,1.25) {$f'(-0.5) = -3$};\n \\end{scope}\n \\end{scope}\n \\end{tikzpicture}\n \\caption{The function $f(x) = x^{2} - 2x$ and its derivative evaluated at,\n different $x$ values.\n }\n\\end{marginfigure}\nWhen evaluated at a particular $x$ value,\nthe derivative of the function will be\nthe slope of the tangent line at that point,\nwhich is indication of the direction and the rate of change.\nNote that in our example above,\nthe derivative of $f(x) = x^{2} - 2x$, $f'(x) = 2 x - 2$, at $x = 1$ is $0$.\nFor a continuous function the derivative of the function is equal to $0$,\nonly on the maximum and minimum points of the function.\nAnd this is the most important reason for all the earlier\nnotes in this introduction.\nIn many methods we see later on,\nwe will be aiming at maximizing or minimizing functions,\nwhere this will be a handy tool.\nIn general,\nderivative evaluated at a particular point will be $0$ for maxima and minima\nof functions,\nit will be a negative value if the function is decreasing (as $x$ increases),\nand a positive value if the function is increasing with $x$.\n\nSo far,\nwe've considered differentiation of functions of a single variables.\nIn machine learning and NLP,\nwe often deal with functions of many variables.\nOne can also differentiate a function of multiple variables with respect to one of them,\nfor which the dependence between the variables should be considered.\nWe will not review how to take (total) derivatives\nof functions of multiple variables.\nHowever, we will introduce \\term{partial derivatives} briefly here.\nA partial derivative is similar to a total derivative,\nbut we assume that except the variable along which we take the derivative,\nall other variables are constants.\nSo, when you evaluate the partial derivative of a function at a particular point,\nit gives you the rate of change along one of the axes.\n\n\\todo[inline]{note the chain rule and the relation between partial/total derivative?}\nThe partial derivative of a function $f$ with respect variable $x$ is denoted by $\\frac{\\partial f}{\\partial x}$.\nFor example, if $f(x, y) = x^{3} + yx$, \n\\[\n \\frac{\\partial f}{\\partial x} = 3 x^{2} + y,\\; \\text{and}\\; \n \\frac{\\partial f}{\\partial y} = x .\n\\]\nThe vector formed by all partial derivatives\nof a (scalar) function of $n$-variables is called its \\term{gradient}.\nGradient of a function $f$ is denoted by $\\nabla f$, or $\\vec{\\nabla} f$.\n\\[\n \\nabla f(x_{1}, \\ldots, x_{n}) = \n \\left(\\frac{\\partial f}{\\partial x_{1}}, \\ldots,\n \\frac{\\partial f}{\\partial x_{n}} \\right)\n\\]\nSimilar to derivative of a function of a single variable,\ngradient points to the direction of greatest change,\nand the magnitude of the gradient indicates the steepness of the change.\nAreas where the gradient is \\num{0} are (local) minima, maxima and saddle points.\nAs a result, it is an important tool\nin finding minimum and maximum values of (objective) functions.\n\n\\todo[inline]{Second derivative, picture with a saddle point, Jacobian matrix(?)}\n\n\n\\begin{marginfigure}[-5\\baselineskip]\n \\tikzsetnextfilename{integration}\n \\begin{tikzpicture}[x=10mm, y=2mm, scale=0.7]\n \\draw[step=1cm,grid] (-4,-2) grid (4,30);\n \\clip(-4,-2) rectangle (4,30);\n \\draw[->] (0,-2) -- (0, 30);\n \\draw[<->] (-4,0) -- (4,0);\n\n \\draw[domain=-4:4,smooth,variable=\\x,blue]\n plot ({\\x},{3*\\x*\\x});\n \\fill[blue,opacity=0.2] \n plot[domain=1:3,smooth,variable=\\x] ({\\x},{3*\\x*\\x}) \n -- (3,0) -- (1,0) -- cycle;\n% plot coordinates {(3,27) (3,0) (1,0) (1,3) };\n \\end{tikzpicture}\n \\caption{\\label{fig:integral}%\n Integral of the function $f(x) = 3 x^{2}$ in range $[1, 3]$.\n }\n\\end{marginfigure}\n\\newthought{Integration} is the inverse of the derivation.\nIn general, the integral of a function in a given range corresponds\nto the (signed) area (or volume) under a function in this range.\nThe notation used for integral of a function $f(x)$ is\n$F(x) = \\int f(x) dx$.\nThis is called an indefinite integral.\nOften we want the integral of a function in an interval $[a,b]$,\nwhich can be calculated by\n\\[\n \\int_{a}^{b} f(x) dx = F(b) - F(a).\n\\]\n\nFor example, if $f(x) = 3 x^{2}$,\nwe know that $F(x) = x^{3}$ (since the integral is the antiderivative,\nand $F'(x) = f(x) = 3 x^{2}$).\nIf we want to know the area under $f(x)$ within range $[1, 3]$,\nwe simply calculate $F(1) - F(0) = 27 - 1 = 26$.\n\n\\begin{marginfigure}\n \\tikzsetnextfilename{numeric-integral1}\n \\begin{tikzpicture}[x=10mm, y=5mm]\n \\draw[step=1cm,grid] (-0.2,-0.2) grid (4,4);\n \\clip(-0.2,-0.2) rectangle (4,4);\n \\draw[->] (0,-0.2) -- (0, 4);\n \\draw[->] (-0.2,0) -- (4,0);\n\n \\draw[domain=-4:4,smooth,variable=\\x,blue]\n plot ({\\x},{0.25*\\x*\\x});\n \\foreach \\x in {1,...,6}{%\n \\fill[draw=black,blue,opacity=0.2]\n (0.5*\\x, 0.25*0.25*\\x*\\x) rectangle ($ (0.5*\\x,0) + (-1*0.5, 0)$);\n }\n \\end{tikzpicture}\\\\\n \\tikzsetnextfilename{numeric-integral2}%\n \\begin{tikzpicture}[x=10mm, y=5mm]\n \\draw[step=1cm,grid] (-0.2,-0.2) grid (4,4);\n \\clip(-0.2,-0.2) rectangle (4,4);\n \\draw[->] (0,-0.2) -- (0, 4);\n \\draw[->] (-0.2,0) -- (4,0);\n\n \\draw[domain=-4:4,smooth,variable=\\x,blue]\n plot ({\\x},{0.25*\\x*\\x});\n \\foreach \\x in {1,...,20}{%\n \\fill[draw=black,blue,opacity=0.2]\n (3*0.05*\\x, 9*0.0025*0.25*\\x*\\x) rectangle ($ (3*0.05*\\x,0) + (-1*3*0.05, 0)$);\n }\n \\end{tikzpicture}\n \\caption{\\label{fig:ingegral-numeric}%\n Demonstration of numerical approximation to an integral.\n Note that as the rectangles get smaller (as in the figure below),\n sum of their areas gets closer to the area under the curve.\n }\n\\end{marginfigure}\nOften integrating functions analytically\n(in closed form) is not easy or possible.\nIn these cases, integrals can be computed with numeric approximation.\nOne way to do this is to sum the areas of rectangles\nas demonstrated in Figure~\\ref{fig:ingegral-numeric}.\nAs we decrease the width of the rectangles,\nor equivalently, increase the number of rectangles in a fixed range,\nthe approximation will be more precise.\nThis also hints at interpreting integrals as a infinite sum.\nThis interpretation will often be more useful\nfor the examples we will see (often in probability theory).\n\n\\section*{Summary}\n\nIn this lecture,\nwe reviewed some concepts from linear algebra and calculus.\nThe aim was to\nprovide a refresher for those who studied these topics,\nfamiliarize the reader with the notation that will be used,\nand also give a feeling of the what mathematical concepts will be useful\nfor studying statistical natural language processing.\nThis overview here is necessarily informal and incomplete.\nBelow, a number of potential sources are listed if you need a better introduction to these concepts.\n\nFor linear algebra,\n\\textcite{cherney2013} and \\textcite{beezer2016} are\ntwo textbooks that are freely available online.\nA classic reference textbook for linear algebra is \\textcite{strang2009}.\nFor a more practical/geometric approach,\nsee \\textcite{farin2014} or \\textcite{shifrin2011}.\n\nFor the concepts we reviewed briefly from calculus,\nany textbook introduction to calculus should be sufficient.\nA well-known (also available online) textbook is \\textcite{strang1991}.\nFor more alternatives on open textbooks on mathematics see\n\\url{http://www.openculture.com/free-math-textbooks}.\n", "meta": {"hexsha": "81807ce880e5fede5d479d32c2ffaaa0d86492fd", "size": 38312, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "math-overview.tex", "max_stars_repo_name": "coltekin/snlp-notes", "max_stars_repo_head_hexsha": "02dddcda0a8ff24f959675f0ad6a9079d573f49b", "max_stars_repo_licenses": ["CC-BY-4.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-08-05T12:58:44.000Z", "max_stars_repo_stars_event_max_datetime": "2020-02-21T14:17:47.000Z", "max_issues_repo_path": "math-overview.tex", "max_issues_repo_name": "coltekin/snlp-notes", "max_issues_repo_head_hexsha": "02dddcda0a8ff24f959675f0ad6a9079d573f49b", "max_issues_repo_licenses": ["CC-BY-4.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "math-overview.tex", "max_forks_repo_name": "coltekin/snlp-notes", "max_forks_repo_head_hexsha": "02dddcda0a8ff24f959675f0ad6a9079d573f49b", "max_forks_repo_licenses": ["CC-BY-4.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 36.9094412331, "max_line_length": 151, "alphanum_fraction": 0.6510492796, "num_tokens": 12241, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8128673087708699, "lm_q2_score": 0.7520125682019722, "lm_q1q2_score": 0.6112864324762073}} {"text": "\\documentclass{article}\n\n\\usepackage{epsfig}\n\\usepackage[export]{adjustbox}% http://ctan.org/pkg/adjustbox\n\\usepackage{caption}\n\\usepackage{subcaption}\n\\usepackage{fullpage}\n\\usepackage{commath}\n\\usepackage{amssymb}\n\\usepackage[space]{grffile}\n\\usepackage{float}\n\\usepackage{float}\n\n\\def\\E{\\mathbb{E}}\n\\def\\R{\\mathbb{R}}\n\\def\\bin{\\text{bin}}\n\\def\\bet{\\text{beta}}\n\\def\\Bin{\\text{Binom}}\n\\def\\diag{\\text{diag}}\n\n\\title{Single Sequence HMM in DNA Methylation Problem}\n\\author{Chicheng Zhang}\n\n\\begin{document}\n\\maketitle\n\n\\section{The Model}\n\nIn the problem of DNA methylation, we are given a sequence $\\cbr{x_t = (c_t, \\mu_t)}_{t=1}^l$, where at each time $t$, the coverage is an nonnegative integer $c_t \\in \\cbr{0,1,\\ldots,N}$ and the methylation count is $\\mu_t \\in \\cbr{0,1,\\ldots,c_t}$. We model the data using a hidden Markov model, that is, the underlying dynamics of the hidden states is driven by a Markov chain $\\cbr{h_t}$, where $h_t \\in [m]$. We leave the consideration of context information to the next stage.\n\nThe HMM can be summarized by parameters $(\\pi, T, O)$, where $\\pi \\in \\R^m$ is the initial probability distribution $\\Pr(h_1)$ and $T \\in \\R^{m \\times m}$ is the transition matrix of the Markov chain $\\Pr(h_{t+1} | h_t)$, $O$ is the observation matrix, namely the conditional probability of methylation and coverage, given hidden state $\\Pr(c_t, \\mu_t | h_t)$. To avoid notation clutter, we drop the time indices $t$ if they are the same among all the variables appeared in a formula.\n\nWe assume the $\\mu_t$ given $c_t$ and $h_t$ is drawn from a binomial distribution, and for each hidden state $h \\in [m]$ there is a methylation probability $p_h \\in [0,1]$ associated with it. Formally,\n\\[ \\mu_t | c_t, h_t \\sim \\Bin(c_t, p_{h_t}) \\]\n\n\\section{Algorithm based on Method of Moments}\n\nGiven an observation $(c, \\mu)$, we map it to a $n$-dimensional vector using feature mapping $\\phi$. Generally speaking, different feature mappings have different properties.\n\nFor mapping $\\phi$, we define the corresponding observation matrices as follows:\n\\[ O_{\\phi}^s := \\E[\\phi(c_s,\\mu_s) | h_2] \\]\nfor $s = 1,2,3$, where $h_2$ is the hidden state at time 2. It is worth pointing out that $O_\\phi$ contains information to recover the $p_h$'s, and since\n\\[ O_{\\phi}^3 = O_{\\phi}^2 T, \\qquad O_{\\phi}^1 = O_{\\phi}^2 \\diag(\\pi) T^T \\diag(T\\pi)^{-1}\\]\nif we can recover $O_{\\phi}^s$, $s = 1,2,3$, then we can recover $T$ and $\\pi$ by algebra.\n\nConditional independence of $(c_1, \\mu_1)$, $(c_2, \\mu_2)$, $(c_3, \\mu_3)$ given $h_2$ implies that\n\\[ \\E[\\phi(c_1, \\mu_1) \\otimes \\phi(c_2, \\mu_2) \\otimes \\phi(c_3, \\mu_3)] = \\sum_{i=1}^m w_i (O_{\\phi}^1)_i \\otimes (O_{\\phi}^2)_i \\otimes (O_{\\phi}^3)_i \\]\nwhere $w_i = \\Pr[h_2 = i] = (T\\pi)_i$.\nThis tensor has a low rank structure.\n\nIts empirical version can be computed by summing the outer products of mapped observation over training data; suppose the training data contains triples $\\cbr{((c_{t,1}, \\mu_{t,1}), (c_{t,2}, \\mu_{t,2}), (c_{t,3}, \\mu_{t,3}))}_{t=1}^{l}$, then\n\\[ \\hat{\\E}[\\phi(c_1, \\mu_1) \\otimes \\phi(c_2, \\mu_2) \\otimes \\phi(c_3, \\mu_3)] = \\frac{1}{l} \\sum_{t=1}^l \\phi(c_{t,1}, \\mu_{t,1}) \\otimes \\phi(c_{t,2}, \\mu_{t,2}) \\otimes \\phi(c_{t,3}, \\mu_{t,3}) \\to \\E[\\phi(c_1, \\mu_1) \\otimes \\phi(c_2, \\mu_2) \\otimes \\phi(c_3, \\mu_3)] \\]\n\nTherefore, applying CP-tensor decomposition on the empirical tensor gives an estimate of $O_{\\phi}^s$, $s = 1,2,3$.\n\n\n\\section{Feature Maps}\nSome examples of feature maps are given as follows.\n\n\\paragraph{Binning mapping} Given observation $(c, \\mu)$, $\\phi_{\\bin, n}(c, \\mu)$ is a $n$-dimensional vector, with its entries as follows:\n\\[ (\\phi_{\\bin, n}(c, \\mu))_i = \\begin{cases} I(\\frac{\\mu}{c} \\in (\\frac{i-1}{n}, \\frac{i}{n}]) & c \\neq 0 \\\\ 0 & c = 0 \\end{cases} \\]\nwhere $n$ is a hyperparameter controlling the number of the bins, and the width of the bins is $\\frac{1}{n}$.\n\n\\paragraph{Beta mapping} Given observation $(c, \\mu)$, $\\phi_{\\bet, n}(c, \\mu)$ is a $n$-dimensional vector, with its entries as follows:\n\\[ (\\phi_{\\bet, n}(c, \\mu))_i = \\frac{1}{B(\\mu+1, c-\\mu+1)} (\\frac{i}{n})^\\mu (1-\\frac{i}{n})^{c-\\mu} \\]\n\n\\section{Experiment Results}\n\\subsection{Setting}\nIn this set of experiments, we focus on the methylation data in cell type E1 and chromosome 1.\n\nLet $s$ be the number of base segments merged to a single segment. Specifically, we aggregate every $s$ observations to a single one. Namely, for all integers $k \\geq 0$, we merge obervations $(c_{ks+1}, \\mu_{ks+1})$, $(c_{ks+2}, \\mu_{ks+2})$, \\ldots, $(c_{ks+s}, \\mu_{ks+s})$ to a single $(\\tilde{c}_k, \\tilde{\\mu}_k)$, where\n\\[ \\tilde{c}_k = \\sum_{i=1}^k c_{ks+i}, \\quad \\tilde{\\mu}_k = \\sum_{i=1}^k \\mu_{ks+i}\\]\nNote that the segments merged are non-overlapping.\n\n\nLet $l$ be the length of the input sequence, that is, we draw $l$ 3-consecutive observations with replacement from the (merged) sequence $\\cbr{(c_t, \\mu_t)}_{t=1}^{L}$, forming a new dataset containing triples $\\cbr{((\\tilde{c}_{t,1}, \\tilde{\\mu}_{t,1}), (\\tilde{c}_{t,2}, \\tilde{\\mu}_{t,2}), (\\tilde{c}_{t,3}, \\tilde{\\mu}_{t,3}))}_{t=1}^{l}$.\n\nLet $n$, the dimensionality of the feature mapping, be 30.\n\nLet $m$ be the number of hidden states assumed. For example if $m = 4$, then the estimated matrix $O^2_{\\phi} = \\E[\\phi(x)|h]$ has 4 columns.\n\n%At this point we do not take into account the contexts, and we further assume that the coverage in different segments are non-uniform.\n\nAlthough our ultimate goal is to recover the transition matrix and observation matrix of the hidden Markov model, at the current stage, we recover $\\E[\\phi(x)|h]$ for each value of $h$. Recall that $x = (c, \\mu)$ is the (coverage, methylation) count pair in each segment. In the figures, the x-axis $t$ corresponds to the $i = (tn)^{\\text{th}}$ coordinate of vector $\\phi$. The $y$ axis corresponds to the values of $(\\E[\\phi(x)|h])_i$.\n\n\nThe value of $p_h$ for each $h$ can be straightforwardly extracted from $\\E[\\phi(x)|h]$. The transition matrix can also be computed subsequently. We leave the full recovery of the parameters to the next stage.\n\n\\paragraph{Recovery of $p_h$ from Expected Beta feature mapping}\nRecall that\n\\[ \\int_0^1 \\frac{t t^{m}t^{c-m}}{B(m+1,c-m+1)} dt = \\int_0^1 \\frac{t t^{(m+1)-1}t^{(c-m+1)-1}}{B(m+1,c-m+1)} dt = \\frac{m+1}{c+2} \\]\nTherefore,\n\\[ \\int_0^1 \\E[\\phi(x,t) | h] dt = \\E[\\frac{m+1}{c+2} | h] = \\E[\\frac{cp_h + 1}{c + 2} | h] = \\E[\\frac{c}{c+2}] p_h + \\E[\\frac{1}{c+2}] \\]\nassuming independence between $h$ and $c$.\nLet $a = \\E[\\frac{1}{c+2}]$. Then, $\\int_0^1 \\E[\\phi(x,t) | h] dt = (1 - 2a) p_h + a$.\nHence,\n\\[ p_h = \\frac{\\int_0^1 \\E[\\phi(x,t) | h] dt - a}{1 - 2 a}. \\]\n\n\n%In subsequent results, we show the entire $\\E[\\phi(x)|h]$, as opposed to a single number $p_h \\in [0,1]$. The reason is that we wish to recover more information about the hidden states learned, to get a better understanding of the HMM model.\n\n%-- what features $\\phi$ are you looking at?\n%-- why are you showing the entire $\\phi$ vector and not a single number?\n%We focus on two types of feature mappings $\\phi$, the first one is binning mapping:\n%\\[ \\phi_{\\bin}(x,y) = \\begin{cases} {I(\\frac{\\mu}{c} \\in (y,y+h])} & c \\neq 0 \\\\ 0 & c = 0\\end{cases} \\]\n%where $h$ is the width of the bins.\n\n%The second one is Beta map:\n%\\[ \\phi(x,y) = \\frac{1}{B(\\mu+1, c-\\mu+1)}y^\\mu (1-y)^{c-\\mu} \\]\n%In this mapping there is no explicit notion of ``bin width\", but the value of $c$ can be thought of as a parameter controlling it: if the coverage $c$ is large, then the bin width is small, and vice versa.\n\n%In experiments, since it is impossible to write down $\\E[\\phi(x_1, \\cdot) \\otimes \\phi(x_2, \\cdot) \\otimes \\phi(x_3, \\cdot)]$ and perform tensor decomposition, we use discretization: compute an empirical version of $\\E[\\phi(x_1, y_1) \\otimes \\phi(x_2, y_2) \\otimes \\phi(x_3, y_3)]$, where $y_1, y_2, y_3$ is in $\\cbr{0,h,2h,\\ldots,1}$. By decomposing this tensor we are able to recover $\\E[\\phi(x, y) | h]$, where $y$ is in $\\cbr{0,h,2h,\\ldots,1}$. By mild regularity assumption on the distribution of $x$, this can give a fairly accurate estimate of $\\E[\\phi(x, y) | h]$ where $y$ ranges in the $[0,1]$ interval.\n\n\n\\subsection{Effect of Sample Size $l$}\nWe fix $m = 4$, $s = 1$, and vary the sample size (number of triples collected) $l =$ 1000, 10000, 20000, 40000, 80000, 160000, 320000. The samples are drawn without replacement from the original dataset. Figure~\\ref{fig:varyl} shows the result. Recall that the value of $m$ is the number of hidden states assumed; We observe that the output of the algorithm stablizes when $l \\geq 80000$. \\footnote{For the last few figures, although the colors of the lines in all subfigures are diffent, we observe that they have the same pattern.}\n\n\\begin{figure}[H]\n\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 10000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 10000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 20000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 20000$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 40000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 40000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 80000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 80000$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 160000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 160000$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$l = 320000$}\n \\end{subfigure}\n \\\\\n \\end{tabular}\n\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for varying sample size $l$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varyl}\n\\end{figure}\n\n\n\\subsection{Effect of Specifying the Number of States}\n\nWe fix $l = 320000$, $s = 1$, and vary the number of states $m = 2,3,4,5,6,7,8,9$. Figure~\\ref{fig:varym} shows the columns of $\\E[\\phi(x)|h]$ recovered. The spectral algorithm provides reasonable results when $m \\leq 4$; when $m \\geq 5$, the algorithm started to recover observation columns that has a lot of negative entries.\n\n\n\\begin{figure}[H]\n\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 1_n = 30_phi = beta_full.pdf}\n \\caption{$m = 1$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 2_n = 30_phi = beta_full.pdf}\n \\caption{$m = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = beta_full.pdf}\n \\caption{$m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$m = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_l/cell = E1_chr = 1_l = 320000_s = 1_m = 5_n = 30_phi = beta_full.pdf}\n \\caption{$m = 5$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for varying number of states $m$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varym}\n\\end{figure}\n\n\\subsection{Binning Feature vs. Beta Feature}\n\nWe compare the experimental results using two types of feature maps $\\phi_{\\bin, n}$ and $\\phi_{\\bet, n}$. Although the two mappings are substantially different, their respective $\\E[\\phi(x)|h]$ have some similarities. In some sense, the beta mapping is performing a ``soft'' binning which takes into account the number count $c$ in $x$: fixing the value of methylation probability $m/c$, if $c$ is larger, then the beta mapping is closer to a binning mapping with smaller bin size. Figure~\\ref{fig:varyphi} shows the columns of the recovered $\\E[\\phi(x)|h]$ recovered. Generally, $\\phi_{\\bet, n}$ produces smoother observation columns, and the results are stabler than $\\phi_{\\bin, n}$.\n\n%-- double check this one experiment -- typo in file naming?\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = beta_full.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 30_phi = binning_igz.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 3$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 4$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_phi/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = binning_igz.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 4$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for $m = 3,4$ and two types of feature map $\\phi$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varyphi}\n\\end{figure}\n\n\\subsection{Effect of Number of Segments Combined}\nWe fix $l = 320000$, $m = 4$, and vary the number of merged segments $s = 1,2,3,4,5,6,7,8$. Figure~\\ref{fig:varys} shows the columns of the recovered $\\E[\\phi(x)|h]$. It can be seen that the recovered result is fairly insensitive to the choice of $s$.\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 1$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 2_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 3_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 4_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.45\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/vary_s/cell = E1_chr = 1_l = 320000_s = 5_m = 4_n = 30_phi = beta_full.pdf}\n \\caption{$s = 5$}\n \\end{subfigure}\n \\end{tabular}\n\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for $m = 4$ and beta feature map $\\phi$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varys}\n\\end{figure}\n\n\\section{Experiments with CG methylations}\nIt is previously known that CG sites has different methylation behavior\ncompared to non-CG sites. We conduct experiments by merging the coverages and\nthe methylation counts only in contexts CGA, CGT, CGC, CGG (rows 13 to 16 of\nthe data).\n\n\\subsection{Effect of Number of Hidden States}\nOur first set of experiments is on investigating the effect of $m$, the number of hidden\nstates. We vary $m$ from 1 to 8, and plot the observation matrices in\nFigure~\\ref{fig:varymcg}.\n\nIt can be seen from the figures below that the range of plausible $m$'s are broader:\nunlike the full dataset where the observation columns start to have\nlarge negative entries starting from $m \\geq 4$, the CG subset still produces\npalusible output when $m = 5,6$. In addition, some hidden states recovered\nhas a large methylation probability, which is not observed in the full dataset\n(see e.g. Figure~\\ref{fig:varym}.)\n\n\\begin{figure}[H]\n\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 1_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 1$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 2_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 5_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 5$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 6_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 6$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 7_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 7$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 8_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$m = 8$}\n \\end{subfigure}\n\n \\end{tabular}\n\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for varying number of states $m$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varymcg}\n\\end{figure}\n\n\\subsection{Effects of Feature Maps}\nWe next conduct a set of comparisons between the beta feature map and the binning\nfeature map, setting the dimension of feature maps $n = 50$.\nIt can be seen that overall, binning features produce ``rougher''\nobservation columns. As with the range of plausible $m$'s, binning feature performs\nbadly when $m \\geq 5$, while beta feature consistently produces reasonable result\neven if $m = 5,6$. Both approaches output nonsensical observation matrices when\n$m = 7, 8$.\n\nIn addition, algorithm using the binning features cannot capture hidden states that has\nlow methylation probability (for all the hidden states, the conditional\nmethylation frequency histograms skew towards right).\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 2_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 2$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 2_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 2$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 3$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 3_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 3$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 4$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 4_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 4$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 5_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 5$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 5_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 5$}\n \\end{subfigure}\n \\end{tabular}\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for $m = 2,3,4,5$ and two types of feature map $\\phi$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varyphicg}\n\\end{figure}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 6_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 6$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 6_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 6$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 7_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 7$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 7_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 7$}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 8_n = 50_phi = beta_full_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bet, n}$ is beta mapping, $m = 8$}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/cg/cell = E1_chr = 1_l = 320000_s = 1_m = 8_n = 50_phi = binning_ctxt = 12131415.pdf}\n \\caption{$\\phi = \\phi_{\\bin, n}$ is binning mapping, $m = 8$}\n \\end{subfigure}\n \\end{tabular}\n \\caption{The observation columns of $\\E[\\phi(x)|h]$recovered, for $m = 6,7,8$ and two types of feature map $\\phi$. The $x$-axis is the value of $t = i/n$, the $y$-axis is the value of $(\\E[\\phi(x)|h])_i$.}\n \\label{fig:varyphicg2}\n\\end{figure}\n\n\n\\section{Experiments with Posterior Decoding}\n\nWith the observation matrix $\\E[\\phi(x)|h]$, transition matrix $T$, and initial probability $\\pi$ recovered,\nwe apply posterior decoding to the original sequence to get estimates of hidden states in each position (every 100 base pairs).\n\nEmpirically we see the result of posterior decoding is similar to that of Viterbi decoding, therefore we present here only the results\nfor posterior decoding. For illustration purposes, we set the length of our test sequence to 100000; in principle it can be set arbitrarily long.\nWe set the size of our training set to be 320000.\n\nBelow we visualize the recovered hidden states using horizontal bar plots of different colors, each color represents a different outcome\nof hidden state. On the right panel, we plot the corresponding expected feature maps of hidden states, and the colors of feature map\ncurves are consistent with those in the bar plot.\n\nIt can be concluded from the plots that all the sequences we are working with are dominated by a certain hidden states. We observe that the\nobservation matrices recovered in the CG contexts are significantly different from that of CC, CT, CA contexts. In particular, the red hidden\nstate in the results of CC, CT, CA contexts represents one with a low methylation probability, whereas the red hidden state in the result of CG\nrepresents one whose methlation probablity is high.\n\n\\subsection{CC Context}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n \\end{figure}\n\n\n \\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\n\\subsection{CT Context}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |4567l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |4567_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |4567l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |4567_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |4567l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |4567_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n \\end{figure}\n\n\n \\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |4567l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |4567_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |4567l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |4567_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\subsection{CA Context}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |891011l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |891011_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |891011l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |891011_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |891011l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |891011_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n \\end{figure}\n\n\n \\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |891011l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |891011_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |891011l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |891011_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\subsection{CG Context}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n \\end{figure}\n\n\n \\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\subsection{All Contexts}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123456789101112131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123456789101112131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123456789101112131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123456789101112131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123456789101112131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123456789101112131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\n \\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123456789101112131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123456789101112131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123456789101112131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123456789101112131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\section{Merging the Contexts using Conditional Independence}\n\nWe concatenate the feature representations in constexts CC, CA, CT and CG. For each\ncontext, the dimension is 10; hence in total the feature map is of 40 dimensions.\nMoreover, when performing posterior decoding, we assume the the observations are\nconditionally independent, that is, for all $i$, the observation probabililty satisfies\n\\begin{eqnarray*}\n &&\\Pr[(m_t^{CC}, m_t^{CA}, m_t^{CT}, m_t^{CG}) | c_t^{CC}, c_t^{CA}, c_t^{CT}, c_t^{CG}, h_t = i] \\\\\n &=&\n \\Pr[m_t^{CC} | c_t^{CC}, h_t = i] \\Pr[m_t^{CA} | c_t^{CA}, h_t = i] \\Pr[m_t^{CT} | c_t^{CT}, h_t = i] \\Pr[m_t^{CG} | c_t^{CG}, h_t = i]\n\\end{eqnarray*}\n\nFrom the experiments below, we can see that the recovered hidden states are diverse,\nand corresponds to different methylation behaviors in different contexts.\n\n\n\\subsection{Cell type E sequences(E1 and E2)}\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell E, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = E_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\n\\subsection{Cell type V (sequences V8 and V9)}\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell V, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell V, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell V, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell V, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell V, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = V_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\n\\subsection{Cell type P (sequences P13P14 and P15P16)}\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell P, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 2_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell P, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 3_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell P, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 4_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\begin{figure}[H]\n \\begin{tabular}{cc}\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell P, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 5_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\\\\n \\begin{subfigure}[t]{0.4\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415l_test = 320000_posterior.pdf}\n \\caption{Posterior Decoding for a substring of methylation of cell P, chromosome 1.}\n \\end{subfigure}\n &\n \\begin{subfigure}[t]{0.6\\textwidth}\n \\includegraphics[width=\\textwidth]{figs/merge_ctxts/cell = P_chr = 1_l = 320000_s = 1_m = 6_n = 40_phi = beta_full_ctxt = |0123|4567|891011|12131415_feature_map.pdf}\n \\caption{Expected Feature map given the hidden states}\n \\end{subfigure}\n \\end{tabular}\n\\end{figure}\n\n\\end{document}\n", "meta": {"hexsha": "eb44b52c7c97a0a8453d2514315e243498d74b6e", "size": 53535, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "kernelbin/expt.tex", "max_stars_repo_name": "anapophenic/knb", "max_stars_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_stars_repo_licenses": ["CC0-1.0"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2016-04-21T21:10:22.000Z", "max_stars_repo_stars_event_max_datetime": "2016-08-27T19:15:23.000Z", "max_issues_repo_path": "kernelbin/expt.tex", "max_issues_repo_name": "anapophenic/knb", "max_issues_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_issues_repo_licenses": ["CC0-1.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "kernelbin/expt.tex", "max_forks_repo_name": "anapophenic/knb", "max_forks_repo_head_hexsha": "0e865a47791bf801c00078a128a4d21e4375e732", "max_forks_repo_licenses": ["CC0-1.0"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 56.5311510032, "max_line_length": 687, "alphanum_fraction": 0.6705706547, "num_tokens": 17846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8740772351648677, "lm_q2_score": 0.6992544335934766, "lm_q1q2_score": 0.6112023819921616}} {"text": "%!TEX root = modelguide.tex\n\n\\chapter{Supporting classes}\n\nArtiSynth uses a large number of supporting classes, mostly defined in\nthe super package {\\tt maspack}, for handling mathematical and\ngeometric quantities. Those that are referred to in this manual are\nsummarized in this section.\n\n\\section{Vectors and matrices}\n\nAmong the most basic classes are those used to implement vectors and\nmatrices, defined in {\\tt maspack.matrix}. All vector classes implement\nthe interface \\javaclass[maspack.matrix]{Vector} and all matrix\nclasses implement \\javaclass[maspack.matrix]{Matrix}, which provide a\nnumber of standard methods for setting and accessing values and\nreading and writing from I/O streams. \n\nGeneral sized vectors and matrices are implemented by\n\\javaclass[maspack.matrix]{VectorNd} and\n\\javaclass[maspack.matrix]{MatrixNd}. These provide all the usual\nmethods for linear algebra operations such as addition, scaling, and\nmultiplication:\n%\n\\begin{lstlisting}[]\n VectorNd v1 = new VectorNd (5); // create a 5 element vector\n VectorNd v2 = new VectorNd (5); \n VectorNd vr = new VectorNd (5); \n MatrixNd M = new MatrixNd (5, 5); // create a 5 x 5 matrix\n\n M.setIdentity(); // M = I\n M.scale (4); // M = 4*M\n\n v1.set (new double[] {1, 2, 3, 4, 5}); // set values\n v2.set (new double[] {0, 1, 0, 2, 0});\n v1.add (v2); // v1 += v2\n M.mul (vr, v1); // vr = M*v1\n\n System.out.println (\"result=\" + vr.toString (\"%8.3f\"));\n\\end{lstlisting}\n%\nAs illustrated in the above example, vectors and matrices both provide\na {\\tt toString()} method that allows their elements to be formated\nusing a C-printf style format string. This is useful for providing\nconcise and uniformly formatted output, particularly for diagnostics.\nThe output from the above example is\n%\n\\begin{verbatim}\n result= 4.000 12.000 12.000 24.000 20.000\n\\end{verbatim}\n%\nDetailed specifications for the format string are provided in the\ndocumentation for \\javamethod[maspack.util]{NumberFormat.set(String)}.\nIf either no format string, or the string {\\tt \"\\%g\"}, is specified,\n{\\tt toString()} formats all numbers using the full-precision output\nprovided by {\\tt Double.toString(value)}.\n\nFor computational efficiency, a number of fixed-size vectors and\nmatrices are also provided. The most commonly used are those defined\nfor three dimensions, including \\javaclass[maspack.matrix]{Vector3d}\nand \\javaclass[maspack.matrix]{Matrix3d}:\n%\n\\begin{lstlisting}[]\n Vector3d v1 = new Vector3d (1, 2, 3);\n Vector3d v2 = new Vector3d (3, 4, 5);\n Vector3d vr = new Vector3d ();\n Matrix3d M = new Matrix3d();\n\n M.set (1, 2, 3, 4, 5, 6, 7, 8, 9);\n\n M.mul (vr, v1); // vr = M * v1\n vr.scaledAdd (2, v2); // vr += 2*v2;\n vr.normalize(); // normalize vr\n System.out.println (\"result=\" + vr.toString (\"%8.3f\"));\n\\end{lstlisting}\n%\n\n\\section{Rotations and transformations}\n\\label{RigidTransform3d:sec}\n\n{\\tt maspack.matrix} contains a number classes that implement rotation\nmatrices, rigid transforms, and affine transforms. \n\nRotations (Section \\ref{Rotations:sec}) are commonly described using a\n\\javaclass[maspack.matrix]{RotationMatrix3d}, which implements a\nrotation matrix and contains numerous methods for setting rotation\nvalues and transforming other quantities. Some of the more commonly\nused methods are:\n%\n\\begin{lstlisting}[]\n RotationMatrix3d(); // create and set to the identity\n RotationMatrix3d(u, angle); // create and set using an axis-angle\n\n setAxisAngle (u, ang); // set using an axis-angle\n setRpy (roll, pitch, yaw); // set using roll-pitch-yaw angles\n setEuler (phi, theta, psi); // set using Euler angles\n invert (); // invert this rotation\n mul (R) // post multiply this rotation by R\n mul (R1, R2); // set this rotation to R1*R2\n mul (vr, v1); // vr = R*v1, where R is this rotation\n\\end{lstlisting}\n%\nRotations can also be described by\n\\javaclass[maspack.matrix]{AxisAngle}, which characterizes a rotation\nas a single rotation about a specific axis.\n\nRigid transforms (Section \\ref{RigidTransforms:sec}) are used by\nArtiSynth to describe a rigid body's pose, as well as its relative\nposition and orientation with respect to other bodies and coordinate\nframes. They are implemented by\n\\javaclass[maspack.matrix]{RigidTransform3d}, which exposes its\nrotational and translational components directly through the fields\n{\\tt R} (a {\\tt RotationMatrix3d}) and {\\tt p} (a {\\tt\nVector3d}). Rotational and translational values can be set and\naccessed directly through these fields. In addition, {\\tt\nRigidTransform3d} provides numerous methods, some of the more commonly\nused of which include:\n%\n\\begin{lstlisting}[]\n RigidTransform3d(); // create and set to the identity\n RigidTransfrom3d(x, y, z); // create and set translation to x, y, z\n\n // create and set translation to x, y, z and rotation to roll-pitch-yaw\n RigidTransfrom3d(x, y, z, roll, pitch, yaw);\n\n invert (); // invert this transform\n mul (T) // post multiply this transform by T\n mul (T1, T2); // set this transform to T1*T2\n mulLeftInverse (T1, T2); // set this transform to inv(T1)*T2\n\\end{lstlisting}\n%\n\nAffine transforms (Section \\ref{AffineTransforms:sec}) are used by\nArtiSynth to effect scaling and shearing transformations on\ncomponents. They are implemented by\n\\javaclass[maspack.matrix]{AffineTransform3d}.\n\nRigid transformations are actually a specialized form of affine\ntransformation in which the basic transform matrix equals a rotation.\n{\\tt RigidTransform3d} and {\\tt AffineTransform3d} hence both derive\nfrom the same base class\n\\javaclass[maspack.matrix]{AffineTransform3dBase}.\n\n\\section{Points and Vectors}\n\nThe rotations and transforms described above can be used to transform\nboth vectors and points in space.\n\nVectors are most commonly implemented using\n\\javaclass[maspack.matrix]{Vector3d}, while points can be implemented\nusing the subclass \\javaclass[maspack.matrix]{Point3d}. The only\ndifference between {\\tt Vector3d} and {\\tt Point3d} is that the former\nignores the translational component of rigid and affine transforms;\ni.e., as described in Sections \\ref{RigidTransforms:sec} and\n\\ref{AffineTransforms:sec}, a vector {\\tt v} has\nan implied homogeneous representation of\n%\n\\begin{equation}\n\\v^* \\equiv \\matl \\v \\\\ 0 \\matr,\n\\end{equation}\n%\nwhile the representation for a point {\\tt p} is\n%\n\\begin{equation}\n\\p^* \\equiv \\matl \\p \\\\ 1 \\matr.\n\\end{equation}\n%\n\nBoth classes provide a number of methods for applying rotational and\naffine transforms. Those used for rotations are\n%\n\\begin{lstlisting}[]\n void transform (R); // this = R * this\n void transform (R, v1); // this = R * v1\n void inverseTransform (R); // this = inverse(R) * this\n void inverseTransform (R, v1); // this = inverse(R) * v1\n\\end{lstlisting}\n%\nwhere {\\tt R} is a rotation matrix and {\\tt v1} is a vector (or a point\nin the case of {\\tt Point3d}).\n\nThe methods for applying rigid or affine transforms include:\n\\begin{lstlisting}[]\n void transform (X); // transforms this by X \n void transform (X, v1); // sets this to v1 transformed by X\n void inverseTransform (X); // transforms this by the inverse of X\n void inverseTransform (X, v1); // sets this to v1 transformed by inverse of X\n\\end{lstlisting}\nwhere {\\tt X} is a rigid or affine transform.\nAs described above, in the case of {\\tt Vector3d}, these methods\nignore the translational part of the transform and apply only the\nmatrix component ({\\tt R} for a {\\tt RigidTransform3d} and {\\tt A} for\nan {\\tt AffineTransform3d}).\nIn particular, that means that for a {\\tt RigidTransform3d} given by {\\tt X}\nand a {\\tt Vector3d} given by {\\tt v},\nthe method calls\n%\n\\begin{lstlisting}[]\n v.transform (X.R)\n v.transform (X)\n\\end{lstlisting}\n%\nproduce the same result.\n\n\\section{Spatial vectors and inertias}\n\\label{SpatialVectors:sec}\n\nThe velocities, forces and inertias associated with 3D coordinate\nframes and rigid bodies are represented using the 6 DOF spatial\nquantities described in Sections \\ref{SpatialVelocitiesAndForces:sec}\nand \\ref{SpatialInertia:sec}. These are implemented by classes in the\npackage {\\tt maspack.spatialmotion}.\n\nSpatial velocities (or twists) are implemented by\n\\javaclass[maspack.spatialmotion]{Twist}, which exposes its\ntranslational and angular velocity components through the publicly\naccessible fields {\\tt v} and {\\tt w}, while spatial forces (or\nwrenches) are implemented by\n\\javaclass[maspack.spatialmotion]{Wrench}, which exposes its\ntranslational force and moment components through the publicly\naccessible fields {\\tt f} and {\\tt m}.\n\nBoth {\\tt Twist} and {\\tt Wrench} contain methods for algebraic\noperations such as addition and scaling. They also contain {\\tt\ntransform()} methods for applying rotational and rigid transforms.\nThe rotation methods simply transform each component by the supplied\nrotation matrix. The rigid transform methods, on the other hand,\nassume that the supplied argument represents a transform between two\nframes fixed within a rigid body, and transform the twist or wrench\naccordingly, using either (\\ref{XvelAB:eqn}) or (\\ref{XforceAB:eqn}).\n\nThe spatial inertia for a rigid body is implemented by\n\\javaclass[maspack.spatialmotion]{SpatialInertia}, which contains a\nnumber of methods for setting its value given various mass, center of\nmass, and inertia values, and querying the values of its components.\nIt also contains methods for scaling and adding, transforming between\ncoordinate systems, inversion, and multiplying by spatial vectors.\n\n\\section{Meshes}\n\\label{Meshes:sec}\n\nArtiSynth makes extensive use of 3D meshes, which are defined in {\\tt\nmaspack.geometry}. They are used for a variety of purposes, including\nvisualization, collision detection, and computing physical properties\n(such as inertia or stiffness variation within a finite element\nmodel).\n\nA mesh is essentially a collection of vertices\n(i.e., points) that are topologically connected in some way. All\nmeshes extend the abstract base class\n\\javaclass[maspack.geometry]{MeshBase}, which supports the vertex\ndefinitions, while subclasses provide the topology.\n\nThrough {\\tt MeshBase}, all meshes provide methods for\nadding and accessing vertices. Some of these include:\n%\n\\begin{lstlisting}[]\n int numVertices(); // return the number of vertices\n Vertex3d getVertex (int idx); // return the idx-th vertex\n void addVertex (Vertex3d vtx); // add vertex vtx to the mesh\n Vertex3d addVertex (Point3d p); // create and return a vertex at position p\n void removeVertex (Vertex3d vtx); // remove vertex vtx for the mesh\n ArrayList getVertices(); // return the list of vertices\n\\end{lstlisting}\n%\nVertices are implemented by \\javaclass[maspack.geometry]{Vertex3d},\nwhich defines the position of the vertex (returned by the method\n\\javamethod*[maspack.geometry.Vertex3d]{getPosition()}), and also\ncontains support for topological connections. In addition, each vertex\nmaintains an index, obtainable via\n\\javamethod*[maspack.geometry.Vertex3d]{getIndex()}, that equals the\nindex of its location within the mesh's vertex list. This makes it\neasy to set up parallel array structures for augmenting mesh vertex\nproperties.\n\nMesh subclasses currently include:\n\n\\begin{description}\n\n\\item[\\protect{\\javaclass[maspack.geometry]{PolygonalMesh}}]\\mbox{}\n\nImplements a 2D surface\nmesh containing faces implemented using half-edges.\n\n\\item[\\protect{\\javaclass[maspack.geometry]{PolylineMesh}}]\\mbox{}\n\nImplements a mesh\nconsisting of connected line-segments (polylines).\n\n\\item[\\protect{\\javaclass[maspack.geometry]{PointMesh}}]\\mbox{}\n\nImplements a point cloud with\nno topological connectivity.\n\n\\end{description}\n\n\\javaclass[maspack.geometry]{PolygonalMesh} is used quite extensively\nand provides a number of methods for implementing faces, including:\n%\n\\begin{lstlisting}[]\n int numFaces(); // return the number of faces\n Face getFace (int idx); // return the idx-th face\n Face addFace (int[] vidxs); // create and add a face using vertex indices\n void removeFace (Face f); // remove the face f\n ArrayList getFaces(); // return the list of faces\n\\end{lstlisting}\n%\nThe class \\javaclass[maspack.geometry]{Face} implements a face as a\ncounter-clockwise arrangement of vertices linked together by\nhalf-edges (class \\javaclass[maspack.geometry]{HalfEdge}).\n{\\tt Face} also supplies a face's (outward facing) normal\nvia \n\\javamethod[maspack.geometry.Face]{getNormal()}.\n\nSome mesh uses within ArtiSynth, such as collision detection, require a\n{\\it triangular} mesh; i.e., one where all faces have three vertices.\nThe method \\javamethod[maspack.geometry.PolygonalMesh]{isTriangular()}\ncan be used to check for this. Meshes that are not triangular can be\nmade triangular using \n\\javamethod[maspack.geometry.PolygonalMesh]{triangulate()}.\n\n\\subsection{Mesh creation}\n\nIt is possible to create a mesh by direct construction. For example,\nthe following code fragment creates a simple closed tetrahedral\nsurface:\n%\n\\begin{lstlisting}[]\n // a simple four-faced tetrahedral mesh \n PolygonalMesh mesh = new PolygonalMesh();\n mesh.addVertex (0, 0, 0);\n mesh.addVertex (1, 0, 0);\n mesh.addVertex (0, 1, 0);\n mesh.addVertex (0, 0, 1);\n mesh.addFace (new int[] { 0, 2, 1 });\n mesh.addFace (new int[] { 0, 3, 2 });\n mesh.addFace (new int[] { 0, 1, 3 });\n mesh.addFace (new int[] { 1, 2, 3 }); \n\\end{lstlisting}\n%\n\nHowever, meshes are more commonly created using either one of the\nfactory methods supplied by \\javaclass[maspack.geometry]{MeshFactory},\nor by reading a definition from a file (Section \\ref{MeshFileIO:sec}).\n\nSome of the more commonly used factory methods for creating polyhedral\nmeshes include:\n%\n\\begin{lstlisting}[]\n MeshFactory.createSphere (radius, nslices, nlevels);\n MeshFactory.createBox (widthx, widthy, widthz);\n MeshFactory.createCylinder (radius, height, nslices);\n MeshFactory.createPrism (double[] xycoords, height);\n MeshFactory.createTorus (rmajor, rminor, nmajor, nminor);\n\\end{lstlisting}\n%\nEach factory method creates a mesh in some standard coordinate\nframe. After creation, the mesh can be transformed using the\n\\javamethodAlt{maspack.geometry.MeshBase.transform(AffineTransform3dBase)}%\n{transform(X)} method, where {\\tt X} is either a rigid transform (\n\\javaclass[maspack.matrix]{RigidTransform3d}) or a more general affine\ntransform (\\javaclass[maspack.matrix]{AffineTransform3d}).\nFor example, to create a rotated box centered on $(5, 6, 7)$,\none could do:\n%\n\\begin{lstlisting}[]\n // create a box centered at the origin with widths 10, 20, 30:\n PolygonalMesh box = MeshFactor.createBox (10, 20, 20);\n\n // move the origin to 5, 6, 7 and rotate using roll-pitch-yaw\n // angles 0, 0, 45 degrees:\n box.transform (\n new RigidTransform3d (5, 6, 7, 0, 0, Math.toRadians(45)));\n\\end{lstlisting}\n%\nOne can also scale a mesh using\n\\javamethodAlt{maspack.geometry.MeshBase.scale(double)}{scale(s)},\nwhere {\\tt s} is a single scale factor, or\n\\javamethodAlt{maspack.geometry.MeshBase.scale(double,double,double)}%\n{scale(sx,sy,sz)}, where {\\tt sx}, {\\tt sy}, and {\\tt sz} are separate\nscale factors for the x, y and z axes. This provides a useful way to\ncreate an ellipsoid:\n%\n\\begin{lstlisting}[]\n // start with a unit sphere with 12 slices and 6 levels ...\n PolygonalMesh ellipsoid = MeshFactor.createSphere (1.0, 12, 6);\n\n // and then turn it into an ellipsoid by scaling about the axes:\n ellipsoid.scale (1.0, 2.0, 3.0);\n\\end{lstlisting}\n%\n\\javaclass[maspack.geometry]{MeshFactory} can also be used to create\nnew meshes by performing boolean operations on existing ones:\n%\n\\begin{lstlisting}[]\n MeshFactory.getIntersection (mesh1, mesh2);\n MeshFactory.getUnion (mesh1, mesh2);\n MeshFactory.getSubtraction (mesh1, mesh2);\n\\end{lstlisting}\n%\n\n\\subsection{Setting normals, colors, and textures}\n\nMeshes provide support for adding normal, color, and texture\ninformation, with the exact interpretation of these quantities\ndepending upon the particular mesh subclass. Most commonly this\ninformation is used simply for rendering, but in some cases normal\ninformation might also be used for physical simulation.\n\n\\begin{sideblock}\nFor polygonal meshes, the normal information described here is used\nonly for smooth shading. When flat shading is requested, normals are\ndetermined directly from the faces themselves.\n\\end{sideblock}\n\nNormal information can be set and queried using\nthe following methods:\n%\n\\begin{lstlisting}[]\n setNormals (\n List nrmls, int[] indices); // set all normals and indices\n\n ArrayList getNormals(); // get all normals\n int[] getNormalIndices(); // get all normal indices\n int numNormals(); // return the number of normals\n Vector3d getNormal (int idx); // get the normal at index idx\n\n setNormal (int idx, Vector3d nrml); // set the normal at index idx\n clearNormals(); // clear all normals and indices\n\\end{lstlisting}\n%\nThe method \\javamethod[maspack.geometry.MeshBase]{setNormals()} takes\ntwo arguments: a set of normal vectors ({\\tt nrmls}), along with a set\nof index values ({\\tt indices}) that map these normals onto the\nvertices of each of the mesh's geometric features. Often, there will\nbe one unique normal per vertex, in which case {\\tt nrmls} will have a\nsize equal to the number of vertices, but this is not always the case,\nas described below. Features for the different mesh subclasses are:\nfaces for {\\tt PolygonalMesh}, polylines for {\\tt PolylineMesh}, and\nvertices for {\\tt PointMesh}. If {\\tt indices} is specified as {\\tt\nnull}, then {\\tt normals} is assumed to have a size equal to the\nnumber of vertices, and an appropriate index set is created\nautomatically using\n\\javamethod[maspack.geometry.MeshBase]{createVertexIndices()}\n(described below). Otherwise, {\\tt indices} should have a size of\nequal to the number of features times the number of vertices per\nfeature. For example, consider a {\\tt PolygonalMesh} consisting of two\ntriangles formed from vertex indices (0, 1, 2) and (2, 1, 3),\nrespectively. If normals are specified and there is one unique normal\nper vertex, then the normal indices are likely to be\n%\n\\begin{verbatim}\n [ 0 1 2 2 1 3 ]\n\\end{verbatim}\n%\nAs mentioned above, sometimes there may be {\\it more} than one normal\nper vertex. This happens in cases when the same vertex uses different\nnormals for different faces. In such situations, the size of the {\\tt\nnrmls} argument will exceed the number of vertices.\n\nThe method {\\tt setNormals()} makes internal copies of the specified\nnormal and index information, and this information can be\nlater read back using \n\\javamethod[maspack.geometry.MeshBase]{getNormals()} \nand\n\\javamethod[maspack.geometry.MeshBase]{getNormalIndices()}.\nThe number of normals can be queried using\n\\javamethod[maspack.geometry.MeshBase]{numNormals()},\nand individual normals can be queried or set using\n\\javamethodAlt{maspack.geometry.MeshBase.getNormal(int)}{getNormal(idx)}\nand\n\\javamethodAlt{maspack.geometry.MeshBase.setNormal(,)}{setNormal(idx,nrml)}.\nAll normals and indices can be explicitly cleared using \n\\javamethod[maspack.geometry.MeshBase]{clearNormals()}.\n\nColor and texture information can be set using analagous methods.\nFor colors, we have\n%\n\\begin{lstlisting}[]\n setColors (\n List colors, int[] indices); // set all colors and indices\n\n ArrayList getColors(); // get all colors\n int[] getColorIndices(); // get all color indices\n int numColors(); // return the number of colors\n float[] getColor (int idx); // get the color at index idx\n\n setColor (int idx, float[] color); // set the color at index idx\n setColor (int idx, Color color); // set the color at index idx\n setColor (\n int idx, float r, float g, float b, float a); // set the color at index idx\n clearColors(); // clear all colors and indices\n\\end{lstlisting}\n%\nWhen specified as {\\tt float[]}, colors are given as RGB or\nRGBA values, in the range $[0,1]$, with array lengths of 3 and 4,\nrespectively. The colors returned by\n\\javamethod[maspack.geometry.MeshBase]{getColors()} are always RGBA\nvalues.\n\n%\nWith colors, there may often be {\\it fewer} colors than the number of\nvertices. For instance, we may have only two colors, indexed by 0 and\n1, and want to use these to alternately color the mesh faces. Using\nthe two-triangle example above, the color indices might then look like\nthis:\n%\n\\begin{verbatim}\n [ 0 0 0 1 1 1 ]\n\\end{verbatim}\n%\n\nFinally, for texture coordinates, we have\n%\n\\begin{lstlisting}[]\n setTextureCoords (\n List coords, int[] indices); // set all texture coords and indices\n\n ArrayList getTextureCoords(); // get all texture coords\n int[] getTextureIndices(); // get all texture indices\n int numTextureCoords(); // return the number of texture coords\n Vector3d getTextureCoords (int idx); // get texture coords at index idx\n\n setTextureCoords (int idx, Vector3d coords);// set texture coords at index idx\n clearTextureCoords(); // clear all texture coords and indices\n\\end{lstlisting}\n\nWhen specifying indices using \n\\javamethodAlt{maspack.geometry.MeshBase.setNormals(,)}{setNormals},\n\\javamethodAlt{maspack.geometry.MeshBase.setColors(,)}{setColors}, or\n\\javamethodAlt{maspack.geometry.MeshBase.setTextureCoords(,)}{setTextureCoords},\nit is common to use the same index set as that which\nassociates vertices with features. For convenience,\nthis index set can be created automatically using\n%\n\\begin{lstlisting}[]\n int[] createVertexIndices();\n\\end{lstlisting}\n%\nAlternatively, we may sometimes want to create a index set\nthat assigns the same attribute to each feature vertex. If\nthere is one attribute per feature, the resulting\nindex set is called a {\\it feature index} set, and\ncan be created using\n%\n\\begin{lstlisting}[]\n int[] createFeatureIndices();\n\\end{lstlisting}\n%\nIf we have a mesh with three triangles and one color per\ntriangle, the resulting feature index set would be\n%\n\\begin{verbatim}\n [ 0 0 0 1 1 1 2 2 2 ]\n\\end{verbatim}\n%\n\n\\begin{sideblock}\nNote: when a mesh is modified by the {\\it addition} of new features\n(such as faces for \\javaclass[maspack.geometry]{PolygonalMesh}), all\nnormal, color and texture information is cleared by default (with\nnormal information being automatically recomputed on demand if\nautomatic normal creation is enabled; see Section \\ref{AutoNormalCreation:sec}).\nWhen a mesh is modified by the {\\it removal} of\nfeatures, the index sets for normals, colors and textures are adjusted\nto account for the removal.\n\nFor colors, it is possible to request that a mesh explicitly maintain\ncolors for either its vertices or features (Section\n\\ref{vertexAndFeatureColoring:sec}). When this is done, colors will\npersist when vertices or features are added or removed, with default\ncolors being automatically created as necessary.\n\\end{sideblock}\n\nOnce normals, colors, or textures have been set, \none may want to know which of these attributes are\nassociated with the vertices of a specific feature. To know this,\nit is necessary to find that feature's offset into the \nattribute's index set. This offset information can\nbe found using the array returned by\n%\n\\begin{lstlisting}[]\n int[] getFeatureIndexOffsets()\n\\end{lstlisting}\n%\nFor example, the three normals associated with a triangle at index\n{\\tt ti} can be obtained using\n%\n\\begin{lstlisting}[]\n int[] indexOffs = mesh.getFeatureIndexOffsets();\n ArrayList nrmls = mesh.getNormals();\n // get the three normals associated with the triangle at index ti:\n Vector3d n0 = nrmls.get (indexOffs[ti]);\n Vector3d n1 = nrmls.get (indexOffs[ti]+1);\n Vector3d n2 = nrmls.get (indexOffs[ti]+2);\n\\end{lstlisting}\n%\nAlternatively, one may use the convenience methods\n%\n\\begin{lstlisting}[]\n Vector3d getFeatureNormal (int fidx, int k);\n float[] getFeatureColor (int fidx, int k);\n Vector3d getFeatureTextureCoords (int fidx, int k);\n\\end{lstlisting}\n%\nwhich return the attribute values for the $k$-th vertex of\nthe feature indexed by {\\tt fidx}.\n\nIn general, the various {\\tt get} methods return references to\ninternal storage information and so should\n{\\bf not} be modified. However, specific values within the lists\nreturned by \n\\javamethod[maspack.geometry.MeshBase]{getNormals()}, \n\\javamethod[maspack.geometry.MeshBase]{getColors()}, or\n\\javamethod[maspack.geometry.MeshBase]{getTextureCoords()}\nmay be modified by the application. This may be\nnecessary when attribute information changes as the simultion\nproceeds. Alternatively, one may use methods such\nas \n\\javamethodAlt{maspack.geometry.MeshBase.setNormal(,)}{setNormal(idx,nrml)}\n\\javamethodAlt{maspack.geometry.MeshBase.setColor(int,float)}%\n{setColor(idx,color)}, or\n\\javamethodAlt{maspack.geometry.MeshBase.setTextureCoords(int,)}%\n{setTextureCoords(idx,coords)}.\n\nAlso, in some situations, particularly with colors and textures, it\nmay be desirable to {\\it not} have color or texture information\ndefined for certain features. In such cases, the corresponding index\ninformation can be specified as -1, and the {\\tt getNormal()}, {\\tt\ngetColor()} and {\\tt getTexture()} methods will return {\\tt null} for\nthe features in question.\n\n\\subsection{Automatic creation of normals and hard edges}\n\\label{AutoNormalCreation:sec}\n\nFor some mesh subclasses, if normals are not explicitly set, they are\ncomputed automatically whenever {\\tt getNormals()} or {\\tt\ngetNormalIndices()} is called. Whether or not this is true\nfor a particular mesh can be queried by the method\n%\n\\begin{lstlisting}[]\n boolean hasAutoNormalCreation();\n\\end{lstlisting}\n%\nSetting normals explicitly, using a call to {\\tt\nsetNormals(nrmls,indices)}, will overwrite any existing normal information,\nautomatically computed or otherwise. The method\n%\n\\begin{lstlisting}[]\n boolean hasExplicitNormals();\n\\end{lstlisting}\n%\nwill return {\\tt true} if normals have been explicitly set, and {\\tt\nfalse} if they have been automatically computed or if there is\ncurrently no normal information. To explicitly remove normals from a\nmesh which has automatic normal generation, one may call {\\tt\nsetNormals()} with the {\\tt nrmls} argument set to {\\tt null}.\n\nMore detailed control over how normals are automatically created may\nbe available for specific mesh subclasses. For example, {\\tt\nPolygonalMesh} allows normals to be created with multiple normals per\nvertex, for vertices that are associated with either open or hard\nedges. This ability can be controlled using the methods\n%\n\\begin{lstlisting}[]\n boolean getMultipleAutoNormals();\n setMultipleAutoNormals (boolean enable);\n\\end{lstlisting}\n%\nHaving multiple normals means that even with smooth shading, open or\nhard edges will still appear sharp. To make an edge hard within\na {\\tt PolygonalMesh}, one may use the methods\n%\n\\begin{lstlisting}[]\n boolean setHardEdge (Vertex3d v0, Vertex3d v1);\n boolean setHardEdge (int vidx0, int vidx1);\n boolean hasHardEdge (Vertex3d v0, Vertex3d v1);\n boolean hasHardEdge (int vidx0, int vidx1);\n int numHardEdges();\n int clearHardEdges();\n\\end{lstlisting}\n%\nwhich control the hardness of edges between individual vertices,\nspecified either directly or using their indices.\n\n\\subsection{Vertex and feature coloring}\n\\label{vertexAndFeatureColoring:sec}\n\nThe method \\javamethod[maspack.geometry.MeshBase]{setColors()} makes\nit possible to assign any desired coloring scheme to a mesh. However,\nit does require that the user explicity reset the color information\nwhenever new features are added.\n\nFor convenience, an application can also request that a mesh\nexplicitly maintain colors for either its vertices or features. These\ncolors will then be maintained when vertices or features are added or\nremoved, with default colors being automatically created as necessary.\n\nVertex-based coloring can be requested with the method\n%\n\\begin{lstlisting}[]\n setVertexColoringEnabled();\n\\end{lstlisting}\n%\nThis will create a separate (default) color for each of the mesh's\nvertices, and set the color indices to be equal to the vertex indices,\nwhich is equivalent to the call\n%\n\\begin{lstlisting}[]\n setColors (colors, createVertexIndices());\n\\end{lstlisting}\n%\nwhere {\\tt colors} contains a default color for each\nvertex. However, once vertex coloring is enabled, the color and index\nsets will be updated whenever vertices or features are added or\nremoved. Meanwhile, applications can query or set the colors\nfor any vertex using {\\tt getColor(idx)}, or any of the\nvarious {\\tt setColor} methods.\nWhether or not vertex coloring\nis enabled can be queried using\n%\n\\begin{lstlisting}[]\n getVertexColoringEnabled();\n\\end{lstlisting}\n%\nOnce vertex coloring is established, the application will typically\nwant to set the colors for all vertices, perhaps using a code fragment\nlike this:\n%\n\\begin{lstlisting}[]\n mesh.setVertexColoringEnabled();\n for (int i=0; i>> from Bio import LogisticRegression\n>>> xs = [[-53, -200.78],\n [117, -267.14],\n [57, -163.47],\n [16, -190.30],\n [11, -220.94],\n [85, -193.94],\n [16, -182.71],\n [15, -180.41],\n [-26, -181.73],\n [58, -259.87],\n [126, -414.53],\n [191, -249.57],\n [113, -265.28],\n [145, -312.99],\n [154, -213.83],\n [147, -380.85],\n [93, -291.13]]\n>>> ys = [1,\n 1,\n 1,\n 1,\n 1,\n 1,\n 1,\n 1,\n 1,\n 1,\n 0,\n 0,\n 0,\n 0,\n 0,\n 0,\n 0]\n>>> model = LogisticRegression.train(xs, ys)\n\\end{verbatim}\n\nHere, \\verb+xs+ and \\verb+ys+ are the training data: \\verb+xs+ contains the predictor variables for each gene pair, and \\verb+ys+ specifies if the gene pair belongs to the same operon (\\verb+1+, class OP) or different operons (\\verb+0+, class NOP). The resulting logistic regression model is stored in \\verb+model+, which contains the weights $\\beta_0$, $\\beta_1$, and $\\beta_2$:\n\n\\begin{verbatim}\n>>> model.beta\n[8.9830290157144681, -0.035968960444850887, 0.02181395662983519]\n\\end{verbatim}\n\nNote that $\\beta_1$ is negative, as gene pairs with a shorter intergene distance have a higher probability of belonging to the same operon (class OP). On the other hand, $\\beta_2$ is positive, as gene pairs belonging to the same operon typically have a higher similarity score of their gene expression profiles.\nThe parameter $\\beta_0$ is positive due to the higher prevalence of operon gene pairs than non-operon gene pairs in the training data.\n\nThe function \\verb+train+ has two optional arguments: \\verb+update_fn+ and \\verb+typecode+. The \\verb+update_fn+ can be used to specify a callback function, taking as arguments the iteration number and the log-likelihood. With the callback function, we can for example track the progress of the model calculation (which uses a Newton-Raphson iteration to maximize the log-likelihood function of the logistic regression model):\n\n\\begin{verbatim}\n>>> def show_progress(iteration, loglikelihood):\n print(\"Iteration:\", iteration, \"Log-likelihood function:\", loglikelihood)\n>>>\n>>> model = LogisticRegression.train(xs, ys, update_fn=show_progress)\nIteration: 0 Log-likelihood function: -11.7835020695\nIteration: 1 Log-likelihood function: -7.15886767672\nIteration: 2 Log-likelihood function: -5.76877209868\nIteration: 3 Log-likelihood function: -5.11362294338\nIteration: 4 Log-likelihood function: -4.74870642433\nIteration: 5 Log-likelihood function: -4.50026077146\nIteration: 6 Log-likelihood function: -4.31127773737\nIteration: 7 Log-likelihood function: -4.16015043396\nIteration: 8 Log-likelihood function: -4.03561719785\nIteration: 9 Log-likelihood function: -3.93073282192\nIteration: 10 Log-likelihood function: -3.84087660929\nIteration: 11 Log-likelihood function: -3.76282560605\nIteration: 12 Log-likelihood function: -3.69425027154\nIteration: 13 Log-likelihood function: -3.6334178602\nIteration: 14 Log-likelihood function: -3.57900855837\nIteration: 15 Log-likelihood function: -3.52999671386\nIteration: 16 Log-likelihood function: -3.48557145163\nIteration: 17 Log-likelihood function: -3.44508206139\nIteration: 18 Log-likelihood function: -3.40799948447\nIteration: 19 Log-likelihood function: -3.3738885624\nIteration: 20 Log-likelihood function: -3.3423876581\nIteration: 21 Log-likelihood function: -3.31319343769\nIteration: 22 Log-likelihood function: -3.2860493346\nIteration: 23 Log-likelihood function: -3.2607366863\nIteration: 24 Log-likelihood function: -3.23706784091\nIteration: 25 Log-likelihood function: -3.21488073614\nIteration: 26 Log-likelihood function: -3.19403459259\nIteration: 27 Log-likelihood function: -3.17440646052\nIteration: 28 Log-likelihood function: -3.15588842703\nIteration: 29 Log-likelihood function: -3.13838533947\nIteration: 30 Log-likelihood function: -3.12181293595\nIteration: 31 Log-likelihood function: -3.10609629966\nIteration: 32 Log-likelihood function: -3.09116857282\nIteration: 33 Log-likelihood function: -3.07696988017\nIteration: 34 Log-likelihood function: -3.06344642288\nIteration: 35 Log-likelihood function: -3.05054971191\nIteration: 36 Log-likelihood function: -3.03823591619\nIteration: 37 Log-likelihood function: -3.02646530573\nIteration: 38 Log-likelihood function: -3.01520177394\nIteration: 39 Log-likelihood function: -3.00441242601\nIteration: 40 Log-likelihood function: -2.99406722296\nIteration: 41 Log-likelihood function: -2.98413867259\n\\end{verbatim}\n\nThe iteration stops once the increase in the log-likelihood function is less than 0.01. If no convergence is reached after 500 iterations, the \\verb+train+ function returns with an \\verb+AssertionError+.\n\nThe optional keyword \\verb+typecode+ can almost always be ignored. This keyword allows the user to choose the type of Numeric matrix to use. In particular, to avoid memory problems for very large problems, it may be necessary to use single-precision floats (Float8, Float16, etc.) rather than double, which is used by default.\n\n\\subsection{Using the logistic regression model for classification}\n\nClassification is performed by calling the \\verb+classify+ function. Given a logistic regression model and the values for $x_1$ and $x_2$ (e.g. for a gene pair of unknown operon structure), the \\verb+classify+ function returns \\verb+1+ or \\verb+0+, corresponding to class OP and class NOP, respectively. For example, let's consider the gene pairs {\\it yxcE}, {\\it yxcD} and {\\it yxiB}, {\\it yxiA}:\n\n\\begin{table}[h]\n\\begin{center}\n\\caption{Adjacent gene pairs of unknown operon status.}\n\\begin{tabular}{|c|c|c|}\n\\hline\nGene pair & Intergene distance $x_1$ & Gene expression score $x_2$ \\\\\n\\hline\n{\\it yxcE} --- {\\it yxcD} & 6 & -173.143442352 \\\\\n{\\it yxiB} --- {\\it yxiA} & 309 & -271.005880394 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\end{table}\n\nThe logistic regression model classifies {\\it yxcE}, {\\it yxcD} as belonging to the same operon (class OP), while {\\it yxiB}, {\\it yxiA} are predicted to belong to different operons:\n\\begin{verbatim}\n>>> print(\"yxcE, yxcD:\", LogisticRegression.classify(model, [6, -173.143442352]))\nyxcE, yxcD: 1\n>>> print(\"yxiB, yxiA:\", LogisticRegression.classify(model, [309, -271.005880394]))\nyxiB, yxiA: 0\n\\end{verbatim}\n(which, by the way, agrees with the biological literature).\n\nTo find out how confident we can be in these predictions, we can call the \\verb+calculate+ function to obtain the probabilities (equations (\\ref{eq:OP}) and \\ref{eq:NOP}) for class OP and NOP. For {\\it yxcE}, {\\it yxcD} we find\n\\begin{verbatim}\n>>> q, p = LogisticRegression.calculate(model, [6, -173.143442352])\n>>> print(\"class OP: probability =\", p, \"class NOP: probability =\", q)\nclass OP: probability = 0.993242163503 class NOP: probability = 0.00675783649744\n\\end{verbatim}\nand for {\\it yxiB}, {\\it yxiA}\n\\begin{verbatim}\n>>> q, p = LogisticRegression.calculate(model, [309, -271.005880394])\n>>> print(\"class OP: probability =\", p, \"class NOP: probability =\", q)\nclass OP: probability = 0.000321211251817 class NOP: probability = 0.999678788748\n\\end{verbatim}\n\nTo get some idea of the prediction accuracy of the logistic regression model, we can apply it to the training data:\n\\begin{verbatim}\n>>> for i in range(len(ys)):\n print(\"True:\", ys[i], \"Predicted:\", LogisticRegression.classify(model, xs[i]))\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\n\\end{verbatim}\nshowing that the prediction is correct for all but one of the gene pairs. A more reliable estimate of the prediction accuracy can be found from a leave-one-out analysis, in which the model is recalculated from the training data after removing the gene to be predicted:\n\\begin{verbatim}\n>>> for i in range(len(ys)):\n model = LogisticRegression.train(xs[:i]+xs[i+1:], ys[:i]+ys[i+1:])\n print(\"True:\", ys[i], \"Predicted:\", LogisticRegression.classify(model, xs[i]))\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 1\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\n\\end{verbatim}\nThe leave-one-out analysis shows that the prediction of the logistic regression model is incorrect for only two of the gene pairs, which corresponds to a prediction accuracy of 88\\%.\n\n\\subsection{Logistic Regression, Linear Discriminant Analysis, and Support Vector Machines}\n\nThe logistic regression model is similar to linear discriminant analysis. In linear discriminant analysis, the class probabilities also follow equations (\\ref{eq:OP}) and (\\ref{eq:NOP}). However, instead of estimating the coefficients $\\beta$ directly, we first fit a normal distribution to the predictor variables $x$. The coefficients $\\beta$ are then calculated from the means and covariances of the normal distribution. If the distribution of $x$ is indeed normal, then we expect linear discriminant analysis to perform better than the logistic regression model. The logistic regression model, on the other hand, is more robust to deviations from normality.\n\nAnother similar approach is a support vector machine with a linear kernel. Such an SVM also uses a linear combination of the predictors, but estimates the coefficients $\\beta$ from the predictor variables $x$ near the boundary region between the classes. If the logistic regression model (equations (\\ref{eq:OP}) and (\\ref{eq:NOP})) is a good description for $x$ away from the boundary region, we expect the logistic regression model to perform better than an SVM with a linear kernel, as it relies on more data. If not, an SVM with a linear kernel may perform better.\n\nTrevor Hastie, Robert Tibshirani, and Jerome Friedman: {\\it The Elements of Statistical Learning. Data Mining, Inference, and Prediction}. Springer Series in Statistics, 2001. Chapter 4.4.\n\n\\section{$k$-Nearest Neighbors}\n\n\\subsection{Background and purpose}\n\nThe $k$-nearest neighbors method is a supervised learning approach that does not need to fit a model to the data. Instead, data points are classified based on the categories of the $k$ nearest neighbors in the training data set.\n\nIn Biopython, the $k$-nearest neighbors method is available in \\verb+Bio.kNN+. To illustrate the use of the $k$-nearest neighbor method in Biopython, we will use the same operon data set as in section \\ref{sec:LogisticRegression}.\n\n\\subsection{Initializing a $k$-nearest neighbors model}\n\nUsing the data in Table \\ref{table:training}, we create and initialize a $k$-nearest neighbors model as follows:\n\n\\begin{verbatim}\n>>> from Bio import kNN\n>>> k = 3\n>>> model = kNN.train(xs, ys, k)\n\\end{verbatim}\n\nwhere \\verb+xs+ and \\verb+ys+ are the same as in Section \\ref{subsec:LogisticRegressionTraining}. Here, \\verb+k+ is the number of neighbors $k$ that will be considered for the classification. For classification into two classes, choosing an odd number for $k$ lets you avoid tied votes. The function name \\verb+train+ is a bit of a misnomer, since no model training is done: this function simply stores \\verb+xs+, \\verb+ys+, and \\verb+k+ in \\verb+model+.\n\n\\subsection{Using a $k$-nearest neighbors model for classification}\n\nTo classify new data using the $k$-nearest neighbors model, we use the \\verb+classify+ function. This function takes a data point $(x_1,x_2)$ and finds the $k$-nearest neighbors in the training data set \\verb+xs+. The data point $(x_1, x_2)$ is then classified based on which category (\\verb+ys+) occurs most among the $k$ neighbors.\n\nFor the example of the gene pairs {\\it yxcE}, {\\it yxcD} and {\\it yxiB}, {\\it yxiA}, we find:\n\\begin{verbatim}\n>>> x = [6, -173.143442352]\n>>> print(\"yxcE, yxcD:\", kNN.classify(model, x))\nyxcE, yxcD: 1\n>>> x = [309, -271.005880394]\n>>> print(\"yxiB, yxiA:\", kNN.classify(model, x))\nyxiB, yxiA: 0\n\\end{verbatim}\nIn agreement with the logistic regression model, {\\it yxcE}, {\\it yxcD} are classified as belonging to the same operon (class OP), while {\\it yxiB}, {\\it yxiA} are predicted to belong to different operons.\n\nThe \\verb+classify+ function lets us specify both a distance function and a weight function as optional arguments. The distance function affects which $k$ neighbors are chosen as the nearest neighbors, as these are defined as the neighbors with the smallest distance to the query point $(x, y)$. By default, the Euclidean distance is used. Instead, we could for example use the city-block (Manhattan) distance:\n\n\\begin{verbatim}\n>>> def cityblock(x1, x2):\n... assert len(x1)==2\n... assert len(x2)==2\n... distance = abs(x1[0]-x2[0]) + abs(x1[1]-x2[1])\n... return distance\n...\n>>> x = [6, -173.143442352]\n>>> print(\"yxcE, yxcD:\", kNN.classify(model, x, distance_fn = cityblock))\nyxcE, yxcD: 1\n\\end{verbatim}\n\nThe weight function can be used for weighted voting. For example, we may want to give closer neighbors a higher weight than neighbors that are further away:\n\n\\begin{verbatim}\n>>> def weight(x1, x2):\n... assert len(x1)==2\n... assert len(x2)==2\n... return exp(-abs(x1[0]-x2[0]) - abs(x1[1]-x2[1]))\n...\n>>> x = [6, -173.143442352]\n>>> print(\"yxcE, yxcD:\", kNN.classify(model, x, weight_fn = weight))\nyxcE, yxcD: 1\n\\end{verbatim}\nBy default, all neighbors are given an equal weight.\n\nTo find out how confident we can be in these predictions, we can call the \\verb+calculate+ function, which will calculate the total weight assigned to the classes OP and NOP. For the default weighting scheme, this reduces to the number of neighbors in each category. For {\\it yxcE}, {\\it yxcD}, we find\n\\begin{verbatim}\n>>> x = [6, -173.143442352]\n>>> weight = kNN.calculate(model, x)\n>>> print(\"class OP: weight =\", weight[0], \"class NOP: weight =\", weight[1])\nclass OP: weight = 0.0 class NOP: weight = 3.0\n\\end{verbatim}\nwhich means that all three neighbors of \\verb+x1+, \\verb+x2+ are in the NOP class. As another example, for {\\it yesK}, {\\it yesL} we find\n\n\\begin{verbatim}\n>>> x = [117, -267.14]\n>>> weight = kNN.calculate(model, x)\n>>> print(\"class OP: weight =\", weight[0], \"class NOP: weight =\", weight[1])\nclass OP: weight = 2.0 class NOP: weight = 1.0\n\\end{verbatim}\nwhich means that two neighbors are operon pairs and one neighbor is a non-operon pair.\n\nTo get some idea of the prediction accuracy of the $k$-nearest neighbors approach, we can apply it to the training data:\n\\begin{verbatim}\n>>> for i in range(len(ys)):\n print(\"True:\", ys[i], \"Predicted:\", kNN.classify(model, xs[i]))\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\n\\end{verbatim}\nshowing that the prediction is correct for all but two of the gene pairs. A more reliable estimate of the prediction accuracy can be found from a leave-one-out analysis, in which the model is recalculated from the training data after removing the gene to be predicted:\n\\begin{verbatim}\n>>> k = 3\n>>> for i in range(len(ys)):\n model = kNN.train(xs[:i]+xs[i+1:], ys[:i]+ys[i+1:], k)\n print(\"True:\", ys[i], \"Predicted:\", kNN.classify(model, xs[i]))\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 1\nTrue: 1 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 1\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 0\nTrue: 0 Predicted: 1\n\\end{verbatim}\nThe leave-one-out analysis shows that $k$-nearest neighbors model is correct for 13 out of 17 gene pairs, which corresponds to a prediction accuracy of 76\\%.\n\n\\section{Na\\\"ive Bayes}\n\nThis section will describe the \\verb|Bio.NaiveBayes| module.\n\n\\section{Maximum Entropy}\n\nThis section will describe the \\verb|Bio.MaximumEntropy| module.\n\n\\section{Markov Models}\n\nThis section will describe the \\verb|Bio.MarkovModel| and/or \\verb|Bio.HMM.MarkovModel| modules.\n\n", "meta": {"hexsha": "1415d5b3e8cd42e3b074dfdb5e05d7d71e299f56", "size": 22295, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Doc/Tutorial/chapter_learning.tex", "max_stars_repo_name": "amblina/biopython", "max_stars_repo_head_hexsha": "5045a7a3e86d5b32e0eaab941ab35daac86c59f8", "max_stars_repo_licenses": ["PostgreSQL"], "max_stars_count": 3, "max_stars_repo_stars_event_min_datetime": "2016-11-21T09:55:56.000Z", "max_stars_repo_stars_event_max_datetime": "2019-04-09T17:39:43.000Z", "max_issues_repo_path": "Doc/Tutorial/chapter_learning.tex", "max_issues_repo_name": "amblina/biopython", "max_issues_repo_head_hexsha": "5045a7a3e86d5b32e0eaab941ab35daac86c59f8", "max_issues_repo_licenses": ["PostgreSQL"], "max_issues_count": 32, "max_issues_repo_issues_event_min_datetime": "2016-11-21T07:38:21.000Z", "max_issues_repo_issues_event_max_datetime": "2017-08-16T13:00:03.000Z", "max_forks_repo_path": "Doc/Tutorial/chapter_learning.tex", "max_forks_repo_name": "amblina/biopython", "max_forks_repo_head_hexsha": "5045a7a3e86d5b32e0eaab941ab35daac86c59f8", "max_forks_repo_licenses": ["PostgreSQL"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2016-11-24T18:57:35.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-16T08:15:25.000Z", "avg_line_length": 53.3373205742, "max_line_length": 1031, "alphanum_fraction": 0.7272034088, "num_tokens": 6478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8289388040954683, "lm_q2_score": 0.7371581568543044, "lm_q1q2_score": 0.6110590009720267}} {"text": "\\documentclass[12pt]{article}\n \\usepackage{mathtools}\n \\usepackage{amsfonts}\n\n \\usepackage{venndiagram}\n \\usepackage{lastpage}\n \\usepackage[hidelinks]{hyperref}\n \\usepackage{color}\n \\definecolor{darkred}{RGB}{200, 0, 0}\n\n \\usepackage{fancyhdr}\n \\pagestyle{fancy}\n \\fancyhf{}\n \\fancyhead[LO]{TMA4140: Homework Set 2}\n \\fancyhead[RO]{Henry S. Sjøen}\n \\fancyfoot[CO]{\\thepage\\ of \\pageref{LastPage}}\n \n \\usepackage{fancyref}\n\n \\author{Henry S. Sjøen}\n \\title{%\n \\textbf{TMA4140 - Homework Set 2}\\\\\n Basic structures: Sets, Functions, Sequences and Sums\\\\\n \\color{darkred}{\\textbf{RETTES}}\n }\n \\begin{document}\n \\maketitle \n \\thispagestyle{empty}\n % \\pagebreak\n \\tableofcontents\n \n \\pagebreak\n \\section{Chapter 2.1 - Sets}\n \\subsection{Exercise 5}\n Determine whether each of these pairs are equal.\\\\\n a) $\\{1,3,3,3,5,5,5,5\\},\\{5,3,1\\} \\Rightarrow True$.\\\\\n b) $\\{\\{1\\}\\},\\{1,\\{1\\}\\} \\Rightarrow False.$\\\\\n c) $\\emptyset,\\{\\emptyset\\} \\Rightarrow False$\n \n \\subsection{Exercise 24}\n Determine wether each of these sets is the power set of a set, where \\emph{a} and \\emph{b} are distinct elements.\\\\\n a) $\\emptyset$, is False. $P(\\emptyset)=\\{\\emptyset\\{\\emptyset\\}\\}$\\\\ \n b) $\\{\\emptyset,\\{a\\}\\}$, is True. $P(\\{a\\})=\\{\\emptyset,\\{a\\}\\}$\\\\\n c) $\\{\\emptyset, \\{a\\},\\{\\emptyset,a\\}\\}$, is False.\\\\\n d) $\\{\\emptyset,\\{a\\},\\{b\\},\\{a,b\\}\\}$, is True.\n\n The powerset of any set $S$ is the set of all subsets of $S$, including the empty set $\\emptyset$ and $S$ itself. \\footnote{Wikipedia, 2018-09-14, 13:54, \\url{https://en.wikipedia.org/wiki/Power_set}}\n If $a$ and $b$ are distinct elements of a set $A$. $A=\\{a,b\\}$\n Then the powerset of $A$ is $p(A)=\\{\\emptyset,\\{a\\},\\{b\\},\\{a,b\\}\\}$.\n\n \\pagebreak\n \\section{Chapter 2.2 - Set Operations}\n \\subsection{Exercise 18c}\n Let A, B and C be sets. Show that:\n $(A \\cap B) \\subseteq (A \\cup B \\cup C)$\n \n % \\begin{equation}\n % \\begin{split}\n % (A \\cap B) \\subseteq (A \\cup B \\cup C)\n % \\end{split}\n % \\end{equation}\n \n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}\n \\fillA\n \\fillB\n \\fillC\n \\end{venndiagram3sets} \n \\caption{$(A \\cup B \\cup C)$}\n \\label{fig:Union}\n \\end{figure}\n \n As illustrated in \\fref{fig:Union} we can se the venndiagram for $(A \\cup B \\cup C)$. Coloring out $(A \\cap B)$, as shown in \\fref{fig:ACapB}, we can see that $(A \\cap B)$ is a subset of $(A \\cup B \\cup C)$.\n \n \\begin{figure}[h]\n \\centering\n \\caption{$A \\cap B$}\n \\begin{venndiagram3sets}\n % [labelOnlyAB=$A \\cap B$]\n \\fillACapB\n \\end{venndiagram3sets}\n \\label{fig:ACapB}\n \\end{figure}\n\n % \\url{http://www3.wolframalpha.com/Calculate/MSP/MSP2891521ibi0a5gd30f95g000037hcb6a7hd034233?MSPStoreType=image/gif&s=44&w=471.&h=279.&cdf=Resizeable}\n\n \\newpage\n \\subsection{Exercise 18d}\n Let A, B and C be sets. Show that:\n $(A - B) - C \\subseteq A - C$\n\n\n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}\n \\fillOnlyA\n \\end{venndiagram3sets}\n \\caption{$(A - B) - C$}\n \\label{fig:onlyA}\n \\end{figure}\n \n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}\n \\fillANotC\n \\end{venndiagram3sets}\n \\caption{$A - C$}\n \\label{fig:ANotC}\n \\end{figure} \n\n We can see that $(A - B)-C$ is a subset of $A-C$, see \\fref{fig:onlyA} and \\ref{fig:ANotC}.\n \n \\newpage\n \\subsection{Exercise 46}\n Show that if A, B, and C are sets, then:\\footnote{This is a special case of the inclusion-exclusion principle, which will be studied in Chapter 8. Also... Wikipedia \\url{https://en.wikipedia.org/wiki/Inclusion-exclusion_principle}}\n \n \\begin{equation}\n |A \\cup B \\cup C| = \n |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|\n \\end{equation}\n\n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}[\n labelOnlyA=1,labelOnlyB=1,labelOnlyC=1,\n labelOnlyAB=2,labelOnlyAC=2,labelOnlyBC=2,\n labelABC=3]\n \\fillA\n \\fillB\n \\fillC\n \\end{venndiagram3sets}\n \\caption{$|A|+|B|+|C|$}\n \\label{fig:u1}\n \\end{figure} \n\n Here we have counted some elements more than once (\\fref{fig:u1}), lets correct that.\n\n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}[\n labelOnlyA=1,labelOnlyB=1,labelOnlyC=1,\n labelOnlyAB=1,labelOnlyAC=1,labelOnlyBC=1,\n labelABC=0]\n \\fillA\n \\fillB\n \\fillC\n \\end{venndiagram3sets}\n \\caption{$|A|+|B|+|C|-(|A \\cap B| - |A \\cap C| - |B \\cap C|)$}\n \\label{fig:u2}\n \\end{figure} \n\n But now, the intersection of A,B and C is not counted. (\\fref{fig:u2}). Lets count the intersection once $|A\\cap B \\cap C|$ as shown in \\fref{fig:u3}.\n\n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}[\n labelOnlyA=1,labelOnlyB=1,labelOnlyC=1,\n labelOnlyAB=1,labelOnlyAC=1,labelOnlyBC=1,\n labelABC=1]\n \\fillA\n \\fillB\n \\fillC\n \\end{venndiagram3sets}\n \\caption{$|A|+|B|+|C|-(|A \\cap B| - |A \\cap C| - |B \\cap C|) + |A \\cap B \\cap C|$}\n \\label{fig:u3}\n \\end{figure} \n\n We have now shown that \n and we can intuitively see that it's the same as counting all elements inside the union of the three sets: $|A \\cup B \\cup C|$. (\\fref{fig:u4})\n\n \\begin{figure}[h]\n \\centering\n \\begin{venndiagram3sets}\n \\fillA\n \\fillB\n \\fillC\n \\end{venndiagram3sets}\n \\caption{Union of the sets A,B and C}\n \\label{fig:u4}\n \\end{figure}\n\n % TODO: Show it with more amazing math.\n\n \\pagebreak \n \\section{Chapter 2.3 - Functions}\n \\subsection{Exercise 12c}\n Determine whether each of these functions from $Z$ to $Z$ is one-to-one.\n\n \\textbf{True.} $f(n) = n^3$, is One-to-One, because it passes both the vertical and horizontal line test.\\footnote{\\url{http://www.mathwords.com/o/one_to_one_function.htm}}\n\n \\subsection{Exercise 38}\n % Todo: Look over later\n Let $f(x) = ax+b$ and $g(x)= cx+d$, where a,b,c, and d are constants. Determine necessary and sufficient conditions on the constants a,b,c, and d so that \n $f \\cdot g = g \\cdot f$.\n \n \\begin{equation}\n \\begin{split}\n f \\cdot g = g \\cdot f = a(cx+d)+b=acx+ad+b\\\\\n g\\cdot f = g(f(x))=c(ax+b)+d=acx+cb+d\\\\\n f \\cdot g = g\\cdot f \\Leftrightarrow ad+b = cb+d\n \\end{split}\n \\end{equation}\n\n \\subsection{Exercise 42}\n % Todo: Look over later\n Let $f$ be the function from $R$ to $R$ defined by $f(x)=x^2$. Find:\\\\\n a) $f^{-1}(\\{1\\}) = \\pm 1$\\\\\n b) $f^{-1}(\\{ x | 0 < x < 1\\}) = \\pm \\{x|-1 4\\}) = \\{x|-2>x \\wedge x>2\\}$\n \n \\pagebreak \n \\section{Chapter 2.4 - Sequences and Summations}\n\n \\subsection{Exercise 12c}\n % Todo: Look over later \n Show that the sequence {$a_n$} is a solution of the recurrence relation $a_n=-3a_{n-1}+4a_{n-2}$ if $a_n = (-4)^n$\n \\begin{equation}\n \\begin{split}\n a_n =& -3a_{n-1}+4a_{n-2} = (-4)^n\\\\\n =&-3a_{n-1} + 4a_{n-2}\\\\\n =& (-4)^{n-1}+4(-4)^{n-2}\\\\\n =& (-4)^{n-2}[(-3)(-4)+4]\\\\\n =& (-4)^{n-2}(16)\\\\\n =& (-4)^{n-2}(-4)^2\\\\\n =& (-4)^{n}\n \\end{split}\n \\end{equation}\n\n \\subsection{Exercise 33d}\n Compute the double sum\n \\begin{equation}\n \\begin{split}\n \\sum_{i=0}^{2}\\sum_{j=1}^{3}ij &= (0*1+0*2+0*3)+(1*1+1*2+1*3)+(2*1+2*2+2*3) \\\\\n &= (0+0+0) + (1+2+3) + (2+4+6)\\\\ \n &= 0 + 6 + 12\\\\ \n &= 18 \n \\end{split}\n \\end{equation}\n\n \\pagebreak \n \\section{Chapter 2.5 - Cardinality of Sets}\n \\subsection{TODO: Exercise 16}\n \\emph{Exercise}: \n Show that a subset of a countable set is also countable.\\\\\n \\emph{Answer}:\n A set is countable if it is finite or is the same size as $\\mathbb{N}$.\n To show that $A$ is countable, it is sufficient to show that there is an injection\n from $A$ to $\\mathbb{N}$.\n\n \\end{document}", "meta": {"hexsha": "d2519098e7a115d9a9cf787363f5065da75b0a87", "size": 8532, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "O2/o2.tex", "max_stars_repo_name": "SjoenH/TMA4140-2018", "max_stars_repo_head_hexsha": "9d2973b94f62f2d30328de3db5f496a42f900ebf", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "O2/o2.tex", "max_issues_repo_name": "SjoenH/TMA4140-2018", "max_issues_repo_head_hexsha": "9d2973b94f62f2d30328de3db5f496a42f900ebf", "max_issues_repo_licenses": ["MIT"], "max_issues_count": 2, "max_issues_repo_issues_event_min_datetime": "2018-09-16T20:12:23.000Z", "max_issues_repo_issues_event_max_datetime": "2018-09-24T09:17:53.000Z", "max_forks_repo_path": "O2/o2.tex", "max_forks_repo_name": "SjoenH/TMA4140-2018", "max_forks_repo_head_hexsha": "9d2973b94f62f2d30328de3db5f496a42f900ebf", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.9920318725, "max_line_length": 235, "alphanum_fraction": 0.5468823254, "num_tokens": 3016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7371581510799253, "lm_q2_score": 0.8289388040954684, "lm_q1q2_score": 0.6110589961854199}} {"text": "\n\\chapter{Preliminary Results} \\label{ch-1}\n\n\\section{Minimal Subspace Polynomial and Multipoint Evaluation}\n\nAn algorithm for multipoint evaluation was presented in \\cite{musschost}. Without loss of generality we may suppose $\\rho$ has skew-degree $d$ a perfect square and let $\\delta = \\sqrt{d}$. The target skew polynomial $\\rho$ can be decomposed as $\\rho = \\rho_0 + \\tau^{\\delta} \\rho_1 + \\ldots + \\tau^{\\delta(\\delta - 1)}\\rho_{\\delta-1}$ where each $\\rho_i$ has skew-degree at most $\\delta - 1$. Set $\\rho_i = \\rho_{i,0} + \\rho_{i,1}\\tau + \\ldots + \\rho_{i,\\delta - 1} \\tau^{\\delta-1}$. For each $0 \\leq i \\leq \\delta -1$ and $1 \\leq j \\leq s$ set $v_{i,j} = \\sigma^i(v_j)$, and each $\\rho_i(v_j) = \\sum_{k=0}^{\\delta - 1} \\rho_{i,k}v_{k,j}$. Horner's scheme can then be used to compute $\\rho(v_j) = \\rho_0(v_j) + \\tau^{\\delta}(\\rho_1(v_j) + \\tau^{\\delta}(\\rho_2(v_j) + \\ldots $.\n\n\\begin{algorithm}\n\\caption{Multipoint Evaluation \\cite{musschost}}\n\\label{alg:mpe2}\n\\hspace*{\\algorithmicindent} \\textbf{Input} A skew polynomial $\\rho \\in \\F_{q^n}\\{\\tau \\}$ of degree $d$, and a set of linearly independent evaluation points $\\{ v_1, \\ldots, v_s \\} \\subset \\F_{q^n}$.\\\\\n \\hspace*{\\algorithmicindent} \\textbf{Output} The set $\\{ \\rho(v_1), \\ldots, \\rho(v_s) \\} \\subset \\F_{q^n}$ \\\\\n \\begin{algorithmic}[1]\n \\State Compute $\\rho_i = \\rho_{i,0} + \\rho_{i,1}\\tau + \\ldots + \\rho_{i,\\delta - 1} \\tau^{\\delta}$ such that $\\rho = \\rho_0 + \\tau^{\\delta} \\rho_1 + \\ldots + \\tau^{\\delta(\\delta - 1)}\\rho_{\\delta-1}$\n \\State Compute $v_{i,j} = \\sigma^i(v_j)$ for each $0 \\leq i \\leq \\delta-1$ and $1 \\leq j \\leq s$\n \\State Compute the $\\delta \\times \\delta$ and $\\delta \\times s$ matrix product:\n $\n \\begin{bmatrix}\n \\rho_{0,0} & \\ldots & \\rho_{0, \\delta - 1} \\\\ \n \\vdots & & \\vdots \\\\\n \\rho_{\\delta - 1, 0} & \\ldots & \\rho_{\\delta - 1, \\delta - 1}\n \\end{bmatrix}\\begin{bmatrix}\n v_{0,0} & \\ldots & v_{0,s} \\\\ \\vdots & & \\vdots \\\\\n v_{\\delta - 1, 0} & \\ldots & v_{\\delta - 1, s}\n \\end{bmatrix} = \\begin{bmatrix}\n \\rho_0(v_1) & \\ldots & \\rho_{0}(v_s) \\\\ \n \\vdots & & \\vdots \\\\\n \\rho_{\\delta - 1}(v_1) & \\ldots & \\rho_{\\delta - 1}(v_s)\n \\end{bmatrix}\n $\n \n \\State For $1 \\leq j \\leq s$ compute $\\rho(v_j) = \\rho_0(v_j) + \\tau^{\\delta}(\\rho_1(v_j) + \\tau^{\\delta}(\\rho_2(v_j) + \\ldots $ via Horner's scheme\n \n\\end{algorithmic}\n\n\\end{algorithm}\n\nSteps 1, 2, and 3 take time equal to the cost of $O(d + d^{1/2}s )$ applications of Frobenius powers. The cost of step 3 is either $O(ds^{\\omega - 2})$ field operations in $\\L$ if $s < \\sqrt{d}$, and $O(d^{(\\omega-1)/2}s)$ $\\L$-operations otherwise. Counting the complexity of modular compositions, the overall bit complexity of the algorithm is either $(ds^{\\omega - 2} n^{\\theta} \\log q)^{1 + o(1)}$ if $s < \\sqrt{d}$, and $(d^{(\\omega-1)/2}sn^{\\theta} \\log q)^{1 + o(1)}$ otherwise.\n\nAlgorithm \\ref{alg:mpe2} may then be leveraged as a subroutine for algorithm \\ref{msp1}; the algebraic runtime of $\\MSP(d)$ satisfies\n\n\\begin{equation*}\n \\MSPr(d) = 2\\MSPr(d/2) + O(d^{(\\omega + 1)/2}).\n\\end{equation*}\n\nApplying the master theorem, $\\MSPr(d) = \\Theta(d^{(\\omega + 1)/2})$. In terms of bit complexity, we have $ \\MSPr(d) = (d^{(\\omega-1)/2}sn^{\\theta} \\log q)^{1 + o(1)}$.\n\n\\section{Narayanan's Algorithm for Rank-two Drinfeld Modules}\n\nWe propose an amendment to an additional algorithm for finding the characteristic polynomial of a rank-two Drinfeld module based on an approach originally due to Narayanan in \\cite{Narayanan18} and later modified in \\cite{musschost}. Recall that in the rank-two case, the characteristic polynomial of $\\phi$ takes the form $X^2 - aX - b$. Moreover, since $\\tau^{2n} - \\phi_a\\tau^n - \\phi_b = \\id - a(\\phi_x) - b(\\phi_x) = 0$, the polynomial $1 - a - b$ is an annihilator of $\\phi_x$; in fact it is the characteristic polynomial of $\\phi_x = \\Delta_2 \\tau^2 + \\Delta_1 \\tau + \\gamma(x)$ \\cite{GEKELE1991187}. If the minimal polynomial $\\Gamma$ of $\\phi_x$ coincides with the characteristic polynomial,then we can once again resort to a Wiedemann style argument: in this case, it suffices to compute the minimal polynomial $\\Gamma_{\\ell, \\alpha}$ of a sequence $\\{r_i\\} = \\{\\ell(\\phi_x^i(\\alpha)) \\}$ for a uniformly random $\\alpha \\in \\L$ and $\\F_q$-linear map $\\ell: \\L \\to \\F_q$ and argue that with probability greater than $\\frac{1}{2}$ we have $\\Gamma_{\\ell, \\alpha} = \\Gamma$. In his original proposal, Narayanan used the automorphism projection algorithm of Kedlaya and Umans to compute $2n$ terms of the sequence $\\{ r_i \\}$, whereas it is unclear to us whether this approach and its complexity bound extends to the case where $\\phi_x$ is not an automorphism of $\\L$. \n\nInstead, we propose the following alternative approach based on a ``block Wiedemann\" style algorithm \\cite{Coppersmith94}. For some positive integer $\\mu$, pick $\\mu$ many $\\F_q$-linear maps $\\ell_1, \\ldots, \\ell_{\\mu}$ as before and $\\mu$ elements $\\alpha_i \\in \\L$. Construct a sequence of $\\mu \\times \\mu$ matrices $R_{k}$ such that the $(i,j)^{th}$ entry of $R_k$ is $ \\ell_i(\\phi_x^k(\\alpha_j))$. The generating series of this sequence can be expressed as a product of the form $Q^{-1}N$ for $\\mu \\times \\mu$ matrices matrices $Q,N$ taking coefficients in $\\F_q[x]$ and with the entries of $Q$ having degree at most $\\lceil n/\\mu \\rceil$. Computing $Q$ can be accomplished by first determining $2n/\\mu$ matrices $R_k$ and using the PM basis algorithm given in \\cite{GiJeVi03} in $(\\mu^{\\omega - 1}n\\log q)^{1 + o(1)}$ bit operations. Fix additional constants $\\frak{b}$and set $K = \\lfloor (n/2\\mu)^{\\frak{b}} \\rfloor$, $K' = \\lceil n/2\\mu K \\rceil$. To compute the $R_k$, we turn to the following procedure:\n\n\\begin{enumerate}\n \\item Compute the $\\F_q$-linear maps $\\ell_{i, \\frak{s}} = \\phi_x^{\\frak{s}}(\\ell_i)$ for $i \\leq \\mu$ and $\\frak{s} < K$.\n \\item Compute $\\phi_x^{K}$ and for $j \\leq \\mu$ and $\\frak{t} \\leq K'$ compute $\\alpha_{\\frak{t}, j} = \\phi_x^{\\frak{t}K}(\\alpha_j)$.\n \\item Let $L$ be the $K\\mu \\times n$ matrix whose rows correspond to the linear map $\\ell_{i, \\frak{s}}$, and $A$ the $n \\times K'\\mu$ matrix whose columns correspond to $\\alpha_{\\frak{t}, j}$. The entries of the matrix product $LA$ are exactly $\\ell_i(\\phi_x^{\\frak{s} + \\frak{t}K}(\\alpha_j))$ for all $i,j \\leq \\mu$, $\\frak{s} < K$, $\\frak{t} < K'$. \n\\end{enumerate}\n\nFor well chosen $\\mu, \\frak{b}$, the procedure outlined above computes all entries of $R_k$ for $k \\leq 2n/\\mu$ in at most $(n^{1.885} \\log q)^{1 + o(1)}$. \n\n%Fix constants, $\\mu$, $\\frak{w}, \\frak{v}$.\n\n\n\n\\section{Minimal Polynomials in Higher Ranks}\n\n\n\\subsection{Schoof-like Algorithms}\n\nOur main results for this section is the following theorem.\n\n\\begin{theorem}\\label{schooflike}\nThere exists a Monte Carlo algorithm such that when $\\gcd(n,r) = 1$ or $r = 3$, then for a generic choice of rank $r$ Drinfeld module $\\phi$ over a degree $n$ extension $\\L$ of $\\F_q$, the algorithm computes the minimal polynomial of $\\phi$ with bit complexity $r^{(\\omega+3)/2}(n^2 \\log q)^{1 + o(1)}$ and returns a correct result when $q > Cnr^2$, for some constant $C > 2$, with probability greater than $\\frac{1}{2}$.\n\\end{theorem}\n\nRecall the \\textit{minimal polynomial} of $\\phi$ is the unique monic polynomial $\\Gamma = T^{d} - a_{d-1}T^{d-1} - \\ldots - a_1T - a_0 \\in \\F_q[x][T]$ of least degree such that:\n\n\\begin{equation} \\label{charpoly1}\n \\tau^{nd} - \\phi_{a_{d-1}}\\tau^{n(d-1)} - \\ldots - \\phi_{a_1} \\tau^n - \\phi_{a_0} = 0\n\\end{equation}\n\n\nThe goal is to compute each $a_i$ modulo a set of polynomials which can then be used to reconstruct the true coefficients. This can be done over the right-quotient space $\\L\\{\\tau \\}/\\phi_{x-e}$ since $\\phi_{a \\bmod b} \\bmod \\phi_b = \\phi_a \\bmod \\phi_b$ for any $a, b \\in \\F_q[x]$. If $b = x - e$ for $e \\in \\F_q$, then in particular $a(c) = \\phi_a \\bmod \\phi_b$. As in the rank 2 case, when $q \\gg nr$, linear polynomials suffice and equation (\\ref{charpoly1}) becomes\n\n\\begin{equation}\\label{linsys}\n\\tau^{n\\hatr} - a_{\\hatr-1}(e)\\tau^{n(\\hatr-1)} - \\ldots - a_1(e)\\tau^n - a_0(e) = 0 \\mod \\phi_{x - e}.\n\\end{equation}\n\n$\\L\\{\\tau \\}/\\phi_{x-e}$ is an $r$-dimensional vector space over $\\L$ spanned by the computational basis $\\{1, \\tau, \\ldots, \\tau^{r-1}\\}$, and let $\\hat{\\varphi}$ denote the column vector of $\\varphi \\in \\L\\{\\tau \\}/\\phi_{x-e}$ in the computational basis. Writing (\\ref{linsys}) as a linear system\n\n\\begin{equation}\\label{system1} \\begin{bmatrix} \\hat{\\tau}^{n(\\hatr-1)} & \\ldots & \\hat{\\tau}^{n} \\end{bmatrix} \\begin{bmatrix} a_{\\hatr-1}(e) \\\\ \\vdots \\\\ a_1(e) \\end{bmatrix} = \\begin{bmatrix} -a_0(e) \\\\ 0 \\\\ \\vdots \\\\ 0 \\end{bmatrix} + \\hat{\\tau}^{n\\hatr}.\n\\end{equation}\n\nLet $ B(\\psi) = \\begin{bmatrix} \\hat{\\tau}^{n(\\hatr-1)} & \\ldots & \\hat{\\tau}^{n} \\end{bmatrix}$ for modulus $\\psi \\in \\L\\{\\tau\\}$. The well-determinedness of the system encoded by $B(\\phi_{x-e})$ can be characterized using the following lemma.\n\n%Let $B_{\\phi}^{n,j} := \\begin{bmatrix} \\tau^{n(j-1)} & \\ldots & \\tau^{n} \\end{bmatrix} \\bmod \\phi_{x-e}$ be the $r \\times (j -1)$ system encoding the coefficients of the $r$-dimensional vector $\\tau^{ni} \\mod \\phi_{x -e}$ in the computational basis. The goal is to show the system given in $(\\ref{system1})$ has a unique solution with high probability when $q \\gg nr$ by applying the Schwartz-Zippel lemma to the polynomial in the coefficients of the Drinfeld module given by $\\det B_{\\phi}^{n,\\hatr}$, where $B_{\\phi}^{n,\\hatr}$ is truncated to the first $\\hatr - 1$ rows, as $\\det B_{\\phi}^{n,\\hatr} \\neq 0$ provides a sufficient condition for solvability.\n\n%Let $ \\Delta_i = \\sum_{j=0}^{n-1}\\alpha_{i,j}x^j$ for $i > 0$ and $ (\\gamma(x) - e) = \\sum_{j=0}^{n-1}\\alpha_{0,j}x^j$. The goal of the next lemma is to bound the degree \n\n\\begin{lemma}\\label{detdeg}\nSuppose $Cn^2r < q$ for a constant $C>2$. Further suppose that there exists at least one skew-polynomial modulus $\\psi$ of skew-degree $r$ such that the system in (\\ref{system1}) is uniquely solvable. Then for choices of Drinfeld coefficients $\\Delta_i$ and evaluation point $e$ except on an asymptotically vanishing set as $q \\to \\infty$, the system given in $(\\ref{system1})$ is well-determined.\n\\end{lemma}\n\n\\begin{proof}\nWe consider general skew-polynomial modulus $\\psi = \\sum_{i=0}^{r} \\Lambda_i \\tau^i$ for equation $(\\ref{system1})$ rather than the special Drinfeld case of $\\Lambda_0 = x - e$. We assume the case $n,d \\geq 3$ we can set $\\mu = n(\\hatr - 1)$, $\\reduc = \\mu - r$, $\\omega = \\mu - n > 0$. We can characterize the linear independence of the set $\\{\\tau^{ni} \\}_{i =1}^{\\hatr} \\bmod \\psi$ as follows: suppose to the contrary that\n\n\\begin{equation}\\label{mult}\n \\sum_{i=1}^{\\hatr - 1} \\beta_i \\tau^{ni} = P \\psi\n\\end{equation}\n\nwith $P = \\sum_{i= 1}^{\\reduc} p_ix^i \\in \\L\\{\\tau \\}$ and $P \\neq 0 $. The choice of $P$ is constrained by the requirement that the coefficient of $\\tau^{j}$ in the product is 0 if $n$ doesn't divide $j$. The linear system representing the product $P\\psi$ can be encoded as an $ (\\omega + 1) \\times (\\omega - r + 1)$ matrix $M$ over $\\L$\n\n\\begin{equation}\n \\label{productsystem}\n \\begin{bmatrix}\n \\Lambda_{r}^{q^{\\reduc}} & 0 & 0 & \\ldots & 0 & \\ldots & 0 \\\\\n \\Lambda_{r - 1}^{q^{\\reduc}} & \\Lambda_{r}^{q^{\\reduc - 1}} & 0 & \\ldots & 0 & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \\\\\n \\Lambda_{0}^{q^{\\reduc}} & \\Lambda_{1}^{q^{\\reduc - 1}} & \\Lambda_{2}^{q^{\\reduc - 2}} & \\ldots & \\Lambda_{r}^{q^{\\reduc - r}} & \\ldots & 0 \\\\\n 0 & \\Lambda_{0}^{q^{\\reduc - 1}} & \\Lambda_{1}^{q^{\\reduc - 2}} & \\ldots & \\Lambda_{r-1}^{q^{\\lambda - r}} & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \\\\\n 0 & 0 & 0 & \\ldots & 0 & \\ldots & \\Lambda_0^{q^{n}}\n \n \n \\end{bmatrix}\n \\begin{bmatrix}\n p_{\\reduc} \\\\ p_{\\reduc - 1} \\\\ \\vdots \\\\ p_{n}\n \\end{bmatrix}\n = \n \\begin{bmatrix}\n \\beta_{\\hatr-1} \\\\ 0 \\\\ \\vdots \\\\ 0 \\\\ \\beta_{1}\n \\end{bmatrix}\n\\end{equation}\n\nDeleting $\\hatr - 1$ rows corresponding to the free parameters $\\beta_i$ leaves the above as a well determined $ (\\omega - \\hatr + 2) \\times (\\omega - r + 1)$ linear system. \n\n\\begin{equation}\n \\label{reduced}\n \\begin{bmatrix}\n \\Lambda_{r - 1}^{q^{\\reduc}} & \\Lambda_{r}^{q^{\\reduc - 1}} & 0 & \\ldots & 0 & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \\\\\n \\Lambda_{0}^{q^{\\reduc}} & \\Lambda_{1}^{q^{\\reduc - 1}} & \\Lambda_{2}^{q^{\\reduc - 2}} & \\ldots & \\Lambda_{r}^{q^{\\reduc - r}} & \\ldots & 0 \\\\\n 0 & \\Lambda_{0}^{q^{\\reduc - 1}} & \\Lambda_{1}^{q^{\\reduc - 2}} & \\ldots & \\Lambda_{r-1}^{q^{\\lambda - r}} & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \\\\\n \n \n \\end{bmatrix}\n \\begin{bmatrix}\n p_{\\reduc} \\\\ p_{\\reduc - 1} \\\\ \\vdots \\\\ p_{n}\n \\end{bmatrix}\n = \n \\begin{bmatrix}\n 0 \\\\ 0 \\\\ \\vdots \\\\ 0 \\\\ 0\n \\end{bmatrix}\n\\end{equation}\n\nDeleting an additional $r - \\hatr + 1$ rows, which can be chosen such that the resulting system derived from the modulus $\\psi$ is invertible, and showing that for generic choices of Drinfeld module parameters the determinant of this resulting square matrix is non-zero is sufficient to establish $P = 0$. To that end, we ``vectorize\" equation $(\\ref{productsystem})$: fix an $\\F_q$-basis of $\\L$, under which elements $a \\in \\L$ can be represented by a length $n$ vector $\\hat{a}$ with entries in $\\F_q$ or by their $n\\times n$ multiplication operator $\\overline{a} \\in \\F_q^{n \\times n}$. Moreover, let $\\overline{\\sigma}$ denote the $n\\times n$ matrix of the Frobenius acting on vectors $\\hat{a}$. This allows a rewriting of the truncated system $\\overline{M}$\n\n\\begin{equation}\n \\label{vectorsystem}\n \\begin{bmatrix}\n \\overline{\\sigma}^{\\lambda}\\overline{\\Lambda}_{r - 1} & \\overline{\\sigma}^{\\lambda-1}\\overline{\\Lambda}_{r} & 0 & \\ldots & 0 & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \\\\\n \\overline{\\sigma}^{\\lambda}\\overline{\\Lambda}_{0} & \\overline{\\sigma}^{\\lambda -1}\\overline{\\Lambda}_{1} & \\overline{\\sigma}^{\\lambda - 2}\\overline{\\Lambda}_{2} & \\ldots & \\overline{\\sigma}^{\\lambda - r}\\overline{\\Lambda}_{r} & \\ldots & 0 \\\\\n 0 & \\overline{\\sigma}^{\\lambda - 1}\\overline{\\Lambda}_{0} & \\overline{\\sigma}^{\\lambda - 2}\\overline{\\Lambda}_{1} & \\ldots & \\overline{\\sigma}^{\\lambda - r}\\overline{\\Lambda}_{r-1} & \\ldots & 0 \\\\\n \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\ddots & \\vdots \n \n \n \\end{bmatrix}\n \\begin{bmatrix}\n \\hat{p}_{\\reduc} \\\\ \\hat{p}_{\\reduc - 1} \\\\ \\vdots \\\\ \\hat{p}_{n}\n \\end{bmatrix}\n = \n \\begin{bmatrix}\n \\hat{0} \\\\ \\vdots \\\\ \\hat{0} \n \\end{bmatrix}\n\\end{equation}\n\nThis system is parametrized by the $\\F_q$-coefficients of $\\Lambda_i = \\sum_{j = 0}^{n-1} \\alpha_{i,j}x^j$. The resulting vectorized system is an $ n(\\omega - r + 1) \\times n(\\omega - r + 1)$ matrix whose entries are degree 1 polynomials in the variables $\\alpha_{i,j}$; therefore when the determinant is not exactly the zero polynomial in the $\\alpha_{i,j}$, then it has degree $n(\\omega - r + 1) = O(n^2r)$. This can't be the case since the system must be invertible for at least one choice of $\\alpha_{i,j}$ corresponding to the choice of skew-polynomial $\\psi$ for which a solution $z$ to $B(\\psi)z = 0$ is uniquely determined. Applying the Schwartz-Zippel lemma to the determinant:\n\n%Set $\\alpha_i = \\frac{\\Delta_i}{\\Delta_r}$ with coefficients $\\alpha_i := \\sum_{j=0}^{n-1}\\alpha_{i,j}x^j$. Furthermore, let $\\overline{\\alpha_i}$ be the $n\\times n$ matrix of the action of left multiplication by $\\alpha_i$ on $\\L/\\F_q$, and let $\\overline{\\tau}$ denote the matrix representation of the Frobenius. If $\\sum_{i=0}^{r-1}c_i \\tau^i$ is any skew polynomial, the action of left multiplication by $\\tau$ on $\\L[\\tau]/\\phi_{x-e}$ is given by the following matrices:\n\n%\\begin{equation*}\n%M := \\begin{bmatrix}\\overline{\\alpha}_{r-1} & I & 0 & \\ldots & 0 \\\\ \\overline{\\alpha}_{r-2} & 0 & I & \\ldots & 0 \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ \\overline{\\alpha}_{0} & 0 & 0 & \\ldots & I \\end{bmatrix} \\hspace{4mm} S := \\begin{bmatrix} \\overline{\\tau} & & \\\\ & \\ddots & \\\\ & & \\overline{\\tau} \\end{bmatrix}\n%\\end{equation*}\n\n%\\begin{equation*}\n%\\begin{bmatrix}\\overline{\\alpha}_{r-1} & I & 0 & \\ldots & 0 \\\\ \\overline{\\alpha}_{r-2} & 0 & I & \\ldots & 0 \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ \\overline{\\alpha}_{0} & 0 & 0 & \\ldots & I \\end{bmatrix} \\begin{bmatrix} \\overline{\\tau} & & \\\\ & \\ddots & \\\\ & & \\overline{\\tau}\\end{bmatrix} \\begin{bmatrix} \\overline{c}_{r-1} \\\\ \\vdots \\\\ \\overline{c}_{0} \\end{bmatrix} = \\begin{bmatrix} \\overline{\\alpha}_{r-1} \\overline{\\tau}(\\overline{c}_{r-1}) + \\overline{\\tau}(\\overline{c}_{r-2}) \\\\ \\overline{\\alpha}_{r-2} \\overline{\\tau} \\overline{c}_{r-1} + \\overline{\\tau} \\overline{c}_{r-3} \\\\ \\vdots \\\\ \\overline{\\alpha}_{0} \\overline{\\tau}(\\overline{c}_{r-1}) \\end{bmatrix}\n%\\end{equation*}\n\n%with each $\\overline{c}_i \\in \\L$ being written as a size $n$ vector over $\\mathbb{F}_q$. Then $\\tau^{in} \\mod \\phi_{x-e}$ is given by \n\n%\\begin{equation*}\n%(MS)^{in-r} \\begin{bmatrix} \\overline{1} \\\\ 0 \\\\ \\vdots \\\\ 0 \\end{bmatrix}.\n%\\end{equation*}\n\n%The entries of this expression have degree $in - r$ in the coefficients $\\alpha_{i,j}$, and the total degree of any term in the square determinant is $2\\sum_{i=1}^{r-1} (in - r) \\leq 2nr^2$.\n\n\\begin{equation*}\n \\Pr\\Big[\\det \\overline{M} = 0\\Big] \\leq \\frac{n^2r}{q} < \\frac{1}{C}.\n\\end{equation*}\n\n\\end{proof}\n\n%If $\\sqdet B_{\\psi}^{n,r} \\neq 0 $, then for generic choices of base field element $e$ and coefficients of $\\phi_x$, the Schwartz-Zippel lemma implies\n\n% $\\{\\alpha_{i,j}\\}_{1 \\leq i \\leq r-1}^{0 \\leq j \\leq n-1}$, the Schwartz-Zippel lemma implies\n\n\n\nThe remainder of the proof of Theorem 1 is classifying pairs of parameters $n,r$ for which there exists at least one choice of modulus $\\psi \\in \\sring$ that guarantees unique solvability of $B(\\psi)$\n\n\\begin{prop}\\label{coprime}\nSuppose $\\gcd(n,r) = 1$. There exists a Drinfeld module $\\phi_x$ such that $ B(\\phi_x)$ is uniquely solvable. %e $\\psi = \\tau^n - 1$.\n\\end{prop}\n\\begin{proof}\nLet $\\phi_x = \\tau^r - \\gamma(x)$. For any $e \\neq -\\gamma(x)$, $\\tau^{ni} = (\\gamma(x) + e)^{\\lfloor ni/r \\rfloor} \\tau^{ni \\bmod r} \\bmod \\phi_{x} - e$. Since $\\gcd(n,r) = 1$, the set $\\{ n \\bmod r, 2n \\bmod r, \\ldots, rn \\bmod r\\}$ is a reordering of $\\{0, 1, \\ldots, r-1\\}$ and so the elements $\\tau^{ni}$ form an $\\L$ basis of $\\L\\{ \\tau\\}/\\phi_{x - e}$.\n\\end{proof}\n\n\n\n\\begin{prop}\\label{rank3}\nIf $r = 3$, then for any $n$ there exists a skew-polynomial $\\psi \\in \\L\\{\\tau\\}$ such that $B(\\psi)$ is uniquely solvable.\n\\end{prop}\n\\begin{proof}\nFor a contradiction, let $\\psi$ be any skew polynomial. The notion of an \\textit{Azumaya algebra} and their connection to skew polynomials was previously studied in \\cite{Ikehata1984AzumayaAA} and \\cite{skewfactor}. In particular, we are interested in a distinguished map sending an Azumaya algebra to its center, the \\textit{reduced norm}, which induces a map $\\cN: \\L\\{\\tau\\} \\to \\comm$ while satisfying the following constraints.\n\n\\begin{enumerate}\n \\item $\\cN(\\psi_1 \\psi_2) = \\cN(\\psi_1) \\cN(\\psi_2)$\n \\item $\\deg_{\\tau^n} \\cN(\\psi) \\leq n \\deg_{\\tau} \\psi$\n \\item $\\cN(\\psi) = \\psi^n$ if $\\psi \\in \\L[\\tau^n]$\n \\item $\\psi$ is irreducible in $\\sring$ if and only if $\\cN(\\psi)$ is irreducible in $\\F_q[\\tau^n]$\n\\end{enumerate}\n\nFor details on the construction of the map and the proof of item (4) we refer to \\cite[Prop. 2.1.17]{skewfactor}. Suppose $\\tau^{2n} - \\lambda \\tau^n = 0$ for some $\\lambda \\in \\F_q$. Then $\\tau^{n} - \\lambda = 0 \\mod \\psi$, and we must have a skew polynomial $D$ such that $\\tau^n - \\lambda = D \\psi$. Taking reduced norms of both sides leads to\n\n\\begin{equation*}\n(\\tau^n - \\lambda)^n = \\cN(D)\\cN(\\psi)\n\\end{equation*}\n\nThe number of irreducible skew-polynomials of any degree was partly characterized in \\cite{odoni_1999}, and in particular they exist for any skew degree. If we choose $\\psi$ to be any degree 3 skew polynomial irreducible in $\\L\\{\\tau \\}$, item (4) implies $\\cN(\\psi)$ is an irreducible central skew-polynomial dividing $(\\tau^n - \\lambda)^n$ and so $\\deg_{\\tau}\\cN(\\psi) = n$. But we must then have $n^2 = \\deg_{\\tau} \\cN(D) + \\deg_{\\tau} \\cN(\\psi) \\leq (n-3)n + n = n^2 - 2n < n^2$, giving a contradiction.\n\n\\end{proof}\n\n\\subsection{Limitations}\n\nOne of the clearest limitations of this approach is the large base field requirement $q > Cn^2 r$. Unlike the case $r = 2$, working with arbitrary degree irreducibles in $\\F_q[x]$ in place of degree 1 polynomials does not appear to repair the issues discussed in lemma (\\ref{detdeg}).\n\nLemma (\\ref{detdeg}) also appears limited to only the parameter cases given in propositions (\\ref{coprime}) and (\\ref{rank3}). As will be seen in the following example, there exist parameter choices for $n, r,$ and Drinfeld module $\\phi_x$ outside these cases where the algorithm will never work.\n\n\\begin{example}\\label{examplefail}\nLet $n = 3$, $r =3$, $\\phi_x = \\tau^3 + x$. For $(\\ref{system1})$ to be uniquely solvable for some $e$, whenever $\\alpha_1 \\tau^3 + \\alpha_2 \\tau^6 = P (\\tau^3 + x - e)$ we must have $P = 0$. However for any $e$ setting $\\alpha_1 = 1$, $\\alpha_2 = x - e $ has the non-zero solution $P = \\tau^3$. \n\\end{example}\n\nIt is possible to detect failure cases such as those in the preceding example by checking dimensionality of the kernel of the linear system in $(\\ref{reduced})$. \n\n\n\n\\subsection{Complexity.}\n\nThe algorithm can be summarized with the following routine:\n\n%\\textit{Subroutine 1}\n\\begin{algorithm}[H]\\label{schoof}\n\\caption{Pseudocode for Schoof-like algorithm}\n\\begin{enumerate}\n \\item Compute $\\tau^{ni} \\mod \\phi_{x-e}$ in the computational basis for $\\lfloor\\frac{r-1}{r}n \\rfloor$ distinct $e \\in \\F_q$ and $1 \\leq i \\leq r$\n \\item Solve the $r \\times (\\hatr-1)$ system given in (\\ref{system1}) for each choice of $e$. If the system $B(\\phi_{x-e})$ is uniquely solvable, apply a uniformly random projection $\\ell : \\L \\to \\F_q$ to the system and solve the system over $\\F_q$.\n \\item Interpolate $\\{a_i\\}_{i=1}^{\\hatr-1}$ using the $\\lfloor\\frac{r-1}{r}n \\rfloor$ evaluation points $\\{a_i(e)\\}_{i=1}^{r-1}$ computed in step (2). \n \\end{enumerate}\n\\end{algorithm}\n\nThe bottleneck of subroutine 1 is step (1). Algorithms for skew-polynomial multiplication can be leveraged to compute the entire collection of $\\tau^{ni} \\mod \\phi_{x - e}$ in $r^{(\\omega+3)/2}(n \\log q)^{1 + o(1)}$ bit operations for a single $e$, with an overall contribution to the algorithm of $r^{(\\omega+3)/2}(n^2 \\log q)^{1 + o(1)}$. Step (2) adds $r^{\\omega}n\\log q$ bit operations, and (3) is of negligible cost. \n\nA generic choice of projection $\\ell$ will preserve unique solvability of $B(\\phi_{x-e})$ with probability at least $1 - \\frac{\\hatr}{q} \\geq 1 - \\frac{1}{Cn^2} $. This holds for $O(n)$ independent choices of projection with probability at least $(1 - \\frac{1}{Cn^2})^n \\to e^{-1/(C)} \\to 1$ for large $n, C$. The remaining probabilistic element is the likelihood of genericity for the choice of Drinfeld module, base field element pair $(\\phi, e)$ for $n$ choices of $e$, which happens with probability $(1 - \\frac{1}{Cn})^{n}$ and which can be made arbitrarily close to 1. This completes the argument for Theorem 1.\n\nWhen working with the prime field case $n = m$ or when $\\gcd(r, s) = 1$, the minimal polynomial is known to have degree $r$ and the routine need only be invoked once. For the general case, including parameter choices not covered by theorem \\ref{schooflike}, we speculate it may be possible to invoke the algorithm for choices of $\\hat{d}$ such that $r = \\hat{d} r'$ with $r' | s$ and invoking probabilistic polynomial testing to verify that the output for a given $\\hat{d}$ is in fact an annihilating polynomial for $\\tau^n$; taking the polynomial corresponding to the least such $\\hat{d}$.\n\n%\\begin{enumerate}\n% \\item Initialize $r_{upper} \\leftarrow r$, $r_{lower} \\leftarrow 1$.\n% \\item If $r_{upper} = r_{lower}$, return $r_{upper}$. Otherwise set $\\hatr \\leftarrow \\frac{r_{upper} + r_{lower}}{2}$.\n% \\item Run Subroutine 1 on the estimate $\\hatr$.\n% \\item If the system in (\\ref{system1}) is inconsistent, $r_{lower} \\leftarrow \\hatr$ and go to step 2.\n% \\item If the system in (\\ref{system1}) returns a solution, use polynomial identity testing to determine if $\\tau^{n \\hatr} - \\sum_{i = 0}^{\\hatr - 1} \\phi_{a_i}\\tau^{ni} = 0$. If this equation holds, set $r_{upper} \\leftarrow \\hatr$, otherwise $r_{lower} \\leftarrow \\hatr$. Go to step 2.\n \n%\\end{enumerate}\n\n\n\n\n\n\\subsection{From Hankel Systems}\nInspired by the Monte Carlo Algorithm given in $\\cite{musschost}$, we will propose an additional algorithm which doesn't depend on the genericity of the chosen Drinfeld module required by the Zippel-Schwartz criteria, and therefore may be usable when the algorithm of theorem \\ref{schooflike} fails, and possesses a similar complexity.\n\\begin{theorem}\\label{hankel}\nLet $\\phi$ be a rank $r$ Drinfeld module over a degree $n$ extension $L/\\F_q$ such that $\\gcd(n, r) = 1$. There is a Monte-Carlo algorithm computing the minimal polynomial of $\\phi_x$ with bit complexity $(r^{3}n^2 \\log q)^{1+o(1)}$ and which returns a correct result with probability greater than $\\frac{1}{2}$ .\n\\end{theorem}\n\nNote that given any algorithm achieving a probability of success greater than $\\frac{1}{2}$, another algorithm that succeeds with probability $\\frac{1}{2} < p < 1$ for any $p$ can be constructed with the same complexity as the original algorithm. We can organize the characteristic equation into a Hankel system.\n\n\\begin{align*}\n\\sum_{i=1}^{r - 1} \\phi_{a_i}\\tau^{ni} = & \\tau^{nr} - \\phi_{a_0}\n%\\end{equation*}\n%\\begin{equation}\n%\\label{chareq}\n\\\\\n\\sum_{i=1}^{r - 1} \\sum_{j=0}^{n(r-i)/r}a_{i,j}\\phi_{x}^j\\tau^{ni} = & \\tau^{nr} - \\phi_{a_0}\n\\end{align*}\n\nWith $a_{i,j} \\in \\F_q$, and with $a_0$ determined via $(\\ref{norm})$. For integers $u, v > 0$, consider the operator Hankel blocks\n\\begin{align*}\nH_{u,v} = & \\begin{bmatrix}\n\\phi_x^u\\tau^{n} & \\phi_x^u\\tau^{2n} & \\ldots & \\phi_x^u\\tau^{n (r - 1 - \\lfloor(ur/n) \\rfloor)} \\\\\n\\phi_x^u\\tau^{2n} & \\phi_x^u\\tau^{3n} & \\ldots & \\phi_x^u\\tau^{(1 + r - 1 - \\lfloor(ur/n) \\rfloor)n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ \\phi_x^u\\tau^{vn} & \\phi_x^u\\tau^{(1+v)n} & \\ldots & \\phi_x^u\\tau^{(v + r - 1 - \\lfloor(ur/n) \\rfloor) n}\n\\end{bmatrix}\n\\\\\n%\\end{equation*}\n%\\begin{equation*}\n G_{v} = &\\begin{bmatrix}\n \\tau^{nr} - \\phi_{a_0} & \\tau^{n(r+1)} - \\phi_{a_0}\\tau^n & \\ldots \\tau^{n(r+v -1)} - \\phi_{a_0}\\tau^{n(v-1)}\n \\end{bmatrix}^T\n\\end{align*}\n\nDefine\n\\begin{align*}\n%\\begin{equation*}\nH:= & \\begin{bmatrix}\nH_{0,v_1} & H_{1, v_1} & \\ldots & H_{\\lfloor n (r- 1)/r \\rfloor, v_1} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ H_{0,v_k} & H_{1, v_k} & \\ldots & H_{\\lfloor n (r-1)/r \\rfloor, v_k}\n\\end{bmatrix}\\\\\nG:= & \\begin{bmatrix}\nG_{v_1} \\\\ \\vdots \\\\ G_{v_k}\n\\end{bmatrix}\n%\\end{equation*}\n\\end{align*}\n\nWe then obtain the system\n\n\\begin{equation*}\nH\n\\begin{bmatrix}\na_{1,0} \\\\ a_{2,0} \\\\ \\vdots \\\\ a_{1,1} \\\\ a_{2,1} \\\\ \\vdots \\\\ a_{1, \\lfloor n (r-1)/r \\rfloor}\n\\end{bmatrix} = G\n\\end{equation*}\n\nIn general, non-trivial relations between the linear maps $\\phi_x^i\\tau^{jn}$ acting on $\\L$ ensure that the space of solutions has dimension larger than 0. To eliminate these relations, we lift to an extension $\\K$ over $\\L$. Let $H_{i, j}(\\alpha_{\\ell})$ denote the entry-wise evaluation of $H_{i, j}$ at $\\alpha_{\\ell}$. The goal is to choose a set of evaluation points $\\alpha_1, \\ldots, \\alpha_k \\in K$ such that the over-determined\n\n\\begin{equation*}\n H(\\alpha_1, \\ldots, \\alpha_k) := \\begin{bmatrix}\nH_{0,v_1}(\\alpha_1) & H_{1, v_1}(\\alpha_1) & \\ldots & H_{\\lfloor n (r-1)/r \\rfloor, v_1}(\\alpha_1) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ H_{0,v_k}(\\alpha_k) & H_{1, v_k}(\\alpha_k) & \\ldots & H_{\\lfloor n (r-1)/r\\rfloor, v_k}(\\alpha_k)\n\\end{bmatrix}\n\\end{equation*}\n\nhas a well-determined solution. Note that $H$ is an $ (v_1 + \\ldots + v_k) \\times \\bigg( \\sum_{i=1}^{n(r-1)/r} r - 1 - \\lfloor(i r/n) \\rfloor \\bigg)$ system, and that the number of columns is a most $nr$. Let $v_1 = v_2 = \\ldots = v_k = r$, $k = n$, and $[\\K:\\L] = r$. So $[\\K:\\F_q] = nr$, and observe that if the following two conditions are satisfied:\n\n\\begin{enumerate}\n \\item The linear maps $\\phi_x^{u}\\tau^{ni}$ on $K$ are independent for $1 \\leq i \\leq r-1$ and $0 \\leq u \\leq \\frac{r-i}{r} n < n$ \n \\item The set $B = \\{\\alpha_1, \\tau^n(\\alpha_1), \\ldots , \\tau^{n(r - 1)}(\\alpha_1), \\ldots , \\alpha_k, \\tau^n(\\alpha_k), \\ldots , $\n $\\tau^{n(r - 1)}(\\alpha_k) \\}$ is a basis for $K/\\F_q$\n\\end{enumerate} \n\nThen the equation $H(\\alpha_1, \\ldots, \\alpha_{n})z = G$ must have a unique solution, corresponding to the coefficients of the minimal polynomial of $phi_x$. The following lemma summarizes the core of this argument.\n\n\\begin{lemma}\\label{charindlem}\nLet $F/G$ be a finite field extension with $[F:G] = m$ and fix a basis $\\alpha_1, \\ldots, \\alpha_m$ for $F$ over $G$. Let $\\sigma_1, \\ldots, \\sigma_m$ be independent linear maps over $F$, fixing $G$. Then the following matrix is has no non-trivial solutions\n\n\\begin{equation*}\n \\begin{bmatrix}\n\\sigma_1(\\alpha_1) & \\ldots & \\sigma_m(\\alpha_1) \\\\ \\vdots & & \\vdots \\\\ \\sigma_1(\\alpha_m) & \\ldots & \\sigma_m(\\alpha_m) \n\\end{bmatrix} \n\\end{equation*}\n\n\n\\end{lemma}\n\\begin{proof}\nSuppose there exist $\\Lambda = \\begin{bmatrix} \\lambda_1 & \\ldots & \\lambda_m \\end{bmatrix}^T \\in F^m$ with not all $\\lambda_i = 0$ such that\n\n\\begin{equation*}\n \\begin{bmatrix}\n\\sigma_1(\\alpha_1) & \\ldots & \\sigma_m(\\alpha_1) \\\\ \\vdots & & \\vdots \\\\ \\sigma_1(\\alpha_m) & \\ldots & \\sigma_m(\\alpha_m) \\end{bmatrix} \\begin{bmatrix} \\lambda_1 \\\\ \\vdots \\\\ \\lambda_m \\end{bmatrix} = \\begin{bmatrix}\n0 \\\\ \\vdots \\\\ 0\n\\end{bmatrix}\n\\end{equation*}\n\nThen for any row-wise re-scaling with $\\ell_1, \\ldots, \\ell_m \\in G$ we must also have\n\n\\begin{equation*}\n \\begin{bmatrix}\n\\ell_1\\sigma_1(\\alpha_1) & \\ldots & \\ell_1\\sigma_m(\\alpha_1) \\\\ \\vdots & & \\vdots \\\\ \\ell_m\\sigma_1(\\alpha_m) & \\ldots & \\ell_m\\sigma_m(\\alpha_m) \\end{bmatrix} \\begin{bmatrix} \\lambda_1 \\\\ \\vdots \\\\ \\lambda_m \\end{bmatrix} = \\begin{bmatrix}\n0 \\\\ \\vdots \\\\ 0\n\\end{bmatrix}\n\\end{equation*}\n\nWe multiply out the left-hand side and sum the rows to obtain\n\n\\begin{equation}\\label{violation}\n \\sum_{i=1}^m\\sum_{j=1}^m\\ell_i\\lambda_j\\sigma_j(\\alpha_i) = \\sum_{j=1}^m\\lambda_j\\sigma_j\\bigg(\\sum_{i=1}^m\\ell_i\\alpha_i\\bigg) = 0\n\\end{equation}\n\nObserve now that (\\ref{violation}) holds for any $\\ell_1, \\ldots, \\ell_m $, violating linear independence.\n\n\\end{proof}\n\nFrom Lemma (\\ref{charindlem}), demonstrating that the system $H(\\alpha_1, \\ldots, \\alpha_{n})z = G$ can be uniquely solved for the coefficients of the minimal polynomial of $\\phi_x$ reduces to establishing the linear independence of $\\phi_x^{u}\\tau^{ni}$ over $\\K$. If $\\gcd(n,r) = 1$, the degree of the lead term of $\\phi_x^u = C\\tau^{ru} + \\ldots$ has a unique residue modulo $n$ for distinct choices of $u$, and so independence of $\\{ \\phi_x^{u}\\tau^{ni} \\}_{0 \\leq u \\leq n(r-i)/r }^{1 \\leq i < r}$ follows from independence of $id, \\tau, \\tau^2, \\ldots, \\tau^{nr-1}$. The values $\\alpha_1, \\ldots, \\alpha_r$ can be constructed from a normal basis for $K/\\F_q$ by selecting $\\alpha_i = \\alpha^{q^i}$ for $0 \\leq i \\leq r - 1$.\n\n\\subsection{Complexity.} Summarizing the preceding results, we have the following procedure:\n\n\\begin{algorithm}[H]\\label{hankel}\n\\caption{Computing the Minimal Polynomial via a Hankel System }\n\n\\begin{enumerate}\n \\item Fix a degree $r$ extension $K$ of $L$ and a normal basis for $K$ over $\\F_q$\n \\item Compute the Hankel Blocks $H_{i,r}(\\alpha_j)$ as well as the right-hand blocks $G_r(\\alpha_j)$\n \\item Fix a random projection $\\ell: K \\to \\F_q$ and apply it to the entries of $H, G$\n \\item Solve the resulting $O(nr) \\times nr$ system over $\\F_q$.\n\\end{enumerate}\n\\end{algorithm}\nDetermining a normal basis of size $nr$ of $K/\\F_q$ can be done in sub-quadratic time in the degree of the extension. Constructing the Hankel system above involves computing $\\phi_x^{u}\\tau^{ni}(\\alpha_j)$ for $O(nr^2)$ choices of $0 \\leq u < \\frac{r-1}{r}n, 1 \\leq i \\leq r, 0 \\leq j \\leq r$. The applications of $\\tau$ can be applied for $O(1)$ cost on the normal basis, while applications of $\\phi_x$ take $O(nr \\log q)$ bit operations. The total bit complexity for the entire procedure to compute the block Hankel system is $O(n^2r^3\\log q)$. We may again apply the uniformly selected linear projection $\\ell$ at an additional cost of $O(n^2r^3\\log q)$. Constructing the normal basis and solving the resulting $nr \\times nr$ Hankel system consisting of $n$ block columns and $r$ block rows can be done in at most $O(n^2r^2\\log q)$.\n\n\\subsection{Comparison and Limitations}\n\nUnlike in the Schoof-like case, there is no dependence on the genericity of the choice of Drinfeld module; the algorithm only fails to return a result when a poor choice of projection is made. For cases other than $\\gcd(n,r) = 1$, lemma (\\ref{charindlem}) can be applied by directly by testing the linear independence of characters $\\{ \\phi_x^{u}\\tau^{ni} \\}_{0 \\leq u \\leq n(r-i)/r }^{1 \\leq i < r}$, which can be done probabilistically using polynomial identity testing. \n\n", "meta": {"hexsha": "e29cd034fdad1779d65b0708dc28fa4e1ca3fc72", "size": 33443, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "MainText/chapter4.tex", "max_stars_repo_name": "DocTrivial/Research-Proposal", "max_stars_repo_head_hexsha": "cf988f64aa400d4c398fe4cb6738a4b90264359d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "MainText/chapter4.tex", "max_issues_repo_name": "DocTrivial/Research-Proposal", "max_issues_repo_head_hexsha": "cf988f64aa400d4c398fe4cb6738a4b90264359d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "MainText/chapter4.tex", "max_forks_repo_name": "DocTrivial/Research-Proposal", "max_forks_repo_head_hexsha": "cf988f64aa400d4c398fe4cb6738a4b90264359d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 79.061465721, "max_line_length": 1374, "alphanum_fraction": 0.656280836, "num_tokens": 11821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8221891392358015, "lm_q2_score": 0.7431680086124811, "lm_q1q2_score": 0.6110246653086806}} {"text": "\\documentclass[a4paper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[english]{babel}\n\\usepackage{graphicx}\n\\usepackage{multicol}\n\\usepackage{amsmath}\n\\usepackage{hyperref}\n\\usepackage{amsthm}\n\\usepackage{geometry}\n\\geometry{a4paper} \n\\usepackage{fancyhdr}\n\\usepackage{xcolor}\n\\begin{document}\n\\author{\\textbf{Elshimaa Ahmed}}\n\\title{\\textbf{Lecture 2}}\n\\date {\\today}\n\\maketitle\n\\theoremstyle{definition}\n\\newtheorem{definition}{Definition}[section]\n\\section{Predicates}\n\\paragraph{}\npredicates are statements involving variables such that \" $x >3$ \" . These statements are neither tru or false where the value of the variable is not specified.\nThe statement \"$x$ is greater than 3\" involving two parts . the variable part $x$ and the predicate(property) it self \" great than 3\" .its denoted by $P(x)$ once a value is assigned to $x$ the $P(x)$ become a preposition\n\\newline\n\\newline\n\\textbf{For example:}\n\\newline\nlet $P(x)$ denote the statement \"$x > 3$\" what are truth values of $P(4)$ and $P(2)$\n\\newline\n\\newline\n\\textbf{Solution: } we obtain $P(4)$ by setting $x = 4$ in the statement, Hence $P(4)$ is the statement \"$3 > 4$\" which is $True$ . $P(2)$ is the statement \"$ 2 >3$\" which is $False$ \n\\section{Quantifiers}\n\\begin{definition}[\\textbf{Universal Quantifier}]\n the universal Quantifier of $P(x)$ is the statement \"$P(X)$ is $True$ for all values of $x$ in the domain\". It is denoted by $\\forall x P(x)$\n\\begin{definition}[\\textbf{Esistential Quantifier}]\n the existential quantifier of $P(x)$ is the preposition \" there exist an element $x$ such that $P(x)$ is $True$\" . It is denoted by $\\exists x P(x)$\n\\end{definition} \n\\end{definition}\n\n\\begin{tabular}{|c|c|c|}\n \\hline\n statement & when true & when false \\\\ \n \\hline\n $\\forall x P(x)$ & P(x) is $True$ for all values of $x$ & there is an $x$ for which $P(x)$ is $False$\\\\\n $\\exists x P(x)$ & there is an $x$ such that $P(x)$ is $True$ & $P(x)$ is $False$ for all $x$ \\\\\n \\hline\n\\end{tabular}\n\\newline\\newline\n\\textbf{English phrases with quatifiers:}\n\\begin{itemize}\n \\item \"no one is P(x)\" $\\longleftrightarrow$ $\\forall x \\neg P(x)$ or $\\neg(\\exists x P(x))$\n \\item \"not every one is P(x)\"$\\longleftrightarrow$ $\\neg(\\forall x P(x))$ or $\\exists x\\neg P(x)$\n \\item \"exactly one is P(x) \" $\\longleftrightarrow$ $\\exists x (P(x) \\wedge \\forall y (P(y) \\rightarrow x = y)) $ \n \\item \"all Q(x) is P(x) \" $\\longleftrightarrow$ $ \\forall x (Q(x) \\rightarrow P(x))$\n \\item \"all Q(x) isnot P(x)\" $\\longleftrightarrow$ $\\forall x (Q(x)\\rightarrow \\neg P(x))$\n \\item \"some Q(x) are P(x)\" $\\longleftrightarrow$ $\\exists x (Q(x)\\wedge P(x))$\n \\item \"some Q(x) are not P(x)\" $\\longleftrightarrow$ $\\exists x (Q(x)\\wedge \\neg P(x))$\n\\newline\\newline\n\\end{itemize}\n\n\\noindent\n\\textbf{Nested Quantifiers}\n\\newline\n\\newline\n\\begin{tabular}{|p{2cm}|p{7cm}|p{7cm}|}\n \\hline\n statement & when true? &when false? \\\\\n \\hline\n $\\forall \\forall y P(x,y)$ & P(x,y) is true for all possible pairs & there is a pair (x,y) such that P(x,y) is false\\\\\n $\\forall y \\forall x P(x,y)$ & &\\\\\n \\hline\n $\\forall x \\exists y P(x,y)$ & for every x there is y (not necessary same for different values of x) such that P(x,y) is true & there is an x such that P(x,y) is false for every y \\\\\n \\hline\n $\\exists y \\forall x P(x,y)$ & there is exist a specific (same) y that P(x,y) is true for all x & for every y there is at leasy one x that P(x,y) fails\\\\\n \\hline\n $\\exists x \\exists y P(x,y)$ & therer is a pair (x,y) for which P(x,y) is true & for all pairs (x,y) P(x,y) is False\\\\\n $\\exists y \\exists x P(x,y) $&&\\\\\n \\hline \n\\end{tabular}\n\n\n\n\\end{document}", "meta": {"hexsha": "452a67a2e820eb764fc286b94cc870994f3fbaed", "size": 3682, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture2/Lecture2.tex", "max_stars_repo_name": "GUC-Notes/discrete-math-", "max_stars_repo_head_hexsha": "f9917c08e9b9706427153a405b03907a26fab2c7", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2021-10-14T03:38:37.000Z", "max_stars_repo_stars_event_max_datetime": "2021-11-01T20:13:25.000Z", "max_issues_repo_path": "Lecture2/Lecture2.tex", "max_issues_repo_name": "GUC-Notes/discrete-math-", "max_issues_repo_head_hexsha": "f9917c08e9b9706427153a405b03907a26fab2c7", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture2/Lecture2.tex", "max_forks_repo_name": "GUC-Notes/discrete-math-", "max_forks_repo_head_hexsha": "f9917c08e9b9706427153a405b03907a26fab2c7", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 44.3614457831, "max_line_length": 221, "alphanum_fraction": 0.659967409, "num_tokens": 1200, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.8221891370573386, "lm_q1q2_score": 0.6110246636897166}} {"text": "\\documentclass[letterpaper,11pt]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amssymb}\n\\usepackage[hmargin=1.25in,vmargin=1in]{geometry}\n\\usepackage{booktabs}\n\\usepackage{graphicx}\n\\usepackage{hyperref}\n\\usepackage{lmodern}\n\\usepackage{microtype}\n\\usepackage{pdflscape}\n\\usepackage{subcaption}\n\n\\title{Coursework 3: STAT 570}\n\\author{Philip Pham}\n\\date{\\today}\n\n\\begin{document}\n\\maketitle\n\n\\begin{enumerate}\n\\item Consider the Poisson-gamma random effects model given by\n \\begin{align}\n Y_i \\mid \\mu_i, \\theta_i\n &\\sim \\operatorname{Poisson}\\left(\\mu_i\\theta_i\\right) \\\\\n \\theta_i\n &\\sim \\operatorname{Gamma}\\left(b, b\\right),\n \\end{align}\n which leads to a negative binomial marginal model with the variance a\n quadratic function of the mean. Design a simulation study, along the lines of\n that which produced Table 2.3 in the book (overdispersed Poisson example) to\n investigate the efficiency and robustness under\n \\begin{itemize}\n \\item a Poisson model;\n \\item quasi-likelihood with $\\mathbb{E}\\left[Y\\right] = \\mu$ and\n $\\operatorname{Var}\\left(Y\\right) = \\alpha\\mu$; and\n \\item sandwich estimation.\n \\end{itemize}\n\n Use a log-linear model\n \\begin{equation}\n \\log \\mu_i = \\beta_0 + \\beta_1x_i,\n \\end{equation}\n with $x_i \\sim_\\mathrm{iid} \\mathcal{N}\\left(0, 1\\right)$ for\n $i = 1,2,\\ldots,n$, and $\\beta_0 = -2$ and $\\beta_1 = \\log 2$.\n\n Simulate for:\n \\begin{itemize} \n \\item $b \\in \\left\\{0.2,1,10,1000\\right\\}$.\n \\item $n \\in \\left\\{10, 20, 50, 100, 250\\right\\}$.\n \\end{itemize}\n\n Summarize what your take away message is after carrying out these simulations.\n\n \\begin{description}\n \\item[Solution:] Note that\n \\begin{align}\n \\mathbb{P}\\left(\n Y_i = y \\mid \\mu_i\n \\right)\n &= \\int_0^\\infty\\mathbb{P}\\left(\n Y_i = y \\mid \\mu_i, \\theta_i = \\theta\n \\right)\\mathbb{P}\\left(\n \\theta_i = \\theta\n \\mid b\n \\right)\n \\,\\mathrm{d}\\theta \\nonumber\\\\\n &= \\int_0^\\infty\n \\left(\n \\frac{\\left(\\mu_i\\theta\\right)^y}{y!}\\exp\\left(-\\mu_i\\theta\\right)\n \\right)\n \\left(\n \\frac{b^b}{\\Gamma(b)}\\theta^{b-1}\n \\exp\\left(-b\\theta\\right)\n \\right) \n \\,\\mathrm{d}\\theta \\nonumber\\\\\n &= \\frac{\\mu_i^y b^b}{y!\\Gamma(b)}\n \\int_0^\\infty\n \\theta^{b + y - 1}\\exp\\left(-\\theta(b + \\mu_i)\\right)\n \\,\\mathrm{d}\\theta \\nonumber\\\\\n &= \\frac{\\Gamma(y + b)}{y!\\Gamma(b)}\n \\frac{\\mu_i^y b^b}{\\left(\\mu_i + b\\right)^{b + y}}\n = \\frac{\\Gamma(y + b)}{y!\\Gamma(b)}\n \\left(\\frac{b}{\\mu_i + b}\\right)^b\n \\left(\\frac{\\mu_i}{\\mu_i + b}\\right)^y\n \\nonumber\\\\\n &\\sim \\operatorname{NegativeBinomial}\\left(\n b,\n \\frac{\\mu_i}{\\mu_i + b}\n \\right).\n \\label{eqn:p1_negative_binomial}\n \\end{align}\n\n By properties of the negative binomial distribution, we have that\n \\begin{align}\n \\mathbb{E}\\left[\n Y_i\n \\mid x_i\n \\right]\n &= \\mu_i = \\exp\\left(\\beta_0 + \\beta_1x_i\\right) \\nonumber\\\\\n \\operatorname{Var}\\left(Y_i \\mid x_i\\right)\n &= \\mu_i\\left(\n 1 + \\frac{\\mu_i}{b}\n \\right).\n \\label{eqn:p1_y_mean_variance}\n \\end{align}\n Thus, variance has a quadratic dependence on the mean, and smaller values of\n $b$ correspond to more dispersion.\n\n \\subsection*{Poisson Model}\n\n In the Poisson model, we assume that\n $\\operatorname{Var}\\left(Y_i \\mid x_i\\right) = \\mu_i$, e.g.\n $b \\rightarrow \\infty$, so we neglect the overdispersion parameter.\n\n In this case, the log-likelihood function is\n \\begin{equation}\n l\\left(\\beta\\right) =\n \\sum_{i=1}^n\\left[\n y_i\\left(\\beta_0 + \\beta_1x_i\\right) -\n \\exp\\left(\\beta_0 + \\beta_1x_i\\right) -\n \\sum_{k=1}^{y_i}\\log k\n \\right],\n \\end{equation}\n which gives us the score function\n \\begin{equation}\n S\\left(\\beta\\right) = \\sum_{i=1}^n\\begin{pmatrix}\n y_i - \\exp\\left(\\beta_0 + \\beta_1x_i\\right) \\\\\n x_iy_i - x_i\\exp\\left(\\beta_0 + \\beta_1x_i\\right)\n \\end{pmatrix}.\n \\label{eqn:p1_score_function}\n \\end{equation}\n We can estimate $\\beta$ by solving for\n $S\\left(\\hat{\\beta}\\right) = \\mathbf{0}$, numerically.\n\n We can estimate the variance of the estimates from the Fisher information,\n \\begin{align}\n \\operatorname{Var}\\left(\n \\hat{\\beta}\n \\right)\n &\\approx\n I_n\\left(\\hat{\\beta}\\right)^{-1} \\label{eqn:p1_beta_hat_variance}\\\\\n &= \\left(\n \\sum_{i=1}^n\\begin{pmatrix}\n \\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right) &\n x_i\\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right) \\\\\n x_i\\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right) &\n x_i^2\\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right)\n \\end{pmatrix} \\right)^{-1} \\nonumber\\\\\n &= \\frac{1}{\n \\left(\\sum_{i=1}^n \\hat{\\mu}_i\\right)\\left(\\sum_{i=1}^n x_i^2\\hat{\\mu}_i\\right)\n - \\left(\\sum_{i=1}^n x_i\\hat{\\mu}_i\\right)^2}\n \\begin{pmatrix}\n \\sum_{i=1}^n x_i^2\\hat{\\mu}_i &\n -\\sum_{i=1}^n x_i\\hat{\\mu}_i \\\\\n -\\sum_{i=1}^n x_i\\hat{\\mu}_i &\n \\sum_{i=1}^n \\hat{\\mu}_i\n \\end{pmatrix}, \\nonumber\n \\end{align}\n where $\\hat{\\mu}_i = \\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right)$.\n \n \\subsection*{Quasi-likelihood}\n\n In a quasi-likelihood model, we specify the mean and variance as\n \\begin{align}\n \\mathbb{E}\\left[\n Y_i\n \\mid x_i\n \\right]\n &= \\mu_i = \\exp\\left(\\beta_0 + \\beta_1x_i\\right) \\nonumber\\\\\n \\operatorname{Var}\\left(Y_i \\mid x_i\\right)\n &= \\alpha \\mu_i\n \\label{eqn:p1_y_quasi_mean_variance}\n \\end{align}\n From Equation \\ref{eqn:p1_y_mean_variance}, we see that this is not quite\n correct, still, but it is closer to the real model than the Poisson model.\n\n Then, by Equation 2.30 of Wakefield's \\emph{Bayesian and Frequentist\n Regression Methods} our estimating function is\n \\begin{align}\n U\\left(\\beta\\right)\n &= D^\\intercal V^{-1}\\left(y - \\mu\\right)/\\alpha \\nonumber\\\\\n &= \\sum_{i=1}^n\\begin{pmatrix}\n \\exp\\left(\\beta_0 + \\beta_1x_i\\right) \\\\\n x_i\\exp\\left(\\beta_0 + \\beta_1x_i\\right)\n \\end{pmatrix}\n \\frac{y_i - \\exp\\left(\\beta_0 + \\beta_1x_i\\right)}{\\alpha\\exp\\left(\\beta_0 + \\beta_1x_i\\right)}\n \\nonumber\\\\\n &= \\frac{1}{\\alpha}\\sum_{i=1}^n\\begin{pmatrix}\n y_i - \\exp\\left(\\beta_0 + \\beta_1x_i\\right) \\\\\n x_iy_i - x_i\\exp\\left(\\beta_0 + \\beta_1x_i\\right)\n \\end{pmatrix} = \\frac{1}{\\alpha}S\\left(\\beta\\right)\n \\label{eqn:p1_quasi_likelihood_score_function}\n \\end{align}\n from Equation \\ref{eqn:p1_score_function}. Thus, the maximum\n quasi-likelihood estimate will be the same as the maximum likelihood\n estimate from the Poisson model.\n\n Having solved for $\\hat{\\beta}$, we have\n \\begin{equation}\n \\hat{\\mu} = \\exp\\left(\\hat{\\beta}_0 + \\hat{\\beta}_1x_i\\right).\n \\label{eqn:p1_mu_hat}\n \\end{equation}\n\n by Equation 2.31 of Wakefield's \\emph{Bayesian and Frequentist Regression\n Methods}, we can then compute\n \\begin{equation}\n \\hat{\\alpha}_n\n = \\frac{1}{n-2}\\sum_{i=1}^n \\frac{\\left(y_i - \\hat{\\mu}_i\\right)^2}{\\hat{\\mu}_i}\n \\label{eqn:p1_alpha_hat}\n \\end{equation}\n\n Then, the variance of our estimates is\n \\begin{align}\n \\operatorname{Var}\\left(\n \\hat{\\beta}\n \\right)\n &\\approx\n \\hat{\\alpha}_n\\left(\\hat{D}^\\intercal \\hat{V}^{-1} \\hat{D}\\right)^{-1} \\nonumber\\\\\n &= \\hat{\\alpha}_n\\left(\\sum_{i=1}^n\n \\begin{pmatrix}\n \\hat{\\mu_i} & x_i\\hat{\\mu_i} \\\\\n x_i\\hat{\\mu_i} & x_i^2\\hat{\\mu_i}.\n \\end{pmatrix}\\right)^{-1}\n \\nonumber\\\\\n &= \\hat{\\alpha}_nI_n\\left(\\hat{\\beta}\\right)^{-1}\n \\label{eqn:p1_beta_hat_quasi_variance}\n \\end{align}\n from Equation \\ref{eqn:p1_beta_hat_variance}.\n\n \\subsection*{Sandwich Estimation}\n\n In sandwich estimation, we only need to specify an estimating function\n $G\\left(\\beta\\right)$. Then, we can apply Equation 2.43 of Wakefield's\n \\emph{Bayesian and Frequentist Regression Methods} to compute the variance\n of our estimates:\n \\begin{align}\n \\operatorname{Var}\\left(\\hat{\\beta}\\right)\n &= \\frac{1}{n}\\hat{A}^{-1}\\hat{B}\\left(\\hat{A}^{-1}\\right)^\\intercal \\nonumber\\\\\n \\hat{A}\n &= -\\frac{1}{n}\\sum_{i=1}^n\n \\frac{\\partial}{\\partial\\beta}G\\left(\\hat{\\beta}, X_i, Y_i\\right) \\nonumber\\\\\n \\hat{B}\n &= \\frac{1}{n}\\sum_{i=1}^n\n G\\left(\\hat{\\beta}, X_i, Y_i\\right)G\\left(\\hat{\\beta}, X_i, Y_i\\right)^\\intercal.\n \\nonumber\n \\end{align}\n\n We can reuse the score function from the quasi-likelihood estimate in\n Equation \\ref{eqn:p1_quasi_likelihood_score_function} without $\\alpha$, so\n \\begin{equation}\n G\\left(\\hat{\\beta}, X_i, Y_i\\right)\n = \\begin{pmatrix}\n Y_i - \\hat{\\mu}_i \\\\\n X_i\\left(Y_i - \\hat{\\mu}_i\\right)\n \\end{pmatrix}\n \\end{equation}\n Thus, our estimate for $\\hat{\\beta}$ will remain the same.\n\n From Equations \\ref{eqn:p1_beta_hat_variance} and\n \\ref{eqn:p1_beta_hat_quasi_variance}, we have that\n \\begin{equation}\n \\hat{A} = \\frac{1}{n}\\hat{D}\\hat{V}^{-1}\\hat{D} = \\frac{1}{n}I_n\\left(\\hat{\\beta}\\right)\n \\end{equation}\n\n From Equation \\ref{eqn:p1_quasi_likelihood_score_function}, we have that\n \\begin{align}\n \\hat{B}\n &= \\frac{1}{n}\n \\hat{D}^\\intercal\\hat{V}^{-1}\\operatorname{diag}\\left(RR^\\intercal\\right)\\hat{V}^{-1}\\hat{D}\n \\nonumber\\\\\n &= \\frac{1}{n}\\hat{D}^\\intercal\\begin{pmatrix}\n \\frac{\\left(y_1 - \\hat{\\mu}_1\\right)^2}{\\hat{\\mu}_1^2} & & \\\\\n & \\ddots &\\\\\n & &\\frac{\\left(y_n - \\hat{\\mu}_n\\right)^2}{\\hat{\\mu}_n^2}\n \\end{pmatrix}\\hat{D} =\n \\frac{1}{n}\\sum_{i=1}^n \\frac{\\left(y_i - \\hat{\\mu}_i\\right)^2}{\\hat{\\mu}_i^2}\n \\begin{pmatrix}\n \\hat{\\mu}_i^2 & x_i\\hat{\\mu}_i^2 \\\\\n x_i\\hat{\\mu}_i^2 & x_i^2\\hat{\\mu}_i^2\n \\end{pmatrix} \\nonumber\\\\\n &= \\frac{1}{n}\\sum_{i=1}^n\\left(y_i-\\hat{\\mu}_i\\right)^2\n \\begin{pmatrix}\n 1 & x_i\\\\\n x_i & x_i^2\n \\end{pmatrix} = \\frac{1}{n}\\sum_{i=1}^n\n G\\left(\\hat{\\beta},x_i,y_i\\right)G\\left(\\hat{\\beta},x_i,y_i\\right)^\\intercal.\n \\end{align}\n\n \\subsection*{Results}\n\n The results of estimating $\\beta_0$ and $\\beta_1$ under various models for\n different $b$ and $n$ can be seen in Tables \\ref{tab:p1_results_beta_0} and\n \\ref{tab:p1_results_beta_1} for $\\beta_0$ and $\\beta_1$, respectively. How\n often the 95\\% confidence interval contains the true value of $\\beta_j$ was\n calculated for each simulation.\n\n When $b \\in \\{10,1000\\}$, there is not much dispersion, and the Poisson\n model does well for all $n$. At smaller values of $b$, the coverage is not\n very good.\n\n The quasi-likelihood model does better than the Poisson model for\n $n \\geq 50$. For smaller values of $n$, its confidence intervals often fail\n to cover the true value. The quasi-likelihood model does a poorer job as\n covering $\\beta_1$, however.\n \n For $\\beta_1$, only sandwich estimation produces good confidence intervals\n for small $b$. However, it needs a lot of samples to do so, typically\n $n = 250$ or $n = 1000$. When $n$ is smaller it often performs worse than\n both the Poisson and quasi-likelihood models.\n \n Code for the simulations can be found at\n \\href{https://nbviewer.jupyter.org/github/ppham27/stat570/blob/master/hw3/poisson\\_gamma\\_random\\_effects.ipynb}{\\texttt{poisson\\_gamma\\_random\\_effects.ipynb}}.\n \\end{description}\n\n \\begin{landscape}\n \\begin{table}\n \\tiny\n \\begin{subtable}{9in}\n \\input{p1_beta_0.tex}\n \\caption{95\\% confidence interval converage for $\\beta_0$.}\n \\label{tab:p1_results_beta_0}\n \\end{subtable}\n \n \\begin{subtable}{9in}\n \\input{p1_beta_1.tex}\n \\caption{95\\% confidence interval converage for $\\beta_1$.}\n \\label{tab:p1_results_beta_1}\n \\end{subtable}\n\n \\caption{The results of fitting various models to the data from a\n Poisson-gamma random effects model. 100,000 simulations were done for\n each $(\\text{model}, b, n)$ tuple.}\n \\label{tab:p1_results} \n \\end{table}\n \\end{landscape}\n \n\n\\item The data in Table \\ref{tab:p2_data} contain data on a typical reliability\n experiment and give the failure stresses (in GPa) of four samples of carbon\n fibers of lengths 1, 10, 20 and 50mm.\n\n \\begin{table}\n \\tiny\n \\centering\n \\input{p2_data.tex}\n \\caption{Failure stress data for four groups of fibers.}\n \\label{tab:p2_data}\n \\end{table}\n\n \\begin{enumerate}\n \\item The exponential distribution\n $Y \\mid \\lambda \\sim_\\mathrm{iid} \\operatorname{Exponential}(\\lambda)$, is a\n simple model for reliability data:\n \\begin{equation}\n p\\left(y \\mid \\lambda \\right) =\n \\lambda \\exp\\left(-\\lambda y\\right),\n \\label{eqn:p2_exponential_pdf}\n \\end{equation}\n with $\\lambda, y > 0$. The hazard function is the probability of imminent\n failure and is given by\n \\begin{equation}\n h\\left(y\\mid\\lambda\\right) =\n \\frac{p\\left(y \\mid \\lambda\\right)}{S\\left(y\\mid\\lambda\\right)},\n \\label{eqn:p2_hazard_definition}\n \\end{equation}\n where\n $S\\left(y\\mid\\lambda\\right) = \\mathbb{P}\\left(Y > y \\mid \\lambda\\right)$ is\n the probability of failure beyond $y$. Derive the hazard function for the\n exponential distribution. Suppose we have a sample $y_1,\\ldots,y_n$, of size\n $n$ from an exponential distribution. Find the form of the MLE of $\\lambda$\n and the asymptotic variance.\n \\begin{description}\n \\item[Solution:] The survival function can be derived with Equation\n \\ref{eqn:p2_exponential_pdf} as\n \\begin{align}\n S\\left(y\\mid\\lambda\\right)\n &= \\mathbb{P}\\left(Y > y \\mid \\lambda\\right) \\nonumber\\\\\n &= \\int_y^\\infty \\lambda \\exp\\left(-\\lambda t\\right)\\,\\mathrm{d}t \\nonumber\\\\\n &= \\left. -\\exp\\left(-\\lambda t\\right) \\right|^\\infty_y \\nonumber \\\\\n &= \\exp\\left(-\\lambda y\\right).\n \\label{eqn:p2_survival}\n \\end{align}\n\n With Equations \\ref{eqn:p2_hazard_definition} and \\ref{eqn:p2_survival},\n the hazard function is\n \\begin{equation}\n h\\left(y\\mid\\lambda\\right) =\n \\frac{p\\left(y \\mid \\lambda\\right)}{S\\left(y\\mid\\lambda\\right)}\n = \\frac{\\lambda \\exp\\left(-\\lambda y\\right)}{\\exp\\left(-\\lambda y\\right)} = \\lambda.\n \\label{eqn:p2_hazard}\n \\end{equation}\n \\end{description}\n\n Given $y_1,\\ldots,y_n$, the log-likelihood function is\n \\begin{equation}\n l\\left(\\lambda\\right) = n\\log\\lambda - \\lambda\\sum_{i=1}^n y_i.\n \\label{eqn:p2_exponential_log_likelihood}\n \\end{equation}\n\n From Equation \\ref{eqn:p2_exponential_log_likelihood}, the score function is\n \\begin{equation}\n S\\left(\\lambda\\right) = \\frac{n}{\\lambda} - \\sum_{i=1}^n y_i.\n \\label{eqn:p2_exponential_score}\n \\end{equation}\n Solving $S\\left(\\hat{\\lambda}\\right) = 0$, gives use the MLE,\n $\\boxed{\\hat{\\lambda} = \\frac{n}{\\sum_{i=1}^n y_i} = \\frac{1}{\\bar{y}}.}$\n\n The asymptotic variance can be derived from the Fisher information\n \\begin{equation}\n I_n\\left(\\hat{\\lambda}\\right)\n = \\operatorname{Var}\\left(S\\left(\\hat{\\lambda}\\right)\\right)\n = \\frac{n}{\\hat{\\lambda}^2}.\n \\end{equation}\n\n Thus, we have that\n $\\boxed{\\operatorname{Var}\\left(\\hat{\\lambda}\\right) =\n \\frac{\\hat{\\lambda}^2}{n}.}$ \n \n \\item For each of the four groups in Table \\ref{tab:p2_data}, estimate a\n separate $\\lambda$, with an associated standard error. Examine the\n appropriateness of the exponential model via Q-Q plots.\n\n \\begin{description}\n \\item[Solution:] The estimates and standard errors for $\\lambda$ are in\n Table \\ref{tab:p2_exponential_estimates}.\n\n \\begin{table}\n \\centering\n \\input{p2_exponential_estimates.tex}\n \\caption{Results of fitting an exponential model for each length.}\n \\label{tab:p2_exponential_estimates}\n \\end{table}\n\n From the Q-Q plots in Figure \\ref{fig:p2_qq_exponential}, we see that\n the exponential model is a very poor fit for the data. The sampled\n quantiles are very different than the theoretical quantiles. Thus, the\n model appears to be misspecified.\n\n \\begin{figure}\n \\centering\n \\includegraphics{p2_qq_exponential.pdf}\n \\caption{Q-Q plots at each length when fitting an exponential model\n using the MLE.}\n \\label{fig:p2_qq_exponential}\n \\end{figure}\n \\end{description}\n\n Code for calculations and plots can be found in\n \\href{https://nbviewer.jupyter.org/github/ppham27/stat570/blob/master/hw3/failure\\_stresses.ipynb}{\\texttt{failure\\_stresses.ipynb}}.\n \n \\item Consider a quasi-likelihood approach to inference for $\\lambda$ under\n the model with\n \\begin{align}\n \\mathbb{E}\\left[Y \\mid \\lambda \\right]\n &= \\lambda^{-1} \\nonumber\\\\\n \\operatorname{Var}\\left(Y \\mid \\lambda\\right)\n &= \\alpha\\lambda^{-2}\n \\label{eqn:p2_quasi_likelihood_model}\n \\end{align}\n with $\\alpha > 0$. Suggest an estimator for $\\alpha$. Estimate $\\lambda$,\n $\\alpha$, and the standard errors, separately for each of the four groups\n in Table \\ref{tab:p2_data}. What do the results suggest to you about the\n fit of the exponential model?\n\n \\begin{description}\n \\item[Solution:] The quasi-likelihood score is\n \\begin{align}\n U\\left(\\lambda\\right)\n &= D^\\intercal V^{-1}\n \\left(Y - \\mathbb{E}\\left[Y \\mid \\lambda \\right]\\right)/\\alpha \\nonumber\\\\\n &= \\begin{pmatrix}\n -\\lambda^{-2} & \\cdots & -\\lambda^{-2}\n \\end{pmatrix} \\begin{pmatrix}\n \\lambda^{2} & & \\\\\n & \\ddots & \\\\\n & & \\lambda^{2}\n \\end{pmatrix} \\begin{pmatrix}\n \\frac{Y_1 - \\lambda^{-1}}{\\alpha} \\\\\n \\vdots \\\\\n \\frac{Y_n - \\lambda^{-1}}{\\alpha}\n \\end{pmatrix} \\nonumber\\\\\n &= -\\frac{1}{\\alpha}\\sum_{i=1}^n\\left(Y_i - \\lambda^{-1}\\right)\n = -\\frac{1}{\\alpha}\\left(\\sum_{i=1}^nY_i - n\\lambda^{-1}\\right).\n \\label{eqn:p2_quasi_score}\n \\end{align}\n \n Solving Equation \\ref{eqn:p2_quasi_score},\n $U\\left(\\hat{\\lambda}\\right) = 0$, we get\n $\\hat{\\lambda} = \\bar{Y}^{-1}$, which is the same as the MLE estimate.\n\n $\\hat{\\alpha}$ is given by Equation 2.31 of Wakefield's \\emph{Bayesian\n and Frequentist Regression Methods}:\n \\begin{equation}\n \\hat{\\alpha}\n = \\frac{1}{n - 1}\\sum_{i=1}^n\\frac{\\left(Y_i - \\hat{\\mu}\\right)^2}{V\\left(\\hat{\\mu}\\right)}\n = \\frac{\\hat{\\lambda}^2}{n - 1}\\sum_{i=1}^n\\left(Y_i - \\hat{\\lambda}^{-1}\\right)^2.\n \\end{equation}\n\n We have that\n \\begin{equation}\n \\operatorname{Var}\\left(U\\left(\\lambda\\right)\\right)\n = \\mathbb{E}\\left[- \\frac{\\partial U}{\\partial \\lambda}(\\lambda)\\right]\n = \\frac{n\\lambda^{-2}}{\\alpha},\n \\end{equation}\n we can estimate\n \\begin{equation}\n \\operatorname{Var}\\left(\\hat{\\lambda}\\right) =\n \\operatorname{Var}\\left(U\\left(\\hat{\\lambda}\\right)\\right)^{-1}\n \\approx \\frac{\\hat{\\alpha}\\hat{\\lambda}^2}{n},\n \\end{equation}\n which is the same as the variance for the MLE estimate multiplied by\n $\\hat{\\alpha}$. The results of fitting the quasi-likelihood model can be\n seen in Table \\ref{tab:p2_quasi_likelihood_estimates}.\n\n \\begin{table}\n \\centering\n \\input{p2_quasi_likelihood_estimates.tex}\n \\caption{Results of fitting a quasi-likelihood model for each length.}\n \\label{tab:p2_quasi_likelihood_estimates}\n \\end{table}\n\n From Equation \\ref{eqn:p2_quasi_score}, we see that a quadratic variance\n function leads to the same score function as Gamma distribution\n with fixed shape parameter $\\alpha^{-1}$ and rate parameter\n $\\lambda\\alpha^{-1}$. $\\alpha = 1$ would correspond to the\n exponential distribution, so it is unsurprising to see that our estimate\n for $\\lambda$ is the same as the MLE estimate.\n\n \\begin{figure}\n \\centering\n \\includegraphics{p2_qq_quasi_likelihood.pdf}\n \\caption{Q-Q plots at each length with a Gamma distribution. The shape\n parameter $\\hat{\\alpha}^{-1}$ and rate parameter\n $\\hat{\\lambda}\\hat{\\alpha}^{-1}$ were estimated using\n quasi-likelihood.}\n \\label{fig:p2_qq_quasi_likelihood}\n \\end{figure}\n \n Standard errors are much smaller than those estimated in Table\n \\ref{tab:p2_exponential_estimates}. From Figure\n \\ref{fig:p2_qq_exponential}, we see that the residuals are\n underdispersed relative to an exponential model, so $\\hat{\\alpha} < 1$,\n which leads to the smaller standard error estimates.\n\n Q-Q plots with the theoretical quantiles derived from\n $\\operatorname{Gamma}\\left(\\hat{\\alpha}^{-1},\n \\lambda\\hat{\\alpha}^{-1}\\right)$ in Figure\n \\ref{fig:p2_qq_quasi_likelihood}. The points lie close the $y = x$\n line. This suggests that the Gamma and quasi-likelihood model are more\n appropriate. They better capture the variance model compared to the\n exponential model.\n\n Code for calculations and plots can be found in\n \\href{https://nbviewer.jupyter.org/github/ppham27/stat570/blob/master/hw3/failure\\_stresses.ipynb}{\\texttt{failure\\_stresses.ipynb}}.\n \\end{description}\n \\item Obtain the form of the sandwich estimate for the variance of\n $\\hat{\\lambda}$. Numerically evaluate sandwich standard errors for the\n estimate of $\\lambda$ in each of the four groups.\n\n \\begin{description}\n \\item[Solution:] We apply Equation 2.43 of Wakefield's \\emph{Bayesian and\n Frequentist Regression Methods} to compute the variance of our\n parameter estimate:\n \\begin{align}\n \\operatorname{Var}\\left(\\hat{\\lambda}\\right)\n &= \\frac{1}{n}\\hat{A}^{-1}\\hat{B}\\left(\\hat{A}^{-1}\\right)^\\intercal \\nonumber\\\\\n \\hat{A}\n &= -\\frac{1}{n}\\sum_{i=1}^n\n \\frac{\\partial}{\\partial\\lambda}G\\left(\\hat{\\lambda}, Y_i\\right) \\nonumber\\\\\n \\hat{B}\n &= \\frac{1}{n}\\sum_{i=1}^n\n G\\left(\\hat{\\lambda}, Y_i\\right)G\\left(\\hat{\\lambda}, Y_i\\right)^\\intercal, \\nonumber\n \\end{align}\n where we resue the quasi-score from Equation \\ref{eqn:p2_quasi_score} to\n specify\n \\begin{equation}\n G\\left(\\lambda, Y_i\\right) = \\frac{1}{\\lambda} - Y_i\n \\label{eqn:p2_sandwich_score}\n \\end{equation}\n as our estimating function.\n\n Thus, we'll have that\n \\begin{align}\n \\hat{A}\n &= \\frac{1}{n}I\\left(\\hat{\\lambda}\\right) = \\frac{1}{\\hat{\\lambda}^2} \\nonumber\\\\\n \\hat{B}\n &= \\frac{1}{n}\\sum_{i=1}^n\\left(Y_i - \\frac{1}{\\hat{\\lambda}}\\right)^2.\n \\end{align}\n\n Thus, our sandwich estimate will be\n \\begin{equation}\n \\boxed{\n \\operatorname{Var}\\left(\\hat{\\lambda}\\right)\n = \\frac{\\hat{\\lambda}^4}{n^2}\\sum_{i=1}^n\\left(Y_i - \\frac{1}{\\hat{\\lambda}}\\right)^2.\n }\n \\label{eqn:p2_sandwich_variance}\n \\end{equation}\n The results of applying Equations \\ref{eqn:p2_sandwich_score} and\n \\ref{eqn:p2_sandwich_variance} can be seen in Table \\ref{tab:p2_sandwich\n estimates}. \n \n \\begin{table}\n \\centering \\input{p2_sandwich_estimates.tex}\n \\caption{Results of fitting a model with sandwich estimation for each\n length.}\n \\label{tab:p2_sandwich estimates}\n \\end{table}\n\n The estimates for $\\hat{\\lambda}$ are of course the same as in Tables\n \\ref{tab:p2_exponential_estimates} and\n \\ref{tab:p2_quasi_likelihood_estimates}, since we reused the same score\n function. The standard errors are smaller than the exponential model since\n the data is underdispersed in that model. However, they are quite similar to\n those in the quasi-likelihood model despite not specifying a variance\n model. From Figure \\ref{fig:p2_qq_quasi_likelihood}, we have evidence that\n the quasi-likelihood model fits the data well, so it is unsurprising that an\n empirical estimate would yield similar results.\n \\end{description}\n \n\\item The Weibull distribution is a common model for survival or reliability\n data:\n $Y \\mid \\eta,\\alpha \\sim_\\mathrm{iid}\n \\operatorname{Weibull}\\left(\\eta,\\alpha\\right)$, with $\\eta > 0$, and\n $\\alpha > 0$. The random variable $Y$ has a Weibull distribution if its\n density can be written in the form\n \\begin{equation}\n p\\left(y \\mid \\eta,\\alpha\\right) = \\eta\\alpha^{-\\eta}y^{\\eta-1}\\exp\\left[\n -\\left(\\frac{y}{\\alpha}\\right)^\\eta\n \\right].\n \\label{eqn:p2_weibull_pdf}\n \\end{equation}\n \n Find the mean, variance and hazard function of a Weibull distribution. For\n what value of the parameters does the exponential distribution result?\n\n \\begin{description}\n \\item[Solution:] The mean can be calculated in terms of the Gamma function\n \\begin{align}\n \\mathbb{E}\\left[\n Y \\mid \\eta,\\alpha\n \\right] = \\int_0^\\infty y p\\left(y \\mid \\eta,\\alpha\\right)\\,\\mathrm{d}y\n &= \\eta\\int_0^\\infty \\left(\\frac{y}{\\alpha}\\right)^{\\eta}\n \\exp\\left[\n -\\left(\\frac{y}{\\alpha}\\right)^\\eta\n \\right]\\,\\mathrm{d}y\\nonumber\\\\\n &= \\alpha\\int_0^\\infty u^{1 + 1/\\eta - 1}\\exp\\left(-u\\right)\\,\\mathrm{d}u \\nonumber\\\\\n &= \\alpha\\Gamma\\left(1 + 1/\\eta\\right).\n \\label{eqn:p2_weibull_mean}\n \\end{align}\n\n We can calculate the second moment similarly,\n \\begin{align}\n \\mathbb{E}\\left[\n Y^2 \\mid \\eta,\\alpha\n \\right] = \\int_0^\\infty y^2 p\\left(y \\mid \\eta,\\alpha\\right)\\,\\mathrm{d}y\n &= \\eta\\int_0^\\infty y\\left(\\frac{y}{\\alpha}\\right)^{\\eta}\n \\exp\\left[\n -\\left(\\frac{y}{\\alpha}\\right)^\\eta\n \\right]\\,\\mathrm{d}y\\nonumber\\\\\n &= \\alpha^2\\int_0^\\infty u^{2 + 1/\\eta - 1}\\exp\\left(-u\\right)\\,\\mathrm{d}u \\nonumber\\\\\n &= \\alpha^2\\Gamma\\left(2 + 1/\\eta\\right).\n \\end{align}\n\n Thus, we have that\n \\begin{equation}\n \\operatorname{Var}\\left(\n Y \\mid \\eta,\\alpha\n \\right) = \\alpha^2\\left(\n \\Gamma\\left(2 + 1/\\eta\\right) - \\left[\\Gamma\\left(1 + 1/\\eta\\right)\\right]^2\n \\right).\n \\end{equation}\n\n The survival function is\n \\begin{align}\n S\\left(y \\mid \\eta,\\alpha\\right)\n &= \\int_y^\\infty p\\left(t \\mid \\eta,\\alpha\\right)\\,\\mathrm{d}t =\n \\int_{\\left(y/\\alpha\\right)^\\eta}^\\infty\\exp\\left(-u\\right)\\,\\mathrm{d}u \\nonumber\\\\\n &= \\exp\\left[-\\left(\\frac{y}{\\alpha}\\right)^\\eta\\right].\n \\label{eqn:p2_weibull_survival}\n \\end{align}\n\n Thus, the hazard function simplifies to\n \\begin{equation}\n h\\left(y \\mid \\eta,\\alpha\\right) =\n \\frac{p\\left(y \\mid \\eta,\\alpha\\right)}{S\\left(y \\mid \\eta,\\alpha\\right)} = \n \\eta\\alpha^{-\\eta}y^{\\eta-1}.\n \\label{eqn:p2_weibull_hazard}\n \\end{equation}\n\n When $\\eta = 1$, this is just the exponential distribution with rate\n parameter $\\lambda = \\alpha^{-1}$. \n \\end{description}\n\\item Is the Weibull distribution with unknown parameters $\\eta$, $\\alpha$ a\n member of the exponential family? What are the implications for inference?\n \\begin{description}\n \\item[Solution:] No, the Weibull distribution with unknown parameters $\\eta$,\n $\\alpha$ is not a member of the exponential family. Mainly, if take the\n $\\log$ of the probability density function, we have the term\n $\\left(\\frac{y}{\\alpha}\\right)^\\eta$. This can not be written in the form\n $\\theta^\\intercal T(y)$, where $\\theta$ are parameters and $T(y)$ is a\n transformation of $y$ into a finite-dimensional vector.\n\n There are implications in inference. The\n \\href{https://en.wikipedia.org/wiki/Sufficient_statistic#Exponential\\_family}{Pitman-Koopman-Darmois\n theorem} states that only in exponential families is there a sufficient\n statistic whose dimension remains bounded as sample size increases. Thus,\n when inferring the parameters from a sample with maximum likelihood, all the\n data must be used, which may make the computation intractable for large\n datasets. Morever, when doing a Bayesian inference, the posterior must be\n conditioned on all the data rather than a finite set of sufficient\n statistics, so no conjugate prior will exist.\n\n In particular, generalized linear models require that the response be\n generated from a distribution in the exponential family. The special\n structure of the exponential family make it so the dispersion estimate is an\n ancillary statistic: its estimation is independent of the estimation of the\n mean.\n \\end{description}\n\n\\item For the Weibull model and a random sample of size $n$ obtain: the\n log-likelihood, the score, and the observed information matrix.\n\n \\begin{description}\n \\item[Solution:] The log-likelihood function is\n \\begin{equation}\n l\\left(\\eta,\\alpha\\right)\n = \\sum_{i = 1}^n\\left(\n \\log\\eta - \\eta\\log\\alpha + \\left(\\eta - 1\\right)\\log y_i\n - \\exp\\left[\\eta\\left(\\log y_i - \\log\\alpha\\right)\\right]\\right)\n \\end{equation}\n\n From which, we have the score function\n \\begin{equation*}\n S\\left(\\eta,\\alpha\\right) = \\sum_{i=1}^n\n \\begin{pmatrix}\n \\frac{1}{\\eta} - \\log\\alpha + \\log y_i -\n \\left(\\log y_i - \\log\\alpha\\right)\n \\exp\\left[\\eta\\left(\\log y_i - \\log\\alpha\\right)\\right]\n \\\\\n -\\frac{\\eta}{\\alpha}\n +\\frac{\\eta}{\\alpha}\\exp\\left[\\eta\\left(\\log y_i - \\log\\alpha\\right)\\right]\n \\end{pmatrix}.\n \\end{equation*}\n\n The observed information for a single observation is\n \\begin{align*}\n I_{y_i}\\left(\\eta, \\alpha\\right)\n &= -\\nabla \\nabla^\\intercal l\\left(\\eta,\\alpha\\right) = \\nabla S\\left(\\eta,\\alpha\\right) \\\\\n &= \\begin{pmatrix}\n \\frac{1}{\\eta^2} + \\left(\\log \\frac{y_i}{\\alpha}\\right)^2\n \\left(\\frac{y_i}{\\alpha}\\right)^\\eta\n &\n \\frac{1}{\\alpha} - \\frac{1}{\\alpha}\\left(\\frac{y_i}{\\alpha}\\right)^\\eta -\n \\frac{\\eta}{\\alpha}\\left(\\log \\frac{y_i}{\\alpha}\\right)\\left(\\frac{y_i}{\\alpha}\\right)^\\eta\n \\\\\n \\frac{1}{\\alpha} - \\frac{1}{\\alpha}\\left(\\frac{y_i}{\\alpha}\\right)^\\eta -\n \\frac{\\eta}{\\alpha}\\left(\\log \\frac{y_i}{\\alpha}\\right)\\left(\\frac{y_i}{\\alpha}\\right)^\\eta\n &\n -\\frac{\\eta}{\\alpha^2} + \\frac{\\eta}{\\alpha^2}\\left(\\frac{y_i}{\\alpha}\\right)^\\eta\n + \\left(\\frac{\\eta}{\\alpha}\\right)^2\\left(\\frac{y_i}{\\alpha}\\right)^\\eta\n \\end{pmatrix}.\n \\end{align*}\n\n The total observed information is then $\\displaystyle\n I_n\\left(\\eta,\\alpha\\right) = \\sum_{i=1}^nI_{y_i}\\left(\\eta,\\alpha\\right)$.\n \\end{description}\n \n\\item Solve the score equations in order to obtain the maximum likelihood\n estimators (MLEs). You should obtain a single equation that needs to be\n numerically solved.\n\n \\begin{description}\n \\item[Solution:] We want to solve the two equations given by\n $S\\left(\\hat{\\eta}, \\hat{\\alpha}\\right) = \\mathbf{0}$.\n\n We can solve for $\\hat{\\alpha}$ in terms of $\\hat{\\eta}$ from the second\n entry:\n \\begin{equation}\n \\hat{\\alpha} = \\left(\\frac{1}{n}\\sum_{i=1}^n y_i^{\\hat{\\eta}}\\right)^{1/\\hat{\\eta}}.\n \\label{eqn:p2_alpha_hat}\n \\end{equation}\n\n Substituting $\\hat{\\alpha}$ with Equation \\ref{eqn:p2_alpha_hat}, we can\n numerically solve for $\\hat{\\eta}$ in the first entry of the score function.\n \\end{description}\n\\item Obtain the MLEs and standard errors for the parameters of the Weibull\n model, for each of the groups in Table \\ref{tab:p2_data}.\n \\begin{description}\n \\item[Solution:] The estimates and standard errors can be found in Table\n \\ref{tab:p2_weibull estimates}. One can see that mean of each estimated\n distribution $\\hat{\\alpha}\\Gamma\\left(1 + 1/\\hat{\\eta}\\right)$ is quite\n close to corresponding mean for the exponential distribution\n $\\hat{\\lambda}^{-1}$.\n\n \\begin{table}\n \\centering \\input{p2_weibull_estimates.tex}\n \\caption{Results of fitting a Weibull model to each length by maximizing\n likelihood.}\n \\label{tab:p2_weibull estimates}\n \\end{table}\n\n \\begin{figure}\n \\centering\n \\includegraphics{p2_qq_weibull.pdf}\n \\caption{Q-Q plots at each length with a Weibull distribution with\n parameters in Table \\ref{tab:p2_weibull estimates}.}\n \\label{fig:p2_qq_weibull}\n \\end{figure}\n\n The Q-Q plots in Figure \\ref{fig:p2_qq_weibull} indicate the Weibull\n distribution fits the data quite well. However, it appears slightly worse\n than the Gamma distribution: compared to Figure\n \\ref{fig:p2_qq_quasi_likelihood}, the lower tail is further away from the\n $y = x$ line. That is, the data are underdispered when fitted to the Weibull\n distribution.\n\n Code for calculations and plots can be found in\n \\href{https://nbviewer.jupyter.org/github/ppham27/stat570/blob/master/hw3/failure\\_stresses.ipynb}{\\texttt{failure\\_stresses.ipynb}}.\n \\end{description}\n\\end{enumerate}\n\\end{enumerate}\n\n\\end{document}\n", "meta": {"hexsha": "497a2a21a65c832fae535d59afbdf71671543f15", "size": 33461, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "hw3/solutions.tex", "max_stars_repo_name": "ppham27/stat570", "max_stars_repo_head_hexsha": "859832aed3ae172abc8b6fbcd2221eb552291a00", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-04-22T11:05:54.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-25T22:46:11.000Z", "max_issues_repo_path": "hw3/solutions.tex", "max_issues_repo_name": "ppham27/stat570", "max_issues_repo_head_hexsha": "859832aed3ae172abc8b6fbcd2221eb552291a00", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "hw3/solutions.tex", "max_forks_repo_name": "ppham27/stat570", "max_forks_repo_head_hexsha": "859832aed3ae172abc8b6fbcd2221eb552291a00", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 40.5587878788, "max_line_length": 165, "alphanum_fraction": 0.6275365351, "num_tokens": 10583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7431680086124811, "lm_q2_score": 0.822189134878876, "lm_q1q2_score": 0.6110246620707529}} {"text": "%mainfile: ../lecture_notes.tex\n\\chapter{Recursive Least-Squares Adaptive Filters}\n\\label{ch:RLS}\n\\section{Review of the Basics}\n\\subsection{The Matrix Inversion Lemma}\n\\label{sec:woodbury}\nLet $\\vect{X}$, $\\vect{Y}$, $\\vect{X}+\\vect{UYV}$, and $\\vect{Y}^{-1}+\\vect{VX}^{-1}\\vect{U}$ all be non-singular matrices. By equating element $(1,1)$ of the two block matrices in \\eq{eq:block_inv1} and \\eq{eq:block_inv2} and setting\n\\begin{align}\n \\vect{X} &= \\vect{A}\\\\\n \\vect{U} &= \\vect{B}\\\\\n \\vect{V} &= \\vect{C}\\\\\n \\vect{Y} &= -\\vect{D}^{-1}\\ ,\n\\end{align}\nwe obtain the matrix inversion lemma\n\\bmath\n (\\vect{X}+\\vect{UYV})^{-1} = \\vect{X}^{-1}-\\vect{X}^{-1}\\vect{U}(\\vect{Y}^{-1}+\\vect{VX}^{-1}\\vect{U})^{-1}\\vect{V}\\vect{X}^{-1}\\ .\n \\label{eq:woodbury}\n\\emath\nThe matrix inversion lemma is also sometimes called the \\textit{Woodbury matrix identity} or the \\textit{Woodbury's identity}.\n\n\\section{Method of Least-Squares}\nThe adaptive filtering problem is shown in Fig.~\\ref{fig:nonwss_block_diagram2}.\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/nonwss_block_diagram}\n \\caption{Block diagram of adaptive filtering in a non-WSS environment.}\n \\label{fig:nonwss_block_diagram2}\n\\end{figure}\nFrom the figure, we have that\n\\begin{itemize}\n \\item[$u(n)$:] zero-mean input signal\n \\item[$w_m(n)$:] $M$-tap FIR-filter with impulse response $w_0(n),w_1(n),\\ldots,w_{M-1}(n)$\n \\item[$y(n)$:] output signal given by $y(n) = \\sum_{m=0}^{M-1}w_m(n) u(n-m)$\n \\item[$d(n)$:] zero-mean desired signal\n \\item[$e(n)$:] error signal\n\\end{itemize}\nDefine\n\\begin{align}\n \\vect{w}(n) &= \\bbmtx w_0(n) & w_{1}(n) & \\cdots & w_{M-1}(n)\\ebmtx^T\\\\\n \\vect{u}(n)&= \\bbmtx u(n) & u(n-1) & \\cdots & u(n-M+1)\\ebmtx^T\\ .\n\\end{align}\nFor $i=1,2,\\ldots,n$ with $n\\geq M$, define\n\\begin{align}\n \\vect{A}(n) &= \\bbmtx \\vect{u}(1) & \\vect{u}(2) & \\cdots & \\vect{u}(n)\\ebmtx^T\\\\\n \\vect{d}(n) &= \\bbmtx d(1) & d(2) & \\cdots & d(n)\\ebmtx^T\\\\\n \\vect{e}(n) &= \\bbmtx e(1) & e(2) & \\cdots & e(n)\\ebmtx^T\\ .\n\\end{align}\nThen\n\\bmath\n \\vect{e}(n) = \\vect{d}(n)-\\vect{A}(n)\\vect{w}(n)\\quad\\iff\\quad\\vect{d}(n)=\\vect{A}(n)\\vect{w}(n)+\\vect{e}(n)\\ .\n\\emath\nNote that we have made $\\vect{A}(n)$, $\\vect{d}(n)$, and $\\vect{e}(n)$ time-dependent in order to emphasise that we are here concerned with an online algorithm. Therefore, we also formulate the squared error cost function $J_2(\\vect{w}(n))$ in a slightly different way compared to \\eq{eq:cf_squared_error}. Here, we define it as\n\\bmath\n J_2(\\vect{w}(n)) = \\sum_{i=1}^{n}e^2(i)=\\vect{e}^T(n)\\vect{e}(n)\\ .\n\\emath\n\nIn the method of least-squares, we wish to minimise $J_2(\\vect{w}(n))$ which we can write as\n\\begin{align}\n J_2(\\vect{w}(n)) &= \\vect{e}^T(n)\\vect{e}(n) = (\\vect{d}(n)-\\vect{A}(n)\\vect{w}(n))^T(\\vect{d}(n)-\\vect{A}(n)\\vect{w}(n))\\notag\\\\\n &= \\vect{d}^T(n)\\vect{d}(n)+\\vect{w}^T(n)\\vect{A}^T(n)\\vect{A}(n)\\vect{w}(n)-2\\vect{w}^T(n)\\vect{A}^T(n)\\vect{d}(n)\\ .\n\\end{align}\nThe minimiser\n\\bmath\n \\vect{w}_o(n) = (\\vect{A}^T(n)\\vect{A}(n))^{-1}\\vect{A}^T(n)\\vect{d}(n)\n\\emath\nis referred to as the least-squares solution, and it is the unique solution to the normal equations\n\\bmath\n \\vect{A}^T(n)\\vect{A}(n)\\vect{w}(n) = \\vect{A}^T(n)\\vect{d}(n)\\ ,\n\\emath\nprovided that $\\vect{A}(n)$ has full rank.\n\n\\subsection{Weighted Least-Squares}\nWhen the statistics of $u(n)$ and/or $d(n)$ is time dependent, the minimisation of the squared error $J_2(\\vect{w}(n))$ may fail to give a good estimate at time $n$ since all data affect the value of $J_2(\\vect{w}(n))$ with the same weight. Ideally, we would like that the new data are assigned a larger weight than the old data. In order to do this, we reformulate $J_2(\\vect{w}(n))$ as a weighted cost function\n\\bmath\n J_\\beta(\\vect{w}(n)) = \\sum_{i=1}^{n}\\beta(n,i) e^2(i) = \\vect{e}^T(n)\\vect{B}(n) \\vect{e}(n)\n\\emath\nwhere $\\beta(n,i)$ contains the weight pertaining to the $i$'th error at time $n$, and $\\vect{B}(n)$ is a diagonal matrix given by\n\\bmath\n \\vect{B}(n) = \\diag{\\beta(n,1), \\beta(n,2), \\ldots , \\beta(n,n)}\\ .\n\\emath\nWe use the five step recipe in Sec.~\\ref{ssec:optimisation} to minimise $J_\\beta(\\vect{w}(n))$ w.r.t $\\vect{w}(n)$.\n\\begin{enumerate}\n \\item Construct the cost function\n \\begin{align}\n J_\\beta(\\vect{w}(n)) &= \\vect{e}^T(n)\\vect{B}(n) \\vect{e}(n) = (\\vect{d}(n)-\\vect{A}(n)\\vect{w}(n))^T\\vect{B}(n)(\\vect{d}(n)-\\vect{A}(n)\\vect{w}(n))\\notag\\\\\n &= \\vect{d}(n)^T\\vect{B}(n)\\vect{d}(n)+\\vect{w}^T(n)\\vect{\\Phi}(n)\\vect{w}(n)-2\\vect{w}^T(n)\\vect{\\varphi}(n)\n \\end{align}\n where we have defined $\\vect{\\Phi}(n)$ and $\\vect{\\varphi}(n)$ as\n\\begin{align}\n \\vect{\\Phi}(n) &= \\vect{A}^T(n)\\vect{B}(n)\\vect{A}(n)\\\\\n \\vect{\\varphi}(n) &= \\vect{A}^T(n)\\vect{B}(n)\\vect{d}(n)\\ .\n\\end{align}\nWe refer to $\\vect{\\Phi}(n)$ and $\\vect{\\varphi}(n)$ as the correlation matrix and the cross-correlation vector, respectively, since they are scaled and weighted estimates of $\\vect{R}_u$ and $\\vect{r}_{ud}$.\n \\item Find the gradient\n \\bmath\n \\vect{g}(\\vect{w}(n)) = (\\vect{\\Phi}(n)+\\vect{\\Phi}^T(n))\\vect{w}(n)-2\\vect{\\varphi}(n)=2\\vect{\\Phi}(n)\\vect{w}(n)-2\\vect{\\varphi}(n)\n \\emath\n \\item Solve $\\vect{g}(\\vect{w}(n))=\\vect{0}$ for $\\vect{w}(n)$\n \\begin{alignat}{3}\n &&\\qquad \\vect{g}(\\vect{w}(n)) &= \\mathrlap{2\\vect{\\Phi}(n)\\vect{w}(n)-2\\vect{\\varphi}(n)}\\notag\\\\\n \\ArrowBetweenLines\n &&\\qquad \\alignedbox{\\vect{\\Phi}(n)\\vect{w}(n)}{=\\vect{\\varphi}(n)} &\\qquad &\\text{(Weighted Normal Equations)}\\\\\n \\ArrowBetweenLines\n &&\\qquad \\vect{w}_o(n) &= \\vect{\\Phi}^{-1}(n)\\vect{\\varphi}(n) &\\qquad &\\text{(If }\\vect{\\Phi}(n)\\text{ is invertible)}\n \\end{alignat}\n \\item Find the Hessian\n \\bmath\n \\vect{H}(\\vect{w}) = 2\\vect{\\Phi}(n)\n \\emath\n which is p.d. for all $\\vect{w}(n)$ if $\\vect{A}(n)$ has full rank and $\\beta(n,i)>0$ for all $n\\geq i > 0$.\n \\item This implies that\n \\begin{itemize}\n \\item $J_\\beta(\\vect{w}(n))$ is a convex function, and\n \\item $\\vect{w}_o(n) = \\vect{\\Phi}^{-1}(n)\\vect{\\varphi}(n)$ is the global minimiser.\n \\end{itemize}\n\\end{enumerate}\nThe solution $\\vect{w}_o(n)$ is often referred to as the weighted least-squares solution.\n\n\\subsubsection{Estimation of the (Cross-)Correlation}\nComparing the weighted normal equations with the Wiener-Hopf equations, we see that\n\\begin{align}\n \\hat{\\vect{R}}_u(n) &= c(n,\\beta)\\vect{\\Phi}(n) = c(n,\\beta)\\vect{A}^T(n)\\vect{B}(n)\\vect{A}(n) = c(n,\\beta)\\sum_{i=1}^{n}\\beta(n,i)\\vect{u}(i)\\vect{u}^T(i)\\label{eq:wls_Ru_est}\\\\\n \\hat{\\vect{r}}_{ud}(n) &= c(n,\\beta)\\vect{\\varphi}(n) = c(n,\\beta)\\vect{A}^T(n)\\vect{B}(n)\\vect{d}(n) = c(n,\\beta)\\sum_{i=1}^{n}\\beta(n,i)\\vect{u}(i)d(i)\\label{eq:wls_rud_est}\n\\end{align}\nare the estimates of the correlation matrix and the cross-correlation vector, respectively. The constant $c(n,\\beta)$ depends on $n$ and the weighting function $\\beta(n,i)$, and it can be selected such that $\\hat{\\vect{R}}_u(n)$ and $\\hat{\\vect{r}}_{ud}(n)$ are unbiased estimates of $\\vect{R}_u(n)$ and $\\vect{r}_{ud}(n)$.\n\n\\subsection{Weight Functions}\nWe consider three simple weight functions.\n\n\\subsubsection{Growing Window Weight Function}\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/growing_window}\n \\caption{The growing window weight function.}\n \\label{fig:growing_window}\n\\end{figure}\n\\noindent If we select the weight function as\n\\bmath\n \\beta(n,i) =\n \\begin{cases}\n 1 & 0 < i \\leq n\\\\\n 0 & \\text{otherwise}\n \\end{cases}\\ ,\n\\emath\nwe obtain the growing window weight function, and it is sketched in Fig.~\\ref{fig:growing_window}. Selecting the growing window weight function reduces the weighted least-squares problem to the standard least-squares problem. In order to obtain unbiased estimates of $\\vect{R}_u(n)$ and $\\vect{r}_{ud}(n)$ in \\eq{eq:wls_Ru_est} and \\eq{eq:wls_rud_est}, respectively, we have to use\n\\bmath\n c(n,\\beta) = \\frac{1}{n}\\ .\n\\emath\n\n\\subsubsection{Sliding Window Weight Function}\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/sliding_window}\n \\caption{The sliding window weight function.}\n \\label{fig:sliding_window}\n\\end{figure}\n\\noindent If we select the weight function as\n\\bmath\n \\beta(n,i) =\n \\begin{cases}\n 1 & n-L < i \\leq n\\\\\n 0 & \\text{otherwise}\n \\end{cases}\n\\emath\nfor $0 < L \\leq n$, we obtain the sliding window weight function, and it is sketched in Fig.~\\ref{fig:sliding_window}. If we select $L=n$, the sliding window weight function reduces to the growing window weight function. In order to obtain unbiased estimates of $\\vect{R}_u(n)$ and $\\vect{r}_{ud}(n)$ in \\eq{eq:wls_Ru_est} and \\eq{eq:wls_rud_est}, respectively, we have to use\n\\bmath\n c(n,\\beta) = \\frac{1}{L}\\ .\n\\emath\n\n\\subsubsection{Exponential Weight Function}\n\\begin{figure}[htbp]\n \\centering\n \\inputTikZ{figures/exponential_window}\n \\caption{The exponential weight function.}\n \\label{fig:exponential_window}\n\\end{figure}\n\\noindent If we select the weight function as\n\\bmath\n \\beta(n,i) =\n \\begin{cases}\n \\lambda^{n-i} & 0 < i \\leq n\\\\\n 0 & \\text{otherwise}\n \\end{cases}\n\\emath\nfor $0<\\lambda\\leq 1$, we obtain the exponential weight function, and it is sketched in Fig.~\\ref{fig:exponential_window}. The parameter $\\lambda$ is called \\textit{the forgetting factor}. If we select $\\lambda=1$, the exponential weight function reduces to the growing window weight function. In order to obtain unbiased estimates of $\\vect{R}_u(n)$ and $\\vect{r}_{ud}(n)$ in \\eq{eq:wls_Ru_est} and \\eq{eq:wls_rud_est}, respectively, we have to use\n\\bmath\n c(n,\\beta) = \n \\begin{cases}\n \\displaystyle\\frac{1-\\lambda}{1-\\lambda^n} & 0<\\lambda< 1\\\\\n \\displaystyle\\frac{1}{n} & \\lambda=1\n \\end{cases}\\ .\n\\emath\n\n\n\\section{The Recursive Least-Squares Algorithm with an Exponential Weight Function}\nIn an online algorithm, we have to solve the weighted normal equations\n\\bmath\n \\vect{\\Phi}(n)\\vect{w}(n) = \\vect{\\varphi}(n)\n\\emath\nfor $\\vect{w}(n)$ at every time index $n$. However, solving this equation directly as\n\\bmath\n \\vect{w}(n) = \\vect{\\Phi}^{-1}(n)\\vect{\\varphi}(n)\n\\emath\nyields a high computational complexity of the algorithm for the following two reasons.\n\\begin{enumerate}\n \\item The matrices $\\vect{A}(n)$ and $\\vect{B}(n)$, and the vector $\\vect{d}(n)$ grows with $n$. Computing $\\vect{\\Phi}(n)$ and $\\vect{\\varphi}(n)$ directly would therefore be infeasible for an online algorithm, unless we use a weight function with a finite duration window.\n \\item For an $M$-tap FIR filter, it requires in the order of $\\mathcal{O}(M^3)$ operations to solve the normal equations for the filter coefficient vector $\\vect{w}(n)$.\n\\end{enumerate}\nThe recursive least-squares (RLS) algorithm bypasses these two problems. Below, we consider how this is obtained for an exponential weight function, which is the most common weight function.\n\n\\subsubsection{Recursive Computation of $\\vect{\\Phi}(n)$ and $\\vect{\\varphi}(n)$}\nWe may compute the correlation matrix recursively by rewriting it as\n\\begin{align}\n \\vect{\\Phi}(n) &= \\vect{A}^T(n)\\vect{B}(n)\\vect{A}(n) = \\sum_{i=1}^{n}\\lambda^{n-i}\\vect{u}(i)\\vect{u}^T(i)\\\\\n &= \\lambda^0 \\vect{u}(n)\\vect{u}^T(n) + \\sum_{i=1}^{n-1}\\lambda^{n-i}\\vect{u}(i)\\vect{u}^T(i) = \\vect{u}(n)\\vect{u}^T(n) + \\lambda\\sum_{i=1}^{n-1}\\lambda^{n-1-i}\\vect{u}(i)\\vect{u}^T(i)\\notag\\\\\n &= \\vect{u}(n)\\vect{u}^T(n) + \\lambda\\vect{\\Phi}(n-1)\\ .\n \\label{eq:rls_corr_recurs}\n\\end{align}\nSimilarly, we may compute the cross-correlation vector recursively by rewriting it as\n\\begin{align}\n \\vect{\\varphi}(n) &= \\vect{A}^T(n)\\vect{B}(n)\\vect{d}(n) = \\sum_{i=1}^{n}\\lambda^{n-i}\\vect{u}(i)d(i)\\\\\n &= \\lambda^0 \\vect{u}(n)d(n) + \\sum_{i=1}^{n-1}\\lambda^{n-i}\\vect{u}(i)d(i) = \\vect{u}(n)d(n) + \\lambda\\sum_{i=1}^{n-1}\\lambda^{n-1-i}\\vect{u}(i)d(i)\\notag\\\\\n &= \\vect{u}(n)d(n) + \\lambda\\vect{\\varphi}(n-1)\\ .\n\\end{align}\nThese recursive formulations of the correlation matrix and the cross-correlation vector clearly reduce the computational complexity and are suitable to use in an online algorithm.\n\n\\subsubsection{Inversion of $\\vect{\\Phi}(n)$}\nFor the inversion of the correlation matrix, we use the matrix inversion lemma stated in Sec.~\\ref{sec:woodbury}. Comparing the recursive formulation of the correlation matrix in \\eq{eq:rls_corr_recurs} with the left side of \\eq{eq:woodbury}, we obtain\n\\begin{align}\n \\vect{X} &= \\lambda\\vect{\\Phi}(n-1)\\\\\n \\vect{U} &= \\vect{u}(n)\\\\\n \\vect{V} &= \\vect{u}^T(n)\\\\\n \\vect{Y} &= 1\\ .\n\\end{align}\nThus, invoking the matrix inversion lemma, we have that\n\\bmath\n \\vect{\\Phi}^{-1}(n) = \\lambda^{-1}\\vect{\\Phi}^{-1}(n-1)-\\lambda^{-2}\\frac{\\vect{\\Phi}^{-1}(n-1)\\vect{u}(n)\\vect{u}^T(n)\\vect{\\Phi}^{-1}(n-1)}{1+\\lambda^{-1}\\vect{u}^T(n)\\vect{\\Phi}^{-1}(n-1)\\vect{u}(n)}\\ .\n \\label{eq:rls_inv_phi}\n\\emath\nNote that the computational complexity of the right side of the equation is much lower than that of the left side of the equation when $\\vect{\\Phi}^{-1}(n-1)$ is known. In order to simplify the notation, we define\n\\begin{alignat}{2}\n \\vect{P}(n) &= \\vect{\\Phi}^{-1}(n)&\\quad&\\text{(inverse correlation matrix)}\\\\\n \\vect{k}(n) &= \\frac{\\vect{P}(n-1)\\vect{u}(n)}{\\lambda+\\vect{u}^T(n)\\vect{P}(n-1)\\vect{u}(n)}&\\quad&\\text{(gain vector)}\n\\end{alignat}\nwhich leads to that we may write \\eq{eq:rls_inv_phi} as\n\\bmath\n \\vect{P}(n) = \\lambda^{-1}\\left[\\vect{P}(n-1)-\\vect{k}(n)\\vect{u}^T(n)\\vect{P}(n-1)\\right]\\ .\n\\emath\nBy rearranging the expression for the gain vector, we obtain\n\\begin{align}\n \\vect{k}(n) &= \\lambda^{-1}\\left[\\vect{P}(n-1)-\\vect{k}(n)\\vect{u}^T(n)\\vect{P}(n-1)\\right] \\vect{u}(n)\\notag\\\\\n &= \\vect{P}(n)\\vect{u}(n)\\ .\n\\end{align}\n\n\\subsubsection{Recursive computation of $\\vect{w}(n)$}\nWe can now develop the recursive update equation of the filter coefficient vector. We have that\n\\begin{align}\n \\vect{w}(n) &= \\vect{P}(n)\\vect{\\varphi}(n)\\\\\n &= \\vect{P}(n)\\left[\\vect{u}(n)d(n) + \\lambda\\vect{\\varphi}(n-1)\\right]=\\lambda\\vect{P}(n)\\vect{P}^{-1}(n-1)\\vect{w}(n-1)+\\vect{k}(n)d(n)\\notag\\\\\n &= \\left[\\vect{P}(n-1)-\\vect{k}(n)\\vect{u}^T(n)\\vect{P}(n-1)\\right]\\vect{P}^{-1}(n-1)\\vect{w}(n-1)+\\vect{k}(n)d(n)\\notag\\\\\n &= \\vect{w}(n-1)-\\vect{k}(n)\\vect{u}^T(n)\\vect{w}(n-1)+\\vect{k}(n)d(n)\\notag\\\\\n &= \\vect{w}(n-1)+\\vect{k}(n)\\xi(n)\n\\end{align}\nwhere we have defined the a priori error as\n\\bmath\n \\xi(n) = d(n)-\\vect{u}^T(n)\\vect{w}(n-1)\\ .\n\\emath\n\n\\subsubsection{The RLS Algorithm}\nThe RLS algorithm may now be formulated as the following set of equations\n\\begin{align}\n \\vect{\\pi}(n) &= \\vect{P}(n-1)\\vect{u}(n)\\\\\n \\vect{k}(n) &= \\frac{\\vect{\\pi}(n)}{\\lambda+\\vect{u}^T(n)\\vect{\\pi}(n)}\\\\\n \\xi(n) &= d(n)-\\vect{u}^T(n)\\vect{w}(n-1)\\\\\n \\vect{w}(n) &= \\vect{w}(n-1)+\\vect{k}(n)\\xi(n)\\\\\n \\vect{P}(n) &= \\lambda^{-1}\\left[\\vect{P}(n-1)-\\vect{k}(n)\\vect{\\pi}^T(n)\\right]\\ .\n\\end{align}\nFor an $M$-tap FIR filter, it requires in the order of $\\mathcal{O}(M^2)$ operations to run one iteration of the RLS algorithm.\n\n\\subsubsection{Initialisation}\nIn order to start the RLS algorithm, we need to select values for the initial inverse correlation matrix $\\vect{P}(0)$, the initial filter coefficient vector $\\vect{w}(0)$, and the input samples $u(n)$ for $n=-M+1,-M+2,\\ldots,1$. Typically, we assume that\n\\begin{align}\n \\vect{P}(0) &= \\delta^{-1}\\vect{I}\\\\\n \\vect{w}(0) &= \\vect{0}\\\\\n u(n) & = 0\\ ,\\quad\\text{for } -M+1t+s|X>s] = \\mathbb{P}[X \\ge t] \\]\n \\begin{remark}\n If we have waited for $s$ units of time, then the probability that we have to wait for $t$ more units is the same as that of we have not waited at all.\\footnote{白~{}等~{}了}\n \\end{remark}\n\n \\subsubsection{Exponential Races}\\label{subs:ExponentialRace}\n Let $X_1 \\sim Exp(\\lambda_1)$, $X_2 \\sim Exp(\\lambda_2)$ be two independent random variables. Let $Y = \\min(X_1, X_2)$, then\n \\[ Y \\sim Exp(\\lambda_1 + \\lambda_2) \\]\n \\begin{proof}\n \\begin{align*}\n \\mathbb{P}[Y \\ge k] &= \\mathbb{P}[\\min\\{X_1,X_2\\} \\ge k]\\\\\n &= \\mathbb{P}[X_1 \\ge k, X_2 \\ge k]\\\\\n &= \\mathbb{P}[X_1 \\ge k]\\cdot \\mathbb{P}[X_2 \\ge k]\\\\\n &= e^{-(\\lambda_1 + \\lambda_2)k}\n \\end{align*}\n \\end{proof}\n\n If we consider the problem of ``Who finishes first between $X_1$ and $X_2$'', then\n \\[ \\mathbb{P}[Y=X_1]=\\sum_{s=0}^\\infty \\mathbb{P}[X_1=s,X_2>s] = \\int_0^{+\\infty}p_1(s)\\mathbb{P}[X_2>s]\\mathrm{d}s = \\frac{\\lambda_1}{\\lambda_1 + \\lambda_2} \\]\n And an exponential race among $n$ random variables is\n \\[ \\mathbb{P}[Y=X_i] = \\frac{\\lambda_i}{\\sum_j \\lambda_j} \\]\n\n\n\\section{Poisson Processes}\n\n \\subsection{Definition of Poisson Processes}\n \\begin{definition}[Poisson Process]\\label{def:PoissonProcess}\n $\\{N(s)|s \\ge 0\\}$ is a \\textbf{Poisson Process}, if\n \\begin{enumerate}\n \\item $N(0)=0$\n \\item $\\forall t, s \\ge 0 \\quad N(t+s) - N(s) \\sim Poisson(\\lambda t)$\n \\item $n$ has \\textbf{independent increments}: $\\forall t_0 \\le t_1 \\le \\dots \\le t_n$,\n \\[ N(t_1)-N(t_0), N(t_2)-N(t_1), \\dots, N(t_n)-N(t_{n-1}) \\]\n are \\textbf{mutually independent}.\n \\end{enumerate}\n \\end{definition}\n\n \\subsection{Alternative Interpretation of Poisson Process}\n We can construct a Poisson process as the sum of multiple random variables with Exponential distributions.\n \\begin{proposition}\\label{prop:AltdefOfPoissonProcess}\n Let $\\tau_1,\\tau_2,\\dots,\\tau_n$ be independent random variables with exponential distribution $Exp(\\lambda)$. Let $T_n=\\sum_{i=1}^n\\tau_i$, $N(s)=\\max\\{n|T_n \\le s\\}$\n\n Then $N(s)$ is a Poisson process with rate $\\lambda$.\n \\end{proposition}\n \\begin{remark}\n If we think of $\\tau_i$ as the time interval between the arrival time of customers in a store, then $N(s)$ is the number of customer arrivals before time $s$.\n \\end{remark}\n\n To prove Proposition \\ref{prop:AltdefOfPoissonProcess}, we first introduce a theorem.\n\n \\begin{theorem}\\label{thm:SumOfExpHasGammaDistribution}\n Let $\\tau_1, \\tau_2, \\dots, \\tau_n$ be independent random variables with Exponential distribution $Exp(\\lambda)$. Let $T_n = \\sum_{i=1}^n\\tau_n$, then $T_n$ has a \\emph{Gamma distribution} $\\Gamma(n, \\lambda)$\n \\[ f_{T_n}(t) = \\lambda e^{-\\lambda t}\\frac{(\\lambda t)^{n-1}}{(n-1)!} \\quad \\forall t \\ge 0 \\]\n \\end{theorem}\n \\begin{proof}\n Prove by induction on $n$.\n \\begin{itemize}\n \\item \\textbf{Base. } $n=1$. $f_{T_1}(t) = \\lambda e^{-\\lambda t}$ obviouly holds.\n \\item \\textbf{Hypothesis. } Suppose it holds for $n$.\n \\item \\textbf{Step. }\n \\begin{align*}\n f_{T_{n+1}}(t) &= \\int_0^t f_{T_n}(s)\\cdot\\lambda e^{-\\lambda(t-s)}\\mathrm{d}s\\\\\n &= \\int_0^t \\lambda e^{-\\lambda s}\\frac{(\\lambda s)^{n-1}}{(n-1)!}\\lambda e^{-\\lambda(t-s)}\\mathrm{d}s \\quad \\text{(Plug in the hypothesis)}\\\\\n &= \\frac{\\lambda^{n+1}}{(n-1)!}e^{-\\lambda t}\\int_0^t s^{n-1}\\mathrm{d}s \\quad \\text{(Move the constants out)}\\\\\n &= \\frac{\\lambda^{n+1}}{n!}e^{-\\lambda t}t^n\n \\end{align*}\n which is exactly $\\Gamma(n+1, \\lambda(n+1))$, so we are done.\n \\end{itemize}\n \\end{proof}\n \\begin{remark}\n Theorem \\ref{thm:SumOfExpHasGammaDistribution} states that the arrival time of the $n$-th customer has a Gamma distribution.\n \\end{remark}\n\n We can now prove Proposition \\ref{prop:AltdefOfPoissonProcess}.\n \\subsection{Proof of Equivalence of the two definitions}\n \\begin{proof}\n To show that the two definitions are equivalent, we need to show that the three requirements in definition \\ref{def:PoissonProcess} can be satisfied by the alternative definition \\ref{prop:AltdefOfPoissonProcess}.\n \\begin{enumerate}\n \\item $N(0)=0$ is trivial.\n \\item We begin proving (2) of definition \\ref{def:PoissonProcess} from the case $s=0$. Notice that $N(t)=n \\Leftrightarrow T_n \\le t \\wedge T_{n+1} > t$, i.e. the $n$-th customer has arrived, but the $(n+1)$-th has not. So we only need to prove the latter has Poisson distribution.\n \\begin{align*}\n \\mathbb{P}[N(t)=n] &= \\mathbb{P}[T_n \\le t, T_{n+1} > t]\\\\\n &= \\int_0^t\\mathbb{P}[T_n = s]\\cdot\\mathbb{P}[\\tau_{n+1} > t-s]\\mathrm{d}s \\quad \\text{(Enumerate all possible values of $t$)}\\\\\n &= \\int_0^t\\lambda e^{-\\lambda s}\\frac{(\\lambda s)^{n-1}}{(n-1)!}\\cdot e^{-\\lambda(t-s)}\\mathrm{d}s \\quad \\text{(Theorem \\ref{thm:SumOfExpHasGammaDistribution})}\\\\\n &= \\frac{\\lambda^n}{(n-1)!}e^{-\\lambda t}\\int_0^t s^{n-1}\\mathrm{d}s \\quad \\text{(Move out the constants)}\\\\\n &= \\frac{(t\\lambda)^n}{n!}\\cdot e^{-\\lambda t}\n \\end{align*}\n which is exactly the expression of $Poisson(\\lambda t)$, so we are done.\n\n When $s > 0$, by the lack of memory property, $N(t+s)-N(s)$ must have the same distribution as $N(t) - N(0) = N(t)$. Of course this can also be verified by some moderate calculations.\n \\item (3) of definitiion \\ref{def:PoissonProcess} can be proved using a similar argument. Notice that the lack-of-memory property and (2) implies that the ``number of arrivals'' after $s$ is independent of the arrivals before $s$. Therefore $N(t_n) - N(t_{n-1})$ is independent of $N(r)$ for all $r < t_{n-1}$. So the result can be proved by induction.\n \\end{enumerate}\n \\end{proof}\n\n\n\\section{Thinning}\n Thinning is associating another random variable with a Poisson Process.\n\n \\subsection{Thinning}\n We can associate some i.i.d. random variables $Y_i$ with each arrival. For example, $Y_i$ can be the gender of the $i$-th customer arriving at some restaurant; or if customers arrive in cars, $Y_i$ can be the number of passengers in a car. Let $P_j = \\mathbb{P}[Y_i = j]$. Let $N_j(t)$ be the total number of $i \\le N(t)$ with $Y_i = j$.\n\n \\begin{theorem}\\label{thm:ThinningOfPoissonProcess}\n $N_j(t)$ are \\emph{independent} Poisson Processes with rate $\\lambda P_j$.\n \\end{theorem}\n \\begin{proof}\n We will prove the result using the first definition of Poisson Distribution \\ref{def:PoissonProcess}. We only need to consider the simplest cases where $Y_i \\in \\{0,1\\}$.\n \\begin{align*}\n \\mathbb{P}[N_0(t) = j, N_1(t) = k] &= e^{-\\lambda t}\\frac{(\\lambda t)^{j+k}}{(j+k)!}\\cdot C^{j+k}_j P_0^j P_1^k \\\\\n &= e^{-\\lambda t}\\frac{(\\lambda t)^{j+k}}{(j+k)!}\\frac{(j+k)!}{j! \\cdot k!}P_0^j P_1^k\\\\\n &= e^{-P_0\\lambda t}\\frac{(P_0\\lambda t)^j}{j!} \\cdot e^{-P_1\\lambda t} \\frac{(P_1 \\lambda t)^k}{k!}\n \\end{align*}\n This is exactly the product of two Poisson Processes, with rate $P_0\\lambda$ and $P_1\\lambda$ respectively. Therefore we have completed our proof.\n \\end{proof}\n \\begin{remark}\n One tricky fact behind this theorem is that, suppose we are interested in the gender of customers arriving at a restaurant. Assume that the probability that the customer is male or female are equal. If one day 40 males came to the restaurant, this does not give any implication to the number of females coming to the restaurant because they are independent.\n \\end{remark}\n\n \\subsection{Examples}\n \\subsubsection{Finding typos in a book.} Two editors read a 300-page manuscript. Editor $A$ finds $100$ typos, Editor $B$ finds $120$ typos, and $80$ of these typos are the same. Suppose $A$ and $B$ have probability $P_A$ and $P_B$ of discovering a typo, and suppose the typos in the book is a rate $\\lambda$ Poisson Process. How can we estimate $\\lambda$, $P_A$ and $P_B$?\n \n We can associate a random variable with each typos appearing in the manuscript.\n \\begin{enumerate}\n \\item Neither $A$ or $B$ found it. $(1-P_A)(1-P_B)$\n \\item Only $A$ found it. $P_A(1-P_B)$\n \\item Only $B$ found it. $P_B(1-P_A)$\n \\item Both $A$ and $B$ found it. $P_AP_B$\n \\end{enumerate}\n From Theorem \\ref{thm:ThinningOfPoissonProcess}, we know that the four cases follow four independent Poisson Processes. So we can estimate the parameters by solving\n \\[\n \\begin{cases}\n 300P_A(1-P_B)\\lambda &= 20\\\\\n 300P_B(1-P_B)\\lambda &= 40\\\\\n \\lambda P_AP_B &= 80\n \\end{cases}\n \\quad \\Longrightarrow \\quad\n \\begin{cases}\n P_A &= 2/3\\\\\n P_B &= 4/5\\\\\n \\lambda &= 1/2\n \\end{cases}\n \\]\n \n \\subsubsection{Coupon Collector Once Again.}\n We revisit the Coupon Collector's problem, but this time we assume the probability of the $j$-th coupon is $P_j$.\n \\begin{itemize}\n \\item Let $N_j$ be the number of gachas before we first get a coupon of the $j$-th category.\n \\item Let $N = \\max_{1 \\le j \\le n}\\{N_j\\}$ is the total number of gachas we need to collect all categories of coupons.\n \\end{itemize}\n We want to find $\\mathbb{E}[N]$.\n\n We re-formulate the problem into the following: suppose we owns a restaurant, and the customers coming to the restaurant is a Poisson Process of rate $\\lambda = 1$. Each customer brings a coupon, let $Y_j$ be the category of the coupon brought by the $j$-th customer. Therefore $N_j(t)$ is a Poisson Process with rate $P_j$. Let $X_j$ be the random variable \\emph{in this Poisson Process setting} denoting the first time to meet a customer with a category $j$ coupon, and we now alternatively consider $X = \\max \\{X_j\\}$.\n\n We are interested in $\\mathbb{P}[X \\le t]$, i.e. the probability that we have got all categories of coupons before time $t$.\n \\begin{align*}\n \\mathbb{P}[X \\le t] &= \\mathbb{P}[X_1 \\le t, X_2 \\le t, \\dots X_n \\le t]\\\\\n &= \\mathbb{P}[X_1 \\le t] \\cdot \\mathbb{P}[X_2 \\le t] \\cdot \\cdots \\cdot \\mathbb{P}[X_n \\le t]\n \\end{align*}\n By the alternative definition of Poisson Process \\ref{prop:AltdefOfPoissonProcess}, $X_j$ has an Exponential Distribution with rate $P_j$. Therefore\n \\[ \\mathbb{P}[X \\le t] = \\prod_{i=1}^n (1-e^{-P_jt}) \\]\n \\[ \\mathbb{P}[X > t] = 1 - \\prod_{i=1}^n (1-e^{-P_jt}) \\]\n \\begin{proposition}\\label{prop:AltComputationOfExpectation}\n If $X \\ge 0$, then\n \\[ \\mathbb{E}[X] = \\sum_{t=0}^{\\infty} \\mathbb{P}[X \\ge t] = \\sum_{i=0}^{\\infty}i\\cdot\\mathbb{P}[X=i] \\]\n \\[ \\mathbb{E}[X] = \\int_0^{\\infty} \\mathbb{P}[X > t]\\mathrm{d}t\\]\n \\end{proposition}\n By Proposition \\ref{prop:AltComputationOfExpectation},\n \\[ \\mathbb{E}[X] = \\int_0^{\\infty} \\mathbb{P}[X > t]\\mathrm{d}t = \\int_0^{\\infty}1 - \\prod_i (1-e^{-P_it})\\mathrm{d}t \\]\n We have figured out $\\mathbb{E}[X]$, but how is $\\mathbb{E}[X]$ related to $\\mathbb{E}[N]$?\n\n Notice that $X$ denotes the exact time at which the customer carrying the $N$-th coupon arrives. If $\\tau_i$ denotes the time interval between the arrival of the $(i-1)$-th and the $i$-th customers, then\n \\[ X = \\sum_{i=1}^N \\tau_i \\]\n Take expectation on both sides\n \\[ \\mathbb{E}[X] = \\mathbb{E}\\left[ \\sum_{i=1}^N \\tau_i \\right] = \\mathbb{E}[N]\\cdot\\mathbb{E}[\\tau_1] = \\mathbb{E}[N] \\]\n Notice that $N$ is a random variable, but we somehow still exchanged the summation and expectation in the second equation, this holds because of Wald's Equation, which will be covered in the future. For now, notice that once $N$ is determined, $\\mathbb{E}[X|N]$ is determined.\n \\[ \\mathbb{E}[X|N] = N\\mathbb{E}[\\tau_1] = N \\]\n Take expectations on both sides again\n \\[ \\mathbb{E}[\\mathbb{E}[X|N]] = \\mathbb{E}[X] = \\mathbb{E}[N] \\]\n \\begin{remark}\n The original Coupon Collector's problem is a special case where $P_j$ are uniform. It can be proved\\footnote{“我也不会积，但是我用Mathematica跑了一下他们确实相等” -- Chihao} that the integral equals to the sum of harmonic series.\\footnote{“这个积分我会积” -- Kuan}\\footnote{“不愧是常州数学帝” -- Chihao}\n \\end{remark}\n\n \\subsection{Nonhomogeneous Thinning}\n Now consider another case. Suppose $P_i$ now depends on the arrival time $s$.\n \\begin{theorem}[Nonhomogeneous Thinning]\n \\label{thm:NonhomogeneousThinning}\n Let $Y_1,\\dots,Y_k$ be random variables, and let $p_1,\\dots,p_k$ be nonnegative functions of time $s$ such that for all $s$, $\\sum_{j=1}^k p_j(s)=1$. Assume $\\mathbb{P}[Y_i=j]=p_j(s)$ at $s$. Then $N_i(t)$'s have \\emph{independent} Poisson distribution with mean $\\lambda\\int_0^t P_i(s)\\mathrm{d}s$.\n \\end{theorem}\n \\begin{proof}\n Suppose there are $k$ categories. Suppose until time $t$, each type $i$ have total number $n_i$.\n \\begin{align*}\n &\\mathbb{P}[N_1(t)=n_1,\\dots,N_k(t)=n_k]\\\\\n &=\\mathbb{P}\\left[ N_1(t)=n_1,\\dots,N_k(t)=n_k|N(t)=\\sum_i n_i \\right]\\mathbb{P}\\left[ N(t) = \\sum_i n_i \\right] \\quad \\text{Always holds}\\\\\n &= \\frac{\\left(\\sum_{i=1}^kn_i\\right)!}{n_1!\\cdots n_k!}p_1^{n_1}\\cdots p_k^{n_k} \\cdot e^{-\\lambda t}\\frac{(\\lambda t)^{\\sum_in_i}}{(\\sum_i n_i)!}\\\\\n &= \\prod_{i=1}^k e^{-\\lambda t P_i} \\frac{(\\lambda t P_i)^{n_i}}{n_i!}\n \\end{align*}\n where\n \\[ P_i = \\frac{1}{t}\\int_0^t P_i(s)\\mathrm{d}s \\]\n \\end{proof}\n \\begin{remark}\n Actually it is a non-homogeneous Poisson Process with rate\n \\[ \\lambda(t) = \\lambda\\cdot P_i(t) \\]\n \\end{remark}\n\n \\begin{definition}[Nonhomogeneous Poisson Process]\\label{def:NonhomogeneousPoissonProcess}\n $\\{ N(t) | t \\ge 0 \\}$ is a \\textbf{nonhomogeneous Poisson Process} with rate $\\lambda(s)$ if\n \\begin{enumerate}\n \\item $N(0)=0$\n \\item $N(t)$ has independent increments.\n \\item $N(t)-N(s)$ has distribution $Poisson\\left(\\int_s^t \\lambda(\\tau)\\mathrm{d}\\tau\\right)$\n \\end{enumerate}\n \\end{definition}\n \\begin{remark}\n In this case, the time intervals are not exponential distributions, and they are not independent.\n \\end{remark}\n\n \\subsection{Applications of Nonhomogeneous Thinning}\n\n \\subsubsection{M/$G$/$\\infty$ Queue}\n Consider a queue at some counters in a bank. M stands for a Poisson Process representing the incoming flow of customers. $G$ is a distribution of the time to serve the customers, $G(t) = \\mathbb{P}[ServiceTime \\le t]$. $\\infty$ is the number of counters. Let\n \\begin{itemize}\n \\item $X(t)$: the number of customers completed services before time $t$.\n \\item $Y(t)$: the number of customers being served at time $t$.\n \\end{itemize}\n\n Consider the customers arriving before $t$, there are two types\n \\begin{itemize}\n \\item Those who have completed their services before $t$. The number is $X(t)$.\n \\item Those who are being served at $t$. The number is $Y(t)$.\n \\end{itemize}\n\n Notice that\n \\[ p_1(s) = G(t-s) \\quad p_2(s) = 1 - G(t-s) = \\bar{G}(t-s) \\]\n Therefore\n \\[ \\mathbb{E}[X(t)] = \\mathbb{E}[N_1(t)] = \\lambda\\int_0^t G(t-s)\\mathrm{d}s = \\lambda\\int_0^t G(s)\\mathrm{d}s \\]\n \\[ \\mathbb{E}[Y(t)] = \\mathbb{E}[N_2(t)] = \\lambda\\int_0^t \\bar{G}(s)\\mathrm{d}s \\]\n\n \\subsubsection{Cars in a Tunnel}\n Consider a tunnel\\footnote{“刚才Kuan提醒我隧道内不能超车” -- Chihao} of length $l$. Suppose we are driving a car, once the car enters the tunnel, its speed cannot change. Assume other NPC cars enter the tunnel with a Poisson distribution of rate $\\lambda$. Suppse the speeds of NPC cars have a distribution $G$, and are also fixed once an NPC car enters the tunnel.\n\n We want to choose an initial speed $x$ to minimize the time we 超车 or 被超车。\n\n Let $t_0 = l/x$. If we enter the tunnel at time $s$, then we must exit the tunnel at time $s + t_0$. For an NPC car with speed $X$, the time interval that the NPC car stays in the tunnel is $T=l/X$. Let $F$ be the distribution of $T$.\n\n \\[ F(t) = \\mathbb{P}[T < t] = \\mathbb{P}[l/X < t] = \\mathbb{P}[X > l/t] = \\bar{G}[l/t] \\]\n\n Let $t$ denote the time when an NPC car enters the tunnel. We divide NPC cars into $3$ categories:\n \\begin{enumerate}\n \\item 超车。$t > s$ and $t + T < s + t_0$\n \\item 被超车。$t < s$ and $t + T > s + t_0$\n \\item No thing happens. Omitted.\n \\end{enumerate}\n At any given time $t$, let $p$ be the probability of overtaking or being overtaken, we have\n \\[\n p(t) =\n \\begin{cases}\n \\mathbb{P}[t + T > s + t_0] = \\bar{F}(s+t_0-t) \\quad &(t < s)\\\\\n \\mathbb{P}[t + T < s + t_0] = \\bar{F}(s+t_0-t) \\quad &(s < t < s+t_0)\\\\\n 0 \\quad &(t > s + t_0)\n \\end{cases}\n \\]\n Notice that only $T$ in the above equation is a random variable.\n \\begin{align*}\n \\mathbb{E}[N_{encountered}(s+t_0)] &= \\lambda \\left( \\int_0^s \\bar{F}(s-t_0-t)\\mathrm{d}t + \\int_s^{s+t_0} F(s+t_0-t)\\mathrm{d}t \\right)\\\\\n &= \\lambda \\left( \\int_{t_0}^{s+t_0} \\bar{F}(t)\\mathrm{d}t + \\int_0^{t_0} F(t)\\mathrm{d}t \\right)\n \\end{align*}\n To maximize the expectation, we take the derivative w.r.t. $t_0$.\n \\[ \\frac{\\mathrm{d}}{\\mathrm{d}t_0} = \\lambda \\left( \\bar{F}(s+t_0) - \\bar{F}(t_0) + F(t_0) \\right) = 0 \\]\n If the system has been running for a long time, i.e. $s$ is sufficiently large, $\\bar{F}(s+t_0) \\approx 0$, so\n \\[ F(t_0) = \\bar{F}(t_0) = \\frac{1}{2} \\]\n and\n \\[ G\\left(\\frac{l}{t_0}\\right) = \\frac{1}{2} \\]\n So we choose an initial speed $x$ such that\n \\[ G(x) = \\frac{1}{2} \\]\n\n \\subsubsection{HIV Patients}\n We want to estimate the number of HIV patients, suppose a patient may or may not show symptoms. Assume the number of patients with HIV virus has a Poisson distribution with an unknown rate $\\lambda$. Suppose we know a distribution $G$ representing ``how long a patient with HIV virus shows symptoms''.\n\n We divide people into 2 types by whether they show symptoms or not. Suppose the patient is infected at time $s$ and we observe if they have shown symptoms at time $t$.\n \\begin{equation}\\label{eq:NumOfPatientsWithSymptoms} \\mathbb{E}[N_1(t)] = \\lambda \\int_0^t G(t-s)\\mathrm{d}s = \\lambda\\int_0^t G(s)\\mathrm{d}s \\end{equation}\n \\[ \\mathbb{E}[N_2(t)] = \\lambda \\int_0^t \\hat{G}(t-s)\\mathrm{d}s = \\lambda\\int_0^t \\hat{G}(s)\\mathrm{d}s \\]\n where $\\lambda$ is unknown.\n\n However, we do know the number of patients who have already shown symptoms. Therefore we can estimate $\\lambda$ with Eq (\\ref{eq:NumOfPatientsWithSymptoms}), and estimate $\\mathbb{E}[N_2(t)]$ with $\\hat{\\lambda}$.\n\n\\section{Conditioning}\n\n \\subsection{Conditioning}\n Let $T_1, T_2, \\dots, T_n$ be the arrival times of a Poisson Process with rate $\\lambda$. Let $U_1, U_2, \\dots, U_n$ be independent uniform random variables on $[0,t]$. Let $V_1, V_2, \\dots, V_n$ be sequence $U_i$ re-arranged in increasing order, then\n \\begin{theorem}[Conditioning]\\label{thm:ConditioningOfPoissonProcess}\n If we condition on $N(t)=n$, then the distribution of $T_1, \\dots, T_n$ is the same as the distribution of $V_1, \\dots, V_n$.\n \\end{theorem}\n \\begin{sketchproof}~{}\n \\begin{itemize}\n \\item The joint distribution of $T_1,\\dots,T_n$ given $N(t)=n$ is $n!/t^n$ (by brutal force calculation).\n \\item The resulting distribution is uniform over $[0,t]$ because the space has volume $t^n$ and $n!$ possible orderings.\n \\end{itemize}\n \\end{sketchproof}\n \\begin{remark}\n This property is useful when computing $\\mathbb{P}[N(s)=m|N(t)=n]$\n \\[ \\mathbb{P}[N(s)=m|N(t)=n] = C^n_m \\left(\\frac{s}{t}\\right)^m\\left(1-\\frac{s}{t}\\right)^{n-m} \\]\n \\end{remark}\n\n\n\\section{Poisson Approximation}\n \\subsection{Motivating Example: Max Load}\\label{subs:m-balls-in-n-bins}\n Consider a m-ball-into-n-bin problem. Suppose we throw balls randomly into several bins, and we want to know how many balls there are in the bin with the most balls.\n\n If we consider the number of balls in each bin, the number follows a Bernoulli distribution. However, since the number of balls in different bins are not independent, analyzing the problem in this way can be complicated.\n\n Alternatively, we can analyze the problem by constructing a Poisson Process.\n\n \\begin{itemize}\n \\item Let $X_i$ be the number of balls in the $i$-th bin.\n \\item $X_i \\sim Bernoulli(m, \\frac{1}{n})$.\n \\item $\\sum_i X_i = m$.\n \\item We are interested in $X = \\max X_i$.\n \\end{itemize}\n\n \\subsection{Poisson Approximation}\n \\begin{theorem}[Poisson Approximation]\\label{thm:PoissonApproximation}\n Let $(X_1,\\dots,X_n)$ be a sequence of random variables where each $X_i$ has a Bernoulli distribution, then its distribution is the same as $(Y_1, \\dots, Y_n)$ conditioned on $\\sum_i Y_i = m$, where $Y_i \\sim Poisson(\\lambda)$ are \\emph{independent} Poisson variables with rate $\\lambda$.\n \\end{theorem}\n \\begin{proof}\n Let $a_1, \\dots, a_n \\ge 0$ be $n$ positive integers s.t. $\\sum_i a_i = m$. We calculate $\\mathbb{P}[X_1 = a_1, \\dots, X_n = a_n]$ and $\\mathbb{P}[Y_1 = a_1, \\dots, Y_n = a_n|\\sum_i Y_i=m]$ to show that they are equal to each other.\n\n Consider throwing $m$ balls into these $n$ bins. There are $n^m$ different kind of throws and $m!$ different permutations of different balls. However, the order of balls in the same bin does not matter, so we shoud divide $m!$ by $a_1!\\cdot \\cdots \\cdot a_n!$.\n \\begin{align*}\n \\mathbb{P}[X_1 = a_1, \\dots, X_n = a_n]\n &= \\frac{1}{n^m} \\cdot \\frac{m!}{a_1! \\dots a_n!}\n \\end{align*}\n\n \\begin{align*}\n \\mathbb{P}[Y_1 = a_1, \\dots, Y_n = a_n | \\sum_iY_i = m]\n &= \\frac{\\mathbb{P}[Y_1=a_1,\\dots,Y_n=a_n]}{\\mathbb{P}[\\sum_i Y_i = m]}\\\\\n &= \\frac{\\prod_{i=1}^n \\mathbb{P}[Y_i = a_i]}{\\mathbb{P}[\\sum_i Y_i = m]} \\quad \\text{(Independence of $Y_i$)}\\\\\n &= \\frac{\\prod_{i=1}^n e^{-\\lambda}\\frac{\\lambda^{a_i}}{a_i!}}{e^{-\\lambda n} \\cdot \\frac{(\\lambda n)^m}{m!}} \\quad \\text{(Denominator is a sum of Poisson)}\\\\\n &= \\frac{1}{n^m} \\cdot \\frac{m!}{a_1! \\dots a_n!} \\quad \\text{(Mafs)}\n \\end{align*}\n\n \\end{proof}\n\n \\subsection{m-balls-in-n-bins: MaxLoad Revisited}\n With Theorem \\ref{thm:PoissonApproximation}, we can solve the motivating example in Section \\ref{subs:m-balls-in-n-bins}.\n \\begin{theorem}\n Let $m=n$. Let $X = \\max_i X_i$. Then there exist constants $c_1$, $c_2$ such that\n \\[ \\mathbb{P}\\left[\\frac{c_1\\log n}{\\log\\log n} < X < \\frac{c_2\\log n}{\\log\\log n}\\right] = 1 - O\\left(\\frac{1}{n}\\right) \\]\n \\end{theorem}\n \\begin{proof}\n We first proof the \\emph{upper bound}. Proving the upper bound is equivalent to proving\n \\[\\mathbb{P}\\left[ X \\ge \\frac{c_2\\log n}{\\log\\log n} \\right] = O\\left(\\frac{1}{n}\\right) \\]\n \\begin{align*}\n \\mathbb{P}\\left[ X \\ge \\frac{c_2\\log n}{\\log\\log n} \\right] &= \\mathbb{P}[\\exists i: X_i \\ge k]\\\\\n &\\le n\\mathbb{P}[X_1 \\ge k] \\quad \\text{(Union Bound)}\\\\\n &\\le n C^k_n\\left(\\frac{1}{n}\\right)^k\\\\\n &\\le n\\cdot \\left(\\frac{en}{k}\\right)^k n^{-k} \\quad \\text{(Mafs)}\\\\\n &= n\\cdot\\left(\\frac{e}{k}\\right)^k\n \\end{align*}\n Therefore it is sufficient to show that $n\\cdot e^k / k^k < 1/n^{1+\\epsilon}$ for some $\\epsilon > 0$. Take logarithm on both sides,\n \\[ (2+\\epsilon)\\log n + k - k\\log k < 0 \\]\n Hence $k\\log k > 3\\log n$ suffices. Plug in $k = \\frac{c_2\\log n}{\\log\\log n}$.\n \\begin{align*}\n k\\log k &= \\frac{c_2\\log n}{\\log\\log n} (\\log\\log n - \\log\\log\\log n + \\log c_2)\\\\\n &> c_2\\log n\\left(1- \\frac{\\log\\log\\log n}{\\log\\log n}\\right)\\\\\n &> \\frac{c_2}{2}\\log n\n \\end{align*}\n So we can choose $c_2 = 6$.\n \n The proof of \\emph{lower bound} requires Theorem \\ref{thm:BadEventBound}. Let $X < \\frac{c_1\\log n}{\\log\\log n}$ be a ``bad event''. This is equivalent to\n \\[ \\forall i, \\quad X_i < \\frac{c_1\\log n}{\\log\\log n} \\]\n Denote this bad event by $\\mathcal{B}$, let $f = \\mathbb{I}[\\mathcal{B}|X_1,\\dots,X_n]$ be an indicator of the bad event. By Theorem \\ref{thm:BadEventBound},\n \\begin{align*}\n \\mathbb{E}[f] &= \\mathbb{P}[\\mathcal{B}]\\\\\n &\\le e\\sqrt{n} \\mathbb{E}[\\mathbb{I}[\\mathcal{B}|(Y_1,\\dots,Y_n)]]\\\\\n &= e\\sqrt{n} \\mathbb{P}[\\forall i \\quad Y_i < \\frac{c_1\\log n}{\\log\\log n}] \\quad \\text{(Independence of $Y_i$)}\\\\\n &= e\\sqrt{n}\\prod_{i=1}^n \\mathbb{P}[Y_i < RHS]\\\\\n &= e\\sqrt{n}(1-\\mathbb{P}[Y_i \\ge RHS])^n\\\\\n &\\le e\\sqrt{n}(1-\\mathbb{P}[Y_i = RHS])^n\\\\\n &= e\\sqrt{n}\\left(1-\\frac{1}{ek!}\\right)^n\\\\\n &\\le e\\sqrt{n} e^{-\\frac{n}{ek!}} < \\frac{1}{n} \\quad \\text{(Mafs)}\n \\end{align*}\n \\end{proof}\n \\begin{theorem}\\label{thm:BadEventBound}\n Let $X_i$ be Bernoulli random variables, and let $Y_i$ be independent Poisson random variables with rate $\\lambda = m/n$. For any function $f$ from $\\mathbb{N}^n$ to $\\mathbb{N}$,\n \\[ \\mathbb{E}[f(X_1, \\dots, X_n)] \\le e\\sqrt{m}\\cdot\\mathbb{E}[f(Y_1,\\dots,Y_m)] \\]\n \\end{theorem}\n \\begin{proof}\n \\begin{align*}\n \\mathbb{E}[f(Y_1,\\dots,Y_n)] &= \\sum_{k=0}^{\\infty}\\mathbb{E}[f(Y_1,\\dots,Y_n) | \\sum_i Y_i = k] \\cdot \\mathbb{P}[\\sum_i Y_i = k] \\quad \\text{(Total Probability)}\\\\\n &\\ge \\mathbb{E}[f(Y_1,\\dots,Y_n)|\\sum Y_i = m]\\cdot \\mathbb{P}[\\sum Y_i = m]\\\\\n &= \\mathbb{E}[f(X_1,\\dots,X_m)]\\cdot\\mathbb{P}[\\sum Y_i = m] \\quad \\text{(thm \\ref{thm:PoissonApproximation})}\n \\end{align*}\n If we let each $Y_i$ has a Poisson distribution of rate $m/n$, by additivity of Poisson variables we have $\\sum Y_i \\sim Poisson(m)$, and then\n \\begin{align*}\n \\mathbb{E}[f(Y_1,\\dots,Y_n)] &= \\mathbb{E}[f(X_1,\\dots,X_n)]\\cdot e^{-m}\\frac{m^m}{m!}\\\\\n &\\ge \\mathbb{E}[f(X_1,\\dots,X_n)]\\cdot\\frac{1}{e\\sqrt{m}}\n \\end{align*}\n \\end{proof}\n \\begin{remark}~{}\n \\begin{itemize}\n \\item $f$ is usually chosen to be some indicator function $\\mathbb{I}$ of some ``bad event''. (In order to show that the probability that this ``bad event'' happens is very small.)\n \\item Choosing $\\lambda = m/n$ is to maximize $\\mathbb{P}[\\sum_i Y_i=m]$.\n \\end{itemize}\n \\end{remark}\n", "meta": {"hexsha": "2495c58a5f8a33606ec0b031d0c6933ae5e5b2b8", "size": 30990, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Stochastic Processes/PoisonProcess.tex", "max_stars_repo_name": "YBRua/CourseNotes", "max_stars_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 8, "max_stars_repo_stars_event_min_datetime": "2021-03-20T10:40:58.000Z", "max_stars_repo_stars_event_max_datetime": "2022-03-25T08:15:15.000Z", "max_issues_repo_path": "Stochastic Processes/PoisonProcess.tex", "max_issues_repo_name": "YBRua/CourseNotes", "max_issues_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Stochastic Processes/PoisonProcess.tex", "max_forks_repo_name": "YBRua/CourseNotes", "max_forks_repo_head_hexsha": "58a4ccb6b8f8d1de9ec10b627a45442519855dfc", "max_forks_repo_licenses": ["MIT"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2021-12-14T11:31:00.000Z", "max_forks_repo_forks_event_max_datetime": "2021-12-14T11:31:00.000Z", "avg_line_length": 66.5021459227, "max_line_length": 529, "alphanum_fraction": 0.5963536625, "num_tokens": 10165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506635289835, "lm_q2_score": 0.803173791645582, "lm_q1q2_score": 0.6109346775443014}} {"text": "\\newpage\n\\subsection{Goal}\\label{subsec:goal}\n\nThe use of stochastic and metaheuristic algorithms (Simulated Annealing, Differential Evolution, Particle Swarm Optimization) in the tasks of unconstrained nonlinear optimization and the experimental comparison of them with Nelder-Mead and Levenberg-Marquardt algorithms.\n\n\\subsection{Formulation of the problem}\\label{subsec:formulation-of-the-problem}\n\nGenerate the noisy data $(x_k, y_k)$, where $k = 0, \\dots, 1000$, according to the rule:\n\n\\begin{equation*}\ny = \\begin{cases}\n-100 + \\delta_k, & f(x_k) < -100, \\\\\nf(x_k) + \\delta_k & -100 \\leq f(x_k) \\leq 100, \\\\\n100 + \\delta_k & f(x_k) > 100,\n\\end{cases}\n\\end{equation*}\n\n\\begin{equation*}\n x_k = \\frac{3k}{1000},\n\\end{equation*}\n\n\\begin{equation*}\n f(x) = \\frac{1}{x^2 - 3x + 2},\n\\end{equation*}\n\nwhere $\\delta_x \\approx N(0, 1)$ are values of a random variable with standard normal distribution.\nApproximate the data by the rational function\n\n\\begin{equation*}\n F(x, a, b, c, d) = \\frac{ax + b}{x^2 + cx + d},\n\\end{equation*}\n\nby means of least squares through the numerical minimization of the following function:\n\n\\begin{equation*}\n D(a, b, c, d) = \\sum^{1000}_{k=0}(F(x_k, a, b, c, d) - y_k)^2.\n\\end{equation*}\n\nTo solve the minimization problem, use Nelder-Mead algorithm, Levenberg-Marquardt algorithm and at least two of the methods among Simulated Annealing, Differential Evolution and Particle Swarm Optimization.\nIf necessary, set the initial approximations and other parameters of the methods.\nUse $\\varepsilon = 0.001$ as the precision;\nat most 1000 iterations are allowed.\nVisualize the data and the approximants obtained in a single plot.\nAnalyze and compare the results obtained (in terms of number of iterations, precision, number of function evaluations, etc.).\n\n\\subsection{Brief theoretical part}\\label{subsec:brief-theoretical-part}\n\nOptimization methods are numerical methods for finding optimal (in some sense) values of objective functions, for example, in the framework of mathematical models of certain processes.\n\n\\textit{Stochastic algorithms} are a broad class of algorithms that rely on repeated random sampling to solve an optimization problem.\nThese methods are useful when it is impossible or difficult to apply others.\n\n\\textit{Metaheuristic algorithms} are algorithms inspired by natural phenomena that solve the optimization problem by trial and error.\nMetaheuristic methods do not guarantee that a solution to the optimization problem will be found.\n\n\\paragraph{Nelder-Mead.}\n\nIf the function $f(x)$ is a ravine function, the efficiency of the simplex method in solving problem:\n\\begin{equation*}\n \\min_{x \\in R^n} f(x) = f(x*) = f*\n\\end{equation*}\nis significantly reduced due to the fact that the regular simplex cannot be \"extended\" along the ravine.\nThe Nelder-Mead method (the deformable polyhedron method) is a development of the simplex method and uses the deformation of the current simplex (not necessarily regular) in the search process.\nThe method uses the following operations on simplexes:\n\n\\begin{itemize}\n \\item reflection;\n \\item reduction;\n \\item compression;\n \\item stretching.\n\\end{itemize}\n\n\\paragraph{Levenberg-Marquardt algorithm}\n\nThe \\textit{Levenberg-Marquardt algorithm} (\\textit{LMA}) is a popular trust region algorithm that is used to find a minimum of a function (either linear or nonlinear) over a space of parameters.\nEssentially, a trusted region of the objective function is internally modeled with some function such as a quadratic.\nWhen an adequate fit is found, the trust region is expanded.\nAs with many numerical techniques, the Levenberg-Marquardt method can be sensitive to the initial starting parameters.\n\n\\paragraph{Simulated annealing}\n\\textit{Simulated annealing} is a random-search technique which exploits an analogy between the way in which a metal cools and freezes into a minimum energy crystalline structure (the annealing process) and the search for a minimum in a more general system;\nit forms the basis of an optimisation technique for combinatorial and other problems.\n\n\\paragraph{Differential evolution}\n\\textit{Differential evolution} is one of the most influential optimization algorithms up-to-date.\nDifferential evolution works through analogous computational steps as used by a standard evolutionary algorithm.\n\n\\subsection{Results}\\label{subsec:results}\n\nIn the case of a \\textit{rational approximation}, significant nonlinearities occur, so the result depends on the initial approximations (Figure~\\ref{ris:plot}).\n\n\\begin{figure}[H]\n \\center\n \\includegraphics[width=\\textwidth]{img/plot.png}\n \\caption{Stochastic and metaheuristic algorithms.}\n \\label{ris:plot}\n\\end{figure}\n\nFrom the table below (Table,~\\ref{tbl:stohmeta}) it is evident that methods \\textit{Nelder-Mead} and \\textit{Levenberg-Marquardt} require an order of magnitude less number of computations.\nHowever, despite the fact that metaheuristic methods do not guarantee finding the result, they allow you to get a solution of fairly good quality in an acceptable time without knowing the search space.\n\nNote that both metaheuristic methods produced extremely similar results for different numbers of iterations.\n\n\\begin{table}[ht]\n\\caption{Stochastic and metaheuristic algorithms}\n\\begin{tabular}{l|l|c|c|}\n\\cline{2-4}\n & \\textbf{method} & \\multicolumn{1}{l|}{\\textbf{iterations}} & \\multicolumn{1}{l|}{\\textbf{function eval}} \\\\ \\hline\n\\multicolumn{1}{|l|}{\\multirow{4}{*}{\\textbf{rational}}} & \\textit{Nelder-Mead} & 480 & 795 \\\\ \\cline{2-4}\n\\multicolumn{1}{|l|}{} & \\textit{Levenberg-Marquardt} & nan & 3 \\\\ \\cline{2-4}\n\\multicolumn{1}{|l|}{} & \\textit{Simulated Annealing} & 1000 & 9111 \\\\ \\cline{2-4}\n\\multicolumn{1}{|l|}{} & \\textit{Differential Evolution} & 149 & 9105 \\\\ \\hline\n\\end{tabular}\n\\label{tbl:stohmeta}\n\\end{table}\n\n\\paragraph{Levenberg-Marquardt algorithm gives poor results}\n\n\\begin{theorem}\n \\label{theorem:conv_newton}\n For doubly continuously differentiable functions (i.e.\\ for $f \\in C^2(D)$) with a non-degenerate matrix $\\nabla^2 f(y^*)$, there exists a $\\varepsilon$-neighborhood of the stationary point $y^*$ of the function $f(y)$ such that for any initial point $y^0$ from this neighborhood, the Newton method will converge superlinearly, and if the Lipschitz condition is met in this neighborhood for Hesse matrices, it will converge quadratically.\n\\end{theorem}\n\nSince the Levenberg-Marquardt algorithm is a hybrid of gradient descent and the Newton method, it therefore has the disadvantages of the Newton method.\nBased on the theorem above [1], we can conclude that if the Hesse matrix is not degenerate, but not sign-positive, then the Newton method can converge to a stationary point that is not a minimum point, but a maximum point or a saddle point.\nAccordingly, a more careful choice of initial approximations is necessary.\n\n\\subsection{Conclusion}\\label{subsec:conclusion}\n\nIn the course of the laboratory work, stochastic and metaheuristic algorithms were implemented and analyzed within the problem of the unconstrained optimization problem.\nThis paper also provides a comparative analysis of stochastic and metaheuristic algorithms.\n\n\\subsection{References}\\label{subsec:references}\n\\begin{enumerate}[label={[\\arabic*]}]\n \\item С.Ю. Городецкий, Лабораторный практикум по методам локальной оптимизации в программной системе LocOpt, 2007.\n\\end{enumerate}\n\n\\subsection{Appendix}\\label{subsec:appendix}\n\nThe source code is located \\href{https://github.com/vanSultan/anal_dev_algo/tree/lab_04}{here}: \\url{https://github.com/vanSultan/anal_dev_algo/tree/lab_04}.\n", "meta": {"hexsha": "969b254a9bf96e6af179673e619ff4012f34bf58", "size": 8120, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "lab_04/report/lab_04_body.tex", "max_stars_repo_name": "vanSultan/anal_dev_algo", "max_stars_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "lab_04/report/lab_04_body.tex", "max_issues_repo_name": "vanSultan/anal_dev_algo", "max_issues_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "lab_04/report/lab_04_body.tex", "max_forks_repo_name": "vanSultan/anal_dev_algo", "max_forks_repo_head_hexsha": "e9d6382103080e6f885b1456cc0a3ce64fbe1863", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 57.5886524823, "max_line_length": 442, "alphanum_fraction": 0.7235221675, "num_tokens": 2006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772884, "lm_q2_score": 0.8031737940012418, "lm_q1q2_score": 0.6109346706203386}} {"text": "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% Written By Michael Brodskiy\n% Class: Analytic Geometry & Calculus III (Math-292)\n% Professor: V. Cherkassky\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\documentclass[12pt]{article} \n\\usepackage{alphalph}\n\\usepackage[utf8]{inputenc}\n\\usepackage[russian,english]{babel}\n\\usepackage{titling}\n\\usepackage{amsmath}\n\\usepackage{graphicx}\n\\usepackage{enumitem}\n\\usepackage{amssymb}\n\\usepackage[super]{nth}\n\\usepackage{everysel}\n\\usepackage{ragged2e}\n\\usepackage{geometry}\n\\usepackage{fancyhdr}\n\\geometry{top=1.0in,bottom=1.0in,left=1.0in,right=1.0in}\n\\newcommand{\\subtitle}[1]{%\n \\posttitle{%\n \\par\\end{center}\n \\begin{center}\\large#1\\end{center}\n \\vskip0.5em}%\n\n}\n\\usepackage{hyperref}\n\\hypersetup{\ncolorlinks=true,\nlinkcolor=blue,\nfilecolor=magenta, \nurlcolor=blue,\ncitecolor=blue,\n}\n\n\\urlstyle{same}\n\n\n\\title{Lecture XII Notes}\n\\date{July 2, 2020}\n\\author{Michael Brodskiy\\\\ \\small Professor: V. Cherkassky}\n\n% Mathematical Operations:\n\n% Sum: $$\\sum_{n=a}^{b} f(x) $$\n% Integral: $$\\int_{lower}^{upper} f(x) dx$$\n% Limit: $$\\lim_{x\\to\\infty} f(x)$$\n\n\\begin{document}\n\n\\maketitle\n\n\\section{The Multivariable Chain Rule $-$ 14.5}\n\nIn three dimensions, the formula for the chain rule looks as follows:\n\n$$\\frac{dz}{dt}=\\frac{\\partial f}{\\partial x}\\frac{dx}{dt} + \\frac{\\partial f}{\\partial y}\\frac{dy}{dt}$$\n\nConditions $z=f(x,y); x=x(t), y=y(t)$\\\\\n\n\\textit{Example:}\n\n$$z=x^2y+3xy^4; x=\\sin2t,y=\\cos t$$\n$$\\frac{dz}{dt}=(2xy+3y^4)(2\\cos2t)+(x^2+12xy^3)(-\\sin t)$$\n\nAnother case looks as follows:\n\nIf $z=f(x,y); x=g(s,t), y=h(s,t)$, then:\n\n$$\\frac{\\partial z}{\\partial s}=\\frac{\\partial z}{\\partial x}\\frac{\\partial x}{\\partial s} + \\frac{\\partial z}{\\partial y}\\frac{\\partial y}{\\partial s}$$\n\n$$\\frac{\\partial z}{\\partial t}=\\frac{\\partial z}{\\partial x}\\frac{\\partial x}{\\partial t} + \\frac{\\partial z}{\\partial y}\\frac{\\partial y}{\\partial t}$$\n\nThe general case of the chain rule, where $u$ is a multivariable function of $n$ variables $x_1$, $x_2$,\\dots $x_n$, and each $x$ is a function of $m$ variables $t_1$,\\dots$t_m$, then:\n\n$$\\frac{\\partial u}{\\partial t_i}=\\frac{\\partial u}{\\partial x_1}\\frac{\\partial x_1}{\\partial t_i}+\\frac{\\partial u}{\\partial x_2}\\frac{\\partial x_2}{\\partial t_i}+\\dots+\\frac{\\partial u}{\\partial x_n}\\frac{\\partial x_n}{\\partial t_i}$$\n\n\\subsection{Implicit Differentiation}\n\nIf $\\frac{\\partial f}{\\partial y}\\neq 0$:\n\n$$\\frac{dy}{dx}=-\\frac{\\frac{\\partial f}{\\partial x}}{\\frac{\\partial f}{\\partial y}}$$\n\nIn addition to this:\n\n$$\\frac{\\partial z}{\\partial x}=-\\frac{\\frac{\\partial f}{\\partial x}}{\\frac{\\partial f}{\\partial z}}$$\n\nand\n\n$$\\frac{\\partial z}{\\partial y}=-\\frac{\\frac{\\partial f}{\\partial y}}{\\frac{\\partial f}{\\partial z}}$$\n\n\\section{The Directional Derivative $-$ 14.6}\n\nThe definition of a directional derivative is:\n\n$$ D_{\\hat{u}}=\\lim_{h\\to0} \\frac{f(x_o+ha, y_o + hb)-f(x_o,y_o)}{h} $$\n\nIn more simplified terms:\n\n$$D_{\\hat{u}}=\\frac{\\partial f}{\\partial x} a + \\frac{\\partial f}{\\partial y} b$$\n\n\\subsection{The Gradient Vector}\n\nThe gradient vector is written $\\overrightarrow{\\nabla} f$, and:\n\n$$\\overrightarrow{\\nabla} f = \\frac{\\partial f}{\\partial x}\\hat{\\bold{i}} + \\frac{\\partial f}{\\partial y}\\hat{\\bold{j}}$$\n\nIn the case that $u = f(x,y,z)$:\n\n$$\\overrightarrow{\\nabla} f = \\frac{\\partial f}{\\partial x}\\hat{\\bold{i}} + \\frac{\\partial f}{\\partial y}\\hat{\\bold{j}}+\\frac{\\partial f}{\\partial z}\\hat{\\bold{k}}$$\n\nand\n\n$$D_{\\hat{u}}=\\overrightarrow{\\nabla}f(x,y,z)\\cdot\\hat{u}$$\n\nThe greatest value of the gradient, written as $|\\overrightarrow{\\nabla}f|$, occurs in the same direction for $\\hat{\\bold{u}}$ and $\\overrightarrow{\\nabla}f$\n\n\\end{document}\n", "meta": {"hexsha": "23dd87ab83c283cbce39d6e490bc14e59c0c1dcf", "size": 3969, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Lecture Notes/Lecture12.tex", "max_stars_repo_name": "MDBrodskiy/Vector_Calculus", "max_stars_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 1, "max_stars_repo_stars_event_min_datetime": "2021-01-15T15:51:52.000Z", "max_stars_repo_stars_event_max_datetime": "2021-01-15T15:51:52.000Z", "max_issues_repo_path": "Lecture Notes/Lecture12.tex", "max_issues_repo_name": "MDBrodskiy/Vector_Calculus", "max_issues_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Lecture Notes/Lecture12.tex", "max_forks_repo_name": "MDBrodskiy/Vector_Calculus", "max_forks_repo_head_hexsha": "d4820f31c0c585ae65e6d61249d8c725077005eb", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 32.8016528926, "max_line_length": 236, "alphanum_fraction": 0.6069538927, "num_tokens": 1288, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7606506526772883, "lm_q2_score": 0.8031737869342624, "lm_q1q2_score": 0.610934665244836}} {"text": "\\documentclass{amsart}\n\n\\usepackage{amssymb}\n\n\\def\\vec#1{\\mathchoice{\\mbox{\\boldmath$\\displaystyle#1$}}\n\t{\\mbox{\\boldmath$\\textstyle#1$}}\n\t{\\mbox{\\boldmath$\\scriptstyle#1$}}\n\t{\\mbox{\\boldmath$\\scriptscriptstyle#1$}}}\n\\newcommand{\\floor}[1]{\\left\\lfloor #1\\right\\rfloor}\n\\newcommand{\\ceil}[1]{\\left\\lceil #1\\right\\rceil}\n\\newcommand{\\F}{\\mathbb{F}}\n\\newcommand{\\N}{\\mathbb{N}}\n\n\\newcommand{\\Input}{\\textbf{Input:}}\n\\newcommand{\\Output}{\\textbf{Output:}}\n\n\\newcommand{\\field}{\\normalfont\\texttt{gf256}}\n\\newcommand{\\transforms}{\\normalfont\\texttt{transforms}}\n\\newcommand{\\hermite}{\\normalfont\\texttt{hermite}}\n\\newcommand{\\encode}{\\normalfont\\texttt{encode}}\n\n\\newcommand{\\XtoN}{LCHToNewton}\n\\newcommand{\\NtoX}{NewtonToLCH}\n\\newcommand{\\XtoL}{LCHEval}\n\\newcommand{\\LtoX}{LCHInterp}\n\\newcommand{\\lsXtoL}{LowSpaceLCHEval}\n\\newcommand{\\lsLtoX}{LowSpaceLCHInterp}\n\\newcommand{\\XtoM}{LCHToMonomial}\n\\newcommand{\\MtoX}{MonomialToLCH}\n\\newcommand{\\Taylor}{TaylorExpansion}\n\\newcommand{\\InverseTaylor}{InverseTaylorExpansion}\n\n\\makeatletter\n\\newcommand*{\\bdiv}{%\n\t\\nonscript\\mskip-\\medmuskip\\mkern5mu%\n\t\\mathbin{\\operator@font div}\\penalty900\\mkern5mu%\n\t\\nonscript\\mskip-\\medmuskip\n}\n\\makeatother\n\n\\let\\originalleft\\left\n\\let\\originalright\\right\n\\renewcommand{\\left}{\\mathopen{}\\mathclose\\bgroup\\originalleft}\n\\renewcommand{\\right}{\\aftergroup\\egroup\\originalright}\n\n\\newenvironment{inputs}{\\Input\\par\\begin{tabular}{@{}rcl}}{\\end{tabular}}\n\\newenvironment{outputs}{\\Output\\par}{}\n\n\\setlength{\\parindent}{0pt}\n\\setlength{\\parskip}{1ex plus 0.5ex minus 0ex}\n\n\\begin{document}\n\t\n\\title{\\lowercase{ftoc2}: Fast transforms over characteristic 2}\n\\maketitle\n\n\\section{\\field{}: Implementation of arithmetic in $\\F_{2^8}$}\n\nThe module \\field{} implements field arithmetic for $\\F_{2^8}$. We let\n$\\F_{2^8}=\\F_2(\\alpha)$ for a root $\\alpha$ of $x^8+x^4+x^3+x^2+1$, and\n$\\{\\beta_0,\\dotsc,\\beta_7\\}$ be the basis of the extension obtained by letting\n$\\beta_7=\\alpha^5$ and $\\beta_i=\\beta^2_{i+1}-\\beta_{i+1}$ for\n$i\\in\\{0,\\dotsc,6\\}$. This construction yields a so-called Cantor basis (see\n\\cite[Appendix~A]{gao2010}). Define maps $[{}\\cdot{}]_k:\\N\\rightarrow\\{0,1\\}$\nfor $k\\in\\N$ by\n\\begin{equation*}\n\ti=\\sum_{k\\in\\N}2^k[i]_k\n\t\\quad\\text{for $i\\in\\N$}.\n\\end{equation*}\nThe field $\\F_{2^8}$ is enumerated as $\\{\\omega_0,\\dotsc,\\omega_{255}\\}$, with\n\\begin{equation*}\n\t\\omega_i=\\sum^7_{k=0}[i]_k\\beta_k\n\t\\quad\\text{for $i\\in\\{0,\\dotsc,255\\}$}.\n\\end{equation*}\nThen $\\F_{2^d}=\\{\\omega_0,\\dotsc,\\omega_{2^d-1}\\}$ for $d\\in\\{1,2,4,8\\}$.\n\nAn element $\\omega_i\\in\\F_{2^8}$ is represented by the byte\n$[i]_7[i]_6\\dotsb[i]_0$. Thus, addition of field elements corresponds to\ncomputing the XOR of bytes. Logarithm and exponent tables are used for\nmultiplication. The array $\\texttt{LOG}$ contains logarithms base $\\alpha$:\n$\\texttt{LOG}[i]$ is the byte that corresponds to $\\log_\\alpha\\omega_i$ for\n$i\\in\\{1,\\dotsc,255\\}$, and $\\texttt{LOG}[0]$ is arbitrarily set to zero.\nSimilarly, $\\texttt{EXP}[i]$ corresponds to $\\alpha^i$ for\n$i\\in\\{0,\\dotsc,254\\}$. \n\n\\subsection{Functions}\\\n\n\\texttt{unsigned char mul(unsigned char x, unsigned char y)}\n\n\\begin{inputs}\n\t\\texttt{x} & - & field element, \\\\\n\t\\texttt{y} & - & field element.\n\\end{inputs}\n\n\\begin{outputs}\n\tThe product of the two elements $x,y\\in\\F_{2^8}$ that correspond to the inputs.\n\\end{outputs}\n\n\\section{\\transforms{}: Transforms over $\\F_{2^8}$}\n\nThe module \\transforms{} implements several of the fast transforms proposed\nin~\\cite{gao2010,lin2014,lin2016a,coxon2018b}. Define polynomials\n\\begin{equation*}\n\tN_i\n\t=\n\t\\prod^{i-1}_{j=0}\n\t\\left(x-\\omega_j\\right)\n\t\\quad\\text{and}\\quad\n\tX_i\n\t=\n\t\\prod^7_{k=0}\n\t\\prod^{2^k[i]_k-1}_{j=0}\n\t\\left(x-\\omega_j\\right)\n\\end{equation*}\nfor $i\\in\\{0,\\dotsc,255\\}$. The polynomials $N_i$ are the Newton polynomials\nassociated with the enumeration $\\{\\omega_0,\\dotsc,\\omega_{255}\\}$ of\n$\\F_{2^8}$. The polynomials $X_i$ were introduced by Lin, Chung and Han\n\\cite{lin2014}, and referred to hereafter as the LCH basis polynomials.\n\n\\subsection{Functions}\\\n\n\\texttt{unsigned int log2(unsigned int n)}\n\nCeiling of base two logarithm.\n\n\\begin{inputs}\n\t\\texttt{n} & - & an integer.\n\\end{inputs}\n\n\\begin{outputs}\n\tThe ceiling of $\\log_2 n$.\n\\end{outputs}\n\n\\texttt{void \\NtoX(int l, unsigned char[:]\\ a)}\n\nNewton basis to LCH basis conversion.\n\n\\begin{inputs}\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat\n\t\\begin{equation}\\label{eqn:newton-lch}\n\t\t\\sum^{\\ell-1}_{i=0}f_iN_i\n\t\t=\\sum^{\\ell-1}_{i=0}h_iX_i.\n\t\\end{equation}\n\\end{outputs}\n\n\\texttt{void \\XtoN(int l, unsigned char[:]\\ a)}\n\nLCH basis to Newton basis conversion.\n\n\\begin{inputs}\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat \\eqref{eqn:newton-lch} holds.\n\\end{outputs}\n\n\\texttt{void \\LtoX(unsigned char shift, int c, int l, unsigned char[:]\\ a)}\n\nInterpolation on the LCH basis.\n\n\\begin{inputs}\n\t\\texttt{shift} & - & shift parameter, \\\\\n\t\\texttt{c} & - & number of evaluations provided, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array of length $2^{\\ceil{\\log_2\\ell}}$.\n\\end{inputs}\n\n\\begin{outputs}\n\tGiven $\\texttt{shift}=\\lambda\\in\\F_{2^8}$, $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for\n\t$i\\in\\{0,\\dotsc,c-1\\}$, and $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for\n\t$i\\in\\{c,\\dotsc,\\ell-1\\}$, the function terminates with\n\t$\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t\\begin{equation}\\label{eqn:lagrange-lch}\n\t\t\\sum^{\\ell-1}_{i=0}\n\t\th_i\n\t\tX_i\\left(\\omega_j+\\lambda\\right)\n\t\t=\n\t\tf_j\n\t\t\\quad\n\t\t\\text{for $j\\in\\{0,\\dotsc,c-1\\}$}.\n\t\\end{equation}\n\\end{outputs} \n\n\\texttt{void \\XtoL(unsigned char shift, int c, int l, unsigned char[:]\\ a)}\n\nEvaluation on the LCH basis.\n\n\\begin{inputs}\n\t\\texttt{shift} & - & shift parameter, \\\\\n\t\\texttt{c} & - & number of evaluations to compute, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array of length\n\t$\\max(\\ell,2^{\\ceil{\\log_2c}})$.\n\\end{inputs}\n\n\\begin{outputs}\n\tGiven $\\texttt{shift}=\\lambda\\in\\F_{2^8}$ and $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for\n\t$i\\in\\{0,\\dotsc,\\ell-1\\}$, the function terminates with\n\t$\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t\\eqref{eqn:lagrange-lch} holds.\n\\end{outputs}\n\n\\texttt{void \\lsXtoL(unsigned char shift, int c, int l, unsigned char[:]~a)}\n\nIn-place evaluation on the LCH basis.\n\n\\begin{inputs}\n\t\\texttt{shift} & - & shift parameter, \\\\\n\t\\texttt{c} & - & number of evaluations to compute, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tGiven $\\texttt{shift}=\\lambda\\in\\F_{2^8}$, $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for\n\t$i\\in\\{0,\\dotsc,\\ell-1\\}$, and $\\texttt{a}[i]\\in\\F_{2^8}$ for\n\t$i\\in\\{c,\\dotsc,\\ell-1\\}$ (i.e, their value does not matter), the function terminates with\n\t$\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t\\eqref{eqn:lagrange-lch} holds, while $\\texttt{a}[c],\\dotsc,\\texttt{a}[\\ell-1]$\n\tretain their initial values if $c<\\ell$.\n\\end{outputs}\n\n\\texttt{void \\lsLtoX(unsigned char shift, int c, int l, unsigned char[:]~a)}\n\nIn-place interpolation on the LCH basis.\n\n\\begin{inputs}\n\t\\texttt{shift} & - & shift parameter, \\\\\n\t\\texttt{c} & - & number of evaluations provided, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tGiven $\\texttt{shift}=\\lambda\\in\\F_{2^8}$, $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for\n\t$i\\in\\{0,\\dotsc,c-1\\}$, and $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for\n\t$i\\in\\{c,\\dotsc,\\ell-1\\}$, the function terminates with\n\t$\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t\\eqref{eqn:lagrange-lch} holds, while $\\texttt{a}[c],\\dotsc,\\texttt{a}[\\ell-1]$ \n\tretain their initial values.\n\\end{outputs} \n\n\\texttt{void \\Taylor(int t, int l, unsigned char[:]\\ a)}\n\nGeneralised Taylor expansion at $x^{2^t}-x$.\n\n\\begin{inputs}\n\t\\texttt{t} & - & exponent, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat\n\t\\begin{equation}\\label{eqn:taylor}\n\t\t\\sum^{\\ell-1}_{i=0}f_ix^i\n\t\t=\n\t\t\\sum^{\\ell-1}_{i=0}\n\t\th_i\n\t\tx^{i-2^t\\floor{i/2^t}}\n\t\t\\left(x^{2^t}-x\\right)^{\\floor{i/2^t}}.\n\t\\end{equation}\n\\end{outputs}\n\n\\texttt{void InverseTaylorExpansion(int t, int l, unsigned char[:]\\ a)}\n\nInverse generalised Taylor expansion at $x^{2^t}-x$.\n\n\\begin{inputs}\n\t\\texttt{t} & - & exponent, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat \\eqref{eqn:taylor} holds.\n\\end{outputs}\n\n\\texttt{void \\MtoX(int l, unsigned char[:]\\ a)}\n\nMonomial basis to LCH basis conversion.\n\n\\begin{inputs}\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat $\\sum^{\\ell-1}_{i=0}f_ix^i=\\sum^{\\ell-1}_{i=0}h_iX_i$.\n\\end{outputs}\n\n\\texttt{void \\XtoM(int l, unsigned char[:]\\ a)}\n\nLCH basis to monomial basis conversion.\n\n\\begin{inputs}\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$ such\n\tthat $\\sum^{\\ell-1}_{i=0}h_iX_i=\\sum^{\\ell-1}_{i=0}f_ix^i$.\n\\end{outputs}\n\n\\texttt{void TaylorExpansionHighCoeffs(int t, int c, int l, unsigned char[:]\\\n\ta)}\n\nHigh coefficients of the generalised Taylor expansion at $x^{2^t}-x$.\n\n\\begin{inputs}\n\t\\texttt{t} & - & exponent, \\\\\n\t\\texttt{c} & - & cut-off, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for ${i\\in\\{c,\\dotsc,\\ell-1\\}}$,\n\tthen on termination $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{c,\\dotsc,\\ell-1\\}$\n\tsuch that\n\t\\begin{equation}\\label{eqn:taylor}\n\t\t\\sum^{\\ell-1}_{i=c}f_ix^i\n\t\t=\n\t\t\\sum^{\\ell-1}_{i=0}\n\t\th_i\n\t\tx^{i-2^t\\floor{i/2^t}}\n\t\t\\left(x^{2^t}-x\\right)^{\\floor{i/2^t}},\n\t\\end{equation}\n\tfor some $h_0,\\dots,h_{c-1}\\in\\F_{2^8}$, while\n\t$\\texttt{a}[0],\\dotsc,\\texttt{a}[c-1]$ retain their initial values.\n\\end{outputs}\n\n\\texttt{void \\MtoX HighCoeffs(int c, int l, unsigned char[:]\\ a)}\n\nMonomial basis to LCH basis conversion.\n\n\\begin{inputs}\n\t\\texttt{c} & - & cut-off, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for ${i\\in\\{c,\\dotsc,\\ell-1\\}}$,\n\tthen on termination $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{c,\\dotsc,\\ell-1\\}$\n\tsuch that\n\t$\\sum^{\\ell-1}_{i=c}f_ix^i=\\sum^{\\ell-1}_{i=0}h_iX_i$ for some\n\t$h_0,\\dots,h_{c-1}\\in\\F_{2^8}$, while $\\texttt{a}[0],\\dotsc,\\texttt{a}[c-1]$\n\tretain their initial values.\n\\end{outputs}\n\n\\section{\\hermite{}: Hermite interpolation and evaluation over $\\F_{2^8}$}\n\nThe module \\hermite{} is an implementation of the Hermite interpolation and\nevaluation algorithms proposed in \\cite{coxon2018}, specialised to $\\F_{2^8}$\nand its subfields. These algorithms reduce instances of the Hermite problems to\ninstances of regular (i.e., without the presence of derivatives) interpolation\nand evaluation problems. We implement these reduction, and use the \\transforms{}\nmodule to solve the problems they admit. The algorithms used in the module are\nmore general than their counterparts in~\\cite{coxon2018}, since they include an\nadditional parameter $\\ell$ that accounts for polynomial length.\n\nFor $i\\in\\N$, let $D^i:\\F_{2^8}[x]\\rightarrow\\F_{2^8}[x]$ be the map that sends\n$f\\in\\F_{2^8}[x]$ to the coefficient of $y^i$ in $f(x+y)\\in\\F_{2^8}[x][y]$,\ncalled the $i$th Hasse derivative on $\\F_{2^8}[x]$.\n\n\\subsection{Functions}\\\n\n\\texttt{void Evaluate(int c, int l, unsigned char[:]\\ a)}\n\nAlgorithm 1.\n\n\\begin{inputs}\n\t\\texttt{c} & - & number of evaluations to compute, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination $\\texttt{a}[i]=f(\\omega_i)\\in\\F_{2^8}$ for\n\t$f=\\sum^{\\ell-1}_{i=0}f_ix^i$ and $i\\in\\{0,\\dotsc,c-1\\}$.\n\\end{outputs}\n\n\\texttt{void Interpolate(int c, int l, unsigned char[:]\\ a)}\n\nAlgorithm 3.\n\n\\begin{inputs}\n\t\\texttt{c} & - & number of evaluations provided, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$, and\n\t$\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{c,\\dotsc,\\ell-1\\}$, then on\n\ttermination $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t$f=\\sum^{\\ell-1}_{i=0}f_ix^i$ satisfies $f(\\omega_i)=h_i$ for\n\t$i\\in\\{0,\\dotsc,c-1\\}$.\n\\end{outputs}\n\n\\texttt{void PrepareLeft(int d, int n, int c, int l, unsigned char[:]\\ a)}\n\nThe function \\textsf{PrepareLeft} from Algorithm 2.\n\n\\texttt{void PrepareRight(int d, int n, int c, int l, unsigned char[:]\\ a)}\n\nThe function \\textsf{PrepareRight} from Algorithm 4.\n\n\\texttt{void HermiteEvaluate(int d, int n, int c, int l, unsigned char[:]\\ a)}\n\nAlgorithm 2.\n\n\\begin{inputs}\n\t\\texttt{d} & - & field degree, \\\\\n\t\\texttt{c} & - & number of evaluations to compute, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array of length\n\t$2^{\\lceil\\log_2\\lceil\\ell/2^d\\rceil\\rceil+d}$.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,\\ell-1\\}$, then\n\ton termination\n\t$\\texttt{a}[i]=(D^{i\\bdiv{2^d}}f)(\\omega_{i\\bmod{2^d}})\\in\\F_{2^8}$ for\n\t$f=\\sum^{\\ell-1}_{i=0}f_ix^i$ and $i\\in\\{0,\\dotsc,c-1\\}$.\n\\end{outputs}\n\n\\texttt{void HermiteInterpolate(int d, int n, int c, int l, unsigned char[:]\\\n\ta)}\n\nAlgorithm 4.\n\n\\begin{inputs}\n\t\\texttt{d} & - & field degree, \\\\\n\t\\texttt{c} & - & number of evaluations provided, \\\\\n\t\\texttt{l} & - & length of the polynomial, \\\\\n\t\\texttt{a} & - & coefficient array of length\n\t$2^{\\lceil\\log_2\\lceil\\ell/2^d\\rceil\\rceil+d}$.\n\\end{inputs}\n\n\\begin{outputs}\n\tIf initially $\\texttt{a}[i]=h_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$, and\n\t$\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{c,\\dotsc,\\ell-1\\}$, then on\n\ttermination $\\texttt{a}[i]=f_i\\in\\F_{2^8}$ for $i\\in\\{0,\\dotsc,c-1\\}$ such that\n\t$f=\\sum^{\\ell-1}_{i=0}f_ix^i$ satisfies\n\t$(D^{i\\bdiv{2^d}}f)(\\omega_{i\\bmod{2^d}})=h_i$ for $i\\in\\{0,\\dotsc,c-1\\}$.\n\\end{outputs}\n\n\\section{\\encode{}: Systematic encoding of multiplicity codes over $\\F_{2^8}$}\n\nThe module \\encode{} is an implementation of the systematic encoding algorithm\nfor multiplicity codes proposed in \\cite{coxon2017}, specialised to codes over\n$\\F_{2^8}$ and its subfields. To encode, the algorithm solves a multivariate\nHermite interpolation problem, followed by a multivariate Hermite evaluation\nproblem. Both these problems are solved by algorithms that reduce the\nmultivariate problems to instances of the univariate problems. The module\n\\hermite{} is used to solve the univariate problems. The reduction also relies\non the use of a multivariate Newton-like basis. The module \\transforms{} is used to\nperform conversions between this basis and the monomial basis.\n\nThe ring of polynomials over $\\F_q$ in indeterminates $x_1,\\dotsc,x_m$ is\ndenoted by $\\F_q[\\vec{x}]=\\F_q[x_1,\\dotsc,x_m]$. Define\n$\\vec{x}^{\\vec{i}}=x^{i_1}_1\\dotsm x^{i_m}_m$ for\n$\\vec{i}=(i_1,\\dotsc,i_m)\\in\\N^m$. For $\\vec{i}\\in\\N^m$, let\n$D^{\\vec{i}}:\\F_q[\\vec{x}]\\rightarrow\\F_q[\\vec{x}]$ be the map that sends\n$f\\in\\F_q[\\vec{x}]$ to the coefficient of $\\vec{y}^{\\vec{i}}$ in\n$f(\\vec{x}+\\vec{y})\\in\\F_q[\\vec{x}][\\vec{y}]$, called the $\\vec{i}$th Hasse\nderivative on $\\F_q[\\vec{x}]$. Let $\\F_q[\\vec{x}]_\\ell$ denote the space of\npolynomials in $\\F_q[\\vec{x}]$ of total degree less than $\\ell$. The weight of a\nvector $\\vec{i}\\in\\N^m$, denoted $\\left|\\vec{i}\\right|$, is defined to be the\nsum of its entries. Define\n$S_{m,s}=\\{\\vec{s}\\in\\N^m\\mid\\left|\\vec{s}\\right|j}^{}\\frac{q\\textsubscript{i}}{(j-i)\\textsuperscript{2}}$ \\bigg)\n\\\\\nF\\textsubscript{j} $=$ C q\\textsubscript{j} c\\textsubscript{j}\n\\\\\\\\\n\nWhere, \\\\\nc\\textsubscript{k} $=$ $\\sum\\limits_{ij}^{}\\frac{q\\textsubscript{i}}{(j-i)\\textsuperscript{2}}$ for all k $\\in$ [1,n] \\hfill(1) \n\\\\\\\\\n\nConsider the two polynomials, \\\\\nA(x) $=$ q\\textsubscript{1}x + q\\textsubscript{2}x\\textsuperscript{2} + . . . + q\\textsubscript{n}x\\textsuperscript{n}\\\\\n\nB(x) $=$ $ - \\frac{1}{(n-1)\\textsuperscript{2}}x - . . . - \\frac{1}{4}x\\textsuperscript{n-2} - x\\textsuperscript{n-1} - 0$.$x\\textsuperscript{n} + x\\textsuperscript{n+1} + \\frac{1}{4}x\\textsuperscript{n+2} + . . . + \\frac{1}{(n-1)\\textsuperscript{2}}x\\textsuperscript{2n-1} $\n\\\\\n\nConsider the product, P(x) = A(x)$*$B(x). \n\\\\Co-efficient of x\\textsuperscript{k} for k in range (n,2n+1) $=$ $\\sum\\limits_{l = 1}^n a\\textsubscript{l}$b\\textsubscript{k-l} \\hfill(2)\\\\\\\\\n\nFrom equation (1) and equation (2), it is observable that coefficient of $x^{n+j}$ in P(x) = $c_j$. So, to obtain $c_j$, we need P(x) which can be calculated using Fast Fourier Transform discussed in the lectures. The time complexity here is $O(nlogn)$.\n\\\\\\\\\nNow, using c\\textsubscript{j} and equation (1), we can obtain F\\textsubscript{j} for all the particles $(j \\in [1,n])$ within $O(nlogn)$ time complexity.\n\\\\\n\n\n{\\centering\n\\begin{minipage}{\\linewidth}\n\\begin{algorithm}[H]\n\\caption{Particle Interaction Algorithm}\\label{pia}\n\\begin{algorithmic}[1]\n\\Procedure{ParticleInteraction}{}\n\\State{$A$ $\\gets$ $[q\\textsubscript{1}, q\\textsubscript{2}, ... , q\\textsubscript{n}]$}\n\\State{$B$ $\\gets$ $[ -$ $\\frac{1}{(n-1)\\textsuperscript{2}}$, - $\\frac{1}{(n-2)\\textsuperscript{2}}$, $-1, 0, 1,$ $\\frac{1}{(n-2)\\textsuperscript{2}}$, $\\frac{1}{(n-1)\\textsuperscript{2}}$]}\n\n\\State{$P \\gets$ MULTIPLY(A, B)}\n\\State{$TotalForce \\gets$ Array of size n}\n\\For{$i \\gets 1$ to $n$}\n \\State{$TotalForce[i] \\gets$ $C * A[i] * P[n+i]$}\n\\EndFor\n\\EndProcedure\n\\Function{Multiply}{$A, B$}\n \\State{$dft_A \\gets DFT(A)$}\n \\State{$dft_B \\gets DFT(B)$}\n \\State{$dft_P \\gets$ Array of size 2n+1}\n \\For{$i \\gets 1$ to $len(dft_A)$}\n \\State{$dft_P[i] \\gets$ $dft_A[i] * dft_B[i]$}\n \\EndFor\n \\State{$P \\gets$ $InvDFT(dft_P)$}\n \\State\\Return{$P$}\n\\EndFunction\n\\end{algorithmic}\n\\end{algorithm}\n\\end{minipage}\n\\par}\n\n\\textbf{Runtime Analysis:}\\\\\nThe above algorithm returns $TotalForce$ array which has $j^{th}$ element as the final totalforce acting on $j^{th}$ particle. The \\textsc{Multiply} function invloves calculating DFT for 2 vectors, each has time complexity of O(nlogn) as discussed in the lecture. \\\\\\\\\nNext we have a for loop which has O(n) time complexity. Further, there is InverseDFT which again takes O(2nlog2n). Next, after the \\textsc{Multiply} function we calculate the $TotalForce$ using elements of P and A. This for loop takes O(2n) time complexity. So,\n\\begin{center}\n T(n) = 2*O(nlogn) + O(n) + O(2nlog2n) + O(2n) = O(nlogn)\n\\end{center}\n\n %\\begin{algorithm}[H] %\n\n%\\end{algorithm}%\n\\newpage\n\\section{Distance Computation using Matrix Multiplication}\n\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item \\textbf{\nProve that the graph H = (V, E\\textsubscript{H}) can be computed from G in O(n\\textsuperscript{$\\omega$}) time, where $\\omega$ is the exponent of matrix-multiplication.}\n\\\\\\\\\nSolution: Using the property of transitive closure(discussed in lecture notes). Assume A is the adjacency matrix of graph G. Then, (A\\textsuperscript{k})\\textsubscript{ij} $>$ 0 iff there is a walk of length exactly k from i to j. Here edges exist between vertices in H, if there is a walk of length 2 or 1 in graph G. (A\\textsuperscript{2}) gives the adjacency matrix of edges who had walks of length of 2 in G. So, Adjacency matrix of H is given by (A\\textsuperscript{2}) + A. Matrix Addition is O(n\\textsuperscript{2}) and Matrix Multiplication is O(n\\textsuperscript{$\\omega$}). So, overall time complexity is O(n\\textsuperscript{$\\omega$}) if $\\omega > 2$\\\\\n\n\\item \\textbf{\nArgue that for any x, y $\\in$ V , D\\textsubscript{H}(x, y) = $\\ceil{ \\textbf{D\\textsubscript{G}(x, y)/2} }$. }\n\\\\\\\\\nSolution: Assume for some vertice x,y the distance in graph G is k. The Path is (x,$a_1$, $a_2$, $a_3$ ........ $a_(k-1)$, y). As all the vertices who have walk of distance 2 between them get edges in the graph H, x and $a_2$, $a_2$ and $a_4$.... all have edges between them. So the path between x and y becomes (x, $a_2$, $a_4$ ........ $a_(k-2)$, y) if k is even and (x, $a_2$, $a_4$ ........ $a_(k-1)$, y) if k is odd. The distance is k/2 if k is even and k/2 + 1 if k is odd.\n$\\implies $ D\\textsubscript{H}(x, y) = $\\ceil{ \\textbf{D\\textsubscript{G}(x, y)/2} }$\nProved. \n\n\n\\item \\textbf{\nLet A\\textsubscript{G} be adjacency matrix of G, and M $=$ D\\textsubscript{H}$\\star$A\\textsubscript{G}. Prove that for any x, y $\\in$ V, the following holds.\\\\\\\\\n$D\\textsubscript{G}(x, y) = \n \\begin{cases} \n 2D\\textsubscript{H}(x, y) & M(x, y) $ \\geqslant$ degree\\textsubscript{G}(y) · D\\textsubscript{H}(x, y) \\\\\n 2D\\textsubscript{H}(x, y) - 1 & M(x, y) $ \\lneq$ degree\\textsubscript{G}(y) · D\\textsubscript{H}(x, y) \\\\\n \\end{cases}\n$}\n\\\\\n\\item \\textbf{\nUse (c) to argue that D\\textsubscript{G} is computable from D\\textsubscript{H} in O(n\\textsuperscript{$\\omega$}) time.}\n\\\\\nFrom (c) we have \n$D\\textsubscript{G}(x, y) = \n \\begin{cases} \n 2D\\textsubscript{H}(x, y) & M(x, y) $ \\geqslant$ degree\\textsubscript{G}(y) · D\\textsubscript{H}(x, y) \\\\\n 2D\\textsubscript{H}(x, y) - 1 & M(x, y) $ \\lneq$ degree\\textsubscript{G}(y) · D\\textsubscript{H}(x, y) \\\\\n \\end{cases}\n$\nwhere A\\textsubscript{G} is the adjacency matrix of G and M = D\\textsubscript{H}$\\star$A\\textsubscript{G}.\\\\\nAs, $D_G$ is computable using $D_H$ which can be computed in O(n\\textsuperscript{$\\omega$}) time. M can be computed in O(n\\textsuperscript{$\\omega$}) time too. So, the overall time complexity is O(n\\textsuperscript{$\\omega$}).\n\n\\item \\textbf{\nProve that all-pairs-distances in n-vertex unweighted undirected graph can be computed in O(n\\textsuperscript{$\\omega$}log n) time, if $\\omega$ is larger than two.}\n\\\\\n\n{\\centering\n\\begin{minipage}{\\linewidth}\n\\begin{algorithm}[H]\n\\caption{Pair Distance}\\label{pia}\n\\begin{algorithmic}[1]\n\\Procedure{shortestdistance}{$A_G$}\n\n\\State \\If{$A_G$ elements are non-empty}\n\\State \\Return $A_G$\n\\State \\Else\n\\State $A_H$ = ($A_G$\\textsuperscript{2}) + $A_G$\n\\State $D_H$ = SHORTESTDISTANCE($A_H$)\n\\State \\Return dpart($D_H$)\n\\EndIf\n\\EndProcedure\n\n\\end{algorithmic}\n\\end{algorithm}\n\\end{minipage}\n\\par}\n\n\\textbf{Runtime Analysis:}\\\\\nThis is a recursive algorithm, where T(n) = T(n/2) + O(n\\textsuperscript{$\\omega$}), as each time part d is called, the graph's distance halves. This when solved, the time complexity becomes O(n\\textsuperscript{$\\omega$}log n) time\n\n\\textbf{Correctness of Algorithm:}\\\\\nAs, $D_H$ gives $D_G$ using computation done in part D, using that we can say that, if $D_H$ is correct, then $D_G$ calculated will also be correct. In the base case, when the all the vertices are connected, that itself is returned. So, the base case is correct. In the induction step, when until n paths are calculated, the next step would give the correct distance too. Therefor, by induction, proved.\n\n\\end{enumerate}\n\\newpage\n\\section{Hashing}\n\n\\begin{enumerate}[label=(\\alph*)]\n\\item The approximate probability can be calculated by:\n\n\\begin{center}\n $P &= {\\frac{n*n^{n - \\log_2 n}}{n^n}} * {n \\choose \\log_2 n}$ \n\\end{center}\nIn the denominator, we have $n^n$ because, n numbers are being assigned and each number has n possible positions that it can go into.\\\\\nIn the numerator, we have n, as the maximum chain, could be any one of the n positions in the hash table. and $n\\textsuperscript{(n-logn)}$ because $n-logn$ numbers can be assigned to any position and the rest is assigned to the single position. It is also multiplied with n choose ($logn$) because from the n numbers, logn numbers are choosen to be used in the max chain. This final calculation is actually an approximation which is greater than the actual probability, because the expression above actually recounts the same cases multiple times due to the $n\\textsuperscript{(n-logn)}$ factor. And below, we prove that this bigger approximation is lesser than 1/n and thus the probability is also lesser than 1/n.\n\\begin{center}\n $P &= {\\frac{n}{n^\\log_2 n}} * {\\frac{n!}{(\\log_2 n)!(n-\\log_2 n)!}}$\n\\end{center}\n\n\\begin{center}\n $P &= \\frac{n}{(\\log_2 n)!} * {\\frac{n(n-1)(n-2)....(n-\\log_2 n+1)}{n^{\\log_2 n}}}$\n\\end{center}\n\n\\begin{center}\n $P \\leq \\frac{n}{(\\log_2 n)!}$\n\\end{center}\n\n\\begin{center}\n Using Sterling Approximation,\n\\end{center}\n \n\\begin{center}\n $\\log(k!) = k*log(k) - k$\n\\end{center} \n\n\\begin{center}\n So, $\\log(\\log_2 n!) = (\\log_2 n)(\\log (n/e))$\n\\end{center}\n \n\\begin{center}\n $ \\frac{\\log(\\log_2 n!)}{\\log n} = \\frac{(\\log_2 n)(\\log (n/e))}{\\log n}$\n\\end{center}\n\n\\begin{center}\n $ \\frac{\\log(\\log_2 n!)}{\\log n} = \\frac{(\\log (n/e))}{\\log 2}$\n\\end{center}\n\n\\begin{center}\n For large n, \n\\end{center}\n\n\\begin{center}\n $\\frac{(\\log (n/e))}{\\log 2} > 2$ \n\\end{center}\n\n\\begin{center}\n $\\frac{\\log(\\log_2 n!)}{\\log n} > 2$\n\\end{center}\n\n\\begin{center}\n $\\log(\\log_2 n!) > 2*\\log n$\n\\end{center}\n\n\\begin{center}\n $\\log(\\log_2 n!/n^2) > 1$\n\\end{center}\n\n\\begin{center}\n $\\log(\\log_2 n!/n^2) > 1$\n\\end{center}\n\n\\begin{center}\n So, $\\frac{n}{(\\log_2 n)!} < \\frac{1}{n}$\n\\end{center}\n\\item Prove that for any given r $\\in$ [1,p-1], there exists at least $M/n \\choose n$ subsets of U of size n in which maximum chain length in hash-table corresponding to $H_r(x)$ is $\\Theta(n)$.\\\\\n\\textbf{Solution:}\\\\\nAs x is invertible with rx mod p with the given constraints(given in lecture), rx mod p $\\in$ [0,M-1].\\\\\nTake y = rx mod p, then $H_r(x) = H(y)$. \\\\\nThe aim is to have a set S such that, when hashed using H(y), all the numbers go to the same spot and the maximum-chain length is n, which is $\\Theta(n)$.\\\\\nAssume S = $\\{a_1, a_2..... a_n \\}$, which implies that, $a_1$ mod n = $a_2$ mod n, $a_2$ mod n = $a_3$ mod n ..... \\\\\n$\\implies a_1-a_2$ mod n = 0 ....\\\\\n$\\implies a_1 - a_2 = cn$ where c is an integer.\\\\\n$\\implies a_1 = b + c_1n$ where b (fixed for a specific set) $\\in$ [0,n-1] and $c_1$ $\\in$ [0, $(M-1)/n$] as $a_1$ $\\in$ [0,M]. \\\\ \nAll elements in S can be represented in the above form. And all the elements with above form can be put in a set S which satisfies the set constraints we are looking for.\\\\\nAs, $c_1$ can have M/n unique values, a set S can choose from M/n numbers while forming the set of n numbers.\\\\\n$\\implies$ the possible number of subsets with a fixed b, is {$M/n\\choose n$}, whose maximum length is n $\\in \\Theta(n)$.\\\\\n$\\implies$ there exists atleast {$M/n\\choose n$} subsets that whose maximum chain length is $\\Theta(n)$. {\\hfill\\qed}\n\n\\item The graph plotted is shown below:\n\n\\begin{figure}[H]\n \\centering\n \t\\includegraphics[scale=0.65]{plot5.png}\n \t\\caption{Graph of Maxchain length. \\label{fig:1}}\n \\end{figure}\n \n We see that the H() hashing function gives a linear line whereas the Hr() hashing function's max chain seems to be almost constant. This is because, as we increase k, the numbers in the set whose hashing table position is same increases linearly. Due to this, the max chain length too increases. But in the Hr() hashing function, the rx mod p part, helps \"randomize\" the number before applying the hashing function, thus reducing the max chain length and almost keeping it at a constant level.\n\\end{enumerate}\n\nThe code used to plot is pasted below:\n\\verbatiminput{4c.py}\n\n\n%------------------------------------------------\n\n\\end{document}\n", "meta": {"hexsha": "0283d10d457de4b7ea6fe569f1e08f0586bf24ea", "size": 17480, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Assignment-3/A3_sol_LaTeX/main.tex", "max_stars_repo_name": "GauravJain28/ADA-Assignments", "max_stars_repo_head_hexsha": "81783248c2850d8963531c74bbf711f37c34b07f", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Assignment-3/A3_sol_LaTeX/main.tex", "max_issues_repo_name": "GauravJain28/ADA-Assignments", "max_issues_repo_head_hexsha": "81783248c2850d8963531c74bbf711f37c34b07f", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Assignment-3/A3_sol_LaTeX/main.tex", "max_forks_repo_name": "GauravJain28/ADA-Assignments", "max_forks_repo_head_hexsha": "81783248c2850d8963531c74bbf711f37c34b07f", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 46.4893617021, "max_line_length": 717, "alphanum_fraction": 0.6811212815, "num_tokens": 5713, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7154239836484143, "lm_q2_score": 0.8539127510928476, "lm_q1q2_score": 0.6109096620750218}} {"text": "\\subsection{Hash-to-curve Functions}\n\\label{ssec:hash-to-curve}\n\nThe contents of this section should not be considered a full proof\nof security but are for implementation reference.\nFormal treatment of the security proofs for hash-to-curve\nfunctions are handled in referenced works.\n\nWe turn our attention to the hash-to-curve functions\nimplemented in our system.\nThe contents of this section are described hereafter.\nFirst, we review a mechanism often employed to solve this problem.\nThen, we briefly discuss why this solution is not ideal in our setting.\nFinally, we introduce the basis of our implementation and proceed into the\ndefinition of the algorithms used in our system after a brief review of\nsome required mathematical operations.\nAll algorithms may be found at the end of this section.\n\nThe seemingly standard, non-deterministic method to perform a hash-to-curve\noperation in the elliptic curve setting is to use the ``MapToGroup''\nmethod as presented in the original BLS short signature\npaper~\\cite{boneh2001short}.\nIn this method values are hashed into\n$\\F_{p}$ by modular arithmetic with a concatenated counter.\nThis counter starts at a fixed value, and is incremented until\na value is found that creates a valid point on the specified\nelliptic curve.\nWhile this may be sufficient in some instances, we prefer \ndeterministic methods due to the need for bounded\ncomputational overhead.\nIn our system, the need for bounded computational\noverhead arises from a desire to allow an Ethereum smart contract to\nperform the hash-to-curve operations with bounded gas consumption.\nDeterministic methods also have the benefit of minimizing\nside-channel attacks in those\nalgorithms that require such protection.\n\nThe hash-to-curve implementation selected allows the problem space to be\ndivided into independent problems such that their solutions may be composed.\nSpecifically, we first hash to the base field\n(hash map $\\mathfrak{h}:\\braces{0,1}^{*}\\to \\F$)\nand then find a deterministic map from the base field\nto the elliptic curve (function $f:\\F\\to E(\\F)$).\nThis approach allows for a separation of concerns and has become a\nstandard approach to the problem of hashing into an elliptic curve\n~\\cite{icart2009hash,ft2012bnhashtocurve,boneh2019h2cBLS12}.\nAlthough this strategy does offer a simplified view of the problem,\nmapping from the base field to the elliptic curve is nontrivial.\nAdditionally, it is frequently the case that $f$ is not surjective, but\nwe can overcome this limitation to obtain a surjective hash-to-curve algorithm\nunder easily-satisfied\nconditions~\\cite{tibouchi2014elligator,ft2012bnhashtocurve}.\nSpecifically, we use domain-separation in order to obtain independent hash\nfunctions.\nThese independent hash functions allow us overcome a non-surjective\nmapping.\nWe will fully address this concern later in this document.\n\nOur discussion of hashing functions follow\n\\cite{ft2012bnhashtocurve,boneh2019h2cBLS12}.\nWe ask the reader to note that in the remaining work of this section,\nwe view $E(\\F_{p})$ as an additive group unless otherwise specified.\nWe highlight this fact because this is different from the multiplicative\nnotation used previously in this document.\nBefore we define our implementation of the hash-to-curve algorithms employed,\nwe feel a review of the mathematics would benefit the reader.\nThus, we first\nreview the mathematics necessary to perform the hash-to-curve operations.\nAfter this review, we present the algorithm for hashing to $\\G_{1}$\nand then present the algorithm for hashing to $\\G_{2}$.\n\nFor reference, the full \\textsc{HashToG1} algorithm can be\nfound in Alg.~\\ref{alg:hash-to-G1} and the full algorithm for \\textsc{HashToG2}\ncan be found in Alg.~\\ref{alg:hash-to-G2}.\n\n\n\n\\subsubsection{Inverses, Square Roots, and Legendre Symbols\n in $\\F_{p}$}\n\\label{sssec:finite_math_Fp}\n\nIn this section we review some finite field mathematics that are\nimportant in our hash-to-curve setting.\n\nWe begin by reviewing inversion in $\\F_{p}$.\nFirst, we recall $\\F_{p}^{*} \\equiv \\F_{p}\\setminus\\braces{0}$,\nthe nonzero elements of our finite field.\nFor $a\\in\\F_{p}^{*}$, we have Euler's formula:\n\n\\begin{equation}\n a^{p-1} = 1 \\mod p.\n\\end{equation}\n\n\\noindent\nThis implies $a^{-1} \\mod p = a^{p-2}$.\nWe acknowledge more efficient methods for computing modular inverses are\npossible, but exponentiation can easily be performed in constant time, which\nis a goal of our implementation. This concludes our discussion of computing\ninverses in $\\F_{p}$.\n\nWe now review the mechanisms for\ncomputing Legendre symbols.\nWe recall that $a\\in\\F_{p}^{*}$ is a \\emph{quadratic residue}\nif there is $x\\in\\F_{p}$ such that $x^{2} = a \\mod p$;\notherwise, $a$ is a called a \\emph{quadratic nonresidue}.\nThis allows us to define the Legendre symbol~\\cite{imc2008}:\n\n\\begin{equation}\n \\chi_{p}(a) \\equiv \\begin{cases}\n 1 \\quad& \\text{if there $a$ is a quadratic residue modulo $p$.} \\\\\n -1 \\quad& \\text{if there $a$ is a quadratic nonresidue modulo $p$.} \\\\\n 0 \\quad& \\text{if $p \\mid a$}\n \\end{cases}\n\\end{equation}\n\n\\noindent\nThe Legendre symbol has a simple formula:\n\n\\begin{equation}\n \\chi_{p}(a) = a^{\\frac{p-1}{2}}.\n\\end{equation}\n\n\\noindent\nThis formula holds even when $a=0$.\nThis concludes our discussion of computing the Legendre Symbol in $\\F_{p}$.\n\nComputing square roots is more challenging.\nWe now review the mechanism of computing square roots.\nIn our case $p = 3\\mod 4$, so there is a simple\nformula to solve $x^{2} = a$:\n\n\\begin{equation}\n x = a^{\\frac{p+1}{4}}.\n\\end{equation}\n\n\\noindent\nThis relies on the assumption that $a$ is a quadratic residue,\nso $a^{\\frac{p-1}{2}} = 1$.\nThis concludes our discussion of computing the square roots in $\\F_{p}$.\n\nTaking inverses, Legendre symbols, and square roots\ngives us the necessary tools to hash to $\\G_{1}$.\nThis ends the preliminary work necessary to develop\n\\textsc{HashToG1}.\n\n\n\\subsubsection{Inverses, Square Roots, and Legendre Symbols\n in $\\F_{p^{2}}$}\n\\label{sssec:finite_math_Fp2}\n\nThe previous section reviewed the mathematics of inverses, square roots, and\nLegendre Symbols in $\\F_{p}$.\nWe must be able to perform the same operations\nin $\\F_{p^{2}}$ if we wish to hash to $\\G_{2}$.\nWhile computing square roots and\ninverses may be familiar to the reader in $\\F_{p}$, the operations require\nspecial handling in $\\F_{p^{2}}$.\nWe review these mechanisms now.\n\nWe first address the problem of computing inverses in $\\F_{p^{2}}$.\nOur discussion and methods follow~\\cite{adj2012sqrtEvenExt},\nand we present their general results applied to our particular case.\nAs mentioned above, the setting of our elliptic curve is $p = 3\\mod 4$.\nThis implies $-1$ is a quadratic nonresidue.\nThus, we have the following isomorphism:\n\n\\begin{equation}\n \\F_{p^{2}} \\simeq \\F_{p}[i]/\\parens{i^{2}+1}.\n\\end{equation}\n\n\\noindent\nThis is the analogous to constructing the complex numbers $\\C$ from\nthe real numbers $\\R$.\nOur implementation uses this construction. We may use\nthis construction to compute an efficient inversion as follows:\n\n\\begin{equation}\n \\parens{a_{0} + a_{1}i}^{-1} = \\frac{a_{0}-a_{1}i}{a_{0}^{2}+a_{1}^{2}}.\n\\end{equation}\n\n\\noindent\nThe main computational cost of this operation is the inversion of an element\nin $\\F_{p}$.\nThis concludes our discussion of computing inverses in $\\F_{p^{2}}$.\n\nWe now begin our discussion of computing the Legendre Symbol in $\\F_{p^{2}}$\nIf $a = a_{0} + a_{1}i\\in\\F_{p^{2}}$, then $a$ is a\nquadratic residue in $\\F_{p^{2}}$ if and only if $a_{0}^{2} + a_{1}^{2}$\nis a quadratic residue in $\\F_{p}$.\nThe main computational cost of this operation is the computation of the Legendre\nsymbol of an element in $F_{p}$.\nFrom the above we have the Legendre symbol in $\\F_{p^{2}}$,\nwhich we denote as \\textsc{LegendreFP2} or $\\chi_{p^{2}}(\\cdot)$;\nthis particular algorithm is presented in\nAlg.~\\ref{alg:legendre-fp2}.\nThis concludes our discussion of computing the Legendre Symbol in\n$\\F_{p^{2}}$.\n\nWe will now look at computing square roots in $\\F_{p^{2}}$.\nThe main idea is to find $b\\in\\F_{p^{2}}$ and odd $s$ such\n$b^{2}a^{s} = 1$.\nIn this case, we see\n\n\\begin{align}\n \\brackets{ba^{\\frac{s+1}{2}}}^{2} &= b^{2}a^{s+1} \\nonumber\\\\\n &= a.\n\\end{align}\n\n\\noindent\nIf we set\n\n\\begin{align}\n b &= \\parens{1 + a^{\\frac{p-1}{2}}}^{\\frac{p-1}{2}} \\nonumber\\\\\n s &= \\frac{p-1}{2},\n\\end{align}\n\n\\noindent\nthen when $b\\ne0$, we have $b^{2}a^{s} = 1$.\nWhen $b=0$, we have $a^{\\frac{p-1}{2}}=-1$.\nIn this case, our square root is $ia^{\\frac{p+1}{4}}$.\nThis procedure is formally presented in~Alg.~\\ref{alg:sqrt-fp2}.\nThe main computational cost of this operation is two exponentiations\nin $\\F_{p^{2}}$.\nThis concludes our discussion of computing square roots in $\\F_{p^{2}}$\nand the preliminary review of those operations\nnecessary to develop \\textsc{HashToG2}.\n\n\n\\subsubsection{Hashing to Base}\n\\label{sssec:hash-to-base}\n\nIn this section we will describe the hash-to-base operation.\nWe begin by discussing the construction of a random oracle into\n$\\F_{p}$ using a single cryptographic hash function with\ndomain separation.\nWe then bound the deviation from uniformity in this operation.\n\nIn the following $H$ is a $256$-bit hash function.\nIn our implementation we use \\textsc{Keccak256}.\nNote this is the \\textsc{Sha3} variant used by Ethereum\nthat differs from the NIST approved \\textsc{Sha3} hash function due to a\ndifference in the handling of padding.\n\nLet $H$ act as a random oracle.\nWe map from $H$ to $\\Z$ by interpreting the output of a\ncryptographic hash function as a big endian unsigned integer.\nThis can be seen in lines 4 and 5 of $\\textsc{HashToBase}$\nas the function named $\\texttt{b2u}$.\nAlthough we may naively map from $\\Z$ to $\\Z_{p}$ by simply taking the\noutput of $H$ modulo $p$, this would be insecure in our setting due to the\nnonuniformity of this operation.\nIn order to ensure the mapping from $\\Z$ to $\\Z_{p}$ is more\nuniformly distributed, domain separation is utilized.\nSpecifically, we use domain separation in order obtain independent hash\nfunctions from $H$ through concatenation of constants with the message being\nhashed.\nThese independent hash functions allow us to create a secure 512-bit random\nnumber from a single cryptographic hash function.\nThe full explanation of this operation is below.\n\nLet $\\textsc{HashToBase}:\\braces{0,1}^{*}\\times\\braces{0,1}^{8}\\times\n\\braces{0,1}^{8}\\to\\F_{p}$ denote our random oracle into the\nunderlying field $\\F_{p}$.\nIn the $\\textsc{HashToBase}$ operation the first component corresponds to the\nunderlying message being hashed, while the last two elements provide the\nnecessary domain separation.\nSee Alg.~\\ref{alg:hash-to-fp} for the full implementation.\n\nDue to the fact $p$ is prime and not a power of 2, there will be some\nnonuniformity in the resulting distribution of \\textsc{HashToBase}.\nWe investigate this nonuniformity now.\n\nFirst, we assume that $H:\\braces{0,1}^{*}\\to\\Z_{N}$\nis a random oracle and $p\\in\\braces{1,2,\\cdots,N-1}$.\nWe want to determine how much $H(m)\\mod p$ deviates from uniformity.\nWe restrict ourselves to the case when $p\\nmid N$.\nLet\n\n\\begin{equation}\n N = qp + r\n\\end{equation}\n\n\\noindent\nwith $0\\le r< p$ and $q\\ge1$.\nBecause $p\\nmid N$, we have $r\\ge1$.\nLet $X$ be uniformly distributed on $\\Z_{N}$ and set $X_{p} = X \\mod p$.\nFurthermore, we let $U_{p}$ be the uniform distribution modulo $p$.\nThen\n\n\\begin{equation}\n \\mathcal{P}\\parens{X_{p}\\in\\braces{0,\\cdots,r-1}} = \\frac{q+1}{N}\n\\end{equation}\n\n\\noindent\nand\n\n\\begin{equation}\n \\mathcal{P}\\parens{X_{p}\\in\\braces{r,\\cdots,p-1}} = \\frac{q}{N}.\n\\end{equation}\n\n\\noindent\nHere, $\\mathcal{P}$ denotes the probability of an event occurring.\nWe now determine the deviation of $X_{p}$ from the uniform\ndistribution $U_{p}$:\n\n\\begin{align}\n \\Delta\\parens{X_{p},U_{p}} &\\equiv\n \\sum_{k=0}^{p-1} \\abs{\\mathcal{P}(X_{p}=k) - \\mathcal{P}(U_{p}=k)}\n \\nonumber\\\\\n &= \\sum_{u=0}^{r-1} \\abs{\\frac{q+1}{N} - \\frac{1}{p}} +\n \\sum_{u=r}^{p-1} \\abs{\\frac{q}{N} - \\frac{1}{p}} \\nonumber\\\\\n &= r\\frac{qp + p - N}{pN} + \\parens{p-r}\\frac{N-qp}{pN}\n \\nonumber\\\\\n &\\le \\frac{p}{N}.\n\\end{align}\n\n\\noindent\nFrom this, we see that if $p$ is a $k$-bit prime and\n$N = 2^{k+\\ell}$, then the deviation from uniformity is $\\le2^{-\\ell}$.\n\nIn our case, $p$ is a 254-bit prime and we concatenate the output\nof independent hash functions in order to have a uniformly distributed\n512-bit output.\nFrom the above it may be seen that \\textsc{HashToBase} produces outputs whose\ndeviation from uniformity is less than $2^{-258}$.\nThis deviation is sufficiently small as to not be of concern.\nFurther work is required to more formally prove this assumption, but\nthis work is not included at this time.\n\n\n\\subsubsection{Base to $\\G_{1}$}\n\\label{sssec:base-to-G1}\n\nIn this section we discuss the construction of a hash-to-G1 algorithm.\nWe begin by noting the non-surjective nature for many functions\n$f:\\F_{p}\\to E(\\F_{p})$.\nThen, we cite a known solution to this problem and provide necessary\nmathematics to understand the fundamental operation that\novercomes the problem.\nNext, we discuss the actual implementation.\nLastly, we discuss specific exclusion of three points from the allowable\noutputs of this algorithm for security purposes.\n\nLet us now suppose that we have a hash function\n$\\mathfrak{h}:\\braces{0,1}^{*}\\to\\F_{p}$ and a\ndeterministic map $f:\\F_{p}\\to E(\\F_{p})$.\nAs has been previously stated, in many\nsituations~\\cite{icart2009hash,ft2012bnhashtocurve,boneh2019h2cBLS12}\n$f$ is not surjective.\nThat is, there are points on the elliptic curve $E(\\F_{p})$\n(sometimes a nontrivial fraction) which cannot be reached by $f$.\nEven so, if we use domain separation to construct independent hash functions,\n$\\mathfrak{h}_{1}, \\mathfrak{h}_{2}:\\braces{0,1}^{*}\\to\\F$,\nthen\n\n\\begin{equation}\n F(\\texttt{m}) = f(\\mathfrak{h}_{1}(\\texttt{m}))\n + f(\\mathfrak{h}_{2}(\\texttt{m}))\n\\end{equation}\n\n\\noindent\nis indistinguishable from a random oracle on $E(\\F_{p})$\nunder certain restrictions on $f$.\nSee~\\cite{ft2012bnhashtocurve,tibouchi2014elligator} for details.\n\nWe now turn our attention to determining $f:\\F_{p}\\to E(\\F_{p})$.\nFinding a map $f:\\F_{p}\\to E(\\F_{p})$ is involved.\nOur BN curve has the form\n\n\\begin{align}\n E:y^{2} &= g(x) \\nonumber\\\\\n &= x^{3} + b.\n\\end{align}\n\n\\noindent\nFrom~\\cite{ft2012bnhashtocurve}, it possible to show there\nare $x_{1},x_{2},x_{3},y\\in\\F_{p}$ such that\n\n\\begin{equation}\n g(x_{1})g(x_{2})g(x_{3}) = y^{2}.\n\\end{equation}\n\n\\noindent\nWhen $y\\ne0$, quadratic reciprocity implies that $g(x_{i})$ is square\nfor some $i$; that is, for some $i$ we have\n$(x_{i},\\sqrt{g(x_{i})})\\in E(\\F_{p})$.\nFor uniqueness, we choose the smallest $i$.\nWe will use the construction of~\\cite{ft2012bnhashtocurve}\nto determine such points with some modifications based\non work in~\\cite{boneh2019h2cBLS12}.\nThe exact algorithm can be found in Alg.~\\ref{alg:base-to-G1}.\n\nOne of the main difficulties is determining $i$ in such a way\nas to not leak information; because of this, we do not wish to\nrely upon if statements.\nIn~\\cite{ft2012bnhashtocurve}, they suggested the function\n\n\\begin{equation}\n \\psi(r_{1},r_{2}) = \\brackets{\\parens{r_{1}-1}r_{2}\\mod3} + 1.\n \\label{eq:ftAB}\n\\end{equation}\n\n\\noindent\nThis function works under the assumption that modular\narithmetic always returns positive integers.\nThis is not always the case in programming languages;\nin particular, it does not hold in \\textsc{Go} (Golang), the language\nwe use to implement these algorithms.\nTo circumvent this, we use the following function,\nwhich is implemented in Line~\\ref{alg_line:new_coef_func}\nof Alg.~\\ref{alg:base-to-G1}:\n\n\\begin{equation}\n \\parens{r_{1}-1}\\parens{r_{2}-3}/4 + 1.\n \\label{eq:newAB}\n\\end{equation}\n\nIn~\\cite{ft2012bnhashtocurve}, the authors encountered an issue when $t=0$.\nTheir solution was to define\n\n\\begin{equation}\n \\textsc{BaseToG1}(0) = \\parens{\\frac{-1+\\sqrt{-3}}{2}, \\sqrt{1+b}}.\n \\label{eq:baseToG1zero}\n\\end{equation}\n\n\\noindent\nWahby and Boneh~\\cite{boneh2019h2cBLS12} suggest another\nmethod, which we implement, in order to have a more efficient\nalgorithm.\nThis implementation also affords the benefit of not needing to\nhandle the case of $t=0$ separately.\nThis can be seen in Line~\\ref{alg_line:alpha_def}\nof Alg.~\\ref{alg:base-to-G1} where we define $\\alpha$ as the inverse\nof a value which depends on $t\\in\\F_{p}$.\nWhen $t=0$, we compute $\\alpha = 0$\nbecause we compute inversions via exponentiation.\nThus, the computational convention $0^{-1} = 0 \\mod p$ is established.\nThis leads to the same result as in Eq.~\\eqref{eq:baseToG1zero}\nwithout special handling.\n\nWe have cryptographic insecurity when\n$\\textsc{HashToG1}(\\texttt{m}) = g_{1}^{\\alpha}$ for known $\\alpha$.\nNote we briefly switch back to multiplicative notation at this time.\nIn practice, it is likely this insecurity will only be known when\n$\\alpha\\in\\braces{-1,0,1}$ or is sufficiently close to these values.\nWe assume we may not fix simple proximity to these values, and thus only\nexplicitly address the case of $\\alpha\\in\\braces{-1,0,1}$.\nFor clarity, this notation specifies the hash function outputs of\nthe identity element, the generator, or the generator's\ninverse.\nDue to the concerns around the use of these points, we will not allow these\nindividual cases.\nIn order to enforce this requirement we include a\n\\textsc{SafePointCheck} in \\textsc{HashToG1};\nsee Line~\\ref{alg_line:safe_point_check} in Alg.~\\ref{alg:hash-to-G1}.\nThis function checks the point returned from the hash function for\nequivalence with the identity element OR equivalence of the point's\n$x$ coordinate with $1$.\nIf either of these conditions is true, an error is raised.\nAlthough the probability of a hash mapping to these points is small, this\nerror must be handled appropriately to prevent an attacker from causing\nunexpected errors in a remote system due to specially crafted messages.\n\n\n\\subsubsection{Base to $\\G_{2}$}\n\\label{sssec:base-to-G2}\n\nWe now take up the slightly more challenging possibility\nof computing a hash function to $\\G_{2}$.\nFortunately, we may utilize many of\nthe previously described operations.\nThus, we do not repeat those explanations and only discuss\nthose operations that differ.\nWe would like to remind the reader that the results in\nSec.~\\ref{sssec:finite_math_Fp2} allow\nus to compute Legendre symbols and square roots in $\\F_{p^{2}}$.\n\nUsing the derivation of~\\cite[Section 3]{boneh2019h2cBLS12},\nwe set $u_{0}=1$ as in $\\G_{1}$ to obtain\n\n\\begin{align}\n x_{1} &= \\frac{\\sqrt{-3}-1}{2} - \\frac{t^{2}\\sqrt{-3}}{t^{2}+g'(1)}\n \\nonumber\\\\\n x_{2} &= -1 - x_{2} \\nonumber\\\\\n x_{3} &= 1 - \\frac{\\parens{t^{2}+g'(1)}^{2}}{3t^{2}},\n\\end{align}\n\n\\noindent\nfor $t\\in\\F_{p^{2}}$.\nHere, we have\n\n\\begin{equation}\n E':y^{2} = g'(x) = x^{3} + b',\n\\end{equation}\n\n\\noindent\nwhere $b' = b/\\xi = 3/\\parens{i+9}$.\nWith this choice, $-g'(1)$ is a residue in $\\F_{p^{2}}$.\nThis ensures that $t\\in\\{0,\\pm\\sqrt{-g'(1)}\\}$\nimplies $x_{1}$ is a valid point on the curve;\nthus, all inputs result in valid outputs.\nSee Alg.~\\ref{alg:base-to-twist} for the algorithm.\nThus, our hash functions to $\\G_{1}$ and $\\G_{2}$ are\nessentially the same.\n\nAt this point, we have successfully mapped into $E'$;\nhowever, our goal is to map into $\\G_{2}$.\nFrom~\\cite{bnCurves}, we know $\\abs{E'(\\F_{p^{2}})} = q\\parens{2p-q}$.\nThis gives a cofactor $r = 2p-q$ because $p\\nmid q$.\nWe take care of this by clearing the cofactor.\nSee Alg.~\\ref{alg:hash-to-G2} for the full hash algorithm.\n\n\\clearpage\n\n\\input{algs/hash_to_G1.tex}\n\n\\input{algs/gfp2_algs.tex}\n\n\\input{algs/hash_to_G2.tex}\n", "meta": {"hexsha": "48b7c50883dfd55a302b18ce814ff134c8a3c375", "size": 19681, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "tex/tcrypt_h2c.tex", "max_stars_repo_name": "chgorman/MadNet-Whitepaper", "max_stars_repo_head_hexsha": "a4cd4946db20713aa0573674c6e1fcc17a76a48a", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "tex/tcrypt_h2c.tex", "max_issues_repo_name": "chgorman/MadNet-Whitepaper", "max_issues_repo_head_hexsha": "a4cd4946db20713aa0573674c6e1fcc17a76a48a", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "tex/tcrypt_h2c.tex", "max_forks_repo_name": "chgorman/MadNet-Whitepaper", "max_forks_repo_head_hexsha": "a4cd4946db20713aa0573674c6e1fcc17a76a48a", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 37.8480769231, "max_line_length": 80, "alphanum_fraction": 0.7304506885, "num_tokens": 5875, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8615382094310357, "lm_q2_score": 0.7090191337850932, "lm_q1q2_score": 0.6108470749735531}} {"text": "\\section{SOLVE Operator}\\ttindex{SOLVE}\nSOLVE is an operator for solving one or more simultaneous algebraic\nequations. It is used with the syntax:\n\\begin{verbatim}\n SOLVE(EXPRN:algebraic[,VAR:kernel|,VARLIST:list of kernels])\n :list.\n\\end{verbatim}\n{\\tt EXPRN} is of the form {\\tt } or\n\\{ {\\tt },{\\tt }, \\dots \\}. Each expression is an\nalgebraic equation, or is the difference of the two sides of the equation.\nThe second argument is either a kernel or a list of kernels representing\nthe unknowns in the system. This argument may be omitted if the number of\ndistinct, non-constant, top-level kernels equals the number of unknowns,\nin which case these kernels are presumed to be the unknowns.\n\nFor one equation, {\\tt SOLVE}\\ttindex{SOLVE} recursively uses\nfactorization and decomposition, together with the known inverses of\n{\\tt LOG}, {\\tt SIN}, {\\tt COS}, {\\tt \\verb|^|}, {\\tt ACOS}, {\\tt ASIN}, and\nlinear, quadratic, cubic, quartic, or binomial factors. Solutions\nof equations built with exponentials or logarithms are often\nexpressed in terms of Lambert's {\\tt W} function.\\index{Lambert's W}\nThis function is (partially) implemented in the special functions package.\n\nLinear equations are solved by the multi-step elimination method due to\nBareiss, unless the switch {\\tt CRAMER}\\ttindex{CRAMER} is on, in which\ncase Cramer's method is used. The Bareiss method is usually more\nefficient unless the system is large and dense.\n\nNon-linear equations are solved using the Groebner basis package.\n\\index{Groebner} Users should note that this can be quite a\ntime consuming process.\n\n{\\it Examples:}\n\\begin{verbatim}\n solve(log(sin(x+3))^5 = 8,x);\n solve(a*log(sin(x+3))^5 - b, sin(x+3));\n solve({a*x+y=3,y=-2},{x,y});\n\\end{verbatim}\n\n{\\tt SOLVE} returns a list of solutions. If there is one unknown, each\nsolution is an equation for the unknown. If a complete solution was\nfound, the unknown will appear by itself on the left-hand side of the\nequation. On the other hand, if the solve package could not find a\nsolution, the ``solution'' will be an equation for the unknown in terms\nof the operator {\\tt ROOT\\_OF}\\ttindex{ROOT\\_OF}. If there\nare several unknowns, each solution will be a list of equations for the\nunknowns. For example,\n\\begin{verbatim}\n solve(x^2=1,x); -> {X=-1,X=1}\n\n solve(x^7-x^6+x^2=1,x)\n 6\n -> {X=ROOT_OF(X_ + X_ + 1,X_,TAG_1),X=1}\n\n solve({x+3y=7,y-x=1},{x,y}) -> {{X=1,Y=2}}.\n\\end{verbatim}\nThe TAG argument is used to uniquely identify those particular solutions.\nSolution multiplicities are stored in the global variable {\\tt\nROOT\\_MULTIPLICITIES} rather than the solution list. The value of this\nvariable is a list of the multiplicities of the solutions for the last\ncall of {\\tt SOLVE}. \\ttindex{SOLVE} For example,\n\\begin{verbatim}\n solve(x^2=2x-1,x); root_multiplicities;\n\\end{verbatim}\ngives the results\n\\begin{verbatim}\n {X=1}\n\n {2}\n\\end{verbatim}\n\nIf you want the multiplicities explicitly displayed, the switch\n{\\tt MULTIPLICITIES}\\ttindex{MULTIPLICITIES} can be turned on. For example\n\\begin{verbatim}\n on multiplicities; solve(x^2=2x-1,x);\n\\end{verbatim}\nyields the result\n\\begin{verbatim}\n {X=1,X=1}\n\\end{verbatim}\n\n\\subsection{Handling of Undetermined Solutions}\nWhen {\\tt SOLVE} cannot find a solution to an equation, it normally\nreturns an equation for the relevant indeterminates in terms of the\noperator {\\tt ROOT\\_OF}.\\ttindex{ROOT\\_OF} For example, the expression\n\\begin{verbatim}\n solve(cos(x) + log(x),x);\n\\end{verbatim}\nreturns the result\n\\begin{verbatim}\n {X=ROOT_OF(COS(X_) + LOG(X_),X_,TAG_1)} .\n\\end{verbatim}\n\nAn expression with a top-level {\\tt ROOT\\_OF} operator is implicitly a\nlist with an unknown number of elements (since we don't always know how\nmany solutions an equation has). If a substitution is made into such an\nexpression, closed form solutions can emerge. If this occurs, the {\\tt\nROOT\\_OF} construct is replaced by an operator {\\tt ONE\\_OF}.\\ttindex{ONE\\_OF}\nAt this point it is of course possible to transform the result of the\noriginal {\\tt SOLVE} operator expression into a standard {\\tt SOLVE}\nsolution. To effect this, the operator {\\tt EXPAND\\_CASES}\n\\ttindex{EXPAND\\_CASES} can be used.\n\nThe following example shows the use of these facilities:\n\\extendedmanual{\\newpage}\n\\begin{verbatim}\nsolve(-a*x^3+a*x^2+x^4-x^3-4*x^2+4,x);\n 2 3\n{X=ROOT_OF(A*X_ - X_ + 4*X_ + 4,X_,TAG_2),X=1}\n\nsub(a=-1,ws);\n\n{X=ONE_OF({2,-1,-2},TAG_2),X=1}\n\nexpand_cases ws;\n\n{X=2,X=-1,X=-2,X=1}\n\\end{verbatim}\n\n\\subsection{Solutions of Equations Involving Cubics and Quartics}\n\nSince roots of cubics and quartics can often be very messy, a switch\n{\\tt FULLROOTS}\\ttindex{FULLROOTS} is available, that, when off (the\ndefault), will prevent the production of a result in closed form. The\n{\\tt ROOT\\_OF} construct will be used in this case instead.\n\nIn constructing the solutions of cubics and quartics, trigonometrical\nforms are used where appropriate. This option is under the control of a\nswitch {\\tt TRIGFORM},\\ttindex{TRIGFORM} which is normally on.\n\nThe following example illustrates the use of these facilities:\n\\begin{verbatim}\nlet xx = solve(x^3+x+1,x);\n\nxx;\n 3\n{X=ROOT_OF(X_ + X_ + 1,X_)}\n\non fullroots;\n\nxx;\n\n - SQRT(31)*I\n ATAN(---------------)\n 3*SQRT(3)\n{X=(I*(SQRT(3)*SIN(-----------------------)\n 3\n\\end{verbatim}\n\\newpage\n\\begin{verbatim}\n - SQRT(31)*I\n ATAN(---------------)\n 3*SQRT(3)\n - COS(-----------------------)))/SQRT(3),\n 3\n\n - SQRT(31)*I\n ATAN(---------------)\n 3*SQRT(3)\n X=( - I*(SQRT(3)*SIN(-----------------------)\n 3\n\n - SQRT(31)*I\n ATAN(---------------)\n 3*SQRT(3)\n + COS(-----------------------)))/SQRT(\n 3\n\n 3),\n\n - SQRT(31)*I\n ATAN(---------------)\n 3*SQRT(3)\n 2*COS(-----------------------)*I\n 3\n X=----------------------------------}\n SQRT(3)\n\noff trigform;\n\nxx;\n 2/3\n{X=( - (SQRT(31) - 3*SQRT(3)) *SQRT(3)*I\n\n 2/3 2/3\n - (SQRT(31) - 3*SQRT(3)) - 2 *SQRT(3)*I\n\n 2/3 1/3 1/3\n + 2 )/(2*(SQRT(31) - 3*SQRT(3)) *6\n\n 1/6\n *3 ),\n\n 2/3\n X=((SQRT(31) - 3*SQRT(3)) *SQRT(3)*I\n\n 2/3 2/3\n - (SQRT(31) - 3*SQRT(3)) + 2 *SQRT(3)*I\n\n 2/3 1/3 1/3\n + 2 )/(2*(SQRT(31) - 3*SQRT(3)) *6\n\n 1/6\n *3 ),\n\n 2/3 2/3\n (SQRT(31) - 3*SQRT(3)) - 2\n X=-------------------------------------}\n 1/3 1/3 1/6\n (SQRT(31) - 3*SQRT(3)) *6 *3\n\\end{verbatim}\n\n\\subsection{Other Options}\n\nIf {\\tt SOLVESINGULAR}\\ttindex{SOLVESINGULAR} is on (the default setting),\ndegenerate systems such as {\\tt x+y=0}, {\\tt 2x+2y=0} will be solved by\nintroducing appropriate arbitrary constants.\nThe consistent singular equation 0=0 or equations involving functions with\nmultiple inverses may introduce unique new indeterminant kernels\n{\\tt ARBCOMPLEX(j)}, or {\\tt ARBINT(j)}, ($j$=1,2,...), % {\\tt ARBREAL(j)},\nrepresenting arbitrary complex or integer numbers respectively. To\nautomatically select the principal branches, do {\\tt off allbranch;} .\n\\ttindex{ALLBRANCH} To avoid the introduction of new indeterminant kernels\ndo {\\tt OFF ARBVARS}\\ttindex{ARBVARS} -- then no equations are generated for the free\nvariables and their original names are used to express the solution forms.\nTo suppress solutions of consistent singular equations do\n{\\tt OFF SOLVESINGULAR}.\n\nTo incorporate additional inverse functions do, for example:\n\\begin{verbatim}\n put('sinh,'inverse,'asinh);\n put('asinh,'inverse,'sinh);\n\\end{verbatim}\ntogether with any desired simplification rules such as\n\\begin{verbatim}\n for all x let sinh(asinh(x))=x, asinh(sinh(x))=x;\n\\end{verbatim}\nFor completeness, functions with non-unique inverses should be treated as\n{\\tt \\verb|^|}, {\\tt SIN}, and {\\tt COS} are in the {\\tt SOLVE}\n\\ttindex{SOLVE} module source.\n\nArguments of {\\tt ASIN} and {\\tt ACOS} are not checked to ensure that the\nabsolute value of the real part does not exceed 1; and arguments of\n{\\tt LOG} are not checked to ensure that the absolute value of the imaginary\npart does not exceed $\\pi$; but checks (perhaps involving user response\nfor non-numerical arguments) could be introduced using\n{\\tt LET}\\ttindex{LET} statements for these operators.\n\n\\subsection{Parameters and Variable Dependency}\n\nThe proper design of a variable sequence\nsupplied as a second argument to {\\tt SOLVE} is important\nfor the structure of the solution of an equation system.\nAny unknown in the system\nnot in this list is considered totally free. E.g.\\ the call\n\\begin{verbatim}\n solve({x=2*z,z=2*y},{z});\n\\end{verbatim}\nproduces an empty list as a result because there is no function\n$z=z(x,y)$ which fulfills both equations for arbitrary $x$ and $y$ values.\nIn such a case the share variable {\\tt requirements}\\ttindex{requirements}\ndisplays a set of restrictions for the parameters of the system:\n\\begin{verbatim}\n requirements;\n\n {x - 4*y}\n\\end{verbatim}\nThe non-existence of a formal solution is caused by a \ncontradiction which disappears only if the parameters\nof the initial system are set such that all members\nof the requirements list take the value zero.\nFor a linear system the set is complete: a solution\nof the requirements list makes the initial\nsystem solvable. E.g.\\ in the above case a substitution\n$x=4y$ makes the equation set consistent. For a non-linear\nsystem only one inconsistency is detected. If such a system\nhas more than one inconsistency, you must reduce them\none after the other. \n\\footnote{\nThe difference between linear and non--linear\ninconsistent systems is based on the algorithms which\nproduce this information as a side effect when attempting\nto find a formal solution; example:\n$solve(\\{x=a,x=b,y=c,y=d\\},\\{x,y\\}$ gives a set $\\{a-b,c-d\\}$\nwhile $solve(\\{x^2=a,x^2=b,y^2=c,y^2=d\\},\\{x,y\\}$ leads to $\\{a-b\\}$.\n}\nThe set shows you also the dependency among the parameters: here\none of $x$ and $y$ is free and a formal solution of the system can be\ncomputed by adding it to the variable list of {\\tt solve}. \nThe requirement set is not unique -- there may be other such sets.\n\n\nA system with parameters may have a formal solution, e.g.\\\n\\begin{verbatim}\n solve({x=a*z+1,0=b*z-y},{z,x});\n\n y a*y + b\n {{z=---,x=---------}}\n b b\n\\end{verbatim}\nwhich is not valid for all possible values of the parameters.\nThe variable {\\tt assumptions}\\ttindex{assumptions} contains then a list of\nrestrictions: the solutions are valid only as long\nas none of these expressions vanishes. Any zero of one of them\nrepresents a special case that is not covered by the\nformal solution. In the above case the value is\n\\extendedmanual{\\newpage}\n\\begin{verbatim}\n assumptions;\n\n {b}\n\\end{verbatim}\nwhich excludes formally the case $b=0$; obviously this special \nparameter value makes the system singular. The set of assumptions\nis complete for both, linear and non--linear systems.\n \n\n{\\tt SOLVE} rearranges the variable sequence\nto reduce the (expected) computing time. This behavior is controlled\nby the switch {\\tt varopt}\\ttindex{varopt}, which is on by default.\nIf it is turned off, the supplied variable sequence is used\nor the system kernel ordering is taken if the variable\nlist is omitted. The effect is demonstrated by an example:\n\\begin{verbatim}\n s:= {y^3+3x=0,x^2+y^2=1};\n\n solve(s,{y,x});\n\n 6 2\n {{y=root_of(y_ + 9*y_ - 9,y_),\n\n 3\n - y\n x=-------}}\n 3\n \n off varopt; solve(s,{y,x});\n\n 6 4 2\n {{x=root_of(x_ - 3*x_ + 12*x_ - 1,x_),\n\n 4 2\n x*( - x + 2*x - 10)\n y=-----------------------}}\n 3\n\n\\end{verbatim}\nIn the first case, {\\tt solve} forms the solution as a set of\npairs $(y_i,x(y_i))$ because the degree of $x$ is higher -- \nsuch a rearrangement makes the internal computation of the Gr\\\"obner basis\ngenerally faster. For the second case the explicitly given variable sequence\nis used such that the solution has now the form $(x_i,y(x_i))$.\nControlling the variable sequence is especially important if\nthe system has one or more free variables.\nAs an alternative to turning off {\\tt varopt}, a partial dependency among\nthe variables can be declared using the {\\tt depend}\\index{depend}\nstatement: {\\tt solve} then rearranges the variable sequence but keeps any\nvariable ahead of those on which it depends.\n\\extendedmanual{\\newpage}\n\\begin{verbatim}\n on varopt;\n s:={a^3+b,b^2+c}$\n solve(s,{a,b,c});\n\n 3 6\n {{a=arbcomplex(1),b= - a ,c= - a }}\n\n depend a,c; depend b,c; solve(s,{a,b,c});\n\n {{c=arbcomplex(2),\n\n 6\n a=root_of(a_ + c,a_),\n\n 3\n b= - a }}\n\\end{verbatim}\nHere {\\tt solve} is forced to put $c$ after $a$ and after $b$, but\nthere is no obstacle to interchanging $a$ and $b$.\n\n", "meta": {"hexsha": "e9543d7154745faa4a29dc891f9aa331b5bc5835", "size": 13682, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "atomic_Decomp/Redlog/reduce.doc/solve.tex", "max_stars_repo_name": "Korosensei42/AtomicDecomposition", "max_stars_repo_head_hexsha": "ca10f97c2cef1a258a4e9fade0a3133d1389d08e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "atomic_Decomp/Redlog/reduce.doc/solve.tex", "max_issues_repo_name": "Korosensei42/AtomicDecomposition", "max_issues_repo_head_hexsha": "ca10f97c2cef1a258a4e9fade0a3133d1389d08e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "atomic_Decomp/Redlog/reduce.doc/solve.tex", "max_forks_repo_name": "Korosensei42/AtomicDecomposition", "max_forks_repo_head_hexsha": "ca10f97c2cef1a258a4e9fade0a3133d1389d08e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 35.9107611549, "max_line_length": 85, "alphanum_fraction": 0.6385762315, "num_tokens": 3798, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8354835534888478, "lm_q2_score": 0.7310585844894971, "lm_q1q2_score": 0.6107874239778122}} {"text": "\n\\begin{questions}\n\n\\question{ Modify the code \\texttt{fluidflowFD.m} to solve the equations\n\\begin{align*}\n w_t + \\psi_y w_x - \\psi_xw_y &= Pr \\Delta w + RaPrT_x, \\\\\n T_t + \\psi_y T_x - \\psi_xT_y &= \\Delta T, \\\\\n\\Delta \\psi &= -w,\n\\end{align*}\nwhere $(x,y) \\in (0,1) \\times (0,1)$ and $t>0$. For this exercise, set $Ra = 2 \\times 10^5$ and $Pr = 0.71$ (air).\n\nThe fluid is at rest at $t=0$, with $T = \\psi = w = 0$. The boundary conditions for the stream function is $\\psi_{Gamma} = 0$, which implies that there is no mass transfer through the boundary $\\Gamma$. The value of the vorticity at the walls is expressed as $w_{\\Gamma} = -\\Delta \\psi_{\\Gamma}$ and the temperature at $\\Gamma$ is defined by\n\\begin{align*}\nT(t,x,y) = \\begin{cases}\n2^9\\tanh^4(100t)x^5(x-1)^4,& y =0,~0\\leq x \\leq 1,~t>0, \\\\\n0, & (x,y) \\in \\Gamma,~y\\neq 0,~t>0.\n\\end{cases}\n\\end{align*}\n\n}\n\n\\begin{solution}\n\n\\begin{figure}[H]\n\\center{\\includegraphics[scale=.5]{p4_f1.eps}}\n\\caption{Solution at $ t=1$ second}\n\\end{figure}\n\n\\end{solution}\n\n\\end{questions}\n\n\\subsubsection*{MATLAB}\n\n \\begin{verbatim}\nfunction HW6P4\n% Parameters\nN = 150;\nRa = 2e5;\nPr = 0.71;\ndt = 1e-5;\n\n\nh = 1/N; % N+1 points in each direction\n[xx,yy] = meshgrid(0:h:1); % 2D gridpoints\n\n% Laplacian with zero BCs\no = ones(N-1,1);\nD2 = (diag(-2*o) + diag(o(1:N-2),-1)+ diag(o(1:N-2),1))/h^2;\n% We need evals and evects to solve Poisson equation\n[eV,eval] = eig(D2); lam = diag(eval);\n\n% Initial condition & pre-allocate memory (start with zeros at t = 0)\nu = 0*xx; v = u; w = u; str = u; T = u;\n\n% boundary condition\n%u(N+1,:) = 1;\nii = 2:N; jj = 2:N; % index for interior nodes\nwi = zeros(N-1,N-1); % interior values of vorticity\nTi = zeros(N-1,N-1); % interior values of temperature\n\ncount = 0;\nt = 0;\nNplot = round(0.1/dt);\n% main loop\nwhile t<50\n if mod(count,1000) == 0\n sprintf('%d -- %f -- %f -- %f -- %f',count,t,norm(str,inf),...\n norm(w,inf),norm(T,inf));\n end\n\n % update temperature at boundary\n T = BCT(T,xx,t);\n\n % update vorticity at boundary\n w = BCw(u,v,w);\n\n % advance temperature with forward Euler\n [Trh,Tx] = Trhs(u,v,T);\n Ti = Ti + dt*Trh;\n T(ii,jj) = Ti;\n\n % advance vorticity with forward Euler\n wi = wi + dt*wrhs(u,v,w,Pr,Ra,Tx);\n w(ii,jj) = wi;\n\n\n % compute stream function with conjugate gradient\n str(2:end-1,2:end-1) = SolvePoisson(wi);\n\n % update velocity (interior nodes only)\n u(ii,jj) = (str(ii+1,jj)-str(ii-1,jj))/(2*h);\n v(ii,jj) = -(str(ii,jj+1)-str(ii,jj-1))/(2*h);\n\n t = t+dt;\n count = count + 1;\n\n % Plot results every now and then\n if count == Nplot\n plotresults(u,v,w,T,str);\n shg\n count = 0;\n disp(t)\n sprintf('%f -- %f -- %f',norm(T,inf),norm(w,inf),norm(str,inf));\n end\n\nend\n\n % Solve Poisson equation (Sylvester equation)\n function S = SolvePoisson(wi)\n ff = -eV'*wi*eV;\n S = ff;\n for j = 1:N-1\n S(:,j) = ff(:,j)./(lam(j)+lam);\n end\n S = eV*S*eV';\n end\n\n % Right-hand side of vorticity equation\n function wrh = wrhs(u,v,w,Pr,Ra,Tx)\n wx = (w(ii,jj+1)-w(ii,jj-1))/(2*h);\n wy = (w(ii+1,jj)-w(ii-1,jj))/(2*h);\n Lw = (w(ii,jj+1)+w(ii,jj-1)+w(ii+1,jj)+w(ii-1,jj)-4*w(ii,jj))/h^2;\n wrh = -u(ii,jj).*wx - v(ii,jj).*wy + (Pr)*Lw + Ra*Pr*Tx;\n end\n\n % Right-hand side of Temperature equation\n function [Trh,Tx] = Trhs(u,v,T)\n Tx = (T(ii,jj+1)-T(ii,jj-1))/(2*h);\n Ty = (T(ii+1,jj)-T(ii-1,jj))/(2*h);\n LT = (T(ii,jj+1)+T(ii,jj-1)+T(ii+1,jj)+T(ii-1,jj)-4*T(ii,jj))/h^2;\n Trh = -u(ii,jj).*Tx - v(ii,jj).*Ty + LT;\n end\n\n function w = BCw(u,v,w)\n % three point sided FD formula (second order accurate)\n w(1,:) = -(-3*u(1,:)+4*u(2,:)-u(3,:))/(2*h); % = -u_y @ y = 0\n w(N+1,:) = (-3*u(N+1,:)+4*u(N,:)-u(N-1,:))/(2*h); % = -u_y @ y = 1\n w(:,1) = (-3*v(:,1)+4*v(:,2)-v(:,3))/(2*h); % = v_x @ x = 0\n w(:,N+1) = -(-3*v(:,N+1)+4*v(:,N)-v(:,N-1))/(2*h);% = v_x @ x = 1\n end\n % Enforce boundary condition (Temperature)\n function T = BCT(T,xx,t)\n %T0 = @(x,y,t) 2^9 * tanh(100*t)^4 * x.^5 .* (x-1).^4 .* (y==0);\n %T(1,:) = T0(x(1,:),y(1,:),t);\n T(1,:) = 2^9*(tanh(100*t))^4*(xx(1,:).^5).*(xx(1,:)-1).^4;\n %T(N+1,:) = 0;\n %T(:,1) = 0;\n %T(:,N+1) = 0;\n end\n\n % Plot results\n function plotresults(u,v,w,T,str)\n % Use a subset of gridpoints\n ip = 1:3:N+1;\n up = u(ip,ip); vp = v(ip,ip);\n xp = xx(ip,ip); yp = yy(ip,ip);\n\n % Velocity\n subplot(2,2,1)\n quiver(xp,yp,up,vp,10)\n axis([0 1 0 1]), axis square\n title('velocity field','fontsize',16)\n\n % Temperature\n subplot(2,2,2)\n contourf(xx,yy,T);\n H = colorbar; set(H,'fontsize',16)\n axis square, title('Temperature','fontsize',16)\n\n % Stream function\n subplot(2,2,3)\n mp = max(max(str));\n mm = min(min(str));\n contour(xx,yy,str,-logspace(-10,log10(-mm),30),'r'), hold on\n contour(xx,yy,str, logspace(-10,log10(mp),20),'k'), hold off\n axis square, title('streamlines (logscale)','fontsize',16)\n\n % Vorticity\n subplot(2,2,4)\n logw = log10(abs(w)); logw(abs(w)==0) = nan;\n contourf(xx,yy,logw,-3:0.5:2),\n H = colorbar; set(H,'fontsize',16)\n set(gca,'position',[0.5703 0.1100 0.3347 0.3412])\n axis square, title('vorticity (log(abs(w))','fontsize',16)\n drawnow\n\n end\nend\n\\end{verbatim}", "meta": {"hexsha": "d5c0f0ac34ff9c3de93a7e7e4bfcbeff34008eab", "size": 5503, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Year_1/ComputationalMethods/Homework 6/Latex/texfiles/problem4.tex", "max_stars_repo_name": "fjcasti1/Courses", "max_stars_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "Year_1/ComputationalMethods/Homework 6/Latex/texfiles/problem4.tex", "max_issues_repo_name": "fjcasti1/Courses", "max_issues_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "Year_1/ComputationalMethods/Homework 6/Latex/texfiles/problem4.tex", "max_forks_repo_name": "fjcasti1/Courses", "max_forks_repo_head_hexsha": "12ab3e86a4a44270877e09715eeab713da45519d", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 28.9631578947, "max_line_length": 341, "alphanum_fraction": 0.5340723242, "num_tokens": 2111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.849971190859164, "lm_q2_score": 0.7185943925708562, "lm_q1q2_score": 0.6107845315981683}} {"text": "\n\\section{Bound-Preserving Methods Using First-Order DG Scheme}\n\n\\subsection{Cartesian Coordinates}\nThis section closely follows \\citet{Qin2016}.\n\n\\subsubsection{Set of Admissible States}\nWe consider a one-dimensional system of conservation laws:\n\\begin{equation}\n \\pd{\\,\\bU}{t} + \\pd{\\,\\bF\\left(\\bU\\right)}{x} = \\vect{0},\n\\end{equation}\nwhere $\\bU$ is a vector of conserved variables, defined as:\n\\begin{equation}\\label{Eq:ConservedVariables}\n \\bU\\longrightarrow\\begin{pmatrix} D\\\\ S\\\\ \\tau\\end{pmatrix}=\\begin{pmatrix} \\rho\\,W\\\\ \\rho\\,h\\,W^{2}\\,v\\\\ \\rho\\,W\\left(h\\,W-1\\right)-p\\end{pmatrix},\n\\end{equation}\nand $\\bF\\left(\\bU\\right)$ are the fluxes of those conserved quantities:\n\\begin{equation}\\label{Eq:FluxVector}\n \\bF\\left(\\bU\\right)\\longrightarrow\\begin{pmatrix}\\rho\\,W\\,v\\\\ \\rho\\,h\\,W^{2}\\,v^{2}+p\\\\\\rho\\,h\\,W^{2}\\,v-D\\,v\\end{pmatrix}.\n\\end{equation}\n\n\nThe physics leads us to define a set of admissible states, $\\cG_{p}$ (the subscript $p$ stands for primitive), as:\n\\begin{equation}\n \\cG_{p}\\equiv\\left\\{\\bU\\Big|\\rho>0,\\,p>0,\\,v^{2}<1\\right\\}.\n\\end{equation}\n\nIt is shown in \\citet{Mignone2005} that $\\cG$ is a convex set\\footnote{Convex in the sense that if $\\bU_{1}\\in\\cG$ and $\\bU_{2}\\in\\cG$, then $\\alpha_{1}\\,\\bU_{1}+\\alpha_{2}\\,\\bU_{2}\\in\\cG$, where $\\alpha_{1},\\,\\alpha_{2}\\in\\left[0,1\\right]$ and $\\alpha_{1}+\\alpha_{2}=1$.} and can equivalently be written in terms of the conserved variables as:\n\\begin{equation}\\label{Eq:SetOfAdmissibleStates}\n \\cG\\equiv\\left\\{\\bU\\Big|D>0,\\,\\tau+D>\\sqrt{D^{2}+S^{2}}\\right\\}.\n\\end{equation}\n\n\\subsubsection{Time-Step Derivation/CFL Condition}\nFor the first-order DG method using forward-Euler time-stepping, we evolve the vector of conserved variables as:\n\\begin{equation}\\label{Eq:1stOrderDG}\n \\ol{\\bU}^{n+1}_{i}=\\ol{\\bU}^{n}_{i}-\\eta_{i}\\left[\\hat{\\bF}\\left(\\ol{\\bU}^{n}_{i},\\ol{\\bU}^{n}_{i+1}\\right)-\\hat{\\bF}\\left(\\ol{\\bU}^{n}_{i-1},\\ol{\\bU}^{n}_{i}\\right)\\right],\n\\end{equation}\nwhere\n\\begin{equation}\n \\ol{\\bU}_{i}\\equiv\\f{1}{\\Delta x_{i}}\\int_{x_{\\imh}}^{x_{\\iph}}\\bU_{i}\\,dx,\n\\end{equation}\n$\\eta_{i}\\equiv\\Delta t_{i}/\\Delta x_{i}$, and $\\hat{\\bF}$ is the numerical flux. In this document we use the local Lax-Friedrichs flux, defined as:\n\\begin{equation}\\label{Eq:LLF}\n \\hat{\\bF}\\left(a,b\\right)=\\f{1}{2}\\left[\\bF\\left(a\\right)+\\bF\\left(b\\right) - \\alpha_{ab}\\left(b-a\\right)\\right],\n\\end{equation}\nwhere $a$ and $b$ represent the state of the fluid in two different elements, $\\alpha_{ab}$ is an estimate for the wave-speed:\n\\begin{equation}\n \\alpha_{ab}=\\text{max}\\left[\\alpha\\left(a\\right),\\alpha\\left(b\\right)\\right],\n\\end{equation}\nand $\\alpha$ is the largest (in absolute value) eigenvalue of the flux-Jacobian:\n\\begin{equation}\n \\alpha=\\left|\\left|\\pderiv{\\bF}{\\bU}\\right|\\right|.\n\\end{equation}\nUsing this we define the following variables:\n\\begin{equation}\\label{Eq:EigVals}\n \\alpha_{\\iph}=\\text{max}\\left[\\alpha\\left(\\ol{\\bU}_{i}\\right),\\alpha\\left(\\ol{\\bU}_{i+1}\\right)\\right],\\hspace{3em} \\alpha_{\\imh}=\\text{max}\\left[\\alpha\\left(\\ol{\\bU}_{i-1}\\right),\\alpha\\left(\\ol{\\bU}_{i}\\right)\\right].\n\\end{equation}\n\nSubstituting \\eqref{Eq:LLF} with \\eqref{Eq:EigVals} into \\eqref{Eq:1stOrderDG}:\n\\begin{align}\n \\ol{\\bU}^{n+1}_{i}&=\\ol{\\bU}^{n}_{i}-\\f{\\eta_{i}}{2}\\left[\\bF\\left(\\ol{\\bU}^{n}_{i}\\right)+\\bF\\left(\\ol{\\bU}^{n}_{i+1}\\right)-\\alpha_{\\iph}\\left(\\ol{\\bU}^{n}_{i+1}-\\ol{\\bU}^{n}_{i}\\right)\\right.\\nonumber\\\\\n &\\left.\\hspace{7em}-\\bF\\left(\\ol{\\bU}^{n}_{i}\\right)-\\bF\\left(\\ol{\\bU}^{n}_{i-1}\\right)+\\alpha_{\\imh}\\left(\\ol{\\bU}^{n}_{i}-\\ol{\\bU}^{n}_{i-1}\\right)\\right]\\nonumber\\\\\n &=\\left[1-\\f{\\eta_{i}}{2}\\left(\\alpha_{\\iph}+\\alpha_{\\imh}\\right)\\right]\\ol{\\bU}^{n}_{i}+\\f{\\eta_{i}}{2}\\,\\alpha_{\\iph}\\left[\\ol{\\bU}^{n}_{i+1}-\\f{1}{\\alpha_{\\iph}}\\bF\\left(\\ol{\\bU}^{n}_{i+1}\\right)\\right]\\nonumber\\\\\n &\\hspace{15em}+\\f{\\eta_{i}}{2}\\,\\alpha_{\\imh}\\left[\\ol{\\bU}^{n}_{i-1}+\\f{1}{\\alpha_{\\imh}}\\bF\\left(\\ol{\\bU}^{n}_{i-1}\\right)\\right]\\nonumber\\\\\n &=\\left[1-\\f{\\eta_{i}}{2}\\left(\\alpha_{\\iph}+\\alpha_{\\imh}\\right)\\right]\\ol{\\bU}^{n}_{i}+\\f{\\eta_{i}}{2}\\,\\alpha_{\\iph}\\,\\ol{\\bH}^{-}\\left(\\ol{\\bU}^{n}_{i+1},\\alpha_{\\iph}\\right)+\\f{\\eta_{i}}{2}\\,\\alpha_{\\iph}\\,\\ol{\\bH}^{+}\\left(\\ol{\\bU}^{n}_{i-1},\\alpha_{\\imh}\\right),\\label{Eq:ConvComb}\n\\end{align}\nwhere\n\\begin{equation}\\label{Eq:Hpm}\n \\ol{\\bH}^{\\pm}\\left(\\ol{\\bU},\\alpha\\right)\\equiv\\ol{\\bU}\\pm\\f{1}{\\alpha}\\,\\bF\\left(\\ol{\\bU}\\right).\n\\end{equation}\n\nThe proof that $\\ol{\\bH}^{\\pm}\\in\\cG$ is given in \\citet{Qin2016}. Therefore, we see that with a restriction on $\\alpha_{i\\pm\\f{1}{2}}$ that \\eqref{Eq:ConvComb} is a convex combination. The restriction is (recalling that $\\eta_{i}=\\Delta t_{i}/\\Delta x_{i}$):\n\\begin{equation}\n 1-\\f{\\eta_{i}}{2}\\left(\\alpha_{\\iph}+\\alpha_{\\imh}\\right)>0\\implies\\f{\\eta_{i}}{2}\\left(\\alpha_{\\iph}+\\alpha_{\\imh}\\right)<1\\implies \\Delta t_{i}<\\f{2\\,\\Delta x_{i}}{\\alpha_{\\iph}+\\alpha_{\\imh}}\\leq\\f{\\Delta x_{i}}{\\text{max}\\left(\\alpha_{i\\pm\\f{1}{2}}\\right)}.\n\\end{equation}\n\nWe want a time-step that is the same for all elements at a given time, so we tighten the restriction to:\n\\begin{equation}\n \\Delta t<\\text{min}_{i}\\left(\\f{\\Delta x_{i}}{\\text{max}\\left(\\alpha_{i\\pm\\f{1}{2}}\\right)}\\right)=\\f{\\Delta x}{\\text{max}_{i}\\left(\\alpha_{i\\pm\\f{1}{2}}\\right)},\n\\end{equation}\nwhere the equality follows for a uniform mesh, i.e. $\\Delta x_{i}=\\Delta x\\,\\forall i$.\n\n\\newpage\n\n\n\\subsection{Curvilinear Coordinates in 1-D}\nNOTE: We assume a conformally-flat, time-independent spatial three-metric:\n\\begin{equation}\n \\gamma_{ij}\\left(x^{k},t\\right)\\longrightarrow\\psi^{4}\\left(x^{k}\\right)\\,\\ol{\\gamma_{ii}}\\left(x^{k}\\right),\n\\end{equation}\nwhere $\\psi\\left(x^{k}\\right)$ is the conformal factor and $\\ol{\\gamma}_{ii}$ is the flat-space metric.\n\n\\subsubsection{Set of Admissible States}\nWe again consider a one-dimensional system of conservation laws, but this time with a curvilinear metric:\n\\begin{equation}\\label{Eq:1DCurvilinearConsLaw}\n \\pd{\\left(\\sqrtgm\\,\\bU\\right)}{t}+\\pd{\\left(\\sqrtgm\\,\\bF^{i}\\right)}{i}=\\sqrtgm\\,\\bQ,\\hspace{1em}\\text{(no sum on $i$)},\n\\end{equation}\nwhere $\\bU$ is given by:\n\\begin{equation}\n \\bU\\longrightarrow\\begin{pmatrix} D\\\\ S_{j}\\\\ \\tau\\end{pmatrix}=\\begin{pmatrix} \\rho\\,W\\\\ \\rho\\,h\\,W^{2}\\,v_{j}\\\\ \\rho\\,W\\left(h\\,W-1\\right)-p\\end{pmatrix}=\\begin{pmatrix} \\rho\\,W\\\\ \\rho\\,h\\,W^{2}\\,\\gamma_{jk}\\,v^{k}\\\\ \\rho\\,W\\left(h\\,W-1\\right)-p\\end{pmatrix},\n\\end{equation}\n$\\bF^{i}\\left(\\bU\\right)$ are the fluxes in the $x^{i}$-direction of those conserved quantities:\n\\begin{equation}\n \\bF^{i}\\left(\\bU\\right)\\longrightarrow\\begin{pmatrix}D\\,v^{i}\\\\ S^{i}\\,v_{j}+p\\,\\delta^{i}_{~j}\\\\ S^{i}-D\\,v^{i}\\end{pmatrix}=\\begin{pmatrix}\\rho\\,W\\,v^{i}\\\\ \\rho\\,h\\,W^{2}\\,v^{i}\\,v_{j}+p\\,\\delta^{i}_{~j}\\\\\\rho\\,h\\,W^{2}\\,v^{i}-D\\,v^{i}\\end{pmatrix}=\\begin{pmatrix}\\rho\\,W\\,v^{i}\\\\ \\rho\\,h\\,W^{2}\\,\\gamma_{jk}\\,v^{i}\\,v^{k}+p\\,\\delta^{i}_{~j}\\\\\\rho\\,h\\,W^{2}\\,v^{i}-D\\,v^{i}\\end{pmatrix},\n\\end{equation}\nand $\\bQ$ is a source term:\n\\begin{align}\n \\bQ\\longrightarrow\\begin{pmatrix}0\\\\\\f{1}{2}\\,P^{ik}\\,\\pd{\\,\\gamma_{ik}}{j}\\\\0\\end{pmatrix}&=\\begin{pmatrix}0\\\\ \\f{1}{2}\\left[P^{11}\\,\\pd{\\,\\gamma_{11}}{j}+P^{22}\\,\\pd{\\,\\gamma_{22}}{j}+P^{33}\\,\\pd{\\,\\gamma_{33}}{j}\\right] \\\\0\\end{pmatrix}\\\\\n &=\\begin{pmatrix}0\\\\ P^{11}\\,h_{1}\\,\\pd{\\,h_{1}}{j}+P^{22}\\,h_{2}\\,\\pd{\\,h_{2}}{j}+P^{33}\\,h_{3}\\,\\pd{\\,h_{3}}{j} \\\\0\\end{pmatrix},\n\\end{align}\nwhere we have used the fact that $\\gamma_{kk}=\\left(h_{k}\\right)^{2}$. The $P^{ik}$ are components of the pressure tensor:\n\\begin{equation}\n P^{ik}=S^{i}\\,v^{k}+p\\,\\gamma^{ik}=\\gamma^{i\\ell}\\,S_{\\ell}\\,v^{k}+p\\,\\gamma^{ik}=\\gamma^{i\\ell}\\,S_{\\ell}\\,v^{k}+p\\,\\gamma^{i\\ell}\\,\\delta^{k}_{~\\ell}=\\gamma^{i\\ell}\\left(S_{\\ell}\\,v^{k}+p\\,\\delta^{k}_{~\\ell}\\right).\n\\end{equation}\nSince the spatial three-metric is diagonal the only non-zero term is that with $\\ell=i$. We can therefore simplify further:\n\\begin{equation}\n P^{ik}=\\gamma^{ii}\\left(S_{i}\\,v^{k}+p\\,\\delta^{k}_{~i}\\right)=\\f{1}{\\gamma_{ii}}\\,\\left(S_{i}\\,v^{k}+p\\,\\delta^{k}_{~i}\\right),\\hspace{1em}\\text{(no sum on $i$)}.\n\\end{equation}\nFor the source-term sum we then have:\n\\begin{equation}\\label{Eq:PressureTensorSum}\n Q_{j}=\\f{1}{2}\\,P^{ik}\\,\\p_{j}\\,\\gamma_{ik}=\\f{1}{2}\\,P^{kk}\\,\\p_{j}\\,\\gamma_{kk}=\\f{1}{2}\\,P^{kk}\\,\\p_{j}\\left(h_{k}\\right)^{2}=P^{kk}\\,h_{k}\\,\\p_{j}\\,h_{k}.\n\\end{equation}\n\nThese definitions lead us to define the same set of admissible states as before, namely:\n\\begin{equation}\n \\cG_{p}\\equiv\\left\\{\\bU\\Big|\\rho>0,\\,p>0,\\,v^{2}<1\\right\\},\n\\end{equation}\nthe only difference being that $v^{2}$ now involves the metric:\n\\begin{equation}\n v^{2}=v^{j}\\,v_{j}=\\gamma_{kj}\\,v^{k}\\,v^{j}.\n\\end{equation}\n\nBefore continuing, we show that the introduction of the metric doesn't affect the translation between $\\cG_{p}$ and $\\cG$...\\sd{I've shown this, just need to TeX it up}\n\n\\subsubsection{Time-Step Derivation/CFL Condition}\nWe start by integrating both sides of \\eqref{Eq:1DCurvilinearConsLaw} over $dx^{i}$ and dividing by the volume of the $K^{th}$ element, $\\Delta V_{K}$ (recalling that there is no sum on $i$):\n\\begin{equation}\n \\f{1}{\\Delta V_{K}}\\int_{x^{i}_{L}}^{x^{i}_{H}}\\pd{\\left(\\sqrtgm\\,\\bU\\right)}{t}dx^{i}+\\f{1}{\\Delta V_{K}}\\int_{x^{i}_{L}}^{x^{i}_{H}}\\pd{\\left(\\sqrtgm\\,\\bF^{i}\\left(\\bU\\right)\\right)}{i}dx^{i}=\\f{1}{\\Delta V_{K}}\\int_{x^{i}_{L}}^{x^{i}_{H}}\\sqrtgm\\,\\bQ\\,dx^{i},\n\\end{equation}\nwhere:\n\\begin{equation}\n \\Delta V_{K}=\\int_{x^{i}_{L}}^{x^{i}_{H}}dV=\\int_{x^{i}_{L}}^{x^{i}_{H}}\\sqrtgm\\,dx^{i}.\n\\end{equation}\n\nBy defining the cell-average as:\n\\begin{equation}\n \\bW_{K}\\equiv\\f{1}{\\Delta V_{K}}\\int_{x^{i}_{L}}^{x^{i}_{H}}\\bW\\,dV,\n\\end{equation}\nwe have:\n\\begin{equation}\n \\f{d\\,\\bU_{K}}{dt}+\\f{1}{\\Delta V_{K}}\\left.\\left(\\sqrtgm\\,\\hat{\\bF}\\left(\\bU\\right)\\right)\\right|^{x^{i}_{H}}_{x^{i}_{L}}=\\bQ_{K}.\n\\end{equation}\nNow, using the common notation of the time step being represented as a superscript $n$:\n\\begin{equation}\n \\bU^{n+1}_{K}=\\bU^{n}_{K}-\\f{\\Delta t^{n}_{K}}{\\Delta V_{K}}\\left[\\sqrtgm_{H}\\,\\hat{\\bF}^{n}_{H}-\\sqrtgm_{L}\\,\\hat{\\bF}^{n}_{L}\\right]+\\Delta t^{n}_{K}\\,\\bQ^{n}_{K}.\n\\end{equation}\n\nNow we define a parameter a la \\citet{ZS2011b}: $\\ve\\in\\left(0,1\\right)$, such that (NOTE: \\citet{ZS2011b} set $\\ve=1/2$):\n\\begin{equation}\n \\bU^{n}_{K}=\\ve\\,\\bU^{n}_{K}+\\left(1-\\ve\\right)\\bU^{n}_{K}.\n\\end{equation}\nWe can use the first term to balance out the term in the square brackets and the second term to balance out the source term.\n\nSo, we get:\n\\begin{align}\n \\bU^{n+1}_{K}&=\\ve\\left\\{\\bU^{n}_{K}-\\f{\\Delta t^{n}_{K}}{\\ve\\,\\Delta V_{K}}\\left[\\sqrtgm_{H}\\,\\hat{\\bF}^{n}_{H}-\\sqrtgm_{L}\\,\\hat{\\bF}^{n}_{L}\\right]\\right\\}+\\left(1-\\ve\\right)\\bU^{n}_{K}+\\Delta t^{n}_{K}\\,\\bQ^{n}_{K}\\\\\n &=\\ve\\left\\{\\bU^{n}_{K}-\\eta^{n}_{K}\\left[\\sqrtgm_{H}\\,\\hat{\\bF}\\left(\\bU^{n}_{K+1},\\bU^{n}_{K}\\right)-\\sqrtgm_{L}\\,\\hat{\\bF}\\left(\\bU^{n}_{K},\\bU^{n}_{K-1}\\right)\\right]\\right\\}+\\left(1-\\ve\\right)\\bU^{n}_{K}+\\Delta t^{n}_{K}\\,\\bQ^{n}_{K}\\\\\n &=\\ve\\,\\bH_{K,1}+\\left(1-\\ve\\right)\\bH_{K,2},\n\\end{align}\nwhere\n\\begin{equation}\n \\bH_{K,1}\\equiv \\bU^{n}_{K}-\\eta^{n}_{K}\\left[\\sqrtgm_{H}\\,\\hat{\\bF}\\left(\\bU^{n}_{K+1},\\bU^{n}_{K}\\right)-\\sqrtgm_{L}\\,\\hat{\\bF}\\left(\\bU^{n}_{K},\\bU^{n}_{K-1}\\right)\\right],\n\\end{equation}\n\\begin{equation}\n \\bH_{K,2}\\equiv\\bU^{n}_{K}+\\f{\\Delta t^{n}_{K}}{1-\\ve}\\,\\bQ^{n}_{K},\n\\end{equation}\nand\n\\begin{equation}\n \\eta^{n}_{K}\\equiv\\f{\\Delta t^{n}_{K}}{\\ve\\,\\Delta V_{K}}.\n\\end{equation}\n\nWe proceed by focusing on each term individually, starting with the numerical flux term, $\\bH_{K,1}$.\n\n\\subsubsection{Numerical flux term}\nWe have to show that $\\bH_{K,1}\\in\\cG$. We again we use the Local-Lax-Friedrichs flux, \\eqref{Eq:LLF}, yielding for $\\bH_{K,1}$:\n\\begin{align}\n \\bU^{n}_{K}-\\f{\\eta^{n}_{K}}{2}\\Big\\{&\\sqrtgm_{H}\\left[\\bF\\left(\\bU^{n}_{K+1}\\right)+\\bF\\left(\\bU^{n}_{K}\\right)-\\alpha^{n}_{H}\\left(\\bU^{n}_{K+1}-\\bU^{n}_{K}\\right)\\right]\\\\\n &-\\sqrtgm_{L}\\left[\\bF\\left(\\bU^{n}_{K}\\right)+\\bF\\left(\\bU^{n}_{K-1}\\right)-\\alpha^{n}_{L}\\left(\\bU^{n}_{K}-\\bU^{n}_{K-1}\\right)\\right]\\Big\\}\\\\\n &\\hspace{-10em}=\\left(1-\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}-\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\right)\\bU^{n}_{K}\\\\\n &\\hspace{-8em}-\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\bF\\left(\\bU^{n}_{K}\\right)+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\bF\\left(\\bU^{n}_{K}\\right)\\\\\n &\\hspace{-8em}+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\left[\\bU^{n}_{K-1}+\\f{1}{\\alpha^{n}_{L}}\\bF\\left(\\bU^{n}_{K-1}\\right)\\right]+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}\\left[\\bU^{n}_{K+1}-\\f{1}{\\alpha^{n}_{H}}\\bF\\left(\\bU^{n}_{K+1}\\right)\\right].\n\\end{align}\nNow we add and subtract $\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}\\,\\bU^{n}_{K}$ and $\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\,\\bU^{n}_{K}$, yielding:\n\\begin{align}\n &\\left(1-\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}-\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\right)\\bU^{n}_{K}\\\\\n &+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}\\left[\\bU^{n}_{K}-\\f{1}{\\alpha^{n}_{H}}\\bF\\left(\\bU^{n}_{K}\\right)\\right]+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\left[\\bU^{n}_{K}+\\f{1}{\\alpha^{n}_{L}}\\bF\\left(\\bU^{n}_{K}\\right)\\right]\\\\\n &+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\left[\\bU^{n}_{K-1}+\\f{1}{\\alpha^{n}_{L}}\\bF\\left(\\bU^{n}_{K-1}\\right)\\right]+\\f{1}{2}\\,\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}\\left[\\bU^{n}_{K+1}-\\f{1}{\\alpha^{n}_{H}}\\bF\\left(\\bU^{n}_{K+1}\\right)\\right].\n\\end{align}\nAll of the terms in square brackets are similar to the $\\bH_{K}$ quantities in \\citet{Qin2016}, and are therefore in $\\cG$. It can easily be seen that the sum of the coefficients is unity. The final condition is that the coefficient of $\\bU^{n}_{K}>0$, or (recalling that $\\eta^{n}_{K}=\\Delta t_{K}/\\left(\\ve\\,\\Delta V_{K}\\right)$):\n\\begin{align}\n 1&-\\eta^{n}_{K}\\,\\sqrtgm_{H}\\,\\alpha^{n}_{H}-\\eta^{n}_{K}\\,\\sqrtgm_{L}\\,\\alpha^{n}_{L}>0\\implies\\eta^{n}_{K}\\left(\\sqrtgm_{H}\\,\\alpha^{n}_{H}+\\sqrtgm_{L}\\,\\alpha^{n}_{L}\\right)<1\\\\\n &\\implies\\Delta t^{n}_{K}<\\f{\\ve\\,\\Delta V_{K}}{\\sqrtgm_{H}\\,\\alpha^{n}_{H}+\\sqrtgm_{L}\\,\\alpha^{n}_{L}}\\leq\\f{\\ve\\,\\Delta V_{K}}{2\\,\\text{max}\\left(\\sqrtgm_{K\\pm\\f{1}{2}}\\,\\alpha^{n}_{K\\pm\\f{1}{2}}\\right)}.\n\\end{align}\nAgain we want a time-step that is the same for all elements at a given time, so:\n\\begin{equation}\n \\Delta t^{n}<\\text{min}_{K}\\left(\\f{\\ve\\,\\Delta V_{K}}{2\\,\\text{max}\\left(\\sqrtgm_{K\\pm\\f{1}{2}}\\,\\alpha^{n}_{K\\pm\\f{1}{2}}\\right)}\\right).\n\\end{equation}\n\nWe close the numerical flux section by writing the explicit form of the time-step for spherical-polar coordinates.\n\n\\subsubsubsection{Time-step for Spherical-Polar Coordinates}\nFor spherical-polar coordinates in 1-D we have that $\\Delta V_{K}=1/3\\left(r_{H}^{3}-r_{L}^{3}\\right)$, and (assuming $\\alpha_{K\\pm\\f{1}{2}}=1\\ \\forall\\ i$) $\\text{max}\\left(\\sqrtgm_{K\\pm\\f{1}{2}}\\,\\alpha_{K\\pm\\f{1}{2}}\\right)=r_{H}^{2}$, so:\n\\begin{align}\n \\Delta t&<\\text{min}_{i}\\left\\{\\f{\\ve\\,1/3\\left[r_{H}^{3}-r_{L}^{3}\\right]}{2\\,r_{H}^{2}}\\right\\}\\\\\n &=\\text{min}_{i}\\left\\{\\f{\\ve}{6}\\,r_{H}\\left[1-\\f{r_{L}^{3}}{r_{H}^{3}}\\right]\\right\\}\\\\\n &=\\text{min}_{i}\\left\\{\\f{\\ve}{6}\\,r_{H}\\left[1-\\left(1-\\f{\\Delta r_{i}}{r_{H}}\\right)^{3}\\right]\\right\\}\\\\\n &=\\text{min}_{i}\\left\\{\\f{\\ve}{6}\\,r_{H}\\left[1-\\left(1+\\left(\\f{\\Delta r_{i}}{r_{H}}\\right)^{2}-2\\frac{\\Delta r_{i}}{r_{H}}\\right)\\left(1-\\f{\\Delta r_{i}}{r_{H}}\\right)\\right]\\right\\}\\\\\n &=\\text{min}_{i}\\left\\{\\f{\\ve}{6}\\,r_{H}\\left[\\left(\\f{\\Delta r_{i}}{r_{H}}\\right)^{3}-3\\left(\\f{\\Delta r_{i}}{r_{H}}\\right)^{2}+3\\f{\\Delta r_{i}}{r_{H}}\\right]\\right\\}\\\\\n &=\\text{min}_{i}\\left\\{\\f{\\ve}{6}\\,\\Delta r_{i}\\left[\\left(\\f{\\Delta r_{i}}{r_{H}}\\right)^{2}-3\\left(\\f{\\Delta r_{i}}{r_{H}}\\right)+3\\right]\\right\\}.\n\\end{align}\nWe know that $\\Delta r_{i}/r_{H}\\in\\left[0,1\\right]$; the minimum value of the quadratic function in this domain is unity. So, we have that for spherical-polar coordinates:\n\\begin{equation}\n \\Delta t<\\f{\\ve}{6}\\,\\text{min}\\left(\\Delta r_{i}\\right).\n\\end{equation}\n\nNext we handle the source term.\n\n\\subsubsection{Source term}\nFor this section we drop the subscript $K$ and the superscript $n$ (but keep in mind that all quantities are still cell-averages). We have to show that $\\bH_{2}\\in\\cG$, where\n\\begin{equation}\n \\bH_{2}=\\begin{pmatrix}D\\\\ S_{j}+\\f{\\Delta t}{1-\\ve}\\,Q_{j} \\\\ \\tau\\end{pmatrix},\\hspace{1em}\\left(H_{2}\\right)_{1}>0,\\hspace{1em}\\left(H_{2}\\right)_{5}+\\left(H_{2}\\right)_{1}>\\sqrt{\\left(H_{2}\\right)_{1}\\left(H_{2}\\right)_{1}+\\left(H_{2}\\right)_{j}\\left(H_{2}\\right)^{j}}.\n\\end{equation}\n\nIt is clear that the first requirement for $\\bH_{2}$ is met, i.e. $D>0$. The second requirement is:\n\\begin{align}\n D+\\tau&>\\sqrt{D^{2}+\\left[S_{j}+\\f{\\Delta t}{1-\\ve}Q_{j}\\right] \\left[S^{j}+\\f{\\Delta t}{1-\\ve}Q^{j}\\right]}\\\\\n &=\\sqrt{D^{2}+S_{j}\\,S^{j}+\\f{\\Delta t}{1-\\ve}\\left(S_{j}\\,Q^{j}+S^{j}\\,Q_{j}\\right)+\\left(\\f{\\Delta t}{1-\\ve}\\right)^{2}\\,Q_{j}\\,Q^{j}}\n\\end{align}\nNow we square both sides:\n\\begin{align}\n &D^{2}+\\tau^{2}+2\\,D\\,\\tau>D^{2}+S_{j}\\,S^{j}+\\f{\\Delta t}{1-\\ve}\\left(S_{j}\\,Q^{j}+S^{j}\\,Q_{j}\\right)+\\left(\\f{\\Delta t}{1-\\ve}\\right)^{2}\\,Q_{j}\\,Q^{j}\\\\\n \\implies&\\tau\\left(\\tau+2\\,D\\right)>\\gamma^{jk}\\,S_{j}\\,S_{k}+\\f{2\\,\\Delta t}{1-\\ve}\\,\\gamma^{jk}\\,S_{j}\\,Q_{k}+\\left(\\f{\\Delta t}{1-\\ve}\\right)^{2}\\,\\gamma^{jk}\\,Q_{j}\\,Q_{k}\\\\\n \\implies&\\tau\\left(\\tau+2\\,D\\right)>\\f{S_{j}\\,S_{k}}{\\gamma_{jk}}+\\f{2\\,\\Delta t}{1-\\ve}\\,\\f{S_{j}\\,Q_{k}}{\\gamma_{jk}}+\\left(\\f{\\Delta t}{1-\\ve}\\right)^{2}\\,\\f{Q_{j}\\,Q_{k}}{\\gamma_{jk}}\\\\\n \\implies&a\\left(\\Delta t\\right)^{2}+b\\,\\Delta t+c<0,\n\\end{align}\nwhere:\n\\begin{align}\n a&=\\f{1}{\\left(1-\\ve\\right)^{2}}\\,\\vv{Q}\\cdot\\vv{Q}=\\f{1}{\\left(1-\\ve\\right)^{2}}\\,\\f{Q_{j}\\,Q_{k}}{\\gamma_{jk}}=\\f{1}{\\left(1-\\ve\\right)^{2}}\\sum\\limits_{k=1}^{3}\\f{\\left(Q_{k}\\right)^{2}}{\\gamma_{kk}}\\\\\n b&=\\f{2}{1-\\ve}\\,\\vv{S}\\cdot\\vv{Q}=\\f{2}{1-\\ve}\\,\\f{S_{j}\\,Q_{k}}{\\gamma_{jk}}=\\f{2}{1-\\ve}\\sum\\limits_{k=1}^{3}\\f{S_{k}\\,Q_{k}}{\\gamma_{kk}}\\\\\n c&=-\\tau\\left(\\tau+2\\,D\\right)+\\vv{S}\\cdot\\vv{S}=-\\tau\\left(\\tau+2\\,D\\right)+\\sum\\limits_{k=1}^{3}\\f{\\left(S_{k}\\right)^{2}}{\\gamma_{kk}}.\n\\end{align}\nWe want to make sure that our function has at least one real root, which means we must have that $b^{2}-4\\,a\\,c\\geq0$:\n\\begin{align}\n b^{2}-4\\,a\\,c&=\\f{4}{\\left(1-\\ve\\right)^{2}}\\left(\\vv{S}\\cdot\\vv{Q}\\right)^{2}-\\f{4}{\\left(1-\\ve\\right)^{2}}\\,\\vv{Q}\\cdot\\vv{Q}\\left[-\\tau\\left(\\tau+2\\,D\\right)+\\vv{S}\\cdot\\vv{S}\\right]\\\\\n &=\\f{4}{\\left(1-\\ve\\right)^{2}}\\left[\\left(\\vv{S}\\cdot\\vv{Q}\\right)^{2}-\\left(\\vv{Q}\\cdot\\vv{Q}\\right)\\left(\\vv{S}\\cdot\\vv{S}\\right)+\\tau\\left(\\tau+2\\,D\\right)\\vv{Q}\\cdot\\vv{Q}\\right]\\\\\n &=\\f{4}{\\left(1-\\ve\\right)^{2}}\\left[\\left|\\vv{S}\\right|^{2}\\left|\\vv{Q}\\right|^{2}\\,\\cos^{2}\\theta_{SQ}-\\left|\\vv{Q}\\right|^{2}\\left|\\vv{S}\\right|^{2}+\\tau\\left(\\tau+2\\,D\\right)\\left|\\vv{Q}\\right|^{2}\\right]\\\\\n &=\\f{4}{\\left(1-\\ve\\right)^{2}}\\left|\\vv{Q}\\right|^{2}\\left[\\tau\\left(\\tau+2\\,D\\right)-\\left|\\vv{S}\\right|^{2}\\left(1-\\cos^{2}\\theta_{SQ}\\right)\\right]\\\\\n &=\\f{4}{\\left(1-\\ve\\right)^{2}}\\left|\\vv{Q}\\right|^{2}\\left[\\tau\\left(\\tau+2\\,D\\right)-\\left|\\vv{S}\\right|^{2}\\sin^{2}\\theta_{SQ}\\right],\n\\end{align}\nwhere $\\theta_{SQ}$ is the angle between the momentum-density vector and the source-term vector. To guarantee at least one real root we must have that\n\\begin{equation}\n \\tau\\left(\\tau+2\\,D\\right)\\geq\\left|\\vv{S}\\right|^{2}\\,\\sin^{2}\\theta_{SQ}.\n\\end{equation}\n\n\\newpage\n\\begin{equation}\n b^{2}-4\\,a\\,c'=\\f{4\\left(S_{1}\\right)^{2}}{\\gamma_{11}}a-4\\,a\\left(S_{1}\\,S^{1}-\\tau^{2}-2\\,D\\,\\tau\\right)=4\\,a\\,\\tau\\left(\\tau+2\\,D\\right).\n\\end{equation}\nSince $\\tau\\geq0$, we must have that $\\tau>-2\\,D$. But, from condition two for $\\bH_{K,2}$ we have that $\\tau>-D$, so this condition is automatically satisfied.\n\nThe solutions to this quadratic equation are:\n\\begin{align}\n \\Delta t&=\\f{-b}{2\\,a}\\pm\\f{1}{2\\,a}\\sqrt{b^{2}-4\\,a\\,c'}=\\f{-S_{1}}{\\sqrt{\\gamma_{11}}\\,\\sqrt{a}}\\pm\\f{1}{2\\,a}\\sqrt{\\f{4S_{1}^{2}}{\\gamma_{11}}a-4\\,a\\,c'}\\\\\n &=\\f{-S_{1}}{\\sqrt{\\gamma_{11}}\\,\\sqrt{a}}\\pm\\f{1}{\\sqrt{\\gamma_{11}}\\,\\sqrt{a}}\\sqrt{S_{1}^{2}-c'\\,\\gamma_{11}}=\\f{1}{\\sqrt{\\gamma_{11}}\\,\\sqrt{a}}\\left[-S_{1}\\pm\\sqrt{S_{1}^{2}-c'\\,\\gamma_{11}}\\right]\\\\\n &=\\f{1}{\\sqrt{\\gamma_{11}}\\,\\sqrt{a}}\\left[-S_{1}\\pm\\sqrt{\\gamma_{11}\\left(\\tau^{2}+2\\,D\\,\\tau\\right)}\\right]\\\\\n &=\\f{2\\left(1-\\ve\\right)}{P^{kk}\\,\\p_{1}\\,\\gamma_{kk}}\\left[-S_{1}\\pm\\sqrt{\\gamma_{11}\\left(\\tau^{2}+2\\,D\\,\\tau\\right)}\\right]\\\\\n &=\\f{2\\left(1-\\ve\\right)}{P^{kk}\\,\\p_{1}\\,\\gamma_{kk}}\\left[-S_{1}\\pm\\sqrt{\\gamma_{11}\\,\\tau\\left(\\tau+2\\,D\\right)}\\right].\n\\end{align}\n\nSo, we end up with:\n\\begin{equation}\n \\Delta t<\\text{min}\\left\\{\\text{min}_{i}\\left(\\f{\\ve\\,\\Delta V_{K}}{2\\,\\text{max}\\left(\\sqrtgm_{i\\pm\\f{1}{2}}\\,\\alpha_{i\\pm\\f{1}{2}}\\right)}\\right),\\text{min}^{n}_{i}\\left(\\f{2\\left(1-\\ve\\right)}{P^{kk}\\,\\p_{1}\\,\\gamma_{kk}}\\left[-S_{1}\\pm\\sqrt{\\gamma_{11}\\,\\tau\\left(\\tau+2\\,D\\right)}\\right]\\right)\\right\\}.\n\\end{equation}\n\n\\subsection{Demanding that $q>0$}\n\nSometimes it happens that the cell-average of $q$, $q_{K}\\equiv q\\left(\\bU_{K}\\right)<0$, so our positivity limiter will fail. To get around this we modify the conserved energy, $\\tau$, to demand that $q=\\ve$, where $0<\\ve\\ll1$. The transformation we make is:\n\\begin{equation}\n \\tau_{K}\\longrightarrow\\alpha\\,\\tau_{K},\\hspace{1em}\\alpha>1.\n\\end{equation}\nThis modifies the definition of $q_{K}$ from:\n\\begin{equation}\n q_{K}=\\tau_{K}+D_{K}-\\sqrt{D_{K}^{2}+S_{K}^{2}+\\ve}<0,\n\\end{equation}\nto:\n\\begin{equation}\n \\ve=\\alpha\\,\\tau_{K}+D_{K}-\\sqrt{D_{K}^{2}+S_{K}^{2}+\\ve}.\n\\end{equation}\nSolving this for $\\alpha$, we get:\n\\begin{equation}\n \\alpha=\\tau_{K}^{-1}\\left[\\ve-D_{K}+\\sqrt{D_{K}^{2}+S_{K}^{2}+\\ve}\\right].\n\\end{equation}\n", "meta": {"hexsha": "66f3dec705f6e2762a2e5ac810b138df4ca288cc", "size": 21767, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "Documents/Euler/SamsTexFiles/BP_Methods_FirstOrder_DGScheme.tex", "max_stars_repo_name": "srichers/thornado", "max_stars_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 6, "max_stars_repo_stars_event_min_datetime": "2019-12-08T16:16:55.000Z", "max_stars_repo_stars_event_max_datetime": "2020-07-24T19:31:21.000Z", "max_issues_repo_path": "Documents/Euler/SamsTexFiles/BP_Methods_FirstOrder_DGScheme.tex", "max_issues_repo_name": "srichers/thornado", "max_issues_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": 9, "max_issues_repo_issues_event_min_datetime": "2019-07-10T20:13:15.000Z", "max_issues_repo_issues_event_max_datetime": "2021-11-11T13:21:00.000Z", "max_forks_repo_path": "Documents/Euler/SamsTexFiles/BP_Methods_FirstOrder_DGScheme.tex", "max_forks_repo_name": "srichers/thornado", "max_forks_repo_head_hexsha": "bc6666cbf9ae8b39b1ba5feffac80303c2b1f9a8", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 8, "max_forks_repo_forks_event_min_datetime": "2018-11-14T01:13:40.000Z", "max_forks_repo_forks_event_max_datetime": "2022-01-24T02:08:20.000Z", "avg_line_length": 72.0761589404, "max_line_length": 393, "alphanum_fraction": 0.6074792117, "num_tokens": 9484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8902942173896131, "lm_q2_score": 0.6859494550081926, "lm_q1q2_score": 0.6106968332153505}} {"text": "\\chapter{Velocity reconstruction using regression} \n\\label{chap_linearregression} \n\nThis chapter presents regression models to address the problem of estimating high resolution fields from low resolution measurements as defined in section \\ref{sec:probdef1}. It reviews the basic ordinary least squares model, regularized linear regression with the common L2 penalty as well as the L1 penalty, and kernel regression. These models are presented in matrix forms, which are more compact than previous works on turbulence. The problem of selecting model complexity and optimizing model parameters, a fundamental topic in learning theory, are also discussed. \n\nThe models are used to reconstruct high-resolution fields from low-resolution measurements. The DNS database of isotropic turbulence presented in section \\ref{sec:data_isotropic} are used for illustration. Only partial results on selecting the models and optimizing parameters are shown in this chapter. Optimal models will be used in chapter \\ref{chap_comparisons} when comparing performances of all proposed models. \n\n\\section{Regression in turbulence studies}\nRegression is probably the earliest method in statistics for prediction. It aims at learning the relation between input variables and corresponding output targets. The model is learned from training samples where both input and output variables are available. the learned model will be used for prediction in situations where only input variables are given. Nice reviews can be found in \\citet{bishop2006pattern} and \\citet{hastie2009elements}. \n\nLeast squares regression has been applied widely in turbulence studies under the name ``\\textit{linear stochastic estimation}'' (LSE) \\citep{adrian1977role,adrian1979conditional}. It has been further investigated \\citep{guezennec1989stochastic,adrian1992stochastic,ewing1999examination} to estimate conditional eddies from the measurements. Later works introduced various extensions such as multi-time, nonlinear or higher-order LSE \\citep{mokhasi2009predictive,durgesh2010multi,nguyen2010proper,meyer2014provide}, which reconstruct velocity fields from pressure or shear-stress measurements. LSE can be also linked to proper orthogonal decomposition (POD), known also as ``principle component analysis'', to reduce the order of reconstruction problems \\citep{bonnet1994stochastic}. \n\nThe idea of combining LR velocity measurements to obtain HR fields using regression has not been addressed until recently \\citep{melnick2012experimental,tu2013integration}. \\citet{melnick2012experimental} used a POD-LSE model to get fully resolved 3D velocities of a flow over a flat plate by combining 3D smoke intensity and 2D PIV measurements. The POD-LSE model has been developed further by \\citet{tu2013integration} with a multi-time LSE reconstruction model. Kalman filter or Kalman smoother are used as real time estimation or data post-processing respectively. The model is tested using time-resolved PIV measurements of a bluff-body wake at a low $ Re $.\n\nRegression models in previous works remains rather simple, with the use of POD to reduce the order of the model. However, simple linear regression models suffer from certain limitations. The use of POD, acting as a low-pass filter, neglects certain small scales \\textit{a priori}. This work discusses further extensions of simple linear regression based on different regularizations or kernel methods. We aim at maximizing the amount of information that each method can recover, so POD is left aside throughout the thesis. Detailed analysis of reconstruction results such as spectra and errors will assess the performances of regression models as mapping functions between large and small scales. \n\n\\section{Ordinary least squares (OLS)}\nGiven a set of $ \\dimtl $ training samples $ \\left\\lbrace \\left(\\y_t, \\z_t\\right)\\right\\rbrace, t= 1,2,...,\\dimtl $ corresponding to pairs of inputs $ \\y_t \\in \\R^{\\dimsl}$ and outputs $ \\z_t \\in \\R^{\\dimsh}$, regression models estimate $ \\z_t $ as a function of $\\y_t $ via a mapping function $ f $: \n\\begin{equation}\n\tf: \\y_t \\mapsto \\myhat{\\z }_t = f(\\y_t) + \\n_t\n\\end{equation}\nwhere $ \\n_t \\sim \\mathscr{N}(0,\\sigma^2_{\\n_t})$ is usually a white Gaussian noise. $ f $ is learned from the $ \\dimtl $ pairs of training samples. This function should minimize the loss $ \\loss(f) $, which measures how far the prediction $ f(\\y) $ and the reference $ \\z $ are different within the training data.\n\nThe ordinary least squares model (OLS) assumes that $ f $ is a empirical linear function of $ \\y_t $. This model is widely used for its simplicity yet efficiency in situations where only a small training set with low signal-to-noise ratio is given. Assuming that $ \\{\\z_t\\} $ and $ \\{\\y_t\\} $ are mean-free and pre-normalized to one-standard deviation, OLS is expressed as:\n\\begin{equation}\n\\myhat{\\z }_t = f(\\y_t) = \\B^{\\mytrans}\\y_t\n\\end{equation}\nwhere $ \\B $ is the coefficient matrix of size $ \\dimsl \\times \\dimsh $. The $ i- $th column of $ \\B $ tells how to weight each measurement $ \\y_{t,j} $ to get an estimate of $ \\z_{t,i} $. The unknown $ \\B $ is learned from training data by minimizing the empirical loss:\n\\begin{equation}\n\\loss(\\B) = \\sum^{\\dimtl}_{t=1}{ \\normtwo{\\z_t-\\B^{\\mytrans}\\y_t}\t}\n\\end{equation}\nor rearranged in the matrix form as:\n\\begin{equation}\n\\loss(\\B) = \\normtwo{\\Y\\B-\\Z} = \\left(\\Y\\B-\\Z\\right)^{\\mytrans}\\left(\\Y\\B-\\Z\\right)\n\\label{eq:LSE2}\n\\end{equation}\nwhere $ \\Y^{\\mytrans} \\mydef \\{\\y_t\\} $ of size $ \\dimsl \\times \\dimtl $ and $ \\Z^{\\mytrans} \\mydef \\{\\z_t \\} $ of size $\\dimsh \\times \\dimtl$. $ \\normtwo{.} $ is the Euclidean distance, also called ``\\textit{L2 norm}''. $ \\mydef $ is the notation of ``\\textit{define as}''. The loss $ \\loss(\\B) $ is the square of errors, which is purely data-dependent. The optimization problem is to finds $ \\B $ such that:\n\\begin{equation}\n\\B= \\argmin_B{ \\left\\lbrace \\loss(\\B) \\right\\rbrace} = \\argmin_B{ \\left\\lbrace\\Vert \\Y\\B-\\Z\\Vert^2_2 \\right\\rbrace}\n\\end{equation}\n$ \\argmin{ \\{ . \\}} $ is the argument of the minimum seeking $ \\B $ such that $ \\loss(\\B) $ attains its minimum. The least square error solution of $ \\B $ is by differentiating equation \\ref{eq:LSE2} with respect to $\\B$ and set to zero:\n\\begin{equation}\n\\B= \\pinv { \\Y } \\Z= \\left( \\Y^{\\mytrans}\\Y \\right)^{-1} \\Y^{\\mytrans}\\Z\n\\label{eq:LSE5}\n\\end{equation}\nwhere $ (.)\\pinv{} $ is Moore-Penrose pseudo inverse. Then, the output $ \\z^\\ext $ of any new input vector $ \\y^\\ext$ out of the training set is estimated as:\n\\begin{equation}\n\\z^\\ext = \\B^{\\mytrans}\\y^\\ext\n\\label{eq:LSE6}\n\\end{equation} \n\n\\section{Regularized linear regression}\n\\label{sec:regularized_linear_regression}\nOLS suffers from critical problems. It requires the inverse of matrix $ \\Y^{\\mytrans}\\Y $ that can be singular or almost singular. The iterative solver can overcome this problem, but may result in a high-variance model with many large coefficients: a small change of $\\y^\\ext$ leads to very different predictions of $\\z^\\ext$. Since learned purely from the training data, it often fits very well the training data but poorly performs otherwise. This phenomena is called \\textit{overfitting} and will be discussed later in this chapter. Regularized least squares are proposed to overcome this problem. The idea is to add a regularization term $ g(\\B) $ on $ \\B $ to the data-dependent error term:\n\\begin{equation}\n\t\\loss(\\B, \\lambda) = \\Vert \\Y\\B-\\Z\\Vert^2_2 +\\lambda g(\\B)\n\t\\label{eq:regularized_linear_regression1}\n\\end{equation}\nwhere the regularization parameter $ \\lambda $ controls the balance between the two terms. Different terms lead to different regression models.\n\n\\subsection{Ridge Regression (RR): L2 penalty}\n\\label{sec:l2_penalty}\nThe most common and simple regularization is L2 penalty, also called ``weight decay'', which leads to the formulation of ridge regression (RR). The regularization term is on the norm of OLS coefficients, i.e. $ g(\\B) = \\Vert \\B \\Vert^2_2$. The loss function becomes:\n\\begin{equation}\n\\loss(\\B, \\lambda) = \\Vert \\Y\\B-\\Z\\Vert^2_2 +\\lambda\\Vert \\B \\Vert^2_2 \n\\label{eq:RR1}\n\\end{equation}\nThe parameter $\\lambda $ controls the balance between the data misfit and regularization term, which is the sum of squares of all coefficients. It controls how far $ \\B $ are shrunk towards zero, the larger $ \\lambda $ the further. Setting the derivative with respect to $ \\B $ to zero, the closed form to estimate $ \\B $ is:\n\\begin{equation}\n\\B=\\left( \\Y ^{\\mytrans}\\Y+\\lambda \\I \\right)^{-1} \\Y ^{\\mytrans}\\Z \n\\label{eq:RR2}\n\\end{equation}\nThis formula is similar to that of OLS except that some positive value $ \\lambda $ is added to the diagonal elements of $ \\Y^{\\mytrans}\\Y $ to ensure $ \\Y^{\\mytrans}\\Y +\\lambda \\I$ is always invertible.\n\nThe mechanism of RR can be analyzed via the singular value decomposition (SVD) of $ \\Y $: \n\\begin{equation}\n\\Y= \\Umat\\Diag\\Vmat^{\\mytrans}\n\\label{eq:RR3}\n\\end{equation}\nwhere $ \\Umat=(\\mybold{u}_1, \\mybold{u}_2,... ,\\mybold{u}_{\\dimsl}) $ is a $ \\dimtl \\times \\dimsl $ orthogonal matrix, $ \\Diag = diag(d_1, d_2, ... , d_{\\dimsl} )$ is a $ \\dimsl \\times \\dimsl $ diagonal matrix $ (d_1 \\geq d_2 \\geq ... \\geq d_{\\dimsl}) $, and $ \\Vmat^{\\mytrans}=(\\mybold{v}_1^{\\mytrans}, \\mybold{v}_2^{\\mytrans},... ,\\mybold{v}_{\\dimsl}^{\\mytrans}) $ is a $ \\dimsl \\times \\dimsl $ orthogonal matrix. The orthogonality implies that $ \\Umat^{\\mytrans}\\Umat = \\I $ or $ \\Umat^{\\mytrans}=\\Umat^{-1} $, similarly for $ \\Vmat $ and $ \\Vmat^{\\mytrans} $. The formula to estimate RR coefficients becomes:\n\\begin{equation}\n\\B =\\left(\\Y^{\\mytrans} \\Y + \\lambda \\I\\right)^{-1}\\Y^{\\mytrans}\\Z = \\Vmat \\, diag \\left( \\frac{d_j^2}{d_j^2+\\lambda} \\right) \\, \\Umat^{\\mytrans} \\Z\n\\label{eq:RR6}\n\\end{equation}\nRe-estimating training output variables $ \\Z $ as a function of inputs, one obtains:\n\\begin{equation}\n\\myhat{\\Z} = \\Y\\B = \\sum\\limits_{j=1}^{\\dimsl} \\left( \\mybold{u}_j \\frac{d_j^2}{d_j^2 + \\lambda} \\mybold{u}^{\\mytrans}\\right)\\Z\n\\label{eq:RR8}\n\\end{equation}\nThis implies that RR projects $ \\Z $ onto the principal components of $ \\Y^{\\mytrans}\\Y $ with large energy content (large $ d_j $) and shrinks the coefficients of low energy (small $ d_j $). This makes RR close to the \\textit{principal component regression} model discussed in \\citet{jolliffe1982note}, where one decomposes $ \\Y^{\\mytrans}\\Y $ using SVD and then set components with low energy content (small $ d_j $) to zeros in a handy manner. The approach also avoids matrix inversion and plays a role similar to regularization.\n\n\\subsection{LASSO: L1 penalty}\n\\label{sec:l1_penalty}\nThe nature of L2 penalty is to reduce model variance by shrinking coefficients corresponding to irrelevant events toward zero. However, they are not exactly zero. L1 penalty will precisely force some coefficients to zero. The model is called least absolute selection and shrinkage operator (LASSO) \\citep{tibshirani1996regression} in statistics, or basis pursuit denoising (BPDN) \\citep{chen1998atomic} in signal processing. It is considered as an implicit subset selection step, where irrelevant events are neglected from the reconstruction. This property favors sparsity - output is estimated as a combination of some input variables only- that is beneficial in many applications.\n\n\\begin{algorithm}[t]\n\\caption{Iterative shrinkage-thresholding algorithm (ISTA)}\\label{algo_ISTA}\n\\begin{algorithmic}[1]\n\\State Set k=0 and initialize $ \\B^{(0)} $; \n\\While{not convergence}\n\t\\State $ \\bigtriangledown f(\\B^{(k)}) = \\Y^{\\mytrans}(\\Z-\\Y\\B^{(k)})$ \\Comment{Residual from step k}\n\t\\State $\\B^{(k+1)} \\gets \\prox_{\\lambda \\normone{.}} \\left[ \\B^{(k)} - t^{(k)} \\bigtriangledown f(\\B^{(k)}) \\right] $\n\t\\State k = k + 1 \n\\EndWhile\n\\State \\textbf{return} $\\B^{(k+1)}$\n\\end{algorithmic}\n\\end{algorithm}\n \nThe LASSO cost function is similar to that of RR with a subtle but very important modification:\n\\begin{equation}\nF(\\B)= \\Vert \\Y\\B-\\Z\\Vert^2_2 +\\lambda \\normone{\\B}\n\\label{eq: LASSO1}\n\\end{equation}\nThe data-dependent term is the same as OLS or RR. The different penalty term $ \\lambda \\normone{\\B}$ imposes a constraint on the sum of absolute values of coefficients. This modification leads to sparsity, the key difference between LASSO and RR. It also makes the problem nonlinear and there exits no closed-form solution. Gradient-based methods are usually used to solve this optimization problem. Iterative shrinkage-thresholding algorithms (ISTA) is one of them, where $ \\B $ is solved iteratively by \\textit{soft-thresholding} as in Algorithm \\ref{algo_ISTA} \\citep{daubechies2004iterative}. In the pseudo code, $ prox $- the shrinkage operator- is the element-wise soft-thresholding: \n\\begin{equation}\n\\left(\\prox_{\\lambda \\normone{.}}[\\B]\\right)_i =\n\\begin{cases}\nb_i-\\lambda \\mysign{b_i} & \\text{if } |b_i|>\\lambda,\n\\\\\n0 & \\text{otherwise }\n\\end{cases}\n\\end{equation}\nwhere $ t^{(k)} $ is the gradient step size, $ b_i $ is the $ i- $th coefficient of $ \\B $. The \\textit{sign function} $ \\mysign{b_i} $ is defined as:\n\\begin{equation}\n \\mysign{b_i} =\n\\begin{cases}\n-1 & \\text{if } b_i<0,\n\\\\\n0 & \\text{if } b_i=0,\n\\\\\n1 & \\text{if } b_i>0.\n\\end{cases}\n\\end{equation}\nFaster schemes have been proposed to accelerate the convergence time \\citep{vonesch2008fast,beck2009fast}, but further discussions are outside the scope of this work.\n\nLASSO often outperforms RR and subset selection methods for several reasons. Compared to RR, LASSO takes most of the advantages, including the stability of the solution and the shrinkage feature. These features give a low-variance model compared to subset selection method. Moreover, LASSO favors sparsity, which can be considered as an implicit subset selection \\citep{hastie2005elements, hastie2009unsupervised}. By forcing some of the coefficients to zeros, the irrelevant predictors are suppressed in final estimation.\n\n\\section{Nonlinear regression}\n\\label{sec:nonlinear_regression}\nLinear regression aims at mapping the output as a linear combination of input variables. This implies a real constraint on performances of this family of models. In many cases including turbulence, linear functions are too simple to describe the undergoing phenomenon. Nonlinear regression models are beneficial, and kernel feature mapping is often used.\n\n\\subsection{Feature mapping}\nKernel methods are used to introduce nonlinearity into the model. The idea is to project the original input vector $ \\y $ onto a fixed feature space:\n\\begin{equation}\n\\y_t \\mymapto \\mybold{\\phi}_t = \\featmap{\\y_t}\n\\end{equation}\nand perform least square regression in this space. $ \\y_t \\in \\R^{\\dimsl}$ is the $ t- $ input vector. The feature vector $ \\mybold{\\phi}_t \\in \\R^{\\Df} $, where $ \\Df \\gg \\dimsl $ is the dimension in feature space. The least square problem becomes: \n\\begin{equation}\n\\B = \\argmin_B{ \\left\\lbrace \\normtwo{ \\Z- \\mathbf{\\Phi}\\B} \\right\\rbrace}\n\\end{equation}\nwhere $ \\mathbf{\\Phi} = \\{\\mybold{\\phi}_t \\} $ ($ t = 1, ..., \\dimtl $), the so-called \\textit{design matrix}, is of size $ \\dimtl \\times \\Df $. Adding L2 regularization term $ \\normtwo{\\B} $ and deriving analogously as RR model, the solution is:\n\\begin{equation}\n\\B=\\left(\\mathbf{\\Phi}^{\\mytrans}\\mathbf{\\Phi} + \\lambda \\I\\right)^{-1}\\mathbf{\\Phi}^{\\mytrans}\\Z\n\\end{equation}\nThen the prediction of a new input variable $ \\y^\\ext $ is:\n\\begin{equation}\n\\z^\\ext=\\B^{\\mytrans} \\featmap{\\y^\\ext}\n\\end{equation} \n\n\\subsection{Kernel ridge regression}\nInvolving nonlinearity via feature mapping significantly increases computational costs. The kernel trick is proposed \\citep{saunders1998ridge} to overcome this problem, resulting in the so-called kernel ridge regression (KRR) model. This trick appears when solving ridge regression using Lagrange dual optimization.\n\n\\subsubsection*{Dual form of ridge regression}\nRR in equation \\ref{eq:RR1} can be re-expressed as a dual Lagrangian optimization problem:\n\\begin{equation}\n\\B=\\argmin_B{ \\left\\lbrace \\sum\\limits_{t=1}^{\\dimtl}\\normtwo{\\mybold{e}_t} + \\lambda \\normtwo{\\B} \\right\\rbrace } \\subjectto \\z_t - \\B^{\\mytrans}\\y_t = \\mybold{e}_t, \\:\\:\\:\\:\\:\\: t = 1, 2, ..., \\dimtl\n\\label{eq:RRdual1}\n\\end{equation}\nwhere $ s.t $ stands for ``\\textit{subject to}''. Introducing Lagrange multipliers $ A[\\dimtl \\times \\dimsh] \\mydef \\{\\mybold{a}_t^{\\mytrans} \\} (t = 1, 2, ..., \\dimtl), \\mybold{a}_t \\in \\R^{\\dimsh}$, the optimization problem \\ref{eq:RRdual1} is equivalent to the problem of finding the saddle point of the function:\n\\begin{equation}\n\\sum\\limits_{t=1}^{\\dimtl}\\normtwo{\\mybold{e}_t} + \\lambda \\normtwo{\\B} + \\sum\\limits_{t=1}^{\\dimtl} \\mybold{a}_t^{\\mytrans}\\left( \\z_t - \\B^{\\mytrans}\\y_t - \\mybold{e}_t\\right)\n\\end{equation}\nThe solution as shown in \\citet{saunders1998ridge} is:\n\\begin{equation}\n\\mathbf{A} = \\left(\\mathbf{K} + \\lambda \\I\\right)^{-1}\\Z\n\\end{equation}\nwhere $ \\mathbf{K} \\mydef \\Y\\Y^{\\mytrans} $ is the matrix of dot products, $ \\mathbf{K}_{m,n} = \\y _m^{\\mytrans} \\y _n $. The prediction for a new input variables $ \\y^\\ext $ is:\n\\begin{equation}\n\\myhat{\\z }^\\ext = \\left(\\sum\\limits_{t=1}^{\\dimtl}\\mybold{a}_t^{\\mytrans}\\y_t\\right)\\y^\\ext = \\mathbf{A}^{\\mytrans} \\mybold{k}^{\\ext}\n\\end{equation}\nwhere $ \\mybold{k}^{\\ext} \\mydef \\{k_t\\} = \\{ \\y_t^{\\mytrans} \\y^\\ext \\} \\in \\R^{\\dimtl}$.\n \n\\subsubsection*{Kernel trick}\n\n\\begin{table}\n\\centering\n\\begin{tabular}{lc} \\toprule\n\tFunction name & $ k (\\mybold{u}, \\mybold{v}) $ \\\\ \\midrule\n\tLinear & $ \\mybold{u}^{\\mytrans} \\mybold{v} $ \\\\ \\midrule\n\tPolynomial & $ (r + \\mybold{u}^{\\mytrans} \\mybold{v})^d $ for $ r,d \\geq 0 $\\\\ \\midrule\n\tRadial basis function (RBF) & $ \\myexp{-\\gamma \\normtwo{\\mybold{u} - \\mybold{v}}}$, $ \\gamma > 0 $ \\\\\n\t\\bottomrule\n\\end{tabular}\n\\caption{Common basis functions for kernel methods.}\n\\label{tab_basisfunctions}\n\\end{table}\n\nSolving RR in the dual form offers no improvement of accuracy, speed or stability. However, the beauty of this approach is the presence of only dot products among input variables in the final prediction step. This implies an analogous procedure when working in a nonlinear feature space. The transformation into such a space is unnecessary if the dot products of the transformed variables can be estimated using the kernel: \n\\begin{equation}\nk (\\mybold{u}, \\mybold{v}) = \\featmap{\\mybold{u}}^{\\mytrans} \\featmap{\\mybold{v}}\n\\end{equation}\nThis explicit transformation is computed in time $ \\mathcal{O}(\\dimsl^2) $, where $ \\dimsl $ is the dimension of the input vectors. The \\textit{kernel trick} permits to estimate directly $ k (\\mybold{u}, \\mybold{v}) $ without computing $ \\featmap{\\mybold{u}} $ and $ \\featmap{\\mybold{v}} $, reducing the computational time into $ \\mathcal{O}(\\dimsl) $ only. Not every kernel has this property. Table \\ref{tab_basisfunctions} gathers three common functions.\n\n\\section{A framework to select model and parameters}\nA model is assessed via its generalization performance, i.e. the prediction capability on a new dataset independent from the training set. In all above models, there exists one or more hyper-parameters that are directly linked to model performances. The arising question is how to optimize those parameters using training data only in a systematic manner. This step is the so-called \\textit{parameter optimization}, while the step to test performances of the model on independent data is the so-called \\textit{model assessment}. \n\nIt is worth also mentioning the definition of different datasets: \\textit{training}, \\textit{validation} and \\textit{test} sets. The training set contains data from which a model is learned. The validation set is used to estimate prediction errors, and from which the best model is chosen. This model is the one that gives the most accurate prediction on the validation set. The testing set is finally used to give an estimate of the prediction error of this optimal model. This error is approximately the generalization error on independent datasets.\n\nTo optimize the generalization capability of a model, one needs to understand the idea of \\textit{bias} and \\textit{variance}. Next sections will discuss this topic, with the technique called ``\\textit{cross-validation}'' to find the trade-off between bias and variance using the training data only. \n\n\\subsection{Bias-variance trade-off}\nThe above regression models can be interpreted as seeking a mapping function $ f $: \n\\begin{equation}\n\tf: \\y \\mapsto \\myhat{\\z } = f(\\y ) + \\n\n\\end{equation}\nwhere $ \\n \\sim \\mathscr{N}(0,\\sigma^2_{\\n})$ is assumed to be a white Gaussian noise. The expected prediction error of this model for an input vector $ \\y_\\knot $ is:\n\\begin{equation}\n\\begin{split}\n\\epsilon(\\y_\\knot) & =\\E{\\left(\\z_\\knot - f(\\y_\\knot)\\right)^2} \\\\\n & = \\underbrace{\\sigma_{\\n}^2}_{\\text{Irreducible Error}} + \\underbrace{\\left(\\E{\\myhat{\\z }_\\knot} - \\z_\\knot\\right)^2}_{\\text{Bias}^2} + \\underbrace{\\E{\\left( \\myhat{\\z }_\\knot - \\E{\\myhat{\\z }_\\knot} \\right)^2}}_{Variance}\n\\end{split}\n\\end{equation}\nwhere $ \\E{.} $ is the expectation of a variable. The first term comes from the noise, which depends only on the data and is irreducible. The second term is the square bias, showing how far the average of the estimates is different from the true mean. The last term is the expected variance of the estimate around its mean. We can interpret the bias as the average prediction error over different data sets from the true mean, and the variance as how this error is sensitive to a particular choice of data set.\n\nAfter the decomposition, a model is said to be \\textit{underfitting} or \\textit{overfitting} depending on the contribution of each term in the total error. An underfitting model is too simple to capture all details of the underlying phenomenon. This model is high-bias and low-variance: it poorly fits the training data and performs similarly in the testing data. An overfitting model is over-complex: it closely fits the training data, including noises, and gives poor performance on new testing data. This model has low bias and high variance. Ideal models should have both low bias and variance. In practice, a good model must satisfy its \\textit{bias-variance trade-off} when it does not suffer from either the problems of overfitting or underfitting. \n\n\\begin{figure}\n\\centering\n\t\\includegraphics[width=0.7\\columnwidth]{./images/regression/bias-variance_tradeoff.png}\n\t\\caption{\\label{fig:bias-variance_tradeoff} Typical behavior of prediction error for testing (red) and training (blue) dataset as a function of a hyper-parameter \\citep{hastie2009elements}. Different curves are for various datasets. The solid ones are expected errors (average of all curves). From left to right can be the direction of decreasing regularization or increasing model complexity.}\n\\end{figure}\n\nThe idea of underfitting and overfitting can be visualized in figure \\ref{fig:bias-variance_tradeoff} as modified from \\citet{hastie2009elements}. It shows the prediction errors for training and testing datasets as a function of a parameter. This parameter can be either the model complexity in an increasing order or the regularization parameter $ \\lambda $ in a decreasing order. Models toward the left (low complexity model, or regression with a strong regularization) are underfitting. It does not fit well both the training and testing data. Toward the right, errors on the training set decrease, while those of the testing set decrease and then increase again. Models in the far right are overfitting. They fit well the training set but not the testing set. The desired model is the one such that the error on the testing set is the lowest. At this point, the model reaches its bias-variance trade-off: bias and variance do not necessarily reach their minima, but a compromise ensures the minimal error on testing data.\n\nThe idea of bias-variance trade-off can be further visualized schematically as in figure \\ref{fig:bias-variance} \\citep{hastie2009elements}. The blue-shaded region shows the irreducible error $ \\sigma_{\\n}^2 $ (due to random noise) of the training set, where the truth is at the center and all realizations are within. To fit the model, the non-regularized model space (red curve) contains all possible models, while the magenta one shows the restricted space of regularized models. Model variances are depicted as yellow curves centered at the so-called ``closest fit''. The model bias is the distance of the closest fit and the truth, which can only be reduced as increasing the model complexity or adding more features. Regularized models add estimation bias on top of the model bias but reduce the variance. The benefit of the trade-off is only when the variance is reduced more than the squared estimation bias.\n\n\\begin{figure}\n\\centering\n\t\\includegraphics[width=0.6\\columnwidth]{./images/regression/bias-variance.png}\n\t\\caption{\\label{fig:bias-variance} Schematic view of the behavior of bias and variance \\citep{hastie2009elements}. The blue-shaded, centered at the truth, is the realization space, where maximum distance is the irreducible error. All possible models are bounded by the red/magenta model space curves. Bias is shown by the black lines as the distance to the truth. Best models are shown as black dots, and circled by model variances. }\n\\end{figure}\n\nThe bias-variance trade-off is essentially estimated from the training data. If this set is large enough, it can be virtually divided into training, cross-validation and testing sets. A general advice for the size of these three sets is $ 50 \\% $, $ 25 \\% $ and $ 25 \\% $ respectively \\citep{hastie2009elements}. However, learned models are usually improved when using more data. Cross-validation is another idea to find the trade-off while keeping the whole training data for learning the model. \n\n\\subsection{Cross-validation}\nIn many cases, the training data is limited. Setting aside $ 50 \\% $ the data to quantify the model might lose all the benefits from regularization and nonlinearity compared to OLS. Cross validation (CV) is commonly used to select the model and optimize parameters using all given data. \n\nThe most popular CV technique is \\textit{k-fold} CV. The dataset is randomly split into $ K $ subsets of approximately equal size. For each $ K $-th subset to estimate the prediction error, the model is trained on the remaining $ K-1 $ subsets. This procedure is repeated $ K $ times for all the subset, and the model error for the current setting is the average of its $ K $ estimates. A summary of k-fold CV is presented in algorithm \\ref{algo_kfold}. $ K $ is at most the number of training samples (\\textit{leave one out cross validation}). When $ K $ is high, the selected model tends to have lower bias but higher variance, since many training samples are similar. A much heavier computation is also required, since the training/predicting is repeated $ K $ times. The popular choice of $ K $ is from 5 to 10 \\citep{breiman1992submodel, kohavi1995study}. \n\n\\begin{algorithm}[t]\n\\caption{K-fold cross-validation for set of L parameters $ \\gamma_1, \\gamma_2, ..., \\gamma_L $} \\label{algo_kfold}\n\\begin{algorithmic}[1]\n\t\\State Divide the training samples into k folds randomly\n\t\\For {$ i = 1, 2, ..., L$}\n\t\t\\For {$ j = 1,2,... K $}\n\t\t\t\\State Train the model with $ \\gamma_i $ using all data set except the $ j^{th} $\n\t\t\t\\State Estimate prediction error $ \\epsilon(i,j) $ of the model on the $ j^{th} $ set\n\t\t\\EndFor\n\t\\EndFor\n\t\\State Estimate averaged error $ \\epsilon(i) = \\frac{1}{K}\\sum\\limits_{j=1}^{K}\\epsilon(i,j)$\n\t\\State Return optimal parameter $ \\gamma_m $ where $ \\epsilon(m) = min \\{\\epsilon(i)\\}$.\n\\end{algorithmic}\n\\end{algorithm}\n\n\\section{Regression models for reconstruction of isotropic turbulence}\nWe now apply regression models to reconstruct HR velocity fields from LR measurements. The data from DNS isotropic turbulence discussed in section \\ref{sec:data_isotropic} is used. Streamwise velocities from $ 37 $ data cubes of fully resolved data $ 96^3 $ are used as the reference. In each data cube, only the streamwise velocity component is considered. The training examples are the HR fields (in spanwise-vertical directions) and equivalent LR ones subsampled by a factor of $ \\sqrt{\\dimsh/\\dimsl} = 3 $ in both vertical and spanwise directions. This ratio is equivalent to an energy loss of $ \\Delta \\kappa_s = 1.03 \\% $ (see table \\ref{tab:energyloss_isotropic}). In streamwise direction, the training planes are selected every 4 snapshots, corresponding to an energy loss of $ 1.23 \\% $. This configuration is chosen to mimic other problems investigated later in chapters \\ref{chap_NLM} and \\ref{chap_BayesianFusion}. \n\nRegression models are learned from $ 37 \\times 24 $ pairs of LR ($ \\dimsl = 32 \\times 32 $) and corresponding HR ($\\dimsh = 96 \\times 96 $) fields. The learned models are then used to reconstruct all $ 37 \\times 96 $ HR fields from $ 37 \\times 96 $ measured LR planes. These results are presented latter in chapter \\ref{chap_comparisons} when comparing various methods. Following sections only discuss the optimization of model parameters.\n\n\\subsection{Regularization parameter and shrinkage effect}\nOrdinary least squares model inverses directly $ \\Y^{\\mytrans}\\Y $, which is usually ill-conditioned, leading to high variance models. A slight change of input variables can lead to very different predictions. Regularization is introduced to reduce this effect by imposing different penalty terms on regression coefficients. RR seeks a model with small sum-of-square coefficients, LASSO imposes a penalty of their absolute values (see section \\ref{sec:regularized_linear_regression}). \n\n\\begin{figure}\n\\centering\n\t\\includegraphics[width=0.8\\columnwidth]{./images/regression/RR_LSE_coefficients.png}\n\t\\caption{\\label{fig:RR_LSE_coefficients} Shrinkage effect: coefficients of OLS (left) vs RR (right) models correspond to an input LR measurement at the center of the field. The coefficients, which are rearranged in a 2D field of the size $ 96 \\times 96 $, can be interpreted as the contributions of this input to the reconstruction of all other high-resolution points. The higher the coefficient, the stronger the impact.}\n\\end{figure}\n\nFigure \\ref{fig:RR_LSE_coefficients} shows the weight of the input variable at the center of the field in reconstructing all other HR points by OLS and RR. These are the coefficients $ \\mybold{b}_j \\mydef [b_{j1}, b_{j2}, ..., b_{j\\dimsh}]^{\\mytrans}$ computed in equation \\ref{eq:LSE5} for OLS or equation \\ref{eq:RR2} for RR. Recall that $ \\B $ is of size $ \\dimsl \\times \\dimsh $. The figures are $ \\mybold{b}_j$ of OLS and RR models, where index $ j $ corresponds to the central point. $ \\mybold{b}_j $ is then rearranged to recover the 2D shape of size $ \\sqrt{\\dimsh} \\times \\sqrt{\\dimsh} $ of the 2D velocity fields. In both plots, the weights remain significant in a neighborhood of several pixels. The coefficients drop very quickly when moving away from the center. This is due to the rapid decay of correlation between the central point and its neighbors. The difference between the two methods is that OLS (using direct inversion) gives large coefficients almost everywhere whereas RR shrinks the contributions of irrelevant points to small values.\n\n\\begin{figure}[t]\n\\centering\n\t\\includegraphics[width=0.8\\columnwidth]{./images/regression/RR_coeffs_lambda.eps}\n\t\\includegraphics[width=0.8\\columnwidth]{./images/regression/Lasso_coeffs_lambda.eps}\n\t\\caption{\\label{fig:coeffs_lambda} Shrinkage effects of L2 (top) and L1 (bottom) penalty: coefficients corresponding to eight high-resolution outputs (in eight different colors) as functions of regularization parameters $ \\lambda $. Higher $ \\lambda $ shrink the coefficients toward zeros differently depending on the penalty terms.}\n\\end{figure}\n\nThe different shrinkage behaviors of RR and LASSO are visualized in figure \\ref{fig:coeffs_lambda}. This figure shows the eight different coefficients corresponding to eight HR outputs as a function of the regularization parameter presented in equation \\ref{eq:regularized_linear_regression1}. It confirms the effect of the penalty term, which shrinks coefficients associated with irrelevant inputs to zeros. RR coefficients do not reach exact zeros even for extremely high $ \\lambda $, while LASSO rapidly shrinks some coefficients to exact zeros.\n\n\\subsection{Optimizing regularization parameters and model complexity via ten-fold cross-validation}\t\n\nAll regression models (except OLS) are parametric, i.e. at least one hyper-parameter controls the construction of the model and its performances. With RR, LASSO or KRR, the regularization parameter $ \\lambda $ controls the balance between data misfit and regularization term. KRR model is also a function of the kernel parameter, which is the standard deviation of the distribution when using RBF kernel, or the polynomial order. All hyper-parameters are chosen from the training samples using ten-fold CV (see algorithm \\ref{algo_kfold}). The plot of the error as a function of model parameters is called the ``validation curve'', from which the best model or parameter is selected.\n\nAnother important test is to see whether the training dataset is sufficient or not. This is examined via the so-called ``learning curve''. To estimate this curve, a small subset of data is drawn out randomly from the training set. This set is retained to estimate the error at the last step. For the remaining data, models are optimized using different portions. Trained models are assessed by computing prediction errors on the testing data. The curve of errors as a function of different data size, the ``learning curve'', will tell whether the current training data is sufficiently large to guarantee an accurate learning.\n\n\\begin{figure}\n\\centering\n\t\\includegraphics[height=6.75cm]{./images/regression/RR_validationcurve.eps}\n\t\\hspace{0.3cm}\n\t\\includegraphics[height=6.75cm]{./images/regression/RR_learningcurve.eps}\t\n\t\\caption{\\label{fig:RR_validationcurve} RR validation curve: errors as functions of regularization parameter $ \\lambda $ (left), and learning curve: errors as functions of training data size (right). Red curves are for the prediction, while blue ones are for the training data}\n\\end{figure}\n\\begin{figure}\n\t\\centering\n\t\\includegraphics[height=6.75cm]{./images/regression/KRR_validationcurve_lambda.eps}\n\t\\hspace{0.3cm}\n\t\\includegraphics[height=6.75cm]{./images/regression/KRR_validationcurve_gamma.eps}\t\n\t\\caption{\\label{fig:KRR_validationcurve} KRR validation curve (RBF kernel), with parameters are kernel parameter $ \\gamma $ and regulartization parameter $ \\lambda $. Red curves are for the prediction, while blue ones are for the training data}\n\\end{figure}\n\nTo plot validation and learning curves, the average normalized root mean square error (NRMSE) $ \\bar{\\epsilon} $ is used as the prediction error. $ \\bar{\\epsilon} $ represents an overall measure of the reconstruction accuracy. NRMSE is estimated between the reconstructed ($ \\hat{\\z}_t $) and reference ($ \\z_t $) fields\n\\begin{equation}\n\\bar{\\epsilon} = \\left(\\frac{ \\sum\\limits_{t} \\sum\\limits_{j\\in \\varmathbb{J}}(\\hat{\\mybold{z}}_{t,j}-\\mybold{z}_{t,j})^2}{\\sum\\limits_{t} \\sum\\limits_{j\\in \\varmathbb{J}}\\mybold{z}_{t,j}^2}\\right)^{1/2}\n\\label{eq:NRMSE}\n\\end{equation} \n$ \\varmathbb{J} $ is the considered set of points used to estimate the error. In the present case of isotropic turbulence, $ \\varmathbb{J} $ contains all points in each plane.\n\nFigure \\ref{fig:RR_validationcurve} shows the validation and learning curve of RR model on the training set (blue) and validation set (red). The solid lines are the means, while a band of colors show the standard deviations of ten estimates of $ \\bar{\\epsilon} $ from ten folds. The validation curve (left) shows the model behavior for different $ \\lambda $. For the training set, the model fits accurately the data when no or small regularization is imposed. This effect of overfitting is shown on the validation set, where higher errors are obtained. The model has also high variance, i.e. prediction errors vary strongly from one set to the other. The optimal value of $ \\lambda \\approx 100 $ is reached where the validation error is minimum. When a very strong regularization is imposed, errors on both training and validation sets increase, since all coefficients are shrunk toward zero. Errors will grow till their maximum value, which is the variance of the HR data. The learning curve (right) shows the effect of the training data size on the performance of the model (with optimal $ \\lambda $). The figure shows that with larger training data, the prediction error on the validation set decrease sharply while this error on the training set gradually increase. With a good model and sufficiently large training data, these two curves will be very close. The curves also suggest that a larger dataset will lead to a better model in this case. \n\nKRR model parameters are optimized using the same validation curve. Different models come with different kernel parameters (recall table \\ref{tab_basisfunctions}), together with the regularization parameter $ \\lambda $. For RBF kernel, $ \\gamma $ represents the kernel shape. Figure \\ref{fig:KRR_validationcurve} shows that the optimal values of both $ \\lambda $ and $ \\gamma $ are both of the order of $ 10^{-5} $. For polynomial kernel, the polynomial order is to be tuned. The model with higher polynomial order will have lower bias but higher variance. An analogous procedure is used to find the optimal order of the polynomial fit, which is either 2 or 3 in general. \n\n\\section{Concluding remarks}\nIn this chapter, regression models are used to reconstruct high resolution fields of an isotropic turbulence from low resolution measurements. The models seek a mapping function between low and high resolved fields. This function is learned from given training samples and used for prediction when only low-resolution measurements are available. \n\nBoth linear and nonlinear models have been discussed. The ordinary least squares model is the simplest one and is widely used in the literature of turbulence. Since suffering from some critical limitations, notably overfitting and ill-conditioning, various regularization methods have been introduced. Both L2 and L1 penalty have been discussed. While L2 penalty reduces model variance by shrinking coefficients of irrelevant events, L1 forces them to zeros by slightly change the penalty term and offers some potential benefits. Nonlinear regression is another approach to improve OLS. Instead of assuming the underlying phenomenon is linear, it introduces nonlinearity by projecting input vectors to a fixed feature space before performing standard OLS. This nonlinearity helps in improving reconstruction accuracy, since assuming a linear relation between large and small scales of turbulence implies a big constraint on model performances. \n\nWe have presented the procedure to optimize hyper-parameters of regression models. This step is crucial to ensure accurate predictions, since performances of all regression models (except OLS) depend on at least one parameter. The models are learned from the training data, but the performances are assessed on independent data only. Cross-validation is used to seek a compromise, the so-called bias-variance trade-off, from training data only. The dataset is virtually split into many subsets. The models are trained within some subsets, and their prediction generalized errors are estimated using the remaining sets. \n\nThe optimal set of parameters is found using the validation curve and ten-fold cross validation technique. This curve shows the prediction errors on the validation set as a function of model parameters. It gives an idea of the optimal parameter range. Through all the tests, it is shown that the gradient of this curve near its minima is not very sharp, implying that a good estimate of these parameters can be achieved from the training data only. \n\nAll model parameters are investigated. RR model is optimal when its regularization parameter $ \\lambda $ is about $ 100 $. KRR model with RBF kernel is constructed with the optimal values of both $ \\lambda $ and $ \\gamma $ are of the order of $ 10^{-5} $. For polynomial KRR takes 2 or 3 as the optimal order of the kernel.\nAll such models will be used later in chapter \\ref{chap_comparisons} to reconstruct high-resolution fields from low-resolution measurements and compared to other proposed methods.", "meta": {"hexsha": "54e69a011ea05fd8147dcca37e13fc3f14b0e601", "size": 40446, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "thesis/manuscripts/final_ver1/corps/linearregression.tex", "max_stars_repo_name": "linhvannguyen/PhDworks", "max_stars_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_stars_repo_licenses": ["MIT"], "max_stars_count": 2, "max_stars_repo_stars_event_min_datetime": "2019-01-01T14:41:09.000Z", "max_stars_repo_stars_event_max_datetime": "2022-02-12T07:08:06.000Z", "max_issues_repo_path": "thesis/manuscripts/final_ver1/corps/linearregression.tex", "max_issues_repo_name": "vanlinhnguyen/PhDworks", "max_issues_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "thesis/manuscripts/final_ver1/corps/linearregression.tex", "max_forks_repo_name": "vanlinhnguyen/PhDworks", "max_forks_repo_head_hexsha": "9336e5257f5ddc3c899a6fb68b1028c905d13ff9", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 121.8253012048, "max_line_length": 1451, "alphanum_fraction": 0.7628937349, "num_tokens": 10454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.7490872243177518, "lm_q2_score": 0.8152324826183822, "lm_q1q2_score": 0.6106802375782737}} {"text": "%* glpk04.tex *%\n\n\\chapter{Advanced API Routines}\n\n\\section{Background}\n\\label{basbgd}\n\nUsing vector and matrix notations the LP problem (1.1)---(1.3) (see\nSection \\ref{seclp}, page \\pageref{seclp}) can be stated as follows:\n\n\\noindent\n\\hspace{.5in} minimize (or maximize)\n$$z=c^Tx_S+c_0\\eqno(3.1)$$\n\\hspace{.5in} subject to linear constraints\n$$x_R=Ax_S\\eqno(3.2)$$\n\\hspace{.5in} and bounds of variables\n$$\n\\begin{array}{l@{\\ }c@{\\ }l@{\\ }c@{\\ }l}\nl_R&\\leq&x_R&\\leq&u_R\\\\\nl_S&\\leq&x_S&\\leq&u_S\\\\\n\\end{array}\\eqno(3.3)\n$$\nwhere:\n\n$x_R=(x_1,\\dots,x_m)$ is the vector of auxiliary variables;\n\n$x_S=(x_{m+1},\\dots,x_{m+n})$ is the vector of structural variables;\n\n$z$ is the objective function;\n\n$c=(c_1,\\dots,c_n)$ is the vector of objective coefficients;\n\n$c_0$ is the constant term (``shift'') of the objective function;\n\n$A=(a_{11},\\dots,a_{mn})$ is the constraint matrix;\n\n$l_R=(l_1,\\dots,l_m)$ is the vector of lower bounds of auxiliary\nvariables;\n\n$u_R=(u_1,\\dots,u_m)$ is the vector of upper bounds of auxiliary\nvariables;\n\n$l_S=(l_{m+1},\\dots,l_{m+n})$ is the vector of lower bounds of\nstructural variables;\n\n$u_S=(u_{m+1},\\dots,u_{m+n})$ is the vector of upper bounds of\nstructural variables.\n\nFrom the simplex method's standpoint there is no difference between\nauxiliary and structural variables. This allows combining all these\nvariables into one vector that leads to the following problem\nstatement:\n\n\\noindent\n\\hspace{.5in} minimize (or maximize)\n$$z=(0\\ |\\ c)^Tx+c_0\\eqno(3.4)$$\n\\hspace{.5in} subject to linear constraints\n$$(I\\ |-\\!A)x=0\\eqno(3.5)$$\n\\hspace{.5in} and bounds of variables\n$$l\\leq x\\leq u\\eqno(3.6)$$\nwhere:\n\n$x=(x_R\\ |\\ x_S)$ is the $(m+n)$-vector of (all) variables;\n\n$(0\\ |\\ c)$ is the $(m+n)$-vector of objective\ncoefficients;\\footnote{Subvector 0 corresponds to objective\ncoefficients at auxiliary variables.}\n\n$(I\\ |-\\!A)$ is the {\\it augmented} constraint\n$m\\times(m+n)$-matrix;\\footnote{Note that due to auxiliary variables\nmatrix $(I\\ |-\\!A)$ contains the unity submatrix and therefore has full\nrank. This means, in particular, that the system (3.5) has no linearly\ndependent constraints.}\n\n$l=(l_R\\ |\\ l_S)$ is the $(m+n)$-vector of lower bounds of (all)\nvariables;\n\n$u=(u_R\\ |\\ u_S)$ is the $(m+n)$-vector of upper bounds of (all)\nvariables.\n\nBy definition an {\\it LP basic solution} geometrically is a point in\nthe space of all variables, which is the intersection of hyperplanes\ncorresponding to active constraints\\footnote{A constraint is called\n{\\it active} if at a given point it is satisfied as equality, otherwise\nit is called {\\it inactive}.}. The space of all variables has the\ndimension $m+n$, therefore, to define some basic solution we have to\ndefine $m+n$ active constraints. Note that $m$ constraints (3.5) being\nlinearly independent equalities are always active, so remaining $n$\nactive constraints can be chosen only from bound constraints (3.6).\n\nA variable is called {\\it non-basic}, if its (lower or upper) bound is\nactive, otherwise it is called {\\it basic}. Since, as was said above,\nexactly $n$ bound constraints must be active, in any basic solution\nthere are always $n$ non-basic variables and $m$ basic variables.\n(Note that a free variable also can be non-basic. Although such\nvariable has no bounds, we can think it as the difference between two\nnon-negative variables, which both are non-basic in this case.)\n\nNow consider how to determine numeric values of all variables for a\ngiven basic solution.\n\nLet $\\Pi$ be an appropriate permutation matrix of the order $(m+n)$.\nThen we can write:\n$$\\left(\\begin{array}{@{}c@{}}x_B\\\\x_N\\\\\\end{array}\\right)=\n\\Pi\\left(\\begin{array}{@{}c@{}}x_R\\\\x_S\\\\\\end{array}\\right)=\\Pi x,\n\\eqno(3.7)$$\nwhere $x_B$ is the vector of basic variables, $x_N$ is the vector of\nnon-basic variables, $x=(x_R\\ |\\ x_S)$ is the vector of all variables\nin the original order. In this case the system of linear constraints\n(3.5) can be rewritten as follows:\n$$(I\\ |-\\!A)\\Pi^T\\Pi x=0\\ \\ \\ \\Rightarrow\\ \\ \\ (B\\ |\\ N)\n\\left(\\begin{array}{@{}c@{}}x_B\\\\x_N\\\\\\end{array}\\right)=0,\\eqno(3.8)$$\nwhere\n$$(B\\ |\\ N)=(I\\ |-\\!A)\\Pi^T.\\eqno(3.9)$$\n\n\\newpage\n\nMatrix $B$ is a square non-singular $m\\times m$-matrix, which is\ncomposed from columns of the augmented constraint matrix corresponding\nto basic variables. It is called the {\\it basis matrix} or simply the\n{\\it basis}. Matrix $N$ is a rectangular $m\\times n$-matrix, which is\ncomposed from columns of the augmented constraint matrix corresponding\nto non-basic variables.\n\nFrom (3.8) it follows that:\n$$Bx_B+Nx_N=0,\\eqno(3.10)$$\ntherefore,\n$$x_B=-B^{-1}Nx_N.\\eqno(3.11)$$\nThus, the formula (3.11) shows how to determine numeric values of basic\nvariables $x_B$ assuming that non-basic variables $x_N$ are fixed on\ntheir active bounds.\n\nThe $m\\times n$-matrix\n$$\\Xi=-B^{-1}N,\\eqno(3.12)$$\nwhich appears in (3.11), is called the {\\it simplex\ntableau}.\\footnote{This definition corresponds to the GLPK\nimplementation.} It shows how basic variables depend on non-basic\nvariables:\n$$x_B=\\Xi x_N.\\eqno(3.13)$$\n\nThe system (3.13) is equivalent to the system (3.5) in the sense that\nthey both define the same set of points in the space of (primal)\nvariables, which satisfy to these systems. If, moreover, values of all\nbasic variables satisfy to their bound constraints (3.3), the\ncorresponding basic solution is called {\\it (primal) feasible},\notherwise {\\it (primal) infeasible}. It is understood that any (primal)\nfeasible basic solution satisfy to all constraints (3.2) and (3.3).\n\nThe LP theory says that if LP has optimal solution, it has (at least\none) basic feasible solution, which corresponds to the optimum. And the\nmost natural way to determine whether a given basic solution is optimal\nor not is to use the Karush---Kuhn---Tucker optimality conditions.\n\n\\def\\arraystretch{1.5}\n\nFor the problem statement (3.4)---(3.6) the optimality conditions are\nthe following:\\footnote{These conditions can be appiled to any solution,\nnot only to a basic solution.}\n$$(I\\ |-\\!A)x=0\\eqno(3.14)$$\n$$(I\\ |-\\!A)^T\\pi+\\lambda_l+\\lambda_u=\\nabla z=(0\\ |\\ c)^T\\eqno(3.15)$$\n$$l\\leq x\\leq u\\eqno(3.16)$$\n$$\\lambda_l\\geq 0,\\ \\ \\lambda_u\\leq 0\\ \\ \\mbox{(minimization)}\n\\eqno(3.17)$$\n$$\\lambda_l\\leq 0,\\ \\ \\lambda_u\\geq 0\\ \\ \\mbox{(maximization)}\n\\eqno(3.18)$$\n$$(\\lambda_l)_k(x_k-l_k)=0,\\ \\ (\\lambda_u)_k(x_k-u_k)=0,\\ \\ k=1,2,\\dots,\nm+n\\eqno(3.19)$$\nwhere:\n$\\pi=(\\pi_1,\\pi_2,\\dots,\\pi_m)$ is a $m$-vector of Lagrange\nmultipliers for equality constraints (3.5);\n$\\lambda_l=[(\\lambda_l)_1,(\\lambda_l)_2,\\dots,(\\lambda_l)_n]$ is a\n$n$-vector of Lagrange multipliers for lower bound constraints (3.6);\n$\\lambda_u=[(\\lambda_u)_1,(\\lambda_u)_2,\\dots,(\\lambda_u)_n]$ is a\n$n$-vector of Lagrange multipliers for upper bound constraints (3.6).\n\n\\newpage\n\nCondition (3.14) is the {\\it primal} (original) system of equality\nconstraints (3.5).\n\nCondition (3.15) is the {\\it dual} system of equality constraints.\nIt requires the gradient of the objective function to be a linear\ncombination of normals to the planes defined by constraints of the\noriginal problem.\n\nCondition (3.16) is the primal (original) system of bound constraints\n(3.6).\n\nCondition (3.17) (or (3.18) in case of maximization) is the dual system\nof bound constraints.\n\nCondition (3.19) is the {\\it complementary slackness condition}. It\nrequires, for each original (auxiliary or structural) variable $x_k$,\nthat either its (lower or upper) bound must be active, or zero bound of\nthe corresponding Lagrange multiplier ($(\\lambda_l)_k$ or\n$(\\lambda_u)_k$) must be active.\n\nIn GLPK two multipliers $(\\lambda_l)_k$ and $(\\lambda_u)_k$ for each\nprimal (original) variable $x_k$, $k=1,2,\\dots,\\linebreak m+n$, are\ncombined into one multiplier:\n$$\\lambda_k=(\\lambda_l)_k+(\\lambda_u)_k,\\eqno(3.20)$$\nwhich is called a {\\it dual variable} for $x_k$. This {\\it cannot} lead\nto the ambiguity, because both lower and upper bounds of $x_k$ cannot be\nactive at the same time,\\footnote{If $x_k$ is a fixed variable, we can\nthink it as double-bounded variable $l_k\\leq x_k\\leq u_k$, where\n$l_k=u_k.$} so at least one of $(\\lambda_l)_k$ and $(\\lambda_u)_k$ must\nbe equal to zero, and because these multipliers have different signs,\nthe combined multiplier, which is their sum, uniquely defines each of\nthem.\n\n\\def\\arraystretch{1}\n\nUsing dual variables $\\lambda_k$ the dual system of bound constraints\n(3.17) and (3.18) can be written in the form of so called {\\it ``rule of\nsigns''} as follows:\n\n\\medskip\n\n\\begin{center}\n\\begin{tabular}{|@{\\,}c@{$\\,$}|@{$\\,$}c@{$\\,$}|@{$\\,$}c@{$\\,$}|\n@{$\\,$}c|c@{$\\,$}|@{$\\,$}c@{$\\,$}|@{$\\,$}c@{$\\,$}|}\n\\hline\nOriginal bound&\\multicolumn{3}{c|}{Minimization}&\\multicolumn{3}{c|}\n{Maximization}\\\\\n\\cline{2-7}\nconstraint&$(\\lambda_l)_k$&$(\\lambda_u)_k$&$(\\lambda_l)_k+\n(\\lambda_u)_k$&$(\\lambda_l)_k$&$(\\lambda_u)_k$&$(\\lambda_l)_k+\n(\\lambda_u)_k$\\\\\n\\hline\n$-\\infty= {\\tt piv\\_tol}\\cdot\\max|u_{i*}|$, i.e. if it is not very\nsmall in the magnitude among other elements in the same row. Decreasing\nthis parameter may lead to better sparsity at the expense of numerical\naccuracy, and vice versa.\n\n\\medskip\n\n{\\tt int piv\\_lim} (default: {\\tt 4})\n\nThis parameter is used on computing $LU$-factorization of the basis\nmatrix and specifies how many pivot candidates needs to be considered\non choosing a pivot element, \\verb|piv_lim| $\\geq$ 1. If \\verb|piv_lim|\ncandidates have been considered, the pivoting routine prematurely\nterminates the search with the best candidate found.\n\n\\newpage\n\n{\\tt int suhl} (default: {\\tt GLP\\_ON})\n\nThis parameter is used on computing $LU$-factorization of the basis\nmatrix. Being set to {\\tt GLP\\_ON} it enables applying the following\nheuristic proposed by Uwe Suhl: if a column of the active submatrix has\nno eligible pivot candidates, it is no more considered until it becomes\na column singleton. In many cases this allows reducing the time needed\nfor pivot searching. To disable this heuristic the parameter\n\\verb|suhl| should be set to {\\tt GLP\\_OFF}.\n\n\\medskip\n\n{\\tt double eps\\_tol} (default: {\\tt 1e-15})\n\nEpsilon tolerance, \\verb|eps_tol| $\\geq$ 0, used on computing\n$LU$-factorization of the basis matrix. If an element of the active\nsubmatrix of factor $U$ is less than \\verb|eps_tol| in the magnitude,\nit is replaced by exact zero.\n\n%\\medskip\n%\n%{\\tt double max\\_gro} (default: {\\tt 1e+10})\n%\n%Maximal growth of elements of factor $U$, \\verb|max_gro| $\\geq$ 1,\n%allowable on computing $LU$-factorization of the basis matrix. If on\n%some elimination step the ratio $u_{big}/b_{max}$ (where $u_{big}$ is\n%the largest magnitude of elements of factor $U$ appeared in its active\n%submatrix during all the factorization process, $b_{max}$ is the\n%largest magnitude of elements of the basis matrix to be factorized),\n%the basis matrix is considered as ill-conditioned.\n\n\\medskip\n\n{\\tt int nfs\\_max} (default: {\\tt 100})\n\nMaximal number of additional row-like factors (entries of the eta\nfile), \\verb|nfs_max| $\\geq$ 1, which can be added to\n$LU$-factorization of the basis matrix on updating it with the\nForrest--Tomlin technique. This parameter is used only once, before\n$LU$-factorization is computed for the first time, to allocate working\narrays. As a rule, each update adds one new factor (however, some\nupdates may need no addition), so this parameter limits the number of\nupdates between refactorizations.\n\n\\medskip\n\n{\\tt double upd\\_tol} (default: {\\tt 1e-6})\n\nUpdate tolerance, 0 $<$ \\verb|upd_tol| $<$ 1, used on updating\n$LU$-factorization of the basis matrix with the Forrest--Tomlin\ntechnique. If after updating the magnitude of some diagonal element\n$u_{kk}$ of factor $U$ becomes less than\n${\\tt upd\\_tol}\\cdot\\max(|u_{k*}|, |u_{*k}|)$, the factorization is\nconsidered as inaccurate.\n\n\\medskip\n\n{\\tt int nrs\\_max} (default: {\\tt 100})\n\nMaximal number of additional rows and columns, \\verb|nrs_max| $\\geq$ 1,\nwhich can be added to $LU$-factorization of the basis matrix on\nupdating it with the Schur complement technique. This parameter is used\nonly once, before $LU$-factorization is computed for the first time, to\nallocate working arrays. As a rule, each update adds one new row and\ncolumn (however, some updates may need no addition), so this parameter\nlimits the number of updates between refactorizations.\n\n%\\medskip\n%\n%{\\tt int rs\\_size} (default: {\\tt 0})\n%\n%The initial size of the Sparse Vector Area, in non-zeros, used to\n%store non-zero elements of additional rows and columns introduced on\n%updating $LU$-factorization of the basis matrix with the Schur\n%complement technique. If this parameter is set to 0, the initial SVA\n%size is determined automatically.\n\n\\subsection{glp\\_get\\_bhead --- retrieve the basis header information}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_get_bhead(glp_prob *P, int k);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_get_bhead| returns the basis header information\nfor the current basis associated with the specified problem object.\n\n\\returns\n\nIf basic variable $(x_B)_k$, $1\\leq k\\leq m$, is $i$-th auxiliary\nvariable ($1\\leq i\\leq m$), the routine returns $i$. Otherwise, if\n$(x_B)_k$ is $j$-th structural variable ($1\\leq j\\leq n$), the routine\nreturns $m+j$. Here $m$ is the number of rows and $n$ is the number of\ncolumns in the problem object.\n\n\\para{Comments}\n\nSometimes the application program may need to know which original\n(auxiliary and structural) variable correspond to a given basic\nvariable, or, that is the same, which column of the augmented\nconstraint matrix $(I\\ |-\\!A)$ correspond to a given column of the\nbasis matrix $B$.\n\n\\def\\arraystretch{1}\n\nThe correspondence is defined as follows:\\footnote{For more details see\nSubsection \\ref{basbgd}, page \\pageref{basbgd}.}\n$$\\left(\\begin{array}{@{}c@{}}x_B\\\\x_N\\\\\\end{array}\\right)=\n\\Pi\\left(\\begin{array}{@{}c@{}}x_R\\\\x_S\\\\\\end{array}\\right)\n\\ \\ \\Leftrightarrow\n\\ \\ \\left(\\begin{array}{@{}c@{}}x_R\\\\x_S\\\\\\end{array}\\right)=\n\\Pi^T\\left(\\begin{array}{@{}c@{}}x_B\\\\x_N\\\\\\end{array}\\right),$$\nwhere $x_B$ is the vector of basic variables, $x_N$ is the vector of\nnon-basic variables, $x_R$ is the vector of auxiliary variables\nfollowing in their original order,\\footnote{The original order of\nauxiliary and structural variables is defined by the ordinal numbers\nof corresponding rows and columns in the problem object.} $x_S$ is the\nvector of structural variables following in their original order, $\\Pi$\nis a permutation matrix (which is a component of the basis\nfactorization).\n\nThus, if $(x_B)_k=(x_R)_i$ is $i$-th auxiliary variable, the routine\nreturns $i$, and if $(x_B)_k=(x_S)_j$ is $j$-th structural variable,\nthe routine returns $m+j$, where $m$ is the number of rows in the\nproblem object.\n\n\\subsection{glp\\_get\\_row\\_bind --- retrieve row index in the basis\nheader}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_get_row_bind(glp_prob *P, int i);\n\\end{verbatim}\n\n\\returns\n\nThe routine \\verb|glp_get_row_bind| returns the index $k$ of basic\nvariable $(x_B)_k$, $1\\leq k\\leq m$, which is $i$-th auxiliary variable\n(that is, the auxiliary variable corresponding to $i$-th row),\n$1\\leq i\\leq m$, in the current basis associated with the specified\nproblem object, where $m$ is the number of rows. However, if $i$-th\nauxiliary variable is non-basic, the routine returns zero.\n\n\\para{Comments}\n\nThe routine \\verb|glp_get_row_bind| is an inversion of the routine\n\\verb|glp_get_bhead|; that is, if \\linebreak\n\\verb|glp_get_bhead|$(P,k)$ returns $i$,\n\\verb|glp_get_row_bind|$(P,i)$ returns $k$, and vice versa.\n\n\\subsection{glp\\_get\\_col\\_bind --- retrieve column index in the basis\nheader}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_get_col_bind(glp_prob *P, int j);\n\\end{verbatim}\n\n\\returns\n\nThe routine \\verb|glp_get_col_bind| returns the index $k$ of basic\nvariable $(x_B)_k$, $1\\leq k\\leq m$, which is $j$-th structural\nvariable (that is, the structural variable corresponding to $j$-th\ncolumn), $1\\leq j\\leq n$, in the current basis associated with the\nspecified problem object, where $m$ is the number of rows, $n$ is the\nnumber of columns. However, if $j$-th structural variable is non-basic,\nthe routine returns zero.\n\n\\para{Comments}\n\nThe routine \\verb|glp_get_col_bind| is an inversion of the routine\n\\verb|glp_get_bhead|; that is, if \\linebreak\n\\verb|glp_get_bhead|$(P,k)$ returns $m+j$,\n\\verb|glp_get_col_bind|$(P,j)$ returns $k$, and vice versa.\n\n\\subsection{glp\\_ftran --- perform forward transformation}\n\n\\synopsis\n\n\\begin{verbatim}\n void glp_ftran(glp_prob *P, double x[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_ftran| performs forward transformation (FTRAN),\ni.e. it solves the system $Bx=b$, where $B$ is the basis matrix\nassociated with the specified problem object, $x$ is the vector of\nunknowns to be computed, $b$ is the vector of right-hand sides.\n\nOn entry to the routine elements of the vector $b$ should be stored in\nlocations \\verb|x[1]|, \\dots, \\verb|x[m]|, where $m$ is the number of\nrows. On exit the routine stores elements of the vector $x$ in the same\nlocations.\n\n\\subsection{glp\\_btran --- perform backward transformation}\n\n\\synopsis\n\n\\begin{verbatim}\n void glp_btran(glp_prob *P, double x[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_btran| performs backward transformation (BTRAN),\ni.e. it solves the system $B^Tx=b$, where $B^T$ is a matrix transposed\nto the basis matrix $B$ associated with the specified problem object,\n$x$ is the vector of unknowns to be computed, $b$ is the vector of\nright-hand sides.\n\nOn entry to the routine elements of the vector $b$ should be stored in\nlocations \\verb|x[1]|, \\dots, \\verb|x[m]|, where $m$ is the number of\nrows. On exit the routine stores elements of the vector $x$ in the same\nlocations.\n\n\\subsection{glp\\_warm\\_up --- ``warm up'' LP basis}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_warm_up(glp_prob *P);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_warm_up| ``warms up'' the LP basis for the\nspecified problem object using current statuses assigned to rows and\ncolumns (that is, to auxiliary and structural variables).\n\nThis operation includes computing factorization of the basis matrix\n(if it does not exist), computing primal and dual components of basic\nsolution, and determining the solution status.\n\n\\returns\n\n\\begin{retlist}\n0 & The operation has been successfully performed.\\\\\n\n\\verb|GLP_EBADB| & The basis matrix is invalid, because the number of\nbasic (auxiliary and structural) variables is not the same as the\nnumber of rows in the problem object.\\\\\n\n\\verb|GLP_ESING| & The basis matrix is singular within the working\nprecision.\\\\\n\n\\verb|GLP_ECOND| & The basis matrix is ill-conditioned, i.e. its\ncondition number is too large.\\\\\n\\end{retlist}\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\newpage\n\n\\section{Simplex tableau routines}\n\n\\subsection{glp\\_eval\\_tab\\_row --- compute row of the tableau}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_eval_tab_row(glp_prob *P, int k, int ind[], double val[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_eval_tab_row| computes a row of the current\nsimplex tableau (see Subsection 3.1.1, formula (3.12)), which (row)\ncorresponds to some basic variable specified by the parameter $k$ as\nfollows: if $1\\leq k\\leq m$, the basic variable is $k$-th auxiliary\nvariable, and if $m+1\\leq k\\leq m+n$, the basic variable is $(k-m)$-th\nstructural variable, where $m$ is the number of rows and $n$ is the\nnumber of columns in the specified problem object. The basis\nfactorization must exist.\n\nThe computed row shows how the specified basic variable depends on\nnon-basic variables:\n$$x_k=(x_B)_i=\\xi_{i1}(x_N)_1+\\xi_{i2}(x_N)_2+\\dots+\\xi_{in}(x_N)_n,$$\nwhere $\\xi_{i1}$, $\\xi_{i2}$, \\dots, $\\xi_{in}$ are elements of the\nsimplex table row, $(x_N)_1$, $(x_N)_2$, \\dots, $(x_N)_n$ are non-basic\n(auxiliary and structural) variables.\n\nThe routine stores column indices and corresponding numeric values of\nnon-zero elements of the computed row in unordered sparse format in\nlocations \\verb|ind[1]|, \\dots, \\verb|ind[len]| and \\verb|val[1]|,\n\\dots, \\verb|val[len]|, respectively, where $0\\leq{\\tt len}\\leq n$ is\nthe number of non-zero elements in the row returned on exit.\n\nElement indices stored in the array \\verb|ind| have the same sense as\nindex $k$, i.e. indices 1 to $m$ denote auxiliary variables while\nindices $m+1$ to $m+n$ denote structural variables (all these variables\nare obviously non-basic by definition).\n\n\\returns\n\nThe routine \\verb|glp_eval_tab_row| returns \\verb|len|, which is the\nnumber of non-zero elements in the simplex table row stored in the\narrays \\verb|ind| and \\verb|val|.\n\n\\para{Comments}\n\nA row of the simplex table is computed as follows. At first, the\nroutine checks that the specified variable $x_k$ is basic and uses the\npermutation matrix $\\Pi$ (3.7) to determine index $i$ of basic variable\n$(x_B)_i$, which corresponds to $x_k$.\n\nThe row to be computed is $i$-th row of the matrix $\\Xi$ (3.12),\ntherefore:\n$$\\xi_i=e_i^T\\Xi=-e_i^TB^{-1}N=-(B^{-T}e_i)^TN,$$\nwhere $e_i$ is $i$-th unity vector. So the routine performs BTRAN to\nobtain $i$-th row of the inverse $B^{-1}$:\n$$\\varrho_i=B^{-T}e_i,$$\nand then computes elements of the simplex table row as inner products:\n$$\\xi_{ij}=-\\varrho_i^TN_j,\\ \\ j=1,2,\\dots,n,$$\nwhere $N_j$ is $j$-th column of matrix $N$ (3.9), which (column)\ncorresponds to non-basic variable $(x_N)_j$. The permutation matrix\n$\\Pi$ is used again to convert indices $j$ of non-basic columns to\noriginal ordinal numbers of auxiliary and structural variables.\n\n\\subsection{glp\\_eval\\_tab\\_col --- compute column of the tableau}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_eval_tab_col(glp_prob *P, int k, int ind[], double val[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_eval_tab_col| computes a column of the current\nsimplex tableau (see Subsection 3.1.1, formula (3.12)), which (column)\ncorresponds to some non-basic variable specified by the parameter $k$:\nif $1\\leq k\\leq m$, the non-basic variable is $k$-th auxiliary\nvariable, and if $m+1\\leq k\\leq m+n$, the non-basic variable is\n$(k-m)$-th structural variable, where $m$ is the number of rows and $n$\nis the number of columns in the specified problem object. The basis\nfactorization must exist.\n\nThe computed column shows how basic variables depends on the specified\nnon-basic variable $x_k=(x_N)_j$:\n$$\n\\begin{array}{r@{\\ }c@{\\ }l@{\\ }l}\n(x_B)_1&=&\\dots+\\xi_{1j}(x_N)_j&+\\dots\\\\\n(x_B)_2&=&\\dots+\\xi_{2j}(x_N)_j&+\\dots\\\\\n.\\ \\ .&.&.\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\\\\n(x_B)_m&=&\\dots+\\xi_{mj}(x_N)_j&+\\dots\\\\\n\\end{array}\n$$\nwhere $\\xi_{1j}$, $\\xi_{2j}$, \\dots, $\\xi_{mj}$ are elements of the\nsimplex table column, $(x_B)_1$, $(x_B)_2$, \\dots, $(x_B)_m$ are basic\n(auxiliary and structural) variables.\n\nThe routine stores row indices and corresponding numeric values of\nnon-zero elements of the computed column in unordered sparse format in\nlocations \\verb|ind[1]|, \\dots, \\verb|ind[len]| and \\verb|val[1]|,\n\\dots, \\verb|val[len]|, respectively, where $0\\leq{\\tt len}\\leq m$ is\nthe number of non-zero elements in the column returned on exit.\n\nElement indices stored in the array \\verb|ind| have the same sense as\nindex $k$, i.e. indices 1 to $m$ denote auxiliary variables while\nindices $m+1$ to $m+n$ denote structural variables (all these variables\nare obviously basic by definition).\n\n\\returns\n\nThe routine \\verb|glp_eval_tab_col| returns \\verb|len|, which is the\nnumber of non-zero elements in the simplex table column stored in the\narrays \\verb|ind| and \\verb|val|.\n\n\\para{Comments}\n\nA column of the simplex table is computed as follows. At first, the\nroutine checks that the specified variable $x_k$ is non-basic and uses\nthe permutation matrix $\\Pi$ (3.7) to determine index $j$ of non-basic\nvariable $(x_N)_j$, which corresponds to $x_k$.\n\nThe column to be computed is $j$-th column of the matrix $\\Xi$ (3.12),\ntherefore:\n$$\\Xi_j=\\Xi e_j=-B^{-1}Ne_j=-B^{-1}N_j,$$\nwhere $e_j$ is $j$-th unity vector, $N_j$ is $j$-th column of matrix\n$N$ (3.9). So the routine performs FTRAN to transform $N_j$ to the\nsimplex table column $\\Xi_j=(\\xi_{ij})$ and uses the permutation matrix\n$\\Pi$ to convert row indices $i$ to original ordinal numbers of\nauxiliary and structural variables.\n\n\\subsection{glp\\_transform\\_row --- transform explicitly specified row}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_transform_row(glp_prob *P, int len, int ind[], double val[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_transform_row| performs the same operation as the\nroutine \\verb|glp_eval_tab_row| with exception that the row to be\ntransformed is specified explicitly as a sparse vector.\n\nThe explicitly specified row may be thought as a linear form:\n$$x=a_1x_{m+1}+a_2x_{m+2}+\\dots+a_nx_{m+n},$$\nwhere $x$ is an auxiliary variable for this row, $a_j$ are coefficients\nof the linear form, $x_{m+j}$ are structural variables.\n\nOn entry column indices and numerical values of non-zero coefficients\n$a_j$ of the specified row should be placed in locations \\verb|ind[1]|,\n\\dots, \\verb|ind[len]| and \\verb|val[1]|, \\dots, \\verb|val[len]|, where\n\\verb|len| is number of non-zero coefficients.\n\nThis routine uses the system of equality constraints and the current\nbasis in order to express the auxiliary variable $x$ through the current\nnon-basic variables (as if the transformed row were added to the problem\nobject and the auxiliary variable $x$ were basic), i.e. the resultant\nrow has the form:\n$$x=\\xi_1(x_N)_1+\\xi_2(x_N)_2+\\dots+\\xi_n(x_N)_n,$$\nwhere $\\xi_j$ are influence coefficients, $(x_N)_j$ are non-basic\n(auxiliary and structural) variables, $n$ is the number of columns in\nthe problem object.\n\nOn exit the routine stores indices and numerical values of non-zero\ncoefficients $\\xi_j$ of the resultant row in locations \\verb|ind[1]|,\n\\dots, \\verb|ind[len']| and \\verb|val[1]|, \\dots, \\verb|val[len']|,\nwhere $0\\leq{\\tt len'}\\leq n$ is the number of non-zero coefficients in\nthe resultant row returned by the routine. Note that indices of\nnon-basic variables stored in the array \\verb|ind| correspond to\noriginal ordinal numbers of variables: indices 1 to $m$ mean auxiliary\nvariables and indices $m+1$ to $m+n$ mean structural ones.\n\n\\returns\n\nThe routine \\verb|glp_transform_row| returns \\verb|len'|, the number of\nnon-zero coefficients in the resultant row stored in the arrays\n\\verb|ind| and \\verb|val|.\n\n\\newpage\n\n\\subsection{glp\\_transform\\_col --- transform explicitly specified\ncolumn}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_transform_col(glp_prob *P, int len, int ind[], double val[]);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_transform_col| performs the same operation as the\nroutine \\verb|glp_eval_tab_col| with exception that the column to be\ntransformed is specified explicitly as a sparse vector.\n\nThe explicitly specified column may be thought as it were added to\nthe original system of equality constraints:\n$$\n\\begin{array}{l@{\\ }c@{\\ }r@{\\ }c@{\\ }r@{\\ }c@{\\ }r}\nx_1&=&a_{11}x_{m+1}&+\\dots+&a_{1n}x_{m+n}&+&a_1x \\\\\nx_2&=&a_{21}x_{m+1}&+\\dots+&a_{2n}x_{m+n}&+&a_2x \\\\\n\\multicolumn{7}{c}\n{.\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .}\\\\\nx_m&=&a_{m1}x_{m+1}&+\\dots+&a_{mn}x_{m+n}&+&a_mx \\\\\n\\end{array}\n$$\nwhere $x_i$ are auxiliary variables, $x_{m+j}$ are structural variables\n(presented in the problem object), $x$ is a structural variable for the\nexplicitly specified column, $a_i$ are constraint coefficients at $x$.\n\nOn entry row indices and numerical values of non-zero coefficients\n$a_i$ of the specified column should be placed in locations\n\\verb|ind[1]|, \\dots, \\verb|ind[len]| and \\verb|val[1]|, \\dots,\n\\verb|val[len]|, where \\verb|len| is number of non-zero coefficients.\n\nThis routine uses the system of equality constraints and the current\nbasis in order to express the current basic variables through the\nstructural variable $x$ (as if the transformed column were added to the\nproblem object and the variable $x$ were non-basic):\n$$\n\\begin{array}{l@{\\ }c@{\\ }r}\n(x_B)_1&=\\dots+&\\xi_{1}x\\\\\n(x_B)_2&=\\dots+&\\xi_{2}x\\\\\n\\multicolumn{3}{c}{.\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .}\\\\\n(x_B)_m&=\\dots+&\\xi_{m}x\\\\\n\\end{array}\n$$\nwhere $\\xi_i$ are influence coefficients, $x_B$ are basic (auxiliary\nand structural) variables, $m$ is the number of rows in the problem\nobject.\n\nOn exit the routine stores indices and numerical values of non-zero\ncoefficients $\\xi_i$ of the resultant column in locations \\verb|ind[1]|,\n\\dots, \\verb|ind[len']| and \\verb|val[1]|, \\dots, \\verb|val[len']|,\nwhere $0\\leq{\\tt len'}\\leq m$ is the number of non-zero coefficients in\nthe resultant column returned by the routine. Note that indices of basic\nvariables stored in the array \\verb|ind| correspond to original ordinal\nnumbers of variables, i.e. indices 1 to $m$ mean auxiliary variables,\nindices $m+1$ to $m+n$ mean structural ones.\n\n\\returns\n\nThe routine \\verb|glp_transform_col| returns \\verb|len'|, the number of\nnon-zero coefficients in the resultant column stored in the arrays\n\\verb|ind| and \\verb|val|.\n\n\\newpage\n\n\\subsection{glp\\_prim\\_rtest --- perform primal ratio test}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_prim_rtest(glp_prob *P, int len, const int ind[], const double val[],\n int dir, double eps);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_prim_rtest| performs the primal ratio test using\nan explicitly specified column of the simplex table.\n\nThe current basic solution associated with the LP problem object must\nbe primal feasible.\n\nThe explicitly specified column of the simplex table shows how the\nbasic variables $x_B$ depend on some non-basic variable $x$ (which is\nnot necessarily presented in the problem object):\n$$\n\\begin{array}{l@{\\ }c@{\\ }r}\n(x_B)_1&=\\dots+&\\xi_{1}x\\\\\n(x_B)_2&=\\dots+&\\xi_{2}x\\\\\n\\multicolumn{3}{c}{.\\ \\ .\\ \\ .\\ \\ .\\ \\ .\\ \\ .}\\\\\n(x_B)_m&=\\dots+&\\xi_{m}x\\\\\n\\end{array}\n$$\n\nThe column is specifed on entry to the routine in sparse format.\nOrdinal numbers of basic variables $(x_B)_i$ should be placed in\nlocations \\verb|ind[1]|, \\dots, \\verb|ind[len]|, where ordinal number\n1 to $m$ denote auxiliary variables, and ordinal numbers $m+1$ to $m+n$\ndenote structural variables. The corresponding non-zero coefficients\n$\\xi_i$ should be placed in locations\n\\verb|val[1]|, \\dots, \\verb|val[len]|. The arrays \\verb|ind| and\n\\verb|val| are not changed by the routine.\n\nThe parameter \\verb|dir| specifies direction in which the variable $x$\nchanges on entering the basis: $+1$ means increasing, $-1$ means\ndecreasing.\n\nThe parameter \\verb|eps| is an absolute tolerance (small positive\nnumber, say, $10^{-9}$) used by the routine to skip $\\xi_i$'s whose\nmagnitude is less than \\verb|eps|.\n\nThe routine determines which basic variable (among those specified in\n\\verb|ind[1]|, \\dots, \\verb|ind[len]|) reaches its (lower or upper)\nbound first before any other basic variables do, and which, therefore,\nshould leave the basis in order to keep primal feasibility.\n\n\\returns\n\nThe routine \\verb|glp_prim_rtest| returns the index, \\verb|piv|, in the\narrays \\verb|ind| and \\verb|val| corresponding to the pivot element\nchosen, $1\\leq$ \\verb|piv| $\\leq$ \\verb|len|. If the adjacent basic\nsolution is primal unbounded, and therefore the choice cannot be made,\nthe routine returns zero.\n\n\\para{Comments}\n\nIf the non-basic variable $x$ is presented in the LP problem object,\nthe input column can be computed with the routine\n\\verb|glp_eval_tab_col|; otherwise, it can be computed with the routine\n\\verb|glp_transform_col|.\n\n\\newpage\n\n\\subsection{glp\\_dual\\_rtest --- perform dual ratio test}\n\n\\synopsis\n\n\\begin{verbatim}\n int glp_dual_rtest(glp_prob *P, int len, const int ind[], const double val[],\n int dir, double eps);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_dual_rtest| performs the dual ratio test using\nan explicitly specified row of the simplex table.\n\nThe current basic solution associated with the LP problem object must\nbe dual feasible.\n\nThe explicitly specified row of the simplex table is a linear form\nthat shows how some basic variable $x$ (which is not necessarily\npresented in the problem object) depends on non-basic variables $x_N$:\n$$x=\\xi_1(x_N)_1+\\xi_2(x_N)_2+\\dots+\\xi_n(x_N)_n.$$\n\nThe row is specified on entry to the routine in sparse format. Ordinal\nnumbers of non-basic variables $(x_N)_j$ should be placed in locations\n\\verb|ind[1]|, \\dots, \\verb|ind[len]|, where ordinal numbers 1 to $m$\ndenote auxiliary variables, and ordinal numbers $m+1$ to $m+n$ denote\nstructural variables. The corresponding non-zero coefficients $\\xi_j$\nshould be placed in locations \\verb|val[1]|, \\dots, \\verb|val[len]|.\nThe arrays \\verb|ind| and \\verb|val| are not changed by the routine.\n\nThe parameter \\verb|dir| specifies direction in which the variable $x$\nchanges on leaving the basis: $+1$ means that $x$ goes on its lower\nbound, so its reduced cost (dual variable) is increasing (minimization)\nor decreasing (maximization); $-1$ means that $x$ goes on its upper\nbound, so its reduced cost is decreasing (minimization) or increasing\n(maximization).\n\nThe parameter \\verb|eps| is an absolute tolerance (small positive\nnumber, say, $10^{-9}$) used by the routine to skip $\\xi_j$'s whose\nmagnitude is less than \\verb|eps|.\n\nThe routine determines which non-basic variable (among those specified\nin \\verb|ind[1]|, \\dots,\\linebreak \\verb|ind[len]|) should enter the\nbasis in order to keep dual feasibility, because its reduced cost\nreaches the (zero) bound first before this occurs for any other\nnon-basic variables.\n\n\\returns\n\nThe routine \\verb|glp_dual_rtest| returns the index, \\verb|piv|, in the\narrays \\verb|ind| and \\verb|val| corresponding to the pivot element\nchosen, $1\\leq$ \\verb|piv| $\\leq$ \\verb|len|. If the adjacent basic\nsolution is dual unbounded, and therefore the choice cannot be made,\nthe routine returns zero.\n\n\\para{Comments}\n\nIf the basic variable $x$ is presented in the LP problem object, the\ninput row can be computed\\linebreak with the routine\n\\verb|glp_eval_tab_row|; otherwise, it can be computed with the routine\n\\linebreak \\verb|glp_transform_row|.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n\\newpage\n\n\\section{Post-optimal analysis routines}\n\n\\subsection{glp\\_analyze\\_bound --- analyze active bound of non-basic\nvariable}\n\n\\synopsis\n\n\\begin{verbatim}\n void glp_analyze_bound(glp_prob *P, int k, double *limit1, int *var1,\n double *limit2, int *var2);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_analyze_bound| analyzes the effect of varying the\nactive bound of specified non-basic variable.\n\nThe non-basic variable is specified by the parameter $k$, where\n$1\\leq k\\leq m$ means auxiliary variable of corresponding row, and\n$m+1\\leq k\\leq m+n$ means structural variable (column).\n\nNote that the current basic solution must be optimal, and the basis\nfactorization must exist.\n\nResults of the analysis have the following meaning.\n\n\\verb|value1| is the minimal value of the active bound, at which the\nbasis still remains primal feasible and thus optimal. \\verb|-DBL_MAX|\nmeans that the active bound has no lower limit.\n\n\\verb|var1| is the ordinal number of an auxiliary (1 to $m$) or\nstructural ($m+1$ to $m+n$) basic variable, which reaches its bound\nfirst and thereby limits further decreasing the active bound being\nanalyzed. if \\verb|value1| = \\verb|-DBL_MAX|, \\verb|var1| is set to 0.\n\n\\verb|value2| is the maximal value of the active bound, at which the\nbasis still remains primal feasible and thus optimal. \\verb|+DBL_MAX|\nmeans that the active bound has no upper limit.\n\n\\verb|var2| is the ordinal number of an auxiliary (1 to $m$) or\nstructural ($m+1$ to $m+n$) basic variable, which reaches its bound\nfirst and thereby limits further increasing the active bound being\nanalyzed. if \\verb|value2| = \\verb|+DBL_MAX|, \\verb|var2| is set to 0.\n\nThe parameters \\verb|value1|, \\verb|var1|, \\verb|value2|, \\verb|var2|\ncan be specified as \\verb|NULL|, in which case corresponding information\nis not stored.\n\n\\subsection{glp\\_analyze\\_coef --- analyze objective coefficient at\nbasic variable}\n\n\\synopsis\n\n\\begin{verbatim}\n void glp_analyze_coef(glp_prob *P, int k,\n double *coef1, int *var1, double *value1,\n double *coef2, int *var2, double *value2);\n\\end{verbatim}\n\n\\description\n\nThe routine \\verb|glp_analyze_coef| analyzes the effect of varying the\nobjective coefficient at specified basic variable.\n\nThe basic variable is specified by the parameter $k$, where\n$1\\leq k\\leq m$ means auxiliary variable of corresponding row, and\n$m+1\\leq k\\leq m+n$ means structural variable (column).\n\nNote that the current basic solution must be optimal, and the basis\nfactorization must exist.\n\nResults of the analysis have the following meaning.\n\n\\verb|coef1| is the minimal value of the objective coefficient, at\nwhich the basis still remains dual feasible and thus optimal.\n\\verb|-DBL_MAX| means that the objective coefficient has no lower\nlimit.\n\n\\verb|var1| is the ordinal number of an auxiliary (1 to $m$) or\nstructural ($m+1$ to $m+n$) non-basic variable, whose reduced cost\nreaches its zero bound first and thereby limits further decreasing the\nobjective coefficient being analyzed.\nIf \\verb|coef1| = \\verb|-DBL_MAX|, \\verb|var1| is set to 0.\n\n\\verb|value1| is value of the basic variable being analyzed in an\nadjacent basis, which is defined as follows. Let the objective\ncoefficient reach its minimal value (\\verb|coef1|) and continue\ndecreasing. Then the reduced cost of the limiting non-basic variable\n(\\verb|var1|) becomes dual infeasible and the current basis becomes\nnon-optimal that forces the limiting non-basic variable to enter the\nbasis replacing there some basic variable that leaves the basis to keep\nprimal feasibility. Should note that on determining the adjacent basis\ncurrent bounds of the basic variable being analyzed are ignored as if\nit were free (unbounded) variable, so it cannot leave the basis. It may\nhappen that no dual feasible adjacent basis exists, in which case\n\\verb|value1| is set to \\verb|-DBL_MAX| or \\verb|+DBL_MAX|.\n\n\\verb|coef2| is the maximal value of the objective coefficient, at\nwhich the basis still remains dual feasible and thus optimal.\n\\verb|+DBL_MAX| means that the objective coefficient has no upper\nlimit.\n\n\\verb|var2| is the ordinal number of an auxiliary (1 to $m$) or\nstructural ($m+1$ to $m+n$) non-basic variable, whose reduced cost\nreaches its zero bound first and thereby limits further increasing the\nobjective coefficient being analyzed.\nIf \\verb|coef2| = \\verb|+DBL_MAX|, \\verb|var2| is set to 0.\n\n\\verb|value2| is value of the basic variable being analyzed in an\nadjacent basis, which is defined exactly in the same way as\n\\verb|value1| above with exception that now the objective coefficient\nis increasing.\n\nThe parameters \\verb|coef1|, \\verb|var1|, \\verb|value1|, \\verb|coef2|,\n\\verb|var2|, \\verb|value2| can be specified as \\verb|NULL|, in which\ncase corresponding information is not stored.\n\n%* eof *%\n", "meta": {"hexsha": "2367db5738fb51b054c8e6a26837bcc0c14533ce", "size": 53063, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "sdk/mmitss/src/glpk-4.55/doc/glpk04.tex", "max_stars_repo_name": "OSADP/MMITSS_AZ_FIELD", "max_stars_repo_head_hexsha": "b4c870061c518eddfa0152938ab60abc2a31023f", "max_stars_repo_licenses": ["Apache-2.0"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "sdk/mmitss/src/glpk-4.55/doc/glpk04.tex", "max_issues_repo_name": "OSADP/MMITSS_AZ_FIELD", "max_issues_repo_head_hexsha": "b4c870061c518eddfa0152938ab60abc2a31023f", "max_issues_repo_licenses": ["Apache-2.0"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "sdk/mmitss/src/glpk-4.55/doc/glpk04.tex", "max_forks_repo_name": "OSADP/MMITSS_AZ_FIELD", "max_forks_repo_head_hexsha": "b4c870061c518eddfa0152938ab60abc2a31023f", "max_forks_repo_licenses": ["Apache-2.0"], "max_forks_count": 2, "max_forks_repo_forks_event_min_datetime": "2017-05-04T16:41:23.000Z", "max_forks_repo_forks_event_max_datetime": "2020-04-01T20:32:44.000Z", "avg_line_length": 38.2298270893, "max_line_length": 80, "alphanum_fraction": 0.7319412773, "num_tokens": 15544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.8152324983301568, "lm_q2_score": 0.7490872075132152, "lm_q1q2_score": 0.610680235648159}} {"text": "\\section{Evaluation}\n\\label{sec:evaluation}\n%\nIn this section, we introduce several pairs of coding solutions,\ntranslated to functional programs.\n%\nFor each pair of programs $P_0$ and $P_1$, we identify an inductive\nset of pairs of expressions $I$, and give a map from all pairs of\nexpressions in $I$ to relational invariants that prove the partial\nequivalence of $P_0$ and $P_1$.\n\n\\paragraph{Solutions to Add Digits}\n%\n\\NV{complete}\n\n\\paragraph{Solutions to Climbing Stairs}\n%\n\\begin{figure}[t]\n \\begin{minipage}{0.48\\linewidth}\n \\input{code/ClimbStairs0.ml}\n \\end{minipage}\n \\begin{minipage}{0.48\\linewidth}\n \\input{code/ClimbStairs1.ml} \n \\end{minipage}\n %\n \\caption{\\cc{ClimbStairs0} and \\cc{ClimbStairs1}: two solutions to\n the Climbing Stairs problem hosted on LeetCode, manually translated\n to OCaml.\n %\n All subexpressions are labeled with indices in comments.\n %\n }\n \\label{fig:climb-stairs}\n\\end{figure}\n%\n\\autoref{fig:climb-stairs} contains the pseudocode for two solutions\nto the Climbing Stairs Problem hosted on the coding platform the\nLeetCode.\n% problem statement:\nThe Climbing Stairs Problem is to take an integer $n$ and return the\nnumber of distinct sequences of steps that can be taken to climb $n$\nsteps, where in each step either one or two steps can be climbed. If\n$n$ is less than or equal to one, then the solution is one.\n%\n\\BH{walk through how the two solutions work}\n\n%\n\\cc{climbStairs0} and \\cc{climbStairs1} have the following proof of\npartial equivalence:\n%\n\\BH{put the pairs of subexpressions in a set}\n%\n\\begin{align*}\n % pair of main subexpressions:\n (\\cc{0}, \\cc{0}) \\mapsto & %\n \\alpha_0^0 = \\alpha_1^0 \\implies \\nu_0 = \\nu_1 \\\\\n % pair of function bodies:\n (\\cc{1}, \\cc{1}) \\mapsto & %\n \\cc{n}_0 = \\cc{n}_1 \\implies \\nu_0 = \\nu_1 \\\\\n % pair of if guards:\n (\\cc{2}, \\cc{2}) \\mapsto & %\n \\cc{n}_0 = \\cc{n}_1 \\implies \\nu_0 = \\nu_1 \\\\\n % pair of then expressions:\n (\\cc{3}, \\cc{3}) \\mapsto & \\nu_0 = \\nu_1 \\\\\n % pair of else expressions:\n (\\cc{4}, \\cc{4}) \\mapsto & %\n \\cc{n}_0 = \\cc{n}_1 \\implies \\nu_0 = \\nu_1 \\\\\n % pair of fixpoint functions:\n (\\cc{5}, \\cc{5}) \\mapsto & %\n % inductive relational invariant (indices in step):\n (\\cc{n}_0 = \\cc{n}_1 \\land %\n \\alpha_0^0 = \\alpha_1^1 \\land %\n \\alpha_0^1 = \\alpha_1^1 \\land %\n \\alpha_0^3 + 1 = \\alpha_1^2 \\implies %\n \\nu_0 = \\nu_1) \\land \\\\\n % for establishing final property:\n & (\\cc{n}_0 = \\cc{n}_1 \\land %\n \\alpha_0^0 = \\alpha_0^1 + \\alpha_1^1 \\land %\n \\alpha_0^1 = \\alpha_1^1 \\land %\n \\alpha_0^3 = \\alpha_1^2 \\implies %\n \\nu_0 = \\nu_1) \\\\\n % pair of argument vectors:\n (\\cc{6}, \\cc{6}) \\mapsto & %\n \\pi_0(\\nu_0) = \\pi_0(\\nu_1) + \\pi_1(\\nu_1) \\land %\n \\pi_1(\\nu_0) = \\pi_1(\\nu_1) \\land %\n \\pi_3(\\nu_0) = \\pi_3(\\nu_3) \\\\\n % fixpoint body:\n (\\cc{7}, \\cc{7}) \\mapsto & \n % inductive relational invariant (indices in step):\n (\\cc{n}_0 = \\cc{n}_1 \\land %\n \\cc{sum}_0 = \\cc{count2}_1 \\land %\n \\cc{prev}_1 = \\cc{count2}_1 \\land %\n \\cc{i}_3 + 1 = \\cc{i}_1 \\implies %\n \\nu_0 = \\nu_1) \\land \\\\\n % for establishing final property:\n & (\\cc{n}_0 = \\cc{n}_1 \\land %\n \\cc{sum}_0 = \\cc{count1}_1 + \\cc{count2}_1 \\land %\n \\cc{prev}_1 = \\cc{count2}_1 \\land %\n \\cc{i}_3 = \\cc{i}_1 \\implies %\n \\nu_0 = \\nu_1) \\\\\n % fixpoint body guards:\n (\\cc{8}, \\cc{8}) \\mapsto & 0 \\leq \\cc{i}_1 - \\cc{i}_0 \\leq 1 \\implies %\n \\nu_0 = \\nu_1 \\\\\n % fixpoint body then expressions:\n (\\cc{9}, \\cc{9}) \\mapsto & \n (\\cc{n}_0 = \\cc{n}_1 \\land %\n \\cc{sum}_0 = \\cc{count2}_1 \\land %\n \\cc{prev}_0 = \\cc{count2}_1 \\land %\n \\cc{i}_0 + 1 = \\cc{i}_1 \\implies %\n \\nu_0 = \\nu_1) \\land \\\\\n % for establishing final property:\n & (\\cc{n}_0 = \\cc{n}_1 \\land \\cc{i}_0 \\land %\n \\cc{sum}_0 = \\cc{count1}_1 + \\cc{count2}_1 \\land %\n \\cc{prev}_1 = \\cc{count2}_1 \\land %\n \\cc{i}_0 = \\cc{i}_1 \\implies %\n \\nu_0 = \\nu_1) \\\\\n % fixpoint body else expressions:\n (\\cc{10}, \\cc{10}) \\mapsto & \\cc{sum}_0 = \\cc{count2}_1 \\implies %\n \\nu_0 = \\nu_1 \\\\\n % argument steps preserve relative invariants over arguments:\n % subcase: CS1 is one step ahead of CS0:\n (\\cc{11}, \\cc{11}) \\mapsto & \n (\\cc{sum}_0 = \\cc{count2}_1 \\land %\n \\cc{prev}_1 = \\cc{count2}_1 \\land %\n \\cc{i}_3 + 1 = \\cc{i}_1 \\implies \\\\\n & \\pi_0(\\nu_0) = \\pi_1(\\nu_1) \\land %\n \\pi_1(\\nu_0) = \\pi_0(\\nu_1) \\land %\n \\pi_3(\\nu_0) = \\pi_2(\\nu_1)) \\land \\\\\n %\n & (\\cc{sum}_0 = \\cc{count1}_1 + \\cc{count2}_1 \\land %\n \\cc{prev}_1 = \\cc{count2}_1 \\land %\n \\cc{i}_0 = \\cc{i}_1 \\implies \\\\ %\n & \\pi_0(\\nu_0) = \\pi_0(\\nu_1) + \\pi_1(\\nu_1) \\land %\n \\pi_1(\\nu_0) = \\pi_1(\\nu_1) \\land %\n \\pi_3(\\nu_0) = \\pi_2(\\nu_1)) \\\\\n\\end{align*}\n%\n\\BH{define notation for logical variables}\n%\n\\BH{refactor entries to use common invariant meta-variables}\n\n\\BH{walk through relational invariants}\n\n\\NV{Question: Why 9 has n0 equals tp n1 precondition but 10 does not?}\n\n\\paragraph{Solutions to Reverse}\n%\n\\NV{complete}\n\n\\paragraph{Solutions to Trailing Zeroes}\n\\begin{hcode}\ntrailingZeroes0 {- 0 -} n \n = {- 1 -} go0 ((0, n) {- 2 -}) {- 3 -}\n where\n go0 {- 4 -} (x, y) {- 5 -}\n | 5 <= y {- 6 -}\n = go0 ((x + (y `div` 5), y `div` 5) {- 9-}) {- 7 -}\n | otherwise \n = x {- 8 -}\n\n\ntrailingZeroes1 {- 0 -} n \n = {- 1 -} go1 ((0, n `div` 5) {- 2 -}) {- 3 -}\n where \n go1 {- 4 -} (s, n) {- 5 -}\n | 0 < n {- 6 -} \n = go1 ((s + n, n `div` 5) {- 9 -}) {- 7 -}\n | otherwise \n = s {- 8 -}\n\\end{hcode}\n\n\n\\newcommand\\rinv[3]{\\ensuremath{(\\cc{#1}, \\cc{#2}) }}\n\\newcommand\\iinv[2]{\\ensuremath{\\mathcal{I}(#1, #2)}}\n\\newcommand\\precondition[2]{\\ensuremath{\\mathcal{P}(#1, #2)}}\n\\newcommand\\postcondition[2]{\\ensuremath{\\mathcal{Q}(#1, #2)}}\n\n$$\n\\text{Invariant: }\\iinv{x_0}{x_1} = x_0^0 = x_1^0 \\land x_0^1 = x_0^1 / 5\n$$\n$$\n\\text{Precondition: }\\precondition{x_0}{x_1} = x_0^0 = x_1^0 \\land x_0^1 = x_0^1 / 5\n$$\n$$\n\\text{Postcondition: }\\postcondition{x_0}{x_1} = x_0 = x_1\n$$\n\n$$\n\\begin{array}{ r c l}\n % pair of main subexpressions:\n \\rinv{0}{0} &\\mapsto& \\alpha_0 = \\alpha_1 \\implies \\nu_0 = \\nu_1 \\\\\n % pair of body \n \\rinv{1}{1} &\\mapsto& n_0 = n_1 \\implies \\nu_0 = \\nu_1 \\\\\n % pair on arguments \n \\rinv{2}{2} &\\mapsto& \\precondition{\\nu_0}{\\nu_1} \\\\\n % pair of fixpoint functions (like step 5 above):\n \\rinv{3}{3} &\\mapsto& n_0 = n_1 \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n % pair of fixpoint functions (like step 5 above):\n \\rinv{4}{4} &\\mapsto& \\precondition{\\alpha_0}{\\alpha_1} \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n \\rinv{5}{5} &\\mapsto& \\precondition{(x_0, y_0)}{(s_1, n_1)} \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n % pair for condition\n \\rinv{6}{6} &\\mapsto& \\precondition{(x_0, y_0)}{(s_1, n_1)} \n \\implies 5 \\leq y_0 \\Leftrightarrow 0 < n_1 \n \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n % pair for then\n \\rinv{7}{7} &\\mapsto& \\precondition{(x_0, y_0)}{(s_1, n_1)} \n \\implies \\iinv{(x_0, y_0)}{(s_1, n_1)} \n \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n % pair for else\n \\rinv{8}{8} &\\mapsto& \\precondition{(x_0, y_0)}{(s_1, n_1)} \n \\implies x_0 = s_1 \n \\implies \\postcondition{\\nu_0}{\\nu_1} \\\\\n % pair for rec call\n \\rinv{9}{9} &\\mapsto& \\iinv{(x_0, y_0)}{(s_1, n_1)} \\implies \\iinv{\\nu_0}{\\nu_1}\\\\\n\\end{array}\n$$\n\n%%% Local Variables: \n%%% mode: latex\n%%% TeX-master: t\n%%% End: \n", "meta": {"hexsha": "b9d9c4d12b5e00934508bf3dcbac97ec030002fd", "size": 7371, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "papers/paper/evaluation.tex", "max_stars_repo_name": "qizhou92/icfp", "max_stars_repo_head_hexsha": "2f84c30e8f564f4bec2933a8b736ae58fd91821e", "max_stars_repo_licenses": ["MIT"], "max_stars_count": null, "max_stars_repo_stars_event_min_datetime": null, "max_stars_repo_stars_event_max_datetime": null, "max_issues_repo_path": "papers/paper/evaluation.tex", "max_issues_repo_name": "qizhou92/icfp", "max_issues_repo_head_hexsha": "2f84c30e8f564f4bec2933a8b736ae58fd91821e", "max_issues_repo_licenses": ["MIT"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "papers/paper/evaluation.tex", "max_forks_repo_name": "qizhou92/icfp", "max_forks_repo_head_hexsha": "2f84c30e8f564f4bec2933a8b736ae58fd91821e", "max_forks_repo_licenses": ["MIT"], "max_forks_count": null, "max_forks_repo_forks_event_min_datetime": null, "max_forks_repo_forks_event_max_datetime": null, "avg_line_length": 33.2027027027, "max_line_length": 102, "alphanum_fraction": 0.5928639262, "num_tokens": 3025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.815232480373843, "lm_q2_score": 0.7490872187162397, "lm_q1q2_score": 0.6106802313303835}} {"text": "\\documentclass{article}\n\\input{../homework-problems}\n\n\\toggletrue{solutions}\n%\\togglefalse{solutions}\n\\toggletrue{answers}\n\\newtheorem{problem}{Problem}\n\n\\newcommand{\\hide}[1]{}\n\n\\renewcommand{\\fcProblemRef}{\\theproblem.\\theenumi}\n\\renewcommand{\\fcSubProblemRef}{\\theenumi.\\theenumii}\n\n\\begin{document}\n\\begin{center}\n\\Large\nMaster Problem Sheet \\\\ Calculus I \\\\ \n\\end{center}\n\n%\\noindent \\textbf{Name:} \\hfill{~}\n%\\begin{tabular}{c|c|c|c|c|c|c|c|c||c}\n%Problem&1 &2&3&4&5&6&7&8& $\\sum$\\\\ \\hline\n%Score&&&&&&&&&\\\\ \\hline\n%Max&20&20&20&20&20&10&20&20&150\n%\\end{tabular}\n\n\n\n\nThis master problem sheet contains all freecalc problems on the topics studied in Calculus I. For a list of contributors/authors of the freecalc project (and in particular, the present problem collection) see file contributors.tex.\n\n\n\\fcLicenseContent\n\n\n\n\\tableofcontents\n\n\\section{Functions, Basic Facts}\\label{secMPSfunctionBasics}\n\\subsection{Understanding function notation}\n\\begin{problem}\n\\input{\\freecalcBaseFolder/modules/precalculus/homework/functions-evaluate-simplify}\n\\end{problem}\n\\subsection{Domains and ranges}\n\\begin{problem}\n\\input{../../modules/precalculus/homework/functions-domains-ranges-1}\n\\end{problem}\n\\subsection{Piecewise Defined Functions}\n\\begin{problem}\n\\input{../../modules/functions-basics/homework/piecewise-linear-formula-from-plot-1}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/functions-basics/homework/piecewise-linear-piecewise-circular-formula-from-plot-1}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/precalculus/homework/piecewise-defined-plot-from-formula}\n\\end{problem}\n\n\\subsection{Function composition}\n\\begin{problem}\n\\input{../../modules/functions-basics/homework/function-composition-domains-and-ranges-1}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/functions-basics/homework/functions-composing-fractional-linear-1}\n\\end{problem}\n\\subsection{Linear Transformations and Graphs of Functions}\n\\begin{problem}\n\\input{../../modules/precalculus/homework/functions-plot-transformations}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/function-graph-linear-transformations/homework/use-known-graph-to-sketch-its-linear-transformation-1}\n\\end{problem}\n\\input{../../modules/function-graph-linear-transformations/homework/use-known-graph-to-sketch-its-linear-transformation-1-solutions}\n\n\n\\section{Trigonometry}\\label{secMPStrigonometry}\n\\subsection{Angle conversion}\n\\begin{problem}\n\\input{../../modules/trigonometry/homework/trigonometry-angle-conversion-degree-to-radian}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/trigonometry/homework/trigonometry-angle-conversion-radian-to-degree}\n\\end{problem}\n\\subsection{Trigonometry identities}\n\\begin{problem}\n\\input{../../modules/trigonometry/homework/trigonometry-identities}\n\\end{problem}\n\n\\subsection{Trigonometry equations}\n\\begin{problem}\n\\input{../../modules/trigonometry/homework/trigonometry-equations}\n\\end{problem}\n\\input{../../modules/trigonometry/homework/trigonometry-equations-solutions}\n\n\\section{Limits and Continuity}\n\\subsection{Limits as $x$ tends to a number}\\label{secMPSlimitsXtendsToNumer}\n\\begin{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-a-common-factor}\n\\end{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-a-common-factor-solutions}\n\\begin{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-a-common-factor2}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-a-direct-substitution}\n\\end{problem}\n\\subsection{Limits involving $\\infty $}\n\\subsubsection{Limits as $x\\to\\pm \\infty$}\\label{secMPSlimitsXtoInfty}\n\\begin{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-infinity-1}\n\\end{problem}\n\\input{../../modules/limits/homework/limit-x-tends-to-infinity-1-solutions}\n\n\\subsubsection{Limits involving vertical asymptote}\\label{secMPSlimitsVerticalAsymptote}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/asymptotes-vertical}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/asymptotes-vertical2}\n\\end{problem}\n\\input{../../modules/curve-sketching/homework/asymptotes-vertical2-solutions}\n\\subsubsection{Find the Horizontal and Vertical Asymptotes}\\label{secMPShorAndVertAsymptotes}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/asymptotes-vertical-horizontal}\n\\end{problem}\n\\input{../../modules/curve-sketching/homework/asymptotes-vertical-horizontal-solutions}\n\n\\subsection{Limits - All Cases - Problem Collection}\n\\begin{problem}\n\\input{../../modules/limits/homework/limit-problem-collection-1}\n\\end{problem}\n\\subsection{Continuity}\n\\subsubsection{Continuity to evaluate limits}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-to-evaluate-limits-1}\n\\end{problem}\n\n\\subsubsection{Conceptual problems} \\label{secMPScontinuityConceptual}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-problems-involving-understanding}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-problems-involving-understanding-2}\n\\end{problem}\n\\subsubsection{Continuity and Piecewise Defined Functions} \\label{secMPScontinuityPiecewise}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-with-piecewise-defined-functions}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-with-piecewise-defined-functions-2}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/continuity/homework/continuity-with-piecewise-defined-functions-3}\n\\end{problem}\n\\subsection{Intermediate Value Theorem}\\label{secMPSintermediateValueTheorem}\n\\begin{problem}\n\\input{../../modules/continuity/homework/IVT-problems}\n\\end{problem}\n\\input{../../modules/continuity/homework/IVT-problems-solutions}\n\\begin{problem}\n\\input{../../modules/continuity/homework/IVT-problems2}\n\\end{problem}\n\\input{../../modules/continuity/homework/IVT-problems2-solutions}\n\\section{Inverse Functions}\\label{secMPSInverseFunctions}\n\n\\subsection{Problems Using Rational Functions Only}\n\\begin{problem}\n\\input{../../modules/inverse-functions/homework/inverse-functions-no-exponents-logarithms-1}\n\\end{problem}\n\\input{../../modules/inverse-functions/homework/inverse-functions-no-exponents-logarithms-1-solutions}\n\n\\begin{problem}\n\\input{../../modules/inverse-functions/homework/inverse-functions4}\n\\end{problem}\n\n\\subsection{Problems Involving Exponents, Logarithms}\n\\begin{problem}\n\\input{../../modules/inverse-functions/homework/inverse-functions1}\n\\end{problem}\n\\input{../../modules/inverse-functions/homework/inverse-functions1-solutions}\n\n\\section{Logarithms and Exponent Basics}\\label{secMPSLogarithmsExponentsBasics}\n\\subsection{Exponents Basics}\n\\begin{problem}\n\\input{../../modules/exponential-functions/homework/exponent-simplify}\n\\end{problem}\n\\input{../../modules/exponential-functions/homework/exponent-simplify-solutions}\n\n\\subsection{Logarithm Basics}\n\\begin{problem}\n\\input{../../modules/logarithms/homework/logarithms-basic-no-properties-1}\n\\end{problem}\n\\input{../../modules/logarithms/homework/logarithms-basic-properties-1-solutions}\n\\begin{problem}\n\\input{../../modules/logarithms/homework/logarithms-combine-1}\n\\end{problem}\n\\input{../../modules/logarithms/homework/logarithms-combine-1-solutions}\n\\begin{problem}\n\\input{../../modules/logarithms/homework/logarithms-basic-properties-1}\n\\end{problem}\n\\input{../../modules/logarithms/homework/logarithms-basic-properties-1-solutions}\n\n\\subsection{Some Problems Involving Logarithms}\n\\begin{problem}\n\\input{../../modules/logarithms/homework/logarithms-equations}\n\\end{problem}\n\\input{../../modules/logarithms/homework/logarithms-equations-solutions}\n\n\\section{Derivatives}\n\\subsection{Derivatives and Function Graphs: basics}\\label{secMPSderivativesFunGraphsBasics}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/match-graph-to-derivative-graph-2}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/match-graph-to-derivative-graph-2-solution}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/match-graph-to-derivative-graph}\n\\end{problem}\n\\subsection{Product and Quotient Rules}\\label{secMPSproductQuotientRules}\n\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-1}\n\\end{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-1-solutions}\n\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-2}\n\\end{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-2-solutions}\n\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-3}\n\\end{problem}\n\\input{../../modules/derivatives/homework/derivatives-product-quotient-rule-3-solutions}\n\n\\subsection{Basic Trigonometric Derivatives}\\label{secMPStrigDerivatives}\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivatives-trig}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivatives-trig-2}\n\\end{problem}\n\\subsection{Natural Exponent Derivatives}\n\\input{../../modules/exponential-functions/homework/exponent-derivative}\n\n\\subsection{The Chain Rule}\\label{secMPSchainRule}\n\n\\begin{problem}\n\\input{../../modules/chain-rule/homework/chain-rule1}\n\\end{problem}\n\\input{../../modules/chain-rule/homework/chain-rule1-solutions}\n\n\\begin{problem}\n\\input{../../modules/chain-rule/homework/chain-rule2}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/chain-rule/homework/chain-rule3}\n\\end{problem}\n\\input{../../modules/chain-rule/homework/chain-rule3-solutions}\n\n\\begin{problem}\n\\input{../../modules/chain-rule/homework/chain-rule4}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/chain-rule/homework/chain-rule5}\n\\end{problem}\n\\subsection{Problem Collection All Techniques}\n\\begin{problem}\n\\input{../../modules/derivatives/homework/derivative-problem-collection-1}\n\\end{problem}\n\\input{../../modules/derivatives/homework/derivative-problem-collection-1-solutions}\n\n\\subsection{Implicit Differentiation}\\label{secMPSImplicitDifferentiation}\n\\begin{problem}\n\\input{../../modules/implicit-differentiation/homework/implicitly-differentiate}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/implicit-differentiation/homework/implicit-tangent}\n\\end{problem}\n\\input{../../modules/implicit-differentiation/homework/implicit-tangent-solutions}\n\n\\subsection{Implicit Differentiation and Inverse Trigonometric Functions}\n\\begin{problem}\n\\input{../../modules/implicit-differentiation/homework/implicit-inverse-trig1}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/implicit-differentiation/homework/implicit-inverse-trig2}\n\\end{problem}\n\n\\subsection{Derivative of non-Constant Exponent with non-Constant Base}\\label{secMPSDerivativeNonConstExponent}\n\\begin{problem}\n\\input{../../modules/logarithms/homework/derivatives-non-const-exponent-2}\n\\end{problem}\n\\input{../../modules/logarithms/homework/derivatives-non-const-exponent-2-solutions}\n\n\\begin{problem}\n\\input{../../modules/logarithms/homework/derivatives-non-const-exponent}\n\\end{problem}\n\n\\subsection{Related Rates}\\label{secMPSrelatedRates}\n\\begin{problem}\n\\input{../../modules/related-rates/homework/related-rates-text-problems-2}\n\\end{problem}\n\\input{../../modules/related-rates/homework/related-rates-text-problems-2-solutions}\n\\begin{problem}\n\\input{../../modules/related-rates/homework/related-rates-text-problems}\n\\end{problem}\n\\input{../../modules/related-rates/homework/related-rates-text-problems-solutions}\n\n\\section{Graphical Behavior of Functions}\n\\subsection{Mean Value Theorem}\\label{secMPS-MVT}\n\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/find-the-mvt-points-1}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/mvt-to-estimate-fmax-fmin-1}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/mvt-show-only-one-root-1}\n\\end{problem}\n\\input{../../modules/maxima-minima/homework/mvt-show-only-one-root-1-solutions}\n\n\\subsection{Maxima, Minima}\\label{secMPSoneVariableMinMax}\n\\subsubsection{Closed Interval method}\\label{secMPSclosedInterval}\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/maxima-minima}\n\\end{problem}\n\n\\begin{problem}\n\\input{../../modules/logarithms/homework/logarithm-physics}\n\\end{problem}\n\n\\subsubsection{Derivative tests}\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/local-maxima-minima-derivative-tests-1}\n\\end{problem}\n\n\\subsubsection{Optimization}\\label{secMPSoptimization}\n\\begin{problem}\n\\input{../../modules/maxima-minima/homework/maxima-minima-text-problems-1}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/optimization/homework/optimization-problem-collection-1}\n\\end{problem}\n\\input{../../modules/optimization/homework/optimization-problem-collection-1-solutions}\n\n\\subsection{Function Graph Sketching}\\label{secMPSfunctionGraphSketching}\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/sketch-graph-with-all-the-details}\n\\end{problem}\n\\input{../../modules/curve-sketching/homework/sketch-graph-with-all-the-details-solutions}\n\n\\begin{problem}\n\\input{../../modules/curve-sketching/homework/sketch-graph-with-all-the-details-2}\n\\end{problem}\n\n\\section{Linearizations and Differentials} \\label{secMPSLinearizationAndDifferentials}\n\\begin{problem}\n\\input{../../modules/differentials/homework/linearization-problems-1}\n\\end{problem}\n\\input{../../modules/differentials/homework/linearization-problems-1-solutions}\n\n\\section{Integration Basics}\n\\subsection{Riemann Sums}\\label{secMPSRiemannSums}\n\\begin{problem}\n\\input{../../modules/integration/homework/riemann-sum-problems-1}\n\\end{problem}\n\\input{../../modules/integration/homework/riemann-sum-problems-1-solutions}\n\n\\subsection{Antiderivatives}\\label{secMPSantiderivatives}\n\\begin{problem}\n\\input{../../modules/antiderivatives/homework/antiderivatives-basic-integrals-1}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/antiderivatives/homework/antiderivatives-basic-integrals-initial-condition-1}\n\\end{problem}\n\\begin{problem}\n\\input{../../modules/antiderivatives/homework/antiderivatives-verify-by-differentiation-1}\n\\end{problem}\n\n\\subsection{Basic Definite Integrals} \\label{secMPSBasicDefiniteIntegrals}\n\\begin{problem}\n\\input{../../modules/integration/homework/basic-definite-integrals-1}\n\\end{problem}\n\\input{../../modules/integration/homework/basic-definite-integrals-1-solutions}\n\\begin{problem}\n\\input{../../modules/integration/homework/basic-definite-integrals-2}\n\\end{problem}\n\\subsection{Fundamental Theorem of Calculus Part I}\\label{secMPSFTCpart1}\n\\begin{problem}\n\\input{../../modules/integration/homework/FTC-part1-problems-1}\n\\end{problem}\n\\input{../../modules/integration/homework/FTC-part1-problems-1-solutions}\n\n\\subsection{Integration with The Substitution Rule}\n\\label{secMPSintegrationSubstitutionRule}\n\\subsubsection{Substitution in Indefinite Integrals}\n\\label{secMPSintegrationSubstitutionRuleIndefinite}\n\\begin{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-1}\n\\end{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-1-solutions}\n\\begin{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-2}\n\\end{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-2-solutions}\n\n\\subsubsection{Substitution in Definite Integrals}\n\\label{secMPSintegrationSubstitutionRuleDefinite}\n\\begin{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-definite-1}\n\\end{problem}\n\\input{../../modules/substitution-rule/homework/substitution-rule-definite-1-solutions}\n\n\\section{First Applications of Integration}\n\\subsection{Area Between Curves}\\label{secMPSareaBetweenCurves}\n\\begin{problem}\n\\input{../../modules/area-between-curves/homework/area-between-curves-problems-1}\n\\end{problem}\n\\input{../../modules/area-between-curves/homework/area-between-curves-problems-1-solutions}\n\\subsection{Volumes of Solids of Revolution}\\label{secMPSvolumesSolidsRevolution}\n\\subsubsection{Problems Geared towards the Washer Method}\\label{secMPSvolumesSolidsRevolutionWashers}\n\\begin{problem}\n\\input{../../modules/volumes/homework/solids-of-revolution-problems-1}\n\\end{problem}\n\\input{../../modules/volumes/homework/solids-of-revolution-problems-1-solutions}\n\n\\subsubsection{Problems Geared towards the Cylindrical Shells Method}\\label{secMPSvolumesSolidsRevolutionShells}\n\\begin{problem}\n\\input{../../modules/volumes/homework/solids-of-revolution-cylindrical-shells-1}\n\\end{problem}\n\\end{document}", "meta": {"hexsha": "877bb9e8fa1199bf0098eabf7c60732ddcebce4a", "size": 16461, "ext": "tex", "lang": "TeX", 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YES\n2. YES", "lm_q1_score": 0.7490872131147275, "lm_q2_score": 0.8152324848629214, "lm_q1q2_score": 0.61068023012656}} {"text": "\\section{Implementation} \\label{sec:implementation}\n\n\\subsection{Parameters of the Implementation}\n\n\n\\begin{table}[tbh]\n\\begin{center}\n\\begin{tabular}{p{.2 \\textwidth} | p{.6 \\textwidth} | p{.2 \\textwidth}}\\hline\nField & Description & Default value\\\\ \\hline\n$P\\in\\mathbb{Z}$ & Number of populations. & 1 \\\\\n$n\\in\\mathbb{Z}$ & Maximum number of pure strategies per population. & - \\\\\n$S\\in\\mathbb{Z}^{P}$ & Vector of pure strategies in each population, such that $1 SM target@\nwhich does the string matching and is indeed a monoid morphism from\n@RString@ to @SM target@ for a given @target@.\n\n\\subsubsection{String Matching Monoid}\n\nWe define the data type\n@SM target@ to contain a refined string field @input@ and\na list of all the indices in @input@ where the\n@target@ appears.\n%\n\\begin{code}\n data SM (target :: Symbol) where\n SM :: input:RString\n -> indices:[GoodIndex input target]\n -> SM target\n\\end{code}\n%\nWe use the string type literal~\\footnote{\\texttt{Symbol} is a kind and\ntarget is effectively a singleton type.} to parameterize the monoid\nover the target being matched. This encoding allows the type checker\nto statically ensure that only searches for the same target can be\nmerged together. The input field is a refined string, and the indices\nfield is a list of good indices. For simplicity we present lists as\nHaskell's built-in lists, but our implementation uses the reflected\nlist type, @L@, defined in \\S~\\ref{sec:haskell-proofs}.\n\nA @GoodIndex input target@ is a refined type alias for a natural\nnumber @i@ for which @target@ appears at position @i@ of @input@. As\nan example, the good indices of @\"abcab\"@ on @\"ababcabcab\"@ are\n@[2,5]@.\n%\n\\begin{code}\n type GoodIndex Input Target\n = {i:Nat | isGoodIndex Input (fromString Target) i }\n\n isGoodIndex :: RString -> RString -> Int -> Bool\n isGoodIndex input target i\n = (subString i (lenStr target) input == target)\n && (i + lenStr target <= lenStr input)\n\n subString :: Int -> Int -> RString -> RString\n subString o l = takeStr l . dropStr o\n\\end{code}\n%\n\\ignore{\n\\begin{code}\ngoodSM :: SM \"abcab\"\ngoodSM = SM \"ababcabcab\" [2, 5]\n\nbadSM :: SM \"abcab\"\nbadSM = SM \"ababcabcab\" [0, 7]\n\\end{code}\n\\ignore{\n\\NV{Liquid Haskell actually will reject both the above, as the lenStr and subString functions are uninterpreted}\n}\n}\n\n\\subsubsection{Monoid Methods for String Matching}~\\label{subsec:monoid:methods}\nNext, we define the mappend and identity elements for string matching.\n\nThe \\textit{identity element} @mempty@ of @SM t@, for each target @t@, is\ndefined to contain the identity @RString@ (@stringMempty@) and the\nidentity @List@ (@listMempty@).\n\\begin{code}\n mempty:: forall (t :: Symbol). SM t\n mempty = SM stringMempty listMempty\n\\end{code}\n%\n\n\\ignore{\nThe associative operator, @(mappend)@,\nappends the two input strings.\nThe appended indices, as depicted in Figure~\\ref{fig:mappend:indices},\nare the concatenations of three list indices:\n\\begin{enumerate}\n\\item The indices @xis@ of the first input, casted to good indices in the new structure,\n\\item the new indices @xyis@ created when concatenating the two strings, and\n\\item the indices @yis@ of the second input, shifted right @lenStr y@ units.\n\\end{enumerate}\n%\n}\n\n\\begin{figure}[t]\n\\centering\n\\captionsetup{justification=centering}\n\\includegraphics[scale=0.5]{text/stringmatcher/makeIndices}\n\\caption{Mappend indices of String Matcher.}\n\\label{fig:mappend:indices}\n\\end{figure}\n%\nThe Haskell definition of @<>@, the monoid operation for @SM t@, is as follows.\n\\begin{code}\n (mappend)::forall (t::Symbol). KnownSymbol t => SM t -> SM t -> SM t\n (SM x xis) mappend (SM y yis)\n = SM (x stringMappend y) (xis' listMappend xyis listMappend yis')\n where\n tg = fromString (symbolVal (Proxy :: Proxy t))\n xis' = map (castGoodIndexLeft tg x y) xis\n xyis = makeNewIndices x y tg\n yis' = map (shiftStringRight tg x y) yis\n\\end{code}\nNote again that capturing target as a type parameter is critical,\notherwise there is no way for the Haskell's type system to specify\nthat both arguments of @(mappend)@ are string matchers on the same target.\n\nThe action of @(<>)@ on the two @input@ fields is straightforward;\nhowever, the action on the two @indices@ is complicated by the need to\nshift indices and the possibility of new matches arising from the\nconcatenation of the two @input@\nfields. Figure~\\ref{fig:mappend:indices} illustrates the three pieces\nof the new @indices@ field which we now explain in more detail.\n\n\\mypara{1. Casting Good Indices}\nIf @xis@ is a list of good indices for the string @x@ and the target\n@tg@, then @xis@ is also a list of good indices for the string\n@x stringMappend y@ and the target @tg@, for each @y@.\n%\nTo prove this property we need to invoke the property\n@subStrAppendRight@ on Refined Strings that establishes\nsubstring preservation on string right appending.\n%\n\\begin{code}\n assume subStrAppendRight\n :: sl:RString -> sr:RString -> j:Int\n -> i:{Int | i + j <= lenStr sl }\n -> {subString sl i j = subString (sl stringMappend sr) i j}\n\\end{code}\n%\nThe specification of @subStrAppendRight@ ensures that for each\nstring @sl@ and @sr@ and each integer @i@ and @j@ whose sum is within @sl@,\nthe substring from @i@ with length @j@ is identical in @sl@ and in @(sl stringMappend sr)@.\n%\nThe function @castGoodIndexLeft@ applies the above property to an index @i@\nto cast it from a good index on @sl@ to a good index on @(sl stringMappend sr)@\n%\n\\begin{code}\n castGoodIndexLeft\n :: tg:RString -> sl:RString -> sr:RString\n -> i:GoodIndex sl tg\n -> {v:GoodIndex (sl stringMappend sr) target | v = i}\n\n castGoodIndexLeft tg sl sr i\n = cast (subStrAppendRight sl sr (lenStr tg) i) i\n\\end{code}\n%\nWhere @cast p x@ returns @x@, after enforcing the properties of @p@ in the logic\n\\begin{code}\n cast :: b -> x:a -> {v:a | v = x}\n cast _ x = x\n\\end{code}\n%\nMoreover, in the logic, each expression @cast p x@\nis reflected as @x@,\nthus allowing random (\\ie non-reflected) Haskell expressions to appear in @p@.\n\n\\mypara{2. Creation of new indices}\nThe concatenation of two input strings @sl@ and @sr@\nmay create new good indices.\n%\nFor instance, concatenation of\n@\"ababcab\"@ with @\"cab\"@\nleads to a new occurence of @\"abcab\"@ at index @5@ which\ndoes not occur in either of the two input strings.\n%\nThese new good indices can appear only at the last @lenStr tg@ positions\nof the left input @sl@.\n%\n@makeNewIndices sl sr tg@ detects all such good new indices.\n%\n\\begin{code}\n makeNewIndices\n :: sl:RString -> sr:RString -> tg:RString\n -> [GoodIndex {sl stringMappend sr} tg]\n \n makeNewIndices sl sr tg\n | lenStr tg < 2 = []\n | otherwise = makeIndices (sl stringMappend sr) tg lo hi\n where\n lo = maxInt (lenStr sl - (lenStr tg - 1)) 0\n hi = lenStr sl - 1\n\\end{code}\nIf the length of the @tg@ is less than 2, then no new good indices are created.\n%\nOtherwise,\nthe call on @makeIndices@ returns all the good indices of the input\n@sl stringMappend sr@ for target @tg@\nin the range from @maxInt (lenStr sl-(lenStr tg-1)) 0@ to @lenStr sl-1@.\n%\n\nGenerally, @makeIndices s tg lo hi@ returns the good indices\nof the input string @s@ for target @tg@ in the range from @lo@ to @hi@.\n%\n\\begin{code}\n makeIndices :: s:RString -> tg:RString -> lo:Nat\n -> hi:Int -> [GoodIndex s tg]\n \n makeIndices s tg lo hi \n | hi < lo = []\n | isGoodIndex s tg lo = lo:rest\n | otherwise = rest\n where\n rest = makeIndices s tg (lo + 1) hi\n\\end{code}\n%Note the similarity to the functional specification of string matching\n%at the beginning of this section.\n\nIt is important to note that\n@makeNewIndices@ does not scan all the input,\ninstead only searching at most @lenStr tg@ positions for new good indices.\n%\nThus, the time complexity to create the new indices is linear\non the size of the target but independent of the size of the input.\n\n\\mypara{3. Shift Good Indices}\nIf @yis@ is a list of good indices on the string @y@ with target @tg@,\nthen we need to shift each element of @yis@ right @lenStr x@ units to\nget a list of good indices for the string @x stringMappend y@.\n\n%\nTo prove this property we need to invoke the property\n@subStrAppendLeft@ on Refined Strings that establishes\nsubstring shifting on string left appending.\n%\n\\begin{code}\n assume subStrAppendLeft\n :: sl:RString -> sr:RString\n -> j:Int -> i:Int\n -> {subStr sr i j = subStr (sl stringMappend sr) (lenStr sl+i) j}\n\\end{code}\n%\nThe specification of @subStrAppendLeft@ ensures that for each\nstring @sl@ and @sr@ and each integers @i@ and @j@,\nthe substring from @i@ with length @j@ on @sr@\nis equal to the substring from @lenStr sl + i@\nwith length @j@ on @(sl stringMappend sr)@.\n%\nThe function @shiftStringRight@ both shifts the input index @i@ by @lenStr sl@\nand applies the @subStrAppendLeft@ property to it,\ncasting @i@ from a good index on @sr@ to a good index on @(sl stringMappend sr)@\n\nThus, @shiftStringRight@ both appropriately shifts the index\nand casts the shifted index using the above theorem:\n\\begin{code}\n shiftStringRight\n :: tg:RString -> sl:RString -> sr:RString\n -> i:GoodIndex sr tg\n -> {v:(GoodIndex (sl stringMappend sr) tg) | v = i + lenStr sl}\n \n shiftStringRight tg sl sr i\n = subStrAppendLeft sl sr (lenStr tg) i `cast` i + lenStr sl\n\\end{code}\n\n\\subsubsection{String Matching is a Monoid}\nNext we prove that the monoid methods @mempty@ and @(mappend)@ satisfy\nthe monoid laws.\n%\n\\begin{theorem}[SM is a Monoid]\\label{theorem:stringmatchers}\n(@SM t@, @mempty@, @mappend@)\nis a monoid.\n\\end{theorem}\n%\n\\begin{proof}\nAccording to the Monoid Definition~\\ref{definition:monoid},\nwe prove that string matching is a monoid,\nby providing safe implementations for the monoid law functions.\n%\nFirst, we prove \\textit{left identity}.\n\\begin{code}\n idLeft :: x:SM t -> {mempty mappend x = xs}\n \n idLeft (SM i is)\n = (mempty :: SM t) mappend (SM i is)\n =. (SM stringMempty listMempty) mappend (SM i is)\n =. SM (stringMempty <+> i) (is1 ++ isNew ++ is2)\n ? idLeftStr i\n =. SM i ([] ++ [] ++ is)\n ? (mapShiftZero tg i is && newIsNullRight i tg)\n =. SM i is\n ? idLeftList is\n ** QED\n where\n tg = fromString (symbolVal (Proxy :: Proxy t))\n is1 = map (castGoodIndexRight tg i stringMempty) []\n isNew = makeNewIndices stringMempty i tg\n is2 = map (shiftStringRight tg stringMempty i) is\n\\end{code}\n\nThe proof proceeds by rewriting, using left identity of the monoid strings and lists,\nand two more lemmata.\n\\begin{itemize}\n\\item Identity of shifting by an empty string.\n\\begin{code}\n mapShiftZero :: tg:RString -> i:RString\n -> is:[GoodIndex i target]\n -> {map (shiftStringRight tg stringMempty i) is = is}\n\\end{code}\nThe lemma is proven by induction on @is@ and\nthe assumption that empty strings have length 0.\n\\item No new indices are created.\n\\begin{code}\n newIsNullLeft :: s:RString -> t:RString \n -> {makeNewIndices stringMempty s t = []}\n\\end{code}\nThe proof relies on the fact that @makeIndices@\nis called on the empty range from @0@ to @-1@\nand returns @[]@.\n\\end{itemize}\n\nNext, we prove \\textit{right identity}.\n\\begin{code}\n idRight :: x:SM t -> {x mappend mempty = x}\n \n idRight (SM i is)\n = (SM i is) mappend (mempty :: SM t)\n =. (SM i is) mappend (SM stringMempty listMempty)\n =. SM (i stringMappend stringMempty) (is1 listMappend isNew listMappend is2)\n ? idRightStr i\n =. SM i (is listMappend N listMappend N)\n ? (mapCastId tg i stringMempty is && newIsNullLeft i tg)\n =. SM i is\n ? idRightList is\n ** QED\n where\n tg = fromString (symbolVal (Proxy :: Proxy t))\n is1 = map (castGoodIndexRight tg i stringMempty) is\n isNew = makeNewIndices i stringEmp tg\n is2 = map (shiftStringRight tg i stringMempty) []\n\\end{code}\nThe proof proceeds by rewriting,\nusing right identity on strings and lists and two more lemmata.\n\\begin{itemize}\n\\item Identity of casting is proven\n\\begin{code}\n mapCastId \n :: tg:RString -> x:RString -> y:RString\n -> is:[GoodIndex x tg] ->\n -> {map (castGoodIndexRight tg x y) is = is}\n\\end{code}\nWe prove identity of casts by induction on @is@ and\nidentity of casting on a single index.\n\\item No new indices are created.\n\\begin{code}\n newIsNullLeft :: s:RString -> t:RString\n -> {makeNewIndices s stringMempty t = listMempty}\n\\end{code}\nThe proof proceeds by case splitting\non the relative length of @s@ and @t@.\nAt each case we prove by induction that all\nthe potential new indices would be out of bounds and thus\nno new good indices would be created.\n\\end{itemize}\n\n- Finally we prove \\textit{associativity}.\nFor space, we only provide a proof sketch.\nThe whole proof is available online~\\cite{implementation}.\n%\nOur goal is to show equality of the left and right associative string matchers.\n%\n\\begin{code}\n assoc :: x:SM t -> y:SM t -> z:SM t\n -> {x mappend (y mappend z) = (x mappend y) mappend z}\n\\end{code}\nTo prove equality of the two string matchers we show\nthat the input and indices fields are respectively equal.\n%\nEquality of the input fields follows by associativity of RStrings.\n%\nEquality of the index list proceeds in three steps.\n%\n\\begin{enumerate}\n\\item Using list associativity and distribution of index shifting,\nwe group the indices in the five lists shown in\nFigure~\\ref{fig:mappend:assoc}: the indices of the input @x@, the new\nindices from mappending @x@ to @y@, the indices of the input @y@, the\nnew indices from mappending @x@ to @y@, and the indices of the input\n@z@.\n\\item The representation of each group depends on the order of appending.\nFor example, if @zis1@ (resp. @zis2@) is the group @zis@ when\nright (resp. left) mappend happened first, then we have\n\\begin{code}\n zis1 = map (shiftStringRight tg xi (yi stringMappend zi))\n (map (shiftStringRight tg yi zi) zis)\n\n zis2 = map (shiftStringRight tg (xi stringMappend yi) zi) zis\n\\end{code}\nThat is, in right first, the indices of @z@ are first shifted\nby the length of @yi@ and then by the length of @xi@,\nwhile in the left first case, the indices of @z@ are shifted by the\nlength of @xi stringMappend yi@.\nIn this second step of the proof we prove, using lemmata,\nthe equivalence of the different group representations.\n%\nThe most interesting lemma we use is called @assocNewIndices@ and proves\nequivalence of all the three middle groups together\nby case analysis on the relative lengths of the target @tg@ and the middle string @yi@.\n\\item After proving equivalence of representations,\nwe again use list associativity and distribution of casts to wrap the\nindex groups back in string matchers.\n\\end{enumerate}\n\\begin{figure}\n\\centering\n\\captionsetup{justification=centering}\n\\includegraphics[scale=0.5]{text/stringmatcher/AssociativeIndices}\n\\caption{Associativity of String Matching.}\n\\label{fig:mappend:assoc}\n\\end{figure}\nWe now sketch the three proof steps, while the whole proof\nis available online~\\cite{implementation}.\n\\begin{code}\n assoc x@(SM xi xis) y@(SM yi yis) z@(SM zi zis)\n -- Step 1: unwrapping the indices\n = x <> (y <> z)\n =. (SM xi xis) <> ((SM yi yis) <> (SM zi zis))\n ...\n -- via list associativity and distribution of shifts\n =. SM i (xis1 ++ ((xyis1 ++ yis1 ++ yzis1) ++ zis1))\n -- Step 2: Equivalence of representations\n =. SM i (xis2 ++ ((xyis1 ++ yis1 ++ yzis1) ++ zis1))\n ? castConcat tg xi yi zi xis\n =. SM i (xis2 ++ ((xyis1 ++ yis1 ++ yzis1) ++ zis2))\n ? mapLenFusion tg xi yi zi zis\n =. SM i (xis2 ++ ((xyis2 ++ yis2 ++ yzis2) ++ zis2))\n ? assocNewIndices y tg xi yi zi yis\n -- Step 3: Wrapping the indices\n ...\n -- via list associativity and distribution of casts\n =. (SM xi xis <> SM yi yis) <> SM zi zis\n =. (x <> y) <> z\n ** QED\n where\n i = xi stringMappend (yi stringMappend zi)\n\n yzis1 = map (shiftStringRight tg xi (yi <+> zi)) yzis\n yzis2 = makeNewIndices (xi <+> yi) zi tg\n yzis = makeNewIndices yi zi tg\n ...\n\\end{code}\n\\cqed\\end{proof}\n", "meta": {"hexsha": "ee9422f0b1d0aefc8139d0c981261806d53de4a2", "size": 16132, "ext": "tex", "lang": "TeX", "max_stars_repo_path": "text/stringmatcher/string-matcher-is-monoid.tex", "max_stars_repo_name": "nikivazou/thesis", "max_stars_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_stars_repo_licenses": ["BSD-3-Clause"], "max_stars_count": 11, "max_stars_repo_stars_event_min_datetime": "2016-12-02T00:46:41.000Z", "max_stars_repo_stars_event_max_datetime": "2021-02-20T07:04:01.000Z", "max_issues_repo_path": "text/stringmatcher/string-matcher-is-monoid.tex", "max_issues_repo_name": "nikivazou/thesis", "max_issues_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_issues_repo_licenses": ["BSD-3-Clause"], "max_issues_count": null, "max_issues_repo_issues_event_min_datetime": null, "max_issues_repo_issues_event_max_datetime": null, "max_forks_repo_path": "text/stringmatcher/string-matcher-is-monoid.tex", "max_forks_repo_name": "nikivazou/thesis", "max_forks_repo_head_hexsha": "a12f2e857a358e3cc08b657bb6b029ac2d500c3b", "max_forks_repo_licenses": ["BSD-3-Clause"], "max_forks_count": 1, "max_forks_repo_forks_event_min_datetime": "2016-12-02T00:46:51.000Z", "max_forks_repo_forks_event_max_datetime": "2016-12-02T00:46:51.000Z", "avg_line_length": 35.9287305122, "max_line_length": 112, "alphanum_fraction": 0.697929581, "num_tokens": 4701, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.8652240860523328, "lm_q2_score": 0.7057850216484838, "lm_q1q2_score": 0.6106622003052353}} {"text": "\\documentclass{subfile}\n\n\\begin{document}\n\t\n\t\\section{SgMO}\\label{sec:sgmo}\n\t\n\t\t\\begin{problem}[Singapore Team Selection Test $2009$, problem $2$]\n\t\t\tIf $a,b,c$ are three positive real numbers such that $ab+bc+ca=1$, prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\sqrt[3]{\\dfrac{1}{a}+6b}+\\sqrt[3]{\\dfrac{1}{b}+6c}+\\sqrt[3]{\\dfrac{1}{c}+6a}\n\t\t\t\t\t\t& \\leq\\dfrac{1}{abc}\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\t\n\t\t\\begin{problem}[Singapore Team Selection Test $2008$, problem $2$]\n\t\t\tLet $x_{1},\\ldots,x_{n}$ be positive real numbers such that $x_{1}\\cdots x_{n}=1$. Prove that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\sum_{i=1}^{n}\\dfrac{1}{n-1+x_{i}}\n\t\t\t\t\t\t& \\leq1\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\t\n\t\t\\begin{problem}[Singapore Team Selection Test $2007$, problem $2$]\n\t\t\tProve the inequality\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\sum_{i1$ be an integer and $x_{1},\\ldots,x_{n}$ be real numbers such that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t|x_{1}|+\\ldots+|x_{n}|\n\t\t\t\t\t\t& = 1\\quad\\mbox{and}\\\\\n\t\t\t\t\tx_{1}+\\ldots+x_{n}\n\t\t\t\t\t\t& = 0\n\t\t\t\t\\end{align*}\n\t\t\tProve that\n\t\t\t\t\\begin{align*}\n\t\t\t\t\t\\left|\\dfrac{x_{1}}{1}+\\ldots+\\dfrac{x_{n}}{n}\\right|\n\t\t\t\t\t\t& \\leq \\dfrac{1}{2}\\left(1-\\dfrac{1}{n}\\right)\n\t\t\t\t\\end{align*}\n\t\t\\end{problem}\n\t\t\n\t\t\\begin{problem}[Singapore $2004$, problem $2$]\n\t\t\tLet $a,b,c$ be real numbers such that $ab+bc+ca=1$ and $0